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ECEX Air Intake Screens can help protect against Legionnaires disease
Over the last 30 years there has been a series of distressing incidents involving Legionnaires’ disease. These have been costly, both in terms of lives destroyed and finances decimated. Legionnaires’ disease, a potentially fatal form of pneumonia, has resulted in manslaughter charges for individuals and companies as well as generating misery for many and tragedy […]Continue Reading
| 24,101
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From Their Home to Yours: Music and Musings in the Time of Quarantine
We joined Tricia Edwards, jazz pianist and keyboardist, on Wednesday, June 24 for a special live streamed performance of some of her favourite tunes and some thoughts about the role of art and music in this time of isolation.
Watch the concert now on Stagestream Live.
Tricia Edwards; Performer | Composer | Bandleader. Studies in jazz resulted in the release in 2008 of her debut jazz recording “Joyspring.”
Tricia now maintains a full schedule based in Calgary,.
Tricia@triciaedwards.ca |
| 220,138
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Here's my dilemma; I'll try to be short and sweet. I currently have one XFX GTX 295 Single PCB (works great) installed in my system. I have another GTX 295 on its way to me in the mail. I am trying to decide if I should unwrap it when it arrives and run quad SLI (my system is completely capable), or if I should sell it on eBay, pocket the cash, and buy a GTX 300 when it comes out (I have no idea when these cards are coming out, or if the 1st gen model will even be stable enough to bother with). Any thoughts/ideas/suggestions/morbid comments? :nerd:
| 324,401
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TITLE: Well-foundedness is not a first order property.
QUESTION [1 upvotes]: In the book 'Logic, Induction and Sets' by Thomas Foster I read the following in page 100 (Section 'The language of predicate logic'):
"We can show that well-foundedness is not a first-order property (there is no first-order theory whose models are precisely those binary structures $\langle X,R\rangle$ where $R$ is a well-founded relation)".
A proof of this can be found here: Why isn't there a first-order theory of well order?. It uses the compactness theorem.
Now, I'm willing to think there are still theories whose models possess a well-founded relation. Is Pearno Arithmetic one of them? What about ZF? In this case however well-foundedness could not be solely derived from the axiom of foundation/induction.
REPLY [4 votes]: The standard model of Peano Arithmetic -- that is, the usual natural numbers whose Platonic existence most of us believe in instinctively -- is well-founded.
However, thanks to Gödel (or just compactness) we know that PA has non-standard models, and a non-standard model of PA is not well-founded. In particular, since arithmetic on elements that can be written as numerals is fully given by the defining axioms for addition and multiplication, any non-standard model must have an element $c$ that does not equal any numeral. And then
$$ c, c-\bar1, c-\bar2, c-\bar3, c-\bar4, \ldots, c-\bar n, \ldots $$
is an infinitely descending sequence. (It can't ever stop unless $c$ actually equals one of the $\bar n$s, which by assumption it doesn't).
The situation with ZF and its various extensions is much the same: While the "standard" model (which we believe in mostly by intuition, if we believe it at all) is supposedly well-founded, one can easily use compactness to produce a model that is not well-founded, by essentially the same reasoning as for PA. (Assuming, of course, that ZF is consistent at all).
In general, it is easy to write a theory whose models are all finite. For such a theory, of course, all models are well-founded.
However, as long as a theory allows for models where there are infinite ascending chains of the relation in question (or merely if there are models with arbitrarily long finite chains), then the compactness argument you link to shows that this theory must have an ill-founded model.
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BY ELI KABERON
Everyone has songs that they associate with specific moments, places and people in their life. For example, I always think of seventh-grade school dances when Juvenile’s “Back Dat Azz Up” comes on, because, well, that’s what we listened to when juking at seventh-grade school dances. We even had an assembly because of that song and the “salacious” dancing it provoked. No matter what takes place for the rest of my life, I’ll always connect that song with that time in my life.
In 2011, two friends and I took a road trip up to a fourth friend’s wedding in northern Michigan. On that trip, another musical connection was made. During the five-hour drive there, and the five-hour drive home, we probably listened to the Wu-Tang Clan’s classic 1993 album Enter the Wu-Tang (36 Chambers) four times, from start to finish. We’d all heard it before. But there was something about that summer afternoon, driving along the highway, en route to a weekend full of fun and shenanigans, that made us want to blast one of the best rap albums ever made.
In honor of that (saw all those buddies last weekend, which got me thinking about it), and because this is the song where the title of this column derives, my favorite verse from an underappreciated track on that album.
Song: “Wu-Tang: 7th Chamber”
Album: Enter the Wu-Tang (36 Chambers)
Artist: Wu-Tang Clan
Verse: Seventh (GZA)
My my my
My Clan is thick like plaster
Bust ya, slash ya
Slit a nigga back like a Dutch Masta Killaas seein pink hearts, yellow moons
Orange stars and green clovers
Analyzing verses off of Wu-Tang songs is nearly impossible, for several reasons. One, this isn’t a full 16 bars; it’s only eight. Second, GZA — along with the other members of the Clan — don’t always speak, you know, English. They use sound effects and made-up terms to get their points across. Part of that is what makes the songs so awesome, and this is an example of that.
This verse comes in the final 65 seconds of a track that lasts 6:07 and includes a long skit about Raekwon looking for his Killer tape, somebody named Shymeek getting killed, and six different Wu members spitting verses before GZA goes and the subjects vary from the Tribe of Shabazz to running up “in spots like Fort Knox.” GZA uses his time to both talk about himself and his crew, saying he’s more dangerous than the most famous boxing match of all time. He also ends the verse – and the song – by implying that after the Wu-Tang Clan is done, people will be seeing the marshmallow objects in Lucky Charms cereal, which is just awesome.
It’s hard to say this verse is the best of all the songs on 36 Chambers. In actuality, it probably isn’t. GZA isn’t the best MC in the Clan in my opinion, but for some reason, this verse really sticks with me. I love how it starts, with the “My my my…” I love how the first four lines all rhyme. I love the Lucky Charms reference. I love the “Blaow!” right in the middle of the verse, something that no rapper today would do.
But most of all, I love what this song reminds me of. It takes me back to a specific moment in time, on an open road with two close friends, which is exactly what music is meant to do.
Respect, Eli. GZA is my favorite in the Wu and one of, if not THE, top emcees on my personal list. Lyrically no one can touch him. Period. However much more goes into emceeing like delivery, and he is pretty much the same always so I understand why some may take some points away from him for not switching it up in that regard.
I went to see him at The Metro in the Fall by myself after a friend bailed last second. Best show I’ve ever seen.
7th Chamber Pt. 2 is one of my favorite Wu songs of all time, but Deck’s verse on it is my favorite – without question. Not to diminish GZA’s or any of the others on that song, but I think his goes the hardest.
What are some of your other favorite GZA verses, not necessarily from that album?
Thanks for writing these. I thoroughly enjoyed the Jay Z one, and this as well.
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\begin{document}
\title{Outage Performance of Two-Hop OFDM Systems with Spatially Random Decode-and-Forward Relays}
\author{Shuping Dang, \IEEEmembership{Student Member, IEEE}, Justin P. Coon, \IEEEmembership{Senior Member, IEEE}, and Gaojie Chen, \IEEEmembership{Member, IEEE}
\thanks{
This work was supported by the SEN grant (EPSRC grant number EP/N002350/1) and the grant from China Scholarship Council (No. 201508060323).
The authors are with the Department of Engineering Science, University of Oxford, Parks Road, Oxford, U.K., OX1 3PJ; tel: +44 (0)1865 283 393, (e-mail: \{shuping.dang, justin.coon, gaojie.chen\}@eng.ox.ac.uk).}}
\maketitle
\begin{abstract}
In this paper, we analyze the outage performance of different multicarrier relay selection schemes for two-hop orthogonal frequency-division multiplexing (OFDM) systems in a Poisson field of relays. In particular, special emphasis is placed on decode-and-forward (DF) relay systems, equipped with bulk and per-subcarrier selection schemes, respectively. The exact expressions for outage probability are derived in integrals for general cases. In addition, asymptotic expressions for outage probability in the high signal-to-noise ratio (SNR) region in the finite circle relay distribution region are determined in closed forms for both relay selection schemes. Also, the outage probabilities for free space in the infinite relay distribution region are derived in closed forms. Meanwhile, a series of important properties related to cooperative systems in random networks are investigated, including diversity, outage probability ratio of two selection schemes and optimization of the number of subcarriers in terms of system throughput. All analysis is numerically verified by simulations. Finally, a framework for analyzing the outage performance of OFDM systems with spatially random relays is constructed, which can be easily modified to analyze other similar cases with different forwarding protocols, location distributions and/or channel conditions.
\end{abstract}
\begin{IEEEkeywords}
Spatially random relay, Poisson point process, multicarrier relay selection, OFDM, outage performance.
\end{IEEEkeywords}
\section{Introduction}
\IEEEPARstart{C}{ooperative} communications have become an important topic for research and industry in recent years \cite{6211487,7060199,1362898,1341264}. It is well known that relay-assisted cooperative communications are capable of providing extra diversity and thus a better system performance in terms of energy efficiency, outage performance and network coverage extension \cite{6182883,4745938,6189409}. In particular, multicarrier relay systems are of high importance, because it fits a number of applications in practice \cite{4342860}. A number of representative and useful multicarrier relay systems have been proposed and analyzed. For example, a block-based orthogonal frequency-division multiplexing (OFDM) decode-and-forward (DF) relay system has been analyzed in \cite{5340982}. Bulk and per-subcarrier relay selection schemes for OFDM systems have been proposed and compared in \cite{4489212} and \cite{7390520}, respectively. However, all of the above achievements regarding cooperative OFDM systems do not consider the location distribution of relays. The conventional network model employed in these previous OFDM-related works assumes the locations of all nodes to be deterministic and stationary, which form a stationary network topology. In practice, however, the dynamic nature of communication nodes is common and should be considered in order to provide a more general and meaningful analysis \cite{5226957}. Hence, a more realistic way to model a communication network is to assume the location of a node to be a random variable. To perform the analysis effectively, Poisson point processes (PPPs) have been used to analyze the location distribution of communication nodes for a large number of applications in wireless communications \cite{haenggi2013stochastic,6515339,7797167,7828097}. Pioneering work related to cooperative transmission in Poisson distributed networks was published in \cite{1705945}, in which an upper bound on outage probability is derived. Then, generalized analyses of DF and AF cooperative systems with spatially random relays distributed within a finite region have been given in \cite{6042306,5706435} and \cite{6328206,6831748}, respectively. The system with relays distributed over an infinite space is analyzed in \cite{6096785,7037462,6515497}. Opportunistic relaying with different combining techniques can be found in \cite{7248672,razi2015outage,6123785}.
However, to the best of the authors' knowledge, the link between OFDM systems and randomly distributed networks is lacking. This motivates us to construct a framework for analyzing the outage performance of two-hop OFDM systems with spatially random relays. In this paper, we analyze the outage performance of the two-hop OFDM system with spatially random DF relays and investigate a series of important properties of cooperative systems in random networks related to the outage performance. Specifically, the contributions of this paper are summarized infra:
\begin{itemize}
\item The exact expressions for outage probability for bulk and per-subcarrier selections are derived in integral forms for general cases. Meanwhile, the asymptotic expressions for finite-region-based outage probability in the high signal-to-noise (SNR) region for bulk and per-subcarrier selections are determined in closed forms. Furthermore, the exact expressions for infinite-region-based outage probability are determined in closed forms for the case of free space.
\item It is proved that the cooperative diversity gain in Poisson random networks can either be zero, one or infinite, which is termed the \textit{ternary property}.
\item An approximate relation between the outage probability ratio of two selection schemes and the relay node density is determined, which can be used to evaluate the performance advantage of per-subcarrier selection over bulk selection in sparse networks.
\item The relation among system throughput, the number of subcarriers and relay node density is investigated and a concave problem is formulated and proved to be capable of producing the optimal number of subcarriers, so that the system throughput can be maximized. Meanwhile, a special optimization case with reliability requirement is discussed and an approximation of the cut-off relay node density above which the formulated problem is solvable, is also derived.
\end{itemize}
All analysis is numerically verified by simulations. The results provided in this paper can be easily modified to analyze other similar cases with different forwarding protocols, location distributions and/or channel conditions.
The rest of this paper is organized as follows. The system model is detailed in Section \ref{sm}. We subsequently analyze the outage performance and discuss a series of related system properties in Section \ref{opa} and Section \ref{disp}. After that, the analysis is numerically verified by simulations in Section \ref{nr}. Finally, Section \ref{c} concludes the paper.
\section{System Model}\label{sm}
\subsection{System configurations and channel model}
In this paper, we consider a network with a single source located at the origin denoted by $\mathbf{p}_S=(0,0)$ and a destination node located at $\mathbf{p}_D=(r_{SD},0)$ in a two-dimensional polar coordinate system. The locations of source and destination are deterministic and stationary. Then, we assume the relays are homogeneously Poisson distributed over a two-dimensional region $\mathcal{C}\subseteq\mathbb{R}^2$ with a constant density $\lambda$, which form a homogeneous PPP denoted as $\Pi(\mathcal{C})$. In particular, a finite circle distribution region centered at the source node with a radius $\varsigma$ and an infinite distribution region are considered and employed to analyze the system performance in this paper, which are denoted by $\mathcal{C}_{\varsigma}$ and $\mathcal{C}_{\mathrm{inf}}$, respectively. Besides, for a typical OFDM system, we assume the number of subcarriers is $K$, which is deterministic. The set of all subcarriers is denoted as $\mathcal{K}$. Furthermore, it is assumed that the channel state information (CSI) can be perfectly estimated without any delay and overhead by the source node, so that relay selection can be effectively performed. We further suppose that the entire network operates in a half-duplex protocol and there is not a direct transmission link between source and destination due to deep fading, so that two orthogonal phases are required for one complete transmission from source to destination. In particular, the source broadcasts the signal to all relays at the first phase and relays decode and forward the received signal to the destination\footnote{We choose DF forwarding protocol in this paper due to its low CSI estimation complexity and satisfactory outage performance \cite{7504171}.}. For the noise, it is assumed to be independent and identically distributed (i.i.d.) at all nodes with noise power $N_0$.
Meanwhile, two signal degradation mechanisms encountered in transmission are considered, which are signal attenuation and multipath fading. Assuming randomly distributed relays are organized in the set $\mathcal{M}$, for the $m$th relay located at $\mathbf{p}_m=(r_{Sm},\theta_{m})$, $\forall~m\in\mathcal{M}$, if equal power allocation scheme is applied over all communication nodes with transmit power $P_t$, the received instantaneous SNR on the $k$th subcarrier is
\begin{equation}\small
\gamma_{1}(m,k)={P_t G_{1}(m,k)r_{Sm}^{-\alpha}}/{N_0},
\end{equation}
where $\alpha$ is the path loss exponent; $G_{i}(m,k)$ is the $i$th hop channel gain on the $k$th subcarrier due to multipath fading and is modeled as an i.i.d. exponentially distributed random variable with unit mean. Therefore, for $i\in\{1,2\}$, $m\in\mathcal{M}$ and $k\in\mathcal{K}$, the probability density function (PDF) and cumulative distribution function (CDF) of $G_i(m,k)$ are given by
\begin{equation}\label{dsad4sad211}\small
f_{G}(s)=e^{-s}~\Leftrightarrow~F_{G}(s)=1-e^{-s}.
\end{equation}
For the second phase, because of the DF forwarding protocol, the received instantaneous SNR at the destination is
\begin{equation}\small
\gamma_{2}(m,k)={P_t G_{2}(m,k)r_{mD}^{-\alpha}}/{N_0},
\end{equation}
where $r_{mD}$ is the distance between the $m$th relay and destination; however, it should be noted that because we assume the locations of all relays are unchanged during a complete transmission process, $r_{mD}$ is a dependent random variable on $r_{Sm}$ and $\theta_m$, which can be expressed by the law of cosines as
\begin{equation}\label{dsakjekl12coslawwww}\small
r_{mD}=\sqrt{r_{SD}^2+r_{Sm}^2-2r_{SD}r_{Sm}\cos\theta_{m}}.
\end{equation}
Finally, the equivalent end-to-end SNR in DF relaying network can be regarded as\footnote{An outage in DF relaying networks depends on the minimum channel coefficient among the source-relay and the relay-destination links. Hence, we can employ the minimum single-hop channel SNR as the equivalent end-to-end SNR here \cite{7445895}.}
\begin{equation}\label{dsakjhe1dfdfdf}\small
\gamma(m,k)=\min\{\gamma_{1}(m,k),\gamma_{2}(m,k)\}.
\end{equation}
\begin{figure}[!t]
\centering
\includegraphics[width=4.0in]{sys.png}
\caption{Illustration of (a) bulk and (b) per-subcarrier relay selection schemes for single source, single destination and multiple spatially randomly distributed relays, given $K = 8$. The numbers in boxes are the sequence numbers of subcarriers.}
\label{sys}
\end{figure}
\subsection{Relay selection schemes and outage probability}
Two selection schemes are considered in the paper. First, only \textit{one} relay among $\mathcal{M}$ is selected by the selection criterion below:
\begin{equation}\label{bulkdsak2s}\small
\mathcal{L}^{bulk}=\arg\max_{m\in\mathcal{M}}\min_{k\in\mathcal{K}}\gamma(m,k).
\end{equation}
This selection scheme is termed \textit{bulk selection}, because all subcarriers will be forwarded by the only one selected relay in bulk. Obviously, this selection scheme is easy to implement for OFDM systems, since only one relay is involved in the entire transmission process. However, its outage performance is obviously not optimal for each individual subcarrier. To obtain the optimal outage performance, we can apply another selection scheme termed \textit{per-subcarrier selection}, in which multiple relays are selected in a per-subcarrier manner, so that all subcarriers can be forwarded by their optimal relays. The per-subcarrier selection criterion is given as follows\footnote{Note that, here it is allowed that $\arg\max_{m\in\mathcal{M}}\gamma(m,k)=\arg\max_{m\in\mathcal{M}}\gamma(m,n)$ for $k\neq n$. In other words, the relay is capable of forwarding two or more subcarriers simultaneously. Because we do not consider a transmit power limit in all nodes, the power imbalance problem among relays is out of the scope of this paper.}
\begin{equation}\label{psscheme1}\small
\mathcal{L}^{ps}=\bigcup_{k=1}^K \left\lbrace \arg\max_{m\in\mathcal{M}}\gamma(m,k)\right\rbrace.
\end{equation}
For clarity, these two selection schemes are illustrated in Fig. \ref{sys}.
To consider the outage performance of multiple subcarriers as an entity, let the superscript $\Xi\in\{bulk, ps\}$ standing for different relay selection schemes and define the \textit{a posteriori} outage probability after selection as\footnote{$\mathcal{M}=\varnothing$ is possible when the area of relay distribution region $|\mathcal{C}|<\infty$ and this special scenario should be regarded as outage as well \cite{6042306}.}
\begin{equation}\label{outagedefdhe89514sas}\small
\Phi^{\Xi}(s)= \mathbb{P}\left\lbrace\left\lbrace\min_{k\in\mathcal{K}} \max_{m_k\in\mathcal{L}^{\Xi}} \gamma(m_k, k)<s\right\rbrace\bigcup\left\lbrace\mathcal{M}=\varnothing\right\rbrace\right\rbrace,
\end{equation}
where $\mathbb{P}\{\cdot\}$ denotes the probability of the random event enclosed; $s$ is a predefined and fixed target SNR threshold ; $m_k$ is the index of the selected relay forwarding the $k$th subcarrier. In this paper, we will take the outage probability defined above as a metric to evaluate the outage performance.
\section{Outage Performance Analysis}\label{opa}
\subsection{Bulk selection}
By (\ref{dsad4sad211}) and (\ref{dsakjhe1dfdfdf}), the CDF of the end-to-end SNR $\gamma(m,k)$ can be derived as
\begin{equation}\label{54651243111}\small
F(s)=1-\mathrm{exp}\left(-\frac{sN_0}{P_t}\left(r_{Sm}^{\alpha}+r_{mD}^\alpha\right)\right),
\end{equation}
and by (\ref{bulkdsak2s}) the a posteriori outage probability after performing bulk selection can be determined by\footnote{The average over $\Pi(\mathcal{C})$ includes the case of $\mathcal{M}=\varnothing$ when $|\mathcal{C}|$ is finite; The area integral given in the second line of the equation can be converted to a double integral by adopting a certain coordinate system (Cartesian, polar or biangular) and thus numerically calculated.}
\begin{equation}\label{phissbulk}\small
\begin{split}
\Phi^{bulk}(s)&=\underset{\Pi(\mathcal{C})}{\mathbb{E}}\left\lbrace\prod_{m\in\mathcal{M}}\left[ 1-(1-F(s))^K\right]\right\rbrace=\mathrm{exp}\left(-\lambda\int_{\mathcal{C}}(1-F(s))^K\mathrm{d}\mathbf{p}_m\right),
\end{split}
\end{equation}
where ${\mathbb{E}}\left\lbrace\cdot\right\rbrace$ denotes the average of the enclosed.
\subsubsection{Finite relay distribution region}
Due to the symmetry of the finite circle region $\mathcal{C}_{\varsigma}$, we can derive the outage probability in the finite circle region by (\ref{phissbulk}) and obtain:
\begin{equation}\label{dsa524651423}\small
\Phi_{\varsigma}^{bulk}(s)=\mathrm{exp}\left(-2\lambda u(\varsigma)\right),
\end{equation}
where $u(\varsigma)=\int_{0}^{\pi}\int_{0}^{\varsigma}\mathcal{H}(K)\mathrm{d}r_{Sm}\mathrm{d}\theta_m$ and $\mathcal{H}(K)=r_{Sm}\mathrm{exp}\left(-\frac{KsN_0}{P_t}\left(r_{Sm}^{\alpha}+r_{mD}^{\alpha}\right)\right)$.
Although there is no closed-form expression of (\ref{dsa524651423}) because of the double integral in $u(\varsigma)$, we can employ a power series expansion at $\frac{P_t}{N_0}\rightarrow \infty$ and obtain the asymptotic expressions of $\Phi^{bulk}_{\varsigma}(s)$ for an arbitrary $\alpha$:
\begin{equation}\label{jianjinhaoc}\small
\Phi_{\varsigma}^{bulk}(s)\sim\tilde\Phi_{\varsigma}^{bulk}(s)=\mathrm{exp}\left(-\lambda\pi\varsigma^2\left(1-\frac{KsN_0\tau_\alpha}{P_t}\right)\right),
\end{equation}
where
\begin{equation}\small
\tau_{\alpha}=
\begin{cases}
r_{SD}^2+\varsigma^2,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\alpha=2\\
r_{SD}^4+2r_{SD}^2\varsigma^2+\frac{2}{3}\varsigma^4,~~~~~~~~~~~~~~~~~~~~~~~\alpha=4\\
\frac{1}{2}\left(2r_{SD}^2+\varsigma^2\right)\left(r_{SD}^4+4r_{SD}^2\varsigma^2+\varsigma^4\right),~~~~\alpha=6
\end{cases}.
\end{equation}
From (\ref{jianjinhaoc}), it can be seen that with an increasing $\frac{P_t}{N_0}$, $\Phi_{\varsigma}^{bulk}(s)$ will converge to the outage floor given by
\begin{equation}\small
\underline \Phi_{\varsigma}^{bulk}(s)=\mathrm{exp}(-\lambda\pi\varsigma^2),
\end{equation}
which is caused by the scenario where there is no relay existing in the finite circle distribution region (c.f. (\ref{outagedefdhe89514sas})).
\subsubsection{Infinite relay distribution region}
Due to the symmetry of the infinite region $\mathcal{C}_{\mathrm{inf}}$, we can obtain
\begin{equation}\label{haochangya1}\small
\begin{split}
\Phi_{\mathrm{inf}}^{bulk}(s)=\mathrm{exp}\left(-2\lambda\int_{0}^{\pi}\int_{0}^{\infty}\mathcal{H}(K)\mathrm{d}r_{Sm}\mathrm{d}\theta_m\right)
\end{split}.
\end{equation}
To the best of the authors' knowledge, there does not exist a closed form of (\ref{haochangya1}) for an arbitrary $\alpha$. However, for a special case when $\alpha=2$ (free space), we can obtain
\begin{equation}\label{haochangya2}\small
\Phi_{\mathrm{inf}}^{bulk}(s)\vert_{\alpha=2}=\mathrm{exp}\left(-\frac{\lambda\pi P_t}{2KsN_0}\mathrm{exp}\left(-\frac{r_{SD}^2KsN_0}{2P_t}\right)\right).
\end{equation}
\subsection{Per-subcarrier selection}
Similarly as the case of bulk selection, by (\ref{psscheme1}) and the binomial theorem, the a posteriori outage probability after performing per-subcarrier selection can be determined by
\begin{equation}\label{phissps}\small
\begin{split}
&\Phi^{ps}(s)=\underset{\Pi(\mathcal{C})}{\mathbb{E}}\left\lbrace1-\left(1-\prod_{m\in\mathcal{M}}F(s)\right)^K\right\rbrace\\
&=\sum_{k=1}^{K}\left[\binom{K}{k}(-1)^{k+1}\underset{\Pi(\mathcal{C})}{\mathbb{E}}\left\lbrace\prod_{m\in\mathcal{M}}F^k(s)\right\rbrace\right]=\sum_{k=1}^{K}\left[\binom{K}{k}(-1)^{k+1}\mathrm{exp}\left(-\lambda\int_{\mathcal{C}}\left(1-F^k(s)\right)\mathrm{d}\mathbf{p}_m\right)\right].
\end{split}
\end{equation}
\subsubsection{Finite relay distribution region}
We can obtain the outage probability of per-subcarrier selection scheme over finite relay distribution region by
\begin{equation}\label{54d65sa465d1}\small
\begin{split}
\Phi_{\varsigma}^{ps}(s)=\sum_{k=1}^{K}\left[\binom{K}{k}(-1)^{k+1} \mathrm{exp}\left(-2\lambda\sum_{n=1}^{k}\binom{k}{n} (-1)^{n+1} u(\varsigma,n)\right)\right],
\end{split}
\end{equation}
where
$u(\varsigma,n)=\int_{0}^{\pi}\int_{0}^{\varsigma} \mathcal{H}(n)\mathrm{d}r_{Sm}\mathrm{d}\theta_{m}$.
Meanwhile, by power series expansion on (\ref{54d65sa465d1}) at $\frac{P_t}{N_0}\rightarrow \infty$, we also obtain the asymptotic expression of $\Phi_{\varsigma}^{ps}(s)$ as
\begin{equation}\label{jianjinhaocpsssss}\small
\begin{split}
&\Phi_{\varsigma}^{ps}(s)\sim\tilde\Phi_{\varsigma}^{ps}(s)=
\mathrm{exp}\left(-\lambda\pi\varsigma^2\right)\left[1-K\left(1-\mathrm{exp}\left(\frac{\lambda\pi\varsigma^2sN_0\tau_\alpha}{P_t}\right)\right)\right]
\end{split},
\end{equation}
from which we can observe that the outage floor given by
\begin{equation}\small
\begin{split}
\underline\Phi_{\varsigma}^{ps}(s)=\mathrm{exp}(-\lambda\pi\varsigma^2)=\underline\Phi_{\varsigma}^{bulk}(s),
\end{split}
\end{equation}
is exactly the same as the case of bulk selection. As expected, the outage event at high SNR within a finite circle distribution region is dominated by the case where there is no relay candidate for selection (i.e. $\mathcal{M}=\varnothing$). As a result, it is irrelevant to the selection schemes and/or forwarding protocols and only dependent on the relay node density $\lambda$ and the radius $\varsigma$. In general, the outage floor for an arbitrary finite distribution region $\mathcal{C}$ can be derived by
\begin{equation}\small
\underline\Phi(\mathcal{C})=\mathrm{exp}\left(-\lambda|\mathcal{C}|\right).
\end{equation}
\subsubsection{Infinite relay distribution region}
Similarly, substituting $\mathcal{C}=\mathcal{C}_{\mathrm{inf}}$ into (\ref{phissps}) yields
\begin{equation}\label{haochangyapsdf}\small
\begin{split}
&\Phi_{\mathrm{inf}}^{ps}(s)=\sum_{k=1}^{K}\left[\binom{K}{k}(-1)^{k+1} \mathrm{exp}\left(-2\lambda\sum_{n=1}^{k}\binom{k}{n} (-1)^{n+1} \int_{0}^{\pi}\int_{0}^{\infty} \mathcal{H}(n) \mathrm{d}r_{Sm}\mathrm{d}\theta_{m}\right)\right].
\end{split}
\end{equation}
Again, there is no closed form of (\ref{haochangyapsdf}) for an arbitrary $\alpha$. When $\alpha=2$, we can obtain
\begin{equation}\label{218sjsyrtysy3}\small
\begin{split}
&\Phi_{\mathrm{inf}}^{ps}(s)\vert_{\alpha=2}=\sum_{k=1}^{K}\left[\binom{K}{k}(-1)^{k+1} \mathrm{exp}\left(-\lambda\pi\sum_{n=1}^{k}\binom{k}{n} (-1)^{n+1}\frac{ P_t}{2 nsN_0}\mathrm{exp}\left(-\frac{r_{SD}^2nsN_0}{2P_t}\right) \right)\right].
\end{split}
\end{equation}
\section{Discussion of Important System Properties}\label{disp}
\subsection{Cooperative diversity analysis}
For deterministic networks in which the number of relays and/or their locations are stationary, diversity gain is an all-important metric to measure the cooperative performance advantage \cite{1362898}. However, we show as follows that this metric is not appropriate anymore for cooperative systems in random networks where relays are Poisson distributed. Subsequently we prove the \textit{ternary property} of diversity gain in Poisson random networks.
For cooperative systems in finite Poisson random networks, the binary property of diversity gain has been proved and it states that the diversity gain $d_o(\mathcal{C})$ can only be either one or zero depending on whether there is a direct transmission link between source and destination or not \cite{6042306}. Following the discussion of outage floor in Section \ref{opa}, this binary property related to diversity gain is straightforward. Because without a direct transmission link, an increasing power would not affect the outage performance at high SNR and this leads to a \textit{zero-diversity-order system}.
However, when considering an infinite distribution region, there does not exist such an outage floor, because if the region is infinite, as long as the relay node density $\lambda$ is positive, there are always an infinite number of relays distributed over region $\mathcal{C}_{\mathrm{inf}}$. As a result, with a \textit{sufficiently} large transmit power (considering the asymptotic region), the two-hop transmission can always be successful, because there must exist a `satisfactory' two-hop link among an infinite number of relays. This indicates the diversity gain of cooperative systems in infinite random networks is infinity. Mathematically, for an arbitrary and bounded $\alpha$, we can prove that\footnote{See Appendix \ref{eppenproxgaindiv}.}
\begin{equation}\label{dsakj12837223gmis}\small
d_o^{\Xi}(\mathcal{C}_{\inf})=-\lim_{\frac{P_t}{N_0}\rightarrow\infty}\frac{\log (\Phi_{\mathrm{inf}}^{\Xi}(s))}{\log (P_t/N_0)}=\infty.
\end{equation}
Therefore, unlike the situations in deterministic networks, it would be impossible to derive a linear relation among $d_o(\mathcal{C})$, $\lambda$ and $\alpha$ in random networks. Instead, the diversity gain can only be either zero, one or infinity. We term this \textit{ternary property} of diversity gain in Poisson random networks. In the meantime, (\ref{dsakj12837223gmis}) also indicates that any high but bounded path loss attenuation does not counter the constructive effects of infinite distribution region on asymptotic outage performance.
\subsection{Comparison of outage performances between bulk and per-subcarrier selections}\label{sa65d6142321312}
By (\ref{phissbulk}) and (\ref{phissps}), we can quantify the outage performance advantage of per-subcarrier selection over bulk selection via the outage probability ratio given by
\begin{equation}\label{45philamd65dsa451}\small
\phi(\lambda)=\frac{\Phi^{ps}(s)}{\Phi^{bulk}(s)}=\sum_{k=1}^{K}\left[\binom{K}{k}(-1)^{k+1}\mathrm{exp}\left(-\lambda\Delta(k)\right)\right],
\end{equation}
where $\Delta(k)=\int_{\mathcal{C}}\left[1-F^k(s)-\left(1-F(s)\right)^K\right]\mathrm{d}\mathbf{p}_m$.
Considering the case of sparse networks with small $\lambda$, the outage probability ratio $\phi(\lambda)$ can be approximated by\footnote{See Appendix \ref{5464218penglua}.}
\begin{equation}\label{jinsidephilamd}\small
\phi(\lambda)\approx 1+\frac{\lambda^2}{2}\sum_{k=1}^{K}\left[\binom{K}{k}(-1)^{k+1}\Delta^2(k)\right].
\end{equation}
Eq. (\ref{jinsidephilamd}) indicates that the outage performance advantage brought by per-subcarrier selection over bulk selection will become negligible with a decreasing $\lambda$, because the number of relays for selection is small and employing multiple relays to perform per-subcarrier selection becomes less likely. Meanwhile, if we would like to ensure the performance advantage by letting the outage probability ratio $\phi(\lambda)\leq\epsilon$ (i.e. the outage probability of per-subcarrrier selection is $\epsilon$ times lower than that of bulk selection), an approximation of the required minimum density $\lambda$ can be derived as
\begin{equation}\label{solutlamndsa2}\small
\lambda\geq\underline{\lambda}(\epsilon)\approx\sqrt{{2(\epsilon-1)}/{\sum_{k=1}^{K}\left[\binom{K}{k}(-1)^{k+1}\Delta^2(k)\right]}},
\end{equation}
which would provide a guidance for whether bulk or per-subcarrier selection should be employed or a switching criterion for a dynamic switching mechanism between these two selection schemes when the relay node density $\lambda$ is known\footnote{Although per-subcarrier selection scheme has the optimal outage performance, it is not always preferable if the performance advantage is not significant, because the selection and synchronization processes among multiple relays and destination node are complicated and thus will require more overheads \cite{6356932}.}.
\subsection{Optimization of the number of subcarriers for bulk selection}
With the development of index modulation and frequency resource allocation, OFDM systems with a varying number of active subcarriers become common in practice \cite{6587554,7809043}. Therefore, it is meaningful to investigate the effects of adopting different numbers of subcarriers $K$ on the system throughput. For simplicity, we assume that equal numbers of bits are carried by each subcarrier. Hence, the average system throughput can be characterized by the average number of successfully decoded subcarriers per transmission at the destination \cite{1343893}, which can be determined by
\begin{equation}\label{213112kappa}\small
\kappa(K,\lambda)=K\left(1-\Phi^{\Xi}(s)\right).
\end{equation}
According to (\ref{outagedefdhe89514sas}), a large $K$ will lead to a large $\Phi^{\Xi}(s)$, since all signals transmitted on these $K$ subcarriers need to be successfully decoded at the destination, or an outage will occur otherwise. Consequently, for a given $\lambda$, there exists an optimal number of subcarriers denoted by $K_{\mathrm{opt}}$, by which the system throughput can be maximized. Also, because relay node density $\lambda$ is varying in practice due to the dynamic on/off switching mechanisms and node mobility \cite{6502479,565669}, it is intuitive that we can dynamically adjust the number of subcarriers $K$ to offset the adverse effects of relay node density $\lambda$, so that the system throughput can always be maximized. To be clear, the relation among $\kappa(K,\lambda)$, $K$ and $\lambda$ is plotted in Fig. \ref{fujia21} for the case of bulk selection in free space.
Now, let us focus on the method to determine $K_{\mathrm{opt}}$. First, we relax the integer $K$ to a positive real number $\tilde{K}$\footnote{Currently, this method is only applicable for bulk selection, since $K$ can be regarded as a product factor in $\Phi^{bulk}(s)$ and thus can be relaxed to $\tilde{K}$. On the other hand, because $K$ is the upper limit of the summation in $\Phi^{ps}(s)$, this relaxing relation is not feasible, and thus the following optimization is not possible for per-subcarrier selection. However, it is possible to propose a sub-optimal solution for per-subcarrier selection in which the $\Phi^{ps}(s)$ is replaced by a certain approximation with $K$ as a product factor.}, so that some optimization tools can be applied. Then, it can be proved that $\kappa(\tilde K,\lambda)$ is a concave function of $\tilde K$\footnote{See Appendix \ref{2311profconcav}}. Therefore, the optimal solution to $\tilde K$ can be obtained via a concave problem formulated by
\begin{equation}\label{dask2caveskaps}\small
\begin{split}
&\tilde K_{\mathrm{opt}}=\arg\max_{\tilde K} \kappa(\tilde K,\lambda)\\
&~~~~~~~~\mathrm{s.t.}~~\tilde K>0,
\end{split}
\end{equation}
which can be solved efficiently using standard optimization techniques (e.g. CVX in MATLAB). Then, $K_{\mathrm{opt}}$ is determined by
\begin{equation}\small
K_{\mathrm{opt}}=\begin{cases}
\lceil\tilde K_{\mathrm{opt}}\rceil,~~\mathrm{if}~\kappa(\lceil\tilde K_{\mathrm{opt}}\rceil,\lambda)\geq \kappa(\lfloor\tilde K_{\mathrm{opt}}\rfloor,\lambda)\\
\lfloor\tilde K_{\mathrm{opt}}\rfloor,~~\mathrm{if}~\kappa(\lceil\tilde K_{\mathrm{opt}}\rceil,\lambda)< \kappa(\lfloor\tilde K_{\mathrm{opt}}\rfloor,\lambda),
\end{cases}
\end{equation}
where $\lceil\cdot\rceil$ and $\lfloor\cdot\rfloor$ denote ceiling and floor functions, respectively.
However, the optimization of $\kappa(K,\lambda)$ given above would not be always applicable, since the chosen $K_{\mathrm{opt}}$ would lead to an inappropriate outage probability beyond a threshold $\Psi$, which should be maintained for a prescribed quality of service (QoS). Unlike networks for real-time streaming media in which throughput is the key \cite{7362038}, this reliability requirement is in particular crucial for some special networks, e.g. the Internet of things (IoT) \cite{atzori2010internet} and military wireless sensor networks (WSNs) \cite{5379900}. In this context, $K_{\mathrm{opt}}$ is dependent on $\lambda$ and $\Psi$. Therefore, with this constraint on outage probability $\Psi$, the concave optimization problem formulated in (\ref{dask2caveskaps}) becomes
\begin{equation}\label{dask2caves232322kaps}\small
\begin{split}
&\tilde K_{\mathrm{opt}}=\arg\max_{\tilde K} \kappa(\tilde K,\lambda)\\
&~~~~~~~~\mathrm{s.t.}~~\tilde K>0~~\mathrm{and}~~\Phi^{\Xi}(s)\vert_{K=\tilde K}\leq\Psi.
\end{split}
\end{equation}
Then $K_{\mathrm{opt}}$ will be determined by
\begin{equation}\small
K_{\mathrm{opt}}=\lfloor\tilde K_{\mathrm{opt}}\rfloor.
\end{equation}
On the other hand, because $K\geq 1$ must be ensured in order to provide transmission service, it is impossible to always maintain a given outage probability $\Psi$ by reducing $K$ when $\lambda$ keeps decreasing. Consequently, there is a cut-off relay node density $\lambda_c$ below which the outage probability $\Psi$ cannot be maintained by reducing $K$. By substituting $K=1$ into (\ref{phissbulk}), the cut-off density can be numerically evaluated by
\begin{equation}\small
\lambda_c=-\frac{\ln \Psi}{\int_{\mathcal{C}}(1-F(s))\mathrm{d}\mathbf{p}_m}.
\end{equation}
By (\ref{haochangya2}), the cut-off relay node density for bulk selection in free space can be approximated by
\begin{equation}\small
\lambda_c\vert_{\alpha=2}\approx-\frac{2sN_0\ln \Psi}{\pi P_t \mathrm{exp}\left(-\frac{r_{SD}^2sN_0}{2P_t}\right)}.
\end{equation}
\begin{figure}[!t]
\centering
\includegraphics[width=5.5in]{fujia21.eps}
\caption{Relation among $\kappa(K,\lambda)$, $K$ and $\lambda$ for bulk selection in free space ($\alpha=2$), when $s=1$, $P_t/N_0=100$, $r_{SD}=5$ and $\varsigma=5$.}
\label{fujia21}
\end{figure}
\begin{figure}[!t]
\centering
\includegraphics[width=5.0in]{tupian1.eps}
\caption{Bulk selection case: outage probability vs. $P_t/N_0$, when $\lambda=1$, $\varsigma=5$ and $r_{SD}=5$.}
\label{tupian1}
\end{figure}
\begin{figure}[!t]
\centering
\includegraphics[width=5.0in]{tupian2.eps}
\caption{Per-subcarrier selection case: outage probability vs. $P_t/N_0$, when $\lambda=1$, $\varsigma=5$ and $r_{SD}=5$.}
\label{tupian2}
\end{figure}
\begin{figure}[!t]
\centering
\includegraphics[width=5.0in]{tupian3.eps}
\caption{Connection probability vs. relay node density $\lambda$, when $P_t/N_0=100$, $\varsigma=5$, $r_{SD}=5$ and $K=4$.}
\label{tupian3}
\end{figure}
\begin{figure}[!t]
\centering
\includegraphics[width=5.0in]{fujia1.eps}
\caption{$\bar\epsilon$ vs. relay node density $\lambda$, when $\varsigma=5$, $r_{SD}=5$ and $K=4$.}
\label{fujia1}
\end{figure}
\begin{figure}[!t]
\centering
\includegraphics[width=5.0in]{kappa.eps}
\caption{Relation between relay node density $\lambda$ and the optimal number of subcarriers $K_{\mathrm{opt}}$ as well as the maximum system throughput characterized by $\kappa(K_{\mathrm{opt}},\lambda)$, when $s=1$, $P_t/N_0=100$, $\varsigma=5$ and $r_{SD}=5$.}
\label{kappa}
\end{figure}
\begin{figure}[!t]
\centering
\includegraphics[width=5.0in]{fujia22.eps}
\caption{Relation between relay node density $\lambda$ and the optimal number of subcarriers $K_{\mathrm{opt}}$ for a given outage probability $\Psi\in\{10^{-2},10^{-3},10^{-5}\}$, when $\alpha=2$, $s=1$, $P_t/N_0=100$, $\varsigma=5$ and $r_{SD}=5$.}
\label{fujia22}
\end{figure}
\section{Numerical Results}\label{nr}
\subsection{Outage performance of bulk and per-subcarrier selections}
First of all, we need to verify the correctness of the analytical results of outage probabilities. To do so, we normalize $s=1$, $N_0=1$ and set $\lambda=1$, $\varsigma=5$ and $r_{SD}=5$\footnote{The lengths given here are relative and thus dimensionless, since other parameters have been normalized.}. Then, we simulate the relation between the outage probability and the transmit power with spatially random relays distributed over a finite circle region. The simulation results for bulk and per-subcarrier selections are presented in Fig. \ref{tupian1} and Fig. \ref{tupian2}, respectively. By the results shown in these two figures, our analysis of the outage performance corresponding to finite circle region has been verified. Meanwhile, we can also see that the variation of the number of subcarriers $K$ has a significant impact on the outage performance for bulk selection, while it is relatively trivial for per-subcarrier selection. This is because bulk selection can only employ one relay to satisfy transmission requirements of all subcarriers, which is more `$K$-sensitive' than the case of per-subcarrier selection which is capable of employing multiple relays. Both selection schemes are affected by path loss exponent in a large but similar scale, which aligns with our expectation.
\subsection{Relation between relay node density and multi-relay performance advantage}
In order to investigate the effects of the relay node density $\lambda$ on outage probability ratio $\phi(\lambda)$, we define the \textit{connection probability} as $\bar \Phi^{\Xi}(s)= 1- \Phi^{\Xi}(s)$, which eases the illustration of some crucial properties of spatially random networks in logarithmic plots when $\lambda$ is small\footnote{Because when $\lambda$ is small, the outage probability will approach to one and the logarithmic scale is a nonlinear scale in terms of the power of ten, which cannot show the details around one.}. Then, we set $P_t/N_0=100$ and $K=4$ and plot the relation between connection probability and $\lambda$ in Fig. \ref{tupian3}. From Fig. \ref{tupian3}, it is obvious that when $\lambda$ is small (i.e. in sparse networks), the multi-relay performance advantage brought by per-subcarrier selection over bulk selection will become negligible, because the number of relays for selection is small and employing multiple relays to perform per-subcarrier selection becomes less likely. Also, in order to show the properties for small $\lambda$ in logarithmic plots, we specify $\bar\epsilon=1-\epsilon$ and plot the relation between $\bar\epsilon$ and $\lambda$ to verify the proposed approximation (\ref{solutlamndsa2}) in Fig. \ref{fujia1}\footnote{As the same reason that we define connection probability, when $\lambda$ is small, $\epsilon$ will approach to one and the details cannot be shown by the logarithmic scale based on the power of ten. In Fig. \ref{fujia1}, the exact $\lambda$ is produced by numerically inversely solving (\ref{45philamd65dsa451}) for $\phi(\lambda)=1-\bar\epsilon$, while the approximate $\lambda$ is produced by calculating $\underline{\lambda}(1-\bar{\epsilon})$, as given in (\ref{solutlamndsa2}).}. From this figure, it is verified that the proposed approximation on $\underline{\lambda}(\epsilon)$ is tight for small $\lambda$ and can thereby provide an efficient metric for evaluating the performance advantage brought by per-subcarrier selection over bulk selection in sparse networks.
\subsection{Relation between the number of subcarriers and system throughput}
To verity the solutions to the optimal number of subcarriers produced by the concave problem formulated in (\ref{dask2caveskaps}), we set $r_{SD}=5$, $\varsigma=5$ and $P_t/N_0=100$ and plot the simulation results in Fig. \ref{kappa}. The jagged nature of $K_{\mathrm{opt}}$ reflects $K$, as the number of subcarriers can only take discrete values in practice. Meanwhile, from this figure, the solutions produced by the concave problem exactly match the numerical results, which verifies the accuracy of the proposed solution. It also aligns with our expectation that with an increasing $\lambda$, both $K_{\mathrm{opt}}$ and the maximum $\kappa(K,\lambda)$ will increase, since the signal propagation environment has been improved with more `appropriate' relays. Also, a higher path loss exponent will significantly degrade the system throughput, which can be observed by comparing the results shown for $\alpha=2$ and $\alpha=4$. The degradation due to high path loss exponent is obvious especially when $\lambda$ is large.
Besides, to verify the constrained solution to the optimal number of subcarriers $K_{\mathrm{opt}}$ for a given outage probability $\Psi$, the simulation results are presented in Fig. \ref{fujia22}. Here, we specify $K_{\mathrm{opt}}=0$ when $\lambda$ is lower than the cut-off density $\lambda_c$ for simplicity and frequency resource saving purposes. Again, both numerical and analytical results match each other and these results verify the feasibility of the proposed solution for the scenario with an outage constraint. Also, in comparison with the results illustrated in Fig. \ref{kappa}, the constraint on outage probability will result in a lower optimal number of subcarriers. Furthermore, by comparing the cases with different $\Psi$, it can be found that the optimal number of subcarriers reduces with the constraint on outage.
\section{Conclusion}\label{c}
In this paper, we analyzed the outage performance of two-hop OFDM systems adopting bulk and per-subcarrier selection schemes, respectively. Also, a series of important properties related to cooperative systems in random networks were investigated, including diversity, outage probability ratio of two selection schemes and the optimization of the number of subcarriers. All analysis has been verified by simulations and some key properties of cooperative OFDM systems over finite and infinite random networks have been revealed and discussed. Moreover, by (\ref{phissbulk}) and (\ref{phissps}), an analytical framework for OFDM systems over random networks has been constructed in this paper, which can be easily modified to analyze other similar cases with different forwarding protocols, location distributions and/or channel conditions.
\appendices
\section{Proof of Diversity Gain in Infinite Poisson Random Networks}\label{eppenproxgaindiv}
To determine the diversity gain in infinite Poisson random networks for bulk selection, we first have the relation
\begin{equation}\label{dachangesdas2}\small
\begin{split}
&d_o^{bulk}(\mathcal{C}_{\mathrm{inf}})=-\lim_{\frac{P_t}{N_0}\rightarrow\infty}\frac{\log\left(\Phi_{\mathrm{inf}}^{bulk}(s)\right)}{\log\left(P_t/N_0\right)}\\
&\overset{(\mathrm{a})}{>}\lim_{\frac{P_t}{N_0}\rightarrow\infty} \frac{2\pi\lambda\int_{0}^{\infty}r_{Sm}\mathrm{exp}\left(-\frac{KsN_0}{P_t}\left(r_{Sm}^\alpha+(r_{Sm}+r_{SD})^\alpha\right)\right)\mathrm{d}r_{Sm}}{\ln 10 \log(P_t/N_0)}\\
&>\mathcal{T}_2=\lim_{\frac{P_t}{N_0}\rightarrow\infty} \frac{2\pi\lambda\mathcal{T}_1}{\ln 10 \log(P_t/N_0)},
\end{split}
\end{equation}
where $(\mathrm{a})$ holds because of the trigonometric relation among $r_{mD}$, $r_{Sm}$ and $\theta_m$ expressed in (\ref{dsakjekl12coslawwww}); $\mathcal{T}_1$ is given by
\begin{equation}\label{1123changdes}\small
\begin{split}
\mathcal{T}_1&=\int_{0}^{\infty}r_{Sm}\mathrm{exp}\left(-\frac{KsN_0}{P_t}\left(r_{Sm}^\alpha+(2\max(r_{Sm},r_{SD}))^\alpha\right)\right)\mathrm{d}r_{Sm}\\
&=\int_{0}^{r_{SD}}r_{Sm}\mathrm{exp}\left(-\frac{KsN_0}{P_t}\left(r_{Sm}^\alpha+(2r_{SD})^\alpha\right)\right)\mathrm{d}r_{Sm}+\int_{r_{SD}}^{\infty}r_{Sm}\mathrm{exp}\left(-\frac{KsN_0}{P_t}\left(r_{Sm}^\alpha+(2r_{Sm})^\alpha\right)\right)\mathrm{d}r_{Sm}\\
&=\left(\frac{P_t}{KsN_0}\right)^{\frac{2}{\alpha}}\mathrm{exp}\left(-\frac{KsN_0(2r_{SD})^\alpha}{P_t}\right)\underline{\Gamma}\left(\frac{2}{\alpha},\frac{KsN_0r_{SD}^\alpha}{P_t}\right)+\frac{r_{SD}^2}{\alpha}{E}_{\frac{\alpha-2}{\alpha}}\left(\frac{(1+2^\alpha)KsN_0r_{SD}^\alpha}{P_t}\right),
\end{split}
\end{equation}
where $\underline{\Gamma}(n,x)=\int_{0}^{x}t^{n-1}e^{-t}\mathrm{d}t$ is the lower incomplete gamma function and ${E}_{n}(x)=\int_{1}^{\infty}{e^{xt}}/{t^n}\mathrm{d}t$ is the exponential integral function.
Substituting (\ref{1123changdes}) into the last line of (\ref{dachangesdas2}) yields
\begin{equation}\small
\mathcal{T}_2=\lim_{\frac{P_t}{N_0}\rightarrow\infty} \frac{2\pi\lambda\mathcal{T}_1}{\ln 10 \log(P_t/N_0)}=\infty.
\end{equation}
As a consequence of $d_o^{bulk}(\mathcal{C}_{\mathrm{inf}})>\mathcal{T}_2$, we have $ d_o^{bulk}(\mathcal{C}_{\mathrm{inf}})=\infty$. Meanwhile, as different selection schemes will not affect diversity gain, which is only related to relay's distribution \cite{4489212}, we can also have $d_o^{ps}(\mathcal{C}_{\mathrm{inf}})=d_o^{bulk}(\mathcal{C}_{\mathrm{inf}})=\infty$. Therefore, considering an infinite distribution region $\mathcal{C}_{\mathrm{inf}}$, for $\Xi\in\{bulk,ps\}$, (\ref{dsakj12837223gmis}) has been proved.
\section{Proof of the Approximation of $\phi(\lambda)$ for Small $\lambda$}\label{5464218penglua}
First, we propose a lemma below and this is the prerequisite for applying power series expansion on $\mathrm{exp}\left(-\lambda\Delta(k)\right)$ for small $\lambda$:
\begin{lemma}\label{das4541652lemma}
$\Delta(k)=\int_{\mathcal{C}}\left[1-F^k(s)-\left(1-F(s)\right)^K\right]\mathrm{d}\mathbf{p}_m$ is positive and bounded for $k\in\mathbb{N}^+$ and $k\leq K$.
\end{lemma}
\begin{IEEEproof}
Because $k\leq K$ and $F(s)$ is a CDF and thus satisfies $0<F(s)<1$, there exists a relation of the integrand in $\Delta(k)$:
\begin{equation}\label{yizhongxs}\small
1-F^k(s)-\left(1-F(s)\right)^K\geq 1-F^k(s)-\left(1-F(s)\right)^k.
\end{equation}
For $1-F^k(s)-\left(1-F(s)\right)^k$, we can employ mathematical induction as below to prove $1-F^k(s)-\left(1-F(s)\right)^k\geq0$:
For $k=1$, $1-F^1(s)-\left(1-F(s)\right)^1=0$ and the statement is true. Then for $k=n>1$, assuming $1-F^n(s)-\left(1-F(s)\right)^n\geq 0$ holds, we can have $1-F^n(s)\geq\left(1-F(s)\right)^n$. Because $0<F(s)<1$, we can further obtain
\begin{equation}\small
1-F^{n+1}(s)>1-F^n(s)\geq\left(1-F(s)\right)^{n}>\left(1-F(s)\right)^{n+1}.
\end{equation}
Therefore, the statement for $k=n+1$ is true and we thus prove the statement $1-F^k(s)-\left(1-F(s)\right)^k\geq 0$ for $k\in\mathbb{N}^+$. Due to (\ref{yizhongxs}) and $K\geq2$ (basic assumption of multicarrier systems), we thereby prove $1-F^k(s)-\left(1-F(s)\right)^K>0$ for $k\in\mathbb{N}^+$ and $k\leq K$. Meanwhile, because $F(s)$ will decrease exponentially with an increasing transmission distance, the area integral of the integrand $1-F^k(s)-\left(1-F(s)\right)^K$ over the region $\mathcal{C}$ is positive and bounded. As a result, the lemma is proved.
\end{IEEEproof}
By Lemma \ref{das4541652lemma}, $\mathrm{exp}\left(-\lambda\Delta(k)\right)$ is proved to be expandable for small $\lambda$ \cite{thomson2008elementary}. we can employ a power series expansion on $\mathrm{exp}\left(-\lambda\Delta(k)\right)$ for small $\lambda$ and obtain
\begin{equation}\small
\mathrm{exp}\left(-\lambda\Delta(k)\right)=1-\lambda\Delta(k)+\frac{\lambda^2\Delta^2(k)}{2}+O(\lambda^3).
\end{equation}
We can truncate $\mathrm{exp}\left(-\lambda\Delta(k)\right)$ by its second order term and substitute $\mathrm{exp}\left(-\lambda\Delta(k)\right)\approx 1-\lambda\Delta(k)+{\lambda^2\Delta^2(k)}/{2}$ into (\ref{45philamd65dsa451}), which yields
\begin{equation}\small
\begin{split}
\phi(\lambda)&\approx \sum_{k=1}^{K}\left[\binom{K}{k}(-1)^{k+1}\left(1-\lambda\Delta(k)+\frac{\lambda^2\Delta^2(k)}{2}\right)\right]\\
&=\sum_{k=1}^{K}\left[\binom{K}{k}(-1)^{k+1}\right]-\lambda\sum_{k=1}^{K}\left[\binom{K}{k}(-1)^{k+1}\Delta(k)\right]+\frac{\lambda^2}{2}\sum_{k=1}^{K}\left[\binom{K}{k}(-1)^{k+1}\Delta(k)^2\right].
\end{split}
\end{equation}
Furthermore, for $K\in\mathbb{N}^+$ and $K\geq 2$, by the binomial theorem, we can derive
\begin{equation}\label{dsa45d51542432hehesum}\small
\begin{split}
\sum_{k=1}^{K}\left[\binom{K}{k}(-1)^{k+1}\right]=1-\sum_{k=0}^{K}\binom{K}{k}(-1)^k=1-(1-1)^K=1.
\end{split}
\end{equation}
For the first order term, we can use the additivity property of integrals and swap the order of summation and integral by
\begin{equation}\label{54d65sa4561heszcs}\small
\begin{split}
&\sum_{k=1}^{K}\left[\binom{K}{k}(-1)^{k+1}\Delta(k)\right]=\int_{\mathcal{C}}\sum_{k=1}^{K}\left\lbrace\binom{K}{k}(-1)^{k+1}\left[1-F^k(s)-\left(1-F(s)\right)^K\right]\right\rbrace\mathrm{d}\mathbf{p}_m.
\end{split}
\end{equation}
Subsequently, the integrand can be determined by
\begin{equation}\small
\begin{split}
&\sum_{k=1}^{K}\left\lbrace\binom{K}{k}(-1)^{k+1}\left[1-F^k(s)-\left(1-F(s)\right)^K\right]\right\rbrace\\
&=\sum_{k=1}^{K}\left[\binom{K}{k}(-1)^{k+1}\right]-\sum_{k=1}^{K}\left[\binom{K}{k}(-1)^{k+1}F^{k}(s)\right]-(1-F(s))^K\sum_{k=1}^{K}\left[\binom{K}{k}(-1)^{k+1}\right]\\
&=1-[1-(1-F(s))^K]-(1-F(s))^K=0.
\end{split}
\end{equation}
As a result, the corresponding area integral of this zero integrand given in (\ref{54d65sa4561heszcs}) is also zero. Finally, the approximation of $\phi(\lambda)$ for small $\lambda$ given in (\ref{jinsidephilamd}) is proved.
\section{Proof of the Concavity of $\kappa(\tilde K,\lambda)$ in Terms of $\tilde K$}\label{2311profconcav}
According to the expressions of outage probability for bulk and per-subcarrier selections given in (\ref{phissbulk}) and (\ref{phissps}), it is obvious that $\Phi^{\Xi}(s)$ is a monotonically deceasing function of $\tilde K$. Therefore, $\forall~\beta\in[0,1]$, $\tilde K_1>0$ and $\tilde K_2>0$, we have
\begin{equation}\small
\Phi^{\Xi}(s)\vert_{K=(1-\beta)\tilde{K}_1+\beta\tilde{K}_2}\geq\Phi^{\Xi}(s)\vert_{K=(1-\beta)\tilde{K}_1}
\end{equation}
and
\begin{equation}\small
\Phi^{\Xi}(s)\vert_{K=(1-\beta)\tilde{K}_1+\beta\tilde{K}_2}\geq\Phi^{\Xi}(s)\vert_{K=\beta\tilde{K}_2}.
\end{equation}
As a result, it can be derived that
\begin{equation}\small
\begin{split}
\kappa((1-\beta)\tilde{K}_1+\beta\tilde{K}_2,\lambda)&\geq (1-\beta)\kappa((1-\beta)\tilde{K}_1,\lambda)+\beta\kappa(\beta\tilde{K}_2,\lambda).
\end{split}
\end{equation}
This proves the concavity of $\kappa(\tilde K,\lambda)$ in terms of $\tilde K$, according to the definition of a concave function \cite{arrow1961quasi}.
\bibliographystyle{IEEEtran}
\bibliography{bib}
\end{document}
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Imagine a Scenario...
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Je t’aime, asshole: Notes Towards a European Ghazal
Typed up my post, then it vanished. An account of meeting Agha Shahid Ali. I wrote it on my home blog, pressed the wrong key and it was gone. Then I came here, wrote it out again, from scratch – what the poet said at night, on a loading dock at Wells College, about the ghazal – and now, once again, it is gone. Where did it go?
Perhaps this accompanies the question of the poet’s body. Where is he?
“He is in the thunder and lightning. He is in the branches of the trees.” — Ghazal 2.b.
I think of typing up my second book, Incubation. And how it vanished. That first version. I pressed something: what? The computer whirred then stopped. It broke. Renee Gladman wrote something along the lines of: “Dear Bhanu, not to worry. We can publish your work the year after next. It’s clear you will not be able to get it to us in time.” I re-wrote the book – a new book – in a two week sitting. Numb, the single mother of a kinetic four year old, and newly divorced/stunned: I’d come home from dropping my son at his preschool, close the door behind me, take off my clothes, lay my clothes over the arm of the sofa, walk through the house to the alcove with the table, strip it of its red tablecloth, wrap myself in the tablecloth, climb on a chair, get down the Maker’s Mark, and pour out a ritualistic tablespoon. To drink.
Then write.
Write then go.
Did you write a book that vanished? Did language return to you? How? Where do words go when they are deleted? Or disappear. Do they recirculate? Are they wet? Are they dead? Can you communicate with language that has gone in the same way that it is possible to communicate with a person who is no longer there? Is it possible to retrieve words that have been lost, to the processes of time, culture or fate?
Describe the pathway of touch – light touch – that allows them to return. Return as vibration. Not vibration. At first, you cannot hear a thing.
The phonemes of Zong! that are the matter of the page.
Though it’s my aunts’ and mother’s stories of epic, colonial and daily life that persist, it’s only my grandfather’s Urdu, Farsi and Persian couplets that have a physical, written presence in his yellowing notebooks on a shelf in India, now Colorado, next to his orchard notes. His list of the seeds and the year he bought them. A pencil drawing of the space and its boundary lines. A mathematics of the mango grove. I want to give this notebook to Matias Viegener and David Burns for their Fallen Fruit project.
Or, on the other side of the family, Kapil Muni, my ancestor, chanting OM NAMAO SHIVAI a thousand years before Christ, off the coast of Bengal. The flame set upon the waves of the sea in a boat of banana leaves and tiny roses tied in bundles with red string, the tendons of a plant that grows near the sea. When I meditate, he sometimes meditates with me. Does he?
A dialect. Language 52.
Ghazal 3:
I touched the bark of an oak tree with my fingertips a thousand times or more on the way to school. One day, a robin flew down out of the lowest branches and landed where I stood upon on my arm. My shirt. My dark green sleeve. Its beak. And sang. Since that time, I have felt great sorrow, in early spring, to see a robin dead on the street. But joy when I see the blue eggs. In the nest by the door. In the garden. In the rain.
The risk of the ghazal, in English, is that, as above, it ends on a lyric note when, in reality, it should be torn from the body like a sob. It should never be written down. It should never be read aloud in a person’s ordinary speaking voice. It is more like a scream than a song. Lisa Robertson: “Form is improvisational.” She said it at CCA, giving, as I recall, one of Trish Salah’s ghazals as an example, to a class of students that included Erin Morrill. I had accompanied Erin to the class because I loved Lisa Robertson’s R’s Boat but had never met her (Lisa R.) properly. Similarly, I liked the example, but began to cough. I coughed and coughed. I left the room. In the bathroom, leaning over the sink, unable to catch my breath, choking, I threw up. When I returned to the class, my face was streaked with mascara and Lisa Robertson asked me if I was okay. I said yes. Then, outside, on the sidewalk, I met Dodie Bellamy and felt so weak I could not greet her like a normal person. My arms were limp. All I could think was: “I just barfed and now I am meeting the author of the BARF MANIFESTO.”
A?
“You never belonged even to yourself though/as you abandoned me your cry was I’m for time,” wrote Ali.
He wrote it down.
Then went.
Posted in Uncategorized on Wednesday, April 20th, 2011 by Bhanu Kapil.
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\begin{document}
\title{Transcendental equations of the running coupling}
\author{Juuso Österman\thanks{juuso.s.osterman@helsinki.fi}\\
University of Helsinki, Helsinki Institute of Physics}
\maketitle
\begin{abstract}
The running coupling of a generic field theory can be described through a separable differential equation involving the corresponding $\beta$-function. Only the first loop order can be solved analytically in terms of well-known functions, all further loop orders lead to transcendental equations. While obscure nowadays, many analytical methods have been devised to study them, most specifically the Lagrange-Bürmann formula. In this article we discuss the structure of transcendental equations that take place at various loop orders. Beyond the first two loop orders, these equations are simplified by applying an optimal Pade approximant on the $\beta$-function. In general, these lead to generalizations of Lambert's equation, the solutions of which are presented in terms of a power series.
\end{abstract}
\section{Introduction}
A renormalization group (RG) equation describes the scale dependence of a quantum field theory through the scale dependence of its parameters, e.g. mass or interaction strength (the running coupling). The relevant differential equations are constructed perturbatively in weak coupling expansion, and can in turn be integrated into transcendental equations \cite{deur,running06}. Each ascending loop order leads to more convoluted equations, with only the one-loop solution given in terms of common (yet non-algebraic) functions.
The transcendental equations can be approached iteratively, as that leads to a unique solution with explicit scale dependence \cite{running06}. However, the iterative method only approximates the correct form of the solution \cite{nesterenko}, which motivates attempting an exact analytical evaluation process. This has been achieved up to three loops in \cite{gardi1}, by using the known solution to Lambert's equation in combination with the Padé approximation. Two further loop orders have been calculated in \cite{Kondrashuk}, where the differential equations are written in terms of an ansatz involving the Lambert W function.
In order to consider a general loop order, we introduce a recipe utilizing (nearly) diagonal Padé approximants, which simplify the transcendental equations significantly while retaining the original accuracy. Additionally, we solve the generic transcendental equation arising from an arbitrary loop order, using the Lagrange inversion theorem. In particular, we explicitly demonstrate the procedure on the four-loop running coupling.
\section{Background}
Coupling constants describe the strength of particular interactions in quantum field theories. The perturbative expansions in QCD and QED, in particular, are described by a single constant (respectively). The generic coupling, $x$, is defined by the renormalization process such that
\begin{equation}
x (\mu^2) Z_x(\mu^2) = x^{ur}
\end{equation}
where the right-hand side refers to the unrenormalized coupling constant, and $Z_x$ to the renormalization counterterm \cite{schwarz, running06, deur}. Through differentiation (with respect to the scale $\mu$) we find the renormalization group equation of the coupling, $x$, to be given by
\begin{equation}
\mu^2 \frac{\partial x}{\partial \mu^2} = \beta (x),
\end{equation}
where the right-hand side is called the $\beta$-function of the coupling. It can be written in terms of a perturbative power series expansion of the coupling. This explicitly reads
\begin{equation}
\label{eq:3}
\begin{split}
\frac{\partial x}{\partial t} &= - \beta_0 x^2\sum_{n \geq 0} c_n x^n,
\end{split}
\end{equation}
where we use the following shorthands: $t \equiv \text{ln} \mu^2$ and $c_0 = 1$. The coefficients have been evaluated up to five loops in Yang-Mills theories \cite{luthe, herzog} , QED \cite{qedbeta, Kataev}, and further in scalar theories \cite{scalarbeta}. However, the differential equations have been solved beyond these loop orders only iteratively.
\subsection{One-loop order and iterative solutions }
As the differential equation given in equation (\ref{eq:3}) is separable, we can easily re-write it in terms of of the following integral:
\begin{equation}
\label{eq:onel}
\begin{split}
\text{ln}\frac{\mu^2}{\mu_0^2} = \int_{x_0}^{x} \frac{d \xi}{\beta( \xi)}.
\end{split}
\end{equation}
We start with the one-loop renormalization group equation (in the weak coupling expansion). Aside from the obvious purpose, this also enables us to describe the standard iterative method to evaluate further loop orders. Thus, the leading order is given by
\begin{equation}
\begin{split}
\beta_0\text{ln}\frac{\mu^2}{\mu_0^2} &= -\int_{x_0}^{x} \frac{d \xi}{\xi^2}. \\
&= \frac{1}{x} -\frac{1}{x_0},\\
\end{split}
\end{equation}
Using trivial algebra, we can further simplify the above to
\begin{equation}
x = \frac{1}{\beta_0 \text{ln} \frac{\mu^2}{\Lambda^2}},
\end{equation}
where we have hidden all $\mu_0$-dependence such that
\begin{equation}
\Lambda^{2 \beta_0} = \frac{\mu_0^{2\beta_0}}{e^{x_0}}.
\end{equation}
To evaluate weak coupling at further loop levels and simultaneously avoid dealing with the non-trivial transcendental equations, one viable approach is to expand the inverse of $\beta$-function \cite{shirkov, running06, nesterenko} such that
\begin{equation}
\begin{split}
\beta_0 \text{ln}\frac{\mu^2}{\mu_0^2} &= -\int \frac{dx}{x^2(1+c_1 x+ c_2 x^2 + c_3 x^3+... )}\\
&= - \int dx \left[\frac{1}{x^2} - c_1 \frac{1}{x} -(c_2-c_1^2) - (c_3-2 c_2 c_1+c_1^3)x \right] + \mathcal{O}(x^3)\\
&= \frac{1}{x}+ c_1 \text{ln} x + (c_2-c_1^2)x + \frac{c_3-2c_1 c_2 + c_1^3}{2}x^2 + C(\mu_0)+\mathcal{O}(x^3).
\end{split}
\end{equation}
By denoting $u = \beta_0 \text{ln} \frac{\mu^2}{\Lambda^2}$ and only considering the leading order $\mathcal{O}\left(\frac{1}{x} \right)$, we obviously find the familiar one-loop solution, $x_1 = \frac{1}{u}$. However, instead of considering further cases explicitly, we write the two-loop equation such that all terms but the inverse coupling are replaced with the one loop solution. This yields
\begin{equation}
\begin{split}
\frac{1}{x_2} &= u -c_1 \text{ln} x_1\\
&= u+c_1 \text{ln}u.
\end{split}
\end{equation}
This process can be extended to the three-loop equation, by inserting the two-loop solution in place of the coupling such that
\begin{equation}
\begin{split}
\frac{1}{x_3} &= u -c_1 \text{ln} x_2 -(c_2-c_1^2)x_2\\
&= u + c_1 \text{ln}(u-c_1\text{ln}u)+\frac{c_1^2-c_2}{u-c_1\text{ln} u}\\
&\simeq u + c_1 \text{ln} u + c_1^2 \frac{\text{ln} u}{u} + \frac{c_1^2-c_2}{u} + \left(\frac{c_1^3}{2}-c_2 c_1^2 \right) \frac{\text{ln}^2 u}{u^2}+ \mathcal{O} \left( \frac{\text{ln}^3 u}{u^3}\right),
\end{split}
\end{equation}
where we performed a simple Taylor expansion on the expression at the limit $u \rightarrow \infty$. The expression at four loops can in turn be written in terms of the three-loop solution such that
\begin{equation}
\begin{split}
\frac{1}{x_4} = u - c_1 \text{ln} x_3 + (c_1^2-c_2)x_3- \frac{c_3-2c_1 c_2 + c_1^3}{2}x_3^2,
\end{split}
\end{equation}
and the process can be continued to an arbitrary loop level. This offers a computationally efficient method with which to consider each loop order. However, it is not exact. It is our motivation to re-write the transcendental equation of each loop order such that we can analytically pursue a power series solution. To this end we introduce two methods, the Padé approximation and the Lagrange inversion theorem.
\subsection{Padé approximation}
The approximation method is named after Henri Padé, but the idea dates back to the work of Georg Frobenius \cite{bakerpade}. In particular, the technique expresses a polynomial of order (M+N) in terms of a rational function, a Padé approximant. The corresponding approximant is denoted [N/M], using the order of the polynomials in numerator and denominator respectively:
\begin{equation}
\frac{1+\sum_{n=1}^N a_n x^n}{1+ \sum_{k= 1}^M b_k x^k} = 1+ \sum_{j=1}^{N+M} c_j x^j + \mathcal{O}(x^{N+M+1}).
\end{equation}
This approximant is unique owing to linear independence of the set $\{x^k\}$. The coefficients are determined by the following $N+M$ linear equations :
\begin{equation}
\left(1+ \sum_{j=1}^{N+M} c_j x^j \right)\left(1+ \sum_{k= 1}^M b_k x^k \right) = 1+\sum_{n=1}^N a_n x^n + \mathcal{O}(x^{N+M+1}).
\end{equation}
Order by order, these are given by
\begin{eqnarray}
c_1+ b_1 &=& a_1,\\
c_2+b_2+c_1 b_1 &=& a_2,\\
\vdots \nonumber\\
\sum_{n = 0}^{\text{min} \{M, N\}} c_{N-n} b_n &=& a_N,\\
\sum_{n=0}^{\text{min}\{N+1, M\}}c_{N+1-n}b_n &=& 0,\\
\vdots \nonumber \\
\sum_{n=0}^M c_{M+N-n} b_n &=& 0.
\end{eqnarray}
Padé approximants can be applied to simplify the integrands corresponding to each loop order. This clearly yields us more manageable transcendental equations, while the price is new higher order terms generated by the approximant. However, each loop order already obeys a similar kind of approximation, which renders the simplification viable. The use of diagonal $[N/N]$ Padé approximants clearly simplifies the integrand most, and is additionally motivated by the reduced renormalization scale dependence. Diagonal approximants have even been shown to be scale free in the limit of large $\beta_0$ \cite{gardi2}. Also, Padé approximants have been shown to yield increasingly accurate preductions to higher order coefficients of the beta function in QCD \cite{ellis}.
This type of approximation becomes relevant at three loops and beyond, which has been known since the late 1990s. However, in terms of direct computation, it has only been extended to consider the lowest relevant order, which reduces to a widely known classic transcendental equation, Lambert's equation \cite{gardi1}. We aim to apply the Padé approximation to find the the generic structure of the transcendental equations that appear beyond three loops, in a sense a generalization of Lambert's equation. The actual evaluation takes place through the Lagrange-Bürmann formula, which enables us to write all sought-after solutions in terms of a power series.
\subsection{Lagrange Inversion theorem}
A holomorphic function can be reversed using the theorem by Joseph Lagrange (and its generalization due to Hans Bürmann), also known as the Lagrange-Bürmann theorem or the reversion of series \cite{annals8, stegun}. Let $f$ be analytic at $a$ such that $f'(a) \neq 0$, and
\begin{equation}
z = f(r).
\end{equation}
The theorem states that this equation can be solved in terms of a power series given by
\begin{equation}
r(z) = a + \sum_{n=1}^\infty r_n \frac{[z-f(a)]^n}{n!},
\end{equation}
where
\begin{equation}
\label{eq:Lagrange}
r_n = \left.\frac{\partial^{n-1}}{\partial r^{n-1}} \left(\frac{r-a}{f(r)-f(a)} \right)^n\right\vert_{r = a}.
\end{equation}
The theorem can be proved using only the standard results of complex analysis \cite{whitaker}. Let us outline the steps for the expansion around origin. To any analytic $h(x)$, we can use the residual theorem to write
\begin{equation}
\frac{1}{2 \pi i} \oint_\Psi h(x) \frac{f'(x)}{f(x)-z}dx = \sum_{k=1}^K h(x_k)
\end{equation}
where $\{x_k\}$ is the set of solutions to the equation $f(x) = z$, which are counted with multiplicity (contained in the closed integration path $\Psi$). Supposing that $f(0) = 0 \neq f'(0)$, we can consider $f(x)$ to be a bijection within a small circle around origin, $S$, which enables us to find a unique inverse function, $r(z) = x $ (which is also analytic). Thus, we can write
\begin{equation}
\begin{split}
r(z)&= \frac{1}{2 \pi i} \oint_S x \frac{f'(x)}{f(x)-z}dx \\
&= \frac{1}{2 \pi i} \sum_{n = 0}^\infty \oint_S x \frac{f'(x)}{f(x)} \left(\frac{z}{f(x)}\right)^n dx\\
&\equiv \sum_{n \geq 1} r_n z^n.
\end{split}
\end{equation}
It is trivial to extract the corresponding coefficient, which simplifies to
\begin{equation}
\begin{split}
r_n &= \frac{1}{2 \pi i}\oint_S x \frac{f'(x)}{f(x)^{n+1}} dx\\
&= \frac{1}{n} \frac{1}{2 \pi i} \oint_S \frac{1}{f(x)^{n}} dx\\
&= \left.\frac{\partial^{n-1}}{\partial x^{n-1}} \left(\frac{x}{f(x)} \right)\right\vert_{x = 0},
\end{split}
\end{equation}
where we applied the residual theorem. The explicit form given in equation (\ref{eq:Lagrange}) is found by re-defining the target function around $a$ instead. Specifically this means that we write for another holomorphic function $g(x)$ the following shift $g(x) = f(x)-f(a)$. This corresponds to the same type of equation $z = g(r)$. By setting $f'(a) \neq 0$, we can again find $g(x)$ to be bijection, but unlike earlier, in a small circle around $a$. Following exactly the same steps as above, apart from defining the contour around $a$, we find the formula of interest.
\section{Exact solution through the Lambert W function}
The Lambert function dates back to 1758 when Johann Lambert solved the trinomial equation
\begin{equation}
x = q+x^m,
\end{equation}
in the sense of finding a power series corresponding to it \cite{corless, lambertorig}. This equation was in 1783 transformed to
\begin{equation}x^{\alpha-\beta}-1 = v(\alpha-\beta) x^{\alpha},
\end{equation}
which in the limit $\beta \rightarrow \alpha =1$ becomes
\begin{equation}
\text{ln} x = vx.
\end{equation}
Leonhard Euler calculated the power series expansion corresponding to this equation \cite{corless, eulerorig}. The solution of the symmetrized equation relates clearly to Lambert's equation,
\begin{equation}
\label{eq:lambert1}
W(z)e^{W(z)} = z,
\end{equation}
where the $W(z)$ notation is due to George Polya and Gabor Szegö \cite{szego}. By algebraic manipulation the equation by Euler becomes
\begin{equation}
-v x e^{-v x} = -v,
\end{equation}
which in turn can be re-expressed in terms of (\ref{eq:lambert1}) such that
\begin{equation}
x = -\frac{W(-v)}{v}.
\end{equation}
The actual power series representation to the Lambert W function can be derived by using the Lagrange inversion theorem. By applying equation (\ref{eq:Lagrange}) on equation (\ref{eq:lambert1}) we find
\begin{equation}
\begin{split}
w_n &= \left.\frac{\partial^{n-1}}{\partial w^{n-1}} \left(\frac{w}{we^{w}} \right)^n\right\vert_{w = 0}\\
&= (-n)^{n-1}.
\end{split}
\end{equation}
Thus, the power series expansion of the Lambert W function around origin is given by
\begin{equation}
W(z) = -\sum_{n=1}^\infty \frac{(-nz)^{n}}{n!n}.
\end{equation}
The generic renormalization group equation can be solved using the Lambert W function up to three loops. In the following subsections we explicitly demonstrate how the structure arises naturally from the transcendental equations of the inverse coupling \cite{gardi1}.
\subsection{Two loops}
Let us return to the renormalization group equation of the running coupling, eq. (\ref{eq:onel}). However, this time we consider an additional term in the polynomial of the denominator. This new integral equation can be solved explicitly somewhat easily, only requiring one additional change of variables such that
\begin{equation}
\begin{split}
\beta_0\text{ln}\frac{\mu^2}{\mu_0^2} &= -\int_{x_0}^{x} \frac{d \xi}{\xi^2(1+c_1 \xi)} \\
&= \int_{x_0^{-1}}^{x^{-1}} \frac{y dy}{y + c_1}\\
&= \frac{1}{x} - c_1 \text{ln} \left(c_1+ \frac{1}{x} \right) + C(\mu_0),
\end{split}
\end{equation}
where we have combined all the terms arising from the lower limit of integration inside the term $C (\mu_0)$. Following the convention we used for the one-loop approximation, we choose to contain all $\mu_0$ dependence inside the logarithm on the right-hand side of the equation. Next, we write the transcendental equation in terms of the scaled inverse coupling by denoting $y = -\frac{1}{ c_1 x}$. This leads to
\begin{equation}
\begin{split}
\text{ln} \Omega &\equiv -\frac{\beta_0}{c_1}\text{ln} \frac{\mu^2}{\Lambda^2}\\
&= -(1-y) +\text{ln}(1-y),
\end{split}
\end{equation}
where we denote
\begin{equation}
\frac{\beta_0}{c_1} \text{ln} \Lambda^2 = \frac{\beta_0}{c_1} \text{ln} \Lambda^2 + \frac{C(\mu_0) }{c_1} - \text{ln} c_1-1.
\end{equation}
This leads to the following transcendental equation
\begin{equation}
\begin{split}
-\Omega &= (1-y)\text{exp}[-(1-y)]\\
&\equiv - W\left(-\Omega\right)e^{W\left(-\Omega\right)},
\end{split}
\end{equation}
where we recognize the Lambert W function. Using this expression we can re-write the inverse two-loop coupling as
\begin{equation}
\frac{1}{x} = - c_1 \left[1+ W\left(-\Omega\right)\right].
\end{equation}
The relevant branch of Lambert W function and its behaviour has been discussed at length in literature \cite{gardi1, corless}.
\subsection{Three loops and Padé approximation}
The three-loop $\beta$ function can be described in terms of a fourth order polynomial such that
\begin{equation}
\beta(x) = \beta_0 x^2(1+c_1 x+ c_2 x^2) + \mathcal{O}(x^5).
\end{equation}
We can achieve the same level of accuracy by writing a $[1/1]$ Padé approximant of the polynomial, following the steps given in section 2. Isolating the relevant polynomial we can write
\begin{equation}
\begin{split}
1+ c_1 x + c_2 x^2 = \frac{1+a_1 x}{1+a_2 x} + \mathcal{O}(x^3),
\end{split}
\end{equation}
where the new coefficients correspond to the following set of linear equations
\begin{eqnarray}
c_1 + a_2 &=& a_1,\\
c_2 + c_1 a_2 &=& 0,
\end{eqnarray}
which in turn yield
\begin{eqnarray}
a_1 &=& c_1-\frac{c_2}{c_1},\\
a_2 &=& -\frac{c_2}{c_1}.
\end{eqnarray}
Through this procedure we have reduced the integrand to the same general structure as the one we found at the two-loop order. In particular we avoid a more convoluted variant of the transcendental equation with a higher order polynomials. The relevant structures are discussed at more length in section 4.
By inserting the Padé approximant to the integral representation of the three-loop equation, we find
\begin{equation}
\begin{split}
\beta_0\text{ln}\frac{\mu^2}{\mu_0^2} &= -\int_{x_0}^{x} \frac{(1+a_2 \xi) d \xi }{\xi^2(1+a_1 \xi)} \\
&= \int_{x_0^{-1}}^{x^{-1}} \frac{(y+a_2) dy}{y + a_1}\\
&= \frac{1}{x} +(a_2-a_1) \text{ln} \left(a_1+ \frac{1}{x} \right) + C(\mu_0).
\end{split}
\end{equation}
We continue by moving the integration constant inside the logarithm and writing the transcendental equation in terms of the scaled inverse coupling $y = -\frac{1}{(a_2-a_1)x}$. Here we have assumed that the coefficients have non-trivial values, as $a_1 = a_2$ would lead to the one-loop solution. Thus, the transcendental equation is given by
\begin{equation}
\begin{split}
\text{ln} \Omega &\equiv \frac{\beta_0}{a_2-a_1}\text{ln}\frac{\mu^2}{\Lambda^2}\\
&= \left(\frac{a_1}{a_2-a_1} - y \right) + \text{ln} \left( \frac{a_1}{a_2-a_1}-y\right).\\
\end{split}
\end{equation}
Similar to the two-loop transcendental equation, we can easily discern the logarithmic version of Lambert's equation. Thus, we are able to write the inverse coupling in terms of the Lambert W function such that
\begin{equation}
\begin{split}
\frac{1}{x} &= - \{a_1 - (a_2-a_1)W(\Omega)\}\\
&= -\left\{c_1-\frac{c_2}{c_1}+c_1 W(\Omega) \right\}.
\end{split}
\end{equation}
\section{General structure of transcendental equations}
Already at the two-loop order we witnessed the appearance of a highly non-trivial transcendental equation. At the three-loop level, we applied the Padé approximation to retain the same level of complexity. However, upon considering further loop orders, or even the three-loop order without algebraic manipulation, it is obvious that we will encounter even more involved transcendental equations, which do not reduce to Lambert's equation. We choose to demand a small amount of regularity from the rational function integrand, in particular binding the coefficients of the $\beta$-function in a specific manner, if need be considering a different number of active fermions. In order to demonstrate the additional condition we write the (N+1)-loop integral equation in terms of the inverse coupling $y$ such that
\begin{equation}
\label{eq:pureb}
\begin{split}
\text{ln} \Omega &= \int dx \frac{1}{\beta_N(x)}\\
&= \int dy \frac{y^N}{y^N + \sum_{k=1}^{N} b_k y^{N-k}}\\
&= \int dy \frac{y^N}{\prod_{k=1}^N (y+a_k)}.
\end{split}
\end{equation}
Here we write the denominator as a product of $n$ monomials, which in turn implies that no two roots are allowed to be equal. This condition is applied to all integrals in this section, to pedagogically motivate the use of diagonal or nearly diagonal Padé approximants. However, as far as results are concerned, the coefficients of the $\beta$-function are limited only such that these Padé approximants contain a denominator that decomposes in the manner described above.
The integral given in equation (\ref{eq:pureb}) is rather straightforward to solve; the relevant steps of the derivation are given in the appendix A. Next we absorb the lower limit on the right-hand side to $\text{ln}\tilde{\Omega} = \text{ln} \Omega - C$. Then we write the logarithmic representation of the corresponding transcendental equation such that
\begin{equation}
\text{ln} \tilde{\Omega} = y +\sum_{n=1}^N \tilde{c}_n \text{ln}(y+a_n),
\end{equation}
the preferable form of which is given by
\begin{equation}
\label{eq:proood}
\tilde{\Omega}e^{-y} = \prod_{n=1}^N (y+a_n)^{\tilde{c}_n} .
\end{equation}
This result corresponds to the [N/0] Padé approximant of the denominator, i.e. the original polynomial. The structural opposite of this being the [0/N] Padé approximant, which leads to the following integral equation
\begin{equation}
\begin{split}
\text{ln} \Omega &= -\int \frac{dx}{x^2} \sum_{k=0}^N c_k x^k\\
&= \int dy \sum_{k=0}^N \frac{c_k}{y^k}.
\end{split}
\end{equation}
Following similar steps to those given above, we find the corresponding transcendental equation to be
\begin{equation}
\tilde{\Omega}\text{exp}\left( -y + \sum_{k=1}^{N-1} \frac{c_{k+1}}{k y^k} \right) = y^{c_1},
\end{equation}
which looks very uninviting to solve for any $N \geq 2$. The former is obviously the preferable form to consider analytically, as the inversion theorem operates through derivatives. However, we are able to reduce the transcendental equation further. This is achieved through diagonal, or nearly diagonal Padé approximants. Supposing that $N = 2n$, we can obviously write instead the $[n/n]$ Padé approximant to achieve
\begin{equation}
\text{ln} \Omega = \int dy \frac{y^n + \sum_{k=0}^{n-1} c_k y^k}{y^n + \sum_{j=0}^{n-1} b_j y^j}.
\end{equation}
Using again the appendix A, we can immediately write the result in the following form
\begin{equation}
\label{eq:easy1}
\tilde{\Omega}e^{-y} = \prod_{n=1}^n (y+a_n)^{\tilde{c}_n},
\end{equation}
which halves the generalized polynomial structure on the right-hand side of equation (\ref{eq:proood}).
In case the loop order is even (the polynomial is of odd order), we can instead write $m = \lfloor N/2 \rfloor$ and instead consider the $[m/m+1]$ Padé approximant to find
\begin{equation}
\begin{split}
\text{ln} \Omega = \int dy \left(\frac{y^m + \sum_{k=0}^{m-1} c_k y^k}{y^m + \sum_{j=0}^{m-1} b_j y^j}+\frac{c_m}{y\left(y^m + \sum_{j=0}^{m-1} b_j y^j \right)} \right),
\end{split}
\end{equation}
which in turn yields a variant of the diagonal result of the order $m+1$ such that
\begin{equation}
\label{eq:easy2}
\tilde{\Omega}e^{-y} = y^{\hat{c}_0}\prod_{k=1}^m (y+a_k)^{\tilde{c}_k}.
\end{equation}
The results given in equations (\ref{eq:easy1}) and (\ref{eq:easy2}) generalize Lambert's equation, and can be solved through the Lagrange inversion theorem (with added combinatorial difficulty).
\section{Four loops and Padé approximation}
After having established the types of transcendental equations we are interested in, it is logical to apply the tools we have to the higher order cases, which have so far been left ambiguous. In particular, we start with the four-loop approximation of the $\beta$-function. As indicated in the earlier section, we aim to avoid higher order contributions by initiating our approach with the $[1/2]$ Padé approximant of the following polynomial
\begin{equation}
1 + c_1 x + c_2 x^2 + c_3 x^3 = \frac{1+ a_1 x}{1+ a_2 x + a_3 x^2} + \mathcal{O}(x^4),
\end{equation}
where we find the following linear relations for the parameters
\begin{eqnarray}
a_2+c_1 &=& a_1,\\
c_1 a_2 + c_2 + a_3 &=& 0,\\
c_1 a_3 + c_3 &=& 0,
\end{eqnarray}
which define the Padé approximant uniquely such that
\begin{eqnarray}
a_1 &=& c_1- \frac{c_2}{c_1}+\frac{c_3}{c_1^2},\\
a_2 &=& -\frac{c_2}{c_1}+\frac{c_3}{c_1^2},\\
a_3 &=& -\frac{c_3}{c_1}.
\end{eqnarray}
These can be used to re-write the four-loop approximation of the renormalization group equation as
\begin{equation}
\begin{split}
\beta_0\text{ln}\frac{\mu^2}{\mu_0^2} &= -\int_{x_0}^{x} \frac{(1+a_2 \xi + a_3 \xi^2) d \xi }{\xi^2(1+a_1 \xi)} \\
&= \int_{x_0^{-1}}^{x^{-1}}dy \left(1 +\frac{a_2-a_1 - \frac{a_3}{a_1}}{y+ a_1} + \frac{a_3}{a_1 y} \right)\\
&= y +\left(a_2-a_1-\frac{a_3}{a_1} \right) \text{ln} \left(a_1+ y \right) + \frac{a_3}{a_1}\text{ln} y + D(\mu_0),
\end{split}
\end{equation}
where we denote $y \equiv \frac{1}{x}$. Let us simplify the notation further by writing
\begin{eqnarray}
A &=& a_2 - a_1 -\frac{a_3}{a_1},\\
B &=& \frac{a_3}{A a_1},\\
C &=& \frac{a_1}{A},\\
z &=& \frac{y}{A}.
\end{eqnarray}
Thus, we are able to write the logarithmic representation of the transcendental equation as
\begin{equation}
\begin{split}
\label{eq:abbreviated}
\text{ln} \Omega &\equiv \frac{\beta_0}{A}\text{ln}\frac{\mu^2}{\Lambda^2}\\
&= z + \text{ln} \left(C + z\right)+ B \text{ln} z.
\end{split}
\end{equation}
This result indeed resembles the type of transcendental equation taking place upon applying an almost diagonal Padé approximation, given in equation (\ref{eq:easy2}):
\begin{equation}
\label{eq:abbreviated2}
\begin{split}
(C+z)z^B = \Omega e^{-z}.
\end{split}
\end{equation}
\subsection{Inversion at four loops}
The equation of interest can be rephrased conveniently such that
\begin{equation}
(v-a)v^b = c e^{-v}.
\end{equation}
We approach this new problem by denoting $f(v) = \frac{e^{-v}}{v^b}$, using which the equation transforms to
\begin{equation}
\label{eq:standard}
v = a +c f(v).
\end{equation}
By re-organizing these terms, we re-write the equation in terms of a constant valued function
\begin{equation}
F(v) = v- c f(v) \equiv a.
\end{equation}
We choose to expand around the bound value $a$, and obviously $F'(a) = -c f'(a) \neq 0$, which explicitly justifies the use of the Lagrange inversion theorem. In order to simplify the power series solution, we introduce three helpful algebraic relations:
\begin{eqnarray}
F(a)-a &=& -c f(a),\\
v-a &=& c f(v),\\
F(v)-F(a) &=& F(v)-a + cf(a)= c f(a).
\end{eqnarray}
The sought after power series solution via the inversion formula is given by
\begin{equation}
v(a) = a + \sum_{k \geq 1} \frac{v_k}{k!} [a-F(a)]^k,
\end{equation}
where we again define the power series coefficient as
\begin{equation}
\begin{split}
v_k &= \left.\partial_v^{k-1} \left( \frac{v-a}{F(v)-F(a)}\right)^k \right|_{v= a}\\
&= \left. \left[\frac{1}{ f(a)}\right]^k \partial_v^{k-1} f(v)^k \right|_{v= a}.
\end{split}
\end{equation}
Thus, we find the following simpler form of the sought after solution to the transcendental equation
\begin{equation}
\label{eq:simple}
v = a + \sum_{k \geq 1} \left. \frac{c^k}{k!} \partial_v^{k-1} f(v)^k \right|_{v= a},
\end{equation}
which is in agreement with the result presented in \cite{Mugnaini}.
Let us next consider the derivative more explicitly by writing
\begin{equation}
\begin{split}
\partial_v^{k-1} f(v)^k &= \partial_v^{k-1} \frac{e^{-kv}}{v^{kb}}\\
&= \sum_{j = 0}^{k-1} {k-1 \choose j} \left[ \frac{\partial^j}{\partial v^j} e^{-k v} \right] \left[ \frac{\partial^{k-1-j}}{\partial v^{k-1-j}} v^{-kb}\right]\\
&= (-1)^{k-1} \sum_{j = 0}^{k-1} {k-1 \choose j} \frac{k^j}{kb-1} \prod_{l=0}^{k-1-j} (kb-1+l) \frac{e^{-kv}}{v^{kb+k-1-j}}.
\end{split}
\end{equation}
Thus, we find the following solution to our transcendental equation
\begin{equation}
\begin{split}
v = a - \sum_{k \geq 1}\sum_{j = 0}^{k-1} \frac{(-c)^k k^j}{(k-1-j)! j! k (kb-1) } \prod_{l=0}^{n-1-j} (kb-1+l) \frac{e^{-ka}}{a^{kb+k-1-j}}.
\end{split}
\end{equation}
By introducing the Pochhammer symbol such that
\begin{equation}
(v)_k = \frac{\Gamma(v+k)}{\Gamma (k)},
\end{equation}
we can simplify the solution given above to the more aesthetic
\begin{equation}
\begin{split}
v = a - \sum_{k \geq 1}\sum_{j = 0}^{k-1} \frac{(-c)^k k^j}{(k-1-j)! j! k } (kb)_{k-1-j} \frac{e^{-ka}}{a^{kb+k-1-j}}.
\end{split}
\end{equation}
\subsection{Full expression at four loops}
Let us return to the relation given in equation (\ref{eq:abbreviated2}). Applying this to the previous identity, we find
\begin{equation}
\begin{split}
z = -C - \sum_{k \geq 1}\sum_{j = 0}^{k-1} \frac{(-\Omega)^k k^j}{(k-1-j)! j! k } (kB)_{k-1-j} \frac{e^{kC}}{(-C)^{kB+k-1-j}},
\end{split}
\end{equation}
or in terms of the inverse coupling
\begin{equation}
\label{eq:fullfour}
y = - AC -A\sum_{k \geq 1}\sum_{j = 0}^{k-1} \frac{\left[-\left(\frac{\mu^2}{\Lambda^2} \right)^\frac{\beta_0}{A}\frac{e^{C}}{(-C)^{B+1}}\right]^k }{(k-1-j)! j! k^2 } \left(kB \right)_{k-1-j} (-kC)^{j+1}.
\end{equation}
It is noteworthy that (\ref{eq:abbreviated}) is not a unique representation, and therefore, we could have instead written
\begin{eqnarray}
A &=& \frac{a_3}{a_1},\\
B &=& \frac{ a_2 - a_1 -\frac{a_3}{a_1}}{A},\\
C &=& \frac{a_1}{A},\\
\tilde{z} &=& \frac{y}{A},
\end{eqnarray}
to introduce instead
\begin{equation}
\frac{\beta_0}{A}\text{ln}\frac{\mu^2}{\Lambda^2} = \tilde{z} + B \text{ln} \left(C + \tilde{z}\right)+ \text{ln} \tilde{z}.
\end{equation}
This, in turn, would lead us to find a solution around $z=0$ and to consider the function $f(x) = \frac{e^{-x}}{(x+a)^b}$. The resulting power series solution is given by
\begin{equation}
y = -A\sum_{k \geq 1}\sum_{j = 0}^{k-1} \frac{\left[-\left(\frac{\mu^2}{\Lambda^2} \right)^\frac{\beta_0}{A}\frac{1}{(-C)^{B+1}}\right]^k }{(k-1-j)! j! k^2 } \left(kB \right)_{k-1-j} (-kC)^{j+1}.
\end{equation}
Also, we note that the four-loop approximation did not demand any new imposed conditions on the denominator structure of the integrand. Rather, the only notable difference to the lower loop order cases was the well defined additional logarithmic term. However, the arising power series solution is somewhat more involved. Also, the former result, given in equation (\ref{eq:fullfour}), is more attractive as the proper parameter values would make it exponentially suppressed (in particular $C < 0$).
\section{Discussion on the higher loop orders}
As elaborated in section 4, the preference for diagonal Padé approximants leads us to consider the higher order integrands in terms of rational functions, the denominators of which decompose to a product of n separate monomials. Specifically, in order to optimize our efforts, we limit to Padé approximants of types $[n-1/n]$ or $[n/n]$, which lower the degree of the sought after polynomial and thus greatly simplify the transcendental equation of interest. In this manner we can find the five-loop coupling structure given by the $[2/2]$ Padé approximant such that
\begin{equation}
\begin{split}
\text{ln} \Omega &= \int dy \frac{1+\frac{a_1}{y}+ \frac{a_2}{y^2}}{1+ \frac{a_3}{y}+\frac{a_4}{y^2}}\\
&= \int dy \left[1 + \frac{(a_1-a_3)y}{y^2+a_3 y+ a_4}+ \frac{a_2-a_4}{y^2+a_3 y + a_4 } \right].
\end{split}
\end{equation}
By recognizing the decomposition of the denominator in terms of the roots of any second order polynomial, we find $y^2+a_3 y + a_4 = (y+a_{+})(y+a_{-})$ and $a_{\pm} = \frac{a_3\pm \sqrt{a_3^2-4a_4}}{2}$. This leads to the following logarithmic representation of the transcendental equation
\begin{equation}
\text{ln} \tilde{\Omega} = y + \frac{a_1-a_3}{a_+ - a_-} \left[\text{ln}\frac{(y+a_+)^{a_+}}{(y+a_-)^{a_-}}\right] +\frac{a_2-a_4}{a_+ - a_-} \text{ln} \left( \frac{y+a_-}{y+a_+} \right)
\end{equation}
These lead to the following type of transcendental equation under sufficient conditions
\begin{equation}
\Xi e^{-y} = (y+a_+)^A (y+a_-)^B,
\end{equation}
or given in a similar manner as in the previous section
\begin{equation}
v = a + b \frac{e^{-v}}{(v+c)^d},
\end{equation}
which leads us to the realization that the five-loop transcendental equation generalizes the four-loop case, something that we recognize taking place also with three and two loops. This corresponds to the pattern we described in section 4, with each consecutive even and odd loop orders, the approximants $[n-1/n]$ and $[n/n]$, being effectively the same (apart from parameter values). Thus, let us establish the duality between the transcendental equations and the n-loop order approximations of the coupling. This allows us to clearly state that all structural information is demonstrated by the odd loop orders, i.e. the diagonal approximants.
Let us consider an arbitrary $(N+1)$-loop (odd) order transcendental equation and simplify it, denoting $n = \frac{N}{2}$, to the familiar structure such that
\begin{equation}
v = a + b \frac{e^{-v}}{\Pi_{k= 1}^{n-1} (v+a_k)^{c_k}}.
\end{equation}
This type of equation can be solved following the exact same steps as before. However, this time we encounter a multinomial coefficient upon derivating instead of the binomial. Explicitly we can formulate the transcendental equation of interest such that the function $f$ from the previous section becomes
\begin{equation}
f(v) = \frac{e^{-v}}{\Pi_{k= 1}^{n-1} (v+a_k)^{c_k}},
\end{equation}
and therefore we find the following derivative in the power series coefficient
\begin{equation}
\begin{split}
\partial_v^{k-1} f(v) &= \sum_{j_1+j_2+...+j_n =k-1} {k-1 \choose j_1, j_2,....j_n} \left[ \frac{\partial^{j_n}}{\partial v^{j_n}} e^{-kv}\right]\prod_{l=1}^{n-1} \left[ \frac{\partial^{j_l}}{\partial v^{j_l}} (v+a_l)^{-kc_l} \right] \\
&= (-1)^{k-1} \sum_{j_1+j_2+...+j_n =k-1} {k-1 \choose j_1, j_2,....j_n} \left[ k^{j_n} e^{-kv}\right]\prod_{l=1}^{n-1} \left(kc_l \right)_{j_l} (v+a_l)^{-kc_l-j_l}.
\end{split}
\end{equation}
Thus, applying above results to equation (\ref{eq:simple}), we find the generic solution type to the arbitrary (diagonal) loop order inverse coupling
\begin{equation}
\label{eq:final}
v = a - \sum_{k=1}^\infty \sum_{j_1+j_2+...+j_n =k-1} \frac{(-1)^k}{k!} {k-1 \choose j_1, j_2,....j_n} \left[ k^{j_n} e^{-ka}\right]\prod_{l=1}^{n-1} \left(kc_l \right)_{j_l} (a+a_l)^{-kc_l-j_l}.
\end{equation}
The resulting series converges very fast with sufficiently large values of the parameters, and thus offers a viable alternative to the iterative approach. Of course, it is noteworthy that some of the uniqueness is lost by approaching the problem through transcendental equations in earnest, as the diagonal structure we described can be chosen to be expanded around multiple separate root values. However, this ambiguity is problematic in appearance only, as a suitable physical value, $a \gg 1$, can be chosen (if any such exists) from the set of roots, leading to correct physical behaviour at the weak coupling limit.
\section{Conclusions}
In this work, we have evaluated the inverse four-loop running coupling analytically for a generic quantum field theory. The result is given in terms of a power series, which solves a generalized variant of Lambert's equation. The components of the power series solution, given in equation (\ref{eq:fullfour}), are characterized by an exponential dependence on the parameters of the transcendental equation. This implies fast convergence for a suitable set of physical parameter values.
We have also discussed the general structure of the n-loop transcendental equations. The equations of interest are generated by the subset of (nearly) diagonal Padé approximants, the denominators of which decompose into a product of discrete roots. We have noted that these transcendental equations are generalizations of Lambert's equation, with each ascending loop order adding new generalized polynomials into the structure. We also established that the forms of the transcendental equations are fully described by the odd loop orders, given in equation (\ref{eq:easy1}).
This generic transcendental equation of an arbitrary loop order has been solved analytically using the Lagrange inversion theorem. The power series solution is akin to the four-loop solution, in terms of its exponential behaviour, as seen in equation (\ref{eq:final}). However, it adds a combinatorially increasing number of components corresponding to the loop order.
\section*{Acknowledgements}
The author wishes to thank Antti Kupiainen, Anca Tureanu and Aleksi Vuorinen for enlightening discussions and feedback. This work has been supported by the European Research Council, grant no.~725369, and by the Academy of Finland, grant no.~1322507.
\appendix
\section{Reduction identities}
In this appendix we give a more detailed description about the integral identities used in sections 4-6. The rational integrals are structured such that the denominators decompose into a product of $n$ monomials, whereas the numerators are taken to be polynomials of arbitrary orders. The integrals are carried out in terms of the inverse coupling, $y$, following the notation in section 4. Also, throughout this appendix we neglect to write the integration constants explicitly.
Let us start with the case of the numerator being a monomial of lower order. In the extreme case we note that
\begin{equation}
\begin{split}
\int dy \frac{y^{n-1}}{\prod_{k=1}^n(y+a_k)} &= \int dy \left[\frac{1}{y+a_1}- \frac{p_{n-2}(y)}{\prod_{k=1}^n(y+a_k)} \right]\\
&= \text{ln} (y+a_1) - \sum_{j=1}^{n-2} a_j \int dy \frac{y^j}{\prod_{k=1}^n(y+a_k)},
\end{split}
\end{equation}
where we denote $p_k(y)$ as a $k$th order polynomial with the following condition
\begin{equation}
y^{n-1}+ p_{n-2}(y) = \prod_{k=2}^n(y+a_k).
\end{equation}
Next, we consider a monomial of the order $0 <m < n $. Following similar steps to those given above, we find
\begin{equation}
\label{eq:ym}
\begin{split}
\int dy \frac{y^{m}}{\prod_{k=1}^n(y+a_k)} &= \int dy \left[\frac{1}{\prod_{k=1}^{n-m}(y+a_k)}- \frac{p_{m-1}(y)}{\prod_{k=1}^n(y+a_k)} \right].\\
\end{split}
\end{equation}
In order to solve the first part of equation (\ref{eq:ym}), we wish to decompose it such that each root is separately summed. This leads to $n-m$ linear equations which uniquely determine the values of the coefficients (up to scaling) such that
\begin{eqnarray}
\sum_{k = 0}^{n-m} A_k &=& 0,\\
\sum_{j=0}^{n-m} \sum_{k=0}^{n-m} A_j (1 -\delta_{jk}) a_k &=& 0,\\
\vdots \nonumber \\
\sum_{j=0}^{n-m} \sum_{l \neq j}^{n-m} A_j \prod_{k \neq j, l} a_k &=& 0.
\end{eqnarray}
This provides us with the following decomposition
\begin{equation}
\label{eq:decomp}
\begin{split}
\int dy \frac{1}{\prod_{k=1}^{n-m}(y+a_k)} &= \frac{1}{\sum_{j=0}^{n-m} A_j \prod_{k \neq j} a_k}\sum_{l=1}^{n-m} \int dy \frac{A_l}{y+a_l}\\
&=\sum_{l=1}^{n-m} \frac{A_l \text{ln} (y+a_l)}{\sum_{j=0}^{n-m} A_j \prod_{k \neq j} a_k}.
\end{split}
\end{equation}
Thus, we can write for any $y^m$
\begin{equation}
\begin{split}
\int dy \frac{y^{m}}{\prod_{k=1}^n(y+a_k)} &= \sum_{j=1}^{n-m} c_j \text{ln}(y+a_j) - \int dy \frac{p_{m-1}(y)}{\prod_{k=1}^n(y+a_k)} \\
&= \sum_{j=1}^n \tilde{c}_j \text{ln}(y+a_j),
\end{split}
\end{equation}
where we iterated the earlier result $m$ times and hid all contributions from the lower order polynomials within the set of new constants $\{\tilde{c}_j \}$. Using this result, we can consider any polynomial $p_n(y)= \sum_{k=0}^n c_{k}y^{n-k}$ in the numerator:
\begin{equation}
\begin{split}
\int dy \frac{p_n (y)}{\prod_{k=1}^n(y+a_k)} &=c_0 \int dy + \int dy \frac{\tilde{p}_{n-1}}{\prod_{k=1}^n(y+a_k)}\\
&= c_0 y + \sum_{k=1}^n \tilde{c}_k \text{ln} (y+a_k).
\end{split}
\end{equation}
This result describes a slight generalization of the integral we need to consider with the diagonal $[n/n]$ Padé approximants.
Additionally, we want to consider the integrals related to the non-diagonal Padé approximants. Thus, we introduce an integrand of the original coupling, in which the numerator is a higher order polynomial than the denominator. It is trivial to re-write this kind of integral in terms of the inverse coupling and to isolate the unfamiliar parts:
\begin{equation}
\begin{split}
\int dy \frac{p_{n+m}(y)}{y^m \prod_{k=1}^n(y+a_k)} &= \sum_{j = 1}^{m} \int dy \frac{d_j}{y^j\prod_{k=1}^n(y+a_k)} + \int dy \frac{\tilde{p}_n (y)}{\prod_{k=1}^n(y+a_k)}. \\
\end{split}
\end{equation}
The unknown integrals, characterised by the denominator of the order $n+j$, can be decomposed in similar manner to equation (\ref{eq:decomp}). By applying some linear algebra, we find
\begin{equation}
\begin{split}
\int dy \frac{1}{y^j\prod_{k=1}^n(y+a_k)} &= \sum_{k = 1}^n \int dy \frac{c_k}{y^j(y+a_k)}\\
&= \sum_{k = 1}^n \int dy \frac{c_k}{a_k y^{j-1}}\left[\frac{1}{y}-\frac{1}{y+a_k} \right]\\
&= \sum_{k = 1}^n \frac{c_k}{a_k} \int dy \left[\sum_{l=0}^{j-1}\frac{1}{(-a_k)^{l} y^{j-l}} -(-a_k)^{1-j}\frac{1}{y+a_k}\right]\\
&= \sum_{k = 1}^n c_k\left[\sum_{l=1}^{j-1} \frac{(-a_k)^{-l}}{l y^{l}} -(-a_k)^{-j} \text{ln} y + (-a_k)^{-j} \text{ln} (y+a_k) \right],
\end{split}
\end{equation}
where we use the convention that the summation vanishes completely if the maximum index is smaller than the minimum. This allows us to identify the structure of the full integral as
\begin{equation}
\int dy \frac{p_{n+m}(y)}{y^m \prod_{k=1}^n(y+a_k)} = \sum_{j=1}^{m-1} \frac{\tilde{c}_j}{y^j} + \tilde{c}_0 \text{ln} y+\tilde{e}_0 y + \sum_{k = 1}^n \tilde{v}_k \text{ln} (y + a_k).
\end{equation}
In particular we find the $[n/n+1]$ Padé approximants to be described by the following special case
\begin{equation}
\int dy \frac{p_{n+1}(y)}{y\prod_{k=1}^n(y+a_k)} = \tilde{c}_0 \text{ln} y+\tilde{e}_0 y + \sum_{k = 1}^n \tilde{v}_k \text{ln} (y + a_k).
\end{equation}
And as a last case of minor interest, we list the extreme case of $[0/n]$ Padé approximant, which yields the following type of integral
\begin{equation}
\begin{split}
\int dy \left(1 + \sum_{k=1}^n a_k y^{-k} \right) = y - \sum_{k=1}^{n-1} \frac{a_{k+1}}{k y^k} + a_1\text{ln} y.
\end{split}
\end{equation}
\bibliographystyle{unsrt}
\bibliography{referencespade}
\end{document}
| 81,644
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QuoteYou are in favor of pre-empting democracy and human rights, on the grounds that things might get worseyou have applied your prejudice against conservative views to what i have said. i have applied my memory of how happy people (especially the left) in the US were for the Iranian students in '79.
You are in favor of pre-empting democracy and human rights, on the grounds that things might get worse
Fear-mongering is the order of the day when it comes to interpreting events in the middle east among most conservatives in the US. Certainly, there is no guarantee that any uprising in Egypt or anywhere else will not be hijacked. Just as there is no guarantee that my firearms or your firearms will not be misused or stolen. Both types of things happen frequently. But we should not shut down rights and institute gun control and preserve dictators and kings because "It might turn out bad if we don't!!"
...some military women are suing because they were raped...
In Muslim countries, they won't touch a girl they don't know.
I suspect Lara Logan thought she could export her foreign morals (or immorality, depending on how you look at it) to Egypt, without taking the time to think about 'doing as Romans do'.
... Try going ...to a muslim country... Why do you think they have resorted to using women suicide bombers?... Because muslim men won't touch a non-family woman to search her for explosives...
A reputation for sleeping with anyone who she thinks can help her somehow. 'Whatever it takes' to get the extra edge on a). we are already going to put them on subs. understanding that the above is a fact, i can't help but wonder about the wisdom of putting more women out there so that more can be violated....and more mens careers tossed in the tank.more guys than girls on here. thoughts?
I'm talking about casual sex. I do bristle at those here who imply that soldiers and sailors in todays forces would resort to rape and pillage upon vanquishing the enemy. Our military does not operate that way.
| 83,704
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\section{Introduction}
We fix an algebraic closure $ \Qbar $ of $ \Q $. If $ k $ is a subfield of $ \Qbar $
then we write $ G_k $ for the Galois group $ \Gal(\Qbar/k) $.
We view all number fields as subfields of $ \Qbar $,
hence we have $ G_k \subseteq G_\Q $.
We also fix a prime number $ p $ and an algebraic closure $ \Qpbar $
of~$ \Q_p $, with absolute value $ |\cdot|_p $ and valuation $v_p$
normalized by $ |p|_p = p^{-1} $ and $ v_p(p) = 1 $.
Let $ k $ be a number field,
$ \eta : G_k \to \Qpbar $ an Artin character.
We write $ \Qp(\eta) $ for the subfield of $ \Qpbar $ obtained
by adjoining all values of $ \eta $ to $ \Qp $, and $ \Z_{p,\eta} $
for the valuation ring of $ \Qp(\eta) $.
Fix an integer $ m\le 0 $.
Let $\sigma: \Qpbar \to \C$ be an embedding.
If $ \eta $ is 1-dimensional, then it follows from \cite[VII Corollary 9.9]{Neu99}
that the value $L(k,\sigma \circ \eta, m)$ of the classical Artin
$ L $-function is in $\sigma(\Qpbar)\subseteq \C$ and that
\begin{equation*}
L^\ast(k,\eta,m) = \sigma^{-1}(L(k,\sigma \circ \eta,m))
\end{equation*}
is independent of the choice of $\sigma$. By the discussion
at the beginning of~\cite[\S~3]{dJNa} this also holds if~$ \eta $
is not 1-dimensional.
In particular, if $ \eta $ takes values in a finite extension
$ E $ of $ \Q $, then all those values are in $ E $.
Clearly,
if $ v $ is a finite place of $ k $, then
$ \Eul_v^\ast(k,\eta,m) = \sigma^{-1}(\Eul_v(k,\sigma \circ \eta,m)) $,
with $ \Eul_v(k,\sigma\circ\eta,s) $ the (reciprocal of the) Euler factor
for~$ v $, is independent of $ \sigma $ and lies in~$ E $ (see ~\eqref{recEulerfactor}
for the expression of this Euler factor).
Write $ q = p $ if $ p \ne 2 $, and $ q = 4 $ if $ p = 2 $.
Let $ \o_p : G_k \to \Qpbar $ be the Teichm\"uller character
for $ p $, which is the composition
\begin{equation*}
G_k \to G_\Q \to \Gal(\Q(\mu_{p^\infty})/\Q) \rightiso \Zp^* \to \mu_{\phi(q)}
\,,
\end{equation*}
where the last homomorphism maps
an element of $\Zp^*$ to the unique element of the torsion subgroup
$\mu_{\phi(q)} $ of $ \Zp^* $ in its coset modulo $1+q\Zp$.
Note that for $ k = \Q $, $ \o_p $ gives an isomorphism $ \Gal(\Q(\mu_q)/\Q) \simeq \Zp^*/(1+q\Zp) \simeq \mu_{\phi(q)} $.
We shall also use the projection $ \Zp^* = \mu_{\phi(q)} \times (1+q\Zp) \to 1+ q \Zp$,
mapping $x$ to $ \langle x \rangle$, so that $ x^{-1} \langle x \rangle $
is in~$ \mu_{\phi(q)} $.
Now assume that $ k $ is totally real. If $ E $ is any extension of $ \Q $,
and $ \eta : G_k \to E $ an Artin character of~$ k $, then $ \eta $ is called even if $ \eta(c) = \eta(1) $ for each $ c $
in the conjugacy class of complex conjugation in $ G_k $. This is equivalent with the fixed field
$ k_\eta $ of the kernel of the underlying representation of $ \eta $ being totally real.
We call $ \eta $ odd if $ \eta(c) = -\eta(1) $ for those $ c $,
so that $ \eta $ is odd if and only if all these $ c $ act as multiplication by $ -1 $ in the underlying representation.
For $ \chi : G_k \to \Qpbar $ a 1-dimensional even Artin character
of the totally real number field $ k \subset \Qbar $, there is
a unique continuous function $ \Lp p k {\chi} s $ on $ \Zp $,
with values in $ \Qp(\chi) $, such that for all integers
$ m \le 0 $ we have the interpolation formula
\begin{equation} \label{interpol}
\Lp p k {\chi} m = L^*(k , \chi\o_p^{m-1} , m) \prod_{v \in P} \Eul_v^\ast(k,\chi\o_p^{m-1},m)
\,,
\end{equation}
where $ P $ consists of the places of $ k $ lying above~$ p $.
The right-hand side here is never zero if $ m < 0 $, and $ L^*(k , \chi\o_p^{-1} , 0) \ne 0 $
as well.
Such continuous functions in particular cases (see, e.g.,
\cite{Kub-Leo,Ser73}) were the starting point, but, in fact,
they have much nicer properties as they can also be described as follows
(see, e.g., \cite{Bar78, Cas-Nog79, Del-Rib, Rib79}, although we
shall mostly follow \cite{greenberg83} here).
Let $ r \ge 1 $ be the largest integer such that $ \mu_{p^r} $ is in $ k(\mu_q) $
and put
\begin{equation*}
\D_k = \{ z \text{ in } \Qpbar \text{ with } |z|_p < p^r p^{-1/(p-1)} \}
\,.
\end{equation*}
Fix a topological generator $\gamma$ of $ \Gal(k_\infty/k) $,
where $ k_\infty $ is the cyclotomic $ \Zp $-extension of $ k $.
(We shall briefly review in Remark~\ref{indeprem} how the various
notions in this introduction depend on this choice.)
Let $ \gamma' $ in $ \Gal(k_\infty(\mu_q)/k) $ be the unique
element that restricts to $ \gamma $ on $ k_\infty $ and to the identity
on $ k(\mu_q) $.
Since the composition
\begin{equation*}
\Gal(k_\infty(\mu_q)/k) \to \Gal(\Q_\infty(\mu_q)/\Q) \rightiso \Z_p^*
\end{equation*}
is injective, there is a unique $ u $ in $ 1+q\Zp $ such that $ \gamma'(\xi) = \xi^u $
for all roots of unity~$ \xi $ of $ p $-power order.
Let $H_\chi(T)= \chi(\gamma)(1+T) - 1$
if $k_\chi \subset k_\infty $ (i.e., $\chi$ is of type $W$) and $ H_\chi(T) = 1 $ otherwise.
There exists a unique power series $G_\chi(T)$, called the Iwasawa power series of $\chi$,
with coefficients in $\Z_{p,\chi}$, such that,
for all $s $ in $ \Zp $, with $s \ne 1$ if $\chi$ is trivial, we have
\begin{equation} \label{iwasawaseries}
\Lp p k {\chi} {s} = \frac{G_\chi(u^{1-s} - 1)}{H_\chi(u^{1-s} - 1)}
\,.
\end{equation}
Note that $ | u-1 |_p = p^{-r} $, so $ u^t = \exp(t\log(u)) $
exists for all $ t $ in~$ \D_k $. Therefore $ \Lp p k {\chi} s $
is a meromorphic function on $ \D_k $,
with values in $ \Qpbar $,
with at most a pole of
order~1 at $ s = 1 $ if $ \chi $ is trivial, and no poles otherwise.
We let $ \zeta_p(k,s) $ denote this function if $ \chi $ is the trivial character.
If $ \chi : G_k \to \Qpbar $ is an even irreducible Artin character
of degree greater than~1, we let $ H_\chi(T) = 1 $.
We then define $ H_\chi(T) $ for all characters $ \chi $
by demanding $ H_{\chi_1+\chi_2}(T) = H_{\chi_1}(T) H_{\chi_2}(T) $.
Using the Galois action on $ \Qpbar $ one sees that $ H_\chi(T) $ has coefficients
in $ \Z_{p,\chi} $.\footnote{In \cite[\S2]{greenberg83}
a ring $ \Zp[\chi] $ is used for all $ \chi $,
but there it is only defined for $ \chi $ 1-dimensional, as
the ring obtained by adjoing the values of $ \chi $ to $ \Zp $.
At the bottom of page 82 of loc.\ cit.\ it is clear that $ \Z_{p,\chi} $
is meant. Note that $ \Zp[\chi] = \Z_{p,\chi} $ if $ \chi $ is
1-dimensional, and that $ \Zp[\textup{values of } \chi] $ is contained
in $ \Z_{p,\chi} $ for all $ \chi $ but may be smaller even for Abelian $ \chi $.}
For $ \chi : G_k \to \Qpbar $ any even Artin character of $ G_k $, one
uses Brauer induction to find a unique $ G_\chi(T) $ in the fraction
field of $ \Z_{p,\chi}[[T]] $
such that~\eqref{interpol} holds for all integers $ m < 0 $ if
we define $ \Lp p k {\chi} s $ by~\eqref{iwasawaseries} using
this $ G_\chi(T) $ and the $ H_\chi(T) $ defined above.
(We shall briefly review this in Section~\ref{Lpsection}.)
It is known for $ p $ odd (and conjectured for $ p = 2 $) that
$ G_\chi $ is in $ p^{-l} \Z_{p,\chi}[[T]] $ for some integer $ l $,
in which case it converges on~$ \D_k $.
This means that $ \Lp p k {\chi} s $
is defined on $ \D_k $ for $ p $ odd, but for $ p=2 $ it might
have finitely many poles if $ \chi $ is not Abelian.
We can now state our main conjecture,
some special cases of which appeared in the literature before
(see Remark~\ref{earlier}).
The equivalence of the
various parts will be proved in Section~\ref{nozeroessection}.
The `missing' case $ m = 0 $ in this conjecture is the subject
of a conjecture by Gross, Conjecture~\ref{gross conj} below.
Note that the statement here always holds for $ m < 0 $ by~\eqref{interpol}
as the right-hand side there is never zero, but we included this
case in the formulation of the conjecture anyway for the sake of uniformity.
\begin{conjecture} \label{mainconjecture}
Fix a prime number $ p $ and an integer $ m \ne 0 $.
Then the following equivalent statements hold, where in the last
three parts the characters always take values in~$ \Qpbar $.
\begin{enumerate}
\item
For every totally real number field $ k $ with $ k/\Q $ Galois
the function $ \zeta_{k,p}(s) $ is non-zero at $ m $ if $ m \ne 1 $ and has a pole of order~1 at $ m = 1 $.
\item
For every totally real number field $ k $ the function $ \zeta_{k,p}(s) $
is non-zero at $ m $ if $ m \ne 1 $ and has a pole of order~1 at $ m = 1 $.
\item
For every totally real number field $ k $ and every 1-dimensional
even Artin character $ \chi \ne {\bf 1}_{G_k} $ of $ G_k $ we have $ \Lp p k {\chi} m \ne 0 $.
\item
For every totally real number field $ k $ and every irreducible
even Artin character $ \chi \ne {\bf 1}_k $ of $ G_k $ we have that $ \Lp p k {\chi} m $ is defined and non-zero.
\item
For every irreducible even Artin character
$ \chi \ne {\bf 1}_{\Q} $ of $ G_\Q $ we have that $ \Lp p {\Q} {\chi} m $ is defined and non-zero.
\end{enumerate}
\end{conjecture}
For $ p $ odd, one can state the conjecture also in terms of
the power series $ G_\chi(T) $ associated to $ k $ and $ \chi $ that we introduced above,
and for $ p=2 $ using fractions of power series. It is more natural
in the sense that $ m = 1 $ is no longer a special case; see Remark~\ref{Gremark}.
We prove this version in a number of cases (see Lemma~\ref{vlemma} and
Proposition~\ref{vprop}).
Our numerical calculations prove Conjecture~\ref{mainconjecture}(3) for many tuples $ (p,k,\chi,m)$ with
$ [k : \Q ] = 1 $ or~2, and in fact, for most tuples $ (p,k,\chi) $
considered the statement holds for all $ m \ne 0 $; see Theorem~\ref{th:mainconj}.
Because Conjecture~\ref{mainconjecture}
has important consequences for
(generalisations of) conjectures made by Coates and Lichtenbaum and by Schneider
(see Section~\ref{consequences}), our calculations prove many
instances of these conjectures.
The conjecture
was itself inspired mostly by the wish
that certain $ p $-adic regulators in algebraic $ K $-theory
should be non-zero, and our calculations also prove this in many
cases (see the end of Section~\ref{consequences}). We can also prove it in certain
cases without relying on computer calculations (see Corollary~\ref{regcor}).
\smallskip
In order to state our second conjecture we need more notation.
\smallskip
If $ \O $ is the valuation ring in a finite extension of $ \Qp $,
then according to~\cite[Theorem~7.3]{wash} any non-zero $ S(T) $
in $ \O[[T]] $ can be written uniquely as $ c D(T) U(T) $,
with $ c \ne 0 $ in $ \O $, $ U(T) $ a unit of $ \O[[T]] $ with $ U(0)=1 $, and $ D(T) $
a distinguished polynomial, i.e., a monic polynomial in $ \O(T) $
such that all non-leading coefficients are in the valuation ideal
of~$ \O $. If we extend the field then this decomposition remains
the same.
For $ S(T) = c D(T) U(T) $ as above we define $ \mu(S) = v_p(c) $
and we let $ \l(S) $ be the degree of $ D(T) $.
We extend those definitions to the field of fractions of $ \O[[T]] $
in the obvious way.
\newpage
We can now state our second conjecture.
\begin{conjecture} \label{Grootsconjecture}
Let $\chi : G_k \to \Qpbar $ be a 1-dimensional even Artin character
of a totally real number field~$ k $.
Assume that $\Ind {G_k} {G_\Q} (\chi)$ is irreducible.
Then any root of the distinguished polynomial $ D_\chi(T) $ of $ G_\chi(T) $ distinct from $u-1$ is simple.
\end{conjecture}
Note that in~\eqref{iwasawaseries} we have $ |u^{1-s}-1|_p < p^{-1/(p-1)} $,
so this conjecture is also about roots of $ D_\chi(T) $ that
are not detected by the zeroes of $ \Lp p k {\chi} s $ on~$ \D_k $.
The cases in which we prove this conjecture numerically are described
in Theorem~\ref{th:simple}.
The potential root $ u-1 $ of $ G_\chi(T) $ corresponds to the
potential root~0 of $ \Lp p k {\chi} s $. For
this root, Gross formulated the following as part of Conjecture~2.12 in~\cite{Gross81}.
\begin{conjecture} \label{gross conj}
Let $\chi : G_k \to \Qpbar $ be a 1-dimensional even Artin character
of a totally real number field~$ k $,
$ P $ the set of places of $ k $ lying above~$ p $,
and $\sigma : \Qpbar \to \mathbb{C}$ any embedding. Then
the $p$-adic $L$-function $\Lp p k {\chi} s $
and the truncated complex $L$-function
$ L(k , \sigma \circ\chi\o_p^{-1} , s) \prod_{v \in P} \Eul_v(k,\sigma \circ \chi\o_p^{-1},s) $
have the same order of vanishing at $ s=0 $.
\end{conjecture}
\begin{remark}
One can also consider the case of truncated \padic \Lfunctions
for a finite set~$ S $ of places of $ k $ that contains~$ P $,
and compare the order of vanishing at $ s = 0 $ to the corresponding
truncated complex \Lfunction, but, as noted in loc.~cit., the correctness of the
conjecture is independent of~$ S $.
\end{remark}
Because of the construction of $ \Lp p k {\chi} s $
for arbitrary $ \chi $, and the behaviour of the classical \Lfunctions
for Brauer induction, the same conjecture would then hold for
all even Artin characters $ \chi $ of~$ k $.
Theoretical evidence for it will be discussed in Remark~\ref{zeroremark},
but we also prove this conjecture in many cases by means of our
calculations; see Theorem~\ref{th:proofgoss=0} and Remark~\ref{grossremark}.
We now formulate the last conjecture that our calculations prove in many cases.
For an even Artin character $ \chi $ of $ G_k $, we define
$ \lambda(\chi) = \lambda(G_\chi/H_\chi) $ and
$ \mu(\chi) = \mu(G_\chi/H_\chi) $.
(Note that $ H_\chi $ is often ignored when defining $ \lambda(\chi) $
and $ \mu(\chi) $, but we follow \cite{Sinnott}.)
Clearly, $ \mu(H_\chi) = 0 $,
so $\mu(\chi) \geq 0$. Deligne and Ribet \cite{Del-Rib} proved
(see \cite[(4.8) and (4.9)]{Rib79})
that $\mu(\chi) \geq [k:\Q]$ if $p=2$.
The so-called ``$\mu = 0$'' conjecture, formulated in~\cite{Iwa73b}, states this
bound should be sharp, and that $ \mu(\chi) $ should be~0 if $ p $ is odd.
\begin{conjecture}[``$\mu = 0$''] \label{muconjecture}
Let $\chi : G_k \to \Qpbar $ be a 1-dimensional even Artin character
of a totally real number field~$ k $.
Then $\mu(\chi) = 0$ if $p$ is odd, and $\mu(\chi) = [k:\Q]$ if $p = 2$.
\end{conjecture}
This conjecture was proved for $k = \Q$ by Ferrero and Washington \cite{Fer-Wash}.
Theorem~\ref{th:proofmu=0} describes for which real quadratic
number fields $ k $ and $ \chi $ our calculations prove it for
more than 6000 conjugacy classes over $\Qp$ of \padic
characters not covered by the result of Ferrero and Washington.
There do not seem to be many conjectures or results on the $ \lambda $-invariant
(but see \cite[Conjecture~1.3]{EJV11} for a very special case).
Based on the (extensive) data provided by our calculations we formulate
the following conjecture.
\begin{conjecture} \label{Qconjecture}
Let $p$ be an odd prime and let $d \geq 1$. Define $\mathfrak{X}_d$ to be the set of even $1$-dimensional Artin characters $\chi : G_\Q \to \Qpbar $ of the form~$\chi = \o_p^i\psi$ with $\psi$ of conductor and order prime to $p$, both $\psi$ and $i$ even, and such that~$[\Q_p(\chi):\Q_p] = d$. For $N \geq 1$, let $\mathfrak{X}_d(N)$ be the subset of those characters in $\mathfrak{X}_d$ whose conductor is at
most~$N$. Then we have
\begin{equation*}
\lim_{N \to +\infty} \frac{\#\{\chi \in \mathfrak{X}_d(N) : \lambda(\chi) > 0\}}{\#\mathfrak{X}_d(N)} = p^{-d}.
