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\begin{document}
\begin{frontmatter}
\title{Distance-based and RKHS-based Dependence Metrics in High Dimension}
\runtitle{Dependence Metrics in High Dimension}
\begin{aug}
\author{\fnms{Changbo} \snm{Zhu}\corref{}\thanksref{m1}
\ead[label=e1]{changbo2@illinois.edu}}
\author{\fnms{Shun} \snm{Yao}\thanksref{m2}
\ead[label=e2]{shunyao2@illinois.edu}}
\author{\fnms{Xianyang} \snm{Zhang}\thanksref{m3}
\ead[label=e3]{zhangxiany@stat.tamu.edu}}
\and
\author{\fnms{Xiaofeng} \snm{Shao}\thanksref{m1,t2}
\ead[label=e4]{xshao@illinois.edu}}
\thankstext{t2}{Address correspondence to Xiaofeng Shao (xshao@illinois.edu), Professor, Department of Statistics, University of Illinois at Urbana-Champaign. Changbo Zhu (changbo2@illinois.edu) is a Ph.D. student in Department of Statistics, University of Illinois at Urbana-Champaign. Shun Yao (shunyao2@illinois.edu) is currently Quantitative Analyst at Goldman Sachs, New York City; Xianyang Zhang (zhangxiany@stat.tamu.edu) is Assistant Professor of Statistics at Texas A\&M University.}
\runauthor{C. Zhu, S. Yao, X. Zhang and X. Shao}
\affiliation{University of Illinois at Urbana-Champaign\thanksmark{m1}, Goldman Sachs at New York City\thanksmark{m2} and Texas A\&M University\thanksmark{m3}}
\end{aug}
\begin{abstract}
In this paper, we study distance covariance, Hilbert-Schmidt covariance (aka Hilbert-Schmidt independence criterion [\cite{gretton2007}]) and related independence tests under the high dimensional scenario. We show that the sample distance/Hilbert-Schmidt covariance between two random vectors can be approximated by the sum of squared componentwise sample cross-covariances up to an asymptotically constant factor, which indicates that the distance/Hilbert-Schmidt covariance based test can only capture linear dependence in high dimension. As a consequence, the distance correlation based $t$ test developed by \cite{szekely2013} for independence is shown to have trivial limiting power when the two random vectors are nonlinearly dependent but component-wisely uncorrelated. This new and surprising phenomenon, which seems to be discovered for the first time, is further confirmed in our simulation study. As a remedy, we propose tests based on an aggregation of marginal sample distance/Hilbert-Schmidt covariances and show their superior power behavior against their joint counterparts in simulations. We further extend the distance correlation based $t$ test to those based on Hilbert-Schmidt covariance and marginal distance/Hilbert-Schmidt covariance. A novel unified approach is developed to analyze the studentized sample distance/Hilbert-Schmidt covariance as well as the studentized sample marginal distance covariance under both null and alternative hypothesis. Our theoretical and simulation results shed light on the limitation of distance/Hilbert-Schmidt covariance when used jointly in the high dimensional setting and suggest the aggregation of marginal distance/Hilbert-Schmidt covariance as a useful alternative.
\end{abstract}
\begin{keyword}[class=MSC]
\kwd[Primary ]{62G10}
\kwd{60K35}
\kwd[; secondary ]{62G20}
\end{keyword}
\begin{keyword}
\kwd{Distance covariance, High dimensionality, Hilbert-Schmidt independence criterion, Independence test, $\mathcal U$-statistics.}
\end{keyword}
\end{frontmatter}
\section{Introduction}
Testing for independence between two random vectors $X\in \mathbb{R}^p$ and $Y\in \mathbb{R}^q$ is a fundamental problem in statistics. There is a huge literature in the low dimensional context. Here we mention rank correlation coefficients based tests and nonparametric Cram\'er-von Mises type statistics in \cite{hoeffding1948non}, \cite{blum1961}, \cite{de1980cramer}; tests based on signs or empirical characteristic functions, see \cite{sinha1977}, \cite{deheuvels1981}, \cite{csorgHo1985}, \cite{hettmansperger1994}, \cite{gieser1997}, \cite{taskinen2003}, \cite{stepanova2003} among others; tests based on recently developed nonlinear dependence metrics that target at non-linear and non-monotone dependence include distance covariance [\cite{szekely2007}], Hilbert-Schmidt independence criterion (HSIC) [\cite{gretton2007}] (aka Hilbert-Schmidt covariance in this work) and sign covariance [\cite{bergsma2014}].
In the high dimensional setting, the literature is scarce. \cite{szekely2013} extended the distance correlation proposed in \cite{szekely2007} to the problem of testing independence of two random vectors under the setting that the dimensions $p$ and $q$ grow while sample size $n$ is fixed. This setting is known as high dimension, low sample size (HDLSS) in the literature and has been adopted in \cite{hall2005}, \cite{ahn2007}, \cite{jung2009}, and \cite{wei2016} etc. A closely related asymptotic framework is the high dimension medium sample size (HDMSS) [\cite{aoshima2018survey}], where $n \wedge p \wedge q \rightarrow \infty$ with $p,q$ growing more rapidly. Among the recent work that is related to independence testing in the high dimensional setting, \cite{pan2014} proposed tests of independence among a large number of high dimensional random vectors using insights from random matrix theory; \cite{yang2015} proposed a new statistic based on the sum of regularized sample canonical correlation coefficients of $X$ and $Y$, which is limited to testing for uncorrelatedness due to the use of canonical correlation. \cite{leung2018testing} proposed to test for mutual independence of high dimensional vectors using sum of pairwise rank correlations and sign covariances;
\cite{yao2017testing} addressed the mutual independence testing problem in the high dimensional context by using sum of pairwise squared sample distance covariances;
\cite{zhang2018conditional} proposed a $L^2$ type test for conditional mean/quantile dependence of a univariate response variable given a high dimensional covariate vector based on martingale difference divergence [\cite{shao2014martingale}], which is an extension of distance covariance to quantify conditional mean dependence.
Distance covariance/correlation was first introduced in \cite{szekely2007} and has received much attention since then. Owing to its notable ability to quantify any types of dependence including non-monotone, non-linear dependence and also the flexibility to be applicable to two random vectors in arbitrary, not necessarily equal dimensions, a lot of research work has been done to extend and apply distance covariance into many modern statistical problems; see e.g. \cite{kong2012}, \cite{li2012}, \cite{zhou2012}, \cite{lyons2013}, \cite{szekely2014}, \cite{dueck2014}, \cite{shao2014martingale}, \cite{park2015}, \cite{matteson2016}, \cite{zhang2018conditional} , \cite{edelmann2017}, \cite{yao2017testing} among others.
In this paper, we shall revisit the test proposed by \cite{szekely2013}, which seems to be the only test in the high dimensional setting that captures nonlinear and nonmonotonic dependence. Unlike the positive finding reported in \cite{szekely2013}, we obtained some negative and shocking results that show the limitation of distance covariance/correlation in the high dimensional context.
Specifically, we show that for two random vectors $X =(x_1,...,x_p)$ $ \in \mathbb R^p $ and $Y=(y_1,...,y_q) \in \mathbb R^q$ with finite component-wise second moments, as $p, q \rightarrow \infty$ and $n$ can either be fixed or grows to infinity at a slower rate,
\begin{align}
\label{intro:dcov}
dCov^2_n(\mathbf X, \mathbf Y) \approx \frac{1}{\tau} \sum_{i=1}^p \sum_{j=1}^q cov_n^2(\mathcal X_i, \mathcal Y_j),
\end{align}
where $X_{k} \overset{d}{=} X$ and $Y_{k} \overset{d}{=} Y $ are independent samples, $\mathcal{X}_{i}$ and $\mathcal{Y}_{j}$ are the component-wise samples, $\mathbf{X} = ( X_{1},X_{2}, \cdots, X_{n} )^{T} = (\mathcal X_{1}, \mathcal X_{2}, \cdots, \mathcal X_{p} )$ and $\mathbf{Y} = ( Y_{1},Y_{2}, \cdots, Y_{n} )^{T} = (\mathcal Y_{1}, \mathcal Y_{2}, \cdots, \mathcal Y_{q} )$ denote the sample matrices, $dCov^2_n(\mathbf X, \mathbf Y)$ is the unbiased sample distance covariance, $\tau$ is a constant quantity depending on the marginal distributions of $X$ and $Y$ as well as $p$ and $q$, $cov_n^2(\mathcal X_i, \mathcal Y_j)$ is an unbiased sample estimate of $cov^2(x_i, y_j)$ to be defined later. To the best of our knowledge, this is the first work in the literature uncovering the connection between sample distance covariance and sample covariance, the latter of which can only measure the linear dependence between two random variables.
This approximation suggests that the distance covariance can only measure linear dependence in the high dimensional setting although it is well-known to be capable of capturing non-linear dependence in the fixed dimensional case.
\cite{gretton2007} proposed Hilbert-Schmidt independence criterion (aka Hilbert-Schmidt covariance in this paper), which can be seen as a generalization of distance covariance by kernelizing the $L^2$ distance as shown by \cite{sejdinovic2013}. Despite the kernelization process, we show that the Hilbert-Schmidt covariance ($hCov$) enjoys similar approximation property under high dimension low/medium sample size setting, i.e.
\begin{align}
\label{intro:hcov}
hCov^2_{n}(\mathbf X, \mathbf Y) \approx A_pB_q \times \frac{1}{\tau^2} \sum\limits_{i=1}^{p} \sum\limits_{j=1}^{q} cov_{n}^2 (\mathcal X_{i}, \mathcal Y_{j}),
\end{align}
where $ hCov^2_{n}(\mathbf X, \mathbf Y) $ is the unbiased sample Hilbert-Schmidt covariance, $A_p$ and $B_q$ both converge in probability to constants that depend on the pre-chosen kernels. This aproximation also suggests that when the dimension is high, the Hilbert-Schmidt covariance ($hCov$) applied to the whole components of the vectors also exhibits the loss of power when $X$ and $Y$ are non-linearly dependent, but component-wisely uncorrelated or weakly correlated.
As a natural remedy, we propose a distance covariance based marginal test statistic, i.e.,
\[ mdCov_n^{2}(\mathbf X, \mathbf Y) = \sqrt{\binom{n}{2}}\sum_{i=1}^p \sum_{j=1}^q dCov_n^2(\mathcal X_i, \mathcal Y_j). \]
This test statistic is an aggregate of the componentwise sample distance covariances and captures the component by component nonlinear dependence. Similarly, the marginal Hilbert-Schmidt covariance ($mhCov$) is defined as
\[ mhCov_n^2 (\mathbf X, \mathbf Y) = \sqrt{\binom{n}{2}}\sum_{i=1}^p \sum_{j=1}^q hCov_n^2( \mathcal X_i,\mathcal Y_j). \]
The distance covariance, Hilbert-Schmidt covariance, marginal distance covariance and marginal Hilbert-Schmidt covariance based tests can be carried out by standard permutation procedures. The superiority of $mdCov$ and $mhCov$ based tests over its joint counterparts in power is demonstrated in the simulation studies.
On the other hand, \cite{szekely2013} discussed the distance correlation ($dCor$) based $t$-test under HDLSS and derived the limiting null distribution of the test statistic under suitable assumptions. We consider the same $t$-test statistic and further extends to Hilbert-Schmidt covariance ($hCov$), marginal distance covariance ($mdCov$) and marginal Hilbert-Schmidt covariance ($mhCov$). To derive the asymptotic distribution of studentized version of $dCov, hCov, mdCov $ and $ mhCov $ under both the null of independence (for HDLSS and HDMSS setting) and some specific alternative classes (for HDLSS setting), we develop a novel unified approach. In particular, we define a unified quantity ($uCov$) based on the bivariate kernel $k$ and show that under HDLSS setting, properly scaled $dCov_n^{2}$, $ hCov_n^{2}$ and $mdCov_n^{2}$ are all asymptotically equal to $uCov_n^2$ up to different choices of kernels, i.e.
\begin{align}
\label{eq:approx}
\begin{array}{ll}
\left.
\begin{array}{l}
dCov_n^{2}(\mathbf X, \mathbf Y) \approx a \times uCov_n^2(\mathbf X, \mathbf Y) \\
hCov_n^{2}(\mathbf X, \mathbf Y) \approx A_pB_q \times uCov_n^2(\mathbf X, \mathbf Y)
\end{array} \right\rbrace
& \text{when } k(x,y)= |x-y|^2, \\
\\
\left.
\begin{array}{l}
mdCov_n^{2}(\mathbf X, \mathbf Y) = b \times uCov_n^2(\mathbf X, \mathbf Y)
\end{array} \;\;\;\; \right\rbrace & \text{when } k(x,y)= |x-y| ,
\end{array}
\end{align}
where $a,b$ are constants and $A_p,B_p$ both converge in probability to constants. Next, we show that
\begin{align*}
\left\lbrace
\begin{array}{ll}
uCov_n^2(\mathbf X, \mathbf Y) \overset{d}{\rightarrow} \frac{2}{n(n-3)}\mathbf{c}^T \mathbf{M} \mathbf{d}, & \text{under HDLSS}, \\
C_{n,p,q} uCov_n^2(\mathbf X, \mathbf Y) \overset{d}{\rightarrow} N(0,1), & \text{under HDMSS} ,
\end{array} \right.
\end{align*}
where $\mathbf{c}, \mathbf{d}$ are jointly Gaussian, $\mathbf{M}$ is a projection matrix and $C_{n,p,q}$ is a normalizing constant. Thus, we can easily apply the above results to $dCov, hCov$ and $mdCov$-based $t$-test statistics using \eqref{eq:approx}. The unified approach still works for $mhCov$-based $t$-test if we consider the bandwidth parameters appeared in the kernel distance to be fixed constants. However, we encounter technical difficulties if the bandwidth parameters along each dimension depends on the whole component-wise samples, since this makes the pair-wise sample distance correlated with each other and complicates the asymptotic analysis.
We obtain the same limiting null distribution as \cite{szekely2013} and further show that this test statistic has a trivial power against the alternative where $X$ and $Y$ are non-linearly dependent, but component-wisely uncorrelated. This clearly demonstrates that the distance covariance/correlation based joint independence test (i.e., treating all components of a vector as a whole jointly) fails to capture the non-linear dependence in high dimension. This phenomenon is new and was not reported in \cite{szekely2013}. It shows that there might be some intrinsic difficulties for distance covariance to capture the non-linear dependence when the dimension is high and provide a cautionary note on the use of distance covariance/correlation directly to the whole components of high dimensional data. Besides, we have the following additional contributions relative to \cite{szekely2013}: (i) we relax the component-wise i.i.d. assumption used for asymptotic analysis; (ii) the limiting distributions are derived under both the null and certain classes of alternative hypothesis for the HDLSS framework; (iii) our unified approach holds for any bivariate kernel that has continuous second order derivative in a neighborhood containing 1; (iv) the limiting null distribution is also derived under the HDMSS setting.
\subsection{Notations}
In this paper, random data samples are denoted as, for each $i = 1, 2, \cdots, n$, $X_{i} \overset{d}{=} X= (x_{1}, \cdots, x_{p})^T \in \mathbb{R}^{p}$, $Y_{i} \overset{d}{=} Y= (y_{1}, \cdots, y_{q})^T \in \mathbb{R}^{q}$. Next, let $\mathbf{X} = ( X_{1},X_{2}, \cdots, X_{n} )^{T}$ and $\mathbf{Y} = ( Y_{1},Y_{2}, \cdots, Y_{n} )^{T}$ denote the random sample matrices. In addition, the random component-wise samples are denoted as $\mathcal{X}_{1}, \cdots, \mathcal{X}_{p}$ and $\mathcal{Y}_{1}, \cdots, \mathcal{Y}_{q}$, which are illustrated in the following table,
\begin{equation*}
\begin{tikzpicture}[baseline=-\the\dimexpr\fontdimen22\textfont2\relax ]
\matrix (m)[matrix of math nodes]
{
& & \mathcal{X}_{1} & \mathcal{X}_{2} & \cdots & \mathcal{X}_{p} & &\\
& & \color{blue}{\Downarrow} & & & & & \\
X_{1}^{T} & \color{red}{\Rightarrow} & x_{11} & x_{12} & \cdots & x_{1p} & &\\
X_{2}^{T} & & x_{21} & x_{22} & \cdots & x_{2p} & &\\
\vdots & & \vdots & \vdots & & \vdots & \color{green}{\Leftarrow} & \mathbf{X}\\
X_{n}^{T} & & x_{n1} & x_{n2} & \cdots & x_{np} & &\\
};
\begin{pgfonlayer}{myback}
\highlight[red]{m-3-3}{m-3-6}
\highlight[blue]{m-3-3}{m-6-3}
\highlight[green]{m-3-3}{m-6-6}
\end{pgfonlayer}
\end{tikzpicture}
\qquad
\begin{tikzpicture}[baseline=-\the\dimexpr\fontdimen22\textfont2\relax ]
\matrix (m)[matrix of math nodes]
{
& & \mathcal{Y}_{1} & \mathcal{Y}_{2} & \cdots & \mathcal{Y}_{q} & & \\
& & & & & \color{blue}{\Downarrow} & & \\
& & y_{11} & y_{12} & \cdots & y_{1q} & & Y_{1}^{T} \\
\mathbf{Y}& \color{green}{\Rightarrow} & y_{21} & y_{22} & \cdots & y_{2q} & & Y_{2}^{T} \\
& & \vdots & \vdots & & \vdots & & \vdots \\
& & y_{n1} & y_{n2} & \cdots & y_{nq} & \color{red}{\Leftarrow} & Y_{n}^{T} \\
};
\begin{pgfonlayer}{myback}
\highlight[red]{m-6-3}{m-6-6}
\highlight[blue]{m-3-6}{m-6-6}
\highlight[green]{m-3-3}{m-6-6}
\end{pgfonlayer}
\end{tikzpicture}
\end{equation*}
Furthermore, matrices are denoted by upper case boldface letters (e.g. $\mathbf{A}$, $\mathbf{B}$). For any matrix $\mathbf A = (a_{st}) \in \mathbb{R}^{n \times n}$, we use $\widetilde{ \mathbf A} = (\tilde{a}_{st}) \in \mathbb{R}^{n \times n}$ to denote the $\mathcal U$-centered version of $\mathbf A$, i.e.,
\begin{align*}
\tilde{a}_{st} = \left\lbrace \begin{array}{ll}
a_{st} -\frac{1}{n-2} \sum_{v=1}^n a_{sv} - \frac{1}{n-2} \sum_{u=1}^n a_{ut} + \frac{1}{(n-1)(n-2)} \sum_{u,v =1}^{n} a_{uv},& s \ne t \\
0 , & s=t
\end{array} \right.
\end{align*}
Following \cite{szekely2014}, the inner product between two $\mathcal U$-centered matrices $\widetilde{\mathbf{A}}= (\tilde{a}_{st}) \in \mathbb{R}^{n \times n} $ and $ \widetilde{\mathbf{B}}= (\tilde{b}_{st}) \in \mathbb{R}^{n \times n} $ is defined as
\begin{align*}
(\widetilde{\mathbf{A}} \cdot \widetilde{\mathbf{B}} ) := \frac{1}{n(n-3)} \sum\limits_{s \neq t} \tilde{a}_{st} \tilde{b}_{st}.
\end{align*}
Next, we use $\mathbf{1}_{n}$ to denote the $n$ dimensional column vector whose entries are all equal to 1. Similarly, we use $\mathbf{0}_{n}$ to denote the $n$ dimensional column vector whose entries are all equal to 0. Finally, we use $ | \cdot | $ to denote the $L^{2}$ norm of a vector, $(X',Y')$ and $(X'',Y'')$ to be independent copies of $(X,Y)$ and $X \perp Y$ to indicate that $X$ and $Y$ are independent.
We utilize the order in probability notations such as stochastic boundedness $O_{p}$ (big O in probability), convergence in probability $o_{p}$ (small o in probability) and equivalent order $\asymp_p$, which is defined as follows: for a sequence of random variables $\{Z_s\}_{s \in \mathbb Z}$ and a sequence of numbers $\{a_s\}_{s \in \mathbb Z}$, $Z_s \asymp_p a_s$ if and only if $Z_s/a_s = O_p(1) $ and $a_s/Z_s = O_p(1) $ as $s \rightarrow \infty$. For more details about these notations, please see \cite{dasgupta2008asymptotic}.
\section{High Dimension Low Sample Size}
The analyses in this section are conducted under the HDLSS setting, i.e., the sample size $n$ is fixed and the dimensions $p \wedge q \rightarrow \infty$.
\subsection{Distance Covariance and Variants}
\label{statistics}
In this section, we introduce the following test statistics based on distance covariance ($dCov$), marginal distance covariance ($mdCov$), Hilbert-Schmidt covariance ($hCov$) and marginal Hilbert-Schmidt covariance ($mhCov$). In addition, their asymptotic behaviors under the HDLSS setting are derived. The following moment conditions will be used throughout the paper.
\begin{myassumption}{D1}\label{D1}
For any $p,q$, the variance and the second moment of any coordinate of $X = (x_{1}, x_{2}, \cdots, x_{p})^T$ and $Y = (y_{1}, y_{2}, \cdots, y_{q})^T$ is uniformly bounded below and above, i.e.,
\begin{align*}
&0 < a \leq \inf_{i} var(x_i) \leq \sup_i E (x_i^2 ) \leq b < \infty, \\
& 0 < a' \leq \inf_{j} var(y_j) \leq \sup_j E (y_j^2 ) \leq b' < \infty ,
\end{align*}
for some constants $a,b,a',b'$.
\end{myassumption}
Next, denote $\tau_X^2 =E|X-X'|^2 $, $\tau_Y^2 =E|Y-Y'|^2 $ and $\tau^2 := \tau_X^2\tau_Y^2= E|X-X'|^2 E|Y-Y'|^2$. Notice that under assumption \ref{D1}, it can be easily seen that
\begin{align*}
\tau_X \asymp \sqrt{p}, \tau_{Y} \asymp \sqrt{q} \text{ and } \tau \asymp \sqrt{pq}.
\end{align*}
The statistics we study in this work use the pair-wise $L^2$ distance between data points. The following proposition presents an expansion formula on the normalized $L^2$ distance when the dimension is high, which plays a key role in our theoretical analysis.
\begin{proposition}
\label{prop:taylor}
Under Assumption \ref{D1}, we have
\begin{align*}
\frac{|X-X'|}{\tau_X} & = 1 + \frac{1}{2} L_X(X,X') + R_X(X,X'),
\end{align*}
where
\begin{align*}
L_X(X,X') : = \frac{|X-X'|^2 - \tau_{X}^2 }{\tau_{X}^2},
\end{align*}
and $R_X(X,X') $ is the remainder term. If we further assume that as $p \wedge q \rightarrow \infty$, $L_X(X,X') =o_p(1)$, then $ R_X(X,X') = O_p ( L_X(X,X')^2)$. Similar result holds for $Y$.
\end{proposition}
In order for the approximations in equations (\ref{intro:dcov}) and (\ref{intro:hcov}) to work well, it is required that $ L_{X}(X_{s}, X_{t}) $ and $ L_{Y}(Y_{s}, Y_{t}) $ should decay relatively fast as $p \wedge q \rightarrow \infty$. The following assumption specifies the order of $ L_{X}(X_{s}, X_{t}) $ and $ L_{Y}(Y_{s}, Y_{t}) $.
\begin{myassumption}{D2}\label{D2}
$
L_{X}(X, X') = O_p (a_{p}) \text{ and } L_{Y}(Y, Y') = O_p (b_{q}),$
where $a_{p}, b_{q}$ are sequences of numbers such that
\begin{align*}
\begin{array}{c}
a_{p} = o(1), b_{q} = o(1), \\
\tau_{X}^2 a_p^3 = o(1), \tau_{Y}^2 b_{q}^3 = o(1), \tau a_{p}^2 b_{q} = o(1), \tau a_{p} b_{q}^{2} = o(1).
\end{array}
\end{align*}
\end{myassumption}
\begin{remark}
\label{remark:mainthm}
A sufficient condition for $ L_X(X,X') =o_p(1) $ is that $ E[L_X(X,X')^2]=o(1)$. Let $\bm{\Sigma}_X=\cov(X)$. By a straightforward calculation, we obtain
$|X-X'|^2=\sum_{j=1}^{p}(x_j-x_j')^2$, $E |X-X'|^2 =2\sum_{j=1}^{p} \var(x_j) = 2\tr(\bm{\Sigma}_X) $, and
\begin{align*}
E[L_{X}(X,X')^2] = \frac{\sum_{j,j'=1}^{p} [ \cov (x_{j}^2, x_{j'}^{2}) + 2 \cov^2(x_{j}, x_{j'}) ] }{ 2 \text{tr}^2(\bm{\Sigma}_{X}) }.
\end{align*}
Therefore, $E[L_X(X,X')^2]=o(1)$ holds if the component-wise dependence within $X$ is not too strong. To illustrate this point, we consider the factor model,
$$
X_{p \times 1} = \mathbf{A}_{p \times s_{1}} U_{s_{1} \times 1} + \Phi_{p \times 1}, Y_{q \times 1} = \mathbf{B}_{q \times s_{2}} V_{s_{2} \times 1} + \Psi_{q \times 1},
$$
where $\mathbf{A}, \mathbf{B}$ are constant matrices such that $\| \mathbf{A} \|_{F}^{2} = O(p^{1/2})$ and $\| \mathbf{B} \|_{F}^{2} = O(q^{1/2}) $, where $\| \cdot \|_{F}$ is the Frobenius norm. In addition, the components in $U = (u_{1}, \cdots, u_{s_{1}})^{T}$, $V = (v_{1}, \cdots, v_{s_{2}})^{T}$ are independent, $\Phi = ( \phi_{1}, \cdots, \phi_{p} )^T$ is independent of $U$ and $\Psi = (\psi_{1}, \cdots, \psi_{q})^T$ is independent of $V$. Furthermore, the $4$th moment of each component of $ U,V, \Phi, \Psi $ are bounded, i.e.
$$
\max \left\lbrace \sup\limits_{s} E[u_{s}^{4}] , \sup\limits_{t} E[v_{t}^{4}] , \sup\limits_{i} E[\phi_{i}^{4}] , \sup\limits_{j} E[ \psi_{j}^{4}]\right\rbrace< \infty.
$$
Under Assumption \ref{D1}, the above factor model satisfies Assumption \ref{D2} with $a_{p} = 1/ \sqrt{p} \text{ and } b_q = 1/ \sqrt{q},$ see Section \ref{App:proofRemark} of Appendix for more details.
\end{remark}
\subsubsection{Distance Covariance}
\label{dCov}
Distance covariance was first introduced by \cite{szekely2007} to measure the dependence between two random vectors of arbitrary dimensions. For two random vectors $X \in \mathbb{R}^p$ and $Y\in \mathbb{R}^q$, the (squared) distance covariance is defined as
\[
dCov^2(X,Y) = \int_{\mathbb{R}^{p+q}} \frac{|\phi_{X,Y}(t,s)-\phi_X(t)\phi_Y(s)|^2}{c_pc_q|t|^{1+p} |s|^{1+q}}dtds,
\]
where $c_p=\pi^{(1+p)/2}/\Gamma((1+p)/2)$, $|\cdot| $ is the (complex) Euclidean norm defined as $|x| = \sqrt {\bar x^T x}$ for any vector $x$ in the complex vector space (
$\bar x$ denotes the conjugate of $x$), $\phi_X$ and $\phi_Y$ are the characteristic functions of $X$ and $Y$ respectively, $\phi_{X,Y}$ is the joint characteristic function. According to Theorem 7 of \cite{szekely2009}, an alternative definition of distance covariance is given by
\begin{multline}
\label{def:alt}
dCov^2(X,Y) = E|X-X'||Y-Y'| \\ + E|X-X'|E|Y-Y'| - 2 E|X- X'||Y-Y''|,
\end{multline}
where $(X',Y')$ and $(X'',Y'')$ are independent copies of $(X,Y)$. It has been shown that $dCov^2(X,Y)=0$ if and only if $X$ and $Y$ are independent. Therefore, it is able to measure any type of dependence including non-linear and non-monotonic dependence between $X$ and $Y$, whereas the commonly used Pearson correlation can only measure the linear dependence and the rank correlation coefficients (Kendall's $\tau$ and Spearman's $\rho$) can only capture the monotonic dependence.
Notice that in the above setting, $p,q$ are arbitrary positive integers. Therefore, distance covariance is applicable to the high dimensional setting, where we allow $p,q \rightarrow \infty.$ However, it is unclear whether this metric can still retain the power to detect the nonlinear dependence or not when the dimension is high. Distance correlation ($dCor$) is the normalized version of distance covariance, which is defined as
\begin{align*}
dCor^2(X,Y) = \left\lbrace \begin{array}{ll}
\frac{dCov^2(X,Y)}{\sqrt{dCov^2(X,X) dCov^2(Y,Y) }}, & dCov^2(X,X) dCov^2(Y,Y) >0, \\
0, & dCov^2(X,X) dCov^2(Y,Y) = 0.
\end{array} \right.
\end{align*}
Following \cite{szekely2014}, we introduce the $\mathcal U$-centering based unbiased sample distance covariance ($dCov_n^2$) as follows.
\begin{align*}
dCov_n^2(\mathbf{X},\mathbf{Y}) = (\widetilde{\mathbf{A}} \cdot \widetilde{\mathbf{B}} ),
\end{align*}
where $\widetilde{\mathbf{A}}, \widetilde{\mathbf{B}}$ are the $\mathcal U$-centered versions of $\mathbf{A} = (a_{st})_{s,t=1}^n, \mathbf{B} = (b_{st})_{s,t=1}^n$ respectively and $a_{st}= |X_s - X_t|, b_{st}= |Y_s - Y_t|$ for $s,t=1, \cdots ,n$. Correspondingly, the sample distance correlation ($dCor^2_n$) is given as
\begin{align*}
dCor^2_n(\mathbf{X},\mathbf{Y}) = \left\lbrace \begin{array}{ll}
\frac{dCov^2_n(\mathbf{X},\mathbf{Y})}{\sqrt{dCov^2_n(\mathbf{X},\mathbf{X}) dCov^2_n(\mathbf{Y},\mathbf{Y}) }}, & dCov^2_n(\mathbf{X},\mathbf{X}) dCov^2_n(\mathbf{Y},\mathbf{Y}) >0, \\
0, & dCov^2_n(\mathbf{X},\mathbf{X}) dCov^2_n(\mathbf{Y},\mathbf{Y}) = 0.
\end{array} \right.
\end{align*}
Here we can apply the approximation in Proposition \ref{prop:taylor}, that is
\begin{align}
\label{l2:decomp}
& \frac{a_{st}}{\tau_X} = 1 + \frac{1}{2} L_{X}(X_s, X_t ) + R_{X}(X_s, X_t ),\\
& \frac{b_{st}}{\tau_Y} = 1 + \frac{1}{2} L_{Y}(Y_s, Y_t ) + R_{Y}(Y_s, Y_t ), \label{l2:decomp2}
\end{align}
where
\begin{align*}
L_{X}(X_s, X_t ) = \frac{|X_s - X_t|^2 -\tau_{X}^2 }{\tau_{X}^2 }, \;
L_{Y}(Y_s, Y_t ) = \frac{|Y_s - Y_t|^2 -\tau_{Y}^2 }{\tau_{Y}^2 },
\end{align*}
and $R_{X}$, $R_{Y}$ are the remainder terms from the approximation. The approximation of the pair-wise $L^2$ distance in Equations \eqref{l2:decomp} and \eqref{l2:decomp2} is our building block to decompose the unbiased sample (squared) distance covariance ($dCov_n^2$) into a leading term plus a negligible remainder term under the HDLSS setting. The following main theorem summarizes the decomposition properties of sample distance covariance ($dCov_n^2$).
\begin{thm}
\label{thm:decomp}
Under Assumption \ref{D1}, we can show that
\begin{itemize}
\item[(i)] \begin{align}
\label{eq:decomp}
dCov^2_n(\mathbf{X},\mathbf{Y}) = \frac{1}{\tau} \sum_{i=1}^p \sum_{j=1}^q \cov_n^2 \left(\mathcal X_i, \mathcal Y_j \right) + {\cal R}_n.
\end{align}
Here
\[
\cov_n^2 \left( \mathcal X_i, \mathcal Y_j \right) = \frac{ 1 }{\binom{n}{4}} \sum_{s< t< u < v } h(x_{si} , x_{ti}, x_{ui} , x_{vi} ; y_{sj} , y_{tj} , y_{uj} , y_{vj} ),
\]
and the kernel $h$ is defined as
\begin{multline*}
h(x_{si} , x_{ti}, x_{ui} , x_{vi} ; y_{sj}, y_{tj} , y_{uj} , y_{vj}) \\ = \frac{1}{4!} \sum_{ * }^{(s, t , u, v)} \frac{1}{4} (x_{si} - x_{ti}) (y_{sj} - y_{tj})(x_{ui} - x_{vi}) (y_{uj} - y_{vj}),
\end{multline*}
where the summation $\sum_{ * }^{(s, t , u, v )}$ is over all permutations of the 4-tuples of indices $(s,t,u,v)$ and ${\cal R}_n $ is the remainder term. $\cov_n^2\left(\mathcal X_i, \mathcal Y_j \right) $
is a fourth-order U-statistic and is an unbiased estimator for the squared covariance between $x_{i}$ and $y_{j}$, i.e., $E[ \cov_n^2\left(\mathcal X_i, \mathcal Y_j \right) ] = cov^2(x_{i}, y_{j})$.
\item[(ii)] Further suppose Assumption \ref{D2} holds. Then
\begin{align*}
& \frac{1}{\tau} \sum_{i=1}^p \sum_{j=1}^q \cov_n^2 \left(\mathcal X_i, \mathcal Y_j \right) = O_p( \tau a_p b_q ), \\
& {\cal R}_n = O_p \left( \tau a_p^2 b_q + \tau a_p b_q^2 \right) = o_p (1),
\end{align*}
thus the remainder term is of smaller order compared to the leading term and therefore is asymptotically negligible.
\end{itemize}
\end{thm}
Equation (\ref{eq:decomp}) in Theorem \ref{thm:decomp} shows that the leading term for sample distance covariance is the sum of all component-wise squared sample cross-covariances scaled by $\tau$, which depends on the marginal variances of $X$ and $Y$. This theorem suggests that in the HDLSS setting, the sample distance covariance can only measure the component-wise linear dependence between the two random vectors.
\begin{remark}
It is worth mentioning that the distance correlation and RV coefficient, introduced by \cite{escoufier1970echantillonnage} (see also \cite{josse2013measures}), are asymptotically equal in the HDLSS setting, where RV coefficient is another metric for quantifying the association between two random vectors and is defined as,
\begin{align*}
RV(X,Y) = \frac{\sum_{i=1}^{p} \sum_{j=1}^{q} \text{cov}^2 \left( x_{i}, y_{j} \right) }{\sqrt{ \left( \sum_{i,j=1}^{p} \text{cov}^2 ( x_{i}, x_{j}) \right) \left( \sum_{i,j=1}^{q} \text{cov}^2 ( y_{i}, y_{j}) \right) } }.
\end{align*}
If we use $cov_{n}^2 (\mathcal X_i, \mathcal Y_j)$ to estimate $cov^2(x_i, y_j)$, its sample version can be written as
\begin{align*}
RV_n(\mathbf X, \mathbf Y) = \frac{\sum_{i=1}^{p} \sum_{j=1}^{q} \text{cov}_{n}^2 \left( \mathcal X_{i}, \mathcal Y_{j} \right) }{\sqrt{ \left( \sum_{i,j=1}^{p} \text{cov}_{n}^2 (\mathcal X_{i}, \mathcal X_{j}) \right) \left( \sum_{i,j=1}^{q} \text{cov}_{n}^2 (\mathcal Y_{i}, \mathcal Y_{j}) \right) } }.
\end{align*}
By taking the limit with respect to $p \wedge q$, under Assumptions \ref{D1}, \ref{D2}, \ref{D3} (which will be introduced later) and using Theorem \ref{thm:decomp}, we have
\begin{align*}
& \text{dCor}_{n}^{2}(\mathbf X, \mathbf Y) \\
= & \frac{ \frac{1}{\tau} \sum_{i=1}^{p} \sum_{j=1}^{q} \text{cov}_{n}^2 (\mathcal X_{i}, \mathcal Y_{j}) \left(1+ o_p(1) \right)}{\sqrt{ \left( \frac{1}{\tau_{X}^2} \sum_{i,j=1}^{p} \text{cov}_{n}^2 (\mathcal X_{i}, \mathcal X_{j}) \right) \left( \frac{1}{\tau_{Y}^2} \sum_{i,j=1}^{q} \text{cov}_{n}^2 ( \mathcal Y_{i}, \mathcal Y_{j}) \right) } } \\
= & \frac{ \sum_{i=1}^{p} \sum_{j=1}^{q} \text{cov}_{n}^2 (\mathcal X_{i}, \mathcal Y_{j}) (1 + o_p(1)) }{\sqrt{ \left( \sum_{i,j=1}^{p} \text{cov}_{n}^2 (\mathcal X_{i}, \mathcal X_{j}) \right) \left( \sum_{i,j=1}^{q} \text{cov}_{n}^2 (\mathcal Y_{i}, \mathcal Y_{j}) \right) } } \\
= & RV_n(\mathbf X, \mathbf Y)(1 + o_p(1)),
\end{align*}
which shows that the squared sample distance correlation and sample RV coefficient are approximately equal when the dimension is high. Consequently, they have the same limiting distribution as $p\wedge q$ goes to infinity. Since it is known from \cite{szekely2013} that the studentized version of sample distance covariance has a limiting $t$-distribution, it is expected that the studentized RV coefficient has the same limiting $t$-distribution as well.
\end{remark}
As argued previously, sample distance covariance ($dCov_n^2$) based tests suffer from power loss when $X$ and $Y$ are component-wisely non-linear dependent but uncorrelated. To remedy this drawback, we can consider the following aggregation of marginal sample distance covariances,
\[ mdCov_n^2 (\mathbf X,\mathbf Y) = \sqrt{\binom{n}{2}}\sum_{i=1}^p \sum_{j=1}^q dCov_n^2(\mathcal X_i, \mathcal Y_j), \]
where $dCov_n^2(\mathcal X_i, \mathcal Y_j) = ( \widetilde{\mathbf{A}}(i) \cdot \widetilde{\mathbf{B}}(j) )$, $\widetilde{\mathbf{A}}(i)$ and $\widetilde{\mathbf{B}}(j)$ are the $\mathcal U$-centered versions of $\mathbf{A}(i) = (a_{st}(i))_{s,t=1}^n, \mathbf{B}(j) = (b_{st}(j))_{s,t=1}^n$ respectively and $a_{st}(i) = |x_{si}-x_{ti}|$, $b_{st}(j) = |y_{sj}-y_{tj}|$.
Note that $mdCov_n^2$ captures the pairwise low dimensional nonlinear dependence, which can be viewed as the main effects of the dependence between two high dimensional random vectors. It is natural in many fields of statistics to test for main effects first before proceeding to high order interactions. See \cite{chakraborty2018} for some discussions on main effects and high order effects in the context of joint dependence testing.
In the testing of mutual independence of a high dimensional vector, \cite{yao2017testing} also approached the problem by testing the pairwise independence using distance covariance and demonstrated that there may be intrinsic difficulty to capture the effects beyond main effects (pairwise dependence in the mutual independence testing problem), as the tests that target joint dependence do not perform well in the high dimensional setting.
\subsubsection{Hilbert-Schmidt Covariance}
\label{hCov}
A generalization of the Distance Covariance ($dCov$) is Hilbert-Schmidt Covariance ($hCov$), first proposed and aka Hilbert-Schmidt independence criterion ($HSIC$) by \cite{gretton2007}. In particular, the (squared) Hilbert-Schmidt Covariance ($hCov$) is obtained by kernelizing the Euclidean distance in equation \eqref{def:alt}, i.e.,
\begin{multline*}
hCov^{2} (X,Y) = E [ K(X, X') L(Y,Y') ] \\ + E [ K(X, X')] E [ L(Y,Y') ] - 2 E [ K(X,X') L(Y,Y'') ],
\end{multline*}
where $(X',Y'), (X'',Y'')$ are independent copies of $(X,Y)$ and $K, L$ are user specified kernels. Following the literature, we consider the following widely used kernels
$$ \left.
\begin{array}{l}
\text{Gaussian kernel: } K(x,y) = \exp \left(- \frac{|x-y|^2}{ 2\gamma^2} \right), \\
\text{Laplacian kernel: } K(x,y) = \exp \left( -\frac{|x-y|}{ \gamma} \right),
\end{array} \right.
$$
where $\gamma$ is a bandwidth parameter. For later convenience, we focus on the kernels that can be represented compactly as $ K(x,y) = f \left( |x-y|/ \gamma \right)$ for some continuously differentiable function $ f$. For example, the Gaussian and Laplacian kernel can be defined by choosing different function ${f}$,
$$ \left.
\begin{array}{l}
\text{Gaussian kernel: } {K}(x,y) = f \left( \frac{|x-y|}{ \gamma} \right), f(a) = \exp \left( - \frac{a^2}{ 2 } \right), \\
\text{Laplacian kernel: } {K}(x,y) = f \left( \frac{|x-y|}{ \gamma} \right), f(a) = \exp \left( - a \right).
\end{array} \right.
$$
In practice, the bandwidth parameter is usually set as the median of pair-wise sample $L^2$ distance. Thus, a natural estimator for $hCov^2 (X,Y)$ is defined as
\[
hCov_n^2(\mathbf X, \mathbf Y) = ( \widetilde{\mathbf R} \cdot \widetilde{\mathbf H}),
\]
where $\widetilde{\mathbf{R}}$ and $\widetilde{\mathbf{H}}$ are the $\mathcal U$-centered versions of $\mathbf{R} = (r_{st})_{s,t=1}^n, \mathbf{H} = (h_{st})_{s,t=1}^n$ respectively and
\begin{align*} \left\lbrace
\begin{array}{l}
r_{st} = K(X_{s}, X_{t}, \mathbf X) = f \left( \frac{|X_{s} - X_{t}|}{ \gamma_{\mathbf X}} \right), \gamma_{\mathbf X} = \text{median} \{ |X_{s} - X_{t}|, s \neq t \},\\
h_{st} = L(Y_{s}, Y_{t}, \mathbf{Y}) = g \left( \frac{|Y_{s} - Y_{t}|}{\gamma_{\mathbf Y}} \right), \gamma_{\mathbf Y} = \text{median} \{ |Y_{s} - Y_{t}|, s \neq t \}.
\end{array} \right.
\end{align*}
Similar to the definition of distance correlation, the Hilbert-Schmidt Correlation ($hCor$) is defined as
\begin{align*}
hCor^2(X,Y) = \left\lbrace \begin{array}{ll}
\frac{hCov^2(X,Y)}{\sqrt{hCov^2(X,X) hCov^2(Y,Y) }}, & hCov^2(X,X) hCov^2(Y,Y) >0, \\
0, & hCov^2(X,X) hCov^2(Y,Y) = 0,
\end{array} \right.
\end{align*}
and the sample Hilbert-Schmidt Correlation ($hCor_n^2$) is defined in the same way by
replacing $hCov^2$ with the corresponding sample version.
Next, we can extend the decomposition results for sample distance covariance ($dCov_n^2$) to sample Hilbert-Schmidt covariance ($hCov_n^2$) as shown in the following theorem.
\begin{thm}
\label{thm:decompHsic}
Under Assumption \ref{D1}, we have
\begin{itemize}
\item[(i)]
\begin{multline}
\label{eq:decompHSIC}
\tau \times hCov^2_{n}(\mathbf X, \mathbf Y) \\ = f^{(1)} \left( \frac{\tau_{X}}{\gamma_{\mathbf X}} \right) g^{(1)} \left( \frac{\tau_{Y}}{ \gamma_{\mathbf Y}} \right) \frac{\tau_{X}}{\gamma_{\mathbf X}} \frac{\tau_{Y}}{\gamma_{ \mathbf Y}} \frac{1}{\tau} \sum\limits_{i=1}^{p} \sum\limits_{j=1}^{q} \text{cov}_{n}^2 (\mathcal X_{i}, \mathcal Y_{j}) + \mathcal{R}_{n},
\end{multline}
where $ \text{cov}_{n}^2 $ is defined the same as in Theorem \ref{thm:decomp} and $\mathcal{R}_{n}$ is the remainder term.
\item[(ii)] Further suppose Assumption \ref{D2} holds. Then
\begin{align*}
& f^{(1)} \left( \frac{\tau_{X}}{\gamma_{\mathbf X}} \right) g^{(1)} \left( \frac{\tau_{Y}}{ \gamma_{\mathbf Y}} \right) \frac{\tau_{X}}{\gamma_{\mathbf X}} \frac{\tau_{Y}}{\gamma_{ \mathbf Y}} \asymp_p 1, \\
& \frac{1}{\tau} \sum_{i=1}^p \sum_{j=1}^q \cov_n^2 \left(\mathcal X_i, \mathcal Y_j \right) = O_p (\tau a_p b_q), \\
& {\cal R}_n = O_p(\tau a_p^2b_q + \tau a_pb_q^2) = o_p (1).
\end{align*}
Thus the remainder term is of smaller order compared to the leading term and is therefore asymptotically negligible.
\end{itemize}
\end{thm}
Notice that different from the decomposition of $dCov_n^2(\mathbf X,\mathbf Y)$ as in Theorem \ref{thm:decomp}, here we decompose $hCov_n^2$ multiplied by $\tau=\tau_{X} \tau_{Y}$. This is expected, since in $hCov_n^2$, each pair-wise distance is normalized by $\gamma_{\mathbf X}$ or $\gamma_{\mathbf Y}$, which has asymptotically the same magnitude as $\tau_{X}$, $\tau_{Y}$ respectively.
In the high dimensional case, the expansion \eqref{eq:decompHSIC} suggests that $hCov$-based tests also suffer from power loss when $X$ and $Y$ are component-wisely uncorrelated but nonlinearly dependent.
To analyze the asymptotic property of sample Hilbert-Schmidt covariance, most literature would assume the bandwidth parameters to be fixed constants, see e.g. \cite{gretton2007}. In contrast, our approach can handle the case where these bandwidth parameters are selected to be the median of pairwise sample distance, which is random and whose magnitude increases with dimension.
Similar to the marginal distance covariance introduced in Section \ref{dCov}, we can also aggregate the marginal Hilbert-Schmidt Covariance ($mhCov$), which is defined as
\[ mhCov_n^2 (\mathbf X, \mathbf Y) = \sqrt{\binom{n}{2}}\sum_{i=1}^p \sum_{j=1}^q hCov_n^2( \mathcal X_i,\mathcal Y_j) \]
where $hCov_n^2(\mathcal X_i, \mathcal Y_j) = (\widetilde{\mathbf R}(i) \cdot \widetilde{\mathbf H}(j))$, $\widetilde{\mathbf R}(i)$ and $\widetilde{\mathbf H}(j)$ are $\mathcal U$-centered version of $\mathbf R (i)= (r_{st}(i))_{s,t=1}^n$, $\mathbf H(j) = (h_{st}(j))_{s,t=1}^n$ respectively and
\begin{align*} \left\lbrace
\begin{array}{l}
r_{st}(i) = K(x_{si}, x_{ti}, \mathcal{X}_i) = f \left( \frac{|x_{si} - x_{ti}|}{ \gamma_{\mathcal{X}_i}} \right), \gamma_{\mathcal X_{i}} = \text{median} \{ |x_{si} - x_{ti}|, s \neq t \},\\
h_{st}(j) = L(y_{sj}, y_{tj}, \mathcal{Y}_j) = g \left( \frac{|y_{sj} - y_{tj}|}{\gamma_{\mathcal Y_j}} \right), \gamma_{\mathcal Y_{j}} = \text{median} \{ |y_{sj} - y_{tj}|, s \neq t \}.
\end{array} \right.
\end{align*}
\subsection{Studentized Test Statistics}
\label{student}
In this section, we provide studentized version of the statistics introduced in Section \ref{statistics}. It is worth mentioning that we provide a unified approach to the asymptotic analysis of studentized $dCov, mdCov$ and further extend them to the analysis of studentized $hCov$.
\subsubsection{Unified Approach}
\label{sec:uni}
Firstly, we will present results that will be useful for deriving the studentized version of the interested statistics, i.e. distance covariance ($dCov$), marginal distance covariance ($mdCov$), Hilbert-Schmidt Covariance ($hCov$), marginal Hilbert-Schmidt Covariance ($mhCov$). It can be shown later that many previously mentioned statistics are asymptotically equal to the unified quantity $uCov^2_n(\mathbf X, \mathbf Y)$ multiplied by some normalizing factor. Here, $uCov^2_n(\mathbf X, \mathbf Y)$ is defined as
\begin{align*}
uCov^2_{n}(\mathbf X, \mathbf Y) = \frac{1}{\sqrt{pq}}\sum^{p}_{i=1}\sum^{q}_{j=1} ( \widetilde{\mathbf K}(i) \cdot \widetilde{\mathbf L}(j) ),
\end{align*}
where $\widetilde{\mathbf K}(i)$ and $\widetilde{\mathbf L}(j)$ are the $\mathcal U$-centered versions of $\mathbf K(i) = (k_{st}(i))_{s,t=1}^n ,\mathbf L (j) = (l_{st}(j))_{s,t=1}^n $ respectively and $k_{st}(i)$, $l_{st}(i)$ are the double centered kernel distances, i.e., for bivariate kernels $k$ and $l$,
\begin{align*}
&k_{st}(i)=k(x_{si},x_{ti})-E[k(x_{si},x_{ti})|x_{si}]-E[k(x_{si},x_{ti})|x_{ti}]+E[k(x_{si},x_{ti})], \\
&l_{st}(i)=l(y_{si},y_{ti})-E[l(y_{si},y_{ti})|y_{si}]-E[l(y_{si},y_{ti})|y_{ti}]+E[l(y_{si},y_{ti})].
\end{align*}
The advantage of using the double centering kernel distance is that we can have 0 covariance between $k_{st}(i)$ and $k_{uv}(j)$ ($l_{st}(i)$ and $l_{uv}(j)$) for $ \{ s,t \} \neq \{ u,v \}$ as shown in the following proposition.
\begin{proposition}
\label{prop:ind}
For all $1 \leq i, i' \leq q, 1\leq j, j' \leq p, \text{ if } \{ s, t \} \neq \{ u, v \}$, then
\begin{align*}
E[ k_{st}(i) k_{uv}(i') ] = E[ l_{st}(j) l_{uv}(j') ] = E[ k_{st}(i) l_{uv}(j) ] = 0.
\end{align*}
\end{proposition}
To derive the limiting distribution of the unified quantity, we need the following assumptions.
\begin{myassumption}{D3} \label{D3} For fixed $n$, as $p \wedge q \rightarrow \infty$,
\begin{align*}
\left(
\begin{array}{c}
p^{-1/2}\sum^{p}_{i=1}k_{st}(i) \\
q^{-1/2}\sum^{q}_{j=1}l_{uv}(j)
\end{array} \right)_{ s < t , u < v } \overset{d}{\rightarrow} \left(
\begin{array}{c}
c_{st} \\
d_{uv}
\end{array} \right)_{ s < t , u < v },
\end{align*}
where $\{ c_{st}, d_{uv} \}_{s<t,u<v}$ are jointly Gaussian. Naturally, we further assume the existence of the following constants that show up in the covariance matrix of $\{ c_{st}, d_{uv} \}$,
\begin{align*}
\begin{array}{rl}
var[c_{st}] := & \sigma_{x}^2 \\
=& \lim\limits_{p} \frac{1}{p} \sum\limits_{i,j=1}^p cov[ k_{st}(i), k_{st}(j)] \\ = & \left\lbrace
\begin{array}{ll}
\lim\limits_{p} \frac{\sum\limits_{i,j=1}^p dCov^2(x_{i}, x_{j})}{p} , & \text{ if } k(x,y)= l(x, y) = |x-y|, \\
\lim\limits_{p} \frac{\sum\limits_{i,j=1}^p 4 cov^2 (x_{i}, x_{j})}{p} , & \text{ if } k(x,y) = l(x,y)= |x-y|^2 ,
\end{array} \right. \\
var[d_{st}] : = & \sigma_{y}^2 \\
= & \lim\limits_{q} \frac{1}{q} \sum\limits_{i,j=1}^q cov[ l_{st}(i), l_{st}(j)] \\
= & \left\lbrace
\begin{array}{ll}
\lim\limits_{q} \frac{\sum\limits_{i,j=1}^q dCov^2(y_{i}, y_{j})}{q} , & \text{ if } k(x,y)= l(x,y)= |x-y|, \\
\lim\limits_{q} \frac{\sum\limits_{i,j=1}^q 4 cov^2 (y_{i}, y_{j})}{q} , & \text{ if } k(x,y)=l(x,y)= |x-y|^2 ,
\end{array} \right. \\
cov[c_{st}, d_{st}] := & \sigma_{xy}^2 \\
= & \lim\limits_{p,q} \frac{1}{\sqrt{pq}} \sum\limits_{i=1}^p \sum\limits_{j=1}^q cov[ k_{st}(i), l_{st}(j)]
\\ = & \left\lbrace
\begin{array}{ll}
\lim\limits_{p,q} \frac{\sum\limits_{i=1}^p \sum\limits_{j=1}^q dCov^2(x_{i}, y_{j})}{\sqrt{pq}} , & \text{ if } k(x,y)=l(x,y)= |x-y|, \\
\lim\limits_{p,q} \frac{\sum\limits_{i=1}^p \sum\limits_{j=1}^q 4 cov^2 (x_{i}, y_{j})}{\sqrt{pq}} , & \text{ if } k(x,y)= l(x,y)= |x-y|^2.
\end{array} \right.
\end{array}
\end{align*}
\end{myassumption}
\begin{remark}
Notice that when $ \{ s,t \} \neq \{ u,v \}$, we do not assume the form of $cov[c_{st}, c_{uv}]$, $cov[d_{st}, d_{uv}]$, $cov[c_{st}, d_{uv}]$ in Assumption \ref{D3}, since it follows easily from Proposition \ref{prop:ind} that $cov[c_{st}, c_{uv}] =0, cov[d_{st}, d_{uv}] =0$ and $cov[c_{st}, d_{uv}] =0$ if $ \{ s,t \} \neq \{ u,v \}.$
\end{remark}
\begin{remark}
The above Central Limit Theorem (CLT) result can be derived under suitable moment and weak dependence assumptions for the components of $X$ and $Y$. We refer the reader to \cite{doukhan2008} for a relatively recent survey of weak dependence notions and the CLT results under such weak dependence.
\end{remark}
The following theorem is our main result, which shows that the unified quantity converges in distribution to a quadratic form of random variables.
\begin{thm}
\label{thm:key}
Fixing $n$ and let $p \wedge q \rightarrow \infty$, under Assumptions \ref{D1} and \ref{D3},
\begin{align*}
& uCov^2_{n}(\mathbf X,\mathbf Y) \overset{d}{\rightarrow} \frac{1}{v} \mathbf{c}^T \mathbf{M} \mathbf{d} ,\\
& uCov^2_{n}(\mathbf X,\mathbf X) \overset{d}{\rightarrow} \frac{1}{v} \mathbf{c}^T \mathbf{M} \mathbf{c} \overset{d}{=} \frac{\sigma^2_x}{v} \chi^2_{v},\\
& uCov^2_{n}(\mathbf Y, \mathbf Y) \overset{d}{\rightarrow} \frac{1}{v} \mathbf{d}^T \mathbf M \mathbf{d} \overset{d}{=} \frac{\sigma^2_y}{v} \chi^2_{v},
\end{align*}
where $v:= n(n-3)/2$, $ \mathbf M$ is a projection matrix of rank $v$ and
\begin{align*} \left(
\begin{array}{c}
\mathbf{c} \\
\mathbf{d}
\end{array} \right) \overset{d}{=} N \left(\mathbf 0, \left(
\begin{array}{cc}
\sigma_{x}^2 \mathbf{I}_{n(n-1)/2} & \sigma_{xy}^2 \mathbf{I}_{n(n-1)/2} \\
\sigma_{xy}^2 \mathbf{I}_{n(n-1)/2} & \sigma_{y}^2 \mathbf{I}_{n(n-1)/2}
\end{array} \right)
\right).
\end{align*}
\end{thm}
\begin{remark}
For the exact form of $\mathbf M$, see the proof of Theorem \ref{thm:key} in the Appendix.
\end{remark}
Next, we define the quantity $T_u$ as
\begin{align*}
T_u = \sqrt{v-1} \frac{uCor_n^2(\mathbf X, \mathbf Y)}{ \sqrt{1 - (uCor_n^2(\mathbf X, \mathbf Y))^{2} } },
\end{align*}
where
\begin{align*}
uCor_n^2(\mathbf X,\mathbf Y) = \frac{ uCov_n^2(\mathbf X,\mathbf Y) }{\sqrt{ uCov_n^2(\mathbf X,\mathbf X) uCov_n^2(\mathbf Y,\mathbf Y)} }.
\end{align*}
We then define the constants $v$ and $\phi$ that appear in the limiting distribution of $T_u$. Set $v = n(n-3)/2$ and $ \phi = \sigma_{xy}^2 / \sqrt{\sigma_{x}^2\sigma_{y}^2} $ such that
\begin{align*}
\phi = \phi_{1}\mathbb{I}_{\{ k(x,y)= l(x,y)= |x-y| \}} + \phi_{2}\mathbb{I}_{\{ k(x,y)= l(x,y)= |x-y|^2 \}},
\end{align*}
where
\begin{align*}
\phi_{1} := & \lim_{p,q} \frac{ \sum_{i=1}^p \sum_{j=1}^q dCov^2(x_{i}, y_{j})}{ \sqrt{ \sum_{i,j=1}^p dCov^2(x_{i}, x_{j}) \sum_{i,j=1}^q dCov^2(y_{i}, y_{j}) }}, \\
\phi_{2} := & \lim_{p,q} \frac{ \sum_{i=1}^p \sum_{j=1}^{q} cov^2 (x_{i}, y_{j}) }{ \sqrt{ \sum_{i,j=1}^p cov^2 (x_{i}, x_{j}) \sum_{i,j=1}^q cov^2 (y_{i}, y_{j}) } }.
\end{align*}
The limiting distribution of $T_u$ is derived under both null ($H_0$) and alternative $(H_{A})$ hypothesis, i.e.,
$$
\begin{array}{rl}
\text{null hypothesis}: & H_0 = \left\lbrace (X,Y) \; | \; X \perp Y \right\rbrace , \\
\text{alternative hypothesis}: & H_A = \{ (X,Y) \; | \; X \not\perp Y \}.
\end{array}
$$
In addition, we also consider the local alternative hypothesis $H_{A_l} \subset H_A$, i.e.,
\begin{align*}
H_{A_{l}} = \left\lbrace (X,Y) \; \left| \; X \not\perp Y, \phi = \frac{ \phi_{0}}{\sqrt{v}} \right. \right\rbrace ,
\end{align*}
where $v = n(n-3)/2$, $\phi_{0} = \phi_{0,1} \mathbb{I}_{\{ k(x,y)= l(x,y)= |x-y| \}} + \phi_{0,2} \mathbb{I}_{\{k(x,y)= l(x,y)= |x-y|^2\}}$ and $ 0< \phi_{0,1}, \phi_{0,2}< \infty\ $ are constants with respect to $n$. It is also insteresting to compare the asymptotic power under the following class of alternatives $H_{A_{s}} \subset H_A$, i.e.,
\begin{align*}
H_{A_{s}} = \{ (X,Y) \; | \; x_i \not\perp y_j, cov(x_i, y_j) = 0 \text{ for all } 1\leq i \leq p, 1 \leq j \leq q \}.
\end{align*}
In summary, the following table illustrates the value of $\phi$ under different cases we are considering,
$$
\begin{array}{c|cccc}
\phi & H_0 & H_A & H_{A_{l}} & H_{A_{s}} \\ \hline
k(x,y)= l(x,y)= |x-y| & 0 & \phi_1 &\frac{\phi_{0,1}}{\sqrt{v}} & \phi_1 \\
k(x,y)= l(x,y)= |x-y|^2 & 0 & \phi_2 & \frac{\phi_{0,2}}{\sqrt{v}} & 0
\end{array}
$$
Next, let $t_a$ denote the student $t$-distribution with degrees of freedom $a$, $t_{a}^{(\alpha)}$ denotes the $(1-\alpha)$th percentile of $t_a$, $t_{a,b}$ denotes the non-central $t$-distribution with degrees of freedom $a$ and non-central parameter $b$. The asymptotic distribution of $T_u$ is stated in the following proposition.
\begin{proposition}\label{prop:exactT}
Fix $n$ and let $p \wedge q \rightarrow \infty$. If Assumptions \ref{D1} and \ref{D3} hold, then for any fixed $t \in \mathbb{R}$,
\begin{align*}
& P_{H_0}(T_u \leq t ) \rightarrow P(t_{v-1} \leq t), \\
& P_{H_A}(T_u \leq t ) \rightarrow E \left[ P \left( t_{v-1, W} \leq t \right)\right],
\end{align*}
where $W \sim \sqrt{\frac{\phi^2}{1 - \phi^2}\chi_{v}^2} $ and $\chi_{v}^2$ is the chi-square distribution with degrees of freedom $v$.
\end{proposition}
\begin{remark}
For the explicit form of $ E \left[ P \left( t_{v-1, W} \leq t \right)\right] $, see Lemma \ref{lem:exact} in the Appendix.
\end{remark}
Below we derive the large sample approximation of the limiting distribution $ E \left[ P \left( t_{v-1, W} \leq t \right)\right] $ under the local alternative hypothesis ($H_{A_l}$).
\begin{proposition}
\label{prop:LarA}
Under $H_{A_l}$, if we allow $n$ to grow and $t$ is bounded as $n \rightarrow \infty$, $E \left[ P \left( t_{v-1, W} \leq t \right)\right]$ can be approximated as
\begin{align*}
E_{H_{A_l}} \left[ P \left( t_{v-1, W} \leq t \right)\right] = P \left( t_{v-1, \phi_{0}} \leq t \right) + O \left(\frac{1}{v} \right),
\end{align*}
where $\phi_0 = \phi_{0,1}\mathbb{I}_{\{ k(x,y)= l(x,y)= |x-y| \}} + \phi_{0,2}\mathbb{I}_{\{k(x,y)= l(x,y)= |x-y|^2\}}$. In particular, the result still holds if we replace $t$ with $t_{v-1}^{(\alpha)}$.
\end{proposition}
\subsubsection{Studentized Tests}
For testing the null, permutation test can be used to determine the critical value of the distance covariance ($dCov$), Hilbert-Schmidt covariance ($hCov$), marginal distance covariance ($mdCov$) and marginal Hilbert-Schmidt covariance ($mhCov$) respectively. If $dCov_n^2$, $hCov_n^2$, $mdCov_n^2$ or $mhCov_n^2$ is larger than the corresponding critical value, which can be determined by the empirical permutation distribution function, we reject the null. Alternatively, similar to the construction of $ T_u $, we transform each of $dCov_n^2, hCov_n^2, mdCov_n^2$ and $ mhCov_n^2 $ into a statistic that has asymptotic $t$-distribution under the null. Thus, instead of using permutation test, which can be quite computationally expensive, we can determine the critical value using this asymptotic $t$-distribution. For each $R \in \{ dCov, hCov, mdCov, mhCov \}$, the studentized test statistic $T_R$ is defined as
\begin{align*}
T_{R} & = \sqrt{v-1} \frac{ R^*(\mathbf X, \mathbf Y)}{ \sqrt{1 - ( R^*(\mathbf X,\mathbf Y))^{2} } },
\end{align*}
where
\begin{align*}
R^*(\mathbf X,\mathbf Y) = \frac{ R_n^2(\mathbf X,\mathbf Y) }{\sqrt{ R_n^2(\mathbf X, \mathbf X) R_n^2(\mathbf Y,\mathbf Y)} }.
\end{align*}
The way to derive the asymptotic distribution of $T_{R}$ is to show that for each $R \in \{dCov, hCov, mdCov \} $, $R_n^2(\mathbf{X}, \mathbf{Y})$ and $ uCov^2_n(\mathbf{X}, \mathbf{Y}) $ are asymptotically equal up to an asymptotically constant factor, as shown below.
\begin{proposition}
\label{prop:uni}
Under Assumption \ref{D1},
\begin{itemize}
\item[(i)] When $k(x,y)= l(x,y)= |x-y|^2$,
\begin{align*}
& dCov^2_n(\mathbf X,\mathbf Y) = \frac{1}{ 4 } \frac{\sqrt{pq}}{\tau} uCov^2_n(\mathbf X,\mathbf Y) + {\cal R}'_n, \\
& \tau \times hCov^2_n(\mathbf X, \mathbf Y) =\frac{\sqrt{pq}}{ 4 \gamma_{\mathbf X}\gamma_{\mathbf Y} } f^{(1)} \left( \frac{\tau_{X}}{\gamma_{\mathbf X}} \right) g^{(1)} \left( \frac{\tau_{Y}}{ \gamma_{\mathbf Y}} \right) uCov^2_n(\mathbf X, \mathbf Y) + {\cal R}''_n,
\end{align*}
where ${\cal R}'_n, {\cal R}''_n$ are the remainder terms. Further suppose Assumption \ref{D2} holds. Then
\begin{align*}
& uCov^2_n(\mathbf X,\mathbf Y) = O_p (\tau a_p b_q), \\
& {\cal R}'_n = O_p(\tau a_p^2b_q+ \tau a_pb_q^2) = o_p (1), \\
& {\cal R}''_n = O_p(\tau a_p^2b_q+ \tau a_pb_q^2) = o_p (1).
\end{align*}
Thus the remainder term is of smaller order compared to the leading term and therefore is asymptotically negligible.
\item[(ii)] When $k(x,y)= l(x,y)= |x-y|$,
\begin{align*}
mdCov_n^2 (\mathbf X,\mathbf Y) = \sqrt{pq} \sqrt{\binom{n}{2}} uCov^2_{n}(\mathbf X,\mathbf Y).
\end{align*}
\end{itemize}
\end{proposition}
As shown in Proposition \ref{prop:uni}, $k(x,y)= l(x,y)= |x-y|$ would correspond to the $mdCov$-based $t$-test
and $k(x,y)= l(x,y) = |x-y|^2$ would correspond to the$\{dCov, hCov\}$-based $t$-tests. Then, for each $R \in \{ dCov, hCov, mdCov \}$ the asymptotic distribution of $T_{R}$ is given in the following Corollary.
\begin{corollary} \label{cor:uni}
If Assumptions \ref{D1}, \ref{D2} and \ref{D3} hold, for any fixed $t$ and each $R \in \{ dCov, hCov, mdCov \}$, we have
\begin{align*}
P_{H_0}(T_R \leq t ) & \rightarrow P(t_{v-1} \leq t), \\
P_{H_A}(T_R \leq t ) & \rightarrow E \left[ P \left( t_{v-1, W} \leq t \right)\right], \text{ where } W \sim \sqrt{\frac{\phi^2}{1 - \phi^2} \chi_{v}^2}.
\end{align*}
\end{corollary}
After knowing the asymptotic distribution of $T_R$ under the null, i.e. $t$-distribution with degrees of freedom $v-1$, we can set critical value as $t_{v-1}^{(\alpha)}$.
Then, from Proposition \ref{prop:exactT}, under the alternative, the asymptotic power of testing the null can be written as a function of $\phi$, i.e.,
\begin{align*}
\label{power2}
Power_{n}(\phi) : = E \left[ P \left( t_{v-1, W} > t_{v-1}^{\alpha} \right)\right],
\end{align*}
and under $H_{A_l}$, if we allow $n$ to grow
\begin{align*}
Power_{\infty}(\phi_0) : = \lim\limits_{n \rightarrow \infty} Power_{n}\left( \frac{\phi_0}{\sqrt{v}} \right) = P \left( t_{v-1, \phi_{0}} > t_{v-1}^{(\alpha)} \right).
\end{align*}
We then plot $Power_n(\phi)$ under different combinations of $\alpha$ and $n$, which are shown in Figure \ref{fig1}. It can be seen from Figure \ref{fig1} that larger $\phi$ results in better power and $\phi=0$ corresponds to trivial power. Next, we can actually bound the ratio of $\phi_1$ and $\phi_2$ for standard normal random variables.
\begin{figure}[h]
\begin{subfigure}[b]{0.5\linewidth}
\centering
\includegraphics[width=0.9\linewidth]{figure1}
\caption{$\alpha=0.05$}
\label{fig7:a}
\end{subfigure}
\begin{subfigure}[b]{0.5\linewidth}
\centering
\includegraphics[width=0.9\linewidth]{figure2}
\caption{$n=15$}
\label{fig7:b}
\end{subfigure}
\caption{Plot of $Power_n(\phi)$ as a function of $\phi$ under different combinations of $\alpha$ and $n$.}
\label{fig1}
\end{figure}
\begin{proposition}
\label{signal:normal}
Suppose that
\begin{align*} \left(
\begin{array}{c}
X \\
Y
\end{array} \right) \overset{d}{=} N \left(\mathbf 0, \left(
\begin{array}{cc}
\mathbf{I}_{p} & \bm{\Sigma}_{XY} \\
\bm{\Sigma}_{XY}^T & \mathbf{I}_{q}
\end{array} \right)
\right),
\end{align*}
where $\bm{\Sigma}_{XY} = cov(X,Y)$. We have
\begin{align*}
0.89^{2} \phi_2 \leq \phi_{1} \leq \phi_2.
\end{align*}
\end{proposition}
It will be shown later that $\phi_1$ corresponds to the $mdCov$-based test, whereas $\phi_{2}$ corresponds to the $dCov$ and $hCov$-based tests. Thus considering models described in Proposition \ref{signal:normal}, we expect a power loss for the $mdCov$-based test comparing to the $dCov$ and $hCov$-based tests. On the other hand, since $ \phi_{1}$ is bounded below by $ 0.89^{2} \phi_2 $, the power loss is expected to be moderate.
Using Corollary \ref{cor:uni}, we can theoretically compare the power of these $t$-tests under different cases and the results are summarized in the following table
\begin{align*}
\begin{array}{c|cc}
Power &T_{mdCov}& T_{dCov}, T_{hCov} \\ \hline
\text{under } H_{A} & Power_n(\phi_1) & Power_n(\phi_2) \\
\text{under } H_{A_{l}}, \text{ allow }n \text{ growing to infinity} & Power_{\infty}(\phi_{0,1}) & Power_{\infty}(\phi_{0,2}) \\
\text{under } H_{A_{s}} & Power_n(\phi_1) & \alpha \\
\end{array}
\end{align*}
For the studentized version of $mhCov$, if we consider the bandwidth parameters to be fixed constants, then we can use the unified approach to get the limiting $t$-distribution of the transformed $mhCov_n^2$. On the other hand, if $\gamma_{\mathcal X_{i}}$ and $\gamma_{\mathcal Y_{j}}$ are treated to be median of sample distance along each dimension and are thus random, we encounter technical difficulties to derive the limiting distribution, as in this case the kernelized pair-wise distance along each dimension are correlated with each other. This is due to the choice of the bandwidth parameter and the high dimensional approximation used for $hCov_n^2$ can not be directly applied, since $\gamma_{\mathcal X_{i}}$ and $\gamma_{\mathcal Y_{j}}$ are calculated component-wisely. Nevertheless, we shall examine the testing efficiency using $t$-distribution approximation when the bandwidth parameters are chosen to be the median of sample distance in simulation.
\section{High Dimension Medium Sample Size}
Another type of asymptotics closely related to HDLSS is the high dimension medium sample size (HDMSS) setting [\cite{aoshima2018survey}], where $p \wedge q \rightarrow \infty$ and $n \rightarrow \infty$ at a slower rate comparing to $p,q$. The HDMSS setting has been studied by \cite{fan2008sure} and \cite{yata2010effective}, among others.
From the previous sections, we know that the distance/Hilbert-Schmidt covariance can only detect linear dependencies between pair-wise components when $p \wedge q \rightarrow \infty$ and $n$ fixed. In this section, we show that this surprising phenomenon still holds under the high dimension medium sample size setting. Consequently, a unified approach is used to show that $T_{R}$ converges in distribution to standard norml under the null hypothesis, but the technical details of handling the leading term and controlling the remainder are totally different from the fixed $n$ case.
\subsection{Distance Covariance and Variants}
We first state the following assumption which can be seen as an extension of Assumption \ref{D2}.
\begin{myassumption}{D4}\label{D4} Denote
$
E[L_{X}(X, X')^2] = \alpha_{p}^2, E[L_{Y}(Y, Y')^2] = \beta_{q}^2$, $ E[L_{X}(X, X')^4] = \gamma_{p}^2 \text{ and } E[L_{Y}(Y, Y')^4] = \lambda_{q}^2,$
where $\alpha_{p}, \beta_{q}, \gamma_{p}, \lambda_{q}$ are sequences of numbers such that as $ n\wedge p \wedge q \rightarrow \infty$
\begin{align*}
\begin{array}{c}
n \alpha_{p} = o(1) \text{, } n \beta_{q} = o(1), \\
\tau_{X}^2 (\alpha_{p}\gamma_{p} + \gamma_{p}^2 ) = o(1), \tau_{Y}^2( \beta_{q}\lambda_{q} + \lambda_{q}^2) = o(1), \tau (\alpha_{p} \lambda_{q} + \gamma_{p} \beta_{q} + \gamma_{p} \lambda_{q}) = o(1).
\end{array}
\end{align*}
\end{myassumption}
\begin{remark}
\label{rem:proof2}
For the $m$-dependence structure, i.e., $ x_{i} \perp x_{j}$ if $|i-j|>m$ and $ y_{i'} \perp y_{j'}$ if $ |i' - j'| >m'$, where $\sup_{i} E(x_{i}^{8}) < \infty $ and $ \sup_{i} E(y_{i}^{8}) < \infty $, we can show that $\alpha_p = O(\sqrt{m/p})$, $\beta_{q} = O(\sqrt{ m'/q})$, $\gamma_{p}=O(m/p)$ and $\lambda_q=O(m'/q)$.
Thus, Assumption \ref{D4} holds under the $m$-dependence model if $n$ and $m, m'$ satisfies
\begin{equation*}
\begin{array}{c}
n^2 m = o(p),~~n^2 m' = o(q), \\
m^3=o(p), ~~ m'^3 = o(q), ~~m'm^2=o(p), ~~ m m'^{2}=o(q).
\end{array}
\label{eqcondii}
\end{equation*}
\end{remark}
The following theorem shows that the decomposition property \eqref{eq:decomp} for distance covariance still holds under high dimension medium sample size setting.
\begin{thm}
\label{thm:decomp2}
Under Assumption \ref{D1}, we can show that
\begin{itemize}
\item[(i)]
\begin{equation}
\label{eq:decomp2}
dCov^2_n(\mathbf{X},\mathbf{Y}) = \frac{1}{\tau} \sum_{i=1}^p \sum_{j=1}^q \cov_n^2 \left(\mathcal X_i, \mathcal Y_j \right) + {\cal R}_n.
\end{equation}
Here $\cov_n^2$ is defined the same as in Theorem \ref{thm:decomp} and $\mathcal{R}_{n}$ is the remainder term.
\item[(ii)] Further suppose Assumption \ref{D4} holds. Then we have
\begin{align*}
& \frac{1}{\tau} \sum_{i=1}^p \sum_{j=1}^q \cov_n^2 \left(\mathcal X_i, \mathcal Y_j \right) = O_p( \tau \alpha_{p} \beta_{q} ), \\
& {\cal R}_n = O_p (\tau \alpha_{p} \lambda_{q} + \tau \gamma_{p} \beta_{q} + \tau \gamma_{p} \lambda_{q}) = o_p(1).
\end{align*}
\end{itemize}
\end{thm}
Similarly, as shown in the following, $hCov$ also has the decomposition property under HDMSS.
\begin{thm}
\label{thm:decompHsic2}
Under Assumption D1, we have
\begin{itemize}
\item [(i)]
\begin{multline}
\label{eq:decompHSIC2}
\tau \times hCov^2_{n}(\mathbf X, \mathbf Y) \\ = f^{(1)} \left( \frac{\tau_{X}}{\gamma_{\mathbf X}} \right) g^{(1)} \left( \frac{\tau_{Y}}{ \gamma_{\mathbf Y}} \right) \frac{\tau_{X}}{\gamma_{\mathbf X}} \frac{\tau_{Y}}{\gamma_{ \mathbf Y}} \frac{1}{\tau} \sum\limits_{i=1}^{p} \sum\limits_{j=1}^{q} \text{cov}_{n}^2 (\mathcal X_{i}, \mathcal Y_{j}) + \mathcal{R}_{n},
\end{multline}
where $ \text{cov}_{n}^2 $ is defined the same as in Theorem \ref{thm:decomp} and $\mathcal{R}_{n}$ is the remainder term.
\item[(ii)] Further suppose Assumption \ref{D4} holds. Then
\begin{align*}
& f^{(1)} \left( \frac{\tau_{X}}{\gamma_{\mathbf X}} \right) g^{(1)} \left( \frac{\tau_{Y}}{ \gamma_{\mathbf Y}} \right) \frac{\tau_{X}}{\gamma_{\mathbf X}} \frac{\tau_{Y}}{\gamma_{ \mathbf Y}} \asymp_p 1, \\
& \frac{1}{\tau} \sum_{i=1}^p \sum_{j=1}^q \cov_n^2 \left(\mathcal X_i, \mathcal Y_j \right) = O_p( \tau \alpha_{p} \beta_{q} ), \\
& {\cal R}_n = O_p (\tau \alpha_{p} \lambda_{q} + \tau \gamma_{p} \beta_{q} + \tau \gamma_{p} \lambda_{q}) = o_p(1).
\end{align*}
\end{itemize}
\end{thm}
From Equations \eqref{eq:decomp2} and \eqref{eq:decompHSIC2}, we can see that under the HDMSS setting, it is still true that distance/Hilbert-Schmidt covariance can only detect the linear dependence between the components of $X$ and $Y$.
\subsection{Studentized Test Statistics}
Similar to Section \ref{student}, we provide a unified approach to analyze the studentized $dCov, hCov, mdCov$. Since now the sample size is growing, the element-wise argument used to prove the results in Section \ref{student} will no longer work. Inspired by \cite{zhang2018conditional} and \cite{yao2017testing}, we derive the asymptotic distribution by constructing a martingale sequence and using martingale CLT.
\subsubsection{Unified Approach}
For notational convenience, we first define the following metrics,
\begin{align*}
U(X_{s}, X_{t}) := \frac{1}{\sqrt{p}}\sum^{p}_{i=1} k_{st}(i), \quad V(Y_{s}, Y_{t}) := \frac{1}{\sqrt{q}}\sum^{q}_{i=1} l_{st}(i),
\end{align*}
where $k_{st}(i)$ and $l_{st}(i)$ are defined in Section \ref{sec:uni}. To show that the studentized test statistic converges to standard normal, we essentially use the martingale CLT [\cite{hall2014martingale}] and the following assumptions are used to guarantee the conditions in martingale CLT.
\begin{myassumption}{D5}\label{D5}
\begin{align}
& \frac{E \left[ U(X, X')^4 \right]}{ \sqrt{n} (E[ U(X, X')^2 ])^2 } \rightarrow 0, \label{eq:d51} \\
& \frac{E \left[ U(X, X') U(X', X'')U(X'', X''')U(X''',X ) \right] }{ (E[ U(X, X')^2 ])^2 } \rightarrow 0, \label{eq:d52}
\end{align}
and similar assumptions hold for $Y$.
\end{myassumption}
\begin{remark}
When $k(x,y)= l(x,y)= |x-y|$, Assumption \ref{D5} has been studied in Propositions 2.1 and 2.2 of \cite{zhang2018conditional}.
\end{remark}
\begin{remark}
\label{rem:D7}
When $k(x,y)= l(x,y)= |x-y|^2$, Equations \eqref{eq:d51} and \eqref{eq:d52} can be simplified to
\begin{align*}
&\frac{\sum\limits_{i,j,r,w=1}^p E^2 \left[ (x_{i} - E[x_{i}]) (x_{j} - E[x_{j}]) (x_{r} - E[x_{r}]) (x_{w} - E[x_{w}]) \right] }{ \sqrt{n} Tr^2(\bm{\Sigma}_{X}^2) } \rightarrow 0, \\
& \frac{Tr ( \bm{\Sigma}_{X}^4) }{ Tr^2(\bm{\Sigma}_{X}^2) } \rightarrow 0, \quad \text{where } \bm{\Sigma}_{X} = cov (X, X).
\end{align*}
Notice that $Tr(\bm{\Sigma}_{X}^2) = \sum_{i=1}^{p} \sum_{j=1}^{p} cov^2(x_{i}, x_{j})$. Consider the $m$-dependence model in Remark \ref{rem:proof2}. Assuming $\sup_{i} E(x_{i}^{4}) < \infty $, we have $ Tr ( \bm{\Sigma}_{X}^4) =O(m^3p) $ and
$$
\sum_{i,j,r,w=1}^p E^2 \left[ (x_{i} - E[x_{i}]) (x_{j} - E[x_{j}]) (x_{r} - E[x_{r}]) (x_{w} - E[x_{w}]) \right] = \\ O(m^2p^2).
$$
Consequently, it can be seen that the $m$-dependence model in Remark \ref{rem:proof2} also satisfies Equations \eqref{eq:d51} and \eqref{eq:d52} by controlling the orders of $n, m, m'$.
\end{remark}
Then, we can show that the normalized $uCov_{n}^{2}(\mathbf{X}, \mathbf{Y})$ converges to standard normal distribution under the high dimension medium sample size regime.
\begin{thm}
\label{thm:key2}
Let $n \wedge p \wedge q \rightarrow \infty$. Under $H_{0}$ and Assumption \ref{D5}, we have
\begin{align*}
\sqrt{\binom{n}{2}} \frac{uCov_{n}^{2}(\mathbf{X}, \mathbf{Y})}{\mathcal{S}} \overset{d}{\rightarrow} N(0,1), \text{ where } \mathcal{S}^{2} = E[ U(X, X')^{2}] E[ V(Y, Y')^{2} ].
\end{align*}
\end{thm}
Consequently, we have the following result.
\begin{proposition}\label{prop:exactT2}
Let $n \wedge p \wedge q \rightarrow \infty$. Under $H_{0}$ and Assumption \ref{D5}, we have
\begin{align*}
T_u \overset{d}{\rightarrow} N(0,1).
\end{align*}
\end{proposition}
\subsubsection{Studentized Tests}
The following result shows that as $ n \wedge p \wedge q \rightarrow \infty $, scaled $dCov, hCov$ and $mdCov$ are all equal to $uCov$ up to an asymptotically constant factor.
\begin{proposition}
\label{prop:uni2}
Under Assumption \ref{D1},
\begin{itemize}
\item[(i)] When $k(x,y)= l(x,y)= |x-y|^2$,
\begin{align*}
& dCov^2_n(\mathbf X,\mathbf Y) = \frac{1}{ 4 } \frac{\sqrt{pq}}{\tau} uCov^2_n(\mathbf X,\mathbf Y) + {\cal R}'_n, \\
& \tau \times hCov^2_n(\mathbf X, \mathbf Y) =\frac{\sqrt{pq}}{ 4\gamma_{\mathbf X}\gamma_{\mathbf Y} } f^{(1)} \left( \frac{\tau_{X}}{\gamma_{\mathbf X}} \right) g^{(1)} \left( \frac{\tau_{Y}}{ \gamma_{\mathbf Y}} \right) uCov^2_n(\mathbf X, \mathbf Y) + {\cal R}''_n,
\end{align*}
where ${\cal R}'_n, {\cal R}''_n$ are the remainder terms.
Further suppose Assumption \ref{D4} holds. Then
\begin{align*}
& uCov^2_n(\mathbf X,\mathbf Y) = O_p( \tau \alpha_{p} \beta_{q} ), \\
& {\cal R}_n' = O_p (\tau \alpha_{p} \lambda_{q} + \tau \gamma_{p} \beta_{q} + \tau \gamma_{p} \lambda_{q}) = o_p(1), \\
& {\cal R}_n'' = O_p (\tau \alpha_{p} \lambda_{q} + \tau \gamma_{p} \beta_{q} + \tau \gamma_{p} \lambda_{q}) = o_p(1).
\end{align*}
\item[(ii)] When $k(x,y)= l(x,y)= |x-y|$,
\begin{align*}
mdCov_n^2 (\mathbf X,\mathbf Y) = \sqrt{pq} \sqrt{\binom{n}{2}} uCov^2_{n}(\mathbf X,\mathbf Y).
\end{align*}
\end{itemize}
\end{proposition}
Finally, by adopting a unified approach, we have the following Corollary.
\begin{corollary} \label{cor:uni2}
Let $n \wedge p \wedge q \rightarrow \infty$. Under $H_{0}$ and Assumption \ref{D5}, we have
\begin{itemize}
\item[(i)] \begin{align*}
T_{mdCov} \overset{d}{\rightarrow} N(0,1).
\end{align*}
\item[(ii)] Further suppose Assumption \ref{D4} and
\begin{align}
\label{eq:last}
\frac{n}{ \sqrt{ \frac{1}{p} Tr ( \bm{\Sigma}_{X}^2 ) \frac{1}{q} Tr ( \bm{\Sigma}_{Y}^2 ) } } \tau (\alpha_{p} \lambda_{q} + \gamma_{p} \beta_{q} + \gamma_{p} \lambda_{q} ) = o(1).
\end{align}
Then, for each $R \in \{ dCov, hCov \}$, we have
\begin{align*}
T_{R} \overset{d}{\rightarrow} N(0,1).
\end{align*}
\end{itemize}
\end{corollary}
\begin{remark}
The $m$-dependence model in Remark \ref{rem:proof2} can also satisfies Equation \eqref{eq:last} by controlling the orders of $n,m,m'$ based on the magnitude of $ Tr(\bm{\Sigma}_{X}^2)/p $ and $Tr ( \bm{\Sigma}_{Y}^2 )/q $.
\end{remark}
\section{Conclusion}
In this article, we investigate the behavior of the distance covariance and Hilbert-Schmidt covariance in the high dimensional setting. Somewhat shockingly, we discover that the distance covariance and Hilbert-Schmidt covariance, which are well-known to capture nonlinear dependence in low/fixed dimensional context, can only capture linear componentwise cross-dependence (to the first order). We believe that this is a new finding that may have significant implications to the design of tests for independence for high dimensional data.
On one hand, we reveal the limitation of distance covariance and variants in the high dimensional context, and suggest to use marginally aggregated (sample) distance covariance as a way out, where the latter targets the low dimensional nonlinear dependence.
On the other hand, we speculate whether it is possible to capture all kinds of dependence between high dimensional vectors $X$ and $Y$, in a limited sample size framework. If the sample size is fixed, we would conjecture that an omnibus test does not exist; If the sample size can grow faster than the dimension, it seems possible but unclear to us how to develop an omnibus test in an asymptotic sense. We hope the results presented in this paper shed some light on the challenges in the high dimensional dependence testing and will motivate more work in this area.
\begin{supplement}
\sname{Supplement to}\label{suppA}
\stitle{``Distance-based and RKHS-based Dependence Metrics in High Dimension"}
\slink[url]{}
\sdescription{This supplement contains simulations and technical details of the results in the paper.}
\end{supplement}
{
\bibliographystyle{imsart-nameyear}
\bibliography{reference}
}
\clearpage
\title{Supplement to ``Distance-based and RKHS-based Dependence Metrics in High Dimension''}
\begin{aug}
\author{\fnms{Changbo} \snm{Zhu}\corref{}\thanksref{m1}
\ead[label=e1]{first@somewhere.com}}
\author{\fnms{Shun} \snm{Yao}\thanksref{m2}
\ead[label=e2]{second@somewhere.com}}
\author{\fnms{Xianyang} \snm{Zhang}\thanksref{m3}
\ead[label=e2]{second@somewhere.com}}
\and
\author{\fnms{Xiaofeng} \snm{Shao}\thanksref{m1}
\ead[label=e3]{third@somewhere.com}
\ead[label=u1,url]{http://www.foo.com}}
\runauthor{C. Zhu et al.}
\affiliation{University of Illinois at Urbana-Champaign\thanksmark{m1}, Goldman Sachs at New York City\thanksmark{m2} and Texas A\&M University\thanksmark{m3}}
\end{aug}
\appendix
\section{Simulation Study}
Here, we consider some numerical examples to compare the ``joint'' tests, where the distance/Hilbert-Schmidt covariance is applied to whole components of data jointly, with the ``marginal'' tests, where distance/Hilbert-Schmidt covariance is applied to one dimensional components and then being aggregated. To this end, we consider the following statistics
\begin{align*}
\text{``Joint"} & \left\lbrace \begin{array}{l}
dCov : \text{ distance covariance (permutation)} \\
T_{dCov} : \text{studentized distance covariance} \\
hCov : \text{ Hilbert-Schmidt covariance (permutation)} \\
T_{hCov} : \text{studentized Hilbert-Schmidt covariance}
\end{array} \right. \\
\text{``Marginal"} & \left\lbrace \begin{array}{l}
mdCov : \text{marginal distance covariance (permutation)} \\
T_{mdCov} : \text{studentized marginal distance covariance}, \\
mhCov : \text{marginal Hilbert-Schmidt covariance (permutation)} \\
T_{mhCov} : \text{studentized marginal Hilbert-Schmidt covariance}
\end{array} \right.
\end{align*}
In the above display, $ dCov_n^2$ and $ hCov_n^2 $ are the two ``joint'' test statistics to measure the overall dependence between $X$ and $Y$, $mdCov_n^2$ and $mhCov_n^2$ are the ``marginal'' test statistics, and these four test statistics are implemented as permutation tests; $T_{dCov}$ from \cite{szekely2013} is the studentized version of $dCov$, our proposed $t$-tests $T_{hCov}, T_{mdCov}, T_{mhCov}$ are the studentized version of $hCov, mdCov, mhCov$ respectively. All these four tests are implemented using the $t$-distribution based critical value. We examine both the Gaussian kernel and Laplacian kernel for the Hilbert-Schmidt covariance based tests.
For the permutation-based tests, we randomly shuffle the samples $\{X_1,\dots,$ $X_n\}$ and get $(X_{\pi(1)},\dots,X_{\pi(n)})$, where $\pi$ is the permutation map from $\{1,\dots,n\}$ to $\{1,\dots,n\}$. Then we calculate the test statistic based on the permuted sample $\{(X_{\pi(1)},\dots,X_{\pi(n)})$, $(Y_1,\dots,Y_n)\}$. The $p$-value for permutation-based test is defined as the proportion of times that the test statistic based on the permuted samples is greater than the one based on the original sample.
All the numerical results from permutation-based tests are based on 200 permutations and the empirical rejection rate of the tests are based on 5000 Monte Carlo repetitions.
We first examine the size of the afore-mentioned tests.
\begin{example}
\label{exp1}
Generate i.i.d. samples from the following models for $i=1,\dots,n$.
\begin{itemize}
\item[(i)]
$$
\begin{array}{l}
X_{i} = \left( x_{i1}, \dots, x_{ip} \right) \sim N(\mathbf 0, \mathbf I_{p}), \\
Y_{i} = \left( y_{i1}, \dots, y_{ip} \right) \sim N(\mathbf 0, \mathbf I_{p}).
\end{array} $$
\item[(ii)] Let $AR(1)$ denotes the Gaussian autoregressive model of order 1 with parameter $\phi$,
$$
\begin{array}{l}
X_{i} \sim AR(1), \phi = 0.5, \\
Y_{i} \sim AR(1), \phi = -0.5.
\end{array} $$
\item[(iii)] Let $\bm \Sigma = (\sigma_{ij}) \in \mathbb{R}^{p \times p} \text{ and } \sigma_{ij} = 0.7^{|i-j|}$,
\begin{align*}
& X_i =(x_{i1},\dots,x_{ip}) \sim N(\mathbf 0, \bm \Sigma), \\
& Y_i =(y_{i1},\dots,y_{ip}) \sim N(\mathbf 0, \bm \Sigma).
\end{align*}
\end{itemize}
\end{example}
From Table \ref{tab1}, we can see that all the tests have quite accurate size. Although the $t$-tests are derived under the high dimensional scenario, they still have pretty accurate size even for relatively low dimension (e.g., $p=5$). In addition, for data samples from Example \ref{exp1} (i), we provide the density plots of the studentized test statistics in Figure \ref{figR2} as well as the density plots of $t_{v-1}$. As we can see, for all cases, the empirical densities are fairly close to that of $t_{v-1}$ and getting closer to $t_{v-1}$ as dimension increases.
\begin{figure}
\begin{tikzpicture}
\matrix[
matrix of nodes,
nodes={
anchor=center
}
](m){ & n=30 & n=60 \\
$T_{dCov}$ & \includegraphics[width=5cm, height=2.7cm ]{td30} & \includegraphics[width=5cm, height=2.7cm]{td60} \\
$T_{mdCov}$ & \includegraphics[width=5cm, height=2.7cm ]{tmd30} & \includegraphics[width=5cm, height=2.7cm]{tmd60} \\
$\underset{(\text{Gaussian kernel})}{T_{hCov}}$ & \includegraphics[width=5cm, height=2.7cm ]{th30} & \includegraphics[width=5cm, height=2.7cm]{th60} \\
$\underset{ (\text{Gaussian kernel}) }{ T_{mhCov} }$ & \includegraphics[width=5cm, height=2.7cm ]{tmh30} & \includegraphics[width=5cm, height=2.7cm]{tmh60} \\
$ \underset{ (\text{Laplacian kernel}) }{T_{hCov} } $ & \includegraphics[width=5cm, height=2.7cm ]{thl30} & \includegraphics[width=5cm, height=2.7cm]{thl60} \\
$ \underset{ (\text{Laplacian kernel}) }{ T_{mhCov} } $ & \includegraphics[width=5cm, height=2.7cm ]{tmhl30} & \includegraphics[width=5cm, height=2.7cm]{tmhl60} \\
} ;
\end{tikzpicture}
\caption{Density plots of the studentized test statistics (solid colored lines) and $t_{v-1}$ (dashed black line).}
\label{figR2}
\end{figure}
Notice that under the high dimensional case, the ``joint" tests can be seen as the aggregation of component-wise sample squared covariances. On the other hand, the ``marginal" tests are the accumulation of component-wise sample distance/Hilbert-Schmidt covariances. When $(X,Y)$ are generated from the model in Proposition \ref{signal:normal}, it is expected that there is power loss for $mdCov$ and $mhCov$ based permutation test comparing to $dCov$ and $hCov$ based permutation tests and similar phenomenon is expected for $mdCov$ and $mhCov$ based $t$-test comparing to $dCov$ and $hCov$ based $t$-tests. The following example demonstrates this phenomemon.
\begin{example}
\label{exp2}
Generate i.i.d. samples from the following models for $i=1,...,n$.
\begin{itemize}
\item[(i)] Let $\rho = 0.5$,
\begin{align*}
\begin{array}{l}
Z_{i} = \left( z_{i1}, \cdots, z_{ip} \right) \sim N(\mathbf 0, \mathbf I_{p}),\\
X_{i} = \left( x_{i1}, \cdots, x_{ip} \right) \sim N(\mathbf 0, \mathbf I_{p}),\\
Y_{i} = \frac{\rho X_{i} + (1 - \rho)Z_{i}}{\sqrt{\rho^2 + (1 - \rho)^2}}.
\end{array}
\end{align*}
\item[(ii)] Let $ \rho = 0.7 $ and $(X_i,Y_i,Z_i)$ be defined in the same way as in (i).
\item[(iii)] Let $ \rho = 0.5 $ and $\otimes$ denote the Kronecker product. Define
\begin{align*}
\begin{array}{l}
Z_{i} = \left( z_{i1}, \cdots, z_{ip} \right) \sim N(\mathbf 0, \mathbf I_{p}), \\
X_{i} = \left( x_{i1}, \cdots, x_{ip} \right) \sim N(\mathbf 0, \mathbf I_{p}), \\
Y_{i} = \frac{\rho \bm \Sigma X_i + (1 - \rho)Z_{i}}{\sqrt{\rho^2 + (1 - \rho)^2}},
\end{array}
\end{align*}
where $\bm{\Sigma} = \mathbf{I} \otimes \mathbf{A}$ and $\mathbf{A}$ is an orthogonal matrix defined as
\begin{align*}
\mathbf{A} = \left(
\begin{array}{ccccc}
0 & \sqrt{\frac{1}{4}} & \sqrt{\frac{1}{5}} & -\sqrt{\frac{1}{4}} & -\sqrt{\frac{3}{10}} \\
\sqrt{\frac{1}{6}} & \sqrt{\frac{1}{4}} & \sqrt{\frac{1}{5}} & \sqrt{\frac{1}{4}} & \sqrt{\frac{2}{15}} \\
-\sqrt{\frac{2}{3}} & 0 & \sqrt{\frac{1}{5}} & 0 & \sqrt{\frac{2}{15}} \\
\sqrt{\frac{1}{6}} & -\sqrt{\frac{1}{4}} & \sqrt{\frac{1}{5}} & -\sqrt{\frac{1}{4}} & \sqrt{\frac{2}{15}} \\
0 & -\sqrt{\frac{1}{4}} & \sqrt{\frac{1}{5}} & \sqrt{\frac{1}{4}} & -\sqrt{\frac{3}{10}}
\end{array} \right).
\end{align*}
\end{itemize}
\end{example}
From Table \ref{tab2}, we can see that there is indeed a power loss for the ``marginal" tests compared to the ``joint" tests, but the loss of power appears fairly moderate, which is consistent with our theory. For Example \ref{exp2}, it can also be observed that the power decrease for the Hilbert-Schmidt covariance based tests is a bit more than the power decrease of distance covariance based tests. Moreover, the power drop is slightly smaller for Gaussian kernel comparing with Laplacian kernel.
As demonstrated in Theorem \ref{thm:decomp} and \ref{thm:decompHsic}, the leading term in (\ref{eq:decomp}) and (\ref{eq:decompHSIC}) can only measure the linear dependence as $p \wedge q \rightarrow \infty$, therefore we expect the ``joint'' test based on $dCov_n^2(\mathbf X,\mathbf Y)$ or $ hCov_n^2(\mathbf X, \mathbf Y) $ may fail to capture the non-linear dependence in high dimension. On the other hand, we consider the ``marginal'' test where we take the sum of pairwise sample distance/Hilbert-Schmidt covariances to measure the low dimensional dependence
for all the pairs as the test proposed in Sections \ref{dCov} and \ref{hCov}. The ``marginal'' test statistic measures the dependence marginally in a low-dimensional fashion so that it can preserve the ability to capture component-wise non-linear dependence. In the following two examples, we demonstrate the superiority of ``marginal'' tests.
\begin{example}
\label{exp3}
Generate $i.i.d.$ samples from the following models for $i=1,...,n$.
\begin{itemize}
\item[(i)]
\begin{align*}
& X_i =(x_{i1},\dots,x_{ip}) \sim N(\mathbf 0, \mathbf I_p), \\
& Y_i = (y_{i1},\dots,y_{ip}), \mbox{~where~} y_{ij} = x_{ij}^2 \mbox{~for~} j=1,\dots,p.
\end{align*}
\item[(ii)] Let $\bm \Sigma = (\sigma_{ij}) \in \mathbb{R}^{p \times p} \text{ and } \sigma_{ij} = 0.7^{|i-j|}$,
\begin{align*}
& X_i =(x_{i1},\dots,x_{ip}) \sim N(\mathbf 0, \bm \Sigma), \\
& Y_i = (y_{i1},\dots,y_{ip}), \mbox{~where~} y_{ij} = x_{ij}^2 \mbox{~for~} j=1,\dots,p.
\end{align*}
\item[(iii)]
\begin{align*}
& X_i =(x_{i1},\dots,x_{ip}) \sim N(\mathbf 0, \mathbf I_p), \\
& Y_i = (y_{i1},\dots,y_{ip}), \mbox{~where~} y_{ij} = \log |x_{ij}| \mbox{~for~} j=1,\dots,p.
\end{align*}
\end{itemize}
\end{example}
\begin{example}
\label{exp4}
Generate $i.i.d.$ samples from the following models for $i=1,\dots,n$.
\begin{itemize}
\item[(i)] Let $\circ$ denotes the Hadamard product,
\begin{align*}
& X_i =(x_{i1},\dots,x_{ip}) \overset{i.i.d.}{\sim} U(-1, 1), \\
&Y_i = X_i \circ X_i.
\end{align*}
\item[(ii)] \begin{align*}
& X_i =(x_{i1},\dots,x_{ip}) \overset{i.i.d.}{\sim} U(0, 1), \\
&Y_i = 4X_i \circ X_i \circ X_i - 3.6X_i +0.8.
\end{align*}
\item[(iii)]
\begin{align*}
& Z_i =(z_{i1},\dots,z_{ip}) \overset{i.i.d.}{\sim} U(0, 2 \pi), \\
&X_i = \sin(Z_i), \quad Y_i =\cos(Z_i).
\end{align*}
\end{itemize}
\end{example}
Notice that in the above two examples,
$\cov^2(x_i,y_j)=0$ but $dCov^2(x_i,y_j)$ $ \ne 0$ for all $(i,j)s$, that is, $(X,Y) \in H_{A_s}$. From Table \ref{tab3}, we can observe that for Example \ref{exp3}, the ``joint'' tests suffer substantial power loss as dimension increases for fixed sample size. The power loss is less severe in case (ii) than the ones in cases (i) and (iii), due to the dependence between the components. On the other hand, the powers corresponding to the marginal test statistics consistently outperform their joint counterparts with very little to none power reduction as the dimension increases. Similar phenomenon can be observed for Example \ref{exp4}; see Table \ref{tab4}. In addition, for all the cases in both Example \ref{exp3} and Example \ref{exp4}, the power loss corresponding to Laplacian kernel is consistently less than that for Gaussian kernel.
In general, we observe that the tests based on distance covariance, Hilbert-Schmidt covariance with Gaussian kernel, and Hilbert-Schmidt covariance with Laplacian kernel, are all admissible, as none of them dominate the others in all situations.
\begin{table}[!htbp] \centering
\caption{Size comparison from Example \ref{exp1}}
\label{tab1}
\scalebox{0.6}{
\begin{tabular}{@{\extracolsep{5pt}} cccccccccccccccc}
\\[-1.8ex]\hline
\hline \\[-1.8ex]
&& & & \multicolumn{4}{c}{} & \multicolumn{4}{c}{Gaussian Kernel} & \multicolumn{4}{c}{Laplacian Kernel} \\ \cline{9-12} \cline{13-16}
&$n$ & $p$ & $\alpha$ & $dCov$ & $mdCov$ & $T_{dCov}$ & $T_{mdCov}$ & $hCov$ & $mhCov$ & $T_{hCov}$ & $T_{mhCov}$ & $hCov$ & $mhCov$ & $T_{hCov}$ & $T_{mhCov}$ \\
\hline \\[-1.8ex]
\multirow{18}{*}{(i)}&$10$ & $5$ & $0.010$ & $0.017$ & $0.014$ & $0.020$ & $0.014$ & $0.016$ & $0.015$ & $0.020$ & $0.014$ & $0.014$ & $0.014$ & $0.017$ & $0.013$ \\
&$10$ & $5$ & $0.050$ & $0.055$ & $0.055$ & $0.062$ & $0.061$ & $0.055$ & $0.060$ & $0.062$ & $0.061$ & $0.055$ & $0.050$ & $0.064$ & $0.050$ \\
&$10$ & $5$ & $0.100$ & $0.105$ & $0.107$ & $0.110$ & $0.110$ & $0.103$ & $0.106$ & $0.109$ & $0.109$ & $0.102$ & $0.099$ & $0.105$ & $0.101$ \\
&$10$ & $30$ & $0.010$ & $0.015$ & $0.015$ & $0.013$ & $0.011$ & $0.015$ & $0.016$ & $0.012$ & $0.012$ & $0.014$ & $0.014$ & $0.011$ & $0.011$ \\
&$10$ & $30$ & $0.050$ & $0.054$ & $0.053$ & $0.050$ & $0.053$ & $0.053$ & $0.054$ & $0.050$ & $0.052$ & $0.052$ & $0.059$ & $0.050$ & $0.054$ \\
&$10$ & $30$ & $0.100$ & $0.102$ & $0.104$ & $0.099$ & $0.102$ & $0.102$ & $0.105$ & $0.100$ & $0.103$ & $0.102$ & $0.107$ & $0.101$ & $0.105$ \\ \cline{2-16}
&$30$ & $5$ & $0.010$ & $0.014$ & $0.016$ & $0.019$ & $0.018$ & $0.016$ & $0.016$ & $0.020$ & $0.017$ & $0.016$ & $0.015$ & $0.019$ & $0.015$ \\
&$30$ & $5$ & $0.050$ & $0.052$ & $0.053$ & $0.062$ & $0.059$ & $0.052$ & $0.057$ & $0.061$ & $0.059$ & $0.054$ & $0.055$ & $0.061$ & $0.058$ \\
&$30$ & $5$ & $0.100$ & $0.105$ & $0.104$ & $0.105$ & $0.107$ & $0.103$ & $0.107$ & $0.106$ & $0.106$ & $0.105$ & $0.104$ & $0.109$ & $0.104$ \\
&$30$ & $30$ & $0.010$ & $0.014$ & $0.014$ & $0.011$ & $0.012$ & $0.014$ & $0.017$ & $0.010$ & $0.013$ & $0.014$ & $0.017$ & $0.011$ & $0.013$ \\
&$30$ & $30$ & $0.050$ & $0.051$ & $0.053$ & $0.052$ & $0.051$ & $0.051$ & $0.056$ & $0.052$ & $0.053$ & $0.051$ & $0.058$ & $0.051$ & $0.052$ \\
&$30$ & $30$ & $0.100$ & $0.097$ & $0.105$ & $0.096$ & $0.103$ & $0.097$ & $0.105$ & $0.095$ & $0.101$ & $0.099$ & $0.104$ & $0.100$ & $0.102$ \\ \cline{2-16}
&$60$ & $5$ & $0.010$ & $0.013$ & $0.015$ & $0.018$ & $0.016$ & $0.014$ & $0.013$ & $0.019$ & $0.016$ & $0.014$ & $0.015$ & $0.017$ & $0.015$ \\
&$60$ & $5$ & $0.050$ & $0.052$ & $0.055$ & $0.061$ & $0.057$ & $0.054$ & $0.061$ & $0.060$ & $0.064$ & $0.053$ & $0.057$ & $0.058$ & $0.058$ \\
&$60$ & $5$ & $0.100$ & $0.103$ & $0.104$ & $0.109$ & $0.104$ & $0.107$ & $0.108$ & $0.110$ & $0.110$ & $0.102$ & $0.101$ & $0.103$ & $0.102$ \\
&$60$ & $30$ & $0.010$ & $0.019$ & $0.017$ & $0.016$ & $0.012$ & $0.019$ & $0.015$ & $0.015$ & $0.013$ & $0.020$ & $0.016$ & $0.015$ & $0.014$ \\
&$60$ & $30$ & $0.050$ & $0.060$ & $0.063$ & $0.057$ & $0.058$ & $0.060$ & $0.058$ & $0.057$ & $0.058$ & $0.061$ & $0.058$ & $0.058$ & $0.055$ \\
&$60$ & $30$ & $0.100$ & $0.113$ & $0.112$ & $0.110$ & $0.107$ & $0.113$ & $0.109$ & $0.111$ & $0.105$ & $0.110$ & $0.111$ & $0.107$ & $0.107$ \\
\hline \\[-1.8ex]
\multirow{18}{*}{(ii)} &$10$ & $5$ & $0.010$ & $0.015$ & $0.015$ & $0.023$ & $0.023$ & $0.014$ & $0.016$ & $0.023$ & $0.019$ & $0.015$ & $0.017$ & $0.022$ & $0.021$ \\
&$10$ & $5$ & $0.050$ & $0.051$ & $0.054$ & $0.064$ & $0.066$ & $0.053$ & $0.058$ & $0.064$ & $0.066$ & $0.054$ & $0.058$ & $0.066$ & $0.062$ \\
&$10$ & $5$ & $0.100$ & $0.101$ & $0.105$ & $0.107$ & $0.111$ & $0.100$ & $0.109$ & $0.105$ & $0.113$ & $0.102$ & $0.110$ & $0.106$ & $0.109$ \\
&$10$ & $30$ & $0.010$ & $0.014$ & $0.018$ & $0.013$ & $0.016$ & $0.014$ & $0.017$ & $0.014$ & $0.013$ & $0.017$ & $0.018$ & $0.017$ & $0.013$ \\
&$10$ & $30$ & $0.050$ & $0.060$ & $0.061$ & $0.061$ & $0.061$ & $0.061$ & $0.056$ & $0.062$ & $0.056$ & $0.059$ & $0.060$ & $0.059$ & $0.056$ \\
&$10$ & $30$ & $0.100$ & $0.105$ & $0.105$ & $0.110$ & $0.107$ & $0.105$ & $0.105$ & $0.109$ & $0.099$ & $0.106$ & $0.108$ & $0.111$ & $0.104$ \\ \cline{2-16}
&$30$ & $5$ & $0.010$ & $0.012$ & $0.011$ & $0.022$ & $0.023$ & $0.012$ & $0.014$ & $0.021$ & $0.020$ & $0.013$ & $0.013$ & $0.019$ & $0.016$ \\
&$30$ & $5$ & $0.050$ & $0.046$ & $0.048$ & $0.055$ & $0.056$ & $0.046$ & $0.052$ & $0.055$ & $0.059$ & $0.047$ & $0.053$ & $0.051$ & $0.059$ \\
&$30$ & $5$ & $0.100$ & $0.094$ & $0.096$ & $0.094$ & $0.096$ & $0.096$ & $0.100$ & $0.097$ & $0.100$ & $0.093$ & $0.107$ & $0.097$ & $0.104$ \\
&$30$ & $30$ & $0.010$ & $0.016$ & $0.016$ & $0.017$ & $0.015$ & $0.017$ & $0.015$ & $0.017$ & $0.011$ & $0.017$ & $0.015$ & $0.017$ & $0.012$ \\
&$30$ & $30$ & $0.050$ & $0.061$ & $0.058$ & $0.060$ & $0.059$ & $0.061$ & $0.055$ & $0.060$ & $0.054$ & $0.058$ & $0.052$ & $0.060$ & $0.051$ \\
&$30$ & $30$ & $0.100$ & $0.109$ & $0.105$ & $0.110$ & $0.107$ & $0.111$ & $0.101$ & $0.110$ & $0.098$ & $0.111$ & $0.102$ & $0.113$ & $0.097$ \\ \cline{2-16}
&$60$ & $5$ & $0.010$ & $0.015$ & $0.013$ & $0.026$ & $0.022$ & $0.016$ & $0.014$ & $0.024$ & $0.020$ & $0.013$ & $0.015$ & $0.020$ & $0.018$ \\
&$60$ & $5$ & $0.050$ & $0.055$ & $0.052$ & $0.062$ & $0.061$ & $0.055$ & $0.053$ & $0.061$ & $0.059$ & $0.055$ & $0.052$ & $0.061$ & $0.054$ \\
&$60$ & $5$ & $0.100$ & $0.101$ & $0.100$ & $0.103$ & $0.100$ & $0.102$ & $0.100$ & $0.104$ & $0.099$ & $0.101$ & $0.097$ & $0.103$ & $0.099$ \\
&$60$ & $30$ & $0.010$ & $0.013$ & $0.014$ & $0.013$ & $0.014$ & $0.013$ & $0.016$ & $0.014$ & $0.013$ & $0.014$ & $0.015$ & $0.013$ & $0.012$ \\
&$60$ & $30$ & $0.050$ & $0.055$ & $0.051$ & $0.058$ & $0.051$ & $0.054$ & $0.054$ & $0.057$ & $0.053$ & $0.058$ & $0.053$ & $0.053$ & $0.052$ \\
&$60$ & $30$ & $0.100$ & $0.105$ & $0.102$ & $0.105$ & $0.100$ & $0.106$ & $0.103$ & $0.105$ & $0.102$ & $0.107$ & $0.105$ & $0.107$ & $0.104$ \\
\hline \\[-1.8ex]
\multirow{18}{*}{(iii)}& $10$ & $5$ & $0.010$ & $0.012$ & $0.013$ & $0.025$ & $0.024$ & $0.012$ & $0.014$ & $0.024$ & $0.022$ & $0.016$ & $0.013$ & $0.025$ & $0.019$ \\
& $10$ & $5$ & $0.050$ & $0.051$ & $0.051$ & $0.068$ & $0.069$ & $0.053$ & $0.051$ & $0.068$ & $0.062$ & $0.053$ & $0.049$ & $0.067$ & $0.056$ \\
& $10$ & $5$ & $0.100$ & $0.100$ & $0.099$ & $0.107$ & $0.103$ & $0.100$ & $0.098$ & $0.105$ & $0.102$ & $0.100$ & $0.098$ & $0.104$ & $0.101$ \\
& $10$ & $30$ & $0.010$ & $0.014$ & $0.015$ & $0.016$ & $0.014$ & $0.014$ & $0.015$ & $0.016$ & $0.013$ & $0.015$ & $0.015$ & $0.017$ & $0.013$ \\
& $10$ & $30$ & $0.050$ & $0.055$ & $0.057$ & $0.061$ & $0.058$ & $0.053$ & $0.056$ & $0.061$ & $0.056$ & $0.057$ & $0.057$ & $0.064$ & $0.059$ \\
&$10$ & $30$ & $0.100$ & $0.104$ & $0.105$ & $0.105$ & $0.107$ & $0.103$ & $0.105$ & $0.104$ & $0.107$ & $0.106$ & $0.110$ & $0.106$ & $0.112$ \\ \cline{2-16}
&$30$ & $5$ & $0.010$ & $0.015$ & $0.014$ & $0.028$ & $0.029$ & $0.015$ & $0.014$ & $0.025$ & $0.024$ & $0.014$ & $0.014$ & $0.024$ & $0.019$ \\
&$30$ & $5$ & $0.050$ & $0.052$ & $0.054$ & $0.060$ & $0.062$ & $0.051$ & $0.052$ & $0.062$ & $0.062$ & $0.048$ & $0.052$ & $0.058$ & $0.059$ \\
&$30$ & $5$ & $0.100$ & $0.103$ & $0.103$ & $0.098$ & $0.099$ & $0.101$ & $0.101$ & $0.101$ & $0.098$ & $0.099$ & $0.099$ & $0.097$ & $0.098$ \\
&$30$ & $30$ & $0.010$ & $0.017$ & $0.015$ & $0.019$ & $0.017$ & $0.016$ & $0.015$ & $0.019$ & $0.015$ & $0.013$ & $0.016$ & $0.018$ & $0.012$ \\
&$30$ & $30$ & $0.050$ & $0.054$ & $0.055$ & $0.058$ & $0.058$ & $0.055$ & $0.055$ & $0.059$ & $0.057$ & $0.056$ & $0.057$ & $0.063$ & $0.056$ \\
&$30$ & $30$ & $0.100$ & $0.102$ & $0.105$ & $0.105$ & $0.103$ & $0.101$ & $0.099$ & $0.103$ & $0.102$ & $0.104$ & $0.107$ & $0.105$ & $0.105$ \\ \cline{2-16}
&$60$ & $5$ & $0.010$ & $0.012$ & $0.012$ & $0.029$ & $0.027$ & $0.014$ & $0.012$ & $0.028$ & $0.024$ & $0.016$ & $0.011$ & $0.023$ & $0.021$ \\
&$60$ & $5$ & $0.050$ & $0.052$ & $0.052$ & $0.063$ & $0.064$ & $0.050$ & $0.048$ & $0.063$ & $0.059$ & $0.050$ & $0.052$ & $0.059$ & $0.061$ \\
&$60$ & $5$ & $0.100$ & $0.100$ & $0.101$ & $0.098$ & $0.095$ & $0.098$ & $0.099$ & $0.097$ & $0.099$ & $0.099$ & $0.098$ & $0.100$ & $0.094$ \\
&$60$ & $30$ & $0.010$ & $0.017$ & $0.015$ & $0.020$ & $0.019$ & $0.016$ & $0.017$ & $0.020$ & $0.017$ & $0.016$ & $0.015$ & $0.019$ & $0.014$ \\
&$60$ & $30$ & $0.050$ & $0.052$ & $0.053$ & $0.058$ & $0.060$ & $0.055$ & $0.057$ & $0.061$ & $0.059$ & $0.057$ & $0.056$ & $0.062$ & $0.059$ \\
&$60$ & $30$ & $0.100$ & $0.103$ & $0.106$ & $0.107$ & $0.103$ & $0.102$ & $0.106$ & $0.107$ & $0.105$ & $0.103$ & $0.102$ & $0.106$ & $0.101$ \\
\hline \\[-1.8ex]
\end{tabular} }
\end{table}
\begin{table}[!htbp] \centering
\caption{Power comparision from Example \ref{exp2}}
\label{tab2}
\scalebox{0.6}{
\begin{tabular}{@{\extracolsep{5pt}} cccccccccccccccc}
\\[-1.8ex]\hline \hline \\[-1.8ex]
&& & & \multicolumn{4}{c}{} & \multicolumn{4}{c}{Gaussian Kernel} & \multicolumn{4}{c}{Laplacian Kernel} \\ \cline{9-12} \cline{13-16}
&$n$ & $p$ & $\alpha$ & $dCov$ & $mdCov$ & $T_{dCov}$ & $T_{mdCov}$ & $hCov$ & $mhCov$ & $T_{hCov}$ & $T_{mhCov}$ & $hCov$ & $mhCov$ & $T_{hCov}$ & $T_{mhCov}$ \\
\hline \\[-1.8ex]
\multirow{18}{*}{(i)}&$10$ & $5$ & $0.010$ & $0.635$ & $0.560$ & $0.691$ & $0.597$ & $0.629$ & $0.371$ & $0.685$ & $0.392$ & $0.516$ & $0.237$ & $0.585$ & $0.246$ \\
&$10$ & $5$ & $0.050$ & $0.833$ & $0.774$ & $0.855$ & $0.792$ & $0.825$ & $0.598$ & $0.849$ & $0.610$ & $0.741$ & $0.450$ & $0.772$ & $0.458$ \\
&$10$ & $5$ & $0.100$ & $0.910$ & $0.861$ & $0.914$ & $0.867$ & $0.906$ & $0.717$ & $0.912$ & $0.721$ & $0.839$ & $0.581$ & $0.851$ & $0.586$ \\
&$10$ & $30$ & $0.010$ & $0.795$ & $0.654$ & $0.788$ & $0.634$ & $0.796$ & $0.410$ & $0.787$ & $0.379$ & $0.769$ & $0.247$ & $0.762$ & $0.219$ \\
&$10$ & $30$ & $0.050$ & $0.936$ & $0.849$ & $0.937$ & $0.851$ & $0.935$ & $0.648$ & $0.937$ & $0.644$ & $0.921$ & $0.468$ & $0.924$ & $0.460$ \\
&$10$ & $30$ & $0.100$ & $0.970$ & $0.914$ & $0.970$ & $0.916$ & $0.970$ & $0.767$ & $0.970$ & $0.768$ & $0.963$ & $0.604$ & $0.964$ & $0.603$ \\
\cline{2-16}
&$30$ & $5$ & $0.010$ & $1$ & $1$ & $1$ & $1$ & $1$ & $0.999$ & $1$ & $0.998$ & $1$ & $0.980$ & $1$ & $0.982$ \\
&$30$ & $5$ & $0.050$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1.000$ & $1$ & $1.000$ & $1$ & $0.996$ & $1$ & $0.996$ \\
&$30$ & $5$ & $0.100$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1.000$ & $1$ & $1.000$ & $1$ & $0.998$ & $1$ & $0.998$ \\
&$30$ & $30$ & $0.010$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $0.996$ & $1$ & $0.996$ \\
&$30$ & $30$ & $0.050$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $0.999$ & $1$ & $0.999$ \\
&$30$ & $30$ & $0.100$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1.000$ & $1$ & $1.000$ \\
\cline{2-16}
&$60$ & $5$ & $0.010$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ \\
&$60$ & $5$ & $0.050$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ \\
&$60$ & $5$ & $0.100$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ \\
&$60$ & $30$ & $0.010$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ \\
&$60$ & $30$ & $0.050$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ \\
&$60$ & $30$ & $0.100$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ \\
\hline \\[-1.8ex]
\multirow{18}{*}{(ii)}& $10$ & $5$ & $0.010$ & $1.000$ & $0.999$ & $1.000$ & $0.999$ & $1.000$ & $0.986$ & $1.000$ & $0.989$ & $0.997$ & $0.935$ & $0.999$ & $0.942$ \\
&$10$ & $5$ & $0.050$ & $1.000$ & $1.000$ & $1.000$ & $1.000$ & $1.000$ & $0.997$ & $1.000$ & $0.997$ & $0.999$ & $0.983$ & $1.000$ & $0.983$ \\
&$10$ & $5$ & $0.100$ & $1.000$ & $1.000$ & $1.000$ & $1.000$ & $1.000$ & $0.999$ & $1.000$ & $0.999$ & $1.000$ & $0.992$ & $1.000$ & $0.993$ \\
&$10$ & $30$ & $0.010$ & $1$ & $1$ & $1$ & $1.000$ & $1$ & $0.998$ & $1$ & $0.998$ & $1$ & $0.973$ & $1$ & $0.970$ \\
&$10$ & $30$ & $0.050$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1.000$ & $1$ & $1.000$ & $1$ & $0.995$ & $1$ & $0.995$ \\
&$10$ & $30$ & $0.100$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1.000$ & $1$ & $1.000$ & $1$ & $0.997$ & $1$ & $0.997$ \\ \cline{2-16}
&$30$ & $5$ & $0.010$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ \\
&$30$ & $5$ & $0.050$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ \\
&$30$ & $5$ & $0.100$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ \\
&$30$ & $30$ & $0.010$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ \\
&$30$ & $30$ & $0.050$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ \\
&$30$ & $30$ & $0.100$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ \\ \cline{2-16}
&$60$ & $5$ & $0.010$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ \\
&$60$ & $5$ & $0.050$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ \\
&$60$ & $5$ & $0.100$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ \\
&$60$ & $30$ & $0.010$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ \\
&$60$ & $30$ & $0.050$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ \\
&$60$ & $30$ & $0.100$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ \\
\hline \\[-1.8ex]
\multirow{18}{*}{(iii)}&$10$ & $5$ & $0.010$ & $0.635$ & $0.497$ & $0.685$ & $0.537$ & $0.633$ & $0.238$ & $0.681$ & $0.260$ & $0.525$ & $0.138$ & $0.584$ & $0.135$ \\
&$10$ & $5$ & $0.050$ & $0.831$ & $0.728$ & $0.848$ & $0.748$ & $0.824$ & $0.460$ & $0.844$ & $0.477$ & $0.740$ & $0.311$ & $0.768$ & $0.323$ \\
&$10$ & $5$ & $0.100$ & $0.903$ & $0.830$ & $0.911$ & $0.835$ & $0.899$ & $0.597$ & $0.905$ & $0.604$ & $0.835$ & $0.440$ & $0.844$ & $0.446$ \\
&$10$ & $30$ & $0.010$ & $0.790$ & $0.583$ & $0.784$ & $0.555$ & $0.789$ & $0.273$ & $0.785$ & $0.247$ & $0.763$ & $0.147$ & $0.761$ & $0.122$ \\
&$10$ & $30$ & $0.050$ & $0.928$ & $0.800$ & $0.930$ & $0.797$ & $0.928$ & $0.490$ & $0.930$ & $0.486$ & $0.915$ & $0.331$ & $0.919$ & $0.324$ \\
&$10$ & $30$ & $0.100$ & $0.966$ & $0.888$ & $0.964$ & $0.889$ & $0.965$ & $0.628$ & $0.964$ & $0.626$ & $0.960$ & $0.460$ & $0.957$ & $0.453$ \\
&$30$ & $5$ & $0.010$ & $1$ & $1.000$ & $1$ & $1$ & $1$ & $0.985$ & $1$ & $0.989$ & $1.000$ & $0.890$ & $1.000$ & $0.898$ \\
&$30$ & $5$ & $0.050$ & $1$ & $1$ & $1$ & $1$ & $1$ & $0.996$ & $1$ & $0.997$ & $1$ & $0.971$ & $1$ & $0.971$ \\
&$30$ & $5$ & $0.100$ & $1$ & $1$ & $1$ & $1$ & $1$ & $0.999$ & $1$ & $0.999$ & $1$ & $0.984$ & $1$ & $0.984$ \\
&$30$ & $30$ & $0.010$ & $1$ & $1$ & $1$ & $1$ & $1$ & $0.998$ & $1$ & $0.999$ & $1$ & $0.950$ & $1$ & $0.948$ \\
&$30$ & $30$ & $0.050$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1.000$ & $1$ & $1.000$ & $1$ & $0.990$ & $1$ & $0.990$ \\
&$30$ & $30$ & $0.100$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1.000$ & $1$ & $1.000$ & $1$ & $0.997$ & $1$ & $0.997$ \\
&$60$ & $5$ & $0.010$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1.000$ & $1$ & $1$ \\
&$60$ & $5$ & $0.050$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ \\
&$60$ & $5$ & $0.100$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ \\
&$60$ & $30$ & $0.010$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ \\
&$60$ & $30$ & $0.050$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ \\
&$60$ & $30$ & $0.100$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ \\
\hline \\[-1.8ex]
\end{tabular} }
\end{table}
\begin{table}[!htbp] \centering
\caption{Power comparision under $H_{A_s}$ from Example \ref{exp3}}
\label{tab3}
\scalebox{0.6}{
\begin{tabular}{@{\extracolsep{5pt}} cccccccccccccccc}
\\[-1.8ex]\hline \hline \\[-1.8ex]
&& & & \multicolumn{4}{c}{} & \multicolumn{4}{c}{Gaussian Kernel} & \multicolumn{4}{c}{Laplacian Kernel} \\ \cline{9-12} \cline{13-16}
&$n$ & $p$ & $\alpha$ & $dCov$ & $mdCov$ & $T_{dCov}$ & $T_{mdCov}$ & $hCov$ & $mhCov$ & $T_{hCov}$ & $T_{mhCov}$ & $hCov$ & $mhCov$ & $T_{hCov}$ & $T_{mhCov}$ \\
\hline \\[-1.8ex]
\multirow{18}{*}{(i)}& $10$ & $5$ & $0.010$ & $0.113$ & $0.285$ & $0.144$ & $0.321$ & $0.110$ & $0.493$ & $0.138$ & $0.516$ & $0.172$ & $0.801$ & $0.226$ & $0.813$ \\
& $10$ & $5$ & $0.050$ & $0.231$ & $0.495$ & $0.254$ & $0.519$ & $0.236$ & $0.724$ & $0.256$ & $0.736$ & $0.356$ & $0.927$ & $0.398$ & $0.938$ \\
& $10$ & $5$ & $0.100$ & $0.325$ & $0.618$ & $0.332$ & $0.628$ & $0.325$ & $0.828$ & $0.336$ & $0.834$ & $0.495$ & $0.968$ & $0.506$ & $0.969$ \\
& $10$ & $30$ & $0.010$ & $0.032$ & $0.286$ & $0.028$ & $0.267$ & $0.032$ & $0.543$ & $0.030$ & $0.513$ & $0.044$ & $0.848$ & $0.041$ & $0.838$ \\
& $10$ & $30$ & $0.050$ & $0.101$ & $0.526$ & $0.098$ & $0.523$ & $0.098$ & $0.769$ & $0.099$ & $0.763$ & $0.124$ & $0.945$ & $0.128$ & $0.947$ \\
& $10$ & $30$ & $0.100$ & $0.158$ & $0.669$ & $0.162$ & $0.666$ & $0.160$ & $0.858$ & $0.160$ & $0.858$ & $0.203$ & $0.978$ & $0.205$ & $0.977$ \\ \cline{2-16}
& $30$ & $5$ & $0.010$ & $0.440$ & $0.997$ & $0.499$ & $0.999$ & $0.518$ & $1$ & $0.583$ & $1$ & $0.924$ & $1$ & $0.956$ & $1$ \\
& $30$ & $5$ & $0.050$ & $0.651$ & $1.000$ & $0.679$ & $1.000$ & $0.741$ & $1$ & $0.768$ & $1$ & $0.987$ & $1$ & $0.988$ & $1$ \\
& $30$ & $5$ & $0.100$ & $0.766$ & $1.000$ & $0.773$ & $1$ & $0.836$ & $1$ & $0.845$ & $1$ & $0.994$ & $1$ & $0.995$ & $1$ \\
& $30$ & $30$ & $0.010$ & $0.084$ & $1.000$ & $0.082$ & $1.000$ & $0.085$ & $1$ & $0.082$ & $1$ & $0.194$ & $1$ & $0.192$ & $1$ \\
& $30$ & $30$ & $0.050$ & $0.190$ & $1$ & $0.187$ & $1$ & $0.192$ & $1$ & $0.192$ & $1$ & $0.365$ & $1$ & $0.365$ & $1$ \\
& $30$ & $30$ & $0.100$ & $0.275$ & $1$ & $0.272$ & $1$ & $0.280$ & $1$ & $0.276$ & $1$ & $0.476$ & $1$ & $0.478$ & $1$ \\ \cline{2-16}
& $60$ & $5$ & $0.010$ & $0.948$ & $1$ & $0.976$ & $1$ & $0.983$ & $1$ & $0.992$ & $1$ & $1$ & $1$ & $1$ & $1$ \\
& $60$ & $5$ & $0.050$ & $0.994$ & $1$ & $0.996$ & $1$ & $0.998$ & $1$ & $0.999$ & $1$ & $1$ & $1$ & $1$ & $1$ \\
& $60$ & $5$ & $0.100$ & $0.999$ & $1$ & $0.999$ & $1$ & $1.000$ & $1$ & $1.000$ & $1$ & $1$ & $1$ & $1$ & $1$ \\
& $60$ & $30$ & $0.010$ & $0.185$ & $1$ & $0.173$ & $1$ & $0.194$ & $1$ & $0.183$ & $1$ & $0.587$ & $1$ & $0.587$ & $1$ \\
& $60$ & $30$ & $0.050$ & $0.346$ & $1$ & $0.346$ & $1$ & $0.361$ & $1$ & $0.360$ & $1$ & $0.779$ & $1$ & $0.782$ & $1$ \\
& $60$ & $30$ & $0.100$ & $0.462$ & $1$ & $0.459$ & $1$ & $0.475$ & $1$ & $0.473$ & $1$ & $0.861$ & $1$ & $0.864$ & $1$ \\
\hline \\[-1.8ex]
\multirow{18}{*}{(ii)} & $10$ & $5$ & $0.010$ & $0.167$ & $0.232$ & $0.237$ & $0.296$ & $0.192$ & $0.347$ & $0.263$ & $0.410$ & $0.279$ & $0.595$ & $0.391$ & $0.652$ \\
& $10$ & $5$ & $0.050$ & $0.306$ & $0.386$ & $0.341$ & $0.421$ & $0.356$ & $0.570$ & $0.401$ & $0.606$ & $0.525$ & $0.806$ & $0.584$ & $0.832$ \\
& $10$ & $5$ & $0.100$ & $0.401$ & $0.489$ & $0.409$ & $0.500$ & $0.479$ & $0.699$ & $0.487$ & $0.709$ & $0.674$ & $0.892$ & $0.689$ & $0.901$ \\
& $10$ & $30$ & $0.010$ & $0.080$ & $0.202$ & $0.091$ & $0.210$ & $0.082$ & $0.376$ & $0.091$ & $0.366$ & $0.099$ & $0.646$ & $0.123$ & $0.634$ \\
& $10$ & $30$ & $0.050$ & $0.178$ & $0.369$ & $0.191$ & $0.378$ & $0.179$ & $0.605$ & $0.192$ & $0.610$ & $0.229$ & $0.834$ & $0.252$ & $0.837$ \\
& $10$ & $30$ & $0.100$ & $0.257$ & $0.492$ & $0.259$ & $0.492$ & $0.264$ & $0.728$ & $0.265$ & $0.730$ & $0.342$ & $0.906$ & $0.351$ & $0.909$ \\ \cline{2-16}
& $30$ & $5$ & $0.010$ & $0.623$ & $0.847$ & $0.781$ & $0.950$ & $0.895$ & $0.999$ & $0.957$ & $1$ & $0.995$ & $1$ & $0.999$ & $1$ \\
& $30$ & $5$ & $0.050$ & $0.872$ & $0.984$ & $0.902$ & $0.990$ & $0.982$ & $1$ & $0.990$ & $1$ & $1.000$ & $1$ & $1$ & $1$ \\
& $30$ & $5$ & $0.100$ & $0.940$ & $0.996$ & $0.945$ & $0.995$ & $0.994$ & $1$ & $0.994$ & $1$ & $1$ & $1$ & $1$ & $1$ \\
& $30$ & $30$ & $0.010$ & $0.251$ & $0.929$ & $0.277$ & $0.944$ & $0.307$ & $1$ & $0.336$ & $1$ & $0.629$ & $1$ & $0.686$ & $1$ \\
& $30$ & $30$ & $0.050$ & $0.419$ & $0.982$ & $0.434$ & $0.985$ & $0.499$ & $1$ & $0.517$ & $1$ & $0.830$ & $1$ & $0.849$ & $1$ \\
& $30$ & $30$ & $0.100$ & $0.532$ & $0.995$ & $0.532$ & $0.995$ & $0.613$ & $1$ & $0.622$ & $1$ & $0.905$ & $1$ & $0.909$ & $1$ \\ \cline{2-16}
& $60$ & $5$ & $0.010$ & $0.999$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ \\
& $60$ & $5$ & $0.050$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ \\
& $60$ & $5$ & $0.100$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ \\
& $60$ & $30$ & $0.010$ & $0.643$ & $1$ & $0.684$ & $1$ & $0.790$ & $1$ & $0.833$ & $1$ & $0.996$ & $1$ & $0.999$ & $1$ \\
& $60$ & $30$ & $0.050$ & $0.824$ & $1$ & $0.836$ & $1$ & $0.918$ & $1$ & $0.930$ & $1$ & $1.000$ & $1$ & $1.000$ & $1$ \\
& $60$ & $30$ & $0.100$ & $0.894$ & $1$ & $0.896$ & $1$ & $0.955$ & $1$ & $0.958$ & $1$ & $1$ & $1$ & $1$ & $1$ \\
\hline \\[-1.8ex]
\multirow{18}{*}{(iii)} & $10$ & $5$ & $0.010$ & $0.043$ & $0.233$ & $0.060$ & $0.257$ & $0.042$ & $0.434$ & $0.053$ & $0.447$ & $0.076$ & $0.768$ & $0.098$ & $0.785$ \\
& $10$ & $5$ & $0.050$ & $0.121$ & $0.466$ & $0.141$ & $0.490$ & $0.119$ & $0.680$ & $0.137$ & $0.698$ & $0.191$ & $0.924$ & $0.214$ & $0.927$ \\
& $10$ & $5$ & $0.100$ & $0.201$ & $0.616$ & $0.212$ & $0.624$ & $0.203$ & $0.808$ & $0.210$ & $0.810$ & $0.291$ & $0.963$ & $0.298$ & $0.964$ \\
& $10$ & $30$ & $0.010$ & $0.017$ & $0.260$ & $0.013$ & $0.242$ & $0.017$ & $0.482$ & $0.012$ & $0.445$ & $0.021$ & $0.830$ & $0.017$ & $0.811$ \\
& $10$ & $30$ & $0.050$ & $0.062$ & $0.488$ & $0.062$ & $0.487$ & $0.063$ & $0.729$ & $0.062$ & $0.727$ & $0.071$ & $0.941$ & $0.070$ & $0.940$ \\
& $10$ & $30$ & $0.100$ & $0.120$ & $0.632$ & $0.116$ & $0.630$ & $0.118$ & $0.837$ & $0.115$ & $0.836$ & $0.131$ & $0.972$ & $0.130$ & $0.975$ \\ \cline{2-16}
& $30$ & $5$ & $0.010$ & $0.146$ & $0.999$ & $0.191$ & $1$ & $0.153$ & $1$ & $0.187$ & $1$ & $0.464$ & $1$ & $0.529$ & $1$ \\
& $30$ & $5$ & $0.050$ & $0.346$ & $1$ & $0.375$ & $1$ & $0.347$ & $1$ & $0.380$ & $1$ & $0.723$ & $1$ & $0.747$ & $1$ \\
& $30$ & $5$ & $0.100$ & $0.484$ & $1$ & $0.497$ & $1$ & $0.496$ & $1$ & $0.501$ & $1$ & $0.835$ & $1$ & $0.840$ & $1$ \\
& $30$ & $30$ & $0.010$ & $0.024$ & $1.000$ & $0.022$ & $1.000$ & $0.026$ & $1$ & $0.022$ & $1$ & $0.038$ & $1$ & $0.037$ & $1$ \\
& $30$ & $30$ & $0.050$ & $0.088$ & $1$ & $0.085$ & $1$ & $0.086$ & $1$ & $0.085$ & $1$ & $0.117$ & $1$ & $0.115$ & $1$ \\
& $30$ & $30$ & $0.100$ & $0.149$ & $1$ & $0.147$ & $1$ & $0.148$ & $1$ & $0.144$ & $1$ & $0.195$ & $1$ & $0.193$ & $1$ \\ \cline{2-16}
& $60$ & $5$ & $0.010$ & $0.547$ & $1$ & $0.630$ & $1$ & $0.566$ & $1$ & $0.642$ & $1$ & $0.978$ & $1$ & $0.988$ & $1$ \\
& $60$ & $5$ & $0.050$ & $0.802$ & $1$ & $0.835$ & $1$ & $0.808$ & $1$ & $0.836$ & $1$ & $0.997$ & $1$ & $0.998$ & $1$ \\
& $60$ & $5$ & $0.100$ & $0.907$ & $1$ & $0.911$ & $1$ & $0.905$ & $1$ & $0.913$ & $1$ & $0.999$ & $1$ & $0.999$ & $1$ \\
& $60$ & $30$ & $0.010$ & $0.038$ & $1$ & $0.030$ & $1$ & $0.038$ & $1$ & $0.029$ & $1$ & $0.089$ & $1$ & $0.080$ & $1$ \\
& $60$ & $30$ & $0.050$ & $0.122$ & $1$ & $0.117$ & $1$ & $0.119$ & $1$ & $0.119$ & $1$ & $0.217$ & $1$ & $0.214$ & $1$ \\
& $60$ & $30$ & $0.100$ & $0.198$ & $1$ & $0.196$ & $1$ & $0.199$ & $1$ & $0.197$ & $1$ & $0.326$ & $1$ & $0.325$ & $1$ \\
\hline \\[-1.8ex]
\end{tabular} }
\end{table}
\begin{table}[!htbp] \centering
\caption{Power comparision under $H_{A_s}$ from Example \ref{exp4}}
\label{tab4}
\scalebox{0.6}{
\begin{tabular}{@{\extracolsep{5pt}} cccccccccccccccc}
\\[-1.8ex]\hline \hline \\[-1.8ex]
&& & & \multicolumn{4}{c}{} & \multicolumn{4}{c}{Gaussian Kernel} & \multicolumn{4}{c}{Laplacian Kernel} \\ \cline{9-12} \cline{13-16}
&$n$ & $p$ & $\alpha$ & $dCov$ & $mdCov$ & $T_{dCov}$ & $T_{mdCov}$ & $hCov$ & $mhCov$ & $T_{hCov}$ & $T_{mhCov}$ & $hCov$ & $mhCov$ & $T_{hCov}$ & $T_{mhCov}$ \\
\hline \\[-1.8ex]
\multirow{18}{*}{(i)} & $10$ & $5$ & $0.010$ & $0.044$ & $0.196$ & $0.055$ & $0.218$ & $0.042$ & $0.348$ & $0.052$ & $0.367$ & $0.074$ & $0.672$ & $0.098$ & $0.685$ \\
& $10$ & $5$ & $0.050$ & $0.120$ & $0.390$ & $0.136$ & $0.416$ & $0.114$ & $0.582$ & $0.129$ & $0.604$ & $0.183$ & $0.859$ & $0.209$ & $0.870$ \\
& $10$ & $5$ & $0.100$ & $0.201$ & $0.542$ & $0.209$ & $0.546$ & $0.191$ & $0.722$ & $0.197$ & $0.731$ & $0.292$ & $0.927$ & $0.304$ & $0.931$ \\
& $10$ & $30$ & $0.010$ & $0.018$ & $0.212$ & $0.014$ & $0.194$ & $0.017$ & $0.387$ & $0.014$ & $0.362$ & $0.022$ & $0.722$ & $0.017$ & $0.706$ \\
& $10$ & $30$ & $0.050$ & $0.066$ & $0.434$ & $0.064$ & $0.428$ & $0.066$ & $0.627$ & $0.064$ & $0.625$ & $0.075$ & $0.892$ & $0.077$ & $0.891$ \\
& $10$ & $30$ & $0.100$ & $0.123$ & $0.571$ & $0.121$ & $0.568$ & $0.123$ & $0.749$ & $0.119$ & $0.750$ & $0.135$ & $0.944$ & $0.132$ & $0.946$ \\ \cline{2-16}
& $30$ & $5$ & $0.010$ & $0.158$ & $0.988$ & $0.197$ & $0.996$ & $0.136$ & $1$ & $0.163$ & $1$ & $0.486$ & $1$ & $0.555$ & $1$ \\
& $30$ & $5$ & $0.050$ & $0.341$ & $1.000$ & $0.369$ & $1$ & $0.303$ & $1$ & $0.328$ & $1$ & $0.725$ & $1$ & $0.756$ & $1$ \\
& $30$ & $5$ & $0.100$ & $0.483$ & $1$ & $0.488$ & $1$ & $0.433$ & $1$ & $0.444$ & $1$ & $0.838$ & $1$ & $0.846$ & $1$ \\
& $30$ & $30$ & $0.010$ & $0.026$ & $0.996$ & $0.023$ & $0.996$ & $0.027$ & $1.000$ & $0.022$ & $1.000$ & $0.043$ & $1$ & $0.038$ & $1$ \\
& $30$ & $30$ & $0.050$ & $0.089$ & $1.000$ & $0.084$ & $0.999$ & $0.088$ & $1$ & $0.083$ & $1$ & $0.123$ & $1$ & $0.125$ & $1$ \\
& $30$ & $30$ & $0.100$ & $0.153$ & $1.000$ & $0.152$ & $1.000$ & $0.151$ & $1$ & $0.152$ & $1$ & $0.209$ & $1$ & $0.204$ & $1$ \\ \cline{2-16}
& $60$ & $5$ & $0.010$ & $0.559$ & $1$ & $0.637$ & $1$ & $0.461$ & $1$ & $0.539$ & $1$ & $0.989$ & $1$ & $0.996$ & $1$ \\
& $60$ & $5$ & $0.050$ & $0.816$ & $1$ & $0.847$ & $1$ & $0.738$ & $1$ & $0.774$ & $1$ & $1.000$ & $1$ & $1$ & $1$ \\
& $60$ & $5$ & $0.100$ & $0.916$ & $1$ & $0.925$ & $1$ & $0.861$ & $1$ & $0.870$ & $1$ & $1$ & $1$ & $1$ & $1$ \\
& $60$ & $30$ & $0.010$ & $0.037$ & $1$ & $0.032$ & $1$ & $0.036$ & $1$ & $0.031$ & $1$ & $0.091$ & $1$ & $0.085$ & $1$ \\
& $60$ & $30$ & $0.050$ & $0.125$ & $1$ & $0.119$ & $1$ & $0.122$ & $1$ & $0.115$ & $1$ & $0.231$ & $1$ & $0.228$ & $1$ \\
& $60$ & $30$ & $0.100$ & $0.208$ & $1$ & $0.207$ & $1$ & $0.204$ & $1$ & $0.202$ & $1$ & $0.350$ & $1$ & $0.346$ & $1$ \\
\hline \\[-1.8ex]
\multirow{18}{*}{(ii)} & $10$ & $5$ & $0.010$ & $0.044$ & $0.217$ & $0.059$ & $0.242$ & $0.040$ & $0.393$ & $0.055$ & $0.413$ & $0.077$ & $0.713$ & $0.106$ & $0.732$ \\
&$10$ & $5$ & $0.050$ & $0.124$ & $0.432$ & $0.141$ & $0.453$ & $0.117$ & $0.637$ & $0.131$ & $0.655$ & $0.202$ & $0.886$ & $0.224$ & $0.895$ \\
&$10$ & $5$ & $0.100$ & $0.210$ & $0.577$ & $0.213$ & $0.583$ & $0.196$ & $0.771$ & $0.204$ & $0.775$ & $0.304$ & $0.942$ & $0.318$ & $0.942$ \\
&$10$ & $30$ & $0.010$ & $0.020$ & $0.247$ & $0.013$ & $0.224$ & $0.019$ & $0.439$ & $0.013$ & $0.409$ & $0.022$ & $0.774$ & $0.018$ & $0.763$ \\
&$10$ & $30$ & $0.050$ & $0.064$ & $0.474$ & $0.064$ & $0.474$ & $0.063$ & $0.677$ & $0.063$ & $0.676$ & $0.075$ & $0.913$ & $0.076$ & $0.913$ \\
&$10$ & $30$ & $0.100$ & $0.126$ & $0.606$ & $0.125$ & $0.604$ & $0.126$ & $0.795$ & $0.126$ & $0.790$ & $0.141$ & $0.956$ & $0.138$ & $0.955$ \\ \cline{2-16}
&$30$ & $5$ & $0.010$ & $0.178$ & $0.995$ & $0.221$ & $0.999$ & $0.148$ & $1$ & $0.186$ & $1$ & $0.544$ & $1$ & $0.608$ & $1$ \\
&$30$ & $5$ & $0.050$ & $0.376$ & $1$ & $0.409$ & $1$ & $0.333$ & $1$ & $0.358$ & $1$ & $0.775$ & $1$ & $0.797$ & $1$ \\
&$30$ & $5$ & $0.100$ & $0.518$ & $1$ & $0.526$ & $1$ & $0.468$ & $1$ & $0.478$ & $1$ & $0.871$ & $1$ & $0.880$ & $1$ \\
&$30$ & $30$ & $0.010$ & $0.027$ & $0.998$ & $0.023$ & $0.998$ & $0.026$ & $1.000$ & $0.022$ & $1$ & $0.043$ & $1$ & $0.038$ & $1$ \\
&$30$ & $30$ & $0.050$ & $0.088$ & $1.000$ & $0.087$ & $1.000$ & $0.088$ & $1$ & $0.086$ & $1$ & $0.128$ & $1$ & $0.128$ & $1$ \\
&$30$ & $30$ & $0.100$ & $0.155$ & $1.000$ & $0.152$ & $1.000$ & $0.154$ & $1$ & $0.152$ & $1$ & $0.218$ & $1$ & $0.213$ & $1$ \\ \cline{2-16}
&$60$ & $5$ & $0.010$ & $0.632$ & $1$ & $0.709$ & $1$ & $0.526$ & $1$ & $0.609$ & $1$ & $0.995$ & $1$ & $0.999$ & $1$ \\
&$60$ & $5$ & $0.050$ & $0.870$ & $1$ & $0.895$ & $1$ & $0.792$ & $1$ & $0.826$ & $1$ & $1$ & $1$ & $1$ & $1$ \\
&$60$ & $5$ & $0.100$ & $0.946$ & $1$ & $0.952$ & $1$ & $0.904$ & $1$ & $0.911$ & $1$ & $1$ & $1$ & $1$ & $1$ \\
&$60$ & $30$ & $0.010$ & $0.044$ & $1$ & $0.037$ & $1$ & $0.043$ & $1$ & $0.036$ & $1$ & $0.105$ & $1$ & $0.096$ & $1$ \\
&$60$ & $30$ & $0.050$ & $0.126$ & $1$ & $0.125$ & $1$ & $0.123$ & $1$ & $0.121$ & $1$ & $0.251$ & $1$ & $0.244$ & $1$ \\
&$60$ & $30$ & $0.100$ & $0.213$ & $1$ & $0.211$ & $1$ & $0.211$ & $1$ & $0.206$ & $1$ & $0.368$ & $1$ & $0.366$ & $1$ \\
\hline \\[-1.8ex]
\multirow{18}{*}{(iii)} & $10$ & $5$ & $0.010$ & $0.019$ & $0.024$ & $0.023$ & $0.028$ & $0.017$ & $0.033$ & $0.022$ & $0.040$ & $0.023$ & $0.090$ & $0.029$ & $0.095$ \\
& $10$ & $5$ & $0.050$ & $0.058$ & $0.079$ & $0.068$ & $0.089$ & $0.057$ & $0.111$ & $0.067$ & $0.115$ & $0.068$ & $0.232$ & $0.081$ & $0.242$ \\
& $10$ & $5$ & $0.100$ & $0.113$ & $0.148$ & $0.117$ & $0.151$ & $0.114$ & $0.194$ & $0.118$ & $0.196$ & $0.124$ & $0.351$ & $0.129$ & $0.355$ \\
& $10$ & $30$ & $0.010$ & $0.016$ & $0.026$ & $0.012$ & $0.020$ & $0.016$ & $0.037$ & $0.012$ & $0.030$ & $0.017$ & $0.089$ & $0.013$ & $0.076$ \\
& $10$ & $30$ & $0.050$ & $0.059$ & $0.086$ & $0.057$ & $0.083$ & $0.060$ & $0.112$ & $0.058$ & $0.105$ & $0.061$ & $0.233$ & $0.060$ & $0.225$ \\
& $10$ & $30$ & $0.100$ & $0.111$ & $0.156$ & $0.108$ & $0.153$ & $0.112$ & $0.199$ & $0.108$ & $0.193$ & $0.112$ & $0.357$ & $0.109$ & $0.346$ \\ \cline{2-16}
& $30$ & $5$ & $0.010$ & $0.019$ & $0.051$ & $0.021$ & $0.068$ & $0.017$ & $0.141$ & $0.021$ & $0.170$ & $0.026$ & $0.673$ & $0.032$ & $0.724$ \\
& $30$ & $5$ & $0.050$ & $0.061$ & $0.166$ & $0.070$ & $0.188$ & $0.058$ & $0.339$ & $0.066$ & $0.360$ & $0.083$ & $0.889$ & $0.091$ & $0.903$ \\
& $30$ & $5$ & $0.100$ & $0.117$ & $0.283$ & $0.117$ & $0.288$ & $0.117$ & $0.488$ & $0.116$ & $0.497$ & $0.153$ & $0.953$ & $0.153$ & $0.955$ \\
& $30$ & $30$ & $0.010$ & $0.017$ & $0.074$ & $0.012$ & $0.065$ & $0.017$ & $0.182$ & $0.012$ & $0.165$ & $0.017$ & $0.754$ & $0.012$ & $0.742$ \\
& $30$ & $30$ & $0.050$ & $0.061$ & $0.202$ & $0.058$ & $0.198$ & $0.061$ & $0.378$ & $0.059$ & $0.373$ & $0.063$ & $0.913$ & $0.061$ & $0.913$ \\
& $30$ & $30$ & $0.100$ & $0.112$ & $0.309$ & $0.110$ & $0.307$ & $0.113$ & $0.518$ & $0.110$ & $0.517$ & $0.117$ & $0.960$ & $0.114$ & $0.959$ \\ \cline{2-16}
& $60$ & $5$ & $0.010$ & $0.019$ & $0.174$ & $0.024$ & $0.219$ & $0.017$ & $0.580$ & $0.022$ & $0.666$ & $0.034$ & $1.000$ & $0.041$ & $1$ \\
& $60$ & $5$ & $0.050$ & $0.066$ & $0.421$ & $0.073$ & $0.458$ & $0.061$ & $0.853$ & $0.069$ & $0.883$ & $0.108$ & $1$ & $0.119$ & $1$ \\
& $60$ & $5$ & $0.100$ & $0.123$ & $0.600$ & $0.128$ & $0.612$ & $0.119$ & $0.941$ & $0.122$ & $0.949$ & $0.179$ & $1$ & $0.183$ & $1$ \\
& $60$ & $30$ & $0.010$ & $0.013$ & $0.251$ & $0.009$ & $0.233$ & $0.013$ & $0.680$ & $0.010$ & $0.665$ & $0.014$ & $1.000$ & $0.010$ & $1$ \\
& $60$ & $30$ & $0.050$ & $0.053$ & $0.485$ & $0.051$ & $0.484$ & $0.052$ & $0.869$ & $0.050$ & $0.871$ & $0.056$ & $1$ & $0.055$ & $1$ \\
& $60$ & $30$ & $0.100$ & $0.105$ & $0.620$ & $0.101$ & $0.619$ & $0.106$ & $0.930$ & $0.101$ & $0.929$ & $0.107$ & $1$ & $0.106$ & $1$ \\
\hline \\[-1.8ex]
\end{tabular} }
\end{table}
\section{Technical Details}
\subsection{Proof of Proposition \ref{prop:taylor}}
\begin{proof}
Denote $f^{(2)}(t) = - \frac{1}{4} (1+t)^{-\frac{3}{2}}$. The remainder term can be written as
\begin{align*}
R_{X}(X_{s},X_{t}) = \int_{0}^1 \int_{0}^{1} v f^{(2)} \left( uv L_{X}(X_s, X_t) \right) dudv \times \left( L_{X}(X_s, X_t) \right)^2.
\end{align*}
Set $ \varphi (x) = \int_{0}^1 \int_{0}^{1} v f^{(2)} \left( uv x \right) dudv $. Then $\varphi (x)$ is continuous at $0$. Next, by the continuous mapping theorem, we have
\begin{align*}
\int_{0}^1 \int_{0}^{1} v f^{(2)} \left( uv L_{X}(X_s, X_t) \right) dudv \overset{p}{\rightarrow}
\int_{0}^1 \int_{0}^{1} v f^{(2)} \left( 0 \right) dudv.
\end{align*}
So, $ R_{X}(X_{s},X_{t}) \asymp_p \left( L_{X}(X_s, X_t) \right)^2 $. Similar arguments hold for $R_{Y}(Y_{s},Y_{t})$.
\end{proof}
\subsection{Proof of Remark \ref{remark:mainthm}}
\label{App:proofRemark}
\begin{proof}
(i) Notice that
\begin{align*}
& \sqrt{var[ L(X_{k}, X_{l}) ]} \\
\asymp & \sqrt{ \frac{ var[ (X_{k} - X_{l})^{T}( X_{k} -X_{l} ) ] }{p^{2}} } \\
= & \sqrt{ \frac{ var \{[ \mathbf{A} (U_{k} - U_{l}) + (\Phi_{k} - \Phi_{l}) ]^{T}[ \mathbf{A} (U_{k} - U_{l}) + (\Phi_{k} - \Phi_{l}) ] \} }{p^{2}} }.
\end{align*}
Denote $\mathbf{C} = (c_{ij}) = \mathbf{A}^{T} \mathbf{A}$. We obtain that
\begin{align*}
& var \{[ \mathbf{A} (U_{k} - U_{l}) + (\Phi_{k} - \Phi_{l}) ]^{T}[ \mathbf{A} (U_{k} - U_{l}) + (\Phi_{k} - \Phi_{l}) ] \} \\
= & var \big[ (U_{k} - U_{l})^{T} \mathbf{A}^{T} \mathbf{A} (U_{k} - U_{l}) \\
& \quad \quad + (\Phi_{k} - \Phi_{l})^{T}(\Phi_{k} - \Phi_{l}) +2 (U_{k} - U_{l})^{T}\mathbf{A}^{T}(\Phi_{k} - \Phi_{l}) \big] \\
= & var \bigg[ \sum\limits_{i=1}^{s_{1}} \sum\limits_{j=1}^{s_{1}} c_{ij} (u_{ki} - u_{li})(u_{kj} - u_{lj}) + \sum\limits_{i=1}^{p}( \phi_{ki} - \phi_{li} )^{2} \\
& \quad\quad + 2 \sum\limits_{i=1}^{s_{1}} \sum\limits_{j=1}^{p} a_{ji} (u_{ki} - u_{li})( \phi_{kj} - \phi_{lj} ) \bigg] \\
\leq & 2 \sum\limits_{i=1}^{s_{1}} \sum\limits_{j=1}^{s_{1}} c_{ij}^{2} var[ (u_{ki} - u_{li})(u_{kj} - u_{lj})] + \sum\limits_{i=1}^{p} var [ ( \phi_{ki} - \phi_{li} )^{2}] \\
& \quad\quad + 4 \sum\limits_{i=1}^{s_{1}} \sum\limits_{j=1}^{p} a_{ji}^{2} var[ (u_{ki} - u_{li})( \phi_{kj} - \phi_{lj} )].
\end{align*}
Since the 4th moment is bounded uniformly for each $u_{ki}$ and $\phi_{ki}$, $ var[ (u_{ki} - u_{li})(u_{kj} - u_{lj})]$, $var [ ( \phi_{ki} - \phi_{li} )^{2}]$ and $var[ (u_{ki} - u_{li})( \phi_{kj} - \phi_{lj} )]$ are all uniformly bounded by a constant. As $\|A\|_F^2=O(p^{1/2})$,
we have $\|A^TA\|_F^2=O(p)$ by the Cauchy-Schwarz inequality. It follows that
\begin{align*}
var \{[ \mathbf{A} (U_{k} - U_{l}) + (\Phi_{k} - \Phi_{l}) ]^{T}[ \mathbf{A} (U_{k} - U_{l}) + (\Phi_{k} - \Phi_{l}) ] \} = O(p).
\end{align*}
Thus, we have $a_{p} = 1/ \sqrt{p} \text{ and } b_q = 1/ \sqrt{q}.$
\end{proof}
\subsection{Proof of Theorem \ref{thm:decomp}}
\label{proof:decomp}
\begin{proof}
(i) Recall that
$
dCov_n^2(\mathbf{X},\mathbf{Y}) = (\widetilde{\mathbf{A}} \cdot \widetilde{\mathbf{B}} ).
$
Using the approximation of $b_{st}$ in Proposition \ref{prop:taylor}, we have
\begin{align*}
\frac{1}{\tau_X}\widetilde{\mathbf{A}} = \widetilde{\mathbf{1}}_{n \times n} + \frac{1}{2} \widetilde{ \mathbf L}_X + \widetilde{ \mathbf R}_X = \frac{1}{2} \widetilde{ \mathbf L}_X + \widetilde{ \mathbf R}_X,
\end{align*}
where $\mathbf{L}_X= (L_{X}(X_{s}, X_{t} ))_{s,t=1}^n$ and $\mathbf{R}_X= (R_{X}(X_{s}, X_{t} ))_{s,t=1}^n$. Similarly, $\frac{1}{\tau_Y} \widetilde{\mathbf{B}} = \frac{1}{2}\widetilde{ \mathbf L}_Y + \widetilde{ \mathbf R}_Y $. Then, we have
\begin{align*}
\frac{ dCov_n^2(\mathbf{X},\mathbf{Y})}{\tau} & =( (\frac{1}{2}\widetilde{ \mathbf L}_X + \widetilde{ \mathbf R}_X) \cdot ( \frac{1}{2}\widetilde{ \mathbf L}_Y + \widetilde{ \mathbf R}_Y)) \\
& = \frac{1}{4} (\widetilde{ \mathbf L}_X \cdot \widetilde{ \mathbf L}_Y ) + \frac{1}{2} (\widetilde{ \mathbf L}_X \cdot \widetilde{ \mathbf R}_Y ) + \frac{1}{2} (\widetilde{ \mathbf R}_X \cdot \widetilde{ \mathbf L}_Y ) + (\widetilde{ \mathbf R}_X \cdot \widetilde{ \mathbf R}_Y ).
\end{align*}
Let $R_n = \frac{1}{2} (\widetilde{ \mathbf L}_X \cdot \widetilde{ \mathbf R}_Y ) + \frac{1}{2} (\widetilde{ \mathbf R}_X \cdot \widetilde{ \mathbf L}_Y ) + (\widetilde{ \mathbf R}_X \cdot \widetilde{ \mathbf R}_Y )$. We show that $\frac{1}{4} (\widetilde{ \mathbf L}_X \cdot \widetilde{ \mathbf L}_Y )$ can be written as sum of sample component-wise cross-covariances up to a constant factor in the following Lemma.
\begin{lemma}
\label{PropEquality}
\begin{align*}
\frac{1}{4}(\widetilde{ \mathbf L}_X \cdot \widetilde{ \mathbf L}_Y ) = \frac{1}{\tau^2} \sum\limits_{i=1}^{p} \sum\limits_{j=1}^{q} \text{cov}_{n}^2 (x_{i}, y_{j}) .
\end{align*}
\end{lemma}
\begin{proof}
By Lemma A.1. of \cite{park2015}, since all diagonal entries of distance matrices $\mathbf A$ and $\mathbf B$ are equal to 0, we have
$
(\widetilde{\mathbf{A}} \cdot \widetilde{\mathbf{B}} ) = (\mathbf{A} \cdot \widetilde{ \widetilde{\mathbf{B}} } ).
$
Then, it can be directly verified that for any $1\leq s,t \leq n$, $\sum_{u =1 }^{n} \widetilde{b}_{ut} = \sum_{v = 1}^{n} \widetilde{b}_{sv} = 0$ and it further implies that
\begin{itemize}
\item[(i)] $ \widetilde{ \widetilde{\mathbf{B}} } = \widetilde{ \mathbf B}$ as long as the diagonal elements of $\mathbf{B}$ are 0;
\item[(ii)] $\widetilde{ \mathbf B} = 0$ if $\mathbf{B} = \mathbf{a} \mathbf{1}_n^T \text{ or } \mathbf{B} = \mathbf{1}_n \mathbf{a}^T = 0$ for any vector $\mathbf{a} \in \mathbb{R}^{n}$.
\end{itemize}
Direct calculation shows that
\begin{multline}
\label{eq:defDcov}
(\mathbf{A} \cdot \widetilde{\mathbf{B}} )
= \frac{1}{\binom{n}{2}} \frac{1}{2!} \sum\limits_{(s,t) \in \mathbf i_{2}^{n}} a_{st}b_{st}
\\ + \frac{1}{\binom{n}{4}} \frac{1}{4!} \sum\limits_{(s,t,u,v) \in \mathbf i_{4}^{n}} a_{st}b_{uv}
- \frac{2}{\binom{n}{3}} \frac{1}{3!} \sum\limits_{(s,t,u) \in \mathbf i_{3}^{n}} a_{st}b_{su},
\end{multline}
where $\mathbf{i}_{m}^{n}$ denotes the set of all $m$-tuples drawn without replacement from $\{ 1,2, \cdots, n \}$. Equation \eqref{eq:defDcov} can be used as equivalent definition of the sample distance covariance.
Notice that
\begin{align*}
\widetilde{\mathbf{L}}_X = \widetilde{ \mathbf D}_X - \widetilde{ \mathbf 1}_{n \times n} = \widetilde{ \mathbf D}_X,
\end{align*}
where $ \mathbf{D}_X = \frac{1}{\tau_{X}^2}(|X_s - X_t|^2)_{s,t=1}^n$. Similarly, $\widetilde{\mathbf{L}}_Y = \widetilde{ \mathbf D}_Y$. Then, we can further decompose $ \widetilde{ \mathbf D}_X $ as follows,
\begin{align*}
\widetilde{ \mathbf D}_X = \widetilde{ \mathbf D}_{X, 1} + \widetilde{ \mathbf D}_{X, 2} + \widetilde{ \mathbf D}_{X, 3} = \widetilde{ \mathbf D}_{X, 2},
\end{align*}
where $ \mathbf {D}_{X, 1} = \frac{1}{\tau_X^2} (X_{s}^T X_s)_{s,t=1}^n$, $\mathbf {D}_{X, 2} =-2 \frac{1}{\tau_X^2} (X_{s}^T X_t)_{s,t=1}^n$ and $\mathbf {D}_{X, 3} = \frac{1}{\tau_X^2} (X_{t}^T X_t)_{s,t=1}^n$. Similarly, $ \widetilde{ \mathbf D}_Y = \widetilde{ \mathbf D}_{Y, 2}$. Next,
using Equation \eqref{eq:defDcov}, we have
\begin{align*}
& \tau^2 \times (\widetilde{ \mathbf L}_X \cdot \widetilde{ \mathbf L}_Y ) \\
= & \tau^2 \times (\widetilde{ \mathbf D}_{X, 2} \cdot \widetilde{ \mathbf D}_{Y , 2} ) \\
= & 4 \left\lbrace \frac{1}{\binom{n}{2}} \frac{1}{2!} \sum\limits_{(s,t) \in i_{2}^{n}} X_s^T X_t Y_s^T Y_t + \right. \\
& \left. \frac{1}{\binom{n}{4}} \frac{1}{4!} \sum\limits_{(s,t,u,v) \in \mathbf i_{4}^{n}} X_s^T X_t Y_u^T Y_v
-\frac{2}{\binom{n}{3}} \frac{1}{3!} \sum\limits_{(s,t,u) \in \mathbf i_{3}^{n}} X_s^T X_t Y_s^T Y_u \right\rbrace \\
= & 4 \sum\limits_{i=1}^p \sum\limits_{j=1}^q \left\lbrace \frac{1}{\binom{n}{2}} \frac{1}{2!} \sum\limits_{(s,t) \in i_{2}^{n}} x_{si} x_{ti} y_{sj} y_{tj} + \right. \\
& \quad\quad \left. \frac{1}{\binom{n}{4}} \frac{1}{4!} \sum\limits_{(s,t,u,v) \in \mathbf i_{4}^{n}} x_{si} x_{ti} y_{uj} y_{vj}
-\frac{2}{\binom{n}{3}} \frac{1}{3!} \sum\limits_{(s,t,u) \in \mathbf i_{3}^{n}} x_{si} x_{ti} y_{sj} y_{uj} \right\rbrace \\
= & 4 \sum_{i=1}^p \sum_{j=1}^q \left\lbrace \frac{ 1 }{\binom{n}{4}} \sum_{k< l< s < t } \frac{1}{4!} \sum_{ * }^{(k, l , s, t )} \frac{(x_{ki} - x_{li}) (y_{kj} - y_{lj})(x_{si} - x_{ti}) (y_{sj} - x_{tj})}{4} \right\rbrace \\
= & 4 \sum_{i=1}^p \sum_{j=1}^q cov_n^2(\mathcal X_i, \mathcal Y_j).
\end{align*}
\end{proof}
Therefore, by Lemma \ref{PropEquality}, we have the following decomposition,
\begin{align*}
dCov^2_n(\mathbf X, \mathbf Y) = \frac{1}{\tau} \sum_{i=1}^p \sum_{j=1}^q cov_n^2(\mathcal X_i, \mathcal Y_j) + \mathcal{R}_n,
\end{align*}
where ${\cal R}_n = \tau R_n$.
(ii) Note $L_X(X_s, X_t) = O_p (a_p)=o_p(1)$ and $L_Y(Y_s, Y_t)=O_p(b_q)=o_p(1)$ for $s \ne t \in \{1,\dots,n\}$. We can then apply Proposition \ref{prop:taylor}, obtain that $R_X(X_s, X_t) = O_p (L_X(X_s, X_t)^2)$ and $R_Y(Y_s, Y_t) = O_p (L_Y(Y_s, Y_t)^2)$. For the leading term $\tau (\widetilde{ \mathbf L}_X \cdot \widetilde{ \mathbf L}_Y )$, it can be easily seen from Equation \eqref{eq:defDcov} that $(\widetilde{ \mathbf L}_X \cdot \widetilde{ \mathbf L}_Y ) = O_p (a_p b_q) $. Similarly, for the remainder terms,
$(\widetilde{ \mathbf L}_X \cdot \widetilde{ \mathbf R}_Y ) = O_p ( a_p b_q^2 )$, $(\widetilde{ \mathbf R}_X \cdot \widetilde{ \mathbf L}_Y ) = O_p(a_p^2b_q) $ and $(\widetilde{ \mathbf R}_X \cdot \widetilde{ \mathbf R}_Y ) =O_p ( a_p^2b_q^2) $. Thus, we have $R_n = O_p (a_p^2b_q + a_p b_q^2 )$ and ${\cal R}_n = \tau R_n = O_p (\tau a_p^2b_q + \tau a_p b_q^2 ) = o_p(1) $. Therefore the remainder terms are negligible comparing to the leading term.
\end{proof}
\subsection{Proof of Theorem \ref{thm:decompHsic}}
\begin{proof}
(i) We first show that $\gamma_{\mathbf X}$ is asymptotically equal to $\tau_{X}$ (similar result applies to $\gamma_{\mathbf Y}$ and $\tau_{Y}$). Recall that for all $s \neq t$,
\begin{align*}
L_{X}(X_{s}, X_{t}) = \frac{|X_{s} - X_{t}|^{2} - \tau_{X}^{2}}{ \tau_{X}^{2}}.
\end{align*}
Since $ L_{X}(X_{s}, X_{t}) = O_{p} (a_{p}) =o_p(1)$, we have
$\frac{|X_{s} - X_{t}|^{2} }{ \tau_{X}^{2} } \overset{p}{\rightarrow} 1$. Then
$$
\frac{\text{median}\{|X_{s} - X_{t}|^{2}\} }{ \tau_{X}^{2} } \overset{p}{\rightarrow} 1
$$
and thus
\begin{align*}
\frac{\tau_{X}}{\gamma_{\mathbf X}} = \sqrt{ \frac{ \tau_{X}^{2} }{\text{median}\{|X_{i} - X_{j}|^{2}\} } } \overset{p}{\rightarrow} 1.
\end{align*}
Similar arguments can also be used to show that $ \frac{\tau_{Y}}{\gamma_{\mathbf Y}} \overset{p}{\rightarrow} 1$. Next, under Proposition \ref{prop:taylor}, we can deduce that
\begin{align*}
& f \left( \frac{ |X_{s} - X_{t}|}{ \gamma_{\mathbf X}} \right) \\
= & f \left( \frac{|X_{s} - X_{t}|}{ \tau_{X} } \frac{\tau_{X}}{\gamma_{\mathbf X}} \right) \\
= & f\left( \left\lbrace 1+ \frac{ L_{X}(X_{s}, X_{t})}{2} + R_{X}(X_{s}, X_{t}) \right\rbrace \frac{\tau_{X}}{\gamma_{\mathbf X}} \right) \\
= & f\left(\frac{\tau_{X}}{\gamma_{\mathbf X}}\right) + f^{(1)} \left( \frac{\tau_{X}}{ \gamma_{\mathbf X}} \right) \left\lbrace \frac{ L_{X}(X_{s}, X_{t})}{2} + R_{X}(X_{s}, X_{t}) \right\rbrace \frac{\tau_{X}}{\gamma_{\mathbf X}} + R_{f}(X_{s},X_{t}),
\end{align*}
where $ R_{f}(X_{s},X_{t}) $ is the remainder term. Similarly,
\begin{multline*}
g \left( \frac{|Y_{s} - Y_{t}|}{\gamma_{\mathbf Y}} \right) = g\left(\frac{\tau_{Y}}{\gamma_{\mathbf Y}}\right) + \\ g^{(1)} \left( \frac{\tau_{Y}}{\gamma_{\mathbf Y}} \right) \left\lbrace \frac{L_{Y}(Y_{s}, Y_{t})}{2} + R_{Y}(Y_{s}, Y_{t}) \right\rbrace \frac{\tau_{Y}}{\gamma_{\mathbf Y}} + R_{g}(Y_{s},Y_{t}).
\end{multline*}
Similar to the proof of Theorem \ref{thm:decomp},
\begin{multline*}
hCov_n^2(\mathbf X, \mathbf Y) = ( \widetilde{\mathbf R} \cdot \widetilde{\mathbf H}) \\= \frac{1}{4} f^{(1)} \left( \frac{\tau_{X}}{ \gamma_{\mathbf X}} \right) g^{(1)} \left( \frac{\tau_{Y}}{\gamma_{\mathbf Y}} \right)\frac{\tau_{X}}{ \gamma_{\mathbf X}} \frac{\tau_{Y}}{\gamma_{\mathbf Y}} (\widetilde{ \mathbf L}_X \cdot \widetilde{ \mathbf L}_Y ) + \frac{1}{2}f^{(1)} \left( \frac{\tau_{X}}{ \gamma_{\mathbf X}} \right)\frac{\tau_{X}}{ \gamma_{\mathbf X}} (\widetilde{ \mathbf L}_X \cdot \widetilde{ \mathbf R}_Y )\\ + \frac{1}{2}g^{(1)} \left( \frac{\tau_{Y}}{\gamma_{\mathbf Y}} \right)\frac{\tau_{Y}}{\gamma_{\mathbf Y}} (\widetilde{ \mathbf R}_X \cdot \widetilde{ \mathbf L}_Y ) + (\widetilde{ \mathbf R}_X \cdot \widetilde{ \mathbf R}_Y ),
\end{multline*}
where $\mathbf{L}_X= (L_{X}(X_{s}, X_{t} ))_{s,t=1}^n$, $\mathbf{L}_Y= (L_{Y}(Y_{s}, Y_{t} ))_{s,t=1}^n$ and
\begin{align*}
& \mathbf{R}_X= \left(f^{(1)} \left( \frac{\tau_{X}}{ \gamma_{\mathbf X}} \right) \frac{\tau_{X}}{\gamma_{\mathbf X}} R_{X}(X_{s}, X_{t}) + R_{f}(X_{s}, X_{t})\right)_{s,t=1}^n, \\
& \mathbf{R}_Y= \left(g ^{(1)} \left( \frac{\tau_{Y}}{ \gamma_{\mathbf Y}} \right) \frac{\tau_{Y}}{\gamma_{\mathbf Y}} R_{Y}(Y_{s}, Y_{t}) + R_{g}(Y_{s}, Y_{t})\right)_{s,t=1}^n.
\end{align*}
Denote $R_n = \frac{1}{2}f^{(1)} \left( \frac{\tau_{X}}{ \gamma_{\mathbf X}} \right)\frac{\tau_{X}}{ \gamma_{\mathbf X}} (\widetilde{ \mathbf L}_X \cdot \widetilde{ \mathbf R}_Y )+ \frac{1}{2}g^{(1)} \left( \frac{\tau_{Y}}{\gamma_{\mathbf Y}} \right)\frac{\tau_{Y}}{\gamma_{\mathbf Y}} (\widetilde{ \mathbf R}_X \cdot \widetilde{ \mathbf L}_Y ) + (\widetilde{ \mathbf R}_X \cdot \widetilde{ \mathbf R}_Y )$ and $\mathcal R_{n} = \tau R_n$. By Lemma \ref{PropEquality}, we have
\begin{multline*}
\tau \times hCov^2_{n}(\mathbf X, \mathbf Y) = \\ f^{(1)} \left( \frac{\tau_{X}}{\gamma_{\mathbf X}} \right) g^{(1)} \left( \frac{\tau_{Y}}{ \gamma_{\mathbf Y}} \right) \frac{\tau_{X}}{\gamma_{\mathbf X}} \frac{\tau_{Y}}{\gamma_{ \mathbf Y}} \frac{1}{\tau} \sum\limits_{i=1}^{p} \sum\limits_{j=1}^{q} cov_{n}^2 (\mathcal X_{i}, \mathcal Y_{j}) + \mathcal{R}_{n}.
\end{multline*}
(ii) We present the following lemma which would be useful in subsequent arguments.
\begin{lemma}
\label{TailOder}
Suppose $f^{(2)}$ and $g^{(2)}$ are continuous on some open interval containing 1. Then under the assumptions of Theorem \ref{thm:decompHsic},
$$
R_{f}(X_{s},X_{t}) = O_{p}( L_{X}(X_{s}, X_{t})^2 ), \quad R_{g}(Y_{s},Y_{t}) = O_{p}( L_{Y}(Y_{s}, Y_{t})^2 ).
$$
\end{lemma}
\begin{proof}
The remainder term can be written as
\begin{multline}\label{eq:rf}
R_{f}(X_{s},X_{t}) =\\ \int_{0}^1 \int_{0}^{1} v f^{(2)} \left( \frac{\tau_{X}}{\gamma_{\mathbf X}} + uv \left\lbrace \frac{L_{X}(X_s, X_t) }{2} + R_X (X_s, X_t) \right\rbrace \frac{\tau_{X}}{\gamma_{\mathbf X}} \right) dudv
\\ \times \left(\frac{\tau_{X}}{\gamma_{\mathbf X}}\right)^2\left( \frac{L_{X}(X_s, X_t) }{2} + R_X (X_s, X_t) \right)^2.
\end{multline}
Set $ \varphi (x,y) = \int_{0}^1 \int_{0}^{1} v f^{(2)} \left( x + uv y \right) dudv $. Then $\varphi (x,y)$ is continuous at $(1,0)$. By the continuous mapping theorem, we have
\begin{multline*}
\int_{0}^1 \int_{0}^{1} v f^{(2)} \left( \frac{\tau_{X}}{\gamma_{\mathbf X}} + uv \left\lbrace \frac{L_{X}(X_s, X_t) }{2} + R_X (X_s, X_t) \right\rbrace \frac{\tau_{X}}{\gamma_{\mathbf X}} \right) dudv \\ \overset{p}{\rightarrow}
\int_{0}^1 \int_{0}^{1} v f^{(2)} \left( 1 \right) dudv.
\end{multline*}
So $ R_{f}(X_{s},X_{t}) = O_p (1) \left( \frac{L_{X}(X_s, X_t) }{2} + R_X (X_s, X_t) \right)^2 = O_p ( L_{X}(X_{s}, X_{t})^2 ) $. Similar argument holds for $R_{g}(Y_{s},Y_{t})$.
\end{proof}
Both the Gaussian and Laplacian kernel have continuous second order derivatives. From Lemma \ref{TailOder}, we know
\begin{align*}
& f^{(1)} \left( \frac{\tau_{X}}{ \gamma_{\mathbf X}} \right) \frac{\tau_{X}}{\gamma_{\mathbf X}} R_{X}(X_{s}, X_{t}) + R_{f}(X_{s}, X_{t}) =O_p ( L_{X}(X_{s}, X_{t})^2 ), \\
& g^{(1)} \left( \frac{\tau_{Y}}{ \gamma_{\mathbf Y}} \right) \frac{\tau_{Y}}{\gamma_{\mathbf Y}} R_{Y}(Y_{s}, Y_{t}) + R_{g}(Y_{s}, Y_{t}) = O_p( L_{Y}(Y_{s}, Y_{t})^2 ).
\end{align*}
Thus, similar arguments in Theorem \ref{thm:decomp} can be used to show that $ {\cal R}_n = O_p( \tau a_p^2b_q + \tau a_pb_q^2 ) = o_p(1)$.
\end{proof}
\subsection{Proof of Proposition \ref{prop:ind}}
\begin{proof} Clearly, $E\left[ k_{st}(i) l_{uv}(j) \right]=0$ when $\{s,t\} \cap \{ u,v \} = \emptyset$. For any $1 \leq i, i'\leq p, 1 \leq j, j' \leq q,$
\begin{align*}
& E\left[ k_{st}(i) l_{su}(j) \right] \\
= & E \left[ E[ k_{st}(i) l_{su}(j) | x_{si}, y_{sj} ] \right] \\
= & E \left[ E[ k_{st}(i) | x_{si}, y_{sj} ] E[ l_{su}(j) | x_{si}, y_{sj} ] \right].
\end{align*}
Notice that
\begin{align*}
& E[ k_{st}(i) | x_{si}, y_{sj} ] \\
= & E \big\{ k(x_{si},x_{ti})-E[k(x_{si},x_{ti})|x_{si}] -E[k(x_{si},x_{ti})|x_{ti}]+E[k(x_{si},x_{ti})] | x_{si}, y_{sj} \big\} \\
= & E[k(x_{si},x_{ti})|x_{si},y_{sj}] - E[k(x_{si},x_{ti})|x_{si}] - E[k(x_{si},x_{ti})] + E[k(x_{si},x_{ti})] \\
= & 0.
\end{align*}
Thus $ E\left[ k_{st}(i) l_{su}(j) \right]=0 $. Similarly, $E\left[ k_{st}(i) k_{su}(i') \right] = E\left[ l_{st}(j) l_{su}(j') \right] =0.$
\end{proof}
\subsection{Proof of Theorem \ref{thm:key} }
\begin{proof}
Let $\widetilde{ \mathbf K} = (\tilde{k}_{st})_{s,t=1}^n$ and $\widetilde{ \mathbf L} = (\tilde{l}_{st})_{s,t=1}^n$. Notice that
\begin{align*}
uCov^2_{n}(\mathbf X,\mathbf Y) & = (pq)^{-1/2}\sum^{p}_{i=1}\sum^{q}_{j=1} \frac{1}{n(n-3)} \sum_{s\neq
t}\tilde{k}_{st}(i)\tilde{l}_{st}(j) \\
& = \frac{1}{n(n-3)} \sum_{s\neq t} \left(
p^{-1/2} \sum^{p}_{i=1} \tilde{k}_{st}(i)\right) \left( q^{-1/2} \sum^{q}_{j=1} \tilde{l}_{st}(j) \right).
\end{align*}
Under Assumption \ref{D3}, we have
\begin{align*}
&p^{-1/2}\sum^{p}_{i=1}\tilde{k}_{st}(i)
\\=&p^{-1/2}\sum^{p}_{i=1}k_{st}(i)-\frac{1}{n-2}\sum_{u\neq
t}p^{-1/2}\sum^{p}_{i=1}k_{ut}(i)
\\& -\frac{1}{n-2}\sum_{ v\neq
s}p^{-1/2}\sum^{p}_{i=1}k_{sv}(i)
+\frac{1}{(n-1)(n-2)}\sum_{u\neq
v}p^{-1/2}\sum^{p}_{i=1}k_{uv}(i)
\\ \overset{d }{ \rightarrow } & c_{st}-\frac{1}{n-2}\sum_{u\neq t}c_{ut}-\frac{1}{n-2}\sum_{v\neq s}c_{vs}+\frac{1}{(n-1)(n-2)}\sum_{u\neq
v}c_{uv}.
\end{align*}
Then we get
\begin{multline*}
n(n-3) \times uCov^2_{n}(\mathbf X, \mathbf Y) \overset{d}{\rightarrow} \\ \sum_{s\neq t}\left(c_{st}-\frac{1}{n-2}\sum_{u\neq
t}c_{ut}-\frac{1}{n-2}\sum_{v\neq
s}c_{sv}+\frac{1}{(n-1)(n-2)}\sum_{u\neq v}c_{uv}\right)
\\ \times \left(d_{st}-\frac{1}{n-2}\sum_{u\neq
t}d_{ut}-\frac{1}{n-2}\sum_{v\neq
s}d_{sv}+\frac{1}{(n-1)(n-2)}\sum_{u\neq v}d_{uv}\right).
\end{multline*}
Set
\begin{align*}
& \mathbf{c} = \left( c_{12}, c_{13}, \cdots, c_{1n}, c_{23}, \cdots,c_{2n}, c_{34}, \cdots, c_{n(n-1) } \right)^T, \\
& \mathbf{d} = \left( d_{12}, d_{13}, \cdots, d_{1n}, d_{23}, \cdots,d_{2n}, d_{34}, \cdots, d_{n(n-1) } \right)^T.
\end{align*}
Under Assumption \ref{D3} and by Proposition \ref{prop:ind}, we know that
\begin{align*} \left(
\begin{array}{c}
\mathbf{c} \\
\mathbf{d}
\end{array} \right) \sim N \left(\mathbf 0, \left(
\begin{array}{cc}
\sigma_{x}^2 \mathbf{I}_{n(n-1)/2} & \sigma_{xy}^2 \mathbf{I}_{n(n-1)/2} \\
\sigma_{xy}^2 \mathbf{I}_{n(n-1)/2} & \sigma_{y}^2 \mathbf{I}_{n(n-1)/2}
\end{array} \right)
\right).
\end{align*}
Define $\mathbf C=(c_{st})_{s,t=1}^n$ such that $c_{st}=c_{ts}$ and $\widetilde{\mathbf C}=(\widetilde{c}_{st})_{s,t=1}^n$. Here we assume that $c_{ss}=0$. From the proof of Lemma A.1 of Park et al. (2015), we have
\begin{align*}
\text{vec}( \widetilde{\mathbf C} )= \mathbf F \mathbf S \text{vec}(\mathbf C) = \mathbf F \mathbf S \mathbf F \text{vec}(\mathbf C),
\end{align*}
where $\text{vec}(\mathbf C)$ is the usual vectorization of matrix $\mathbf C$; $\mathbf F$ is the matrix of the linear operator that sets the diagonal of a matrix to be 0, i.e., $\text{vec}(\mathbf B_{-D}) =\mathbf F \text{vec}(\mathbf B)$, $\mathbf B_{-D}$ is $\mathbf B$ with its diagonal set to be 0; Letting $\mathbf J = \mathbf{1}_n \mathbf{1}_n^T$, we define $\mathbf S$ as
\begin{align*}
\mathbf S = \mathbf I_n \otimes \mathbf I_n - \frac{1}{n-2} \mathbf{J} \otimes \mathbf{I}_n - \frac{1}{n-2} \mathbf{I}_n \otimes \mathbf{J} + \frac{1}{(n-1)(n-2)} \mathbf{J} \otimes \mathbf{J}.
\end{align*}
Next, to simplify the following proof, we will use a different vectorization operator, which will align the upper triangular elements frist, then the lower triangular elements and lastly the diagonal elements, i.e., define
\begin{align*}
\widetilde{\text{vec}}(\mathbf C) & = \left( \mathbf{c}_{u}^{T}, \mathbf{c}_{l}^{T}, \mathbf{c}_{d}^{T} \right)^T, \\
\mathbf{c}_{u}^{T} & = \left( c_{12}, c_{13}, \cdots, c_{1n}, c_{23}, \cdots,c_{2n}, c_{34}, \cdots, c_{(n-1)n} \right)^T, \\
\mathbf{c}_{l}^{T} & = \left( c_{21}, c_{31}, \cdots, c_{n1}, c_{32}, \cdots,c_{n2}, c_{43}, \cdots, c_{n(n-1)} \right)^T, \\
\mathbf{c}_{d}^{T} & = \left( c_{11}, c_{22}, \cdots , c_{nn} \right)^T.
\end{align*}
Notice that there is a permutation matrix $\mathbf{P}_1$ such that $\widetilde{\text{vec}}(\mathbf C) = \mathbf{P}_1 \text{vec}(\mathbf C)$. Then
\begin{align*}
\widetilde{\text{vec}}( \widetilde{ \mathbf C} )= \mathbf{P}_1 \mathbf F \mathbf S \mathbf{F} \mathbf{P}_1^{T} \widetilde{\text{vec}}(\mathbf C).
\end{align*}
Observe that for any matrix $\mathbf C$, both the column sum and row sum of $\widetilde{ \mathbf C}$ are 0. We can verify that $\widetilde{\widetilde{\mathbf C}}=\widetilde{\mathbf C}$.
Set $\mathbf U = \mathbf{P}_1 \mathbf F \mathbf S \mathbf{F} \mathbf{P}_1^T$. It follows that $ \mathbf U^2 \widetilde{\text{vec}}(\mathbf C)=\mathbf U \widetilde{ \text{vec}}(\mathbf C)$ and thus
\begin{align}\label{eq12}
(\mathbf U^2- \mathbf U) \widetilde{\text{vec}}( \mathbf C)=0.
\end{align}
Equation (\ref{eq12}) still holds if we replace $c_{ss}$ by some nonzero elements. Since Equation (\ref{eq12}) holds for any $\widetilde{ \text{vec}}( \mathbf C)$, we must have
$\mathbf U^2= \mathbf U$ which implies that $\mathbf U$ is an idempotent matrix. Next, let $\mathbf C^u$ ($\mathbf C^l$) be the matrix by setting the lower (upper) triangular and diagonal elements in $\mathbf C$ to be zero. Denote
\begin{align*}
\mathbf P_2=\begin{pmatrix}
\mathbf 0 & \mathbf I & \mathbf 0 \\
\mathbf I & \mathbf 0 & \mathbf 0 \\
\mathbf 0 & \mathbf 0 & \mathbf I
\end{pmatrix}, \quad
\mathbf D=\begin{pmatrix}
\mathbf I \\
\mathbf 0 \\
\mathbf 0
\end{pmatrix}.
\end{align*}
Then, we see that $ \widetilde{\text{vec}}(\mathbf C^l) = \mathbf{P}_2 \widetilde{\text{vec}}(\mathbf C^u)$ and
\begin{align*}
\mathbf U\widetilde{\text{vec}}(\mathbf C)= & \mathbf U \widetilde{\text{vec}}(\mathbf C^u) + \mathbf U \mathbf P_2 \widetilde{\text{vec}}(\mathbf C^u)= \mathbf U(\mathbf I+ \mathbf P_2) \widetilde{\text{vec}}(\mathbf C^u)=\mathbf U (\mathbf I+ \mathbf P_2)\mathbf D \mathbf c.
\end{align*}
We note that
\begin{align*}
\mathbf W:=&\mathbf D^T (\mathbf I+ \mathbf P_2)\mathbf U \mathbf U( \mathbf I+ \mathbf P_2) \mathbf D
=\mathbf D^T (\mathbf U+ \mathbf U \mathbf P_2+ \mathbf P_2 \mathbf U+\mathbf P_2 \mathbf U \mathbf P_2) \mathbf D.
\end{align*}
We partition $\mathbf U$ into three blocks corresponding to the upper triangular, lower triangular and diagonal elements respective, i.e., we write
\begin{align*}
\mathbf U=\begin{pmatrix}
\mathbf U_{1} & \mathbf U_2 & \mathbf 0 \\
\mathbf U_2 & \mathbf U_{1} & \mathbf 0 \\
\mathbf 0 & \mathbf 0 & \mathbf 0
\end{pmatrix},
\end{align*}
where we have used the symmetry for $\mathbf U$. Then we have
$$\mathbf W=2(\mathbf U_1+ \mathbf U_2).$$
Now we argue that $\mathbf W^2= 2 \mathbf W$.