\end{equation*}
\end{conjecture}
During the calculations we discovered this conjecture does not apply
if we allow~$ p $ to divide the order of the character. In fact, in
that case the $ \lambda $-invariant can be quite large (see
Examples~\ref{firstexample} through~\ref{lastexample}). This phenomenon
can be explained by a result of Sinott~\cite[Theorem~2.1]{Sinnott}; see Corollary~\ref{lambdappowercase}.
The conjecture does also not apply if we allow both
$ i $ and $ \psi $ to be odd; see the beginning of Section~\ref{lambdamodel}.
\medskip
The structure of this paper is as follows.
In Section~\ref{Lpsection} we
review and slightly extend the description of $ \Lp p k {\chi} s $ if
$ \chi $ is an even $ \Qpbar $-valued Artin character of higher
dimension. We also review truncated \padic \Lfunctions, which play a
role in some of the conjectures or their consequences.
Section~\ref{nozeroessection} proves the equivalence of the various
parts of Conjecture~\ref{mainconjecture}, and gives a formulation
using the power series~$ G_\chi(T) $. It also discusses some earlier cases
of this conjecture and some theoretical evidence for it.
We conclude this section with a review of
what is known about Conjecture~\ref{gross conj},
and a discussion of our motivation for making Conjecture~\ref{Grootsconjecture}.
In Section~\ref{consequences} we discuss some important consequences for a
generalisation of a conjecture by Coates and Lichtenbaum
(\cite[Conjecture~1]{Co-Li}, but see also \cite[\S1]{dJNa}) and
for a conjecture by Schneider \cite{schn79}.
We also discuss the relation with a conjecture
made in~\cite{BBDJR}. In fact, this last conjecture inspired the
current paper as it is about the non-vanishing of \padic regulators in $ K $-theory,
in analogy with the Leopoldt conjecture, and our calculations
that are described in Section~\ref{zeronumevidence} prove this
non-vanishing in many cases.
In Section~\ref{NPsection} we review and slightly refine the theory of
Newton polygons, in order to help us rule out by computations
the existence of multiple factors in the corresponding distinguished polynomials in Section~\ref{zeronumevidence},
thus proving Conjecture~\ref{Grootsconjecture} in many cases.
In that section we also prove, by means of calculations, Conjectures~\ref{mainconjecture},
\ref{gross conj} and~\ref{muconjecture} for many characters.
Finally, in Section~\ref{lambdamodel}, we discuss the behaviour
of the $ \l $-invariants of the Iwasawa power series, and our
(substantial) numerical data leads us to make and corroborate Conjecture~\ref{Qconjecture}.
| 33,410
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TITLE: Associate elements in non-integral domains.
QUESTION [2 upvotes]: Elements $a$ and $b$ of an integral domain are associates if
$a\mathrel{\vdots}b$ and $b\mathrel{\vdots}a$
I have proved this fact. Then I tried to find out $a$ and $b$ which divide each other and does not associate. This may happen only in non-integral domains of course. But I couldn't find an example. Can someone suggest one?
Definition of associates.
$a$ and $b$ are associates if $a=b\epsilon$ where $\epsilon$ is
invertible element of the ring.
$a\mathrel{\vdots}b$ means $a$ is divided by $b$. ($a=bc$). It is the
same as $b|a$.
REPLY [3 votes]: Here is an example due to Kaplansky.
Let $R=C([0,3])$ be the ring of continuous funtions $f:[0,3]\rightarrow\mathbb{R}$. I claim that
$$R^{\times}=\left\lbrace f(t)\in R \hspace{2.5mm} | \hspace{2.5mm} f(t)\neq 0 \hspace{2.5mm} \forall t\in[0,3]\right\rbrace.$$
Indeed, if $f(t)g(t)=1$ for all $t\in[0,3]$ then $f(t)\neq 0$ because if there exists $s\in [0,3]$ such that $f(s)=0$ then $1=f(s)g(s)=0\cdot g(s)=0,$ a contradiction. Conversely, if $f(t)\neq 0$ then $\frac{1}{f(t)}\neq 0$ is a continuous function.
Now define the three elements
$$a(t)=\begin{cases}1-t&t\in[0,1]\\0&t\in[1,2]\\t-2&t\in[2,3]\end{cases},$$
$$b(t)=\begin{cases}1-t&t\in[0,1]\\0&t\in[1,2]\\2-t&t\in[2,3]\end{cases},$$
and
$$c(t)=\begin{cases}1&t\in[0,1]\\3-2t&t\in[1,2]\\-1&t\in[2,3]\end{cases}.$$
Then $a(t),b(t),c(t)\in R$ and it's clear that $c(t)a(t)=b(t)$ and $c(t)b(t)=a(t)$, so $a(t)\mid b(t)$ and $b(t)\mid a(t)$.
Let $u(t)\in R^{\times}$ be given. If $a(t)=b(t)u(t)$ then
$$a(t)=\begin{cases}1-t&t\in[0,1]\\0&t\in[1,2]\\t-2&t\in[2,3]\end{cases}=\begin{cases}(1-t)u(t)&t\in[0,1]\\0&t\in[1,2]\\(2-t)u(t)&t\in[2,3]\end{cases}=b(t)u(t),$$
which implies that
$$u(t)=\begin{cases}1&t\in[0,1]\\\star&t\in[1,2]\\-1&t\in[2,3]\end{cases}.$$
But due to the Intermediate Value Theorem we see that there exists $s\in[0,3]$ such that $u(s)=0$ because it attains all the values between $-1$ and $1$. Thus we conclude that $u(t)$ is not a unit and therefore we see that $a(t)$ and $b(t)$ are not associates.
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Previous Questions and Answers
Is adoption Biblical?
Is
adoption Biblical?
A reference is made to it in
Proverbs 29:21, in regard to a man sometimes adopting a servant, "He that delicately
bringeth up his servant from a child SHALL HAVE HIM BECOME HIS SON at the length."
In a sense, we get adopted by
God when we get saved.." Also, in
Ephesians."
Based upon these Scriptures,
adoption is Biblical.
Mordecai
adopted Esther. Esther 2:5-7 says, HER FOR HIS OWN
DAUGHTER."
Previous Questions and Answers
Ask A Question
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What Is The Omega 3 Omega 6 Ratio And Why Do We Need To Know?
It may pay to consider what your Omega 3 Omega 6 ratio is.
We talk a lot about Omega 3 essential fatty acids on our website, and for good reason, they’re extremely important to your health, as long as you get enough. But today we wanted to look at Omega 6 fatty acids, and what the Omega 3 Omega 6 ratio is. And how it affects you.
Omega 6 fats are also essential fatty acids. Omega 6 is like Omega 3, we can’t make it in our bodies and so we need to get Omega 6 from our diet. However unlike Omega 3 fatty acids, which 90% if us are deficient in, we are not generally deficient in Omega 6 fats. Omega 6 fats are important for a number of reasons, including for maintenance of growth and for good general health. They are important for maintaining the health of the cells in the brain.
Omega 6 fatty acids can also go by the names n-6 fatty acids, w6 fatty acids and linoleic acids.
So whilst we need Omega 6 EFAs in our diet just like we need Omega 3, most of us get enough, if not too much, or way too much.
Omega 6 fats come from completely different sources than the Omega 3 fatty acids. Omega 6 fats are found in plant sources like many of our oils, particularly corn oil, safflower oil and sunflower oil, sesame oil and peanut oil. Soybean oil is almost all Omega 6 fat.
And because many of these oils are used in the production of many of the processed foods we eat, like margarine, they are also found extensively in a wide range of foods we eat every day.
On top of that our meat is now higher in Omega 6 than it was. In the past our beef came from grass fed animals. Cows that eat grass are higher in Omega 3, because it’s found in grass. But now our animals are mainly grain fed, and as you can see the oils from grains are high in Omega 6. So our meat is now much higher in Omega 6 than it was.
Now on to the Omega 3 Omega 6 ratio. It is thought that some time ago we ate roughly as much Omega 3 as we ate Omega 6, That makes an Omega 3 Omega 6 ratio of 1:1. (And you need to change that.)
However our diets have swung much more in favor of a higher intake of Omega 6, and a much lower intake of Omega 3, from eating less fish and from less Omega 2 in meat.
Some scientists suggest that some of us may have an Omega 6 Omega 3 ratio that is as high as 10 times as much 6 as 3. So as you can see we don’t need any more Omega 6. Some estimates are even that some of us can be eating more than 20 times as much Omega 6 as 3, and we have even seen estimates of a ratio of 30 times as much Omega 6 as 3.
Although not fully understood it is thought that a diet high in 6 and low in 3 can lead to some health problems, and that our Omega 3 Omega 6 ratio should be somewhere much lower than it is now, perhaps around 1:3.
A high ratio, it is thought, can lead to thickening of the blood possibly leading to blood clots, a worsening of some autoimmune diseases and an increase in some inflammatory diseases, and more, including cancer. A lower ratio can is associated with a lower breast cancer risk. And there is a wide range of lifestyle diseases that are thought to be susceptible to a high ratio between the 2, including many that are related to inflammation in the body.
So there are 2 things you need to do if you want to improve your ratio and lessen your risk of various diseases, some serious. You need to lower your intake of Omega 6 by lowering your intake of processed foods and oils, for instance by changing to good oils such as Olive Oil.
And you need to increase your intake of Omega 3 EFAs, by eating more fish or by daily supplementation with high quality fish oil supplements.
Remember, although all essential fatty acids are essential, that doesn’t mean you don’t get enough of them. It seems clear that we get quite enough Omega 6 fats, and that we need to be getting less of them.
Update: A study was undertaken in 2011 to actually identify how our consumption of Omega 6 has changed in the 20th century. Amongst the authors of the study was Joseph Hibbeln, who was have referred to on this site before. Dr Hibbeln is one of the more well known experts in the field of the health benefits of the Omega 3 fatty acids.
The study confirmed we have said here, that there has been a noticeable increase in the intake of Omega 6 fats over the last century.
In fact it was found that since 1909, where the omega 6 fats (as linoleic acide, the form found in the oils from seeds, which are high on our diet now) provided around 2.3 percent of daily calories, that figure has now increased, since 1999, to 7.2 percent, making a 213 percent increase over that 90 year period.
And there was a decrease in the intake of DHA and EPA.
And the study also noted that the consumption of oil from soybeans increased more than 1000 times, which, according to the authors has “likely decreased tissue concentrations of EPA and DHA during the 20th century”.
To read the study click here.To find out more about Peter Click Here
Tagged with: omega 3 omega 6 • omega 3 omega 6 ratio • omega 6
Filed under: Omega 3 And Omega 6
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Weather Journal: Sunny Week, Rising Water
- Getty Images
- A relaxing scene in the sunshine at Manhattan’s Bryant Park on Tuesday.
Sun will continue across Greater New York for the rest of the week and into the weekend, but water levels in area rivers are expected to keep climbing.
High pressure that helped drive Irene out to sea after landfall will remain in place for the next several days, with highs expected to be in the low- to mid-80s. It’ll feel like near perfect weather for those who felt minor impacts from the weekend storm. For everyone else, well… there are other things to worry about, but hopefully the sun helps bring some cheer. The NYC Department of Heath also has some Post-Irene safety tips for those doing cleanup.
There’s too much weather news still happening, and I can’t possibly cover it all, but a quick rundown follows. For the latest up-to-the-minute coverage of Irene’s lingering impacts, keep following Metropolis.
Area flooding update As Governor Christie has been good to keep emphasizing, New Jersey remains at a critical risk of river flooding this week following record rains. But the same holds true across the tri-state.
In New York City, August 2011 is now our wettest month (of any month and any year) since at least 1869. Central Park has officially recorded 18.95″ of rain this month, beating the previous record (Sept 1882) by more than two inches, and dumping nearly half a year’s worth of precipitation. A bit farther away, one meteorologist points out that the 19.40″ of rain in Philadelphia this month is probably the most rain in any month for any major Northeast city since Newark, N.J. received 22+ inches back in 1843. All that water has to reach the ocean, and it’s doing so in catastrophic fashion.
Irene is proving that the deadliest part of landfalling hurricanes is often freshwater flooding after the storm passes. But because our current measure of hurricane intensity relies only on wind speed, information about these risks is often lost. With at least 37 related deaths, Irene has now become the fourth deadliest hurricane to impact the US since 1980. Damage estimates range from $7B to $20B, making Irene the record 10th billion dollar weather disaster so far this year. And peak hurricane season is still two weeks away.
A look at area flood gauges show that the Ramapo, Passaic and Raritan Rivers in New Jersey are all at or near record flood stage. I’m especially concerned with the Connecticut River near Hartford, as all the flood waters from Vermont will have to make their way through there. Even the Hudson River may experience some flooding as waters from the devastated Catskills make their way downstream.
If you find yourself near moving water, DO NOT attempt to enter it, no matter how shallow it may seem, especially if you are in a vehicle. One unfortunate fatality this weekend in Westchester county was from a man who accidentally stepped in a submerged street drain, and was washed away.
Watching the tropics (again) Don’t look now, but there is also another tropical threat looming out in the Atlantic. The National Hurricane Center on Tuesday will likely announce the birth of Tropical Storm Katia off the coast of West Africa, expected to grow into the season’s second hurricane by the end of the week.
The storm’s name is the first time a replacement name for “Katrina” has been used, and coincidentally formed almost exactly six years to the day after its predecessor made landfall in New Orleans. One reliable weather model is already pointing to a possible risk to the East Coast from future Katia about 10-12 days from now. It’s still VERY early and forecasts are notoriously uncertain this far out, but at this point I’d give it about a 25-35% chance of US landfall. Enough to pay close attention, but not enough to take any actions quite yet.
Keep following Weather Journal this week as we keep you informed of Katia’s progress westward across the Atlantic Ocean.
Monday’s Greater New York forecast roundup:
WSJ Headquarters in Manhattan:
- Brooklyn: 83/68
- Queens: 84/69
- Bronx: 85/66
- Staten Island: 84/65
- Poughkeepsie, N.Y.: 81/56
- Trenton, N.J.: 84/64
- Islip, Long Island: 83/65
- New Haven, Conn.: 84/59
Actual High: 84; RealFeel High: 86
Actual Low: 67; RealFeel Low: 65
Weather: Another sunny day
Meteorologist Eric Holthaus contributes daily weather reports and analysis on Metropolis. For the latest on conditions in New York and elsewhere, follow his updates (@wxrisk) on Twitter.
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Babies Are Not Toys. And You Wouldn't Treat Toys Like That, Either
This woman, a real estate agent in La Jolla, California, is going to jail for violating her probation. What was she on probation for? she was convicted of scratching and pinching eight babies after befriending their mothers. They still don't know why she did that. Sure are some weird people out there. (KSWB-TV (Fox 5/69)/San Diego)
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Have an opinion? Add your comment below.
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Professor Melanie Welham selected for the role of Executive Chair of the Biotechnology and Biological Sciences Research Council.
UKRI will be the main mechanism to promote the UK's unrivalled strengths in research and innovation both at home and around the world. It is at the heart of our modern Industrial Strategy and will ensure that we continue to make the most of our world-leading R&D sector and provide support for our researchers and scientists.
Professor Welham said:
"The excellent research, people and infrastructure that BBSRC invests in makes a real difference to how we understand the world around us.
Creatively exploring the frontiers of bioscience yields remarkable insights into how life works and will, in turn, improve our lives; giving us secure and nutritious food, renewable resources, and better health for us and our animals.
I’m delighted to have the opportunity to continue to lead BBSRC in UK Research and Innovation – to work in partnership to keep the UK at the global forefront of research and innovation."
Sir Mark Walport, UKRI CEO Designate, said:
"I'm delighted with the appointment of Melanie as the first BBSRC Executive Chair. She has an excellent track record as a research leader and as interim BBSRC Chief Executive since 2016. Melanie has played a central role in developing and delivering the Industrial Strategy Challenge Fund and I look forward to continuing to work closely with her as we now embark on establishing UK Research and Innovation."
The post of BBSRC Executive Chair is potentially subject to a pre-appointment hearing by the House of Commons Science and Technology Committee. The Committee will consider this in due course.
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aaahhhh, it feels good to be blogging again. i feel so guilty when i take too many days off; i don't know why? it makes me feel so out of touch.
anyway, let's rewind back to friday. i'm all about tgif and usually count down the seconds for the work day to end and start getting my weekend on. i look so forward to the weekend rides and that helps get me thru the work week. so friday i got home around 6:30pm, i'm chilling out, waiting for the j-man to call me around 7 (when he gets off from work). yay, the phone rings, it's the j-man (yes, i still get excited when he calls me after all these years) and the first thing he says is, "you aren't going to believe this." i'm thinking, he got laid off but no, someone stole the catalytic converter off of his truck. better said, someone cut it off. for those of you who don't know what that is, here is a picture of one:
harley davidson motorcycle!! it will make you crap your pants because you are not expecting that at all. apparently, this is the thing to steal off cars to make a quick buck. salvage yards will pay about $50 a piece for these cat converter's and then they melt them down and extract the metal inside. many of these cat converter's have platinum and 2 other metals inside of them and the salvage yard can make about $6000 per ounce of platinum. these thieves are scum!
Caratunk Girl to help me out so we both put the Old Caratunk Curse on these thieves. she did her version(click on her link to read her version) and here's the one i said:
ok, on to bigger and better things. i had a good swim and bike yesterday and a great run this morning (7.5 miles before work).
when i left work last night it was like a monsoon and the lightning was absolutely incredible but really scary too. i made it home safely thank goddess and reaped the benefits this morning as it was not nearly as hot or humid. i'm very happy with my training up to this point. i'm following the iron distance schedule on BT. there are several others who are following it as well and they are getting ready to do IMCDA this weekend. i can't wait to get their feedback on it. i like the schedule so far but have tweaked things here and there to fit my needs and level. if no rain tonight, i'll be biking.
*****************
in other news, welcome to some new readers. i enjoy reading all the comments and following your blogs. i've added a sidebar (finally!) for current giveaways around the blogosphere and will do my best to keep that up to date. lots of cool stuff out there to win so go check em' out!
much love and peace out!
Those jerks. As I told you before, CARATUNK CURSE to those jerks.
I am following BT schedule as well, I would love to know if it worked for other folks or not - I mean, I am assuming it did, but I would like to know if they thought, "Gee, I wish I did a little more time on the bike" or whatever.
That's a good curse. remind me never to cross you!
No monsoon here today. Just a gusty Northerly wind and 7 deg C. Just right for a 50k ITT.
That is just crazy about your hubby's truck! I can't believe that people do that but I am pretty naive. Congrats on adpating and overcoming!
NEVER heard of them stealing the catalytic converter... man would I BE PISSED! My truck would crush them if they touched it!
Unbelievable these people... sigh, I said the curse for you too... hopefully it works!
Good for you two overcoming and being the better people!
Have a great week training!
Loving the caratunk curse! Excellent.I can totally relate to hating having the schedule thrown into chaos...such creatures of habit are we type A's, huh?
waht comes around goes around, and I am sure 'they' will have it coming to them. It's only a truck....good choice on not letting it ruin the weekend! Way to go on the training, you will knock that race out of the park!
That's so much worse than coming out and finding your spare tire gone, bummer. Good thing you didn't let it get you down!
OMG! That is just insane! I can't even believe that! I'm glad it didn't end up ruining your weekend. Love the curse! :)
Although I can't comment on the BT ironman plan, I'm following the BT HIM program and really like it. I may try a different one next year just to switch things up, but we'll see! Let us know what you think about it as you get more into it!
I literally went "cata-what? what does it do?" I admit, I dont know much about cars. then my anger set in because I HATE people like that!!! ticks me off. Between you and mandy, karma will get him.
I like how you enjoyed the weekend after it though
Sucks about the truck, but glad you rose above and didn't let it get to you. They will get what's coming to them one of these days.
I have heard great things about the BT plan. Not following that one for IMCDA, but have some friends that are.
Let me know what you think after your race.
catalytic converter ?!?! CRAZY you are now the 5th YES, FIFTH person I know who has had theirs stolen. so crazy!!!
For 50Bucks! unreal ... totally get the schedule interruption ruining the rhythm of planned training...way to overcome it!
have a great week
-Derek
That sucks about the catalytic converter. Way to roll with the punches though.
Part of the problem with catalytic converter thefts is that junk yards accept them no questions asked. There really isn't any reason for them to accept them (or at least not to accept more than one per person). I've heard some cities have started to require identification when turned into a junk yard because of the increasing problem.
Way to "rise up against the man" and keep the positive attitude!
there was a rash of cc's stolen around her a few months back and sprinkler head copper. People suck but when the economy gets bad people suck even worse!
This is your onsite IMCDA reporter.. gathering info for a full report of pre race activities, celebrity interviews, and general IM hoopla
Dude, that sucks to hear about J's truck, but I love the friendship prayer you said for those jerks...I'll have to remember it!
I have never heard of cutting out stuff like that out of a truck. Shame on the junk yards as well for turning a blind eye!
That just sucks about the truck. No curse would be too harsh.
In Memphis, this happens alot! Sux!
Wow, that sucks, but there is no way in hell I am going to crap my pants, I am just sorry KC. That is definitely where I draw the line (unless I get REALLY fast at Ironman and the only way I can podium is to ..... okay... never mind.)
What wa the first thing that went through your mind when you heard "my catalytic converter has been stolen"? I ask because I don't hear that every day. Surprised that other people are like "oh yeah, those things...happens in Memphis".
In LA we just steal the whole car.
Glad you got it all back on track...
I think that it is pretty crappy that someone took the cat converter for a quick buck! I just had one replaced on my Honda van. It cost me $800! Ouch! I know that some of it was for labor but the tech told me that it was made of platinum...worth a chunk of money:) I am a firm believer in karma too!
Nice work on your bike, swim and run! Keep it up Kristine:)
Oh my that is crazy!!!! Glad you got it figured out!
that sucks! we had a string of those thieves a year or so ago... people at work had theirs cut off/stolen from the parking lot! some people are so stupid and worthless. glad you were able to brush it off and have a good weekend regardless - better to not let them get the best of you.
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14k Two Toned Gold Pave Textured Earrings
by RC Jewelry
Sale
Original price $316.16
Current price $197.99
SKU 68560
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Waves of polished 14k yellow gold and contrasting white pave texture swirl together to create these unique and sophisticated hoop earrings. Earrings are 24mm long and are held in place by hinged clasps.
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\begin{document}
\maketitle
\begin{abstract}
Every 4-connected graph with minimum degree $\delta$ and connectivity $\kappa$ either has a cycle of length at least $4\delta-2\kappa$ or has a dominating cycle.\\
\noindent\textbf{Keywords:} Circumference, dominating cycle, connectivity, Dirac-type result, endfragment.
\end{abstract}
\section{Introduction}
Long and large cycles (paths) are the main general research objects in hamiltonian graph theory basing on two different initial conceptions: Hamilton and dominating cycles (paths). A cycle $C$ of a graph $G$ is called a Hamilton cycle if it contains every vertex of $G$. A cycle $C$ is called a dominating cycle if $V(G-C)$ is an independent set.
In 1952, Dirac [2] obtained the first sufficient condition for a graph to be hamiltonian by showing that every graph $G$ of order $n$ and minimum degree $\delta$ at least $n/2$, is hamiltonian. Although this bound $n/2$ is tight in Dirac's theorem, it was essentially lowered in several ways by direct incorporation of some additional graph invariants (namely connectivity $\kappa$ and independence number $\alpha$) due to the author [6, 7], Nash-Williams [5] and the author [8]. The latter recently was improved by Yamashita [15]. The evolution of these developments can be demonstrated by the following order.\\
\noindent\textbf{Theorem A} [2]. Every graph with $\delta\geq\frac{1}{2}n$ is hamiltonian.\\
\noindent\textbf{Theorem B} [6, 7]. Every 2-connected graph with $\delta \geq \frac{1}{3}(n+\kappa)$ is hamiltonian.\\
\noindent\textbf{Theorem C} [5]. Every 2-connected graph with $\delta \geq \max\{\frac{1}{3}(n+2),\alpha\}$ is hamiltonian.\\
\noindent\textbf{Theorem D} [8]. Every 3-connected graph with $\delta\geq\max\lbrace \frac{1}{4}(n+2\kappa),\alpha\rbrace$ is hamiltonian.\\
\noindent\textbf{Theorem E} [15]. Every 3-connected graph with $\delta\geq\max\lbrace \frac{1}{4}(n+\kappa+3),\alpha\rbrace$ is hamiltonian.\\
A short proof of Theorem B was given in [3] due to H\"{a}ggkvist. Theorems A-D have their reverse versions due to Dirac [2], the author [7], Voss and Zuluaga [14] and the author [9], respectively, concerning the circumference and Hamilton cycles.\\
\noindent\textbf{Theorem F} [2]. Every 2-connected graph has a cycle of length at least $\min\lbrace n,2\delta\rbrace$.\\
\noindent\textbf{Theorem G} [7]. Every 3-connected graph has a cycle of length at least $\min\lbrace n,3\delta -\kappa\rbrace$.\\
\noindent\textbf{Theorem H} [14]. Every 3-connected graph with $\delta \geq \alpha$ has a cycle of length at least $\min \lbrace n, 3\delta -3\rbrace$.\\
\noindent\textbf{Theorem I} [9, 11]. Every 4-connected graph with $\delta\geq\alpha$ has a cycle of length at least $\min\lbrace n,4\delta-2\kappa\rbrace$.\\
Theorem I was firstly announced in 1985 [9]. The detailed proof of this theorem recently was given in [11]. In view of Theorem E, the following reverse version is reasonable.\\
\noindent\textbf{Conjecture 1}. Every 4-connected graph with $\delta\geq\alpha$ has a cycle of length at least $\min\lbrace n,4\delta-\kappa-4\rbrace$.\\
It is natural to look for analogous results for dominating cycles. Observe that Theorems C, D, E, H and I can not have dominating versions due to the condition $\delta\geq\alpha$. In 1971, Nash-Williams [5] obtained the first Dirac-type sufficient condition for dominating cycles, i.e. the dominating version of Theorem A. The dominating version of Theorem B is due to Sun, Tian and Wei [12], which recently was improved by Yamashita [15].\\
\noindent\textbf{Theorem J} [5]. Every 2-connected graph with $\delta \geq \frac{1}{3}(n+2)$ has a dominating cycle.\\
\noindent\textbf{Theorem K} [12]. Every 3-connected graph with $\delta \geq \frac{1}{4}(n+2\kappa)$ has a dominating cycle.\\
\noindent\textbf{Theorem L} [15]. Every 3-connected graph with $\delta \geq \frac{1}{4}(n+\kappa+3)$ has a dominating cycle.\\
The reverse version of Theorem J is due to Voss and Zuluaga [14]. \\
\noindent\textbf{Theorem M} [14]. Every 3-connected graph either has a cycle of length at least $3\delta-3$ or has a dominating cycle.\\
In this paper we prove the reverse version of Theorem K, recently conjectured in [10]. \\
\noindent\textbf{Theorem 1}. Every 4-connected graph either has a cycle of length at least $4\delta-2\kappa$ or has a dominating cycle.\\
The limit example $4K_2+K_3$ shows that the 4-connectivity condition in Theorem 1 can not be replaced by 3-connectedness. Further, the limit example $5K_2+K_4$ shows that the conclusion "has a cycle of length at least $4\delta-2\kappa$" in Theorem 1 can not be replaced by "has a cycle of length at least $4\delta-2\kappa$+1". Finally, the limit example $H(1,n-2\delta,\delta,\kappa)$ shows that the conclusion "has a dominating cycle" can not be replaced by "has a Hamilton cycle", where $H(1,n-2\delta,\delta,\kappa)$ is defined as follows. Given four integers $a,b,t,\kappa$ with $\kappa\leq t$, we denote by $H(a,b,t,\kappa)$ the graph obtained from $tK_a+\overline{K}_t$ by taking any $\kappa$ vertices in subgraph $\overline{K}_t$ and joining each of them to all vertices of $K_b$.\\
In view of Theorem E and Conjecture 1, the following reverse version of Theorem L is reasonable.\\
\noindent\textbf{Conjecture 2}. Every 4-connected graph either has a cycle of length at least $4\delta-\kappa-4$ or has a dominating cycle.\\
Let $G$ be a graph. For $X\subset V(G)$, we denote by $N(X)$ the set of all vertices of $G-X$ having neighbors in $X$. Furthermore, $\hat{X}$ is defined as $V(G)-(X\cup N(X))$. Following Hamidoune [4], we define a subset $X$ of $V(G)$ to be a fragment of $G$ if $N(X)$ is a minimum cut-set and $\hat{X}\neq\emptyset$. If $X$ is a fragment then $\hat{X}$ is a fragment too and $\hat{\hat{X}}=X$. For convenience, we will use $X^{\uparrow}$ and $X^{\downarrow}$ to denote $X$ and $\hat{X}$, respectively. Throughout the paper, we suppose w.l.o.g. that $|A^{\uparrow}|\geq|A^{\downarrow}|$. An endfragment is a fragment that contains no other fragments as a proper subset.
To prove Theorem 1, we present four more general Dirac-type results for dominating cycles centered around a lower bound $c\geq4\delta-2\kappa$ under four alternative conditions in terms of endfragments.\\
\noindent\textbf{Theorem 2}. Let $G$ be a 3-connected graph. If $|A^{\uparrow}|\leq 3\delta-\kappa-4$ and $|A^{\downarrow}|\leq 3\delta-3\kappa+1$ for an endfragment $A^{\downarrow}$ of $G$, then either $G$ has a cycle of length at least $4\delta-2\kappa$ or has a dominating cycle. \\
\noindent\textbf{Theorem 3}. Let $G$ be a 4-connected graph. If $|A^{\uparrow}|\leq 3\delta-\kappa-4$ and $|A^{\downarrow}|\geq 3\delta-3\kappa+2$ for an endfragment $A^{\downarrow}$ of $G$, then either $G$ has a cycle of length at least $4\delta-2\kappa$ or has a dominating cycle.\\
\noindent\textbf{Theorem 4}. Let $G$ be a 4-connected graph. If $|A^{\uparrow}|\geq 3\delta-\kappa-3$ and $|A^{\downarrow}|\leq 3\delta-3\kappa+1$ for an endfragment $A^{\downarrow}$ of $G$, then either $G$ has a cycle of length at least $4\delta-2\kappa$ or has a dominating cycle. \\
\noindent\textbf{Theorem 5}. Let $G$ be a 4-connected graph. If $|A^{\uparrow}|\geq 3\delta-\kappa-3$ and $|A^{\downarrow}|\geq 3\delta-3\kappa+2$ for an endfragment $A^{\downarrow}$ of $G$, then either $G$ has a cycle of length at least $4\delta-2\kappa$ or has a dominating cycle. \\
\section{Definitions and notations}
By a graph we always mean a finite undirected graph $G$ without loops or multiple edges. A good reference for any undefined terms is [1]. For $H$ a subgraph of $G$ we will denote the vertices of $H$ by $V(H)$ and the edges of $H$ by $E(H)$. For every $S\subset V(G)$ we use $G-S$ short for $\langle V(G)-S\rangle$, the subgraph of $G$ induced by $V(G)-S$. In addition, for a subgraph $H$ of $G$ we use $G-H$ short for $G-V(H)$.
Let $\delta$ denote the minimum degree of vertices of $G$. The connectivity $\kappa$ of $G$ is the minimum number of vertices whose removal from $G$ results in a disconnected or trivial graph. We say that $G$ is $s$-connected if $\kappa\ge s$. A set $S$ of vertices is independent if no two elements of $S$ are adjacent in $G$. The cardinality of maximum set of independent vertices is called the independence number and denoted by $\alpha$.
Paths and cycles in a graph $G$ are considered as subgraphs of $G$. If $Q$ is a path or a cycle of $G$, then the length of $Q$, denoted by $|Q|$, is $|E(Q)|$. Throughout this paper, the vertices and edges of $G$ can be interpreted as cycles of lengths 1 and 2, respectively. A graph $G$ is hamiltonian if it contains a Hamilton cycle. The length of a longest cycle (the circumference) will be denoted by $c$.
Let $Q$ be a cycle of $G$ with a fixed cyclic orientation. For $x\in V(Q)$, we denote the successor and predecessor of $x$ on $Q$ by $x^+$ and $x^-$, respectively. For $X\subseteq V(Q)$, we define $X^+=\{x^+|x\in X\}$ and $X^-=\{x^-|x\in X\}$. For two vertices $x$ and $y$ of $Q$, let $x\overrightarrow{Q}y$ denote the segment of $Q$ from $x$ to $y$ in the chosen direction on $Q$ and $x\overleftarrow{Q}y$ denote the segment in the reverse direction. We also use similar notation for a path $P$ of $G$. We call $x\overrightarrow{Q}y$ an $m$-segment if $|x\overrightarrow{Q}y|\geq m$. For $P$ a path of $G$, denote by $F(P)$ and $L(P)$ the first and the last vertices of $P$, respectively.
Let $Q$ be a cycle of a graph $G$, $r\geq 2$ a positive integer and $Z_1,Z_2,...,Z_p$ are subsets of $V(Q)$ with $p\geq 2$. A collection $(Z_{1},...,Z_{p})$ is called a $(Q,r)$-scheme if $x\overrightarrow{Q}y$ is a 2-segment for each distinct $x,y\in Z_{i}$ (where $i\in\lbrace1,...,p\rbrace)$ and is an $r$-segment for each distinct $x\in Z_{i}$ and $y\in Z_{j}$ (where $i,j\in\{1,...,p\}$ and $i\neq j)$. A $(Q,r)$-scheme is nontrivial if $(Z_{1},...,Z_{p})$ has a system of distinct representatives. The definition of $(Q,r)$-scheme was first introduced by Nash-Williams [5] for $p=2$. \\
\noindent\textbf{Definition A $\{Q^{\uparrow}_{1},...,Q^{\uparrow}_{m};V^{\uparrow}_{1},...,V^{\uparrow}_{m};V^{\uparrow}\}$}. Let $A^{\uparrow}$ be a fragment of $G$ with respect to a minimum cut-set $S$. Define $Q^{\uparrow}_{1},...,Q^{\uparrow}_{m}$ as a collection of vertex disjoint paths in $\langle A^{\uparrow}\cup S\rangle$ with terminal vertices in $S$ such that
$(*1)$\quad $|V(Q^{\uparrow}_i)|\geq 2$ $(i=1,...,m)$ and $\sum_{i=1}^{m}|V(Q^{\uparrow}_{i})|$ is as great as possible.
Abbreviate $V^{\uparrow}_{i}=V(Q^{\uparrow}_{i})$ $(i=1,...,m)$ and $V^{\uparrow}=\bigcup^{m}_{i=1}V^{\uparrow}_{i}$. \\
\noindent\textbf{Definition B $\{Q^{\downarrow}_{1},...,Q^{\downarrow}_{m};Q^{\downarrow}_{*};V^{\downarrow}_{1},...,V^{\downarrow}_{m};V^{\downarrow}\}$}. Let $A^{\uparrow}$ be a fragment of $G$ with respect to a minimum cut-set $S$ and let $Q^{\uparrow}_{1},...,Q^{\uparrow}_{m}$ be as defined in Definition A.
If $|A^{\downarrow}|\geq2(\delta-\kappa+1)$, then we denote by $Q^{\downarrow}_{1},...,Q^{\downarrow}_{m}$ a collection of paths (if exist) in $\langle A^{\downarrow}\cup S\rangle$ with
$(*2)$ combining $Q^{\uparrow}_{1},...,Q^{\uparrow}_{m}$ with $Q^{\downarrow}_{1},...,Q^{\downarrow}_{m}$ results in a simple cycle such that $\sum_{i=1}^{m}|V(Q^{\downarrow}_{i})|$ is as great as possible.
Abbreviate $V^{\downarrow}_{i}=V(Q^{\downarrow}_{i})$ $(i=1,...,m)$ and $V^{\downarrow}=\bigcup^{m}_{i=1}V^{\downarrow}_{i}$. For the special case $|V^{\downarrow}\cap S|=2$ and $S\not\subseteq V^{\uparrow}$, we will use $Q^{\downarrow}_{*}$ to denote a longest path in $\langle A^{\downarrow}\cup S\rangle$ such that combining $Q^{\uparrow}_{1}$ and $Q^{\downarrow}_{*}$ results in a simple cycle with $|V(Q^{\downarrow}_{*})\cap S|\geq3$.
For the reverse case $|A^{\downarrow}|\leq2\delta-2\kappa+1$, we use the same notation $Q^{\downarrow}_{1},...,Q^{\downarrow}_{m}$ (if no ambiguity can arise) to denote a collection of paths in $\langle A^{\downarrow}\cup S\rangle$ with
$(*3)$ combining $Q^{\uparrow}_{1},...,Q^{\uparrow}_{m}$ with $Q^{\downarrow}_{1},...,Q^{\downarrow}_{m}$ results in a simple cycle such that $|(\cup_{i=1}^{m}V^{\downarrow}_{i})\cap S|$ is as great as possible,
$(*4)$ $|\cup_{i=1}^{m}V^{\downarrow}_{i}|$ is as great as possible, subject to $(*3)$.
In this case, $V^{\downarrow}_{i}$ and $V^{\downarrow}$ are as defined in the previous case.\\
\noindent\textbf{Definition C}. Throughout the paper we will use $C$ to denote the cycle obtained by combining $Q^{\uparrow}_{1},...,Q^{\uparrow}_{m}$ and $Q^{\downarrow}_{1},...,Q^{\downarrow}_{m}$. Assume w.l.o.g. that
$(*5)$\quad $Q^{\uparrow}_{1},...,Q^{\uparrow}_{m}$ is chosen such that $|C|$ is as great as possible.
\section{Preliminaries}
In [5], Nash-Williams proved the following result concerning $(Q,r)$-schemes for a cycle $Q$ and a pair $(Z_1 ,Z_2 )$ of subsets of $V(Q)$.\\
\noindent\textbf{Lemma A [5]}. Let $Q$ be a cycle and $(Z_{1},Z_{2})$ be a nontrivial $(Q,r)$-scheme. Then
$$
|Q|\geq\min\Big\{ 2(|Z_{1}|+|Z_{2}|)+2r-6,\frac{1}{2}r(|Z_{1}|+|Z_{2}|)\Big\}.
$$
By some modification of Lemma A, we obtain the following.\\
\noindent\textbf{Lemma 1}. Let $Q$ be a cycle and $(Z_{1},Z_{2})$ be a nontrivial $(Q,r)$-scheme. Then $|Q|\geq |Z_1|+|Z_2|+|Z_1\cup Z_2|$. Moreover,
$(a1)$ if $r=4$, then $|Q|\geq 2(|Z_1|+|Z_2|)$,
$(a2)$ if $r\geq5$, then $|Q|\geq 2(|Z_1|+|Z_2|)+r-3$,
$(a3)$ if $|Z_2|=1$, then $|Q|\geq2|Z_1|+2r-4$,
$(a4)$ if $|Z_1|=|Z_2|=1$, then $|Q|\geq2r$,
$(a5)$ if $|Z_1|+|Z_2|\geq4$ and $r\geq4$, then $|Q|\geq2(|Z_1|+|Z_2|)+2r-8$.\\
For $(Z_1,Z_2,Z_3)$-schemes we need a result proved in [11, Lemma 1].\\
\noindent\textbf{Lemma B [11]}. Let $Q$ be a cycle and $(Z_{1},Z_{2},Z_3)$ be a nontrivial $(Q,r)$-scheme with $|Z_3|=1$. Then
$$
|Q|\geq\min\{ 2(|Z_1|+|Z_2|)+3r-10,\frac{1}{2}r(|Z_1|+|Z_2|)\}.
$$
For the special cases $|Z_i|=1$ $(i=2,3)$ and $|Z_i|=1$ $(i=1,2,3)$, we need the following result.\\
\noindent\textbf{Lemma 2}. Let $Q$ be a cycle and $(Z_{1},Z_{2},Z_3)$ be a nontrivial $(Q,r)$-scheme.
(b1) If $r\geq4$, $|Z_1|+|Z_2|\geq 6$ and $|Z_3|=1$, then $|Q|\geq2(|Z_1|+|Z_2|)+3r-12$.
(b2) If $|Z_i|=1$ $(i=2,3)$, then $|Q|\geq2|Z_1|+3r-6$.
(b3) If $|Z_i|=1$ $(i=1,2,3)$, then $|Q|\geq3r$.\\
For $(Z_1,Z_2,Z_3,Z_4)$-schemes we need the next result proved in [11, Lemma 2].\\
\noindent\textbf{Lemma C [11]}. Let $Q$ be a cycle and $(Z_{1},Z_{2},Z_3,Z_4)$ be a nontrivial $(Q,r)$-scheme with $|Z_3|=|Z_4|=1$. Then
$$
|Q|\geq\min\{ 2(|Z_1|+|Z_2|)+4r-14,\frac{1}{2}r(|Z_1|+|Z_2|)\}.
$$
For the special cases $|Z_i|=1$ $(i=2,3,4)$ and $|Z_i|=1$ $(i=1,2,3,4)$, we prove the following.\\
\noindent\textbf{Lemma 3}. Let $Q$ be a cycle and $(Z_{1},Z_{2},Z_3,Z_4)$ be a nontrivial $(Q,r)$-scheme.
(c1) If $r\geq4$, $|Z_1|+|Z_2|\geq8$ and $|Z_3|=|Z_4|=1$, then $|Q|\geq2(|Z_1|+|Z_2|)+4r-16$.
(c2) If $|Z_i|=1$ $(i=2,3,4)$, then $|Q|\geq2|Z_1|+4r-8$.
(c3) If $|Z_i|=1$ $(i=1,2,3,4)$, then $|Q|\geq4r$.\\
We will show that the main conclusion of Theorem 1 easily follows when $\delta\leq 2\kappa-3$. \\
\noindent\textbf{Lemma 4}. Every 3-connected graph with $\delta\leq 2\kappa-3$ either has a cycle of length at least $4\delta-2\kappa$ or has a dominating cycle.\\
Our constructions of dominating cycles are based on a combination of $Q^{\uparrow}_{1},...,Q^{\uparrow}_{m}$ and $Q^{\downarrow}_{1},...,Q^{\downarrow}_{m}$ (see Definitions A-B). For the case $\delta>3\kappa/3-1$, the existence of $Q^{\downarrow}_{1},...,Q^{\downarrow}_{m}$ follows from the next lemma proved in [11, Lemma 10].\\
\noindent\textbf{Lemma D [11]}. Let $G$ be a 2-connected graph, $A^{\downarrow}$ be an endfragment of $G$ with respect to a minimum cut-set $S$ and let $L$ be a set of independent edges in $\langle S\rangle$. If $\delta > 3\kappa /2 -1$, then $\langle A^{\downarrow}\cup V(L)\rangle$ contains a cycle that uses all the edged in $L$.\\
We need also the following result that occurs in [13].\\
\noindent\textbf{Lemma E [13]}. Let $G$ be a hamiltonian graph with $\{v_1,...,v_r\}\subseteq V(G)$ and $d(v_i)\geq r$ $(i=1,...,r)$. Then any two vertices of $G$ are connected by a path of length at least $r$.\\
To prove Theorems 2-5, we need a number of lemmas on total lengths of $Q^{\uparrow}_{1},...,Q^{\uparrow}_{m}$ and $Q^{\downarrow}_{1},...,Q^{\downarrow}_{m}$ by covering some alternative conditions. In particular, the alternative conditions $|A^{\uparrow}|\leq 3\delta-\kappa-4$ and $|A^{\uparrow}|\geq 3\delta-\kappa-3$ are covered by the following two lemmas.\\
\noindent\textbf{Lemma 5}. Let $G$ be a 3-connected graph and $|A^{\uparrow}|\leq 3\delta-\kappa-4$ for a fragment $A^{\uparrow}$ of $G$. Then $A^{\uparrow}-V^{\uparrow}$ is an independent set.\\
\noindent\textbf{Lemma 6}. Let $G$ be a 4-connected graph and $|A^{\uparrow}|\geq 3\delta-\kappa-3$ for a fragment $A^{\uparrow}$ of $G$. Then either $A^{\uparrow}-V^{\uparrow}$ is an independent set or $|V^{\uparrow}|\geq3\delta-5$.\\
The next lemma insures $A^{\downarrow}-V^{\downarrow}=\emptyset$ when $|A^{\downarrow}|\leq 2\delta-2\kappa+1$.\\
\noindent\textbf{Lemma 7}. Let $G$ be a 2-connected graph with $|A^{\downarrow}|\leq 2\delta-2\kappa+1$ for an endfragment $A^{\downarrow}$ of $G$. If $\{ Q^{\downarrow}_{1},...,Q^{\downarrow}_{m}\}$ exists, then $A^{\downarrow}-V^{\downarrow}=\emptyset$ and $A^{\downarrow}-V(Q^{\downarrow}_*)=\emptyset$.\\
The alternative conditions $|A^{\downarrow}|\leq 3\delta - 3\kappa+1$ and $|A^{\downarrow}|\geq 3\delta - 3\kappa+2$ are covered by the next two lemmas.\\
\noindent\textbf{Lemma 8}. Let $G$ be a 3-connected graph with $\delta \geq2\kappa -2$ and let $|A^{\downarrow}|\leq 3\delta - 3\kappa+1$ for an endfragment $A^{\downarrow}$ of $G$ with respect to a minimum cut-set S.
$(d1)$ If $f\geq3$, then $A^{\downarrow}-V^{\downarrow}$ is an independent set, where $f=|V^{\downarrow}\cap S|$.
$(d2)$ If $f=2$ and $S\not\subseteq V^{\uparrow}$, then $A^{\downarrow}-V(Q^{\downarrow}_*)$ is an independent set.
$(d3)$ If $f=2$ and $S\subseteq V^{\uparrow}$, then either $A^{\downarrow}-V^{\downarrow}=\emptyset$ or $|V^{\downarrow}|\geq 2\delta - 2\kappa+3$.\\
\noindent\textbf{Lemma 9}. Let $G$ be a 3-connected graph with $\delta \geq2\kappa -2$ and let $|A^{\downarrow}|\geq 3\delta - 3\kappa+2$ for an endfragment $A^{\downarrow}$ of $G$ with respect to a minimum cut-set S.
$(e1)$ If $f\geq3$, then either $A^{\downarrow}-V^{\downarrow}$ is an independent set or $|V^{\downarrow}|\geq 3\delta -3\kappa +f$, where $f=|V^{\downarrow}\cap S|$.
$(e2)$ If $f=2$ and $S\not\subseteq V^{\uparrow}$, then either $A^{\downarrow}-V(Q^{\downarrow}_*)$ is an independent set or $|V^{\downarrow}|\geq 3\delta -3\kappa +3$.
$(e3)$ If $f=2$ and $S\subseteq V^{\uparrow}$, then $|V^{\downarrow}|\geq 2\delta -2\kappa +3$.\\
Some extremal cases are considered in the following lemma.\\
\noindent\textbf{Lemma 10}. Let $G$ be a 2-connected graph and $A^{\uparrow}$ is a fragment of $G$.
$(f1)$ If $A^{\uparrow}-V^{\uparrow}$ is an independent non-empty set, then $|V^{\uparrow}|\geq 2\delta +m-2$.
$(f2)$ If $A^{\downarrow}-V^{\downarrow}$ is an independent non-empty set, then $|V^{\downarrow}|\geq 2\delta-2\kappa+2f-m$.
$(f3)$ If $A^{\downarrow}-V(Q^{\downarrow}_*)$ is an independent non-empty set, then $|V^{\downarrow}|\geq 2\delta-2\kappa+5$.\\
For the special case $|A^{\downarrow}|\leq 3\delta - 3\kappa$, we need the following more simple result proved in [11, Lemma 14].\\
\noindent\textbf{Lemma F [11]}. Let $G$ be a 3-connected graph with $\delta\geq2\kappa-2$ and let $|A^{\downarrow}|\leq 3\delta-3\kappa$ for a fragment $A^{\downarrow}$ of $G$. Then $A^{\downarrow}-V^{\downarrow}$ is an independent set.\\
\section{Proofs of lemmas}
\noindent\textbf{Proof of Lemma 1}. Let $x_1,...,x_t$ be the elements of $Z_1\cup Z_2$, occuring on $\overrightarrow{Q}$ in a consecutive order. Consider the segments $I_i=x_i\overrightarrow{Q}x_{i+1}$ $(i=1,...,t)$, where $x_{t+1}=x_1$ and $t=|Z_1\cup Z_2|$. Since $(Z_1,Z_2)$ is a nontrivial $(Q,r)$-scheme, a segment $I_i$ is an $r$-segment if $x_i\in Z_1\cap Z_2$ and is a 2-segment, otherwise. Hence,
$$
|Q|\geq 3|Z_1\cap Z_2|+2(t-|Z_1\cap Z_2|)=|Z_1|+|Z_2|+|Z_1\cup Z_2|.
$$
\textbf{Case 1.} $r=4$.
By Lemma A, $|Q|\geq2(|Z_1|+|Z_2|)$.\\
\textbf{Case 2}. $r\geq5$.
By the hypothesis, $(r-4)(|Z_1|+|Z_2|)\geq2$ which is equivalent to
$$
\frac{1}{2}r(|Z_1|+|Z_2|)\geq2(|Z_1|+|Z_2|)+r-3
$$
and the result follows from Lemma A immediately.\\
\textbf{Case 3}. $|Z_2|=1$.
Clearly at least two of segments $I_1,...,I_t$ are $r$-segments. If $Z_2\subseteq Z_1$, then $t=|Z_1|$ and $|C|\geq2r+2(t-2)=2|Z_1|+2r-4$. If $Z_2\not\subseteq Z_1$, then $t=|Z_1|+1$ and $|Q|\geq2r+2(t-1)>2|Z_1|+2r-4$.\\
\textbf{Case 4}. $|Z_1|=|Z_2|=1$.
The result is trivial.\\
\textbf{Case 5}. $r\geq4$ and $|Z_1|+|Z_2|\geq4$.
By the hypothesis, $(r-4)(|Z_1|+|Z_2|-4)\geq0$ which is equivalent to
$$
\frac{1}{2}r(|Z_1|+|Z_2|)\geq2(|Z_1|+|Z_2|)+2r-8
$$
and the result follows from Lemma A, immediately. \qquad $\Delta$\\
\noindent\textbf{Proof of Lemma 2}. Let $x_1,...,x_t$ be the elements of $Z_1\cup Z_2\cup Z_3$, occuring on $\overrightarrow{Q}$ in a consecutive order. Consider the segments $I_i=x_i\overrightarrow{Q}x_{i+1}$ $(i=1,...,t)$, where $x_{t+1}=x_1$ and $t=|Z_1\cup Z_2\cup Z_3|$. \\
\textbf{Case 1}. $r\geq4$, $|Z_1|+|Z_2|\geq6$ and $|Z_3|=1$.
By the hypothesis, $(r-4)(|Z_1|+|Z_2|-6)\geq0$ which is equivalent to
$$
\frac{1}{2}r(|Z_1|+|Z_2|)\geq2(|Z_1|+|Z_2|)+3r-12
$$
and the result follows from Lemma B, immediately. \\
\textbf{Case 2}. $|Z_2|=|Z_3|=1$.
Clearly at least three of segments $I_1,...,I_t$ are $r$-segments. If $Z_2\cup Z_3\subseteq Z_1$, then $t=|Z_1|$ and $|Q|\geq3r+2(t-3)\geq2|Z_1|+3r-6$. If $Z_2\subseteq Z_1$ and $Z_3\not\subseteq Z_1$ or $Z_3\subseteq Z_1$ and $Z_2\not\subseteq Z_1$, then $t=|Z_1|+1$ and
$$
|Q|\geq3r+2(t-3)=2|Z_1|+3r-4>2|Z_1|+3r-6.
$$
Finally, if $Z_2\not\subseteq Z_1$ and $Z_3\not\subseteq Z_1$, then $t=|Z_1|+2$ and
$$
|Q|\geq3r+2(t-3)=2|Z_1|+3r-2>2|Z_1|+3r-6.
$$
\textbf{Case 3}. $|Z_1|=|Z_2|=|Z_3|=1$.
The result is trivial. \qquad $\Delta$\\
\noindent\textbf{Proof of Lemma 3}. Let $x_1,...,x_t$ be the elements of $Z_1\cup Z_2\cup Z_3\cup Z_4$, occuring on $\overrightarrow{Q}$ in a consecutive order. Consider the segments $I_i=x_i\overrightarrow{Q}x_{i+1}$ $(i=1,...,t)$, where $x_{t+1}=x_1$ and $t=|Z_1\cup Z_2\cup Z_3\cup Z_4|$. \\
\textbf{Case 1}. $r\geq4$, $|Z_1|+|Z_2|\geq8$ and $|Z_3|=|Z_4|=1$.
By the hypothesis, $(r-4)(|Z_1|+|Z_2|-8)\geq0$ which is equivalent to
$$
\frac{1}{2}r(|Z_1|+|Z_2|)\geq2(|Z_1|+|Z_2|)+4r-16
$$
and the result follows from Lemma C, immediately. \\
\textbf{Case 2}. $|Z_i|=1$ (i=2,3,4).
We can argue as in proof of Lemma 2 (Case 2).\\
\textbf{Case 3}. $|Z_i|=1$ (i=1,2,3,4).
The result is trivial. \qquad $\Delta$\\
\noindent\textbf{Proof of Lemma 4}. If $G$ has a dominating cycle, then we are done. Otherwise, using Theorem M and the fact that $\delta\leq2\kappa-3$, we obtain $c\geq 3\delta-3\geq4\delta-2\kappa$. \quad $\Delta$\\
\noindent\textbf{Proofs of Lemmas 5-6}. Let $V^{\uparrow}$ be as defined in Definition A and let $P$ be a longest path in $\langle A^{\uparrow}-V^{\uparrow}\rangle$. If $|V(P)|\leq1$, then clearly $A^{\uparrow}-V^{\uparrow}$ is an independent set. Let $|V(P)|\geq2$. To prove Lemma 5, we can argue exactly as in [11, proof of Lemma 12]. To prove Lemma 6, we can argue exactly as in [11, proof of Lemma 13].\qquad $\Delta$\\
\noindent\textbf{Proof of Lemma 7}. Let $A^{\downarrow}$ is defined with respect to a minimum cut-set $S$ and let $Q^{\downarrow}_{1},...,Q^{\downarrow}_{m}$, $V^{\downarrow}$ and $C$ be as defined in Definition B. Put $f=|V^{\downarrow}\cap S|$. Assume w.l.o.g. that $F(Q^{\uparrow}_i)=v_i$ and $L(Q^{\uparrow}_i)=w_i$ $(i=1,...,m)$. Form a cycle $C^{\downarrow}$ by deleting $Q^{\uparrow}_{1},...,Q^{\uparrow}_{m}$ from $C$ and adding extra edges $v_1w_1,v_2w_2,...,v_mw_m$, respectively. Clearly $V^{\downarrow}=V(C^{\downarrow})$. Conversely, suppose that $A^{\downarrow}-V^{\downarrow}\neq\emptyset$, and choose a connected component $H$ of $\langle A^{\downarrow}-V^{\downarrow}\rangle$. Put $M=N(V(H))$. If $|M|=\kappa$, then $M$ is a minimum cut-set of $G$ with $M\subset A^{\downarrow}\cup S$ and $M\neq S$, contradicting the definition of $A^{\downarrow}$. Let $|M|\geq\kappa+1$. Clearly $|M\cap (S-V^{\downarrow})|\leq \kappa-f$. Since $M\subseteq V(C^{\downarrow})\cup S$, we have
$$
|M\cap V(C^{\downarrow})|=|M|-|M\cap (S-V^{\downarrow})|\geq f+1.
$$
Recalling that $|V(C^{\downarrow})\cap S|=f$, we have $x^+\notin S$ for some $x\in M\cap V(C^{\downarrow})$. Then by deleting $xx^+$ from $C^{\downarrow}$ and adding an edge $xy$ with $y\in V(H)$, we get a path $R=x^+\overrightarrow{C}^{\downarrow}xy$ with endvertices in $A^{\downarrow}$. Choose a maximal path $R^*=\xi R^*\eta$ in $\langle A^{\downarrow}\cup V^{\downarrow}\rangle$ containing $R$ as a subpath with $\xi ,\eta\in A^{\downarrow}$ and put
$$
d_1=|N(\xi)\cap V(R^*)|, \quad d_2=|N(\eta)\cap V(R^*)|.
$$
Since $R^{*}$ is extreme, we have $d_i\geq\delta-\kappa+f$ $(i=1,2)$ and
$$
|V(R^*)|\leq|A^{\downarrow}|+f\leq2\delta-2\kappa+f+1\leq d_1+d_2-f+1,
$$
\noindent implying that $d_1+d_2\geq|V(R^*)|+f-1\geq|V(R^*)|+m$. By standard arguments, there is a vertex $\lambda\in V(\overrightarrow{R}^*)$ such that $\xi\lambda^+\in E(G)$, $\eta\lambda\in E(G)$ and $\lambda\lambda^+\not\in\{v_1w_1,...,v_mw_m\}$. Then by deleting $\lambda\lambda^+$ and adding $\xi\lambda^+$, $\eta\lambda$, we can form a new collection of paths instead of $Q^{\downarrow}_{1},...,Q^{\downarrow}_{m}$, contrary to $(*2)$. So, $A^{\downarrow}-V^{\downarrow}=\emptyset$. By the same way it can be shown that $A^{\downarrow}-V(Q^{\downarrow}_*)=\emptyset$. \qquad $\Delta$ \\
\noindent\textbf{Proof of Lemma 8}. Let $Q^{\downarrow}_{1},...,Q^{\downarrow}_{m}$ and $V^{\downarrow}_{1},...,V^{\downarrow}_{m},V^{\downarrow}$ are as defined in Definition B. The existence of $Q^{\downarrow}_{1},...,Q^{\downarrow}_{m}$ follows from Lemma D. Clearly $f\geq2m$. If $|A^{\downarrow}|\leq2\delta-2\kappa+1$, then by Lemma 7, we are done. Let $|A^{\downarrow}|\geq2\delta-2\kappa+2$\\
\textbf{Case 1}. $f\geq3$.
Suppose first that $\delta-\kappa\leq1$. Combining this with $\delta\geq2\kappa-2$, we get $f=\kappa=3$, $\delta=4$, $m=1$ and $|A^{\downarrow}|\leq4$. If $A^{\downarrow}-V^{\downarrow}$ is an independent set, then we are done. Otherwise, $|V^{\downarrow}|\leq |A^{\downarrow}|+f-2\leq5$. Choose a vertex $z\in A^{\downarrow}-V^{\downarrow}$. If $z$ and $Q^{\downarrow}_{1}$ are connected by at most three paths having no vertex other than $z$ in common, then $G$ has a cut-set $S^{\prime}$ of order three with $S^{\prime}\subset A^{\downarrow}\cup S$ and $S^{\prime}\neq S$, contradicting the definition of $A^{\downarrow}$. Otherwise, it is easy to see that $|V^{\downarrow}|\geq7$, a contradiction. So, we can assume that $\delta-\kappa\geq2$. Let $P=y_1...y_p$ be a longest path in $\langle A^{\downarrow}-V^{\downarrow}\rangle$. By the hypothesis,
$$
p+|V^{\downarrow}|-f\leq|A^{\downarrow}|\leq3\delta-3\kappa+1,
$$
implying that
$$
|V^{\downarrow}|\leq3\delta-3\kappa+f-p+1. \eqno{(1)}
$$
Put
$$
Z_1=N(y_1)\cap V^{\downarrow},\quad Z_2=N(y_p)\cap V^{\downarrow},
$$
$$
Z_{1,i}=Z_1\cap V^{\downarrow}_{i},\quad Z_{2,i}=Z_2\cap V^{\downarrow}_{i}\quad (i=1,...,m).
$$
Clearly $Z_1=\cup^{m}_{i=1}Z_{1,i}$ and $Z_2=\cup^{m}_{i=1}Z_{2,i}$. If $p\leq1$, then $A^{\downarrow}-V^{\downarrow}$ is independent and we are done. Let $p\geq2$.\\
\textbf{Case 1.1}. $p=2$.
Clearly $|Z_i|\geq\delta-\kappa+f-1$ $(i=1,2)$. For each $i\in \{1,...,m\}$, form a cycle $C^{\downarrow}_i$ by adding to $Q^{\downarrow}_i$ an extra path of length 3 connecting $F(Q^{\downarrow}_i)$ and $L(Q^{\downarrow}_i)$. Since $\{Q^{\downarrow}_{1},...,Q^{\downarrow}_{m}\}$ is $(*2)$-extreme, $(Z_{1,i},Z_{2,i})$ is a $(C^{\downarrow}_{i},3)$-scheme. If $(Z_{1,i},Z_{2,i})$ is a nontrivial $(C^{\downarrow}_{i},3)$-scheme, then by Lemma 1, $|C^{\downarrow}_{i}|\geq2|Z_{1,i}|+|Z_{2,i}|$. Otherwise, it can be checked easily. Hence,
$$
|V^{\downarrow}|=\sum^{m}_{i=1}|V^{\downarrow}_{i}|=\sum^{m}_{i=1}(|C^{\downarrow}_{i}|-2)\geq\sum^{m}_{i=1}|C^{\downarrow}_{i}|-2m
$$
$$
\geq2|Z_1|+|Z_2|-2m\geq(3\delta-3\kappa+f)+2f-2m-3.
$$
Since $f\geq2m$ and $f\geq3$, we have $2f-2m-3\geq0$, implying that $|V^{\downarrow}|\geq3\delta-3\kappa+f$, contrary to (1).\\
\textbf{Case 1.2.} $p=3$.
Clearly $|Z_i|\geq\delta-\kappa+f-2$ $(i=1,2)$. For each $i\in \{1,...,m\}$ form a cycle $C^{\downarrow}_i$ by adding to $Q^{\downarrow}_i$ an extra path of length 4, connecting $F(Q^{\downarrow}_i)$ and $L(Q^{\downarrow}_i)$. Since $\{Q^{\downarrow}_{1},...,Q^{\downarrow}_{m}\}$ is $(*2)$-extreme, $(Z_{1,i},Z_{2,i})$ is a $(C^{\downarrow}_{i},4)$-scheme. If $(Z_{1,i},Z_{2,i})$ is a nontrivial $(C^{\downarrow}_{i},4)$-scheme, then by $(a1)$ (see Lemma 1), $|C^{\downarrow}_{i}|\geq2(|Z_{1,i}|+|Z_{2,i}|)$. Otherwise, it can be checked easily. Hence,
$$
|V^{\downarrow}|=\sum^{m}_{i=1}|V^{\downarrow}_{i}|=\sum^{m}_{i=1}(|C^{\downarrow}_{i}|-3)\geq\sum^{m}_{i=1}|C^{\downarrow}_{i}|-3m
$$
$$
\geq2(|Z_1|+|Z_2|)-3m\geq(3\delta-3\kappa+f)+3f-3m-6.
$$
If $f=3$, then $m=1$ and $3f-3m-6=0$. If $f=4$, then $m\leq2$ and $3f-3m-6\geq0$. Finally, if $f\geq5$, then again $3f-3m-6\geq f+m-6\geq0$. So, in any case, $|V^{\downarrow}|\geq3\delta-3\kappa+f$, contrary to (1).\\
\textbf{Case 1.3.} $p\geq4$.
Let $w_1,w_2,...,w_s$ be the elements of $(N(y_p)\cap V(P))^+$ occuring on $\overrightarrow{P}$ in a consecutive order, where $w_s=y_p$. Put $P_0=w^-_1\overrightarrow{P}w_s$ and $p_0=|V(P_0)|$. For each $w_i$ $(i\in \{1,...,s\})$ there is a path $y_1\overrightarrow{P}w^{-}_{i}w_s\overleftarrow{P}w_i$ in $\langle V(P)\rangle$ of length $p$ connecting $y_1$ and $w_i$. Hence, we can assume w.l.o.g. that for each $i\in \{1,...,s\}$, $P$ is chosen such that
$$
|Z_1|\geq|N(w_i)\cap V^{\downarrow}|,\quad N(w_i)\cap V(P)\subseteq V(P_0).
$$
Choose $w\in\lbrace w_1,...,w_s\rbrace$ as to maximize $|N(w_i)\cap V^{\downarrow}|$, $i=1,...,s$. Set
$$
Z_3=N(w)\cap V^{\downarrow}, \quad Z_{3,i}=Z_3\cap V^{\downarrow}_{i} \quad (i=1,...,m).
$$
Clearly
$$
|Z_1|\geq|Z_3|\geq|Z_2|\geq\delta-\kappa+f-p_0+1. \eqno{(2)}
$$
\textbf{Claim 1.} $|V^{\downarrow}_{i}|\geq2(|Z_{1,i}|+|Z_{3,i}|)-2$ $(i=1,...,m)$.
Proof. For each $i\in \{1,...,m\}$, form a cycle $C^{\downarrow}_{i}$ by adding to $Q^{\downarrow}_{i}$ an extra path of length $p+1$ connecting $F(Q^{\downarrow}_i)$ and $L(Q^{\downarrow}_i)$. Since $\{Q^{\downarrow}_{1},...,Q^{\downarrow}_{m}\}$ is $(*2)$-extreme, $(Z_{1,i},Z_{3,i})$ is a $(C^{\downarrow}_{i},p+1)$-scheme. If $(Z_{1,i},Z_{3,i})$ is a nontrivial $(C^{\downarrow}_{i},p+1)$-scheme, then by $(a2)$ (see Lemma 1), $|C^{\downarrow}_{i}|\geq2(|Z_{1,i}|+|Z_{3,i}|)+p-2$. Otherwise, it can be checked easily. Observing that $|V^{\downarrow}_{i}|=|C^{\downarrow}_{i}|-p$ $(i=1,...,m)$, we get the desired result immediately. \qquad $\Delta$\\
By Claim 1,
$$
|V^{\downarrow}|=\sum^{m}_{i=1}|V^{\downarrow}_{i}|\geq2(|Z_1|+|Z_3|)-2m\geq4(\delta-\kappa+f-p_0+1)-2m
$$
$$
=(3\delta-3\kappa+f)+(\delta-\kappa-2)+4(m-p_0+1)+3(f-2m)+2
$$
$$
>(3\delta-3\kappa+f)+4(m-p_0+1).
$$
If $p_0\leq m+1$, then clearly $|V^{\downarrow}|\geq3\delta-3\kappa+f$, contrary to (1). So, we can assume that
$$
p_0 \geq m+2. \eqno{(3)}
$$
Further, assume that $\delta-\kappa+f-|Z_3|\leq2$, implying that $|Z_1|\geq|Z_3|\geq\delta-\kappa+f-2$. By Claim 1,
$$
|V^{\downarrow}|=\sum^{m}_{i=1}|V^{\downarrow}_{i}|\geq2(|Z_1|+|Z_3|)-2m\geq4(\delta-\kappa+f-2)-2m
$$
$$
=(3\delta-3\kappa+f)+(\delta-\kappa-2)+3f-2m-6\geq3\delta-3\kappa+f,
$$
contrary to (1). Now let $\delta-\kappa+f-|Z_3|\geq3$. By the choice of $w$,
$$
|N(w_i)\cap V(P_0)|\geq\delta-\kappa+f-|Z_3|\geq3 \quad (i=1,...,s).
$$
In particular, for $i=s$, we have $s\geq\delta-\kappa+f-|Z_3|$. By Lemma E, in $\langle V(P_0)\rangle$ any two vertices are joined by a path of length at least $\delta-\kappa+f-|Z_3|$. Observing that $p\geq s+1\geq \delta-\kappa+f-|Z_3|+1$ and combining it with (1), we get
$$
|V^{\downarrow}|\leq2\delta-2\kappa+|Z_3|. \eqno{(4)}
$$
For each $i\in \{1,...,m\}$, form a cycle $C^{\downarrow}_{i}$ by adding to $Q^{\downarrow}_{i}$ an extra path of length $r$, where $r=\delta-\kappa+f-|Z_3|+2\geq5$, connecting $F(Q^{\downarrow}_i)$ and $L(Q^{\downarrow}_i)$. Let $G^{\prime}$ be the subgraph of $G$ obtained from $G$ by deleting all vertices of $S-V^{\downarrow}$. Clearly $G^{\prime}$ is $f$-connected. If $m=1$, then $f\geq3=m+2$. If $m\geq2$, then again $f\geq2m\geq m+2$. So, in any case, $G^{\prime}$ is $(m+2)$-connected. Since $|V^{\downarrow}|>f\geq m+2$ and $|V(P_0)|=p_0\geq m+2$ (by (3)), there are vertex disjoint paths $R_1,...,R_{m+2}$ in $G^{\prime}$ connecting $V(P_0)$ and $V^{\downarrow}$. Let $F(R_i)\in V(P_0)$ and $L(R_i)\in V^{\downarrow}$ $(i=1,...,m+2)$. Then we can assume w.l.o.g. that either $L(R_i)\in V^{\downarrow}_{1}$ $(i=1,2)$ and $L(R_i)\in V^{\downarrow}_{2}$ $(i=3,4)$ or $L(R_i)\in V^{\downarrow}_{1}$ $(i=1,2,3)$. \\
\textbf{Claim 2.} Let $\tau\in\{1,...,m\}$ and $a,b\in\{1,...,m+2\}$. If $L(R_i)\in V^{\downarrow}_{\tau}$ $(i=a,b)$, then $|V^{\downarrow}_{\tau}|\geq2(|Z_{1,\tau}|+|Z_{3,\tau}|)+r-7$.
Proof. Suppose first that $|Z_{1,\tau}|\leq2$, $|Z_{3,\tau}|\leq2$. Clearly $(\{L(R_a)\},\{L(R_b)\})$ is a nontrivial $(C^{\downarrow}_{\tau},r)$-scheme. By $(a4)$ (see Lemma 1), $|C^{\downarrow}_{\tau}|\geq2r$, implying that
$$
|V^{\downarrow}_{\tau}|=|C^{\downarrow}_{\tau}|-r+1\geq r+1\geq 2(|Z_{1,\tau}|+|Z_{3,\tau}|)+r-7.
$$
Now let $|Z_{1,\tau}|\geq3$ and $|Z_{3,\tau}|\leq2$. If either $R_a$ or $R_b$, say $R_a$, has no common vertices with $y_1\overrightarrow{P}w^-_1$, then $(Z_{1,\tau},\{L(R_a)\})$ is a nontrivial $(C^{\downarrow}_{\tau},r)$-scheme. Otherwise, let $t$ be the smallest integer such that $y_t\in V(R_a\cup R_b)$. Assume w.l.o.g. that $y_t\in V(R_b)$. Then due to $y_1\overrightarrow{P}y_tR_bF(R_b)$, we again can state that $(Z_{1,\tau},\{L(R_a)\})$ is a nontrivial $(C^{\downarrow}_{\tau},r)$-scheme. By $(a3)$ (see Lemma 1), $|C^{\downarrow}_{\tau}|\geq2|Z_{1,\tau}|+2r-4$, implying that
$$
|V^{\downarrow}_{\tau}|=|C^{\downarrow}_{\tau}|-r+1\geq 2|Z_{1,\tau}|+r-3\geq 2(|Z_{1,\tau}|+|Z_{3,\tau}|)+r-7.
$$
Finally, we suppose that $|Z_{1,\tau}|\geq3$ and $|Z_{3,\tau}|\geq3$. It means that $(Z_{1,\tau},Z_{3,\tau})$ is a nontrivial $(C^{\downarrow}_{\tau},r)$-scheme. By $(a5)$ (see Lemma 1), $|C^{\downarrow}_{\tau}|\geq2(|Z_{1,\tau}|+|Z_{3,\tau}|)+2r-8$, implying that
$$
|V^{\downarrow}_{\tau}|=|C^{\downarrow}_{\tau}|-r+1\geq 2(|Z_{1,\tau}|+|Z_{3,\tau}|)+r-7. \qquad \Delta\\
$$
\textbf{Claim 3.} Let $\tau\in\{1,...,m\}$ and $a,b,c\in\{1,...,m+2\}$. If $L(R_i)\in V^{\downarrow}_{\tau}$ $(i=a,b,c)$, then $|V^{\downarrow}_{\tau}|\geq2(|Z_{1,\tau}|+|Z_{3,\tau}|)+2r-11$.
Proof. Suppose first that $|Z_{1,\tau}|\leq3$ and $|Z_{3,\tau}|\leq3$. Clearly $\{L(R_a)\}$, $\{L(R_b)\}$ and $\{L(R_c)\}$ form a nontrivial $(C^{\downarrow}_{\tau},r)$-scheme. By $(b3)$ (see Lemma 2), $|C^{\downarrow}_{\tau}|\geq3r$ and therefore,
$$
|V^{\downarrow}_{\tau}|=|C^{\downarrow}_{\tau}|-r+1\geq 2r+1\geq 2(|Z_{1,\tau}|+|Z_{3,\tau}|)+2r-11.
$$
Now let $|Z_{1,\tau}|\geq4$ and $|Z_{3,\tau}|\leq3$. If at least two of $R_a$, $R_b$, $R_c$, say $R_a$ and $R_b$, have no common vertices with $y_1\overrightarrow{P}w^-_1$, then $Z_{1,\tau}$, $\{L(R_a)\}$ and $\{L(R_b)\}$ form a nontrivial $(C^{\downarrow}_{\tau},r)$-scheme. Otherwise, let $t$ be the smallest integer such that $y_t\in V(R_a\cup R_b\cup R_c)$. Assume w.l.o.g. that $y_t\in V(R_c)$. Then due to $y_1\overrightarrow{P}y_tR_cF(R_c)$, we again can state that $Z_{1,\tau}$, $\{L(R_a)\}$ and $\{L(R_b)\}$ form a nontrivial $(C^{\downarrow}_{\tau},r)$-scheme. By $(b2)$ (see Lemma 2), $|C^{\downarrow}_{\tau}|\geq2|Z_{1,\tau}|+3r-6$, implying that
$$
|V^{\downarrow}_{\tau}|=|C^{\downarrow}_{\tau}|-r+1\geq 2|Z_{1,\tau}|+2r-5\geq 2(|Z_{1,\tau}|+|Z_{3,\tau}|)+2r-11.
$$
Finally, let $|Z_{1,\tau}|\geq4$ and $|Z_{3,\tau}|\geq4$. If at least two of $R_a$, $R_b$, $R_c$, say $R_a$ and $R_b$, have no common vertices with $y_1\overrightarrow{P}w^-_1$, then choosing one of $R_a$, $R_b$, say $R_a$, satisfying $F(R_a)\neq w$, we conclude that $Z_{1,\tau}$, $Z_{3,\tau}$ and $\{L(R_a)\}$ form a nontrivial $(C^{\downarrow}_{\tau},r)$-scheme. Otherwise, let $t$ be the smallest integer such that $y_t\in V(R_a\cup R_b\cup R_c)$. Assume w.l.o.g. that $y_t\in V(R_c)$. Choosing one of $R_a$, $R_b$, say $R_a$, satisfying $F(R_a)\neq w$, we again can state that $(Z_{1,\tau},Z_{3,\tau},\{L(R_a)\})$ is a nontrivial $(C^{\downarrow}_{\tau},r)$-scheme. By $(b1)$ (see Lemma 2), $|C^{\downarrow}_{\tau}|\geq2(|Z_{1,\tau}|+|Z_{3,\tau}|)+3r-12$, implying that
$$
|V^{\downarrow}_{\tau}|=|C^{\downarrow}_{\tau}|-r+1\geq 2(|Z_{1,\tau}|+|Z_{3,\tau}|)+2r-11. \qquad \Delta
$$
\textbf{Claim 4.} Let $\tau\in\{1,...,m\}$ and $a,b,c,d\in\{1,...,m+3\}$. If $L(R_i)\in V^{\downarrow}_{\tau}$ $(i=a,b,c,d)$, then $|V^{\downarrow}_{\tau}|\geq2(|Z_{1,\tau}|+|Z_{3,\tau}|)+3r-15$.
Proof. Suppose, that $|Z_{1,\tau}|\leq4$, $|Z_{3,\tau}|\leq4$. Clearly $\{L(R_a)\}$, $\{L(R_b)\}$, $\{L(R_c)\}$ and $\{L(R_d)\}$ form a nontrivial $(C^{\downarrow}_{\tau},r)$-scheme. By $(c3)$ (see Lemma 3), $|C^{\downarrow}_{\tau}|\geq4r$, implying that
$$
|V^{\downarrow}_{\tau}|=|C^{\downarrow}_{\tau}|-r+1\geq 3r+1\geq 2(|Z_{1,\tau}|+|Z_{3,\tau}|)+3r-15.
$$
Now let $|Z_{1,\tau}|\geq5$ and $|Z_{3,\tau}|\leq4$. If at least three of $R_a$, $R_b$, $R_c$, $R_d$ say $R_a$, $R_b$, $R_c$, do not intersect $P_0$, then
$Z_{1,\tau}$, $\{L(R_a)\}$, $\{L(R_b)\}$ and $\{L(R_c)\}$ form a nontrivial $(C^{\downarrow}_{\tau},r)$-scheme. Otherwise, let $t$ be the smallest integer such that $y_t\in V(R_a\cup R_b\cup R_c\cup R_d)$. Assume w.l.o.g. that $y_t\in V(R_d)$. Then due to $y_1\overrightarrow{P}y_tR_dF(R_d)$, we again can state that $Z_{1\tau}$, $\{L(R_a)\}$, $\{L(R_b)\}$ and $\{L(R_c)\}$ form a nontrivial $(C^{\downarrow}_{\tau},r)$-scheme. By $(c2)$ (see Lemma 3), $|C^{\downarrow}_{\tau}|\geq2|Z_{1,\tau}|+4r-8$, implying that
$$
|V^{\downarrow}_{\tau}|=|C^{\downarrow}_{\tau}|-r+1\geq 2|Z_{1,\tau}|+3r-7\geq 2(|Z_{1,\tau}|+|Z_{3,\tau}|)+3r-15.
$$
Finally, let $|Z_{1,\tau}|\geq5$ and $|Z_{3,\tau}|\geq5$. If at least three of $R_a$, $R_b$, $R_c$, $R_d$ say $R_a$, $R_b$, $R_c$, do not intersect $P_0$, then choosing two of $R_a$, $R_b$, $R_c$, say $R_a$, $R_b$, such that $F(R_a)\neq w$ and $F(R_b)\neq w$, we conclude that $Z_{1,\tau}$, $Z_{3,\tau}$, $\{L(R_a)\}$ and $\{L(R_b)\}$ form a nontrivial $(C^{\downarrow}_{\tau},r)$-scheme. Otherwise, let $t$ be the smallest integer such that $y_t\in V(R_a\cup R_b\cup R_c\cup R_d)$. Assume w.l.o.g. that $y_t\in V(R_d)$. Then choosing two of $R_a$, $R_b$, $R_c$, say $R_a$, $R_b$, satisfying $F(R_a)\neq w$ and $F(R_a)\neq w$, we again can state that $Z_{1\tau}$, $Z_{3\tau}$, $\{L(R_a)\}$ and $\{L(R_b)\}$ form a nontrivial $(C^{\downarrow}_{\tau},r)$-scheme. By $(c1)$ (see Lemma 3), $|C^{\downarrow}_{\tau}|\geq2(|Z_{1,\tau}|+|Z_{3,\tau}|)+4r-16$, implying that
$$
|V^{\downarrow}_{\tau}|=|C^{\downarrow}_{\tau}|-r+1\geq 2(|Z_{1,\tau}|+|Z_{3,\tau}|)+3r-15. \qquad \Delta
$$
\textbf{Case 1.3.1}. $L(R_i)\in V^{\downarrow}_{1}$ $(i=1,2)$ and $L(R_i)\in V^{\downarrow}_{2}$ $(i=3,4)$.
It follows that $f\geq4$. By Claim 2, $|V^{\downarrow}_{i}|\geq2(|Z_{1,i}|+|Z_{3,i}|)+r-7$ $(i=1,2)$. By Claim 1, $|V^{\downarrow}_{i}|\geq2(|Z_{1,i}|+|Z_{3,i}|)-2$ $(i=3,...,m)$. By summing, we get \\
$$
|V^{\downarrow}|=|V^{\downarrow}_{1}|+|V^{\downarrow}_{2}|+\sum^{m}_{i=3}|V^{\downarrow}_{i}|\geq2(|Z_1|+|Z_3|)+2r-14-2(m-2)
$$
$$
=(2\delta-2\kappa+|Z_3|+1)+f+|Z_3|-7\geq2\delta-2\kappa+|Z_3|+1,
$$
contrary to (4).\\
\textbf{Case 1.3.2}. $L(R_i)\in V^{\downarrow}_{1}$ $(i=1,2,3)$.
By Claim 3, $|V^{\downarrow}_{1}|\geq2(|Z_{1,1}|+|Z_{3,1}|)+2r-11$. By Claim 1, $|V^{\downarrow}_{i}|\geq2(|Z_{1,i}|+|Z_{3,i}|)-2$ $(i=2,...,m)$. By summing, we get
$$
|V^{\downarrow}|\geq2(Z_1|+|Z_3|)+2r-11-2(m-1)
$$
$$
\geq(2\delta-2\kappa+|Z_3|+1)+f+|Z_3|-6\geq2\delta-2\kappa+|Z_3|+1,
$$
contrary to (4).\\
\textbf{Case 2}. $f=2$ and $S\not\subseteq V^{\uparrow}$.
Choose a vertex $z\in S-V^{\uparrow}$. Put $F(Q^{\downarrow}_1)=v_1$ and $L(Q^{\downarrow}_1)=w_1$. If there is a cut-vertex $x$ in $\langle A^{\downarrow}\cup\{v_1,w_1,z\}\rangle$ that separates $z$ and $Q^{\downarrow}_{1}$, then $S^{\prime }=\{S-z\}\cup \{x\}$ is a minimum cut-set of $G$ other than $S$ with $S^{\prime}\subset A^{\downarrow}\cup S$, contradicting the definition of $A^{\downarrow}$. Otherwise, the existence of $Q^{\downarrow}_{*}$ (see Definition B) follows easily. Put $g=|V(Q^{\downarrow}_{*})\cap S|$. By the definition, $g\geq3$. As in Case 1, we can assume that $\delta-\kappa\geq2$. Let $P=y_1...y_p;Z_1;Z_2$ be as defined in Case 1. If $p\leq1$, then $A^{\downarrow}-V(Q^{\downarrow}_{*})$ is independent. Let $p\geq2$. Put $S_0=(V(Q^{\downarrow}_{*})\cap S)-\{v_1,w_1\}$.\\
\textbf{Case 2.1.} $p=2$.
Clearly $|Z_i|\geq\delta-\kappa+g-1\geq4$ $(i=1,2)$. Form a cycle $C^{\downarrow}_{1}$ by adding to $Q^{\downarrow}_{*}$ an extra path of length 3 with endvertices $v_1,w_1$. Let $I_1,...,I_t$ be the segments of $C^{\downarrow}_{1}$ having only their ends in common with $Z_1\cup Z_2$ and having at least one inner vertex in common with $S$. \\
\textbf{Case 2.1.1.} $S_0\not\subseteq V(I_i)-\{F(I_i),L(I_i)\}$ $(i=1,...,t)$.
It is easy to see that $(Z_1,Z_2)$ is a nontrivial $(C^{\downarrow}_{1},3)$-scheme. By Lemma 1,
$$
|V^{\downarrow}|\geq|V(Q^{\downarrow}_{*})|=|C^{\downarrow}_{1}|-2\geq2|Z_1|+|Z_2|-2\geq3\delta-3\kappa+3g-5.
$$
Hence,
$$
|A^{\downarrow}|\geq|V^{\downarrow}|-g+2\geq(3\delta-3\kappa+2)+2g-5>3\delta-3\kappa+2,
$$
contradicting the hypothesis.\\
\textbf{Case 2.1.2.} $S_0\subseteq V(I_i)-\{F(I_i),L(I_i)\}$ for some $i\in \{1,...,t\}$, say $i=1$.
Let $I_1=x\overrightarrow{I_1}y$. Denote by $R$ a longest path connecting $x$ and $y$ and passing through $V(P)$. Clearly $|R|=3$ if $x$ and $y$ belong to different $Z_1,Z_2$ and $|R|=2$, otherwise. Form a new cycle $C^{\downarrow}_{2}$ by deleting $I_1$ from $C^{\downarrow}_{1}$ and adding $R$. Clearly $(Z_1,Z_2)$ is a nontrivial $(C^{\downarrow}_{2},3)$-scheme and we can reach a contradiction as in Case 2.1.1.\\
\textbf{Case 2.2.} $p=3$.
Clearly $|Z_i|\geq\delta-\kappa+g-2\geq3$ $(i=1,2)$. Form a cycle $C^{\downarrow}_{1}$ by adding to $Q^{\downarrow}_{*}$ an extra path of length 4 with endvertices $v_1,w_1$. Let $I_1,...,I_t$ be as defined in Case 2.1.\\
\textbf{Case 2.2.1.} $S_0\not\subseteq V(I_i)-\{F(I_i),L(I_i)\}$ $(i=1,...,t)$.
It is easy to see that $(Z_1,Z_2)$ is a nontrivial $(C^{\downarrow}_{1},4)$-scheme. By Lemma 1,
$$
|V^{\downarrow}|\geq|V(Q^{\downarrow}_{*})|=|C^{\downarrow}_{1}|-3\geq2|Z_1|+|Z_2|-3\geq4\delta-4\kappa+4g-11.
$$
Hence,
$$
|A^{\downarrow}|\geq|V^{\downarrow})|-g+3\geq(3\delta-3\kappa+2)+\delta-\kappa+3g-10>3\delta-3\kappa+2,
$$
contradicting the hypothesis.\\
\textbf{Case 2.2.2.} $S_0\subseteq V(I_i)-\{F(I_i),L(I_i)\}$ for some $i\in \{1,...,t\}$, say $i=1$.
Let $I_1=x\overrightarrow{I_1}y$. Denote by $R$ a longest path connecting $x$ and $y$ and passing through $V(P)$. Clearly $|R|=4$ if $x$ and $y$ belong to different $Z_1,Z_2$ and $|R|\geq2$, otherwise. Form a new cycle $C^{\downarrow}_{2}$ by deleting $I_1$ from $C^{\downarrow}_{1}$ and adding $R$. Clearly $(Z_1,Z_2)$ is a nontrivial $(C^{\downarrow}_{2},4)$-scheme and we can reach a contradiction as in Case 2.2.1.\\
\textbf{Case 2.3.} $p\geq4$.
Let $P_0,p_0,w$ and $Z_3$ are as defined in Case 1.3. By the definition, $|Z_1|\geq|Z_3|\geq|Z_2|$ and $p_0\geq2$. If $p_0=2$, then $|Z_1|\geq|Z_2|\geq\delta-\kappa+g-1$ and we can argue as in case $p=2$. If $p_0=3$, then $|Z_1|\geq|Z_2|\geq\delta-\kappa+g-2$ and we can argue as in case $p=3$. Let $p_0\geq4$. Further, if $\delta-\kappa+g-|Z_3|\leq1$, then $|Z_1|\geq|Z_3|\geq\delta-\kappa+g-1$ and we can argue as in case $p=2$. So, we can assume that $\delta-\kappa+g-|Z_3|\geq2$. Since $p_0\geq4$, there are vertex disjoint paths $R_1,R_2,R_3,R_4$ in $\langle A^{\downarrow}\cup V(Q^{\downarrow}_{*}) \rangle$ connecting $P_0$ and $Q^{\downarrow}_{*}$ (otherwise, there exist a cut-set of $\langle A^{\downarrow}\cup V(Q^{\downarrow}_{*}) \rangle$ of order at most three, contradicting the definition of $A^{\downarrow}$). Let $F(R_i)\in V(P)$ and $L(R_i)\in V(Q^{\downarrow}_{*})$ $(i=1,2,3,4)$. By the choice of $w$, $|N(w_i)\cap V(P_0)|\geq\delta-\kappa+g-|Z_3|$ for each $i\in\{1,...,s\}$. In particular, for $i=s$, we have $s\geq\delta-\kappa+g-|Z_3|$. By Lemma E, in $\langle V(P_0)\rangle$ any two vertices are joined by a path of length at least $\delta-\kappa+g-|Z_3|$. Let $I_1,...,I_t$ be the segments of $Q^{\downarrow}_{*}$ having only their ends in common with $Z_1\cup Z_3\cup Z_4\cup Z_5 $, where $Z_4=\{L(R_1)\}$ and $Z_5=\{L(R_2)\}$. Form a cycle $C^{\downarrow}_{1}$ by adding to $Q^{\downarrow}_{*}$ an extra path of length $r$, where $r=\delta-\kappa+g-|Z_3|+2$, connecting $v_1$ and $w_1$. \\
\textbf{Case 2.3.1.} At least two of segments $I_1,...,I_t$ intersect $S_0$.
In this case we can apply Claim 4, which gives
$$
|V^{\downarrow}|\geq|V(Q^{\downarrow}_{*})|\geq2(|Z_1|+|Z_3|)+3r-15
$$
$$
=(3\delta-3\kappa+2)+(|Z_1|-|Z_3|)+3g+|Z_1|-11.
$$
Since $|A^{\downarrow}|\geq|V^{\downarrow}|-g+p$, we have
$$
|A^{\downarrow}|\geq(3\delta-3\kappa+2)+(|Z_1|-|Z_3|)+2g+p+|Z_1|-11>3\delta-3\kappa+2,
$$
contradicting the hypothesis.\\
\textbf{Case 2.3.2}. Only one of segments $I_1,...,I_t$, say $I_1$, intersects $S_0$.
In this case, each vertex of $S_0$ is an inner vertex for $I_1$. Form a path $L$ by deleting $I_1$ from $Q^{\downarrow}_{*}$ and adding a longest path connecting $F(I_1)$ and $L(I_1)$ and passing through $V(P)$. Then we can apply Claim 4 with respect to $L$ and can reach a contradiction as in Case 2.3.1.\\
\textbf{Case 3.} $f=2$ and $S\subseteq V^{\uparrow}$.
If $A^{\downarrow}\subseteq V^{\downarrow}$, then we are done. Let $A^{\downarrow}\not\subseteq V^{\downarrow}$ and let $P=y_1...y_p$; $P_0$, $p_0$, $w$ and $Z_1$, $Z_2$, $Z_3$ are as defined in Case 1. If $p=1$, then $|Z_1|\geq\delta-\kappa+2$ and by standard arguments, $|V^{\downarrow}|\geq2|Z_1|-1\geq2\delta-2\kappa+3$. Let $p\geq2$, implying in particular that $p_0\geq2$. If $p_0=2$, then $|Z_1|\geq|Z_2|\geq\delta-\kappa+1$ and we can argue as in Case 1.1. Let $p_0\geq3$. As above, there are vertex disjoint paths $R_1,R_2,R_3$ connecting $V^{\downarrow}$ and $V(P_0)$. Let $F(R_i)\in V(P_0)$ and $L(R_i)\in V^{\downarrow}$ $(i=1,2,3)$. If $\delta-\kappa-|Z_3|+2\leq1$, then $|Z_1|\geq|Z_3|\geq\delta-\kappa+1$ and we can argue as in the case $p_0\leq2$. Let $\delta-\kappa-|Z_3|+2\geq2$. By the choice of $w$, $|N(w^-_i)\cap V(P_0)|\geq\delta-\kappa+2-|Z_3|$ $(i=1,...,s)$. In particular, for $i=s$, we have $s\geq\delta-\kappa-|Z_3|+2$. By Lemma E, in $\langle V(P_0)\rangle$ any two vertices are joined by a path of length at least $\delta-\kappa-|Z_3|+2$. By Claim 3,
$$
|V^{\downarrow}|\geq2(|Z_1|+|Z_3|)+2r-11\geq2\delta-2\kappa+3.\qquad \Delta
$$
\noindent\textbf{Proof of Lemma 9}. Let $Q^{\downarrow}_{1},...,Q^{\downarrow}_{m}$ and $V^{\downarrow}_{1},...,V^{\downarrow}_{m}$, $V^{\downarrow}$ be as defined in Definition B. The existence of $Q^{\downarrow}_{1},...,Q^{\downarrow}_{m}$ follows from Lemma D. Clearly $f\geq2m$. Further, let $P=y_1...y_p$, $Z_1$, $Z_2$ and $Z_{1,i}$, $Z_{2,i}$ $(i=1,...,m)$ be as defined in proof of Lemma 8 (Case 1). \\
\textbf{Case 1}. $f\geq3$.
Assume first that $\delta-\kappa\leq1$. By combining this with $\delta\geq2\kappa-2$, we get $f=\kappa=3$, $\delta=4$ and $m=1$. If $A^{\downarrow}\subseteq V^{\downarrow}$, then we are done. Let $A^{\downarrow}\not\subseteq V^{\downarrow}$ and let $z\in A^{\downarrow}-V^{\downarrow}$. If $z$ and $Q^{\downarrow}_1$ are connected in $\langle A^{\downarrow}\cup S\rangle$ by at most three paths having no vertex other than $z$ in common, then $G$ has a cut-set $S^{\prime}$ of order three with $S^{\prime}\subset A^{\downarrow}\cup S$ and $S^{\prime}\neq S$, contradicting the definition of $A^{\downarrow}$. Otherwise, it is easy to see that $|V^{\downarrow}|\geq7>3\delta-3\kappa+f$. So, we can assume that $\delta-\kappa\geq2$. If $p\leq3$, then we can argue as in proof of Lemma 8 (Cases 1.1-1.2). Let $p\geq4$ and let $P_0,p_0,w,Z_3$ and $Z_{3,i}$ $(i=1,...,m)$ are as defined in proof of Lemma 8 (Case 1.3). Clearly
$$
|Z_1|\geq|Z_3|\geq|Z_2|\geq\delta-\kappa+f-p_0+1. \eqno {(5)}
$$
By Claim 1 (see the proof of Lemma 8),
$$
|V^{\downarrow}|=\sum^{m}_{i=1}|V^{\downarrow}_{i}|\geq2(|Z_1|+|Z_3|)-2m\geq4(\delta-\kappa+f-p_0+1)-2m,
$$
implying that
$$
|V^{\downarrow}|\geq(3\delta-3\kappa+f)+(\delta-\kappa-2)+3(f-2m)+4(m-p_0+1)+2. \eqno{(6)}
$$
If $p_0\leq m+1$, then $|V^{\downarrow}|\geq3\delta-3\kappa+f$ and we are done. Now let $p_0\geq m+2$. Further, if $\delta-\kappa+f-|Z_3|\leq2$, then $|Z_1|\geq|Z_3|\geq\delta-\kappa+f-2$ and by Claim 1,\\
$|V^{\downarrow}|=\sum^{m}_{i=1}|V^{\downarrow}_{i}|\geq2(|Z_1|+|Z_3|)-2m\geq4(\delta-\kappa+f-2)-2m$\\
$=(3\delta-3\kappa+f)+(\delta-\kappa-2)+3f-2m-6\geq3\delta-3\kappa+f.$\\
Now let $\delta-\kappa+f-|Z_3|\geq3$. By the choice of $w$,
$$
|N(w_i)\cap V(P_0)|\geq\delta-\kappa+f-|Z_3|\quad (i=1,...,s).
$$.
In particular, for $i=s$, we have $s\geq\delta-\kappa+f-|Z_3|$. By Lemma E, in $\langle V(P_0)\rangle$ any two vertices are joined by a path of length at least $\delta-\kappa+f-|Z_3|$. For each $i\in \{1,...,m\}$, form a cycle, denoted by $C^{\downarrow}_{i}$, by adding to $Q^{\downarrow}_{i}$ an extra path of length $r$, where $r=\delta-\kappa+f-|Z_3|+2\geq5$, connecting $F(Q^{\downarrow}_{i})$ and $L(Q^{\downarrow}_{i})$. Combining $\delta-\kappa+f-|Z_3|\geq3$ with (5), we get, $p_0\geq4$. \\
\textbf{Case 1.1}. $p_0=m+2$.
Substituting $p_0=m+2$ in (6), we get $|V^{\downarrow}|\geq(3\delta-3\kappa+f)+\delta-\kappa-4$. If $\delta-\kappa\geq4$, then we are done. Let $\delta-\kappa\leq3$. By the hypothesis, $\delta-\kappa\geq\kappa-2$, implying that $\kappa\leq5$, as well as $f\leq5$ and $m\leq2$. Since $m+2=p_0\geq4$, we have $m=2$ and $p_0=4$. If $f=5$, then the desired result follows from (6). If $f=4$, then it is easy to see that $|V^{\downarrow}|\geq13\geq3\delta-3\kappa+f$.\\
\textbf{Case 1.2}. $p_0\geq m+3$.
It is easy to see that $|V^{\downarrow}|\geq f+2\geq m+3$. Recalling also that $p_0\geq m+3$, we conclude that there are vertex disjoint paths $R_1,...,R_{m+3}$ in $G-(S-V^{\downarrow})$ connecting $V(P_0)$ and $V^{\downarrow}$. Since $p_0\geq m+3$, we can assume w.l.o.g. that either $L(R_i)\in V^{\downarrow}_{1}$ $(i=1,2,3,4)$ or $L(R_i)\in V^{\downarrow}_{1}$ $(i=1,2,3)$ and $L(R_i)\in V^{\downarrow}_{2}$ $(i=4,5)$ or $L(R_i)\in V^{\downarrow}_{1}$ $(i=1,2)$, $L(R_i)\in V^{\downarrow}_{2}$ $(i=3,4)$ and $L(R_i)\in V^{\downarrow}_{3}$ $(i=5,6)$.
\textbf{Case 1.2.1.} $L(R_i)\in V^{\downarrow}_{1}$ $(i=1,2,3,4)$.
By Claim 4 (see the proof of Lemma 8), $|V^{\downarrow}_1|\geq2(|Z_{1,1}|+|Z_{3,1}|)+3r-15$. By Claim 1, $|V^{\downarrow}_i|\geq2(|Z_{1,i}|+|Z_{3,i}|)-2$ for each $i\in \{2,...,m\}$. By summing, we get\\
$|V^{\downarrow}|=|V^{\downarrow}_1|+\sum^{m}_{i=2}|V^{\downarrow}_{i}|\geq \sum^{m}_{i=1}2(|Z_{1,i}|+|Z_{3,i}|)+3r-15-2(m-1)$\\
$= 2(|Z_1|+|Z_3|)+3(\delta-\kappa+f-|Z_3|+2)-2m-13$\\
$=(3\delta-3\kappa+f)+(|Z_1|-|Z_3|)+|Z_1|+2f-2m-7>3\delta-3\kappa+f$.\\
\textbf{Case 1.2.2.} $L(R_i)\in V^{\downarrow}_{1}$ $(i=1,2,3)$ and $L(R_i)\in V^{\downarrow}_{2}$ $(i=4,5)$.
Clearly $f\geq4$. By Claim 3, $|V^{\downarrow}_1|\geq2(|Z_{1,1}|+|Z_{3,1}|)+2r-11$. By Claim 2, $|V^{\downarrow}_2|\geq2(|Z_{1,2}|+|Z_{3,2}|)+r-7$. By Claim 1, $|V^{\downarrow}_i|\geq2(|Z_{1,i}|+|Z_{3,i}|)-2$ for each $i\in \{3,...,m\}$. By summing, we get\\
$|V^{\downarrow}|=|V^{\downarrow}_1|+|V^{\downarrow}_2|+\sum^{m}_{i=3}|V^{\downarrow}_{i}|$\\
$\geq \sum^{m}_{i=1}2(|Z_{1,i}|+|Z_{3,i}|)+3r-18-2(m-2)$\\
$= 2(|Z_1|+|Z_3|)+3(\delta-\kappa+f-|Z_3|+2)-2m-14$\\
$\geq(3\delta-3\kappa+f)+|Z_1|+2f-2m-8\geq3\delta-3\kappa+f$.\\
\textbf{Case 1.2.3.} $L(R_i)\in V^{\downarrow}_{1}$ $(i=1,2)$, $L(R_i)\in V^{\downarrow}_{2}$ $(i=3,4)$, $L(R_i)\in V^{\downarrow}_{3}$ $(i=5,6)$.
Clearly $f\geq6$. By Claim 2, $|V^{\downarrow}_i|\geq2(|Z_{1,i}|+|Z_{3,i}|)+r-7$ $(i=1,2,3)$. By Claim 1, $|V^{\downarrow}_i|\geq2(|Z_{1,i}|+|Z_{3,i}|)-2$ for each $i\in \{4,...,m\}$. By summing, we get\\
$|V^{\downarrow}|=|V^{\downarrow}_1|+|V^{\downarrow}_2|+|V^{\downarrow}_3|+\sum^{m}_{i=4}|V^{\downarrow}_{i}|$\\
$\geq \sum^{m}_{i=1}2(|Z_{1,i}|+|Z_{3,i}|)+3r-21-2(m-3)$\\
$= 2(|Z_1|+|Z_3|)+3(\delta-\kappa+f-|Z_3|+2)-2m-15$\\
$\geq(3\delta-3\kappa+f)+|Z_1|+2f-2m-9>3\delta-3\kappa+f$.\\
\textbf{Case 2}. $f=2$ and $S\not\subseteq V^{\uparrow}$.
Choose a vertex $z\in S-V^{\uparrow}$. Put $F(Q^{\downarrow}_1)=v_1$ and $L(Q^{\downarrow}_1)=w_1$. If there is a cut-vertex $x$ in $\langle A^{\downarrow}\cup\{v_1,w_1,z\}\rangle$ that separates $z$ and $Q^{\downarrow}_{1}$, then $S^{\prime }=\{S-z\}\cup \{x\}$ is a cut-set of $G$ other than $S$ with $S^{\prime}\subset A^{\downarrow}\cup S$, contradicting the definition of $A^{\downarrow}$. Otherwise, the existence of $Q^{\downarrow}_{*}$ (see Definition B) follows easily. Put $g=|V(Q^{\downarrow}_{*})\cap S|$. By the definition, $g\geq3$. Assume first that $\delta-\kappa\leq1$. Combining this with $\delta\geq2\kappa-2$, we obtain $\delta=4$ and $\kappa=3$. If $A^{\downarrow}\subseteq V(Q^{\downarrow}_{*})$, then we are done. Let $A^{\downarrow}\not\subseteq V(Q^{\downarrow}_{*})$ and choose a vertex $z\in A^{\downarrow}- V(Q^{\downarrow}_{*})$. As in Case 1, there are at least 4 paths connecting $z$ and $Q^{\downarrow}_{*}$, implying that $|V^{\downarrow}|\geq|V(Q^{\downarrow}_{*})|\geq7>3\delta-3\kappa+3$. Now let $\delta-\kappa\geq2$. Let $P=y_1...y_p;Z_1;Z_2$ be as defined in proof of Lemma 8 (Case 1). If $p\leq1$, then $A^{\downarrow}-V(Q^{\downarrow}_{*})$ is independent. Let $p\geq2$. Put $S_0=(V(Q^{\downarrow}_{*})\cap S)-\{v_1,w_1\}$.\\
\textbf{Case 2.1.} $p=2$.
Clearly $|Z_i|\geq\delta-\kappa+g-1\geq4$ $(i=1,2)$. Form a cycle $C^{\downarrow}_{1}$ by adding to $Q^{\downarrow}_{*}$ an extra path of length 3 with endvertices $v_1,w_1$. Let $I_1,...,I_t$ be the segments of $C^{\downarrow}_{1}$ having only their ends in common with $Z_1\cup Z_2$ and having at least one inner vertex in common with $S$. \\
\textbf{Case 2.1.1.} $S_0\not\subseteq V(I_i)-\{F(I_i),L(I_i)\}$ $(i=1,...,t)$.
It is easy to see that $(Z_1,Z_2)$ is a nontrivial $(C^{\downarrow}_{1},3)$-scheme. By Lemma 1,\\
$|V^{\downarrow}|\geq|V(Q^{\downarrow}_{*})|=|C^{\downarrow}_{1}|-2\geq|Z_1|+|Z_2|+|Z_1\cup Z_2|-2$\\
$\geq3(\delta-\kappa+g-1)-2=3\delta-3\kappa+3g-5>3\delta-3\kappa+3$.\\
\textbf{Case 2.1.2.} $S_0\subseteq V(I_i)-\{F(I_i),L(I_i)\}$ for some $i\in \{1,...,t\}$, say $i=1$.
Let $I_1=x\overrightarrow{I_1}y$. Denote by $R$ a longest path connecting $x$ and $y$ and passing through $V(P)$. Clearly $|R|=3$ if $x$ and $y$ belong to different $Z_1,Z_2$ and $|R|=2$, otherwise. Form a new cycle $C^{\downarrow}_{2}$ by deleting $I_1$ from $C^{\downarrow}_{1}$ and adding $R$. Clearly $(Z_1,Z_2)$ is a nontrivial $(C^{\downarrow}_{2},3)$-scheme and as in Case 2.1.1, $|V^{\downarrow}|>3\delta-3\kappa+3$.\\
\textbf{Case 2.2.} $p=3$.
Clearly $|Z_i|\geq\delta-\kappa+g-2\geq3$ $(i=1,2)$. Form a cycle $C^{\downarrow}_{1}$ by adding to $Q^{\downarrow}_{*}$ an extra path of length 4 with endvertices $v_1,w_1$. Let $I_1,...,I_t$ and $S_0$ be as defined in Case 2.1.\\
\textbf{Case 2.2.1.} $S_0\not\subseteq V(I_i)-\{F(I_i),L(I_i)\}$ $(i=1,...,t)$.
It is easy to see that $(Z_1,Z_2)$ is a nontrivial $(C^{\downarrow}_{1},4)$-scheme. By Lemma 1,\\
$|V^{\downarrow}|\geq|V(Q^{\downarrow}_{*})|=|C^{\downarrow}_{1}|-3\geq2(|Z_1|+|Z_2|)-3$\\
$\geq4(\delta-\kappa+g-2)-3=3\delta-3\kappa+3+(\delta-\kappa)+4g-14\geq3\delta-3\kappa+3$.\\
\textbf{Case 2.2.2.} $S_0\subseteq V(I_i)-\{F(I_i),L(I_i)\}$ for some $i\in \{1,...,t\}$, say $i=1$.
Let $I_1=x\overrightarrow{I_1}y$. Denote by $R$ a longest path connecting $x$ and $y$ and passing through $V(P)$. Clearly $|R|=4$ if $x$ and $y$ belong to different $Z_1,Z_2$ and $|R|\geq2$, otherwise. Form a new cycle $C^{\downarrow}_{2}$ by deleting $I_1$ from $C^{\downarrow}_{1}$ and adding $R$. Clearly $(Z_1,Z_2)$ is a nontrivial $(C^{\downarrow}_{2},4)$-scheme and as in Case 2.2.1, $|V^{\downarrow}|\geq3\delta-3\kappa+3$.\\
\textbf{Case 2.3.} $p\geq4$.
Let $P_0,p_0,w$ and $Z_3$ are as defined in proof of Lemma 8 (Case 1.3). By the definition, $|Z_1|\geq|Z_3|\geq|Z_2|$ and $p_0\geq2$. If $p_0=2$, then $|Z_1|\geq|Z_2|\geq\delta-\kappa+g-1$ and we can argue as in case $p=2$. If $p_0=3$, then $|Z_1|\geq|Z_2|\geq\delta-\kappa+g-2$ and we can argue as in case $p=3$. Let $p_0\geq4$. Further, if $\delta-\kappa+g-|Z_3|\leq1$, then $|Z_1|\geq|Z_3|\geq\delta-\kappa+g-1$ and we can argue as in case $p=2$. Let $\delta-\kappa+g-|Z_3|\geq2$. Since $p_0\geq4$, there are vertex disjoint paths $R_1,R_2,R_3,R_4$ in $\langle A^{\downarrow}\cup V(Q^{\downarrow}_{*}) \rangle$ connecting $P_0$ and $Q^{\downarrow}_{*}$ (otherwise, there exist a cut-set of $\langle A^{\downarrow}\cup V(Q^{\downarrow}_{*}) \rangle$ of order 3 contradicting the definition of $A^{\downarrow}$). Let $F(R_i)\in V(P)$ and $L(R_i)\in V(Q^{\downarrow}_{*})$ $(i=1,2,3,4)$. Clearly, $|N(w_i)\cap V(P_0)|\geq\delta-\kappa+g-|Z_3|$ for each $i\in\{1,...,s\}$. In particular, for $i=s$, we have $s\geq\delta-\kappa+g-|Z_3|$. By Lemma E, in $\langle V(P_0)\rangle$ any two vertices are joined by a path of length at least $\delta-\kappa+g-|Z_3|$. Assume first that $|Z_3|\leq3$ and assume w.l.o.g. that $L(R_1),L(R_2),L(R_3),L(R_4)$ occur on $v_1\overrightarrow{Q}^{\downarrow}_{*}w_1$ in a consecutive order. Consider the segments $J_i=L(R_i)\overrightarrow{Q}^{\downarrow}_{*}L(R_{i+1})$, $i=1,2,3$. If at least two of $J_1,J_2,J_3$ intersect $S_0$, then by the definition of $Q^{\downarrow}_{*}$, $|J_i|\geq\delta-\kappa+g-|Z_3|+2\geq4$ $(i=1,2,3)$ and hence,\\
$|V^{\downarrow}|\geq|V(Q^{\downarrow}_{*})|\geq|J_1|+|J_2|+|J_3|+1\geq3(\delta-\kappa+g-|Z_3|+2)+1$\\
$=(3\delta-3\kappa+3)+3g-3|Z_3|+4>3\delta-3\kappa+3$.\\
Now let only one of $J_1,J_2,J_3$, say $J_1$, intersects $S_0$. Form a new path $L$ by deleting $J_1$ from $Q^{\downarrow}_{*}$ and adding a path of length $\delta-\kappa+g-|Z_3|+2$, connecting $L(R_1)$ and $L(R_2)$ and passing through $V(P_0)$. Then as above, $|V^{\downarrow}|>|L|\geq3\delta-3\kappa+3$. Now let $|Z_3|\geq4$, implying that $|Z_1|\geq|Z_3|\geq4$. Assume w.l.o.g. that $R_1\cup R_2$ does not intersect $y_1\overrightarrow{P}w^{-}_{1}$ (see the proof of Lemma 8, Case 1.3) and does not contain $w$. Let $I_1,...,I_t$ be the segments of $Q^{\downarrow}_{*}$ having only their ends in common with $Z_1\cup Z_3\cup Z_4\cup Z_5 $, where $Z_4=\{L(R_1)\}$ and $Z_5=\{L(R_2)\}$. Form a cycle $C^{\downarrow}_{1}$ by adding to $Q^{\downarrow}_{*}$ an extra path of length $r$, where $r=\delta-\kappa+g-|Z_3|+2$, connecting $v_1$ and $w_1$. \\
\textbf{Case 2.3.1.} At least two of segments $I_1,...,I_t$ intersect $S_0$.
It is easy to see that $(Z_1,Z_3,Z_4,Z_5)$ is a nontrivial $(C^{\downarrow}_{1},r$-scheme. By $(c1)$ ( see Lemma 3), \\
$|V^{\downarrow}|\geq|V(Q^{\downarrow}_{*})|=|C^{\downarrow}_{1}|-r+1\geq2(|Z_1|+|Z_3|)+3r-15$\\
$=(3\delta-3\kappa+3)+(|Z_1|-|Z_3|)+3g+|Z_1|-12>3\delta-3\kappa+3$.\\
\textbf{Case 2.3.2}. Only one of segments $I_1,...,I_t$, say $I_1$, intersects $S_0$.
In this case, each vertex of $S_0$ is an inner vertex for $I_1$. Let $I_1=xI_1y$. Form a cycle $C^{\downarrow}_{2}$ by deleting $I_1$ from $C^{\downarrow}_{1}$ and adding a longest path connecting $x$ and $y$ and passing through $V(P)$. It is easy to see that $(Z_1,Z_3,Z_4,Z_5)$ is a nontrivial $(C^{\downarrow}_{1},r)$-scheme. Then we can argue as in Case 2.2.1.\\
\textbf{Case 3.} $f=2$ and $S\subseteq V^{\uparrow}$.
If $A^{\downarrow}\subseteq V^{\downarrow}$, then $|V^{\downarrow}|\geq|A^{\downarrow}|+2\geq3\delta-3\kappa+4>2\delta-2\kappa+3$. Let $A^{\downarrow}\not\subseteq V^{\downarrow}$. Then we can argue exactly as in proof of Lemma 8 (Case 3). \qquad $\Delta$\\
\noindent\textbf{Proof of Lemma 10}. To prove $(f1)$, choose a vertex $z\in A^{\uparrow}-V^{\uparrow}$. Clearly $N(z)\subseteq V^{\uparrow}\cup S$. Put\\
$Z_i=N(z)\cap V^{\uparrow}_i \quad (i=1,...,m), \quad Z=\cup^{m}_{i=1}Z_{i},$\\
$W=N(z)\cap \{v_1,...,v_m,w_1,...,w_m\}.$\\
Since $\{Q^{\uparrow}_{1},...,Q^{\uparrow}_{m}\}$ is $(*1)$-extreme, we have $|W|\leq2$ and $|Z|\geq\delta-1.$ If $|W|=0$, then by standard arguments, $|V^{\uparrow}_{i}|\geq2|Z_i|+1$ $(i=1,...,m)$ and
$$
|V^{\uparrow}|=\sum^{m}_{i=1}|V^{\uparrow}_{i}|\geq\sum^{m}_{i=1}(2|Z_i|+1)=2|Z|+m\geq2\delta+m-2.
$$
Further, let $|W|=1$. Assume w.l.o.g. that $W=\{v_1\}$. By $(*1)$, $N(z)\subseteq V^{\uparrow}$, i.e. $|Z|\geq\delta$. Then clearly $|V^{\uparrow}_{1}|\geq2|Z_1|$ and $|V^{\uparrow}_{i}|\geq2|Z_i|+1$ $(i=2,...,m)$, implying that\\
$|V^{\uparrow}|= |V^{\uparrow}_{1}|+\sum^{m}_{i=2}|V^{\uparrow}_{i}|\geq 2|Z_1|+\sum^{m}_{i=2}(2|Z_i|+1)$\\
$\geq2|Z|+m-1>2\delta+m-2.$\\
Finally, we assume that $|W|=2$. By $(*1)$, $W\subseteq V^{\uparrow}_{i}$ for some $i\in \{1,...,m\}$, say $i=1$, and $N(z)\subseteq V^{\uparrow}$ implying that $|Z|\geq\delta$. Then clearly $|V^{\uparrow}_{1}|\geq2|Z_1|-1$ and $|V^{\uparrow}_{i}|\geq2|Z_i|+1$ $(i=2,...,m)$, whence\\
$|V^{\uparrow}|= |V^{\uparrow}_{1}|+\sum^{m}_{i=2}|V^{\uparrow}_{i}|\geq 2|Z_1|-1+\sum^{m}_{i=2}(2|Z_i|+1)$\\
$=2|Z|+m-2\geq2\delta+m-2.$\\
To prove $(f2)$, choose a vertex $z\in A^{\downarrow}-V^{\downarrow}$ and put
$$
Z_i=N(z)\cap V^{\downarrow}_i \quad (i=1,...,m), \quad Z=\cup^{m}_{i=1}Z_{i},
$$
Clearly $|Z|\geq\delta-(\kappa-f)$. Since $\{Q^{\downarrow}_{1},...,Q^{\downarrow}_{m}\}$ is $(*2)$-extreme, by standard arguments, $|V^{\downarrow}_{i}|\geq2|Z_i|-1$ $(i=1,...,m)$, implying that
$$
|V^{\downarrow}|=\sum^{m}_{i=1}|V^{\downarrow}_{i}|\geq\sum^{m}_{i=1}(2|Z_i|-1)=2|Z|-m\geq2\delta-2\kappa+2f-m.
$$
To prove $(f3)$, choose a vertex $z\in A^{\downarrow}-V(Q^{\downarrow}_{*})$. Put $Z=N(z)\cap V(Q^{\downarrow}_{*})$ and $g=|V(Q^{\downarrow}_{*})\cap S|$. Clearly $|Z|\geq\delta-\kappa+g\geq\delta-\kappa+3$. By standard arguments, $|V(Q^{\downarrow}_{*})|\geq2|Z|-1\geq2\delta-2\kappa+5$. \qquad $\Delta$
\section{Proofs of theorems}
\textbf{Proof of Theorem 2}. If $G$ has a dominating cycle, then we are done. Otherwise, by Lemma M, $c\geq 3\delta-3$. If $\delta\leq 2\kappa-3$, then $c\geq 3\delta-3\geq4\delta-2\kappa$ and again we are done. So, we can assume that $\delta\geq 2\kappa-2>3\kappa/2-1$. Let $Q^{\uparrow}_{1},...,Q^{\uparrow}_{m}$; $Q^{\downarrow}_{1},...,Q^{\downarrow}_{m}$ and $C$ be as defined in Definitions A-C. The existence of $Q^{\downarrow}_{1},...,Q^{\downarrow}_{m}$ follows from Lemma D. Put $f=|V^{\downarrow}\cap S|$. Assume w.l.o.g. that $F(Q^{\uparrow}_{i})=v_i$ and $L(Q^{\uparrow}_{i})=w_i$ $(i=1,...,m)$. Form a cycle $C^{\uparrow}$ consisting of $Q^{\uparrow}_{1},...,Q^{\uparrow}_{m}$ and extra edges $w_1v_2, w_2v_3,...,w_{m-1}v_m,w_mv_1$. By Lemma 5, $A^{\uparrow}-V^{\uparrow}$ is independent.\\
\textbf{Case 1}. $|A^\downarrow|\leq2\delta-2\kappa+1.$
By Lemma 7, $A^{\downarrow}\subseteq V^{\downarrow}$. By $(*1)$, $S-V(C)$ is independent. If $V(G-C)$ is independent, then clearly $C$ is a dominating cycle. Otherwise, each edge of $G-C$ has one end in $A^{\uparrow}-V^{\uparrow}$ and the other in $S-V(C)$. Let $xy\in E(G)$, where $x\in A^{\uparrow}-V^{\uparrow}$ and $y\in S-V(C)$. By $(*1)$, $N(x)\subseteq V^{\uparrow}\cup\{y\}$.\\
\textbf{Case 1.1.} $N(y)\not\subseteq V(C)\cup \{x\}$.
Let $yz\in E(G)$ for a vertex $z\in A^{\uparrow}-V^{\uparrow}$ other than $x$. By $(*1)$ and the fact that $A^{\uparrow}-V^{\uparrow}$ is independent, we have $N(z)\subseteq V^{\uparrow}\cup\{y\}$. Set $Z_1=N(x)\cap V^{\uparrow}$ and $Z_2=N(z)\cap V^{\uparrow}$. Clearly $|Z_i|\geq\delta-1$ $(i=1,2)$. Basing on $(*1)$, it is easy to see that $(Z_1,Z_2)$ is a nontrivial $(C^{\uparrow},4)$-scheme. By $(a1)$ (see Lemma 1), $|C|>|C^{\uparrow}|\geq4(\delta-1)>4\delta-2\kappa$.\\
\textbf{Case 1.2.} $N(y)\subseteq V(C)\cup \{x\}$.
\textbf{Case 1.2.1.} $|N(y)\cap V^{\downarrow}|\geq f-m+1.$
Set $Z_{i}=N(y)\cap V^{\downarrow}_{i}$ $(i=1,...,m)$ and $Z=\cup^{m}_{i=1}Z_{i}$. By the hypothesis, $|Z|=|N(y)\cap V^{\downarrow}|\geq f-m+1$. By $(*1)$, $y$ is not adjacent to $F(Q^{\downarrow}_{i})$ and $L(Q^{\downarrow}_{i})$ $(i=1,...,m)$. Since $\{Q^{\downarrow}_{1},...,Q^{\downarrow}_{m}\}$ is $(*4)$-extreme, by standard arguments, $|V^{\downarrow}_{i}\cap S|\geq|Z_{i}|+1$ $(i=1,...,m)$ and
$$
f=|V^{\downarrow}\cap S|=\sum^{m}_{i=1}|V^{\downarrow}_{i}\cap S|\geq \sum^{m}_{i=1}(|Z_i|+1)\geq|Z|+m\geq f+1,
$$
a contradiction. \\
\textbf{Case 1.2.2.} $|N(y)\cap V^{\downarrow}|\leq f-m\leq f-1$.
Since $N(y)\subseteq V(C)\cup \{x\}$, we have $|N(y)\cap V^{\uparrow}|\geq \delta-|N(y)\cap V^{\downarrow}|-1\geq\delta-f.$ Set $Z_1=N(x)\cap V^{\uparrow}$ and $Z_2=N(y)\cap V^{\uparrow}$. Clearly $|Z_1|\geq\delta-1$ and $|Z_2|\geq\delta-f$. By $(*1)$, it is easy to see that $(Z_1,Z_2)$ is a nontrivial $(C^{\uparrow},3)$-scheme. By Lemma 1, $|V^{\uparrow}|=|C^{\uparrow}|\geq2(\delta-1)+\delta-f=3\delta-f-2$. Observing that $|A^{\downarrow}|\geq\delta-\kappa+1$, we obtain
$$
|C|\geq|V^{\uparrow}|+|A^{\downarrow}|\geq (4\delta-2\kappa)+\kappa-f-1\geq4\delta-2\kappa.
$$.
\textbf{Case 2}. $|A^\downarrow|\geq2\delta-2\kappa+2.$
\textbf{Case 2.1.} $f\geq3$.
By Lemma 8 and $(*1)$, $A^{\downarrow}-V^{\downarrow}$ and $S-V(C)$ both are independent sets. If $A^{\downarrow}\subseteq V^{\downarrow}$, then we can argue exactly as in Case 1. Let $A^{\downarrow}\not\subseteq V^{\downarrow}$.
By $(f2)$ (see Lemma 10), $|V^{\downarrow}|\geq2\delta-2\kappa+2f-m.$ If $A^{\uparrow}\not\subseteq V^{\uparrow}$, then by $(f1)$ (see Lemma 10), $|V^{\uparrow}|\geq2\delta+m-2$, whence
$$
|C|\geq|V^{\uparrow}|+|V^{\downarrow}|-2m\geq(4\delta -2\kappa)+2f-2m-2>4\delta-2\kappa.
$$
Now let $A^{\uparrow}\subseteq V^{\uparrow}$. Since $S-V(C)$ is independent, either $C$ is a dominating cycle (then we are done) or each edge in $G-C$ has one end in $S-V(C)$ and the other in $A^{\downarrow}-V^{\downarrow}$.\\
\textbf{Case 2.1.1.} There is a path $xyz$ in $G-C$ with $x,z\in A^{\downarrow}$ and $y\in S$.
Put $Z_1=N(x)\cap V^{\downarrow}$, $Z_2=N(z)\cap V^{\downarrow}$ and
$$
Z_{1,i}=N(x)\cap V^{\downarrow}_{i}, \quad Z_{2,i}=N(z)\cap V^{\downarrow}_{i}\quad (i=1,...,m).
$$
Clearly $|Z_i|\geq\delta-\kappa+f$ $(i=1,2)$. For each $i\in \{1,...,m\}$, form a cycle $C^{\downarrow}_{i}$ by adding to $Q^{\downarrow}_{i}$ an extra path of length 4 with endvertices $F(Q^{\downarrow}_{i})$ and $L(Q^{\downarrow}_{i})$. If $(Z_{1,i},Z_{2,i})$ is a nontrivial $(Q^{\downarrow}_{i},4)$-scheme, then by $(a1)$ (see Lemma 1), $|V^{\downarrow}_{i}|=|C^{\downarrow}_{i}|-3\geq2(|Z_{1,i}|+|Z_{2,i}|)-3$. Otherwise, it can be checked easily. By summing, we get\\
$|V^{\downarrow}|=\sum^{m}_{i=1}|V^{\downarrow}_{i}|\geq\sum^{m}_{i=1}(2(|Z_{1,i}|+|Z_{2,i}|)-3)=2(|Z_1|+|Z_2|)-3m$\\
$\geq 4(\delta-\kappa+f)-3m \geq 3\delta-3\kappa+f+2,$\\
\noindent whence $|A^{\downarrow}|\geq|V^{\downarrow}|-f > 3\delta-3\kappa+2$, contrary to the hypothesis.\\
\textbf{Case 2.1.2.} There is a path $xyz$ in $G-C$ with $x,z\in S$ and $y\in A^{\downarrow}$.
Clearly $\kappa\geq4$. If $xy^{\prime}\in E(G)$ or $zy^{\prime}\in E(G)$ for some $y^{\prime}\in A^{\downarrow}-V^{\downarrow}$ other than $y$, then we can argue as in Case 2.1.1. So, we can assume that $N(x)\subseteq V(C)\cup \{y\}$ and $N(z)\subseteq V(C)\cup \{y\}$. Put $Z_1=N(x)\cap V(C)$ and $Z_2=N(z)\cap V(C)$. \\
\textbf{Case 2.1.2.1.} $m=1$.
Assume w.l.o.g. that $Q^{\uparrow}_{1}=v_1\overrightarrow{C}w_1$. Let $I=\xi\overrightarrow{C}\eta$ be a segment on $C$ having only $\xi,\eta$ in common with $Z_1\cup Z_2$. By $(*1)$, $\{\xi,\eta\}\cap \{v_1,w_1\}=\emptyset.$ If $\xi,\eta\in Z_1$ or $\xi,\eta\in Z_2$, then by $(*1)$ and $(*2)$, $|I|\geq2$. Let $\xi\in Z_1$ and $\eta\in Z_2$. If $\xi,\eta\in V^{\downarrow}$, then by $(*2)$, $|I|\geq4$. Let $\xi,\eta\in V^{\uparrow}$. If $|I|\leq2$, then the collection of paths $v_1\overrightarrow{C}\xi x$, $z\eta\overrightarrow{C}w_1$ contradicts $(*1)$. If $|I|=3$, then the cycle $v_1\overrightarrow{C}\xi xyz\eta\overrightarrow{C}v_1$ contradicts $(*4)$. So, $|I|\geq4$ if $\xi,\eta\in V^{\uparrow}$. As for the cases $\xi\in V^{\uparrow}$, $\eta\in V^{\downarrow}$ or $\xi\in V^{\downarrow}$, $\eta\in V^{\uparrow}$, there exists exactly one segment $I_1=\xi_1\overrightarrow{C}\eta_1$ with $\xi_1\in V^{\uparrow}$, $\eta_1\in V^{\downarrow}$, $w_1\in I_1$ and exactly one segment $I_2=\xi_2\overrightarrow{C}\eta_2$ with $\xi_2\in V^{\downarrow}$, $\eta_2\in V^{\uparrow}$, $v_1\in I_2$, each of length at least 2. Form a new cycle $C^{\prime}$ by deleting $I_1$ and $I_2$ from $C$ and adding appropriate extra paths each of length 4. Clearly $(Z_1,Z_2)$ is a nontrivial $(C^{\prime},4)$-scheme. By $(a1)$ (see Lemma 1),
$$
|C|\geq |C^{\prime}|-4\geq 2(|Z_1|+|Z_2|)-4\geq 4(\delta-1)-4\geq 4\delta-2\kappa.
$$
\textbf{Case 2.1.2.2.} $m\geq2$.
By arguing as in Case 2.1.2.1, we obtain
$$
|C|\geq |C^{\prime}|-4m\geq 4(\delta-1)-4m=(4\delta-2\kappa)+2(\kappa-2m-2).
$$
Since $\kappa\geq f+2\geq2m+2$, we have $|C|\geq4\delta-2\kappa$.\\
\textbf{Case 2.1.3.} $xy\in E(G)$ for some $x\in S-V(C)$ and $y\in A^{\downarrow}-V^{\downarrow}.$
Form a new cycle $C^{\downarrow}$ by deleting $Q^{\uparrow}_{1},...,Q^{\uparrow}_{m}$ from $C$ and adding extra paths $v_iz_iw_i$ $(i=1,...,m)$. Put $V_{0}=\{v_1,...,v_m,w_1,...,w_m\}$ and
$$
Z_0=N(y)\cap V_0, \quad Z_1=N(x)\cap V^{\uparrow}, \quad Z_2=N(x)\cap V^{\downarrow},
$$
$$
Z_3=N(y)\cap (V^{\uparrow}-V_0), \quad Z_4=N(y)\cap (V^{\downarrow}-V_0).
$$
If $N(x)\not\subseteq V(C)\cup\{y\}$ or $N(y)\not\subseteq V(C)\cup\{x\}$, then we can argue as in Cases 2.1.1-2.1.2. Let $N(x)\subseteq V(C)\cup\{y\}$ and $N(y)\subseteq V(C)\cup\{x\}$. Abbreviate $|Z_0|=t$, $|Z_2|=d$ and $|Z_3|=h$. Clearly $|Z_1|\geq\delta-d-1$ and $|Z_4|\geq \delta-h-t-1.$ If $d=0$, then $|Z_1|\geq\delta-1$ and arguing exactly as in proof of $(f1)$ (see Lemma 10), we get $|V^{\uparrow}|\geq2\delta+m-2$. By $(f2)$ (see Lemma 10), $|V^{\downarrow}|\geq2\delta-2\kappa+2f-m$. Hence,
$$
|C|\geq |V^{\uparrow}|+|V^{\downarrow}|-2m\geq(4\delta-2\kappa)+2f-2m-2> 4\delta-2\kappa.
$$
Let $d\geq1$. If $h=0$, then it is not hard to see that $(Z_2,Z_0\cup Z_4)$ is a nontrivial $(C^{\downarrow},3)$-scheme. By Lemma 1,
$$
|V^{\downarrow}|\geq |C^{\downarrow}|-m\geq 2|Z_0\cup Z_4|+|Z_2|-m\geq2(\delta-1)+d-m\geq2\delta-1-m.
$$
Further, $|A^{\downarrow}|\geq |V^{\downarrow}|-f \geq 2\delta-1-f-m$, implying that $|A^{\uparrow}|\geq2\delta-1-f-m$ and
$$
|C|\geq |A^{\uparrow}|+|V^{\downarrow}|\geq (4\delta-2\kappa)+2\kappa-f-2m-2.
$$
Observing that $\kappa\geq f+1$ and $f\geq2m$, we get $|C|\geq 4\delta-2\kappa.$ Now let $h\geq1$. Basing on $(*1)$ and $(*4)$, it is not hard to see that $(Z_1,Z_0\cup Z_3)$ is a nontrivial $(C^{\uparrow},3)$-scheme. Since $|Z_1|\geq\delta-d-1$ and $|Z_0\cup Z_3|=t+h$, we have by Lemma 1,
$$
|V^{\uparrow}|=|C^{\uparrow}|\geq 2|Z_1|+|Z_0\cup Z_3|\geq2\delta-2d-2+t+h.
$$
On the other hand, by $(*2)$, $(Z_0\cup Z_4,Z_2)$ is a nontrivial $(C^{\downarrow},3)$-scheme. By Lemma 1,
$$
|V^{\downarrow}|=|C^{\downarrow}|-m\geq 2|Z_0\cup Z_4|+|Z_2|-m\geq 2\delta-2h+d-m-2.
$$
Note that for the special case $m=1$ and $t=0$, by the same arguments, we can obtain a slightly better estimation, namely $|V^{\downarrow}|\geq2\delta-2h+d-2$. So,
$$
|C|\geq |V^{\uparrow}|+|V^{\downarrow}|-2m\geq(4\delta-2\kappa)+2\kappa-d-h-3m+t-4.
$$
If $d\leq 2\kappa-h-3m+t-4$, then clearly $c\geq 4\delta-2\kappa$. Let $d\geq 2\kappa-h-3m+t-3$. Therefore,
$$
|V^{\downarrow}|\geq2\delta-2h+d-m-2\geq 2\delta+2\kappa-3h-4m+t-5
$$
whence
$$
|A^{\uparrow}|\geq |A^{\downarrow}|\geq |V^{\downarrow}|-f+1\geq2\delta+2\kappa-3h-4m+t-f-4.
$$
Further,
$$
|C|\geq |A^{\uparrow}|+|V^{\downarrow}|\geq(4\delta-2\kappa)+6\kappa-6h-8m+2t-f-9.
$$
Observing that $\kappa\geq f+h+1$, we obtain $6\kappa\geq6(f+h+1)\geq6h+f+10m+6$, whence $|C|\geq(4\delta-2\kappa)+2m+2t-3$. If $m+t\geq2$, then clearly $|C|\geq4\delta-2\kappa$. Otherwise, we have $m=1$ and $t=0$. Recalling that in this special case, $|V^{\downarrow}|\geq 2\delta-2h+d-2$, we can obtain the desired result by the same arguments used above.\\
\textbf{Case 2.2.} $f=2$ and $S\not\subseteq V^{\uparrow}$.