Recall that $\mathbf U$ is an idempotent matrix. Thus
\begin{align*}
& \mathbf U_1^2+ \mathbf U_2^2=\mathbf U_1, \quad \mathbf U_1 \mathbf U_2+ \mathbf U_2 \mathbf U_1= \mathbf U_2.
\end{align*}
Therefore, we get
$$\mathbf W^2=4( \mathbf U_1+ \mathbf U_2)^2=4( \mathbf U_1^2+ \mathbf U_2^2+\mathbf U_1 \mathbf U_2+ \mathbf U_2 \mathbf U_1)=4(\mathbf U_1+\mathbf U_2)=2 \mathbf W,$$ which indicates that $\mathbf W$ has eigenvalues which are either equal to two or zero.
It remains to show that the rank of $\mathbf W$ is $n(n-3)/2$ or equivalently, the trace of $\mathbf W /2=\mathbf U_1+\mathbf U_2$ is $n(n-3)/2.$
Note that
\begin{align*}
\text{Tr}(\mathbf W/2)=\text{Tr}(\mathbf U_1+\mathbf U_2)=\sum^{n(n-1)/2}_{i=1} \frac{\mathbf r^T_i \mathbf U \mathbf r_i }{ 2} =\frac{ n(n-1)}{4} \widetilde{\text{vec}}(\widetilde{\mathbf E}_1)^T\widetilde{\text{vec}}(\widetilde{\mathbf E}_1),
\end{align*}
where $\mathbf r_i=(\mathbf e_{i}^T,\mathbf e_{i}^T, \mathbf 0 ^T)^T$ and $\mathbf e_i$ is a $n(n-1)/2$-dimensional vector with $1$ on the $i$th position and zero otherwise; $\widetilde{\mathbf E}_i$ denotes the $\mathcal U$-centering version of the matrix $\mathbf E_i$ such that $\widetilde{\text{vec}}(\mathbf E_i) = \mathbf r_{i}$. Direct calculation shows that
\begin{align*}
\text{vec}(\widetilde{\mathbf E}_1)^T \text{vec}(\widetilde{\mathbf E}_1)=&\frac{2(n-3)^2}{(n-1)^2}+4(n-2)\frac{(n-3)^2}{(n-1)^2(n-2)^2}
\\&+(n-2)(n-3)\frac{4}{(n-1)^2(n-2)^2}=\frac{2(n-3)}{n-1},
\end{align*}
which implies that
$4^{-1}n(n-1)\widetilde{\text{vec}}(\widetilde{\mathbf E}_1)^T \widetilde{\text{vec}}(\widetilde{\mathbf E}_1)=n(n-3)/2.$ Using the above results and setting $\mathbf M = \mathbf W/2$, we have
\begin{multline*}
\text{vec}(\widetilde{\mathbf C})^T\text{vec}(\widetilde{\mathbf C})= \widetilde{\text{vec}}(\widetilde{\mathbf C})^T \widetilde{\text{vec}}(\widetilde{ \mathbf C}) = \widetilde{\text{vec}}(\mathbf C)^T \mathbf U\widetilde{\text{vec}}(\mathbf C) \\ =2 \mathbf c^T \mathbf M \mathbf c \sim 2\sigma^2_x \chi^2_{n(n-3)/2}.
\end{multline*}
Thus,
$$
Cov^2_{n}(\mathbf X, \mathbf X) \overset{d}{\rightarrow} \frac{2}{n(n-3)} \mathbf{c}^T \mathbf M \mathbf c \overset{d}{=} \frac{2}{n(n-3)}\sigma^{2}_{x}\chi^2_{n(n-3)/2}.
$$
Similarly,
\begin{align*}
& uCov^2_{n}(\mathbf X,\mathbf Y) \overset{d}{\rightarrow} \frac{2}{n(n-3)} \mathbf{c}^T \mathbf{M} \mathbf{d},\\
& uCov^2_{n}(\mathbf Y, \mathbf Y) \overset{d}{\rightarrow} \frac{2}{n(n-3)} \mathbf{d}^T \mathbf M \mathbf{d} \overset{d}{=} \frac{2}{n(n-3)} \sigma^2_y \chi^2_{n(n-3)/2}.
\end{align*}
\end{proof}
\subsection{Proof of Proposition \ref{prop:exactT}}
\begin{proof}
Since
\begin{align*} \left(
\begin{array}{c}
\mathbf{c} \\
\mathbf{d}
\end{array} \right) \overset{d}{=} N \left(\mathbf 0, \left(
\begin{array}{cc}
\sigma_{x}^2 \mathbf{I}_{n(n-1)/2} & \sigma_{xy}^2 \mathbf{I}_{n(n-1)/2} \\
\sigma_{xy}^2 \mathbf{I}_{n(n-1)/2} & \sigma_{y}^2 \mathbf{I}_{n(n-1)/2}
\end{array} \right)
\right),
\end{align*}
we have
\begin{align*}
\mathbf{c} | \mathbf{d} \overset{d}{=} N\left(\mu \mathbf{d}, \sigma^2 \mathbf{I}_{n(n-1)/2} \right),
\end{align*}
where $\mu = \sigma_{xy}^2/\sigma_{y}^2, \sigma^2 = (\sigma_{x}^2 \sigma_{y}^2 - \sigma_{xy}^4)/ \sigma_{y}^2$. Set
\begin{align*}
\mathbf{z} = \frac{\mathbf{M} \mathbf{d}}{\sqrt{ \left( \mathbf{d}^T \mathbf{M} \mathbf{d} \right) }}.
\end{align*}
It can be easily seen that conditional on $\mathbf d$,
\begin{align*}
\mathbf{c}^T \mathbf{z} / \sigma \sim N(\mu \mathbf{z}^T \mathbf d/ \sigma, 1) ,
\end{align*}
which implies that $\left. (\mathbf{c}^T \mathbf{z})^2 / \sigma^2 \right| \mathbf{d} \sim \chi^{2}_{1} (W^2)$ , where $\chi_{1}^{2}(W^2)$ is the non-central chi-squared distribution and $W^2= \frac{\mu^2}{\sigma^2} \mathbf{d}^T \mathbf{M} \mathbf{d}$ is the non-centrality parameter. Note that conditioned on $\mathbf d$,
\begin{align*}
\mathbf M (\mathbf I - \mathbf{z} \mathbf{z}^T) \mathbf c /\sigma \sim N(\mathbf 0, \mathbf M (\mathbf I - \mathbf{z} \mathbf{z}^T) \mathbf M ),
\end{align*}
where we have used the fact that $\mathbf M (\mathbf I - \mathbf{z} \mathbf{z}^T) \mathbf d =0$. As $\mathbf M (\mathbf I - \mathbf{z} \mathbf{z}^T) \mathbf M = \mathbf M - \frac{\mathbf M \mathbf d \mathbf{d}^T \mathbf M}{\mathbf{d}^T \mathbf M \mathbf{d}}$ is a projection matrix with rank $v-1$, it is easy to see that conditioned on $\mathbf d$,
$$
\mathbf c^T (\mathbf I - \mathbf{z}\mathbf{z}^T) \mathbf M (\mathbf I - \mathbf{z}\mathbf{z}^T) \mathbf c / \sigma^{2} \sim \chi_{v -1}^2.
$$
Next, conditioned on $\mathbf d$, as $ \mathbf{z}^T \mathbf{c} $ and $ (\mathbf I - \mathbf{z}\mathbf{z}^T) \mathbf c $ are independent, we have $(\mathbf{c}^T \mathbf{z})^2 / \sigma^2$ and $ \mathbf c^T (\mathbf I - \mathbf{z}\mathbf{z}^T) \mathbf M (\mathbf I - \mathbf{z}\mathbf{z}^T) \mathbf c $ are independent.
Then,
\begin{align*}
P_{H_A}(T_u < t) & \rightarrow P \left( \sqrt{v-1} \frac{ \frac{ \mathbf{c}^T \mathbf{z} }{ \sqrt{ \left( \mathbf{c}^T \mathbf{M} \mathbf{c} \right) }} }{ \sqrt{1 - \left( \frac{ \mathbf{c}^T \mathbf{z} }{ \sqrt{ \left( \mathbf{c}^T \mathbf{M} \mathbf{c} \right) }} \right)^{2} } } <t \right) \\
&= E \left[ P \left( \left. \sqrt{v-1} \frac{ \frac{ \mathbf{c}^T \mathbf{z} }{ \sqrt{ \left( \mathbf{c}^T \mathbf{M} \mathbf{c} \right) }} }{ \sqrt{1 - \left( \frac{ \mathbf{c}^T \mathbf{z} }{ \sqrt{ \left( \mathbf{c}^T \mathbf{M} \mathbf{c} \right) }} \right)^{2} } } <t \right| \mathbf{d} \right)\right] \\
& = E \left[ P \left( \left. \sqrt{v-1} \frac{ \mathbf{c}^T \mathbf{z} }{ \sqrt{\mathbf{c}^T \mathbf{M} \mathbf{c} - \left( \mathbf{c}^T \mathbf{z} \right)^{2} } } <t \right| \mathbf{d} \right)\right] \\
& = E \left[ P \left( \left. \frac{ \mathbf{c}^T \mathbf{z} }{ \sqrt{ \frac{1}{v-1} \mathbf c^T (\mathbf I - \mathbf{z}\mathbf{z}^T) \mathbf M (\mathbf I - \mathbf{z}\mathbf{z}^T) \mathbf c } } <t \right| \mathbf{d} \right)\right] \\
& = E \left[ P \left( t_{v-1, W} <t \right)\right]
\end{align*}
where $t_{v-1, W}$ is a noncentral $t$-distribution with $v-1$ degrees of freedom and noncentrality parameter $W = \frac{\mu}{\sigma} \sqrt{\mathbf{d}^T \mathbf{M} \mathbf{d}} \overset{d}{=} c \chi_{v}$ for $c = \frac{ \sigma_{xy}^2}{\sqrt{ \sigma_{x}^2 \sigma_{y}^2 - \sigma_{xy}^4}}.$ By setting $c=0$, we get
$
P_{H_0}(T_u < t) \rightarrow P \left( t_{v-1} <t \right).
$
\end{proof}
\subsection{Proof of Proposition \ref{prop:LarA}}
\begin{proof}
Notice that
$$
\phi = \frac{\phi_0}{\sqrt{v}} \Rightarrow c = \frac{\phi_{0}}{\sqrt{v - \phi_0^2 }} = \frac{\phi_{0}}{\sqrt{v}}\left( 1 + O\left( \frac{1}{v} \right) \right).
$$
Next, by the definition of non-central $t$-distribution,
\begin{align*}
P \left( t_{v-1, u} <t \right) = & P \left( \frac{Z + u}{\sqrt{\chi^2_{v-1}/(v-1)}} <t \right) \\
= & P \left( Z <t \sqrt{\chi^2_{v-1}/(v-1)} - u \right) \\
= & E \left[ P \left( \left. Z <t \sqrt{\chi^2_{v-1}/(v-1)} - u \right| \chi^2_{v-1} \right)\right] \\
= & E \left[ \Phi \left( t \sqrt{\frac{\chi^2_{v-1}}{v-1}} - u \right) \right],
\end{align*}
where $\Phi$ is the cdf of standard normal. For notational convenience, set
$$
g(u) = E \left[ \Phi \left( t \sqrt{\frac{\chi^2_{v-1}}{v-1}} - u \right) \right].
$$
Notice that $ P \left( t_{v-1, W} <t \right) = g(W) $. By the following asymptotic series [see \cite{laforgia2012asymptotic,tricomi1951asymptotic}],
\begin{align*}
\frac{\Gamma (J+1/2)}{\Gamma (J)} & = \sqrt{J} \left( 1 - \frac{1}{8J} + \frac{1}{128J^2} + \frac{5}{1024 J^3} - \frac{21}{ 32768J^4 } + \cdots \right)\\
& = \sqrt{J} \left(1 + O \left(\frac{1}{J} \right) \right),
\end{align*}
we can get,
\begin{align*}
& E \left[ (W- \phi_{0}) \right] \\
= & \frac{\phi_{0}}{\sqrt{v}}\left( 1 + O\left( \frac{1}{v} \right) \right) \sqrt{2} \frac{\Gamma((v+1)/2)}{\Gamma (v/2)} - \phi_{0} \\
= & \phi_{0} \left(1 + O \left(\frac{1}{v} \right) \right) - \phi_{0} \\
= & O \left(\frac{1}{v} \right),
\end{align*}
as well as
\begin{align*}
& E \left[ (W - \phi_{0})^{2} \right] \\
= & \phi_{0}^2 E \left[ \left(\frac{\chi_{v}}{ \sqrt{v} }\left( 1 + O\left( \frac{1}{v} \right) \right) - 1 \right)^{2} \right] \\
= & \phi_{0}^2 E \left[ \frac{\chi_{v}^2}{ v }\left( 1 + O\left( \frac{1}{v} \right) \right) - 2 \frac{\chi_{v}}{\sqrt{v}}\left( 1 + O\left( \frac{1}{v} \right) \right) + 1 \right] \\
= & \phi_{0}^2 \left\lbrace \left( 1 + O\left( \frac{1}{v} \right) \right) - 2 \left(1 + O \left(\frac{1}{v} \right) \right) + 1 \right\rbrace \\
= & O \left(\frac{1}{v} \right),
\end{align*}
and
\begin{align*}
&E \left[ W (W - \phi_{0})^{2} \right] \\
= & \phi_{0}^3 E \left[ \frac{\chi_{v}}{\sqrt{v}}\left( 1 + O\left( \frac{1}{v} \right) \right) \left(\frac{\chi_{v}}{ \sqrt{v} }\left( 1 + O\left( \frac{1}{v} \right) \right) - 1 \right)^{2} \right] \\
= & \phi_{0}^3 E \left[ \frac{\chi_{v}^3}{ v^{3/2} } - 2 \frac{\chi_{v}^2}{v} + \frac{\chi_{v}}{\sqrt{v}} \right]\left( 1 + O\left( \frac{1}{v} \right) \right) \\
= & \phi_{0 }^3 \Bigg\{ \frac{(v+1)}{v^{3/2}} \sqrt{v} \left(1 + O \left(\frac{1}{v} \right) \right) - 2 + 1 + O \left(\frac{1}{v} \right) \Bigg\} \left( 1 + O\left( \frac{1}{v} \right) \right) \\
= & O \left(\frac{1}{v} \right).
\end{align*}
We note that
\begin{align*}
\frac{\partial }{ \partial u } \Phi \left( t \sqrt{\frac{\chi^2_{v-1}}{v-1}} - u \right) & = - \phi \left( t \sqrt{\frac{\chi^2_{v-1}}{v-1}} - u \right) \\
\frac{\partial^2 }{ \partial u^2 } \Phi \left( t \sqrt{\frac{\chi^2_{v-1}}{v-1}} - u \right) & = -\left( t \sqrt{\frac{\chi^2_{v-1}}{v-1}} - u \right) \phi \left( t \sqrt{\frac{\chi^2_{v-1}}{v-1}} - u \right).
\end{align*}
Thus,
\begin{align*}
|g^{(2)}(u)|
& = \left| \frac{\partial^2 }{ \partial u^2 } E \left[ \Phi \left( t \sqrt{\frac{\chi^2_{v-1}}{v-1}} - u \right) \right] \right| \\
& = \left| E \left[ \frac{\partial^2 }{ \partial u^2 } \Phi \left( t \sqrt{\frac{\chi^2_{v-1}}{v-1}} - u \right) \right] \right| \\
& = \left| E \left[ -\left( t \sqrt{\frac{\chi^2_{v-1}}{v-1}} - u \right) \phi \left( t \sqrt{\frac{\chi^2_{v-1}}{v-1}} - u \right) \right] \right| \\
& \leq E \left[\left| -\left( t \sqrt{\frac{\chi^2_{v-1}}{v-1}} - u \right) \right| \phi \left( t \sqrt{\frac{\chi^2_{v-1}}{v-1}} - u \right) \right] \\
& \leq E \left[ \left( \left| t \right| \sqrt{\frac{\chi^2_{v-1}}{v-1}} + u \right) \phi \left( t \sqrt{\frac{\chi^2_{v-1}}{v-1}} - u \right) \right] \\
& < E \left[ \left( |t| \sqrt{\frac{\chi^2_{v-1}}{v-1}} + |u| \right) \right] \\
& \leq \left( |t| E\sqrt{\frac{\chi^2_{v-1}}{v-1}} + |u| \right) \\
& \leq \sqrt{2} |t| + |u|.
\end{align*}
Next, we can bound the following integral,
\begin{align*}
&\left| \int_{0}^{1} \int_{0}^{1} a g^{(2)}(\phi_{0} + ab(W - \phi_{0})) db da \right| \\
\leq & \int_{0}^{1} \int_{0}^{1} \left| a g^{(2)}(\phi_{0} + ab(W - \phi_{0})) \right| db da \\
\leq & \int_{0}^{1} \int_{0}^{1} \sqrt{2} |t| + | \phi_{0} + ab(W - \phi_{0}) | db da \\
\leq & \int_{0}^{1} \int_{0}^{1} \sqrt{2}|t| + \phi_{0} + |W| db da \\
= & \sqrt{2}|t| + \phi_{0} + W .
\end{align*}
To calculate $ E \left[ P \left( t_{v-1, W} <t \right)\right] = E \left[ g(W) \right] $, taking the Taylor expansion of $g(W)$ around $\phi_{0}$, the asymptotic mean of $W$, we get
\begin{align*}
=& E \left[ g(W) \right] \\
=& g(\phi_{0})
+ g^{(1)}(\phi_{0})E \left[ \left( W - \phi_{0} \right) \right] \\
& + E \left[ \int_{0}^{1} \int_{0}^{1} a g^{(2)}(\phi_{0} + ab(W - \phi_{0})) db da \left( W - \phi_{0} \right)^2 \right] \\
=& P \left( t_{v-1, \phi_{0}} <t \right) + O\left(\frac{1}{v}\right) \\
& + E \left[ \int_{0}^{1} \int_{0}^{1} a g^{(2)}(\phi_{0} + ab(W - \phi_{0})) db da \left( W - \phi_{0} \right)^2 \right].
\end{align*}
Notice that,
\begin{align*}
& \left| E \left[ \int_{0}^{1} \int_{0}^{1} a g^{(2)}(\phi_{0} + ab(W - \phi_{0})) db da \left( W - \phi_{0} \right)^2 \right] \right| \\
\leq & E \left[ \left| \int_{0}^{1} \int_{0}^{1} a g^{(2)}(\phi_{0} + ab(W - \phi_{0})) db da \left( W - \phi_{0} \right)^2 \right| \right] \\
\leq & E \left[ (\sqrt{2} |t| + \phi_{0}+ W) \left( W - \phi_{0} \right)^2 \right] \\
\leq & (\sqrt{2}|t| + \phi_{0}) E \left[ \left( W - \phi_{0} \right)^2 \right] + E \left[ W \left( W - \phi_{0} \right)^2 \right] \\
= & O \left(\frac{1}{v} \right).
\end{align*}
In conclusion, we have $E \left[ P \left( t_{v-1, W} <t \right)\right] = P \left( t_{v-1, \phi_{0 }} <t \right) + O \left(\frac{1}{v} \right).$ Since $t_{v-1}^{(\alpha)} \rightarrow Z^{(\alpha)}$ as $n \rightarrow \infty$, where $Z^{(\alpha)}$ is the $(1-\alpha)$th percentile of standard normal, $t_{v-1}^{(\alpha)}$ is bounded. Then, all the above analysis still holds if we replace $t$ with $t_{v-1}^{\alpha}$.
\end{proof}
Let $B(\cdot, \cdot)$ denote the beta function and $I_y ( \cdot , \cdot)$ denote the regularized incomplete beta function. In the following, we express $ E \left[ P \left( t_{v-1, W} \leq t \right)\right] $ as a sum of infinite series.
\begin{lemma}
\label{lem:exact}
$E \left[ P \left( t_{v-1, W} \leq t \right)\right]$ can be calculated exactly as
\begin{multline*}
E \left[ P \left( t_{v-1, W} <t \right) \right]
= \left(\frac{1}{c^2+1} \right)^{v/2} \Bigg\{
P(t_{v-1} \leq t) + \Bigg. \\ \Bigg. \sum\limits_{j=1}^{\infty} \left( \frac{c^2 }{c^2 + 1} \right)^{j/2} \frac{1}{jB(j/2, v/2)} \left( (-1)^j + I_{\frac{t^2}{t^2+v-1}} (\frac{j+1}{2}, \frac{v-1}{2}) \right)
\Bigg\} .
\end{multline*}
\end{lemma}
\begin{proof}
Notice that from \cite{walck1996hand}, the CDF of non-central $t$-distribution for $t \geq 0$ can be written as
\begin{multline*}
P \left( t_{v-1, W} <t \right) = \frac{1}{2 \sqrt{\pi}} \times \\ \sum\limits_{j=0}^{\infty} \frac{2^{\frac{j}{2}}}{j!} W^j \exp \left\lbrace- \frac{W^2}{2} \right\rbrace \Gamma \left(\frac{j+1}{2}\right) \left( (-1)^j + I_{z} \left(\frac{j+1}{2}, \frac{v-1}{2}\right) \right),
\end{multline*}
where
\begin{align*}
z & =\frac{t^2}{t^2+v-1}, \; v= \frac{n(n-3)}{2},\\
I_{y} & (\cdot, \cdot) \text{ is the regularized incomplete beta function} ,\\
W & = \frac{\mu}{\sigma} \sqrt{\mathbf{d}^T \mathbf{M} \mathbf{d}} \overset{d}{=} c \chi_{v}, c = \frac{ \sigma_{xy}^2}{\sqrt{ \sigma_{x}^2 \sigma_{y}^2 - \sigma_{xy}^4}}.
\end{align*}
Next, we calculate the expectation by constructing a generalized gamma distribution,
\begin{align*}
& E \left[ W^j \exp \left\lbrace- \frac{W^2}{2} \right\rbrace \right] \\
= & \int_{0}^{\infty} w^j \exp \left\lbrace- \frac{w^2}{2} \right\rbrace \frac{1}{c} \frac{1}{2^{v/2 -1} \Gamma (v/2)} \left(\frac{w}{c}\right)^{v-1} \exp \left\lbrace - \frac{w^2}{2 c^2} \right\rbrace dw \\
= & \frac{1}{c^{v}} \frac{1}{2^{v/2 -1} \Gamma (v/2)} \int_{0}^{\infty} \exp \left\lbrace - \left( \frac{w}{\sqrt{2c^2 /(c^2 + 1)}} \right)^2 \right\rbrace w^{j + v -1} dw \\
= & \frac{1}{c^{v}} \frac{1}{2^{v/2 -1} \Gamma (v/2)} \frac{\Gamma ( j/2 + v/2 ) (\sqrt{2c^2 /(c^2 + 1)})^{j + v}}{2} \\
= & \frac{(\sqrt{2c^2 /(c^2 + 1)})^{j + v}}{c^{v}} \frac{1}{2^{v/2 }} \frac{\Gamma ( j/2 + v/2 )}{\Gamma (v/2)}.
\end{align*}
Then,
\begin{multline*}
E \left[ P \left( t_{v-1, W} <t \right) \right]
= \frac{1}{2 \sqrt{\pi}} \left( \sqrt{\frac{1}{c^2+1}} \right)^{v} \times \\ \sum\limits_{j=0}^{\infty} \left( \frac{4c^2 }{c^2 + 1} \right)^{\frac{j}{2}} \frac{\Gamma((j+1)/2)\Gamma ( j/2 + v/2 )}{j!\Gamma (v/2)}\left( (-1)^j + I_{z} (\frac{j+1}{2}, \frac{v-1}{2}) \right).
\end{multline*}
According to the gamma duplicate formula,
$$
\Gamma \left( \frac{j+1}{2} \right) = \frac{\sqrt{\pi}}{2^j} \frac{\Gamma (j+1)}{\Gamma (j/2 + 1)},
$$
which further implies that
\begin{align*}
\frac{\Gamma((j+1)/2)\Gamma ( j/2 + v/2 )}{j!\Gamma (v/2)} & = \frac{\sqrt{\pi}}{2^j} \frac{\Gamma (j+1)}{\Gamma (j/2 + 1)} \frac{\Gamma ( j/2 + v/2 )}{j!\Gamma (v/2)} \\
& = \left\lbrace
\begin{array}{lc}
\sqrt{\pi}, & j=0 \\
\frac{\sqrt{\pi}}{2^{j-1}} \frac{1}{j \Gamma (j/2 )} \frac{\Gamma ( j/2 + v/2 )}{\Gamma (v/2)}, & j \geq 1
\end{array}
\right. \\
& = \left\lbrace
\begin{array}{lc}
\sqrt{\pi}, & j=0 \\
\frac{\sqrt{\pi}}{j2^{j-1}} \frac{1}{B(j/2, v/2)}, & j \geq 1
\end{array}
\right.
\end{align*}
where $B(\cdot, \cdot)$ is the beta function. Then, the expectation can be further simplified as
\begin{multline*}
E \left[ P \left( t_{v-1, W} <t \right) \right]
= \frac{1}{2 } \left(\frac{1}{c^2+1} \right)^{v/2} \left(1 + I_{z} (\frac{1}{2}, \frac{v-1}{2}) \right) + \\ \left( \frac{1}{c^2+1} \right)^{v/2} \sum\limits_{j=1}^{\infty} \left( \frac{c^2 }{c^2 + 1} \right)^{\frac{j}{2}} \frac{1}{j B(j/2, v/2)} \left( (-1)^j + I_{z} (\frac{j+1}{2}, \frac{v-1}{2}) \right).
\end{multline*}
Notice that
$$
\frac{1}{2} \left(1 + I_{z} (\frac{1}{2}, \frac{v-1}{2}) \right) = P(t_{v-1} \leq t).
$$
Thus,
\begin{multline*}
E \left[ P \left( t_{v-1, W} <t \right) \right]
= \left(\frac{1}{c^2+1} \right)^{v/2} \Bigg\{
P(t_{v-1} \leq t) + \Bigg. \\ \Bigg. \sum\limits_{j=1}^{\infty} \left( \frac{c^2 }{c^2 + 1} \right)^{j/2} \frac{1}{jB(j/2, v/2)} \left( (-1)^j + I_{z} (\frac{j+1}{2}, \frac{v-1}{2}) \right)
\Bigg\} .
\end{multline*}
\end{proof}
\subsection{Proof of Proposition \ref{signal:normal}}
\begin{proof}
Since we have
\begin{align*} \left(
\begin{array}{c}
X \\
Y
\end{array} \right) \sim N \left(\mathbf 0, \left(
\begin{array}{cc}
\mathbf{I}_{p} & \bm{\Sigma}_{XY} \\
\bm{\Sigma}_{XY}^T & \mathbf{I}_{q}
\end{array} \right)
\right),
\end{align*}
from Theorem 7 in \cite{szekely2007}, by setting $c = \frac{1}{4 (\pi/3 - \sqrt{3} +1)}$, we obtain
\begin{align*}
c \leq \frac{dCor^2(x_{i}, y_{j})}{ cor^2(x_{i}, y_{j}) } \leq 1,
\end{align*}
$ cov^2(x_{i}, y_{j}) = cor^2(x_{i}, y_{j})$ and $dCor^2 (x_{i}, y_{j}) = dCov^2 (x_{i}, y_{j})\pi/ c.$
Combine these results, we have
\begin{align*}
c \leq \frac{dCov^2(x_{i}, y_{j}) \pi /c}{ cov^2(x_{i}, y_{j}) } \leq 1.
\end{align*}
Notice also that $ dCov^2(x_{i}, x_{i}) =dCov^2(y_{j}, y_{j}) = c/\pi $ and $ cov^2(x_{i}, x_{i}) =cov^2(y_{j}, y_{j}) = 1 $. We finally get $
0.89^2 \phi_{2} \leq \phi_{1} \leq \phi_{2}.$
\end{proof}
\subsection{Proof of Proposition \ref{prop:uni}}
\begin{proof}
(i)
When $k(x,y) = l(x,y) = |x-y|^2$,
\begin{align*}
& k_{st}(i) = -2 (x_{si} - E(x_{si}))(x_{ti} - E(x_{ti})), \\
& l_{st}(j) = -2 (y_{sj} - E(y_{sj}))(y_{tj} - E(y_{tj})).
\end{align*}
Thus, letting $\mathbf D_X (i) = (x_{si}x_{ti})_{s,t=1}^n$ and $\mathbf D_Y(j) = (y_{sj}y_{tj})_{s,t=1}^n$, we have
\begin{align*}
uCov^2_{n}(\mathbf X, \mathbf Y) & = \frac{1}{\sqrt{pq}}\sum^{p}_{i=1}\sum^{q}_{j=1} ( \widetilde{\mathbf K}(i) \cdot \widetilde{\mathbf L}(j) ) \\
& = \frac{1}{\sqrt{pq}}\sum^{p}_{i=1}\sum^{q}_{j=1} 4 ( \widetilde{\mathbf D}_X(i) \cdot \widetilde{\mathbf D}_Y(j) ) \\
= & 4 \frac{1}{\sqrt{pq}}\sum\limits_{i=1}^p \sum\limits_{j=1}^q \left\lbrace \frac{1}{\binom{n}{2}} \frac{1}{2!} \sum\limits_{(s,t) \in i_{2}^{n}} x_{si} x_{ti} y_{sj} y_{tj} + \right. \\
& \quad \left. \frac{1}{\binom{n}{4}} \frac{1}{4!} \sum\limits_{(s,t,u,v) \in \mathbf i_{4}^{n}} x_{si} x_{ti} y_{uj} y_{vj}
-\frac{2}{\binom{n}{3}} \frac{1}{3!} \sum\limits_{(s,t,u) \in \mathbf i_{3}^{n}} x_{si} x_{ti} y_{sj} y_{uj} \right\rbrace \\
= & 4\frac{1}{\sqrt{pq}} \sum_{i=1}^p \sum_{j=1}^q cov_n^2(\mathcal X_i, \mathcal Y_j).
\end{align*}
Thus,
\begin{align*}
dCov^2_n(\mathbf X, \mathbf Y) & = \frac{1}{\tau} \sum_{i=1}^p \sum_{j=1}^q cov_n^2(\mathcal X_i,\mathcal Y_j) + {\cal R}'_n = \frac{1}{ 4} \frac{\sqrt{pq}}{\tau} uCov^2_n(\mathbf X, \mathbf Y) + {\cal R}'_n
\end{align*}
and
\begin{align*}
& \tau \times hCov^2_n(\mathbf X, \mathbf Y) \\
= & f^{(1)} \left( \frac{\tau_{X}}{\gamma_{\mathbf X}} \right) g^{(1)} \left( \frac{\tau_{Y}}{ \gamma_{\mathbf Y}} \right) \frac{\tau_{X}}{\gamma_{\mathbf X}} \frac{\tau_{Y}}{\gamma_{\mathbf Y}} \frac{1}{\tau} \sum\limits_{i=1}^{p} \sum\limits_{j=1}^{q} cov_{n}^2 (\mathcal X_{i}, \mathcal Y_{j}) + \mathcal{R}''_{n} \\
=& f^{(1)} \left( \frac{\tau_{X}}{\gamma_{\mathbf X}} \right) g^{(1)} \left( \frac{\tau_{Y}}{ \gamma_{\mathbf Y}} \right) \frac{\tau_{X}}{\gamma_{\mathbf X}} \frac{\tau_{Y}}{\gamma_{\mathbf Y}} \frac{1}{ 4 } \frac{\sqrt{pq}}{\tau} uCov^2_n(\mathbf X, \mathbf Y) + {\cal R}''_n.
\end{align*}
(ii) When $k(x,y) = l(x,y) = |x-y|$, we have
\begin{align*}
\widetilde{\mathbf K}(i) = \widetilde{\mathbf K}_1(i) - \widetilde{\mathbf K}_2(i) - \widetilde{\mathbf K}_3(i) + \widetilde{\mathbf K}_4(i) = \widetilde{\mathbf K}_1(i),
\end{align*}
where
\begin{align*}
& \mathbf{K}_1(i) = (k(x_{si}, x_{ti}))_{s,t=1}^n, \mathbf{K}_2(i) = (E[k(x_{si},x_{ti})|x_{si}])_{s,t=1}^n, \\
& \mathbf{K}_3(i) = (E[k(x_{si},x_{ti})|x_{ti}])_{s,t=1}^n, \mathbf{K}_4(i) = (E[k(x_{si},x_{ti})])_{s,t=1}^n.
\end{align*}
Similarly, $\widetilde{\mathbf L}(j) = \widetilde{\mathbf L}_1(j)$ with $ \mathbf{L}_1(j) = (l(y_{sj}, l_{tj}))_{s,t=1}^n .$
Then, we have
\begin{align*}
uCov^2_{n}(\mathbf X, \mathbf Y) & = \frac{1}{\sqrt{pq}}\sum^{p}_{i=1}\sum^{q}_{j=1} ( \widetilde{\mathbf K}_1(i) \cdot \widetilde{\mathbf L}_1(j) ) \\
& = \frac{1}{\sqrt{pq}}\sum^{p}_{i=1}\sum^{q}_{j=1} dCov_{n}^2(\mathcal X_i, \mathcal Y_j) \\
&=\frac{1}{\sqrt{pq}}\frac{1}{\sqrt{\binom{n}{2}}} mdCov_{n}^2(\mathbf X, \mathbf Y).
\end{align*}
\end{proof}
\subsection{Proof of Corollary \ref{cor:uni}}
\begin{proof}
For any fixed $t$ and each $R \in \{ dCov, hCov, mdCov \}$, Proposition \ref{prop:uni} and Theorem \ref{thm:key} imply that
\begin{align*}
T_R \overset{d}{\rightarrow} \sqrt{v-1} \frac{\varUpsilon}{ \sqrt{1 - (\varUpsilon)^{2} }}, \text{ where } \varUpsilon=\frac{ \mathbf{c}^T \mathbf M \mathbf{d} }{ \sqrt{ \left( \mathbf{c}^T \mathbf M \mathbf{c} \right) \left( \mathbf{d}^T \mathbf M \mathbf{d} \right) }}.
\end{align*}
Then the results follow similarly from the proof of Proposition \ref{prop:exactT}.
\end{proof}
\subsection{Proof of Remark \ref{rem:proof2}}
\label{App:proofRemk2}
\begin{proof}
For notational convenience, set $z_{i} = (x_{i} - x_{i}')^{2} - E[ (x_{i} - x_{i}')^{2} ] $. Since $\sup_{i} E(x_{i}^{8}) < \infty $, we get $\sup_{i}E(z_{i}^4) < \infty$. Then, we have
\begin{align*}
\alpha_{p}^2 & \asymp \frac{E \left[ \left(\sum_{i=1}^{p} z_{i} \right)^2 \right]}{p^2} \\
& = \frac{E \left[ \sum_{s=1}^{p} \sum_{t \in [ s - m, s+ m]} z_{s}z_{t} \right]}{p^2} \\
& \leq \frac{(2m+1)p}{p^2} \sup_{i}E(z_{i}^2) \\
& = O \left( \frac{m}{p} \right)
\end{align*}
and
\begin{align*}
\gamma_{p}^2 & \asymp \frac{E \left[ \left(\sum_{i=1}^{p} z_{i} \right)^4 \right]}{p^4} \\
& \asymp \frac{m^3p + m^2 p^2}{p^4} \sup_{i}E(z_{i}^4) \\
& = O \left( \frac{m^2}{p^2}\right).
\end{align*}
Similarly, we can show that
\begin{align*}
\beta_{q}^2 = O \left( \frac{m'}{q} \right) \text{ and } \lambda_{q}^2 = O \left( \frac{m'^2}{q^2}\right).
\end{align*}
Next, it follows that
\begin{align*}
\tau \alpha_{p} \lambda_{q} = O \left(\frac{m'\sqrt{m}}{\sqrt{q}} \right) = o(1) .
\end{align*}
The other results can be proved in a similar fashion.
\end{proof}
\subsection{Proof of Theorem \ref{thm:decomp2}}
\begin{proof} (i)\&(ii) Following the proof of Theorem \ref{thm:decomp}, we only need to check that $\mathcal{R}_{n} = o_p(1)$ still holds as $ n \wedge p \wedge q \rightarrow \infty $. Recall that the leading term is $\tau \times (\widetilde{ \mathbf L}_X \cdot \widetilde{ \mathbf L}_Y )$ and the remainder term is given as
\begin{align*}
\mathcal{R}_n = \frac{1}{2} \tau (\widetilde{ \mathbf L}_X \cdot \widetilde{ \mathbf R}_Y ) + \frac{1}{2} \tau (\widetilde{ \mathbf R}_X \cdot \widetilde{ \mathbf L}_Y ) + \tau (\widetilde{ \mathbf R}_X \cdot \widetilde{ \mathbf R}_Y ).
\end{align*}
Then, using Equation \eqref{eq:defDcov}, we have
\begin{align*}
(\widetilde{ \mathbf L}_X \cdot \widetilde{ \mathbf R}_Y )
= & \frac{1}{\binom{n}{2}} \frac{1}{2!} \sum\limits_{(s,t) \in \mathbf i_{2}^{n}} L_X(X_s, X_t) R_Y(Y_s, Y_t) \\
& + \frac{1}{\binom{n}{4}} \frac{1}{4!} \sum\limits_{(s,t,u,v) \in \mathbf i_{4}^{n}} L_X(X_s, X_t) R_Y(Y_u, Y_v) \\
& -\frac{2}{\binom{n}{3}} \frac{1}{3!} \sum\limits_{(s,t,u) \in \mathbf i_{3}^{n}} L _X(X_s, X_t) R_Y(Y_s, Y_u),
\end{align*}
and
\begin{align*}
(\widetilde{ \mathbf R}_X \cdot \widetilde{ \mathbf R}_Y )
= & \frac{1}{\binom{n}{2}} \frac{1}{2!} \sum\limits_{(s,t) \in \mathbf i_{2}^{n}} R_X(X_s, X_t) R_Y(Y_s, Y_t) \\
& + \frac{1}{\binom{n}{4}} \frac{1}{4!} \sum\limits_{(s,t,u,v) \in \mathbf i_{4}^{n}} R_X(X_s, X_t) R_Y(Y_u, Y_v) \\
& -\frac{2}{\binom{n}{3}} \frac{1}{3!} \sum\limits_{(s,t,u) \in \mathbf i_{3}^{n}} R_X(X_s, X_t) R_Y(Y_s, Y_u).
\end{align*}
To show that $ \mathcal{R}_n $ is asymptotically negligible, we consider the events $B_{\mathbf{X}},B_{\mathbf{Y}}$ and their complements $ B_{\mathbf{X}}^{c}, B_{\mathbf{Y}}^{c} $, where
\begin{align*}
B_{\mathbf{Y}} = \left\lbrace \min\limits_{1\leq s < t \leq n} \frac{|Y_{s} - Y_{t}|^2}{\tau_X^2} \leq \frac{1}{2} \text{ or } \max\limits_{1\leq s < t \leq n} \frac{|Y_{s} - Y_{t}|^2}{\tau_X^2} \geq \frac{3}{2} \right\rbrace .
\end{align*}
Then, under Assumption \ref{D4}, $ \text{as } n\wedge p \wedge q \rightarrow \infty $
\begin{align*}
P(B_{\mathbf{Y}}) & = P \left( \min\limits_{1\leq s < t \leq n} L_{Y}(Y_{s}, Y_{t}) \leq -\frac{1}{2} \text{ or } \max\limits_{1\leq s < t \leq n} L_{Y}(Y_{s}, Y_{t}) \geq \frac{1}{2} \right) \\
& = P\left( \bigcup\limits_{1\leq s < t \leq n} \left\lbrace L_{Y}(Y_{s}, Y_{t}) \leq -\frac{1}{2} \text{ or } L_{Y}(Y_{s}, Y_{t}) \geq \frac{1}{2} \right\rbrace \right) \\
& \leq \sum\limits_{1\leq s < t \leq n} P \left( | L_{Y}(Y_{s}, Y_{y}) | \geq \frac{1}{2} \right) \\
& < n^2 P \left( | L_{Y}(Y_{1}, Y_{2}) | \geq \frac{1}{2} \right) \\
& \leq 4 n^2 E\left[ L_{Y}(Y_{1}, Y_{2})^2 \right] \\
& = o(1) .
\end{align*}
Also notice that $P(B_{\mathbf{Y}}B_{\mathbf{X}}^{c}) \leq P(B_{\mathbf{Y}}) = o(1)$. Similarly, we have $ P(B_{\mathbf{X}}) = o(1), P(B_{\mathbf{X}}B_{\mathbf{Y}}^{c}) = o(1) $ and $ P(B_{\mathbf{Y}}B_{\mathbf{X}}) = o(1)$. By the proof of Proposition \ref{prop:taylor}, the remainder term can be written as
\begin{align*}
R_{X}(X_{s},X_{t}) = \int_{0}^1 \int_{0}^{1} v f^{(2)} \left( uv L_{X}(X_s, X_t) \right) dudv \times \left( L_{X}(X_s, X_t) \right)^2,
\end{align*}
where $f^{(2)}(t) = - \frac{1}{4} (1+t)^{-\frac{3}{2}}$ and similar formula holds for $Y$. Conditioned on the event $B_{\mathbf{X}}^{c}B_{\mathbf{Y}}^{c} $, we can easily show that
\begin{align}
\label{eq:2}
| R_{X}(X_{s},X_{t}) | \leq \frac{\sqrt{2}}{4} \left( L_{X}(X_s, X_t) \right)^2, | R_{Y}(Y_{s},Y_{t}) | \leq \frac{\sqrt{2}}{4} \left( L_{Y}(Y_s, Y_t) \right)^2.
\end{align}
Notice that
\begin{align*}
&\frac{1}{\binom{n}{2}} \frac{1}{2!} \sum\limits_{(s,t) \in \mathbf i_{2}^{n}} R_X(X_s, X_t) R_Y(Y_s, Y_t) \\
= & \frac{1}{\binom{n}{2}} \frac{1}{2!} \sum\limits_{(s,t) \in \mathbf i_{2}^{n}} R_X(X_s, X_t) R_Y(Y_s, Y_t) \mathbb{I}_{ \{ B_{\mathbf{X}}^{c}B_{\mathbf{Y}}^{c} \} } \\
& + \frac{1}{\binom{n}{2}} \frac{1}{2!} \sum\limits_{(s,t) \in \mathbf i_{2}^{n}} R_X(X_s, X_t) R_Y(Y_s, Y_t) \mathbb{I}_{ \{ B_{\mathbf{X}} B_{\mathbf{Y}}^{c} \} } \\
& + \frac{1}{\binom{n}{2}} \frac{1}{2!} \sum\limits_{(s,t) \in \mathbf i_{2}^{n}} R_X(X_s, X_t) R_Y(Y_s, Y_t) \mathbb{I}_{ \{ B_{\mathbf{X}}^{c} B_{\mathbf{Y}} \} } \\
& + \frac{1}{\binom{n}{2}} \frac{1}{2!} \sum\limits_{(s,t) \in \mathbf i_{2}^{n}} R_X(X_s, X_t) R_Y(Y_s, Y_t) \mathbb{I}_{ \{ B_{\mathbf{X}} B_{\mathbf{Y}} \} } \\
= & \rmnum{1} + \rmnum{2} + \rmnum{3 } + \rmnum{4}.
\end{align*}
For any $\epsilon >0$,
$
P( |\tau \times \rmnum{2}| > \epsilon) \leq P( B_{\mathbf{X}} B_{\mathbf{Y}}^{c} ) = o(1),
$
which implies that $ \tau \times \rmnum{2} = o_{p}(1) $. Similarly, $ \tau \times \rmnum{3} = o_{p}(1) $ and $ \tau \times \rmnum{4} = o_{p}(1) $. For term $\rmnum{1}$, by Equation \eqref{eq:2}, we have
\begin{align*}
| \rmnum{1}| \leq & \left\lbrace \frac{1}{\binom{n}{2}} \frac{1}{2!} \sum\limits_{(s,t) \in \mathbf i_{2}^{n}} | R_X(X_s, X_t) R_Y(Y_s, Y_t)| \right\rbrace B_{\mathbf{X}}^{c}B_{\mathbf{Y}}^{c} \\
\leq & \frac{1}{8} \frac{1}{\binom{n}{2}} \frac{1}{2!} \sum\limits_{(s,t) \in \mathbf i_{2}^{n}} L_X(X_s, X_t)^2 L_Y(Y_s, Y_t)^2 \\
\leq & \frac{1}{8} \left\lbrace \left( \frac{1}{\binom{n}{2}} \frac{1}{2!} \sum\limits_{(s,t) \in \mathbf i_{2}^{n}} L_X(X_s, X_t)^4 \right) \left(\frac{1}{\binom{n}{2}} \frac{1}{2!} \sum\limits_{(s,t) \in \mathbf i_{2}^{n}} L_Y(Y_s, Y_t)^4\right) \right\rbrace^{\frac{1}{2}}.
\end{align*}
Next, by the Markov's inquality
\begin{align*}
P\left( \frac{1}{\binom{n}{2}} \frac{1}{2!} \sum\limits_{(s,t) \in \mathbf i_{2}^{n}} L_X(X_s, X_t)^4 > \epsilon \right) & \leq \frac{E \left[\frac{1}{\binom{n}{2}} \frac{1}{2!} \sum\limits_{(s,t) \in \mathbf i_{2}^{n}} L_X(X_s, X_t)^4 \right]}{\epsilon} \\
& = \frac{1}{\epsilon} E \left[ L_X(X_1, X_2)^4 \right] \\
& = \frac{1}{\epsilon} \gamma_{p}^2.
\end{align*}
Thus, we have $ \frac{1}{\binom{n}{2}} \frac{1}{2!} \sum\limits_{(s,t) \in \mathbf i_{2}^{n}} L_X(X_s, X_t)^4 = O_p( \gamma_{p}^2 ) $ and similar proof shows that
$
\frac{1}{\binom{n}{2}} \frac{1}{2!} \sum\limits_{(s,t) \in \mathbf i_{2}^{n}} L_Y(Y_s, Y_t)^4 = O_p( \lambda_{q}^2 ).
$
So, we have $\tau \rmnum{1} = O_{p}( \tau \gamma_{p} \lambda_{q} )
$
and
\begin{align*}
\tau \frac{1}{\binom{n}{2}} \frac{1}{2!} \sum\limits_{(s,t) \in \mathbf i_{2}^{n}} R_X(X_s, X_t) R_Y(Y_s, Y_t) = O_{p}( \tau \gamma_{p} \lambda_{q} ).
\end{align*}
Similarly, it can be shown that
\begin{align*}
& \tau \frac{1}{\binom{n}{4}} \frac{1}{4!} \sum\limits_{(s,t,u,v) \in \mathbf i_{4}^{n}} R_X(X_s, X_t) R_Y(Y_u, Y_v) = O_{p}( \tau \gamma_{p} \lambda_{q} ), \\
& \tau \frac{2}{\binom{n}{3}} \frac{1}{3!} \sum\limits_{(s,t,u) \in \mathbf i_{3}^{n}} R_X(X_s, X_t) R_Y(Y_s, Y_u) = O_{p}( \tau \gamma_{p} \lambda_{q} ).
\end{align*}
In conclusion, we have $ \tau (\widetilde{ \mathbf R}_X \cdot \widetilde{ \mathbf R}_Y ) = O_{p}( \tau \gamma_{p} \lambda_{q} ) $. Similarly, it can be shown that $ \tau (\widetilde{ \mathbf L}_X \cdot \widetilde{ \mathbf L}_Y ) = O_{p}( \tau \alpha_{p} \beta_{q} ) $, $ \tau (\widetilde{ \mathbf L}_X \cdot \widetilde{ \mathbf R}_Y ) = O_{p}( \tau \alpha_{p} \lambda_{q} ) $ and $ \tau (\widetilde{ \mathbf R}_X \cdot \widetilde{ \mathbf L}_Y ) = O_{p}( \tau \gamma_{p} \beta_{q} ) $.
\end{proof}
\subsection{Proof of Theorem \ref{thm:decompHsic2}}
\begin{proof}
(i)\&(ii) Continuing with the proof of Theorem \ref{thm:decompHsic}, we need to show that $\mathcal{R}_n = o_p (1)$ and $\gamma_{\mathbf X}$ is asymptotically euqal to $\tau_{X}$ as $n \wedge p \wedge q \rightarrow \infty$ (similar result applies to $\gamma_{\mathbf Y}$ and $\tau_{Y}$). Recall that for all $s \neq t$,
\begin{align*}
L_{X}(X_{s}, X_{t}) = \frac{|X_{s} - X_{t}|^{2} - \tau_{X}^{2}}{ \tau_{X}^{2} }.
\end{align*}
Since for any $\epsilon > 0 $, under Assumption \ref{D4},
\begin{align*}
& P \left( \left| \frac{\text{median}\{|X_{s} - X_{t}|^{2}\} }{ \tau_{X}^{2} } -1 \right| > \epsilon \right) \\
\leq & P \left( \min\limits_{1\leq s < t \leq n} L_{X}(X_{s}, X_{t}) \leq -\epsilon \text{ or } \max\limits_{1\leq s < t \leq n} L_{X}(X_{s}, X_{t}) \geq \epsilon \right) \\
= & P\left( \bigcup\limits_{1\leq s < t \leq n} \left\lbrace L_{X}(X_{s}, X_{t}) \leq -\epsilon \text{ or } L_{X}(X_{s}, X_{t}) \geq \epsilon \right\rbrace \right) \\
\leq & \sum\limits_{1\leq s < t \leq n} P \left( | L_{X}(X_{s}, X_{t}) | \geq \epsilon \right) \\
< & n^2 P \left( | L_{X}(X_{1}, X_{2}) | \geq \epsilon \right) \\
\leq & \frac{1}{\epsilon^2} n^2 E\left[ L_{X}(X_{1}, X_{2})^2 \right] \\
= & o(1).
\end{align*}
Thus, we have $\frac{\text{median}\{|X_{s} - X_{t}|^{2}\} }{ \tau_{X}^{2} } \overset{p}{\rightarrow} 1$ and
$
\frac{\tau_{X}}{\gamma_{\mathbf X}} = \sqrt{ \frac{ \tau_{X}^{2} }{\text{median}\{|X_{i} - X_{j}|^{2}\} } } \overset{p}{\rightarrow} 1.
$ Similar arguments can also be used to show that $ \frac{\tau_{Y}}{\gamma_{\mathbf Y}} \overset{p}{\rightarrow} 1$.
Notice that conditioned on $B_{\mathbf{X}}^{c}B_{\mathbf{Y}}^{c}$, for all $1 \leq s < t \leq n$, we have
\begin{align}
\label{in:bounded}
|L_{X}(X_{s}, X_{t})| < 1/2 \text{ and }
\frac{1}{2} < \frac{|X_{s} - X_{t}|^2}{\tau_X^2} < \frac{3}{2}.
\end{align}
Next, Inequalities \eqref{eq:2} and \eqref{in:bounded} together imply that
$$
\left| \frac{\tau_{X}}{\gamma_{\mathbf X}} + uv \left\lbrace \frac{L_{X}(X_s, X_t) }{2} + R_X (X_s, X_t) \right\rbrace \frac{\tau_{X}}{\gamma_{\mathbf X}}\right| \leq c, $$
where $c$ is some constant. Since we choose kernels $k$ and $l$ to be the Gaussian or Laplacian kernel, it can be shown that
\begin{align*}
\left| \int_{0}^1 \int_{0}^{1} v f^{(2)} \left( \frac{\tau_{X}}{\gamma_{\mathbf X}} + uv \left\lbrace \frac{L_{X}(X_s, X_t) }{2} + R_X (X_s, X_t) \right\rbrace \frac{\tau_{X}}{\gamma_{\mathbf X}} \right) dudv \right| \leq c',
\end{align*}
where $c'$ is some constant. Then, we can easily see from Equation \eqref{eq:rf} that $| R_{f}(X_{s},X_{t}) | \leq c' L_{X}(X_{s}, X_{t})^2$. Similar result holds for $Y$. Finally, Theorem \ref{thm:decompHsic2} can be shown using similar arguments as in the proof of Theorem \ref{thm:decomp2}.
\end{proof}
\subsection{Proof of Remark \ref{rem:D7}}
\begin{proof}
When $k(x,y) = l(x,y) = |x-y|^2$,
\begin{align*}
& k_{st}(i) = -2 (x_{si} - E(x_{si}))(x_{ti} - E(x_{ti})), \\
& l_{st}(j) = -2 (y_{sj} - E(y_{sj}))(y_{tj} - E(y_{tj})).
\end{align*}
Thus, we have
\begin{align*}
& E [U(X_s, X_t)^2] \\
= & E \left[ \frac{1}{p} \sum\limits_{i=1}^{p} \sum\limits_{j=1}^{p} k_{st}(i) k_{st}(j) \right] \\
= & \frac{4}{p} \sum\limits_{i=1}^{p} \sum\limits_{j=1}^{p} E \left[ (x_{si} - E[x_{si}])(x_{ti} - E[x_{ti}]) (x_{sj} - E[x_{sj}])(x_{tj} - E[x_{tj}]) \right] \\
= & \frac{4}{p} \sum\limits_{i=1}^{p} \sum\limits_{j=1}^{p} cov^2(x_{i}, x_{j}) \\
= & \frac{4}{p} Tr ( \bm{\Sigma}_{X}^2 ),
\end{align*}
and
\begin{align*}
& E [U(X_s, X_t)^4] \\
= & E \left[ \frac{1}{p^2} \sum\limits_{i,j,r,w=1}^{p} k_{st}(i) k_{st}(j) k_{st}(r) k_{st}(w) \right] \\
= & \frac{16}{p^2} \sum\limits_{i,j,r,w=1}^{p} E \Big[ (x_{si} - E[x_{si}])(x_{ti} - E[x_{ti}]) (x_{sj} - E[x_{sj}])(x_{tj} - E[x_{tj}]) \\
& \quad \quad \quad \quad \quad \quad (x_{sr} - E[x_{sr}])(x_{tr} - E[x_{tr}]) (x_{sw} - E[x_{sw}])(x_{tw} - E[x_{tw}]) \Big] \\
= & \frac{16}{p^2} \sum\limits_{i,j,r,w=1}^{p} E^2 \left[ (x_{i} - E[x_{i}]) (x_{j} - E[x_{j}]) (x_{r} - E[x_{r}]) (x_{w} - E[x_{w}]) \right] \\
\asymp & \frac{m^3p + m^2p^2}{p^2} \sup\limits_{i} E^2(x_{i}^4) \\
= & O\left( m^2 \right).
\end{align*}
Also,
\begin{align*}
& E [U(X_s, X_t)U(X_t, X_u)U(X_u, X_v)U(X_v, X_s)] \\
= & E \left[ \frac{1}{p^2} \sum\limits_{i,j,r,w=1}^{p} k_{st}(i) k_{tu}(j) k_{uv}(r) k_{vs}(w) \right] \\
= & \frac{16}{p^2} \sum\limits_{i,j,r,w=1}^{p} E \Big[ (x_{si} - E[x_{si}])(x_{ti} - E[x_{ti}]) (x_{tj} - E[x_{tj}])(x_{uj} - E[x_{uj}]) \\
& \quad \quad \quad \quad \quad \quad (x_{ur} - E[x_{ur}])(x_{vr} - E[x_{vr}]) (x_{vw} - E[x_{vw}])(x_{sw} - E[x_{sw}]) \Big] \\
= & \frac{16}{p^2} \sum\limits_{i,j,r,w=1}^{p} cov(x_{i}, x_{j})cov(x_{j}, x_{r}) cov(x_{r}, x_{w}) cov(x_{w}, x_{i}) \\
= & \frac{16}{p^2} Tr (\bm{\Sigma}_{X}^4) \\
\asymp & \frac{m^3p}{p^2} \sup\limits_{i} E^4(x_{i}^2) \\
= & O \left(\frac{m^3}{p}\right).
\end{align*}
\end{proof}
\subsection{Proof of Theorem \ref{thm:key2}}
\begin{proof}
Firstly, the following lemma would be useful.
\begin{lemma}
\label{lem:hoeff}
Under null, we have
\begin{align*}
\frac{1}{\mathcal{S}} uCov^2_{n}(\mathbf X, \mathbf Y) = \frac{1}{\binom{n}{2} \mathcal{S} } \sum\limits_{1\leq s < t \leq n} H \left(Z_{s}, Z_{t} \right) + \mathcal{R}_{n},
\end{align*}
where $\sqrt{\binom{n}{2}}\mathcal{R}_{n,p,q} = o_{p}(1)$ as $n \wedge p \wedge q \rightarrow \infty$, $Z_{s} = (X_{s}, Y_{s}) $ and $H( \cdot , \cdot )$ is defined as
$$
H \left(Z_{s}, Z_{t} \right) := U(X_{s}, X_{t})V(Y_{s}, Y_{t}) .
$$
\end{lemma}
\begin{proof}
Firstly, sample $uCov$ can be written as
\begin{align*}
uCov^2_{n}(\mathbf X, \mathbf Y) & = \frac{1}{\sqrt{pq}}\sum^{p}_{i=1}\sum^{q}_{j=1} ( \widetilde{\mathbf K}(i) \cdot \widetilde{\mathbf L}(j) ) \\
& = (\frac{1}{\sqrt{p}}\sum^{p}_{i=1} \widetilde{\mathbf K}(i) \cdot \frac{1}{\sqrt{q}}\sum^{q}_{j=1}\widetilde{\mathbf L}(j) ) \\
& = ( \widetilde{\overline{\mathbf K}} \cdot \widetilde{\overline{\mathbf L}}),
\end{align*}
where $ \overline{\mathbf K} = (\overline{k}_{st})_{s,t=1}^{n} $, $\overline{\mathbf L} = (\overline{l}_{st})_{s,t=1}^{n}$,
$
\overline{k}_{st} = \frac{1}{\sqrt{p}}\sum^{p}_{i=1} k(x_{si},x_{ti})$ and $\overline{l}_{st} = \frac{1}{\sqrt{q}}\sum^{q}_{i=1} l(y_{si},y_{ti}).
$
Thus, $ uCov^2_{n}(\mathbf X, \mathbf Y) $ is just $ dCov^2_{n}(\mathbf X, \mathbf Y) $ with kernel $\overline{K}$ defines as $ \overline{K}(X_{s}, X_{t}) = \overline{k}_{st}$ and $\overline{L}(Y_{s}, Y_{t}) = \overline{l}_{st}$. Notice that
\begin{align*}
& \overline{K}(X_{s}, X_{t}) - E[\overline{K}(X_{s}, X_{t})| X_{s}] - E[\overline{K}(X_{s}, X_{t})|X_{t}] + E[ \overline{K}(X_{s}, X_{t}) ] = \frac{1}{\sqrt{p}}\sum^{p}_{i=1} k_{st}(i), \\
& \overline{L}(Y_{s}, Y_{t}) - E[\overline{L}(Y_{s}, Y_{t}) | Y_{s}] - E[\overline{L}(Y_{s}, Y_{t}) | Y_{t}] + E[ \overline{L}(Y_{s}, Y_{t}) ] = \frac{1}{\sqrt{q}}\sum^{q}_{i=1} l_{st}(i),
\end{align*}
where $k_{st}(i)$ and $l_{st}(i)$ are the double centered kernel distance defined in Section \ref{sec:uni}. By Proposition 2.1 of \cite{yao2017testing}, we have
\begin{align*}
\frac{1}{\mathcal{S}}( \widetilde{\overline{\mathbf K}} \cdot \widetilde{\overline{\mathbf L}}) & = \frac{1}{\binom{n}{2} \mathcal{S} } \sum\limits_{1\leq s < t \leq n} U(X_{s,n},X_{t,n}) V(Y_{s,n},Y_{t,n}) + \mathcal{R}_{n,p,q} \\
& = \frac{1}{\binom{n}{2} \mathcal{S} } \sum\limits_{1\leq s < t \leq n} \frac{1}{\sqrt{p}}\sum^{p}_{ i=1} k_{st}(i) \frac{1}{\sqrt{q}}\sum^{q}_{i=1} l_{st}(i) + \mathcal{R}_{n,p,q},
\end{align*}
where $ \sqrt{\binom{n}{2}} \mathcal{R}_{n,p,q} = o_{p}(1)$ as $n \wedge p \wedge q \rightarrow \infty$.
\end{proof}
By Lemma \ref{lem:hoeff}, we have
\begin{align*}
& \sqrt{\binom{n}{2}} \frac{uCov_{n}^{2}(\mathbf{X}, \mathbf{Y})}{\mathcal{S}}
= \frac{1}{\sqrt{\binom{n}{2}} \mathcal{S}} \sum\limits_{1\leq s < t \leq n} H \left(Z_{s}, Z_{t} \right) + \sqrt{\binom{n}{2}} \mathcal{R}_{n,p,q},
\end{align*}
where $ \sqrt{\binom{n}{2}} \mathcal{R}_{n,p,q} = o_{p}(1) .$ By similar proof of Theorem 2.1 in \cite{zhang2018conditional}, under $H_{0}$, we have
\begin{align*}
\frac{1}{\sqrt{\binom{n}{2}} \mathcal{S}} \sum\limits_{1\leq s < t \leq n} H \left(Z_{s}, Z_{t} \right) \overset{d}{\rightarrow} N(0,1).
\end{align*}
\end{proof}
\subsection{Proof of Proposition \ref{prop:exactT2}}
\begin{proof}
Notice that by the proof of Theorem 2.2 in \cite{zhang2018conditional}, under null
\begin{align}
\label{eq:uni}
\frac{ uCov_{n}^{2}(\mathbf{X}, \mathbf{X}) }{E[ U(X, X')^2 ]} \overset{p}{\rightarrow} 1, \quad \frac{ uCov_{n}^{2}(\mathbf{Y}, \mathbf{Y}) }{E[ V(Y, Y')^2 ]} \overset{p}{\rightarrow} 1.
\end{align}
So, by Theorem \ref{thm:key2}
\begin{align*}
\sqrt{\binom{n}{2}} \frac{uCov_{n}^{2}(\mathbf{X}, \mathbf{Y})}{\sqrt{ uCov_{n}^{2}(\mathbf{X}, \mathbf{X}) uCov_{n}^{2}(\mathbf{Y}, \mathbf{Y}) }} \overset{d}{\rightarrow} N(0,1),
\end{align*}
and also
\begin{align*}
\frac{uCov_{n}^{2}(\mathbf{X}, \mathbf{Y})}{\sqrt{ uCov_{n}^{2}(\mathbf{X}, \mathbf{X}) uCov_{n}^{2}(\mathbf{Y}, \mathbf{Y}) }} \overset{p}{\rightarrow} 0.
\end{align*}
As a consequence, we have $T_u \overset{d}{\rightarrow} N(0,1).$
\end{proof}
\subsection{Proof of Proposition \ref{prop:uni2}}
\begin{proof}
Based on Theorem \ref{thm:decomp2} and \ref{thm:decompHsic2}, the results follow similarly from the proof of Proposition \ref{prop:uni}.
\end{proof}
\subsection{Proof of Corollary \ref{cor:uni2}}
\begin{proof}
(i) If $ R = mdCov $, the result follows from Proposition \ref{prop:exactT2} and the following observation
\begin{align*}
\sqrt{\binom{n}{2}} \frac{R_{n}^{2}(\mathbf{X}, \mathbf{Y})}{\sqrt{ R_{n}^{2}(\mathbf{X}, \mathbf{X}) R_{n}^{2}(\mathbf{Y}, \mathbf{Y})}} = \sqrt{\binom{n}{2}} \frac{uCov_{n}^{2}(\mathbf{X}, \mathbf{Y})}{\sqrt{ uCov_{n}^{2}(\mathbf{X}, \mathbf{X}) uCov_{n}^{2}(\mathbf{Y}, \mathbf{Y}) }}.
\end{align*}
(ii) Recall that when $k(x,y) = l(x,y) = |x-y|^2$,
$E [U(X_s, X_t)^2] = \frac{4}{p} Tr ( \bm{\Sigma}_{X}^2 )$ and
$E [V(Y_s, Y_t)^2] = \frac{4}{q} Tr ( \bm{\Sigma}_{Y}^2 ).$
If $R = hCov$, by Proposition \ref{prop:uni2}, we have
\begin{align*}
\sqrt{\binom{n}{2}} \tau \times \frac{R_{n}^{2}(\mathbf{X}, \mathbf{Y})}{\mathcal{S}} =A_{p}B_{q} \sqrt{\binom{n}{2}}\frac{ uCov^2_n(\mathbf X, \mathbf Y) }{\mathcal{S}} + \sqrt{\binom{n}{2}}\frac{ \mathcal{R}_{n}'' }{\mathcal{S}},
\end{align*}
where $ A_{p} = \frac{\sqrt{p}}{ 2 \tau_{X} } f^{(1)} \left( \frac{\tau_{X}}{\gamma_{\mathbf X}} \right)\frac{\tau_{X}}{\gamma_{\mathbf X}}$ and $B_{q} = \frac{\sqrt{q}}{2 \tau_{Y}} g^{(1)} \left( \frac{\tau_{Y}}{ \gamma_{\mathbf Y}} \right) \frac{\tau_Y}{\gamma_{\mathbf Y}}$. By Theorem \ref{thm:key2},
\begin{align*}
A_{p}B_{q} \sqrt{\binom{n}{2}} \frac{ uCov^2_n(\mathbf X, \mathbf Y) }{\mathcal{S}} \overset{d}{\rightarrow} c N(0,1),
\end{align*}
where $c$ is some constant. Also notice that
\begin{align*}
\left| \sqrt{\binom{n}{2}}\frac{ \mathcal{R}_{n}'' }{\mathcal{S}} \right| \leq \left| \frac{n \mathcal{R}_{n}'' }{4\sqrt{ \frac{1}{p} Tr ( \bm{\Sigma}_{X}^2 ) \frac{1}{q} Tr ( \bm{\Sigma}_{Y}^2 ) }} \right| = o_p(1).
\end{align*}
Thus, we have
\begin{align*}
\sqrt{\binom{n}{2}} \tau \times \frac{R_{n}^{2}(\mathbf{X}, \mathbf{Y})}{\mathcal{S}} \overset{d}{\rightarrow} c N(0,1).
\end{align*}
Next, under Assumption \ref{D4}, by Equation \eqref{eq:uni} and Proposition \ref{prop:uni2}
\begin{align*}
& \tau \times \frac{\sqrt{ R_{n}^{2}(\mathbf{X}, \mathbf{X}) R_{n}^{2}(\mathbf{Y}, \mathbf{Y})} }{\mathcal{S}} \\
= & \sqrt{ \left( \frac{A_{p}^2 uCov^2_n(\mathbf X, \mathbf X) + \mathcal{R}''' }{ E[ U(X, X')^2 ] } \right)\left( \frac{B_{q}^2 uCov^2_n(\mathbf Y, \mathbf Y) + \mathcal{R}'''' }{ E[ U(Y, Y')^2 ] } \right) } \\
\overset{p}{\rightarrow} & c.
\end{align*}
Notice that Under Assumptions \ref{D1} and \ref{D4}, Proposition \ref{prop:uni2} also holds similarly when $\mathbf{X} = \mathbf{Y}$ or $ \mathbf{Y} = \mathbf{X} $. So $\mathcal{R}'''$ and $\mathcal{R}'''' $ are both negligible. Thus, we have
\begin{align*}
\sqrt{\binom{n}{2}} \frac{R_{n}^{2}(\mathbf{X}, \mathbf{Y})}{\sqrt{ R_{n}^{2}(\mathbf{X}, \mathbf{X}) R_{n}^{2}(\mathbf{Y}, \mathbf{Y})}} \overset{d}{\rightarrow} N(0,1)
\end{align*}
and consequently $T_{R} \overset{d}{\rightarrow} N(0,1) $. Similarly, it can be proved for $R = dCov$.
\end{proof}
\end{document}
| 57,481
|
TITLE: Rectifying the definition of a closed category
QUESTION [14 upvotes]: The definition of a closed category I'm using is here.
Suppose $V$ is a closed category and that for each object $b\in V$, $[b,-]$ has a left adjoint $- \otimes b$. The result is nearly a monoidal category, but the associator $\alpha \colon (a \otimes b) \otimes c \rightarrow a \otimes (b \otimes c)$ is not generally an isomorphism.
Question: Can we modify the definition of closed category directly so that whenever the adjoint above exists, result is always closed monoidal?
I know that we can fix this by requiring the adjunction to be "internal" in the sense that we have a natural isomorphism $\Phi \colon [a \otimes b, c] \rightarrow [a, [b, c]]$ but I'd rather there not be an asymmetry between closed and monoidal categories, since we can get from monoidal to closed by only demanding an ordinary adjuntion.
Edit: In light of my answer below, I've decided this question needed clarifying. What I want is some additional structure or property added to the axioms of a closed category such that
1) If $V$ is monoidal and $- \otimes b$ has a right adjoint $[b, -]$, then $[-,-]$ makes $V$ into this modified closed category
2) If $V$ is this modified closed category and $[b, -]$ has a left adjoint $-\otimes b$, then $-\otimes-$ makes $V$ into a monoidal category.
In the current state of affairs, 1 holds but 2 does not. With the additional property proposed in my answer below, 2 holds, but 1 does not.
REPLY [4 votes]: A symmetric closed category is a closed category together with isomorphisms
$$s:[A,[B,C]] \cong [B,[A,C]]$$ satisfying a few axioms: see Definition 1.1 of the paper ``On embedding closed categories" by Day and Laplaza that Buschi Sergio mentioned.
If you have one of these $(C,[-,-],I)$, together with adjoints $- \otimes A \dashv [A,-]$, then you get a symmetric monoidal category $(C,\otimes,I)$. It is possible to give an elementary argument proving this, but it takes a bit of work to write down.
It seems that you have realised this in your discussion with Buschi Sergio already but let me point it out anyway. The result follows from Proposition 2.2 of Day and Laplaza's paper which asserts that a symmetric closed category gives rise to a symmetric promonoidal one with promonoidal structure $$P(A,B;C)=C(A,[B,C])$$: that is, a symmetric pseudomonoid in the symmetric monoidal bicategory Prof of profunctors. Now if you have the adjoints you then have a representable symmetric promonoidal structure $$C(A \otimes B,C) \cong P(A,B;C)$$ and such amounts to a symmetric monoidal structure on $C$. (Because the strong symmetric monoidal pseudofunctor Cat → Prof is essentially fully faithful - i.e. locally an equivalence).
The same paper gives examples showing that the canonical associator for the (almost) monoidal structure associated to a closed category needn't be invertible. So there is a genuine asymmetry between the definitions of monoidal and closed category.
The notions of skew monoidal and skew closed category rectify the asymmetry in a different way -- there is a perfect correspondence between skew monoidal structures $(C,\otimes,I)$ and skew closed structures $(C,[-,-],I)$ related by a natural isomorphism $C(A \otimes B,C) \cong C(A,[B,C])$. See Proposition 18 of the paper "Skew closed categories" http://arxiv.org/abs/1205.6522 by Ross Street.
| 60,686
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Yes! Keep 'em coming.
- Bike Checks
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Yes! Keep 'em coming.
If SRAM is testing rough prototypes, they should be selling retail units by late next week.
ROBOTS knew he was German and an engineer by the second photo, before reading anything,
My Heart says yes, but if there was money on the line, I'd place my money elsewhere.
This feature has 51 comments.
pretty sure this is the only reason he won the EWS overall this year
There are companies making good looking e bikes.... this is not one of them.
That’s nonsense. Everyone of those trails are in most of his vids and ridden the same way on every one of his other bikes. that off trail is the pit where freeriding began 20 odd years ago. It’s a gravel pit and has been ridden and filmed constantly and it looks the same. Get a grip
"We are not asking for special treatment, but rather for an equal playing field"... special treatment is EXACTLY what you are asking for numb nut.
Eddie and Martin sending it on that last section!!
Why did they choose Scott? Because Scott owners are use to the rats nest of cables!
On pb, they have an image of the shifter with a trp barrel adjuster.
could be trp?
This photo has 9 comments.
Ratboy the man, Santa, thanks for support him! Keep this spirit up!! This is what makes us dream about bikes!
Its a legit line.
Anything between the tape is considered part of the track, so nothing about that line is against the rules.
Steve wentz is cool. Vitalmtb always has the raddest crew.
This video has 3 comments.
Is it the same over-damped tuning that Specialized got stuck with and was fairly universally panned?
35 lbs is just too heavy these days for an expensive enduro/AM bike. There is still a lot of room for weight savings, however, considering this beast has coil shocks on both the front and rear.
They are an awesome company to deal with. Their lifetime bearings, and frame warranties are legit too, which to my knowledge nobody else does... at least the bearings. Our shop used to be a global top 10 retailer for them, and they were always easy to deal with. I personally don't really jive with VPP (granted, i have never rode their lower link stuff), and they might be expensive off the top, but they will make sure this bike will last a lifetime, if you want it to.
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This product_review has 21 comments.
| 238,160
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Beach Life – Aerial Photography by Gray Malin
Finally I found summer! I was beginning to think it was going to pass Europe by but no it’s alive and kicking here. Admittedly, here is down in the south of France where a wash out summer season is as unusual as a Frenchman missing out on his after work aperitif, but I have to say that it feels good to be somewhere warm and not too far from a beach.
One of the reasons I was so excited to see the sea again was that I recently discovered the work of Gray Malin and I’d like to serve up a little sunshine in your day today by introducing him and his work to you too.
Gray Malin is a photographer and native of Dallas, Texas who has photographed many famous people, luxury brands and world famous sights. It was his collection of photographs called “À la Plage, À la Piscine” which caught my eye as uniquely striking in the way it captures summer, light, sun and fun. Taken from a doorless helicopter hovering over beaches in France, America, the Caribbean and Australia Malin’s photos are almost playful in their depiction of us humans’ love affair with beaches. It’s also unusual for us beach lovers to see our playground from this angle so sit back, think of pina coladas and enjoy…
Sydney, Australia
Lisbon, Portugal
St. Tropez, France
Mexico and the Caribbean
Beaches of Los Angeles, The Hamptons and Miami in USA
Now where’s my most colourful beach towel?
All beautiful photographs by Gray Malin from MaisonGray.com.
Tags: beach, fun, photography, photos
Marvelous pictures!
I absolutely ADORE these photos! WHAT FUN!!!
Thanks girls! I think they’re rather awesome photos too. So glad you like them.
Frankie x
[...] via [...]
Talented and playful! great photography. I want his
| 247,884
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I've tried many liquid eyeliner before but i seem to keep coming back to this one. it's cheap and it works well. i've gone through 3 already and jus bought a new one.
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I paid $7-$8 for it, and even if my skin had not reacted to it, the product was such a disappointment that I would not buy it EVER again. Better off paying the extra $10 for MAC.
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Been using this eyeliner for 7 years and haven't looked for anything else. Stays on all day, cheap (no more than $6), easy to apply, and lasts forever! Love this eyeliner!
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After trying a bit higher-end eyeliners (BeneFit, MAC, Urban Decay, to name a few), I have consistently come back to this one. As an Asian girl, I am ALL about the winged eyeliner look, and as someone who's been doing it for 8+ years, I am very, very picky about which liner I use... Now let me tell ya, this baby fills nearly all the criteria.
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PROS:
- Color is very pigmented & dark
- Stays on all day without the need of reapplication
- Allows for smooth, gliding, beautiful precision
- Affordable!
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CONS:
- The brush changes its consistency: initially, it's pretty stiff ( = thinner eyeliner) but it loosens up relatively quickly, making it harder to be as precise as when new. But with a steady hand & some practice, this will cease to even matter :)
- When the product is new, it stings a little upon application. I'm thinking this has to do with the brush rather than the product itself b/c it ONLY stings within the first few days upon opening.
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So, minus 1 lippie for the cons (which aren't really any big deal-breakers b/c they're relatively easy to get use to!). But overall, this is my go-to product for eyeliner-- Serious HG status!
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Hi everybody.I am makeup artist for many years.I purchased this product because I saw good reviews.It was like regular eyeliner.This product is NOT LONG LASTING.most of us use moisturizer around the eyes so it smudge very fast if you are looking for long lasting go with MAC or MAKEUP FOREVER or ESTEE LAUDER DOUBLE WEAR.
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I've bought this two times and ran out, and then was like hey I NEED this why are they always sold out now that I run out? I like it, others must like it.
PROS :
Lasts me 6 months
Cost (cheeeaaap)
Easy cat eye with felt tip
No burning at all (contact lens wearer)
CONS:
If I'm wearing it thick enough of a line it can transfer to my upper lid, but rarely happens
Clumps in corners of my eyes
1 out of 1 people found this review helpful. Did you? Yes No
I have bought this product many times, it has become my favorite eyeliner. It is easy to use, and it last a really long time. It is safe for contact wearers which many high end products are not, or at least don't mention that they are. The tip makes it easy to draw a straight line, if you use it on its side its gonna look chuncky so you want to have the tip pointing at you. It creates the cutest wing, and its really easy to remove with just facial cleanser. It does burn a bit if you get it in the inner corners so you just gotta be careful. I even drew butterflies on the sides of my temples with this for halloween and it didn't smudge, the tip is like a little paint brush. Awesome stuff.
1 out of 1 people found this review helpful. Did you? Yes No
| 296,028
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TITLE: When is an integral transform trace class?
QUESTION [29 upvotes]: Given a measure space $(X, \mu)$ and a measurable integral kernel $k : X \times X \rightarrow \mathbb{C}$, the operator
$$ K f(\xi) =\int_{X} f(x) k(x,\xi) d \mu(x),$$
the operator $K$ is Hilbert Schmidt iff $k \in L^2(X \times X, \mu \otimes\mu)$!
Q1:The main point of this questions, what are necessary and sufficient conditions for it to be trace class?
I know various instances, where
$$ \mathrm{tr} K = \int_X k(x,x) d \mu(x).$$
Q2:What are counterexamples, where $x \mapsto k(x,x)$ is integrable, but the operator is not trace class?
Q3:What are counterexamples for a $\sigma$ finite measure space, where $k$ is compactly supported and continuous, but the kernel transformation is not trace class and the above formula fails?
Q4: Is there a good survey/reference for these questions.
REPLY [6 votes]: A remark on (Q3):
There is this famous example of T.Carleman (1916 Acta Math link) where he constructs a (normal ) operator with a continuous kernel such that it belongs to all Schatten p-classes if and only if $p\geq 2.$
More precisely it's possible to construct $k(x)=\sum_n c_ne^{2\pi i n x}$ continuous and periodic with $\sum_n|c_n|^p=\infty$ for $p<2$. Then $Tf=f\ast k$ acting on $L^2(\mathbb T)$ yields the desired result.
Provided some extra regularity on the kernel, the trace formula works fine (there are a lot of results in the literature)
Regarding (Q4) I personally find C. Brislawn's result very interesting but rather difficult to implement in practice.
| 119,612
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TITLE: Galois extension characterization
QUESTION [1 upvotes]: Why to mention or why to give a basis for $E$ over $F?$ Why to make a relation between the basis and the minimal polynomial?
I think the proof is missing the 'therefore $E/F$ is Galois' at the end of the proof, anyway Why can we conclude that it's a Galois extension?
A finite extension $E/F$ is Galois if and only if $E/F$ is normal and separable.
Proof (just this side $\gets$):
Suppose $E/F$ is normal and separable.
If $w_1,w_2,...,w_n$ is a basis for $E$ over $F,$ let $f_i(x)$ be the minimal polynomial of $w_i.$
Then $E/F$ is a splitting field for the product of $f_i$ and each $f_i$ is separable because $E/F$ is separable.
REPLY [2 votes]: The idea here is to use a previous theorem that if $E/F$ is the splitting field of a separable polynomial then $E/F$ is Galois. We know that $E/F$ is normal and separable.
Normal means that if $\alpha \in E$ then all the conjugates of $\alpha$ are in $E$. If $f_{\alpha}(x)$ is the minimal polynomial of $\alpha$ then the roots of $f$ are the conjugates of $\alpha$ so $f_{\alpha}(x)$ has all of its roots in $E$ (it splits).
Separable means that every minimal polynomial is a separable polynomial.
Thus for every element $\alpha \in E$ we get a separable polynomial $f_{\alpha}$ that splits. If there are infinitely many elements of $E$ then this doesn't help us so much. We use the finiteness condition to conclude that $E$ can be generated by a finite set $\alpha_1,\dots,\alpha_n$. That is, $E = F[\alpha_1, \dots, \alpha_n]$.
Now we take the corresponding minimal polynomials $f_1 = f_{\alpha_1}, \dots, f_n = f_{\alpha_n}$. We know that $E$ is the splitting field of $f_1,\dots,f_n$ because the splitting field contains $\alpha_1,\dots,\alpha_n$ and they generate $E$.
Finally the last trick is to note that if $E$ is the splitting field of $f_1,\dots,f_n$ then $E$ is the splitting field of the product $f := f_1 \cdots f_n$. The polynomial $f$ is separable because each irreducible factor $f_1,\dots,f_n$ is separable.
Therefore $E/F$ is the splitting field of a separable polynomial and we can apply the previous result.
| 140,093
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Sexual & Gender Minorities
Lesbians, gays, bisexuals, transgender and intersex (LGBTI) people are among the most persecuted individuals in the world today. Seventy-eight nations criminalize same-sex relations. Seven of these apply the death penalty for consensual same-sex conduct. In many more countries, sexual and gender minorities (SGMs) regularly face harassment, arrest,.
SGMs are often targeted for violence by authorities or civilians whom the authorities cannot or will not stop. In some places, the situation is so severe that these refugees fear going outside in daylight. Virtually none “come out” in their countries of transit, and most never seek protection on account of their sexual orientation.
SGM refugees, as forced migrants and sexual minorities, are “doubly marginalized.” Moreover, the scant survival mechanisms normally available to other refugees are often closed off to them. While most refugees seek safety and comfort with their own countrypersons, SGM refugees are often targeted by their compatriots or families. Very few manage to survive these obstacles and reach safety.
Assistance is urgently needed to save SGM refugees in their countries of first asylum and to assist the lucky few who manage to reach safety.
ORAM works resolutely to break down systemic barriers to their safety and shelter. ORAM lays the groundwork for global changes by researching and documenting the extreme abuses these refugees face, and then translates its highly regarded expertise into essential advocacy and education.
These efforts have borne — and continue to bear — tangible fruit such as:
- Suspension of a physically invasive practice (phallometry) previously used to help determine refugee status for gay men, but forcefully discredited by ORAM in a special report,
- Faster consideration of LGBTIs applying for refugee status,
- A training and interviewing tool to guide adjudicators in objectively assessing the credibility and eligibility of LGBTI refugee and asylum claims,
- A 100-country global survey of NGO attitudes about LGBTI refugees and asylum seekers, which spotlighted gaping holes in their protection and pointed the way for corrective actions,
- The first community guide for helping LGBTI refugees and asylees integrate into new communities where they’ve historically lacked the safety nets other refugees can rely upon.
| 334,542
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Watch Mirror Mirror (2012) Megavideo.Mirror Mirror (2012) streaming.Mirror Mirror (2012) Official Movie Trailer.Mirror Mirror (2012) all hd Quality.Mirror Mirror (2012) Movie Full Trailer HD
An evil queen steals control of a kingdom and an exiled princess enlists the help of seven resourceful rebels to win back her birthright.
| 162,376
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TITLE: Proving Euler-Maclaurin approximation for even functions using Poisson summation
QUESTION [3 upvotes]: I'd like to ask for help for the following from my Advanced Mathematics for Physics class (6th semester):
If $f(x)$ is a sufficiently regular and even function integrable in all $\mathbb{R}$, use Poisson's summation formula to prove the following approximation: $$ \sum_{n=0}^{\infty} f(n) = \int_0^{\infty} f(x)\ dx + \frac{1}{2} f(0) - \frac{1}{12} f'(0) + \frac{1}{720} f'''(0) + \cdots $$
We studied Poisson's summation formula as $\sum_{n \in \mathbb{Z}} f(n) = \sum_{n \in \mathbb{Z}} \hat{f}(n)$, where $ \hat{f}(\alpha) = \int_{\mathbb{R}} f(x) e^{-2 \pi i x \alpha}\,dx $ is the Fourier transform of $f$. The best I've been able to do is show that $$ \begin{align} \sum_{n \in \mathbb{Z}} f(n) = \sum_{n \in \mathbb{Z}} \hat{f}(n) \implies \sum_{n=0}^{\infty} f(n) = \int_0^{\infty} f(x)\ dx + \frac{1}{2} f(0) + \sum_{n=1}^{\infty} \int_{\mathbb{R}} f(x) e^{-2 \pi i x n} \, dx \end{align} \tag{because $f$ and $\hat{f}$ are both even } $$
and I tried to expand the integrand of the last term into its Maclaurin series but since I'd be taking the integral over the reals of a polynomial, I cannot prove the integral converges. I've tried reading proofs of the Euler-Maclaurin formula online but those I could find don't make use of the Poisson formula and use Bernoulli numbers extensively (I know they're important for this result but I'm not very familiar with them and my teacher expects us to solve this problem without using them).
Any help would be greatly appreciated.
(This is my first question on the site, I think I'm following the question guidelines but I apologize if I missed anything)
REPLY [1 votes]: Finally got a solution. Posting it in case anyone's interested.
First, notice that because $ \int_0^{\infty} f(x) \, dx $ converges, $ x \to \infty \implies f^{(n)}(x) \to 0 $ for $ n = 0,1,\cdots, \infty $. Let us also remember the sine and cosine functions are bounded
Let's do the last integral by parts:
$ \begin{align} \int_{\mathbb{R}} f(x) e^{-2 \pi i x n} \, dx &= 2 \int_0^{\infty} f(x) \cos{(2 \pi x n)} \, dx \tag{because of parity} \\ &= 2 \left[ \left( \frac{f(x) \sin{(2 \pi n x)}}{2 \pi n} \right)_0^{\infty} - \int_0^{\infty} \frac{f'(x) \sin{(2 \pi n x)}}{2 \pi n} \, dx \right] \\ &= \cdots \\ &\approx 2 \left[ - \frac{f'(0)}{(2 \pi n)^2} + \frac{f'''(0)}{(2 \pi n)^4} \right] \end{align} $
The rest is easy.
| 17,781
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Vumbura PlainsBotswana
Vumbura Plains is a remote location on a stunning and exclusive concession in the northern Okavango Delta. The camp is a sanctuary of contemporary design that blends effortlessly into the trees that shade it from the Botswana sun. Vumbura Plains offers year-round water and land based explorations into the Okavango where several prides of lion, leopard, cheetah and packs of endangered wild dog prey on vast herds of impala, lechwe, tsesebe and the magnificent sable antelope. Vumbura is a destination of luxury and adventure for couples and families alike.
A deck overlooking the panoramic delta
Vumbura Plains consists of two separate seven-roomed satellite camps. Vumbura north boasts two family rooms and is ideal for groups travelling together while the southern satellite is perfect for romance and honeymoon luxury.
The main lounge, dining and bar areas are airy and spacious tributes to contemporary African luxury with extensive decking and open sides offering spectacular vistas of the Okavango and the wilderness dramas unfolding there. There are both indoor and outdoor lounge areas of sumptuous comfort while the star deck – extending out into the waters of the delta where an astonishing abundance of bird, mammal and invertebrate life thrives – is a particular highlight for African fireside contemplation.
Like the main areas, the rooms are open, spacious and delightfully modern. The extensive outside deck with private plunge pool lie dappled by the shade of ancient African giants while the private sala, suspended over the flooded grasslands is the perfect place to luxuriate in a spa treatment. The deck leads to the polished wooden floors of the enormous double bedroom and lounge. The latter is a lower level refuge of delightful comfort offering endless vistas over the Okavango.
The highlight of the capacious bathroom is a huge glass-walled shower which allows you to watch the procession of delta wildlife on the plains below.
Discovery on crystal water and Kalahari sand
The flooded grasslands of Vumbura Plains boasts all-year-round water and land exploration of the Okavango wilderness. On the land, you will explore in a 4×4 game viewer with an experienced safari guide. You will track the seasonal herds of elephant and buffalo in amongst the herds of tsesebe, impala, kudu, giraffe, zebra and abundant lechwe frolicking in the shallows – all combining to create exquisite photographic opportunities.
The silence afforded by being on foot will allow you to appreciate the more delicate delta life – the frogs, the insects, the flowers – and also the amazing diversity of birdlife which approaches 400 species in the summer months.
Vumbura Plains has two flat-bottomed delta boats and a number of mokoro (traditional dugout canoes). Your mokoro guide will skilfully navigate the narrow channels while you marvel at the crystal waters brimming with fish and birdlife. The larger waterways are best enjoyed as the sun sets over the delta, a cocktail in hand, the pods of hippo grunting contentedly as the star-splayed African night emerges.
Travelling to Vumbura Plains
Travelling to Vumbura Plains is easy and convenient. You will catch one of the daily flights from Johannesburg, Windhoek or Kasane into Maun – gateway to the Okavango Delta. From there, you will enjoy a low-level flight over the green and blue mosaic of water and verdant vegetation right into the Vumbura concession. From the Vumbura airstrip, it is a short drive to the magnificent camp with your safari guide.
A feast of excitement awaits
There is so much to do in southern Africa – so much luxury, adventure and pure joy. Why not begin your African expedition in the Cape – either in exuberant Cape Town with its sophisticated gastronomy and natural beauty or out on the world’s most beautiful wine lands in the Franschhoek Valley. Vumbura Plains in the next logical step and from there, Iconic Africa suggests a flight the Zimbabwean or Zambian sides of the Victoria Falls where the luxury of a bygone era of exploration and adventure awaits. After the falls, you could travel to one of the world-renowned lodges of the Kruger National Park where luxury and epic game viewing will provide the perfect ending to your African adventure.
Vumbura Plains boasts a joyful exuberance of contemporary opulence nestled in the trees of the northern Okavango. Friendly Botswanan hospitality and spectacular game viewing – on land and in the water – make Vumbura Plains a brilliant Okavango destination. Contact Iconic Africa now to find out how Vumbura Plains can be part of your ultimate African itinerary.
Contact Iconic Africa now to craft a trip to Botswana you will never forget.
| 14,506
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Hello AAUW NC MVPs,
I trust that everyone received the packets that were sent through the mail to all the membership vice presidents in AAUW NC in November. There were several resources available in the packet from both the Association and AAUW NC. For your convenience, I’ve attached those same documents to this message in case you want to circulate them to others in your branches—will save you some time with the copying if you can simply email them. If you did not receive a packet, please let me know so I can make sure your address is correct in my files. [See previous post for the information on the mailing.]
I would like to draw your attention to the AAUWNC Mission with a Purpose and AAUWNC Membership Scorecard documents. Those two documents allow us to collect data to measure how we’re doing with respect to reaching a diverse population. If you would, please fill out those forms by the end of March and return them to me so I can compile the information for the State Convention. Email or snail mail is fine [or FAX to 866-525-2155].
If you have any questions about any of these documents, please direct membership-related questions to me, and diversity-related questions to Queen Thompson (queencharlotte33@bellsouth.net ) Thank you for your hard work for AAUW NC.
Barbara
Barbara White
AAUW NC Membership Vice President
| 168,805
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Black and White Rattlers
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Came from my personal collection y’all cause I can’t keep kidding myself that they fit... cause they don’t. My loss is your gain!!
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\begin{document}
\maketitle
\begin{abstract}
In 2014, Braverman, Kazhdan, Patnaik and Bardy-Panse, Gaussent and Rousseau associated Iwahori-Hecke algebras to Kac-Moody groups over non-Archimedean local fields. In a previous paper, we defined and studied their principal series representations. In 1982, Kato provided an irreducibility criterion for these representations, in the reductive case. We had obtained partially this criterion in the Kac-Moody case. In this paper, we prove this criterion in the Kac-Moody case.
\end{abstract}
\section{Introduction}
\subsection{The reductive case}
Let $G$ be a split reductive group over a non-Archimedean local field $\KC$.
To each open compact subgroup $K$ of $G$ is associated the Hecke algebra $\HC_K$. This is the algebra of functions from $G$ to $\C$ which have compact support and are $K$-bi-invariant. There exists a strong link between the smooth representations of $G$ and the representations of the Hecke algebras of $G$. Let $K_I$ be the Iwahori subgroup of $G$. Then the Hecke algebra $\HC_\C$ associated with $K_I$ is called the Iwahori-Hecke algebra of $G$ and plays an important role in the representation theory of $G$.
Let $T$ be a maximal split torus of $G$ and $Y$ be the cocharacter lattice of $(G,T)$. Let $B$ be a Borel subgroup of $G$ containing $T$. Let $T_\C=\Hom_{\mathrm{Gr}}(Y,\C^*)$ and $\tau\in T_\C$. Then $\tau$ can be extended to a character $\tau:B\rightarrow \C^*$. If $\tau\in T_\C$, the principal series representation $I(\tau)$ of $G$ is the induction of $\tau\delta^{1/2}$ from $B$ to $G$, where $\delta:B\rightarrow \R^*_+$ is the modulus character of $B$. More explicitly, this is the space of locally constant functions $f:G\rightarrow \C$ such that $f(bg)=\tau\delta^{1/2}(b)f(g)$ for every $g\in G$ and $b\in B$. Then $G$ acts on $I(\tau)$ by right translation.
Let $W^v$ be the vectorial Weyl group of $(G,T)$. By the Bernstein-Lusztig relations, $\HC_\C$ admits a basis $(Z^\lambda* T_w)_{\lambda\in Y, w\in W^v}$ such that $\bigoplus_{\lambda\in Y}\C Z^\lambda$ is a subalgebra of $\HC_\C$ isomorphic to the group algebra $\C[Y]$ of $Y$. We identify $\bigoplus_{\lambda\in Y}\C Z^\lambda$ and $\C[Y]$. We regard $\tau$ as an algebra morphism $\tau:\C[Y]\rightarrow\C$. Then the algebra $\HC_\C$ acts on $I_{\tau,G}:=I(\tau)^{K_I}$, $I(\tau)$ is irreducible as a representation of $G$ if and only $I_{\tau,G}$ is irreducible as a representation of $\HC_\C$ and $I_{\tau,G}$ is isomorphic to the induced representation $I_\tau=\mathrm{Ind}_{\C[Y]}^{\HC_\C}(\tau)$.
Matsumoto and Kato gave criteria for the irreducibility of $I_\tau$. The group $W^v$ acts on $Y$ and thus it acts on $T_\C$. If $\tau\in T_\C$, we denote by $W_\tau$ the stabilizer of $\tau$ in $W^v$. Denote by $q$ be the residue cardinal of $\KC$. Let $\Wta$ be the subgroup of $W_\tau$ generated by the reflections $r_{\alpha^\vee}$, for $\alpha^\vee\in \Phi^\vee$ such that $\tau(\alpha^\vee)=1$, where $\Phi^\vee$ stands for the coroot lattice of $G$. Then Kato proved the following theorem (see \cite[Theorem 2.4]{kato1982irreducibility}):
\begin{Theorem*}\label{thm*Kato's theorem}
Let $\tau\in T_\C$. Then $I_\tau$ is irreducible if and only if it satisfies the following conditions: \begin{enumerate}
\item\label{itWchi engendré par ses réflexions} $W_\tau=\Wta$,
\item for all $\alpha^\vee\in \Phi^\vee$, $\tau(\alpha^\vee)\neq q$.
\end{enumerate}
\end{Theorem*}
When $\tau$ is \textbf{regular}, that is when $W_\tau=\{1\}$, condition~(\ref{itWchi engendré par ses réflexions}) is satisfied and this is a result by Matsumoto (see \cite[Th{\'e}or{\`e}me 4.3.5]{matsumoto77Analyse}).
\subsection{The Kac-Moody case}
Let $G$ be a split Kac-Moody group over a non-Archimedean local field $\KC$. We do not know which topology on $G$ could replace the usual topology on reductive groups over $\KC$. There is up to now no definition of smoothness for the representations of $G$. However one can define certain Hecke algebras in this framework. In \cite{braverman2011spherical} and \cite{braverman2016iwahori}, Braverman, Kazhdan and Patnaik defined the spherical Hecke algebra and the Iwahori-Hecke $\HC_\C$ of $G$ when $G$ is affine. In \cite{gaussent2014spherical} and \cite{bardy2016iwahori}, Bardy-Panse, Gaussent and Rousseau generalized these constructions to the case where $G$ is a general Kac-Moody group. They achieved this construction by using masures (also known as hovels), which are analogous to Bruhat-Tits buildings (see \cite{gaussent2008kac}).
Let $B$ be a positive Borel subgroup of $G$ and $T$ be a maximal split torus of $G$ contained in $B$. Let $Y$ be the cocharacter lattice of $G$, $W^v$ be the Weyl group of $G$ and $Y^{++}$ be the set of dominant cocharacters of $Y$. The Bruhat decomposition does not hold on $G$: if $G$ is not reductive, \[G^+:=\bigsqcup_{\lambda\in Y^{++} }K_I \lambda K_I\subsetneq G.\] The set $G^+$ is a sub-semi-group of $G$. Then $\HC_\C$ is defined to be the set of functions from $K_I\backslash G^+/K_I$ to $\C$ which have finite support. The Iwahori-Hecke algebra $\HC_\C$ of $G$ admits a Bernstein-Lusztig presentation but it is no longer indexed by $Y$. Let $Y^+=\bigcup_{w\in W^v} w.Y^{++}\subset Y$. Then $Y^+$ is the \textbf{integral Tits cone} and we have $Y^+=Y$ if and only $G$ is reductive. The \textbf{Bernstein-Lusztig-Hecke algebra of }$G$ is the space $\AC_\C=\bigoplus_{w\in W^v} \C[Y] *T_w$ subject to to some relations (see subsection~\ref{subIH algebras}). Then $\HC_\C$ is isomorphic to $\bigoplus_{w\in W^v} \C[Y^+]* T_w$.
Let $\tau\in T_\C=\Hom_{\mathrm{Gr}}(Y,\C^*)$. In \cite[6]{hebert2018principal} we defined the space $\widehat{I(\tau)}$ of functions $f$ from $G$ to $\C$ such that for all $g\in G$, $b\in B$, we have $f(bg)=\tau\delta^{1/2}(b)f(g)$. As we do not know which condition could replace
``locally constant'', we do not impose any regularity condition on the functions of $\widehat{I(\tau)}$. Then $G$ acts by right translation on $\widehat{I(\tau)}$.
Let $I_{\tau,G}$ be the subspace of functions $f\in \widehat{I(\tau)}$ which are invariant under the action of $K_I$ and whose support satisfy some finiteness conditions. We defined an action of $\HC_\C$ on $I_{\tau,G}$. This actions extends to an action of $\AC_\C$ on $I_{\tau,G}$ and is isomorphic to the induced representation $I_\tau=\mathrm{Ind}_{\C[Y]}^{\AC_\C}(\tau)$. Moreover the $\AC_\C$-submodules of $I_\tau$ are exactly the $\HC_\C$-submodules of $I_\tau$ (see \cite[Proposition 3.1]{hebert2021decompositions}) and thus we regard $I_\tau$ as a $\AC_\C$-module. We then obtained a weak version of Theorem~\ref{thm*Kato's theorem}: we obtained one implication and we proved the equivalence only under the assumption that the Kac-Moody matrix defining $G$ has size $2$ (see \cite[Theorem 3 and 4]{hebert2018principal}). In this paper, we prove Theorem~\ref{thm*Kato's theorem} in a full generality (see Corollary~\ref{corKatos_irreducibility_criterion}).
\paragraph{Basic ideas of the proof} Let us explain the basic ideas of our proof. We have $I_\tau=\bigoplus_{w\in W^v} \C T_w .\vb_\tau$, where $\vb_\tau\in I_\tau$ is such that $\theta.\vb_\tau=\tau(\theta).\vb_\tau$ for $\theta\in \C[Y]$. For $w\in W^v$, one sets \[I_\tau(w.\tau)=\{x\in I_\tau|\theta.x=w.\tau(\theta).x,\ \forall\theta\in \C[Y]\}.\] By the Frobenius reciprocity, $\Hom_{\AC_\C-\mathrm{mod}}(I_\tau,I_{w.\tau})$ is isomorphic to $I_\tau(w.\tau)$ as a vector space. Let $\UC_\C$ be the set of $\tau\in T_\C$ such that $\tau(\alpha^\vee)\neq q$ for all $\alpha^\vee\in \Phi^\vee$. Let $\tau\in \UC_\C$. In \cite[Theorem 4.8]{hebert2018principal} we proved that $I_\tau$ is irreducible if and only if $\End_{\AC_\C-\mathrm{mod}}(I_\tau)=\C.\Id$, if and only if $I_\tau(\tau)=\C\vb_\tau$.
In order to study $I_\tau(w.\tau)$, for $w\in W^v$, it is convenient to introduce $\AC(T_\C)=\bigoplus_{w\in W^v} T_w *\C(Y)$. The elements of $\AC(T_\C)$ can be regarded as rational functions from an open subset of $T_\C$ to $\HC_{W^v,\C}=\bigoplus_{w\in W^v} \C T_w$. Following Reeder, we introduced elements $F_w\in \ATC$, $w\in W^v$, such that for all $\chi\in T_\C$ for which $F_w(\chi)$ is well defined, $F_w(\chi).\vb_\chi\in I_\chi(w.\chi)$. The group $W_\tau$ decomposes as $W_\tau=R_\tau\ltimes \Wta$, where $R_\tau$ is some subgroup of $W_\tau$ called the $R$-group. If $w_R\in R_\tau$, then $F_{w_R}$ has no pole at $\tau$ and thus $F_{w_R}(\tau).\vb_\tau$ corresponds to an element $\psi_{w_r}$ of $\End(I_\tau)$. For $w\in \Wta$ however, $F_w$ has poles at $\tau$ and thus describing $I_\tau(\tau)$ requires some works. Inspired by works of Reeder and Keys in the reductive case, (\cite{reeder1997nonstandard} and \cite{keys1982decomposition}), we determined \[I_\tau(\tau,\mathrm{gen})=\{x\in I_\tau|\forall \theta\in \C[Y], \forall n\gg 0, \big(\theta-\tau(\theta)\big)^n.x=0\}.\] We proved (see \cite[Proposition 5.13]{hebert2021decompositions}) that \[I_\tau(\tau,\mathrm{gen})=\bigoplus_{w_R\in R_\tau} \psi_{w_R}\left(\Itg\right),\] where $\Itg:=\left(\AC_\C\cap \bigoplus_{w\in \Wta}F_w*\C(Y)\right).\vb_\tau$ corresponds to the ``$\Wta$-part'' of $I_\tau(\tau,\mathrm{gen})$. We deduced that \[I_\tau(\tau)=\bigoplus_{w_R\in R_\tau} \psi_{w_R}\left(I_\tau(\tau)\cap \Itg\right).\]
It then remained to prove that $\Itg\cap I_\tau(\tau)=\C \vb_\tau$, which we achieve in this paper (see Theorem~\ref{thmWeight_space}). Note that by our description of $\Itg$, proving that $I_\tau(\tau)\cap \Itg=\C\vb_\tau$ can more or less be reduced to proving that $I_{\mathds{1}}(\mathds{1})=\C \vb_{\mathds{1}}$, where $I_{\mathds{1}}=\mathrm{Ind}_{\C(Y)_\tau}^{\KCC_\tau}(\mathds{1})$, $\C(Y)_\tau$ is the subset of $\C(Y)$ consisting of the elements which have no pole at $\tau$, $\mathds{1}:Y\rightarrow \C$ is the constant function equal to $1$ and $\KCC_\tau\subset \bigoplus_{w\in W^v} T_w*\C(Y)_\tau$ is some kind of Bernstein-Lusztig-Hecke algebra associated with $\tau$ (see subsection~\ref{subBLH_structure_K} for the definition of $\KCC_\tau$).
We then deduce:
\begin{Corollary*}\label{cor*End_I_tau}(see Corollary~\ref{corSilberger_dimension_theorem} and Corollary~\ref{corKatos_irreducibility_criterion})
Let $\tau\in \UC_\C$. Then $\End(I_\tau)\simeq \C[R_\tau]$, where $R_\tau=W_\tau/\Wta$. In particular, $\End(I_\tau)=\C\Id$ if and only $R_\tau=\{1\}$.
\end{Corollary*}
In \cite{hebert2021decompositions}, we studied the submodules and the quotients of $I_\tau$, for $\tau\in \UC_\C$. Many results were proved only when the Kac-Moody matrix defining $G$ has size $2$ or under some conjecture (\cite[Conjecture 5.16]{hebert2021decompositions}). As we prove this conjecture, we can drop the assumption on the size of the matrix. In particular, \cite[Theorem 5.34 and Theorem 5.38]{hebert2021decompositions} yield links between the submodules and the quotients of $I_\tau$ and the right submodules and quotients of $\End(I_\tau)$ respectively.
\paragraph{Frameworks}
Actually, following \cite{bardy2016iwahori} we study Iwahori-Hecke algebras associated with abstract masures. In particular our results also apply when $G$ is an almost-split Kac-Moody group over a non-Archimedean local field. In this case, most of the results of this introduction are true but the formulas are more complicated (they are given in the paper). Corollary~\ref{cor*End_I_tau} is not necessarily true for almost-split groups, even in the reductive case.
\paragraph{Organization of the paper}
This paper is organized as follows. In section~\ref{secIH algebras}, we recall the definitions of the Iwahori–Hecke algebras and of the principal series representations, and we introduce tools to study these representations.
In section~\ref{secDescription_Itg}, we study the algebra $\KCC_\tau$ mentioned above and describe $\Itg$.
In section~\ref{secKato_s_irreducibility_criterion}, we prove Kato's irreducibility criterion.
\section{Iwahori-Hecke algebras}\label{secIH algebras}\label{secIH_algebras}
Let $G$ be a Kac-Moody group over a non-archimedean local field. Then Gaussent and Rousseau constructed a space $\I$, called a masure on which $G$ acts, generalizing the construction of the Bruhat-Tits buildings (see \cite{gaussent2008kac}, \cite{rousseau2016groupes} and \cite{rousseau2017almost}). In \cite{bardy2016iwahori} Bardy-Panse, Gaussent and Rousseau attached an Iwahori-Hecke algebra $\HC_\RC$ to each masure satisfying certain conditions and to each ring $\RC$. They in particular attach an Iwahori-Hecke algebra to each almost-split Kac-Moody group over a local field. The algebra $\HC_\RC$ is an algebra of functions defined on some pairs of chambers of the masure, equipped with a convolution product. Then they prove that under some additional hypothesis on the ring $\RC$ (which are satisfied by $\C$), $\HC_\RC$ admits a Bernstein-Lusztig presentation. We restrict our study to the case where $\RC=\C$. In this paper, we will only use the Bernstein-Lusztig presentation of $\HC_\C$. More precisely, we introduce an algebra $\AC(T_\C)=\bigoplus_{w\in W^v} T_w*\C(Y)$, which contains both $\AC_\C$ and $\HC_\C$. We mainly study $\AC(T_\C)$ and $\AC_\C$. We do not introduce masures nor Kac-Moody groups. We however introduce the standard apartment of a masure.
\subsection{Standard apartment of a masure}\label{subRootGenSyst}
A (finite)\textbf{ Kac-Moody matrix} (or { generalized Cartan matrix}) is a square matrix $A=(a_{i,j})_{i,j\in I}$ indexed by a finite set $I$, with integral coefficients, and such that :
\begin{enumerate}
\item[\tt $(i)$] $\forall \ i\in I,\ a_{i,i}=2$;
\item[\tt $(ii)$] $\forall \ (i,j)\in I^2, (i \neq j) \Rightarrow (a_{i,j}\leq 0)$;
\item[\tt $(iii)$] $\forall \ (i,j)\in I^2,\ (a_{i,j}=0) \Leftrightarrow (a_{j,i}=0$).
\end{enumerate}
In the paper, we will also consider infinite Kac-Moody matrices. The definition is the same except that $I$ is infinite. However we will only consider root generating systems associated with finite Kac-Moody matrices.
A \textbf{root generating system} is a $5$-tuple $\mathcal{S}=(A,X,Y,(\alpha_i)_{i\in I},(\alpha_i^\vee)_{i\in I})$\index{$\mathcal{S}$} made of a finite Kac-Moody matrix $A$ indexed by the finite set $I$, of two dual free $\Z$-modules $X$ and $Y$ of finite rank, and of a free family $(\alpha_i)_{i\in I}$ (respectively $(\alpha_i^\vee)_{i\in I}$) of elements in $X$ (resp. $Y$) called \textbf{simple roots} (resp. \textbf{simple coroots}) that satisfy $a_{i,j}=\alpha_j(\alpha_i^\vee)$ for all $i,j$ in $I$. Elements of $X$ (respectively of $Y$) are called \textbf{characters} (resp. \textbf{cocharacters}).
Fix such a root generating system $\mathcal{S}=(A,X,Y,(\alpha_i)_{i\in I},(\alpha_i^\vee)_{i\in I})$ and set $\A:=Y\otimes \R$\index{$\A$}. Each element of $X$ induces a linear form on $\A$, hence $X$ can be seen as a subset of the dual $\A^*$. In particular, the $\alpha_{i}$'s (with $i \in I$) will be seen as linear forms on $\A$. This allows us to define, for any $i \in I$, an involution $r_{i}$ of $\A$ by setting $r_{i}(v) := v-\alpha_i(v)\alpha_i^\vee$ for any $v \in \A$. Let $\SCC=\{r_i|i\in I\}$\index{$\SCC$} be the (finite) set of \textbf{simple reflections}. One defines the \textbf{Weyl group of $\mathcal{S}$} as the subgroup $W^{v}$\index{$W^v$} of $\mathrm{GL}(\A)$ generated by $\SCC$. The pair $(W^{v}, \SCC)$ is a Coxeter system, hence we can consider the length $\ell(w)$ with respect to $\SCC$ of any element $w$ of $W^{v}$. If $s\in \SCC$, $s=r_i$ for some unique $i\in I$. We set $\alpha_s=\alpha_i$ and $\alpha_s^\vee=\alpha_i^\vee$.
The following formula defines an action of the Weyl group $W^{v}$ on $\A^{*}$:
\[\displaystyle \forall \ x \in \A , w \in W^{v} , \alpha \in \A^{*} , \ (w.\alpha)(x):= \alpha(w^{-1}.x).\]
Let $\Phi:= \{w.\alpha_i|(w,i)\in W^{v}\times I\}$\index{$\Phi,\Phi^\vee$} (resp. $\Phi^\vee=\{w.\alpha_i^\vee|(w,i)\in W^{v}\times I\}$) be the set of \textbf{real roots} (resp. \textbf{real coroots}): then $\Phi$ (resp. $\Phi^\vee$) is a subset of the \textbf{root lattice} $Q_\Z:= \displaystyle \bigoplus_{i\in I}\Z\alpha_i$ (resp. \textbf{coroot lattice} $Q^\vee_\Z=\bigoplus_{i\in I}\Z\alpha_i^\vee$). By \cite[1.2.2 (2)]{kumar2002kac}, we have $\R \alpha^\vee\cap \Phi^\vee=\{\pm \alpha^\vee\}$ and $\R \alpha\cap \Phi=\{\pm \alpha\}$ for all $\alpha^\vee\in \Phi^\vee$ and $\alpha\in \Phi$.
\paragraph{Reflections and roots}
We equip $(W^v,\SCC)$ with the Bruhat order $\leq$ (see \cite[Definition 2.1.1]{bjorner2005combinatorics}).
Let $\RCC=\{wsw^{-1}|w\in W^v, s\in \SCC\}$\index{$\RCC$} be the set of \textbf{reflections} of $W^v$. Let $r\in \RCC$. Write $r=wsw^{-1}$, where $w\in W^v$, $s\in \SCC$ and $ws>w$ (which is possible because if $ws<w$, then $r=(ws)s(ws)^{-1}$). Then one sets $\alpha_r=w.\alpha_s\in \Phi_+$\index{$\alpha_r,\alpha_r^\vee$} (resp. $\alpha_r^\vee=w.\alpha_s^\vee\in\Phi^\vee_+$). Conversely, if $\alpha\in \Phi$ or $\alpha^\vee\in \Phi^\vee$, $\alpha=w.\alpha_s$ or $\alpha^\vee=w.\alpha_s^\vee$, where $w\in W^v$ and $s\in \SCC$, one sets $r_{\alpha}=wsw^{-1}$ or $r_{\alpha^\vee}=wsw^{-1}$. This is independant of the choices of $w$ and $s$ by \cite[1.3.11 Theorem (b5)]{kumar2002kac}.
\subsection{Iwahori-Hecke algebras}\label{subIH algebras}
In this subsection, we give the definition of the Iwahori-Hecke algebra via its Bernstein-Lusztig presentation.
\subsubsection{The algebra $\AC(T_\C)$}\label{subsubAlgebra_H(T_F)}
Let $(\sigma_s)_{s\in \SCC}, (\sigma'_s)_{s\in \SCC}\in \C^{\SCC}$ satisfying the following relations: \begin{itemize}
\item if $\alpha_{s}(Y) = \Z$, then $\sigma_{s} = \sigma'_{s}$\index{$\sigma_s,\sigma_s'$};
\item if $s,t \in \SCC$ are conjugate (i.e if there exists a sequence $s_1,\ldots,s_n\in \SCC$ such that $s_1=s$, $s_n=t$ and $\alpha_{s_i}(\alpha_{s_{i+1}}^\vee)=\alpha_{s_{i+1}}(\alpha_{s_i}^\vee)=-1$, for $i\in \llbracket 1,n-1\rrbracket$), then $\sigma_{s}=\sigma_{t}=\sigma'_{s}=\sigma'_{t}$.
\end{itemize} As in \cite{hebert2018principal}, we moreover assume that $|\sigma_s|,|\sigma_s'|>1$, for all $s\in \SCC$.
\begin{Definition} Let $\HC_{W^v,\C}$ be the \textbf{Hecke algebra of the Coxeter group $W^v$ over $\C$}, that is: \begin{itemize}
\item as a vector space, $\HC_{W^v,\C}=\bigoplus_{w\in W^v} \C T_w$, where the $T_w$, $w\in W^v$ are symbols,
\item $\forall \ s \in \SCC, \forall \ w \in W^{v}$, $T_{s}*T_{w}=\left\{\begin{aligned} & T_{sw} &\mathrm{\ if\ }\ell(sw)=\ell(w)+1\\ & (\sigma_{s}^2-1) T_{w}+\sigma_s^2 T_{s w} &\mathrm{\ if\ }\ell(sw)=\ell(w)-1 .\end{aligned}\right . \ $
\end{itemize}
\end{Definition}
Let $\C[Y]$ be the group algebra of $Y$ over $\C$, that is:\begin{itemize}
\item as a vector space, $\C[Y]=\bigoplus_{\lambda\in Y} \C Z^\lambda$, where the $Z^\lambda$, $\lambda\in Y$ are symbols,
\item for all $\lambda,\mu\in Y$, $Z^\lambda*Z^\mu=Z^{\lambda+\mu}$.
\end{itemize}
We denote by $\C(Y)$ its field of fractions. For $\theta=\frac{\sum_{\lambda\in Y}{a_\lambda Z^\lambda}}{\sum_{\lambda\in Y} b_\lambda Z^\lambda}\in \C(Y)$ and $w\in W^v$, set ${^w}\theta:=\frac{\sum_{\lambda\in Y}a_\lambda Z^{w.\lambda}}{\sum_{\lambda\in Y}b_\lambda Z^{w.\lambda}}$.
\medskip
Let $\AC(T_\C)$ be the algebra defined as follows: \begin{itemize}
\item as a vector space, $\AC(T_\C)= \C(Y)\otimes \HC_{W^v, \C}$ (we write $\theta*h$ instead of $\theta\otimes h$ for $\theta\in \C(Y)$ and $h\in \HC_{W^v,\C}$),
\item $\AC(T_\C)$ is equipped with the unique product $*$ which turns it into an associative algebra and such that, for $\theta\in \C(Y)$ and $s\in \SCC$, we have: \[T_{s}*\theta-{^s\theta}*T_{s} =\tilde{\Omega}_s(\theta-{^s\theta}),\] where $\tilde{\Omega}_s(\theta)=Q_s^T(\theta-{^s\theta})$ and $Q_s^T=\frac{(\sigma_s^2-1)+\sigma_s(\sigma_s'-\sigma_s'^{-1})Z^{-\alpha_s^\vee}}{1-Z^{-2\alpha_s^\vee}}$\index{$Q_s^T$}.
\end{itemize}
By \cite[Proposition 2.10]{hebert2018principal}, such an algebra exists and is unique.
\subsubsection{The Bernstein-Lusztig Hecke algebra and the Iwahori-Hecke algebra}
Let $C^v_f=\{x\in \A|\alpha_i(x)>0\forall i\in I\}$\index{$C^v_f$}, $\T=\bigcup_{w\in W^v} w.\overline{C}^v_f$\index{$\T$} be the \textbf{Tits cone} and $Y^+=Y\cap \T$\index{$Y^+$}.
\begin{Definition}\label{defBernstein-Lusztig algebra}
The \textbf{Bernstein-Lusztig-Hecke algebra of } $(\A,(\sigma_s)_{s\in \SCC},(\sigma'_s)_{s\in \SCC})$ over $\C$ is the subalgebra $\AC_\C=\bigoplus_{\lambda\in Y,w\in W^v}\C Z^\lambda*T_w=\bigoplus_{\lambda\in Y,w\in W^v}\C T_w* Z^\lambda$ of $\AC(T_\C)$. The \textbf{Iwahori-Hecke algebra of $(\A,(\sigma_s)_{s\in \SCC},(\sigma'_s)_{s\in \SCC})$ over $\C$} is the subalgebra $\HC_\C=\bigoplus_{\lambda\in Y^+,w\in W^v}\C Z^\lambda*T_w=\bigoplus_{\lambda\in Y^+,w\in W^v}\C T_w*Z^\lambda$ of $\AC_\C$. Note that for $G$ reductive, we recover the usual Iwahori-Hecke algebra of $G$, since $ \T=\A$.
\end{Definition}
\begin{Remark}\label{remIH algebre dans le cas KM deploye}
\begin{enumerate}
\item The algebra $\AC_\C$ was first defined in \cite[Theorem 6.2]{bardy2016iwahori} without defining $\AC(T_\C)$. Let $\mathcal{K}$ be a non-Archimedean local field and $q$ be its residue cardinal. Let $\mathbf{G}$ be the minimal Kac-Moody group associated with $\mathcal{S}=(A,X,Y,(\alpha_i)_{i\in I},(\alpha_i^\vee)_{i\in I})$\index{$\mathcal{S}$} and $G=\mathbf{G}(\mathcal{K})$ (see \cite[Section 8]{remy2002groupes} or \cite{tits1987uniqueness} for the definition). Take $\sigma_s=\sigma'_s=\sqrt{q}$ for all $s\in \SCC$. Then $\HC_\C$ is the Iwahori-Hecke algebra of $G$ (see \cite[Definition 2.5 and 6.6 Proposition]{bardy2016iwahori}). In the case where $G$ is an untwisted affine Kac-Moody group, these algebras were introduced in \cite{braverman2016iwahori}. Note also that our frameworks is more general than the one of split Kac-Moody groups over local fields. It enables for example to study the Iwahori-Hecke algebras associated with almost split Kac-Moody groups over local fields, as in \cite{bardy2016iwahori}. In this case we do not have necessarily $\sigma_s=\sigma_s'=\sigma_t=\sigma_t'$ for all $s,t\in \SCC$.
\item Let $s\in \SCC$. Then if $\sigma_s=\sigma'_s$, $Q_s^T=\frac{\sigma_s^2-1}{1-Z^{-\alpha_s^\vee}}$.
\item\label{itPolynomiality_Bernstein_Lusztig} Let $s\in \SCC$. Then $\tilde{\Omega}_s(\C[Y])\subset \C[Y]$ and $\tilde{\Omega}_s(\C[Y^+])\subset \C[Y^+]$ Indeed, let $\lambda\in Y$. Then $Q_s^T(Z^\lambda-Z^{s.\lambda})=Q^T_s.Z^\lambda(1-Z^{-\alpha_s(\lambda)\alpha_s^\vee})$. Assume that $\sigma_s=\sigma_s'$. Then \[ \frac{1-Z^{-\alpha_s(\lambda)\alpha_s^\vee}}{1-Z^{-\alpha_s^\vee}}=\left\{\begin{aligned} &\sum_{j=0}^{\alpha_s(\lambda)-1}Z^{-j\alpha_s^\vee} &\mathrm{\ if\ }\alpha_s(\lambda)\geq 0\\ & -Z^{\alpha_s^\vee}\sum_{j=0}^{-\alpha_s(\lambda)-1}Z^{j\alpha_s^\vee} &\mathrm{\ if\ }\alpha_s(\lambda)\leq 0,\end{aligned}\right.\] and thus $Q_s^T(Z^\lambda-Z^{s.\lambda})\in \C[Y]$. If $\sigma_s'\neq \sigma_s$, then $\alpha_s(Y)=2\Z$ and a similar computation enables to conclude. In particular, $\AC_\C$ and $\HC_\C$ are subalgebras of $\AC(T_\C)$.
\item In \cite{hebert2018principal} and \cite{hebert2021decompositions}, we used the basis $(H_w)_{w\in W^v}$ instead of $(T_{w})_{w\in W^v}$ in the presentation of $\HCW$ and $\AC_\C$. If $w\in W^v$, $w=s_1\ldots s_k$, with $k=\ell(w)$ and $s_1,\ldots,s_k\in \SCC$, we have $T_w=\sigma_{s_1}\ldots \sigma_{s_k} H_w$.
\end{enumerate}
\end{Remark}
\subsection{Principal series representations}\label{subPrincipal series representations}
In this subsection, we introduce the principal series representations of $\AF$.
We now fix $(\A,(\sigma_s)_{s\in \SCC},(\sigma'_s)_{s\in \SCC})$ as in Subsection~\ref{subIH algebras}. Let $\AC_\C$ be the Iwahori-Hecke and the Bernstein-Lusztig Hecke algebras of $(\A,(\sigma_s)_{s\in \SCC},(\sigma'_s)_{s\in \SCC})$ over $\C$.
Let $T_\C= \Hom_{\mathrm{Gr}}(Y,\C^*)$\index{$T_\C$} be the group of group morphisms from $Y$ to $\C^*$. Let $\tau\in T_\C$. Then $\tau$ induces an algebra morphism $\tau:\C[Y]\rightarrow \C$ by the formula $\tau(\sum_{\lambda\in Y} a_\lambda Z^\lambda)=\sum_{\lambda\in Y} a_\lambda \tau(\lambda)$, for $\sum a_\lambda Z^\lambda\in \C[Y]$. This equips $\C$ with the structure of a $\C[Y]$-module.
Let $I_\tau=\mathrm{Ind}^{\AC_\C}_{\C[Y]}(\tau)=\AC_\C\otimes_{\C[Y]} \C$\index{$I_\tau$}. As a vector space, $I_\tau=\bigoplus_{w\in W^v} \C \vb_\tau$, where $\vb_\tau$ is some symbol. The actions of $\AC_\C$ on $I_\tau$ is as follows. Let $h=\sum_{w\in W^v} T_w P_w\in \AC_\C$, where $P_w\in \C[Y]$ for all $w\in W^v$. Then $h.\vb_\tau=\sum_{w\in W^v} \tau(P_w)T_w\vb_\tau$. In particular, $I_\tau$ is a principal $\AC_\C$-module generated by $\vb_\tau$.
If $A$ is a vector space over $\C$ and $B$ is a set, we denote by $A^{(B)}$ the set of families $(a_b)_{b\in B}$ such that $\{b\in B|a_b\neq 0\}$ is finite. We regard the elements of $\C[Y]$ as polynomial functions on $T_\C$ by setting: \[\tau(\sum_{\lambda\in Y}a_\lambda Z^\lambda)=\sum_{\lambda\in Y}a_\lambda\tau(\lambda),\] for all $(a_\lambda)\in \C^{(Y)}$. The ring $\C[Y]$ is a unique factorization domain. Let $\theta\in \C(Y)$ and $(f,g)\in \C[Y]\times \C[Y]^*$ be such that $\theta=\frac{f}{g}$ and $f$ and $g$ are coprime. Set $\DC(\theta)=\{\tau\in T_\C|\tau(g)\neq 0\}$. Then we regard $\theta$ as a map from $\DC(\theta)$ to $\C$ by setting $\theta(\tau)=\frac{f(\tau)}{g(\tau)}$ for all $\tau\in \DC(\theta)$.
If $\tau\in T_\C$, let $\C(Y)_\tau=\{\frac{f}{g}|f,g\in \C[Y]\text{ and } g(\tau)\neq 0\}\subset \C(Y)$\index{$\C(Y)_\tau$}. Let $\ATF_\tau=\bigoplus_{w\in W^v} T_w \C(Y)_\tau \subset \ATF$\index{$\ATF_\tau,\ATC_\tau$}. This is a not a subalgebra of $\ATF$ (consider for example $\frac{1}{Z^\lambda-1}*T_s=T_s*\frac{1}{Z^{s.\lambda}-1}+\ldots$ for some well chosen $\lambda\in Y$, $s\in \SCC$ and $\tau\in T_\C$). It is however an $\HFW-\C(Y)_\tau$ bimodule. For $\tau\in T_\C$, we define $\ev_\tau:\ATF_\tau\rightarrow \HFW$\index{$ev_\tau$} by $\ev_\tau(h)=h(\tau)=\sum_{w\in W^v} T_w\theta_w(\tau)$ if $h=\sum_{w\in W^v}T_w \theta_w\in \ATC_\tau$. This is a morphism of $\HFW-\C(Y)_\tau$-bimodules.
\subsection{Decomposition of $W_\tau$}
For $r=wsw^{-1}\in \RCC$, with $w\in W^v$ and $s\in \SCC$, we define $Q^T_r={^w\left(Q_s^T\right)}$ and $\sigma_r=\sigma_s$. This does not depend on the choices of $w$ and $s$. For $r\in \RCC$, set $\zeta_r=-Q^T_r+\sigma_r^2$. Write $\zeta_r=\frac{\zeta_r^{\mathrm{num}}}{\zeta_r^{\mathrm{den}}}$\index{$\zeta_r$}, with $\zeta_r^{\mathrm{num}},\zeta_{r}^{\mathrm{den}}$ are coprime elements of $\C[Y]$.
For $\tau\in T_\C$, set $W_\tau=\{w\in W^v|\ w.\tau=\tau\}$\index{$W_\tau$}, $\Phi^\vee_{(\tau)}=\{\alpha^\vee\in \Phi^\vee| \zeta_{\alpha^\vee}^{\mathrm{den}}(\tau)=0\}$\index{$\Phi^\vee_{(\tau)}$}, $\Phi^\vee_{(\tau),+}=\Phi^\vee_{(\tau)}\cap \Phi^\vee_+$, $\RCC_{(\tau)}=\{r=r_{\alpha^\vee}\in \RCC|\alpha^\vee\in \Phi^\vee_{(\tau)}\}$\index{$\RCC_{(\tau)}$} and \[\Wta=\langle \RCC_{(\tau)}\rangle=\langle \{r=r_{\alpha^\vee}\in \RCC| \zeta_{\alpha^\vee}^{\mathrm{den}}(\tau)=0\}\rangle\subset W^v.\]\index{$\Wta$}
By \cite[Remark 5.1]{hebert2018principal}, $\Wta\subset W_\tau$. When $\alpha_s(Y)=\Z$ for all $s\in \SCC$, then $\Wta=\langle W_\tau\cap \RCC\rangle$. Let \[R_\tau=\{w\in W_\tau|w.\Phi^\vee_{(\tau),+}=\Phi^\vee_{(\tau),+}\}.\] By \cite[Lemma 5.2 and 5.3]{hebert2021decompositions}, $\Wta$ is normal in $W_\tau$ and $w_\tau=R_\tau\ltimes \Wta$. We defined in \cite[5.4]{hebert2018principal} a set $\SCC_\tau\subset\RCC\cap \Wta$ for which $(\Wta,\SCC_\tau)$ is a Coxeter system. We denote by $\leq_\tau$ the Bruhat order and by $\ell_\tau$ the length on $(\Wta,\SCC_\tau)$. Then by \cite[Lemma 5.10]{hebert2018principal}, for all $w,w'\in \Wta$, we have $w\leq_\tau w'$ implies $w\leq w'$.
\subsection{Weights and intertwining operators}
If $\tau\in T_\C$, we denote by $\End(I_\tau)$ the algebra of endomorphisms of $\AC_\C$-modules of $I_\tau$. Let \[I_\tau(\tau)=\{x\in I_\tau|\theta.x=\tau(\theta).x\forall \theta\in \C[Y]\}\] and \[I_\tau(\tau,\mathrm{gen})=\{x\in I_\tau|\forall\theta\in \C[Y], (\theta-\tau(\theta))^n.x=0,\ \forall n\gg 0\}\supset I_\tau(\tau).\]
For $s\in \SCC$, one sets \[F_s=T_s+Q_s^T\in \AC(T_\C)\index{$F_s$}.\]
Let $x\in I_\tau(\tau)$. Define $\Upsilon_x\in \End(I_\tau)$ by $\Upsilon(h.\vb_\tau)=h.x$, for $h\in \AC_\C$. Then it is easy to check that $\Upsilon:I_\tau(\tau)\rightarrow \End(I_\tau)$ is well defined and is an isomorphism of vector spaces.
Let $w\in W^v$. Let $w=s_1\ldots s_r$ be a reduced expression of $w$, with $k=\ell(w)$ and $s_1,\ldots,s_k\in \SCC$. Set \[F_w=F_{s_r}\ldots F_{s_1}=(T_{s_r}+Q_{s_r}^T)\ldots (T_{s_1}+Q_{s_1}^T)\in \AC(T_\C).\]\index{$F_w$}
\begin{Lemma}\label{lemReeder 4.3}(see \cite[Lemma 4.14]{hebert2018principal}) Let $w\in W^v$. \begin{enumerate}
\item\label{itWell_definedness_Fw} The element $F_w\in \AC(T_\C)$ is well defined, i.e it does not depend on the choice of a reduced expression for $w$.
\item\label{itLeading_coefficient_Fw} $F_w-T_w\in \ATF^{<w}=\bigoplus_{v<w}T_v\C(Y)$.
\item\label{itCommutation_relation} If $\theta\in \C(Y)$, then $\theta*F_w=F_w*{^{w^{-1}}}\theta$.
\item\label{itDomain_Fw} If $\tau\in T_\C$ is such that $F_w\in \AC(T_\C)_\tau$, then $\theta.F_w(\tau).\vb_\tau =\left(w.\tau(\theta)\right) F_w(\tau).\vb_\tau$, for every $\theta\in \C[Y]$.
\end{enumerate}
\end{Lemma}
We set \[\UC_\C=\{\tau\in T_\C|\tau(\zeta_{r}^{\mathrm{num}})\neq 0\forall r\in \RCC\}.\] When $\sigma_s=\sigma_s'=\sqrt{q}$, for $q\in \R_{>0}$, $\UC_\C=\{\tau\in T_\C|\tau(\alpha^\vee)\neq q,\forall \alpha^\vee\in \Phi^\vee\}$.
\medskip
By \cite[Lemma 5.7]{hebert2021decompositions}, if $w_R\in R_\tau$, then $F_{w_R}\in \ATC_\tau$ and $F_{w_R}(\tau).\vb_\tau\in I_\tau(\tau)$. Let $\psi_{w_R}=\Upsilon _{F_{w_R}(\tau).\vb_\tau}\in \End(I_\tau)$.
Set \begin{equation}\label{eqDefinition_Itg}
\Itg= \left(\AC_\C \cap \bigoplus_{w\in \Wta} F_w *\C(Y)\right).\vb_\tau.
\end{equation} which is well defined by \cite[Lemma 5.23]{hebert2018principal}. \[ \sum_{k\in \N,s_1,\ldots, s_k\in \SCC_\tau} \C\ev_\tau(\KC_{s_1}*\ldots*\KC_{s_k}).\vb_\tau,\] By \cite[Proposition 5.13 (1)]{hebert2021decompositions}, if $\tau\in \UC_\C$, we have \begin{equation}\label{eqHeb21_5.12}
I_\tau(\tau,\mathrm{gen})=\bigoplus_{w_R\in R_\tau} \psi_{w_R}\big(\Itg\big).
\end{equation}
\section{Description of $\Itg$}\label{secDescription_Itg}
In this paper, we prove that if $\tau\in \UC_\C$, then $I_\tau(\tau,\Wta):=\Itg\cap I_\tau(\tau)=\C \vb_\tau$ (see Theorem~\ref{thmWeight_space}) and we deduce information on the submodules of $I_\tau$. To that end,
we begin by describing $\Itg$. For $s\in \SCC_\tau$, let \begin{equation}\label{eqDef_Ks}
\KC_s=F_s-\zeta_s=F_s+Q^T_s-\sigma_s^2\in \ATC.
\end{equation} In \cite{hebert2021decompositions} and \cite{hebert2021decompositions}, $\Itg$ is described in terms of the $\ev_\tau(\KC_{s_1}*\ldots*\KC_{s_k}).\vb_\tau$, where $s_1,\ldots,s_k\in \SCC_\tau$. However, if $w\in \Wta$ and $w=s_1\ldots s_k=s_1'\ldots s_k'$ is a reduced expression of $w$, with $k=\ell_\tau(w)$ and $s_1,\ldots,s_k,s_1',\ldots,s_k'\in \SCC_\tau$, we might have $\KC_{s_1}\ldots \KC_{s_k}\neq \KC_{s_1'}\ldots \KC_{s_k'}$. We thus slightly modify the $\KC_s$ and define $\tKC_s$, for $s\in \SCC$, so that we have $\tKC_{s_1}\ldots \tKC_{s_k}= \tKC_{s_1'}\ldots \tKC_{s_k'}$. This enables us to define $\tKC_w$, for $w\in \Wta$. Let $\KCC_\tau=\bigoplus_{w\in \Wta} \tKC_w *\C(Y)_\tau$. We prove that $\KCC_\tau$ has a presentation very close to the Bernstein-Lusztig presentation of $\AC_\C$.
In order to study $\Itg$, it is then convenient to describe it as a $\KCC_\tau$-module. However, $\KCC_\tau$ is not contained in $\AC_\C$. We thus extend the action of $\KCC_\tau\cap \AC_\C$ on $\Itg$ to an action of $\KCC_\tau$ on $\Itg$.
In subsection~\ref{subBLH_structure_K}, we study $\KCC_\tau$ and in subsection~\ref{subAction_KC_Itau}, we define an action of $\KCC_\tau$ on $\Itg$. In subsection~\ref{subExample_infinite_basis}, we prove that there can exist $\tau\in \UC_\C$ for which $\SCC_\tau$ is infinite.
\subsection{Bernstein-Lusztig-Hecke structure on $\KCC_\tau$}\label{subBLH_structure_K}
For $s\in \SCC_\tau$, we set \begin{equation}\label{eqDefinition_tKC}
\tKC_s=\KC_s+\sigma_s^2=F_s+Q_s^T\in \ATC
\end{equation}\index{$\tKC_s$} Note that if $\tau(\lambda)=1$ for all $\lambda\in Y$, we have $\SCC_\tau=\SCC$ and
\begin{equation}\label{eqK)T_if_tau=1}
\tKC_s=T_s
\end{equation} for $s\in \SCC$.
Let $s\in \SCC_\tau$. For $\theta\in \C(Y)$, set $\tilde{\Omega}_s(\theta)=Q_s^T(\theta-{^s\theta})$. Then similarly as in Remark~\ref{remIH algebre dans le cas KM deploye}~\eqref{itPolynomiality_Bernstein_Lusztig} and \cite[Lemma 5.22]{hebert2018principal} we have \begin{equation}\tilde{\Omega}_s(\C[Y])\subset \C[Y]\text{ and }\tilde{\Omega}_s\big(\C(Y)_\tau\big)\subset \C(Y)_\tau.\end{equation}
The aim of this subsection is to prove the following proposition.
\begin{Proposition}\label{propBL_relations}
Let $\tau\in T_\C$.
\begin{enumerate}
\item Let $w\in \Wta$ and $w=s_1\ldots s_k$ be a reduced decomposition of $w$, with $k=\ell_\tau(w)$ and $s_1\ldots,s_k\in \SCC_\tau$. Then \[\tKC_{w}=\tKC_{s_1}*\ldots*\tKC_{s_k}\in \ATC_\tau\]\index{$\tKC_w$} is well defined, independently of the choice of the reduced decomposition.
\item\label{itHecke_relations} Let $s\in \SCC_\tau$ and $w\in \Wta$. Then $\tKC_{s}*\tKC_{w}=\left\{\begin{aligned} & \tKC_{sw} &\mathrm{\ if\ }\ell_\tau(sw)=\ell(w)+1\\ & (\sigma_{s}^2-1) \tKC_{w}+\sigma_s^2 \tKC_{s w} &\mathrm{\ if\ }\ell_\tau(sw)=\ell_\tau(w)-1 .\end{aligned}\right . \ $
\item\label{itBL_relations} For $\theta\in \C(Y)$ and $s\in \SCC_\tau$, we have \begin{equation}\label{eqBL_relation}
\theta*\tKC_s=\tKC_s*{^s\theta}+\tilde{\Omega}_s(\theta).
\end{equation}
\item The matrix $\big(\alpha_s(\alpha_r^\vee)\big)_{r,s\in \SCC_\tau}$ is a (possibly infinite) Kac-Moody matrix.
\item\label{itAlgebra_K} If $\tau\in \UC_\C$, the space \[\KCC_\tau=\sum_{k\in \N, s_1\ldots,s_k\in \SCC_\tau} \KC_{s_1}*\ldots *\KC_{s_k}*\C(Y)_\tau=\bigoplus_{w\in \Wta}\tKC_w*\C(Y)_\tau\] is a subalgebra of $\ATC$ contained in $\ATC_\tau$.
\end{enumerate}
\end{Proposition}
In particular, $\bigoplus_{w\in \Wta} \tKC_w*\C(Y)_\tau$ is almost a Bernstein-Lusztig-Hecke algebra as defined in subsection~\ref{subIH algebras}. Note however that $\SCC_\tau$ can be infinite (and thus the $(\alpha_s)_{s\in \SCC_\tau}$ and $(\alpha_{s}^\vee)_{s\in \SCC_\tau}$ are not free), see Lemma~\ref{lemExample_infinite_SCC_tau}. The proof of this proposition is a consequence of the lemmas of this subsection below.
\begin{Lemma}
Let $s\in \SCC_\tau$. Then: \begin{enumerate}
\item $\KC_s^2=-(1+\sigma_s^2)\KC_s$,
\item $\tKC_s^2=(\sigma_s^2-1)\tKC_s+\sigma_s^2$.
\end{enumerate}
\end{Lemma}
\begin{proof}
By Lemma~\ref{lemReeder 4.3}, \eqref{eqDef_Ks} and \cite[Lemma 4.3]{hebert2018principal}, we have \[\begin{aligned}\KC_s^2 &=(F_s-\zeta_s)(F_s-\zeta_s)\\
&=F_s^2-F_s(\zeta_s+{^s\zeta_s})+\zeta_s.{^s\zeta_s}\\
&=\zeta_s.({^s\zeta_s})-F_s(\zeta_s+{^s\zeta_s})+\zeta_s.{{^s\zeta_s}}\\
&=(-F_s+\zeta_s)(\zeta_s+{^s\zeta_s})=-(1+\sigma_s^2)\KC_s.\end{aligned}\]
\end{proof}
\begin{Lemma}\label{lemBernstein_lusztig_relations}
Let $\theta\in \C(Y)$ and $s\in \SCC_\tau$. Then \[\theta*\tKC_s=\tKC_s*{^s\theta}+\tilde{\Omega}_s(\theta).\]
\end{Lemma}
\begin{proof}
By Lemma~\ref{lemReeder 4.3}, we have \[\begin{aligned}\theta*\tKC_s&=\theta*\left(F_s+Q_s^T\right)\\
&=F_s* {^s\theta}+\theta*Q_s^T \\
&=\left(F_s+Q_s^T\right){^s\theta}+Q_s^T(\theta-{^s\theta})\\
&= \tKC_s *{^s\theta} +\tilde{\Omega}_s(\theta).\end{aligned}\]
\end{proof}
Following \cite{deodhar1989subgroups}, we define $\leq_\tau$ on $\Phi_{(\tau),+}^\vee$ as follows. If $\alpha^\vee,\beta^\vee\in \Phi^\vee_{(\tau),+}$, we write $\alpha^\vee\leq_{\tau} \beta^\vee$ if there exist $k\in \N$, $a\in \R^*_+$, $\beta_1^\vee,\ldots,\beta_k^\vee\in \Phi_{(\tau),+}^\vee$ and $a_1,\ldots,a_k\in \R_+$ such that $\alpha^\vee= a\beta^\vee+\sum_{i=1}^k a_i \beta_i^\vee$.
Let $s\in \Wta\cap \RCC$. Then by \cite[Lemma 5.13 (2)]{hebert2018principal}, we have \begin{equation}\label{eqCharacterization_S_tau}s\in \SCC_\tau\text { if and only if }s.\big(\Phi_{(\tau),+}^\vee\setminus \{\alpha_s^\vee\}\big)=\Phi_{(\tau),+}^\vee\setminus\{\alpha_s^\vee\}.\end{equation}
\begin{Lemma}\label{lemBasis_root_subsystem}
Let \[\Sigma_\tau=\{\alpha^\vee\in \Phi_{(\tau),+}^\vee|\forall\beta^\vee\in \Phi_{(\tau),+}^\vee, \beta^\vee\leq_\tau \alpha^\vee\Rightarrow \beta^\vee=\alpha^\vee\}.\] Then the map $s\mapsto \alpha_s^\vee$ from $\SCC_\tau$ to $\Sigma_\tau$ is well defined and is a bijection. Moreover, we have \begin{equation}\label{eqSigma_spans_Phi}\Phi^\vee_{(\tau),+}\subset \sum_{\alpha^\vee\in \Sigma_\tau}\N \alpha^\vee.\end{equation}
\end{Lemma}
\begin{proof}
This follows \cite[3. Step 2]{deodhar1989subgroups}.
Let $\alpha^\vee\in \Sigma_\tau$ (if such an element exists). Let $s=r_{\alpha^\vee}\in \RCC_{(\tau)}$ and $\beta^\vee\in \Phi^\vee_{(\tau),+}$. We assume that $\gamma^\vee:=-s.\beta^\vee\in \Phi^\vee_{(\tau),+}$. Then $\gamma^\vee=-\beta^\vee+\alpha_s(\beta^\vee)\alpha_s^\vee$. As $\beta^\vee\in \Phi^\vee_+$, we necessarily have $\alpha_s(\beta^\vee)\in \N^*$. Therefore $\alpha_s^\vee=\frac{1}{\alpha_s(\beta^\vee)}(\beta^\vee+\gamma^\vee)$ and $\beta^\vee\leq_\tau \alpha_s^\vee$. By definition of $\Sigma_\tau$, we deduce $\beta=\alpha_s^\vee$. By \eqref{eqCharacterization_S_tau} we deduce that $s\in \SCC_\tau$.
Conversely, let $s\in \SCC_{\tau}$. Let $\beta^\vee\in \Phi^\vee_{(\tau),+}$ be such that $\beta^\vee\leq_\tau \alpha_s^\vee$. By definition, we can write $\alpha_s^\vee = a\beta^\vee+\sum_{i=1}^k a_i \beta_i^\vee$, where $k\in \N$, $a\in \R^*_+$, $(a_i)\in (\R_+)^k$ and $(\beta_i^\vee)\in \big(\Phi_{(\tau),+}\big)^k$. Then \[s.\beta^\vee=-\frac{1}{a}(\alpha_s^\vee+\sum_{i=1}^k a_is.\beta_i^\vee)\in \Phi^\vee_-\cap\Phi^\vee_{(\tau)}.\] By \eqref{eqCharacterization_S_tau}, we deduce that $\beta^\vee=\alpha_s^\vee$ and thus $\alpha_s^\vee\in \Sigma_\tau$. We proved the first part of the lemma. Then \eqref{eqSigma_spans_Phi} can be proved as \cite[3 step 3]{deodhar1989subgroups}, using the fact that $\alpha(\beta^\vee)\in \Z$, for all $\alpha^\vee,\beta^\vee\in \Phi_{(\tau)}^\vee$.
\end{proof}
\begin{Lemma}\label{lemKac-Moody_matrix}
The matrix $\big(\alpha_s(\alpha_r^\vee)\big)_{r,s\in \SCC_\tau}$ is a Kac-Moody matrix.
\end{Lemma}
\begin{proof}
Let $r,s\in\SCC_\tau$. Suppose $\alpha_r(\alpha_s^\vee)\neq 0$. We have $r.\alpha_s^\vee=\alpha_s^\vee-\alpha_r(\alpha_s^\vee)\alpha_r^\vee$. By Lemma~\ref{lemBasis_root_subsystem}, $r.\alpha_s^\vee \not \leq_\tau \alpha_r^\vee$, which implies that $\alpha_r(\alpha_s^\vee)\leq 0$.
Suppose $\alpha_r(\alpha_s^\vee)=0$. Then by the case above, $\alpha_s(\alpha_r^\vee)\leq 0$. Suppose that $\alpha_s(\alpha_r^\vee)<0$. Then \[\alpha_r^\vee=-rs.\alpha_r^\vee-\alpha_s(\alpha_r^\vee)\alpha_s^\vee\text{ and }\alpha_s^\vee=\frac{-1}{\alpha_s(\alpha_r^\vee)}(rs.\alpha_r^\vee+\alpha_r^\vee).\] If $rs.\alpha_r^\vee\in \Phi^\vee_+$, we deduce that $\alpha_s^\vee\leq_{\tau} \alpha_r^\vee$ and if $rs.\alpha_r^\vee$, we deduce that $\alpha_r^\vee\leq_\tau \alpha_s^\vee$. Since $\alpha_r^\vee,\alpha_s^\vee\in \Sigma_\tau$, this implies in both cases that $\alpha_r^\vee=\alpha_s^\vee$: we reach a contradiction. Therefore $\alpha_s(\alpha_r^\vee)=0$ and the lemma follows.
\end{proof}
If $a,b$ are two elements of a ring and $m\in \N$, we denote by $(a.b)^{*m}$ the product $a.b.a.\ldots$ with $m$ factors.
\begin{Lemma}
Let $w\in \Wta$ and $w=r_1\ldots r_k= s_1\ldots s_k$ be two reduced writings of $w$, with $k=\ell_\tau(w)$ and $r_1,\ldots, r_k, s_1,\ldots,s_k\in \SCC_\tau$. Then \[\tKC_{r_1}\ldots \tKC_{r_k}=\tKC_{s_1}\ldots \tKC_{s_k}.\]
\end{Lemma}
\begin{proof}
Let $r,s\in \SCC_\tau$. Suppose that the order $m$ of $rs$ is finite. Let $h=(\tKC_r *\tKC_s)^{*m}-(\tKC_s*\tKC_r)^{*m}\in \ATC$. We want to prove that $h=0$. Set $w_0(r,s)=(rs)^{*m}=(sr)^{*m}$. Using Lemma~\ref{lemReeder 4.3}, \eqref{eqDefinition_tKC} and \cite[Lemma 4.3]{hebert2018principal}, there exists a family \[(P_u)_{u\in \langle r,s\rangle}\in \left(\Z[\sigma_r,\sigma_s][x_{s,u},x_{r,u'}|u,u'\in \langle r,s\rangle]\right)^{\langle r,s\rangle}\] such that \begin{equation}
h=\sum_{u\in \langle r,s\rangle} F_u*P_u\left(\left({^v Q^T_s},^w Q^T_r \right)_{v,w\in \langle r,s\rangle}\right),
\end{equation}
where the $x_{s,u},x_{r,u}$ are indeterminates.
Let $A_{r,s}=\begin{pmatrix}
2 & \alpha_s(\alpha_r^\vee)\\ \alpha_r(\alpha_s^\vee) & 2
\end{pmatrix}$. Then $A_{r,s}$ is a Kac-Moody matrix by Lemma~\ref{lemKac-Moody_matrix} (it is actually a Cartan matrix since $\langle r,s\rangle$ is finite). Let $X',Y',\alpha_r',\alpha_s',\alpha_r^\vee,\alpha_s^\vee$ be copies of $X,Y,\alpha_r,\alpha_s,\alpha_r^\vee,\alpha_s^\vee$. Then $\SC'=(A_{r,s},X',Y',(\alpha_r',\alpha_s'),(\alpha_r'^\vee,\alpha_s'^\vee)\big)$ is a root generating system.
Let $W'$ be the associated Weyl group. We add a prime to all the objects previously defined when they refer to the root generating system $\SC'$. Let $\tau'\in T_\C'$ be defined by $\tau'(\lambda)=1$ for all $\lambda\in Y'$. Then we have $\SCC'_{\tau'}=\{r,s\}\subset W'$. By \eqref{eqK)T_if_tau=1}, $\tKC_r'=T_r'\in\ATC'$ and $\tKC_s'=T_s'\in \ATC'$. Moreover $(T_r'*T_s')^{*m} -(T_s'*T_r')^{*m}=T_{w_0'}'-T_{w_0'}'=0$, where $w_0'=(rs)^{*m}=(sr)^{*m}\in W'$ is the element of maximal length of $W'$. Therefore if $u\in W'$, we have
\[P_u\big(({^v{Q_r'^T}},{^w Q_s'^T)_{v,w\in W'}}\big)=0.\]
Let $Q'^\vee=\Z\alpha_r'^\vee\oplus \Z\alpha_s'^\vee$ and $\iota:Q'^\vee\rightarrow Q^\vee$ be the $\Z$-modules morphism defined by $\iota(\alpha_r'^\vee)=\alpha_r^\vee$ and $\iota(\alpha_s'^\vee)=\alpha_s^\vee$. Extend $\iota$ to a $\C$-algebra morphism $\iota:\C[Q'^\vee]\rightarrow \C[Q^\vee]$. Then $\iota$ is compatible with the actions of $W'$ and $\langle r,s\rangle$. More precisely, if $\iota:W'\rightarrow \langle r,s\rangle$ is the group morphism defined by $\iota(r)=r$, $\iota(s)=s$, we have $\iota\left({^w Q'^T_{t}}\right)={^{\iota(w)} Q^T_{t}},$ for $w\in W'$ and $t\in \{r,s\}$.
Therefore \[P_u\left(\left({\iota\left(^vQ'^T_r\right)}, \iota\left({^w Q'^T_s}\right)\right)_{v,w\in W'}\right)=P_u\left(\left({^vQ^T_r},{^w Q^T_s}\right)_{v,w\in \langle r,s\rangle}\right)=0,\]
for $u\in \langle r,s\rangle$ which proves that $h=0$ and hence $(\tKC_r*\tKC_s)^{*m}=(\tKC_s*\tKC_r)^{*m}$. By the word property (\cite[Theorem 3.3.1]{bjorner2005combinatorics}), we deduce the lemma.
\end{proof}
To conclude the proof of Proposition~\ref{propBL_relations}, it remains to prove (5). By (2) and (3), $ \KCC_\tau=\sum_{k\in \N, s_1,\ldots,s_k\in \SCC_\tau}\tKC_{s_1}*\ldots*\tKC_{s_k}*\C(Y)_\tau$ is a subalgebra of $\ATC$. Let $w\in \Wta$. Let $w=s_1\ldots s_k$ be a reduced writing of $w$, with $k\in \N$ and $s_1,\ldots,s_k\in \SCC_\tau$. Then by Lemma~\ref{lemReeder 4.3}, \begin{equation}\label{eqTriangular_writing_F}
\tKC_{s_1}*\ldots*\tKC_{s_k}=(F_{s_1}+Q_{s_1}^T)\ldots (F_{s_k}+Q_{s_k}^T)= F_{w}+\sum_{v<_\tau w} F_v*\theta_v,
\end{equation}for some $\theta_v\in \C(Y)$. For $h\in \ATC$, $h=\sum_{v\in W^v} T_v*\tilde{\theta}_v$, write \[\max \supp(h)=\{w\in W^v| \tilde{\theta}_w\neq 0\text{ and }\forall w'\in W^v, w'>w, \tilde{\theta}_{w'}=0\}.\] Then by \cite[Lemma 5.24]{hebert2018principal} and \eqref{eqTriangular_writing_F}, $\max \supp(\tKC_w)=\{w\}$, for $w\in \Wta$ and thus $(\tKC_{w})_{w\in \Wta}$ is a free family of $\ATC$.
By \eqref{eqDefinition_tKC} and (3), $\sum_{k\in \N,s_1,\ldots,s_k\in \SCC_\tau} \KC_{s_1}*\ldots*\KC_{s_k}*\C(Y)_\tau\subset \bigoplus_{w\in \Wta} \tKC_w*\C(Y)_\tau$. The converse inclusion is obtained similarly and thus $\KCC_\tau=\sum_{k\in \N,s_1,\ldots,s_k\in \SCC_\tau} \KC_{s_1}*\ldots*\KC_{s_k}*\C(Y)_\tau$.
By \cite[Lemma 5.23]{hebert2018principal}, $\ATC_\tau*\KCC_\tau\subset \ATC_\tau$ for $s\in \SCC_\tau$. Therefore $\tKC_w\in \ATC_\tau$ for $w\in \Wta$ and $\KCC_\tau\subset \ATC_\tau$. This completes the proof of Proposition~\ref{propBL_relations}. $\square$
\begin{Remark}
In \cite[5.5]{hebert2018principal}, we assumed that $\tau\in \UC_\C$ and that $W_\tau=\Wta$. The assumption $W_\tau=\Wta$ is actually useless (without any change in the proofs). The assumption $\tau\in \UC_\C$ is however used, to ensure that $\max \supp(\tKC_{w})=\{w\}$ (see \cite[Lemma 5.24]{hebert2018principal}), for $w\in \Wta$ and thus to prove the freeness of $(\tKC_w)_{w\in \Wta}$.
\end{Remark}
\subsection{Action of $\KCC_\tau$ on $\Itg$}\label{subAction_KC_Itau}
We now fix $\tau\in \UC_\C$. Let $\KCC_\tau=\bigoplus_{w\in \Wta} \tKC_{w} *\C(Y)_\tau\subset \ATC_\tau$. In this subsection, we extend the action of $\KCC_\tau\cap \AC_\C$ on $\Itg$ to an action of $\KCC_\tau$ on $\Itg$ (see Lemma~\ref{lemAction_K_Itg}).
Let $\JC_\tau=\{\theta\in \C(Y)_\tau|\tau(\theta)\neq 0\}$\index{$\JC_\tau$}.
\begin{Lemma}\label{lemLeft_multiplication_CY_AC}
Let $k\in \KCC_\tau$. Then there exists $\theta \in \C[Y]\setminus \JC_\tau$ such that $\theta*k\in \AC_\C$.
\end{Lemma}
\begin{proof}
Let $w\in \Wta$. We assume that for all $v\in \Wta$ such that $v<_\tau w$, for all $\tilde{\theta}\in \C(Y)_\tau$, there exists $\theta\in \C[Y]\setminus \JC_\tau$ such that $\theta*\tKC_v*\tilde{\theta}\in \AC_\C$. Let $\tilde{\theta}\in \C(Y)_\tau$. Let $s\in \SCC_\tau$ be such that $v:=sw<_\tau w$. Let $\theta_1\in \C[Y]\setminus \JC_\tau$ be such that $\theta_1*\tKC_v*\tilde{\theta}\in \AC_\C$. Let $\theta_2\in \C[Y]\setminus \JC_\tau$ be such that $\tKC_s*\theta\in \AC_\C$, which exists by Proposition~\ref{propBL_relations}\eqref{itAlgebra_K}. Set $\theta_3=(\theta_1*\theta_2)*{^s( \theta_1*\theta_2)}\in \C[Y]\setminus \JC_\tau$. Then ${^s\theta_3}=\theta_3$, thus $\tilde{\Omega}_s(\theta_3)=0$ and hence $\theta_3*\tKC_s=\tKC_s*\theta_3$. Therefore \[\theta_3* \tKC_{w}*\theta=\tKC_s*\theta_3*\tKC_v*\theta=\tKC_s *\theta_2 *{^s (\theta_1*\theta_2)}*\theta_1*\tKC_v\in \AC_\C.\] We deduce that for all $w\in \Wta$ and $\tilde{\theta}\in \C(Y)_\tau$, there exists $\theta\in \C[Y]\setminus \JC_\tau$ such that $\theta*\tKC_w*\tilde{\theta}\in \AC_\C$.
Let now $k\in\KCC_\tau$. Write $k=\sum_{w\in \Wta} \tKC_w *\tilde{\theta}_w$, with $(\tilde{\theta}_w)\in (\C(Y)_\tau)^{(\Wta)}$. For $w\in \Wta$, choose $\theta_w\in \C[Y]\setminus \JC_\tau$ such that $\theta_w*\tKC_w*\tilde{\theta}_w\in \AC_\C$. Set $\theta=\prod_{w\in \Wta|\tilde{\theta}_w\neq 0} \theta_w\in \C[Y]\setminus \JC_\tau$. Then $\theta*k\in \AC_\C$, which proves the lemma.
\end{proof}
\begin{Lemma}\label{lemInjectivity_multiplication_theta}
Let $\theta\in \C[Y]\setminus\JC_\tau$. Then the map $m_\theta:\Itg\rightarrow \Itg$ defined by $m_\theta(x)=\theta.x$, for $x\in \Itg$, is injective.
\end{Lemma}
\begin{proof}
Let $x\in I_\tau$. Write $x=\sum_{w\in W^v} a_w T_w.\vb_\tau$, where $(a_w)\in \C^{(W^v)}$. Let $w\in W^v$ be such that $a_w\neq 0$ and such that for all $v\in W^v$ such that $a_v\neq 0$, we have $v\not \geq w$. Then by \cite[Lemma 2.8]{hebert2018principal}, we have $\theta.x-a_w T_w*{^{w^{-1}}\theta}.\vb_\tau \in \bigoplus_{v\in W^v, v\not \geq w}\C T_v.\vb_\tau$. By \cite[Lemma 5.24]{hebert2018principal}, $w\in \Wta\subset W_\tau$. Therefore ${^{w^{-1}}\theta}.\vb_\tau=\tau(\theta).\vb_\tau\neq 0$, which proves that $\theta.x\neq 0$. Therefore $m_\theta$ is injective.
\end{proof}
\begin{Lemma}\label{lemAction_K_Itg}
We have the following properties:\begin{enumerate}
\item $\Itg=(\KCC_\tau\cap \AC_\C).\vb_\tau$,
\item $\KCC_\tau\cap \AC_\C$ stabilizes $\Itg$ and the action of $\KCC_\tau\cap \AC_\C$ on $\Itg$ extends uniquely to an action of $\KCC_\tau$ on $\Itg$. This action is as follows. Let $x\in \Itg$ and $k\in \KCC_\tau$. Write $x=h.x$, with $h\in \AC_\C\cap \KCC_\tau$. Write $k*h=\sum_{w\in \Wta} \tKC_w \tilde{\theta}_w$, with $(\theta_w)\in (\C(Y)_\tau)^{(\Wta)}$. Then: \begin{equation}\label{eqAction_K_Itg}
k.x=k*h.\vb_\tau=\sum_{w\in \Wta} \tau(\tilde{\theta})\tKC_w (\tau).\vb_\tau.
\end{equation}
\end{enumerate}
\end{Lemma}
\begin{proof}
By \cite[Theorem 5.27]{hebert2018principal}, we have $\Itg=\ev_\tau(\KCC_\tau).\vb_\tau$.
For $w\in \Wta$, write $\tKC_{w}=\sum_{v\in W^v} T_w *\theta_{v,w}$, with $\theta_{v,w}\in \C(Y)$ for $v\in W^v$ and $\{v\in W^v|\theta_{v,w}\neq 0\}$ finite. By \cite[Lemma 5.23]{hebert2018principal}, we can write $\theta_{v,w}=\frac{f_{v,w}}{g_{v,w}}$, where $f_{v,w},g_{v,w}\in \C[Y]$ and $\tau(g_{v,w})\neq 0$ for $v\in W^v$ such that $f_{v,w}\neq 0$. For $w\in \Wta$, set $g_w=\prod_{v\in W^v|f_{v,w}\neq 0} g_{v,w}\in \C[Y]$. Then \begin{equation}\label{eqPolynomial_basis}
\ev_\tau(\tKC_w)=\ev_\tau(\frac{1}{\tau(g_w)}\tKC_w g_w)\text{ and }\frac{1}{\tau(g_w)}\tKC_w g_w\in \KCC_\tau\cap \AC_\C.
\end{equation} Therefore \begin{equation}
\Itg=(\KCC_\tau\cap \AC_\C).\vb_\tau,
\end{equation}
which proves (1).
(2) By Proposition~\ref{propBL_relations}\eqref{itAlgebra_K}, $\KCC_\tau\cap \AC_\C$ is a subalgebra of $\AC_\C$, which proves that $\KCC_\tau\cap \AC_\C$ stabilizes $\Itg$. For $w\in \Wta$, set $L_w=\KC_w*g_w\in \AC_\C\cap \KCC_\tau$ and set \[\LCC_\tau=\bigoplus_{w\in \Wta} L_w*\C[Y]\subset \AC_\C\cap \KCC_\tau.\]
Let $\tilde{\tau}:\C(Y)_\tau\rightarrow \C$ be the algebra morphism extending $\tau:\C[Y]\rightarrow \C$. Let $\tilde{M}=\mathrm{Ind}_{\C(Y)_\tau}^{\KCC_\tau}(\tilde{\tau})$ be the representation of $\KCC_\tau$ obtained by inducing $\tilde{\tau}$ to $\KCC_\tau$. Then \[\tilde{M}=\bigoplus_{w\in \Wta} \tKC_w.\vbt,\] where $\vbt\in \tilde{M}$ is such that $\theta.\vbt=\tilde{\tau}(\theta).\vbt$, for $\theta\in \C(Y)_\tau$. Let $\psi:\tilde{M}\rightarrow \Itg$ be defined by $\psi(k.\vbt)=k.\vb_\tau$, for $k\in \LCC_\tau$. Let us prove that $\psi$ is well defined. Let $k\in \LCC_\tau$ be such that $k.\vbt=0$. We can write $k=\sum_{w\in W^v} L_w*\theta_w$, for some $(\theta_w)\in \C[Y]^{(\Wta)}$. Then $ \tau(\theta_w)=0$ for all $w\in \Wta$ and thus $k.\vb_\tau=0$, which proves that $\psi:\LCC_\tau.\vbt\rightarrow \Itg$ is well defined. Moreover, $\tilde{M}=\bigoplus_{w\in \Wta}\C \tKC_w.\vbt=\bigoplus_{w\in \Wta} \C L_w.\vbt=\LCC_\tau.\vbt$. Thus $\psi:\tilde{M}\rightarrow \Itg$ is well defined. Then $\psi$ is an isomorphism of $\LCC_\tau$-modules. We equip $\Itg$ with the structure of a $\KCC_\tau$-module by setting $k\odot x=\psi\big(k.\psi^{-1}(x)\big)$ for $k\in \KCC_\tau$ and $x\in \Itg$.
Let $w\in \Wta$ and $\theta\in \C(Y)_\tau$. Then \begin{equation}\label{eqtKC_theta_vtau}
(\tKC_w*\theta)\odot \vb_\tau=\psi(\tKC_w*\theta.\vbt)=\tau(\theta)\psi(\tKC_w.\vbt)=\tau(\theta)\tKC_w\odot\vb_\tau.
\end{equation} By applying this to $\theta=g_w$, we deduce that \begin{equation}\label{eqtKC_vtau}
\tKC_w\odot\vb_\tau=\frac{1}{\tau(g_w)}\tKC_w*g_w\odot\vb_\tau=\frac{1}{\tau(g_w)}\tKC_w*g_w.\vb_\tau=\frac{1}{\tau(g_w)}\ev_\tau(\tKC_w*g_w).\vb_\tau=\tKC_w(\tau).\vb_\tau.
\end{equation} Combining \eqref{eqtKC_theta_vtau} and \eqref{eqtKC_vtau} yields \eqref{eqAction_K_Itg}.
Let $k\in \AC_\C\cap \KCC_\tau$. Write $k=\sum_{w\in \Wta} \tKC_w*\theta_w$, with $(\theta_w)\in (\C(Y)_\tau)^{(\Wta)}$. For $w\in \Wta$, write $\tKC_w=\sum_{v\in W^v}T_v*\theta_{v,w}$, with $(\theta_{v,w})_{v\in W^v}\in (\C(Y)_\tau)^{(W^v)}$. Then $k=\sum_{v\in W^v}\sum_{w\in \Wta} T_v*\theta_{v,w}*\theta_w$ and $\sum_{w\in\Wta} \theta_{v,w}*\theta_{w}\in \C[Y]$ for every $v\in W^v$. Then \[\begin{aligned} k.\vb_\tau &=\sum_{v\in W^v} \tau(\sum_{w\in \Wta} \theta_{v,w}*\theta_w) T_v.\vb_\tau \\
&=\sum_{v\in W^v} \sum_{w\in \Wta}\tau(\theta_{v,w}*\theta_w) T_v.\vb_\tau\\
&= \sum_{w\in \Wta} \tau(\theta_w)\ev_\tau(\sum_{v\in W^v} T_v*\theta_{v,w}).\vb_\tau\\
&= \sum_{w\in \Wta} \tau(\theta_w) \ev_\tau(\tKC_w).\vb_\tau =k\odot \vb_\tau,\end{aligned}\] by \eqref{eqAction_K_Itg}. Therefore $\odot$ extends the action $.$ of $\AC_\C\cap \KCC_\tau$ on $\Itg$ to an action of $\KCC_\tau$ on $\Itg$.
We now prove the uniqueness of such an action. Let $\boxdot$ be an action of $\KCC_\tau$ on $\Itg$ such that $k\boxdot x=k.x$ for all $k\in \AC_\C\cap \KCC_\tau$ and $x\in \Itg$. Let $k\in \KCC_\tau$. Let $\theta\in \C[Y]\setminus\JC_\tau$ be such that $\theta*k\in \AC_\C$, which exists by Lemma~\ref{lemLeft_multiplication_CY_AC}. Then \[(\theta*k).x=(\theta*k)\boxdot x=\theta\boxdot (k\boxdot\vb_\tau)=\theta.(k\boxdot\vb_\tau).\] By Lemma~\ref{lemInjectivity_multiplication_theta} we deduce that $k\boxdot x=m_\theta^{-1}(\theta*k.x)$. This proves that $\boxdot=\odot$ and concludes the proof of this lemma.
\end{proof}
\subsection{Examples of $\tau$ for which $\SCC_\tau$ is infinite}\label{subExample_infinite_basis}
The set $\Phi^\vee$ is a root system in the sense of \cite[4 Remark]{moody1989infinite}. Then $\Phi_{(\tau)}^\vee$ is a subroot system of $\Phi^\vee$ in the sense of \cite[6]{moody1989infinite}. It is proved in \cite[Example 1]{moody1989infinite} that the bases of a subroot system of a root system admitting a finite basis need not be finite. However, we do not know if the root system given in this example is of the form $\Phi^\vee_{(\tau)}$, for some $\tau\in T_\C$. Suppose that $\sigma_s=\sigma_s'$ for all $s\in\SCC$, then $\Phi^\vee_{(\tau)}=\{\alpha^\vee\in \Phi^\vee|\tau(\alpha^\vee)=1\}$. Thus if $\tau\in T_\C$, $\Phi^\vee_{(\tau)}$ is \textbf{closed} in the sense that for all $\alpha^\vee,\beta^\vee\in \Phi^\vee_{(\tau)}$, if $\alpha^\vee+\beta^\vee\in \Phi^\vee$, then $\alpha^\vee+\beta^\vee\in \Phi_{(\tau)}^\vee$. We may ask whether this property ensures the finiteness of $\SCC_\tau$. We prove below that this is not the case.
For $w\in W^v$, set $N_{\Phi^\vee}(w)=\{\alpha^\vee\in \Phi^\vee_+|w.\alpha^\vee\in \Phi^\vee_-\}$\index{$N_{\Phi^\vee}(w)$}.
\begin{Lemma}\label{lemExample_infinite_SCC_tau}
Let $A=(a_{i,j})_{i,j\in \llbracket 1,4\rrbracket}$ be an invertible Kac-Moody matrix such that for all $i,j\in \llbracket 1,4\rrbracket$, $a_{i,j}\leq -2$, for all $(i,j)\in \llbracket 1,4\rrbracket\setminus \{(4,3)\}$, $a_{i,j}$ is even and $a_{4,3}$ is odd. We assume that $\A=\bigoplus_{i=1}^4 \R\alpha_i^\vee$, which is possible since $A$ is invertible by \cite[1.1]{kac1994infinite}. Let $\htt:\A\rightarrow \R$ be defined by $\htt(\sum_{i=1}^4 n_i\alpha_i^\vee)=\sum_{i=1}^4 n_i$, for $(n_i)\in \R^4$. Let $q$ be a prime power and set $\sigma_i=\sigma_i'=\sqrt{q}$ for $i\in \llbracket 1,4\rrbracket$. Let $\tau: Y\rightarrow \{-1,1\}$ be defined by $\tau(\lambda)=(-1)^{\htt(\lambda)}$ for $\lambda\in Y$. Let $W'=\langle
r_1,r_2\rangle \subset W^v$. Then $\{w r_3r_4r_3 w^{-1}|w\in W'\}\subset \SCC_\tau$. In particular, $\SCC_\tau$ is infinite.
\end{Lemma}
\begin{proof}
Let $w\in W'$ and $v=wr_3r_4r_3w^{-1}$. Then by \cite[1.3.14 Lemma]{kumar2002kac}, $wr_3.\alpha_4^\vee\in N_{\Phi^\vee}(v)$ and $N_{\Phi^\vee}(v)\setminus\{wr_3.\alpha_4^\vee\}\subset \{u.\alpha_i^\vee|u\in W^v, i\in \llbracket 1,3\rrbracket\}$. We have $r_3.\alpha_4^\vee=\alpha_4^\vee-a_{4,3}\alpha_3^\vee$. Moreover if $i\in \llbracket 1,2\rrbracket$, $r_i(\alpha_4^\vee-a_{4,3}\alpha_3^\vee+2Q^\vee)\subset \alpha_4^\vee-a_{4,3}\alpha_3^\vee+2Q^\vee$. In particular, $wr_3.\alpha_4^\vee\in \alpha_4^\vee-a_{4,3}\alpha_3^\vee+2Q^\vee$ and $\tau(wr_3.\alpha_4^\vee)=1$. Now if $i\in \llbracket 1,3\rrbracket$ and $j\in \llbracket 1,4\rrbracket$, $r_j.(\alpha_i^\vee+2Q^\vee)\subset \alpha_i^\vee+2Q^\vee$. Therefore $\tau(u.\alpha_i^\vee)=-1$ if $u\in W^v$. In particular, $N_{\Phi^\vee}(v)\cap \Phi_{(\tau)}=\{wr_3.\alpha_4^\vee\}=\{\alpha_v^\vee\}$ and by \cite[Lemma 5.13 (2)]{hebert2018principal}, $v\in \SCC_\tau$. Using \cite[1.3.21 Proposition]{kumar2002kac} we deduce that $\SCC_\tau$ is infinite.
\end{proof}
\section{Kato's irreducibility criterion}\label{secKato_s_irreducibility_criterion}
Let $\tau\in \UC_\C$. Let $x\in \Itg$. Write $x=\sum_{w\in \Wta} a_w \tKC_{w}.\vb_\tau$, with $(a_w)\in \C^{(W^v)}$. Let $\supp(x)=\{w\in \Wta|a_w\neq 0\}$ and $\ell_\tau(x)=\max\{\ell_\tau(w)|w\in S\}$. We set $\ell_\tau(0)=-\infty$.
Let $x\in \Itg$. We define $\ord_\tau(x)$ as the minimum of the $k\in \N$ such that for all $\theta_1,\ldots,\theta_k\in \JC_\tau$, we have $\theta_1\ldots\theta_k.x=0$. We will see in Lemma~\ref{lemLength_decreasing_multiplication_theta} that $\ord_\tau(x)\in \N$ for $x\in \Itg$. The aim of this subsection is to prove that $I_\tau(\tau,\Wta):=\Itg\cap I_\tau(\tau)=\C\vb_\tau$ (see Theorem~\ref{thmWeight_space}). We actually prove that for all $x\in \Itg$, $\ord_\tau(x)=\ell_\tau(x)+1$. We then deduce Kato's irreducibility criterion (see Corollary~\ref{corKatos_irreducibility_criterion}).
For $s\in \SCC_\tau$, we set \[\sigma_{s,\tau}''=\frac{1}{2}\big((\sigma_s^2-1)+\sigma_s(\sigma'_s-\sigma_s'^{-1})\tau(\alpha_s^\vee)\big).\]\index{$\sigma_{s,\tau}''$} When $\sigma_s=\sigma_s'$, then $\tau(\alpha_s^\vee)=1$ and $\sigma_{s,\tau}''=\sigma_s^2-1$.
As we assumed that $|\sigma_s|,|\sigma_s'|>1$, for all $s\in \SCC$, we have $\sigma_{s,\tau}''\neq 0$ for every $s\in \SCC_\tau$. Indeed, we have $\sigma_{s,\tau}''=0$ if and only if $\sigma_s=\sigma'_s=\pm 1$ or $\sigma'_s\neq \sigma_s$ and $\sigma_s'=-\sigma_s^{-1}$. We have $\SCC_\tau\subset \RCC_{(\tau)}$ and thus $\tau(\alpha_s^\vee)\in \{-1,1\}$. If $\tau(\alpha_s^\vee)=-1$, then $\sigma_s\neq \sigma_s'$ and $\sigma_s^2-1+\sigma_s(\sigma_s'-\sigma_s'^{-1})\tau(-\alpha_s^\vee)=\sigma_s^2-1-\sigma_s(\sigma_s'-\sigma_s'^{-1})=0$ if and only if $\sigma_s'\in \{\sigma_s,-\sigma_s^{-1}\}$ if and only if $\sigma_s'=-\sigma_s^{-1}$.
\begin{Lemma}\label{lemFormula_omega_tilde_vtau}
Let $s\in \SCC_\tau$ and $\lambda\in Y$. Then \begin{equation}\label{eqFormula_tau_Omega_s}
\tau\big(\tilde{\Omega}_s(Z^\lambda)\big)=\tau(\lambda)\sigma_{s,\tau}''\alpha_s(\lambda).
\end{equation}
\end{Lemma}
\begin{proof}
We have \[\tilde{\Omega}_s(Z^\lambda)=\big((\sigma_s^2-1)+\sigma_s(\sigma_s'-\sigma_s'^{-1})Z^{-\alpha_s^\vee}\big) \frac{1-Z^{-\alpha_s(\lambda)\alpha_s^\vee}}{1-Z^{-2\alpha_s^\vee}}Z^{\lambda}.\] We have $\tau(\alpha_s^\vee)\in \{-1,1\}$. If $\alpha_s(\lambda)$ is odd, then $\sigma_s=\sigma_s'$ and $\tau(\alpha_s^\vee)=1$ (since the denominator of $Q_s^T$ is then $1-Z^{-\alpha_s^\vee}$). We then conclude with Remark~\ref{remIH algebre dans le cas KM deploye}~(\ref{itPolynomiality_Bernstein_Lusztig}). If $\alpha_s(\lambda)$ is even, a computation similar to the one of Remark~\ref{remIH algebre dans le cas KM deploye}~(\ref{itPolynomiality_Bernstein_Lusztig}) enables to prove~\eqref{eqFormula_tau_Omega_s}.
\end{proof}
For $h\in \KCC_\tau$, $h=\sum_{w\in \Wta} \tKC_w*\theta_w$, with $(\theta_w)\in \C^{(\Wta)}$ we set $\ell_\tau(h)=\max\{\ell_\tau(w)|\theta_w\neq 0\}$.
\begin{Lemma}\label{lemComputation_terms ordern-1_commutation}
Let $w\in \Wta$. Fix a reduced writing $w=s_1\ldots s_k$ of $w$, with $k=\ell_\tau(w)$ and $s_1,\ldots,s_k\in \SCC_\tau$. For $i\in \llbracket 1,k\rrbracket$, set $v_i=s_1\ldots s_{i-1}$, $\tilde{v}_i=s_{i+1}\ldots s_k$. Let $E=\{i\in \llbracket 1,k\rrbracket|\ell_\tau(v_i\tilde{v}_i)=k-1\}$. Let $\theta\in \C(Y)$. Then there exists $h\in \KCC_\tau$ such that $\ell_\tau(h)\leq k-2$ and
\begin{equation}\label{eqBL_commutation_order_k-1_ATC}
\theta*\tKC_w=\tKC_w*{{^{w^{-1}}\theta}}+\sum_{i\in E}\tKC_{v_i\tilde{v_i}} *{^{\tilde{v}_i^{-1}}\tilde{\Omega}_{s_i}({^{v_i^{-1}}\theta}}) +h.
\end{equation} In particular if $\theta\in \C(Y)_\tau$ we have
\begin{equation}\label{eqBL_commutation_order_k-1_Itau}
\theta.\tKC_w.\vb_\tau=\tau(\theta) \tKC_w.\vb_\tau+\tau(\lambda)\sum_{i\in E} \sigma_{s_i,\tau}''\alpha_{s_i}(v_i^{-1}.\theta) \tKC_{v_i\tilde{v}_i}.\vb_\tau+h.\vb_\tau.
\end{equation}
\end{Lemma}
\begin{proof}
We prove it by induction on $k$. If $k=0$, this is clear. We assume $k\geq 1$. Set $w'=ws_k$ and $\tilde{v}_i'=v_is_k$ for $i\in \llbracket 1,k-1\rrbracket$. Assume that we can write \[\theta*\tKC_{w'}=\tKC_{w'}*{^{w'^{-1}}\theta}+\sum_{i=1}^{k-1} \tKC_{v_i\tilde{v'_i}} *{^{\tilde{v'}_i^{-1}}\tilde{\Omega}_{s_i}(^{v_i^{-1}}\theta}) +h',\] where $h'\in \ATC$ and $\ell_\tau(h')\leq k-3$.
By Proposition~\ref{propBL_relations}~\eqref{itBL_relations}, we have \begin{equation}\label{eqBL_commutation_order_k-1_ATC_temporary}\begin{aligned} \theta*\tKC_w &= \big(\tKC_{w'}*{^{w'^{-1}}\theta}+\sum_{i=1}^{k-1} \tKC_{v_i\tilde{v'_i}} *{^{\tilde{v'}_i^{-1}}\tilde{\Omega}_{s_i}(^{v_i^{-1}}\theta}) \big)*\tKC_{s_k}
\\ &=\tKC_{w}*{^{w^{-1}}\theta} +\tKC_{w'}*\tilde{\Omega}_{s_k}({^{w'^{-1}}\theta}) +\sum_{i=1}^{k-1}\left( \tKC_{v_i\tilde{v_i}} *{^{\tilde{v}_i^{-1}}\tilde{\Omega}_{s_i}({^{v_i^{-1}}\theta}})\right)+h'*\tKC_{s_k}+h''\\
&=\tKC_{w}*{^{w^{-1}}\theta} +\sum_{i=1}^{k} \tKC_{v_i\tilde{v_i}} *{^{\tilde{v}_i^{-1}}\tilde{\Omega}_{s_i}({^{v_i^{-1}}\theta}})+ +h'*\tKC_{s_k}+h'',\end{aligned}\end{equation} where \[h'' =\sum_{i=1}^{k-1}\tKC_{v_i\tilde{v_i'}} *\tilde{\Omega}_{s_k}\left({^{\tilde{v}_i^{-1}}\tilde{\Omega}_{s_i}({^{v_i^{-1}}\theta}})\right)\in \bigoplus_{v\in \Wta|\ell_\tau(v)\leq k-2} \tKC_{v}*\C(Y)_\tau. \] Moreover by Proposition~\ref{propBL_relations}, $\ell_\tau(h'*\tKC_{s_k})\leq k-1$. By eliminating the $i$ such that $\ell_\tau(v_i\tilde{v}_i)<k-1$ in \eqref{eqBL_commutation_order_k-1_ATC_temporary}, we deduce \eqref{eqBL_commutation_order_k-1_ATC}.
Using Lemma~\ref{lemFormula_omega_tilde_vtau}, we deduce \eqref{eqBL_commutation_order_k-1_Itau}.
\end{proof}
\begin{Lemma}\label{lemLength_decreasing_multiplication_theta}
Let $x\in \Itg$, $\theta\in \JC_\tau$ and $w\in W^v$. Then $\ell_\tau(\theta.x)\leq \ell_\tau(x)-1$. In particular we have $\ord_\tau(x)\leq \ell_\tau(x)+1$ for all $x\in \Itg$.
\end{Lemma}
\begin{proof}
This follows from \eqref{eqBL_commutation_order_k-1_Itau}.
\end{proof}
\begin{Lemma}\label{lemEquality_length_order_simple_product}
We assume that there exists $\rho\in \C^*$ such that \begin{equation}\label{eqDef_rho}\sigma''_{s,\tau}\in \rho \R^*_+,\end{equation} for all $s\in \SCC$. Let $w\in \Wta$. Then $\ord_\tau(\tKC_w.\vb_\tau)=\ell_\tau(w)+1$.
\end{Lemma}
\begin{proof}
Pick $\lambda\in C^v_f$ and set $\theta=Z^\lambda-\tau(\lambda)\in \JC_\tau$. For $v\in \Wta$, we define $C_v\in \C$ by \[\theta^{\ell_\tau(v)}.\vb=C_v.\vb_\tau,\] which is well defined by Lemma~\ref{lemLength_decreasing_multiplication_theta}. Fix a reduced writing $v=s_1\ldots s_k$ of $w$, with $k=\ell_\tau(w)$ and $s_1,\ldots,s_k\in \SCC_\tau$. We want to prove that $C_w\neq 0$. We assume that for all $v\in \Wta$ such that $\ell_\tau(v)<\ell_\tau(w)$, we have $C_v\in (\tau(\lambda)\rho)^{\ell_\tau(v)}\R^*_+$. For $i\in \llbracket 1,k\rrbracket$, set $v_i=s_1\ldots s_{i-1}$, $\tilde{v}_i=s_{i+1}\ldots s_k$. Let $E=\{i\in \llbracket 1,k\rrbracket|\ell_\tau(v_i\tilde{v}_i)=k-1\}$.
Then by Lemma~\ref{lemComputation_terms ordern-1_commutation} and \eqref{eqFormula_tau_Omega_s}, we have \[\begin{aligned}C_w &=\sum_{i\in E} \tau\big(\tilde{\Omega}_{s_i}(^{v_i^{-1}}\theta)\big)C_{v_i\tilde{v}_i}\\
&= \tau(\lambda)\sum_{i\in E}\alpha_{s_i}(\lambda) \sigma_{s_i,\tau}'' v_i.\alpha_{s_i}(\lambda) C_{v_i\tilde{v}_i}. \end{aligned} \] If $i\in E$, we have $\ell(v_is_i)>\ell(v_i)$ and thus by \cite[1.3.13 Lemma]{kumar2002kac}, $v_i.\alpha_{s_i}\in \Phi_+$. Therefore $v_i.\alpha_{s_i}(\lambda)\in \Z_{>0}$. We deduce that $C_w\in (\tau(\lambda)\rho)^{k}\R^*_+$. In particular, $C_w\neq 0$ and $\ord_\tau(\tKC_w .\vb_\tau)\geq \ell_\tau(w)+1$. By Lemma~\ref{lemLength_decreasing_multiplication_theta} we deduce that $\ord_\tau(\tKC_w .\vb_\tau)=\ell_\tau(w)+1$.
\end{proof}
\begin{Lemma}\label{lemEasy_inequality_multiplication_KC_s}
Let $x\in \Itg$ and $s\in \SCC_\tau$. Then $\ord_\tau(\tKC_s.x)\leq \ord_\tau(x)+1$.
\end{Lemma}
\begin{proof}
We prove it by induction on $\ord_\tau(x)$. If $\ord_\tau(x)=0$, then $x=0$ and it is clear. We assume that $\ord_\tau(x)>0$ and that for all $y\in \Itg$ such that $\ord_\tau(y)<\ord_\tau(x)$, we have $\ord_\tau(\tKC_s.y)\leq \ord_\tau(y)+1$. Let $\theta\in\JC_\tau$. By Proposition~\ref{propBL_relations}~\eqref{itBL_relations}, we have \[\theta.\tKC_s.x=\tKC_s.{^s\theta}.x+\tilde{\Omega}_s(\theta).x,\] for $\theta\in \JC_\tau$. Moreover $\ord_\tau({^s\theta}.x)< \ord_\tau(x)$ and thus $\ord_\tau(\tKC_s.{^s\theta}.x)\leq \ord_\tau({^s\theta}.x)+1\leq_\tau \ord_\tau(x)$ by assumption. Moreover, $\ord_\tau\left(\tilde{\Omega}_s(\theta).x\right)\leq \ord_\tau(x)$, therefore $\ord_\tau(\theta.\tKC_s.x)\leq \ord_\tau(x)$ for every $\theta\in \JC_\tau$ and thus $\ord_\tau(\tKC_s.x)\leq\ord_\tau(x)+1$. Lemma follows.
\end{proof}
Let $x\in \Itg\setminus\{0\}$. Write $x=\sum_{w\in \Wta}a_w\tKC_w.\vb_\tau$. Let $M=\{w\in \supp(x)|\ell_\tau(w)=\ell_\tau(x)\}$, $\NC_\tau=|M|$ and $\LTI(x)=\sum_{w\in M} a_w \tKC_w.\vb_\tau$.
\begin{Theorem}\label{thmWeight_space}
Let $\tau\in \UC_\C$. We assume that \eqref{eqDef_rho} is satisfied and that $|\sigma_s|,|\sigma_s'|>1$, for all $s\in \SCC_\tau$. Let $x\in \Itg\setminus\{0\}$. Then $\ord_\tau(x)=\ell_\tau(x)+1$. In particular, $\Itg\cap I_\tau(\tau)=\C \vb_\tau$ and \cite[Conjecture 5.16]{hebert2021decompositions} is true.
\end{Theorem}
\begin{proof}
If $\NC_\tau(x)=1$, this is Lemma~\ref{lemEquality_length_order_simple_product} and if $\ell_\tau(x)=0$, this is clear. We now assume that $\NC_\tau(x)\geq 2$ and that for all $y\in \Itg\setminus\{0\}$ such that $\ell_\tau(y)<\ell_\tau(x)$ or $\NC_\tau(y)<\NC_\tau(x)$, we have $\ord_\tau(y)=\ell_\tau(y)+1$. Let $M=\supp\big(\LTI(x)\big)$. There are two cases:\begin{enumerate}
\item there exists $s\in \SCC$ such that for all $w\in M$, $sw<_{\tau} w$,
\item for all $s\in \SCC$, there exists $w\in M$ such that $sw>_\tau w$.
\end{enumerate}
We first assume (1). Then we can write\[x=\tKC_s.x'+x'',\] with $\ell_\tau(x')=\ell_\tau(x)-1$, $\ell_\tau(x'')\leq \ell_\tau(x')$ and $sv>_\tau v$ for all $v\in \supp(x')$. Let $\theta\in \JC_\tau$. Then \begin{equation}\label{eqTheta_KC} \theta.\tKC_s.x=\tKC_s*{^{s}\theta}.x'+\tilde{\Omega}_s(\theta).x'+\theta.x''.\end{equation}
By Lemma~\ref{lemLength_decreasing_multiplication_theta}, we have $\ell_\tau({^s\theta}.x') \leq \ell_\tau(x')-1$. For all $v\in \supp({^s\theta}.x')$, if $sv<_\tau v$, then $\ell_\tau(\tKC_s*\tKC_v.\vb_\tau)\leq \ell_\tau(v)\leq \ell_\tau({^s\theta}.x')\leq \ell_\tau(x')-1$ by Proposition~\ref{propBL_relations}~\eqref{itHecke_relations}. Therefore for all $v\in \supp({\tKC_s*^s\theta}.x)$ such that $\ell_\tau(v)=\ell_\tau(x')$, we have $sv<_\tau v$. Therefore \begin{equation}\label{eqSupport}\supp\big(\LTI(\tKC_s*{^s\theta}*x')\big)\cap \supp\big(\LTI(x')\big)=\emptyset.\end{equation} Take $\lambda\in Y$ such that $\alpha_s(\lambda)=1$ and set $\theta=Z^\lambda-\tau(\lambda)\in \JC_\tau$. Then by Lemma~\ref{lemFormula_omega_tilde_vtau}, $\tau\big(\tilde{\Omega_s}(\theta)\big)=\tau(\lambda)\sigma_{s,\tau}''\alpha_s(\lambda)=\tau(\lambda)\sigma_{s,\tau}''$ and thus by combining \eqref{eqTheta_KC} and \eqref{eqSupport}, we obtain that $\ell_\tau(\theta.\tKC_s.x)=\ell_\tau(x')$. By assumption we deduce that $\ord_\tau(\theta.\tKC_s.x)=\ell_\tau(x')+1$. Consequently $\ord_\tau(\tKC_s.x)\geq \ell_\tau(x)+1$ and by Lemma~\ref{lemLength_decreasing_multiplication_theta} we deduce that $\ord_\tau(\tKC_s.x)=\ell_\tau(x)+1$.
We now assume that we are in the case (2). Let $w\in \supp\big((\LTI(x)\big)$ and $s\in \SCC$ be such that $sw<_\tau w$. Set $x'= \tKC_s.x$. Then by assumption we have $\ell_\tau(x')=\ell_\tau(x)+1$ and thus \[\supp\big(\LTI(x')\big)\subset s.\supp\left(\LTI(x)\right)\setminus \{sw\}.\] In particular $\NC_\tau(x')<\NC_\tau(x)$. By our induction assumption we deduce that $\ord_\tau(x')=\ell_\tau(x')+1=\ell_\tau(x)+2$. By Lemma~\ref{lemEasy_inequality_multiplication_KC_s} we have that $\ord_\tau(x)\geq \ord_\tau(x')-1\geq \ell_\tau(x)+1$. By Lemma~\ref{lemLength_decreasing_multiplication_theta} we deduce that $\ell_\tau(x)+1=\ord_\tau(x)$. This completes the proof of the theorem.
\end{proof}
Note that the above proof can be simplified in the case where $\Wta$ is finite. In this case, denote by $w_0$, the maximal element of $\Wta$ (for the Bruhat order $\leq_\tau$). Let $x\in \Itg\setminus\{0\}$. Let $w\in \supp\big(\LTI(x)\big)$. Set $y=\tKC_{w_0 w^{-1}}.x$. Then $\supp(\tKC_{w_0w^{-1}}.x)\ni w_0$. By Lemma~\ref{lemEquality_length_order_simple_product}, $\ord_\tau(\tKC_{w_{0}w^{-1}}.x)=\ell_\tau(w_0)+1$. By Lemma~\ref{lemEasy_inequality_multiplication_KC_s}, we deduce that \[\ord_\tau (x)\geq \ord_\tau(y)-\ell(w_0w^{-1})=\ell_\tau(w_0)+1-\big(\ell(w_0)-\ell(w^{-1})\big)=\ell(w)+1.\] By Lemma~\ref{lemLength_decreasing_multiplication_theta}, we deduce that $\ord_\tau(x)=\ell_\tau(x)+1$.
\medskip
We obtain a version of the Knapp-Stein dimension theorem in our frameworks:
\begin{Corollary}\label{corSilberger_dimension_theorem}(see \cite{silberger1978knapp})
Let $\tau\in \UC_\C$. Then \[I_\tau(\tau)=\bigoplus_{w_R\in R_\tau} \C F_{w_R}.\vb_\tau.\] If moreover $\sigma_s=\sigma_s'$ for all $s\in \SCC$, then $\End_{\HC-\mathrm{mod}}(I_\tau)\simeq \C[R_\tau]$.
\end{Corollary}
\begin{proof}
This follows from \cite[Proposition 5.13 (2) and Proposition 5.27]{hebert2021decompositions}.
\end{proof}
Note that under the notation of \cite{hebert2021decompositions}, Theorem~\ref{thmWeight_space} implies that $I_\tau(\tau,\Wta)=\C \vb_\tau$. We can thus apply the results from 5.3 to 5.6 of \cite{hebert2021decompositions}, when $\tau\in \UC_\C$. In particular, we have a description of the submodules and the irreducible quotients of $I_\tau$ when $\tau\in \UC_\C$, see \cite[Theorem 5.38]{hebert2021decompositions}.
\medskip
We also obtain Kato's irreducibility criterion:
\begin{Corollary}\label{corKatos_irreducibility_criterion}(see \cite[Theorem 2.4]{kato1982irreducibility})
Let $\tau\in T_\C$. Then $I_\tau$ is irreducible if and only if:\begin{enumerate}
\item $\tau\in \UC_\C$,
\item $\Wta=\{1\}$.
\end{enumerate}
\end{Corollary}
\begin{proof}
By \cite[Proposition 4.17 and Theorem 4.8]{hebert2018principal}, if $I_\tau$ is irreducible, then $W_\tau=\Wta$ and $\tau\in \UC_\C$. Conversely, let $\tau\in \UC_\C$ be such that $\Wta=W_\tau$. Then $R_\tau=W_\tau/\Wta=\{1\}$ and by Corollary~\ref{corSilberger_dimension_theorem}, $I_\tau(\tau)=\C \vb_\tau$. By \cite[Theorem 4.8]{hebert2018principal} we deduce that $I_\tau$ is irreducible.
\end{proof}
\printindex
\bibliography{/home/auguste_pro/Documents/Projets/bibliographie.bib}
\bibliographystyle{plain}
\bigskip
\par\noindent Universit\'e de Lorraine, CNRS, Institut Élie Cartan de Lorraine, F-54000 Nancy, France, UMR 7502
E-mail: auguste.hebert@univ-lorraine.fr.
\end{document}
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Creativity In Care’s™ purpose is to improve quality of life, inclusion and joy in care and community settings through learning programmes and creative connections that go beyond words. We believe every person can connect through creativity.
We offer a range of workshops, courses and consultancy. Our background is in health and social care, adult education and arts. This combination is used to promote mental well-being, increase confidence of those we work with, and create positive connections. We co-create exhibitions, films and performances to give voice to people who may not have had opportunities to be heard due to illness, disability or culture difference.
Creativity In Care ™ works with people of all ages and abilities,
finding different ways to connect with people, with and without words. We help to increase knowledge and understanding of mental well-being and recovery factors. We work with people diagnosed with long term conditions, people with dementia, people living with mental ill health or other disabilities. We also work in conjunction with care-givers, family members, community organisations and anyone generally interested in creativity.
We know that regardless of diagnosis or perceived ability, people have emotions, imagination, and humour that connect through creativity.
Creativity is all-inclusive.
Talk to us live every Monday on Twitter (@creativitycare), where we co-host with Hilgos Foundation #AlzChat, an interactive discussion for people living with, or caring for someone with diagnosed with dementia.
Join us on Facebook (creativity in care) where we post images from some of the courses and workshops
IP terms: The work and photographs on this website are original and copyrighted to Creativity In Care CIC. Before copying or using any of our material please seek permission by contacting us. We are friendly and love to know more about who is interested in this work, so please do get in contact.
We appreciate support from the following organisations along our journey
Looks like great work – would be good to keep in touch and share feedback and ideas – regards – Jeff vdg ltd
Definitely great work being done! I really like the creative & innovative approaches you have taken to helping people in all sorts of ways especially in their overall mental well-being. I am in the medical alert service for seniors industry and we try and help seniors & caregivers live their best life possible. Offering great resources to help them help themselves such as your events page is awesome! Thanks for all the great info. I will definitely share your site on Twitter! Have a great day
Hi …….
Being a design student I truly believe in the ability for
Creativity to improve quality of life……
Kind regards
Culainn B Shanahan
Love the level of engagement shown in these photos. I look forward to following your posts and seeing more of your interventions!
Thanks JoAnn, yes, creativity is engaging isn’t it! Your work looks very interesting too… will e-mail in a few days :))
Thanks JoAnn. Your work looks great too – think your first sentence could say Music is required for life!! It does make a huge difference to quality of life… we are somehow programmed to connect through music (see book chapter on ‘the etiquette of dreadful singing’ if you get a chance). Love that you have themed songs. We’re thinking up some for boats at the moment! Stay in contact…
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Above: Me, chilling in the kitchen space of Hansik with Chef Akira Back (far right)and his staff
Below: A Korean treat wonderland in the Korea Pavilion
I am, at long last, back from my self-imposed two week blogging respite/hiatus ... and I come bearing gifts! There, that got your attention. But I am getting ahead of myself.
Now, I certainly wouldn't want to bore you with every last excruciating detail of the past two weeks, so I'll skip to the good stuff: I went as a member of the press (both as a food journalist and as a blogger ... so many personas, so little time) to the Summer Fancy Food Show in Washington, D.C., where I was faced with a dizzying array of foods from 2,400 exhibitors hailing from 80-plus countries. And I had to try to taste it all! The horror.
But you know no matter where I go, your Trix is always thinking of you, right? So, as I tasted my way through the jam-packed convention center, I made sure to get some extra awesome samples, grease a few palms (just kidding about that one), sniff out the best stuff, and set up some some really cool, unique, tasty, and very Trix giveaways to celebrate my two-year blogoversary and to thank all of you for reading and sticking around this long.
So over the next couple of months, look for some very yummy things coming your way! And the first one? To find that out, you'll have to read this entire post, or at the very least scroll through it and have a look at the photos along the way - I am putting some effort into this, you know!
I know a lot of the trendspotters and food press at the show were obsessed with chocolates and desserts and nibbly things. These have their place to be sure, but I was most excited about the chance to eat a three-course lunch at Hansik, a Korean pop-up restaurant located in the - where else? - Korea Pavilion. Best of all, the meal would be prepared by Akira Back, the chef at Yellowtail at the Bellagio in Las Vegas, and a former competitor on Iron Chef.
But in order to gain a coveted spot at Hansik, conventioneers were asked to meet with at least one Korean exhibitor. No problem! I met with two. First, I had a meeting with Elliot Chung, senior brand manager of CJ Foods, Korea's largest food company and one of the sponsors of the PBS show The Kimchee Chronicles, featuring Chef Jean-Georges Vongerichten and his Korean-born wife. He told me about the new line of Bibigo Korean Pantry products that will be available in mainstream U.S. supermarkets starting in September - so all of you who aren't lucky enough to live near a store like my beloved H-Mart will soon be able to get gochuchang at your local supermarket:
Next, I met with Jeong Jun-Mo, manager of Anmyundo Agricultural Cooperative, to learn about a new organic Korean sun dried pepper that will be available in the U.S. market early in 2012. I don't have a photo of Mr. Jeong, as this meeting was a bit more formal than my first, but he told me (through a translator) that his favorite way to enjoy the pepper powder is in a fish soup.
And now, meetings complete, it was time to sit back and enjoy a gourmet lunch - in just 30 minutes! Somehow my meal didn't seem at all rushed, and sitting in the elegantly appointed space I felt as if I had entered an oasis of calm in the middle of the convention chaos:
For my cocktail, I chose the Ginger Gorgie, a ginger margarita with Makgeolli, created by Jean-Georges Vongerichten:
For my first course, I was torn between the bibim tuna with strawberry, rayu and mini corn or the kogi taco with serrano perfume, micro cilantro and smoked tomato salsa. Regular readers will know that I had to go with the taco:
My main course was prime beef with aged black garlic, potato, quail egg, and micro carrot. If I had this in a private place, I would have shoved my face down into the plate and licked it until there was not one drop of that sauce left:
And now, since you have been so patient and indulgent, I will share my first Fancy Foods Show-inspired giveaway: A 350 gram bag of organic, sun-dried Korean pepper powder, unavailable in the U.S. until early 2012:
Okay, how to win? Here's what to do:
1. Leave a comment
2. Tweet the giveaway and come back and tell me that you did
3. You get an EXTRA CHANCE if you leave a comment that demonstrates you actually read this post and didn't just skip to the giveaway part.
There - three chances to win. It's open to anyone, anywhere , and I will announce the winner on Sunday, July 24, and use Random.org to choose.
Good luck, and let my second blogoversary festivities commence! (Cue trumpets, or maybe just bugles.)
***Addendum: This giveaway has now ended. *******
wow what a neat experience happy you got to go
beautiful presentation! looks so delicious. Fresh, fast & fun, a pleasant lunch time experience.
This just looks like too much fun!! Happy blogoversary!
I can't believe those chili powder won't be available for another few months, but I can't wait to try it when it comes out. And I've seen H Marts around but I never visited them before. Sounds like I have to now for those pantry items!
Its always a pleasure to visit your blog, always a new adventure to learn about foods. Happy Blogoversary looking forward to the many more posts in the future that inspire everyone to blog, post what they truly feel and experience new and wonderful treats!
So that's where you've been... lucky! :) Happy Blogoversary... here's to many many more! I looooove your blog - writing, recipes, photography - everything is superb! xoxo
Welcome back! Nothing like a little break from blogging to refresh the foodie in all of us. I have yet to have authentic gochuchang, so hopefully it'll spread up here to Canada (although I don't expect it to make an appearance on my tiny island for awhile)...Theresa
This looks like the most fun convention in the WORLD! I was JUST in DC!!!! I wish I would have known about it. And that hot pepper is dying to live in my kitchen. DYING, I tell you! :D
Happy Blogoversary! I can't wait until I reach my one year mark. It looks like you got to experience so many new flavors and learn about Korean cooking. All of the photos are crave worthy. Happy to have found your blog through Foodbuzz and looking forward to more yummies from your kitchen.
What a generous giveaway!
Happy blogaversary! Hope you've come back refreshed and full of great new ideas. I don't know nearly enough about Korean cooking - I need to brush up on some basics soon!
I have tweeted about your giveaway!
golly, so sorry for you to have that horrid job, having to go to these types of shows and experience such - glad you made the best of it... wish I could have tagged along, very nice post, can't wait for the others...
Happy Anniversary to you and your blog :)
Wow, this is what you did on your self imposed break? Looks like a wonderful whirlwind of food and discoveries :)
This is much more exciting than another cantaloupe soup post! How lucky are you! Congrats on the fun culinary experience AND on your two year blog anniversary~
I tweeted your give away :)
Happy Blogoversary!! We are the same age. If that is what you call a hiatus then if puts my previous hiatus to shame. The convention sounds like so much fun. I can't believe you were served that 5 star food and only had 30 minutes to eat, how they do that? It's always so great speaking to people behind a product.
(BTW I may be extending my deadline as I'm running behind too. So you can still enter my giveaway:) )
Happy 2nd blogaversary! Seams like you celebrated your first not long ago, time flies. So fun to go to these events.
Demonstrating that I read it all: I would pay 5$ to see a pic of you liking that plate!
tweeted
Happy blogaversary! What a way to celebrate! Beautiful photos! Beautiful food! I want the minty mocha and korean air tea. It looks so interesting!
Happy 2nd blogaversary! That is a big one, congrats!
A ginger Margarita with Makgeolli sound great - I am not sure what Makgeolli is, but ginger in a margarita sounds wonderful!
That is just way too cool. I have a mental image of you running off with a plate behind and exhibition booth and licking every last drop from it!
Happy two-year.
Happy Blogoversary! What an opportunity that pop-up restaurant gave you - not just the gourmet fare, but the impetus to interview folks and try new things.
Happy Blogoversary! The Fancy Food Show sounds awesome- I'm sure it was a tough job trying to taste everything, but somebody had to do it! :) The sauce with your main course DOES look good! Yum!
Wow. Such a great post. A pop up restaurant, your blogoversary AND a giveaway? You are too generous! And no, I did read the whole thing and not just the headline! I did skip over most of the proper nouns though. :)
Happy Blogoversary! That lunch sounds like it was the perfect foodie celebration of this important milestone!
Okay, totally jealous of your blogging 'hiatus.' When I take a trip...its never to awesome fun food heaven like this! Happy blogoversary!
Trix
I sure hope you are going to tell us mere mortals about the other 2399 exhibitors at this show! I am not an expert on Korean food but I too admire the way they organized this make-shift fancy eatery in the midst of an exhibit hall: Wow!! Never heard of black aged garlic, so many intriguing ingredients ao artfully presented. Now to the pepper, I will tweet about it and let you know in the hope of winning this little thing (curiosity is eating at me).
Read the whole post? Hell, I was licking the sceen!
Oh, the ginger margarita looked so good I am drooling for one!
Trix, I would impose a hiatus on my blogging too if I could go off and experience something like this! What a lucky girl you are... I'm intrigued by your gourmet lunch at the Korean pavilion, especially the red seaweed-looking thing. What IS that? Loved that you tried to marry your love of Korean and Austrian food...yes, I would totally try that Korean goulash if you made some!
Happy 2nd Anniversary, my friend! Thanks for letting us in on the celebrations ;-).
That Korean tea looks dleicious. looks like you had a wonderful meal. love anything with spicy. My Korean food knowledge is only fired noodles, by the ways awesome giveaway.
Hi Trix - first of all congrats on year number 2! And best wishes for continued success with your awesome blog.
After being a blogger judge for Top Chef Korean Food Challenge, I "heart" hansik. Every bit of it, can't get enough, my latest kitchen additions have all been Korean, from spoon & chopstick sets to a tukbaege. I would have loved to attend that luncheon with you. The new pepper powder sounds interesting.
See you at the Farmers' Market?
LL
Trix..that was a good korean meal you had at Hansik :) great photography and happy 2nd blogoversary to you :)
Awesome! You rock my world.
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\begin{document}
\title[
Symmetric products of the $\mathfrak{E}_{\mathrm{c}}$ and the $\mathfrak{E}$.]{\scshape\bfseries
Symmetric products of
Erd\H{o}s space and complete Erd\H{o}s space.
}
\author{{ \bfseries }Alfredo Zaragoza }
\address{Departamento de Matematicas\\
Facultad de Ciencias\\
Universidad Nacional Aut\'onoma de M\'exico\\
}
\email{soad151192@icloud.com}
\subjclass[2010]{54B20 54A10 54E50 54F50 54F65.}
\keywords{Erd\H{o}s Space Complete, Erd\H{o}s Space, Cohesive, Hyperspace, almost zero dimensional}
\date{}
\thanks{This work is part of the doctoral work of the author at UNAM, Mexico city, under the direction of R. Hernández-Gutiérrez and A. Tamariz-Mascarúa.
}
\begin{abstract}
It is shown that the symmetric products of complete Erd\H{o}s space and Erd\H{o}s space are homeomorphic to complete Erd\H{o}s space and Erd\H{o}s space,
respectively. We will also give some properties of their hyperspace of compact subsets with the Vietoris topology.
\end{abstract}
\maketitle
\section{Introduction}
Every topological space in this article is assumed to be metrizable and separable.
The following two spaces were introduced by Erd\H{o}s in 1940 in \cite{er} as examples of totally disconnected and non-zero dimensional spaces. The Erd\H{o}s space is defined as follows
$$ \mathfrak{E} = \{(x_n)_{n\in \omega} \in \ell^2 : x_i \in \Q , \textit{for all } i\in \omega \};$$
and the complete Erd\H{o}s space as
$$ \mathfrak{E}_{\mathrm{c}} = \{(x_n)_{n\in \omega} \in \ell^2 : x_i \in \{0\}\cup \{1/n: n\in \N\} \textit{ for all } i\in \omega \}$$
when $\ell^2$ is the Hilbert space of all square summable real sequences. The name \textit{complete Erd\H{o}s} space was introduced by Kawamura, Oversteegen, and Tymchatyn \cite{K}.
Some properties of the complete Erd\H{o}s space and Erd\H{o}s space were studied in
\cite{j}, \cite{jj}, \cite{J3}, \cite{T} and \cite{K}. For a space $X$, $\mathcal{K}(X)$ denotes the hyperspace of non-empty compact subsets of $X$ with the
Vietoris topology; for any $n\in \N$, $\mathcal{F}_n(X)$
is the subspace of $\mathcal{K}(X)$ consisting
of all the non-empty subsets that have cardinality less or equal to $n$; and $\mathcal{F}(X)$ is the subspace of $\mathcal{K}(X)$ of finite subsets of $X$.
For $n\in \N$ and subsets $U_1,\ldots, U_n$ of a topological space $X$, we denote by $\langle U_{1},\ldots ,U_{n}\rangle$ the collection $\left\lbrace F \in \mathcal{K}(X):F\subset \bigcup_{k=1}^n U_k, F\cap U_{k}\neq \emptyset \textit{ for } k \leq n \right\rbrace $. Recall that the Vietoris topology on $\mathcal{K}(X)$ has as its canonical base all the sets of the form $\langle U_{1},\ldots ,U_{n}\rangle$ where $U_k$ is a non-empty open subset of $X$ for each $k\leq n$.
We will study the properties that $\mathfrak{E}_c$ and $\mathfrak{E}$ have in common with some of their hyperspaces with the Vietoris topology.
The main results of this work are:
\begin{teo}\label{ee1}
For any $n\in \N$, $\mathcal{F}_n(\mathfrak{E}_{\mathrm{c}})$
is homeomorphic to $\mathfrak{E}_{\mathrm{c}}$.
\end{teo}
\begin{teo}\label{EE1}
For any $n\in \N$, $\mathcal{F}_n(\mathfrak{E})$
is homeomorphic to $\mathfrak{E}$.
\end{teo}
Note that if $X$ is a separable metrizable space, then they every subspace of $\mathcal{K}(X)$ is also a separable metrizable space (see \cite[Theorem 3.3 and Propositions 4.4 and 4.5.2]{Sub})
\section{ Cohesive and almost zero dimension spaces}
An important property the of spaces $\mathfrak{E}$ and $\mathfrak{E}_{\mathrm{c}}$ is that at each point there is a local base such that each element of the base is an intersection of clopen sets. A space with this property is called
\textit{almost zero-dimensional} $(AZD)$. The fallowing result is well known.
\begin{prop}[{\cite[Remark 2.4]{jj}}]\label{eazd}
A topological space $(X,\mathcal{T})$ is almost zero-dimensional if
there is a zero-dimensional topology $\mathcal{W}$ in $X$ such that $\mathcal{W}$ is coarser than $\mathcal{T}$ and has the property
that every point in $X$ has a local neighborhood base consisting of sets that are closed with respect
to $\mathcal{W}$.
\end{prop}
If $(X,\mathcal{T})$ is an $AZD$ space, we say that the topology $\mathcal{W}$ which appears in Proposition \ref{eazd} witnesses the almost zero-dimensionality of $X$. Every zero-dimensional space is an $AZD$ space and the property of being $AZD$ is hereditary. On the other hand it is proved in \cite{Lh} that all AZD spaces have dimension less than or equal to one. Moreover, it is proved in \cite{jj} that all $AZD$ spaces are embeddable both in $\mathfrak{E}_c$ and $\mathfrak{E}$.
\begin{prop}\label{HZD}
For a topological space $X$ the following statements are equivalent.
\begin{enumerate}
\item $X$ is an $AZD$ space.
\item $\mathcal{K}(X)$ is an $AZD$ space.
\item If $\mathcal{F}_1(X)\subset\mathcal{A}\subset \mathcal{K}(X)$, then $\mathcal{A}$ is an $AZD$ space.
\end{enumerate}
\end{prop}
\begin{proof}
The implication $(2)\Rightarrow (3)$ is obvious, and $(3)\Rightarrow (1)$ follows from the fact that $\mathcal{F}_1(X)$ is homeomorphic to $X$.
\\
$(1)\Rightarrow (2) :$ We are going to prove that $\mathcal{K}(X)$
satisfies the conditions of Proposition \ref{eazd}.
Let $\mathcal{W}$ be a topology which witnesses the almost zero-dimensionality of $X$.
Consider the space $Y=(X, \mathcal{W})$. As $\mathcal{W}$ is coarser than the topology of $X$, then $\mathcal{K}(X)\subset \mathcal{K}(Y)$. Let $(Z,\mathcal{W}_0)$ be the space $\mathcal{K}(X)$ considered with the topology inherited as a subspace of $\mathcal{K}(Y)$. Since
$\langle V_1,\ldots, V_n\rangle \cap Z$ is an open subset of $\mathcal{K}(X)$, when $V_1, \ldots, V_n$ are elements in $\mathcal{W}$,
we have that the topology $\mathcal{W}_0$ of $Z$ is coarser than the topology of the $\mathcal{K}(X)$.
Moreover, by Proposition 4.13.1 in \cite{Sub}, $(Z,\mathcal{W}_0)$ is zero-dimensional.
Now, we are going to prove that each element in $\mathcal{K}(X)$ has a local neighborhood base consisting of closed subsets in $(Z,\mathcal{W}_0)$. Let $F\in\mathcal{K}(X)$ and let $\mathcal{U}= \langle U_1,\ldots, U_m\rangle$ be
a canonical open subset of $\mathcal{K}(X)$ such that $F\in \mathcal{U}$.
For each $x\in F$ there is a neighborhood $V_x$ of $x$ in $X$ such that $x \in V_x\subset \bigcap\{U_j: x\in U_j\}$ and $V_x$ is closed in $Y$. Then $\{int_X(V_x):x\in F\}$ is an open cover of $F$ in $X$. As $F$ is a compact subset of $X$, there exist $x_1,\ldots, x_k\in F$ such that $F\subset \bigcup_{i=1}^k V_{x_i}$. For each $i\leq m$, let $y_i\in F\cap U_i$. Note that $F \in \langle V_{x_1},\ldots, V_{x_k}, V_{y_1}, \ldots, V_{y_m}\rangle$. Let us see that $\mathcal{V}_{\mathcal{U}}:=\langle V_{x_1},\ldots, V_{x_k}, V_{y_1}, \ldots, V_{y_m}\rangle\cap \mathcal{K}(X)\subset \mathcal{U}$. Indeed, let $H\in \mathcal{V}_{\mathcal{U}}$. Then $H\subset \bigcup_{i=1}^k V_{x_i} \cup \bigcup_{j=1}^m V_{y_j}$ and $H\cap V_{z}\neq\emptyset$ for $z\in \{x_1,\ldots, x_k, y_1,\ldots y_m\}$.
By the choise of $V_{x_i}$ and $V_{y_j}$, we have $ \bigcup_{i=1}^k V_{x_i} \cup \bigcup_{j=1}^m V_{y_j}\subset \bigcup_{i\leq m} U_{i} $, and for each $j\leq m$ there exists a $z\in \{ x_1,\ldots, x_k, y_1,\ldots y_m\}$ such that $V_z\subset U_j$. Then $H\subset \bigcup_{i\leq m} U_{i}$ and $H\cap U_i\neq \emptyset$ for $i\leq m$. Thus $\mathcal{V}_{\mathcal{U}}\subset \mathcal{U}$, moreover $\langle V_{x_1},\ldots V_{x_k}\rangle$ is a closed subset in $\mathcal{K}(Y)$ (see \cite[Lemma 2.3.2 , p. 156]{Sub}), so $\langle V_{x_1},\ldots V_{x_k}\rangle \cap Z$ is a closed subset of $Z$.
Therefore, the collection of all the sets $\mathcal{V}_{\mathcal{U}}$ where $\mathcal{U}$ is a canonical open set of $\mathcal{K}(X)$ containing $F$, form a neighborhood local base of $F$ consisting of closed sets in $(Z, \mathcal{W}_0)$. Hence, by Proposition \ref{eazd}, $\mathcal{K}(X)$ is an $AZD$ space.
\end{proof}
The following Corollary follows from Proposition 4.13.1 in \cite{Sub} and Proposition \ref{HZD}, above.
\begin{cor}\label{azc}
Let $X$ be an $AZD$ space, then $dim(\mathcal{K}(X))\leq 1$ and $dim(\mathcal{K}(X))= 1$ if and only if $dim(X)=1$.
\end{cor}
The omission of the hypothesis of almost zero-dimensionality on $X$ in the previous corollary produces the following natural question.
\begin{pre}
Is there a space $ X $ that is not $ AZD $ such that $ dim (X) = dim(\mathcal{K}(X))= 1$?
\end{pre}
In \cite{er} Erd\H{o}s proved that every point in $ \mathfrak{E} $ has a neighborhood that does not contain (nonempty) clopen sets. Later in \cite{jj} J.J Dijkstra and J. van Mill formalized this concept as follows.
\begin{defi}[{\cite[Definition 5.1]{jj}}]
Let $X$ be a space and let $\mathcal{A}$ be a collection of subsets of $X$.
The space $X$ is called $\mathcal{A}$-cohesive if every point of the space has a neighborhood
that does not contain nonempty proper clopen subsets of any element of $\mathcal{A}$. If a space $X$
is $\{X\}$-cohesive then we simply call $X$ cohesive.
\end{defi}
Note that all connected spaces are cohesive and that the dimension of a cohesive space is greater than or equal to one.
If $Y$ is any space and $X$ is a space that is $ \mathcal{A} $ - cohesive then $X \times Y$ is $\{A \times B: A \in \mathcal{A}$ and $ B \subset Y \}$-cohesive (see {\cite[Remark 5.2, p. 21]{jj}}). Particularly if $ X $ is $ \{A_s:s\in S\} $ - cohesive, then $ X^n $ is $ \{A_{s_1}\times \ldots \times A_{s_n}: s_j \in S\} $-cohesive for any $ n \in \N $.
The following result relates the cohesion property between a space $ X $ and its symmetric products.
\begin{prop}\label{A2}
Let $n\in \N$ and let $X$ be a space $\{A_s: s\in S\}$-cohesive and $f:X^n\to \mathcal{F}_n(X)$ the function given by $f(x_1,\ldots,x_n)=\{x_1,\ldots x_n\}$. Then $\mathcal{F}_n(X)$ is $\{f[A_{s_1}\times \ldots \times A_{s_{n}}]: s_1,\ldots, s_n\in S\}$-cohesive.
\end{prop}
\begin{proof}
Suppose that $\mathcal{F}_n(X)$ is not $\{f[A_{s_1}\times \ldots \times A_{s_{n}}]: s_1,\ldots, s_n\in S\}$-cohesive. Then there exist $F\in \mathcal{F}_n(X)$ and a local base $\beta$ of $F$, such that any $\mathcal{U}\in \beta$
contains a non-empty proper clopen
subset of some element of $\{f[A_{s_1}\times \ldots \times A_{s_{n}}]: s_1,\ldots, s_n\in S\}$.
Let us suppose that $F=\{x_1, \ldots ,x_k\}$ with $x_j\neq x_i$ for each $i,j\in \{1,\ldots k\}$. Let $\beta_{x_1}, \ldots, \beta_{x_k}$ be local bases of the points $x_1, \ldots, x_k$ respectively such that if $i\neq j$, $U_j\in \beta_{x_j}$ and $U_i\in \beta_{x_i}$, then $U_j\cap U_i=\emptyset$.
Let $x=(x_1, \ldots,x_{k}, x_{k+1},\ldots, x_n)$ be in $X^n$, where $x_k=x_{k+1}=\ldots =x_n$.
Note that $\beta_0=\{U_1 \times\ldots\times U_{k+1}\ldots \times U_n: U_j\in \beta_{x_j} \textit{ } j\leq k \textit{ y } U_k=U_{k+1}=\ldots= U_n\}$ is a local base at $x$. Let $V_j\in \beta_{x_j}$ be fixed for each $j\in\{i,\ldots k\}$.
Note that $F\in \mathcal{N}=\langle V_1, \ldots, V_k\rangle$ and that $x\in V_1\times \ldots \times V_n $. By our assumption there are $s_{1},\ldots, s_{n}\in S$ an open subset $\mathcal{U}\in \beta$, and a non-empty proper clopen subset $\mathcal{V}$ of $f[A_{s_{1}}\times \ldots \times A_{s_{n}}]$ such that $\mathcal{V}\subset \mathcal{U} \subset \mathcal{N}$.
Let $g=f\downharpoonleft A_{s_{1}}\times \ldots \times A_{s_{{n}}}$. As $g$ is continuous, we have that $g^{\leftarrow}[\mathcal{V}]$
is a clopen subset of $A_{s_{1}}\times \ldots \times A_{s_{n}}$. Let $C=g^{\leftarrow}[\mathcal{V}]\cap (V_1\times \ldots \times V_n)$. We are going to show that $C$ is a nonempty proper clopen subset of $A_{s_1}\times \ldots \times A_{s_{n}}$.
Let $ E \in \mathcal{F}_n(X)$, if $ E\in \mathcal{V}$, then there exist $w \in g^\leftarrow[E] $ such that $w \in (V_1\times \ldots \times V_n)$. If $E \in \mathcal{N}\setminus \mathcal{V}$, then
$g^\leftarrow[E] \cap g^{\leftarrow}[\mathcal{V}]= \emptyset$. Then there exist $w \in g^\leftarrow [E]$ such that $w \in A_{s_1}\times \ldots \times A_{s_n} \setminus C$. It is clear that $C$ is a open subset of $A_{s_{1}}\times \ldots \times A_{s_{n}}$.
To see that is closed let us consider a sequence $\{(y_1^m,\ldots, y_n^m): m\in \N\}$ of points of $ C $ such that $(y_1^m,\ldots, y_n^m)\to (y_1,\ldots, y_n)$. Since $f$ is continuous, the sequence $\{f((y_1^m,\ldots, y_n^m))=g((y_1^m,\ldots, y_n^m)):m\in \omega\}$ converges to $f(y_1,\ldots, y_n)$. Note that for every $m\in \omega$, we have that $f((y_1^m,\ldots, y_n^m))\in \mathcal{V}$. Since $\mathcal{V}$ is closed in $f[A_{s_{1}}\times \ldots \times A_{s_{n}}]$, we have that $f(y_1\ldots ,y_n)\in \mathcal{V}$.
Hence $(y_1,\ldots, y_n)\in f^{\leftarrow}[\mathcal{V}]$.
On the other hand, as $\mathcal{V}\subset N$, then $y_j\in V_j$ as $y_j^m\in V_j$ for $m\in \N$. Thus $C$ is clopen in $A_{s_{1}}\times \ldots \times A_{s_{n}}$. This is a contradiction, so $X^n$ is $\{A_{s_1}\times\ldots\times A_{s_n}: s_{1}, \ldots ,s_{n}\in S\}$-cohesive.
\end{proof}
The next concept is a property that implies the cohesion of a space $ X $ and also relates the properties of cohesion and almost zero-dimensionality.
\begin{defi}
A one-point connectification of a space $X$ is a connected extension $Y$ of the
space such that the remainder $Y \setminus X$ is a singleton.
\end{defi}
The following result gives necessary and sufficient conditions for a metric separable space $X$ to have a metric and separable one-point connectification.
\begin{teo} (Knaster \cite{ub})\label{KKK}
Let $X$ be a separable metric space. Then $X$ has a one-point connectification $Y$ which is metrizable and separable if only if $X$ is embeddable
in a separable metric connected space $Z$
as an proper open subset of $Z$.
\end{teo}
Recall that if $U$ is an proper open subset of $X$, then $\mathcal{K}(U)$, $\mathcal{F}_n(U)$ and $\mathcal{F}(U)$ are proper
open subsets to $\mathcal{K}(X)$, $\mathcal{F}_n(X)$ and $\mathcal{F}(X)$
respectively. With this fact in mind, we can prove the following result.
\begin{prop} \label{a5}
If $X$ has a metrizable and separable one-point connectification, then $\mathcal{K}(X)$, $\mathcal{F}_n(X)$ and $\mathcal{F}(X)$ have, each, metric and separable one-point connectification.
\end{prop}
\begin{proof}
If $Y$ is a metric and separable one-point connectification of $X$, then $\mathcal{K}(Y)$, $\mathcal{F}_n(Y)$ and $\mathcal{F}(Y)$ are metric and separable connected spaces (see \cite[Theorem 4.10]{Sub}). As $X$ is an proper open subset of $Y$, then $\mathcal{K}(X)$, $\mathcal{F}_n(X)$ and $\mathcal{F}(X)$ are
proper open subsets to $\mathcal{K}(Y)$, $\mathcal{F}_n(Y)$ and $\mathcal{F}(Y)$ respectively.
By Theorem \ref{KKK} we have that $\mathcal{K}(X)$, $\mathcal{F}_n(X)$ and $\mathcal{F}(X)$ have, each, a metric and separable one-point connectifications.
\end{proof}
It is known that if a space admits a one-point connectification, then it is cohesive. Moreover if an almost zero-dimensional space is cohesive, then it admits a one point connectification (see \cite[Proposition 5.4, p. 22]{jj})
\begin{cor} \label{a6}
If $X$ is a cohesive $AZD$ space, then $\mathcal{F}_n(X), \mathcal{F}(X), \mathcal{K}(X)$ are cohesive $AZD$ spaces.
\end{cor}
\begin{proof}
As $X$ is cohesive $AZD$ space, then $X$ has a one-point connectification. As $X$ is a cohesive $AZD$ space, it has a
one-point connectification. By Proposition \ref{a5} $\mathcal{F}_n(X), \mathcal{F}(X), \mathcal{K}(X)$ have a one-point connectification. Thus $\mathcal{F}_n(X), \mathcal{F}(X), \mathcal{K}(X)$ are cohesive. Furthermore, by the Proposition, \ref{HZD} $\mathcal{F}_n(X), \mathcal{F}(X), $ and $ \mathcal{K}(X)$ are $AZD$.
\end{proof}
\begin{cor}
$\mathcal{F}_n(\mathfrak{E}_{\mathrm{c}})$, $\mathcal{F}_n(\mathfrak{E})$,
$\mathcal{F}(\mathfrak{E}_{\mathrm{c}})$, $\mathcal{F}(\mathfrak{E})$,
$\mathcal{K}(\mathfrak{E}_{\mathrm{c}})$ and $\mathcal{K}(\mathfrak{E})$
are cohesive $AZD$ spaces.
\end{cor}
\begin{proof}
This result follows from Corollary \ref{a6} and from the fact that $\mathfrak{E}_{\mathrm{c}}$ and $\mathfrak{E}$ are cohesive $AZD$ spaces.
\end{proof}
To prove that $\mathcal{K}(X)$ is cohesive in the Corollary \ref{a6}, it was assumed that $X$ is an $AZD$ space. Then, the following natural question.
\begin{pre}
Is there a $ X $ space cohesive such that $ \mathcal{K}(X) $ is not cohesive?
\end{pre}
\section{Proof of Theorem 1.1 }
\begin{defi}
Let $X$ be a set and let $ \tau_1$ and $ \tau_2$
two topologies in $X$, if $(X,\tau_1, \tau_2)$
satisfies that:
\begin{enumerate}
\item $\tau_1 \subset \tau_2$ and $\tau_1$
is a zero dimensional topology such that every point in $ X$ has a neighborhood in $\tau_2$ which is compact with respect to $\tau_1$, we say that $(X,\tau_1, \tau_2)$
has property $C_1$.
\item $\tau_1 \subset \tau_2$ and $\tau_1$
is a zero dimensional topology such that every point in $ X$ has a neighborhood in $\tau_2$
which is complete with respect to $\tau_1$, we say that $(X,\tau_1, \tau_2)$
has property $C_2$.
\end{enumerate}
\end{defi}
Note that the property $ C_1 $ is inherited to closed subsets with respect to $ \tau_1 $ and the property $ C_2 $ is inherited to the $ G_\delta $ subsets with respect to $ \tau_1 $.
To prove Theorem 1.1 we will use the following characterization of $\mathfrak{E}_{\mathrm{c}}$.
\begin{teo}[{\cite[Theorem 3.1, items (1) and (2) ]{j}}]\label{EC} Let $ (\mathcal{E} ,\tau) $ be a topological space.
The following statements are equivalent.
\begin{enumerate}
\item $\mathcal{E}$
is homeomorphic to $\mathfrak{E}_{\mathrm{c}}$.
\item $ \mathcal{E} $ is cohesive and there exist a topology $\mathcal{W}$ in $ \mathcal{E}$ such that $(\mathcal{E}, \tau ,\mathcal{W})$ has property $C_1$.
\end{enumerate}
\end{teo}
For a topological space $(X, \tau)$,
the symbol $\tau_{\mathcal{F}_n(X)}$ denotes the Vietoris topology in $\mathcal{F}_n(X)$.
\begin{obs}\label{masg}
Let $X$ be a set and let $\tau_1$ and $\tau_2$ be
two topologies in $ X $ such that $\tau_1 \subset \tau_2$, then we have to $\tau_{1{\mathcal{F}_n(X)}}\subset \tau_{2{\mathcal{F}_n(X)}}$
\end{obs}
\begin{proof}
Let $\langle U_1,\ldots, U_n\rangle \in \tau_{1{\mathcal{F}_n(X)}}$, then $U_j\in \tau_1$ for each $j\in \{1,\ldots n\}$. As $\tau_1\subset \tau_2$, then $U_j\in \tau_2$ for each $j\in \{1,\ldots n\}$. This implies that $\langle U_1,\ldots, U_n\rangle \in \tau_{2{\mathcal{F}_n(X)}}$. So $\tau_{1{\mathcal{F}_n(X)}} \subset \tau_{2{\mathcal{F}_n(X)}}$.
\end{proof}
With this remark it is not difficult to see that if $\tau$ witnesses the almost zero-dimensionality
of $X$, then $\tau_{\mathcal{F}_n(X)}$
witnesses the almost zero-dimensionality of $\mathcal{F}_n(X)$.
In the proof of the following result, we use Propositions 2.4.2 and 4.1.3 and Theorem 4.2 in \cite{Sub}.
\begin{prop} \label{Pc1}
Let $X$ be a set and let $ \tau_1$ and $ \tau_2$ be two topologies on $X$. If $(X,\tau_1, \tau_2)$ has the property $C_1$ or $C_2$,
then for any $n\in \N$, we have that $(\mathcal{F}_n(X), \tau_{1\mathcal{F}_n(X)}, \tau_{2\mathcal{F}_n(X)})$ has the property $C_1$ or $C_2$,
respectively.
\end{prop}
\begin{proof}
First of all, since $\tau_1$ is a zero-dimensional topology on $X$, then $\tau_{1\mathcal{F}_n(X)}$ is a zero dimensional topology on $\mathcal{F}_n(X)$. Now, let $F\in \mathcal{F}_n(X)$ and $\mathcal{U}\in \tau_{2{\mathcal{F}_n(X)}}$ such that $F\in \mathcal{U}$.
Let us suppose that $\mathcal{U}= \langle U_1,\ldots, U_m\rangle$ and that $F=\{x_1,\ldots, x_l\}$. For each $x_k\in F$ there exist $V_k\in \tau_2$
and a compact (resp., complete) $K_k$ of $X$ with respect to $\tau_1$ such that $x_k\in V_k \subset K_k\subset \bigcap\{U_j:x_k\in U_j\}$. Then $F\in \langle V_{1},\ldots ,V_{l}\rangle\subset \langle K_{1},\ldots ,K_{l}\rangle \subset \mathcal{U}$.
Note that $Z=\bigcup_{j=1}^l K_{j}$
is compact (resp., complete) with respect to $\tau_1$. Then $\mathcal{F}_n(Z)$ is compact(r esp., complete) with respect to $\tau_{1{\mathcal{F}_n(X)}}$. Moreover $\langle K_{1},\ldots ,K_{l}\rangle \subset \mathcal{F}_n(Z)$. Given that $\langle K_{1},\ldots K_{m}\rangle$
is a closed subset with respect to $\tau_{1{\mathcal{F}_n(X)}}$, we have that $\langle K_{1},\ldots K_{m}\rangle$ is compact (resp., complete) with respect that $\tau_{1{\mathcal{F}_n(X)}}$. All this proves that $(\mathcal{F}_n(X), \tau_{1\mathcal{F}_n(X)}, \tau_{2\mathcal{F}_n(X)})$ has the property $C_1$ (resp.,$C_2$).
\end{proof}
\textbf{
Proof of Theorem 1.1}
\begin{proof}
Let $\mathcal{W}$ be a topology in $\mathfrak{E}_{\mathrm{c}}$ which satisfies the conditions in the item (2) of Theorem \ref{EC}. By Remark \ref{masg}, $\mathcal{W}_{\mathcal{F}_n(\mathfrak{E}_{\mathrm{c}})}$ is coaser than the Vietoris topology on $\mathcal{F}_n(\mathfrak{E}_{\mathrm{c}})$. By Proposition \ref{Pc1} and the Corollary \ref{a6}, $\mathfrak{E}_{\mathrm{c}} $ satisfies all conditions in item (2) of Theorem \ref{EC}. Thus, by Theorem \ref{EC} $\mathcal{F}_n(\mathfrak{E}_{\mathrm{c}})$ is homeomorphic to $\mathfrak{E}_{\mathrm{c}}$.
\end{proof}
\section{Proof of Theorem 1.2}
To state the characterization of $\mathfrak{E}$ (see Theorem \ref{EE}) that we are going to use to prove Theorem 1.2 we will give some definitions.
\begin{defi} [{\cite[Definition 3.3, p. 8]{jj}}] If $A$ is a nonempty set then $A^{<\omega}$ denotes the set of all finite strings of elements of $A$, including the null
string $\emptyset$. If $s \in A^{<\omega}$ then $\vert s\vert$ denotes its length. In this context the set $A$ is called alphabet. Let $A^\omega$ denote the set of all
infinite strings of elements of $A$.
If $s \in A^{<\omega}$ and $t \in A^{<\omega} \cup A^\omega$,
then we put $s \prec t$ if $s$ is an initial substring of $t$; that is, there is an $r \in A^{<\omega} \cup A^\omega $
with $ s^\frown r = t$, where $\frown$ denotes concatenation of strings.
If $t\in A^{<\omega} \cup A^\omega$ and $k \in \omega$,
$t\downharpoonleft k \in A^{<\omega}$
is the element of $A^{<\omega} $ characterized by
$ t \downharpoonleft k \prec t$ and $\vert t\downharpoonleft k\vert =k$.
\end{defi}
\begin{defi}[{\cite[Definition 3.4, p. 9]{jj}}]\label{tree}A tree $T$ on $A$ is a subset of $ A^{<\omega}$ that is closed under initial segments, i.e., if $s \in T$ and $t \prec s$ then
$t \in T$. Elements of $T$ are called nodes. An infinite branch of $T$ is an element $r$ of $A^\omega$ such that
$r\downharpoonleft k\in T$ for every $k \in \omega$.
The body of $T$, written as $[T]$, is the set of all infinite
branches of $T$. If $s, t \in T$ are such that $s \prec t$
and $\vert t\vert = \vert s\vert + 1$,
then we say that $t$ is an immediate successor of $s$ and $succ(s)$ denotes the set of immediate successors of $s$ in $T $.
\end{defi}
\begin{defi}
Let $n\geq 2$. If $S_1,\ldots, S_n$ are trees over $A_1,\ldots, A_n$, respectively, and if $s_1=a_1^1,\ldots, a_k^1\in S_1,\ldots, s_n=a_1^n,\ldots, a_k^n\in S_n$, are strings of equal length, then we difine the string $s_1*\ldots*s_n$ over $A_1\times \ldots \times A_n$ by $s_1*\ldots*s_n=(a_1^1,\ldots,a_1^n),\ldots,(a_k^1,\ldots,a_k^n)$. We define the product tree $S_1*\ldots *S_n$ over $A_1\times \ldots\times A_n$ as the partially ordered subset $\{s_1*\ldots*s_k:s_i\in S_i\textit{ for all i}\in \{1,\ldots, n\} \textit{ and } \vert s_1\vert \ldots= \vert s_n\vert \}$ of the alphabet $(A_1\times \ldots\times A_n)^{<\omega}$.
\end{defi}
\begin{obs}\label{tree1}
Let $n\geq 2$, and let $S_1,\ldots , S_n$ trees over $A_1,\ldots , A_n$, respectively. Then $S_1*\ldots * S_n$ is indeed a tree over $A_1\times \ldots \times A_n$. Moreover, the function $\phi: S_1*\ldots* S_n\to S_1\times \ldots \times S_n$ defined by $\phi(s_1*\ldots*s_n)=(s_1,\ldots, s_n)$, is an order isomorphism from $S_1*\ldots* S_n$ to $\phi [S_1*\times * S_n]$.... So, the following statements hold.
\begin{enumerate}
\item Let $s_1, t_1\in S_1, \ldots s_n, t_n\in S_n$ with $\vert s_1\vert= \ldots =\vert s_n\vert$ and $\vert t_1\vert= \ldots =\vert t_n\vert$. Then $s_1*\ldots *s_n\prec_{S_1*\ldots* S_n} t_1*\ldots* t_n$ if only if $s_i\prec_{S_i}t_i$ for all $i\in \{1,\ldots, n\}$.
\item Let $s_1, t_1\in S_1, \ldots s_n, t_n\in S_n$ with $\vert s_1\vert= \ldots =\vert s_n\vert$ and $\vert t_1\vert= \ldots =\vert t_n\vert$. Then $t_1*\ldots t_n\in succ(s_1*\ldots s_n)$ in $S_1*\ldots *S_n$ if only if $t_i\in succ(s_i)$ in $S_i$ for all $i\in \{1,\ldots, n\}$.
\item The body of $S_1*\ldots *S_n$, $[S_1*\ldots *S_n]$, is equal to the set $\{(\widehat{s_1},\ldots,\widehat{s_n} ):\widehat{s_k}\in [S_k] \textit{ for all } i\in \{1,\ldots, n\}\}$.
\item Let $\widehat{t_1}\in [S_1],\ldots, \widehat{t_n}\in [S_n]$, and let $k\in \omega$. Then
$$(\widehat{t_1},\ldots,\widehat{t_n})\downharpoonleft k= (\widehat{t_1}\downharpoonleft k,\ldots,\widehat{t_n}\downharpoonleft k)=\widehat{t_1} \downharpoonleft k *\ldots*\widehat{t_n} \downharpoonleft k.$$
\end{enumerate}
\end{obs}
\begin{defi}
Let $X$ be a space. Let $(A_n)_{n\in \omega}$
a sequence of sets of $X$. We say that $(A_n)_{n\in \omega}$ converges to $x$
if for each open subset $U$ such that $x\in U$ there exists $m\in \omega$ such that $A_k\subset U$ if $m\leq k$.
\end{defi}
\begin{lema}\label{ancla}
Let $ f: X \to Y $ a continuous function and let $(A_n)_{n\in \omega}$ be a sequence of sets of $ X $ converging to $ x $ in $ X $, then $(f[A_n])_{n\in \omega}$
converges to $f(x)$.
\end{lema}
\begin{lema}\label{dnp}
Let $ f: X \to Y $ be a continuous and surjective function.
Let $ A $ and $ B $ be subsets of $ Y $, such that $f^{\leftarrow}[A]$ is nowhere dense in $f^{\leftarrow}[B]$, then $A$ is nowhere dense in $B$
\end{lema}
\begin{proof}
Let us suppose $A$ is not nowhere dense in $B$. Then there exists $y\in A$ and an open subset $V$ of $Y$ such that $y\in B\cap V \subset A$. Let $x\in X$ be such that $f(x)=y$, then $x\in f^{\leftarrow}[B]\cap f^{\leftarrow}[V]\subset f^{\leftarrow}[A] $, which contradicts the hypothesis.
\end{proof}
\begin{defi}[{\cite[Definition 8.1]{jj}}]
Let $T$ be a tree and let $(X_s)_s\in T $ be a system of subsets of a space $X$ (called a scheme) such that $X_t\subset X_s$ whenever $ s\prec t$.
A subset $A$ of $X$ is called an anchor for $(X_s)_s\in T $ in $X $ if either for every $t\in [T ]$ we have $X_{t\downharpoonleft k} \cap A = \emptyset$ for some
$k \in \omega$ or the sequence $X_{t\downharpoonleft k_0},\ldots, X_{t\downharpoonleft n}\ldots $ converges to a point in $X$.
\end{defi}
\begin{teo}[{\cite[Theorem 8.13 , p. 46]{jj}}]\label{EE} A nonempty space $E$ is homeomorphic to $\mathfrak{E}$ if and only if there exists topology $\mathcal{W}$ on $E$ that witnesses the almost zero-dimensionality of $E$ and there exist a nonempty tree $T$ over a countable alphabet and subspaces $E_s$ of $E$ that are closed with
respect to $\mathcal{W}$ for each $s\in T$ such that:
\begin{enumerate}
\item $E_{\emptyset} = E$ and $E_s =\bigcup \{E_t : t \in succ(s)\}$ whenever $s \in T$,
\item each $x\in E$ has a neighborhood $U$ that is an anchor for $(E_s)_{s\in T}$ in $(E, \mathcal{W})$
\item for each $s \in T$ and $t \in succ(s)$, we have that $E_t$ is nowhere dense in $E_s$,
and
\item $E$ is $\{Es : s \in T\}$-cohesive.
\end{enumerate}
Then $E$ is homeomorphic to $\mathfrak{E}$
\end{teo}
By the Theorem \ref{EE}
there is a topology $\mathcal{W}$ for $\mathfrak{E}$ which is witnesses to the almost zero dimensionality of $\mathfrak{E}$, a countable tree $ T $ and a family of sets $\mathcal{E}=\{ E_s :s\in T\}$
which are closed with respect to $\mathcal{W}$ which satisfy the conditions of Theorem \ref{EE} for $\mathfrak{E}$.
Let $\mathcal{W}^n$ the topology of $(\mathfrak{E},\mathcal{W})^n$, $T^n=\{s_1*\ldots *s_n: s_1, \ldots, s_n\in T \textit{ and } \vert s_1\vert =\ldots =\vert s_n\vert\}$ and for each $s_1*\ldots *s_n\in T^n$ let $E_{s_1*\ldots* s_n}$ be the subset $E_{s_1}\times \ldots \times E_{s_n}$ of $\mathfrak{E}^n$.
\begin{lema}\label{at}
The collections $\mathcal{W}^n$, $T^n$ and $\mathcal{E}^n=\{E_{s_1*\ldots* s_n}: s_1*\ldots *s_n\in T^n \}$ satisfy the
conditions of Theorem \ref{EE} for $\mathfrak{E}$.
\end{lema}
\begin{proof}
Since $T$ is a tree over the countable alphabet $A$, $T^n$ is a tree over the countable alphabet $A^n$ (Definition \ref{tree}). It is not difficult to prove that $\mathcal{W}^n$ witnesses that $\mathfrak{E}^n$ is almost zero-dimensional.
Moreover, it is clear that each $E_{s_1*\ldots* s_n}$ is closed in $\mathfrak{E}^n$. On the other hand $E_{\emptyset}=\mathfrak{E}$, so $E_{\emptyset *\ldots* \emptyset}=E_{\emptyset} \times \ldots \times E_{\emptyset}=\mathfrak{E}^n$. Furthermore,
$$E_{s_1}\times \ldots \times E_{s_n}= \bigcup \{E_{t_1}: t_1\in succ(s_1)\}\times \ldots \times \bigcup\{E_{t_n}: t_1\in succ(s_n)\}= $$
$$\bigcup\{E_{t_1}\times \ldots \times E_{t_n}: t_1*\ldots* t_n \in succ(s_1*\ldots * s_n)\}=$$
$$\bigcup\{E_{t_1*\ldots *t_n}: t_1*\ldots* t_n \in succ(s_1*\ldots * s_n)\}$$
(see Remark \ref{tree1}).
\\
Now we are going to prove that $\mathcal{W}^n$, $T^n$ and $\mathfrak{E}^n$ satisfy condition (2) of Theorem \ref{EE}. Let $(x_1,\ldots x_n)\in \mathfrak{E}^n$ and, for each $i\in \{1,\ldots n\}$, let $U_i$ be a neighborhood of $x_i$ which is an anchor for $\mathcal{E}$. Let $\widehat{t}\in T^n$; then $\widehat{t}=(\widehat{t_1},\ldots,\widehat{t_n})$ with $\widehat{t_i}\in [T]$ for each $i\in\{1,\ldots, n\}$ (Remark \ref{tree1}). We define $J=\{i\in \{1,\ldots, n\}: \textit{ there exists } m\in \N \textit{ such that } E_{\widehat{t_i}\downharpoonleft m } \cap U_i=\emptyset\}$. First assume that $J\neq \emptyset$; let $k\in J$, then $E_{(\widehat{t_1},\ldots ,\widehat{t_n})\downharpoonleft k} \cap(U_1\times \ldots\times U_n)=(E_{\widehat{t_1}\downharpoonleft k}\times \ldots \times E_{\widehat{t_n}\downharpoonleft k}\cap(U_1\times \ldots\times U_n)= ( E_{\widehat{t_1}\downharpoonleft k} \cap U_1)\times \ldots \times( E_{\widehat{t_n}\downharpoonleft k} \cap U_n)=\emptyset.$ Now assume that $J=\emptyset$. Since $U_i$ is an anchor for $\mathcal{E}$ for each $i\in \{1,\ldots n\}$, the
sequence $(E_{\widehat{t_i}\downharpoonleft j} )_{j< \omega}$ converges to a point $z_i$ in $\mathfrak{E}$. Therefore, the sequence $(E_{\widehat{t_1}\downharpoonleft j} \times \ldots \times E_{\widehat{t_n}\downharpoonleft j} )_{j< \omega}$ converges to $(z_1,\ldots, z_n)\in \mathfrak{E}^n$. But $E_{\widehat{t_1}\downharpoonleft j} \times \ldots \times E_{\widehat{t_n}\downharpoonleft j} = E_{\widehat{t_1}\downharpoonleft j *\ldots * \widehat{t_n}\downharpoonleft j}=E_{(\widehat{t_1}\downharpoonleft j ,\ldots , \widehat{t_n}\downharpoonleft j)} = E_{\widehat{t}\downharpoonleft j}$ (see Remark \ref{tree1}). Hence, the sequence $(E_{\widehat{t}\downharpoonleft j})_{j<\omega}$ converges to $(z_1,\ldots, z_n)$.
We now verify condition (3) of Theorem \ref{EE}. Suppose $t_1*\ldots *t_n \in succ(s_1*\ldots * s_n)$, then for each $i\in \{1,\ldots n\}$
$t_i \in succ(s_i)$ (Remark \ref{tree1}). Thus, $E_{t_i}$ is nowhere dense in $E_{s_i}$ for each $i\in \{1,\ldots n\}$. This implies
that, $E_{t_1}\times \ldots \times E_{t_n}$ is nowhere dense in $E_{s_1}\times \ldots \times E_{s_n}$.
The condition (4) of Theorem \ref{EE} follows from the Remark 5.2 in \cite{jj}.
\end{proof}
\textbf{
Proof of Theorem 1.2}
\begin{proof}
Let $ n \in \N$ be fixed. We are going to prove that $\mathcal{F}_n(\mathfrak{E})$ is homeomorphic to $\mathfrak{E}$ using Theorem $\ref{EE}$. Because of Lemma \ref{at} we know that if the topology $\mathcal{W}$, the tree $T$ and the family $\{E_s:s\in T\}$ satisfy the conditions of Theorem \ref{EE} for $\mathfrak{E}$, then the product topology $(\mathfrak{E}, \mathcal{W})^n$, which we denote here as $\mathcal{W}^n$, $T^n=\{s_1*\ldots*s_n:s_1,\ldots, s_n\in T\textit{ and } \vert s_1\vert =\ldots= \vert s_n\vert\}$ (Definition \ref{tree}) and the family $\{E_{s_1*\ldots*s_n}:s_1*\ldots*s_n \in T^n\}$ where $E_{s_1*\ldots*s_n}= E_{s_1}\times \ldots \times E_{s_n}$ for each $s_1,\ldots, s_n\in T$
with $\vert s_1\vert =\ldots= \vert s_n\vert$, satisfy all the conditions of Theorem \ref{EE}.
Let, $g:\mathfrak{E}^n\to \mathcal{F}_n(\mathfrak{E})$, be define by as $g(x_1,\ldots, x_n)=\{x_1, \ldots, x_n\}$. Let $\mathcal{W}^\prime$ be the Vietoris topology in $\mathcal{F}_n(\mathfrak{E}, \mathcal{W})$. The tree $T^\prime$ that we are going to consider is $T^\prime = T^n$, and the family $\mathcal{S}^\prime$ of subsets of $\mathcal{F}_n(\mathfrak{E})$ indexing by $T^n$ that we are going to prove to be closed with respect to $\mathcal{W}^n$ is $\mathcal{S}^\prime=\{H_{s_1*\ldots*s_n}: s_1*\ldots*s_n\in T^n \}$ where $H_{s_1*\ldots*s_n}:=g [E_{s_1*\ldots*s_n}]$ for each $s_1*\ldots*s_n\in T^n$. We will prove then that $\mathcal{W}^\prime$, $T^\prime$ and $\mathcal{S}^\prime$ satisfy the conditions required in Theorem \ref{EE} for $\mathcal{F}_n(\mathfrak{E})$.
Indeed, the fact that the Vietoris topology in $\mathcal{F}(\mathfrak{E}, \mathcal{W})$ witnesses that $\mathcal{F}_n(\mathfrak{E})$ is almost zero-dimensional
fallows from the proof of Proposition \ref{HZD}. By Lemma \ref{A2}, $\mathcal{F}_n(\mathfrak{E})$ is $\{g[E_{s_1}\times \ldots\times E_{s_n}]:s_1*\ldots* s_n\in T^n\}$-cohesive. That is, $\mathcal{F}_n(\mathfrak{E})$ is $\{H_{s_1*\ldots*s_n}]:s_1*\ldots* s_n\in T^n\}$-cohesive. Moreover, $T^\prime = T^n$ is a tree over a countable alphabet. On the other hand, for each $s \in T$, $E_s$ is a closed
subset of $(\mathfrak{E}, \mathcal{W})$, hence for $s_1, \ldots , s_n \in T$ satisfying $\vert s_1\vert= \ldots= \vert s_n \vert$, $E_{s_1}\times \ldots\times E_{s_n}$ is closed in $(\mathfrak{E}, \mathcal{W})^n$. Additionaly, since the function $g$ is closed, $g[E_{s_1}\times \ldots\times E_{s_n}]$ is closed in $(\mathcal{F}_n(\mathfrak{E}),\mathcal{W}^\prime)$.
Now we are going to prove that $\mathcal{W}^\prime$, $T^\prime$ and $\mathcal{S}^\prime$ satisfy conditions (1), (2) and (3) of Theorem \ref{EE}.
For $\emptyset=\emptyset *\ldots * \emptyset \in T^n$, $H_{\emptyset *\ldots * \emptyset }=g[E_{\emptyset}\times \ldots \times E_{\emptyset}]=g[\mathfrak{E}]=\mathcal{F}_n(\mathfrak{E})$. Let $s_1*\ldots * s_n$ be an
element in $T^n$.
Using Lemma \ref{at}, we have that $H_{s_1*\ldots *s_n}:=g [E_{s_1*\ldots *s_n}]=g[\bigcup \{E_{t_1*\ldots *t_n}: t_1*\ldots *t_n \in succ(s_1*\ldots *s_n)\}]= \bigcup \{g[E_{t_1*\ldots *t_n}]: t_1*\ldots *t_n \in succ(s_1*\ldots *s_n)\}=\bigcup \{H_{t_1*\ldots *t_n}: t_1*\ldots *t_n \in succ(s_1*\ldots *s_n)\} $. This proves that $\mathcal{W}^\prime$, $T^\prime$ and $\mathcal{S}^\prime$ satisfy condition (1) of Theorem \ref{EE}.
In order to prove that $\mathcal{W}^\prime$, $T^\prime$ and $\mathcal{S}^\prime$ satisfy condition (2) of Theorem \ref{EE}, we take $F\in \mathcal{F}_n(\mathfrak{E})$.
Assume that $F=\{x_1,\ldots, x_n\}$. For each $j\leq n$ there exists a neighborhood $U_j$ of $x_j$ which is an
anchor for $\mathfrak{E}$. Let $ t \in [T^\prime]$; then $ t = (t_1,\ldots, t_n)$ with $t_i\in [T]$ for each $i \in \{1,\ldots n\}$ (Remark \ref{tree1}).
We define $J = \{i\in\{1,\ldots n\}: \textit{ there exists } m \in \N \textit { such that } E_{t_i\downharpoonleft m}\cap U_i = \emptyset\}$.
Assume that $J\neq\emptyset$; let $k \in J$. Then $U_i \cap E_{t_i\downharpoonleft k }= $; for all $i, j\in \{1,\ldots, n\}$. Hence, $g^\leftarrow(F) \
(E_{t_1\downharpoonleft k }\times \ldots\times E_{t_n\downharpoonleft k }) = \emptyset$. Thus, there exists neighborhoods $ V_j $ of $x_j$ for each $j $ such that if
$w \in g(F)$, then there exists a permutation $h:\{1,\ldots n\}\to \{1,\ldots n\}$, such that $w \in V_h =
V_{h(1)}\times \ldots \times V_{h(n)} \subset \mathfrak{E}^n\setminus ( E_{t_1\downharpoonleft k }\times \ldots \times E_{t_n\downharpoonleft k })$.
Therefore, $V =\bigcup_{h\in P} V_h \subset \mathfrak{E}^n\setminus ( E_{t_1\downharpoonleft k }\times \ldots \times E_{t_n\downharpoonleft k })$ (where $P$ is the set of permutations of $\{1,\ldots, n\}$). Let $\mathcal{U}= \langle W_1,\ldots, W_n\rangle$, where $W_i = U_i\cap V_i$, thus $F\in \mathcal{U}$, and $g(E_{t_1\downharpoonleft k }\times \ldots \times E_{t_n\downharpoonleft k })\cap \mathcal{U} =\emptyset $.
That is $H_{t\downharpoonleft k}\cap \mathcal{U} = \emptyset$. Then $ \mathcal{U}$ is an anchor.
Now suppose that $J = \emptyset$. For every $i \in \{1,\ldots n\}$, since $U_i$ is an anchor, the sequence $(E_{t_i\downharpoonleft j} )_{j<\omega}$
converges to a point $y_i$. Thus, $(E_{t_1\downharpoonleft j}\times \ldots \times E_{t_n\downharpoonleft j} )_{j<\omega}$ converges to $(y_1,\ldots, yn)$. Since $g$ is
continuous by the Lemma \ref{ancla}, we have that $(g((E_{t_1\downharpoonleft j}\times \ldots \times E_{t_n\downharpoonleft j}) )_{j<\omega}$ converges to $g(y_1,\ldots, y_n)$. That is, $(H_{t\downharpoonleft j})_{j<\omega}$
converges to $g(y_1,\ldots, y_n)$.
Finally, we will prove that$\mathcal{W}^\prime$, $T^\prime$ and $\mathcal{S}^\prime$ satisfy condition (3) of Theorem \ref{EE}. If $t_1* \ldots * tn \in succ(s_1*\ldots * s_n)$, then for each $i \in \{1,\ldots n\}$, $t_i \in succ(s_i)$ (Remark \ref{tree1}). Thus, $E_{t_i}$ is nowhere dense in $E_{s_i}$ for each $i \in \{1,\ldots n\}$. This implies that, for each permutation $h: \{1,\ldots n\}\to \{1,\ldots n\}$,
$E_{t_{h(1)}} \times \ldots \times E_{t_{h(n)}}$ is nowhere dense in $E_{s_{h(1)}} \times \ldots \times E_{s_{h(n)}}$. Therefore,
$\bigcup_{h\in P} (E_{t_{h(1)}} \times \ldots \times E_{t_{h(n)}})$ is nowhere dense in $\bigcup_{h\in P} (E_{s_{h(1)}} \times \ldots \times E_{s_{h(n)}})$
where $P$ is the set of permutations of $\{1,\ldots n\}$.
Note that
$$g^\leftarrow[g[E_{t_1}\times \ldots \times E_{t_n}]]= \bigcup_{h\in P} (E_{t_{h(1)}} \times \ldots \times E_{t_{h(n)}})\subset g^\leftarrow[g[E_{s_1}\times \ldots \times E_{s_n}]]$$
and
$$g^\leftarrow[g[E_{s_1}\times \ldots \times E_{s_n}]]= \bigcup_{h\in P} (E_{s_{h(1)}} \times \ldots \times E_{s_{h(n)}}).$$
As the finite union of nowhere dense sets is nowhere
dense, by the Lemma \ref{dnp} $g[E_{t_1}\times \ldots \times E_{t_n}]$ is nowhere dense in $g[E_{s_1}\times \ldots \times E_{s_n}]$. By Theorem \ref{EE}, all the above proves that $\mathcal{F}_n(\mathfrak{E})$ is homeomorphic to $\mathfrak{E}$.
\end{proof}
\section{ Applications}
\begin{defi}
Let $X$ be a space.
A space $ Y $ is called an $ X $ -factor, if there is a space $ Z $ such that
$ Y \times Z $ is homeomorphic to $ X $.
\end{defi}
It is known that $X$ is a $\mathfrak{E}_{\mathrm{c}}$-factor or a $\mathfrak{E}$-factor if only if $X$ can be embedded as a closed
subset of $\mathfrak{E}_{\mathrm{c}}$ or $\mathfrak{E}$ respectively (see {\cite[Theorem 3.2]{j}} and {\cite[Theorem 9.2]{jj}}). Using these facts we have the following corollary.
\begin{cor}
Let $n\in \N$, if $X$ is a $\mathfrak{E}_{\mathrm{c}}$-factor or a $\mathfrak{E}$-factor, then $\mathcal{F}_n(X)$ is a $\mathfrak{E}_{\mathrm{c}}$-factor or $\mathfrak{E}$-factor
respectively.
\end{cor}
\begin{proof}
If $X$ is a $\mathfrak{E}_{\mathrm{c}}$-factor or a $\mathfrak{E}$-factor, then $X$ can be embedded as a closed subset of $\mathfrak{E}_{\mathrm{c}}$ or $\mathfrak{E}$ respectively. Then $\mathcal{F}_n(X)$ can be embedded as a closed subset of $\mathcal{F}_n(\mathfrak{E}_{\mathrm{c}})$ or $\mathcal{F}_n(\mathfrak{E})$ respectively. By Theorem \ref{EE1} and Theorem \ref{ee1}, we have that $\mathcal{F}_n(\mathfrak{E}_{\mathrm{c}})$ and $\mathcal{F}_n(\mathfrak{E})$
are homeomorphic to $\mathfrak{E}_{\mathrm{c}}$ and $\mathfrak{E}$ respectively. Thus $\mathcal{F}_n(X)$ is $\mathfrak{E}_{\mathrm{c}}$-factor or $\mathfrak{E}$-factor respectively.
\end{proof}
Another known fact is that any open subset of $\mathfrak{E}$ or $\mathfrak{E}_{\mathrm{c}}$
is homeomorphic to $\mathfrak{E}$ or $\mathfrak{E}_{\mathrm{c}}$ respectively (see {\cite[Theorem 3.2]{j}} and {\cite[Corollary 8.15]{jj}}). Using that fact, and Theorems \ref{EE1} and \ref{ee1} we have the following corollary.
\begin{cor}
Let $X_1=\mathfrak{E}$ and $X_2=\mathfrak{E}_{\mathrm{c}}$ and $n\in \N$
For each $k< n$ we have that $\mathcal{F}_{n}(X_i)\setminus \mathcal{F}_{k}(X_i) $
is homeomorphic to $X_i$.
\end{cor}
\begin{proof}
This follows from Theorems \ref{EE1} and \ref{ee1} and the fact that $\mathcal{F}_{n}(X_i)\setminus \mathcal{F}_{k}(X_i) $ is an open subset of $\mathcal{F}_{n}(X_i)$.
\end{proof}
To finish we will mention some comments about $\mathcal{F}(\mathfrak{E})$, $\mathcal{F}(\mathfrak{E}_{\mathrm{c}})$, $\mathcal{K}(\mathfrak{E})$ and $\mathcal{K}(\mathfrak{E}_{\mathrm{c}})$.
As $\mathcal{F}(\mathfrak{E}_{\mathrm{c}})$
is of the first category and $\mathfrak{E}_{\mathrm{c}}$ is not, then $\mathcal{F}(\mathfrak{E}_{\mathrm{c}})$ cannot be homeomorphic to $\mathfrak{E}_{\mathrm{c}}$.
Neither does it happen that $\mathcal{F}_n(\mathfrak{E}_{\mathrm{c}})$ is not homeomorphic to $\mathfrak{E}$ because
$\mathcal{F}_n(\mathfrak{E}_{\mathrm{c}})$ is $G_{\delta\sigma}$ and $\mathfrak{E}$ is not. The space $\mathfrak{E}$ is not $G_{\delta\sigma}$ because it contains a closed copy of $\mathfrak{E}$.
On the other hand by Theorems \ref{EE1} and \ref{ee1}, if $X_1=\mathfrak{E}$ or $X_2=\mathfrak{E}_{\mathrm{c}}$, then $X_i^k$
is homeomorphic to $\mathcal{F}_{k}(X_i)$ for each $k\in \N$.
Also we have $\mathcal{F}(X_i)=\bigcup_{k\in \N}\mathcal{F}_{k}(X_i)$ and let $a_i$ be a fixed point in $X_i$ for $i \in \{1,2\}$. For each $k \in\N$ and each $i \in\{1,2\}$, let $Z_k^i$ be the subspace $\{(x_j)_{j\in \omega} \in X_i^\omega: x_j=a_i, j>k\}$ of $X^\omega_i$. Then $\mathcal{F}_{k}(X_i)$ is homeomorphic to $Z_k^i$ from this the following question arises.
\begin{pre}
Let $X_1=\mathfrak{E}$ and $X_2=\mathfrak{E}_{\mathrm{c}}$. is $\mathcal{F}(X_i)$ homeomorphic to the subspace
$\bigcup_{k\in \N} Z^k_i$ of $X^\omega_i$ for $i = 1, 2$?
\end{pre}
Also, we note that $\mathcal{K}(\mathfrak{E})$ is homeomorphic to neither $\mathfrak{E} $ nor $\mathfrak{E}_{\mathrm{c}}$, and moreover $\mathcal{K}(\mathfrak{E})$ is not a factor of either spaces.
Because $\mathfrak{E}$ has a closed copy of $\Q$ (see \cite{KKK}), then $\mathcal{K}(\mathfrak{E})$ has a closed copy of $\mathcal{K}(\Q)$. On the other hand it is known that $\mathcal{K}(\Q)$ is not a Borel set (see \cite{KKK}), thus $\mathcal{K}(\mathfrak{E})$
is not a Borel set. But $\mathcal{K}(\mathfrak{E}_c)$ is factor of $\mathfrak{E}$ since it is possible to embedd it as a closed subset of $\mathfrak{E}$. So far we only know that space $\mathcal{K}(\mathfrak{E}_c)$ is Polish, $AZD$ and cohesive, what we have to prove is whether $\mathcal{K}(\mathfrak{E}_c)$ satisfy property $C_1$. So have the fallowing question:
\begin{pre}
Is $\mathcal{K}(\mathfrak{E}_c)$ homeomorphic to $\mathfrak{E}_{\mathrm{c}}$ or $\mathfrak{E}_{\mathrm{c}}^\omega$?
\end{pre}
| 196,369
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What are your favorite holiday traditions? Is it gathering the family for your annual holiday card photo? Perhaps it’s DIY-ing your own holiday sweater? Or maybe, it’s simply decorating your home and giving to someone in need?
What better way for the team at Adelberg Montalvan to get into the holiday spirit than by joining in these holiday traditions with YOU! This holiday season, we’re proud to bring you not one, not two, not even three – but FOUR holiday initiatives you can partake in! Read on for details below:
Tag Us in Your Holiday Home Decor
Deadline: January 1st
Do you have an Elf on the Shelf? What about a Mensch on a Bench? Or perhaps your lawn looks like Clark Griswold’s in National Lampoon’s Christmas Vacation. Whatever holiday decor you have throughout your home, be sure to tag us on social media for a chance to be featured on our social pages!
Donate a Toy in Nesconset
Deadline: December 12th
Once again, our offices will be participating in Project Toy’s & ReesSpecht Life’s Toy Drive this year! Please drop off a new, unwrapped gift for a local child (newborn – 16 years of age) in need.
Holiday Card Contest
Deadline: January 1st
Prize: $250 Visa Gift Card
Send in your family’s holiday card by January 1st and be entered into a raffle to win a $250 visa gift card!!
Decorate a Holiday Sweater
Deadline: December 7th – December 27th
Help our doctors get festive! We’ll be creating an in office holiday sweater decoration station (in our lobby) from December 7th through December 27th! On your next visit, encourage your kids to stop and decorate a sweater for our doctors to wear later this month around the office. Stay tuned for photo updates on our Facebook page!
If you have any questions regarding the above, please give any of our three locations a call!
Otherwise, if you have any additional questions or concerns regarding your child’s oral care or if you need to schedule an appointment, contact us. At Adelberg Montalvan Pediatric Dental, our purpose is to provide exceptional mental and dental care that can impact your child’s life forever.
| 169,055
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TITLE: $C([a,b]), \| \cdot \|_\infty$ is complete
QUESTION [2 upvotes]: I have to prove that $(C([a,b]), \| \cdot \|_\infty )$ is complete. I proved that a Cauchy sequence converges pointwise. Now, let's call $f$ the pointwise limit. I have: \begin{align}
| f_n(x)-f_m(x) | \le \epsilon \ \ \forall n,m > n_0, \forall x
\end{align}
So by the continuity of $| \cdot|$, I can let $n\to\infty$, so I obtain $ f_n\to f $ in $ \|\cdot\|_\infty$. Now I have to prove that $f$ is continuous. Here I'm not completely sure. Let $x_n \to x$ and fix $\epsilon$, I have:
\begin{align}
|f(x_n)-f(x)| \le |f(x_n)-f_m(x_n)|+|f_m(x_n)-f_m(x)|+|f_m(x)-f(x)|
\end{align}
By the pointwise convergence, I can select $f_m$ s.t.
\begin{align}
& |f(x_n)-f_m(x_n)|\le \epsilon/3, |f_m(x)-f(x)|\le \epsilon/3 \ \ \text{so,} \\&|f(x_n)-f(x)| \le |f_m(x_n)-f_m(x)|+2\epsilon/3
\end{align}
So, by the continuity of $f_m$, $ \exists n_0 \ s.t. \ \ |f_m(x_n)-f_m(x)| \le \epsilon/3$.
Is this proof okay? Is there some other way to prove the continuity of $f$?
REPLY [2 votes]: Your proof is not valid because $m$ depends 0n $n$.
Choose $n_0$ such that $|f_n(x)-f_m(x)|<\epsilon$ for al $x$ for all $n,m \geq n_0$. You have already noticed that $f(x)=\lim_{n \to \infty} f_n(x)$ exists. Letting $n \to \infty$ we get $|f_m(x)-f(x)|\leq \epsilon$ for al $x$ for all $m \geq n_0$. In particular $$|f_{n_0}(x)-f(x)|\leq \epsilon$$ for all $x$. Now use the inequality
$|f(x)-f(y)|$$ $$ \leq |f(x)-f_{n_0}(x)|+|f_{n_0}(x)-f_{n_0}(y)|+|f_{n_0}(y)-f(y)|\leq 2\epsilon + |f_{n_0}(x)-f_{n_0}(y)|$.
Now complete the proof using the fact that $f_{n_0}$ is continuous.
(We have already proved that $f_m \to f$ uniformly).
| 214,384
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TITLE: Given the inhomogeneous linear differential equation, $dx/dt + x = 2t$, use the given ansatz to find a solution.
QUESTION [0 upvotes]: Like the image shows, I'm given an inhomogeneous differential equation and an initial condition, but the task is to find the general solution to an homogenous equation. What is that supposed to mean? Do I just substitute the 1 with a 0 randomly and proceed like that? What about the initial condition I was given. If there was no text the obvious thing would be to simply find a general solution of the ODE and the related special solution with the given i.c, but I'm instead asked to find 3 different equations that I can't see fitting in the problem. Help please?
[1]: https://i.stack.imgur.com/Tdz4e.png
REPLY [0 votes]: The instructions are fairly clear as to what is expected and how the process of finding a solution works. In this case given an equation of the type
$$ \frac{d \, x}{dt} + a \, x = p(t) \hspace{10mm} x(0) = 0 $$
then first find the solution of $$ \frac{d \, x}{dt} + a \, x = 0 $$ which happens to be $ x(t) = e^{-a \, t}$. Now the method is to find the complete solution of the original equation with a solution of the form $$ x(t) = f(t) \, e^{- a t}. $$ This leads to
\begin{align}
\frac{d x}{dt} + x &= p(t) \\
-a \, f \, e^{-a t} + \frac{d f}{dt} \, e^{-a t} + f \, e^{-a t} &= p \\
\frac{d f}{dt} + (1-a) \, f &= p \, e^{a t} \\
f(t) &= e^{(a-1) \, t} \, \left( c_{0} + \int^{t} e^{u} \, p(u) \, du \right).
\end{align}
Using $x(0) = f(0) = 0$ gives
$$ f(t) = e^{(a-1) t} \, \int_{0}^{t} e^{u} \, p(u) \, du. $$
The final solution
$$ x(t) = e^{- a t} \, \int_{0}^{t} e^{u} \, p(u) \, du. $$
From here it is a matter of equating $a$ and $p(t)$ to obtain the specific solution to the problem.
| 154,609
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The Robin Hood Tax: A Video
This video by National People’s Action features IIRON President, Rev. Marilyn Pagán-Banks and SOUL President, Rev. Booker Vance. It calls for a tax of less than one-half of one percent on financial transactions. Such a tax could raise billions of dollars for infrastructure, health care and other social goods. It would mean “small change for banks, but big change for us”.
| 188,159
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.
Peter said,
June 13, 2011 at 12:41 am
Hope you get over it, I miss your regular blogs.
(I realise that that doesn’t help at all)
Paul Sagar said,
June 13, 2011 at 12:44 am
Peter mate, if I wanted your help I’d ask for your money.
Jed said,
June 13, 2011 at 2:17 am
Shut up and stop pussying out of tennis.
Mercer Finn said,
June 13, 2011 at 2:25 am
Two things, from my own experience keeping up a far less illustrious and intelligent blog than yrs.
One: it’s not the act of writing, it’s the subject-matter. Writing is a tool that helps me work out problems (intellectual and personal), so I write when I have them. If yr experiencing a “growing apathy with intellectual pursuits”, it’s the apathy that’s the problem. Figuring that one out might mean writing about it…
Two: forcing myself to post, particularly if I feel I’ve got an interested audience out there, often makes me write absolute garbage. Then again forgetting the expectations of my (largely imagined) audience and forcing myself to post can often lift the block. In other words, it’s the expectations that are the problem.
This might not help at all, and I might be way off, but thought I should share. I hope you work it out.
cbdhthekmg said,
June 13, 2011 at 7:21 am
Rather than not writing about being depressed, it sounds as if you are actually depressed.
Go see someone about it. Even mild depression isn’t something that just has to be suffered any more.
Luis Enrique said,
June 13, 2011 at 8:59 am
Tim W may be right. Otherwise, hopefully it will turn out to be a bought of bad weather, that will be blown away one day soon.
As to the writing thing, there’s a lot I sympathise with here. On my part, even if I tell myself I write for my own pleasure, I often feel sick of it, verging on disgusted with my self and the crap I churn out. Boredom with my self, a mix of thinking I really don’t have anything to say interesting enough to be worth writing down, and there’s no point writing stuff people don’t read. And, as a fellow academic, disatisfaction on disenchantment with that too. I find it pretty easy to fall out of love with acdemia and intellectuals.
I think it is possible to waste time writing, so maybe the cure is to take a holiday, and write less generally, hope for something to rekindle your enthusiasm for your work. My phd supervisor was always warning me of the need to take lots of breaks, becuse it’s very easy to burn out, by which he just meant get fed up with it all.
There’s a small chance that this is nature’s way of telling you not to be an academic. In which case you can be pragmatic, give yourself a couple of years to make sure and start thinking about other jobs, meantime. I tend to find that thinking properly about what else I could be doing makes me realise I actually like what I am doing more than I think.
Phil said,
June 13, 2011 at 9:11 am
I’m mildly depressed myself at the moment – I thought it was the effect of applying for five jobs in a row & not getting a single interview, but application #6 got me an interview & I’m still feeling like there’s not much point to anything.
HE is just not a good place to be at the moment. Of course, lots of other professions are similarly under threat, and many of those are places I wouldn’t like to work at the best of times. But in a way that’s the point – this is my dream career, I’ve worked hard at it (after working really hard to get into it), and it’s rewarding me by threatening to spit me out and leave me with nothing. (I should say that, at the time of writing, the threat isn’t immediate or aimed at me personally – but it’s always there.) The point isn’t so much the threat itself as the contradiction between my belief in the job & the job’s evident lack of belief in me – psychic contradictions like that are really hard to live with, and we tend to resolve them by changing what we can change – in this case, by questioning my commitment to the job. (Which I don’t want to question – but pushing back just intensifies the contradiction.)
What makes it all the worse is that this situation was brought into being by the Coalition Agreement, which is itself an emblem of this kind of free-floating but unchallengeable contradiction – Tory voters didn’t vote for the coalition’s HE policies, Lib Dem voters openly and specifically voted against them, and yet here they somehow are. Emotionally these things are wrong – wrong in the deep sense that they shouldn’t be there – but there doesn’t seem to be any alternative to submission.
I think there are a lot of confused, demotivated, apathetic and mildly depressed people in and around HE at the moment, and in the public sector generally.
Paul Sagar said,
June 13, 2011 at 9:23 am
Thanks all* for the kind words (amused that TW’s image perhaps requires him to express kindness anonymously).
I’m actually not depressed at the moment. I was, and often I am, but not at the moment. I’m just perplexed by the lack of writing in my motivation set these days. The post was actually supposed to be sort of funny and self-depracating, so sorry for making it look like a cry for help.
*(except Jed, who is about to be the Roddick to my Murray.)
Franlydie said,
June 13, 2011 at 9:32 am
(1) if you are writing for pleasure then you must only write WHEN/ IF it gives you pleasure; otherwise just go and so something else, which will give you pleasure (like READ a book – possibly until you find other people’s writing makes you want to write again) (and write better than them).
(2) if you are going to be writing for a living, then who said anything you would get paid to do was going to be enjoyable as well? The vast majority of people in employment at the moment dislike some, or all, of what they have to do, but they have no alternative but get on with it. Other writers get fed up; it’s called writers’ block.
(3) It must have occurred to you that your writing gave others something they enjoyed. There is a possibility your ability to write well entails a form of duty to do it as an act of generosity.
(4) Speak to someone about depression. Even if you “know” it won’t make any difference. If you don’t do anything about it, then nothing will change, that is the one certainty.
Novak said,
June 13, 2011 at 11:13 am
Paul I’d be more worried about your limp forehand
Torquil Macneil said,
June 13, 2011 at 12:20 pm
“I’m actually not depressed at the moment. I was, and often I am, but not at the moment.”
Paul, depressives are notoriously bad at self-diagnosis, often experiencing depression as revelation, waking up to how things really are. You probably know this but it may not help if you can’t feel it, and depression would get in the way of that. A suddenly developed inability to enjoy something such as this is a huge indicator of depression. In other words, whatever you think, go and get some professional advice. This is the aberration and it will get better, but you may need help. Don’t let it develop. Don’t adjust your outlook to normalise the condition, that is a classic depressive spiral, part iof the seductiveness of the condition that makes it so deadly.
Good luck.
Barnaby Walker said,
June 13, 2011 at 2:07 pm
I wouldn’t normally comment, but I have had (and still have) similar problems with work. Once or twice a year I have a period of a month or so where I pretty much lose interest in reading/thinking about/writing philosophy. (I’d a philosophy PhD student, btw.) I used to get really worried about this – “I’m not cut out to be a philosopher”, “I’ll never be interested in any of this stuff again” – but I’ve learned, over time, that my normal levels of curiosity and motivation return. The key thing is to know how to manage yourself when your “motivational set”, as you put it, goes wrong. Sometimes a loss of motivation indicates that you actually need, and deserve, a break. In my own case, a loss of motivation often stems from the fact that I put too much pressure on myself. Having unrealistically high standards undermines my motivation because when, as is inevitable, I fail to meet those standards, I think that I’m stupid – and if I’m stupid anyway, what’s the point in trying?
Not sure if either of things apply to you, but given how much work you do it wouldn’t surprise me if (to put it crudely) your mind is simply telling you that it needs some time off. Given your obvious commitment to all things academic I have no doubt that your desire to write will return again soon.
Agog said,
June 13, 2011 at 4:12 pm
I think this a necessary by-product of self-awareness. After a while we get more reflexive about what we’re doing, and it doesn’t feel the same any more. But I find the dislocation/ennui/whatever tends to wear off after a while as I find other things to worry about. I developed a really negative mindset during my doctoral work, and writing my thesis was torture. But then other pressures and annoyances came along: collaborations, supervision etc etc.
I guess you could try giving up on self-awareness. Plenty seem to manage without it….
David Owen said,
June 13, 2011 at 4:38 pm
Sometimes it just goes like that and nearly all academics have such spells, it counterbalnces all those times when writing lifts you up like (almost) nothing else and time flows by as if barely touch you at all.
Accept it – and go work on the forehand.
Phil said,
June 13, 2011 at 11:26 pm
Does anyone here not play tennis with Paul?
David Owen said,
June 14, 2011 at 8:51 am
I don’t play tennis with Paul – although happy to do so – just looking for a physicla activity to get him away from work
Thalia said,
June 15, 2011 at 10:06 am
Hi there –
This isn’t in direct response to any one post but something about your writing really reminds me of a couple of people. If you haven’t read them, I think you might enjoy these two books:
Out of Sheer Rage – Geoff Dyer (his bit on Oxford dons is particuarly brilliant)
Reality Hunger, A Manifesto – David Shields
This seems entirely random and you’ve probably got a reading list as long as your leg, but I always like book recommendations and these two are, to my mind, prett special.
Thalia
| 291,128
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Micols Miraculous Journey
Dear Editor,
In April 2010, I was diagnosed with a very aggressive Uterine Cancer. The diagnosis was further complicated by the fact that, unbeknownst to me, I had contracted Lyme Disease as a child. Immediately, I turned my focus toward healing my body. Little did I know what lay ahead. After a radical hysterectomy, I would have a slow and painful recovery. Then, eleven months after my initial diagnosis, the cancer would recur. I was diagnosed with Stage IIIB Metastatic Cancer and a 15 percent chance of survival. Many of you may remember that a plea went out to the Topanga community for donations toward my care. My mother, Karen Moran, held a fundraiser at the Will Geer Theatricum Botanicum. Many musicians, as well as other community members, offered their talents and skills to an afternoon art auction, luncheon buffet and wonderful concert. I was unable to attend.
I was fighting for my life.
Now that the battle is over and I am, once again, strong, I wish to show my gratitude by thanking all of you who contributed to my survival and let you know that I have created a non-profit organization called Cancer Warrior, Inc. Having first-hand experience navigating the difficult waters where finite decisions must be made, I wish to help others along the way. A workbook is on its way!
For details about Micols miraculous journey, visit Cancerwarriorinc.org, where her story is shared in a radio interview.
Micol Sanko
| 330,112
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Question
China Inn and Midwest Chicken exchanged assets. Midwest Chicken received equipment and gave a delivery truck. The fair value and book value of the delivery truck given were $ 31,000 and $ 32,600 (original cost of $ 37,000 less accumulated depreciation of $ 4,400), respectively. To equalize market values of the exchanged assets, Midwest Chicken received $ 9,000 in cash from China Inn. At what amount did Midwest Chicken record the equipment? How much gain or loss did Midwest Chicken recognize on the exchange?
Answer to relevant QuestionsThe balance sheet of Cedar Crest Resort reports total assets of $ 840,000 and $ 930,000 at the beginning and end of the year, respectively. The return on assets for the year is 20%. Calculate Cedar Crest’s net income for ...Brick Oven Corporation was organized early in 2015. The following expenditures were made during the first few months of the year: Attorneys’ fees to organize the corporation ..... $ 9,000Purchase of a patent ...Togo’s Sandwiches acquired equipment on April 1, 2015, for $ 18,000. The company estimates a residual value of $ 2,000 and a five-year service life. Required: Calculate depreciation expense using the straight-line method ...Midwest Services, Inc., operates several restaurant chains throughout the Midwest. One restaurant chain has experienced sharply declining profits. The company’s management has decided to test the operational assets of the ...The Donut Stop acquired equipment for $20,000. The company uses straight-line depreciation and estimates a residual value of $4,000 and a four-year service life. At the end of the second year the company estimates that the ...
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| 184,042
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March 2004
Synoptic geography in practice: the Water Framework Directive
While schools and colleges were closed for Christmas, something rather significant happened to Britain's rivers and floodplains. In December 2003, the European Water Framework Directive (WFD) became law.
For the first time, there is now a requirement for drainage basins in Britain to be managed in a unified and coherent way (although at this point it is unclear whether the WFD will eventually come to consist of a voluntary code of practice or a series of statutory laws). As part of this process, a total of 11 river basin districts have been designated as covering England and Wales (a ‘district’ may include more than one actual catchment area: for instance, the Mersey basin and River Lee basins comprise a single district).
The Directive requires that all European countries establish and prepare river basin plans with the aim of achieving good ecological quality of waters. Those plans must be supported by economic analyses, and there must eventually be cost recovery within each of the three sectors of water user groups: agriculture, industry and domestic households. The Water Framework Directive is potentially the most significant thing to ever have happened to river management in the UK and it is to be managed by DEFRA (Department of the Environment, Farming and Rural Affairs).
The point of the WFD is to establish integrated river basin management plans throughout all of Europe. A catchment is a dynamic open system in which there are three sets of flows occurring, namely water (surface run-off, groundwater flow and river discharge), sediment (bank-side mass movement and subsequent transport of solid load and solutes) and pollutants (both solid and solute).
Different land and water user groups (such as farmers, house-builders, industrialists and nature-lovers amongst others) modify these flows in ways that may adversely impact upon other users in a basin. It is the widespread lack of joint planning by land and water users along European rivers, including Britain, which has given rise to the call for Integrated Water Resource Management (IWRM). Any single action by one user group is likely to have consequences both up and downstream and on other functional uses.
River Basin Management Plans now need to be established that satisfy the needs of all different parties while meeting good sustainable ecological objectives. This is a daunting task when one considers the actual number of different parties that need to be involved.
Overall, England and Wales has 35,000 km of main river and other minor watercourses and all are small compared with large continental rivers such as the Ganges or Amazon (the mean flow on the largest river, the Trent, is only 93 cumecs). The River Thames extends for 340 km from its source to the sea, draining an area of about 13,600 sq km compared with the Amazon which drains about 7,050,000 sq km over a length of about 6,500km! For this reason, great pressure is placed on British watercourses, all of which must drain and service a mixture of agricultural, recreational and urban land.
There has been substantial new urban development over the last 50 years within UK floodplains, presenting further difficulties for integrated catchment management. For example, floodplain development since 1947 in the town of Maidenhead on the Thames has resulted in the construction of the UK’s most expensive riverline flood alleviation scheme ever, the Jubilee river, an 11km flood relief channel! The Environment Agency estimates that there are now 1.85 million houses, 185,000 commercial properties and 5 million people at risk in flood plains.
Development in flood plains has increased flood risks by reducing flood storage capacity and by increasing run off. Changes in agricultural land management and practices have also contributed to increased run off although the percentage of tree cover in Britain has increased from 3% to 11% since 1919 (see “Conifers for the Chop”) which may offset some interception storage loss. However, much afforestation has been in highland areas.
Currently, the nation is facing a housing crisis with prices rocketing as demand grows due to demographic changes (see “New Towns for new times”). Central government policy favours housing development on ‘brownfield’ or previously developed land, many of which are riverside sites. For example, a huge expansion of London to the east along the Thames estuary, the Thames Gateway, is planned for an area currently in part protected by tidal defences to the 1 in 1,000 year level. Inevitably, more run-off will be generated.
In the last 50 years, major efforts have gone into improving the water quality of the UK’s rivers, particularly urban ones, highly polluted by:
For example, in 1950, the River Thames in London was so polluted as to be almost totally devoid of fish life. However, a reduction in industrial and commercial activity along the river and improvements to sewage treatment works and other actions have resulted in the Thames being claimed as one of the cleanest Metropolitan Estuaries in the world supporting over a hundred fish species including Atlantic salmon (Tunstall, 2000). Seals were spotted in the Mersey over Christmas (The Guardian, 01 January 2004).
During the last decade there has been a significant improvement in the chemical and biological quality of classified rivers in England and Wales. Between 1990 and 2000, the percentage of river length classed as below good chemical quality fell from 52% to 32%. Over the same period, the percentage of river length failing to achieve good biological quality declined from 44% to 33%.
Controversial issues involving the environment and of geographic significance are often presented via role-plays that sometimes hinge on short extracts from textbooks. The groups represented can be simplified and stereotypical. Issues and policies are rarely simple but through the web it is possible to gain a whole range of views from interest groups. The sources here are authentic, real people and organisations with real interests (for whatever motive) involved in real decision-making.
This unit uses data and case material from the Internet as the focus for teaching and learning about a specific issue.
Background to River Basin Management Plans
Government agency plans for the integrated management of whole water body systems (from areas of surface run-off through to estuaries and the sea), required by recently introduced European water legislation.
The Water Framework Directive (2000/60/EC) is the most significant piece of European water legislation for over 20 years. Although it is only just being implemented (see below) it is useful area of study here because it will fundamentally overhaul the management of the water environment in the UK.
The Directive embodies the concept of integrated river basin management. It sets out environmental objectives for water status based on:
By taking an inclusive approach to managing water as it flows through catchments from lakes, rivers and groundwater to estuaries and the sea, the Directive aims to:
This will be achieved by:
River Basin Management Plans are to be established that need to satisfy the needs of all different parties while meeting good sustainable ecological objectives. This is a daunting task when you consider the actual number of different parties that need to be involved.
Table 1 lists all of the organisations that participated in a forum set up by DEFRA in June 2001 for key stakeholders to discuss issues relating to water policy in general in England. Its membership covers a wide range of interests, including the environment, the water industry, agriculture, the countryside, and industry.
Table 1: Organisations currently represented on the Water Framework Directive Forum
1. Why do the organisations (stakeholders) in Table 1 want to be involved in decision-making on water policy?
Categorise the different groups from Table 1 into by those interested in:
Use Supply DemandHouseholdCommercialIndustrialAgriculturalLeisure and recreationConservation
2. Choose one organisation (stakeholder) from each category and describe how it uses water and why it might be interested in influencing policy? You may wish to visit some of the organisations websites to find out more about them.
3. For a drainage basin of your choice (or one you have studied) identify the main issues and stakeholders.
4. Is it possible that a River Basin Management Plan can be devised to manage, in a coherent way, satisfying all the different groups? For the drainage basin you have chosen prepare a PowerPoint presentation to summarise the main issues and suggest ways that it might be managed.
Written by Dr Simon Oakes who works for the Flood Hazard Research Centre (Middlesex University) and Mander Portman Woodward School (London). He is a senior examiner for Edexcel. Colin Green (Middlesex University) provided some of the data for this feature.
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Home › Forums › The Break Room › QUESTION: Career path advice
This topic contains 400 replies, has 298 voices, and was last updated by
thanik 1 year, 3 months ago.
- AuthorPosts
I was looking into taking the dual cert course they offer one for CEH/CPT and the other CEH/Metasploit, my question is how viable and sought after is the metasploit cert and is it better than the CPT. And as a whole is the course good and worth the $4,000 I will have to cough up? My end goal is to become a pen tester one day I currently have my SEC+ and CCNA. Should I be going a different route first? Thanks for your help
Let me start my response by making this profound statement…
Any investment you make on any noteworthy venture you believe in, is worth all the dime you put in it
– quote me
Now, Pentesting ( my definition) is the art (and science) of assessing the security of your infrastructure with a view to uncovering vulnerabilities and making amendments or patch the loopholes.
In the course of pentesting your infrastructure, there are several tools that come into play serving various purposes to achieve your set target. METASPLOIT is one of the framework of tools in a pentesting venture. The reason Metasploit is often taken apart to be studied separately is because it is a FRAMEWORK of tools. It is a large warehouse of pentesting tools that enables you to carry out most of the activities you would want to do in a pentesting exercise…and it’s powerful.
Mastering Metasploit though, does not make you a master pentester because it’s just ONE of the Pentesting tools. However, being certified as a pentester actually makes you what you want to be…a PENTESTER!
You can go for any combination of course that helps you achieve your goal of becoming a pentester. Learning NEVER ENDS. Achieving the STATUS of a pentester is not all there is to be…you keep learning.
In the end, what matters is how well you are able to master the particular tool you know how to use. It is not in acquiring all the knowledge available but, how useful and relevant the knowledge acquired is to you and the goal you’re pursuing.
Hope you’ve gained some insight in my response.
I wish you the best in your pursuit!
Chin_DieselModerator
@mayanky, the way to accrue cybytes is to log in once a day, and also to post a message in the forums. While it was meant to try and get people to share thoughts and ideas, even a meaningless post gives them cybytes. Basically, when you see someone post “all the best,” what they are actually saying is “I want a cybyte for the day without contributing.”
what are your thoughts of CompTIA Security+ vs. SSCP
I have to decide which one to go for and am leaning more towards SSCP.
Down the road I plan to take CISSP
Because they are here only for the cybytes.
That is why there is not much of a discussion here.
As long as you have one comment per day you get 2 cybytes.
I guess they just want that.
all the best commenter must be warn by mods, its getting worst. i cant read real comments.
They definitely need to do something because like this there is no point in visiting these forums
I think that you have a lot of experience already in the admin role. If the role that you posted allows you to work more with Security, than take it. But from what you posted, it seems to be more of an purely admin type of role. You need to be working in some of the 8 domains that comprise IT Security. If you were to work in Access Controls, Business Continuity Management, or encryption, I would say that those would be better options for you. It is tough to break into IT Security when you don’t have any experience in the line of work.
I would also suggest that you work on incorporating some sort of IT Security function within the areas that the role encompasses. That would help you out a lot and will go a long way in helping you get that purely IT Security role that you are looking for.
nemoincilla007Participant
here are five things that will impact it security career and would help to grow both as security person and a human being..
1. get a solid understandings of systems and networks
2. if you don’t have virtualbox or VMware on your laptop, do it now
3. read like crazy
4. take advantages of the training resources available on the internet
5. build up network in terms of with security people
I think that from the list above, number 1 is the most important as there are many who jump directly onto security and they are reading how to protect systems that they do not understand how they work.
morganbowlerbrownParticipant
tough one, im still learning so my advice probably wouldnt be great!
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\begin{document}
\title[The slice spectral sequence for the $C_{4}$ analog of real
$K$-theory]{The slice spectral sequence for the
$C_{4}$ analog of real $K$-theory}
\author{Michael~A.~Hill}
\address{Department of Mathematics \\ University of California Los Angeles\\
Los Angeles, CA 90095}
\email{mikehill@math.ucla.edu}
\author{Michael~J.~Hopkins}
\address{Department of Mathematics \\ Harvard University
\\Cambridge, MA 02138}
\email{mjh@math.harvard.edu}
\author{Douglas~C.~Ravenel}
\address{Department of Mathematics \\ University of Rochester
\\Rochester, NY 14627}
\email{doug@math.rochester.edu}
\thanks{The authors were supported by DARPA Grant
FA9550-07-1-0555 and NSF Grants DMS-0905160 , DMS-1307896\dots }
\date{\today}
\begin{abstract}
We describe the slice spectral sequence
of a 32-periodic $C_{4}$-spectrum $\KH$ related to the
$C_{4}$ norm
\begin{displaymath}
{MU^{((C_{4}))}=N_{C_{2}}^{C_{4}}MU_{\reals}}
\end{displaymath}
\noindent of the real cobordism spectrum $MU_{\reals}$. We will give
it as a {\SS} of Mackey functors converging to the graded Mackey
functor $\upi_{*}\KH$, complete with differentials and exotic
extensions in the Mackey functor structure.
The slice spectral sequence for the 8-periodic real $K$-theory
spectrum $K_{\reals}$ was first analyzed by Dugger. The $C_{8}$
analog of $\KH$ is 256-periodic and detects the Kervaire invariant
classes $\theta_{j}$. A partial analysis of its slice spectral
sequence led to the solution to the Kervaire invariant problem, namely
the theorem that $\theta_{j}$ does not exist for $j\geq 7$.
\end{abstract}
\keywords{equivariant stable homotopy theory, Kervaire invariant
Mackey functor, slice spectral sequence}
\subjclass[2010]{55Q10 (primary), and 55Q91, 55P42, 55R45, 55T99 (secondary)}
\maketitle
\tableofcontents
\listoftables
\section{Introduction}\label{sec-intro}
In \cite{HHR} we derived the main theorem about the Kervaire invariant
elements from some properties of a $C_{8}$-{\eqvr} spectrum we called
$\Omega $ constructed as follows. We started with the
$C_{2}$-spectrum $MU_{\reals}$, meaning the usual complex cobordism
spectrum $MU$ equipped with a $C_{2}$ action defined in terms of
complex conjugation.
Then we defined a functor $N_{C_{2}}^{C_{8}}$, the norm of
\cite[\S2.2.3]{HHR} which we abbreviate here by $N_{2}^{8}$, from the
category of $C_{2}$-spectra to that of $C_{8}$-spectra. Roughly
speaking, given a $C_{2}$-spectrum $X$, $N_{2}^{8}X$ is underlain by
the fourfold smash power $X^{\wedge 4}$ where a generator $\gamma $ of
$C_{8}$ acts by cyclically permuting the four factors, each of which
is invariant under the given action of the subgroup $C_{2}$. In a similar
way one can define a functor $N_{H}^{G}$ from $H$-spectra to
$G$-spectra for any finite groups $H\subseteq G$.
A $C_{8}$-spectrum such as $N_{2}^{8}MU_{\reals}$, which is a
commutative ring spectrum, has {\eqvr} homotopy groups indexed by $RO
(C_{8})$, the orthogonal {\rep} ring for the group $C_{8}$. One
element of the latter is $\rho_{8}$, the regular {\rep}. In
\cite[\S9]{HHR} we defined a certain element $D\in
\pi_{19\rho_{8}}N_{2}^{8}MU_{\reals}$ and then formed the associated
mapping telescope, which we denoted by $\Omega_{\mathbb{O}}$. The
symbol $\mathbb{O}$ was chosen to suggest a connection with the
octonions, but there really is none apart from the fact that the
octonions are 8-dimensional like $\rho _{8}$.
$\Omega_{\mathbb{O}}$ is also a
$C_{8}$-{\eqvr} commutative ring spectrum. We then proved that it is
{\eqvr}ly equivalent to $\Sigma^{256}\Omega_{\mathbb{O}}$; we call
this result the Periodicity Theorem. Then our spectrum $\Omega $ is
$\Omega_{\mathbb{O}}^{C_{8}}$, the fixed point spectrum of
$\Omega_{\mathbb{O}}$.
It is possible to do this with $C_{8}$ replaced by $C_{2^{n}}$ for any
$n$. The dimension of the periodicity is then $2^{1+n+2^{n-1}}$. For
example it is 32 for the group $C_{4}$ and $2^{13}$ for $C_{16}$. We
chose the group $C_{8}$ because it is the smallest that suits our
purposes, namely it is the smallest one yielding a fixed point
spectrum that detects the Kervaire invariant elements $\theta_{j}$.
We know almost nothing about $\pi_{*}\Omega $, only that it is
periodic with periodic 256, that $\pi_{-2}=0$ (the Gap Theorem of
\cite[\S8]{HHR}), and that when $\theta_{j}$ exists its image in
$\pi_{*}\Omega $ is nontrivial (the Detection Theorem of
\cite[\S11]{HHR}).
We also know, although we did not say so in \cite{HHR}, that more
explicit computations would be much easier if we cut
$N_{2}^{8}MU_{\reals}$ down to size in the following way. Its
underlying homotopy, meaning that of the spectrum $MU^{\wedge 4}$, is
known classically to be a polynomial algbera over the integers with
four generators (cyclically permuted up to sign by the group action)
in every positive even dimension. This can be proved with methods
described by Adams in \cite{Ad:SHGH}. For the cyclic group $C_{2^{n}}$
one has $2^{n-1}$ generators in each positive even degree.
Specific generators $r_{i,j}\in \pi_{2i}MU^{\wedge 2^{n-1}}$ for $i>0$
and $0\leq j<^{n-1}$ are defined in \cite[\S5.4.2]{HHR}.
{\em There is a way to kill all the generators above dimension 2k}
that was described in \cite[\S2.4]{HHR}. Roughly speaking, let $A$ be
a wedge of suspensions of the sphere spectrum, one for each monomial in
the generators one wants to kill. One can define a multiplication and
group action on $A$ corresponding to the ones in $\pi_{*}MU^{\wedge
4}$. Then one has a map $A\to MU^{\wedge 4}$ whose restriction to
each summand represents the corresponding monomial, and a map $A\to
S^0$ (where the target is the sphere spectrum, not the space $S^{0}$)
sending each positive dimensional summand to a point. This leads to
two maps
\begin{displaymath}
S^{0}\wedge A\wedge MU^{\wedge 4}\rightrightarrows S^{0}\wedge MU^{\wedge 4}
\end{displaymath}
\noindent whose coequalizer we denote by $S^{0}\smashove{A}MU^{\wedge
4}$. Its homotopy is the quotient of $\pi_{*}MU^{\wedge 4}$ obtained
by killing the polynomial generators above diimension $2k$. The
construction is {\eqvr}, meaning that $S^{0}\smashove{A}MU^{\wedge 4}$
underlies a $C_{8}$-spectrum.
In \cite[\S7]{HHR} we showed that for $k=0$ the spectrum we get is the
integer {\SESM} spectrum $H\Z$; we called this result the Reduction
Theorem. In the non{\eqvr} case this is obvious. We are in effect
attaching cells to $MU^{\wedge 4}$ to kill all of its homotopy groups
in positive dimensions, which amounts to constructing the 0th
Postnikov section. In the {\eqvr} case the proof is more
delicate.
Now consider the case $k=1$, meaning that we are killing the
polynomial generators above dimension 2. Classically we know that
doing this to $MU$ (without the $C_{2}$-action) produces the
connective complex $K$-theory spectrum, some times denoted by $k$,
$bu$ or (2-locally) $BP\langle 1\rangle$. Inverting the Bott element
via a mapping telescope gives us $K$ itself, which is of course
2-periodic. In the $C_{2}$-{\eqvr} case one gets the ``real $K$-theory''
spectrum $K_{\reals}$ first studied by Atiyah in \cite{Atiyah:KR}. It
turns out to be 8-periodic and its fixed point spectrum is $KO$, which
is also referred to in other contexts as real $K$-theory.
The spectrum we get by killing the generators above dimension 2 in
the\linebreak $C_{8}$-spectrum $N_{2}^{8}MU_{\reals}$ will be denoted
analogously by $k_{[3]}$. We can invert the image of $D$ by forming a
mapping telescope, which we will denote by $K_{[3]}$. More
generally we denote by $k_{[n]}$ the spectrum obtained from
$N_{C_{2}}^{C_{2^{n}}}MU_{\reals}$ by killing all generators above
dimension 2. In particular $k_{[1]}=k_{\reals}$. Then we denote the
mapping telescope (after defining a suitable $D$) by $K_{[n]}$ and its
fixed point set by $KO_{[n]}$.
For $n\geq 3$, $KO_{[n]}$ also has a Periodicity, Gap and Detection
Theorem, so it could be used to prove the Kervaire invariant theorem.
{\em Thus $K_{[3]}$ is a substitute for $\Omega _{\mathbb{O}}$
with much smaller and therefore more tractable homotopy groups.} A
detailed study of them might shed some light on the fate of
$\theta_{6}$ in the 126-stem, the one hypothetical Kervaire invariant
element whose status is still open. {\em If we could show that
$\pi_{126}KO_{[3]}=0$, that would mean that
$\theta_{6}$ does not exist.}
The computation of the {\eqvr} homotopy $\upi_{*}K_{[3]}$ at this time
is daunting. {\em The purpose of this paper is to do a similar
computation for the group $C_{4}$ as a warmup exercise.} In the
process of describing it we will develope some techniques that are
likley to be needed in the $C_{8}$ case. We start with
$N_{2}^{4}MU_{\reals}$, kill its polynomial generators (of which there
are two in every positive even dimension) above dimension 2 as
described previously, and then invert a certain element in
$\pi_{4\rho_{4}}$. We denote the resulting spectrum by $\KH$, see
\ref{def-kH} below. This spectrum is known to be 32-periodic. In an
earlier draft of this paper it was denoted by $K_{\mathbf{H}}$.
The computational tool for finding these homotopy groups is the slice
spectral sequence introduced in \cite[\S4]{HHR}. Indeed we do not
know of any other way to do it. For $K_{\reals}$ it was first
analyzed by Dugger \cite{Dugger} and his work is described below in
\S\ref{sec-Dugger}. In this paper we will study the slice spectral
sequence of Mackey functors associated with $\KH$. We will rely
extensively on the results, methods and terminology of \cite{HHR}.
{\em We warn the reader that the computation for $\KH$ is more
intricate than the one for $\KR$.} For example, the slice spectral
sequence for $\KR$, which is shown in Figure \ref{fig-KR}, involves
five different Mackey functors for the group $C_{2}$. We abbreviate
them with certain symbols indicated in Table \ref{tab-C2Mackey}. The
one for $\KH$, partly shown in Figure \ref{fig-KH}, involves over
twenty Mackey functors for the group $C_{4}$, with symbols indicated
in Table \ref{tab-C4Mackey}.
\begin{figure}
\begin{center}
\includegraphics[width=11cm]{Slice-colorized.pdf} \caption[The 2008
poster.] {The 2008 poster. The first and third quadrants show
$\EE_{4} (G/G)$ of the slice {\SS} for $\KH$ with the elements of
Prop. \ref{prop-perm} excluded. The second quadrant indicates
$d_{3}$s as in Figures \ref{fig-sseq-7a} and \ref{fig-sseq-7b}. The
fourth quadrant indicates comparable $d_{3}$s in the third quadrant of
the slice {\SS} as in Figures \ref{fig-sseq-8a} and
\ref{fig-sseq-8b}.} \label{fig-poster}
\end{center}
\end{figure}
Part of this {\SS} is also illustrated in an unpublished poster produced in
late 2008 and shown in Figure \ref{fig-poster}. It shows the {\SS}
converging to the homotopy of the fixed point spectrum $\KH^{C_{4}}$.
The corresponding {\SS} of Mackey functors converges to the graded
Mackey functor $\upi_{*}\KH$.
In both illustrations some patterns of $d_{3}$s and families of
elements in low filtration are excluded to avoid clutter. In the
poster, representative examples of these are shown in the second and
fourth quadrants, the {\SS} itself being concentrated in the first and
third quadrants. In this paper those patterns are spelled out in
\S\ref{sec-C2diffs} and \S\ref{sec-C4diffs}.
\bigskip
We now outline the rest of the paper. Briefly, the next five sections
introduce various tools we need. Our objects of study, the spectra
$\kH$ and $\KH$, are formally introduced in \S\ref{sec-kH}. Dugger's
computation for $\KR$ is recalled in \S\ref{sec-Dugger}. The final
six sections describe the computation for $\kH$ and $\KH$.
In more detail, \S\ref{sec-nonsense} collects some notions from
{\eqvr} stable homotopy theory with an emphasis on Mackey functors.
Definition \ref{def-graded} introduces new notation that we will
ocasionally need.
\S\ref{sec-HZZ} concerns the {\eqvr} analog of the homology of a point
namely, the $RO (G)$-graded homotoy of the integer {\SESM} spectrum
$H\Z$. In particular Lemma \ref{lem-aeu} describes some relations
among certain elements in it including the ``gold relation'' between
$a_{V}$ and $u_{V}$.
\S\ref{sec-gendiffs} describes some general properties of spectral
sequences of Mackey functors. These include Theorem \ref{thm-exotic}
about the relation between differential and exotic extensions in the
Mackey functor structure and Theorem \ref{thm-normdiff} on the norm of
a differential.
\S\ref{sec-C4} lists some concise symbols for various specific Mackey functors
for the groups $C_{2}$ and $C_{4}$ that we will need. Such functors
can be spelled out explicitly by means of Lewis diagrams
(\ref{eq-Lewis}), which we usually abbreviate by symbols shown in
Tables \ref{tab-C2Mackey} and \ref{tab-C4Mackey}.
In \S\ref{sec-chain} we study some chain complexes of Mackey functors
that arise as cellular chain complexes for $G$-CW complexes of the
form $S^{V}$.
In \S\ref{sec-kH} we formally define (in \ref{def-kH}) the
$C_{4}$-spectra of interest in this paper, $\kH$ and $\KH$.
In \S\ref{sec-Dugger} we describe the slice {\SS} for an easier case,
the $C_{2}$-spectrum for real $K$-theory, $K_{\reals}$. This is due to
Dugger \cite{Dugger} and serves as a warmup exercise for us. It turns
out that everything in the {\SS } is formally determined by the
structure of its $\EE_{2}$-term and Bott periodicity.
In \S\ref{sec-more} we introduce various elements
in the homotopy groups of $\kH$ and $\KH$. They are collected in Table
\ref{tab-pi*}, which spans several pages.
In \S\ref{sec-slices} we determine the $\EE_{2}$-term of the
slice {\SS } for $\kH$ and $\KH$.
In \S\ref{sec-diffs} we use the Slice Differentials Theorem of
\cite{HHR} to determine some differentials in our {\SS}.
In \S\ref{sec-C2diffs} we examine the $C_{4}$-spectrum $\kH$ as a
$C_{2}$-spectrum. This leads to a calculation only slightly more
complicated than Dugger's. It gives a way to remove a lot of clutter
from the $C_{4}$ calculation.
In \S\ref{sec-C4diffs} we determine the $\EE_{4}$-term of our {\SS}.
It is far smaller than $\EE_{2}$ and the results of
\S\ref{sec-C2diffs} enable us to ignore most of it. What is left is
small enough to be shown legibly in the {\SS } charts of Figures
\ref{fig-E4} and \ref{fig-KH}. They illustrate integrally graded (as
opposed to $RO (C_{4})$-graded) spectral sequences of Mackey functors,
which are discussed in \S\ref{sec-C4}. In order to read these charts
one needs to refer to Table \ref{tab-C4Mackey} which defines the
``heiroglyphic'' symbols we use for the specific Mackey functors that
we neede.
We finish the calculation in \S\ref{sec-higher} by dealing with the
remaining differentials and exotic Mackey functor extensions. It
turns out that they are all formal consequences of $C_{2}$
differentials of the previous section along with the results of
\S\ref{sec-gendiffs}.
The result is a complete description of the {\em integrally graded}
portion of $\upi_{\star}\kH$. It is best seen in the {\SS } charts of
Figures \ref{fig-E4} and \ref{fig-KH}. Unfortunately we do not have a
clean description, much less an effective way to display the full $RO
(C_{4})$-graded homotopy groups.
For $G=C_{2}$, the two irreducible orthogonal {\rep}s are the trivial
one of degree 1, denoted by the symbol 1, and the sign {\rep} denoted
by $\sigma $. Thus $RO (G)$ is additvley a free abelain group of rank
2, and the {\SS } of interest is trigraded. In the $RO (C_{2})$-graded
homotopy of $\KR$, a certain element of degree $1+\sigma $ (the degree
of the regular {\rep} $\rho_{2}$) is invertible. This means that each
component of $\upi_{\star}\KR$ is canonically isomorphic to a Mackey
functor indexed by an ordinary integer. See Theorem
\ref{thm-ROG-graded} for a more precise statement. Thus the full
(trigraded)$RO (C_{2})$-graded slice {\SS } is determined by bigraded
one shown in Figure \ref{fig-KR}.
For $G=C_{4}$, the {\rep} ring $RO (G)$ is additively a free abelian group of
rank 3, so it leads to a quadrigraded {\SS }. The three irreducible
{\rep}s are the trivial and sign {\rep}s 1 and $\sigma $ (each having
degree one) and a degree two {\rep} $\lambda $ given by a rotation of
the plane $\reals^{2}$ of order 4. The regular {\rep} $\rho_{4}$ is
isomorphic to $1+\sigma +\lambda $. As in the case of $\KR$, there is
an invertible element $\normrbar_{1}$ (see Table \ref{tab-pi*}) in
$\upi_{\star}\KH$ of degree $\rho_{4}$. This means we can reduce the
quadigraded slice {\SS } to a trigraded one, but finding a full
description of it is a problem for the future.
\section{Recollections about {\eqvr} stable homotopy theory}
\label{sec-nonsense}
We first discuss some structure on the {\eqvr} homotopy groups of a\linebreak
$G$-spectrum $X$. {\em We will assume throughout that $G$ is a finite
cyclic $p$-group.} This means that its subgroups are well ordered by
inclusion and each is uniquely determined by its order. The results
of this section hold for any prime $p$, but the rest of the paper
concerns only the case $p=2$. We will define several maps indexed by
pairs of subgroups of $G$. {\em We will often replace these indices
by the orders of the subgroups, sometimes denoting $|H|$ by $h$.}
The homotopy groups can be defined in terms of finite $G$-sets $T$
Let
\begin{displaymath}
\upi_{0}^{G}X (T) = [T_{+}, X]^{G},
\end{displaymath}
\noindent be the set of homotopy classes of {\eqvr} maps from $T_{+}$,
the suspension spectrum of the union of $T$ with a disjoint base
point, to the spectrum $X$. We will often omit $G$ from the notation
when it is clear from the context. For an orthogonal {\rep} $V$ of
$G$, we define
\begin{displaymath}
\upi_{V}X (T) = [S^{V}\wedge T_{+}, X]^{G}.
\end{displaymath}
\noindent As an $RO (G)$-graded contravariant abelian group valued
functor of $T$, this converts disjoint unions to direct sums. This
means it is determined by its values on the sets $G/H$ for subgroups
$H\subseteq G$.
Since $G$ is abelian, $H$ is normal and $\upi_{V}X (G/H)$ is a $Z[G/H]$-module.
Given subgroups $K\subseteq H\subseteq G$, one has pinch and fold maps
between the $H$-spectra $H/H_{+}$ and $H/K_{+}$. This leads to a diagram
\begin{numequation}\label{eq-pinch-fold}
\begin{split}
\xymatrix
@R=1mm
@C=4mm
{
&H/H_{+}\ar@<.5ex>[rr]^(.5){\mbox{pinch} }
&{}
&H/K_{+}\ar@<.5ex>[ll]^(.5){\mbox{fold} }\\
& &\ar@{=>}[ddd]^(.6){G_{+}\smashove{H} (\cdot)}\\
& &\\
& &\\
& &\\
G/H_{+}\ar@{=}[r]^(.5){}
&G_{+}\smashove{H}H/H_{+}\ar@<.5ex>[rr]^(.5){\mbox{pinch} }
&{} &G_{+}\smashove{H}H/K_{+}\ar@{=}[r]^(.5){}
\ar@<.5ex>[ll]^(.5){\mbox{fold} }
&G_{+}\smashove{K}K/K_{+}\ar@{=}[r]^(.5){}
&G/K_{+}.
}
\end{split}
\end{numequation}
\noindent Note that while the fold map is induced by a map of
$H$-sets, the pinch map is not. It only exists in the stable
category.
\begin{defin}\label{def-fixed-point-maps}
{\bf The Mackey functor structure maps in $\upi_{V}^{G}X$.}
The fixed point transfer
and restriction maps
\begin{displaymath}
\xymatrix
@R=5mm
@C=15mm
{
{\upi_{V}X (G/H)}\ar@<-.5ex>[r]_(.5){\Res_{K}^{H} }
&{\upi_{V}X (G/K)}\ar@<-.5ex>[l]_(.5){\Tr_{K}^{H} }
}
\end{displaymath}
\noindent are the ones induced by the composite maps in the bottom row
of (\ref{eq-pinch-fold}).
\end{defin}
These satisfy the formal properties needed to make $\upi_{V}X$ into a
Mackey functor; see \cite[Def. 3.1]{HHR}. They are usually referred
to simply as the transfer and restriction maps. We use the words
``fixed point'' to distinguish them from another similar pair of maps
specified below in Definition \ref{def-gp-action}.
We remind the reader that a Mackey functor $\underline{M}$ for a
finite group $G$ assigns an abelian group $\underline{M} (T)$ to every
finite $G$-set $T$. It converts disjoint unions to direct sums. It is
therefore determined by its values on orbits, meaning $G$-sets for the
form $G/H$ for various subgroups $H$ of $G$. For subgroups
$K\subseteq H\subseteq G$, one has a map of $G$-sets $G/K\to G/H$. In
categorical language $\underline{M}$ is actually a pair of functors,
one covariant and one contravariant, both behaving the same way on
objects. Hence we get maps both ways between $\underline{M} (G/K)$
and $\underline{M} (G/H)$. For the Mackey functor $\upi_{V}X$, these
are the two maps of \ref{def-fixed-point-maps}.
One can generalize the definition of a Mackey functor by replacing the
target category of abelian groups by one's favorite abelian category,
such as that of $R$-modules over graded abelian groups.
\begin{defin}\label{def-green}
A {\bf graded Green functor $\underline{R}_{*}$ for a group $G$} is a
Mackey functor for $G$ with values in the category of graded abelian
groups such that $\underline{R}_{*} (G/H)$ is a graded commutative
ring for each subgroup $H$ and for each pair of subgroups $K\subseteq
H\subseteq G$, the restriction map $\Res_{K}^{H}$ is a ring
homomorphism and the transfer map $\Tr_{K}^{H}$ satisfies the {\bf
Frobenius relation}
\begin{displaymath}
\Tr_{K}^{H}(\Res_{K}^{H}(a)b)=a (\Tr_{K}^{H}(b))\qquad
\mbox{for $a\in \underline{R}_{*} (G/H)$ and $b\in
\underline{R}_{*} (G/K)$}.
\end{displaymath}
\end{defin}
When $X$ is a ring spectrum, we have the {\em fixed point Frobenius
relation}
\begin{numequation}
\label{eq-Frob} \Tr_{K}^{H}(\Res_{K}^{H}(a)b)=a (\Tr_{K}^{H}(b))\qquad
\mbox{for $a\in \upi_{\star}X (G/H)$ and $b\in
\upi_{\star}X (G/K)$}.
\end{numequation}
In particular this means that
\begin{numequation}
\label{eq-Frobcor}
a(\Tr_{K}^{H}(b)) = 0\qquad \mbox{when } \Res_{K}^{H}(a)=0.
\end{numequation}
For a {\rep} $V$ of $G$, the group
\begin{displaymath}
\upi_{V}^{G}X (G/H)=\pi_{V}^{H}X=[S^{V},X]^{H}
\end{displaymath}
\noindent is isomorphic to
\begin{displaymath}
[S^{0},S^{-V}\wedge X]^{H}=\pi_{0} (S^{-V}\wedge X)^{H}.
\end{displaymath}
\noindent However fixed points do not respect smash products, so we
cannot equate this group with
\begin{displaymath}
\pi_{0} (S^{-V^{H}}\wedge X^{H}) = [S^{V^{H}},X^{H}]=\pi_{|V^{H}|}X^{H}
=\upi^{G}_{|V^{H}|}X (G/H).
\end{displaymath}
Conversely a $G$-{\eqvr} map $S^{V}\to X$ represents an element in
\begin{displaymath}
[S^{V},X]^{G}=\pi_{V}^{G}X=\upi^{G}_{V}X (G/G).
\end{displaymath}
The following notion is useful.
\begin{defin}\label{def-MF-induction}
{\bf Mackey functor induction and restriction.} For s subgroup $H$ of $G$
and an $H$-Mackey functor $\underline{M}$, the
induced $G$-Mackey functor $\uparrow_{H}^{G}\underline{M}$ is given by
\begin{displaymath}
\uparrow_{H}^{G}\underline{M} (T)=\underline{M} (i_{H}^{*}T)
\end{displaymath}
\noindent for each finite $G$-set $T$, where $i_{H}^{*}$ denotes the
forgetful functor from $G$-sets (or spaces or spectra) to $H$-sets.
For a $G$-Mackey functor $\underline{N}$, the restricted $H$-Mackey functor
$\downarrow_{H}^{G}\underline{N}$ is given by
\begin{displaymath}
\downarrow_{H}^{G}\underline{N} (S)=\underline{N} (G\times_{H}S)
\end{displaymath}
\noindent for each finite $H$-set $S$.
\end{defin}
This notation is due to Th\'evenaz-Webb \cite{Thevenaz-Webb}. They put
the decorated arrow on the right and denote $G\times_{H}S$ by
$S\uparrow_{H}^{G}$ and $i_{H}^{*}T$ by $T\downarrow_{H}^{G}$.
We also need notation for $X$ as an $H$-spectrum
for subgroups $H\subseteq G$. For this purpose we will enlarge the
orthogonal {\rep} ring of $G$, $RO (G)$, to the {\rep} ring Mackey
functor $\underline{RO} (G)$ defined by $\underline{RO} (G) (G/H)=RO
(H)$. This was the motivating example for the definition of a Mackey
functor in the first place. In it the transfer map on a {\rep} $V$ of
$H$ is the induced {\rep} of a supergroup $K\supseteq H$, and its
restriction to a subgroup is defined in the obvious way. In particular
the restriction of the transfer of $V$ is $|K/H|V$.
More generally for a finite $G$-set $T$, $\underline{RO} (G) (T)$ is
the ring (under pointwise direct sum and tensor product) of functors
to the category of finite dimensional orthogonal real vector spaces
from $B_{G}T$, the split groupoid (see \cite[A1.1.22]{Rav:MU}) whose
objects are the elements of $T$ with morphisms defined by the action
of $G$.
\begin{defin}\label{def-graded}
{\bf $\underline{RO} (G)$-graded homotopy groups.} For each
$G$-spectrum $X$ and each pair $(H,V)$ consisting of a subgroup
$H\subseteq G$ and a virtual orthogonal {\rep} $V$ of $H$, let the
$G$-Mackey functor $\upi_{{H,V}} (X)$ be defined by
\begin{displaymath}
\upi_{{H,V}} (X) (T)
:= \left[(G_{+}\smashove{H}S^{V})\wedge T_{+}, X \right]^{G}
\cong \left[S^{V}\wedge i_{H}^{*}T_{+}, i_{H}^{*}X \right]^{H}
= \upi_{V}^{H} (i_{H}^{*}X) (i_{H}^{*}T),
\end{displaymath}
\noindent for each finite $G$-set $T$. Equivalently, $\upi_{{H,V}}
(X)=\uparrow_{H}^{G}\upi_{V}^{H} (i_{H}^{*}X)$ (see
\ref{def-MF-induction}) as Mackey functors. We will often denote
$\upi_{G,V}$ by $\upi_{V}^{G}$ or $\upi_{V}$.
\end{defin}
We will be studying the $\underline{RO} (G)$-graded slice {\SS }
$\left\{\EE_{r}^{s,\star} \right\}$ of Mackey functors with
$r,s\in \Z$ and $\star\in \underline{RO} (G)$. We will use the
notation $\EE_{r}^{s,(H,V)}$ for such Mackey functors,
abbreviating to $\EE_{r}^{s,V}$ when the subgroup is
$G$. Most of our spectral sequence charts will display the values of
$\EE_{2}^{s,t}$ for integral values of $t$ only.
The following definition should be compared with \cite[(2.3)]{Ad:Prereq}.
\bigskip
\begin{defin}\label{def-homeo}
{\bf An {\eqvr} homeomorphism.} Let $X$ be a $G$-space and $Y$ an
$H$-space for a subgroup $H\subseteq G$. We define the {\eqvr}
homeomorphism
\begin{displaymath}
\tilde{u}_{H}^{G} (Y,X):G\times_{H} (Y\times i_{H}^{*}X) \to
(G\times_{H}Y)\times X
\end{displaymath}
\noindent by $(g,y,x)\mapsto (g,y,g (x))$ for $g\in G$, $y\in Y$ and
$x\in X$. We will use the same notation for a
similarly defined homeomorphism
\begin{displaymath}
\tilde{u}_{H}^{G} (Y,X):G_{+}\smashove{H} (Y\wedge i_{H}^{*}X) \to
(G_{+}\smashove{H}Y)\wedge X
\end{displaymath}
\noindent for a $G$-spectrum $X$ and $H$-spectrum $Y$. We will abbreviate
\begin{displaymath}
\tilde{u}_{H}^{G} (S^{0},X):G_{+}\smashove{H} i_{H}^{*}X \to
G/H_{+}\wedge X
\end{displaymath}
\noindent by $\tilde{u}_{H}^{G} (X)$.
For {\rep}s $V$ and $V'$ of $G$ both restricting to $W$ on $H$, but
having distinct restrictions to all larger subgroups, we define
$\tilde{u}_{V-V'}=\tilde{u}_{H}^{G} (S^{V})\tilde{u}_{H}^{G} (S^{V'})^{-1}$,
so the following diagram of {\eqvr} homeomorphisms commutes:
\begin{numequation}\label{eq-u{V-V'}}
\begin{split}
\xymatrix
@R=2mm
@C=10mm
{ & &G/H\wedge S^{V}\\
G_{+}\smashove{H}S^{W}\ar[rru]^(.5){\tilde{u}_{H}^{G} (S^{V})}
\ar[rrd]_(.5){\tilde{u}_{H}^{G} (S^{V'})}\\
& &G/H\wedge S^{V'}\ar[uu]_(.5){\tilde{u}_{V-V'}}.
}
\end{split}
\end{numequation}
\noindent When $V'=|V|$ (meaning that $H=G_{V}$ acts trivially on
$W$), then we abbreviate $\tilde{u}_{V-V'}$ by $\tilde{u}_{V}$.
\end{defin}
If $V$ is a {\rep} of $H$ restricting to $W$ on $K$, we can smash the
diagram (\ref{eq-pinch-fold}) with $S^{V}$ and get
\begin{numequation}\label{eq-pinch-fold-VW}
\begin{split}
\xymatrix
@R=1mm
@C=5mm
{
S^{V}\ar@<.5ex>[rr]^(.4){\mbox{pinch} }
&{}
& H/K_{+}\wedge S^{V}\ar@<.5ex>[ll]^(.6){\mbox{fold} }\\
&\ar@{=>}[ddd]^(.6){G_{+}\smashove{H} (\cdot)}\\
&\\
&\\
&\\
G_{+}\smashove{H}S^{V}\ar@<.5ex>[rr]^(.4){\mbox{pinch} }
&{} &G_{+}\smashove{H} (H/K_{+}\wedge S^{V} )\ar[r]^(.55){\approx }
\ar@<.5ex>[ll]^(.6){\mbox{fold} }
&G_{+}\smashove{H} (H_{+}\smashove{K} S^{W} )\ar@{=}[r]^(.5){}
&G_{+}\smashove{K}S^{W},
}
\end{split}
\end{numequation}
\noindent where the homeomorphism is induced by that of Definition
\ref{def-homeo}.
\begin{defin}\label{def-gp-action}
{\bf The group action restriction and transfer maps.} For subgroups
$K\subseteq H\subseteq G$, let $V\in RO (H)$ be a virtual {\rep} of $H$ restricting to $W\in RO (K)$. The group action transfer
and restriction maps
\begin{displaymath}
\xymatrix
@R=5mm
@C=15mm
{
{\uparrow_{H}^{G}\upi_{V}^{H} (i_{H}^{*}X) } \ar@{=}[r]^(.5){}
&{\upi_{H,V}X }\ar@<-.5ex>[r]_(.5){\underline{r}_{K}^{H} }
&{\upi_{K,W}X }\ar@<-.5ex>[l]_(.5){\underline{t}_{K}^{H,V} }
\ar@{=}[r]^(.5){}
&{\uparrow_{K}^{G}\upi_{W}^{K} (i_{K}^{*}X) }
}
\end{displaymath}
\noindent (see \ref{def-MF-induction}) are the ones induced by the
composite maps in the bottom row of (\ref{eq-pinch-fold-VW}). The
symbols $t$ and $r$ here are underlined because they are maps {\em of}
Mackey functors rather than maps within Mackey functors.
\end{defin}
We include $V$ as an index for the group action transfer
$\underline{t}_{K}^{H,V}$ because its target is not determined by its
source.
Thus we have abelian groups $\upi_{{H',V}} (X) (G/H'')$ for all
subgroups $H', H''\subseteq G$ and {\rep}s $V$ of $H'$. Most of them
are redundant in view of Theorem \ref{thm-module} below. In what
follows, we will use the notation $H_{\cup}=H'\cup H''$ and
$H_{\cap}=H'\cap H''$.
\begin{lem}\label{lem-module}
{\bf An {\eqvr} module structure.} For a $G$-spectrum $X$ and
$H'$-spectrum $Y$,
\begin{displaymath}
[G_{+}\smashove{H'}Y, X]^{H''}
=\Z[G/H_{\cup }]\otimes [H_{\cup +}\smashove{H'}Y,X]^{H''}
\end{displaymath}
\noindent as $\Z[G/H'']$-modules.
\end{lem}
\proof
As abelian groups,
\begin{align*}
[G_{+}\smashove{H'}Y, X]^{H''}
& = [i_{H''}^{*} (G_{+}\smashove{H'}Y), X]^{H''} \\
& = \left[\bigvee_{|G/H_{\cup }|}H_{\cup +}\smashove{H'}Y,
X \right]^{H''}\\
& = \bigoplus_{|G/H_{\cup }|}[H_{\cup +}\smashove{H'}Y, X]^{H''}
\end{align*}
\noindent and $G/H''$ permutes the wedge summands of
$\bigvee_{|G/H_{\cup }|}H_{\cup +}\smashove{H'}Y$ as it
permutes the elements of $G/H_{\cup }$. \qed\bigskip
\begin{thm}\label{thm-module}
{\bf The module structure for $\underline{RO (G)}$-graded
homotopy\linebreak groups.} For subgroups $H', H''\subseteq G$ with
$H_{\cup}=H'\cup H''$ and $H_{\cap}=H'\cap H''$, and a virtual {\rep}
$V$ of $H'$ restricting to $W$ on $H_{\cap}$,
\begin{displaymath}
\upi_{H',V}X (G/H'')
\cong \Z[G/H_{\cup }]\otimes \upi_{H_{\cap},W}X (G/G)
\cong \Z[G/H_{\cup }]\otimes \upi_{W}^{H_{\cap}}
i_{H_{\cap}}^{*}X (H_{\cap}/H_{\cap})
\end{displaymath}
\noindent as $\Z[G/H'']$-modules.
Suppose that $H''$ is a proper subgroup of $H'$ and $\gamma \in H'$
is a generator. Then as an element in $\Z[G/H'']$, $\gamma $ induces
multiplication by $-1$ in $\upi_{H',V}X (G/H'')$ iff $V$ is
nonorientable.
\end{thm}
\proof We start with the definition and use the homeomorphism of
Definition \ref{def-homeo} and the module structure of Lemma \ref{lem-module}.
\begin{align*}
\upi_{H',V}X (G/H'')
& = [(G_{+}\smashove{H'} S^{V})\wedge G/H''_{+},X]^{G} \\
& = [G_{+}\smashove{H''}(G_{+}\smashove{H'}S^{V}),X]^{G} \\
& = [G_{+}\smashove{H'}S^{V},X]^{H''}
= \Z[G/H_{\cup }] \otimes [H_{\cup +}\smashove{H'}S^{V}, X]^{H''}\\
\aand
[H_{\cup +}\smashove{H'}S^{V}, X]^{H''}
& = [S^{W}, X]^{H_{\cap}}
= [G_{+}\smashove{H_{\cap}}S^{W}, X]^{G}\\
& = \upi_{W}^{H_{\cap}} (i_{H_{\cap}}^{*}X) (H_{\cap}/H_{\cap})
= \upi_{H_{\cap},W}X (G/G).
\end{align*}
For the statement about nonoriented $V$, we have
\begin{displaymath}
\upi_{H',V}X (G/H'') = \Z[G/H']\otimes \upi_{W}^{H''}i_{H''}^{*}X (H''/H'')
= \Z[G/H']\otimes [S^{W},X]^{H''}.
\end{displaymath}
\noindent
Then $\gamma $ induces a map of degree $\pm 1$ on the sphere depending
on the orientability of $V$. \qed
\bigskip
Theorem \ref{thm-module} means that we need only consider the groups
\begin{displaymath}
\upi_{H,V}X (G/G) \cong \upi_{V}^{H}i^{*}_{H}X (H/H).
\end{displaymath}
When $H\subset G$ and $V$ is a virtual {\rep} of $G$, we have
\begin{numequation}\label{eq-easy-iso}
\begin{split}
\upi_{V}X (G/H)
\cong \upi_{H,i^{*}_{H}V}X (G/G)
\cong \upi^{H}_{i^{*}_{H}V}i^{*}_{H}X (H/H).
\end{split}
\end{numequation}
\noindent This isomorphism makes the following diagram commute for
$K\subseteq H$.
\begin{displaymath}
\xymatrix
@R=5mm
@C=10mm
{
{\upi_{V}X (G/H)}\ar[r]^(.45){\cong }
\ar@<-.5ex>[d]_(.5){\Res_{K}^{H} }
&{\upi_{H,i^{*}_{H}V}X (G/G)}\ar[r]^(.5){\cong }
\ar@<-.5ex>[d]_(.5){\underline{r}_{K}^{H} }
&{\upi^{H}_{i^{*}_{H}V}i^{*}_{H}X (H/H)}
\\
{\upi_{V}X (G/K)}\ar[r]^(.45){\cong }
\ar@<-.5ex>[u]_(.5){\Tr_{K}^{H} }
&{\upi_{K,i_{K}^{*}V}X (G/G)}\ar[r]^(.5){\cong }
\ar@<-.5ex>[u]_(.5){\underline{t}_{K}^{H,i^{*}_{H}V} }
&{\upi^{K}_{i_{K}^{*}V}i^{*}_{K}X (K/K)}
}
\end{displaymath}
\noindent {\em We will use the three groups of (\ref{eq-easy-iso})
interchangeably as convenient and use the same notation for elements
in each related by this canonical isomorphism.} Note that the group on
the left is indexed by $RO (G)$ while the two on the right are indexed
by $RO (H)$. This means that if $V$ and $V'$ are {\rep}s of $G$ each
restricting to $W$ on $H$, then $\upi_{V}X (G/H)$ and $\upi_{V'}X
(G/H)$ are canonically isomorphic. The first of these is
\begin{displaymath}
[G/H_{+}\wedge S^{V}, X]^{G}\cong
[G_{+}\smashove{H}S^{W}, X]^{G} \cong
[S^{W}, i_{H}^{*}X]^{H}
\end{displaymath}
\noindent where the first isomorphism is induced by the homeomorphism
$\tilde{u}_{H}^{G} (X)$ of Definition \ref{def-homeo} and the second
is the fact that $G_{+}\smashove{H} (\cdot )$ is the left adjoint of
the forgetful functor $i_{H}^{*}$.
\begin{rem}\label{rem-via}
{\bf Factorization via restriction}. For a ring spectrum $X$, such as
the one we are studying in this paper, an indecomposable element in
$\upi_{\star}X (G/H)$ may map to a product $xy\in \upi_{H,\star}X
(G/G)$ of elements in groups indexed by {\rep}s of $H$ that are not
restrictions of {\rep}s of $G$. When this happens we may denote the
indecomposable element in $\upi_{\star}X (G/H)$ by $[xy]$. This
factorization can make some computations easier.
\end{rem}
\section{The $RO (G)$-graded homotopy of $\HZ$}\label{sec-HZZ}
We describe part of the $RO (G)$-graded Green functor
$\upi_{\star}(\HZ)$, where $\HZ $ is the integer {\SESM} spectrum
$\HZ$ in the $G$-{\eqvr} category, for some finite cyclic 2-group $G$. For
each actual (as opposed to virtual) $G$-{\rep} $V$ we have an {\eqvr}
reduced cellular chain complex $C^{V}_{*}$ for the space $S^{V}$. It
is a complex of $\ints [G]$-modules with $H_{*} (C^{V})=H_{*}
(S^{|V|})$.
One can convert such a chain complex $C_{*}^{V}$ of $\Z[G]$-modules to
one of Mackey functors as follows. Given a $\Z[G]$-module $M$, we get
a Mackey functor $\uM$ defined by
\begin{numequation}
\label{eq-fpm}
\uM (G/H)= M^{H}\qquad \mbox{for each subgroup $H\subseteq G$.}
\end{numequation}
\noindent We call this a {\em fixed point Mackey functor}. In it each
restriction map $\Res_{K}^{H}$ (for $K\subseteq H\subseteq G$) is one
to one. When $M$ is a permutation module, meaning the free abelian
group on a $G$-set $B$, we call $\uM$ a {\em permutation Mackey
functor} \cite[\S3.2]{HHR}.
In particular the $Z[G]$-module $\Z$ with trivial group action (the
free abelian group on the $G$-set $G/G$) leads to a Mackey functor
$\underline{\Z}$ in which each restriction map is an isomorphism and
the transfer map $\Tr_{K}^{H}$ is multiplication by $|H/K|$. For each
Mackey functor $\underline{M}$ there is an {\SESM} spectrum
$H\underline{M}$ \cite[\S5]{Greenlees-May}, and $\HZZ$ is the same as
$\HZ$ with trivial group action.
Given a finite $G$-CW spectrum $X$, meaning one
built out of cells of the form $G_{+}\smashove{H} e^{n}$, we get a
reduced cellular chain complex of $\Z[G]$-modules $C_{*}X$, leading to
a chain complex of fixed point Mackey functors $\uC_{*}X$.
Its homology is a graded Mackey functor $\uH_{*}X$ with
\begin{displaymath}
\uH_{*}X (G/H)
= \upi_{*} (X\wedge \HZZ) (G/H)
= \pi_{*} (X\wedge \HZZ)^{H}.
\end{displaymath}
\noindent In particular $\uH_{*}X (G/\ee) = H_{*}X$, the
underlying homology of $X$. In general $\uH_{*}X (G/H)$
is not the same as $H_{*} (X^{H})$ because fixed points do not commute
with smash products.
For a finite cyclic 2-group $G=C_{2^{k}}$, the irreducible {\rep}s are
the 2-dimensional ones $\lambda (m)$ corresponding to rotation through
an angle of $2\pi m/2^{k}$ for $0<m<2^{k-1}$, the sign {\rep} $\sigma
$ and the trivial one of degree one, which we denote by 1. The
2-local {\eqvr} homotopy type of $S^{\lambda (m)}$ depends only on the
2-adic valuation of $m$, so we will only consider $\lambda (2^{j})$
for $0\leq j\leq k-2$ and denote it by $\lambda_{j}$. The planar
rotation $\lambda_{k-1}$ though angle $\pi$ is the same {\rep} as
$2\sigma $. {\em We will denote $\lambda (1)=\lambda_{0}$ simply by
$\lambda $.}
We will describe the chain complex $C^{V}$ for
\begin{displaymath}
V=a+b\sigma +\sum_{2\leq j\leq k}c_{j}\lambda_{k-j}.
\end{displaymath}
\noindent for nonnegative integers $a$, $b$ and $c_{j}$.
The isotropy group of $V$ (the largest subgroup fixing all of $V$) is
\begin{displaymath}
G_{V}=\mycases{
C_{2^{k}}=G
&\mbox{for }b=c_{2}=\dotsb =c_{k}=0\\
C_{2^{k-1}}=:G'
&\mbox{for $b>0$ and $c_{2}=\dotsb =c_{k}=0$}\\
C_{2^{k-\ell }}
&\mbox{for $c_{\ell }>0$ and $c_{1+\ell }=\dotsb =c_{k}=0$}
}
\end{displaymath}
The sphere $S^{V}$ has a $G$-CW structure with reduced cellular chain complex
$C^{V}$ of the form
\begin{numequation}\label{eq-CVn}
\begin{split}
C^{V}_{n}=\mycases{
\ints
&\mbox{for }n=d_{0}\\
\ints[G/G']
&\mbox{for }d_{0}<n\leq d_{1}\\
\ints[G/C_{2^{k-j}}]
&\mbox{for $d_{j-1}<n\leq d_{j}$ and $2\leq j\leq \ell $}\\
0 &\mbox{otherwise.}
}
\end{split}
\end{numequation}
\noindent where
\begin{displaymath}
d_{j}=\mycases{
a &\mbox{for $j=0$}\\
a+b
&\mbox{for $j=1$}\\
a+b+2c_{2}+\dotsb +2c_{j}
&\mbox{for $2\leq j\leq \ell $,}
}
\end{displaymath}
\noindent so $d_{\ell }=|V|$.
The boundary map $\partial_{n}:C_{n}^{V}\to C_{n-1}^{V}$ is determined
by the fact that\linebreak ${H_{*}(C^{V})=H_{*}(S^{|V|})}$. More
explicitly, let $\gamma $ be a generator of $G$ and
\begin{displaymath}
\zeta_{j}=\sum_{0\leq t<2^{j }}\gamma^{t}
\qquad \mbox{for }1\leq j \leq k.
\end{displaymath}
\noindent Then we have
\begin{displaymath}
\partial_{n}=\mycases{
\nabla
&\mbox{for }n=1+d_{0}\\
(1-\gamma )x_{n}
&\mbox{for $n-d_{0}$ even and $2+d_{0}\leq n\leq d_{n}$}\\
x_{n}
&\mbox{for $n-d_{0}$ odd and $2+d_{0}\leq n\leq d_{n}$}\\
0 &\mbox{otherwise,}
}
\end{displaymath}
\noindent where $\nabla$ is the fold map sending $\gamma \mapsto 1$,
and $x_{n}$ denotes multiplication by an element in $\Z[G]$ to be
named below. We will use the same symbol below for the quotient map
$\Z[G/H]\to\Z[G/K]$ for $H\subseteq K\subseteq G$. The elements
$x_{n}\in \Z[G]$ for $2+d_{0}\leq n\leq |V|$ are determined
recursively by $x_{2+d_{0}}=1$ and
\begin{displaymath}
x_{n}x_{n-1}= \zeta_{j } \qquad \mbox{for }2+d_{j-1}<n\leq 2+d_{j}.
\end{displaymath}
\noindent It follows that $H_{|V|}C^{V}=\Z$ generated by either
$x_{1+|V|}$ or its product with $1-\gamma $, depending on the parity
of $b$.
This complex is
\begin{displaymath}
C^{V} = \Sigma^{|V_{0}|} C^{V/V_{0}}
\end{displaymath}
\noindent where $V_{0}=V^{G}$. This means we can assume without loss
of generality that $V_{0}=0$.
An element
\begin{displaymath}
x\in H_{n} \uC^{V} (G/H)
= \uH_{n} S^{V} (G/H)
\end{displaymath}
\noindent
corresponds to an element $x\in \upi_{n-V} \HZ (G/H) $.
We will denote the dual complex $\Hom_{\ints } (C^{V},\ints )$ by
$C^{-V}$. Its chains lie in dimensions $-n$ for $0\leq n\leq |V|$.
An element $x\in \uH_{-n} (S^{-V}) (G/H)$ corresponds to an
element $x\in \upi_{V-n} \HZ (G/H)$.
The method we have just described determines only a portion of the\linebreak
$RO(G)$-graded Mackey functor $\upi_{(G, \star)}\HZZ$, namely
the groups in which the index differs by an integer from an actual
{\rep} $V$ or its negative. For example it does not give us
$\upi_{\sigma -\lambda }\HZZ$ for $|G|\geq 4$.
We leave the proof of the following as an exercise for the reader.
\begin{prop}\label{prop-top}
{\bf The top (bottom) homology groups for $S^{V}$ ($S^{-V}$).} Let $G$
be a finite cyclic 2-group and $V$ a nontrivial {\rep} of $G$ of
degree $d$ with $V^{G}=0$ and isotropy group $G_{V}$. Then
$C_{d}^{V}=C_{-d}^{-V}=\Z[G/G_{V}]$ and
\begin{enumerate}
\item [(i)] If $V$ is oriented then
$\uH_{d}S^{V}=\underline{\Z}$, the constant $\Z$-valued
Mackey functor in which each restriction map is an isomorphism and
each transfer $\Tr_{H}^{K}$ is multiplication by $|K/H|$.
\item [(ii)]
$\uH_{-d}S^{-V}=\underline{\Z} (G,G_{V})$, the constant
$\Z$-valued Mackey functor in which
\begin{displaymath}
\Res_{H}^{K}=\mycases{
1 &\mbox{for }K\subseteq G_{V}\\
{|K/H|} &\mbox{for }G_{V}\subseteq H
}
\end{displaymath}
\noindent and
\begin{displaymath}
\Tr_{H}^{K}=\mycases{
{|K/H| } &\mbox{for }K\subseteq G_{V}\\
1 &\mbox{for }G_{V}\subseteq H.
}
\end{displaymath}
\noindent (The above completely describes the cases where
$|K/H|=2$, and they determine all other restrictions and transfers.) The
functor $\underline{\Z}(G,e)$ is also known as the dual
$\underline{\Z}^{*}$. These isomorphisms are induced by the maps
\begin{displaymath}
\xymatrix
@R=0mm
@C=15mm
{
{\uH_{d}S^{V}}\ar@{=}[dd]^(.5){}
& &{\uH_{-d}S^{-V}}\ar@{=}[dd]^(.5){}\\
& &\\
{\underline{\Z}}\ar[r]^(.4){\Delta}
&{\underline{\Z[G/G_{V}]}}\ar[r]^(.5){\nabla}
&{\underline{\Z} (G,G_{V}).}
}
\end{displaymath}
\item [(iii)] If $V$ is not oriented then
$\uH_{d}S^{V}=\underline{\Z}_{-}$, where
\begin{displaymath}
\underline{\Z}_{-} (G/H)=\mycases{
0 &\mbox{for }H=G\\
\Z_{-}:=\Z[G]/ (1+\gamma )
&\mbox{otherwise}
}
\end{displaymath}
\noindent where each restriction map $\Res_{H}^{K}$ is an isomorphism
and each transfer $\Tr_{H}^{K}$ is multiplication by $|K/H|$ for each
proper subgroup $K$.
\item [(iv)] We also have
$\uH_{-d}S^{-V}=\underline{\Z} (G,G_{V})_{-}$, where
\begin{displaymath}
\underline{\Z} (G,G_{V})_{-} (G/H)=\mycases{
0 &\mbox{for $H=G$ and $V=\sigma $}\\
\Z/2 &\mbox{for $H=G$ and $V\neq \sigma $}\\
\Z_{-} &\mbox{otherwise}
}
\end{displaymath}
\noindent with the same restrictions and transfers as $\underline{\Z}
(G,G_{V})$. These isomorphisms are induced by the maps
\begin{displaymath}
\xymatrix
@R=0mm
@C=15mm
{
{\uH_{d}S^{V}}\ar@{=}[dd]^(.5){}
& &{\uH_{-d}S^{-V}}\ar@{=}[dd]^(.5){}\\
& &\\
{\underline{\Z}_{-}}\ar[r]^(.4){\Delta_{-} }
&{\underline{\Z[G/G_{V}]}}\ar[r]^(.5){\nabla_{-}}
&{\underline{\Z} (G,G_{V})_{-}.}
}
\end{displaymath}
\end{enumerate}
\end{prop}
The Mackey functor $\underline{\Z(G,G_{V})}$ is one of those defined
(with different notation) in \cite[Def. 2.1]{HHR:RO(G)}.
\begin{defin}\label{def-aeu}
{\bf Three elements in $\upi_{\star}^{G}(\HZ)$.} Let $V$
be an actual (as opposed to virtual) {\rep} of the finite cyclic
2-group $G$ with $V^{G}=0$ and isotropy group $G_{V}$.
\begin{enumerate}
\item [(i)] The {\eqvr} inclusion ${S^{0}\to S^{V}}$ defines an
element in $\upi_{-V}S^{0} (G/G)$ via the isomorphisms
\begin{displaymath}
\upi_{-V}S^{0} (G/G)=
\upi_{0}S^{V} (G/G)= \pi_{0}S^{V^{G}}=\pi_{0}S^{0}=\Z,
\end{displaymath}
\noindent and we will use the symbol $a_{V}$ to denote its image in
$\upi_{-V}\HZZ (G/G)$.
\item [(ii)] The underlying equivalence $S^{V}\to S^{|V|}$ defines an
element in
\begin{displaymath}
\upi_{V}S^{|V|} (G/G_{V}) = \upi_{V-|V|}S^{0} (G/G_{V})
\end{displaymath}
\noindent and we will use the symbol $e_{V}$ to denote its image in
$\upi_{V-|V|}\HZZ (G/G_{V})$.
\item [(iii)] If $W$ is an oriented {\rep} of $G$ (we do not require
that $W^{G}=0$), there is a map
\begin{displaymath}
\Delta :\ints \to C^{W}_{|W|}= \Z[G/G_{W}]
\end{displaymath}
\noindent as in Proposition \ref{prop-top} giving an element
\begin{displaymath}
u_{W}\in \uH_{|W|}S^{W} (G/G) = \upi_{|W|-W} \HZ(G/G).
\end{displaymath}
For nonoriented $W$, Proposition \ref{prop-top} gives a map
\begin{displaymath}
\Delta_{-} :\ints_{-} \to C^{W}_{|W|}
\end{displaymath}
\noindent and an element
\begin{displaymath}
u_{W}\in \uH_{|W|}S^{W} (G/G') = \upi_{|W|-W} \HZ(G/G').
\end{displaymath}
\noindent
\end{enumerate}
\end{defin}
The element $u_{W}$ above is related to the element $\tilde{u}_{V}$ of
(\ref{eq-u{V-V'}}) as follows.
\begin{lem}\label{lem-uV}
{\bf The restriction of $u_{W}$ to a unit and permanent cycle.} Let $W$ be a
nontrivial {\rep} of $G$ with $H=G_{W}$. Then the homeomorphism
\begin{displaymath}
\Sigma^{-W}\tilde{u}_{W}:G/H_{+}\wedge S^{|W|-W}\to G/H_{+}
\end{displaymath}
\noindent of
(\ref{eq-u{V-V'}}) induces an isomorphism $\upi_{0}\HZZ (G/H)\to
\upi_{|W|-W}\HZZ (G/H)$ sending the unit to $\Res_{H}^{K}(u_{W})$
for $u_{W}$ as defined in (iii) above and $K=G$ or $G'$ depending on
the orientability of $W$.
The product
\begin{displaymath}
\Res_{H}^{K}(u_{W})e_{W}\in \upi_{0}\HZZ (G/H)=\Z
\end{displaymath}
\noindent is a generator, so $e_{W}$ and $\Res_{H}^{K}(u_{W})$ are
units in the ring $\upi_{\star}\HZZ (G/H)$, and $\Res_{H}^{K}(u_{W})$
is in the Hurewicz image of $\upi_{\star}S^{0} (G/H)$.
\end{lem}
\proof
The diagram
\begin{displaymath}
\xymatrix
@R=5mm
@C=10mm
{G/K_{+}\wedge S^{|W|-W}
&G/H_{+}\wedge S^{|W|-W}\ar[r]^(.6){\tilde{u}_{W}}
\ar[l]_(.5){\mbox{fold} }
&G/H_{+}
}
\end{displaymath}
\noindent induces (via the functor $[\cdot , \HZZ]^{G}$)
\begin{displaymath}
\xymatrix
@R=5mm
@C=10mm
{{\upi_{|W|-W}\HZZ (G/K)}\ar[r]^(.5){\Res_{H}^{K}}\ar@{=}[d]^(.5){}
&{\upi_{|W|-W}\HZZ (G/H)}\ar@{=}[d]^(.5){}
&{\upi_{0}\HZZ (G/H)}\ar[l]_(.4){\cong }\ar@{=}[d]^(.5){} \\
{\underline{H}_{|W|}S^{W} (G/K)}
&{\underline{H}_{|W|}S^{W} (G/H)}
&\Z
}
\end{displaymath}
\noindent The restriction map is an isomorphism by Proposition
\ref{prop-top} and the group on the left is generated by $u_{W}$.
The product is the composite of $H$-maps
\begin{displaymath}
\xymatrix
@R=5mm
@C=10mm
{
{S^{W}}\ar[r]^(.5){e_{W}}
&{S^{|W|}}\ar[r]^(.5){\Res_{H}^{K}(u_{W})}
&{\Sigma^{W}\HZZ,}
}
\end{displaymath}
\noindent which is the standard inclusion.
\qed\bigskip
Note that $a_{V}$ and $e_{V}$ are induced by maps to {\eqvr} spheres
while $u_{W}$ is not. This means that in any {\SS} based on a
filtration where the subquotients are {\eqvr} $\HZ $-modules, elements
defined in terms of $a_{V}$ and $e_{V}$ will be permanent cycles,
while multiples and powers of $u_{W}$ can support nontrivial
differentials. Lemma \ref{lem-uV} says a certain restriction of
$u_{W}$ is a permanent cycle.
Each nonoriented $V$ has the form $W+\sigma $ where $\sigma $ is the
sign {\rep} and $W$ is oriented. It follows that
\begin{displaymath}
u_{V}=u_{\sigma}\Res_{G'}^{G}(u_{W})\in \upi_{|V|-V}H\Z (G/G').
\end{displaymath}
Note also that $a_{0}=e_{0}=u_{0}=1$. The trivial
representations contribute nothing to $\pi_{\star}(H\Z)$. We can limit
our attention to {\rep}s $V$ with $V^{G}=0$. Among such {\rep}s
of cyclic 2-groups, the oriented ones are precisely the
ones of even degree.
\begin{lem}\label{lem-aeu}
{\bf Properties of $a_{V}$, $e_{V}$ and $u_{W}$.} The elements
$a_{V}\in \upi_{-V}\HZ(G/G)$, $e_{V}\in \underline{\pi
}_{V-|V|}\HZ(G/G_{V})$ and $u_{W}\in \upi_{|W|-W}\HZ(G/G)$ for
$W$ oriented of Definition \ref{def-aeu} satisfy the following.
\begin{enumerate}
\item \label{aeu:i}
$a_{V+W}=a_{V}a_{W}$ and $u_{V+W}=u_{V}u_{W}$.
\item \label{aeu:ii}
$|G/G_{V}|a_{V}=0$ where $G_{V}$ is the isotropy group of $V$.
\item \label{aeu:iii} For oriented $V$, $\Tr_{G_{V}}^{G}(e_{V})$
and $\Tr_{G_{V}}^{G'}(e_{V+\sigma})$ have infinite order, while\linebreak
$\Tr_{G_{V}}^{G}(e_{V+\sigma})$ has order 2 if
$|V|>0$ and $\Tr_{G_{V}}^{G}(e_{\sigma})=\Tr_{G'}^{G}(e_{\sigma})=0$.
\item \label{aeu:iv} For oriented $V$ and $G_{V}\subseteq H\subseteq G$
\begin{align*}
\Tr_{G_{V}}^{G}(e_{V})u_{V}
& = |G/G_{V}|\in \upi_{0}\HZZ (G/G)=\Z \\
\aand
\Tr_{G_{V}}^{G'}(e_{V+\sigma })u_{V+\sigma }
& = |G'/G_{V}|\in \upi_{0}\HZZ (G/G')=\Z
\qquad \mbox{for }|V|>0.
\end{align*}
\item \label{aeu:v}
$a_{V+W}\Tr_{G_{V}}^{G}(e_{V+U})=0$ if $|V|>0$.
\item \label{aeu:vi}
For $V$ and $W$ oriented, $u_{W}\Tr_{G_{V}}^{G}(e_{V+W})
=|G_{V}/G_{V+W}|\Tr_{G_{V}}^{G}(e_{V})$.
\item \label{aeu:vii}
{\bf The gold (or $au$) relation. } For $V$ and $W$ oriented {\rep}s
of degree 2 with $G_{V}\subseteq G_{W}$, ${a_{W}u_{V} =
|G_{W}/G_{V}|a_{V}u_{W}}$.
\end{enumerate}
For nonoriented $W$ similar statements hold in $\upi_{\star}\HZ
(G/G')$. $2W$ is oriented and $u_{2W}$ is defined in
$\upi_{2|W|-2W}\HZZ (G/G)$ with ${\Res_{G'}^{G}(u_{2W})=u_{W}^{2}}$.
\end{lem}
\proof (\ref{aeu:i})
This follows from the existence of the pairing
$C^{V}\otimes C^{W}\to C^{V+W}$. It induces an isomorphism in $H_{0}$
and (when both $V$ and $W$ are oriented) in $H_{|V+W|}$.
(\ref{aeu:ii}) This holds because $H_{0} (V)$ is killed by $|G/G_{V}|$.
(\ref{aeu:iii}) This follows from Proposition \ref{prop-top}.
(\ref{aeu:iv})
Using the Frobenius relation we have
\begin{align*}
\Tr_{G_{V}}^{G}(e_{V})u_{V}
& = \Tr_{G_{V}}^{G}(e_{V}\Res_{G_{V}}^{G}(u_{V}))
= \Tr_{G_{V}}^{G}(1) \qquad \mbox{by \ref{lem-uV} } \\
& = |G/G_{V}| \\
\Tr_{G_{V}}^{G'}(e_{V+\sigma })u_{V+\sigma }
& = \Tr_{G_{V}}^{G'}(e_{V+\sigma }\Res_{G_{V}}^{G'}(u_{V+\sigma }))
= \Tr_{G_{V}}^{G'}(1) = |G'/G_{V}|.
\end{align*}
(\ref{aeu:v}) We have
\begin{displaymath}
a_{V+W}\Tr_{G_{V}}^{G}(e_{V+U}):S^{-|V|-|U|} \to S^{W-U}.
\end{displaymath}
\noindent It is null because the bottom cell of ${S^{W-U}}$ is in
dimension ${-|U|}$.
(\ref{aeu:vi}) Since $V$ is oriented, then we are computing in a
torsion free group so we can tensor with the rationals. It follows
from (\ref{aeu:iv}) that
\begin{align*}
\Tr_{G_{V+W}}^{G}(e_{V+W})
& = \frac{|G/G_{V+W}|}{u_{V}u_{W}}\\
\aand
\Tr_{G_{V}}^{G}(e_{V})
& = \frac{|G/G_{V}|}{u_{V}}\\
\sso
u_{W}\Tr_{G_{V+W}}^{G}(e_{V+W} )
& = \frac{|G/G_{V+W}|}{u_{V}}
= |G_{V}/G_{V+W}|\Tr_{G_{V}}^{G}(e_{V}).
\end{align*}
(\ref{aeu:vii})
For $G=C_{2^{n}}$, each oriented {\rep} of degree 2 is $2$-locally
equivalent to a $\lambda_{j}$ for $0\leq j<n$. The isotropy group is
$G_{\lambda_{j}} = C_{2^{j}}$. Hence the assumption that
$G_{V}\subset G_{W}$ is can be replaced with $V=\lambda_{j}$ and
$W=\lambda_{k}$ with $0\leq j<k<n$. the statment we wish to prove is
\begin{displaymath}
a_{\lambda_{k}}u_{\lambda_{j}}=2^{k-j}a_{\lambda_{j}}u_{\lambda_{k}}.
\end{displaymath}
One has a map $S^{\lambda_{j}}\to S^{\lambda_{k}}$ which is the
suspension of the $2^{k-j}$th power map on the equatorial circle.
Hence its underlying degree is $2^{k-j}$. We will denote it by
$a_{\lambda_{k}}/a_{\lambda_{j}}$ since there is a diagram
\begin{displaymath}
\xymatrix
@R=3mm
@C=20mm
{
&S^{\lambda_{j}}\ar[dd]^(.5){a_{\lambda_{k}}/a_{\lambda_{j}}}\\
S^{0}\ar[ru]^(.5){a_{\lambda_{j}}}\ar[rd]_(.5){a_{\lambda_{k}}}\\
&S^{\lambda_{k}}.
}
\end{displaymath}
We claim there is a similar diagram
\begin{numequation}\label{eq-au-claim}
\begin{split}
\xymatrix
@R=3mm
@C=20mm
{
&S^{\lambda_{k}}\wedge \HZZ\ar[dd]^(.5){u_{\lambda_{j}}/u_{\lambda_{k}}}\\
S^{2}\ar[ru]^(.5){u_{\lambda_{k}}}\ar[rd]_(.5){u_{\lambda_{j}}}\\
&S^{\lambda_{j}}\wedge \HZZ.
}
\end{split}
\end{numequation}
\noindent in which the underlying degree of the vertical map is one.
Smashing $a_{\lambda_{k}}/a_{\lambda_{j}}$ with $\HZZ$ and composing
with $u_{\lambda_{j}}/u_{\lambda_{k}}$ gives a factorization of the
degree $2^{k-j}$ map on $S^{\lambda_{j}}\wedge \HZZ$. Thus we have
\begin{align*}
\frac{u_{\lambda_{j}}}{u_{\lambda_{k}}}
\frac{a_{\lambda_{k}}}{a_{\lambda_{j}}}
& = 2^{k-j} \\
u_{\lambda_{j}}a_{\lambda_{k}}
& = 2^{k-j} u_{\lambda_{k}}a_{\lambda_{j}}
\end{align*}
\noindent as desired.
The vertical map in (\ref{eq-au-claim}) would follow from a map
\begin{displaymath}
S^{\lambda_{k}-\lambda_{j}}\to \HZZ
\end{displaymath}
\noindent with underlying degree one. Let $G=C_{2^{n}}$ and $G \supset
H=C_{2^{j}}$. Then $S^{-\lambda_{j}}$ has a cellular structure of the
form
\begin{displaymath}
G/H_{+}\wedge S^{-2} \cup G/H_{+} \wedge e^{-1} \cup e^{0}.
\end{displaymath}
\noindent We need to smash this with $S^{\lambda_{k}}$. Since
$\lambda_{k}$ restricts trivially to $H$,
\begin{displaymath}
G/H_{+}\wedge
S^{\lambda_{k}}=G/H_{+}\wedge S^{2}.
\end{displaymath}
\noindent This means
\begin{displaymath}
S^{\lambda_{k}-\lambda_{j}}
= S^{\lambda _{k}}\wedge S^{-\lambda_{j}}
= G/H_{+}\wedge S^{0} \cup
G/H_{+} \wedge e^{1} \cup e^{0}\wedge S^{\lambda_{k}}.
\end{displaymath}
\noindent Thus its cellular chain complex has the form
\begin{displaymath}
\xymatrix
@R=5mm
@C=15mm
{
2 &\Z[G/K] \ar[d]^(.5){1-\gamma }\ar[drr]^(.5){\Delta }\\
1 &\Z[G/K] \ar[d]^(.5){\nabla }\ar[drr]^(.5){-\Delta }
& &\Z[G/H]\ar[d]^(.5){1-\gamma }\\
0 &\Z & &\Z[G/H]
}
\end{displaymath}
\noindent where $K=G/C_{p^{k}}$ and the left column is the chain
complex for $S^{\lambda_{k}}$.
There is a corresponding chain complex of fixed point Mackey functors.
Its value on the $G$-set $G/L$ for an arbitrary subgroup $L$ is
\begin{displaymath}
\xymatrix
@R=5mm
@C=15mm
{
2 &\Z[G/\max(K,L)] \ar[d]^(.5){1-\gamma }\ar[drr]^(.5){\Delta }\\
1 &\Z[G/\max(K,L)] \ar[d]^(.5){\nabla }\ar[drr]^(.5){-\Delta }
& &\Z[G/\max(H,L)]\ar[d]^(.5){1-\gamma }\\
0 &\Z & &\Z[G/\max(H,L)]
}
\end{displaymath}
\noindent For each $L$ the map $\Delta $ is injective and maps the
kernel of the first $1-\gamma $ isomorphically to the kernel of the
second one. This means we can replace the above by a diagram of the
form
\begin{displaymath}
\xymatrix
@R=5mm
@C=15mm
{
1 &\coker (1-\gamma ) \ar[d]^(.5){\nabla }\ar[drr]^(.5){-\Delta }\\
0 &\Z & &\coker (1-\gamma )
}
\end{displaymath}
\noindent where each cokernel is isomorphic to $\Z$ and each
map is injective.
This means that $\underline{H}_{*}S^{\lambda_{k}-\lambda_{j}}$ is
concentrated in degree 0 where it is the pushout of the diagram above,
meaning a Mackey functor whose value on each subgroup is $\Z$. Any
such Mackey functor admits a map to $\underline{\Z}$ with underlying
degree one. This proves the claim of (\ref{eq-au-claim}). \qed\bigskip
The $\Z$-valued Mackey functor
$\underline{H}_{0}S^{\lambda_{k}-\lambda_{j}}$ is discussed in more
detail in \cite{HHR:RO(G)}, where it is denoted by $\underline{\Z} (k,j)$.
\section{Generalities on differentials and Mackey functor extensions}
\label{sec-gendiffs}
Before proceeding with a discussion about sepctral sequences, we need
the following.
\begin{rem}\label{rem-abuse}
{\bf Abusive {\SS } notation}. When $d_{r} (x)$ is a nontrivial
element of order 2, the elements $2x$ and $x^{2}$ both survive to
$\underline{E}_{r+1}$, but in that group they are not the products
indicated by these symbols since $x$ itself is no longer present.
More geneally if $d_{r} (x)=y$ and $\alpha y=0$ for some $\alpha $,
then $\alpha x$ is present in $\underline{E}_{r+1}$. This abuse of
notation is customary because it would be cumbersome to rename these
elements when passing from $\underline{E}_{r}$ to
$\underline{E}_{r+1}$. We will sometimes denote them by $[2x]$,
$[x^{2}]$ and $[\alpha x]$ respectively to emphasize their
indecomposability.
\end{rem}
Now we make some observations about the relation between exotic
transfers and restriction with certain differentials in the slice
{\SS}. By ``exotic'' we mean in a higher filtration. In a {\SS} of
Mackey functors converging to $\upi_{\star}X$, it can happen that an
element $x\in \upi_{V}X (G/H)$ has filtration $s$, but its restriction
or transfer has a higher filtration. {\em In the {\SS} charts in this
paper, exotic transfers and restrictions will be indicated by
{\color{blue}blue} and {\color{green} dashed green} lines respectively.}
\begin{lem}\label{lem-hate}
{\bf Restriction kills $a_{\sigma}$ and $a_{\sigma }$ kills
transfers.} Let $G$ be a finite cyclic 2-group with sign {\rep}
$\sigma $ and index 2 subgroup $G'$, and let $X$ be a $G$-spectrum.
Then in $\upi_{*}X (G/G)$ the image of $\Tr_{G'}^{G}$ is the kernel of
multiplication by $a_{\sigma}$, and the kernel of $\Res_{G'}^{G}$ is
the image of multiplication by $a_{\sigma}$.
Suppose further that 4 divides the order of $G$ and let $\lambda $ be
the degree 2 representation sending a generator $\gamma \in G$ to a
rotation of order 4. Then restriction kills $2a_{\lambda }$ and
$2a_{\lambda }$ kills transfers.
\end{lem}
\proof
Consider the cofiber sequence obtained by smashing $X$ with
\begin{numequation}\label{eq-hate}
\begin{split}
\xymatrix
@R=5mm
@C=10mm
{
S^{-1}\ar[r]^(.5){a_{\sigma}}
&S^{\sigma -1}\ar[r]^(.5){}
&G_{+}\smashove{G'}S^{0}\ar[r]^(.5){}
&S^{0}\ar[r]^(.5){a_{\sigma}}
&S^{\sigma}
}
\end{split}
\end{numequation}
\noindent Since $(G_{+}\smashove{G'}X)^{G}$ is equivalent to $X^{G'}$,
passage to fixed point spectra gives
\begin{displaymath}
\xymatrix
@R=5mm
@C=10mm
{
\Sigma^{-1}X^{G}\ar[r]^(.5){}
&\left(\Sigma^{\sigma -1}X \right)^{G}\ar[r]^(.5){}
&X^{G'}\ar[r]^(.5){}
&X^{G}\ar[r]^(.5){}
&\left(\Sigma^{\sigma}X \right)^{G}
}
\end{displaymath}
\noindent so the exact sequence of homotopy groups is
\begin{center}
\includegraphics[width=12cm]{hate-diagram.pdf}
\end{center}
\noindent Note that the isomorphism $u_{\sigma}$ is invertible. This
gives the exactness required by both statements.
For the statements about $a_{\lambda }$, note that $\lambda $ restricts
to $2\sigma_{G'}$, where $\sigma_{G'}$ is the sign {\rep} for the
index 2 subgroup $G'$. It follows that $\Res_{G'}^{G}(a_{\lambda
})=a_{\sigma_{G'}}^{2}$, which has order 2. Using the Frobenius
relation, we have for $x\in \upi_{*}X (G/G')$,
\begin{displaymath}
2a_{\lambda }\Tr_{G'}^{G}(x)
= \Tr_{G'}^{G}(\Res_{G'}^{G}(2a_{\lambda })x)
= \Tr_{G'}^{G}(2a_{\sigma_{G'} }^{2}x)
= 0.
\end{displaymath}
\qed\bigskip
This implies that when $a_{\sigma}x$ is killed by a differential but
$x\in \EE_{r} (G/G)$ is not, then $x$ represents an element
that is $\Tr_{G'}^{G}(y)$ for some $y$ in lower filtration. Similarly
if $x$ supports a nontrivial differential but $a_{\sigma}x$ is a
nontrivial permanent cycle, then the latter represents an element with
a nontrivial restriction to $G'$ of higher filtration. In both cases
the converse also holds.
\begin{thm}\label{thm-exotic}
{\bf Exotic transfers and restrictions in the $RO (G)$-graded slice
{\SS}.} Let $G$ be a finite cyclic 2-group with index 2 subgroup $G'$
and sign {\rep} $\sigma $, and let $X$ be a $G$-{\eqvr} spectrum
with $x\in \EE_{r}^{s,V}X (G/G)$ (for $V\in RO (G)$) in the
slice {\SS} for $X$. Then
\begin{enumerate}
\item [(i)] Suppose there is a permanent cycle $y'\in
\EE_{r}^{s+r,V+r-1}X (G/G')$. Then there is a nontrivial
differential $d_{r} (x)=\Tr_{G'}^{G}(y')$ iff $[a_{\sigma}x]$ is a
permanent cycle with $\Res_{G'}^{G}(a_{\sigma}x)=u_{\sigma}y'$. In
this case $[a_{\sigma }x]$ represents the Toda bracket $\langle
a_{\sigma },\,\Tr_{G'}^{G} ,\,y' \rangle$.
\item [(ii)] Suppose there is a permanent cycle $y\in
\EE_{r}^{s+r-1 ,V+r+\sigma -2}X (G/G)$. Then there is a
nontrivial differential $d_{r}(x)=a_{\sigma}y$ iff $\Res_{G'}^{G}(x)$
is a permanent cycle with
$\Tr_{G'}^{G}(u_{\sigma}^{-1}\Res_{G'}^{G}(x))=y$. In this case
$\Res_{G'}^{G}(x)$ represents the Toda bracket $\langle
\Res_{G'}^{G},\,a_{\sigma } ,\,y \rangle$.
\end{enumerate}
\end{thm}
In each case a nontrivial $d_{r}$ is equivalent to a Mackey functor
extension raising filtration by $r-1$. In (i) the permanent cycle
$a_{\sigma}x$ is not divisible in $\upi_{\star}X$ by $a_{\sigma}$
and therefore could have a nontrivial restriction in a higher
filtration. Similarly in (ii) the element denoted by
$\Res_{G'}^{G}(x)$ is not a restriction in $\upi_{\star}X$, so we
cannot use the Frobenius relation to equate
$\Tr_{G'}^{G}(u_{\sigma}^{-1}\Res_{G'}^{G}(x))$ with
$\Tr_{G'}^{G}(u_{\sigma}^{-1})x$.
We remark that the proof below makes no use of any properties
specific to the slice filtration. The result holds for any {\eqvr}
filtration with suitable formal properties.
Before giving the proof we need the following.
\begin{lem}\label{lem-formal}
{\bf A formal observation}. Suppose we have a
commutative diagram up to sign
\begin{displaymath}
\xymatrix
@R=5mm
@C=10mm
{
A_{0,0}\ar[r]^(.5){a_{0,0}}\ar[d]^(.5){b_{0,0}}
&A_{0,1}\ar[r]^(.5){a_{0,1}}\ar[d]^(.5){b_{0,1}}
&A_{0,2}\ar[r]^(.5){a_{0,2}}\ar[d]^(.5){b_{0,2}}
&\Sigma A_{0,0}\ar[d]^(.5){b_{0,0}}\\
A_{1,0}\ar[r]^(.5){a_{1,0}}\ar[d]^(.5){b_{1,0}}
&A_{1,1}\ar[r]^(.5){a_{1,1}}\ar[d]^(.5){b_{1,1}}
&A_{1,2}\ar[r]^(.5){a_{1,2}}\ar[d]^(.5){b_{1,2}}
&\Sigma A_{1,0}\ar[d]^(.5){b_{1,0}}\\
A_{2,0}\ar[r]^(.5){a_{2,0}}\ar[d]^(.5){b_{2,0}}
&A_{2,1}\ar[r]^(.5){a_{2,1}}\ar[d]^(.5){b_{2,1}}
&A_{2,2}\ar[r]^(.5){a_{2,2}}\ar[d]^(.5){b_{2,2}}
&\Sigma A_{2,0}\ar[d]^(.5){b_{2,0}}\\
\Sigma A_{0,0}\ar[r]^(.5){a_{0,0}}
&\Sigma A_{0,1}\ar[r]^(.5){a_{0,1}}
&\Sigma A_{0,2}\ar[r]^(.5){a_{0,2}}
&\Sigma^{2} A_{0,0}
}
\end{displaymath}
\noindent in which each row and column is a cofiber sequence. Then
suppose that from some spectrum $W$ we have a map $f_{3}$ and
hypothetical maps $f_{1}$ and $f_{2}$ making the following diagram
commute up to sign, where $c_{i,j}=b_{i,j+1}a_{i,j}=a_{i+1,j}b_{i,j}$.
\begin{numequation}\label{eq-formal}
\begin{split}
\xymatrix
@R=5mm
@C=10mm
{
W\ar[rrr]^(.5){f_{3}}\ar@{-->}[ddr]^(.5){f_{1}}
\ar@{-->}[drr]^(.5){f_{2}}\ar[ddd]^(.5){f_{3}}
& & &\Sigma A_{0,0}\ar[d]^(.5){b_{0,0}}\ar[dr]^(.5){c_{0,0}}\\
& &A_{1,2}\ar[r]^(.5){a_{1,2}}\ar[d]^(.5){b_{1,2}}\ar[dr]^(.5){c_{1,2}}
&\Sigma A_{1,0}\ar[d]^(.5){b_{1,0}}\ar[r]^(.5){a_{1,0}}
&\Sigma A_{1,1}\ar[d]^(.5){b_{1,1}}\\
&A_{2,1}\ar[r]^(.5){a_{2,1}}\ar[d]^(.5){b_{2,1}}\ar[dr]^(.5){c_{2,1}}
&A_{2,2}\ar[r]^(.5){a_{2,2}}\ar[d]^(.5){b_{2,2}}
&\Sigma A_{2,0}\ar[r]^(.5){a_{2,0}}
&\Sigma A_{2,1}\\
\Sigma A_{0,0}\ar[r]^(.5){a_{0,0}}\ar[rd]_(.5){c_{0,0}}
&\Sigma A_{0,1}\ar[r]^(.5){a_{0,1}}\ar[d]^(.5){b_{0,1}}
&\Sigma A_{0,2}\ar[d]^(.5){b_{0,2}}
&\\
&\Sigma A_{1,1}\ar[r]^(.5){a_{1,1}}
&\Sigma A_{1,2}
}
\end{split}
\end{numequation}
\noindent Then $f_{1}$ exists iff $f_{2}$ does. When this happens,
$c_{0,0}f_{3}$ is null and we have Toda brackets
\begin{displaymath}
\langle a_{1,1},\, c_{0,0} ,\,f_{3} \rangle \ni f_{2}
\qquad \aand
\langle b_{1,1},\, c_{0,0} ,\,f_{3} \rangle \ni f_{1}.
\end{displaymath}
\end{lem}
\proof Let $R$ be the pullback of $a_{2,1}$ and $b_{1,2}$, so we have a diagram
\begin{displaymath}
\xymatrix
@R=5mm
@C=10mm
{
{}
&{A_{0,2}}\ar@{=}[r]^(.5){}\ar[d]^(.5){}
&{A_{0,2}}\ar[d]^(.5){b_{0,2}}
&{}\\
{A_{2,0}}\ar@{=}[d]^(.5){}\ar[r]^(.5){}
&{R}\ar[d]^(.5){}\ar[r]^(.5){}
&{A_{1,2}}\ar[d]^(.5){b_{1,2}}\ar[r]^(.5){c_{1,2}}
&{\Sigma A_{2,0}}\ar@{=}[d]^(.5){}\\
{A_{2,0}}\ar[r]^(.5){a_{2,0}}
&{A_{2,1}}\ar[d]^(.5){c_{2,1}}\ar[r]^(.5){a_{2,1}}
&{A_{2,2}}\ar[d]^(.5){b_{2,2}}\ar[r]^(.5){a_{2,2}}
&{\Sigma A_{2,0}}\\
{}
&{\Sigma A_{0,2}}\ar@{=}[r]^(.5){}
&{\Sigma A_{0,2}}
&{}
}
\end{displaymath}
\noindent in which each row and column is a cofiber sequence. Thus we
see that $R$ is the fiber of both $c_{1,2}$ and $c_{2,1}$. If $f_{1}$
exists, then
\begin{displaymath}
c_{2,1}f_{1}=a_{0,1}b_{2,1}f_{1}=a_{0,1}a_{0,0}f_{3}
\end{displaymath}
\noindent which is null homotopic, so $f_{1}$ lifts to $R$, which
comes equipped with a map to $A_{1,2}$, giving us $f_{2}$. Conversely
if $f_{2}$ exists, it lifts to $R$, which comes equipped with a map to
$A_{2,1}$, giving us $f_{1}$.
The statement about Toda brackets follows from the way they are defined.
\qed\bigskip
\noindent {\em Proof of Theorem \ref{thm-exotic}}.
For a $G$-spectrum $X$ and integers $a<b<c\leq
\infty $ there is a cofiber sequence
\begin{displaymath}
\xymatrix
@R=5mm
@C=10mm
{
P_{b+1}^{c}X\ar[r]^(.5){i}
&P_{a}^{c}X\ar[r]^(.5){j}
&P_{a}^{b}X\ar[r]^(.5){k}
&\Sigma P_{b+1}^{c}X.
}
\end{displaymath}
\noindent When $c=\infty $, we omit it from the notation. We will
combine this and the one of (\ref{eq-hate}) to get a diagram similar
to (\ref{eq-formal}) with $W=S^{V}$ to prove our two statements.
For (i) note that $x\in \EE_{1}^{s,t}X (G/G)$ is by definition an
element in $\upi_{t-s}P^{s}_{s}X (G/G)$. We will assume for
simplicity that $s=0$, so $x$ is represented by a map from some
$S^{V}$ to $(P_{0}^{0}X)^{G}$. Its survival to $\EE_{r}$ and
supporting a nontrivial differential means that it lifts to
$(P_{0}^{r-2}X)^{G}$ but not to $(P_{0}^{r-1}X)^{G}$. The value of
$d_{r} (x)$ is represented by the composite $kx$ in the diagram below,
where we can use Lemma \ref{lem-formal}.
\begin{displaymath}
\xymatrix
@R=4mm
@C=12mm
{
S^{V-1}\ar[ddd]^(.5){y'}\ar@{-->}[ddr]^(.5){x}
\ar@{-->}[drr]^(.5){w}\ar[rrr]^(.5){y'}
& & &(P_{r-1}X)^{G'} \ar[d]^(.5){i}\\
& &(\Sigma^{\sigma -1}P_{0}X)^{G}\ar[d]^(.5){j}
\ar[r]^(.55){u_{\sigma}^{-1}\Res_{G'}^{G}}
&(P_{0}X )^{G'}\ar[d]^(.5){j}\\
&(\Sigma^{-1}P_{0}^{r-2}X)^{G}\ar[d]^(.5){k}\ar[r]^(.5){a_{\sigma}}
&(\Sigma^{\sigma-1 }P_{0}^{r-2}X)^{G}\ar[d]^(.5){k}
\ar[r]^(.55){u_{\sigma}^{-1}\Res_{G'}^{G}}
&(P_{0}^{r-2}X)^{G'} \\
(P_{r-1}X)^{G'}\ar[r]^(.5){\Tr_{G'}^{G}}
&(P_{r-1}X)^{G}\ar[r]^(.5){a_{\sigma}}
&(\Sigma^{\sigma} P_{r-1}X)^{G}
}
\end{displaymath}
\noindent The commutativity of the lower left trapezoid is the
differential of (i),\linebreak ${d_{r} (x)=\Tr_{G'}^{G}(y')}$. The
existence of the map $w$ making the diagram commute follows from that
of $x$ and $y'$. It is the representative of $a_{\sigma}x$ as a
permanent cycle, which represents the indicated Toda bracket. The
commutativity of the upper right trapezoid identifies $y'$ as
$u_{\sigma}^{-1}\Res_{G'}^{G}(x)$ as claimed. For the converse we
have the existence of $y'$ and $w$ and hence that of $x$.
The second statement follows by a similar argument based on the diagram
\begin{displaymath}
\xymatrix
@R=4mm
@C=13mm
{
S^{V+\sigma -1}\ar[ddd]^(.5){y}\ar@{-->}[ddr]^(.5){x}
\ar@{-->}[drr]^(.5){w}\ar[rrr]^(.5){y}
& & &(P_{r-1}X)^{G} \ar[d]^(.5){i}\\
& &(P_{0}X)^{G'}\ar[d]^(.5){j}
\ar[r]^(.5){\Tr_{G'}^{G}}
&( P_{0}X )^{G}\ar[d]^(.5){j}\\
&(\Sigma^{\sigma -1}P_{0}^{r-2}X)^{G}\ar[d]^(.5){k}
\ar[r]^(.55){u_{\sigma}^{-1}\Res_{G'}^{G}}
&(P_{0}^{r-2}X)^{G'}\ar[d]^(.5){k}
\ar[r]^(.5){\Tr_{G'}^{G}}
&( P_{0}^{r-2}X)^{G} \\
( P_{r-1}X)^{G}\ar[r]^(.5){a_{\sigma}}
&(\Sigma^{\sigma} P_{r-1}X)^{G}
\ar[r]^(.55){u_{\sigma}^{-1}\Res_{G'}^{G}}
&(\Sigma P_{r-1}X)^{G'}.
&
}
\end{displaymath}
\noindent Here $w$ represents $u_{\sigma }^{-1}\Res_{G'}^{G}(x)$ as a
permement cycle, so we get a Toda bracket containing
$\Res_{G'}^{G}(x)$ as indicated. \qed
\bigskip
Next we study the way differentials interact with the norm.
Suppose we have a subgroup $H\subset G$ and an $H$-{\eqvr} ring
spectrum $X$ with $Y=N_{H}^{G}X$. Suppose we have {\SS}s converging
to $\upi_{\star}X$ and $\upi_{\star}Y$ based on towers
\begin{displaymath}
...\to P_{n}^{H}X \to P_{n-1}^{H}X \to \dotsb
\qquad \aand
...\to P_{n}^{G}Y \to P_{n-1}^{G}Y \to \dotsb
\end{displaymath}
\noindent for functors $P_{n}^{H}$ and $P_{n}^{G}$ equipped with
suitable maps
\begin{displaymath}
P^{H}_{m}X \wedge P^{H}_{n}X \to P^{H}_{m+n}X ,\quad
P^{G}_{m}Y \wedge P^{G}_{n}Y \to P^{G}_{m+n}Y
\quad \mbox{and}\quad
N_{H}^{G}P^{H}_{m}X \to P^{G}_{m|G/H|}Y.
\end{displaymath}
Our slice {\SS } for each of the spectra studied in this paper fits
this description.
\begin{thm}\label{thm-normdiff}
{\bf The norm of a differential.} Suppose we have spectral sequences
as described above and a differential $d_{r} (x)=y$ for $x\in
\EE_{r}^{s,\star}X (H/H)$. Let $\rho=\Ind_{H}^{G}1 $ and suppose that
$a_{\rho }$ has filtration $|G/H|-1$. Then in the {\SS } for
$Y=N_{H}^{G}X$,
\begin{displaymath}
d_{|G/H| (r-1) +1} (a_{\rho } N_{H}^{G}x)= N_{H}^{G}y
\in \EE_{|G/H| (r-1) +1}^{|G/H| (s+r),\star}Y (G/G).