By Lemma 8, $A^{\downarrow}-V(Q^{\downarrow}_{*})$ is independent. If $A^{\uparrow}\subseteq V^{\uparrow}$ and $A^{\downarrow}\subseteq V(Q^{\downarrow}_{*})$, then clearly $C$ is a dominating cycle. Next, if $A^{\uparrow}\subseteq V^{\uparrow}$ and $A^{\downarrow}\not\subseteq V(Q^{\downarrow}_{*})$, then we can argue exactly as in Case 2.1.3. Let $A^{\uparrow}\not\subseteq V^{\uparrow}$ and $A^{\downarrow}\subseteq V(Q^{\downarrow}_{*})$. By $(f1)$ (see Lemma 10), $|V^{\uparrow}|\geq2\delta-1$. Recalling that $|A^{\downarrow}|\geq 2\delta-2\kappa+2$, we get
$$
|C|\geq |V^{\uparrow}|+|A^{\downarrow}|\geq (4\delta-2\kappa)+1>4\delta-2\kappa.
$$
If $A^{\uparrow}\not\subseteq V^{\uparrow}$ and $A^{\downarrow}\not\subseteq V(Q^{\downarrow}_{*})$, then by $(f1)$ and $(f3)$ (see Lemma 10), $|V^{\uparrow}|\geq2\delta-1$ and $|V^{\downarrow}|\geq2\delta-2\kappa+5$, whence
$$
|C|\geq |V^{\uparrow}|+|V^{\downarrow}|-2\geq 4\delta-2\kappa+2>4\delta-2\kappa.
$$
\textbf{Case 2.3.} $f=2$ and $S\subseteq V^{\uparrow}.$
By Lemma 8, either $A^{\downarrow}-V^{\downarrow}=\emptyset$ or $|V^{\downarrow}|\geq2\delta-2\kappa+3$. If $A^{\downarrow}-V^{\downarrow}=\emptyset$, then $C$ is a dominating cycle, since $A^{\uparrow}-V^{\uparrow}$ is independent (by Lemma 5). So, we can assume that $|V^{\downarrow}|\geq2\delta-2\kappa+3$. If $A^{\uparrow}\not\subseteq V^{\uparrow}$, then by $(f1)$ (see Lemma 10), $|V^{\uparrow}|\geq2\delta-1$ and $|C|\geq |V^{\uparrow}|+|V^{\downarrow}|-2\geq 4\delta-2\kappa$. Now let $A^{\uparrow}\subseteq V^{\uparrow}$. If $|A^{\downarrow}|\leq3\delta-3\kappa$, then by Lemma F, $A^{\downarrow}-V^{\downarrow}$ is independent and $C$ is a dominating cycle. Let $|A^{\downarrow}|=3\delta-3\kappa+1\geq2\delta-\kappa-1$, implying that $|A^{\uparrow}|\geq2\delta-\kappa-1$. Then $|V^{\uparrow}|\geq |A^{\uparrow}|+\kappa\geq 2\delta-1$ and
$$
|C|\geq |V^{\uparrow}|+|V^{\downarrow}|-2\geq 4\delta-2\kappa. \qquad \Delta
$$
\noindent\textbf{Proof of Theorem 3}. If $\delta\leq2\kappa-3$, then we are done by Lemma 4. Let $\delta\geq2\kappa-2$ and let $Q^{\uparrow}_{1},...,Q^{\uparrow}_{m}$; $Q^{\downarrow}_{1},...,Q^{\downarrow}_{m}$ and $C$ be as defined in Definitions A-C. The existence of $Q^{\downarrow}_{1},...,Q^{\downarrow}_{m}$ follows from Lemma D. Assume w.l.o.g. that $F(Q^{\uparrow}_{i})=v_i$ and $L(Q^{\uparrow}_{i})=w_i$ $(i=1,...,m)$. Form a cycle $C^{\uparrow}$ consisting of $Q^{\uparrow}_{1},...,Q^{\uparrow}_{m}$ and extra edges $w_1v_2, w_2v_3,...,w_{m-1}v_m,w_mv_1$. By Lemma 5, $A^{\uparrow}-V^{\uparrow}$ is independent. Put $f=|V^{\downarrow}\cap S|$. By the hypothesis, $|A^{\uparrow}|\geq|A^{\downarrow}|\geq3\delta-3\kappa+2\geq2\delta-\kappa$. \\
\textbf{Case 1}. $f\geq3$.
By Lemma 9, either $A^{\downarrow}-V^{\downarrow}$ is independent or $|V^{\downarrow}|\geq3\delta-3\kappa+f$. If $A^{\downarrow}-V^{\downarrow}$ is independent, then we can argue as in proof of Theorem 2 (Case 2.1). Let $|V^{\downarrow}|\geq3\delta-3\kappa+f\geq2\delta-2\kappa+1$. If $A^{\uparrow}\subseteq V^{\uparrow}$, then
$$
|C|\geq |A^{\uparrow}|+|V^{\downarrow}|\geq(2\delta-\kappa)+(2\delta-\kappa+1)>4\delta-2\kappa.
$$
Otherwise, by $(f1)$ (see Lemma 10), $|V^{\uparrow}|\geq2\delta+m-2$, whence
$$
|C|\geq |V^{\uparrow}|+|V^{\downarrow}|-2m\geq(4\delta-2\kappa)+\delta-\kappa+f-m-2\geq4\delta-2\kappa.
$$
\textbf{Case 2}. $f=2$ and $S\not\subseteq V^{\uparrow}$.
By Lemma 9, either $A^{\downarrow}-V(Q^{\downarrow}_{*})$ is independent or $|V^{\downarrow}|\geq3\delta-3\kappa+3$. If $A^{\downarrow}-V(Q^{\downarrow}_{*})$ is independent, then we can argue as in proof of Theorem 2 (Case 2.2). If $|V^{\downarrow}|\geq3\delta-3\kappa+3$, then we can argue as in case $f\geq3$. \\
\textbf{Case 3}. $f=2$ and $S\subseteq V^{\uparrow}$.
By Lemma 9, $|V^{\downarrow}|\geq2\delta-2\kappa+3$. If $A^{\uparrow}\subseteq V^{\uparrow}$, then
$$
|C|\geq |A^{\uparrow}|+|V^{\downarrow}|+\kappa-2\geq(4\delta-2\kappa)+1>4\delta-2\kappa.
$$
Otherwise, by $(f1)$ (see Lemma 10), $|V^{\uparrow}|\geq2\delta-1$, whence
$$
|C|\geq |V^{\uparrow}|+|V^{\downarrow}|-2\geq4\delta-2\kappa. \qquad \Delta
$$
\noindent\textbf{Proof of Theorem 4}. If $\delta\leq2\kappa-3$, then we are done by Lemma 4. Let $\delta\geq2\kappa-2$ and let $Q^{\uparrow}_{1},...,Q^{\uparrow}_{m}$; $Q^{\downarrow}_{1},...,Q^{\downarrow}_{m}$ and $C$ be as defined in Definitions A-C. The existence of $Q^{\downarrow}_{1},...,Q^{\downarrow}_{m}$ follows from Lemma D. Assume w.l.o.g. that $F(Q^{\uparrow}_{i})=v_i$ and $L(Q^{\uparrow}_{i})=w_i$ $(i=1,...,m)$. Form a cycle $C^{\uparrow}$ consisting of $Q^{\uparrow}_{1},...,Q^{\uparrow}_{m}$ and extra edges $w_1v_2, w_2v_3,...,w_{m-1}v_m,w_mv_1$. By Lemmas 6, either $A^{\uparrow}-V^{\uparrow}$ is independent or $|V^{\uparrow}|\geq3\delta-5$. If $A^{\uparrow}-V^{\uparrow}$ is independent, then we can argue as in proof of Theorem 2. Let $|V^{\uparrow}|\geq3\delta-5$. \\
\textbf{Case 1.} $f\geq3.$
By Lemma 8, $A^{\downarrow}-V^{\downarrow}$ is independent. If $A^{\downarrow}\subseteq V^{\downarrow}$, then
$$
|C|\geq |V^{\uparrow}|+|A^{\downarrow}|\geq (3\delta-5)+(\delta-\kappa+1)=4\delta-\kappa-4\geq4\delta-2\kappa.
$$
Otherwise, by $(f2)$ (see Lemma 10), $|V^{\downarrow}|\geq 2\delta-2\kappa+2f-m$, whence
$$
|C|\geq |V^{\uparrow}|+|V^{\downarrow}|-2m\geq (4\delta-2\kappa)+\delta+2f-3m-5\geq4\delta-2\kappa.
$$
\textbf{Case 2.} $f=2$ and $S\not\subseteq V^{\uparrow}.$
By Lemma 8, $A^{\downarrow}-V(Q^{\downarrow}_*)$ is independent. If $A^{\downarrow}\subseteq V(Q^{\downarrow}_*)$, then we can argue as in Case 1. Otherwise, by $(f3)$ (see Lemma 10), $|V(Q^{\downarrow}_*)|\geq2\delta-2\kappa+5$, whence
$$
|C|\geq |V^{\uparrow}|+|V(Q^{\downarrow}_*)|-2\geq(4\delta-2\kappa)+\delta-2>4\delta-2\kappa.
$$
\textbf{Case 3.} $f=2$ and $S\subseteq V^{\uparrow}.$
By Lemma 6, either $A^{\uparrow}-V^{\uparrow}$ is independent or $|V^{\uparrow}|\geq3\delta-5$. If $A^{\uparrow}-V^{\uparrow}$ is independent, then we can argue as in proof of Theorem 2 (Case 2.3). So, we can assume that $|V^{\uparrow}|\geq3\delta-5$. By Lemma 8, either $A^{\downarrow}-V^{\downarrow}=\emptyset$ or $|V^{\downarrow}|\geq2\delta-2\kappa+3$. If $A^{\downarrow}-V^{\downarrow}=\emptyset$, then
$$
|C|\geq |V^{\uparrow}|+|A^{\downarrow}|\geq(3\delta-5)+(\delta-\kappa+1)\geq4\delta-2\kappa.
$$
Let $|V^{\downarrow}|\geq2\delta-2\kappa+3$. Then
$$
|C|\geq |V^{\uparrow}|+|V^{\downarrow}|-2\geq(3\delta-5)+(2\delta-2\kappa+3)-2\geq4\delta-2\kappa. \qquad \Delta
$$
\noindent\textbf{Proof of Theorem 5}. If $\delta\leq2\kappa-3$, then we are done by Lemma 4. Let $\delta\geq2\kappa-2$ and let $Q^{\uparrow}_{1},...,Q^{\uparrow}_{m}$; $Q^{\downarrow}_{1},...,Q^{\downarrow}_{m}$ and $C$ be as defined in Definitions A-C. The existence of $Q^{\downarrow}_{1},...,Q^{\downarrow}_{m}$ follows from Lemma D. Assume w.l.o.g. that $F(Q^{\uparrow}_{i})=v_i$ and $L(Q^{\uparrow}_{i})=w_i$ $(i=1,...,m)$. By Lemma 6, either $A^{\uparrow}-V^{\uparrow}$ is independent or $|V^{\uparrow}|\geq3\delta-5$.\\
\textbf{Case 1.} $f\geq3$.
By Lemma 9, either $A^{\downarrow}-V^{\downarrow}$ is independent or $|V^{\downarrow}|\geq3\delta-3\kappa+f.$\\
\textbf{Case 1.1.} $A^{\uparrow}-V^{\uparrow}$ and $A^{\downarrow}-V^{\downarrow}$ both are independent sets.
If $A^{\uparrow}\subseteq V^{\uparrow}$ and $A^{\downarrow}\subseteq V^{\downarrow}$, then clearly $C$ is a dominating cycle. If $A^{\uparrow}\subseteq V^{\uparrow}$ and $A^{\downarrow}\not\subseteq V^{\downarrow}$, then by $(f2)$ (see Lemma 10), $|V^{\downarrow}|\geq2\delta-2\kappa+2f-m$, whence
$$
c\geq|A^{\uparrow}|+|V^{\downarrow}|\geq(4\delta-2\kappa)+\delta-\kappa+f+m-3>4\delta-2\kappa.
$$
Next, if $A^{\uparrow}\not\subseteq V^{\uparrow}$ and $A^{\downarrow}\subseteq V^{\downarrow}$, then by $(f1)$ (see Lemma 10), $|V^{\uparrow}|\geq2\delta+m-2$, whence
$$
c\geq|V^{\uparrow}|+|A^{\downarrow}|\geq(4\delta-2\kappa)+\delta-\kappa+m>4\delta-2\kappa.
$$
Finally, if $A^{\uparrow}\not\subseteq V^{\uparrow}$ and $A^{\downarrow}\not\subseteq V^{\downarrow}$, then by $(f1)$ (see Lemma 10), $|V^{\uparrow}|\geq2\delta+m-2$ and by $(f2)$ (see Lemma 10), $|V^{\downarrow}|\geq2\delta-2\kappa+2f-m$. Hence,
$$
c\geq|V^{\uparrow}|+|V^{\downarrow}|-2m\geq(4\delta-2\kappa)+2f-2m-2>4\delta-2\kappa.
$$
\textbf{Case 1.2.} $A^{\uparrow}-V^{\uparrow}$ is independent and $|V^{\downarrow}|\geq3\delta-3\kappa+f$.
If $A^{\uparrow}\subseteq V^{\uparrow}$, then
$$
c\geq|A^{\uparrow}|+|V^{\downarrow}|\geq(4\delta-2\kappa)+2\delta-2\kappa+f-3\geq4\delta-2\kappa.
$$
Otherwise, by $(f1)$ (see Lemma 10), $|V^{\uparrow}|\geq2\delta+m-2$ and
$$
c\geq|V^{\uparrow}|+|V^{\downarrow}|-2m\geq(4\delta-2\kappa)+\delta-\kappa+f+m-2>4\delta-2\kappa.
$$
\textbf{Case 1.3.} $|V^{\uparrow}|\geq3\delta-5$ and $A^{\downarrow}-V^{\downarrow}$ is independent.
If $A^{\downarrow}\subseteq V^{\downarrow}$, then
$$
c\geq|V^{\uparrow}|+|A^{\downarrow}|\geq(4\delta-2\kappa)+2\delta-\kappa-3>4\delta-2\kappa.
$$
Otherwise, by $(f2)$ (see Lemma 10), $|V^{\downarrow}|\geq2\delta-2\kappa+2f-m$ and
$$
c\geq|V^{\uparrow}|+|V^{\downarrow}|-2m\geq(4\delta-2\kappa)+\delta+2f-3m-5\geq4\delta-2\kappa.
$$
\textbf{Case 1.4.} $|V^{\uparrow}|\geq3\delta-5$ and $|V^{\downarrow}|\geq3\delta-3\kappa+f$.
Clearly
$$
c\geq|V^{\uparrow}|+|V^{\downarrow}|-2m\geq(4\delta-2\kappa)+2\delta-\kappa-5\geq4\delta-2\kappa.
$$
\textbf{Case 2.} $f=2$ and $S\not\subseteq V^{\uparrow}$.
By Lemma 9, either $A^{\downarrow}-V(Q^{\downarrow}_*)$ is independent or $|V^{\downarrow}|\geq3\delta-3\kappa+3.$\\
\textbf{Case 2.1.} $A^{\uparrow}-V^{\uparrow}$ and $A^{\downarrow}-V(Q^{\downarrow}_*)$ both are independent sets.
If $A^{\uparrow}\subseteq V^{\uparrow}$ and $A^{\downarrow}\subseteq V^{\downarrow}$, then clearly $C$ is a dominating cycle. If $A^{\uparrow}\subseteq V^{\uparrow}$ and $A^{\downarrow}\not\subseteq V^{\downarrow}$, then by $(f3)$ (see Lemma 10), $|V^{\downarrow}|\geq2\delta-2\kappa+5$ and
$$
c\geq|A^{\uparrow}|+|V^{\downarrow}|\geq(4\delta-2\kappa)+\delta-\kappa+2>4\delta-2\kappa.
$$
Further, if $A^{\uparrow}\not\subseteq V^{\uparrow}$ and $A^{\downarrow}\subseteq V^{\downarrow}$, then by $(f1)$ (see Lemma 10), $|V^{\uparrow}|\geq2\delta-1$, whence
$$
c\geq|V^{\uparrow}|+|A^{\downarrow}|\geq(4\delta-2\kappa)+\delta-\kappa+1>4\delta-2\kappa.
$$
Finally, if $A^{\uparrow}\not\subseteq V^{\uparrow}$ and $A^{\downarrow}\not\subseteq V^{\downarrow}$, then by $(f1)$ (see Lemma 10), $|V^{\uparrow}|\geq2\delta-1$ and by $(f3)$ (see Lemma 10), $|V^{\downarrow}|\geq2\delta-2\kappa+5$. Then
$$
c\geq|V^{\uparrow}|+|V^{\downarrow}|-2\geq(4\delta-2\kappa)+2>4\delta-2\kappa.
$$
\textbf{Case 2.2.} $A^{\uparrow}-V^{\uparrow}$ is independent and $|V^{\downarrow}|\geq3\delta-3\kappa+3$.
If $A^{\uparrow}\subseteq V^{\uparrow}$, then
$$
c\geq|A^{\uparrow}|+|V^{\downarrow}|\geq(4\delta-2\kappa)+2\delta-2\kappa\geq4\delta-2\kappa.
$$
Otherwise, by $(f1)$ (see Lemma 10), $|V^{\uparrow}|\geq2\delta-1$ and
$$
c\geq|V^{\uparrow}|+|V^{\downarrow}|-2\geq(4\delta-2\kappa)+\delta-\kappa\geq4\delta-2\kappa.
$$
\textbf{Case 2.3.} $|V^{\uparrow}|\geq3\delta-5$ and $A^{\downarrow}-V(Q^{\downarrow}_*)$ is independent.
If $A^{\downarrow}\subseteq V(Q^{\downarrow}_*)$, then
$$
c\geq|V^{\uparrow}|+|A^{\downarrow}|\geq(4\delta-2\kappa)+2\delta-\kappa-3>4\delta-2\kappa.
$$
Otherwise, by $(f3)$ (see Lemma 10), $|V^{\downarrow}|\geq2\delta-2\kappa+5$ and
$$
c\geq|V^{\uparrow}|+|V^{\downarrow}|-2\geq(4\delta-2\kappa)+\delta-2>4\delta-2\kappa.
$$
\textbf{Case 2.4.} $|V^{\uparrow}|\geq3\delta-5$ and $|V^{\downarrow}|\geq3\delta-3\kappa+3$.
Clearly
$$
c\geq|V^{\uparrow}|+|V^{\downarrow}|-2\geq(4\delta-2\kappa)+2\delta-\kappa-4\geq4\delta-2\kappa.
$$
\textbf{Case 3.} $f=2$ and $S\subseteq V^{\uparrow}$.
By Lemma 6, either $A^{\uparrow}-V^{\uparrow}$ is independent or $|V^{\uparrow}|\geq3\delta-5$. By Lemma 9, $|V^{\downarrow}|\geq2\delta-2\kappa+3$.\\
\textbf{Case 3.1.} $A^{\uparrow}-V^{\uparrow}$ is independent.
If $A^{\uparrow}\subseteq V^{\uparrow}$, then
$$
c\geq|A^{\uparrow}|+|V^{\downarrow}|\geq(4\delta-2\kappa)+\delta-\kappa\geq4\delta-2\kappa.
$$
Otherwise, by $(f1)$ (see Lemma 10), $|V^{\uparrow}|\geq2\delta-1$ and
$$
c\geq|V^{\uparrow}|+|V^{\downarrow}|-2\geq4\delta-2\kappa.
$$
\textbf{Case 3.2.} $|V^{\uparrow}|\geq3\delta-5$.
Clearly
$$
c\geq|V^{\uparrow}|+|V^{\downarrow}|-2\geq(4\delta-2\kappa)+\delta-4\geq4\delta-2\kappa. \qquad \Delta
$$
\noindent\textbf{Proof of Theorem 1}. If $\delta\leq2\kappa-3$, then we are done by Lemma 4. Let $\delta\geq2\kappa-2$. Then the desired result follows from Theorems 2-5, immediately. \qquad $\Delta$
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Archive
Naaman’s Healing
Naaman’s master considered him an extraordinary man. He was the military commander of Aram’s army, and he had won many important battles for Aram by the power of the Eternal. Naturally he was greatly esteemed by his king. Naaman was a fierce warrior, but he also had a skin disease. Now one time, the Arameans went out in raiding parties and took a little girl from Israel as their prisoner. The little girl became a servant to Naaman’s wife.
Girl (to Naaman’s wife): If only my master could be near the prophet in Samaria, the prophet there could heal my master’s disease.
Naaman became hopeful, and he went and told his king what the little girl from Israel said.
King of Aram: I am going to write a letter to Israel’s king, and I want you to take it to him immediately.
Naaman left with the king’s letter in his hand, plus 750 pounds of silver, 150 pounds of gold, and 10 sets of fineclothing. Naaman handed the letter to Israel’s king, and the king read it.
King of Aram’s Message: The man carrying this letter is my servant, Naaman. He has a skin disease, and I request that you heal him.
King of Israel (ripping his clothing): Who does he think I am—God? Why does Aram’s king think I have the power to kill and make alive again? What in the world makes him think that I can heal you of your disease? It is obvious that Aram’s king is trying to create trouble between us.
Elisha, the man of God, received word that Israel’s king had ripped his clothing, so he sent a message to Israel’s king.
Elisha’s Message: What has caused you to rip your clothing? Tell the man who has come to you for healing to come to me. Then he will be assured that a prophet lives in Israel.
The king told Naaman to go find Elisha, so Naaman showed up at Elisha’s door with his horses and chariots. Elisha did not show his face to Naaman, but instead sent instructions: “Wash yourself in the Jordan River seven times. The waters will heal you, and your skin will be back to normal. You will be cleansed.” Naaman boiled with anger as he left Elisha. He had come to his house expecting something much different.
Naaman: What is this! I came here thinking that Elisha would come outside and call upon the name of the Eternal One his God, and that Elisha’s hand would pass over my sores and heal my skin disease, not the waters of the Jordan River. The Abanah and Pharpar Rivers in Damascus are greater rivers than all the rivers of Israel combined, so why couldn’t I just go bathe in those and be healed?
Naaman then stormed away, boiling with anger. Later his servants approached and spoke to him with respect.
Naaman’s Servants: Father, if the prophet had told you to do some important thing, wouldn’t you have done what he asked? Why is it difficult for you to follow his instructions when he tells you, “Bathe yourself in the Jordan River, and be cleansed”?
So Naaman swallowed his pride, walked down to the Jordan River, and washed himself seven times, just as the man of God had instructed him to do. There, the miracle occurred. Naaman’s disease was healed: his skin was as new as an infant’s, and he was clean from the disease (2 Kings 5:1-14, The Voice).
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Montgomery County nixes proposed billboard plan
Company wanted to put digital signs on top of traffic signals
Published 5:30 am, Tuesday, August 2, 2011
Billboards attached to stop lights? Not in Montgomery County.
Montgomery County Commissioners said no to the idea of placing digital advertising monitors on top of traffic signals at various intersections.
Lucky Srinivasan, owner of Sterling Systems in Houston, recently presented a revenue sharing plan to commissioners that projected the county would make $42 million over 10 years through the sale of advertising space on the monitors.
All the county would have to do is allow Srinivasan's company to set up monitors and the county would collect 10 percent of what each monitor generated.
Srinivasan said the 50 to 72-inch flat screen monitors would be perched directly above traffic lights. Once the light turned red, the monitor would come on showing an advertising image. Just before the light changes to green, the monitor would shut off.
He asked commissioners for a chance to test out the monitors at an intersection, but with little fan fare.
Citing safety concerns, commissioners quickly shot down any interest for the program.
"I don't think people need another distraction from the roadway, especially in this day and age," Precinct 2 Commissioner Craig Doyle said.
Precinct 3 Commissioner Ed Chance felt that having monitors was an accident waiting to happen.
"I agree with Commissioner Doyle. We've got enough as it is with people using their cell phone. The last thing we want is people watching TVs as well," Chance said.
Srinivasan tried to quash the commissioners' concerns by explaining that the advertisements would only turn on after the traffic stops. He said his company is in discussions with CBS Outdoor to help with the instillation of the monitors.
"The screen comes on two seconds after the light changes red, then the (image) disappears two seconds before the light changes back to green," he said.
Srinivasan, a long-time developer in the Houston area, is the husband of the late Dee Srinivasan, the former mayor of Hedwig Village who died from heart complications in 2005 while in office.
He said that his advertising concept is new and he has yet to contract with a municipality.
Judge Alan B. Sadler pointed out the revenue sharing was a bit on the low side.
The projections Srinivasan submitted to the county was for Harris County, but he said Montgomery County could stand to make a similar amount.
While the municipalities are projected to take in revenues of $42 million over 10 years, the projected revenue for Srinivasan's company is $251 million.
"The revenue sharing for municipalities is kind of on the light side at 10 percent," Sadler said. "This is really an issue for the Commissioners since they're the ones with the red lights."
Srinivasan said that the company would be willing to work with the county to determine the right amount for revenue sharing.
"Initially there would be no cost to the municipality," Srinivasan said.
The commissioners voted unanimously not to pursue an agreement with Sterling Systems.
Srinivasan could not be reached for additional comment.
Nancy Crecelius, a resident from The Woodlands who watched the presentation, said she could not imagine billboards at intersections, and sees it as a public safety issue.
"I think it's the most ridicules thing I've ever heard," Crecelius said. "Are we so information-starved that we can't have a minute or two to ourselves at the stop light without being bombarded with advertisement? There's no need for it. I get enough advertisement in my life."
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Charge, parity, and time reversal symmetry is a fundamental symmetry of physical laws under the simultaneous transformations of charge conjugation (C), parity transformation (P), and time reversal (T). CPT is the only combination of C, P, and T that is observed to be an exact symmetry of nature at the fundamental level. The CPT theorem says that CPT symmetry holds for all physical phenomena, or more precisely, that any Lorentz invariant local quantum field theory with a Hermitian Hamiltonian must have CPT symmetry.
The CPT theorem appeared for the first time, implicitly, in the work of Julian Schwinger in 1951 to prove the connection between spin and statistics. In 1954, Gerhart Lüders and Wolfgang Pauli derived more explicit proofs, so this theorem is sometimes known as the Lüders–Pauli theorem. At about the same time, and independently, this theorem was also proved by John Stewart Bell. These proofs are based on the principle of Lorentz invariance and the principle of locality in the interaction of quantum fields. Subsequently, Res Jost gave a more general proof in the framework of axiomatic quantum field theory.
Efforts during the late 1950s revealed the violation of P-symmetry by phenomena that involve the weak force, and there were well-known violations of C-symmetry as well. For a short time, the CP-symmetry was believed to be preserved by all physical phenomena, but in the 1960s that was later found to be false too, which implied, by CPT invariance, violations of T-symmetry as well.
Consider a Lorentz boost in a fixed direction z. This can be interpreted as a rotation of the time axis into the z axis, with an imaginary rotation parameter. If this rotation parameter were real, it would be possible for a 180° rotation to reverse the direction of time and of z. Reversing the direction of one axis is a reflection of space in any number of dimensions. If space has 3 dimensions, it is equivalent to reflecting all the coordinates, because an additional rotation of 180° in the x-y plane could be included.
This defines a CPT transformation if we adopt the Feynman–Stueckelberg interpretation of antiparticles as the corresponding particles traveling backwards in time. This interpretation requires a slight analytic continuation, which is well-defined only under the following assumptions:
#The theory is Lorentz invariant;
#The vacuum is Lorentz invariant;
#The energy is bounded below.
When the above hold, quantum theory can be extended to a Euclidean theory, defined by translating all the operators to imaginary time using the Hamiltonian. The commutation relations of the Hamiltonian, and the Lorentz generators, guarantee that Lorentz invariance implies rotational invariance, so that any state can be rotated by 180 degrees.
Since a sequence of two CPT reflections is equivalent to a 360-degree rotation, fermions change by a sign under two CPT reflections, while bosons do not. This fact can be used to prove the spin-statistics theorem.
The implication of CPT symmetry is that a "mirror-image" of our universe — with all objects having their positions reflected through an arbitrary point (corresponding to a parity inversion), all momenta reversed (corresponding to a time inversion) and with all matter replaced by antimatter (corresponding to a charge inversion) — would evolve under exactly our physical laws. The CPT transformation turns our universe into its "mirror image" and vice versa. CPT symmetry is recognized to be a fundamental property of physical laws.
In order to preserve this symmetry, every violation of the combined symmetry of two of its components (such as CP) must have a corresponding violation in the third component (such as T); in fact, mathematically, these are the same thing. Thus violations in T symmetry are often referred to as CP violations.
The CPT theorem can be generalized to take into account pin groups.
In 2002 Oscar Greenberg published an apparent proof that CPT violation implies the breaking of Lorentz symmetry. If correct, this would imply that any study of CPT violation also includes Lorentz violation. However, Chaichian et al later disputed the validity of Greenberg's result. Greenberg replied that the model used in their paper meant that their "proposed objection was not relevant to my result".
The overwhelming majority of experimental searches for Lorentz violation have yielded negative results. A detailed tabulation of these results was given in 2011 by Kostelecky and Russell.
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{\bf Problem.} Let $f(x)$ be a function defined for all positive real numbers satisfying the conditions $f(x) > 0$ for all $x > 0$ and
\[f(x - y) = \sqrt{f(xy) + 2}\]for all $x > y > 0.$ Determine $f(2009).$
{\bf Level.} Level 4
{\bf Type.} Intermediate Algebra
{\bf Solution.} First, we claim there exist positive real numbers $x$ and $y$ so that $x - y = xy = 2009.$ From these equations,
\[x^2 - 2xy + y^2 = 2009^2,\]so $x^2 + 2xy + y^2 = 2009^2 + 4 \cdot 2009.$ Then $x + y = \sqrt{2009^2 + 4 \cdot 2009},$ so by Vieta's formulas, $x$ and $y$ are the roots of
\[t^2 - (\sqrt{2009^2 + 4 \cdot 2009}) t + 2009 = 0.\](The discriminant of this quadratic is $2009^2,$ so it does have real roots.)
Then for these values of $x$ and $y,$
\[f(2009) = \sqrt{f(2009) + 2}.\]Let $a = f(2009),$ so $a = \sqrt{a + 2}.$ Squaring both sides, we get $a^2 = a + 2,$ so $a^2 - a - 2 = 0.$ This factors as $(a - 2)(a + 1) = 0.$ Since $a$ is positive, $a = \boxed{2}.$
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Results of the 2014 NSERC Strategic Project Grants Competition
McGill received 10 awarded projects of 35 submitted applications. The success rate is 28.57% and a total amount awarded to McGill over 3-year period is $4,673,796.
Please find below – 6 successful awardees of 2014 Strategic Grants from the Faculty of Engineering:
Discovery Accelerator Grants awarded to Two Researchers
Denis Mitchell (Civil Engineering) and Andrew Higgins (Mechanical Engineering) were awarded grants from NSERC’s 2014 Discovery Accelerator Supplements Program (DAS).
The program provides “substantial and timely resources to researchers who have a superior research program that is highly rated in terms of originality and innovation, and who show strong potential to become international leaders within their field". Each award has a total value of $120,000 ($40,000 annually) and is normally paid over three years.
Previous Recipients
2013
- Marco Amabili (Mechanical Engineering)
- Raynald Gauvin (Mining and Materials Engineering)
- Alejandro Rey (Chemical Engineering)
- Thomas Szkopek (Electrical and Computer Engineering)
2012
- Damiano Pasini (Mechanical Engineering)
- Jeffrey Bergthorson (Mechanical Engineering)
- Francois Barthelat (Mechanical Engineering)
2011
- Showan Nazhat (Mining and Materials Engineering)
2010
- Nathalie Tufenkji (Chemical Engineering)
- Tal Arbel (Electrical & Computer Engineering)
2009
- Warren Gross (Electrical & Computer Engineering)
- Srikar Vengallatore (Mechanical Engineering)
2008
- Alejandro Rey (Chemical Engineering)
2007
- Meyer Nahon (Mechanical Engineering)
Winners of the 2013 William and Rhea Seath Awards Competition
James McGoff and Charles A. Vincent, both undergraduate students in Mining and Materials Engineering for their “Novel Insulation Inserts for Improving Large-Scale BioPharma Logistics Operations”. LIFEPACK, the company that James and Charles co-founded, is commercializing a novel insulation product for companies seeking to improve their large scale cold chain operations without disrupting their current shipping logistics. Adopting the LIFEPACK product is the quickest and easiest way for companies to improve their cold chain packaging by 40% to 80%. The product’s development was heavily based on a continuous dialogue with key partners in the biopharma industry. These industrial partners included pharmaceutical developers, centralized laboratory services, contract research organizations (CRO’s), blood banks, hospitals, and third party logistics providers (3PL). The award will go towards scaling and sustaining the growth of the company.
Professor Andrew Kirk, Electrical and Computer Engineering, Dr. Philip Roche of the Lady Davis Institute for Medical Research, and Professor Mark Trifiro, McGill Department of Medicine and Chief of Endocrinology at the Jewish General Hospital, for their “Multiplex Measurement of PCR Reactions by a Label Free Plasmonic Thermocycler”. This new, better, cheaper, and faster method in the performance of the polymerase chain reaction (PCR) has been demonstrated and patented at McGill. PCR is an essential tool in molecular biological investigations and dominates the field of molecular diagnostics, providing DNA fingerprinting for crime scene analysis as well as identification of disease processes (cancer resistance and susceptibility, hereditary illness and microbial infection). The award will meet the specific aim of delivering the prototype of a miniaturized, multiplex, energy efficient, and rapid real-time PCR platform that outperforms market leaders.
Professor Jun Song publishes in the Journal of Nature Materials
A recent article authored by Professor Jun Song (Multiscale Modeling of Materials Group, Mining and Materials Engineering) in collaboration with Professor William Curtin (Director of the Institute of Mechanical Engineering at École polytechnique fédérale de Lausanne in Switzerland) was published in the journal of Nature Materials (impact factor 32.84). Song's study, entitled “Atomic mechanism and prediction of hydrogen embrittlement in iron”, reveals clues towards fundamental understanding of hydrogen embrittlement, and provides a framework for interpreting experiments and designing next-generation embrittlement-resistant structural materials. For details, please read the report at the McGill Newsroom.
Xinyu Liu Receives the CMC Award
Professor Xinyu Liu of the Biomedical Microsystems Laboratory in Mechanical Engineering received the 2012 Douglas R. Colton Medal for Research Excellence from the Canadian Microelectronics Corporation (CMC). He is seen here with Dr. Ian L. McWalter, President and CEO of CMC Microsystems. This is awarded annually to a Canadian researcher to recognize excellence in research leading to new understanding and novel developments in microsystems and related technologies. Professor Liu was honoured for developing “innovative microrobotic and bioMEMS technologies that enable high-throughput, automated manipulation and characterization of single cells for biological and medical applications”.
Tom Gleeson Publishes Article in "Nature"
Professor Tom Gleeson of Civil Engineering and Applied Mechanics recently published in Nature (Article: 488, 197–200; 2012). His paper entitled “Water balance of global aquifers revealed by groundwater footprint” received extensive international media coverage, including articles in the Los Angeles Times, Vancouver Sun, La Presse, Chicago Tribune, le Nouvel Observateur, Huffington Post, Nature Geoscience, Sciences & Vie, Sciences & Avenir, CBCnews, Nature News, Science News, CNBC, Agence Science-Press, Tahiti Info, Orlando Sentinel, Boise Weekly, Hong Kong Daily News, The Nation (Pakistan), Newsweek (Pakistan), the Daily Star (Bangladesh), the Saudi Gazette (Saudi Arabia), Terra (Spain) and Yesil Gazette (Turkey) following news wire stories by Reuters and Agence France-Presse. Blog coverage includes the New York Times, National Geographic, NPR , Scientific American, Washington Post, Kennislink, Business Insider, Deutsche Welle, Medical Daily, rtbf.be, pieuvre.ca, revmodo.com and landfood.ubc.
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TITLE: Some basic properties of Pisot number
QUESTION [1 upvotes]: A Pisot number is an algebraic integer $>1 $ and all of whose conjugates have modulus $<1$. First assume that $q$ is a Pisot number. Denote by $q_1,\ldots, q_d$ the algebraic conjugates of $q$. Let $P(x)=\sum_{i=0}^n\epsilon_i x^i$ be a polynomial with coefficients in $\{0, \pm 1,\ldots, \pm m\}$. Suppose that $P(q)\neq 0$. Then $P(q_j)\neq 0$ for $1\leq j\leq d$, and $P(q)\prod_{j=1}^dP(q_j)$ is a non-zero integer
How can we prove this result?
REPLY [1 votes]: The fundamental theorem of symmetric polynomials says that a symmetric polynomial $Q(x_1,x_2,\ldots,x_n)\in \mathbb Z[x_1,x_2,\ldots,x_n]$ is a polynomial in the elementary symmetric polynomials $\sigma_1,\sigma_2,\ldots,\sigma_n$, i.e. $Q(x_1,x_2,\ldots,x_n)=R(\sigma_1,\sigma_2,\ldots,\sigma_n)$ for a polynomial $R\in\mathbb Z[x_1,x_2,\ldots,x_n]$.
The polynomial $P(q)\prod_{i=1}^dP(q_i)$ is symmetric in $q,q_1,\ldots,q_d$, therefore $P(q)\prod_{i=1}^dP(q_i)=R(\sigma_1,\sigma_2,\ldots,\sigma_n,\sigma_{n+1})$ for a polynomial $R\in\mathbb Z[x_1,x_2,\ldots,x_n,x_{n+1}]$. Since $\sigma_i$, the elementary symmetric polynomials on $q,q_1,\ldots,q_d$, are the coefficients of the minimal polynomial of $q$ (here we use the fact that $q$ is an algebraic integer) it follows that $\sigma_i$ are integers (see this). Therefore $P(q)\prod_{i=1}^dP(q_i)=R(\sigma_1,\sigma_2,\ldots,\sigma_n,\sigma_{n+1})\in\mathbb Z$.
As Gerry Myerson said this has nothing to do with Pisot numbers. It is true for any algebraic integer.
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She used our problems to get closer to you. I'm not suppose to be upset. I was just jealous n crazy becomes I said she wasn't just trying to be your friend. Now I'm the crazy ex. Ya'll have fun with that. She'll need another friend soon enough. If she did it with you she will do it to you.
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A long time ago when we met you were lonely
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Description
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| 360,269
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TITLE: ODE problem with an integral on one side.
QUESTION [0 upvotes]: $$(f(x))^2=\int_{0}^xf(t)dt$$
I have this ODE problem in my textbook, the question is to find all $f(x), (x>0)$. I do realize that the left side of the equation is the derivation of the right side (obviously squared). However, I found my self stuck after this.
I tried to assign the integral as $U(x)$, and so the ODE that I got is :
$(U(x)')^2=u(x)$.
REPLY [2 votes]: Firstly we find all the absolutely continuous functions $f$. By taking the derivative at both sides we have
$$
2f(x)f'(x)=f(x)
$$
hence we have $f(x)(1-2f'(x))=0$, so if $f(x)\neq 0$ then $2f'(x)=1$, which means $f(x)=\frac{1}{2} x+\mathrm{constant}$ for some region $\{x\geq C\} $. Therefore the space of solutions can be described by the set
$$
S=\left\{f\in\mathrm{AC} (0,\infty )\vert f(x)=0\ \text{for }x\leq C,\ f(x)=\frac{1}{2} x-\frac{1}{2} C\ \text{for } x\geq C\right\}
$$
Since $S$ is closed under the norm $\Vert\cdot\Vert_{\infty } $, this is the set of all solutions to this integral equation.
| 161,048
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Stats & Rankings for Xu4N F. Gr3G0Ri0
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All the news that shouldn't be news for November 20, 2015 including: - Commentary: Stop Being an As… Paul Tourville Nov 19, 2015 35 views
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Lawrence Krauss talks about how science is being abused and misused in politics.... and why it is h… Morgan Matthew Oct 1, 2008 65 views
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TITLE: keeping track of proper classes vs sets when showing ZFC has a countable model (assuming ZFC is consistent)
QUESTION [8 upvotes]: I've heard the claim that ZFC has a countable model in ZFC if ZFC is consistent. As far as I can tell, this is a straightforward consequence of Gödel's completeness theorem and the Löwenheim–Skolem theorem.
I'm curious, though, how careful one needs to be about bookkeeping details related to sets and proper classes when formulating the argument ... or if the issue just doesn't come up for an obvious reason that I'm missing.
So, my question is twofold
Does the completeness theorem give us a model over a set anyway?
If we pick a model over a proper class initially, can we use Löwenheim–Skolem?
When I nonstandardly say a model over X, I mean a model whose universe is X. I nonstandardly say a model in X when the fundamental set theory used to construct it is the set theory X.
ZFC is a first order theory with one relation symbol $\in$, by Gödel's completeness theorem it has a model. I'm not sure whether this theorem promises us the existence of a model over a set or if the promised model could be a model over a proper class instead.
Also, ZFC does not have any finite models, since $\emptyset, P(\emptyset), P(P(\emptyset)), \cdots$ are all distinct, where $P$ denotes the powerset.
Suppose for the sake of argument that we conclude that ZFC has a model over a proper class (using ZFC as the raw material for building the model). In that case, we interpret $\in$ as $\in$ in ZFC. So, we've built a trivial model, let's call it $M$. The universe of $M$ is the class of all sets in ZFC. It's possible that by applying the completeness theorem, we could have gotten a model over a set, which would make this problem go away. I'm not sure. Suppose we decided to pick $M$ as our model even if it's an inconvenient choice.
The second prong of the argument, as far as I can tell, is an application of the (downward) Löwenheim-Skolem theorem. The downward Löwenheim-Skolem theorem does not address the [proper class]-set boundary explicitly, so I'm wondering how the argument works. This question invokes Löwenheim-Skolem as an explanation for the existence of countable models.
In the proof of the Löwenheim-Skolem theorem sketched in the Wikipedia article, we use the axiom of choice repeatedly to select elements of the universe that have to be included in the new universe for the model we're constructing.
A strict reading of the axiom of choice promises nothing about the ability to make arbitrary choices over proper classes (and a choice function cannot have a proper class as a domain), so I'm not sure how to get from my trivial self-model of ZFC $M$ to any model whose universe is a set, let alone a countable set.
REPLY [10 votes]: Does the completeness theorem give us a model over a set anyway?
Yes. The completeness theorem says that every consistent theory has a model, and in this context, "model" is defined to mean a model over a set.
If we pick a model over a proper class initially, can we use Löwenheim–Skolem?
This is a subtle issue--what does it even mean to have a model over a proper class? The issue is that the usual recursive definition of what it means for a structure to satisfy a formula does not work when your structure is a proper class. So, if you have a structure $(M,\epsilon)$ where $M$ is a proper class, you cannot even state $(M,\epsilon)\models ZFC$ in the language of ZFC.
However, for any individual sentence $\varphi$, you can state $(M,\epsilon)\models \varphi$, by carrying out the recursive definition "by hand" in the specific case of $\varphi$. So typically, when people talk about "class models", what they mean instead is that there is a theorem scheme which says that for each axiom $\varphi$ of ZFC, you can prove that $(M,\epsilon)\models\varphi$. Alternatively, you could assume that your class model has the special property that you actually can define a formula satisfaction relation for it, but this is rarely actually true in practice.
As for carrying out a Löwenheim–Skolem argument for class models, you can do this as long as you can define satisfaction with respect to all the formulas you care about. (This means you can carry it out as usual if your class model has a satisfaction relation, or if it doesn't, you can only carry it out to get a submodel that satisfies finitely many particular axioms of ZFC, rather than all of ZFC.) The issue with the axiom of choice that you point out can be avoided using Scott's trick. For instance, every time the proof of Löwenheim–Skolem would have you make a choice of an element with some property, instead of choosing just one element, you can pick all the elements with the desired property that have minimal rank. There may be uncountably many such elements so this may not give you a countable model, but it will at least get you down to a set-sized model. Then, you can use ordinary Löwenheim–Skolem to get down to a countable model.
Note that in this weak (theorem scheme) sense of "class model", you do not even need to assume that ZFC is consistent in order to get a class model of ZFC: $(V,\in)$ is a class model of ZFC. The argument above then shows that for any finite subset of the axioms of ZFC, you can prove there is a (set) model of those axioms. This is known as the reflection principle. (Crucially, though, this is a separate theorem for each finite set of axioms, rather than a single theorem with a quantified variable representing an arbitrary finite set of axioms. As a result, you cannot combine it with compactness to conclude that ZFC is consistent.)
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TITLE: Probability on a Die
QUESTION [1 upvotes]: The normal probability of a number in a regular die (6 faces) is $\dfrac{1}{6}$. Let in an addicted [that is, "loaded"] die, the probability of a even number (2, 4 and 6) be twice the normal probability;
I've got such outcome: $regular\space probability \space on \space evens \space is \space \dfrac{3}{6} $, doubling it, it would become $\dfrac{6}{6}$, in other words, a certain event, it sounds strange to me, is that right ?
Thanks in advance;
REPLY [2 votes]: You might check the wording of the question. If the probability of each even number is twice that of a normal die, you are correct. If the probability of each even number is twice that of each odd number, the result is different-then the evens come up $2/9$ each for a total probability of $6/9=2/3$
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Written By:
- Date published:
6:00 am, January 10th, 2017 - 144 comments
Categories: open mike - Tags:
Open mike is your post.
For announcements, general discussion, whatever you choose. The usual rules of good behaviour apply (see the Policy).
Step up to the mike …
This from RadioNZ – haven’t looked at all the details but sounds interesting.
9 Jan 2017
RNZ helping launch new digital innovation for Radio Stations
“Vox Populi” – latin for ‘voice of the people’ – takes on a whole new meaning as RNZ helps the launch of the diigital innovation VoxPop. It’s a new way of giving listeners the chance to give us feedback on stories – and have your voice on air. 2,45m
Try these for voxpop. Don’t know about the other one. Seems about Oz – as so often they spoil the proceedings.
or
The future of manufacturing employment – robots are becoming cheap enough that even third world wages aren’t low enough to compete. And yet, New Zealand still lags in using them. In terms of robots per 10,000 manufacturing workers, Russia 3, Indonesia 6, Brazil 11, NZ 41, China 49, USA 176, Germany 301, South Korea 531.
And automation does the one thing that eliminates trade – it removes economies of scale. And that will mean that we’ll have to be economic.
Right we will have to write shorter quicker comments I can see. Perhaps a guide beside us with common words matching each letter of the alphabet. Then a lot more phrases like WTF and LOL and IIRR. There will be a little guide with newest acronyms that people can have in a small window or print off. Much more efficient and save fingertip skin.
Andre you sound as if you are welcoming low cost competition for the few jobs available now on random part-time basis. The people are going to have to form a parallel government called WGAD (We Give a Damn) with slogan JUNABAGPCI (Join us now and bring a good practical costed idea).
And one idea will be to start guilds in each town and tell people of the value when they commit to the producers in their town first before looking at the tempting stuff made overseas.
Then there are the NZ labels and designs made overseas China, Vietnam.
They will get a look in after buying locally made. Shopping will have to be to build one’s own economy. Guilds will be started and take on apprenticeships and the locals will support this by spending strategically on local goods. Any sneers, go blow your nose.
On religion’s importance to those in Europe and USA.
<iLTwo…
I think you will find religion is a taboo subject greywarshark, just like that other highly divisive subject of Israel.
Too much emotion and antagonism involved.
Try it as a post.
“Religion is the sigh of the oppressed creature, the heart of a heartless world, and the soul of soulless conditions.”
Conversely one could say “Religion is a powerful, moneymaking method of controlling people”.
I chimed in with my Marx quote to support greywarshark’s thesis that Americans’ religiosity is connected with the lack of social solidarity in the US, and the high levels of social precariousness. In other words; religion is a kind of mythical security blanket for people who do genuinely lack real security. I don’t see how your statement relates to that — are you suggesting that Americans religious feelings are actually foisted on them by powerful financial interests?
Strangely enough, history generally shows religion closely connected with repressive regimes. Among the great, cruel Tsars of Russia, only Stalin was an atheist, and he all but replaced the Greek Orthodox religion with Marxist dogma. Putin has restored the Greek Orthodox.. The American oligarchy have their silly fundamentalist Bible Belt Christianity.
By and large, religion has largely functioned as a blunt instrument of social control. The few who get transports of spiritual delight out of religion are the lucky but deluded ones.
If you are one of those few, enjoy it for as long as you can.
‘Out-sourcing’ = ‘contracting out’ = PRIVATISATION.
‘Inefficient’ is corporate speak for we haven’t yet got our filthy hands on it!
‘Inefficient’ was the unsubstantiated, unproven corporate mantra behind the ‘commercialise, corporatise’ PRIVATISE Neo-Liberal Rogernomics agenda.
In my considered opinion as an anti-privatisation / anti-corruption campaigner – the only ones who have benefited from running public services in a more ‘business-like’ way – are those businesses which have been awarded the contracts.
And how much corruption has been involved in the awarding of contracts across central and local government?
Locally, nationally and internationally?
Penny Bright
Independent candidate
Mt Albert by-Election
#PennyBrightNZ
Mário Soares has died.
Portugal’s former president and prime minister, Mário Soares, a central figure in the country’s return to democracy in the 1970s after decades of rightwing dictatorship, has died aged 92.
[…].
Also, an obit from a Portuguese newspaper – google translate
Mario Soares left us and left us everything
He was a bourgeois revolutionary. The bourgeois criticized him for being revolutionary, and the revolutionaries criticized him for being bourgeois. That is why he is so refreshingly modern: we have not yet come close to what he wanted for us.
Mário Soares took nothing with him. Left everything with us. This is the greatest generosity a person can have: wanting everything to others and dedicating his life to fighting for it – and for us.
The list of reasons why Putin might have wanted Trump just keeps growing. There’s Junior telling us back in 2008 Russians made up a disproportionate part of the business, there’s just the general principle that shit-stirring, mayhem and loss of credibility in the US is good for Russia, and then there’s Russia and Big Oil wanting to pump out and burn vastly more fossil fuels…
Ahhhh the liberal tears keep being so salty. Don’t forget the Russians didn’t want warhawk Hillary Clinton starting a nuclear conflict over Syria, either.
That’s real motivation. Detente.
Let’s see how former Exxon Mobil CEO Tillerson’s confirmation goes. That’s going to be a rough one and a major test of Trump’s political management on the Hill.
Don’t forget the Russians didn’t want warhawk Hillary Clinton starting a nuclear conflict over Syria, either.
Thing is, Andre’s reasons are actual ones, as opposed to fantasy ones like the above in Colonial Viper’s head.
the threat was real ie clintons solution for syria was a no fly zone,which was an act of war.
The definition of the word “real” is no longer useful if we allow it to encompass possibilities at the end of a tenuous chain of “ifs.”
The garden of forking paths is well known,as is the problem of future contingents.
The problem at the time was real as the pathway was to an illegal act of war if enacted ( a syrian no fly zone) .
Well, according to that theory any US personnel on the ground and US airstrikes on Syrian territory (even ones in territory not in the direct control of the Syrian government) has already been an “act of war”, yet not precipitated a nuclear war.
Correct. Just as it would be an act of war if Syrian troops deployed in, say, North Dakota.
Yes. It’s truly amazing how many acts of war that the US commits and never gets called on.
Indeed.
And yet still no nuclear conflict.
It’s almost as if international relations are complex interactions betweeen state, non-state and substate actors, rather than just a simple “OMG, that’s technically an act of war, press the fucking button!!!”
Which is why a person with a brain is preferable to an oompah-loompah with poor impulse control as head of state.
I’m more concerned with them not being called on them when they call anything that anyone else does that’s exactly the same as what they do such as fast as they can.
Well, you’re not the only one who prefers to call moral equivalence rather than avoid geopolitics being dominated by an orangutan with a twitter account.
Well, according to that theory any US personnel on the ground and US airstrikes on Syrian territory (even ones in territory not in the direct control of the Syrian government) has already been an “act of war”, yet not precipitated a nuclear war.
Um no.
Airstrikes against isil and Nustra are legitimate targets ,as authorized by the UN sc resolution. US coalition bombing of syrian soldiers ,they used the get out of jail card of a mistake undertaken in good faith to legitimize the fact.
So wouldn’t that also legitimise a no fly zone?
Nope.That constrains the ability of syria (and russia as an invited party) to use the self defence mechanisms of the UN charter.
The no fly zone would need a separate un resolution,which would not get through the SC.
Hence it always was a binary outcome,either Clinton was full of shit or as PONTUS would have invoked an unlawful act of war,
No.
It depends on the extent of the proposed no fly zone, but if say Syrian airstrikes on non-ISIL groups take the pressure of ISIL (because non-ISIL are being bombed), then the no-fly zone satisfies the current UN request.
Russia and Syria might have arguable legal justification to defend themselves (just as they had the arguable justification when the US accidentally bombed that outpost), but even if the current airstrikes bombed something the Syrians didn’t want bombed, that’s still an arguable act of war.
There are very few binary situations in law or international relations. The no-fly confluence of both is not such a situation.
The key phrase here is “in compliance with international law, in particular with the United Nations Charter”. The Security Council has asked member states to take on ISIL, but notice they require that it is done within the bounds of the UN charter, which forbids states from attacking other countries. The Syrian government has asked the Russians and Iranians to provide military support, which means that support conforms to the UN charter. If the Syrian government (as a member state of the UN) had asked the US to bomb the ISIS rebels besieging government forces at Deir ez-Zor, in the famous incident there, and the US had committed an honest mistake, and blown up the Syrian government troops instead, that would have been merely a nasty diplomatic incident, but the Syrian government had not authorised that bombing (still less had they authorised a “mistaken” bombing of Syrian Army positions), and that meant it was an act of war and a breach of the UN charter.
In practice, the US gets away with it not because international law says it’s OK, but because they have the military and political clout to get away with it, whatever its legal status might be. This is the norm for US military action: they have invaded or attacked countless countries over the years without even a pretence of legal justification. There are exceptions, of course, where the law has been on their side, but in practical terms that’s of no consequence: the law is for the weak; the strong can rely instead on force.
Pro tip: you can’t learn about coding or the UN charter with a couple couple hours trolling Google. We’ve got unilateral deployments and legal exemptions from prosecutions now, because:
US: motion to bomb Syria
Russia: lol nope, we veto that
Russia: motion to bomb Syria
US: lol nope, we veto that
If the US goes all gung-ho on oil extraction…fracking etc…,wouldn’t that actually hurt the Russian economy (its oil sector revenues)?
And if Trump increases the US’s nuclear arsenal (as he’s sign-posted) then wouldn’t that also have the potential to hurt the Russian economy (ie – a ‘new’ arms race bleeding resources/budgets)?
Clinton would probably have been more rational on the extraction front and, war monger as she is, less inclined to increase the US’s nuclear stockpiles.
edit – the prospect of more cordial relations with one President as opposed to the other is a genuine reason to prefer one to the other. Nothing suspicious about that, is there?
Fiat Chrysler announces US$1B investment in USA, 2,000 new American jobs to be created
But denies it has anything to do with Trump
A few hours later:
Fiat Chrysler may have to abandon Mexico production if Trump tariff is high
The truth comes out. Nothing to do with Trump eh??? LOL
Again this is the brilliance of Trump as a business man. He knows how big business makes decisions. He doesn’t need to individually talk to the CEO of Fiat Chrysler to signal to them what they need to do.
“the brilliance of Trump as a business man”
Pffft. The guy has been bankrupt so many times that only financiers outside the US will continue doing business with him. But keep wearing out those kneepads, sir.
I think Trump has been bankrupt once. Not unusual for an entrepreneur. Thereafter some of his companies have been bankrupt or liquidated. Again, not unusual in the entrepreneurial world.
Many times and quite cynically. As mentioned, US banks et al refuse to touch the guy.
Really? How many times has Donald Trump, personally i.e. not his business entities, been bankrupt?
I can only recall one occasion.
Well, there’s his ongoing complete moral and ethical bankruptcy. And I understand daddy bailed him out several times, which may have taught him to make sure it’s a company that goes under, not him. But I haven’t found where he’s actually filed for personal financial bankruptcy. Care to educate me?
greater love hath no man than to lay down his creditors for his ego… half a dozen times or more.
Not to mention settle fraud investigations and so on.
I was wrong. The truth is that Donald Trump has NEVER filed for bankruptcy himself. He has however for companies that he is associated with.
Sacha
CV was being sarcastic. Trump is a brilliant businessman at doing whatever and still holding onto plenty of dosh.
Someone who can go bankrupt and just go around the barriers, is a Grand Master of Chicanery. That reminds of University of Chicago, the place where Milton Friedman et al and his theory came from. He’s partly Milt’s creation.
” the brilliance of Trump as a business man”
Uh, yeah, he’s done astoundingly well out of being an obnoxious loudmouth buffoon on TV. And he’s done mediocre-to-middling in real estate, a business he learned at his daddy’s knee. Considering what he was given to start with, he’s way underperformed the general New York real estate market and the general stock market.
On the flipside, he’s been an abject failure at everything else he’s tried, apart from fleecing investors and stiffing contractors. FFS, how do you lose money running casinos?
“he’s been an abject failure at everything else he’s tried, apart from fleecing investors and stiffing contractors”
Yet this “abject failure” has many millions (billions?) of dollars, a loving tight knit family and become the President of the United States.
So Im guessing you are more of a failure than he.
Any verification on what he is worth?
Forbes Sept 2016: Donald Trump’s value falls US$800M to US$3.7 Billion
What a financial failure
That just proves that you have a warped values system as you celebrate his stealing from others. Not that Trump is successful.
What a ridiculously callow comment James. It’s as good as saying that no one can comment on anyone/anything unless they enjoy (?) more or less equivalence with the players and involvement in the things. Almost as ridiculous as the never-endingly malignant CV, self-proclaimed as the purest leftie in the whole of New Zealand (hahaha), suddenly expert in ‘mega-business’. And adoring of the sharpest practices associated therewith.
Have a listen to Meryl Streep re the NYT reporter. The foulness she identifies is all swept away because (however questionably or by virtue only of the accident of birth) Trump got his “millions (billions?)” ? Yeah I know…….success/failure is ALL about money and the surplus/deficit thereof. I understand how that’s your buzz James but in New Zealand’s purest leftie……WTF ?
Socialist intervener. Wouldn’t have thought that was your cuppa.
Hmmm.
Was the main motivation a theoretical 35% import duty that Trump might or might not be able to push through congress, vs $1.7Billion in Michigan tax credits exchanged for $1billion investment by Fiat-Chrysler that was announced a little over a year ago, I wonder.
BMW is staying in mexico. I guess detroit didn’t promise them tax credits.
Ah, so massive tax payers subsidies to line private pockets.
Yup. The art of the deal /sarc
My pick: at this rate, in 2020 Trump will win the Rust Belt with bigger majorities than 2016.
If he does, it’ll be by claiming credit for shit he had nothing to do with.
Like manufacturers taking tax breaks that states had negotiated when Trump had barely announced his candidacy.
Oh, and blaming Obama for things Trump fucked up.
Ain’t politics grand. Watch out for Toyota to fold soon and put new big money in the US.
11 days to 16 years of Trump rule.
16 years…?
two terms for each face.
Junior, Ivanka, Eric and beyond……
What about Jared?
nah, our champion who railed against dynastic presidencies when hillary was running wouldn’t be so hypocritical, surely /sarc
For one who thinks he is so knowledgable, how come you think Trump will be in power for 16 years, given the two terms that all other presidents legally have in office. Guess he could try to change the law and become an octogenarian dictator for life or family could become de facto president.
I think you are getting carried away CV on your predictions.
I always backed you on your reasoning for Trump to beat Hillary, but I think it would be safer for you to predict that Trump will be an abject failure than claiming there’s going to be a Trump dynasty. Better still why don’t you just buy a lotto ticket instead! Cheers.
Well my prediction is 8 years of Trump; the 16 years of Trump rule thing is tongue in cheek.
But I am also indicating that (IMO) by looking at the tea leaves, the Trump family is planning ahead at least that far.
BTW I’d bet anyone real money that Hillary Clinton is trying to figure out a way that she can run again in 2020
My name is CV and it’s “Life” I say, “Prez for Life !” The US and The World really does ‘owe’ Trump 16 years to Life and stuff the Constitution. Why though CV do you presage it being all over by 31 January potentially ?
To think all this fucked-up hubris of months now started with a tatty internecine dispute in Dunedin. 150 km south-west of Gore, the country music capital of New Zealand ??? “Stand By Your Man……”
The multiverse works in truly mysterious ways
It is not funny to tease people unduly, CV – you may have pushed taking the piss a bit far, if I have managed to understand correctly (always a highly debatable point..)
True true. Though I have friends who live in Gore so I took exception 🙂
Either way Americans will pay more for their Chryslers
Bit weird, these lefty Trump fans. Normally you’d think someone inheriting huge amonuts to get started, exploiting loopholes to dodge tax, rip off investors and get away scot-frree (oh it wasn’t me going bankrupt, it was my company!) would be the natural enemy of the left. Basically capital personified. But the alt-left are so desperate to convince themselves that senpai is going to fix ecerything they’re praising Trump for being a typical business psychopath.
[Kindly (I thought) I decided to make sure you had seen this before deciding what to do. You got quotes or links? Or is it an apology? Or…] – Bill
This?
Link in my above comment is about –
Malcolm Roberts, Australian senator, wanting to bash NZs because we voted in the UN on Israeli settlements!
Just a piece for those following Brexit. Ongoing thoughts and doubts.
Ah OK. Thanks, greywarshark. The link in your above comment @ 9.45am, like the earlier one on voxpops, goes to an audio file with not content. when I click on the play button for both, I just get the same audio I listened to a couple of weeks back.
So I thought the second link was an attempt to post the correct link about voxpops.
carolyn-nth
I am having a bit of trouble with Radionz new web site. I am a bit of a moaner about this so have tried to nut it out for myself as well as get help, and may have the answer sorted eventually. I liked the old setup before they glamourised and filled up the pages with photos for those who need visual affirmation to get context for what they are hearing and reading.
I was reading a blogpost the other day and an older person, I feel, said he had just updated his computer and spent the morning closing down the apps he didn’t need so that it was just like his olde one except faster.
I agree.
Ah. Yes, I see your point, greywarshark, on RNZ changes.
The vox pop is an interesting thing. I’ll be interested how it develops.
The Brexit article is interesting. I can see a bit of both sides, there.
But focusing on eastern Europe is an interesting way to go. There are problems with the way Germany has dominated the EU – works more for them than southern Europe.
Roberts is a climate denialist and one nation senator.
Oz voted with us as well but he doesnt have a go at turnbull does he ?
Oz has its own little tea party banging a drum for support wherever it can get it. Best we all ignore them as the xenophobic dinosaurs they are voted in by queenslanders.
In the Guardian, a piece from Brother Cornel.
Though I think his statement that ‘The age of Barack Obama may have been our last chance to break from our neoliberal soulcraft.’, is overly polite. Obama wasn’t financed into power to change anything.
As with despots and tyrants – nepotism.
/
Robert Reich’s 15 signs of impending tyranny.
As a follow-up, an analysis supporting the idea that racism, not economics, was the main driver for Trump’s success. Race, not Rust!
I read that article and it struck me how little the logico-mathematical core made sense. But the headline is a tell: an idea that “needs to die” … so that the neoliberal juggernaut can continue to roll!
Expand your thought processes I would say, Andre
Your country has been a tyranny for an age, already. Do you not understand that, or will narrow parameters continue to discombobulate you?
There is nothing ‘impending’ about it!
The mind is beyond incredible abstract beauty
Long time since I’ve seen that word used One Two. Thanks.
By “New Zealanders” do they mean “ex-Labour MPs-cum-National stooges”?
heh
That’s opaque.
Might have something to do with this.
It was illegally occupied East Jerusalem, and it was a massacre, not a “war” in 2014
—but Eric Frykberg’s dishonest and loaded report leaves all that out.
We have had several looks at the mediocrity and lack of professionalism of Radio New Zealand’s political commentary, from Jim Mora’s light chat vehicle The Panel through Kim Hill’s tendency to indulge nasty attack dogs like Alex Gibney and A.A. Gill, to the dismal naïveté of Bryan Crump, Jesse Mulligan, Anusha Bradley and John Campbell.
This morning we must, sadly, add one more to this unedifying list. Long term sufferers of RNZ’s steadily deteriorating news service will be familiar with the name of Eric Frykberg. In the following item about a courageous New Zealand teacher in the besieged enclave of Gaza, Frykberg—or perhaps it was some nervous higher-up—manages to undermine it by tagging on three final paragraphs which are pure black propaganda. If you can read this without gnashing your teeth in fury, then you are either an ACT cultist or you simply have no clue….'t-quit-gaza
Not sure how you tag those paragraphs as propaganda Moz ?
Okay, mullet, here they are, in italics, with my comments after each one. Of course, those three paragraphs are there for no other reason than to distract from and undermine the bravery of Julie Webb Pullman. They certainly are not relevant, even slightly, to her story.
1. The Foreign Affairs Ministry warnings came after a Palestinian man rammed his truck into a group of Israeli soldiers in Jerusalem, killing four of them and wounding 17, raising tensions throught the region.
The truck attack occurred in illegally occupied East Jerusalem. They were IDF soldiers, and as such were legitimate targets for resistance. International law recognizes the right of occupied people to resist with force. And, no, I do not endorse such actions; I do not support Palestinians using terror tactics against Israelis. I think they should resist this brutal occupation actively but nonviolently, as they do 99 per cent of the time. I don’t support people shooting other people either, even if they are provoked beyond reason as the Palestinians are. But international law does recognize the right to defend yourself militarily if attacked, and that’s what some desperate Palestinians occasionally feel driven to do. Russian soldiers in Chechnya suffered similarly, so did U.S. soldiers in Iraq, and so did German soldiers in France. It is worthwhile considering what would happen to, say, any heavily armed Iranian soldiers who walked through Tel Aviv, or Houston, or London, routinely cowering the population.
2. The truck attack was praised by the Hamas rulers of Gaza.
This banal sentence at least is undisputed, bearing out the old adage that even the most egregious propaganda usually has at least some truth to it.
3. The kidnap and murder of three Israeli teenagers in the West Bank in 2014 led to a full scale war in Gaza that year.
Again, Eric Frykberg—whether unwittingly or by self-censoring—neglects to mention that those teenagers were kidnapped in the illegally occupied West Bank. He—or some nervous sub-editor—then compounds this misinformation by adding another, even nastier, piece of disinformation, by labeling the massacre of the trapped, unarmed population of Gaza as a “full scale war.” It’s worth contrasting the shoddy work of people like Erik Frykberg with the words of Israeli soldiers who actually take part in Israel’s brutal repression of the Palestinians….
Those bastards with their facts! I’m positively livid!
Sorry, Gabby, but you’ve lost me. Which bastards, and which facts?
Anti-smoking campaigner says tax hikes aren’t working–antismoking-campaigner-and-former-nz-tobacco-tax-supporter-says-it-isnt-working
They kind of work out quite nicely for the pharmaceutical industry that gets $x?? of public money by way of subsidy for all the patches and gums and what not they supply.
They work out quite nicely as a revenue stream too. (The article covers that).
The only area the price increase has an effect is in youngsters not taking up smoking. I stumbled across all this when compiling the Chematistic Camel post back when. Oddly. Smoking rates have increased among the oldest of us…which I’m waiting for someone to spin as a sign of the health benefits of smoking 🙂
Anyway. Vaping would have a huge impact. I’m an ex-smoker and know many people who have only been able to quit by switching to vaping. I don’t know of a single person who smoked and took up vaping who then defaulted back to smoking.
But here’s a thing – there’s no money in vaping for either the government nor the pharmaceutical industry. So it’s ‘dangerous’ and a ‘gateway’.
“They kind of work out quite nicely for the pharmaceutical industry that gets $x?? of public money by way of subsidy for all the patches and gums and what not they supply.”
Indeed.
“Vaping would have a huge impact”
One of the concerns with vaping is they are known for blowing up in your face.
Kiwi entrepreneurs call for legalisation of cannabis, following worldwide success
Fantastico, may it happen.
Did you know that nitrogen is one of the nutrients that cannabis thrives on? Dairy farming diversification springs to mind.
There is massive public support for cannabis legalisation throughout NZ and across many social groups including right wingers.
Watch the alcohol lobbyists continue to fight against the legalisation of cannabis, they will be super concerned that it may eat into their profits.
Playing nicey-nicey with people outside the “regular commentariat” is over-rated. We’ve been doing that for decades and racism hasn’t subsided in any meaningful way at all. In fact if you measure it on outcomes, it’s got worse.
“I’m just saying what you’re all thinking.”
[TheStandard: A moderator moved this comment to Open Mike as being off topic or irrelevant in the post it was made in. Be more careful in future.]
If you think I’m suggesting playing nice-nicey you’re really not paying attention and you’ve missed the point.
Besides, the massive testosterone-fest that TS commenting often is regularly bleeds commenters and authors, so I don’t see how your approach is designed to work.
edit, and while I’m at it, how about you take a look at your contributions to TS being hostile to women. Male supremacy, the dynamics are remarkably similar. See, not particularly nice-nicey.
It’s designed to work by bringing the issue of white supremacy front and centre and putting its advocates on the defensive.
Hostile to women? Can you illustrate that with an example please?
Have you thought about why there are no regular feminist authors writing on TS? Or why the sole current regular woman author won’t write from a feminist perspective? What’s happened to all those women? Have you even noticed that there is a problem?
Like I said, the dynamics are remarkably similar. How about I start calling you a misogynist then and attacking you every time to you do shit that makes this place worse for women. I’m not actually comparing you to James, I just want you to pay more attention to what is going on here and the fact that you might be missing significant parts of the picture.
“It’s designed to work by bringing the issue of white supremacy front and centre and putting its advocates on the defensive.”
That much I understand. What happens after that?
Yes I’ve noticed there’s a problem. I’ve also noticed you calling certain people on their misogyny – TRP (rightly or wrongly) also made a point of doing so to those same people while he was here. This is the first time you’ve characterised my comments that way (I think); time for some introspection I guess.
See my response to Carolyn_nth above for “what happens next”.
Despite his on the surface pro-feminism position, TRP is part of the problem (go look at what happened on the one post I’ve made on a feminist topic, assuming the evidence is still there because TRP was deleting it). He’s had his moderation privileges reduced so he can now only moderate his own posts, but he’s also caused problems with some of the posts he writes on gender issues too.
There is a huge problem on TS for women. Mostly it gets ignored, but this is a very hostile place for feminists, and the macho nature of the debate culture is a big part of the problem.
I don’t think you are a misogynist, and I don’t think you are one of the main problems in terms of individuals (although the whole soundbite zen thing is fucking annoying and counterproductive to good communication, and poor communication is part of the problem). And all things being equal, your approach to racism probably wouldn’t matter. But in the culture that exists here, it’s like holy fuck, another dude setting fires when we can’t even keep up with the existing ones and meanwhile does it even matter what is happening to women here? The irony of seeing you argue against white supremacy while taking part in the male supremacy of this site was just too much.
At the very least then, “casual” misogyny. Food for thought. Thanks.
cheers OAB.
A necessary first step in that direction requires the development of a more detailed and transparent exploration of the concept known as “white supremacy.”
I think a far more pressing priority is challenging it directly wherever it rears its head, and forcing its mouthpieces onto the defensive, rather than allowing their rhetoric free reign to hurt people while we search for a nuanced response.
This is a pākehā problem. Pākehā hand-wringing is just another way of enabling it.
[I’m shifting this and the conversation below to OM, because while it’s broadly on topic, I see micky attempting to get people to address the post and I don’t want this to detract from that. – weka]
[TheStandard: A moderator moved this comment to Open Mike as being off topic or irrelevant in the post it was made in. Be more careful in future.]
Oh fuck off. It’s only in your head that the only two options are hard out aggression or Pākehā hand-wringing.
It’s possible to challenge directly and not allow free reign to racism without turning every conversation into a war. Maybe consider that you aren’t the only person in the world with a strategy, and have a listen to your peers from time to time on what might be best.
Agree, weka. I do find an aggressively combative approach to politics tends to result in reinforcing polarised views and superficial point scoring – much like the farce Question Time has often become in parliament.
There are multiple lines of evidence that false beliefs are reinforced by exposure to facts.
There are also multiple lines of evidence that creating dissonance is a useful tool against racism. So we see James and Newsflash (yesterday) getting all hot and bothered about my negative characterisations of their behaviour, and then attempting to defend it.
There is also evidence that change is impossible without negative consequences for a person holding racist views.
I would rather they get hurt than their targets.
I agree that facts are helpful. My experience is that when people get called “racist”, or accused of expressing “racism”, they tend to get very defensive & then are not so open to attending to the facts. Then discussion is shut down.
I think it’s better to go straight to the facts and reasoned arguments rather than (over)using accusatory terms like “racism”.
I think facts are un-helpful, because they harden false beliefs. This is well-documented.
Emotive arguments, on the other hand, elicit defensive responses, forcing the antagonist (in this case the racist) onto the back foot, and diverting their attention from the actual targets of their hate speech.
That’s the theory anyway.
This is well-documented.
eg?
Example.
Thanks, OAB
But then that article ends thus:
My bold.
So they are saying including facts in a debate is helpful to some extent – Aand necesary as part of a wider strategy.
I think that, behind all facts and arguments are some basic assumptions that are value-based. Including facts, critiques and reasoned arguments into the discussion, does help expose the underlying biases and related values.
Cognitive dissonance can be achieved by exposing such biases and evidence based arguments. IMO it doesn’t require a combative approach.
Actually, I think that being aggressively combative is more likely to close down the discussion and result in strengthening of biases, with no way to usefully expose those biases.
And then there is the collateral damage that weka mentions.
Aggressive approaches may have their uses, if used very sparingly, but as a regular and persistent strategy, I think it only reinforces polarisation and entrenched positions.
well put Carolyn.
Your point is well taken, however – IMO – it’s the double standard infested passive aggressive behaviour which flavours The Standard at the moment.
“There are multiple lines of evidence that false beliefs are reinforced by exposure to facts.”
No-one that I can see is objecting to you posting facts.
There are also multiple lines of evidence that creating dissonance is a useful tool against racism. So we see James and Newsflash (yesterday) getting all hot and bothered about my negative characterisations of their behaviour, and then attempting to defend it.
That may well be, but there is still collateral damage.
There is also evidence that change is impossible without negative consequences for a person holding racist views.
I would rather they get hurt than their targets.
Are you familiar with theories of horizontal and lateral abuse?
To labour the metaphor, if you never respond to the opening shots of a war (clue: I didn’t fire them, Bill English did) you just get picked off one by one.
It’s not a war OAB, at least not one in which that metaphor works. After all these years I understand your rationale, and I have some sympathy for it and can the usefulness of the strategy when applied with discernment. But there is so much more going on than that. I’m suggesting that you look at the collateral damage. You’ve now got two feminists calling you on that.
And I’m addressing it directly in the context of gender because of all the shit that goes down here regarding women and where their place is it’s close to intolerable to see a progressive man arguing for an end to white supremacy and using the very tools that exclude women.
(apologies micky, we can shift this to OM if you prefer).
Yawn the best response to OABs tough man, key board warrior dribble
🙄
Thanks for providing an example of exactly what Weka is talking about.
Pleasure 😀, don’t agree with much what weka says but I respect the way she communicates , so glad to help, you in turn….. of the highest order
Given the general tone of comment the past couple of days (on a variety of topics) the following appears pertinent.
“O con noi o contro di not” Benito Mussolini
So we shouldn’t argue with one another then?
argue all you like….if you believe black and white thinking is debate.
I guess it depends on the purpose of your argument…argument for arguments sake or to seek common ground. I think the previous links indicate where the former leads
if you believe black and white thinking is debate.
I don’t think like that.
I guess it depends on the purpose of your argument…argument for arguments sake or to seek common ground.
I think we all do a bit of both. Arguing purely for argument’s sake is nothing more than simple contrarianism—it’s what a lot of talkback hosts do in the absence of having read anything substantial.
….. just to put it out there
There are still one or two journalists (often with vast experience) still interested.
Some of them still have mortgages to pay, so they’re signed up to the corporate machine – whether Granny and her peons, of Fearfex, or even 3 – worse still the state owned commercial machine.
More O’Keefe/Project Veritas fun and games.
This was a great interview, and very insightful if you have the time. Anthony Flaccavento is a farmer who is highly critical of trickle down economics, a Green and a supporter of bottom up economics.
High income earners getting “free ride”
And the rich keep stealing from the rest of us.
Really, we need to change the tax system so that this simply cannot happen.
Tax on income is so terribly passé
Tax on land and financial capital, that’s where you go to really get at true wealth
People making $100K to $200K pa aren’t “rich” – unless they have millions in assets.
Many of those declaring personal income at just less than the $70k threshold are the capital-rich who can hide the rest of their wealth through trusts, companies, etc. No accident that the bulge moved when the tax threshold did.
Sure, but if you want to get them you need taxes on capital, not taxes on income
Agree – wealth/asset tax is the way forward there.
and that should encompass a financial transactions tax
Total agreement. Which (I think) means you are serious this time.
Although I agree with you I don’t think that we can get rid of an income tax just yet.
“Greed and curiosity were teamed up against motivated ignorance,” they explained in the LA Times – and it was a clear victory for staying in political comfort zones. Most conservatives, 61%, chose to stay in their bubble and forgo the extra cash”
Forward looking thoughts by Gareth Morgan”s TOP Party.
Making NZ fair again requires an investment by somebody, there’s no free lunch here. The somebody is those of us who have enjoyed a tremendous rise in our wealth that a tax loophole has generated over the last few decades. Yes, us the Babyboomers are the ones who have to first acknowledge what’s happened and then step up and deal with it… Read more
TOP’s policy to make New Zealand fair again; Some numbers
Is there a simple way for me to work out how TOP’s tax package might affect me? Yes there is, it’s crude but gives you an idea at least. Take 8% of your gross income, and that’s your tax cut
link?
Spent quite a bit of time today mulling this over…from a personal POV…
well, we couldn’t afford a yearly tax on imputed rent at .5-1.5%..
.1% maybe, but not any higher on our fixed and very limited income.
Our plan C looking betterer and betterer…
Another example of a Republican voter being fooled into voting against their interests.
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Security forces kill seven Somalis during food aid handout: police
At least seven people were killed and 12 others wounded Monday when a gunfight broke out in Somalia between rival security forces as they waited for cards for food aid, police said.
"There was exchange of gunfire between policemen who were guarding a distribution site and members from the military," police officer Mohamed Burhan said, adding that seven civilians were killed in the crossfire in the clashes, just outside the capital Mogadishu.
Witnesses said the dead included children as well as elderly men and women.
Shooting between rival security forces broke out in a camp for displaced people in the Afgoye corridor, where thousands of people live in dire conditions ©Mohamed Abdiwahab (AFP/File)
United Nations envoy to Somalia, Nick Kay, said he was "shocked at shooting of civilians queuing for food rations" and added that the "perpetrators must face justice."
Shooting broke out in a camp for displaced people in the Afgoye corridor, where thousands of people live in dire conditions along a key route leading northwest from Mogadishu.
"Many people were waiting in lines receiving cards for food aid distribution when the gunfight broke out," said witness Mumino Dinow.
"It was horrible and I saw several dead people including a woman, child and two elderly men."
It was not immediately clear what sparked the fighting between the rival forces.
The area is under the control of the government, after the Al-Qaeda-linked Shebab were driven out of Afgoye in 2012.
Tens of thousands of people who have fled war as well as hunger -- amid poor rains in some areas and floods elsewhere -- live in basic camps in and around the capital.
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TITLE: determine the locations of all the poles and zeros (including zeros at s = infinite). Make an S-Plane plot of the infinite poles and zeros
QUESTION [1 upvotes]: Determine the locations of all the poles and zeros (including zeros at $S = \infty$). Make an $S$-Plane plot of the infinite poles and zeros.
$$G(s) = \dfrac{5S^2 + 20S + 15}{S(S + 3)(S^2 + 4S + 4)}$$
REPLY [1 votes]: We can find the non-infinite poles and zeros as follows:
First of all, factor the top and bottom, canceling out any common factors. We have
$$
G(s) = \frac{5s^2 + 20s + 15}{s(s+3)(s^2 + 4s + 4)}
= \frac{5(s+1)(s+3)}{s(s+3)(s+2)^2}
= \frac{5(s+1)}{s(s+2)^2}
$$
The zeros are the values of $s$ for which $G(s)=0$. $G(s)$ will be zero whenever our simplified numerator is zero. That is to say $s$ will be a zero if
$$
5(s+1) = 0
$$
The poles are the values of $s$ for which $G(s)$ has an infinite discontinuity. $G(s)$ will be undefined whenever our simplified denominator is zero, so we have to find the zeros of the bottom, which is to say that $s$ will be a pole if
$$
s(s+2)^2 = 0
$$
Now, to find the behavior of $G$ at infinity, you need to find
$$
\lim_{s\to \infty} G(s) =
\lim_{s \to \infty} \frac{5s^2 + 20s + 15}{s(s+3)(s^2 + 4s + 4)}
$$
If this limit is $0$, then $G$ has a zero at $\infty$. If this limit is $\pm \infty$, then $G$ has a pole at $\infty$. If this limit is any non-zero constant, then $G$ has neither a zero nor a pole at $\infty$.
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TITLE: Why was this set $N_{n,m}$ defined?
QUESTION [0 upvotes]: I'm trying to understand the proof below.
Why was the set $N_{n,m}$ of measure zero defined in this proof?
Why is the sequence uniformly convergent?
Prove that the normed space $L^{\infty}$ equipped with $\lVert\cdot\rVert_{\infty}$ is complete.
REPLY [1 votes]: How are the sets $N_{n,m}$ defined?
By definition of the $L^\infty$ norm, for any $f\in L^\infty(X,\mathcal{A},\mu)$, there exists a set $B$ with $\mu(B)=0$ such that for every $x\in B^c$,
$$
|f(x)|\le\|f\|_\infty.
$$
By what we have above, one has for any fixed positive integers $m$ and $n$, there exists $N_{n,m}$ with $\mu(N_{n,m})=0$, such that for every $x\in N_{n,m}^c$,
$$
|f_n(x)-f_m(x)|\le \|f_{n}-f_m\|_{\infty}.
$$
Why does the sequence $(g_n)$ with $g_n:=\tilde{f}_n$ converge uniformly on $N^c$?
Given $\epsilon>0$, there exists $M>0$ such that for all $x\in N^c$ and for all $n,m\geq M$, one has
$$
|f_{n}(x)-f_m(x)|\le \|f_{n}-f_m\|_{\infty}<\epsilon,\tag{1}
$$
which implies that for every (fixed) $x\in N^c$, $\{f_n(x)\}_{n=1}^\infty$ is a Cauchy sequence in $\mathbb{R}$ and hence the pointwise limit $\lim_nf_n(x)$ exists (in $N^c$). Thus, we can define:
$$
f(x):=\lim_{n\to\infty}f_n(x),\quad x\in N^c.
$$
On the other hand, (1) implies that given $\epsilon>0$, there exists $M>0$, such that for all $n>M$,
$$
\sup_{x\in N^c}|f(x)-f_n(x)|<\epsilon,
$$
which, by the definition of uniform convergence, implies that $g_n$ converges uniformly to $f$ in $N^c$.
| 252
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XX1 Eagle Trigger Shifter | Single Click
$170
Ships after .
Currently out of stock until ..
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- E-MTB specific. Limits the pull lever to a single engagement to help eliminate premature chain wear and chain breaking.
- Carbon trigger and cover reduce weight.
- Master XX1 Eagle™ group
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September 26, 2008
Revamped Parks Web Site Drawing Raves
By Jeanine Benca
LIVERMORE -- The park district's newly improved Web site has a host of user-friendly functions, including a key that can translate the entire site into more than 20 languages.Livermore Area Recreation and Park District officials say the renovated site will help members of Livermore's diverse ethnic community, especially non-English speakers, have better access to current information about parks and programs.
A click from the homepage will activate Google Translate software, which almost instantly translates the site into Spanish, Chinese or another language of the user's choice.
"The board members are really excited about the site translation," said Lea Blevins, a district spokeswoman. "There are a lot of Spanish-speaking people in Livermore, but why stop at Spanish?"
In addition to the language feature, the site includes slide-out menu options for easier navigation. The more streamlined Web page also enables park district officials to post frequent updates about programs and notify the community of building closures or other issues in the parks, Blevins said.
"(The old site) had a lot of decent information, but it just wasn't as user-friendly. We as a staff didn't have the ability to update it ourselves -- we had to rely on a Webmaster," she added.
Since the site went online several weeks ago, the park district has received scores of positive feedback, according to a district news release.
Park district officials said the total cost of the Web site improvements was $18,000.
Visit the Web site at
Reach Jeanine Benca at [email protected] or 925-847- 2125.NEW LIVERMORE PARKS WEB SITE
Visit the new Livermore Area Recreation and Park District Web site at
Originally published by Jeanine Benca, Valley Times.
(c) 2008 Oakland Tribune. Provided by ProQuest LLC. All rights Reserved.
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What It Is
Schleich introduces all-new animal figures in its Farm Life and Wild Life figure lines. These cold-cast resin animals (including pandas, gorillas, lambs, and sheep) have beautiful detailing and wonderful painting, making them look very realistic. Each is sold separately.
Why It’s Fun
The great detail on these figures makes them great for collectors. Of course, kids love playing with them, too, and making up all kinds of different narrative-based scenes.
Who It’s For
Schleich Figurines are for ages 3 and up.
What To Be Aware Of
Each figure differs in price between $2.99 and $7.99.
What do you think about Schleich Figurines?
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ST LOUIS, MO–Today at the annual ComicsPRO meeting in Portland, the St. Louis based publisher of “comics for everyone” announces!
.”.
About Lion Forge Comics:
Lion Forge Comics strives to publish comics that everyone—regardless of background or ethnicity—can identify with. The original graphic novel Andre the Giant: Closer to Heaven was nominated for three Glyph Awards, as well as the Dwayne McDuffie Award for Diversity. Lion Forge Comics also includes the Roar imprint for young adults and the CubHouse imprint for readers twelve and under. It is also home to the Magnetic Collection of premium and critically acclaimed comics and graphic novels, boasting multiple Eisner Award nominees. From licensed properties, including Dreamworks Voltron Legendary Defender, original work from top independent creators, and the original and all-inclusive Catalyst Prime superhero universe, there is something for every level of comic fan, young and old, across multiple imprints and formats. Lion Forge Comics—Comics for Everyone. Visit for more information.
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Advent is upon us and we have developed our very special Advent calendar just for you. Each day right up to the 25th of December the windows will reveal something significant that you can download and keep.
To our subscribers we are offering information, insight and a special offer as part of our celebrations, as we approach Christmas. Thank you for your continued support.
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HBR | Petraeus and the Rise of Narcissistic Leaders
Forbes | 9 Personality Types & How to Inspire & Motivate
Infographic | Equality in Leadership
CJ | Do Sweat the Small Stuff
Podcast | How to Succeed at Anything by Really Really Trying: Lyman MacInnis
TED | Let’s Raise Kids to be Entrepreneurs; Cameron Herold
Both my nine- and seven-year-olds have a stockbroker already.Cameron Herold
Leading, along the road with you!
Alan
P.S. New employees? The 90-Day New Employee Success Program provides a strategy to get your new hires engaged and up to speed quickly and effectively.
P.P.S. Did you know over 80% of positions are found through networking? Increase your chances in connecting professionals through my LinkedIn network. Expand your network by 3800 professionals by joining mine. Click here & mention you are a friend.
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Restaurant Week Review: Bud & Marilyn's January 30, 2016 Erik Anderson "Just when I thought it couldn't get better." That should be the title of my miniseries of Restaurant Week Reviews here on Mogblog. Because honestly, I'm amazed. How this city's restaurant selection continues to top itself, I'll never know. Then again, I'd like to think my good taste and knowing how to pick 'em really comes in handy. Maybe. A little. Ok, I should mostly be thanking the chefs. Now, to preface this, I had high hopes for Bud & Marilyn's going in. Not only has it been named one of Philly's top restaurants of 2015, but Zagat recently featured their Chef-de-Cuisine, Dan Giorgio, in their famous "30 under 30" article. (Dan also just so happens to be a dear friend of mine, so I was expecting him to pull out all of the stops.) And oh, how the stops were pulled. Learning from my experience at Zahav earlier this week, a couple friends and I abandoned the prix fixe and surrendered to the chefs tasting menu. (This was just another in a long string of excellent decisions I've made regarding restaurant week.) And from the first plate of fried cheese curds - yes, it's true - I was hooked. Inspired by American comfort food of yesteryear, the chefs tasting took us through a veritable pu pu platter of treats. Oh wait, there was a actually a pu pu platter in there. With spareribs and shrimp toast and poke... But that was after the potato roll fried chicken sandwiches, and the kale salad with hush puppies. Though before the pork n' pickles platter with homemade pâté and biscuits. I think.At this point, we were slowing down, because all of that rich food will get to you after a while. (To be fair, Dan warned me of this going in. But I couldn't stop. I wanted more.) And in what became one of my favorite moments of the night, our server came over to say "The chef knows you're getting full, so he's ONLY going to send over two more entrees." I actually LOLed. Especially when those entrees turned out to be meatloaf n mashed potatoes, and campanelli with meatballs. I mean come on. Both were marvelous, by the way. But that goes without saying at this point, right? Finally, we gathered our strength and marched bravely on into dessert. I had been eyeing this marvelous triple layer chocolate cake all evening, and I wasn't prepared to leave without getting to know it up close and personal. Covered in malted milk balls (How retro and charming! And fat.) and every bit as rich and tasty as the rest of our meal, it was the real cherry on the sundae of our Bud and Marilyn's experience. It would be a sin to do a writeup on Bud & Marilyn's without mentioning the mid century ambiance. Every detail, from the lightbulbs, to the wood paneling, to the tiki-inspired glassware that held our potent cocktails, was meticulously designed to feel like your quirky grandparents rec room. That hasn't changed since 1976. At least. I loved it. But I'm not surprised, owners Val Safran and Marcie Turney are no stranger to a beautiful restaurant experience - the pair owns 7 other successful businesses on 13th street alone! I can't wait to try them all. 10/10Oh and Dan, if you're reading this: what was that third entree you might've sent if we weren't too full? Asking for a friend. Cheers! x Subscribe to #MOGBLOG via EmailFollow #MOGBLOG on Bloglovin'
| 384,652
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Rick.
DIY Shibori – HonestlyWTF
Today marks HonestlyWTF’s four year anniversary. Four years! To celebrate, we’re revisiting the very first tutorial we ever featured on the site: shibori tie dye....
to experience pearltrees activate javascript.
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Just an avid reader. Mostly SF/Fantasy, some hobbies, paranormal, urban fantasy and lighter, fluffier things.
The.
Thiss.
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Lakers vs. Celtics Game Five: The Keys to Victory for Both TeamsJune 12, 2010
Breaking Down Game 5 of the NBA Finals
Up to this point, the NBA Finals between the Boston Celtics and the Los Angeles Lakers have lived up to the hype. After four games the series is all tied up at two with one game left in Boston and then the rest of the series playing out in LA. The series now boils down to a best two out of three for the NBA Championship.
Both teams have had their good and bad moments throughout the finals, but there are certain keys to victory that both teams need to achieve to win the ring.
NBA Finals Game Five Odds: The opening Line for Game Five at BETUS.com was Boston -3 / 188. Since the opening, there has been very little line movement. If you want to bet on Boston, the best line is Boston -3 -109 at Bet Phoenix.com , and if you want to bet on the Lakers, the best line is +3 found at several of the Top Ten sportsbooks.
Here is a breakdown of keys to victory for each team.
Boston Celtics
Go Deep: The Celtics depth is a strong point, and it was obvious in game four when the bench basically won the game and let the Celtics starters rest for almost the entire fourth quarter. Glenn Davis and Nate Robinson led the charge and they have to bring that same energy and determination to game five.
The Truth shall set you free: Paul Pierce sets the tone for this team. He started off game five by scoring double digits in the first quarter and when the Lakers made it close late he made the shot that sealed the win. He has to maintain that focus and set the tone again for the Celtics to win game five.
Save the Best For Last: Ray Allen doesn’t have to break records to be a player in game five. In game four, he was the leader on the court playing as the only starter. He yelled at some of the players when they got technical fouls and kept them focused on the win. He only scored 12 points, but if he shows he can be a scoring threat early, it takes the pressure off some of the others. Game Five could be his last game in Boston.
Los Angeles Lakers
Kobe’s Killer Instinct: You could see it in game four; Kobe has his killer instinct back. He was draining shots from all over the court and willing his team to play harder. This is the kind of heart and drive that will motivate his team to play well.
Two Trick Ponies: Someone else has to stand up and make a difference. In game four, Pau Gasol and Kobe Bryant both had good games yet it wasn’t enough. Lamar Odom was invisible and Derrick Fisher could not do it again, the Lakers need more output from other players to win game five.
Andrew Bynum’s injury: Bynum’s injury has seemed to have gotten worse. He is sitting out more and more and the Lakers need him. He provides an offensive threat on the inside and defends against Glenn Davis keeping him at bay. His health could be a difference maker in game five.
Regardless of the outcome of game five and the series, this is setting up to be another classic match up between the Celtics and the Lakers as they battle to see who is the best in the NBA
Report: Pelicans Extend Steven Adams
New Orleans' big man agrees to 2-year, $35M extension
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TITLE: Perfect sets of a complete metric space
QUESTION [3 upvotes]: So I am aware something like this question had been asked, so I would appreciate if you could direct me to a duplicate, if there is any.
In particular, I've read the post
Proof that a perfect set is uncountable,
And of course there is a classical proof when the metric space is $\mathbf{R}^n$ endowed with the usual Euclidean metric, the crux of the proof of which is very well surmised here:
Rudin 2.43 Every nonempty perfect set in $\mathbb{R}^k$ is uncountable.
My question is, in seeing Rudin's proof, it seems the only place where he actually used the fact that the metric space is $\mathbf{R}^n$ (and not just any old complete metric space) is when he invoked the Heine-Borel theorem to conclude that the closure of each $V_n$ is compact. But we know in general that in a complete metric space, if $(F_n)$ is a nested sequence of closed sets whose diameters converge to $0$, then the intersection $\bigcap_{n=1}^{\infty}F_n$ is nonempty, and that is all we need! (Of course, Rudin's proof doesn't explicitly mention that the diameter of $V_n$ tends to $0$, nor did he have to, but we can always choose the nested sequence so that the radii decay exponentially.)
Am I missing a critical detail here?
REPLY [0 votes]: Preliminaries:
$\textbf{Theorem:1}$ If $\{K_{\alpha} :\}$ is a family of non empty
compact subsetsets of a metric space $(X, d) $ such that the
intersection of any finite subfamily is non empty , then
$\bigcap_{\alpha} K_{\alpha}\neq \emptyset$
$\text{Theorem} \space 2.36$ (Principles of mathematical analysis,
Rudin)
$\textbf{Corollary:1} $ If $(K_n) $ be any descending sequence of non empty compact subsets in a metric space $(X, d) $ , then $\bigcap_{n} K_n\neq \emptyset$
$\bigstar$ Note 1: The same is not true for any sequence of non empty closed sets in a metric space.
Ex: Consider $(\Bbb{Q}, d_{\text{sub}}) $ as a metric subspace of $(\Bbb{R}, d_{\text{eucliden}}) $ .Let $\Bbb{Q}=(r_n)_{n\in\Bbb{N}}$ be the enumeration and $C_n=\Bbb{Q}\setminus \{r_n\}$.Then
$\emptyset \neq C_n$ closed set.
$C_{n+1}\subset C_n$
But $\bigcap_{n} C_n=\emptyset$
$\bigstar $ Note 2 : $(X, d) $ be a metric space and $A\subset X$ . Let $x\in X$ is a limit point of $A$ and $x\in U$ be a open set. Is it always possible to find an open set $V$ such that $x\in \overline{V}\subset U$ where $\overline{V}$ is compact?
Hint: Consider the metric space $(\Bbb{Q}, d_{\text{sub}}) $ .
$\textbf{Definition :}$ A topological space $(X, \tau) $ is called a Locally compact Haussdoff space if $\forall x\in X$ and forall open sets $U_x$ containing $x$ there exists an open set $V_x$ such that
$x\in V\subset \overline{V}\subset U$
Topological closure $\overline{V}$ is closed.
Example 1 : $\Bbb{R^k}$ is locally compact Hausdorff space.For any point $x\in
\Bbb{R^k}$ and a open set $U$ containing $x$ , there is a basic open set i.e an open ball $B(x, r) $ such that $x\in \overline{B(x, \frac{r}{2})}\subset B(x, r) \subset U$
Now by Heine-Borel theorem, $B(x, \frac{r}{2}) $ is compact.
Example 2 : Let $(X,d)$ be a metric space where $|X|\ge \aleph_{0}$ i.e $X$ is an infinite set and $d$ is the discrete metric. Then $(X, d) $ is locally compact Hausdorff space as for any $x\in X$ and a open set $U$ containing $x$ , $x\in \{x\}\subset U$ where $\{x\}$ is compact. But $(X, d) $ doesn't have Heine-Borel property as $X$ is closed , bounded but not compact.
$\textbf{Theorem:}$ Let $\emptyset \neq P$ perfect subset of $\Bbb{R}^k$.
Claim : $P$ is uncountable.
Proof:(Rudin's way) $P\neq \emptyset$ implies $\exists x\in P$ .Since $\Bbb{R^k} $ is Hausdorff space , any open set $U$ containing $x$ contains an infinite subset of $P$. $($ Infact this result is also true for any $T_1$-space $)$ . Hence $P$ is an infinite set.
To show $P$ is not countably infinite . Suppose $P$ is countably infinite and $P=\{x_1,x_2,\ldots\}$.
Let $V_0$ is an open set containing $x_1$. Then by Local compactness of $\Bbb{R}^k$ , $\exists V_1\subset V_0$ open set such that $x_1\in \overline{V_1}\subset V_0$ and since $x_1$ is a limit point of $P$ , $V_1\setminus \{x_1\} \cap P\neq \emptyset$ and $\overline{V_1}$ is compact.
Let $x_2\in V_1\setminus \{x_1\} \cap P$
Again by Local compactness, we can find an open set $V_2\subset V_1$ such that $x_2\in \overline{V_2}\subset V_1$ and $\overline{V_2}$ is compact. Again $V_2\setminus \{x_2\} \cap P\neq \emptyset$
Continuing in that way, we produce a sequence of open sets $(V_n) $ with the properties :
$\overline{V_{n+1}}\subset {V_n}$
$x_n\notin V_{n+1}$
$V_{n+1}\cap P\neq \emptyset$
Put $K_n=\overline{V_n}\cap P$
As $2$ implies $x_n \notin{V_{n+1}}$ implies $\bigcap_{n}K_n=\emptyset$.
But $(K_n) $ is a descending sequence of non empty compact sets and $\bigcap_{n}K_n=\emptyset$ directly contradict $\textbf{Corollary:1}$.
Note : A compact subset of a metric space (infact in any Hausdorff space) is closed.
We can use completeness ,by choosing $V_n$ as the basic open sets i.e open balls in $\Bbb{R^n}$ , $V_n=B(x_n, \frac{r}{2^n})$ and then the Cantor intersection theorem for complete metric space.
The theorem is still true in every complete metric space as a closed subset of a complete metric space is complete and a complete metric space with no isolated point is uncountable. (an application of Baire Category theorem) .
$\textbf{Theorem}$ A normed linear space is finite dimensional iff closed unit ball is compact iff closed bounded sets are compact i.e Heine-Borel property is true.
Choose an infinite dimensional Banach space, then it doesn't have the Heine-Borel property but every non empty perfect set is uncountable.
| 217,077
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\chapter{States and observables}\label{chap:QM}
\blfootnote{This manuscript will be published by Cambridge University Press in the series ``Cambridge Studies in Advanced Mathematics''. This version is free to view and download for personal use only. Not for re-distribution, re-sale or use in derivative works. \newline \noindent {\copyright} Jan van Neerven}
\noindent
In this final chapter we apply some of the ideas developed in the preceding chapters to set up a functional analytic framework for Quantum Mechanics. More specifically, we will show how replacing the role of Borel sets in classical mechanics by orthogonal projections in Quantum Mechanics leads, in a natural way, to the Quantum Mechanics formalism for states and observables. This chapter is not a course on Quantum Mechanics. Rather, it aims at explaining, on a conceptual and mathematically rigorous level, the mathematical structure of the theory which, in most physicist's accounts, is simply taken for granted or introduced axiomatically.
\section{States and observables in classical mechanics}
Before entering the realm of Quantum Mechanics, we start by taking a brief look at the notions of state and observable in classical mechanics from a rather abstract measure theoretic point of view.
\subsection{States}
In classical mechanics, the {\em state space} of a physical system is a measurable space
$(X,\X)$, typically a manifold with its Borel $\sigma$-algebra. For example, the state space of an ensemble of $k$ free moving point particles in $\R^3$ is $\R^{3k}\times \R^{3k}$ (three position coordinates $x_j$ and three momentum coordinates $p_j$ for each particle) and that of a harmonic oscillator is the submanifold of $\R\times \R$ given by $x^2+p^2 =$ constant.
\begin{definition}[States, pure states]\label{def:clas-state} Let $(X,\X)$ be a measurable space.
\begin{enumerate}[\rm(1)]
\item
A {\em state}\index{state} is a probability measure $\nu$ on $(X,\X)$.
\item
A {\em pure state}\index{pure state}\index{state!pure} is an extreme point of the set of probability measures on $(X,\X)$.
\end{enumerate}
For a measurable set $B\in\X$, the number $\nu(B)$ is thought of as ``the probability that the state is described by a point in $B$''.
\end{definition}
Thus we identify the ``state'' of a system with the ensemble of truth probabilities of certain questions about the system.
For example, the exact positions and momenta of all particles in a gas container at a given time cannot be known with complete precision, but one might ask about the probability of finding a certain portion of the gas in a certain subset of the container.
Recall that a measure $\nu$ on $(X,\X)$ is said to be {\em atomic}\index{atomic}\index{measure!atomic} if, whenever we have $\nu(B)>0$ and $B = B_0\cup B_1$ with disjoint $B_0,B_1\in\X$, it follows that either $\nu(B_0)=0$ or $\nu(B_1)=0$.
The following was shown in Example \ref{ex:extreme-points-K}:
\begin{proposition} The pure states are precisely the atomic probability measures.
\end{proposition}
\subsection{Observables}
\begin{definition}[Observables] Let $(\Om,\calF)$ be a measurable space.
An {\em $\Om$-valued observable}\index{observable} is a measurable function $f: X\to \Om$.
An {\em elementary observable}\index{observable!elementary} is a $\{0,1\}$-valued observable.
\end{definition}
For example, the three position coordinates $x_j$ and momentum coordinates $p_j$
of a free moving particle in $\R^3$ are real-valued observables on the state space $X = \R^3\times \R^3$, and so are its kinetic energy $|p|^2/2m$ (where the mass $m$ is treated as a constant) and potentials $V(x)$.
If $\nu$ is a state on $(X,\X)$ and $f:X\to\Om$ is an observable, then for $F\in\calF$ the number
$$\nu(f^{-1}(F)) = \nu(\{x\in X:\, f(x)\in F\})$$ belongs to the interval $[0,1]$ and is interpreted as
``the probability that measuring $f$ results in a value in $F$ when the system is
in state $\nu$.''
\subsection{From classical to quantum}\label{subsec:class-quantum}
An elementary observable is of the form $\one_B$ with $B\in\X$.
Its range equals $\{0,1\}$ unless $B = \emptyset$ or $B = \X$, in which case one has $\one_\emptyset \equiv 0$ and $\one_X \equiv 1$. Orthogonal projections in a complex Hilbert space enjoy similar
properties {\em spectrally}: if $P$ is an orthogonal projection in
a Hilbert space $H$, its spectrum equals
$\sigma(P) = \{0,1\}$ unless $P=0$ or $P=I$; in these cases one has
$\sigma(0) = \{0\}$ and $\sigma(I) = \{ 1\}$.
The basic idea that underlies Quantum Mechanics is to {\em replace elementary
observables by orthogonal projections}.
The set of all orthogonal projections in a complex Hilbert space $H$ is denoted by $\PP(H)$.\index{$P$@$\calP(H)$} This set is partially ordered in
a natural way by declaring $P_1\le P_2$ to mean that the range of $P_1$ is contained in the range of $P_2$; this is equivalent to the statement that $P_2-P_1$ is positive.
With respect to this partial ordering, $\PP(H)$ is a {\em lattice}\index{lattice}
in the sense that for all $P_1$ and $P_2$ in $\PP(H)$ there exists a greatest lower bound $$P_1\wedge P_2$$ in $\PP(H)$, given as the orthogonal projection
onto $\Ran(P_1)\cap \Ran(P_2)$,
and a least upper bound $$P_1\vee P_2$$ in $\PP(H)$, given as the orthogonal projection
onto the closed subspace spanned by $\Ran(P_1)$ and $\Ran(P_2)$. The `negation' of an orthogonal projection
$P$ is the projection $$\neg P = I-P$$ onto the orthogonal complement of $\Ran(P)$.
One has the associative laws $$(P_1\wedge P_2)\wedge P_3 = P_1\wedge (P_2\wedge P_3), \quad
(P_1\vee P_2)\vee P_3 = P_1\vee (P_2\vee P_3)$$ and the identities
$$ \neg(P_1\wedge P_2) = \neg P_1\vee \neg P_2, \quad \neg(P_1\vee P_2) = \neg P_1\wedge \neg P_2,$$
but the important difference with the classical setting is that the distributive laws
\begin{align*}P_1\wedge (P_2\vee P_3) & = (P_1\wedge P_2)\vee (P_1\wedge P_3)\\
P_1\vee (P_2\wedge P_3) & = (P_1\vee P_2)\wedge (P_1\vee P_3)
\end{align*}
generally fail.
\begin{example}
In $\C^2$ consider the orthogonal projections $P_1$, $P_2$, and $P_3$ onto the first and second coordinate axes and the diagonal, respectively. Then $P_2\vee P_3 = I$, $P_1\wedge P_2 = P_1\wedge P_3 =0$, and
$$ P_1\wedge (P_2\vee P_3) = P_1, \quad (P_1\wedge P_2)\vee (P_1\wedge P_3) =0.$$
\end{example}
\section{States and observables in Quantum Mechanics}\label{sec:states-observables}
From now on $H$ is a {\em separable complex Hilbert space.}
\subsection{States}\label{subsec:states}
Upon replacing indicator functions of measurable sets by orthogonal projections in $H$, one is led to the idea to define a state as a mapping $\nu:\calP(H)\to [0,1]$ that satisfies $\nu(0)=0$, $\nu(I)=1$, and is countably additive in the sense that $$\sum_{n\ge 1} \nu(P_n) = \nu(P)$$ whenever $(P_n)_{n\ge 1}$ is a (finite or infinite) sequence of pairwise disjoint orthogonal projections and $P$ is their least upper bound, that is, $P$ is the orthogonal projection onto the closure of the span of the ranges of $P_n$, $n\ge 1$.
Although this definition is quite satisfactory in many ways, it suffers from the defect
that it does not present an obvious way to extend $\nu$ to positive linear combinations of pairwise disjoint orthogonal projections.
In the classical picture, the expected value of a nonnegative simple function
$f = \sum_{n=1}^N c_n \one_{B_n}$ in state $\nu$ is given by its integral
$\int_X f\ud \nu = \sum_{n=1}^N c_n \nu(B_n)$. The desideratum
\begin{align}\label{eq:welldef-mu}
\nu\Bigl(\sum_{n=1}^N c_n P_n\Bigr)= \sum_{n=1}^N c_n \nu(P_n)
\end{align}
can be thought of as a quantum analogue of this and constitutes the first step toward defining the expected value for more general classes of observables. However, if one attempts to take \eqref{eq:welldef-mu} as a definition, a problem of well-definedness arises (that such a problem indeed may arise is demonstrated by the example at the end of this section). The next definition proposes a way around this difficulty.
The {\em convex hull}\index{convex hull} of a subset $S$ of a vector space $V$ is the smallest convex set in $V$ containing $S$ and is denoted by ${\rm co}(S)$\index{$C$@${\rm co}(S)$}.
\begin{definition}[Affine mappings] Let $S$ be a subset of a vector space $V$.
A mapping $\nu:S \to [0,1]$ is called
{\em affine}\index{affine} if it extends to a mapping $\nu: {\rm co}(S) \to [0,1]$ satisfying
$$\nu\Bigl(\sum_{n=1}^N \la_n v_n\Bigr) = \sum_{n=1}^N \la_n \nu(v_n)$$ for all $N\ge 1$, $v_1,\dots,v_N\in S$,
and scalars $\la_1,\dots,\la_N\ge 0$ satisfying $\sum_{n=1}^N \la_n = 1$.
\end{definition}
\begin{definition}[States, preliminary definition]\label{def:state-preliminary}
A {\em state}\index{state} is a mapping $\nu:\calP(H)\to [0,1]$ that is affine and satisfies $\nu(0)=0$
and $\nu(I)=1$.
\end{definition}
In what follows we call two orthogonal projections {\em disjoint}\index{disjoint!projections} if their ranges are mutually orthogonal. A family of projections said to be {\em disjoint} if every two distinct members of this family are disjoint.
Let us denote by $\calP_{\rm fin}(H)$\index{$P$@$\calP_{\rm fin}(H)$} the set of all finite rank projections in $\calP(H)$.
\begin{theorem}[States as positive trace class operators with unit trace]\label{thm:Gleason-light} Suppose
$\nu:\calP_{\rm fin}(H)\to [0,1]$ is affine and satisfies $\nu(0)=0$. Then there exists a unique positive trace class operator $T$ on $H$ such that
for all $P\in \calP_{\rm fin}(H)$ we have
\begin{align}\label{eq:nuP} \nu(P) = \tr(PT),
\end{align}
and we have $$\tr(T) = \sup_{P\in \calP_{\rm fin}(H)} \nu(P).$$
The identity \eqref{eq:nuP} defines an extension $\nu:\calP(H)\to [0,1]$ which is countably additive. This extension is a state if and only if $\nu(I)=1$.
\end{theorem}
\begin{remark}[Density functions]
In the physics literature, the operator $T$ is called the {\em density operator}\index{density operator} associated with the state $\nu$.
\end{remark}
\begin{proof}[Proof of Theorem \ref{thm:Gleason-light}]
To prove uniqueness, suppose that $T,\wt T\in \calL(H)$ are such that
$ \tr(PT) =\tr(P\wt T)$ for all $P\in \PP_{\rm fin}(H)$. Taking $P$ to be the rank one projection $ h\,\bar\otimes\, h$, with $h\in H$ of norm one, gives
$\iprod{Th}{h} = \iprod{\wt Th}{h}$. By scaling, this extends to arbitrary $x\in H$, and this implies $T=\wt T$ by Proposition \ref{prop:polarisation}.
The existence proof proceeds in several steps.
\smallskip
{\em Step 1} --
Fix orthogonal projections $P_1,\dots,P_N\in \calP_{\rm fin}(H)$ and scalars $c_1,\dots,c_N\ge 0$.
If $\sum_{n=1}^N c_n < 1$, then with $c_{N+1}:= 1- \sum_{n=1}^N c_n$ and $P_{N+1}:=0$ the affinity assumption implies
$$ \nu\Bigl(\sum_{n=1}^N c_n P_n\Bigr)
= \nu\Bigl(\sum_{n=1}^{N+1} c_n P_n\Bigr) = \sum_{n=1}^{N+1} c_n \nu(P_n) = \sum_{n=1}^N c_n \nu(P_n),$$
where we used that $\nu(0)=0$.
Suppose now that the scalars $c_1,\dots,c_N\ge 0$ are arbitrary and consider the operator $S:= \sum_{n=1}^N c_n P_n.$
Fix an arbitrary integer $k\ge \sum_{n=1}^N c_n$. Then, by what we just proved, the number
$$k \nu(\frac1k S) = k\nu\Bigl(\sum_{n=1}^N \frac{c_n}{k} P_n\Bigr) = k\sum_{n=1}^N \frac{c_n}{k} \nu(P_n) = \sum_{n=1}^N c_n \nu(P_n)$$
is independent of $k$. Hence we may define an extension of $\nu$, again denoted by $\nu$, by
\begin{align*}
\nu(S):= k \nu(\frac1k S)
\end{align*}
and obtain, for arbitrary $c_1,\dots,c_N\ge 0$,
\begin{align*}
\nu\Bigl(\sum_{n=1}^N c_n P_n\Bigr)= \sum_{n=1}^N c_n \nu(P_n).
\end{align*}
The extension just defined is finitely additive on the set of operators $S$ of the form just described. Indeed, this follows by induction from the fact that if $S = \sum_{n=1}^N c_n P_n$ and $S' = \sum_{n=N+1}^{N'} c_n P_n$ are as intended,
then
\begin{align*} \nu(S+S') & = \nu\Bigl(\sum_{n=1}^{N'} c_n P_n\Bigr) = \sum_{n=1}^{N'} c_n \nu(P_n)
\\ & = \sum_{n=1}^{N} c_n \nu(P_n) + \sum_{n=N+1}^{N'} c_n \nu(P_n) = \nu(S)+\nu(S').
\end{align*}
\smallskip
{\em Step 2} -- Consider now an operator of the form
$S = \sum_{n=1}^N c_n P_n$ with coefficients $c_n\in \R$.
Then we may write $S = S_+-S_-$ by considering positive and negative coefficients and define $$\nu(S) := \nu(S_+) - \nu(S_-).$$ To see that this is well defined, suppose that
$S = S_+-S_- = S_+'-S_-'$, where $S = \sum_{n=1}^N c_n P_n = \sum_{n=1}^{N'} c_n' P_n'$.
Then, by the finite additivity observed at the end of Step 1,
$$\nu(S_+)+\nu(S_-') = \nu(S_++S_-') =\nu(S_+'+S_-) = \nu(S_+')+\nu(S_-),$$
so $\nu(S_+) - \nu(S_-) = \nu(S_+') - \nu(S_-')$ as desired.
Similarly it is checked that
$c\nu(S) = \nu (cS)$ for all $c\in\R$ and that $\nu(S+S') = \nu(S)+\nu(S')$.
If $S = \sum_{n=1}^N c_n P_n$ with coefficients $c_n\in \C$, we set
\begin{align}\label{eq:nuS} \nu(S) := \frac12\nu(S+S^\star) + \frac1{2i}\nu(i(S-S^\star)).
\end{align}
Then $\nu$ is easily seen to be additive and real-linear, and from
\begin{align*}
\nu(iS) & = \frac12\nu(iS-iS^\star) + \frac1{2i}\nu(i(iS+iS^\star))
\\ & = i\Bigl(\frac1{2i}\nu(iS-iS^\star) -\frac1{2}\nu(-(S+S^\star))\Bigr)
= i\nu(S)
\end{align*}
it follows that $\nu$ is in fact complex-linear.
\smallskip
{\em Step 3} --
If $S = \sum_{n=1}^N c_n P_n$ with $c_1,\dots,c_N\in \C$ and with mutually orthogonal projections $P_1,\dots,P_N \in \calP_{\rm fin}(H)$, then
\begin{equation}\label{eq:absnuS}
\begin{aligned}
|\nu(S)| = \Bigl|\nu\Big(\sum_{n=1}^N c_n P_n\Bigr) \Bigr|
& = \Bigl|\sum_{n=1}^N c_n \nu(P_n)\Bigr| \le \max_{1\le n \le N} |c_n| \sum_{n=1}^N \nu(P_n)
\\ & = \max_{1\le n \le N} |c_n| \nu\Big(\sum_{n=1}^N P_n\Bigr) \le \max_{1\le n \le N} |c_n|
\\ & = \Big\n \sum_{n=1}^N c_n P_n\Big\n = \n S\n.
\end{aligned}
\end{equation}
Here we used that $\nu(\sum_{n=1}^N P_n)\le 1$ since $\sum_{n=1}^N P_n$ is an orthogonal projection.
By the spectral theorem (Theorem \ref{thm:spect-thm-comp}), every compact self-adjoint operator $S\in \calL(H)$ can be approximated, in the norm of
$\calL(H)$, by operators $S_n$ of the form just considered and the estimate \eqref{eq:absnuS} entails that the limit
$ \nu(S):= \limn \nu(S_n)$ exists. Moreover, if the operators $S_n'$ form another approximating sequence, then
by finite additivity we have
$|\nu(S_n) - \nu(S_n')| = |\nu(S_n - S_n')| \le \n S_n-S_n\n \to 0$, proving that the number $\nu(S)$ is independent
of the choice of approximating sequence.
For general compact operators $S\in \calL(H)$ we define
$\nu(S)$ by \eqref{eq:nuS} and find
$$|\nu(S)| \le \frac12\n S+S^\star\n + \frac12 \n i(S-S^\star)\n \le 2\n S\n.$$
This extension is again linear.
\smallskip
{\em Step 4} --
The argument of Step 3 proves that we may identify $\nu$ with an element in $({\mathscr{K}}(H))^*$, the dual of the space ${\mathscr{K}}(H)$ of compact operators on $H$.
By trace duality (Theorem \ref{thm:traceduality})
there exists a unique trace class operator $T\in \calL_1(H)$ such that for all $S\in {\mathscr{K}}(H)$
we have $\nu(S) = \tr(ST)$.
By considering the orthogonal projection $P = h\,\bar\otimes\, h$ onto the span of the norm one vector $h$, we obtain $\iprod{Th}{h} = \tr(PT) = \nu(P) \ge 0$. This implies that $T$ is positive.
If $P_n$ is an increasing sequence of finite rank projections converging to the identity operator strongly, then
$$\tr(T) = \limn \tr (P_nT) = \limn \tr (TP_n) =\limn \nu(P_n) \le \sup_{P\in \calP_{\rm fin}(H)} \nu(P).$$
In the opposite direction, for any $P\in \calP_{\rm fin}(H)\nu(P)$ we have
$$ \nu(P) = \tr (PT) = \tr(TP) \le \tr (T).$$
Taking the supremum over all $P\in \calP_{\rm fin}(H)$ we obtain $M\le \tr(T)$.
This proves the identity $\tr (T) = \sup_{P\in \calP_{\rm fin}(H)} \nu(P)$.
\smallskip
{\em Step 5} --
It remains to prove that the extended mapping $\nu:\calP(H)\to [0,1]$ is countably additive, the final assertion in the statement of the theorem being clear. Let $(P_n)_{n\ge 1}$ be a sequence of disjoint orthogonal projections and let $P$ be the orthogonal projection onto the closure of the span of their ranges.
If $(h_j^{(n)})_{j\ge 1}$ is an orthonormal basis for the range of $P_n$, then the union of these sequences can be relabelled into an orthonormal basis for the range of $P$. Then, since $P^\star = P$, $P_n^\star = P_n$, and $Ph_j^{(n)} =P_nh_j^{(n)} = h_j^{(n)}$,
\begin{align*} \nu(P) = \tr(PT) & = \sum_{j, n\ge 1} \iprod{Th_j^{(n)}}{h_j^{(n)}}
\\ & =\sum_{n\ge 1} \sum_{j\ge 1} \iprod{P_n T h_j^{(n)}}{h_j^{(n)}} = \sum_{n\ge 1}\tr(P_nT) = \sum_{n\ge 1} \nu(P_n),
\end{align*}
the third identity being justified by the nonnegativity of the summands.
\end{proof}
In what follows we denote by $\mathscr{S}(H)$\index{$S$@$\mathscr{S}(H)$} the set of all positive trace class operators with unit trace on $H$ corresponding to a pure state. We will see below (Proposition \ref{prop:states-SH}) that
this set is the closed convex hull of its set of extreme points.
Theorem \ref{thm:Gleason-light} establishes a one-to-one correspondence between the set of states $\nu:\calP(H)\to [0,1]$ and the set of positive trace class operators $T$ on $H$ with unit trace. The following theorem establishes two further one-to-one correspondences, namely, with the set of positive normal functionals $\phi:\calL(H)\to \C$ satisfying $\phi(I)=1$, and with the set of all positive functionals $\phi:\mathscr{K}(H)\to \C$ satisfying a similar normalisation. Here, {\em positivity}\index{positive!functional, Hilbertian}\index{functional!positive, Hilbertian} means that $T\ge 0$ implies $\phi(T)\ge 0$ and {\em normality}\index{normal!functional}\index{functional!normal} means that $$\sum_{n\ge 1}\phi(P_n) = \phi(P)$$ whenever $(P_n)_{n\ge 1}$ is a sequence of disjoint orthogonal projections in $H$ and $P$ is their least upper bound.
\begin{theorem}[States as positive normal functionals]\label{thm:state-posfc} Let $H$ be a separable Hilbert space.
\begin{enumerate}[\rm(1)]
\item \label{it:state-posfc1}
If $T$ is a positive trace class operator on $H$ with unit trace, then
\begin{align}\label{eq:phiSstate}\phi(S):= \tr(ST), \quad S\in \calL(H),
\end{align}
defines a bounded linear functional $\phi:\calL(H)\to \C$ that is positive, normal, and satisfies $\phi(I)=1$.
\item \label{it:state-posfc3} If $\phi:\mathscr{K}(H)\to \C$ is a positive bounded linear functional satisfying
$$\sup_{P\in \calP_{\rm fin}(H)}\phi(P)=1,$$ then there exists a unique positive trace class operator $T$ on $H$ with unit trace such that
\begin{align}\label{eq:phiSstateK}\phi(S):= \tr(ST), \quad S\in \mathscr{K}(H).
\end{align}
Moreover, upon replacing in \eqref{eq:phiSstateK} the compact operators by arbitrary bounded operators, an extension of $\phi$ to a state is obtained.
\item \label{it:state-posfc2} If $\phi:\calL(H)\to \C$ is a bounded linear functional that is positive, normal, and satisfies $\phi(I)=1$ and if there exists a positive trace class operator on $H$ with unit trace such that
\begin{align}\label{eq:phiSstateP}\phi(P)= \tr(PT), \quad P\in \calP_{\rm fin}(H),
\end{align}
then \eqref{eq:phiSstateP} holds with the projections $P$ replaced by arbitrary bounded operators on $H$, that is, \eqref{eq:phiSstate} holds, and we have $ \sup_{P\in \calP_{\rm fin}(H)}\phi(P)=\phi(I)=\tr(T) =1$.
\end{enumerate}
\end{theorem}
\begin{proof} \eqref{it:state-posfc1}:\ Suppose that $T$ is a positive trace class operator on $H$ with unit trace
and define $\phi$ as in the statement of the theorem. It is clear that $\phi(I) = \tr(T) = 1$. To prove the
positivity of $\phi$, let $S\ge 0$. If $(h_n)_{n\ge 1}$ is an orthonormal basis for $H$, the positivity of $T$ implies
$$\phi(S) = \tr(ST) = \tr(S^{1/2}TS^{1/2}) = \sum_{n\ge 1} \iprod{TS^{1/2} h_n}{S^{1/2}h_n} \ge 0.$$
The normality of $\phi$ follows from the countable additivity of the mapping $\nu:\calP(H)\to [0,1]$
given by $\nu(P) := \tr(PT)$.
\smallskip
\eqref{it:state-posfc3}:\ Let $\phi:\mathscr{K}\to \C$ be a positive functional satisfying the normalisation condition in the statement of the theorem.
Define the mapping $\nu:\calP_{\rm fin}(H)\to \C$ by $\nu(P):= \phi(P)$. Then $P\ge 0$ implies $\nu(P)\ge 0$ and
$I-P\ge 0$ implies that if $ P\le Q\le I$ with $Q\in \calP_{\rm fin}(H)$, then
$$1-\nu(P) \ge \phi(Q) -\nu(P) = \phi(Q)-\phi(P) = \phi(Q-P)\ge 0$$
by the positivity of $\phi$. Together this gives $\nu(P)\in [0,1]$. The linearity of $\phi$ implies that $\nu(0)=0$ and that $\nu$ is affine. It follows that $\nu$ satisfies the assumptions of Theorem \ref{thm:Gleason-light}.
Therefore $$\phi(P) = \nu(P) = \tr(PT)$$ for a unique positive trace class operator $T$ with unit trace,
and this formula extends $\nu$ to a state.
\smallskip
\eqref{it:state-posfc2}:\
As was established in the proof of Theorem \ref{thm:Gleason-light}, the identity $ \nu(S) = \tr(ST)$, which is assumed to hold for all $S\in \calP_{\rm fin}(H)$, extends uniquely to arbitrary compact operators $S\in \mathscr{K}(H)$ by continuity. Since $\phi$ is continuous, we obtain that $\phi(S) = \tr(ST)$ for all compact operators $S$.
Now let $S\in \calL(H)$ be an arbitrary bounded operator. Let $(h_n)_{n\ge 1}$ be an orthonormal basis for $H$ and let $P_n$ denote the orthogonal projection onto the span of the set $\{h_1,\dots,h_n\}$. We claim that
$$ \phi(S) = \limn \phi(P_nS).$$
Indeed, since $\phi$ is positive we may apply the Cauchy--Schwarz inequality to the sesqui\-linear mapping $(A,B)\mapsto \phi(AB^\star)$ and obtain, with $Q_n:= I-P_n$,
\begin{align*} |\phi(S) - \phi(P_n S)| & = |\phi(Q_nS)| \le |\phi(Q_n^2)||\phi(SS^\star)| \\ & = |\phi(I-P_n)||\phi(SS^\star)| = |\phi(I)-\phi(P_n)||\phi(SS^\star)| \to 0,
\end{align*}
where the last step follows from the normality of $\phi$. This proves the claim.
Putting things together we obtain
$$\phi(S) = \limn \phi(P_n S) = \limn \tr(P_n S T) = \tr (ST).$$
\end{proof}
The following corollary summarises what we have proved:
\begin{corollary}\label{cor:states-3defs} The following four sets are in one-to-one correspondence:
\begin{itemize}
\item states $\nu:\calP(H)\to [0,1]$;
\item positive trace class operators $T$ on $H$ satisfying $\tr(T)=1$, via
$$ \nu(P) = \tr(PT), \quad P\in \calP(H);$$
\item positive functionals $\phi:\mathscr{K}(H)\to \C$ satisfying $\sup_{P\in \calP_{\rm fin}(H)}\phi(P)=1$, via
$$ \phi(S) = \tr(ST), \quad S\in \calL(H);$$
\item positive normal functionals $\phi:\calL(H)\to \C$ satisfying $\phi(I)=1$, via
$$ \phi(S) = \tr(ST), \quad S\in \calL(H).$$
\end{itemize}
\end{corollary}
Here, and below, it is understood that a {\em functional} is a bounded linear functional.
In most treatments the following definition is adopted. It replaces the preliminary Definition \ref{def:state-preliminary}.
\begin{definition}[States]\label{def:state}
A {\em state}\index{state} is a positive normal functional $\phi:\calL(H)\to \C$ satisfying $\phi(I)=1$.
\end{definition}
This definition captures what is generally called a {\em normal state} in the mathematical literature on Quantum Mechanics; the term {\em state} is usually reserved for general positive functionals $\phi:\calL(H)\to \C$ satisfying $\phi(I)=1$. The small abuse of terminology committed by omitting the adjective `normal' from our terminology may be justified by the third item in the above list, which does not involve normality.