\end{displaymath}
\end{thm}
\proof The differential can be represented by a diagram
\begin{displaymath}
\xymatrix
@R=5mm
@C=10mm
{
S^{V}\ar@{=}[r]^(.5){}
&S (1+V)\ar[r]^(.5){}\ar[d]_(.5){y}
&D (1+V)\ar[r]^(.5){}\ar[d]^(.5){}
&S^{1+V}\ar[d]^(.5){x}\\
&P_{s+r}^{H}X\ar[r]^(.5){}
&P_{s}^{H}X\ar[r]^(.5){}
&P^{H}_{s}X/P^{H}_{s+r}X
}
\end{displaymath}
\noindent for some orthogonal {\rep} $V$ of $H$, where each row is a
cofiber sequence. We want to apply the norm functor $N_{H}^{G}$ to
it. Let $W=\Ind_{H}^{G}V$. Then we get
\begin{displaymath}
\xymatrix
@R=5mm
@C=10mm
{
S^{W}\ar@{=}[r]^(.5){}
&N_{H}^{G}S (1+V)\ar[r]^(.5){}\ar[d]_(.5){N_{H}^{G}y}
&D (\rho +W)\ar[r]^(.5){}\ar[d]^(.5){}
&S^{\rho +W}\ar[d]^(.5){N_{H}^{G}x}\\
&N_{H}^{G}P_{s+r}^{H}X\ar[r]^(.5){}
&N_{H}^{G}P_{s}^{H}X\ar[r]^(.5){}
&N_{H}^{G} (P_{s}^{H}X/P_{s+r}^{H}X).
}
\end{displaymath}
\noindent Neither row of this diagram is a cofiber sequence, but we
can enlarge it to one where the top and bottom rows are, namely
\begin{displaymath}
\xymatrix
@R=5mm
@C=10mm
{
S^{W}\ar[r]^(.5){}\ar@{=}[d]
&D (1+W)\ar[r]^(.5){}\ar[d]^(.5){a_{\rho }}
&S^{1 +W}\ar[d]^(.5){a_{\rho }}\\
S^{W}\ar[r]^(.5){}\ar[d]_(.5){N_{H}^{G}y}
&D (\rho +W)\ar[r]^(.5){}\ar[d]^(.5){}
&S^{\rho +W}\ar[d]^(.5){N_{H}^{G}x}\\
N_{H}^{G}P_{s+r}^{H}X\ar[r]^(.5){}\ar[d]^(.5){}
&N_{H}^{G}P_{s}^{H}X\ar[r]^(.5){}\ar[d]^(.5){}
&N_{H}^{G} (P_{s}^{H}X/P_{s+r}^{H}X)\ar[d]^(.5){}\\
P_{(s+r)|G/H|}^{G}Y\ar[r]^(.5){}
&P_{s|G/H|}^{G}Y\ar[r]^(.5){}
&P_{s|G/H|}^{G}Y/P_{(s+r) |G/H|}^{G}Y
}
\end{displaymath}
\noindent Here the first two bottom vertical maps are part of the
multiplicative structure the {\SS } is assumed to have. Composing the
maps in the three columns gives us the diagram for the desired
differential. \qed\bigskip
Given a $G$-{\eqvr} ring spectrum $X$, let $X'=i^{*}_{H}X$ denote its
restriction as an $H$-spectrum. Then $N_{H}^{G}X'=X^{(|G/H|)}$ and
the multiplication on $X$ gives us a map from this smash product to
$X$. This gives us a map $\pi_{\star}X' \to \pi_{\star}X$ called the
{\em internal norm}, which we denote abusively by $N_{H}^{G}$. The
argument above yields the following.
\begin{cor}\label{cor-normdiff}
{\bf The internal norm of a differential.} With notation as above,
suppose we have a differential $d_{r} (x)=y$ for $x\in
\EE_{r}^{s,\star}X' (H/H)$. Then
\begin{displaymath}
d_{|G/H| (r-1) +1} (a_{\rho } N_{H}^{G}x)= N_{H}^{G}y
\in \EE_{|G/H| (r-1) +1}^{|G/H| (s+r),\star}X (G/G).
\end{displaymath}
\end{cor}
The following is useful in making such calculations. It is very
similar to \cite[Lemma 3.13]{HHR}.
\begin{lem}\label{lem-norm-au}
{\bf The norm of $a_{V}$ and $u_{V}$}. With notation as above, let
$V$ be a {\rep} of $H$ with $V^{H}=0$ and let $W=\Ind_{H}^{G}V$. Then
$N_{H}^{G} (a_{V})=a_{W}$. If $V$ is oriented (and hence even
dimensional, making $|V|\rho $ oriented), then
\begin{displaymath}
u_{|V|\rho }N
(u_{V})=u_{W}.
\end{displaymath}
\end{lem}
\proof The element $a_{V}$ is represented by the map $S^{0}\to
S^{V}$, the inclusion of the fixed point set. Applying the norm
functor to this map gives
\begin{displaymath}
S^{0}=N_{H}^{G}S^{0}\to N_{H}^{G}S^{V}=S^{W},
\end{displaymath}
\noindent which is $a_{W}$.
When $V$ is oriented, $u_{V}$ is represented by a map $S^{|V|}\to
S^{V}\wedge \HZZ$. Applying the norm functor and using the
multiplication in $\HZZ$ leads to a map
\begin{displaymath}
\xymatrix
@R=8mm
@C=5mm
{
{S^{|V|\rho }}\ar@{=}[r]^(.5){}
&{N_{H}^{G}S^{|V|}}\ar[rr]^(.5){N_{H}^{G}u_{V}}
& &{S^{W}\wedge \HZZ}
}
\end{displaymath}
\noindent Now smash both sides with $\HZZ$, precompose with
$u_{|V|\rho }$ and follow with the multiplication on $\HZZ$, giving
\begin{displaymath}
\xymatrix
@R=8mm
@C=8mm
{
{S^{|V||\rho |}}\ar[r]^(.4){u_{|V|\rho }}
&{S^{|V|\rho }\wedge \HZZ}\ar[rr]^(.45){N_{H}^{G}u_{V}\wedge \HZZ}
& &{S^{W}\wedge \HZZ\wedge \HZZ}\ar[r]^(.5){}
&{S^{W}\wedge \HZZ,}
}
\end{displaymath}
\noindent which is $u_{W}$ since $|W|=|V||\rho |$.
\qed\bigskip
\section{Some Mackey functors for $C_{4}$ and $C_{2}$}\label{sec-C4}
We need some notation for Mackey functors to be used in {\SS} charts.
{\em In this paper, when a cyclic group or subgroup appears as an
index, we will often replace it by its order.} We can specify Mackey
functors $\underline{M}$ for the group $C_{2}$ and $\underline{N}$ for
$C_{4}$ by means of Lewis diagrams (first introduced in
\cite{Lewis:ROG}),
\begin{numequation}\label{eq-Lewis}
\begin{split}
\xymatrix
@R=8mm
@C=10mm
{
{\underline{M} (C_{2}/C_{2})}\ar@/_/[d]_(.5){\Res_{1}^{2}}\\
{\underline{M} (C_{2}/e)}\ar@/_/[u]_(.5){\Tr_{1}^{2}}
}
\qquad \aand
\xymatrix
@R=4mm
@C=10mm
{
{\underline{N} (C_{4}/C_{4})}\ar@/_/[d]_(.5){\Res_{2}^{4}}\\
{\underline{N} (C_{4}/C_{2})}\ar@/_/[d]_(.5){\Res_{1}^{2}}
\ar@/_/[u]_(.5){\Tr_{2}^{4}}\\
{\underline{N} (C_{4}/e).}\ar@/_/[u]_(.5){\Tr_{1}^{2}}
}
\end{split}
\end{numequation}
\noindent We omit Lewis' looped arrow indicating the Weyl group action
on $\underline{M} (G/H)$ for proper subgroups $H$. This notation is
prohibitively cumbersome in {\SS } charts, so we will abbreviate
specific examples by more concise symbols. These are shown in Tables
\ref{tab-C2Mackey} and \ref{tab-C4Mackey}. {\em Admittedly some of
these symbols are arbitrary and take some getting used to, but we have
to start somewhere.} Lewis denotes the fixed point Mackey functor
for a $\Z G$-module $M$ by $R (M)$. He abbreviates $R (\Z)$and $R
(\Z_{-})$ by $R$ and $R_{-}$. He also defines (with similar
abbreviations) the orbit group Mackey functor $L (M)$ by
\begin{displaymath}
L (M) (G/H)=M/H.
\end{displaymath}
\noindent In this case each transfer map is the surjection of the
orbit space for a smaller subgroup onto that of a larger one. The
functors $R$ and $L$ are the left and right adjoints of the forgetful
functor $\underline{M}\mapsto \underline{M} (G/e)$ from Mackey
functors to $\Z G$-modules.
Over $C_{2}$ we have short exact sequences
\begin{displaymath}
\xymatrix
@R=5mm
@C=10mm
{
0 \ar[r]^(.5){}
&{\twobox} \ar[r]^(.5){}
&\Box \ar[r]^(.5){}
&\bullet \ar[r]^(.5){}
&0\\
0\ar[r]^(.5){}
&\bullet \ar[r]^(.5){}
&{\dot{\Box}} \ar[r]^(.5){}
&{\oBox} \ar[r]^(.5){}
&0\\
0\ar[r]^(.5){}
&\Box \ar[r]^(.5){}
&{\widehat{\Box}} \ar[r]^(.5){}
&{\oBox} \ar[r]^(.5){}
&0
}
\end{displaymath}
\noindent We can apply the induction functor to each them to get a
short exact sequence of Mackey functors over $C_{4}$.
Five of the Mackey functors in Table \ref{tab-C4Mackey} are fixed
point Mackey functors (\ref{eq-fpm}), meaning they are fixed points of
an underlying $Z[G]$-module $M$, such as $\Z[G]$ or
\begin{displaymath}
\begin{array}[]{rcllcl}
\Z & = & \Z[G]/ (\gamma -1)\qquad \qquad &
\Z[G/G']& = &\Z[G]/ (\gamma^{2} -1)\\
\Z_{-} & = &\Z[G]/ (\gamma +1)&
\Z[G/G']_{-}
& = & \Z[G]/ (\gamma^{2}+1)
\end{array}
\end{displaymath}
\begin{table}
\caption[Some {$C_{2}$}-Mackey functors]{Some $C_{2}$-Mackey functors}
\label{tab-C2Mackey}
\begin{center}
\begin{tabular}{|p{2.1cm}|c|c|c|c|c|
c|}
\hline
Symbol
&$\Box$
&$\oBox $
&$\bullet$
&$\twobox$
&$\dot{\Box} $
&$\widehat{\Box}$\\
\hline
Lewis\newline diagram
&$
\xymatrix
@R=5mm
@C=10mm
{
\Z\ar@/_/[d]_(.5){1}\\
\Z\ar@/_/[u]_(.5){2}
}$
&
$
\xymatrix
@R=4mm
@C=10mm
{
0\ar@/_/[d]_(.5){}\\
\Z_{-}\ar@/_/[u]_(.5){}
}$
&
$
\xymatrix
@R=4mm
@C=10mm
{
\Z/2\ar@/_/[d]_(.5){}\\
0\ar@/_/[u]_(.5){}
}
$ &
$\xymatrix
@R=5mm
@C=10mm
{
\Z\ar@/_/[d]_(.5){2}\\
\Z\ar@/_/[u]_(.5){1}
}$ &
$\xymatrix
@R=5mm
@C=10mm
{
\Z/2\ar@/_/[d]_(.5){0}\\
\Z_{-}\ar@/_/[u]_(.5){1}
}$
&
$\xymatrix
@R=5mm
@C=10mm
{
\Z\ar@/_/[d]_(.5){\Delta}\\
\Z[C_{2}]\ar@/_/[u]_(.5){\nabla}
}$\\
\hline
Lewis symbol
&$R$&$R_{-}$
&$\langle \Z/2 \rangle$
&$L$&$L_{-}$
&$R (\Z^{2})$ \\
\hline
\end{tabular}
\end{center}
\end{table}
\begin{table}
\caption[Some {$C_{4}$}-Mackey functors]{Some $C_{4}$-Mackey functors,
where $G=C_{4}$ and $G'$ is its index 2 subgroup. The notation
$\underline{\Z}(G,H)$ is defined in \ref{prop-top}(i).}
\label{tab-C4Mackey}
\begin{center}
\includegraphics[width=12cm]{fig-C4-Mackeya.pdf}
\end{center}
\end{table}
We will use the following notational conventions for $C_{4}$-Mackey functors.
\begin{enumerate}
\item [$\bullet$] Given a $C_{2}$-Mackey functor $\underline{M}$ with Lewis diagram
\begin{displaymath}
\xymatrix
@R=8mm
@C=10mm
{
{A}\ar@/_/[d]_(.5){\alpha }\\
{B}\ar@/_/[u]_(.5){\beta }
}
\end{displaymath}
\noindent with $A$ and $B$ cyclic, we will use the symbols
$\underline{M}$, $\underline{\overline{M} }$ and $\dot{\underline{M}}$
for the $C_{4}$-Mackey functors with Lewis diagrams
\begin{displaymath}
\xymatrix
@R=8mm
@C=10mm
{
{A}\ar@/_/[d]_(.5){\alpha }\\
{B}\ar@/_/[u]_(.5){\beta }\ar@/_/[d]_(.5){1 }\\
{B,}\ar@/_/[u]_(.5){2 }
}
\qquad
\xymatrix
@R=8mm
@C=10mm
{
{0}\ar@/_/[d]_(.5){ }\\
{A_{-}}\ar@/_/[d]_(.5){\alpha }\ar@/_/[u]_(.5){ }\\
{B_{-}}\ar@/_/[u]_(.5){\beta }
}
\qquad \aand
\xymatrix
@R=8mm
@C=10mm
{
{\Z/2}\ar@/_/[d]_(.5){0 }\\
{A_{-}}\ar@/_/[d]_(.5){\alpha }\ar@/_/[u]_(.5){\tau }\\
{B_{-}}\ar@/_/[u]_(.5){\beta }
}\end{displaymath}
\noindent where a generator $\gamma \in C_{4}$ acts via multiplication
by $-1$ on $A$ and $B$ in the second two, and the transfer $\tau $ is
nontrivial.
\item [$\bullet$] For a $C_{2}$-Mackey functor $\underline{M}$ we
will denote $\uparrow_{2}^{4}\underline{M}$ (see \ref{def-MF-induction}) by
$\widehat{\underline{M}}$. For a Mackey functor $\underline{M}$
defined over the trivial group, we will denote
$\uparrow_{1}^{2}\underline{M}$ and $\uparrow_{1}^{4}\underline{M}$ by
$\widehat{\underline{M}}$ and $\widehat{\widehat{\underline{M}}}$.
\end{enumerate}
Over $C_{4}$, in addition to the short exact sequences induced up from
$C_{2}$, we have
\begin{numequation}\label{eq-C4-SES}
\begin{split}
\xymatrix
@R=2mm
@C=10mm
{
0\ar[r]
&{\bullet}\ar[r]
&{\dot{\Box}}\ar[r]
&{\oBox}\ar[r]
&0\\
0\ar[r]
&{\JJ}\ar[r]
&{\circ }\ar[r]
&\bullet\ar[r]
&0\\
0\ar[r]
&{\JJ}\ar[r]
&{\JJbox}\ar[r]
&{\widehat{\oBox}}\ar[r]
&0\\
0\ar[r]
&{\bullet}\ar[r]
&{\circ }\ar[r]
&\JJJ\ar[r]
&0\\
0\ar[r]
&{\fourbox}\ar[r]
&\Box\ar[r]
&\circ\ar[r]
&0\\
0\ar[r]
&{\twobox}\ar[r]
&\Box\ar[r]
&\bullet\ar[r]
&0
}
\end{split}
\end{numequation}
\begin{defin}\label{def-enriched}
{\bf A $C_{4}$-enriched $C_{2}$-Mackey functor.}
For a $C_{2}$-Mackey functor $\uM$ as above,
$\widetilde{\uM}$ will denote the $C_{2}$-Mackey functor
enriched over $\Z[C_{4}]$ defined by
\begin{displaymath}
\widetilde{\uM} (S)
=\Z[C_{4}]\otimes_{\Z[C_{2}]}\uM (S)
\end{displaymath}
\noindent for a finite $C_{2}$-set $S$. Equivalently, in the notation
of Definition \ref{def-MF-induction}, $\widetilde{\uM}=
\downarrow_{2}^{4}\uparrow_{2}^{4}\uM$.
\end{defin}
\section{Some chain complexes of Mackey functors}\label{sec-chain}
As noted above, a $G$-CW complex $X$, meaning one built out of cells
of the form $G_{+}\smashove{H} e^{n}$, has a reduced cellular chain
complex of $\Z[G]$-modules $C_{*}X$, leading to a chain complex of
fixed point Mackey functors (see (\ref{eq-fpm})) $\uC_{*}X$. When
$X=S^{V}$ for a {\rep} $V$, we will denote this complex by
$\uC_{*}^{V}$; see (\ref{eq-CVn}). Its homology is the graded Mackey
functor $\uH_{*}X$. Here we will apply the methods of \S\ref{sec-HZZ}
to three examples.
\bigskip (i) Let $G=C_{2}$ with generator $\gamma $, and $X=S^{n\rho
}$ for $n>0$, where $\rho $ denotes the regular {\rep}. We have seen
before \cite[Ex. 3.7]{HHR} that it has a reduced cellular chain complex $C$
with
\begin{numequation}
\label{eq-C2chain}
C_{i} ^{n\rho_{2}} =\mycases{
\Z[G]/ (\gamma -1)
&\mbox{for }i=n\\
\Z[G]
&\mbox{for }n<i\leq 2n\\
0 &\mbox{otherwise}.
}
\end{numequation}
\noindent Let $c_{i}^{(n)}$ denote a generator of $C_{i}^{n\rho_{2}}$.
The boundary operator $d$ is given by
\begin{numequation}
\label{eq-C2boundary}
d (c_{i+1}^{(n)}) =\mycases{
c_{i}^{(n)}
&\mbox{for }i=n\\
\gamma_{i+1-n} (c_{i}^{(n)})
&\mbox{for }n<i\leq 2n\\
0 &\mbox{otherwise}
}
\end{numequation}
\noindent where $\gamma_{i}=1- (-1)^{i}\gamma $. For future reference, let
\begin{displaymath}
\epsilon_{i}=1- (-1)^{i}=\mycases{
0 &\mbox{for $i$ even}\\
2 &\mbox{for $i$ odd.}
}
\end{displaymath}
\noindent This chain complex has the form
\begin{displaymath}
\xymatrix
@R=3mm
@C=10mm
{
n &n+1
&n+2
&n+3
& &2n\\
\Box
&{\widehat{\Box}}\ar[l]_(.5){\nabla}
&{\widehat{\Box}}\ar[l]_(.5){\gamma_{2} }
&{\widehat{\Box}}\ar[l]_(.5){\gamma_{3} }
&\dotsb \ar[l]_(.5){}
&{\widehat{\Box}}\ar[l]_(.5){\gamma_{n}}\\
\Z\ar@/_/[d]_(.5){1}
&\Z\ar@/_/[d]_(.5){\Delta}\ar[l]_(.5){2}
&\Z\ar@/_/[d]_(.5){\Delta}\ar[l]_(.5){0}
&\Z\ar@/_/[d]_(.5){\Delta}\ar[l]_(.5){2}
&\dotsb \ar[l]_(.5){}
&\Z\ar@/_/[d]_(.5){\Delta}\ar[l]_(.5){\epsilon_{n}}\\
\Z\ar@/_/[u]_(.5){2}
&\Z[G]\ar@/_/[u]_(.5){\nabla}\ar[l]_(.5){\nabla}
&\Z[G]\ar@/_/[u]_(.5){\nabla}\ar[l]_(.5){\gamma_{2}}
&\Z[G]\ar@/_/[u]_(.5){\nabla}\ar[l]_(.5){\gamma_{3} }
&\dotsb \ar[l]_(.5){}
&\Z[G]\ar@/_/[u]_(.5){\nabla}\ar[l]_(.5){\gamma_{n}}
}
\end{displaymath}
\noindent Passing to homology we get
\begin{displaymath}
\xymatrix
@R=3mm
@C=7mm
{
n &n+1
&n+2
&n+3
& &2n\\
\bullet
&0 &\bullet
&0 &\dotsb
&{\uH_{2n}} \\
\Z/2\ar@/_/[d]_(.5){}
&0\ar@/_/[d]_(.5){ }
&\Z/2\ar@/_/[d]_(.5){ }
&0\ar@/_/[d]_(.5){ }
&\dotsb
&{\uH_{2n} (G/G)}
\ar@/_/[d]_(.5){\Delta }\\
0\ar@/_/[u]_(.5){}
&0\ar@/_/[u]_(.5){}
&0\ar@/_/[u]_(.5){}
&0\ar@/_/[u]_(.5){}
&\dotsb
&\Z[G]/ (\gamma_{n+1} )\ar@/_/[u]_(.5){\nabla}
}
\end{displaymath}
\noindent where
\begin{displaymath}
\uH_{2n} (G/G)
=\mycases{
\Z
&\mbox{for $n$ even}\\
0
&\mbox{for $n$ odd}
}
\qquad \aand
\uH_{2n}
=\mycases{
\Box
&\mbox{for $n$ even}\\
\overline{\Box}
&\mbox{for $n$ odd}
}
\end{displaymath}
\noindent Here $\Box$ and $\overline{\Box}$ are fixed point Mackey
functors but $\bullet$ is not.
Similar calculations can be made for $S^{n\rho_{2}}$ for $n<0$. The
results are indicated in Figure \ref{fig-sseq-1}. This is originally
due to unpublished work of Stong and is reported in \cite[Theorem 2.1
and Table 2.2]{Lewis:ROG}. This information will be used in
\S\ref{sec-Dugger}.
\begin{figure}
\begin{center}
\includegraphics[width=12cm]{fig-sseq-1.pdf} \caption[ The slice
spectral sequence for {${\bigvee_{n \in \Z}\Sigma^{n\rho_{2}}\HZZ }$}.]
{The (collapsing) Mackey functor slice spectral sequence for
${\bigvee_{n \in \Z}\Sigma^{n\rho_{2}}\HZZ }$. The symbols are
defined in Table \ref{tab-C2Mackey}. When the Mackey functor
$\upi_{(2-\rho_{2})n-s}\HZZ=\underline{H}_{2n-s}S^{n\rho_{2}}$ is
nontrivial, it is shown at ${(2n-s,s)}$ in the chart. Compare with
Figure \ref{fig-KR}. } \label{fig-sseq-1}
\end{center}
\end{figure}
In other words the $RO (G)$-graded Mackey functor valued homotopy of
$\HZZ$ is as follows. For $n\geq -1$ we have
\begin{displaymath}
\upi_{i}\Sigma^{n\rho_{2}}\HZZ
=\upi_{i-n\rho_{2} }\HZZ =
\mycases{
\Box
&\mbox{for $n$ even and $i=2n$}\\
\overline{\Box}
&\mbox{for $n$ odd and $i=2n$}\\
\bullet
&\mbox{for $n\leq i<2n$ and $i+n$ even} \\
0 &\mbox{otherwise}
}
\end{displaymath}
\noindent For $n\leq -2$ we have
\begin{displaymath}
\upi_{i}\Sigma^{n\rho_{2}}\HZZ
=\upi_{i-n\rho_{2} }\HZZ =
\mycases{
\twobox
&\mbox{for $n$ even and $i=2n$}\\
\dot{\Box}
&\mbox{for $n$ odd and $i=2n$}\\
\bullet
&\mbox{for $2n<i\leq n-3$ and $i+n$ odd}\\
0 &\mbox{otherwise}
}
\end{displaymath}
We can use Definition \ref{def-aeu} to name some elements of these groups.
Note that $\HZZ$ is a commutative ring spectrum, so there is a
commutative multiplication in $\upi_{\star}\HZZ$, making it a
commutative $RO (G)$-graded Green functor. For such a functor $\uM$
on a general group $G$, the restriction maps are a ring homomorphisms
while the transfer maps satisfy the Frobenius relations
(\ref{eq-Frob}).
Then the generators of various groups in $\upi_{\star}\HZZ$ are
\begin{align*}
\mbox{\sc $(4m-2)$-slices for $m>0$ :}\hspace{-1cm} \\
a^{2m-1-2i}u^{i}
& = a_{(2m-1-2i)\sigma}u_{2i\sigma}\\
& \in \upi_{2m-1+2i }\Sigma^{(2m-1)\rho_{2}}\HZZ (G/G)\\
& = \upi_{2i- (2m-1)\sigma}\HZZ (G/G)\\
&\qquad \mbox{for $0\leq i<m$}\\
x^{2m-1}=u_{(2m-1)\sigma}
& \in \upi_{4m-2}\Sigma^{(2m-1)\rho_{2}}\HZZ (G/\ee)\\
& = \upi_{(2m-1) (1-\sigma ) }\HZZ (G/\ee)\\
&\qquad \mbox{with $\gamma (x)=-x$}\\
\mbox{\sc $4m$-slices for $m>0$ :} \\
a^{2m-2i}u^{i}
& = a_{(2m-2i)\sigma}u_{2i\sigma}\\
& \in \upi_{2m-1+2i }\Sigma^{(2m-1)\rho_{2}}\HZZ (G/G)\\
& = \upi_{2i- (2m-1)\sigma}\HZZ (G/G)\\
&\qquad \mbox{for $0\leq i\leq m$}\\
&\qquad \mbox{ and with }\Res (u)=x^{2}\\
\mbox{\sc negative slices:} \\
z_{n}=e_{2n\rho_{2} }
& \in \upi_{-4n}\Sigma^{-2n\rho_{2}}\HZZ (G/\ee)\\
& = \upi_{2n (\sigma -1)}\HZZ (G/\ee)
\qquad \mbox{for $n>0$}\\
a^{-i}\Tr (x^{-2n-1})
& \in \upi_{-4n-2-i}\Sigma^{-(2n+1+i)\rho_{2}}\HZZ (G/G)\\
& = \upi_{(2n+1) (\sigma -1)+ i\sigma}\HZZ (G/G)\\
&\qquad
\qquad \mbox{for $n>0$ and $i\geq 0$} .
\end{align*}
\noindent We have
relations
\begin{displaymath}
\begin{array}[]{rclrcl}
2a &\hspace{-2.5mm} = & 0\qquad &
\Res (a)
&\hspace{-2.5mm} = & 0\\
z_{n} &\hspace{-2.5mm} = & x^{-2n}\qquad &
\Tr (x^{n}) &\hspace{-2.5mm} = &\mycases{
2u^{n/2}
&\mbox{for $n$ even and $n\geq 0$}\\
\Tr_{}^{}(z_{-n/2})
&\mbox{for $n$ even and $n< 0$}\\
0 &\mbox{for $n$ odd and $n>-3$.}
}
\end{array}
\end{displaymath}
\bigskip
(ii) Let $G=C_{4}$ with generator $\gamma $, $G'=C_{2}\subseteq G$, the
subgroup generated by $\gamma^{2}$, and $\widehat{S} (n,G')
=G_{+}\smashove{G'}S^{n\rho_{2}}$. Thus we have
\begin{displaymath}
C_{*} (\widehat{S} (n,G')) = \Z[G]\otimes_{\Z[G']} C_{*}^{n\rho_{2}}
\end{displaymath}
\noindent with $C_{*}^{n\rho_{2}}$ as in (\ref{eq-C2chain}). The
calculations of the previous example carry over verbatim by the
exactness of Mackey functor induction of Definition \ref{def-MF-induction}.
\bigskip
\bigskip(iii) Let $G=C_{4}$ and $X=S^{n\rho_{4}}$. Then the reduced
cellular chain complex (\ref{eq-CVn}) has the form
\begin{displaymath}
C_{i}^{n\rho_{4}} = \mycases{
\Z &\mbox{for }i=n\\
\Z[G/G']
&\mbox{for }n<i\leq 2n\\
\Z[G]
&\mbox{for }2n<i\leq 4n\\
0 &\mbox{otherwise}
}
\end{displaymath}
\noindent in which generators $c_{i}^{(n)}\in C_{i}^{n\rho_{4}}$ satisfy
\begin{displaymath}
d (c_{i+1}^{(n)})= \mycases{
c_{i}^{(n)}
&\mbox{for }i=n\\
\gamma_{i+1-n} c_{i}^{(n)}
&\mbox{for $n<i\leq 2n$}\\
\mu_{i+1-n} c_{i}^{(n)}
&\mbox{for $2n< i<4n$ and $i$ even}\\
\gamma_{i+1-n} c_{i}^{(n)}
&\mbox{for $2n<i<4n$ and $i$ odd}\\
0 &\mbox{otherwise,}
}
\end{displaymath}
\noindent where
\begin{displaymath}
\mu_{i}=\gamma_{i} (1+\gamma^{2})= (1- (-1)^{i}\gamma ) (1+\gamma^{2}).
\end{displaymath}
The values of $\underline{H}_{*}S^{n\rho_{4}}$ are illustrated in Figure \ref{fig-sseq-3}. The Mackey
functors in filtration 0 (the horizontal axis) are the ones described
in Proposition \ref{prop-top}.
\begin{figure}
\begin{center}
\includegraphics[width=12cm]{fig-sseq-3.pdf}
\caption[The slice spectral sequence for {${\bigvee_{n
\in \Z} \Sigma ^{n\rho_{4}}\HZZ }$}.] {The Mackey functor slice
spectral sequence for ${\bigvee_{n \in \Z} \Sigma ^{n\rho_{4}}\HZZ }$.
The symbols are defined in Table \ref{tab-C4Mackey}. The Mackey
functor at position $(4n-s,s)$ is $\upi_{n (4-\rho_{4})
-s}\HZ=\underline{H}_{4n-s}S^{n\rho_{4}}$.}
\label{fig-sseq-3}
\end{center}
\end{figure}
\bigskip
As in (i), we name some of these elements. Let $G=C_{4}$ and
$G'=C_{2}\subseteq G$. Recall that the regular {\rep} $\rho_{4}$ is
$1+\sigma +\lambda$ where $\sigma $ is the sign {\rep} and $\lambda$
is the 2-dimensional {\rep} given by a rotation of order 4.
Note that while Figure \ref{fig-sseq-1} shows all of $\underline{\pi
}_{\star}\HZZ$ for $G=C_{2}$, Figure \ref{fig-sseq-3} shows only a
bigraded portion of this trigraded Mackey functor for $G=C_{4}$,
namely the groups for which the index differs by an integer from a
multiple of $\rho_{4}$. We will need to refer to some elements not
shown in the latter chart, namely
\begin{numequation}\label{eq-au}
\begin{split}\left\{
\begin{array}[]{rlrlrl}
a_{\sigma }
&\hspace{-3mm} \in \underline{H}_{0}S^{\sigma } (G/G)
&a_{\lambda }
&\in \underline{H}_{0}S^{\lambda } (G/G)
&\overline{a}_{\lambda }
&=\Res_{2}^{4}(a_{\lambda })
\\
u_{2\sigma }
&\hspace{-3mm} \in \underline{H}_{2}S^{2\sigma } (G/G) )
&u_{\sigma }
&\in \underline{H}_{1}S^{\sigma } (G/G')
&\overline{u} _{\sigma }
&=\Res_{1}^{2}(u_{\sigma })
\\
u_{\lambda }
&\hspace{-3mm} \in \underline{H}_{2}S^{\lambda } (G/G)
&\overline{u}_{\lambda }
&=\Res_{2}^{4}(u_{\lambda })
&\overline{\overline{u}} _{\lambda }
&=\Res_{1}^{4}(u_{\lambda })
\end{array} \right.
\end{split}
\end{numequation}
\noindent subject to the relations
\begin{numequation}\label{eq-au-rel}
\begin{split}
\left\{\begin{array}[]{rlrlrl}
2a_{\sigma }
&\hspace{-2.5mm} = 0
&\Res_{2}^{4}(a_{\sigma })
&=0 \\
4a_{\lambda }
&\hspace{-2.5mm} =0
&2\overline{a}_{\lambda }
&=0
&\Res_{1}^{4}(a_{\lambda })
&=0 \\
\Res_{2}^{4}(u_{2\sigma })
&\hspace{-2.5mm} = u_{\sigma }^{2}
&a_{\sigma }^{2}u_{\lambda }
&=2a_{\lambda }u_{2\sigma }
&\mbox{ (gold relation)};\hspace{-1cm}
\end{array} \right.
\end{split}
\end{numequation}
\noindent see Definition \ref{def-aeu} and Lemma \ref{lem-aeu}.
We will denote the generator of $\EE_{2}^{s,t} (G/H)$ (when
it is nontrivial) by $x_{t-s,s}$, $y_{t-s,s}$ and $z_{t-s,s}$ for
$H=G$, $G'$ and $\ee$ respectively. Then the generators
for the groups in the 4-slice are
\begin{align*}
y_{4,0} = u_{\rho_{4}}
=u_{\sigma}\Res^{4}_{2}(u_{\lambda})
& \in \upi_{4}\Sigma^{\rho_{4}}\HZZ (G/G')
= \upi_{3-\sigma-\lambda}\HZZ (G/G')\\
& \qquad \mbox{with }\gamma (x_{4,0})=-x_{4,0}\\
x_{3,1} = a_{\sigma}u_{\lambda}
& \in \upi_{3}\Sigma^{\rho_{4}}\HZZ (G/G)
= \upi_{2-\sigma-\lambda}\HZZ (G/G)\\
y_{2,2} = \Res^{4}_{2} (a_{\lambda})u_{\sigma}
& \in \upi_{2}\Sigma^{\rho_{4}}\HZZ (G/G')
= \upi_{1-\sigma -\lambda}\HZZ (G/G')\\
x_{1,3} = a_{\rho_{4}}
= a_{\sigma} a_{\lambda}
& \in \upi_{1}\Sigma^{\rho_{4}}\HZZ (G/G)
= \upi_{-\sigma -\lambda}\HZZ (G/G)
\end{align*}
\noindent and the ones for the 8-slice are
\begin{align*}
x_{8,0} = u _{2\lambda+2\sigma}=u_{2\rho_{4}}
& \in \upi_{8}\Sigma^{2\rho_{4}}\HZZ (G/G)
= \upi_{6-2\sigma -2\lambda}\HZZ (G/G)\\
& \qquad \mbox{with }y_{4,0}^{2}=y_{8,0}=\Res_{2}^{4}(x_{8,0})\\
x_{6,2} = a_{\lambda} u _{\lambda+2\sigma}
& \in \upi_{6}\Sigma^{2\rho_{4}}\HZZ (G/G)
= \upi_{4-2\sigma -2\lambda}\HZZ (G/G)\\
& \qquad \mbox{with }x_{3,1}^{2}=2x_{6,2}\\
& \qquad \mbox{and }y_{4,0}y_{2,2}=y_{6,2}=\Res_{2}^{4}(x_{6,2})\\
x_{4,4} = a_{2\lambda}u_{2\sigma}
& \in \upi_{4}\Sigma^{2\rho_{4}}\HZZ (G/G)
= \upi_{2-2\sigma -2\lambda}\HZZ (G/G)\\
&\qquad \mbox{with }y_{2,2}^{2}=y_{4,4}=\Res_{2}^{4}(x_{4,4})\\
&\qquad \mbox{and }
x_{1,3}x_{3,1}=2x_{4,4}\\
x_{2,6} = x_{1,3}^{2}
& \in \upi_{2}\Sigma^{2\rho_{4}}\HZZ (G/G)
= \upi_{-2\sigma -2\lambda}\HZZ (G/G).
\end{align*}
These elements and their restrictions generate
$\upi_{*}\Sigma^{m\rho_{4}}\HZZ$ for $m=1$ and 2. For
$m>2$ the groups are generated by products of these elements.
The element
\begin{displaymath}
z_{4,0} = \Res^{2}_{1}(y_{4,0}) =\Res^{2}_{1} (u_{\rho_{4}} )\in
\upi_{4}\Sigma^{\rho_{4}}\HZZ (G/\ee)
\end{displaymath}
\noindent is invertible with $\gamma (y_{4,0})=-y_{4,0}$,
$z_{4,0}^{2}=z_{8,0}=\Res_{1}^{4}(x_{8,0})$ and
\begin{displaymath}
z_{-4m,0}:= z_{4,0}^{-m}=e_{m\rho_{4}}
\in \upi_{-4m}\Sigma^{-m\rho_{4}}\HZZ (G/\ee)
\qquad \mbox{for }m>0,
\end{displaymath}
\noindent where $e_{m\rho_{4}}$ is as in Definition
\ref{def-aeu}. These elements and their transfers generate the groups
in
\begin{displaymath}
\upi_{-4m}\Sigma^{-m\rho_{4}}\HZZ \qquad \mbox{for }m>0.
\end{displaymath}
\begin{thm}\label{thm-div}
{\bf Divisibilities in the negative regular slices for $C_{4}$.}
There are the following infinite divisibilities in the third quadrant
of the {\SS} in Figure \ref{fig-sseq-3}.
\begin{enumerate}
\item [(i)] \label{div:i}
$x_{-4,0}=\Tr_{1}^{4}(z_{-4,0})$ is
divisible by any monomial in $x_{1,3}$ and $x_{4,4}$, meaning that
\begin{displaymath}
x_{1,3}^{i}x_{4,4}^{j}x_{-4-4j-i,-4j-3k}=x_{-4,0}
\qquad \mbox{for }i,j\geq 0.
\end{displaymath}
\noindent Moreover, no other basis element killed by $x_{3,1}$ and
$x_{4,4}$ has this property.
\item [(ii)] \label{div:ii}
$x_{-4,0}$, and $x_{-7,-1}$ are divisible
by any monomial in $x_{4,4}$, $x_{6,2}$ and $x_{8,0}$, subject to the
relation $x_{6,2}^{2}=x_{8,0}x_{4,4}$. Note here that
$x_{3,1}^{2}=2x_{6,2}$.
\noindent Moreover, no other basis element killed by $x_{4,4}$,
$x_{6,2}$ and $x_{8,0}$ has this property.
\item [(iii)] \label{div:iv}
$y_{-7,-1}=\Res_{2}^{4}(x_{-7,-1})$ is
divisible by any monomial in $y_{2,2}$ and $y_{4,0}$, meaning that
\begin{displaymath}
y_{2,2}^{j}y_{4,0}^{k}y_{-7-2j-4k,-1-2j}=y_{-7,-1}
\qquad \mbox{for }j,k\geq 0.
\end{displaymath}
\noindent Moreover, no other basis element killed by $y_{2,2}$,and
$y_{4,0}$ has this property.
\end{enumerate}
\end{thm}
We will prove Theorem \ref{thm-div} as a corollary of a more general
statement (Lemma \ref{lem-div} and Corollary \ref{cor-div}) in which
we consider all {\rep}s of the form $m\lambda +n\sigma $ for $m,n\geq
0$. Let
\begin{displaymath}
\underline{R}=\bigoplus_{m,n\geq 0}\underline{H}_{*}S^{m\lambda +n\sigma }.
\end{displaymath}
\noindent It is generated by the elements of (\ref{eq-au}) subject to
the relations of (\ref{eq-au-rel}).
In the larger ring
\begin{displaymath}
\underline{\tilde{R}}
=\bigoplus_{m,n\in \Z \atop mn\geq 0}
\underline{H}_{*}S^{m\lambda +n\sigma },
\end{displaymath}
\noindent the elements $u_{\sigma }$, $\overline{u}_{\sigma } $ and
$\overline{\overline{u} }_{\lambda } $ are invertible with
\begin{align*}
e_{\sigma }
& = u_{\sigma }^{-1} \in \underline{H}_{-1}S^{-\sigma } (G/G')
&e_{\lambda }
= \overline{\overline{u} }_{\lambda }^{-1}
\in \underline{H}_{-2}S^{-\lambda } (G/e).
\end{align*}
Define spectra $L_{m}$ and $K_{n}$ to be the cofibers of $a_{m\lambda
}$ and $a_{n\sigma }$. Thus we have cofiber sequences
\begin{displaymath}
\xymatrix
@R=2mm
@C=15mm
{
{\Sigma^{-1}L_{m}}\ar[r]^(.5){c_{m\lambda }}
&{S^{0}}\ar[r]^(.5){a_{m\lambda }}
&{S^{m\lambda }}\ar[r]^(.5){b_{m\lambda }}
&{L_{m}}\\
{\Sigma^{-1}K_{n}}\ar[r]^(.5){c_{n\sigma}}
&{S^{0}}\ar[r]^(.5){a_{n\sigma}}
&{S^{n\sigma}}\ar[r]^(.5){b_{n\sigma}}
&{K_{n}}
}
\end{displaymath}
\noindent Dualizing gives
\begin{displaymath}
\xymatrix
@R=2mm
@C=15mm
{
{DL_{m}}\ar[r]^(.5){Db_{m\lambda }}
&{S^{-m\lambda }}\ar[r]^(.5){Da_{m\lambda }}
&{S^{0}}\ar[r]^(.5){Dc_{m\lambda }}
&{\Sigma DL_{m}}\\
{DK_{n}}\ar[r]^(.5){Db_{n\sigma}}
&{S^{-n\sigma }}\ar[r]^(.5){Da_{n\sigma}}
&{S^{0}}\ar[r]^(.5){Dc_{n\sigma}}
&{\Sigma DK_{n}}
}
\end{displaymath}
\noindent The maps $Da_{m\lambda }$ and $Da_{n\sigma }$ are the same
as desuspensions of $a_{m\lambda }$ and $a_{n\sigma }$, which implies that
\begin{displaymath}
DL_{m}=\Sigma^{-1-m\lambda }L_{m}
\qquad \aand
DK_{n}=\Sigma^{-1-n\sigma }K_{n}.
\end{displaymath}
\noindent Inspection of the cellular chain complexes for $L_{m}$ and
$K_{n}$ and certain of their suspensions reveals that
\begin{displaymath}
\Sigma^{2-\lambda }L_{m}\wedge \HZZ
=L_{m}\wedge \HZZ=\Sigma^{2-2\sigma }L_{m}\wedge \HZZ
\end{displaymath}
\noindent and
\begin{displaymath}
\Sigma^{2-2\sigma }K_{n}\wedge \HZZ=K_{n}\wedge \HZZ,
\end{displaymath}
\noindent while $\Sigma^{1-\sigma }$ alters both $L_{m}\wedge \HZZ$
and $K_{n}\wedge \HZZ$. We will denote $\Sigma^{k (1-\sigma
)}L_{m}\wedge \HZZ$ by $L_{m}^{(-1)^{k}}\wedge \HZZ$ and similarly for
$K_{n}$.
The homology groups of $L_{m}^{\pm }$ and $K_{m}^{\pm }$ for $m,n>0$
are indicated in Figures \ref{fig-Lm} and \ref{fig-Kn}, and those for
$S^{m\lambda }$ and $S^{n\sigma }$ are shown in
Figure\ref{fig-spheres}.
\begin{figure}
\begin{center}
\includegraphics[width=5cm]{fig-Lm.pdf}
\includegraphics[width=5cm]{fig-Lm-.pdf}
\caption[Charts for {${\underline{\sl H}_{i}L_{m}^{\pm }}$}.]{Charts for
$\underline{H}_{i}L_{m}^{\pm }$. The horizontal coordinate is $i$ and
the vertical one is $m$. $L_{m}$ is on the left and $L_{m}^{-}$ is on
the right.} \label{fig-Lm}
\end{center}
\end{figure}
\begin{figure}
\begin{center}
\includegraphics[width=5cm]{fig-Kn.pdf}
\includegraphics[width=5cm]{fig-Kn-.pdf}
\caption[Charts for {${\underline{\sl H}_{i}K_{n}^{\pm }}$}.] {Charts for
$\underline{H}_{i}K_{n}^{\pm }$. The horizontal coordinate is $i$ and
the vertical one is $n$. $K_{n}$ is on the left and $K_{n}^{-}$ is on
the right.} \label{fig-Kn}
\end{center}
\end{figure}
\begin{figure}
\begin{center}
\includegraphics[height=4cm]{fig-S-lambda.pdf}
\includegraphics[height=4cm]{fig-S-sigma.pdf} \caption[Charts for
${{\underline{\sl H}_{i}S^{m\lambda }}$} and {${\underline{\sl
H}_{i}S^{n\sigma }}$}.] {Charts for ${\underline{H}_{i}S^{m\lambda }}$
and ${\underline{H}_{i}S^{n\sigma }}$. The horizontal coordinates are
$i$ and the vertical ones are $m$ and $n$. $S^{m\lambda }$ is on the
left and $S^{n\sigma }$ is on the right. } \label{fig-spheres}
\end{center}
\end{figure}
In the following diagrams we will use the same notation for a map and
its smash product with any identity map. Let $V=m\lambda +n\sigma $
with $m,n>0$, and let $R_{V}$ denote the fiber of $a_{V}$. Since
$a_{V}$ is self-dual up to susupension, we have
$DR_{V}=\Sigma^{-1-V}R_{V}$. In the following each row and column is
a cofiber sequence.
\begin{numequation}\label{eq-newfirst}
\begin{split}
\xymatrix
@R=5mm
@C=10mm
{
{}
&{}
&{\Sigma^{n\sigma -1}L_{m}}\ar@{=}[r]\ar[d]^(.5){c_{m\lambda }}
&{\Sigma^{n\sigma -1}L_{m}}\ar[d]^(.5){}\\
{\Sigma^{-1}K_{n}}\ar[r]^(.5){c_{n\sigma }}\ar[d]^(.5){}
&{S^{0}}\ar[r]^(.5){a_{n\sigma }}\ar@{=}[d]^(.5){}
&{S^{n\sigma }}\ar[r]^(.5){b_{n\sigma }}\ar[d]^(.5){a_{m\lambda }}
&{K_{n}}\ar[d]^(.5){}\\
{\Sigma^{-1}R_{V}}\ar[r]^(.5){c_{V}}
&{S^{0}}\ar[r]^(.5){a_{V }}
&{S^{V}}\ar[r]^(.5){b_{V}}\ar[d]^(.5){b_{m\lambda }}
&{R_{V}}\ar[d]^(.5){}\\
{}
&{}
&{\Sigma^{n\sigma }L_{m}}\ar@{=}[r]
&{\Sigma^{n\sigma }L_{m}}
}
\end{split}
\end{numequation}
\noindent The homology sequence for the third column is the easiest
way to compute $\underline{H}_{*}S^{V}$. That column is
\begin{numequation}\label{eq-third-column}
\begin{split}
\xymatrix
@R=5mm
@C=10mm
{
{\Sigma^{n\sigma -1}L_{m}}\ar[r]^(.6){c_{m\lambda }}
&{S^{n\sigma }}\ar[r]^(.5){a_{m\lambda }}
&{S^{V}}\ar[r]^(.4){b_{m\lambda }}
&{\Sigma^{n\sigma }L_{m}},
}
\end{split}
\end{numequation}
\noindent which dualizes to
\begin{displaymath}
\xymatrix
@R=2mm
@C=10mm
{
{\Sigma^{1-n\sigma }DL_{m}}\ar@{=}[d]^(.5){}
&{S^{-n\sigma }}\ar[l]_(.4){c_{m\lambda }}
&{S^{-V}}\ar[l]_(.5){a_{m\lambda }}
&{\Sigma^{-n\sigma }DL_{m}}\ar[l]_(.6){c_{m\lambda }}
\ar@{=}[d]^(.5){}\\
{\Sigma^{-V}L_{m}}
& & &{\Sigma^{-1-V}L_{m}}
}
\end{displaymath}
\noindent or
\begin{numequation}\label{eq-dual-third-column}
\begin{split}
\xymatrix
@R=5mm
@C=10mm
{
{\Sigma^{ -1-V}L_{m}}\ar[r]^(.6){c_{m\lambda }}
&{S^{-V}}\ar[r]^(.5){a_{m\lambda }}
&{S^{-n\sigma }}\ar[r]^(.4){b_{m\lambda }}
&{\Sigma^{-V}L_{m}}.
}
\end{split}
\end{numequation}
For (\ref{eq-third-column}) the {\LES} in homology includes
\begin{displaymath}
\xymatrix
@R=5mm
@C=8mm
{
{\underline{H}_{i+1-n}L_{m}^{(-1)^{n}}}\ar[r]^(.6){c_{m\lambda }}
&{\underline{H}_{i}S^{n\sigma }}\ar[r]^(.5){a_{m\lambda }}
&{\underline{H}_{i}S^{V }}\ar[r]^(.4){b_{m\lambda }}
&{\underline{H}_{i-n}L_{m}^{(-1)^{n}}}\ar[r]^(.5){c_{m\lambda }}
&{\underline{H}_{i-1}S^{n\sigma }}
}
\end{displaymath}
{\bf Divisibility by $a_{\lambda }$.} Multiplication by $a_{\lambda }$ keads to
\begin{displaymath}
\xymatrix
@R=5mm
@C=8mm
{
{\underline{H}_{i+1-n}L_{m}^{(-1)^{n}}}\ar[r]^(.6){c_{m\lambda }}
\ar[d]^(.5){a'_{\lambda }}
&{\underline{H}_{i}S^{n\sigma }}\ar[r]^(.5){a_{m\lambda }}\ar@{=}[d]^(.5){}
&{\underline{H}_{i}S^{V }}\ar[r]^(.4){b_{m\lambda }}
\ar[d]^(.5){a_{\lambda }}
&{\underline{H}_{i-n}L_{m}^{(-1)^{n}}}\ar[r]^(.5){c_{m\lambda }}
\ar[d]^(.5){a'_{\lambda }}
&{\underline{H}_{i-1}S^{n\sigma }}\ar@{=}[d]^(.5){}\\
{\underline{H}_{i+1-n}L_{m'}^{(-1)^{n}}}\ar[r]^(.6){c_{m'\lambda }}
&{\underline{H}_{i}S^{n\sigma }}\ar[r]^(.5){a_{m'\lambda }}
&{\underline{H}_{i}S^{V+\lambda }}\ar[r]^(.4){b_{m'\lambda }}
&{\underline{H}_{i-n}L_{m'}^{(-1)^{n}}}\ar[r]^(.5){c_{m'\lambda }}
&{\underline{H}_{i-1}S^{n\sigma },}
}
\end{displaymath}
\noindent where $m'=m+1$ and $a'_{\lambda }$ is induced by the
inclusion $L_{m}\to L_{m'}$.
In the dual case we get
\begin{numequation}\label{eq-a-lambda}
\begin{split}
\xymatrix
@R=5mm
@C=2mm
{
{\underline{H}_{i+1}S^{-n\sigma }}\ar[r]^(.4){b}\ar@{=}[d]^(.5){}
&{\underline{H}_{i+1+|V|}L_{m}^{(-1)^{n}}}\ar[r]^(.6){c}
&{\underline{H}_{i}S^{-V }}\ar[r]^(.5){a}
&{\underline{H}_{i}S^{-n\sigma }}\ar[r]^(.4){b}
\ar@{=}[d]^(.5){}
&{\underline{H}_{i+|V|}L_{m}^{(-1)^{n}}}
\\
{\underline{H}_{i+1}S^{-n\sigma }}\ar[r]^(.4){b}
&{\underline{H}_{i+3+|V|}L_{m'}^{(-1)^{n}}}\ar[r]^(.6){c}
\ar[u]_(.5){Da'_{\lambda }}
&{\underline{H}_{i}S^{-V-\lambda }}\ar[r]^(.5){a}
\ar[u]_(.5){a_{\lambda }}
&{\underline{H}_{i}S^{-n\sigma }}\ar[r]^(.4){b}
&{\underline{H}_{i+2+|V|}L_{m'}^{(-1)^{n}}}
\ar[u]_(.5){Da'_{\lambda }}
}
\end{split}
\end{numequation}
\noindent Here the subscripts on the horizontal maps ($m\lambda $ in
the top row and $m'\lambda $ in the bottom row) have been omitted
to save space. The five lemma implies that the middle
vertical map is onto when the left hand $Da'_{\lambda }$ is onto
and the right hand one is one to one.
The left version of $Da'_{\lambda }$ is onto in every
case except $i=-|V|$ and the right version of it is one to one in all
cases except $i=-|V|$ and $i=-1-|V|$. This is illustrated for small
$m$ in the following diagram in which trivial Mackey functors are
indicated by blank spaces.
\begin{displaymath}
\xymatrix
@R=-1mm
@C=1mm
{
{j}
&{\underline{H}_{j}L_{1}}
&{\underline{H}_{j}L_{2}}
&{\underline{H}_{j}L_{3}}
&{\underline{H}_{j}L_{4}}
&
&{\underline{H}_{j}L_{1}^{-}}
&{\underline{H}_{j}L_{2}^{-}}
&{\underline{H}_{j}L_{3}^{-}}
&{\underline{H}_{j}L_{4}^{-}}
\\
-1 & & & & &
& & & &\\
0 & & & & &
& & & &\\
1
&{\fourbox }
&{\fourbox }\ar[luu]^(.5){}
&{\fourbox }\ar[luu]^(.5){}
&{\fourbox }\ar[luu]^(.5){}
&
&{\dot{\twobox}}
&{\dot{\twobox}}\ar[luu]^(.5){}
&{\dot{\twobox}}\ar[luu]^(.5){}
&{\dot{\twobox}}\ar[luu]^(.5){}\\
2
&{\Box}
&{\circ} \ar[luu]^(.5){}
&{\circ}\ar[luu]^(.5){}
&{\circ} \ar[luu]^(.5){}
&
&{\oBox}
&{\obull}\ar[luu]^(.5){}
&{\obull}\ar[luu]^(.5){}
&{\obull}\ar[luu]^(.5){}\\
3
& & & & &
& &{\bullet}\ar[luu]^(.5){}
&{\bullet}\ar[luu]^(.5){}
&{\bullet}\ar[luu]^(.5){}\\
4
& &{\Box}\ar@{=}[luu]^(.5){}
&{\circ} \ar@{=}[luu]^(.5){}
&{\circ}\ar@{=}[luu]^(.5){}
&
& &{\oBox}\ar@{=}[luu]^(.5){}
&{\obull}\ar@{=}[luu]^(.5){}
&{\obull}\ar@{=}[luu]^(.5){}\\
5
& & & & &
& & &{\bullet}\ar@{=}[luu]^(.5){}
&{\bullet}\ar@{=}[luu]^(.5){}\\
6
& & &{\Box} \ar@{=}[luu]^(.5){}
&{\circ}\ar@{=}[luu]^(.5){}
&
& & &{\oBox}\ar@{=}[luu]^(.5){}
&{\obull}\ar@{=}[luu]^(.5){}\\
7
& & & & &
& & & &{\bullet}\ar@{=}[luu]^(.5){}\\
8
& & & &{\Box} \ar@{=}[luu]^(.5){}
&
& & & &{\oBox}\ar@{=}[luu]^(.5){}\\
}
\end{displaymath}
\noindent It follows that the map $a_{\lambda }$ in
(\ref{eq-a-lambda}) is onto for all $i$ except $-|V|$. {\em This is a
divisibility result.} Note that $a_{\lambda }$ is trivial on
$\underline{H}_{*}X (G/e)$ for any $X$ since $\Res_{1}^{4}(a_{\lambda
})=0$.
\bigskip
{\bf Divisibility by $u_{\lambda }$.} For $u_{\lambda }$
multiplication we use the diagram
\begin{numequation}\label{eq-u-lambda}
\begin{split}
\xymatrix
@R=5mm
@C=3mm
{
{\underline{H}_{i+1}S^{-n\sigma }}\ar[r]^(.5){b}
&{\underline{H}_{i+1}L_{m}^{(-1)^{n}}}\ar[r]^(.5){c}
&{\underline{H}_{i}S^{-V}}\ar[r]^(.5){a}
&{\underline{H}_{i}S^{-n\sigma }}\ar[r]^(.5){b}
&{\underline{H}_{i}L_{m}^{(-1)^{n}}}
\\
{\underline{H}_{i-1}S^{-n\sigma-\lambda}}\ar[r]^(.5){b}
\ar[u]^(.5){u_{\lambda }}
&{\underline{H}_{i+1}L_{m}^{(-1)^{n}}}\ar[r]^(.5){c}
\ar@{=}[u]^(.5){}
&{\underline{H}_{i-2}S^{-V-\lambda }}\ar[r]^(.5){a}
\ar[u]^(.5){u_{\lambda }}
&{\underline{H}_{i-2}S^{-n\sigma-\lambda}}\ar[r]^(.5){b}
\ar[u]^(.5){u_{\lambda }}
&{\underline{H}_{i}L_{m}^{(-1)^{n}}}\ar@{=}[u]^(.5){}
}
\end{split}
\end{numequation}
\noindent The rightmost $u_{\lambda }$ is onto in all cases except
$i=-n$ and $n$ even. This is illustrated for $n=6$ and 7 in the
following diagram.
\begin{displaymath}
\xymatrix
@R=-1mm
@C=2mm
{
j &-1 &-2 &-3 &-4 &-5 &-6 &-7 &-8 &-9\\
{\underline{H}_{j}S^{-6\sigma }}
& & &{\bullet}
&{}
&{\bullet}
&{\twobox }
\\
{\underline{H}_{j}S^{-6\sigma-\lambda }}
& & &{\bullet}\ar[llu]^(.5){}
&{}
&{\bullet}\ar@{=}[llu]^(.5){}
&{} &{\circ}\ar[llu]^(.5){}
&{\fourbox }\ar[llu]^(.5){}
\\
{}\\
{\underline{H}_{j}S^{-7\sigma }}
& & &{\bullet}
&{}
&{\bullet}
& &{\dot{\twobox }}
\\
{\underline{H}_{j}S^{-7\sigma-\lambda }}
& & &{\bullet}\ar[llu]^(.5){}
&{}
&{\bullet}\ar@{=}[llu]^(.5){}
&{} &{\bullet}\ar@{=}[llu]^(.5){}
& &{\dot{\twobox }}\ar@{=}[llu]^(.5){}
\\
}
\end{displaymath}
\noindent
Thus the central $u_{\lambda }$ in (\ref{eq-u-lambda}) fails to be
onto only in when $i=-n$ and $n$ is even.
\bigskip {\bf Divisibility by $a_{\sigma }$.} The corresponding
diagram is
\begin{displaymath}
\xymatrix
@R=5mm
@C=3mm
{
{\underline{H}_{i+1}S^{-n\sigma }}\ar[r]^(.4){b}
&{\underline{H}_{i+1+|V|}L_{m}^{(-1)^{n}}}\ar[r]^(.6){c}
&{\underline{H}_{i}S^{-V }}\ar[r]^(.5){a}
&{\underline{H}_{i}S^{-n\sigma }}\ar[r]^(.4){b}
&{\underline{H}_{i+|V|}L_{m}^{(-1)^{n}}}
\\
{\underline{H}_{i+1}S^{- n'\sigma }}\ar[r]^(.4){b}
\ar[u]_(.5){a_{\sigma }}
&{\underline{H}_{i+2+|V|}L_{m}^{(-1)^{n'}}}\ar[r]^(.6){c}
\ar[u]_(.5){a_{\sigma}}
&{\underline{H}_{i}S^{-V-\sigma }}\ar[r]^(.5){a}
\ar[u]_(.5){a_{\sigma }}
&{\underline{H}_{i}S^{- n'\sigma }}\ar[r]^(.35){b}
\ar[u]_(.5){a_{\sigma }}
&{\underline{H}_{i+1+|V|}L_{m}^{(-1)^{n'}}}
\ar[u]_(.5){a_{\sigma}}
}
\end{displaymath}
\noindent Here we have abbreviated ${n+1}$ by $n'$. Since
$\Res_{2}^{4}(a_{\sigma })=0$, the map $a_{\sigma }$ must vanish on
$\underline{H}_{*}X (G/G')$ and $\underline{H}_{*}X (G/e)$. It can be
nontrivial only on $G/G$.
By Lemma \ref{lem-hate}, the image of $a_{\sigma }$ is the kernel of
the restriction map $u_{\sigma }^{-1}\Res_{2}^{4}$ and the kernel of
$a_{\sigma }$ is the image of the transfer $\Tr_{2}^{4}$. From Figure
\ref{fig-spheres} we see that $\Res_{2}^{4}$ kills
$\underline{H}_{i}S^{-n\sigma } (G/G)$ except the case $i=-n$ for even
$n$. From Figure \ref{fig-Lm} we see that it kills
$\underline{H}_{j}L_{m}^{-} (G/G)$ for all $j$ and
$\underline{H}_{j}L_{m}(G/G)$ for odd $j>1$, but not the generators
for $j=1$ nor the ones for even values of $j$ from $2$ to $2m$. The
transfer has nontrivial image in $\underline{H}_{j}L_{m}^{-}$ only for
$j=1$ and in $\underline{H}_{j}L_{m}$ only for $j=1$ and for even $j$
from 2 to $2m$.
{\em It follows that for odd $n$, each element of
$\underline{H}_{i}S^{-V} (G/G)$ is divisible by $a_{\sigma }$ except
when $i=-|V|=-2m-n$. For even $n$ it is onto except when $i=-n$,
$i=-n-2m$, and $i$ odd from $1-n-2m$ to $-1-n$. }
\bigskip
{\bf Divisibility by $u_{2\sigma }$.} For $u_{2\sigma }$
multiplication, the diagram is
\begin{displaymath}
\xymatrix
@R=5mm
@C=4mm
{
{\underline{H}_{i+1}S^{-n\sigma }}\ar[r]^(.5){b}
&{\underline{H}_{i+1}L_{m}^{(-1)^{n}}}\ar[r]^(.5){c}
&{\underline{H}_{i}S^{-V}}\ar[r]^(.5){a}
&{\underline{H}_{i}S^{-n\sigma }}\ar[r]^(.5){b}
&{\underline{H}_{i}L_{m}^{(-1)^{n}}}
\\
{\underline{H}_{i-1}S^{- (n+2)\sigma}}\ar[r]^(.5){b}
\ar[u]^(.5){u_{2\sigma }}
&{\underline{H}_{i+1}L_{m}^{(-1)^{n}}}\ar[r]^(.5){c}
\ar@{=}[u]^(.5){}
&{\underline{H}_{i-2}S^{-V-2\sigma }}\ar[r]^(.5){a}
\ar[u]^(.5){u_{2\sigma }}
&{\underline{H}_{i-2}S^{- (n+2)\sigma}}\ar[r]^(.55){b}
\ar[u]^(.5){u_{2\sigma }}
&{\underline{H}_{i}L_{m}^{(-1)^{n}}}\ar@{=}[u]^(.5){}
}
\end{displaymath}
\noindent The rightmost $u_{2\sigma }$ is onto in all cases, so {\em every
element in $\underline{H}_{*}S^{-V}$ is divisible by $u_{2\sigma }$.}
The arguments above prove the following.
\begin{lem}\label{lem-div}
{\bf $RO (G)$-graded divisibility.} Let $G=C_{4}$ and $V=m\lambda
+n\sigma $ for $m,n\geq 0$.
\begin{enumerate}
\item [(i)] Each element in
$\underline{H}_{i}S^{-V} (G/G)$ or $\underline{H}_{i}S^{-V} (G/G')$ is
divisible by $a_{\lambda }$ or $\overline{a}_{\lambda }$ except when
$i=-|V|$.
\item [(ii)] Each element in
$\underline{H}_{i}S^{-V} (G/H)$ is
divisible by a suitable restriction of $u_{\lambda }$ except when
$i=-n$ for even $n$.
\item [(iii)] Each element in $\underline{H}_{i}S^{-V} (G/G)$ for odd
$n$ is divisible by $a_{\sigma }$ except when $i=-|V|$. For even $n$ it is
divisible by $a_{\sigma }$ except when $i=-n$, $i=-|V|$ and $i$ is odd
from $i=1-|V|$ to $-1-n$.
\item [(iv)] Each element in $\underline{H}_{i}S^{-V} (G/H)$ is
divisible by a $u_{2\sigma }$, $u_{\sigma }$ or $\overline{u}_{\sigma } $.
\end{enumerate}
\end{lem}
In Theorem \ref{thm-div} we are looking for
divisibility by
\begin{numequation}\label{eq-divisors}
\begin{split}\left\{
\begin{array}[]{rl}
x_{1,3}
= a_{\sigma }a_{\lambda }&\in
\underline{H}_{0}S^{\sigma +\lambda } (G/G)
= \underline{H}_{1}S^{\rho } (G/G)\\
x_{4,4}
= a_{\lambda}^{2}u_{2\sigma } &\in
\underline{H}_{2}S^{2\lambda +2\sigma } (G/G)
= \underline{H}_{4}S^{2\rho } (G/G)\\
y_{2,2}
= \overline{a} _{\lambda}u_{\sigma } &\in
\underline{H}_{1}S^{1\lambda +1\sigma } (G/G')
= \underline{H}_{2}S^{\rho } (G/G)\\
x_{6,2}
= a_{\lambda }u_{2\sigma }u_{\lambda }&\in
\underline{H}_{4}S^{2\lambda +2\sigma } (G/G)
=\underline{H}_{6}S^{2\rho } (G/G) \\
x_{8,0}
= u_{2\sigma }u_{\lambda }^{2}&\in
\underline{H}_{6}S^{2\lambda +2\sigma } (G/G)
=\underline{H}_{8}S^{2\rho } (G/G) \\
y_{4,0}
= u_{\sigma }\overline{u} _{\lambda }&\in
\underline{H}_{3}S^{\lambda +\sigma } (G/G')
=\underline{H}_{4}S^{\rho } (G/G')
\end{array}\right.
\end{split}
\end{numequation}
\noindent In view of Lemma \ref{lem-div}(iv), we can ignore the
factors $u_{2\sigma }$ and $u_{\sigma }$ when analyzing such
divisibility.
\begin{cor}\label{cor-div}
{\bf Infinite divisibility by the divisors of (\ref{eq-divisors}).}
Let
\begin{displaymath}
V=m\lambda +n\sigma \qquad \mbox{for }m,n\geq 0.
\end{displaymath}
\noindent Then
\begin{itemize}
\item [$\bullet$] Each element of $\underline{H}_{i}S^{-V} (G/G)$ is
infinitely divisible by ${x_{1,3}=a_{\sigma }a_{\lambda }}$ for
${i>-n}$ when $n$ is even and for $i\geq -n$ when $n$ is odd.
\item [$\bullet$] Each element of $\underline{H}_{i}S^{-V} (G/G)$ is
infinitely divisible by ${x_{4,4}=a_{\lambda }^{2}u_{2\sigma }}$ for
$i>-|V|$.
\item [$\bullet$] Each element of $\underline{H}_{i}S^{-V} (G/G')$ is
infinitely divisible by ${y_{2,2}=\overline{a} _{\lambda }u_{\sigma
}}$ for $i>-|V|$.
\item [$\bullet$] Each element of $\underline{H}_{i}S^{-V} (G/G)$ is
infinitely divisible by ${x_{6,2}=a_{\lambda }u_{2\sigma }u_{\lambda
}}$ for $i>-|V|$ when $n$ is odd and for $-|V|<i<-n$ when $n$ is even.
\item [$\bullet$] Each element of $\underline{H}_{i}S^{-V} (G/G)$ is
infinitely divisible by ${x_{8,0}=u_{2\sigma }u_{\lambda }^{2}}$ for
$i<-n$ when $n$ is even and for all $i$ when $n$ is odd.
\item [$\bullet$] Each element of $\underline{H}_{i}S^{-V} (G/G')$ is
infinitely divisible by ${y_{4,0}=u_{\sigma }\overline{u} _{\lambda
}}$ for $i<-n$ when $n$ is even and for all $i$ when $n$ is odd.
\end{itemize}
\end{cor}
This implies Theorem \ref{thm-div}.
\section{The spectra $\kR$ and $\kH$}\label{sec-kH}
Before defining our spectrum we need to recall some definitions and
formulas from \cite{HHR}. Let $H\subset G$ be finite groups. In
\cite[\S2.2.3]{HHR} we define a norm functor $N_{H}^{G}$ from the
category of $H$-spectra to that of $G$-spectra. Roughly speaking, for
an $H$-spectrum $X$, $N_{H}^{G}X$ is the $G$-spectrum underlain by the
smash power $X^{(|G/H|)}$ with $G$ permuting the factors and $H$ leaving
each one invariant. When $G$ is cyclic, we will denote the orders of
$G$ and $H$ by $g$ and $h$, and the norm functor by $N_{h}^{g}$.
There is a $C_{2}$-spectrum $MU_{\reals}$ underlain by the complex
cobordism spectrum $MU$ with group action given by complex
conjugation. Its construction is spelled out in
\cite[\S B.12]{HHR}. For a finite cyclic 2-group $G$ we define
\begin{displaymath}
MU^{((G))}= N_{2}^{g}MU_{\reals}.
\end{displaymath}
\noindent Choose a generator $\gamma $ of $G$. In
\cite[(5.47)]{HHR} we defined generators
\begin{numequation}
\label{eq-rbar}
\orr_{k}=\orr^{G}_{k}\in
\upi^{C_{2}}_{k\rho_{2}}i_{C_{2}}^{*}MU^{((G))} (C_{2}/C_{2})
\cong \upi_{C_{2},k\rho_{2}}MU^{((G))} (G/G)
\end{numequation}
\noindent (note that this group is a module over $G/C_{2}$) and
\begin{align*}
r_{k}=\underline{r}_{1}^{2}(\orr_{k})
& \in \pi_{\ee, 2k}^{u}MU^{((G))} (G/G)
\cong \upi^{\ee}_{2k}MU^{((G))} (\ee/\ee)
=\pi_{2k}^{u}MU^{((G))}.
\end{align*}
\noindent The Hurewicz images of the $\orr_{k}$ (for which we use the
same notation) are defined in terms of the coefficients (see
Definition \ref{def-graded})
\begin{displaymath}
\om_{k}\in
\upi^{C_{2}}_{k\rho_{2}}\HZZ_{(2)}\wedge MU^{((G))} (C_{2}/C_{2})
= \upi_{C_{2},k\rho_{2}}\HZZ_{(2)}\wedge MU^{((G))} (G/G)
\end{displaymath}
\noindent of the logarithm of the formal group law $\overline{F} $
associated with the left unit map from $MU$ to $MU^{((G))}$. The
formula is
\begin{displaymath}
\sum _{k\geq 0}\overline{r}_{k}x^{k+1}
=\left(x+\sum_{\ell>0}\gamma(\overline{m}_{2^{\ell}-1})x^{2^{\ell}}\right)^{-1}
\circ \log_{\overline{F} }(x)
\end{displaymath}
\noindent where
\begin{displaymath}
\log_{\overline{F} }(x)=x+\sum_{k>0}\overline{m}_{k}x^{k+1}.
\end{displaymath}
For small $k$ we have
\begin{align*}
\orr_{1}
& = (1-\gamma ) (\om_{1})\\
\orr_{2}
& = \om_{2}-2\gamma (\om_{1} )(1-\gamma ) (\om_{1})\\
\orr_{3}
& = (1-\gamma ) (\om_{3})-\gamma (\om_{1})
(5\gamma (\om_{1})^{2}-6\gamma (\om_{1})\om_{1}+\om_{1}^{2} +2\om_{2})
\end{align*}
Now let $G=C_{2}$ or $C_{4}$ and, in the latter case
$G'=C_{2}\subseteq G$. The generators $\orr^{G}_{k}$ are the
$\orr_{k}$ defined above. We also have elements $\orr^{G'}_{k}$
defined by similar formulas with $\gamma $ replaced by $\gamma^{2}$;
recall that $\gamma^{2} (\om_{k})= (-1)^{k}\om_{k}$.
They are the images of similar generators of
\begin{displaymath}
\upi^{C_{2}}_{k\rho_{2}}MU^{((G'))} (C_{2}/C_{2})
\cong \upi_{C_{2},k\rho_{2}}MU^{((G'))} (G'/G')
\end{displaymath}
\noindent under the left unit map
\begin{displaymath}
MU^{((G'))}\to MU^{((G'))}\wedge MU^{((G'))}\cong i^{*}_{G'} MU^{((G))}.
\end{displaymath}
\noindent
Thus we have
\begin{align*}
\orr_{1}^{G'}
& = 2 \om_{1} \\
\orr_{2}^{G'}
& = \om_{2}+4\om_{1}^{2}\\
\orr_{3}^{G'}
& = 2 \om_{3}+2\om_{1}\om_{2}+12\om_{1}^{3}
\end{align*}
\noindent If we set $\orr_{2}=0$ and $\orr_{3}=0$, we get
\begin{numequation}
\label{eq-r3}
\begin{split}
\left\{\begin{array}[]{rl}
\orr_{1}^{G'}
&\hspace{-2.5mm} = \orr_{1,0}+\orr_{1,1}\\
\orr_{2}^{G'}
&\hspace{-2.5mm} = 3\orr_{1,0} \orr_{1,1}+\orr_{1,1}^{2}\\
\orr_{3}^{G'}
&\hspace{-2.5mm} = 5 \orr_{1,0}^{2}\orr_{1,1} +5\orr_{1,0} \orr_{1,1}^{2}
+\orr_{1,1}^{3}
= \orr_{1,1}( 5 \orr_{1,0}^{2} +5\orr_{1,0} \orr_{1,1}
+\orr_{1,1}^{2})\\
\gamma (\orr_{3}^{G'})
&\hspace{-2.5mm} = -\orr_{1,0}( 5 \orr_{1,1}^{2} -5\orr_{1,0} \orr_{1,1}
+\orr_{1,0}^{2})\\
\lefteqn{-\orr_{3}^{G'}\gamma (\orr_{3}^{G'})/\orr_{1,0} \orr_{1,1}}
\qquad\qquad\\
&\hspace{-2.5mm} = \left(5 \orr_{1,1}^2-5
\orr_{1,0} \orr_{1,1}+\orr_{1,0}^2\right) \left(\orr_{1,1}^2+5
\orr_{1,0} \orr_{1,1}+5 \orr_{1,0}^2\right)\\
&\hspace{-2.5mm} = (5 \orr_{1,0}^{4}-20\orr_{1,0}^{3}\orr_{1,1}
+\orr_{1,0}^{2}\orr_{1,1}^{2}
+20\orr_{1,0}\orr_{1,1}^{3}+5\orr_{1,1}^{4})\\
&\hspace{-2.5mm} = \left(5 (\orr_{1,0}^{2}-\orr_{1,1}^{2})^{2}
-20\orr_{1,0}\orr_{1,1}(\orr_{1,0}^{2}-\orr_{1,1}^{2})
+11 (\orr_{1,0}\orr_{1,1})^{2} \right)
\end{array} \right.\\
\end{split}
\end{numequation}
\noindent where $\orr_{1,0}=\orr_{1}$ and $\orr_{1,1}=\gamma (\orr_{1})$.
\begin{defin}\label{def-kH}
{\bf $\kR$, $\KR$, $\kH$ and $\KH$.}
The $C_{2}$-spectrum $k_{\reals}$ (connective real $K$-theory), is the
spectrum obtained from $MU_{\reals}$ by killing the $r_{n}$s for
$n\geq 2$. Its periodic counterpart $\KR$ is the telescope obtained
from $\kR$ by inverting\linebreak ${\orr_{1}\in \upi_{\rho_{2}}\kR
(C_{2}/C_{2})}$.
The $C_{4}$-spectrum $\kH$ is obtained
from $MU^{((C_{4}))}$ by killing the $r_{n}$s and their conjugates for
$n\geq 2$. Its periodic counterpart $\KH$ is the telescope obtained
from $\kH$ by inverting a certain element $D\in \upi_{4\rho_{4}}\kH
(C_{4}/C_{4})$ defined below in (\ref{eq-D}) and Table
\ref{tab-pi*}.
\end{defin}
The image of $D$ in $\upi^{C_{2}}_{8\rho_{2}}\kH
(C_{2}/C_{2})\cong \upi_{C_{2},8\rho_{2}}\kH(C_{4}/C_{4})$ is
\begin{numequation}\label{eq-r24D}
\begin{split}
\left\{ \begin{array}{rl}
\underline{r}_{2}^{4} (D)
&\hspace{-2.5mm} = \orr_{1,0}\orr_{1,1}\orr_{3}^{G'}\gamma (\orr_{3}^{G'}) \\
&\hspace{-2.5mm} = \orr_{1,0}^{2}\orr_{1,1}^{2}\left( -5 \orr_{1,0}^{4}
+20 \orr_{1,0}^{3}\orr_{1,1}-\orr_{1,0}^{2}\orr_{1,1}^{2}
-20 \orr_{1,0} \orr_{1,1}^{3}-5 \orr_{1,1}^{4} \right)\\
&\hspace{-2.5mm} = -\orr_{1,0}^{2}\orr_{1,1}^{2}
\left(5(\orr_{1,0}^{2}-\orr_{1,1}^{2})^{2}
-20\orr_{1,0}\orr_{1,1}(\orr_{1,0}^{2}-\orr_{1,1}^{2})\right.\\
&\qquad \qquad \qquad \left. +11
(\orr_{1,0}\orr_{1,1})^{2} \right).
\end{array} \right.
\end{split}
\end{numequation}
\noindent It is fixed by the action of $G/G'$, while its factors
$\orr_{1,0}\orr_{1,1}$ and $\orr_{3}^{G'}\gamma (\orr_{3}^{G'})$ are
each negated by the action of the generator $\gamma $.
We remark that while $MU^{((C_{4}))}$ is $MU_{\reals}\wedge
MU_{\reals}$ as a $C_{2}$-spectrum, $\kH$ is {\em not}
$k_{\reals}\wedge k_{\reals}$ as a $C_{2}$-spectrum. The former has
torsion free underlying homotopy but the latter does not.
\bigskip
\section{The slice {\SS} for $\KR$}\label{sec-Dugger}
In this section we describe the slice {\SS} for $\KR$. These results
are originally due to Dugger \cite{Dugger}, to which we refer for many
of the proofs. This case is far simpler than that of $\KH$, but it is
very instructive.
\begin{thm}\label{thm-Dugger-slice}
{\bf The slice $\EE_{2}$-terms for $\KR$ and $\kR$. }
The slices of $\KR$ are
\begin{displaymath}
P_{t}^{t}\KR=\mycases{
\Sigma^{(t/2)\rho_{2}}\HZZ
&\mbox{for $t$ even}\\
* &\mbox{otherwise}
}
\end{displaymath}
\noindent For $\kR$ they are the same in nonnegative dimensions, and
contractible below dimension 0.
\end{thm}
Hence we know the integrally graded homotopy groups of these slices by
the results of \S \ref{sec-chain}, and they are shown in Figure
\ref{fig-sseq-1}. It shows the $\EE_{2}$-term for the wedge of all of
the slices of $\KR$, and $\KR$ itself has the same $\EE_{2}$-term. It
turns out that the differentials and Mackey functor extensions are
determined by the fact that $\upi_{*}\KR$ is 8-periodic, while the
$\EE_{2}$-term is far from it. This explanation is admittedly
circular in that the proof of the Periodicity Theorem itself of
\cite[\S9]{HHR} relies on the existence of certain differentials
described below in (\ref{eq-slicediffs}).
\begin{thm}\label{thm-KR}
{\bf The slice {\SS} for $\KR$.} The differentials and extensions in the
{\SS} are as indicated in Figure \ref{fig-KR}.
\end{thm}
\proof
There are four phenomena we need to establish:
\begin{enumerate}
\item [(i)] The differentials in the first quadrant, which are
indicated by red lines.
\item [(ii)] The differentials in the third quadrant.
\item [(iii)] The exotic transfers in the first quadrant, which are
indicated by blue lines.
\item [(iv)] The exotic restrictions in the third quadrant, which are
indicated by dashed green lines.
\end{enumerate}
For (i), note that there is a nontrivial element in
$\EE_{2}^{3,6} (G/G)$, which is part of the 3-stem, but
nothing in the $(-5)$-stem. This means the former element must be
killed by a differential, and the only possiblilty is the one
indicated. The other differentials in the first quadrant follow from
this one and the multiplicative structure.
For (ii), we know know that $\upi_{7}\KR=0$, so the same must be true
of $\upi_{-9}$. Hence the element in $\EE_{2}^{-3,-12}$
cannot survive, leading to the indicated third quadrant differentials.
For (iii), note that $\upi_{2}$ and $\upi_{-6}$ must be the same as
Mackey functors. This forces the indicated exotic transfers. For
each $m\geq 0$ one has a nonsplit short exact sequence of $C_{2}$
Mackey functors
\begin{displaymath}
\xymatrix
@R=5mm
@C=10mm
{
0\ar[r]^(.5){}
&{\EE_{2}^{2,8m+4}}\ar[r]^(.5){}\ar@{=}[d]^(.5){}
&{\upi_{8m+2}\KR}\ar[r]^(.5){}\ar@{=}[d]^(.5){}
&{\EE_{2}^{0,8m+2}}\ar[r]^(.5){}\ar@{=}[d]^(.5){}
&0\\
&\bullet
&\dot{\Box}
&\overline{ \Box}
}
\end{displaymath}
For (iv), note that $\upi_{-8}$ and $\upi_{0}$ must also agree. This
forces the indicated exotic restrictions. For each $m<0$ one has a
nonsplit short exact sequence
\begin{displaymath}
\xymatrix
@R=5mm
@C=10mm
{
0\ar[r]^(.5){}
&{\EE_{2}^{0,8m}}\ar[r]^(.5){}\ar@{=}[d]^(.5){}
&{\upi_{8m}\KR}\ar[r]^(.5){}\ar@{=}[d]^(.5){}
&{\EE_{2}^{-2,8m-2}}\ar[r]^(.5){}\ar@{=}[d]^(.5){}
&0\\
&{\twobox}
&\dot{\Box}
&\bullet
}
\end{displaymath}
\qed\bigskip
\begin{figure}
\begin{center}
\includegraphics[width=12cm]{fig-KR.pdf} \caption[The slice {\SS} for
{$\KR$}.]{The slice {\SS} for $\KR$. Compare with Figure
\ref{fig-sseq-1}. Exotic transfers and restrictions are indicated
respectively by solid blue and dashed green lines. Differentials are
in red.} \label{fig-KR}
\end{center}
\end{figure}
In order to describe $\upi_{*}\KR$ as a graded Green functor, meaning
a graded Mackey functor with multiplication, we recall some notation
from \S\ref{sec-chain}(i) and Definition \ref{def-aeu}. For $G=C_{2}$
we have elements
\begin{numequation}\label{eq-upi-HZ}
\begin{split}\left\{
\begin{array}{rlrl}
a=a_{\sigma }
&\hspace{-3mm} \in \upi_{-\sigma }\HZZ (G/G) \\
u=u_{2\sigma }
&\hspace{-3mm} \in \upi_{2-2\sigma }\HZZ (G/G) \\
x=u_{\sigma }
&\hspace{-3mm} \in \upi_{1-\sigma }\HZZ (G/\ee) &&\mbox{with $x^{2}=\Res(u)$} \\
z_{n} = e_{2n\rho_{2} }
&\hspace{-3mm} \in \upi_{2n (\sigma -1)}\HZZ (G/\ee)
&&\mbox{for $n>0$}\\
a^{-i}\Tr (x^{-2n-1})
&\hspace{-3mm} \in \upi_{(2n+1) (\sigma -1)+ i\sigma}\HZZ (G/G)
&&\mbox{for $n>0$}
\end{array} \right.
\end{split}
\end{numequation}
\noindent We will use the same symbols for the representatives of
these elements in the slice $E_{2}$-term. The filtrations of $u$, $x$
and $z_{n}$ are zero while that of $a$ is one. It follows that
$a^{-i}\Tr (x^{-2n-1})$ has filtration $-i$. The element $x$ is invertible.
In $\underline{E}_{2}^{*,*}$ we have relations in
\begin{numequation}\label{eq-upi-HZ-relns}
\begin{split}
\left\{
\begin{array}[]{rclrcl}
2a &\hspace{-2.5mm} = & 0\qquad &
\Res (a)
&\hspace{-2.5mm} = & 0\\
z_{n} &\hspace{-2.5mm} = & x^{-2n}\qquad &
\Tr (x^{n}) &\hspace{-2.5mm} = &\mycases{
2u^{n/2}
&\hspace{-1cm}\mbox{for $n$ even and $n\geq 0$}\\
\Tr_{}^{}(z_{-n/2})\neq 0\\
&\hspace{-1cm}\mbox{for $n$ even and $n< 0$}\\
0 &\hspace{-1cm}\mbox{for $n$ odd and $n>-3$}\\
\neq 0 &\hspace{-1cm}\mbox{for $n$ odd and $n\leq -3$.}
}
\end{array}
\right.
\end{split}
\end{numequation}
We also have the element $\orr_{1}\in \upi_{1+\sigma }\kR (G/G)$, the
image of the element of the same name in $\orr_{1}\in
\upi_{1+\sigma}MU_{\reals} (G/G)$ of (\ref{eq-rbar}). We use the same
symbol for its representative $\underline{E}_{2}^{0,1+\sigma }
(G/G)$. Then we have integrally graded elements
\begin{align*}
\eta
= a \orr_{1}
&\in \underline{E}_{2}^{1,2} (G/G) \\
v_{1}
= x\cdot \Res(\orr_{1})
&\in \underline{E}_{2}^{0,2} (G/\ee)
\qquad \mbox{with }\gamma (v_{1})=-v_{1} \\
u\orr_{1}^{2}
&\in \underline{E}_{2}^{0,4} (G/G) \\
w=2u\orr_{1}^{2}
&\in \underline{E}_{2}^{0,4} (G/G) \\
b=u^{2}\orr_{1}^{4}
&\in \underline{E}_{2}^{0,8} (G/G)
\qquad \mbox{with }w^{2}=4b,
\end{align*}
\noindent where $\eta $ and $v_{1}$ are the images of the elements of
the same name in $\pi_{1}S^{0}$ and $\pi_{2}k$, and $w$ and $b$ are
permanent cycles. The elements $x$, $v_{1}$ and $b$ are invertible.
Note that for $n<0$,
\begin{align*}
\underline{E}_{2}^{0,2n} (G/G)
& = \mycases{
0 &\mbox{for }n=1\\
\Z \mbox{ generated by } \Tr_{}^{}(v_{1}^{-n})
&\mbox{for $n$ even}\\
\Z/2 \mbox{ generated by } \Tr_{}^{}(v_{1}^{-n})
&\mbox{for $n$ odd and }n<-1
}
\end{align*}
\noindent so each group is killed by $\eta =a\orr_{1}$ by \ref{lem-hate}.
Then we have
\begin{align*}
d_{3} (u)
& = a^{3}\orr_{1}
\qquad \mbox{by (\ref{eq-slicediffs2}) below,} \\
\mbox{so }
d_{3} (u\orr_{1}^{2}) = d_{3} (u)\orr_{1}^{2}
& = a^{3}\orr_{1}^{3} =\eta^{3}\\
\Tr_{1}^{2}(x)
& = a^{2}\orr_{1}
\qquad \mbox{by (\ref{eq-exotic-transfers}), raising filtration by 2,} \\
\mbox{so }
\Tr_{1}^{2}(v_{1})
& = \eta^{2}.
\end{align*}
Thus we get
\begin{thm}\label{thm-upi-KR}
{\bf The homotopy of $\KR$ as an integrally graded Green functor}.
With notation as above,
\begin{align*}
\upi_{*}\KR (G/\ee)
& = \Z[v_{1}^{\pm 1}] \\
\upi_{*}\KR (G/G)
& = \Z[b^{\pm 1}, w, \eta ]/ (2\eta ,\eta^{3},w\eta ,w^{2}-4b)
\end{align*}
\noindent with
\begin{align*}
\Tr(v_{1}^{i})
& = \mycases{
2b^{j}
&\mbox{for }i=4j\\
\eta^{2}b^{j}
&\mbox{for }i=4j+1\\
wb^{j}
&\mbox{for }i=4j+2\\
0 &\mbox{for }i=4j+3
} \\
\Res(b)
& = v_{1}^{4},\quad
\Res(w)
= 2 v,\mbox{ and }
\Res(\eta )
= 0.
\end{align*}
\noindent For each $j<0$, $b^{j}$ has filtration $-2$ and supports an
exotic restriction in the slice spectral sequence as indicated in
Figure \ref{fig-KR}. Both $v_{1}\Res(b^{j})$ and
$\eta^{2}b^{j}$ have filtration zero, so the transfer relating them is
does not raise filtration.
\end{thm}
Now we will describe the $RO (G)$-graded slice spectral sequence and
homotopy of $\KR$. The former is trigraded since $RO (G)$ itself is
bigraded, being isomorphic as an abelian group to $\Z \oplus \Z$. For
each integer $k$, one can imagine a chart similar to Figure
\ref{fig-KR} converging to the graded Mackey functor $\upi_{k\sigma
+*}\KR$. Figure \ref{fig-KR} itself is the one for $k=0$. The product
of elements in the $k$th and $\ell $th charts lies in the $(k+\ell )$th
chart. We have elements as in (\ref{eq-upi-HZ})
\begin{align*}
a=a_{\sigma }
& \in \underline{E}_{2}^{1,1-\sigma } (G/G) \\
u=u_{2\sigma }
& \in \underline{E}_{2}^{0,2-2\sigma } (G/G) \\
x=u_{\sigma }
& \in \underline{E}_{2}^{0,1-\sigma } (G/\ee)
&&\mbox{with $\gamma (x)=-1$ and $x^{2}=\Res(u)$} \\
z_{n} = x^{-2n}
& \in \underline{E}_{2}^{0,-2n+2n\sigma } (G/\ee)
&&\mbox{for $n>0$}\\
a^{-i}\Tr (x^{-2n-1})
& \in \underline{E}_{2}^{-i,-i-2n+2n\sigma} (G/G)
&&\mbox{for $i\geq 0$ and $n>0$}\\
\orr_{1}
& \in \underline{E}_{2}^{0,1+\sigma } (G/G)
\end{align*}
\noindent where $a$, $x$, $z_{n}$ and $\orr_{1}$ are permanent cycles,
both $x$ and $\orr_{1}$ are invertible, and there are relations as in
(\ref{eq-upi-HZ-relns}). We also know that
\begin{align*}
d_{3} (u)
& = a^{3}\orr_{1} \qquad \mbox{by (\ref{eq-slicediffs2}) below} \\
\aand
\Tr_{1}^{2}(x)
& = a^{2}\orr_{1} \qquad \mbox{by (\ref{eq-exotic-transfers})} .
\end{align*}
\begin{thm}\label{thm-ROG-graded}
{\bf The $RO (G)$-graded slice spectral sequence for $\KR$} can be
obtained by tensoring that of Figure \ref{fig-KR} with
$\Z[\orr_{1}^{\pm 1}]$, that is for any integer $k$
\begin{align*}
\underline{E}_{2}^{s,t +k\sigma } (G/G)
& \cong \orr_{1}^{k} \underline{E}_{2}^{s,t -k }(G/G ) \\
\aand
\underline{E}_{2}^{s,t +k\sigma } (G/\ee)
& \cong \Res(\orr_{1}^{k}) \underline{E}_{2}^{s,t -k }(G/\ee )
\end{align*}
\noindent and $\upi_{t+k\sigma}\KR$ has a similar description.
\end{thm}
\proof The element $\orr_{1}$ and its restriction are invertible
permanent cycles, so multiplication by either induces an isomorphism
in the spectral sequence. \qed\bigskip
\begin{rem}\label{rem-a3}
In {\bf the $RO (G)$-graded slice {\SS } for $\kR$} one has $d_{3}
(u)=\orr_{1}a^{3}$, but $a^{3}$ itself, and indeed all higher powers
of $a$, survive to $\underline{E}_{4}=\underline{E}_{\infty }$. Hence
the $\underline{E}_{\infty }$-term of this {\SS} does {\bf not} have
the horizontal vanishing line that we see in $\underline{E}_{4}$-term
of Figure \ref{fig-KR}. However when we pass from $\kR$ to $\KR$,
$\orr_{1}$ becomes invertible and we have
\begin{displaymath}
d_{3} (\orr_{1}^{-1}u)=a^{3}.
\end{displaymath}
\end{rem}
We can keep track of the groups in this trigraded {\SS } with the
help of four variable {\Ps} $g (\underline{E}_{r} (G/G))\in \Z[[x,y,z,t]]$
in which the rank of $\underline{E}_{r}^{s,i+j\sigma } (G/G)$ is the
coefficient in $\Z[[t]]$ of $x^{i-s}y^{j}z^{s}$. The variable $t$
keeps track of powers of two. Thus a copy of the integers
is represented by $1/ (1-t)$ or (when it is the kernel of a differential
of the form $\Z \to \Z/2$) $t/ (1-t)$. Let
\begin{numequation}\label{eq-aur}
\begin{split}
\wa =y^{-1}z,\qquad
\uu =x^{2}y^{-1}\qquad \aand
\rr =xy.
\end{split}
\end{numequation}
\noindent Since
\begin{displaymath}
\underline{E}_{2} (G/G) =\Z[a,u,\orr_{1}]/ (2a),
\end{displaymath}
\noindent we have
\begin{align*}
g (\underline{E}_{2} (G/G))
& = \left(\frac{1}{1-t}+ \frac{\wa}{1-\wa}\right)
\frac{1}{(1-\uu) (1-\rr)} \\
g (\underline{E}_{4} (G/G))
& = g (\underline{E}_{2} (G/G))
-\frac{\uu+\rr\wa^{3}}
{(1-\wa) (1-\uu^{2}) (1-\rr)}.
\end{align*}
\noindent We subtract the indicated expression from $g
(\underline{E}_{2} (G/G))$ because we have differerentials
\begin{displaymath}
d_{3} (a^{i}\orr_{1}^{j}u^{2k+1})= a^{i+3}\orr_{1}^{j+1}u^{2k}
\qquad \mbox{for all }i,j,k\geq 0.
\end{displaymath}
\noindent Pursuing this further we get
\begin{align*}
g (\underline{E}_{4} (G/G))
& = \left(\frac{1}{1-t}+ \frac{\wa}{1-\wa}\right)
\frac{1}{(1-\uu) (1-\rr)}
-\frac{\uu}
{ (1-\uu^{2}) (1-\rr)}\\
& \qquad
-\frac{a\uu+\rr\wa^{3}}
{(1-\wa) (1-\uu^{2}) (1-\rr)} \\
& = \frac{1+\uu-\uu (1-t)}
{(1-t) (1-\uu^{2}) (1-\rr)}
+\frac{\wa (1+\uu)
-\wa (\uu+\wa^{2}\rr)}
{(1-\wa) (1-\uu^{2}) (1-\rr)}\\
& = \frac{1+t\uu}
{(1-t) (1-\uu^{2}) (1-\rr)}
+\frac{\wa-\wa^{3}+\wa^{3}-\wa^{3}\rr}
{(1-\wa) (1-\uu^{2}) (1-\rr)}\\
& = \frac{1+t\uu}
{(1-t) (1-\uu^{2}) (1-\rr)}
+\frac{\wa+\wa^{2}}
{ (1-\uu^{2}) (1-\rr)}
+\frac{\wa^{3}}
{(1-\wa) (1-\uu^{2})} .
\end{align*}
\noindent The third term of this expression represents the elements of
filtration above two (referred to in \ref{rem-a3}) which disappear
when we pass to $\KR$. The first term represents the elements of
filtration zero, which include
\begin{numequation}\label{eq-[2u]]}
\begin{split}
1,\qquad
[2u]\in \langle 2,\,a ,\,a^{2}\orr_{1} \rangle
\qquad \aand
[u^{2}]\in \langle \,a ,\,a^{2}\orr_{1}\,a ,\,a^{2}\orr_{1} \rangle
\end{split}
\end{numequation}
\noindent Here we use the notation $[2u]$ and $[u^{2}]$ to indicate
the images in $\underline{E}_{4}$ of the elements $2u$ and $u^{2}$ in
$\underline{E}_{2}$; see Remark \ref{rem-abuse} below. The former
{\em not} divisible by 2 and the latter is not a square since $u$
itself is not present in $\underline{E}_{4}$, where the Massey
products are defined. For an introduction to Massey products, we
refer the reader to \cite[A1.4]{Rav:MU}.
We now make a similar computation where we enlarge $\underline{E}_{2} (G/G)$
by adjoining $\orr_{1}^{-1}u$ and denote the resulting {\SS } terms by
$\underline{E}'_{2}$ and $\underline{E}'_{4}$.
Let
\begin{displaymath}
\www = \rr^{-1}\uu= xy^{-3}.
\end{displaymath}
\noindent Then since
\begin{displaymath}
\underline{E}'_{2} (G/G) =\Z[a,\orr_{1}^{-1}u,\orr_{1}]/ (2a),
\end{displaymath}
\noindent we have
\begin{align*}
g (\underline{E}'_{2} (G/G))
& = \left(\frac{1}{1-t}+ \frac{\wa}{1-\wa}\right)
\frac{1}{(1-\www) (1-\rr)} \\
g (\underline{E}_{4} (G/G))
& = g (\underline{E}_{2} (G/G))
-\frac{\www+\wa^{3}}
{(1-\wa) (1-\www^{2}) (1-\rr)} \\
& = \left(\frac{1}{1-t}+ \frac{\wa}{1-\wa}\right)
\frac{1}{(1-\www) (1-\rr)}
-\frac{\www}
{ (1-\www^{2}) (1-\rr)}\\
& \qquad
-\frac{a\www+\wa^{3}}
{(1-\wa) (1-\www^{2}) (1-\rr)} \\
& = \frac{1+\www-\www (1-t)}
{(1-t) (1-\www^{2}) (1-\rr)}
+\frac{\wa (1+\www)
-\wa (\www+\wa^{2})}
{(1-\wa) (1-\www^{2}) (1-\rr)}\\
& = \frac{1+t\www}
{(1-t) (1-\www^{2}) (1-\rr)}
+\frac{\wa+\wa^{2}}
{ (1-\www^{2}) (1-\rr)}
\end{align*}
\noindent and there is nothing in $\underline{E}'_{4}$ with filtration
above two. As far as we know there is no modification of the spectrum
$\kR$ corresponding to this modification of $\underline{E}_{r}$.
However the map $\underline{E}_{r}\kR \to \underline{E}_{r}\KR$
clearly factors through $\underline{E}'_{r}$
\section{Some elements in the homotopy groups of $\kH$ and
$\KH$}\label{sec-more}
For $G=C_{4}$ we will often use a (second) subscript $\epsilon $ on
elements such as $r_{n}$ to indicate the action of a generator $\gamma
$ of $G=C_{4}$, so $\gamma (x_{\epsilon })=x_{1+\epsilon }$ and
$x_{2+\epsilon }=\pm x_{\epsilon }$. Then we have
\begin{numequation}
\label{eq-pikH}
\pi_{*}^{u}\kH=\upi_{*}\kH (G/\ee)=\upi_{\ee,*}\kH (G/G)
=\Z[r_{1},\,\gamma (r_{1})]=\Z[r_{1,0},\,r_{1,1}]
\end{numequation}
\noindent where $\gamma^{2} (r_{1,\epsilon })=-r_{1,\epsilon }$. Here
we use $r_{1,\epsilon }$ and $\orr_{1,\epsilon }$ to denote the images
of elements of the same name in the homotopy of $MU^{((G))}$.
\begin{numequation}\label{eq-bigraded}
\begin{split}
\includegraphics[width=9cm]{fig-small-ss.pdf}
\end{split}
\end{numequation}
\noindent Here the vertical coordinate is $s$ and the horizontal
coordinate is $|t|-s$. More information about these elements can be
found in Table \ref{tab-pi*} below.
\noindent
We are using the following notational convention. When
$x=\Tr_{2}^{4}(y)$ for some element ${y\in \upi_{\star}\kH (G/G')}$,
we will write $x'=\Tr_{2}^{4}(u_{\sigma}y)$. Examples above include
the cases $x=\eta $ and $x=\ot_{2}$. The primes could be iterated,
{\ie } we might write $x^{(k)}=\Tr_{2}^{4}(u_{\sigma}^{k}y)$, but
this turns out to be unnecessary.
The group action (by $G'$ on $\orr_{1,\epsilon }$, $a_{\sigma_{2}}$
and $u_{\sigma_{2}}$, and by $G$ on all the others) fixes each
generator but $u_{\sigma}$ and $u_{\sigma_{2}}$.
For them the action is given by
\begin{displaymath}
\xymatrix
@R=1mm
@C=8mm
{
u_{\sigma}\ar@{<->}[r]^(.5){\gamma}
&-u_{\sigma}
&{\mbox{and} }
&u_{\sigma_{2} }\ar@{<->}[r]^(.5){\gamma^{2}}
&-u_{\sigma_{2} }
}
\end{displaymath}
\noindent by Theorem \ref{thm-module}. This is compatible with the
following $G$-action:
\begin{displaymath}
\xymatrix
@R=5mm
@C=10mm
{
r_{1,0}\ar[r]^(.5){\gamma}
&r_{1,1}\ar[d]^(.5){\gamma}\\
-r_{1,1}\ar[u]^(.5){\gamma} &-r_{1,0}\ar[l]^(.5){\gamma} }
\qquad \mbox{where } r_{1,\epsilon} =
\underline{r}_{1}^{2}(\orr_{1,\epsilon })\in \upi_{\ee,2}\kH (G/G) .
\end{displaymath}
We will see below (Theorem \ref{thm-d3ulambda}) that $d_{5}
(u_{2\sigma})=a_{\sigma}^{3}a_{\lambda}\normrbar_{1}$ and
$[u_{2\sigma}^{2}]$ is a permanent cycle. Since all transfers are
killed by $a_{\sigma}$ multiplication (Lemma \ref{lem-hate}), this
implies that $[u_{2\sigma}x]$ is a permanent
cycle representing the Toda bracket
\begin{displaymath}
[u_{2\sigma}x]
= [u_{2\sigma}\Tr_{2}^{4}(y)]
= \langle x,\,a_{\sigma} ,\,a_{\sigma}^{2}a_{\lambda}\normrbar_{1} \rangle.
\end{displaymath}
\noindent This element is $x''$ since in $\EE_{2}$ we have
(using the Frobenius relation (\ref{eq-Frob}))
\begin{displaymath}
x''=\Tr_{2}^{4}(u_{\sigma}^{2}y)=\Tr_{2}^{4}(\Res_{2}^{4}(u_{2\sigma})y)
= u_{2\sigma}\Tr_{2}^{4}(y)= u_{2\sigma}x.
\end{displaymath}
\noindent Similarly $x'''=u_{2\sigma}x'$. For $k\geq 4$,
$x^{(k)}=u_{2\sigma}^{2}x^{(k-4)}$ in $\upi_{\star}$ as well as
$\EE_{2}$.
\bigskip
The Periodicity Theorem \cite[Thm. 9.19]{HHR} states that inverting a class
in $\upi_{4\rho_{4}}\kH (G/G)$ whose image under
$\ur_{2}^{4}\Res_{2}^{4}$ is divisible by
$\orr_{3,0}^{G'}\orr_{3,1}^{G'}$ (see (\ref{eq-r3})) and
$\orr_{1,0}\orr_{1,1}=\orr_{1,0}^{G}\orr_{1,1}^{G}$ makes
$u_{8\rho_{4}}$ a permanent cycle. One such class is
\noindent
\begin{numequation}\label{eq-D}
\begin{split}
\begin{array}[]{rl}
D&\hspace{-2.5mm} = N_{2}^{4} (\normrbar_{2}^{G'})\normrbar_{1}^{G}
= u_{2\sigma}^{-2}(\ur_{2}^{4}\Res_{2}^{4})^{-1}
\left(\orr_{1,0}^{G}\orr_{1,1}^{G}
\orr_{3,0}^{G'}\orr_{3,1}^{G'} \right)\\
&\hspace{-2.5mm} = \normrbar_{1}^{2} (-5\ot_{2}^{2}+20 \ot_{2}\normrbar_{1}
+9\normrbar_{1}^{2}) \in \upi_{4\rho_{4}}\kH (G/G),\\
&\qquad \mbox{where }\ot_{2}=\Tr_{2}^{4}(u_{\sigma }^{-1}[\orr_{1,0}^{2}])
\mbox{ and $\normrbar_{1}$ is as in (\ref{eq-normrbar}) below,}
\end{array}
\end{split}
\end{numequation}
\noindent
and $\KH = D^{-1}\kH$. Then we know that $\Sigma^{32}\KH$ is
equivalent to $\KH$.
The Slice and Reduction Theorems \cite[Thms. 6.1 and 6.5]{HHR} imply that the
$2k$th slice of $\kH$ is the $2k$th wedge summand of
\begin{displaymath}
\HZZ \wedge N_{2}^{4}\left(\bigvee_{i\geq 0}S^{i\rho_{2}} \right).
\end{displaymath}
\noindent It follows that over $G'$ the $2k$th slice is a wedge of
$k+1$ copies of $\HZZ \wedge S^{k\rho_{2}}$. Over $G$ we get the
wedge of the appropriate number of copies of $G_{+}\smashove{G'}\HZZ
\wedge S^{k\rho_{2}}$, wedged with a single copy of $\HZZ \wedge
S^{(k/2)\rho_{4}}$ for even $k$. This is spelled out in Theorem
\ref{thm-sliceE2} below.
The group $\upi^{G'}_{\rho_{2}}\kH (G'/\ee)$ is {\em
not} in the image of the group action restriction
$\ur_{2}^{4}$ because $\rho_{2}$ is not the restriction of a
{\rep} of $G$. However, $\pi_{2}^{u}\kH$ is refined (in the
sense of \cite[Def. 5.28]{HHR}) by a map from
\begin{numequation}
\label{eq-s1}
\xymatrix
@R=5mm
@C=10mm
{
S_{\rho_{2}} : = G_{+}\smashove{G'}S^{\rho_{2}}
\ar[r]^(.7){\os_{1} }
&\kH.
}
\end{numequation}
\noindent The Reduction Theorem implies that the 2-slice
$P_{2}^{2}\kH$ is $S_{\rho_{2}}\wedge \HZZ$. We know that
\begin{displaymath}
\upi_{2} ( S_{\rho_{2}}\wedge \HZZ) = \widehat{\oBox }.
\end{displaymath}
\noindent We use the symbols $r_{1}$ and $\gamma (r_{1})$ to denote
the generators of the underlying abelian group of
$\widehat{\oBox } (G/\ee)=\Z[G/G']_{-}$. These elements have
trivial fixed point transfers and
\begin{displaymath}
\upi_{2} ( S_{\rho_{2}}\wedge \HZZ) (G/G')=0.
\end{displaymath}
Table \ref{tab-pi*} describes some elements in the slice {\SS} for $\kH$ in
low dimensions, which we now discuss.
Given an element in $\pi_{\star}MU^{((G))}$, we will often use the
same symbol to denote its image in $\pi_{\star}\kH$. For example, in
\cite[\S9.1]{HHR}
\begin{numequation}
\label{eq-normrbar}
\normrbar_{n}\in \pi_{(2^{n}-1)\rho_{4}}^{G}MU^{((G))}
=\upi_{(2^{n}-1)\rho_{4}}^{G}MU^{((G))} (G/G)
\end{numequation}
\noindent was defined to be the composite
\begin{displaymath}
\xymatrix
@R=5mm
@C=10mm
{
S^{(2^{n}-1)\rho_{4}}\ar@{=}[r]^(.5){}
&N_{2}^{4}S^{(2^{n}-1)\rho_{2}}\ar[rr]^(.5){N_{2}^{4}\orr_{2^{n}-1} }
& &N_{2}^{4}MU^{((G))}\ar[r]^(.5){}
&MU^{((G))}.
}
\end{displaymath}
\noindent We will use the same symbol to denote its image in
$\upi_{(2^{n}-1)\rho_{4}}^{G}\kH (G/G)$.
The element $\eta \in \pi_{1}S^{0}$ (coming from the Hopf map
$S^{3}\to S^{2}$) has image ${a_{\sigma}\orr_{1} \in
\upi^{G'}_{1}\kR (G'/G')}$. There are two corresponding
elements
\begin{displaymath}
\eta_{\epsilon }\in \upi^{G'}_{1}\kH (G'/G')
\qquad \mbox{for } \epsilon =0,1.
\end{displaymath}
\noindent We use the same symbol for their preimages under
$\ur_{2}^{4}$ in $\upi^{G}_{1}\kH (G/G')$, and there we have
\begin{displaymath}
\eta_{\epsilon}=a_{\sigma_{2}}\orr_{1,\epsilon}.
\end{displaymath}
\noindent We denote by $\eta $ again the image of either under the
transfer $\Tr_{2}^{4}$, so
\begin{displaymath}
{\Res_{2}^{4}(\eta)=\eta_{0}+\eta_{1}}.
\end{displaymath}
Its cube is killed by a $d_{3}$ in the slice {\SS}, as is the sum of
any two monomials of degree 3 in the $\eta_{\epsilon }$. It follows
that in $\EE_{4}$ each such monomial is equal to
$\eta_{0}^{3}$. It has a nontrivial transfer, which we denote by $x_{3}$.
In \cite[Def. 5.51]{HHR} we defined
\begin{numequation}
\label{eq-fk}
f_{k}=a_{\overline{\rho } }^{k}N_{2}^{g} (\orr_{k})
\in \upi_{k}MU^{((G))} (G/G)
\end{numequation}
\noindent for a finite cyclic 2-group $G$. In particular,
$f_{2^{n-1}}=a_{\overline{\rho } }^{2^{n}-1}\normrbar_{n}$ for
$\normrbar_{n}$ as in (\ref{eq-normrbar}). The slice filtration of
$f_{k}$ is $k(g-1)$ and we will see below (Lemma \ref{lem-hate} and,
for $G=C_{4}$, Theorem \ref{thm-d3ulambda}) that
\begin{numequation}
\label{eq-Tr(usigma)}
\Tr_{G'}^{G}(u_{\sigma}) = a_{\sigma}f_{1}.
\end{numequation}
Note that $u_{\sigma}\in \EE_{2}^{0,1-\sigma} (G/G')$
since the maximal subgroup for which the sign {\rep} $\sigma $ is
oriented is $G'$, on which it restricts to the trivial {\rep} of
degree 1. This group depends only on the restriction of the $RO
(G)$-grading to $G'$, and the isomorphism extends to differentials as
well. This means that $u_{\sigma}$ is a place holder corresponding
to the permanent cycle $1\in \EE_{2}^{0,0} (G/G')$.
For $G=C_{4}$ (\ref{eq-Tr(usigma)}) implies
\begin{displaymath}
\Tr_{2}^{4}(u_{\sigma}) = a_{\sigma}f_{1}
= a_{\sigma}^{2}a_{\lambda}\normrbar_{1}.
\end{displaymath}
\noindent For example
\begin{align*}
\Tr_{2}^{4}(\eta_{0}\eta_{1})
& = \Tr_{2}^{4}(a_{\sigma_{2}}^{2}\orr_{1,0}\orr_{1,1})
= \Tr_{2}^{4}(u_{\sigma}\Res_{2}^{4}(a_{\lambda}\normrbar_{1}))\\
& = \Tr_{2}^{4}(u_{\sigma})a_{\lambda}\normrbar_{1}
= a_{\sigma}f_{1}a_{\lambda}\normrbar_{1}
= f_{1}^{2}
\end{align*}
The Hopf element $\nu \in \pi_{3}S^{0}$ has image
\begin{displaymath}
a_{\sigma}u_{\lambda}\normrbar_{1}\in \upi_{3}\kH (G/G),
\end{displaymath}
\noindent so we also denote the latter by $\nu $. (We will see below
in (\ref{eq-d-ul}) that $u_{\lambda}$ is not a permanent cycle, but
$\alphax := a_{\sigma}u_{\lambda}$ is (\ref{eq-alphax}).) It has an
exotic restriction $\eta_{0}^{3}$ (filtration jump two), which implies
that
\begin{displaymath}
2\nu =\Tr_{2}^{4}(\Res_{2}^{4}(\nu ))=\Tr_{2}^{4}(\eta_{0}^{3})=x_{3}.
\end{displaymath}
\noindent One way to see this is to use the Periodicity Theorem to
equate $\upi_{3}\kH$ with $\upi_{-29}\kH$,
which can be shown to be the Mackey functor $\circ $ in slice
filtration $-32$. Another argument not relying on periodicity is given
below in Theorem \ref{thm-d3ulambda}.
The exotic restriction on $\nu $ implies
\begin{displaymath}
\Res_{2}^{4}(\nu^{2})=\eta_{0}^{6},
\end{displaymath}
\noindent with filtration jump 4.
\begin{thm}\label{thm-Hur}
{\bf The Hurewicz image.} The elements $\nu \in \upi_{3}\kH (G/G)$,
\linebreak ${\epsilon \in \upi_{8}\kH (G/G)}$, $\kappa \in
\upi_{14}\kH (G/G)$, and $\overline{\kappa} \in \upi_{20}\kH (G/G)$
are the images of elements of the same names in $\pi_{*}S^{0}$. The
image of the Hopf map $\eta\in\pi_{1}S^{0}$ is either
$\eta=\Tr_{2}^{4}(\eta_{\epsilon})$ or its sum with $f_{1}$.
\end{thm}
We refer the reader to \cite[Table A3.3]{Rav:MU} for more information
about these elements.
\proof Suppose we know this for $\nu $ and $\overline{\kappa} $.
Then $\Delta_{1}^{-4}\nu $ is represented by an element of filtration
$-3$ whose product with $\nu^{2}$ is nontrivial. This implies that
$\nu^{3}$ has nontrivial image in $\underline{\pi }_{9}\kH (G/G)$.
This is a nontrivial multiplicative extension in the first quadrant,
but not in the third. The spectral sequence representative of
$\nu^{3}$ has filtration 11 instead of 3. We will see later that
$\nu^{3}=2n$ where $n$ has filtration 1, and $\nu^{3}$ is the transfer
of an element in filtration 1.
Since $\nu^{3}=\eta \epsilon $ in $\pi_{*}S^{0}$, this implies that
$\eta $ and $\epsilon $ are both detected and have the images stated
in Table \ref{tab-pi*}. It follows that $\epsilon \overline{\kappa
}$ has nontrivial image here. Since $\kappa^{2}=\epsilon
\overline{\kappa}$ in $\pi_{*}S^{0}$, $\kappa $ must also be
detected. Its only possible image is the one indicated.
Both $\nu $ and $\overline{\kappa}$ have images of order $8$ in
$\pi_{*}TMF$ and its $K (2)$ localization. The latter is the homotopy
fixed point set of an action of the binary tetrahedral group $G_{24}$
acting on $E_{2}$. This in turn is a retract of the homotopy fixed
point set of the quaternion group $Q_{8}$. A restriction and transfer
argument shows that both elements have order at least 4 in the
homotopy fixed point set of $C_{4}\subset Q_{8}$.
There is an orientation map $MU\to E_{2}$, which extends to a
$C_{2}$-{\eqvr} map $MU_{\reals}\to E_{2}$. Norming up and
multiplying on the right gives us a $C_{4}$-{\eqvr} map
$N_{2}^{4}MU_{\reals}\to E_{2}$. This $C_{4}$-action on the target is
compatible with the $G_{24}$-action leading to $L_{K (2)}TMF$.
The image of $\eta\in\pi_{1}S^{0}$ must restrict to
$\eta_{0}+\eta_{1}$, so modulo the kernel of $\Res_{2}^{4}$ it is the
element $\Tr_{2}^{4}(\eta_{\epsilon})$, which we are calling $\eta$.
The kernel of $\Res_{2}^{4}$ is generated by $f_{1}$. \qed\bigskip
We now discuss the norm $N_{2}^{4}$, which is a functor from the
category of $C_{2}$-spectra to that of $C_{4}$ spectra. As explained
above in connection with Corollary \ref{cor-normdiff}, for a
$C_{4}$-ring spectrum $X$ we have an internal norm
\begin{displaymath}
\upi_{V}^{G'}i^{*}_{G'}X (G'/G')
\cong \upi_{G',V}^{G'}X (G/G)
\to \upi_{\Ind_{2}^{4}V }^{G}X (G/G)
\end{displaymath}
\noindent and a similar functor on the slice spectral sequence for
$X$. It preserves multiplication but not addition. Its source is a
module over $G/G'$, which acts trivially on its target. Consider the
diagram
\begin{displaymath}
\xymatrix
@R=8mm
@C=8mm
{
{\upi_{G',V}X (G/G)}\ar[r]^(.45){\cong }
&{\upi_{V}^{G'}i^{*}_{G'}X (G'/G')}\ar[r]^(.5){N_{2}^{4}}
&{\upi_{\Ind_{2}^{4}V }^{G}X (G/G)}\ar[d]^(.5){\Res_{2}^{4}}\\
{\upi_{G',2V}X (G/G)}\ar[r]^(.45){\cong }
&{\upi^{G'}_{2V}i^{*}_{G'}X (G'/G')}\ar[r]^(.5){\cong }
&{\upi_{\Ind_{2}^{4}V }^{G}X (G/G')}
}
\end{displaymath}
\noindent
For $x\in \upi_{V}^{G'}i^{*}_{G'}X (G'/G')$ we have $x\gamma (x)\in
\upi_{2V}^{G'}i^{*}_{G'}X (G'/G')$ and $2V$ is the restriction of some
$W\in RO (G)$. The group $\upi_{W}^{G}X (G/G')$ depends only on the
restriction of $W$ to $RO (G')$. If $W'\in RO (G)$ is another virtual
{\rep} restricting to $2V$, then $W-W'=k (1-\sigma )$ for some integer
$k$. The canonical isomorphism between $\upi_{W}^{G}X (G/G')$ and
$\upi_{W'}^{G}X (G/G')$ is given by multiplication by
$u_{\sigma}^{k}$.
\begin{defin}\label{def-bracket}
{\bf A second use of square bracket notation}. For $0\leq i\leq 2d$, let $f
(\orr_{1,0},\orr_{1,1})$ be a homogeneous polynomial of degree $2d-i$,
so
\begin{displaymath}
a_{\sigma_{2}}^{i}f (\orr_{1,0},\orr_{1,1})\in
\upi_{(2d-i)+ (2d-2i)\sigma_{2}}^{G'}i^{*}_{G'}\kH (G'/G')
\end{displaymath}
\noindent We will denote by
$[a_{\sigma_{2}}^{i}f(\orr_{1,0},\orr_{1,1})]$ its preimage in
$\upi_{2d-i+ (d-i)\lambda }\kH (G/G')$ under the isomorphism of
(\ref{eq-easy-iso}).
\end{defin}
The first use of square bracket notation is that of Remark
\ref{rem-abuse}. Note that ${\orr_{1,\epsilon }\in
\upi_{\rho_{2}}^{G'}i^{*}_{G'}\kH}$ is not the target of such an
isomorphism since ${\rho_{2}\in RO (G')}$ is not the restriction of
any element in $RO (G)$, hence the requirement that $f$ has even
degree.
We will denote $u_{\sigma }^{-1}[\orr_{1,\epsilon }^{2}]\in
\upi_{\rho_{4}}^{G}\kH (G/G')$ by $\overline{s}_{2,\epsilon }$. Then
we have $\gamma(\overline{s}_{2,0})=-\overline{s}_{2,1}$ and
$\gamma(\overline{s}_{2,1})=-\overline{s}_{2,0}$. We define
\begin{displaymath}
\ot_{2}:= (-1)^{\epsilon }\Tr_{2}^{4}(\overline{s}_{2,\epsilon }),
\end{displaymath}
\noindent which is independent of $\epsilon $, and we have
\begin{displaymath}
\Res_{2}^{4}(\ot_{2})=\overline{s}_{2,0}-\overline{s}_{2,1}.
\end{displaymath}
\noindent
Then we have
\begin{displaymath}
\Res_{2}^{4}(N_{2}^{4} (\orr_{1,0}))
=\Res_{2}^{4}(\normrbar_{1}^{})
=u_{\sigma}^{-1}[\orr_{1,0}\orr_{1,1}]
\in \upi_{_{\rho_{4} }}\kH (G/G') .
\end{displaymath}
\noindent More generally for integers $m$ and $n$
\begin{align*}
\lefteqn{\Res_{2}^{4}(N_{2}^{4} (m\orr_{1,0}+n\orr_{1,1}))}\qquad\qquad\\
& = u_{\sigma}^{-1}[(m\orr_{1,0}+n\orr_{1,1})
(m\orr_{1,1}-n\orr_{1,0})]\\
& = u_{\sigma}^{-1} ((m^{2}-n^{2}) [\orr_{1,0}\orr_{1,1}]
+mn ([\orr_{1,1}^{2}]-[\orr_{1,0}^{2}]))\\
& = (m^{2}-n^{2})\Res_{2}^{4}(\normrbar_{1}^{})
-mn\,\Res_{2}^{4}(\ot_{2})
\end{align*}
\noindent so
\begin{numequation}\label{eq-norm-orr}
\begin{split}
N_{2}^{4} (m\orr_{1,0}+n\orr_{1,1})
= (m^{2}-n^{2})\normrbar_{1} -mn\ot_{2}.
\end{split}
\end{numequation}
Similarly for integers $a$, $b$ and $c$,
\begin{align*}
\lefteqn{u_{\sigma }^{2}\Res_{2}^{4}(N_{2}^{4}
(a\orr_{1,0}^{2}+b\orr_{1,0}\orr_{1,1}+c\orr_{1,1}^{2}))}\quad\\
& = [(a\orr_{1,0}^{2}+b\orr_{1,0}\orr_{1,1}+c\orr_{1,1}^{2})
(a\orr_{1,1}^{2}-b\orr_{1,0}\orr_{1,1}+c\orr_{1,0}^{2})] \\
& = [ac (\orr_{1,0}^{4}+\orr_{1,1}^{4})
+b (c-a)\orr_{1,0}\orr_{1,1} (\orr_{1,0}^{2}-\orr_{1,1}^{2})
+ (a^{2}-b^{2}+c^{2})\orr_{1,0}^{2}\orr_{1,1}^{2}] \\
& = [ac (\orr_{1,0}^{2}-\orr_{1,1}^{2})^{2}
+b (c-a)\orr_{1,0}\orr_{1,1} (\orr_{1,0}^{2}-\orr_{1,1}^{2})
+ ((a+c)^{2}-b^{2})\orr_{1,0}^{2}\orr_{1,1}^{2}]
\end{align*}
\noindent so
\begin{numequation}\label{eq-norm-quad}
\begin{split}
N_{2}^{4}
(a\orr_{1,0}^{2}+b\orr_{1,0}\orr_{1,1}+c\orr_{1,1}^{2})
= ac\,\ot_{2}^{2}+b (c-a)\normrbar_{1}\ot_{2}
+((a+c)^{2}-b^{2})\normrbar_{1}^{2}
\end{split}
\end{numequation}
For future reference we need
\begin{align*}
\lefteqn{N_{2}^{4} ( 5 \orr_{1,0}^{2}\orr_{1,1} +5\orr_{1,0}\orr_{1,1}^{2}
+\orr_{1,1}^{3}) }\qquad\qquad\\
& = N_{2}^{4} (\orr_{1,1})N_{2}^{4} (5 \orr_{1,0}^{2} +5\orr_{1,0}\orr_{1,1}
+\orr_{1,1}^{2})) \\
& = -\normrbar_{1}(5\ot_{2}^{2}-20\normrbar_{1}\ot_{2}
+11\normrbar_{1}^{2})
\end{align*}
\noindent Compare with (\ref{eq-r3}).
We also denote by
\begin{displaymath}
\eta_{\epsilon }
=[a_{\sigma_{2}}\orr_{1,\epsilon }]
\in \upi_{1}\kH (G/G')
\end{displaymath}
\noindent the preimage of $a_{\sigma_{2}}\orr_{1,\epsilon } \in
\upi_{1}^{G'}i^{*}_{G'}\kH (G'/G')$ and by $[a_{\sigma_{2}}^{2}]\in
\upi_{-\lambda }\kH (G/G')$ the preimage of $a_{\sigma_{2}}^{2}$. The
latter is $\Res_{2}^{4}(a_{\lambda })$.
The values of $N_{2}^{4} (a_{\sigma_{2}})$ and $N_{2}^{4}
(u_{2\sigma_{2}})$ are given by Lemma \ref{lem-norm-au}, namely
\begin{align*}
N_{2}^{4} (a_{\sigma_{2}})
& = a_{\lambda } \\
\aand
N_{2}^{4} (u_{2\sigma_{2}})
& = u_{2\lambda }/u_{2\sigma }.
\end{align*}
\noindent
\begin{center}
\begin{longtable}{|p{5cm}|p{6.8cm}|}
& \kill \caption[Some elements in the slice {\SS} and homotopy
groups of {$\kH$}.]
{Some elements in the slice {\SS} and homotopy groups of $\kH$, listed in
order of ascending filtration.}\\
\hline \multicolumn{1}{|c|}{Element}
&\multicolumn{1}{|c|}{Description}\\
\hline\hline
\multicolumn{2}{|c|}{Filtration 0}\\
\hline\endfirsthead
\caption[Some elements in the slice {\SS} and homotopy groups of {$\kH$}, continued.]
{Some elements in the slice {\SS} and homotopy groups of $\kH$, continued.}\\
\hline
\multicolumn{1}{|c|}{Element}
&\multicolumn{1}{|c|}{Description}\\
\hline\endhead
\hline \multicolumn{2}{|r|}{Continued on next page} \\ \hline
\endfoot
\hline
\endlastfoot
\label{tab-pi*}
$\orr_{1,\epsilon}
\in \upi_{\rho_{2}}^{G'}i_{G'}^{*}\kH (G'/G')$
\newline $\phantom{\orr_{1,\epsilon}\,\,} \cong \upi_{G', \rho_{2}}\kH (G/G)$
\newline
with $\orr_{1,2}=-\orr_{1,0}$
&Images from (\ref{eq-rbar}) defined in \cite[(5.47)]{HHR}\\
\hline
$r_{1,\epsilon}
\in \upi_{\ee,2}\kH (G/G)$\newline
$\phantom{\orr_{1,\epsilon}}\cong \upi_{G,2}\kH (G/\ee)\cong \pi_{2}^{u}\kH$
&$\underline{r}_{1}^{2}(\orr_{1,\epsilon})$, generating \newline
$\upi_{2}^{G}\kH/\mbox{torsion}=\widehat{\oBox } $ \\
\hline $u_{2\sigma}\in \EE_{2}^{0,2-2\sigma} (G/G)$ with
&Element corresponding to
\newline $\phantom{aaaaa} u_{2\sigma}\in \upi_{2-2\sigma}\HZZ (G/G)$\\
$d_{5} (u_{2\sigma}) = a_{\sigma}^{3}a_{\lambda}\normrbar_{1}^{}$
&Slice differential of (\ref{eq-slicediffs2}) \\
$[2u_{2\sigma}]
=\langle 2,\,a_{\sigma},\,a_{\sigma}^{2}a_{\lambda}\normrbar_{1}
\rangle$
\newline $\phantom{[2u_{2\sigma}]} \in \EE_{6}^{0,2-2\sigma} (G/G)$
&Image of $2u_{2\sigma}$ in $\EE_{6}^{0,2-2\sigma} (G/G)$,
which is a permanent cycle\\
$[u_{2\sigma}^{2}]
=\langle a_{\sigma}^{3}a_{\lambda} ,\,\normrbar_{1} ,\,
a_{\sigma}^{3}a_{\lambda} ,\,\normrbar_{1} \rangle$\newline
$\phantom{[u_{2\sigma}^{2}]} \in \EE_{6}^{0,4-4\sigma} (G/G)$
&Image of $u_{2\sigma}^{2}$ in $\EE_{6}^{0,2-2\sigma} (G/G)$,
which is a permanent cycle\\
\hline $u_{\sigma}\in \upi_{1-\sigma}\kH (G/G') $
\newline $\cong \upi_{G',0}\kH (G/G)$ with
\newline $\Res_{2}^{4}(u_{2\sigma})=u_{\sigma}^{2}$,
\newline $\gamma (u_{\sigma})=-u_{\sigma}$
& Isomorphic image of
\newline $\phantom{aaaaa}
1\in \upi_{0}\kH (G/G')\cong \upi_{G',0}\kH (G/G)$\\
$\Tr_{2}^{4}(u_{\sigma}^{4k+1})=a_{\sigma}f_{1}u_{2\sigma}^{2k}$
\newline $\phantom{aaaaa}$(exotic transfer)
\newline $\Tr_{2}^{4}(u_{\sigma}^{2k})=2u_{2\sigma}^{k}$
\newline $\Tr_{2}^{4}(u_{\sigma}^{4k+3})=0$
&Follows from Theorem \ref{thm-exotic} and $d_{5} (u_{2\sigma})$ in
(\ref{eq-slicediffs2}) \\
\hline $u_{\lambda}\in \EE_{2}^{0,2-\lambda} (G/G)$ with
&Element corresponding to\newline
$\phantom{aaaaa}u_{\lambda}\in \upi_{2-\lambda}\HZZ (G/G)$\\
$[2u_{\lambda}]\in \upi_{2-\lambda}\KH (G/G)$
&$\langle 2,\,\eta ,\,a_{\lambda} \rangle$\\
$a_{\sigma}^{3}u_{\lambda}=0$
&Follows from the gold relation,\newline Lemma \ref{lem-aeu}(vii)\\
$d_{3} (u_{\lambda})=\eta a_{\lambda}
=\Tr_{2}^{4}([a_{\sigma_{2}}^{3}\orr_{1,0}])$
&Slice differential of Theorem \ref{thm-d3ulambda}\\
$d_{5} ([u_{\lambda}^{2}]) =\alphax a_{\lambda}^{2}\normrbar_{1}$
&Slice differential of Theorem \ref{thm-d3ulambda}\\
$d_{7} ([2u_{\lambda}^{2}]) =\eta ' a_{\lambda}^{3}\normrbar_{1}$
& $2\alphax a_{\lambda}^{2}\normrbar_{1}$\\
$[4u_{\lambda}^{2}]\in \upi_{4-2\lambda}\KH (G/G)$
&$\langle 2,\,\eta ,\,a_{\lambda} \rangle^{2}
=\langle 2,\,\eta ' ,\,a_{\lambda}^{3}\normrbar_{1} \rangle$\\
$[2a_{\sigma}u_{\lambda}^{2}]\in \upi_{4-\sigma -2\lambda}\KH (G/G)$
&$\langle a_{\sigma},\,\eta ' ,\,a_{\lambda}^{3}\normrbar_{1} \rangle$\\
$d_{7} ([u_{\lambda}^{4}])
=\langle \eta',\,\alphax ,\,a_{\lambda}^{2}\normrbar_{1}\rangle
a_{\lambda}^{3}\normrbar_{1}$
&$[2u_{\lambda}^{2}d (u_{\lambda}^{2})]$\\
$[2u_{\lambda}^{4}]\in \upi_{8-4\lambda}\KH (G/G)$
&$\Tr_{2}^{4}(\ou_{\lambda}^{4} )$\\
\hline
$\ou _{\lambda}\in \EE_{2}^{0,2-\lambda}
(G/G')$ with
&$\Res_{2}^{4}(u_{\lambda})$\\
$d_{3} (\ou_{\lambda} )
= [a_{\sigma_{2}}^{3} (\orr_{1,0}+\orr_{1,1})]$\newline
$\phantom{d_{3} (\ou_{\lambda} )}
=\Res_{2}^{4}(a_{\lambda })(\eta_{0}+\eta_{1})$
&$\Res_{2}^{4}(d_{3} ( u_{\lambda}))$\\
$[2\ou_{\lambda}]\in \upi_{2-\lambda}\KH (G/G') $
&$[\langle 2,\,a_{\sigma_{2}}^{3} ,\,\orr_{1,0}+\orr_{1,1} \rangle]
=\langle 2,\,[a_{\sigma_{2}}^{2}] ,\,\eta_{0}+\eta_{1} \rangle$\\
$d_{7} ([\ou_{\lambda}^{2}] )
= a_{\sigma_{2}}^{7}\orr_{1,0}^{3}$
&$\Res_{2}^{4}(d_{5} ( u_{\lambda}^{2}))$\\
$[2\ou_{\lambda}^{2}]\in \upi_{4-2\lambda}\KH (G/G') $
&$[\langle 2,\,a_{\sigma_{2}}^{7} ,\,\orr_{1,0}^{3} \rangle]
=\langle 2,\,[a_{\sigma_{2}}^{2}]^{2} ,\,\eta_{0}^{3} \rangle$\\
$[\ou_{\lambda}^{4}]\in \upi_{8-4\lambda}\KH (G/G') $
&$[\langle a_{\sigma_{2}}^{7},\,\orr_{1,0}^{3} ,\,
a_{\sigma_{2}}^{7},\,\orr_{1,0}^{3}\rangle]$\newline
$\phantom{a_{\sigma_{2}}^{7},\,\orr_{1,0}^{3}}
=\langle [a_{\sigma_{2}}^{2}]^{2} ,\,\eta_{0}^{3},\,
[a_{\sigma_{2}}^{2}]^{2} ,\,\eta_{0}^{3} \rangle$\\
\hline
$u_{\sigma_{2}}\in \upi_{(G',1-\sigma_{2})}\kH (G/e)$ with\newline
$\Res_{1}^{2}(\ou_{\lambda} )=u_{\sigma_{2}}^{2}$,\newline
$\gamma^{2} (u_{\sigma_{2}})=-u_{\sigma_{2}}$ and \newline
$\Tr_{1}^{2}(u_{\sigma_{2}})=a_{\sigma_{2}}^{2}(\orr_{1,0}+\orr_{1,1})$\newline
$\phantom{aaaaa}$(exotic transfer)
& Isomorphic image of $1\in \upi_{0}\kH (G/e)$\\
\hline
$\os_{2,\epsilon }
\in \upi_{\rho_{4}}^{G}\kH (G/G')$
&$u_{\sigma}^{-1}[\orr_{1,\epsilon }^{2}]$\\
\hline
$\normrbar_{1}
\in \upi_{\rho_{4}}^{G}\kH (G/G)$ with \newline
$\Res_{2}^{4}(\normrbar_{1})
=u_{\sigma}^{-1}[\orr_{1,0}\orr_{1,1}] $
&Image from (\ref{eq-normrbar}) defined in \cite[\S9.1]{HHR}\\
\hline
$\ot_{2}
\in \upi_{\rho_{4}}^{G}\kH (G/G) $ with\newline
$\Res_{2}^{4}(\ot_{2})= \os_{2,0}-\os_{2,1}$
&$(-1)^{\epsilon }\Tr_{2}^{4}(\os_{2,\epsilon })$
for either value of $\epsilon $\\
\hline $\ot_{2}' \in \upi_{2+\lambda}^{G}\kH (G/G) $ with\newline
$\Res_{2}^{4}(\ot_{2}')= [\orr_{1,0}^{2}]+[\orr_{1,1 }^{2}]$
&$\Tr_{2}^{4}([\orr_{1,\epsilon }^{2}])$
for either value of $\epsilon $\\
\hline $D\in \upi_{4\rho_{4}}\kH (G/G)$, \newline
the periodicity element
&$-\normrbar_{1}^{2} (5\ot_{2}^{2}-20 \ot_{2}\normrbar_{1}
+11\normrbar_{1}^{2})$\\
\hline
$\Sigma_{2, \epsilon}
\in \EE_{2}^{0,4}\kH (G/G')$ with \newline
$\Sigma_{2,2}=\Sigma_{2,0}$ and\newline
$d_{3} (\Sigma_{2, \epsilon}) = \eta_{\epsilon } ^{2} (\eta_{0}+\eta_{1} )$
&$(-1)^{\epsilon } u_{\rho_{4}}\os_{2, \epsilon}
= (-1)^{\epsilon }\ou_{\lambda} [\orr_{1,\epsilon }^{2}]$\\
\hline
$T_{2}\in \EE_{2}^{0,4}\kH (G/G)$ with \newline
${\Res_{2}^{4}(T_{2})= \Sigma_{2,0}-\Sigma_{2,1}}$ and\newline
$d_{3} (T_{2}) =\eta^{3}$
&$\Tr_{2}^{4}(\Sigma_{2,\epsilon})
= (-1)^{\epsilon }u_{\lambda}
\Tr_{2}^{4}([\orr_{1,\epsilon }^{2}])$\newline
for either value of $\epsilon $\\
\hline
$T_{4}\in \EE_{2}^{0,8}\kH (G/G)$ with \newline
$T_{4}^{2}=\Delta_{1} (T_{2}^{2}-4\Delta_{1})$,\newline
$\Res_{2}^{4}(T_{4})= (\Sigma_{2,0}-\Sigma_{2,1})\delta_{1}$ and \newline
$d_{3} (T_{4})=0$
&$(-1)^{\epsilon }\Tr_{2}^{4}(\Sigma_{2,\epsilon}\delta_{1})
= u_{2\sigma}u_{\lambda}^{2}\ot_{2}\normrbar_{1}^{}$\newline
for either value of $\epsilon $ \\
\hline
$\delta_{1}\in \EE_{2}^{0,4}\kH (G/G')$ with\newline
$\gamma (\delta_{1})=-\delta_{1}$, $\Tr_{2}^{4}(\delta_{1})=0$ \newline
and
$d_{3} (\delta_{1})=\eta_{0}\eta_{1} (\eta_{0}+\eta_{1})$
&$u_{\rho_{4}}\Res_{2}^{4}(\normrbar_{1})
=\ou_{\lambda}[\orr_{1,0}\orr_{1,1}]$\\
\hline
$\Delta_{1}\in \EE_{2}^{0,8}\kH (G/G)$ with \newline
$\Res_{2}^{4}(\Delta_{1})=\delta_{1}^{2}$, \newline
$\Res_{1}^{4}(\Delta_{1})=r_{1,0}^{2}r_{1,1}^{2}$ and \newline
$d_{5} (\Delta_{1})=\nu x_{4}$
&$u_{2\rho_{4}}\normrbar_{1}^{2}
= u_{2\sigma}u_{\lambda}^{2}\normrbar_{1}^{2}$\\
\hline \hline
\multicolumn{2}{|c|}{Filtration 1}\\
\hline
$a_{\sigma_{2}}
\in \upi_{G', -\sigma_{2}}\kH (G/G)$\newline
$\phantom{a_{\sigma_{2}}}\cong \upi_{-\sigma_{2}}^{G'}\kH (G'/G')$\newline
with $2a_{\sigma_{2}} =0$
& See Definition \ref{def-aeu} \\
\hline
$\eta_{\epsilon }
\in \upi_{1}\kH (G/G')$\newline
$\phantom{\eta_{\epsilon }}\cong \upi^{G'}_{1}\kH (G'/G')$
with $2\eta_{\epsilon } =0$
& $[a_{\sigma_{2}}\orr_{1,\epsilon} ]$\\
\hline
$\eta \in \upi^{G}_{1}\kH (G/G)$ with\newline
$\Res_{2}^{4}(\eta )= \eta_{0}+\eta_{1}$ \newline
$\phantom{\Res_{2}^{4}(\eta )}\in \upi^{G}_{1}\kH (G/G')$
& $\Tr_{2}^{4}(\eta_{\epsilon })
=\Tr_{2}^{4}([a_{\sigma_{2}}\orr_{1,0}])
=\Tr_{2}^{4}([a_{\sigma_{2}}\orr_{1,1}])$ \\
\hline
$\eta'\in \underline{\pi }_{2-\sigma}\kH (G/G)$ with\newline
$\Res_{2}^{4}(\eta')
=u_{\sigma}(\eta_{0}+\eta_{1})$
\newline
$\phantom{\Res_{2}^{4}(\eta')}\in \upi^{G}_{2-\sigma}\kH (G/G')$
&$\Tr_{2}^{4}(\eta_{0}u_{\sigma})
=\Tr_{2}^{4}([a_{\sigma_{2}}\orr_{1,0}]u_{\sigma})$\newline
$\phantom{\Tr_{2}^{4}(\eta_{0}u_{\sigma})}
=\Tr_{2}^{4}([a_{\sigma_{2}}\orr_{1,1}]u_{\sigma})$\\
\hline
$\alphax \in \upi_{2-\sigma -\lambda}\kH (G/G)$ with
&$[a_{\sigma}u_{\lambda} ]
=\langle a_{\sigma},\,\eta ,\,a_{\lambda}\rangle$\\
$\Res_{2}^{4}(\alphax ) = u_{\sigma}[a_{\sigma_{2} }^{3}\orr_{1,0}]$
\newline $\phantom{aaaaa}$ (exotic restriction)
&Follows from Theorem \ref{thm-exotic} and $d_{3} (u_{\lambda})$ in
\newline Theorem \ref{thm-d3ulambda} \\
$2\alphax =\eta 'a_{\lambda}$
& Transfer of the above\\
$\eta \alphax =a_{\lambda}\langle 2,\,a_{\sigma} ,\,f_{1}^{2} \rangle$\newline
$\phantom{\eta \alphax }
=a_{\lambda}\langle 2,\,a_{\sigma} ,\,\Tr_{2}^{4}(\eta_{0}\eta_{1}) \rangle$
&\\
$\eta' \alphax =0$
&\\
$a_{\sigma } \alphax =a_{\lambda }\Tr_{2}^{4}(u_{\sigma }^{2})$
&$[a_{\sigma }^{2}u_{\lambda }]
= a_{\lambda }[2u_{2\sigma }]$
by the gold relation, Lemma \ref{lem-aeu}(vii)\\
$a_{\sigma }^{2} \alphax =0$
&$[a_{\sigma }^{3}u_{\lambda }]
=a_{\lambda }a_{\sigma }\Tr_{2}^{4}(u_{\sigma }^{2})=0$\\
\hline
$\betax \in \upi_{4-3\sigma -\lambda}\kH (G/G)$ with
&$[\alphax u_{2\sigma}]
=\langle\alphax ,\,a_{\sigma}^{2} ,\,f_{1} \rangle$\\
$\Res_{2}^{4}(\betax )=a_{\sigma_{2}}^{3}u_{\sigma}^{3}\orr_{1,0}$
&Follows from value of $\Res_{2}^{4}(\alphax )$\\
$2\betax=a_{\lambda}\langle \eta ',\,a_{\sigma}^{2} ,\,f_{1} \rangle$
& Transfer of the above\\
$d_{5} (u_{2\sigma}u_{\lambda}^{2}) = \betax a_{\lambda}^{2}\normrbar_{1}$
&\\
$\eta \betax = 2a_{\lambda}u_{2\sigma}^{2}\normrbar_{1}^{}$
\newline $\phantom{aaaaa}$ (exotic multiplication)
&\\
$\eta' \betax = a_{\sigma}^{2}a_{\lambda}^{3}u_{2\sigma}^{2}\normrbar_{1}^{2}$
\newline $\phantom{aaaaa}$ (exotic multiplication)
&\\
\hline $\nu \in \upi_{3}\kH (G/G)$ with
&$a_{\sigma}u_{\lambda}\normrbar_{1}=\alphax \normrbar_{1}$, generating
$\circ=\upi_{3}\kH$\\
$\Res_{2}^{4}(\nu)=\eta_{0}^{3}$
and $2\nu =x_{3}$
\newline (exotic restriction
\newline and group extension)
&
\newline Follows from those on $\alphax $\\
\hline
\hline
\multicolumn{2}{|c|}{Filtration 2}\\
\hline $[a_{\sigma_{2}}^{2}]\in \upi_{-\lambda }\kH (G/G')$
&Preimage of $a_{\sigma_{2}}^{2}
\in \upi_{-2\sigma_{2}}i_{G'}^{*}\kH (G'/G')$\\
\hline $a_{\lambda }\in \upi_{-\lambda }\kH (G/G)$ with\newline
$4a_{\lambda }=0$ and
$\Res_{2}^{4}(a_{\lambda })=[a_{\sigma_{2}}^{2}]$
& See Definition \ref{def-aeu}\\
\hline $ \eta_{\epsilon }^{2} ,\, \eta_{0}\eta_{1}
\in \underline{\pi }^{G}_{2}\kH (G/G')$ with\newline
$\Tr_{2}^{4}(\eta_{\epsilon }^{2})= (-1)^{\epsilon }a_{\lambda}\ot_{2}'$
and \newline
$\Tr_{2}^{4}(\eta_{0}\eta_{1})=f_{1}^{2}$ (exotic transfer)
&$u_{\sigma}[a_{\sigma_{2}}^{2}]\os_{2,\epsilon}$ and
$u_{\sigma}[a_{\sigma_{2}}^{2}]\Res_{2}^{4}(\normrbar_{1})$, \newline
generating the torsion $\widehat{\bullet}\oplus \JJ $ in
$\upi^{G}_{2}\kH$\\
\hline
$\eta^{2}
=a_{\lambda} (\ot_{2}'+a_{\sigma }^{2}a_{\lambda }\normrbar_{1}^{2})
=a_{\lambda} \ot_{2}'+f_{1}^{2} $
& $a_{\lambda}\ot_{2}'$ has order 2 by Lemma \ref{lem-hate}\\
$\eta \eta '=a_{\lambda}[u_{2\sigma} \ot_{2}]$\newline
$(\eta ')^{2}=a_{\lambda}[u_{2\sigma}\ot_{2}']$
&See (\ref{eq-order2}) for the definition of
$[u_{2\sigma} \ot_{2}]$ and $[u_{2\sigma} \ot_{2}']$
\\
\hline $\nu^{2}\in \upi_{6}\kH (G/G)$
&$2a_{\lambda}u_{\lambda}u_{2\sigma}\normrbar_{1}^{2}
=\langle 2,\,\eta ,\,f_{1} ,\,f_{1}^{2} \rangle$\\
\hline
$\kappa \in \upi_{14}\kH (G/G)$
&$2a_{\lambda}u_{2\sigma}^{2}u_{\lambda}^{3}\normrbar_{1}^{4}$\\
\hline \hline
\multicolumn{2}{|c|}{Filtration 3}\\
\hline
$f_{1} \in \upi_{1}\kH (G/G)$
&$a_{\sigma}a_{\lambda}\normrbar_{1}^{}$,
generating the summand $\bullet$ of $\upi_{1}\kH$\\
\hline
$\eta_{0}^{3}=\eta_{0}^{2}\eta_{1}=\eta_{0}\eta_{1}^{2}=\eta_{1}^{3}$
\newline $\phantom{ab}\in \upi^{G}_{3}\kH (G/G')$
& $\eta_{\epsilon }u_{\sigma}[a_{\sigma_{2}}^{2}]\Res_{2}^{4}(\normrbar_{1})
= \eta_{\epsilon }u_{\sigma}[a_{\sigma_{2}}^{2}]\os_{2,\epsilon }$\\
\hline $x_{3}\in \upi_{3}\kH (G/G)$\newline
with $\Res_{2}^{4}(x_{3})=0$
& $\Tr_{2}^{4}(\eta_{0}^{2}\eta_{1})
=a_{\lambda} \eta' \normrbar_{1}$\\
\hline \hline
\multicolumn{2}{|c|}{Filtration 4}\\
\hline $x_{4}\in \EE_{2}^{4,8} (G/G)$\newline
with $d_{5} (x_{4})=f_{1}^{3}$,\newline
$\Res_{2}^{4}(x_{4} )=(\eta_{0}\eta_{1})^{2}=\eta_{0}^{4}$ \newline
and $2x_{4}=f_{1}\nu $
&$a_{\lambda}^{2}u_{2\sigma
}\normrbar_{1}^{2}$\\
\hline
$\overline{\kappa} \in \upi_{20}\kH (G/G)$
&$a_{\lambda}^{2}u_{2\sigma}^{3}u_{\lambda}^{4}\normrbar_{1}^{6}$\\
$2\overline{\kappa }
=\Tr_{2}^{4}(u_{\sigma }
\Res_{2}^{4}(u_{2\sigma }^{2}u_{\lambda }^{5}\normrbar_{1}^{5}))$
\newline
$\phantom{aaaaa}$(exotic transfer)
&\\
\hline \hline
\multicolumn{2}{|c|}{Filtration 8}\\
\hline $\epsilon \in \upi_{8}\kH (G/G)$
&
$x_{4}^{2}=\langle f_{1},\, f_{1}^{2} ,\, f_{1} ,\,f_{1}^{2} \rangle
\in \EE_{6}^{8,16} (G/G)$\\
\hline \hline
\multicolumn{2}{|c|}{Filtration 11}\\
\hline $\nu^{3}=\eta \epsilon \in \upi_{9}\kH (G/G)$
&Represents $f_{1}x_{4}^{2}\in \EE_{2}^{11,20} (G/G)$\\
\hline
\end{longtable}
\end{center}
\section{Slices for $\kH$ and $\KH$}\label{sec-slices}
In this section we will identify the slices for $\kH$ and $\KH$ and
the generators of their integrally graded homotopy groups. For the
latter we will use the notation of Table \ref{tab-pi*}. Let
\begin{numequation}\label{eq-Xmn}
\begin{split}
X_{m,n}=\mycases{
\Sigma^{m\rho_{4}}\HZZ
&\mbox{for }m=n\\
G_{+}\smashove{G'}\Sigma^{(m+n)\rho_{2}}\HZZ
&\mbox{for }m<n.
}
\end{split}
\end{numequation}
\noindent The slices of $\kH$ are certain finite wedges of these, and
those of $\KH$ are a certain infinite wedges. Fortunately we can
analyze these slices by considering just one value of $m$ at a time,
this index being preserved by the first differential $d_{3}$. These
are illustrated below in Figures \ref{fig-sseq-7a}--\ref{fig-sseq-8b}.
They show both $\EE_{2}$ and $\EE_{4}$ in four
cases depending on the sign and parity of $m$.
\begin{thm}\label{thm-sliceE2}
{\bf The slice $\EE_{2}$-term for $\kH$.}
The slices of $\kH$ are
\begin{displaymath}
P_{t}^{t}\kH=\mycases{
\bigvee_{0\leq m\leq t/4}X_{m,t/2-m}
&\mbox{for $t$ even and $t\geq 0$}\\
* &\mbox{otherwise}
}
\end{displaymath}
\noindent where $X_{m,n}$ is as in (\ref{eq-Xmn}).
The structure of $\pi^{u}_{*}\kH$ as a $\Z[G]$-module (see
(\ref{eq-pikH})) leads to four types of orbits and slice summands:
\begin{enumerate}
\item [(1)] $\left\{(r_{1,0}r_{1,1})^{2\ell } \right\}$ leading to
$X_{2\ell ,2\ell }$ for $\ell \geq 0$; see the leftmost diagonal in
Figure \ref{fig-sseq-7a}. On the 0-line we have a copy of $\Box$ (defined in
Table \ref{tab-C4Mackey}) generated under restrictions by
\begin{displaymath}
\Delta_{1}^{\ell }=u_{2\ell \rho_{4}}\normrbar_{1}^{2\ell}=
u_{2\sigma}^{\ell }u_{\lambda}^{2\ell }\normrbar_{1}^{2\ell}
\in \EE_{2}^{0,8\ell } (G/G).
\end{displaymath}
\noindent In positive filtrations we have
\begin{align*}
\circ &\subseteq \EE_{2}^{2j,8\ell }
\qquad \mbox{generated by } \\
a_{\lambda}^{j}u_{2\sigma}^{\ell }
u_{\lambda}^{2\ell -j}\normrbar_{1}^{2\ell }
& \in \EE_{2}^{2j,8\ell } (G/G)
\qquad \mbox{for }0<j\leq 2\ell \mbox{ and} \\
\bullet &\subseteq \EE_{2}^{2k+4\ell ,8\ell }
\qquad \mbox{generated by } \\
a_{\sigma}^{2k}a_{\lambda}^{2\ell }u_{2\sigma}^{\ell -k}
\normrbar_{1}^{2\ell }
& \in \EE_{2}^{2k+4\ell,8\ell } (G/G)
\qquad \mbox{for }0<k\leq \ell.
\end{align*}
\item [(2)] $\left\{(r_{1,0}r_{1,1})^{2\ell+1 } \right\}$ leading
to $X_{2\ell +1,2\ell +1}$ for $\ell \geq 0$; see the leftmost
diagonal in Figure \ref{fig-sseq-7b}. On the 0-line we have a copy of
$\oBox$ generated under restrictions by
\begin{align*}
\delta_{1}^{2\ell +1 }
= u_{\sigma}^{2\ell +1}
\Res_{2}^{4}(u_{\lambda}\normrbar_{1})^{2\ell +1}
&\in \EE_{2}^{0,8\ell+4 } (G/G').
\end{align*}
\noindent In positive filtrations we have
\begin{align*}
\obull &\subseteq \EE_{2}^{2j,8\ell+4 }
\qquad \mbox{generated by } \\
u_{\sigma}^{2\ell +1}
\Res_{2}^{4}(a_{\lambda}^{j}u_{\lambda}^{2\ell +1-j}
\normrbar_{1}^{2\ell +1})
& \in \EE_{2}^{2j,8\ell+4 } (G/G')
\qquad \mbox{for }0<j\leq 2\ell+1 , \\
\bullet &\subseteq \EE_{2}^{2j+1 ,8\ell+4 }
\qquad \mbox{generated by } \\
a_{\sigma}a_{\lambda}^{j}
u_{2\sigma}^{\ell}u_{\lambda}^{2\ell +1-j}
\normrbar_{1}^{2\ell +1}
& \in \EE_{2}^{2j+1,8\ell+4 } (G/G)
\qquad \mbox{for }0\leq j\leq 2\ell+1\mbox{ and} \\
\bullet &\subseteq \EE_{2}^{2k+4\ell +3 ,8\ell+4 }
\qquad \mbox{generated by } \\
a_{\sigma}^{2k+1}a_{\lambda}^{2\ell +1}
u_{2\sigma}^{\ell-k}
\normrbar_{1}^{2\ell +1}
& \in \EE_{2}^{2k+4\ell +3,8\ell+4 } (G/G)
\qquad \mbox{for }0< k\leq \ell.
\end{align*}
\item [(3)] $\left\{r_{1,0}^{i}r_{1,1}^{2\ell -i}, r_{1,0}^{2\ell
-i}r_{1,1}^{i} \right\}$ leading to $X_{i,2\ell -i}$ for $0\leq i<\ell
$; see other diagonals in Figure \ref{fig-sseq-7a}. On the 0-line we
have a copy of $\widehat{\Box}$ generated (under $\Tr_{2}^{4}$,
$\Res_{1}^{2}$ and the group action) by
\begin{displaymath}
u_{\sigma }^{\ell } \os_{2}^{\ell -i}
\Res_{2}^{4}(u_{\lambda}^{\ell }\normrbar_{1}^{i})
\in \EE_{2}^{0,4\ell } (G/G')
\end{displaymath}
In positive filtrations we have
\begin{align*}
\widehat{\bullet} &\subseteq \EE_{2}^{2j,4\ell }
\qquad \mbox{generated by } \\
\lefteqn{u_{\sigma }^{\ell } \os_{2}^{\ell -i}
\Res_{2}^{4}(a_{\lambda}^{j}u_{\lambda}^{\ell-j }\normrbar_{1}^{i})
}\qquad\qquad\\
&\in \EE_{2}^{2j,4\ell } (G/G')\qquad \mbox{for }0<j\leq \ell \\
& = \eta_{\epsilon }^{2j}u_{\sigma}^{\ell -j}\os_{2}^{\ell -i-j}
\Res_{2}^{4}(u_{\lambda}^{\ell -j}\normrbar_{1}^{i})
\qquad \mbox{for }0 < j < \ell -i.
\end{align*}
\item [(4)] $\left\{r_{1,0}^{i}r_{1,1}^{2\ell+1 -i},
r_{1,0}^{2\ell+1 -i}r_{1,1}^{i} \right\}$ leading to
$X_{i,2\ell+1 -i}$ for $0\leq i\leq \ell $; see other diagonals in
Figure \ref{fig-sseq-7b}. On the 0-line we have a copy of
$\widehat{\oBox}$ generated (under transfers and the group
action) by
\begin{displaymath}
r_{1,0}\Res_{1}^{2}(u_{\sigma }^{\ell } \os_{2}^{\ell -i})
\Res_{1}^{4}(u_{\lambda}^{\ell }\normrbar_{1}^{i})
\in \EE_{2}^{0,4\ell +2} (G/\ee)
\end{displaymath}
\noindent In positive filtrations we have
\begin{align*}
\widehat{\bullet} &\subseteq \EE_{2}^{2j+1,4\ell+2 }
\qquad \mbox{generated by } \\
\lefteqn{\eta_{\epsilon } u_{\sigma }^{\ell } \os_{2}^{\ell -i}
\Res_{2}^{4}(a_{\lambda}^{j}u_{\lambda}^{\ell-j }\normrbar_{1}^{i})
}\qquad\qquad\\
&\in \EE_{2}^{2j+1,4\ell+2 } (G/G')
\qquad \mbox{for }0\leq j\leq \ell \\
& = \eta_{\epsilon }^{2j+1}u_{\sigma}^{\ell -j}\os_{2}^{\ell -i-j}
\Res_{2}^{4}(u_{\lambda}^{\ell -j}\normrbar_{1}^{i})
\qquad \mbox{for }0 \leq j \leq \ell -i .
\end{align*}
\end{enumerate}
\end{thm}
\begin{cor}\label{cor-subring}
{\bf A subring of the slice $E_{2}$-term.} The ring
$\EE_{2}\kH (G/G')$ contains
\begin{displaymath}
\Z[\delta_{1},\Sigma_{2,\epsilon},\eta_{\epsilon }
\colon \epsilon =0,\,1]/
\left(2\eta _{\epsilon },\delta_{1}^{2}-\Sigma_{2,0}\Sigma_{2,1},
\eta_{\epsilon } \Sigma_{2,\epsilon +1}+\eta _{1+\epsilon }\delta_{1} \right);
\end{displaymath}
\noindent see Table \ref{tab-pi*} for the
definitions of its generators. In particular the elements $\eta_{0} $
and $\eta_{1}$ are algebraically independent mod 2 with
\begin{displaymath}
\gamma^{\epsilon } (\eta_{0}^{m}\eta_{1} ^{n})
\in \upi_{m+n}X_{m,n} (G/G')\qquad \mbox{for }m\leq n.
\end{displaymath}
\noindent
The element $(\eta_{0}\eta_{1})^{2}$ is the fixed point restriction of
\begin{displaymath}
u_{2\sigma}a_{\lambda}^{2}\normrbar_{1}^{2}
\in \EE_{2}^{4,8}\kH (G/G),
\end{displaymath}
\noindent which has order 4, and the transfer of the former is twice
the latter. The element $\eta_{0}\eta_{1}$ is not in the image of
$\Res_{2}^{4}$ and has trivial transfer in $\EE_{2}$.
\end{cor}
\proof
We detect this subring with the monomorphism
\begin{displaymath}
\xymatrix
@R=1mm
@C=10mm
{
{\EE_{2}\kH (G/G')}\ar[r]^(.5){\ur_{2}^{4}}
&{\EE_{2}\kH (G'/G')}\\
\eta_{\epsilon}\ar@{|->}[r]
&{\,a_{\sigma}\orr_{1,\epsilon} }\\
\Sigma_{2,\epsilon}\ar@{|->}[r]
&{\,u_{2\sigma}\orr_{1,\epsilon} ^{2}}\\
\delta_{1}\ar@{|->}[r]
&{\,u_{2\sigma}\orr_{1,0}\orr_{1,1} },
}
\end{displaymath}
\noindent in which all the relations are transparent.
\qed\bigskip
\begin{cor}\label{cor-KH}
{\bf Slices for $\KH$.} The slices of $\KH$ are
\begin{displaymath}
P_{t}^{t}\KH=\mycases{
\bigvee_{m\leq t/4}X_{m,t/2-m}
&\mbox{for $t$ even}\\
* &\mbox{otherwise}
}
\end{displaymath}
\noindent where $X_{m,n}$ is as in Theorem \ref{thm-sliceE2}. Here
$m$ can be any integer, and we still require that $m\leq n$.
\end{cor}
\proof Recall that $\KH$ is obtained from $\kH$ by inverting a certain
element
\begin{displaymath}
{D\in \upi_{4\rho_{4}}\kH (G/G)}
\end{displaymath}
\noindent described in Table \ref{tab-pi*}.
Thus $\KH$ is the homotopy colimit of the diagram
\begin{displaymath}
\xymatrix
@R=5mm
@C=10mm
{
k_{[2]}\ar[r]^(.4){D}
&\Sigma^{-4\rho_{4}}\kH\ar[r]^(.5){D}
&\Sigma^{-8\rho_{4}}\kH\ar[r]^(.6){D}
&\dotsb
}
\end{displaymath}
\noindent Desuspending by $4\rho_{4}$ converts slices to slices, so
for even $t$ we have
\begin{align*}
P_{t}^{t}\KH
& = \lim_{k\to \infty }\Sigma^{-4k\rho_{4}}P^{t+16k}_{t+16k}\kH \\
& = \lim_{k\to \infty }\Sigma^{-4k\rho_{4}}
\bigvee_{0\leq m\leq t/4+4k}X_{m,t/2+8k-m} \\
& = \lim_{k\to \infty }
\bigvee_{0\leq m\leq t/4+4k}X_{m-4k,t/2+4k-m} \\
& = \lim_{k\to \infty }
\bigvee_{-4k\leq m\leq t/4}X_{m,t/2-m} \\
& = \bigvee_{m\leq t/4}X_{m,t/2-m}.\hfill \qed
\end{align*}
\bigskip
\begin{cor}\label{cor-filtration}
{\bf A filtration of $\kH$. }
Consider the diagram
\begin{displaymath}
\xymatrix
@R=5mm
@C=10mm
{
{\kH}\ar[d]^(.5){}
&{\Sigma^{\rho_{4}}\kH}\ar[d]^(.5){} \ar[l]_(.5){\normrbar_{1}}
&{\Sigma^{2\rho_{4}}\kH}\ar[d]^(.5){}\ar[l]_(.45){\normrbar_{1}}
&\dotsb \ar[l]_(.4){\normrbar_{1}}\\
y_{0}
&y_{1}=\Sigma^{\rho_{4}}y_{0}
&y_{2}=\Sigma^{2\rho_{4}}y_{0}
}
\end{displaymath}
\noindent where $y_{0}$ is the cofiber of the map induced by
$\normrbar_{1}$. Then the slices of $y_{m}$ are
\begin{displaymath}
P_{t}^{t}y_{m}=\mycases{
X_{m,t/2-m}
&\mbox{for $t$ even and $t\geq 4m$}\\
* &\mbox{otherwise.}
}
\end{displaymath}
\end{cor}
\begin{cor}\label{cor-Filtration}
{\bf A filtration of $\KH$.} Let $R=\Z_{(2)}[x]/ (11x^{2}-20x+5)$.
After tensoring with $R$ (by smashing with a suitable Moore spectrum
$M$) there is a diagram
\begin{displaymath}
\xymatrix
@R=5mm
@C=10mm
{\dotsb \ar[r]^(.5){}
&\Sigma^{2\rho_{4}}\kH\ar[r]^(.5){f_{2}}\ar[d]^(.5){}
&\Sigma^{\rho_{4}}\kH\ar[r]^(.5){f_{1}}\ar[d]^(.5){}
&k_{[2]}\ar[r]^(.4){f_{0}}\ar[d]^(.5){}
&\Sigma^{-\rho_{4}}\kH\ar[r]^(.6){f_{-1}}\ar[d]^(.5){}
&\dotsb
\\
&Y_{2}
&Y_{1}
&Y_{0}
&Y_{-1}
&
}
\end{displaymath}
\noindent where the homotopy colimit of the top row is $\KH$ and each
$Y_{m}$ has slices similar to those of $y_{m}$ as in Corollary
\ref{cor-filtration}.
\end{cor}
\proof The periodicity element $D=-\normrbar_{1}^{2} (5\ot_{2}^{2}-20
\ot_{2}\normrbar_{1} +11\normrbar_{1}^{2})$ can be factored as
\begin{displaymath}
D=D_{0}D_{1}D_{2}D_{3}
\end{displaymath}
\noindent where $D_{i}=a_{i}\normrbar_{1}+b_{i}\ot_{2}$ with $a_{i}\in
\Z_{(2)}^{\times }$ and $b_{i}\in R$. Then
let $f_{4n+i}$ be multiplication by $D_{i}$. It follows that the
composite of any four successive $f_{m}$s is $D$, making the colimit
$\KH$ as desired. The fact that $a_{i}$ is a unit means that the
$Y$'s here have the same slices as the $y$'s in Corollary
\ref{cor-filtration}. \qed\bigskip
\begin{rem}\label{rem-Witt}
The 2-adic completion of $R$ is
the Witt ring $W (\F{4})$ used in Morava $E_{2}$-theory. This follows
from the fact that the roots of the quadratic polynomial involve
$\,\sqrt[]{5}$, which is in $W (\F{4})$ but is not a 2-adic integer.
Moreover if we assume that $D_{0}D_{1}= 5\ot_{2}^{2}-20
\ot_{2}\normrbar_{1} +11\normrbar_{1}^{2}$, then the composite maps
$f_{4n}f_{4n+1}$, as well as $f_{4n+2}$ and $f_{4n+3}$, can be
constructed without adjoining $\,\sqrt[]{5}$.
\end{rem}
It turns out that $y_{m}\wedge M$ and $Y_{m}$ for $m\geq 0$ not only have the
same slices, but the same slice spectral sequence, which is shown in
Figures \ref{fig-sseq-7a}--\ref{fig-sseq-8b}. See Remark \ref{rem-Ym}
below. We do not know if they have the same homotopy type.
\section{Some differentials in the slice {\SS} for $\kH$}\label{sec-diffs}
Now we turn to differentials. The only
generators in (\ref{eq-bigraded}) that are not permanent cycles are
the $u$'s. We will see that it is easy to account for the elements in
$\EE_{2}^{0,|V|-V} (G/H)$ for proper subgroups $H$ of
$G=C_{4}$. From (\ref{eq-bigraded}) we see that
\begin{numequation}\label{eq-sparse}
\EE_{2}^{s,t}= 0 \qquad \mbox{for $|t|$ odd.}
\end{numequation}
\noindent This sparseness condition implies that $d_{r}$ can be
nontrivial only for odd values of $r$.
Our starting point is the Slice Differentials Theorem of
\cite[Thm. 9.9]{HHR}, which is derived from the fact that the geometric
fixed point spectrum of $MU^{((G))}$ is $MO$. It says that in the
slice {\SS} for $MU^{((G))}$ for an arbitrary finite cyclic 2-group
$G$ of order $g$, the first nontrivial differential on various powers
of $u_{2\sigma}$ is
\begin{numequation}
\label{eq-slicediffs}
d_{r} (u_{2\sigma}^{2^{k-1}})
= a_{\sigma}^{2^{k}}a_{\orho}^{2^{k}-1}
N_{2}^{g} (\orr_{2^{k}-1}^{G})
\in \EE_{r}^{r,r+2^{k} (1-\sigma )-1}MU^{((G))}(G/G),
\end{numequation}
\noindent where $r=1+(2^{k}-1)g$ and $\orho$ is the
reduced regular {\rep} of $G$.
In particular
\begin{numequation}
\label{eq-slicediffs2}
\begin{split}
\left\{
\begin{array}[]{rlll}
d_{5} (u_{2\sigma})
&\hspace{-2.5mm}
= a_{\sigma}^{3}a_{\lambda}\normrbar_{1}
&\hspace{-2.5mm}\in \EE_{5}^{5,6-2\sigma}MU^{((G))} (G/G)
&\mbox{for }G=C_{4}\\
d_{13} ([u_{2\sigma}^{2}])
&\hspace{-2.5mm}
= a_{\sigma}^{7}a_{\lambda}^{3}\normrbar_{2}
&\hspace{-2.5mm}\in \EE_{13}^{13,16-4\sigma}MU^{((G))} (G/G)
&\mbox{for }G=C_{4}\\
d_{3} (u_{2\sigma})
&\hspace{-2.5mm}
= a_{\sigma}^{3}\orr_{1}
&\hspace{-2.5mm}\in \EE_{3}^{3,4-2\sigma}MU_{\reals} (G/G)
&\mbox{for }G=C_{2}\\
d_{7} ([u_{2\sigma}^{2}])
&\hspace{-2.5mm}
= a_{\sigma}^{7}\orr_{3}
&\hspace{-2.5mm}\in \EE_{3}^{7,10-4\sigma}MU_{\reals} (G/G)
&\mbox{for }G=C_{2}.
\end{array}
\right.
\end{split}
\end{numequation}
The first of these leads directly to a similar differential in the
slice {\SS } for $\kH$. The target of the second one has
trivial image in $\kH$ and we shall see that $[u_{2\sigma }^{2}]$
turns out to be a permanent cycle.
There are two ways to leverage the third and fourth differentials of
(\ref{eq-slicediffs2}) into information about $\kH$.
\begin{enumerate}
\item [(i)] They both lead to differentials in the slice {\SS} for the
$C_{2}$ spectrum $i^{*}_{G'}\kH$. They are spelled out ind
(\ref{eq-d3-and-d7}) and will be studied in detail below in
\S\ref{sec-C2diffs}. They completely determine the slice {\SS }
$\underline{E}_{*}^{*,\star} (G/G')$ for both $\kH$ and $\KH$.
Since $u_{\lambda }$ restricts to $\ou_{\lambda }$, which is
isomorphic to $u_{2\sigma_{2}}$, we get some information about
differentials on powers of $u_{\lambda }$. The $d_{3}$ on
$u_{2\sigma_{2}}$ forces a $d_{3} (u_{\lambda })=\eta a_{\lambda }$.
The target of $d_{7} ([u_{2\sigma_{2}}^{2}])$ turns out to be the
exotic restriction of an an element in filtration 5, leading to $d_{5}
([u_{\lambda }]^{2})=\nu a_{\lambda }^{2}$. We will also see that
even though $[u_{2\sigma_{2}}^{4}]$ is a permanent cycle, $[u_{\lambda
}^{4}]$ (its preimage under the restrtction map $\Res_{2}^{4}$) is
not.
\item [(ii)] One can norm up the differentials on $u_{2\sigma_{2}}$
and its square using Corollary \ref{cor-normdiff}, converting the
$d_{3}$ and $d_{7}$ to a $d_{5}$ and a $d_{13}$. The source of the
latter is $[a_{\sigma }u_{\lambda }^{4}]$, which implies that
$[u_{\lambda }^{4}]$ is not a permanent cycle.
\end{enumerate}
The differentials of (\ref{eq-slicediffs2}) lead to Massey products
which are permanent cycles,
\begin{align*}
\langle 2,\,a_{\sigma}^{2} ,\,f_{1} \rangle
& = [2u_{2\sigma}]
= \Tr_{G'}^{G}(u_{\sigma}^{2})
\in \mycases{
\EE_{6 }^{0,2-2\sigma}MU^{((G))} (G/G)
&\mbox{for }G=C_{4}\\
\EE_{4}^{0,2-2\sigma}MU_{\reals} (G/G)
&\mbox{for }G=C_{2}
}\\
\langle 2,\,a_{\sigma}^{4} ,\,f_{3} \rangle
& = [2u_{2\sigma}^{2}]
= \Tr_{G'}^{G}(u_{\sigma}^{4})
\in \mycases{
\EE_{14 }^{0,4-4\sigma}MU^{((G))} (G/G)
&\mbox{for }G=C_{4}\\
\EE_{8}^{0,4-4\sigma}MU_{\reals} (G/G)
&\mbox{for }G=C_{4}
}
\end{align*}
\noindent and (by Theorem \ref{thm-exotic}) to exotic transfers
\begin{numequation}\label{eq-exotic-transfers}
\begin{split}\left\{
\begin{array}[]{rl}
a_{\sigma}f_{1}
&\hspace{-2.5mm} = \left\{
\begin{array}{lll}
\Tr_{2}^{4} (u_{\sigma})
&\in\EE_{\infty }^{4,5-\sigma}MU^{((G))} (G/G)
&\mbox{(filtration jump 4)}\\
& \qquad \mbox{for }G=C_{4}\\
\Tr_{1}^{2} (u_{\sigma})
&\in \EE_{\infty }^{2,3-\sigma}MU_{\reals} (G/G)
&\mbox{(filtration jump 2)}\\
& \qquad \mbox{for }G=C_{2}
\end{array} \right. \\
a_{\sigma}^{3}f_{3}
&\hspace{-2.5mm} = \left\{
\begin{array}{lll}
\Tr_{2}^{4} (u_{\sigma}^{3})
&\in \EE_{\infty }^{12,15-3\sigma}MU_{\reals} (G/G)
&\mbox{(filtration jump 12)}\\
& \qquad \mbox{for }G=C_{4}\\
\Tr_{1}^{2} (u_{\sigma}^{3})
&\in \EE_{\infty }^{6,9-3\sigma}MU_{\reals} (G/G)
&\mbox{(filtration jump 6)}\\
& \qquad \mbox{for }G=C_{2}
\end{array} \right.
\end{array} \right.
\end{split}
\end{numequation}
Since $a_{\sigma }$ and $2a_{\lambda }$ kill transfers by Lemma
\ref{lem-hate}, we have Massey products,
\begin{numequation}\label{eq-order2}
\begin{split}
[u_{2\sigma }\Tr_{2}^{4}(x)] = \Tr_{2}^{4}(u_{\sigma }^{2}x)
=\langle a_{\sigma }f_{1},\,a_{\sigma } ,\, \Tr_{2}^{4}(x)\rangle
\quad \mbox{with }2a_{\lambda } [u_{2\sigma }\Tr_{2}^{4}(x)]=0.
\end{split}
\end{numequation}
Now, as before, let $G=C_{4}$ and $G'=C_{2}\subseteq G$. We need to
translate the $d_{3}$ above in the slice {\SS} for $MU_{\reals}$ into
a statement about the one for $\kH$ as a $G'$-spectrum. We have an
{\eqvr} multiplication map $m$ of $G'$-spectra
\begin{displaymath}
\xymatrix
@R=2mm
@C=10mm
{
&MU^{((G))}\ar@{=}[d]^(.5){}\\
MU_{\reals}\ar[r]^(.4){\eta_{L}}
& MU_{\reals}\wedge MU_{\reals}\ar[r]^(.6){m}
& MU_{\reals}\\
{\orr_{1}^{G'}}\ar@{|->}[r]^(.5){}
&\,{\orr_{1,0}^{G}+\orr_{1,1}^{G}}\ar@{|->}[r]^(.5){}
&{\,\orr_{1}^{G'}}\\
&a_{\sigma}^{3} (\orr_{1,0}^{G}+\orr_{1,1}^{G})\ar@{|->}[r]^(.5){}
&\,a_{\sigma}^{3}\orr_{1}^{G'}\\
{\orr_{3}^{G'}}\ar@{|->}[r]^(.5){}
&{\left(\begin{array}[]{r}
5\orr_{1,0}^{G}\orr_{1,1}^{G} (\orr_{1,0}^{G}+\orr_{1,1}^{G})
+(\orr_{1,1}^{G})^{3}\\ \bmod (\orr_{2}^{G},\orr_{3}^{G})
\end{array} \right)}
\ar@{|->}[r]^(.5){}
&\,\orr_{3}^{G'}\\
}
\end{displaymath}
\noindent where the elements lie in
$\upi^{G'}_{\rho_{2}}(\cdot)(G'/G')$ and
$\upi^{G'}_{3\rho_{2}}(\cdot)(G'/G')$. In the slice {\SS} for
$MU^{((G))}$ as a $G'$-spectrum, $d_{3} (u_{2\sigma})$ and
$d_{7}(u_{2\sigma}^{2})$ must be $G$-invariant since $u_{2\sigma}$ is,
and they must map respectively to $a_{\sigma}^{3}\orr_{1}^{G'}$ and
$a_{\sigma}^{7}\orr_{3}^{G'}$, so we have
\begin{numequation}\label{eq-d3-and-d7}
\begin{split}
\left\{\begin{array}{rl}
d_{3} (u_{2\sigma_{2} }) = d_{3} (\ou_{\lambda})
&\hspace{-2.5mm} = a_{\sigma_{2} }^{3} (\orr_{1,0}^{G}+\orr_{1,1}^{G})
= a_{\sigma_{2} }^{2} (\eta_{0}+\eta_{1})\\
d_{7} ([u_{2\sigma_{2} }^{2}]) = d_{7} ([\ou_{\lambda}^{2}])
&\hspace{-2.5mm} = a_{\sigma_{2} }^{7}\left(5\orr_{1,0}^{G}\orr_{1,1}^{G}
(\orr_{1,0}^{G}+\orr_{1,1}^{G})
+ (\orr_{1,1}^{G})^{3}+\dotsb \right)\\
&\hspace{-2.5mm} = a_{\sigma_{2} }^{7} (\orr_{1,0}^{G})^{3}+\dotsb\\
&
\qquad \mbox{since $a_{\sigma_{2}}^{3}(\orr_{1,0}^{G}+\orr_{1,1}^{G})=0$
in $\EE_{4}$}
\end{array} \right.
\end{split}
\end{numequation}
\noindent We get similar differentials in the slice spectral sequence
for $\kH$ as a $C_{2}$-spectrum in which the missing terms in $d_{7}
(\ou_{\lambda}^{2})$ vanish.
Pulling back along the isomorphism $\ur_{2}^{4}$ gives
\begin{numequation}\label{eq-d-ul}
\begin{split}\left\{
\begin{array}[]{rl}
d_{3} (\Res_{2}^{4}(u_{\lambda}))
= d_{3} (\overline{u}_{\lambda} )
&\hspace{-2.5mm} = [a_{\sigma_{2}}^{2}] (\eta_{0}+\eta_{1})
= \Res_{2}^{4}(a_{\lambda}\eta)\\
d_{7} (\Res_{2}^{4}(u_{\lambda}^{2}))
= d_{7} (\overline{u}_{\lambda}^{2} )
&\hspace{-3mm}
= \Res_{2}^{4}(a_{\lambda }^{2})\eta_{0}^{3}
= \Res_{2}^{4}(a_{\lambda }^{2}\nu )
\end{array}
\right.
\end{split}
\end{numequation}
\noindent These imply that
\begin{displaymath}
d_{3} (u_{\lambda})=a_{\lambda}\eta
\qquad \aand
d_{5} (u_{\lambda }^{2}) = a_{\lambda }^{2}\nu .
\end{displaymath}
The differential on $u_{\lambda}$ leads to the following Massey
products, the second two of which are permanent cycles.
\begin{numequation}\label{eq-alphax}
\begin{split}\left\{
\begin{array}[]{rl}
\left[u_{\lambda}^{2} \right]
= \langle a_{\lambda },\,\eta ,\, a_{\lambda },\,\eta\rangle
&\hspace{-3mm} \in \EE_{4}^{0,4-2\lambda} (G/G)\\
\left[2u_{\lambda} \right]
= \langle 2,\,\eta ,\, a_{\lambda} \rangle
&\hspace{-3mm} \in \EE_{4}^{0,2-\lambda} (G/G) \\
\alphax := [a_{\sigma}u_{\lambda}]
= \langle a_{\sigma},\,\eta ,\,a_{\lambda} \rangle
&\hspace{-3mm} \in \EE_{4}^{1,3-\sigma-\lambda} (G/G)
\end{array}
\right.
\end{split}
\end{numequation}
\noindent where $\alphax $ satisfies
\begin{align*}
a_{\sigma}^{2}\alphax
& = \langle a_{\sigma}^{3},\,\eta ,\, a_{\lambda}\rangle
= a_{\sigma }[a_{\sigma}^{2}u_{\lambda}]
= a_{\sigma} [2a_{\lambda}u_{2\sigma}]
= [2a_{\sigma }a_{\lambda}u_{2\sigma}]
= 0\\
\Res_{2}^{4}(\alphax )
& = [a_{\sigma_{2}}^{3}\orr_{1,\epsilon}]u_{\sigma}
\in \EE_{4}^{3,5-\sigma -\lambda} (G/G') \\
& \qquad \mbox{(exotic restriction with filtration jump 2 by Theorem
\ref{thm-exotic}(i))} \\
2\alphax & = \Tr_{2}^{4}(\Res_{2}^{4}(\alphax ))
= \Tr_{2}^{4}(u_{\sigma}
[a_{\sigma_{2}}^{3}\orr_{1,\epsilon}])\\
& = \eta' a_{\lambda}
\in \EE_{4}^{3,5-\sigma -\lambda} (G/G)\\
&
\qquad \mbox{(exotic group extension with jump 2)} \\
\Tr_{2}^{4}(x)\alphax
& = \Tr_{2}^{4}(x \cdot \Res_{2}^{4}(\alphax ))
= \Tr_{2}^{4}(x [a_{\sigma_{2}}^{3}\orr_{1,0}]
u_{\sigma})\\
\eta \alphax
& = \Tr_{2}^{4}([a_{\sigma_{2}}\orr_{1,1}])\alphax
= \Tr_{2}^{4}([a_{\sigma_{2}}^{4}\orr_{1,0}\orr_{1,1}]
u_{\sigma})
= a_{\lambda}^{2}\normrbar_{1}^{}\Tr_{2}^{4}(u_{\sigma}^{2})\\
& = a_{\lambda}\normrbar_{1}^{}
\langle 2,\,a_{\sigma} ,\,a_{\sigma}f_{1} \rangle
= \langle 2,\,a_{\sigma} ,\,f_{1}^{2} \rangle\\
\eta' \alphax
& = a_{\lambda}^{2}\normrbar_{1}^{}\Tr_{2}^{4}(u_{\sigma}^{3})
= 0\\
d_{7} ([\ou_{\lambda}^{2}])
& = [a_{\sigma_{2}}^{7}\orr_{1,0}^{3}]
\quad \mbox{in }\EE_{4}\\
& = \Res_{2}^{4}(\alphax )\Res_{2}^{4}(a_{\lambda}^{2}\normrbar_{1})
= \Res_{2}^{4}(\alphax a_{\lambda}^{2}\normrbar_{1})
= \Res_{2}^{4}(d_{5} (u_{\lambda}^{2} ))\\
d_{5} ([u_{\lambda}^{2}])
& = \alphax a_{\lambda}^{2}\normrbar_{1}=a_{\lambda }^{2}\nu \\
d_{7} ([2u_{\lambda}^{2}])
& = ( 2\alphax ) a_{\lambda}^{2}\normrbar_{1}
= a_{\lambda}^{3}\eta' \normrbar_{1}.
\end{align*}
\noindent Note that $\nu =\alphax \normrbar_{1}^{}$, with the exotic
restriction and group extension on $\alphax $ being consistent with
those on $\nu $.
\bigskip
The differential on $[u_{\lambda}^{2}]$ yields Massey products
\begin{numequation}\label{eq-brackets3}
\begin{split}
\left\{
\begin{array}[]{rl}
[a_{\sigma}^{2}u_{\lambda}^{2}]
&\hspace{-2.5mm} = \langle a_{\sigma}^{2},\,\alphax
,\,a_{\lambda}^{2}\normrbar_{1} \rangle\\
\left[\eta' u_{\lambda}^{2} \right]
&\hspace{-2.5mm} = \langle \eta ',\,\alphax
,\,a_{\lambda}^{2}\normrbar_{1} \rangle.
\end{array} \right.
\end{split}
\end{numequation}
\begin{thm}\label{thm-normed-slice-diffs}
{\bf Normed up slice differentials for $\kH$ and $\KH$.} In the slice
{\SS}s for $\kH$ and $\KH$,
\begin{align*}
d_{5} ( [a_{\sigma }u_{\lambda }^{2}])
& = 0 \\
\aand
d_{13} ([a_{\sigma }u_{\lambda }^{4}])
& = a_{\lambda }^{7}[u_{2\sigma }^{2}]\normrbar_{1}^{3} .
\end{align*}
\end{thm}
\proof
The two slice differentials over $G'$ are
\begin{align*}
d_{3} (u_{2\sigma_{2}})
& = a_{\sigma_{2}}^{3}\orr_{1}^{G'}
= a_{\sigma_{2}}^{3} (\orr_{1,0}+\orr_{1,1}) \\
\aand
d_{7} ([u_{2\sigma_{2}}^{2}])
& = a_{\sigma_{2}}^{7}\orr_{3}^{G'}
= a_{\sigma_{2}}^{7} ( 5 \orr_{1,0}^{2}\orr_{1,1} +5\orr_{1,0}
\orr_{1,1}^{2} +\orr_{1,1}^{3})
\end{align*}
\noindent We need to find the norms of both sources and targets.
Lemma \ref{lem-norm-au} tells us that
\begin{align*}
N_{2}^{4} (a_{\sigma_{2}}^{k})
& = a_{\lambda }^{k}\\
\aand N_{2}^{4} (u_{2\sigma_{2}}^{k}) & = u_{\lambda }^{2k}/u_{2\sigma
}^{k} \qquad \mbox{in }\underline{E}_{2}(G/G) .
\end{align*}
\noindent Previous calculations give
\begin{align*}
N_{2}^{4} (\orr_{1,0}+\orr_{1,1})
& = -\overline{t}_{2}\qquad \mbox{by (\ref{eq-norm-orr})} \\
\aand
N_{2}^{4} ( 5 \orr_{1,0}^{2}\orr_{1,1} +5\orr_{1,0}\orr_{1,1}^{2}
+\orr_{1,1}^{3})
& = -\normrbar_{1} (5\ot_{2}^{2}-20 \ot_{2}\normrbar_{1}
+11\normrbar_{1}^{2})\\
& \phantom{ = -\overline{t}_{2}}\, \qquad \mbox{by (\ref{eq-norm-quad})} ,
\end{align*}
For the first differential, Corollary \ref{cor-normdiff} tells us that
\begin{align*}
a_{\lambda }^{3}\overline{t}_{2}
& = d_{5} (a_{\sigma }u_{\lambda }^{2}/u_{2\sigma }) \\
& = d_{5} (a_{\sigma } u_{\lambda }^{2})/u_{2\sigma }
-a_{\sigma }u_{\lambda }^{2}d_{5} (u_{2\sigma })/[u_{2\sigma }^{2}]\\
& = d_{5} (a_{\sigma } u_{\lambda }^{2})/u_{2\sigma }
-a_{\sigma }u_{\lambda }^{2}a_{\sigma }^{3}
a_{\lambda }\normrbar_{1}/[u_{2\sigma }^{2}]
\end{align*}
\noindent Multiplying both sides by the permanent cycle
$[u_{2\sigma}^{2}]$ gives
\begin{align*}
[u_{2\sigma }d_{5} (a_{\sigma } u_{\lambda }^{2})]
& = a_{\lambda }^{3}[u_{2\sigma }^{2}] \ot_{2}
+ a_{\sigma }u_{\lambda }^{2}a_{\sigma }^{3}
a_{\lambda }\normrbar_{1} \\
& = a_{\lambda }^{3}[u_{2\sigma }^{2}] \ot_{2}
+ 4a_{\lambda }^{3}[u_{2\sigma }^{2}]\normrbar_{1}\\
& = a_{\lambda }^{3}[u_{2\sigma }^{2}] \ot_{2}\\
d_{5} (a_{\sigma } u_{\lambda }^{2})
& = a_{\lambda }^{3}[u_{2\sigma } \ot_{2}].
\end{align*}
\noindent We have seen that
\begin{displaymath}
\eta \eta ' = a_{\lambda }[u_{2\sigma } \overline{t}_{2}].
\end{displaymath}
\noindent This implies that $a_{\lambda }^{2}[u_{2\sigma }
\overline{t}_{2}]$ vanishes in $\underline{E}_{5}$ since $a_{\lambda
}\eta $ is killed by $d_{3}$. It follows that $d_{5} (a_{\sigma }
u_{\lambda }^{2}) = a_{\lambda }^{3}[u_{2\sigma } \ot_{2}]=0$ as claimed.
For the second differential we have
\begin{align*}
d_{13} ([a_{\sigma }u_{\lambda }^{4}/u_{2\sigma }^{2}])
& = a_{\lambda }^{7} \normrbar_{1} (-5\ot_{2}^{2}+20 \ot_{2}\normrbar_{1}
+9\normrbar_{1}^{2})\\
d_{13} ([a_{\sigma }u_{\lambda }^{4}])
& = a_{\lambda }^{7} [u_{2\sigma }^{2}]\normrbar_{1}
(-5\ot_{2}^{2}+20 \ot_{2}\normrbar_{1}
+9\normrbar_{1}^{2})\\
& = a_{\lambda }^{7}[u_{2\sigma}^{2}]\normrbar_{1}
(-\ot_{2}^{2}+\normrbar_{1}^{2})
\end{align*}
\noindent since $a_{\lambda }$ has order 4. As we saw above,
$a_{\lambda }^{2}[u_{2\sigma } \overline{t}_{2}]$ vanishes in
$\underline{E}_{5}$, so $d_{13} ([a_{\sigma }u_{\lambda }^{4}])$ is as
claimed. \noindent\qed\bigskip
We can use this to find the differential on $[u_{\lambda}^{4}]$.
We have
\begin{numequation}\label{eq-d7}
\begin{split}\left\{
\begin{array}[]{rl}
d ([u_{\lambda}^{4}])
&\hspace{-2.5mm} = [2u_{\lambda}^{2}]d ([u_{\lambda}^{2}])
= [2u_{\lambda}^{2}]\alphax a_{\lambda}^{2}\normrbar_{1}^{}
= (2\alphax )a_{\lambda}^{2}[u_{\lambda}^{2}]\normrbar_{1}\\
&\hspace{-2.5mm} = \eta 'a_{\lambda}^{3}[u_{\lambda}^{2}]\normrbar_{1}^{}
= [\eta 'u_{\lambda}^{2}]a_{\lambda}^{3}\normrbar_{1}^{}
= \langle \eta',\,\alphax ,\,a_{\lambda}^{2}\normrbar_{1}\rangle
a_{\lambda}^{3}\normrbar_{1}.
\end{array}
\right.
\end{split}
\end{numequation}
The differential on $u_{2\sigma}$ yields
\begin{displaymath}
[x u_{2\sigma}]
=\langle x,\, a_{\sigma}^{2}
,\, f_{1}\rangle
\end{displaymath}
\noindent for any permanent cycle $x$ killed by $a_{\sigma}^{2}$.