We conclude with an example of a finitely additive mapping $\nu:\calP(H)\to [0,1]$ which is not affine (and therefore does not define a state).
\begin{example}[Failure of affineness in two dimensions]\label{ex:cex-Gleason}
Let $H= \C^2$ and let $S$ denote its unit sphere. Let $f:S\to [0,1]$ be a function with the following two properties:
\begin{enumerate}[\rm(i)]
\item\label{it:cex-Gleason1} $f(h_1) = f(h_2)$ whenever ${\rm span}(h_1) = {\rm span}(h_2)$;
\item\label{it:cex-Gleason2} $f(h_1)+f(h_2) = 1$ whenever $h_1\perp h_2$.
\end{enumerate}
Apart from these restrictions, $f$ can be completely arbitrary.
Define $\nu:\mathscr{P}(H)\to [0,1]$ by $ \nu(0) :=0$, $\nu(I) = 1$, and
$$ \nu(P_h):= f(h), \quad h\in S,$$
where $P_h$ is the orthogonal projection onto ${\rm span}(h)$.
It is clear that $\nu$ is finitely additive: if the orthogonal projections $P_1,\dots,P_N$ are pairwise disjoint, then all but at most two must be zero. If there are zero or one nonzero projections, then additivity is trivial, and if there are
two nonzero projections they must be of the form $P_{h_1}$ and $P_{h_2}$ with $h_1\perp h_2$; in that case additivity
follows from
$$ \nu(P_{h_1}) + \nu(P_{h_2}) = f(h_1)+f(h_2) = 1 = \nu(I) = \nu(P_{h_1}+P_{h_2}).$$
If there exists a positive operator $T$ on $H$ with unit trace such that
for all $P\in \mathscr{P}(H)$ we have $\nu(P) = \tr(PT)$, then $$f(h) = \nu(P_h) = \tr(P_h T) = \iprod{Th}{h}$$ depends
continuously on $h$. It is however easy to construct discontinuous functions $f$ satisfying the conditions \eqref{it:cex-Gleason1} and \eqref{it:cex-Gleason2}. Indeed, once the value of $f$ at a given point $h_0\in S$ is fixed, the conditions \eqref{it:cex-Gleason1} and \eqref{it:cex-Gleason2} fix the values of $f$ only on the points $e^{i\theta}h_0$ and all points orthogonal to them. If we identify $S_H$ with the unit sphere $S^3$ in $\R^4$, these points define a `great circle' incident with $h_0$ and an `equator' relative to the `north pole' $h_0$. Therefore, in a sufficiently small neighbourhood of $h_0$, $f$ is only determined on a submanifold of dimension $1$. This leaves enough room to construct functions $f$ satisfying \eqref{it:cex-Gleason1} and \eqref{it:cex-Gleason2} but discontinuous at $h_0$.
If $\nu$ were affine we could represent it by a positive operator $T$. This would contradict the discontinuity of $f$.
\end{example}
It is not a coincidence that this counterexample lives in two dimensions: A celebrated theorem due to Gleason asserts that if $\dim(H)\ge 3$, then every countably additive mapping $\nu:\calP(H)\to [0,1]$ is affine and hence defines a state.
\subsection{Pure states}
Corollary \ref{cor:states-3defs} establishes three equivalent ways of looking at the convex set of all states.
Since the correspondence between them preserves convex combinations and hence extreme points, the following definition makes sense from each of the three points of view:
\begin{definition}[Pure states]
A {\em pure state}\index{pure state}\index{state!pure} is an extreme point of the convex set of states.
\end{definition}
\begin{proposition}\label{prop:pure} A state $\phi:\calL(H)\to \C$ is pure if and only if it is a {\em vector state},\index{vector state}\index{state!vector} that is,
there exists a unit vector $h\in H$ such that
$$\phi(S) = \iprod{Sh}{h}, \quad S\in \calL(H). $$
The unit vector $h$ is unique up to a scalar multiple of modulus one.
\end{proposition}
The first assertion can be equivalently stated as saying that the extreme points of the set of all positive trace class operators with unit trace are precisely the orthogonal projections of rank one.
\begin{proof}
`Only if': \ Let $\phi$ be a state and let $T$ be the associated positive trace class operator on $H$ with unit trace.
By the singular value decomposition (Theorem \ref{thm:tc-ell1}) we have
$ T =\sum_{n\ge 1} \la_n h_n\,\bar\otimes\, h_n$ for some orthonormal basis $(h_n)_{n\ge 1}$ of $H$ and nonnegative scalar sequence $(\la_n)_{n\ge 1}$ such that $\sum_{n\ge 1} \la_n = \tr(T) = 1$. This allows us to write $T$ as a convex combination of distinct states unless all but one $\la_n$ vanish, in which case we have $T = h_\nu\,\bar\otimes\, h_\nu$ for some unit vector $h_\nu\in H$ and $\nu(P) = \tr(P\circ (h_\nu\,\bar\otimes\, h_\nu)) = \iprod{Ph_\nu}{h_\nu}$ for all orthogonal projections $P\in \calP(H)$.
\smallskip `If': \ If $\phi$ is a vector state, then the associated positive trace class operator is of the form $T = h\,\bar\otimes\, h$ with $\n h\n=1$. If $T = (1-\la) T_0 + \la T_1$ is a convex combination of positive trace class operators $T_0$ and $T_1$ with unit trace, then the unit vector $h = Th = (1-\la) T_0 h+ \la T_1 h$ is a convex combination of
two vectors of norm at most one. Hence we must have either $h = (1-\la) T_0 h$ or $h = \la T_1 h$.
Since $T_0$ and $T_1$ are contractive, this is only possible if $\la=0$ (in the first case) or $\la=1$ (in the second case). This means that either $T = T_0$ or $T = T_1$, so $T$ is an extreme point of the convex set of
positive trace class operator on $H$ with unit trace. Since the correspondence between states and the associated
positive trace class operators preserves convex combinations, it follows that $\phi$ is an extreme point of the convex set of states.
The uniqueness assertion follows by observing that for all $\theta\in\R$ we have $$(e^{i\theta}h_\nu) \,\bar\otimes\, (e^{i\theta}h_\nu) = h_\nu \,\bar\otimes\, h_\nu.$$
\end{proof}
\begin{remark}[Bras, Kets, and wave functions] In the physics literature, the pure state corresponding to a unit vector $h\in H$ is commonly denoted by\index{$H$@$\vert h\rb$}
$$|h\rb$$ and referred to as the {\em ket}\index{ket} or {\em wave function}\index{wave!function} associated with $h$.
It is common to identify $|h\rb$ with $h$. The addition in $H$ can be used to define, for orthogonal unit vectors $h_1,h_2\in H$ and scalars $\al_1,\al_2\in \C$ satisfying $|\al_1|^2+|\al_2|^2$, the pure state
$$\al_1|h_1\rb + \al_2 |h_2\rb := | \al_1 h_1+\al_2 h_2\rb.$$
Such states are referred to as (coherent) {\em superpositions}\index{superposition} of the states corresponding to $h_1$ and $h_2$.
Such states should be carefully distinguished from states that can be built by using the addition of $\calL_1(H)$.
Indeed, for $h_1,h_2\in H$ are linearly independent unit vectors in $H$ and $\la\in [0,1]$ the convex combination
$ (1-\la) h_1\,\bar\otimes\,h_1 + \la h_2\,\bar\otimes\,h_2$, or, in physics notation,
\begin{align*}
(1-\la) |h_1\rb\lb h_1| + \la |h_2\rb\lb h_2|,
\end{align*}
defines a state in $\calL_1(H)$.
Such states, which are not pure unless either $\la=0$ or $\la=1$, are called {\em mixed states}\index{state!mixed}\index{mixed state} or, more precisely, {\em mixtures} of the states corresponding to $h_1$ and $h_2$.
\end{remark}
We recall that $\mathscr{S}(H)$ denotes the convex set of all positive trace class operators with unit trace on $H$.
As we have seen in Corollary \ref{cor:states-3defs}, the elements of this set are in one-to-one correspondence with states.
By Proposition \ref{prop:pure}, the extreme points of $\mathscr{S}(H)$ are the rank one projections of the form $h\,\bar\otimes\,h$ with $h\in H$ of norm one.
\begin{proposition}\label{prop:states-SH} The set $\mathscr{S}(H)$ is the closed convex hull of its extreme points.
The extreme points of this set are precisely the rank one projections of the form $h\,\bar\otimes\,h$ with $h\in H$ of norm one.
\end{proposition}
\begin{proof} By the singular value decomposition of Theorem \ref{thm:tc-ell1}, every element of $T\in \mathscr{S}(H)$
is of the form $ T = \sum_{n\ge 1} \la_n h_n\,\bar\otimes\,h_n$, with convergence in trace norm, with $(h_n)_{n\ge 1}$ an orthonormal bases in $H$
and $(\la_n)_{n\ge 1}$ a nonnegative sequence satisfying $\sum_{n\ge 1}\la_n = 1$. This gives the first assertion. The second follows from Corollary \ref{cor:states-3defs}, which informs us that the operators of the form $h\,\bar\otimes\, h$ with $h\in H$ of norm one are in one-to-one correspondence with the vector states, which are the extreme points of the convex set of all states by Proposition \ref{prop:pure}.
\end{proof}
\subsection{Observables}
Let $(\Om,\calF)$ be a measurable space. Classically, an $\Om$-valued observable is a measurable function $f:X\to \Om$,
where $(X,\X)$ is another measurable space. By definition of measurability,
$f$ induces a mapping from $\calF$ to $\X$ given by $$F \mapsto f^{-1}(F), \quad F\in \calF,$$ and this mapping is countably additive in the sense that if the sets $F_n\in\calF$, $n\ge 1$ are pairwise disjoint, then
$f^{-1}(\bigcup_{n\ge 1} F_n) =\bigcup_{n\ge 1} f^{-1}(F_n)$.
Identifying sets in $\X$ by their indicator functions and replacing them by orthogonal projections in a Hilbert space $H$, we arrive at the following definition of an observable in Quantum Mechanics.
\begin{definition}[Observables]\label{def:observable}\index{observable} Let $(\Om,\calF)$ be a measurable space and $H$ a Hilbert space.
An {\em $\Om$-valued observable} is a countably additive mapping $P:\calF\to \mathscr{P}(H)$ satisfying $P(\Om) = I$.
An {\em elementary observable}\index{elementary observable}\index{observable!elementary} is a $\{0,1\}$-valued observable.
\end{definition}
Observables defined in this way are sometimes called {\em sharp observables}\index{sharp observable}\index{observable!sharp}, as opposed to unsharp observables which will be introduced in Section \ref{subsec:POVM}.
Following notation introduced in Chapter \ref{chap:spectral-theorem} we write $P_F := P(F)$ for $F\in\calF$. For vectors $h\in H$, we denote by $P_h$ the nonnegative probability measure on $\Om$ given by $$P_h(F) := \iprod{P_F h}{h}, \quad F\in\calF.$$
In the operator-theoretic language of Chapter \ref{ch:unbdd}, a real-valued observable is nothing but
a projection-valued measure on $\R$ and
we can associate a unique self-adjoint operator $(A,\Dom(A))$ with $P$
determined by
$$ \Dom(A) = \Big\{h\in H: \ \int_\R |\la|^2 \ud P_h(\la)< \infty\Big\}$$
and, for $h\in\Dom(A)$,
$$ \iprod{Ah}{h} = \int_\R \lambda\ud P_h(\la)$$
(see Theorem \ref{thm:Borel-FC-unbddnormal}).
In the converse direction, the spectral theorem asserts that every self-adjoint operator
$(A,\Dom(A))$ arises from a projection-valued measure on $\R$ by the above procedure and hence defines an observable.
In conclusion, {\em real-valued} observables are in one-to-one correspondence with self-adjoint operators. This is the point of view taken in most treatments of Quantum Mechanics. While being more direct in many ways, the advantage of the present treatment is that it provides a deeper motivation as being a direct consequence of the point of view that, on the mathematical level, the classical-to-quantum transition is the transition from the Boolean algebra of sets to the lattice of orthogonal projections. A further advantage of the present approach is that, in the same vein,
the spectral theorem for normal operators can be reinterpreted as establishing a one-to-one correspondence between {\em complex-valued observables} and normal operators, and observables with values in the unit circle correspond to unitary operators.
If $P:\calF\to \calP(H)$ is an observable and $\phi:\calL(H)\to\C$ is a pure state represented by the unit vector $h\in H$, then
$$\phi(P_F) = \iprod{P_F h}{h} = P_h (F), \quad F\in \cal F,$$
so the assignment $F\mapsto \phi(P_F)$ defines a probability measure.
The following proposition is an immediate consequence of the fact that states are normal. It generalises this observation to arbitrary states and allows us to interpret
the number $\phi(P_F)$
as ``the probability that measuring $P$ results in a value contained in $F\in\F$ when the system is
in state $\phi$''. If $P$ is a real-valued (or complex-valued) observable represented by a self-adjoint (or normal) operator $A$,
then $P$ is supported on the spectrum $\sigma(A)$ and $P$ can be thought of as an $\sigma(A)$-valued observable. The physical interpretation is that ``with probability one,
a measurement of $A$ produces a value belonging to $\sigma(A)$''.
\begin{proposition}\label{prop:observ-mapping}
If $\phi:\calL(H)\to\C$ is a state and $P:\calF\to \calP(H)$ an $\Om$-valued observable, the mapping $$ F \mapsto \phi(P_F), \quad F\in \calF,$$
defines a probability measure on $(\Om,\calF)$.
\end{proposition}
\subsection{The uncertainty principle}
If $P$ is an observable represented by a bounded self-adjoint operator $A$, the {\em expected value of $P$ in state $\phi$}\index{expected value!of an observable} is defined as the number
$$ \lb A\rb_\phi := \phi(A).$$
This definition extends to unbounded self-adjoint operators as long as $TA$ defines a bounded operator of trace class.
If $\phi$ is a pure state associated with the unit vector $h\in H$ with $h\in \Dom(A)$, we have
$$ \lb A\rb_\phi = \iprod{Ah}{h}.$$
In this situation, if $P$ is a real-valued observable represented by a self-adjoint operator $A$ with $h\in \Dom(A)$, we can define the {\em variance}\index{variance!of an observable} of $A$ in state $\nu$ by
$$\var_\phi(A) := \big\lb (A - \lb A\rb_\phi)^2\big\rb_\phi = \n (A - \iprod{Ah}{h})h\n^2.$$
The {\em uncertainty}\index{uncertainty!of an observable} of $A$ in state $\phi$ is defined by
$$\Delta_\phi(A) := \sqrt{\var_\phi(A)}.$$
\begin{theorem}[Uncertainty principle]\label{thm:uncertainty}\index{uncertainty!principle} Let $\phi$ be a pure state associated with the unit vector $h\in H$,
and consider two real-valued observables with associated self-adjoint operators
$A$ and $B$.
If $h\in \Dom([A,B]):= \{h\in \Dom(A)\cap \Dom(B): \, Ah\in \Dom(B), \, Bh\in \Dom(A)\}$, then
$$ \Delta_\phi(A)\Delta_\phi(B) \ge \frac12 |\iprod{[A,B])h}{h}|,$$
where $[A,B] = AB - BA$ is the commutator of $A$ and $B$.
\end{theorem}
\begin{proof} The operators $\wt A:= A - \iprod{Ah}{h}$ and $\wt B:= B - \iprod{Bh}{h}$
are self-adjoint and satisfy $\Dom(\wt A) = \Dom(A)$ and $\Dom(\wt B) = \Dom(B)$. In particular
we note that $\wt A h\in \Dom(\wt B)$ and $\wt B h\in \Dom(\wt A)$.
The Cauchy--Schwarz inequality implies
\begin{align*}
\Delta_\phi(A)\Delta_\phi(B)
& = \n \wt A h\n\n \wt B h\n
\ge |\iprod{\wt A h}{\wt B h}|
\\ & \ge |\Im \iprod{\wt A h}{\wt B h}|
= \frac12 |\iprod{\wt A h}{\wt B h} - \iprod{\wt B h}{\wt A h}|
\\ & = \frac12 |\iprod{[\wt A, \wt B]h}{h}| = \frac12 |\iprod{[A, B]h}{h}|.
\end{align*}
\end{proof}
\begin{proposition}\label{prop:pure-state} Let $P$ be a real-valued observable, represented by the self-adjoint operator $A$, and let $h\in \Dom(A)$ satisfy $\n h\n=1$. The following assertions are equivalent:
\begin{enumerate}[\rm(1)]
\item\label{it:pure-state1} $A$ has zero uncertainty in the pure state $\phi$ associated with $h$;
\item\label{it:pure-state2} $h$ is an eigenvector for $A$.
\end{enumerate}
If these equivalent conditions hold, then for the corresponding eigenvalue $\la$ have $$
\la = \lb A\rb_\phi \ \ \hbox{and} \ \ \phi(P_{\{\la\}}) = 1.$$
\end{proposition}
The physical interpretation is that a measurement of $A$ in a pure state $\phi$ gives the expected value $\lb A\rb_\phi$ with probability one if and only if the representing unit vector $h$ is an eigenvector of $A$, and in this case the eigenvalue equals $\lb A\rb_\phi$.
\begin{proof}
\eqref{it:pure-state1}$\Rightarrow$\eqref{it:pure-state2}: \ If $ \var_\phi(A) =0$, then $Ah = \iprod{Ah}{h}h$, so $h$ is an eigenvector of $A$ with eigenvalue $\la = \iprod{Ah}{h} = \lb A\rb_\phi$.
\eqref{it:pure-state2}$\Rightarrow$\eqref{it:pure-state1}: \ $Ah = \lambda h$, then
$$ \var_\phi(A) = \n (A - \iprod{Ah}{h})h\n^2 = \n (A - \la)h\n^2 = 0.$$
If the equivalent conditions hold, then by
Corollary \ref{cor:normal-eigen-mapping} for all measurable functions $f:\sigma(A)\to \C$
we have $f(A)h = f(\la)h$ and consequently
$$ \int_{\sigma(A)} f\ud P_h = \iprod{f(A)h}{h} = f(\la).$$
This forces $P_h = \delta_{\{\la\}}$ and therefore
$$\phi(P_{\{\la\}}) = \iprod{P_{\{\la\}} h}{h} = \iprod{h}{h} = 1.$$
\end{proof}
\subsection{The qubit}\label{subsec:qubit}
It is instructive to take a closer look at the simplest genuinely quantum mechanical system, the qubit. It is the quantum version of the {\em bit}\index{bit} $\{0,1\}$ and should be thought of as equipped with the counting measure $\mu$ giving mass $1$ to each of the two elements of $\{0,1\}$.
We write
$\one_{\{0\}}$ and $\one_{\{1\}}$ for the unit basis vectors of the Hilbert space $L^2(\{0,1\})$, which we isometrically identify with $\C^2$.
Identifying pure states with unit vectors $h\in \C^2$, every pure state is of the form $|\alpha\one_{\{0\}}+\beta\one_{\{1\}}\rb$ where $\alpha,\beta\in\C$ satisfy $|\alpha|^2 + |\beta|^2 = 1$;
here we used the ket notation $|h\rb$ to denote the pure state represented by a unit vector $h$.
Since pure states are defined up to a complex number of modulus one, every pure state can be uniquely written in the form $|\cos \left(\theta /2\right) \one_{\{0\}} + e^{i\phi }\sin \left(\theta /2\right)\one_{\{1\}}\rb$ or, in the
physicist's notation,
\begin{align}\label{eq:Bloch1}\cos \left(\theta /2\right)\ket{0} +e^{i\phi }\sin \left(\theta /2\right)\ket{1}
\end{align}
for suitable $0\leq \theta \leq \pi$ and $0\leq \phi <2\pi$ (in
In spherical coordinates, the variables $\theta $ and $\phi$
uniquely determine a point
\begin{align}\label{eq:Bloch3}
(\sin \theta \cos \phi ,\,\sin \theta \sin \phi ,\,\cos \theta)
\end{align}
on the unit sphere $S^2$ of $\R^3$. This representation of pure states is frequently referred to as the {\em Bloch sphere}.\index{Bloch!sphere}
\begin{wrapfigure}{r}{5cm}
\begin{center}
\includegraphics[scale=0.16]{Bloch_sphere.png}
\footnotesize{The Bloch sphere \\ (Source: Wikipedia)}
\end{center}
\end{wrapfigure}
Arbitrary states can be identified with points in the closed unit ball of $\R^3$ as follows. Any self-adjoint operator $T = (t_{ij})_{i,j=1}^2$ on $\C^2$ with unit trace $\tr(T) = t_{11}+t_{22} = 1$ is of the form
\begin{align}\label{eq:T-qubit} T = \frac12\begin{pmatrix}
1+c_3 & \, c_1-ic_2 \\ \, c_1+ic_2 & \, 1-c_3
\end{pmatrix}
\end{align}
with $c_1,c_2,c_3\in \R$. The eigenvalues of $T$ are $\frac12(1\pm |c|)$, where $c = (c_1,c_2,c_3)$ is called the {\em Bloch vector}\index{Bloch!vector} of $T$. From this we see that $T\ge 0$ if and only if $|c|\le 1$, that is, $c\in \ov B_{\R^3}$.
A routine computation shows that the pure state
associated with the unit vector $h:= \cos \left(\theta /2\right)\ket{0} +e^{i\phi }\sin \left(\theta /2\right)\ket{1}$
corresponds to the operator
\begin{align}\label{eq:Bloch2} T = h\,\bar\otimes\, h
= \begin{pmatrix}
1+ \cos \theta & \, e^{-i\phi }\sin\theta \\ \, e^{i\phi }\sin \theta & \, 1-\cos\theta
\end{pmatrix}
\end{align}
with Bloch vector $$(\sin\theta\cos\phi, \,\sin\theta\sin\phi, \,\cos\theta).$$
Thus the Bloch sphere representation of the pure state associated with a unit vector $h\in\C^2$ equals the Bloch vector of the associated operator $h\,\bar\otimes\, h$.
Equation \eqref{eq:T-qubit} can be written as
\begin{align*} T & = \frac12 \begin{pmatrix}\, 1 & \, 0\, \\ \,0 & 1 \end{pmatrix}
+ \frac{c_1}2 \begin{pmatrix}\, 0 & \, 1\, \\ 1 & 0 \end{pmatrix}
+ \frac{c_2}2 \begin{pmatrix}\, 0 & \,-i\, \\ i & 0 \end{pmatrix}
+ \frac{c_3}2 \begin{pmatrix}\, 1 & \, 0\, \\ 0 & -1 \end{pmatrix}
\\ & = \frac12(I + c_1\sigma_1 + c_2 \sigma_2+ c_3 \sigma_3),
\end{align*}
where
$$\sigma_1 = \begin{pmatrix}\, 0 & \, 1\, \\ 1 & 0 \end{pmatrix}, \quad
\sigma_2 = \begin{pmatrix}\, 0 & \,-i\, \\ i & 0 \end{pmatrix}, \quad
\sigma_3 = \begin{pmatrix}\, 1 & \, 0\, \\ 0 & -1 \end{pmatrix}$$
are the three {\em Pauli matrices}.\index{Pauli matrices}
These matrices are self-adjoint and their spectra equal $\{\pm 1\}$. Therefore they are associated with $\pm 1$ valued observables, also denoted by $\sigma_1$, $\sigma_2$, and $\sigma_3$. The corresponding eigenstates of $\sigma_j$ are called the `spin up/down states along the $j$-th axis'.
Every self-adjoint operator $A$ on $\C^2$ is of the form
$$ A = \begin{pmatrix}\, a & c_1-ic_2\, \\ \,c_1+ic_2 & b \end{pmatrix}
= \begin{pmatrix}\, c_0+c_3 & c_1-ic_2\, \\ \,c_1+ic_2 & c_0-c_3 \end{pmatrix}
= c_0 I + c_1\sigma_1 + c_2 \sigma_2+ c_3 \sigma_3
$$
for certain $a,b,c_0,c_1,c_2,c_3\in \R$ with $a = c_0+c_3$ and $b = c_0-c_3$. It follows that
the quadruple $\{I,\sigma_1,\sigma_2,\sigma_3\}$ is a basis for the real-linear vector space of self-adjoint operators on $\C^2$.
\section{Entanglement}\label{sec:entaglement}
The natural choice for the state space of a system of $N$ classical point particles in $\R^3$ is $\R^{6N}$, the idea being that six coordinates (three for position, three for momentum) are needed to describe the state of each particle.
In Quantum Mechanics, the natural choice of Hilbert space is $L^2(\R^{3N})$. Labelling the points of $\R^{3N}$ as
$x = (x_j^{(n)})_{j,n=1}^{3,N}$ this choice suggests the following natural definition of observables $\wh x_j^{(n)}$ describing the $j$-th coordinate of the $n$-th particle:
$$ \wh x_j^{(n)} f(x) := x_j^{(n)}f(x),\quad f\in \Dom(\wh x_j^{(n)}),$$
where $\Dom(\wh x_j^{(n)})= \{f\in L^2(\R^{3N}):\, \wh x_j^{(n)} f\in L^2(\R^{3N})\}$.
The space $L^2(\R^{3N}$ is isometric in a natural way to the $N$-fold Hilbert space tensor product (see Definition \ref{def:Hilbert-tensor} and the discussion following it)
$$ L^2(\R^{3N}) \simeq \underbrace{L^2(\R^3)\ot\cdots\ot L^2(\R^3)}_{n \ \rm{times}}.$$
This suggests that if the Hilbert spaces $H_1,\hdots,H_N$ describe the states of $N$ quantum mechanical systems, then
its Hilbert space tensor product $$H_1\ot\cdots\ot H_N$$ serves to describe the $N$-partite system composed of these $N$ subsystems. In what follows we focus on the case $N=2$, but everything we say extends to general $N$ without difficulty.
Let $H$ and $K$ be Hilbert spaces, and let $H\otimes K$ be their Hilbert space tensor product.
For unit vectors $h \in H$ and $k\in K$ we write $|h\rb$ and $|k\rb$ for the pure states in $H$ and $K$ represented by these vectors, and
$$ |h\rb |k\rb:= |h\otimes k\rb $$ for the pure state represented by the unit vector $h\otimes k$ in $H\otimes K$.
Suppose now that orthonormal vectors $h_1, h_2 \in H$ and orthonormal vectors $k_1,k_2\in K$ are given.
Then the unit vectors $h_1\otimes k_1$ and $h_2\otimes k_2$ are orthogonal in $H\ot K$. Hence,
for scalars $\al_1,\al_2\in \C$ satisfying $|\al_1|^2+|\al_2|^2 = 1$, the superposition
$\al_1 h_1\otimes k_1 + \al_2 h_2\otimes k_2 $ defines a unit vector in $H\ot K$. To this unit vector corresponds the pure state
$$ \al_1 |h_1\rb|k_1\rb + \al_2 |h_2\rb|k_2\rb := |\al_1 h_1\otimes k_1 + \al_2 h_2\otimes k_2\rb. $$
Unless $\al_1=0$ or $\al_2=0$, this state cannot be written in the form $|h\rb|k\rb$. States with this property are called {\em entangled states}\index{entangled state}\index{state!entangled}.
The partial trace (see Section \ref{subsec:partialtrace}) can be used to define states of subsystems
starting from the state of a composite system. More concretely, suppose that $T\in \calL_1(H\ott K)$ is a positive trace class operator with unit trace. Then the operators
$ \tr_K(T)$ and $\tr_H(T)$
are positive trace class operators with unit trace in $\calL_1(H)$ and $\calL_1(K)$, respectively. If we think of $T$ as describing the state of a bipartite system with Hilbert space $H\ott K$, $\tr_K(T)$ and $\tr_H(T)$ can be thought of as describing the states of the two constituent subsystems with Hilbert spaces $H$ and $K$, respectively. For example, by the result of Example \ref{ex:partialtrace}, in the case of a pure states $T$ represented by the unit vector $h\ot k$ in $H\ott K$, the states $\tr_K(T)$ and $\tr_H(T)$ are the pure states represented by the unit vectors $h$ and $k$, that is, we have
$$ \tr_K(T) = |h\rb\lb h|, \quad \tr_H(T) = |k\rb\lb k|.$$
\section{Positive operator-valued measures}
We discuss next a natural extension of the notion of an observable.
\subsection{Effects}
Let $(X,\X)$ be a measurable space. As a warm-up we show:
\begin{proposition}\label{prop:convexhullBbM} The closed convex hull in $B_{\rm b}(X)$ of the set of elementary observables
$\{\one_B:\,B\in \X\}$ equals
$$ \calE(X):= \{f\in B_{\rm b}(X): \ 0\le f \le \one \ \hbox{pointwise}\}.$$
The extreme points of $\calE(X)$ are precisely the elementary observables.
\end{proposition}
\begin{proof}
Denote by $E(X)$ the closed convex hull of the set of elementary observables.
The inclusion $E(X)\subseteq \calE(X)$ is trivial. To prove the inclusion $\calE(X)\subseteq E(X)$,
let $f\in \calE(X)$ be given. Given $\eps>0$, select a simple function
$g =\sum_{j=1}^k c_j \one_{C_j}$ such that $\n f - g\n_\infty<\eps$; this function may be chosen in such a way that
the measurable sets $C_j$ are disjoint and the coefficients satisfy $0\le c_j\le 1$.
After relabelling we may assume that $0\le c_1\le \dots\le c_k\le 1$.
If $k=1$, then $g = (1-c_1)\one_{\emptyset} + c_1 \one_{C_1}$ belongs to $E(X)$. If $k\ge 2$ we
set $$L_0:=\emptyset \quad \hbox{and} \quad L_i:= \bigcup_{j=i}^k C_j \quad (j=1,\dots,k)$$ and
$$\la_0:= 1-c_k, \quad \la_1:= c_1, \quad \hbox{and} \quad \la_i:= c_i-c_{i-1} \quad (i=2,\dots,k).$$
Then $0\le \la_i\le 1$, $ \sum_{i=0}^k \la_i = 1$, and
$$ g =\sum_{j=1}^k c_j \one_{C_j} = \sum_{i=0}^k \la_i \one_{L_i}.$$
It follows that $g$ belongs to $E(X)$.
Since $\eps>0$ was arbitrary, this proves that $f\in {E(X)}$.
If $g\in \calE(X)$ is an elementary observable and $g = \lambda f_0+(1-\lambda)f_1$ with $0< \la< 1$ and $0\le f_j\le \one$ for $j=0,1$, then $0 = g(\xi) = \lambda f_0(\xi) +(1-\lambda)f_1(\xi)$ implies $f_0(\xi) =f_1(\xi) = 0$
and $1 = g(\xi') = \lambda f_0(\xi') +(1-\lambda)f_1(\xi')$ implies $f_0(\xi') =f_1(\xi') = 1$, that is,
$f_0 = f_1 = g$ pointwise. It follows that every elementary observable is an extreme point
of $\calE(X)$. If $g\in \calE(X)$ is not an elementary
observable, then the set $\{\eps\le g \le 1-\eps\}$ is nonempty for sufficiently small $\eps>0$, and then
it is easy to produce measurable $f_0\not=f_1$ satisfying $0\le f_j\le \one$ for $j=0,1$ and $g = \frac12f_0+\frac12 f_1$. It follows that $g$ is not an extreme point of $\calE(X)$.
\end{proof}
The closed convex hull of $\calP(H)$ in $\calL(H)$ is identified in the next proposition.
We write $S\le T$ to express that $T-S$ is a positive operator.
\begin{proposition}\label{prop:effect-convexhull} The closed convex hull in $\calL(H)$ of $\calP(H)$ equals \index{$E$@$\calE(H)$}
$$ \calE(H):= \{ E\in \calL(H): \, 0\le E\le I\}.$$
The extreme points of $\calE(H)$ are precisely the orthogonal projections.
\end{proposition}
\begin{proof}
Every element of the convex hull of $\calP(H)$ belongs to $\calE(H)$, and this passes on to the closed convex hull.
Since elements of $\calE(H)$ are positive and hence self-adjoint, every $E\in \calE(H)$ admits a representation as
$$ E =\int_{\sigma(E)} \l\ud P(\l)$$
where $P$ is the projection-valued measure of $E$.
Let $$f_n := \sum_{j=1}^{2^n}\frac{j}{2^{n}} \one_{I_j}
= \frac{1}{2^{n}}\sum_{j=1}^{2^n} \one_{(\frac{j-1}{2^n}, 1]} $$
where $I_j:=(\frac{j-1}{2^n}, \frac{j}{2^n}]$ for $1\le j\le 2^n$. Set
$$ E_n :=\int_{\sigma(E)} f_n(\l)\ud P(\l)
=\frac1{2^n}\sum_{j=1}^{2^n} P_{((j-1)/2^n, 1]}.
$$
Then $E_n$ is in the convex hull of $\calP(H)$ and
$$\limn \n E-E_n \n \le \limn \sup_{\l\in [0,1]}|\l - f_n(\l)| = 0.$$
This proves that $E$ is in the closed convex hull of $\calP(H)$.
If $E\in \calE(H)$ is an orthogonal projection and $E = \lambda E_0+(1-\lambda)E_1$ with $0< \la< 1$ and $0\le E_j\le I$ for $j=0,1$, then for all $x\in \ker(E)$ we have $ \lambda \iprod{E_0 x}{x} +(1-\lambda)\iprod{E_1 x}{x}=0$
with $\iprod{E_i x}{x} \ge 0$ for $i=0,1$, and this is possible only if $\iprod{E_0 x}{x} = \iprod{E_1 x}{x} = 0$.
For all norm one vectors $x\in \ran(E)$ we have $\iprod{Ex}{x} = \iprod{x}{x} = 1$ and consequently $\lambda \iprod{E_0 x}{x} +(1-\lambda)\iprod{E_1 x}{x} = 1$.
Since $\iprod{E_i x}{x} \le 1$ for $i=0,1$, this is possible only if $\iprod{E_0 x}{x} = \iprod{E_1 x}{x} = 1$.
It follows that $E_0=E_1 = 0$ on $\ker(E)$ and $E_0=E_1=I$ on $\ran(E)$,
and therefore $E_0=E_1=E$.
It follows that $E$ is an extreme point of $\calE(H)$.
If $E\in \calE(H)$ is not an orthogonal projection, then the spectral theorem for bounded self-adjoint operators implies that the spectrum $\sigma(E)$ cannot be equal to $\{0,1\}$. Since $\sigma(E)$ is contained in $[0,1]$ it follows that
$[\eps,1-\eps]\cap\sigma(E)$ is nonempty for all sufficiently small $\eps>0$ and then, again by the spectral theorem,
it is easy to produce operators $E_0\not=E_1$ in $\calE(H)$ such that $E = \frac12E_0+\frac12 E_1$. It follows that $E$ is not an extreme point of $\calE(H)$.
\end{proof}
\begin{definition}[Effects]
The elements of the set $\calE(H)$ are called {\em effects}.\index{effect}
\end{definition}
Effects are self-adjoint, and it follows from Theorem \ref{thm:spect-sa} that
a self-adjoint operator on $H$ is an effect if and only if
its spectrum is contained in the unit interval $[0,1]$.
If $T$ is an arbitrary nonzero positive operator, then for all $0\le c \le \n T\n^{-1}$ the operator $cT$ is an effect. Indeed, this is clear for $c=0$, and if $c>0$ the operator
$c^{-1} I - T$ positive since it is self-adjoint and has positive spectrum.
A mapping $\nu: \calE(H)\to [0,1]$ is said to be {\em finitely additive}\index{finitely additive} if
$$\sum_{n=1}^N \nu(E_n) = \nu(E)$$ whenever $E_1,\dots,E_N, E\in \calE(H)$ satisfy $E_1+\cdots+E_N = E$.
\begin{theorem}[Busch]\index{theorem!Busch}\label{thm:Busch} Every finitely additive mapping $\nu: \calE(H)\to [0,1]$ satisfying $\nu(I)=1$ restricts to an affine mapping $\nu:\calP(H)\to [0,1]$ and hence defines a state.
\end{theorem}
\begin{proof} By assumption we have $\nu(I)=1$ and
from $1 = \nu(I) = \nu(I+0) = \nu(I)+\nu(0) = 1 + \nu(0)$ it follows that $\nu(0)=0$. We must show that
$\nu$, as a mapping from $\calP(H)$ to $[0,1]$, is affine.
If $E_1,E_2$ are effects satisfying $E_1\le E_2$, then $E_2-E_1$ is an effect
and $\nu(E_1)\le \nu(E_1)+\nu(E_2-E_1) = \nu(E_2)$.
If $E$ is an effect, then so is $\frac1nE$, and by finite additivity we obtain $\nu(E) = n\nu(\frac1n E)$ for all $n=1,2,\dots$
If $0\le q = m/n\le 1$ is a rational number, then $qE$ is an effect and $q\nu(E) = nq \nu(\frac1n E) = m \nu(\frac1n E) = \nu(\frac{m}n E) = \nu(q E)$.
If $0\le \alpha\le 1$ is a real number, then $\alpha E$ is an effect, and if $p_j,q_j$ are rational numbers satisfying $p_j\uparrow \alpha$ and $q_j\downarrow \alpha$, then
$$ p_j\nu(E) = \nu(p_jE) \le \nu(\alpha E) \le \nu(q_jE) = q_j\nu(E),$$
and therefore $ \alpha \nu(E) = \nu(\alpha E)$.
We have already observed that if $T\in\calL(H)$ is a positive operator, then for all $0< c\le \n T\n^{-1}$ the operator $cT$ is an effect. We now define $$\nu(T):= c^{-1}\nu(cT).$$
This quantity is well defined: if $0<c_1\le c_2\le \n T\n^{-1}$, then
$c_2^{-1}\nu(c_2 T) = c_1^{-1} \frac{c_1}{c_2}\nu(c_2 T) = c_1^{-1} \nu({c_1}T)$.
Suppose now that positive operators $T_1,\dots, T_N$ and $T$ satisfy $\sum_{n= 1}^N T_n = T$.
Choose $c>0$ such that $c T$ is an effect.
Then each operator $cT_n$ is an effect, and $\nu(T)
= c^{-1}\nu(cT) = c^{-1}\sum_{n= 1}^N \nu(cT_n) =\sum_{n= 1}^N \nu(T_n)$.
In particular this applies to convex combinations of orthogonal projections, from which it follows that
$\nu$, as a mapping from $\calP(H)$ to $[0,1]$, is affine.
\end{proof}
\subsection{Positive operator-valued measures}\label{subsec:POVM}
The next definition generalises the notion of a projection-valued measure by replacing the role
of orthogonal projections by effects.
\begin{definition}[Positive operator-valued measures]\label{def:POVM} A {\em positive op\-erator-valued measure} (POVM)\index{POVM}
on a measurable space $(\Om,\calF)$
is a mapping $Q: \calF\to \calE(H)$ that assigns to every set $F\in \calF$ an effect $Q_F:=Q(F) \in \calE(H)$ with
the following properties:
\begin{enumerate}[\rm(i)]
\item\label{it:POVM1} $Q_\Om = I$;
\item\label{it:POVM2} for all $x\in H$ the mapping $$F\mapsto \iprod{Q_F x}{x}, \quad F\in \calF,$$
defines a finite measure $Q_{x}$ on $(\Om,\calF)$.
\end{enumerate}
\end{definition}
The measures defined by \eqref{it:POVM2} is denoted by $Q_{x}$. Thus, for all $F\in \calF$ and $x\in H$, by definition we have
$$ \iprod{Q_F x}{x} = Q_{x}(F) =\int_{\Om} \one_F \ud Q_{x}.$$
Note that
$$ Q_x(\Om) = \iprod{Q_\Om x}{x} = \iprod{x}{x} =\n x\n^2.$$
Every projection-valued measure is a POVM. In the converse direction we have the following simple result.
\begin{proposition}\label{prop:POVM-PVM} A POVM $Q: \calF\to \calE(H)$ is a projection-valued measure
if and only if $Q_FQ_{F'} = Q_{F\cap F'}$ for all $F,F'\in \calF$.
\end{proposition}
\begin{proof}
The `only if' part has already been established in Section \ref{sec:PVM}. The `if' part is evident from $Q_F^2 = Q_{F\cap F} = Q_F$, which shows that $Q_F$ is a projection. Since it is also positive, $Q_F$ is an orthogonal projection.
\end{proof}
A POVM which is not projection-valued is sometimes called an {\em unsharp observable}.\index{unsharp!observable}\index{observable!unsharp} An example will be discussed in
Section \ref{subsec:number-phase}.
We have seen in Proposition \ref{prop:observ-mapping} that if $\phi$ is a state and $P:\calF\to \calP(H)$ is a projection-valued measure, the mapping $$ F \mapsto \phi(P_F), \quad f\in \calF,$$
defines a probability measure on $(\Om,\calF)$. Identifying a state $\phi$ with its associated positive trace class operator $T$, we obtain a mapping which assigns a probability measure to every positive trace class operator with unit trace which preserves convex combinations.
Inspection of this argument shows that it extends to POVMs. The following proposition shows that in the converse direction, every POVM arises in this way.
We recall that $\mathscr{S}(H)$ denotes the convex set of all positive trace class operators with unit trace on $H$. As we have seen in Proposition \ref{prop:states-SH}, this set is the closed convex hull of its the extreme points, which are precisely the rank one projections of the form $h\,\bar\otimes\,h$ with $h\in H$ of norm one.
\begin{theorem}[POVMs as unsharp observables]\label{thm:POVM-unsharp} Let $(\Om,\calF)$ be a measurable space.
If $\Phi: \mathscr{S}(H) \to M(\Om)$ is an affine mapping, then there exists a unique POVM $Q:\calF\to\calE(H)$ such that
for all $T\in \mathscr{S}(H)$ we have
$$ (\Phi(T))(F) = \tr(Q_F T), \quad F\in \calF.$$
\end{theorem}
\begin{proof} The proof consists of two steps.
\smallskip {\em Step 1} -- We claim that
$\Phi$ extends to a bounded linear operator from $\calL_1(H)$ into $M(\Om)$. The proof of this claim is accomplished in three steps.
First, we set $\Phi(0):=$ and, for an arbitrary nonzero positive operator $T\in \calL_1(H)$,
$$ \Phi(T):= \n T\n_1 \Phi(T/\n T\n_1),$$
where $\n T\n_1 = \tr(T) >0$ since $T$ is positive and nonzero. Note that for all $c\ge 0$ we have $$\Phi(cT) = c\Phi(T).$$ The identity
$$ S+T = (\n S\n_1+\n T\n_1) \Bigl(\la \frac{S}{\n S\n_1} + (1-\la) \frac{T}{\n T\n_1}\Bigr),$$
where $\la = \n S\n_1/(\n S\n_1+\n T\n_1)$,
implies that if $S,T\in \calL_1(H)$ are positive, then so is $S+T$ and
\begin{align*}\Phi(S+T) & = \n S+T\n_1 \Phi\Bigl(\frac{S+T}{\n S+T\n_1}\Bigr)
\\ & = \Phi\Bigl((\n S\n_1+\n T\n_1) \Bigl(\la \frac{S}{\n S\n_1} + (1-\la) \frac{T}{\n T\n_1}\Bigr)\Bigr)
\\ & = (\n S\n_1+\n T\n_1)\Phi\Bigl(\la \frac{S}{\n S\n_1} + (1-\la) \frac{T}{\n T\n_1}\Bigr)
\\ & = (\n S\n_1+\n T\n_1)\Bigl(\la\Phi\Bigl(\frac{S}{\n S\n_1}\Bigr) + (1-\la)\Phi\Bigl( \frac{T}{\n T\n_1}\Bigr)\Bigr)
\\ & = \n S\n_1\Phi\Bigl(\frac{S}{\n S\n_1}\Bigr) + \n T\n_1\Phi\Bigl( \frac{T}{\n T\n_1}\Bigr)
\\ & = \Phi(S)+\Phi(T),
\end{align*}
where we used the assumption that $\Phi$ preserves convex combinations of states.
Applying this to $aS$ and $bT$ with $a,b\ge 0$ we find that $aS+bT$ is positive and
$$\Phi(aS+bT) = \Phi(aS)+\Phi(bT) = a\Phi(S)+b\Phi(T).$$
Next, for an arbitrary self-adjoint $T\in \calL_1(H)$ write $T = T_1-T_2$ with $T_1,T_2$ positive operators in $\calL_1(T)$. Such decompositions always exist; one could take for instance $T_1 = \frac12(T+|T|)$ and $T_2 = T-T_1$.
We then set
$$ \Phi(T):= \Phi(T_1) - \Phi(T_2).$$
To see that this is well defined, suppose that we also have $T = T_1'-T_2'$ with $T_1',T_2'$ positive operators in $\calL_1(T)$. Then $T_1+T_2' = T_2+T_1'$ and hence, by what we just proved,
$$ (\Phi(T_1) - \Phi(T_2)) - (\Phi(T_1') - \Phi(T_2'))
= \Phi(T_1+T_2') - \Phi(T_2+T_1') = 0.
$$
Finally, for an arbitrary $T\in \calL_1(T)$ we set
$$ \Phi(T) := \phi(A)+i\Phi(B),$$
where $A:= \frac12(T+T^\star)$ and $B:=\frac12i(T^\star - T)$ are the unique self-adjoint operators such that $T = A+iB$. The easy proof that $\Phi$ is linear and bounded is left to the reader.
\smallskip
{\em Step 2} -- We now turn to the proof of the theorem. Using the extension provided by Step 1,
for every fixed $F\in \calF$ the mapping $T\mapsto (\Phi(T))(F) $ defines a bounded linear functional
on $\calL_1(H)$ and therefore by Theorem \ref{thm:traceduality} it defines a bounded operator $Q_F\in \calL(H)$ such that
$$ (\Phi(T))(F) = \tr(Q_F^\star T), \quad T\in \calL_1(H).$$
In particular, for all $h\in H$ we have
$$ \iprod{Q_F h}{h} = \tr(Q_F\circ (h\,\bar\otimes\, h)) = (\Phi(h\,\bar\otimes\, h))(F) \in [0,1],$$
which gives the operator inequality $0\le Q_F\le I$, that is, we have $Q_F \in \calE(H)$. In particular we see that $Q_F$ is self-adjoint and therefore $\tr(Q_F^\star T) = \tr(Q_F T)$.
It is clear that $Q_{\Om} = I$, and for every $h\in H$ the measure
$F\mapsto \iprod{Q_F h}{h} = \Phi(h\,\bar\otimes\, h) \in M(\Om)$ is a probability measure. This proves that
$Q: F\mapsto Q_F$ is a POVM.
Uniqueness is clear from the fact that $\tr(Q_F T) = 0$ for all $T\in \calL_1(H)$
implies $Q_F = 0$.
\end{proof}
\begin{remark}\label{rem:convex-reasonable}
The assumption that $\Phi$ should preserve convex combinations is reasonable one in the light of the following argument. Suppose we have two quantum mechanical systems at our disposal, represented by the operators $T_1$ and $T_2$ in $\mathscr(H)$ describing their states. We use a classical coin to decide which state is going to be observed: if, with probability $p$, `heads' comes up we observe the system corresponding to $T_1$; otherwise we measure the system corresponding to $T_2$. This experiment can be described as observing the state corresponding to the convex combination $pT_1 + (1-p)T_2$. If $\Phi$ is the observable to be measure, we expect the probability distribution of the outcomes,
$\Phi(pT_1 + (1-p)T_2)$,
to be given by
$p\Phi(T_1)+(1-p)\Phi(T_2)$.
\end{remark}
POVM's admit a bounded functional calculus, but an important difference with the bounded functional calculus for projection-valued measures of Theorem \ref{thm:Borel-FC} is that the calculus for POVM's fails to be multiplicative. A partial result in the positive direction is given at the end of Section \ref{subsec:Naimark}.
\begin{proposition}[Bounded functional calculus for POVM's]\label{prop:POVM-calculus} \ Let \ $Q: \calF\to \calL(H)$ be a POVM.
There exists a unique linear mapping $\Psi_Q:B_{\rm b}(\Om)\to \calL(H)$
satisfying
$$ \Psi_Q(\one_F) = Q_F, \quad F\in \calF,$$
and
$$ \n \Psi_Q(f)\n \le \n f\n_\infty, \quad f\in B_{\rm b}(\Om).$$
It satisfies $$\Psi_Q(f)^\star = \Psi_Q(\ov f), \quad f\in B_{\rm b}(\Om).$$
\end{proposition}
\begin{proof}
For $x,y\in H$ consider the complex measure $Q_{x,y}$ defined by
$$ Q_{x,y} (F):= \iprod{Q_F x}{y}, \quad F\in \calF.$$
That this indeed defines a measure follows by a polarisation argument from the countable additivity of the measures $Q_h$, $h\in H$.
For any measurable partition $\Om = F_1\cup\cdots\cup F_k$ we have, by the Cauchy--Schwarz inequality applied twice,
\begin{align*} \sum_{j=1}^k |Q_{x,y}(F_k)| = \sum_{j=1}^k |\iprod{Q_{F_j} x}{y}|
& \le \sum_{j=1}^k \iprod{Q_{F_j} x}{x}^{1/2}\iprod{Q_{F_j} y}{y}^{1/2}
\\ & = \sum_{j=1}^k Q_x(F_j)^{1/2}Q_y(F_j)^{1/2}
\\ & \le \Bigl(\sum_{j=1}^k Q_x(F_j)\Bigr)^{1/2} \Bigl(\sum_{j=1}^k Q_y(F_j)\Bigr)^{1/2}
\\ & = Q_x(\Om)^{1/2}Q_y(\Om)^{1/2}
\\ & = \n x\n\n y\n,
\end{align*}
from which it follows that $$|Q_{x,y}|(\Om) \le \n x\n\n y\n.$$
For $f\in B_{\rm b}(\Om)$ define
$$\aa_f(x,y):= \int_\Om f\ud Q_{x,y}, \quad x,y\in H.$$
The form $\aa$ is sesquilinear
and bounded and defines a bounded operator $\Psi_Q(f)$ on $H$ by Proposition \ref{prop:bilin-T}.
It is clear that $\Psi_Q(\one_F) = Q_F$ for all $F\in \calF$ and
$$ |\iprod{\Psi_Q(f)x}{y}| = \Big|\int_\Om f\ud Q_{x,y}\Big| \le \int_\Om |f|\ud |Q_{x,y}| \le \n f\n_\infty\n x\n\n y\n.$$
The identity $(\Psi_Q(f))^\star = \Psi_Q(\ov f)$ is a consequence of $\ov{Q_{y,x}} = Q_{x,y}$,
from which it follows that
\begin{align*}
\iprod{(\Psi_Q(f))^\star x}{y} & =\iprod{x}{\Psi_Q(f) y} = \ov{\iprod{\Psi_Q(f) y}{x}}
= \ov{\aa_f (y,x)} \\ & = \ov{\int_\Om f\ud Q_{y,x}} = \int_\Om \ov f \ud Q_{x,y} = \aa_{\ov f}(x,y) =\iprod{\Psi_Q(\ov f)x}{y}.
\end{align*}
Uniqueness is clear from the fact that $\Psi_Q(\one_F) = Q_F$ and the simple functions are dense in $B_{\rm b}(\Om)$.
\end{proof}
\subsection{Naimark's theorem}\label{subsec:Naimark}
The following result gives a method of producing POVMs from projection-valued measures:
\begin{proposition}[Compression]\label{prop:POVM-compression}
Let $H$ be a closed subspace of a larger Hilbert space $\wt H$ and let $J:H\to \wt H$ be the inclusion mapping. If
$P: \calF\to \calE(\wt H)$ is a projection-valued measure, then
$Q:= J^\star P J: \calF\to \calE(H)$ is a POVM.
\end{proposition}
\begin{proof}
For all $x\in H$ and $F\in \calF$ we have $\iprod{Q_F x}{x} = \iprod{P_FJx}{Jx}$ and
$$0\le \iprod{P_FJx}{Jx} = \n P_F Jx\n^2 \le \n x\n^2 = \iprod{x}{x}$$ and therefore $0\le Q_F \le I$. Furthermore, from $Q_x(F) = \iprod{Q_F x}{x} = \iprod{P_FJx}{Jx} = P_{Jx}(F)$ it follows that
$Q_x$ is a finite measure on $(\Om,\calF)$.
\end{proof}
The main result of this section is {\em Naimark's theorem} which asserts that, conversely, every POVM arises in this way, that is, every POVM dilates to a projection-valued measure.
\begin{theorem}[Naimark]\label{thm:Naimark}\index{theorem!Naimark} Let $(\Om,\calF)$ be a measurable space and let
$Q: \calF\to \calL(H)$ be a POVM. There exists a Hilbert space $\wt H$,
a projection-valued measure $P: \calF\to \calE(\wt H)$, and an isometry $J:H\to \wt H$ such that
$$ Q_F = J^\star P_F J, \quad F\in \calF.$$
\end{theorem}
\begin{proof}
Let
$$S:= \calF \times H = \{(F,x): \, F\in \calF, \,x\in H\}$$
and consider the function $\wt Q:S\times S \to \C$ by
$$ \wt Q(p,p') := \iprod{Q_{F\cap F'}x}{x'} \ \ \hbox{for} \ \ p=(F,x), \ p'=(F',x').$$
We claim that this function is {\em positive definite}\index{positive!definite} in the sense
that for all finite choices of
$p_1,\dots,p_N\in S$ and $z_1,\dots,z_N\in \C$ we have
\begin{align}\label{eq:posdef-tildeQ} \sum_{n,m=1}^N \wt Q(p_n,p_m)z_n\ov z_m \ge 0.
\end{align}
First assume that $p_n = (F_n,x_n)$ with the sets $F_n$ disjoint.
In that case,
\begin{align*}
\sum_{n,m=1}^N \wt Q(p_n,p_m)z_n\ov z_m & \sum_{n,m=1}^N \iprod{Q_{F_n\cap F_m}x_n}{x_m}z_n\ov z_m
= \sum_{n=1}^N \iprod{ Q_{F_n}z_nx_n}{z_nx_n} \ge 0
\end{align*}
by the positivity of the operators $Q_{F_n}$.
For general $F_1,\dots,F_N \in \calF$ we write their union $\bigcup_{n=1}^N F_n$ as a union of $2^N$ disjoint sets $C_\sigma$ in $\calF$, indexed by the elements $\sigma\in 2^N$, the power set of $\{1,\dots,N\}$, as follows.
For $\sigma\in 2^N$ we set
$$ C_\sigma := \bigcap_{n\in\sigma}F_n\setminus\bigcup_{m\not\in \sigma}F_m.$$
It is straightforward to check that the sets $C_\sigma$ are pairwise disjoint and
that for all $n=1,\dots,N$ we have
$$ F_n = \bigcup_{\substack{\sigma\in 2^N \\ n\in \sigma}}C_\sigma, \quad F_n\cap F_m =\bigcup_{\substack{\sigma\in 2^N \\ \{n,m\}\subseteq \sigma}}C_\sigma .$$
Then, by the additivity of $Q$ and the positivity of the operators $Q_{C_\sigma}$,
\begin{align*}
\sum_{n,m=1}^N \wt Q(p_n,p_m)z_n\ov z_m
& = \sum_{n,m=1}^N (Q_{F_n\cap F_m}x_n | x_m)z_n\ov z_m
\\ & = \sum_{n,m=1}^N \Bigl( \sum_{\substack{\sigma\in 2^N \\ \{n,m\}\subseteq \sigma}}Q_{C_\sigma}x_n\Big| x_m\Bigr) z_n\ov z_m
\\ & = \sum_{\sigma\in 2^N}\sum_{\substack{1\le n,m\le N \\ \{n,m\}\subseteq \sigma}}\iprod{Q_{C_\sigma}x_n}{ x_m} z_n\ov z_m
\\ & = \sum_{\sigma\in 2^N}\Bigl(Q_{C_\sigma}\sum_{\substack{1\le n\le N \\ n\in \sigma}}z_nx_n\Big|\sum_{\substack{1\le m\le N \\ m\in \sigma}} z_mx_m\Bigr) \ge 0.
\end{align*}
This completes the proof of \eqref{eq:posdef-tildeQ}.
Let $V$ be the vector space of finitely supported complex-valued functions defined on $S$. The elements
of $V$ are functions $h:S\to \C$ such that $f(p) = 0$ for all but at most finitely many
pairs $p=(F,x)\in S$. The function $v\in V$ that maps $p\in S$ to the complex number $z$ and is identically zero otherwise
will be denoted as $v = z \one_{p}.$ For two functions $v,v'\in V$, say
$v = \sum_{n=1}^N z_n \one{p_n}$ and $v' = \sum_{n=1}^N z_n' \one{p_n}$ (allowing some of the $z_n$ and $z_n'$ to be zero) we define
\begin{align}\label{eq:pos-def-sesqS}\iprod{v}{v'}:= \sum_{n,m=1}^N\wt Q(p_n,p_m)z_n\ov{z_m'}.
\end{align}
Arguing as in the proof of Theorem \ref{thm:dilation-G}, this uniquely defines a sesquilinear mapping from $V\times V$ to $\C$ which satisfies $\iprod{v}{v'} = \ov{\iprod{v'}{v}}$
for all $v,v'\in V$ and $\iprod{v}{v}\ge 0$ for all $v\in V$, and
$$ N = \{v\in V:\, \iprod{v}{v}= 0\}$$
is a subspace of $V$. It follows that \eqref{eq:pos-def-sesqS} induces an inner product on the vector space quotient $V/N$.
Let $\wt H$ denote the Hilbert space completion $\wt H$ of $V/N$ with respect to this inner product.
Consider elements in $\wt H$ of the form $p+N=(\Om,x)+N$ and $p'+N=(\Om,x')+N$ with $x,x'\in H$. Then
$$ \iprod{p+N}{p'+N}_{\wt H} = \iprod{Q_\Om x}{x'} = \iprod{x}{x'}.$$
Taking $x'=x$. in particular we may identify $x\in H$ isometrically with the element $p+N$ in $\wt H$, where $p=(\Om,x)$.
In this way we obtain an isometric embedding $J$ of $H$ into $\wt H$.
To ease up notation we use the notation $p=(F,X)$ for general elements of $\wt H$, rather than the more precise notation $p+N=(F,X)+N$. With this notation, $Jx = (\Om,x)$.
The mapping $\pi:\wt H\to \wt H$ defined by
$$\pi(F,x):= (\Omega, Q_F x)$$
satisfies $\pi^2(F,x)= \pi(\Omega, Q_F x)= (\Omega, Q_\Omega Q_F x) = (\Omega, Q_F x) = \pi(F,x).$
We extend $\pi$ by linearity and check that this results in a self-adjoint, hence orthogonal, projection
in $\wt H$ whose range equals $H$. From
$$\iprod{\pi(F,x)}{x'}_{\wt H} =\iprod{Q_F x} {x'}_{H} = \iprod{(F,x)} {(\Om,x')}_{\wt H} =\iprod{(F,x)} {Jx'}_{\wt H} $$
it follows that $\pi$, viewed as a mapping from $\wt H$ to $H$, equals $ J^\star$.
Finally set $$ P_F (F',x): = (F\cap F', x).$$ Again it is routine to check that $P_F $ is an orthogonal projection
in $\wt H$. By the properties \eqref{it:POVM1} and \eqref{it:POVM2} in Definition \ref{def:POVM} the mapping $P: \calF\to \calL(\wt H)$ is a projection valued measure. Finally, from
$$ J^\star P_F Jx = \pi P_F (\Om,x) = \pi (\Om\cap F, x) = \pi(F,x) = Q_F x$$
we conclude that $Q_F = J^\star P_F J$.
\end{proof}
\begin{theorem}\label{thm:contr-POVM}\index{contraction!represented by a POVM} For every contraction $T\in\calL(H)$, there exists a unique POVM $Q:\calB(\T)\to\calE(H)$ on the unit circle
such that $$ T^n = \int_{\mathbb{T}} z^n \ud Q(z), \quad n\in\N.$$
If $Q$ is a POVM with the above property, then $T$ is unitary if and only if $Q$ is a projection-valued measure.
\end{theorem}
\begin{proof}
Considering the projection-valued measure $P$ of a unitary dilation $U$
of $T$ as in Theorem \ref{thm:Nagy} and compressing to $H$ as in Proposition \ref{prop:POVM-compression},
we obtain the existence of a POVM $Q = J^\star P J$ on the unit circle
such that $$ T^n = J^\star U^n J = J^\star\Bigl( \int_{\mathbb{T}} z^n \ud P(z)\Bigr) J= \int_{\mathbb{T}} z^n\ud Q(z), \quad n\in\N.$$
To prove uniqueness, suppose that for all $n\in\N$ and $x\in H$ we have
$$ \iprod{T^n x}{x} = \int_{\mathbb{T}} z^n \ud Q_x^{(1)}(z) = \int_{\mathbb{T}} z^n \ud Q_x^{(2)}(z).$$
This means that the nonnegative Fourier coefficients of the probability measures
$Q_x^{(1)}$ and $Q_x^{(2)}$ agree. Now Theorem \ref{thm:uniq-FT-T} (and the simple observation following it) can be applied to see that $Q_x^{(1)} = Q_x^{(2)}$.
\end{proof}
If $Q$ is as in Theorem \ref{thm:contr-POVM}, then for all trigonometric polynomials $f\in C(\mathbb{T})$ of the form
$f(z) =\sum_{n=0}^N c_n z^n$ we have
$$\Psi_Q(f) = \int_{\mathbb{T}} f \ud Q = f(T),$$ where $ T = \int_{\mathbb{T}} z \ud Q(z)$.
By the continuity of the bounded functional calculus with respect to the supremum norm, this identity
persists for functions $f$ in the disc algebra $A(\mathbb{D})$, the Banach space of all functions $f\in C(\mathbb{T})$
which have continuous extension to $\ov{\mathbb{D}}$ which is holomorphic on ${\mathbb{D}}$; these are precisely the functions belonging to the closure in $f\in C(\mathbb{T})$ of the trigonometric polynomials of the above form.
An easy consequence is that the bounded functional calculus
of a POVM $Q$ is multiplicative on the disc algebra, that is, for all $f,g\in A(\mathbb{D})$ we have $$\Psi_Q(f) \Psi_Q(g)= \Psi_Q(fg).$$
\begin{remark}
The POVM $Q$ obtained from Theorem \ref{thm:contr-POVM} has the property that
$$ T^n = \int_{\mathbb{T}} z^n\ud Q(z), \quad n\in\N.$$
This property, in turn, can be used to obtain a dilation of $Q$ to a projection-valued measure $P$ as follows.
By Theorem \ref{thm:Nagy} there exist a Hilbert space $\wt H$, a unitary operator $U\in \calL(\wt H)$, and an isometry $J:H\to\wt H$ such that
$$ T^n = J^\star U^n J, \quad n\in\N.$$
Using the spectral theorem for bounded normal operators, let $P:\mathscr{B}(\mathbb{T})\to \mathscr{P}(\wt H)$ be its associated projection-valued measure, that is, $U = \int_{\mathbb{T}}\la\ud P(\la)$.
We claim that $P$ has the desired properties. To see this, first note that by multiplicativity of the bounded measurable calculus of $U$ we have
$$U^n = \int_{\mathbb{T}}\la^n\ud P(\la), \quad n\in\Z.$$
Combining things,
for all $x\in H$ we obtain
\begin{align*} \int_{\mathbb{T}}\la^n\ud Q_x(\la) = \iprod{T^nx}{x} = \iprod{U^nJx}{Jx} = \int_{\mathbb{T}}\la^n\ud P_{Jx}(\la), \quad n\in\N.
\end{align*}
This means that the nonnegative Fourier coefficients of the probability measures
$Q_x$ and $P_{Jx}$ agree. As in the proof of Theorem \ref{thm:contr-POVM} this implies that $Q_x = P_{Jx}$. But this implies, for all Borel subsets $F$,
$$ \iprod{Q_F x}{x} = Q_x(F) = P_{Jx}(F) = \iprod{P_FJx}{Jx} = \iprod{J^\star P_F Jx}{x}.$$
This being true for all $x\in H$, we conclude that $Q_F = J^\star P_F J$.
\end{remark}
\subsection{The phase/number pair}\label{subsec:number-phase}
A convenient model for the {\em number operator}, the self-adjoint operator in quantum optics which corresponds to the observable of counting the number of photons, can be given
on the Hilbert space $\ell^2(\N)$ whose elements consist of the square summable complex sequences $c=(c_n)_{n\in\N}$.
On $\ell^2(\N)$ we consider the unbounded self-adjoint operator $\wh n$ on $\ell^2(\N)$ given by
$$ \wh ne_n = ne_n, \quad n\in\N,$$
where $e_n = (0,\dots,0,1,0\dots)$ is the $n$-th standard unit vector, with maximal domain
$$\Dom(\wh n) = \Bigl\{c \in \ell^2(\N):\, \sum_{n\in\N} n^2|c_n|^2 <\infty\Bigr\}.$$
The sequence $(e_n)_{n\in \N}$ is an orthonormal basis of eigenvectors for $\wh n$ satisfying $\wh n e_n = ne_n$ for all $n\in\N$. In particular, every $n\in\N$ is an eigenvalue of $\wh n$.
On the other hand if $\la\in\C\setminus\N$, then for every $c = (c_n)_{n\in \N}\in \ell^2(\N)$ equation $(\la - \wh n )u = c$
is uniquely solved by $u = (u_n)_{n\in \N}\in \ell^2(\N)$ with coefficients $u_n = c_n/(\la-n)$, $n\in \N$. This implies that
$\la\in \varrho(\wh n)$. We conclude that
$$\sigma(\wh n) = \N.$$
We think of $e_n$ as the pure state describing an $n$-photon state of the electromagnetic field.
Its associated projection-valued measure $N$ is given by $N(\{n\}) = \pi_n$, the orthogonal projection in $\ell^2(\N)$
onto the one-dimensional span of $e_n$, so that
$$ \iprod{\wh n x}{x} = \int_{\N} n \ud N_x(n) = \sum_{n\in \N} n \iprod{N(\{n\})x}{x}, \quad x\in \Dom(\wh n). $$
To define {\em phase}\index{phase} as a $\T$-valued observable in the sense of POVM's we proceed as follows.
Let $V$ be the left shift on $\ell^2(\N)$,
$$ Ve_0 := 0, \quad Ve_n := e_{n-1} \ \ (n\ge 1).$$
By Theorem \ref{thm:contr-POVM} there exists a unique POVM $\Phi:\calB(\T)\to \calE(\ell^2(\N))$
such that
$$ V^n = \int_{\mathbb T} z^n\ud \Phi(z), \quad n\in\N.$$
As will be argued in Section \ref{sec:symmetries}, the covariance property expressed in the following theorem identifies the POVM $\Phi$
as the complementary observable to the number observable $\wh n$.
The notions of covariance and complementarity will be developed more systematically in Section \ref{sec:symmetries}.
\begin{theorem}[Phase and number]\label{thm:quantum-phase}
Let $(U(t))_{t\ge 0}$ be the unitary $C_0$-group generated by $i\wh n$.
The phase observable $\Phi$ is {\em covariant} with respect to this group, that is,
for all Borel subsets $B\subseteq\mathbb{T}$ we have
\begin{align*}
U(t)\Phi_B U^\star(t) = \Phi_{e^{it}B}, \quad t\in \R,
\end{align*}
where $e^{it}B = \{e^{it}z: \, z\in B\}$ is the rotation of $B$ over $t$.
\end{theorem}
\begin{proof} The properties of the projection-valued measure $P$ obtained from Nai\-mark's theorem (Theorem \ref{thm:Naimark}) imply that for the trigonometric functions $e_k$, $k\in\N$, we have
\begin{align*}
(\Phi_B U^\star(t)e_n|e_m) = (P_B J U^\star(t)e_n|Je_m)= e^{-int}({\bf 1}_B e_n|e_m)
\end{align*}
while at the same time
\begin{align*}
(U^\star(t) \Phi_{e^{it}B}e_n|e_m) &= (P_{e^{it}B}Je_n|JU(t)e_m) = e^{-itm}({\bf 1}_{e^{it}B}e_n|e_m)
\\ & = e^{-itm}\int_{e^{it}B}z^{n-m}\ud z = e^{-itm}\int_{B}(e^{-it}z)^{n-m}\ud z
\\ & = e^{-itn}\int_{B}z^{n-m}\ud z = e^{-int}({\bf 1}_B e_n|e_m).
\end{align*}
Since the span of the functions $e_n$, $n\in\N$, is dense in $\ell^2(\N)$, this completes the proof.
\end{proof}
\section{Hidden variables}\label{sec:hidden}
The one-to-one correspondence of Theorem \ref{thm:POVM-unsharp} between the set of POVMs and the set of affine mappings from $\mathscr{S}(H)$
to $M(\Om)$ is particularly satisfying from a philosophical point of view, as it characterises unsharp observables in an operational way: an unsharp observable is nothing but a rule of assigning probability distributions to states in such a way that convex combinations are respected. The rationale of this assumption has been discussed in Remark \ref{rem:convex-reasonable}.
Thinking of quantum mechanical unsharp observables as convexity preserving mappings from $\mathscr{S}(H)$ to $M(\Om)$, analogously we can define classical unsharp observables as convexity-preserving mappings from $M(X)$ to $M(\Om)$, where $X$ is the state space of the classical system. Indeed, in Section \ref{sec:states-observables} we have defined
observables as measurable functions from $X$ to $\Om$, and such functions $f$ induce a mapping from the set of probability measures $X$ to $M(\Om)$
by $$(f(\mu))(B):= \mu(x\in X:\,f(x)\in B).$$ This mapping preserves convex combinations. Thus every classical observable is a classical unsharp observable.
The following interesting result shows that every family of quantum observables with values in a locally compact Hausdorff spaces admits a classical model.
\begin{theorem}[Hidden variables]\label{thm:hidden}\index{hidden variables} Let $H$ be a separable Hilbert space and $\Om$ be a locally compact Hausdorff space.
Suppose $\Phi=\{\Phi_i:\, i\in I\}$ is a family of unsharp quantum mechanical observables, each of which maps $\mathscr{S}(H)$ to $M(\Om)$ in an affine way, that is, preserving convex combinations. Then there exists a compact Hausdorff space $X$
and a family $f=\{f_i:\, i\in I\}$ of unsharp classical observables, each of which maps $M(X)$ to $M(\Om)$ preserving convex combinations, such that the following conditions hold:
\begin{enumerate}[\rm(1)]
\item The elements of $X$ are the equivalence classes of the extreme points of $\mathscr{S}(H)$ modulo indiscernibility under $\Phi$; here, two extreme points $T_1,T_2$ of $\mathscr{S}(H)$ are said to be {\em indiscernible}\index{indiscernible} under $\Phi$
if $$ \Phi_i(T_1) = \Phi_i(T_2), \quad i\in I;$$
\item the quotient mapping $q:T\mapsto [T]$ from the set of extreme points of $\mathscr{S}(H)$ to $X$ is continuous;
\item The elements of $f$ are related to those of $\Phi$ by passing to equivalence classes; more precisely, if $[T]$ denotes the equivalence class of $T$ modulo indiscernibility under $\Phi$, then
$$ f_i(\delta_{\{[T]\}}) = \Phi_i(T),$$
where $\delta_{\{[T]\}}\in M(X)$ is the Dirac measure supported on the singleton $\{[T]\}$.
\end{enumerate}
\end{theorem}
The classical observables in $f$ can be thought of as {\em hidden variables}: by their very definition they cannot be observed using the measurements of the quantum observables in $\Phi$.
\begin{proof}[Sketch of the proof]
In what follows we think of the set ${\rm Extr}(\mathscr{S}(H))$ of extreme points of $\mathscr{S}(H)$ as being equipped with the
coarsest topology $\tau$ such that all mappings $T\mapsto \lb g, \Phi_i(T)\rb$ with $i\in I$ and $g\in C_0(\Om)$ are continuous; the brackets refer between the duality of $C_0(\Om)$ and $M_{\rm R}(\Om)$
(see Theorem \ref{thm:CK-dual}). Adapting the proof of the Banach--Alaoglu in a straightforward way, it is seen that $\tau$ turns ${\rm Extr}(\mathscr{S}(H))$ into a compact Hausdorff space.
The definitions of $X$ has already been given in the statement of the theorem.
We declare a subset of $X$ to be open if its pre-image under the mapping $T\mapsto [T]$ belongs to $\tau$.
This defines a topology $\upsilon$ on $X$ which renders the quotient mapping from ${\rm Extr}(\mathscr{S}(H))$ to $X$ continuous. As a result, the space $X$ is compact with respect to $\upsilon$.
We extend the mapping $\delta_{\{[T]\}}\mapsto \Phi_i(T)$ to the convex hull in $M(X)$ of the Dirac measures by convexity:
$$ f_i\Bigl(\sum_{n=1}^N \la_n \delta_{\{[T]\}}\Bigr) := \sum_{n=1}^N \la_n\Phi_i(T)$$
for scalars $0\le \la_n\le 1$ such that $\sum_{n=1}^N \la_n = 1$. Clearly, $f_i$ preserves convex combinations. The functions $f_i$ are continuous
with respect to the weak$^*$ topologies of $M(X)$ and $M(\Om)$. The span of the Dirac measures is dense in
$M(X)$ with respect to the weak$^*$ topology. Leaving aside some topological subtleties, these functions admit unique extensions to all of $M(X)$, and these extensions again preserve convex combinations.
\end{proof}
It is of some interest to work through the details of this construction for the qubit. Accordingly let $H = \C^2$.
As we have seen in Section \ref{subsec:qubit}, the set of extreme points of $\mathscr{S}(H)$ then corresponds to the Bloch sphere $S^2$ in $\R^3$. Under this correspondence the Bloch vector
$\xi = (\sin\theta\cos\phi, \,\sin\theta\sin\phi, \,\cos\theta) \in S^2$ corresponds to the operator $T_\xi\in \mathscr{S}(\C^2)$ given in matrix form as
\begin{align*} T_\xi = \begin{pmatrix}
1+ \cos \theta & \, e^{-i\phi }\sin\theta \\ \, e^{i\phi }\sin \theta & \, 1-\cos\theta
\end{pmatrix}.
\end{align*}
Let us take the Pauli matrices as the set of observables of interest:
$$\Phi:= \{\sigma_1, \sigma_2,\sigma_3\}.$$ Let
$Q_j$ be the projection-valued measure associated with $\sigma_j$. For example,
$(Q_1)_{\{1\}}$ and $(Q_1)_{\{-1\}}$ are the orthogonal projections onto the one-dimensional subspaces
spanned by the eigenvectors corresponding to the eigenvalues $1$ and $-1$ of $\sigma_1 = \begin{pmatrix}\, 0 & \, 1\, \\ 1 & 0 \end{pmatrix}$:
$$ (Q_1)_{\{1\}}= \begin{pmatrix}\, \frac12 & \, \frac12\, \\ \frac12 & \frac12 \end{pmatrix},
\quad (Q_1)_{\{-1\}}=
\begin{pmatrix} \phantom{-}\frac12 &\, -\frac12\, \\ -\frac12\, & \phantom{-}\frac12\, \end{pmatrix}.$$
We view these as observables with values in $\Om:= \{\pm1\}$. Since $\Phi$ separates the points of $S^2$, the space $X$ constructed in the proof of Theorem \ref{thm:hidden}
can be identified with $S^2$. The corresponding family classical variables $f = \{f_1,f_2,f_3\}$
is given by the mappings $f_j: M(S^2)\to M(\{\pm 1\})$,
\begin{align*} (f_1(\delta_\xi))(\{1\})
& = \tr((Q_1)_{\{1\}}T_\xi)
\\ & = \tr\left(\begin{pmatrix} \frac12 & \, \frac12\, \\ \frac12 & \frac12 \end{pmatrix}\begin{pmatrix}
1+ \cos \theta & \, e^{-i\phi }\sin\theta \\ \, e^{i\phi }\sin \theta & \, 1-\cos\theta
\end{pmatrix}\right)
\\ & =\frac12 \tr
\begin{pmatrix}
1 + \cos\theta+e^{i\phi }\sin \theta & \ 1-\cos\theta+e^{-i\phi }\sin \theta \\
1 + \cos\theta+e^{i\phi }\sin \theta & \ 1-\cos\theta+e^{-i\phi }\sin \theta
\end{pmatrix}
\\ & = \frac12(1 + \cos\theta+e^{i\phi }\sin \theta + 1-\cos\theta+e^{-i\phi }\sin \theta)
\\ & = 1 + \cos\phi\sin\theta
\\ & = 1 + \xi_1
\end{align*}
and likewise
$$(f_1(\delta_\xi))(\{-1\}) = 1 - \xi_1.
$$
Similar computations give
$$ (f_2(\delta_\xi))(\{\pm 1\}) = 1 \pm \xi_2, \quad (f_3(\delta_\xi))(\{\pm 1\}) = 1 \pm \xi_3. $$
By considering convex combinations of Dirac measures and a limiting argument, it follows that the hidden variables $f_j: M(S^2)\to M(\{\pm 1\})$ are given by
$$ \ud f_j(\mu) = \frac12\Bigl(1+p\int_{S^2}x_j\ud \mu(x)\Bigr)\ud p,\quad j = 1,2,3.$$
where $\ud p$ is the probability measure on $\{\pm 1\}$ giving each point mass $\frac12$.
\section{Symmetries}\label{sec:symmetries}
In order to motivate our definition of a symmetry we need to introduce some notation and terminology.
A real-linear mapping $T:H\to H$ is said to be {\em antilinear}\index{antilinear}\index{operator!antilinear}
if $T(cx) = \ov c Tx$ for all $c\in \C$ and $x\in H$. The {\em adjoint} of an antilinear mapping $T:H\to H$
is the unique antilinear mapping $T^\star:H\to H$ defined by
$$ \iprod{x}{T^\star y} = \iprod{Tx}{y}, \quad x,y\in H.$$
A mapping $T:H\to H$ is called {\em antiunitary}\index{antiunitary}\index{operator!antiunitary} if it is antilinear and satisfies $TT^\star = T^\star T = I.$
It is straightforward to check that if $U:H\to H$ is unitary or antiunitary, then the mapping
$$ \mathscr{U}(T):= U TU^\star$$
is well defined as a mapping from $\mathscr{S}(H)$ to $\mathscr{S}(H)$ and satisfies
\begin{align}\label{eq:U-trans-prob}\tr(\mathscr{U}(T_1)\mathscr{U}(T_2)) = \tr(T_1T_2), \quad T_1,T_2\in \mathscr{S}(H). \end{align}
Here, as before, $\mathscr{S}(H)$ denotes the set of all positive trace class operators with unit trace on $H$ corresponding to a pure state. As we have seen, the elements of this set are precisely the rank one projections $ h\,\bar\otimes\,h$ with $h\in H$ of norm one.
The physical interpretation of \eqref{eq:U-trans-prob} is that $\mathscr{U}$ {\em preserves transition probabilities} between pure states.\index{transition probability}
To see this, recall that if $A$ is the self-adjoint operator in $H$ representing a real-valued observable and $h\in H$ has norm one (and belongs to $\Dom(A)$ in case $A$ is unbounded),
then the quantity $$\tr(A\circ (h\,\bar\otimes\,h)) = \iprod{Ah}{h}$$ is interpreted as the expected value of the observable
in the pure state $\ket{h}$. In particular, if $\ket{g}$ is another pure state, then
$$ \tr((g\,\bar\otimes\,g)\circ (h\,\bar\otimes\,h)) = \iprod{(g\,\bar\otimes\,g)h}{h}
= |\iprod{g}{h}|^2$$
is the expected value of the observable $g\,\bar\otimes\,g$ in state $\ket{h}$. In physics parlance, this the probability of ``finding a system with state $|h\rb$ in state $|g\rb$ when measuring it against an orthonormal basis containing $g$'',
that is, the transition probability between $|h\rb$ and $|g\rb$.
A remarkable theorem due to Wigner provides a converse to
\eqref{eq:U-trans-prob}:
\begin{theorem}[Wigner]\label{thm:Wigner}\index{theorem!Wigner} If $\mathscr{U}:\mathscr{S}(H)\to \mathscr{S}(H)$ is a bijection with the property that $$\tr(\mathscr{U}(T_1)\mathscr{U}(T_2)) = \tr(T_1T_2), \quad T_1,T_2\in \mathscr{S}(H),$$
there exists a mapping $U:H\to H$ which is either unitary or antiunitary such that
$$ \mathscr{U}(T) = UTU^\star, \quad T\in \mathscr{S}(H).$$
This mapping is unique up to a complex scalar of modulus one.
\end{theorem}
We sketch a proof of the theorem only for the case of qubits, that is, for $H = \C^2$, and refer to the Notes for some missing details and references to the general case.
\begin{proof}[Proof of Theorem \ref{thm:Wigner} for qubits]
We begin by recalling \eqref{eq:Bloch1} and \eqref{eq:Bloch2}, which say that
if we write unit vectors $h\in \C^2$ as $ \cos(\theta/2)|0\rb + e^{i\phi} \sin(\theta/2)|1\rb$ with $0\leq \theta \leq \pi$ and $0\leq \phi <2\pi$, then the rank one projection $h\,\bar\otimes\,h$ in $\C^2$ onto the span of $h$ is given as a matrix by
$$ h\,\bar\otimes\,h = \begin{pmatrix}
1+ \cos \theta & \, e^{-i\phi }\sin\theta \\ \, e^{i\phi }\sin \theta & \, 1-\cos\theta
\end{pmatrix}.
$$
By elementary computation,
\begin{align*}
\tr((h\,\bar\otimes\,h)(h'\,\bar\otimes\,h')) & = |\iprod{h}{{h'}}|^2
= \frac12(1+ x_h\cdot x_{h'}),
\end{align*}
where, as in \eqref{eq:Bloch3}, $$x_h = (\sin\theta\cos\phi, \,\sin\theta\sin\phi, \,\cos\theta)$$
is the Bloch vector of $h$.
Under the bijective correspondence $h\,\bar\otimes\,h\leftrightarrow x_h$ between the elements of $\mathscr{S}(\C^2)$ and the points of the unit sphere $S^2$ of $\R^3$, the assumption of the theorem implies that
$\mathscr{U}$ induces a mapping $\mathscr{R}:S^2\to S^2$ satisfying
$\frac12(1+ \mathscr{R}x_h\cdot \mathscr{R}x_{h'}) = \frac12(1+ \mathscr{R}x_h\cdot \mathscr{R}x_{h'})$, that is,
$$ \mathscr{R}x_h\cdot \mathscr{R}x_{h'} = x_h\cdot x_{h'}.$$
This identity implies that the $3\times 3$ matrix $R$ defined by
$$ R_{ij} = \mathscr{R}u_i \cdot u_j, \quad i,j\in\{1,2,3\},$$
with $u_1,u_2,u_3$ the standard unit vectors of $\R^3$, is orthogonal. Now we use the algebraic fact, taken for granted here, that for every
orthogonal $3\times 3$ matrix $R$ with real coefficients there exists mapping $U:\C^2\to \C^2$ which is either unitary or antiunitary, and which is unique up to a complex scalar of modulus one, such that
$$ U (x\cdot \sigma) U^\star = (Rx)\cdot \sigma,\quad x\in \R^3,$$
where $x\cdot\sigma:= x_1\sigma_1+x_2\sigma_2+x_3\sigma_3$, where $x\in \R^3$ and $\sigma_1,\sigma_2,\sigma_3$ are the Pauli matrices. The mapping $U$ has the required properties.
\end{proof}
Informally speaking, Wigner's theorem informs us that symmetries $\mathscr{U}$ of quantum mechanical systems are given by operators $U$ acting on the underlying Hilbert space that are either unitary or antiunitary.
In practice on is primarily interested in one-parameter groups of symmetries indexed by time. Suppose $(\mathscr{U}(t))_{t\in \R}$ is such a group. By the uniqueness part of Wigner's theorem, for all $s,t\in \R$ we the identity $\mathscr{U}(t)\mathscr{U}(s) = \mathscr{U}(t+s)$ implies the existence of a scalar $c(t,s)$ of modulus one
such that the corresponding (anti)unitary operators satisfy $$U(t)U(s) = c(t,s)U(t+s).$$
From the associativity law $U(t)(U(s)U(r)) = (U(t)U(s))U(r)$
we obtain the {\em cocycle identity}\index{cocycle identity}
$$c(s,r) c(t,s+r) = c(t,s) c(t+s,r).$$
In this situation, a theorem of Bargmann implies that there exists function $d$, taking values in the scalars of modulus one, such that $$ c(t,s) = \frac{d(t)d(s)}{d(t+s)}$$
and the operators $V(t):= d(t)^{-1}U(t)$ are unitary. They satisfy
$$ \mathscr{U}(t)(T) = V(t)^\star TV(t), \quad V(t)V(s) = V(t+s).$$
The unitary group $(V(t))_{t\in \R}$ can be shown to be strongly continuous. Hence, by Stone's theorem,
it follows that there exists a self-adjoint operator $\mathscr{H}$, the {\em Hamiltonian} associated with the family $\mathscr{U}$, such that $V(t) = e^{it\mathscr{H}}$ for $t\in \R$.
The action of the unitary $C_0$-group $(e^{it\mathscr{H}})_{t\in\R}$ on pure states is given by
$ \mathscr{U}(t) (h\,\bar\otimes\, h)
= V(t)h \,\bar\otimes\, V(t)h$.
The equation
$$ \frac{{\rm d}}{{\rm d}t} V(t)h = i\mathscr{H}V(t)$$
is an abstract version of the Schr\"odinger equation\index{Schr\"odinger!equation, abstract}\index{equation!Schr\"odinger, abstract} \eqref{eq:Schrodinger} (which corresponds to the special case $H = L^2(\R^d,m)$ and $\mathscr{H} = -\Delta +V$).
These considerations motivate the following definition.
\begin{definition}[Symmetry, of a Hilbert space]
A {\em symmetry}\index{symmetry} of $H$ is a unitary operator on $H$.
\end{definition}
\subsection{Covariance}
\begin{definition}[Symmetry, of a measure space] A {\em symmetry} of the measure space $(\Om,\calF,\mu)$ is a measurable bijective mapping $g:\Om\to \Om$
with measurable inverse that leaves $\mu$ invariant, that is,
$$ (g(\mu))(F):= \mu(g^{-1}(F)) = \mu(F), \quad F\in \calF.$$
\end{definition}
If $g$ is a symmetry of $(\Om,\calF,\mu)$, then for all $F\in \calF$ the set $g(F)$ is measurable and
$\mu(F) = \mu(g^{-1}(g(F))) = \mu(g(F))$, that is, $\mu$ is also invariant under $g^{-1}$.
\begin{definition}[Conservation and covariance]\index{covariant}\index{conserved} Let $(\Om,\F,\mu)$ be a measure space and let $U$ be a symmetry of a Hilbert space $H$.
\begin{enumerate}[\rm (i)]
\item An observable $P:\calF\to \PP(H)$ is said to be {\em conserved} under $U$ if
$U P_F U^\star = P_F$ for all $F\in \calF.$
\item An observable $P:\calF\to \PP(H)$ is said to be {\em covariant} under the pair $(g,U)$, where $g$ is a symmetry of $(\Om,\calF,\mu)$, if
$U P_F U^\star = P_{g(F)}$ for all $F\in \calF,$
that is, if the following diagram commutes:
\begin{figure}[ht]
\begin{center}
\begin{tikzcd}[scale cd=1.1, sep=huge]
\calF \arrow[r, "F\mapsto g(F)"] \arrow[d, "P"]
& \calF \arrow[d, "P"] \\
\PP(H) \arrow[r, " P\mapsto UPU^\star"]
& \PP(H)
\end{tikzcd}
\end{center}
\end{figure}
\end{enumerate}
\end{definition}
As we will
see shortly, position is covariant with respect to translation (an object at position $x$ appears at position $x-x'$ if the origin is translated over $x'$) and momentum is covariant with respect to boosts (an object with momentum $p$
appears with momentum $p-p'$ if a boost of size $p'$ is applied, that is, if the origin `in momentum space' is translated over $p'$).
\paragraph{Locally compact abelian groups}
An interesting special case arises when observables take values in a locally compact abelian (LCA) group $G$.
Our treatment borrows some results from the theory of LCA groups that will not proved here. For the special cases $\R^d$ and $\T$ the presentation is self-contained, as all missing details can be filled in with the help of the results of Chapter \ref{ch:operators}.
A fuller treatment of symmetries should also cover the case of (noncommutative) Lie groups such as $SO(3)$ and $SU(2)$, but thus would take us too far afield.
Every LCA group $G$ admits a {\em Haar measure}\index{Haar!measure}, that is, a Borel measure $\mu$ such that $\mu(B) = \mu (g^{-1}(B))$ for all $B\in \calB(G)$ and $g\in G$. This measure is unique up to a scalar multiple.
With respect to a Haar measure $\mu$, every $g\in G$ induces a symmetry on $G$ by
$$ g: g'\mapsto gg', \quad g'\in G.$$
This induces a symmetry $U_g$ on $L^2(G) := L^2(G,\mu)$ given by
$$U_g f = f \circ g^{-1}.$$
We refer to $U_g$ as the {\em translation over $g$}.\index{translation}\index{$U_g$}
We have
$$U(g_1)U(g_2)f = (f\circ g_2^{-1})\circ g_1^{-1} = f \circ (g_1 g_2)^{-1} = U(g_1 g_2)f,$$
so $U(g_1)U(g_2) = U(g_1 g_2)$. This means that the mapping $U:G\to\calL(L^2(G))$, $g\mapsto U_g$, is multiplicative and hence defines a unitary representation. It is easily checked that this representation is strongly continuous.
A {\em character}\index{character} of $G$ is a continuous group homomorphism $\gamma:G\to \T$.
The set $\Gamma$ of all characters of $G$ is called the {\em Pontryagin dual of $G$}.\index{Pontryagin dual}\index{dual!Pontryagin} It has the structure of a locally compact abelian group in a natural way by endowing it with the
weak$^*$ topology inherited from $L^\infty(G)$ (local compactness being a consequence of the fact that
$\Gamma\cup\{0\}$ is a weak$^*$ closed subset of the closed unit ball $\ov B_{L^\infty(G)}$ which is weak$^*$ compact by the Banach--Alaoglu theorem).
It follows that $\Gamma$ carries a Haar measure which is again unique up to a normalisation.
Every $g\in G$ defines a character $\gamma\mapsto \gamma(g)$ on $\Gamma$, and the {\em Pontryagin duality theorem} asserts that these are the only ones and the Pontryagin dual of $\Gamma$ equals $G$ both as a set and as an LCA group.
Every character $\gamma\in\Gamma$ induces a
symmetry on $L^2(G)$ via
$$V_\gamma f(g') = \gamma(g')f(g'), \quad g'\in G, \ f\in L^2(G).$$
We refer to $V_\gamma$ as the {\em boost over $\gamma$}.\index{boost}\index{$V_\gamma$}
In the language of Chapter \ref{ch:operators}, $V_\gamma$ is the pointwise multiplier with $\gamma$.
It is immediate from the above definitions that the so-called {\em Weyl commutation relation}\index{Weyl commutation relation}\index{commutation relation!Weyl} holds:
\begin{proposition}[Weyl commutation relation]\label{prop:WeylCR}
For all $ g\in G$ and $\gamma\in \Gamma$
we have $$ V_\gamma U_g = \gamma(g) U_g V_\gamma .$$
\end{proposition}
\begin{proof} For $f\in L^2(G)$ and $g'\in G$ we have
\begin{align*}
V_\gamma U_g f(g') &= \gamma(g')U_g f(g') = \gamma(g') f(g^{-1}g')
\intertext{and}
\gamma(g) U_g V_\gamma f(g') & = \gamma(g) V_\gamma f(g^{-1}g') \\ & = \gamma(g) \gamma(g^{-1}g') f(g^{-1}g')
= \gamma(g') f(g^{-1}g').
\end{align*}
\end{proof}
We now turn to definition of a pair of canonical observables that can be associated with LCA groups.
For the statement of the second part of the theorem we need the {\em Plancherel theorem for LCA groups}, which asserts that the Fourier--Plancherel transform $\calF:L^1(G)\to L^\infty(\Gamma)$ defined by
$$ \F f(\gamma):= \int_G f(g)\ov{\gamma(g)}\ud \mu(g), \quad \gamma\in\Gamma,$$
where $\mu$ is a Haar measure on $G$,
maps $L^1(G)\cap L^2(G)$ into $L^2(\Gamma)$ and there is a unique normalisation of the Haar measure of $\Gamma$ such that $\F$ extends to a unitary operator from $L^2(G)$ onto $L^2(\Gamma)$.
For later reference we
note that $\ov{\gamma(g)} = \gamma(g^{-1})$ and hence
\begin{equation}\label{eq:transl-Fourier}\begin{aligned}
\F U_g f(\gamma) &= \int_G f(g^{-1}g')\ov{\gamma(g')}\ud \mu(g')
\\ & = \int_G f(g')\ov{\gamma(g g')}\ud \mu(g') =\gamma(g^{-1})\calF f(\gamma),
\end{aligned}
\end{equation}
where the second identity follows by substitution and invariance of $\mu$.
\begin{theorem}[Position and momentum]\label{thm:Ugamma} Let $G$ be an LCA group, let $\Gamma$ be its Pontryagin dual, and let $\calB(G)$ and $\calB(\Gamma)$ be their Borel $\sigma$-algebras.
\begin{enumerate}[\rm(1)]
\item\label{it:Ugamma1} There exists a unique $G$-valued observable $X:\calB(G)\to \calP(L^2(G))$ such that for all $\gamma\in \Gamma$ we have
$$ V_\gamma =\int_G \gamma \ud X.$$
It is given by $X(B)f = \one_B f$ for $f\in L^2(G)$ and $B\in \calB(G)$.
\item\label{it:Ugamma2} There exists a unique $\Gamma$-valued observable $\Xi:\calB(\Gamma)\to \calP(L^2(G))$ such that for all
$g\in G$ we have
$$ U_g =\int_\Gamma g \ud \Xi.$$
It is given by $\Xi(B) f = \calF^{-1}\one_B\calF f$ for $f\in L^2(\Gamma)$ and $B\in \calB(G)$.
\end{enumerate}
\end{theorem}
\begin{proof}
\eqref{it:Ugamma1}: \
Consider the $G$-valued observable $X:\calB(G)\to \calL(L^2(G))$ defined by
$$ X_B f:= \one_B f, \quad B\in \calB(G), \ f\in L^2(G).$$
For Borel sets $B\subseteq G$ the operator $T_{\one_B} := \int_G \one_B \ud X$ on $L^2(G)$ is the pointwise multiplier
$T_{\one_B}f = \one_B f$. By linearity, for $\mu$-simple functions $\phi$ on $G$ the operator $T_{\phi} := \int_G \phi \ud X$ on $L^2(G)$ is the pointwise multiplier
$T_{\phi}f = \phi f$. By approximation, the operator $T_\gamma:= \int_G \gamma \ud X$ is the pointwise multiplier $T_{\gamma}f = \gamma f = V_\gamma f$. This proves existence.
We only sketch the proof of uniqueness; for the special cases $G = \R^d$ and $G=\T$ the missing details are easily filled in by using the properties of the Fourier transform proved in Chapter \ref{ch:operators}.
If $\wt X$ is an observable satisfying $ V_\gamma =\int_G \gamma \ud \wt X$,
then for all $f\in L^2(G)$ we have
$\int_G \gamma \ud \wt X_f = \int_G \gamma \ud X_f$. This can be interpreted as saying that the finite Borel measures
$\wt X_f$ and $X_f$ have the same Fourier transforms. Their equality therefore follows from the injectivity
of the Fourier transform as a mapping from $M(G)$ to $L^\infty(\Gamma)$.
\smallskip
\eqref{it:Ugamma2}: \ We begin with the existence part.
Applying the construction of the preceding part to $\Gamma$ we obtain the dual position operator $X^\Gamma:\calB(\Gamma)\to \calL(L^2(\Gamma))$ given by
$$ X_B^\Gamma \phi := \one_B \phi, \quad B\in \calB(\Gamma), \ f\in L^2(\Gamma).$$
By conjugation with the Fourier--Plancherel transform it induces an observable $\Xi: \calB(\Gamma)\to \calL(L^2(G))$:
\begin{align*}
\Xi_B f := \F^{-1}X_B^\Gamma \F f, \quad B\in \calB(\Gamma), \ f\in L^2(G).
\end{align*} This observable has the desired property. Uniqueness is proved in the same way as in part \eqref{it:Ugamma1}.
\end{proof}
\begin{definition}[Position and momentum]\label{def:posmom} The $G$-valued observable $X$ and
the $\Gamma$-valued observable $\Xi$ of the theorem are called the {\em position}\index{position operator} and {\em momentum}\index{momentum operator} observables of $G$, respectively.
\end{definition}
\begin{proposition}\label{prop:LCA-pos-mom} Let $G$ be an LCA group and let $X$ and $\Xi$ be the position and momentum observables of $G$.
\begin{enumerate}[\rm(1)]
\item\label{it:LCA-covar1} $X$ is covariant with respect to every $(g,U_g)$ and conserved under every $V_\gamma$,
$$ U_g X_B U_g^\star f(g')= X_{g B}f(g'), \quad V_\gamma X_B V_\gamma^\star f = \one_B f = X_B f.$$
\item\label{it:LCA-covar2} $\Xi$ is covariant with respect to every $(\gamma,V_\gamma)$ and conserved under every $U_g$,
$$ V_\gamma \Xi_B V_\gamma^\star f(\gamma')= \Xi_{\gamma B}f(\gamma'), \quad U_g \Xi_B U_g^\star f = \one_B f = \Xi_B f.$$
\end{enumerate}
\end{proposition}
\begin{proof}
\eqref{it:LCA-covar1}: \ For all $g\in G$ and $B\in \calB(G)$ we have
\begin{align*}
U_g X_B U_g^\star f(g')
= [\one_BU_{g^{-1}}f](g^{-1}g')
= \one_B(g^{-1}g')f(g') = X_{g B}f(g'),
\end{align*}
proving covariance with respect to $(g,U_g)$. Conservation under $V_\gamma$ is even simpler:
\begin{align*}
V_\gamma X_B V_\gamma^\star f = \gamma(\cdot) \one_B(\cdot) \gamma^{-1}(\cdot)f(\cdot) = \one_B f = X_B f.
\end{align*}
The proof of \eqref{it:LCA-covar2} is entirely similar.
\end{proof}
\begin{remark}[Complementarity]
In the physics literature, the `duality' between the position and momentum observables is referred to as {\em complementarity}.\index{complementarity} As we will see in the next two sections, this captures the complementarity of the position and momentum observables in $\R^d$ as well as that of the angle and angular momentum observables in $\T$.
\end{remark}
\subsection{The case $G = \R^d$: position and momentum}\label{subsec:pos-mom}
We now specialise to $G = \R^d$ with normalised Lebesgue measure ${\rm d}m(x) = (2\pi)^{-d/2}\ud x$ as the Haar measure.
Every character $\gamma:\R^d\to \T$ is of the form
$$\gamma(x) = e_\xi(x) := e^{ ix\cdot\xi}$$ for some $\xi\in \R^d$.
Under the identification $\gamma \leftrightarrow e_\xi$ we have $\Gamma = \R^d$, the Haar measure being again the normalised Lebesgue measure $m$.
To distinguish $G=\R^d$ from its dual $\Gamma = \R^d$ we use roman letters for elements of $G$ and greek letters for elements of its dual $\Gamma$.
For every $y\in \R^d$, the translation $y: x\mapsto x+y$ is a symmetry of $\R^d$. The induced symmetry $U_y$ on $L^2(\R^d,m)$ is right translation over $y$:
$$ U_y f(x) = f(x-y), \quad x\in \R^d, \ f\in L^2(\R^d,m).$$
For every $\xi\in \R^d$ the boost $V_\xi$ on $L^2(\R^d,m)$ is given by
$$ V_\xi f(x) = e^{i x\cdot\xi}f(x), \quad x\in \R^d,\ f\in L^2(\R^d,m).$$
The Weyl commutation relation takes the form
$$ V_\xi U_x = e^{2\pi i x\cdot \xi} U_x V_\xi, \quad x,\xi\in \R^d. $$
Applying this identity to functions in $C_{\rm c}^1(\R^d)$ and
differentiating this relation with respect to $x_j$ and $\xi_k$, we obtain the
{\em Heisenberg commutation relation}\index{Heisenberg!commutation relation}\index{commutation relation!Heisenberg}
\begin{align}\label{eq:Heisenberg}
\wh x_j\wh \xi_k - \wh\xi_k\wh x_j = i \delta_{jk} I,
\end{align}
where
\begin{equation}\label{eq:defxjxik}
\begin{aligned}
\wh x_jf(x) & := x_j f(x), \\
\wh \xi_k f(x) & := \frac1{ i} \frac{\partial f}{\partial x_k}(x).
\end{aligned}
\end{equation}
The rigorous interpretation of \eqref{eq:Heisenberg} is that for all $f\in C_{\rm c}^1(\R^d)$ the equality $\wh x_j\wh \xi_k f - \wh\xi_k\wh x_j f = \frac1{2\pi} i \delta_{jk} f
$ holds in $L^2(\R^d,m)$. Of course \eqref{eq:Heisenberg} could also be derived directly from the definitions given in \eqref{eq:defxjxik}.
When considered as densely defined operators in $L^2(\R^d,m)$ with domain $C_{\rm c}(\R^d)$, the operators
$\wh x_j$ and $\wh \xi_k$ are closable and their closures are self-adjoint on $L^2(\R^d,m)$. Indeed, by Stone's theorem (Theorem \ref{thm:Stone}), the generators of the unitary $C_0$-groups $(U_{te_j})_{t\in \R}$ and
$(V_{te_k})_{t\in \R}$ are of the form $iA_j$ and $iB_k$ with $A_j, B_k$ self-adjoint. Moreover,
$C_{\rm c}^1(\R^d)$ is contained in their domains and we have $A_j f = \wh x_jf$ and $B_k f = \wh \xi_k f$ for all $f\in C_{\rm c}^1(\R^d)$. The space $C_{\rm c}^1(\R^d)$ is
dense in $L^2(\R^d,m)$ and invariant under both groups, and therefore by Proposition \ref{prop:core} it is dense in both $\Dom(A_j)$ and $\Dom(B_k)$
with respect to their respective graph norms.
With slight abuse of notation we write $\wh x_j$ and $\wh\xi_k$ for the self-adjoint closures and denote their domains by $\Dom(\wh x_j)$ and $\Dom(\wh \xi_k)$. For pure states $\phi$ represented by a norm one function $f\in \Dom(\wh x_j)\cap \Dom(\wh\xi_j)$ such that $\wh x_jf, \wh\xi_jf\in \Dom(\wh x_j)\cap \Dom(\wh\xi_j)$, the uncertainty principle of Theorem \ref{thm:uncertainty} takes the form
$$ \Delta_\phi(\wh x_j)\Delta_\phi(\wh\xi_j) \ge \frac1{2}.$$
It follows from Proposition \ref{prop:LCA-pos-mom} that the position observable
$X$ is covariant with respect to translations and conserved under boosts, and that the momentum observable $\Xi$ is conserved under translations and covariant with respect to boosts. This essentially characterises these observables:
\begin{theorem}[Covariance characterisation of position and momentum]\label{thm:pos-mom} Up to conjugation with a translation, respectively a boost, position and momentum are characterised by their covariance and conservation properties. More precisely, denoting by $X$ and $\Xi$ the position and momentum observables of Definition \ref{def:posmom} the following assertions hold.
\begin{enumerate}[\rm(1)]
\item If the observable $P: \calB(\R^d)\to \PP(L^2(\R^d,m))$ is covariant with respect to all pairs $(x,U_x)$, $x\in \R^d$, and conserved under all boosts $V_\xi$, $\xi\in \R^d$, then there exists a unique $y\in \R^d$ such that
$$P_B = U_{y} X_B U_y^\star, \quad B\in \calB(\R^d). $$
\item If the observable $P: \calB(\R^d)\to \PP(L^2(\R^d,m))$ invariant under all translations $U_x$, $x\in \R^d$, and covariant with respect to all pairs $(\xi,V_\xi)$, $\xi\in \R^d$, then there exists a unique $\eta\in \R^d$ such that
$$P_B = V_{\eta} \Xi_B V_{\eta}^\star, \quad B\in \calB(\R^d).$$
\end{enumerate}
\end{theorem}
\begin{proof}
Let the projection-valued measure
$P: \calB(\R^d)\to \PP(L^2(\R^d,m))$ be covariant with respect to all pairs $(x,U_x)$ and conserved under all boosts $V_\xi$.
The boost invariance means that every projection $P_B$ commutes with pointwise multiplication with every trigonometric exponential $x\mapsto \exp(i x\cdot\xi)$. By Lemma \ref{lem:char-mult}, this implies that $P_B $ is a pointwise multiplier of the form $P_B f(x) = m_{B}(x)f(x)$ with $m_B\in L^\infty(\R^d)$. Since $P_B$ is a projection, $m_{B}$ must be an indicator function, say of the set $C_{B}$: $$ P_B f = \one_{C_B} f.$$
Substituting this into the covariance with respect to translations, we arrive at the identity
$ \one_{C_{B}}(x-t) = \one_{C_{B+t}}$ as elements of $L^\infty(\R^d)$, that is, we have $$C_{B}+t = C_{B+t}$$
up to a null set. Similarly one sees that $C_{\R^d} = \R^d$ and $C_{\complement B} = \complement C_B$ up to null sets. Finally, if $B$ and $B'$ are disjoint, then so are $P_B $ and $P_{B'}$, and therefore $ C_{B}$ and $C_{B'}$ are disjoint up to a null set. It follows that the mapping $B\mapsto C_B$ commutes up to null sets with
translations and the Boolean set operations.
Let $B := [0,1)^d$ be the half-open unit cube. Suppose $x,y\in \R^d$ are Lebesgue points of $\one_{C_B}$ satisfying $\max_{1\le j\le d}|x_j-y_j|>1$. Then $C_B$ and $C_B+y-x$ intersect in a set of positive measure. This is only possible if $B$ and $B+y-x$ intersect in a set of positive measure, but these sets are disjoint. This contradiction proves that all Lebesgue points $x,y$ of $\one_{C_B}$ satisfy $\max_{1\le j\le d}|x_j-y_j|\le 1$.
Since almost every point of $\one_{C_B}$ is a Lebesgue point, it follows that up to a null set, $C_B$ is contained in the rectangle $\prod_{j=1}^d [a_j,b_j]$, where
\begin{align*}a_j &:= \inf\{x_j:\, x \ \hbox{is a Lebesgue point of} \ \one_{C_B}\}, \\
b_j & := \sup\{x_j:\, x \ \hbox{is a Lebesgue point of} \ \one_{C_B}\}.
\end{align*}
In particular, since $0\le b_j-a_j\le 1$, up to a null set we have $C_B \subseteq \prod_{j=1}^d [a_j,a_j+1) = a+B$, where $a = (a_1,\dots,a_d)$.
We next claim that up to a null set we have equality $$C_B= a+B.$$ Indeed, the sets $k+B$ with $k\in \Z^d$ are pairwise disjoint and their union is $\R^d$. Hence, up to null sets, the sets $C_{k+B} = k+C_B$ are disjoint and their union
is $\R^d$. This is only possible if $(a+B)\setminus C_B$ is a null set. This proves the claim.
Let $n\in \N$ and consider the set $B^{(n)} := [0, 2^{-n})^d$. The same argument as above proves that there exists
an $a^{(n)}\in \R^d$ such that $C_{B^{(n)}} = a^{(n)}+B^{(n)}$. Now $B$ is the disjoint union of $2^{nd}$ translates $k+B^{(n)}$, $k\in \{0,2^n-1\}^d$. Therefore, up to a null set, $C_B = a+B$ is the disjoint union of the $2^{nd}$ sets $C_{k+B^{(n)}} = a^{(n)} + k + B^{(n)}$. This union equals $a^{(n)} + B$. This shows that $a^{(n)} = a$ for all $n\in \N$.
Summarising what we have proved, we find that for all sets $B$ of the form $y+B^{(n)}$ with $y\in \R^d$ and $n\in \Z$,
we have
$$P_B f = \one_{a+B} f.$$
Equivalently, this can be expressed as
$$P_B = U_{a} X_B U_{-a}= U_{a} X_B U_{a}^\star.$$
This proves the first part of the theorem.
To prove the second part, suppose that the projection-valued measure
$P: \calB(\R^d)\to \PP(L^2(\R^d,m))$ is conserved under all $U_y$ and covariant with respect to all pairs $(\eta,V_\eta)$. Then $\wt P_B := \F^{-1} P_B \F$ defines a projection-valued measure that is covariant with respect to all pairs $(y,U_y)$ and conserved under all $V_\eta$. It follows from the previous step that $\wt P = U_aXU_{a}^\star$ for some $a\in \R^d$,
and therefore $P = \F \wt P \F^{-1} = V_a \Xi V_{a}^\star$ by \eqref{eq:transl-Fourier}.
\end{proof}
\subsection{The case $G = \T$: angle and angular momentum}
The results of the preceding section have natural analogues for the unit circle $\T$.
We identify $\T$ with the unit circle of $\C$ and take the normalised Lebesgue measure on $\T$ as Haar measure. Every character $\gamma:\T\to \T$ is of the form
$$ \gamma(z) = z^k, \quad z\in \T,$$ for some $k\in \Z$.
Under this identification we have $\Gamma = \Z$, its normalised Haar measure being the counting measure.
For every $w\in \mathbb{T}$ the rotation
$z\mapsto wz$ is a symmetry of $\T$.
The induced symmetry $U_w$ on $L^2(\R^d,m)$ is given by
\begin{align}\label{def-Uw} U_w f(z) = f(w^{-1}z) , \quad z\in \T, \ f\in L^2(\T).
\end{align}
For every $k\in \Z$ the boost $V_k$ on $L^2(\T)$ is given by
\begin{align}\label{def-Vk} V_k f(z) = z^{k}f(z), \quad z\in \T, \ f\in L^2(\T).
\end{align}
The Weyl commutation relation takes the form
$$ V_k U_z = z^k U_z V_k, \quad z\in \T, \ k\in\Z. $$
The position and momentum observables in $\T$ associated with the symmetries $U_z$ and $V_k$ are denoted by $\Theta$ and $L$ and are called the
{\em angle}\index{angle}
and (orbital) {\em angular momentum}\index{angular momentum}
observables. They take values in $\T$ and $\Z$ respectively; in particular, angular momentum can only assume discrete values. In the physics literature one speaks about `quanta' of angular momentum.
\begin{remark}
By viewing $\Z$ as a subset of the real line, we may identify $L$ with a real-valued observable and thus associate with $L$ a self-adjoint operator $\wh l$ on $L^2(\R)$. There is no natural way, however, to do the same with $\Theta$. One could identify $\T$ with the interval $(-\pi,\pi]$ contained in the real line and thus identify $\Theta$ with a real-valued observable. The choice of the interval $(-\pi,\pi]$ is somewhat arbitrary, however, and entails a nonuniqueness issue that cannot be resolved satisfactorily.
The associated self-adjoint operator $\wh\theta$ appears not to be very useful. For instance, it does not satisfy the `continuous variable' Weyl commutation relation $$ e^{is\wh \theta}e^{it\wh l} = e^{ist} e^{it\wh l}e^{is\wh \theta}.$$
This will be further discussed in Problem \ref{prob:angle}.
\end{remark}
By Proposition \ref{prop:LCA-pos-mom}, $\Theta$ is covariant under every pair $(z,U_z)$ and conserved under every $V_k$,
and $L$ is conserved under every $U_z$ and covariant under every pair $(k,V_k)$.
Repeating the proof of Theorem \ref{thm:pos-mom} almost {\em verbatim} one arrives at the following theorem.
\begin{theorem}[Covariance characterisation of angle and angular momentum]\label{thm:angle} Up to conjugation with a translation, respectively a boost, angle and angular momentum are characterised by their covariance and conservation properties. More precisely, denoting by
$\Theta$ and $L$ the position and momentum operators associated with the rotations $U_z$ and boosts $V_k$ given by \eqref{def-Uw} and \eqref{def-Vk}, the following assertions hold.
\begin{enumerate}[\rm(1)]
\item If the observable $P: \calB(\T)\to \PP(L^2(\T))$ is covariant with respect to all pairs $(z,U_z)$, $z\in\T$, and conserved under all $V_k$, $k\in \Z$, then there exists a unique $w\in \T$ such that
$$P_B = U_w \Theta_B U_w^\star, \quad B\in \calB(\T). $$
\item If the observable $P: \calB(\Z)\to \PP(L^2(\T))$ is conserved under all $U_z$, $z\in \T$, and covariant with respect to all pairs $(k,V_k)$, $k\in\Z$, then there exists a unique $j\in\Z $ such that
$$P_B = V_j L_B V_j^\star, \quad B\in \calB(\Z).$$
\end{enumerate}
\end{theorem}
\subsection{The Stone--von Neumann theorem}\label{subsec:StonevonNeumann}
We have seen in Theorem \ref{thm:pos-mom} that the $\R^d$-valued position and momentum observables are uniquely determined, up to conjugation with a translation and a boost respectively, by their transformation properties under translations and boosts. It is interesting to observe that both the covariance relation for position
$$
U_x X_B U_{x}^\star f= X_{x B}f, \quad x\in \R^d, \quad f\in L^2(\R^d,m),
$$
and the covariance relation for momentum
$$
V_\xi \Xi_B V_{\xi}^\star f= \Xi_{\xi B}f\quad \xi\in \R^d, \quad f\in L^2(\R^d,m),
$$ imply the Weyl commutation relation. Here, as before, ${\rm d}m(x) = (2\pi)^{-d/2}\ud x$ is the normalised Lebesgue measure.
Indeed, approximating $e^{i x\cdot \xi}$ by simple functions, as in the proof of Theorem \ref{thm:Ugamma} we find that the covariance relation for position implies the identity $U_x V_\xi U_{-x}f= e^{- i x\cdot \xi}V_\xi f$, which is the Weyl commutation relation. In the same way
the covariance relation for momentum implies the Weyl commutation relation. In view of this it is reasonable to ask to what extent position and momentum are determined by the Weyl commutation relation. The answer to this question is
provided by a theorem due to Stone and von Neumann (Theorem \ref{thm:Stone-vonNeumann} and its corollary). Proving this theorem is the main objective of the present section.
We start with some preparation. Suppose that $\wt U, \wt V:\R^d\to \calL(H)$ are strong\-ly continuous unitary representations of $\R^d$ on some Hilbert space $H$ such that the Weyl commutation relation holds, that is,
\begin{align}\label{eq:WCR}
\wt V_\xi \wt U_x = e^{ ix\cdot \xi} \wt U_x \wt V_\xi, \quad x,\xi\in \R^d.
\end{align}
The relation \eqref{eq:WCR} states that $\wt U$ and $\wt V$ `commute up to a multiplicative scalar of modulus one'. This suggests to interpret \eqref{eq:WCR} as a `projective' unitary representation\index{representation!projective} of $\R^d\times\R^d$ on $H$. There is a quick way to extend \eqref{eq:WCR} to a unitary representation as follows.
Consider the unitary operators
\begin{align}\label{eq:WWeyl} \wt W(x,\xi):= e^{\frac12 ix\cdot \xi}\wt U_x \wt V_\xi = e^{-\frac12 ix\cdot \xi}\wt V_\xi \wt U_x, \quad x,\xi\in \R^d.
\end{align}
The operators $\wt W(x,\xi)$ defined by \eqref{eq:WWeyl} satisfy
\begin{equation}\label{eq:WW}\begin{aligned}
\wt W(x,\xi)\wt W(x',\xi')
& = e^{-\frac12 i(x\cdot\xi+x'\cdot\xi')} \wt V_\xi \wt U_x\wt V_{\xi'} \wt U_{x'}
\\ & = e^{-\frac12 i(x\cdot\xi+x'\cdot\xi') - ix\cdot\xi'}\wt V_\xi \wt V_{\xi'}\wt U_x \wt U_{x'}
\\ & = e^{\frac12 i(x'\cdot\xi - x\cdot\xi')} e^{-\frac12 i(x+x')(\xi+\xi')}\wt V_{\xi+\xi'} \wt U_{x+x'}
\\ & = e^{\frac12 i(x'\cdot\xi - x\cdot\xi')} \wt W(x+x',\xi+ \xi').
\end{aligned}
\end{equation}
From this it follows that the unitary operators defined by
\begin{align}\label{eq:Wxxit} \wt W(x,\xi,t) := e^{it} \wt W(x,\xi)
\end{align}
satisfy
\begin{equation}\label{eq:Heisenberg-repr}
\begin{aligned} \wt W(x,\xi,t) \wt W(x',\xi',t')
& = e^{i(t+t')} \wt W(x,\xi)\wt W(x',\xi')
\\ & = e^{i(t+t' + \frac12(x'\cdot\xi-x\cdot\xi'))} \wt W(x+x',\xi+\xi')
\\ & = \wt W((x,\xi,t)\circ (x',\xi',t')),
\end{aligned}
\end{equation}
where
$$ (x,\xi,t)\circ (x',\xi',t'):= (x+x', \xi+\xi', t+t' + \frac12(x'\cdot\xi-x\cdot\xi')).$$
This prompts the following definition. One easily checks that the operation $\circ$ turns $\H^d:=\R^d\times\R^d\times \R$\index{$H$@$\H^d$} into a group:
\begin{definition}[Heisenberg group]
The {\em Heisenberg group}\index{Heisenberg!group}\index{group!Heisenberg} in dimension $d$ is the group $\H^d:=\R^d\times\R^d\times \R$ with composition law
$$ (x,\xi,t)\circ (x',\xi',t'):= (x+x', \xi+\xi', t+t' + \frac12(x\cdot\xi'-x'\cdot\xi)).$$
\end{definition}
The identity \ref{eq:Heisenberg-repr} informs us that $\wt W$ defines a
unitary representation of the Heisenberg group $\H^d$ on $H$. It is strongly continuous and it satisfies
\begin{align}\label{eq:W00t} \wt W(0,0,t) = e^{it}I, \quad t\in \R.
\end{align}
\begin{definition}[Schr\"odinger representation] The {\em Schr\"odinger representation}\index{representation!Schr\"odinger}\index{Schr\"odinger!representation} is the unitary representation $W: \H^d\to \calL(L^2(\R^d,m))$ arising
in the special case where the unitary representations $U,V:\R^d\to \calL(L^2(\R^d,m))$ are given by translations and boosts, respectively.
\end{definition}
Explicitly, the Schr\"odinger representation is given by
$$ W(x,\xi,t)f(x') =
e^{ it} e^{-\frac12 ix\cdot \xi} e^{i x'\cdot \xi}f(x'-x).
$$
\begin{proposition}\label{prop:Schrod-irred}
The\, Schr\"odinger\, representation\, is\, {\em irreducible},\,\index{representation!irreducible}\index{irreducible} that is,\, the\, only\, closed subspaces of $L^2(\R^d,m)$ invariant under the action of $W$ are the trivial subspaces $\{0\}$ and $L^2(\R^d,m)$.
\end{proposition}
The proof of this proposition will be given at the end of the section, for it uses elements of the proof of the following theorem which says that the Schr\"odinger representation is essentially the only irreducible unitary representation of $\H^d$ satisfying \eqref{eq:W00t}:
\begin{theorem}[Stone--von Neumann]\label{thm:Stone-vonNeumann}\index{theorem!Stone--von Neumann}
Let $\wt W: \H^d\to \calL(H)$ be a strongly continuous unitary representation of $\H^d$ on a separable Hilbert space $H$. If $\wt W$ is irreducible and satisfies $\wt W(0,0,t) = e^{it}I$ for all $t\in \R$, then
$\wt W$ is unitarily equivalent to the Schr\"odinger representation $W$.
More precisely, there exists a unitary operator $S: L^2(\R^d,m)\to H$ such that
$$\wt W(x,\xi,t) = SW(x,\xi,t)S^\star, \quad (x,\xi,t)\in \H^d.$$
The operator $S$ is unique up to a multiplicative scalar of modulus one.
\end{theorem}
Here, we use the term {\em unitary operator}\index{unitary!operator}\index{operator!unitary} for an operator $S: H\to K$, where $H$ and $K$ are Hilbert spaces, such that $S^\star S = I$ (the identity operator on $H$) and $SS^\star = I$ (the identity operator on $K$).
We have the following immediate corollary for representations arising from pairs of unitary representations satisfying the Weyl commutation relation.
\begin{corollary} Let $\wt U, \wt V:\R^d\to\calL(H)$ be strongly continuous unitary representations on a separable Hilbert space $H$ satisfying the Weyl commutation relation
$$ \wt V_\xi \wt U_x = e^{ix\cdot \xi} \wt U_x \wt V_\xi, \quad x,\xi\in \R^d.$$
Suppose furthermore that the family $\{\wt U_x, \wt V_\xi: x\in \R^d,\, \xi\in\R^d\} $ acts irreducibly on $H$
in the sense that the only closed subspace invariant under all operators $ U_x$ and $\wt V_\xi$, $x\in \R^d,\, \xi\in\R^d$, are the trivial subspaces $\{0\}$ and $H$.
Then there exists a unitary operator $S:L^2(\R^d,m)\to H$ such that
\begin{align*}
\wt U_x & = S U_x S^\star,\quad x\in\R^d,
\\ \wt V_\xi & = S V_\xi S^\star, \quad \xi\in\R^d,
\end{align*}
where $U$ and $V$ are the translation and boost representations on $L^2(\R^d,m)$, respectively.
The operator $S$ is unique up to a multiplicative constant of modulus one.
\end{corollary}
\begin{proof} Defining the operators $\wt W:\H^d\to \calL(H)$ by \eqref{eq:WWeyl} and \eqref{eq:Wxxit},
the irreducibility assumption of the corollary
translates into the irreducibility of the representation $\wt W$.
\end{proof}
In what follows we fix a strongly continuous unitary representation $\wt W: \H^d\to \calL(H)$
and define $$\wt U_x := W(x,0,0), \quad \wt V_\xi:= W(0,\xi,0), \quad \wt W(x,\xi):= \wt W(x,\xi,0).$$
Then \eqref{eq:WCR}--\eqref{eq:Heisenberg-repr} hold again.
We write $m$ for both the normalised Lebesgue measures on $\R^d$ and $\R^{2d}$.
\begin{definition}[Weyl transform] For $a\in L^1(\R^{2d},m)$
we define the operator $\wt W(a)\in \calL(H)$ by
\begin{align*}
\wt W(a)h
& : = \int_{\R^{2d}} a(x,\xi) \wt W(x,\xi)h \ud m(x)\ud m(\xi), \quad h\in H,
\end{align*}
where the integral is a Bochner integral in $H$.
\end{definition}
The next two lemmas state some properties for the Weyl transform associated with the Schr\"odinger representation $W:\H^d\to \calL(L^2(\R^d,m))$.
\begin{lemma}\label{lem:Moyal-injective}
For all $a\in L^1(\R^{2d})\cap L^2(\R^{2d},m)$ the operator $W(a)$ is Hilbert--Schmidt and
$$ \n W(a)\n_{\calL_2(L^2(\R^{d}))} = \n a\n_{L^2(\R^{2d},m)}.$$
\end{lemma}
\begin{proof}
For the Schr\"odinger representation $W:\H^d\to \calL(L^2(\R^d,m))$ we have the explicit formula
\begin{align}\label{eq:Schrod-repr}
W(x,\xi) f(x') = e^{-\frac12 ix\cdot \xi}e^{ix'\cdot\xi} f(x'-x), \quad f\in L^2(\R^d,m),
\end{align}
where $W(x,\xi) := W(x,\xi,0)$ as in \eqref{eq:Wxxit}.
By a change of variables and Fubini's theorem we obtain
\begin{align*}
W(a)f & = \int_{\R^{2d}} a(x,\xi) e^{-\frac12 ix\cdot \xi}e^{i(\cdot)\cdot\xi} f(\cdot-x)\ud m(x)\ud m(\xi)
\\ & = \int_{\R^{d}} \Bigl(\int_{\R^d} a(\cdot-x,\xi) e^{-\frac12 i(\cdot-x)\cdot \xi}e^{i(\cdot)\cdot\xi}\ud m(\xi)\Bigr) f(x) \ud m(x)
\\ & := \int_{\R^d} k(x,\cdot)f(x)\ud m(x),
\end{align*}
where
$$ k(x,x') = \int_{\R^d} a(x'-x,\xi) e^{\frac12 i(x+x')\cdot\xi}\ud m(\xi).$$
By Plancherel's theorem,
\begin{align*}
\ & \int_{\R^{2d}} \Big|k\Bigl(\frac{y-z}{2},\frac{y+z}{2}\Big)\Big|^2\ud m(y)\ud m(z)
\\ & \qquad = \int_{\R^{2d}} \Big|\int_{\R^d} a(z,\xi) e^{\frac12 iy\cdot\xi}\ud m(\xi)\Big|^2\ud m(y)\ud m(z)
\\ & \qquad = 2\int_{\R^{d}}\int_{\R^d}| a(z,\xi)|^2\ud m(\xi) \ud m(z)
= 2\n a\n_2^2
\end{align*}
and hence
$$ \int_{\R^{2d}} |k(x,x')|^2 \ud m(x)\ud m(x') = \frac12 \int_{\R^{2d}}\Big|k\Bigl(\frac{y-z}{2},\frac{y+z}{2}\Big)\Big|^2\ud m(y)\ud m(z) = \n a\n_2^2.$$
The result now follows from Example \ref{ex:kernel-HS}, which says that an integral operator with square integrable kernel is Hilbert--Schmidt, with Hilbert--Schmidt norm equal to the $L^2$-norm of the kernel.
\end{proof}
Since $L^1(\R^{2d},m)\cap L^2(\R^{2d},m)$ is dense in $L^2(\R^{2d},m)$, the lemma implies that the mapping $W: a\mapsto W(a)$ has a unique extension to an isometry from $L^2(\R^{2d},m)$ into the space of Hilbert--Schmidt operators $\calL_2(L^2(\R^{d}))$. This extension is again denoted by $W$.
A special role is played by the functions
\begin{align*}a_0 (x,\xi) & := \exp\Bigl(-\frac14|x|^2/2\Bigr) \exp\Bigl(-\frac14|\xi|^2\Bigr),\\
\phi_0(x) &:= \pi^{d/4} \exp\Bigl(-\frac12|x|^2\Bigr).
\end{align*}
\begin{lemma}\label{lem:Schrod-rankone}
The operator $W(a_0)$ equals the rank one projection $\phi_0\,\bar\otimes\, \phi_0$.
\end{lemma}
\begin{proof}
Using \eqref{eq:Schrod-repr}
and the elementary identity (which follows from Lemma \ref{lem:Gauss})
$$ \int_{\R^d} e^{-\frac12 ix\cdot\xi} e^{iu\cdot \xi} \exp\Bigl(-\frac14|\xi|^2\Bigr) \ud m(\xi)
= \pi^{d/2} \exp\Bigl(-|u-\frac12x|^2\Bigr)$$
we obtain, for $f\in L^2(\R^d,m)$,
\begin{align*}
W(a_0) f
& = \int_{\R^{2d}} \exp\Bigl(-\frac14|x|^2\Bigr) \exp\Bigl(-\frac14|\xi|^2\Bigr) e^{-\frac12 ix\cdot\xi}e^{ i(\cdot)\cdot \xi} f(\cdot-x) \ud m(x)\ud m(\xi)
\\ & = \pi^{d/2} \int_{\R^{d}} \exp\Bigl(-|(\cdot)-\frac12x|^2\Bigr) \exp\Bigl(-\frac14|x|^2\Bigr) f(\cdot-x) \ud m(x)
\\ & = \pi^{d/2} \exp\Bigl(-\frac12|\cdot|^2\Bigr) \int_{\R^{d}} \exp\Bigl(-\frac12|x|^2\Bigr)f(x)\ud m(x)
= (\phi_0\,\bar\otimes\, \phi_0)f.
\end{align*}
\end{proof}
Returning to a general strongly continuous unitary representation $\wt W: \H^d\to \calL(H)$,
we note the following important multiplicativity property.
\begin{lemma}\label{lem:Moyal}
For all $a,b\in L^1(\R^{2d})$ we have
$$ \wt W(a)\wt W(b) = \wt W(a\#b), $$
where the so-called {\em twisted convolution}\index{twisted convolution}\index{convolution!twisted} $a\# b \in L^1(\R^{2d})$ is defined by
$$ {a \# b}(x,\xi) := \int_{\R^{2d}} e^{\frac12 i(x'\cdot\xi - x\cdot \xi' )} a(x-x', \xi-\xi') b(x',\xi')\ud m(x')\ud m(\xi'). $$
\end{lemma}
Young's inequality implies that $a \# b$ does indeed belong to $L^1(\R^d)$.
\begin{proof} Fix $h\in H$.
By \eqref{eq:WW}, a change of variables, and Fubini's theorem,
\begin{align*}
& \wt W(a)\wt W(b)h\\ & = \int_{\R^{4d}} a(x,\xi)b(x',\xi') \wt W(x,\xi)\wt W(x',\xi')h \ud m(x)\ud m(\xi)\ud m(x')\ud m(\xi')
\\ & = \int_{\R^{4d}} e^{\frac12 i(x'\cdot\xi - x\cdot \xi')} a(x,\xi)b(x',\xi')
\\ & \hskip3cm \times \wt W(x+x',\xi+ \xi')h \ud m(x)\ud m(\xi)\ud m(x')\ud m(\xi')
\\ & = \int_{\R^{4d}} e^{\frac12 i(x'\cdot\xi - x\cdot \xi')} a(x-x',\xi-\xi')b(x',\xi')
\\ & \hskip3cm \times \wt W(x,\xi) h \ud m(x)\ud m(\xi)\ud m(x')\ud m(\xi')
\\ & = \int_{\R^{2d}} a\#b(x,\xi) \wt W(x,\xi)h \ud m(x)\ud m(\xi)
= \wt W(a\#b)h.
\end{align*}
\end{proof}
This lemma is used to establish the following technical fact.
\begin{lemma}\label{lem:WWW} For $x,\xi\in \R^d$ we have
$$\wt W(a_0)\wt W(x,\xi)\wt W(a_0) = a_0(x,\xi)\wt W(a_0).$$
\end{lemma}
\begin{proof}
Repeating the steps in the proof of Lemma \ref{lem:Moyal}, for all $h\in H$ we obtain
\begin{equation}\label{eq:Waxi}
\begin{aligned}
\wt W(x,\xi)\wt W(a_0)h & = \int_{\R^{2d}} a_0(x',\xi')\wt W(x,\xi)\wt W(x',\xi')h\ud m(x')\ud m(\xi')
\\ & = \int_{\R^{2d}} e^{\frac12 i(x'\cdot\xi - x\cdot \xi')} a_0(x-x',\xi-\xi')\wt W(x,\xi)h\ud m(x')\ud m(\xi')
\\ & = \wt W(a_{x,\xi})h,
\end{aligned}
\end{equation}
where $ a_{x,\xi}(x',\xi') := e^{\frac12 i(x'\cdot\xi - x\cdot \xi')} a_0(x-x',\xi-\xi').$
Hence, by Lemma \ref{lem:Moyal}, the lemma is equivalent to the statement that
$$\wt W(a_0 \# a_{x,\xi}) = a_0(x,\xi) \wt W(a_0).$$
For this it suffices to show that $$a_0 \# a_{x,\xi} = a_0(x,\xi)a_0.$$
By the injectivity of the Schr\"odinger representation $W$, which follows from Lemma \ref{lem:Moyal-injective},
this in turn is equivalent to showing that
$$W(a_0)W(x,\xi)W(a_0) = W(a_0 \# a_{x,\xi}) = a_0(x,\xi) W(a_0).$$
The verification of this identity proceeds by explicit calculation. By
Lemma \ref{lem:Schrod-rankone},
\begin{align*}
W(a_0)W(x,\xi)W(a_0)f
& = (\phi_0\,\bar\otimes\, \phi_0) W(x,\xi)(\phi_0\,\bar\otimes\, \phi_0)f
\\ & = \iprod{W(x,\xi)\phi_0}{\phi_0} \iprod{f}{\phi_0}\phi_0
= \iprod{W(x,\xi)\phi_0}{\phi_0} W(a_0)f.
\end{align*}
Moreover, by \eqref{eq:Schrod-repr}, Lemma \ref{lem:Schrod-rankone}, and an elementary computation,
\begin{align*}
\iprod{W(x,\xi)\phi_0}{\phi_0}
& = e^{-\frac12 ix\cdot \xi}\iprod{e^{ i(\cdot)\cdot\xi} \phi_0(\cdot-x)}{\phi_0}
\\ & = \pi^{d/2} e^{-\frac12 ix\cdot \xi}\int_{\R^d} e^{iy\cdot\xi}\exp\Bigl(-\frac12|y-x|^2\Bigr)\exp\Bigl(-\frac12|y|^2\Bigr)\ud m(y)
\\ & = \exp \Bigl(-\frac14|x|^2\Bigr)\exp\Bigl(-\frac14|\xi|^2\Bigr) = a_0(x,\xi).
\end{align*}
\end{proof}
We are now ready for the proof of the Stone--von Neumann theorem.
\begin{proof}[Proof of Theorem \ref{thm:Stone-vonNeumann}]
We split the proof into three steps.
\smallskip
{\em Step 1} --
We begin by showing that $\wt W(a_0)$ is a rank one projection.
By Lemmas \ref{lem:Schrod-rankone} and \ref{lem:Moyal} (applied to $W$),
$$W(a_0 \# a_0) = W(a_0)W(a_0) = (\phi_0\,\bar\otimes\,\phi_0)^2 = \phi_0\,\bar\otimes\,\phi_0 = W(a_0).$$
By the injectivity of $W$ (which follows from Lemma \ref{lem:Moyal-injective}), this implies that $ a_0 \# a_0 = a_0$. Another application of Lemma \ref{lem:Moyal}, this time to $\wt W$,
gives $$\wt W(a_0)\wt W(a_0) = \wt W(a_0\#a_0) = \wt W(a_0).$$ This means that $\wt W(a_0)$ is a projection.
We will use the assumption of irreducibility of $\wt W$ to prove that this projection has rank one.
We begin by showing that $\wt W(a_0) \not=0$. Indeed, if we had $\wt W(a_0) = 0$, then
for all $x,\xi\in \R^d$ and $h,h'\in H$ we would have, by \eqref{eq:WW} and \eqref{eq:Waxi},
\begin{align*}
0 & = \iprod{\wt W(x,\xi)\wt W(a_0) \wt W(-x,-\xi)h}{h'}
\\ & = \iprod{\wt W(a_{x,\xi}) \wt W(-x,-\xi)h}{h'}
\\ & = \int_{\R^{2d}} e^{\frac12 i(x'\xi-x\xi')}a_0(x-x',\xi-\xi')
\\ & \qquad\qquad\qquad\qquad\times \iprod{\wt W(x',\xi')\wt W(-x,-\xi)h}{h'} \ud x'\ud\xi'
\\ & = \int_{\R^{2d}} e^{-\frac12 i(x'\xi-x\xi')}a_0(x',\xi')
\\ & \qquad\qquad\qquad\qquad\times \iprod{\wt W(x-x',\xi-\xi')\wt W(-x,-\xi)h}{h'} \ud x'\ud\xi'
\\ & = \int_{\R^{2d}} e^{- i(x'\xi-x\xi')}a_0(x',\xi')\iprod{\wt W(-x',-\xi')h}{h'} \ud x'\ud\xi'
\\ & = \int_{\R^{2d}} e^{2 i(x'\xi-x\xi')}a_0(x',\xi')\iprod{\wt W(x',\xi')h}{h'} \ud x'\ud\xi'.
\end{align*}
This being true for all $x,\xi\in \R^d$, the Fourier inversion theorem would then imply that $\iprod{\wt W(x',\xi')h}{h'}=0$ for almost all $x',\xi'\in\R^d$.
Since $h,h'\in H$ were arbitrary, it would follow that $W(x',\xi') = 0$ for almost all $x',\xi'\in\R^d$,
contradicting the fact that all these operators are unitary.
Fix any nonzero $h\in \Ran(\wt W(a_0))$; this is possible by the preceding argument. Let
$\wt Y_h$ be the closed linear span of the set $\{\wt W(x,\xi)h:\, x,\xi\in \R^d\}$.
From \eqref{eq:WW} we see that $\wt Y_h$ is invariant under each operator $\wt W(x,\xi)$ and hence under the representation $\wt W$. Since $\wt W$ is assumed to be
irreducible it follows that $\wt Y_h = H$.
By \eqref{eq:WCR} and \eqref{eq:WWeyl},
$$ \wt W(x,\xi)^\star= e^{\frac12 ix\cdot \xi} \wt U_x^\star \wt V_\xi^\star
= e^{\frac12 ix\cdot \xi} \wt U_{-x} \wt V_{-\xi}
= e^{-\frac12 ix\cdot \xi} \wt V_{-\xi} \wt U_{-x}
= \wt W(-x,-\xi).
$$
This identity implies that $\wt W(a_0)$ is self-adjoint, and then we infer from Lemma \ref{lem:WWW} that if $h,h'\in \Ran(\wt W(a_0))$, then
\begin{equation}\label{eq:iprodWW}
\begin{aligned}
\iprod{\wt W(x,\xi)h}{\wt W(x',\xi')h'}
& = \iprod{\wt W(-x',-\xi')\wt W(x,\xi)\wt W(a_0)h}{\wt W(a_0)h'}
\\ & = e^{\frac12 i (x'\cdot\xi - x\cdot \xi')}\iprod{\wt W(x-x',\xi-\xi')\wt W(a_0)h}{\wt W(a_0)h'}
\\ & = e^{\frac12 i (x'\cdot\xi - x\cdot \xi')} a_0(x-x',\xi-\xi') \iprod{h}{h'}.
\end{aligned}
\end{equation}
In particular, if $h\perp h'$ with $h,h'\in \Ran(\wt W(a_0))$, then $\wt Y_h\perp \wt Y_{h'}$. Since $\wt Y_h = H$ this implies $\wt Y_{h'} = \{0\}$.
In particular $\wt W(x,\xi)h'=0$ for all $x,\xi\in \R^d$, and therefore $h'=0$ by Lemma \ref{lem:Moyal-injective}.
This proves that $\Ran(\wt W(a_0))$ equals the one-dimensional span of $h$.
\smallskip
{\em Step 2} -- Define
$$ S \sum_{n=1}^N c_n W(x_n,\xi_n)\phi_0 := \sum_{n=1}^N c_n \wt W(x_n,\xi_n)h_0,$$
where $h_0\in \Ran(\wt W(a_0))$ has norm $\n h_0\n = 1 = \n \phi_0\n$.
It follows from \eqref{eq:iprodWW} (applied to both $W$ and $\wt W$) that $S$ is well defined and isometric
on the linear span
of the functions $ W(x,\xi)\phi_0$, $x,\xi\in\R^d$,
and hence extends to an isometry from $Y_{\phi_0}$ onto $\wt Y_{h_0}$, the former being defined as the
closed linear span of the functions $W(x,\xi)\phi_0$, $x,\xi\in \R^d$. But $\wt Y_{h_0} = H$, and by applying this to
$W$ we see that likewise $Y_{\phi_0} = L^2(\R^d,m)$. This proves that $S$ is isometric from
$L^2(\R^d,m)$ onto $H$, and hence
unitary. Since $S\phi_0 = h_0$,
this proves that $S$ has the desired properties.
\smallskip
{\em Step 3} -- If $T:L^2(\R^d,m)\to H$ is another unitary operator with the property
that $\wt W(x,\xi,t) = TW(x,\xi,t)T^{-1}$ for all $(x,\xi,t)\in \H^d$,
then
$ S^{-1}TW(x,\xi) = W(x,\xi)S^{-1}T$ for all $x,\xi\in \R^d$.
From this it follows that $S^{-1}T$ commutes with $W(a_0)$,
and therefore it maps the one-dimensional range of this operator onto itself.
This implies that $S^{-1}Tf = e^{i\theta}f$ for some $\theta\in \R$ and all $f\in \Ran(W(a_0))$.
Then,
\begin{align*} T \sum_{n=1}^N c_n W(x_n,\xi_n)f
& = S S^{-1}T\sum_{n=1}^N c_n W(x_n,\xi_n)f
\\ & = S\sum_{n=1}^N c_n W(x_n,\xi_n)S^{-1}T f
= e^{i\theta}S \sum_{n=1}^N c_n W(x_n,\xi_n)f
\end{align*}
and therefore $T = e^{i\theta}S$.
\end{proof}
\begin{proof}[Proof of Proposition \ref{prop:Schrod-irred}]
Reasoning by contradiction, suppose that $Y$ is a nontrivial closed subspace invariant under $W$ and let $Y^\perp$ be its orthogonal complement. The identity $W(x,\xi)^\star = W(-x,-\xi)$ implies that $Y^\perp$ is invariant under $W$ as well. By restriction we thus obtain two strongly continuous unitary representations $W_Y: \H^d\to \calL(Y)$
and $W_{Y^\perp}:\H^d\to \calL(Y^\perp)$, and they satisfy $$W_{Y}(0,0,t) = e^{ it}I_Y, \quad W_{Y^\perp}(0,0,t) =e^{ it} I_{Y^\perp}.$$ Lemma \ref{lem:Schrod-rankone} and Step 1 of the proof of Theorem \ref{thm:Stone-vonNeumann} imply that $W(a_0)$, $W_{Y}(a_0)$ and $W_{Y^\perp}(a_0)$ are orthogonal projections of rank one in $L^2(\R^d.m)$, $Y$, and $Y^\perp$, respectively.
This leads to the contradiction
$$ y_0\,\bar\otimes\, y_0 = W(a_0) = W_{Y}(a_0) + W_{Y^\perp}(a_0),$$
as it represents the rank one projection $y_0\,\bar\otimes\, y_0$ as a sum of two disjoint rank one projections.
\end{proof}
The final result of this section describes the Ornstein--Uhlenbeck semigroup in terms of the Weyl calculus.
Let us first recall some notation from Theorem \ref{thm:Hermite} (where a different normalisation of Lebesgue measure was used). The mapping $E: f\mapsto ef$, where
$$e(x) :=
\exp\Bigl(-\frac14|x|^2\Bigr),$$ is unitary from $L^2(\R^d,\gamma)$ to $L^2(\R^d,m)$, and
the dilation $D: L^2(\R^d,m)\to L^2(\R^d,m)$,
$$ D f(x) := (\sqrt 2)^{d/2} f\bigl({\sqrt 2}x\bigr)$$
is unitary on $L^2(\R^d,m)$. Consequently the operator
$$ U:= D \circ E$$
is unitary from $L^2(\R^d,\gamma)$ to $L^2(\R^d,m)$.
\begin{theorem}\label{thm:exptL} For all $t > 0$ we have,
with $s:= \frac{1-e^{-t}}{1+e^{-t}}$,
\begin{equation*}
OU(t) = (1+s)^d\, U^{-1} \wt W(a_{s})U,
\end{equation*}
where $\wt W$ is the Schr\"odinger representation and $$ a_s(x,\xi):= \exp(-s(|x|^2+|\xi|^2)), \quad x,\xi\in\R^d.$$
\end{theorem}
\begin{proof}
By the definition of $\wt W$ and a change of variables one has
\begin{align*}
\wt W(a)f(y)
= \int_{\R^{2d}}
a\Bigl(\frac12(v+y),\xi\Bigr)\exp(-i\xi(v-y))f(v)\ud m(v)\ud m(\xi).
\end{align*}
By the definition of $U$, this
gives the explicit formula
\begin{align*}
& U^{-1}\wt W(a)U f(y)
\\ & =\int_{\R^{2d}} a\Bigl(\frac12(x+\frac{y}{\sqrt
2}),\xi\Bigr) \\ & \hskip1cm \times \exp\Bigl(-i\xi(x-\frac{y}{\sqrt{2}}
)\Bigr)\exp\Bigl(-\frac12|x|^2+\frac14|y|^2\Bigr)f(x\sqrt{2})\ud m(\xi)\ud m(x)
\\ & =\frac{1}{(\sqrt{2})^{d}}\int_{\R^{2d}} a\Bigl(\frac{x+y}{2\sqrt
2},\xi\Bigr)
\\ & \hskip1cm \times \exp\Bigl(-i\xi(\frac{x-y}{\sqrt{2}}
)\Bigr)\exp\Bigl(-\frac14|x|^2+\frac14|y|^2\Bigr)f(x)\ud m(\xi)\ud m(x)
\\ & = \int _{\R^{d}} K_{a}(y,x)f(x)\ud m(x),
\end{align*}
where
\begin{align*}
\ & K_{a}(y,x) \\ & \ \ = \frac1{(\sqrt 2)^d}{\exp\Bigl(-\frac14|x|^2+\frac14|y|^2\Bigr)}\int_{\R^{d}}
a\Bigl(\frac{x+y}{2\sqrt
2},\xi\Bigr)\exp\Bigl(-i\xi(\frac{x-y}{\sqrt{2}})\Bigr)\ud m(\xi).
\end{align*}
Applying this to the function $a_s$ we obtain
\begin{align*}
& K_{a_s}(y,x) \\ & = \frac1{(\sqrt 2)^d}{\exp\Bigl(-\frac14|x|^2+\frac14|y|^2\Bigr)}
\\ & \hskip1cm \times\int_{\R^{d}}
\exp\Bigl(-s(|\xi|^2+\frac18|x+y|^2)\Bigr)\exp\Bigl(-i\xi(\frac{x-y}{\sqrt{2}})\Bigr)\ud m(\xi)
\\ & = \frac1{(\sqrt 2)^d}{\exp\Bigl(-\frac{s}{8}|x+y|^2\Bigr)\exp\Bigl(-\frac14|x|^2+\frac14|y|^2\Bigr)}
\\ & \hskip1cm \times \int_{\R^d} \exp\Bigl(-s(|\xi|^2
+i\xi(\frac{x-y}{\sqrt 2}))\Bigr) \ud m(\xi)
\\ & = \frac1{(\sqrt 2)^d}{\exp\Bigl(-\frac{1}{8s}|x-y|^2\bigr)\exp\Bigl(-\frac{s}{8}
|x+y|^2\Bigr)\exp\Bigl(-\frac14|x|^2+\frac14|y|^2\Bigr)}
\\ & \hskip1cm \times \int_{\R^d}\! \exp(-s|\eta|^2) \ud m(\eta)
\\ & = \frac1{2^d s^{d/2}} \exp\Bigl(-\frac{1}{8s}|x-y|^2\Bigr)
\exp(-\frac{s}{8}|x+y|^2)\exp\Bigl(-\frac14|x|^2+\frac14|y|^2\Bigr)
\\ & = \frac1{2^d
s^{d/2}}\exp\Bigl(-\frac{1}{8s}(1-s)^2(|x|^2+|y|^2)+\frac{1}{4}(\frac1s -
s)xy\Bigr)\exp\Bigl(-\frac12|x|^2\Bigr).
\end{align*}
Therefore,
\begin{align*}
& U^{-1} \wt W(a_{s})Uf(y)\\ & = \int _{\R^{d}} K_{a_s}(y,x)f(x)\ud m(x)
\\ & = \frac1{2^d (2\pi s)^{d/2}}\!
\int_{\R^d} \exp\Bigl(-\frac{1}{8s}(1-s)^2(|x|^2+|y|^2)+\frac{1}{4}(\frac1s - s)xy\Bigr) f(x)
e^{-\frac12|x|^2}\!\ud x.
\end{align*}
With $s= \frac{1-e^{-t}}{1+e^{-t}}$, this identity implies
\begin{align*}
& (1+s)^d\, U^{-1} \wt W(a_{s})U f(y)
\\ & \ = \frac1{2^d(2\pi)^{d/2}} \Bigl(\frac{2}{1+e^{-t}}\Bigr)^d\Bigl(\frac{1+e^{-t}}{1-e^{-t}}\Bigr)^{d/2}
\\ & \phantom{=}\ \times
\int_{\R^d} \exp\Bigl(-\frac{1}{2}\frac{e^{-2t}}{1-e^{-2t}}(|x|^2+|y|^2)+{\frac{e^{-t}}{1-e^{-2t}}}xy\Bigr) f(x)\exp\Bigl(-\frac12|x|^2\Bigr)\ud x
\\ & \ = \frac1{(2\pi)^{d/2}} \Bigl(\frac{1}{1-e^{-2t}}\Bigr)^{d/2}
\int_{\R^d} \exp\Bigl(-\frac12\dfrac{|e^{-t}y - x|^2}{1- e^{-2 t}} \Bigr)f(x)\ud x
\\ & \ = \int_{\R^d} M_t(y,x)f(x)\ud x = OU(t)f(y),
\end{align*}
where
$$M_t(y,x) = \frac1{(2\pi)^{d/2}} \Bigl(\frac{1}{1-e^{-2t}}\Bigr)^{d/2}
\exp\Bigl(-\frac12\dfrac{|e^{-t}y - x|^2}{1- e^{-2 t}} \Bigr)
$$
is the Mehler kernel;
the last step uses the Mehler formula \eqref{eq:OU-Mehler} for $OU(t)$.
\end{proof}
\section{Second quantisation}\label{sec:SQ}
Up to this point we have been concerned with what physicists call the problem of {\em first quantisation}\index{first quantisation}\index{quantisation!first}, namely, how to define the quantum analogues of classical observables. In order to arrive at a version of Quantum Mechanics that is consistent with Special Relativity, one must be able to describe systems with
a variable number of particles. This is due to the fact that the equivalence of mass and energy makes it possible that particles are created and annihilated. If one uses a Hilbert space $H$ to describe the pure states of a single particle,
one postulates that the $n$-fold Hilbert space tensor product $H^{\hot n}:= H\otimes\cdots\ot H$
describes the pure states of a system of $n$ such particles. As explained in Appendix \ref{sec:tensor} have a direct sum decomposition $H^{\hot n} = \Gamma^n(H)\oplus \Lambda^n(H)$
into symmetric and antisymmetric tensor products; a {\em boson}\index{boson} is a particle whose $n$ particle states are given by elements of $\Gamma^n(H)$ and a {\em fermion}\index{fermion} is a particle whose $n$ particle states are given by elements of $H^{\aot n} = \Lambda^n(H)$. In this section we will discuss the bosonic theory only, the reason being its deep connections with stochastic analysis.
The elements of the Hilbert space direct sum
$$\Gamma(H):= \bigoplus_{n=0}^\infty \Gamma^n(H) $$
correspond to superpositions of states carrying different numbers of bosons. The process of passing from $H$ to $ \bigoplus_{n=0}^\infty \Gamma^n(H)$ is called {\em (bosonic) second quantisation}.\index{second quantisation}\index{quantisation!second}
The elementary observation that every contraction $T$ on $H$ extends to a contraction $\Gamma(T) := \bigoplus_{n=0}^\infty \Gamma^n(T)$ on $\Gamma(H)$
allows us to establish a beautiful connection, for the special case $H = \R^d$, with the Ornstein--Uhlenbeck semigroup discussed in Section \ref{subsec:OU}, namely,
$$ OU(t) = \Gamma(e^{-t}I) .$$
Under this correspondence, the negative generator $-L$ of this semigroup corresponds to the number operator of Section \ref{subsec:number-phase} (where a unitarily equivalent model of it was studied)
which counts the number of photons. Perhaps even more profoundly, our study will uncover a deep connection between second quantisation and the Fourier transform: We will show that the Fourier--Plancherel transform is unitarily equivalent to the second quantisation of the operator $-i I$.
In view of what has just been said, this relates the Fourier--Plancherel transform to the Ornstein--Uhlenbeck semigroup. Some connections of second quantisation with Number Theory will be discussed in the Notes.
For simplicity we will limit our treatment to the case where the Hilbert space describing the pure states of a single particle is $H = \R^d$. {\em Mutatis mutandis}, the theory generalises to arbitrary Hilbert spaces if one replaces the Gaussian measure on $\R^d$ by a so-called $H$-isonormal process, a central object in Malliavin calculus.\index{Malliavin calculus} Pursuing this would take us too far afield, however.
As was already implicit in the above discussion, unless otherwise stated {\em we work over the real scalar field.} This might seem at odds with the formalism of Quantum Mechanics but has certain mathematical advantages. As we will see, the main mathematical results can be complexified afterwards.
\subsection{The Wiener--It\^o chaos decomposition}\label{subsec:WI}
For $h\in \R^d$ we define $\phi_h\in L^2(\R^d,\gamma)$\index{$F$@$\phi_h$} by
$$ \phi_h (x):= x\cdot h = \sum_{j=1}^d x_j h_j , \quad x\in \R^d.$$
Let $(H_n)_{n\in\N}$ be the sequence of Hermite polynomials introduced in Section \ref{subsec:Hermite}.
For $n\in\N$
we define\index{$H$@$\HH_n$} $$\HH_n:=\overline{\hbox{span}}\,\{{H_n}(\phi_h):\, |h|=1\},$$
the closure being taken in $L^2(\R^d,\gamma)$. Here, $({H_n}(\phi_h))(x):= {H_n}(\phi_h(x)) = H_n(x\cdot h)$.
The space $\HH_n$ is sometimes referred to as the
{\em Gaussian chaos} of order $m$.\index{chaos!Gaussian} Note that $\HH_0 = \R\one$ is the one-dimensional space of constant functions.
We are going to prove that the
subspaces $\HH_n$ are pairwise orthogonal and induce a direct sum decomposition
$$\bigoplus_{n=0}^\infty \HH_n = L^2(\R^d,\gamma).$$
In a second step we will identify orthonormal bases for the summands $\HH_n$.
To start our analysis, for $h\in \R^d$ we define the functions\index{$K$@$K_h$}
\begin{align*}K_h := \exp\Bigl(\phi_h-\frac12 |h|^2\Bigr).
\end{align*}
From $\exp(\phi_h)\in L^2(\R^d,\gamma)$ we see that
$K_h$ is well defined as an element of $L^2(\R^d,\gamma)$.
From \eqref{eq:Hermite-gen-fc} we see that
\begin{align}\label{eq:Kh-powerseries}
K_h = \sum_{n=0}^\infty \frac{|h|^n}{n!}{H_n}( \phi_{h/|h|}).
\end{align}
In particular,
\begin{align}\label{eq:Kh-powerseries-1}
K_h = \sum_{n=0}^\infty \frac1{n!}{H_n}(\phi_h), \quad |h|=1.
\end{align}
\begin{lemma}\label{lem:Kh} The functions $K_h$, $h\in \R^d$, span a dense subspace in $L^2(\R^d,\gamma)$.
\end{lemma}
\begin{proof}
Suppose that $f\in L^{2}(\R^d,\gamma)$ is such that $\int_{\R^d}
f K_h\ud\gamma
= 0$ for all $h\in \R^d$. Then $\int_{\R^d}
f \exp(\phi_h) \ud\gamma= 0$ for all $h\in \R^d$. Taking $h:=-\sum_{k=1}^d t_k e_k$, with $e_k$ the $k$-th standard unit vector of $\R^d$, we see that
\begin{align*}\int_{\R^d} f(x)\exp\Bigl(-\sum_{k=1}^d t_k x_k \Bigr)\ud \gamma(x) = 0
\end{align*}
for all $t_1,\dots ,t_d\in\R$. By analytic continuation, the same holds for all $t_1,\dots ,t_d\in\{z\in\C\}$.
Taking $t_k = iy_k$ with $y_k\in \R$, this means that the Fourier transform of the function $x\mapsto f(x)\exp(-\frac12|x|^2)$ vanishes.
By the injectivity of the Fourier transform (Theorem \ref{thm:FT-inversion}) we
conclude that $f(x) \exp(-\frac12|x|^2)=0$ for almost all $x\in\R^d$, that is, $f(x)=0$ for almost all $x\in\R^d$.
\end{proof}
\begin{theorem}[Wiener--It\^o decomposition]\label{thm:WI}
The following orthogonal decomposition holds:
$$L^2(\R^d,\gamma)=\bigoplus_{n=0}^\infty \HH_n.$$
\end{theorem}
\begin{proof}
Fix $h,h'\in\R^d$ with $|h| = |h'| = 1$ and $s,t\in\R$. Repeating the steps of
\eqref{it:Hermite-c},
for all $s,t\in\R$ we have
\begin{equation*}
\begin{aligned}
\ & \int_{-\infty}^\infty H(s,\phi_h(x)) H(t,\phi_{h'}(x)) \ud \gamma(x)
= \exp(st h\cdot h').
\end{aligned}
\end{equation*}
Taking the $(n+m)$-th partial derivative
$\frac{\partial^{n+m}}{\partial s^n \partial t^m}$
at $s=t=0$
on both sides of this identity
we obtain
\begin{align*}
\int_{\R^d} \frac{H_m(\phi_h)}{\sqrt{m!}} \frac{H_n(\phi_{h'})} {\sqrt{n!}}\ud\gamma= \delta_{mn} (h\cdot h')^n.
\end{align*}
For $m\not=n$ this implies $\HH_m \perp\HH_n$.
If $f\perp \HH_n$ for all $n\in\N$, then $\int_{\R^d}f H_n(\phi_h)\ud \gamma = 0$ for all $h\in \R^d$ with $|h|=1$,
and therefore \eqref{eq:Kh-powerseries-1}
implies that
$$ \int_{\R^d} f(x) K_h \ud \gamma =0, \quad h\in \R^d.$$
By Lemma \ref{lem:Kh}, this implies that $f=0$.
\end{proof}
The Wiener--It\^o decomposition diagonalises the Ornstein--Uhlenbeck semigroup on $L^2(\R^d,\gamma)$ introduced in Section \ref{subsec:OU}:
\begin{theorem}\label{thm:OU-diag}
The following identities hold:
\begin{enumerate}[\rm(1)]
\item\label{it:OU-diag1} For all $h\in \R^d$ and $t\ge 0$,
$$ OU(t)K_h = K_{e^{-t}h};$$
\item\label{it:OU-diag2} For all $n\in \N$, $F\in \HH_n$, and $t\ge 0,$ $$OU(t)F = e^{-n t}F.$$
\end{enumerate}
\end{theorem}
\begin{proof}
Completing squares in the exponential, for $y,h\in \R^d$ and $t\ge 0$ we have
\begin{align*} \ & \int_{\R^d}\exp(\sqrt{1-e^{-2t}}y\cdot h)\ud \gamma(y)
\\ & \qquad = \prod_{j=1}^d \frac1{\sqrt{2\pi}}\int_\R \exp\Bigl(\sqrt{1-e^{-2t}}y_j h_j - \frac12|y_j|^2\Bigr)\ud y_j
\\ & \qquad = \prod_{j=1}^d \exp\Bigl(\frac12(1-e^{-2t})|h_j|^2\Bigr)
= \exp\Bigl(\frac12(1-e^{-2t})|h|^2\Bigr).
\end{align*}
Hence, from the definitions of $OU(t)$, $K_h$, and $\phi_h$,
\begin{align*}
OU(t)K_h(x) & = \int_{\R^d} \exp\Bigl(\phi_h(e^{-t}x+ \sqrt{1-e^{-2t}}y)-\frac12 |h|^2\Bigr)\ud \gamma(y)
\\ & = \exp\Bigl(e^{-t}x\cdot h -\frac12 |h|^2\Bigr)\int_{\R^d}\exp(\sqrt{1-e^{-2t}}y\cdot h)\ud \gamma(y)
\\ & = \exp\Bigl(e^{-t}x\cdot h -\frac12 |h|^2\Bigr) \exp\Bigl(\frac12(1-e^{-2t})|h|^2\Bigr)
\\ & = \exp\Bigl(x\cdot e^{-t}h -\frac12 |e^{-t}h|^2\Bigr) \
= K_{e^{-t}h}(x).
\end{align*}
This gives \eqref{it:OU-diag1}. If $|h|=1$ and $s\ge 0$, it follows from \eqref{eq:Kh-powerseries} and \eqref{eq:Kh-powerseries-1} and the preceding calculation that
\begin{align*}
OU(t)\sum_{n=0}^\infty \frac{s^n}{n!}{H_n}( \phi_{h})= OU(t)K_{sh} = K_{e^{-t}sh} = \sum_{n=0}^\infty \frac{s^n e^{-nt}}{n!}{H_n}( \phi_{h}).
\end{align*}
Taking $n$-th derivatives in $s$ and evaluating at $s=0$, we obtain the identity
$$OU(t) {H_n}( \phi_{h}) = e^{-nt}H_n( \phi_{h}).$$ By linearity and taking limits, this gives \eqref{it:OU-diag2}.
\end{proof}
\subsection{The Wiener--It\^o isometry}
Our next aim is to find an orthonormal basis for
each summand $\HH_n$. This will be achieved by means of multivariate Hermite polynomials in Theorem \ref{thm:3.1.9b}.
By dominated convergence we obtain:
\begin{lemma}\label{lem:2.1.1}
The mapping $h\mapsto \exp(\phi_h)$ is continuous from $\R^d$ to
$L^2(\R^d,\gamma)$.
\end{lemma}
For $\bfn = (n_1,\dots,n_d)\in \N^d$ we write $$ |{\bf n}| :=\sum_{i=1}^d n_i, \quad \bfn!:= \prod_{i=1}^d n_i!,$$
and define $\Phi_{\bfn}\in L^2(\R^d,\gamma)$ by
\begin{align*}
H_{\bfn} :=\prod_{i=1}^d H_{ n_i}(\phi_{e_i}),
\end{align*}
where $(e_i)_{i=1}^d$ is the standard unit basis of $\R^d$ (so $\phi_{e_i}(x) = x_i$).
\begin{theorem}\label{thm:3.1.9b}
For each $n\in\N$ the system
$\{\frac1{\sqrt{{\bfn}!}} H_{\bfn}: \, {\bfn}\in {\N^d},\,|{\bfn}|=n\}$
is an orthonormal basis for $\HH_n$. As a consequence,
the family
$\{\frac1{\sqrt{{\bfn}!}} H_{\bfn}: \, {\bfn}\in {\N^d}\}$ is an orthonormal basis for
$L^2(\R^d,\gamma)$.
\end{theorem}
\begin{proof} The proof is divided into three steps.
\smallskip
{\em Step 1} --
First we prove that $\{\frac1{\sqrt{{\bfn}!}} H_{\bfn}: \, {\bfn}\in {\N^d}\}$ is an orthonormal
system in $L^2(\R^d,\gamma)$.
Let ${\bfm},{\bfn}\in {\N^d}$. By separation of variables and the orthonormality of the scaled Hermite polynomials $H_n/\sqrt{n!}$ in one variable,
\begin{align*}\Bigl(\frac1{\sqrt{{\bfn}!}}H_{\bfm}\Big|\frac1{\sqrt{{\bfn}!}} H_{\bfn}\Bigr) &=
\frac1{ \sqrt{{\bfm}! {\bfn}!}}\Bigl( \prod_{i=1}^d H_{ m_i}(\phi_{e_i})\Big|
\prod_{j=1}^d H_{n_j}(\phi_{e_j})\Bigr)
\\ &=\prod_{i=1}^d\prod_{j=1}^d\Bigl(\frac1{\sqrt{m!}}H_{ m_i}(\phi_{e_i})\Big| \frac1{\sqrt{n!}}H_{n_j}(\phi_{e_j})\Bigr)
=\delta_{{\bfm} {\bfn}}.
\end{align*}
\smallskip
{\em Step 2} -- Next we prove completeness of this system in $L^2(\R^d,\gamma)$.
Suppose $f\in L^2(\R^d,\gamma)$ is such that $\iprod{f}{H_{\bfn}}=0$ for all
${\bfn}\in{\N^d}$. Fix an arbitrary $h\in \R^d$ and
put $g_k := \sum_{j=1}^k \frac1{j!} \phi_{h}^j$.
Then $\lim_{k\to\infty} g_{k} =
\exp(\phi_h)$ in $L^2(\R^d,\gamma)$ by Lemma \ref{lem:2.1.1}.
Each $g_{k}$ is a polynomial in
$\phi_{e_{1}},\dots ,\phi_{e_{d}}$. Clearly, any such polynomial is
a linear combination of $H_{\bfn}$'s for appropriate $\bfn\in{\N^d}$.
It follows that $\iprod{f}{g_{k}}=0$ for all $n\ge 1$.
Passing to the limit $n\to\infty$ it follows that also
$\iprod{f}{\exp(\phi_h)}=0$, and therefore $\iprod{f}{K_h}=0$.
Since $h\in \R^d$ was arbitrary, Lemma \ref{lem:Kh} implies that $f=0$.
Together with Step 1, this proves that $\{H_{\bfn}: \, \bfn\in {\N^d}\}$ is an orthonormal basis of
$L^2(\R^d,\gamma)$.
\smallskip
{\em Step 3} --
The final step is to prove that $\{\frac1{\sqrt{{\bfn}!}}H_{\bfn}: \, \bfn\in {\N^d},\,|{\bfn}|=n \}$
is an orthonormal basis for $\HH_n$.
Denote by $\GG_n$ the closed linear span
of the set $\{H_{\bfn}: \, \bfn\in {\N^d},\,|{\bfn}|=n\}$. By Step 1,
$\GG_m\perp \GG_n$ if $m\not=n$.
If $|h|=1$ and $0\le m\le n$, then $H_m(\phi_h)\in \bigoplus_{j=1}^m
\GG_j\subseteq \bigoplus_{j=1}^n
\GG_j$.
Hence $\HH_m\subseteq \bigoplus_{j=1}^n
\GG_j$ and therefore
\begin{align}\label{eq:3.1.9}\bigoplus_{j=1}^n \HH_j\subseteq \bigoplus_{j=1}^n
\GG_j.
\end{align}
Also by Step 1, $${\mathscr H}_n\perp \bigoplus_{j=1}^{n-1}\GG_j.$$
Projecting \eqref{eq:3.1.9} onto $\HH_n$,
it follows that $\HH_n\subseteq\GG_n$.
But then by Step 2,
$$L^2(\R^d,\gamma)= \bigoplus_{n=0}^\infty \HH_n\subseteq \bigoplus_{n=0}^\infty \GG_n=
L^2(\R^d,\gamma),$$
and from this we infer that $\HH_n=\GG_n$ for all $n\in\N$.
\end{proof}
The orthogonal projection in $L^2(\R^d,\gamma)$ onto $\HH_n$ will be denoted by
$J_n$.\index{$J$@$J_n$}
\begin{corollary}\label{cor:3.1.10}
For all $n\in\N$ and $h\in \R^d$ with $|h|=1$ we have
$$J_n(\phi_h^n) = H_n(\phi_h).$$
More generally, if $h_1,\dots,h_k$ are orthonormal in $\R^d$ and ${\bf n} = (n_1,\dots,n_k)\in \N^k$,
then
$$J_{|{\bf n}|}(\phi_{h_1}^{n_1}\cdot\hdots \cdot \phi_{h_k}^{n_k}) =
H_{n_1}(\phi_{h_1})\cdot\hdots \cdot H_{n_k}(\phi_{h_k}).$$
\end{corollary}
\begin{proof}
By Theorem \ref{thm:3.1.9b},
$$H_{n_1}(\phi_{h_1})\cdot\hdots \cdot H_{n_k}(\phi_{h_k})
=\phi_{h_1}^{n_1}\cdot\hdots \cdot \phi_{h_k}^{n_k} + q(h_1,\dots ,h_k),$$
where $q$ is a polynomial of
degree less than $|{\bf n}|$. Hence Theorem \ref{thm:3.1.9b} implies that
$q(h_1,\dots ,h_k)\in \bigoplus_{j=0}^{|{\bf n}|-1} \HH_j$,
and the corollary follows
by projecting onto $\HH_{|{\bf n}|}$.
\end{proof}
\begin{corollary}\label{cor:3.1.11}
For all $h\in \R^d$ with $|h|=1$,
$$K_h= \sum_{n=0}^\infty
\frac{1}{n!} J_n(\phi_h^n).$$
\end{corollary}
\begin{proof}
This follows from the previous corollary and \eqref{eq:Kh-powerseries-1}.
\end{proof}
\begin{theorem}\label{thm:3.1.12}
For all $f_1,\dots ,f_n\in \R^d$ and $g_1,\dots ,g_n\in \R^d$
we have
$$\iprod{J_n(\phi_{f_1}\cdot\hdots \cdot \phi_{f_n})}{J_n(\phi_{g_1}\cdot\hdots \cdot
\phi_{g_n})}
=\!\sum_{\pi\in S_n} \iprod{f_1}{g_{\pi(1)}}\cdot\hdots \cdot \iprod{f_n}{g_{\pi(n)}},$$
where $S_n$ is the group of permutations of $\{1,\dots ,n\}$.
\end{theorem}
\begin{proof}
Choose nonnegative integers
$l_1,\dots ,l_j$ and $m_1,\dots ,m_k$ such that $l_1+\cdots +l_j = m_1+\cdots +m_k=n$.
By adding extra zero's to the smallest of these index sets we may assume that
$j=k$.
By Corollary \ref{cor:3.1.10},
\begin{align*}J_n(\phi_{e_{i_1}}^{l_1}\cdot\hdots \cdot \phi_{e_{i_k}}^{l_k})
& = H_{l_1}(\phi_{e_{i_1}})\cdot\hdots \cdot H_{l_k}(\phi_{e_{i_k}}), \\
J_n(\phi_{e_{i_1}}^{m_1}\cdot\hdots \cdot \phi_{e_{i_k}}^{m_k})
& = H_{m_1}(\phi_{e_{i_1}})\cdot\hdots \cdot H_{m_k}(\phi_{e_{i_k}}).
\end{align*}
Hence,
\begin{align*}
\iprod{J_n (\phi_{e_{i_1}}^{l_1}\cdot\hdots \cdot \phi_{e_{i_k}}^{l_k})}{J_n(\phi_{e_{i_1}}^{m_1}\cdot\hdots \cdot \phi_{e_{i_k}}^{m_k})}
= \bfn!\prod_{j=1}^j\delta_{l_j,m_j}.
\end{align*}
On the other hand, with
\begin{equation}\label{eq:f1dotsfm}
\begin{aligned}
f_1 = \dots = f_{m_1}
:=e_{i_1},\quad & \dots \,,\quad f_{m_1+\cdots+m_{k-1}+1}=\dots =f_{m_1+\cdots+m_k} = e_{i_k},\\
g_1 = \hdots = g_{l_1}
:=e_{i_1},\quad & \dots \,,\quad g_{l_1+\cdots+l_{k-1}+1}=\dots =g_{l_1+\cdots+l_k} =e_{i_k},
\end{aligned}
\end{equation}
we have
\begin{align*}
& \sum_{\pi\in S_n}
\iprod{f_1}{g_{\pi(1)}}\cdot\hdots \cdot \iprod{f_{n}}{g_{\pi(n)})}
\\ & = \sum_{\pi\in S_n}
\iprod{e_{i_1}}{g_{\pi(1)}}\cdot\hdots \cdot \iprod{e_{i_1}}{g_{\pi(l_1)}}\cdot\hdots \\
& \qquad \hdots \cdot
\iprod{e_{i_k}}{g_{\pi(l_1+\hdots +l_{k-1}+1)}}\cdot\hdots \cdot \iprod{e_{i_k}}{g_{\pi(l_1+\hdots +l_{k-1}+l_k)}}
\\ &=m_1!\cdot\hdots \cdot
m_k!\cdot\delta_{m_1,l_1}\cdot\hdots \cdot\delta_{m_k,l_k}\\ &
= \bfn!\cdot\delta_{m_1,l_1}\cdot\hdots \cdot\delta_{m_k,l_k}.
\end{align*}
This proves the corollary in the special case
\eqref{eq:f1dotsfm}.
By $n$-linearity, the corollary then also follows if each of the $f$'s and
$g$'s are finite linear combinations of such expressions, and finally the
general case follows by density.
\end{proof}
The final result of this section relates the spaces $\HH_n$ to the $n$-fold symmetric tensor product of $\R^d$, which we will define next.
The $n$-fold tensor product\index{$R$@$(\R^d)^{\ot n}$}
$$ (\R^d)^{\otimes n} := \underbrace{\R^d \ot\cdots \otimes \R^d }_{n \ \mathrm{ times}},$$
is a real Hilbert space with respect to the inner product
\begin{align*}
\Bigl(\sum_{j=1}^\ell g_1^{(j)}\ot\cdots\ot g_n^{(j)}\Big| \sum_{k=1}^m h_1^{(k)}\ot\cdots\ot h_n^{(k)}
\Bigr)
:= \sum_{j=1}^\ell \sum_{k=1}^m (g_1^{(j)}| h_1^{(k)})\cdot\hdots\cdot (g_n^{(j)}| h_n^{(k)}).
\end{align*}
For $n=0$ we identify $(\R^d)^{\ot 0}$ with the scalar field $\R$.
From Section \ref{sec:tensor} we recall that the {\em $n$-fold symmetric tensor product}\index{symmetric!tensor product}\index{tensor product!symmetric} of $\R^d $, denoted by $\Gamma^n(\R^d)$,\index{$R$@$\Gamma^n(\R^d)$} is defined
as the range of the orthogonal projection $P_\Sigma \in \calL((\R^d)^{\ot
n})$ given by\index{$P$@$P_\Sigma$}
\begin{align*}
P_\Sigma(h_1 \ot \cdots \ot h_n) =
\frac{1}{n!} \sum_{\pi \in S_n} h_{\pi(1)} \ot \cdots \ot
h_{\pi(n)}, \quad h_1, \ldots, h_n \in \R^d ,
\end{align*}
where $S_n$ is the group of permutations of $\{1,\ldots,n\}$,
or equivalently, as the subspace of those elements of $(\R^d)^{\ot n}$ that are invariant
under the action of $S_n$.
For the formulation of the next theorem we introduce the following notation.
Let ${\bf h} = (h_i)_{i=1}^d $ be an orthonormal basis in $\R^d $.
For $\bfn\in \N^d$ let
\begin{align*} {\bf h}_\Sigma^{\bfn}:= P_\Sigma (h_{i_1}^{\ot n_{i_1}} \ot \cdots \ot h_{i_k}^{\ot n_{i_k}}),
\end{align*}
where $\{i_1,\dots,i_k\} = \{1\le i\le d:\, n_i\not = 0\}$. This is well defined since the right-hand side does not depend on the order of the elements in the set $\{i_1,\dots,i_k\}$.
We further define
$$H_{\bfn}(\phi_{\bf h}) :=\prod_{i=1}^d H_{n_i}(\phi_{h_i}).$$
\begin{theorem}[Wiener-It\^o isometry]\label{thm:Wiener-Ito-isometry}\index{theorem!Wiener--It\^o isometry}
For each $n\in \N$ the mapping
\begin{align*}
\Phi_n: {\bf h}_\Sigma^{\bfn}
\mapsto \frac{1}{\sqrt{n!}}H_{\bfn}(\phi_{\bf h}), \quad \bfn\in\N^d,\ |\bfn|=n,
\end{align*}
where ${\bf h} = (h_i)_{i=1}^d$ is any orthonormal basis in $\R^d$,
is well defined and extends by linearity to an isometry $\Gamma^n(\R^d)\eqsim \HH_n$.
\end{theorem}
Taking direct sums, this results in an isometry
$$ \bigoplus_{n=0}^\infty \Gamma^n(\R^d) \eqsim L^2(\R^d,\gamma).$$
\begin{proof} We split the proof into two parts.
\smallskip
{\em Step 1} -- First let ${\bf e} = (e_i)_{i=1}^d $ be the standard basis in $\R^d $
and consider the mapping $W: \Gamma^n(\R^d)\to \HH_n$ defined by
$$W({\bf e}_\Sigma^{\bfn}):=\frac{1}{\sqrt{n!}}H_{\bfn}(\phi_{\bf e}), \quad \bfn\in\N^d,\ |\bfn|=n,$$
and linearity.
To prove well-definedness of the mapping in the statement of the theorem we will show that
$$ W({\bf h}_\Sigma^{\bfn}) = \frac{1}{\sqrt{n!}}H_{\bfn}(\phi_{\bf h}), \quad \bfn\in\N^d,\ |\bfn|=n.$$
Suppose $h_i = \sum_{j=1}^d c_{ij} e_j$, $i=1,\dots,d$,
and let $\bfn\in\N^d$ satisfy $|\bfn|=n$.
Then
\begin{align*}
W({\bf h}_\Sigma^{\bfn})
& = W\left(P_\Sigma \Biggl(\Bigl(\sum_{j=1}^d c_{i_1j}e_j\Bigr)^{\ot n_{i_1}} \ot \cdots \ot \Bigl(\sum_{j=1}^d c_{i_kj}e_j\Bigr)^{\ot n_{i_k}}\Biggr)\right)
\\ & = W\Bigl(P_\Sigma\sum_{|\bfm|=n}^d a_{\bfm} {\bf e}^\bfm\Bigr)
= W\Bigl(\sum_{|\bfm|=n}^d a_{\bfm} {\bf e}_\Sigma^\bfm\Bigr)
\\ & = \sum_{|\bfm|=n}^d a_{\bfm} \frac{1}{\sqrt{n!}}H_{\bfm}({\bf e})
= \frac{1}{\sqrt{n!}}\sum_{|\bfm|=n}^d a_{\bfm} \prod_{i=1}^d H_{m_i}(\phi_{e_i})
\\ & = \frac{1}{\sqrt{n!}}\sum_{|\bfm|=n}^d a_{\bfm} J\Bigl(\prod_{i=1}^k \phi_{e_i}^{m_i}\Bigr)
= \frac{1}{\sqrt{n!}}J_n\Bigl(\sum_{|\bfm|=n}^d a_{\bfm} \prod_{i=1}^k \phi_{e_i}^{m_i}\Bigr)
\\ & = \frac{1}{\sqrt{n!}}J_n\Bigl(\sum_{|\bfm|=n}^d a_{\bfm} \phi_{\bf_{\bf e} }^{\bfm}\Bigr)
= \frac{1}{\sqrt{n!}}J_n\Bigl(\prod_{\ell=1}^k\Bigl(\sum_{j=1}^d c_{i_\ell j}\phi_{e_j}\Bigr)^{n_{i_\ell}} \Bigr)
\\ & = \frac{1}{\sqrt{n!}}J_n\Bigl(\prod_{\ell=1}^k \phi_{h_\ell}^{n_{i_\ell}}\Bigr)
= \frac{1}{\sqrt{n!}}\prod_{i=1}^k H_{n_i}(\phi_{h_i}) = \frac{1}{\sqrt{n!}}H_{\bfn}(\bf h),
\end{align*}
where the coefficients $a_{\bfm}$, $\bfm\in\N^d$, are determined by the identity
$$ \sum_{|\bfm|=n}^d a_{\bfm} \xi^\bfm = \prod_{\ell=1}^k\Bigl(\sum_{j=1}^d c_{i_\ell j}\xi_j\Bigr)^{n_{i_1}}$$
in the formal variables $\xi_1,\dots,\xi_d$ and $\xi^\bfm = \xi_1^{\bf m_1}\cdot\hdots\cdot\xi_d^{m_d}$.
\smallskip
{\em Step 2} -- In this step we show that the mapping is an isometric isomorphism.
First let $h_1,\dots,h_n\in \R^d $ be arbitrary.
By Theorem \ref{thm:3.1.12} we have
\begin{align*}
\n J_n(\phi_{h_1}\cdot\hdots \cdot \phi_{h_n})\n^2
& =\sum_{\pi\in S_n} \iprod{h_1}{h_{\pi(1)}}\cdot\hdots \cdot \iprod{h_n}{h_{\pi(n)}}
\\ & =\sum_{\pi\in S_n} \iprod{h_1\ot\cdots\ot h_n}{ h_{\pi(1)}\ot\cdots\ot h_{\pi(n)}}
\\ & = n! \iprod{h_1\ot\cdots\ot h_n}{ P_\Sigma(h_1 \ot \cdots \ot h_n)}
\\ & = n!\n P_\Sigma(h_1 \ot \cdots \ot h_n)\n^2.
\end{align*}
Specialising to the standard basis $(e_i)_{i=1}^d$ of $\R^d$
and let $${\bf e}_\Sigma^{\bfn}:= P_\Sigma (e_{i_1}^{\ot n_{i_1}} \ot \cdots \ot e_{i_k}^{\ot n_{i_k}}),$$
where $\{i_1,\dots,i_k\} = \{1\le i\le d:\, n_i\not = 0\}$,
using Corollary \ref{cor:3.1.10} we obtain
\begin{align*}
\n {\bf e}_\Sigma^{\bfn} \n & =
\n P_\Sigma(e_{i_1}^{\ot n_{i_1}}\ot\cdots\ot e_{i_k}^{\ot n_{i_k}})\n
= \frac1{\sqrt{n!}} \n J_n (\phi_{e_{i_1}}^{ n_{i_1}}\cdot\hdots\cdot \phi_{e_{i_k}}^{ n_{i_k}})\n
\\ & = \frac1{\sqrt{n!}} \n H_{ n_{i_1}}(\phi_{e_{i_1}})\cdot\hdots\cdot H_{ n_{i_k}}(\phi_{e_{i_k}})\n
= \frac{1}{\sqrt{n!}}\n H_{\bfn}(\phi_{\bf e})\n.
\end{align*}
This identity extends to finite linear combinations by the Pythagorean identity, noting that
both on the left and on the right the contributing terms in the sums are orthogonal.
This proves that the mapping in the statement of the proposition is an isometry.
Since the multivariate Hermite polynomials of degree $n$ form an orthonormal basis in $\HH_n$, this isometry is onto.
\end{proof}
The Wiener--It\^o isometry behaves well under complexification. Indeed, the
functions $H_{ n_i}(\phi_{e_i})\in L^2(\R^d,\gamma)$ can be interpreted as functions in $L^2(\R^d,\gamma;\C)$.
Under these identifications, by redoing the proof of Theorem \ref{thm:Wiener-Ito-isometry}
for complex scalars we obtain:
\begin{theorem}[Wiener-It\^o isometry, complex version]\index{theorem!Wiener--It\^o isometry}\label{thm:WI-compl}
The Wiener--It\^o isometry extends to an isometry
$$ \bigoplus_{n=0}^\infty \Gamma^n(\C^d) \eqsim L^2(\R^d,\gamma;\C).$$
\end{theorem}
\subsection{Second quantisation}
Let $H$ be a real or complex Hilbert space. The completion of the $n$-fold tensor product $H^{\ot n}$
will be denoted by $H^{\hot n}$.\index{$H$@$H^{\hot n}$}
For a bounded operator $T\in \calL(H)$
we define $T^{\ot n}\in \calL(H^{\ot n})$ by
$$ T^{\ot n}(h_1\otimes \cdots\otimes h_n):= (Th_1\otimes \cdots\otimes Th_n).$$
For $n=0$ we understand that $T^{\ot 0} = I_\K$ on $(\R^d)^{\ot 0} = \K$, where $\K$ is the scalar field $\R$ or $\C$.
\begin{proposition}
If $T$ is a contraction on $H$, then $T^{\ot n}$ is a contraction on $H^{\ot n}$ and
$\n T^{\ot n}\n = \n T\n^n$.
\end{proposition}
\begin{proof}
Both statements will follow once we have shown that $\n T\n =1$ implies $\n T^{\ot n}\n = 1$.
The inequality $\n T^{\ot n}\n \ge 1$ being obvious, we concentrate on the inequality $\n T^{\ot n}\n \le 1$.
Let $H_\C$ denote the complexification of $H$. Endowed with the norm $\n h+ih'\n_{H_\C}^2:= \n h\n^2+\n h'\n^2$, the space $H_\C$ is a Hilbert space. If $T$
is a contraction on $H$, then $T_\C(h+ih'):= Th+iTh'$ defines a contraction on $H_\C$ of the same norm.
Hence by Theorem \ref{thm:WI-compl} (and noting that $(T_\C)^{\ot n} = T^{\ot n})_\C$) it suffices to prove the proposition in the case of complex scalars.
If $T$ is a unitary operator on the complex Hilbert space $H$, then
\begin{align*}\Big\n T^{\ot n} \sum_{j=1}^k c_j h_1^{(j)}\otimes \cdots\otimes h_n^{(j)}\Big\n^2
& = \sum_{i=1}^k\sum_{j=1}^k c_ic_j\prod_{m=1}^n \iprod{Th_m^{(i)}}{Th_m^{(j)}}
\\ & = \sum_{i=1}^k\sum_{j=1}^k c_ic_j\prod_{m=1}^n \iprod{h_m^{(i)}}{h_m^{(j)}}
\\ & = \Big\n \sum_{j=1}^k c_j h_1^{(j)}\otimes \cdots\otimes h_n^{(j)}\Big\n^2
\end{align*}
and therefore $T^{\ot n}$ is an isometry on $H$.
The corresponding result for contractions then follows from the fact that every contraction $T$
can be represented as a convex combination of four unitaries by Lemma \ref{lem:sum-of-unitaries}.
\end{proof}
As a consequence of this proposition we see that if $T$ is a contraction on $H$, then\index{$G$@$\Gamma(T)$}
$$ \Gamma(T) := \bigoplus_{n=0}^\infty \Gamma^n(T)$$
defines a contraction on $\bigoplus_{n=0}^\infty \Gamma^n(H)$.
\begin{definition}[Symmetric second quantisation]
The Hilbert space $$\Gamma(H):= \bigoplus_{n=0}^\infty \Gamma^n(H)$$ is called the
{\em symmetric Fock space}\index{symmetric!Fock space} over $H$. When $T$ is a contraction on $H$,
the contraction $\Gamma(T)$ on $\bigoplus_{n=0}^\infty \Gamma^n(T)$ is called the {\em symmetric second quantisation}\index{symmetric!second quantisation} of $T$.
\end{definition}
{\em Antisymmetric second quantisation} can be defined similarly but will not be studied here. Because if this, we will omit the adjective `symmetric' from now on and simply talk about {\em second quantisation}.
If $S$ and $T$ are contractions on $H$, their second quantisations satisfy
$$ \Gamma(I) = I, \quad \Gamma(S) = \Gamma(S)\Gamma(T), \quad \Gamma(T^\star) = \Gamma(T)^\star.$$
Moreover, if $T$ is a contraction on a real Hilbert space, then
$$\Gamma(T_\C) = \Gamma(T)_\C.$$
We now return to the setting where $H = \R^d$ and work in what follows over the real scalar field. If $T$ is a contraction on $\R^d$, via the isometry of Theorem \ref{thm:Wiener-Ito-isometry} the operator $\Gamma(T)$ induces a contraction on $L^2(\R^d,\gamma)$ which, by a slight abuse of notation, will be denoted by $\Gamma(T)$ as well. It is easily checked that the above four identities hold again.
\begin{lemma}\label{lem:Mall-Kh}
If $T$ is a contraction on $\R^d$, then for all orthonormal systems ${\bf h} = (h_j)_{j=1}^k$ in $\R^d$ and $\bfn = (n_1,\dots,n_k)\in\N^d$ with $|\bfn|=n$ we have
$$ \Gamma(T) \prod_{j=1}^k H_{n_j}(\phi_{h_j}) = \prod_{j=1}^k H_{n_j}(\phi_{Th_j}).$$
As a consequence, for all $h\in \R^d$,
$$ \Gamma(T) K_h = K_{Th}.$$
\end{lemma}
\begin{proof}
This is immediate from Theorem \ref{thm:WI} and the definition of $\Gamma(T)$.
\end{proof}
As a special case, for the Ornstein--Uhlenbeck semigroup we obtain:
\begin{theorem}[Ornstein--Uhlenbeck semigroup and second quantisation]\label{thm:OU-2Q} Under the Wiener--It\^o isometry, for all $t\ge 0$ we have $$ OU(t) = \Gamma(e^{-t} I).$$
\end{theorem}
\begin{proof} This follows from Theorem \ref{thm:OU-diag}, where it is shown that $OU(t)K_h = K_{e^{-t}h}$,
and the density of the span of the functions $K_h$ in $L^2(\R^d,\gamma)$, which has been shown in Lemma \ref{lem:Kh}.
\end{proof}
\begin{theorem}[Positivity]\label{thm:sec-quant-pos}
If $T$ is a contraction on $\R^d$, then $\Gamma(T)$
is a positivity preserving contraction on $L^2(\R^d,\gamma)$.
\end{theorem}
Here, {\em positivity preserving} means that $f\ge 0$ implies $\Gamma(T)f\ge 0$.
\begin{proof}
By Lemma \ref{lem:Mall-Kh}, for all $h\in \R^d$ we have $\Gamma(T)K_h = K_{Th}\ge 0$.
Moreover, for $c \in\R$,
\begin{align*}
\Gamma(T)(\exp(c \phi_h))
= \Gamma(T)(\exp(\phi_{ch}))
& = \exp \bigl(\frac12c ^2|h|^2\bigr)\Gamma(T)K_{c h}
\\ & = \exp \bigl(\frac12c ^2|h|^2\bigr)K_{c Th}
\\ & = \exp\Bigl(c \phi_{Th} + \frac12c ^2(|h|^2 -|Th|^2)\Bigr).
\end{align*}
By analytic continuation this identity extends to arbitrary $c \in \C$.
Let $0\le f\in \calF^2(\R^d):= \big\{f\in L^1(\R^d)\cap L^2(\R^d): \ \wh f\in L^1(\R^d)\cap L^2(\R^d)\big\}.$
By Fourier inversion, for almost all $x\in \R^d$ we have
\begin{align*}
\Gamma(T)f (x)&= \frac1{(2\pi)^{d/2}}
\int_{\R^d} \wh f(\xi_1,\dots,\xi_n) \Gamma(T)
\exp\bigl( i \phi_\xi(x)\bigr)\ud\xi
\\ &= \frac1{(2\pi)^{d/2}}
\int_{\R^d} \wh f(\xi_1,\dots,\xi_n)
\exp\bigl( i\phi_{T\xi}(x) -\frac12(|\xi|^2 -|T\xi|^2)\bigr)\ud\xi.
\end{align*}
By dominated convergence, it follows that
$\Gamma(T)f = \lim_{\eps\downarrow 0}
\widecheck{\!F}_\eps$, where
$$F_\eps(\xi) :=\wh f(\xi)
g_\eps(\xi) \quad\hbox{with} \ \
g_\eps(\xi):=\exp\Bigl(-\frac12(|\xi|^2 -(1-\eps)|T\xi|^2)\Bigr).$$
By taking inverse Fourier transforms and Lemma \ref{lem:Gauss},
$ (2\pi)^{d/2}\widecheck{\!F}_\eps = f *\,\widecheck{\!g}_\eps.$
If we can prove that the $\widecheck g_\eps$ is nonnegative almost everywhere,
it follows that $\Gamma(T)f\ge 0$ almost everywhere, it
being limit almost everywhere of convolutions of nonnegative functions.
Since $T$ is a contraction we may write
$$|\xi|^2 -(1-\eps)|T\xi|^2 = \iprod{(I - (1-\eps)T^\star T)\xi}{\xi}= \n (I - (1-\eps)T^\star T)^{1/2}\xi\n^2,$$
where $D_\eps:= (I - (1-\eps)T^\star T)^{1/2}$ is invertible. Hence,
\begin{align*}
\widecheck g_\eps(x) &
= \int_{\R^d} \exp\Bigl(-\frac12\n D_\eps \xi\n^2\Bigr) \exp(- i x\cdot\xi)\ud \xi.
\end{align*}
After a change of variables, the right-hand side can be evaluated as a Fourier transform of a Gaussian and is therefore strictly positive.
\end{proof}
\subsection{The Segal--Plancherel transform}
In this section we discuss a Gaussian analogue of the Fourier--Plancherel transform $\calF$,
the so-called {\em Segal--Plancherel transform}\index{Segal--Plancherel transform}\index{transform!Segal--Plancherel} $\mathscr{W}$ on $L^2(\R^d,\gamma)$.
As before we denote by $$ \ud m(x):= \frac1{(2\pi)^{d/2}}\ud x$$ the normalised Lebesgue measure.
If we reinterpret the Fourier transform as an operator from $L^1(\R^d, m)$ to $L^\infty(\R^d, m)$,
$$ \calF f(\xi) = \int_{\R^d} \exp(-ix\cdot \xi) f(x)\ud m(x), \quad \xi\in \R^d, \ f\in L^1(\R^d, m),$$
its restriction to
$L^1(\R^d,m)\cap L^2(\R^d,m)$ extends to an isometry on $L^2(\R^d,m)$. In the present section, the term Fourier--Plancherel transform will refer to this operator.
As in Section \ref{subsec:StonevonNeumann} we let $ U:= D\circ E$, where $D: L^2(\R^d,m)\to L^2(\R^d,m)$ and $E:L^2(\R^d,\gamma)\to L^2(\R^d,m)$
are the unitary operators given by
\begin{align*} Df(x) &:= (\sqrt 2)^{d/2} f\bigl({\sqrt 2}x\bigr), \\ Ef(x) &:= e(x)f(x),
\end{align*}
with $e(x) :=\exp(-\frac14|x|^2).$
\begin{theorem}\label{thm:Segal}
The mapping $\mathscr{W}: f\mapsto \mathscr{W}f$, defined for multivariate polynomials $f:\R^d\to\C$ by
$$ \mathscr{W} f(x):= \int_{\R^d} f(-ix+\sqrt 2 y)\ud \gamma(y), \quad x \in \R^{d},
$$
extends to a unitary operator on $L^2(\R^d,\gamma;\C)$ and we have
\begin{align}\label{eq:FWT}
\mathscr{W} = U^{-1} \circ \calF \circ U.
\end{align}
\end{theorem}
\begin{proof}
Since $D$, $E$, and $\calF$ are unitary, the unitarity of $\mathscr{W}$ will follow from
the operator identity \eqref{eq:FWT}. To prove this identity,
we substitute $\eta = \sqrt 2 y$ and $\xi=-ix+ \eta$ to obtain
\begin{align*}\mathscr{W}f(x)
& = \frac1{(2\pi)^{d/2}} \int_{\R^d} f(-ix+ \sqrt 2 y)\exp\Bigl(-\frac12|y|^2\Bigr)\ud y
\\ & = \frac1{(4\pi)^{d/2}} \int_{\R^d} f(-ix+ \eta)\prod_{k=1}^d\exp\Bigl(-\frac14\eta_k^2\Bigr)\ud \eta
\\ & = \frac1{(4\pi)^{d/2}} \int_{-ix+\R^d} f(\xi)\prod_{k=1}^d \exp\Bigl(-\frac14(\xi_k +ix_k)^2\Bigr)\ud \xi
\\ & \stackrel{(*)}{=} \frac1{(4\pi)^{d/2}} \int_{\R^d} f(\xi)\prod_{k=1}^d \exp\Bigl(-\frac14(\xi_k+ix_k)^2\Bigr)\ud \xi
\\ & = \frac1{(4\pi)^{d/2}} \int_{\R^d} f(\xi) \exp\Bigl(-\frac14(|\xi|^2 - \frac12i\xi\cdot x + \frac14|x|^2)\Bigr)\ud \xi.\end{align*}
To justify $(*)$ it suffices, by writing $f$ as a linear combination of monomials and
separating variables, to show that for any $k\in \N$ and $x\in\R$,
$$ \int_{-ix+\R} \xi^k\exp\Bigl(-\frac14(\xi+ix)^2\Bigr)\ud \xi= \int_{\R} \xi^k\exp\Bigl(-\frac14(\xi+ix)^2\Bigr)\ud \xi.
$$
But this is clear by Cauchy's integral formula and a limiting argument using the decay at infinity.
Hence,
\begin{align*}
& (E \circ\mathscr{W}\circ E^{-1})f (x)
\\ &\ = \frac{1}{(4\pi)^{d/2}}\! \exp\Bigl(-\frac14|x|^2\Bigr)
\\ & \qquad\qquad \times\int_{\R^d} \exp\Bigl(\frac14|\xi|^2\Bigr)f(\xi)
\exp\Bigl(-\frac14|\xi|^2 -\frac12i\xi\cdot x + \frac14|x|^2\Bigr)\ud \xi
\\ & \ = \frac1{(4\pi)^{d/2} }\int_{\R^d} f(\xi) \exp\Bigl(-\frac12 i\xi\cdot x\Bigr)\ud \xi
\\ & \ = \int_{\R^d} 2^{d/2}f(2\xi) \exp(- i\xi\cdot x)\ud m(\xi)
\\ & \ = (\calF \circ D^{2})f(x) = (D^{-1}\circ\calF \circ D)f(x),
\end{align*}
where we used that $\calF \circ D = D^{-1}\circ\calF$.
\end{proof}
We have the following representation of $\mathscr{W}$ in terms of second quantisation:
\begin{theorem}[Segal]\label{thm:WT}\index{theorem!Segal} On $L^2(\R^d,\gamma)$ we have
$\mathscr{W} = \Gamma(-i I)$.
\end{theorem}
\begin{proof} Let $f:\R^d\to\C$ be a multivariate polynomial.
Mehler's formula and Theorem \ref{thm:OU-2Q} tell us that for all $t>0$,
\begin{align*}
\Gamma(e^{-t}I)f(x) = OU(t)f(x) = \int_{\R^d} f(e^{-t}x + \sqrt{1-e^{-2t}}y)\ud \gamma(y).
\end{align*}
By analytic continuation we may replace $t>0$ by any $z\in \C$ with $\Re z>0$. Now let $z\to \frac12\pi i$.
\end{proof}
\begin{corollary} The Fourier--Plancherel transform is unitarily equivalent to $\Gamma(-i I)$.
\end{corollary}
This gives a neat ``explanation'' of the identity $\calF^4 = I$: by the multiplicativity of second quantisation it follows from the identity $(-i)^4 =1$!
\subsection{Creation and annihilation}\label{subsec:annih-creat}
Using Theorem \ref{thm:WI-compl} we will identify $L^2(\R^d,\gamma;\C)$ with the symmetric Fock space
$$\Gamma(\C^d) := \bigoplus_{n=0}^\infty \Gamma^n(\C^d)$$
via the Wiener-It\^o isometry (Theorem \ref{thm:Wiener-Ito-isometry}). Here $\Gamma^n(\C^d)$
denotes the $n$-fold symmetric tensor product of $\C^d$. To shorten notation we write $H = \R^d$ and $H_\C = \C^d$.
For $h\in H_\C$ the {\em creation operator}
$a^\dagger(h): \Gamma^n(H_\C)\to \Gamma^{n+1}(H_\C)$ is defined by
$$
\begin{aligned}
\ & a_n^\dagger(h)\sum_{\sigma\in S_n} g_{\sigma(1)}\otimes\cdots
\otimes g_{\sigma(n)}
\\ & := \frac{1}{\sqrt{n+1}}
\sum_{\sigma\in S_n} \sum_{\ell=1}^{n+1} g_{\sigma(1)}
\otimes\cdots\otimes g_{\sigma(\ell-1)}\otimes h\otimes g_{\sigma(\ell)}\otimes \cdots \otimes g_{\sigma(n)},
\end{aligned}
$$
and the {\em annihilation operator}
$a(h): \Gamma^{n+1}(H_\C)\to \Gamma^n(H_\C)$ by
$$
\begin{aligned}
\ & a_{n+1}(h)\sum_{\sigma\in S_{n+1}} g_{\sigma(1)}\otimes\cdots
\otimes g_{\sigma(n+1)}
\\ & := \frac{1}{\sqrt{n+1}}
\sum_{\sigma\in S_{n+1}} \sum_{\ell=1}^{n+1} \iprod{g_{\sigma(\ell)}}{h} \
g_{\sigma(1)}\otimes\cdots\otimes \wh{g_{\sigma(\ell)}}\otimes\cdots \otimes
g_{\sigma(n+1)}
\end{aligned}
$$
where $\ \wh{}\ $ signifies that this term is omitted.
These operators are well defined and bounded, and their operator norms
are bound\-ed by
\begin{equation*}
\n a_n^\dagger\n_{\calL(\Gamma^n(H_\C), \Gamma^{n+1}(H_\C)}
= \n a_{n+1}(h)\n_{\calL(\Gamma^{n+1}(H_\C), \Gamma^n(H_\C))} \le C_n |h|
\end{equation*}
with constants $C_n$ depending on $n$ only.
The first equality follows from the duality relations
\begin{equation*}
a_{n}^{\dagger \star}(h) = a_{n+1}(h).
\end{equation*}
Furthermore, we have the commutation relations
\begin{equation*}
a_{n+2}(h) a_{n+1}^{\dagger}(h) - a_{n}^{\dagger}(h)a_{n+1}(h) =
|h|^2 I.
\end{equation*}
Extending the operators $a^\dagger(h)$ and $a(h)$ by linearity to the span of the spaces
$\Gamma^n(H_\C)$, $n\in\N$, and $\Gamma^{n+1}(H_\C)$, $n\in\N$, the resulting operators are easily seen to be densely defined and closable. In what follows, we use the same notation for their closures.
The Wiener-It\^o isometry allows us to transfer them to densely defined closed
operators in $L^2(\R^d,\gamma)$.
Let $(e_j)_{j=1}^d$ denote the standard basis of $H_\C$ and write
$$ a_j := a(e_j), \quad a_j^\dagger = a^\dagger(e_j); \quad j=1,\dots,d.$$
The operators $$A: f\mapsto (a_1 f, \dots, a_d f)$$ and
$$A^\dagger: g\mapsto a_1^\dagger g_1+\cdots+a_d^\dagger g_d$$
are densely defined and closed as operators from $L^2(\R^d,\gamma)$ to $L^2(\R^d,\gamma;\C^d)$
and from $L^2(\R^d,\gamma;\C^d)$ to $L^2(\R^d,\gamma)$, respectively.
The operators $a_j(h)$ and $a_j^\dagger(h)$ are dual to each other for each $j=1,\dots,d$, and so are the operators $A$ and $A^\dagger$.
Let $\nabla = (\partial_1,\dots,\partial_d)$ be the gradient, viewed as a densely defined closed operator from
$L^2(\R^d,\gamma)$ to $L^2(\R^d,\gamma;\C^d)$, and let $\nabla^\star$ denote its adjoint.
\begin{proposition}\label{prop:grad} $A = \nabla $ and $A^\dagger = \nabla^\star$.
\end{proposition}
\begin{proof}
We only need to prove the first identity, as the second follows from it.
For $j=1,\dots,d$ let $\pi_j(x) := x_j$ be the coordinate projections.
For the exponential functions $$\eta_k(x) := \exp\Bigl(\pi_k(x) - \frac12|x|^2\Bigr), \quad x\in \R^d,$$
it is easily checked that
we have
\begin{align*} \iprod{ A \eta_k}{e_j} = a_j \eta_k = \delta_{jk} \eta_k.
\end{align*}
On the other hand,
\begin{align*}
\iprod{ \nabla\eta_k}{ e_j} = \partial_j \eta_k & = \frac{{\rm d}}{{\rm d}t} \exp\Bigl(\pi_k(\cdot+te_j) - \frac12|\cdot|^2\Bigl)\Big|_{t=0}
\\ & = \delta_{jk} \exp\Bigl(\pi_k(\cdot) - \frac12|\cdot|^2\Bigr)= \delta_{jk} \eta_k.
\end{align*}
Since the exponential functions are dense in $\Dom(A)$ and $\Dom(\nabla)$ with their graph norms, this gives the result.
\end{proof}
Define the {\em position operator}\index{position operator}\index{operator!position} $Q = (q_1,\dots,q_d)$ by
\begin{align*}
q_j :=\frac1{\sqrt2}(a_j + a^\dagger_j) .
\end{align*}
In the light of the next claim,
the choice of the normalising constant $1/\sqrt 2$ may appear unnatural.
The reason for this choice will become apparent in \eqref{reason:1}, \eqref{reason:2}, \eqref{reason:3}.
We claim that for
all $j=1,\dots,d$ we have, for almost all $x \in \R^d$,
$$q_jf(x) = \frac1{\sqrt 2} x_j f(x).$$
Indeed, since by Proposition \ref{prop:grad} we have $a_j = \partial_j$, the directional derivative in the direction of $e_j$, it follows that
$$ \sqrt{2}\iprod{ q_jf}{g} = \iprod{ f}{ \partial_j g} + \iprod{ \partial_j f}{g}.$$
Suppose now that $f,g \in C_{\rm c}^1(\R^d)$ are test functions.
Then, with $x = (x_1,\dots,x_d)$,
\begin{align*}\iprod{\partial_j^\star f}{g} = \int_{\R^d} f\ov{\partial_j g}\ud \gamma
& = \frac1{(2\pi)^{d/2}} \int_{\R^d} f(x) \ov{\partial_j g(x)}\exp(-\frac12|x|^2)\ud x
\\ & = \frac1{(2\pi)^{d/2}}\int_{\R^d} [x_jf(x)-\partial_j f(x)] \ov{g(x)}\exp(-\frac12|x|^2)\ud x
\\ & = -\iprod{\partial_j f}{g} + \iprod{ x_jf}{g},\end{align*}
It follows that $(\partial_j f+\partial_j^\star )f (x)= (x\cdot h ) f(x)$.
In a similar way we define the {\em momentum operator}\index{momentum operator}\index{operator!momentum} $P = (p_1,\dots,p_j)$ by
\begin{align*}
p_j :=\frac1{i\sqrt{2}}(a_j - a_j^\dagger).
\end{align*}
From the commutation relation $[a_j,a_j^\dagger] = I$ we have
\begin{equation}\label{reason:1}
\begin{aligned}
\ [p_j,q_j] & = \frac1{2i}\Bigl((a_j - a^\dagger_j)(a_j +a^\dagger_j) - (a_j + a^\dagger_j)(a_j - a^\dagger_j)\Bigr)
\\ & = \frac1{i}(a_ja^\dagger_j - a^\dagger_ja_j ) = \frac1{i}I.
\end{aligned}
\end{equation}
and the identity
\begin{equation}\label{reason:2}
\begin{aligned}
\frac12(p_j^2 + q_j^2)
& = -\frac14(a_j^2 - a_ja_j^\dagger - a_j^\dagger a_j + a_j^{\dagger2}) + \frac14 (a_j^2 + a_ja_j^\dagger + a_j^\dagger a_j + a_j^{\dagger2})
\\ & = \frac12(a_j^\dagger a_j + a_ja_j^\dagger)
= a_j^\dagger a_j + \frac12[a_j,a_j^\dagger]
= a_j^\dagger a_j + \frac12 I.
\end{aligned}
\end{equation}
As is checked by an easy computation, in terms of the annihilation and creation operators, the Ornstein--Uhlenbeck operator
is given by
\begin{align}\label{eq:QHO1} - L & = \nabla^\star\nabla = A^\dagger A = \sum_{j=1}^d a^\dagger_ja_j,
\intertext{so that by \eqref{reason:2},}
\label{eq:QHO2}
-L & = -\frac{d}{2} + \frac12(P^2+Q^2) = -\frac{d}{2} + \sum_{j=1}^d \frac12(p_j^2+q_j^2)
\end{align}
The operators $P$ and $Q$ intertwine with the momentum operator $D = \frac1i \nabla$ and the position operator $X$, in the sense that
\begin{equation}\label{reason:3}
\begin{aligned}
U \circ q_j \circ U ^{-1} & = x_j,\\
U \circ p_j \circ U ^{-1} & = \frac1{i} \partial_j,
\end{aligned}
\end{equation}
with $U$ as before. These relations are easy to check by explicit computation and justify the terminology `position' and `momentum' for $q_j$ and $p_j$.
\begin{problems}
\item
Prove the assertions about orthogonal projections in Section \ref{subsec:class-quantum}.
\item\label{prob:orthomodular}
Prove that if $P$ and $Q$ are orthogonal projection on a Hilbert space such that $\ran(P)$ is contained in $\Ran(Q)$, then $ Q = P \vee (Q \wedge \neg P).$
\item
Give an alternative proof of Proposition \ref{prop:observ-mapping} based on the equivalence of states and positive trace class operators with unit trace.
\noindent{\em Hint:}\ One may use Theorem \ref{thm:tc-ell1} to
write $T = \sum_{n\ge 1} \la_n h_n\,\bar\otimes\, h_n$
with $h_1,h_2,\dots$ an orthonormal basis and $\la_1,\la_2,\hdots \ge 0$ such that $\sum_{n\ge 1} \la_n = 1$.
\item
Consider a qubit in state $a\ket{0} + b\ket{1}$, where $a,b\in\C $ satisfy $|a|^2+|b|^2 =1$. Compute the probabilities
that upon measuring the spin in direction $j\in \{1,2,3\}$ we find `up', respectively `down'.
\item
Prove that if $U$ is a symmetry of a complex Hilbert space $H$, then
the mapping $\tau_U:\calP(H)\to\calP(H)$ given by $\tau_U(P) := U^\star PU$ enjoys the following properties:
\begin{enumerate}[\rm(i)]
\item\label{it:tau-symm1a} $\tau_U(I)=I$;
\item\label{it:tau-symm1b} for all $P\in \calP(H)$ we have $\tau_U(\neg P) = \neg \tau_U(P);$
\item\label{it:tau-symm2} for all $P_1,P_2\in \calP(H)$ we have
\begin{align*}
\tau_U(P_1\wedge P_2) &= \tau_U(P_1)\wedge \tau_U(P_2);
\\ \tau_U(P_1\vee P_2) &= \tau_U(P_1)\vee \tau_U(P_2).
\end{align*}
\end{enumerate}
\item
Show that position and momentum are covariant with respect to rotations $R_\rho$ on $L^2(\R^d,m)$ given by $ R_\rho f(x) = f(\rho^{-1} x)$, where $\rho\in SO(d)$, the group of orthogonal transformations on $\R^d$ with determinant $1$.
\item
For the group $G=\Z/2\Z$, determine the position and momentum operators on $L^2(G)\simeq \C^2$.
\item
Find the projection-valued measures associated with
the self-adjoint operators $\wh x_j$ and $\wh \xi_j$ discussed in Section \ref{subsec:pos-mom}.
\item
In this problem we prove {\em Wintner's theorem}:\index{theorem!Wintner}
There exist no pair of {\em bounded} operators $S,T\in \calL(H)$ satisfying the Heisenberg commutation relation $ST-TS = iI$. We may absorb the imaginary constant $i$ into one of the two operators and consider the identity $ST-TS = I$ instead. Assuming that $S,T\in \calL(H)$ satisfy $ST-TS = I$, obtain a contradiction by completing the following steps.
\begin{enumerate}[\rm(a)]
\item Show that for all $n=1,2,\hdots$ we have $S^n T -TS^n = n S^{n-1}$.
\item Deduce that $S^{n-1}\not=0$ and $n\n S^{n-1}\n \le2\n S^{n-1}\n \n S\n\n T\n.$
\end{enumerate}
\item\label{prob:phi-state}
Prove that for a linear mapping $\phi: \calL(H)\to \C$ the following assertions are equivalent:
\begin{enumerate}[\rm(1)]
\item\label{it:lem:weakoperator1} there exist sequences $(x_n)_{n\ge 1}$ and $(y_n)_{n\ge 1}$ satisfying
$$\sum_{n\ge 1} \n x_n\n^2<\infty \ \hbox{ and } \ \sum_{n\ge 1} \n y_n\n^2<\infty$$ such that for all $T\in \calL(H)$ we have
$$\phi(T) = \sum_{\ge 1} \iprod{Tx_n}{y_n};$$
\item\label{it:lem:weakoperator3} $\phi$ is continuous on $\ov B_{\calL(H)}$ with respect to the weak topology of $\calL(H)$;
\item\label{it:lem:weakoperator2} $\phi$ is continuous on $\ov B_{\calL(H)}$ with respect to the strong topology of $\calL(H)$;
\item $\phi$ is normal.
\end{enumerate}
If $\phi$ is positive and satisfies $\phi(I)=1$, these conditions are equivalent to:
\begin{enumerate}[\rm(1)]\setcounter{enumii}{4}
\item there exists an orthogonal sequence $(x_n)_{n\ge 1}$ satisfying $\sum_{n\ge 1} \n x_n\n^2=1$ such that for all $T\in \calL(H)$ we have
$$\phi(T) = \sum_{\ge 1} \iprod{Tx_n}{x_n}.$$
\end{enumerate}
\item\label{prob:angle}
Using the bounded functional calculus for projection-valued measures on $\T$ we may define
$$\wh\theta := \int_{\mathbb{T}} \arg(z)\ud \Theta(z),$$
where $\Theta:\calB(\T)\to\PP(L^2(\T))$ is the projection-valued measure defining the angle observable.
There is some ambiguity here as to how to take the argument; for the sake of definiteness we take it in $(-\pi,\pi]$.
\begin{enumerate}[\rm(a)]
\item Show that $\wh\theta$ is bounded and self-adjoint on $L^2(\T)$.
\item Show that for all $f,g\in L^2(\T)$ we have
\begin{align*} \iprod{\wh\theta f}{g} = \frac1{2\pi} \int_{-\pi}^\pi \theta f(e^{i\theta}) \ov{g(e^{i\theta})}\ud \theta.
\end{align*}
\end{enumerate}
Define the {\em angular momentum operator}\index{operator!angular momentum}
$$ \wh l := \sum_{n\in\Z} n L(\{n\}).$$
\begin{enumerate}[\rm(a)]\setcounter{enumi}{2}
\item Show that, with an appropriate choice of domain, this operator is self-adjoint on $L^2(\T)$.
\item Prove that $\wh\theta$ and $\wh l$ satisfy the Heisenberg commutation relation
$$\wh l\,\wh\theta - \wh\theta \,\wh l = iI$$
on $\Dom(\wh l\,\wh \theta)\cap \Dom(\wh\theta\, \wh l)$ and show that this domain is dense in $L^2(\T)$.
\end{enumerate}
Encouraging as this may seem, the operator $\wh\theta$ appears to be of little use in Physics. This is related to
the failure of the `continuous variable' Weyl commutation relation for $\wh\theta$ and $\wh l$:
\begin{enumerate}[\rm(a)]\setcounter{enumi}{4}
\item Show that there exists no bounded operator $T$ such that the following identity holds for all $s,t\in \R$:
\begin{align}\label{eq:Weyl-phase}
e^{isT}e^{it\wh l} = e^{ist} e^{it\wh l}e^{isT}.
\end{align}
Show that the same conclusion holds if we assume that $T$ is a (possibly unbounded) self-adjoint operator.
\noindent {\em Hint:}\ Show that if an $s\in \R$ exists such that the identity in \eqref{eq:Weyl-phase} holds for all $t\in\R$, then $s \in \Z$.
\item Prove a similar result for the phase operator of Section \ref{subsec:number-phase}.
\end{enumerate}
\item
Show that if $T$ is a contraction on $\R^d$, then for every $1\le p<\infty$
the second quantised operator $\Gamma(T)$ extends to a contraction on $L^p(\R^d,\gamma)$.
\item
Show that if $U$ is unitary on $\C^d$, then for all
$f\in L^2(\R^d,\gamma;\C)$ we have $$\Gamma_\C(U)f(x) = f(U^\star x)$$ for almost all $x\in \R^d$.
\item
Show that the position and momentum operators $q_j$ and $p_j$ introduced in Section \ref{subsec:annih-creat} satisfy the relations
\begin{align*}
q_j\circ \mathscr{W} = \mathscr{W} \circ p_j,\\
p_j\circ \mathscr{W} = -\mathscr{W}\circ q_j,
\end{align*}
consistent (modulo the difference in normalisations of the Fourier transform) with the relations $x_j\circ \calF = \calF\circ (\frac1i \partial_j)$ and $(\frac1i \partial_j)\circ \calF = -\calF \circ x_j$ for position and momentum operators of Section \ref{subsec:pos-mom}.
\end{problems}
\endinput
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TITLE: Proving arg(z/w)=arg(z)-arg(w)
QUESTION [2 upvotes]: I need to prove that
$$arg\left(\frac{z}{w}\right)=arg(z)-arg(w)$$
However, I am a little stuck as to how to go about this.
I know the proof for $arg(zw)=arg(z)+arg(w)$ happens by letting $z=r(cos\theta+isin\theta)$ and $w=s(cos\phi+isin\phi)$ and then multiplying them together and expanding out the brackets, combining the arguments using the double angle formulae.
I have been told that the proof of this is analogous, but I have absolutely no idea how I would simplify
$$\frac{r(cos\theta+isin\theta)}{s(cos\phi+isin\phi)}$$
Is there some identity I don't know about?
EDIT: We have not covered $e^{ix}=cos(x)+isin(x)$ yet, so the proof will not need to use that fact.
REPLY [6 votes]: You could multiply by $\frac{\cos\phi-i\sin\phi}{\cos\phi-i\sin\phi}$ and use:
$$
\sin(x-y) = \sin x \cos y - \cos x \sin y
$$
$$
\cos(x-y) = \cos x \cos y + \sin x \sin y
$$
REPLY [1 votes]: If we interpret $\arg$ as the imaginary part of a branch of the logarithm function, then this is not true: If we choose the branch so that $\arg \zeta \in [0, 2 \pi)$, then
$$\arg\left(\frac{1}{-1}\right) = \arg(-1) = \pi$$
but
$$\arg 1 - \arg (-1) = 0 - \pi = - \pi.$$
That said, the identity can hold for certain choices of branch and restricted values of $z, w$ (for example if we choose the branch so that $\arg \zeta \in [-\pi, \pi)$ and restrict to $z, w \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right)$).
It's anyway more natural (though often not more convenient) to view the argument as a function that takes values modulo $2\pi$ (equivalently, as elements of $\mathbb{S^1}$), which corresponds to the fact that the above argument function $\arg$ of any branch of $\log$ (with domain $\mathbb{C} - \{0\}$ anyway) satisfies.
$$\arg \frac{z}{w} \equiv (\arg z - \arg w) \bmod 2\pi.$$
REPLY [0 votes]: What you need to use is not an identity but (it can be called at most) a method: You multiply both the numerator and the denominator by the complex conjugate of the latter one:
$\dfrac{a+b\imath}{c+d\imath}=\dfrac{(a+b\imath)(c-d\imath)}{c^2+d^2}$
Using this you will get the cosine/sine of the difference of angles formulae.
Observe that this is actually the way to find a simpler form of the (multiplicative) inverse of a nonzero complex number.
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French.
That biker wasn’t even trying, but Rollerman is still very cool. Totally insane, but very cool.
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We chose the Marriott Myrtle Beach Resort & Spa at Grande Dunes for our home away from home during the Baseball Youth Nationals! It was a lovely resort, located directly on the beach in the quieter North Myrtle Beach area.
The pool and beach areas were just lovely and had a wonderful resort feel. We spent almost all of our time off the field here, not having much desire to leave outside of team events!
There was a great waterslide, ping pong, all the beach things and two large pools with plenty of seating. The boys and adults alike had plenty to do (or not do, for the adults).
I think I heard “meet me in the game room” approximately 497 times during the week. The boys would literally finish a game, wave their hats to the other team and say “meet me in the game room”! They never got tired. It was located steps from the pool, which was steps from the beach, and the kids easily moved from one thing to the next.
As for me, I couldn’t go down to the beach (no sand in the wound!) and couldn’t swim. But, I managed to eek out a good relaxing time in spite of that. I found shady chairs on the pool deck and tore through some books.
The other parents graciously took Whit to the beach when his teammates headed down. I did take a few peeks from the safety of the ramp. :-)
We ended up getting an ocean view room (booked resort view despite my own best advise to every client – if you’re going to the beach you’re going to want to see and hear the ocean), so this was a welcome sight day in and day out.
Aside from the resort life, we headed out for a few Myrtle Beach-y things with the team. Go Karts are not on my usual vacation agenda, but he LOVED them! :-)
Unlimited putt putt? Yes please, said every 12 year old boy ever!
As for food, well, Whit and I love a good meal. We found a few gems during our stay, mostly in the North Myrtle Beach area. We enjoyed a delicious brunch at Blueberry’s Grill. His Dublin omelet was AMAZING! Oh, and we both loved the unique blueberry hush puppies.
Croissants was right at the entrance to our resort and was a lovely pit stop for an afternoon coffee and lemon cookie! Both were excellent.
Cafe Gelato was one of our favorites! Their chicken salad was divine. I honed my order to the half and half – salad with scoop of chicken salad and the balsamic fig dressing. YUM! They also have gelato, which was delicious. We enjoyed dining in, but on one rainy day just grabbed our lunch in the drive through to enjoy on our balcony.
Whit loves sushi and pho, and “googled up” CO Sushi in Market Common, which was steps from the baseball fields. This was a really cute area that we dined in a couple of times. The food here was fine and it made for a yummy lunch after playing ball.
Our team dined at Nacho Hippo, also in Market Common, one evening. They have a back room great for a large party. The food here was delicious, as were the margs.
Whit and I did a couple dinners off on our own as well. One favorite was Sea Captain’s Quarters. We had a reservation for inside, but when we arrived and saw the outdoor seating and live country music we asked to change. We both LOVED our meal here. The hush puppies with honey butter reminded me of Pearl’s growing up in Savannah. I had the shrimp and grits, which photographed as well as slop but were some of the best I’ve tasted. Whit had the bbq fried chicken and a side of grits and said his chicken was excellent and agreed that the grits were the “best ever”.
Our other favorite “date night” was Hook & Barrel in North Myrtle just outside of our resort. We dined on the porch under palms and beside hydrangeas. It was lovely! The food was excellent and I fully appreciated little details like the South Carolina shaped board the starter of pimento cheese and lavash crackers was served on. I enjoyed the scallops and Whit had the shrimp alfredo. For dessert we splurged on this INCREDIBLE dish of a chocolate bowl filled with salted caramel ice cream, drizzled with chocolate and caramel, and topped the crumbles of their chocolate chip skillet cookie. It was memorable!
There were many parts of Myrtle Beach we didn’t see and there were tons of touristy things we didn’t do. The things we did do, see and eat were perfect for the two of us and we had a fun, enjoyable week together. I treasure time with that baby of mine.
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7 thoughts on “Myrtle Beach {Stay, See, Eat, Do}”
Looks like the boys had a great time at the Baseball Nationals and what better place to be at than the beach. Sorry you didn’t get to enjoy the amazing pools and beach at the resort. I know you will be beyond excited to finally be able to get back to water activities. We haven’t been to Myrtle Beach in a while since we typically prefer Hilton Head but when we go we always choose the North Myrtle Beach area as well as its a little less crowded and the Sea Captains House is one of our favorite restaurants.
Long-time follower..and I was shocked at how grown-up Whit looks in these pics!! Enjoy all of your posts..and enjoy all of your posts. I travel through MB for work and will make stops at some of those lunch places you suggested. Have a happy summer!
He has grown up SO much in the past year!
Oh my, the resort looks amazing and your culinary adventures sound so fun! I spent many happy summers in the MB area, some February school vacations as well. There sure are loads of dining options and plenty to do. And, let us not forget the golfers for whom Myrtle Beach is a mecca. So glad that you and Whit had such a memorable time to kick off the summer. (Hope the wound is healing well and that you will be in the pool and at the beach very soon.)
Your doc is lucky you’re such a good rule follower. I would have been bandaging up that wound with some sort of plastic barrier and hanging out on the beach FOR SURE, haha.
Glad this post didn’t happen before my husband worked in Myrtle Beach without me, I would have had even more FOMO.
I want to get past this so badly that I literally do NOTHING but will my body to heal. :-)
What fun…..my gosh Whit is growing up so fast. I’ve been following you since you posted his darling baby pictures. He’s an amazing young man….but you already knew that! Praying for continued healing for you and hopeful you can make a big splash in your pool soon. God bless…
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TITLE: How to recognize a finite dimensional algebra is Koszul or quadratic?
QUESTION [8 upvotes]: I have a family of finite dimensional algebras that are directed quasihereditary. I think they might be Koszul algebras and I am wondering what approaches there are to check Koszulness or even quadraticity. I know the quivers of these algebras and can compute Ext^n between simple modules for all n, but I do not have a quiver presentation. I know that there are paths of length 2 and of higher lengths between all vertices of the quiver with nonvanishing Ext^2 so I cannot prove or eliminate quadraticity for trivial reasons. Any thoughts?
I should add that I do not have explicit minimal projective resolutions of the simple modules.
REPLY [4 votes]: One incredibly useful fact is that it suffices to find linear resolutions of standard modules, not simples. These are usually much easier to find "by hand." Strictly speaking this is stronger than Koszul (the term is "standard Koszul") but in practice it seems rare for a quasi-hereditary algebra to be Koszul and not standard Koszul.
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\begin{document}
\title[Torsion points of elliptic curves]{Torsion of elliptic curves and unlikely intersections}
\author{Fedor Bogomolov}
\address{Courant Institute of Mathematical Sciences, N.Y.U. \\
251 Mercer str. \\
New York, NY 10012, U.S.A.}
\address{National Research University Higher School of Economics,
Russian Federation \\
AG Laboratory, HSE \\
7 Vavilova str., Moscow, Russia, 117312}
\email{bogomolo@cims.nyu.edu}
\author{Hang Fu}
\address{Courant Institute of Mathematical Sciences, N.Y.U. \\
251 Mercer str. \\
New York, NY 10012, U.S.A.}
\email{fu@cims.nyu.edu}
\author{Yuri Tschinkel}
\address{Courant Institute of Mathematical Sciences, N.Y.U. \\
251 Mercer str. \\
New York, NY 10012, U.S.A.}
\address{Simons Foundation\\
160 Fifth Av. \\
New York, NY 10010, U.S.A.}
\email{tschinkel@cims.nyu.edu}
\keywords{Elliptic curves, torsion points, fields}
\begin{abstract}
We study effective versions of unlikely intersections of images of
torsion points of elliptic curves on the projective line.
\end{abstract}
\maketitle
\begin{dedication}
To Nigel Hitchin, with admiration.
\end{dedication}
\setcounter{section}{0}
\section*{Introduction}
\label{sect:introduction}
Let $k$ be a field of characteristic $\neq 2$ and $\bar{k}/k$ an algebraic closure of $k$.
Let $E$ be an elliptic curve over $k$, presented as a double cover
$$
\pi: E\ra \P^1,
$$
ramified in 4 points, and
$E[\infty]\subset E(\bar{k})$ the set of its torsion points. In \cite{BT-small} we proved:
\begin{thm}
If $E_1, E_2$ are nonisomorphic elliptic curves over $\bar{\Q}$, then
$$
\pi_1(E_1[\infty]) \cap \pi_2(E_2[\infty])
$$
is finite.
\end{thm}
Here, we explore effective versions of this theorem, specifically,
the size and structure of such intersections (see \cite{zannier} for
an extensive study of related problems).
We expect the following universal bound:
\begin{conj}[Effective Finiteness--EFC-I]
\label{conj:main}
There exists a constant $c>0$ such that
for every pair of nonisomorphic elliptic curves $E_1,E_2$ over $\C$ we have
$$
\pi_1(E_1[\infty])\cap \pi_2(E_2[\infty])<c.
$$
\end{conj}
We say that two subsets of the projective line
$$
S=\{ s_1,\ldots, s_n\},\quad S':= \{ s_1',\ldots, s_n'\} \subset \P^1(\bar{k})
$$
are projectively equivalent, and write $S\sim S'$, if there is a $\gamma\in \PGL_2(\bar{k})$
such that (modulo permutation of the indices) $s_i=\gamma(s_i')$, for all $i$.
Let $E$ be an elliptic curve over $k$, $e\in E$ the identity, and
$$
\begin{array}{rcc}
E & \stackrel{\iota}{\lra} & E\\
x & \mapsto & -x
\end{array}
$$
the standard involution. The corresponding quotient map
$$
\pi: E\ra E/\iota =\bP^1
$$
is ramified in the image of the 2-torsion points of $E(\bar{k})$.
Conversely, for
$$
r:=\{ r_1,r_2,r_3, r_4\} \subset \bP^1(\bar{k}),
$$
the double cover
$$
\pi_r: E_r\ra \bP^1
$$
with ramification in $r$ defines an elliptic curve;
given another such $r'$, the curves $E_r$ and $E_{r'}$ are isomorphic (over $\bar{k})$
if and only if $r\sim r'$, in particular, the image of 2-torsion determines the elliptic curve, up to isomorphism.
Let $E_r[n]\subset E_r(\bar{k})$ be the set of elements of order {\em exactly} $n$, for $n\in \N$.
The behavior of torsion points of other small orders is also
simple:
$$
\pi_r(E_r[3])\sim \{ 1,\zeta_3,\zeta_3^2,\infty\},
$$
where $\zeta_3$ is a nontrivial third root of 1, and
$$
\pi_r(E_r[4]) \sim \{ 0,1,-1,i,-i,\infty \}.
$$
In particular, up to projective equivalence, these are {\em independent} of $E_r$.
However, for all $n\ge 5$,
the sets $\pi_r(E_r[n])$, modulo $\PGL_2(\bar{k})$, do depend on $E_r$, and it is
tempting to inquire into the nature of this dependence.
In this note, we study $\pi_r(E_r[n])$, for varying curves $E_r$ and varying $n$.
Our goal is to establish effective and uniform finiteness results for intersections
$$
\pi_r(E_r[n])\cap \pi_{r'}(E_{r'}[n']), \quad n, n'\in \N,
$$
for elliptic curves $E_r,E_{r'}$, defined over $k$.
We formulate several conjectures in this direction and provide evidence for them.
The next step is to ask: given elliptic curves $E_r,E_{r'}$ over $\bar{k}$,
when is
$$
r \subset \pi_{r'}(E_{r'}[\infty])?
$$
We modify this question as follows:
Which minimal subsets $\tilde{L}\subset \P^1(\bar{k})$ have the property
$$
r\subset \tilde{L} \quad \Rightarrow \pi_r(E_r[\infty])\subseteq \tilde{L}?
$$
The sets $\tilde{L}$ carry involutions, obtained from
the translation action of the 2-torsion points of $E$ on $E$,
which
descends, via $\pi$, to an action on $\P^1$
and defines an embedding of $\Z/2\oplus \Z/2\hookrightarrow \PGL_2(\bar{k})$.
It is conjugated to the standard embedding
of $\Z/2\oplus \Z/2$, generated by involutions
$$
z\mapsto -z\quad \text{ and } \quad z\mapsto 1/z,
$$
acting on $\tilde{L}$.
This observation is crucial for the discussion in Section~\ref{sect:cyclo}, where we prove
that, modulo projectivities, $L:=\tilde{L}\setminus \{ \infty\} $ are fields.
\bigskip
\noindent {\bf Acknowledgments:}
The first author was partially supported by the Russian Academic
Excellence Project `5-100' and by Simons Fellowship and
by EPSRC programme grant EP/M024830.
The second author was supported by the MacCracken Program offered
by New York University.
The third author was partially supported by NSF grant 1601912.
\section{Generalities}
\label{sect:general}
Let $j:\mathcal E\ra \bP^1$ be the standard universal elliptic curve, with $j$ the $j$-invariant morphism.
Consider the diagram
\
\centerline{
\xymatrix{
E_{\lambda} \ar^{\iota}[r]
\ar@{}[d]|-*[@]{\subset} & P_{\lambda} \ar@{}[d]|-*[@]{\subset}
\\
\mathcal E \ar^{\iota}[r] \ar[d]_{j} & \mathcal P \ar[d]_j \\
\bP^1 \ar@{=}[r] & \bP^1
}
}
\
\noindent
assigning to each fiber $E_{\lambda}:=j^{-1}(\lambda)$ the quotient $P_{\lambda}= \pi(E_{\lambda})\simeq \bP^1$,
by the involution $\iota: x\mapsto -x$ on $E_{\lambda}$. (This is well-defined even for singular fibers of $j$.)
Note that $\mathcal P\ra \bP^1$ is a $\PGL_2$-torsor. Taking fiberwise $n$-symmetric product:
$$
P_{\lambda} \mapsto \Sym^n(P_{\lambda})
$$
we have associated $\PGL_2$-torsors
$$
j_n: \mathcal P_n=\Sym^n(\mathcal P)\ra \bP^1.
$$
Taking $\PGL_2$-invariants, we have a canonical projection
$$
\Sym^n(P_\lambda) \ra \mathcal M_{0,n}(P_{\lambda}) \simeq \mathcal M_{0,n},
$$
to the moduli space of $n$-points on $\bP^1$.
The associated $\PGL_2$-torsor is trivial; fixing a trivialization we obtain a morphism
$$
\mu_n: \mathcal P_n \ra \overline{\mathcal M}_{0,n}
$$
For every $N\in \N$, we have the modular curve $X(N)\ra \bP^1$, parametrizing pairs of elliptic curves together with $N$-torsion subgroups.
The involution $\iota$ induces an involution on every $X(N)$, we have the induced quotient
$$
X(N)\ra Y(N):=X(N)/\iota.
$$
Since the family $j:\mathcal E\ra \P^1$ has maximal monodromy $\SL_2(\Z)$, the curves $X(N)$ and $Y(N)$ are irreducible.
We have a natural embedding $Y(N)\hookrightarrow \mathcal P$. Put
$$
Y:=\cup_{N\in \N} Y(N)
$$
and consider
$$
\Sym^n(Y)\hookrightarrow \mathcal P_n \ra \overline{\mathcal M}_{0,n}.
$$
Note that $\Sym^n(Y)$ is a union of infinitely many irreducible curves, each corresponding to
an orbit of the action of the monodromy group $\PGL_2(\Z)$ on the generic fiber of the restriction of $j_n$ to $\Sym^n(Y)$.
Let $Y_{n,\omega}\subset \Sym^n(Y)$
be an irreducible component corresponding to a $\PGL_2(\Z)$-orbit $\omega$ (for the monodromy action, as above).
We now formulate conjectures
about $\mu_n$, for small $n$, which guide our approach to the study of images of torsion points.
\begin{conj}
\label{conj:1}
The map
$$
\mu_4 : Y_{4,\omega} \to \overline{\mathcal M}_{0,4} =\P^1
$$
is finite surjective, for all but finitely many $\omega$.
\end{conj}
\begin{conj}
\label{conj:2}
The map
$$
(\mu_4,j): Y_{4,\omega}\to \overline{\mathcal M}_{0,4}\times \P^1
$$
is
a rational embedding, for all but finitely many $\omega$.
\end{conj}
\begin{conj}
\label{conj:3}
The map
$$
\mu_5 : Y_{5,\omega }\to \overline{\mathcal M}_{0,5}
$$
is a rational
embedding, for all but finitely many $\omega$. Moreover, if for some distinct orbits $\omega$ and $\omega'$
the corresponding images $\mu_5(Y_{5,\omega })$ and $\mu_5(Y_{5,\omega'})$ are curves, then they are different.
\end{conj}
\begin{conj}
\label{conj:4}
The map
$$
\mu_6 : Y_{6,\omega}\to \overline{\mathcal M}_{0,6}
$$
is a rational embedding, for all but finitely many $\omega$.
Moreover, if $\mu_6 (Y_{6,\omega})$ is a curve then there exist
at most finitely many $\omega'$ such that
\begin{itemize}
\item $\mu_6 (Y_{6,\omega'})$ is a curve and
\item $\mu_6(Y_{6,\omega})\cap \mu_6 (Y_{6,\omega'})\neq \emptyset$.
\end{itemize}
\end{conj}
\section{Examples and evidence}
\label{sect:ex}
We now discuss examples and evidence for Conjectures in Section~\ref{sect:general}.
\begin{exam}
We have
\begin{itemize}
\item $\mu_4(\Sym^4(Y(2))\simeq \overline{\mathcal M}_{0,4}=\bP^1$,
\item $\mu_4(\Sym^4(Y(3))$ is a point in $\overline{\mathcal M}_{0,4}$.
\end{itemize}
Consider $\Sym^4(Y(4))$. Note that $\pi(E[4])=\{ 0,1,-1, i,-i,\infty\} $
is an orbit of the symmetric group $\mathfrak S_4$, acting on $\P^1$.
The pairs
$$
(0, \infty),(1,-1),(i,-i)
$$
are pairs of stable points for $3$ even involutions in $\mathfrak S_4$,
and the action of $\mathfrak S_4$ is transitive on pairs and inside
each pair. There are two different $\mathfrak S_4$-orbits
of $4$-tuples: either the orbit contains two pairs
of vertices such as $(0, \infty),(1,-1)$, or a pair and two points
from different pairs $(0, \infty),(1, i)$.
Thus $\Sym^4(Y(4))$ has two components which project to different
points modulo $\PGL_2$; therefore, there exist
exceptional orbits $\omega$ such that $\mu_4(Y_{4,\omega})$ is a point.
\end{exam}
\begin{lemm}
\label{lemm:mu-cn}
If $\mu_4(Y_{4,\omega})$ is a point
then all cross ratios of 4-tuples of points parametrized by $Y_{4,\omega}$
are constant.
\end{lemm}
\begin{proof}
The map $\mu_4 $ can be viewed
as a composition
$$
(\P^1)^4\stackrel{\mathit{cr}}{\longrightarrow} (\Z/2\oplus \Z/2) \backslash (\P^1)^4/\PGL_2 = \P^1_1\to \mathfrak S_3 \backslash \P^1_1.
$$
Thus we have a diagram
\centerline{
\xymatrix{
(\P^1)^4 \ar[r]^{cr\quad\quad\quad }\ar[d] & (\Z/2\oplus \Z/2) \backslash (\P^1)^4 /\PGL_2 \ar[d]^{\mathfrak S_3} \\
\mathfrak S_4\backslash (\P^1)^4 \ar[r] & \mathfrak S_4\backslash (\P^1)^4/\PGL_2
}
}
\
\no
Note that any irreducible $Y_{4,\omega}$ lifts
to a union of connected components $Y_{4,\omega,i}\subset (\Z/2\oplus \Z/2)\backslash Y^4$,
where cross-ratio is well defined.
Thus if $\mu_4$ is a rational function of cross-ratio
on any four-tuple of points and if $\mu_4$ is constant
then the cross-ratio is also constant.
\end{proof}
\begin{prop}
\label{prop:sur}
There exist orbits $\omega$ such that
$$
\mu_4: Y_{4,\omega}\ra \P^1
$$
is surjective.
\end{prop}
\begin{proof}
The singular fiber $\mathcal E_{\infty}:=j^{-1}(\infty)$
is an irreducible rational curve with one node $p_{\infty}$. The group scheme $\cup _{d\mid n} \mathcal E[d]$, whose generic fiber is isomorphic to
$\Z/n\oplus\Z/n$, specializes to $\{\zeta_n^i\}\subset \mathbb G_m=\mathcal E_{\infty}\setminus p_{\infty}$.
Let $\mathcal E_{\infty}[n]$ be the specialization of $\mathcal E[n]$; then
\begin{itemize}
\item $\mathcal E_{\infty}[n]\subset \{ \zeta_n^i\}$,
\item there exists a subgroup scheme $\mathcal W_n\simeq \Z/n\subset\Z/n\oplus \Z/n$ in the group scheme
of points killed by $n$, specializing to $\mathcal E_{\infty}$, while the complemenary branches specialize to
$p_{\infty}$.
\end{itemize}
Taking the quotient by $\iota$, we find that
$((\Z/n\oplus \Z/n)\setminus \Z/n)/\iota $ specializes to $0$ in the fiber $\P^1_{\infty}$ and
all other points specialize to subset in $(\Z/n)/\iota$; the limit depends
on the selected direction of specialization.
Assume that we have distinct points $\{ z_1,z_2,z_3,z_4\} \subset \pi(E[n])$, for a smooth fiber $E$ of $\mathcal E$, such that
$$
z_1,z_2\in W_n/\iota \quad \text{ and } \quad z_3,z_4\notin W_n/\iota.
$$
The $z_1, z_2$ can be specialized to different nonzero points in $\mathcal E_{\infty}/\iota$, and
$z_3,z_4$ will specialize to $0$.
Assume that $\mu_4$ is constant, i.e., the cross-ratio is constant.
Since $z_3,z_4$ will specialize to $0$, the cross-ratio equals 1.
Then
$$
(z_1- z_3)(z_2-z_4)= (z_2- z_3)(z_1-z_4),
$$
and
$$
z_1(z_3-z_4)=z_2(z_3-z_4).
$$
Near the special fiber, $z_3\neq z_4$, thus $z_1=z_2$, contradiction.
Thus on orbits of this type, $\mu_4$ is not constant, hence surjective.
\end{proof}
\section{Geometric approach to effective finiteness}
\label{sect:geom}
Let $E:=E_r,E':=E_{r'}$ be elliptic curves.
Consider the diagram
\centerline{
\xymatrix{
C\ar[d] \ar[r]& E\times E' \ar[d] \\
\Delta \ar[r] & \P^1\times \P^1
}
}
\no
where
$C\subset E\times E'$ be the fiberwise product over the diagonal
$\Delta\subset \P^1\times \P^1$. If $r\neq r'$ then $C$ has genus $\ge 2$.
By Raynaud's theorem \cite{raynaud},
$$
C(\bar{k})\cap E[\infty]\times E'[\infty]
$$
is finite, since it is
the preimage of $\pi(E[\infty])\cap\pi(E'[\infty])\subset \Delta$, the latter set is also finite.
This finiteness argument appeared in \cite{BT-small}.
Consider the curves $C$ occurring in this construction. We have a diagram
\centerline{
\xymatrix{
C \ar[d]_{\sigma'}\ar[r]^{\sigma} & E\\
E' &
}
}
\
\noindent
where $\sigma, \sigma'$ are involutions with fixed points $c_1,c_2$ and $c_1', c_2'$, respectively. Assume that
$$
r\cap r' = \{ 0, 1, \infty\}.
$$
Then the product involution $\sigma\sigma'$ on $C\subset E\times E'$ has fixed points in the 6
preimages of the points $\{ 0,1,\infty \}\subset \Delta_{\P^1}\subset \P^1\times \P^1$ (diagonally), i.e., is the hyperelliptic involution.
Thus there is an action of $ \Z/2\oplus \Z/2$ on $C$, induced by the covering maps $\pi$ and $\pi'$.
The curve $C\subset A=E\times E'$ has self-intersection $C^2=8$
since it is a double cover of both $E$ and $E'$ and its class is equal to $2(E+E')$.
\begin{itemize}
\item
If the genus $\mathsf g(C)=2$ (three such points) then the image of $C$ in its Jacobian $J(C)$ has self-intersection 2.
Consider the map
$$
\nu: J(C)\to A= E\times E'.
$$
and let $n$ be its degree.
The preimage $\nu^{-1}(C)\subset J(C)$ has self-intersection $8n$.
On the other hand, its homology class is equal to $n$ translations of $C$, hence has self-intersection $2n^2$, thus $n=4$.
Moreover, $\ker(\nu) = \Z/2\oplus \Z/2$, generated by the pairwise differences of preimages of points $\{ 0,1,\infty \}$.
Thus, $J(C)$ is 4-isogenous to $A:=E\times E'$ and
$\nu(C)$ is singular, with nodes exactly at the preimages of $\{ 0,1,\infty \}\subset \Delta_{\P^1}$.
Consider a point $c\in C\subset J(C)$ and assume that $\nu(c)$ has order $m$ with respect to $0\in A$.
Then $c$ has order $m$ or $2m$ in $J(C)$, with respect to $0 \in J(C)$.
Hence the corresponding curve $Y(m)\subset \P^1\times \P^1$ (viewed as a moduli
space of pairs $E,E'$)
is given as an intersection of genus $2$ curves containing a point of order $m$ or $2m$, respectively. This is a locus in the moduli space
$\mathcal M_2$ of genus 2 curves.
\item
If $\mathsf g(C)=3$ (two such points) then
there are three quotients of $C$ which are
elliptic curves $E_1,E_2,E_3$, with involutions $\sigma_i\in \Z/2\oplus \Z/2$
fixing $4$ points on $E_i$ which are invariant under the hyperelliptic
involution given by complement to $\sigma_j$.
The kernel of
$$
\nu_i: J(C)\to E_j\times E_k
$$
contains
$E_i$, for $i,j,k\in \{ 1,2,3\} $.
\item
If $\mathsf g(C)= 4$ then $C$ is $C/\sigma_i = E_i, i=1,2$
and $C/\sigma_1\sigma_2 = C'$ where $\mathsf g(C')=2$ and
there are exactly two ramification points on $C'$.
\item
If $\mathsf g(C)= 5$ then $C/\sigma_1\sigma_2 = C'$ is a hyperelliptic
curve of genus $3$ and the covering is an unramified double cover.
\end{itemize}
\begin{rema}
\label{rema:stand}
Assume that there is $b\in \P^1$
and a subset $S\subset C(\bar{k})$ such that $S+b \subset C\subset E\times E'$.
Then
$$
\# S\le 8 = C^2 = C \cap (C + b);
$$
hence we have
at most $8$ points $c_i\in \P^1$ such that for $x$-coordinates
$ c_i+_1 b = c_i +_2 b$, where the summation $+_1$ corresponds to the
summation on the first curve and $+_2$ on the second.
\end{rema}
\begin{rema}
The construction can be extended to products of more than two elliptic curves.
We may consider
$$
\pi:=\prod_{i=1}^r \pi_i : \mathcal A:=\prod E_i\to \mathcal P :=\prod \P^1_i.
$$
The ramification divisor of $\pi :\mathcal A \ra \mathcal P$
is a union of products of projective lines. Let $\Delta=\P^1\subset \mathcal P$ be the diagonal,
there exists canonical identifications $\delta_i: \P_i^1 \simeq \Delta$.
If $p\in \Delta$ is contained in $\delta_i(\pi_i(E_i[\infty]))$, for all $i$,
then the preimage of $p$ in $\mathcal A$ is contained
in the preimage of the diagonal. This is a curve of genus at least 2, provided
there exist $E_i, E_j$ with $r_i\neq r_j$. Then the set of such $p$ is finite.
In particular, if $E$ is defined over a number field $k$ and $p$ is defined over a proper subfield, then
$p$ is also in the image torsion points of $\gamma(E)$, where $\gamma$ is a Galois conjugation. Hence,
the existence of torsion points with $x$-coordinate in a smaller field has a geometric implication.
\end{rema}
We expect the following version of Conjecture~\ref{conj:main}:
\begin{conj}[Effective Finiteness--EFC-II]
\label{conj:main2}
There exists a constant $c>0$ such that for every elliptic curve $E_r$ over a number field
and every $\gamma\in \PGL_2(\bar{\Q})$ with $\gamma(r)\neq r$ we have
$$
\pi_r(E_r[\infty])\cap \pi_\gamma(E_\gamma[\infty]) < c.
$$
\end{conj}
\section{Fields generated by elliptic division}
\label{sect:cyclo}
In this section, we explore properties of subsets of $\P^1(\bar{k})$ generated by images of torsion points, following closely \cite{BT-small}.
For
$$
r:=\{ r_1, r_2,r_3,r_4\}\subset \P^1(\bar{k}),
$$
a set of four distinct points, let $E_r$ be the corresponding elliptic curve defined in the Introduction. Let
$$
\tilde{L}_r\subset \P^1(\bar{k})
$$
be the smallest subset such that for every $E_{r'}$ with $r'\subseteq \tilde{L}_r$ we have $\pi_{r'}(E_{r'}[\infty])\subseteq \tilde{L}_r$.
\begin{thm}\cite{BT-small}
\label{thm:field}
Let $k$ be a number field.
For every $a\in k\setminus \{0,\pm1,\pm i\}$, and
\begin{equation}
\label{eqn:ra}
r=r_a:=\{ a,-a,a^{-1} , -a^{-1}\} \subset \P^1(k)
\end{equation}
the set
$$
L_{a}:=\tilde{L}_{r_a}\backslash\{\infty\}
$$
is a field.
\end{thm}
At first glance, it is rather surprising that such a simple and natural construction, inspired by comparisons of $x$-coordinates of torsion points of
elliptic curves, produces a field. The conceptual reason for this is the rather peculiar structure of 4-torsion points of elliptic curves:
translations by 2-torsion points yields, upon projection to $\P^1$, {\em two} standard commuting involutions on $\P^1(\bar{k})$, which allow to define
addition and multiplication on $L_a$.
We may inquire about arithmetic and geometric properties of the fields $L_a$. For $a\in \bar{k}$ we let $k(a)\subseteq \bar{k}$
denote the smallest subfield containing $a$. We have:
\begin{itemize}
\item
For every $a\in \bar{k}$, the field $L_a$ is a Galois extension of $\Q(a)$.
\item
For every $k$ of characteristic zero, $L_a$ contains $\Q^{ab}$, the maximal abelian extension of $\Q$.
\item
The field $L_\zeta$, where $\zeta$ is a primitive
root of order $8$, is contained in any field $L_a$.
Indeed, the corresponding elliptic curve $E$ has ramification subset
$$
\{ \zeta,\zeta^3,\zeta^5,\zeta^7\},
$$
which is projectively equivalent to $\{ 1,-1, i,-i\} \subset \pi(E[4])$.
Since $\pi(E[4])$ projectively does not depend on the curve $E$, we obtain
that $L_\zeta\subset L_a$, for all $a$.
The same holds for $L_a$ where $E_a$ is isomorphic
to $E_3$ (elliptic curve with an automorphism of order $3$).
\item
The field $L_a$ is {\em contained} in a field obtained as an iteration of Galois extensions with Galois groups either abelian or
$\PGL_2(\mathbb F_q)$, for various prime powers $q$. Is $L_a$ equal to such an extension?
As soon as the absolute Galois group is not equal to a group of this type, e.g., for
a number field $k$, we have
$$
L_a\subsetneq \bar{k}.
$$
\item
Let $a,a'\in \bar{\Q}$ be algebraic numbers such that $\Q(a)=\Q(a')$. Then $L_a=L_{a'}$.
Varying $a\in \bar{\Q}$, we obtain a supply of interesting infinite extensions $L_a/\Q$.
\end{itemize}
The rest of this section is devoted to the proof of Theorem~\ref{thm:field}.
\begin{proof}
Let $r_0:=\{ 0, \infty, 1, -1\}$ and put $L:=\tilde{L}_{r_0}\setminus \{ \infty\}$. Let
$$
\pi=\pi_{r_a}:E_{r_a}\ra \P^1
$$
be the elliptic curve
with ramification in $r_a$. Since
$$
\{0,\infty,\pm1\}\subseteq\pi(E_{r_a}[4]),
$$
we have
$L\subseteq L_{a}$, for all $a$.
We first show that $L$ is a field.
\
{\em Step 1.}
$L\backslash\{0\}$ is a multiplicative group.
Indeed, for any $b\in L\backslash\{0\}$, we have
$$
r_0:=\{ 0,1,-1,\infty\} = b^{-1} \cdot \{0,b,-b,\infty\} =:r_b
$$
and hence
$$
L_{r_b}= b\cdot L_{r_0}= b\cdot L.
$$
Since $b^{-1},-b^{-1}\in L$ we also
have $\{ 0,1,-1,\infty\} \subset b\cdot L$.
Thus $L\subseteq b L$.
Similarly, $ L\subseteq b^{-1}\cdot L$ or $b \cdot L\subseteq L$,
which implies $L=bL$.
Thus for any $a,b\in L$ we have $ ab \in L$,
and since the same holds for $ ab^{-1}$, $b\neq 0$, we obtain
$L\setminus \{ 0\} \subseteq \bar{k}^\times$.
\
{\it Step 2.}
Let
$$
\Aut_{L}:=\{ \gamma \in \PGL_2(\bar{k}) \, | \, \gamma(\tilde{L})\subseteq \tilde{L} \}
$$
be the subgroup preserving $\tilde{L}$.
It is nontrivial, since it contains $L\setminus \{ 0\} $ as a multiplicative subgroup, together
with the involution $x\mapsto x^{-1}$.
Consider
$$
\gamma_1: x\mapsto (x-1)/(x+1).
$$
It is an involution with
$\gamma_1(\infty)= 1, \gamma_1(0)= -1$
and hence $\gamma_1$ is coming from
$r:=\{0,1,-1,\infty\}$.
Thus it maps $L$ into $L$ and
$\gamma_1\in \Aut_L$
Consider any pair of distinct elements
$\{ b,c\} \subset L$: it can be transformed into $\{ 0,1\}$
by an element from $\Aut_L$.
If $b\neq 0,\infty$ then,
dividing on $b$, we obtain
$\{ 1,c/b\}$ and $\gamma_1(\{ 1,c/b\})=\{0,1\}$.
If $b=0$ and $c\neq \infty$ then, dividing
on $c$, we obtain $\{0,1\}$.
If $b=0,c= \infty$ then $\gamma_1(\{0,\infty\})= \{-1,1\}$
and we reduce to the first case.
\
{\it Step 3.}
$L$ is closed under addition.
We show that $\gamma: x\mapsto x+1$ is contained in
$\Aut_L$: by Step 2, there exists a $g\in \Aut_L$
which maps $\{-1,\infty\}$ to $\{0,\infty\}$ and hence
$\{-1,0,\infty\}$ to $\{0,b,\infty\}$,
for some $b\in L\setminus \{0\}$.
Then $b^{-1}g\in \Aut_L$ maps $\{ -1,0,\infty\}$ to
$\{0,1,\infty\}$ and hence $b^{-1}g(x)= \gamma(x)=x+1$.
Thus for any $a\in L$ we have $a+b = b(a/b + 1)\in L$, which shows that
$L$ is an abelian group.
\
Now let us turn to the general $L_{a}$.
\
{\it Step 4.}
Note that $L\subset L_a$ and that
$L_a$ is closed under taking square roots.
Indeed for any $a\in L$ and $E_r$ with $r:=\{ 0,1,a,\infty\}$, we have
$\sqrt{a}\in\pi_r(E_r[4])$ and hence $\sqrt{a}\in L_a$.
Furthermore, for any $a,b\in L_a$ we have $\sqrt{ab}\in L_a$.
Indeed, consider the curve $E_r$ with $r= \{0,a,b,\infty\}$.
Then $\sqrt{ab}\in \pi(E_r[4])$, since
the involution $z\to ab/z$ is contained in the subgroup $\Z/2\oplus \Z/2$ corresponding to the two-torsion on $E_r$,
its invariant points are in $\pi_r(E_r[4])$.
Iterating, we obtain that
$$
\sqrt[2^{m-1}]{b_1\cdots b_m}\in \tilde{L}_a\setminus \{ \infty\} \text{ for all } \, b_i\in
\tilde{L}_a\setminus \{ \infty\}
$$
\
{\em Step 5.}
For all $b\in L_a, c\in L$ we have $\sqrt{b+ c}\in L_a$.
Indeed, for $c\in L$ we know that there is a solution $d\in L$
of the quadratic equation $d^2+d + c =0$.
Consider the curve $E_r$ for $r:=\{\infty,b, d , d+1\}$.
Then
$$
d\pm \sqrt{b-d}\in \pi(E_r[4])
$$
and hence $d\pm\sqrt{b-d}\in L_a$.
Thus
$$
\sqrt{(\sqrt{b-d} + d)(\sqrt{b-d}- d)}= \sqrt{b -d^2 -d}= \sqrt{b+c} \in L_a.
$$
\
{\em Step. 6.}
Let $P_m\in L[x]$ be a monic
polynomial of degree $m$ and let $b\in L_a$.
Then there is an $N(m)\in \N$ such that
$$
\sqrt[4^{N(m)}]{P_m(b)}\in L_a.
$$
Indeed, we have
$$
P_m(b)= c_m + b(c_{m-1} + b(c_{m-2}+\cdots )\cdots ).
$$
The statement holds for $m=1$ by Step 4.
Assume that it holds for $m-1$.
Then $c_{m-1} + b(c_{m-2}+ \cdots )= d^{4^{N(m-1)}}$ for
some $d\in L_a$.
We can then write
$$
P^m(b)=c_m + bd^{4^{N(m-1)}},
$$
by taking $t= \sqrt[4^{N(m-1)}]{b}$ and $u_m= \sqrt[4^{N(m-1)}]{c_m}$
we obtain
$$
P^m(b)= \prod (t+ \zeta^i u_m),
$$
where
$t\in L_a, u_m \in L$ and $\zeta^i$ runs through the roots of unity
of order $4^{N(m-1)}$.
By Steps 4 and 5, we obtain that $4^m 4^{N(m-1)}$-th
root of $P_m(b)$
is contained in $L_a$, thus the result holds for
$N(m)= 4^{N(m-1)}$
\
{\em Step 7.}
Let $b\in L_a$ be any algebraic element over $L$.
Then the field $L(b)$ is a finite extension of $L$ and
there is an $n\in \N$ such that any $x\in L(b)$
can be represented as a monic polynomial of $b$ with coefficients in $L$
of degree $ \leq n$. For such $n$ we define
a power $4^N$ such $\sqrt[4N]{x}\in L_a$, but then
any element in $L(b)$ is in $L_a$.
\end{proof}
\
\begin{rema}
In the proof we have only used
points in $\pi(E[4])$. Therefore, for any subset $D\subset \N$
containing $4$ we can define $L_{a,D}$, as the smallest subset containing all $\pi(E[n])$ for all $n\in D$ and
all elliptic curves obtained as double covers with ramification in $L_{a,D}$.
It will also be a field.
For example, if $D=\{3,4\} $ then $L_{a,D}$
is exactly the closure of $L_a$ under abelian degree $2$ and $3$ extensions, since
$\PGL_2(\mathbb F_2) = \mathfrak S_3$ and $\PGL_2(\mathbb F_3)= \mathfrak S_4$ and both groups
are solvable with abelian quotients of exponent $3,2$.
\end{rema}
On $(\Sym^4 (\P^1(\bar \Q))\setminus \Delta)/ \PGL_2(\bar \Q)$
we can define a directed graph structure $DGS$, postulating that
$$
r_z= \{ z_1,z_2,z_3,z_4\} \ra r_w= \{ w_1,w_2,w_3,w_4\}
$$
if there is an elliptic curve $E'$ isogeneous to $E_{r_z}$ such that $r_w$ is projectively
equivalent to a subset in $\pi(E'[\infty])$.
Any path in the graph is equivalent to a path contained
in $(\Sym^4 (\P^1(L(E)))\setminus \Delta)/ \PGL_2(\bar \Q)$, for some $E$.
The graph contains cycles, periodic orbits, and preperiodic orbits, i.e.,
paths which at some moment end in periodic orbits.
\begin{ques}
Consider the field $L_0= L_{r_0}$ for $r_0=\{ 0,1,-1,\infty\} $.
Does
$$
(\Sym^4(\P^1(L(E)))\setminus \Delta)/ \PGL_2(L(E))
$$
consist
of one cycle in $DGS$?
Note that any path beginning from $r_0$
extends
to a cycle (in many different ways) since $r_0$
is $\PGL_2$-equivalent to a four-tuple of points of order $4$
on any elliptic curve.
\end{ques}
\begin{rema}
In Step 7, we have used algebraicity of $L_a/L$, and we do not know how to
extend the proof to geometric fields. What are the properties of $L_a$ in
geometric situations, when $a$ is transcendental over $k$?
\end{rema}
We have seen in the proof that the field $L_a$ is closed under extensions of degree 2. We also have:
\begin{lemm}
For any $b\in L_a$, we have $\sqrt[3]{b}\in L_a$.
\end{lemm}
\begin{proof}
Consider a curve $E_r$ with $r:= \{ b, \sqrt b, -\sqrt b, \infty\}$.
Its 3-division polynomial takes the form:
$$
f_3(x) = 3x^4 - 4b x^3 - 6b x^2 + 12 b^2 x - 4 b^3 - b^2.
$$
We can represent it as a product:
$3\prod_{i=1}^4 (x-x_i)$, where the set $\{ x_i\} \subset L_a$ is equal
$\pi(E_r[3])$. The corresponding cubic resolvent
$$
rc(x):=\prod (x- (x_ ix_j+ x_k x_l)),
$$
where $(i,j),(k,l)$ is any splitting
into pairs of indices among $1,2,3,4$.
In terms of $b$, we have
$$
rc(x)= x^3 + 2bx^2 + 4b^2 x/3 + 8b^3/3 - 128 b^4/27 + 64 b^5/27.
$$
Since the set $\{ x_i\}$ is projectively equivalent to $\{ 0,1,\zeta_3,\zeta_3^2\}$,
we can see that the cubic polynomial above
has the form $C (x^3 + B)$, for some constants $C,B$.
It can be checked that
$$
rc( 2b(2x-1)/ 3 ) = (4b/3)^3 (x^3 + (b-1)^2).
$$
After a projective map in $\PGL_2(L_a)$ we can transform the
the elements $x_ix_j+ x_kx_l$ into $-\sqrt[3]{(b-1)^2}$.