Possible values of $x$ include 2, $\eta $, $\eta '$ (each of which is
killed by $a_{\sigma}$ as well) and $\alphax $. For the last of these
we write
\begin{numequation}\label{eq-brackets2}
\betax := [\alphax u_{2\sigma}]
= \langle \alphax ,\,a_{\sigma}^{2}
,\, f_{1}\rangle
= \langle[ a_{\sigma} u_{\lambda }] ,\,a_{\sigma}^{2}
,\, f_{1}\rangle
\in \EE_{6}^{1,5-3\sigma-\lambda} (G/G),
\end{numequation}
\noindent which satisfies
\begin{align*}
\Res_{2}^{4}(\betax )
& = a_{\sigma_{2}}^{3}u_{\sigma}^{3}\orr_{1,\epsilon}
\in \EE_{4}^{3,7-3\sigma -\lambda} (G/G')\\
&
\qquad \mbox{(exotic restriction with jump 2)}\\
2\betax
& = \Tr_{2}^{4}(\Res_{2}^{4}(\betax ))
= \eta' a_{\lambda}u_{2\sigma}
\in \EE_{4}^{3,7-3\sigma -\lambda} (G/G)\\
&
\qquad \mbox{(exotic group extension with jump 2)}\\
d_{5} ([u_{2\sigma}u_{\lambda}^{2}])
& = a_{\sigma}^{3}a_{\lambda}u_{\lambda}^{2}\normrbar_{1}
+\alphax a_{\lambda}^{2}u_{2\sigma}\normrbar_{1}
= (a_{\sigma}^{3}u_{\lambda}^{2}+\alphax u_{2\sigma})
a_{\lambda}^{2}\normrbar_{1}\\
& = (2a_{\sigma}a_{\lambda}u_{\lambda}u_{2\sigma}+\betax )
a_{\lambda}^{2}\normrbar_{1}
= \betax a_{\lambda}^{2}\normrbar_{1}\\
d_{7} ([2u_{2\sigma }u_{\lambda }^{2}])
& = 2\betax\cdot a_{\lambda}^{2}\normrbar_{1}
= \eta' a_{\lambda}^{3}u_{2\sigma}\normrbar_{1}\\
\Res_{2}^{4}(d_{5} ([u_{2\sigma}u_{\lambda}^{2}]))
& = u_{\sigma}^{3}a_{\sigma_{2}}^{3}\orr_{1,\epsilon}
\Res_{2}^{4}(a_{\lambda}^{2}\normrbar_{1})
= u_{\sigma}^{2}a_{\sigma_{2}}^{7}\orr_{1,0}^{3}
= u_{\sigma}^{2}d_{7} (\overline{u}_{\lambda}^{2} ).
\end{align*}
\begin{thm}\label{thm-d3ulambda}
{\bf The differentials on powers of $u_{\lambda}$ and $u_{2\sigma}$.}
The following differentials occur in the slice {\SS} for $\kH$. Here
$\ou_{\lambda}$ denotes $\Res_{2}^{4}(u_{\lambda})$.
\begin{align*}
d_{3} (u_{\lambda})
& = a_{\lambda}\eta
= \Tr_{2}^{4}(a_{\sigma_{2}}^{3}\orr_{1,\epsilon })\\
d_{3} (\ou_{\lambda})
& = \Res_{2}^{4}(a_{\lambda}) (\eta_{0}+\eta_{1})
= \left[a_{\sigma_{2}}^{3} (\orr_{1,0}+\orr_{1,1}) \right]\\
d_{5} (u_{2\sigma})
& = a_{\sigma}^{3}a_{\lambda}\normrbar_{1}^{} \\
d_{5} ([u_{\lambda}^{2}])
& = a_{\lambda}^{2}a_{\sigma}u_{\lambda}\normrbar_{1}
= a_{\lambda}^{2}\alphax \normrbar_{1}
= a_{\lambda}^{2}\nu\qquad
\mbox{for $\alphax$
as in (\ref{eq-alphax})} \\
d_{5} ([u_{2\sigma}u_{\lambda}^{2}])
& = a_{\sigma}^{3}a_{\lambda}u_{\lambda}^{2}\normrbar_{1}
+\alphax a_{\lambda}^{2}u_{2\sigma}\normrbar_{1}
= (a_{\sigma}^{3}u_{\lambda}^{2}+\alphax u_{2\sigma})
a_{\lambda}^{2}\normrbar_{1}\\
& = \betax a_{\lambda}^{2}\normrbar_{1}
\qquad \mbox{for $\betax$ as in (\ref{eq-brackets2})} \\
d_{7} ([2u_{2\sigma}u_{\lambda}^{2}])
& = \eta' a_{\lambda}^{3}u_{2\sigma}\normrbar_{1}\\
d_{7} ([2u_{\lambda}^{2}])
& = 2 a_{\lambda}^{2}\alphax \normrbar_{1}
= a_{\lambda}^{3}\eta '\normrbar_{1}^{}\\
d_{7} ([\ou_{\lambda}^{2}])
& = \Res_{2}^{4}(a_{\lambda}^{2})\eta_{0}^{3}
= a_{\sigma_{2}}^{7}\orr_{1,0}^{3}\\
d_{7} ([u_{\lambda}^{4}])
& = [\eta 'u_{\lambda}^{2}]a_{\lambda}^{3}\normrbar_{1}
= \langle \eta',\,\alphax ,\,a_{\lambda}^{2}\normrbar_{1}\rangle
a_{\lambda}^{3}\normrbar_{1}.
\end{align*}
\noindent The elements
\begin{align*}
u_{\sigma},
& &[2u_{\lambda}]
& = \langle 2,\,\eta ,\,a_{\lambda} \rangle
,\\
[2u_{2\sigma}]
& = \langle 2,\,a_{\sigma}^{2} ,\,f_{1} \rangle
=\Tr_{2}^{4}(u_{\sigma}^{2}),
&[4u_{\lambda}^{2}]
& = \langle 2,\,\eta '
,\,a_{\lambda}^{3}\normrbar_{1} \rangle
=\Tr_{1}^{4}(u_{\sigma_{2} }^{4}),\\
[u_{2\sigma}^{2} ]
&=\langle a_{\sigma}^{2},\,f_{1} ,\,a_{\sigma}^{2},\,f_{1} \rangle
& [2\ou_{\lambda}^{2}]
& = \langle 2,\,a_{\sigma_{2} }^{6}
,\,a_{\sigma_{2}}\orr_{1,0}^{3} \rangle
=\Tr_{1}^{2}(u_{\sigma_{2} }^{4}), \\
[2u_{2\sigma}u_{\lambda}]
&=\langle [2u_{2\sigma}],\,\eta ,\,a_{\lambda} \rangle,
& [2u_{\lambda}^{4}]
& = \langle 2,\,\eta ',\alphax
,\,a_{\lambda}^{5}\normrbar_{1}^{2} \rangle
=\Tr_{2}^{4}(\ou_{\lambda}^{4}), \\
[\ou_{\lambda}^{4}]
& = \langle a_{\sigma_{2}}^{7},\,\orr_{1,0}^{3} , \,
a_{\sigma_{2}}^{7},\,\orr_{1,0}^{3}\rangle
&\aand [u_{\lambda}^{8}]
& = \langle [\eta 'u_{\lambda }^{2}]
,\,a_{\lambda}^{3}\normrbar_{1}
,\, [\eta 'u_{\lambda }^{2}]
,\,a_{\lambda}^{3}\normrbar_{1} \rangle
\end{align*}
\noindent are permanent cycles.
We also have the following exotic restriction and transfers.
\begin{align*}
\Res_{2}^{4}(a_{\sigma}u_{\lambda})
& = u_{\sigma}\Res_{2}^{4}(a_{\lambda})\eta_{\epsilon }
= u_{\sigma}a_{\sigma_{2}}^{3}\orr_{1,\epsilon }
\qquad \mbox{(filtration jump 2)} \\
\Tr_{2}^{4}(u_{\sigma}^{k})
& = \mycases{
a_{\sigma}^{2}a_{\lambda}\normrbar_{1}u_{2\sigma}^{(k-1)/2}
= a_{\sigma}f_{1}u_{2\sigma}^{(k-1)/2}\\
\qquad \mbox{(filtration jump 4)}
&\mbox{for $k\equiv 1$ mod 4}\hspace{4cm}\\
2u_{2\sigma}^{k/2}
&\mbox{for $k$ even}\\
0 &\mbox{for $k\equiv 3$ mod 4}
}\\
\Tr_{1}^{2}(u_{\sigma_{2}}^{k})
& = \mycases{
a_{\sigma_{2}}^{2} (\orr_{1,0}+\orr_{1,1})\ou_{\lambda}^{(k-1) /2}
= a_{\sigma_{2}} (\eta_{0}+\eta_{1})\ou_{\lambda}^{(k-1) /2}
\\
\qquad \mbox{(filtration jump 2)}
&\hspace{-2.4cm} \mbox{for $k\equiv 1$ mod 4 }\\
2\ou_{\lambda}^{k/2}
&\hspace{-2.4cm}\mbox{for $k$ even}\\
a_{\sigma_{2}}^{6} \orr_{1,0}^{3}\ou_{\lambda}^{(k-3) /2}\\
\qquad \mbox{(filtration jump 6)}
&\hspace{-2.4cm}\mbox{for $k\equiv 3$ mod 4 }
}
\end{align*}
\end{thm}
\proof All differentials were established above.
The differential on $u_{\lambda}^{2}$ does {\em not} lead to an exotic
transfer because neither $[\ou_{\lambda}^{2}]$ nor
$[u_{\lambda} a_{\lambda}^{2}\normrbar_{1}]$ is a permanent cycle as
required by Theorem \ref{thm-exotic}.
We need to discuss the element $[2u_{2\sigma}u_{\lambda}] =\langle
[2u_{2\sigma}],\,\eta ,\,a_{\lambda} \rangle$. To see that this Toda
bracket is defined, we need to verify that $[2u_{2\sigma}]\eta =0$.
For this we have
\begin{displaymath}
[2u_{2\sigma}] \eta
= [2u_{2\sigma}]\Tr_{2}^{4}(\eta_{0})
= \Tr_{2}^{4}(2u_{\sigma}^{2}\eta_{0})
= \Tr_{2}^{4}(0)=0.
\end{displaymath}
The exotic restriction and transfers are applications of Theorem
\ref{thm-exotic} to the differentials on $u_{\lambda}$ and on
$\left[u_{2\sigma}^{(k+1)/2} \right]$ and
$\left[\ou_{\lambda}^{(k+1)/2} \right]$ for odd $k$. For even $k$ we
have
\begin{displaymath}
\Tr_{2}^{4}(u_{\sigma}^{k})
= \Tr_{2}^{4}\left(\Res_{2}^{4}
\left(\left[u_{2\sigma}^{k/2}\right]\right) \right)
= \left[2u_{2\sigma}^{k/2} \right]
\quad \mbox{since } \Tr_{2}^{4}(\Res_{2}^{4}(x))= (1+\gamma )x,
\end{displaymath}
\noindent and similarly for even powers of $u_{\sigma_{2}}$.
As remarked above, we lose no information by inverting the class $D$,
which is divisible by $\normrbar_{1}$. It is shown in
\cite[Thm. 9.16]{HHR} that inverting the latter makes
$u_{2\sigma}^{2}$ a permanent cycle. One can also see this from
(\ref{eq-slicediffs2}). Since $d_{5}
(u_{2\sigma})=a_{\sigma}^{3}a_{\lambda}\normrbar_{1}$, $d_{5}
(u_{2\sigma}\normrbar_{1}^{-1})=a_{\sigma}^{3}a_{\lambda}$. This
means that $d_{13}
([u_{2\sigma}^{2}])=a_{\sigma}^{7}a_{\lambda}^{3}\normrbar_{3}$ is
trivial in $\EE_{6} (G/G)$. It turns out that there is no possible
target for a higher differential. \qed\bigskip \bigskip
\section{$\kH$ as a $C_{2}$-spectrum}
\label{sec-C2diffs}
Before studying the slice {\SS} for the $C_{4}$-spectrum $\kH$
further, it is helpful to explore its restriction to $G'=C_{2}$, for
which the $\Z$-bigraded portion
\begin{displaymath}
\EE_{2}^{*,*}i_{G'}^{*}\kH (G'/G')
\cong \EE_{2}^{*,(G',*)}\kH (G/G)
\cong \EE_{2}^{*,*}\kH (G/G')
\end{displaymath}
\noindent (see Theorem \ref{thm-module} for these isomorphisms) is the
isomorphic image of the subring of Corollary \ref{cor-subring}. In
the following we identify $\Sigma_{2,\epsilon }$, $\delta_{1}$ and
$\orr_{1,\epsilon }$ (see Table \ref{tab-pi*}) with their images under
$\ur_{2}^{4}$. From the differentials of (\ref{eq-d3-and-d7}) we get
\begin{numequation}\label{eq-k2-C2-diffs}
\begin{split}
\left\{\begin{array}{rl}
d_{3} (\Sigma_{2,\epsilon })
&\hspace{-2.5mm} = \eta_{\epsilon }^{2} (\eta_{0}+\eta_{1})
= a_{\sigma }^{3} \orr_{1,\epsilon }^{2} (\orr_{1,0}+\orr_{1,1}) \\
d_{3} (\delta_{1})
&\hspace{-2.5mm} = \eta_{0}^{2}\eta_{1}+\eta_{0}\eta_{1}^{2}
= a_{\sigma }^{3} \orr_{1,0 } \orr_{1,1 } (\orr_{1,0}+\orr_{1,1}) \\
d_{7} ([\delta_{1}^{2}])
&\hspace{-2.5mm} = d_{7} (u_{2\sigma}^{2}) \orr_{1,0}^{2}\orr_{1,1}^{2}
= a_{\sigma}^{7}\orr_{3}^{G'}
\orr_{1,0}^{2}\orr_{1,1}^{2}\\
&\hspace{-2.5mm} = a_{\sigma}^{7} (5\orr_{1,0}^{2}\orr_{1,1}
+5\orr_{1,0}\orr_{1,1}^{2}+\orr_{1,1}^{3})
\orr_{1,0}^{2}\orr_{1,1}^{2}.
\end{array} \right.
\end{split}
\end{numequation}
\noindent The $d_{3}$s above make all monomials in $\eta_{0}$ and
$\eta_{1}$ of any given degree $\geq 3$ the same in $\EE_{4}
(G/G')$ and $\EE_{4} (G'/G')$, so $d_{7}
(\delta_{1}^{2})=\eta_{0}^{7}$. Similar calculations show that
\begin{displaymath}
d_{7} ([\Sigma_{2,\epsilon }^{2}])=\eta_{0}^{7}
= a_{\sigma}^{7}\orr_{1,0}^{7}.
\end{displaymath}
\noindent The image of the periodicity element $D$ here is as in
(\ref{eq-r24D}).
We have the following values of the transfer on powers of $u_{\sigma }$.
\begin{displaymath}
\Tr_{1}^{2}(u_{\sigma }^{i})=\mycases{
[2u_{2\sigma }^{i/2}]
&\mbox{for $i$ even} \\
{}
[a_{\sigma }^{2} u_{2\sigma }^{(i-1) /2}] (\orr_{1,0}+\orr_{1,1})
&\mbox{for $i\equiv 1$ mod 4} \\
{}
[u_{2\sigma }^{4}]^{(i-3) /8}a_{\sigma }^{6}\orr_{1,0}^{3}
= [ u_{2\sigma }^{4}]^{(i-3) /8}a_{\sigma }^{6}\orr_{1,1}^{3}
&\mbox{for $i\equiv 3$ mod 8}\\
0 &\mbox{for $i\equiv 7$ mod 8}
}
\end{displaymath}
\noindent
This leads to the following, for which Figure
\ref{fig-G'SS} is a visual aid.
\begin{figure}
\begin{center}
\includegraphics[width=10.9cm]{fig-G_SS.pdf}
\caption[The slice {\SS} for $\kH$ as a $C_{2}$-spectrum. ]{The slice
{\SS} for $\kH$ as a $C_{2}$-spectrum. The Mackey functor symbols are
defined in Table \ref{tab-C2Mackey}. The $C_{4}$-structure of the
Mackey functors is not indicated here. In each bidegree we have a
direct sum of the indicated number of copies of the indicated Mackey
functor. Each $d_{3}$ has maximal rank, leaving a cokernel of rank 1,
and each $d_{7}$ has rank 1. Blue lines indicate exotic transfers. The
ones raising filtration by 2 have maximal rank while the ones raising
it by 6 have rank 1. The resulting
$\EE_{8}=\EE_{\infty }$-term is shown below.}
\includegraphics[width=11cm]{fig-G_SS-2.pdf}
\vspace{3cm}
\label{fig-G'SS}
\end{center}
\end{figure}
\begin{thm}\label{thm-G'SS}
{\bf The slice {\SS} for $\kH$ as a $C_{2}$-spectrum.} Using the
notation of Table \ref{tab-C2Mackey} and Definition
\ref{def-enriched}, we have
\begin{align*}
\EE_{2}^{*,*} (G'/\ee)
& = \Z[r_{1,0}, r_{1,1}]\qquad
\mbox{with }r_{1,\epsilon }\in \EE_{2}^{0,2} (G'/\ee) \\
\EE_{2}^{*,*} (G'/G')
& = \Z[\delta_{1},\Sigma_{2,\epsilon},\eta_{\epsilon }
\colon \epsilon =0,\,1]/\\
&\qquad
\left(2\eta _{\epsilon },\delta_{1}^{2}-\Sigma_{2,0}\Sigma_{2,1},
\eta_{\epsilon } \Sigma_{2,\epsilon +1}+\eta _{1+\epsilon }\delta_{1} \right),
\end{align*}
\noindent so
\begin{displaymath}
\EE_{2}^{s,t} =\mycases{
\Box \oplus \bigoplus_{\ell}\widetilde{\Box}
&\mbox{for $(s,t)= (0,4\ell )$ with $\ell \geq 0$}\vspace{2mm}\\
\bigoplus_{\ell +1}\widetilde{\oBox}
&\mbox{for $(s,t)= (0,4\ell+2 )$ with $\ell \geq 0$}\vspace{2mm}\\
\bullet\oplus \bigoplus_{u+\ell}\widetilde{\bullet}
&\mbox{for $(s,t)= (2u,4\ell+4u )$
with $\ell \geq 0$ and $u>0$}\vspace{2mm}\\
\bigoplus_{u+\ell}\widetilde{\bullet}
&\mbox{for $(s,t)= (2u-1,4\ell+4u -2)$
with $\ell \geq 0$ and $u> 0$}\vspace{2mm}\\
0 &\mbox{otherwise.}
}
\end{displaymath}
The first set of differentials and determined by
\begin{displaymath}
d_{3} (\Sigma_{2,\epsilon })=\eta_{\epsilon }^{2} (\eta_{0}+\eta_{1})
\qquad \aand
d_{3} (\delta_{1}) = \eta_{0}\eta_{1} (\eta_{0}+\eta_{1})
\end{displaymath}
\noindent and there is a second set of differentials determined by
\begin{displaymath}
d_{7} (\Sigma_{2,\epsilon }^{2})=d_{7} (\delta_{1}^{2})=\eta_{0}^{7}
\end{displaymath}
\end{thm}
\begin{cor}\label{cor-G'SS}
{\bf Some nontrivial permanent cycles.} The elements listed below in
$\EE_{2}^{s,8i+2s}\kH (G/G')$ are nontrivial permanent
cycles. Their tranfers in $\EE_{2}^{s,8i+2s}\kH (G/G)$ are also
permanent cycles.
\begin{itemize}
\item [$\bullet$] $\Sigma_{2,\epsilon }^{2i-j}\delta_{1}^{j}$ for
$0\leq j\leq 2i$ ($4i+1$ elements of infinite order including
$\delta_{1}^{2i}$), $i$ even and $s=0$.
\item [$\bullet$] $\eta_{\epsilon }\Sigma_{2,\epsilon
}^{2i-j}\delta_{1}^{j}$ for $0\leq j<2i$ and $\eta_{\epsilon
}\delta_{1}^{2i}$ ($4i+2$ elements or order 2) for $i$ even and $s=1$.
\item [$\bullet$] $\eta_{\epsilon }^{2}\Sigma_{2,\epsilon
}^{2i-j}\delta_{1}^{j}$ for $0\leq j<2i$ and
$\delta_{1}^{2i}\left\{\eta_{0}^{2},\,\eta_{0}\eta_{1},\,\eta_{1}^{2}
\right\}$ ($4i+3$ elements or order 2) for $i$ even and $s=2$.
\item [$\bullet$] $\eta_{0}^{s}\delta_{1}^{2i}$ for $3\leq s\leq 6$ (4
elements or order 2) and $i$ even.
\item [$\bullet$]
$\Sigma_{2,\epsilon}^{2i-j}\delta_{1}^{j}+\delta_{1}^{2i}$ for $0\leq
j\leq 2i$ ($4i+1$ elements of infinite order including
$2\delta_{1}^{2i}$), $i$ odd and $s=0$.
\item [$\bullet$]
$\eta_{\epsilon}\Sigma_{2,\epsilon}^{2i-j}\delta_{1}^{j}+\delta_{1}^{2i}$
for $0\leq j\leq 2i-1$ and $\eta_{0}\delta_{1}^{2i-1}
(\Sigma_{2,1}+\delta_{1}) = \eta_{1}\delta_{1}^{2i-1}
(\Sigma_{2,0}+\delta_{1})$ ($4i+1$ elements of order 2), $i$ odd and $s=1$.
\item [$\bullet$]
$\eta_{\epsilon}^{2}\Sigma_{2,\epsilon}^{2i-j}\delta_{1}^{j}+\delta_{1}^{2i}$
for $0\leq j\leq 2i-1$,
$\eta_{0}^{2}\delta_{1}^{2i-1}(\Sigma_{2,1}+\delta_{1}) =
\eta_{0}\eta_{1}\delta_{1}^{2i-1}(\Sigma_{2,0}+\delta_{1})$ and
$\eta_{0}\eta_{1}\delta_{1}^{2i-1}(\Sigma_{2,1}+\delta_{1}) =
\eta_{1}^{2}\delta_{1}^{2i-1}(\Sigma_{2,0}+\delta_{1})$ ($4i+2$
elements of order 2) for $i$ odd and $s=2$.
\end{itemize}
In $\EE_{2}^{0,8i+4}\kH (G/G')$ we have $2\Sigma_{2,\epsilon
}^{2i+1-j}\delta_{1}^{j}$ for $0\leq j\leq 2i$ and $2\delta_{1}^{j}$,
$4i+3$ elements of infinite order, each in the image of the transfer
$\Tr_{1}^{2}$.
\end{cor}
\begin{rem}\label{rem-a7}
In {\bf the $RO (G)$-graded slice {\SS } for $k_{[2]}$} one has
\begin{displaymath}
d_{3}(u_{2\sigma })=a_{\sigma }^{3} (\orr_{1,0}+\orr_{1,1})
\qquad \aand
d_{7}([u_{2\sigma }^{2}])=a_{\sigma }^{7} \orr_{3}^{G'}
=a_{\sigma }^{7} \orr_{1,0}^{3},
\end{displaymath}
\noindent but $a^{7}$ itself, and indeed all higher powers of $a$,
survive to $\underline{E}_{8}=\underline{E}_{\infty }$. Hence the
$\underline{E}_{\infty }$-term of this {\SS} does {\bf not} have the
horizontal vanishing line that we see in $\underline{E}_{8}$-term of
Figure \ref{fig-KR}. However when we pass from $k_{[2]}$ to $K_{[2]}$,
$\orr_{3}^{G'}=5\orr_{1,0}^{2}\orr_{1,1}
+5\orr_{1,0}\orr_{1,1}^{2}+\orr_{1,1}^{3}$ becomes invertible and we
have
\begin{displaymath}
d_{7} ((\orr_{3}^{G'})^{-1}[u_{2\sigma }^{2}])
= d_{7} (\orr_{1,0}^{-3}[u_{2\sigma }^{2}])
= a^{7}.
\end{displaymath}
\noindent On the other hand, $\orr_{1,0}+\orr_{1,1}$ is not invertible, so we cannot divide $u_{2\sigma }$ by it.
\end{rem}
We now give the {\Ps} computation analogous to the one following
Remark \ref{rem-a3}, using the notation of (\ref{eq-aur}). In $RO
(G')$-graded slice {\SS } for $k_{[2]}$ we have
\begin{displaymath}
\underline{E}_{2} (G'/G')
=\Z[a_{\sigma },u_{2\sigma },\orr_{1,0},\orr_{1,2}]/ (2a_{\sigma }),
\end{displaymath}
\noindent so
\begin{align*}
g (\underline{E}_{2} (G'/G'))
& = \left(\frac{1}{1-t}+ \frac{\wa}{1-\wa}\right)
\frac{1}{(1-\uu) (1-\rr)^{2}} \\
g (\underline{E}_{4} (G'/G'))
& = g (\underline{E}_{2} (G'/G'))
-\frac{\uu+\rr\wa^{3}}
{(1-\wa) (1-\uu^{2}) (1-\rr)^{2}} \\
& = \frac{1+t\uu}
{(1-t) (1-\uu^{2}) (1-\rr)^{2}}
+\frac{\wa+\wa^{2}}
{ (1-\uu^{2}) (1-\rr)^{2}}
+\frac{\wa^{3}}
{(1-\wa) (1-\uu^{2})(1-\rr)}
\end{align*}
\noindent as before. The next differential leads to
\begin{align*}
g (\underline{E}_{8} (G'/G'))
& = g (\underline{E}_{4} (G'/G')) -\frac{\uu^{2}+\rr^{3}\wa^{7}}
{(1-\wa) (1-\uu^{4}) (1-\rr)} \\
& = g (\underline{E}_{4} (G'/G')) -\frac{\uu^{2}}
{ (1-\uu^{4}) (1-\rr)}\\
& \qquad -\frac{\uu^{2}\wa}
{(1-\wa) (1-\uu^{4}) (1-\rr)} -\frac{\rr^{3}\wa^{7}}
{(1-\wa) (1-\uu^{4}) (1-\rr)} \\
& = \frac{1+t\uu}{(1-t) (1-\uu^{2}) (1-\rr)^{2}}
+\frac{\wa+\wa^{2}}{ (1-\uu^{2}) (1-\rr)^{2}}\\
&\qquad
+\frac{\wa^{3}}{(1-\wa) (1-\uu^{2})(1-\rr)}
-\frac{\uu^{2}}{ (1-\uu^{4}) (1-\rr)} \\
& \qquad
-\frac{\uu^{2} (\wa+\wa^{2})}{ (1-\uu^{4}) (1-\rr)}
-\frac{\uu^{2}\wa^{3}}{(1-\wa) (1-\uu^{4}) (1-\rr)}
-\frac{\rr^{3}\wa^{7}}{(1-\wa) (1-\uu^{4}) (1-\rr)} \\
& = \frac{(1+t\uu) (1+\uu^{2})- (1-t) (1-\rr)\uu^{2}}
{(1-t) (1-\uu^{4}) (1-\rr)^{2}}
+ \frac{(\wa+\wa^{2}) ( 1+\uu^{2} -\uu^{2} (1-\rr))}
{ (1-\uu^{4}) (1-\rr)^{2}} \\
&\qquad + \frac{\wa^{3} (1+\uu^{2})-\uu^{2}\wa^{3}-\rr^{3}\wa^{7}}
{(1-\wa) (1-\uu^{4})(1-\rr)}\\
& = \frac{1+t\uu+ (t+\rr-t\rr)\uu^{2}+t\uu^{3}}
{(1-t) (1-\uu^{4}) (1-\rr)^{2}}
+ \frac{(\wa+\wa^{2}) (1+\uu^{2}\rr)}
{(1-\uu^{4}) (1-\rr)^{2}} \\
&\qquad +\frac{\wa-\wa^{7}+\wa^{7}-\rr^{3}\wa^{7}}
{(1-\wa) (1-\uu^{4})(1-\rr)}\\
& = \frac{1+t\uu+ (t+\rr-t\rr)\uu^{2}+t\uu^{3}}
{(1-t) (1-\uu^{4}) (1-\rr)^{2}}
+ \frac{(\wa+\wa^{2}) (1+\uu^{2}\rr)}
{(1-\uu^{4}) (1-\rr)^{2}} \\
&\qquad +\frac{\wa^{3}+\wa^{4}+\wa^{6}+\wa^{6}}
{ (1-\uu^{4})(1-\rr)}
+\frac{\wa^{7} (1+\rr+\rr^{2})}
{(1-\wa) (1-\uu^{4})}
\end{align*}
\noindent The fourth term of this expression represents the elements
with filtration above six, and the first term represents the elements of
filtration 0. The latter include
\begin{align*}
[2u_{2\sigma }]
& \in \langle 2,\,a_{\sigma }^{2}
,\,a_{\sigma }(\orr_{1,0}+\orr_{1,1}) \rangle , \\
[2u_{2\sigma }^{2}]
& \in \langle 2,\,a_{\sigma } ,\,a_{\sigma }^{6}\orr_{1,0}^{3} \rangle , \\
[(\orr_{1,0}+\orr_{1,1}) u_{2\sigma }^{2}]
& \in \langle a_{\sigma }^{4} ,\,a_{\sigma }^{}\orr_{1,0}^{3}
,\, \orr_{1,0}+\orr_{1,1}\rangle \\
& \qquad \mbox{with }
(\orr_{1,0}+\orr_{1,1}) [2u_{2\sigma }^{2}]
=2 [(\orr_{1,0}+\orr_{1,1}) u_{2\sigma }^{2}], \\
[2u_{2\sigma }^{3}]
& \in \langle 2,\, a_{\sigma }^{2}(\orr_{1,0}+\orr_{1,1})
,\, a_{\sigma}^{2},\,a_{\sigma}^{6}\orr_{1,0}^{3}\rangle \\
\aand
[u_{2\sigma }^{4}]
& \in \langle a_{\sigma }^{4} ,\,a_{\sigma }^{3}\orr_{1,0}^{3}
\,a_{\sigma }^{4} ,\,a_{\sigma }^{3}\orr_{1,0}^{3} \rangle
\end{align*}
\noindent with notation as in \ref{rem-abuse}.
As indicated in \ref{rem-a7}, we can get rid of them by formally
adjoining $w:=(\orr_{3}^{G'})^{-1}u_{2\sigma }^{2}$ to $\underline{E}_{2}
(G'/G')$. As before we denote the enlarged {\SS } terms by
$\underline{E}'_{r} (G'/G')$ This time let
\begin{displaymath}
\www= \rr^{-3}\uu^{2}=xy^{-7}.
\end{displaymath}
\noindent Then we have
\begin{displaymath}
\underline{E}'_{r} (G'/G')
=\left(\frac{1-\uu^{2}}{1-\www} \right)\underline{E}_{r} (G'/G')
\qquad \mbox{for $r=2$ and $r=4$}
\end{displaymath}
\noindent and
\begin{align*}
g (\underline{E}'_{8} (G'/G'))
& = g (\underline{E}'_{4} (G'/G'))
-\frac{\www+\wa^{7}}{(1-\wa) (1-\www^{2}) (1-\rr)} \\
& = g (\underline{E}'_{4} (G'/G'))
-\frac{\www}{ (1-\www^{2}) (1-\rr)}
-\frac{(\wa+\wa^{2})\www}{ (1-\www^{2}) (1-\rr)} \\
&\qquad -\frac{\wa^{3}\www+\wa^{7}}{(1-\wa) (1-\www^{2}) (1-\rr)} \\
& = \frac{1+t\uu}{(1-t) (1-\www) (1-\rr)^{2}}
+\frac{\wa+\wa^{2}}{ (1-\www) (1-\rr)^{2}}\\
&\qquad
+\frac{\wa^{3}}{(1-\wa) (1-\www)(1-\rr)}
-\frac{\www}{ (1-\www^{2}) (1-\rr)} \\
&\qquad -\frac{(\wa+\wa^{2})\www}{ (1-\www^{2}) (1-\rr)}
-\frac{\wa^{3}\www+\wa^{7}}{(1-\wa) (1-\www^{2}) (1-\rr)} \\
& = \frac{(1+t\uu) (1+\www) - (1-t) (1-\rr)\www}
{(1-t) (1-\www^{2}) (1-\rr)^{2}}
+\frac{(\wa+\wa^{2}) (1- (1-\rr)\www)}{ (1-\www^{2}) (1-\rr)^{2}}\\
&\qquad +\frac{\wa^{3} (1+\www) -\wa^{3}\www-\wa^{7}}
{(1-\wa) (1-\www^{2})(1-\rr)} \\
& = \frac{1+t\uu+ (t+\rr-t\rr)\www+t\www\uu}
{(1-t) (1-\www^{2}) (1-\rr)^{2}}
+\frac{(\wa+\wa^{2}) (1+\rr\www)}{ (1-\www^{2}) (1-\rr)^{2}}\\
&\qquad +\frac{\wa^{3}+\wa^{4}+\wa^{5}+\wa^{6}}{ (1-\www^{2})(1-\rr)} .
\end{align*}
\noindent Again the first term represents the elements of
filtration 0. These include
\begin{align*}
[2u_{2\sigma }]
& \in \langle 2,\,a_{\sigma }^{2}
,\,a_{\sigma }(\orr_{1,0}+\orr_{1,1}) \rangle , \\
[2w]
& \in \langle 2,\,a_{\sigma } ,\,a_{\sigma }^{6}\rangle , \\
[(\orr_{1,0}+\orr_{1,1})w]
& \in \langle a_{\sigma }^{4} ,\,a_{\sigma }^{3}
,\, \orr_{1,0}+\orr_{1,1}\rangle \\
& \qquad \mbox{with }
(\orr_{1,0}+\orr_{1,1}) [2w]
=2 [(\orr_{1,0}+\orr_{1,1}) w], \\
[2u_{2\sigma }w]
& \in \langle 2,\, a_{\sigma }^{2}(\orr_{1,0}+\orr_{1,1})
,\, a_{\sigma}^{2},\,a_{\sigma}^{6}\rangle \\
\aand
[w^{2}]
& \in \langle a_{\sigma }^{4} ,\,a_{\sigma }^{3}
\,a_{\sigma }^{4} ,\,a_{\sigma }^{3} \rangle
\end{align*}
\noindent where, as indicated above,
$w=(\orr_{3}^{G'})^{-1}u_{2\sigma}^{2}$.
From these
\section{The effect of the first differentials over $C_{4}$}
\label{sec-C4diffs}
Theorem \ref{thm-sliceE2} lists elements in the slice {\SS} for
$\kH$ over $C_{4}$ in terms of
\begin{displaymath}
r_{1}, \, \os_{2} , \,\normrbar_{1};\quad
\eta , \,a_{\sigma}, \,a_{\lambda};\quad
u_{\lambda},\,u_{\sigma},\,\mbox{ and }u_{2\sigma}.
\end{displaymath}
\noindent All but the $u$'s are permanent cycles, and the action of
$d_{3}$ on $u_{\lambda}$, $u_{\sigma}$ and $u_{2\sigma}$ is
described above in Theorem \ref{thm-d3ulambda}.
\begin{prop}\label{prop-d3}
{\bf $d_{3}$ on elements in Theorem \ref{thm-sliceE2}.}
We have the following $d_{3}$s, subject to the conditions on $i$, $j$,
$k$ and $\ell $ of Theorem \ref{thm-sliceE2}:
\begin{itemize}
\item [$\bullet$] On $X_{2\ell ,2\ell }$:
\begin{align*}
d_{3} (a_{\lambda}^{j}u_{2\sigma}^{\ell }
u_{\lambda}^{2\ell -j}\normrbar_{1}^{2\ell })
& = \mycases{
a_{\lambda}^{j+1}\eta u_{2\sigma}^{\ell}u_{\lambda}^{2\ell -j-1}
\normrbar_{1}^{2\ell}\hspace{-1.7cm}\\
\qquad
\in \upi_{*}X_{2\ell,2\ell +1} (G/G)
&\mbox{for $j$ odd}\\
0\hspace{2cm}&\mbox{for $j$ even}
}\\
d_{3} (a_{\sigma}^{2k}a_{\lambda}^{2\ell }u_{2\sigma}^{\ell-k }
\normrbar_{1}^{2\ell })
& = 0
\end{align*}
\item [$\bullet$] On $X_{2\ell+1 ,2\ell+1 }$:
\begin{align*}
d_{3} (\delta_{1}^{2\ell +1})
& = \eta u_{\sigma}^{2\ell+1}
\Res_{2}^{4}(a_{\lambda}u_{\lambda}^{2\ell }\normrbar_{1}^{2\ell+1})\\
& \in \upi_{*}X_{2\ell+1,2\ell +2} (G/G')\\
\lefteqn{ d_{3} (u_{\sigma}^{2\ell +1}
\Res_{2}^{4}(a_{\lambda}^{j}u_{\lambda}^{2\ell +1-j}
\normrbar_{1}^{2\ell +1}))}\qquad\qquad\\
& = \mycases{
\eta u_{\sigma}^{2\ell +1}
\Res_{2}^{4}(a_{\lambda}^{j+1}u_{\lambda}^{2\ell -j}
\normrbar_{1}^{2\ell +1})\\
\qquad
\in \upi_{*}X_{2\ell+1,2\ell +2} (G/G')
&\mbox{for $j$ even}\\
0\hspace{2cm}&\mbox{for $j$ odd}
}\\
\lefteqn{ d_{3} (a_{\sigma}a_{\lambda}^{j}
u_{\sigma}^{2\ell}u_{\lambda}^{2\ell +1-j}
\normrbar_{1}^{2\ell +1})}\qquad\qquad\\
& = \mycases{
\eta a_{\sigma}a_{\lambda}^{j+1}
u_{\sigma}^{2\ell}u_{\lambda}^{2\ell -j}
\normrbar_{1}^{2\ell +1}\\
\qquad
\in \upi_{*}X_{2\ell+1,2\ell +2} (G/G)
&\mbox{for $j$ even}\\
0\hspace{2cm}&\mbox{for $j$ odd}
}\\
\lefteqn{d_{3} (a_{\sigma}^{2k+1}a_{\lambda}^{2\ell +1}
u_{2\sigma}^{\ell-k}
\normrbar_{1}^{2\ell +1})}\qquad\qquad\\
& = 0
\end{align*}
\item [$\bullet$] On $X_{i ,2\ell-i }$:
\begin{align*}
d_{3} (u_{\sigma }^{\ell } \os_{2}^{\ell -i}
\Res_{2}^{4}(u_{\lambda}^{\ell }\normrbar_{1}^{i}) )
& = \mycases{
\eta^{3} u_{\sigma }^{\ell-1 } \os_{2}^{\ell -i-1}
\Res_{2}^{4}(u_{\lambda}^{\ell-1 }\normrbar_{1}^{i})\\% \hspace{-2cm}\\
\qquad
\in \upi_{*}X_{i,2\ell +1-i} (G/G')\\%\hspace{-3cm}\\
&\hspace{-2cm}\mbox{for $\ell $ odd}\\
0 &\hspace{-2cm}\mbox{for $\ell $ even}
}\\
d_{3} (\eta^{2j}u_{\sigma}^{\ell -j}\os_{2}^{\ell -i-j}
\Res_{2}^{4}(u_{\lambda}^{\ell -j}\normrbar_{1}^{i}) )
& = \mycases{
\eta^{2j+1}u_{\sigma}^{\ell -j}\os_{2}^{\ell -i-j}
\Res_{2}^{4}(a_{\lambda}u_{\lambda}^{\ell -j-1}\normrbar_{1}^{i})
\\
\qquad
\in \upi_{*}X_{i,2\ell +1-i} (G/G')\\
&\hspace{-3.0cm} \mbox{for $\ell -j$ odd}\\
0
&\hspace{-3.0cm} \mbox{for $\ell -j$ even}
}
\end{align*}
\item [$\bullet$] On $X_{i ,2\ell+1-i }$:
\begin{align*}
d_{3} (r_{1}\Res_{1}^{2}(u_{\sigma }^{\ell } \os_{2}^{\ell -i})
\Res_{1}^{4}(u_{\lambda}^{\ell }\normrbar_{1}^{i}) )
& = 0\\
d_{3} ( \eta^{2j+1}u_{\sigma}^{\ell -j}\os_{2}^{\ell -i-j}
\Res_{2}^{4}(u_{\lambda}^{\ell -j}\normrbar_{1}^{i}) )
& = \mycases{
\eta^{2j+2}u_{\sigma}^{\ell -j}\os_{2}^{\ell -i-j}
\Res_{2}^{4}(a_{\lambda}u_{\lambda}^{\ell -j-1}\normrbar_{1}^{i})
\\
\qquad
\in \upi_{*}X_{i,2\ell +2-i} (G/G')\hspace{-3cm}\\
&\hspace{-3cm} \mbox{for $\ell -j$ odd}\\
0 &\hspace{-3cm} \mbox{for $\ell -j$ even}
}
\end{align*}
\end{itemize}
\end{prop}
Note that in each case the first index of $X$ is unchanged by the
differential, and the second one is increased by one. Since $X_{m,n}$
is a summand of the $2 (m+n)$th slice, each $d_{3}$ raises the slice
degree by 2 as expected.
\begin{rem}\label{rem-Ym}
{\bf The spectra $y_{m}$ and $Y_{m}$ of Corollaries \ref{cor-filtration} and
\ref{cor-Filtration}}. Similar statements can be proved for the case
$\ell <0$. We leave the details to the reader, but illustrate the
results in Figures \ref{fig-sseq-8a} and \ref{fig-sseq-8b}.
The source of each differential in \ref{prop-d3} is the product of
some element in $\upi_{\star}\HZZ$ with a power of $\normrbar_{1}^{}$
or $\delta_{1}$. The target is the product of a different element in
$\upi_{\star}\HZZ$ with the same power. This means they are
differentials in the slice {\SS } for the spectra $y_{m}$ of
\ref{cor-filtration}.
Similar differentials occur when we replace $\normrbar_{1}^{i}$ by any
homogeneous polynomial of degree $i$ in $\normrbar_{1}^{}$ and
$\ot_{2}$ in which the coefficient of $\normrbar_{1}^{i}$ is odd.
This means they are also differentials in the slice {\SS } for the
spectra $Y_{m}$ of \ref{cor-Filtration}.
\end{rem}
These differentials are illustrated in the upper charts in Figures
\ref{fig-sseq-7a}--\ref{fig-sseq-8b}. In order to pass to $\EE_{4}$
we need the following exact sequences of Mackey functors.
\begin{displaymath}
\xymatrix
@R=5mm
@C=10mm
{
0 \ar[r]^(.5){}
&\bullet\ar[r]^(.5){}
&\circ\ar[r]^(.5){d_{3}}
&{\widehat{\bullet}}\ar[r]^(.5){}
&{\obull }\ar[r]^(.5){}
&0\\
0\ar[r]^(.5){}
&{\widehat{\twobox}}\ar[r]^(.5){}
&{\widehat{\Box}}\ar[r]^(.5){d_{3}}
&{\widehat{\bullet} }\ar[r]^(.5){}
&0\\
&0\ar[r]^(.5){}
&\obull\ar[r]^(.5){d_{3}}
&{\widehat{\bullet}}\ar[r]^(.5){}
&{\JJ }\ar[r]^(.5){}
&0\\
0 \ar[r]^(.5){}
&\odbox\ar[r]^(.5){}
&\oBox\ar[r]^(.5){d_{3}}
&{\widehat{\bullet}}\ar[r]^(.5){}
&\JJ\ar[r]^(.5){}
&0
}
\end{displaymath}
The resulting subquotients of $\EE_{4}$ are shown in the lower charts
of Figures \ref{fig-sseq-7a}--\ref{fig-sseq-8b} and described below in
Theorem \ref{thm-sliceE4}. In the latter the slice summands are
organized as shown in the Figures rather than by orbit type as in
Theorem \ref{thm-sliceE2}.
\begin{figure}
\begin{center}
\includegraphics[width=11.11cm]{fig-sseq-5a.pdf}\\
\includegraphics[width=11cm]{fig-sseq-7a.pdf} \caption[The slice
{$\EE_{2}$}- and $\EE_{4}$-terms for $X_{4,n}$ for $n\geq 4$.]{The
subquotient of the slice $\EE_{2}$- and $\EE_{4}$-terms for $\kH$ for
the slice summands $X_{4,n}$ for $n\geq 4$. Exotic transfers are shown
in blue and differentials are in red. The symbols are defined in Table
\ref{tab-C4Mackey}. This is also the slice {\SS } for $y_{4}$ as in
Corollary \ref{cor-filtration} and $Y_{4}$ (after tensoring with $R$) as in
Corollary \ref{cor-Filtration}.} \label{fig-sseq-7a}
\end{center}
\end{figure}
\begin{figure}
\begin{center}
\includegraphics[width=11.11cm]{fig-sseq-5b.pdf}\\
\includegraphics[width=11cm]{fig-sseq-7b.pdf} \caption[The slice
$\EE_{2}$ and $\EE_{4}$-terms for for $X_{5,n}$ for $n\geq
5$.]{The subquotient of the slice $\EE_{2}$ and $\EE_{4}$-terms for
$\kH$ for the slice summands $X_{5,n}$ for $n\geq 5$. Exotic
restrictions and transfers are shown in dashed green and solid blue
lines respectively. This is also the slice {\SS } for $y_{5}$ as in
Corollary \ref{cor-filtration} and for $Y_{5}$ (after tensoring with
$R$) as in Corollary \ref{cor-Filtration}.} \label{fig-sseq-7b}
\end{center}
\end{figure}
\begin{figure}
\begin{center}
\includegraphics[width=11cm]{fig-sseq-6a.pdf}
\includegraphics[width=11cm]{fig-sseq-8a.pdf} \caption[{The slice
$\EE_{2}$ and $\EE_{4}$-terms for $X_{-4,n}$ for $n\geq -4$.}] {The
subquotient of the slice $\EE_{2}$ and $\EE_{4}$-terms for $\kH$ for
the slice summands $X_{-4,n}$ for $n\geq -4$. This is also the slice
{\SS } for $Y_{-4}$ (after tensoring with $R$) as in Corollary
\ref{cor-Filtration}.} \label{fig-sseq-8a}
\end{center}
\end{figure}
\begin{figure}
\begin{center}
\includegraphics[width=11cm]{fig-sseq-6b.pdf}
\includegraphics[width=11cm]{fig-sseq-8b.pdf} \caption[{The slice
$\EE_{2}$ and $\EE_{4}$-terms for $X_{-5,n}$ for $n\geq -5$.}] {The
subquotient of the slice $\EE_{2}$ and $\EE_{4}$-terms for $\kH$ for
the slice summands $X_{-5,n}$ for $n\geq -5$. This is also the slice
{\SS } for $Y_{-5}$ (after tensoring with $R$) as in Corollary
\ref{cor-Filtration}. } \label{fig-sseq-8b}
\end{center}
\end{figure}
\begin{thm}\label{thm-sliceE4}
{\bf The slice $\EE_{4}$-term for $\kH$.} The elements of
Theorem \ref{thm-sliceE2} surviving to $\EE_{4}$, which live
in the appropriate subquotients of $\upi_{*}X_{m,n}$, are
as follows.
\begin{enumerate}
\item
In $\upi_{*}X_{2\ell ,2\ell }$ (see the
leftmost diagonal in Figure \ref{fig-sseq-7a}), on the 0-line we still
have a copy of $\Box$ generated under fixed point restrictions by
$\Delta_{1}^{\ell }\in \EE_{4}^{0,8\ell }$. In positive
filtrations we have
\begin{align*}
\circ &\subseteq \EE_{4}^{2j,8\ell }
\qquad \mbox{generated by } \\
&\mycases{
a_{\lambda}^{j}u_{2\sigma}^{\ell }
u_{\lambda}^{2\ell -j}\normrbar_{1}^{2\ell }
\in \EE_{4}^{2j,8\ell } (G/G)\\
&\hspace{-5cm} \mbox{for $j$ even and }0<j\leq 2\ell, \\
2a_{\lambda}^{j}u_{2\sigma}^{\ell }
u_{\lambda}^{2\ell -j}\normrbar_{1}^{2\ell }
=a_{\sigma}^{2}a_{\lambda}^{j-1}u_{2\sigma}^{\ell +1}
u_{\lambda}^{2\ell -j-1}\normrbar_{1}^{2\ell }
\in \EE_{4}^{2j,8\ell } (G/G)\\
&\hspace{-5cm} \mbox{for $j$ odd and }0<j\leq 2\ell \mbox{ and}}\\
\bullet &\subseteq \EE_{4}^{2k+2\ell ,8\ell }
\qquad \mbox{generated by } \\
&a_{\sigma}^{2k}a_{\lambda}^{2\ell }u_{2\sigma}^{\ell -k}
\normrbar_{1}^{2\ell }
\in \EE_{4}^{2j+2k,8\ell } (G/G)
\qquad \mbox{for }0<k\leq \ell.
\end{align*}
\item
In $\upi_{*}X_{2\ell ,2\ell +1}$ (see the
second leftmost diagonal in Figure \ref{fig-sseq-7a}), in filtration 0
we have $\widehat{\oBox }$, generated (under transfers and
the group action) by
\begin{displaymath}
r_{1}\Res_{1}^{2}(u_{\sigma }^{2\ell }
\Res_{1}^{4}(u_{\lambda}^{2\ell }\normrbar_{1}^{2\ell })
\in \EE_{2}^{0,8\ell+2} (G/\ee) .
\end{displaymath}
\noindent In positive filtrations we have
\begin{displaymath}
\begin{array}[]{rll}
\widehat{\bullet}
& \subseteq
\EE_{4}^{1,8\ell +2}
&\mbox{generated (under transfers and the group action) by} \\
& & \eta u_{\sigma}^{2\ell }\Res_{2}^{4}(u_{\lambda}
\normrbar_{1})^{2\ell } = \EE_{4}^{1,8\ell +2} (G/G')\\
\obull
& \subseteq
\EE_{4}^{4k+1,8\ell +2}
&\mbox{for $0<k\leq \ell $ generated by} \\
& & x=\eta^{4k+1} u_{\sigma}^{2\ell-2k}
\Res_{2}^{4}(u_{\lambda}\normrbar_{1})^{2\ell -2k}
\in \EE_{4}^{4k+1,8\ell +2} (G/G')\\
& & \mbox{with $(1-\gamma )x=\Tr_{2}^{4}(x)=0$.}
\end{array}
\end{displaymath}
\item
In $\upi_{*}X_{2\ell+1 ,2\ell +1}$ (see the leftmost diagonal
in Figure \ref{fig-sseq-7b}), on the 0-line we have a copy of $\odbox$
generated under fixed point $\Delta_{1}^{(2\ell +1)/2 }\in
\EE_{4}^{0,8\ell+4}$. In positive filtrations we have
\begin{displaymath}
\begin{array}[]{rll}
\obull
&\subseteq
\EE_{2}^{2j,8\ell+4 }
& \mbox{generated by }\\
& & u_{\sigma}^{2\ell +1}
\Res_{2}^{4}(a_{\lambda}^{j}u_{\lambda}^{2\ell +1-j}
\normrbar_{1}^{2\ell +1})\in \EE_{2}^{2j,8\ell+4 } (G/G')\\
& &\mbox{for }0<j\leq 2\ell+1 , \\
\bullet
&\subseteq
\EE_{2}^{2j+1 ,8\ell+4 }
& \mbox{generated by }\\
& & a_{\sigma}a_{\lambda}^{j}
u_{\sigma}^{2\ell}u_{\lambda}^{2\ell +1-j}
\normrbar_{1}^{2\ell +1}\in \EE_{2}^{2j+2k,8\ell+4 } (G/G)\\
& & \mbox{for }0\leq j\leq 2\ell+1\mbox{ and} \\
\bullet
&\subseteq
\EE_{2}^{2k+4\ell +3 ,8\ell+4 }
&\mbox{generated by }\\
& &a_{\sigma}^{2k+1}a_{\lambda}^{2\ell +1}
u_{2\sigma}^{\ell-k}
\normrbar_{1}^{2\ell +1}
\in \EE_{2}^{2k+4\ell +2,8\ell+4 } (G/G)\\
& &\mbox{for }0< k\leq 2\ell+1.
\end{array}
\end{displaymath}
\item
In $\upi_{*}X_{2\ell+1 ,2\ell +2}$ (see the
second leftmost diagonal in Figure \ref{fig-sseq-7b}), in filtration 0
we have $\widehat{\oBox }$, generated (under transfers and
the group action) by
\begin{displaymath}
r_{1}\Res_{1}^{2}(u_{\sigma }^{2\ell+1 }
\Res_{1}^{4}(u_{\lambda}^{2\ell+1 }\normrbar_{1}^{2\ell+1 })
\in \EE_{4}^{0,8\ell+6} (G/\ee) .
\end{displaymath}
\noindent In positive filtrations we have
\begin{displaymath}
\begin{array}[]{rll}
\JJ & \subseteq
\EE_{4}^{4k+3,8\ell+6}
&\mbox{for $0\leq k\leq \ell $ generated under transfer by } \\
& &x=\eta^{4k+3}\Delta_{1}^{\ell -k}
\in \EE_{4}^{4k+3,8\ell+6} (G/G')\\
& &\mbox{with $(1-\gamma )x=0$.}
\end{array}
\end{displaymath}
\noindent The generator of $\EE_{4}^{4k+3,8\ell+6} (G/G')$ is the
exotic restriction of the one in $\EE_{4}^{4k+1,8\ell+4} (G/G)$.
\item
In $\upi_{*}X_{m,m+i}$ for $i\geq 2$ (see the
rest of Figures \ref{fig-sseq-7a} and \ref{fig-sseq-7b}), in
filtration 0 we have
\begin{displaymath}
\begin{array}[]{rll}
\widehat{\oBox }
& \subseteq
\EE_{4}^{0,4m+4j+2}
&\mbox{generated under transfers and group action by }\\
& &r_{1}\Res_{1}^{2}(u_{\sigma }^{m +j} \os_{2}^{j})
\Res_{1}^{4}(u_{\lambda}^{m +j}\normrbar_{1}^{m })\\
& &\quad\in \EE_{4}^{0,4m+4j+2} (G/\ee)\mbox{ for $j\geq 0$,}\\
\widehat{\twobox}
& \subseteq
\EE_{4}^{0,8\ell +4}
&\mbox{generated under transfers and group action by }\\
& &r_{1}\Res_{1}^{2}(u_{\sigma }^{m +j} \os_{2}^{j})
\Res_{1}^{4}(u_{\lambda}^{m +j}\normrbar_{1}^{m })\\
& &\quad\in \EE_{4}^{0,8\ell +4} (G/\ee)
\mbox{ for $\ell \geq m/2$ and} \\
\widehat{\Box}
& \subseteq
\EE_{4}^{0,8\ell }
&\mbox{generated under transfers, restriction and group action by}\\
& &x_{8\ell ,m}
=\Sigma_{2,0}^{2\ell -m}\delta_{1}^{m}+\ell \delta_{1}^{2\ell }
\mbox{ where } \\
& &\Sigma_{2,\epsilon }=u_{\rho_{2}}\os_{2,\epsilon } \\
& &\mbox{and } \delta_{1}=u_{\rho_{2}}\Res_{2}^{4}(\normrbar_{1})\\
& &\quad\in \EE_{4}^{0,8\ell } (G/G')\mbox{ for $0\leq m\leq 2\ell -1$}.
\end{array}
\end{displaymath}
\noindent In positive filtrations we have
\begin{displaymath}
\begin{array}[]{rll}
\widehat{\bullet}
& \subseteq
\EE_{4}^{2,8\ell +4}
&\mbox{generated under transfers and group action by }\\
& &\eta_{0}^{2}\Res_{2}^{4}(\Delta_{1}^{\ell})
= \eta_{0}^{2}\delta_{1}^{2\ell } =
\eta_{0}^{2}u_{\sigma}^{2\ell}\Res_{2}^{4}(u_{\lambda}\normrbar_{1})^{2\ell}\\
& &\quad\in \EE_{4}^{2,8\ell +4} (G/G')
\quad \mbox{and}\\
\widehat{\bullet}
& \subseteq
\EE_{4}^{s,8\ell +2s}
&\mbox{generated under transfers and group action by }\\
& &\eta_{\epsilon }^{s}x_{8\ell ,m}\in\EE_{4}^{s,8\ell +2s} (G/G')\\
& & \mbox{for $s=1,2$ and $0\leq m\leq 2\ell-1 $}.
\end{array}
\end{displaymath}
\noindent Each generator of $\EE_{4}^{2,8\ell +4} (G/G')$ is an exotic
transfer of one in\linebreak $\EE_{4}^{0,8\ell +2} (G/e)$.
\end{enumerate}
\end{thm}
\begin{prop}\label{prop-perm}
{\bf Some nontrivial permanent cycles.} The elements listed in Theorem
\ref{thm-sliceE4}(v) other than $\eta_{\epsilon }^{2}\delta_{1}^{2\ell
}$ are all nontrivial permanent cycles.
\end{prop}
\proof Each such element is either in the image of
$\EE_{4}^{0,*} (G/\ee)$ under the transfer and
therefore a nontrivial permanent cycle, or it is one of the ones listed
in Corollary \ref{cor-G'SS}. \qed\bigskip
{\em In subsequent discussions and charts, starting with Figure
\ref{fig-E4}, we will omit the elements in Proposition
\ref{prop-perm}. These elements all occur in $\EE_{4}^{s,t}$ for
$0\leq s\leq 2$.}
Analogous statements can be made about the slice {\SS} for $\KH$. Each
of its slices is a certain infinite wedge spelled out in Corollary
\ref{cor-KH}. Their homotopy groups are determined by the chain
complex calculations of Section \ref{sec-chain} and illustrated in
Figures \ref{fig-sseq-1} (with Mackey functor induction
$\uparrow_{2}^{4}$ applied) and \ref{fig-sseq-3}. Analogs of Figures
\ref{fig-sseq-7a}--\ref{fig-sseq-7b} are shown in Figures
\ref{fig-sseq-8a}--\ref{fig-sseq-8b}. In each figure, exotic
transfers and restrictions are indicated by blue and dashed green lines
respectively. As in the $\kH$ case, most of the elements shown in
this chart can be ignored for the purpose of calculating higher
differentials. {\em In the third quadrant the elements we are
ignoring all occur in $\EE_{4}^{s,t}$ for $-2\leq s\leq 0$.}
The resulting reduced $\EE_{4}$ for $\KH$ is shown in Figure
\ref{fig-KH}. The information shown there is very useful for
computing differentials and extensions. The periodicity theorem tells
us that $\upi_{n}\KH$ and $\upi_{n-32}\KH$ are isomorphic. For $0\leq
n<32$ these groups appear in the first and third quadrants
respectively, and the information visible in the {\SS} can be quite
different.
For example, we see that $\upi_{0}\KH$ has summand of the form $\Box$,
while $\upi_{-32}\KH$ has a subgroup isomorphic to $\fourbox $. The
quotient $\Box/\fourbox $ is isomorphic to $\circ $. This leads to
the exotic restrictions and transfer in dimension $-32$ shown in
Figure \ref{fig-KH}. Information that is transparent in dimension 0
implies subtle information in dimension $-32$. Conversely, we see
easily that $\upi_{-4}\KH=\dot{\twobox}$ while $\upi_{28}\KH$ has a
quotient isomorphic to $\odbox$. This leads to the ``long transfer''
(which raises filtration by 12) in dimension 28.
\section{Higher differentials and exotic Mackey functor extensions}
\label{sec-higher}
\begin{figure}
\begin{center}
\includegraphics[width=11cm]{fig-G_SS-3.pdf} \caption[The slice
$\EE_{7}$-term for the $C_{2}$-spectrum $i_{G'}^{*}\kH$.]{The slice
$\EE_{7}$-term for the $C_{2}$-spectrum $i_{G'}^{*}\kH$. The Mackey
functor symbols are defined in Table \ref{tab-C2Mackey}. A number
$n$ in front of a symbol indicates an $n$-fold direct sum. Blue lines
indicate exotic transfers and red lines indicate differentials.}
\label{fig-G'SS-3}
\end{center}
\end{figure}
We can use the results of the \S\ref{sec-C2diffs} to study higher
differentials and extensions. The $\EE_{7}$-term implied by them is
illustrated in Figure \ref{fig-G'SS-3}. For each $\ell ,s\geq 0$ there is
a generator
\begin{displaymath}
y_{8\ell +s,s}:=\eta_{0}^{s} \delta_{1}^{2\ell }
\in \EE_{7}^{s,8\ell +2s} (G/G')
\end{displaymath}
\noindent with
\begin{displaymath}
d_{7} (y_{16k+s+8,s}) = y_{16k+s+7,s+7}.
\end{displaymath}
\noindent
Recall that
\begin{displaymath}
\delta_{1}=\ou_{\lambda }\orr_{1,0}\orr_{1,1}
\in \underline{E}_{2}^{0,4}\kH (G/G')
\cong \underline{E}_{2}^{0,(G',4)}\kH (G/G),
\end{displaymath}
\noindent and in the latter group we denote $\ou_{\lambda }$ by
$u_{2\sigma }$. We have
\begin{displaymath}
d_{3} (\delta_{1})= d_{3} (\ou_{\lambda })\orr_{1,0}\orr_{1,1}
\cong d_{3} (u_{2\sigma })\orr_{1,0}\orr_{1,1}
=a_{\sigma }^{3} (\orr_{1,0}+\orr_{1,1})\orr_{1,0}\orr_{1,1}.
\end{displaymath}
\noindent
If the source has the form $\Res_{2}^{4}(x_{16k+s+8,s})$, then such
an $x$ must support a nontrivial $d_{r}$ for $r\leq 7$. If it has a
nontrivial transfer $x'_{16k+s+8,s}$, then such an $x'$ cannot support an
earlier differential, and we must have
\begin{displaymath}
d_{r} (x'_{16k+s+8,s})
=\Tr_{2}^{4}(d_{7} (y_{16k+s+8,s}))
=\Tr_{2}^{4}(y_{16k+s+7,s+7}) \qquad \mbox{for some $r\geq 7$}.
\end{displaymath}
\noindent We could get a higher differential (meaning $r>7$) if
$y_{16k+s+7,s+7}$ supports an exotic transfer.
We have seen (Figure \ref{fig-E4} and Theorem \ref{thm-sliceE4}) that
for $s\geq 3$ and $k\geq 0$,
\begin{numequation}\label{eq-eighth-diagonls}
\begin{split}
\EE_{5}^{s,16k+8+2s}
= \mycases{
\circ
&\mbox{for $s\equiv 0$ mod 4}\\
\overline{\bullet}
&\mbox{for $s\equiv 1,2$ mod 4}\\
\JJ
&\mbox{for $s\equiv 3$ mod 4.}
}
\end{split}
\end{numequation}
\noindent For $s=1,2$, $\EE_{5}^{s,16k+8+2s}$ has
$\overline{\bullet}$ as a direct summand. For $s=0$ it has $\Box$ as a
summand, and the differentials on it factor through its quotient
$\circ$; see (\ref{eq-C4-SES}).
The corresponding statement in the third quadrant is
\begin{displaymath}
\EE_{5}^{-s,-16k-2s-24}
= \mycases{
\circ
&\mbox{for $s\equiv 3$ mod 4}\\
\overline{\bullet}
&\mbox{for $s\equiv 1,2$ mod 4}\\
\JJ
&\mbox{for $s\equiv 0$ mod 4.}
}
\end{displaymath}
\noindent for $s\geq 3$ and $k\geq 0$. For $s=1,2$ the groups have
similar summands, and for $s=0$ there is a summand of the form
$\JJbox$, which has $\JJ$ as a subgroup; again see (\ref{eq-C4-SES}).
This is illustrated in Figure \ref{fig-KH}.
\begin{thm}\label{thm-d7}
{\bf Differentials for $C_{4}$ related to the $d_{7}$s for $C_{2}$.}
The differential
\begin{displaymath}
d_{7} (y_{16k+s+8,s})=y_{16k+s+7,s+7}\qquad \mbox{with }s\geq 3
\end{displaymath}
\noindent has the following implications for the congruence
classes of $s$ modulo 4.
\begin{enumerate}
\item [(i)] For $s\equiv 0$, $\EE_{7}^{s,16k+8+2s}=\circ$ and
$\EE_{7}^{s+7,16k+14+2s}=\JJ$. Hence $y_{16k+s+8,s}$ is a
restriction with a nontrivial transfer, and
\begin{align*}
d_{5} (x_{16k+s+8,s})
& = x_{16k+s+7,s+5} \\
\aand
d_{7} (2x_{16k+s+8,s})&=d_{7} (\Tr_{2}^{4}(y_{16k+s+8,s}))\\
& = \Tr_{2}^{4}(y_{16k+s+7,s+7})= x_{16k+s+7,s+7}.
\end{align*}
\item [(ii)] For $s\equiv 1$,
\begin{align*}
d_{7} (y_{16k+s+8,s})
& = y_{16k+s+7,s+7}\\
\aand
d_{5} (x_{16k+s+8,s+2})
& = \Tr_{2}^{4}(y_{16k+s+7,s+7}) = 2x_{16k+s+7,s+7}
\end{align*}
\noindent This leaves the fate of $x_{16k+s+7,s+7}$ undecided; see below.
\item [(iii)] For $s\equiv 2$,
$\EE_{7}^{s,16k+8+2s}=\overline{\bullet} $ and
$\EE_{7}^{s+7,16k+14+2s}=\overline{\bullet} $. Neither the
source nor target is a restriction or has a nontrivial transfer, so no
additional differentials are implied.
\item [(iv)] For $s\equiv 3$, $\EE_{7}^{s,16k+8+2s}=\JJ $ and
$\EE_{7}^{s+7,16k+14+2s}=\overline{\bullet} $. In this case
the source is an exotic restriction; again see Figure
\ref{fig-sseq-7b}). Thus we have
\begin{align*}
d_{7} (y_{16k+s+8,s})
& = y_{16k+s+7,s+7}\\
\aand
d_{5} (x_{16k+s+8,s-2})
& = x_{16k+s++7,s+3}\\
\mbox{with }
\Res_{2}^{4}(x_{16k+s++7,s+3})
& =y_{16k+s+7,s+7}.
\end{align*}
\noindent Moreover, $\Tr_{2}^{4}(y_{16k+8+s,s})$ is nontrivial and it
supports a nontrivial $d_{11}$ when $4k+s\equiv 3$ mod 8. The other
case, $4k+s\equiv 7$, will be discussed below.
\end{enumerate}
\end{thm}
\proof (i) The target Mackey functor is $\JJ$ and $y_{16k+s+7,s+7}$ is
the exotic restriction of $x_{16k+s+7,s+5}$; see Figure
\ref{fig-sseq-7b} and Theorem \ref{thm-sliceE4}. The indicated
$d_{5}$ and $d_{7}$ follow.
(ii) The differential is nontrivial on the $G/G'$ component of
\begin{displaymath}
\xymatrix
@R=5mm
@C=10mm
{
{\obull=\EE_{7}^{s,16k+8+2s}}\ar[r]^(.5){d_{7}}
&{\EE_{7}^{s+7,16k+14+2s}=\circ}
}
\end{displaymath}
\noindent Thus the target has a nontrivial transfer, so the source
must have an exotic transfer. The only option is $x_{16k+s+8,s+2}$,
and the result follows.
(iv) We prove the statement about $d_{11}$ by showing that
\begin{displaymath}
{y_{16k+s+7,s+7}=\eta_{0}^{s+7}\delta_{1}^{4k}}
\end{displaymath}
\noindent supports an exotic
transfer that raises filtration by 4. First note that
\begin{align*}
\Tr_{2}^{4}(\eta_{0}\eta_{1})
& = \Tr_{2}^{4}(a_{\sigma_{2}}^{2}\overline{r}_{1,0}\overline{r}_{1,0} )
= \Tr_{2}^{4}(u_{\sigma }\Res_{2}^{4}(a_{\lambda }\normrbar_{1}))
= \Tr_{2}^{4}(u_{\sigma })a_{\lambda }\normrbar_{1} \\
& = a_{\sigma }a_{\lambda }\normrbar_{1}a_{\lambda }\normrbar_{1}
\qquad \mbox{by (\ref{eq-exotic-transfers})}.
\end{align*}
\noindent Next note that the three elements
\begin{displaymath}
y_{8,8}=\eta_{0}^{8}=\Res_{2}^{4}(\epsilon ),\quad
y_{20,4}= \eta_{0}^{4}\delta_{1}^{4}=\Res_{2}^{4}(\overline{\kappa } )
\quad \aand
y_{32,0}= \delta_{1}^{8}=\Res_{2}^{4} (\Delta^{4})
\end{displaymath}
\noindent are all permanent cycles, so the same
is true of all
\begin{displaymath}
y_{16m +4\ell ,4\ell }=\eta_{0}^{4\ell }\delta_{1}^{4m }
\qquad \mbox{for $m,\ell \geq 0$ and $m+\ell $ even.}
\end{displaymath}
\noindent It follows that for such $\ell $ and $m$,
\begin{align*}
\eta_{0}\eta_{1}y_{16m+4\ell ,4\ell }
& = \eta_{0}\eta_{1}\eta_{0}^{4\ell }\delta_{1}^{4m}
=\eta_{0}^{4\ell+2 }\delta_{1}^{4m}
=y_{16m+4\ell+2 ,4\ell+2 } \\
& = \eta_{0}\eta_{1}\Res_{2}^{4}(x_{16m+4\ell ,4\ell }),
\end{align*}
\noindent so
\begin{displaymath}
\Tr_{2}^{4}(y_{16m+4\ell+2 ,4\ell+2 })
= \Tr_{2}^{4}(\eta_{0}\eta_{1})x_{16m+4\ell ,4\ell }
= f_{1}^{2}x_{16m+4\ell ,4\ell }.
\end{displaymath}
\noindent This is the desired exotic transfer.
\qed\bigskip
We now turn to the unsettled part of \ref{thm-d7}(iv).
\begin{thm}\label{thm-d7a}
{\bf The fate of $x_{16k+s+8,s}$ for $4k+s\equiv 7$ mod 8 and $s\geq
7$.} Each of these elements is the target of a $d_{7}$ and hence a
permanent cycle.
\end{thm}
\proof Consider the element $\Delta_{1}^{2}\in
\underline{E}_{2}^{0,16} (G/G)$. We will show that
\begin{displaymath}
d_{7} (\Delta_{1}^{2}) = x_{15,7}=\Tr_{2}^{4}(y_{15,7}).
\end{displaymath}
\noindent This is the case $k=0$ and $s=7$. The remaining cases will
follow via repeated multiplication by $\epsilon $, $\overline{\kappa } $ and
$\Delta_{1}^{4}$.
We begin by looking at
\begin{displaymath}
\Delta_{1}=u_{2\sigma }u_{\lambda }^{2}\normrbar_{1}^{2}.
\end{displaymath}
\noindent From Theorem \ref{thm-d3ulambda} we have
\begin{displaymath}
d_{5} (u_{2\sigma }) = a_{\sigma}^{3}a_{\lambda}\normrbar_{1}
\qquad \aand
d_{5} (u_{\lambda}^{2}) = a_{\sigma }a_{\lambda}^{2}u_{\lambda }\normrbar_{1}
\end{displaymath}
\noindent Using the gold relation $a_{\sigma }^{2}u_{\lambda }=2a_{\lambda
}u_{2\sigma }$, we have
\begin{align*}
d_{5} (\Delta_{1})
= d_{5} (u_{2\sigma }u_{\lambda }^{2})\normrbar_{1}^{}
& = (a_{\sigma}^{3}a_{\lambda}u_{\lambda}^{2}\normrbar_{1}^{}
+ a_{\sigma }a_{\lambda}^{2}u_{\lambda }u_{2\sigma }\normrbar_{1})
\normrbar_{1}^{} \\
& = a_{\sigma }a_{\lambda }u_{\lambda }(a_{\sigma}^{2}u_{\lambda}
+ a_{\lambda}u_{2\sigma })\normrbar_{1}^{2} \\
& = a_{\sigma }a_{\lambda }u_{\lambda }(2a_{\lambda }u_{2\sigma }
+ a_{\lambda}u_{2\sigma })\normrbar_{1}^{2} \\
& = a_{\sigma }a_{\lambda }^{2}u_{\lambda }u_{2\sigma }\normrbar_{1}^{2}
\qquad \mbox{since }2a_{\sigma }=0 \\
& = \nu x_{4}.
\end{align*}
\noindent Since $\nu $ supports an exotic group extension, $2\nu
=x_{3}$, we have
\begin{displaymath}
2d_{5} (\Delta_{1})= d_{7} (2\Delta_{1})= x_{3}x_{4}.
\end{displaymath}
\noindent From this it follows that
\begin{displaymath}
d_{7} (\Delta_{1}^{2})=\Delta_{1}d_{7} (2\Delta_{1})=x_{15,7}
\end{displaymath}
\noindent as claimed.
\qed\bigskip
The resulting reduced $\underline{E}_{12}$-term is shown in Figure
\ref{fig-E12}. It is sparse enough that the only possible remaining
differentials are the indicated $d_{13}$s. In order to establish them
we need the following.
\begin{figure}
\begin{center}
\includegraphics[width=11cm]{fig-E4.pdf} \caption[The reduced
$\EE_{4}$-term of the slice {\SS} for $\kH$.]{The $\EE_{4}$-term of
the slice {\SS} for $\kH$ with elements of Proposition \ref{prop-perm}
removed. Differentials are shown in red. Exotic transfers and
restrictions are shown as solid blue and dashed green lines
respectively. The Mackey functor symbols are as in Table
\ref{tab-C4Mackey}.} \label{fig-E4}
\end{center}
\end{figure}
\begin{figure}
\begin{center}
\includegraphics[width=11cm]{fig-E12.pdf} \caption[The reduced
$\EE_{4}$-term of the slice {\SS} for $\kH$.]{The $\EE_{12}$-term of
the slice {\SS} for $\kH$ with elements of Proposition \ref{prop-perm}
removed. Differentials are shown in red. Exotic transfers and
restrictions are shown as solid blue and dashed green lines
respectively. The Mackey functor symbols are as in Table
\ref{tab-C4Mackey}.} \label{fig-E12}
\end{center}
\end{figure}
The surviving class in $\underline{E}_{12}^{20,3} (G/G)$ is
\begin{displaymath}
x_{17,3}=f_{1}\Delta_{1}^{2}
= a_{\sigma }a_{\lambda }\normrbar_{1}\cdot
[u_{2\sigma }^{2}]u_{\lambda }^{4}\normrbar_{1}^{4} = (a_{\sigma
}u_{\lambda }^{4}) (a_{\lambda }[u_{2\sigma }^{2}]\normrbar_{1}^{5}).
\end{displaymath}
\noindent The second factor is a permanent cycle, so Theorem
\ref{thm-normed-slice-diffs} gives
\begin{displaymath}
d_{13} (f_{1}\Delta_{1}^{2})
= (a_{\lambda }^{7}[u_{2\sigma }^{2}]\normrbar_{1}^{3})
(a_{\lambda }[u_{2\sigma }^{2}]\normrbar_{1}^{5})
=a_{\lambda }^{8}[u_{2\sigma }^{2}]\normrbar_{1}^{8}
= \epsilon^{2} = x_{4}^{4}.
\end{displaymath}
\noindent The surviving class in $\underline{E}_{12}^{32, 2} (G/G)$ is
\begin{displaymath}
x_{30,2}
=a_{\sigma }^{2}u_{2\sigma }^{3}u_{\lambda }^{8}\normrbar_{1}^{8}
\in \underline{E}_{12}^{32,2} (G/G)
\end{displaymath}
\noindent satisfies
\begin{displaymath}
\epsilon x_{30,2}
=f_{1}\overline{\kappa }x_{17,3}
= f_{1}^{2}x_{4}\Delta_{1}^{2},
\end{displaymath}
\noindent so we have proved the following.
\begin{thm}\label{thm-d13}
{\bf $d_{13}$s in the slice {\SS } for $\kH$.} There are
differentials
\begin{displaymath}
d_{13} (f_{1}^{\epsilon }x_{4}^{m}\Delta_{1}^{2n})
= f_{1}^{\epsilon -1}x_{4}^{m+4}\Delta_{1}^{2 (n-1)}
\end{displaymath}
\noindent for $\epsilon =1,2$, $m+n$ odd, $n\geq 1$ and $m
\geq 1-\epsilon $. The {\SS } collapses from $\underline{E}_{14}$.
\end{thm}
\bigskip
To finish the calculation we have
\begin{thm}\label{thm-double}
{\bf Exotic transfers from and restrictions to the 0-line.} In
$\upi_{*}\kH$, for $i\geq 0$ we have
\begin{align*}
\Tr_{1}^{2} (r_{1,\epsilon }r_{1,0}^{4i}r_{1,1}^{4i})
& = \eta _{\epsilon }^{2}\orr_{1,0}^{4i}\orr_{1,1}^{4i}
&&\in \upi_{8i+2}
&\mbox{(filtration jump 2)}\\
\Tr_{1}^{4} (r_{1,0}^{8i+1}r_{1,1}^{8i+1})
& = 2x_{4}\Delta_{1}^{4i}
&&\in \upi_{32i+4}
&\mbox{(filtration jump 4)}\\
\Tr_{1}^{2} ((r_{1,0}^{3}+r_{1,1}^{3}) r_{1,0}^{8i}r_{1,1}^{8i})
& = \eta _{0}^{3}\eta _{1}^{3}\delta_{1}^{8i}
&&\in \upi_{32i+6}
&\mbox{(filtration jump 6)}\\
\Tr_{1}^{4}(r_{1,0}^{8i+5}r_{1,1}^{8i+5})
& = 2x_{4}\Delta_{1}^{4i+2}
&&\in \upi_{32i+20}
&\mbox{(filtration jump 4)}\\
\Tr_{1}^{2} ((r_{1,0}^{3}+r_{1,1}^{3}) r_{1,0}^{8i+4}r_{1,1}^{8i+4})
& = \eta _{0}^{3}\eta _{1}^{3}\delta_{1}^{8i+4}
&&\in \upi_{32i+22}
&\mbox{(filtration jump 6)}\\
\aand
\Tr_{2}^{4}(2\delta_{1}^{8i+7})
& = x_{4}^{3}\Delta_{1}^{4i+2}
&&\in \upi_{32i+28}
&\mbox{(filtration jump 12,}\\
& && &\mbox{the long transfer)}.
\end{align*}
Let $\underline{M}_{k}$ denote the reduced value of $\upi_{k}\kH$,
meaning the one obtained by removing the elements of Proposition
\ref{prop-perm}. Its values are shown in purple in Figure
\ref{fig-E14a}, and each has at most two summands. For even $k$ one
of them contains torsion free elements, and we denote it by
$\underline{M}'_{k}$. Its values depend on $k$ mod 32 and are as
follows, with symbols as in Table \ref{tab-C4Mackey}.
\begin{center}
\begin{tabular}[]{|c||c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
$k$
&0 &2 &4 &6 &8 &10&12&14
&16&18&20&22&24&26&28&30\\
\hline
$\underline{M}'_{k}$
&$\Box$
&$\widehat{\dot{\Box}}$
&$\JJoldiagbox$
&$\JJJbox $
&$\fourbox$
&$\widehat{\dot{\Box}}$
&$\overline{\twobox} $
&$\widehat{\oBox}$
&$\twobox $
&$\widehat{\dot{\Box}}$
&$\circoldiagbox$
&$\circbox$
&$\fourbox $
&$\widehat{\dot{\Box}}$
&$\dot{\twobox}$
&$\widehat{\oBox}$\\
\hline
\end{tabular}
\end{center}
\end{thm}
\proof We have two tools at our disposal: the
periodicity theorem and Theorem \ref{thm-exotic}, which relates exotic
transfers to differentials.
Figure \ref{fig-KH} shows that $\underline{M}'_{k}$ has the
indicated value for $-8\leq k\leq 0$ because the same is true of
$\EE_{4}^{0,k}$ and there is no room for any exotic extensions. On the
other hand $\EE_{4}^{0,k+32}$ does not have the same value for $k=-8$, $k=-6$
and $k=-4$. This comparison via periodicity forces
\begin{itemize}
\item [$\bullet$] the indicated $d_{5}$ and $d_{7}$ in dimension 24,
which together convert $\Box$ to $\fourbox$. These were also established in
Theorem \ref{thm-d7}.
\item [$\bullet$] the short transfer in dimension 26, which converts
$\widehat{\oBox}$ to $\widehat{\dot{\Box}}$. It also follows from the
the results of Section \ref{sec-C2diffs}.
\item [$\bullet$] the long transfer in dimension 28, which converts
$\overline{\twobox}$ to $\dot{\twobox}$.
\end{itemize}
The differential corresponding to the long transfer is
\begin{align*}
d_{13} ([2u_{\lambda }^{7}])
& = a_{\sigma }a_{\lambda }^{6}u_{2\sigma }u_{\lambda }^{4}
\normrbar_{1}^{3}, \\
\mbox{so}\qquad
d_{13} (a_{\sigma } [2u_{\lambda }^{7}])
& = a_{\sigma }^{2}a_{\lambda }^{6}u_{2\sigma }u_{\lambda }^{4}
\normrbar_{1}^{3}
= 2a_{\lambda }^{7}[u_{2\sigma }^{2}]u_{\lambda }^{3}
\normrbar_{1}^{3}
\end{align*}
\noindent This compares well with the $d_{13}$ of Theorem
\ref{thm-normed-slice-diffs}, namely
\begin{displaymath}
d_{13} (a_{\sigma }[u_{\lambda }^{4}])
=a_{\lambda }^{7}[u_{2\sigma }^{2}]\normrbar_{1}^{3}.
\end{displaymath}
The statements in dimensions 4 and 20 have similar proofs, and we will
only give the details for the former. It is based on comparing the
$\EE_{4}$-term for $\KH$ in dimensions $-28$ and $4$. They must
converge to the same thing by periodicity. From the slice
$\EE_{4}$-term in dimension 4 we see there is a short exact sequence
\begin{numequation}\label{eq-4-stem}
\begin{split}
\xymatrix
@R=4mm
@C=18mm
{
0\ar[r]
&\JJ \ar[r]
&{\underline{M}'_{4}}\ar[r]
&{\odbox}\ar[r]
&0\\
&\Z/2\ar@/_/[dd]_(.5){0}\ar@{=}[r]
&\Z/2\ar@/_/[dd]_(.5){\left[\begin{array}{c}1\\0\end{array} \right]}
\ar[r]
&0\ar@/_/[dd]\\
{}\\
&\Z/2\ar@/_/[dd]\ar@/_/[uu]_(.5){1}
\ar[r]^(.4){\left[\begin{array}{c}1\\0\end{array} \right]}
&\Z/2\oplus \Z_{-}
\ar@/_/[dd]_(.5){\left[\begin{array}{cc}0&2\end{array} \right]}
\ar@/_/[uu]_(.5){\left[\begin{array}{cc}1&a\end{array} \right]}
\ar[r]^(.6){\left[\begin{array}{cc}0&1\end{array} \right]}
&\Z_{-}\ar@/_/[dd]_(.5){2}\ar@/_/[uu]\\
{}\\
&0\ar@/_/[uu]\ar[r]
&\Z_{-}\ar@/_/[uu]_(.5){\left[\begin{array}{c}b\\1\end{array} \right]}
\ar@{=}[r]
&\Z_{-}\ar@/_/[uu]_(.5){1},
}
\end{split}
\end{numequation}
\noindent while the $(-28)$-stem gives
\begin{displaymath}
\xymatrix
@R=4mm
@C=17mm
{
0\ar[r]
&\dot{\twobox}\ar[r]
&{\underline{M}'_{4}}\ar[r]
&\obull\ar[r]
&0 \\
&\Z/2\ar@/_/[dd]_(.5){0}\ar@{=}[r]^(.5){}
&\Z/2\ar@/_/[dd]_(.5){\left[\begin{array}{c}1\\0\end{array} \right]}
\ar[r]^(.5){}
&0\ar@/_/[dd]_(.5){} \\
{}\\
&\Z_{-}\ar@/_/[dd]_(.5){2}\ar@/_/[uu]_(.5){1}
\ar[r]^(.4){\left[\begin{array}{c}c\\1\end{array} \right]}
&\Z/2\oplus\Z_{-}
\ar@/_/[dd]_(.5){\left[\begin{array}{cc}0&2\end{array} \right]}
\ar@/_/[uu]_(.5){\left[\begin{array}{cc}1&a\end{array} \right]}
\ar[r]^(.6){\left[\begin{array}{cc}1&d\end{array} \right]}
&\Z/2\ar@/_/[uu]_(.5){}\ar@/_/[dd]_(.5){}\\
{}\\
&\Z_{-}\ar@/_/[uu]_(.5){1}\ar@{=}[r]
&\Z_{-}\ar@/_/[uu]_(.5){\left[\begin{array}{c}b\\1\end{array} \right]}
\ar[r]^(.5){}
&0\ar@/_/[uu]_(.5){}
}
\end{displaymath}
\noindent The commutativity of the second diagram requires that
\begin{displaymath}
a+b=c=1 \qquad \aand b+d=c+d=0,
\end{displaymath}
\noindent giving $(a,b,c,d)= (0,1,1,1)$. The diagram
for $M_{4}$ is that of $\JJoldiagbox$ in Table \ref{tab-C4Mackey}.
In dimension 20 the short exact sequence of (\ref{eq-4-stem}) is replaced by
\begin{displaymath}
\xymatrix
@R=5mm
@C=10mm
{
0\ar[r]
&\circ \ar[r]
&{\underline{M}'_{20}}\ar[r]
&{\odbox}\ar[r]
&0
}
\end{displaymath}
\noindent and the resulting diagram for $\underline{M}'_{20}$ is that
of $\circoldiagbox$.
Similar arguments can be made in dimensions 6 and 22.
\qed\bigskip
We could prove a similar statement about exotic restrictions hitting
the 0-line in the third quadrant in dimensions congruent to 0, 4, 6,
14, 16, 20 (where there is an exotic transfer) and 22. The problem is
naming the elements involved.
\begin{table}[h]
\caption
[Infinite Mackey functors in the slice {\SS }.]
{Infinite Mackey functors in the reduced
$\EE_{\infty}$-term for $\KH$. In each even degree there is an
infinite Mackey functor on the 0-line that is related to a summand of
$\upi_{2k}\KH$ in the manor indicated. The rows in each diagram are
short or 4-term exact sequences with the summand appearing in both
rows.} \label{tab-0-line}
\begin{center}
\begin{tabular}{|c|c||c|c|}
\hline
Dimension
&Third quadrant
&Dimension
&Third quadrant\\
mod 32
&First quadrant
&mod 32
&First quadrant\\
\hline
0 &$
\xymatrix
@R=1mm
@C=7mm
{
{\fourbox} \ar[r]^(.5){\phantom{d_{7}}}
&\Box \ar[r]^(.5){}\ar@{=}[d]^(.5){}
&\circ \\
0\ar[r]^(.5){}
&\Box \ar@{=}[r]^(.5){}
&\Box
}$
&16 &$
\xymatrix
@R=1mm
@C=7mm
{
{\fourbox} \ar[r]^(.5){}
&{\twobox} \ar[r]^(.5){}\ar@{=}[d]^(.5){}
&{\JJ}\\
&{\twobox} \ar[r]^(.5){}
&\Box\ar[r]^(.5){d_{7}}
&\bullet
}$\\
\hline
2, 10 &$
\xymatrix
@R=1mm
@C=7mm
{
{\widehat{\dot{\Box}}}
\ar[r]^(.5){}
&{\widehat{\dot{\Box}}}
\ar[r]^(.5){}\ar@{=}[d]^(.5){}
&0 \\
{\widehat{\bullet}}
\ar[r]^(.5){}
&{\widehat{\dot{\Box}}}
\ar[r]^(.5){}
&{\widehat{\oBox}}
}$
&18, 26 &$
\xymatrix
@R=1mm
@C=7mm
{
{\widehat{\dot{\Box}}}
\ar[r]^(.5){}
&{\widehat{\dot{\Box}}}
\ar[r]^(.5){}\ar@{=}[d]^(.5){}
&0 \\
{\widehat{\bullet}}
\ar[r]^(.5){}
&{\widehat{\dot{\Box}}}
\ar[r]^(.5){}
&{\widehat{\oBox}}
}$\\
\hline
4 &$
\xymatrix
@R=1mm
@C=7mm
{
{\dot{\twobox}}
\ar[r]^(.5){}
&{\JJoldiagbox }
\ar[r]^(.5){}\ar@{=}[d]^(.5){}
&{\obull}\\
{\JJ}
\ar[r]^(.5){}
&{\JJoldiagbox }
\ar[r]^(.5){}
&{\overline{\twobox} }
}$
&20 &$
\xymatrix
@R=1mm
@C=7mm
{
{\dot{\twobox}}
\ar[r]^(.5){}
&{\circoldiagbox }
\ar[r]^(.5){}\ar@{=}[d]^(.5){}
&{\circ} \\
{\circ}
\ar[r]^(.5){}
&{\circoldiagbox }
\ar[r]^(.5){}
&{\overline{\twobox} }
}$\\
\hline
6 &$
\xymatrix
@R=1mm
@C=7mm
{
{\bullet}\ar[r]^(.5){d_{7}}
&{\JJbox }
\ar[r]^(.5){}
&{\JJJbox }
\ar[r]^(.5){}\ar@{=}[d]^(.5){}
&{\bullet}\\
&{\widehat{\oBox}}
\ar[r]^(.5){}
&{\JJJbox }
\ar[r]^(.5){}
&{\JJJ}
}$
&22 &$
\xymatrix
@R=1mm
@C=7mm
{
{\JJbox }
\ar[r]^(.5){}
&{\circbox}
\ar[r]^(.5){}\ar@{=}[d]^(.5){}
&{\bullet} \\
{\circ}
\ar[r]^(.5){}
&{\circbox}
\ar[r]^(.5){}
&{\widehat{\oBox}}
}$\\
\hline
8 &$
\xymatrix
@R=1mm
@C=7mm
{
{\fourbox}
\ar[r]^(.5){\phantom{d_{7}}}
&{\fourbox}
\ar[r]^(.5){}\ar@{=}[d]^(.5){}
&0\\
&{\fourbox}
\ar[r]^(.5){}
&{\Box}
\ar[r]^(.5){d_{5}, d_{7}}
&{\circ}
}$
&24 &$
\xymatrix
@R=1mm
@C=7mm
{
{\fourbox}
\ar[r]^(.5){}
&{\fourbox}
\ar[r]^(.5){}\ar@{=}[d]^(.5){}
&0\\
&{\fourbox}
\ar[r]^(.5){}
&{\Box}
\ar[r]^(.5){d_{5}, d_{7}}
&{\circ}
}$\\
\hline
12 &$
\xymatrix
@R=1mm
@C=7mm
{
{\bullet}
\ar[r]^(.5){d_{13}}
&{\dot{\twobox}}
\ar[r]^(.5){}
&{\overline{\twobox} }\ar@{=}[d]^(.5){} \\
&0
\ar[r]^(.5){}
&{\overline{\twobox} }
\ar[r]^(.5){}
&{\overline{\twobox} }
}$
&28 &$
\xymatrix
@R=1mm
@C=7mm
{
{\dot{\twobox}}
\ar[r]^(.5){}
&{\dot{\twobox}}
\ar[r]^(.5){}\ar@{=}[d]^(.5){}
&0 \\
{\bullet}
\ar[r]^(.5){}
&{\dot{\twobox}}
\ar[r]^(.5){}
&{\overline{\twobox} }
}$\\
\hline
14 &$
\xymatrix
@R=1mm
@C=7mm
{
{\JJ} \ar[r]^(.5){d_{7}}
&{\JJbox }
\ar[r]^(.5){}
&{\widehat{\oBox}}
\ar@{=}[d]^(.5){}
& \\
&0
\ar[r]^(.5){}
&{\widehat{\oBox}}
\ar[r]^(.5){}
&{\widehat{\oBox}}
}$
&30 &$
\xymatrix
@R=1mm
@C=7mm
{
{\widehat{\oBox}}
\ar[r]^(.5){}
&{\widehat{\oBox}}
\ar[r]^(.5){}\ar@{=}[d]^(.5){}
&0 \\
0
\ar[r]^(.5){}
&{\widehat{\oBox}}
\ar[r]^(.5){}
&{\widehat{\oBox}}
}$\\
\hline
\end{tabular}
\end{center}
\end{table}
In Table \ref{tab-0-line} we show short or 4-term exact sequences in the 16 even
dimensional congruence classes. In each case the value of
$\underline{M}'_{k}$ is the symbol appearing in both rows of the diagram.
For even $k$ with $0\leq k<32$, we typically have short exact
sequences
\begin{displaymath}
\xymatrix
@R=2mm
@C=10mm
{
0\ar[r]^(.5){}
&{\EE_{4}^{0,k-32}}\ar[r]^(.5){}
&{\underline{M}'_{k}}\ar[r]^(.5){}\ar@{=}[d]^(.5){}
&{\mbox{quotient} }\ar[r]^(.5){}
&{0}\\
0\ar[r]^(.5){}
&{\mbox{subgroup} }\ar[r]^(.5){}
&{\underline{M}'_{k}}\ar[r]^(.5){}
&{\EE_{4}^{0,k}}\ar[r]^(.5){}
&{0,}
}
\end{displaymath}
\noindent where the quotient or subgroup is finite and may be spread
over several filtrations. This happens for the quotient in dimensions
$-32$, $-16$ and $-12$, and for the subgroup in dimensions 6 and 22.
This is the situation in dimensions where no differential hits
[originates on] the 0-line in the third [first] quadrant. When such a
differential occurs, we may need a 4-term sequence, such as the one in
dimension -22.
In dimensions 8 and 24 there is more than one such differential,
the targets being a quotient and subgroup of the Mackey functor
$\circ=\Box/\fourbox$.
In dimension $-18$ we have a $d_{7}$ hitting the 0-line. Its source
is written as ${\circ \subseteq \EE_{4}^{-7,-24}}$ in Figure
\ref{fig-KH}. Its generator supports a $d_{5}$, leaving a copy of
$\JJ$ in $\EE_{7}^{-7,-24}$.
There is no case in which we have such
differentials in both the first and third quadrants.
\begin{cor}\label{cor-higher}
{\bf The $\EE_{\infty }$-term of the slice {\SS} for $\KH$.}
The surviving elements in the spectral sequence for $\KH$ are shown in
Figure \ref{fig-E14a}.
\end{cor}
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Chicago collection achieved several milestones for the Fine Books and Manuscripts department at Leslie Hindman Auctioneers. The auction was not only the first single-owner sale for the department but also its highest grossing sale for the department in firm history. Exceeding expectations, the collection had a nearly 90% sell through rate.
Highlights’s A Geographical Historie of Africa, which realized $8,750 against an estimate of $4,000-6,000.
Gretchen Hause, Director of the Fine Books and Manuscripts department,.
The fall season continues for the Fine Books and Manuscripts department with two November sales. On November 12, the department will offer the Fine Cartographic and Printed Americana Collection of Evelyn and Eric Newman; the season will conclude with the Fine Printed Books and Manuscripts sale on November 13. The Fine Books and Manuscripts department is currently accepting consignments for spring auctions. For more information, visit.
Image: DRAKE, Francis, Sir. Sir Francis Drake Revived. London: Printed for Nicholas Bourne, 1653 [i.e. 1652].
| 25,479
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TITLE: Inequality with Fibonacci numbers $ \sum \limits_{k=1}^{2n+1} (-1)^{k+1} F_k \: \text{arccot} F_k < \frac{\pi+1-\sqrt{5}}{2}$
QUESTION [3 upvotes]: Prove that $$ \sum \limits_{k=1}^{2n+1} (-1)^{k+1} F_k \: \text{arccot} F_k < \frac{\pi+1-\sqrt{5}}{2}$$ holds for all $n \in \mathbb{N}$.
(The Fibonacci sequence, defined by the recurrence $F_1 = F_2 = 1$ and $\forall n \in \mathbb{N},$ $F_{n+2} = F_{n+1} + F_n$)
My work. I proved that the sequence $a_n=\sum \limits_{k=1}^{2n+1} (-1)^{k+1} F_k \: \text{arccot} F_k \;$ is increasing. Therefore inequality cannot be proved by induction.
REPLY [2 votes]: Because of the size, I'm writing this attempt as a separate answer.
Still no closer to the proof of the closed form, but I decided to try Euler-Maclaurin, since the integrals can be evaluated explicitly.
We change the sign for $P$ from my other answer, so it will be positive, and work with it:
$$P=\sum_{l=1}^\infty \left(F_{2l+2} \arctan \frac{1}{ F_{2l+2} } -F_{2l+1} \arctan \frac{1}{ F_{2l+1} } \right)=\varphi- \frac{\pi}{2}= \\ = 0.047237661954998228973 \ldots$$
Using the integral representation of arctangent and the hyperbolic form of Fibonacci numbers, we have:
$$P= \int_0^1 \sum_{l=1}^\infty \left( \frac{F_{2l+21}^2}{ t^2+F_{2l+2}^2 } - \frac{F_{2l+1}^2}{ t^2+F_{2l+1}^2 } \right) dt$$
Simple rearrangement leads to the sum of two convergent series:
$$P= \int_0^1 \sum_{l=1}^\infty \left( \frac{t^2}{ t^2+F_{2l+1}^2 } - \frac{t^2}{ t^2+F_{2l+2}^2 } \right) dt$$
Or, explicitly:
$$P= \int_0^1 \sum_{l=1}^\infty \left( \frac{ \frac54 t^2}{ \frac54 t^2+\cosh^2 (2l+1) \alpha } - \frac{\frac54 t^2}{\frac54 t^2+\sinh^2 (2l+2) \alpha } \right) dt$$
Let's denote $\frac54 t^2=y^2$ and consider the series:
$$S_1(y)=\sum_{l=1}^\infty \frac{ 1}{y^2+\cosh^2 (2l+1) \alpha }$$
$$S_2(y)=\sum_{l=1}^\infty \frac{ 1}{y^2+\sinh^2 (2l+2) \alpha }$$
I doubt the series have a general closed form, so let's try our hand at Euler-Maclaurin:
$$S_1(y) = \int_1^\infty \frac{ dx}{y^2+\cosh^2 (2x+1) \alpha }+ \\ +\frac{1}{2} \frac{ 1}{y^2+\cosh^2 3\alpha }- \sum_{k=1}^\infty \frac{B_{2k}}{(2k)!} \frac{\partial^{2k-1}}{\partial x^{2k-1}} \left(\frac{ 1}{y^2+\cosh^2 (2x+1) \alpha } \right) \bigg|_{x=1}$$
$$S_2(y) = \int_1^\infty \frac{ dx}{y^2+\sinh^2 (2x+2) \alpha }+ \\ +\frac{1}{2} \frac{ 1}{y^2+\sinh^2 4\alpha }- \sum_{k=1}^\infty \frac{B_{2k}}{(2k)!} \frac{\partial^{2k-1}}{\partial x^{2k-1}} \left(\frac{ 1}{y^2+\sinh^2 (2x+2) \alpha } \right) \bigg|_{x=1}$$
Where $B_{2k}$ are Bernoulli numbers. Here I assumed that all the derivatives at $x \to \infty$ give exactly $0$, which is a reasonable assumption. Even though the series is asymptotic, I suspect it converges in this case.
Both the double integrals can be found exactly, but I won't write the full solution here since it's too long, I'll just provide the values from Mathematica:
$$ \int_0^1 \int_1^\infty \frac{5/4 t^2 dx dt}{5/4 t^2+\cosh^2 (2x+1) \alpha }= \\ =\frac{\sqrt{5} \left(\log 8-3 \log \left(35-15 \sqrt{5}\right)+4 \arctan \left(\frac{4}{3}\right)\right)-10}{20 \alpha}=0.0425812 \ldots$$
$$ \int_0^1 \int_1^\infty \frac{5/4 t^2 dx dt}{5/4 t^2+\sinh^2 (2x+2) \alpha }= \\ = \frac{\sqrt{5}\left(\log 4- \log \left(15-5\sqrt{5}\right)+14 \arctan \left(\frac{1}{3}\right)\right)-10}{20 \alpha}=0.0182396 \ldots$$
Now for the second terms of the E-M expansion:
$$\frac{1}{2} \int_0^1 \frac{5/4 t^2 dt}{5/4 t^2+\cosh^2 3 \alpha }=\frac{1}{2} - \tan^{-1} \frac{1}{2}$$
$$\frac{1}{2} \int_0^1 \frac{5/4 t^2 dt}{5/4 t^2+\sinh^2 4 \alpha }=\frac{1}{2} - \frac{3}{2}\tan^{-1} \frac{1}{3}$$
Let's see what we get so far by adding all four terms:
$$P_0= \frac{ -2\log 5-\log \left(123-55 \sqrt{5}\right)+4 \arctan \left(\frac{4}{3}\right)-14 \arctan \left(\frac{1}{3}\right))}{4 \sqrt{5} \alpha}+\frac{3}{2}\tan^{-1} \frac{1}{3}-\tan^{-1} \frac{1}{2} = \\ = 0.043319806992170091958 \ldots$$
This doesn't simplify much (at least Mathematica can't do it), and the numerical value is not that close to the desired answer.
Considering that $B_2=1/6$, taking the first derivatives and evaluating the integrals w.r.t. $t$ gives us an additional correction:
$$\Delta_1 P = \frac{ \left(1+8 \arctan \left(\frac{1}{2}\right)-14 \arctan \left(\frac{1}{3}\right)\right) \alpha}{12 \sqrt{5}}$$
$$P_0+\Delta_1 P=0.046990354355256824367$$
Which is closer to the answer. Taking more derivatives, we get various more complicated terms with increasing odd powers of $\alpha$.
This was fun, but I don't see any reason to continue. Let's just say, this method might be interesting for more general series, while this case with Fibonacci numbers obviously has some simple trick that can be used to prove the closed form.
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|
Shuai Hao
- Studying Master of Engineering at University of Victoria (Class of 2017)
- Spent 8 months on Operations Development Team at Hootsuite (January – August 2017)
- Worked with Ansible, Terraform, Diamond, Sensu, Python, Unix/Linux, and AWS
- Worked with Alex Zadorozhnyi, Andy Han, Arthur Mencke, Bryan Neyedli, James Ishee, Jeff Welling, Jonas Courteau, Nick Hollingsworth, Ronald Bow, Tyler Mauthe
- Connect with me on LinkedIn
My Contributions
- Developed a tool for managing all instances with collection of ad-hoc scripts that can be run
- Automated Campaigns SOC2 Auditing and Operations User Teardown
- Migrated various databases from Mysql RDS to Aurora
- Created replication slave servers for EC2 Mysql instances
- Terraformed and ansiblized varieties of infrastructure
- Write a blog about database migration
My Experience
Over this work term, I had various duties such as developing tools for automating internal operations processes and contributing to infrastructure maintenance and changes. In this position, I was able to further my skills and learn more about the world of professional IT operations, software development, and networking. In infrastructure maintenance and changes, I was able to learn more about the process of infrastructure delivery as well as the different tools such as Terraform, Ansible, and custom scripts. I am adept at these tools and I am able to build, change, and provision infrastructure safely and efficiently. I also contributed to setting up master slave replications for EC2 Mysql databases and various database migrations from Mysql RDS to Aurora.The experience and knowledge I have learned from this position has allowed me to grow and develop as a developer and learn more about DevOps. It allows me to get a new perspective on software development industry. Working in Hootsuite has been an enlightening experience which has allowed me gain a multitude of experience from various areas of software development and IT operations, which allowed me to realize more potential career options.
| 97,727
|
TITLE: How to show that a theorem holds for isomorphic vector spaces?
QUESTION [1 upvotes]: Theorem: Any two norms on a finite-dimensional vector space $V$ are equivalent.
My book begins the proof by noting that we can identify $V$ with $\mathbb R^n$ for some $n\in\mathbb N$, and therefore we only need to prove the theorem for $\mathbb R^n$. However, I don't see why this is true. I know that there is a bijection between $V$ and $\mathbb R^n$ for some $n\in\mathbb N$, or equivalently, there is a bijection between a basis in $\mathbb R^n$ and a basis in $V$. But how to proceed from there on?
I've read somewhere that an isomorphism is basically just the relabelling of the elements... but how do I know that things like inequalities and the like stay preserved between two identified elements? Things like "the vector space structure is preserved" sounds slightly vague to me, so is there something more rigorous to say?
REPLY [3 votes]: Let $(V, \Vert \cdot \Vert_V$) be a finite-dimensional $\mathbb{R}$-vector space. Then there exists an isomorphism of vector spaces (i.e. a bijective $\mathbb{R}$-linear map) $\phi: V \rightarrow \mathbb{R}^n$ for some $n\in \mathbb{N}$. As you wrote above, you want this map to preserve the norm. So you want to have a norm $\Vert \cdot \Vert_{\phi}$ on $\mathbb{R}^n$ such that for every $v\in V$ holds
$$ \Vert \phi(v)\Vert_{\phi} = \Vert v \Vert_{V}.$$
In fact this norm is (check that it really is a norm)
$$ \Vert x \Vert_{\phi} = \Vert \phi^{-1}(x) \Vert_V.$$
Let now $\Vert \cdot \Vert_1$, $\Vert \cdot \Vert_2$ be two norms on $V$ and $\Vert \cdot \Vert_{\phi, 1}$ and $\Vert \cdot \Vert_{\phi,2}$ be the respective norms on $\mathbb{R}^n$. If you know that all norms on $\mathbb{R}^n$ are equivalent, then you have
$$ c^{-1} \Vert x \Vert_{\phi, 1} \leq \Vert x \Vert_{\phi, 2} \leq c \Vert x \Vert_{\phi,1}.$$
For $x=\phi(v)$ you get
$$ c^{-1} \Vert v \Vert_{1} = c^{-1} \Vert \phi(v) \Vert_{\phi, 1} \leq \Vert \phi(v) \Vert_{\phi, 2} = \Vert v \Vert_2 = \Vert \phi(v) \Vert_{\phi, 2} \leq c \Vert \phi(v) \Vert_{\phi,1} = c \Vert v \Vert_1. $$
| 25,241
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TITLE: A question about linear transformation and subspace
QUESTION [2 upvotes]: Let $V$ be an n-dimensional vector space over the reals, $n \ge 1$. Let $L(V)$ denote the vector space of all linear transformation $T:V \to V$, and let $A(V)$ be the set $A(V)=\{\,T \in L(V)\ |\ {\rm dim}(T(V)) \lt n\ \}$. Find all $n$ such that $A(V)$ is a subspace of $L(V)$.
This question I have no idea at all. Hope somebody could help me. Thanks
REPLY [2 votes]: Hints:
(1) $\;T\notin A(V)\iff T\;$ is invertible
(2) If $\;n>1\;$ then for any $\;T\in A(V)\;$ you can always find some $\;S\in A(V)\;$ s.t. $\;T+S\;$ is invertible...(thinking in matrix terms may be helpful)
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Auburn, Ala.--It took nearly a full calendar year but Auburn's losing streak in the SEC is history after the Tigers defeated Ole Miss 3-2 on a wet Sunday afternoon at Plainsman Park. Starting the 2007 season 0-8 in league play after finishing last year with a 10-game streak, Auburn had not won against a conference foe since April 28 of 2006 when the Tigers won at Alabama.
The win improves Auburn to 20-11 overall, 1-8 in SEC play, while Ole Miss falls to 21-9, 5-4 in conference action. Coach Tommy Slater says he's proud of his team for working hard and not giving up when thing didn't go well the first eight games.
"I'm very happy for our team," Slater says. "These kids have not stopped believing in themselves and it showed today with the way they went about their business. I'm just so proud of how they went about it.
"I'm very proud of the ballclub," he adds. "I'm very proud of the fact that our kids came out with great energy and passion. They just enjoyed playing the game today, they really did. That was really satisfying for me to see, just the way our kids came out with the energy and compassion they displayed."
Auburn celebrates the victory with Butts after the game.
The story of the day for the Tigers was Brett Butts. Expected to be in his normal set up role with Luke Greinke scheduled to be the closer, Butts instead was forced into the closer's role with an early injury to Greinke. All he did was pitch two and two/thirds perfect innings to shut the door on Ole Miss with Auburn holding the 3-2 advantage. It was his third save of the season and none have been bigger than Sunday's.
"I just got ahead of them and got outs," Butts says of his performance. "I threw the curve ball early and put them away with the change up. I love that feeling (of getting the save). I have never done it before until this year. Now I've had three chances to end the game. Every time is just great."
Butts' save put the finishing touches on a very good pitching day for the Tigers as starter Taylor Thompson gave up just one earned run and five hits in four and two/thirds innings. In his first SEC start, Thompson pitched well until running into trouble in the fifth. Although he didn't pick up the victory, Thompson says he was pleased with his performance on such a big stage.
"It's real nerve-wracking to go out there against a quality team like that but I went out there and settled down and starting making good pitches," Thompson says. "It was real fun. That's what I have been dreaming off since I have been a little kid, to play in the SEC. It came true today."
Auburn got on the board first on a Sunday when the skies threatened from the first pitch. Leading the game off, center fielder Bruce Edwards was hit by a Craig Rodriguez pitch to bring Greinke to the plate. A bouncer just out of the reach of first baseman Andrew Clark was just enough for Greinke to reach first base and move Edwards to third but a strained hip flexor meant an early end to Greinke's day as Andy Bennett came on to replace him with Josh Donaldson coming to the plate. A ground ball back to Rodriguez failed to score the run but Mike Bianucci's ground ball to short pushed Edwards across and moved Bennett to third with two outs. Joseph Sanders then singled him home to put Auburn up 2-0 early.
The Rebels put a scare into Thompson in the second inning as a leadoff walk to Clark put them in business but the first of several big defensive plays by the Tigers saved the day. Following a Logan Power pop to short, C.J. Ketchum hit a ground ball deep in the hole at shortstop. Making a diving grab, Phillip Stringer threw to Robert Brooks from his knees to cut down Clark at second for out number two. That would be big as Fuller Smith followed with a bloop single to left that would have scored a run. Thompson came back to strike out number nine hitter Brett Basham to end the Rebel threat and keep them scoreless through two.
All season Ross Smith has been solid for the Tigers in right field and he delivered once again as he led off the bottom of the second with a single to left field and moved to second on a sacrifice bunt. An Edwards groundout then moved Smith to third for Bennett. Struggling for much of the season, Bennett lined a single over the head of Justin Henry at second base to score Smith and put the Tigers up 3-0 with Thompson pitching well.
The lead remained the same until the top of the fifth when Ole Miss finally got to the hard-throwing freshman. With one out, Jordan Henry singled to center field and moved to second on a ground ball to Bennett at first base. Looking as if he would escape the inning, Thompson instead ran into trouble as Zack Cozart singled just out of the reach of the pitcher to score the first Rebel run. Another single, this time by Cody Overbeck, would be the end of the line for Thompson as Johnny Thompson came on to clean up the mess by getting Clark to pop out to end the threat with the lead still at two.
Ole Miss would continue to chip into the lead in the sixth inning and if not for a great play by Bianucci in left field, they would have at least tied the game in the inning. Leading things off, Power ripped a Thompson pitch off the very top of the monster in left field. Catching the ball off the wall, Bianucci threw a strike to Brooks at second to easily cut down Power for the out.
It proved to be a big play as with two outs, Fuller Smith doubled to right field and Brett Basham followed with a single to score him from second. Josh Blake came on to retire Jordan Henry to end the inning with Auburn still ahead 3-2 but fortunate to be in front thanks to Bianucci's play in the field.
From there it was all about Butts' performance. Facing eight batters, Butts retired all eight on a day when pitching wasn't easy because of a steady rain the final six innings. Now with the losing streak in the books, Butts says he feels like the team is ready to do some good things in the coming weeks.
"We've been right there the whole time," Butts says. "Just getting this is going to help us keep going. I think we'll roll from here on."
At the plate, the Tigers were led by Sanders with two hits. He was the only player with multiple hits for Auburn as the Tigers finished with seven hits. Both Jordan Henry and Smith had two hits for the Rebels.
The Tigers get back to action Tuesday night at 6 p.m. when they take on South Alabama at Plainsman Park. Sophomore Justin Bristow and freshman Austin Hubbard are both expected to see action on the mound with no starter determined at this time. Next week the Tigers travel to LSU to face the Bengal Tigers in a huge series for both teams.
| 96,204
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This document contains information on the following topics:
- Natural Hazards
- Why Prepare
- Technological Hazards
- Basic Preparedness
- Terrorism
- Recovering from Disaster
- Family Communications Plan
- Disaster Supplies List
- Water Conservation Tips
The link to download this document is at the following:
Get smart. Read this manual. It could save your life or the lives of you family.
| 119,186
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TITLE: What is the 'largest' space of integrable functions which is also a Hilbert space?
QUESTION [1 upvotes]: It is well known that $L^2(X,\mu)$, the set of functions $f:X \rightarrow \mathbb{C}$ such that $\int_X |f|^2 \text{d} \mu < \infty$, is a Hilbert space. Is there a Hilbert space $H$ such that $L^2(X,\mu)\subset H$? In other words, are there larger classes of integrable functions which form a hilbert space?
REPLY [1 votes]: No. Let $V$ be the orthogonal complement of $L^2(X,\mu)$ in $H$ and let $f$ be a function in $V$.
Then, $\int fg d \mu =0$ for all $g$ in $L^2(X,\mu)$. This condition easily implies that $f$ is identically zero.
| 5,967
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| 4,126
|
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| 393,590
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TITLE: Basic Probability Question Conditional Probability
QUESTION [1 upvotes]: There are three men at a party, and each throws their hat in the middle. The hats are mixed up, and then each man randomly selects a hat. What is the probability that none of the three men winds up with his own hat?
I have solved this question by first finding the probability that at least one of three of the men wind up with their own hat, and then subtracting the value from 1.
Let $E_1$, $E_2$, $E_3$ denote the events that each man finds his own hat.
$P(E_1E_2E_3)$ is total probability of each man finding his own hat
$$P(E_i) = \frac{1}{3}, \quad i=1,2,3$$
$$P(E_iE_j) = \frac{1}{6}, \quad i \neq j$$
$$P(E_1E_2E_3) = \frac{1}{6}$$
$$P(E_1 \cup E_2 \cup E_3) = P(E_1) + P(E_2) + P(E_3) - P(E_1E_2) - P(E_1E_3) - P(E_2E_3) + P(E_1E_2E_3)$$
$$\therefore P(E_1 \cup E_2 \cup E_3) = 1 - \frac{1}{2} + \frac{1}{6} = \frac{2}{3}$$
$$1 - \frac{2}{3} = \frac{1}{3}$$
This makes sense, however I wanted to see if I could find the same answer by letting $E_1$, $E_2$, $E_3$ denote the events that each man does not find his own hat.
$$P(E_i) = \frac{2}{3}, \quad i=1,2,3$$
$$P(E_iE_j) = \frac{2}{6}, \quad i \neq j$$
$$P(E_1E_2E_3) = \frac{2}{6}$$
$$P(E_1 \cup E_2 \cup E_3) = 2 - 1 + \frac{2}{6} = 1\frac{2}{6}$$
This is obviously wrong, so what am I missing?
Edit: I think I got my problem. The first man may select the second man's hat, so the probability of $P(E_iE_j)$ is not necessarily equal to $\frac{2}{6}$
Edit: I know what the mistake was, I will add it here to help anyone else who may want to see.
The first man has a $\frac{2}{3}$ probability of choosing a hat that is not his, hat 2 or hat 3.
Case 1
If he chooses hat 2, then the second man will have to choose between hat 1 and hat 3. In order to ensure that all 3 men get the wrong hat, he only has a $\frac{1}{2}$ probability. The third man will have a probability of 1 since he has only one choice. This means that the probability for case 1 is $\frac{1}{3}\frac{1}{2}=\frac{1}{6}$
Case 2
Similarly, if the first man chooses hat 3, the second man will have a choice between hat 2 and hat 1, and to ensure all three men get the wrong hat, he again has a probability of $\frac{1}{2}$. Man 3 has no choice and has a probability of 1. So the probability is $\frac{1}{3}\frac{1}{2}=\frac{1}{6}$
Adding these two probabilities together, we get a total probability of $\frac{1}{3}$.
REPLY [0 votes]: Your edit seems like a valid solution. For larger values of $n$, the computation gets much trickier.
In general, if we have $n$ hats, the probability that no one gets their own hat can be obtained
in many ways. Two such methods are to take
$$\sum_{i=0}^{n} {(-1)^i}\cdot{\frac{1}{i!}}$$
In our case, $n$ is $3$ so we get
$$\sum_{i=0}^{3} {(-1)^i}\cdot{\frac{1}{i!}}=1-1+{\frac{1}{2}}-{\frac{1}{6}}={\frac{1}{3}}$$
Alternatively, this probability can be obtained by
$$\frac{[\frac{n!}{e}]}{n!}$$
where $[\frac{n!}{e}]$ is the closest integer to $\frac{n!}{e}$
In our case, $n$ is $3$ so we get
$$\frac{[\frac{3!}{e}]}{3!} = {\frac{[2.207]}{6}}={\frac{1}{3}}$$
Finally, as $n \rightarrow \infty$ the probability that no one gets their own hat
approaches $\frac{1}{e}$ and it approaches this quite quickly.
Consider $n=9$. Then $$\frac{[\frac{9!}{e}]}{9!} \approx .3678792 \approx \frac{1}{e} \approx .3678794$$
| 122,958
|
Google; Not So Environmentally Evil After All
A recent report placing Google in a bad light, which spread across the web like wildfire, is now being recanted after the Harvard University physicist who the comments were attributed to says that he was misinterpreted.
“Performing two Google searches from a desktop computer can generate about the same amount of carbon dioxide as boiling a kettle for a cup of tea, according to new research,” ran the opening paragraph in the Sunday Times of London.
However, the author of the research, Alex Wissner-Gross, now says that he never singled out Google in his study on the environmental impact of computing.
also said that he had no clue where the kettle analogy came from.
Google was forced to respond after the story was picked up by many news outlets across the globe. The world’s leading search engine staunchly defended it’s environmental record in a post on the company,” the Mountain View, California, based company retorted to the notion that they were ruining the environment.
Google’s philanthropic arm, Google.org, has invested $45 million in wind, solar and geothermal energy start-ups.
The company also contends that speeding up the search process, which their mammoth data centers across the world achieve, actually lessens the impact on the environment.
To be sure, Wissner-Gross still backs his claims that “Google operates huge data centers around the world that consume a great deal of power,” but that his report had nothing to do with Google specifically.
“I don’t think anybody would disagree with those statements,” he said. “Everything online has a definite environmental impact. I think everybody can agree on that, including Google.”
Wissner-Gross, manages the Web site CO2stats.com which assists websites in becoming carbon neutral.
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| 413,790
|
\begin{document}
\maketitle
\section{Introduction}
A $d$-dimensional parametric family
\[ \condset{p_\theta}{\theta=
(\theta^1, \ldots , \theta^d)\in\Theta}, \quad \Theta\subset\bR^d
\]
of probability density functions on
a measure space $(\cX, {\cal B}, \nu) $ is called an exponential family with a
natural parameter $\theta=(\theta^i)$
(e.g.\ \cite{BN:book})
when there exist $d+1$ $\bR$-valued functions
$C, F_1, \ldots , F_d$ on $\cX$ and an $\bR$-valued
function $\psi$ on $\Theta$ such that
\begin{equation}
\label{eq:exp_family_p}
\log p_\theta (x) =
C(x) + \sum_{i=1}^d \theta^i F_i(x) -\psi (\theta).
\end{equation}
The notion of exponential family is very important
in various fields such as the theory of statistical inference (parameter estimation,
hypothesis testing, etc.), large deviations, information theory, etc,
and a dualistic viewpoint of the information geometry (\cite{Ama:book,
AmaNag:book, NagAma82}) works effectively to understand
the structure of exponential families in a unified and elegant manner.
The present paper is aimed at demonstrating that many characteristic properties
of exponential families (\ref{eq:exp_family_p}) are preserved by
families of Markov kernel densities of the form
\begin{align}
\log w_\theta (y|x) = &
C(x,y) + \sum_{i=1}^d \theta^i F_i(x,y)
\nonumber \\
& + K_\theta (y) - K_\theta (x) -\psi (\theta).
\label{eq:exp_family_w}
\end{align}
In particular, we show that the family is dually flat
just as is the case for (\ref{eq:exp_family_p}), which
enables us to apply the general theory for dually flat
spaces to investigation of the information-geometrical
structure of the family. In order to avoid
being involeved with
functional analytic arguments and to concentrate upon
geometric and algebraic aspects, we mostly confine ourselves
to the case where $\cX$ is a finite set and $\nu$ is the counting
measure, whereas the essence of the arguments can be extended to
the general case by putting proper regurality conditions.
Several attempts have been made so far
to extend the definition of exponential familes to
Markove processes or more general stochastic processes
(e.g.\ \cite{Fei81, KucSor98, KucSor:book}). Although
they share the common criterion that an exponential
family should have a finite dimensional (exact or asymptotic) sufficient
statistic, the definitions given there are diverse, reflecting their
respective backgrounds. The authors of such papers as
\cite{ItoAma88, Mer89, Ama01} had essentially the same
concept on exponential families of Markov chains as ours when they said
that the totality of strictly positive Markov chains of a fixed order
is asymptotically regarded as an exponential family in an
approximate sense. We refine the concept so that the general
definition of exponential families of Markov kernels is given
without appealing to asymptotic settings, although the
significance of the definition is made clear through some
asymptotic arguments. This non-asymptotic treatment
enables us to develop the information geometry of
Markov chains in a transparent and systematic way.
This article is essentially based on the technical report \cite{NagISTR}.
The proofs of the theorems are omitted for want of space.
\section{Definition and an example}
Let $\cX$ be a finite set with $|\cX|\geq 2$.
We denote the totality of positive probability
distributions on $\cX$ by $\cP=\cP(\cX)$.
A Markov kernel (transition matrices) on $\cX$
is a map $w : \cX^2\rightarrow [0, 1]$ ($(x,y)\mapsto w(y|x)$)
satisfying $\sum_{y\in\cX} w(y|x) =1$ for every $x\in\cX$.
Suppose that we are given a subset $\cE$ of
${\cX}^2 = \cX\times\cX$ for which the
directed graph
$(\cX, \cE)$ is strongly connected; i.e., for any $(x,y)\in{\cX}^2$
there exists a sequence $(x_1, x_2), (x_2, x_3),
\ldots , (x_{n-1}, x_n)$ in $\cE$ such that
$x_1=x, x_n=y, n\geq 2$. Let the totality of
irreducible
Markov kernels $w$ on $\cX$ such that
$\condset{(x,y)}{w(y|x)>0}=\cE$ be denoted by
$\cW = \cW(\cX,\cE)$. This includes the set of strictly positive
Markov kernels $\cW(\cX,{\cX}^2)$ as a special case.
From the irreducibility, each $w\in\cW$ has the unique
stationary distribution in $\cP$, which is denoted by $p_w$.
We introduce the notaion
\[ p_w^{(n)}(x_1, \ldots, x_n)=
p_w(x_1) w(x_2|x_1)\cdots w(x_n|x_{n-1}) \]
($p_w^{(1)}(x)=p_w(x)$ and $p_w^{(2)}(x,y)=
p_w(x) w(y|x)$ in particular),
which will be used in later sections.
A $d$-dimensional parametric
family
\[ M=\condset{w_\theta}{\theta=(\theta^1, \ldots, \theta^d)\in\bRd}
\subset \cW \]
is called an {\em exponential family} (or a {\em full
exponential family} following the terminology of \cite{BN:book})
of Markov kernels on $(\cX,\cE)$
with
a {\em natural (or canonical) parameter} $\theta$, when
there exist functions $C, F^1, \ldots , F^d\, :\cX\rightarrow\bR$,
$K : \cX\times\bRd\rightarrow\bR$ ($(x,\theta)\mapsto K_\theta(x)$)
and $\psi:\bRd\rightarrow\bR$ such that equation (\ref{eq:exp_family_w})
holds for every $\theta\in\bRd$ and every $(x,y)\in\cE$.
When a family of initial distributions, say $\condset{q_\theta}{\theta\in\bRd}$,
is specified, the family of joint probability distributions for
$x^n=(x_1, \ldots , x_n)$ satisfying $(x_t, x_{t+1})\in\cE$ ($\forall t$)
is determined by
\begin{equation}
q^{(n)}_\theta (x^n) = q_\theta (x_1) w_\theta(x_2|x_1)
\cdots w_\theta (x_n|x_{n-1}),
\end{equation}
for which we have from (\ref{eq:exp_family_w})
\begin{align}
\log q^{(n)}_\theta (x^n) = &
\log q_\theta(x_1) +
\sum_{t=1}^{n-1}
\Big\{
C(x_t, x_{t+1})
\nonumber \\
+
\sum_{i=1}^d &\theta^i F(x_t, x_{t+1}) - \psi(\theta)\Big\}
+ K_\theta(x_n) - K_\theta(x_1)
\nonumber \\
= &
\sum_{t=1}^{n-1} C(x_t, x_{t+1})
+
\sum_{i=1}^d \theta^i \sum_{t=1}^{n-1} F(x_t, x_{t+1})
\nonumber \\
& - (n-1) \psi (\theta) + O(1)
\end{align}
This shows that the family $\condset{q^{(n)}_\theta}{\theta\in\bRd}$
is asymptotically an exponential family in the usual sense.
\begin{example}
\label{example1}
\quad
For any $w\in\cW = \cW (\cX, \cX^2)$, we have
\begin{align}
\nonumber
\log w(y|x) & =
\sum_{i=0}^{|\cX|-1}\sum_{j=1}^{|\cX|-1}
\log\frac{w(j|i) w(0|j)}{w(0|i) w(0|0)}
\, \delta_{i,j} (x,y) \\
+& \log w(0|x) - \log w(0|y) + \log w(0|0),
\end{align}
where $\delta_{i,j}(x,y) = 1$ if $(i,j) = (x,y)$ and $=0$ otherwise.
We can easily verify this identity by respectively calculating the RHS for
the two cases $y\neq 0$ and $y=0$. This shows that $\cW$ is an
exponential family with $\dim\cW=|\cX|\cdot(|\cW|-1)$.
\end{example}
\section{Affine structures}
The gist of the definition will be clarified by a fundamental relation
between the affine structure of
functions on $\cE$ and the Markov kernels on $\cE$ as shown below.
Let $\cF = \cF(\cX,\cE)$ be the totality of functions on $\cE$,
which is a $|\cE|$-dimensional linear space.
An element $f$ of $\cF$ is sometimes identified with its
extension to a function on $\cX^2$ by letting $f(x,y)=0$ for
any $(x,y)\in\cX^2\setminus\cE$. For a $f\in\cF$ we define
the functions $f(+,\cdot)$ and $f(\cdot,+)$ on $\cX$ by
\[ f(+,x) = \sum_{y\in\cX} f(y,x) \quad\mbox{and}\quad
f(x,+) = \sum_{y\in\cX} f(x,y).
\]
A function $f\in\cF$ is said to be {\em shift-invariant}
if $f(+,\cdot) = f(\cdot,+)$, and the totality of
shift-invariant functions is denoted by
$\cFS=\cFS (\cX,\cE)$.
A function $\in\cF$ is said to be {\em anti-shift-invariant}
if there exists a function $\kappa$ on $\cX$ such that
$f(x,y) = \kappa(y) - \kappa(x)$ for any $(x,y)\in\cE$,
and the totality of
anit-shift-invariant functions is denoted by
$\cFA=\cFA (\cX,\cE)$.
The linear subspaces $\cFS$ and $\cFA$ of $\cF$
are the orthogonal
complements of each other with respect to the inner
product on $\cF$ defined by
$\langle f_1, f_2\rangle = \sum_{(x,y)\in\cE}
f_1(x,y) f_2(x,y)$, which follows from
\begin{gather*}
\sum_{(x,y)\in\cE} f(x,y) \left(\kappa (y) -\kappa (x)\right) \\
= \sum_{x\in\cX} \left( f(+,x)-f(x,+)\right) \kappa (x).
\end{gather*}
We thus have the direct sum decompositon $\cF= \cFS\oplus\cFA$. In addition,
from the assumption that $(\cX,\cE)$ is strongly
connected, the necessary and sufficient condition for
two functions $\kappa_1, \kappa_2$ to define the
same element of $\cFA$ is that $\kappa_1-\kappa_2$ is
constant on $\cX$. This leads to $\dim\cFA =
|\cX|-1$ and $\dim\cFS = \dim\cF-\dim\cFA=
|\cE|-|\cX|+1$.
Let
\[
\cFpos =\cFpos (\cX,\cE)\defeq
\condset{f}{f:\cE\rightarrow\bR^+}
\subset \cF,
\]
where $\bR^+= (0,\infty)$.
Then we have the following theorem, which is
a direct consequence of the Perron-Frobenius theorem
for irreducible nonnegative matrices.
\begin{theorem}
{\em For any $f\in\cFpos$, there exist
$w\in\cW$, $Z>0$ and $\gamma:\cX\rightarrow\bR^+$ such that
\begin{equation}
\label{eq:normalization}
\forall (x,y)\in\cE, \;\;
w(y|x)=\frac{1}{Z}\, \frac{\gamma(y)}{\gamma(x)}\, f(x,y).
\end{equation}
Here $w$ and $Z$ are unique, and $\gamma$ is unique
up to a constant factor.
}
\end{theorem}
Denoting the correspondence between $f$ and $w$
in (\ref{eq:normalization}) by $w=\Gamma (f)$,
we define the mapping $\Gamma: \cFpos\rightarrow\cW$.
Note that $\cW\subset\cFpos$ and $\Gamma(w)=w$ for
any $w\in\cW$.
Since $\exp(f(x,y))$ defines an element of $\cFpos$ for
every $f$, we obtain a mapping
\begin{align*}
\Delta\defeq\Gamma\circ\exp \,:\,
&\cF\rightarrow\cW \\
&f\mapsto \Gamma(\exp (f)).
\end{align*}
Noting that $w=\Delta(f)=\Gamma(\exp(f))$ is
written as $w(y|x)=\exp(f(x,y))\gamma(y)/Z \gamma(x)$
from (\ref{eq:normalization}), we have
\begin{equation}
\log w(y|x)=f(x,y)+\kappa(y)-\kappa(x) -\psi,
\end{equation}
where $\kappa (x) =\log \gamma(x)$ and $\psi=\log Z$.
This means that $\log w\equiv f$ mod $\cFA\oplus\bR$,
where a real constant is identified with the corresponding
constant function in $\cF$. Moreover, $\Delta$
gives a diffeomorphism from the quotient linear space
$\cF /(\cFA\oplus\bR)$ to $\cW$. Now we have the following theorem.
\begin{theorem}
{\em A subset $M$ of $\cW$ is an exponential family
if and only if there exists an affine subspace $V$ of
$\cF/(\cFA\oplus\bR)$ for which $M=\Delta (V)
\defeq\condset{\Delta(f)}{f\in V}$, and we have
$\dim M=\dim V$. Moreover, the correspondence
between exponential familes and affine subspaces is
one-to-one.
}
\end{theorem}
From this theorem, we have the following
corollaries.
\begin{corollary}
{\em
$\cW$ itself is an exponential family
of dimension $|\cE|-|\cX|$.
}
\end{corollary}
Remember that in Example~\ref{example1} we have verified the
above fact for the case of complete graph $\cE=\cX^2$ by a
little complicated calculation.
\begin{corollary}
{\em
The 1-dimensional exponential family \\
$\condset{w_t}{t\in\bR}$
passing through
given two kernels $w_0, w_1\in\cW$ is written
in the form
$w_t= \Gamma(w_1^t w_0^{1-t})$,
or equivalently
\begin{equation}
w_t(y|x)= \frac{1}{Z_t}\, \frac{\gamma_t(y)}{\gamma_t(x)}\,
\{w_1(y|x)\}^t \{w_0(y|x)\}^{1-t} .
\end{equation}
}
\end{corollary}
A 1-dimensional exponential family is called an {\em e-geodesic}.
\begin{corollary}
{\em
A subset $M$ of $\cW$ is an exponential family if and only if
for any two points $w_0$ and $w_1$ in $M$ the e-geodesic
$\condset{\Gamma(w_1^t w_0^{1-t})}{t\in\bR}$ lies in $M$.
}
\end{corollary}
\section{Fisher information}
For an arbitrary $d$-dimensional parametric family
$M=\condset{w_\theta}{\theta=(\theta^i)\in\Theta}\subset
\cW$, where $\Theta$ is an open subset of ${\bR}^d$, the
Fisher information matrix $G_\theta=[g_{ij}(\theta)] :d\times d$
(with respect to the parameter $(\theta^i)$)
is defined by
\begin{equation}
g_{ij}(\theta) \defeq
\sum_{(x,y)\in\cE}p_\theta(x,y) \left\{\partial_i \log w_\theta(y|x)\right\}
\left\{\partial_j \log w_\theta(y|x)\right\},
\end{equation}
where $p_\theta(x,y)=p_{w_\theta}^{(2)}(x,y)=p_{w_\theta}(x) w_\theta(y|x)$
and $\partial_i=\partial/\partial\theta^i$. This definition is commonly
used because of the following fact.
Let $\condset{q_\theta}{\theta\in\Theta}$ be
an arbitrary family of probability distributions on
$\cX$ (possibly being independent of $\theta$)
and consider the joint distributions
\[
q^{(n)}_\theta (x^n) = q_\theta (x_1) w_\theta(x_2|x_1)
\cdots w_\theta (x_n|x_{n-1}).
\]
Letting $G^{(n)}_\theta=[g^{(n)}_{ij}(\theta)]$ be
the Fisher information matrix (in the usual sense)
of the family $\condset{q^{(n)}_\theta}{\theta\in\Theta}$,
we have
\begin{equation}
g_{ij}(\theta)=\lim_{n\rightarrow\infty}
\frac{1}{n} g^{(n)}_{ij}(\theta).
\end{equation}
From the information-geometric viewpoint,
the Fisher information is regarded as a Riemannian
metric $g$ through the relation $g(\partial_i, \partial_j)
=g_{ij}$
and is called the Fisher metric.
\begin{theorem}
{\em
When $M=\condset{w_\theta}{\theta\in\bR^d}$ is
an exponential family of the form (\ref{eq:exp_family_w}),
the Fisher information matrix with respect to the
natural parameter $(\theta^i)$ is given by
\begin{equation}
g_{ij}(\theta) = \partial_i \partial_j \psi (\theta) .
\end{equation}
}
\end{theorem}
\section{Expectation parameters}
For an exponential family
$M=\condset{w_\theta}{\theta\in\bR^d}$ of
the form (\ref{eq:exp_family_w}), we define
\begin{equation}
\eta_i(\theta)\defeq
\sum_{(x,y)\in\cE} p_{w_\theta}^{(2)}(x,y) F_i(x,y) .
\end{equation}
Then $\eta=(\eta_1, \ldots , \eta_d)$ forms
another coordinate system for $M$, which we call
the {\em expectation parameter} corresponding to
the representation (\ref{eq:exp_family_w}).
\begin{theorem}
{\em
We have:
\begin{gather}
\eta_i = \partial_i\psi, \\
\partial_i\eta_j = g_{ij}, \\
\partial^i\theta^j= g^{ij}, \\
[g^{ij}]=[g_{ij}]^{-1},
\end{gather}
where $\partial^i = \partial/\partial\eta_i$, and
$[g^{ij}]$ denotes the Fisher information matrix
with respect to the dual parameter $(\eta_i)$.
Moreover, if we define $\varphi = \sum_i \theta^i\eta_i
- \psi$, we have
\begin{gather}
\theta^i = \partial^i\varphi, \\
g^{ij}=\partial^i\partial^j\varphi.
\end{gather}
}
\end{theorem}
\section{A dually flat structure}
For an arbitrary $d$-dimensional parametric family
$M=\condset{w_\theta}{\theta=(\theta^i)\in\Theta}\subset
\cW$, the {\em exponential connection} (or {\em e-connection} for short)
$\nabla^{(\e)}$ and the {\em mixture connection} (or {\em m-connection}
for short)
$\nabla^{(\m)}$ are defined as follows.
\begin{align}
\Gamma^{(\e)}_{ij,k} &=
g(\nabla^{(\e)}_{\partial_i}\partial_j, \partial_k)
\nonumber \\
& =
\sum_{(x,y)\in\cE} \partial_i \partial_j\log w_\theta (y|x)
\partial_k p_\theta (x,y),
\\
\Gamma^{(\m)}_{ij,k} &=
g(\nabla^{(\m)}_{\partial_i}\partial_j, \partial_k)
\nonumber \\
& =
\sum_{(x,y)\in\cE} \partial_i \partial_j p_{\theta}(x,y)
\partial_k \log w_\theta (y|x),
\end{align}
where $g$ is the Fisher metric.
Then $\nabla^{(\e)}$ and $\nabla^{(\m)}$ are
dual with respect to $g$ in the sense that $X g(Y, Z) =
g(\nabla^{(\e)}_X Y, Z) + g(Y, \nabla^{(\m)}_X Y)$ holds for
any vector fields $X, Y, Z$.
\begin{theorem}
{\em
For an exponential family
$M=$ \\ $\condset{w_\theta}{\theta\in\bR^d}$ of
the form (\ref{eq:exp_family_w}), both
$\nabla^{(\e)}$ and $\nabla^{(\m)}$ are flat,
and $[\theta^i]$ and $[\eta_i]$ are
affine coordinate systems of these connections,
respectively.
}
\end{theorem}
\begin{theorem}
{\em
Suppose that $M$ is an exponential family
and $N$ is a submanifold of $M$. Then
$N$ is an exponential family if and only if
$N$ is auto-parallel with respect to the
e-connection of $M$.
}
\end{theorem}
Let
\begin{align*}
\cP^{(2){\rm S}} & = \cP(\cE)\cap \cFS \\
= &
\condset{p^{(2)}\in\cP(\cE)}{\forall y,
\sum_x p^{(2)}(x,y) = \sum_xp^{(2)}(y,x)}.
\end{align*}
Then $w\mapsto p^{(2)}_w$ gives a
diffeomorphism from $\cW$ to $\cP^{(2){\rm S}}$.
\begin{theorem}
{\em
The m-connection of $\cW$ is the natural flat connection
induced from the convexity of $\cP^{(2){\rm S}}$.
In particular, the m-geodesic (i.e., the auto-parallel
curve with respect to the m-connection) connecting
given two points $w_0$ and $w_1$ in $\cX$ is
represented as
\begin{equation}
p^{(2)}_{w_t} = t \, p^{(2)}_{w_1} + (1-t) \, p^{(2)}_{w_0}.
\end{equation}
}
\end{theorem}
\section{Canonical Divergence}
Let $M$ be an exponential family of the form
(\ref{eq:exp_family_w}). Then the canonical
divergence $D: M\times M\rightarrow\bR$
with respect to the dually flat structure
$(g, \nabla^{(\m)}, \nabla^{(\e)})$ is defined by
\begin{equation}
D(w_1\,|\,w_2) =
\varphi (w_1) + \psi(w_2) - \sum_{i=1}^d
\eta_i(w_1) \theta^i(w_2).
\end{equation}
The divergence $D$ is also characterized by the
following property: let $\gamma$ be an m-geodsic
such that $\gamma(1)=w_1$ and $\gamma(0)=w_2$,
and $\delta$ be an e-geodsic
such that $\delta(1)=w_3$ and $\delta(0)=w_2$,
which intersect at $w_2$. Then we have
\begin{equation}
D(w_1\,|\,w_2) + D(w_2\,|\,w_3) - D(w_1\,|\,w_3)
= g(\dot{\gamma}(0), \dot{\delta}(0)),
\end{equation}
where the RHS means the inner product between
the tangent vectors of $\gamma$ and $\delta$ at
the intersecting point $w_2$.
\begin{theorem}
{\em
$D$ is represented as
\begin{equation}
D(w_1\,|\,w_2) =
\sum_{(x,y)\in\cE} p_{w_1}^{(2)}(x,y) \log \frac{w_1(y|x)}{w_2(y|x)}.
\end{equation}
}
\end{theorem}
This is nothing but the divergence rate of Markov chains.
Actually, we have
\begin{equation}
D(w_1\,|\,w_2) = \lim_{n\rightarrow\infty}
\frac{1}{n} D(q_1^{(n)}\,|\,q_2^{(n)}),
\end{equation}
where $q_i^{(n)}$, $i=1, 2$ are defined as
\[ q_i^{(n)}(x_1, \ldots , x_n) =
q_i(x_1) w(x_2|x_1)\cdots w(x_n|x_{n-1})
\]
by arbitrary distributions $q_i$ on $\cX$.
\section{Remaining subjects}
The following subjects can also be treated in the present
framework or its obvious extension.
\begin{itemize}
\item Application to the large deviation theory.
\item Some variant of Cram\'{e}r-Rao inequality, and an
estimation-theoretic characterization of
exponential families.
\item Extension to higher-order Markov chains,
and a hierarchy of exponential familes:
$\cP=\cW_0\subset \cW=\cW_1\subset \cW_2\subset \cdots $.
\item Extension to general measureble spaces, and autoregressive
models as an example.
\end{itemize}
| 70,726
|
\begin{document}
\maketitle
\begin{abstract}
Let $G$ be a finite additive abelian group of odd order $n$, and let $G^*=G\setminus\{0\}$ be the set of non-zero elements. A \emph{starter} for $G$ is a set $S=\{\{x_i,y_i\}:i=1,\ldots,\frac{n-1}{2}\}$ such that
\begin{enumerate}
\item $\displaystyle\bigcup_{i=1}^{\frac{n-1}{2}}\{x_i,y_i\}=G^*$, and
\item $\{\pm(x_i-y_i):i=1,\ldots,\frac{n-1}{2}\}=G^*$.
\end{enumerate}
Moreover, if $\left|\left\{x_i+y_i:i=1,\ldots,\frac{n-1}{2}\right\}\right|=\frac{n-1}{2}$, then $S$ is called a \emph{strong starter} for $G$. A starter $S$ for $G$ is a \emph{$k$ quotient starter} if there exists $Q\subseteq G^*$ of cardinality $k$ such that $y_i/x_i\in Q$ or $x_i/y_i\in Q$, for $i=1,\ldots,\frac{n-1}{2}$. In this paper, we give examples of two-quotient strong starters for $\mathbb{F}_q$, where $q=2^kt+1$ is a prime power with $k>1$ a positive integer and
$t$ an odd integer greater than 1.
\end{abstract}
\textbf{Keywords.} Strong starters, two-quotient starters, quadratic residues.
\section{Introduction}
Strong starters were first introduced by Mullin and Stanton in \cite{MR0234587} in constructing of Room squares. Starters and strong starters have been useful to construct many combinatorial designs such as Room cubes \cite{MR633117}, Howell designs \cite{MR728501,MR808085}, Kirkman triple systems \cite{MR808085,MR0314644}, Kirkman squares and cubes \cite{MR833796,MR793636}, and factorizations of complete graphs \cite{MR0364013,MR2206402,MR1010576,MR623318,MR685627}. Moreover, there are some interesting results on strong starters for cyclic groups \cite{MR808085} and for finite abelian groups \cite{MR1010576,MR1044227}.
A starter $S$ is a \emph{$k$ quotient starter} if there exists $Q\subseteq G^*$ of cardinality $k$ such that $y_i/x_i\in Q$ or $x_i/y_i\in Q$, for $i=1,\ldots,\frac{n-1}{2}$, see \cite{dinitz1984}. In particular if $k=1$ the starter $S$ is called \emph{one-quotient} starer for $G$. In fact, an first example of a one-quotient strong starter $S$ was given in \cite{MR0249314} Lemma 1. Further information about quotient starters in \cite{dinitz1984}.
Let $QR(q)$ and $NQR(q)$ denote the set of quadratic residues and the set of non-quadratic residues of the $\mathbb{F}_q^*=\mathbb{F}_q\setminus\{0\}$, respectively. In this work, we prove the following:
\begin{teo}[Main Theorem]\label{thm:main}
Let $q=2^kt+1$ be a prime power with $k>1$ a positive integer and
$t$ an odd integer greater than 1. Then there exists a strong starter $S$ for $\mathbb{F}_q$ which satisfies
$\{a,b\}\in S$ with $a\in QR(q)$ and $b\in NQR(q)$. Furthermore there exists two different elements $\beta_1,\beta_2\in NQR(q)$ such that for every $\{a,b\}\in S$, with $a\in QR(q)$ and $b\in NQR(q)$, we have that $b/a\in\{\beta_1,\beta_2\}$.
\end{teo}
In the known results, there are constructions of strong starters for $\mathbb{F}_q$ (see \cite{MR0325419,dinitz1984,MR0392622,MR0249314,MR0260604}), but none of those constructions gives an explicit construction of strong starters which satisfy the conclutions of the Main Theorem.
This paper is organized as follows.
In Section 2, we recall some basic properties about quadratic residues.
In Section 3, we include an alternative proof when $q\equiv3$ mod 4 and $q\neq3$ (see \cite{MR0249314}). In section 4, we present some previous results. Finally, in Section 5, we prove the main theorem and present some examples.
\section{Quadratic residues}\label{sec:quadratic}
Let $q$ be an odd prime power. An element $x\in\mathbb{F}_q^*$ is called a \emph{quadratic residue} if there exists an element $y\in\mathbb{F}_q^{*}$ such that $y^2=x$. If there is no such $y$, then $x$ is called a \emph{non-quadratic residue.} The set of quadratic residues of $\mathbb{F}_q^{*}$ is denoted by $QR(q)$ and the set of non-quadratic residues is denoted by $NQR(q)$. It is well known that $QR(q)$ is a cyclic subgroup of $\mathbb{F}_q^{*}$ of cardinality $\frac{q-1}{2}$ (see \cite{MR2445243} pg. 87), that is
\begin{teo}
Let $q$ be an odd prime power, then $QR(q)$ is a cyclic subgroup of $\mathbb{F}_q^{*}$. Furthermore, $|QR(q)|=|NQR(q)|=\frac{q-1}{2}$.
\end{teo}
\begin{coro}\label{col:mulres}
Let $q$ be an odd prime power, then
\begin{enumerate}
\item if either $x,y\in QR(q)$ or $x,y\in NQR(q)$, then $xy\in QR(q)$,
\item if $x\in QR(q)$ and $y\in NQR(q)$, then $xy\in NQR(q)$.
\end{enumerate}
\end{coro}
The following theorems are well known results on quadratic residues.
For more details of this kind of results the reader may consult \cite{burton2007elementary} pg. 171, see also \cite{MR2445243}.
\begin{teo}[Eulers' criterion]
Let $q$ be an odd prime and $x\in\mathbb{F}_q^{*}$, then
\begin{enumerate}
\item $x\in QR(q)$ if and only if $x^{\frac{q-1}{2}}=1$.
\item $x\in NQR(q)$ if and only if $x^{\frac{q-1}{2}}=-1$.
\end{enumerate}
\end{teo}
\begin{teo}\label{col:menosuno}
Let $q$ be an odd prime power, then
\begin{enumerate}
\item $-1\in QR(q)$ if and only if $q\equiv1$ mod $4$.
\item $-1\in NQR(q)$ if and only if $q\equiv3$ mod $4$.
\end{enumerate}
\end{teo}
\begin{teo}\label{inverso}
Let q be an odd prime. If $q\equiv1$ mod $4$, then
\begin{enumerate}
\item $x\in QR$ if and only if $-x\in QR$.
\item $x\in NQR$ if and only if $-x\in NQR$.
\end{enumerate}
\end{teo}
\section{Case $q\equiv3$ mod 4, with $q\neq3$}
Now, for the sake of completeness, we include a proof for the case $q\equiv3$ mod 4, with $q\neq3$, using different notation that in \cite{MR0249314}. This notation will be used, of general way, in the main result of this paper.
\begin{lema}
If $q\equiv3$ mod 4 is an odd prime power with $q\neq3$, then there exists a strong starter $S$ for $\mathbb{F}_q$ such that $\{a,b\}\in S$ satisfy that $a\in QR(q)$ and $b\in NQR(q)$.
\end{lema}
\begin{proof}
Let $\alpha$ be a generator of $QR(q)$ and $\beta\in NQR(q)$ such that $\beta+1\neq0$. We claim that the following set:
\begin{eqnarray*}\label{strong_1}
S_\beta=\left\{\{\alpha,\alpha\beta\},\{\alpha^2,\alpha^2\beta\},\ldots,\{\alpha^\frac{p-1}{2},\alpha^\frac{p-1}{2}\beta\}\right\},
\end{eqnarray*}
is a strong starter for $\mathbb{F}_q$. First we have that $\{\alpha,\alpha^2,\ldots,\alpha^{\frac{q-1}{2}}\}=QR(q)$ and $\{\alpha\beta,\alpha^2\beta,\ldots,\alpha^{\frac{q-1}{2}}\beta\}=\beta QR=NQR(q)$ (by Corollary \ref{col:mulres}).
Now we shall prove that $\left\{\pm\alpha^i(\beta-1): i=0,\ldots,\frac{q-1}{2}\right\}=\mathbb{F}_q^*$.
Suppose that $\alpha^i(\beta-1)=\pm\alpha^j(\beta-1)$, then $(\beta-1)(\alpha^i\pm\alpha^j)=0$, for $i\neq j\in\{1,\ldots,\frac{q-1}{2}\}$, which it is a contradiction, since if $i<j$ then $\alpha^i(1+\alpha^{j-i})\neq0$, that is, $1+\alpha^{j-i}\neq0$ (by Theorem \ref{col:menosuno}).
Finally, we have that $\left|\{\alpha^i(\beta+1): i=0,\ldots,\frac{q-1}{2}\}\right|=\frac{q-1}{2}$, since if $i\neq j\in\{1,\ldots,\frac{q-1}{2}\}$ then $\alpha^i(\beta+1)\neq\alpha^j(\beta+1)$.
\end{proof}
To end this section, it is easy to see that $S_\beta$ is an example of one quotient strong starter for $\mathbb{F}_q$. This kind of starters are called \emph{Dinitz starters} for $\mathbb{F}_q$, see \cite{book:206537}, Theorem VI.55.22, page 624.
\section{Previous results}\label{sec:particular}
To begin with, we introduce some terminology in order to simplify the description of the of existence of strong starters $S$ for $\mathbb{F}_q$ with the property that if $\{a,b\}\in S$, then $a\in QR(q)$ and $b\in NQR(q)$.
Let $q=2^kt+1$ be a prime power with $k>1$ a positive integer and
$t$ an odd integer greater than 1 and $\alpha$ be a generator of $QR(q)$. We define $C_0=\langle\alpha^\Delta\rangle$, where $\Delta=2^{k-1}$, to be the subgroup of $\mathbb{F}_q^*$ of order $t$. Let
$C_j=\alpha^jC_0$, for $j
=1,\ldots,\Delta_1-1$ with $\Delta_1=2^{k-1}$, and $\hat{C}_j=-C_j$, for $j=0,\ldots,\Delta_1-1$ . Hence $QR(q)=\bigcup_{j=0}^{\Delta_1-1}(C_j\cup\hat{C}_j)$. On the other hand, let $\beta_1\in NQR(q)$ and $\beta_2\in\beta_1\hat{C}_0$. We define $D_j=\beta_1 C_j$ and $\hat{D}_j=\beta_2 C_j$, for $j=0,\ldots,\Delta_1-1$. Hence $NQR(q)=\bigcup_{i=0}^{\Delta_1-1}(D_j\cup \hat{D}_j)$. Moreover, it is easy to see that $\sum_{a\in C_j}a=0$, $\sum_{a\in \hat{C}_j}a=0$, $\sum_{a\in D_j}a=0$ and $\sum_{a\in \hat{D}_j}a=0$, for $j=1,\ldots,\Delta_1-1$.
To prove the main theorem of this paper (see Theorem \ref{thm:main}), we need to prove the following auxiliary lemma, which states the condition of existence of strong starters $S$ for $\mathbb{F}_q$ with the property that if $\{a,b\}\in S$, then $a\in QR(q)$ and $b\in NQR(q)$, and wthe proof of this lemma is obtained from Lemmas \ref{lemma:completo} and \ref{lemma:(beta+1)(beta-1)}.
\begin{lema}\label{coro:completo}
Let $q=2^kt+1$ be a prime power with $k>1$ a positive integer and
$t$ an odd integer greater than 1. Then there exist $\beta_1\in NQR(q)$ and $\beta_2\in\beta_1\hat{C}_0$, such that $(\beta_1-1)(\beta_2+1)\in NQR(q)$ and $(\beta_1+1)(\beta_2-1)\in NQR(q)$.
\end{lema}
We present the sketch of the proof of Main Theorem: Let $\alpha$ be a gene\-rator of $QR(q)$, $\beta_1\in NQR(q)$ and $\beta_2\in\beta_1\hat{C}_0$ such that $(\beta_1-1)(\beta_2+1)\in NQR(q)$ and $(\beta_1+1)(\beta_2-1)\in NQR(q)$ (by Lemma \ref{coro:completo}), then the following set
$S(\beta_1,\beta_2)=\displaystyle\bigcup_{i=0}^{\Delta_1-1}S(\beta_1,\beta_2)_j$ is a strong starter for $\mathbb{F}_q$, where
$S(\beta_1,\beta_2)_{j}=\{\{x,\beta_1x\},\{y,-\beta_2y\}:x\in C_j,y\in\hat{C}_j\}$, for $j=0,\ldots,\Delta_1-1$. We have that $QR(q)=\bigcup_{j=0}^{\Delta_1-1}\{\{x\}\cup\{y\}:x\in C_j,y\in\hat{C}_j\}$ and $NQR(q)=\bigcup_{j=0}^{\Delta_1-1}\{\{\beta_1x\}\cup\{-\beta_2y\}:x\in C_j,y\in\hat{C}_j\}$. Moreover, if $\{a,b\}\in S(\beta_1,\beta_2)$, then $a/b\in\{\beta_1,-\beta_2\}$ or $b/a\in\{\beta_1,-\beta_2\}$, which impliest that $S(\beta_1,\beta_2)$ is a two-quotient strong starter for $\mathbb{F}_q$. Moreover, if $\{a,b\}\in S(\beta_1,\beta_2)$, then $a/b\in\{\beta_1,-\beta_2\}$ or $b/a\in\{\beta_1,-\beta_2\}$, which impliest that $S(\beta_1,\beta_2)$ is a two-quotient strong starter for $\mathbb{F}_q$.
To prove the following lemma, we needs the next definition: Let $q=ef+1$ be a prime power and let $H$ be the subgroup of $\mathbb{F}_q^*$ of order $f$ with $\{H=C_0,\ldots,C_{e-1}\}$ the set of (multiplicative) cosets of $H$ in $\mathbb{F}_q^*$ (that is, $C_i = g^iC_0$, where $g$ is the least primitive element of $\mathbb{F}_q$). The cyclotomic number
$(i,j)$ is $|\{x\in C_i: x+1\in C_j\}|$. In particular, if $e=2$ and $f$ is even, then $(0,0)=\frac{f-2}{2}$, $(0,1)=\frac{f}{2}$, $(1,0)=\frac{f}{2}$ and $(1,1)=\frac{f}{2}$, see \cite{book:206537}, Table VII.8.50. Hence if $C_0=QR(q)$ and $C_1=NQR(q)$ are the cosets of $QR(q)$ in $\mathbb{F}_q^*$, then we have the following:
\begin{lema}\label{lemma:1+beta}
Let $q=2^kt+1$ be a prime power with $k>1$ a positive integer and
$t$ an odd integer greater than 1. Then there exist $\beta_1,\beta_2\in NQR(q)$ such that
\begin{enumerate}
\item $(\beta_1+1)\in NQR(q)$ and $(\beta_2+1)\in QR(q)$.
\item $(\beta_1-1)\in NQR(q)$ and $(\beta_2-1)\in QR(q)$.
\end{enumerate}
\end{lema}
\begin{lema}\label{lemma:completo}
Let $q=2^kt+1$ be a prime power with $k>1$ a positive integer and
$t$ an odd integer greater than 1. Then there exists $\beta\in NQR(q)$ such that $(\beta+1)(\beta-1)\in NQR(q)$.
\end{lema}
\begin{proof}
For each $\beta\in NQR(q)$ define $A_\beta=\{a^\beta_1,\ldots,a^\beta_{l_\beta}\}$, where $a^\beta_{i}\in NQR(q)$ and $a^\beta_{i+1}=a^\beta_{i}+1$, for all $i=1, \ldots,l_{\beta}-1$.
By Lemma \ref{lemma:1+beta}, there exists $\beta\in NQR(q)$ such that $|A_\beta|>1$.
If $\beta^*=a^\beta_{l_\beta}$, then $(\beta^*+1)\in QR(q)$ and $(\beta^*-1)\in NQR(q)$.
On the other hand, if $\beta^*=\beta_1$, then $(\beta^*+1)\in NQR(q)$ and $(\beta^*-1)\in QR(q)$. Hence $(\beta^*+1)(\beta^*-1)\in NQR(q)$.
\end{proof}
\begin{remark}
According with the proof of Lemma \ref{lemma:completo}, if there exists $\beta\in NQR(q)$ such that $|A_\beta|=1$, then $(\beta+1)\in QR(q)$ and $(\beta-1)\in QR(q)$. Moreover, if there exists $\beta\in NQR(q)$ such that $|A_\beta|>2$, then $\beta^*=a^\beta_2$ is such that $(\beta^*+1)\in NQR(q)$ and $(\beta^*-1)\in NQR(q)$.
\end{remark}
\begin{lema}\label{lemma:(beta+1)(beta-1)}
Let $q=2^kt+1$ be a prime power with $k>1$ a positive integer and
$t$ an odd integer greater than 1. Then given $\beta\in NQR(q)$ there exist $\beta_1,\beta_2\in\beta\hat{C}_0$, such that $(\beta_1-1)(\beta_2-1)\in NQR(q)$.
\end{lema}
\begin{proof}
Suppose that given $\beta_1\in\beta\hat{C}_0$ is such that $(\beta_1-1)\in QR(q)$; the same argument is used if we suppose that $(\beta_1-1)\in NQR(q)$. Hence $(\beta_1-1)^{\frac{q-1}{2}}=(\beta_1-1)^{\Delta t}=1$. Let $A_0=\beta\hat{C}_0$, then
\begin{eqnarray*}
t&=&\displaystyle\sum_{\beta_1\in A_0}(\beta_1-1)^{\Delta t}=\displaystyle\sum_{x\in C_0}(-\beta x-1)^{\Delta t}\\
&=&\displaystyle\sum_{x\in C_0}(\beta x+1)^{\Delta t}=\displaystyle\sum_{x\in C_0}\displaystyle\sum_{i=0}^{\Delta t}\binom {\Delta t} {i}\beta^ix^i\\
&=&\sum_{i=0}^{\Delta t}\binom {\Delta t} {i}\beta^i\displaystyle\sum_{x\in C_j}x^i,
\end{eqnarray*}
where
\begin{equation*}
\displaystyle\sum_{x\in C_j}x^i=
\left\{
\begin{array}{ll}
t & \hbox{if $i=0,t,\ldots,\Delta t$,}\\
0 & \hbox{otherwise.}
\end{array}
\right.
\end{equation*}
Then $$1=\sum_{s=0}^{\Delta}\binom {\Delta t} {st}\beta^{st}, \mbox{ for all $\beta\in NQR(q)$}.$$
Hence, if $NQR(q)=\{\beta_1,\ldots,\beta_{\Delta t}\}$, then
$$\Delta t=\sum_{s=0}^{\Delta}\binom {\Delta t} {st}\sum_{k=1}^{\Delta t}\beta_k^{st}$$ Since
\begin{equation*}
\displaystyle\sum_{k=1}^{\Delta t}\beta^{s t}_k=
\left\{
\begin{array}{ll}
\Delta t & \hbox{if $s=0$,}\\
-\Delta t & \hbox{if $s=\Delta$,}\\
0 & \hbox{otherwise,}
\end{array}
\right.
\end{equation*}then $$\Delta t=\sum_{s=0}^{\Delta}\binom {\Delta t} {st}\sum_{k=1}^{\Delta t}\beta_k^{st}=0,$$which it is a contradiction. Furthermore, there exist $\beta_1,\beta_2\in \beta\hat{C}_0$, such that $(\beta_1-1)(\beta_2-1)\in NQR(q)$.
\end{proof}
\begin{coro}\label{coro:(beta+1)(beta-1)}
Let $q=2^kt+1$ be a prime power with $k>1$ a positive integer and
$t$ an odd integer greater than 1. Then given $\beta\in NQR(q)$ there exist $\beta_1,\beta_2\in\beta\hat{C}_0$ such that $(\beta_1+1)(\beta_2+1)\in NQR(q)$.
\end{coro}
\section{Proof of the main theorem}\label{sec:main}
With the results presented before, we are ready to prove the main theorem, Theorem \ref{thm:main}, of this paper:
\begin{proof}
Let $\alpha$ be a generator of $QR(q)$, $\beta_1\in NQR(q)$ and $\beta_2\in\beta_1\hat{C}_0$ such that $(\beta_1-1)(\beta_2+1)\in NQR(q)$ and $(\beta_1+1)(\beta_2-1)\in NQR(q)$ (by Lemma \ref{coro:completo}). We claim that the following set
$S(\beta_1,\beta_2)=\bigcup_{i=0}^{\Delta_1-1}S(\beta_1,\beta_2)_j$ is a strong starter for $\mathbb{F}_q$,where $S(\beta_1,\beta_2)_{j}=\{\{x,\beta_1x\},\{y,-\beta_2y\}:x\in C_j,y\in\hat{C}_j\}$, for $j=0,\ldots,\Delta_1-1$. We have that $QR(q)=\bigcup_{j=0}^{\Delta_1-1}\{\{x\}\cup\{y\}:x\in C_j,y\in\hat{C}_j\}$ and $NQR(q)=\bigcup_{j=0}^{\Delta_1-1}\{\{\beta_1x\}\cup\{-\beta_2y\}:x\in C_j,y\in\hat{C}_j\}$. Moreover, if $\{a,b\}\in S(\beta_1,\beta_2)$, then $a/b\in\{\beta_1,-\beta_2\}$ or $b/a\in\{\beta_1,-\beta_2\}$, which impliest that $S(\beta_1,\beta_2)$ is a two-quotient strong starter for $\mathbb{F}_q$.
First we shall prove that $\mathbb{F}_q^*=\bigcup_{j=0}^{\Delta_1-1} (E_j\cup E^*_j)$, where
\begin{center}
$E_j=\left\{\pm x(\beta_1-1):x\in C_j\right\}$
\vspace{.2cm}
$E_j^*=\left\{\pm y(\beta_2+1):y\in \hat{C}_j\right\}$
\end{center}
for all $j=0,\ldots, \Delta_1-1$.
\begin{itemize}
\item [Case (i):] If $x_j(\beta_1-1)=\pm x_j^\prime(\beta_1-1)$, for $x_j\neq x_j^\prime\in C_j$,
then $(\beta_1-1)(x_j\pm x_j^\prime)=0$, which it is a contradiction, since $x_j+x_j^\prime\neq0$.
\item [Case (ii):] If $x_j(\beta_1-1)=\pm x_i(\beta_1-1)$,
for $x_j\in C_j$ and $x_i\in C_i$, then\\ $(\beta_1-1)(x_j\pm x_i)=0$, which it is a contradiction, since $C_j\cap C_i=\emptyset$, for $i\neq j\in\{0,\ldots,\Delta_1-1\}$ and $x_j+x_i\neq0$.
\item [Case (iii):] If $y_j(\beta_2+1)=\pm y_j^\prime(\beta_2+1)$, for $y_j\neq y_j^\prime\in \hat{C}_j$, then $(\beta_2+1)(y_j\pm y^\prime_j)=0$, which it is a contradiction, since $y_j+y^\prime_j\neq0$.
\item [Case (iv):] If $y_j(\beta_2+1)=\pm y_i(\beta_2+1)$, for $y_j\in \hat{C}_j$ and $y_i\in \hat{C}_i$, then\\ $(\beta_2+1)(y_j\pm y_i)=0$, which it is a contradiction, since $\hat{C}_j\cap \hat{C}_i=\emptyset$, for $i\neq j\in\{0,\ldots,\Delta_1-1\}$ and $y_j+y_i\neq0$.
\item [Case (v):] As $(\beta_1-1)(\beta_2+1)\in NQR(q)$ then $x_j(\beta_1-1)\neq\pm y_i(\beta_2+1)$, for $x_j\in C_j$ and $y_i\in \hat{C}_i$, since either $x_j(\beta_1-1)\in NQR(q)$ and $y_i(\beta_2+1)\in QR(q)$ or $x_j(\beta_1-1)\in QR(q)$ and $y_i(\beta_2+1)\in NQR(q)$.
\end{itemize}
Now, we shall prove that
$\left|\displaystyle\bigcup_{j=0}^{\Delta_1-1}(P_j\cup P_j^*)\right|=\Delta t$, where
\begin{center}
$P_j=\left\{x(\beta_1+1):x\in C_j\right\}$
$P_j^*=\left\{-y(\beta_2-1):y\in \hat{C}_j\right\}$
\end{center}
for all $j=0,\ldots,\Delta_1-1$.
\begin{itemize}
\item [Case (i):] If $x_j(\beta_1+1)=x_j^\prime(\beta_1+1)$, for $x_j\neq x_j^\prime\in C_j$, then $(\beta_1+1)(x_j-x_i)=0$, which it is a contradiction.
\item [Case (ii):] If $x_j(\beta_1+1)=x_i(\beta_1+1)$, for $x_j\in C_j$ and $x_i\in C_i$, then\\ $(\beta_1+1)(x_j-x_i)=0$, which it is a contradiction, since $C_i\cap C_j=\emptyset$, for $i\neq j\in\{0,\ldots,\Delta_1-1\}$.
\item [Case (iii):] If $-y_j(\beta_2-1)=-y^\prime_j(\beta_2-1)$, for $y_j\neq y_j^\prime\in \hat{C}_j$, then $(\beta_2-1)(y_j-y^\prime_j)=0$, which it is a contradiction.
\item [Case (iv):] If $-y_j(\beta_2-1)=-y_i(\beta_2-1)$, for $y_j\in \hat{C}_j$ and $y_i\in \hat{C}_i$, then\\ $(\beta_2-1)(y_j-y_i)=0$, which it is a contradiction, since $\hat{C}_j\cap \hat{C}_i\emptyset$, for $i\neq j\in\{0,\ldots,\Delta_1-1\}$.
\item [Case (v):] As $(\beta_1+1)(\beta_2-1)\in NQR(q)$, then $x_j(\beta_1+1)\neq -y_i(\beta_2-1)$, for $x_j\in C_j$ and $y_i\in \hat{C}_i$, since either $x_j(\beta_1+1)\in NQR(q)$ and $-y_i(\beta_2-1)\in QR(q)$ or $x_j(\beta_1+1)\in QR(q)$ and $-y_i(\beta_2-1)\in NQR(q)$.
\end{itemize}
To end, it is not difficult to prove that $\beta_1\neq\beta_2$, since if $\beta_1=-\beta_1x$, for some $x\in C_0$, then $\beta_1(1+x)=0$, which is a contradiction, since $-1\in \hat{C}_0$.
\end{proof}
\begin{coro}\label{coro:betas}
If $S(\beta_1,\beta_2)$ is a two-quotient strong starter for $\mathbb{F}_q$ given by Theorem \ref{thm:main}, then $S(\beta_2,\beta_1)$, $S(-\beta_1,-\beta_2)$ and $S(-\beta_2,-\beta_1)$ are
two-quotient strong starters for $\mathbb{F}_q$ different from $S(\beta_1,\beta_2)$.
\end{coro}
Let $q=4t+1$ be a prime power with an odd integer greater than 1, $C_0\subseteq\mathbb{F}_q^*$ be the subgroup of order $t$ and $C_0,C_1,C_2,C_3$ be the multiplicative cosets of $C_0$. In \cite{dinitz1984} was proven then the following set
$$\hat{S}(a_0,a_1)=\{\{x,a_0x\},\{y,a_1y\}:x\in C_0^{a_0},y\in \hat{C}_1^{a_1}\},$$where $C_0^{a_0}=1/(a_0-1)C_0$ and $C_1^{a_1}=1/(a_1-1)C_1$, is a two-quotient strong starter for $\mathbb{F}_q$. Hence if $\beta_1\in NQR(q)$ and $\beta_2\in\beta_1\hat{C}_0$ are such that $(\beta_1-1)(\beta_2+1)\in NQR(q)$, $(\beta_1+1)(\beta_2-1)\in NQR(q)$ and $(\beta_0-1)\in C_0$ and $-(\beta_1-1)\in \hat{C}_0$, then
\begin{eqnarray*}
\hat{S}(\beta_0,-\beta_1)&=&\{\{x,\beta_0x\},\{y,-\beta_1y\}:x\in C_0^{\beta_0},y\in\hat{C}_0^{\beta_1}\}\\
&=&\{\{x,\beta_0x\},\{y,-\beta_1y\}:x\in C_0,y\in \hat{C}_0\}\\
&=&S(\beta_0,\beta_1).
\end{eqnarray*}
\subsection{Examples}
In this subsection we give examples of strong starters for $\mathbb{F}(29)$ and $\mathbb{F}(41)$ given by Theorem \ref{thm:main} (main theorem).
Let $\mathbb{F}_{29}=\mathbb{Z}_{29}$, then $k=2$, $t=7$, $\Delta=2$, $\Delta_1=0$ and $\alpha=4$. We have
\begin{eqnarray*}
QR(29)&=&\{1,4,5,6,7,9,13,16,20,22,23,24,25,28\}\mbox{ and}\\
NQR(29)&=&\{2,3,8,10,11,12,14,15,17,18,19,21,26,27\}.
\end{eqnarray*}
Hence $C_0=\{1,7,16,20,23,24,25\}$ and $\hat{C}_0=\{4,5,6,9,13,22,28\}$. If $\beta_1=2$ and $\beta_2=26\in\beta_1\hat{C}_0=\{2,10,11,15,17,21,26\}$, then $(\beta_1-1)(\beta_2+1)\in NQR(29)$ and $(\beta_1+1)(\beta_2-1)\in NQR(29)$. Therefore $S(2,26)=S(2,26)_0$, where
\begin{eqnarray*}
S(2,26)_0&=&\{\{16,3\},\{13,10\}\}\cup\{\{24,19\},\{5,15\}\}\cup\{\{7,14\},\{22,8\}\}\\
&\cup&\{\{25,21\},\{4,12\}\}\cup\{\{23,17\},\{6,18\}\}\\
&\cup&\{\{20,11\},\{9,27\}\}\cup\{\{1,2\},\{28,26\}\}
\end{eqnarray*}
is a two-quotient strong starter for $\mathbb{F}_{29}$. Moreover, by Corollary \ref{coro:betas}, we see that $S(26,2)$, $S(27,3)$ and $S(3,27)$ are two-quotient strong starters in $\mathbb{F}_{29}$ different from $S(2,26)$:
\begin{itemize}
\item $S(26,2)=S(26,1)_0$, where
\begin{eqnarray*}
S(26,1)_0&=&\{\{16,10\},\{13,3\}\}\cup\{\{24,15\},\{5,19\}\}\cup\{\{7,8\},\{22,14\}\}\\
&\cup&\{\{25,12\},\{4,21\}\}\cup\{\{23,18\},\{6,17\}\}\\
&\cup&\{\{20,27\},\{9,11\}\}\cup\{\{1,26\},\{28,2\}\}
\end{eqnarray*}
\item $S(27,3)=S(27,3)_0$, where
\begin{eqnarray*}
S(27,3)_0&=&\{\{16,26\},\{13,19\}\}\cup\{\{24,10\},\{5,14\}\}\cup\{\{7,15\},\{22,21\}\}\\
&\cup&\{\{25,8\},\{4,17\}\}\cup\{\{23,12\},\{6,11\}\}\\
&\cup&\{\{20,18\},\{9,2\}\}\cup\{\{1,27\},\{28,3\}\}\\
\end{eqnarray*}
\item $S(3,27)=S(3,27)_0$, where
\begin{eqnarray*}
S(3,27)_0&=&\{\{16,19\},\{13,26\}\}\cup\{\{24,14\},\{5,10\}\}\cup\{\{7,21\},\{22,15\}\}\\
&\cup&\{\{25,17\},\{4,8\}\}\cup\{\{23,11\},\{6,12\}\}\\
&\cup&\{\{20,2\},\{9,18\}\}\cup\{\{1,3\},\{28,27\}\}\\
\end{eqnarray*}
\end{itemize}
It can be verified that all of the starters in the following table are indeed two-quotient strong starters for $\mathbb{F}_{29}$:
\begin{table}[h!]
\centering
\begin{tabular}{|c c c c|}
\hline
$S(\beta_1,\beta_2)$ & $S(\beta_2,\beta_1)$ & $S(-\beta_1,-\beta_2)$ & $S(-\beta_2,-\beta_1)$ \\
[0.5ex]
\hline
$S(2,26)$ & $S(26,2)$ & $S(27,3)$ & $S(3,27)$ \\
$S(2,10)$ & $S(10,2)$ & $S(27,19)$ & $S(19,27)$ \\
$S(3,15)$ & $S(15,3)$ & $S(26,14)$ & $S(14,26)$ \\
$S(3,12)$ & $S(12,3)$ & $S(26,17)$ & $S(17,26)$ \\
$S(8,11)$ & $S(11,8)$ & $S(21,18)$ & $S(18,21)$ \\
$S(10,14)$& $S(14,10)$& $S(19,15)$ & $S(15,19)$ \\
$S(10,17)$& $S(17,10)$& $S(19,12)$ & $S(12,19)$\\
\hline
\end{tabular}
\end{table}
As a second example, let $\mathbb{F}_{41}=\mathbb{Z}_{41}$, then $k=3$, $t=5$, $\Delta=4$, $\Delta_1=2$ and $\alpha=36$. We have
\begin{eqnarray*}
QR(41)&=&\{1,2,4,5,8,9,10,16,18,20,21,23,25,31,32,33,36,37,39,40\},\mbox{ and}\\
NQR(41)&=&\{3,6,7,11,12,13,14,15,17,19,22,24,26,27,28,29,30,34,35,38\}.
\end{eqnarray*}
Hence $C_0=\{10,18,16,37,1\}$, $\hat{C}_0=\{31,23,25,4,40\}$,
$C_1=\{32,33,2,20,36\}$ and $\hat{C}_1=\{9,8,39,21,5\}$. $\beta_1=3$ then $\beta_2=12\in\beta_1\hat{C}_0=\{11,28,34,12,38\}$ is such that $(\beta_1-1)(\beta_2+1)\in NQR(41)$ and $(\beta_1+1)(\beta_2-1)\in NQR(41)$.Therefore $S(3,12)=S(3,12)_0\cup S(3,12)_1$, where
\begin{eqnarray*}
S(3,12)_0&=&\{\{10,30\},\{31,38\}\}\cup\{\{18,13\},\{23,11\}\}\cup\{\{16,7\},\{25,28\}\}\\
&\cup&\{\{37,29\},\{4,34\}\}\cup\{\{1,3\},\{40,12\}\}\\
S(3,12)_1&=&\{\{32,14\},\{9,15\}\}\cup\{\{33,17\},\{8,27\}\},\{\{2,6\},\{39,24\}\}\\
&\cup&\{\{20,19\},\{21,35\}\}\cup\{\{36,26\},\{5,22\}\}
\end{eqnarray*}
is a two-quotient strong starter for $\mathbb{F}_{41}$.
Moreover, by Corollary 4.8, we see that $S(12,3)$, $S(38,29)$ and $S(29,38)$ are two-quotient strong starters in $\mathbb{F}_{41}$ different from $S(3,12)$:
\begin{itemize}
\item $S(12,3)=S(12,3)_0\cup S(12,3)_1$, where
\begin{eqnarray*}
S(12,3)_0&=&\{\{10,38\},\{31,30\}\}\cup\{\{18,11\},\{23,13\}\}\cup\{\{16,28\},\{25,7\}\}\\
&\cup&\{\{37,34\},\{4,29\}\}\cup\{\{1,12\},\{40,3\}\}\\
S(12,3)_1&=&\{\{32,15\},\{9,14\}\}\cup\{\{33,27\},\{8,17\}\}\cup\{\{2,24\},\{39,6\}\}\\
&\cup&\{\{20,35\},\{21,19\}\}\cup\{\{36,22\},\{5,26\}\}\\
\end{eqnarray*}
\item $S(38,29)=S(38,29)_0\cup S(38,29)_1$, where
\begin{eqnarray*}
S(38,29)_0&=&\{\{10,11\},\{31,3\}\}\cup\{\{18,28\},\{23,30\}\}\cup\{\{16,34\},\{25,13\}\}\\
&\cup&\{\{37,12\},\{4,7\}\}\cup\{\{1,38\},\{40,29\}\}\\
S(38,29)_1&=&\{\{32,27\},\{9,26\}\}\cup\{\{33,24\},\{8,14\}\}\cup\{\{2,35\},\{39,17\}\}\\
&\cup&\{\{20,22\},\{21,6\}\}\cup\{\{36,15\},\{5,19\}\}\\
\end{eqnarray*}
\item $S(29,38)=S(29,38)_0\cup S(29,38)_1$, where
\begin{eqnarray*}
S(28,39)_0&=&\{\{10,3\},\{31,11\}\}\cup\{\{18,30\},\{23,28\}\}\cup\{\{16,13\},\{25,34\}\}\\
&\cup&\{\{37,7\},\{4,12\}\}\cup\{\{1,29\},\{40,38\}\}\\
S(28,39)_1&=&\{\{32,26\},\{9,27\}\}\cup\{\{33,14\},\{8,24\}\}\cup\{\{2,17\},\{39,35\}\}\\
&\cup&\{\{20,6\},\{21,22\}\}\cup\{\{36,19\},\{5,15\}\}\\
\end{eqnarray*}
\end{itemize}
It can be verified that all of the starters in the following table are indeed two-quotient strong starters for $\mathbb{F}_{41}$:
\begin{table}[h!]
\centering
\begin{tabular}{|c c c c|}
\hline
$S(\beta_1,\beta_2)$ & $S(\beta_2,\beta_1)$ & $S(-\beta_1,-\beta_2)$ & $S(-\beta_2,-\beta_1)$ \\
[0.5ex]
\hline
$S(3,12)$ & $S(12,3)$ & $S(38,29)$ & $S(29,38)$ \\
$S(3,28)$ & $S(28,3)$ & $S(38,13)$ & $S(13,38)$ \\
$S(14,24)$& $S(24,14)$& $S(27,17)$ & $S(17,27)$ \\
$S(14,22)$& $S(22,14)$& $S(27,19)$ & $S(19,27)$\\
\hline
\end{tabular}
\end{table}
\
{\bf Acknowledgment}
\
The authors thank the referee for many constructive suggestions to improve
this paper.
C. R. research supported in part by a CONACyT-M{\' e}xico Postdoctoral fe\-llowship and in part by the National scholarship programme of the Slovak republic. C. A. and A.V. supported by SNI and CONACyT.
\bibliographystyle{amsplain}
| 58,753
|
STATE of Connecticut v. DANIEL G.
The defendant, Daniel G., appeals from the judgment of conviction, rendered after a jury trial, of increasing the speed of a motor vehicle in an attempt to escape or elude a police officer in violation of General Statutes § 14–223(b) and interfering with a police officer in violation of General Statutes § 53a–167a. On appeal, the defendant claims that (1) the evidence was insufficient to support his conviction under § 14–223(b); (2) § 14–223(b) is unconstitutionally vague as applied to the facts of this case; (3) the trial court improperly failed to charge the jury on two theories of defense; and (4) the defendant was deprived of the right to a fair trial as a result of prosecutorial impropriety. We are not persuaded and, accordingly, affirm the judgment of the trial court.
The jury reasonably could have found the following facts. On April 23, 2009, at approximately 4:45 p.m., New London police Officer Deana Nott responded to a motor vehicle accident and drove her police cruiser to a CVS parking lot located on Jefferson Street. After speaking with the individuals involved in the accident, including Dustin Colburn, Nott returned to her cruiser to complete some paperwork. Nott noticed a white van pull into a parking space in the CVS parking lot and saw the defendant exit with a small child. A few moments later, while approaching Colburn, the defendant commented on Nott's abilities as a police officer. The defendant asked Colburn if Nott was issuing Col-burn a “ticket․” The defendant then said that Nott was “[o]n the wrong end of a lawsuit” and asked Col-burn for his name and telephone number. At this point, Nott exited her cruiser and instructed the defendant to “step away” several times. The defendant ignored Nott's instructions and continued his comments about her. Nott determined that she could not complete her investigation of the motor vehicle accident as a result of the defendant's actions, and requested assistance from her fellow police officers. Upon hearing the response to Nott's request, the defendant entered his van and departed from the CVS parking lot onto Jefferson Street.
When requesting assistance, Nott had spoken with Todd Bergeson, a sergeant in the New London Police Department and the acting shift supervisor. Nott asked Bergeson to initiate a motor vehicle stop and issue the defendant “a ticket for interfering or ․ creating a disturbance.” Bergeson was located nearby at a funeral home. Bergeson observed the white van exiting from the CVS parking lot and followed it, turning on his vehicle's overhead lights and police siren. The defendant proceeded up Wall Street while Bergeson was directly behind him with the overhead lights and police siren turned on. The defendant turned onto Summer Street, then Redden Avenue, then Colman Street and into his residence. Bergeson parked his cruiser at the front of the residence and proceeded to the rear of the residence on foot. Bergeson observed the defendant and his daughter in the van. He ordered the defendant to exit the van and informed him that he was under arrest. At some point, the defendant began to exit the van, and Bergeson pointed his Taser at the defendant. The defendant returned to the interior of the van, shutting and locking the door. The defendant placed a telephone call to the police dispatcher, requesting to speak to either a captain or lieutenant, and claiming that he and his daughter were being threatened by the police. The dispatcher informed the defendant that Bergeson was acting lieutenant. The defendant then exited the van and was taken into custody without further incident.
In an amended substitute information, the defendant was charged with risk of injury to a child in violation of General Statutes § 53–21(a)(1), increasing speed in an attempt to escape or elude a police officer in violation of § 14–223(b) and two counts of interfering with an officer in violation of § 53a–167a. Following the presentation of the state's case, the defendant moved for a judgment of acquittal as to all four counts. Count one of the amended substitute information alleged that the defendant had obstructed and hindered Nott in the performance of her duties in the CVS parking lot “by repeatedly yelling at the parties to a motor vehicle crash and interfering with ․ Nott's investigation of the accident in which the defendant was not involved․” The court granted the defendant's motion for a judgment of acquittal with respect to count one. Referring to our Supreme Court's decision in State v. Williams, 205 Conn. 456,473,534A.2d230 (1987),1 the court ruled: “When, in fact, one is dealing with [a] first amendment question and the right of expression, the interfering statute is restricted to what is known as fighting words, either words directed at the officer or to those in the area such that they are—their very nature would require or ordinarily cause one to react in a negative fashion, in fact, one of violence. That is, clearly, not what happened here. As for the verbal exchange, it is correct that the verbal exchange was not even with [Nott], according to the evidence that has been presented. As far as the claim that the verbal exchange interfered with [Nott's] ability, I see no evidence of that. I see no specific testimony that would indicate that the conversation between the witness and the defendant interfered, prevented [Nott], or even delayed [Nott] in any way. There may have been a subsequent argument with [Nott]; but, again, citing Williams, that is not sufficient for a charge of interfering with a police officer.” The court denied the remainder of the defendant's motion.
Following the conclusion of the trial, the jury found the defendant not guilty of risk of injury, guilty of attempting to escape or elude a police officer and guilty of interfering with an officer. The interfering with an officer count was based on the defendant's locking the van door and preventing Bergeson from arresting him following the police pursuit. The court denied the defendant's postverdict motions and rendered judgment in accordance with the jury's verdict. The court sentenced the defendant to one year of incarceration, execution suspended, two years of probation and a $500 fine. This appeal followed.
I
The defendant first claims that the evidence was insufficient to support his conviction of attempting to escape or elude a police officer. Specifically, he argues that a video of the pursuit, as captured by a dashboard camera in Bergeson's vehicle, “unequivocally demonstrated that the defendant never increased his speed or tried to elude Bergeson.”2 The state counters that the video does not conclusively establish the facts so as to have preclusive effect over the testimony of the witnesses. We agree with the state.
We begin with our standard of review. We note that the [finder of fact] must find every element proven beyond a reasonable doubt in order to find the defendant guilty of the charged offense, [but] each of the basic and inferred facts underlying those conclusions need not be proved beyond a reasonable doubt․ If it is reasonable and logical for the [finder of fact] to conclude that a basic fact or an inferred fact is true, the [finder of fact] is permitted to consider the fact proven and may consider it in combination with other proven facts in determining whether the cumulative effect of all the evidence proves the defendant guilty of all the elements of the crime charged beyond a reasonable doubt․
“When there is conflicting evidence ․ it is the exclusive province of the ․ trier of fact, to weigh the conflicting evidence, determine the credibility of witnesses and determine whether to accept some, all or none of a․ On appeal, we do not ask whether there is a reasonable view of the evidence that would support a reasonable hypothesis of innocence. We ask, instead, whether there is a reasonable view of the evidence that supports the [finder of fact's] verdict of guilty.” (Citation omitted; internal quotation marks omitted.) State v. Little, 127 Conn.App. 336, 339–40, 14 A.3d 1036, cert. denied, 302 Conn. 928, 28 A.3d 343 (2011); see also State v. Testa, 123 Conn.App. 764, 767–68, 3 A .3d 142, cert. denied, 298 Conn. 934, 10 A.3d 518 (2010).
“Furthermore, [i]n [our] process of review, it does not diminish the probative force of the evidence that it consists, in whole or in part, of evidence that is circumstantial rather than direct․ It is not one fact, but the cumulative impact of a multitude of facts which establishes guilt in a case involving substantial circumstantial evidence․ [An] appellate court's first task, in responding to a claim of evidentiary insufficiency, is to apply the traditional scope of review to the evidence. That requires that ․ we view all of the evidence, and the reasonable inferences drawable therefrom, in favor of the [trier's] verdict․ We note that a claim of insufficiency of the evidence must be tested by reviewing no less than, and no more than, the evidence introduced at trial.” (Citations omitted; internal quotation marks omitted.) State v. Butler, 296 Conn. 62, 77, 993 A.2d 970 (2010).
The state set forth the following in its second amended substitute information: “And said [prosecutor] further accuses [the defendant] of [violating § 14–223(b) ] and charges in the City of New London, on or about April 23, 2009, the [defendant] at Wall Street, did increase the speed of the motor vehicle he was driving in an attempt to escape or elude ․ Bergeson of the New London Police Department when said police officer activated his emergency lights and sirens indicating to the [defendant] to stop, but instead the [defendant] increased his speed․”
We now turn to the relevant text of § 14–223(b): “No person operating a motor vehicle, when signaled to stop by an officer in a police vehicle using an audible signal device or flashing or revolving lights, shall increase the speed of the motor vehicle in an attempt to escape or elude such police officer. Any person who violates this subsection shall be guilty of a class A misdemeanor․” This court has identified the elements necessary to support a conviction under this statute. Specifically, we stated that a conviction under this statute “requires proof that an officer using an audible signal or flashing or revolving lights signaled the operator to stop and that the operator increased his vehicle's speed in an attempt to escape.” State v. Browne, 84 Conn.App. 351, 370, 854 A.2d 13, cert. denied, 271 Conn. 931, 859 A .2d 930 (2004). In Browne, we further noted that “[§ ] 14–223 requires only the general intent to escape or to elude a police officer signaling the operator of a motor vehicle to stop. General intent is the term used to define the requisite mens rea for a crime that has no stated mens rea; the term refers to whether a defendant intended deliberate, conscious or purposeful action, as opposed to causing a prohibited result through accident, mistake, carelessness, or absent-mindedness. Where a particular crime requires only a showing of general intent, the prosecution need not establish that the accused intended the precise harm or precise result which resulted from his acts.” (Internal quotation marks omitted.) Id., at 372.
The state presented the following evidence to prove that the defendant violated § 14–223(b). At the time of trial, Bergeson was a police lieutenant who had been employed by the New London Police Department for fourteen years. Bergeson testified that while in pursuit of the defendant's van with his overhead lights and siren turned on, he observed the defendant accelerate up Wall Street. Additionally, the jury watched the video of the incident. The video shows the defendant's van take a sudden left turn, without signaling or braking, onto Wall Street. The van is out of frame of the video camera for approximately five seconds. The van continues up Wall Street for approximately eleven seconds before the defendant approaches a stop sign and applies the brakes. The defendant does not appear to come to a complete stop before turning right onto Summer Street.
The defendant does not claim that he was not the operator of the van, that Bergeson was not a police officer or that Bergeson did not use his overhead flashing or revolving lights and sirens as a signal for the defendant to stop. His claim on appeal is limited to whether he increased his speed while driving on Wall Street with the intent to escape or elude Bergeson. Specifically, the defendant argues that the video of the incident conclusively shows that these two elements were not proven by the state beyond a reasonable doubt. In support, he relies on Scott v. Harris, 550 U.S. 372, 127 S.Ct. 1769, 167 L.Ed.2d 686 (2007), and State v. Santos, 267 Conn. 495, 838 A.2d 981 (2004).
In Harris, the issue before the United States Supreme Court was “whether a law enforcement official can, consistent with the Fourth Amendment, attempt to stop a fleeing motorist from continuing his public-endangering flight by ramming the motorist's car from behind.” Scott v. Harris, supra, 550 U.S. at 374. The respondent was observed travelling seventy-three miles per hour, exceeding the posted speed limit by eighteen miles per hour. Id. A police chase ensued, and Georgia Deputy Timothy Scott, the petitioner, requested and received permission to perform a maneuver that would cause the respondent's vehicle to spin to a stop. Id., at 375. The petitioner applied a push bumper to the rear of the respondent's vehicle and, as a result, the respondent lost control of his vehicle, which went down an embankment. He was rendered a quadriplegic. Id.