Hence $-\sqrt[3]{(b-1)^2}\in L_a$, for any $b\in L_a$; since
$L_a$ is a field closed under $2$-extensions we obtain the claim.
\end{proof}
This raises a natural
\begin{ques}
Is $L_a$ is closed under taking roots of arbitrary degree?
\end{ques}
If we add $\mathbb G_m$ to the set of allowed elliptic curves then the answer
is affirmative. However, there may exist a purely {\em elliptic} substitute
for obtaining roots of prescribed order.
\begin{coro}
If the $j(E)\in L_a$ then any set $\{ b,-b,b^{-1},-b^{-1}\} $ with $\mu_4((b,-b,b^{-1},-b^{-1}))=j(E)$
is contained in $L_a$. Note that such $b$ are solutions
of a cubic equation. Thus $L_a$ depends only
on the curve $E$ and we will write $L(E)$.
\end{coro}
It is also easy to see that $L(E)=L(E')$ if $E$ and $E'$ are isogenous.
\section{Intersections}
\label{sect:inter}
In this section we present further results concerning intersections
$$
\pi_1(E_1[\infty])\cap \pi_2(E_2[\infty])
$$
for
different elliptic curves $E_1,E_2$ and
provide evidence for the Effective Finiteness Conjecture \ref{conj:main}.
\begin{prop}
\label{prop:ar}
Assume that
\begin{equation}
\label{eqn:eqb}
\pi_1(E_1[4])=\pi_2(E_2[4])= \{ 0,1, -1, i, -i, \infty\}
\end{equation}
and that
$$
\#\{ \pi_1(E_1[3])\cap \pi_2(E_2[3])\} \ge 2.
$$
Then $r_1=r_2$ and $E_1 = E_2$.
\end{prop}
\begin{proof}
By our assumption \eqref{eqn:eqb},
$E_i$ are given by the equation
$$
y^2=x^4- t_ix^2 + 1.
$$
With $a_i$ defined by
$$
r_i = \{ a_i,-a_i,a_i^{-1},-a_i^{-1} \},
$$
we have
$$
t_i= a^2_i + a^{-2}_i.
$$
We assume that $\pi_i(e_i)=a_i$.
In this case, points $\pi_i(E_i[3])\subset \bar{\Q}\subset \P^1$ are
the roots of
\begin{equation}
\label{eqn:aaa}
x^4+2a x^3-(2/a)x-1=0
\end{equation}
or, equivalently,
$$
2x^{3}a^{2}+(x^{4}-1)a-2x.
$$
If $x,y\in\pi_{a_{1}}(E_{a_{1}}[3])\cap\pi_{a_{2}}(E_{a_{2}}[3])$,
where $x\neq y$ and $a_{1}\neq a_{2}$, then $a_{1}$
and $a_{2}$ are the roots of $2x^{3}a^{2}+(x^{4}-1)a-2x$
and of $2y^{3}a^{2}+(y^{4}-1)a-2y$, that means that their coefficients are proportional
\[
\frac{2x^{3}}{2y^{3}}=\frac{x^{4}-1}{y^{4}-1}=\frac{-2x}{-2y}.
\]
Then, on the one hand,
$x^{3}/y^{3}=x/y$ implies $x^{2}=y^{2}$, and hence $x=-y$, by our assumption that $x\neq y$.
On the other hand,
$$
x/y=-1 = (x^{4}-1)/(y^{4}-1)=1,
$$
a contradiction.
\end{proof}
Given any $x\in \bar{\Q}$
we obtain $a_i=a_i(x)$, $i=1,2$, which satisfy \eqref{eqn:aaa}.
Then the resulting elliptic curves $E_i$ satisfy
\eqref{eqn:eqb} and we have
$$
\#\{ \pi_1(E_1[3])\cap \pi_2(E_2[3])\} = 1.
$$
unless
$$
(x^4-1)^2 + 16x^4 = x^8 + 14x^4 + 1= 0
\quad \text{ or }\quad x^4= -7 \pm 4\sqrt{3}.
$$
Moreover,
\begin{equation}
\label{eqn:in}
\#\{ \pi_{a_1}(E_{a_1}[\infty])\cap \pi_{a_2}(E_{a_2}[\infty])\} = 6 + 4 n\ge 10,
\end{equation}
where $6$
is the number of images of common points of order $4$ (from Equation~\ref{eqn:eqb})
and $4$ stands for the size of $(\Z/2\oplus \Z/2)$-orbit of a point in $\P^1$.
However, it may happen that the inequality in \eqref{eqn:in} is strict.
\begin{exam}
Consider the polynomial $f_5(x ,a )$ defined in \cite[Theorem 18]{BF}). Its roots are exactly
$\pi_a(E_a[5])$. It has degree $12$ with respect to $x$ and $6$ with respect to $a$.
The polynomial $f_3(x,a)$ has degree $2$ with respect to $a$ and generically has exactly
two solutions $a_1(z),a_2(z)$, for any given $z$.
We want also $f_5(v, a_i(z))= 0$ for some $v$ and $z$.
This is equivalent to $f_5(v, a)$ being divisible by $f_3(z, a)$, as polynomials in $a$.
Writing division with remainder
$$
f_5(v,a)= g(a)f_3(z,a) + C(v,z)a + C'(v,z)
$$
for some explicit polynomials $C$, and $C'$, which have to vanish.
This condition is gives an explicit polynomial in $u$, which is divisible by a high power of $u$ and $(u-1)$.
Excluding the trivial solutions $u=0,1$, and substituting $t=u^4$ we obtain the equation
\begin{eqnarray*}
& & 32u^{24}+ 1369u^{20}+18812u^{16}+90646u^{12}+18812u^{8}+1369u^{4}+32\\
& = & 32t^{6}+1369t^{5}+18812t^{4}+90646t^{3}+18812t^{2}+1369t+32\\
& = & t^{3}\left[32\left(t^{3}+\frac{1}{t^{3}}\right)+1369\left(t^{2}+\frac{1}{t^{2}}\right)+18812\left(t+\frac{1}{t}\right)+90646\right]
\end{eqnarray*}
Since $t\neq 0$, we have
\begin{eqnarray*}
& =& 32\left(t+\frac{1}{t}\right)^{3}+1369\left(t+\frac{1}{t}\right)^{2}+18716\left(t+\frac{1}{t}\right)+87908\\
&=& 32r^{3}+1369r^{2}+18716r+87908\\
&=:& f(r)
\end{eqnarray*}
Computing the discriminant of this cubic polynomial, we find that it has no multiple roots.
Its solutions give rise to pairs $u,v$ such that
for $a_1:=a_1(u),a_2:=a_2(u)$ we have
$$
f_5(v, a_i) = f_3(u, a_i) =0
$$
and hence
$$
\#\{ \pi_{a_1}(E_{a_1}[\infty])\cap \pi_{a_2}(E_{a_2}[\infty])\} \geq 14.
$$
The symmetry of the above equation reduced the problem to a cubic equation with coefficients in $\Q$,
followed by a quadratic equation. The roots can be expressed in closed form and hence we get explicit
description for the 24 roots $u$.
\end{exam}
The same scheme can be applied to points of higher order.
Indeed we have a polynomial $f_n(u,x)= 0$ which has
increasing degree with respect to $u$, and the
existence of a pair $u,v$ such that $f_n(v,x)= 0$ is divisible
by $f_3(u,x)$ depend on the divisibilty of $f_n(v,x)$ by $f_3(u,x)$.
Using long division we obtain two polynomials
$C_{0,n}( u,v)$ and $C_{1,n}( u,v)$ so that their common
zeroes $(u,v)$ correspond to pairs $( u,v)$
with $f_3(u,x)= 0$ and $f_n(v,x)= 0$ simulaneously.
\begin{exam}
Applying this scheme to points of order $3$ and $7$ (or 3 and 11, 3 and 13, 3 and 17)
we obtain that the corresponding resultant has
roots of multiplicity three which implies the existence
of three points $v$ for a given $u$ with
$f_3(u,x)= 0$ and $f_7(v,x)= 0$ and hence
$$
\#\{ \pi_1(E_1[\infty])\cap \pi_2( E_2[\infty])\} \ge 6 + 16 = 22.
$$
\end{exam}
Since we have every reason to expect
polynomials $C_{0,n}( u,v)$ and $C_{1,n}( u,v)$ to have increasing
number of intersection points with the growth of $n$ we
are led to the following conjecture:
\begin{conj}
There is an infinite dense subset
of points $a\in \P^1$ such that
$$
\pi_a(E_{a}[\infty])\cap \pi_{a_2}(E_{a_2}[\infty]) \ge 22
$$
with
$$
\pi_a(E_{a}[3])\cap \pi_{a_2}(E_{a_2}[3]) \neq 0.
$$
\end{conj}
Note that in all such cases the fields $L_{a} = L_{a_2}$.
\section{General Weierstrass families}
\label{sect:gen-fam}
The family of elliptic curves considered in Section~\ref{sect:inter}
is the most promising for obtaining large intersections of torsion points.
In this section, we consider other families where the intersections
tend to be smaller, following \cite{BF}.
We consider elliptic curves $E_a$ with the same
$$
\pi_a(e_a)=\infty\in \P^1.
$$
These are given by their Weierstrass form
\begin{equation}
\label{eqn:wei}
y^2 = x^3 + a_2x^2+ a_4 x + a_6.
\end{equation}
Using formulas in, e.g., \cite[III, Section 2]{knapp}, we write down
(modified) division polynomials
$f_{n, a}$, whose zeroes are {\em exactly} $\pi_a(E_a[n])$:
$$
f_{n,a}(x) =\sum_{0\le r,s,t, r+ 2s +3t \leq d(n)} c_{r,s,t}(n) a_2^r a_4^s a_6^t x^{d(n)- (r+ 2s +3t)},
$$
where $d(n)$ and the coefficients $c_{r,s,t}(n)$
can be expressed via totient functions
$J_k(n)$, with $d(n) = J_2(n)/2$, if $n> 2$, and $d(2)=3$ (see \cite{BF}).
\begin{lemm}
\label{lemm:fn}
Let $E_1, E_2$ be elliptic curves in generalized
Weierstrass form \eqref{eqn:wei} such that, for some $n>1$ we have
$$
\pi_1(E_1[n])=\pi_2(E_2[n]).
$$
Then $E_1\simeq E_2$.
\end{lemm}
\begin{proof}
The statement is trivial for $n=2$.
For $n > 2$, we have $d(n) \geq 4$, the comparison of division polynomials
implies that the terms
$$
a_2^r a_4^s a_6^t, \quad r+ 2s +3t \leq 3
$$
must be equal. For
$$
(r,s,t)=(0,0,1), (0,1,0), (1,0,0)
$$
we find equality of coefficients $a_i$ for both curves.
\end{proof}
Often, already the existence of nontrivial
intersections
\begin{equation}
\label{eqn:eee}
\pi_1(E_1[n]) \cap \pi_2(E_2[n]) \ge 1
\end{equation}
leads to the
isomorphism of curves $E_1,E_2$.
For example, if both curves are defined over a number field $k$ and
the action of the absolute Galois group $G_k$ on $\pi_1(E_1[n])$
and $\pi(E_2[n])$ is transitive then \eqref{eqn:eee} implies that $E_1\simeq E_2$.
For many, but not all, $n\in \N,$ the equality of totient functions $J_2(n)= J_2(m)$, for some $m\in \N$, implies $n=m$.
\begin{exam}
There exist many tuples $(m,n)$ for which
$$
J_2(m)=J_2(n)\quad \text{ and } \quad J_1(m)\neq J_1(n).
$$
For example,
$$
J_2(5)=J_2(6)\quad \text{ but } \quad J_1(5) = 4, \quad J_1(6)= 2.
$$
We also have
$$
J_2(35)=J_2(40)= J_2(42), \text{ while } J_1(35)= 24, J_1(40)=16, J_1(42)=12.
$$
On the other hand, we have
$$
J_2(15)=J_2(16)=192 \quad \text{ and } \quad J_1(15)=J_1(16)=8.
$$
\end{exam}
These results indicate a relation of our question
to Serre's conjecture. He considered the action of the Galois group
on torsion points of an elliptic curve $E$ defined over a number field $k$.
If $E$ does not have complex multiplication, then
the image of the absolute Galois group $G_k$ is an open subgroup of $\GL_2(\hat \Z)$, i.e., of finite
index.
\begin{conj}[Serre]
For any number field $k$ there exists a constant $c=c(k)$ such that for every
non-CM elliptic curve $E$
over $k$ the index of the image of the Galois group $G_k$ in $\GL_2(\hat \Z)$
is smaller than $c$.
\end{conj}
In particular, for $k= \Q$ he conjectured that for primes $\ell \geq 37$ the
image of $G_k$ surjects onto $\PGL_2(\Z_\ell)$.
Thus, modulo Serre's conjecture, our conjecture holds for curves
defined over $\Q$.
\begin{prop}
Assume that
$$
\pi_1( E_1[n])=\pi_2( E_2[m]), \quad n\neq m.
$$
Then $k(E[n])$ contains $\Q(\zeta_d)$, where $d = \mathrm{lcm}(m,n)$,
the least common multiple of $m,n$.
\end{prop}
\begin{proof}
By Serre, we have
$$
\Q(\zeta_n)\subset k(E[n]) \quad \text{ and } \quad
\Q(\zeta_m)\subset k(E[m])
$$
as subfields
of index at most $2$. \end{proof}
\begin{coro}
Assume that $k$ does not contain roots of $1$
of order divisible by $n,m$. Then
$k(E[n]), k(E(m))$ contain a cyclotomic subfield
of index at most $2$.
\end{coro}
This provides a strong restriction on intersections of images of torsion points
for elliptic curves over $\Q$, or over more general number fields $k$ with this property.
This yields a restriction on fields $k(E[n])$,
since $(n,m) > 4$, for all $(n,m)$ with $J_2(n)= J_2(m)$.
\bibliographystyle{plain}
\bibliography{elliptic}
\end{document}
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\begin{document}
\begin{abstract}
Over complex numbers, the Fourier-Mukai partners of abelian varieties are well-understood. A celebrated result is Orlov's derived Torelli theorem. In this note, we study the FM-partners of abelian varieties in positive characteristic. We notice that, in odd characteristics, two abelian varieties of odd dimension are derived equivalent if their associated Kummer stacks are derived equivalent, which is Krug and Sosna's result over complex numbers. For abelian surfaces in odd characteristic, we show that two abelian surfaces are derived equivalent if and only if their associated Kummer surfaces are isomorphic. This extends the result \cite{HosonoLianEtAl2003} to odd characteristic fields, which solved a classical problem originally from Shioda. Furthermore, we establish the derived Torelli theorem for supersingular abelian varieties and apply it to characterize the quasi-liftable birational models of supersingular generalized Kummer varieties.
\end{abstract}
\maketitle
\section{Introduction}
Let $A$ be an abelian variety over an algebraically closed field $k$. It is desirable to have a description of the derived category of $A$ via its associated Kummer stack $\mathrm{K}(A)\coloneqq [A/\iota]$, where $\iota$ acts on $A$ by the involution $-1 \colon A \to A$. When $A$ is an abelian surface, the singular Kummer variety $A/\iota$ admits a crepant resolution, denoted by $\km(A)$.
A classical problem raised by Shioda is
\begin{question}[\cite{Shioda1977}]\label{Q1}
{For two abelian surfaces $A_1$ and $A_2$, if $\km(A_1) \cong \km(A_2)$, then can we conclude that $A_1 \cong A_2$?}
\end{question}
Over complex numbers, Question \ref{Q1} was solved in \cite{HosonoLianEtAl2003,orlov97}: two abelian surfaces have isomorphic Kummer surfaces if and only if they are derived equivalent. Their proof relies on the use of Hodge theory and global Torelli theorem for abelian surfaces, which are missing in positive characteristic fields. More generally, Stellari has investigated this problem for abelian varieties of arbitrary dimension in \cite{Stellari2007}.
In this paper, we are interested in above questions over positive characteristic fields. In particular, we would like to extend the result of \cite{HosonoLianEtAl2003} and \cite{Stellari2007} to fields with odd characteristic. The following result gives an answer of Shioda's question over positive characteristic fields.
\begin{thm}\label{mainthm1}
Assume $\mathrm{char}(k)\neq 2$. Let $A_1$ and $A_2$ be abelian varieties over $k$ of dimension $n$. Then the following holds:
\begin{enumerate}
\item If $A_1$ is derived equivalent to $A_2$, then Kummer stack $\rK(A_1)$ is derived equivalent to $\rK(A_2)$.
\item If $n$ is odd, then the converse of (i) holds.
\item If $n=2$, then $A_1$ and $A_2$ are derived equivalent if and only if $\km(A_1)\cong \km(A_2)$. If $A_1$ is supersingular, then $A_2$ is derived equivalent to $A_1$ if and only if $A_2\cong A_1$.
\end{enumerate}
\end{thm}
The statements (i) and (ii) will be proved with techniques around equivariant derived categories, which is already known in \cite{KrugSosna2015,ploog07} with $k=\mathbb{C}$. In the same way, we will generalize them to the case that $\Char(k)>2$ .
The proof of statement (iii) will be divided into two cases.
\begin{itemize}
\item For the finite height case, we can prove a lifting theorem for Kummer structures (see \S \ref{subsec:liftingKummerstructure}). Then the specialization argument of derived equivalences will imply this statement.
\item For the supersingular case, it can be concluded by supersingular Torelli theorem for abelian varieties (see \S \ref{subsec:supersingularderivedtorelli}).
\end{itemize}
Notice that Theorem \ref{mainthm1} shows that supersingular abelian surfaces do not have any non-trivial Fourier-Mukai partners. We expect that there is a similar characterization for higher dimension supersingular abelian varieties.
As an application, we can characterize the quasi-liftably birational class of irreducible symplectic varieties which come from the moduli space of sheaves on supersingular abelian surfaces. Consider the moduli space of Gieseker-Maruyama $H$-stable sheaves on $A$ with Mukai vector $v$ such that $p\nmid v^2$, denoted by $M_H(A,v)$. Recall that the fiber $K_v(A)$ of the Albanese morphism
\[
(\det, c_{2})\colon M_{H}(A,v)\to \Pic^{0}(A)\times A,
\]
is an irreducible symplectic variety of dimension $v^2-2$ in the sense of \cite{FL18}. A consequence in \loccit asserts that the generalized Kummer type variety $K_v(A)$ is quasi-liftably birational to some generalized Kummer variety $K_n(A')$, where $A'$ is a Fourier-Mukai partner of $A$. Here we verify that $A \cong A'$.
\begin{thm}\label{thm:supersingulargeneralizedKummer}Let $n = \frac{v^2}{2}$. Suppose that $p \nmid n$ and $p \nmid n+1$. Let $A$ be a supersingular abelian surface and let $K_v(A)$ be the generalized Kummer type variety. Then
\begin{enumerate}
\item $K_v(A)$ is quasi-liftably birational equivalent to $K_n(A)$.
\item if $K_v(A)$ is quasi-liftably birational to $K_v(A')$ for some abelian surface $A'$ and $v'$, then $A\cong A'$.
\end{enumerate}
\end{thm}
\noindent\textbf{Acknowledgement:} We are grateful to Lie Fu for helpful discussions and comments. The authors are supported by NSFC grants for General Program (11771086), Key Program (11731004), the Shu Guang Program (17SG01) of Shanghai Education Commission and NKRD Program of China (Grant No. 2020YFA0713200).
\section{Equivariant derived category of schemes}
\label{equi-dc}
\subsection{Equivariant quasi-coherent sheaves} \label{sec:eqcoh}
Let $G$ be a constant finite group scheme over $k$ and let $X$ be a quasi-compact and quasi-separated scheme over a field $k$ with a $G$-action $$\mu \colon G \times_k X \to X.$$
A {\em $G$-equivariant quasi-coherent sheaf} on $X$ is a pair $(\cF,\lambda)$ such that $\cF$ is a quasi-coherent sheaf on $X$ and $\lambda$ is a family $ \left\{\lambda_g \colon \cF \to g^*\cF \right\}_{g \in G}$ of isomorphisms satisfying the following cocycle condition:
\begin{equation}
\label{eq:cocyclecondition}
\lambda_{fg} = \lambda_f \circ \lambda_g \colon \cF \to g^* \cF \to (fg)^*\cF.
\end{equation}
We call $\lambda$ a \emph{$G$-linearization} of $\cF$. For instance, $\cO_{X}^{\rm can}= (\cO_X, \id_{\cO_{X \times G} })$ is a $G$-equivariant quasi-coherent sheaf. In this paper, we denote by $\Qcoh_G(X)$ the category of $G$-equivariant quasi-coherent $\cO_X$-modules.
\begin{rmk}
If the cocyle condition \eqref{eq:cocyclecondition} for $\gamma$ is missing, then $\cF$ will be called \emph{$G$-invariant}.
\end{rmk}
Let $[X/G]$ be the quotient stack given by the $G$-action on $X$ and denote by $\Qcoh([X/G])$ the category of quasi-coherent sheaves on $[X/G]$ (\cf\cite[\S 9]{Olsson2016}). There is a well-known stacky description for $G$-equivariant quasi-coherent sheaves on $X$ as follows.
\begin{lemma}
\label{lemma:stackydescrip}
There is a canonical equivalence of categories \[
\Qcoh([X/G]) \simeq \Qcoh_{G}(X).
\]
Moreover, if $X$ is locally noetherian, then we also have $\coh([X/G]) \cong \coh_G(X)$.
\end{lemma}
\begin{proof}
Assume that $X$ is locally noetherian.
Consider the functor
\[
\begin{aligned}
\Phi \colon \Qcoh_G(X) &\to \Qcoh([X/G])\\
(\cF,\lambda) &\mapsto \widetilde{\cF}
\end{aligned}
\]
where $\widetilde{\cF}$ is the sheaf on $[X/G]$ defined as follows.
For any object
\[
\begin{tikzcd}
\mathscr{P} \ar[r,"\pi"] \ar[d] & X_T \\
T
\end{tikzcd}
\]
lying on a $k$-scheme $T$, the pull-back $\pi^* \cF_T$ is a $G$-equivariant quasi-coherent sheaf on $\mathscr{P}$ as $\pi$ is $G$-equivariant.
The descent theory along $G$-torsor for the stack $\Qcoh_k$ of quasi-coherent sheaves establishes a canonical equivalence:
\begin{equation}\label{eq:descentalongtorsor}
\Qcoh(T) \simeq \Qcoh_G(\mathscr{P}),
\end{equation}
see \cite[Theorem 4.46]{vistolinote} for example. Take $\widetilde{F}(T, \mathscr{P}, \pi)$ to be one in the isomorphism class in $\Qcoh(T)$ corresponding to $\pi^* \cF_T$.
It remains to show that $\widetilde{F}$ is quasi-coherent (resp.\ coherent).
Consider the smooth covering $(G_X,\mu) \colon X \to [X/G]$ where $G_X$ is the trivial torsor on $X$ defined by $p_X \colon X \times_k G \to X$ and $\mu$ is the group action of $G$ on $X$. Since the $\lambda$ is an isomorphism
\[
p_X^* \cF \xrightarrow{\sim} \mu^* \cF,
\]
we can see $\widetilde{\cF}(X,G_X,\mu) \cong \cF$
by the previous construction, which is quasi-coherent (resp.\ coherent). Thus by \cite[Proposition 9.1.15]{Olsson2016}, we can see $\widetilde{\cF}$ is quasi-coherent (resp.~coherent).
The converse is similar. We just take $\Phi^{-1} \widetilde{\cF}$ to be the quasi-coherent sheaf (resp.\ coherent sheaf) $\widetilde{\cF}(X,G_X,\mu)$. The linearization $\lambda$ is from the definiton of quasi-coherent sheaves (resp.\ coherent sheaves) on $[X/G]$.
\end{proof}
The Lemma \ref{lemma:stackydescrip} implies that the category $\Qcoh_G(X)$ is a Grothendieck category (\cf\cite[\href{https://stacks.math.columbia.edu/tag/0781}{Tag 0781}]{stacks-project}). This promises a nice homological algebraic theory on $G$-equivariant quasi-coherent sheaves. In the following literature, the \emph{$G$-equivariant derived category} of $X$ means the bounded derived category of $\coh_{G}(X)$, denoted by $\D^b_G(X)$. A useful fact for $G$-equivariant derived category is the derived McKay correspondence established by Bridgeland--King--Reid \cite{BKR}. Let $n= \lvert G \rvert$. Let $X\!\sslash\!G \subset X^{[n]}$ be the closure of the set of $G$-clusters in $X$, i.e.~$G$-invariant $0$-dimensional subscheme $Z \subset X$ such that $H^0(Z,\cO_{Z}) \cong k[G]$. By composing the projection $ X\!\sslash\!G \to X$ and the quotient $X \to X/G$, we can get a birational morphism
\[
\tau\colon X\sslash G \to X/G.
\]
\begin{thm}[Bridgeland--King--Reid]
\label{thm:BKR}Assume that $(|G|,p)=1$.
Suppose the following two conditions hold
\begin{enumerate}[label=\tt{(BKR\arabic*)},leftmargin=3pc]
\item $\omega_X$ is locally trivial as a $G$-bundle,
\item $X\sslash G \times_{\tau} X\sslash G$ has dimension $d \leq \dim X+1$.
\end{enumerate}
Then there is a derived equivalence between $\D^b(X\!\sslash\!G)$ and $\D^b_G(X)$.
\end{thm}
\begin{proof}
The original proof in \loccit is for $k =\mathbb{C}$. However, this also proceeds for general case that $(|G|,p) =1$ (\cf\cite[Theorem 2.4.5]{dolgachevnote}).
\end{proof}
The following consequence is well-known over complex numbers (\cf\cite[\S 10.2]{BKR} or \cite[\S 3.2]{ploog07}).
\begin{cor}\label{cor:equivariantabelian}
Suppose $\Char(k)\neq 2$. Let $A$ be an abelian surface over $k$. Let $\iota$ be the involution on $A$ and $\langle \iota \rangle$ the finite group scheme over $k$ generated by $\iota$. Then $\D^b([A/\langle\iota\rangle])\simeq \D^b(\km(A))$.
\end{cor}
\begin{proof}
We can view $\km(A)$ as the closure of the subset of reduced $\langle\iota\rangle$-clusters in the Hilbert scheme $\hilb^{2}(A)$ of two points on $A$. In this case {\texttt{(BKR2)}} is satisfied and the canonical sheaf is trivial as an $\langle \iota \rangle$-bundle. Thus Theorem \ref{thm:BKR} implies that there is a Fourier-Mukai transform
\[
\Phi^{P} \colon \D^b(\km(A)) \to \D^b_{\langle \iota \rangle}(A) \simeq \D^b([A/\langle\iota\rangle])
\]
as $\Char(k) \neq 2$.
\end{proof}
Let us recall some general duality theorem for the $G$-equivariant derived category.
The $G$-action on $X$ also induces $G$-action on the triangulated category $\D^b(X)$. Thus we can also consider the $G$-equivariant category $\D^b(X)^{G}$ (\cf\cite[\S 2]{Elagin2014}). Under the assumption that $(|G|,p)=1$, we have an exact equivalence
\begin{equation}\label{eq:equicat}
\D^b(X)^{G} \simeq \D^b_{G}(X)
\end{equation}
by \loccit Theorem 7.1. Therefore, we will not distinguish $\D^b_{G}(X)$ and $\D^b(X)^{G}$ in the rest of the paper if $G$ acts on $X$ and $(\lvert G \rvert , p)=1$.
Let $\widehat{G}= \Hom(G, k^*)$ be the character group of $G$. There exists an induced $\widehat{G}$-action on the $G$-equivariant derived category $\D^b_G(X)$ as follows. For each $\chi \in \widehat{G}$, one can define a line bundle on $[X/G]$ twisted by $\chi$ as
\[
\cL_{\chi} \coloneqq \cO_{X}^{can} \otimes_k \chi\cong(\cO_X, \id_{\cO_{G\times X}} \otimes \chi) \quad \text{ for $\chi \in \widehat{G}$ }.
\]
The action of $\chi$ on $\D^b([X/G])$ is given by tensoring $\cL_{\chi}$. Under the identification $\D^b([X/G]) \simeq \D^b(X)^{G}$, the $\widehat{G}$-action on a $G$-equivariant object $(E,\lambda)$ is given by twisting the linearization:
\[
\lambda \otimes_k \chi \coloneqq \left\{ E \xrightarrow{\lambda_g \cdot \chi(g)} g^* E \right\}_{g \in G} \quad \text{ for any $\chi \in \widehat{G}$}.
\]
\begin{prop}[A. Elagin]\label{prop:dualities}
Assume $X$ is noetherian. There are exact equivalences
\[ \D^b_G(X)^{\widehat{G}} \simeq (\D^b(X)^{G})^{\widehat{G}} \simeq \D^b(X).
\]
\end{prop}
\begin{proof}
The first exact equivalence is given by \eqref{eq:equicat}. For the second exact equivalence, we shall note that $\D^b_G(X)$ is idempotent complete for any noetherian scheme or algebraic stack $X$ as $\coh(X)$ is a Grothendieck category and admits all products. Now we can apply the duality theorem \cite[Theorem 4.2]{Elagin2014} to conclude it.
\end{proof}
With the same notations in Corollary \ref{cor:equivariantabelian}, we have $\widehat{\langle \iota \rangle} \cong \langle \iota^* \rangle$, where $\iota^*$ is the character dual to $\iota$, acting naturally on $\D^b([A/\langle \iota \rangle])$.
\begin{cor}
If $\Char(k) >2$, then $\D^b\left([A/\iota]\right)^{\langle \iota^* \rangle} \simeq \D^b(A)$. In particular, if $A$ is an abelian surface, then $\D^b(\km(A))^{\langle \iota^* \rangle} \simeq \D^b(A)$. \qed
\end{cor}
\section{Lifting of derived equivalences}\label{sec:descend}
\subsection{Equivariant derived equivalences}
In this part, we will recollect some preliminary facts on lifting theory and descent theory of equivariant equivalences in \cite{KrugSosna2015,ploog07} and extend them to all algebraically closed fields. With the notations as in \S 2, we will always assume the order of $\lvert G \rvert$ is coprime to $p$. Let $X_1$ and $X_2$ be two projective $k$-schemes or quotient stacks equipped with $G$ actions. Let $\Phi \colon \D^b(X_1) \to \D^b(X_2)$ be an exact functor.
\begin{defn}\label{defn:liftdescent}
An exact functor $\widetilde{\Phi} \colon \D^b_G(X_1) \to \D^b_G(X_2)$ is called a {\it descent} of $\Phi$ if it fits into the following 2-commutative diagrams
\[ \begin{tikzcd}
\D^b(X_1) \ar[r,"\Phi"] \ar[d,"\pi_{1,*}"] & \D^b(X_2) \ar[d,"\pi_{2,*}"] \\
\D^b_G(X_1) \ar[r, "\widetilde{\Phi}"] & \D^b_G(X_2)
\end{tikzcd}
\quad
\begin{tikzcd}
\D^b_G(X_1) \ar[r,"\widetilde{\Phi}"] \ar[d,"\pi_{1}^*"] & \D^b_G(X_2) \ar[d,"\pi_{2}^{*}"] \\
\D^b(X_1) \ar[r, "\Phi"] & \D^b(X_2),
\end{tikzcd}
\]
where $\pi_i \colon X_i \to [X_i/G]$ are structure morphisms of quotients. We may also call $\Phi$ {\it a lift} of $\widetilde{\Phi}$.
\end{defn}
\begin{rmk}
The push-forward and pull-back functors $\pi^*_i, \pi_{i,*}$ are both exact under the assumption that $(|G|,p)=1$ (\cf\cite[Lemma 3.8]{Elagin2014}). Thus the functors in Definition \ref{defn:liftdescent} are all exact.
\end{rmk}
Consider the $k$-linear triangluated category $\cT = \D^b_G(X)$. The non-trivial line bundles $\cL_{\chi}$ on $[X/G]$ induce a natural $\widehat{G}$-action on $\cT$ by taking tensor product (see also \S\ref{sec:eqcoh}). This leads to the following definition.
\begin{defn}\label{defn:equivariantfunctor}
Let $\cT_1$ and $\cT_2$ be two $k$-linear triangulated categories with $G_1$ and $G_2$ actions respectively. Let $\gamma \colon G_1 \xrightarrow{\cong} G_2$ be a group isomorphism. An exact functor $\Phi\colon \cT_1 \to \cT_2$ is called $\gamma$-twisted equivariant if
\[
\Phi \circ g^* \simeq \gamma(g)^* \circ \Phi \quad \text{for all } g \in G_1 .
\]
When $\gamma = \id_G$, the $\Phi$ will be called $G$-equivariant for simplicity.
If a descent $\widetilde{\Phi}$ of $\Phi$ is $\widehat{G}$-equivariant as an exact functor, then it will be called a \emph{$\widehat{G}$-equivariant descent} of $\Phi$.
\end{defn}
The following examples of equivariant functors is the most frequently used throughout this paper.
Let $\gamma\colon G \xrightarrow{\cong} G$ be an abstract group isomorphism, then there is an action of $G$ on $X_1 \times_k X_2$ (or $\D^b(X_1 \times X_2)$) by
\[
g \cdot (x_1,x_2) = (g\cdot x_1,\gamma(g) \cdot x_2).
\]
For instance, if $\gamma= \id$, then this action is just the diagonal action of $G$. The $G$-equivariant derived category of $X_1 \times_k X_2$ under the action given by $\gamma$ is denoted by $\D^b_{G_{\gamma}}(X_1 \times X_2)$. If there is an object $\cP=(P,\lambda)$ in $\D^b_{G_\gamma}(X_1 \times X_2)$ such that
\[\Phi \simeq \mathbf{R}\!p_{2,*}(p_1^*(-)\otimes \cP),
\] then $\Phi$ is a $\gamma$-twisted $G$-equivariant exact functor. A $\gamma$-twisted $G$-equivariant functor of this form is called {\it integral} and will be denoted by $\Phi^{\cP}$, where the object $\cP$ is called its \emph{kernel}. In this case, we can forget the linearization of the kernel $\cP$, which will defines an integral functor
\[
\Phi^{P} \colon \D^b(X_1) \to \D^b(X_2).
\]
It is not hard to check that $\Phi^{P}$ is a lift of $\Phi^{\cP}$.
A well-known fact is that a $G$-equivariant descent of a Fourier-Mukai transform is still an exact equivalence (\cf\cite[Proposition 3.10]{KrugSosna2015} or \cite[Lemma 5]{ploog07}).
\begin{prop}\label{prop:equivalencelift}
Let $\cP=(P,\lambda)$ be a $G_{\gamma}$-equivariant object in $\D^b_{G_{\gamma}}(X_1 \times X_2)$ for some abstract isomorphism $\gamma\colon G \xrightarrow{\cong} G$. If $\Phi^{P}_{X_1 \to X_2}$ is an exact equivalence, then the integral functor
\[\Phi\coloneqq\Phi^{\cP}\colon \D^b_G(X_1) \to \D^b_G(X_2)\]
will also be an exact equivalence.
\end{prop}
The proof in \loccit also works for any algebraically closed field $k$ such that $(|G|,p)=1$. We sketch it here for reader's convenience.
\begin{proof}
Take the kernel of right adjoint inverse of $\Phi$: $$Q= P^{\vee} \otimes p_{X_1}^* \omega_{X_1}[\dim X_1].$$ It admits a natural $G$-lineaization $\lambda'$ induced by $\lambda$. Denote $(Q,\lambda')$ by $\cQ$.
Note that the pull-back functor $\pi^*$ is just the forgetful functor of linearization. Therefore $\pi^*(\cQ \circ \cP) \cong Q \circ P\cong \cO_{\Delta_{X_1}}$. On the other hand, we have $\Aut(\cO_{\Delta_{X_1}}) = k^*$ in $\D^b(X_1 \times X_1)$.
Via computing the cohomology of $G$, we can show that the set of linearizations of $\cO_{\Delta_{X_1}}$ forms a principal homogeneous space under the action of $\widehat{G}=\Hom(G,k^*)$ (\cf\cite[Lemma 1]{ploog07}). In particular, since $\cO_{\Delta_{X_1}}$ admits the trivial linearization, the set of its linearizations is equal to $\widehat{G}$. It means that there is some $\chi \in \widehat{G}$ such that $\cQ \circ \cP \cong \Delta_*\cL_{\chi}$. Since tensoring line bundles will induce a derived equivalence, we can conclude that $\Phi$ is an exact equivalence.
\end{proof}
Suppose $G$ is a cyclic group. Then $\rH^2(G,k^*) =\rH^2(\widehat{G},k^*)=0$ by a direct computation. This implies that any Fourier-Mukai kernel $P$ (resp.~$\cP=(P,\lambda)$) admits a $G$-linearization (resp.~$\widehat{G}$-linearization). Now we can lift a $G$-equivariant derived equivalence along cyclic Galois coverings, i.e.~morphisms of smooth $k$-stacks $Y_i \to X_i=Y_i/G$\footnote{The quotient here is in stack-theoretical sense (\cf\cite{romagny05}).}.
\begin{cor}[{\cite[Proposition 4.3]{KrugSosna2015}}]\label{cor:cyclicdescent}
Let $Y_1 \to X_1$ and $Y_2 \to X_2$ be two cyclic Galois coverings. Suppose that there is an isomorphism $\gamma \colon \widehat{G} \xrightarrow{\cong} \widehat{G}$ . Then any $\gamma$-twisted equivariant Fourier-Mukai transform
\[
\Phi^{\cP} \colon \D^b(X_1) \xrightarrow{\sim} \D^b(X_2)
\]
lifts to a derived equivalence
\[
\widetilde{\Phi} \colon \D^b(Y_1) \xrightarrow{\sim} \D^b(Y_2).
\] \qed
\end{cor}
\subsection{Proof of Theorem 1.2 (i) and (ii)}
\begin{prop}[Proposition 3.1, \cite{Stellari2007}]\label{prop:stellari1}
If there is a Fourier-Mukai transform
\[
\Phi^{P} \colon \D^b(A_1) \xrightarrow{\sim} \D^b(A_2),
\]
then $\D^b([A_1/\iota]) \simeq\D^b([A_2/\iota])$.
\end{prop}
\begin{proof}
We repeat Stellari's proof for this statement here.
Let $\gamma$ be the abstract isomorphism $ [-1]_{A_1} \mapsto[-1]_{A_2}$.
Let $\tau$ be the generator of $G_{\gamma}$. Let $T_{(x,y)}$ be the translation of a point $(x,y)$ on $A_1 \times A_2$. As $\Phi^{\tau^* P} \in \text{Eq}(\D^b(A_1),\D^b(A_2))$, by \cite[Corollary 3.4]{orlov02}, we have
\[
\tau^* P \cong T_{(a,0),*} P \otimes p_{A_1}^* \cL_{\alpha}
\]
for some $a \in A_1(k)$, line bundle $\cL_{\alpha} \in \Pic^0(A_1) \cong \widehat{A}_1$.
It is well-known that the Fourier-Mukai kernel $P$ is isomorphic to $\cE[i]$ for some semi-homogeneous sheaf $\cE$ on $A_1 \times A_2$, which implies
\[
T_{(a,0),*}P \cong T_{(a,0),*} \cE [i] \cong P \otimes \cL_{\beta}
\]
for some $\cL_{\beta} \in \Pic^0(A_1 \times A_2)(k)$. Thus $\tau^* P \cong P \otimes \cL$, where $\cL= \cL_{\beta} \otimes p_A^*\cL_{\alpha}[i]$ . As $\Pic^0(A_1\times A_2)(k)$ is divisible, we can find a line bundle $\cN$ on $A_1 \times A_2$ such that $\cN^2 = \cL$. Let $\widetilde{P} = P \otimes \cN$. We can see
\begin{equation}\label{eq:tildeP}
\tau^* \widetilde{P} \cong \tau^* P \otimes \tau^* N \cong P \otimes \cL \otimes N^{\vee} \cong \widetilde{P}.
\end{equation}
On the other hand, since $\widetilde{P}$ is obtained from $P$ by tensoring line bundles, it also induces a Fourier-Mukai transform from $A_1$ to $A_2$. The isomorphism \eqref{eq:tildeP} ensures that $\widetilde{P}$ is $G_{\Delta}$-invariant, which also implies $\widetilde{P}$ admits a $G_{\gamma}$ linearization $\lambda$ as $\rH^2(G_{\gamma}, k^*)=0$. Therefore $\D^b([A/\iota]) \simeq \D^b([B/\iota])$ by Proposition \ref{prop:equivalencelift}.
\end{proof}
Combining the identification in Corollary \ref{cor:equivariantabelian}, the statement (i) in Theorem \ref{mainthm1} can be concluded.
Krug and Sosna proved the converse direction for odd dimensional abelian varieties (\cf\cite[Proposition 5.13]{KrugSosna2015}). Their statement there is over $\mathbb{C}$, while we extend it to algebraically closed fields in positive characteristic.
\begin{prop}
Let $A_1$ and $A_2$ are two abelian varieties in dimension $2n+1$, $n \in \mathbb{Z}_{>0}$. Any Fourier-Mukai transform between Kummer stacks
\[
\Phi^{\cP} \colon \D^b([A_1/\iota]) \xrightarrow{\sim} \D^b([A_2/\iota])
\]
has a lift
\[
\widetilde{\Phi} \colon \D^b(A_1) \xrightarrow{\sim} \D^b(A_2).
\]
\end{prop}
\begin{proof}
The stack $[A/\iota]$ has non-trivial torsion canonical bundle when $\dim A$ is odd ({\it cf}.~\cite[see][]{KrugSosna2015}).
Let $X_i = [A_i /\iota]$.
We can take the abstract isomorphism $\langle K_{X_1}\rangle \xrightarrow[\sim]{\gamma} \langle K_{X_2} \rangle$, whose actions are given by tensoring. Notice that $K_{X_i} \otimes (-)$ is just the shift of Serre functor $S_i[-\dim_{X_i}]$ on $\D^b(X_i)$. Therefore $\Phi^{\cP}$ is $\gamma$-twisted equivariant as Serre functor commutes with all Fourier-Mukai transforms. Thus any $\D^b([A_1/\iota]) \xrightarrow{\sim} \D^b([A_2/\iota])$ can be lifted to an exact equivalence $\D^b(A_1) \simeq \D^b(A_2)$ by the Corollary \ref{cor:cyclicdescent}.
\end{proof}
This proves Theorem \ref{mainthm1} (ii).
\begin{rmk}
We wonder if $\rK(A_1) \cong \rK(A_2)$ will imply $\D^b(A_1) \cong \D^b(A_2)$ (which doesn't satisfy \texttt{(BKR2)}). The same question can be asked for two pairs $(X_1,G)$ and $(X_2,G)$ satisfying \texttt{(BKR1)} and \texttt{(BKR2)} such that their quotients have isomorphic crepant resolution.
\end{rmk}
\begin{comment}
Let $(X,G)$ and $(X',G)$ be two pairs which satisfy assumptions of the BKR's theorem. Assume that we have an isomorphism $\varphi \colon X\sslash G \to X'\sslash G$. Consider the following commutative diagram from the BKR's theorem
\begin{equation}
\begin{tikzcd}[row sep = small, column sep =small]
& & \cZ \times_{\varphi} \cZ' \ar[rd] \ar[ld] & & \\
& \cZ \ar[rd] \ar[ld] & & \cZ' \ar[ld] \ar[rd] & \\
X & & X\sslash G \Cong^{\varphi} X'\sslash G & & X'
\end{tikzcd}
\end{equation}
where $\cZ$ and $\cZ'$ are the restrictions of universal $[n]$-points on $X\sslash G$ and $X'\sslash G$ respectively.
We can see the kernel of \[\Phi \coloneqq \Phi^{BKR}_{X'\sslash G \to [X'/G]} \circ \Phi^{\cO_{\Gamma_{\varphi}}}_{X\sslash G \to X'\sslash G} \circ\Phi^{BKR}_{[X/G] \to X\sslash G} \] is just $\cP \coloneqq(p_{13}^*i_* \cO_{Z \times_{\varphi} Z'},\lambda) \in \D^b_{G\times G}(X \times X)$. We can apply \ref{cor:cyclicdescent} to achieve our main result in this part.
Let $\widetilde{\Phi}\coloneqq\Phi^{p_{13}^*i_* \cO_{Z \times_{\varphi}Z}}_{X \to X'}$. We need to show that it is a lift of $\Phi$
\begin{equation}\label{eq:inflatcompatibility}
\begin{tikzcd}
D^b(X) \ar[r,"\widetilde{\Phi}"] \ar[d,"\pi'_*"] & D^b(X') \ar[d,"\pi_*"] \\
D^b_G(X) \ar[r,"\Phi","\sim"'] & D^b_G(X')
\end{tikzcd}
\end{equation}
such that the $\widetilde{\Phi}$ is a derived equivalence. In fact, we have the following statement.
\begin{thm}\label{thm:liftBKR}
Suppose that $G$ is cyclic. Let $(X_1,G)$ and $(X_2,G)$ be two pairs such that they satisfy \texttt{(BKR1)} and \texttt{(BKR2)}. If their quotients have isomorphic crepant resolution $Y$, then there is a derived equivalence
\[
\widetilde{\Phi} \colon \D^b(X_1) \xrightarrow{\sim} \D^b(X_2).
\]
\end{thm}
\begin{proof}
Under the assumption, we can identify $X_1\sslash G$ and $X_2\sslash G$, denoted by $Y$. We have the following diagram
\[
\begin{tikzcd}[column sep=small]
& \D^b(Y) \ar[ld,"\Phi_1"'] \ar[rd,"\Phi_2"] & \\
\D^b_G(X_1) & & \D^b_G(X_2)
\end{tikzcd}
\]
where $\Phi_i$ is the BKR equivalence of $X_i$. Let $\sigma$ be the generator of $G$. In this case, $\widehat{G}$ is generated by $\widehat{\sigma}$. Let $T_i$ be the autoequivalence of $\D^b(Y)$ given by
\[
T_i \coloneqq \Phi_1^{-1} \circ (\cL_{\widehat{\sigma}} \otimes -) \circ \Phi_1.
\]
Clearly, since $\Phi_i$ are $G$-equivariant, $T_1$ commutes with $T_2$.
Suppose there is some $K \in \Aut(\D^b(Y))$ such that $K \circ T_1 \simeq T_2 \circ K$, then the equivalence
\[
\Phi \coloneqq \Phi_2 \circ K \circ \Phi_1 \colon \D^b_G(X_1) \to \D^b_G(X_2)
\]
satisfies
\[
(\cL_{\widehat{\sigma}} \otimes -) \circ \Phi \simeq \Phi \circ (\cL_{\widehat{\sigma}} \otimes -).
\]
Then by applying Corollary \ref{cor:cyclicdescent}, we can see there is a lift $\widetilde{\Phi}$. Therefore, it is sufficient to find such autoequivalence $K$.
Let $K= \sum_{i=0}^{n-1} T_1^{-i} \circ T_2^i$. As $T_1$ commutes with $T_2$, we can see $(\id-T_1^{-1}\circ T_2) \circ K = \id$. Thus $K \in \Aut(\D^b(Y))$. Moreover, we can see $T_1^{-1} \circ K \circ T_2 \simeq K$.
\end{proof}
\begin{cor}
If $A$ and $A'$ are two abelian surfaces over $k$ such that $\km(A) \cong \km(A')$, then $\D^b(A) \simeq \D^b(A')$.
\end{cor}
\end{comment}
\subsection{Lifting Kummer structures}\label{subsec:liftingKummerstructure}
To prove the Theorem (iii), we need to use the lifting of Kummer surfaces. A technical lemma is
\begin{lemma}\label{lemma:liftingkummer}
Let $X$ be a K3 surface over $k$. Suppose $\cX$ is a relative K3 surface, lifting $X$ to some finite extension $V$ of $W$ such that the specialization map is an isomorphism
\begin{equation}\label{lift-ns}
\NS(\cX_{F}) \xrightarrow{\sim} \NS(X).
\end{equation}
For any abelian surface $A$ satisfying $\km(A) \cong X$, there is an abelian scheme $\cA$ over $V$ such that $\cA_{k} \cong A$ and $\cX \cong \km(\cA)$.
\end{lemma}
\begin{proof}
We may follow the idea of \cite[Proposition 1.1]{shioda78} to prove the existence of the lifting. Let $E_a$ be the exceptional curve on $X$ with respect to the $2$-torsion point $a \in A[2]$. It has a unique Cartier divisor extension $\cE_a \in \Pic(\cX)$ by the construction. Moreover, the divisor
\[
\cE \coloneqq \sum_{a \in A[2]} \cE_{a} \in \Pic(\cX)
\]
is $2$-divisible as $\cE$ is equal to the extension of $\frac{1}{2} \sum_{a \in A[2]} E_a \in \NS(X)$ via \eqref{lift-ns}. Therefore, we have a double covering of $V$-schemes
\[
\pi \colon \cY \longrightarrow \cX,
\]
which is branched along $\cE \subset \cX$. We can contract the curves $\cC_a \subset \cY$ lying over $\cE_a$ to points:
\[
f \colon \cY \longrightarrow \cA,
\]
where $\cA$ is a smooth proper scheme over $V$. Thus we can see $\cA$ is the required abelian scheme since $\cX \cong \cY/ \iota \cong \km(\cA)$ where $\iota$ is the involution of $\cY$ induced by the double covering $\pi$.
\end{proof}
\subsection{Derived Torellli theorem and specialization}
Let $A_1$ and $A_2$ be two abelian varieties. Consider the set of \emph{symplectic isomorphisms}
\[
U(A_1,A_2)=\left\{ f = \begin{psmallmatrix}
f_1 & f_2\\
f_3& f_4
\end{psmallmatrix}\colon A_1 \times \widehat{A}_1 \xrightarrow{\sim} A_2 \times \widehat{A}_2 \Big| f^{-1}= \widetilde{f} \coloneqq \begin{psmallmatrix}
\hat{f_4} & -\hat{f_2} \\
-\hat{f_3} & \hat{f_1}
\end{psmallmatrix}\right\}.
\]
Let $\text{Eq}(\D^b(A_1), \D^b(A_2))$ be the set of exact equivalences from $\D^b(A_1)$ to $\D^b(A_2)$. Then the derived Torelli theorem for abelian varieties asserts
\begin{thm}[Orlov--Polishchuk]\label{thm:OrlovPolischuk}
There is a morphism
\begin{equation}
\label{eq:symplectic}
\gamma_{A_1,A_2}\colon \text{Eq}\left(\D^b(A_1),\D^b(A_2)\right) \to U(A_1,A_2).
\end{equation}
If $U(A_1,A_2) \neq \emptyset$, then $A_1$ and $A_2$ are Fourier-Mukai partners (\cf\cite{orlov02} or \cite{Polishchuk2003}).
\end{thm}
The following lemma shows that derived equivalence is preserved under smooth specialization.
\begin{lemma}\label{lemma:specializationofDE}
For two abelian schemes $\cA_1$ and $\cA_2$ over $V$, any derived equivalence $\D^b(\cA_{1,F}) \xrightarrow{\sim} \D^b(\cA_{2,F})$ has a specialization $\D^b(\cA_{1,k}) \to \D^b(\cA_{2,k})$, which is also a derived equivalence.
\end{lemma}
\begin{proof}
It is known that any derived equivalence $\D^b(\cA_{1,F}) \xrightarrow{\sim} \D^b(\cA_{2,F})$ induces a symplectic isomorphism $f_F \colon \cA_{1,F} \times \widehat{\cA}_{1,F} \xrightarrow{\sim}\cA_{2,F} \times \widehat{\cA}_{2,F}$.
By using the Matsusaka-Mumford theorem (\cf\cite[Chapter I. Corollary 1]{MatsusakaMumford}), $f_F$ can be extended to an isomorphism $f \colon \cA_1 \times \widehat{\cA}_1 \xrightarrow{\sim}\cA_2 \times \widehat{\cA_2}$ such that the restriction $f_k$ is a specialization of $f_F$.
On the other hand, we can also take the specialization of $\widetilde{f}_F$. As $(\widetilde{f} \circ f)_F = \widetilde{f}_F \circ f_F = \id_{\cA_{1,F} \times \widehat{\cA}_{1,F}}$. The graph of $\widetilde{f} \circ f$ is the diagonal of $\cA \times \widehat{\cA}$. Thus the specialization $f_k$ is also a symplectic isomorphism. Hence the derived equivalence $\D^b(\cA_{1,k}) \xrightarrow{\sim} \D^b(\cA_{2,k})$ is from Theorem \ref{thm:OrlovPolischuk}.
\end{proof}
\subsection{Proof of Theorem \ref{mainthm1} (iii): finite height case}
Given two abelian surfaces $A_1$ and $A_2$, if there is an exact derived equivalence $\D^b(A_1) \simeq \D^b(A_2)$, then we have $\D^b(\km(A_1)) \cong \D^b(\km(A_2))$ by Proposition \ref{prop:stellari1}. Then we have
$$\km(A_1)\cong \km(A_2)$$
because the Kummer surface does not have non-trivial Fourier-Mukai partners ({\it cf.}~\cite[Theorem 1.1]{LO15}).
Recall that an abelian surface is called of finite height if it is not supersingular. Assume that $A_1$ and $A_2$ are of finite heights. In this case, if $X$ is a Kummer surface isomorphic to $\km(A_1)\cong \km(A_2)$, then $X$ is a K3 surface with finite height. According to \cite[(5.6)]{TateK3finiteh}, there is a N\'eron-Severi lattice preserving lifting $\cX$ of $X$ over some $V$. By Lemma \ref{lemma:liftingkummer}, there exist relative abelian surfaces $\cA_1$ and $\cA_2$ over $V$ such that $$\km(\cA_1)\cong \cX\cong \km(\cA_2).$$
Due to \cite[Theorem 0.1 (1)]{HosonoLianEtAl2003}, the generic fibers of $\cA_1$ and $\cA_2$ are geometrically derived equivalent. By using the standard spreading out argument and \cite[Lemma 2.12]{orlov02}, we can find a finite extension $V'$ of $V$ such that $\D^b(A_{1,F'}) \xrightarrow{\sim} \D^b(A_{2,F'})$, where $F'$ is the fraction field of $V'$. The assertion follows from Lemma \ref{lemma:specializationofDE}.
\section{Supersingular derived Torelli theorem}\label{sec:FMpartnersabelian}
In this section, we will focus on derived categories of supersingular abelian varieties. There is a cohomological realization of the derived Torelli theorem for supersingular abelian varieties using supersingular abelian crystals and supersingular K3 crystals.
\subsection{Abelian crystals and K3 crystals} For supersingular abelian varieties, we can give a cohomological realization of Theorem \ref{thm:OrlovPolischuk} via Ogus' supersingular Torelli theorem (\cite{Og79}).
Let $A$ be an abelian variety of dimension $g$ over $k$. Recall that $A$ is {\it supersingular} if the $\rH_{\crys}^1(A/W)$ is purely of slope $\frac{1}{2}$ as a weight one $F$-crystal with natural Frobenius. Over an algebraically closed field, it is also equivalent to say that $A$ is isogenous to a $g$-fold product of supersingular elliptic curves (\cf\cite[Theorem (4.2)]{oort74}).
\begin{defn}
An abelian crystal of genus $g$ is a $F$-crystal $(H,\varphi)$ of rank $2g$ endowed with an isomorphism between $F$-crystals $\tr\colon \Lambda^{2g}H \to W(-n)$, whose Hodge numbers of Hodge polygon are both equal to $g$. The isomorphism $\tr$ is called the trace of abelian crystal $(H,\varphi)$.
\end{defn}
As its name indicates, for a $g$-dimensional abelian variety $A$, the $F$-crystal $\rH^1_{\crys}(A/W)$ is an example of abelian crystal. The trace map of $\rH^1_{\crys}(A/W)$ is given by
\[
\Lambda^{2g} \rH^1_{\crys}(A/W) \cong \rH^{2g}_{\crys}(A/W) \cong W(-n).
\]
Here the first isomorphism is coming from the multiplication of $A$ and the cup-product of its crystalline cohomology.
The crystalline Torelli theorem of supersingular abelian varieties states that, for any integer $g \geq 2$, there is a bijection
\begin{equation}
\left\lbrace \begin{matrix}\text{isomorphism classes of } \\
\text{ supersingular abelian varieties}\\
\text{of genus $g$}\end{matrix} \right\rbrace \xrightarrow[H^1_{\crys}(-/W)]{\sim} \left\lbrace \begin{matrix}
\text{isomorphism classes of} \\
\text{supersingular abelian crystals} \\
\text{ of genus $g$}\end{matrix} \right\rbrace_.
\end{equation}
See \cite[Theorem 6.2]{Og79}.
The second crystalline cohomology of an abelian surface is also equipped with a structure called \emph{K3-crystal} which is defined as follows.
\begin{defn}
We say a $F$-crystal $(H,\varphi)$ is K3-crystal with rank $n$, if there is a symmetric bilinear form $\langle, \rangle_H$ on $H$, satisfying:
\begin{enumerate}
\item $H$ is of weight $2$, that means $p^2H \subset \im(\varphi)$;
\item the Hodge number $h^0(H)=1$, that means $\varphi \otimes \id_k$ is of rank 1;
\item $\langle, \rangle_H$ is perfect;
\item $\langle \varphi x, \varphi y \rangle_H = p^2 \sigma\langle x, y \rangle_H$ for any $x,y \in H$.
\end{enumerate}
Inside a K3-crystal $(H,\varphi)$, there is a natural $\mathbb{Z}_p$-lattice defined as
\[
T_H \coloneqq H^{\varphi =p} =\left\{ x \in H \big| \varphi(x) = px \right\}
\]
equipped with bilinear form $\langle,\rangle_{T_H}$ induced from $H$.
A K3-crystal is called \emph{supersinguar} if it is purely of slope 1. If $(H,\varphi)$ is supersingular of rank $n$, then its Tate module $T_H$ is a free $\Z_p$-module of rank $n$, which comes from the definition of pure of slope $1$. Its discriminant is equal to $p^{-2\sigma_0}$ for some integer $\sigma_0 \geq 1$, which is called the \emph{Artin invariant} of $H$.
\end{defn}
Let $A$ be an abelian surface over $k$. Then $\rH^2_{\crys}(A/W)$ is naturally endowed with a K3-crystal structure. Consider the canonical isomorphism
\[\rH^2_{\crys}(A/W) \cong \Lambda^2\rH^1_{\crys}(A/W),
\]
one can see that $A$ is supersingular if and only if $\rH^2_{\crys}(A/W)$ is supersingular as a K3-crystal. An important observation due to Ogus is that the functor $\Lambda^2$ forms an equivalence from the category of supersingular abelian crystals of genus 2 to the category of supersingular K3 crystals of rank 6 (cf.\ \cite[Proposition 6.9]{Og79}). Thus the crystalline Torelli theorem for supersingular abelian surfaces can be rephrased in terms of K3-crystals, that means
two supersingular abelian surfaces $A_1$ and $A_2$ are isomorphic if and only if
\[
\rH^2_{\crys}(A_1/W)\cong \rH^2_{\crys}(A_2/W).
\]
\begin{rmk}\label{rmk:pp}
This also implies that any supersingular abelian surface $A$ is principal polarized, since there is a natural isomorphism of K3-crystals $\rH^2_{\crys}(A/W) \xrightarrow{\sim} \rH^2_{\crys}(\widehat{A}/W)$. This can also deduced from the classification of N\'eron-Severi lattices of supersingular abelian surfaces (\cf\cite[Lemma 6.2]{FL18}).
\end{rmk}
One can characterize $K3$-crystals via the characteristic space. Let $H$ be a supersingular K3 crystal with Artin invariant $\sigma_0$. We have the an orthogonal decomposition (not unique!) for its Tate module:
\begin{equation}\label{eq:decompositonTatemodule}
(T_H, \langle,\rangle_{T_H}) \cong (T_0, p \langle, \rangle_{T_0}) \oplus (T_1, \langle, \rangle_{T_1}),
\end{equation}
such that $T_0$ is of rank $2\sigma_0$, $\langle, \rangle_{T_0}$ and $\langle, \rangle_{T_1}$ are both perfect, since the cokernel of $T_H \hookrightarrow T_H^{\vee}$ is killed by $p$. The kernel
\[
\overline{H} \coloneqq \ker (T_H \otimes_{\mathbb{Z}_p} k \to H\otimes_W k)
\]
forms a $\sigma_0$-dimensional $\mathbb{F}_p$-vector space which is totally isotropic with respect ot $\langle,\rangle_{T_H}$ and it is isomorphic to the image of
\[
H \cong H^{\vee} \to T_H^{\vee}\otimes_{\mathbb{Z}_p} W
\]
It also forms a \emph{strictly characteristic subspace} of $T_0 \otimes_{\mathbb{Z}_p} k$ (\cf\cite[Definition 3.19]{Og79} for the definition and {\loccit~Remark 3.16} for the explanation). Let $\mathbb{K}_H = \varphi^{-1}(\overline{H}) \subset T_0 \otimes_{\mathbb{Z}_p} k$, which is another strictly characteristic subspace of $T_0\otimes_{\mathbb{Z}_p} k$. We also call it the characteristic space of the $K3$-crystal $H$.
The classification of $K3$-crystals (see \cite[Theorem 3.20]{Og79}) asserts that
\begin{thm}[Ogus]\label{thm:characteristicspace}
For two supersingular $K3$-crystals $H$ and $H'$ of the same rank, $H \cong H'$ if and only if there is an isomorphism of pairs $(T_0 \otimes k, \mathbb{K}_H) \cong (T_0' \otimes k, \mathbb{K}_{H'})$.
\end{thm}
\subsection{Supersingular derived Torelli theorem} \label{subsec:supersingularderivedtorelli}
Let $\cP$ be the Poincar\'e line bundle on $A \times \widehat{A}$ associated to the polarization $\varphi_{\cL}$. There is a canonical isomorphism between Dieudonn\'e modules:
\begin{equation}\label{eq:dieudual}
\Phi_A \colon \rH^1_{\crys}(A/W)^* \xrightarrow{} \rH^1_{\crys}(\widehat{A}/W),
\end{equation}
which corresponds to the first crystalline Chern class
\[
c_1(\cP) \in \rH^1_{\crys}(A/W) \otimes_W \rH^1_{\crys}(\widehat{A}/W) \subseteq \rH^2_{\crys}(A \times \widehat{A}/W);
\]
see \cite[Th\'eor\`em 5.1.2]{BBM2006}. We have a natural quadratic form on the $F$-crystal $\rH^1_{\crys}(A \times \widehat{A})$:
\[
q_A(a, \alpha) = 2 \Phi_A^{-1}(\alpha)(a) \quad \text{for } (a,\alpha) \in \rH^1_{\crys}(A/W) \oplus \rH^1_{\crys}(\widehat{A}/W)
\]
With abelian crystals, we are able to provide a cohomological description of Theorem \ref{thm:OrlovPolischuk} for supersingular abelian varieties.
It is clear that any symplectic isomorphism $f \colon A_1\times \widehat{A}_1 \xrightarrow{\sim} A_2 \times \widehat{A}_2$ will induce an isomorphism of abelian crystals
\[
\varphi \colon \rH^1_{\crys}(A_1/W) \oplus \rH^1_{\crys}(\widehat{A}_1/W) \xrightarrow{\sim}\rH^1_{\crys}(A_2/W) \oplus \rH^1_{\crys}(\widehat{A}_2/W),
\]
by taking K\"unneth decomposition.
Conversely, we can write the isomorphism $\varphi$ into a $2 \times 2$-matrix
\[
\rH^1_{\crys}(A_1/W) \oplus \rH^1_{\crys}(\widehat{A}_1/W) \xrightarrow{
\begin{pmatrix}
\varphi_1 & \varphi_2 \\
\varphi_3 & \varphi_4
\end{pmatrix}}\rH^1_{\crys}(A_2/W) \oplus \rH^1_{\crys}(\widehat{A}_2/W)
\]
in which $\varphi_i$ are all morphisms between $F$-crystals. Then $\varphi$ being isometry in terms of the quadratic form $q_A$ is equivalent to satisfying matrix equation
\[
\begin{pmatrix}
\varphi_1^t & \varphi_3^t \\
\varphi_2^t & \varphi_4^t
\end{pmatrix}
\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}
\begin{pmatrix}
\varphi_1 & \varphi_2 \\
\varphi_3 & \varphi_4
\end{pmatrix}
= \begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}.
\]
It implies that
\[
\varphi^{-1}= \begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}
\begin{pmatrix}
\varphi_1^t & \varphi_3^t \\
\varphi_2^t & \varphi_4^t
\end{pmatrix}
\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}=
\begin{pmatrix}
\varphi_4^t & \varphi_2^t \\
\varphi_3^t & \varphi_1^t
\end{pmatrix}
\]
Now suppose $A_1$ or $A_2$ is supersingular. By Ogus's crystalline Torelli theorem for supersinsgular abelian varieties (cf.\ \cite[Theorem 6.2]{Og79}), we can find an isomorphism $f \in \Hom(A_1\times \widehat{A}_1, A_2\times \widehat{A}_2)$ such that $\rH^1_{\crys}(f) = \varphi$ . It also satisfies $H^1_{\crys}(f_i)= \varphi_i$ if we rewrite $f$ into the following matrix form uniquely:
\[
f=\begin{pmatrix}
f_1 & f_2 \\
f_3 & f_4
\end{pmatrix}.
\]
Note that
\[
c_1(\widehat{\cP}) = - c_1(\cP) \in \rH^1_{\crys}(\widehat{A}/W) \otimes \rH^1_{\crys}(A/W)
\]
by the identification $\rH^1_{\crys}(\widehat{\widehat{A}}/W) \cong \rH^1_{\crys}(A/W)$. Thus we have
\[
\varphi_1^t = \rH^1_{\crys}(\widehat{f}_1) \quad \varphi_2^t = - \rH^1_{\crys}(\widehat{f}_2) \quad \varphi_3^t = - \rH^1_{\crys}(\widehat{f}_3) \quad \varphi_4^t = \rH^1_{\crys}(\widehat{f}_4).
\]
Thus $f \in U(A_1,A_2)$. Therefore, we have the following
\begin{prop}
For arbitrary supersingular abelian varieties $A_1$ and $A_2$ over $k$, the following statements are equivalent.
\begin{enumerate}
\item $\D^b(A_1) \simeq \D^b(A_2)$.
\item There is a isometry of abelian crystals
\[
\rH^1_{\crys}(A_1 \times \widehat{A}_1/W) \xrightarrow{\sim} \rH^1_{\crys}(A_2 \times \widehat{A}_2/W).
\]
with respect to $q_{A_i}$.
\end{enumerate}
\end{prop}
For supersingular abelian surfaces, we have the following consequence via K3-crystals.
\begin{thm}\label{thm:supersingularDerivedTorelli}
Let $A_1$ and $A_2$ be two supersingular abelian surfaces over $k$. Then
$\D^b(A_1)\cong \D^b(A_2)$ if and only if
$A_1 \cong A_2$.
\end{thm}
\begin{proof}
It is clear that when $\dim A_1 = 2$ then $\dim A_2 =2$ if they are derived equivalent. The given derived equivalence $\Phi^P$ induces an isometry between K3 crystals $\widetilde{\rH}(A_1/W) \cong \widetilde{\rH}(A_2/W)$, where
\[
\widetilde{\rH}(A_i/W) \coloneqq W(-1) \oplus \rH^2_{\crys}(A_i/W) \oplus W(-1)
\]
is the Mukai K3-crystal of $A_i$ equipped with the Mukai pairing.
Since $\rH^2_{\crys}(A_i/W)^{\varphi_i =p} \cong \NS(A_i) \otimes \ZZ_p$ (\cf \cite[(1.6)]{Og79}), we have
\[
\widetilde{\rH}(A_i/W)^{\varphi_i =p} \cong \mathbb{Z}_p^{\oplus 2} \oplus \NS(A_i) \otimes \ZZ_p.
\]
Therefore, the characteristic subspace of $\widetilde{\rH}(A_i/W)$ is equal to
\[
\ker \left( \NS(A_i) \otimes k \to \rH^2_{\dR}(A_i/k) \right),
\]
which is isomorphic to the characteristic subspace of $\rH^2_{\crys}(A_i/W)$. Let
\[
\NS(A_i) \otimes \mathbb{Z}_p \cong \left(T_0^{(i)}, p\langle, \rangle \right)\oplus \left(T_1^{(i)}, \langle, \rangle\right)
\]
be decomposition of $\mathbb{Z}_p$-lattice as in \eqref{eq:decompositonTatemodule}. Then there is a decomposition of the Tate module of $\widetilde{\rH}(A_i/W)$:
\[
\left(T_0^{(i)},p \langle, \rangle\right) \oplus \left(T_1^{(i)} \oplus \mathbb{Z}_p^{\oplus 2},\langle,\rangle \right),
\]
where the second inner product is from the restriction of Mukai pairing.
Therefore
\[
(T_0^{(1)}, \mathbb{K}_{A_1}) \cong (T_0^{(2)}, \mathbb{K}_{A_2}),
\]
which implies an isomorphism between K3-crystals:
\[
\rH^2_{\crys}(A_1/W) \cong \rH^2_{\crys}(A_2/W).
\]
Then the isomorphism $A_1 \cong A_2$ follows from the crystalline Torelli theorem.
\end{proof}
Now we can give a summary of the known equivalence relations between supersingular abelian surfaces.
\begin{cor}\label{cor3}
Let $A_1, A_2$ be two abelian surfaces. If $A_1$ is supersingular, then the following statements are equivalent:
\begin{TFAE}[leftmargin=1.5pc]
\item There is an isomorphism $A_1 \cong A_2$;
\item There is an isomorphism between K3-crystals $\rH^2_{\crys}(A_1/W) \cong \rH^2_{\crys}(A_2/W)$;
\item There is an isomorphism between Kummer surfaces $\km(A_1) \cong \km(A_2)$;
\item There is a derived equivalence $\D^b(\km(A_1)) \xrightarrow{\sim} \D^b(\km(A_2))$;
\item There is a derived equivalence $\D^b(A_1) \xrightarrow{\sim} \D^b(A_2)$.
\end{TFAE}
\end{cor}
\begin{proof}
We firstly note that the supersingularity is invariant under the relations listed in the statements. Hence $A_2$ is supersingular in each statement.
Then the equivalence between (a), (b) is just the Ogus's crysalline Torelli theorem for supersingular abelian surfaces. The statements (c), (d) and (e) are equivalent by the Theorem \ref{mainthm1}. The (a) and (e) are equivalent by Theorem \ref{thm:supersingularDerivedTorelli}.
\end{proof}
Therefore, the statement (iii) in Theorem \ref{mainthm1} is true for supersingular abelian surfaces.
\begin{rmk}
We suspect whether every supersingular abelian variety $A$ has only trivial (in the strong sense) Fourier-Mukai partners, i.e. $\mathrm{FM}(A)=\{A\}$. This requires that $A$ admits a principal polarization, which holds if $\dim A=2$ (\cf Remark \ref{rmk:pp}). But it may fail in higher dimensional case. See \cite[ \S 10]{LOo98} for the discussion of non-principal polarizations on supersingular abelian varieties of any genus. It will be very interesting to know if $\mathrm{FM}(A)=\{A, \widehat{A} \}$.
\end{rmk}
\subsection{Proof of Theorem \ref{thm:supersingulargeneralizedKummer}}
We can see there is a quasi-liftably birational map
\begin{equation}\label{eq:birationalKumtype}
K_{v}(A)\dashrightarrow K_{n}(A'),
\end{equation}
with $n=\frac{v^{2}}{2}-1$, $p \nmid n$ and some (supersingular) abelian surface $A'$ derived equivalent to $A$ (\cf\cite[Theorem 6.12]{FL18}). Then (i) follows from the fact $A$ does not have non-trivial Fourier-Mukai partners (\cf\ref{cor3}).
To prove (ii), our strategy is to endow $\rH^2_{\crys}(K_n(A)/W)$ with a natural\footnote{Here the ``natural" means the structure should be at least functorial with respect to algebraic correpsondences.} K3-crystal structure, which is closely related to that of $A$. Combining the algebraic correspondence given by the birational map \eqref{eq:birationalKumtype}, one can show that there is an isomorphism between K3-crystals of $A$ and $A'$. In fact,
\begin{lemma}\label{lemma:BBforms}
Assume that $p \nmid n+1$.
\begin{enumerate}
\item For any abelian surface $A$ and positive integer $n$, we can endow the $F$-crystal $\rH^2_{\crys}(K_n(A)/W)$ with the Beauville-Bogomolov form $q$, which makes $(\rH^2_{\crys}(K_n(A)/W),q)$ a K3 crystal of rank 7.
\item There is an orthogonal decomposition with respect to $q$:
\begin{equation}\label{eq:FcrystalDecom}
\rH^2_{\crys}(K_n(A)/W) \cong \rH^2_{\crys}(A/W) \oplus W(-1).
\end{equation}
\end{enumerate}
\end{lemma}
\begin{proof}
Let $\cA$ be a lifting of abelian surface $A$ to the Witt vector ring $W$, which is an abelian scheme over $W$ (\cf\cite[(11.1)]{oort85}). Consider the summation of points $s \colon \cA^{[n]}\to \cA $. The fiber of $s$ at the origin point $\spec(W) \to \cA$ is a lifting of generalized Kummer variety $K_n(A)$, which will be denoted by $\cK_n(\cA)$ as a $W$-scheme.
For the geometric generic fiber $\cK_n(\cA)_{\bar{K}}$, we have the Beauville-Bogomolov form $q_{\bar{K}}$ on $p$-adic \'etale cohomology $\rH^2_{\et}(\cK_n(\cA)_{\bar{K}},\ZZ_p)$. There is an orthogonal decomposition
\begin{equation}
\rH^2_{\et}(\cK_n(\cA)_{\bar{K}}, \ZZ_p) \cong \rH^2_{\et}(\cA_{\bar{K}},\ZZ_p) \oplus \ZZ_p\delta,
\end{equation}
such that $q_{\bar{K}}(\delta) = -2(n+1)$.
The Beauville-Bogomolov form on $\rH^2_{\crys}(K_n(A)/W)$ can be defined by applying integral crystalline-\'etale comparison (\cf\cite[Theorem 1]{BMS1}) and denoted by $q$. It is followed by a decomposition of $F$-crystals
\[
\rH^2_{\crys}(K_n(A)/W) \cong \rH^2_{\crys}(A/W) \oplus W(-1),
\]
with respect to $q$.
The bilinear form induced by $q$ is perfect since $p \nmid n+1$. A direct computation shows that
\[
q(\varphi^*{\alpha})= p^2 \sigma q(\alpha).
\]
The canonical decomposition \eqref{eq:FcrystalDecom}
implies the Hodge number $h^0(\rH^2_{\crys}(K_n(A)/W)) =1$. Therefore, $(\rH^2_{\crys}(K_n(A)/W), q)$ is a K3-crystal.
\end{proof}
By the decomposition \eqref{eq:FcrystalDecom}, we have
\[
\rH^2_{\crys}(K_n(A)/W)^{\varphi =p} \cong \rH^2_{\crys}(A/W)^{\varphi =p} \oplus \mathbb{Z}_p \delta.
\]
Thus there is an isomorphism between Tate modules (as $\ZZ_p$-lattices)
\begin{equation}\label{eq:decomposeNS}
\NS(K_n(A)) \otimes \mathbb{Z}_p \cong \left(\NS(A) \otimes \mathbb{Z}_p \right) \oplus \langle -2(n+1) \rangle.
\end{equation}
\begin{lemma}\label{lemma:abelian-Kummer}
Under the same assumption in \ref{lemma:BBforms}, if $A$ is supersingular, then we have an isomorphism of characteristic subspaces
\[
\mathbb{K}_{\rH^2_{\crys}(A/W)} \xrightarrow{\sim} \mathbb{K}_{\rH^2_{\crys}(K_n(A)/W)}.
\]
\end{lemma}
\begin{proof}
Consider the following commutative diagram
\[
\begin{tikzcd}
\NS(A) \otimes \mathbb{Z}_p \ar[r,"c_1"] \ar[d,hook] & \rH^2_{\crys}(A/W) \ar[d,hook] \\
\NS(K_n(A)) \otimes \mathbb{Z}_p\ar[r,"c_1"] & \rH^2_{\crys}(K_n(A)/W)
\end{tikzcd}
\]
The horizontal injective morphisms are isometric embeddings from \eqref{eq:decomposeNS}. Thus it is not hard to see that they induces isomorphic characteristic subspaces.
\end{proof}
\begin{rmk}
The Lemma \ref{lemma:abelian-Kummer} can be viewed as a higher dimensional analogue to the relationship between supersingular abelian surfaces and their associated supersingular Kummer surfaces, which is a miracle used in Ogus's proof of crystalline Torelli theorem for supersingular Kummer surfaces.
\end{rmk}
\begin{lemma}\label{lemma:isocrystalH2}
If $K_n(A)$ and $K_n(A')$ are quasi-liftably birational equivalent, then there is an isomorphism of $F$-crystals $\rH^2_{\crys}(K_n(A)/W) \xrightarrow{\sim} \rH^2_{\crys}(K_n(A')/W)$ preserving Beauville-Bogomolov forms.
\end{lemma}
\begin{proof}
We may assume that $K_n(A)$ and $K_n(A')$ are liftable as $\cK$ and $\cK'$ over some finite extension $V$ of $W$, whose geometric generic fibers are birational equivalent irreducible symplectic varieties. Then there is a correspondence $Z_{F} \subset (\cK_F \times \cK_F')$ for some totally ramified finite field extension $F$ of $K=\Frac(V)$, such that $[Z_F]_{*} \colon \rH^2_{\et}(\cK_{\bar{K}},\ZZ_p) \xrightarrow{\sim} \rH^2_{\et}(\cK'_{\bar{K}},\ZZ_p)$ is a $G_F$-equivariant isomorphism, compatible with Beauville-Bogomolov forms (\cf\cite[Lemma 2.6]{Huybrechts99}). The construction in Lemma \ref{lemma:BBforms} implies that there is an isomorphism of $F$-crystals as we required.
\end{proof}
If $K_v(A)$ and $K_{v'}(A')$ are quasi-liftably birational equivalent, then $K_n(A)$ and $K_n(A')$ are also quasi-liftably birational equivalent by combining \eqref{eq:birationalKumtype}.
This implies that there exists isomorphisms between characteristic subspaces
\[
\bK_{A} \xrightarrow[\text{Lemma \ref{lemma:abelian-Kummer}}]{\sim} \bK_{K_n(A)} \xrightarrow[\text{Lemma \ref{lemma:isocrystalH2}}]{\sim} \bK_{K_n(A')} \xrightarrow[\text{Lemma \ref{lemma:abelian-Kummer}}]{\sim} \bK_{A'}.
\]
Therefore, by Theorem \ref{thm:characteristicspace} there is an isomorphism of K3-crystals
\[
\rH^2_{\crys}(A/W) \cong \rH^2_{\crys}(A'/W).
\]
Now we can conclude that $A \cong A'$ by Ogus's crystalline Torelli theorem for supersingular abelian surfaces.
\begin{comment}
\section{FM partners of twisted abelian surface}
\label{abelian-surface}
In this section, we study the derived category of twisted abelian surfaces and characterize Fourier-Mukai partners of twisted abelian surfaces.
\subsection{Twisted abelian surface}
We adopt Lieblich's definition for twisted scheme(\cf \cite{Lie07}). Let $A$ be an abelian surface over $k$ and let $\mathscr{A}\rightarrow A$ be a $\mu_n$-gerbe over $A$. This corresponds to a pair $(A,\alpha)$ for some $\alpha\in H^2_{\fl}(A,\mu_n)$, where the cohomology group is with respect to the flat topology. There is a bijection of sets
\[
\left\{ \text{$\mu_n$-gerbes on $A$ } \right\} \xrightarrow{\sim} H^2_{\fl}(A,\mu_n).
\]
The corresponding cohomology class $\alpha$ is also denoted by $[\srA]$. The gerbe $\srA$ is called a \emph{twisted abelian surface} with the underlying space $A$. The Kummer exact sequence induces a surjective map
\begin{equation}\label{braumap}
H^2_{\fl}(A,\mu_n)\rightarrow \Br(A)[n]
\end{equation}
where the right-hand side is the \emph{cohomological Brauer group} $\br(A) \coloneqq H^2(A,\mathbb{G}_m)_{\rm tor}$. For any $\mu_n$ gerbe on $A$, there is an associated $\G_m$-gerbe on $A$ via \eqref{braumap}, denoted by $\srA_{\G_m}$. We will denote $\srA^{(i)}$ by the gerbe corresponding to cohomological class $i[\srA]$. Then the $\srA^{(i)}$-twisted sheaves are called $i$-fold $\srA$-twisted sheaves on $A$.
If $k$ has characteristic $p\neq 2$, for a twisted abelian surface, there is an associated twisted K3 Kummer surface constructed as follows: consider the following commutative diagram
\begin{equation}
\label{eq:kummerconstruction}
\begin{tikzcd}
\widetilde{A} \ar[r,"\widetilde{\sigma}"] \ar[d,"\pi"] & A \ar[d] \\
\km(A) \ar[r,"\sigma"] & A/{\iota}
\end{tikzcd}
\end{equation}
where
\begin{itemize}
\item $\iota$ is the involution of $A$,
\item $\sigma$ is the crepant resolution of quotient singularities and
\item $\widetilde{\sigma}$ is the blow-up of $A$ along the points in $A[2]$. Its birational inverse is denoted by $\widetilde{\sigma}^{-1}$.
\end{itemize}
\begin{prop}\label{eq:Kummerbrauer}
Let $A$ be an abelian surface over $k$. The $(\widetilde{\sigma}^{-1})^*\pi^*$ induces an isomorphism between cohomological Brauer groups
\[
\Theta \colon \br(\km(A))\xrightarrow{\sim} \br(A).
\]
In particular, when $A$ is supersingular, then $\br(A)$ is isomorphic to additive group $k$.
\end{prop}
\begin{proof}
The comparison between these two $\ell$-torsion parts is almost same as \cite[Proposition 1.3]{Skorobogatov2009}. As $\widetilde{\sigma}$ is a birational morphism, the pull-back $\widetilde{\sigma}^*$ induces an isomorphism
\[
\br(A) \xrightarrow{\sim}\br(\widetilde{A} )
\]
since the Brauer group is a birational invariant. Let $U= A\backslash A[2]$. There is an isomorphism $\widetilde{\sigma}^{-1}(U) \cong U$. Thus the restriction $\br(A) \hookrightarrow \br(U)$
can be identified with the restriction \[\br(\widetilde{A}) \hookrightarrow \br(\widetilde{\sigma}^{-1}(U))\] by the following commutative diagram
\[
\begin{tikzcd}[column sep=small]
\br(A) \ar[r,hook, "res"] \ar[d,"\cong"] & \br(U) \ar[d,equal] \\
\br(\widetilde{A}) \ar[r,hook,"res"] & \br(\widetilde{\sigma}^{-1}(U))
\end{tikzcd}
\]
The Grothendieck's purity theorem for Brauer groups implies that the top arrow is an isomorphism as $A$ is of dimension 2 and the complement $A-U$ is of codimension 2 in $A$.
Thus $\br(\widetilde{A}) \cong \br(\widetilde{\sigma}^{-1}(U))$. The restriction of $\pi$ on $\widetilde{\sigma}^{-1}(U)$ is a $\mathbb{Z}/2\mathbb{Z}$-torsor on $\km(A)\backslash E$ where $E$ is the ramified locus of $\pi$. Consider the Hochschild-Serre spectral sequence
\[
E^{p,q}_2=H^p\!\left(\mathbb{Z}/2\mathbb{Z}, H^q_{\et}(\widetilde{\sigma}^{-1}(U),\mathbb{G}_m) \right) \Rightarrow H^{p+q}_{\et}(\km(A)\backslash E,\mathbb{G}_m).
\]
The elements in the $E_2$-page are as follows.
\begin{itemize}
\item $E_2^{0,2}= \left( H^2_{\et}(\widetilde{\sigma}^{-1}(U),\mathbb{G}_m) \right)^{\mathbb{Z}/2\mathbb{Z}} \cong \left(\br(A) \right)^{\mathbb{Z}/2\mathbb{Z}}$.
\item $E_2^{1,1} = H^1(\mathbb{Z}/2\mathbb{Z}, \Pic(A))$ since $H^1_{\et}(\widetilde{\sigma}^{-1}(U),\mathbb{G}_m) \cong \Pic(\widetilde{\sigma}^{-1}(U)) \cong \Pic(A)$.
\item $E_2^{2,0} = H^2(\mathbb{Z}/2\mathbb{Z},k^*)=0$ since $k$ is algebraically closed and $\Char(k)\neq 2$.
\end{itemize}
Since the Neron-Severi group $N(A)$ is torsion-free and the involution on $N(A)$ is trivial, we can see
\[
H^1(\mathbb{Z}/2\mathbb{Z},\Pic(A)) \cong H^1(\mathbb{Z}/2\mathbb{Z},\Pic^0(A)).
\]
Thus $E_2^{1,1}=0$ as $\Pic^0(A)$ is divisible. As a consequence, we have an isomorphism
\begin{equation}\label{eq:brauer:kummer-abelian}
\br(\km(A)\backslash E) \cong \ker \left( (\br(A)^{\mathbb{Z}/2\mathbb{Z}} \xrightarrow{d_2^{0,2}} H^2(\mathbb{Z}/2\mathbb{Z}, \Pic(A)) \right).
\end{equation}
The $(\widetilde{\sigma}^*)^{-1} \pi^*$ induces an injective map $\br(\km(A)\backslash E) \hookrightarrow \br(\widetilde{\sigma}^{-1}(U))$. Now the restriction diagram
\[
\begin{tikzcd}
\br(\km(A)) \ar[r,hook,"res"] \ar[d,"(\widetilde{\sigma}^*)^{-1}\pi^*"]& \br(\km(A)\backslash E) \ar[d,hook] \\
\br(A) \ar[r,"\sim"] & \br(U)
\end{tikzcd}
\]
implies that $(\widetilde{\sigma}^*)^{-1}\pi^*$ is injective. For any smooth surface $X$ over $k$, whose Neron-Severi group $\NS(X)$ is torsion-free, we can characterize its Brauer group as
\begin{equation}\label{eq:grobrauer1}
\br(X)[\ell] \cong (\mathbb{Q}_{\ell}/\mathbb{Z}_{\ell})^{b_2 -\rho},
\end{equation}
where $\ell$ is any prime not equal to $p$, $b_2$ is the Betti number and $\rho$ is the Picard number. Since the Neron-Severi groups of both abelian surface and Kummer surfaces are torsion-free, we can conclude that $\widetilde{\sigma}^{-1} \pi^*$ is surjective from \eqref{eq:grobrauer1} as \[b_2(\km(A)) - b_2(A) = \rho(\km(A)) - \rho(A) = 16\] by the construction of Kummer surface.
For the $p$-primary torsions, we have $$\br(\km(A))[p^{\infty}]\cong \br(A)^{\mathbb{Z}/2\mathbb{Z}}[p^{\infty}]$$ from \eqref{eq:brauer:kummer-abelian} since $H^2(\mathbb{Z}/2\mathbb{Z},\Pic(A))$ is a $2$-torsion group. Thus it suffices to prove that the involution on $\br(A)$ is trivial. In fact, $H^2_{\fl}(A,\mu_p)$ can be embedded to $H^2_{dR}(A/k)$ by de-Rham-Witt theory; \cite[see][Proposition 1.2]{Og79}. Thus the involution on $H^2_{\fl}(A,\mu_p)$ is trivial. Then by the exact sequence
\[
0 \to \NS(A) \otimes \mathbb{Z}/p \to H^2_{\fl}(A,\mu_p) \to \br(A)[p] \to 0
\]
we can deduce that $\br(A)[p]$ is invariant under the involution. Furthermore, for $p^n$-torsions that $n \geq 2$, we can run induction. Assume that elements in $\br(A)[p^k]$ are $\iota$-invariant for any $1 \leq k < n$. For any $\alpha \in \br(A)[p^{n}]$, $p \alpha \in \br(A)[p^{n-1}]$, so $p \alpha$ is $\iota$-invariant. Thus
\[
p\alpha = \iota(p \alpha) = p \iota(\alpha),
\]
which means $\alpha - \iota(\alpha) \in \br(A)[p]$. Applying $\iota$ on $\alpha - \iota(\alpha)$, we obtain equation
\[
\alpha - \iota(\alpha) = \iota(\alpha) - \alpha.
\]
It implies $\alpha- \iota(\alpha)$ is also a $2$-torsion. Since $p$ is coprime to $2$, we can conclude that $\alpha = \iota(\alpha)$.
If $A$ is supersingular, then $\km(A)$ is also supersingular. We have already known that the Brauer group of a supersingular K3 surface over $k$ is isomorphic to $k$ which is a $p$-torsion group; \cite[see][]{Ar74}. Thus $\br(A)\cong k$.
\end{proof}
\subsection{Twisted Fourier-Mukai partner of abelian surfaces}Let us introduce the $B$-field theory on twisted abelian surfaces. We mainly follow the content of \cite[Section 5.1]{FL18} and {\cite[Section 3.4]{BL18}}.
\begin{defn}[$B$-fields]
For a prime $\ell\neq p$, an {\it $\ell$-adic $B$-field} on $A$ is an element $B\in H^2_{\et}(A,\QQ_\ell(1))$. It can be written as $\alpha/ \ell^n$ for some $\alpha\in H^2_{\et}(A,\ZZ_\ell(1))$ and $n\in\ZZ$. The associated Brauer class $[B_\alpha]$ is the image of $\alpha$ under the following composition of natural maps
\begin{equation*}
H^2_{\et}(A,\ZZ_\ell(1)) \rightarrow H^2(A,\mu_{\ell^n})\rightarrow \Br(S)[\ell^n],~\hbox{if $\ell\nmid p$}.
\end{equation*}
A {\it crystalline B-field} is an element $B=\frac{\alpha}{p^n}\in H^2_{\cris}(A/W)\otimes K$ with $\alpha\in H^2_{\cris}(A/W)$, so that the projection of $\alpha$ in $H^2_{\cris}(A/W_n(k))$ lies in the image of the map
\begin{equation}\label{dlog}
H^2_{\fl}(A,\mu_{p^n})\xrightarrow{d\log} H^2_{\cris}(A/W_n(k))
\end{equation}
(See \cite[I.3.2, II.5.1]{Il79} for the details of the map \eqref{dlog}). Then we can associate a $p^n$-torsion Brauer class $[B_\alpha]$ via the map \eqref{braumap}.
\end{defn}
The \emph{Mukai lattice} $\widetilde{H}(A)$ of for integral $\ell$-adic cohomology and crystalline cohomology of $A$ is defined as follows.
\begin{itemize}
\item ($\ell$-adic): $\widetilde{H}(A,\mathbb{Z}_{\ell}) \coloneqq H^0_{\et}(A,\mathbb{Z}_{\ell})(-1) \oplus H^2_{\et}(A,\mathbb{Z}_{\ell}) \oplus H^4_{\et}(A, \mathbb{Z}_{\ell})(1)$;
\item (crystalline): $\widetilde{H}_{\crys}(A/W) \coloneqq H^0_{\crys}(A/W)(-1) \oplus H^2_{\crys}(A/W) \oplus H^4_{\crys}(A/W)(1)$
\end{itemize}
endowed with the Mukai pairing
\begin{equation}
\langle (r,c,\chi),(r',c',\chi')\rangle_{\widetilde{H}} = c \cdot c' - r\chi'-r'\chi.
\end{equation}
Moreover, for any $\mu_m$-gerbe $\srA \to A$, we can associate it with the \emph{twised Mukai lattice} as follows. Write $B=\frac{\alpha}{m} $ as either a $\ell$-adic or crystalline $B$-field of $\srA\rightarrow A$. Let
\begin{equation}\label{Mlat}
\widetilde{H}(\srA)=\begin{cases}
e^{a/m} (\widetilde{H}(A,\ZZ_\ell)) &\text{ if } p\nmid m\\ ~\\
e^{a/m}(\widetilde{H}_{\cris}(A/W)) & \text{ if }m=p^n
\end{cases}
\end{equation}
where $e^{a/m}$ acts on the Mukai lattices by cup-product. We can see the definition of $\widetilde{H}(\srA)$ is independent of the choice of $B$-field for the Brauer class $[\srA_{\mathbb{G}_m}]$. The $\widetilde{A}$ is the required twised Mukai lattice of $\srA$. Sometimes, we will also write it as $\widetilde{H}(A, B)$ to emphasize the $B$-field. For Neron-Severi group, we also have the \emph{extended (twised) Neron-Severi lattice} defined as
\[
\widetilde{N}(\srA) = \widetilde{H}(\srA) \cap \widetilde{N}(A)[\frac{1}{\ell}].
\]
Inspired by \cite[Theorem 1.2]{Stellari2007}, we have the following observation shows that there is also a close relationship between $\mu_m$-gerbes on a abelian surface and those on its Kummer surface.
\begin{lemma}\label{lemma:twistedabelian-kummer}
Let $A$ be an abelian surface and $\km(A)$ its associated Kummer surface. There is a bijection
\[
\left\{ \widetilde{H}(\srA)\Big|\srA \to A \text{ is a $\mu_m$-gerbe} \right\}/_{\cong} \xrightarrow{\sim} \left\{ \widetilde{H}(\mathscr{X})\Big| \mathscr{X} \to \km(A) \text{ is a $\mu_m$-gerbe} \right\}/_{\cong}
\]
\end{lemma}
\begin{proof}
For any $[\srA] \in H^2(X,\mu_m)$ and a B-field lift $B$ of $[\srA]$, we can embed $B$ as an element $B'$ in
\begin{equation}\label{eq:kummerdecompose}
H^2(\km(A)) \cong \pi_* H^2(A) \oplus \Pi_A
\end{equation}
where $H^2(-)$ can be $H^2_{\et}(- ,\mathbb{Z}_{\ell}(1)$ or $H^2_{\crys}(-/W_n(k))$ and $\Pi_A$ is the Kummer lattice associated to $A$. We claim that $B'$ is a B-field on $\km(A)$. For instance, when $\srA$ is a $\mu_p$-gerbe, then $B$ is a $B$-field if and only if $B -\varphi(B) \in H^2_{\crys}(A/W) + \varphi(H^2_{\crys}(A/W)$. As the decomposition \eqref{eq:kummerdecompose} is compatible with Frobenius, hence $B'$ also satisfies this relation on $\km(A)$. It implies that $B'$ is also a $B$-field and $\widetilde{H}(\km(A),B')$
is a well-defined twisted Mukai lattice on $\km(A)$.
Let $\alpha$ be the image of $\srA$ in $\br(A)[m]$ under \eqref{braumap}. Let $\beta$ the inverse image $\Theta^{-1}(\alpha)$. We can see $\widetilde{H} \cong \widetilde{H}(\srX)$ for any $\mu_m$-gerbe whose associated Brauer class is equal to $\beta$. Thus we establish the required map, which is bijection since $\Theta$ is an isomorphism and a twisted Mukai lattice only depends on the Brauer class.
\end{proof}
Let $\Phi^\mathscr{P} \colon \D^b(\srA_1) \to \D^b(\srA_2)$ be a Fourier-Mukai transform such that $\mathscr{P}$ is a $\srA_1^{(-1)}\times \srA_2$-twisted perfect complex. We can identify $\mathscr{P}$ with a sheaf on $\srA_1^{(-1)} \times \srA_2$ by the following fact.
\begin{lemma}\label{lemma:kernel-sheaf}
There exists an $\srA_1^{(-1)}\times \srA_2^{(1)}$-twisted sheaf $\cE$ such that the Fourier-Mukai kernel $\mathscr{P} \cong \cE[n]$ for some integer $n$.
\end{lemma}
\begin{proof}
The proof is same to that of \cite[Proposition 3.8]{Huybrechts2008} as it is well-known that the derived category of a twisted abelian surface is generic, i.e.\ it contains no spherical objects.
\end{proof}
For the Fourier-Mukai transform $\Phi^{\mathscr{P}}$, there is a \emph{cohomological Fourier-Mukai transform} $\phi^{\mathscr{P}}$ which fits into the following commutative diagram
\[
\begin{tikzcd}
\widetilde{N}(\srA_1) \ar[r, "\simeq"] \ar[d] & \widetilde{N}(\srA_2)\ar[d] \\
\widetilde{H}(\srA_1)_{\mathbb{Q}} \ar[r,"\phi^{\mathscr{P}}","\sim"'] &\widetilde{H}(\srA_2)_{\mathbb{Q}}
\end{tikzcd}
\]
This cohomological realization enables us to have the following deifinition.
\begin{defn}
A Fourier-Mukai transform is called \emph{filtered} if its cohomological Fourier-Mukai transform sends Mukai vector $(0,0,1)$ to $(0,0,1)$.
\end{defn}
\begin{prop}[filtered Torelli theorem for twisted abelian surfaces]
\label{thmfilter}
Let $\srA_1,\srA_2$ be a pair of twisted abelian surfaces. Then following statements are equivalent
\begin{TFAE}
\item There is an isomorphism between associated $\G_m$-gerbes $\srA_{1,{\G_m}}$ and $\srA_{2,{\G_m}}$.
\item There is filtered Fourier-Mukai transform $\Phi^{\mathscr{P}}_{\srA_1 \to \srA_2}$.
\end{TFAE}
\end{prop}
\begin{proof} The theorem for the untwisted abelian surfaces in positive characteristic is given in \cite[Proposition 3.1]{Honigs2019}. Here we extend it to the twisted case.
The (a) induces (b) obviously.
For the converse, we have seen $\Phi^{\mathscr{P}}_{\srA_1 \to \srA_2}(\sheaf_x) = \mathscr{P}_{x} \coloneqq \mathscr{P}|_{\{x\}\times A_2}$ is an $\srA_{2}$-twisted sheaf up to shift in Lemma \ref{lemma:kernel-sheaf}. The $\mathscr{P}_x$ is just a skyscraper sheaf (up to shift) on $A_2$ as $\Phi^{\mathscr{P}}_{\srA_1 \to \srA_2}$ sends $(0,0,1)$ to $(0,0,1)$ by the assumption. The induced integral functor $\Phi^{P}_{A_1 \to A_2}$ also maps skyscraper sheaves to skyscraper sheaves since skyscraper sheaves are naturally twisted as they are supported at closed points. Hence $\Phi^{P}_{A_1 \to A_2}$ is fully faithful as skyscraper sheaves form a spanning class of the bounded derived category of a smooth projective variety. Moreover, it is an equivalence as abelian surfaces are with trivial canonical bundle. It is well-known that any derived equivalence between smooth projective varieties, which maps closed points to closed points, induces an isomorphism between varieties (cf.\ \cite[Corollary 6.14]{Huybrechts2008}). Thus the underlying abelian surfaces $A_1$ and $A_2$ are isomorphic as proved in the untwisted case. Let $f \colon A_1 \to A_2$ be the induced isomorphism. Since $\Phi^{P}_{\srA_1 \to \srA_2}$ sends $(0,B,0)$ to $(0,B',0)$ as it is filtered, we have $B' = f^*B$. Whence, $f$ is an isomorphism between $\G_m$-gerbe $\srA_{1,\G_m}$ and $\srA_{2,\G_m}$.
\end{proof}
\begin{thm}
For $\mathbb{G}_m$-gerbes $\srA_1 \to A_1$ and $\srA_2 \to A_2$, the following statements are equivalent:
\begin{TFAE}
\item There is a Fourier-Mukai transform $\D^b(\srA_1) \simeq \D^b(\srA_2)$.
\item There is an isomorphism between K3 crystals $\rho \colon \widetilde{H}(\srA_1/W) \cong \widetilde{H}(\srA_2/W)$.
\end{TFAE}
\end{thm}
\begin{proof} Consider the codimension filtration on $\widetilde{H}(A)[\frac{1}{\ell}]$ (the prime $\ell$ depends on the choice of $\ell$-adic or crystalline twisted Mukai lattice) given by $F^i = \oplus_{r \geq i} H^{2r}(A)[\frac{1}{\ell}]$. The filtraion on $\widetilde{H}(\srA)$ is defined as
\[
F^0\widetilde{H}(\srA) \coloneqq \widetilde{H}(\srA) \quad F^1\widetilde{H}(\srA) \coloneqq \widetilde{H}(\srA) \cap F^1\widetilde{H}(A)[\frac{1}{\ell}] \quad F^2\widetilde{H}(\srA) \coloneqq \widetilde{H}(\srA) \cap F^2 \widetilde{H}(A)[\frac{1}{\ell}].
\]
Let $B$ be a B-field of $[\srA]$. A direct computation shows that the graded pieces of $F^*$ are
\[
\begin{aligned}
&\Gr^0_F \widetilde{H}(\srA) = \left\{ (r, rB, \frac{rB^2}{2}) \Big| r \in H^0(A) \right\},\\
&\Gr^1_F \widetilde{H}(\srA)\cong \left\{
(0, c, c \cdot B) | c \in H^2(A)\right\} \cong H^2(A),\\
& \Gr_F^2 \widetilde{H}(\srA) = \left\{(0,0, \chi)| \chi \in H^4(A) \right\} \cong H^4(A)(1).
\end{aligned}
\]
If a morphism between twisted Mukai lattices preserves this filtration, we also call it filtered. Note that there is a filtered isomorphism
\begin{equation}\label{eq:abeliantomoduli}
\gamma\colon \widetilde{H}(\srA_1) \xrightarrow{\rho} \widetilde{H}(\srA_2) \xrightarrow{\phi^{\cE}} \widetilde{H}(\srM_H(\srA_2,v))
\end{equation}
where $\phi^{\cE}$ is the cohomological Fourier-Mukai transform given by universal sheaf $\cE$. Then $A_1 \cong M_H(\srA_2,v)$ abstractly since \eqref{eq:abeliantomoduli} induces an isomorphism between supersingular K3 crystals.
Let $B$ be a B-field of $[\srA_1]$ and $B'$ a B-field of $[\srM_H(\srA_2,v)]$. Let $\Aut_{K3}(H^2_{\crys}(A_1/W))$ the automorphism group of K3 crystal $H^2_{\crys}(A_1/W)$. Then $\gamma^1 \coloneqq \Gr^1_F(\gamma)$ can be viewed as an element in $\Aut_{K3}(H^2_{crys}(A_1))$ such that $\gamma^1(B)= B'$ since $\gamma$ sends $(0,B,0)$ to $(0,B',0)$ as it is filtered. Note that there is a surjection
\[
\Aut(A_1) \twoheadrightarrow \Aut(H^1_{crys}(A_1),\tr) \xrightarrow{\sim} \Aut_{K3}(H^2_{\crys}(A_1)).
\]
The left-hand surjective map is obtained from Ogus's Torelli theorem for supersingular varieties by strong approximation (cf. \cite[Remark 6.8]{Og79}). The right-hand isomorphism is from the fact that the category of supersingular abelian crystals of genus 4 is equivalent to the category of supersingular K3 crystals of rank 6. Thus $\gamma^1$ extends to an automorphism $f \in \Aut(A_1)$ such that $f^*[\srA_{1,\mathbb{G}_m}] = [\srM_H(\srA_2,v)]$.
\end{proof}
\begin{rmk}
The $0$-graded piece $\Gr_F^0 \widetilde{H}(\srA)$ depends on the transcendental part of the B-field. More precisely, consider the exact sequence
\[
H^1(A,\mathbb{G}_m) \xrightarrow{c} H^2_{\fl}(A, \mu_m) \to H^2(A, \mathbb{G}_m)
\]
the cohomology class $[\srA]$ can be uniquely written as
\[
[\srA]= c(\cL) + \alpha.
\]
Then $B= B_{\alpha} + \frac{c_1({\cL})}{m}$ is a B-field lift of $[\srA]$. As $\cL$ is equal to a $m$-th power for some line bundle $\cL^{\frac{1}{m}} \in \Pic(A)$, we can see
\[
e^{\cL^{\frac{1}{m}}} \colon e^{B_{\alpha}}\widetilde{H}(A) \to e^{B}\widetilde{H}(A)
\]
is an isometry. Thus $\widetilde{H}(\srA) \cong e^{B_\alpha}\widetilde{H}(A)$.
\end{rmk}
\subsection{Fourier-Mukai partners are moduli of twisted sheaves}\label{subsec:twistedab}
Let $\srA \to A$ be a $\mu_n$-gerbe on $A$. Consider the moduli stack $\srM^{ss}_{H}(\srA,v)$ (resp.\ $\srM^{s}_H(\srA,v)$) of $H$-semistable (resp.\ $H$-stable) $\srA$-twisted sheaves with fixed Mukai vector $v \in N(\srA)$. If $H$ is $v$-generic, then $\srM^{ss}_H(\srA,v) = \srM^{s}_H(\srA,v)$. We will also denote them by $\srM_H(\srA,v)$ if $H$ is $v$-generic.
The following is Yoshioka's criterion on non-emptiness of moduli space of sheaves.
\begin{prop}
\label{criteria}
Let $\srA \to A$ be a $\mu_n$-gerbe on an abelian surface $A$. Let $v=(r,\ell,s) \in N(\srA)$ be a primitive Mukai vector such that $v^2=0$. Fix a $v$-generic ample divisor $H$. If one of the following holds:
\begin{enumerate}
\item $r>0$.
\item $r=0$ and $\ell$ is effective.
\item $r=\ell=0$ and $s>0$.
\end{enumerate}
then coarse moduli space $\moduli_H(\srA,v) \neq \emptyset$ and the moduli stack $\srM_H(\srA, v)$ is a $\G_m$-gerbe on $\moduli_H(\srA,v)$. Moreover, $\moduli_H(\srA,v)$ is an abelian surface.
\end{prop}
\begin{proof}
If $\srA \to A$ is a $\mu_n$-gerbe such that $p \nmid n$, then the statements are proven in \cite[Proposition A.2.1]{MMY11}. It is based on a statement of lifting a Brauer classes on $A$ to characteristic 0 which really needs the condition $p \nmid n$. Hence we just need to prove this for the case $\srA$ being a $\mu_p$-gerbe.
Let $B$ be an artinian local $W(k)$-algebra which fits in the following exact sequence
\[
0 \to I \to R \to k.
\]
Suppose that there is a thickening $A_R$ of $A$ to $\spec(R)$. There is a functorial obstruction class for lifting the associated Brauer class $\alpha \coloneqq [\srA_{\mathbb{G}_m}]$ to $\br(A_R)$:
\[
o(\alpha) \in H^3(A,\cO_A)\otimes I, \]
see \cite[Proposition 2.1]{Bragg2019}. If $\alpha= \Theta(\beta)$ for some $\beta \in \br(\km(A))$. Then the functorial property implies
\[
o(\alpha) = (\widetilde{\sigma}^{-1})^*\pi^*(o(\beta)) =0
\]
since $o(\beta) \in H^3\left(\km(A),\cO_{\km(A)}\right) \otimes I =0$. Thus, as we have seen in Lemma \ref{eq:Kummerbrauer} that $\Theta$ is surjective, the Brauer class $\alpha$ is unobstructed. Hence the forgetful functor
\[
\Def_{(A,\alpha)} \to \Def_{A}
\]
is smooth of relative dimension 1 as the tangent space of each fiber is isomorphic to $H^2(A,\cO_A)$. Furthermore, we can consider the deformation problem as follows. Suppose that $\alpha_R \in \br(A_R)$ is the deformed Brauer class of $\alpha$. The morphism $\eta \colon \Def_{(A,[\srA])} \to \Def_{(A,\alpha)}$ maps a pair of $(A_R, [\srA_R])$ to $H^2_{\fl}(A_R,\mu_p)$ whose associated Brauer class is $\alpha_R$. It is shown that there $\eta$ is a closed immersion of formal schemes, see \cite[Proposution 3.3]{Bragg2019}.
Now we can follow a standard argument to show that there is a finite extension (possibly ramified) $W \subset W'$ such that there is a lifting of triple $(\cA, \cL, \widetilde{\alpha})$ in which
\begin{itemize}
\item $(\cA,\cL)$ is a lifting of polarized abelian surface $(A,L)$ to $W'$,
\item $\widetilde{\alpha} \in H^2(\cA,\mu_p)$ with $\widetilde{\alpha}|_{A} = [\srA]$.
\end{itemize}
Then by taking the relative moduli space $\srM_H(\srA,\widetilde{\alpha},v) \to W'$ to show that the special fiber $\srM_H(\srA,v)$ is non-empty under the assumption (i), (ii) or (iii).
\end{proof}
One can characterize the Fourier-Mukai partners of twisted abelian surfaces as moduli spaces of twisted sheaves.
\begin{thm}\label{thm1}
Let $\srA \to A$ be $\mu_m$-gerbe on an abelian surface $A$. Then the associated $\mathbb{G}_m$-gerbe of any Fourier-Mukai partner of $\srA$ is isomorphic to a $\G_m$-gerbe on the moduli space of $\srX$-twisted sheaves $M_{H}(\srX,v)$ with $\srX$ being $\srA$ or $\widehat{\srA}$. In particular, if $A$ is principally polarized then $\srX=\srA$.
\end{thm}
\begin{proof}
Let $\mathscr{M}$ be a Fourier-Mukai partner of $\srA$. Let $\Phi^{\mathscr{P}}_{\mathscr{M}\to \srA}$ be the Fourier-Mukai transform. Denote $v$ the Mukai vector of image of $(0,0,1)_{\mathscr{M}}$ under $\Phi^{\mathscr{P}}_{\mathscr{M} \to \srA}$. We can assume $v$ satisfying one of the conditions in \ref{criteria} by changing $\srA$ to $\widehat{\srA}$ if it is necessary. Denote $\srX$ by $\srA$ or $\widehat{\srA}$. It is proved that the moduli stack $\srM_{H}(\srX,v)$ is a $\mathbb{G}_m$-gerbe on $M_H(A,v)$ in Proposition \ref{criteria}.
There is a Fourier-Mukai transform
\begin{equation}
\Phi^{\mathscr{P}} \colon \D^b(\srM_{H}(\srX,v)^{(-1)}) \to \D^b(\srX^{(1)})
\end{equation}
induced by the universal sheaf $\mathscr{P}$ on $\srM_{H}(\srX,v) \times \srX$, whose image of Mukai vector $(0,0,1)$ is $v$. Combine with the derived equivalence to its Fourier-Mukai partner $\srM$
\[
\Phi \colon \D^b(\srX) \to \D^b(\srM)
\]
we will obtain a filtered derived equivalence from $\srM_H(\srX,v)^{(-1)}$ to $\srM^{(1)}$. This induces an isomorphism from $\srM_H(\srX,v)^{(-1)}$ to $\srM^{(1)}_{\mathbb{G}_m}$ by the Theorem \ref{thmfilter}.
\end{proof}
\begin{cor}
Let $A$ be the a superspecial supersingular abelian surface with Artin invariant $1$. For any Brauer class $[\srA] \in \br(A)$ , there is non-superspecial supersingular abelian surface $A'$ such that $\D^b(\srA)\simeq \D^b(A')$.
\end{cor}
\begin{proof}
The (iii) implies that it suffices to find an supersingular abelian surface $A'$ whose extended K3 crystal is isomorphic to $\widetilde{H}(\srA)$. Let $\widetilde{H}(\mathscr{X})$ be the image of $\widetilde{H}(\srA)$ under the map in Lemma \ref{lemma:twistedabelian-kummer}. We can see the Artin invariant of $\widetilde{H}(\srX)$ is 2. According to \cite{BL18}, we can find a supersingular Kummer surface $\km(A')$ such that
\[
\widetilde{H}(\km(A')) \cong \widetilde{H}(\mathscr{X}).
\]
Finally we get an abelian surface $A$ such that $\widetilde{H}(A') \cong \widetilde{H}(\srA)$ by Lemma \ref{lemma:twistedabelian-kummer}.
\end{proof}
\end{comment}
\printbibliography
\end{document}
| 209,290
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TITLE: Proof of Hilbert's nullstellensatz,
QUESTION [6 upvotes]: Let $k$ be an algebraically closed field and $$K=\frac{k[x_1,\dots,x_n]}{m}$$
be a finitely generated $k$-algebra, where $m$ is a maximal ideal.
$K$ is algebraic over $k$. Then why is $k$ isomorphic to $K$? Sorry if this is obvious.
REPLY [3 votes]: This answer from a book: Elementary algebtaic geometry, Klaus Hulek, p25:
Theorem: Let $k$ be a field with infinitely many elements, and let $A=k[a_1,...,a_n]$ be a finitely generated $k-$algebra. If $A$ is a field, then $A$ algebraic over $k$.
In proof of Hilbert's Nullstellensatz we have $K$ is a field and finitely generated $k-$ algebra, then by theorem implies that $K$ is algebraic over $k$ then we have $k \hookrightarrow k[x_1,...x_n] \to k[x_1,...x_n]/m =K $
then $K$ is an algebraic extension of $k$ so $K\cong k$ where $k$ is algebraic closed.
| 145,376
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Again, I'm linking to Gillian of Crafting a Rainbow in the top 5 of 2016
These are in no way New Year's Resolutions. Those are just too easy to break!
These are those goals and plans that relate to sewing.
Strangely, though not so when I tell you more, my first goals concern weight and health.
- Many of my clothes have become too tight
- My bras were strangling me and I had to buy new ones (I do not have any plans to make bras)
- I saw a couple of RTW items that I liked but I was too big for the biggest size (regular range, not plus sized range) or they are too short (one of the reasons I started making my own)
- I was becoming breathless far too easily
- I felt unwell
- I was sent for heart tests because of some enzyme and chest X-ray abnormalities and need to get a 24 hour monitor in January. There isn't a major problem. My heart rate is fast, probably because I am unfit. I have a minor ECG issue. I had heart surgery in 2000 to repair a congenital hole in the heart, found by chance after I'd had a couple of brief transient ischaemic attacks which can be stroke precursors. I've been fine since.
- I have been getting no exercise since not playing golf, that's since the end of September. I can only imagine increasing my fitness will benefit my heart!
- My knees hurt more than they did. They're having to carry more weight.
- I didn't like what I was seeing in the mirror.
- I had put on weight, but not as much I thought as would account for the way I was feeling and looking. I'm out of shape even more.
Today, I went back to Slimming World. I've been before and did well. David doesn't need to lose weight beyond the 2 or 3 pounds he's put on over Christmas but will be supportive. He doesn't see why people need to attend such things as Slimming World but I need that input. I was heavier this time than when I started last time. That time I lost 2 stones; this time it needs to be nearer 3.
In addition, I will restart golf. I find that type of exercise more enjoyable than a gym.
Alison gave me a watch that measures steps etc and I've set some goals on that for daily activity. Even on a day to day basis, I'd become less active. Unfortunately, sewing is not a particularly active hobby. So far, I've met my goals on daily activity, but I did start easy!
Being slimmer will allow me to wear some clothes that I can't currently. Both those that I have that are currently too tight and those I might buy.
Blocks
I will sort out my blocks after I've got about half way with my weight loss. I have to do this. I will sew for where I am, though, as well as thinking about where I want to be. So I won't delay until I reach a weight goal - that was one of my previous u=issues with making trousers and I didn't ever finish.
Sewing Plans:.
I have issues with RTW relating to fit (sure - quality too but I'm referring only to fit here). Some of the issues are related to my height (5’11”), some to my basic shape (pear), some to my ‘maturing’ body shape - more around the middle than I ever had and some because I'm overweight.
These are the items I find hardest to buy:
Trousers
RTW issues
- crotch length and depth too short;
- waist gaping as my waist is much smaller than my hips.
- My thighs are quite chunky, too.
- I don't tend to have problems getting the legs long enough - and there are some specialist companies catering for tall women, should I need that.
- In addition, I don't like low rise. I prefer my waistband at my natural waist, or just below.
I intend 2017 to be the year of trousers. I have consistently failed to get trousers to fit, either RTW or me-made. I attended a jeans making class; I can sew them, but they don't fit. In class we drew up a pattern using Dennic Chunman Lo rather than say Aldrich and it just didn't work. I wasn't the only one in this position. We tried to make changes but again the term ended before we had got far enough and I had no desire to continue.
My tutor is going to help me get a decent trouser block. I believe she is thinking about draping though I'm not entirely sure.
Blouses/shirts
RTW issues
- Sleeve and body length. Most of my blouses and shirts have sleeves that are too short. While I'm happy to wear some styles over a skirt or trousers, I want to be able to tuck my blouses in securely.
- Being tall, I sometimes find that bust darts are too high as there isn't enough length between shoulder and bust.
- I don't like shirts with pockets as the positioning of them is usually not flattering on me.
- More recently I have found that my bust is bigger because of weight gain I assume and a couple of the shirts I have pull a bit at the front, not an issue I had previously. I am quite wide shouldered. I am having to do FBAs on virtually all patterns and now feel that some of the. RTW that fitted me previously could do with the same!
As yet, I haven't chosen what I might like to make.
Jackets
RTW issues similar to blouses and shirts.
Last term my class was on jacket tailoring. We made up a standard size jacket - 10 or 12 (UK). It would have fitted Helen widthwise but would have been too short in body and arms. No chance of it fitting me! I didn't actually finish it as after we made the practice jacket, the idea was to make our own fitted jacket. We all ran out of time. I have made an initial fitting toile in calico and will make a few changes to the pattern and make up another toile, this time in boiled wool. I'll give more details at a later stage.
.
.
After Christmas, I plan to finish this blazer, too. I have had to abandon the plans I had to use striped jersey as it is too thin. I may use a plain boiled wool.
SWAP (Sewing with a Plan)
The Stitchers Guild runs a kind of contest each year. I have never entered. It involves producing a number of garments (11 this year)by sewing or knitting which work together, by end April. There are guidelines for these garments which are detailed elsewhere so I don't need to go into detail, particularly as I’m not sure which path I’ll take. There should be a combination of ‘bottoms’, ‘tops’, dresses and ‘overlayers’ some of which can be outerwear. They can include one made previously, one bought and although the official start date is Boxing Day, you can start early. I tried to do one of the tops early (hacking from a RTW top) but ran out of time before Christmas.
Provided I manage the jackets, above, and the trouser block, I hope to enter the SWAP. I would put in 2 or 3 pairs of trousers, plus or minus jeans, the two jackets mentioned above, some tops and a cardigan. Even though my class next term is outerwear, I don't think I'll be trying any of that for this. If I decide to do this, I may not finish it in the time available but would continue until I had completed it. I would include a couple of blouses/shirts and try to make a pair of close fitting trousers. Obviously, leggings would be easier and so faster, but I don't wear them.
I'm not yet absolutely committed to this so haven't made any detailed plan. You'll be the first to know if I go ahead! It could work for my reunion weekend.
Helen's wedding dress
I need to reassemble it! I took it apart to try to clean it, but unsuccessfully, and Helen is quite distressed that it has remained that way (disassembled more than the dirty part). So I'll put it together and get it back to her.
Helen's red silk dress
I should get that done relatively quickly as the toile only needed minor alteration. I'm likely to find the pockets most difficult. Interestingly, the pockets are actually gusseted and are like those in outerwear. We'll be looking at those next term. I don't see them as being very appropriate in a silk dress but Helen loves them. I have some photos of the dress but Ill wait until things are back to normal here (Just escaping for a few minutes)
Level 3 Sewing Class
Next term is outerwear and I think the third term is lingerie (can't quite remember offhand). I'm carrying on with the course.
40 year reunion
For our reunion 40 years after graduating from Glasgow University, we will have a long weekend in a spa hotel in Scotland, in November 2017. Some of my friends from uni don't intend to go, which is sad, so I don't know how popular it will be. I plan to make a few garments for this long weekend, a travel wardrobe, maybe. Not sure what yet. I will include an evening dress, though. Maybe.
I hope you all had a good Christmas break - and I wish you a Happy New Year 2017
I hope you all had a good Christmas break - and I wish you a Happy New Year 2017
| 204,784
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TITLE: Sum of the square distances from a point to the sides/faces of a regular polygon/polyhedra
QUESTION [0 upvotes]: This is a variant of the result discussed in this link: On a constant associated to equilateral triangle and its generalization.
Consider any regular polygon and an arbitrary point, $P$, on an arbitrary circle with center at the centroid of the polygon. The sum of the square distances from $P$ to the sides of the polygon is a constant.
This can be extended to a tetrahedron. Consider a tetrahedron $ABCD$ and an arbitrary point, $P$, on any sphere with center at the centroid of the tetrahedron. The sum of the square distances from $P$ to the planes containing the faces of the tetrahedron is a constant. I suspect this can be generalized to any regular polyhedra.
These two similar results are probably particular cases of a more general theorem, so my question is: which whould be this general theorem for which these two results are just particular cases?
REPLY [1 votes]: Let $A_1$, $A_2$, $A_3$, … , $A_n$ be a set of points, and let $G=\displaystyle\frac{A_1+A_2+\ldots+A_n}{n}$.
Now take any circle centered at $G$. (If you're working in 3 dimensions, take a sphere centered at $G$ instead.) For any point $P$ on this circle, the value $A_1P^2+A_2P^2+\ldots+A_nP^2$ is constant.
To prove this, first note that, for any point $X$, we have $\vec{A_1X}+\vec{A_2X}+\ldots+\vec{A_nX}=n\vec{XG}$.
From this, you can conclude that $\vec{A_1G}+\vec{A_2G}+\ldots+\vec{A_nG}=0$.
So, for any point $X$, we have the following:
$$A_1X^2+A_2X^2+\ldots+A_nX^2=\left|\vec{A_1X}\right|^2+\left|\vec{A_2X}\right|^2+\ldots+\left|\vec{A_nX}\right|^2$$
$$=\left|\vec{A_1G}+\vec{XG}\right|^2+\left|\vec{A_2G}+\vec{XG}\right|^2+\ldots+\left|\vec{A_nG}+\vec{XG}\right|^2$$
$$=n\cdot\left|\vec{XG}\right|^2+2\cdot\vec{XG}\cdot\left(\vec{A_1G}+\vec{A_2G}+\ldots+\vec{A_nG}\right)+\left|\vec{A_1G}\right|^2+\left|\vec{A_2G}\right|^2+\ldots+\left|\vec{A_nG}\right|^2$$
$$=n\left|\vec{XG}\right|^2+\left|\vec{A_1G}\right|^2+\left|\vec{A_2G}\right|^2+\ldots+\left|\vec{A_nG}\right|^2$$
$$=n\cdot {XG}^2+A_1G^2+A_2G^2+\ldots+A_nG^2$$
So, when $XG$ is constant, $A_1X^2+A_2X^2+\ldots+A_nX^2$ is constant.
Sources: http://www.cut-the-knot.org/triangle/medians.shtml
Side note if you like physics:
Imagine $n$ Hookean springs, all with the same spring constant. The first spring connects $A_1$ to $X$, the second spring connects $A_2$ to $X$, and etc. In this setup, points $A_1$ through $A_n$ are fixed, but point $X$ is free to move. The potential energy of this setup is proportional to $A_1X^2+A_2X^2+\ldots+A_nX^2$. Since nature likes to minimize potential energy, this setup will come to rest when $A_1X^2+A_2X^2+\ldots+A_nX^2$ is at it's minimum. This setup will also come to rest when all the forces sum to $0$. At any point $X$, the sum of the forces will be $k\cdot\left(\vec{A_1}+\vec{A_2}+\ldots+\vec{A_n}\right)=k\cdot n\cdot XG$, where $k$ is the spring constant of each spring. This vector is sort of the opposite of a gradient vector, in the sense that it points to the direction of greatest decrease. If point $X$ moves perpendicular to this vector, there will be no change in the value of $A_1X^2+A_2X^2+\ldots+A_nX^2$. Ergo, you can move $X$ along a circle centered at $G$, and the value $A_1X^2+A_2X^2+\ldots+A_nX^2$ will remain constant.
| 77,093
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\begin{document}
\title[non-asymptotic $\ell_1$ spaces with unique $\ell_1$ asymptotic model]{ non-asymptotic $\ell_1$ spaces with unique $\ell_1$ asymptotic model }
\author[S. A. Argyros]{Spiros A. Argyros}
\address{National Technical University of Athens, Faculty of Applied Sciences,
Department of Mathematics, Zografou Campus, 157 80, Athens, Greece}
\email{sargyros@math.ntua.gr}
\author[A. Georgiou]{Alexandros Georgiou}
\address{National Technical University of Athens, Faculty of Applied Sciences,
Department of Mathematics, Zografou Campus, 157 80, Athens, Greece}
\email{ale.grgu@gmail.com}
\author[P. Motakis]{Pavlos Motakis}
\address{Department of Mathematics, University of Illinois at Urbana-Champaign, Urbana, IL 61801, U.S.A.}
\email{pmotakis@illinois.edu}
\thanks{{\em 2010 Mathematics Subject Classification:} Primary 46B03, 46B06, 46B25, 46B45.}
\thanks{The third author's research was supported by the National Science Foundation under Grant Numbers NSF DMS-1600600 and NSF DMS-1912897.}
\begin{abstract}
A recent result of Freeman, Odell, Sari, and Zheng \cite{FOSZ} states that whenever a separable Banach space not containing $\ell_1$ has the property that all asymptotic models generated by weakly null sequences are equivalent to the unit vector basis of $c_0$ then the space is Asymptotic $c_0$. We show that if we replace $c_0$ with $\ell_1$ then this result is no longer true. Moreover, a stronger result of B. Maurey - H. P. Rosenthal \cite{MR} type is presented, namely, there exists a reflexive Banach space with an unconditional basis admitting $\ell_1$ as a unique asymptotic model whereas any subsequence of the basis generates a non-Asymptotic $\ell_1$ subspace.
\end{abstract}
\maketitle
\section{Introduction}
In this paper we study the question whether the uniqueness of asymptotic models, or equivalently, the uniform uniqueness of joint spreading models in a given Banach space implies that the space must be Asymptotic $\ell_p$. This is a coordinate free version from \cite{MMT} of the notion of asymptotic $\ell_p$ spaces with a Schauder basis by Milman and Tomczak-Jaegermann from \cite{MT}. The question draws its motivation from the following Problem of Halbeisen and Odell from \cite{HO} and a subsequent remarkable result from \cite{FOSZ}. Given a Banach space $X$, let $\mathscr{F}_0(X)$ denote the class of normalized weakly null sequences and $\mathscr{F}_b(X)$ denote the class of all normalized block sequences of a fixed basis, if $X$ has one.
\begin{pr}[\cite{HO}]
\label{asmodelasellp}
Let $X$ be a Banach space that admits a unique asymptotic model with respect to $\mathscr{F}_0(X)$, or with respect to $\mathscr{F}_b(X)$ if $X$ has a basis. Does $X$ contain an Asymptotic $\ell_p$, $1\le p<\infty$, or an Asymptotic $c_0$ subspace?
\end{pr}
An asymptotic model is a notion which describes the asymptotic behavior of an array of sequences $(x_j^{i})_j$, $i\in\N$. On the contrary a space is Asymptotic $\ell_p$, for $1\le p<\infty$, (resp. Asymptotic $c_0$) if the asymptotic behavior of the whole space resembles that of $\ell_p$ (resp. $c_0$). Remarkably, in some cases unique asymptotic array structure implies that a space is Asymptotic $c_0$.
\begin{thm}[\cite{FOSZ}]\label{theorem FOSZ}
Let $X$ be a separable Banach space that does not contain $\ell_1$ and admits a unique $c_0$ asymptotic model with respect to $\mathscr{F}_0(X)$. Then the space $X$ is Asymptotic $c_0$.
\end{thm}
It was observed by Baudier, Lancien, Kalton, the third author, and Schlumprecht in \cite[Section 9.2]{BLMS} that Theorem \ref{theorem FOSZ} no longer holds if we replace $c_0$ with $\ell_p$ for any $1<p<\infty$. The counterexamples are spaces very similar to the space defined by Szlenk in \cite{Sz}. The main purpose of this paper is to provide an answer for the remaining case $p=1$. Note that the main obstruction in this case is the fact that the $\ell_1$-norm is the largest one and hence, assuming that the space admits a unique $\ell_1$ asymptotic model which means a very strong presence of asymptotic $\ell_1$ structure, it is not obvious how to preserve a tree structure in the space which has norm smaller than $\ell_1$.
\begin{thm}
\label{main}
There exists a reflexive Banach space $X$ with an unconditional basis that admits a unique $\ell_1$ asymptotic model with respect to $\mathscr{F}_0(X)$, whereas it is not an Asymptotic $\ell_1$ space.
\end{thm}
In fact, for every countable ordinal $\xi$, there is a space $T^\xi_{inc}$, as in Theorem \ref{main}, that contains a weakly null $\ell_2$-tree of height $\omega^\xi$. An easy modification of $T^\xi_{inc}$ can yield a space containing a weakly null $\ell_p$-tree, for any $1<p<\infty$ with $p\neq2$, or a weakly null $c_0$-tree of height $\omega^\xi$. Furthermore, the following analogue of the classical B. Maurey - H. P. Rosenthal \cite{MR} result is proved, which yields a stronger separation of the two properties than Theorem \ref{main}.
\begin{thm}
There exists a reflexive Banach space $X$ with an unconditional basis that admits a unique $\ell_1$ asymptotic model with respect to $\mathscr{F}_0(X)$, whereas any subsequence of the basis generates a non-Asymptotic $\ell_1$ subspace.
\end{thm}
More specifically, for every countable ordinal $\xi$, there is a space $T^\xi_{ess\-inc}$ as in the theorem above such that the space generated by any infinite subsequence of its basis contains a block $c_0$-tree of height $\omega^\xi$. It is possible to modify $T^\xi_{ess\-inc}$ to contain $\ell_p$-trees for any $1<p<\infty$, instead of $c_0$-trees.
In the final part of this paper we show that, for $1<r<p<\infty$, certain spaces ${JT}^\xi_{r,p}$, similar to those defined by Odell and Schlumprecht in \cite[Example 4.2]{OS3} (see also \cite[page 66]{O2}), admit a unique $\ell_p$ asymptotic model but are not Asymptotic $\ell_p$. These are spaces with an unconditional Schauder basis $(e_t)_{t\in\mathcal{\T_\xi}}$ indexed over a well-founded and infinite branching countable tree $\T_\xi$ of height $\omega^\xi$. The norm of ${JT}^\xi_{r,p}$ is defined as follows: if $x = \sum_{t\in\mathcal{T}_\xi}a_te_t$ and $S$ is a segment of $\mathcal{T}_\xi$ define $\|S(x)\|^r_r = \sum_{t\in S}|a_t|^r$ and
\begin{equation}\label{jtqp}\tag{1.8}
\|x\|_{{JT}_{r,p}^\xi} = \sup\bigg\{\Big(\sum_{i=1}^n\|S_i(x)\|_r^p\Big)^{1/p}\!\!\!\!: (S_i)_{i=1}^n \text{ disjoint segments of }\T_\xi\bigg\}.\end{equation}
The space $T^\xi_{inc}$ from Theorem \ref{main} is defined on the same tree. We say that two segments $S_1$, $S_2$ of $\mathcal{T}_\xi$ are incomparable if any node of $S_1$ is incomparable to any node of $S_2$. We relabel the basis of the Tsirelson space $T$ as $(e_t)_{t\in\mathcal{T}_\xi}$ so that the order is compatible with the initial one and define the norm of $T^\xi_{inc}$ as follows : for $x = \sum_{t\in\mathcal{T}_\xi}a_te_t$ define $\|S(x)\|^2_2 = \sum_{t\in S}|a_t|^2$ and
\[\|x\|_{T_{inc}^{\xi}} = \sup\bigg\{\Big\|\sum_{i=1}^n\|S_i(x)\|_{2}e_{\min{S_i}}\Big\|_T\!\!\!: (S_i)_{i=1}^n \text{ incomparable segments of } \T_\xi\bigg\}.\]
However, we will not use the above description of the norms. Instead we revert to the notion of norming sets and norming functionals. This makes some parts of the proof easier and it can also be potentially useful to show similar results on more complicated spaces based on these norms.
Finally, we should mention that Problem \ref{asmodelasellp} is only one of several concerning the separation of different asymptotic structures in Banach space theory. For example, in \cite{AM3} the first and third author showed that there exist spaces with a uniformly unique spreading model, which can be chosen to be any $\ell_p$ or $c_0$, that have no Asymptotic $\ell_p$ or $c_0$ subspace. This answers a question by Odell in \cite{O1} and Junge, Kutzarova, and Odell in \cite{JKO}. Moreover, in \cite{KM} Kutzarova and the third author showed that certain spaces by Beanland, the first author, and the third author from \cite{ABM} are asymptotically symmetric and have no Asymptotic $\ell_p$ or $c_0$ subspaces, answering a question from \cite{JKO}.
\begin{notn*}
By $\N=\{1,2,\ldots\}$ we denote the set of all positive integers. We will use capital letters as $L,M,N,\ldots$ (resp. lower case letters as $s,t,u,\ldots$) to denote infinite subsets (resp. finite subsets) of $\N$. For every infinite subset $L$ of $\N$, the notation $[L]^\infty$ (resp. $[L]^{<\infty}$) stands for the set of all infinite (resp. finite) subsets of $L$. For every $s\in[\N]^{<\infty}$, by $|s|$ we denote the cardinality of $s$. For $L\in[\N]^\infty$ and $k\in\N$, $[L]^k$ (resp. $[L]^{\le k}$) is the set of all $s\in[L]^{<\infty}$ with $|s|=k$ (resp. $|s|\le k$). For every $s,t\in[\N]^{<\infty}$, we write $s<t$ if either at least one of them is the empty set, or $\max s<\min t$. Also for $\emptyset\neq M\in[\N]^\infty$ and $n\in\N$ we write $n<M$ if $n<\min M$. For $s=\{n_1<\ldots<n_k\}\in[\N]^{<\infty}$ and for each $1\le i\le k$, we set $s(i)=n_i$.
Moreover, we follow \cite{LT} for standard notation and terminology concerning Banach space theory.
\end{notn*}
\section{Asymptotic structures}
Let us recall the definitions of the asymptotic notions that appear in the results of this paper and were mentioned in the introduction. Namely, asymptotic models, joint spreading models and the notions of Asymptotic $\ell_p$ and Asymptotic $c_0$ spaces. For a more thorough discussion, including several open problems and known results, we refer the reader to \cite[Section 3]{AM3}.
\begin{dfn}[\cite{HO}]
An infinite array of sequences $(x^{i}_j)_j$, $i\in\N$, in a Banach space $X$, is said to generate a sequence $(e_i)_i$, in a seminormed space $E$, as an asymptotic model if for every $\varepsilon>0$ and $n\in\N$, there is a $k_0\in\N$ such that for any natural numbers $k_0\leq k_1<\cdots<k_n$ and any choice of scalars $a_1,\ldots,a_n$ in $[-1,1]$ we have that
\[\Bigg|\Big\|\sum_{i=1}^na_ix_{k_i}^{i}\Big\| - \Big\|\sum_{i=1}^na_ie_{i}\Big\|\Bigg| < \varepsilon.\]
\end{dfn}
A Banach space $X$ is said to admit a unique asymptotic model with respect to a family $\mathscr{F}$ of normalized sequences in $X$ if whenever two infinite arrays, consisting of sequences from $\mathscr{F}$, generate asymptotic models then those must be equivalent. Typical families under consideration are those of normalized weakly null sequences, denoted $\mathscr{F}_0(X)$, normalized Schauder basic sequences, denoted $\mathscr{F}(X)$, or the family all normalized block sequences of a fixed basis of $X$, if it has one, denoted $\mathscr{F}_b(X)$.
The notion of plegma families was first introduced by Kanellopoulos, Tyros, and the first author in \cite{AKT}. We will use the slightly modified definition of \ from \cite{AGLM}.
\begin{dfn}[\cite{AGLM}]
Let $M\in[\N]^\infty$ and $k\in\N$. A {plegma} (resp. {strict plegma}) family in $[M]^k$ is a finite sequence $(s_i)_{i=1}^l$ in $[M]^k$ satisfying the following.
\begin{enumerate}
\item[(i)] $s_{i_1}(j_1)<s_{i_2}(j_2)$ for every $1\le j_1<j_2\le k$ and $1\le i_1,i_2\le l$.
\item[(ii)] $s_{i_1}(j)\le s_{i_2}(j)$ $\big($resp. $s_{i_1}(j)< s_{i_2}(j)\big)$ for all $1\le i_1<i_2\le l$ and $1\le j\le k$ .
\end{enumerate}
For each $l\in \N$, the set of all sequences $(s_i)^l_{i=1}$ which are plegma families in $[M]^k$ will be denoted by $Plm_l([M]^k)$ and that of the strict plegma ones by $S$-$Plm_l([M]^k)$.
\end{dfn}
\begin{dfn}[\cite{AGLM}]
A finite array of sequences $(x^{i}_j)_j$, $1\leq i\leq l$, in a Banach space $X$, is said to generate another array of sequences $(e_j^{i})_j$, $1\leq i\leq l$, in a seminormed space $E$, as a joint spreading model if for every $\varepsilon>0$ and $n\in\N$, there is a $k_0\in\N$ such that for any $(s_i)_{i=1}^l\in S$-$Plm([\N]^n)$ with $k_0\le s_1(1)$ and any $l\times n$ matrix $A=(a_{ij})$ with entries in $[-1,1]$ we have that
\[\Bigg|\Big\|\sum_{i=1}^l\sum_{j=1}^na_{ij}x_{s_i(j)}^{i}\Big\| - \Big\|\sum_{i=1}^l\sum_{j=1}^na_{ij}e_j^{i}\Big\|\Bigg|<\varepsilon.\]
\end{dfn}
A Banach space $X$ is said to admit a uniformly unique joint spreading model with respect to a family of normalized sequences $\mathscr{F}$ in $X$, if there exists a constant $C$ such that whenever two arrays $(x_j^{i})_j$ and $(y_j^{i})_j$, $1\leq i\leq l$, of sequences from $\mathscr{F}$ generate joint spreading models then those must be $C$-equivalent. Moreover, a Banach space admits a uniformly unique joint spreading model with respect to a family $\mathscr{F}$ if and only if it admits a unique asymptotic model with respect to $\mathscr{F}$ (see, e.g., \cite[Remark 4.21]{AGLM} or \cite[Proposition 3.12]{AM3}).
It was proved in \cite{AGLM} that whenever a Banach space admits a uniformly unique joint spreading model with respect to some family satisfying certain stability conditions, then it satisfies a property concerning its bounded linear operators, called the Uniform Approximation on Large Subspaces property (see \cite[Theorem 5.17]{AGLM} and \cite[Theorem 5.23]{AGLM}).
\begin{dfn}[\cite{MT} and \cite{MMT}]
A Banach space $X$ is called Asymptotic $\ell_p$, $1\leq p<\infty$, (resp. Asymptotic $c_0$) if there exists a constant $C$ such that in a two-player $n$-turn game $G(n,p,C)$, where in each turn $k=1,\ldots,n$ player (S) picks a finite codimensional subspace $Y_k$ of $X$ and then player (V) picks a normalized vector $x_k\in Y_k$, player (S) has a winning strategy to force player (V) to pick a sequence $(x_k)_{k=1}^n$ that is $C$-equivalent to the unit vector basis of $\ell^n_p$ (resp. $\ell^\infty_n)$.
\end{dfn}
The typical example of a non-classical Asymptotic $\ell_p$ space is the Tsirelson space as defined by Figiel and Johnson in \cite{FJ}. This is a reflexive Asymptotic $\ell_1$ space and it is the dual of Tsirelson's original space from \cite{T} that is Asymptotic $c_0$. Finally, whenever a Banach space is Asymptotic $\ell_p$ or Asymptotic $c_0$, then it admits a uniformly unique joint spreading model with respect to $\mathscr{F}_0(X)$ (see, e.g., \cite[Corollary 4.12]{AGLM}).
\section{A family of non-asymptotic $\ell_1$ spaces admitting uniformly unique $\ell_1$ joint spreading models}
\label{ell1section}
In this section we define the spaces $T_{inc}^\xi$, for each countable ordinal $\xi$, and we prove that they admit a uniformly unique $\ell_1$ joint spreading model with respect to $\mathscr{F}_b(T_{inc}^\xi)$ and are not Asymptotic $\ell_1$. The spaces are defined in terms of norming sets and norming functionals as this is more convenient to prove the desired result.
\subsection{Measures on Well-Founded Countable Compact Trees} We start with a key result that will be used later to prove that $T^\xi_{inc}$ admits a uniformly unique joint spreading model equivalent to the unit vector basis of $\ell_1$.
Let $\preceq$ be a partial order on some infinite subset $M$ of the naturals, which is compatible with the standard order, i.e. $n\preceq m$ implies $n\le m$, for all $n,m\in M$. Assume that, for each $n\in M$, the set $S_n=\{m\in M: m\preceq n \}$ is finite and totally ordered with respect to $\preceq$, that is, $\T=(M,\preceq)$ is a tree. Let us also assume that the tree $\T$ is well-founded, i.e., it contains no infinite totally ordered sets, and infinite branching, i.e., every non-maximal node has infinitely many immediate successors.
Observe that $\tT=(\{S_t :{t\in \T}\},\subset)$ is also a tree and is in fact isomorphic to $\T$ via the mapping $t\mapsto S_t$. Given $t\in\T$, we will denote $S_t$ by $\tt$. Moreover, two nodes $\tt_1,\tt_2$ are incomparable in $\tT$ if and only if the nodes $t_1,t_2$ are incomparable in $\T$, i.e. not comparable in the respective order. For $\tt\in\T$, we denote by $V_{\tt}$ the set consisting of $\tt$ and all of its successors.
Note that $\tT$ is a countable compact space when equipped with the pointwise convergence topology and hence $\mathcal{M}(\tT)$, the set of all regular measures on $\tT$, is isometric to $\ell_1(\tT)$. In particular, each $\mu\in\mathcal{M}(\tT)$ is of the form $\mu=\sum_{\tt\in \tT}a_\tt\delta_\tt$, where $\delta_\tt$ is the Dirac measure centered on $\tt$, and $\|\mu\|=\sum_{\tt\in \tT}|a_\tt|$. Finally, the support of $\mu$ is defined as $\supp\mu=\{\tt\in \tT:a_\tt\neq0 \}$. We will prove the following proposition, starting with Lemma \ref{lemma incomparable}
\begin{prop}\label{proposition incomparable}
Let $(\mu_j)_j$ be a sequence of positive regular measures on $\tT$ with disjoint finite supports and let $c>0$ be such that $\mu_j(\tT)<c$ for all $j\in\N$. Then, for every $\varepsilon>0$, there is an $L\in[\N]^\infty$ and a subset $G_j$ of $\supp\mu_{j}$ for each $j\in L$, satisfying the following.
\begin{itemize}
\item[(i)] $\mu_{j}(\tT\setminus G_j)\le \varepsilon$ for every $j\in L$.
\item[(ii)] The sets $G_j$, $j\in\N$, are pairwise incomparable.
\end{itemize}
\end{prop}
\begin{lem}\label{lemma incomparable}
Let $(\mu_j)_j$ be a sequence of positive regular measures on $\tT$ with disjoint finite supports and let $c>0$ be such that $\mu_j(\tT)<c$ for all $j\in\N$. Assume that $w^*\-\lim_j\mu_j=\mu=\sum_{\tt\in \tT}a_\tt\delta_\tt$. Then, for every $\tt\in\supp\mu$ and $\varepsilon>0$, there is an $L\in[\N]^\infty$ and a subset $G^\tt_j$ of $\supp\mu_{j}$ for each $j\in L$, satisfying the following.
\begin{itemize}
\item[(i)] $G^\tt_j\subset V_\tt$ for every $j\in L$.
\item[(ii)] $|\mu_{j}(G^\tt_j)-a_\tt|<\varepsilon$ for every $j\in L$.
\item[(iii)] The sets $G^\tt_j$, $j\in L$, are pairwise incomparable.
\end{itemize}
\end{lem}
\begin{proof} Recall that the nodes of $\T$ are in fact naturals numbers. Hence identifying $\{t:\tt\in\supp \mu_j \}$, $j\in\N$, as subsets of the naturals and passing to a subsequence, we may assume that they are successive.
Let $(\tt_j)_j$ be an enumeration of the immediate successors of $\tt$ and for each $j\in \N$ define $W^\tt_j=V_\tt\setminus\cup_{i=1}^jV_{\tt_i}$. Observe that $(W^\tt_j)_j$ is a decreasing sequence of clopen subsets of $\tT$ with $\cap_{j}W^\tt_j=\{\tt\}$ and hence $\lim_j\mu(W^\tt_j)=a_\tt$ and $\lim_j\mu_j(W^\tt_i)=\mu(W^\tt_i)$ for all $i\in\N$. We can thus find $N\in[\N]^\infty$ and pass to a subsequence of $(\mu_{j})_j$, which we relabel for convenience, so that $\lim_{j\in N}|\mu_{j}(W^\tt_j)-\mu(W^\tt_j)|=0$ and define $G^\tt_j=\supp\mu_{j}\cap W^\tt_j$ for each $j\in N$. Note then that $\lim_{j\in N}\mu_{j}(G^\tt_j)=a_\tt$ and $\mu_{j}|_{G^\tt_j}(\cup_{i=1}^j V_{\tt_i})=0$ for all $j\in N$.
There is at most one $j\in N$ such that $\tt\in G^\tt_j$ and hence, passing to a subsequence, we may assume that $\tt\notin G^\tt_j$ for all $j\in N$. Moreover, since $\lim_{j\in N}\mu_{j}(G^\tt_j)=a_\tt$, we may even pass to a further subsequence such that $|\mu_{j}(G^\tt_j)-a_\tt|<\varepsilon$ for all $j\in N$. For the remaining part of the proof we will choose, by induction, an $L\in[N]^\infty$ such that $G^\tt_{j}$, $j\in L$, are pairwise incomparable. Set $l_1=\min N$ and suppose that we have chosen $l_1<\ldots<l_k$ in $N$, for some $k\in\N$, such that $G^\tt_{l_i}$ and $G^\tt_{l_j}$ are incomparable, $1\le i<j\le k$. Pick $l_k<l_{k+1}\in N$ such that $\mu_{l_{k+1}}|_{G^\tt_{l_{k+1}}}(\cup\{ V_\ts:\ts\in\cup_{i=1}^k G^\tt_{l_i}\} )=0$. Then, if for some $1\le i \le k$ the nodes $\ts_1\in G^\tt_{l_i}$ and $\ts_2\in G^\tt_{l_{k+1}}$ are comparable, we have that $\ts_2\in V_{\ts_1}$ whereas $\mu_{l_{k+1}}(V_{\ts_{1}})=0$, which is a contradiction. Hence $G^\tt_{l_{1}},\ldots,G^\tt_{l_{k+1}}$ are pairwise incomparable.
\end{proof}
\begin{proof}[Proof of Proposition \ref{proposition incomparable}]
Passing to a subsequence, since $\tT$ is compact with respect to the pointwise convergence topology and $(\mu_j)_j$ are uniformly bounded, we may assume that $(\mu_j)_j$ $w^*$-converges to some measure $\mu=\sum_{\tt\in\tT} a_\tt\delta_\tt$ in $\mathcal{M}(\tT)$.
Let $\delta>0$ be such $(1-\delta)(\mu(\tT)-\delta)> \mu(\tT)-{\varepsilon}/{2}$ and pick $n_0\in\N$ such that $\sum_{i=1}^{n_0}a_{\tt_i}\ge \mu(\tT)-\delta$. Applying the previous lemma successively for each $\tt_i$, $i=1,\ldots,n_0$, we obtain an $L\in[\N]^\infty$ and, for each $j\in L$ and $i=1,\ldots,n_0$, a subset $G^i_j$ of $\supp\mu_{j}$ satisfying items (i) - (iii) of Lemma \ref{lemma incomparable} for $\tt_i$ and $\delta a_{\tt_i}$. Note that if $\tt_{i_1},\tt_{i_2}$ are incomparable for some $1\le i_1,i_2\le n_0$, then by item (i), the sets $G^{i_1}_{j_1}$ and $G_{j_2}^{i_2}$ are also incomparable for any $j_1,j_2\in L$. If the nodes $\tt_{i_1},\tt_{i_2}$ are comparable, say $\tt_{i_1}\subset\tt_{i_2}$, then there exists at most one $j\in L$ such that $\tt_{i_2}\in G^{i_1}_j$. Hence by a finite induction argument, we may pass to a subsequence such that the sets $G^i_j$, for $i=1,\ldots,n_0$ and $j\in L$, are pairwise incomparable. Define $G_j=\cup_{i=1}^{n_0}G^i_j$, $j\in L$, and conclude that
\[
\mu_{j}(G_j)=\sum_{i=1}^{n_0}\mu_{j}(G^i_j)\ge\sum_{i=1}^{n_0}a_{\tt_i}-\delta a_{\tt_i}\ge(\mu(\tT)-\delta)(1-\delta)> \mu(\tT)-\frac{\varepsilon}{2}.
\]
Finally, passing to a further subsequence if necessary, we may also assume that $|\mu_j(\tT)-\mu(\tT)|<{\varepsilon}/{2}$ and hence $\mu_j(\tT\setminus G_j)<\varepsilon$ for every $j\in L$.
\end{proof}
\subsection{Tsirelson Extension of a Ground Set.} In order to define $T_{inc}^\xi$, we first introduce some necessary concepts used in the construction of Tsirelson type spaces.
\begin{dfn}
A subset $W$ of $c_{00}(\N)$ is called a norming set if it satisfies the following conditions.
\begin{itemize}
\item[(i)] $W$ is symmetric and $ e^*_i\in W$ for every $i\in\N$.
\item[(ii)] $\|f\|_\infty\le1$ for every $f\in W$.
\item[(iii)] $W$ is closed under the restriction of its elements to intervals of $\N$.
\end{itemize}
\end{dfn}
A norming set $W$ induces a norm $\|\cdot\|_W$ on $c_{00}(\N)$ defined as
\[
\|x\|_W=\sup\{f(x):f\in W\}.
\]
\begin{dfn}Let $G$ be a norming set on $c_{00}(\N)$. The Tsirelson extension of $G$, denoted by $W_G$, is the minimal subset of $c_{00}(\N)$ that contains $G$ and is closed under the $(\mathcal{S},1/2)$-operation, i.e., if $f_1,\ldots,f_n$ are in $W_G$ and $n\le \supp f_1<\ldots<\supp f_n$, then $1/2\sum_{i=1}^nf_i$ is also in $W_G$. We call $G$ the ground set of $W_G$.
\end{dfn}
Note that $W_G$ is a norming set on $c_{00}(\N)$. Moreover, the induced norm $\|\cdot\|_{W_G}$ satisfies the following implicit equation
\[
\|x\|_{W_G}=\max\Big\{\|x\|_G,\frac{1}{2}\sup\sum_{i=1}^n\|E_ix\|_{W_G} \Big\}
\]
where the supremum is taken over all finite collections $E_1,\ldots,E_n$ of successive intervals of $\N$ with $n\le E_1$.
\begin{dfn}
Let $f\in W_G$. For a finite tree $\mathcal{A}$, a family $(f_\alpha)_{\alpha\in \mathcal{A}}$ is said to be a tree analysis of $f$ if the following are satisfied.
\begin{itemize}
\item[(i)] $\mathcal{A}$ has a unique root denoted by 0 and $f_0=f$.
\item[(ii)] For every maximal node $\alpha\in\mathcal{A}$ we have that $f_\alpha\in G$.
\item[(iii)] Let $\alpha$ be a non-maximal node of $\mathcal{A}$ and denote by $S(\alpha)$ set of immediate successors of $\alpha$. Then $f_\alpha\in W_G$ and the ranges of $f_s$, $s\in S(\alpha)$, are disjoint and $f_\alpha=1/2\sum_{s\in S(\alpha)}f_s$.
\end{itemize}
\end{dfn}
It follows, by minimality, that every $f\in W_G$ admits a tree analysis.
\begin{prop}\label{proposition tree analysis}
Let $f\in W_G$ with a tree analysis $(f_\alpha)_{\alpha\in \mathcal{A}}$ and denote by $\mathcal{M}$ the set of all maximal nodes of $\mathcal{A}$. Then the following hold.
\begin{itemize}
\item[(i)] For every $\alpha\in\mathcal{M}$, there is a $k_\alpha\in\N\cup\{0\}$ such that $f=\sum_{\alpha\in \mathcal{M}}f_\alpha/2^{k_\alpha}$.
\item[(ii)] If $\mathcal{N}\subset\mathcal{M}$, then $g=\sum_{\alpha\in \mathcal{N}}f_\alpha/2^{k_\alpha}$ is in $W_G$ and $g=f|_{\cup\{\supp f_\alpha:\alpha\in \mathcal{N}\}}$.
\end{itemize}
\end{prop}
For an extensive review on Tsirelson's space we refer the reader to \cite{CS}.
\subsection{Definition of the space $T^\xi_{inc}$}
We define the space $T^\xi_{inc}$ as the completion of $c_{00}(\N)$ with respect to the norm induced by a norming set $W_\xi$. This norming set is a subset of the Tsirelson extension of a ground set $G^\xi_2$, the functionals of which satisfy a certain property. Both this property and $G^\xi_2$ are defined via an infinite branching well-founded tree $\T_\xi$ on the natural numbers.
We start by fixing a partition of the naturals $\N=\cup_{j=0}^\infty N_j$ into infinite sets and an injection $\phi:[\N]^{<\infty}\to\N$. Recall the definition of the Schreir families $(\S_\xi)_{\xi<\omega_1}$.
\begin{dfn}
Let $\xi$ be a countable ordinal. We define, by transfinite induction, the Schreier family $\mathcal{S}_\xi\subset[\N]^{<\infty}$ as follows.
\begin{itemize}
\item[(i)] If $\xi=0$, then $\mathcal{S}_0=\{\{n\}:n\in\N \}\cup\{\emptyset\}$.
\item[(ii)] If $\xi=\alpha+1$, then \[\mathcal{S}_\xi =\{\cup_{j=1}^nE_j:n\in\N,\; E_1<\ldots<E_n\in\S_\alpha\text{ and }n\le E_1 \}.\]
\item[(iii)] If $\xi$ is a limit ordinal we choose a fixed sequence $(\alpha(\xi,j))_j\subset[1,\xi)$ which increases to $\xi$ and set \[\mathcal{S}_\xi=\{E\subset \N:\text{ there exists } j\in\N \text{ such that }E\in S_{\alpha(\xi,j)}\text{ and }j\le E\}.\]
\end{itemize}
\end{dfn}
We now define the tree $\T_\xi$, by defining a partial order $\preceq_\xi$ on $\N$.
\begin{dfn}
Fix a countable ordinal $\xi$ and define the partial order $\preceq_\xi$ on $\N$ as follows: $n\preceq_\xi m$ if there exists $\{n_0,\ldots,n_k \}\in \mathcal{S}_\xi$ such that
\begin{itemize}
\item[(i)] $n_0\in N_0$ and $n_i\in N_{\phi(n_0,\ldots,n_{i-1})}$ with $n_{i-1}<n_i$ for every $1\le i \le k$,
\item[(ii)] $n=n_i$ and $m=n_j$ for some $0\le i\le j\le k$.
\end{itemize}
\end{dfn}
\begin{rem} Note that $\T_\xi=(\N,\preceq_\xi)$ is an infinite branching tree and it is also well-founded since $\mathcal{S}_\xi$ is a compact family, i.e., $\{\rchi_E:E\in\S_\xi\}$ is a compact subset of $\{0,1\}^\N$. Moreover, the partial order $\preceq_\xi$ is compatible with the standard order on the naturals and finally, standard inductive arguments yield that $\T_\xi$ is of height $\omega^\xi$.
\end{rem}
\begin{dfn} Define the following norming set on $c_{00}(\N)$
\[
{G_2^{\xi}}=
\Big\{ \sum_{i\in S}a_ie^*_i:S\text{ is a segment of }\T_\xi\text{ and }\sum_{i\in S}a_i^2\le 1\Big\}
\]
and denote by $W_\xi$ the subset of $W_{G^\xi_2}$ containing all $f$ with tree analysis $(f_\alpha)_{\alpha\in\mathcal{A}}$ such that there exist pairwise incomparable segments $S_{\alpha}$ of $\T_\xi$ with $\supp f_\alpha\subset S_{\alpha}$ for every maximal node $\alpha\in\mathcal{A}$.
Denote by $T_{inc}^\xi$ the completion of $c_{00}(\N)$ with respect to the norm $\|\cdot\|_{W_\xi}$ induced by the norming set $W_\xi$.
\end{dfn}
\begin{rem}
The unit vector basis $(e_j)_j$ of $c_{00}(\N)$ forms a 1-unconditional Schauder basis for the space $T^\xi_{inc}$. Moreover it is boundedly complete, since $T^\xi_{inc}$ admits $\ell_1$ as a uniformly unique spreading model as shown in Proposition \ref{first space unique as model}.
\end{rem}
First, we show that $T_{inc}^\xi$ admits a uniformly unique joint spreading model with respect to $\mathscr{F}_b(T_{inc}^\xi)$, that is equivalent to the unit vector basis of $\ell_1$.
\begin{prop}\label{first space unique as model}
The space $T_{inc}^\xi$ admits a uniformly unique joint spreading model with respect to $\mathscr{F}_b(T_{inc}^\xi)$, which is equivalent to the unit vector basis of $\ell_1$.
\end{prop}
\begin{proof}
Let $(x^i_j)_j$, $1\le i\le l$, be an array of normalized block sequences in $T_{inc}^\xi$ and $\varepsilon>0$. Passing to a subsequence, we assume that $\supp x^{i_1}_j <\supp x^{i_2}_{j+1}$ for every $i_1,i_2=1,\ldots, l$ and $j\in\N$. For every $i=1,\ldots,l$ and $j\in\N$, pick a functional $f^i_j=\sum_{\alpha\in\mathcal{M}_{j}^i }f^i_{j,\alpha}/2^{k^i_{j,\alpha}}$ in $W_\xi$ such that $f^i_j(x^i_j)\ge1-\varepsilon$ and $f^i_{j,\alpha}(x^i_j)>0$ for every $\alpha\in \mathcal{M}^i_j$, where $\mathcal{M}^i_j$ denotes the set of all maximal nodes of a fixed tree analysis of $f^i_j$. For every $i=1,\ldots,l$, $j\in\N$ and $\alpha\in \mathcal{M}^i_j$, define $\lambda^i_{j,\alpha}=f^i_{j,\alpha}(x^i_j)/2^{k^i_{j,\alpha}}$ and $t^i_{j,\alpha}=\min\supp f^i_{j,\alpha}$. Moreover, for each $j\in\N$, define the probability measure \[\mu_j=\frac{1}{l}\sum_{i=1}^l\frac{1}{f^i_j(x^i_j)}\sum_{\alpha\in \mathcal{M}_{j}^i}\lambda^i_{j,\alpha}\delta_{\tt^i_{j,\alpha}}.\] Then, Proposition \ref{proposition incomparable} yields an $L\in[\N]^\infty$ and a sequence $(G_j)_{j\in L}$ of pairwise incomparable subsets of ${\tT_\xi}$ such that $\mu_{j}(G_j)\ge 1-\delta$ for every $j\in L$ and for some $\delta$ sufficiently small such that for any $i=1,\ldots,l$ and $j\in L$
\begin{equation}\label{each measure on Gj}\tag{2.8.1}
\frac{1}{f_j^i(x_j^i)}\sum_{\alpha\in \mathcal{M}_{j}^i}\lambda^i_{j,\alpha}\delta_{\tt^i_{j,\alpha}}(G_j)\ge (1-\varepsilon)^2.
\end{equation}
Let $k\in \N$ and $(s_i)_{i=1}^l\in S$-$Plm_l([L]^k)$ with $kl\le x_{s_1(1)}$. Then, for $i=1,\ldots,l$ and $j\in L$, if $\mathcal{N}^i_j=\{\alpha \in\mathcal{M}^i_{s_i(j)}:\tt^i_{j,\alpha}\in G_{s_i(j) }\}$, item (ii) of Proposition \ref{proposition tree analysis} yields that
\[g^i_j\;=\sum_{\alpha\in\mathcal{N}^i_j } \frac{1}{2^{k^i_{s_i(j),\alpha}}}f^i_{s_i(j),\alpha}\in W_{\xi}.\] Moreover, \eqref{each measure on Gj} implies $g^i_j(x^i_{s_i(j)})\ge (1-\varepsilon)^2$ for all $i=1,\ldots,l$ and $j\in L$, and since $G_j$, $j\in L$, are pairwise incomparable, we have that $g=1/2\sum_{i=1}^l\sum_{j=1}^kg^i_j$ is in $W_{{\xi}}$. Then for any choice of scalars $(a_{ij})_{i=1,j=1}^{l,k}$, due to unconditionality, we conclude that
\[
\Big\|\sum_{i=1}^l\sum_{j=1}^ka_{ij}x^i_{s_i(j)}\Big\|\ge \Big\|\sum_{i=1}^l\sum_{j=1}^k|a_{ij}|x^i_{s_i(j)}\Big\|\ge\frac{(1-\varepsilon)^2}{2}\sum_{i=1}^l\sum_{j=1}^k|a_{ij}|.\]
\end{proof}
\begin{prop}\label{section 1 reflexivity}
The space $T_{inc}^\xi$ is reflexive.
\end{prop}
\begin{proof}
Since $T_{inc}^\xi$ admits a boundedly complete unconditional Schauder basis, it does not contain $c_0$ (see \cite[Theorem 1.c.10]{LT}) and hence it suffices to show that it does not contain $\ell_1$ as follows from \cite[Theorem 2]{J1}.
Fix $n\in\N$. Let $(x_j)_j$ be a normalized block sequence in $T_{inc}^\xi$ and $f=\sum_{i\in S}b_ie^*_i$ in $G^\xi_2$. For each $j=1,\ldots,n$, define $I_j=\{i\in S: i\in\supp x_j \}$ and note that
\[
\Big(\sum_{i\in I_j}b_jx_j(i)\Big)^2\le\sum_{i\in I_j}b_i^2.
\]
Then, for any choice of scalars $a_1,\ldots,a_n$, we have that
\begin{align*}
f\Big(\sum_{j=1}^na_jx_j\Big)&=\sum_{j=1}^na_j\sum_{i\in S}b_ix_j(i)\le\Big(\sum_{j=1}^na_j^2\Big)^{\frac{1}{2}}\Big(\sum_{j=1}^n\Big(\sum_{i\in I_j}b_ix_j(i)\Big)^2\Big)^{\frac{1}{2}}\\
&\le \Big(\sum_{j=1}^na_j^2\Big)^{\frac{1}{2}}\Big(\sum_{j=1}^n\sum_{i\in I_j}b_i^2\Big)^{\frac{1}{2}}\le \Big(\sum_{j=1}^na_j^2\Big)^{\frac{1}{2}}
\end{align*}
and hence
\[
\Big\|\sum_{j=1}^{n}a_jx_j\Big\|_{G^\xi_2}\le \Big(\sum_{j=1}^na_j^2\Big)^{\frac{1}{2}}.
\]
That is, for any normalized block sequence $(x_j)_j$ in $T_{inc}^\xi$, there exists a block subsequence $(y_j)_j$ with $\|y_j\|_{G_2^\xi}\to0$.
We show that $T^\xi_{inc}$ does not contain $\ell_1$ in a similar manner as in the proof of the reflexivity for the classical Tsirelson space \cite{FJ}. Suppose that $T^\xi_{inc}$ contains $\ell_1$. Then James' $\ell_1$ distortion theorem \cite{J2} implies that, for $\varepsilon<{1}/{4}$, there exists a normalized block sequence $(x_j)_j$ in $T^\xi_{inc}$ such that
\[
\Big\|\sum_{j=1}^na_jx_j\Big\|\ge (1-\varepsilon)\sum_{j=1}^n|a_j|
\]
for any $n\in\N$ and any choice of scalars $a_1,\ldots,a_n$. Applying the result of the previous paragraph, we may also assume that $\|x_j\|_{G^\xi_2}<{1}/{2}$ for every $j\in\N$ and hence, for any $n\ge2$, we have that
\begin{equation}\tag{2.9.1}\label{reflexivity they are not normed by ground set}
\Big\|x_1+\frac{1}{n}\sum_{i=2}^{n+1}x_i\Big\|> \Big\|x_1+\frac{1}{n}\sum_{i=2}^{n+1}x_i\Big\|_{G^\xi_2}.
\end{equation}
Moreover, for any $n\in\N$, we have that
\[
\Big\|x_1+\frac{1}{n}\sum_{i=2}^{n+1}x_i\Big\|\ge 2(1-\varepsilon).
\]
Observe that (\ref{reflexivity they are not normed by ground set}) implies that there exists $f={1}/{2}\sum_{j=1}^kf_j\in W_\xi\setminus G^\xi_2$ such that
\[
f\Big(x_1+\frac{1}{n}\sum_{i=2}^{n+1}x_i\Big)> \Big\|x_1+\frac{1}{n}\sum_{i=2}^{n+1}x_i\Big\|-{\varepsilon} \ge\frac{5}{4}
\]
and that $\min\supp f_1\le \max\supp x_1$, since otherwise
\[
f\Big(x_1+\frac{1}{n}\sum_{i=2}^{n+1}x_i\Big)=\frac{1}{n}\sum_{i=2}^{n+1}f(x_i)\le 1.
\]
Therefore, $k\le \max\supp x_1$. Note that there are at most $k$ $i$'s such that the support of $x_i$ intersects the supports of at least two $f_j$'s and hence
\[
f\Big(x_1+\frac{1}{n}\sum_{i=2}^{n+1}x_i\Big)\le 1 +\frac{k}{n}+\frac{n-k}{2n}\le 1+\frac{n+\max\supp x_1}{2n}\xrightarrow[n \to \infty]{}\frac{3}{2}.
\]
This yields a contradiction for sufficiently large $n$ since ${3}/{2}<2(1-\varepsilon)$.
\end{proof}
\begin{prop}\label{NOT Asymptotic l1}
The space $T_{inc}^\xi$ is not Asymptotic $\ell_1$.
\end{prop}
\begin{proof}
Suppose that $T_{inc}^\xi$ is $C$-Asymptotic $\ell_1$ and let $n\in\N$ be such that $n> C^2$. Since $T_{inc}^\xi$ is reflexive, we may assume that player (S) chooses tail subspaces (see \cite[Lemma 5.18]{AGLM}) throughout any winning strategy in the game $G(n,1,C)$. Let us assume the role of player (V) and let $Y_1$ be the tail subspace with which player (S) initiates the game. Then, as player $(V)$, we choose an element of the basis $e_{j_1}\in Y_1$, such that $|S|\ge n$ for every maximal segment $S$ of $\T_\xi$ with $\min S={j_1}$. Suppose that in the $k+1$ turn of the game, for $k<n$, player $(S)$ chooses the subspace $Y_{k+1}$. Then, again as player (V), we choose a vector $e_{j_{k+1}}\in Y_{k+1}$ with $j_{k+1}$ an immediate successor of $j_k$. Note that, in the final outcome of the game, we have chosen elements of the basis $e_{j_1},\ldots,e_{j_n}$ such that $\{j_1,\ldots,j_n\}$ is a segment of $\T_\xi$ and hence $\{e_{j_1},\ldots,e_{j_n}\}$ is isometric to the standard basis of $\ell_2^n$. We calculate
\[
\Big\| \frac{1}{n}\sum_{i=1}^n e_{j_i}\Big\|= \Big\| \frac{1}{n}\sum_{i=1}^n e_{j_i}\Big\|_{G^\xi_2}=n^{-\frac{1}{2}}
\]
whereas, since $T_{inc}^\xi$ is $C$-Asymptotic $\ell_1$, we have that
\[
\frac{1}{C}\le\Big\| \frac{1}{n}\sum_{i=1}^n e_{j_i}\Big\|
\]
and this is a contradiction.
\end{proof}
\begin{rem}
For any $1<p<\infty$, we may replace the norming set $G^\xi_2$ with
\[
G^\xi_p=\bigg\{\sum_{i\in S}a_ie^*_i:\text{ S is a segment of }\T_\xi\text{ and }\sum_{i\in S}|a_i|^q\le1 \bigg\}
\]
where $p^{-1}+q^{-1}=1$, to obtain a reflexive Banach space admitting a uniformly unique $\ell_1$ joint spreading model, that contains a weakly null $\ell_p$-tree of height $\omega^\xi$ or a weakly null $c_0$-tree if we replace $G^\xi_2$ with $G=\{\pm e^*_i:i\in\N\}$.
\end{rem}
\section{A stronger separation of the two properties}
The spaces $T^\xi_{inc}$ constructed in the previous section, yield a separation between the properties of being an Asymptotic $\ell_1$ space and admitting a unique $\ell_1$ asymptotic model. It is easy however to see that these spaces contain subsequences of their bases that generate Asymptotic $\ell_1$ subspaces. For example, consider any subspace generated by a subsequence $(e_j)_{j\in M}$ of the basis of some $T^\xi_{inc}$, such that the elements of $M$ are pairwise incomparable in $\T_\xi$. In this section we show that, for any countable ordinal $\xi$, there is a reflexive Banach space $T^\xi_{ess\-inc}$ that admits a unique $\ell_1$ asymptotic model with respect to $\mathscr{F}_b(T^\xi_{ess\-inc})$ and any subsequence of its basis generates a non-Asymptotic $\ell_1$ subspace. To some extent, this family of spaces is the Maurey - Rosenthal \cite{MR} analogue of the two aforementioned properties.
Start by fixing a countable ordinal $\xi$ and let $(m_j)_{j\ge0}$, $(n_j)_{j\ge 0}$ be increasing sequences of natural numbers such that :
\begin{itemize}
\item[(i)] $m_0=2$, $m_1=4$ and $m_j\ge m_{j-1}^2$ for every $j\ge 2$ and
\item[(ii)] $n_0=1$, $n_1=6$ and $n_j>\log_2m^2_{j}+n_{j-1}$ for every $j\ge2$.
\end{itemize}
Let $\mathcal{Q}$ denote the collection of all finite sequences $((g_1,m_{j_1}),\ldots,(g_{k},m_{j_k}))$, where $g_i:\N\to\{-1,0,1\}$ has finite support and $j_i\in\N$ for $1\leq i\leq k$, and $m_{j_1}<\cdots<m_{j_k}$. Let $\sigma :\mathcal{Q}\to \{m_j:j\in\N\}$ be an injection so that each sequence $((g_1,m_{j_1}),\ldots,(g_{k},m_{j_k}))$ is mapped to some $m_j$ with $m_{j_k}<m_j$.
\begin{dfn}\label{definition full tree support+weight}
Let $\tT_\xi$ be the set of all finite sequences $((g_1,m_{j_1}),\ldots,(g_k,m_{j_k}))$ satisfying the following conditions.
\begin{itemize}
\item[(i)] $g_i:\N\to\{-1,0,1\}$ for $i=1,\ldots,k$ with $\supp g_1<\ldots<\supp g_k$.
\item[(ii)] $\supp g_i\in \S_{n_{j_i}}$ for $i=1,\ldots,k$, where $n_{j_1}=n_1$ and $n_{j_1}<\ldots<n_{j_k}$.
\item[(iii)] $m_{j_1}=m_1$ and $m_{j_i}=\sigma ((g_1,m_{j_1}),\ldots,(g_{i-1},m_{j_{i-1}}))$ for every $i=2,\ldots,k$.
\item[(iv)] $\{\min \supp g_i:i=1,\ldots,k\}\in\S_\xi$.
\end{itemize}
\end{dfn}
Note that item (iii) of the above definition implies that $\tT_\xi$, equipped with the partial order $\le_{\tT_\xi}$ where $\tt_1\le_{\tT_\xi}\tt_2$ if $\tt_1$ is an initial segment of $\tt_2$, is a tree. Moreover, it is easy to see that it is infinite-branching, and as follows from item (iv) and standard inductive arguments, it is also well founded and of height $\omega^\xi$. In particular, the above remain true if for an infinite subset of the naturals $M$ we additionally require that $\supp g_i\subset M$ for every $i=1,\ldots,k$, in Definition \ref{definition full tree support+weight}.
We may also identify $\tT_\xi$ as a closed subset, with respect to the pointwise convergence topology, of $\big\{ \{{\pm m_j^{-1}}\}_{j\in\N}\cup\{0\} \big\}^\N$ via the mapping
\[
((g_1,m_{j_1}),\ldots,(g_k,m_{j_k}))\mapsto {m^{-1}_{j_1}}\;g_1+\cdots+{m^{-1}_{j_k}}\;g_k.
\]
The fact that $\lim_jm_j^{-1}=0$ implies that $\big\{ \{{\pm m_j^{-1}}\}_{j\in\N}\cup\{0\} \big\}^\N$ is compact with respect to the pointwise convergence topology of $[-1,1]^\mathbb{N}$.
Observe that, as a consequence of item (iii), any $\tt=((g_1,m_{j_1}),\ldots,(g_k,m_{j_k}))$ in $\tT_\xi$ is uniquely determined by the pair $(g_k,m_{j_k})$, which we will denote by $(g_t,m_{j_t})$ or just by $t$ (i.e., $t = (g_t,m_{j_t})$. Taking advantage of this we may define $\T_\xi=\{(g_t,m_{j_t}):\tt\in\tT_\xi \}$, which is in bijection with $\tT_\xi$ via the mapping $t=(g_t,m_{j_t})\mapsto \tt$. Note that $\le_{\tT_\xi}$ induces a natural order, denoted by $\le_{\T_\xi}$, on $\T_\xi$, where $(g_{t_1},m_{j_{t_1}})\le_{\T_\xi}(g_{t_2},m_{j_{t_2}})$ if $\tt_1\le_\tT \tt_2$. Clearly, the tree $(\T_\xi,\le_{\T_\xi})$ is isomorphic to $(\tT_\xi,\le_{\tT_\xi})$ via the mapping $t=(g_t,m_{j_t})\mapsto \tt$.
\begin{dfn}
Let $\tW_\xi$ be the set of all finite sequences $(m_{j_1},m_{j_2},\ldots,m_{j_k})$\linebreak for which there exist $g_1,\ldots,g_k:\N\to\{-1,0,1\}$ with $((g_1,m_{j_1}),\ldots,(g_k,m_{j_k}))\in\tT_\xi$.
\end{dfn}
The initial segment order $\le_{\tW_\xi}$ is a partial order on $\tW_\xi$ and is in fact naturally induced by the order $\le_{\tT_\xi}$. Moreover, it is easy to verify that $(\tW_\xi,\le_{\tW_\xi})$ is a well founded infinite-branching tree of height $\omega^\xi$. It is also isomorphic to the tree $(\W_\xi,\le_{\W_\xi})$, where $\W_\xi=\{m_{j_t}:\tt\in\tT_\xi \}$ and $m_{j_{t_1}}\le_{\W_\xi} m_{j_{t_2}}$ if $\tt_1\le_{\tT_\xi}\tt_2$. This correspondence between $\tW_\xi$ and $\W_\xi$ is identical to that of $\tT_\xi$ and $\T_\xi$.
\begin{rem}
\begin{itemize}
\item[(i)] If $m_{j_{t_1}}$, $m_{j_{t_2}}$ are incomparable nodes in $\W_\xi$, then for every $g_1,g_2:\N\to\{-1,0,1\}$ such that $(g_1,m_{j_{t_1}})$ and $(g_2,m_{j_{t_2}})$ are in $\T_\xi$, these are also incomparable.
\item[(ii)] Note that there exist nodes $\tt_1$ and $\tt_2$ which are incomparable in $\tT_\xi$, whereas $m_{j_{t_1}}$ and $m_{j_{t_2}}$ are comparable in $\W_\xi$. To see this, consider any node\linebreak $\tt=((g_1,m_{j_1}),\ldots,(g_k,m_{j_k}))$ in $\tT_\xi$ with $k>1$ and, for each $i=1,\ldots,k-1$, let $h_i:\N\to\{-1,0,1\}$ be such that $h_i\neq g_i$ and $t_i=(h_i,m_{j_i})$ is in $\T_\xi$. Then, item (iii) of Definition \ref{definition full tree support+weight} implies that the nodes $\tt_i$ and $\tt$ are incomparable whereas $m_{j_{t_i}}$ and $m_{j_{t}}$ are comparable for every $i=1,\ldots,k-1$, since $\tt\in\tT_\xi$.
\end{itemize}
\end{rem}
\begin{dfn}
We say that a subset $X$ of $\T_\xi$ is essentially incomparable if whenever $(g_{t_1},m_{j_{t_1}})$, $(g_{t_2},m_{j_{t_2}})$ are in $X$ with $m_{j_{t_1}}<_{\W_\xi} m_{j_{t_2}}$ and $g:\N\to\{-1,0,1\}$ is the unique sequence such that $(g,m_{j_{t_1}})\le _{\T_\xi} (g_{t_2},m_{j_{t_2}})$, then $\supp g<\supp g_{t_1}$.
\end{dfn}
\begin{rem}
Let $X$ be an essentially incomparable subset of $\T_\xi$ and $h_t:\N\to\{-1,0,1\}$ with $\supp h_t\subset\supp g_t$ for every $t\in X$. Then $\{(h_t,m_{j_t}):t\in X\}$ is also an essentially incomparable subset of $\T_\xi$.
\end{rem}
The following lemma is an extension of Proposition \ref{proposition incomparable} and is the main ingredient of the proof that the space $T^\xi_{ess\- inc}$ admits a uniformly unique joint spreading model.
\begin{lem}\label{combinatorial lemma essentially incomparable}
Let $(\mu_i)_i$ be a sequence of positive regular measures on $\tT_\xi$ with finite supports and let $C>0$ be such that $\mu_i(\tT_\xi)<C$ for all $i\in\N$. Assume that the sets $\cup\{\supp g_t:\tt\in\supp\mu_i\}$, $i\in\N$, are disjoint. Then, for every $\varepsilon>0$, there exists an $M\in[\N]^\infty$ and $G^1_i$, $G^2_i$ subsets of $\tT_\xi$ for each $i\in M$, such that
\begin{itemize}
\item[(i)] $G^1_i$, $G^2_i$ are disjoint subsets of $\supp\mu_i$ for every $i\in M$,
\item[(ii)] $\mu_i(\tT_\xi\setminus G^1_i\cup G^2_i)<\varepsilon$ for every $i\in M$,
\item[(iii)] $\{t\in\T_\xi:\tt\in \cup_{i\in M}G^1_i\}$ is essentially incomparable and
\item[(iv)] for every $i_1\neq i_2$ in $M$, every $\tt_1\in G^2_{i_1}$ and $\tt_2\in G^2_{i_2}$, the nodes $m_{j_{t_1}}$ and $m_{j_{t_2}}$ are incomparable in $\W_\xi$.
\end{itemize}
\end{lem}
Before we are able to prove this Lemma it is necessary to introduce the notion of successor limits of measures. We find this limit notion to be of independent interest and therefore we use broader terminology to define it and prove its properties.
\begin{notn}
Let $\mathcal{T}$ be a countably branching well founded tree. For each $t\in\mathcal{T}$ we denote $\mathrm{succ}_{\mathcal{T}}(t)$ the set of immediate successors of $t$. In particular, if $t$ is maximal then $\mathrm{succ}_{\mathcal{T}}(t)$ is empty. For $t\in\mathcal{T}$ we denote $V_t = \{s\in\mathcal{T}:t\leq s\}$. We view $\mathcal{T}$ as topological space with the topology generated by the sets $V_t$ and $\mathcal{T}\setminus V_t$, $t\in\mathcal{T}$. This is a compact metric topology for which the sets of the form $V_t\setminus(\cup_{s\in F}V_s)$, $t\in\mathcal{T}$ and $F\subset\mathrm{succ}_\mathcal{T}(t)$ finite, form a base of clopen sets. We denote by $\mathcal{M}_+(\mathcal{T})$ the cone of all bounded positive measures $\mu:\mathcal{P}(\mathcal{T})\to[0,+\infty)$. For $\mu\in\mathcal{M}_+(\mathcal{T})$ we define the support of $\mu$ to be the set $\mathrm{supp}(\mu) = \{t\in\mathcal{T}:\mu(\{t\})>0\}$. A set $A$ in $\mathcal{M}_+(\mathcal{T})$ is called bounded if $\sup_{\mu\in\mathcal{A}}\mu(\mathcal{T})<\infty$.
\end{notn}
Recall that a sequence $(\mu_i)$ in $\mathcal{M}_+(\mathcal{T})$ converges in the $w^*$-topology to a $\mu\in\mathcal{M}_+(\mathcal{T})$ if and only if for all clopen sets $V\subset \mathcal{T}$ we have $\lim_i\mu_i(V) = \mu(V)$ if and only if for all $t\in\mathcal{T}$ we have $\lim_i\mu_i(V_t) = \mu(V_t)$.
\begin{dfn}
Let $\mathcal{T}$ be a countably branching well founded tree, $(\mu_i)$ be a disjointly supported sequence in $\mathcal{M}_+(\mathcal{T})$ and $\nu\in\mathcal{M}_+(\mathcal{T})$. We say that $\nu$ is the successor-determined limit of $(\mu_i)$ if for all $t\in\mathcal{T}$ we have $\mu(\{t\}) = \lim_i\mu_i(\mathrm{succ}_\mathcal{T}(t))$. In this case we write $\nu = \mathrm{succ}\-\!\lim_i\mu_i$.
\end{dfn}
\begin{rem}
It is possible for a disjointly supported and bounded sequence $(\mu_i)\in\mathcal{M}_+(\mathcal{T})$ to satisfy $w^*\-\lim_i\mu_i\neq \mathrm{succ}\-\!\lim_i\mu_i$. Take for example $\mathcal{T} = [\mathbb{N}]^{\leq 2}$ (all subsets of $\mathbb{N}$ with at most two elements with the partial order of initial segments). Define $\mu_i = \delta_{\{i,i\}}$. Then, $w^*\-\lim_i\mu_i = \delta_\emptyset$ whereas $\mathrm{succ}\-\!\lim_i\mu_i = 0$.
\end{rem}
Although these limits are not the necessarily the same, there is an explicit formula relating $\mathrm{succ}\-\!\lim_i\mu_i$ to $w^*\-\lim_i\mu_i$.
\begin{lem}
\label{formula w-succ}
Let $\mathcal{T}$ be a countable well founded tree, $(\mu_i)$ be a bounded and disjointly supported sequence in $\mathcal{M}_+(\mathcal{T})$ so that $w^*\-\lim_i\mu_i = \mu$ exists and for all $t\in\mathcal{T}$ the limit $\nu(\{t\}) = \lim_i\mu_i(\mathrm{succ}_\mathcal{T}(t)) $ exists as well. Then, for every $t\in\mathcal{T}$ and enumeration $(t_j)$ of $\mathrm{succ}_\mathcal{T}(t)$ we have
\begin{equation}
\label{formula w-succ formula}
\mu(\{t\}) = \nu(\{t\}) + \lim_j\lim_i\mu_i\Big(\cup_{k\geq j}(V_{t_k}\setminus\{t_{k}\})\Big).
\end{equation}
In particular, $\mu(\{t\}) = \nu(\{t\})$ if and only if the double limit in \eqref{formula w-succ formula} is zero.
\end{lem}
\begin{proof}
For $j\in\mathbb{N}$ we have $\{t\}\cup(\cup_{k\geq j}V_{t_k}) = V_t\setminus(\cup_{k<j}V_{t_k})$ which is clopen and thus
\begin{equation}
\label{thankfully its clopen}
\lim_i\mu_i\big(\{t\}\cup(\cup_{k\geq j}V_{t_k})\big) = \mu\big(\{t\}\cup(\cup_{k\geq j}V_{t_k})\big).
\end{equation}
Because $(\mu_i)$ is disjointly supported we observe that for all $j\in\N$
\begin{equation}
\label{formula w-succ eq1}
\lim_i\mu_i(\{t_{k}:k\geq j\}) = \lim_i\mu_i(\mathrm{succ}_\mathcal{T}(t)) = \nu(\{t\}).
\end{equation}
We calculate
\begin{equation*}
\begin{split}
\mu(\{t\}) &= \lim_{j\to\infty}\mu\Big(\{t\}\cup(\cup_{k\geq j}{V}_{t_j})\Big) \stackrel{\eqref{thankfully its clopen}}{=} \lim_j\lim_i\mu_i\Big(\{t\}\cup(\cup_{k\geq j}V_{t_j})\Big)\\
&= \lim_j\lim_i\mu_i(\cup_{k\geq j}V_{t_j}) = \lim_j\lim_i\mu_i(\{t_{k}:k\geq j\}\cup(\cup_{k\geq j}(V_{t_k}\setminus\{t_{k}\})))\\
&= \lim_j\lim_i\mu_i(\{t_{k}:k\geq j\}) + \lim_j\lim_i\mu_i\Big(\cup_{k\geq j}(V_{t_k}\setminus\{t_{k}\})\Big).
\end{split}
\end{equation*}
Thus, \eqref{formula w-succ eq1} yields the conclusion.
\end{proof}
\begin{cor}
Let $\mathcal{T}$ be a countable well founded tree and $(\mu_i)$ be a bounded and disjointly supported sequence in $\mathcal{M}_+(\mathcal{T})$. Then, there exist a subsequence $(\mu_{i_n})$ of $(\mu_i)$ and $\nu\in\mathcal{M}_+(\mathcal{T})$ with $\nu = \mathrm{succ}\-\!\lim_n\mu_{i_n}$.
\end{cor}
\begin{proof}
By passing to a subsequence, $\mu = w^*\-\lim_i\mu_i$ exists and for all $t\in\mathcal{T}$ the limit $\nu(\{t\}) = \lim_i\mu_i(\mathrm{succ}_\mathcal{T}(t)) $ exists as well. By \eqref{formula w-succ formula} for all $t\in\mathcal{T}$ we have $\nu(\{t\})\leq \mu(\{t\})$. Thus $\sum_{t\in\mathcal{T}}\nu(\{t\}) \leq \mu(\mathcal{T})$, i.e., $\nu$ defines a bounded positive measure
\end{proof}
\begin{lem}
\label{splitting lemma}
Let $\mathcal{T}$ be a countable well founded tree and $(\mu_i)$ be a bounded and disjointly supported sequence in $\mathcal{M}_+(\mathcal{T})$ so that $\mathrm{succ}\-\!\lim_i\mu_i = \nu$ exists. Then, there exist an infinite $L\subset\mathbb{N}$ and partitions $A_i$, $B_i$ of $\mathrm{supp}(\mu_i)$, $i\in L$, so that the following are satisfied.
\begin{itemize}
\item[(i)] If for all $i\in L$ we define the measure $\mu_i^1$ given by $\mu_i^1(C)= \mu_i(C\cap A_i)$, then $\nu = w^*\-\lim_{i\in L}\mu_i^1 = \mathrm{succ}\-\!\lim_{i\in L}\mu_i^1$.
\item[(ii)] If for all $i\in L$ we define the measure $\mu_i^2$ given by $\mu_i^2(C)= \mu_i(C\cap B_i)$ then for all $t\in\mathcal{T}$ the sequence $(\mu_i^2(\mathrm{succ}_\mathcal{T}(t)))_i$ is eventually zero. In particular, $\mathrm{succ}\-\!\lim_{i\in L}\mu_i^2 = 0$.
\end{itemize}
\end{lem}
\begin{proof}
Enumerate $\mathcal{T} = \{s_n:n\in\mathbb{N}\}$ and assume, passing if necessary to a subsequence, that for all $n\in\mathbb{N}$ and $i> n$ we have
\begin{equation}
\label{sufficiently close}
|\mu_{i}(\mathrm{succ}_\mathcal{T}(s_n)) - \nu(s_n)| < \frac{1}{2^n}.
\end{equation}
Let us point out that for $m\neq n$ the sets $\mathrm{succ}_\mathcal{T}(s_m)$ and $\mathrm{succ}_\mathcal{T}(s_n)$ are disjoint and $\cup_n\mathrm{succ}_\mathcal{T}(s_n) = \mathcal{T}\setminus\{t_0\}$, where $t_0$ denotes the root of the tree $\mathcal{T}$. We may, and will, assume that for all $i\in\mathbb{N}$, $t_0\not\in\mathrm{supp}(\mu_i)$. Define for each $i\in \mathbb{N}$ the sets
\begin{equation*}
A_i = \mathrm{supp}(\mu_i)\cap\Big(\cup_{n=1}^i\mathrm{succ}_\mathcal{T}(s_n)\Big)\text{ and } B_i = \mathrm{supp}(\mu_i)\cap\Big(\cup_{n=i+1}^\infty\mathrm{succ}_\mathcal{T}(s_n)\Big).
\end{equation*}
We point out that for all $i\in\mathbb{N}$, $A_i$, $B_i$ forms a partition of $\mathrm{supp}(\mu_i)$ and we will show that it has the desired properties.
Statement (ii) follows directly from the fact that for every $t\in\mathcal{T}$ the sequence of sets $(B_i\cap\mathrm{succ}_\mathcal{T}(t))_i$ is eventually empty. To show that (i) holds we fix $t\in\mathcal{T}$ and let $(t_j)$ be an enumeration of $\mathrm{succ}_\mathcal{T}(t)$. Define $L_j = \cup_{k=j}^\infty\{n\in\mathbb{N}:t_k\leq s_n\}$, for each $j\in\mathbb{N}$, and observe that $\cap_j L_j = \emptyset$. Also observe that for all $j\in\mathbb{N}$ we have $\cup_{k\geq j}(V_{t_k}\setminus\{t_k\}) = \cup_{n\in L_j}\mathrm{succ}_\mathcal{T}(s_n)$. Therefore we have
\begin{equation*}
\begin{split}
\mu_i&\Big(A_i \cap\big(\cup_{k\geq j}(V_{t_k}\setminus\{t_k\})\big) \Big) = \mu_i\Big(\big(\cup_{n=1}^i\mathrm{succ}_\mathcal{T}(s_n)\big)\cap\big(\cup_{n\in L_j}\mathrm{succ}_\mathcal{T}(s_n)\big)\Big)\\
&= \mu_i\Big(\cup_{n\in L_j\cap[1,i]}\mathrm{succ}_\mathcal{T}(s_n)\Big) = \sum_{n\in L_j\cap[1,i]} \mu_i\big(\mathrm{succ}_\mathcal{T}(s_n)\big)\\
&\stackrel{\eqref{sufficiently close}}{\leq} \sum_{n\in L_j\cap[1,i]}\nu(s_n) + \sum_{n\in L_j\cap[1,i]}\frac{1}{2^n} \leq \nu(\{s_n :n\in L_j\}) + 2^{-\min(L_j)+1}\\
&= \nu(\cup_{k\geq j}V_{t_k}) + 2^{-\min(L_j)+1}.
\end{split}
\end{equation*}
Therefore, $\lim_j\sup_i \mu_i\Big(A_i \cap\big(\cup_{k\geq j}(V_{t_k}\setminus\{t_k\})\big) \Big) = 0$ and by Lemma \ref{formula w-succ}, (i) is satisfied.
\end{proof}
\begin{proof}[Proof of Lemma \ref{combinatorial lemma essentially incomparable}]
Apply Lemma \ref{splitting lemma} so that, by passing to a subsequence of $(\mu_i)$, there are, for each $i\in\mathbb{N}$, partitions $A_i$, $B_i$ of $\mathrm{supp}(\mu_i)$ so that the conclusion of that Lemma it satisfied. Define, for each $i\in\mathbb{N}$, the measures $\mu_i^1$, $\mu_i^2$ given by $\mu_i^1(C) = \mu_i(A_i\cap C)$ and $\mu_i^2(C) = \mu_i(B_i\cap C)$. Let $\nu = w^*\-\lim_i\mu^1_i = \mathrm{succ}\-\!\lim_i\mu^1_i$. Pick a finite subset $F$ of $\tT_\xi$ so that $\nu(\tT_\xi\setminus F) <\varepsilon/2$. Then, because $\nu = w^*\-\lim_i\mu^1_i$ we have $\lim_i\mu_i^1(\tT_\xi) = \nu(\tT_\xi)$ and because $\nu = \mathrm{succ}\-\!\lim_i\mu^1_i$
\begin{equation*}
\lim_i\Big|\mu^1_i(\tT_\xi) - \mu^1_i(\cup_{\tilde t\in F}\mathrm{succ}(\tilde t))\Big| = \Big|\nu(\tT_\xi) - \lim_i\sum_{\tilde t\in F}\mu_i^1(\mathrm{succ}(\tilde t))\Big| = \nu(\tT_\xi\setminus F)<\frac{\varepsilon}{2}.
\end{equation*}
We can find $i_0\in\mathbb{N}$ so that for all $i\geq i_0$ we have
\begin{equation}
\label{successor is close enough}
\Big|\mu_i(A_i) - \mu_i\Big(A_i\cap\big(\cup_{\tilde t\in F}\mathrm{succ}(\tilde t)\big)\Big)\Big| = \Big|\mu^1_i(\tT_\xi) - \mu^1_i(\cup_{\tilde t\in F}\mathrm{succ}(\tilde t))\Big| <\frac \varepsilon 2.
\end{equation}
We may, using the fact that the sets $\cup\{\supp g_t:\tt\in\supp\mu_i\}$ for $i\in\N$ are disjoint, find $j_0\geq i_0\in\N$ such that
\begin{equation}
\label{avoid all supports}
\cup_{\tilde s\in F}\supp g_{s} < \supp g_t\text{ for every }\tilde t\in\cup_{i\geq j_0}\mathrm{supp}(\mu_i^1).
\end{equation}
We define $G_i^1 = A_i\cap(\cup_{\tt\in F}\mathrm{succ}(\tt))$, $i\geq j_0$. By \eqref{successor is close enough} we have that for all $i\geq j_0$, $|\mu_i(A_i) - \mu_i(G_i^1)|<\varepsilon/2$. Additionally, $\{t\in\T_\xi:\tt\in\cup_{i\geq j_0}G_1^i\}$ is essentially incomparable. Indeed, let $\ts_1,\ts_2\in \cup_{i\geq j_0}G^1_i$ with $m_{j_{s_1}}<_{\W} m_{j_{s_2}}$ and $(h,m_{j_{s_1}})\in\T_\xi$ be such that and $(h,m_{j_{s_1}})\le_{\T_\xi} s_2$. Then \eqref{avoid all supports} implies that $\supp h<\supp g_{s_1}$.
For the remaining part of the proof, since for all $i\in\N$ the set $B_i = \supp\mu_i^2$ is finite (as a subset of the finite support of $\mu_i$) and for each $\tt\in\tT_\xi$ the sequence $(\mu_i^2(\mathrm{succ}(\tt)))_i$ is eventually zero, we may pass to a subsequence so that for all $i<j$ we have $\{m_{j_t}:\tt\in \supp\mu_i^2\}\cap \{m_{j_t}:\tt\in\supp\mu_j^2\} = \emptyset$. We can therefore define the bounded sequence of disointly supported measures $(\nu_i)$ on $\widetilde{\mathcal{W}}_\xi$ with $\nu_i(\{(w_1,\ldots,w_k)\}) = \mu_i^2(\{\tilde t\in\tT_\xi: m_{j_t} = w_k\})$. Hence, applying Proposition \ref{proposition incomparable} and passing to a subsequence, we obtain a subset $E_i$ of $\supp\nu_i$ such that $\nu_i(\tW_\xi\setminus E_i)<\varepsilon/2$ and the sets $E_i$, $i\in\N$, are pairwise incomparable. It is easy to verify that $G^2_i=\{\tt\in B_i:m_{j_t}\in 2_i \}$, $i\in\N$, are pairwise incomparable and $|\mu_i(B_i) - \mu_i(G_i^2)| = \mu^2_i(\tT_\xi\setminus G^2_i)<\varepsilon/2$ for every $i\in\N$.
\end{proof}
We now define the space $T^\xi_{ess\-inc}$ in a similar way to $T^\xi_{inc}$, that is, using the notion of the Tsirelson extension $W_G$ of a ground set $G$.
\begin{dfn}
Define the following norming sets on $c_{00}(\N)$.
\[
G_0=\big\{\pm e^*_n:n\in\N \big\}
\]
\[
G_1=\Big\{\frac{1}{m_j}\sum_{n\in \N}g(n)e^*_n:j\in\N\;\text{ and }g:\N\to\{-1,0,1\}\text{ with }\supp g\in \S_{n_j} \Big\}.
\]
For each $f=m_j^{-1}\sum_{n\in \N}g(n)e^*_n$ in $G_1$, set $t_f=(g,m_j)$. Moreover, if $G=G_1\cup G_0$ and $f$ is in $W_G$ with a tree analysis $(f_\alpha)_{\alpha\in \mathcal{A}}$, define
\[
\M^1_f=\{\alpha:\alpha\text{ is a maximal node of }\mathcal{A}\text{ and }f_a\in G_1 \}.
\]
Let $W$ be the subset of $W_G$ containing all functionals $f$ such that $\{t_{f_\alpha}:\alpha\in\M^1_f \}$ is an essentially incomparable subset of $\T_\xi$. Denote by $T^\xi_{ess\-inc}$ the completion of $c_{00}(\N)$ with respect to the norm $\|\cdot\|_W$ induced by $W$.
\end{dfn}
\begin{rem}
\begin{itemize}
\item[(i)] The standard basis $(e_j)_j$ of $c_{00}(\N)$ forms a $1$-unconditional basis for the space $T^\xi_{ess\-inc}$ and it is also boundedly complete since $T^\xi_{ess\-inc}$ admits a uniformly unique $\ell_1$ spreading model as shown in Proposition \ref{first space unique}.
\item[(ii)] If $f\in W_G$ with a tree analysis $(f_\alpha)_{\alpha\in \mathcal{A}}$ and $m_{j_{t_{f_\alpha}}}$, for $\alpha\in\M^1_f$, are pairwise incomparable nodes in $\W_\xi$ , then $f\in W$.
\item[(iii)] The norming set of $T^\xi_{ess\- inc}$ contains the norming set of Tsirelson's original space, i.e., the Tsirelson extension of $G_0$.
\end{itemize}
\end{rem}
\begin{prop}\label{first space unique}
The space $T^\xi_{ess\-inc}$ admits $\ell_1$ a uniformly unique joint spreading model with respect to $\mathscr{F}_b(T^\xi_{ess\-inc})$.
\end{prop}
\begin{proof}
Let $(x^i_j)_j$, $1\le i \le l$, be an array of normalized block sequences in $T^\xi_{ess\-inc}$ and fix $\varepsilon>0$. Passing to a subsequence, we may assume that $\supp x^{i_1}_j<\supp x^{i_2}_{j+1}$ for all $i_1, i_2=1,\ldots,l$ and $j\in\N$. For each $i=1,\ldots,l$ and $j\in\N$, pick a functional $f^i_j=\sum_{\alpha\in \M^i_j}f^i_{j,\alpha}/2^{k^i_{j,\alpha}}$ in $W$ with $f^i_j(x^i_j)\ge 1-\varepsilon$ and $f^i_{j,\alpha}(x^i_j)>0$ for every $\alpha\in\M^i_j$, where $\mathcal{M}^i_j$ denotes the set of all maximal nodes of a fixed tree analysis of $f^i_j$. Moreover, for each $\alpha\in \M^1_{f^i_j}=\{\alpha\in\M^i_j:f^i_{j,\alpha}\in G_1\}$, define $t^i_{j,\alpha}=t_{f^i_{j,\alpha}}$ and, for each $j\in\N$, the measure $\mu_j$ as follows:
\[
\mu_j = \sum_i\sum_{\alpha\in \M^1_{f^i_j}}\frac{f^i_{j,\alpha}(x^i_j)}{2^{k^i_{j,\alpha}}}\delta_{\tt^i_{j,\alpha}}.
\]
Passing to a subsequence assume that $\lim_j\mu_j(\tT_\xi)=c$. If $c=0$, then we may assume that $f^i_{j,\alpha}\in G_0$ for every $i=1,\ldots,l$, $j\in\N$ and $\alpha\in\M^i_j$, in which case the desired result is immediate. Hence, if $c>0$, applying Lemma \ref{combinatorial lemma essentially incomparable} and passing to a subsequence, we obtain $(G^1_j)_j$, $(G^2_j)_j$ satisfying items (i) - (iv) with $\mu_j(\tT_\xi\setminus G^1_j\cup G^2_j)<1/8$. Then, for each pair $(i,j)$, set
\[
\mathcal{M}^1_{i,j}=\{\alpha\in\M^i_{f^i_j}:t^i_{j,\alpha}\in G^1_j \}\quad\text{ and }\quad\mathcal{M}^2_{i,j}=\mathcal{M}_{i}^j\setminus \mathcal{M}^1_{i,j}
\]
and
\[
f^k_{i,j}=\sum_{\alpha\in \mathcal{M}^k_{i,j}}f^i_{j,\alpha}/2^{k^i_{j,\alpha}},\quad k=1,2.
\]
Note in particular that, for every pair $(i,j)$, the fact that $\mu_j(\tT_\xi\setminus G^1_j\cup G^2_j)<1/8$ implies that $|f^i_j(x^i_j)-(f^1_{i,j}(x^i_j)+f^2_{i,j}(x^i_j))|<1/8$ and hence that there exists $k=1,2$ such that $f^k_{i,j}(x^i_j)\ge (7-\varepsilon)/16$. Set
\[
A_k=\{(i,j) :f^{k}_{i,j}(x^i_j)\ge (7-8\varepsilon)/16 \},\quad k=1,2.
\]
Let $n\in\N$, $\{\lambda_{ij}\}_{i=1,j=1}^{l,n}\subset[-1,1]$ with $\sum_{i,j}|\lambda_{ij}|=1$ and $s=(s_i)_{i=1}^l\in S\- Plm_l([\N]^k)$ with $ln\le\min\supp x^1_{s_1(1)}$. Then let $k=1,2$ be such that $\sum_{(i,s_i(j))\in A_k}|\lambda_{ij}|\ge 1/2$ and observe that $f=1/2\sum_{(i,s_i(j))\in A_k}f^k_{i,s_i(j)}$ is in $W$. Hence, we calculate
\[
\bigg\|\sum_{i=1}^l\sum_{j=1}^n|\lambda_{ij}|x^i_{s_i(j)} \bigg\|\ge f\bigg(\sum_{i=1}^l\sum_{j=1}^n|\lambda_{ij}|x^i_{s_i(j)}\bigg)=\frac{1}{2}\sum_{(i,s_i(j))\in A_k}|\lambda_{ij}|f^k_{i,s_i(j)}\big(x^i_{s_i(j)}\big)\ge\frac{7-8\varepsilon}{32}
\]
and due to unconditionality this yields that
\[
\bigg\|\sum_{i=1}^l\sum_{j=1}^n\lambda_{ij}x^i_{s_i(j)}\bigg\|\ge\frac{7-8\varepsilon}{32}.
\]
\end{proof}
It remains to show that for every $M\in[\N]^\infty$, the space $T^\xi_{ess\-inc}$ contains a $c_0$-tree of height $\omega^\xi$ supported by $(e_j)_{j\in M}$. To this end, let us recall the following definition.
\begin{dfn}
Let $n\in\N$ and $\varepsilon>0$. We say that a convex combination $x=\sum_{i\in \Delta}\lambda_ie_i$ in $c_{00}(\N)$ is an $(n,\varepsilon)$-special convex combination if
\begin{itemize}
\item[(i)] $\Delta\in \S_n$ and
\item[(ii)] $\sum_{i\in \Delta'}\lambda_i<\varepsilon$ for every $\Delta'\in \S_m$ with $m<n$.
\end{itemize}
\end{dfn}
The main ingredient in the proof of the following proposition is the notion of repeated averages, first defined by Argyros, Mercourakis, and Tsarpalias. in \cite{AMT}. We refer the reader to \cite[Chapter 2]{AT} for further details.
\begin{prop}
For every $n\in\N$ and $\varepsilon>0$, there is a $k\in\N$ such that, for every maximal subset $F$ of $\S_n$ with $k<F$, there exists an $(n,\varepsilon)$-special convex combination $x$ in $c_{00}(\N)$ with $\supp x=F$.
\end{prop}
For a functional $f$ in $W$ with tree analysis $(f_\alpha)_{\alpha\in \mathcal{A}}$, we define the height of $f$, denoted by $h(f)$, as the maximum of $|a|$ over all maximal nodes $\alpha\in\mathcal{A}$. Moreover, if $f=m^{-1}_j\sum_{n\in\N}g(n)e^*_n$ is in $W$, we say that $f$ is a weighted functional and define the weight of $f$ as $w(f)=m_j$.
\begin{lem}\label{admissibility height of tree}
Let $j\in \N$ and $f$ be a functional in $W$ with a tree analysis $(f_\alpha)_{\alpha\in \mathcal{A}}$ such that $w(f_\alpha)<m_j$ for every $\alpha\in\mathcal{M}^1_f$. Then $\supp f\in \S_{k}$, where $k\le n_{j-1}+h(f)$.
\end{lem}
\begin{proof}
For each $\alpha\in\mathcal{A}$, let $k_\alpha\in\N$ be such that $\supp f_\alpha\in\S_{k_\alpha}$. Then, since $w(f_\alpha)<m_j$, we have that $k_\alpha\le n_{j-1}$ for every $\alpha\in\M^1_f$. Note then that, as follows from the definition of $W_G$, this implies that $k_\alpha\le n_{j-1}+1$ for every $\alpha\in\mathcal{A}$ with $|\alpha|=h(f)-1$. In particular, a finite induction argument yields that $k_\alpha\le n_{j-1}+i$ whenever $|\alpha|=h(f)-i$ and this proves the desired result.
\end{proof}
\begin{prop}\label{special convex combination seminormalized}
Let $j\in\N$ and $x=\sum_{i\in\Delta}\lambda_ie_i$ be an $(n_j,m^{-2}_j)$-special convex combination, then
\[
\frac{1}{m_j}\le\|x\|_W\le\frac{1}{m_j}+\frac{1}{m^2_j}.
\]
\end{prop}
\begin{proof}
Pick an $f$ in $W$ and define $\Delta_1=\{i\in\Delta:|f(e_i)|>m_j^{-1} \}$ and $\Delta_2 =\Delta\setminus \Delta_1$. Consider the tree analysis $(f^1_\alpha)_{\alpha\in\mathcal{A}}$ of $f_1=f|_{\Delta_1}$ and note that $w(f^1_\alpha)<m_j$ for every $\alpha\in\mathcal{M}^1_{f_{1}}$. Indeed, if $w(f^1_\alpha)=m_{j'}\ge m_j$ for some $\alpha$, then for any $i\in \supp f^1_\alpha$ we have that $|f(e_i)|\le m^{-1}_{j'}$ and this is a contradiction. Moreover, the fact that $|f_1(e_i)|>m^{-1}_j$ for every $i\in \Delta_1=\supp f_1$ implies that $h(f_1)<\log_2m_{j-1}$ and hence the previous proposition yields that $\supp f_1\in\S_l$, where $l\le\log_2m_j+n_{j-1}<n_j$. Therefore, since $x=\sum_{i\in\Delta}\lambda_ie_i$ is an $(n_j,m^{-2}_j)$\-special convex combination, we have that
\[
|f|_{\Delta_1}(\sum_{i\in\Delta}\lambda_ie_i)|\le\sum_{i\in \Delta_1}\lambda_i<\frac{1}{m^{2}_j}.
\]
We also calculate
\[
|f|_{\Delta_2}(\sum_{i\in\Delta}\lambda_ie_i)|\le \frac{1}{m_j}\sum_{i\in \Delta_2}\lambda_i\le \frac{1}{m_j}
\]
and conclude that $\|x\|_W\le m^{-1}_j+m^{-2}_j$. For the remaining part notice that the functional $f=m_j^{-1}\sum_{i\in \Delta}e^*_i$ is in $W$.
\end{proof}
\begin{prop}\label{special convex combination upper bound}
Let $j\in\N$ and $x=\sum_{i\in\Delta}\lambda_ie_i$ be an $(n_j,m^{-2}_j)$-special convex combination. Then $|f(x)|<2m_j^{-2}$, for every $f\in W$ with a tree analysis $(f_\alpha)_{\alpha\in\mathcal{A}}$ such that $w(f_\alpha)\neq m_j$ for all $\alpha\in \M^1_f$.
\end{prop}
\begin{proof}
Define $\Delta_1=\{i\in\Delta:|f(e_i)|>m^{-2}_{j} \}$ and $\Delta_2=\Delta\setminus \Delta_1$ and let $(f^1_\alpha)_{\alpha\in\mathcal{A}_1}$ be the tree analysis of $f_1=f|_{\Delta_1}$. Similar arguments as in the previous proof yield that $w(f^1_\alpha)<m^2_j<m_{j+1}$ and hence $w(f^1_\alpha)<m_j$ for all $\alpha\in\M^1_{f_1}$, since $w(f_\alpha)\neq m_j$ for all $\alpha\in M^1_f$. Moreover, since $|f_1(e_i)|>m^2_j$ for all $i\in \supp f_1$, we have that $h(f_1)<\log_2m_j^{2}$ and therefore Proposition \ref{admissibility height of tree} yields that $\Delta_1=\supp f_1\in\S_l$ with $l\le \log_2m_j^{2}+n_{j-1}<n_j$. The fact that $x=\sum_{i\in\Delta}\lambda_ie_i$ is an $(n_j,m^{-2}_j)$\-special convex combination implies that
\[
|f_1(\sum_{i\in\Delta}\lambda_ie_i)|\le\sum_{i\in \Delta_1}\lambda_i<\frac{1}{m_j^2}.
\]
We also calculate
\[
|f|_{\Delta_2}(\sum_{i\in\Delta}\lambda_ie_i)|\le\frac{1}{m^2_j}\sum_{i\in\Delta_2}\lambda_i\le\frac{1}{m^2_j}
\]
and this completes the proof.
\end{proof}
Let $M$ be an infinite subset of the naturals and consider the collection $T_\xi(M)$ of all finite sequences $(x_1,\ldots,x_k)$ of vectors in $c_{00}(\N)$ such that
\begin{itemize}
\item[(i)] $x_l=m_{j_l}\sum_{i\in \Delta_l}\lambda_ie_i$, where $\sum_{i\in \Delta_l}\lambda_ie_i$ is an $(n_{j_l},m^{-2}_{j_l})$\-special convex combination for every $l=1,\ldots,k$,
\item[(ii)] $\Delta_l$ is a subset of $M$ for every $l=1,\ldots,k$ and
\item[(iii)] $((\rchi_{\Delta_1},m_{j_1}),\ldots,(\rchi_{\Delta_k},m_{j_k}))\in\tT_\xi$.
\end{itemize}
Note that $T_\xi(M)$, equipped with the initial segment order, is a well-founded infinite branching tree of height $\omega^\xi$.
\begin{prop}
Let $M$ be an infinite subset of the naturals and $(x_1,\ldots,x_k)$ be any node of $T_\xi(M)$, then $\|x_1+\ldots+x_k\|_W\le 3$.
\end{prop}
\begin{proof}
Let $f\in W$ with a tree analysis $(f_\alpha)_{\alpha\in \mathcal{A}}$. Observe that there exists at most one $1\le l_0\le k$ such that there is an $\alpha\in \M^1_f$ with $w(f_\alpha)=m_{j_{l_0}}$ and $\supp f_a\cap \Delta_{l_0}$ is non-empty. Indeed, suppose that there exist $1\le l_1<l_2\le k$ and $\alpha_1,\alpha_2\in \M^1_f$ with $w(f_{\alpha_1})=m_{j_{l_1}}$, $w(f_{\alpha_2})=m_{j_{l_2}}$, $\supp f_{\alpha_1}\cap \Delta_{l_1}\neq\emptyset$ and $\supp f_{\alpha_2}\cap \Delta_{l_2}\neq\emptyset$. Then since $\{t_{f_{\alpha}}:\alpha\in \M^1_f\}$ is essentially incomparable and $m_{j_{l_1}}<_{\W_\xi} m_{j_{l_2}}$ we have that $\Delta_{l_1}<\Delta_{t_{f_{\alpha_1}}}=\supp f_{\alpha_1}$ which is a contradiction.
Therefore, for any $l\neq l_0$, we have that $w(f_\alpha)\neq m_{j_l}$ for every $\alpha\in\M^1_f$ and the previous proposition yields that $|f(x_l)|<2m^{-1}_{j_l}$. Moreover, Proposition \ref{special convex combination seminormalized} yields that $|f(x_{l_0})|\le 1+m^{-1}_{j_{l_0}}$ and hence we conclude that
\[
|f(x_1+\cdots+x_k)|\le 1+2\sum_{l=1}^k\frac{1}{m_{j_l}}\le 3.
\]
\end{proof}
The previous proposition and the fact that the tree $T_\xi(M)$ is of height $\omega^\xi$ yield the following result.
\begin{thm}
For every $M\in[\N]^\infty$, the space $T^\xi_{ess\-inc}$ contains a $c_0$-tree of height $\omega^\xi$, supported by $(e_j)_{j\in M}$. In particular, the space generated by $(e_j)_{j\in M}$ is not Asymptotic $\ell_1$.
\end{thm}
\begin{rem}
There exist modifications of the ground set $G$ that yield, for any $1<p<\infty$, a space, as in the previous theorem, that contains $\ell_p$-trees instead of $c_0$-trees.
\end{rem}
\begin{thm}
The space $T^\xi_{ess\-inc}$ is reflexive.
\end{thm}
\begin{proof}
Note that since $\T_\xi$ is a countable compact space with respect to the pointwise convergence topology, the completion of $c_{00}(\N)$ with respect to $\|\cdot\|_G$ is embedded in $C[\T_\xi]$, i.e., the space of all continuous real functions on $\T_\xi$, and hence is $c_0$-saturated. Furthermore, $T^\xi_{ess\-inc}$ admits a boundedly complete basis and therefore does not contain $c_0$. The above imply that the identity operator $Id: (c_{00}(\N),\|\cdot\|_W)\to (c_{00}(\N),\|\cdot\|_G)$ is strictly singular and hence for any normalized block sequence $(x_j)_j$ in $T^\xi_{ess\-inc}$ there exists a subsequence $(x_j)_{j\in M}$ such that $\lim_{j\in M}\|x_j\|_G=0$. The remainder of the proof is identical to the last paragraph of Proposition \ref{section 1 reflexivity}.
\end{proof}
\section{More non-asymptotic $\ell_p$ spaces with uniformly unique $\ell_p$ joint spreading models}
In this final section we show that, for every $1<p<\infty$, there is a reflexive Banach space that admits a uniformly unique $\ell_p$ asymptotic model whereas it is not Asymptotic $\ell_p$. This was also observed in \cite[Section 7.2]{BLMS} for a slightly different type of spaces. We show this for a class of spaces very similar to those defined in \cite[Example 4.3]{OS3}.
\begin{dfn}
Let $1<p<\infty$ and denote its conjugate by $q$, i.e., $p^{-1}+q^{-1}=1$. Fix a countable ordinal $\xi$ and define the following norming sets on $c_{00}(\N)$.
\[
G_1^{\xi}=\Big\{\sum_{i\in S}\epsilon_ie^*_i:S\text{ is a segment of }\T_\xi\text{ and }\epsilon_i=\pm1 \Big\}
\]
\begin{align*}
G_{1,p}^{\xi}=\Big\{\sum_{i=1}^mb_if_{i}:m\in\N,\; \sum_{i=1}^m|b_i|^q\le1,&\; f_i\in G^{\xi}_1\text{ for }i=1,\ldots,m\text{ and} \\\supp f_1,&\ldots,\supp f_m\text{ are pairwise disjoint} \Big\}.
\end{align*}
Denote by $JT^\xi_{1,p}$ the completion of $c_{00}(\N)$ with respect to the norm induced by the norming set ${G^{\xi}_{1,p}}$.
\end{dfn}
We start with some necessary remarks on the above norming sets and a Ramsey type result.
\begin{rem}\label{remark comb lem pointwise limit}
Let $(f_j)_j$ be a sequence in $G_{1}^{\xi}$ with $f_j=\sum_{i \in S_j}\epsilon^j_ie^*_i$, $j\in\N$, and for each $i,j\in\N$, define $\epsilon_j(i)=\epsilon^i_j$ if $i\in S_j$ and $\epsilon_j(i)=0$ otherwise. Passing to a subsequence, we may assume that $(S_j)_j$ converges pointwise to a segment $S$, since $\T_\xi$ is well-founded, and that $(\epsilon_j)_j$ also converges to some $\epsilon$ in $\{-1,1\}^\N$. Then, clearly, $(f_j)_j$ converges pointwise to $f=\sum_{i\in S}\epsilon(i)e^*_i$ and $f$ is in $G^\xi_1$.
\end{rem}
\begin{rem}\label{remark inequality norm <= 1}
Let $x$ be a normalized vector in $JT^\xi_{1,p}$ with finite support.
\begin{itemize}
\item[(i)] If for some $\varepsilon>0$ there is a family $\{f_i\}_{i\in I}$ in $G_{1}^{\xi}$ whose members have pairwise disjoint supports and $|f_{i}(x)|\ge\varepsilon$ for all $i\in I$, then $\#I\le\varepsilon^{-p}$.
\item[(ii)] Let $f_1,\ldots,f_m\in G_1^{\xi}$ have pairwise disjoint supports and $\supp f_i \subset\range (x)$ for $i=1,\ldots,m$. Then, for any choice of scalars $b_1,\ldots,b_m$, we have that
\[
\Big|\sum_{i=1}^mb_if_{i}(x)\Big|^q\le\sum_{i=1}^m|b_i|^q.
\]
\end{itemize}
\end{rem}
\begin{dfn}
We call a family $(F_j)_j$ of finite subsets of $JT^\xi_{1,p}$ a normalized block family if for any choice of $x_j\in F_j$, $j\in\N$, the sequence $(x_j)_j$ is block and $\|x\|=1$ for any $x\in F_j$ and $j\in\N$. Moreover, for such a family, define $M(F_j)=\max\{\supp x:x\in F_j\}$ and $r(F_j)=\#(M(F_{j-1}),M(F_j)]$, where $M(F_0)=0$.
\end{dfn}
\begin{lem}\label{comb lem ell2 subsequence}
Let $(F_j)_j$ be a normalized block family in $JT^\xi_{1,p}$ with $\sup_j\# F_j<\infty$. Then, for every $\varepsilon>0$ and $n_0\in \N$, there is an $L\in[\N]^\infty$ such that, for any segment $S$ of $\T_\xi$ with $\min S\le n_0$ and any $f\in G_1^{\xi}$ with $\supp f=S$, there is at most one $j\in L$ with the property that $|f(x)|\ge\varepsilon$ for some $x\in F_j$.
\end{lem}
\begin{proof}
For a segment $S$ of $\T_\xi$, let $G_S$ denote the set of all $f\in G_1^{\xi}$ with $\supp f=S$. If the conclusion is false for some $\varepsilon>0$ and $n_0\in\N$, then using Ramsey Theorem from \cite{Ra}, there exists an $L\in[\N]^\infty$ such that, for any $i<j$ in $L$, there is a segment $S_{ij}$ with $\min S_{ij}\le n_0$, a functional $f_{ij}\in G_{S_{ij}}$ and $x_{ij}\in F_i$, $y_{ij}\in F_j$ such that $|f_{{ij}}(x_{ij})|\ge\varepsilon$ and $|f_{{ij}}(y_{ij})|\ge\varepsilon$. Assume for convenience that $L=\N$. Since $\#\sup_j F_j<\infty$, using the pigeon hole principle and a diagonal argument we may assume that there exist sequences $(x_j)_j$, $(y_j)_j$ such that $x_j,y_j\in F_j$ and, for every $i<j\in\N$, a segment $S_{ij}$ of $\T_\xi$ with $\min S_{ij}\le n_0$ and an $f_{ij}\in G_{S_{ij}}$ such that $|f_{{ij}}(x_i)|\ge\varepsilon$ and $|f_{{ij}}(y_j)|\ge\varepsilon$.
For each $i<j<k$ in $\N$, define $S_{ijk}= S_{ik}\cap S_{jk}\cap \range(y_k)$. Once more, using Ramsey theorem and passing to a further infinite subset, we may assume that $S_{ijk}$ is either always empty or always non-empty for every $i<j<k$ in $\N$. Item (i) of Remark \ref{remark inequality norm <= 1} and the fact that $\|y_k\|=1$ for all $k\in\N$ rules out the first case and hence $S_{ijk}\neq\emptyset$ for all $i<j<k$ in $\N$. This in particular implies that if we fix $i<j_1<k$ and $i<j_2<k$, then $S_{ij_1k}|_{[n_0,m(x_k))}=S_{ij_2k}|_{[n_0,m(x_k))}$. For any $j\in\N$, take an arbitrary $i$ with $1<i<j$ and set $S_{j}=S_{1ij}|_{[n_0,m(x_j))}$. Then we conclude that, for any $j\in \N$, there is an $f_j\in G_{S_j}$ such that $|f_j(x_i)|\ge\varepsilon$ for all $i<j$, where $\min S_j\le n_0$. This is a contradiction, since Remark \ref{remark comb lem pointwise limit} implies that there exists an $f\in G^{\xi}_1$ with the property that $|f(x_j)|\ge\varepsilon$ for all $j\in\N$, whereas $\supp f$ is finite since $\T_\xi$ is well-founded.
\end{proof}
\begin{lem}\label{no2 comb lem ell2 subsequence}
Let $\varepsilon>0$ and $(F_j)_j$ be a normalized block family in $JT^\xi_{1,p}$ with $\sup_j\# F_j<\infty$. Then there exists a strictly increasing sequence $(n_j)_j$ of naturals and a decreasing sequence $(\varepsilon_j)_j$ of positive reals such that
\begin{enumerate}
\item[(i)] for every $j\in \N$, every segment $S$ of $\T_\xi$ with $\min S\le M(F_{n_j})$ and $f\in G^{\xi}_1$ with $\supp f=S$, there exists at most one $j'>j$ such that $|f(x)|\ge\varepsilon_j$ for some $x\in F_{n_{j'}}$ and
\item[(ii)] $\sum_{j=1}^\infty r(F_{n_j})\sum_{i=j}^\infty(i+1)\varepsilon_i<\varepsilon$.
\end{enumerate}
\end{lem}
\begin{proof}
Let $(\delta_j)_j$ be a sequence of positive reals such that $\sum_{j=1}^\infty\delta_j<\varepsilon$. We will construct $(n_j)_j$ and $(\varepsilon_j)_j$ by induction, along with a decreasing sequence $(L_j)_j$ of infinite subsets of $\N$. Set $n_1=1$ and $L_1=\N$ and choose $\varepsilon_1>0$ such that $2r(F_1)\varepsilon_1<\delta_1$. Suppose that $n_1,\ldots,n_j$, $\varepsilon_1,\ldots,\varepsilon_j$ and $L_1,\ldots,L_j$ have been chosen for some $j$ in $\N$. Then, the previous lemma yields an $L_{j+1}\in[L_{j}]^\infty$ such that, for every segment $S$ of $\T_\xi$ with $\min S\le M(F_{n_j})$ and every $f\in G_1^{\xi}$ with $\supp f=S$, there is at most one $j'>j$ such that $|f(x)|\ge\varepsilon_{j}$ for some $x\in F_{n_{j'}}$. Choose $n_{j+1}\in L_{j+1}$ with $n_{j+1}>n_j$ and $\varepsilon_{j+1}<\varepsilon_j$ such that
\begin{enumerate}
\item[(a)] $r(F_{n_{j+1}})(j+2)\varepsilon_{j+1}<\delta_{j+1}$ and
\item[(b)] $r(F_{n_{k}})\sum_{i=k}^{j+1}(i+1)\varepsilon_i<\delta_k$ for all $k\le j$.
\end{enumerate}
It follows quite easily that $(n_j)_j$ and $(\varepsilon_j)_j$ are as desired.
\end{proof}
\begin{prop}\label{upper ell2 for ground set}
Let $\varepsilon>0$ and $(F_j)_j$ be a normalized block family in $JT^\xi_{1,p}$ with $\sup_j\# F_j<\infty$ satisfying the following.
\begin{enumerate}
\item[(i)] For every $j\in \N$, every segment $S$ of $\T_\xi$ with $\min S\le M(F_{n})$ and $f\in G^{\xi}_{1}$ with $\supp f =S$, there exists at most one $j'>j$ such that $|f(x)|\ge\varepsilon_j$ for some $x\in F_{j'}$ and
\item[(ii)] $\sum_{j=1}^\infty r(F_j)\sum_{i=j}^\infty(i+1)\varepsilon_i<\varepsilon$.
\end{enumerate}
Then, for every $n\in\N$, every choice of $x_1,\ldots,x_n$ with $x_j\in F_{j}$ and scalars $a_1,\ldots,a_n$, we have that
\[
\Big(\sum_{j=1}^n|a_j|^p\Big)^{\frac{1}{p}}\le\Big\|\sum_{j=1}^na_jx_j\Big\|\le (2^{\frac{1}{q}}+\varepsilon)\Big(\sum_{j=1}^n|a_j|^p\Big)^{\frac{1}{p}}.
\]
\end{prop}
\begin{proof} The lower inequality follows easily from the definition of $G_{1,p}^{\xi}$. Let us first observe that if $(x_j)_j$ is a sequence with each $x_j\in F_j$, then, for any $j\in\N$ and any segment $S$ of $\T_\xi$ with $M(x_{j-1})<\min S\le M(x_j)$ and $f\in G_1^{\xi}$ with $\supp f = S$, the following hold due to (i).
\begin{enumerate}
\item[(a)] $\#\{i>j:|f(x_i)|\ge\varepsilon_j\}\le 1$.
\item[(b)] $\#\{i>j:\varepsilon_{k-1}>|f(x_i)|\ge\varepsilon_k\}\le k$ for all $k>j$.
\end{enumerate}
Let $f=\sum_{i=1}^mb_if_i$ be in $G_{1,p}^{\xi}$ with $\supp f_i ={S_i}$, for $i=1,\ldots,m$. For each $i$, we will denote by $j_{i,1}$ the unique $1\le j\le n$ such that $M(x_{j_{i,1}-1})<\min S_i\le M(x_{j_{i,1}})$ and by $j_{i,2}$ the unique, if there exists, $j_{i,1}<j\le n$ such that $|f_{i}(x_{j_{i,2}})|\ge\varepsilon_{j_{i,1}}$. Denote by $f_{i,1}$ the restriction of $f_i$ to ${\range(x_{j_{i,1}})}\cap\range(x_{j_{i,2}})$ and set $f_{i,2}=f_i-f_{i,1}$ for $i=1,\ldots,m$, and $I_j=\{i:j=j_{i,1}\text{ or }j=j_{i,2}\}$ for $j=1,\ldots,n$. Note that, due to (a), each $i$ appears in $I_j$ for at most two $j$ and hence $\sum_{j=1}^n\sum_{i\in I_j}|b_i|^q\le 2$.\linebreak We thus calculate applying item (ii) of Remark \ref{remark inequality norm <= 1}
\begin{align*}
\sum_{i=1}^mb_if_{i,1}\Big(\sum_{j=1}^na_jx_j\Big)&=\sum_{j=1}^na_j\sum_{i\in I_j}b_if_{i,1}(x_j)\le\Big(\sum_{j=1}^n|a_j|^p \Big)^{\frac{1}{p}}\Big(\sum_{j=1}^n\big|\sum_{i\in I_j}b_if_{i,1}(x_j)\big|^q \Big)^{\frac{1}{q}}\\&\le\Big(\sum_{j=1}^n|a_j|^p \Big)^{\frac{1}{p}}\Big(\sum_{j=1}^n\sum_{i\in I_j}|b_i|^q \Big)^{\frac{1}{q}}\le2^{\frac{1}{q}}\Big(\sum_{j=1}^n|a_j|^p \Big)^{\frac{1}{p}}.
\end{align*}
Finally, for each $j\in\N$, set $G_j=\{i:M(x_{j_{i,1}-1})<\min S_i\le M(x_{j_{i,1}})\}$. Note that, as follows from (b), $\#G_j\le r(F_j)$ and $|f_{i,2}(\sum_{k=1}^nx_k)|<\sum_{k=i}^\infty(k+1)\varepsilon_k$ for any $i\in G_j$. Hence (ii) yields that $\sum_{i=1}^m|f_{i,2}(\sum_{k=1}^nx_k)|<\varepsilon$ and we conclude that
\[
\Big|\sum_{i=1}^mb_if_{i,2}\Big(\sum_{j=1}^na_jx_j\Big)\Big|=\Big|\sum_{j=1}^na_j\sum_{i=1}^mb_if_{i,2}(x_j)\Big|<\varepsilon\Big(\sum_{j=1}^n|a_j|^p \Big)^{\frac{1}{p}}
\]
which along with the above calculation yield the desired result.
\end{proof}
\begin{prop}
The space $JT^\xi_{1,p}$ admits a uniformly unique joint spreading model with respect to $\mathscr{F}_b(JT^\xi_{1,p})$, equivalent to the unit vector basis of $\ell_p$.
\end{prop}
\begin{proof}
Let $(x^1_j)_j,\ldots,(x^l_j)_j$ be normalized block sequences in $JT^\xi_{1,p}$ and let $\varepsilon>0$.\linebreak Applying Lemma \ref{no2 comb lem ell2 subsequence} and passing to a subsequence, we may assume that $F_j=\{x^i_j:i=1,\ldots,l\}$ is a normalized block family in $JT^\xi_{1,p}$ satisfying items (i) and (ii) of Proposition \ref{upper ell2 for ground set}. Then, for every $k\in\N$, every $s=(s_i)_{i=1}^l$ in $S$-$Plm_l([L]^k)$ and any choice of scalars $(a_{ij})_{i=1,j=1}^{l,k}$, we calculate
\[
\Big(\sum_{i=1}^l\sum_{j=1}^k|a_{ij}|^p\Big)^{\frac{1}{p}}\le\Big\|\sum_{i=1}^l\sum_{j=1}^ka_{ij}x^i_{s_i(j)}\Big\|\le (2^{\frac{1}{q}}+\varepsilon)\Big(\sum_{i=1}^l\sum_{j=1}^k|a_{ij}|^p\Big)^{\frac{1}{p}}.
\]
A diagonal argument then yields that there exists $L\in[\N]^\infty$ such that $((x^i_j)_{j\in L})_{i=1}^l$ generates a joint spreading model $2^\frac{1}{q}$-equivalent to the unit vector basis of $\ell_p$.
\end{proof}
\begin{prop}
The space $JT^\xi_{1,p}$ is reflexive.
\end{prop}
\begin{proof}
Note that the unit vector basis of $c_{00}(\N)$ forms a boundedly complete unconditional Schauder basis for $JT^\xi_{1,p}$, that is, it does not contain $c_0$. Moreover, Proposition \ref{upper ell2 for ground set} yields that it does not contain $\ell_1$ and hence Theorem 2 from \cite{J1} yields the desired result.
\end{proof}
\begin{prop}
The space $JT^\xi_{1,p}$ is not Asymptotic $\ell_p$.
\end{prop}
\begin{proof}
Suppose that $JT^\xi_{1,p}$ is $C$-Asymptotic $\ell_p$ and let $n\in\N$ be such that $C\le n^\frac{1}{q}$. Then, following the same arguments as in Proposition \ref{NOT Asymptotic l1}, in the final outcome of $G(n,p,C)$ we, as player (V), have chosen elements of the basis $e_{j_1},\ldots,e_{j_n}$ such that $\{j_1,\ldots,j_n\}$ is a segment of $\T_\xi$ and hence $\{e_{j_1},\ldots,e_{j_n}\}$ is isometric to $\ell_1^n$. We then calculate
\[
\Big\| {n^{-\frac{1}{p}}}\sum_{i=1}^n e_{j_i}\Big\|=\Big\| {n^{-\frac{1}{p}}}\sum_{i=1}^n e_{j_i}\Big\|_{G_1^\xi}=n^\frac{1}{q}
\]
whereas, since $JT^\xi_{1,p}$ is $C$-Asymptotic $\ell_p$, we have that
\[
\Big\| {n^{-\frac{1}{p}}}\sum_{i=1}^n e_{j_i}\Big\|\le C
\]
and this is a contradiction.
\end{proof}
\begin{rem}
We may also define a conditional version of $JT^\xi_{1,p}$, denoted as $JT^\xi_{p}$, by replacing the norming set $G_1^\xi$ with
\[
G^\xi_{sum}=\Big\{\sum_{i \in S}e^*_i:S\text{ is a segment of } \T_\xi \Big\}.
\]
Note that the above results hold for $JT^\xi_{p}$. For the reflexivity part, notice that it suffices to show that $(e_j)_j$ is shrinking for $JT^\xi_{p}$. If not, then there is an $x^*\in (JT^\xi_{p})^{*}\setminus\overline{\text{span}}\{e^*_j \}_{j=1}^\infty$ and an $x^{**}\in (JT^\xi_{p})^{**}$ with $x^{**}(e^*_j)=0$ for all $j\in\N$ and $x^{**}(x^*)=1$. Then, from Odell-Rosenthal Theorem \cite{OR} and the fact that $x^{**}(e^*_j)=0$, $j\in\N$, we may find a seminormalized block sequence $(x_j)_j$ in $JT^\xi_{p}$ with $w^*$-$\lim_jx_j=x^{**}$ and, passing to a subsequence, we may assume that it also satisfies items (i) and (ii) of Proposition \ref{upper ell2 for ground set} for some $\varepsilon>0$. Since $x^{**}(x^*)=1$, there exists $n_0\in\N$ such that $x^*(x_n)\ge1/2$ for all $n\ge n_0$. Then, for $k\in\N$ such that $(2^\frac{1}{q}+\varepsilon)k^{-\frac{1}{q}}<1/2$, Proposition \ref{upper ell2 for ground set} yields that
\[
x^*\Big( \frac{x_{n_0+1}+\ldots+x_{n_0+k}}{k}\Big)\le(2^\frac{1}{q}+\varepsilon)k^{-\frac{1}{q}}
\]
which is a contradiction.
\end{rem}
\begin{rem}
Note that by replacing the norming set $G_1^\xi$ with
\[
G_r^{\xi}=\Big\{\sum_{i\in S}b_ie^*_i:S\text{ is a segment of }\T_\xi\text{ and }\sum_{i \in S}|b_i|^{r'}\le1 \Big\}
\]
where $r^{-1}+r'^{-1}=1$ and $1<r<p$, we define the spaces $JT^\xi_{r,p}$ whose norm is described in \eqref{jtqp}. These spaces are also reflexive, admit a unique $\ell_p$ asymptotic model and are not Asymptotic $\ell_p$.
\end{rem}
\begin{rem}
The approach used in \cite{BLMS} can be used to show that the spaces ${JT}^\xi_{r,p}$ and ${JT}^\xi_{p}$ have the property that any joint spreading model generated by an array of weakly null sequences is isometrically equivalent to the unit vector basis of $\ell_p$. That approach provides less insight and has no potential to apply to cases with a non-isometric result, e.g., the space from Section \ref{ell1section}.
\end{rem}
\section*{Acknowledgement}
The authors would like to thank the anonymous referees whose useful remarks helped us identify and correct a mistake in the original proof of Lemma \ref{combinatorial lemma essentially incomparable}.
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\section{Dependence Transform}
\label{transformability}
We provide the dependence manipulation techniques for both the spatial dependence and temporal dependence of a stochastic process.
The manipulation of the spatial dependence means the dependence manipulation of the random parameters of the stochastic process at some time epochs, while the manipulation of the temporal dependence means the manipulation of some random parameters on the time line.
Primary results of temporal dependence manipulation are shown in \cite{sun2018hidden}.
We define the unconditionally increasingly functions
\begin{multline}
\mathfrak{F}_{UI} = \{ f: \mathbb{R}^n \rightarrow \mathbb{R}; f(x_i| \qty{\bm{x}\setminus x_i }) \text{ is increasing at } x_i, \\
\forall \bm{x} \in \mathbb{R}^{n}, \forall 1\le i\le n \},
\end{multline}
and if the function $f$ is strictly increasing, we denote $f \in \mathfrak{F}_{USI}$.
We define the unconditionally increasingly affine functions
\begin{multline}
\mathfrak{F}_{UIA} = \{ f: \mathbb{R}^n \rightarrow \mathbb{R};
f(x_i| \qty{\bm{x}\setminus x_i }) \text{ is affine function of } x_i, \\
\text{ and is increasing at } x_i,
\forall \bm{x} \in \mathbb{R}^{n}, \forall 1\le i\le n \},
\end{multline}
and if the function $f$ is strictly increasing, we denote $f \in \mathfrak{F}_{USIA}$.
These function classes are sufficiently general in some scenarios, e.g., the wireless channel capacity process.
\subsection{Identical Marginals}
We present a sufficient condition that the composition of a supermodular function with some multivariate functions is a supermodular function.
\begin{lemma}\label{lemma-sm-function}
Let $f_t: \mathbb{R}^n \rightarrow \mathbb{R}$, $\forall t\ge 1$.
If $g: \mathbb{R}^t \rightarrow \mathbb{R}$ is supermodular, $g^\prime := g \qty(f_1, \ldots, f_t) : \mathbb{R}^{t\times n} \rightarrow \mathbb{R}$, and
\begin{multline}
g\qty( \qty( f_1(\bm{x}_1), \ldots, f_t(\bm{x}_t) ) \wedge \qty( f_1(\bm{y}_1), \ldots, f_t(\bm{y}_t) ) ) \\
+ g\qty( \qty( f_1(\bm{x}_1), \ldots, f_t(\bm{x}_t) ) \vee \qty( f_1(\bm{y}_1), \ldots, f_t(\bm{y}_t) ) )\\
=
g^\prime \qty( (\bm{x}_1, \ldots, \bm{x}_t) \wedge (\bm{y}_1, \ldots, \bm{y}_t) ) \\
+ g^\prime \qty( (\bm{x}_1, \ldots, \bm{x}_t) \vee (\bm{y}_1, \ldots, \bm{y}_t) ),
\end{multline}
then $g^\prime$ is supermodular.
\end{lemma}
\begin{proof}
Considering
\begin{multline}
g\qty( \qty( f_1(\bm{x}_1), \ldots, f_t(\bm{x}_t) ) \wedge \qty( f_1(\bm{y}_1), \ldots, f_t(\bm{y}_t) ) ) \\
+ g\qty( \qty( f_1(\bm{x}_1), \ldots, f_t(\bm{x}_t) ) \vee \qty( f_1(\bm{y}_1), \ldots, f_t(\bm{y}_t) ) ) \\
\ge g^\prime\qty( \bm{x}_1, \ldots, \bm{x}_t ) + g^\prime\qty( \bm{y}_1, \ldots, \bm{y}_t ),
\end{multline}
the proof follows directly.
\end{proof}
\begin{remark}
The implicit function theorem \cite{spivak1965calculus}\cite{rudin1976principles} gives a sufficient condition on the functions for the existence of their inverse in general.
\end{remark}
We investigate the scenario, where the multidimensional process is temporally independent and spatially dependent.
\begin{theorem}\label{theorem-sm-sp-dependent}
Assume the random parameters are spatially dependent and temporally independent.
Assume $f_t : \mathbb{R}^n \rightarrow \mathbb{R}$,
\begin{equation}
f_t\qty(\bm{x}_t) \diamond f_t\qty(\bm{y}_t) = f_t\qty( \bm{x}_t \hat{\diamond} \bm{y}_t ),\ \forall \bm{x}_t, \bm{y}_t \in \mathbb{R}^{n},\ \forall t\ge 0,
\end{equation}
where $\diamond, \hat{\diamond} \in \qty{\wedge, \vee}$ preserves one of the following three relations for all $t\ge 0$$:$ $\diamond = \hat{\diamond}$, $\qty{\diamond = \wedge|\hat{\diamond}=\vee} $, and $\qty{\diamond = \vee|\hat{\diamond}=\wedge}$.
If, for any $1\le j\le t,$
\begin{equation}
\left( X_j^1, X_j^2, \ldots, X_j^n \right) \le_{sm} \left( \widetilde{X}_j^1, \widetilde{X}_j^2, \ldots, \widetilde{X}_j^n \right),
\end{equation}
and $\qty( X_j^1, \ldots, X_j^n )$ and $\qty( \widetilde{X}_j^1, \ldots, \widetilde{X}_j^n )$, and $\qty( X_k^1, \ldots, X_k^n )$ are independent for all $j\neq k$,
then
\begin{equation}
\left( X_1, X_2, \ldots, X_t \right) \le_{sm} \left( {{X}}_1, {{X}}_2,\ldots, \widetilde{X}_j, \ldots, {{X}}_t \right),
\end{equation}
where ${{X}}_i = f_i \qty( {X}_i^1, \ldots, {X}_i^n )$, $\forall 1\le i\le t$, and ${\widetilde{X}}_j = f_j \qty( \widetilde{X}_j^1, \ldots, \widetilde{X}_j^n )$, $1\le j\le t$.
If
\begin{equation}
\left( X_j^1, X_j^2, \ldots, X_j^n \right) \le_{sm} \left( \widetilde{X}_j^1, \widetilde{X}_j^2, \ldots, \widetilde{X}_j^n \right),\ \forall 1\le j \le t,
\end{equation}
and $\qty( X_j^1, \ldots, X_j^n )$ and $\qty( X_k^1, \ldots, X_k^n )$ are independent for all $j\neq k$, so are $\qty( \widetilde{X}_j^1, \ldots, \widetilde{X}_j^n )$ and $\qty( \widetilde{X}_k^1, \ldots, \widetilde{X}_k^n )$,
then
\begin{multline}
\left( \widetilde{X}_1,\ldots, \widetilde{X}_k, {X}_{k+1}, \ldots, X_t \right) \\
\le_{sm} \left( {\widetilde{X}}_1, \ldots, {\widetilde{X}}_j, X_{j+1}, \ldots, {{X}}_t \right),\ \forall 1\le k\le j\le t,
\end{multline}
where ${{X}}_j = f_j \qty( {X}_j^1, \ldots, {X}_j^n )$ and ${\widetilde{X}}_j = f_j \qty( \widetilde{X}_j^1, \ldots, \widetilde{X}_j^n )$, $\forall 1\le j\le t$.
\end{theorem}
\begin{proof}
Considering the temporal independence assumption and the conjunction property of supermodular order \cite{shaked2007stochastic},
we have
$
\qty( X_1^1, \ldots, X_1^n, \ldots, X_t^1, \ldots, X_t^n )
\le_{sm}
\qty( {X}_1^1, \ldots, {X}_1^n, \ldots, \widetilde{X}_j^1, \ldots, \widetilde{X}_j^n, \ldots, {X}_t^1, \ldots, {X}_t^n )
$.
Letting $g: \mathbb{R}^t \rightarrow \mathbb{R}$ be supermodular and denote $g^\prime := g \qty(f_1, \ldots, f_t) : \mathbb{R}^{t\times n} \rightarrow \mathbb{R}$,
we have $g^\prime$ is supermodular, which follows Lemma \ref{lemma-sm-function}.
Thus, it directly implies
$
\qty( f_1 \qty( X_1^1, \ldots, X_1^n ), \ldots, f_t\qty( X_t^1, \ldots, X_t^n ) )
\le_{sm}
\qty( f_1 \qty({X}_1^1, \ldots, {X}_1^n ), \ldots, f_j \qty( \widetilde{X}_j^1, \ldots, \widetilde{X}_j^n ), \ldots, f_t \qty( {X}_t^1, \ldots, {X}_t^n ) )
$.
The proof of the other result follows the reflexivity and transitivity property of supermodular order \cite{muller2002comparison}.
\end{proof}
\begin{remark}
The results indicate that the spatial dependence of the random parameters also influences the dependence of the stochastic process and more manipulations of the spatial dependence has more strength to transform the dependence of the stochastic process.
\end{remark}
\begin{remark}
It is interesting to investigate the relationship between the requirement of the functional and the spatial dependence of the random parameters.
An example is the comonotonicity dependence structure with identical marginal distribution, i.e., the random parameters are equal almost surely, thus the requirement of the function reduces to the scenario of the requirement of the univariate functional scenario, i.e., decreasing or increasing for each variate on the function domain.
\end{remark}
We present a result without specification on the spatial and temporal dependence. Note the relaxation of the specification on dependence is replaced by the additional conditions on the functionals.
\begin{theorem}\label{theorem-sm-dependence}
Assume $f_t : \mathbb{R}^n \rightarrow \mathbb{R}$ and $f_t\qty(\bm{X}_t^i | \bm{Z}_t^i = \bm{z}_t^i )$ are all increasing or all decreasing at each component of $\bm{X}^i = \qty(X^i_1, \ldots, X^i_t)$, for any $\bm{z}_t^i = \qty(x_t^1,\ldots, x_t^{i-1}, x_t^{i+1}, \ldots, x_t^n)$ in the support of $\bm{Z}_t^i = \qty(X^1_t,\ldots, X_t^{i-1}, X_t^{i+1}, \ldots, X_t^n) $, $\forall 1\le i\le n$, $\forall t\ge 1$.
Denote $\bm{z}^i = \qty{ \bm{z}_1^i, \ldots, \bm{z}_t^i }$, $\bm{Z}^i = \qty{ \bm{Z}_1^i, \ldots, \bm{Z}_t^i }$, and $\qty(\bm{X}^i | \bm{Z}^i) \le_{sm} \qty(\widetilde{\bm{X}}^i | \bm{Z}^i) \iff \qty(\bm{X}^i | \bm{Z}^i = \bm{z}^i ) \le_{sm} \qty(\widetilde{\bm{X}}^i | \bm{Z}^i =\bm{z}^i ),\ \forall \bm{z}^i \in \bm{Z}^i$.
If, for any $1\le i \le n,$
\begin{equation}
\left( X_1^i, X_2^i, \ldots, X_t^i | \bm{Z}^i \right)
\le_{sm} \left( \widetilde{X}_1^i, \widetilde{X}_2^i, \ldots, \widetilde{X}_t^i | \bm{Z}^i \right),
\end{equation}
then
\begin{equation}
\left( X_1, X_2, \ldots, X_t \right) \le_{sm} \left( {\widetilde{X}}_1, {\widetilde{X}}_2, \ldots, {\widetilde{X}}_t \right),
\end{equation}
where ${\widetilde{X}}_j = f_j ( X_j^1, \ldots, X_j^{i-1}, \widetilde{X}_j^i, X_j^{i+1}, \ldots, X_j^n )$, $\forall 1\le j\le t$.
If, $\forall 1\le j \le i$,
\begin{equation}
\left( X_1^j, X_2^j, \ldots, X_t^j | \bm{Z}^j \right)
\le_{sm} \left( \widetilde{X}_1^j, \widetilde{X}_2^j, \ldots, \widetilde{X}_t^j | \bm{Z}^j \right),
\end{equation}
then
\begin{equation}
\widetilde{\bm{X}}_{t}^{k} \le_{sm} \widetilde{\bm{X}}_{t}^{j},\ \forall 0\le k \le j \le i,
\end{equation}
with $\widetilde{{X}}_{{t}_m}^{l} = f_m ( \widetilde{X}_m^1, \ldots, \widetilde{X}_m^l, {X}_m^{l+1}, \ldots, {X}_m^n )$, $\ 1\le m \le t$, and $\widetilde{\bm{X}}_{t}^{l} = ( \widetilde{{X}}_{t_1}^{l}, \ldots, \widetilde{{X}}_{t_t}^{l} )$, $l\in\{k,j\}$.
\end{theorem}
\begin{proof}
The proof follows analogically to the proof of the independence scenario \cite{sun2018hidden}, by using the conditional probability.
\end{proof}
\begin{remark}
The stochastic orders $\qty(\bm{X}|\bm{Z} = \bm{z}) \le_{sm} \qty(\bm{Y}|\bm{Z} = \bm{z})$ and $\mathbb{E}\qty(\bm{X}|\bm{Z}) \le_{sm} \mathbb{E}\qty(\bm{Y}|\bm{Z})$ correspond to the conditional supermodular order in the sense of the uniform conditional ordering
\cite{whitt1980uniform}\cite{ruschendorf1991conditional}.
On the one hand, the conditional formulation influences the stochastic ordering of the probability measures, moreover, it influences the property of the functions of the random variables, e.g., the monotonicity.
\end{remark}
\begin{remark}
There is an implicit condition that the spatial dependence must not influence the temporal dependence ordering, or the temporal dependence ordering is conditional on the spatial dependence.
Specifically, if spatial independence is assumed, the conditional event disappears.
On the other hand, it is interesting to investigate what type of spatial dependence sufficiently imply the conditional ordering.
\end{remark}
\begin{remark}
It is interesting to investigate the conditional probability and conditional stochastic order expression of the spatial dependence manipulation scenario, e.g., Theorem \ref{theorem-sm-sp-dependent}.
\end{remark}
\begin{remark}
As an example of conditional probability, the function of two random variables $f(X,Y)$,
the independence assumption implies $\mathbb{E}\qty[f(X,Y)] = \mathbb{E}_{X}\mathbb{E}_{Y}\qty[f(X,Y)]$, while the absence of independence implies that $\mathbb{E}\qty[f(X,Y)] = \mathbb{E}_{X}\mathbb{E}_{Y|X=x}\qty[f(X,Y)| X=x]$.
\end{remark}
\begin{remark}
An an example of conditional monotonicity of functions, let $f(x,y) = x^y$, $x>0$, then $f(x,y|y>0)$ is increasing at $x$, $f(x,y|y<0)$ is decreasing at $x$, and $f(x,y|y=0)=1$ is constant.
\end{remark}
\begin{remark}
It is interesting to extend the results in Theorem \ref{theorem-sm-dependence} and the corresponding results without conditional probability to the increasing supermodular order $\le_{ism}$ and the symmetric supermodular order $\le_{symsm}$.
\end{remark}
\subsection{Different Marginals}
We present a result about the manipulation of stochastic process based on the marginals.
\begin{theorem}\label{theorem-marginal-copula}
Assume the random parameters are spatially independent and temporally dependent.
If, for any $1\le i\le n$,
$
\left( X_1^i, X_2^i, \ldots, X_t^i \right) \le_{dcx} \left( \widetilde{X}_1^i, \widetilde{X}_2^i, \ldots, \widetilde{X}_t^i \right),
$
$f_j\qty( X_j^1\ldots, X_j^n ) \in \mathfrak{F}_{USIA}$, $\forall 1\le j\le t$,
and $\left( X_1^i, X_2^i, \ldots, X_t^i \right)$ and $\left( {\widetilde{X}}_1^i, {\widetilde{X}}_2^i, \ldots, {\widetilde{X}}_t^i \right)$ have the common conditionally increasing copula
$C^i_t \qty(u_1^i,\ldots, u_t^i)$,
then
\begin{equation}
\left( X_1, X_2, \ldots, X_t \right) \le_{dcx} \left( {\widetilde{X}}_1, {\widetilde{X}}_2, \ldots, {\widetilde{X}}_t \right),
\end{equation}
where ${\widetilde{X}}_j = f_j ( X_j^1, \ldots, X_j^{i-1}, \widetilde{X}_j^i, X_j^{i+1}, \ldots, X_j^n )$, $\forall 1\le j\le t$,
and
\begin{equation}
\sum_{j=1}^{t} \alpha_j X_j \le_{cx} \sum_{j=1}^{t} \alpha_j \widetilde{X}_j.
\end{equation}
where $\alpha_j \in \mathbb{R}_{\ge 0}$, $\forall 1\le j\le t$.
\end{theorem}
\begin{proof}
Without loss of generality, we consider the first variate.
We have
$\qty(X_1^1, \ldots, X_t^1) \le_{dcx} \qty(\widetilde{X}_1^1, \ldots, \widetilde{X}_t^1)
\implies X_j^1 \le_{cx} \widetilde{X}_j^1,\ \forall 1\le j\le t
$.
Considering that the composition $g\circ f$ of a convex function $g$ and an affine function $f$ is a convex function \cite{simchi2005logic}, we have
$
f_j \qty( X_j^1, x_j^2, \ldots, x_j^n ) \le_{cx} f_j \qty( \widetilde{X}_j^1, {x}_j^2, \ldots, x_j^n ), \ \forall \qty(x_j^2, \ldots, x_j^n) \in \qty(X_j^2,\ldots, X_j^n),\ \forall 1\le j\le t
$.
Since the functional $\qty{f_j}_{1\le j\le t}$ is unconditionally increasing, the copula of the sequence, $\qty{ f_j \qty( X_j^1, x_j^2, \ldots, x_j^n ) }_{1\le j\le t}$, equals the copula of the sequence $\qty{ X_j^1}_{1\le j\le t}$.
Thus, we obtain $\qty( f_1\qty( X_1^1, x_1^2, \ldots, x_1^n ), \ldots, f_t\qty( X_t^1, x_t^2, \ldots, x_t^n ) ) \le_{dcx} \qty( f_1\qty( \widetilde{X}_1^1, x_1^2, \ldots, x_1^n ), \ldots, f_t\qty( \widetilde{X}_t^1, x_t^2, \ldots, x_t^n ) )$, $\forall \qty(x_j^2, \ldots, x_j^n) \in \qty(X_j^2,\ldots, X_j^n)$, $\forall 1\le j\le t$.
This conditional case follows the proof of the one dimensional result in \cite{muller2001stochastic}.
By taking the expectation, we obtain
$ \left( X_1, X_2, \ldots, X_t \right) \le_{dcx} \left( {\widetilde{X}}_1, {\widetilde{X}}_2, \ldots, {\widetilde{X}}_t \right) $, which further implies the convex order of the weighted sum \cite{muller2001stochastic}.
\end{proof}
\begin{remark}
It is interesting to consider the copula construction of the functional sequence based on the copulas of each sub-sequence, especially for some special cases, e.g., without Granger causality .
\end{remark}
\begin{remark}
The invariance of copula under strictly increasing transformation of random variables requires that the functionals are strictly increasing.
\end{remark}
\begin{remark}
This result is extensible to any functions $f_j\qty( X_j^1\ldots, X_j^n ) \in \mathfrak{F}$, $\forall 1\le j\le t$, which are componentwisely and strictly increasing and preserves the convexity under the composition $g\circ f_j\qty(X_j^i| \qty{\bm{x}_j \setminus x_j^i} )$, $\forall 1\le i \le n$, with a convex function $g: \mathbb{R} \rightarrow \mathbb{R}$.
\end{remark}
\begin{remark}
The manipulation of marginal distributions is more complicated in the sense that there is a more involved requirement on the functionals.
\end{remark}
We present the result of the strength of the marginal distribution manipulation.
\begin{theorem}\label{theorem-dcx-manipulation-strength}
Assume the random parameters are spatially independent and temporally dependent.
If, for all $1\le i\le n$,
$\left( X_1^i, X_2^i, \ldots, X_t^i \right) \le_{dcx} \left( \widetilde{X}_1^i, \widetilde{X}_2^i, \ldots, \widetilde{X}_t^i \right)$,
$f_j\qty( X_j^1\ldots, X_j^n ) \in \mathfrak{F}_{USIA}$, $\forall 1\le j\le t$,
and $\left( X_1^i, X_2^i, \ldots, X_t^i \right)$ and $\left( {\widetilde{X}}_1^i, {\widetilde{X}}_2^i, \ldots, {\widetilde{X}}_t^i \right)$ have the common conditionally increasing copula
$C^i_t \qty(u_1^i,\ldots, u_t^i)$,
then
\begin{equation}
\widetilde{\bm{X}}_{t}^{k} \le_{dcx} \widetilde{\bm{X}}_{t}^{k^\prime},\ \forall 0\le k \le k^\prime \le n,
\end{equation}
with $\widetilde{{X}}_{{t}_m}^{l} = f_m ( \widetilde{X}_m^1, \ldots, \widetilde{X}_m^l, {X}_m^{l+1}, \ldots, {X}_m^n )$, $\ 1\le m \le t$, and $\widetilde{\bm{X}}_{t}^{l} = ( \widetilde{{X}}_{t_1}^{l}, \ldots, \widetilde{{X}}_{t_t}^{l} )$, $l\in\{k,k^\prime \}$,
and
\begin{equation}
\sum_{j=1}^{t} \alpha_j X_j^k \le_{cx} \sum_{j=1}^{t} \alpha_j \widetilde{X}_j^{k^\prime}.
\end{equation}
where $\alpha_j \in \mathbb{R}_{\ge 0}$, $\forall 1\le j\le t$.
\end{theorem}
\begin{proof}
The proof follows the proof of Theorem \ref{theorem-marginal-copula} and the transitivity of the directionally convex order.
\end{proof}
\begin{remark}
The results show that the manipulation of the marginal distributions of one dimension is able to transform the distribution ordering properties of the overall stochastic process, the more dimensions the more manipulation strength.
In addition, the marginal distribution manipulation is feasible only for positive dependence, while the dependence structure manipulation has no dependence bias.
\end{remark}
\begin{remark}
The results have a dependence control utility.
On the one hand, it means that the negative dependence endows the advantage of reducing the power or capacity cost while attaining a higher performance, which is a physical perspective, on the other hand, it means that the advantage of negative dependence maps to the property of the negative dependence and the convex order, which is a mathematical perspective.
Thus, the mathematical property corresponds to a physical resource, which can be taken advantage of and exploited.
In addition, it indicates that the marginal distribution manipulation is invalid for negative dependence and should be avoided in practice.
\end{remark}
\begin{corollary}
The results in Theorem \ref{theorem-marginal-copula} and Theorem \ref{theorem-dcx-manipulation-strength} extend to the increasing directionally convex order $\le_{idcx}$ of the random vectors and the corresponding increasing convex order $\le_{icx}$ of the weighted partial sums.
\end{corollary}
\begin{proof}
The proof follows the proofs of Theorem \ref{theorem-marginal-copula} and Theorem \ref{theorem-dcx-manipulation-strength}, and the result that the composition of an increasing convex function and an increasing affine function is an increasing convex function \cite{shaked2007stochastic}, and the result that \cite{balakrishnan2012increasing}, letting $\bm{X}$ and $\bm{Y}$ be random vectors with a common conditionally increasing copula and assuming that $X_i \le_{icx} Y_i$, $\forall i$, then $\bm{X} \le_{idcx} \bm{Y}$.
\end{proof}
\begin{corollary}
Assume the random parameters are spatially dependent and temporally independent.
If $f_j\qty( X_j^1\ldots, X_j^n )$, $\forall 1\le j\le t$, are increasing and directionally convex,
and
$
\left( X_j^1, X_j^2, \ldots, X_j^n \right) \le_{idcx} \left( \widetilde{X}_j^1, \widetilde{X}_j^2, \ldots, \widetilde{X}_j^n \right),
$
$\forall 1\le j\le t$,
then the same results hold as in Theorem \ref{theorem-sm-sp-dependent}, but in the sense of the increasing and directionally convex order $\le_{idcx}$.
\end{corollary}
\begin{proof}
Note the composition of an increasing and convex function $g: \mathbb{R} \rightarrow \mathbb{R}$ and an increasing and directionally convex function $f: \mathbb{R}^n \rightarrow \mathbb{R}$ is an increasing and directionally convex function $g\circ f$ \cite{shaked2007stochastic}.
Then, the proof follows that the increasing and convex order of each elements implies the increasing and directionally convex order of the random vector with independent elements.
\end{proof}
\begin{remark}
It is interesting to extend the temporal and spatial manipulation results to the conditional (increasing) directionally convex order.
\end{remark}
Assuming independence among the random vectors, we present the directionally convex order result for random sums, which are not necessarily independent.
\begin{theorem}\label{theorem-random-multiplexing-dcx}
Let $\bm{X}_j = ( X_{j,1}, \ldots, X_{j,m} )$ and $\bm{Y}_j = ( Y_{j,1}, \ldots, Y_{j,m} )$, $j = 1,2,\ldots$, be two sequences of non-negative random vectors with independence among components, and let $\bm{M} = \left(M_1, M_2, \ldots , M_m \right)$ and $\bm{N} = \left( N_1, N_2, \ldots, N_m \right)$ be two vectors of non-negative integer-valued random variables. Assume that both $\bm{M}$ and $\bm{N}$ are independent of the $\bm{X}_j$'s and $\bm{Y}_j$'s.