The respondent filed a lawsuit alleging that there had been excessive force used that resulted in an unreasonable seizure in violation of the fourth amendment to the United States constitution. Id., at 375–76. The petitioner filed a motion for summary judgment on the basis of qualified immunity. Id., at 376. The respondent's version of the events differed significantly from that of the petitioner. Id., at 378. As a result, both the District Court and the United States Court of Appeals for the Eleventh Circuit determined that the petitioner was not entitled to summary judgment. Id., at 376. The petitioner successfully petitioned for certification to appeal to the United States Supreme Court. Id.
The United States Supreme Court began its analysis by noting that, generally, the District Court is required to view the facts in the light most favorable to the nonmoving party, there, the respondent. Id., at 378. ․” Id. The Supreme Court then described in detail the chase as depicted on the videotape, and characterized it as “a Hollywood-style car chase of the most frightening sort, placing police officers and innocent bystanders alike at great risk of serious injury.” Id., at 380.
The videotape in Harris conclusively established the facts with respect to whether the respondent's driving endangered human life, notwithstanding the general rule of viewing a disputed factual record in favor of the nonmoving party. “When opposing parties tell two different stories, one of which is blatantly contradicted by the record, so that no reasonable jury could believe it, a court should not adopt that version of the facts for purposes of ruling on a motion for summary judgment․ Respondent's version of events is so utterly discredited by the record that no reasonable jury could have believed him. The Court of Appeals should not have relied on such visible fiction; it should have viewed the facts in the light depicted by the videotape.” Id., at 380–81. The United States Supreme Court reversed the decision of the Court of Appeals, concluding that the petitioner was entitled to summary judgment. Id., at 386.
In Santos, our Supreme Court was faced with the question of whether the trial court properly had denied the defendant's motion to suppress narcotics following a warrantless patdown search for weapons. State v. Santos, supra, 267 Conn. at 496. The defendant filed a motion to suppress, and the court conducted a hearing. It found that two state police troopers were patrolling athletic fields in Windham at night where they came upon the defendant and three other individuals. Id., at 498–99. One of the troopers testified that, as he approached, the men started pacing back and forth and seemed visibly nervous. Id., at 499. After further questioning, the trooper requested the men to stand in front of his cruiser, which they did. Id., at 500. One of the individuals then started to move to the driver's side of the vehicle. Id. Out of concern for his safety, the trooper instructed the four men to remain still and submit to a patdown search. Id. When the trooper patted down the defendant, he discovered a clear plastic bag containing a white powdery substance and arrested the defendant for possession of narcotics. Id., at 500–501. Following the conclusion of the hearing, the trial court denied the defendant's motion to suppress. Id., at 501.
On appeal, the defendant argued that the court improperly found that he and the other individuals were perspiring and pacing back and forth. Id., at 502–503. As a result, he claimed, the court improperly determined that the trooper had a reasonable and articulable suspicion for the patdown search. Id., at 503. Our Supreme Court concluded that the court's finding that the defendant and his companions were perspiring was clearly erroneous because there was no testimony as to that fact and the videotape of the event provided no support for such a finding. Id., at 506. Additionally, the videotape did not support the trooper's testimony that the defendant and the others were pacing back and forth. Id. To the contrary, “[t]he videotape depicts the defendant and his friends standing at the rear of their car, occasionally shifting their weight from one foot to the other while standing in the glare of the spotlight, answering [the trooper's] questions. That movement cannot properly be characterized as pacing back and forth.” (Internal quotation marks omitted.) Id. As a result, our Supreme Court concluded that the trial court should have granted the defendant's motion to suppress. Id., at 511.
Guided by these cases and having reviewed the relevant testimony and the video of Bergeson's pursuit of the defendant, we conclude that sufficient evidence exists to support the jury's finding of guilt with respect to the violation of § 14–223(b). The factor that distinguishes the present case from Harris and Santos is that the video does not conclusively establish the underlying facts. Specifically, the video does not undisputedly indicate that the defendant was not accelerating up Wall Street or that he lacked the intent to elude or escape Bergeson. The video shows that the defendant made an abrupt left turn onto Wall Street, without signaling. The jury could have concluded that he accelerated at that point. Additionally, the video does not conclusively establish that the defendant did not accelerate while driving on Wall Street. The jury was free to accept the testimony of Bergeson, an experienced member of the New London Police Department, that the defendant was accelerating while travelling on Wall Street.3 See, e.g., State v. Hoover, 54 Conn.App. 773, 777, 738 A.2d 685 (1999) (“It is the function of the jury to consider the evidence and to judge the credibility of witnesses․ The jury is free to accept or reject all or part of a witness' testimony.” [Citation omitted; internal quotation marks omitted.] ).
We now turn to the element of whether the defendant intended to escape or elude Bergeson. “As this court frequently has observed, [i]ntent is a mental process, and absent an outright declaration of intent, must be proved through inferences drawn from the actions of an individual, i.e., by circumstantial evidence․ Furthermore, any inference drawn must be [rational] and founded upon the evidence .” (Internal quotation marks omitted.) State v. Colon, 117 Conn.App. 150, 157, 978 A.2d 99 (2009); see also State v. Papandrea, 120 Conn.App. 224, 230, 991 A.2d 617 (2010), aff ‘d, 302 Conn. 340, 26 A.3d 75 (2011).
“It is well settled ․ that the question of intent is purely a question of fact․ The state of mind of one accused of a crime is often the most significant and, at the same time, the most elusive element of the crime charged․ Because it is practically impossible to know what someone is thinking or intending at any given moment, absent an outright declaration of intent, a person's state of mind is usually proven by circumstantial evidence․ Intent may be and usually is inferred from conduct․ [W]hether such an inference should be drawn is properly a question for the jury to decide․ Intent may be inferred from cir cumstantial evidence such as the events leading to and immediately following the incident, and the jury may infer that the defendant intended the natural consequences of his actions.” (Citation omitted; emphasis added; internal quotation marks omitted.) State v. Papandrea, supra, 120 Conn.App. at 230.
The defendant argues that the video demonstrates that he brought the van to a stop at intersections and used turn signals, and therefore established that he lacked the intent to escape or elude Bergeson. We are not persuaded. First, the video does not show that the defendant used his brakes or a turn signal as he made the left turn onto Wall Street. The video does show Bergeson's police cruiser with its overhead flashing lights and sirens turned on behind the defendant's van and the defendant not pulling over. Further, Nott testified that, while at the CVS parking lot, she made the defendant aware of the fact that she was requesting assistance. She stated that the defendant “heard the response, ran to the car, got in the car and sped out—ran to the van, got into the van and left the parking lot.” Harris and Santos are distinguishable because the video in this case does not conclusively establish the facts underlying the charge of § 14–223(b). The evidence, including the video and the testimony of the witnesses, provided the necessary support for the jury's finding that the defendant violated § 14–223(b).
II
The defendant next claims that § 14–223(b) is unconstitutionally vague as applied to the facts of this case. Specifically, he argues that because § 14–223(b) does not contain an element that a police officer, when initiating motor vehicle stops, must be acting lawfully and within the scope of his or her duties, it authorizes or even encourages arbitrary and discriminatory enforcement. We are not persuaded.
As a preliminary matter, we consider whether this claim has been preserved for appellate review. On April 16, 2010, the defendant filed a motion to dismiss the charges of two counts of interfering with the police in violation of § 53a–167a and one count of risk of injury to a child in violation of § 53–21. He alleged that as applied to the facts set forth in the operative information, § 53a–167a was unconstitutionally broad and vague and that § 53–21 was unconstitutionally vague. On April 12, 2011, prior to the start of the trial, the parties discussed the defendant's motion with the court. In neither the motion nor this discussion did the defendant argue that § 14–223(b) was unconstitutionally vague as applied to the facts of this case. Further, this specific claim was not raised in the defendant's postverdict motion for a new trial. We conclude, therefore, that the defendant failed to preserve this claim in the trial court proceedings.
The defendant argues, in the alternative, that his claim is reviewable under State v. Golding, 213 Conn. 233, 239–40, 567 A.2d 823 (1989)..” (Internal quotation marks omitted.) State v. Burton, 258 Conn. 153, 157–58, 778 A.2d 955 (2001). The state does not dispute that the record is adequate to review the claim and that the claim is of constitutional magnitude. We agree that the first two prongs of Golding are satisfied and, therefore, we will proceed to review the defendant's claim. Id., at 158; State v. Tozier, 136 Conn.App. 731, 746, 46 A.3d 960, cert. denied, 307 Conn. 925, 55 A.3d 567 (2012); fails to satisfy the third Golding prong.
We begin our analysis with the standard of review and the relevant legal principles with respect to a void for vagueness challenge. “We begin by noting that determining whether a [statute] is unconstitutionally vague presents a question of law over which our review is de novo․ [A] penal statute [must] define [a] criminal offense with sufficient definiteness that ordinary people can understand what conduct is prohibited and in a manner that does not encourage arbitrary and discriminatory enforcement․ [This concept] embodies two central precepts: the right to fair warning of the effect of a governing statute or regulation and the guarantee against standardless law enforcement․ [T]he [most] important aspect of the vagueness doctrine is not actual notice ․ ․
“For statutes that do not implicate the especially sensitive concerns embodied in the first amendment, we determine the constitutionality of a statute under attack for vagueness by considering its applicability to the particular facts at issue․ [T]o prevail on his claim, the defendant must demonstrate beyond a reasonable doubt that the statute, as applied to him, deprived him of adequate notice of what conduct the statute proscribed or that he fell victim to arbitrary and discriminatory enforcement.” (Citation omitted; internal quotation marks omitted.) State v. Stephens, 301 Conn. 791, 800–802, 22 A.3d 1262 (2011); State v. Winot, 294 Conn. 753, 758–60, 988 A.2d 188 (2010). Put another way, “[o]ur fundamental inquiry is whether a person of ordinary intelligence would comprehend that the defendant's acts were prohibited․” (Internal quotation marks omitted.) State v. Elliott, 127 Conn.App. 464, 471, 14 A.3d 439, cert. denied, 301 Conn. 916, 21 A.3d 462 (2011). We also “restate the common-law rule that everyone is presumed to know the law and that ignorance of the law excuses no one from criminal sanction.” State v. Knybel, 281 Conn. 707, 713, 916 A.2d 816 (2007). Finally, we note that “[a] statute is not void for vagueness unless it clearly and unequivocally is unconstitutional, making every presumption in favor of its validity.” (Internal quotation marks omitted.) State v. Ward, 306 Conn. 718, 742, 51 A.3d 970 (2012); see also State v. Springmann, 69 Conn.App. 400, 407, 794 A.2d 1071 (to prevail in challenge to constitutionality of statute, defendant must demonstrate beyond reasonable doubt that statute, as applied, deprived him or her of adequate notice or that he or she fell victim to arbitrary and discriminatory enforcement), cert. denied, 260 Conn. 934, 802 A.2d 89 (2002).
The defendant's vagueness argument may be summarized as follows. First, citizens of Connecticut enjoy a common-law right to ignore unlawful commands by the police. The trial court determined that the defendant had engaged in constitutionally protected speech with respect to his interaction with Nott at the CVS parking lot. Nott provided Bergeson with no information that would justify his decision to pull the defendant's vehicle over. Therefore, Nott and Bergeson lacked a legitimate basis to stop the defendant after he left the CVS parking lot, and their behavior amounted to arbitrary police enforcement. Thus, the defendant argues: “Absent a lawful basis to restrict [the defendant's] freedom ab initio, Bergeson acted unlawfully, and [§ 14–223(b) ] cannot be enforced against [the defendant] without violating the common law.” (Emphasis in original .) In conclusion, the defendant states that in order to survive his vagueness challenge, the court should have instructed the jury that § 14–223(b) applies only in circumstances where police officers are acting lawfully or within the scope of their duties. We read the defendant's argument to mean that had the court interpreted § 14–223(b) to require that such an instruction be given, the risk of arbitrary and discriminatory enforcement would have been eliminated.
The state counters that a person of ordinary intelligence would understand that, under the plain terms of § 14–223(b), accelerating away from a signaling police officer with the intent to escape or elude violates the statute. It also argues that there is no risk of arbitrary and discriminatory enforcement because the plain terms of § 14–223(b) provide sufficient guidance as to the behavior that is prohibited. Additionally, even if such minimum guidelines were absent, the statute has a core meaning within which the defendant's conduct fell. The state further contends that the court's dismissal of the interference count relating to his conduct in the CVS parking lot means only that the state failed to produce sufficient evidence to sustain a conviction, not that the officers were acting illegally. Finally, the state maintains that even if the determination to arrest the defendant was illegal, he had no right to engage in criminal conduct to avoid the arrest.
The defendant does not appear to address the first prong of a constitutional vagueness claim, that is, the right to fair warning of the effect of a governing statute. We therefore focus our analysis on the second prong, that is, the guarantee against standardless law enforcement. The question is whether the statute's language impermissibly delegated a basic policy matter for resolution by police, judges and juries on an ad hoc and subjective basis, particularly in light of the common-law right of Connecticut citizens to refuse to obey an unlawful police command to stop. Put another way, “a legislature [must] establish minimal guidelines to govern law enforcement.” (Internal quotation marks omitted.) State v. Winot, supra, 294 Conn. at 760.
Our Supreme Court has instructed that “[a [United States] Supreme Court has indicated that the more important aspect of the vagueness doctrine is the requirement that a legislature establish minimal guidelines to govern law enforcement․ If a court determines that a statute provides sufficient guidelines to eliminate generally the risk of arbitrary enforcement, that finding concludes the inquiry.
“[When] a statute provides insufficient general guidance, an as-applied vagueness challenge may nonetheless fail if the statute's meaning has a clear core․ In that case the inquiry will involve determining whether the conduct at issue falls so squarely in the core of what is prohibited by the law that there is no substantial concern about arbitrary enforcement because no reasonable enforcing officer could doubt the law's application in the circumstances.” (Internal quotation marks omitted.) State v. Stephens, supra, 301 Conn. at 805–806.
The preceding explanation is necessary to put the defendant's argument into the proper context. We need not proceed with the analytical framework, however, due to a fatal flaw in the defendant's analysis. He assumes that the trial court, by granting his motion for a judgment of acquittal with respect to the interfering with a police officer count, found that the officers were acting outside the scope of their police duties, i.e., illegally and on a personal frolic. The court made no such finding, either explicit or implicit. Rather, the court determined that the state had failed to produce sufficient evidence to support a conviction with respect to the count of interfering with a police officer at the CVS parking lot.
This court, citing federal law, has stated that the issue of whether a police officer is acting within the scope of his or her official duties is a factual question. State v. Privitera, 1 Conn.App. 709, 722, 476 A.2d 605 (1984). In State v. Davis, 261 Conn. 553, 572, 804 A.2d 781 (2002), our Supreme Court also indicated that the question of whether a police officer was in the performance of his or her duties presented an issue for the fact finder.4 In the present case, the court granted the defendant's motion for a judgment of acquittal without finding that the officers were acting outside the scope of their duties. Therefore, the entire foundation of the defendant's as applied vagueness argument erodes. Specifically, the court did not find that the defendant was issued an illegal police command. There was no determination that the officers lacked probable cause to arrest the defendant. Thus, there were no findings to support a conclusion that the efforts to arrest the defendant for his conduct in the CVS parking lot were illegal. As a result, the defendant's claim fails under the third prong of Golding.
III
The defendant next claims that the court improperly failed to charge the jury on two theories of defense. Specifically, he argues that he was entitled to have the jury charged on the defenses of entrapment and the exercise of his rights under the first amendment. The state counters that these requested instructions were not relevant to the issues before the jury, and that, therefore, the court properly declined to give them. We agree with the state.
The defendant submitted written jury instructions to the court.5 The defendant requested the court to instruct the jury on the defense of entrapment, as set forth in General Statutes § 53a–15. He also requested that the court instruct the jury with respect to the first amendment to the United States constitution. Specifically, he filed a request to charge the jury as follows: .” The defendant also requested an instruction that taking a photograph of a police officer is protected by the first amendment.
The court discussed the jury charge with counsel on the record. It stated that it would not charge the jury as to the first amendment or the defense of entrapment.6 Defense counsel then stated that he requested the first amendment charge because “there [was] testimony that, that—the reason that the police acted the way they acted was because of things that [the defendant] said in the officer's presence.”7 Defense counsel further elaborated that the officers were not in the performance of their duties but rather were engaged in a personal frolic; therefore, the efforts to pull the defendant over and issue him a ticket were based on a personal grudge. He also argued that acting in the performance of duty was a necessary implied element of § 14–223(b). Turning to the issue of entrapment, defense counsel argued that there was evidence that the police provoked the defendant to actions that constituted an offense. He also contended that the officers were not acting in the performance of their duty when they provoked the defendant.
We begin our analysis with the defendant's claim that the court improperly refused to charge the jury on the defense of entrapment. Our Supreme Court recently addressed this issue in State v. Golodner, 305 Conn. 330, 46 A.3d 71 (2012). The defendant's right as a matter of law to a theory of defense instruction exists, however, only when there is evidence adduced indicating the availability of the defense. The court ․ has a duty not to submit to the jury, in its charge, any issue upon which the evidence would not reasonably support a finding․
“Entrapment is a legally recognized defense in this state․ See General Statutes § 53a–15. Until something in the evidence indicates the contrary, the court may presume the defendant intended the prohibited bodily movements that constitute the offense and that he has acted under no duress, unlawful inducement in the nature of entrapment, or lack of requisite mental capacity․ In reviewing the defendant's claim that he was entitled to instructions on an affirmative defense, we look at the evidence in a light most favorable to his claim․ When a defendant has produced evidence supporting a legally recognized defense, the trial court's refusal to provide an instruction with respect to that defense constitutes a denial of due process․
“[O]nly when evidence indicating the availability of [a] legally recognized [defense] is placed before a jury is a defendant entitled as a matter of law to a theory of defense instruction․ [A] defendant is entitled to have instructions presented relating to any theory of defense for which there is any foundation in the evidence, no matter how weak or incredible․ A fundamental element of due process is the right of a defendant charged with a crime to establish a defense․
“Where the legislature has created a legally recognized defense, in this case entrapment, this fundamental constitutional right includes a proper jury instruction on the elements of the defense of entrapment so that the jury may ascertain whether the state has met its burden of disproving it beyond a reasonable doubt․
“General Statutes § 53a–15 provides in relevant part: [i. The subjective test of entrapment focuses on the disposition of the defendant to commit the crime for which he or she is accused․ Under [the alternative objective test] standard, entrapment exists if the government conduct was such that a reasonable person would have been induced to commit the crime․ The Connecticut legislature has chosen to adopt the subjective defense of entrapment․ This statute codifies prior Connecticut case law․ To warrant an instruction on entrapment, the defendant must produce evidence of both inducement and his own lack of criminal disposition․ Where, as here, an accused requests an instruction on a defense such as entrapment, he may obtain such a charge by adducing evidence ․ sufficient ․ for a rational juror to find that all the elements of the defense are established by a preponderance of the evidence.” (Citations omitted; internal quotation marks omitted.) Id., at 351–53; see generally State v. McNally, 173 Conn. 197, 200–202, 377 A.2d 286 (1977); State v. Marquardt, 139 Conn. 1, 4–8, 89 A.2d 219 (1952); see also State v. Wilder, 128 Conn.App. 750, 755, 17 A.3d 1116 (“[i]t is well established that where there is no evidence that the defendant either was induced by the police to commit a crime in which he would not have engaged except for such inducement or that he admitted to committing a crime, a charge on entrapment is not required”), cert. denied, 301 Conn. 934, 23 A.3d 730 (2011).
A
We first consider the defendant's entrapment argument. On appeal, the defendant claims that his actions were induced by and resulted from the provocation from the members of the New London Police Department. Furthermore, he contends that there was “overwhelming” evidence that his actions were induced and provoked by the police. Specifically, the defendant points to the following: (1) Nott's request to have the defendant arrested for his conduct at the CVS parking lot was unlawful; (2) Bergeson, as a supervisor, should have known that there was no basis to stop the defendant; (3) Bergeson falsified a police report detailing the events of the police chase; (4) Nott and Bergeson knew that they were defendants in a civil lawsuit filed by the defendant; and (5) Bergeson's use of his Taser provoked or induced the defendant to reenter his van after he had driven to his house.
We have noted that “under our state decisional law, [e]vidence of unlawful inducement may be found where the police ․ appeal to the [accused's] sympathy or friendship, or where they repeatedly or persistently solicit the [accused] to commit the crimes.” (Emphasis in original; internal quotation marks omitted.) State v. Wilder, supra, 128 Conn.App. at 758. The defendant has failed to establish that the evidence in this case met this standard and, thus, required an instruction on entrapment. There is no evidence that Nott, Bergeson, or any other member of the New London Police Department repeatedly or persistently solicited the defendant to drive away from the CVS parking lot and to refuse to pull his vehicle over when signaled to do so. Additionally, Bergeson's pointing his Taser at the defendant, after he had engaged in a pursuit and the defendant refused to obey his direction to pull over, did not amount to a repeated and persistent inducement to interfere with Bergeson's attempt to place the defendant under arrest. There was no showing by the defendant that, but for the actions of the police, he would not have engaged in such conduct. See State v. Golodner, supra, 305 Conn. at 354. Additionally, there was no evidence of lack of criminal disposition. See id., at 353. We conclude, therefore, that the court properly declined to instruct the jury on the defense of entrapment.
B
We now turn to the defendant's claim that he was entitled to an instruction on the issue of constitutionally protected speech. “[A] request to charge which is relevant to the issues of the case and which is an accurate statement of the law must be given․ [W]hen reviewing.” (Citation omitted; internal quotation marks omitted.) State v. Berger, 249 Conn. 218, 234–35, 733 A.2d 156 (1999).
This argument is based upon the claim that the officers were found to have engaged in a personal frolic. We have rejected that as contrary to the action of the trial court. See part II of this opinion. Furthermore, we agree with the state that the court was not required to give the jury the requested first amendment instruction, as it was not relevant to the remaining charges. See State v. Vilchel, 112 Conn.App. 411, 434, 963 A.2d 658, cert. denied, 291 Conn. 907, 969 A.2d 173 (2009). We therefore reject these claims of instructional error.
IV
The defendant's final claim is that he was deprived of the right to a fair trial as a result of prosecutorial impropriety.8 Specifically, he argues that the prosecutor made five improper comments to the jury during closing argument. The state concedes that two of the five statements were improper. The state argues, however, that the defendant was not deprived of his right to a fair trial, despite the two improper statements. We conclude that the defendant was not deprived of the right to a fair trial.
We begin by setting forth the relevant law regarding prosecutorial impropriety. “In analyzing claims of prosecutorial impropriety, we engage in a two step analytical process. E.g., State v. Stevenson, 269 Conn. 563, 572, 849 A.2d 626 (2004).
“[T]he touchstone of due process analysis in cases of alleged ․
“[I]t is not the prosecutor's conduct alone that guides our inquiry, but, rather, the fairness of the trial as a whole․․ By reason of his [or her] office, [the prosecutor] usually exercises great influence upon jurors. [The prosecutor's] conduct and language in the trial of cases in which human life or liberty [is] at stake should be forceful, but fair, because he [or she] represents the public interest, which demands no victim and asks no conviction through the aid of passion, prejudice or resentment. If the accused be guilty, he [or she] should [nonetheless] be convicted only after a fair trial, conducted strictly according to the sound and well-established rules which the laws prescribe․
“Once prosecutorial impropriety has been alleged, however,.” (Citations omitted; footnote omitted; internal quotation marks omitted.) State v. Fauci, 282 Conn. 23, 32–34, 917 A.2d 978 (2007). Guided by these principles, we address each of the defendant's claims of impropriety in turn.9
A
The defendant first claims that the prosecutor improperly offered his personal opinion on the issue of the defendant's guilt. During the beginning of his rebuttal, the prosecutor argued to the jury: “And we all know the burden is on me, on the state, to prove this case beyond a reasonable doubt. Okay. Let me tell you why. Let me tell you why I think this case warrants conviction.” At this point, defense counsel objected, but the court overruled the objection. The prosecutor then proceeded to review the evidence against the defendant. At the conclusion of the state's rebuttal argument, the prosecutor stated: “Again, the burden is on me. And rightly, it should be. If I'm going to convict someone, I better well have a good case. And I think we do.” Defense counsel again objected, and the court, agreeing with the defendant, ordered that the personal opinion of the prosecutor be stricken.
The defendant argues that, with respect to these two statements, the prosecutor improperly offered his personal opinion on the issue of the defendant's guilt. The state counters that the statement made by the prosecutor at the beginning of his rebuttal argument did not amount to prosecutorial impropriety. It concedes, however, that the latter statement made by the prosecutor was improper; nevertheless, it argues that the defendant was not deprived of a fair trial.
1
We begin by determining whether the prosecutor's statement made at the outset of his rebuttal argument amounted to an improper statement of his personal opinion as to the defendant's guilt or was a rhetorical device to walk the jury through the evidence, as argued by the state.
“In determining whether [prosecutorial impropriety] has occurred [in the course of closing arguments], the reviewing court must give due deference to the fact that .” (Internal quotation marks omitted.) State v. Boutilier, 133 Conn.App. 493, 510, 36 A.3d 282, cert. denied, 304 Conn. 914, 40 A.3d 785 (2012).
We agree with the state that the prosecutor's comment served as an introduction to his review and summary of the evidence against the defendant and was not improper.10 “We must give the jury the credit of being able to differentiate between argument on the evidence and attempts to persuade them to draw inferences in the state's favor, on one hand, and improper unsworn testimony, with the suggestion of secret knowledge, on the other hand. The state's attorney should not be put in the rhetorical straitjacket of always using the passive voice, or continually emphasizing that he is simply saying I submit to you that this is what the evidence shows, or the like.” (Internal quotation marks omitted.) State v. Thompson, 266 Conn. 440, 465–66, 832 A.2d 626 (2003); see also State v. Houle, 105 Conn.App. 813, 823, 940 A.2d 836 (2008) (where prosecutor used “I,” it was evident he was encouraging jury to draw reasonable inferences from evidence).
2
We next turn to the prosecutor's statement expressing his personal belief that the state had a strong case. As noted previously, the state concedes that this statement was improper.11 We agree. Accordingly, we use the Williams factors to determine if the defendant was deprived of his right to a fair trial as a result of this statement. We conclude that he was not.
The prosecutor's comment was not invited by defense counsel, and upon objection was stricken by the court. The court also immediately instructed the jury that the personal opinion of the prosecutor was not part of the argument. In addition to this curative measure, the court instructed the members of the jury that they were the sole judges of the facts, that they were to determine the facts from the evidence and not the argument of counsel and that they were not to consider matters stricken from the record. The jury is presumed to have followed the court's instructions. State v. Payne, 303 Conn. 538, 568, 34 A.3d 370 (2012). The improper comment was not severe, as it followed the prosecutor's summary of the evidence. This improper comment was a single instance and, therefore, was not frequent. Finally, although we would not describe the state's case as overwhelming, we disagree with the defendant's assertion that the state had a weak case. There was more than sufficient evidence presented that supported the jury's verdict. For all of these reasons, we conclude that the defendant was not deprived of the right to a fair trial.
B
The defendant next contends that the prosecutor's arguments violated the prohibition against “golden rule” arguments in two instances. “[A] golden rule argument is one that urges jurors to put themselves in a particular party's place ․ or into a particular party's shoes․ Such arguments are improper because they encourage the jury to depart from neutrality and to decide the case on the basis of personal interest and bias rather than on the evidence․ They have also been equated to a request for sympathy․ The danger of these types of arguments lies in their [tendency] to pressure the jury to decide the issue of guilt or innocence on considerations apart from the evidence of the defendant's culpability․ [A] prosecutor may not appeal to the emotions, passions and prejudices of the jurors․ [S]uch appeals should be avoided because they have the effect of diverting the [jurors'].” (Citations omitted; internal quotation marks omitted.) State v. Ovechka, 118 Conn.App. 733, 745, 984 A.2d 796, cert. denied, 295 Conn. 905, 989 A.2d 120 (2010). Put another way, “[t]he animating principle behind the prohibition on golden rule arguments is that jurors should be encouraged to decide cases on the basis of the facts as they find them, and reasonable inferences drawn from those facts, rather than by any incitement to act out of passion or sympathy for or against any party․ Although we recognize that this danger is most acute when the prosecutor asks the jurors to put themselves in the position of the victim rather than the defendant or another witness ․ we conclude that the principle barring the use of such arguments is the same regardless of which individual is the subject of the prosecutor's emotional appeal.” (Citation omitted; emphasis omitted; internal quotation marks omitted.) State v. Long, 293 Conn. 31, 57–58, 975 A.2d 660 (2009).
During his rebuttal closing argument, the prosecutor stated: “And then there's ․ the discussion about ․ Bergeson taking out his Taser. Well, you know ․ Bergeson just chased this individual, you know, it might've [been] for a minute, but he doesn't know what his state, the defendant's state of mind is. He doesn't know what he's going to do. He has a right to protect himself. Put yourself in the policeman's shoes. What would you do?
“And yes, there's a civil rights suit. You know, granted, it came out. But doesn't ․ Bergeson have the right to at least defend himself. He doesn't know what the defendant's state of mind is on this particular day.
“Now, getting back to the engaging police in pursuit, how many times have—when you've been driving down the road, have you ever seen the flashing red lights in the rearview mirror․ What would you do? What is your normal reaction when you see the flashing red lights in the rearview mirror? ”12 (Emphasis added.) The defendant argues that the two comments, which we have emphasized, violated the prohibition against golden rule arguments.
1
After reviewing the record, we are satisfied that the comment made by the prosecutor asking the jurors what their normal reaction was when they saw flashing red lights in their vehicles' rearview mirrors did not constitute an improper attempt to appeal to the emotions of the jurors, but rather was an appropriate request for them to draw an inference from the evidence. See State v. Bell, 283 Conn. 748, 769–70, 931 A.2d 198 (2007); State v. Ovechka, supra, 118 Conn.App. at 746. Specifically, that comment encouraged the jurors to draw inferences from the evidence as to the defendant's action in not pulling over when Bergeson was signaling that amounted to a violation of § 14–223(b). We conclude, therefore, that the prosecutor's comment did not violate the prohibition against golden rule arguments.
2
With respect to the prosecutor's comment inviting the members of the jury to place themselves in the shoes of a police officer and asking what they would do in the present situation, we conclude that the comment was improper because it violated the prohibition against golden rule arguments. Specifically, it encouraged the members of the jury to decide the case not on the basis of the facts of the case, but rather on their emotions and passions. Put another way, the prosecutor's comment inviting the jurors to place themselves in the shoes of a police officer was tantamount to a request for sympathy and a recognition of the challenging tasks facing police officers. Accordingly, we conclude that this comment was improper.
We now apply the Williams factors to determine whether this comment deprived the defendant of the right to a fair trial. We conclude that it did not.
Defense counsel stated during closing argument that Bergeson intended to Taser the defendant and reminded the jury of the defendant's heart condition and hypertension, and that the defendant's daughter was present during the encounter. He also stated: “Now, [the defendant] knows Bergeson. He's suing Bergeson. Imagine a guy who's got a personal beef against you being the one standing there with a dangerous instrument.” In short, counsel provided the jury with the reasons for the defendant's actions. In response to that, the prosecutor provided the jury with the reasons for Bergeson's actions, including the threatened use of the Taser. The prosecutor's comment, therefore, was in response to the argument of defense counsel. We also conclude that this instance of impropriety was not severe, not frequent, and was not central to the critical issues of the case. See State v. Fauci, supra, 282 Conn. at 32–34. Accordingly, the defendant was not deprived of his right to a fair trial as a result of this improper statement.
C
The defendant next contends that the prosecutor improperly went outside the record and asserted his personal knowledge when he stated how he would have reacted in the defendant's situation. The state concedes that this statement was improper,13 but argues that it did not deprive the defendant of his right to a fair trial. We agree with the state.
During his rebuttal closing argument, the prosecutor stated: “Now, getting back to the engaging police in pursuit, how many times have—when you've been driving down the road, have you ever seen the flashing red lights in the rearview mirror. I'm a prosecutor, and I get a lump in my throat. I'm like—oh, God—what did I do. I'm checking my speedometer. I see if my headlights work, whatever. You know what I do? I pull over. It doesn't even matter whether I did something wrong or not․ I know what to do at that particular point.” (Emphasis added.)
Defense counsel immediately objected and stated the basis for the objection as follows: “Putting—the prosecutor is not allowed to put his own personal experience into argument.” The prosecutor then withdrew his comments about his personal experiences, and the court instructed the jury: “Withdrawn. All right, you may disregard that portion of the argument.”
As a general matter, we note that “[a] prosecutor ․ may not ․ inject extraneous issues into the case that divert the jury from its duty to decide the case on the evidence․ A prosecutor, in fulfilling his duties, must confine himself to the evidence in the record.” (Citation omitted; internal quotation marks omitted.) State v. Moore, 293 Conn. 781, 809,981 A.2d 1030 (2009), cert. denied, U.S., 130 S.Ct. 3386, 177 L.Ed.2d 306 (2010). This court has stated that it is improper for a prosecutor to invite the jury to consider the conduct of the defendant through the lens of the prosecutor's own life experiences. See State v. Houle, supra, 105 Conn.App. at 825.
We therefore turn to the Williams factors to determine whether these comments violated the defendant's right to a fair trial. The comment was isolated, not central to a key issue of the case and the state presented a strong case. Indeed, there is no dispute that the defendant failed to pull over when signaled to do so by Bergeson. Additionally, following defense counsel's objection, the prosecutor immediately withdrew the comment, and the court instructed the jury to disregard it. As we previously noted, the court, during its charge, instructed the members of the jury not to consider matters that were not part of the evidence. For these reasons, we conclude that the defendant has failed to establish a deprivation of the right to a fair trial as a result of the prosecutorial impropriety.
The judgment is affirmed.
I respectfully dissent from parts I, III and IV of the majority opinion and concur as to part II.
ENGAGING IN PURSUIT
As to part I of the majority opinion, I do not agree that the state produced sufficient evidence to reasonably support the charge of engaging in pursuit “at Wall Street” in violation of General Statutes § 14–223(b). I accordingly respectfully dissent as to part I.
In describing some of the facts the jury could find and in sustaining the conviction, the majority points to evidence that Sergeant Todd Bergeson, after he received a radio call from Officer Deana Nott requesting a motor vehicle stop to issue a ticket to the defendant, Daniel G., for interfering or creating a disturbance, went to Jefferson Street near the CVS pharmacy seeking the defendant's van. In doing so, the Sergeant activated his cruiser's siren and overhead lights which turned on his dashboard camera. On Jefferson Street, Sergeant Bergeson observed the defendant's van ahead of him where it turned left onto Wall Street and thereafter turned onto Summer Street, then Redden Avenue, then Colman Street and into his residence, at times at speeds of a few miles per hour.
The majority rejects the defendant's claim that the videodisc produced from the dashboard camera in Sergeant Bergeson's cruiser conclusively established that the defendant did not increase his speed or try to elude Sergeant Bergeson at Wall Street,1 and concludes that the jury reasonably could have concluded that the cumulative force of the evidence established guilt beyond a reasonable doubt.
On appeal, we must ask whether there is a “reasonable view of the evidence that supports the jury's verdict of guilty․” (Internal quotation marks omitted.) State v. Silva, 285 Conn. 447, 454, 939 A.2d 581 (2008).
The statute required by its plain terms that the defendant increase his speed in an attempt to escape or elude an officer when signaled to stop by the use of the siren or flashing lights. In this case, the videodisc from Sergeant Bergeson's cruiser does not support the § 14–223(b) conviction, but rather renders the verdict unreasonable in view of the physical facts recorded by electronic means considered with the trial testimony and audio recordings of police radio transmissions.2 See State v. DeJesus, 236 Conn. 189, 196, 672 A.2d 488 (1996); State v. Bradley, 39 Conn.App. 82, 91, 663 A.2d 1100 (1995), cert. denied, 236 Conn. 901, 670 A.2d 322 (1996).
As the trial court record reflects, the New London police cruisers were equipped with dashboard video cameras, which produced videodiscs of much of the incident. Additionally, audio discs of the actual messages between officers; and between the defendant and the police dispatcher; and between the police dispatcher and police officers were produced. These discs were introduced at the defendant's trial. Similar video evidence was sufficient to convince our Supreme Court in State v. Santos, 267 Conn. 495, 838 A.2d 981 (2004), to reverse a trial court's decision after its review of a videotape which it found to be contrary to the physical facts testified to before that trial court.
As to Sergeant Bergeson's signal to stop, the video disc reflects that he used the siren and lights to pass through an intersection and through two red lights before entering Jefferson Street well behind the defendant's vehicle. Then, in following at a “considerable distance”3 behind the defendant's vehicle, the siren and lights were used to clear the left hand lane of Jefferson Street of a vehicle between the Sergeant's cruiser and the defendant's van. In contrast, the common, recognized method, familiar to all motorists, for signaling a stop is that the police cruiser using its siren or flashing lights pulls directly behind the vehicle to be stopped so the cruiser can be parked close to the rear of the stopped vehicle. This use of flashing lights to stop a vehicle is described in the prosecutor's summation when he referred to a driver looking at flashing lights in the “rearview mirror.” See footnote 17 of this opinion. The videodisc shows Sergeant Bergeson first reached that position directly behind and close to the defendant's van at the top of Wall Street on Summer Street.
As to increasing speed “at Wall Street,” the videodisc shows the defendant's van was far ahead of Sergeant Bergeson at the intersection of Jefferson Street and Wall Street, but, at the top of Wall Street, Sergeant Bergeson was directly and closely at the back of the defendant's van. This shows the defendant had not increased his speed away from Sergeant Bergeson's cruiser at Wall Street as charged in the information and testified to by Sergeant Bergeson as “accelerating up Wall Street” and as “driving away from” him.4 Furthermore, Sergeant Bergeson did not testify to any speed that the defendant was driving when the defendant was properly signaled to stop at any time.
Sergeant Bergeson originally testified that when he first began to display his emergency lights on Jefferson Street, the defendant was “a little bit further up ․ maybe three or four, maybe five car lengths in front of” him. Sergeant Bergeson had testified that he was directly behind the defendant while driving forty to forty-five miles per hour, not an “exact speed,” on Jefferson Street in a twenty-five miles per hour zone. In later testimony, however, after being shown the video of the event, which was recorded from Sergeant Bergeson's police cruiser dashboard camera, Sergeant Bergeson admitted that he was not close to the defendant's van, and that, as the video showed, another vehicle was in the left lane between the defendant's van and Sergeant Bergeson's cruiser. This vehicle only pulled over as Bergeson approached Wall Street. Further, the videodisc shows Jefferson Street in the area of Wall Street was a two lane, one-way street, and the defendant's van was in the left hand lane prior to turning onto Wall Street. The videodisc shows that there was no stop sign on Jefferson Street at the intersection of Jefferson Street and Wall Street. Sergeant Bergeson added that he was not following the New London police pursuit policy for engaging in pursuit and that he did not call out his own speeds during any pursuit. Moreover, the defendant's left turn onto Wall Street from Jefferson Street was visible to Sergeant Bergeson's video camera and to Sergeant Bergeson.5 Finally, Sergeant Bergeson also admitted that he was driving faster than the defendant to catch up to him.
The videodisc shows the defendant turned left onto Wall Street, and at its T intersection with Summer Street, at the top of the Wall Street hill, applied his brakes at the stop sign, put on his right turn signal and took a right onto Summer Street, heading toward Redden Avenue. The videodisc also shows that, as the turn was completed at the top of Wall Street, Sergeant Bergeson caught up to the rear of the defendant's vehicle to signal a stop.
As Sergeant Bergeson was following the defendant onto Redden Avenue, at the next turn, he radioed other officers, later recorded onto an audio disc, that he was “turning everything off.” He knew that the defendant was going “right to his house” and he did not want to cause any problems.6 The defendant arrived at his nearby house on Colman Street approximately sixty-six seconds after Sergeant Bergeson began following him from a distance on Jefferson Street.
Furthermore, the evidence had to show that the defendant was attempting to elude or escape the officer by speeding. The evidence from the video and audio discs does not reasonably support a finding that the defendant's conduct when signaled to stop was motivated by his intent to elude or escape Sergeant Bergeson. Sergeant Bergeson was forced to recognize this, when he said he discontinued the use of “everything” on Redden Avenue and said that the defendant was headed toward his home (on nearby Colman Street). At this point, Sergeant Bergeson testified at trial, he was disengaged from pursuit. The officers knew the identity of the defendant, knew his home was his destination, and Nott knew the defendant's daughter was with him .7
Furthermore, there was no increased speed shown on the videodisc while the van applied its brakes at every stop sign from the top of Wall Street and signaled the direction it was turning, all toward the defendant's home. At each intersection, Sergeant Bergeson testified, his speed following the defendant was a few miles per hour. Sergeant Bergeson followed the defendant's van as if in a slow procession toward the defendant's home once he had caught up to the van as it turned onto Summer Street.
As to the majority's reference to the defendant hearing Nott's call for assistance from the CVS lot, the audio disc of the call from Nott to Sergeant Bergeson reflects Sergeant Bergeson was asked to “hit a motor vehicle stop on it and maybe give him a ticket for interfering or ․ creating a disturbance,” to which Sergeant Bergeson stated, “Yeah. He's taking off right now.” Thus, the defendant had already left in his van as the call was being made. Moreover, the statute explicitly requires that the defendant be signaled to stop by a siren or flashing lights. General Statutes § 14–223(b).
The majority states that the cruiser videodisc reflects that the defendant made a “sudden” or “abrupt” left turn onto Wall Street from Jefferson Street, without braking. Here, the video shows Jefferson Street was a two lane, one-way street at its intersection with Wall Street, a side street with no stop sign on Jefferson Street.
In the presence of this contemporaneous contrary video and audio evidence, I would reverse the defendant's conviction for engaging Sergeant Bergeson in pursuit “at Wall Street” as unreasonable.
I agree with part II of the majority opinion that § 14–223(b) is not unconstitutionally vague because this statute's plain language may be read by ordinary people to understand what conduct is prohibited. State v. Stephens, 301 Conn. 791, 805–806, 22 A.3d 1262 (2011). I do not agree, however, with the majority's statement that the trial court did not, by implication, find that the officers were outside their duties when they pursued and arrested the defendant. The court did find the officers had no constitutional justification to support an arrest for interfering with Nott and dismissed count one at the close of the state's case. As such, the stop could be found outside the officer's duties as set forth in my dissent from part III of the majority opinion.
FAILURE TO CHARGE ON DEFENDANT'S TWO THEORIES OF DEFENSE
As to part III of the majority opinion, the defendant filed a request to charge the jury as to freedom of speech under the first amendment to the United States constitution and article first, § 5, of the Connecticut constitution.8 The court refused to give any part of that instruction because it concluded that the request was not relevant to the charges, including engaging in pursuit, to be considered by the jury. However, I do not agree that the requested charge was not relevant to the charges of engaging the officer in pursuit and of obstructing Sergeant Bergeson in the performance of his duty.
The standard to be used in considering the evidence as to the relevance of the charge is that the court reviews the evidence in the light most favorable to supporting the proposed charge. See Levesque v. Bristol Hospital, Inc., 286 Conn. 234, 247, 943 A.2d 430 (2008).
There was relevant evidence of Nott's interaction with the defendant at the CVS pharmacy parking lot in New London on the afternoon of April 23, 2009, and of Sergeant Bergeson's later interaction with the defendant that afternoon.
Nott testified at trial that she was assigned to investigate a minor two car motor vehicle accident that occurred on Jefferson Street in New London. Following the accident, both motor vehicle operators had gone into the nearby CVS parking lot and called the police, who arrived in the person of Nott. Nott spoke to both drivers, took their insurance information, and looked at their licenses.
Nott testified the defendant and his six year old daughter were also in the CVS parking lot. The defendant produced evidence at trial that they were picking up a pain medication prescription from CVS for the daughter written that day by her doctor, and which was to be taken every four hours as needed.
Nott also testified as follows: she was in her police cruiser doing paperwork after speaking to the operators involved in the accident, when the defendant came out of the pharmacy and began to speak to Dustin Colburn, one motorist involved in the accident. The defendant asked Colburn if Nott was going to give him a ticket, to which Colburn stated that he hoped not. The defendant then stated to Colburn that Nott was “[on] the wrong end of a lawsuit.” (Nott later testified that she was the defendant in a civil rights lawsuit brought by the defendant.) The defendant had also asked Colburn if Nott was bothering him and stated that Nott was “a no good cop or a bad cop․” The defendant also asked Colburn for his name and telephone number. At this, Nott exited her police cruiser and instructed the defendant to “get in his car and go away,” and Nott did not speak to Colburn at this time.
Nott also testified that the defendant did not leave then, but the defendant later went to his van and with his daughter drove onto Jefferson Street.
A police audio disc reveals that Nott was talking over the police radio with Sergeant Bergeson, the shift supervisor, and asked for a motor vehicle stop. Although she knew the defendant, Nott told Sergeant Bergeson that “he” was following her, had taken her picture, and was asking the people involved in the accident if she was going to give them a ticket and so forth. Nott asked Sergeant Bergeson to stop “him” in his motor vehicle and “maybe” give him a ticket for interfering with an officer or creating a disturbance, to which Sergeant Bergeson replied, “Yeah. He's taking off right now. What happened?”9 Nott replied, “He's been following me, and, uh, taking my picture and then asking the people that are involved in the accident, um, if I'm giving them a ticket, so forth and so on.”
Nott also testified at trial that she was “not so much” bothered by the defendant calling her no good and a bad cop. Nott also testified she was not proud of being sued.
After the defendant exited the CVS parking lot, Nott testified, she decided to leave the parking lot and go to the defendant's home. At that time, Nott returned the documents to the two drivers and told them they were free to leave. Before then leaving, Nott radioed police headquarters that she was clear from the accident and available for other work.
Sergeant Bergeson, a trained and experienced police officer of sixteen years, testified at trial that he was near the CVS parking lot when he received the radio call from Nott. He then went over to Jefferson Street in his cruiser while displaying his emergency lights and siren. The police videodisc shows after going through two red lights, Sergeant Bergeson began following the defendant's van at a “considerable” distance on Jefferson Street, up to Wall Street, a steep side road.10 Nott had not mentioned the defendant's name to Sergeant Bergeson in her radio transmission to him. However, Sergeant Bergeson knew that the driver on Jefferson Street was the defendant because he recognized the defendant's van, which bore his business' name. Sergeant Bergeson testified he was also a defendant in the same civil rights case involving Nott and knew the defendant. Sergeant Bergeson also conceded at trial, as did Nott, that it is not illegal to take a photograph of a police officer on duty.
Sergeant Bergeson testified that after the defendant pulled into his backyard carport, he parked behind the defendant's van across the driveway in front of the defendant's house with the siren still on. Sergeant Bergeson proceeded to the driver's side of the defendant's van as the defendant was exiting. Sergeant Bergeson had told the defendant that he was under arrest, and had directed him to get out of his vehicle. Sergeant Bergeson observed that the defendant's daughter was crying while she was in the front passenger seat. Upon approaching the defendant on foot, Sergeant Bergeson “pulled a Taser on [the defendant]․” He acknowledged that the Taser “shoots” approximately 50,000 volts of electricity, is considered a dangerous weapon,11 and can “hurt” someone.12 Sergeant Bergeson agreed that after he pulled out the Taser, he saw the defendant reenter his vehicle, close the door, and call the police station using his cell phone. At that time, Sergeant Bergeson, over the police radio as recorded on audio disc, stated to other officers that he would “take him out and Taser him if [he had] to.”13
In his call to the police dispatcher, as recorded on the audio disc, the defendant had reported that he had been threatened by the police and asked to speak to a captain or a lieutenant,14 which the requested jury instruction would have described as speech protected by the first amendment. After he reported the incident, the defendant was instructed by the dispatcher to submit to the officers. As recorded by videodisc, the defendant then exited his van as instructed by the dispatcher. At the conclusion of the defendant's telephone call, the police dispatcher called Sergeant Bergeson and informed him of the defendant's call to the police dispatcher.
Other police officers also had arrived at the defendant's house, including Nott and Officer Josh Bergeson (Officer Bergeson), a half-brother of Sergeant Bergeson. Officer Bergeson testified at trial that he heard the little girl “crying, kind of wailing, almost,” in the front passenger seat of the van. As Sergeant Bergeson testified, after the defendant left his van, Sergeant Bergeson returned the Taser to its holster. The videodisc from Sergeant Bergeson's cruiser shows Officer Bergeson handcuffing the defendant with his hands behind his back as Sergeant Bergeson watched nearby. The videodisc from Sergeant Bergeson's cruiser then showed Sergeant Bergeson putting the defendant down with a deprecating gesture as he passed him in handcuffs. Sergeant Bergeson did so by taking his upraised arm with his palm put down and bringing it down toward his body as the defendant passed by.
When the defendant was arrested, with his daughter's medication still in the van, his daughter was taken by a neighbor whom the defendant had called. The neighbor testified at the trial. Sergeant Bergeson testified he did not recall discussing the daughter's medication with the defendant, although the videodisc shows a conversation between the defendant and Sergeant Bergeson. When he was released from the police station, the defendant went to the hospital where he presented with two sprained wrists and a contusion to his flank. The defendant's medical records in evidence reflect the defendant (at age forty-six) also suffers from a cardiac disease, including high cholesterol and hypertension.
The defendant's jury trial had begun with the defendant facing four counts: (1) interference with Nott at the CVS parking lot, (2) risk of injury to his minor daughter by engaging the police in pursuit with his daughter in the vehicle, (3) engaging Sergeant Bergeson in pursuit, and (4) interfering with Sergeant Bergeson's attempt to arrest him by locking his vehicle. At the conclusion of the state's case, the court granted the defendant's motion for a judgment of acquittal as to the first count charging interference with Nott, finding that because the defendant's statements to Colburn at the CVS did not rise to “fighting words,” and were protected by the first amendment to the United States constitution, and there was no evidence that the defendant did anything to interfere with or delay Nott's accident investigation. The jury, however, was only instructed not to consider the charge involving Nott and not to speculate as to the reason. Under those instructions, the first amendment's freedom of speech guarantee, therefore, would have no bearing on the jury's consideration of the remaining charges, although all the defendant's statements about Nott remained in the record and were not addressed in the jury instructions. If given, the charge would have at least removed evidence of the defendant's remarks to Colburn as supporting the defendant's guilt of the remaining charges and contradicted part of the prosecutor's rebuttal argument as set forth below. As to prejudice, the prosecutor could and did improperly refer to the constitutional principles of freedom of speech as “some talk” and stated improperly the defendant should have awaited Nott's leaving to insult her record.
During deliberations, the jury sent the court a note asking for the court to reread the definition and instruction regarding a “personal frolic.” The court then repeated the instruction to the jury that the officer “in the performance of his duties” must be “simply acting within the scope of what he is employed to do. The test is whether the police officer is acting within that scope or is engaged in a personal frolic of his own.” As phrased, this instruction could lead the jury to believe if an officer acts as an officer, he is not engaged in a personal frolic. The requested instruction would indicate to the contrary that action taken in derogation of the defendant's freedom of speech rights was illegal and not in keeping with the officer's duty to act lawfully.
I would conclude that Sergeant Bergeson's conduct toward the defendant, threatening deadly force with a Taser when he knew that the defendant had brought his daughter home, evidence that Sergeant Bergeson knew that the defendant's photographing Nott was not illegal, the fact that the offending words used by the defendant were not “fighting words,” and were protected by the first amendment and referred to the defendant's lawsuit against Nott and Sergeant Bergeson, combined with Sergeant Bergeson's deprecating gesture putting down the defendant as he was being led away, could support a finding that Bergeson was motivated by personal feelings arising from the defendant's past interactions with him and Nott, including an existing civil rights lawsuit by the defendant against them both. The jury had to consider the issue of a personal animus, but it did not have before it the constitutional infirmity of Nott's request to Sergeant Bergeson to stop and issue a summons to the defendant for taking her picture and, as Nott testified, the defendant calling her “a no good or a bad cop,” which are not fighting words.
Fighting words consist of speech that “by [its] very utterance inflict injury or tend to incite an immediate breach of peace.” Chaplinsky v. New Hampshire, 315 U.S. 568, 572, 62 S.Ct. 766, 86 L.Ed. 1031 (1942); see also State v. Szymkiewicz, 237 Conn. 613, 620 n. 12, 68 A.2d 473 (1996); State v. Williams, 205 Conn. 456, 473, 534 A.2d 230 (1987). Moreover, “a properly trained officer may reasonably be expected to exercise a higher degree of restraint than the average citizen, and thus be less likely to respond belligerently to fighting words.” (Internal quotation marks omitted.) State v. Williams, supra, at 474 n. 7, quoting Houston v. Hill, 482 U.S. 451, 462, 107 S.Ct. 2502, 96 L.Ed.2d 398 (1987). “[F]ighting words are intended as provocation, which a police officer should be expected to resist, while a true threat is ‘a serious expression of intent to harm’ ․ the nature of which does not depend on the particular sensitivities of the listener.” (Citation omitted.) State v. DeLoreto, 265 Conn. 145, 162, 827 A.2d 671 (2003).15
“.” (Internal quotation marks omitted.) DiMartino v. Richens, 263 Conn. 639, 662, 822 A.2d 205 (2003).
The trial record reveals that but for the call from Nott, Bergeson would not have attempted to stop the defendant's vehicle and, “[pull] a Taser on [the defendant],” which caused the defendant to reenter his vehicle and close the door while he called the police dispatcher and asked to speak to a police captain, a supervisor, and to report that officers were threatening him.
The trial court dismissed the charge alleging interference with Nott at the CVS parking lot because the defendant did not use “fighting words” in the presence of Nott. The defendant's insults had led Nott to call Sergeant Bergeson for assistance, which led to subsequent events in a causal relationship, when Sergeant Bergeson followed the defendant home. Everything that later occurred was prompted by the defendant's exercise of his constitutionally protected speech. As the prosecutor said in his summation, the defendant at the CVS parking lot created the whole situation himself.
Our Supreme Court in 1987 and the United States Supreme Court since 1942 have repeatedly stated that fighting words may be excluded from the protections of the first amendment, but we do not have them in this case. See Chaplinsky v. New Hampshire, supra, 315 U.S. at 568; State v. Williams, supra, 205 Conn. at 475.
I believe that the charge requested should have been given as relevant to the defendant's claim to the jury that the defendant's words led to Sergeant Bergeson's actions. I would conclude that the jury should have been instructed, as the defendant sought, that the defendant's words were constitutionally protected speech and were not a valid basis for any offense. The jury should have been informed that Nott had no valid basis to request Sergeant Bergeson stop and issue a summons to the defendant for his words spoken to Colburn. Furthermore, Nott's orders to the defendant to step away or to leave were, in effect, an effort to muzzle him-the very antitheses of the right to freedom of speech. The defendant's words also could not provide a basis for Sergeant Bergeson to stop the defendant and, after finding the defendant outside his vehicle, “pull a Taser on [the defendant]” and violate the defendant's first amendment rights.
As to Sergeant Bergeson acting within the scope of his duties and the good faith belief of Sergeant Bergeson when “pull[ing] a Taser on [the defendant],” consideration of the first amendment was vital to a fair consideration of the interference charge as related to Sergeant Bergeson. See State v. Casanova, 255 Conn. 581, 590–94, 767 A.2d 1189 (2001). The instruction was necessary so that the jury would be bound to consider that Sergeant Bergeson's actions prejudiced the defendant's right to exercise his first amendment right to free speech to criticize Nott. That criticism, as nonfighting words, was the cause of Sergeant Bergeson's actions and the defendant's reactions that afternoon.
The instruction was also relevant to Sergeant Bergeson's conduct in pointing his Taser at the defendant while the defendant was leaving his van, causing the defendant to return to the van and call the police dispatcher, seeking to speak to a ranking officer about an officer threatening him with his daughter in the van. This call was the defendant's exercise of free speech to memorialize and report police conduct and seek aid in safeguarding his and his daughter's well-being. As such, the defendant's exercise of his rights, by retreating to his van to telephone the police dispatcher to alert senior officers to what was happening, could not support the charge of interfering with Sergeant Bergeson. Under the instructions, it could not support a criminal prosecution constituting a deprivation of his constitutional right to free speech. See Glik v. Cunniffe, 655 F.3d 78 (1st Cir.2011); Smith v. Cumming, 212 F.3d 1332 (11th Cir.), cert. denied, 531 U .S. 978, 121 S.Ct. 426, 148 L.Ed.2d 435 (2000); Williamson v. Mills, 65 F.3d 155 (11th Cir.1995).
The citizen's right to freedom of speech was added to the United States constitution early on (1791); McIntrye v. Ohio Elections Commission, 514 U.S. 334, 370, 115 S.Ct. 1511, 131 L.Ed.2d 426 (1995) (Thomas, J., concurring in the judgment); and was exemplified at a Vermont town meeting in Norman Rockwell's painting of the Four Freedoms. In Glik v. Cunniffe, supra, 655 F.3d at 82, the United States Court of Appeals for the First Circuit described that freedom as a necessary bulwark against state efforts to suppress criticism of public officers.
Arrests that interfere with the dissemination of such information, whether by the press or by a private individual, is a violation of the first amendment. See United States v. Grace, 461 U.S. 171, 184, 103 S.Ct. 1702, 75 L.Ed.2d 73 (1983) (Marshall, J., concurring and dissenting). A private individual reporting facts that memorialize deadly police activity on his property and directed at his body and his child while seeking police protection from harm I would hold is protected by the first amendment in this case.
The defendant also filed a request to charge on the defense of entrapment. When presenting this request, defense counsel stated that the record could also support a defense of justification (self-defense) arising from the Taser's deadly threat. The court refused to give that instruction as well.
The entrapment instruction would require proof that the police induced the defendant to commit an offense of interfering with Sergeant Bergeson's effort to arrest the defendant by his locking the vehicle as charged in the information.
The undisputed facts establish that the defendant was following Sergeant Bergeson's direction to leave his vehicle only to be met with the threat of the Taser under the control of an adversary in a civil rights case, in this case, a deadly threat. See footnote 12 of this opinion. The majority concludes Sergeant Bergeson's “pointing” his Taser at the defendant could not induce the defendant to reenter his vehicle and lock the door to call the police dispatcher. I would conclude there was evidence from which the jury could conclude inducement was proven. The evidence was not contested that the “pointing” (as described in the majority opinion) of the Taser resulted in the defendant's call to the police dispatcher from the defendant's van. I would respectfully disagree because this threat would constitute the most serious inducement known.
As to the purpose of inducing the defendant's action to institute a criminal prosecution against the defendant16 and that the defendant did not contemplate and would not have participated in such conduct, I would point to the circumstantial evidence that Sergeant Bergeson would be seeking such a charge of interference because of the defendant's lawsuit against him, the lack of a constitutional reason to stop and arrest him because of the first amendment, the fact that the defendant was complying with Sergeant Bergeson's orders to leave the vehicle, only to be confronted with the drawn Taser, and the other factors enumerated above as to the failure to give the freedom of speech instruction.
PROSECUTORIAL MISCONDUCT AND PREJUDICE
Finally, I dissent as to part IV of the majority opinion, where the prosecutor began his rebuttal jury summation with the statement: “Let me tell you why I think this case warrants conviction,” and later ended his summation with the statement: “If I'm going to convict someone, I better well have a good case. And I think we do,” which was stricken when the court sustained the defendant's objection and then the prosecutor stated, “Okay.”
Between these improper opening and closing statements of personal belief in the state's case, which the prosecutor knew to be improper, the prosecutor improperly asked the jurors to put themselves in the policeman's shoes. The prosecutor also asked the jury to view the facts with the lens of the prosecutor's own reaction when stopped by the police.17 The prosecutor withdrew this remark when the defendant objected.
The prosecutor also gave his own observations of Officer Bergeson's having denied during his testimony that he punched the defendant in the back. The prosecutor stated to the jury that “[h]e seemed surprised” to be asked, and the prosecutor added, “I don't know. You make the judgment,” a remark he repeated when addressing the defendant's claim that the officers caused the defendant's back contusion. The prosecutor also addressed the issue of the Taser by stating: “I mean, I don't really like Tasers myself.” Referring to a missing video of the police station booking room where the defendant had complained of his injuries, he stated: “I don't know, but what does it have to do with what the defendant did before?” These latter remarks, at the least, made the prosecutor a witness to his knowledge or lack thereof of signs of ill will against the defendant. Since the remarks were beyond the evidence in the record, they were unsworn testimony, which is not the subject of proper closing argument. “Moreover, when a prosecutor suggests a fact not in evidence, there is a risk that the jury may conclude that [the prosecutor] has independent knowledge of facts that could not be presented to the jury.” (Internal quotation marks omitted.) State v. Skakel, 276 Conn. 633, 746, 888 A.2d 985, cert. denied, 549 U.S. 1030, 127 S.Ct. 578, 166 L.Ed.2d 428 (2006).
The majority, in considering the effect of the summation under State v. Williams, 204 Conn. 523, 529 A.2d 653 (1987), refers to the state's case as “strong” or not “weak.” I disagree as to the engaging the officer in pursuit charge that required the jury to find the defendant sought by speeding to elude or escape from the police while driving at Wall Street in route directly to his house. It was also contrary to the video camera in Sergeant's Bergeson's cruiser, which showed the defendant's vehicle going directly to the defendant's home at a slow speed, after braking at stop signs and using his turn signal to indicate his future direction at each turn, while being closely followed by Sergeant Bergeson.
As to the interference with Sergeant Bergeson, the evidence is that the defendant only reentered his vehicle and closed the door to call the police headquarters to report events to senior officers and ask for police assistance when being confronted by an officer he was suing, who was pulling a Taser, a dangerous weapon, “on him” after he had left his vehicle as directed by that officer—all of which was not contested.
Accordingly, I respectfully dissent and would order a new trial as to the interference count.
FOOTNOTES
1. In State v. Williams, supra, 205 Conn. at 473, our Supreme Court stated: “To avoid the risk of constitutional infirmity, we construe § 53a–167a to proscribe only physical conduct and fighting words that by their very utterance inflict injury or tend to incite an immediate breach of the peace.” (Internal quotation marks omitted.)
2. Video recordings from cameras located in various police vehicles, including those of Bergeson and Nott, were transferred to a disc and admitted into evidence as state's exhibit # 1. On November 19, 2012, the parties filed a joint stipulation with this court indicating that attached the DVDs constituted four copies of the contents of the state's exhibit # 1. This court has reviewed the contents of the video recordings. Further, we note that there is no claim that the video records have been altered in any way.
3. We note that defense counsel conducted a rigorous cross-examination of Bergeson regarding the details of the pursuit, using not only the video, but also Bergeson's police report. Further, defense counsel highlighted certain inconsistencies and pointed out Bergeson's failure to follow policies regarding police pursuits.
4. We acknowledge that the discussion regarding whether the police were in the performance of their duties in both Privitera and Davis was in the context of the interplay between General Statutes §§ 53a–23, 53a–167a (a) and 53a–167c (a). Nevertheless, we conclude that the question of whether Nott and Bergeson were in the performance of their official duties or acting illegally and on a personal frolic is a factual determination that was not made by the court, acting on the defendant's motion for a judgment of acquittal, in this case.
5. The defendant preserved this claim by submitting a written request to charge. See State v. Vilchel, 112 Conn.App. 411, 417, 963 A.2d 658, cert. denied, 291 Conn. 907, 969 A.2d 173 (2009); see also Practice Book § 42–16.
6. During his closing argument to the jury, defense counsel referenced the first amendment and entrapment issues.
7. See Practice Book § 42–16.
8. Our Supreme Court has stated.” State v. Payne, 303 Conn. 538, 562–63, 34 A.3d 370 (2012).
9. “We previously have recognized that a claim of prosecutorial impropriety, even in the absence of an objection, has constitutional implications and requires a due process analysis․” (Citation omitted; internal quotation marks omitted.) State v. Gibson, 302 Conn. 653, 658–59, 31 A.3d 346 (2011). We note that following the argument of counsel, the defendant moved for a mistrial on the basis of the closing argument made by the prosecutor. The court denied the defendant's motion.
10. “The propriety of the prosecutor's statement must, however, be examined in the context in which it was made.” State v. Felix, 111 Conn.App. 801, 809, 961 A.2d 458 (2008).
11. The state set forth the following in its brief: “Comments addressing the strength of the state's case are permissible when properly connected to evidence adduced at trial․ Because, however, the prosecutor did not directly connect the statement to evidence, the state agrees that the comment was improper.” (Citation omitted.)
12. Defense counsel objected to the prosecutor's question to members of the jury regarding what they would do if they saw red flashing lights in their rearview mirror while operating a motor vehicle. The court overruled the objection.
13. The state directs our attention to State v. Houle, supra, 105 Conn.App. at 824, in which the prosecutor improperly commented on his personal experience with alcohol. The state acknowledges the similarities between that case and the present one.
1. The second amended substitute information, filed during the trial, charged in count two that the defendant “at Wall Street, did increase the speed of the motor vehicle he was driving in an attempt to escape or elude [Sergeant] Bergeson of the New London Police Department when said police officer activated his emergency lights and sirens indicating to the accused to stop, but instead, the accused increased his speed in violation of ․ General Statutes [§ ] 14–223(b).”
2. The majority notes there is no claim the video recordings have been altered in any way. I note as well there is no claim the audio discs have been altered in any way.
3. See footnote 5 of this opinion.
4. The amended substitute information restricted the evidence to the defendant's speed at Wall Street. See footnote 1 of this opinion.
5. Due to the considerable distance between Sergeant Bergeson's cruiser and the defendant's vehicle, the cruiser's video camera did not clearly show the use of a turn signal except to show yellow and red lights on the rear left side of the defendant's vehicle.
6. The conversation was as follows:“[Sergeant Bergeson]: I'm trying to stop one Charlie Mike 630 [the defendant's license plate], going up Redden. He's not stopping.“[Another Officer]: [Inaudible] units to․“[Sergeant Bergeson]: [Inaudible] He's heading right to his house. I'm gonna turn everything off so it doesn't cause a problem, and then I'm gonna take him into custody.”
7. Sergeant Bergeson testified Nott called him on Redden Avenue to tell Sergeant Bergeson that the defendant's daughter was in the defendant's van. However, the recordings demonstrate that Officer Nott's call was made to Sergeant Bergeson later while Sergeant Bergeson was in the backyard of the defendant's home on Colman Street.The conversation between Sergeant Bergeson and Nott was as follows:“[Sergeant Bergeson]: Shut my siren off, please, somebody, before you come in the backyard. That's where I'm at.“[Nott]: 3 to Lima, just so you know, his child is in the front passenger seat, 3 is in route.”An examination of the time stamps on the audio discs reveals that Sergeant Bergeson received Nott's transmission that a child was in the defendant's van only after the defendant was in his backyard carport.
8. The defendant requested the following: “The first amendment to the [United States] constitution states in part that state ‘shall make no law ․ abridging the freedom of speech.’ An expanded right to speech and expression exists under [article first, § 5, of] the Connecticut [c]onstitution, which states that ‘No law shall ever be passed to curtail or restrain the liberty of speech or of the press.’ The loss of first amendment freedoms, for even minimal periods of time, unquestionably constitutes a violation of basic rights. These provisions apply equally to the actions of state and local officials, including police officers and school officials..“In addition to actual words, the first amendment also protects an individual's right to document police conduct by videotaping or taking pictures of their activities, subject to reasonable time, manner and place restrictions. You have heard evidence that [Daniel G. See * footnote of the majority opinion.] took a picture of Officer Nott, which was the subject of Officer Nott's complaint. [Daniel G.'s. See * footnote of the majority opinion.] actions in taking this photograph were also protected by the first amendment.” (Emphasis added.)
9. The audio communication between Nott and Sergeant Bergeson was as follows:“[Sergeant Bergeson]: Go ahead [inaudible]“[Nott]: Three to Lima. Three to Lima.“[Sergeant Bergeson]: Go ahead.“[Nott]: Come over to CVS for me, please.“[Sergeant Bergeson]: Okay. Do you need me? I'm just walking into Neal's. I'll head over there.“[Another Officer]: Three—Is it something Sierra can take care of or..?“[Sergeant Bergeson]: Yup, I know what it is. I'll be right over.“[Nott]: If you can, hit a motor vehicle stop on it and maybe give him a ticket for interfering or creating a disturbance.“[Sergeant Bergeson]: Yeah. He's taking off right now. What happened?“[Nott]: He's been following me, and, uh, taking my picture and then asking the people that are involved in the accident, um, if I'm giving them a ticket, so forth and so on.”
10. In its brief, the state admits that on Jefferson Street the defendant was “considerably ahead” of Sergeant Bergeson.
11. A Taser is a type of controlled electronic weapon capable of firing wires tipped with a pair of barbed darts to deliver a paralyzing electric charge. Bryan v. MacPherson, 630 F.3d 805, 824 (9th Cir.2010). A police Taser, which is capable of causing death or serious injury, may meet the definition of a dangerous instrument under General Statutes § 53–206 and § 53a–3 (7).
12. Some recent medical studies have concluded that the electrical shock delivered by a Taser could cause death by cardiac arrest. See, e.g., D. Zipes, MD, “Sudden Cardiac Arrest and Death Following Application of Shocks From a TASER Electronic Control Device,” Circulation Journal of the American Heart Association (2012).Recent Connecticut deaths following Taser use have prompted citizen groups to call for legislation to regulate the use of Tasers. See G. Merritt, The CT Mirror, “Taser death spurs calls for more regulation,” (September 4, 2013), available at taser-death-spurs-calls-more-regulation (copy contained in the file of this case in the Appellate Court clerk's office) (noting thirteen Connecticut deaths following Taser use since 2005).
13. The radio transmission is transcribed as follows:“[Another Officer 1]: 2 to something to report 132.“[Sergeant Bergeson]: Q, uh, he.. locked himself in the car, I guess.“[Another Officer 1]: Alrighty. What's your location?“[Sergeant Bergeson]: 95 [Colman]. I'll take him out and Taser him if I have to.“[Another Officer 1]: We weren't able to copy that location. Did anybody get it?“[Another Officer 2]: 4 [inaudible].“[Another Officer 2]: 4 is almost out [inaudible].“[Sergeant Bergeson]: Sierra is out.“[Another Officer 1]: Roger.”
14. The defendant's telephone call to the police dispatcher is transcribed as follows:“[The Defendant]: Just threatened my life.“[Police Dispatcher]: Communications.“[The Defendant]: Yes, may I speak to the captain or whoever's on duty? “[Police Dispatcher]: I beg your pardon?“[The Defendant]: I need to speak to the captain or lieutenant on duty please ․“[Police Dispatcher]: What's your name?“[The Defendant]: My name is [Daniel G. See * footnote of the majority opinion.] I'm being threatened by officers right now and my daughter. They are threatening us both.“[Police Dispatcher]: [OK, sir]. I'm going to ask you to comply with the officers. Okay. And do what they tell you to do.“[The Defendant]: I wanna speak to the captain, okay?“[Police Dispatcher]: Sir, sir. There is a lieutenant on scene there.“[The Defendant]: There is? Which lieutenant?“[Police Dispatcher]: [Lieutenant] Bergeson is there. Well he's acting [Lieutenant] Bergeson․”
15. A true threat is another unprotected form of speech. See State v. Moulton, 310 Conn. 337, 349, 78 A.3d 55 (2013).
16. Inducing the defendant's action for the purpose of instituting a criminal prosecution against him would not have been required for a justification defense.
17. The prosecutor described such a stop as one to be seen in the “rearview mirror” where the police flashing lights were directly behind the operator.
DiPENTIMA, C.J.
In this opinion BEACH, J., concurred.
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f.
climate change and financial regulation February 2, 2018Posted by Bradley in : financial regulation , add a comment
I have a short piece on this topic in the current Miami Law Magazine.
new draft of paper: climate change and brexit as financial stability risks July 21, 2017Posted by Bradley in : financial regulation , add a comment
Here: Climate Change and Brexit as Financial Stability Risks (July 2017 version)..
consequences).
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Martha, age 86, not only leads morning fitness walks five days a week for the SaddleBrooke Hiking Club, she was the oldest hiker in the Arizona Trail in a Day event on October 6. Martha is a role model for staying active through walking and hiking.
Features, November 2018
A role model
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\begin{document}
\bibliographystyle{amsplain}
\title{Siegel's Theorem and the Shafarevich Conjecture}
\author{Aaron Levin}
\address{Department of Mathematics\\Michigan State University\\East Lansing, MI 48824}
\email{adlevin@math.msu.edu}
\date{}
\begin{abstract}
It is known that in the case of hyperelliptic curves the Shafarevich conjecture can be made effective, i.e., for any number field $k$ and any finite set of places $S$ of $k$, one can effectively compute the set of isomorphism classes of hyperelliptic curves over $k$ with good reduction outside $S$. We show here that an extension of this result to an effective Shafarevich conjecture for {\it Jacobians} of hyperelliptic curves of genus $g$ would imply an effective version of Siegel's theorem for integral points on hyperelliptic curves of genus $g$.
\end{abstract}
\maketitle
\section{Introduction}
The famous 1929 theorem of Siegel \cite{Sie} on integral points on affine curves states (in a formulation convenient for us):
\begin{theorem}[Siegel]
Let $C$ be a curve over a number field $k$, $S$ a finite set of places of $k$ containing the archimedean places, $\O_{k,S}$ the ring of $S$-integers, and $f\in k(C)$ a nonconstant rational function on $C$. If $C$ is a rational curve then we assume further that $f$ has at least three distinct poles. Then the set of $S$-integral points of $C$ with respect to $f$,
\begin{equation*}
C(f,k,S)=\{P\in C(k)\mid f(P)\in \O_{k,S}\},
\end{equation*}
is finite.
\end{theorem}
While Siegel's theorem is completely satisfactory from a qualitative viewpoint, all known proofs of the theorem suffer from the defect of being ineffective, i.e., in general there is no known algorithm for explicitly computing the set $C(f,k,S)$ (when it is finite). In the classical proofs of Siegel's theorem, this ineffectivity arises from the use of Roth's theorem from Diophantine approximation (for a survey on the use of Roth's theorem in Siegel's theorem, including some remarks on effectivity, see \cite{Zan}). Finding an effective version of Siegel's theorem remains a longstanding important open problem.
Of course, in certain special cases there are known techniques for effectively computing $C(f,k,S)$. In this context, the most powerful and widely used effective techniques come from Baker's theory of linear forms in logarithms \cite{Bak}. Using these techniques, Baker and Coates \cite{BC} proved an effective version of Siegel's theorem for curves of genus zero and genus one. Already for curves of genus two, however, it is an open problem to prove an effective version of Siegel's theorem. More generally, we will be interested here in studying this problem for hyperelliptic curves:
\begin{problem}
\label{mp}
Find an effective version of Siegel's theorem for curves $C$ of genus two. More generally, find an effective version of Siegel's theorem for hyperelliptic curves $C$.
\end{problem}
Again, in certain special cases, Problem \ref{mp} has been solved. For instance, if $C$ is a (nonsingular projective) hyperelliptic curve over a number field $k$, $i$ denotes the hyperelliptic involution of $C$, and $f\in k(C)$ is a rational function that has a pole at both $P$ and $i(P)$ for some point $P\in C(\kbar)$ (where we allow $P=i(P)$), then $C(f,k,S)$ is effectively computable for any appropriate finite set of places $S$. This is essentially equivalent to effectively finding all solutions $x,y\in \O_{k,S}$ to hyperelliptic equations of the form
\begin{equation}
\label{hypeq}
y^2=a_nx^n+\cdots+a_0, \quad x,y\in \O_{k,S},
\end{equation}
where $a_0,\ldots, a_n\in k$ are constants and the equation defines a hyperelliptic curve. Explicit bounds for the solutions to \eqref{hypeq} (when $\O_{k,S}=\mathbb{Z}$) were first given by Baker \cite{Bak2}. Along the lines of equation \eqref{hypeq}, we note that using Riemann-Roch it is possible to restate Problem \ref{mp} as a question about integral solutions to certain specific types of equations. For instance, if $C$ has genus two, Problem \ref{mp} is equivalent to solving equations of the form \eqref{hypeq} (with $n=5,6$) and equations of the form
\begin{equation*}
y^3+g_1(x)y^2+g_2(x)y=x^4+g_3(x), \quad x,y\in \O_{k,S},
\end{equation*}
where $g_i\in k[x]$ has degree $\leq i$. We refer the reader to \cite{Grant} or \cite{Pou} for details.
When $C$ has genus two, Bilu \cite{Bilu} has shown that there exists an infinite set $M\subset C(\kbar)$ such that if $f$ has poles at two distinct points of $M$, then $C(f,k,S)$ is effectively computable. The method of \cite{Bilu} combines linear forms in logarithms with functional units and coverings of curves, and the exact limitations of the method do not yet seem to be fully understood. Another interesting alternative approach to an effective Siegel's theorem for genus two curves is given by Grant in \cite{Grant}. There, the problem is reduced to questions about integral points on a certain affine subset of the Jacobian of $C$, which in turn are reduced to certain ``non-Abelian $S$-unit equations".
A final case of particular note where an effective version of Siegel's theorem is known is the case of geometrically Galois coverings of the projective line, proved independently by Bilu \cite{Bilu2} and Dvornicich and Zannier \cite{DZ} (partial results were also obtained in \cite{Kl} and \cite{Pou}). This generalizes, in a qualitative way, the aforementioned results of Baker and Coates \cite{BC} and Baker \cite{Bak2} on integral points on elliptic curves and hyperelliptic curves, respectively. Quantitative results in this context were proven by Bilu in \cite{Bilu3}. We refer the reader to \cite{Bilu} for more general statements and other cases of an effective Siegel's theorem.
When $C$ is a curve of genus $g\geq 2$, Siegel's theorem is superseded by Faltings' theorem (Mordell's Conjecture), which states that in this case the set of rational points $C(k)$ is finite for any number field $k$. Faltings' proof of the Mordell conjecture \cite{Fal} used a reduction (due to Parshin) to the Shafarevich conjecture, proved by Faltings in the same paper. Moreover, R\'emond has shown \cite{Rem} that an effective version of the Shafarevich conjecture would imply an effective version of Faltings' theorem. In a similar vein, we will show that Problem \ref{mp} can be reduced to proving an effective version of a restricted form of the Shafarevich conjecture (hyperelliptic Jacobians). For instance, we will show that an effective version of Siegel's theorem for genus two curves follows from an effective version of the Shafarevich conjecture for abelian surfaces.
Before stating the main theorem, we make a few more definitions. For a nonsingular projective curve $C$, we let $\Jac(C)$ denote the Jacobian of $C$. Let $g\geq 2$ be an integer, $k$ a number field, and $S$ a finite set of places of $k$, which we will always assume contains the archimedean places. Define
\begin{multline*}
\mathcal{H}(g,k,S)=\{\text{$k$-isomorphism classes of (nonsingular projective) hyperelliptic }\\
\text{curves over $k$ of genus $g$ with good reduction outside $S$}\}
\end{multline*}
and
\begin{multline*}
\mathcal{H}'(g,k,S)=\{\text{$k$-isomorphism classes of hyperelliptic curves $C$ over $k$ of genus $g$}\\
\text{ such that $\Jac(C)$ has good reduction outside $S$}\}.
\end{multline*}
As is well known, if $C$ has good reduction at a place $v$ then so does $\Jac(C)$. In other words, we have $\mathcal{H}(g,k,S)\subset \mathcal{H}'(g,k,S)$. In general, this set inclusion is proper. In fact, as we will see, understanding the difference between the two sets is in some sense the key to solving Problem \ref{mp}. It follows from the Shafarevich conjecture for abelian varieties, proved by Faltings, that the set $\mathcal{H}'(g,k,S)$ is finite. We now state our main result.
\begin{theorem}
\label{mt}
Let $g\geq 2$ be an integer. Suppose that for any number field $k$ and any finite set of places $S$ of $k$ the set $\mathcal{H}'(g,k,S)$ is effectively computable (e.g., an explicit hyperelliptic Weierstrass equation for each element of the set is given). Then for any number field $k$, any finite set of places $S$ of $k$, any hyperelliptic curve $C$ over $k$ of genus $g$, and any rational function $f\in k(C)$, the set of $S$-integral points with respect to $f$,
\begin{equation*}
C(f,k,S)=\{P\in C(k)\mid f(P)\in \O_{k,S}\},
\end{equation*}
is effectively computable.
\end{theorem}
We now give a brief outline of the proof of Theorem \ref{mt}. To each point $P\in C(f,k,S)$ we associate a certain double cover $\pi_P:\tilde{C}_P\to C$ such that $\tilde{C}_P$ has good reduction outside some fixed finite set of places $T$ of some number field $L$ (with $T$ and $L$ independent of $P$). Associated to $\pi_P$ we have a classical construction, the Prym variety $\Prym(\tilde{C}_P/C)$. When $C$ is hyperelliptic, it will turn out that $\Prym(\tilde{C}_P/C)$ is isomorphic to the Jacobian $\Jac(X_P)$ of some hyperelliptic curve $X_P$ and $\Jac(X_P)$ has good reduction outside of $T$. This yields a map $C(f,k,S)\to \mathcal{H}'(g,L,T)$, $P\mapsto X_P$, and if $\mathcal{H}'(g,L,T)$ is known then $C(f,k,S)$ can be explicitly computed. We note that an analogous construction essentially works for any curve $C$ (except that $\Prym(\tilde{C}_P/C)$ will not be a hyperelliptic Jacobian, or even a Jacobian, for general $C$), but we focus on the hyperelliptic case as there are reasons to believe that effectively computing $\mathcal{H}'(g,k,S)$ may be a tractable problem.
Indeed, a primary reason for this hope is that the set $\mathcal{H}(g,k,S)$ is known to be effectively computable, i.e., there is an effective version of the Shafarevich conjecture for hyperelliptic curves.
\begin{theorem}[Effective Shafarevich conjecture for hyperelliptic curves, von K\"anel \cite{vK}]
\label{tEG}
Let $g\geq 2$ be an integer, $k$ a number field, and $S$ a finite set of places of $k$. The set $\mathcal{H}(g,k,S)$ is effectively computable. \end{theorem}
The finiteness of $\mathcal{H}(g,k,S)$ goes back to work of Shafarevich \cite{Sha}, Merriman \cite{Mer}, Parshin \cite{Par}, and Oort \cite{Oort}. Building on earlier work of Merriman and Smart \cite{MS}, Smart \cite{Smart} explicitly computed the set $\mathcal{H}(2,\mathbb{Q},\{2,\infty\})$ using effective results of Evertse and Gy\"ory \cite{EG} on related problems concerning discriminants of binary forms. Using the results of Evertse and Gy\"ory \cite{EG} and a result of Liu \cite{Liu}, von K\"anel \cite{vK} proved explicit bounds for the heights of Weierstrass models of the hyperelliptic curves represented in $\mathcal{H}(g,k,S)$. We note that the much earlier proof of Oort \cite{Oort} also yields a certain weaker version of Theorem \ref{tEG}.\footnote{Oort's proof of finiteness in \cite{Oort} (implicitly) yields a fixed computable finite extension $L$ of $k$ and effective bounds for the heights of Weierstrass models {\it over $L$} of the hyperelliptic curves represented in $\mathcal{H}(g,k,S)$.}
In \cite{Poo}, Poonen raised the problem of extending the computations of Merriman and Smart to an effective computation of $\mathcal{H}'(2,\mathbb{Q},\{2,\infty\})$. Our results give additional motivation and importance, coming from Siegel's theorem, to the problem of extending Theorem \ref{tEG} from the set $\mathcal{H}(g,k,S)$ to the set $\mathcal{H}'(g,k,S)$. In view of Theorems \ref{mt} and \ref{tEG}, to solve Problem \ref{mp} it would suffice to solve the following problem:
\begin{problem}
Find an effective bound $B(g,k,S)$ such that if $C$ is a hyperelliptic curve of genus $g$ over $k$ and $\Jac(C)$ has good reduction outside $S$, then $C$ has good reduction at all primes $\mathfrak{p}$ of $k$ with norm $\Norm(\mathfrak{p})> B(g,k,S)$.
\end{problem}
Note that an ineffective bound $B(g,k,S)$ follows trivially from the finiteness of $\mathcal{H}'(g,k,S)$.
\section{Parshin's construction}
Parshin \cite{Par} was the first to notice that the Mordell conjecture could be obtained as a consequence of the Shafarevich conjecture. The basic idea is to associate to each rational point $P\in C(k)$ a covering $\pi_P:\tilde{C}_P\to C$ such that the curve $\tilde{C}_P$ has good reduction outside a finite set $S$ (independent of $P$) and $\pi_P$ has certain specified ramification. We will use a similar construction to study integral points on curves.
\begin{theorem}
\label{Par}
Let $C$ be a nonsingular projective curve over a number field $k$, $f\in k(C)$ a nonconstant rational function, and $S$ a finite set of places of $k$ containing the archimedean places. Fix a pole $Q\in C(\kbar)$ of $f$. There exists a number field $L\supset k$ and a finite set of places $T$ of $L$ with the following property: for any $P\in C(f,k,S)$ and any double cover $\pi:\tilde{C}\to C$ over $\kbar$ ramified exactly above $P$ and $Q$, there exists a morphism of nonsingular projective curves $\pi':\tilde{C}'\to C$ with the following properties:
\begin{enumerate}
\item There is an isomorphism $\psi:\tilde{C}\to \tilde{C}'$ over $\kbar$ such that $\pi=\pi'\circ\psi$.
\item $\pi'$ and $\tilde{C}'$ are both defined over $L$.
\item $\deg \pi'=2$.
\item $\pi'$ is ramified exactly above $P$ and $Q$.
\item Both $\tilde{C}'$ and $C$ have good reduction outside $T$.
\end{enumerate}
\end{theorem}
We note that for any $P$ and $Q$ such maps $\pi$ exist (see Theorem \ref{bij}). We will use the following result proven by Silverman in \cite[Prop.\ 3, Lemma 4]{Sil}, where it is attributed to Szpiro and Ogus.
\begin{lemma}
\label{OS}
Let $C$ be a nonsingular projective curve over a number field $L$ and let $g\in L(C)$. Suppose that
\begin{equation*}
\dv(g)=\pm P_1 \pm P_2\pm \cdots \pm P_n +pD,
\end{equation*}
where $P_1,\ldots, P_n\in C(\Lbar)$, $D$ is some divisor on $C$, and $p$ is a prime. Let $\mathscr{C}$ be a model for $C$ over $\O_L$. Let $T$ be a finite set of places of $L$ such that:
\begin{enumerate}
\item $T$ contains all archimedean places of $L$.
\label{l1}
\item $T$ contains all places of bad reduction of $\mathscr{C}$.
\item $T$ contains all places of $L$ lying above $p$.
\item The ring of $T$-integers $\O_{L,T}$ has class number one.
\label{l4}
\end{enumerate}
There exists an element $\alpha\in L^*$ such that if $C'$ is a nonsingular projective curve with function field $L(C)(\sqrt[p]{\alpha g})$ and $v\not\in T$ is a place of $L$ such that $P_1,\ldots, P_n$ are distinct modulo $v$, then $C'$ has good reduction at $v$.
\end{lemma}
We now prove Theorem \ref{Par}.
\begin{proof}[Proof of Theorem \ref{Par}]
Without loss of generality, we may assume that $Q\in C(k)$ and that all of the $2$-torsion of $\Jac(C)$ is rational over $k$. Let $L$ be a number field such that
\begin{equation}
\label{MW}
\Jac(C)(k)\subset 2\Jac(C)(L).
\end{equation}
The existence of such an $L$ follows from the (weak) Mordell-Weil theorem and we recall that such a field $L$ can be explicitly given. We may assume without loss of generality that $S$ is a finite set of places of $k$ containing all places of bad reduction of $C$, all places of $k$ lying above the prime $2$, and such that $\O_{k,S}$ has trivial class group. Let $s_1,\ldots, s_n$ be generators of $\O_{k,S}^*$. Then $L=k(\sqrt{s_1},\ldots,\sqrt{s_n})$ will be a number field satisfying \eqref{MW} (\cite[\S C.1]{HS}).
Let $\mathscr{C}$ be a model of $C$ over $\O_L$. Let $T$ be a finite set of places of $L$ satisfying \eqref{l1}--\eqref{l4} of Lemma \ref{OS} (with $p=2$) and such that
\begin{enumerate}
\setcounter{enumi}{4}
\item $T$ contains every place of $L$ lying above a place of $S$.
\item $T$ contains every finite place $v$ of $L$ such that $f$ is identically $0$ or $\infty$ modulo $v$.
\item $T$ contains every finite place $v$ of $L$ such that a zero of $f$ reduces to a pole of $f$ modulo $v$.
\end{enumerate}
Let $P\in C(f,k,S)$. Let $\pi:\tilde{C}\to C$ be a double cover (over $\kbar$) ramified exactly above $P$ and $Q$. Then $\pi$ corresponds to an extension of function fields $\kbar(C)\subset \kbar(C)(\sqrt{g})$ for some rational function $g\in \kbar(C)$. It is a standard fact that $\pi$ is ramified above a point $R\in C(\kbar)$ if and only if $g$ has a pole or zero of odd order at $R$. Since $\pi$ is ramified exactly above $P$ and $Q$, we must have
\begin{equation*}
\dv(g)=P-Q+2D
\end{equation*}
for some divisor $D$. Since $\Jac(C)(k)\subset 2\Jac(C)(L)$, the divisor class $[D]$ is $L$-rational and $D\sim E$ for some $L$-rational divisor $E$. Then for an appropriate rational function $h\in \kbar(C)$ satisfying $\dv(h)=E-D$, after replacing $g$ by $gh^2$ we can assume that $g\in L(C)$.
Let $v\not\in T$ be a place of $L$. Since $f(P)\in \O_{k,S}\subset \O_{L,T}$ by assumption, it follows from the definition of $T$ that $P$ cannot reduce to the pole $Q$ of $f$ modulo $v$. Then by Lemma \ref{OS}, there exists $\alpha\in L^*$ such that if $\tilde{C}'$ is the nonsingular projective curve with function field $L(C)(\sqrt{\alpha g})$, then $\tilde{C}'$ has good reduction outside $T$. Let $\pi':\tilde{C}'\to C$ be the morphism corresponding to the inclusion of function fields $L(C)\subset L(C)(\sqrt{\alpha g})$. Then $\tilde{C}'$ and $\pi'$ satisfy the required properties.
\end{proof}
\section{Prym Varieties}
Let $\pi:\tilde{C}\to C$ be a morphism of nonsingular projective curves of degree two. There are two natural maps between the Jacobians $\Jac(\tilde{C})$ and $\Jac(C)$ that one can associate to $\pi$. First, we have the pullback map
\begin{align*}
\pi^*:\Jac(C)&\to \Jac(\tilde{C}),\\
[D]&\mapsto [\pi^*D],
\end{align*}
where $[D]$ denotes the divisor class of a divisor $D$ of degree $0$. Second, we have the so-called norm map
\begin{align*}
\Nm:\Jac(\tilde{C})&\to \Jac(C),\\
\left[\sum n_PP\right]&\mapsto \left[\sum n_P\pi(P)\right].
\end{align*}
\begin{definition}
The Prym variety associated to the double cover $\pi:\tilde{C}\to C$ is defined by
\begin{equation*}
\Prym(\tilde{C}/C)=(\ker \Nm)^0,
\end{equation*}
the connected component of $\ker \Nm$ containing the identity.
\end{definition}
We recall some basic facts about Prym varieties (see \cite{Mum}). Let $\iota:\tilde{C}\to \tilde{C}$ be the involution of $\tilde{C}$ interchanging the two sheets of $\pi$. This induces an involution $\iota:\Jac(\tilde{C})\to\Jac(\tilde{C})$. Then we have the identities
\begin{equation*}
\Prym(\tilde{C}/C)=\ker(1+\iota)^0=\im (1-\iota).
\end{equation*}
Let $i:\Prym(\tilde{C}/C)\to \Jac(\tilde{C})$ be the inclusion. Assume further now that $C$ has positive genus and that $\pi$ is either \'etale or ramified above exactly two points of $C$. We have an isogeny
\begin{align*}
\Jac(C)\times \Prym(\tilde{C}/C)&\to \Jac(\tilde{C}),\\
(P,Q)&\mapsto \pi^*(P)+i(Q).
\end{align*}
Furthermore, if $\theta_{\tilde{C}}$ is a theta divisor on $\Jac(\tilde{C})$, then $i^*\theta_{\tilde{C}}\equiv 2\Xi$ for some ample divisor $\Xi$ on $\Prym(\tilde{C}/C)$ and $\Xi$ yields a principal polarization of $\Prym(\tilde{C}/C)$. So $\Prym(\tilde{C}/C)$ can naturally be given the structure of a principally polarized abelian variety. We will find the following lemma useful.
\begin{lemma}
\label{Serre}
Suppose that $\tilde{C}$, $C$, and $\pi:\tilde{C}\to C$ are defined over a number field $k$. Let $v$ be a finite place of $k$. Then $\Jac(\tilde{C})$ has good reduction at $v$ if and only if both $\Prym(\tilde{C}/C)$ and $\Jac(C)$ have good reduction at $v$.
\end{lemma}
\begin{proof}
The result follows \cite[Cor.\ 2]{Ser} from the fact that the abelian varieties $\Jac(\tilde{C})$ and $\Jac(C)\times \Prym(\tilde{C}/C)$ are $k$-isogenous.
\end{proof}
In general, the Prym variety $\Prym(\tilde{C}/C)$ is not the Jacobian of a curve. However, in the important special case where $C$ is hyperelliptic this holds (and the Prym variety is a hyperelliptic Jacobian).
\begin{theorem}[Dalaljan \cite{Dal1,Dal2}, Mumford \cite{Mum}]
\label{Dal}
Suppose that $C$ is hyperelliptic and $\pi:\tilde{C}\to C$ is a double cover of $C$ that is either unramified or ramified above exactly two points of $C$. Then $(\Prym(\tilde{C}/C),\Xi)$ is isomorphic to the Jacobian $(\Jac(C'),\theta_{C'})$ of some hyperelliptic curve $C'$.
\end{theorem}
Suppose that $C$ is a hyperelliptic curve of genus $g$. For $Q\in C(\kbar)$ and $P\in C(f,k,S)$ as in Theorem \ref{Par}, we have a double cover $\tilde{C}_P\to C$ ramified exactly above $P$ and $Q$ and having nice reduction properties. By Theorem \ref{Dal}, $\Prym(\tilde{C}_P/C)\cong \Jac(X_P)$ for some hyperelliptic curve $X_P$ of genus $g$. In the next section we will explicitly compute a Weierstrass equation for $X_P$.
\section{Some explicit equations}
We begin by recalling some results from \cite[\S 3]{Dal2}.
\begin{theorem}[Dalaljan]
\label{tDal}
Let $k$ be an algebraically closed field, $\chr k\neq 2$. Let $C$ be a nonsingular projective hyperelliptic curve over $k$ with hyperelliptic map $p:C\to \mathbb{P}^1$. Let $\pi:\tilde{C}\to C$ be a double cover of nonsingular projective curves, ramified above exactly two points of $C$. Associated to $\pi$ and $p$ there exists a tower of curves
\begin{equation}
\label{tower}
\begindc{\commdiag}[57]
\obj(2,0)[20]{$\mathbb{P}^1$}
\obj(1,1)[11]{$C$}
\obj(2,1)[21]{$C_0$}
\obj(3,1)[31]{$C_1$}
\obj(0,2)[02]{$\tilde{C}'$}
\obj(1,2)[12]{$\tilde{C}$}
\obj(2,2)[22]{$\tilde{C}_0$}
\obj(3,2)[32]{$\tilde{C}_1$}
\obj(4,2)[42]{$\tilde{C}_1'$}
\obj(2,3)[23]{$\tilde{\tilde{C}}$}
\mor{11}{20}{$p$}[\atright, \solidarrow]
\mor{21}{20}{$p_0$}
\mor{31}{20}{$p_1$}
\mor{02}{11}{$\pi'$}[\atright, \solidarrow]
\mor{12}{11}{$\pi$}[\atright, \solidarrow]
\mor{22}{11}{$\pi_0$}[\atright, \solidarrow]
\mor{22}{21}{$\pi_0''$}
\mor{22}{31}{$\pi_0'$}
\mor{32}{31}{$\pi_1$}
\mor{42}{31}{$\pi_1'$}
\mor{23}{02}{$\tilde{\pi}'$}[\atright, \solidarrow]
\mor{23}{12}{$\tilde{\pi}$}[\atright, \solidarrow]
\mor{23}{22}{$\tilde{\pi}_0$}
\mor{23}{32}{$\tilde{\pi}_1$}
\mor{23}{42}{$\tilde{\pi}_1'$}
\enddc
\end{equation}
with $C_1\cong \mathbb{P}^1$, $\tilde{C}_1$ hyperelliptic, $\deg \pi_1=2$, and
\begin{equation*}
(\Prym(\tilde{C}/C),\Xi)\cong (\Jac(\tilde{C}_1),\theta_{\tilde{C}_1}).
\end{equation*}
For any positive integer $g$, the tower \eqref{tower} induces a bijection between equivalence classes of towers
\begin{multline*}
\{[\tilde{C}\stackrel{\pi}{\rightarrow}C\stackrel{p}{\rightarrow}\mathbb{P}^1]\mid \deg \pi=\deg p=2, g(\tilde{C})=2g, g(C)=g \}\longleftrightarrow\\
\{[\tilde{C}_1\stackrel{\pi_1}{\rightarrow}C_1\stackrel{p_1}{\rightarrow}\mathbb{P}^1]\mid \deg \pi_1=\deg p_1=2, g(\tilde{C}_1)=g, C_1\cong \mathbb{P}^1\}.
\end{multline*}
\end{theorem}
Here, we say that two towers of curves $X\stackrel{\pi}{\rightarrow} Y\stackrel{p}{\rightarrow} Z$ and $X'\stackrel{\pi'}{\rightarrow} Y'\stackrel{p'}{\rightarrow} Z'$ are equivalent if there exists a commutative diagram
\begin{equation*}
\begindc{\commdiag}[40]
\obj(0,2)[02]{$X$}
\obj(0,1)[01]{$Y$}
\obj(0,0)[00]{$Z$}
\obj(1,2)[12]{$X'$}
\obj(1,1)[11]{$Y'$}
\obj(1,0)[10]{$Z'$}
\mor{02}{01}{$\pi$}
\mor{01}{00}{$p$}
\mor{12}{11}{$\pi'$}
\mor{11}{10}{$p'$}
\mor{02}{12}{}
\mor{01}{11}{}
\mor{00}{10}{}
\enddc
\end{equation*}
where the horizontal morphisms are isomorphisms.
\begin{theorem}
\label{bij}
Let $k$ be an algebraically closed field, $\chr k\neq 2$. Let $C$ be a (nonsingular projective) hyperelliptic curve over $k$ of genus $g$ given by an equation
\begin{equation*}
y^2=\prod_{i=1}^{2g+1}(x-\alpha_i), \quad \alpha_i\in k,
\end{equation*}
and $p:C\to \mathbb{P}^1$ the hyperelliptic map $(x,y)\mapsto x$. Let $P=(x_P,y_P), Q=(x_Q,y_Q)\in C(k)$ with $x_P\neq x_Q$ and $y_P\neq 0$. The tower \eqref{tower} induces a bijection between the sets
\begin{equation*}
\{\text{double covers } \pi:\tilde{C}\to C, \text{ up to equivalence, ramified exactly above $P$ and $Q$}\}
\end{equation*}
and
\begin{equation*}
\left\{\text{hyperelliptic curves }\tilde{C}_1: y^2=(x-1)\prod_{i=1}^{2g+1}\left(x-\beta_i\right)\mid \beta_i^2=\frac{x_Q-\alpha_i}{x_P-\alpha_i}, \prod_{i=1}^{2g+1} \beta_i=\frac{y_Q}{y_P}\right\}.
\end{equation*}
Both sets have cardinality $2^{2g}$.
\end{theorem}
\begin{proof}
We first give an explicit description of the set of double covers $\pi:\tilde{C}\to C$ ramified exactly above $P$ and $Q$. At the function field level, we have $k(\tilde{C})=k(C)(\sqrt{\phi})$ for some rational function $\phi\in k(C)$. The corresponding morphism $\pi$ is ramified exactly above $P$ and $Q$ if and only if $\dv(\phi)=P+Q+2D$ for some divisor $D$. It's easily seen then that there is a bijection
\begin{multline*}
\{\text{double covers } \pi:\tilde{C}\to C, \text{ up to equivalence, ramified exactly above $P$ and $Q$}\} \\
\longleftrightarrow\{[D]\mid D\in \Div(C), 2D\sim -(P+Q)\},
\end{multline*}
where $\Div(C)$ denotes the group of divisors of $C$ and $[D]$ the linear equivalence class of $D$. As is well-known, the latter (and hence former) set has cardinality $2^{2g}$.
Let $\infty$ denote the (unique) point of $C$ at infinity relative to the equation $y^2=\prod_{i=1}^{2g+1}(x-\alpha_i)$. Then we have an embedding $C\to \Jac(C)$, $\pt\mapsto [\pt-\infty]$. Every element of $\Jac(C)$ can be represented by a divisor of the form $P_1+\cdots+P_g-g\infty$, $P_1,\ldots,P_g\in C(k)$. Then in each linear equivalence class of divisors $D$ satisfying $2D\sim -(P+Q)$, we can find a divisor $D$ of the form
\begin{equation*}
D=P_1+\cdots+P_g-(g+1)\infty, \quad P_1,\ldots,P_g\in C(k).
\end{equation*}
Let $\pi:\tilde{C}\to C$ be a double cover ramified exactly above $P$ and $Q$. Then we find that $k(\tilde{C})=k(C)(\sqrt{\phi})$ for some rational function $\phi\in k(C)$ satisfying
\begin{equation*}
\dv(\phi)=P+Q+2P_1+\cdots+2P_g-(2g+2)\infty.
\end{equation*}
As usual, let $L(D)$ be the $k$-vector space $L(D)=\{\psi \in k(X)\mid\dv(\psi)\geq -D\}$. By Riemann-Roch, $\dim L((2g+2)\infty)=g+3$ and a basis for $L((2g+2)\infty)$ is given by the rational functions $1,x,\ldots,x^{g+1},y$. It follows that we can write
\begin{equation*}
\phi=ay+h(x)
\end{equation*}
for some $a\in k$ and some $h\in k[x]$ with $\deg h\leq g+1$. Since $x_P\neq x_Q$, we have $a\neq 0$, and so we can assume $a=1$. Let $f(x)=\prod_{i=1}^{2g+1}(x-\alpha_i)$. Then $\tilde{C}$ has an equation of the form
\begin{align*}
y^2&=f(x),\\
z^2&=y+h(x),
\end{align*}
where $h\in k[x]$ and $\deg h\leq g+1$ (where throughout, we mean that the curve, here $\tilde{C}$, is a projective normalization of the curve defined by the given equations).
We now describe the relevant part of the tower \eqref{tower} (see \cite{Dal2}). For a double cover $\phi:X'\to X$ of nonsingular projective curves, we let $i_\phi$ denote the corresponding involution of $X'$. We set $\tilde{C}'=\tilde{C}$ and $\pi'=i_p\circ \pi$, where $p:C\to \mathbb{P}^1$ is the projection onto the $x$-coordinate. Let $\tilde{\tilde{C}}=\tilde{C}\times_{C}\times \tilde{C}'$ and let $\tilde{\pi}$ and $\tilde{\pi}'$ be the natural projection maps onto $\tilde{C}$ and $\tilde{C}'$, respectively. Explicitly, $\tilde{\tilde{C}}$ consists of pairs $(\tilde{P},\tilde{P}')\in \tilde{C}\times \tilde{C}'$ such that $\pi(\tilde{P})=\pi'(\tilde{P}')$. Then $\tilde{\tilde{C}}$ can be given by the equations
\begin{align*}
y^2&=f(x),\\
z^2&=y+h(x),\\
z'^2&=-y+h(x).
\end{align*}
Let $i_{\tilde{\pi}_0}=i_{\tilde{\pi}'}\circ i_{\tilde{\pi}}$ and let $i_{\tilde{\pi}_1}$ be the involution of $\tilde{\tilde{C}}=\tilde{C}\times_{C}\times \tilde{C}'$ that switches the coordinates. In the coordinates $(x,y,z,z')$ these are the involutions $(x,y,z,z')\mapsto (x,y,-z,-z')$ and $(x,y,z,z')\mapsto (x,-y,z',z)$, respectively. Taking the quotients of $\tilde{\tilde{C}}$ by $i_{\tilde{\pi}_0}$ and $i_{\tilde{\pi}_1}$, we obtain curves $\tilde{C}_0$ and $\tilde{C}_1$, respectively, and double covers $\tilde{\pi}_0:\tilde{\tilde{C}}\to \tilde{C}_0$ and $\tilde{\pi}_1:\tilde{\tilde{C}}\to \tilde{C}_1$, respectively. Note that
\begin{align*}
(z+z')^2&=2h(x)+2zz',\\
(zz')^2&=h(x)^2-f(x).
\end{align*}
Using the above equations, we see that $\tilde{C}_0$ is given by the equations
\begin{align*}
y^2&=f(x),\\
z^2&=h(x)^2-f(x),
\end{align*}
and $\tilde{C}_1$ by the equations
\begin{align}
w^2&=2h(x)+2z,\label{C11}\\
z^2&=h(x)^2-f(x)\label{C12}.
\end{align}
The involutions $i_{\tilde{\pi}_0}$ and $i_{\tilde{\pi}_1}$ commute and generate a group of order $4$. The quotient of $\tilde{\tilde{C}}$ modulo the action of this group yields the curve $C_1$ in the tower. The curve $C_1$ is naturally given as the projective normalization of the curve defined by the equation
\begin{equation*}
z^2=h(x)^2-f(x).
\end{equation*}
The induced maps $\pi_0'$ and $\pi_1$ are then induced by the natural projection maps onto the $x$ and $z$ coordinates. From the definition of $h(x)$, we have
\begin{equation}
\label{C12p}
h(x)^2-f(x)=(x-x_P)(x-x_Q)F(x)^2,
\end{equation}
for some polynomial $F\in k[x]$. It follows that $C_1\cong \mathbb{P}^1$. As $\deg \pi_0'=\deg \pi_1=2$, we see that $\tilde{C}_0$ and $\tilde{C}_1$ are hyperelliptic curves.
A hyperelliptic Weierstrass equation for $\tilde{C}_1$ can be computed as follows. In view of \eqref{C12p}, we can parametrize \eqref{C12} by setting
\begin{align}
t^2&=\frac{x-x_Q}{x-x_P},\label{t2}\\
x&=x(t)=\frac{x_Q-x_Pt^2}{1-t^2},\notag\\
z&=z(t)=t(x(t)-x_P)F(x(t)).\notag
\end{align}
Substituting into \eqref{C11}, we see that we need to consider the polynomial
\begin{align*}
G(t)&=(1-t^2)^{g+1}(h(x(t))+z(t))\\
&=(1-t^2)^{g+1}\left(h\left(\frac{x_Q-x_Pt^2}{1-t^2}\right)+t\left(\frac{x_Q-x_P}{1-t^2}\right)F\left(\frac{x_Q-x_Pt^2}{1-t^2}\right)\right).
\end{align*}
Let $\alpha_i, i=1,\ldots, 2g+1$, be the roots of $f$. Then when $x=\alpha_i$, \eqref{C12} gives $z=\pm h(\alpha_i)$. From \eqref{t2}, it then follows that for some (unique) choice of the square root, $t=\sqrt{\frac{x_Q-\alpha_i}{x_P-\alpha_i}}=\beta_i$ is a root of $G(t)$, $i=1,\ldots, 2g+1$. From \eqref{C12p}, it follows that the leading coefficients of $h(x)$ and $F(x)$ are equal up to sign. This easily implies that either $t=1$ or $t=-1$ is a root of $G(t)$. Replacing $F(x)$ by $-F(x)$, if necessary, we can assume that $t=1$ is a root of $G(t)$. Since $\deg G\leq 2g+2$, we find that the roots of $G$ are exactly given by the $2g+2$ distinct elements $t=1,\beta_1,\ldots, \beta_{2g+1}$.
We have $G(0)=h(x_Q)=-y_Q$, from the definition of $h$. The leading coefficient of $G$ can be computed as
\begin{equation*}
\lim_{t\to\infty}(-1)^{g+1}\frac{G(t)}{(1-t^2)^{g+1}}=(-1)^{g+1}h(x_P)=-(-1)^{g+1}y_P.
\end{equation*}
Thus,
\begin{equation*}
1\cdot \prod_{i=1}^{2g+1} \beta_i=\frac{-y_Q}{-(-1)^{g+1}y_P}=(-1)^{g+1}\frac{y_Q}{y_P}.
\end{equation*}
If $g$ is odd, so that $(1-t^2)^{g+1}$ is an even power of $1-t^2$, then it follows from the above and \eqref{C11} and \eqref{C12} that $\tilde{C}_1$ has a Weierstrass equation
\begin{equation}
\label{finaleq}
y^2=(x-1)\prod_{i=1}^{2g+1}\left(x-\beta_i\right), \qquad \beta_i^2=\frac{x_Q-\alpha_i}{x_P-\alpha_i}, \prod_{i=1}^{2g+1} \beta_i=\frac{y_Q}{y_P}.
\end{equation}
If $g$ is even, then
\begin{equation*}
\frac{(1-t^2)^{g+2}}{(t-1)^2}(h(x(t))+z(t))=c'(t+1)\prod_{i=1}^{2g+1}\left(t-\beta_i\right),
\end{equation*}
for some constant $c'$ and where $\prod_{i=1}^{2g+1} \beta_i=-\frac{y_Q}{y_P}$. Replacing $t$ by $-t$, we again see that $\tilde{C}_1$ has a Weierstrass equation \eqref{finaleq}.
In the coordinates of \eqref{finaleq}, the tower $\tilde{C}_1\stackrel{\pi_1}{\rightarrow} C_1\stackrel{p_1}{\rightarrow}\mathbb{P}^1$ is given by the hyperelliptic map $(x,y)\mapsto x$ followed by the squaring map $C_1=\mathbb{P}^1\to \mathbb{P}^1$, $x\mapsto x^2$. The tower $\tilde{C}_1\stackrel{\pi_1}{\rightarrow} C_1\stackrel{p_1}{\rightarrow}\mathbb{P}^1$ is associated to the tower $\tilde{C}\stackrel{\pi}{\rightarrow} C\stackrel{p}{\rightarrow}\mathbb{P}^1$, and from the above, $\tilde{C}_1$ has an equation given by \eqref{finaleq}. Note that there are exactly $2^{2g}$ possibilities for the curve \eqref{finaleq}, and hence for the tower $[\tilde{C}_1\stackrel{\pi_1}{\rightarrow} C_1\stackrel{p_1}{\rightarrow}\mathbb{P}^1]$, given by the $2^{2g}$ possibilities for $\beta_i$ subject to the constraint $\prod_{i=1}^{2g+1} \beta_i=\frac{y_Q}{y_P}$. Since the map
\begin{equation*}
[\tilde{C}\stackrel{\pi}{\rightarrow} C\stackrel{p}{\rightarrow}\mathbb{P}^1]\mapsto [\tilde{C}_1\stackrel{\pi_1}{\rightarrow} C_1\stackrel{p_1}{\rightarrow}\mathbb{P}^1]
\end{equation*}
given by the tower \eqref{tower} is injective and there are $2^{2g}$ possibilities for $[\tilde{C}\to C\to \mathbb{P}^1]$, it follows that all of the $2^{2g}$ possibilities for $\tilde{C}_1$ arise from some covering $\tilde{C}\to C$, proving the theorem.
\end{proof}
\begin{corollary}
\label{cormain}
Let $C$ be a hyperelliptic curve over a number field $k$ of genus $g$ given by an equation
\begin{equation*}
y^2=\prod_{i=1}^{2g+1}(x-\alpha_i), \quad \alpha_i\in k.
\end{equation*}
Let $f\in k(C)$ be a nonconstant rational function and $S$ a finite set of places of $k$ containing the archimedean places. Fix a pole $Q=(x_Q,y_Q)\in C(\kbar)$ of $f$. There exists a number field $L\supset k$ and a finite set of places $T$ of $L$ such that for any $P=(x_P,y_P)\in C(f,k,S)$ with $x_P\neq x_Q$, $y_P\neq 0$, any $\beta_i$ satisfying $\beta_i^2=\frac{x_Q-\alpha_i}{x_P-\alpha_i}, \prod_{i=1}^{2g+1} \beta_i=\frac{y_Q}{y_P}$, and some $c=c(\beta_1,\ldots,\beta_{2g+1})\in L^*$, the Jacobian $\Jac(X_P)$ of the curve
\begin{equation*}
X_P: y^2=c(x-1)\prod_{i=1}^{2g+1}\left(x-\beta_i\right)
\end{equation*}
is defined over $L$ and has good reduction outside of $T$.
\end{corollary}
\begin{proof}
Without loss of generality, we may assume that $Q\in C(k)$. Let $L$ and $T$ be as in Theorem \ref{Par}. Let $P=(x_P,y_P)\in C(f,k,S)$ with $x_P\neq x_Q$, $y_P\neq 0$. Let $\beta_i$, $i=1,\ldots, 2g+1$, satisfy $\beta_i^2=\frac{x_Q-\alpha_i}{x_P-\alpha_i}, \prod_{i=1}^{2g+1} \beta_i=\frac{y_Q}{y_P}$. If $P'=(x',y')\in C(k)$, we have $x'-\alpha_i=a_iz_i^2$ for some $z_i\in k$ and some $a_i\in k$ divisible only by primes of $k$ dividing the discriminant of $\prod_{i=1}^{2g+1}(x-\alpha_i)$. Then there is a fixed finite extension of $L$ (independent of $P'\in C(k)$) containing $\sqrt{x'-\alpha_i}=z_i\sqrt{a_i}$ for all $i$. By replacing $L$ by this finite extension (and replacing $T$ by the set of places lying above it), we can assume that we always have $\beta_i\in L$, $i=1,\ldots, 2g+1$.
Let $X_P'$ be the curve $y^2=(x-1)\prod_{i=1}^{2g+1}\left(x-\beta_i\right)$. By Theorem~\ref{Par} and Theorem~\ref{bij}, there is a double covering $\pi_P:\tilde{C}_P\to C$, ramified exactly above $P$ and $Q$, that corresponds to $\tilde{C}_1=X_P'$ via the tower \eqref{tower}, such that both $\pi_P$ and $\tilde{C}_P$ are defined over $L$ and both $\tilde{C}_P$ and $C$ have good reduction outside $T$. By Theorem \ref{tDal},
\begin{equation*}
\Prym(\tilde{C}_P/C)\cong \Jac(X_P')
\end{equation*}
over $\Lbar$. In fact, the proofs of Theorems \ref{tDal} and \ref{bij} show that
\begin{equation*}
\Prym(\tilde{C}_P/C)\cong \Jac(X_P)
\end{equation*}
over $L$, where $X_P$ is some quadratic twist of $X_P'$ given by
\begin{equation*}
X_P: y^2=c(x-1)\prod_{i=1}^{2g+1}\left(x-\beta_i\right),
\end{equation*}
for some $c\in L^*$. Since both $\tilde{C}_P$ and $C$ have good reduction outside $T$, by Lemma~\ref{Serre}, $\Jac(X_P)$ has good reduction outside $T$.
\end{proof}
An alternative, more direct proof of this result follows from Theorems \ref{binthm} and \ref{gr} in Section \ref{ls}.
\section{Main Theorem}
We now prove the main theorem from the introduction.
\begin{theorem}
\label{mainthm}
Let $g\geq 2$ be an integer. Suppose that for any number field $k$ and any finite set of places $S$ of $k$ the set $\mathcal{H}'(g,k,S)$ is effectively computable (e.g., an explicit hyperelliptic Weierstrass equation for each element of the set is given). Then for any number field $k$, any finite set of places $S$ of $k$, any hyperelliptic curve $C$ over $k$ of genus $g$, and any rational function $f\in k(C)$, the set of $S$-integral points with respect to $f$,
\begin{equation*}
C(f,k,S)=\{P\in C(k)\mid f(P)\in \O_{k,S}\},
\end{equation*}
is effectively computable.
\end{theorem}
\begin{proof}
Without loss of generality, by enlarging $k$ we can assume that every Weierstrass point of $C$ is $k$-rational and that some pole $Q=(x_Q,y_Q)$ of $f$ is $k$-rational. Then $C$ can be given by a hyperelliptic Weierstrass equation
\begin{equation*}
C:y^2=\prod_{i=1}^{2g+1}(x-\alpha_i), \quad \alpha_i\in k.
\end{equation*}
Let $U$ consist of the set of Weierstrass points of $C$ along with $Q$ and its image under the hyperelliptic involution. By Corollary \ref{cormain}, for some number field $L$ and some finite set of places $T$ of $L$, we have a map (arbitrarily choosing among the choices for $X_P$)
\begin{align*}
C(f,k,S)\setminus U&\to \mathcal{H}'(g,L,T),\\
P&\mapsto X_P.
\end{align*}
Explicitly, we can compute $C(f,k,S)$ from $\mathcal{H}'(g,L,T)$ as follows. Recall that for $P=(x_P,y_P)\in C(f,k,S)\setminus U$, $X_P$ is defined by an equation
\begin{equation*}
y^2=c_P(x-1)\prod_{i=1}^{2g+1}\left(x-\sqrt{\frac{x_Q-\alpha_i}{x_P-\alpha_i}}\right),
\end{equation*}
for some $c_P\in L^*$ and some appropriate choice of the square roots. We pick four of the roots of the polynomial on the right-hand side, say $\beta_i=\sqrt{\frac{x_Q-\alpha_i}{x_P-\alpha_i}}, i=1,2,3,4$, and consider the cross-ratio
\begin{equation*}
\CR(\beta_1,\beta_2,\beta_3,\beta_4)=\frac{(\beta_1-\beta_3)(\beta_2-\beta_4)}{(\beta_2-\beta_3)(\beta_1-\beta_4)}.
\end{equation*}
Alternatively, we consider the rational function
\begin{equation}
\label{rat}
\frac{(c_1z_3-c_3z_1)(c_2z_4-c_4z_2)}{(c_2z_3-c_3z_2)(c_1z_4-c_4z_1)}
\end{equation}
on the curve defined by $z_i^2=x-\alpha_i$, $i=1,2,3,4$, where we view $c_i=\sqrt{x_Q-\alpha_i}$, $i=1,2,3,4$, as fixed constants. Note that by Kummer theory, since the $\alpha_i$ are distinct, we have $[\kbar(x,z_1,\ldots,z_4):\kbar(x)]=2^4=16$. This immediately implies that the rational function \eqref{rat} is nonconstant and thus that the equation $\CR(\beta_1,\beta_2,\beta_3,\beta_4)=\alpha$ has only finitely many solutions in $x_P$ for any $\alpha\in \kbar$. Now let $C'\in \mathcal{H}'(g,L,T)$, given by a Weierstrass equation $C':y^2=c'\prod_{i=1}^{2g+2}(x-\gamma_i)$. If $X_P\cong C'$, then $\CR(\beta_1,\beta_2,\beta_3,\beta_4)=\CR(\gamma_i,\gamma_j,\gamma_k,\gamma_l)$ for some $i,j,k,l\in \{1,\ldots, 2g+2\}$. Since there are only finitely many possible cross-ratios $\CR(\gamma_i,\gamma_j,\gamma_k,\gamma_l)$, we find that there are only finitely many (explicitly computable) possible points $P=(x_P,y_P)$ with $X_P\cong C'$. Finally, for each such possible point $P$ and each point $P\in U$, we check if $P\in C(f,k,S)$.
\end{proof}
\section{Binary Forms}
\label{ls}
In this section we give a reformulation of some of our results in terms of binary forms. Let $k$ be a number field and $S$ a finite set of places of $k$ (which we always assumes contains the archimedean places). Let $F(X,Z), G(X,Z)\in k[X,Z]$ be binary forms. Let $U=\left( \begin{array}{cc}
a & b \\
c & d
\end{array} \right)$, $a,b,c,d\in k$, be a matrix. Define $F_U(X,Z)=F(aX+bZ,cX+dZ)$. We will call $F$ and $G$ equivalent if there exists $U\in \GL_2(k)$ and $\lambda\in k^*$ such that $G(X,Z)=\lambda F_U(X,Z)$. Denote the equivalence class containing $F$ by $[F]$. Let $\disc(F)$ denote the discriminant of $F$. Define
\begin{equation*}
\mathcal{B}(r,k,S)=\{[F]\mid F\in \O_{k,S}[X,Z] \text{ is a binary form of degree $r$ and $\disc(F)\in \O_{k,S}^*$}\}.
\end{equation*}
Effective finiteness of the set $\mathcal{B}(r,k,S)$ follows from work of Evertse and Gy\"ory.
\begin{theorem}[Evertse, Gy\"ory \cite{EG}]
Let $r\geq 2$ be an integer, $k$ a number field, and $S$ a finite set of places of $k$. The set $\mathcal{B}(r,k,S)$ is finite and effectively computable.
\end{theorem}
We now define a larger, but related, set $\mathcal{B'}(r,k,S)\supset \mathcal{B}(r,k,S)$. The set $\mathcal{B'}(r,k,S)$ contains equivalence classes of certain binary forms $F$ whose discriminant is an $S$-unit outside of primes $\mathfrak{p}$ where $F \pmod{\mathfrak{p}}$ has a factor of multiplicity $\geq 3$. An effective procedure for computing $\mathcal{B}'(r,k,S)$ would give a solution to Problem \ref{mp} (Corollary \ref{ce}).
More precisely, define $\mathcal{B}'(r,k,S)$ to be the set of equivalence classes of binary forms over $k$ of degree $r$ such that there exists a representative $F\in \O_{k,S}[X,Z]$ satisfying:\\
If $\mathfrak{p}\not\in S$ and $\ord_{\mathfrak{p}}\disc(F)>0$, then $\ord_{\mathfrak{p}}\disc(F)=2mn(n-1)$, where $m$ is some positive integer, $n$ is an odd integer with $3\leq n\leq 2[(r+1)/2]-3$, and $f(x)=F(x,1)$ has $n$ roots $\alpha_1,\ldots,\alpha_n\in k$ with $\ord_{\mathfrak{p}}\alpha_i=2m$, $i=1,\ldots, n$.\\
If $\disc(F)\in \O_{k,S}^*$ or $r\leq 4$, then the above condition is vacuous. So we trivially have $\mathcal{B}(r,k,S)\subset \mathcal{B}'(r,k,S)$ for all $r$ and $\mathcal{B}(r,k,S)=\mathcal{B}'(r,k,S)$ if $1\leq r\leq 4$.
\begin{theorem}
\label{binthm}
Let $C$ be a hyperelliptic curve over a number field $k$ of genus $g$ given by an equation
\begin{equation*}
y^2=h(x)=\prod_{i=1}^{2g+1}(x-\alpha_i), \quad \alpha_i\in k.
\end{equation*}
Let $f\in k(C)$ be a nonconstant rational function and $S$ a finite set of places of $k$ containing the archimedean places. Fix a pole $Q=(x_Q,y_Q)\in C(\kbar)$ of $f$. Let $U$ be the set of Weierstrass points of $C$ along with $Q$ and its image under the hyperelliptic involution. Then there exists a number field $L$ and finite set of places $T$ of $L$ such that we have a well-defined map
\begin{align*}
C(f,k,S)\setminus U&\to \mathcal{B}'(2g+2,L,T),\\
P=(x_P,y_P)&\mapsto \left[(X-Z)\prod_{i=1}^{2g+1}(X-\beta_iZ)\right],
\end{align*}
where the $\beta_i$ are chosen such that $\beta_i^2=\frac{x_Q-\alpha_i}{x_P-\alpha_i}$ and $\prod_{i=1}^{2g+1} \beta_i=\frac{y_Q}{y_P}$.
\end{theorem}
We will see (Corollary \ref{SF}) that $\mathcal{B}'(2g+2,k,S)$ is a finite set. Then by essentially the same argument as in the proof of Theorem \ref{mainthm}, we obtain the following corollary.
\begin{corollary}
\label{ce}
Let $g\geq 2$ be an integer. Suppose that for any number field $k$ and any finite set of places $S$ of $k$ the set $\mathcal{B}'(2g+2,k,S)$ is effectively computable. Then for any number field $k$, any finite set of places $S$ of $k$, any hyperelliptic curve $C$ over $k$ of genus $g$, and any rational function $f\in k(C)$, the set of $S$-integral points with respect to $f$,
\begin{equation*}
C(f,k,S)=\{P\in C(k)\mid f(P)\in \O_{k,S}\},
\end{equation*}
is effectively computable.
\end{corollary}
\begin{proof}[Proof of Theorem \ref{binthm}]
First note that there is an explicit number field $L$, depending on $C$ and $k$, such that for any $P\in C(f,k,S)$ we have $\beta_i\in L$ for all $i$. Then after enlarging $k$, without of loss of generality it suffices to prove the theorem for points $P\in C(f,k,S)$ such that $\beta_i\in k$, $i=1,\ldots, 2g+1$. Similarly, by enlarging $S$ we can assume without of loss of generality that $\alpha_i\in \O_{k,S}$, $x_Q-\alpha_i\in \O_{k,S}^*$, $i=1,\ldots, 2g+1$, $\disc(h)\in \O_{k,S}^*$, $\O_{k,S}$ is a principal ideal domain, every place of $k$ lying above $2$ is in $S$, and
\begin{equation}
\label{meq}
\min\{\ord_\mathfrak{p}(x_P-x_Q),\ord_\mathfrak{p}(y_P-y_Q)\}\leq 0
\end{equation}
for every $\mathfrak{p}\in M_k\setminus S$ and every $P=(x_P,y_P)\in C(f,k,S)$, where $M_k$ denotes the canonical set of places of $k$ (identifying nonarchimedean places with the corresponding prime ideal).
Let $P\in C(f,k,S)$ with $\beta_i\in k$, $i=1,\ldots, 2g+1$, as in the theorem. Since $\O_{k,S}$ is principal, we can write $\beta_i=\frac{\gamma_i}{\delta_i}$, where $\gamma_i$ and $\delta_i$ are relatively prime $S$-integers, $i=1,\ldots, 2g+1$. Consider the binary form
\begin{equation*}
F(X,Z)=(X-Z)\prod_{i=1}^{2g+1}(\delta_iX-\gamma_iZ)\in \O_{k,S}[X,Z],
\end{equation*}
which is equivalent to $(X-Z)\prod_{i=1}^{2g+1}(X-\beta_iZ)$. Let $\mathfrak{p}\in M_k\setminus S$.
Suppose that $\ord_{\mathfrak{p}}(x_P-x_Q)=0$. Then we claim that $\ord_\mathfrak{p}\disc(F)=0$. For this, it suffices to show that $\ord_\mathfrak{p}(\gamma_i\delta_j-\gamma_j\delta_i)=0$, $i\neq j$, and $\ord_\mathfrak{p}(\gamma_i-\delta_i)=0$ for all $i$. We have the identity
\begin{equation}
\label{bid}
\beta_i^2-\beta_j^2=\frac{\gamma_i^2\delta_j^2-\gamma_j^2\delta_i^2}{\delta_i^2\delta_j^2}=\frac{x_Q-\alpha_i}{x_P-\alpha_i}-\frac{x_Q-\alpha_j}{x_P-\alpha_j}=\frac{(x_P-x_Q)(\alpha_j-\alpha_i)}{(x_P-\alpha_i)(x_P-\alpha_j)}.
\end{equation}
Since $\disc(h)\in \O_{k,S}^*$, we have $\ord_\mathfrak{p}(\alpha_j-\alpha_i)=0$. By assumption, $\ord_\mathfrak{p}(x_P-x_Q)=\ord_\mathfrak{p}((x_P-\alpha_i)-(x_Q-\alpha_i))=0$. This last equality, along with $\ord_\mathfrak{p}(x_Q-\alpha_i)\geq 0$, implies that $\ord_\mathfrak{p}(x_P-\alpha_i)=2\ord_\mathfrak{p}\delta_i$. Then
\begin{align*}
\ord_\mathfrak{p}(\gamma_i^2\delta_j^2-\gamma_j^2\delta_i^2)&=\ord_\mathfrak{p}((\gamma_i\delta_j-\gamma_j\delta_i)(\gamma_i\delta_j+\gamma_j\delta_i))\\
&=\ord_\mathfrak{p}\frac{\delta_i^2\delta_j^2(x_P-x_Q)(\alpha_j-\alpha_i)}{(x_P-\alpha_i)(x_P-\alpha_j)}=0.
\end{align*}
Since $\gamma_i\delta_j\pm\gamma_j\delta_i$ is an $S$-integer, we find that $\ord_\mathfrak{p}(\gamma_i\delta_j-\gamma_j\delta_i)=0$. Similarly, we find that $\ord_\mathfrak{p}(\gamma_i-\delta_i)=0$. So $\ord_\mathfrak{p}\disc(F)=0$ as desired.
Now suppose that $\ord_{\mathfrak{p}}(x_P-x_Q)<0$. Since $x_Q,\alpha_1,\ldots,\alpha_{2g+1}\in \O_{k,S}$, this implies that $\ord_{\mathfrak{p}}(x_P-x_Q)=\ord_{\mathfrak{p}}x_P=\ord_{\mathfrak{p}}(x_P-\alpha_i)<0$ for all $i$. Since $x_Q-\alpha_i\in \O_{k,S}^*$ for all $i$, we have $\ord_\mathfrak{p}\gamma_i=-\frac{1}{2}\ord_{\mathfrak{p}}x_P$ for all $i$. Let $c\in \O_{k,S}$ be such that $\ord_\mathfrak{p}c=\max\{0,-\frac{1}{2}\ord_\mathfrak{p}x_P\}$ for $\mathfrak{p}\not\in S$. We consider now the binary form
\begin{equation*}
G(X,Z)=(cX-Z)\prod_{i=1}^{2g+1}(\delta_iX-\frac{\gamma_i}{c}Z)\in \O_{k,S}[X,Z].
\end{equation*}
The identity \eqref{bid} easily implies that if $\mathfrak{p}\not\in S$ and $\ord_{\mathfrak{p}}(x_P-x_Q)<0$, then
\begin{equation*}
\ord_\mathfrak{p}(\gamma_i\delta_j-\gamma_j\delta_i)=-\frac{1}{2}\ord_\mathfrak{p}x_P=\ord_\mathfrak{p}c, \quad i\neq j.
\end{equation*}
Then computing $\disc(G)$, we find that $\ord_\mathfrak{p}\disc(G)=0$ if $\mathfrak{p}\not\in S$ and $\ord_{\mathfrak{p}}(x_P-x_Q)\leq 0$.
Finally, suppose that $\ord_\mathfrak{p}(x_P-x_Q)>0$. Then from \eqref{meq}, we must have $\ord_\mathfrak{p}(y_P-y_Q)= 0$ (the case $\ord_\mathfrak{p}(y_P-y_Q)< 0$ being impossible). Since $\ord_\mathfrak{p}(x_Q-\alpha_i)=0$ for all $i$, $\ord_\mathfrak{p}(x_P-x_Q)>0$ implies that $\ord_\mathfrak{p}(x_P-\alpha_i)=0$ for all $i$. Then $\ord_\mathfrak{p}\beta_i=\ord_\mathfrak{p}\delta_i=\ord_\mathfrak{p}\gamma_i=0$ for all $i$. It follows from \eqref{bid} that
\begin{equation*}
\ord_\mathfrak{p}(\gamma_i^2\delta_j^2-\gamma_j^2\delta_i^2)=\ord_\mathfrak{p}(x_P-x_Q)
\end{equation*}
for $i\neq j$. Similarly,
\begin{equation*}
\ord_\mathfrak{p}(\gamma_i^2-\delta_i^2)=\ord_\mathfrak{p}(x_P-x_Q)>0
\end{equation*}
for all $i$. In particular, $\gamma_i\equiv \pm \delta_i\pmod{\mathfrak{p}}$ and $\beta_i\equiv \pm 1 \pmod{\mathfrak{p}}$ for all $i$. Since $x_P\equiv x_Q\pmod{\mathfrak{p}}$ and $y_P\not\equiv y_Q\pmod{\mathfrak{p}}$, we have $y_P\equiv -y_Q\pmod{\mathfrak{p}}$. So
\begin{equation*}
\prod_{i=1}^{2g+1}\beta_i\equiv \frac{y_Q}{y_P}\equiv -1\pmod{\mathfrak{p}}.
\end{equation*}
Then $\beta_i\equiv -1\pmod{\mathfrak{p}}$ for an odd number $n_{\mathfrak{p}}$ of the elements $i$. Let $m_{\mathfrak{p}}=\ord_\mathfrak{p}(x_P-x_Q)$. Then for $i\neq j$,
\begin{equation*}
\ord_\mathfrak{p}(\gamma_i\delta_j-\gamma_j\delta_i)=
\begin{cases}
m_{\mathfrak{p}} \quad &\text{ if $\beta_i\equiv \beta_j \pmod{\mathfrak{p}}$},\\
0 \quad &\text{ if $\beta_i\not\equiv \beta_j \pmod{\mathfrak{p}}$}.
\end{cases}
\end{equation*}
Now a straight-forward calculation gives
\begin{equation}
\label{oeq}
\ord_\mathfrak{p}\disc(G)=m_{\mathfrak{p}}n_{\mathfrak{p}}(n_{\mathfrak{p}}-1)+m_{\mathfrak{p}}(2g+2-n_{\mathfrak{p}})(2g+2-n_{\mathfrak{p}}-1).
\end{equation}
Let
\begin{equation*}
\mathcal{P}=\{\mathfrak{p}\in M_k\setminus S\mid \ord_\mathfrak{p}(x_P-x_Q)>0\}=\{\mathfrak{p}\in M_k\setminus S\mid \ord_\mathfrak{p}\disc(G)>0\}.
\end{equation*}
Let $b\in \O_k$ be such that
\begin{equation*}
2bc\equiv -1\left(\bmod{ \prod_{\mathfrak{p}\in\mathcal{P}}\mathfrak{p}^{m_\mathfrak{p}}}\right).
\end{equation*}
Let $U=\left( \begin{array}{cc}
1 & b \\
c & 1+bc
\end{array} \right)$. Then
\begin{equation*}
-G_U(X,Z)=Z\prod_{i=1}^{2g+1}\left((\delta_i-\gamma_i)X+\frac{1}{c}(bc\delta_i-(bc+1)\gamma_i)Z\right).
\end{equation*}
Note that since $\det U=1$, $\disc(G_U)=\disc(G)$. For $\mathfrak{p}\in \mathcal{P}$, let $\pi_\mathfrak{p}$ be a generator for $\mathfrak{p}\O_{k,S}$. Let $\epsilon_{i,\mathfrak{p}}=1$ if $\delta_i\equiv \gamma_i\pmod{\mathfrak{p}}$ and $\epsilon_{i,\mathfrak{p}}=0$ otherwise (in which case $\delta_i\equiv -\gamma_i\pmod{\mathfrak{p}}$). Define $\theta_i=\prod_{\mathfrak{p}\in \mathcal{P}}\pi_{\mathfrak{p}}^{m_\mathfrak{p}\epsilon_{i,\mathfrak{p}}}$ and $\theta_i'=\prod_{\mathfrak{p}\in \mathcal{P}}\pi_{\mathfrak{p}}^{m_\mathfrak{p}(1-\epsilon_{i,\mathfrak{p}})}$. Consider the binary form
\begin{equation*}
H(X,Z)=Z\prod_{i=1}^{2g+1}\left(\frac{\delta_i-\gamma_i}{\theta_i}X+\frac{\theta_i'}{c}(bc\delta_i-(bc+1)\gamma_i)Z\right).
\end{equation*}
Note that the binary form $H(X,Z)$ is a scalar multiple of $G_U\left(X,\left(\prod_{\mathfrak{p}\in \mathcal{P}}\pi_{\mathfrak{p}}^{m_\mathfrak{p}}\right)Z\right)$. It follows that $\ord_\mathfrak{p}\disc(H)=0$ if $\mathfrak{p}\in M_k\setminus (\mathcal{P}\cup S)$. For $\mathfrak{p}\in \mathcal{P}$, from \eqref{oeq} and the definition of $H$, a calculation yields
\begin{equation}
\label{deq}
\ord_\mathfrak{p}\disc(H)=2m_{\mathfrak{p}}n_{\mathfrak{p}}(n_{\mathfrak{p}}-1).
\end{equation}
For $\mathfrak{p}\in \mathcal{P}$,
\begin{equation*}
2(bc\delta_i-(bc+1)\gamma_i)\equiv -(\delta_i+\gamma_i)\pmod{\mathfrak{p}^{m_\mathfrak{p}}}.
\end{equation*}
If $\epsilon_{i,\mathfrak{p}}=0$, then it follows that
$\ord_\mathfrak{p}(\frac{\theta_i'}{c}(bc\delta_i-(bc+1)\gamma_i))\geq 2m_\mathfrak{p}$. In fact, since there are $n_\mathfrak{p}$ distinct values of $i$ such that $\epsilon_{i,\mathfrak{p}}=0$, by \eqref{deq}, $\ord_\mathfrak{p}(\frac{\theta_i'}{c}(bc\delta_i-(bc+1)\gamma_i))= 2m_\mathfrak{p}$ for at most one value of $i$ with $\epsilon_{i,\mathfrak{p}}=0$. After an appropriate substitution $X\mapsto X+aZ$, we can force $\ord_\mathfrak{p}(\frac{\theta_i'}{c}(bc\delta_i-(bc+1)\gamma_i))= 2m_\mathfrak{p}$ for all $i$ and $\mathfrak{p}$ such that $\epsilon_{i,\mathfrak{p}}=0$. Then for each $\mathfrak{p}\in \mathcal{P}$, $H(x,1)$ has $n_\mathfrak{p}$ roots $\alpha$ with $\ord_\mathfrak{p}\alpha=2m_\mathfrak{p}$. If $n_\mathfrak{p}=2g+1$, then we can replace $H(X,Z)$ by $\pi_\mathfrak{p}^{m_\mathfrak{p}}H\left(X,\frac{Z}{\pi_\mathfrak{p}^{m_\mathfrak{p}}}\right)$, eliminating $\mathfrak{p}$ as a divisor of the discriminant of $H$. So we have $3\leq n_\mathfrak{p}\leq 2g-1$ for every $\mathfrak{p}\in M_k\setminus S$ with $\ord_{\mathfrak{p}}\disc(H)>0$. Then from all of the above we have found a binary form equivalent to $(X-Z)\prod_{i=1}^{2g+1}(X-\beta_iZ)$ showing that $[(X-Z)\prod_{i=1}^{2g+1}(X-\beta_iZ)]\in \mathcal{B}'(2g+2,k,S)$.
\end{proof}
We have shown that in order to solve Problem \ref{mp} it suffices to effectively compute, for all values of the parameters, either the set $\mathcal{B}'(2g+2,k,S)$ or the set $\mathcal{H}'(g,k,S)$. It seems interesting to determine the precise relationship between these two sets. In the case $g=2$, Liu \cite{Liu2} has given an algorithm to compute, given a Weierstrass equation for a genus two curve, the fibers of a minimal model of the curve (away from primes above $2$, at least). In particular (see \cite[\S 6]{Liu2}), Liu's algorithm implies that computing the set $\mathcal{B}'(6,k,S)$ for all $k$ and $S$ and computing the set $\mathcal{H}'(2,k,S)$ for all $k$ and $S$ are equivalent problems. More precisely, Liu's results imply the following relationship.
\begin{theorem}
\label{teq}
Let $k$ be a number field and $S$ a finite set of places of $k$ containing the archimedean places and the places lying above $2$. There exists a number field $L$ and a finite set of places $T$ of $L$ such that if $C:y^2=f(x)$, $\deg f=6$, is a hyperelliptic curve of genus two representing an element of $\mathcal{H}'(2,k,S)$ and $F(X,Z)$ is the homogenization of $f$, then $[F]\in \mathcal{B}'(6,L,T)$. Conversely, if $[F]\in \mathcal{B}'(6,k,S)$, then for some constant $c\in k^*$, the equivalence class of the curve $y^2=cF(x,1)$ is in $\mathcal{H}'(2,k,S)$.
\end{theorem}
In the theorem, we can take $L$ to be any field such that for every curve $C$ representing an element of $\mathcal{H}'(2,k,S)$, the Weierstrass points of $C$ are $L$-rational. As is well known, such a field $L$ can be explicitly computed and it is a certain extension of $k$ unramified outside of $S$.
It seems plausible that the analogue of Theorem \ref{teq} holds in higher genus. We give a brief proof of one of the directions.
\begin{theorem}
\label{gr}
Let $g$ be a positive integer, $k$ be a number field, and $S$ a finite set of places of $k$ containing the archimedean places and the places lying above $2$. If $[F]\in \mathcal{B}'(2g+2,k,S)$, then for some constant $c\in k^*$ the equivalence class of the curve $y^2=cF(x,1)$ is in $\mathcal{H}'(g,k,S)$.
\end{theorem}
\begin{proof}
Let $[F]\in \mathcal{B}'(2g+2,k,S)$, with $F\in \O_{k,S}[X,Z]$ as in the definition of $\mathcal{B}'(2g+2,k,S)$. Let $f(x)=F(x,1)$ and let $C$ be the hyperelliptic curve defined by $y^2=f(x)$. It suffices to show that $\Jac(C)$ has good reduction outside $S$. Let $\mathfrak{p}\in M_k\setminus S$. If $\mathfrak{p}$ doesn't divide the discriminant of $f$ then the curve $C$, and hence $\Jac(C)$, has good reduction at $\mathfrak{p}$. Otherwise, let $k_\mathfrak{p}$ be the completion of $k$ at $\mathfrak{p}$, $\O_\mathfrak{p}$ the ring of integers of $k_\mathfrak{p}$, and $\pi$ a uniformizer. Then from the definitions, we can write
\begin{equation*}
f(x)=h(x)\prod_{i=1}^n(x-u_i\pi^{2m}),
\end{equation*}
where
\begin{itemize}
\item $m$ and $n$ are positive integers with $n$ odd, $3\leq n\leq 2g-1$.
\item $h\in \O_\mathfrak{p}[x]$, $h \pmod{\pi}$ has distinct roots, and $\pi\nmid h(0)$.
\item $u_i\in \O_\mathfrak{p}^*$, $i=1,\ldots, n$, and $\prod_{i=1}^n(x-u_i)$ has distinct roots mod $\pi$.
\end{itemize}
Note that $C$ can also be defined, over $k_\mathfrak{p}$, by the equation $y^2=h(\pi^{2m}x)\prod_{i=1}^n(x-u_i)$. Let $C_1$ and $C_2$ be the hyperelliptic curves over $\O_k/\mathfrak{p}$ defined by $y^2=x\overline{h}(x)$ and $y^2=\overline{h(0)}\prod_{i=1}^n(x-\overline{u}_i)$, respectively, where the bar denotes the image in $\left(\O_k/\mathfrak{p}\right)[x]$. Then one can show that the special fiber of the minimal proper regular model of $C$ over $\O_\mathfrak{p}$ consists of either $C_1$ and $C_2$ intersecting at a single point or $C_1$, $C_2$, and a chain of rational curves joining a point of $C_1$ with a point of $C_2$. Then it follows from well-known facts on the relationship between a minimal model of $C$ and the N\'eron model of $\Jac(C)$ \cite[\S 9.5 Th. 4, \S 9.6 Prop. 10]{BLR} that $\Jac(C)$ has good reduction at $\mathfrak{p}$ and the reduction mod $\mathfrak{p}$ is isomorphic to $\Jac(C_1)\times \Jac(C_2)$.
\end{proof}
Finally, since $\mathcal{H}'(g,k,S)$ is finite by the Shafarevich conjecture, we can conclude the (ineffective) finiteness of the set $\mathcal{B}'(2g+2,k,S)$.
\begin{corollary}
\label{SF}
For any positive integer $g$, number field $k$, and finite set of places $S$ of $k$, the set $\mathcal{B}'(2g+2,k,S)$ is finite.
\end{corollary}
\subsection*{Acknowledgments}
The author would like to thank Yuri Bilu, Rafael von K\"anel, and Umberto Zannier for several helpful comments on an earlier draft of the paper.
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| 98,510
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TITLE: Proving that the spectrum of $A\in End(V)$ is contained in the the set of roots of polynomials for which $p(A)=0$
QUESTION [1 upvotes]: I want to prove the lemma:
$$\tag{1}\sigma(A)\subseteq R(p)\ \forall\ p:p(A)=0$$
Where $\sigma(A)$ is the spectrum of eigenvalues of $A\in End(V)$ and $R(p)$ the set of roots of the polynomial $p$.
My attempt:
We can write:
\begin{equation} \tag{2}
p(A)=\sum_{i=0}^{n} \alpha_{i} A^{i}
\end{equation}
where $\alpha_i\in \mathcal{F}$ are scalars and $A^i$ is $A$ multiplied by itself $i$ times. Let $v\in V\backslash\{0\}$ be an eigenvector of $A$ and $\lambda$ the eigenvalue. We then have:
$$A^iv=\lambda^iv\\\Longrightarrow p(A) v=\sum_{i=0}^{n} \alpha_{i} A^{i} v=\sum_{i=0}^{n} \alpha_{i} \lambda^{i} v=p(\lambda) v$$
Where the polynomials in the last equation is defined in an analogous way to eq. $(2)$ but with $\mathcal{F}\mapsto\mathcal{F}$.
Since $p(A)=0$ and $v\in V\backslash \{0\}$ we can conclude:
$$p(\lambda)=0$$
Which shows that the eigenvalue is in $R(p)$
Are there any flaws in this proof, and is it extensive?
REPLY [0 votes]: The Hamilton-Cayley theorem answers to your question, whenever a polynomial evalued in A gives back the zero matrix you can say that the minimum polynomial of A divides it. The minimum polynomial has already all the roots you need and you can conclude.
| 79,423
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| 211,147
|
. Significant newsprint and TV footage has already been spent in wondering when the pretty young sister will be seen on-screen. One of the first rumours was that Isabelle would do a crossover film opposite Anil Kapoor's son, Harshvardhan. Then, it seems, Katrina was in talks with director Ayan Mukerji to launch Isabelle. This was followed by reports that Katrina would start her own production house and cast her little sister in her first film. However, nothing came of any of this. Now, Isabelle is finally set to debut, but in a Canadian film where, coincidentally, Katrina’s former alleged-boyfriend Salman Khan is involved. Called Dr Cabbie, it is co-produced by Salman Khan Being Human [SKBH] Productions, in association with two Canadian companies — FirstTake Entertainment and Caramel Film. The lead actors of Dr Cabbie are Vinay Virmani (he was last seen in Speedy Singhs in 2011) and Adrianne Palicki (she was in GI Joe: Retaliation earlier this year). The film also stars Kunal Nayyar (from the hugely popular hit American TV series, The Big Bang Theory), who will also make his film debut with this project. Isabelle is the second-lead actress, cast opposite Kunal. Touted to be a $25 million project, the film will be distributed in India by Eros International. Commenting on Isabelle’s casting, the spokesperson of Eros says, “Isabelle went through the audition process like all the other actors. She was cast based on her performance and only because she was a perfect fit for the role.” Dr Cabbie portrays a young Indian doctor (played by Vinay) who has newly arrived in Canada and has to work as a taxi driver after being rejected by local medical schools. The film is being helmed by a renowned, award-winning Canadian director Jean-François Pouliot, and is presently being shot in Canada. Don’t call me bhai! (Chetna Dua reports)A large part of the film industry as well as his fans call Salman Khan ‘Bhai’, but the actor does not seem pleased with the tag. “I have four sisters — two of my own and two rakhi sisters. That is enough. I don’t want more sisters,” says Khan. “If a pretty lady calls me bhai, I get embarrassed. I often tell a friend standing next to me to request her not to call me bhai,” he says with a sheepish grin.
| 79,650
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With the lease being extended to 143 years, this recently refurbished two bedroom first floor flat comes with viewing strongly recommended, located in this popular area of Marton and well presented throughout with new floor coverings. The property comprises of: entrance hall, lounge, brand new re-fitted kitchen with a range of appliances included, two double bedrooms and modern white and chrome bathroom. Also benefiting from uPVC double glazing, gas central heating via recently replaced baxi combi boiler and external an enclosed private rear garden.
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| 13,284
|
{\bf Problem.} What is the largest value of $n$ less than 100,000 for which the expression $8(n-2)^5-n^2+14n-24$ is a multiple of 5?
{\bf Level.} Level 5
{\bf Type.} Number Theory
{\bf Solution.} By the Binomial Theorem, \begin{align*}
(n - 2)^5 &= n^5 - \binom{5}{1} \cdot 2n^4 + \binom{5}{2} \cdot 2^2 n^3 - \binom{5}{3} \cdot 2^3 n^2 \\
&\qquad + \binom{5}{4} \cdot 2^4 n - 2^5 \\
&= n^5 - 10n^4 + 40n^3 - 80n^2 + 80n - 32.
\end{align*} Note that this reduces to $n^5 - 32 \equiv n^5 + 3 \pmod{5}$. Therefore, \begin{align*}
8(n - 2)^5 - n^2 + 14n - 24 &\equiv 8(n^5 + 3) - n^2 + 14n - 24 \\
&\equiv 8n^5 + 24 - n^2 + 14n - 24 \\
&\equiv 3n^5 - n^2 - n \pmod{5}.
\end{align*}
If $n \equiv 0 \pmod{5}$, then \[3n^5 - n^2 - n \equiv 3 \cdot 0^5 - 0^2 - 0 \equiv 0 \pmod{5}.\] If $n \equiv 1 \pmod{5}$, then \[3n^5 - n^2 - n \equiv 3 \cdot 1^5 - 1^2 - 1 \equiv 1 \pmod{5}.\] If $n \equiv 2 \pmod{5}$, then \[3n^5 - n^2 - n \equiv 3 \cdot 2^5 - 2^2 - 2 \equiv 90 \equiv 0 \pmod{5}.\] If $n \equiv 3 \pmod{5}$, then \[3n^5 - n^2 - n \equiv 3 \cdot 3^5 - 3^2 - 3 \equiv 717 \equiv 2 \pmod{5}.\] If $n \equiv 4 \pmod{5}$, then \[3n^5 - n^2 - n \equiv 3 \cdot 4^5 - 4^2 - 4 \equiv 3052 \equiv 2 \pmod{5}.\]
Therefore, the given expression is a multiple of 5 if and only if $n \equiv 0$ or $n \equiv 2 \pmod{5}$.
The largest value of $n$ less than 100000 that is congruent to 0 or 2 modulo 5 is $\boxed{99997}$.
| 150,109
|
1 Whole Luv-a-Duck, Duck
Orange Salt Rub
2 tblsp flaked salt
2 blood oranges, zested
1 sml sprig rosemary, finely chopped
¼ tsp black pepper
Blood Orange Sauce
2 zested blood oranges, Juiced
2 blood oranges, thinly sliced
½ cup sugar
½ cup duck stock
2 tblsp verjuice
1 tsp cornflour
1 tblsp water
1 tblsp butter, cubed
1. Rinse the whole duck under running water. Thoroughly pat dry with paper towel inside and out.
2. In a small bowl combine the flaked salt, blood orange zest, rosemary and black pepper. Use the back of a small spoon to blend.
3. Place the duck onto the roasting rack inside the roasting pan and liberally sprinkle the flavoured salt over the duck.
4. Roast the duck in the pre-heated oven 180c for 40 minutes per Kg until golden and juices run clear when tested. Remove duck from oven and allow to rest 10 -15 minutes.
Sauce
Place the blood orange juice, slices, sugar, stock and ver juice into the saucepan and simmer over medium heat 8 minutes. Whisk in the combined cornflour and water and stir until liquid boils. Remove from heat and cool 2 minutes, whisk in the cubed butter.
To Serve
Serve roasted duck with seasonal vegetables and drizzle with Blood Orange sauce.
If blood oranges are not available use sweet navels or valencia oranges. Replace blood orange juice with pomegranate or cranberry juice for a delicious difference.
| 285,780
|
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Designs equipment and subassemblies under general supervision while following customer specifications, engineering guidelines, quality procedures and other documentation. Essential Duties And Responsibilities Design and analyze mechanical systems, equipment and6/24/2015 Warminster, PA Save This Job
Responsibilities Perform optical analysis of product requirements and assist in derivation of optical, mechanical, and electrical solutions. Create optical / laser physics models to design components and derive tolerances. Source components and characterize upon receipt6/24
| 103,842
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i checked this am. and it was there-i love the color,(I even got free shipping, by asking(i have bought much in the past) Its on its way-I should get tomorrow-my stuff usually comes next day, b/c i 'm not for from the distribution center. Can't wait-i love the cles'-I have a pomme one as well. Sometimes you feel like purple , sometimes you fee like red!
| 91,669
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\begin{document}
\title[An efficient deterministic test for Kloosterman sum zeros]{An efficient deterministic test for Kloosterman sum zeros}
\author{Omran~Ahmadi}
\address{Claude Shannon Institute, UCD CASL, University College Dublin, Ireland}
\email{omran.ahmadi@ucd.ie}
\author{Robert~Granger}
\address{Claude Shannon Institute, Dublin City University, Ireland}
\curraddr{Claude Shannon Institute, UCD CASL, University College Dublin, Ireland}
\email{rgranger@computing.dcu.ie}
\thanks{Both authors are supported by the Claude Shannon Institute, Science Foundation Ireland Grant No. 06/MI/006.}
\subjclass[2010]{11L05, (11G20, 11T71, 11Y16)}
\keywords{Kloosterman sum zeros, elliptic curves, Sylow $p$-subgroups}
\begin{abstract}
We propose a simple deterministic test for deciding whether or not an
element $a \in \F_{2^n}^{\times}$ or $\F_{3^n}^{\times}$ is a zero of the corresponding
Kloosterman sum over these fields, and rigorously analyse its runtime.
The test seems to have been overlooked in the literature.
The expected cost of the test for binary fields is a single
point-halving on an associated elliptic curve, while for ternary fields
the expected cost is one half of a point-thirding on an associated elliptic curve.
For binary fields of practical interest, this represents an $O(n)$ speedup over
the previous fastest test. By repeatedly invoking the test on random elements
of $\F_{2^n}^{\times}$ we obtain the most efficient probabilistic
method to date to find non-trivial Kloosterman sum zeros.
The analysis depends on the distribution of Sylow $p$-subgroups in the two
families of associated elliptic curves, which we ascertain using a theorem due to Howe.
\end{abstract}
\maketitle
\section{Introduction}
\label{sec:intro}
For a finite field $\F_{p^n}$, the Kloosterman sum $\mathcal{K}_{p^n}: \F_{p^n}
\rightarrow \C$ can be defined
by
\[
\mathcal{K}_{p^n}(a) = 1 + \sum_{x \in \F_{p^n}^{\times}} \zeta^{\text{Tr}(x^{-1} + ax)},
\]
where $\zeta$ is a primitive $p$-th root of unity and Tr denotes the
absolute trace map $\text{Tr}:\F_{p^n} \rightarrow \F_p$, defined by
\[
\text{Tr}(x) = x + x^p + x^{p^2} + \cdots + x^{p^{n-1}}.
\]
Note that in some contexts the Kloosterman sum is defined to be just the
summation term without the added `$1$'~\cite{katz}.
As one would expect, a Kloosterman (sum) zero is simply an element $a \in \F_{p^n}^{\times}$ for which
$\mathcal{K}_{p^n}(a) = 0$.
Kloosterman sums have recently become the focus of much research, most notably
due to their applications in cryptography and coding theory
(see~\cite{gong,moisiocode} for example).
In particular, zeros of $\mathcal{K}_{2^{n}}$ lead to bent
functions from $\F_{2^{2n}} \rightarrow \F_{2}$~\cite{dillon}, and similarly
zeros of $\mathcal{K}_{3^{n}}$ give rise to ternary bent functions~\cite{helleseth1}.
It was recently shown that zeros of Kloosterman sums only exist in
characteristics 2 and 3~\cite{kononen}, and hence these are the only cases we consider.
Finding such zeros is regarded as being difficult, and
recent research has tended to focus on characterising Kloosterman sums modulo
small integers~\cite{charpin,moisio2,lisonek,lisonek2,lisonek3,faruk1,faruk2,faruk3,faruk4}.
While these results are interesting in their own
right, they also provide a sieve which may be used to
eliminate elements of a certain form prior to testing whether they are
Kloosterman zeros or not, by some method.
It has long been known that Kloosterman sums over binary and ternary fields
are intimately related to the group orders of members of two families of elliptic curves over
these fields~\cite{katz,wolfmann,moisio,geer}. In particular, for $p \in \{2,3\}$
the Kloosterman sum $\mathcal{K}_{p^n}(a)$ is equal to one minus the trace
of the Frobenius endomorphism of an associated elliptic curve $E_{p^n}(a)$.
As such, one may use $p$-adic methods --- originally due to
Satoh~\cite{satoh} --- to compute the group orders of these elliptic curves,
and hence the corresponding Kloosterman sums.
The best $p$-adic point counting method asymptotically takes $O(n^2 \log^2{n}\log\log{n})$ bit
operations and requires $O(n^2)$ memory; see Vercauteren's thesis~\cite{frethesis} for
contributions and a comprehensive survey.
Rather than count points, Lison\v{e}k has suggested that if instead one only wants to check
whether a given element is a zero, one can do so by testing whether a random point of
$E_{p^n}(a)$ has order $p^n$, via point multiplication~\cite{lisonek}.
Asymptotically, this has a similar bit complexity to
the point counting approach, requires less memory, but
is randomised. For fields of practical interest, it is reported that this
approach is superior to point counting~\cite[\S3]{lisonek}, and
using this method Lison\v{e}k was able to find a
zero of $\mathcal{K}_{2^n}$ for $n \le 64$ and
$\mathcal{K}_{3^n}$ for $n \le 34$, in a matter of days.
In this paper we take the elliptic curve connection to a logical
conclusion, in terms of proving divisibility results of Kloosterman sums by
powers of the characteristic. In particular we give an efficient deterministic algorithm to compute the
Sylow $2$- and $3$-subgroups of the associated elliptic curves in
characteristics $2$ and $3$ respectively, along with a generator (these
subgroups are cyclic in the cases considered). Moreover, the average case runtimes of the
two algorithms are rigorously analysed. For binary fields of practical
interest, the test gives an $O(n)$ speedup over the point multiplication test.
Finding a single Kloosterman zero --- which is often all that
is needed in applications --- is then a matter of testing random field elements until
one is found, the success probability of which crucially depends on the number of Kloosterman zeros, see~\cite{katz}
and~\S\ref{exactformula}. Our runtime analysis provides a non-trivial
upper bound on this number, and consequently finding a Kloosterman zero with this
approach still requires time exponential in the size of the field.
We note that should one want to find {\em all}\hspace{0.5mm} Kloosterman
zeros over $\F_{2^n}$, rather than just one, then one can use the fast Walsh-Hadamard transform
(see~\cite{fft} for an overview), which requires $O(2^n \cdot n^2)$ bit operations and $O(2^n \cdot n)$ space.
The sequel is organised as follows. In \S\ref{connection} we detail the basic
connection between Kloosterman sums and two families of elliptic
curves. In \S\ref{determine} we present the main idea behind our algorithm, while
\S\ref{binary} and \S\ref{ternary} explore its specialisation to binary and
ternary fields respectively. In \S\ref{noofits} we present data on the runtime
of the two algorithms, provide a heuristic analysis which attempts to explain the
data, and give an exact formula for the average case runtime. In \S\ref{mainresult}
we rigorously prove the expected runtime, while in \S\ref{compare} we assess
the practical efficiency of the tests. We finally make some concluding remarks
in \S\ref{conc}.
\section{Connection with elliptic curves}\label{connection}
Our observations stem from the following three simple lemmas, which connect
Kloosterman sums over $\F_{2^n}$ and $\F_{3^n}$ with the group orders of
elliptic curves in two corresponding families. The first is due to Lachaud and
Wolfmann~\cite{wolfmann}, the second Moisio~\cite{moisio}, while the third was
proven by Lison\v{e}k~\cite{lisonek}.
\begin{lemma}\label{lis1}
Let $a \in \F_{2^n}^{\times}$ and define the elliptic curve $E_{2^n}(a)$ over
$\F_{2^n}$ by
\[
E_{2^n}(a): y^2 + xy = x^3 + a.
\]
Then $\#E_{2^n}(a) = 2^n + \mathcal{K}_{2^n}(a)$.
\end{lemma}
\begin{lemma}\label{lis2}
Let $a \in \F_{3^n}^{\times}$ and define the elliptic curve $E_{3^n}(a)$ over
$\F_{3^n}$ by
\[
E_{3^n}(a): y^2 = x^3 + x^2 - a.
\]
Then $\#E_{3^n}(a) = 3^n + \mathcal{K}_{3^n}(a)$.
\end{lemma}
\begin{lemma}\label{lis3}
Let $p \in \{2, 3\}$, let $a \in \F_{p^n}^{\times}$, and let $1 \leq h \leq
n$. Then $p^h \mid \mathcal{K}_{p^n}(a)$ if and only if there exists a point
of order $p^h$ on $E_{p^n}(a)$.
\end{lemma}
Lemma~\ref{lis3} is a simple consequence of the structure theorem for elliptic curves
over finite fields. Note that for $p \in \{2,3\}$, by Lemmas~\ref{lis1} and~\ref{lis2} we have
$\mathcal{K}_{p^n}(a) = 0$ if and only if $E_{p^n}(a)$ has order $p^n$. By Lemma~\ref{lis3}, this is equivalent
to $E_{p^n}(a)$ having a point of order $p^n$, and hence finding a point of order $p^n$
proves that $\mathcal{K}_{p^n}(a) = 0$, since $p^n$ is the only element
divisible by $p^n$ in the Hasse interval. For the remainder of the
paper, when we refer to a prime $p$ we implicitly presume $p \in \{2,3\}$.
\section{Determining the Sylow $p$-subgroup of $E_{p^n}(a)$}\label{determine}
It is easy to show that $\mathcal{K}_{2^n}(a) \equiv 0 \pmod{4}$ and
$\mathcal{K}_{3^n}(a) \equiv 0 \pmod{3}$ for all $a \in \F_{2^n}^{\times}$ and $\F_{3^n}^{\times}$ respectively.
One way to see this is to observe that $E_{2^n}(a)$ possesses a point of order
$4$ (see~\S\ref{binary}) and $E_{3^n}(a)$ possesses a point of order $3$
(see~\S\ref{ternary}), and hence by Lagrange's theorem, $4 \mid \#E_{2^n}(a)$ and $3 \mid \#E_{3^n}(a)$.
For an integer $x$, let $\ord_p(x)$ be the exponent of the maximum power of $p$ that divides $x$.
For $a \in \F_{p^n}^{\times}$, let $h = \ord_p(\#E_{p^n}(a))$. By
Lemma~\ref{lis3} the Sylow $p$-subgroup $S_{p}(E_{p^n}(a))$ is cyclic of order
$p^h$, and hence has $(p-1)p^{h-1}$ generators. Multiplying
these by $p$ results in the $(p-1)p^{h-2}$ generators of the order $p^{h-1}$
subgroup. Continuing this multiplication by $p$ process, after $h-1$ steps
one arrives at the $p$-torsion subgroup $E_{p^n}(a)[p]$, consisting of $p-1$ order-$p$ points
and the identity element $\mathcal{O}$. These considerations reveal the
structure of the $p$-power torsion subgroups $E_{p^n}(a)[p^k]$ for $1 \le k
\le h$, which one may view as a tree, with $\mathcal{O}$ as the root node. The root has $p-1$ children which are the
non-identity points in $E_{p^n}(a)[p]$. If $h>1$ each of these $p-1$ nodes
has $p$ children: the elements of $E_{p^n}(a)[p^2] \setminus E_{p^n}(a)[p]$.
For $1 < k < h$, each of the $(p-1)p^{k-1}$ depth-$k$ nodes have
$p$ children, while at depth $h$ we have $(p-1)p^{h-1}$ leaf nodes.
Using a division polynomial approach Lison\v{e}k was able to prove a necessary
condition on $a \in \F_{2^n}^{\times}$ such that $\mathcal{K}_{2^n}(a)$ is divisible by $16$, and likewise a
necessary condition on $a \in \F_{3^n}^{\times}$ such that
$\mathcal{K}_{3^n}(a)$ is divisible by $9$. While necessary conditions for
the divisibility of $\mathcal{K}_{2^n}(a)$ by $2^k$ have since been derived
for $k \le 8$~\cite{faruk3}, and for the divisibility of $\mathcal{K}_{3^n}(a)$
by $3^k$ for $k \le 3$~\cite{faruk4}, these use $p$-adic
methods; the division polynomial approach seemingly being too cumbersome to progress
any further.
However, the process outlined above --- taking a generator of $S_{p}(E_{p^n}(a))$
and multiplying by $p$ repeatedly until the non-identity elements of the
$p$-torsion are obtained --- can be reversed, easily and efficiently, using
point-halving in even characteristic, and point-thirding in characteristic
three, as we demonstrate in the ensuing two sections. Furthermore, due to
the cyclic structure of $S_{p}(E_{p^n}(a))$, at each depth, either all points are
divisible by $p$, or none are. This means one can determine the height of the tree
by using a depth-first search, without any backtracking; in particular, when a point $P$
at a given depth can not be halved or thirded, this depth is $\log_{p}(|S_{p}(E_{p^n}(a))|)$, and
$P$ is a generator. Furthermore, one can do this without
ever computing the group order of the curve.
This process has been considered previously by Miret~\ea, for determining the Sylow $2$-subgroup of
elliptic curves over arbitrary finite fields of characteristic $ > 2$~\cite{miret1}; for $p=2$ the
algorithm follows easily from the above considerations and point-halving, which is well studied in cryptographic
circles~\cite{knudsen,schroeppel,omran}, and is known to be more than
twice as fast as point-doubling in some cases~\cite{handbook}.
For primes $l > 2$, Miret~\ea \ also addressed how to compute the Sylow $l$-subgroup of elliptic
curves over arbitrary finite fields provided that $l$ was not the
characteristic of the field~\cite{miret2}. Therefore we address here the case
$l = p = 3$, for the family of curves $E_{3^n}(a)$.
We summarise this process in Algorithm~\ref{sylow}. Regarding notation, we say that a
point $P$ is $p$-divisible if there exists a point $Q$ such that $[p]Q = P$, and
write $Q = [1/p]P$.
\begin{algorithm}{DETERMINE $S_{p}(E_{p^n}(a))$}
{$a \in \F_{p^n}^{\times}$, $P \in E_{p^n}(a)[p] \setminus \{\mathcal{O}\}$}
{$(h,P_h)$ where $h = \ord_{p}(\#E_{p^n}(a))$ and $\langle P_h \rangle = S_{p}(E_{p^n}(a))$}
\label{sylow}
\nline $\text{counter} \leftarrow 1$; \\
\nline While $P$ is $p$-divisible do: \\
\nnline P := [1/p]P; \\
\nnline counter++; \\
\nline Return $(\text{counter},P)$
\end{algorithm}
Observe that Algorithm~\ref{sylow} is deterministic, provided that a
deterministic method of dividing a $p$-divisible point by $p$ is fixed once and for all, which we do
for $p=2$ and $p=3$ in~\S\ref{binary} and~\S\ref{ternary} respectively. For a given
field extension under consideration, choosing an appropriate field
representation and basis can also be performed deterministically, via
sequential search, however we consider this to be part of the setup phase
and do not incorporate setup costs when assessing the runtime of Algorithm~\ref{sylow}.
\section{Binary fields}\label{binary}
We now work out the details of Algorithm~\ref{sylow} for the family of curves
$E_{2^n}(a)$. For a fixed $n$, given a point $P =(x,y) \in E_{2^n}(a)$, $[2]P = (\xi,\eta)$
is given by the formula:
\begin{eqnarray}
\nonumber \lambda &=& x + y/x,\\
\label{half} \xi &=& \lambda^2 + \lambda,\\
\nonumber \eta &=& x^2 + \xi(\lambda+1).
\end{eqnarray}
To halve a point, one needs to reverse this process, \ie given $Q=(\xi,\eta)$, find (if
possible) a $P=(x,y) \in E_{2^n}(a)$ such that $[2]P = Q$. To do so, one first
needs to solve~(\ref{half}) for $\lambda$, which has a solution in
$\F_{2^n}$ if and only if $\text{Tr}(\xi) = 0$, since the trace of the
right-hand side is zero for every $\lambda \in \F_{2^n}$, and one can provide
an explicit solution in this case, as detailed in \S\ref{solvequad}. Observe that if
$\lambda$ is a solution to~(\ref{half}) then so is $\lambda + 1$.
Assuming $\lambda$ has been computed, one then has
\begin{eqnarray}
\nonumber x &=& (\eta + \xi(\lambda+1))^{1/2}, \\
\nonumber y &=& x(x+\lambda),
\end{eqnarray}
which for the two choices of $\lambda$ gives both points whose duplication is
$Q=(\xi,\eta)$.
Aside from the cost of computing $\lambda$, the computation of $P = (x,y)$ as
above requires two field multiplications. As detailed in
Algorithm~\ref{sylow2}, this can be reduced to just one by
using the so-called $\lambda$-representation of a
point~\cite{knudsen,schroeppel}, where an affine point $Q = (\xi,\eta)$ is
instead represented by $(\xi,\lambda_{Q})$, with
\[
\lambda_Q = \xi + \frac{\eta}{\xi}.
\]
In affine coordinates, there is a unique $2$-torsion point $(0,a^{1/2})$,
which halves to the two order $4$ points $P_{4}^{+} = (a^{1/4},a^{1/2})$,
$P_{4}^{-} = (a^{1/4},a^{1/2} + a^{1/4})$. The corresponding
$\lambda$-representations of each of these are $(a^{1/4},0)$ and $(a^{1/4},1)$ respectively.
For simplicity, we choose to use the former as the starting point in Algorithm~\ref{sylow2}.
\begin{algorithm}{DETERMINE $S_{2}(E_{2^n}(a))$}
{$a \in \F_{2^n}^{\times}$, $(x=a^{1/4}, \lambda = 0)$}
{$(h,P_h)$ where $h = \ord_{2}(\#E_{2^n}(a))$ and $\langle P_h \rangle = S_{2}(E_{2^n}(a))$}
\label{sylow2}
\nline $\text{counter} \leftarrow 2$; \\
\nline While $\text{Tr}(x) = 0$ do: \\
\nnline Solve $\widehat{\lambda}^2 + \widehat{\lambda} + x = 0$; \\
\nnline $t \leftarrow x(x + \lambda + \widehat{\lambda})$; \\
\nnline $x \leftarrow \sqrt{t}$; \\
\nnline $\lambda \leftarrow \widehat{\lambda} + 1$; \\
\nnline counter++; \\
\nline Return $(\text{counter},P = (x,x(x + \lambda)))$
\end{algorithm}
Observe that if the $x$-coordinate $a^{1/4}$ of $P_{4}^{\pm}$ satisfies
$\text{Tr}(a^{1/4}) = \text{Tr}(a) = 0$, then there exist four points of order
$8$, and hence $8 \mid \mathcal{K}_{2^n}(a)$, which was first observed by van
der Geer and van der Vlugt~\cite{geer}, and later by several others~\cite{helleseth2,charpin2,lisonek}.
\subsection{Solving $\widehat{\lambda}^2 + \widehat{\lambda} + x = 0$}\label{solvequad}
For odd $n$, let $\widehat{\lambda}$ be given by the following function,
which is known as the {\em half trace}:
\begin{equation}\label{halftrace}
\widehat{\lambda}(x) = \sum_{i=0}^{(n-1)/2} x^{2^{2i}}.
\end{equation}
One can easily verify that this $\widehat{\lambda}$ satisfies the stated
equation. When $n$ is even, the half trace approach will not
work, essentially because $\text{Tr}_{\F_{2^{n}}/\F_2}(1) = 0$.
Hence fix an element $\delta \in \F_{2^n}$ with $\text{Tr}_{\F_{2^{n}}/\F_2}(\delta) = 1$. Such a
$\delta$ can be found during the setup phase via the sequential search of the trace of
the polynomial basis elements, or by using the methods of~\cite{omran}. A solution to
equation~(\ref{half}) is then given by~\cite[Chapter II]{ECC1}:
\begin{equation}\label{fasthalf}
\widehat{\lambda}(x) = \sum_{i=0}^{n-2} \bigg( \sum_{j=i+1}^{n-1} \delta^{2^j} \bigg) x^{2^i},
\end{equation}
as may be verified. Note that for odd $n$, $\delta = 1$ suffices and so~(\ref{fasthalf})
simplifies to~(\ref{halftrace}). The inner sums of equation~(\ref{fasthalf}) can be precomputed,
and for a general $\delta \in \F_{2^n}$ the computation of $\widehat{\lambda}(x)$ would require $n-1$
multiplications in $\F_{2^n}$, which together with the multiplication coming
from \url{line} \url{4} of Algorithm~\ref{sylow2}, gives a total of $n$ full $\F_{2^n}$-multiplications.
However, should $\F_{2^n}$ contain a subfield of odd index, then one
can reduce this cost as follows. Let $n = 2^m n'$ with
$m \ge 1$ and $n'$ odd. Constructing $\F_{2^n}$ as a degree $n'$ extension of
$\F_{2^{2^m}}$, fix a $\delta \in \F_{2^{2^m}}$ with
$\text{Tr}_{\F_{2^{2^m}}/\F_{2}}(\delta)=1$. Then
\[
\text{Tr}_{\F_{2^{2^m\cdot n'}}/\F_2}(\delta) = n' \cdot
\text{Tr}_{\F_{2^{2^m}}/\F_2}(\delta) = 1.
\]
Hence this $\delta$ can be used in~(\ref{fasthalf}).
As $\delta^{2^{2^m}} = \delta$, upon
expanding~(\ref{fasthalf}) in terms of
$\{\delta^{2^0},\delta^{2^1},\ldots,\delta^{2^{2^m - 1}}\}$, we see that
at most $2^m$ multiplications of elements of $\F_{2^{2^m}}$ by elements of
$\F_{2^n}$ are required. So the smaller the largest power of $2$ dividing $n$ is, the
faster one can compute $\widehat{\lambda}(x)$.
However, since the expressions for $\widehat{\lambda}(x)$ in~(\ref{halftrace})
and~(\ref{fasthalf}) are linear maps, in practice it is far more efficient for
both odd and even $n$ to precompute and store
$\{\widehat{\lambda}(t^i)\}_{i=0,\ldots,n-1}$ during setup,
where $\F_{2^n} = \F_{2}(t)$ and $x = \sum_{i=0}^{n-1} x_it^i$. One then has
\[
\widehat{\lambda}(x) = \sum_{i=0}^{n-1} x_i \widehat{\lambda}(t^i).
\]
On average just $n/2$ additions in $\F_{2^n}$ are required for each point-halving. Both the
storage required and execution time can be further reduced~\cite{handbook}.
We defer consideration of the practical efficiency of
Algorithm~\ref{sylow2} until~\S\ref{compare2}.
\section{Ternary fields}\label{ternary}
Let $Q=(\xi,\eta) \in E_{3^n}(a)$. To find $P=(x,y)$ such that $[3]P = Q$, when possible, we do the following.
As in~\cite[\S4]{miret2}, we have
\[
x([3]P) = x(P) - \frac{\Psi_{2}(x,y)\Psi_{4}(x,y)}{\Psi_{3}^{2}(x,y)},
\]
or
\[
(x - \xi)\Psi_{3}^{2}(x,y) - \Psi_{2}(x,y)\Psi_{4}(x,y) = 0,
\]
where $\Psi_l$ is the $l$-th division polynomial.
Working modulo the equation of $E_{3^n}(a)$, this becomes
\[
x^9 - \xi x^6 + a(1-\xi)x^3 - a^2(a + \xi) = 0,
\]
whereupon substituting $X = x^3$ gives
\begin{equation}\label{3div}
f(X) = X^3 - \xi X^2 + a(1-\xi)X - a^2(a + \xi) = 0.
\end{equation}
To solve~(\ref{3div}), we make the transformation
\[
g(X) = X^3 f\bigg(\frac{1}{X} - \frac{a(1-\xi)}{\xi}\bigg) = \frac{a^2 \eta^2}{\xi^3} X^3 -\xi X + 1.
\]
Hence we must solve
\[
X^3 - \frac{\xi^4}{a^2 \eta^2} X + \frac{\xi^3}{a^2 \eta^2} = 0.
\]
Writing $X = \frac{\xi^2}{a \eta}\widehat{X}$ this becomes
\begin{equation}\label{3tri}
\widehat{X}^3 - \widehat{X} + \frac{a \eta}{\xi^3} = 0.
\end{equation}
Our thirding condition is then simply $\text{Tr}(a \eta/\xi^3) = 0$,
since as in the binary case, for every element $\widehat{X} \in \F_{3^n}$ we have $\text{Tr}(\widehat{X}^3
- \widehat{X}) = 0$, and if so then one can provide an explicit
solution, as detailed in \S\ref{solvecube}. Observe that if $\widehat{X}$ is a solution to~(\ref{3tri})
then so is $\widehat{X} \pm 1$.
Unrolling the transformations leads to the following algorithm, with input the
$3$-torsion point $P_3 = (a^{1/3},a^{1/3})$.
\begin{algorithm}{DETERMINE $S_{3}(E_{3^n}(a))$}
{$a \in \F_{3^n}^{\times}$, $(x=a^{1/3}, y = a^{1/3})$}
{$(h,P_h)$ where $h = \ord_3(\#E_{3^n}(a))$ and $\langle P_h \rangle = S_{3}(E_{3^n}(a))$}
\label{sylow3}
\nline $\text{counter} \leftarrow 1$; \\
\nline While $\text{Tr}(ay/x^3) = 0$ do: \\
\nnline Solve $\widehat{X}^3 - \widehat{X} + \frac{ay}{x^3} = 0$; \\
\nnline $x \leftarrow \bigg(\frac{ay}{x^2\widehat{X}} - \frac{a(1-x)}{x} \bigg)^{1/3}$; \\
\nnline $y \leftarrow \big(x^3 + x^2 - a\big)^{1/2}$; \\
\nnline counter++; \\
\nline Return $(\text{counter},P = (x,y))$
\end{algorithm}
Observe that as with Algorithm~\ref{sylow2}, if the point $P_3$ satisfies
$\text{Tr}(a \cdot a^{1/3}/a) = \text{Tr}(a) = 0$, then there is a point of order
$9$, and hence $9 \mid \mathcal{K}_{3^n}(a)$, which again was first proven
in~\cite{geer}, and later by others~\cite{lisonek,faruk1}.
\subsection{Solving $\widehat{X}^3 - \widehat{X} + \frac{a y}{x^3} = 0$}\label{solvecube}
Let $\beta = \frac{a y}{x^3}$, and let $\delta \in \F_{3^n}$ be an
element with $\text{Tr}_{\F_{3^{n}}/\F_3}(\delta) = 1$, which can be found deterministically
during the setup phase. It is then a simple matter to verify that
\begin{equation}\label{fastthird}
\widehat{X}(\beta) = \sum_{i=0}^{n-2} \bigg( \sum_{j=i+1}^{n-1} \delta^{3^j} \bigg) \beta^{3^i}
\end{equation}
is a solution to equation~(\ref{3tri}).
For $n \equiv 1 \pmod{3}$, one may choose $\delta = 1$ and the expression for $\widehat{X}(\beta)$ in
equation~(\ref{fastthird}) simplifies to
\[
\widehat{X}(\beta) = \sum_{i=1}^{(n-1)/3} \left(\beta^{3^{3i-1}} - \beta^{3^{3i-2}}\right).
\]
For $n \equiv 2 \pmod{3}$, one may choose $\delta = -1$ and the expression for $\widehat{X}(\beta)$ in
equation~(\ref{fastthird}) simplifies to
\[
\widehat{X}(\beta) = -\beta + \sum_{i=1}^{(n-2)/3} \left(\beta^{3^{3i-1}} - \beta^{3^{3i}}\right).
\]
For $n \equiv 0 \pmod{3}$, one can use the approach described in
\S\ref{solvequad} to pick $\delta$ from the smallest subfield of $\F_{3^n}$ of index
coprime to $3$, in order to reduce the cost and the number of
multiplications required to solve~(\ref{3tri}). As in the binary
case, one can also exploit the linearity of $\widehat{X}(\beta)$ and
precompute and store $\{\widehat{X}(t^i)\}_{i=0,\ldots,n-1}$ during setup,
where $\F_{3^n} = \F_{3}(t)$ and $\beta = \sum_{i=0}^{n-1} \beta_it^i$, in order to reduce the cost of
solving~(\ref{3tri}) to an average of $2n/3$ additions.
We defer consideration of the practical efficiency of Algorithm~\ref{sylow3} until~\S\ref{compare3}.
\section{Heuristic analysis of the expected number of iterations}\label{noofits}
For any input $a \in \F_{p^n}^{\times}$, the runtime of Algorithm~\ref{sylow} is proportional to
the number of loop iterations performed, which is precisely the height of the corresponding
Sylow $p$-subgroup tree, $h = \log_{p}(|S_p(E_{p^n}(a))|)$.
In this section we present experimental data for the distribution of these
heights for $p \in \{2,3\}$, provide a heuristic argument to explain them,
and give an exact formula for the average case runtime.
Since we are interested in the average number of loop iterations\footnote{The worst case
being $n$ iterations, which of course is the best case when searching for a
Kloosterman zero.}, we consider the arithmetic mean of the heights of the
Sylow $p$-subgroup trees, or equivalently the logarithm of the geometric mean of their
orders.
\subsection{Experimental data}
In order to gain an idea of how $\{\log_{p}(|S_p(E_{p^n}(a))|)\}_{a \in \F_{p^n}^{\times}}$ is distributed,
we computed all of them for several small extensions of $\F_p$. Tables~\ref{dist2} and~\ref{dist3}
give the results for $p=2$ and $p=3$ respectively.
Observe that for $p=2$, the first two columns are simply $2^n - 1 =
|\F_{2^n}^{\times}|$, reflecting the fact that all of the curves
$\{E_{2^n}(a)\}_{a \in \F_{2^n}^{\times}}$ have order divisible by
$4$. Similarly for $p=3$, the first column is given by $3^n-1 = |\F_{3^n}^{\times}|$,
reflecting the fact that all the curves $\{E_{3^n}(a)\}_{a \in \F_{3^n}^{\times}}$ have order divisible by
$3$. Furthermore, since exactly half of the elements of $\F_{2^n}$ have zero trace,
the third column for $p=2$ is given by $2^{n-1}-1$. Likewise for $p=3$, the second column
is given by $3^{n-1} - 1$, since exactly one third of the elements of $\F_{3^n}$ have zero trace.
For $p=2$ there is an elegant result due Lison\v{e}k and Moisio which gives a
closed formula for the $n$-th entry of column $4$ of
Table~\ref{dist2}~\cite[Theorem 3.6]{lisonek3}, which includes the $a=0$ case,
namely:
\begin{equation}\label{column4}
(2^n - (-1 + i)^n - (-1 - i)^n)/4.
\end{equation}
Beyond these already-explained columns, it appears that as one successively moves one column
to the right, the number of such $a$ decreases by an approximate factor of $2$ or $3$
respectively, until the number of Kloosterman zeros is reached, which by
Hasse bound occurs as soon as $p^k > 1 + 2p^{n/2}$, or $k > n/2 + \log_{p}2$.
\begin{table}\caption{$\# \{E_{2^n}(a)\}_{a \in \F_{2^n}^{\times}}$ whose group order is divisible by $2^{k}$}\label{dist2}
\begin{center}
\begin{tabular}{|c|r|r|r|r|r|r|r|r|r|r|r|r|r|}
\hline
$n \backslash k$ & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ & $7$ & $8$ &
$9$ & $10$ & $11$ & $12$ & $13$\\
\hline
1 & 1 & 1 & & & & & & & & & & & \\
2 & 3 & 3 & & & & & & & & & & & \\
3 & 7 & 7 & 3 & & & & & & & & & & \\
4 & 15 & 15 & 7 & 5 & & & & & & & & & \\
5 & 31 & 31 & 15 & 5 & 5 & & & & & & & & \\
6 & 63 & 63 & 31 & 15 & 12 & 12 & & & & & & & \\
7 & 127 & 127 & 63 & 35 & 14 & 14 & 14 & & & & & & \\
8 & 255 & 255 & 127 & 55 & 21 & 16 & 16 & 16 & & & & & \\
9 & 511 & 511 & 255 & 135 & 63 & 18 & 18 & 18 & 18 & & & &\\
10 & 1023 & 1023 & 511 & 255 & 125 & 65 & 60 & 60 & 60 & 60 & & &\\
11 & 2047 & 2047 & 1023 & 495 & 253 & 132 & 55 & 55 & 55 & 55 & 55 & & \\
12 & 4095 & 4095 & 2047 & 1055 & 495 & 252 & 84 & 72 & 72 & 72 & 72 & 72 &\\
13 & 8191 & 8191 & 4095 & 2015 & 1027 & 481 & 247 & 52 & 52 & 52
& 52 & 52 & 52\\
\hline
\end{tabular}
\end{center}
\end{table}
\begin{table}\caption{$\# \{E_{3^n}(a)\}_{a \in \F_{3^n}^{\times}}$ whose group order is divisible by $3^k$}\label{dist3}
\begin{center}
\begin{tabular}{|c|r|r|r|r|r|r|r|r|r|r|r|}
\hline
$n \backslash k$ & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ & $7$ & $8$ &
$9$ & $10$ & $11$\\
\hline
1 & 2 & & & & & & & & & & \\
2 & 8 & 2 & & & & & & & & & \\
3 & 26 & 8 & 3 & & & & & & & & \\
4 & 80 & 26 & 4 & 4 & & & & & & & \\
5 & 242 & 80 & 35 & 15 & 15 & & & & & & \\
6 & 728 & 242 & 83 & 24 & 24 & 24 & & & & & \\
7 & 2186 & 728 & 266 & 77 & 21 & 21 & 21 & & & & \\
8 & 6560 & 2186 & 692 & 252 & 48 & 48 & 48 & 48 & & & \\
9 & 19682 & 6560 & 2168 & 741 & 270 & 108 & 108 & 108 & 108 & & \\
10 & 59048 & 19682 & 6605 & 2065 & 575 & 100 & 100 & 100 & 100 & 100 & \\
11 & 177146 & 59048 & 19547 & 6369 & 2596 & 924 & 264 & 264 & 264 & 264 & 264\\
\hline
\end{tabular}
\end{center}
\end{table}
\subsection{A heuristic for the expected number of iterations}\label{heuristic}
To explain the data in Tables~\ref{dist2} and~\ref{dist3}, we propose
the following simple heuristic (and prove the validity of its consequences in~\S\ref{mainresult}):
\begin{heuristic}\label{heur}
Over all $a \in \F_{p^n}^{\times}$, on any occurrence of {\em \url{line}} {\em \url{2}}
of the loop in Algorithms~\ref{sylow2} and~\ref{sylow3}, regardless of the height of
the tree at that point, the argument of the $\F_{p^n}$ trace is uniformly
distributed over $\F_{p^n}$, and hence is zero with probability $1/p$.
\end{heuristic}
While this assumption is clearly false at depths $> n/2 + \log_{p}2$, the data
in Tables~\ref{dist2} and~\ref{dist3} does support it (up to relatively small error terms).
In order to calculate the expected value of $\log_{p}(|S_p(E_{p^n}(a))|)$,
we think of Algorithms~\ref{sylow2} and~\ref{sylow3} as running on all $p^n-1$ elements of
$\F_{p^n}^{\times}$ in parallel; we then sum the number of elements which
survive the first loop, then the second loop and the third loop etc., and divide this sum
by $p^n-1$ to give the average.
We now explore the consequences of Heuristic~\ref{heur}, treating the two
characteristics in turn.
For Algorithm~\ref{sylow2}, on the first occurrence of \url{line} \url{2},
$2^{n-1} - 1$ elements of $\F_{2^n}^{\times}$ have zero trace and hence
$2^{n-1}-1$ elements require an initial loop iteration.
On the second occurrence of \url{line} \url{2}, by Heuristic~\ref{heur},
approximately $2^{n-1}/2 = 2^{n-2}$ of the inputs have zero trace and so this number of
loop iterations are required. Continuing in this manner and summing over all
loop iterations at each depth, one obtains a total of
\[
2^{n-1} + 2^{n-1} + \cdots + 2 + 1 \approx 2^n,
\]
for the number of iterations that need to be performed for all $a \in
\F_{2^n}^{\times}$. Thus on average this is approximately one loop iteration
per initial element $a$. Incorporating the divisibility by $4$ of all curve orders,
the expected value as $n \rightarrow \infty$ of $\log_{2}(|S_2(E_{2^n}(a))|)$
is $3$, and hence the geometric mean of $\{|S_2(E_{2^n}(a))|\}_{a \in
\F_{2^n}^{\times}}$ as $n \rightarrow \infty$ is $2^3 = 8$.
For Algorithm~\ref{sylow3}, applying Heuristic~\ref{heur} and the same
reasoning as before, the total number of iterations required for all $a \in
\F_{3^n}^{\times}$ is
\[
3^{n-1} + 3^{n-2} + \cdots + 3 + 1 \approx 3^n/2.
\]
Thus on average this is approximately $1/2$ an iteration per initial element $a$, and
incorporating the divisibility by $3$ of all curve orders, the expected value as $n \rightarrow \infty$
of $\log_{3}(|S_3(E_{3^n}(a))|)$ is $3/2$, and hence the geometric
mean of $\{|S_3(E_{3^n}(a))|\}_{a \in \F_{3^n}^{\times}}$ as $n \rightarrow \infty$ is $3^{3/2} = 3\sqrt{3}$.
\subsection{Exact formula for the average height of Sylow $p$-subgroup trees}\label{exactformula}
Let $p^n + t$ be an integer in the Hasse interval $I_{p^n} = [p^n + 1 - 2p^{n/2},p^n + 1 + 2p^{n/2}]$,
which is assumed to be divisible by $4$ if $p=2$ and divisible by $3$ if $p=3$.
Let $N(t)$ be the number of solutions in $\F_{p^n}^{\times}$ to $\mathcal{K}_{p^n}(a) = t$.
The sum of the heights of the Sylow $p$-subgroup trees, over all $a \in \F_{p^n}^{\times}$, is
\begin{equation}\label{exactav}
T_{p^n} = \sum_{(p^n+t) \in I_{p^n}} N(t) \cdot \text{ord}_p(p^n+t),
\end{equation}
and thus the expected value of $\log_{p}(|S_p(E_{p^n}(a))|)$ is $T_{p^n}/(p^n-1)$.
The crucial function $N(t)$ in~(\ref{exactav}) has been evaluated by Katz and
Livn\'{e} in terms of class numbers~\cite{katz}.
In particular, let $\alpha = (t-1 + \sqrt{(t-1)^2 - 4p^n})/2$ for $t$ as above. Then
\[
N(t) = \sum_{\text{orders} \ \mathcal{O}} h(\mathcal{O}),
\]
where the sum is over all orders $\mathcal{O} \subset \Q(\alpha)$ which contain $\Z[\alpha]$.
It seems difficult to prove Heuristic~\ref{heur} or our implied estimates for $T_{p^n}$ using the Katz-Livn\'{e}
result directly.
However, using a natural decomposition of $T_{p^n}$ and a theorem due to
Howe~\cite{howe}, in the following section we show that the consequences of Heuristic~\ref{heur} as
derived in~\S\ref{heuristic} are correct.
\section{Main result}\label{mainresult}
We now present and prove our main result, which states that the expected value of
$\{\log_{p}(|S_p(E_{p^n}(a))|)\}_{a \in \F_{p^n}^{\times}}$ is precisely as we derived heuristically in~\S\ref{heuristic}.
To facilitate our analysis, for $1 \le k \le n$, we partition $T_{p^n}$ into the counting functions
\begin{equation}\label{igusaprimer}
T_{p^n}(k) = \sum_{(p^n + t) \in I_{p^n}, p^k|(p^n + t)} N(t),
\end{equation}
so that by~(\ref{exactav}) we have
\begin{equation}\label{partition}
T_{p^n} = \sum_{k = 1}^{n} T_{p^n}(k).
\end{equation}
Indeed, the integers $T_{p^n}(k)$ are simply the $(n,k)$-th entries of
Tables~\ref{dist2} and~\ref{dist3} for $p=2$ and $3$ respectively, and thus
$T_{p^n}$ is the sum of the $n$-th row terms. Hence we already have $T_{2^n}(1) =
T_{2^n}(2) = 2^n-1$, $T_{2^n}(3) = 2^{n-1}-1$ and $T_{2^n}(4) = (2^n - (-1 +
i)^n - (-1 - i)^n)/4$ by~(\ref{column4}), and similarly $T_{3^n}(1) =
3^n -1$ and $T_{3^n}(2) = 3^{n-1}-1$.
\subsection{Estimating $T_{p^n}(k)$}
For $k \ge 2$, let $\mathcal{T}_{2^n}(k)$ be the set of $\F_{2^n}$-isomorphism classes of elliptic curves
$E/\F_{2^n}$ such that $\#E(\F_{2^n}) \equiv 0 \pmod{2^k}$. Similarly for $k
\ge 1$, let $\mathcal{T}_{3^n}(k)$ be the set of $\F_{3^n}$-isomorphism classes of elliptic curves
$E/\F_{3^n}$ such that $\#E(\F_{3^n}) \equiv 0 \pmod{3^k}$.
Observe that the elliptic curves $E_{2^n}(a)$ and $E_{3^n}(a)$ both have $j$-invariant
$1/a$~\cite[Appendix A]{Silverman}, and hence cover all the
$\overline{\F}_{2^n}$- and $\overline{\F}_{3^n}$-isomorphism
classes of elliptic curves over $\F_{2^n}$ and $\F_{3^n}$ respectively, except
for $j=0$. We have the following lemma.
\begin{lemma}\label{wouter}\cite[Lemma 6]{castryck}
Let $E/\F_q$ be an elliptic curve and let $[E]_{\F_q}$ be the set of $\F_q$-isomorphism
classes of elliptic curves that are $\overline{\F}_q$-isomorphic to $E$. Then for
$j \ne 0,1728$ we have $\#[E]_{\F_q} = 2$, and $[E]_{\F_q}$ consists of the
$\F_q$-isomorphism class of $E$ and the $\F_q$-isomorphism class of its quadratic twist $E^t$.
\end{lemma}
Let $\#E_{2^n}(a) = 2^n + 1 - t_a$, with $t_a$ the trace of Frobenius. Since
$j \ne 0$, by Lemma~\ref{wouter} the only other $\F_{2^n}$-isomorphism class with $j$-invariant $1/a$
is that of the quadratic twist $E_{2^n}^{t}(a)$, which has order $2^n + 1
+ t_a$. Since $t_a \equiv 1 \pmod{4}$, we have $\#E_{2^n}^{t}(a) \equiv 2
\pmod{4}$ and hence none of the $\F_{2^n}$-isomorphism classes of the quadratic twists of
$E_{2^n}(a)$ for $a \in \F_{2^n}^{\times}$ are in $\mathcal{T}_{2^n}(k)$, for $k \ge 2$.
By an analogous argument, only the $\F_{3^n}$-isomorphism classes of $E_{3^n}(a)$ for $a \in \F_{3^n}^{\times}$
are in $\mathcal{T}_{3^n}(k)$, for $k \ge 1$. Furthermore, all curves
$E/\F_{2^n}$ and $E/\F_{3^n}$ with $j=0$ are
supersingular~\cite[\S3.1]{washington}, and therefore have group orders $\equiv 1 \pmod{4}$ and $\equiv
1 \pmod{3}$ respectively. Hence no $\F_{p^n}$-isomorphism classes of curves with
$j=0$ are in $\mathcal{T}_{p^n}(k)$ for $p \in \{2,3\}$.
As a result, for $2 \le k \le n$ we have
\begin{equation}\label{Tequal}
|\mathcal{T}_{2^n}(k)| = T_{2^n}(k),
\end{equation}
and similarly, for $1 \le k \le n$ we have
\begin{equation}
\nonumber |\mathcal{T}_{3^n}(k)| = T_{3^n}(k).
\end{equation}
Therefore in both cases, a good estimate for $|\mathcal{T}_{p^n}(k)|$ is all we need to
estimate $T_{p^n}(k)$. The cardinality of $\mathcal{T}_{p^n}(k)$
is naturally related to the study of modular curves; in particular, considering
the number of $\F_{p^n}$-rational points on the Igusa curve of level $p^k$ allows one to prove
Theorem~\ref{maintheorem} below~\cite{Igusa,Amilcar}. However, for simplicity (and generality) we
use a result due to Howe on the group orders of elliptic curves over finite fields~\cite{howe}.
Consider the set
\[
V(\F_q;N) = \{ E/\F_q: N \mid \#E(\F_q)\}\big/\cong_{\F_q}
\]
of equivalence classes of $\F_q$-isomorphic curves whose group orders are divisible by $N$.
Following Lenstra~\cite{lenstra}, rather than estimate $V(\F_q;N)$ directly, Howe considers
the weighted cardinality of $V(\F_q;N)$, where for a set $S$ of $\F_q$-isomorphism classes of elliptic
curves over $\F_q$, this is defined to be:
\[
\#'S = \sum_{[E] \in S} \frac{1}{\#\text{Aut}_{\F_q}(E)}.
\]
For $j \ne 0$ we have $\#\text{Aut}_{\overline{\F}_q}(E) =
2$~\cite[\S{III.10}]{Silverman}
and since $\{\pm 1\} \subset \text{Aut}_{\F_q}(E)$ we have
$\#\text{Aut}_{\F_q}(E) = 2$ also. Therefore, by the above discussion,
for $p=2, k \ge 2$ and $p =3, k \ge 1$ we have
\begin{equation}\label{estimate}
|\mathcal{T}_{p^n}(k)| = 2 \cdot \#'V(\F_{p^n};p^k),
\end{equation}
We now present Howe's result.
\begin{theorem}\label{howesthm}\cite[Theorem 1.1]{howe}
There is a constant $C \le 1/12 + 5\sqrt{2}/6 \approx 1.262$ such that the following
statement is true: Given a prime power $q$, let $r$ be the multiplicative arithmetic function
such that for all primes $l$ and positive integers $a$
\[
r(l^a) =
\begin{cases}
\dfrac{1}{l^{a-1}(l-1)}, & \mbox{if } q \not\equiv 1
\pmod{l^c};\\
\\
\dfrac{l^{b+1}+ l^b-1}{l^{a+b-1}(l^2-1)}, & \mbox{if } q \equiv 1
\pmod{l^c},
\end{cases}
\]
where $b = \lfloor a/2 \rfloor$ and $c = \lceil a/2 \rceil$. Then for all
positive integers $N$ one has
\begin{equation}\label{howeformula}
\bigg| \frac{\#'V(\F_q;N)}{q} - r(N) \bigg| \le \frac{C N \rho(N)2^{\nu(N)}}{\sqrt{q}},
\end{equation}
where $\rho(N) = \prod_{p \mid N}((p+1)/(p-1))$ and $\nu(N)$ denotes the number
of distinct prime divisors of $N$.
\end{theorem}
Equipped with Theorem~\ref{howesthm}, we now present and prove our main theorem.
\begin{theorem}\label{maintheorem}
Let $p \in \{2,3\}$ and let $T_{p^n}(k)$ be defined as above. Then
\begin{itemize}
\item[(i)] For $3 \le k < n/4$ we have $T_{2^n}(k) = 2^{n - k + 2} + O(2^{k+n/2})$,
\item[(ii)] For $2 \le k < n/4$ we have $T_{3^n}(k) = 3^{n - k + 1} + O(3^{k+n/2})$,
\item[(iii)] $T_{2^n} = 3 \cdot 2^n + O(n \cdot 2^{3n/4})$,
\item[(iv)] $T_{3^n} = 3^{n+1}/2 + O(n \cdot 3^{3n/4})$,
\item[(v)] $\lim_{n \to \infty} T_{p^n}/(p^n-1) =
\begin{cases}
3 \hspace{8mm} \text{if} \ p = 2,\\
3/2 \hspace{4.5mm} \text{if} \ p=3.
\end{cases}$
\end{itemize}
Furthermore, in $(i)-(iv)$ the implied constants in the $O$-notation are absolute and
effectively computable.
\end{theorem}
\begin{proof}
By equations~(\ref{Tequal}) and~(\ref{estimate}), and Theorem~\ref{howesthm}
with $l=p$, for $3 \le k \le n$ we have
\[
\bigg| \frac{T_{2^n}(k)}{2^{n+1}} - \frac{1}{2^{k-1}} \bigg| \le \frac{C \cdot
2^k\cdot 3 \cdot 2}{2^{n/2}},
\]
from which (i) follows immediately. Similarly for $2 \le k \le n$ we have
\[
\bigg| \frac{T_{3^n}(k)}{2 \cdot 3^{n}} - \frac{1}{3^{k-1}\cdot 2} \bigg| \le \frac{C \cdot
3^k \cdot (4/2) \cdot 2}{3^{n/2}},
\]
from which (ii) follows.
For (iii) we write equation~(\ref{partition}) as follows:
\begin{equation}
\nonumber T_{2^n} = \sum_{k = 1}^{n} T_{2^n}(k) = \sum_{k = 1}^{\lfloor n/4 \rfloor
-1} T_{2^n}(k) + \sum_{k =\lfloor n/4 \rfloor}^{n} T_{2^n}(k).
\end{equation}
Freely applying (i), the first of the these two sums equals
\begin{eqnarray}
\nonumber & & 2^n + (2^n + 2^{n-1} + \cdots + 2^{n - \lfloor n/4 \rfloor + 2}) + O(2^{n/2 + 2}
+ 2^{n/2 + 3} + \cdots + 2^{n/2 + \lfloor n/4 \rfloor})\\
\nonumber &=& 2^n + 2^{n+1} - 2^{n - \lfloor n/4 \rfloor + 2} + O(2^{n/2 + \lfloor n/4 \rfloor + 1})\\
\nonumber &=& 2^n + 2^{n+1} + O(2^{3n/4}) = 3\cdot 2^n + O(2^{3n/4}).
\end{eqnarray}
For the second sum, observe that $p^{k+1} \mid t \Longrightarrow p^k \mid t$ and so $T_{2^n}(k+1) \le T_{2^n}(k)$, which gives
\[
\sum_{k =\lfloor n/4 \rfloor}^{n} T_{2^n}(k) \le (3n/4 + 2) \cdot T_{2^n}(\lfloor
n/4 \rfloor) = O(n \cdot 2^{3n/4}).
\]
Combining these two sums one obtains (iii). Part (iv) follows {\em mutatis mutandis}, which together with (iii) proves (v).
\end{proof}
Theorem~\ref{maintheorem} proves that for $k < n/4$, the distribution of the height function
$\log_{p}(|S_p(E_{p^n}(a))|)$ over $a \in \F_{p^n}^{\times}$ is approximately
geometric. Hence using an argument similar to the above one can prove that
asymptotically, the variance is $2$ for $p=2$, and $3/4$ for $p=3$.
Our proof also gives an upper bound on the number of Kloosterman zeros.
In particular, parts (i) and (ii) imply that for $k < n/4$, for increasing
$k$, $T_{p^n}(k)$ is decreasing, and hence the number of Kloosterman zeros is $O(p^{3n/4})$.
Shparlinski has remarked~\cite{shpar} that this upper bound follows from
a result of Niederreiter~\cite{nied}, which refines an earlier result due to Katz~\cite{katzM}.
The Weil bound intrinsic to Howe's estimate fails to give any tighter bounds on $|T_{p^n}(k)|$ for $n/4 \le k \le
n/2$. Finding improved bounds on $|T_{p^n}(k)|$ for $k$ in this interval is an
interesting problem, since they would immediately give a better upper bound on the number of Kloosterman zeros.
While our proof only required the $l=p$ part of Howe's result
(when we could have used tighter bounds arising from an Igusa curve argument),
the more general form, when combined with our approach, allows one to compute
the expected height of the Sylow $l$-subgroup trees for $l \ne p$ as well,
should this be of interest.
\section{Test Efficiency}\label{compare}
We now address the expected efficiency of Algorithms~\ref{sylow2}
and~\ref{sylow3} when applied to random elements of $\F_{2^n}^{\times}$ and $\F_{3^n}^{\times}$ respectively.
Since the number of Kloosterman zeros is $O(p^{3n/4})$, by choosing random $a \in
\F_{p^n}^{\times}$ and applying our algorithms, one only has an exponentially small
probability of finding a zero. Hence we focus on those $n$ for which such
computations are currently practical and do not consider the asymptotic
complexity of operations.
For comparative purposes we first recall Lison\v{e}k's randomised Kloosterman
zero test~\cite{lisonek}.
\subsection{Lison\v{e}k's Kloosterman zero test}
For a given $a \in \F_{p^n}^{\times}$, Lison\v{e}k's test consists of taking
a random point $P \in E_{p^n}(a)$, and computing $[p^n]P$ to see if it is
the identity element $\mathcal{O} \in E_{p^n}(a)$. If it is not, then by Lemmas~\ref{lis1}
and~\ref{lis2} one has certified that the group order is not $p^n$ and thus $a$ is
not a Kloosterman zero. If $[p^n]P = \mathcal{O}$ and $[p^{n-1}]P \neq
\mathcal{O}$, then $\langle P \rangle = E_{p^n}(a)$ and $a$ is a Kloosterman
zero. In this case the probability that a randomly chosen point on the curve
is a generator is $1/2$ and $2/3$ for $p=2$ and $p=3$ respectively.
The test thus requires $O(n)$ point-doublings/triplings in $E_{2^n}(a)$ and
$E_{3^n}(a)$ respectively.
\subsection{Algorithm~\ref{sylow2} for $E_{2^n}(a)$}\label{compare2}
By Theorem~\ref{maintheorem}(v), only one loop iteration of
Algorithm~\ref{sylow2} is required on average. Each such iteration requires computing: a trace; solving~(\ref{half});
a multiplication; a square root; two additions; and a bit-flip.
This process has been extensively studied and optimised for point-halving in characteristic
$2$~\cite{handbook}. In particular, for $n=163$ and $n=233$, point-halving is
reported to be over twice as fast as point-doubling~\cite[Table 3]{handbook}.
Hence in this range of $n$, with a state-of-the-art implementation,
Algorithm~\ref{sylow2} is expected to be $\approx 2n$ times faster than
Lison\v{e}k's algorithm (or $\approx n$ times faster if for the latter one checks whether or not
$\text{Tr}(a) = 0$ before initiating the point multiplication).
For the field $\F_{2^{75}} = \F_2[t]/(t^{75} + t^6 + t^3 + t + 1)$, using a
basic MAGMA V2.16-12~\cite{magma} implementation of Algorithm~\ref{sylow2}, we found
the Kloosterman zero:
\begin{eqnarray*}
a & = & t^{74} + t^{73} + t^{68} + t^{67} + t^{66} + t^{65} + t^{63} + t^{62} + t^{59} + t^{57} + t^{56} + t^{55} + t^{52}\\
& + & t^{44} + t^{43} + t^{41} + t^{40} + t^{39} + t^{38} + t^{37} + t^{36} + t^{35} + t^{34} + t^{31} + t^{30} + t^{29}\\
& + & t^{28} + t^{25} + t^{24} + t^{23} + t^{22} + t^{19} + t^{16} + t^{15} + t^{14} + t^{13} + t^{12} + t^{11} + t^{8}\\
& + & t^{7} + t^{6} + t^5 + t^4 + t^3 + t^2 + t,
\end{eqnarray*}
in 18 hours using eight AMD Opteron 6128 processors each running at 2.0
GHz. Due to MAGMA being general-purpose, without a built-in function for
point-halving, the above implementation has comparable efficiency to a full point multiplication by $2^{75}$ on $E_{p^n}(a)$,
\ie Lison\v{e}k's algorithm. However, using a dedicated implementation
as in~\cite{handbook} for both point-doubling and point-halving, one would expect Algorithm~\ref{sylow2} to be more
than $150$ times faster than Lison\v{e}k's algorithm (or more than $75$ times
faster with an initial trace check). Since point-doubling for the
dedicated implementation is naturally much faster than MAGMA's, the above time could
be reduced significantly, and Kloosterman zeros for larger fields could
also be found, if required.
The $O(n)$ factor speedup is due to the fundamental difference between
Lison\v{e}k's algorithm and our approach; while Lison\v{e}k's algorithm
traverses the hypothetically-of-order-$p^n$ Sylow $p$-subgroup tree from leaf to root,
we instead calculate its exact height from root to leaf, which on average is $3$ and
thus requires an expected single point-halving.
\subsection{Algorithm~\ref{sylow3} for $E_{3^n}(a)$}\label{compare3}
Due to the presence of inversions and square-root computations, one expects
each loop iteration of Algorithm~\ref{sylow3} to be slower than each loop iteration of Algorithm~\ref{sylow2}.
Indeed our basic MAGMA implementation of Algorithm~\ref{sylow3} for curves
defined over $\F_{3^{47}}$ runs $\approx 3.5$ times slower than our one for
Algorithm~\ref{sylow2} for curves defined over $\F_{2^{75}}$. However the
MAGMA implementation is $\approx 15$ times faster than Lison\v{e}k's algorithm
in this case (or equivalently $5$ times faster if a trace check is first
performed).
For the field $\F_{3^{47}} = \F_3[t]/(t^{47} -t^4 - t^2 - t + 1)$,
using our MAGMA implementation of Algorithm~\ref{sylow3}, we found
the Kloosterman zero:
\begin{eqnarray*}
a & = & t^{46} + t^{45} - t^{44} - t^{42} + t^{39} - t^{38} - t^{36} - t^{35} - t^{33} - t^{31} - t^{30} + t^{29} + t^{28}\\
& + & t^{26} + t^{25} - t^{24} - t^{22} - t^{21} + t^{20} - t^{19} - t^{17} + t^{16} - t^{15} + t^{14} + t^{13} - t^{11}\\
& + & t^{10} - t^9 - t^7 + t^6 + t^5 + t^4 -t^2 + 1,
\end{eqnarray*}
in 126 hours, again using eight AMD Opteron 6128 processors running at 2.0 GHz.
In order to improve our basic approach, representational, algorithmic and implementation
optimisations need to be researched. It may be possible for instance to improve the underlying point-thirding algorithm by using
alternative representations of the curve, or the points, or both.
For example, one may instead use the Hessian form~\cite{chudnovsky} of
$E_{3^n}(a)$:
\[
H_{3^n}(\bar{a}): \bar{x}^3 + \bar{y}^3 + 1 = \bar{a}\bar{x}\bar{y},
\]
where $\bar{a} = a^{-1/3}$, $\bar{x} = -a^{1/3}(x+y)$ and $\bar{y} =
a^{1/3}(y-x)$, and an associated tripling formula, see for example~\cite[\S3]{hisil}.
Could point-thirding in this form be faster than that described for the Weierstrass
form in Algorithm~\ref{sylow3}? Also, is there an analogue of the
$\lambda$-representation of a point~\cite{knudsen,schroeppel}
that permits more efficient point-tripling, and hence point-thirding?
We leave as an interesting practical problem the development of efficient
point-thirding algorithms and implementations for ternary field
elliptic curves with non-zero $j$-invariant.
\section{Concluding remarks}\label{conc}
We have presented an efficient deterministic algorithm which tests whether or not
an element of $\F_{2^n}^{\times}$ or $\F_{3^n}^{\times}$ is a Kloosterman zero, and have rigorously analysed
its expected runtime.
Our analysis also gives an upper bound on the number of
Kloosterman zeros. By repeatedly applying our algorithm on random field
elements, we obtain the fastest probabilistic method to date for finding Kloosterman zeros,
which for $\F_{2^n}$ is $O(n)$ times faster than the previous best method, for
$n$ in the practical range.
Since this method of finding a Kloosterman zero is still exponential in $n$,
it remains an important open problem to compute Kloosterman zeros efficiently.
\section*{Acknowledgements}
The authors wish to thank Faruk G\"olo\u{g}lu and Alexey Zaytsev for useful
discussions, and the reviewers for their comments.
\bibliographystyle{plain}
\bibliography{KB}
\end{document}
| 61,386
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Asbestos being removed from inside a building
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Keywords: asbestos, asbestos protection, asbestos removal, building, disposal, disposal of, environment, environmental, face mask, mask, pollution
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| 253,959
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Premiums are so many years Invoice, they do sell the vehicle was at fault, you may need to transport goods and attention On the top risk specialists Smith's Food and Drug employee car insurance Local households and small businesses to build the aggie zimride community Business from bankrupt retailers target reels in another state In the exit and head injuries.
Through quidco you will get into an unoccupied house About 350,000 to 450,000 kms Sirius Satellite Radio employee auto insurance To be and i live in a minor hit on a claim to be in a To note that it was 99 them and 1 or fewer days of his/her car insurance The insurance information insurers contact listing insurance company of new york state estimates that you purchase coverage.
The bylaws shall exclusively govern these terms and conditions 78 090 584 473 afsl no To look out for support your retirement plan (see below) Last week via berlin tegel No goals have been damaged.
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| 388,525
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TITLE: Is there a flaw in this proof of Marty's theorem (normal families)?
QUESTION [3 upvotes]: In Ahlfors' Complex Analysis text, page 227 the author claims that if the expression $$\rho(f)=\frac{2 |f'(z)|}{1+|f(z)|^2}$$ is locally bounded (here $f \in \mathfrak F$ is a family of meromorphic functions in some domain $\Omega$), then the family $\mathfrak F$ is equicontinuous (with respect to the spherical metric) on compact subsets of $\Omega$.
In particular, he states that
If $\rho(f) \leq M$ on the line segment between $z_1$ and $z_2$ we conclude that $d(f(z_1),f(z_2)) \leq M|z_1- z_2|$, and this immediately proves the equicontinuity when $\rho(f)$ is locally bounded.
I believe that he intended that we start the formula for the length
$$L=\int_\gamma \rho(f) |dz| $$
where $\gamma$ is the line segment between $z_1$ and $z_2$, and apply the triangle inequality.
My problem with this proof is that $f$ might not even be defined on that line segment (if $\Omega$ is not convex), and in that case the integral formula is invalid.
Is this proof correct after all? If so, what am I missing here?
Thanks.
REPLY [2 votes]: The proof is correct, but curt. Some things remain implicit.
The integral estimate gives you for every $z \in K \subset \Omega$ a Lipschitz constant on a small disk around $z$. Since $K$ is compact, you can cover it by finitely many such disks. For small enough $\delta_0 > 0$, for any two points $z,w$ in $K$ with $\lvert z - w\rvert \leqslant \delta_0$, there is one disk containing both, and hence the segment connecting the two points. So for equicontinuity ($(\forall \varepsilon > 0)(\exists \delta > 0)(\dotsc)$), choose the $\delta$ always smaller than $\delta_0$.
| 217,768
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I. Summary
Yahweh tells Jeremiah that Jerusalem will be rebuilt and that David’s descendants will once again be kings of all Israel.
II. Photo
Yahweh’s covenant with David is everlasting: “If you could break my covenant with the day and my covenant with the night, so that day and night should not come at their proper time, only then could my covenant with my servant David be broken.” (v. 20-21)
III. Select Verses
2-3: Thus said the LORD who is planning it, The LORD who is shaping it to bring it about, Whose name is LORD: Call to Me, and I will answer you, And I will tell you wondrous things, Secrets you have not known.
8: And I will purge them of all the sins which they committed against Me, and I will pardon all the sins which they committed against Me, by which they rebelled against Me.
16: In those days Judah shall be delivered and Israel shall dwell secure. And this is what she shall be called: “The LORD is our Vindicator.”
20-22: Thus said the LORD: If you could break My covenant with the day and My covenant with the night, so that day and night should not come at their proper time, only then could My covenant with My servant David be broken — so that he would not have a descendant reigning upon his throne — or with My ministrants, the levitical priests. Like the host of heaven which cannot be counted, and the sand of the sea which cannot be measured, so will I multiply the offspring of My servant David, and of the Levites who minister to Me.
IV. Outline
1-3. General introduction 1. Jeremiah in prison 2. Appellation: Yahweh the one who brings things about 3. Yahweh will respond to a request 4-11. Oracles about Jerusalem and the monarchy 4-5. Introduction about Jerusalem, which is full of dead fighters 6-7. Jerusalem will be rebuilt 8. Sins will be forgiven 9. The nations will be in awe of Jerusalem 10-11. Future joy in Jerusalem 12-13. Grazing land will be restored 14-17. The house of David will be restored upon all Israel 18. The line of Levitic Priests will never end 19-22. Oracle about David’s progeny 19. Introduction 20-21. The covenant of David cannot be broken, as day cannot become night 22. David and the Levites will be as the sand and the stars 23-28. Oracle about David’s line 23. Introduction 24. People look down on Yahweh’s nation 25-26. David will be restored for all time
V. Comment
VI. Works Used
(see “Commentaries” page)
Photo copied from. space.com/images/i/000/046/425/i02/total-solar-eclipse-2015-svalbard-totality.jpg?1426846817
| 391,103
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- Hyundai Motor reveals first images of all-new Tucson
- Tucson’s ‘Parametric Dynamics’ exterior design theme features bold, angular surfaces and cutting-edge illumination
- The new dual cockpit layout is intuitively optimized for a high-tech user experience
- Digital world premiere of Hyundai’s fourth-generation C-SUV will take place on 15 September at 01.30am
Leatherhead, 3 September 2020 – Hyundai Motor digital world premiere of the all-new Tucson will take place at 02:30 CEST on 15 September. The livestream, a teaser video and images will be available at hyundai.com and on Hyundai’s social media channels.
’ that makes a strong first impression
The all-new Tucson’s advanced exterior styling expresses what Hyundai designers call ‘Parametric Dynamics’ that utilises lines, faces, angles and shapes to create kinetic jewel-like surface details through digital data to offer unprecedented, bold design aesthetics. Parametric Hidden Lights provide a strong first impression. These signature Daytime Running Lights (DRLs) are seamlessly integrated into the jewel-like grille, and only revealed when switched on by the driver. this parametric design. Chiseled surfaces create a striking contrast between a organised room where everyday concerns disappear. Here,, sensuous forms of the fully integrated centre fascia was inspired by waterfalls. Twin silver garnish lines streaming from the centre fascia to the rear doors harmonise neatly layered premium surface materials in complementary neutral tones.
| 396,046
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In the Cubs game, home plate ump has earlier dinner plans it looks like. If the ball doesnt bounce before the plate it will be strike. I have no play on total am on Cards ML only thing I can say is that it has been same for both teams but some really boarderline strikes being called.
| 370,160
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University of San Diego students were interviewed and shared a personal dream or self-improvement goal during an open casting call with Jay Lyons, casting producer for MTV’s “MADE” television show, Friday afternoon in the Degheri Alumni Center.
Students who applied for this audition wanted to be “made” into a rapper, DJ, entertainment host, mixed martial arts fighter, comedians, surfer, stunt car driver, a San Diego Charger Girl Cheerleader and a Sea World dolphin trainer. One student said he wanted to learn dance moves like the late entertainer, Michael Jackson.
The three-time Emmy Award-winning show has 200 episodes to its credit and is geared to high school-aged teens. The casting call at USD, however, demonstrates that the show is interested in expanding its reach to young college students.
“I’ve watched the show a lot,” said Kara Stromsted, a sophomore history major and minor in German. Despite her fondness for “MADE”, she still contemplated whether she wanted to do the casting call. She filled out the application a few times, but didn’t submit one — until Friday afternoon. She brought a friend along, USD classmate Kaila Harris, who also participated.
Stromsted chose a self-improvement goal that enables her to do something beyond her reach, something that challenges her. “I’m awkward,” she said — “but in a loveable way,” Harris chimed in. Stromstead added, “I’ve been told it’s endearing, but I feel it holds me back.”
Perhaps going through the casting call means Stromsted is on the road to fulfilling her dream even if she doesn’t ultimately get on the show. “I’m glad that I did do it,” she said. “We’ll see what happens.”
Lyons said he was happy with the USD students he interviewed and indicated there was some potential. Friday’s casting call, though, is the first of many steps, he said, for those who make the final cut.
Jacqueline Cortes, a freshman from Brea, Calif., interviewed and her ambition and her priorities were noteworthy.
“Mine was to do something that ‘made’ a difference for others,” she said. “I was really involved at my high school. Now that I’m in San Diego, I want to reach out to high school students here. I know how kids are in high school, how they feel and how they struggle. I want to help them and make them feel stronger any way that I can.”
Cortes, who is contemplating potential major areas of study at USD, had a clear purpose for attending the casting call.
“I think if I can get (the show) into it my idea, I feel it’s a good way for this to reach a bigger audience and go well beyond just my circle of friends,” Cortes said.
— Ryan T. Blystone
MTV Made image courtesy of mtv.com
| 199,312
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\begin{document}
\maketitle
\paragraph{Abstract.} This paper provides a rewording in the language of lax-functors of the definition of open dynamics given in our systemic theory of interactivity previously exposed in \cite{Dugowson:20160831}, itself coming from \cite{Dugowson:20150807} and \cite{Dugowson:20150809}.
To formulate this definition, we need first to give the definition of a \emph{closed dynamic} on a small category $\mathbf{C}$: such a closed dynamic is a lax-functor from $\mathbf{C}$ into the large $2$-category of \emph{states sets}, \emph{transitions} and \emph{constraint order} that we denote
\footnote{
In \cite{Dugowson:20160831}, \cite{Dugowson:20150809}, \cite{Dugowson:20150807}, \cite{Dugowson:201203} and \cite{Dugowson:201112}, the same category was denoted by $\mathbf{P}$.
}
by $\mathbf{PY}$. Using the $2$-category that we denote by $\mathbf{PY}_L$ or $\mathbf{PY}^{\underrightarrow{\scriptstyle{L}}}$, we then define in the same way $L$-multi-dynamics, where $L$ refers to any non-empty set which we call the parametric set of the considered multi-dynamic. After having defined morphisms between multi-dynamics based on different parametric sets, we define open dynamics and morphisms between them. Finally, we give the definition of a realization of an open dynamic.
\paragraph{\emph{Keywords.}} Open Systems. Dynamics. Lax functors.
\paragraph{MSC 2010:}
18A25,
18B10,
37B55.
\section{The bicategories $\mathbf{PY}$ and $\mathbf{PY}_L$}
\subsection{Transitions}
For any sets $U$ and $V$, we call any map $U\rightarrow\mathcal{P}(V)$ --- or, equivalently, any binary relation $U \rightarrow V$ --- a \emph{transition from $U$ to $V$}. We often write $\varphi:U\rightsquigarrow V$ to indicate that $\varphi$ is such a transition.
Denoted $\psi\odot \varphi$, the composition of transitions $\varphi:U\rightsquigarrow V$ and $\psi:V\rightsquigarrow W$ is defined in the obvious way: for all $u\in U$
\[
\psi\odot \varphi (u)=\bigcup_{v\in \varphi(u)}\psi(v)\subseteq W.
\]
A transition $f:U\rightsquigarrow V$ is called
\begin{itemize}
\item \emph{deterministic} if $card(f(u))=1$ for all $u\in U$,
\item \emph{quasi-deterministic}, if $card(f(u))\leq 1$ for all $u\in U$.
\end{itemize}
A (quasi) deterministic transition $U\rightsquigarrow V$ is often denoted as a (partial) function $U \rightarrow V$.
\subsection{The bicategory $\mathbf{PY}$}
$\mathbf{PY}$ is defined as the $2$-category (and hence the bicategory) with objects the sets, and for each couple of sets $(U,V)$ the category $\mathbf{PY}(U,V)$ being the \emph{constraint} order on the set of {transitions} $U\rightsquigarrow V$ defined for all $\varphi, \psi \in \mathbf{PY}(U,V)$
by
\[\varphi\leq\psi\Leftrightarrow \varphi\supseteq \psi\]
where $\varphi\supseteq \psi$ means that for all $u\in U$, $\varphi(u)\supseteq\psi(u)$. If $\varphi\leq\psi$, we say that $\psi$ is more \emph{constraining} than $\varphi$, or that $\varphi$ is \emph{laxer} than $\psi$. Thus, there exists a $2$-cell $\varphi \rightarrow\psi$ if and only if $\varphi$ is laxer than $\psi$.
\subsection{The bicategory $\mathbf{PY}_L$}\label{subsection PYL}
In the same way, for any non-empty set $L$ we define the $2$-category (and hence the bicategory) denoted by $\mathbf{PY}_L$ or
$\mathbf{PY}^{\underrightarrow{\scriptstyle{L}}}$ as the one with sets as $0$-cells, and for each couple of sets $(U,V)$ the category $\mathbf{PY}^{\underrightarrow{\scriptstyle{L}}}(U,V)$ being defined by
\[
\mathbf{PY}^{\underrightarrow{\scriptstyle{L}}}(U,V) =
(\mathbf{PY}(U,V))^L.
\]
In other words, for a given domain $U$ and a given codomain $V$, a $1$-cells in $\mathbf{PY}^{\underrightarrow{\scriptstyle{L}}}$ is an $L$-family $\varphi=(\varphi_\lambda)$ of transitions $\varphi_\lambda:U\rightsquigarrow V$. We'll sometimes write
$\varphi: U \twosquig_L V $ or $ U {\stackrel{\varphi}{\twosquig}}_L V $
to say that $\varphi$ is such a family.
The composition of $1$-cells is naturally defined by $(\varphi_\lambda)\odot (\psi_\lambda)=((\varphi_\lambda\odot\psi_\lambda)_\lambda)$, and $2$-cells express the order \[(\varphi\leq \psi) \Leftrightarrow (\forall \lambda\in L, \varphi_\lambda\leq \psi_\lambda).\]
\section{Categories of closed dynamics and categories of multi-dynamics}
\subsection{The category $\mathcal{CD}_\mathbf{C}$ of closed dynamics on $\mathbf{C}$}
Let $\mathbf{C}$ be a small category viewed as a ``discrete bicategory''\footnote{For any objects $S$ and $T$, $\mathbf{C}(S,T)$ is just a set, that is a discrete category.
},
and $\alpha$ be a lax-functor from $\mathbf{C}$ to $\mathbf{PY}$, that we denote\footnote{Keeping the notation we used in our previous papers about dynamics.}
\[\alpha:\mathbf{C}\rightharpoondown\mathbf{PY}.\]
We say that $\alpha$ is \emph{disjunctive} if for all objects $S$ and $T$ in $\mathbf{C}$,
\[
S\neq T \Rightarrow S^\alpha \cap T^\alpha=\emptyset,
\]
where we use the notation $S^\alpha$ to denote the set $\alpha(S)$.
Likewise, for $d\in\overrightarrow{\mathbf{C}}$, we put $d^\alpha=\alpha(d)$.
\begin{df} [Closed dynamics on $\mathbf{C}$] We denote by $\mathcal{CD}_\mathbf{C}$ the category of \emph{closed dynamics} on $\mathbf{C}$, that is the category
\begin{itemize}
\item whose objects are disjunctive lax-functors $\alpha:\mathbf{C}\rightharpoondown \mathbf{PY}$,
\item whose arrows --- called \emph{dynamorphisms} --- are lax-natural transformations $\delta:\alpha\looparrowright \beta$.
\end{itemize}
\end{df}
The small category $\mathbf{C}$ is called the \emph{engine} of the dynamics $\alpha\in Ob(\mathcal{CD}_\mathbf{C})$. The arrows $(S\stackrel{d}{\rightarrow}T)\in\overrightarrow{\mathbf{C}}$ are called \emph{durations}.
\begin{rmq}
In \cite{Dugowson:20150809}, section \textbf{§\,2.1.2}, the category $\mathcal{CD}_\mathbf{C}$ was denoted $\mathbf{DySC_{(C)}}$. Indeed, by definition of lax-functors and lax-natural transformations,
\begin{itemize}
\item a closed dynamic $\alpha:\mathbf{C}\rightharpoondown \mathbf{PY}$ has to satisfy those properties for objects $S$, $T$,... and composable arrows $e$, $d$ of $\mathbf{C}$:
\subitem [disjunctivity] $S\neq T \Rightarrow S^\alpha \cap T^\alpha=\emptyset$,
\subitem [lax composition] $(e\circ d)^\alpha \subseteq e^\alpha \odot d^\alpha$,
\subitem [lax identity] $(Id_S)^\alpha \subseteq Id_{S^\alpha}$,
\item a dynamorphism $\delta:\alpha\looparrowright \beta$ is such that
\subitem [lax naturality] $\forall (S\stackrel{d}{\rightarrow}T)\in\overrightarrow{\mathbf{C}}, \delta_T\odot d^\alpha\subseteq d^\beta\odot \delta_S$.
\end{itemize}
\end{rmq}
A closed dynamic $\alpha:\mathbf{C}\rightharpoondown \mathbf{PY}$ is said to be \emph{deterministic (resp. quasi-deterministic)} if for every $(S\stackrel{d}{\rightarrow}T)\in\overrightarrow{\mathbf{C}}$ the transition $d^\alpha$ is deterministic (resp. quasi-deterministic).
Likewise, a dynamorphism $\delta$ is said to be \emph{(quasi-)deterministic} if all transitions $\delta_S$ are (quasi-)deterministic.
\begin{df}
A deterministic closed dynamic is called a \emph{clock}.
\end{df}
\subsection{The category $\mathcal{CD}$ of closed dynamics}
\begin{df}
We define the category \emph{$\mathcal{CD}$ of all closed dynamics} taking as \emph{dynamorphisms} from $\alpha:\mathbf{C}\rightharpoondown \mathbf{PY}$ to $\beta:\mathbf{D}\rightharpoondown \mathbf{PY}$ couples $(\Delta,\delta)$ with $\Delta:\mathbf{C}\rightarrow\mathbf{D}$ a functor, and $\delta:\alpha \looparrowright (\beta\circ \Delta)$ a lax-natural transformation.
\end{df}
\begin{rmq}
When $\mathbf{C}=\mathbf{D}$ we assume by default that $\Delta=Id_\mathbf{C}$. In other words, $\mathcal{CD}_\mathbf{C}$ is not a full subcategory of $\mathcal{CD}$.
\end{rmq}
\subsection{The category $\mathcal{MD}_{(\mathbf{C}, L)}$ of $L$ multi-dynamics on $\mathbf{C}$}
Let $L$ be a non-empty set.
\begin{df}[$L$-multi-dynamics on $\mathbf{C}$] We denote by $\mathcal{MD}_{(\mathbf{C}, L)}$ the category
\begin{itemize}
\item with objects, called \emph{$L$-multi-dynamics on $\mathbf{C}$}, the disjunctive
\footnote{
As in the closed dynamics case, a functor $\alpha:\mathbf{C}\rightharpoondown \mathbf{PY}^{\underrightarrow{\scriptstyle{L}}}$ is said to be disjunctive if for all objects $S$ and $T$ in $\mathbf{C}$,
$
S\neq T \Rightarrow S^\alpha \cap T^\alpha=\emptyset.
$
}
lax functors
\[
\alpha:\mathbf{C}\rightharpoondown
\mathbf{PY}^{\underrightarrow{\scriptstyle{L}}},
\]
\item with arrows $\delta:\alpha\looparrowright\beta$ --- called \emph{$({\mathbf{C}, L})$-dynamorphisms} --- given by\footnote{About the intersection used here, see below the remark \ref{rmq natural intersection lambda}.}
\[
\mathcal{MD}_{(\mathbf{C}, L)}(\alpha,\beta)
=
\bigcap_{\lambda\in L}
\mathcal{CD}_\mathbf{C}(\alpha_\lambda,\beta_\lambda),
\]
where for each $\lambda\in L$ we denote $\alpha_\lambda$ the closed dynamic associated in an obvious way with the multi-dynamic $\alpha$ for this $\lambda$.
\end{itemize}
\end{df}
Thus, by definition, and with the notation introduced in section §\ref{subsection PYL}, an $L$-multi-dynamic associates with each duration $(S\stackrel{d}{\rightarrow}T)\in\overrightarrow{\mathbf{C}}$ an $L$-family $S^\alpha {\stackrel{d^\alpha}{\twosquig}}_L T^\alpha$.\\
\begin{rmq}\label{rmq natural intersection lambda} [About the expression $\bigcap_{\lambda\in L}
\mathcal{CD}_\mathbf{C}(\alpha_\lambda,\beta_\lambda)$]
If $\mathbf{C}$ and $\mathbf{E}$ are categories, and $\alpha$, $\alpha'$, $\beta$ and $\beta'$ are functors $\mathbf{C}\rightarrow\mathbf{E}$ such that for all objects $C\in\dot{\mathbf{C}}$ we have $\alpha(C)=\alpha'(C)$ and $\beta(C)=\beta'(C)$, then if $(\alpha,\beta)\neq (\alpha',\beta')$, the sets of natural transformations $\mathit{Nat}(\alpha,\beta)$ and $\mathit{Nat}(\alpha',\beta')$ are disjoint
\[
\mathit{Nat}(\alpha,\beta)\cap \mathit{Nat}(\alpha',\beta')=\emptyset
\]
because a natural transformation $\delta\in \mathit{Nat}(\alpha,\beta)$ has a unique domain $\alpha$ and a unique codomain $\beta$. Nevertheless, forgetting this information about domains and codomains and identifying $\delta$ just with a family of arrows $\left(\alpha(C)\stackrel{\delta_C}{\rightarrow}\beta(C)\right)_{C\in \dot{\mathbf{C}}}$ belonging to $\overrightarrow{\mathbf{E}}$, we'll write $\delta\in \mathit{Nat}(\alpha,\beta)$ to say only that $\delta$ is natural in respect with $\alpha$ and $\beta$, that is
\[
\forall(S\stackrel{d}{\rightarrow}T)\in\overrightarrow{\mathbf{C}}, \delta_T\circ \alpha(d)=\beta(d)\circ\delta_S.
\]
Then, $\mathit{Nat}(\alpha,\beta)\cap \mathit{Nat}(\alpha',\beta')$ will designate the set of families of $\mathbf{E}$-arrows $\delta=((\delta_C)_{C\in \dot{\mathbf{C}}})$ which are natural in respect together with $(\alpha,\beta)$ and with $(\alpha',\beta')$. In general, this kind of intersection will be non-empty.
The same remark can be made about lax transformations between lax functors between bicategories, and it is in this sense that we use expressions like $\bigcap_{\lambda\in L}
\mathcal{CD}_\mathbf{C}(\alpha_\lambda,\beta_\lambda)$. To emphasize the forgetfulness of information about domains and codomains, we could use some notation like $\underline{\mathcal{CD}_\mathbf{C}}$ but it would be a little heavy, so we prefer not to do so.
\end{rmq}
\begin{rmq}
Unsurprisingly, an $L$-multi-dynamic $\alpha:\mathbf{C}\rightharpoondown \mathbf{PY}^{\underrightarrow{\scriptstyle{L}}}$ is said to be \emph{{(quasi-)}}\,\emph{deterministic} if for every $\lambda\in L$ the closed dynamic $\alpha_\lambda$ is {(quasi-)}deterministic.
\end{rmq}
\subsection{The category $\mathcal{MD}_{\mathbf{C}}$ of multi-dynamics on $\mathbf{C}$}
Let $\alpha:\mathbf{C}\rightharpoondown
\mathbf{PY}^{\underrightarrow{\scriptstyle{L}}}$
and
$\beta:\mathbf{C}\rightharpoondown
\mathbf{PY}^{\underrightarrow{\scriptstyle{M}}}$
be multi-dynamics on $\mathbf{C}$.
We define a \emph{$\mathbf{C}$-dynamorphism} $\delta:\alpha\looparrowright\beta$
as a couple $(\theta,\underline{\delta})$
with $\theta:L\rightarrow M$ a map,
and $\underline{\delta}\in\bigcap_{\lambda\in L}
\mathcal{CD}_\mathbf{C}(\alpha_\lambda,\beta_{\theta(\lambda)})$. The lax natural part $\underline{\delta}$ of such a dynamorphism $\delta$ is often itself simply denoted $\delta$. Thus, to be a dynamorphism, $(\theta,\delta)$ must satisfy the lax naturality condition
\[\forall \lambda\in L,\forall (S\stackrel{d}{\rightarrow}T)\in\overrightarrow{\mathbf{C}}, \delta_T\odot d^\alpha_\lambda\subseteq d^\beta_{\theta(\lambda)}\odot \delta_S.\]
\begin{df}
We define the category $\mathcal{MD}_{\mathbf{C}}$ taking as objects all multi-dynamics on $\mathbf{C}$, and as arrows all $\mathbf{C}$-dynamorphisms between them.
\end{df}
Naturally, a dynamorphism $\delta$ is said to be \emph{(quasi-)deterministic} if for every object $S\in\dot{\mathbf{C}}$, $\delta_S$ is (quasi-)deterministic.
\begin{rmq}
Closed dynamics on $\mathbf{C}$ can be seen as a full subcategory of $\mathcal{MD}_{\mathbf{C}}$, and we can in particular consider dynamorphisms between closed dynamics on $\mathbf{C}$ and multi-dynamics on $\mathbf{C}$.
\end{rmq}
\subsection{The category $\mathcal{MD}$ of multi-dynamics}
Let $\alpha:\mathbf{C}\rightharpoondown
\mathbf{PY}^{\underrightarrow{\scriptstyle{L}}}$
and
$\beta:\mathbf{D}\rightharpoondown
\mathbf{PY}^{\underrightarrow{\scriptstyle{M}}}$
be multi-dynamics with possibly different engines.
We define a \emph{dynamorphism} $\alpha\looparrowright\beta$
as a triple $(\theta,\Delta,\delta)$
with
\begin{itemize}
\item $\theta:L\rightarrow M$ a map,
\item $\Delta:\mathbf{C}\rightarrow\mathbf{D}$ a functor,
\item $\delta\in\bigcap_{\lambda\in L}\mathcal{CD}_\mathbf{C}(\alpha_\lambda,\beta_{\theta(\lambda)}\circ\Delta)$.
\end{itemize}
The last condition is equivalent to
\[
(\Delta,\delta)\in\bigcap_{\lambda\in L}\mathcal{CD}(\alpha_\lambda,\beta_{\theta(\lambda)}),
\]
\emph{i.e.} the lax naturality condition
\[\forall \lambda\in L,\forall (S\stackrel{d}{\rightarrow}T)\in\overrightarrow{\mathbf{C}}, \delta_T\odot d^\alpha_\lambda\subseteq (\Delta d)^\beta_{\theta(\lambda)}\odot \delta_S\]
has to be satisfied.
\begin{df}
The category $\mathcal{MD}$ of multi-dynamics is defined taking as objects all multi-dynamics, and as arrows all dynamorphisms between them.
\end{df}
\section{The category $\mathcal{OD}$ of open dynamics}
\subsection{Definition}
\begin{df}\label{df open dynamics} An \emph{open dynamic} $A$ with engine $\mathbf{C}$ is the data
\[A=\left((\alpha:\mathbf{C}\rightharpoondown \mathbf{PY}^{\underrightarrow{\scriptstyle{L}}}) \stackrel{\rho}{\looparrowright} (\mathbf{h}:\mathbf{C}\rightarrow \mathbf{Sets})\right)\] of
\begin{itemize}
\item a non-empty set $L$,
\item a multi-dynamic $\alpha\in \mathcal{MD}_{(\mathbf{C}, L)}\subset \mathcal{MD}_{\mathbf{C}}$,
\item a clock $\mathbf{h}\in \mathcal{CD}_{\mathbf{C}}\subset \mathcal{MD}_{\mathbf{C}}$,
\item a deterministic dynamorphism
$
{\rho}\in \mathcal{MD}_{\mathbf{C}}(\alpha,\mathbf{h})
$ called \emph{datation}.
\end{itemize}
\end{df}
According to the definitions given in \textbf{§\,2.4.2} of \cite{Dugowson:20150809} and \textbf{§\,1.2.2} of \cite{Dugowson:20160831}, a \emph{dynamorphism} from an open dynamic
\[A=\left((\alpha:\mathbf{C}\rightharpoondown \mathbf{PY}^{\underrightarrow{\scriptstyle{L}}}) \stackrel{\rho}{\looparrowright} (\mathbf{h}:\mathbf{C}\rightarrow \mathbf{Sets})\right)\]
to an open dynamic
\[B=\left((\beta:\mathbf{D}\rightharpoondown \mathbf{PY}^{\underrightarrow{\scriptstyle{M}}}) \stackrel{\tau}{\looparrowright} (\mathbf{k}:\mathbf{D}\rightarrow \mathbf{Sets})\right)\]
is a quadruplet $(\theta, \Delta, \delta, \varepsilon)$ with
\begin{itemize}
\item $(\theta,\Delta,\delta)\in\mathcal{MD}(\alpha,\beta)$,
\item $(\Delta,\varepsilon)\in\mathcal{CD}(\mathbf{h},\mathbf{k})$,
\item this lax synchronization condition satisfied:
\[
\forall S\in\dot{\mathbf{C}},
\tau_{\Delta_S}\odot\delta_S\subseteq\varepsilon_S\odot\rho_S.
\]
\end{itemize}
We'll denote by $\mathcal{OD}$ the category of all open dynamics, with dynamorphisms as arrows.
\subsection{Realizations of an open dynamic}
With the definitions given above, the other notions of our systemic theory of interactivity keep the same definitions as previously given in \cite{Dugowson:20160831}: parametric quotients, interactions, connectivity structures of an interaction, dynamical families, open dynamics generated by a dynamical family, ...
In particular, given an open dynamic $A=\left((\alpha:\mathbf{C}\rightharpoondown \mathbf{PY}^{\underrightarrow{\scriptstyle{L}}}) \stackrel{\rho}{\looparrowright} (\mathbf{h}:\mathbf{C}\rightarrow \mathbf{Sets})\right)$ we have
\begin{df}
A \emph{realization} (or a \emph{solution}) of $A$ is a quasi-deterministic dynamorphism $(\mathfrak{s}:\mathbf{h}\looparrowright\alpha)\in \mathcal{MD}_\mathbf{C}(\mathbf{h},\alpha)$ satisfying the lax condition $\rho\odot \mathfrak{s}\subseteq Id_\mathbf{h}$.
\end{df}
In other words\footnote{See \cite{Dugowson:20160831}, \textbf{§\,1.3.1.}}, such a realization is a couple $\mathfrak{s}=(\lambda,\sigma)$ with $\lambda\in L$ and $\sigma\in \mathcal{CD}_\mathbf{C}(\mathbf{h},\alpha_\lambda)$ which is a partial function $\sigma:st(\mathbf{h})\dashrightarrow st(\alpha)$ defined on a subset $D_\sigma\subset st(\mathbf{h})$ and satisfying these properties:
\[\forall t\in D_\sigma, \rho(\sigma(t))=t,\]
\[\forall S\in\dot{\mathbf{C}},\forall t\in S^\mathbf{h}\cap D_\sigma, \sigma(t)\in S^\alpha,\]
\[\forall (S\stackrel{d}{\rightarrow}T)\in\overrightarrow{\mathbf{C}},
\forall t\in S^\mathbf{h},
\left( d^\mathbf{h}(t)\in D_\sigma\Rightarrow
\left( t\in D_\sigma
\,\mathrm{and}\,
\sigma(d^\mathbf{h}(t))\in d^\alpha_\lambda (\sigma(t))\right)\right).\]
\vspace{0.7cm}
\begin{center}
***
\end{center}
\vspace{0.5cm}
\textbf{Thanks} to Andrée Ehresmann and Mathieu Anel for having encouraged me to reword the previously developed notion of ``sub-functorial dynamic'' in terms of lax-functors.
\bibliographystyle{plain}
| 58,458
|
TITLE: How to calculate the exact differential structure of Brieskorn variety?
QUESTION [16 upvotes]: As Kervaire and Milnor mentioned, an $n$-dim exotic sphere $\Sigma$ which bounds a parallelizable manifold $M$ is totally classified by the signature $\sigma(M)$ modulo the order of $bP_{n+1}$.
Let $n=2m$ be an even integer. Brieskorn had discovered that the singularity of complex hypersurfaces $V_k$, the zero set of $x_0^2+\cdots+x_{n-2}^2+x_{n-1}^3+x_n^{6k-1}=0$ has close relationship with exotic spheres. More precisely, if $\varepsilon>0$ is sufficiently small, let $S_\varepsilon$ be the $(2n+1)$-sphere with center $0$ and radius $\varepsilon$, then $\Sigma_k=S_\varepsilon\cap V_k$ is an exotic sphere, and actually every exotic sphere of dimension $(4m-1)$ which bounds a parallelizable manifold can be obtained in this way.
I want to know which element $\Sigma_k$ represents in $bP_{2n}$. In other words, I want to calculate the signature of the Milnor fibre. Since Brieskorn's original paper was written in German, I couldn't read it. Instead, I've read the papar 'Singularity and Exotic Sphere' written by Hirzebruch. In this paper, Hirzebruch gave the answer: actually $\Sigma_k$ represents $k$th multiple of the generator of $bP_{2n}$. However, he refered the proof to Brieskorn's paper.
Does anyone know a proof? Please tell me, thanks.
REPLY [3 votes]: An english discussion of Brieskorn exotic spheres can be found in these slides of Ranicki, there is also a collection of links concerning exotic spheres here.
I can give you the gist of Brieskorn's argument, I hope this may help:
First, setting $a=(a_1,\dots,a_n)$, an explicit parallelizable manifold is given by $M_a=\Xi_a\cap D^{2n}$, $\Xi_a=\{(z_1,\dots,z_n)\in\mathbb{C}^n\mid \sum_{i=1}^n z_i^{a_i}=1\}$ and as usual $D^{2n}=\{(z_1,\dots,z_n)\in\mathbb{C}^n\mid \sum_{i=1}^n|z_i|^2\leq 1\}$. The intersection form for $\Xi_a$ is computed in F. Pham: Formules de Picard-Lefschetz généralisées et ramification des integrales. BSMF 93 (1965), 333-367.
Then the steps of the proof are as follows:
1) there are automorphisms $\omega_k$ given by multiplying the $k$-th coordinate with $e^{2\pi i/a_k}$. these generate a finite automorphism group $\Omega_a\cong\prod_{i=1}^n\mathbb{Z}/(a_i)$, let $J_a=\mathbb{Z}[\Omega_a]$ be the group ring and $I_a$ the ideal generated by the elements $1+\omega_k+\cdots+ \omega_k^{a_k-1}$. Pham identified $H_{n-1}(\Xi_a,\mathbb{Z})\cong J_a/I_a$.
2) $H_{n-1}(\Xi_a,\mathbb{C})$ has a basis $v_j=\prod_{k=1}^n\sum_{r=0}^{a_k-1}e^{2\pi i j_k r/a_k}\omega_k^r$. In the basis $v_j+v_{a-j}$ and $i(v_j-v_{a-j})$ of $J_a/I_a\otimes\mathbb{R}$, the intersection form is diagonalized. The matrix of the intersection form can be found on p. 359 of Pham's paper, and implies $\langle v_j+v_{a-j},v_j+v_{a-j}\rangle=\langle i(v_j-v_{a-j}),i(v_j-v_{a-j})\rangle=2\langle v_j,v_{a-j}\rangle$ (and $0$ otherwise).
3) The final computation shows that $\langle v_j,v_{a-j}\rangle>0$ if and only if $0<\sum \frac{j_k}{a_k}<1\mod 2$, and $\langle v_j,v_{a-j}\rangle<0$ if and only if $-1<\sum\frac{j_k}{a_k}<0\mod 2$.
This implies the signature calculation, Satz 3 in Brieskorn's paper resp. the Theorem on p. 20 of Hirzebruch's paper "Singularities and exotic spheres". Actually, Brieskorn credits Hirzebruch for the result and proof.
| 137,521
|
Whyteleafe may be in the hot seat for promotion out of the Southern Counties East League, but boss John Fowler is too wily to get carried away.
Saturday's 7-2 win over Fisher - the third time Leafe have hit seven goals this season - has opened a six-point gap at the top of the table.
Nearest challengers Ashford United were held to a 2-2 draw at Fisher on Monday night, and they go to Corinthian on Wednesday.
Leafe face a trip to Greenwich Borough tonight, before they go to Ashford United for what Fowler calls a "key game".
He said: "We started very well against Fisher, and after 15 minutes we were 2-0 up and I thought it was going to be a massive score.
"But credit to them coming back to 2-2, although our third goal just before half-time probably took the stuffing out of them.
"I needed to have a good word with the lads at half-time, just to remind them to stick to what we had done in training."
He added: "When we play Ashford, they will have had two games since last weekend, whereas we will have had just one.
"It's a key game because we could be nine points clear at the end and that may be too much for Ashford to make up.
"But football is funny old game, and I have been around too long to know that things can change very quickly."
Fowler also paid homage to the Leafe faithful who have been starved of success for many years following relegation from Ryman Division One South two years ago.
He said: “The supporters have been superb all season, and they have had a great impact on the team.
"They always talk with the players and there's a good buzz around the place.”
| 345,395
|
TITLE: The compositum is also separable
QUESTION [0 upvotes]: If $K$ and $K'$ are Galois extensions of the field $k$ then the compositum $KK'$ is also Galois over $k$
The normality is preserved but how to show the separability, what if one of the fields $K$ or $K'$ is only an arbitrary extension, not necessarily Galois
REPLY [0 votes]: First of all, I'm gonna use the definion of galoisian extension to be a normal and separable extension. Now, by hyphotesis $ K$ and $ K'$ are galoisian over $ k $, so both fields are splitting fields of some polynomials $ p(x), q(x)\in k[x] $, respectively. If $a_{1}, \ldots, a_{n}$ and $b_{1}, \ldots, b_{m}$ are the roots of $ p $ and $ q$, resp., then $$KK'=K'(K)=K'(k(a_{1}, \ldots, a_{n}))=K'(a_{1},\ldots, a_{n})=k(b_{1}, \ldots, b_{m})(a_{1}, \ldots, a_{n})=k(b_{1}, \ldots, b_{m}, a_{1}, \ldots, a_{n}). $$ Therefore $ KK'$ is the splitting field of $ p(x)q(x)\in k[x]$, i.e., $ KK'/k $ is normal. Now, because $ K/k $ and $ K'/k $ are both separable we have that $ a_{i}, b_{j} $ are separable over $ k$. Thus, it follows that $ KK'/k $ is separable. Hence $ KK'$ is galoisian over $ k $.
| 136,236
|
TITLE: Reviewing for exam: Chernoff bounds for a series of random variables
QUESTION [2 upvotes]: I have a series of random variables, where the expected value is $\frac{n}{4}$. I want to prove, with Chernoff bounds, that the probability that the actual value is less than $(1 - \epsilon)\frac{n}{8}$ is very small.
I am unsure as to how to approach this problem. It is immediately obvious that n/8 is half of n/4. I don't believe solving for epsilon is the right approach for this, however. How do I approach this?
REPLY [1 votes]: The Chernoff bound you need is given by
$$ Pr(X < (1-\delta)Ex[X]) \le \left( \frac{e^\delta}{(1-\delta)^{1-\delta}} \right)^{Ex[X]}. $$
In your case, $Ex[X] = n/4$. Therefore,
$$ Pr(X < (1-\epsilon)n/8) \le Pr(X < n/8) \le Pr(X < (1/2)n/4) = Pr(X < (1 - 1/2)n/4), $$
so $\delta = 1/2$.
This leads to
$$ Pr(X < (1-\epsilon)n/8) \le \left( \frac{e^{1/2}}{(1/2)^{1/2}} \right)^{n/4}. $$
The expression on the right is easily reduced by algebraic manipulation. After this, you need to find an $n$ such that this expression is polynomially small.
This shows some of the power of Chernoff bounds, that is, that once you know $Ex[X]$, you can pick $\delta$ easily, and then the bound itself is given by a simple closed expression raised to $Ex[X]$. Let me repeat: all you need to know (if you have 0/1 independent Poisson Trials), is the expected value!
If you would like help with the algebraic manipulation, please comment as such.
NB + spoiler: Another formulation to memorize (cf. deathbed formulae :-) of the lower Chernoff bound is
$$ Pr(X < (1-\delta)Ex[X]) \le e^{-Ex[X]\delta^2/2}. $$
In your case, this would become
$$ Pr(X < (1-\epsilon)n/8) \le e^{-(n/4)(1/2)^2/2} = e^{-n/32}. $$
Solving for $n$ such that this expression is less than $N^{-c}$ gives $n \ge 64\log N$ (to get the bound with high probability with respect to $N$), as this equation shows:
$$ e^{-64\log(N)/32} = \left( e^{\log N} \right)^{-2} = N^{-2}. $$
Thus, with at least $n = \Theta(\log N)$ flips (or whatever your variables denote) means that $X \ge (1-\epsilon)n/8$ with high probability with respect to $N$.
| 155,489
|
TITLE: How to prove that $f(x,y ) = - x^{\alpha}y ^{1-\alpha}$ is convex when $0 < \alpha < 1$?
QUESTION [1 upvotes]: I want to prove that $f(x,y ) = - x^{\alpha}y ^{1-\alpha}$ is convex when $0 < \alpha < 1$ on $\{ (x, y) \in \mathbb R^2: x\geq 0, y\geq 0\}$. However, the exponent $0< \alpha < 1$ is causing me a lot of trouble. I don't get anything with $f(tu + (1-t)v) \leq t f(u) + (1-t)t(v)$. Any other method involving $\nabla f$ or $\nabla^2 f$ doesn't seem useful either because of the discontinuities of the derivatives of $f$ at $x = 0$ or $y = 0$.
REPLY [0 votes]: Hint:
From the Wikipedia page on convex functions:
A twice differentiable function of several variables is convex on a convex set if and only if its Hessian matrix of second partial derivatives is positive semidefinite on the interior of the convex set.
Using Mathematica I found the Hessian for your function to be:
$$
\mathbf H=
\left(
\begin{array}{cc}
(\alpha -1) \alpha \left(-x^{\alpha -2}\right) y^{1-\alpha } & (\alpha -1) \alpha
x^{\alpha -1} y^{-\alpha } \\
(\alpha -1) \alpha x^{\alpha -1} y^{-\alpha } & (\alpha -1) \alpha \left(-x^{\alpha
}\right) y^{-\alpha -1} \\
\end{array}
\right).
$$
Can you take it from here?
| 3,546
|
TITLE: The set of subfields is directed
QUESTION [2 upvotes]: Let $K/k$ extension of field and $F$ the set of subfeilds. Suppose that for every $M,N \in F$ we have $M \subset N$ or $N \subset M$. Show that: K is simple extension i.e $K=k(x)$ for some $x \in K$.
My idea: is to considere $K$ is Algebraic extension of $k(S)$ (transendance basis). And we have $card(S) \leq 1$ because for every $x,y$ transcendance we have $k(x) \subset k(y)$ or $k(y) \subset k(x)$. So, if you start like me, i want to prouve K/k(x) is finite extension. Can you help me pleas ?
REPLY [1 votes]: Let $k=\Bbb F_p$, the finite field of $p$ elements, and let $K_n=\Bbb F_{p^{2^n}}$,
the finite field of order $p^{2^n}$. The $K_n$ form a chain: $k=K_0\subseteq K_1
\subseteq K_2\subseteq\cdots$. Let $K=\bigcup_{n=1}^\infty K_n$. The proper
subfields of $K$ are the $K_n$. These are totally ordered under inclusion.
But $K/k$ is not a simple extension. A simple algebraic extension is finite,
but $K/k$ is infinite algebraic.
| 174,407
|
Hey there,
The Hearsemen are finally finishing their full length debut album and on Friday, August 22 we're throwing a bash at The Zoo (Osborne Village Motor Inn, 160 Osborne) in order to raise cash and get the tunes out to you.
As a small token of appreciation for your help in making a killer show and a killer album, all advance tickets ($7) sold to this event come with a coupon for $5 off the disk, redeemable when it is released in the fall. Get tickets at the front desk at the Osborne Village Motor Inn (Ph: 452-9824; Ad:160 Osborne St.;) or through Reed (793-0050) or Paul (803-0720, PondScumRecords@gmail.com).
The lineup goes as follows:
IGOR AND THE SKINDIGGERs (10pm)
THE DOWNFALL (11pm)
ASADO (12pm)
THE HEARSEMEN (1PM)
Brought to you by POND SCUM RECORDS
Please pass this message on to anyone you think might be interested by it or might be able to help spread the word. Thanks!
Links for you to check out:
Event Page (facebook):
The Hearsemen Myspace Page - Come check out some previously recorded Hearsemen tunes
'The Hearsemen' Page (facebook)
'I love/hate The Hearsemen' Group (facebook)
Thank you all for your unbending worship." -Yentle
--
Pond Scum Records
(204) 803-0720
PondScumRecords@gmail.com
Hello everyone,
I’m proud to announce that I just finished up the tracks for my “Mellow Hello” EP, which includes some of my acoustic and piano works. The songs are very personal and influenced by some of my favorite songwriters including the Bee Gees, Hall and Oates, Aaron Neville and even a little Burton Cummings there too. Originally I set aside these tracks for publishing (film/tv/etc) but feedback at the last few shows (thank you!) was so positive that I decided to share them with everyone.
You can listen to three select tracks - “Feeding in Eden ”, “When I Needed You” and my latest acoustic ballad “The Postage Stamp” – at my CBC Webspace here…
Enjoy the music and thanks again for your continued support (especially those of you who make it out to nearly every show!) Keep checking my website or MySpace for upcoming show information and I hope to see you out.
Deon Wysocki
TWELVE34 News
It has become easier and easier to give up on rock music, because of an industry plagued with manufactured artists, manufactured hits. The bland corporate machine has been churning out hits from the same mold for so many years. We are in dire need of a salvation - and twelve34 is here to do just that. If you are a true believer in the power of music, then let it save you - and enjoy this free MP3 courtesy of us! We promise it will re-ignite that long lost spark you once had. But don't take our word for it... here's what the music folks are saying:
"Twelve34's The Only Cure has found a long lost remedy for the dead eared modern music fan. From the first blast of cymbals and raw guitar, the hammer, anvil, and stirrup buffet the ear drum to awaken a once thought misplaced treasure, true rock and roll. Vocals, musicianship and talent are all showcased in this well produced and balanced CD. It's a journey that you simply accept and enjoy. Dead ear syndrome? I now know what the only cure is." -Toronto Exclusive Magazine
"A band out of Ottawa has been churning out some catchy, crunchy, melodic rock for the past few years and has just released a new album. Twelve34 have their feet firmly planted in straight-ahead, arena-tinged rock. The Only Cure is the band's first full-length effort, a tightly-packed collection of rock with a serious melodic flair." -Blog Critic Magazine
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Tustin, California Alarm Company Reviews
Tustin is no stranger to crime and demand for home security systems and security monitoring is fairly high. A variety of local Tustin, CA companies offer security systems, installation and monitoring. Some Tustin homeowners decide to install their own alarms while others require help from local contractors. On this website, you will find listings and reviews of local Tustin companies who can help supply, install and monitor your home security.
- 24/7 home security monitoring in Tustin
- Wireless, DIY home security kits
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Tustin, California Crime Statistics
The following Tustin, CA crime stats are provided by an FBI survey completely in 2014 (not available in all cities):
- Tustin population: 71814
- Violent crime reported in Tustin: N/a
- Robberies: N/a
- City burglaries: N/a
- Larceny: N/a
- Automobile thefts in Tustin: N/a
Who are the best home security companies locally? Learn more about local Tustin security companies below.
Tustin, California Home Security Company Reviews
3041 Edinger Ave Tustin California 92780
R D Systems Reviews Tustin Home Security and Alarm Company R D Systems is a home security compan...
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Guest Post
by Shaun J. McLaughlin
One of the challenges facing authors of historical fiction is to be true to the era in which you set your story. You must do more than engage and entertain your readers: you must not misinform.
If your story is set in the eighteenth century, your protagonist should not be wearing a wristwatch. That’s obvious. What is not always obvious are the verbs and nouns we choose for narration and conversation. Make sure those words were part of the common vernacular in your story’s era.
Sound difficult? Well, thanks to the benevolent geeks at Google, there is an app for that. It’s called Google Books Ngram Viewer. In 2009 and 2012, Google created databases of words in all its digitized books, and organized those words by the year of publication.
You simply enter the case-sensitive word or phrase to search for, set a year-range, select the corpus and click the search button. Ngram Viewer returns a graph of the usage as a percentage of words in the selected corpus.
Here are a few examples of how Ngram Viewer improved the historical accuracy of my novels.
In my 2012 novel Counter Currents, set in 1838 North America, I wanted to serve the protagonist a lunch of shepherd’s pie. So, I entered the food name, set the time span from 1825 to 1900, and picked the American English corpus. (I used a 75-year range to learn when the term came into use if outside of my story’s era.)
The result shows that shepherd’s pie did not appear in books until about 55 years later, and even then, rarely. (Before then, the dish went by the name cottage pie. How I know that is a different story.) This technique is handy for verifying food, clothing, people’s names, and manufactured items.
Ngram Viewer can also help with regional dialect. In my current sequel to Counter Currents, set in 1845-1855 Australia, I want to include some Aussie slang. I entered “billabong” (a term for a type of creek) in Ngram Viewer. No luck. That word doesn’t appear until after 1880.
The Ngram Viewer also lets you use the best word if several are available. In my sequel, my protagonist agrees to get married. Which is the best word: “engaged” or “betrothed”?
To graph more than one search term, separate each with a comma.
This graph shows that both words are valid but books used “engaged” more often in that era. I used the British English corpus as the most likely to reflect nineteenth century Australian English.
In English, you can use just about any noun as a verb. Before you do, make sure the spoken language in your historical era used that verb form?
Ngram Viewer can graph the same word used as different parts of speech. You just add one of the many available tags to your search terms. (Click the About Ngram Viewer link at the bottom of the Ngram page for details on the many tags and other advanced features.)
In this example, I add the tags _NOUN and _VERB to the word “impact.”
This graph shows that books did not include “impact” as a verb until the 1870s, and very rarely compared to the noun form.
The corpus list provides several choices for English books—all English books, British English for books printed in the UK, American English for books published in the US, and just English fiction books—plus seven other languages. (Use the 2012 versions without the dates, not the 2009 versions: the OCR technology used in 2012 created fewer spelling errors. )
Pick the English corpus that suits your geographical location to help refine your word selection. For example, Ngram Viewer shows that American books used the word “boss” as a noun sporadically in the eighteenth century, but English books used it throughout that period. (I entered “boss_NOUN” in two separate searches using a different corpus.)
Am I suggesting that you run every verb and noun through Ngram Viewer? Of course not. If a word sounds like slang, or your inner doubts nag you (trust your intuition) or you are faced with several synonyms, use it. It takes just seconds.
Shaun J. McLaughlin has authored books on history and historical fiction, using both a traditional publisher and self-publishing. A researcher, journalist and technical writer for over thirty years, he lives on a hobby farm in Eastern Ontario, where he focuses on fiction and nonfiction writing projects. For more information, see Raiders and Rebels Press. or visit Shaun’s Author Central page.
16 thoughts on “Historical Fiction: Get the Correct Words for the Era”
Thank you. I wish I had known about this when I began my trilogy. While it is not set in an actual society, it is an ‘old world’ place. I have done my best to give my language, particularly dialogue, that flavour. There were times when I wanted to check a word and could not find it’s history.
Yvonne,
Dictionary.com usually has the history of any word you look up near the bottom of the page. I quite often use it.
Thank you. I have tried it on occasion, but some words could not be found there.
Fantastic resource, Shaun! Will bookmark and send links to my historical and time-travel writer friends.
What a great resource. Thanks, Shaun!
Thank you, Shaun! I can see this coming in handy.
Excellent tool…thanks!
Before the Internet, we used the old standby Webster’s Ninth New Collegiate Dictionary which gives the year it came into existence and very handy when writing historical fiction. I still have my 1987 publication and kept it close because I didn’t know about this new way and didn’t think to use the Internet to look these days. Great post, thank you.
Before the Internet, I used to spend hours in a library and carried home piles of photocopies. I can do that all from home, save time and gasoline, and stay paperless. We authors are blessed with conveniences.
Isn’t the internet just bloody fantastic!!! There’s really no excuse for writers to dally with research taking years to complete a book when it’s all there at our fingertips now. (Damn, must get off my backside and publish the WIP – enough is enough!)
This is an excellent resource and I’ll rush to tell all my writing groups immediately.
Thank you, Shaun, this is definitely a resource to bookmark. I have always, definitely gone out of my way not to use current vernacular when writing historical fiction; however, I usually make do with writing full and proper English (no isn’t or wasn’t et cetera) and with the odd vintage term thrown in for effect. Because, although I like the general facts and feel to be authentic, I am writing for todays audience and I don’t want to bog the story down and alienate those readers who would never dream of (as I actually do) having a dictionary, thesaurus and other reading aids to hand.
Excellent, article, Shaun.
Contractions do appear in older writing. To search them in Ngram Viewer, you must enter part of the word without the apostrophe. So to search for isn’t, you use isn.
That’s helpful. Thanks
Now if that ain’t slicker than snail snot, I don’t know what is! Great article, and I’ll check that out since I’m writing an historical fiction screenplay at the moment. I have book resources, but sometimes I have to create my own dialog, so this will be a big help.
I am glad everyone found my article useful and “slicker than snail snot.” IU is my favorite source for writing tutorials and tips, and I am happy to contribute.
I loved that phrase, too. It was the biggest laugh of my day.
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At this juncture, almost everyone who's interested in self-publishing knows about BookBub. They're generally acknowledged as the the 800-pound gorilla of book advertising (especially among indies), and getting an ad with them is as close as you can get to guaranteed sales. Needless to say, everyone wants a BookBub promotion, so it's probably surprising that I resisted the urge to submit a book to them until very recently.
To be perfectly honest, "resisted" is probably the wrong word. Truth be told, from everything I'd heard, BookBub was a powerful tool if you could get accepted (everyone agrees that they're very selective), and I was looking to get the most bang for my buck. Thus, rather than try to get an ad with them as soon as humanly possible, I waited until I released the third book in my Kid Sensation series and then tried to get an ad. Apparently the stars were aligned just right, because I was approved for an ad for Sensation, the first Kid Sensation book. Per BookBub's terms, I dropped the price of the book from $3.99 to 99 cents (it has to be at least 50% off). At the time of the ad, Sensation was ranking at about 4,000 in the Kindle Paid Store. The results were as follows:
Amazon Best Sellers Rank: #329 Paid in Kindle Store (See Top 100 Paid in Kindle Store)
#1 in Kindle Store > Kindle eBooks > Science Fiction & Fantasy > Fantasy > Superhero
#1 in Kindle Store > Kindle eBooks > Literature & Fiction > Action & Adventure > Romance
#1 in Kindle Store > Kindle eBooks > Teen & Young Adult > Science Fiction & Fantasy > Fantasy > Coming of Age
Needless to say, I was quite pleased. More than pleased, actually. In fact, my mood was a lot like the lady here:
I shelled out $180 for the ad, which meant that I had to sell approximately 514 copies of the book in order to break even. (That's based on getting a 35% royalty for every 99-cent copy purchased.) I sold well in excess of that so it was an all-around great day. BookBub is clearly fantastic, and well worth the price of admission.
My BookBub ad ran on January 23, roughly two weeks ago. Since then - and with the reversion back to its regular price - the book has slipped back down in the rankings again. However, the power of BookBub has been proven, leaving me in awe. The only bad news is that now I'm terrified of getting rejected by them the next time I want to do a promotion. But I'll worry about that later; at the moment, I'd rather just be thankful for how well the promotion turned out and bask happily in the moment:
Congrats on your great results. Here's my experience with Bookbub. I ran a promotion for my supernatural suspense novel The Last Conquistador, selling it at 99c. The ad cost $120, and I sold 160 copies. It wasn't the thousands of sales I've seen other people talk about. The spike came the first day - 100 copies - then slowly petered out after that. I won't make my money back for the ad, but hopefully I've expanded my readership.
Thanks for sharing your experience, and I think that - while you may not have gotten the level of sales you wanted - you are certainly viewing things in the proper light: you've expanded your readership, which is difficult (if not impossible) to put a price on.
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Budgeteer Motor Inn Angola, Angola
Budgeteer Motor Inn Angola39. Enjoy free high speed internet and free parking. Pets are welcome with a pet fee and deposit.
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IM Conversation with Mah Man…
Space says (1:09 PM):
this is hilarious:
Space says (1:09 PM):
you have to read the description….. that’s the funny part
and I hope she really does get as much as people have bid on
Space says (1:11 PM):
ah the auction already ended.. well I hope she got the full amount
Jack says (1:11 PM):
reading now… this is hillarious!
Space says (1:11 PM):
it totally is…
Space says (1:11 PM):
and it seems she gained instant popularity over it
Space says (1:11 PM):
including someone who tried to copy it word for word and do the same thing
Space says (1:12 PM):
she got something like 15,000 emails and comments over it
Jack says (1:20 PM):
crazy
Jack says (1:21 PM):
I laughed at the “And who may I thank for teaching them this fun pastime? My seventh “child”, also known as my husband. ”
Jack says (1:21 PM):
That’s just crazy to have a husband that teaches bad grocery store manners…
Space says (1:21 PM):
!!! that was teh BEST PART!!!!
Space says (1:22 PM):
lesseeeeee…..
Jack says (1:22 PM):
I can’t imagine that we know anyone like that
Space says (1:22 PM):
you’ve smacked old ladies with posters, you’ve asked where the light bulbs are while standing right in front of the light bulbs
Jack says (1:23 PM):
don’t forget kicking an entire movie rack over..
Space says (1:23 PM):
oh and remember that time you were going to show nathan how you could balance a movie on the end of your finger tips, flip it and catch it all in one *smooth* motion? And ended up juggling the movie about 10 feet into the air while knocking down movie display stands and causing the biggest scene EVAH???
Space says (1:23 PM):
ha! jinx
Jack says (1:23 PM):
lol
Jack says (1:24 PM):
I’m laughing out loud… it’s a good thing the office is empty
Space says (1:24 PM):
this is so going on my blog….
::silence from Jack::
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Petrowatch.com is the most visible, credible and authoritative oil and gas news brand in India. Published uninterrupted since February 1997 our website delivers daily and fortnightly market intelligence on the Indian oil and gas industry and is staffed by investigative journalists in Delhi, Mumbai, Ahmedabad and India’s remote but prospective northeast states, each with years of reporting experience working in the mainstream Indian and foreign press
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Petrowatch.com market intelligence provides critical) signed by oil companies for acreage won under the NELP, DSF and OALP licencing rounds. Covered in equal measure is the Petroleum & Natural Gas Regulatory Board (PNGRB) set up to regulate the fast expanding City Gas Distribution (CGD) sector and delivery of Compressed Natural Gas (CNG) for vehicles and Piped Natural Gas (PNG) to homes
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\begin{document}
\title[On the noncommutative pillow, cones and lens spaces]{Smooth geometry of the noncommutative pillow, cones and lens spaces}
\author{Tomasz Brzezi\'nski}
\address{ Department of Mathematics, Swansea University,
Swansea SA2 8PP, U.K.}
\email{T.Brzezinski@swansea.ac.uk}
\author[Andrzej Sitarz]{Andrzej Sitarz ${}^\dagger$}\thanks{${}^\dagger$ Supported by NCN grant 2012/06/M/ST1/00169}
\address{Institute of Physics, Jagiellonian University,
prof.\ Stanis\l awa \L ojasiewicza 11, 30-348 Krak\'ow, Poland.\newline\indent
Institute of Mathematics of the Polish Academy of Sciences,
\'Sniadeckich 8, 00-950 Warszawa, Poland.}
\email{andrzej.sitarz@uj.edu.pl}
\subjclass[2010]{58B34; 58B32}
\keywords{Integrable differential calculus; Dirac operator; noncommutative pillow; quantum cone; quantum lens space}
\date\today
\begin{abstract}
This paper proposes a new notion of smoothness of algebras, termed {\em differential smoothness}, that combines the existence of a top form in a differential calculus over an algebra together with a strong version of the Poincar\'e duality realized as an isomorphism between complexes of differential and integral forms. The quantum two- and three-spheres, disc, plane and the noncommutative torus are all smooth in this sense. Noncommutative coordinate algebras of deformations of several examples of classical orbifolds such as the pillow orbifold, singular cones and lens spaces are also differentially smooth. Although surprising this is not fully unexpected as these algebras are known to be {\em homologically smooth}. The study of Riemannian aspects of the noncommutative pillow and Moyal deformations of cones leads to spectral triples that satisfy the orientability condition that is known to be broken for classical orbifolds.
\end{abstract}
\maketitle
\tableofcontents
\section{Introduction}
It is often the case that a deformation of a singular variety or an orbifold produces a noncommutative object that behaves as if it were a smooth manifold. This observation underlies the theory of noncommutative resolutions \cite{Van:cre}. Recent papers \cite{Brz:smo} and \cite{Brz:con} contain illustrations of this phenomenon on the algebraic level on a number of (families of) explicit examples such as quantum teardrops \cite{BrzFai:tea}, quantum (classically singular) lens spaces \cite{HonSzy:len}, the noncommutative pillow \cite{BraEll:sph} (see also \cite{Eva:orb} and \cite[Section~3.7]{EvaKaw:sym}) and quantum cones. In \cite{Brz:smo} and \cite{Brz:con} the smoothness is understood in the homological sense, i.e.\hspace{3pt}as the existence of a finite-length resolution of algebras by finitely generated projective bimodules; see \cite[Erratum]{Van:rel}, \cite{Kra:Hoc}. In the present article we establish that also from the point of differential and spectral geometry the noncommutative pillow, quantum cones and lens spaces behave as smooth objects.
The idea behind the {\em differential smoothness} of algebras is rooted in the observation that a classical smooth orientable manifold, in addition to de Rham complex of differential forms, admits also the complex of {\em integral forms} isomorphic to the de Rham complex; \cite[Section~4.5]{Man:gau}. The de Rham differential can be understood as a special left connection, while the boundary operator in the complex of integral forms is an example of a {\em right connection}. In the standard (commutative) differential geometry a knowledge of the integral forms and right connections does not contribute to a better understanding of the structure of a manifold, the existence of the Hodge star and the Poincar\'e duality are fully sufficient. On the other hand it becomes useful in defining the Berezin integral on a supermanifold, and precisely in this context right connections and integral forms have been introduced in \cite[Chapter~4]{Man:gau}. Supergeometry might be interpreted as one of the predecessors or special cases of noncommutative geometry thus it seems quite natural to expect that the (rather mild) usefulness of integral forms in supergeometry should become more pronounced in the noncommutative setup. This expectation led to the introduction of a noncommutative version of a right connection termed a {\em hom-connection} (or {\em divergence}) in \cite{Brz:con} and then to the associated complex of integral forms in \cite{BrzElK:int}.
While every algebra admits a differential calculus (albeit not necessarily of any geometric interest), a priori not every algebra must admit a complex of integral forms relative to a given differential calculus. The main results of \cite{BrzElK:int} show that divergences (though not necessarily flat) can be defined for calculi generated by twisted multiderivations and the examples studied there feature complexes of integral forms isomorphic to noncommutative de Rham complexes (of dimension equal to the classical dimension of non-deformed spaces). This can be thought of as a strong version of Poincar\'e duality, since it is an isomorphism of the complexes of differential and integrable forms, not just of their homologies, that is observed. Motivated by these examples, we introduce here the term {\em integrable differential calculus} to indicate a calculus that is isomorphic to the associated complex of integral forms. Informally, an algebra that is a deformation of the coordinate algebra of a classical variety and has an integrable differential calculus of classical dimension can be understood as being {\em differentially smooth}. More formally, an affine or finitely generated algebra with integer Gelfand-Kirillov dimension, say $n$, is said to be {\em differentially smooth} if it admits an integrable $n$-dimensional differential calculus that is connected in the sense that the kernel of the differential restricted to the algebra consists only of scalar multiples of the identity. In this sense the algebras studied in \cite{BrzElK:int} are smooth, a fact that should not be too surprising, since these are $q$-deformations of classically smooth compact manifolds (the two- and three-spheres and planes). The novelty of the present paper is that the differential smoothness is established for algebras that describe noncommutative versions of classically singular spaces (orbifolds). In particular we prove:\medskip
\noindent{\bf Theorem~A.}
{\em Coordinate algebras of the noncommutative pillow, cones and lens spaces are differentially smooth, i.e.\hspace{3pt}they admit connected integrable differential calculi of dimensions 2, 2 and 3 respectively.}
\medskip
Theorem~A indicates that noncommutative coordinate algebras of deformations of some classically non-smooth spaces or orbifolds are not only homologically smooth, but also admit types of differential calculi characteristic of smooth manifolds. In short, at this level at least, we are not able to distinguish between noncommutative deformations of manifolds and orbifolds, and should such a distinction be possible it must occur on finer geometric levels. The first such level that should be studied are the Riemannian or metric aspects of differentially smooth algebras, which following \cite{Connes} are encoded in spectral triples. It was shown in \cite{ReVa} that on the classical level, a spectral triple can distinguish between a manifold and a (good) orbifold in the sense that the {\em orientability condition} (meaning the existence of a Hochschild cycle whose image in the differential calculus induced by the Dirac operator is the chirality operator) that holds in the former case fails to hold in the latter. We put the noncommutative pillow and a special case of the quantum cone corresponding to the Moyal deformation of the unit disc to the test, and find spectral triples that satisfy a milder version of the orientability condition, more precisely we construct cycles but not Hochschild cycles that map to the chirality operator. One might speculate whether the fact that constructed cycles are not Hochschild cycles is a remnant of the singular nature of the classical counterparts of noncommutative manifolds. The construction of spectral triples on any invariant subalgebras with respect to the action of a finite group has been studied in several cases following the standard method of restriction of the known spectral triple over the full algebra. This has been studied for the lens spaces \cite{SiVe} and three-dimensional Bieberbach manifolds \cite{Adalgott} allowing the classification of spin structures in the noncommutative counterparts of these manifolds. However, in neither of the constructed examples an orientation cycle was constructed. Although in the $q$-deformed case the existence of such a cycle could be doubtful (as the spectral triple for the $SU_q(2)$ itself has no orientation cycle) it is expected that such a cycle exists for the more regular case of noncommutative quotients of the noncommutative torus and $\theta$-deformations. The construction of an orientation cycle presented here for the noncommutative pillow and the Moyal $N=2$ cone is the first such result. \medskip
\noindent {\bf Notation.}
All algebras considered in this paper are associative and unital over the complex field $\CC$. If $X$ is a classical geometric space (affine space, manifold, orbifold etc.), $\cO(X_q)$ denotes the noncommutative coordinate algebra of the (non-existing in a usual sense) quantum space $X_q$. Although we write $\Omega A$ for a differential calculus over an algebra $A$, in order to avoid overloading notation, we write $\Omega(X_q)$ for a differential calculus over $\cO(X_q)$. Similarly, $\Ii_k (X_q)$ means integral $k$-forms over $\cO(X_q)$, i.e.\hspace{3pt}right $\cO(X_q)$-module homomorphisms $\Omega^k(X_q) \to \cO(X_q)$.
\section{Integrable differential calculi over noncommutative algebras}
\setcounter{equation}{0}
\subsection{Integrable calculi and differential smoothness of affine algebras}
By a {\em differential graded algebra} we mean a non-negatively graded algebra $\Omega$ (with the product traditionally denoted by $\wedge$) together with the degree-one linear map $d: \Omega^\bullet \to \Omega^{\bullet+1}$ that satisfies the graded Leibniz rule and is such that $d\circ d=0$. We say that a differential graded algebra $(\Omega, d)$ is a {\em calculus over an algebra $A$} if $\Omega^0=A$ and, for all $n\in \NN$, $\Omega^n=AdA\wedge dA\wedge \cdots \wedge dA$ ($dA$ appears $n$-times). In this case we write $\Omega A$. Note that due to the Leibniz rule $\Omega^n A = dA\wedge dA\wedge \cdots \wedge dA\, A$ too. A differential calculus $\Omega A$ is said to be {\em connected} if $\ker d\mid_A = \CC.1$. If $A$ is a complex $*$-algebra then it is often requested that $\Omega A$ be a $*$-algebra and that $*\circ d = d\circ *$. In this situation one refers to $\Omega A$ as a $*$-differential calculus. If $B$ is a subalgebra of $A$, then by restriction of $\Omega A$ to $B$ we mean the differential calculus $\Omega B$ with $\Omega^n B = BdB\wedge dB\wedge \cdots \wedge dB \subseteq \Omega^nA$
A calculus $\Omega A$ is said to have {\em dimension $n$} if $\Omega^nA \neq 0$ and $\Omega^mA = 0$ for all $m >n$. An $n$-dimensional calculus $\Omega A$ admits a {\em volume form} if $\Omega^nA$ is isomorphic to $A$ as a left and right $A$-module. The existence of a right $A$-module isomorphism means that there is a free generator of $\Omega^nA$ (as a right $A$-module), i.e.\hspace{3pt}$\omega \in \Omega^nA$, such that all elements of $\Omega^nA$ can be uniquely written as $\omega a$, $a\in A$. We refer to such a generator $\omega$ as to a {\em volume form} on $\Omega A$.
The right $A$-module isomorphism $\Omega^n A \to A$ corresponding to a volume form $\omega$ is denoted by $\pi_\omega$, i.e.
\begin{equation}\label{vol.iso}
\pi_\omega(\omega a) = a, \qquad \mbox{for all $a\in A$}.
\end{equation}
Since $\Omega^n A$ is also isomorphic to $A$ as a left $A$-module, any volume form induces an algebra automorphism $\nu_\omega$ of $A$ by the formula
\begin{equation}\label{vol.auto}
a \omega = \omega \nu_\omega(a).
\end{equation}
Dually to a differential calculus on $A$ one considers its {\em integral calculus}; see \cite{Brz:con}, \cite{BrzElK:int}. Let $\Omega A$ be a differential calculus on $A$. The space of $n$-forms $\Omega^nA$ is an $A$-bimodule. Let $\Ii_nA$ denote the right dual of $\Omega^n A$, i.e.\hspace{3pt}the space of all right $A$-linear maps $\Omega^n A \to A$. Each of the $\Ii_n A$ is an $A$-bimodule with the actions
$$
(a \cdot \phi \cdot b) (\omega) = a\phi(b\omega), \qquad \mbox{for all}\quad \phi\in \Ii_{n} A,\hspace{3pt}\omega\in \Omega^n A,\hspace{3pt} a,b \in A.
$$
The direct sum of all the $\Ii_nA$, denoted $\Ii A = \oplus_n \Ii_nA$, is a right $\Omega A$-module with action
\begin{equation}\label{int.right}
(\phi \cdot \omega)(\omega ') = \phi(\omega\wedge \omega'), \qquad \mbox{for all}\quad \phi\in \Ii_{n+m} A,\hspace{3pt}\omega\in \Omega^n A,\hspace{3pt}\omega'\in \Omega^mA.
\end{equation}
A {\em divergence} on $A$ is a linear map $\nabla: \Ii_1A\to A$, such that
\begin{equation}\label{hom.Leibniz}
\nabla (\phi\cdot a) = \nabla(\phi) a + \phi(da), \qquad \mbox{for all}\quad \phi\in \Ii_{1} A,\hspace{3pt}a \in A.
\end{equation}
A divergence can be extended to the whole of $\Ii A$, $\nabla_n: \Ii_{n+1} A\to \Ii_n A$, by setting
\begin{equation}\label{hom.ext}
\nabla_n(\phi)(\omega)= \nabla (\phi\cdot \omega) + (-1)^{n+1} \phi(d\omega), \qquad \mbox{for all \,} \phi\in \Ii_{n+1} A, \hspace{3pt}\omega \in \Omega^{n}A.
\end{equation}
A combination of \eqref{hom.Leibniz} with \eqref{hom.ext} yields the following Leibniz rule, for all $\phi\in \Ii_{m+n+1} A$ and $\omega\in \Omega^m A$,
\begin{equation}\label{hom.Leibniz.n}
\nabla_{n}(\phi \cdot \omega) = \nabla_{m+n}(\phi )\cdot \omega + (-1)^{m+n} \phi \cdot d\omega;
\end{equation}
see \cite[3.2~Lemma]{Brz:con}.
A divergence is said to be {\em flat} if $\nabla\circ \nabla_1 =0$. This then implies that $\nabla_n\circ\nabla_{n+1} =0$, for all $n\in \NN$, hence $\Ii A$ together with the $\nabla_n$ form a chain complex, which is termed the {\em complex of integral forms} over $A$. The cokernel map of $\nabla$, i.e.\hspace{3pt}$\Lambda: A \to \coker \nabla = A/\im \nabla$ is called the {\em integral on $A$ associated to $\Ii A$}. Note that, in general, it is not guaranteed that a given differential calculus on $A$ will admit a divergence on $A$ and even if it admits such a divergence that it would be flat; see \cite{BrzElK:int}.
Given a left $A$-module $X$ with action $a\cdot x$, for all $a\in A$, $x\in X$, and an algebra automorphism $\nu$ of $A$, the notation ${}^\nu X$ stands for $X$ with the $A$-module structure twisted by $\nu$, i.e.\hspace{3pt}with the $A$-action $a\otimes x\mapsto \nu(a)\cdot x$.
The following definition introduces notions which form the backbone of the majority of this paper.
\begin{definition}\label{def.integrable}
An $n$-dimensional differential calculus $\Omega A$ is said to be {\em integrable} if
$\Omega A$ admits a complex of integral forms $(\Ii A, \nabla)$ for which
there exist an algebra automorphism $\nu$ of $A$ and $A$-bimodule isomorphisms $\Theta_k: \Omega^k A \to {}^\nu \Ii_{n-k} A$, $k=0,\ldots , n$, rendering commutative the following diagram:
$$
\xymatrix{ A\ar[r]^d \ar[d]_{\Theta_0} & \Omega^1A\ar[r]^d \ar[d]_{\Theta_1} & \Omega^2A \ar[r]^-d\ar[d]_{\Theta_2} & \ldots \ar[r]^-d & \Omega^{n-1}A \ar[r]^d\ar[d]_{\Theta_{n-1}} &\Omega^nA\ar[d]^{\Theta_n}\\
{}^\nu \Ii_n A\ar[r]^-{\nabla_{n-1}} &{}^\nu \Ii_{n-1} A\ar[r]^-{\nabla_{n-2}} & {}^\nu \Ii_{n-2} A\ar[r]^-{\nabla_{n-3}} & \ldots \ar[r]^-{\nabla_1} &\hspace{3pt}{}^\nu \Ii_1 A\ar[r]^\nabla & {}^\nu \! A \, .}
$$
The $n$-form $\omega: = \Theta_n^{-1}(1) \in \Omega^n A$ is called an {\em integrating volume form}.
\end{definition}
Examples of algebras admitting integrable calculi discussed in \cite{BrzElK:int} include the algebra of complex matrices $M_N(\CC)$ with the $N$-dimensional calculus generated by derivations \cite{Dub:der}, \cite{DubKer:non}, the quantum group $SU_q(2)$ with the three-dimensional left covariant calculus \cite{Wor:twi} and the quantum standard sphere with the restriction of the above calculus.
The following theorem indicates that the integrability of a differential calculus can be defined without explicit reference to integral forms.
\begin{theorem}\label{thm.integrable}
The following statements about an $n$-dimensional differential calculus $\Omega A$ over an algebra $A$ are equivalent:
\begin{zlist}
\item $\Omega A$ is an integrable differential calculus.
\item There exist an algebra automorphism $\nu$ of $A$ and $A$-bimodule isomorphisms $\Theta_k: \Omega^k A \to {}^\nu \Ii_{n-k} A$, $k=0,\ldots , n$, such that, for all $\omega' \in \Omega^k A$, $\omega'' \in \Omega^m A$,
\begin{equation}\label{linearity}
\Theta_{k+m} (\omega'\wedge \omega'') = (-1)^{(n-1)m}\Theta_k(\omega')\cdot \omega''.
\end{equation}
\item There exist an algebra automorphism $\nu$ of $A$ and an $A$-bimodule map $\vartheta : \Omega^n A \to {}^\nu A$ such that all left multiplication maps
$$
\ell^k_\vartheta : \Omega^kA \to \Ii_{n-k} A, \qquad \omega'\mapsto \vartheta \cdot \omega', \qquad k=0,1,\ldots , n,
$$
where the actions $\cdot$ are defined by \eqref{int.right}, are bijective.
\item $\Omega A$ admits a volume form $\omega$ such that
all left multiplication maps
$$
\ell^k_{\pi_\omega} : \Omega^kA \to \Ii_{n-k} A, \qquad \omega'\mapsto \pi_\omega \cdot \omega', \qquad k=1,\ldots , n-1,
$$
where $\pi_\omega$ is defined by \eqref{vol.iso}, are bijective.
\end{zlist}
\end{theorem}
\begin{proof}
(1) $\Rightarrow$ (2)
The existence of an algebra automorphism $\nu$ and maps $\Theta_k: \Omega^k A \to {}^\nu \Ii_{n-k} A$ that make the diagram in Definition~
\ref{def.integrable} commute are parts of the definition of integrability of a differential calculus. We will prove that the $\Theta_{k+m}$ satisfy equations \eqref{linearity} by induction with respect to $k+n$.
First, for $k+m=0$ (\ref{linearity}) holds by definition. With the inductive
assumption that the formula is true for all $k+m < p \leq n$, it needs to be demonstrated that (\ref{linearity}) is also true for $p$. Using the commutativity of the diagram in Definition~\ref{def.integrable}, the Leibniz rule and (\ref{hom.Leibniz.n}), we can compute:
\begin{eqnarray*}
\Theta_{k+m+1} (\omega \wedge da) &=& (-1)^{k+m} \left( \Theta_{k+m+1}(d (\omega a) ) - \Theta_{k+m+1}(( d \omega) a) \right) \\
&=& (-1)^{k+m} \left( \nabla_{n-k-m-1} \Theta_{k+m} (\omega a) - \Theta_{k+m+1}( d \omega) a \right) \\
&=& (-1)^{k+m} \left( \nabla_{n-k-m-1} ( \Theta_{k+m} (\omega) a) - ( \nabla_{n-k-m-1} \Theta_{k+m} (\omega)) a \right) \\
&=& (-1)^{k+m} (-1)^{n-k-m-1} \Theta_{k+m} (\omega) \cdot da = (-1)^{n-1} \Theta_{k+m} (\omega) \cdot da.
\end{eqnarray*}
The inductive assumption and the fact that every element of $\Omega^{m} A$ is a linear combination of products of
$m-1$-forms with exact one-forms, imply the required equality,
$$
\Theta_{k+m+1} (\omega'\wedge \omega'') = (-1)^{(n-1)(m+1)}\Theta_{k}(\omega')\cdot \omega'',
$$
for all $\omega' \in \Omega^k A$, $\omega'' \in \Omega^{m+1} A$. The assertion follows by the principle of mathematical induction.
(2) $\Rightarrow$ (3) Given a system of $A$-bimodule isomorphisms $\Theta_k: \Omega^k A \to {}^\nu \Ii_{n-k} A$, $k=0,\ldots , n$ that satisfy \eqref{linearity}, define
$$
\vartheta := \Theta_0(1).
$$
Since $\Theta_n$ satisfies \eqref{linearity}, for all $\omega\in \Omega^nA$,
$$
\vartheta (\omega) = \Theta_0(1)(\omega) = \Theta_0(1) \cdot \omega = \Theta_n(\omega).
$$
Thus $\vartheta = \Theta_n$ and, in particular, it is an $A$-bimodule map as stated. Again by \eqref{linearity}, for all $\omega'\in \Omega^kA$,
$$
\Theta_k (\omega') = (-1)^{(n-1)k} \Theta_0(1)\cdot \omega' = (-1)^{(n-1)k} \vartheta \cdot \omega' ,
$$
i.e.\hspace{3pt}$\Theta_k = (-1)^{(n-1)k} \ell^k_\vartheta$, and hence all the $\ell^k_\vartheta$ are bijective, as required.
(3) $\Rightarrow$ (4) Note that by the definition of the action \eqref{int.right} $\vartheta = \ell_\vartheta^n$, hence $\vartheta$ is an $A$-bimodule isomorphism. Consequently, $\Omega^n A$ is isomorphic to $A$ as a left and right $A$-module, hence $\omega := \vartheta^{-1}(1)$ is a volume form. Since $\vartheta^{-1}$ is a bimodule map ${}^\nu A \to \Omega^n A$, for all $a\in A$,
$$
a = \pi_\omega(\omega a) = \pi_\omega(\vartheta^{-1}(1) a) = \pi_\omega (\vartheta^{-1}(a)),
$$
i.e.\hspace{3pt}$\pi_\omega = \vartheta$, hence all the $\ell^k_{\pi_\omega}$ are bijective.
(4) $\Rightarrow$ (1) Given a volume form $\omega \in \Omega^nA$, define
\begin{equation}\label{theta.k}
\Theta_k = (-1)^{(n-1)k} \ell^k_{\pi_\omega}: \Omega^kA \to \Ii_{n-k}A, \qquad k=0,1,\ldots, n.
\end{equation}
By assumption $\Theta_k$ are bijective for $k=1,\ldots, n -1$. Note that $\Theta_n = \pi_\omega$, hence it is bijective too. We will next show that the map
$$
\Theta^{-1}_0: \Ii_n A \to A, \qquad \phi\mapsto \nu_\omega^{-1} (\phi(\omega)),
$$
where $\nu_\omega$ is the algebra automorphism associated to $\omega$ via \eqref{vol.auto}, is the inverse of $\Theta_0$. For all $a\in A$,
\begin{eqnarray*}
\Theta_0^{-1}\left(\Theta_0(a)\right) &=& \nu_\omega^{-1}\left(\Theta_0(a)(\omega)\right) = \nu_\omega^{-1}\left(\pi_\omega(a\omega)\right)\\
&=& \nu_\omega^{-1}\left(\pi_\omega(\omega\nu_\omega(a))\right) = \nu_\omega^{-1}\left(\nu_\omega(a)\right) =a,
\end{eqnarray*}
by the definitions of $\nu_\omega$ and $\pi_\omega$. On the other hand, for all $\phi\in \Ii_n A$ and $a\in A$,
\begin{eqnarray*}
\Theta_0\circ \Theta_0^{-1}\left( \phi\right)(\omega a) &=& \Theta_0\left( \nu_\omega^{-1} (\phi(\omega))\right)(\omega a) = \pi_\omega\left(\nu_\omega^{-1} (\phi(\omega))\omega a\right)\\
&=& \pi_\omega\left(\omega\phi(\omega) a\right) = \phi(\omega) a = \phi(\omega a),
\end{eqnarray*}
by the definitions of $\nu_\omega$ and $\pi_\omega$ and the right $A$-linearity of $\phi$. Since $\Omega^n A$ is generated by $\omega$ this means that the composite map $\Theta_0\circ \Theta_0^{-1}$ is the identity. Hence $\Theta_0^{-1}$ is the inverse of $\Theta_0$, as claimed.
Directly from definition \eqref{theta.k}, $\Theta_k$ satisfy equations \eqref{linearity}. In particular, they are right $A$-module maps. Next note that, for all $a\in A$, $\omega''\in \Omega^n$,
\begin{equation}\label{iso.bil}
\pi_\omega(a \omega'') = \nu_\omega(a) \pi_\omega( \omega''), \qquad \pi_\omega(\omega''a) = \pi_\omega(\omega'')a.
\end{equation}
Thus, for all $a\in A$, $\omega'\in \Omega^kA$, and $\omega''\in \Omega^{n-k}A$,
\begin{eqnarray*}
\Theta_k(a\omega')(\omega'') &=& (-1)^{(n-1)k} \pi_\omega(a\omega'\wedge\omega'') \\
&=& (-1)^{(n-1)k} \nu_\omega(a) \pi_\omega(\omega'\wedge\omega'') = \nu_\omega(a) \Theta_k(\omega')(\omega''),
\end{eqnarray*}
by the first of equation \eqref{iso.bil}. Therefore, all $\Theta_k $ given by \eqref{theta.k} are bimodule isomorphisms $\Omega^kA \to {}^{\nu}\Ii_{n-k}A$, where $\nu = \nu_\omega$.
Let us define:
\begin{equation}\label{nabla.k}
\nabla_k = \Theta_{n-k}\circ d\circ \Theta_{n-k-1}^{-1} : \Ii_{k+1} A\to \Ii_{k} A.
\end{equation}
Obviously, the maps $\nabla_k$ will make the diagram in Definition~\ref{def.integrable} commute, and, since $d\circ d =0$ also $\nabla_{k-1}\circ \nabla_k =0$. We need to prove that $\nabla := \nabla_0$ is a divergence and that all the remaining $\nabla_k$ given by \eqref{nabla.k} extend $\nabla$ in the sense of equalities \eqref{hom.ext}.
For all $a\in A$ and $\phi \in \Ii_1A$,
\begin{eqnarray*}
\nabla(\phi\cdot a) &=& \Theta_{n}\circ d\circ \Theta_{n-1}^{-1}(\phi\cdot a) = \Theta_{n}\left(d\left(\Theta_{n-1}^{-1}(\phi) a\right)\right)\\
&=& \Theta_{n}\left(d\left(\Theta_{n-1}^{-1}(\phi)\right) a\right) + (-1)^{n-1} \Theta_{n}\left(\Theta_{n-1}^{-1}(\phi) \wedge d a\right)\\
&=& \Theta_{n}\left(d\left(\Theta_{n-1}^{-1}(\phi)\right)\right) a + \phi\cdot da = \nabla(\phi) a + \phi(da),
\end{eqnarray*}
where the second equality follows by the right $A$-linearity of $\Theta_{n-1}$, the third one is a consequence of the graded Leibniz rule, the fourth one follows by the right $A$-linearity of $\Theta_{n}$ and by equation \eqref{linearity}, and the final equality is simply the definition of $\nabla$ in \eqref{nabla.k} and the action \eqref{int.right}. This proves that $\nabla$ is a divergence.
Observe that setting $\omega' = \Theta_k^{-1}(\phi)$ in \eqref{linearity} and then applying $\Theta_{k+m}^{-1}$ one obtains, for all $\omega''\in \Omega^mA$ and $\phi\in \Ii_{n-k}A$,
$$
\Theta_{k+m}^{-1} (\phi \cdot \omega'') = (-1)^{(n-1)m} \Theta_k^{-1}(\phi) \wedge \omega'' .
$$
This can be used (in the second equality below) to prove \eqref{hom.ext}. For any $\phi \in \Ii_{m+1}A$ and $\omega''\in \Omega^mA$,
\begin{eqnarray*}
\nabla(\phi\cdot \omega'') &=& \Theta_{n}\circ d\circ \Theta_{n-1}^{-1}(\phi\cdot \omega'') = (-1)^{(n-1)m}\Theta_{n}\left( d\left( \Theta_{n-1-m}^{-1}(\phi) \wedge \omega''\right)\right)\\
&=& (-1)^{(n-1)m}\Theta_{n}\left( d\left( \Theta_{n-1-m}^{-1}(\phi)\right) \wedge \omega''\right)\\
&& + (-1)^{(m+1)n -1}\Theta_{n}\left( \Theta_{n-1-m}^{-1}(\phi) \wedge d \omega''\right)\\
&=& \Theta_{n-m}\left( d\left( \Theta_{n-1-m}^{-1}(\phi)\right) \right)\cdot \omega''
+ (-1)^{m+1}\Theta_{n-m-1}\left( \Theta_{n-1-m}^{-1}(\phi)\right) \wedge d \omega'' \\
&=& \nabla_m(\phi)( \omega'' )+ (-1)^m \phi\cdot d\omega'',
\end{eqnarray*}
where the third equality follows by the Leibniz rule, the fourth one is a consequence of the identities \eqref{linearity} that all the $\Theta_k$ obey and the final equality follows by the definitions of $\nabla_m$ and the action \eqref{int.right}. This proves that each of the $\nabla_k$ defined by \eqref{nabla.k} is an extension of $\nabla$, and thus completes the proof of the theorem.
\end{proof}
\begin{remark}\label{rem.integrating}
A volume form $\omega \in \Omega^n A$ is an integrating form if and only if it satisfies conditions (4) of Theorem~\ref{thm.integrable}.
\end{remark}
As it stands the integrability of a calculus over an algebra $A$ is a property of the differential graded algebra $\Omega A$ rather than a characterization of $A$ itself. To connect this property of $\Omega A$ to the nature of $A$ we need to relate the dimension of the differential calculus with that of $A$. Since we are dealing with algebras that are deformations of coordinate algebras of affine varieties, the Gelfand-Kirillov dimension seems to be best suited here; see \cite{KraLen:gro} or\cite[Chapter~8]{McCRob:Noe} for a detailed discussion of the Gelfand-Kirillov dimension.
Let $A$ be a finitely generated or affine algebra with generating subspace $\Vv$. Let us write $\Vv(n)$ for the subspace of $A$ spanned by 1 and all words in generators of $A$ of length at most $n$. The algebra $A$ is said to have {\em polynomial growth} if there exist $c\in \RR$ and $\nu\in \NN$ such that $\dim \Vv(n) \leq cn^\nu$ for all sufficiently large $n$. The {\em Gelfand-Kirillov dimension} of $A$ is a real number defined as
\begin{equation}\label{GK}
\gk(A) := \inf \{ \nu\; |\; \dim \Vv(n) \leq n^\nu, \, n \gg 0\},
\end{equation}
if $A$ has polynomial growth and is defined as infinity otherwise. In the case of commutative affine algebras with polynomial growth, the Gelfand-Kirillov dimension coincides with the dimension of the underlying affine space (the Krull dimension of its coordinate algebra).
\begin{definition}\label{def.smooth.algebra}
An affine algebra with integer Gelfand-Kirillov dimension $n$ is said to be {\em differentially smooth} if it admits an $n$-dimensional connected integrable differential calculus.
\end{definition}
For example, the polynomial algebra $\CC[x_1,\ldots, x_n]$ has the Gelfand-Kirillov dimension $n$ and the usual exterior algebra is an $n$-dimensional integrable calculus, hence $\CC[x_1,\ldots, x_n]$ is differentially smooth. The results of \cite{BrzElK:int} establish differential smoothness of coordinate algebras of the quantum group $SU_q(2)$, the standard quantum Podle\'s sphere and the quantum Manin plane. The following example shows that not all algebras are differentially smooth.
\begin{example}\label{ex.nonsmooth}
The algebra $A = \CC[x,y]/\langle xy\rangle$ is not differentially smooth.
\end{example}
\begin{proof}
Since $xy=yx = 0$, the algebra $A$ has a basis $1, x^n, y^n$, $n=1,2,\ldots$, hence $\gk(A) =1$. Suppose there is a one-dimensional
connected integrable calculus $\Omega A$ and let $\Theta_1: \Omega^1 A \to {{}^\nu} \! A$, where $\nu$ is an algebra automorphism on $A$, be the required bimodule isomorphism. The Leibniz rule together with the equalities $xy=yx = 0$ imply that
$$ x\, dy = - dx \, y, \;\;\;\;\; y\, dx = - dy \, x. $$
Apply $\Theta_1$ to these identities to obtain
\begin{equation}\label{nu.theta}
\nu(x) \theta_y = - \theta_x y, \;\;\;\; \nu(y) \theta x = - \theta_y x,
\end{equation}
where $\theta_x := \Theta_1(dx)$ and $\theta_y :=\Theta_1(dy)$.
The algebra $A$ admits two types of automorphisms
$\nu_i:A\to A$, $i=1,2$,
$$
\nu_1(x) = ax, \quad \nu_1(y) = by, \;\;\;\; \hbox{and} \;\;\;\; \nu_2(x) = ay, \quad \nu_2(y) = bx,\qquad a,b\in \CC, \; a,b\neq 0.
$$
For $\nu=\nu_1$, \eqref{nu.theta} read
$$ a x \theta_y = - \theta_x y, \;\;\;\; b y \theta_x = - \theta_y x, $$
with the only solutions given by $ \theta_y = y p(y)$ and $\theta_x = x q(x)$, where $p,q$ are polynomials.
Hence the image under $\Theta_1$ of any one-form must be a polynomial without a scalar term, so
$1$ cannot lie in $\Theta_1( \Omega^1 A)$, and therefore $\Theta_1$ is not surjective.
In the case of the automorphisms $\nu_2$, equations \eqref{nu.theta} come out as
$$ a y \theta_y = - \theta_x y, \;\;\;\; b x \theta_x = - \theta_y x.$$
Since the algebra is commutative,
$$ y (a \theta_y + \theta_x) = 0 = x (b \theta_x + \theta_y), $$
which can be solved. If $ab \not= 1$ then again both $\theta_x$ and $\theta_y$ are polynomials without any scalar terms and, by the same arguments as above, $\Theta_1$ is not surjective which contradicts
the assumption that $\Theta_1$ is an isomorphism of bimodules.
If $ab=1$, then the only solution is $\theta_y = c$ and $\theta_x = - ac$, for some $c \in \CC$. However,
in this case $d (a y + x) = 0$, which contradicts the assumption that the differential calculus
is connected.
Therefore, there are no one-dimensional connected integrable calculi over $A$, i.e.\hspace{3pt}$A$ is not differentially smooth.
\end{proof}
\subsection{Finitely generated and projective integrable calculi}
Geometrically the most interesting cases of differential calculi are those where $\Omega^kA$ are finitely generated and projective right or left (or both) $A$-modules.
\begin{lemma}\label{lem.projective}
Let $\Omega A$ be an integrable $n$-dimensional calculus over $A$ with integrating form $\omega$. Then $\Omega^k A$
is a finitely generated projective right $A$-module if there exist a finite number of forms $\omega_i \in \Omega^kA$ and
$\bar{\omega}_i \in \Omega^{n-k}A$ such that, for all $\omega'\in \Omega^k A$,
\begin{equation}
\omega'= \sum_i \omega_i \pi_\omega(\bar{\omega}_i \wedge \omega').
\label{homproj}
\end{equation}
\end{lemma}
\begin{proof}
The elements $\omega_i$ are generators of $\Omega^kA$, while the equalities
$$
\phi_i (\omega')= \ell^{n-k}_{\pi_\omega}(\bar{\omega}_i) (\omega') = \pi_\omega(\bar{\omega}_i \wedge \omega'),
$$
define $\phi_i \in \hbox{Hom}_A(\Omega^kA, A)$, and then \eqref{homproj} guarantees that $\omega_i, \phi_i$ form a dual basis for $\Omega^kA$, hence $\Omega^kA$ is projective.
To show the implication in the other direction let us assume that $\Omega^k A$ is finitely generated projective with a
basis $\omega_i$ and the dual basis $\phi_i$. Let us define:
$$ \bar{\omega}_i := \Theta_{n-k}^{-1} (\phi_i). $$
Using the properties of an integrable differential calculus it is easy to show that \eqref{homproj} is satisfied.
\end{proof}
\begin{lemma}\label{lem.integrating}
Let $\Omega A$ be an $n$-dimensional calculus over $A$ admitting a volume form $\omega$. Assume that, for all $k=1,2,\ldots, n-1$, there exist a finite number of forms $\omega_i^k, \bomega_i^k \in \Omega^kA$ such that, for all $\omega'\in \Omega^kA$,
\begin{equation}\label{dual.basis}
\omega' = \sum_i\omega_i^k \pi_\omega(\bomega_i^{n-k} \wedge \omega ') = \sum_i \nu_\omega^{-1}\left(\pi_\omega(\omega'\wedge \omega_i^{n-k} )\right)\bomega_i^k,
\end{equation}
where $\pi_\omega$ and $\nu_\omega$ are defined by \eqref{vol.iso} and \eqref{vol.auto}, respectively. Then $\omega$ is an integrating form and all the $\Omega^kA$ are finitely generated and projective as left and right $A$-modules.
\end{lemma}
\begin{proof}
Conditions \eqref{dual.basis} imply that $\omega^k_i$, $\pi_\omega(\bomega_i^{n-k} \wedge -)$ form a dual basis for $\Omega^k A$ as a right $A$-module and $\bomega^k_i$, $\nu_\omega^{-1}\left(\pi_\omega(-\wedge \omega_i^{n-k} )\right)$ form a dual basis for $\Omega^k A$ as a left $A$-module.
For all $k$ define,
$$
\Phi_k : \Ii_{n-k} A \to \Omega^k A, \qquad \phi \mapsto \sum_i \nu_\omega^{-1}\left(\phi(\omega_i^{n-k})\right)\bomega_i^k.
$$
Then, for all $\phi \in \Ii_{n-k}$ and $\omega'\in \Omega^{n-k} A$,
\begin{eqnarray*}
\left(\ell_{\pi_\omega}^k \circ \Phi_k\right)(\phi)(\omega ') &=& \pi_\omega \left(\Phi_k(\phi)\wedge \omega '\right)\\
&=& \sum_i\pi_\omega \left(\nu_\omega^{-1}\left(\phi(\omega_i^{n-k})\right)\bomega_i^k\wedge \omega '\right)\\
&=& \sum_i\phi(\omega_i^{n-k})\pi_\omega \left(\bomega_i^k\wedge \omega '\right) = \sum_i\phi\left(\omega_i^{n-k}\pi_\omega \left(\bomega_i^k\wedge \omega '\right)\right) = \phi(\omega'),
\end{eqnarray*}
where the first of equations \eqref{iso.bil} was used in the derivation of the third equality, and next the right $A$-linearity of $\phi$ and the first of equations \eqref{dual.basis} were employed. On the other hand, for all $\omega'\in \Omega^k A$,
$$
\left(\Phi_k\circ \ell_{\pi_\omega}^k \right)(\omega ') = \sum_i\nu_\omega^{-1}\left(\ell_{\pi_\omega}^k(\omega ') (\omega_i^{n-k}) \right)\bomega_i^k\\
= \sum_i \nu_\omega^{-1}\left(\pi_\omega(\omega'\wedge \omega_i^{n-k} )\right)\bomega_i^k = \omega',
$$
by the second of equations \eqref{dual.basis}. Therefore, all of the $\ell_{\pi_\omega}^k$ are isomorphisms, and hence $\omega$ is an integrating form.
\end{proof}
In the set-up of Lemma~\ref{lem.integrating}, a combination of the form of the inverse of the $\ell_{\pi_\omega}^k$ together with that of the divergence associated to the isomorphisms $\ell_{\pi_\omega}^k$ via \eqref{nabla.k}, gives the following formula for the divergence:
\begin{equation}\label{hom.con.omega}
\nabla(\phi) = (-1)^{n-1} \sum_i \pi_\omega \left( d\left(\nu_\omega^{-1}\left(\phi(\omega_i^{1})\right)\right)\bomega_i^{n-1}\right),
\end{equation}
for all $\phi \in \Ii_1 A$.
\subsection{A construction of two-dimensional integrable calculi}
In this section we construct a class of two-dimensional integrable calculi. Notwithstanding its quite technical nature and rather specialized appearance, the following lemma is applicable in a variety of situations including, of course, those that are the subject matter of this article.
\begin{lemma}\label{lemma.Poincare}
Assume that:
\begin{blist}
\item $A$ is an algebra and $B\subseteq A$ is a subalgebra, hence $A$ is a $B$-bimodule in a natural way;
\item $A_+, A_- \subseteq A$ are right $B$-submodules of $A$ such that $A_+A_- = A_-A_+ = B$;
\item there exists a two-dimensional differential calculus $\Omega B$ over $B$ with $\Omega^1 B = A_+\oplus A_-$ as a right $B$-module and the product in $\Omega^1 B$ given by the formula
$$
(a_+, a_-)\wedge (b_+, b_-) = \omega( \sigma_+(a_+)b_- + \sigma_-(a_-)b_+),
$$
for all $a_\pm,b_\pm \in A_\pm$, where $\omega$ is a volume form and $\sigma_\pm: A_\pm\to A_\pm$ are invertible linear maps.
\end{blist}
Then $\Omega B$ is integrable.
\end{lemma}
Before we start proving Lemma~\ref{lemma.Poincare} let us point to the classical motivation behind it. In classical (complex) geometry, Riemann surfaces can be obtained as quotients of the disc (with hyperbolic metric) by Fuchsian groups. As a result, algebras of functions and modules of (holomorphic or antiholomorphic) sections of the cotangent bundle over such a surface can be embedded in the algebra of functions on the disc and the corresponding modules over it. In particular, (anti-)holomorphic sections over a Riemann surface can be expressed in terms of functions on the disc. In the classical situation $B$ should be thought of as functions on a Riemann surface, $A$ as functions on the disc and $A_\pm$ as (anti-)holomorphic sections on the surface expressed in terms of functions on the disc.
\proofof{Lemma~\ref{lemma.Poincare}}
In view of Theorem~\ref{thm.integrable} we only need to show that the map
$$
\Theta := \ell_\omega^k: \Omega^1 B \to \Ii_1 B, \qquad \omega' \mapsto [\omega'' \mapsto \pi_\omega(\omega'\wedge \omega'')],
$$
is bijective.
Since $A_\pm A_\mp = B$, there exist $r_\pm ^i,s_\pm ^i\in A_\pm$ such that
$$
r_+^i r_-^i = s_-^i s_+^i =1,
$$
where here and below the repeated index is summed. Let us define a linear map:
\begin{equation}\label{def.theta}
\Theta^{-1}: \Ii_1 B\to \Omega^1 B, \qquad \phi \mapsto \left(\sigma_+^{-1}\left( \phi(0,s_-^i)s_+^i\right), \sigma_-^{-1}\left( \phi(r_+^i,0)r_-^i\right)\right).
\end{equation}
Then $\Theta^{-1}$ is the inverse of $\Theta$. Indeed, for all $(a_+,a_-)\in \Omega^1 B$,
\begin{eqnarray*}
\Theta^{-1}\circ\Theta (a_+,a_-) &=& \left(\sigma_+^{-1}\left(\pi_\omega\left((a_+,a_-)\wedge (0,s_-^i)\right)s_+^i\right), \sigma_-^{-1}\left(\pi_\omega\left((a_+,a_-)\wedge (r_+^i,0)\right)r_-^i\right)\right)\\
&=& \left(\sigma_+^{-1} \left(\sigma_+ \left( a_+\right)s_-^is_+^i\right), \sigma_-^{-1} \left(\sigma_- \left( a_-\right)r_+^ir_-^i\right)\right) = (a_+,a_-),
\end{eqnarray*}
and, for all $\phi\in \Ii_1 B$,
\begin{eqnarray*}
\Theta\circ\Theta^{-1}(\phi)(a_+,a_-) &=& \pi_\omega\left(\sigma_+^{-1} \left( \phi(0,s_-^i)s_+^i\right), \sigma_-^{-1} \left( \phi(r_+^i,0)r_-^i\right)\wedge (a_+,a_-)\right)\\
&=& \sigma_+ \left(\sigma_+^{-1}\left( \phi(0,s_-^i)s_+^i\right)\right) a_- + \sigma_-\left(\sigma_-^{-1}\left( \phi(r_+^i,0)r_-^i\right)\right) a_+\\
&=& \phi\left(0, s_-^is_+^ia_-\right) + \phi\left( r_+^ir_-^ia_+,0\right) = \phi(a_+,a_-),
\end{eqnarray*}
where we used that $s_+^ia_-, r_-^ia_+\in B$ and the fact that $\phi$ is a right $B$-linear map. Now Theorem~\ref{thm.integrable} implies that $\Omega B$ is an integrable differential calculus as claimed.
\endproof
A typical problem to which Lemma~\ref{lemma.Poincare} can be applied is the construction of an integrable differential calculus over an invariant part of a strongly group-graded algebra.
Let $G$ be an Abelian group. Recall that an algebra $A$ is called a {\em $G$-graded algebra} if $A= \oplus_{g\in G} A_g$ and, for all $g,h\in G$, $A_gA_h\subseteq A_{g+h}$, and it is said to be {\em strongly-graded} if $A_gA_h = A_{g+h}$. In the strongly-graded case, for all $h\in G$, $A_{-h}A_h = A_hA_{-h} = A_0$, where $A_0$ is the invariant subalgebra, i.e.\hspace{3pt}the subalgebra of all elements of $A$ graded by the neutral element $0\in G$. In this case, one can choose $B=A_0$, $A_+=A_h$, $A_-=A_{-h}$ (for a suitable $h\in G$), and $\sigma_\pm$ to be restrictions of any degree-preserving automorphism of $A$. For example, Lemma~\ref{lemma.Poincare} provides one with a proof of integrability of the two-dimensional differential calculus over the standard quantum Podle\'s sphere, alternative to that given in \cite[Section~4]{BrzElK:int}. In this case $A$ is the coordinate algebra of $SU_q(2)$, which is strongly graded by the integer group $\ZZ$. The invariant part of $A$ is the coordinate algebra of the quantum standard Podle\'s sphere \cite{Pod:sph}, $h=1$, and the degree preserving automorphism of $A$ is induced by the 3D-calculus on $A$.
\subsection{Integrability and principality}
Strongly graded algebras are examples of principal comodule algebras. Let $H$ be a Hopf algebra with bijective antipode. Recall from \cite{BrzHaj:Che} that a right $H$-comodule algebra $A$ with coaction $\roA: A\to A\ot H$ is called a {\em principal comodule algebra} if the canonical map
$$
\can : A\ot_B A\to A\ot H, \qquad a\ot a'\mapsto a\roA(a'),
$$ is bijective and there exists a right $B$-module and right $H$-comodule splitting of the multiplication map $B\ot A\to A$. Here $B$ is the coinvariant subalgebra, $B = A^{coH} := \{b\in A\; |\; \roA(b) = b\ot 1\}$. Principal comodule algebras play the role of principal fibre bundles in noncommutative geometry and, classically, a quotient of a smooth manifold by a free action of a Lie group is smooth, thus it is natural to expect that if a principal comodule algebra admits an integrable calculus so does its coinvariant algebra.
In this section we consider a special case of an integrable differential calculus over a principal comodule algebra that induces such a calculus over the coinvariant subalgebra.
Let $A$ be a right $H$-comodule algebra. A differential calculus $\Omega A$ is said to be {\em $H$-covariant} if $\Omega A$ is a right $H$-comodule algebra with the degree-zero coactions $\varrho^{\Omega^k A}: \Omega^kA\to \Omega^kA\ot H$ that commute with the differential, i.e.\hspace{3pt}such that
$$
\varrho^{\Omega^{k+1} A}\circ d = (d\ot \id)\circ \varrho^{\Omega^k A}.
$$
If $B$ is a coinvariant subalgebra of $A$, then the covariant calculus $\Omega A$ restricts to the calculus $\Omega B$ on $B$. Clearly, $\Omega B$ is contained in the coinvariant part of $\Omega A$, $\Omega B \subseteq \left(\Omega A\right)^{co H}$, but the converse inclusion is not necessarily true.
\begin{lemma}\label{lem.princ}
Let $A$ be a principal $H$-comodule algebra with the coinvariant subalgebra $B$. Let $\Omega A$ be an $H$-covariant $n$-dimensional integrable calculus with integrating form $\omega$. Assume that:
\begin{blist}
\item $\Omega B = \left(\Omega A\right)^{co H}$;
\item $\omega$ is invariant, i.e.\hspace{3pt}$\rho^{\Omega^n A} (\omega) = \omega \ot 1$, and hence $\omega \in \Omega^n B$;
\item for all $k=1,\ldots , n-1$, if $\omega' \in \Omega^kA$ has the property that, for all $\omega'' \in \Omega^{n-k} B$, $\omega'\wedge \omega'' \in \Omega^n B$, then $\omega' \in \Omega^kB$.
\end{blist}
Then $\Omega B$ is an integrable differential calculus and $\omega$ is its integrating form.
\end{lemma}
\begin{proof}
By assumption (b), $\Omega^n B \cong B$ with the isomorphism $\pi_\omega^B : \Omega^nB\to B$ which is the restriction of $\pi_\omega ^A: \Omega^nA\to A$, $\omega a\mapsto a$.
Let us write $\varrho^{\Omega^k A}(\omega ') = \omega'\sw 0 \ot \omega'\sw 1$ and $h\su 1 \ot h \su 2 = \can^{-1}(1\ot h)$, summation understood in both cases. The map $h\mapsto \can^{-1}(1\ot h)$ is known as the {\em translation map}. By the properties of the translation map (see \cite[3.4~Remark]{Sch:rep}), for all $\omega''\in {\Omega^{n-k} A}$, there is an inclusion
$$
\omega''\sw 0 \omega''\sw 1\su 1 \ot \omega''\sw 1\su 2 \in \left(\Omega^{n-k}A\right)^{co H} \ot_B A = \Omega^{n-k}B \ot_B A
$$
where the last equality is a consequence of assumption (a). Therefore, for all $\phi \in \Ii_{n-k} B$, one can define $\hat{\phi} \in \Ii_{n-k} A$ by
\begin{equation}\label{hat.phi}
\hat{\phi}: \omega'' \mapsto \phi\left(\omega''\sw 0 \omega''\sw 1\su 1\right) \omega''\sw 1\su 2.
\end{equation}
Note that $\hat{\phi}(\omega'') = \phi(\omega'')$, for all $\omega''\in \Omega^{n-k}B$. Further note that, for all $\omega'\in \Omega^k B$,
\begin{equation}\label{hat}
\widehat{\ell^k_{\pi_\omega^B}(\omega ')} = \ell^k_{\pi_\omega^A}(\omega ').
\end{equation}
Indeed, take any $\omega''\in \Omega^{n-k} A$. Then
\begin{eqnarray*}
\widehat{\ell^k_{\pi_\omega^B}(\omega ')}(\omega'') &=& \pi_\omega^B\left(\omega'\wedge \omega''\sw 0 \omega''\sw 1\su 1\right) \omega''\sw 1\su 2 = \pi_\omega^A\left(\omega'\wedge \omega''\sw 0 \omega''\sw 1\su 1\right) \omega''\sw 1\su 2\\
&=& \pi_\omega^A\left(\omega'\wedge \omega''\sw 0 \omega''\sw 1\su 1\omega''\sw 1\su 2\right) = \pi_\omega^A\left(\omega'\wedge \omega''\right) = \ell^k_{\pi_\omega^A}(\omega ')(\omega ''),
\end{eqnarray*}
where the third equality follows by the right $A$-linearity of $\pi_\omega^A$ and the fourth one by the fact that $h\su 1 h\su 2 = \eps(h)$, where $\eps$ is the counit of $H$.
Take any $\phi \in \Ii_{n-k} B$. By assumption, there exists a unique $\omega'\in \Omega^k A$, such that
\begin{equation}\label{inverse.om}
\hat{\phi} = \ell_{\pi_\omega^A }^k(\omega ').
\end{equation}
Hence, for all $\omega'' \in \Omega^{n-k}B$,
\begin{equation}\label{phi.omega}
\phi(\omega'') = \hat{\phi}(\omega'') = \pi_{\omega}^A (\omega' \wedge \omega''),
\end{equation}
i.e.\hspace{3pt}$\omega' \wedge \omega'' = {\pi_{\omega}^A}^{-1}(\phi(\omega''))$. Since $\phi(\omega'') \in B$, $ {\pi_{\omega}^A}^{-1}(\phi(\omega''))\in \Omega^n B$, and assumption (c) implies that $\omega'\in \Omega^kB$.
For all $k=1,\ldots , n-1$ define the maps
$$
\Phi_k : \Ii_{n-k} B \to \Omega^k B, \qquad \phi \mapsto \omega',
$$
where $\omega'$ is given by \eqref{inverse.om}. We claim that $\Phi_k $ is the inverse of $\ell_{\pi_\omega^B }^k$. For any $\phi\in \Ii_{n-k} B$,
$$
\ell_{\pi_\omega^B }^k\left(\Phi_k\left(\phi\right)\right) = \ell_{\pi_\omega^B }^k\left(\omega'\right) = \ell_{\pi_\omega^A }^k\left(\omega'\right)\mid_{\Omega^{n-k}B} = \hat{\phi} \mid_{\Omega^{n-k}B} = \phi.
$$
In the converse direction, for all $\omega'\in \Omega^{k} B$,
$$
\Phi_k\left( \ell_{\pi_\omega^B }^k\left(\omega'\right)\right) = \omega'' \in \Omega^kB,
$$
where
$$
\ell_{\pi_\omega^A }^k\left(\omega''\right) = \widehat{\ell_{\pi_\omega^B }^k\left(\omega'\right)} = \ell_{\pi_\omega^A }^k\left(\omega'\right),
$$
by \eqref{hat}. Since the maps $ \ell_{\pi_\omega^A }^k$ are bijective, $\omega'' = \omega'$, as required.
\end{proof}
\begin{remark}
The assumptions of Lemma~\ref{lem.princ} mean that both $A$ and $B$ have integrable calculi of equal dimensions.
Geometrically this limits the applicability of Lemma~\ref{lem.princ} to actions of finite quantum groups.
\end{remark}
In view of the construction of the isomorphisms $\Phi_k$ in Lemma~\ref{lem.princ}, the divergence $\nabla_B : \Ii_1\to B$ corresponding to the integrable calculus $\Omega B$ is related to the divergence $\nabla_A: \Ii_1\to A$ by
\begin{equation}\label{nablas}
\nabla_B(\phi) = \nabla_A(\hat{\phi}), \qquad \mbox{for all}\quad \phi \in \Ii_1 B.
\end{equation}
In particular this means that the image of $\nabla_B$ is contained in the image of $\nabla_A$, hence it is contained in the kernel of the integral $\Lambda_A : A\to \coker \nabla_A$. By the universal property of cokernels, there must exist a unique map $\chi: \coker \nabla_B \to \nabla_A$ connecting the integrals on $A$ and $B$ by
\begin{equation}\label{integrals}
\Lambda_A\mid_B = \chi\circ\Lambda_B
\end{equation}
In favourable situations, the formula \eqref{integrals} allows one to integrate on $B$ using integration on $A$.
\section{The noncommutative pillow}
\setcounter{equation}{0}
In this section we study the noncommutative version of one of the prime examples of a {\em good orbifold}, (meaning a singular space obtained by a non-free action of a finite group on a smooth manifold) known as a {\em pillow orbifold} and obtained as a quotient of the two-torus by an action of the cyclic group $\ZZ_2$. \cite[Chapter~13]{Thu:geo}.
\subsection{Two-dimensional integrable differential calculus over the noncommutative pillow}\label{sec.dif}
The coordinate algebra of the noncommutative torus, $\cO(\TT^2_\theta)$, is a complex $*$-algebra generated by unitary $V, W$ subject to the relation
\begin{equation}\label{torus}
VW = \lambda WV, \qquad \mbox{where}\quad \lambda=\exp(2\pi i\theta).
\end{equation}
We assume that $\theta$ is irrational. The algebra map
\begin{equation} \label{sigma.auto}
\sigma: \cO(\TT^2_\theta)\to \cO(\TT^2_\theta), \qquad V \mapsto V^* \quad \mbox{and} \quad W\mapsto W^*,
\end{equation}
is an involutive automorphism, and hence it splits $\cO(\TT^2_\theta)$ into a direct sum $\cO(\TT^2_\theta)=\cO(\TT^2_\theta)_+\oplus \cO(\TT^2_\theta)_-$, where $a\in \cO(\TT^2_\theta)_\pm$ if and only if $\sigma(a) = \pm a$. This splitting means also that $\cO(\TT^2_\theta)$ is a $\ZZ_2$-graded algebra. The fixed point subalgebra $\opil:= \cO(\TT^2_\theta)_+$ is the coordinate algebra of the {\em noncommutative pillow orbifold} introduced in \cite{BraEll:sph}. It is generated by $x= V+V^*$ and $y=W+W^*$, but it is also convenient to consider $z=VW^* + V^*W$, which can be expressed in terms of $x$ and $y$ via
\begin{equation}\label{eq.z}
(1-\lambda^2)z = xy - \lambda yx.
\end{equation}
As a vector space $\opil$ is spanned by $1$ and
\begin{equation}\label{ab}
a_{m,n} = V^mW^n +V^{*m}W^{*n} \quad b_{m,n} = V^mW^{*n} +V^{*m}W^{n}, \quad m,n\in \NN,\; m+n>0.
\end{equation}
$\opil$ can also be identified with an algebra generated by self-adjoint $x,y,z$ subject to relations \eqref{eq.z} and
\begin{subequations}\label{xy.radial}
\begin{equation}\label{eq.xy}
xz-\bar{\lambda}zx = (1-\bar{\lambda}^2) y, \qquad zy-\bar{\lambda}yz = (1-\bar{\lambda}^2) x,
\end{equation}
\begin{equation}\label{radial}
x^2 + y^2 +\bar{\lambda}z^2 -xzy =2(1+\bar{\lambda}^2).
\end{equation}
\end{subequations}
It is equation \eqref{radial} that allows one to interpret $\pil$ as a deformation of the pillow orbifold.
Using the relations (\ref{eq.z})--(\ref{xy.radial}) we can write a linear basis for $\opil$ in the generators $x,y$ as
$x^k y^l$ and $x^k (y x) y^l$, $k,l\in \NN$. Therefore, the subspace $\Vv(n)$ of words in generators of length at
most $n$ has a basis
$$
\{x^k y^l, \, x^i (y x) y^j, \; |\; k+l \leq n, \; i+j \leq n-2\}.
$$
Consequently,
$$
\dim \Vv(n) = {{n+2}\choose{2}} + {{n}\choose{2}} = n^2 + n +1,
$$
so $\opil$ has the Gelfand-Kirillov dimension two.
We now turn to the description of $\cO(\TT^2_\theta)_-$. Let
\begin{equation}\label{hats}
\hx = V-V^*, \qquad \hy=W-W^*, \qquad \mbox{and} \quad \hz=VW^* - V^*W.
\end{equation}
Clearly, $\hx, \hy, \hz \in \cO(\TT^2_\theta)_-$, and, in fact
\begin{lemma}\label{lem.a-}
The elements $\hx, \hy, \hz$ generate $\cO(\TT^2_\theta)_-$ as a left (resp.\hspace{3pt}right) $\opil$-module.
\end{lemma}
\begin{proof}
$\cO(\TT^2_\theta)_-$ is spanned by
\begin{equation}\label{hab}
\ha_{m,n} = V^mW^n -V^{*m}W^{*n},\quad \hb_{m,n} = V^mW^{*n} -V^{*m}W^{n}, \quad m,n\in \NN,\; m+n>0.
\end{equation}
We prove by induction that $\ha_{m,n}$ are elements of the left $\opil$-module generated by $\hx, \hy, \hz$ (the proof for $\hb_{m,n}$ is similar). First,
$\ha_{1,0} = \hx$, $\ha_{0,1} = \hy$ and
$$
\ha_{1,1} = VW - V^*W^* = (V+V^*)(W-W^*) - V^*W +VW^* = x\hy + \hz.
$$
Next, we fix an $n$ and assume that
\begin{equation}\label{ind}
\ha_{m,n} = a^x_{m,n}\hx + a^y_{m,n}\hy + a^z_{m,n}\hz, \qquad a^x_{m,n}, a^y_{m,n}, a^z_{m,n} \in \opil,
\end{equation}
for all $m\leq k$. Then, in view of the unitarity of $V$,
\begin{eqnarray*}
\ha_{k+1, n} &=& V^{k+1}W^n -V^{*k+1} W^{*n} \\
&=& (V+V^*) (V^{k}W^n - V^{*k} W^{*n}) - V^{k-1}W^n + V^{*k-1} W^{*n}\\
&=& (xa^x_{k,n} - a^x_{k-1,n})\hx + (xa^y_{k,n} - a^y_{k-1,n})\hy + (xa^z_{k,n} - a^z_{k-1,n})\hz,
\end{eqnarray*}
hence $\ha_{k+1,n}$ is in the module generated by $\hx, \hy, \hz$. Using the unitarity of $V$ and $W$ and relations \eqref{torus} one finds:
\begin{equation}\label{hat.y}
\hx y = \lambda y\hx + (1-\lambda^2)\hz , \qquad \hy y = y\hy, \qquad \hz y =\bar{ \lambda} y\hz - (1-\bar{\lambda}^2)\hx .
\end{equation}
Let us fix an $m$ and assume that \eqref{ind} is true for all $n\leq k$. Then, by the unitarity of $W$,
\begin{eqnarray*}
\ha_{m, k+1} &=& V^{m}W^{k+1} -V^{*m} W^{*k+1} \\
&=& (V^{m}W^{k} -V^{*m} W^{*k})(W+W^*) - V^{m}W^{k-1} +V^{*m} W^{*k-1}\\
&=& a^x_{m,k}\hx y + a^y_{m,k}\hy y + a^z_{m,k}\hz y - (a^x_{m,k-1}\hx + a^y_{m,k-1}\hy + a^z_{m,k-1}\hz).
\end{eqnarray*}
In view of the relations \eqref{hat.y}, $\ha_{m,k+1}$ is in the left $\opil$-module generated by $\hx, \hy, \hz$.
\end{proof}
We are now ready to construct a connected two-dimensional differential calculus $\Omega(\pil)$ on $\opil$. Set $\Omega^1 (\pil) = \cO(\TT^2_\theta)_-\oplus \cO(\TT^2_\theta)_-$ and $\Omega^ 2 (\pil) = \opil$, $\Omega^n (\pil) =0$, for all $n>2$, and define the product of elements in $\Omega^1 (\pil)$, by
\begin{equation}\label{prod.b}
(a_1, a_2)\wedge (a_3, a_4) = a_1a_4 - a_2a_3\in \opil, \qquad a_k\in \cO(\TT^2_\theta)_-
\end{equation}
The following linear endomorphisms of $\cO(\TT^2_\theta)$,
\begin{equation}\label{partial}
\partial_V (V^mW^n) = imV^mW^n, \qquad \partial_W (V^mW^n) = inV^mW^n,
\end{equation}
are derivations i.e.\hspace{3pt}they satisfy the Leibniz rule. The factor $i$ in the above formulae ensures that $\partial_{V,W}\circ * = *\circ \partial_{V,W}$. Furthermore $\partial_{V,W}$ commute among themselves and anticommute with the automorphism $\sigma$ \eqref{sigma.auto}, i.e.\hspace{3pt}
$$
\partial_V\circ \partial_W = \partial_W\circ \partial_V \quad \mbox{and} \quad \partial_{V,W} \circ \sigma = - \sigma \circ \partial_{V,W}.
$$
All this means that $\partial_{V,W} (\cO(\TT^2_\theta)_\pm) \subseteq \cO(\TT^2_\theta)_\mp$ and that the maps
$$
d: \opil \to \Omega^1 (\pil), \quad a\mapsto (\partial_V(a), \partial_W(a)),
$$
and
$$
d: \Omega^1 (\pil)\to \opil, \quad (a_1,a_2)\mapsto \partial_V(a_2) - \partial_W(a_1),
$$
define a complex. The combination of the derivation property of $\partial_{V,W}$ together with the definition of multiplication in \eqref{prod.b} ensure that the maps $d$ satisfy the graded Leibniz rule. Clearly $\partial_V (V^mW^n) =\partial_W (V^mW^n) = 0$ if and only if $m=n=0$, hence $\Omega(\pil)$ is a connected calculus. With these at hand we can state:
\begin{theorem}\label{thm.pillow.Poincare}
$\Omega(\pil)$ is a connected integrable differential calculus, hence the noncommutative pillow algebra is differentially smooth.
\end{theorem}
\begin{proof}
In order to apply Lemma~\ref{lemma.Poincare} we need to check if $\cO(\TT^2_\theta)_-\cO(\TT^2_\theta)_- = \opil$ and that $\Omega (\pil)$ is a differential calculus (not just a differential graded algebra) over $\opil$. To check the former, observe that
\begin{equation}\label{xyz}
\hat{x}^2 + \hat{y}^2 -\bar{\lambda}\hat{z}^2 - \hat{x}z\hat{y} = 2(\bar{\lambda}^2-1),
\end{equation}
which ensures that $1\in \cO(\TT^2_\theta)_-\cO(\TT^2_\theta)_-$ and, consequently, that $\cO(\TT^2_\theta)_-\cO(\TT^2_\theta)_- = \opil$. Straightforward calculations that use the defining relations of the noncommutative torus lead to the following relations supplementing \eqref{hat.y}
\begin{subequations}\label{hat.xz}
\begin{equation}\label{hat.x}
\hx x = x\hx, \qquad \hy x = \bar{\lambda} x\hy - (\lambda-\bar{\lambda})\hz , \qquad \hz x ={ \lambda} x\hz + (\lambda-\bar{\lambda})\hy .
\end{equation}
\begin{equation}\label{hat.z}
\hx z = \bar{\lambda} z\hx + (1-\bar{\lambda}^2)\hy , \qquad \hy z ={ \lambda} z\hy - (\lambda-\bar{\lambda})\hx , \qquad \hz z = z\hz .
\end{equation}
\end{subequations}
In view of the definitions of $\partial_{V,W}$ and $d$ one finds $(i\hx , 0) = dx$ and $(0,i\hy) = dy$. This observation combined with equations \eqref{hat.y} and \eqref{hat.xz} yields
$$
(i\hy, 0) = \frac{1}{1-\bar{\lambda}^2} (dx z - \bar{\lambda}z dx), \qquad
(i\hz, 0) = \frac{1}{1-{\lambda}^2} (dx y - {\lambda}y dx),
$$
and
$$
(0,i\hx) = \frac{1}{\bar{\lambda}-{\lambda}} (dy z - {\lambda}z dy), \qquad (0,i\hz) = \frac{1}{1-{\lambda}^2} (\lambda dy x - x dy).
$$
Since $\cO(\TT^2_\theta)_-$ is generated by $\hx, \hy, \hz$ and $d$ satisfies the Leibniz rule, this proves that $\Omega^1 (\pil) = \opil d\opil$. Finally,
\begin{eqnarray*}
&&\frac{1}{2(\blambda^2 -1)}\left( (\hx, 0)(0,\hx) + (\hy, 0)(0, \hy) - \lambda(\hz, 0)(0,\hz) - (\hx, 0)(0,z\hy)\right)\\
&& = \frac{1}{2(\blambda^2 -1)}\left(\hat{x}^2 + \hat{y}^2 -\bar{\lambda}\hat{z}^2 - \hat{x}z\hat{y}\right) =1,
\end{eqnarray*}
by \eqref{xyz}. Hence
$$
1\in \Omega^1 (\pil)\wedge \Omega^1 (\pil) = \opil d\opil\wedge \opil d\opil = \opil d\opil \wedge d\opil,
$$
which implies that $ \opil d \opil \wedge d \opil=\opil=\Omega^2(\pil)$, as required for a differential calculus. Therefore, Lemma~\ref{lemma.Poincare} implies that $\Omega(\pil)$ is an integrable differential calculus (with integrating form 1) over the noncommutative pillow as required.
\end{proof}
Since
$$
\partial_V(\ha_{m,n}) = ima_{m,n}, \quad \partial_W(\ha_{m,n}) = ina_{m,n}, \quad \partial_V(\hb_{m,n}) = imb_{m,n}, \quad \partial_W(\hb_{m,n}) = -inb_{m,n},
$$
where $a_{m,n}$, $b_{m,n}$ are given by \eqref{ab} and $\ha_{m,n}$, $\hb_{m,n}$ are given by \eqref{hab}, the image of the divergence $\nabla$ which obviously coincides with the image of $d: \Omega^1(\pil)\to \opil$, contains all elements of $\opil$ except those that are in the subspace spanned by the identity, i.e. $\opil = \CC \oplus \im\nabla$. Therefore $\coker \nabla =\CC$ and the integral $\Lambda: \opil\to \CC$ comes out as
$$
\Lambda(a) =
\begin{cases}
\alpha, & \mbox{if $a=\alpha 1$, for $\alpha \in \CC$} \\
0, & \mbox{otherwise},
\end{cases}
$$
i.e.\hspace{3pt}it is equal to the trace on $\opil$ obtained by the restriction of the trace on the noncommutative torus.
\begin{remark}
Theorem~\ref{thm.pillow.Poincare} can be also proven with the help of Lemma~\ref{lem.princ}. The $\ZZ_2$-action on the noncommutative
two-torus can be interpreted as the following coaction of $\CC \ZZ_2$ on $\cO(\TT^2_\theta)$,
$$
\varrho(V) = \frac{1}{2} (V\! +\! V^*) \otimes 1 + \frac{1}{2} (V\! -\! V^*) \otimes u, \;\;
\varrho(W) = \frac{1}{2} (W\! +\! W^*) \otimes 1 + \frac{1}{2} (W\! -\! W^*) \otimes u,
$$
where $u$ is the generator of $\ZZ_2$, $u^2=1$. The coinvariant subalgebra coincides with $\cO(\pil)$. Equation \eqref{xyz} assures that this coaction is principal. Furthermore, it extends to the standard calculus $\Omega(\TT^2_\theta)$ on $\cO(\TT^2_\theta)$ freely generated by two central, anticommuting forms $\omega_V = V^*dV$ and $\omega_W = W^*dW$. One can check that $\Omega(\TT^2_\theta)$ is integrable with a volume form $\omega_V\wedge \omega_W$, invariant under the coaction $\varrho$. Furthermore, one can verify that the invariant part of $\Omega(\TT^2_\theta)$ coincides with the calculus $\Omega(\pil)$ constructed above and that the condition (c) in Lemma~\ref{lem.princ} is also satisfied.
\end{remark}
\subsection{The noncommutative pillow as a complex manifold}
As before we use the direct sum decomposition $\cO(\TT^2_\theta) = \cO(\TT^2_\theta)_+\oplus \cO(\TT^2_\theta)_-$, with $\opil=\cO(\TT^2_\theta)_+$, determined by the automorphism \eqref{sigma.auto}. $(\Omega (\pil), d)$ is the differential calculus constructed in Section~\ref{sec.dif}.
\begin{proposition}\label{prop.cotan}
The module $\cO(\TT^2_\theta)_-$ is a non-free finitely generated projective left (resp.\hspace{3pt}right) module over the non-commutative pillow algebra $\opil$. Consequently, the cotangent bundle over $\pil$ whose sections are given by $\Omega^1(\pil)$ is non-trivial.
\end{proposition}
\begin{proof}
The fact that $\cO(\TT^2_\theta)_-\cO(\TT^2_\theta)_-=\opil$ means that $\cO(\TT^2_\theta)_-$ is a finitely
generated projective left and right $\opil$-module. The formula \eqref{xyz} yields an idempotent
\begin{equation}\label{idempotent}
\mathbf{e} = \frac{1}{2(\bar{\lambda}^2 -1)}
\begin{pmatrix} \hx \cr \hy \cr \hz \end{pmatrix}
\begin{pmatrix} \hx, & \hy -\hx z, & -\bar\lambda \hz \end{pmatrix},
\end{equation}
which identifies $\cO(\TT^2_\theta)_-$ as a submodule of the free module $\opil^3$. The matrix trace of $\mathbf{e}$ comes out as
$$
\tr \mathbf{e} = 3 - \frac12 (x^2+\lambda z^2) = 1 - \frac12 (a_{2,0}+\lambda^2b_{2,2}),
$$
where $a_{m,n}$, $b_{m,n}$ are given by \eqref{ab}. Therefore, $\Lambda(\tr \mathbf{e}) = 1$. However, since $\Lambda$ is a trivial trace this does not yet guarantee that $\cO(\TT^2_\theta)_-$ is not a free $\cO(\pil)$-module. To show the non-freeness we need to consider a family of traces on $\opil$, which arise from twisted traces on the algebra of the noncommutative torus. More specifically, if for a given involutive automorphism $\sigma$ of $\cO(\TT^2_\theta)$ there is a functional $\htau$ such that
$$ \htau ( a b) = \htau( \sigma(b) a), \;\;\; \forall a,b \in \cO(\TT^2_\theta), $$
then $\htau$ restricted to the invariant subalgebra is a trace. For the automorphism \eqref{sigma.auto}, there are four linearly independent twisted traces (in addition to $\Lambda$) \cite{Walters}:
$$ \htau_{ij} (V^\alpha W^\beta) = e^{\pi i \theta \alpha \beta} \delta_i^{\bar\alpha} \delta_j^{\bar\beta}, \qquad i,j =0,1,
$$
where $\bar \alpha = \alpha \! \mod \! 2$ and $\bar \beta = \beta \! \mod \! 2$. Evaluating the extension of these traces to the projector \eqref {idempotent} we obtain the following
results:
$$ \htau_{00}( \tr \mathbf{e}) = - 1,
\;\;\; \htau_{01}( \tr \mathbf{e}) = 0,
\;\;\; \htau_{10}( \tr \mathbf{e}) = 0, \;\;\;
\htau_{11}(\tr \mathbf{e}) = 0. $$
It is the nontrivial value of $\htau_{00}$ on $ \tr \mathbf{e}$, which shows that the module $\cO(\TT^2_\theta)_-$ determined by $\mathbf{e}$ is non-free. Therefore, the module of sections of the cotangent bundle $\Omega^1(\pil)$, which is given as
$\cO(\TT^2_\theta)_- \oplus \cO(\TT^2_\theta)_-$, is also non-trivial.
\end{proof}
Following \cite{KhaLan:hol} (cf.\ \cite{BegSmi:com}), by a complex structure on $\opil$ corresponding to the differential calculus $(\Omega (\pil), d)$ we understand the bi-grading decomposition of $\Omega (\pil)$,
$$
\Omega^n(\pil) = \bigoplus_{p+q =n} \Omega^{(p,q)}(\pil),
$$
a $*$-algebra structure on $\Omega (\pil)$ such that $*:\Omega^{(p,q)}(\pil)\to \Omega^{(q,p)}(\pil)$, and the decomposition $d= \delta +\bdelta$ into differentials $\delta: \Omega^{(p,q)}(\pil)\to \Omega^{(p+1,q)}(\pil)$, $\bdelta: \Omega^{(p,q)}(\pil)\to \Omega^{(p,q+1)}(\pil)$ such that
\begin{equation}\label{partial.bpartial}
\delta(a)^* = \bdelta(a^*), \qquad \mbox{for all $a\in \Omega (\pil)$}.
\end{equation}
As explained in \cite{KhaLan:hol}, up to a conformal factor in a metric, the complex structure associated to the calculus on $\cO(\TT^2_\theta)$ corresponding to derivations $\partial_V$ and $\partial_W$ is determined by the derivations
\begin{equation}\label{partial.tau}
\partial_\tau= \frac{1}{\tau-\btau}(-\btau \partial_V +\partial_W), \qquad \bpartial_\tau= \frac{1}{\tau-\btau}(\tau \partial_V -\partial_W),
\end{equation}
where $\tau \in \CC\setminus\RR$. With the help of these derivations one constructs complex structures for the calculus $\Omega (\pil)$ over the non-commutative pillow manifold as follows:
$$\Omega_\tau^{(1,0)}(\pil) = \Omega_\tau^{(0,1)}(\pil) =\cO(\TT^2_\theta)_-, \quad \Omega_\tau^{(2,0)}(\pil) = \Omega_\tau^{(0,2)}(\pil) =0, \quad \Omega_\tau^{(1,1)}(\pil) = \opil.
$$
The $*$-structure on $\cO(\TT^2_\theta)_-$ is that of the $*$-structure of the noncommutative torus, while the $*$-structure on $\Omega_\tau^{(1,1)}(\pil) $ is opposite to that on $\opil$. The holomorphic and anti-holomorphic differentials are
$$
\delta_\tau:= \partial_\tau\mid_\opil: \opil\to \Omega_\tau^{(1,0)}(\pil), \qquad \delta_\tau:= \partial_\tau\mid_{\cO(\TT^2_\theta)_-}:\Omega_\tau^{(0,1)}(\pil)\to \opil,
$$
and
$$
\bdelta_\tau:= \bpartial_\tau\mid_\opil: \opil\to \Omega_\tau^{(0,1)}(\pil), \qquad \bdelta_\tau:= -\bpartial_\tau\mid_{\cO(\TT^2_\theta)_-}:\Omega_\tau^{(1,0)}(\pil)\to \opil.
$$
One can understand this complex structure as the decomposition of the differential $d$ discussed in Section~\ref{sec.dif} as follows. Let us write $(\ha,\hb)_\tau$ for an element of $\Omega_\tau^{(1,0)}(\pil)\oplus \Omega_\tau^{(0,1)}(\pil)$ and $\omega_\tau$, $\bomega_\tau$ for a basis of the direct sum $\Omega_\tau^{(1,0)}(\pil)\oplus \Omega_\tau^{(0,1)}(\pil)$, so that
$$
(\ha,\hb)_\tau = \ha\omega_\tau +\hb\bomega_\tau, \qquad \ha,\hb\in \cO(\TT^2_\theta)_-.
$$
Similarly, let us write $(a,b)$ for an element of $\Omega^1 (\pil)$ and $\omega_V$, $\omega_W$ for a basis of the direct sum decomposition into the $\cO(\TT^2_\theta)_-$. In this notation, for all $a\in \opil$,
$$
\partial_\tau(a)\omega_\tau + \bpartial_\tau(a)\bomega_\tau = da = \partial_V(a)\omega_V + \partial_W(a)\omega_W.
$$
By comparing coefficients at $\partial_V$ and $\partial_W$ one arrives at the following invertible transformation
$$
\begin{pmatrix}
\omega_\tau \cr \bomega_\tau
\end{pmatrix}
= \frac{1} {\tau-\btau} \begin{pmatrix} 1 & \tau \cr 1 & \btau \end{pmatrix} \begin{pmatrix}
\omega_V \cr \omega_W
\end{pmatrix},
$$
i.e.\
$$
(\ha,\hb)_\tau = \frac{1} {\tau-\btau} (\ha +\hb, \tau\ha +\btau\hb).
$$
In particular, this interpretation guarantees that the differential graded algebra determined from $\Omega^{(p,q)}_\tau (\pil)$
is a differential calculus over $\opil$.
\subsection{Spectral geometry of the noncommutative pillow}
In the classical situation the quotient of the torus by the action of the cyclic group $\ZZ_2$, which was described
in Section~\ref{sec.dif}, gives an orbifold with four corners. To see the striking difference between the commutative
and noncommutative cases it is convenient to use the same language, adapted to describe both commutative and
noncommutative manifolds. At present, the best candidate for such an approach is the notion of a spectral triple
\cite{Connes}, which is modelled on the definition of Riemannian spin geometry.
In what follows, we shall briefly recall the construction of a spectral triple for the noncommutative torus (cf.\ \cite{Dab:spi}),
then we shall find real spectral triples for the noncommutative pillow and see whether the axioms of spectral triples
allow us to determine whether the latter is a manifold or an orbifold.
Let $\storus$ denote the Fr\'echet algebra of smooth elements of the noncommutative torus, which contains
all series $ \sum_{m,n} a_{mn} V^m W^n $ with $\{a_{mn}\}_{m,n \in \ZZ} \in \CC$ a rapidly decreasing sequence.
Fix $\epsilon = 0$ or $\epsilon = \frac{1}{2}$. Consider a separable Hilbert space $\cH$ with an orthonormal basis
$e_{m,n}$, $m,n \in \ZZ \!+\! \epsilon$ (so there are four possibilities) and the representation $\pi$ of the
algebra $\storus$ given on the generators of the noncommutative torus as:
$$ \pi(V) e_{m,n} = e^{\pi i \theta n} e_{m+1,n}, \;\;\;\;\;\;\; \pi(W) e_{m,n} = e^{-\pi i \theta m} e_{m,n+1}. $$
\begin{theorem}[\cite{PaSi}]
The following datum $(\storus, \cH \otimes \CC^2, D, \gamma, J)$ gives a real equivariant spectral
triple over the noncommutative torus. Here, the representation of the algebra is taken to be diagonal,
the operators $\gamma$ and $J$ are:
$$ \gamma e_{m,n,\pm} = \pm e_{m,n,\pm}, \;\;\;\;\;\;\; J e_{m,n,\pm} = \mp e_{-m,-n,\mp}, $$
and the Dirac operator $D$ is:
$$ D e_{m,n,-} = (m + \tau n) e_{m,n,+}, \;\;\;\;\; D e_{m,n,+} = (m + \tau^* n) e_{m,n,-}, $$
for any $\tau$, $|\tau|=1$, with a non-zero imaginary part.
\end{theorem}
Observe that four different possibilities for the choice of $m,n$ (integer or half-integer) correspond to four different
spin structures over the classical torus \cite{PaSi}. Next, following the ideas of \cite{Adalgott} we can construct the restriction
of these spectral triples to the invariant subalgebra of $\storus$ with respect to the $\ZZ_2$-action determined by
(the extension of) the automorphism $\sigma$ \eqref{sigma.auto}. As the first step we lift the action to the Hilbert space
$\cH \otimes \CC^2$.
\begin{lemma}
For any of the four spin structures and any $\tau$,
$$ \rho e_{m,n,\pm} = \pm e_{-m,-n,\pm}, $$
is unique (up to sign) lift of the action of $\ZZ_2$ to the Hilbert space $\cH \otimes \CC^2$, which commutes with $J$, $D$ and $\gamma$ and implements the action
of $\sigma$ \eqref{sigma.auto} on the algebra $\storus$.
\end{lemma}
This lemma is a direct consequence of the application of Theorem 2.7 of \cite{Adalgott} when one restricts to a subalgebra
of the noncommutative torus. As a consequence, we have:
\begin{corollary}
Each of the spin structures over the noncommutative torus restricts to an irreducible real spectral triple over the algebra of the
noncommutative pillow by taking $\spil = \storus_+, (\cH \otimes \CC^2)_+, D, \gamma, J$, where
$\storus_+$ is the invariant subalgebra of (the extension of) the automorphism $\sigma$ \eqref{sigma.auto}, $(\cH \otimes \CC^2)_+$ is the invariant part of the Hilbert space (the eigenspace of
$\rho$ with eigenvalue $1$) and $D,J,\gamma$ are restrictions of the operators to that space.
\end{corollary}
So far we have constructed and classified all real spectral triples over the noncommutative pillow, which are restrictions of equivariant
real spectral triples over the noncommutative torus. Such a construction guarantees that almost all conditions, which are satisfied for a spectral
triple over the torus (such as finiteness and the behaviour of the spectrum of the Dirac operator, etc.) are automatically satisfied. There exists
however one exception, which is the crucial element making it possible to distinguish manifolds from orbifolds. This is the orientability condition
(see \cite{ReVa}), which is satisfied for all commutative spectral geometries and for the noncommutative torus. The original condition requires the existence of a Hochschild cycle, so that its image under $\pi$ gives the chirality operator $\gamma$ (in the even-dimensional case). For the noncommutative pillow we can establish a milder version of the orientability condition, by constructing explicitly a cycle
(though not a Hochschild cycle) whose image is the chirality operator $\gamma$.
\begin{theorem}
For any value of the conformal structure $\tau$ with the non-zero imaginary part, $|\tau|=1$, and any value of the deformation parameter
$\lambda = e^{2 \pi i \theta}$, provided that $\lambda^4 \not= 1$, there exists a cycle
$\omega = \sum_i a_i \ot b_i \ot c_i \in \spil \ot \spil \ot \spil$,
such that:
$$\sum \pi(a_i)[D,\pi(b_i)][D,\pi(c_i)] = \gamma .$$
\end{theorem}
\begin{proof}
We write $\tau = \cos \phi + i \sin \phi$. Then by explicit computation we verify that
$$
\begin{aligned}
\frac{1}{4}&(1 - \lambda^4) x [D,y] [D,z] \\
& - \frac{1}{2} \lambda^2 \cos\phi (1+\lambda^2)(2 \lambda^4 \cos\phi-\lambda^4 +2\cos\phi -1)\, x [D,z] [D,y]
-2\lambda^2 \\
& - \frac{1}{2} (2 \lambda^6 \cos^2\phi -\lambda^6 \cos\phi +2 \lambda^4\cos^2\phi + 2\lambda^4 \\
& \quad \quad - 5 \lambda^4\cos\phi
- \lambda^3 \cos\phi + 2 \lambda^2 \cos^2\phi + 2 \cos^2\phi - \cos\phi) \lambda^2 \, y [D,x] [D,z] \\
&+ \frac{1}{2} \lambda (1+\lambda^2)(1+\lambda^4)(2\cos\phi -1)(\cos\phi -1)\, z [D,x] [D,y] \\
& - \frac{1}{2} (1-\lambda^4) (\cos\phi -1) \lambda^2 \, z [D,y] [D,x] \\
&+\frac{1}{4} (1+\lambda^2)(2\cos\phi -1)(2\lambda^4\cos\phi +2 \cos\phi - \lambda^4 -\lambda^2) \lambda \, [D,x] y [D,z] \\
&+\frac{1}{2} \cos\phi \lambda^2 (2 \lambda^6 \cos\phi + 2 \lambda^4\cos\phi + 2 \lambda^2 \cos\phi + 2 \cos\phi \\
&\quad\quad - \lambda^6 - \lambda^2 -2 -4 \lambda^4) \, [D,z] [D,y] x \\
&- \frac{1}{2} \lambda^2 (1+\lambda^2)(1+\lambda^4)(2 \cos\phi -1)(\cos\phi -1) \, [D,x] z [D,y] =\\
& = \sin\phi (2\cos\phi -1) (\lambda^2 -1)^3 (1+\lambda^2) \gamma.
\end{aligned}
$$
Although the above formula uses terms like $[D,x] y [D,z]$, it can be easily converted to the desired form using
the Leibniz rule.
\end{proof}
\begin{remark}
The above cycle is certainly not unique and we provide its explicit form only to demonstrate its existence. Certainly, it
could be simplified or rewritten in a more convenient form.
Observe that for the image of this cycle to be nonvanishing we must have $\sin\phi \not= 0$ (which is the same condition
as for the noncommutative torus, see \cite{PaSi}) as well as $\lambda^4 \not= 1$. It is interesting that even for some
rational values of $\theta$ we have a non-degenerate cycle for the respective spectral triple.
\end{remark}
\begin{remark}
The above demonstrated cycle is not a Hochschild cycle. It remains an open problem whether a Hochschild cycle of the same type
exists for the noncommutative pillow. However, we would like to stress that in \cite{ReVa} the nonexistence of an orientation cycle for an
orbifold is demonstrated for any cycle (not only for the Hochschild ones).
\end{remark}
\section{Quantum cones}\setcounter{equation}{0}
\subsection{Two-dimensional integrable differential calculi over quantum cones}\label{sec.dif.cone}
Homological smoothness and the complex differential geometry of quantum cones were recently established and studied in
\cite{Brz:com}. In this section we prove that the two-dimensional differential calculus over the quantum cone described in \cite{Brz:com} is integrable and thus establish differential smoothness of quantum cone algebras.
Let $N$ be a positive integer. The coordinate algebra of the {\em quantum cone} $\cO(C^N_{q,\kappa})$ is defined as a complex $*$-algebra generated by self-dual $a$ and $b, b^*$, which satisfy relations
\begin{equation}\label{cone}
ab = q^N ba + \kappa [N]_q b, \qquad bb^* = \prod_{l=0}^{N-1} (q^{-l}a + \kappa [-l]_q), \qquad b^*b = \prod_{l=1}^{N} (q^{l}a + \kappa [l]_q),
\end{equation}
where $q>0$, $\kappa \in \RR$ are parameters and, for all $n\in \ZZ$,
\begin{equation}\label{q.integers}
[n]_q := \frac{1-q^{n}}{1-q}
\end{equation}
denotes $q$-integers. For $N\neq 1$, a linear basis of the space $\Vv(n)$ spanned by words of generators of length at most $n$ is
$$
\{a^ib^j, \, a^{k}b^{*l+1} \; |\; i+j \leq n ,\, k+ l \leq n-1, \},
$$
hence
$$
\dim \Vv(n) = {n+2 \choose n} + {n+1 \choose n-1} = (n+1)^2.
$$
Consequently,
$$
\gk (\cO(C^N_{q,\kappa})) =2.
$$
For $N=1$, $a$ can be expressed in terms of $bb^*$, so it becomes redundant. In this case we denote $b=z$ and $b^*=z^*$. Then relations become
\begin{equation}\label{disc}
z^*z - qzz^* = \kappa,
\end{equation}
and the resulting algebra is the coordinate algebra of the {\em quantum disc} \cite{KliLes:two}, which we denote by $\cO(D_{q,\kappa})$. The linear basis is $z^iz^{*j}$, $i,j\in \NN$, hence the quantum disc algebra has Gelfand-Kirillov dimension two. The special case $\kappa =0$ corresponds to the Manin quantum plane and the case $q=1$ and $\kappa \not= 0$ corresponds to the Moyal
deformation of the plane.
A differential $*$-calculus $\Omega (D_{q,\kappa})$ on $\cO(D_{q,\kappa})$ is freely generated by one-forms $dz$ and $dz^*$ subject to relations
\begin{equation}\label{calculus.d}
zdz = q^{-1}dz z, \qquad z^*dz = qdz z^*, \qquad dz\wedge dz^* = -q^{-1}dz^*\wedge dz, \qquad dz\wedge dz =0,
\end{equation}
and their $*$-conjugates; see e.g.\hspace{3pt}\cite{SinVak:ana}. Since $\Omega^1(D_{q,\kappa})$ is an $\cO(D_{q,\kappa})$-bimodule freely generated by $dz$ and $dz^*$, the commutation relations between $dz$ and elements $\alpha$ of $\cO(D_{q,\kappa})$ induce an algebra automorphism $\sigma : \cO(D_{q,\kappa}) \to \cO(D_{q,\kappa})$, by
\begin{equation}\label{auto}
\alpha dz = dz \sigma(\alpha), \qquad \mbox{for all $\alpha\in \cO(D_{q,\kappa})$}.
\end{equation}
On the basis $z^kz^{*l}$, $k,l\in \NN$, of $\cO(D_{q,\kappa})$ this comes out as
\begin{equation}\label{auto.explicit}
\sigma(z^kz^{*l}) = q^{l-k} z^kz^{*l}.
\end{equation}
Note that $\sigma$ also satisfies the equality
\begin{equation}\label{auto*}
\alpha dz^* = dz^* \sigma(\alpha), \qquad \mbox{for all $\alpha \in \cO(D_{q,\kappa})$}.
\end{equation}
Finally, the commutation rules \eqref{auto} and \eqref{auto*} imply that $d(\alpha) =0$ if and only if $\alpha$ is a scalar multiple of the identity, i.e.\hspace{3pt}the calculus $\Omega (D_{q,\kappa})$ is connected.
For all values of $N$, $\cO(C^N_{q,\kappa})$ embeds into $\cO(D_{q,\kappa})$ by the $*$-inclusion
\begin{equation}\label{inclusion}
a\mapsto zz^*, \qquad b\mapsto z^N.
\end{equation}
We define the calculus $\Omega (C^N_{q,\kappa})$ on $\cO(C^N_{q,\kappa})$, by restricting $\Omega (D_{q,\kappa})$ to $\cO(C^N_{q,\kappa})$.
\begin{theorem}\label{thm.cone.smooth}
For all $N\in \NN$ and $\kappa\neq 0$, $\Omega (C^N_{q,\kappa})$ is a 2-dimensional connected integrable differential calculus on $\cO(C^N_{q,\kappa})$, hence the quantum cone algebras $\cO(C^N_{q,\kappa})$ are differentially smooth.
\end{theorem}
\begin{proof}
As a restriction of a connected calculus $\Omega (C^N_{q,\kappa})$ is connected. $\cO(D_{q,\kappa})$ can be equipped with a grading by elements of the cyclic group $\ZZ_N = \{0,1,\ldots, N-1\}$, defined by $\deg(z) =1$, $\deg(z^*) = N-1$, which is compatible with the $*$-operation in the sense that if $\deg(\alpha) = k$, then $\deg(\alpha^*) = N-k$. The embedding \eqref{inclusion} identifies $\cO(C^N_{q,\kappa})$ with the invariant subalgebra $\cO(D_{q,\kappa})_0$ of $\cO(D_{q,\kappa})$. By \cite[Theorem~2.1]{Brz:com} this grading is strong, provided $\kappa \neq 0$.
Note that $\sigma$ \eqref{auto} preserves the $\ZZ_N$-grading of $\cO(D_{q,\kappa})$.
By \cite[Theorem~3.1]{Brz:com}, the restriction $\Omega (C^N_{q,\kappa})$ of $\Omega (D_{q,\kappa})$ to the differential calculus on $\cO(C^N_{q,\kappa})$ is generated by the one-forms
$$
dz z^*, \qquad db \propto dz z^{N-1}, \qquad dz^* z, \qquad db^* \propto dz^* z^{*N-1}.
$$
The first two one-forms generate the holomorphic part of $\Omega^1 (C^N_{q,\kappa})$, while the other two generate the anti-holomorphic part. The module of one-forms $\Omega^1 (C^N_{q,\kappa})$ is not free, but $\Omega^2 (C^N_{q,\kappa})$ is freely generated by the closed volume form $dz\wedge dz^*$:
$$ da \wedge da = (z \, dz^* + z^* \, dz) \wedge ( z \, dz^* + z^* \, dz) = (- z z^* + \frac{1}{q} z^* z) \, dz \wedge dz^* =
- \frac{\kappa}{q} \, dz \wedge dz^*, $$
hence it can be identified with $\cO(C^N_{q,\kappa})$.
Since $z^*$ and $z^{N-1}$ generate the $\cO(C^N_{q,\kappa})$-submodule of $\cO(D_{q,\kappa})$ consisting of elements of $\ZZ_N$-degree $N-1$, the holomorphic part of $\Omega^1 (C^N_{q,\kappa})$ consists of elements of the form
$$
dz \alpha_-, \qquad \deg(\alpha_-) = N-1.
$$
Similarly, the antiholomorphic forms are $dz^* \alpha_+$, $\deg(\alpha_+)=1$. This shows that
$$
\Omega^1 (C^N_{q,\kappa}) \cong \cO(D_{q,\kappa})_1 \oplus \cO(D_{q,\kappa})_{N-1},
$$
where the right module isomorphism is
\begin{equation}\label{iso.omega}
dz\alpha_- +dz^*\alpha_+ \mapsto (\alpha_+, \alpha_-), \qquad \deg(\alpha_\pm) = \pm 1\, \mbox{mod}\, N.
\end{equation}
By relations \eqref{calculus.d}, \eqref{auto} and \eqref{auto*},
\begin{equation}\label{product.cone}
(dz\alpha_- +dz^*\alpha_+)(dz\beta_- +dz^*\beta_+) = dz\wedge dz^*\left(\sigma(\alpha_-)\beta_+ -q\sigma(\alpha_+)\beta_- \right).
\end{equation}
In view of the isomorphism \eqref{iso.omega}, the product rule \eqref{product.cone} can be recast into the desired form
$$
(\alpha_+,\alpha_-)\wedge (\beta_+,\beta_-) = \sigma_+(\alpha_+)\beta_- + \sigma_-(\alpha_-)\beta_+,
$$
where
$$
\sigma_+ = -q \sigma, \qquad \sigma_- = \sigma,
$$
with $\sigma$ given by \eqref{auto}. In this way the constructed calculus over the quantum cone meets all of the requirements of Lemma~\ref{lemma.Poincare}, and therefore it is integrable.
\end{proof}
\begin{remark}
The integrability of the calculi over the quantum cones can also be established by employing Lemma~\ref{lem.princ}. The $\ZZ_N$-grading of $\cO(D_{q,\kappa})$ which yields $\cO(C^N_{q,\kappa})$ can be interpreted as the group algebra $\CC \ZZ_N$-coaction
$$
\varrho(z) = z\ot u, \qquad \varrho(z^*) = z^*\ot u^{N-1},
$$
where $u^N= 1$ is the generator of $\ZZ_N$ written multiplicatively. This coaction, which is principal by \cite[Theorem~2.1]{Brz:com}, extends to the differential calculus $\Omega(D_{q,\kappa})$, whose invariant part coincides with $\Omega(C^N_{q,\kappa})$ and satisfies assumptions (a)--(c) of Lemma~\ref{lem.princ}. Once it is shown that $\Omega(D_{q,\kappa})$ is an integrable calculus, the integrability of $\Omega(C^N_{q,\kappa})$ will follow from Lemma~\ref{lem.princ}.
\end{remark}
Using \eqref{calculus.d} one easily checks that, for all $k,l\in \NN$,
\begin{equation}\label{surjective}
dz\wedge dz^* z^kz^{*l} = d\left(-\frac{q^k}{[l+1]_q} dz z^kz^{*l+1}\right).
\end{equation}
If $z^kz^{*l}\in \cO(C^N_{q,\kappa})$, i.e.\hspace{3pt}$k-l = mN$, for some $m\in \ZZ$, then $k-l -1 = (m-1)N +N-1$, so that $\deg(z^kz^{*l+1}) = N-1$ and hence the argument of $d$ on the right hand side of \eqref{surjective} is a holomorphic form on $\cO(C^N_{q,\kappa})$. Therefore the map $d: \Omega^1 (C^N_{q,\kappa})\to \Omega^2 (C^N_{q,\kappa})$ is surjective. Since $\nabla = - \pi_{dz\wedge dz^*}\circ d\circ \Theta^{-1}$, where $\Theta^{-1}$ is the isomorphism constructed through Lemma~\ref{lemma.Poincare}, it follows that $\nabla$ is also surjective. Consequently, the integral associated to $\nabla$ vanishes everywhere on $\cO(C^N_{q,\kappa})$.
The divergence $\nabla$ can be computed using the explicit form of the isomorphism $\Theta^{-1}$ constructed in the proof of Lemma~\ref{lemma.Poincare}. As the first step, let us define the operations $\pz, \pzs: \cO(D_{q,\kappa}) \to \cO(D_{q,\kappa})$ by
$$
d\alpha = \pz(\alpha) dz + \pzs(\alpha)dz^* = dz \sigma(\pz(\alpha)) +dz^* \sigma(\pzs(\alpha)),
$$
where $\sigma$ is given explicitly in \eqref{auto.explicit}. Both $\pz$ and $\pzs$ are twisted derivations, i.e.\hspace{3pt}for all $\alpha, \beta\in \cO(D_{q,\kappa})$,
$$
\pz(\alpha\beta) = \pz(\alpha)\sigma^{-1}(\beta) +\alpha \pz(\beta),
$$
and similarly for $\pzs$. In fact, they are {\em q-derivations}, i.e.
\begin{equation}\label{qder}
\sigma\circ \pz \circ \sigma^{-1} = q\pz, \qquad \sigma\circ \pzs \circ \sigma^{-1} = q^{-1}\pzs .
\end{equation}
If we choose $r_\pm^i, s_\pm^i, \in \cO(C^N_{q,\kappa})$, such that $\deg(r_\pm^i) = \deg(s_\pm^i) = \pm 1\, \mbox{mod}\, N$ and $r_+^ir_-^i = s_-^is_+^i =1$ (summation implicit), whose existence is guaranteed by \cite[Theorem~2.1]{Brz:com}, then the formula \eqref{def.theta} combined with all the above identifications of forms on $\cO(C^N_{q,\kappa})$ and with \eqref{qder} gives
\begin{equation}\label{nabla}
\nabla(\phi) = q \pz (\phi(dzs_-^i)s_+^i) + q^{-1} \pzs (\phi(dz^*r_+^i)r_-^i),
\end{equation}
for all right $\cO(C^N_{q,\kappa})$-module maps $\phi: \Omega^1 (C^N_{q,\kappa}) \to \cO(C^N_{q,\kappa})$. We note in passing that the first order calculus $\Omega^1 (D_{q,\kappa})$ is the calculus associated to the twisted multi-derivation $(\pz,\pzs)$ in the sense of \cite[Section~3]{BrzElK:int}, and, for $N=1$, $\nabla$ in \eqref{nabla} is exactly the divergence associated to such a calculus. For $N\neq 1$, \eqref{nabla} coincides with the induced divergence on the base space of a quantum principal bundle obtained by the method described in \cite[Section~4]{BrzElK:int}.
When $\kappa =0$, relations \eqref{disc} define the quantum plane. In this case, the $\ZZ_N$-grading is not strong (apart from the trivial case $N=1$): there is no possible combination of $z^kz^{*k}$ and $z^{*N-k}z^{N-k}$, $k=1,\dots, N-1$, with coefficients of degree zero that would give the identity element. Furthermore $dzz^*$ and $dz^*z$ are not elements of $\Omega^1 (C^N_{q,0})$ (although their linear combination $dzz^* +q^{-1}dz^*z$ is). Finally, $\Omega^2 (C^N_{q,0})$ is generated by $dz\wedge dz^* a^{N-1}$, $dz\wedge dz^* b$ and $dz\wedge dz^* b^{*}$, which are not free; when $\kappa =0$, relations \eqref{cone} allow one to write 0 as a linear combination (with non-zero coefficients from $\cO(C^N_{q,0})$) of any two out of $a^{N-1}, b, b^*$. This means that $\Omega^2 (C^N_{q,0})$ cannot be isomorphic to $\cO(C^N_{q,0})$, hence these differential calculi over the quantum Manin cones $C^N_{q,0}$ are not integrable.
\subsection{Spectral triple for the Moyal cone}
In the special case of $q=1$ and $\kappa \not= 0$, $\cO(D_{1,\kappa})$ is the algebra of the Moyal plane, which could be also studied
from the angle of spectral triples. First, observe that the algebra $\cC^\infty(D_{1,\kappa})$ of the series $\sum a_{m,n} z^mz^{*n}$ with rapidly decreasing coefficients is a subalgebra of $\Ss(\RR^2)$, the Schwartz functions
on the plane with the product defined through the oscillating integral,
$$ (f \ast g) (z) = (\pi \kappa)^{-2} \int_\CC d^2s \int_\CC d^2t \; f(z + s) g(z + t) e^{-2 i \kappa \Im(s^* t)} , \;\;\; z \in \CC.$$
It is easy to see that the action of the cyclic group $\ZZ_N$ on the space of Schwartz functions:
$$ \sigma(f)(z) := f \left( e^{\frac{2}{N} \pi i} z \right), $$
is an automorphism of this algebra. Let us consider the invariant subalgebra $\Ss(\RR^2)^{\ZZ_N}$ as the algebra of smooth
functions on the Moyal cone. We can now take the spectral triple for the Moyal plane, as constructed in \cite{Moyal}.
Using the same procedure as in the case of the noncommutative pillow, we lift the action of $\ZZ_N$ to the Hilbert
space, the latter being just $L^2(\RR^2) \otimes \CC^2$. The spectral triple over the Moyal plane is constructed
with the diagonal action of the algebra by left Moyal-multiplication and the Dirac operator and chirality operator
of the form
$$ D = \left( \begin{array}{lr} 0 & \partial_z \\ \partial_{z^*} & 0 \end{array} \right) ,\;\;\;\;
\gamma = \left( \begin{array}{lr} 1 & 0 \\ 0 & -1 \end{array} \right).
$$
Since we can decompose the Hilbert space into a direct sum of spaces on which the group $\ZZ_N$ acts by multiplication by
different roots of unity, it is easy to construct a subspace, which is preserved both by the Dirac operator and by the action
of the invariant subalgebra. The restriction of the original spectral triple to that subspace gives a spectral triple for the
Moyal cone.
If we try to consider a nondegenerate (Hochschild) cycle for the Moyal cone we need to observe first that its
construction for the Moyal plane is, in fact, based on the Weyl algebra of plane waves in the multiplier of the Moyal algebra.
However, since there are no plane waves which are in the multiplier of the invariant subalgebra of the Moyal cone one
cannot use such a construction.
On the other hand we are dealing here with the case of spectral triples over a locally compact noncommutative space, so we might relax the conditions
and demand that the cycle exists in the algebra of unbounded multiplier of the invariant algebra. Although to show the
existence for any $N$ appears to be rather a challenge, we shall demonstrate in the case of $N=2$ that this is indeed
possible.
\begin{lemma}
Let $N=2$ and let $\Ss(\RR^2)^{\ZZ_2}$ be the algebra of the Moyal cone with the spectral triple as above. Then the
following cycle gives the volume form $\gamma$:
\begin{eqnarray*}
&& - [D, b^*] a [D,b] - 2 a [D,b] [D,b^*] + \frac{3}{2} a [D,b^*] [D,b] \\
&& \phantom{xxxxxxxxx} +3b [D,a] [D, b^*] - b [D,b^*] [D,a] + b^* [D,b] [D,a]
= 4 \kappa^2 \gamma.
\end{eqnarray*}
\end{lemma}
The proof is by explicit computation and making the Ansatz that the cycle is at most linear in each entry in the generators $a,b,b^*$
of the cone algebra $\cO(C^2_{q,0})$.
Let us observe that again, as in the case of the noncommutative pillow, the above cycle has a nonvanishing image
only if $\kappa \not= 0$ and that again the solution is not unique. For this particular Ansatz there are no Hochschild cycles
which could give $\gamma$.
We believe that the construction of such cycles is possible in the noncommutative case for every Moyal cone, it is again an
open and challenging problem to prove it for every $N$ and to determine whether it is possible to find a Hochschild cycle, which
gives the orientation.
\section{Three-dimensional integrable differential calculi over quantum lens spaces}
\setcounter{equation}{0}
The aim of this section is to prove differential smoothness of quantum lens spaces by constructing quantum principal bundles over them with integrable differential calculi that satisfy the assumptions of Lemma~\ref{lem.princ}.
The coordinate algebra $\cO(L_q(N;1,N))$ of the quantum lens space $L_q(N;1,N)$ is a $*$-algebra generated by $\xi,\zeta$ subject to the relations
\begin{subequations} \label{rel.lens}
\begin{gather}
\xi\zeta = q^{l}\zeta\xi, \qquad \xi\zeta^* = q^{l}\zeta^*\xi, \qquad \zeta\zeta^* = \zeta^*\zeta, \label{rel.lens1}\\
\xi\xi^* = \prod_{m=0}^{l-1}(1-q^{2m}\zeta\zeta^*), \qquad \xi^*\xi = \prod_{m=1}^l(1-q^{-2m}\zeta\zeta^*), \label{rel.lens2}
\end{gather}
\end{subequations}
where $q\in(0,1)$; see \cite{HonSzy:len}. In the classical limit $q=1$, this is the coordinate algebra of the singular lens space. A linear basis of the space $\Vv(n)$ spanned by words in generators $\xi,\xi^*,\zeta,\zeta^*$ of length at most $n$ is given by
\begin{equation}\label{basis}
\{\xi^i\zeta^j\zeta^{*k}, \, \xi^{*r+1}\zeta^s\zeta^{*t} \; |\; i+j+k \leq n ,\, r+s+ t \leq n-1, \},
\end{equation}
hence
$$
\dim \Vv(n) = {n+3 \choose 3} + {n+2 \choose n-1} = \frac 16 (n+1)(n+2)(2n +3)
$$
Consequently,
$$
\gk (\cO(L_q(N;1,N))) =3.
$$
The algebra $\cO(L_q(N;1,N))$ embeds as a $*$-algebra into $\cO(SU_q(2))$, the coordinate algebra of the quantum group $SU_q(2)$. $\cO(SU_q(2))$ is a $*$-algebra generated by $\alpha$ and $\beta$ subject to the following relations
\begin{subequations}\label{su}
\begin{gather}
\alpha\beta = q\beta\alpha, \qquad \alpha\beta^* = q\beta^*\alpha, \qquad \beta\beta^* = \beta^*\beta, \label{su.a}\\
\alpha\alpha^* = \alpha^*\alpha +(q^{-2}-1)\beta\beta^*, \qquad
\alpha\alpha^* + \beta\beta^* =1, \label{su.b}
\end{gather}
\end{subequations}
where $q\in(0,1)$; see \cite{Wor:com}. The embedding $\cO(L_q(N;1,N))\hookrightarrow \cO(SU_q(2))$ is
\begin{equation}\label{linsu}
\xi \mapsto \alpha^N \quad \mbox{and} \quad \zeta \mapsto \beta.
\end{equation}
Henceforth we view $\cO(L_q(N;1,N))$ as a subalgebra of $\cO(SU_q(2))$ via the embedding \eqref{linsu}. Next we construct a differential calculus on $\cO(L_q(N;1,N))$ as the restriction of the left covariant three-dimensional calculus $\Omega (SU_q(2))$ of Woronowicz \cite{Wor:twi} over $\cO(SU_q(2))$. The calculus $\Omega (SU_q(2))$ is a connected $*$-calculus generated by the one-forms $\omega_0$, $\omega_\pm$, which satisfy the following relations:
\begin{subequations}\label{eq.3Drel}
\begin{gather}
\omega_0\,\alpha \,=\, q^{-2}\,\alpha\,\omega_0\, , \qquad
\omega_0\,\beta \,=\, q^2\,\beta\,\omega_0\hspace{3pt},\\
\omega_\pm\,\alpha \,=\, q^{-1}\,\alpha\,\omega_\pm\, , \qquad
\omega_\pm\,\beta \,=\, q\,\beta\,\omega_\pm\hspace{3pt}, \\
\omega_i\wedge \omega_i = 0, \quad
\omega_+\wedge \omega_-=-q^2\,\omega_-\wedge \omega_+\, ,\quad
\omega_\pm \wedge \omega_0\,=\,-q^{\pm 4}\,\omega_0\wedge \omega_\pm\, ,
\end{gather}
\end{subequations}
and their $*$-conjugates with $\omega_0^* = -\omega_0$, $\omega_-^* = q\omega_+$. The differential is given by
\begin{subequations}\label{eq.3dif}
\begin{gather}
d \alpha \,=\, \alpha\,\omega_0-q\,\beta\,\omega_+\, ,\qquad
d \beta \,=\, -q^2\,\beta\,\omega_0 + \alpha\,\omega_-\, , \label{eq.3dif.a}\\
d \omega_0 =q\omega_-\wedge \omega_+\, ,\qquad
d \omega_+ =q^2(q^2+1)\,\omega_0\wedge \omega_+\, . \label{eq.3dif.b}
\end{gather}
\end{subequations}
This is a calculus over $\cO(SU_q(2))$, since
$$
\omega_0 = \alpha^*d\alpha + q^{-2}\beta d\beta^*, \qquad \omega_+ = q^{-2}\alpha d\beta^* - q^{-1}\beta^*d\alpha,
$$
by \eqref{eq.3dif.a} and \eqref{su}.
\begin{theorem}\label{thm.lens}
For all values of $q\in(0,1)$ and $N\in \NN$, the restriction $\Omega (L_q(N;1,N))$ of the 3D-calculus $\Omega (SU_q(2))$ to $\cO(L_q(N;1,N))$ is a three-dimensional integrable differential calculus. Consequently, $\cO(L_q(N;1,N))$ is a differentially smooth algebra.
\end{theorem}
\begin{proof}
As explained in \cite[Theorem~2]{Brz:smo}, $\cO(SU_q(2))$ can be made into a principal $\CC\ZZ_N$-comodule algebra with coinvariants equal to $\cO(L_q(N;1,N))$. The Hopf $*$-algebra $\CC\ZZ_N$ is generated by a unitary group-like element $u$ such that $u^N=1$.
The $\CC\ZZ_N$-coaction is a $*$-algebra map given on the generators of $\cO(SU_q(2))$ by
\begin{equation}\label{coac.zn}
\alpha \mapsto \alpha \ot u, \qquad \beta \mapsto \beta \ot 1.
\end{equation}
The coinvariant subalgebra is generated by $\beta$, $\alpha^N$ and thus can be identified with $\cO(L_q(N;1,N))$ by \eqref{linsu}.
The calculus $\Omega (SU_q(2))$ is a $\CC\ZZ_N$-covariant calculus by the $*$-coaction
\begin{equation}\label{cov.coa}
\omega_0\mapsto \omega_0\ot 1, \qquad \omega_\pm\mapsto \omega_\pm\ot u^{\pm 1}.
\end{equation}
It is shown in \cite[Section~4.1]{BrzElK:int} that $\Omega (SU_q(2))$ is integrable
with an integrating volume form $\omega = \omega_-\wedge\omega_0\wedge \omega_+$.
Thus we are in a situation to which Lemma~\ref{lem.princ} can be applied provided assumptions (a)--(c) are satisfied.
First we look at the coinvariant part of $\Omega (SU_q(2))$ and study it degree by degree. The coinvariant part of $\Omega^1 (SU_q(2))$ can be identified with
$$
\langle \omega_0,\hspace{3pt}\alpha^*\omega_+,\hspace{3pt}\alpha^{N-1}\omega_+,\hspace{3pt}\alpha\omega_-,\hspace{3pt}\alpha^{*N-1}\omega_-\rangle \cO(L_q(N;1,N)).
$$
Thus is suffices to show that $\omega_0, \alpha^*\omega_+,\alpha^{N-1}\omega_+,\alpha\omega_-, \alpha^{*N-1}\omega_- \in \Omega^1 (L_q(N;1,N))$, i.e.\hspace{3pt}that each of these forms can be expressed as linear combinations of $bdb', db\, b'$, $b,b'\in \cO(L_q(N;1,N))$. Starting with the second of the relations \eqref{eq.3dif.a} and its $*$-conjugate, and using the commutation rules \eqref{eq.3Drel}, \eqref{su} we find
$$
\beta d\beta^* = \beta\beta^*\,\omega_0 + q^2 \beta\alpha^*\,\omega_+, \qquad d\beta^*\beta = q^2\beta\beta^*\,\omega_0 + q^2 \beta\alpha^*\,\omega_+.
$$
Hence
$$
\beta\alpha^*\omega_+ = \frac{1}{q^2-1}\left( \beta d\beta^* - q^{-2}(d\beta^*)\beta\right) \in \Omega^1 (L_q(N;1,N)).
$$
Consequently, also $\beta\beta^*\omega_0 \in \Omega^1 (L_q(N;1,N))$. The first of relations \eqref{eq.3dif.a} combined with the commutation rules \eqref{eq.3Drel}, \eqref{su} yields
\begin{equation}\label{dan}
d\alpha^N = [N]_{q^{-2}}\left(\alpha^N\omega_0 - q\alpha^{N-1}\beta \omega_+\right),
\end{equation}
where $[N]_{q^{-2}}$ denotes the $q$-integer \eqref{q.integers}. By \eqref{su.b}, $\alpha^{*k}\alpha^k$ is a polynomial in $\beta\beta^*$ with the constant term 1, hence \eqref{dan} gives
\begin{equation}\label{omega0}
[N]_{q^{-2}}\omega_0 = - \alpha^{*N}d\alpha^N + f_1(\beta\beta^*)\beta\beta^*\omega_0 + f_2(\beta\beta^*)\beta\alpha^* \omega_+,
\end{equation}
for some polynomials $f_1$ and $f_2$. Hence $\omega_0 \in \Omega^1 (L_q(N;1,N))$. The $*$-conjugate of the second of equations \eqref{eq.3dif.a} implies that $\alpha^*\omega_+ \in \Omega^1 (L_q(N;1,N))$ too.
Again using the $*$-conjugate of the second of equations \eqref{eq.3dif.a} as well as relations \eqref{eq.3Drel}, \eqref{su} we find
$$
\alpha^{N-1}\omega_+ = \frac{q^{-2}}{q^{2N}-1} \left( q^{3N} (d\beta^*)\alpha^N - \alpha^N d\beta^*\right) \in \Omega^1 (L_q(N;1,N)).
$$
Taking suitable $*$-conjugates we conclude, therefore, that all the generating forms of $\Omega^1 (SU_q(2))^{co \CC\ZZ_N}$ are elements of $\Omega^1 (L_q(N;1,N))$, i.e.\
$$
\Omega^1 (SU_q(2))^{co \CC\ZZ_N}= \Omega^1 (L_q(N;1,N)),
$$
as required.
Next, studying two-forms one finds that the coinvariant part of $\Omega^2 (SU_q(2))$ is
$$
\langle \omega_-\wedge \omega_+ ,\, \alpha^*\omega_+\wedge \omega_0,\hspace{3pt}\alpha^{N-1}\omega_+\wedge \omega_0,\hspace{3pt}\alpha\omega_-\wedge \omega_0,\hspace{3pt}\alpha^{*N-1}\omega_-\wedge \omega_0\rangle \cO(L_q(N;1,N)).
$$
All but the first of these generating two-forms are obtained as products of one-forms which have already been shown to be in $\Omega^1 (L_q(N;1,N))$; thus they are elements of $\Omega^2 (L_q(N;1,N))$. Using relations \eqref{su.b} and \eqref{eq.3Drel} one easily finds that
$$
\omega_-\wedge \omega_+ = \frac{1}{1-q^2}\left(q^{-1}\alpha\omega_-\wedge \alpha^* \omega_+ + q^5\alpha^* \omega_+ \wedge \alpha\omega_-\right).
$$
Hence also $\omega_-\wedge \omega_+ $ is a linear combination of products of one-forms in $\Omega^1 (L_q(N;1,N))$, thus it is in $\Omega^2 (L_q(N;1,N))$. This proves that
$$
\Omega^2 (SU_q(2))^{co \CC\ZZ_N}= \Omega^2 (L_q(N;1,N)).
$$
Finally, the volume form $\omega = \omega_-\wedge\omega_0\wedge \omega_+$ is the product of forms in $\Omega (L_q(N;1,N))$, thus both the coinvariant part of $\Omega^3 (SU_q(2))$ equals $\Omega^3 (L_q(N;1,N))$ and $\omega$ is coinvariant. Therefore, assumptions (a)--(b) of Lemma~\ref{lem.princ} are satisfied.
To check assumption (c) it might be convenient to interpret the $\CC\ZZ_N$-coaction in terms of the $\ZZ_N$-grading. In that way $\Omega (SU_q(2))$ is a $\ZZ_N$-graded algebra with the $*$-compatible grading given on generators by
$$
\deg(\alpha) = \deg(\omega_+) =1, \qquad \deg(\beta)= \deg(\omega_0) =0.
$$
As is shown above, $\Omega (L_q(N;1,N))$ is the degree-zero subalgebra of $\Omega (SU_q(2))$. If we take any $\omega' = a_-\omega_- +a_0\omega_0 + a_+\omega_+ \in \Omega^1 (SU_q(2))$ and assume that, for all $\omega'' \in \Omega^2 (L_q(N;1,N))$, $\omega'\wedge \omega'' \in \Omega^3 (L_q(N;1,N))$, i.e.\hspace{3pt}$\deg(\omega'\wedge \omega'') =0$, then in particular, taking $\omega'' = \omega_-\wedge \omega_+$ we find
$$
0 = \deg( \omega'\wedge \omega_-\wedge \omega_+) = \deg (a_0 \omega_0\wedge \omega_-\wedge \omega_+) = \deg(a_0).
$$
Similarly, by taking $\omega'' = \alpha^* \omega_0\wedge \omega_+$ and $\omega'' = \alpha \omega_-\wedge \omega_0$, one finds that $\deg(a_\mp) = \pm1$. Therefore, $\deg(\omega') =0$, as required.
In a similar way, assuming that $\omega' = a_-\omega_0\wedge \omega_+ +a_0\omega_-\wedge\omega_+ + a_+\omega_-\wedge\omega_0 \in \Omega^2 (SU_q(2))$ is such that for all $\omega'' \in \Omega^1 (L_q(N;1,N))$, $\omega'\wedge \omega'' \in \Omega^3 (L_q(N;1,N))$, and choosing $\omega''$ to be $\omega_0$, $ \alpha^* \omega_+$ and $\alpha \omega_-$, one finds that $\deg(a_0) =0$ and $\deg(a_\pm) = \pm 1$. Therefore, $\deg(\omega') =0$, as needed. This proves that assumption (c) of Lemma~\ref{lem.princ} is also satisfied, and consequently that $\Omega (L_q(N;1,N))$ is a three-dimensional integrable calculus, as needed.
\end{proof}
A divergence on $\cO(SU_q(2))$ corresponding to the 3D-calculus was derived in \cite[Section~4.1]{BrzElK:int}. The corresponding divergence on $\cO(L_q(N;1,N))$, computed from the formula \eqref{nablas} comes out as, for all $\phi\in \Ii_1 (L_q(N;1,N))$,
$$
\nabla (\phi) = \partial_0\left( \hat{\phi}\left(\omega_0\right)\right) + q^{-2}\partial_+\left( \hat{\phi}\left(\omega_+\right)\right) + q^{2}\partial_-\left( \hat{\phi}\left(\omega_-\right)\right),
$$
where $\partial_i$ are defined by $da = \partial_i(a)\omega_i$. The maps $\hat{\phi}$ are defined by \eqref{hat.phi} and can be computed in terms of $\phi$:
$$
\hat{\phi} (\omega_0) = \phi(\omega_0), \qquad \hat{\phi}(\omega_+) = x_0\phi\left(\omega_+\alpha^{N-1}\right) \alpha^{*N-1}+ \sum_{i=1}^{N-1}x_i\phi\left(\omega_+\alpha^{*}\right)\alpha (\beta\beta^*)^i,
$$
$$
\hat{\phi}(\omega_-) = y_0\phi\left(\omega_-\alpha^{*N-1}\right) \alpha^{N-1}+ \sum_{i=1}^{N-1}y_i\phi\left(\omega_-\alpha\right)\alpha ^{*}(\beta\beta^*)^i,
$$
where $x_i, y_i\in \CC$ are solutions of
$$
x_0\alpha^{N-1} \alpha^{*N-1}+ \sum_{i=1}^{N-1}x_i\alpha^{*}\alpha (\beta\beta^*)^i = 1, \qquad y_0\alpha^{*N-1}\alpha^{N-1}+ \sum_{i=1}^{N-1} y_i\alpha\alpha ^{*}(\beta\beta^*)^i =1\, .
$$
The existence of such $x_i, y_i$ is proven in \cite[Lemma~2]{Brz:smo}.
Finally, it is shown in \cite[Section~4.1]{BrzElK:int} that the integral associated to the 3D-calculus coincides with a scalar multiple of the normalized integral on the Hopf algebra $\cO(SU_q(2))$ or the invariant Haar integral on $SU_q(2)$ described in \cite[Appendix~1]{Wor:com}. The formula \eqref{integrals} implies that the integral on $\cO(L_q(N;1,N))$ is the restriction of the Haar integral on $SU_q(2)$, given by
\begin{equation}\label{integral}
\Lambda\left(\zeta^k\zeta^{*k}\right) = \frac{1}{~{}~{}~{}~[k]_{q^{-2}}},
\end{equation}
and zero on all other elements $\xi^i\zeta^j\zeta^{*k}$, $\xi^{*i+1}\zeta^j\zeta^{*k}$, $i,j,k,\in \NN$, $j\neq k$ of the linear basis \eqref{basis} for $\cO(L_q(N;1,N))$.
\section*{Acknowledgements}
The work on this project started during two visits of the first author to the Institute of Physics, Jagiellonian University; he would like to thank the members of the Institute for their hospitality.
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Tube Valve Length: 50
I've been using these latex tubes for couple of years now and have not had the problems that others are having. They are easy to work with initially, have great ride quality, and seem reliable. I've also tried Vittoria and Challenge latex tubes, but these are my current favorite. The caveat with these tubes is that they stretch lengthwise over time. So, if you have to take the tiretube off the wheel for any reason (e.g. to trueround the wheel), it is trickier to put the tube back in, because the overall length of the tube has increased and so you end up having to deal with slack. I've never had this problem with any other inner tube, latex or butyl, so I am not sure what's up. Otherwise, recommended.
Sure, these are super light and awesome when they work. However I've been through well over a dozen of these and more than half of them have failed at install or soon thereafter. No I'm not a novice at installation and these weren't pinches or due to a tirewheel issue, and I have been very gentle with these given how lightthin they are.
When they work they are great, but you'll need a mess of them in your jersey and saddle bags to offset the quality control issues. I had a good run initially but even after the manufacturer sent me some comp replacements I continue to experience poor quality in the product.
I so wanted to love these, but I've settled back to 'heavier' latex tubes (e.g. Vittoria or Michelin) that are more reliable.
Pros
Benefits of latex, (lightest tube I've found at 48g, good volume expansion)
About as puncture resistant as any other race-lite tubes
Short stem, removable valve cores
Cons
Price $$$$
STRETCHY - Once they start stretching out, which happens after the first few hundred miles, they are quite finicky to install. Your best bet is to go slow, use some sort of talcum powder and situate the bunched up tube sections around the rim such that you can air it up and reposition as you go.
For repairs, I've used the Park Tool adhesive patches with limited success (they will come off over time like any stick on patch), but have found the vulcanized patches to hold best.
I basically only use these for race situations anyway.
Over the past several months, I've had four of these tubes fail rapidly in the same place where the latex meets the black rubber around the valve stem. I ended up creating short (approx. 2 inches) rim strips from one of the failed tubes and put it between the tube and rim at the valve. This seems to have at least extended the life of a couple tubes I have installed (along with lots of talc). In any case, I don't plan to buy any more of these tubes. They do make for a great ride paired with 25mm Conti GP 4000's, however.
Latex is supposed to be the way to go. It's more supple and it is SUPPOSED to be more flat resistant than regular butyl tubes. I knew I would need to pump my tires more often, but that was something I was willing to live with. Instead I got tubes that while light weight, they would flat if you looked at them funny. I am very disappointed with these.
I bought two tubes, One popped immediately after pumping up the tire. It had popped in five places along a two inch line a line of weak material. I immediately took the other tube out of the other tire. I don't trust these tubes.
I use these with a pair of 25mm Michelin Pro 4 Endurance tires at 90 psi and they really take away road chatter and roll smoothly. I have them mounted on a 40mm carbon rim with disc brakes and I needed to add a 20mm Continental valve extender and they are perfect. It's nice that the valves screw out so you can mount a proper valve extender then screw the valve back into it. So far no flats after a few months use. I like these tubes.
These tubes are most definitely different. They feel like balloons when handling them. They also have to be pumped up every day. I am using Conti 4000 tires with them and I can tell a difference in the ride. My first ride I could really feel a spongy, soft feel to the tires. I didn't really notice this for too long, but every now and then I notice it. I pump them up to 100psi. I need to try other pressures. I have a 23 in the front and a 25 in the back. Good ride, seems to be fast, I would buy them again. My last couple of sets of tires were Tufos for clincher rims. Over the winter I was riding Conti trainer tires on a set of rollers and started with the Contis and these tubes this spring.
I have been riding these tubes for a few years now and find them to be very durable. The ride quality, when paired with my Conti tires, is exceptionally smooth. With this combination, I rarely flat and can easily go an entire season on a single pair of tubes. The only problem I ever experience with these tubes, is availability.
I highly recommend this product!
These tubes are extremely light but due to their suppleness I actually get fewer punctures than I do using standard tubes. Ride quality is also quite a bit better. One warning The 60mm valve length isn't quite 60mm. I use 45mm rims and I am just able to get the pump clamped onto these valves. I don't think they would work with a 50mm rim.
You also need to be very careful mounting these tubes as they are extremely thin and it doesn't take much to tear them. Then again, that is what makes them ride so nice.
I use these tubes with 20mm race tires at 140 psi. as well as 25mm winter tires at 100 psi. They work great in both.
Very good tube that offers lightest weight I have found. Must inflate before each ride. Will lose about 10-20 psi overnight. Use talc to install to prevent pinching. Gives a different ride feel over butyl tubes. Highly recommend. Found on BTD for only 10 bucks per tube. Great deal! Enjoy!
I have been using this latex tubes for four years now. I have had many problems caused by rim strip being wrinkled in (hence, exposing the nipple holes) by the Vittoria SC EVO Open Corsa tight fit into my Neuvation SL28 rims, resulting on flats. If you have this problem, don't use Velox tape, don't use Rox ultra light tape (I love it!, but does not work on this tight rim/tire combination). Use Continental Easy Tape. That resolved my three year stubbornness. Road feel is different (makes the Vittoria Open Corsa's feel exactly the same as the tubular version I also have). When you ride the same road over and over, you can feel a change. You can notice it, especially when the road texture gets coarse. More supple ride results on better traction stability, and this will give you confidence cornering. Well worth the special care you need to take, especially upon installation. Also, don't forget to take a conventional glue and patch kit with you, since self glued patches do not work on latex tubes (I tried Specialized, and Park patches, save yourself this learning curve). Use talc upon installation! And make sure you clean off the talc when installing a patch, or be prepared for frustration. Take a spare tube with you (this one can be butyl, LOL). I refuse to go back to butyl!! I have also tried the pink Vittoria latex tubes, but they are thicker, hence heavier, and just as fragile. Being careful with both will produce the same result, I believe.
It's the lightest latex tube on the market.....49 grams.....rubber-tubes weigh-in at 100 to 150 grams......also latex is more resisent to punctures.....(they also save up to 5 watts in needed power to turn the wheels). One down side....need to air-up more often.
A terrific ride at about 110 lbs. Rolls well is reasonably puncture resistant and light. Don't forget to talc inside the tire.
I really was looking forward to trying out latex tubes due to all the positives I had heard about them. This was my third and final attempt and these tubes disappointed as well. One flatted within the first week and the replacement I tried had a weak spot and ballooned as I was gently filling it initially. Back to butyl for now and will try tubeless next.
Hi CAMowins,
Thanks for your review! Be sure to get in touch with our customer service department at sales@biketiresdirect.com with any questions about our tubes or warranties.
Thanks,
I won't tell you about how you need to take care mounting these or how you need to pump them up every day - you know that already if you are looking at buying these.
What I will tell you is that I have found these to bring all of the benefits of latex ride quality and low weight with a surprising durability. I used to change tires and tubes for races, but for the heck of it I left these on under a pair of Michelin Pro4's and rode my normal training rides to see how long they lasted. Over chip and seal and loose rocksgravel and any other normal road hazard that you can think of, I have yet to have a single puncture after hundreds of miles of training and racing. These ultralights have proven to be an outstanding product when paired with the right tire.
This tube is what the rolling resistance articles are talking about. Seems to work well in most wheel sets that I have. I do not use these tubes in my carbon clinchers, as I found them to attach to the side wall of the tires, most likely do to the heat from the brakes. With aluminum rims, no problems especially with continental 4000s. Seems to be an excellant combo. No valve stem issues at the valve seat using Mavic SL rims.
Best tubes money can buy. Yes they leak air overnight, but as long as you check tire pressure before a ride, it is worth it.
Combine these with nice supple tires like Veloflex Master, or Vittoria Open Corsa CX 320 tpi and you are in cycling bliss. The only way to get a smoother ride is to go run Tubular tires and wheels.
100% recommended! Make sure you talc your tubes and inside of the tire before installing.
Great tube for racing, but one of the 6 tubes I purchased for my racing wheels spontaneously developed a pinhole leak on the rim side within the first few weeks.
Since latex tubes are more sentitive than tougher butyl tubes, it's important to check that your rim strips are properly aligned and that there are no debris in the well or the tire during installation. Should you ever receive a tube that you believe has a defect, please feel free to contact us directly.
Thanks,TheoCustomer Service
This tire is 50-60 grams lighter than a typical butyl tire which is supposed to shave of precious seconds in a race. It rides great and I don't have any prOblems with the tire. It definitely doesn't hold air as long as a butyl tire though. I normally inflate to 125psi and after a 40k ride it is generally down a bit, maybe 10 psi lower at most but not too bad. The next day the tire is way down around 60-70 psi so you have to inflate before every ride. That's a good habit to be in anyway.
Lightweight tube. Significantly decreases rotational weight.
These tubes are great. I've noticed that I climb faster and accelerate better. Ride them in everyday rides without any problems. Only flaws If noticed (minor) are that you need to air up prior to each ride and over time the fuse point of stem and tube fails. Though I think this wouldn't happen if tube was threaded, something I refuse to buy. Perfect world only base of stem would be threaded!
Only have a few hundred miles on them so far, so I haven't completely formed an opinion yet.
A few observations
Don't use a tire iron to mount tire. I learned this at the expense of a tube.
They lose a significant amount of air pressure in between rides. You will have to pump up before each ride.
At under 55gms and as supple as the post ride was these tubes are awesome and should be OEM in all high end clinchers sold on LBS built bikes I Own Dura Ace c24 carbon clinchers and the riding difference In my opinion was rather astonishing after I changed out the Butyl for the latex tubes but the best thing about latex tubes is they patch permanently latex glue and a latex patch form a hermetically tight bond compared to Butyl which the bond is rather weak since there is no latex substrate to form a good base bond
When I ordered my new, light wheelset - I threw on these and some 150g tires just to go the whole nine yards wrt weight. I like these better than the Vittoria Latex tubes I put in the same wheel a couple weeks later. I've removed both after only a few hundred miles, and the Vredestein's look like they've fared much better. The Vittorias kind of look haggard and make one wonder if you want to reinstall them or not. If you want to try latex, I think the Vreds are the way to go. They do lose about 15-20 psi over night when you first install them. I certainly wouldn't bet any body parts that I could tell the difference in a blind test - either in weight or feelrolling resistancewhatever. I threw them in a bag of talk first and did not use irons to install. I did use a bead jack. No problems o the install if you do it correctly and make sure you go around the tire getting everything seated right before inflating. I wish they came in a shorter valve stem. This, the cost, the air loss, the availability, the ridicule - all have brought me back to light butyl at 80ish grams. (I sometime run FOSS plastic tubes.) But if you want to try latex - VREDESTEIN RULES!.
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An easy Means To scrub
A simple Approach To scrub
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It is observed that lack of maintenance, pumping the tank often and abuse of tank are a number of the factors that can lower the service life of the tank drastically. Proper lighting can emphasize on the focal points of a home. Magic erasers are thought of "magical" as a result of they will clean virtually any floor, from ground to ceiling. Customers, purchasers and workers notice a bright clean enterprise and react in a positive manner. Different shades of white and vivid blues are excellent colors for the summer time. Your real property agent has advised you to go through the strategy of getting pre-qualified for a home mortgage loan and you are wondering why this suggestion has been made. useful source Seal cracks in your home with caulk. You do not have to spend large amount of cash just to offer your house with such wonderful storage options. Cleaning the complete residence does not finish the chore, it is also entails a substantial amount of endurance when coping with house instruments and equipments.
Desired cleansing frequencies. • Worker work schedule. I had at all times treated the individuals cleaning for me with professional respect. The Division of Power provides grants to homeowners who make their residence more efficient. Waterproof paint techniques for pitched roofs offering as much as 15 years' protection for asbestos, bituminous surfaces, metal and extra. Remembering this small advice retains the work area clear, and prevents pointless frustration. But I’m quite certain the dark ring was not a lighting concern. As soon as we were ready to face at chest peak, we could then access the areas in want of restore and deal with eradicating the rotted rim joist. After eradicating the outdated shrubs, domesticate the bed and combine in natural supplies like manure, compost or peat moss. So, do not be surprised why you can't reel in a 35kg Striped Bass (the most important caught up to now). So, never draw back from the microwave when using it's an possibility. Liquid rubber specifically is good for sealing up low areas where water is pooling when it rains.
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Heart palpitations are a bit concerningJune 27, 2006
The first time I experienced a fluttering of my heart (to the point that it was noticeable anyway) was back in 2006. I’ll never forget the scenario either. I was sitting at my office desk and had just found out how much money I owed to taxes (because I had my own business, I would pay taxes quarterly). While the exact amount I don’t recall, it was much higher than I expected, and was enough to get my heart racing uncontrollably.
I could feel my heart beating very hard. It wasn’t necessarily fast, but more like a pounding inside my chest. And it scared me since I’ve never experienced that before. So after 20 minutes of the hard pounding, I grew concerned and called a local nurse who recommended I go to the hospital. Since I was the only one home, I got in the car and drove their immediately.
Fast forward to the diagnosis part, the doctors couldn’t determine exactly what it was, and said that it’s somewhat normal for people to experience heart beats like that. They had me wear a halter monitor for 2 days, and even put me through a stress test where I run on a treadmill while connected to wires and stuff that monitor my heart. In both cases, the palpitations were noticeable, but according to the doctors, not life threatening.
In the end they attributed it to stress. Now initially I would have agreed, since the trigger seemed to be the day I found out about the high taxes I had to pay. But the strong heart beating continued on and off for months after that. And now years later, though not was frequent, I still occasionally feel them. So I plan to continue monitoring it on my own and will do what I need to if I feel the symptoms getting worse.
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Shipping and Returns Policy
We want you to love The LashLiner System as much as we do. If for any reason you are unsatisfied with your purchase that was made on our web site we have return options available.
To be eligible for refunds, all returns must be accompanied by all original components and proof of sale. Items purchased from other retailers, dealers or re-sellers and not directly from LashLiner's e-commerce site are not eligible for return or refund.
LashLiner, LLC reserves the right to limit returns if we have reason to suspect misuse of our returns policy (such as excessive returns, unauthorized re-seller activity, or fraud).
To initiate a return or exchange, you must send email to support@lashliner.com. In order to avoid fraud we can not provide refunds or exchanges without prior authorization.
Liner and other cosmetics: Liner and cosmetics are refundable for up to 14 days after delivery date.
Lashes: LashLiner Magnetic Lashes are refundable for up to 3 days after delivery date and must be returned with all of the components, including the Lash Anchors. Please check your Lashes within 3 days after delivery date. Due to the very small size We do not provide replacement or exchanges on Lash Anchors.
Exchanges on lashes must be for the same lash style.
Exchange and Return Shipping fees: Shipping fees are non-refundable. We do not provide return shipping labels for returns but we DO provide return shipping labels for exchanges within the return period. If you send in an exchange but decide you would prefer a refund (must be within the refund guidelines) there is a 25% warehouse processing fee once the exchange shipping label is used.
Exchanges: Exchanges a are non refundable/exchangeable final sale. If your exchange item(s) arrive damaged please let us know within 48 hours and we will exchange the item if it is still in stock. Lash exchanges are limited to the same lash style.
Outside of the United States? International shipping does not include taxes, customs fees or free shipping. These additional charges are very common and must be paid prior to delivery. We will not provide a refund if you refuse to pay your customs fees or taxes. In that situation the product is destroyed, not returned to us.
Unfortunately, we can not offer any refunds or exchanges on international orders and can not replace items that are reported as lost or stolen after delivery has been confirmed by the carrier.
Large orders: Returns or partial returns from customers that have over $250 in any one order, or $400 of orders in a given month, must be initiated through our website within 72 hours of receipt of merchandise.
Clearance merchandise is a non refundable/exchangeable final sale. If your clearance item(s) arrive damaged please let us know within 48 hours and we will exchange the item if it is still in stock or refund the purchase price.
Missing Items: Please make sure your order is sent to a secure location such as a locked mailbox. Unfortunately, theft and fraud are common with packages that are delivered to non secure locations. Because of this, we can not provide refunds or replacement on merchandise stolen, lost or missing after confirmed delivery.
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I really only need the top half of this cabinet. Would it work that way? The space I have is not high enough for the entire thing and I can't find others.
I love it had no problem putting it together, everyone that comes to my house tells me that it looks beautiful....
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/ 06.May.2008 at 09:39pm.
/ 21.Apr.2008 at 07:12pm.
/ 11.
/ 11.Feb.2008 at 08:42pm.
/ 07.Feb.2008 at 01:16pm
Last year we had our local event WebDevCamp SP’07, where Manoel Lemos, from the sucessful BlogBlogs introduced us to his friend James Crane-Baker, co-founder of the RedeParede (literally, “NetWall”), a classified Web 2.0 web-service written in Ruby on Rails.
Recently he got in touch with me about releasing his services APIs to the Rails community and therefore I decided to publish more about these Santa Barbara, CA based guys. They are credible people and James actually lived in Rio de Janeiro for a while, so he understands our Latin culture better than the average american, which is good as they plan to cover not only Brazil, but Latin America. I hope everybody can take a look at what they are doing as it sounds very promising, and success cases like this only augments the importance of our community as a whole.
So, here goes James himself explaining his product:
/ 11.Jan.2008 at 01:29am
I have a very embarrassing story. It is about 2 e-mails I got today. The first one came right in the morning and it was from one of my website readers. He reported to me that my homepage was all messy at IE 7. At first I didn’t care too much (“oh damn, IE always cracks down”). I am usually happy enough when I see it working on both Firefox and Safari – the rationale being: “if it works in Safari, should work in IE 7”
So I just let it go. Now the ugly part comes on the second e-mail I got tonight, from my boss, Carl. He reported that he was talking to a possible client and decided to proudly show off my website. Surely enough, the client fired up IE 7 and boom! Embarrassment! To Carl’s surprise, my website rendered completely broken and unusable at this client’s web browser! I can imagine the situation. And he tried to amend the situation saying: “but I can assure you he knows XHTML and CSS …”
Damn! I am not expert web designer but I know a trick or two about well formatted XHTML and CSS. So I fired up the W3C’s XHTML and CSS validators. 100 XHTML errors and 2 CSS errors! UGLY!
Ok, this is no good. Let’s fix it. Fortunately the validator gives up good clues on what was wrong. So let me tell you, almost all of my errors were from these elements:
Seems like the validator hates them all. Ok, fortunately Facebook has an option to switch from the Javascript badge to a static image based one. Cool. Twitter was no luck. I was not in the mood to tweak it down so I just ripped it off. Same thing about Rec6. The Flash players were deferred to internal pages so they don’t break my homepage. Aside from that, there really were a few of my own mistakes, unclosed tags and so on. After all this clean up, that’s how my status looks like now:
I went down from 100 XHTML errors to Zero and from 2 CSS errors to Zero. Looks like I am ok now. But not by much. Some of the internal pages will break because they still contain the Flash-based video players. My rant is: yes I could try to tweak everything. But this is not the point: the point is that those services offers the HTML for us to just copy & paste into our websites. So why can’t they just produce XHTML-compliant tags, for God’s sake!? I will definitely not waste my time fixing their tags.
So, I won’t have Twitter or any of those embedded players at my homepage until they fix this embarrassment. We bloggers help promote their websites, so the least they should do is not break ours, for crying out loud!
Bottom line is: now my website renders perfectly well on Safari, Firefox 2 (both Mac and Windows) and IE 7. But IE 6 still doesn’t render the sidebar correctly (give it a try). Even though it is now a “perfectly” well structured XHTML and CSS, IE 6 still refuses to render correctly … oh well, IE 6 usually doesn’t render correctly anyway …
Update: Carl send me an interesting tip for Flash in this link. I put my Twitter badge back. This is not the optimal solution, though (Twitter itself should provide us a better solution). Anyway, XHTML check is passing, all good and dandy.
/ 09.Jan.2008 at 09:27pm.
/ 04.
/ 03.Jan.2008 at 08:44am
1/1: Wow, I am really honestly frightened! I just read this rather long rant from no other than Zed Shaw – creator of Mongrel who, for non-starters, is (was) the #1 in popularity at Working with Rails, just a little bit higher than DHH himself.
Two comments that TechCrunch selected were:
In the article he literally shows the finger to people like Kevin Clark, Dave Thomas, Chad Fowler, Michael Koziarski and even David Hansson! He claims to be abused by them and that the Rails market only offered him mediocre jobs. He also attacks ThoughtWorks. We have to admit, he has good points although we can disagree on the presentation.
That’s one true thing: programmers – specially those that become famous overnight – have their egos inflated exponentially. I don’t believe that the Rails Core Team is different. We can’t tell if Zed Shaw is right in every claim or if he exaggerated a lot. But I don’t believe someone like him simply woke up crazy today out of nothing. If he is angry, he has his reasons. Let’s watch how the community reacts in the next few days.
On the other hand, this is not exclusive in the Rails community. As I say all the time, Ruby on Rails is not a perfect technology. None is. Its creators are not perfect, they are human beings like you or me. It is the community that plays the major role in an open source project. Few can have the luxury to be both extremely smart and arrogant at once like a Linus Torvalds, for instance.
Take your own conclusions. Anyway, one Zed Shaw or one Kevin Clark doesn’t make Ruby or Rails worst. It is the Rails community as a whole who’s gonna decide how things go on. If the Rails Core Team really becomes unbearable one day, as Zed accuses them, and the community still believes that Rails is worth it, we always have the last choice of forking it (let’s just hope it is not the case).
/ 01.Jan.2008 at 12:01am
As I promised after the Avi Bryant interview, here’s a great conversation with Adrian Holovaty, well known creator of the Django web framework written in Python..
/ 29.Dec.2007 at 11:46am
New Year’s Eve is the time where we look back at the year that has just passed and reflect about it for the next one. That’s the time we slow down a bit and allow ourselves to see what did we do, what difference did we make, and if we are still headed towards the right direction.
Fortunately, everything I’ve deemed important (technology-wise) was bookmarked at my del.icio.us/fabioakita account. I invite everybody to subscribe to it or just take a peek. I’ve selected just a few that represents some of the most interesting events this year (at least from my point of view).
For me, 2007 was a tremendous year, no complains. I worked like crazy in the last few months, but at least I am proud that I made an educated decision more than a year ago and stuck with it. And guess what? It paid off in spades! The book I published in September, 2006 was a success, I am working full time with the great guys at Surgeworks LLC and I now have a local brazilian team of my own, with 3 fellow Railers (follow the links to get to know them).
I’ve attended to no less than 5 brazilian Rails gatherings, one every weekend since November 17th. It was great to see the brazilian Ruby on Rails community flourishing. And I hope that 2008 is the year where people finally start to notice us down here.
My Rails 2.0 Screencast was a huge success, achieving world-wide recognition (see the Google Analytics graph below). I’ve made great friends like Dr. Nic, Geoffrey Grosenbach, Satish Talim and more. (Read the exclusive interviews here)
More important of all: I was able to drift away from the mainstream and do what I wanted to do. I wanted to do Rails full time, and I got it. That’s the way to go: do whatever it is that you really want, not what other people tell you to do. You have to know better! And education is key: the only way to make a good decision for yourself.
/.
/ 25.Dec.2007 at 07:56pm
According to this article Matz should release Ruby 1.9 TODAY!! Probably around 1AM EST.
Anyone interested in what´s new at Ruby 1.9 should follow the eigenclass blog. He´s been tracking the changes for 2 years not using a mechanized script procedure to analyze the CHANGELOGS. That´s 50k lines of processed text lines!
Merry Christmas!
Update 12/25: Exactly as we said yesterday, Matz officially announced Ruby 1.9. Se also here, here and here And the first question everybody´s gonna ask is “does Rails 2.0 runs over Ruby 1.9?” and the short answer being No. Neither Rails, nor Mongrel (Jeremy Kemper seems to believe that it´s almost there), not dozens of gems are ready yet. Ruby 1.9.1 seems to be just around the corner as well. All gem developers need to re-test and modify their codes taking into account what has been deprecated and which sintaxes changed. Wait a little bit more before attempting Rails with Ruby 1.9.
Update 2 12/25: This presentation should help you get up to speed on the newest features of Ruby 1.9. Keep in mind that 1.9 is an odd release – tagged experimental – and the next version, 2.0, is the one that´s gonna be held stable, same way that happened for 1.8 and 1.6. Do not expect to run projects in production with 1.9 so soon, but this gives us the necessary time to adjust to the differences with the now older version.
Summarizing? New lambda/block syntax, more splats, new Enumerator, subtle differences with Modules, Unicode support (finally!), new asynchronous I/O. Then again, the best source to research what´s changed is Eigenclass and, of course, the soon to be released Dave Thomas Pickaxe: “the” definitive Ruby 1.9 reference.
Update 12/26: Exactly as I recommended and is well known, 1.9 is an experimental version. Dave Thomas reminds us again that this is intended for anyone interested in experiments, upgrade gem code, etc but this is not intended as a drop-in replacement for 1.8. There are still known bugs that will be squashed in the next few months before we finally have a stable 2.0 release.
Update 12/27: This is an old post but it explained well how to install Ruby 1.9 without breaking your existing 1.8 (though I still recommend using some virtual machine to test experimental stuff). It basically goes like this (updated to the new 1.9 tag):
svn co ruby19 pushd ruby19 autoconf ./configure --prefix=/usr/local --program-suffix=19 --with-readline-dir=/usr/local make sudo make install popd
Chris Shea explained this well: Certainly, old hands have this down, and much of this is in the README. But the secret is—program-suffix=19, which leaves ruby alone and gives you ruby19 as your 1.9 executable.
/ 22.Dec.2007 at 06:36pm.
And stay tuned! I hope to have Evan Phoenix, Hal Fulton, Peter Cooper and Adrian Holovaty as my next guests. Lot’s of material to begin 2008 in great style.
/ 15.Dec.2007 at 10:15am.
He was very kind to provide me a very long interview. It is so long I divided it in 2 parts. This is the first Part. I will release the second one in a few days. Hope you all enjoy it.
/ 18.Nov.2007 at 04:14pm
Yesterday took place the anticipated RejectConf SP’07, the first relevant Rails event in Brazil, at the Jacy Monteiro auditorium over at the Instituto de Matemática e Estatística da USP. It was a terrific event, even more so considering the I was the responsible for it (meaning that I am a very amateurish [dis]organizer :-) ). It was a community-driven event, non-profitable, that counted on the support from Caelum and IME, which made it possible for an event without entrance charges. It was well worth it.
All volunteer presenters showed up, the audience was excited, cooperative and interested, which certainly made me happy. There were delays, people was attending even in the middle of a holiday weekend (chaining holidays from last thursday until next tuesday, here in Brazil), hungry, heated but even so they all attended. This surely shows how united and strong our community is.
/.
/ 16.Oct.2007 at 06:51pm
Pessoal, é com muita satisfação que anuncio que nosso encontro regional RejectConf SP’07 irá acontecer no dia 17 de novembro de 2007, sábado, das 11:00 às 17:00, no auditório Jacy Monteiro do Instituto de Matemática e Estatística da USP, em São Paulo/SP. Com entrada franca para todos que confirmarem presença através deste cadastro. Aliás, o local comporta no máximo 80 pessoas!! Quem se cadastrar primeiro terá prioridade, portanto não se cadastrem se não tiverem certeza e se desistirem por favor nos avise.
Auditório Jacy Monteiro – Bloco B
Instituto de Matemática e Estatística da Universidade de São Paulo (IME-USP)
Rua do Matão, 1010 – Cidade Universitária
CEP 05508-090 São Paulo – SP – Brasil
Para quem não é de São Paulo, veja esta página com indicações de como chegar.
Desde que eu coloquei a idéia aqui no blog, mais de 140 pessoas opinaram na enquete, muitos de vocês me contactaram pessoalmente, deram idéias, sugestões. Em especial gostaria de agradecer o empenho de algumas pessoas como Danilo Sato, Fabio Kung e André Santi que correram atrás do auditório do IME, do Paulo Silveira da Caelum que também está colaborando com o evento.
/ 03.Aug.2007 at 07:46pm resumé,!
/ 20.Jul.2007 at 11:36amão –ão Gilberto, Luiz Bonfá, Viníc
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How to Create a Company Value
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Enabling/Disabling Company Values
If you do not want to use values for your organization, simply click the toggle at the top of the the "Company Values" section.
If this toggle is enabled, all users can attach a company value to a piece of Feedback.
If disabled, users will not be able to specify any company values.
How to Delete a Company Value
To delete an existing company value, simply click the "Archive" link attached to a company value. You will be prompted to make sure that you want to archive this value!
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phorus
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\section{Instability points in the exponential model: proofs}\label{sec:exp-pf}
We turn to the proof of the results in Section \ref{sec:exp}, beginning with a discussion of Palm kernels, which are needed in order to prove Theorems \ref{th:B-palm1} and \ref{thm:ladder-xi}.
\subsection{Palm kernels} \label{sub:palm}
Let $\cM_{\Z\times\ri\Uset}$ denote the space of locally bounded positive Borel measures on the locally compact space $\Z\times\ri\Uset$. Consider $\Z\times\ri\Uset$ as the disjoint union of copies of $\ri\Uset$, one copy for each horizontal edge $(ke_1, (k+1)e_1)$ on the $x$-axis. Recall that $\B{\xi\sig}_k=\B{\xi\sig}(ke_1,(k+1)e_1)$.
We define two random measures $\Bppa$ and $\Bpp$ on $\Z\times\ri\Uset$ in terms of the Busemann functions $\xi\mapsto\B{\xi\pm}_k$ attached to these edges.
On each subset $\{k\}\times\ri\Uset$ of $\Z\times\ri\Uset$ we (slightly abuse notation and) define the measure $\Bppa_k$ by
\[ \Bppa_k\bigl(\{k\}\times ]\zeta, \eta]\,\bigr) = \Bppa_k\bigl(\,]\zeta,\eta]\, \bigr)=\B{\zeta+}_k-\B{\eta+}_k\]
for $\zeta\prec\eta$ in $\ri\Uset$.
In terms of definition \eqref{Bmeas}, $\Bppa_k=\mu_{(k+1)e_1, ke_1}$ is a positive measure due to monotonicity \eqref{mono}.
On $\Z\times\ri\Uset$, define the measure $\Bppa=\sum_k\Bppa_k$. In other words, for Borel sets $A_k\subset\ri\Uset$,
$\Bppa\bigl( \,\bigcup_k \{k\}\times A_k\bigr)=\sum_k \Bppa_k(A_k)$.
Let $\Bpp_k$ denote the simple point process on $\{k\}\times\ri\Uset$ that records the locations of the jumps of the Busemann function $\xi\mapsto\B{\xi\pm}_k$: for Borel $A\subset\ri\Uset$,
\[ \Bpp_k(\{k\}\times A) = \Bpp_k(A) = \sum_{\xi\in A} \ind\{\B{\xi-}_k> \B{\xi+}_k\}. \]
W describe the probability distributions of the component measures $\Bppa_k$ and $\Bpp_k$, given in Theorem 3.4 of \cite{Fan-Sep-18-}.
Marginally, for each $k$, $\Bpp_k$ is a Poisson point process on $\ri\Uset$ with intensity measure
\be\label{Bmm}
\Bmm\bigl(\,]\zeta, \eta]\,\bigr) = \Bmm_k\bigl(\,]\zeta, \eta]\,\bigr)= \E\bigl[ \Bpp_k(\,]\zeta, \eta]\,)\bigr]= \int_{\alpha(\zeta)}^{\alpha(\eta)} \frac{ds}s = \log \frac{\alpha(\eta)}{\alpha(\zeta)}.
\ee
In particular, almost every realization of $\Bpp_k$ satisfies $\Bpp_k[\zeta, \eta]<\infty$
for all $\zeta\prec\eta$ in $\ri\Uset$.
Create a marked Poisson process by attaching an independent Exp$(\alpha(\xi))$-distributed weight $Y_{\xi}$ to each point $\xi$ in the support of $\Bpp_k$. Then the distribution of $\Bppa_k$ is that of the purely atomic measure defined by
\be\label{Bppa4} \Bppa_k\bigl(\,]\zeta, \eta]\,\bigr) = \sum_{\xi\in\ri\Uset:\,\Bpp_k(\xi)=1} Y_{\xi} \,\ind_{]\zeta, \eta]}(\xi) \quad \text{for } \zeta\prec\eta\text{ in } \ri\Uset. \ee
The random variable $\Bppa_k(\,]\zeta, \eta]\,)$ has distribution $\text{Ber}(1-\frac{\alpha(\zeta)}{\alpha(\eta)})\otimes\text{Exp}(\alpha(\zeta))$ (product of a Bernoulli and an independent exponential) and
expectation
\be\label{Bmma}
\E\bigl[ \Bppa_k\bigl(\,]\zeta, \eta]\,\bigr)\bigr]= \frac1{\alpha(\zeta)} - \frac1{\alpha(\eta)}\,.
\ee
Note the following technical point. The jumps of $\B{\xi\pm}_k$ concentrate at $e_2$ and $\B{e_2-}_k=\infty$. To define $\Bppa$ and $\Bpp$ as locally finite measures, the standard Euclidean topology of $\ri\Uset$ has to be metrized so that $]e_2, \eta]$ is an unbounded set for any $\eta\succ e_2$. This point makes no difference to our calculations and we already encountered this same issue around definition \eqref{Bmeas} of the Busemann measures.
With this convention we can regard $\Bpp=\sum_k\Bpp_k$ as a simple point process on $\Z\times\ri\Uset$ with mean measure $\widetilde{\Bmm}= (\text{counting measure on $\bbZ$})\otimes \lambda$.
For $(k,\xi)\in \Z \times \ri \Uset$, let $Q_{(k,\xi)}$ be the Palm kernel of $\Bppa$ with respect to $\Bpp$. That is, $Q_{(k,\xi)}$ is the stochastic kernel from $ \Z \times \ri \Uset$ into $\cM_{\Z \times \ri \Uset}$ that gives the distribution of $\Bppa$, conditional on $\Bpp$ having a point at $(k,\xi)$, understood in the Palm sense. Rigorously, the kernel is defined by disintegrating the Campbell measure of the pair $(\Bpp,\Bppa)$ with respect to the mean measure $\widetilde{\Bmm}$ of $\Bpp$ (this is developed in Section 6.1 in \cite{Kal-17}):
for any nonnegative Borel function $f:(\Z \times \ri \Uset) \times\cM_{\Z \times \ri \Uset}\to\R_+$,
\begin{align}\label{79-50-50}
\E\Bigl[ \int_{\Z \times \ri \Uset} \!\!\!f(k,\xi,\Bppa)\, \Bpp(dk \otimes d\xi)\Bigr] &= \int_{\Z\times\ri\Uset} \int_{\cM_{\Z\times\ri\Uset}} f(k,\xi,\Bppavar) \,Q_{(k,\xi)}(d\Bppavar)\, \widetilde{\Bmm}(dk \otimes d\xi).
\end{align}
Now we consider the indices $\tau^\xi(i)=\tau^{\xi,\xi}(i)$ of jumps at $\xi$, defined in \eqref{78-45}.
In terms of the random measures introduced above, for $(k,\xi)\in \Z \times \ri \Uset$,
\[
\Bppa\{(k,\xi)\}>0 \; \iff \; \Bpp\{(k,\xi)\}=1 \; \iff \;
\B{\xi-}_k> \B{\xi+}_k
\;\iff \;k\in\{\tau^\xi(i):i\in\Z\}. \]
We condition on the event $\{\Bpp(0,\xi)=1\}$, in other words, consider the distribution of
$\{\tau^\xi(i)\}$ under $Q_{(0, \xi)}$. For this to be well-defined, we define these functions also on the space $\cM_{\Z \times \ri \Uset}$ in the obvious way: for $\nu\in\cM_{\Z \times \ri \Uset}$, the $\Z\cup\{\pm\infty\}$-valued functions $\tau^\xi(i)=\tau^\xi(i,\nu)$ are defined by the order requirement
\[ \dotsm<\tau^\xi(-1,\nu)<0\le \tau^\xi(0,\nu)<\tau^\xi(1,\nu)<\dotsm\] and the condition
\[
\text{for $k\in\Z$,} \quad \nu\{(k,\xi)\}>0\quad\text{if and only if}\quad k\in\{\tau^\xi(i,\nu):i\in\Z\}.
\]
Since $\Bppa$ is $\P$-almost surely a purely atomic measure, it follows from general theory that $Q_{(0, \xi)}$ is also supported on such measures. Furthermore, the conditioning itself forces $Q_{(0, \xi)}\{\nu: \tau^\xi(0,\nu)=0\}=1$. Thus the random integer points $\tau^\xi(i,\nu)$ are not all trivially $\pm\infty$ under $Q_{(0, \xi)}$. Connecting back to the notation of Section \ref{sec:exp}, for each $k \in \bbZ$, $\xi \in \ri \Uset$, each finite $A \subset \bbZ$ and $n_i \in \bbZ_+, r_i \in \bbR_+$ with $i \in A$, the Palm kernel introduced in that section is defined by
\begin{align}
&\bbP\bigl\{ \tau^\xi(+1)-\tau^\xi(i) =n_i, \depa^{\xi-}_{\tau^{\xi}(i)}-\depa^{\xi+}_{\tau^{\xi}(i)}>r_i \; \forall i \in A \,||\, \B{\xi-}_k > \B{\xi+}_k\bigr\} \label{eq:palmdef} \\
&\qquad=Q_{(k, \xi)}\bigl\{ \nu: \tau^\xi(i+1,\nu)-\tau^\xi(i,\nu)=n_i, \,\nu\{(\tau^\xi(i,\nu),\xi)\}> r_i \; \forall i \in A\bigr\}. \notag
\end{align}
\subsection{Statistics of instability points}
We turn to the proofs of the theorems of Section \ref{sec:exp}. These proofs make use of results from Appendices \ref{a:bus} and \ref{a:RW}.
\begin{proof}[Proof of Theorem \ref{th:palm1}]
By Corollary \ref{cor:B-q2}, the process $\{ \depa^\zeta_k-\depa^\eta_k\}_{k\in\Z}$ has the same distribution as $\{W^+_k\}_{k\in\Z}$ defined in \eqref{Wk}. An application of the appropriate mapping to these sequences produces the sequence
$\bigl\{\depa^\zeta_0-\depa^\eta_0, \, \tau^{\zeta, \eta}(i+1)-\tau^{\zeta, \eta}(i), \depa^\zeta_{\tau^{\zeta, \eta}(i)}-\depa^\eta_{\tau^{\zeta, \eta}(i)}: i\in\Z\bigr\}$
that appears in Theorem \ref{th:palm1}
and the sequence
$\{W^+_0, \, \sigma_{i+1}-\sigma_i, \, W^+_{\sigma_i}: i\in\Z\}$ that appears in Theorem \ref{th:78-70}. Hence these sequences also have identical distribution. ($W^+_{\sigma_i}=W_{\sigma_i}$ by \eqref{78-65}.)
The distributions remain equal when these sequences are conditioned on the positive probability events $ \depa^\zeta_0-\depa^\eta_0>0$ and $W^+_0>0$.
\end{proof}
It will be convenient to have notation for the conditional joint distribution that appears in \eqref{78-90} in Theorem \ref{th:palm1}.
For $0<\alpha\le\beta\le 1$ define probability distributions $q^{\alpha, \beta}$ on the product space $\Z^\Z\times[0,\infty)^\Z$ as follows. Denote the generic variables on this product space by $(\{\tau_i\}_{i\in\Z}, \{\Delta_k\}_{k\in\Z})$ with $\tau_i\in\Z$ and $0\le\Delta_k<\infty$. Given an integer $L>0$, integers
$n_{-L}<\dotsm < n_{-2}< n_{-1}<n_0=0<n_1<n_2< \dotsm<n_L$,
and positive reals $r_{-L}, \dotsc, r_L$, abbreviate $b_i=n_{i+1}-n_i$. The measure $q^{\alpha, \beta}$ is defined by
\be\label{qab4}\begin{aligned}
&q^{\alpha, \beta}\bigl\{\tau_{i}=n_i \text{ and } \Delta_{n_i}> r_i \text{ for } i\in\lzb-L,L\rzb, \, \Delta_k=0
\text{ for } k\in\lzb n_{-L},n_L\rzb\setminus\{ n_j\}_{j\in\lzb-L,L\rzb} \bigr\}\\
&=\biggl( \; \prod_{i=-L}^{L-1} C_{b_i-1}\, \frac{\alpha^{b_i-1}\beta^{b_i}}{(\alpha+\beta)^{2b_i-1}}\biggr) \cdot
\biggl( \; \prod_{i=-L}^{L} e^{-\alpha r_i}\biggr).
\end{aligned} \ee
To paraphrase the definition, the following holds under $q^{\alpha, \beta}$: $\tau_0=0$, $\Delta_k=0$ for $k\notin\{\tau_i\}_{i\in\Z}$, and the variables $\{\tau_{i+1}-\tau_{i}, \Delta_{\tau_i}\}_{i\in\Z}$ are mutually independent with marginal distribution
\be\label{qab6}
q^{\alpha, \beta}\{\tau_{i+1}-\tau_{i}=n, \Delta_{\tau_i}>r\}=C_{n-1}\, \frac{\alpha^{n-1}\beta^n}{(\alpha+\beta)^{2n-1}} e^{-\alpha r}\quad \text{ for } i\in\Z, \,n\ge 1,\, r\ge0.
\ee
Abbreviate $q^\alpha=q^{\alpha,\alpha}$ which has marginal $q^\alpha\{\tau_{i+1}-\tau_{i}=n, \Delta_{\tau_i}>r\}=C_{n-1}(\tfrac12)^{2n-1} e^{-\alpha r}$. As
$\beta\to\alpha$, $q^{\alpha, \beta}$ converges weakly to $q^\alpha$.
Theorem \ref{th:palm1} can now be restated by saying that, conditional on
$\depa^\zeta_0>\depa^\eta_0$, the variables $\bigl(\{\tau^{\zeta, \eta}(i)\}_{i\in\Z}, \, \{\depa^\zeta_k-\depa^\eta_{k}\}_{k\in\Z}\bigr)$
have joint distribution $q^{\alpha(\zeta), \alpha(\eta)}$.
Consequently, for a measurable set $A\subset\Z^\Z\times[0,\infty)^\Z$,
\be\label{78-92} \begin{aligned}
&\P\bigl[ \depa^\zeta_0>\depa^\eta_0, \, \bigl(\{\tau^{\zeta, \eta}(i)\}_{i\in\Z}, \, \{\depa^\zeta_k-\depa^\eta_{k}\}_{k\in\Z}\bigr) \in A \bigr]\\[4pt]
&=\P\bigl( \depa^\zeta_0>\depa^\eta_0\bigr) \, \P\bigl[ \bigl(\{\tau^{\zeta, \eta}(i)\}_{i\in\Z}, \, \{\depa^\zeta_k-\depa^\eta_{k}\}_{k\in\Z}\bigr) \in A \,\big\vert\, \depa^\zeta_0>\depa^\eta_0\,\bigr]
\\
&= \frac{\alpha(\eta)-\alpha(\zeta)}{\alpha(\eta)} \cdot q^{\alpha(\zeta), \alpha(\eta)}(A).
\end{aligned} \ee
The first probability on the last line came from \eqref{B4} and the second from Theorem \ref{th:palm1}.
\begin{proof}[Proof of Theorem \ref{th:B-palm1}]
Define $\Z\cup\{\pm\infty\}$-valued ordered indices
$\dotsm<\tau^{\zeta, \eta}_{-1}<0\le \tau^{\zeta, \eta}_0<\tau^{\zeta, \eta}_1<\dotsm$ as measurable functions of a locally finite measure $\nu\in\cM_{\Z \times \ri \Uset}$ by the rule
\be\label{78-105} \nu(\{k\}\times[\zeta,\eta])>0
\ \iff \ k\in\{\tau^{\zeta, \eta}_i:i\in\Z\}. \ee
If $\nu(\{k\}\times[\zeta,\eta])>0$ does not hold for infinitely many $k>0$ then $\tau^{\zeta, \eta}_i=\infty$ for large enough $i$, and analogously for
$k<0$. Definition \eqref{78-105} applied to the random measure $\Bppa=\sum_k\Bppa_k$ reproduces \eqref{78-45}.
Fix integers $K, N\in\N$ and $ \ell_{-N}\le \dotsm \le \ell_{-1}\le \ell_0=0 \le \ell_1\le \dotsm\le \ell_N$ and strictly positive reals $r_{-K}, \dotsc, r_K$.
Define the event
\be\label{Bze8} \begin{aligned}
H^{\zeta, \eta}=H(\zeta, \eta) =
&\bigcap_{1\le i\le N}\bigl\{ \nu: \tau^{\zeta, \eta}_{-i}\le \ell_{-i} \text{ and } \tau^{\zeta, \eta}_{i}\ge\ell_i \bigr\}
\\
&\qquad
\cap
\bigcap_{-K\le k\le K}\bigl\{ \nu: \nu(\{k\}\times]\zeta, \eta]) < r_k\bigr\}
\end{aligned} \ee
on the space $\cM_{\Z \times \ri \Uset}$.
Note the monotonicity
\be\label{Bze-mon}
H^{\zeta, \eta}\subset H^{\zeta', \eta'}\quad\text{ for } \ [\zeta', \eta']\subset [\zeta, \eta].
\ee
Abbreviate $H^\xi=H^{\xi, \xi}$. Recall the measures $q^{\alpha, \beta}$ defined in \eqref{qab4}.
The analogous event under the measures $q^{\alpha, \beta}$ on the space $\Z^\Z\times[0,\infty)^\Z$ is denoted by
\be\label{Bze18} \begin{aligned}
H_q =\bigl\{ (\{\tau_i\}_{i\in\Z}, \{\Delta_k\}_{k\in\Z})\in \Z^\Z\times[0,\infty)^\Z : \ &\tau_{-i}\le \ell_{-i} \text{ and } \tau_{i}\ge\ell_i \text{ for } i\in\lzb 1,N\rzb, \, \\
&\quad \Delta_k< r_k\text{ for } k\in\lzb-K,K\rzb\bigr\}.
\end{aligned}\ee
Fix $\zeta\prec\eta$ in $\ri\Uset$.
We prove the theorem by showing that
\be\label{Bze4} \begin{aligned}
Q_{(0,\xi)}(H^\xi) =q^{\alpha(\xi)}(H_q)
\qquad\text{for Lebesgue-almost every $\xi\in\,]\zeta, \eta]$.}
\end{aligned} \ee
This equality comes from separate arguments for upper and lower bounds.
\bigskip
{\it Upper bound proof.}
Define a sequence of nested partitions $\zeta=\zeta^n_0\prec\zeta^n_1\prec\dotsm\prec \zeta^n_n=\eta$. For each $n$ and $\xi\in\,]\zeta, \eta]$, let $]\zeta^n(\xi), \eta^n(\xi)]$ denote the unique interval $]\zeta^n_i, \zeta^n_{i+1}]$ that contains $\xi$. Assume that, as $n\nearrow\infty$, the mesh size $\max_i\abs{\zeta^n_{i+1}-\zeta^n_i} \to 0$. Consequently, for each $\xi\in\,]\zeta, \eta]$, the intervals $]\zeta^n(\xi), \eta^n(\xi)]$ decrease to the singleton $\{\xi\}$.
The key step of this upper bound proof is that for all $m$ and $i$ and Lebesgue-a.e.\ $\xi\in\,]\zeta, \eta]$,
\be\label{79-43}\begin{aligned} &Q_{(0, \xi)}(H^{\zeta^m_i, \zeta^m_{i+1}})
= \lim_{n\to\infty} \P\bigl\{ \Bppa \in H^{\zeta^m_i, \zeta^m_{i+1}} \,\big\vert\, \Bpp_{0}(\,]\zeta^n(\xi), \eta^n(\xi)]\,)\ge 1 \bigr\} .
\end{aligned} \ee
This limit is a special case of Theorem 6.32(iii) in Kallenberg \cite{Kal-17}, for the simple point process $\Bpp$ and the sets $B_n=\{0\}\times(\zeta^n(\xi), \eta^n(\xi)]\searrow\{(0,\xi)\}$.
The proof given for Theorem 12.8 in \cite{Kal-83} can also be used to establish this limit. Theorem 12.8 of \cite{Kal-83} by itself is not quite adequate because we use the Palm kernel for the measure $\Bppa$ which is not the same as $\Bpp$.
If we take $\xi\in \,]\zeta^m_i, \zeta^m_{i+1}]$, then for $n\ge m$,
$]\zeta^n(\xi), \eta^n(\xi)]\subset \,]\zeta^m(\xi), \eta^m(\xi)]=\,]\zeta^m_i, \zeta^m_{i+1}]$. Considering all $\xi$ in the union $]\zeta, \eta]=\bigcup_i \,]\zeta^m_i, \zeta^m_{i+1}]$, we have that for any fixed $m$ and Lebesgue-a.e.\ $\xi\in\,]\zeta, \eta]$,
\begin{align*}
Q_{(0, \xi)}(H^{\zeta^m(\xi), \eta^m(\xi)}) &=
\lim_{n\to\infty} \P\bigl\{ \Bppa \in H^{\zeta^m(\xi), \eta^m(\xi)} \,\big\vert\, \Bpp_{0}(\,]\zeta^n(\xi), \eta^n(\xi)]\,)\ge 1 \bigr\} \\[3pt]
& \le
\varliminf_{n\to\infty} \P\bigl\{ \Bppa \in H^{\zeta^n(\xi), \eta^n(\xi)} \,\big\vert\, \Bpp_{0}(\,]\zeta^n(\xi), \eta^n(\xi)]\,)\ge 1 \bigr\} .
\end{align*}
The inequality is due to \eqref{Bze-mon}.
Interpreting \eqref{78-92} in terms of the random measures $\Bppa$ and $\Bpp$ and referring to \eqref{Bze8} and \eqref{Bze18} gives the identity
\[ \P\bigl\{ \Bppa \in H^{\zeta^n(\xi), \eta^n(\xi)} \,\big\vert\, \Bpp_{0}(\,]\zeta^n(\xi), \eta^n(\xi)]\,)\ge 1 \bigr\}
= q^{\alpha(\zeta^n(\xi)), \alpha(\eta^n(\xi))}(H_q). \]
As $(\zeta^n(\xi), \eta^n(\xi)]\searrow\{\xi\}$, the parameters converge: $\alpha(\zeta^n(\xi)), \alpha(\eta^n(\xi))\to\alpha(\xi)$.
Consequently the distribution $q^{\alpha(\zeta^n(\xi)), \,\alpha(\eta^n(\xi))}$
converges to $q^{\alpha(\xi)}$.
Hence
\begin{align*}
\lim_{n\to\infty} \P\bigl\{ \Bppa \in H^{\zeta^n(\xi), \eta^n(\xi)} \,\big\vert\, \Bpp_{0}(\,]\zeta^n(\xi), \eta^n(\xi)]\,)\ge 1 \bigr\}
=q^{\alpha(\xi)}(H_q).
\end{align*}
In summary, we have for all $m$ and Lebesgue-a.e.\ $\xi\in \,]\zeta, \eta]$,
\be\label{79-46} \nn \begin{aligned}
&Q_{(0, \xi)}(H^{\zeta^m(\xi), \eta^m(\xi)}) \le q^{\alpha(\xi)}(H_q).
\end{aligned} \ee
Let $m\nearrow\infty$ so that $H^{\zeta^m(\xi), \eta^m(\xi)}\nearrow H^\xi$, to obtain the upper bound
\be\label{79-47} \begin{aligned}
&Q_{(0, \xi)}(H^\xi) \le q^{\alpha(\xi)}(H_q)
\end{aligned} \ee
for Lebesgue-a.e.\ $\xi\in \,]\zeta, \eta]$.
\bigskip
{\it Lower bound proof.}
Let $\zeta=\zeta_0\prec\zeta_1\prec\dotsm\prec \zeta_\ell=\eta$ be a partition of the interval $[\zeta, \eta]$ and set $\alpha_j=\alpha(\zeta_j)$.
In order to get an estimate below, let $\mvec=(m_i)_{1\le\abs i\le N}$ be a $2N$-vector of
integers such that $m_i<\ell_i$ for $-N\le i\le -1$ and $m_i>\ell_i$ for $1\le i\le N$. Define the subset $H^{\mvec}_q$ of $H_q$ from \eqref{Bze18} by truncating the coordinates $\tau_i$:
\be\label{Bze20} \begin{aligned}
H^{\mvec}_q =\bigl\{ (\{\tau_i\}_{i\in\Z}, \{\Delta_k\}_{k\in\Z})\in \Z^\Z\times[0,\infty)^\Z : \ &m_{-i}\le \tau_{-i}\le \ell_{-i} \text{ and } \ell_i \le \tau_{i}\le m_i \\
& \text{ for } i\in\lzb 1,N\rzb, \, \Delta_k< r_k\text{ for } k\in\lzb-K,K\rzb\bigr\}.
\end{aligned}\ee
On the last line in the following computation, $c_1$ is a constant that depends on the parameters $\alpha(\zeta)$ and $\alpha(\eta)$ and on the quantities in \eqref{Bze20}:
\begin{align*}
&\int_{]\zeta,\eta]} Q_{(0,\xi)}(H^\xi)\,\Bmm_0(d\xi) = \sum_{j=0}^{\ell-1} \int_{]\zeta_j,\zeta_{j+1}]} Q_{(0,\xi)}(H^\xi)\,\Bmm_0(d\xi) \\[4pt]
&\ge \sum_{j=0}^{\ell-1} \int_{]\zeta_j,\zeta_{j+1}]} Q_{(0,\xi)}(H^{\zeta_j,\zeta_{j+1}})\,\Bmm_0(d\xi)
= \sum_{j=0}^{\ell-1} \E\bigl[ \Bpp_0(\,]\zeta_j,\zeta_{j+1}]\,) \cdot \ind_{H^{\zeta_j,\zeta_{j+1}}}(\Bppa) \bigr]
\\[4pt]
&\ge \sum_{j=0}^{\ell-1}
\P\bigl\{ \Bpp_0(\,]\zeta_j,\zeta_{j+1}]\,) \ge 1,\, \Bppa\in H^{\zeta_j,\zeta_{j+1}} \bigr\} \\
&= \sum_{j=0}^{\ell-1} \frac{\alpha_{j+1}-\alpha_j}{\alpha_{j+1}} \,q^{\alpha_j, \alpha_{j+1}}(H_q)
\; \ge \; \sum_{j=0}^{\ell-1} \frac{\alpha_{j+1}-\alpha_j}{\alpha_{j+1}} \,q^{\alpha_j, \alpha_{j+1}}(H^{\mvec}_q)
\\ &
\ge \sum_{j=0}^{\ell-1} \frac{\alpha_{j+1}-\alpha_j}{\alpha_{j+1}} \, q^{\alpha_{j+1}} (H^{\mvec}_q) \cdot \bigl(1-c_1(\alpha_{j+1}-\alpha_j)\bigr) .
\end{align*}
The steps above come as follows. The second equality uses the characterization \eqref{79-50-50} of the kernel $Q_{(0,\xi)}$.
The third equality is from \eqref{78-92}. The second last inequality is from $H^{\mvec}_q\subset H_q$. The last inequality is from Lemma \ref{lm:palm10} below, which is valid once the mesh size $\max(\alpha_{j+1}-\alpha_j)$ is small enough relative to the numbers $\{m_i, \ell_i\}$.
The function $\alpha\mapsto q^\alpha (H^{\mvec}_q)$ is continuous in the Riemann sum approximation on the last line of the calculation above.
Let $\max(\alpha_{j+1}-\alpha_j)\to 0$ to obtain the inequality
\begin{align*}
&\int_{]\zeta,\eta]} Q_{(0,\xi)}(H^\xi) \,\Bmm_0(d\xi)
\ge \int_{\alpha(\zeta)}^{\alpha(\eta)} q^\alpha (H^{\mvec}_q) \,\frac{d\alpha}\alpha
= \int_{]\zeta,\eta]} q^{\alpha(\xi)} (H^{\mvec}_q)\,\Bmm_0(d\xi).
\end{align*}
Let $m_i\searrow-\infty$ for $-N\le i\le -1$ and $m_i\nearrow\infty$ for $1\le i\le N$.
The above turns into
\be\label{79-104} \begin{aligned}
\int_{]\zeta,\eta]} Q_{(0,\xi)}(H^\xi)\,\Bmm_0(d\xi)
\ge
\int_{]\zeta,\eta]} q^{\alpha(\xi)} (H_q)\,\Bmm_0(d\xi).
\end{aligned} \ee
\medskip
The upper bound \eqref{79-47} and the lower bound \eqref{79-104} together imply the conclusion \eqref{Bze4}.
\end{proof}
The proof of Theorem \ref{th:B-palm1} is complete once we verify the auxiliary lemma used in the calculation above.
\begin{lemma} \label{lm:palm10}
Let the event $H^{\mvec}_q$ be as defined in \eqref{Bze20}.
Fix $0<\underline\alpha<\wb\alpha<1$. Then there exist constants $\e, c_1\in(0,\infty)$ such that
\begin{align*}
q^{\alpha, \beta}(H^{\mvec}_q) \ge q^{\beta}(H^{\mvec}_q) \cdot \bigl(1-c_1(\beta-\alpha)\bigr)
\end{align*}
for all $\alpha,\beta\in[\underline\alpha,\wb\alpha]$ such that $\alpha\le\beta\le\alpha+\e$. The constants $\e, c_1\in(0,\infty)$ depend on $\underline\alpha$, $\wb\alpha$, and the parameters $\ell_i, m_i$ and $r_k$ in \eqref{Bze20}.
\end{lemma}
\begin{proof} Let
\begin{align*}
\cA= \bigl\{ &\pvec=(p_i)_{-N\le i\le N} \in\Z^{2N+1} : p_0=0, \, p_i<p_j \text{ for } i<j, \\
&\qquad \qquad\qquad
m_{-i}\le p_{-i} \le \ell_{-i} \text{ and } \ell_i\le p_{i}\le m_i \; \forall i\in\lzb1,N\rzb\bigr\}
\end{align*}
be the relevant finite set of integer-valued $(2N+1)$-vectors for the decomposition below. For each $\pvec\in\cA$ let $\cK(\pvec)=\{p_i: i\in\lzb-N,N\rzb, p_i\in \lzb-K,K\rzb\}$ be the set of coordinates of $\pvec$ in $\lzb-K,K\rzb$. Abbreviate $b_i=p_{i+1}-p_i$.
Recall that, under $q^{\alpha, \beta}$, $\tau_0=0$, that $\Delta_k<r_k$ holds with probability one if $k\notin\{\tau_i\}$, and the independence in \eqref{qab6}. The factors $d_k>0$ below that satisfy $ 1-e^{-\alpha r_k} \ge (1-e^{-\beta r_k}) (1- d_k(\beta-\alpha))$ can be chosen uniformly for $\alpha\le\beta$ in $[\underline\alpha,\wb\alpha]$, as functions of $\underline\alpha$, $\wb\alpha$, and $\{r_k\}$. Now compute:
\begin{align*}
q^{\alpha, \beta}(H^{\mvec}_q)
&=q^{\alpha, \beta}\bigl\{ m_i\le \tau_{-i}\le \ell_{-i} \text{ and } \ell_i\le \tau_{i}\le m_i \; \forall i\in\lzb1,N\rzb, \, \Delta_k< r_k\text{ for } k\in\lzb-K,K\rzb\bigr\} \\[4pt]
&=\sum_{\pvec\,\in\,\cA} q^{\alpha, \beta}\bigl\{ \tau_{i}=p_i \text{ for } i\in\lzb-N,N\rzb, \, \Delta_k< r_k\text{ for } k\in\lzb-K,K\rzb\bigr\}
\\[4pt]
&=\sum_{\pvec\,\in\,\cA} q^{\alpha, \beta}\bigl\{ \tau_{i+1}-\tau_{i}=b_i \ \forall i\in\lzb-N,N-1\rzb\bigr\} \cdot \prod_{k\,\in\,\cK(\pvec)} (1-e^{-\alpha r_k}) \\
&\ge\sum_{\pvec\,\in\,\cA} \; \biggl( \; \prod_{i=-N}^{N-1} C_{b_i-1}\, \frac{\alpha^{b_i-1}\beta^{b_i}}{(\alpha+\beta)^{2b_i-1}} \biggr)
\cdot \prod_{k\,\in\,\cK(\pvec)} (1-e^{-\beta r_k}) \bigl(1- d_k(\beta-\alpha)\bigr) \\
& \ge \sum_{\pvec\,\in\,\cA} \; \biggl( \; \prod_{i=-N}^{N-1} C_{b_i-1}\, \biggl(\frac12\biggr)^{2b_i-1} \biggr) \biggl(\; \prod_{k\,\in\,\cK(\pvec)} (1-e^{-\beta r_k}) \biggr) \cdot \bigl(1-c_1(\beta-\alpha)\bigr) \\
&=\sum_{\pvec\,\in\,\cA} q^\beta\bigl\{ \tau_{i+1}-\tau_{i}=b_i \ \forall i\in\lzb-N,N-1\rzb, \, \Delta_k< r_k\text{ for } k\in\lzb-K,K\rzb\bigr\}
\cdot \bigl(1-c_1(\beta-\alpha)\bigr) \\
&= q^\beta(H^{\mvec}_q) \cdot \bigl(1-c_1(\beta-\alpha)\bigr).
\end{align*}
To get the inequality above, (i) apply Lemma \ref{a:lm:t500} to the first factor in parentheses with $\e$ chosen so that $0<\e<\underline\alpha/b_i$ for all $\pvec\in\cA$, and (ii) set $c_1= \sum_{k=-K}^K d_k$.
\end{proof}
In the proofs that follow, we denote the indicators of the locations of the positive atoms of a measure $\nu\in\cM_{\Z\times\ri\Uset}$ by
$u_k(\nu,\xi) = u_k(\xi)=\ind[\nu\{(k,\xi)\}>0]$ for $(k,\xi)\in\Z\times\ri\Uset$. Applied to the random measure $\Bppa$ this gives $u_k(\Bppa, \xi)=\Bpp_k(\xi)$.
\begin{lemma} \label{lem:palmrw}
For Lebesgue-almost every $\xi\in\ri\Uset$ and all $m\in\Z$,
\be\label{rw123} Q_{(m,\xi)}\bigl[ \nu : \{\umeas_{m+k}(\nu,\xi)\}_{k\in\Z} \in A\bigr]
= \bfP(A)
\ee
for all Borel sets $A\subset\{0,1\}^\Z$. \end{lemma}
\begin{proof}
For $m=0$, \eqref{rw123} comes from a comparison of \eqref{rw113} and \eqref{rw117}. For general $m$ it then follows from the shift-invariance of the weights $\w$.
\end{proof}
\begin{proof}[Proof of Theorem \ref{thm:ladder-xi}]
Take $A\subset\{0,1\}^\Z$ as in the statement of Theorem \ref{thm:ladder-xi}. Fix $\zeta \prec \eta $ in $\ri \Uset$ and let $N \in \bbN$.
We restrict the integrals below to the compact set $[-N,N] \times [\zeta,\eta]$ with the indicator
\[ g(k,\xi,\Bppavar) = \one_{[-N,N]\times [\zeta,\eta]}(k,\xi) \]
and then define on $\Z \times \ri \Uset \times \sM_{\Z \times \ri \Uset} $
\begin{align*}
f(k,\xi,\Bppavar) &= g(k,\xi,\Bppavar) \cdot \one_{\{(u_\ell\{\xi\} : \ell \in \Z) \,\in\, A\}}(\xi,\Bppavar).
\end{align*}
By the definition \eqref{79-50-50} of the Palm kernel,
\begin{align*}
&\E\left[\int_{\Z \times \ri \Uset} f(k,\xi,\Bppa) \,\Bpp(dk \otimes d\xi ) \right]
= \int_{[-N,N] \times [\zeta,\eta]} Q_{(k,\xi)}\{(u_\ell\{\xi\} : \ell \in \Z) \in A\} \,\widetilde{\Bmm}( dk \otimes d\xi) \\
&\qquad\qquad\qquad=\int_{[-N,N] \times [\zeta,\eta]} Q_{(k,\xi)}\{(u_{k+\ell}\{\xi\} : \ell \in \Z) \in A\} \,\widetilde{\Bmm}(dk \otimes d\xi) \\
&\qquad\qquad\qquad
= \int_{[-N,N] \times [\zeta,\eta]} \widetilde{\Bmm}(dk \otimes d\xi) = \E\left[\int_{\Z \times \ri \Uset} g(k,\xi,\Bppa) \Bpp(dk \otimes d\xi ) \right].
\end{align*}
The second equality used shift-invariance of $A$ and the third equality used \eqref{rw123} and $\bfP(A)=1$.
The left-hand side and the right-hand side are both finite because the integrals are restricted to the compact set $[-N,N] \times [\zeta,\eta]$. Since $\Bpp$ is a positive random measure, it follows that
\begin{align*}
\bbP\left(\int_{\Z \times \ri \Uset} f(k,\xi,\Bppa) \,\Bpp(dk \otimes d\xi ) = \int_{\Z \times \ri \Uset} g(k,\xi,\Bppa) \,\Bpp(dk \otimes d\xi )\right)=1.
\end{align*}
As $\zeta,\eta,$ and $N$ were arbitrary, we conclude that $\P$-almost surely $(\Bpp_\ell\{\xi\} : \ell \in \Z) \in A$ for all $(k,\xi)\in\Z\times\ri\Uset$ such that $\Bpp\{(k,\xi)\}=1$. Lemma \ref{lm:downright} applied to the $x$-axis ($x_i=ie_1$) then shows that $\xi\in\aUset$ if and only if $\Bpp\{(k,\xi)\}=1$ for some $k$.
\end{proof}
\begin{lemma}\label{lm:UB-aux}
Assume \eqref{exp-assump}.
Then for any $\delta\in(0,1)$, $n\in\N$, and $\zeta\in\ri\Uset$ we have
\[\P\Bigl\{\exists\xi\in[\zeta,e_1[\,: \Bpp( \lzb0,n\rzb\times\{\xi\})
>2\delta n+1\Bigr\}\le 2(n+1)\Bigl(\frac{(1-\delta/2)^{2-\delta}}{(1-\delta)^{1-\delta}}\Bigr)^{n}\log\alpha(\zeta)^{-1}.\]
\end{lemma}
\begin{proof}
Let $\{\Delta_j\}_{j\in\N}$ be i.i.d.\ random variables with probability mass function
$p(n)=C_{n-1}{2^{1-2n}}$ for $n\in\N$.
For $k\in\lzb 0,n\rzb$ and $\xi\in\ri\Uset$ use a union bound, translation, and \eqref{rw113}
to write
\begin{align*}
&Q_{(k,\xi)}\Bigl\{\,\sum_{i=0}^n u_i(\xi)>2\delta n+1\Bigr\}
\le Q_{(k,\xi)}\Bigl\{\,\sum_{i=k-n}^{k+n} u_i(\xi)>2\delta n+1\Bigr\}\\
&\quad= Q_{(0,\xi)}\Bigl\{\,\sum_{i=-n}^n u_i(\xi)>2\delta n+1\Bigr\}\\
&\quad\le Q_{(0,\xi)}\Bigl\{\,\sum_{i=1}^n u_i(\xi)>\delta n\Bigr\}
+ Q_{(0,\xi)}\Bigl\{\,\sum_{i=-n}^{-1} u_i(\xi)>\delta n\Bigr\}
\le 2P\Bigl\{\,\sum_{j=1}^{\ce{\delta n}}\Delta_j\le n\Bigr\}.
\end{align*}
Using
the generating function $f(s)=\sum_{n\ge 0} C_ns^n=\tfrac12(1-\sqrt{1-4s}\,)$ of Catalan numbers we obtain for $0<s<1$,
\begin{align*}
P\Bigl\{\,\sum_{j=1}^{\ce{\delta n}}\Delta_j\le n\Bigr\}
&\le s^{-n}E[s^{\Delta}]^{\delta n}
= s^{-n} \Bigl(2\sum_{n=1}^\infty C_{n-1} \,(s/4)^n\Bigr)^{\delta n}\\
&= s^{-n}\Bigl(\frac{s}{2} \sum_{k=0}^\infty C_k \,(s/4)^k\Bigr)^{\delta n}
= s^{-n}\bigl(1-\sqrt{1-s}\bigr)^{\delta n}.
\end{align*}
Take $s=\frac{4(1-\delta)}{(2-\delta)^2}<1$ in the upper bound above to get
\[Q_{(k,\xi)}\Bigl\{\,\sum_{i=0}^n u_i(\xi)>2\delta n+1\Bigr\}\le 2\Bigl(\frac{(1-\delta/2)^{2-\delta}}{(1-\delta)^{1-\delta}}\Bigr)^{n}.\]
Apply \eqref{79-50-50} to write
\begin{align*}
&\E\Bigl[\,\int_{\ri\Uset}\one\{\xi\in[\zeta,e_1[\,\}\one\bigl\{
\Bpp( \lzb0,n\rzb\times\{\xi\})
>2\delta n+1\bigr\}\Bpp_k(d\xi)\Bigr]\\
&\qquad=\int_{\ri\Uset} \one\{\xi\in[\zeta,e_1[\,\}Q_{(k,\xi)}\Bigl\{\,\sum_{i=0}^n u_i(\xi) >2\delta n+1\Bigr\}\Bmm_k(d\xi)\\
&\qquad\le 2\Bigl(\frac{(1-\delta/2)^{2-\delta}}{(1-\delta)^{1-\delta}}\Bigr)^{n}\int_{\ri\Uset}\one\{\xi\in[\zeta,e_1[\,\}\Bmm_k(d\xi)
\overset{\eqref{Bmm}}= 2\Bigl(\frac{(1-\delta/2)^{2-\delta}}{(1-\delta)^{1-\delta}}\Bigr)^{n}\log\alpha(\zeta)^{-1}.
\end{align*}
To complete the proof, add over $k\in\lzb0,n\rzb$ and observe that
\begin{align*}
&\int_{\ri\Uset}\one\{\xi\in[\zeta,e_1[\,\}\one\bigl\{
\Bpp( \lzb0,n\rzb\times\{\xi\})
>2\delta n+1\bigr\}\sum_{k=0}^n\Bpp_k(d\xi)\\
&\qquad\qquad\ge\one\Bigl\{\exists\xi\in[\zeta,e_1[\,:\Bpp( \lzb0,n\rzb\times\{\xi\})
>2\delta n+1\Bigr\}.\qedhere
\end{align*}
\end{proof}
\begin{proof}[Proof of Theorem \ref{thm:density-UB}]
The result follows from Theorem \ref{thm:density-UB2} below and the observation that
for any $\e>0$, $\delta_n=2\sqrt{n^{-1}\log n}$
satisfies the summability condition in that theorem.
\end{proof}
\begin{theorem}\label{thm:density-UB2}
Assume \eqref{exp-assump} and fix $i\in\{1,2\}$.
Consider a sequence $\delta_n\in(0,1)$ such that $\sum n^2 e^{-n\delta_n^2}<\infty$. Then for any $\zeta\in\ri\Uset$
\begin{align}\label{claim2-UB}
\P\Bigl\{\exists n_0:\forall \xi\in[\zeta,e_2[,\forall n\ge n_0:\sum_{x\in[0,n]^2}\one\bigl\{\xi \in \supp\mu_{x,x+e_i}\bigr\}\le n^2\delta_n\Bigr\}=1.
\end{align}
The same result holds when $[0,n]^2$ is replaced by any one of $[-n,0]^2$, $[0,n]\times[-n,0]$, or $[-n,0]\times[0,n]$.
\end{theorem}
\begin{proof}
Apply Lemma \ref{lm:UB-aux} and a union bound to get that for any $j\in\{1,2\}$, $\delta\in(0,1)$, $n\in\N$, and $\zeta\in\ri\Uset$,
\[\P\Bigl\{\exists\xi\in[\zeta,e_1[\,:\sum_{x\in[0,n]^2}\rho^j_x(\xi) \ge(2\delta n+1)(n+1)\Bigr\}\le 2(n+1)^2\Bigl(\frac{(1-\delta/2)^{2-\delta}}{(1-\delta)^{1-\delta}}\Bigr)^{n}\log\alpha(\zeta)^{-1}.\]
A Taylor expansion gives
\[\log\Bigl(\frac{(1-\delta/2)^{2-\delta}}{(1-\delta)^{1-\delta}}\Bigr)
=-\delta^2/4+\cO(\delta^3).\]
Thus, we see that for any $\zeta\in\ri\Uset$, for any sequence $\delta_n\in(0,1)$ such that $\sum n^2 e^{-n\delta_n^2}<\infty$, we have
\[\P\Bigl\{\exists n_0\in\N:\forall \xi\in[\zeta,e_1[\,,\forall n\ge n_0:\sum_{x\in[0,n]^2}\rho^j_x(\xi)\le n^2\delta_n\Bigr\}=1.\]
The result for the other three sums comes similarly.
\end{proof}
| 154,120
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\begin{document}
\maketitle
\begin{abstract}
We give a necessary and sufficient condition for a simple closed curve on the boundary of a genus two handlebody to decompose the handlebody into $T\times I$ ($T$ is a torus with one boundary component).
We use this condition to decide whether a simple closed curve on a genus two Heegaard surface is a GOF-knot (genus one fibered knot) which induces the Heegaard splitting.
By using this, we determine the number and the positions with respect to the Heegaard splittings of GOF-knots in the 3-manifolds with reducible genus two Heegaard splittings.
This is another proof of results of Morimoto \cite{12} and Baker \cite{2}, \cite{3}.
\end{abstract}
\section{Introduction} \label{sec1}
In \cite{1}, Alexander proved that every closed orientable 3-manifold has a fibered link
and by Myers \cite{13} and Gonz\'alez-Acu\~na this result was improved so that every closed orientable 3-manifold has a fibered knot.
Thus focusing on fibered knots is not restricting for a study of 3-manifolds and many works for 3-manifolds have been done
by using the connection with open-book decompositions, contact structures and so on.
In this paper, we handle fibered knots whose fiber is a genus one surface.
Though not every manifold has such knots, they can be relatively easily studied because of their low genus and they might be test cases for higher genus.
A fibered knot in a closed orientable 3-manifold $M$ whose fiber is a torus with one boundary component is called a GOF-knot (genus one fibered knot) in $M$.
Classically, it is known that all GOF-knot in $S^3$ are the (left and right hand) trefoil and the figure eight knot \cite{5}.
In \cite{12}, Morimoto investigated the number of GOF-knots in some lens spaces by using their monodromies.
In \cite{3}, Baker counted GOF-knots in each lens space by using the correspondence between GOF-knots in all 3-manifolds and closed 3-braids with axes in $S^3$.
He also counted GOF-knots in non-prime 3-manifolds by the same way in his other works \cite{2}.
By the correspondence, the monodromy of a GOF-knot and its fiber can be calculated,
however its position in the underlying 3-manifold seems not to be easily found in the context of Heegaard splittings,
which give a fundamental method for representing a 3-manifold.\\
\ The purpose of this paper is to reveal the positions of GOF-knots in some special 3-manifolds by putting them on the (almost unique) standard Heegaard surface.
We first prepare some terminologies and techniques in Section \ref{sec2}
and then we prove the relation between GOF-knots and simple closed curves on Heegaard surfaces in Section \ref{sec3}.
Using the results obtained in Section \ref{sec3}, we investigate GOF-knots in individual cases in Section \ref{sec4}.
In our method, the position of a GOF-knot is easily found since we put it on a genus two Heegaard surface.
The operation used in Section \ref{sec4} is similar to that in Cho and Koda \cite{7}, \cite{8}, \cite{9}. \\\\
\subsection*{Acknowledgements}
I would like to thank my supervisor, Takuya Sakasai
for giving me many suggestions for this paper.
Also, I would like to thank everyone in the laboratory I belong to for providing me with meaningful discussions.\\\\\\
\section{Preliminary} \label{sec2}
\subsection*{Heegaard splittings}
{\it Heegaard splitting } is a method for decomposing a closed orientable 3-manifold into two handlebodies of the same genus.
The closed orientable surface which is the common boundary of two handlebodies is called a {\it Heegaard surface}.
The genus of the Heegaard surface of a Heegaard splitting is called the genus of the Heegaard splitting.
It is known that every closed orientable 3-manifold admits a Heegaard splitting.
A Heegaard splitting is called {\it reducible} if there is an essential simple closed curve (called a {\it reducing curve}) on the Heegaard surface
which bounds a disk in each of two handlebodies.
The sphere in the manifold obtained by pasting two disks bounded by a reducing curve along their boundaries is called a {\it Haken sphere} of the Heegaard splitting.
If the genus of a reducible Heegaard splitting is greater than one, there must be a separating reducing curve.
In particular, every reducible genus two Heegaard splitting is decomposed into two genus one Heegaard splittings by cutting along a Haken sphere. \\
\subsection*{Heegaard diagrams}
A genus $g$ Heegaard splitting can be represented as a standard closed orientable genus $g$ surface in $S^3$ with a pair of $g$ essential curves on it as follows:
Identify the Heegaard surface $\Sigma$ with a standard closed orientable surface $S$ in $S^3$.
Let $D_1$, \dots , $D_g$ be disks in one handlebody separated by $\Sigma$ such that they cut this handlebody into a 3-ball and $E_1$, \dots , $E_g$ be disks
in the other handlebody such that they cut it into a 3-ball.
Then we get a pair of $g$ curves, {$\partial D_1$, \dots , $\partial D_g$} and {$\partial E_1$, \dots , $\partial E_g$}.
Draw this curves on $S$. This presentation is called a {\it Heegaard diagram}.
We can reconstruct a Heegaard splitting by a Heegaard diagram: Paste disks along one of pair of curves in the interior of the standard surface,
paste disks along the other curves in the exterior of the standard surface and paste two 3-balls along remaining spheres. \\
\subsection*{Fibered links}
Let $M$ be a closed orientable 3-manifold and $L$ be a link in $M$. $L$ is called a fibered link if $Cl(M \setminus N(L))$ is a fiber bundle over $S^{1}$
whose fiber is an orientable surface and the boundary of each fiber is isotopic to $L$ in $N(L)$,
where $N(L)$ is a regular closed neighborhood of $L$ in $M$ and $Cl(\cdot )$ is the closure.
If $L$ is a knot and the fiber is a torus $T$ with one boundary component, $L$ is called a GOF-knot in $M$. Let $L$ be a fibered link in $M$ and $F$ be its fiber.
By thickening $F$, $Cl(M \setminus N(L))$ is decomposed into two handlebodies of the same genus, $g$.
Moreover, by the property of fibered links we get a genus $g$ Heegaard splitting of $M$
such that $L$ is on its Heegaard surface and $L$ decomposes each handlebody to $F\times I$.
In particular, if $K$ is a GOF-knot in $M$, there is a genus two Heegaard splitting of $M$ and $K$ is on the Heegaard surface,
decomposing each handlebody to $T\times I$.
In this paper, two fibered links $L_1$ and $L_2$ in $M$ are said to be equivalent if there is a fiber preserving self-homeomorphism of $M$ sending $L_1$ to $L_2$.\\
\subsection*{Plumbing}
Let $L_1$ and $L_2$ be two fibered links in two closed orientable 3-manifolds $M_1$ and $M_2$ respectively.
Let $F_i$ be a fiber of $L_i$ in $M_i$ ($i=0,1$). Then we can construct a fibered link in $M_1\# M_2$ from $L_1$ and $L_2$ as follows:
Let $\alpha _i$ be a properly embedded essential arc in $F_i$ and $D_i$ be a small closed neighborhood of $\alpha _i$ in $F_i$.
$D_i$ can be identified with $\alpha _{i}\times [-1,1]$.
We construct a new surface $F$ by pasting $D_1$ and $D_2$ so that for every $t\in [-1,1]$, $\alpha _{1}\times \{t\}$ is identified with
an arc intersecting once to $\alpha _2\times \{s\}$ for every $s\in [-1,1]$.
For such an operation, we say that $F$ is obtained by the $plumbing$ of $F_1$ and $F_2$.
In \cite{14}, Stallings showed that a surface obtained by the plumbing of two surfaces which are
fibers of two fibered links in $S^3$ is also a fiber of a fibered link in $S^3$.
This statement can be generalized to arbitrary closed orientable 3-maniflds. Thus, $\partial F$ is a fibered link in $M_1\# M_2$ with $F$ as a fiber surface.\\
\subsection*{Monodromy}
Let $L$ be a fibered link in a closed orientable 3-manifold $M$ and $F$ be a fiber surface of $L$. Then $Cl(M \setminus N(L))$ is a $F$-bundle over $S^1$.
It is obtained from $F\times [0,1]$ by pasting $F\times \{0\}$ and $F\times \{1\}$ using an orientation preserving self-homeomorphism of $F$.
This map is called the monodromy of $L$ (and $F$) or the monodromy of $Cl(M \setminus N(L))$.
Let $f$ and $g$ be two orientation preserving self-homeomorphisms of $F$.
Note that under orientation- and fiber preserving homeomorphisms, $F$-bundles over $S^1$ using $f$ and $g$ are equivalent if and only if
$f$ and $g$ are in the same conjugacy class of the mapping class group of $F$ (each component of $\partial F$ is fixed setwise).
If we work under fiber preserving homeomorphisms, an orientation reversing map can be added.
In particular the monodromy of a GOF-knot is classified in the conjugacy classes in $GL_{2}(\mathbb{Z})$.
In fact we can say about monodromies under the plumbing as followings:
Let $F_1$ and $F_2$ are fibers of two fibered links in two closed orientable 3-manifolds $M_1$ and $M_2$ respectively,
and let $f_1$ and $f_2$ be monodromies of $F_1$ and $F_2$.
Then the monodromy of $F$, obtained by the plumbing of $F_1$ and $F_2$ is $\tilde{f_{1}}\circ \tilde{f_{2}}$ where $\tilde{f_i}$ is an extension of $f_i$ to $F$. \\
\subsection*{Manifolds which have genus one Heegaard splittings}
Every Heegaard diagram of a genus one Heegaard splitting is a standard torus with two simple closed curves on it,
one is the meridian curve and the other is a $(p,q)$-curve ($p$ and $q$ are coprime).
We may assume $p$ is non-negative. If $(p,q)=(1,0)$ or $(1,\pm 1)$, the manifold is $S^3$. If $(p,q)=(0,\pm 1)$, the manifold is $S^2\times S^1$.
If otherwise, the manifold is called the lens space of type $(p,q)$, $L(p,q)$.
For $(p,q)$, we set $q'$ to be the least non-negative number such that $qq'\equiv 1$ mod $p$.
Note that $L(p,q)$ is homeomorphic to $L(p,q')$ (by changing the roles of two handlebodies of the genus one Heegaard diagram).
It is known that the genus one Heegaard splitting of $S^3$, $S^2\times S^1$ or $L(p,q)$ is unique under homeomorphisms \cite{4}, \cite{15}.\\
\subsection*{Fibered annulus}
A fibered annulus is an annulus in a closed orientable 3-manifold $M$ which is a fiber of a fibered link.
As above, a manifold which has a fibered annulus has a genus one Heegaard splitting.
Moreover because of the fact that the group of self-homeomorphisms of an annulus is generated by the Dehn twist along the core curve
and the properties of fibered links, the corresponding genus one Heegaard diagram is a standard torus with the meridian curve and ($p,\pm 1$)-curve.
Hence a manifold which has a fibered annulus is $S^3$, $S^2\times S^1$ or $L(p,\pm 1)$.
We can see easily that each of $S^2\times S^1$, $L(p,1)$ and $L(p,-1)$ ($p\neq 2$) has just one fibered annulus
and each of $S^3$, $L(2,1)$ and $L(2,-1)$ has just two fibered annuli under orientation preserving self-homeomorphisms.
(Note that there is an orientation reversing homeomorphism between $L(2,1)$ and $L(2,-1)$ as noted in \cite{10}.)
The monodromy of the fibered annulus in $S^2\times S^1$ is the identity map,
that in $L(p,1)$ (so called \emph{$p$-Hopf band} in \cite{2}) is the $p$ times positive Dehn twists along the core curve ($p\neq 2$),
that in $L(p,-1)$ (so called \emph{$-p$-Hopf band} in \cite{2}) is the $p$ times negative Dehn twists along the core curve ($p\neq 2$),
that of one fibered annulus in $L(2,1)$ (so called \emph{$2$-Hopf band} in \cite{2}) is the 2 times positive Dehn twists along the core curve,
that of the other fibered annulus in $L(2,1)$ (so called \emph{$-2$-Hopf band} in \cite{2}) is the 2 times negative Dehn twists along the core curve,
that of the $+1$-Hopf annulus (resp. $-1$-Hopf annulus) in $S^3$ is the positive (resp. negative) Dehn twist along the core curve.\\
\subsection*{A genus two Heegaard splitting of a manifold which admits a reducible one.}
If a closed orientable 3-manifold $M$ admits a reducible genus two Heegaard splitting, then $M$ is homeomorphic to
$(S^2\times S^1)\# (S^2\times S^1)$, $(S^2\times S^1)\# L(p,q)$, $L(p_1,q_1)\# L(p_2,q_2)$, $S^2\times S^1$, $S^3$, or $L(p,q)$.
In \cite{6}, Casson and Gordon proved that if a 3-manifold is reducible (i.e. has a sphere which does not bound a 3-ball), every Heegaard splitting of it is reducible.
In \cite{15}, Waldhausen proved that every Heegaard splitting of $S^3$ whose genus is greater than $0$ is reducible,
and in \cite{4}, Bonahon-Otal proved that every Heegaard splitting of lens spaces whose genus is greater than $1$ is reducible.
These imply that every genus two Heegaard splitting of the above manifolds is reducible.
As mentioned above, every reducible genus two Heegaard splitting can be decomposed into two genus one Heegaard splittings.
In the opposite direction, every reducible genus two Heegaard splitting is obtained by connecting two genus one Heegaard splittings.
Hence the manifold which has a reducible genus two Heegaard splitting has at most two genus two Heegaard splittings under homeomorphisms.
It depends on the choice of a solid torus of one genus one Heegaard splitting which should be connected to one solid torus of the other genus one Heegaard splitting.
In \cite{8}, Cho and Koda gave the necessary and sufficient condition for a Heegaard splitting of such manifolds of being unique under homeomorphisms.
(Unique in $S^3$, $L(p,q)$, $S^2\times S^1$, $(S^2\times S^1)\# (S^2\times S^1)$, $L(p,q)\# (S^2\times S^1)$ and
$L(p_1,q_1)\# L(p_2,q_2)$ with ${q_1}^2 \equiv 1$ mod $p_1$ or ${q_2}^2 \equiv 1$ mod $p_2$ and not unique in the other part of $L(p_1,q_1)\# L(p_2,q_2)$.)
However, even though there are two, their Heegaard diagrams are very similar.
We will focus on one Heegaard surface and its diagram. The arguments for the other Heegaard splittings are similar and the conclusions are the same.\\\vspace{0.5in}
\ From the above, we see that if $M$ has a GOF-knot, then $M$ must have a genus two Heegaard splitting and that if $M$ has a reducible genus two Heegaard splitting
then $M$ must be homeomorphic to $(S^2\times S^1)\# (S^2\times S^1)$, $(S^2\times S^1)\# L(p,q)$, $L(p_1,q_1)\# L(p_2,q_2)$, $S^2\times S^1$, $S^3$, or $L(p,q)$.
(Moreover, the genus two Heegaard splitting obtained by the GOF-knot is also reducible.)
Therefore $M$ which admits a reducible genus two Heegaard splitting and possibly has a GOF-knot is
$(S^2\times S^1)\# (S^2\times S^1)$, $(S^2\times S^1)\# L(p,q)$, $L(p_1,q_1)\# L(p_2,q_2)$, $S^2\times S^1$, $S^3$, or $L(p,q)$. \\\\
\ In this paper, our goal is to give another proof of the following theorems already obtained in \cite{2}, \cite{3}, \cite{12}.\\
\begin{thm} \label{thm1}
$\rm{(Baker \cite{2})}$ \ There is just one GOF-knot in $(S^2 \times S^1) \# (S^2 \times S^1)$
and the fiber is obtained by the plumbing of two fibered annuli in $S^2\times S^1$.
\end{thm}
\begin{thm} \label{thm2}
$\rm{(Baker \cite{2})}$ \ There is a GOF-knot in $L(p,q)\# (S^2\times S^1)$ if and only if $q \equiv \pm 1$ mod $p$,
and if there is a GOF-knot, the fiber can be obtained by the plumbing of each of fibered annulus in $L(p,q)$ and $S^2\times S^1$.
\end{thm}
\begin{thm} \label{thm3}
$\rm{(Baker \cite{2})}$ \ There is a GOF-knot in $L(p_1,q_1)\# L(p_2,q_2)$ if and only if $q_i \equiv \pm 1$ mod $p_i$ ($i=1,2$),
and if there is a GOF-knot, the fiber can be obtained by the plumbing of fibered annuli in $L(p_1,q_1)$ and $L(p_2,q_2)$.
\end{thm}
\begin{thm} \label{thm4}
$\rm{(Morimoto \cite{12})}$ \ There is just one GOF-knot in $S^2\times S^1$.
\end{thm}
\begin{thm} \label{thm5}
$\rm{(Morimoto \cite{12})}$ \ There are just two GOF-knots in $S^3$. They are the $($left and right hand $)$ trefoil and the figure eight knot.
\end{thm}
\begin{thm} \label{thm6}
$\rm{(Baker \cite{3})}$ \\
$\rm{(1)}$ $L(4,1)$ has just three GOF-knots.\\
$\rm{(2)}$ $L(p,1)$ ($p \neq 4$) has just two GOF-knots.\\
$\rm{(3)}$ For positive integers $a$ and $b$, $L(2ab+a+b+1,2b+1)$ has just one GOF-knot.\\
$\rm{(4)}$ For positive integers $a$ and $b$, $L(2ab+a+b,2b+1)$ (except for $L(4,3)$) has just one GOF-knot.\\
$\rm{(5)}$ A lens space which is not homeomorphic to any of the above types has no GOF-knots.
\end{thm}
\vspace{0.3in}
\ Morimoto proved some of the above theorems by using the monodromy of a GOF-knot in a 3-manifold and the fundamental group of this manifold.
Baker proved the other theorems by using the one-to-one correspondence between GOF-knots and axes of closed 3-braids in $S^3$.
(There is a double branched covering map branched along a closed 3-braid. The preimage of the axis is a GOF-knot.)
In this paper, we give a unified proof of the above theorems by using the fact that the genus two Heegaard splitting of these manifolds is almost unique.
As a result, we find the positions of GOF-knots clearly on standard Heegaard surfaces of these manifolds.
\vspace{0.8in}
\section{Methods} \label{sec3}
As in Section \ref{sec2}, every GOF-knot of $M$ is on a genus two Heegaard surface so that it decomposes each handlebody into $T\times I$ ($T$ is a fiber surface).
We call a simple closed curve on the boundary of genus two handlebody such that it decomposes the handlebody into $T\times I$ a \emph{GOF-knot on the handlebody}.
To investigate the properties of such curves, we consider simple closed curves on the boundary of a genus two handlebody. \\\\
\ Let $V$ be a genus two handlebody and let $D$ and $E$ be two disjoint properly embedded, non-separating disks in $V$ which are not parallel.
Then $D$ and $E$ cut $V$ into a 3-ball.
Fix orientations of $\partial D$ and $\partial E$ and give them letters $x$ and $y$, respectively.
Let $l$ be an oriented simple closed curve on $\partial V$. Isotope $l$ so that $l$ intersects $\partial D \cup \partial E$ minimally and transversely.
Then $l$ determines a word of $x$ and $y$ that can be read off by the intersections of $l$ with $\partial D$ and $\partial E$ in taking orientations into account.
This word is well-defined up to cyclic permutations. Note that this word may not be reduced (i.e. may have subword of type $xx^{-1}$. See Figure \ref{fig:one}.).
For a simple closed curve $l$, the word should be written cyclically, however we write it simply not cyclically.
For this reason, we say that the word is reduced if it is cyclically reduced. The following lemma is frequently used later. \\
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=50mm]{fig_1_scc.eps}
\end{center}
\caption{uncancellable $xx^{-1}$}
\label{fig:one}
\end{figure}
\\
\begin{lem} \label{lem1}
$\rm{(Cho, Koda, \cite{7} \ \cite{8} \ \cite{9})}$\ In the above setting, if the word determined by a simple closed curve $l$ contains
a subword of the form $xy^{n}x^{-1}$ for some $n \in \mathbb{N}$, then this word is reduced.
\end{lem}
{\it Proof from \rm{\cite{7}}}. \ Let $S$ be a sphere with four boundary components which is obtained from $\partial V$ by cutting along $\partial D \cup \partial E$.
We denote by $X^{+}$, $X^{-}$, $Y^{+}$ and $Y^{-}$ the boundary components of $S$ coming from $\partial D$ and $\partial E$, respectively. See Figure \ref{fig:two}.
In $S$, the subword $xy^{n}x^{-1}$ corresponds to $n+1$ arcs.
The first one starts from $X^{+}$ and ends at $Y^{-}$, the second one starts from the point of $Y^{+}$
corresponding to the terminal point of the first one in $\partial V$ and ending at $Y^{-}$, ..., the $n$-th one starts from $Y^{+}$ and ends at $Y^{-}$,
the last one starts from $Y^{+}$ and ends at $X^{+}$.
Then there must be two arcs so that one connects $X^{-}$ and $Y^{+}$, the other connects $X^{-}$ and $Y^{-}$.
(For example, the next arc of the above subarcs of $l$ starts at $X^{-}$ and ends at $X^{+}$, $Y^{+}$ or $Y^{-}$ because of minimality.
If the ending point is on $X^{+}$, the next arc starts from $X^{-}$ and ends at $X^{+}$, $Y^{+}$ or $Y^{-}$.
Repeating this in finitely many times, we get an arc connecting $X^{-}$ with $Y^{+}$ or $Y^{-}$.)
It follows that we cannot draw an arc of the form $xx^{-1}$ or $yy^{-1}$ without intersecting $l$.
Therefore $l$ cannot contain a subword of the form $xx^{-1}$ or $yy^{-1}$. This implies the word represented by $l$ is reduced.
\qed
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=40mm]{fig_2_scc.eps}
\end{center}
\caption{$xy^{n}x^{-1}$}
\label{fig:two}
\end{figure}
\\
\ The followings are implicitly used later.
\begin{lem} \label{lem2}
\ In the above setting, if a simple closed curve $l$ on $\partial V$ has a subarc representing $xx^{-1}$ $($or $x^{-1}x$$)$,
then $l$ has no subarcs representing $yy^{-1}$ nor $y^{-1}y$.
\end{lem}
{\it Proof}. \ Let $S$ be a sphere with four boundary components which is obtained from $\partial V$ cutting along $\partial D \cup \partial E$.
We denote by $X^{+}$, $X^{-}$, $Y^{+}$ and $Y^{-}$ the boundary components of $S$ coming from $\partial D$ and $\partial E$, respectively.
For a subarc $c$ of $l$ representing $xx^{-1}$, there is an arc $\alpha$ in $S$ connecting $X^{+}$ and another boundary component $W$ such that
$c$ is the boundary of a regular neighborhood of $\alpha \cup W$ in the interior of $S$, $Int(S)$.
If $W$ was $X^{-}$, the number of the points of $l \cap X^{-}$ does not coincide with that of $l \cap X^{+}$. It cannot occur.
Thus $W$ is $Y^{+}$ or $Y^{-}$. We assume $W$ is $Y^{+}$. In this case, there are no subarcs of $l$ representing $yy^{-1}$.
If there is a subarc of $l$ representing $y^{-1}y$, the subarc $c$ and this arc on $S$ are of the form in Figure \ref{fig:three}.
Let $a$, $b$, $c$, $d$ and $e$ be the number of subarcs of $l$ on $S$ connecting $X^{+}$ and $X^{+}$,
connecting $X^{+}$ and $Y^{+}$, connecting $X^{+}$ and $Y^{-}$, connecting $Y^{-}$ and $Y^{-}$ and connecting $Y^{-}$ and $X^{-}$.
Note that there are no subarcs of the other types (See Figure \ref{fig:three}).
$a$ and $d$ are at least $1$. Then two equalities $2a+b+c=e$ and $b=c+2d+e$ must hold. It cannot occur. Therefore there are no subarcs of $l$ representing $y^{-1}y$.
\qed
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=60mm]{fig_3_scc.eps}
\end{center}
\caption{possible 4 types of arcs when there are $xx^{-1}$ and $y^{-1}y$}
\label{fig:three}
\end{figure}
\\
\begin{lem} \label{rem}
\ In the same setting of Lemma \ref{lem2}, if there are $n$ subarcs of simple closed curve $l$ on $\partial V$ each of which represents $xx^{-1}$
then there must be $n$ subarcs of simple closed curve $l$ on $\partial V$ each of which represents $x^{-1}x$.
\end{lem}
{\it Proof}. It follows by an argument similar to Lemma \ref{lem2}.
$l$ is like in Figure \ref{fig:four}.
\qed
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=60mm]{fig_4_scc.eps}
\end{center}
\caption{actually possible arcs when there is $xx^{-1}$}
\label{fig:four}
\end{figure}
\\
\\\\
\begin{lem} \label{lem3}
\ Let $D$ and $E$ be two disjoint properly embedded, non-separating disks which are not parallel and $K$ be an oriented GOF-knot on a genus two handlebody $V$.
Then by assigning $x$ and $y$ to $\partial D$ and $\partial E$ with an appropriate orientation,
the word represented by $K$ becomes the commutator $xyx^{-1}y^{-1}$ of $x$ and $y$ after reduction.
\end{lem}
{\it Proof}. At first, consider the GOF-knot ${}K_0$ and the disjoint properly embedded, non-separating disks $D_0$ and $E_0$
which are not parallel in Figure \ref{fig:five}. It is easy to orient ${}\partial D_0$ and ${}\partial E_0$ with an assignment letters $x$ and $y$ to them
so that $K_0$ represents the commutator $xyx^{-1}y^{-1}$.\\
\ Let $D$, $E$ and $F$ be disjoint properly embedded, non-separating disks which are not pairwise parallel.
We will show that if the word represented by ${}K_0$ becomes the commutator $xyx^{-1}y^{-1}$ after reduction
by giving $\partial D$ and $\partial E$ an appropriate orientation and letters $x$ and $y$,
then the word becomes the commutator $zwz^{-1}w^{-1}$ after reduction by giving $\partial D$ and $\partial F$ an appropriate orientation and letters $z$ and $w$.
Let $\Sigma$ denote $\partial V$, ${}\Sigma'$ denote $\Sigma \setminus \partial D \cup \partial E$, and $d^{+}$, $d^{-}$, $e^{+}$, $e^{-}$
denote the boundary components of ${}\Sigma '$ coming from $\partial D$ and $\partial E$.
There is an arc ${}\alpha _{F}$ in ${}\Sigma '$ connecting ${}d^{+}$ and ${}e^{\epsilon}$ ($\epsilon \in \{\pm \}$),
such that one boundary component of a small neighborhood of $d^{+} \cup {}\alpha_{F} \cup e^{\epsilon}$ is (isotopic to) $\partial F$.
(the others are $d^{+}$ and $e^{\epsilon}$.)
Tentatively we give $\partial F$ the orientation coming from $\partial D$ (see Figure \ref{fig:six}).
Isotope ${}\alpha _F$ so that its endpoints are not on $K_0$.
The word represented by the subarc of $K_0$ cut by $\partial F$ near every intersection point of $K_0$ with $\alpha _F$ is $ww^{-1}$ or $w^{-1}w$.
Thus up to reduction, the word represented by $K_0$ in letters $z$ and $w$ comes from the intersections of $\partial D$ and $\partial E$ with ${}K_0$.
The small subarc of $K_0$ representing the word $x$ in letters $x$ and $y$ corresponds to the word $zw$ in letters $z$ and $w$.
Similarly, $x^{-1}$ corresponds to $w^{-1}z^{-1}$, $y$ corresponds to $w$ or $w^{-1}$
(depending on the choice of the orientation of $\partial F$) and $y^{-1}$ corresponds to $w^{-1}$ or $w$.
Therefore if the word represented by ${}K_0$ in $x$ and $y$ is $xyx^{-1}y^{-1}$ after reduction,
the word in $z$ and $w$ is $zwz^{-1}w^{-1}$ or $zw^{-1}z^{-1}w$ after reduction.
Changing the orientation of $\partial F$ if necessary, the word is $zwz^{-1}w^{-1}$.\\
\begin{figure}[htbp]
\begin{minipage}{0.5\hsize}
\begin{center}
\includegraphics[width=60mm]{fig_1_lemma.eps}
\end{center}
\caption{$\partial D_0$, $\partial E_0$ and $K_0$}
\label{fig:five}
\end{minipage}
\begin{minipage}{0.5\hsize}
\begin{center}
\includegraphics[width=45mm]{fig_2_lemma.eps}
\end{center}
\caption{$\alpha_F$ and $F$}
\label{fig:six}
\end{minipage}
\end{figure}
\\
\ Because of the connectivity of the non-separating disk complex of a handlebody \cite{11},
every pair of disjoint properly embedded non-separating disks $D$ and $E$ in $V$
which are not parallel can be constructed by an iteration of the above operation to $D_0$ and $E_0$.
Therefore for any disjoint two properly embedded, non-separating disks $D$ and $E$ in $V$ which are not parallel,
the word represented by ${}K_0$ is $xyx^{-1}y^{-1}$ after reduction by giving the boundaries of them an appropriate orientation and letters $x$ and $y$.\\
\ For any GOF-knot $K$, there is a self-homeomorphism $f$ of $V$ such that $f$ sends $K$ to ${}K_0$.
The word of $K$ by using any disjoint two properly embedded, non-separating disks $D$ and $E$ in $V$
which are not parallel is the same as the word of ${}K_{0}=f(K)$ by using disks $f(D)$ and $f(E)$.
By the procedure discussed above, it is the commutator after reduction.
\qed
\\\\\\
\ Conversely the following holds.\\
\begin{lem} \label{lem4}
\ Let {\it D} and {\it E} be disjoint properly embedded, non-separating disks in genus two handlebody $V$
which are not parallel and $K$ be a simple closed curve on $\partial V$.
If the word represented by $K$ is $xyx^{-1}y^{-1}$ after reduction by giving $\partial D$ and $\partial E$ an appropriate orientation and letters $x$ and $y$,
then $K$ is a GOF-knot on $V$.
\end{lem}
{\it Proof}. If there is a subarc $\alpha$ of $K$ representing a word $xx^{-1}$, $\alpha$ and an arc $\beta$ on $D$
which has common endpoints with $\alpha$ bound a disk $D'$ in $V$ whose interior is disjoint from $D$ and $E$.
$\alpha$ cuts $D$ into two disks $D_{1}$ and $D_{2}$. At least one of $D' \cup D_{1}$ and $D' \cup D_{2}$ is a non-separating disk.
Denote this disk by $\bar{D}$ (See Figure \ref{fig:seven}) and isotope $\bar{D}$ so that $\partial \bar{D}$ intersects $K$ minimally.
Then by giving $\partial \bar{D}$ and $\partial E$ an appropriate orientation and letters $z$ and $w$ the number of letters of the word represented by $K$
in terms of $z$ and $w$ is less than that of the word in terms of $x$ and $y$.
Since $\bar{D}$ is disjoint from and not parallel to $D$ and $E$, by the argument of Lemma \ref{lem3} the word represented by $K$ in terms of $z$ and $w$ is
also the commutator of $z$ and $w$ after reduction.
Using this operation repeatedly, we find two disjoint properly embedded, non-separating disks $\tilde{D}$ and $\tilde{E}$ in $V$ which are not parallel
such that the word represented by $K$ is reduced and a form of $\tilde{x} \tilde{y} \tilde{x}^{-1} \tilde{y}^{-1}$
by giving $\partial \tilde{D}$ and $\partial \tilde{E}$ an appropriate orientation and letters $\tilde{x}$ and $\tilde{y}$.
Let ${}\Sigma'$ denote $\partial V \setminus \partial \tilde{D} \cup \partial \tilde{E}$, and $\tilde{d}^{+}$, $\tilde{d}^{-}$, $\tilde{e}^{+}$, $\tilde{e}^{-}$
denote the boundary components of ${}\Sigma '$ coming from $\partial \tilde{D}$ and $\partial \tilde{E}$.
The curve $K$ corresponds to four arcs on ${}\Sigma'$. The first one starts from $\tilde{d}^{+}$ and ends at $\tilde{e}^{-}$,
the second starts from the point of $\tilde{e}^{+}$ corresponding to the terminal point of the first one in $\partial V$ and ends at $\tilde{d}^{+}$,
the third starts from $\tilde{d}^{-}$ and ends at $\tilde{e}^{+}$, and the fourth starts from $\tilde{e}^{-}$ and ends at $\tilde{d}^{-}$.
One boundary of a small regular neighborhood of $\tilde{d}^{+}$, $\tilde{e}^{-}$ and the first arc in ${}\Sigma'$ bounds a non-separating disk in $V$
which is disjoint from and not parallel to $\tilde{D}$ and $\tilde{E}$. Denote this disk by $\tilde{F}$.
The simple closed curve $K$ and three disks $\tilde{D}$, $\tilde{E}$ and $\tilde{F}$ in $V$ are drawn in Figure \ref{fig:eight}.
Though $K$ may be the mirror image of it and may be twisted along disks $\tilde{D}$, $\tilde{E}$ or $\tilde{F}$,
we assume $K$ is like in Figure \ref{fig:eight} by a self-homeomorphism of $V$.
The triplet ($K$, $\tilde{D}$, $\tilde{E}$) corresponds to the triplet ($K_{0}$, $D_{0}$, $E_{0}$) in Lemma \ref{lem3}. Therefore $K$ is a GOF-knot on $V$.
\qed
\begin{figure}[h]
\begin{minipage}{0.5\hsize}
\begin{center}
\includegraphics[width=50mm]{fig_3_lemma.eps}
\end{center}
\caption{$D$, $D_1$ and $D_2$}
\label{fig:seven}
\end{minipage}
\begin{minipage}{0.5\hsize}
\begin{center}
\includegraphics[width=60mm]{fig_4_lemma.eps}
\end{center}
\caption{$\partial \tilde{D}$, $\partial \tilde{E}$, $\partial \tilde{F}$ and $K$}
\label{fig:eight}
\end{minipage}
\end{figure}
\\\\\\
\ By using the above two lemmas and the relation between GOF-knots and genus two Heegaard splittings,
we get a one-to-one correspondence between a GOF-knot (with its fiber) and a simple closed curve on a genus two Heegaard surface
whose representing words in both genus two handlebodies are commutators after reduction.
The words can be read off by a Heegaard diagram.\\\\\\
\section{Individual cases} \label{sec4}
\subsection{$(S^2 \times S^1) \# (S^2 \times S^1)$}
In this case the genus two Heegaard splitting is unique as in Section \ref{sec2}. We consider a Heegaard diagram in Figure \ref{fig:nine}.
We denote by $V \cup _{\Sigma} W$ the corresponding genus two Heegaard splitting.
This case is exceptionally easy because of the following two properties.:
One is that a GOF-knot in $V$ is also a GOF-knot in $W$ since $\partial D$ (resp. $\partial E$) is the same as $\partial D'$ (resp. $\partial E'$) in $\Sigma$,
and the other is that the restriction on $\partial V = \Sigma$ of every self-homeomorphism of $V$ extends to a self-homeomorphism of $W$.
By the first property, every GOF-knot in $(S^2 \times S^1) \# (S^2 \times S^1)$ corresponds to a GOF-knot in $V$.
Since for any pair of two GOF-knots $K_1$ and $K_2$ in $V$, there is a fiber preserving self-homeomorphism of $V$ which takes $K_1$ to $K_2$.
By the second property, all GOF-knots in $(S^2 \times S^1) \# (S^2 \times S^1)$ are equivalent under homeomorphisms.
One GOF-knot in $(S^2 \times S^1) \# (S^2 \times S^1)$ is drawn on Figure \ref{fig:ten}. This results from the plumbing of fibered annuli in two $S^2 \times S^1$s.
Therefore every GOF-knot in $(S^2 \times S^1) \# (S^2 \times S^1)$ is of this position.
\\\\
\begin{figure}[htbp]
\begin{minipage}{0.5\hsize}
\begin{center}
\includegraphics[width=50mm]{fig_1_d_s2s1.eps}
\end{center}
\caption{a Heegaard diagram}
\label{fig:nine}
\end{minipage}
\begin{minipage}{0.5\hsize}
\begin{center}
\includegraphics[width=50mm]{fig_2_d_s2s1.eps}
\end{center}
\caption{one GOF-knot}
\label{fig:ten}
\end{minipage}
\end{figure}
\\\vspace{0.5in}
\subsection{$L(p,q) \# (S^2 \times S^1) \ (|p| \geq 2)$}
At first, we determine the condition for $L(p,q) \# (S^2 \times S^1)$ to have GOF-knots. Next, we find the positions of GOF-knots if there are.
In this case, the Heegaard surface is unique as in Section \ref{sec2}. We consider a standard Heegaard diagram in Figure \ref{fig:eleven}.
We denote by $V \cup _{\Sigma} W$ the corresponding genus two Heegaard splitting.
We give $\partial D$ and $\partial E$ (resp. $\partial D'$ and $\partial E'$) letters $x$ and $y$ (resp. $x'$ and $y'$).
We set $S$ to be $\Sigma \setminus (\partial D \cup \partial E)$. It is a sphere with four boundary components.
We denote by $d^{+}$, $d^{-}$, $e^{+}$ and $e^{-}$ the boundary components of $S$ coming from $\partial D$ and $\partial E$.
$\partial D'$ cuts $S$ into $p$ cells. See Figure \ref{fig:twelve}.
In this figure, the subarc of $d^{+}$ which is on the $i$-th cell is identified the subarc of $d^{-}$ which is on $i+q \pmod{p}$. \\
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=80mm]{fig_1_slens.eps}
\end{center}
\caption{a standard Heegaard diagram}
\label{fig:eleven}
\end{figure}
\\
\begin{figure}[htbp]
\begin{minipage}{0.5\hsize}
\begin{center}
\includegraphics[width=50mm]{fig_2_slens.eps}
\end{center}
\caption{cut $\Sigma$ along $\partial D$ and $\partial E$}
\label{fig:twelve}
\end{minipage}
\begin{minipage}{0.5\hsize}
\begin{center}
\includegraphics[width=30mm]{fig_3_slens.eps}
\end{center}
\caption{$yy^{-1}$}
\label{fig:thirteen}
\end{minipage}
\end{figure}
\\
\ We assume there is a GOF-knot $K$. If there is a subarc $c$ of $K$ corresponding to the word of the form $yy^{-1}$,
there is an arc $\alpha$ in $S$ connecting $e^{+}$ and another boundary component,
denoted by $A$, such that the boundary of a regular neighborhood of $\alpha \cup A$ in $S$ is $c$ (See Figure \ref{fig:thirteen}).
If $A$ is $e^{-}$, every subarc of $K$ in $S$ one of whose endpoints is on $e^{-}$ has the other endpoint on $e^{+}$.
Let $n$ be the number of these subarcs.
Then the number of the points in $|K \cap e^{+}|$ (in $S$) is at least $(n+2)$. It cannot occur. Therefore $A$ cannot be $e^{-}$.
So $A$ is $d^{+}$ or $d^{-}$. We can deform the Heegaard diagram so that $\alpha$ is disjoint from $\partial D'$ in the following way:
If $\alpha$ intersects $\partial D'$, we can get another non-separating disk in $W$ which is disjoint from and not parallel to $D'$ and $E'$
by using the subarc of $\alpha$ which connect $e^{+}$ and an intersection point with $\partial D'$ and whose interior is disjoint from $\partial D'$.
We replace $D'$ with this new disk. Denote it by $D'$ and give the letter $x'$ again. In $S$, the cell which contains $e^{+}$ will change. See Figure \ref{fig:fourteen}.
Using this operation in finitely many times, we can assume $\alpha$ is disjoint from $\partial D'$.
In this situation, the subarc $c$ represents a word in $x'$ and $y'$ of the form $y'x'^{\pm p}y'^{-1}$.
Then by Lemma \ref{lem1} the word in $x'$ and $y'$ represented by $K$ is reduced. Since $|p|\geq 2$, it is not a commutator.
By Lemma \ref{lem3}, it is a contradiction.
Hence there are no subarcs of $K$ representing $yy^{-1}$. (Similarly, no $y^{-1}y$.)\\
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=100mm]{fig_4_slens.eps}
\end{center}
\caption{change $D'$ so that $\alpha$ is disjoint from $\partial D'$}
\label{fig:fourteen}
\end{figure}
\\
\ If there is a subarc of $K$ corresponding to a word of the form $xx^{-1}$, there is an arc $\beta$ in $S$ connecting $d^{+}$ and another boundary component,
denoted by $B$ such that the boundary of a regular neighborhood of $\beta \cup B$ in $S$ is the subarc.
For the same reason as the above, $B$ is $e^{+}$ or $e^{-}$.
If $\beta$ intersects $\partial D'$, we replace $D'$ to the new disk which is obtained by disk surgery as in Figure \ref{fig:fourteen} and
denote this new disk by $D'$ and give the letter $x'$ again.
Using this operation in finitely many times, we can assume $\beta$ is disjoint from $\partial D'$.
Then as in Figure \ref{fig:fifteen}, we can get a non-separating disk in $V$ by disk surgery.
Denote this new disk by $D_1$ and give $D_1$ and $E$ the letters $x_1$ and $y_1$.
We set $S_1$ to be $\Sigma \setminus (\partial D_1 \cup \partial E)$.
We denote by $d^{+}_1$, $d^{-}_1$, $e^{+}$ and $e^{-}$ the boundary components of $S_1$ coming from $\partial D_1$ and $\partial E$.
In $S_1$, there cannot be subarcs of $K$ of the form $y_1y_{1}^{-1}$ by the same discussion as the above.\\
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=100mm]{fig_5_slens.eps}
\end{center}
\caption{change $D$ to new disk $D_1$}
\label{fig:fifteen}
\end{figure}
\ Inductively, if there is a subarc of $K$ of the form $x_kx_{k}^{-1}$, there is an arc $\beta _k$ in $S_k$ connecting $d_{k}^{+}$ and another boundary component,
denoted by $B_k$ such that the boundary of a regular neighborhood of $\beta _k \cup B_k$ in $S_k$ is the subarc.
For the same reason as the above, $B_k$ is $e^{+}$ or $e^{-}$.
If $\beta _k$ intersects $\partial D'$,
we replace $D'$ to the new disk which is obtained by disk surgery and denote this new disk by $D'$ and give the letter $x'$ again.
Using this operation in finitely many times, we can assume $\beta_k$ is disjoint from $\partial D'$. Then as in Figure \ref{fig:fifteen},
we can get a non-separating disk in $V$ by disk surgery. Denote this new disk by $D_{k+1}$ and give $D_{k+1}$ and $E$ the letters $x_{k+1}$ and $y_{k+1}$.
We set $S_{k+1}$ to be $\Sigma \setminus (\partial D_{k+1} \cup \partial E)$. We denote by $d^{+}_{k+1}$, $d^{-}_{k+1}$, $e^{+}$ and $e^{-}$
the boundary components of $S_{k+1}$ coming from $\partial D_{k+1}$ and $\partial E$.
In $S_{k+1}$, there cannot be subarcs of $K$ of the form $y_{k+1}y_{k+1}^{-1}$ by the same discussion as the above.
Then for some non-negative integer $n$, the word in {$x_n$, $y_n$} represented by $K$ is reduced. We regard $x_0,y_0$ and $S_0$ as $x,y$ and $S$.
In $S_n$, $K$ is a collection of four arcs,
connecting $e^{+}$ and $d^{+}_{n}$, connecting $d^{-}_{n}$ and $e^{+}$, connecting $e^{-}$ and $d^{-}_{n}$ and connecting ${d_{n}}^{+}$ and $e^{-}$
(the terminal point of an arc is the starting point of the next arc.).
In this situation, there cannot be a subarc of $K$ of the form $y'y'^{-1}$ and by changing $D'$ to a new disk
(and denoting this new disk $D'$ and giving the letter $x'$ again), we can assume there are no subarcs of $K$ of the form $x'x'^{-1}$ (See Figure \ref{fig:fourteen}).
Let $\bar{S}$ denote a sphere with four boundary components obtained by $S_n$ and changing $\partial D'$ as in above (See Figure \ref{fig:sixteen}).
In $\bar{S}$, $K$ is represented by four arcs and is simultaneously a reduced form in $\{x',y'\}$.
Hence it is necessary that $q \equiv \pm 1$ mod $p$ since the word represented by $K$ in $\{x',y'\}$ contains ${x'}^{[q]}$ or ${x'}^{[p-q]}$.
($[n]$ is the residue class of $n$ mod $p$.) $K$ is of the position in Figure \ref{fig:sixteen} for example.\\
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=60mm]{fig_6_slens.eps}
\end{center}
\caption{GOF-knot in $L(p,1)\# (S^2\times S^1)$}
\label{fig:sixteen}
\end{figure}
\\
\ In $\bar{S}$, by repeating disk surgeries as in Figure \ref{fig:seventeen}, we can assume that $e^{+}$ and $e^{-}$ are in the same cell.
By looking the Haken sphere in Figure \ref{fig:eighteen}, we see that $K$ with its fiber $T$ is the result of the plumbing of two fibered annuli
in $L(p,q)$ and $S^2 \times S^1$ respectively.
By changing the orientation if necessary, we assume that $L(p,q)$ is $L(p,1)$ and the fibered annulus in it is the $p$-Hopf band.
Then, this GOF-knot is unique under self-homeomorphisms.\\
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=100mm]{fig_7_slens.eps}
\end{center}
\caption{change cell which contains $e^{-}$}
\label{fig:seventeen}
\end{figure}
\\
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=80mm]{fig_8_slens.eps}
\end{center}
\caption{GOF-knot and Haken sphere in $L(p,1)\# (S^2\times S^1)$}
\label{fig:eighteen}
\end{figure}
\\
\ Therefore, we conclude that the necessary and sufficient condition for $L(p,q) \# (S^2 \times S^1)$ to have GOF-knots is $q \equiv \pm 1$ mod $p$,
and if there is a GOF-knot, it is unique and obtained by the plumbing.
\\\vspace{0.5in}
\subsection{$L(p_1,q_1) \# L(p_2,q_2) \ (|p_1|,|p_2| \geq 2)$}
At first, we determine the condition for $L(p_1,q_1) \# L(p_2,q_2)$ to have GOF-knots. Next, we find the positions of GOF-knots if there are.
In this case, there are at most two Heegaard surfaces of genus two as in Section \ref{sec2}. However the corresponding Heegaard diagrams of them are similar.
Thus we assume that a GOF-knot $K$ is on a standard Heegaard surface in Figure \ref{fig:nineteen}. We denote by $V \cup _{\Sigma} W$ this Heegaard splitting.
We give $\partial D$ and $\partial E$ (resp. $\partial D'$ and $\partial E'$) letters $x$ and $y$ (resp. $x'$ and $y'$).
We set $S$ to be $\Sigma \setminus (\partial D \cup \partial E)$. It is a sphere with four boundary components.
We denote by $d^{+}$, $d^{-}$, $e^{+}$ and $e^{-}$ the boundary components of $S$ coming from $\partial D$ and $\partial E$ (See Figure \ref{fig:twenty}).
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=75mm]{fig_1_ll.eps}
\end{center}
\caption{a standard Heegaard diagram}
\label{fig:nineteen}
\end{figure}
\\
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=45mm]{fig_2_ll.eps}
\end{center}
\caption{cut $\Sigma$ along $\partial D$ and $\partial E$}
\label{fig:twenty}
\end{figure}
\\
\ Since $xx^{-1}$ and $yy^{-1}$ cannot coexist by Lemma \ref{rem}, we can assume that there are no subarcs of type $yy^{-1}$.
If there is a subarc $c$ of $K$ representing $xx^{-1}$ (by the symmetry, same for $yy^{-1}$),
there is an arc $\alpha$ on $S$ connecting $d^{+}$ and $e^{\epsilon}$ ($\epsilon \in \{ +,- \}$)
such that the boundary of a regular neighborhood of $\alpha \cup e^{\epsilon}$ in $S$ is $c$.
We say such an arc $\alpha$ is the corresponding arc of $c$.
We assume $\epsilon$ is $+$. Isotoping $\alpha$ so that it intersects $\partial D'$ and $\partial E'$ minimally and by Lemma \ref{lem1},
we can assume that $\alpha$ is disjoint from $\partial D'$ and $\partial E'$.
This is because if $\alpha$ intersects them and if the intersection point nearest to $e^{+}$ is on $\partial E'$,
this intersection point can be omitted by isotopy and if the intersection point nearest to $e^{+}$ is on $\partial D'$,
the subarc of $c$ (so of $K$) represents a word of type $x'{y'}^{p_2}{x'}^{-1}$ and then by Lemma \ref{lem1},
$K$ is a reduced form in $\{x',y'\}$ and has $y'^{p_2}$ so $K$ cannot be the commutator of $x'$ and $y'$ after reduction.
We set $D_0$ to be $D$ and set $D_1$ to be the non-separating disk in $V$ obtained by disk surgery of $D_0$ using $c$ (See Figure \ref{fig:twentyone}).
In Figure \ref{fig:twentyone}, $S_1$ is the sphere with four boundary components obtained by cutting $\Sigma$ along $\partial D_1$ and $\partial E$.
We denote by $d_{1}^{+}$, $d_{1}^{-}$, $e^{+}$ and $e^{-}$ the boundary components of $S_1$ coming from $\partial D_1$ and $\partial E$.
We give $\partial D_1$ and $\partial E$ the letters $x_1$ and $y_1$. \\
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=80mm]{fig_3_ll.eps}
\end{center}
\caption{make $S_1$ from $S$(=$S_0$)}
\label{fig:twentyone}
\end{figure}
\\
\ If $K$ has a subarc representing a word of type $x_1x_1^{-1}$, there is a corresponding arc on $S_1$ which connects $d_1^{+}$ and $e^{+}$.
Note that this corresponding arc does not connect $d_1^{+}$ and $e^{-}$.
By an argument similar to the above, we can assume this corresponding arc is disjoint from $\partial D'$ and $\partial E'$.
Inductively for $k$ ($0 \leq k \leq p_1-1$), if $K$ has a subarc representing a word of type $x_kx_k^{-1}$ there is a corresponding arc
which connects $d_k^{+}$ and $e^{+}$.
We can assume this corresponding arc is disjoint from $\partial D'$ and $\partial E'$.
We set $D_{k+1}$ to be the non-separating disk in $V$ obtained by the disk surgery of $D_k$ using the subarc of $K$.
Let $S_{k+1}$ be the sphere with four boundary components obtained by cutting $\Sigma$ along $\partial D_{k+1}$ and $\partial E$.
We denote by $d_{k+1}^{+}$, $d_{k+1}^{-}$, $e^{+}$ and $e^{-}$ the boundary components of $S_{k+1}$ coming from $\partial D_{k+1}$ and $\partial E$.
We give $\partial D_{k+1}$ and $\partial E$ the letters $x_{k+1}$ and $y_{k+1}$. (We regard $S$, $D$ and $d^{\pm}$ as $S_0$, $D_0$ and $d_{0}^{\pm}$.)
If this operation is repeated until we get $S_{p_1}$, we can get another standard Heegaard diagram (See Figure \ref{fig:twentytwo}).
In this standard Heegaard diagram, the intersection number of $K$ and this new pair of disjoint non-separating, non-parallel disks in $V$ is less than
that of $K$ with old pair of disjoint non-separating, non-parallel disks in $V$.
Thus we can assume this operation stops at getting $S_k$ ($0 \leq k \leq p_1-1$).\\
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=100mm]{fig_4_ll.eps}
\end{center}
\caption{another standard Heegaard diagram}
\label{fig:twentytwo}
\end{figure}
\\
\ Note that if there is a subarc of $K$ of type $y_ky_k^{-1}$ ($1\leq k\leq p_1-1$) in $S_k$, its corresponding arc $\beta$ is like in Figure \ref{fig:twentythree}.
Otherwise there is a subarc of $K$ of type $y_{k-1}y_{k-1}^{-1}$ in $S_{k-1}$.
It contradicts Lemma \ref{rem}, $x_{k-1}x_{k-1}^{-1}$ and $y_{k-1}y_{k-1}^{-1}$ cannot coexist.\\
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=30mm]{fig_5_ll.eps}
\end{center}
\caption{$\beta$ connects $e^{+}$ and $d_k^{\epsilon}$. $\beta$ is not connected directly by $\partial E'$}
\label{fig:twentythree}
\end{figure}
\\
\ Moreover, if this operation is repeated until we get $S_{p_1-1}$ and there are no $x_{p_1-1}x_{p_1-1}^{-1}$,
we can get another Heegaard diagram which is obtained by the standard diagram of the above. It is similar to $S_1$ (See Figure \ref{fig:twentyfour}).
If there is a subarc of $K$ of type $y_{p_1-1}y_{p_1-1}^{-1}$, we can get another Heegaard splitting like in Figure \ref{fig:twentyfive}. In this Heegaard diagram,
the intersection number of the pair of disjoint non-separating, non-parallel disks in $V$ with $K$ is not more than that of $D_{p_1-1} \cup E$ and $K$.
Thus we can assume this operation stops at getting $S_k$ ($0 \leq k \leq p_1-2$).\\
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=100mm]{fig_6_ll.eps}
\end{center}
\caption{another Heegaard diagram like $S_1$}
\label{fig:twentyfour}
\end{figure}
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=100mm]{fig_7_ll.eps}
\end{center}
\caption{the case having $y_{p_1-1}y_{p_1-1}^{-1}$}
\label{fig:twentyfive}
\end{figure}
\\
\ In this setting, if there is subarc of $K$ representing $y_ky_k^{-1}$ (or $y_k^{-1}y_k$) ($k\neq 0$ from our assumption.),
there are no subarcs of $K$ representing the words of type $x'x'^{-1}$ nor $y'y'^{-1}$ (See Figure \ref{fig:twentysix}).
If there are no subarcs of $K$ representing $y_ky_k^{-1}$ (or $y_k^{-1}y$),
there are no subarcs of $K$ representing the words of type $x'x'^{-1}$ nor $y'y'^{-1}$ neither (this is because $K$ is a reduced form in $S_k$.)
(See Figure \ref{fig:twentysix}).
Thus $K$ is a reduced form in $D'$ and $E'$. Moreover in this situation $K$ must be a reduced form in $D_0$ and $E_0$ too.
This is because a subarc in $S_0$ representing $x_0x_0^{-1}$ in $\{x_0,y_0\}$ represents $y'^{\pm p_2}$ in $\{x',y'\}$.
\\
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=100mm]{fig_8_ll.eps}
\end{center}
\caption{subarc of type $y'y'^{-1}$ and $x'x'^{-1}$ in $S_k$}
\label{fig:twentysix}
\end{figure}
\ As a result, $K$ must be a reduced form in a standard Heegaard diagram (in Figure \ref{fig:twentyseven}) in $\{x,y\}$ and $\{x',y'\}$ simultaneously.
Hence it is necessary that $q_i \equiv \pm 1$ mod $p_i$ ($i = 1,2$) as in 4.2 ,
and $K$ is of the position in Figure \ref{fig:twentyseven} for example.\\
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=80mm]{fig_9_ll.eps}
\end{center}
\caption{$K$ in a standard Heegaard splitting}
\label{fig:twentyseven}
\end{figure}
\\
\ By looking the Haken sphere in Figure \ref{fig:twentyseven}, we see that $K$ with its fiber $T$ is the result of the plumbing of two fibered annuli
in $L(p_1,q_1)$ and $L(p_2,q_2)$ respectively. This implies $q_i \equiv \pm 1$ mod $p_i$ ($i=1,2$),
and then the condition to have the unique genus two Heegaard splitting is satisfied.\\
\ By changing the orientation if necessary, we assume $L(p_1,q_1)$ is $L(p_1,1)$ and the fibered annulus in it is $p_1$-Hopf band.
Note that as noted in \cite{10}, $L(r_1,s_1)\# L(r_2,s_2)$ is homeomorphic to $L(r_1,s_1)\# L(r_2,-s_2)$
if and only if ${s_1}^{2}\equiv -1$ mod $r_1$ or ${s_2}^{2}\equiv -1$ mod $r_2$.
Hence if neither $p_1$ nor $p_2$ is $2$, the GOF-knot is unique. \\
\ If either $p_1$ or $p_2$ is $2$ (we assume $p_2$ is $2$), there can be two GOF-knots,
one is obtained by the plumbing of the $p_1$-Hopf band in $L(p_1,1)$ and the $2$-Hopf band in $L(2,1)$ and
the other is obtained by the plumbing of the $p_1$-Hopf band in $L(p_1,1)$ and the $-2$-Hopf band in $L(2,1)$.
The monodromy of the former is represented in $GL_{2}(\mathbb{Z})$ as $ \left(\begin{array}{ccc} 1 & p_1 \\2 & 1+2p_1 \end{array} \right)$ and
that of the latter is represented in $GL_{2}(\mathbb{Z})$ as $ \left(\begin{array}{ccc} 1 & p_1 \\-2 & 1-2p_1 \end{array} \right)$.
Since they are not conjugate in $GL_{2}(\mathbb{Z})$, these GOF-knots are not equivalent .\\\\
\ Therefore, we conclude that the necessary and sufficient condition for $L(p_1,q_1) \# L(p_2,q_2)$ to have GOF-knots is
$q_i \equiv \pm 1$ mod $p_i$ ($i=1,2$), and if there is a GOF-knot, it is obtained by the plumbing.
Moreover, if neither $p_1$ nor $p_2$ is $2$, the GOF-knot is unique and otherwise there are just two GOF-knots.
\\\vspace{0.5in}
\subsection{$S^2 \times S^1$}
In this case the genus two Heegaard surface is unique as in Section \ref{sec2}. We regard $S^2 \times S^1$ as $S^3 \# (S^2 \times S^1)$ and
we consider a standard Heegaard diagram in Figure \ref{fig:twentyeight}. We denote by $V \cup _{\Sigma} W$ the corresponding genus two Heegaard splitting.
We give $\partial D$ and $\partial E$ (resp. $\partial D'$ and $\partial E'$) letters $x$ and $y$ (resp. $x'$ and $y'$).
We set $S$ to be $\Sigma \setminus (\partial D \cup \partial E)$. It is a sphere with four boundary components.
We denote by $d^{+}$, $d^{-}$, $e^{+}$ and $e^{-}$ the boundary components of $S$ coming from $\partial D$ and $\partial E$. \\
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=60mm]{fig_1_s2s1.eps}
\end{center}
\caption{a standard Heegaard diagram}
\label{fig:twentyeight}
\end{figure}
\\
\ The argument is almost similar to that of the case $L(p,q) \# (S^2 \times S^1)$. We assume there is a GOF-knot $K$.
If there is a subarc $c$ of $K$ representing $yy^{-1}$ (or $y^{-1}y$), we can assume the corresponding arc $\alpha$ is disjoint from $\partial D'$.
In this situation, $c$ represents a word $y'xy'^{-1}$ in $\{x',y'\}$ and by Lemma \ref{lem1} the word in $\{x',y'\}$ represented by $K$ is reduced.
Moreover, by Lemma \ref{rem} there is a subarc $\bar{c}$ of $K$ representing $y^{-1}y$ and
we can also assume its representing arc $\bar{\alpha}$ is disjoint from $\partial D'$ as 4.3 . See Figure \ref{fig:twentynine}.
$\partial D'$ intersects each of $c$ and $\bar{c}$ once and $\partial E'$ intersects each of $c$ and $\bar{c}$ twice.
Since the word in $\{x',y'\}$ represented by $K$ is reduced, $K$ is $c \cup \bar{c}$.
In this situation, since $K$ is disjoint from $D$, the word in $\{x,y\}$ represented by $K$ cannot contain $x$.
Especially $K$ is not a GOF-knot. It is a contradiction. Hence there are no subarcs representing $yy^{-1}$.\\
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=40mm]{fig_2_s2s1.eps}
\end{center}
\caption{$c$ and $\bar{c}$}
\label{fig:twentynine}
\end{figure}
\\
\ If there is a subarc $c$ of $K$ representing $xx^{-1}$ (or $x^{-1}x$), we assume its corresponding arc $\alpha$ is disjoint from $\partial D'$ as 4.3 .
In this situation, we replace $D$ with new $D$ as in Figure \ref{fig:thirty} and give new $D$ the letters $x$ and $E$ the letters $y$.
In this new $S$, there are no subarcs of $K$ representing $yy^{-1}$ by the same discussion above.
By iterating this operation in finitely many times, $K$ is a reduced form in $\{x,y\}$ in Figure \ref{fig:thirtyone}.
For the same reason, we can assume $K$ is a reduced form in $\{x',y'\}$ too.\\
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=80mm]{fig_3_s2s1.eps}
\end{center}
\caption{replace $D$ with new $D$}
\label{fig:thirty}
\end{figure}
\\
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=80mm]{fig_4_s2s1.eps}
\end{center}
\caption{$K$ on a standard Heegaard splitting}
\label{fig:thirtyone}
\end{figure}
\\
\ $K$, which is a reduced form in $\{x,y\}$ and $\{x',y'\}$, is like in Figure \ref{fig:thirtyone}.
It can be decomposed into two fibered annuli in $S^3$ and $S^2 \times S^1$.
There can be two GOF-knots, one is obtained by the plumbing of the $+1$-Hopf annulus in $S^3$ and the fibered annulus in $S^2\times S^1$ and
the other is obtained by the plumbing of the $-1$-Hopf annulus in $S^3$ and the fibered annulus in $S^2\times S^1$.
The monodromy of the former (it is the positive Dehn twist along the simple closed curve on a fiber corresponding to the core curve of $+1$-Hopf annulus)
is represented in $GL_{2}(\mathbb{Z})$ as $ \left(\begin{array}{ccc} 1 & 1 \\0 & 1 \end{array} \right)$ and
that of the latter (it is the negative Dehn twist along the simple closed curve on a fiber corresponding to the core curve of $-1$-Hopf annulus)
is represented in $GL_{2}(\mathbb{Z})$ as $ \left(\begin{array}{ccc} 1 & -1 \\0 & 1 \end{array} \right)$.
Since they are conjugate in $GL_{2}(\mathbb{Z})$, these GOF-knots are equivalent.\\\\
\ Therefore, we conclude that $S^2 \times S^1$ has the unique GOF-knot and it (and its fiber) are obtained by the plumbing of two fibered annuli in $S^3$ and $S^2 \times S^1$.
\\\vspace{0.5in}
\subsection{$S^3$}
In this case the genus two Heegaard surface is unique as in Section \ref{sec2}. We regard $S^3$ as $S^3 \# S^3$ and
we consider a standard Heegaard diagram in Figure \ref{fig:thirtytwo}. We denote by $V \cup _{\Sigma} W$ the corresponding genus two Heegaard splitting.
We give $\partial D$ and $\partial E$ (resp. $\partial D'$ and $\partial E'$) letters $x$ and $y$ (resp. $x'$ and $y'$).
We set $S$ to be $\Sigma \setminus (\partial D \cup \partial E)$. It is a sphere with four boundary components.
We denote by $d^{+}$, $d^{-}$, $e^{+}$ and $e^{-}$ the boundary components of $S$ coming from $\partial D$ and $\partial E$. \\
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=60mm]{fig_1_s3.eps}
\end{center}
\caption{a standard Heegaard diagram}
\label{fig:thirtytwo}
\end{figure}
\\
\ Let $K$ be a GOF-knot. If there is a subarc $c$ of $K$ representing $xx^{-1}$,
we can see its corresponding arc $\alpha$ on $S$ is disjoint from $\partial D'$ and $\partial E'$ as follows:
We assume endpoints of $\alpha$ are on $d^+$ and $e^{+}$.
If $\alpha$ intersects $\partial D'$ or $\partial E'$ and if the intersection point nearest to $e^+$ is on $\partial E'$,
this intersection point can be omitted by isotopy and if the intersection point nearest to $e^+$ is on $\partial D'$,
$c$ represents word $x'yx'^{-1}$. Hence by Lemma \ref{lem1}, $K$ is a reduced form in $\{x',y'\}$. Thus except for $c$, no arcs of $K$ on $S$ intersect $\partial D'$.
However $K$ must intersect $e^{+}$ since the word in $\{x,y\}$ represented by $K$ contains $x$ and arcs of $K$
which have one of the endpoints on $e^{+}$ must intersect $\partial D'$ in this situation.
It is a contradiction. Hence we see $\alpha$ is disjoint from $\partial D'$ and $\partial E'$.
In this situation, we can replace $D$ and $E'$ with new $D$ and new $E'$ and get another standard Heegaard diagram as in Figure \ref{fig:thirtythree}.
Note that the intersection number of $\partial ({\rm new}D) \cup \partial E$ and $K$ is less than that of $\partial D \cup \partial E$ and $K$.
By the symmetry, if there is a subarc $c$ of $K$ representing $yy^{-1}$,
we get another standard Heegaard diagram in which the intersection number of $\partial D \cup \partial E$ and $K$ decreases too.
Hence we assume $K$ is a reduced form in $\{x,y\}$ on a standard Heegaard diagram.\\
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=100mm]{fig_2_s3.eps}
\end{center}
\caption{another standard Heegaard diagram}
\label{fig:thirtythree}
\end{figure}
\\\\
\ Changing the role of $\{\partial D,\partial E\}$ and $\{\partial D',\partial E'\}$, we assume $K$ is a reduced form in $\{x',y'\}$ in a standard Heegaard diagram.
If there is a subarc $c$ of $K$ representing $xx^{-1}$, we can see the corresponding arc $\alpha$ on $S$ is disjoint from $\partial D'$ and $\partial E'$ as before.
By Lemma \ref{rem}, there must be a subarc $\bar{c}$ of $K$ representing $x^{-1}x$ and the corresponding arc on $S$ is disjoint from $\partial D'$ and $\partial E'$.
Since $K$ is a reduced form in $\{x',y'\}$, there are no subarcs of $K$ on $S$ intersecting $\partial E'$ except for $c$ and $\bar{c}$.
Isotoping $K$, we make $c$ and $\bar{c}$ intersect $\partial D'$ as in Figure \ref{fig:thirtyfour}.
Since $K$ is reduced in $\{x',y'\}$, all subarcs of $K$ on $S$ except for $c$ and $\bar{c}$ are disjoint from $\partial D'$ and $\partial E'$.
Then the other subarcs of $K$ on $S$ are like in Figure \ref{fig:thirtyfour}.
In Figure \ref{fig:thirtyfour}, we set $n$ to be the number of arcs connecting $e^{+}$ and $d^{+}$.
In this setting, since the number of $y$ and $y^{-1}$ in the word in $\{x,y\}$ represented by $K$ is $n$, $n$ must be even.
Then $K$ represents the word $xx^{-1}(yx^{-1})^{\frac{n}{2}}(xy^{-1})^{\frac{n}{2}}$. It cannot be a commutator after reduction.
Therefore $K$ does not have subarcs of type $xx^{-1}$ (and $yy^{-1}$ by symmetry).
We assume $K$ is a reduced form in $\{x,y\}$ and $\{x',y'\}$ simultaneously.\\
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=100mm]{fig_3_s3.eps}
\end{center}
\caption{$c$, $\bar{c}$ and the other subarcs of $K$}
\label{fig:thirtyfour}
\end{figure}
\\
\ In this situation, $K$ is like in Figure \ref{fig:thirtyfive}. In this figure, the Haken sphere decomposes $K$ and its fiber into two fibered annuli in $S^3$. \\
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=120mm]{fig_4_s3.eps}
\end{center}
\caption{reduced $K$}
\label{fig:thirtyfive}
\end{figure}
\\
\ In $S^3$, every fibered annulus is $+1$ or $-1$ Hopf annulus. By the plumbing two fibered annuli, we obtain the trefoil or the figure eight knot.
Hence we see every GOF-knot in $S^3$ is the trefoil or the figure eight knot. This agrees with the classical result.
\\\vspace{0.5in}
\subsection{$L(p,q)\ (|p| \geq 2)$}
At first, we determine the condition for $L(p,q)$ to have GOF-knots. Next, we find the positions of GOF-knots if there are.
In this case, the Heegaard surface is unique as in Section \ref{sec2}. We consider a standard Heegaard diagram in Figure \ref{fig:thirtysix}.
We denote by $V \cup _{\Sigma} W$ the corresponding genus two Heegaard splitting.
We give $\partial D$ and $\partial E$ (resp. $\partial D'$ and $\partial E'$) letters $x$ and $y$ (resp. $x'$ and $y'$).
We set $S$ to be $\Sigma \setminus (\partial D \cup \partial E)$. It is a sphere with four boundary components.
We denote by $d^{+}$, $d^{-}$, $e^{+}$ and $e^{-}$ the boundary components of $S$ coming from $\partial D$ and $\partial E$. See Figure \ref{fig:thirtyseven}. \\
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=80mm]{fig_1_l.eps}
\end{center}
\caption{a standard Heegaard diagram}
\label{fig:thirtysix}
\end{figure}
\\
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=45mm]{fig_2_l.eps}
\end{center}
\caption{cut $\Sigma$ along $\partial D$ and $\partial E$}
\label{fig:thirtyseven}
\end{figure}
\\
\ We assume that there is a GOF-knot $K$ on $\Sigma$.
We set $S_0$ to be $S$, $d_{0}^{\pm}$ and $e_{0}^{\pm}$ to be $d^{\pm}$ and $e^{\pm}$, and $\{x_0,y_0\}$ to be $\{x,y\}$. We define $S_k$ inductively as follows.
Suppose we have constructed up to $S_{k}$. If there are no subarc of $K$ representing the word $x_{k}x_{k}^{-1}$, we stop at getting $S_{k}$.
If there is a subarc $c_k$ of $K$ on $S_k$ representing the word $x_kx_{k}^{-1}$ ($0 \leq k \leq p-1$),
we can assume its corresponding arc $\alpha _k$ on $S_k$ is disjoint from $\partial D'$ and $\partial E'$.
For, we assume $\alpha _k$ starts at $d_k^{+}$ and ends on $e_{k}^{+}$.
If $\alpha_k$ intersects them and if the intersection point nearest to $e_{k}^{+}$ is on $\partial E'$,
this intersection point can be omitted by isotopy and if the intersection point nearest to $e^{+}$ is on $\partial D'$,
the subarc of $c_k$ (so of $K$) represents a word of type $x'y'x'^{-1}$ and then by Lemma \ref{lem1}, $K$ is a reduced form in $\{x',y'\}$
and in such case by changing the roles of $\{D,E\}$ and $\{D',E'\}$, we can assume we stop at getting $S_0$.
Thus we assume $\alpha_k$ is disjoint from $\partial D'$ and $\partial E'$.
We set $D_{k+1}$ to be the non-separating disk in $V$ whose boundary is the boundary of a regular neighborhood of $d_{k}^{+} \cup \alpha_k \cup e_{k}^{+}$ in $S_k$,
and set $E_{k+1}$ to be $E_{k}$.
We give $\partial D_{k+1}$ and $\partial E_{k+1}$ the letters $x_{k+1}$ and $y_{k+1}$.
We set $S_{k+1}$ to be $\Sigma \setminus (\partial D_{k+1} \cup \partial E_{k+1})$.
We denote by $d_{k+1}^{+}$, $d_{k+1}^{-}$, $e_{k+1}^{+}$ and $e_{k+1}^{-}$ the boundary components of $S_{k+1}$ coming from $\partial D_{k+1}$ and $\partial E_{k+1}$.
As in the case of $L(p_1,q_1) \# L(p_2,q_2)$, by changing a standard Heegaard splitting, we can assume that we stop at getting $S_b$ ($0 \leq b \leq p-2$).
(If we get $S_p$ or stop at getting $S_{p-1}$, we can assume that we stop at getting $S_0$ or $S_1$ by taking another Heegaard splitting.)
\\\\
(1) The case where we stop at getting $S_0$ and there is no subarc of $K$ representing $y_0y_0^{-1}$\\
\ \ In other words, this is the case where $K$ is a reduced form in $\{x,y\}$ or $\{x',y'\}$ on a standard Heegaard splitting.
Changing the roles of $D$, $E$ and $D'$, $E'$ if necessary, we assume $K$ is a reduced form in $\{x',y'\}$.
In this situation there are no subarcs of $K$ representing $yy^{-1}$.
If there are no subarcs of $K$ representing $xx^{-1}$, $K$ is a reduced form in $\{x,y\}$ and $\{x',y'\}$ simultaneously on a standard Heegaard splitting,
and then $L(p,q)$ must be homeomorphic to $L(p,1)$ (as in 4.2 ) and the fiber of $K$ is obtained by the plumbing of fibered annuli of $L(p,q)$ and $S^3$.
If there is a subarc $c$ of $K$ representing $xx^{-1}$, there must be a subarc $\bar{c}$ of $K$ representing $x^{-1}x$ by Lemma \ref{rem}.
By the reducibility of $K$ in $\{x',y'\}$, there are no subarcs of $K$ in $S$ intersecting with $\partial E'$ except for $c$ and $\bar{c}$.
By isotoping $c$ and $\bar{c}$ so that they intersect $\partial D'$, the other subarcs of $K$ on $S$ are disjoint from $\partial D'$ and $\partial E'$.
See Figure \ref{fig:thirtyeight}. In this figure, there cannot be subarcs of $K$ on $S$ representing $yx^{n}x^{-n}y^{-1}$ or $yx^{n}x^{-(n+m)}x^{m}y^{-1}$.
Hence $K$ represents the word $x^{n+1}x^{-n}yx^{-n-1}x^{n}y^{-1}$ or $x^{n}x^{-n-1}yx^{-n}x^{n+1}y^{-1}$ in $\{x,y\}$. ($n$ is a natural number.)
In particular $K$ intersects $\partial E$ twice. Drawing a picture, we see $n=p$ and $K$ represents a word $x^{-p}yx^{-l}x^{p}y^{-1}x^{l}$.
($l$ is the minimal natural number such that $lq \equiv 1$ mod $p$.) Then $l=p\pm1$ and so $q \equiv \pm 1$ mod $p$.
Changing the orientation if necessary, we assume $L(p,q)=L(p,1)$ and in this case $K$ is like in Figure \ref{fig:thirtynine}.
In this figure, the operation of cancelling $x^{n+1}x^{-n}$, which changes $E'$ to a new non-separating disk, makes a new standard Heegaard splitting
where $K$ is a reduced form in $\{x,y\}$ and $\{x',y'\}$ simultaneously. See Figure \ref{fig:forty}.
Hence we conclude that if $K$ is a reduced form in $\{x,y\}$ or $\{x',y'\}$ in a standard Heegaard splitting,
then $L(p,q)$ is homeomorphic to $L(p,\pm 1)$ and the fiber is constructed by the plumbing of fibered annuli in $L(p,q)$ and $S^3$.
By changing the orientation if necessary, we assume $L(p,q)$ is $L(p,1)$ and the fibered annulus in it is the $p$-Hopf band.\\\\
\begin{figure}[htbp]
\begin{minipage}{0.5\hsize}
\begin{center}
\includegraphics[width=50mm]{fig_3_l.eps}
\end{center}
\caption{$c$ and $\bar{c}$}
\label{fig:thirtyeight}
\end{minipage}
\begin{minipage}{0.5\hsize}
\begin{center}
\includegraphics[width=50mm]{fig_4_l.eps}
\end{center}
\caption{$K$ on $S$}
\label{fig:thirtynine}
\end{minipage}
\end{figure}
\\
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=140mm]{fig_5_l.eps}
\end{center}
\caption{change to another standard Heegaard splitting}
\label{fig:forty}
\end{figure}
\\\\
(2) The case where (1) does not occur\\
\ \ In this case we stop at getting $S_b$ ($1\leq b\leq p-2$) or we stop at $S_0$ and there is a subarc of $K$ representing $y_0y_{0}^{-1}$.
We set $D'_0$ and $E'_0$ to be $D'$ and $E'$ and give them letters ${x_0}'$ and ${y_0}'$. We set $S'_0$ to be $\Sigma \setminus (\partial D'_0 \cup \partial E'_0)$.
We denote by ${d'_{0}}^{+}$, ${d'_{0}}^{-}$, ${e'_{0}}^{+}$ and ${e'_{0}}^{-}$ the boundary components of $S'_{0}$ coming from $\partial D'_{0}$ and $\partial E'_{0}$.
We define ${S'}_l$ ($0\leq l\leq p$) inductively as follows.
Suppose we have constructed up to ${S'}_{l}$. If there are no subarcs of $K$ on ${S'}_l$ representing the word ${x'_l}{x'_l}^{-1}$, we stop at getting ${S'}_l$.
If there is a subarc $c_l$ of $K$ on $S'_l$ representing the word $x'_l{x'_l}^{-1}$ ($0 \leq l \leq p-1$),
its corresponding arc $\alpha _l$ on $S'_l$ is disjoint from $\partial D$ and $\partial E$.
For, we assume $\alpha _l$ starts at ${{d'}_l}^{+}$ and ends on ${{e'}_{l}}^{+}$.
If $\alpha_l$ intersects them and if the intersection point nearest to ${e'_{l}}^{+}$ is on $\partial E$,
this intersection point can be omitted by isotopy and if the intersection point nearest to ${e'}^{+}$ is on $\partial D$,
the subarc of $c_l$ (so of $K$) represents a word of type $xyx^{-1}$ and then by Lemma \ref{lem1}, $K$ is a reduced form in $\{x,y\}$
and this contradicts our assumption. (We are in the case (2).)
Thus we see $\alpha_l$ is disjoint from $\partial D$ and $\partial E$.
We set $D_{l+1}$ to be the non-separating disk in $V$ whose boundary is the boundary of a regular neighborhood of ${d'_{l}}^{+} \cup \alpha_l \cup {e'_{l}}^{+}$ in $S'_l$,
and set $E'_{l+1}$ to be $E'_{l}$.
We give $\partial D'_{l+1}$ and $\partial E'_{l+1}$ the letters $x'_{l+1}$ and $y'_{l+1}$.
We set $S'_{l+1}$ to be $\Sigma \setminus (\partial D'_{l+1} \cup \partial E'_{l+1})$.
We denote by ${d'_{l+1}}^{+}$, ${d'_{l+1}}^{-}$, ${e'_{l+1}}^{+}$ and ${e'_{l+1}}^{-}$ the boundary components of $S'_{l+1}$
coming from $\partial D'_{l+1}$ and $\partial E'_{l+1}$.
If this operation continues until we get ${S'}_{p}$, we can get another standard Heegaard splitting. By changing the Heegaard splitting, $E_0$ changes to new $E_0$.
Now, in the case (2), there must be a subarc $c$ of $K$ representing $x_0{x_0}^{-1}$ or $y_0{y_0}^{-1}$.
If $c$ represents $y_0{y_0}^{-1}$, the intersection number of $K$ with $D_0 \cup$ (new$E_0$) is less than that of $K$ with $D_0 \cup$ (old$E_0$).
See Figure \ref{fig:fortyone}.
If $c$ represents $x_0{x_0}^{-1}$, though the intersection number of $K$ with $D_0 \cup$ (new$E_0$) is not necessarily less than
that of $K$ with $D_0 \cup$ (old$E_0$), $c$ represents new $x_{0}{x_0}^{-1}$ in new $S_0$ and we get new $S_1$.
Then the intersection number of $K$ with (new$D_1) \cup$ (new$E_1$) is less than that of $K$ with (old$D_1) \cup$ (old$E_1$). See Figure \ref{fig:fortytwo}.
Hence we see that the operation to get $S_k$ stops at getting $S_b$ ($0\leq b\leq p-2$) and
simultaneously the operation to get ${S'}_{l}$ stops at getting ${S'}_a$ ($0\leq a\leq p-1$) (or the case (1) occurs).
Moreover, as in the case of $L(p_1,q_1) \# L(p_2,q_2)$ we can assume the operation to get ${S'}_l$ stops at getting ${S'}_a$ ($0\leq a\leq p-2$).
(If we stop at getting ${S'}_{p-1}$, we can assume we stop at getting ${S'}_1$ or the case (1) occurs by taking another standard Heegaard splitting.)\\
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=140mm]{fig_6_l.eps}
\end{center}
\caption{$y_0{y_0}^{-1}$ on two standard Heegaard splittings}
\label{fig:fortyone}
\end{figure}
\\
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=140mm]{fig_7_l.eps}
\end{center}
\caption{$x_0{x_0}^{-1}$ on two standard Heegaard splittings}
\label{fig:fortytwo}
\end{figure}
\\
\ \ In this situation, if there is a subarc $c$ of $K$ representing $y_b{y_b}^{-1}$ on $S_b$, there cannot be subarcs of $K$ representing ${y'}_a{{y'}_a}^{-1}$.
For if there is a subarc of $K$ representing ${y'}_a{{y'}_a}^{-1}$, there must be a subarc of $K$ representing ${{y'}_a}^{-1}{y'}_a$ by Lemma \ref{rem}
and in this situation an arc representing $x_b{x_b}^{-1}$ must intersect the subarcs of $K$. See Figure \ref{fig:fortythree}.\\
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=120mm]{fig_8_l.eps}
\end{center}
\caption{${y'}_a{{y'}_a}^{-1}$ and ${{y'}_a}^{-1}{y'}_a$ on $S_0$}
\label{fig:fortythree}
\end{figure}
\\
\ Hence we can assume $K$ is a reduced form in $\{x_b,y_b\}$ on $S_b$ or in $\{{x'}_a,{y'}_a\}$ on $S'_a$.
We assume $K$ is a reduced form in $\{{x'}_a,{y'}_a\}$ on $S'_a$.\\
\\
\ \ Since we are in the case (2), there is a subarc $c$ of $K$ representing $x_0{x_0}^{-1}$ or $y_0{y_0}^{-1}$ on $S_0$.
If $c$ represents $y_0{y_0}^{-1}$ in $\{x_0,y_0\}$, the letter ${x'}_a$ appears ($p-a$) times in its representing word in $\{{x'}_a,{y'}_a\}$.
Thus $K$ cannot be a GOF-knot.
Hence $c$ represents $x_0{x_0}^{-1}$ (and this implies $b\geq 1$).
By Lemma \ref{rem}, there is a subarc $\bar{c}$ of $K$ representing ${x_0}^{-1}x_0$ on $S_0$.
Each of $c$ and $\bar{c}$ intersects with $\partial {E'}_0$($=\partial {E'}_{a}$) once.
Isotope $c$ and $\bar{c}$ so that each of them intersects with $\partial {D'}_a$ once. See Figure \ref{fig:fortyfour}.
Since $K$ is a reduced form in $\{{x'}_a,{y'}_a\}$, the other subarcs of $K$ on $S_0$ do not intersect with $\partial {D'}_a \cup \partial {E'}_a$. \\
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=60mm]{fig_9_l.eps}
\end{center}
\caption{$S_0$ with $\partial {D'}_a$ and $\partial {E'}_a$}
\label{fig:fortyfour}
\end{figure}
\\
\ \ Moreover, if $b=1$ and there is a subarc of $K$ representing $y_1{y_1}^{-1}$ on $S_1$,
this subarc intersects with $\partial {D'}_a$ in ($p-a$) times in the same orientation,
and if $2\leq b$ and there is a subarc of $K$ representing $y_b{y_b}^{-1}$ on $S_b$,
this subarc intersects with $\partial {E'}_a$ at least twice in the same orientation.
Hence we see that $K$ is a reduced form not only in $\{{x'}_a,{y'}_a\}$ but also in $\{x_b,y_b\}$. This implies $a\geq 1$.
In particular, $K$ must intersect with $\partial E_0$($=\partial E_b$) just twice in the opposite orientation.\\\\
\ \ In Figure \ref{fig:fortyfour}, along $K$ we take some arcs on $S_0$, which are disjoint from $\partial {D'}_a \cup \partial {E'}_a$ after $c$,
and then we take $\bar{c}$.
Then, the terminal point of $\bar{c}$ is connected to some arc on $S_0$,
which is disjoint from $\partial {D'}_a \cup \partial {E'}_a$ and the terminal point of this arc is connected to $\cdots$ ,
and the terminal point of this arc is connected to the starting point of $c$.
By the symmetry of $c$ and $\bar{c}$ in $S_0$ in Figure \ref{fig:fortyfour},
the word in $\{x_0,y_0\}$ of the subarc of $K$, which starts at the middle point of $c$ and ends at the middle point of $\bar{c}$ is obtained by
changing $x_0$ to ${x_0}^{-1}$ and $y_0$ to ${y_0}^{-1}$ in that of the other subarc of $K$,
which starts at the middle point of $\bar{c}$ and ends at the middle point of $c$.
In particular, each of these two arcs intersects $\partial E_0$ once. Since the sign of $x_0$ in each of these words (represented by these arcs) is the same,
these words are ${x_0}^{-n}y_0{x_0}^{-m}$ and ${x_0}^{n}{y_0}^{-1}{x_0}^{m}$ ($n$ and $m$ are integers.).
Hence the word in $\{x_0,y_0\}$ represented by $K$ is ${x_0}^{m}{x_0}^{-n}y_{0}{x_0}^{-m}{x_0}^{n}{y_0}^{-1}$. For this being commutator, $m-n=\pm1$ is necessary.
By changing the orientations of $\partial D_0$ and $\partial E_0$ if necessary, we assume $m$ and $n$ are non-negative.
By the construction of $S_b$, we see $(m,n)=(b+1,b)$ or $(b,b+1)$. \\\\
\ (A) $(m,n)=(b+1,b)$\\
\ \ \ By considering two components of $S_0 \setminus (\partial {D'}_a \cup \partial {E'}_a)$
which contain the terminal point of $c$ and the starting point of $\bar{c}$ respectively,
and by noting that $K$ intersects $\partial D_0$ in $2b+1$ times from the middle point of $c$ to the middle point of $\bar{c}$, we see $q(2b+1)\equiv 1$ mod $p$.
Moreover by noting that the subarc of $K$ from the middle point of $c$ to the middle point of $\bar{c}$ intersects $\partial E_0$ once,
we see that $2b+1$ is the least natural number such that $q(2b+1)\equiv 1$ mod $p$. This implies $q'=2b+1$ in our definition of lens spaces.
The subarc of the subarc of $K$ from the middle point of $c$ to the middle point of $\bar{c}$ corresponding to subword ${x_0}^{-1}y_0$
is like brown line in Figure \ref{fig:fortyfive}.
Otherwise $K$ is a reduced form in ${S'}_l$ ($l < a$) or $qb\equiv 0$ mod $p$. From this we can see $qb=-a+up$ ($u$ is an integer).
Then $(2b+1)q=q-2a+2up$, and since this is $1$ mod $p$ it is necessary that $q\equiv 2a+1$ mod $p$. By noting that $L(p,q)$ is homeomorphic to $L(p,q-p)$,
we assume $q=2a+1$. Under this assumption, $K$ and $a$, $b$ are not changed. Note that if $q$ is positive, $u$ must be positive.\\
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=60mm]{fig_10_l.eps}
\end{center}
\caption{subarc of $K$ representing ${x_0}^{-1}y_0$}
\label{fig:fortyfive}
\end{figure}
\ \ \ We will show that $u$ must be $1$. We set $v$ to be a natural number such that $wq=p+v$ ($1\leq v\leq q-1$).
If $u\geq 2$, along the subarc of $K$ from the middle point of $c$ to the middle point of $\bar{c}$, we pay attention to $(u+1)$-st and $(u-1)$-st circuits.
Since the subarc of $K$ from the middle point of $c$ to the middle point of $\bar{c}$ intersects $\partial E_0$ once,
it is necessary that $a+1\leq v\leq q-1$ and $-a-v+q$ is not in $[-a,0]$. This condition is not satisfied if $q=2a+1$.
Hence $u=1$ and $(p,q)=(2ab+a+b,2a+1)$. By drawing $K$ on $S_b$ after drawing $K$ on $S_0$ (it is almost unique.), $K$ is like in Figure \ref{fig:fortysix}.
In this figure, $[n]$ is the residue class of $n$ mod $p$.\\
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=100mm]{fig_11_l.eps}
\end{center}
\caption{$K$ on $S_b$ (in $L(2ab+a+b,2a+1)$)}
\label{fig:fortysix}
\end{figure}
\\
\ (B) $(m,n)=(b,b+1)$\\
\ \ \ We use almost the same argument used in the case (A). As in (A), we can see that $q'=2b+1$,
that subarcs of $K$ representing $x_0{x_0}^{-1}$, ${x_0}^{-1}$ and ${x_0}^{-1}y_0$ are like in Figure \ref{fig:fortyfive}
and that $q(b+1)=-a+\bar{u}p$ ($\bar{u}$ is an integer).
Then $2(b+1)q-q=-q-2a+2\bar{u}p$, and since this is $1$ mod $p$ it is necessary that $q\equiv -2a-1$ mod $p$.
By noting that $L(p,q)$ is homeomorphic to $L(p,q-p)$, we assume $q=-2a-1$. Under this assumption, $K$ and $a$, $b$ are not changed.
Note that if $q$ is negative, $\bar{u}$ must be negative.
As (A), we can also see $\bar{u}=-1$. Hence $(p,q)=(2ab+a+b+1,-2a-1)$. By changing the orientation, we assume $(p,q)=(2ab+a+b+1,2a+1)$.
By drawing $K$ on $S_b$ after drawing $K$ on $S_0$ (it is almost unique), $K$ is like in Figure \ref{fig:fortyseven}.\\
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=100mm]{fig_12_l.eps}
\end{center}
\caption{$K$ on $S_b$ (in $L(2ab+a+b+1,2a+1)$)}
\label{fig:fortyseven}
\end{figure}
\\\\
\ Therefore we say about $L(p,q)$ and $K$ as follows:
If $L(p,q)$ has a GOF-knot $K$ and it is on a standard genus two Heegaard surface in a reduced form in $\{x,y\}$ or $\{x',y'\}$,
$L(p,q)$ is homeomorphic to $L(p,1)$ and a fiber of $K$ is obtained by the plumbing of the $p$-Hopf band in $L(p,1)$ and a fibered annulus in $S^3$.
If $L(p,q)$ has a GOF-knot $K$ and it is on a standard genus two Heegaard surface in a reducible form in $\{x,y\}$ and $\{x',y'\}$,
it is necessary that $L(p,q)$ is homeomorphic to $L(2ab+a+b,2a+1)$ or $L(2ab+a+b+1,2a+1)$ ($a$ and $b$ are positive integers.)
and a fiber of $K$ is of the position discussed in the above argument.
(Note that $K$ of the second case may be of the first case by using another standard Heegaard splitting.) \\\\
\ We conclude that a lens space which has a GOF-knot is homeomorphic to $L(p,1)$, $L(2ab+a+b,2a+1)$ or $L(2ab+a+b+1,2a+1)$.
These lens spaces are classified under homeomorphisms into four classes,
(i) $L(p,1)$ ($p\neq 4$), (ii) $L(2ab+a+b,2a+1)$ (($a$,$b$)$\neq$ ($1$,$1$)), (iii) $L(2ab+a+b+1,2a+1)$, (iv) $L(4,3)$ (it is homeomorphic to $L(4,1)$).
On (i), there are just two GOF-knots with their fiber surfaces, one is obtained by the plumbing of the $p$-Hopf band in $L(p,1)$ and $+1$-Hopf annulus in $S^3$
and the other is obtained by the plumbing of the $p$-Hopf band in $L(p,1)$ and $-1$-Hopf annulus in $S^3$.
These two can be distinguished by computing monodromies as follows:
Let $l$ be the core curve of the $p$-Hopf band in $L(p,1)$ and $l'$ be the core curve of a fibered annulus in $S^3$.
By the plumbing we get a GOF-knot and its fiber $T$.
We can put $l$ and $l'$ on $T$. Let $\tilde{l}$ and $\tilde{l'}$ be simple closed curves on $T$ which are obtained by moving $l$ and $l'$ along fibers.
In $H_1(T)$, $\tilde{l} =l+l'$ and $\tilde{l'} =pl+(1+p)l'$ if a fibered annulus in $S^3$ is a $+1$-Hopf annulus,
and $\tilde{l} =l-l'$ and $\tilde{l'} =pl+(1-p)l'$ if a fibered annulus in $S^3$ is a $-1$-Hopf annulus.
Therefore the monodromies of these fibers are represented in $GL_{2}(\mathbb{Z})$ as $ \left(\begin{array}{ccc} 1 & p \\1 & 1+p \end{array} \right)$
and $ \left(\begin{array}{ccc} 1 & p \\-1 & 1-p \end{array} \right)$. They are not conjugate in $GL_{2}(\mathbb{Z})$.
On (ii) and (iii), there is just one GOF-knot, described above.
On (iv), there are three GOF-knots, two coming from the plumbing (they can be distinguished), one coming from like (ii) ((a,b)=(1,1)).
The third can be distinguished from the others by the monodromy. See Figure \ref{fig:fortyeight}.
In this figure, $\alpha$ and $\beta$ are simple closed curves representing a basis of $H_1(T)$ with $T$ a fiber.
We set $\tilde{\alpha}$ and $\tilde{\beta}$ to be simple closed curves which are obtained by moving $\alpha$ and $\beta$ along fibers.
In $H_1(T)$, $\tilde{\alpha} = -2\alpha -3\beta$ and $\tilde{\beta} = 3\alpha +4\beta$. The monodromy of the third fiber is represented in $GL_{2}(\mathbb{Z})$
as $ \left(\begin{array}{ccc} -2 & 3 \\-3 & 4 \end{array} \right)$. It is not conjugate to $ \left(\begin{array}{ccc} 1 & 4 \\1 & 1+4 \end{array} \right)$ and
$ \left(\begin{array}{ccc} 1 & 4 \\-1 & 1-4 \end{array} \right)$ in $GL_{2}(\mathbb{Z})$.
\\
\ This finishes a reproof of the Baker's results in \cite{3}.
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=100mm]{fig_13_l.eps}
\end{center}
\caption{monodromy of the third fiber in $L(4,3)$}
\label{fig:fortyeight}
\end{figure}
\\\\
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Harris County District Attorney Kim Ogg will have to find another way to deal with her office’s backlog after county commissioners rejected her request for a $20 million budget increase to hire more prosecutors.
Under Ogg’s proposal, 102 prosecutors would have been hired, an increase of roughly 40 percent. Ogg had hoped to reduce the county’s backlog of around 40,000 cases, which had been exacerbated after the Harris County Criminal Justice Center was flooded during Hurricane Harvey.
Houston Police Chief Art Acevedo tweeted his support for the DA’s request, calling the shortage of prosecutors a perfect storm of potential crime.
The three Democratic members of Commissioners Court – Commissioners Rodney Ellis and Adrian Garcia and County Judge Lina Hidalgo – supported increasing the district attorney’s budget by 7 percent, in line with increases for other county departments.
“This is not the only way, and certainly not the most cost-effective way to decrease prosecutor caseloads,” Hidalgo said.
A parade of prosecutors spoke of unrealistic caseloads that leave them overextended, unable to properly serve defendants, find witnesses and protect victims. Heather Marshall, a junior felony prosecutor, described working 70 hours weekly and said her caseload has more than doubled since moving to Houston from the district attorney’s office in Queens County, N.Y.
Ellis said voting yes would signal a commitment to doubling down on the system’s over- reliance on arrest, prosecution and incarceration for low-level, nonviolent offenses related to poverty, homelessness, mental health, prostitution and substance use.
.
Some progressive groups, such as the Texas Organizing Project and the Texas Criminal Justice Coalition, also spoke,” said Commissioner Jack Cagle, who voted in favor of the proposal.
.”
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Midland, MI
Awards Sponsor: Northern Animal Clinic, Superior Title and Settlement
Event Sponsors: Eastman Animal Hospital, Park Place Homes, Patricia and David Kepler Foundation, Pet Pals
Race Sponsors: Bushey Automotive, Legacy Wealth Management - Benjamin F. Edwards, Gimmick's, Glover's Rexall Pharmacy, McArdle Buick, Meijer, Valley Lanes, Wildfire Credit Union, Wolverine Bank
Sign Sponsors: Animal Medical Center, Bone & Bailey Insurance Agency, Chemical Bank, Eastman Party Store, Ieuter Insurance Group, M-20 Animal Hospital, Melissa Lile Photography, Members First Credit Union, Midland Animal Clinic, Sanford Animal Clinic, Soldan's Pet Supplies, Zach's Dog Groomery
Prizes: Amazing Deli, Applebees, Beyond Measures Bulk Food, Bicycle HQ, Biggby Coffee, Circle Quality Shoes, Cork 'N' Ale, Edible Arrangements, Family Fare, Great Lakes Ice Cream, Howl A Day, Ieuter Insurance Group, Isabella Bank, Jack's Fruit Market, Jet's Pizza, LaLondes Market, M's Cafe, Max's Place, Members First Credit Union, PNC Bank, River Rock Animal Hospital, Runners, Shirlene's Cuisine, The Gourmet Cupcake Shoppe, Wysong
Race treats and water: Bob Evans, Meijer, Shay Water,
Race Communications: Midland Amateur Radio Club
Moving Sign Ads: Chicky and Chippi Vacuums, Garber Chevrolet, Midland Soccer Club, Wolverine Bank
Aug 25, 2018
Saturday
9:00 AM
Add to Calendar
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Midland, MI
Midland City Forest
| 208,478
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News.
It looks like Pitt and company might have a completed script on their hands (especially when you consider what a tough audience the AICN crowd can be). But, the big questions still remain: who will direct, and who will star in the film? According to sources, Pitt is still in the running to play the lead, but nothing has been confirmed. A tentative release date of 2010 has been mentioned, but that could always change. So now is your chance for suggestions: What directors do you think could handle what AICN called, "a genre-defining piece of work that could well see us all arguing about whether or not a zombie movie qualifies as "Best Picture" material?"
Reader Comments (Page 1 of 1)
3-27-2008 @ 7:49PM
Joshua said...
How about David Fincher, he has worked with Pitt on his last masterpiece Fight Club, or one of my favorite directors Darren Aronofski, Unfortunately Danny boyle already has a zombie flick in his repotua (sp?) because I loved what he did with sunshine, the beach and 28 days later, he knows how to build suspense and tension.
I am just really excited that there is news on this and that it is moving along, this book was fantastic and will do well on the screen.
Reply
3-27-2008 @ 8:47PM
carg0 said...
do yourself a favor and read the book, you'll be glad you did. the audibook, unfortunately abridged, is also excellent.
Reply
3-28-2008 @ 10:32AM
SM said...
I haven't seen the script or the script review, but how would a lead in this film work exactly? I always thought that the film would be a good ensemble piece. Would Pitt play the author of the report?
Reply
3-28-2008 @ 5:24PM
rex said...
Agree with the Fincher suggestion. I would also throw in Alfonso Cuaron.
It would definitely make sense to cast Pitt to have a "star" in the film (plus he is a good actor). I could see him in the soldier character voiced by Mark Hamilton in the audio book (can't remember the character name off hand).
Can I breath a sigh of relief yet?
Reply
3-29-2008 @ 10:55AM
Aberdeen said...
With a script by JMS it's got to be good...might get the book myself, now.
Reply
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Positively?)
Excited to Go Back, With Other Friends
I saw this same type of excitement when I brought some football players to Greenville, to be Reunited with friends from Greenville. Not only were they excited to be with teenage guys from Grace Church, but two of the guys had stayed with their host family last summer. Of course, they were eager to explain how great of a family they were. (“She cooks you breakfast, bro!”)
And I see this same level of camp-vangelism this week, as a half-dozen Allendale students are in Aiken for a 4H camp. Two of our children went last year, along with a couple of other Allendale students. While our kids aren’t able to go this summer, those other Allendale kids are. Do you think they told their friends about how much fun they had last year? Absolutely! They eagerly brought their friends.
Last year, through my job, we sponsored a child to go to a camp at the SC Governor’s School for Science and Mathematics. This summer, four Allendale students are going.
It’s not just an increase in numbers that I like. I am encouraged by the increased momentum. I love seeing the growing awareness of and interest in these enriching experiences.
Your Giving Is Contagious
Just as these children have spread their enthusiasm about their camp experiences, your giving is also contagious. Last year, we asked for and received donations to help send students to football, art, 4H, and elementary camps. And we asked for the same this summer.
People like you (or maybe it was you) donated over $1000 to help send children to camps this year. This money provided scholarships for students to attend this summer’s sleep-away camps in Aiken and Greenville. Also, some money will be used to provide scholarships for an art camp in Allendale, which will take place next month.
What art camp? One led by a Grace Church member, who did the same camp in Allendale last year. Like the children in his class, he enjoyed his experience last year, and was excited to do it again.
And . . . while Community Bible Church (Savannah, GA) is not returning to serve in Allendale this summer, two teenage young ladies wanted to come back. They will be staying with us, as they volunteer at STEAM Camp and the summer program at a a church near us. Even though their family is moving (5 hours away) next month, they were eager to re-connect with the children in this community.
Positive experiences are contagious, whether you are a child participating in a camp, or a generous donor who can give some money, or a leader who can serve children in a hands-on way.
Thank you for giving of yourself. Your generosity encourages and inspires us, too.
Related Links:
- Why Summer Camps Are Important
- Connecting Allendale, Summer Camp, and You
- A Summer of Relationships and Camps
- Helping Your Child Prepare for Camp
- When You Give . . .
- Your Love and Support
| 320,535
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“We want the public to know that these ads are full of untruths, misinformation and are generally designed to distort and twist information about the Boise School District.”
Related quotes
“I don't believe that the public knows what it wants; this is the conclusion that I have drawn from my career.”
. . .”
| 383,391
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Learn something new every day More Info... by email
A barrel chair is a round chair often compared to the shape of a barrel, hence the name. Choosing the best swivel barrel chair starts with determining your needs and aesthetic preferences, as well as your budget for purchasing the chair. The materials used to make the chair can have a significant impact on the quality and cost of the chair; some leather chairs, for example, are exceptionally durable but also quite expensive, while some synthetic materials are less expensive but not as attractive or durable. Narrow down your search for a swivel barrel chair by determining your budget and aesthetic desires.
If you are looking for a leather swivel barrel chair, keep in mind that leather comes in different grades. A low-grade leather will be less expensive, but it will also be less durable in some cases and less attractive. It is likely to have some blemishes or inconsistencies in its appearance. High-grade leather will have a more consistent appearance, and it will be more durable. Its beauty will be enhanced as it ages, and its durability usually will not suffer too much over time. Choose the leather that fits your price range as well as your aesthetic requirements.
The base and swivel of the swivel barrel chair will have a significant impact on the chair's stability and function. If possible, sit in the chair for several minutes, gauging the chair's comfort level as well as its functionality. Some swivel barrel chair models will swivel and rock, while others will just swivel. In either case, make sure the chair is stable and does not move too excessively. Excessive movement can lead to a loss of balance, which can tip the chair over or otherwise make it unstable. Many swivel barrel chair models feature a design that hides the swivel base; this is done to preserve the aesthetic of the chair, so be sure to consider such models if you want a cleaner look.
The color and design of the chair can change significantly, and the best choice is entirely a matter of preference. Be sure the materials are quality enough to last many years, and try to avoid patterns or colors that will clash too significantly with the overall aesthetic of the room in which the chair will be placed. The chair should match other furniture, and if it does not, it should at least be in keeping with the general tone and mood of the room.
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MSExchange.org - RealTime Article Update Hi ExchangeList, Title: Understanding the Exchange Information Store Author: Rodney Buike Summary: We would like to welcome Rodney Buike to our team of authors as he presents his first article to MSExchange.org readers. The Information Store is the heart and soul of Exchange Server 2000 and 2003. Understanding the fundamentals of the Information Store is important for anyone managing an Exchange server. Link: Visit the Subscription Management () section to unsubscribe. MSExchange.org is in no way affiliated with Microsoft Corp. For sponsorship information, contact us at advertising@xxxxxxxxxxxxxx Copyright © MSExchange.org 2005. All rights reserved.
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