Assume that $X_{j,i} \le_{cx} X_{j+1,i}$, $\forall 1\le i \le m$, $\forall j \ge 1$.
If $\bm{M} \le_{dcx} \bm{N}$, then
\begin{equation}
\left( \sum_{j=1}^{M_1} X_{j,1}, \ldots, \sum_{j=1}^{M_m} X_{j,m} \right)
\le_{dcx} \left( \sum_{j=1}^{N_1} X_{j,1}, \ldots, \sum_{j=1}^{N_m} X_{j,m} \right). \nonumber
\end{equation}
If $\bm{X}_j \le_{dcx} \bm{Y}_j $, $\forall j$, then
\begin{equation}
\left( \sum_{j=1}^{N_1} X_{j,1}, \ldots, \sum_{j=1}^{N_m} X_{j,m} \right)
\le_{dcx} \left( \sum_{j=1}^{N_1} Y_{j,1}, \ldots, \sum_{j=1}^{N_m} Y_{j,m} \right). \nonumber
\end{equation}
If $\bm{M} \le_{dcx} \bm{N}$ and $\bm{X}_j \le_{dcx} \bm{Y}_j $, $\forall j$, then
\begin{equation}
\left( \sum_{j=1}^{M_1} X_{j,1}, \ldots, \sum_{j=1}^{M_m} X_{j,m} \right) \le_{dcx}
\left( \sum_{j=1}^{N_1} Y_{j,1}, \ldots, \sum_{j=1}^{N_m} Y_{j,m} \right). \nonumber
\end{equation}
\end{theorem}
\begin{proof}
The first result is available in \cite{pellerey1999stochastic}\cite{shaked2007stochastic}.
For the second result, $\bm{X}_j \le_{dcx} \bm{Y}_j \implies X_{j,i} \le_{cx} Y_{j,i}$, $\forall 1\le i\le m$, the independence assumption implies that the convex order is closed under convolutions \cite{shaked2007stochastic}, i.e., $\sum_{j=1}^{n_i} X_{j,i} \le_{cx} \sum_{j=1}^{n_i} Y_{j,i}$, $\forall 1\le i\le m$, furthermore, it implies
$
\mathbb{E} \qty[ \phi \left( \sum_{j=1}^{N_1} X_{j,1}, \ldots, \sum_{j=1}^{N_m} X_{j,m} \right) | \bm{N} = \qty(n_1, \ldots, n_m) ]
\le \mathbb{E} \qty[ \phi \left( \sum_{j=1}^{N_1} Y_{j,1}, \ldots, \sum_{j=1}^{N_m} Y_{j,m} \right) | \bm{N} = \qty(n_1, \ldots, n_m) ],
$
where $\phi$ is directionally convex. By integrating for expectation, we obtain the final result.
The third result follows the transitivity of the directionally convex order.
\end{proof}
\begin{corollary}
With proper revisions, the results in Theorem \ref{theorem-random-multiplexing-dcx} extend to the increasing directionally convex order $\le_{idcx}$ and (increasing) componentwise convex order ($\le_{iccx}$) $\le_{ccx}$.
Specifically, the corresponding revisions are $X_{j,i} \le_{icx} (\le_{cx}, \le_{icx}) X_{j+1,i}$, $\bm{M} \le_{idcx} (\le_{ccx}, \le_{iccx}) \bm{N}$, and $\bm{X}_j \le_{idcx} (\le_{ccx}, \le_{iccx}) \bm{Y}_j$.
\end{corollary}
\begin{proof}
The proof follows the properties of each stochastic orders and preliminary results in \cite{pellerey1999stochastic}\cite{shaked2007stochastic}.
\end{proof}
\begin{remark}
It is interesting to notice the fact that: Let $\bm{X} = \qty(X_1, \ldots, X_m )$ be a set of independent random variables and let $\bm{Y} = \qty( Y_1, \ldots, Y_m )$ be another set of independent random variables, then, $\bm{X} \le_{dcx} (\le_{idcx}) \bm{Y} \iff X_i \le_{cx} (\le_{icx}) Y_i,\ \forall 1\le i\le m \iff \bm{X} \le_{ccx} (\le_{iccx}) \bm{Y}$.
\end{remark}
We study the ordering property of the partial sums under the ordering condition of the sequences.
\begin{theorem}
For the stochastic process, if $\qty(X_{t_1}, \ldots, X_{t_k}) \le_{idcx} \qty(\widetilde{X}_{t_1}, \ldots, \widetilde{X}_{t_k})$, $\forall t_1, \ldots, t_k \in \mathbb{N}$, $\forall k\in \mathbb{N}$, then, we have
\begin{equation}
\qty( \sum_{j_1 \in \mathcal{T}_1} X_{j_1}, \ldots, \sum_{j_k \in \mathcal{T}_k} X_{j_k} )
\le_{idcx}
\qty( \sum_{j_1 \in \mathcal{T}_1} \widetilde{X}_{j_1}, \ldots, \sum_{j_k \in \mathcal{T}_k} \widetilde{X}_{j_k} ),
\end{equation}
for any disjoint subsets $\mathcal{T}_1, \ldots, \mathcal{T}_k \in \mathbb{N}$.
\end{theorem}
\begin{proof}
The proof follows that, if $f : \mathbb{R}^m \rightarrow \mathbb{R}^k$ is increasing and directionally convex and $g : \mathbb{R}^n \rightarrow \mathbb{R}^m$ is increasing and directionally convex, then the composition $f\circ g$ is increasing and directionally convex \cite{shaked2007stochastic}.
\end{proof}
\begin{remark}
An alternative approach is to treat the functional stochastic process as a random field on $\mathbb{N}^n \times \mathbb{R}$, then the comparison result directly follows the comparison result of random field in \cite{miyoshi2004note}\cite{shaked2007stochastic}.
\end{remark}
\begin{remark}
The result indicates that the $\le_{idcx}$ ordering of the instantaneous values implies the $\le_{idcx}$ ordering of the accumulated values.
\end{remark}
\begin{remark}
It is interesting to study the corresponding property of the supermodular order or the counter examples.
\end{remark}
Since the usual stochastic order has a direct indication on the mean values, i.e., $\bm{X} \le_{st} \bm{Y} \implies \mathbb{E}\bm{X} \le \mathbb{E}\bm{Y} \implies \sum\mathbb{E}{X}_i \le \sum\mathbb{E}{Y}_i$, $X_i \in \bm{X},\ Y_i \in \bm{Y}$, it is interesting to consider the dependence manipulation with respect to the usual stochastic order when the mean value is the objective measure.
\begin{theorem}\label{theorem-marginal-st}
Assume the random parameters are spatially independent and temporally dependent.
If $f_j\qty( X_j^1\ldots, X_j^n )$, $\forall 1\le j\le t$, are increasing,
and
$
\left( X_1^i, X_2^i, \ldots, X_t^i \right) \le_{st} \left( \widetilde{X}_1^i, \widetilde{X}_2^i, \ldots, \widetilde{X}_t^i \right),
$
$\forall 1\le i\le n$,
then
\begin{equation}
\left( X_1, X_2, \ldots, X_t \right) \le_{st} \left( {\widetilde{X}}_1, {\widetilde{X}}_2, \ldots, {\widetilde{X}}_t \right),
\end{equation}
where ${\widetilde{X}}_j = f_j ( X_j^1, \ldots, X_j^{i-1}, \widetilde{X}_j^i, X_j^{i+1}, \ldots, X_j^n )$, $\forall 1\le j\le t$, for any $1\le i \le n$;
and
\begin{equation}
\widetilde{\bm{X}}_{t}^{k} \le_{st} \widetilde{\bm{X}}_{t}^{k^\prime},\ \forall 0\le k \le k^\prime \le n,
\end{equation}
where $\widetilde{{X}}_{{t}_m}^{l} = f_m ( \widetilde{X}_m^1, \ldots, \widetilde{X}_m^l, {X}_m^{l+1}, \ldots, {X}_m^n )$, $\ 1\le m \le t$, and $\widetilde{\bm{X}}_{t}^{l} = ( \widetilde{{X}}_{t_1}^{l}, \ldots, \widetilde{{X}}_{t_t}^{l} )$, $l\in\{k,k^\prime \}$.
\end{theorem}
\begin{proof}
The spatial independence implies the conjunction $\qty( \bm{X}^1, \ldots, \bm{X}^i, \ldots, \bm{X}^n ) \le_{st} \qty( \bm{X}^1, \ldots, \widetilde{\bm{X}}^i, \ldots, \bm{X}^n )$, where $\bm{X}^i = \left( X_1^i, X_2^i, \ldots, X_t^i \right)$, then the first result directly follows the closure property of the usual stochastic order \cite{shaked2007stochastic}.
The second result follows the transitivity of the usual stochastic order.
\end{proof}
\begin{remark}
Particularly, if $\left( X_1^i, X_2^i, \ldots, X_t^i \right)$ and $\left( {\widetilde{X}}_1^i, {\widetilde{X}}_2^i, \ldots, {\widetilde{X}}_t^i \right)$ have the common copula $C^i_t \qty(u_1^i,\ldots, u_t^i)$, the order condition $\left( X_1^i, X_2^i, \ldots, X_t^i \right) \le_{st} \left( \widetilde{X}_1^i, \widetilde{X}_2^i, \ldots, \widetilde{X}_t^i \right)$ can be replaced by $X_j^i \le_{st} \widetilde{X}_j^i$, $\forall 1\le j \le t$.
This result is available in \cite[p. 272]{shaked2007stochastic}.
\end{remark}
\begin{corollary}
Assume the random parameters are spatially dependent and temporally independent.
If $f_j\qty( X_j^1\ldots, X_j^n )$, $\forall 1\le j\le t$, are increasing,
and
$
\left( X_j^1, X_j^2, \ldots, X_j^n \right) \le_{st} \left( \widetilde{X}_j^1, \widetilde{X}_j^2, \ldots, \widetilde{X}_j^n \right),
$
$\forall 1\le j\le t$,
then the same results hold as in Theorem \ref{theorem-sm-sp-dependent}, but in the sense of the usual stochastic order $\le_{st}$.
\end{corollary}
\begin{proof}
The results follow the closure property and transitivity of the usual stochastic order \cite{shaked2007stochastic}.
\end{proof}
\begin{remark}
It is interesting to extend the results to the scenario without spatial and temporal dependence specification and express the results in terms of the conditional probability and conditional stochastic order as in Theorem \ref{theorem-sm-dependence}.
\end{remark}
We have the following results of the random sums with respect to the usual stochastic order.
\begin{remark}
Let $\bm{X}_j = ( X_{j,1}, \ldots, X_{j,m} )$ and $\bm{Y}_j = ( Y_{j,1}, \ldots, Y_{j,m} )$, $j = 1,2,\ldots$, be two sequences of non-negative random vectors, and let $\bm{M} = \left(M_1, M_2, \ldots , M_m \right)$ and $\bm{N} = \left( N_1, N_2, \ldots, N_m \right)$ be two vectors of non-negative integer-valued random variables. Assume that both $\bm{M}$ and $\bm{N}$ are independent of the $\bm{X}_j$'s and $\bm{Y}_j$'s.
If $\qty{ \bm{X}_j, j \in \mathbb{N} } \le_{st} \qty{ \bm{Y}_j, j \in \mathbb{N} }$ and $\bm{M} \le_{st} \bm{N}$, then
\begin{equation}
\left( \sum_{j=1}^{M_1} X_{j,1}, \ldots, \sum_{j=1}^{M_m} X_{j,m} \right)
\le_{st} \left( \sum_{j=1}^{N_1} Y_{j,1}, \ldots, \sum_{j=1}^{N_m} Y_{j,m} \right). \nonumber
\end{equation}
This result is available in \cite{shaked2007stochastic}. Specifically, the proof follows the transitivity of the following results,
\begin{equation}
\left( \sum_{j=1}^{M_1} X_{j,1}, \ldots, \sum_{j=1}^{M_m} X_{j,m} \right)
\le_{st} \left( \sum_{j=1}^{N_1} X_{j,1}, \ldots, \sum_{j=1}^{N_m} X_{j,m} \right), \nonumber
\end{equation}
which is provided in \cite{pellerey1999stochastic}, and
\begin{equation}
\left( \sum_{j=1}^{N_1} X_{j,1}, \ldots, \sum_{j=1}^{N_m} X_{j,m} \right)
\le_{st} \left( \sum_{j=1}^{N_1} Y_{j,1}, \ldots, \sum_{j=1}^{N_m} Y_{j,m} \right), \nonumber
\end{equation}
which follows the closure property of the usual stochastic order, by conditioning on $\qty(N_1, \ldots, N_m) = \qty(n_1, \ldots, n_m)$ and integrating for expectation.
\end{remark}
| 200,248
|
No glass by the pool? Scared of popping that champagne cork? No problem!
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A new canned sparkling wine company called Shamps sells canned sparkling wine with 110 calories and 5 percent ABV per 12-ounce can, and it comes in six-packs.
Mimosas by the pool anyone!?
| 196,869
|
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| 75,473
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{\bf Problem.} Find the sum of the roots of the equation \[(2x^3 + x^2 - 8x + 20)(5x^3 - 25x^2 + 19) = 0.\]
{\bf Level.} Level 3
{\bf Type.} Intermediate Algebra
{\bf Solution.} By Vieta's formulas, the sum of the roots of $2x^3 + x^2 - 8x + 20 = 0$ is $-\tfrac{1}{2}.$ Similarly, the sum of the roots of $5x^3-25x^2+19=0$ is $-\tfrac{-25}{5} = 5.$ Notice that the roots of the given equation consist of the roots of both equations put together (since, in general, $ab = 0$ if and only if $a=0$ or $b=0$). Therefore, the sum of the roots of the given equation is $-\tfrac{1}{2} + 5 = \boxed{\tfrac{9}{2}}.$
| 39,271
|
\begin{document}
\title{Gradings on the Lie algebra $D_4$ revisited}
\author[A. Elduque]{Alberto Elduque${}^\star$}
\address{Departamento de Matem\'{a}ticas
e Instituto Universitario de Matem\'aticas y Aplicaciones,
Universidad de Zaragoza, 50009 Zaragoza, Spain}
\email{elduque@unizar.es}
\thanks{${}^\star$supported by the Spanish Ministerio de Econom\'{\i}a y Competitividad---Fondo Europeo de Desarrollo Regional (FEDER) MTM2010-18370-C04-02 and MTM2013-45588-C3-2-P, and by the Diputaci\'on General de Arag\'on---Fondo Social Europeo (Grupo de Investigaci\'on de \'Algebra)}
\author[M. Kochetov]{Mikhail Kochetov${}^\dagger$}
\address{Department of Mathematics and Statistics,
Memorial University of Newfoundland,
St. John's, NL, A1C5S7, Canada}
\email{mikhail@mun.ca}
\thanks{${}^\dagger$supported by Discovery Grant 341792-2013 of the Natural Sciences and Engineering Research Council (NSERC) of Canada and a grant for visiting scientists by Instituto Universitario de Matem\'aticas y Aplicaciones, University of Zaragoza}
\subjclass[2010]{Primary 17B70; Secondary 17B25, 17C40, 17A75}
\keywords{Graded algebra, graded module, exceptional simple Lie algebra, composition algebra,
cyclic composition algebra, triality, trialitarian algebra, exceptional simple Jordan algebra}
\date{}
\begin{abstract}
We classify group gradings on the simple Lie algebra $\cL$ of type $D_4$ over an algebraically closed field of characteristic different from $2$:
fine gradings up to equivalence and $G$-gradings, with a fixed group $G$, up to isomorphism. For each $G$-grading on $\cL$, we also study graded $\cL$-modules
(assuming characteristic $0$).
\end{abstract}
\maketitle
\section{Introduction}
In the past two decades, there has been much interest in gradings on simple Lie algebras by arbitrary groups --- see our recent monograph \cite{EKmon} and references therein. In particular, the classification of fine gradings (up to equivalence) on all finite-dimensional simple Lie algebras over an algebraically closed field of characteristic $0$ is essentially complete (\cite[Chapters 3--6]{EKmon}, \cite{E14}, \cite{YuExc}). For a given group $G$, the classification of $G$-gradings (up to isomorphism) on classical simple Lie algebras over an algebraically closed field of characteristic different from $2$ was done in \cite{BK10} (see also \cite[Chapter 3]{EKmon}), excluding type $D_4$, which exhibits exceptional behavior due to the phenomenon of triality. Although the case of $D_4$ is included in \cite{E09d} (see also \cite{DMV} and \cite[\S 6.1]{EKmon}), only fine gradings are treated there and the characteristic is assumed to be $0$. Since we do not see how to extend those arguments to positive characteristic, here we use an approach based on affine group schemes, which was also employed in \cite{BK10}.
Let $\FF$ be the ground field. Except in the Preliminaries, we will assume $\FF$ {\em algebraically closed} and $\chr{\FF}\ne 2$. All vector spaces, algebras, tensor products, group schemes, etc. will be assumed over $\FF$ unless indicated otherwise. The superscript $\times$ will indicate the multiplicative group of invertible elements.
Recall that affine group schemes are representable functors from the category $\Alg_\FF$ of unital associative commutative algebras over $\FF$ to the category of groups --- we refer the reader to \cite{Wh}, \cite[Chapter VI]{KMRT} or \cite[Appendix A]{EKmon} for the background. Every (na\"\i ve) algebraic group gives rise to an affine group scheme. These are precisely the {\em smooth algebraic} group schemes, i.e., those whose representing (Hopf) algebra is finitely generated and reduced. In characteristic $0$, all group schemes are reduced, but it is not so in positive characteristic. We will follow the common convention of denoting the (smooth) group schemes corresponding to classical groups by the same letters, but using bold font to distinguish the scheme from the group (which is identified with the $\FF$-points of the scheme). It is important to note that this convention should be used with care: for example, the automorphism group scheme $\AAut_\FF(\cU)$ of a finite-dimensional algebra $\cU$ is defined by $\AAut_\FF(\cU)(\cS)=\Aut_\cS(\cU\ot\cS)$ for every $\cS$ in $\Alg_\FF$, and may be strictly larger than the smooth group scheme corresponding to the algebraic group $\Aut_\FF(\cU)$.
Let $\cL$ be a Lie algebra of type $D_4$. It is well known that the automorphism group scheme $\AAut_\FF(\cL)$ is smooth and we have a short exact sequence
\begin{equation}\label{eq:exact_D4}
\xymatrix{
\mathbf{1}\ar[r] & \PGOs^+_8\ar[r] & \AAut_\FF(\cL)\ar[r]^-{\pi} & \mathbf{S}_3\ar[r] & \mathbf{1}
}
\end{equation}
where $\PGOs^+_8$ is the group scheme of inner automorphisms (which corresponds to the algebraic group $\PGO^+_8$, see e.g. \cite[\S 12.A]{KMRT}) and $\mathbf{S}_3$ is the constant group scheme corresponding to the symmetric group $S_3$. This sequence can be split by identifying $\cL$ with the {\em triality Lie algebra} (see Definition \ref{df:tri} below) of the {\em para-Cayley algebra} $\cC$, i.e., the Cayley algebra equipped with the new product $x\bullet y=\bar{x}\bar{y}$, where juxtaposition denotes the usual product of $\cC$ and bar denotes its standard involution (see e.g. \cite[\S 4.1]{EKmon}). If we use the standard model for Lie algebras of series $D$, i.e., identify $\cL$ with the skew elements in $\cR=M_8(\FF)$ with respect to an orthogonal involution $\sigma$, then the restriction $\AAut_\FF(\cR,\sigma)\to\AAut_\FF(\cL)$ is a closed imbedding whose image coincides with the semidirect product of $\PGOs^+_8$ and the constant group scheme corresponding to one of the subgroups of order $2$ in $S_3$ (see e.g. \cite[\S 3.1]{EKmon}).
Now suppose that we have a grading $\Gamma:\;\cL=\bigoplus_{g\in G}\cL_g$ by a group $G$. Since the elements of the support of a group grading on a simple Lie algebra necessarily commute, we will always assume that $G$ is {\em abelian}. (Since the support is finite, we may also assume $G$ finitely generated.) Then $\Gamma$ is equivalent to a morphism $\eta=\eta_\Gamma\colon G^D\to\AAut_\FF(\cL)$ where $G^D$ is the Cartier dual of $G$ (i.e., the affine group scheme represented by the group algebra $\FF G$, so $G^D(\cS)=\Hom(G,\cS^\times)$ for every $\cS$ in $\Alg_\FF$) and $\eta$ is defined in Equation \eqref{eq:def_eta} below --- the details can be found in e.g. \cite[\S 1.4]{EKmon}. The image $\pi\eta(G^D)$ is an abelian subgroupscheme of $\mathbf{S_3}$. Since the subgroupschemes of a constant group scheme correspond to subgroups, here we have three possibilities: the image has order $1$, $2$ or $3$. The grading $\Gamma$ will be said to have {\em Type I, II or III}, respectively. (We use a capital letter here to distinguish from the other meaning of {\em type} common in the literature on gradings, which refers to the sequence of integers that records the number of homogeneous components of each dimension.) The subgroupscheme $\eta^{-1}(\eta(G^D)\cap\PGOs^+_8)$ of $G^D$ corresponds to a subgroup $H$ of $G$ of order $1$, $2$ or $3$, respectively. We will refer to $H$ as the {\em distinguished subgroup} of $\Gamma$. It is the smallest subgroup of $G$ such that the induced $G/H$-grading is of Type I.
If $\Gamma$ is of Type I then the image $\eta(G^D)$ lies in $\PGOs^+_8$, which is a subgroupscheme of index $2$ in $\PGOs_8=\AAut_\FF(\cR,\sigma)$. In the case of Type II, applying an outer automorphism of order $3$ if necessary, we may assume that $\eta(G^D)$ lies in $\AAut_\FF(\cR,\sigma)$. Therefore, any $G$-grading $\Gamma$ on $\cL$ of Type I or II is isomorphic to the restriction of a $G$-grading $\Gamma'$ on the algebra with involution $(\cR,\sigma)$. For this reason, we will sometimes collectively refer to gradings of Type I and II as {\em matrix gradings} (respectively, ``inner'' and ``outer''). They were classified in \cite{BK10} (see also \cite[\S 2.4]{EKmon}) up to {\em matrix isomorphism}, i.e., up to the action of $\Aut_\FF(\cR,\sigma)$. Note that there is a subtlety here: one isomorphism class of Type I gradings on $\cL$ may correspond to $1$, $2$ or $3$ isomorphism classes of gradings on $(\cR,\sigma)$ --- see Section \ref{s:Type_I}. No such difficulty arises for Type II gradings.
Our main concern in this paper are Type III gradings on $\cL$. We note that, since $\pi\eta(G^D)$ is a diagonalizable subgroupscheme, the group algebra of the corresponding subgroup of $S_3$ must be semisimple (being isomorphic to the Hopf dual of the group algebra of a finite subgroup of $G$). If $\chr\FF=3$ then the group algebra of the cyclic group of order $3$ is not semisimple (in other words, the corresponding constant group scheme is not diagonalizable), hence {\em Type III gradings do not occur in characteristic~$3$}.
An essential ingredient in our approach is the concept of {\em trialitarian algebra} introduced in \cite[Chapter X]{KMRT}, which is a central simple associative algebra over a cubic \'{e}tale algebra $\LL$ equipped with an orthogonal involution and some additional structure. The general definition is quite involved; here we will only need a special case of trialitarian algebras arising as endomorphisms of so-called {\em cyclic composition algebras} (see Definitions \ref{df:cyclic_comp} and \ref{df:tri_alg}). As shown in \cite[\S 45.C]{KMRT}, over any field of characteristic different from $2$, any simple Lie algebra $\cL$ of type $D_4$ is isomorphic to a canonically defined Lie subalgebra of a unique (up to isomorphism) trialitarian algebra $E$.
Moreover, the restriction $\AAut_\FF(E)\to\AAut_\FF(\cL)$ is an isomorphism. This means, in particular, that any $G$-grading on $\cL$ is the restriction of a unique $G$-grading on $E$, hence $\cL$ and $E$ have the same classifications of gradings.
The structure of the paper is the following. In Section~\ref{s:preliminaries}, we introduce the necessary background on gradings and the objects that will be our main tools: composition algebras, cyclic composition algebras and trialitarian algebras. Section~\ref{s:Type_I} discusses Type I gradings, especially the facts that are relevant to the above-mentioned subtlety in their classification and will be crucial in Section~\ref{s:lifting}, where we show how to reduce the classification of Type III gradings on $\cL$ (or, equivalently, on the trialitarian algebra $E$) to the corresponding cyclic composition algebra. A description of Type III gradings on the latter is given in Section~\ref{s:Type_III_fine}, from where we derive the classification of fine gradings on $\cL$ up to equivalence, under the assumption $\chr\FF\ne 2$, thus extending the result known for characteristic $0$. In Section~\ref{s:Type_III}, we obtain the classification of Type III gradings by a fixed group $G$ up to isomorphism, which is new even for characteristic $0$. (If the reader is only interested in characteristic $0$, then there is no need to deal with affine group schemes: it is sufficient to consider the algebraic groups $\Aut_\FF(\cU)$ and $\wh{G}=\Hom(G,\FF^\times)$, which are the $\FF$-points of the schemes $\AAut_\FF(\cU)$ and $G^D$, respectively.) An appendix on graded modules is included with a twofold purpose: to complete the results in \cite{EK14} on graded modules for the classical simple Lie algebras by including Type III gradings for $D_4$, and to show the analogy of the Brauer invariants introduced in \cite{EK14} with Tits algebras.
\section{Preliminaries}\label{s:preliminaries}
\subsection{Group gradings on algebras}
Let $\cU$ be an algebra (not necessarily associative) over a field $\FF$ and let $G$ be a group (written multiplicatively).
\begin{df}\label{df:G_graded_alg}
A {\em $G$-grading} on $\cU$ is a vector space decomposition
\[
\Gamma:\;\cU=\bigoplus_{g\in G} \cU_g
\]
such that
$
\cU_g \cU_h\subset \cU_{gh}\quad\mbox{for all}\quad g,h\in G.
$
If such a decomposition is fixed, $\cU$ is referred to as a {\em $G$-graded algebra}.
The nonzero elements $x\in\cU_g$ are said to be {\em homogeneous of degree $g$}, and one writes $\deg_\Gamma x=g$ or just $\deg x=g$ if the grading is clear from the context. The {\em support} of $\Gamma$ is the set $\supp\Gamma\bydef\{g\in G\;|\;\cU_g\neq 0\}$.
\end{df}
If $(\cU,\sigma)$ is an algebra with involution, then we will always assume $\sigma(\cU_g)=\cU_g$ for all $g\in G$.
There is a more general concept of grading: a decomposition $\Gamma:\;\cU=\bigoplus_{s\in S}\cU_s$ into nonzero subspaces indexed by a set $S$ and having the property that, for any $s_1,s_2\in S$ with $\cU_{s_1}\cU_{s_2}\ne 0$, there exists (unique) $s_3\in S$ such that $\cU_{s_1}\cU_{s_2}\subset\cU_{s_3}$. For such a decomposition $\Gamma$, there may or may not exist a group $G$ containing $S$ that makes $\Gamma$ a $G$-grading. If such a group exists, $\Gamma$ is said to be a {\em group grading}. However, $G$ is usually not unique even if we require that it should be generated by $S$. The {\em universal grading group} is generated by $S$ and has the defining relations $s_1s_2=s_3$ for all $s_1,s_2,s_3\in S$ such that $0\ne\cU_{s_1}\cU_{s_2}\subset\cU_{s_3}$ (see e.g. \cite[Chapter 1]{EKmon} for details).
Here we will deal exclusively with abelian groups, and we will sometimes write them additively. Gradings by abelian groups often arise as eigenspace decompositions with respect to a family of commuting diagonalizable automorphisms. If $\FF$ is algebraically closed and $\chr\FF=0$ then all abelian group gradings on finite-dimensional algebras can be obtained in this way. Over an arbitrary field, a $G$-grading $\Gamma$ on $\cU$ is equivalent to a morphism of affine group schemes $\eta_\Gamma:G^D\to\AAut_\FF(\cU)$ as follows: for any $\cR\in\Alg_\FF$, the corresponding homomorphism of groups
$(\eta_\Gamma)_\cR:\Alg_\FF(\FF G,\cR)\to\Aut_\cR(\cU\ot\cR)$ is defined by
\begin{equation}\label{eq:def_eta}
(\eta_\Gamma)_\cR(f)(x\ot r)=x\ot f(g)r\;\text{for all}\;x\in\cU_g, g\in G, r\in\cR, f\in\Alg_\FF(\FF G,\cR).
\end{equation}
Consequently, if we have two algebras, $\cU$ and $\cV$, and a morphism $\theta\colon\AAut_\FF(\cU)\to\AAut_\FF(\cV)$ then any $G$-grading $\Gamma$ on $\cU$ gives rise to a $G$-grading $\theta(\Gamma)$ on $\cV$ by setting $\eta_{\theta(\Gamma)}\bydef\theta\circ\eta_\Gamma$.
Let $\Gamma:\, \cU=\bigoplus_{g\in G} \cU_g$ and $\Gamma':\,\cU'=\bigoplus_{h\in H} \cU'_h$
be two group gradings, with supports $S$ and $T$, respectively.
We say that $\Gamma$ and $\Gamma'$ are {\em equivalent} if there exists an isomorphism of algebras $\vphi\colon\cU\to\cU'$ and a bijection $\alpha\colon S\to T$ such that $\varphi(\cU_s)=\cU'_{\alpha(s)}$ for all $s\in S$. If $G$ and $H$ are universal grading groups then $\alpha$ extends to an isomorphism $G\to H$. In the case $G=H$, the
$G$-gradings $\Gamma$ and $\Gamma'$ are {\em isomorphic} if $\cU$ and $\cU'$ are isomorphic as $G$-graded algebras, i.e., if there exists an isomorphism of algebras $\vphi\colon\cU\to\cU'$ such that $\varphi(\cU_g)=\cU'_g$ for all $g\in G$. Note that $\theta\colon\AAut_\FF(\cU)\to\AAut_\FF(\cV)$ sends isomorphic gradings on $\cU$ to isomorphic gradings on $\cV$.
If $\Gamma:\,\cU=\bigoplus_{g\in G} \cU_g$ and $\Gamma':\,\cU=\bigoplus_{h\in H} \cU'_h$ are two gradings on the same algebra, with supports $S$ and $T$, respectively, then we will say that $\Gamma'$ is a {\em refinement} of $\Gamma$ (or $\Gamma$ is a {\em coarsening} of $\Gamma'$) if for any $t\in T$ there exists (unique) $s\in S$ such that $\cU'_t\subset\cU_s$. If, moreover, $\cU'_t\ne\cU_s$ for at least one $t\in T$, then the refinement is said to be {\em proper}. A grading $\Gamma$ is said to be {\em fine} if it does not admit any proper refinement.
Given a $G$-grading $\Gamma:\,\cU=\bigoplus_{g\in G} \cU_g$, any group homomorphism $\alpha\colon G\to H$ induces an $H$-grading ${}^\alpha\Gamma$ on $\cU$ whose homogeneous component of degree $h$ is the sum of all $\cU_g$ with $\alpha(g)=h$. Note that $\eta_{{}^\alpha\Gamma}=\eta_\Gamma\circ\alpha^D$ and hence $\theta({}^\alpha\Gamma)={}^\alpha\theta(\Gamma)$. Clearly, ${}^\alpha\Gamma$ is a coarsening of $\Gamma$ (not necessarily proper). If $G$ is the universal group of $\Gamma$ then every coarsening of $\Gamma$ is obtained in this way. If $\Gamma$ and $\Gamma'$ are two gradings, with universal groups $G$ and $H$, then $\Gamma'$ is equivalent to $\Gamma$ if and only if $\Gamma'$ is isomorphic to ${}^\alpha\Gamma$ for some group isomorphism $\alpha\colon G\to H$. It follows that, if universal groups are used, $\theta$ preserves equivalence of gradings.
\subsection{Graded division algebras over algebraically closed fields}
Let $G$ be a group. If $\cR$ is an associative algebra with a $G$-grading such that $\cR$ is graded simple and satisfies the descending chain condition on graded left ideals then, by the graded version of a classical result, there exists a graded division algebra $\cD$ over $\FF$ and a finite-dimensional graded right vector space $W$ over $\cD$ such that $\cR\cong\End_\cD(W)$ as a $G$-graded algebra (\cite[Theorem 2.6]{EKmon}). Here by a {\em graded division algebra} we mean a unital associative algebra with a $G$-grading such that every nonzero homogeneous element is invertible. A {\em graded vector space} over $\cD$ is just a graded $\cD$-module (in the obvious sense), which is automatically free.
Now we collect for future use some general facts about finite-dimensional graded division algebras in the following situation: $\FF$ is algebraically closed and $G$ is abelian.
Let $\cD$ be a graded division algebra and let $T$ be the support of the grading on $\cD$. Then $T$ is a subgroup of $G$ and $\cD$ can be identified with a twisted group algebra $\FF^\tau T$ for some $2$-cocycle $\tau\colon T\times T\to\FF^\times$. Indeed, the identity component $\cD_e$ is a division algebra over $\FF$, so $\cD_e=\FF$ and hence all nonzero homogeneous components have dimension $1$. We fix a basis $X_t$ in each of them and write $X_s X_t=\tau(s,t)X_{st}$ ($s,t\in T$). Define $\beta=\beta_\tau$ by $\beta(s,t)\bydef\frac{\tau(s,t)}{\tau(t,s)}$. This is an alternating bicharacter $T\times T\to\FF^\times$, independent of the scaling of the $X_t$ since $X_s X_t=\beta(s,t)X_t X_s$.
The pair $(T,\beta)$ determines $\cD$ up to isomorphism of graded algebras. Indeed, the graded division algebras with support $T$ are classified by the cohomology class $[\tau]\in \mathrm{H}^2(T,\FF^\times)$ (see e.g. \cite[Theorem 2.13]{EKmon}), and we can use a standard cohomological result: the quotient of the group of symmetric $2$-cocycles $\mathrm{Z}^2_{sym}$ by the $2$-coboundaries $\mathrm{B}^2$ can be identified with $\mathrm{Ext}(T,\FF^\times)$ in the category of abelian groups, but $\FF^\times$ is a divisible group, so $\mathrm{Ext}(T,\FF^\times)$ is trivial, i.e., $\mathrm{B}^2=\mathrm{Z}^2_{sym}$. Since the mapping $\tau\mapsto\beta_\tau$ from $\mathrm{Z}^2$ to the group of alternating bicharacters, $\Hom(T\wedge T,\FF^\times)$, is a homomorphism with kernel $\mathrm{Z}^2_{sym}$, we obtain an injection $[\tau]\mapsto\beta_\tau$ from $\mathrm{H}^2(T,\FF^\times)$ to $\Hom(T\wedge T,\FF^\times)$ (in fact, an isomorphism, see \cite{Yamazaki}).
We will need the $G$-graded Brauer group of $\FF$, which we will denote by $B_G(\FF)$. There are several versions of Brauer group associated to a field (or, more generally, a commutative ring) $\FF$ and an abelian group $G$. They consist of equivalence classes of certain $\FF$-algebras equipped with a $G$-grading, a $G$-action or both. The multiplication is induced by tensor product of algebras or its twisted version. The Brauer group we need here is the one defined in \cite{PP}, where there is only a $G$-grading and the tensor product is not twisted. For a field $\FF$ and an abelian group $G$, the group $B_G(\FF)$ consists of the equivalence classes of finite-dimensional central simple associative $\FF$-algebras equipped with a $G$-grading, where $\cA_1\sim\cA_2$ if and only if there exist finite-dimensional $G$-graded $\FF$-vector spaces $V_1$ and $V_2$ such that $\cA_1\otimes\End(V_1)\cong\cA_2\otimes\End(V_2)$ as graded algebras. Every class contains a unique graded division algebra (up to isomorphism). The classical Brauer group $B(\FF)$ is imbedded in $B_G(\FF)$ as the classes containing a division algebra with trivial $G$-grading. As shown in \cite{PP}, if $G$ is finite and $\FF$ contains enough roots of unity (so that $|\wh{G}|=|G|$) then there is a split short exact sequence
$
1\to B(\FF)\to B_G(\FF)\to \mathrm{H}^2(\wh{G},\FF^\times)\to 1
$. If $\FF$ is algebraically closed (as we will assume in this paper) then, for any abelian group $G$, the Brauer group $B_G(\FF)$ is isomorphic to the group of alternating continuous bicharacters of the pro-finite group $\wh{G_0}$ where $G_0$ is the torsion subgroup of $G$ if $\chr{\FF}=0$ and the $p'$-torsion subgroup of $G$ if $\chr{\FF}=p>0$ (i.e., the set of all elements whose order is finite and coprime with $p$). Namely, the class of a $G$-graded matrix algebra corresponds to its ``commutation factor'' $\hat{\beta}$, which can be seen as an alternating bicharacter $\wh{G}\times \wh{G}\to\FF^\times$ (by means of the canonical homomorphism $\wh{G}\to\wh{G_0}$) and determines the parameters $(T,\beta)$ of the graded division algebra representing the class (see \cite[\S 2]{EK14}). Note that in this case $\beta$ is nondegenerate since the graded division algebra is central.
Let $\cK$ be the center of a graded division algebra $\cD$, so $\cK$ is a graded field over $\FF$. Let $H\subset T$ be the support of the grading on $\cK$. The following result holds for arbitrary $\FF$.
\begin{lemma}\label{lm:not_divisible}
Let $\cK$ be a graded field that is finite-dimensional over its identity homogeneous component $\FF$. If $\cK$ is separable as an algebra over $\FF$ then $\dim_\FF\cK$ is not divisible by $\chr{\FF}$.
\end{lemma}
\begin{proof}
Assume, to the contrary, that $\chr{\FF}=p>0$ and $p$ divides $\dim_\FF\cK=|H|$. Then $H$ contains an element $g$ of order $p$. Pick a nonzero element $x\in\cK_g$. Then $x^p\in\cK_e$ and the powers $1, x, x^2,\ldots,x^{p-1}$ are linearly independent over $\FF$, hence the minimal polynomial of $x$ has the form $X^p-\lambda$, $\lambda\in\FF$, which contradicts separability.
\end{proof}
Let $\wb{G}=G/H$ and $\wb{T}=T/H$. Since $H$ is precisely the radical of the alternating bicharacter $\beta$, we obtain a nondegenerate alternating bicharacter $\bar{\beta}\colon\wb{T}\times\wb{T}\to\FF^\times$, hence $|\wb{T}|$ is not divisible by $\chr{\FF}$. Assume $\cD$ is semisimple as an ungraded algebra. Then $\cK$ is the direct product of $k$ copies of $\FF$ where, by Lemma \ref{lm:not_divisible}, $k=|H|$ is not divisible by $\chr{\FF}$.
Accordingly, we can write $\cD=\cD_1\times\cdots\times\cD_k$ where $\cD_i$ are the ideals generated by the minimal central idempotents of $\cD$, so they are simple and $\wb{G}$-graded. Since $\chr{\FF}$ does not divide $|H|$, the $H$-grading on $\cK$ is equivalent to an action of the group of characters $\wh{H}=\{\chi_1,\ldots,\chi_k\}$, which must permute the factors of $\cK$ transitively since $\cK$ is a graded field. Relabeling if necessary, we may assume that $\chi_i$ sends the first factor to the $i$-th one. We can extend the characters $\chi_i$ to $T$ in some way. Then each $\chi_i$ acts as an automorphism of $\cD$ that preserves the $G$-grading on $\cD$, hence $\chi_i$ yields an isomorphism of $\wb{G}$-graded algebras $\cD_1\to\cD_i$.
Since the identity component of each $\cD_i$ (with respect to the $\wb{G}$-grading) has dimension $1$, they are graded division algebras (see e.g. \cite[Lemma 2.20]{EKmon}). Therefore, we can regard them as elements of the $\wb{G}$-graded Brauer group $B_{\wb{G}}(\FF)$.
\begin{proposition}\label{prop:beta_bar}
Let $\cD$ be a graded division algebra over an algebraically closed field $\FF$ of arbitrary characteristic. Assume that $\cD$ is finite-dimensional and semisimple as an algebra, decomposing into simple components $\cD_1\times\cdots\times\cD_k$. Define $\wb{G}$, $\wb{T}$ and $\bar{\beta}$ as above. Then the $\cD_i$ are $\wb{G}$-graded division algebras, all isomorphic to each other, and the $\wb{G}$-graded Brauer class of $\cD_i$ corresponds to the pair $(\wb{T},\bar{\beta})$.
\end{proposition}
\begin{proof}
The only statement that has not yet been proved is the one about the Brauer class. Indeed, $\beta$ is defined by the equation $X_sX_t=\beta(s,t)X_tX_s$ for all $s,t\in T$. Since the value of $\beta$ depends only on the cosets of $s$ and $t$ in $\wb{T}$, in the $\wb{G}$-grading we have $xy=\bar{\beta}(\bar{s},\bar{t})yx$ for all $x\in\cD_{\bar{s}}$ and $y\in\cD_{\bar{t}}$. Taking the projection onto $\cD_i$, we see that $\bar{\beta}$ satisfies the same property with respect to the $\wb{G}$-grading of $\cD_i$ as $\beta$ with respect to the $G$-grading of $\cD$.
\end{proof}
\subsection{Composition algebras}
Recall that a (finite-dimensional) {\em composition algebra} is a nonassociative algebra $\cA$ with a nonsingular quadratic form $n$ such that $n(xy)=n(x)n(y)$ for all $x,y\in\cA$. It is known that $\dim\cA$ can be $1$, $2$, $4$ or $8$. The unital composition algebras are called {\em Hurwitz algebras} and can be obtained using the Cayley--Dickson doubling process. The ones of dimension $8$ are called {\em octonion}, or {\em Cayley algebras}. If the ground field is algebraically closed then, up to isomorphism, there is only one Hurwitz algebra in each dimension.
Here we will also need another kind of composition algebras.
\begin{df}\label{df:symmetric}
A composition algebra $\cS$, with multiplication $\star$ and norm $n$, is said to be \emph{symmetric} if the polar form of the norm, $n(x,y)\bydef n(x+y)-n(x)-n(y)$, is associative:
\[
n(x\star y,z)=n(x,y\star z),
\]
for all $x,y,z\in\cS$.
\end{df}
As a consequence of this definition, $\cS$ satisfies the following identities:
\[
(x\star y)\star x=n(x)y=x\star(y\star x).
\]
Over an algebraically closed field, there are, up to isomorphism, only two symmetric composition algebras of dimension $8$: the para-Cayley and the Okubo algebras (see \cite{EPI96} or \cite[Theorem 4.44]{EKmon}). Both can be obtained from the Cayley algebra $\cC$ by introducing a new product: in the first case, $x\star y=x\bullet y\bydef \bar{x}\bar{y}$, where $\bar{x}\bydef n(x,1)1-x$ is the standard involution, and in the second case, $x\star y=\tau(x)\bullet\tau^2(y)$, where $\tau$ is a certain automorphism of order $3$. If $\chr\FF\ne 3$ then the Okubo algebra can also be realized as the space of traceless $3\times 3$ matrices with a certain product (see e.g. \cite[\S 4.6]{EKmon} for details).
From now on, we assume $\chr{\FF}\ne 2$.
\begin{df}\label{df:tri}
Let $\cS$ be a symmetric composition algebra of dimension $8$. Its \emph{triality Lie algebra} is defined as
\[
\tri(\cS,\star,n)=\{(d_1,d_2,d_3)\in\frso(\cS,n)^3\;|\; d_1(x\star y)=d_2(x)\star y+x\star d_3(y)\ \forall x,y\in\cS\}.
\]
This is a Lie algebra with componentwise multiplication.
\end{df}
It turns out that this definition is symmetric with respect to cyclic permutations of $(d_1,d_2,d_3)$, and each projection determines an isomorphism $\tri(\cS)\to\frso(\cS,n)$, so $\tri(\cS)$ is a Lie algebra of type $D_4$ (see e.g. \cite[\S 5.5, \S 6.1]{EKmon} or \cite[\S 45.A]{KMRT}, but note that in the latter the ordering of triples differs from ours).
This fact is known as the ``local triality principle''. There is also a ``global triality principle'', as follows.
\begin{df}\label{df:Tri}
Let $\cS$ be a symmetric composition algebra of dimension $8$. Its \emph{triality group} is defined as
\[
\TRI(\cS,\star,n)=\{(f_1,f_2,f_3)\in\Ort(\cS,n)^3\;|\; f_1(x\star y)=f_2(x)\star f_3(y)\ \forall x,y\in\cS\}.
\]
This is an algebraic group with componentwise multiplication.
\end{df}
It turns out that this definition is symmetric with respect to cyclic permutations of $(f_1,f_2,f_3)$, and $\TRI(\cS)$ is isomorphic to $\Spin(\cS,n)$. In fact, this isomorphism can be defined at the level of the corresponding group schemes (see \cite[\S 35.C]{KMRT} for details). The said cyclic permutations determine outer actions of $A_3$ on $\Spin(\cS,n)$ and its Lie algebra $\frso(\cS,n)$. If $\cS$ is a para-Cayley algebra then one can define an outer action of $S_3$ using $(f_1,f_2,f_3)\mapsto(\bar{f}_1,\bar{f}_3,\bar{f}_2)$ and $(d_1,d_2,d_3)\mapsto(\bar{d}_1,\bar{d}_3,\bar{d}_2)$ as the action of the transposition $(2,3)$, where $\bar{f}$ is defined by $\bar{f}(\bar{x})=\overline{f(x)}$. This allows us to split the exact sequence \eqref{eq:exact_D4}.
\subsection{Cyclic composition algebras}
A convenient way to ``package'' triples of maps as above is the following concept due to Springer. Let $\LL$ be a Galois algebra over $\FF$ with respect to the cyclic group of order $3$ (see e.g. \cite[\S 18.B]{KMRT}). Fix a generator $\rho$ of this group. For any $\ell\in\LL$, define the norm $N(\ell)=\ell\rho(\ell)\rho^2(\ell)$, the trace $T(\ell)=\ell+\rho(\ell)+\rho^2(\ell)$, and the adjoint $\ell^\#=\rho(\ell)\rho^2(\ell)$. Our main interest is in the case $\LL=\FF\times\FF\times\FF$ where $\rho(\ell_1,\ell_2,\ell_3)=(\ell_2,\ell_3,\ell_1)$.
\begin{df}\label{df:cyclic_comp}
A {\em cyclic composition algebra} over $(\LL,\rho)$ is a free $\LL$-module $V$ with a nonsingular $\LL$-valued quadratic form $Q$ and an $\FF$-bilinear multiplication $(x,y)\mapsto x*y$ that is $\rho$-semilinear in $x$ and $\rho^2$-semilinear in $y$ and satisfies the following identities:
\begin{align*}
Q(x*y)&=\rho(Q(x))\rho^2(Q(y)),\\
b_Q(x*y,z)&=\rho(b_Q(y*z,x))=\rho^2(b_Q(z*x,y)),
\end{align*}
where $b_Q(x,y)\bydef Q(x+y)-Q(x)-Q(y)$ is the polar form of $Q$.
An {\em isomorphism} from $(V,\LL,\rho,*,Q)$ to $(V',\LL',\rho',*',Q')$ is a pair of $\FF$-linear isomorphisms
$\varphi_0\colon(\LL,\rho)\to(\LL',\rho')$ (i.e., $\varphi_0$ is an isomorphism that satisfies $\varphi_0\rho=\rho'\varphi_0$) and $\varphi_1\colon V\to V'$ such that $\varphi_1$ is $\varphi_0$-semilinear, $\varphi_1(x*y)=\varphi_1(x)*'\varphi_1(y)$ and $\varphi_0(Q(x))=Q'(\varphi_1(x))$ for all $x,y\in V$.
\end{df}
As a consequence, $V$ also satisfies
\begin{equation}\label{eq:cyclic_comp_id}
(x*y)*x=\rho^2(Q(x))y\quad\text{and}\quad x*(y*x)=\rho(Q(x))y.
\end{equation}
One checks that, for $\lambda,\mu\in\LL^\times$, the new product $x\mathbin{\tilde{*}}y=\lambda(x*y)$ and the new quadratic form $\tilde{Q}(x)=\mu Q(x)$ define a cyclic composition algebra if and only if $\mu=\lambda^\#$ (\cite[Lemma 36.1]{KMRT}). We will say that an isomorphism $(V,\LL,\rho,\tilde{*},\tilde{Q})\to(V',\LL',\rho',*',Q')$ is a {\em similitude} from $(V,\LL,\rho,*,Q)$ to $(V',\LL',\rho',*',Q')$ with {\em parameter} $\lambda$ and {\em multiplier} $\lambda^\#$. In particular, for any $\ell\in\LL^\times$, the mappings $\varphi_0=\id$ and $\varphi_1(x)=\ell x$ define a similitude from $(V,\LL,\rho,*,Q)$ to itself with parameter $\ell^{-1}\ell^\#$ and multiplier $\ell^2$.
If $(\cS,\star,n)$ is a symmetric composition algebra then $\cS\ot\LL$ becomes a cyclic composition algebra with $Q(x\ot\ell)=n(x)\ell^2$ (extended to sums in the obvious way using the polar form of $n$) and $(x\ot\ell)*(y\ot m)=(x\star y)\ot\rho(\ell)\rho^2(m)$. With $\LL=\FF\times\FF\times\FF$ and $\rho(\ell_1,\ell_2,\ell_3)=(\ell_2,\ell_3,\ell_1)$, this gives $V=\cS\times\cS\times\cS$ with $Q=(n,n,n)$ and
\begin{equation}\label{df:cyclic_prod}
(x_1,x_2,x_3)*(y_1,y_2,y_3)=(x_2\star y_3,x_3\star y_1,x_1\star y_2).
\end{equation}
It turns out that, with $\LL$ and $\rho$ as above, any cyclic composition algebra is similar to $\cS\ot\LL$ where $\cS$ is a para-Hurwitz algebra (see e.g. \cite[\S 36, \S 36B]{KMRT}). Hence, for any $\LL$, the $\LL$-rank of a cyclic composition algebra can be $1$, $2$, $4$ or $8$. Also, if $\FF$ is algebraically closed then there is only one isomorphism class of cyclic composition algebras in each rank. (In this case, similar cyclic composition algebras are isomorphic because, for any $\lambda\in\LL^\times$, we can find $\ell\in\LL^\times$ such that $\ell^{-1}\ell^\#=\lambda$.)
Assume now that the rank is $8$. The multiplication \eqref{df:cyclic_prod} allows us to interpret $\TRI(\cS,\star,n)$ as $\Aut_\LL(V,*,Q)$, the group of $\LL$-linear automorphisms (i.e., with $\varphi_0=\id$), and $\TRI(\cS,\star,n)\rtimes A_3$ as $\Aut_\FF(V,\LL,\rho,*,Q)$, the group of all automorphisms (see Definition \ref{df:cyclic_comp}). This interpretation can be made at the level of group schemes:
\[
\AAut_\FF(V,\LL,\rho,*,Q)=\TRIs(\cS,\star,n)\rtimes\mathbf{A}_3\cong\Spins(\cS,n)\rtimes\mathbf{A}_3.
\]
Similarly, $\tri(\cS,\star,n)$ can be interpreted as $\Der_\LL(V,*,Q)$.
\begin{remark}
If $\FF$ is algebraically closed then the para-Cayley and Okubo algebras are ``$\FF$-forms'' of the unique cyclic composition algebra $V$ of rank $8$ (in the sense that, for $\cS$ either the para-Cayley or Okubo algebra, we have $V\cong\cS\ot\LL$ as an $\LL$-module with a quadratic form, while the product is extended by semilinearity). Any outer automorphism of order $3$ defines such an ``$\FF$-form'' of $V$, which must be isomorphic either to the para-Cayley or the Okubo algebra; this explains the fact that there are $2$ conjugacy classes of outer automorphisms of the algebraic group $\Spin_8(\FF)$.
\end{remark}
We will later need the group scheme of similitudes $\SIMs_\FF(V,\LL,\rho,*,Q)$, which is defined as follows.
Let $\Gs$ be the group scheme of invertible $\FF$-linear endomorphisms of $V$ that are semilinear with respect to some automorphism of $(\LL,\rho)$. This is a subgroupscheme of $\GLs_\FF(V)\times\AAut_\FF(\LL,\rho)$ whose projection onto the second factor can be split by choosing an ``$\FF$-form'' of $V$, thus identifying $\Gs$ with $\GLs_\LL(V)\rtimes\AAut_\FF(\LL,\rho)$.
\begin{remark}
Here $\GLs_\LL(V)$ is regarded as a group scheme over $\FF$, sending $\cR\in\Alg_\FF$ to the group of invertible elements in $\End_{\LL\ot\cR}(V\ot\cR)$. One could define $\GLs_\LL(V)$ as a group scheme over $\LL$, sending $\cR'\in\Alg_\LL$ to the group of invertible elements in $\End_{\cR'}(V\ot_\LL\cR')$. Our $\GLs_\LL(V)$ is the $\LL/\FF$-corestriction of this latter (i.e., the group scheme over $\FF$ obtained by substituting $\LL\ot\cR$ for $\cR'$). In particular, we will regard the multiplicative group scheme of $\LL$ as a group scheme over $\FF$ and will denote it by $\GLs_1(\LL)$.
\end{remark}
Now, we have natural representations of $\Gs$ in the $\FF$-linear space of $\LL$-bilinear functions $V\times V\to\LL$ and in the $\FF$-linear space of $\FF$-bilinear functions $V\times V\to V$ that are $\rho$-semilinear in the first variable and $\rho^2$-semilinear in the second variable. Then $\AAut_\FF(V,\LL,\rho,*,Q)$ is the intersection of the stabilizers in $\Gs$ of the vectors $b_Q$ and $*$. We define $\SIMs_\FF(V,\LL,\rho,*,Q)$ as the intersection of the stabilizers of the $\LL$-submodules spanned by each of $b_Q$ and $*$.
Using $\FF b_Q$ as an $\FF$-form of the free $\LL$-module $\LL b_Q$, we identify the group scheme of semilinear endomorphisms of $\LL b_Q$ with $\GLs_\LL(\LL b_Q)\rtimes\AAut_\FF(\LL,\rho)=\GLs_1(\LL)\rtimes\mathbf{A}_3$, and similarly for $\LL *$. Thus the representations of $\SIMs_\FF(V,\LL,\rho,*,Q)$ on $\LL *$ and on $\LL b_Q$ give rise to morphisms $\SIMs_\FF(V,\LL,\rho,*,Q)\to\GLs_1(\LL)\rtimes\mathbf{A}_3$, which we denote by $\theta$ and $\theta'$, respectively. We will now obtain a relation between these two morphisms, to be used later. For any $\cR\in\Alg_\FF$, we extend the action of $\rho$ from $\LL$ to $\LL\ot\cR$ by $\cR$-linearity and define $N$ and $\#$ for $\LL\ot\cR$ by the same formulas as before. Hence we obtain an automorphism of the group scheme $\GLs_1(\LL)$ defined by $a\mapsto a^\#$ for all $a\in(\LL\ot\cR)^\times$, $\cR\in\Alg_\FF$. Together with the identity on $\mathbf{A}_3$, this yields an automorphism of $\GLs_1(\LL)\rtimes\mathbf{A}_3$, which we will denote by $(\#,\id)$.
\begin{lemma}\label{lm:lambda_mu}
We have $\theta'=(\#,\id)\theta$.
\end{lemma}
\begin{proof}
We fix $\cR\in\Alg_\FF$, denote $\LL_\cR=\LL\ot\cR$ and $V_\cR=V\ot\cR$, and extend $b_Q$ and $*$ to $V_\cR$ by $\cR$-linearity. Let $(\varphi_1,\varphi_0)$ be an element of $\SIMs_\FF(V,\LL,\rho,*,Q)(\cR)\subset\Gs(\cR)$, so $\varphi_1$ is an invertible $\varphi_0$-semilinear endomorphism of the $\LL_\cR$-module $V_\cR$. For any $a\in\LL_\cR$, the action of $(\varphi_1,\varphi_0)$ sends $ab_Q$ to the bilinear form $V_\cR\times V_\cR\to\LL_\cR$ defined by $(x,y)\mapsto\varphi_0(aQ(\varphi_1^{-1}(x),\varphi_1^{-1}(y)))$, which must be an element of $\LL_\cR b_Q$. We see that this mapping on $\LL_\cR b_Q$ is $\varphi_0$-semilinear and hence $\theta'_\cR(\varphi_1,\varphi_0)=(\mu,\varphi_0)\in (\LL_\cR)^\times\rtimes\Aut_\cR(\LL_\cR,\rho)$, where $\mu b_Q$ is the image of $b_Q$ under the action. Similarly, $\theta_\cR(\varphi_1,\varphi_0)=(\lambda,\varphi_0)$, where $\lambda *$ is the image of $*$. By construction, $(\varphi_1,\varphi_0)$ is an isomorphism $(V_\cR,\LL_\cR,\rho,*,Q)\to(V_\cR,\LL_\cR,\rho,\tilde{*},\tilde{Q})$ where $\tilde{Q}(x)=\mu Q(x)$ and $x\mathbin{\tilde{*}}y=\lambda(x*y)$.
It follows that $\mu=\lambda^\#$.
\end{proof}
\subsection{The trialitarian algebra $\End_\LL(V)$}
Let $V$ be a cyclic composition algebra over $(\LL,\rho)$ of rank $8$. Consider $E=\End_\LL(V)$. This is a central separable associative algebra over $\LL$. We will denote by $\sigma$ the involution of $E$ determined by the quadratic form $Q$. The even Clifford algebra $\Cl_0(V,Q)$ can be defined purely in terms of $(E,\sigma)$ as the quotient $\Cl(E,\sigma)$ of the tensor algebra of $E$ (regarded as an $\LL$-module) by certain relations --- see \cite[\S 8.B]{KMRT}, where the construction is carried out for a central simple algebra over a field, but we can apply it to each simple factor of $E$ if $\LL=\FF\times\FF\times\FF$. The imbedding of $E$ into its tensor algebra yields a canonical $\LL$-linear map $\kappa\colon E\to\Cl(E,\sigma)$, whose image generates $\Cl(E,\sigma)$, but which is neither injective nor a homomorphism of algebras. This construction has the advantage of being functorial for isomorphisms of algebras with involution. Thus, $\sigma$ determines a unique involution $\ul{\sigma}$ on $\Cl(E,\sigma)$ such that $\kappa\sigma=\ul{\sigma}\kappa$. Also, any action (respectively, grading) by a group on the algebra with involution $(E,\sigma)$ gives rise to a unique action (respectively, grading) on the algebra with involution $(\Cl(E,\sigma),\ul{\sigma})$ such that $\kappa$ is equivariant (respectively, preserves the degree). In fact, we can define a morphism of group schemes $\AAut_\FF(E,\sigma)\to\AAut_\FF(\Cl(E,\sigma),\ul{\sigma})$. There is an isomorphism $\Cl_0(V,Q)\to\Cl(E,\sigma)$ sending, for any $x,y\in V$, the element $x\cdot y\in\Cl_0(V,Q)$ to the image of the operator $z\mapsto xb_Q(y,z)$ under $\kappa$.
\begin{df}\label{df:tri_alg}
The multiplication $*$ of $V$ allows us to define an additional structure on $E$, namely, an isomorphism of $\LL$-algebras with involution:
\[
\alpha\colon\Cl(E,\sigma)\stackrel{\sim}{\to}\,{}^{\rho} E\times{}^{\rho^2}E,
\]
where the superscripts denote the twist of scalar multiplication (i.e., ${}^\rho E$ is $E$ as an $\FF$-algebra with involution, but with the new $\LL$-module structure defined by $\ell\cdot a=\rho(\ell)a$). This is done using the Clifford algebra $\Cl(V,Q)$ as follows. Identities \eqref{eq:cyclic_comp_id} imply that the mapping
\[
x\mapsto\begin{pmatrix}0 & l_x \\ r_x & 0\end{pmatrix}\in\End_{\LL}({}^\rho V\oplus{}^{\rho^2}V),\;x\in V,
\]
where $l_x(y)\bydef x*y\defby r_y(x)$, extends to an isomorphism of $\ZZ_2$-graded algebras with involution
\[
\alpha_V\colon(\Cl(V,Q),\tau)\stackrel{\sim}{\to}(\End_{\LL}({}^\rho V\oplus{}^{\rho^2}V),\tilde\sigma),
\]
where $\tau$ is the standard involution of the Clifford algebra and $\tilde\sigma$ is induced by the quadratic form $({}^\rho Q,{}^{\rho^2}Q)$ on ${}^\rho V\oplus{}^{\rho^2}V$, with ${}^\rho Q(x)\bydef\rho^{-1}(Q(x))$. Then $\alpha$ is obtained by restricting $\alpha_V$ to the even part $\Cl_0(V,Q)$ and identifying $\Cl(E,\sigma)$ with $\Cl_0(V,Q)$ as above. Explicitly,
\[
\alpha\colon \kappa\bigl(xb_Q(y,\cdot)\bigr)\mapsto(l_x r_y,r_x l_y),\;x,y\in V.
\]
\end{df}
The structure $(E,\LL,\rho,\sigma,\alpha)$ is an example of {\em trialitarian algebra} (with trivial discriminant) --- the general definition is given in \cite[\S 43.A]{KMRT} but we will not need it here. An {\em isomorphism} $(E,\LL,\rho,\sigma,\alpha)\to(E',\LL',\rho',\sigma',\alpha')$ is defined to be an isomorphism $\varphi\colon (E,\sigma)\to (E',\sigma')$ of $\FF$-algebras with involution such that the following diagram commutes:
\[
\xymatrix{
\Cl(E,\sigma)\ar[r]^\alpha\ar[d]_{\Cl(\varphi)} & {}^{\rho}E\times{}^{\rho^2}E\ar[d]^{\varphi\ot\Delta(\varphi)}\\
\Cl(E',\sigma')\ar[r]^{\alpha'} & {}^{\rho'}E'\times{}^{\rho'^2}E'
}
\]
where we have identified ${}^\rho E\times{}^{\rho^2}E=E\ot(\FF\times\FF)$ as $\FF$-algebras (similarly for $E'$) and $\Delta(\varphi)\in\Aut_\FF(\FF\times\FF)$ is the identity if the restriction $\varphi\colon\LL\to\LL'$ conjugates $\rho$ to $\rho'$ and the flip if it conjugates $\rho$ to the inverse of $\rho'$. Note that this definition does not depend on the identifications of $E$ and $E'$ with algebras of endomorphisms but depends on the choice of ``orientations'' $\rho$ and $\rho'$. Extending scalars to $\cR\in\Alg_\FF$, one defines the automorphism group scheme $\AAut_\FF(E,\LL,\sigma,\alpha)$, which does not depend on the choice of $\rho$. It can be shown (\cite[\S 44]{KMRT}) that, for $V=\cC\ot\LL$ with $\LL=\FF\times\FF\times\FF$ and $(\cC,\bullet,n)$ a para-Cayley algebra, the group scheme $\AAut_\FF(E,\LL,\sigma,\alpha)$ is isomorphic to $\PGOs^+(\cC,n)\rtimes\mathbf{S}_3$. Moreover, the natural morphism $\mathrm{Int}\colon\Gs\to\AAut_\FF(E)$, where $\Gs$ is the group scheme of invertible $\FF$-linear endomorphisms of $V$ that are semilinear over $(\LL,\rho)$ and $\mathrm{Int}_\cR(a)$ is the conjugation by $a\in\Gs(\cR)$ on $E_\cR=\End_{\LL\ot\cR}(V\ot\cR)$, $\cR\in\Alg_\FF$, yields a morphism of short exact sequences:
\[
\xymatrix{
\mathbf{1}\ar[r] & \AAut_\LL(V,\LL,\rho,*,Q)\ar[r]\ar[d] & \AAut_\FF(V,\LL,\rho,*,Q)\ar[r]\ar[d] &
\AAut_\FF(\LL,\rho)\ar[r]\ar[d] & \mathbf{1}\\
\mathbf{1}\ar[r] & \AAut_\LL(E,\LL,\sigma,\alpha)\ar[r] & \AAut_\FF(E,\LL,\sigma,\alpha)\ar[r] &
\AAut_\FF(\LL)\ar[r] & \mathbf{1}
}
\]
where the left vertical arrow is the quotient map $\Spins(\cC,n)\to\PGOs^+(\cC,n)$, the middle vertical arrow has image $\AAut_\FF(E,\LL,\rho,\sigma,\alpha)$ (i.e., the automorphisms that preserve $\rho$), and the right vertical arrow is the injection $\mathbf{A}_3\to\mathbf{S}_3$.
Our interest in the trialitarian algebra $\End_\LL(V)$ comes from the following fundamental connection with Lie algebras of type $D_4$ (see \cite[\S 45]{KMRT}). It turns out that the restriction $\frac12\kappa\colon\mathrm{Skew}(E,\sigma)\to\mathrm{Skew}(\Cl(E,\sigma),\ul{\sigma})$ is an injective homomorphism of Lie algebras over $\LL$, and the $\FF$-subspace
\[
\cL(E,\LL,\rho,\sigma,\alpha)\bydef\{x\in\mathrm{Skew}(E,\sigma)\;|\;\alpha(\kappa(x))=2(x,x)\}
\]
is precisely the Lie subalgebra $\tri(\cC,\bullet,n)=\Der_\LL(V,*,Q)$, which is isomorphic to $\So(\cC,n)$ by the local triality principle. (The equality $\cL(E,\LL,\rho,\sigma,\alpha)=\tri(\cC,\bullet,n)$ follows from the well-known fact that $\tri(\cC,\bullet,n)$ is spanned by the triples of the form $\bigl(xn(y,\cdot)-yn(x,\cdot),\frac12(r_xl_y-r_yl_x),\frac12(l_xr_y-l_yr_x)\bigr)$, $x,y\in\cC$, where $l_x(y)\bydef x\bullet y\defby r_y(x)$ --- see e.g. \cite[\S 5.5]{EKmon}, where this fact is used to prove the local triality principle.) Moreover, the restriction $\mathrm{Res}$ from $E$ to
$\cL(E)\bydef \cL(E,\LL,\rho,\sigma,\alpha)$ allows us to recover the short exact sequence \eqref{eq:exact_D4} as follows:
\[
\xymatrix{
\mathbf{1}\ar[r] & \AAut_\LL(E,\LL,\sigma,\alpha)\ar[r]\ar[d]^[@!-90]{\sim} &
\AAut_\FF(E,\LL,\sigma,\alpha)\ar[r]\ar[d]^[@!-90]{\sim} & \AAut_\FF(\LL)\ar[r]\ar@{=}[d] & \mathbf{1}\\
\mathbf{1}\ar[r] & \AAut_\FF(\cL(E))_0\ar[r] & \AAut_\FF(\cL(E))\ar[r] &
\AAut_\FF(\LL)\ar[r] & \mathbf{1}\\
}
\]
where the subscript $0$ denotes the connected component of identity. Note that the composition $\mathrm{Res}\circ\mathrm{Int}$ is the adjoint representation
\[
\Ad\colon\AAut_\FF(V,\LL,\rho,*,Q)\to\AAut_\FF(\Der(V,*,Q)).
\]
Since there is only one trialitarian algebra (up to isomorphism) over an algebraically closed field, it follows that, over any field, each Lie algebra of type $D_4$ can be realized as $\cL(E)$ for a unique trialitarian algebra (though not necessarily of the sort discussed here). And what is important for our purpose, any $G$-grading on $\cL(E)$ is the restriction of a unique $G$-grading on $E$, so the classification of gradings is the same for $\cL(E)$ as for $E$.
\section{Type I gradings and their Brauer invariants}\label{s:Type_I}
Let $\FF$ be an algebraically closed field, $\chr\FF\ne 2$. Take $\LL=\FF\times\FF\times\FF$ and $\rho\in\Aut_\FF(\LL)$ defined by $\rho(\lambda_1,\lambda_2,\lambda_3)=(\lambda_2,\lambda_3,\lambda_1)$.
Let $(V,\LL,\rho,*,Q)$ be a cyclic composition algebra of rank $8$ and let $E=\End_\LL(V)$ be the corresponding trialitarian algebra. Suppose we have a grading $\Gamma$ on $(E,\LL,\rho,$ $\sigma,\alpha)$ by an abelian group $G$ (which we may assume finitely generated), i.e., a grading on the $\FF$-algebra $E$ such that $\sigma$ and $\alpha$ preserve the degree.
In this section we assume that $\Gamma$ is of Type I, i.e, its restriction to the center $\LL=Z(E)$ is trivial. Then the decomposition $\LL=\FF\times\FF\times\FF$ yields the $G$-graded decomposition $E=E_1\times E_2\times E_3$, where each $E_i$ is isomorphic to $M_8(\FF)$ and equipped with an orthogonal involution $\sigma_i$ (the restriction of $\sigma$). Denote the $G$-grading on $E_i$ by $\Gamma_i$, $i=1,2,3$.
\subsection{Related triples of gradings on $M_8(\FF)$ with orthogonal involution}
Write $V=\cC\otimes\LL$ where $(\cC,\bullet,n)$ is the para-Cayley algebra. Then we can identify each $(E_i,\sigma_i)$ with $\End_\FF(\cC)$, where the
involution is induced by the norm $n$. Thus, we can view all $\Gamma_i$ as gradings on the same algebra $\End_\FF(\cC)\cong M_8(\FF)$. We will say that $(\Gamma_1,\Gamma_2,\Gamma_3)$ is the {\em related triple} associated to $\Gamma$.
Recall that the $G$-grading $\Gamma$ is equivalent to a morphism $G^D\to\AAut_\LL(E,\LL,\sigma,\alpha)$, where $G^D$ is the Cartier dual of $G$. The triality principle implies that each restriction morphism $\pi_i\colon\AAut_\LL(E,\LL,\sigma,\alpha)\to\AAut_\FF(E_i,\sigma_i)$ is a closed imbedding whose image is the connected component of $\AAut_\FF(E_i,\sigma_i)$ (isomorphic to $\PGOs_8^+$). Hence each of the gradings $\Gamma_i=\pi_i(\Gamma)$ uniquely determines $\Gamma$.
Moreover, there exists $\Gamma$ such that $\Gamma_1$ is any given ``inner'' grading on the algebra with involution $\End_\FF(\cC)$.
The outer action of $S_3$ on $\TRIs(\cC,\bullet,n)$ and on its quotient $\AAut_\LL(E,\LL,\sigma,\alpha)$ yields the following action
on related triples: $A_3$ permutes the components of $(\Gamma_1,\Gamma_2,\Gamma_3)$ cyclically and the transposition $(2,3)$ sends
$(\Gamma_1,\Gamma_2,\Gamma_3)$ to $(\wb{\Gamma}_1,\wb{\Gamma}_3,\wb{\Gamma}_2)$ where $\wb{\Gamma}_i$ denotes the image of $\Gamma_i$ under the inner automorphism of $\End_\FF(\cC)$ corresponding to
the standard involution of $\cC$ (which is an improper isometry of $n$).
The classification up to isomorphism of $G$-gradings on the algebra $M_8(\FF)$ with orthogonal involution is known (see e.g. \cite[Theorem 2.64]{EKmon}). It can also be determined explicitly which of these gradings are ``inner'' (see Remark \ref{rem:inner}). Let us see to what extent this allows us to classify Type I gradings on the trialitarian algebra $E$ or, equivalently, on the Lie algebra $\cL(E)$. For two ``inner'' gradings $\Gamma'$ and $\Gamma''$, we will write $\Gamma'\sim\Gamma''$ if there exists an element of $\PGO_8^+(\FF)$ sending $\Gamma'$ to $\Gamma''$. Thus, $\Gamma'$ and $\Gamma''$ are isomorphic if and only if $\Gamma'\sim\Gamma''$ or $\wb{\Gamma'}\sim\Gamma''$.
The stabilizer of a given Type I grading $\Gamma$ under the outer action of $S_3$ can have size $1$, $2$, $3$ or $6$.
Therefore, we have the following three possibilities (the last one corresponding to sizes $3$ and $6$):
\begin{itemize}
\item If $\Gamma_i\nsim\Gamma_j$ for $i\ne j$ and $\Gamma_i\nsim\wb{\Gamma}_j$ for all $i,j$ then the isomorphism class of $\Gamma$ corresponds to $3$ distinct isomorphism classes of ``inner'' gradings (one for each of $\Gamma_i$);
\item If $\Gamma_i\sim\wb{\Gamma}_i$ for some $i$ and $\Gamma_i\nsim\Gamma_j\sim\wb{\Gamma}_k$, where $\{i,j,k\}=\{1,2,3\}$, then the isomorphism class of $\Gamma$ corresponds to $2$ distinct isomorphism classes of ``inner'' gradings (one for $\Gamma_i$ and one for $\Gamma_j$);
\item If $\Gamma_1\sim\Gamma_2\sim\Gamma_3$ then the isomorphism class of $\Gamma$ corresponds to $1$ isomorphism class of ``inner'' gradings.
\end{itemize}
Given $\Gamma_1$, one can --- in principle --- compute $\Gamma_2$ and $\Gamma_3$, but obtaining explicit formulas seems to be difficult.
\subsection{Relation among the Brauer invariants in a related triple}
Recall that the projections $E\to E_i$ define the three irreducible representations of dimension $8$ for the Lie algebra $\cL=\cL(E)$ of type $D_4$: the natural and the two half-spin representations. As before, suppose we are given a Type I grading $\Gamma$ on $E$ and consider its related triple $(\Gamma_1,\Gamma_2,\Gamma_3)$. Since the image of $\cL$ generates each $E_i$, the gradings $\Gamma_i$ on $E_i$ are the unique $G$-gradings such that the three representations $\cL\to E_i$ are homomorphisms of graded Lie algebras.
For each $i$, we can write $E_i=\End_{\cD_i}(W_i)$ where $\cD_i$ is a graded division algebra and $W_i$ is a graded right vector space over $\cD_i$ (\cite[Theorem 2.6]{EKmon}). We are interested in the Brauer classes $[E_i]$, i.e., the isomorphism classes of $\cD_i$, $i=1,2,3$.
Since the even Clifford algebra corresponding to the space of the natural representation of $\cL$ can be defined purely in terms of the corresponding algebra with involution $E_i$ (say, $i=1$) and hence inherits the $G$-grading regardless of the characteristic of $\FF$, we can derive relations among the $G$-graded Brauer classes $[E_i]$ using the same arguments as in \cite[Proposition 39]{EK14}:
\begin{equation}\label{eq:Brauer_relations}
[E_i]^2=1\text{ and }[E_1]=[E_2][E_3].
\end{equation}
Alternatively, extending the proof of \cite[Theorem 9.12]{KMRT} to the graded setting, one arrives at the same relations (see also Proposition 42.7 therein).
\section{Lifting Type III gradings from $\End_\LL(V)$ to $V$}\label{s:lifting}
Let $\FF$ be an algebraically closed field, $\chr\FF\ne 2,3$, and $\LL=\FF\times\FF\times\FF$. Define $\rho\in\Aut_\FF(\LL)$ by $\rho(\lambda_1,\lambda_2,\lambda_3)=(\lambda_2,\lambda_3,\lambda_1)$ and fix a primitive cube root of unity $\omega\in\FF$. The element $\xi=(1,\omega,\omega^2)$ spans the $\omega$-eigenspace for $\rho$ in $\LL$ and satisfies $N(\xi)=1$.
Let $(V,\LL,\rho,*,Q)$ be a cyclic composition algebra of rank $8$ and let $E=\End_\LL(V)$ be the corresponding trialitarian algebra. Suppose we have a Type III grading $\Gamma$ on $(E,\LL,\rho,$ $\sigma,\alpha)$ by an abelian group $G$, i.e, the projection of the image of $\eta=\eta_\Gamma\colon G^D\to\AAut_\FF(E,\LL,\sigma,\alpha)$ in $\AAut_\FF(\LL)$ is $\mathbf{A}_3$. This is equivalent to saying that the $G$-grading on $\LL$ obtained by restricting $\Gamma$ to the center $\LL=Z(E)$ has 1-dimensional homogeneous components: $\LL=\FF\oplus\FF\xi\oplus\FF\xi^2$, so $\LL$ becomes a graded field. Let $h\in G$ be the degree of $\xi$. This is an element of order $3$, which we will call the {\em distinguished element} of the grading. Note that the subgroup $H\bydef\langle h\rangle$ is precisely the distinguished subgroup, i.e., the subgroup of $G$ corresponding to the inverse image $\eta^{-1}(\eta(G^D)\cap\AAut_\LL(E,\LL,\sigma,\alpha))$ in $G^D$.
The purpose of this section is to show that $\eta$ can be ``lifted'' from $E$ to $V$ in the sense that there exists a morphism $\eta'$ such that the following diagram commutes:
\begin{equation}
\begin{aligned}\label{eq:eta_diag}
\xymatrix{
& \AAut_\FF(V,\LL,\rho,*,Q)\ar[dd]^{\mathrm{Int}}\\
G^D\ar@{-->}[ru]^-{\eta'}\ar[rd]^-{\eta} &\\
& \AAut_\FF(E,\LL,\sigma,\alpha)
}
\end{aligned}
\end{equation}
or, in other words, there exists a $G$-grading $\Gamma'$ on $(V,\LL,\rho,*,Q)$, i.e., a grading on the $\FF$-algebra $(V,*)$ making it a graded $\LL$-module with $b_Q$ a degree-preserving map, such that the grading on $E=\End_\LL(V)$ induced by $\Gamma'$ is precisely $\Gamma$.
We will proceed in steps. First we will forget about all extra structure and will treat $E$ simply as a central algebra over $\LL$, so $V$ will be just a graded vector space over $\LL$. Then we will include the involution $\sigma$ and the bilinear form $Q$ in the picture, and finally will take care of $\alpha$ and $*$.
\subsection{Triviality of the graded Brauer invariants}\label{sse:lifting1}
Since $E$ is graded simple, we can write $E=\End_\cD(W)$ as a $G$-graded algebra, where $\cD$ is a graded division algebra over $\FF$ and $W$ is a graded right vector space over $\cD$ (\cite[Theorem 2.6]{EKmon}). Clearly, $\LL$ is the center of $\cD$. We are going to show that $\cD=\LL$.
Let $\wb{G}=G/H$ and consider the $\wb{G}$-grading on $E$ induced by the quotient map $G\to\wb{G}$. Since it is a Type I grading, we can decompose
$E=E_1\times E_2\times E_3$ as a $\wb{G}$-graded algebra. We also have $\cD=\cD_1\times\cD_2\times\cD_3$, $W=W_1\times W_2\times W_3$ and $E_i=\End_{\cD_i}(W_i)$.
Since the $E_i$ are simple algebras, so are the $\cD_i$. Now Proposition \ref{prop:beta_bar} tells us that the $\wb{G}$-graded algebras $\cD_1$, $\cD_2$ and $\cD_3$ are isomorphic graded division algebras, hence $[E_1]=[E_2]=[E_3]$ in the $\wb{G}$-graded Brauer group.
On the other hand, we have relations \eqref{eq:Brauer_relations}. This forces $[E_i]=1$ for all $i$. In other words, $\wb{T}$ and $\bar{\beta}$ are trivial,
which means that $T=H$ and $\beta=1$. We have proved $\cD=\LL$.
\subsection{Construction of $\eta'$}
Since the graded division algebra corresponding to the $G$-graded algebra $E$ is the graded field $\LL$, we can give $V$ a $G$-grading $\Gamma'$ making it into a graded vector space over $\LL$ such that the grading induced on $E=\End_\LL(V)$ is precisely the given grading $\Gamma$. But so far we ignored the quadratic form $Q$ and the multiplication $*$ on $V$. Taking $Q$ into account is easy: by replacing $Q$ with $\mu Q$ for a suitable $\mu\in\LL^\times$, we can make $b_Q\colon V\times V\to\LL$ a homogeneous map of some degree and the possible choices of $\mu$ differ by a homogeneous factor in $\LL^\times$ (\cite[Theorem 2.57]{EKmon}). Since $\FF$ is algebraically closed, we can find $\lambda\in\LL^\times$ such that $\lambda^\#=\mu$. Replacing $*$ with $\lambda*$ and $Q$ with $\mu Q$, we obtain a structure of cyclic composition algebra on $V$ (similar to the original one) that has homogeneous $b_Q$ and induces the original $\sigma$ and $\alpha$ on $E$. We will now show that, by a suitable shift of grading $\Gamma'$, we can make $*$ and $Q$ degree-preserving (i.e., homogeneous maps of degree $e$).
Recall the morphism $\mathrm{Int}\colon\Gs\to\AAut_\FF(E)$, where $\Gs$ is the group scheme of invertible $\FF$-linear endomorphisms of $V$ that are semilinear over $(\LL,\rho)$. The grading $\Gamma'$ on $V$ is equivalent to a morphism $\eta'\colon G^D\to\Gs$ such that $\mathrm{Int}\circ\eta'=\eta$. Since $\eta(G^D)$ is contained in $\mathbf{H}\bydef\AAut_\FF(E,\LL,\rho,\sigma,\alpha)$, we conclude that $\eta'(G^D)$ is contained in $\Gs_0\bydef\mathrm{Int}^{-1}(\mathbf{H})$. Since $\mathrm{Int}$ is a separable morphism (i.e., the kernel is smooth), $\Gs_0$ is smooth as the inverse image of a smooth subgroupscheme. On the other hand, $\mathrm{Int}(\SIMs(V,\LL,\rho,*,Q))$ is contained in $\mathbf{H}$, hence $\SIMs(V,\LL,\rho,*,Q)$ is contained in $\Gs_0$. Since the $\FF$-points of these two group schemes coincide, the schemes have the same dimension, hence $\SIMs(V,\LL,\rho,*,Q)$ is also smooth and coincides with $\Gs_0$. We have shown that $\eta'(G^D)$ is contained in $\SIMs(V,\LL,\rho,*,Q)$. This means that $\eta'(G^D)$ stabilizes the $\LL$-submodules spanned by each of $b_Q$ and $*$.
Now recall the morphisms $\theta'$ and $\theta$ of $\SIMs(V,\LL,\rho,*,Q)$ to $\GLs_1(\LL)\rtimes\mathbf{A}_3$ associated to the actions of this group scheme on the $\LL$-submodules $\LL b_Q$ and $\LL*$, respectively. We know that $b_Q$ is a homogeneous element in the space of bilinear forms (with respect to the grading induced by $\Gamma'$), hence $\eta'(G^D)$ stabilizes the $\FF$-subspace spanned by $b_Q$. This means that, for any $\cR\in\Alg_\FF$, the homomorphism $\theta'_\cR$ sends every element of the group $\eta'(G^D)(\cR)$ to an element of the form $(\mu,\varphi_0)\in(\LL_\cR)^\times\rtimes\Aut_\cR(\LL_\cR,\rho)$, where $\mu$ actually belongs to $\cR^\times$. Let $(\lambda,\varphi_0)$ be the image of the same element of $\eta'(G^D)(\cR)$ under $\theta_\cR$. In view of Lemma \ref{lm:lambda_mu}, we have $\mu=\lambda^\#$. But then $N(\lambda)\lambda=\mu^\#\in\cR^\times$, hence also $\lambda\in\cR^\times$. We have shown that $\eta'(G^D)$ stabilizes the $\FF$-subspace spanned by $*$. In terms of the grading $\Gamma'$, this means that $*$ is a homogeneous element of some degree, say, $g_0$, i.e., we have $x*y\in V_{g_0ab}$ for all $x\in V_a$ and $y\in V_b$ ($a,b\in G$). Shifting the grading $\Gamma'$ by $g_0$, we obtain $V_a*V_b\subset V_{ab}$, i.e., the new $\eta'$ sends $G^D$ to the stabilizer of $*$. Applying Lemma \ref{lm:lambda_mu} again, we see that $\eta'(G^D)$ stabilizes $b_Q$ as well. (Alternatively, one may invoke identities \eqref{eq:cyclic_comp_id} relating $*$ and $Q$.) We have constructed $\eta'$ that fits diagram \eqref{eq:eta_diag} for the cyclic composition algebra similar to the original one.
\subsection{Reduction of the classification of Type III gradings from trialitarian algebras to cyclic composition algebras}
First we summarize the result of the previous subsection:
\begin{theorem}\label{th:tri_to_comp}
Let $(E,\LL,\rho,\sigma,\alpha)$ be a trialitarian algebra over an algebraically closed field $\FF$, $\chr{\FF}\ne 2,3$. Suppose $E$ is given a Type III grading by an abelian group $G$. Then there exists a cyclic composition algebra $(V,\LL,\rho,*,Q)$ with a $G$-grading such that $E$ is isomorphic to $\End_\LL(V)$ as a $G$-graded trialitarian algebra.\qed
\end{theorem}
Since we want to classify $G$-gradings up to isomorphism, there still remains the question of uniqueness of $V$ in the above theorem. To state the answer, it is convenient to introduce some terminology. For a cyclic composition algebra $(V,\LL,\rho,*,Q)$ over $(\LL,\rho)$, consider the same $\LL$-module $V$ with the same quadratic form $Q$ but with the new multiplication $x*^\mathrm{op}y\bydef y*x$. This is a cyclic composition algebra over $(\LL,\rho^2)$, called the {\em opposite} of $V$ and denoted $V^\mathrm{op}$. We will say that an isomorphism $(\varphi_1,\varphi_0)\colon(V,\LL,\rho)\to(V',\LL',\rho')$ and an algebra isomorphism $\varphi\colon\End_\LL(V)\to\End_{\LL'}(V')$ are {\em compatible} if $\varphi_1(ax)=\varphi(a)\varphi_1(x)$ for all $a\in\End_\LL(V)$ and $x\in V$. The next result does not require algebraic closure.
\begin{theorem}\label{th:tri_isomorphism}
Let $(V,\LL,\rho,*,Q)$ and $(V',\LL',\rho',*',Q')$ be two cyclic composition algebras of rank $8$ where $\LL=\FF\times\FF\times\FF$ such that $\chr{\FF}\ne 2,3$ and $\FF$ contains a primitive cube root of unity. Suppose $V$ and $V'$ are given Type III gradings by an abelian group $G$. Let $E=\End_\LL(V)$ and $E'=\End_{\LL'}(V')$ be the corresponding trialitarian algebras with induced $G$-gradings. Then, for any isomorphism $(\varphi_1,\varphi_0)$ of graded cyclic composition algebras from either $V$ or $V^\mathrm{op}$ to $V'$, there exists a unique compatible isomorphism $\varphi\colon E\to E'$ of graded trialitarian algebras. Conversely, for any isomorphism $\varphi\colon E\to E'$ of graded trialitarian algebras, there exists a unique compatible isomorphism of graded cyclic composition algebras $(\varphi_1,\varphi_0)$ from either $V$ or $V^\mathrm{op}$ (but not both) to $V'$.
\end{theorem}
\begin{proof}
Given $(\varphi_1,\varphi_0)$, the only mapping $\varphi\colon E\to E'$ that will satisfy the compatibility condition is the one defined by $\varphi(a)x'=\varphi_1(a\varphi_1^{-1}(x'))$, for all $a\in E$ and $x'\in V'$, and it is an isomorphism of algebras. Since $\sigma$ and $\alpha$ are defined in terms of $Q$ and $*$, it is straightforward to verify that $\varphi$ is actually an isomorphism of trialitarian algebras. The definition of induced grading on the algebra of endomorphisms of a graded module implies that $\varphi$ preserves the degree.
Conversely, suppose $\varphi$ is given. Let $\varphi_0\colon\LL\to\LL'$ be the restriction of $\varphi$. (Since $V'$ is a faithful $\LL'$-module, this is the only possibility to satisfy the compatibility condition.) Then either $\varphi_0\colon(\LL,\rho)\to(\LL',\rho')$ or $\varphi_0\colon(\LL,\rho^2)\to(\LL',\rho')$. The second possibility reduces to the first if we replace $V$ with $V^\mathrm{op}$, so assume $\varphi_0\colon(\LL,\rho)\to(\LL',\rho')$. By \cite[Proposition 44.16]{KMRT}, there exists a similitude of (ungraded) cyclic composition algebras $(\tilde{\varphi}_1,\tilde{\varphi}_0)\colon(V,\LL,\rho,*,Q)\to(V',\LL',\rho',*',Q')$. Let $\tilde{\varphi}$ be the corresponding isomorphism $E\to E'$. Then $\tilde{\varphi}^{-1}\varphi$ is an automorphism of $(E,\LL,\rho,\sigma,\alpha)$, and it follows from \cite[Proposition 44.2]{KMRT} that the group $\Aut_\FF(E,\LL,\rho,\sigma,\alpha)$ is the image of the homomorphism $\mathrm{Int}_\FF\colon\SIM_\FF(V,\LL,\rho,*,Q)\to\Aut_\FF(E,\LL,\sigma,\alpha)$. Therefore, we can find $(\psi_1,\psi_0)\in\SIM_\FF(V,\LL,\rho,*,Q)$ that is sent to $\tilde{\varphi}^{-1}\varphi$, in particular $\psi_0=\tilde{\varphi}_0^{-1}\varphi_0$. Set $\hat{\varphi}_1\bydef\tilde{\varphi}_1\psi_1$. Then $(\hat{\varphi}_1,\varphi_0)$ will satisfy the compatibility condition with $\varphi$.
It remains to take care of the gradings. By \cite[Theorem 2.10]{EKmon}, there exists $u\in G$ and $\varphi_0$-semilinear isomorphism $\varphi_1\colon V^{[u]}\to V'$ of graded spaces over $\LL$ such that $(\varphi_1,\varphi_0)$ is compatible with $\varphi$. Here $V^{[u]}$ denotes a shift of grading, i.e., the new grading $V=\bigoplus_{g\in G}\tilde{V}_g$ where $\tilde{V}_{gu}=V_g$ for all $g\in G$. Now observe that $\hat{\varphi}_1^{-1}\varphi_1$ is an endomorphism of $V$ as an $E$-module, hence it is the multiplication by an element $\ell\in\LL^\times$, i.e., $\varphi_1(x)=\hat{\varphi}_1(\ell x)$ for all $x\in V$. Since $\hat{\varphi}_1$ is a similitude, so is $\varphi_1$. If $\hat{\lambda}\in\LL^\times$ is the parameter of $\hat{\varphi}_1$ then the parameter of $\varphi_1$ is $\lambda=\hat{\lambda}\ell^{-1}\ell^\#$. On the other hand, we have $\varphi_1(V_g)=V'_{gu}$ for all $g\in G$. Pick $s,t\in G$ with $V_s*V_t\ne 0$ and pick $x\in V_s$, $y\in V_t$ such that $z\bydef x*y\ne 0$. Then $z'\bydef\varphi_1(z)$ is a nonzero element of $V'_{stu}$. At the same time, we have $\varphi_0(\lambda)z'=\varphi_1(\lambda(x*y))=\varphi_1(x)*'\varphi(y)\in V'_{stu^2}$. It follows that $\varphi_0(\lambda)$ is a homogeneous element of degree $u$. Replacing $\varphi_1$ by the map $x\mapsto\varphi_1(\lambda^{-1}x)$, which is a similitude with parameter $\lambda_0\bydef\lambda^2(\lambda^\#)^{-1}\in\LL_e^\times=\FF^\times$, we obtain $\varphi_1(V_g)=V'_g$ for all $g\in G$. Finally, since $\lambda_0\in\FF^\times$, the mapping $x\mapsto\lambda_0 x$ is a similitude $V\to V$ with parameter $\lambda_0$ and leaves each $V_g$ invariant, hence replacing $\varphi_1$ by the map $x\mapsto\varphi_1(\lambda_0^{-1}x)$ yields the desired isomorphism of graded cyclic composition algebras.
Finally, if $(\tilde{\varphi}_1,\tilde{\varphi}_0)$ is another isomorphism compatible with $\varphi$ and preserving degree then there exists $\ell\in\LL^\times$ such that $\tilde{\varphi}_1(x)=\varphi_1(\ell x)$ for all $x\in V$. But this is possible only if $\ell=1$.
\end{proof}
\begin{corollary}\label{cor:tri_isomorphism}
Under the conditions of Theorem \ref{th:tri_to_comp}, fix an identification $E=\End_\LL(V)$ as a $G$-graded algebra. Then there exist exactly $4$ gradings on the cyclic composition algebra $V$ that induce the given grading on $E$, and they form an orbit of the subgroup
\[
C\bydef\{(\veps_1,\veps_2,\veps_3)\;|\;\veps_i\in\{\pm 1\},\,\veps_1\veps_2\veps_3=1\}\subset\LL^\times
\]
(the center of the spin group) with respect to its natural action on $V$.
\end{corollary}
\begin{proof}
If $V$ has two $G$-gradings, $\Gamma_1$ and $\Gamma_2$, that induce the same grading $\Gamma$ on $E$ then there exists an algebra automorphism $\varphi_1\colon V\to V$ that sends $\Gamma_1$ to $\Gamma_2$ and induces the identity map on $E$ (so $\varphi_0=\id$). Hence $\varphi_1$ is given by $\varphi_1(x)=\ell x$ for some $\ell\in\LL^\times$. This map is a similitude with multiplier $\ell^{-1}\ell^\#$, which must be equal to $1$. It is easy to see that $\ell^{-1}\ell^\#=1$ if and only if $\ell\in C$.
It remains to observe that, if $\ell\in C$ is different from $1$, then, for any homogeneous component $V_g\ne 0$ of $\Gamma_1$, the image $\varphi_1(V_g)=\ell V_g$ cannot coincide with $V_g$. Indeed, consider any character $\chi\in\wh{G}$ such that $\chi(h)=\omega$. In the action of the group $\wh{G}$ associated to the grading $\Gamma_1$ on $V$, $\chi$ will act differently on $V_g$ and on $\ell V_g$, namely, as the scalar $\chi(g)$ on the former and as $\rho(\ell)\ell^{-1}\chi(g)$ on the latter.
\end{proof}
We now turn to the classification of fine gradings up to equivalence. Clearly, if a Type III grading cannot be refined in the class of Type III gradings then it is fine. It is also clear from Theorems \ref{th:tri_to_comp} and \ref{th:tri_isomorphism} that a fine Type III grading on a cyclic composition algebra $V$ induces a fine Type III grading on the trialitarian algebra $E=\End_\LL(V)$. We will now establish the converse.
\begin{theorem}\label{th:tri_equivalence}
Let $(E,\LL,\rho,\sigma,\alpha)$ be a trialitarian algebra over an algebraically closed field $\FF$, $\chr{\FF}\ne 2,3$. Then every fine Type III grading on $E$ with universal group $G$ is induced from a fine Type III grading on the cyclic composition algebra $(V,\LL,\rho,*,Q)$ with the same universal group. Moreover, two such gradings on $E$ are equivalent if and only if they are induced from equivalent gradings on $V$.
\end{theorem}
\begin{proof}
Let $\Gamma$ be a fine Type III grading on $E$ with universal group $G$. It is determined by a maximal diagonalizable subgroupscheme $\Qs$ of $\AAut_\FF(E,\LL,\rho,\sigma,\alpha)$, which we may identify with $G^D$. By Theorem \ref{th:tri_to_comp}, we can induce $\Gamma$ from a Type III $G$-grading $\Gamma'$ on $V$. Let $\Qs'$ be the corresponding diagonalizable subgroupscheme of $\AAut_\FF(V,\LL,\rho,*,Q)$. Then the morphism $\mathrm{Int}$ restricts to an isomorphism $\Qs'\to\Qs$ (see diagram \eqref{eq:eta_diag}). We claim that $\Gamma'$ is fine. Assume, to the contrary, that $\Qs'$ is not maximal. Then there exists diagonalizable $\tilde{\Qs}'$ properly containing $\Qs'$. The image $\mathrm{Int}(\tilde{\Qs}')$ is necessarily $\Qs$ because the latter is maximal. We will obtain a contradiction if we can show that the intersection of $\tilde{\Qs}'$ with the kernel $\mathbf{K}$ of $\mathrm{Int}$ is trivial. Since $\mathbf{K}$ is isomorphic to $\bmu_2^2$ and every subgroupscheme of $\bmu_2^2$ is smooth, it suffices to show that the intersection has no $\FF$-points different from the identity. But this is clear since $\mathbf{K}(\FF)=C$ (see the Corollary \ref{cor:tri_isomorphism}) and $\Qs'(\FF)$ contains an automorphism that acts as a cyclic permutation on $C$ and hence does not commute with any element of $C$ except the identity. The assertion about equivalence follows from Theorem \ref{th:tri_isomorphism}.
\end{proof}
\section{Gradings on cyclic composition algebras}\label{s:Type_III_fine}
As in the previous section, let $\FF$ be an algebraically closed field, $\chr\FF\ne 2,3$, and let $\LL=\FF\times\FF\times\FF$, with $\rho(\lambda_1,\lambda_2,\lambda_3)=(\lambda_2,\lambda_3,\lambda_1)$.
\subsection{The Albert algebra}
If $(V,\LL,\rho,*,Q)$ is a cyclic composition algebra of rank $8$, then the direct sum
\begin{equation}\label{eq:JLV}
\cJ(\LL,V)\bydef \LL\oplus V,
\end{equation}
is the Albert algebra (i.e., the simple exceptional Jordan algebra), which contains $\LL$ as a subalgebra and whose
norm, trace form and adjoint extend those of $\LL$ in the following way (see \cite[Theorem 38.6]{KMRT}):
\begin{align*}
&N\bigl((\ell,v)\bigr)=N(\ell)+b_Q(v,v*v)-T\bigl(\ell Q(v)\bigr),\\
&T\bigl((\ell_1,v_1),(\ell_2,v_2)\bigr)= T(\ell_1\ell_2)+T\bigl(b_Q(v_1,v_2)\bigr),\\
&(\ell,v)^\#=\bigl(\ell^\#-Q(v),v*v-\ell v).
\end{align*}
In particular, $V$ is the orthogonal complement to $\LL$ relative to the trace form.
Any element $X$ in $\alb\bydef\cJ(\LL,V)$ satisfies the generic degree $3$ equation:
\[
X^3-T(X)X^2+S(X)X-N(X)1=0,
\]
where $T\bigl((\ell,v)\bigr)=T(\ell)$ and $S(X)=\frac{1}{2}\left(T(X)^2-T(X^2)\right)$.
Note that the adjoint is defined by $X^\#=X^2-T(X)X+S(X)1$, hence the commutative multiplication in $\alb$ can be expressed in terms of $\#$ and $T$.
Therefore, any grading $\Gamma$ on the cyclic composition algebra $(V,\LL,\rho,*,Q)$ by an abelian group $G$ extends to a grading $\Gamma_\cJ$ on $\alb$ by the same group $G$, given by $\alb_g=\LL_g\oplus V_g$ for all $g\in G$. The gradings on the Albert algebra have been determined in \cite{EK12a} (see also \cite[Chapter 5]{EKmon}).
\subsection{Type III gradings on cyclic composition algebras}
Recall that, for a symmetric composition algebra $(\cS,\star,n)$, the associated cyclic composition algebra $(V,\LL,\rho,*,Q)$ is given by $V=\cS\ot \LL$,
\begin{align*}
&(x\ot \ell)*(y\ot m)=(x\star y)\ot \rho(\ell)\rho^2(m),\\
&Q(x\otimes\ell)=n(x)\ell^2,\\
&b_Q(x\otimes\ell,y\otimes m)=n(x,y)\ell m,
\end{align*}
for all $x,y\in\cS$ and $\ell,m\in\LL$. If we think of $\cS\ot \LL$ as $\cS\times\cS\times\cS$ then the product expands as in \eqref{df:cyclic_prod}, namely,
\[
(x_1,x_2,x_3)*(y_1,y_2,y_3)=(x_2\star y_3,x_3\star y_1,x_1\star y_2)
\]
and $Q$ becomes $(n,n,n)$. This cyclic composition algebra will be denoted by $(\cS,\star,n)\ot(\LL,\rho)$.
Any pair of gradings, $\Gamma_\cS$ on $\cS$ and $\Gamma_\LL$ on $\LL$ (such that $\rho$ is degree-preserving), by the same abelian group $G$, induces a $G$-grading $\Gamma$ on the cyclic composition algebra $(\cS,\star,n)\ot(\LL,\rho)$ with $\bigl(\cS\ot\LL)_g=\bigoplus_{k\in G}\left(\cS_{gk^{-1}}\ot \LL_k\right)$ for all $g\in G$. This grading will be denoted by $\Gamma_\cS\ot\Gamma_\LL$. We are interested in the case of Type III gradings, where $\LL$ is a graded field: its homogeneous components are $\LL_e=\FF 1$, $\LL_h=\FF\xi$ and $\LL_{h^2}=\FF\xi^2$ where, as before, $\xi=(1,\omega,\omega^2)$ and $h\in G$ is the distinguished element.
\begin{theorem}\label{th:gradings_cyclic}
Let $\Gamma$ be a Type III grading by an abelian group $G$ on the cyclic composition algebra $(V,\LL,\rho,*,Q)$ of rank $8$ over an algebraically closed field $\FF$, $\chr\FF\ne 2,3$, and let $\Gamma_\LL$ be the induced grading on $\LL$.
\begin{enumerate}
\item
If $V_e=0$, then $(V,\LL,\rho,*,Q)$ is isomorphic to $(\cO,\star,n)\ot(\LL,\rho)$ as a graded cyclic composition algebra, where $(\cO,\star,n)$ is the Okubo algebra, endowed with a $G$-grading $\Gamma_\cO$ with $\cO_e=0$, and the grading on $(\cO,\star,n)\ot(\LL,\rho)$ is $\Gamma_\cO\ot\Gamma_\LL$.
\item
Otherwise, $(V,\LL,\rho,*,Q)$ is isomorphic to $(\cC,\bullet,n)\ot(\LL,\rho)$ as a graded cyclic composition algebra, where $(\cC,\bullet,n)$ is the para-Cayley algebra, endowed with a $G$-grading $\Gamma_\cC$, and the grading on $(\cC,\bullet,n)\ot(\LL,\rho)$ is $\Gamma_\cC\ot\Gamma_\LL$.
\end{enumerate}
\end{theorem}
\begin{proof}
Assume first that $V_e=0$ and consider the Albert algebra $\alb=\cJ(\LL,V)$ as in the previous subsection. Then the grading $\Gamma_\cJ$ induced by $\Gamma$ on $\alb$ satisfies the condition $\alb_e=\FF 1$, and hence, by \cite[Theorem 5.12]{EKmon}, all the homogeneous components of $\Gamma_\cJ$ have dimension $1$ and the support is a $3$-elementary abelian subgroup of $G$ (isomorphic to $\ZZ_3^3$). Set $h_1=h$ (the distinguished element) and pick $h_2,h_3\in G$ such that the support of $\Gamma_\cJ$ is generated by $h_1,h_2,h_3$. Then $V=\bigoplus_{g\in \supp\Gamma}V_g$, with $\dim V_g=1$ for any $g\in \supp\Gamma$, and $\supp\Gamma=\supp\Gamma_\cJ\setminus H=\langle h_1,h_2,h_3\rangle\setminus\langle h_1\rangle$.
Consider the graded $\FF$-subalgebra $\cO\bydef\bigoplus_{g\in\langle h_2,h_3\rangle} V_g$ in $(V,*)$. The values of
$Q$ (or $b_Q$) on $\cO$ are contained in
$\bigoplus_{g\in\langle h_2,h_3\rangle}\LL_g=\LL_e=\FF 1$,
and hence $n\bydef Q\vert_\cO$ is a nondegenerate quadratic form on $\cO$. Because of identity \eqref{eq:cyclic_comp_id}, $\cO$ is a symmetric composition algebra of dimension $8$ with norm $n$. Besides, it is graded by $\langle h_2,h_3\rangle\cong\ZZ_3^2$, with one-dimensional homogeneous components and support $\langle h_2,h_3\rangle\setminus\{e\}$. It follows that $(\cO,*,n)$ is the Okubo algebra (see \cite{Eld09} or \cite[Theorems 4.12 and 4.51]{EKmon}). Moreover, $V=\cO\oplus\xi\cO\oplus\xi^2\cO=\cO\ot\LL$ and the first part of the theorem follows.
We proceed to the case $V_e\ne 0$. Then $V_e$ is an $\FF$-subalgebra of $(V,*)$ and, again, the values of $Q$ on $V_e$ are contained in $\FF 1$. Hence $(V_e,*,Q)$ is a symmetric composition algebra. Since $\FF$ is algebraically closed, there exists a nonzero idempotent $\varepsilon$ in $V_e$: $0\ne\varepsilon=\varepsilon*\varepsilon$ (see e.g. \cite[Proposition 4.43]{EKmon}). Substituting $x=y=\varepsilon$ into identity \eqref{eq:cyclic_comp_id} gives $Q(\varepsilon)=1$.
The cyclic composition algebra $(V,\LL,\rho,*,Q)$ is isomorphic to $(\cC,\bullet,n)\ot(\LL,\rho)$, with $(\cC,\bullet,n)$ the para-Cayley algebra, so we may identify these and hence we may identify $\varepsilon$ with a triple $(x_1,x_2,x_3)\in\cC\times\cC\times\cC$ such that $x_i\bullet x_{i+1}=x_{i+2}$ for any $i=1,2,3$ (indices modulo $3$) and $n(x_1)=n(x_2)=n(x_3)=1$. Using \cite[Corollary 5.6 and Lemma 5.25]{EKmon} we conclude that there is a triple $(f_1,f_2,f_3)\in \TRI(\cC,\bullet,n)$ such that $f_1(x_1)=f_2(x_2)=1$ (the unit of the Cayley algebra or, equivalently, the para-unit of the para-Cayley algebra). But then we get $f_3(x_3)=f_3(x_1\bullet x_2)=f_1(x_1)\bullet f_2(x_2)=1\bullet 1=1$. Since $\TRI(\cC,\bullet,n)$ is contained in the group of automorphisms of our cyclic composition algebra, we may assume, without loss of generality, that $\varepsilon=\buno\bydef (1,1,1)$.
Note that, for any $(x_1,x_2,x_3)\in V=\cC\ot \LL$, we have
\[
(x_1,x_2,x_3)*\buno=(\bar x_2,\bar x_3,\bar x_1)=b_Q\bigl((x_2,x_3,x_1),\buno\bigr)\buno-(x_2,x_3,x_1).
\]
For $X=(x_1,x_2,x_3)\in V$, define $\bar{X}\bydef b_Q(X,\buno)\buno-X$. Hence $X*\buno=\bar{X}$ if and only if $x_1=x_2=x_3$, if and only if $X\in \cC\ot 1$. But $\{X\in V\;|\; X*\buno=\bar{X}\}$ is a graded subspace of $V$, so we conclude that $\cC\cong \cC\ot 1$ (with the para-Hurwitz multiplication) is a graded $\FF$-subalgebra of $(V,*)$, and the second part of the Theorem follows.
\end{proof}
\subsection{Application to fine gradings on simple Lie algebras of type $D_4$}
Let $\cL$ be the simple Lie algebra of type $D_4$ over an algebraically closed field $\FF$, $\chr\FF\ne 2$. We are ready to obtain the classification of fine gradings on $\cL$ up to equivalence. Recall that Type III gradings exist only if $\chr\FF\ne 3$. In this case we use the realization $\cL=\cL(E)$ where $E$ is the trialitarian algebra. The following result implies the analogs of Theorems 6.8 and 6.11 in \cite{EKmon}, where only characteristic $0$ was considered.
\begin{corollary}\label{cor:fineIII}
Up to equivalence, there are three fine gradings of Type III on the simple Lie algebra of type $D_4$ over an algebraically closed field $\FF$, $\chr\FF\ne 2,3$.
Their universal groups are $\ZZ^2\times\ZZ_3$, $\ZZ_2^3\times\ZZ_3$ and $\ZZ_3^3$.
\end{corollary}
\begin{proof}
Recall that, since the restriction from $E$ to $\cL$ yields an isomorphism of automorphism group schemes, $E$ and $\cL$ have the same classification of gradings. By Theorem \ref{th:tri_equivalence}, the classification of fine gradings of Type III on $E$ is the same as that on $V$, the cyclic composition algebra of rank $8$. Finally, Theorem \ref{th:gradings_cyclic} implies that any fine grading of Type III on $V$ comes from a fine grading on either the para-Cayley algebra $\cC$ or the Okubo algebra $\cO$, with $\cO_e=0$. On $\cC$, there are only two fine gradings, up to equivalence; their universal groups are $\ZZ^2$ and $\ZZ_2^3$ (see e.g. \cite[Theorem 4.51]{EKmon}). On $\cO$, there is a unique fine grading with trivial identity component; its universal group is $\ZZ_3^2$ (see e.g. \cite[Corollary 4.54]{EKmon}). Conversely, these gradings on $\cC$ and $\cO$ give rise to three gradings on $V$, whose universal groups are $\ZZ^2\times\ZZ_3$, $\ZZ_2^3\times\ZZ_3$ and $\ZZ_3^3$. By looking at $V_e$ and the universal groups, we see that none of these three can be a coarsening of another, hence they are fine.
\end{proof}
We now turn to Types I and II (so $\chr\FF=3$ is allowed) and use the matrix realization $\cL=\mathrm{Skew}(\cR,\sigma)$ where $\cR=M_8(\FF)$ and $\sigma$ is an orthogonal involution. Up to equivalence, there are $15$ fine gradings on $(\cR,\sigma)$, which restrict to $15$ gradings of Type I or II on $\cL$ (see e.g. \cite[Example 3.44]{EKmon}).
\begin{remark}\label{rem:inner}
Since we assume $\chr\FF\ne 2$, a $G$-grading on $(\cR,\sigma)$ restricts to a Type~I grading on $\cL$ if and only if the group of characters $\wh{G}$ acts by inner automorphisms of $\cL$. Hence Lemma 33 in \cite{EK14} allows us to determine which restrictions are of Type I and which of Type II and to compute the generator of the distinguished subgroup (see Definition 34 in \cite{EK14}). A direct computation shows that, out of the above $15$ gradings on $\cL$, $8$ are of Type I and $7$ are of Type II.
\end{remark}
\begin{theorem}\label{th:D4_fine}
Let $\FF$ be an algebraically closed field and let $\cL$ be the simple Lie algebra of type $D_4$ over $\FF$.
\begin{enumerate}
\item If $\chr\FF\ne 2,3$ then there are, up to equivalence, $17$ fine gradings on $\cL$. Their universal groups and types are given in Theorem 6.15 of \cite{EKmon}.
\item If $\chr\FF=3$ then there are, up to equivalence, $14$ fine gradings on $\cL$. They correspond to cases (1)---(14) in Theorem 6.15 of \cite{EKmon}.
\end{enumerate}
\end{theorem}
\begin{proof}
The Type II gradings on $\cL$ obtained from fine gradings on $(\cR,\sigma)$ cannot be refined in the class of Type II gradings, hence they are fine. The Type I gradings, on the other hand, could fail to be fine because of the possibility of a Type III refinement (if $\chr\FF\ne 3$). However, such a grading $\Gamma$ would then be the coarsening of one of the gradings in Corollary \ref{cor:fineIII} obtained by taking the universal group modulo the distinguished subgroup of order $3$, so the universal group of $\Gamma$ would be $\ZZ^2$, $\ZZ_2^3$ or $\ZZ_3^2$, but none of these occurs on the list (cf. \cite[Corollary 6.12]{EKmon}).
Out of the $15$ fine gradings on $\cL$ coming from $(\cR,\sigma)$, there are only two that share the same universal group (namely, $\ZZ_2^3\times\ZZ_4$) and the same type ($24$ components of dimension $1$ and $2$ components of dimension $2$) --- see the discussion in \cite[\S 6.1]{EKmon} following Corollary 6.12. It turns out that these two are actually equivalent. This can be shown as in \cite{EKmon}, since the proofs of Lemma 6.13 and Proposition 6.14 are valid under the assumption $\chr\FF\ne 2$. Alternatively, we can consider the graded Brauer invariants $(T_i,\beta_i)$ of the related triple $(\Gamma_1,\Gamma_2,\Gamma_3)$ where $\Gamma_1$ is one of these two gradings (which are of Type I). For one of them, we have $T_1\cong\ZZ_2^2$, so Equation (30) and Remark 43 in \cite{EK14} apply. For the other, we have $T_1\cong\ZZ_2^4$, so Equation (32) and Remark 44 apply. Whichever we choose, it is easy to see, using the results in \cite{EK14} mentioned above, that the $T_i$ are not all isomorphic to each other (in fact, two of them are $\ZZ_2^4$ and one is $\ZZ_2^2$), hence the $\Gamma_i$ are not all equivalent as gradings on $M_8(\FF)$ and the result follows.
\end{proof}
\section{Type III gradings up to isomorphism}\label{s:Type_III}
We continue to assume that the ground field $\FF$ is algebraically closed and that $\chr\FF\ne 2,3$.
Let $\Gamma$ be a Type III grading by an abelian group $G$ on the cyclic composition algebra $(V,\LL,\rho,*,Q)$ of rank $8$.
Define the \emph{rank} of $\Gamma$ as the dimension of the neutral homogeneous component $V_e$. Recall from the proof of Theorem \ref{th:gradings_cyclic}
that either $V_e=0$, or $V_e$ is a symmetric composition algebra, and hence its dimension is restricted to $1$, $2$, $4$ or $8$.
Given two such Type III gradings $\Gamma$ and $\Gamma'$, they will be said to be \emph{similar} if they induce isomorphic gradings
on the trialitarian algebra $E=\End_\LL(V)$ or, equivalently, on the Lie algebra $\cL(E)$, which is simple of type $D_4$.
According to Theorem \ref{th:tri_isomorphism}, $\Gamma$ and $\Gamma'$ are similar if and only if the graded cyclic composition algebras $(V,\Gamma)$ and $(V,\Gamma')$
are isomorphic or anti-isomorphic, i.e., there exists an isomorphism $(\varphi_1,\varphi_0)$ from either $V$ or $V^\mathrm{op}$, endowed with the grading $\Gamma$,
onto $V$, endowed with the grading $\Gamma'$. Thus, the classification of $G$-gradings on the simple Lie algebra of type $D_4$ up to isomorphism is the same as
the classification of $G$-gradings on the cyclic composition algebra $(V,\LL,\rho,*,Q)$ up to similarity.
Clearly, the rank is an invariant of the similarity class of a given Type III grading $\Gamma$.
\subsection{Construction of Type III gradings}
Fix an abelian group $G$. For each possible rank $r$, we will define a list of Type III gradings by $G$ on the unique cyclic composition algebra $(V,\LL,\rho,*,Q)$,
which can be realized as $(\cC,\bullet,n)\otimes(\LL,\rho)$ or as $(\cO,\star,n)\otimes(\LL,\rho)$, where $(\cC,\bullet,n)$ and $(\cO,\star,n)$
are the para-Cayley and the Okubo algebra, respectively. As before, $h$ will denote the distinguished element of the grading
(the degree of $\xi=(1,\omega,\omega^2)\in\LL$, which spans the $\omega$-eigenspace of $\rho$), so the distinguished subgroup is $H=\langle h\rangle$.
Note that the distinguished elements of $V$ and $V^\mathrm{op}$ are inverses of each other.
\begin{itemize}
\item[$\boxed{r=0}$] By \cite[Corollaries 4.54 and 4.55]{EKmon}, given a subgroup $K$ of $G$ isomorphic to $\ZZ_3^2$, there are, up to isomorphism,
exactly two $G$-gradings, $\Gamma_\cO^+$ and $\Gamma_\cO^-$, with support $K\setminus\{e\}$ on the Okubo algebra $(\cO,\star,n)$.
Pick an order $3$ element $h\in G\setminus K$ and let $\Gamma_\LL$ be the grading on $\LL$ with $\deg \xi=h$.
Denote by $\GIII_{0}(G,K,h,\pm)$ the grading $\Gamma_\cO^\pm\otimes\Gamma_\LL$ on $(\cO,\star,n)\otimes(\LL,\rho)$.
Note that the support of this grading is $KH\setminus H$ and the subgroup generated by the support is
the direct product $KH$, where $H=\langle h\rangle$.
\item[$\boxed{r=1}$] By \cite[Theorems 4.21 and 4.51]{EKmon}, given a subgroup $K$ of $G$ isomorphic to $\ZZ_2^3$, there is, up to isomorphism, a unique grading $\Gamma_\cC$ with support $K$
on the Cayley algebra, or equivalently on the para-Cayley algebra $(\cC,\bullet,n)$. Pick an order $3$ element $h\in G\setminus K$ and let $\Gamma_\LL$ be as above.
Denote by $\GIII_{1}(G,K,h)$ the grading $\Gamma_\cC\otimes\Gamma_\LL$ on $(\cC,\bullet,n)\otimes(\LL,\rho)$.
\item[$\boxed{r=2}$] Pick an order $3$ element $h\in G$ and elements $g_1,g_2,g_3\in G\setminus H$, $H=\langle h\rangle$, with $g_1g_2g_3=e$.
Write $\gamma=(g_1,g_2,g_3)$. Consider the grading $\Gamma_\cC(G,\gamma)$ on the para-Cayley algebra $(\cC,\bullet,n)$ induced from the Cartan grading
(see \cite[Theorem 4.21]{EKmon}) by the homomorphism $\ZZ^2\rightarrow G$ sending $(1,0)$ to $g_1$ and $(0,1)$ to $g_2$.
Denote by $\GIII_{2}(G,\gamma,h)$ the grading $\Gamma_\cC(G,\gamma)\otimes\Gamma_\LL$ on $(\cC,\bullet,n)\otimes(\LL,\rho)$.
The restrictions on $h$ and $\gamma$ ensure that the rank of $\GIII_{2}(G,\gamma,h)$ is $2$.
\item[$\boxed{r=4}$] Pick again an order $3$ element $h\in G$ and another element $g\in G\setminus H$, $H=\langle h\rangle$.
Set $\gamma=(e,g,g^{-1})$ and consider the grading $\Gamma_\cC(G,\gamma)$ on $\cC$ as in the previous case.
Denote by $\GIII_{4}(G,g,h)$ the grading $\Gamma_\cC(G,\gamma)\otimes\Gamma_\LL$ on $(\cC,\bullet,n)\otimes(\LL,\rho)$. Its rank is easily checked to be $4$.
\item[$\boxed{r=8}$] Consider the trivial gradings $\Gamma_\cC^\mathrm{triv}$ and $\Gamma_\cO^\mathrm{triv}$ on the para-Cayley algebra and the Okubo algebra,
respectively. Pick an order $3$ element $h\in G$ and denote by $\GIII_{8}(G,h,\mathrm{p})$ the grading
$\Gamma_\cC^\mathrm{triv}\otimes\Gamma_\LL$ on $(\cC,\bullet,n)\otimes(\LL,\rho)$ and
by $\GIII_{8}(G,h,\mathrm{o})$ the grading $\Gamma_\cO^\mathrm{triv}\otimes\Gamma_\LL$ on $(\cO,\star,n)\otimes(\LL,\rho)$.
\end{itemize}
\subsection{Classification up to isomorphism}
The next result classifies Type III gradings on the simple Lie algebra of type $D_4$ up to isomorphism by classifying
Type III gradings on the cyclic composition algebra of rank $8$ up to similarity.
\begin{theorem}\label{th:class_isomIII}
Let $\Gamma$ be a Type III grading by an abelian group $G$ on the cyclic composition algebra $(V,\LL,\rho,*,Q)$ of rank $8$
over an algebraically closed field $\FF$, $\chr\FF\ne 2,3$. Then $\Gamma$ is similar to one of the gradings
$\GIII_{0}(G,K,h,\delta)$, $\GIII_{1}(G,K,h)$, $\GIII_{2}(G,\gamma,h)$, $\GIII_{4}(G,g,h)$, or $\GIII_{8}(G,h,\mathrm{t})$,
where $\delta$ is $+$ or $-$ and $\mathrm{t}$ is $\mathrm{p}$ or $\mathrm{o}$.
Moreover, the gradings with different ranks on the list above are not similar, and for gradings of the same rank we have:
\begin{itemize}
\item If $\GIII_{0}(G,K,h,\delta)$ is similar to $\GIII_{0}(G,K',h',\delta')$, then $K\langle h\rangle=K'\langle h'\rangle$
and also $\langle h\rangle=\langle h'\rangle$. Assuming the subgroups $H=\langle h\rangle$ and $KH$ are fixed, there are exactly two similarity classes:
the gradings $\GIII_{0}(G,K,h,\delta)$ and $\GIII_{0}(G,K',h',\delta')$ are similar if and only if either $\delta'=\delta$ and $h'=h$
or $\delta'=-\delta$ and $h'=h^{-1}$.
\item $\GIII_{1}(G,K,h)$ is similar to $\GIII_{1}(G,K',h')$ if and only if $K'=K$ and $\langle h'\rangle=\langle h\rangle$.
\item $\GIII_{2}(G,\gamma,h)$ is similar to $\GIII_{2}(G,\gamma',h')$ if and only if $\langle h'\rangle =\langle h\rangle$ and there exists
a permutation $\pi\in S_3$ and $1\leq j\leq 3$ such that either $g'_i=g_{\pi(i)}h^j$, for all $i=1,2,3$, or $g'_i=g_{\pi(i)}^{-1}h^j$, for all $i=1,2,3$.
\item $\GIII_{4}(G,g,h)$ is similar to $\GIII_{4}(G,g',h')$ if and only if $\langle h'\rangle=\langle h\rangle$ and $g'$ equals either $g$ or $g^{-1}$.
\item $\GIII_{8}(G,h,\mathrm{t})$ is similar to $\GIII_{8}(G,h',\mathrm{t}')$ if and only if $\langle h'\rangle=\langle h\rangle$ and $\mathrm{t}'=\mathrm{t}$.
\end{itemize}
\end{theorem}
\begin{proof}
We start with the most difficult case, which is the case of rank $0$.
The fact that any grading of Type III and rank $0$ is isomorphic to a grading of the form $\GIII_{0}(G,K,h,\pm)$ follows from
Theorem \ref{th:gradings_cyclic} and its proof. Now, if $\GIII_{0}(G,K,h,\delta)$ is similar to $\GIII_{0}(G,K',h',\delta')$ then their supports and distinguished subgroups coincide,
and hence $K\langle h\rangle=K'\langle h'\rangle$ and $\langle h\rangle=\langle h'\rangle$.
In particular, $K$ is a subgroup in the support of $\GIII_{0}(G,K',h',\delta')$. Hence, as in the proof of Theorem \ref{th:gradings_cyclic},
we may consider the $\FF$-subalgebra $\bigoplus_{k\in K} V_k$ in $V=\cO\otimes \LL$, endowed with the grading
$\GIII_{0}(G,K',h',\delta')=\Gamma_\cO(G,K,\delta')\otimes \Gamma_\LL'$, where $\Gamma_\LL'$ is the grading on $\LL$ with $\deg \xi=h'$.
This is a symmetric composition algebra, graded by a group isomorphic to $\ZZ_3^2$ with one-dimensional homogeneous components, so it is the Okubo algebra.
This shows that $\GIII_{0}(G,K',h',\delta')$ is isomorphic to the grading $\Gamma_\cO^{\delta''}\otimes\Gamma_\LL'=\GIII_{0}(G,K,h',\delta'')$,
where $h'$ equals $h$ or $h^{-1}$ and $\delta''\in\{+,-\}$.
Therefore, once we fix the subgroups $\langle h\rangle$ and $K\langle h\rangle$, we get at most four similarity classes.
Now consider the case $K'=K$, $h'=h^{-1}$, $\delta=+$ and $\delta'=-$.
Fix two generators $k_1$ and $k_2$ of $K$ (recall that $K$ is isomorphic to $\ZZ_3^2$) and pick elements
$x,y\in\cO$ with $\deg x=k_1$, $\deg y=k_2$ (with respect to $\Gamma_\cO^+$) and $n(x,x\star x)=1=n(y,y\star y)$.
By \cite[Lemma 4.47]{EKmon}, we have either $x\star y=0$ or $y\star x=0$ but not both. Interchanging $k_1$ and $k_2$ if necessary, we will assume that $x\star y=0$.
The grading $\Gamma_\cO^-$ is given by $\deg x=k_2$ and $\deg y=k_1$. Because of \cite[Lemma 4.47]{EKmon}, there is a unique involution $\sigma$ on $\cO$
such that $\sigma(x)=y$. Also consider the automorphism $\tau\colon (x_1,x_2,x_3)\mapsto (x_1,x_3,x_2)$ of $\LL$, which takes $\xi$ to $\xi^2$.
Then the map $\sigma\otimes \tau$ is a graded isomorphism from the opposite of $(\cO,\star,n)\otimes(\LL,\rho)$,
endowed with the grading $\Gamma_\cO^+\otimes\Gamma_\LL$, onto $(\cO,\star,n)\otimes(\LL,\rho)$, endowed with the grading $\Gamma_\cO^-\otimes\Gamma_\LL'$.
This shows that we get at most two similarity classes, with representatives $\Gamma_\cO^+\otimes\Gamma_\LL$ and $\Gamma_\cO^+\otimes\Gamma_\LL'$.
Finally, there is no graded isomorphism or anti-isomorphism from $\Gamma_\cO^+\otimes\Gamma_\LL$ onto $\Gamma_\cO^+\otimes\Gamma_\LL'$.
Indeed, any such (anti-)isomorphism $(\varphi_1,\varphi_0)$ takes the $\FF$-subalgebra $\cO\otimes 1=\bigoplus_{k\in K}V_k$ of $V=\cO\otimes\LL$ onto itself.
Since $\cO_{k_1}=\FF x$ and $\cO_{k_2}=\FF y$, we obtain $\varphi_1(x\otimes 1)*\varphi_1(y\otimes 1)=0=\varphi_1\bigl((x\otimes 1)*(y\otimes 1)\bigr)$,
whereas $0\ne \varphi_1\bigl((y\otimes 1)*(x\otimes 1)\bigr)$. It follows that $(\varphi_1,\varphi_0)$
is necessarily an isomorphism, so $\varphi_0\rho=\rho\varphi_0$ and hence $\varphi_0$ must respect the grading on $\LL$, which is a contradiction
because it takes $\Gamma_\LL$ to $\Gamma_\LL'$. The proof of the rank $0$ case is complete.
\smallskip
If the rank is $8$, then $V_e$ is a symmetric composition algebra of dimension $8$ and hence isomorphic to $(\cC,\bullet,n)$ or $(\cO,\star,n)$.
This isomorphism class is an invariant of the grading, and the result follows.
\smallskip
If the rank is $1$, then $V_e=\FF \varepsilon$, for a unique idempotent $\varepsilon$. As in the proof of Theorem \ref{th:gradings_cyclic},
we may identify $V$ with $\cC\otimes\LL$, where $\cC=\{X\in V\,|\, X*\varepsilon=\bar X\}$. This is the para-Cayley algebra with para-unit $\varepsilon$, and
it is an invariant of the grading. Besides, $\cC$ is $G$-graded with neutral homogeneous component of dimension $1$.
The only possibility is that the support of the grading on $\cC$ is isomorphic to $\ZZ_2^3$, and this support is also an invariant of the grading.
Thus $\Gamma$ is isomorphic to $\GIII_{1}(G,K,h)$. The standard involution $\sigma\colon x\mapsto \bar x$ on $\cC$ preserves the grading on $\cC$,
and the map $\sigma\otimes\tau$ ($\tau$ as above interchanging $\xi$ and $\xi^2$) gives a graded anti-isomorphism from $(\cC,\bullet,n)\otimes(\LL,\rho)$,
endowed with the grading $\GIII_{1}(G,K,h)$, onto itself, but endowed with the grading $\GIII_{1}(G,K,h^2)$.
\smallskip
If the rank is $4$, then $V_e$ is a symmetric composition algebra of dimension $4$, so it is a para-quaternion algebra and hence contains a unique
para-unit $\varepsilon$ (\cite[Proposition 4.43 and Theorem 4.44]{EKmon}), which is thus an invariant of the grading.
The graded para-Cayley algebra $\cC=\{X\in V \,|\,X*\varepsilon=\bar X\}$ is then also an invariant of the grading. Since any grading on the Cayley algebra
with neutral homogeneous component of dimension $\geq 2$ is induced from the Cartan grading (\cite[Corollary 4.13]{EKmon}), the grading on $\cC$ is of the form $\Gamma_\cC(G,\gamma)$ with $\gamma=(e,g,g^{-1})$ for some $g\in G$. Moreover, the condition $g\not\in \langle h\rangle$ follows from the fact $\dim V_e=4$. The pair $\{g,g^{-1}\}$ is an invariant of the grading, and the gradings $\GIII_{4}(G,g,h)$ and $\GIII_{4}(G,g,h^2)$ are proved to be similar with the same argument as for rank $1$.
\smallskip
Finally, if the rank is $2$, then $V_e$ is the para-quadratic composition algebra $(\cK,\bullet,n)$, and this contains exactly three para-units.
Actually, $\cK$ is isomorphic to $\FF\times \FF$ but with the para-Hurwitz product: $(x_1,x_2)\bullet (y_1,y_2)=(x_2y_2,x_1y_1)$.
The para-units are $(1,1)$, $(\omega,\omega^2)$ and $(\omega^2,\omega)$.
If $\varepsilon$ is a para-unit of $V_e$, we may take, as in the proof of Theorem \ref{th:gradings_cyclic}, the para-Cayley algebra
$\cC=\{ X\in V\,|\, X*\varepsilon=\bar X\}$ with para-unit $\varepsilon$, and then identify $V$ with $\cC\otimes\LL$.
This para-Cayley subalgebra is graded with $\dim \cC_e=2$, so the grading is induced from the Cartan grading, and hence is of the form
$\Gamma_\cC(G,\gamma)$ with $\gamma=(g_1,g_2,g_3)$ satisfying $g_1g_2g_3=e$. By \cite[Theorem 4.21]{EKmon}, two such gradings
$\Gamma_\cC(G,\gamma)$ and $\Gamma_\cC(G,\gamma')$ are isomorphic if and only if there is a permutation $\pi\in S_3$ such that
either $g_i'=g_{\pi(i)}$ for all $i=1,2,3$, or $g'_i=g_{\pi(i)}^{-1}$ for all $i=1,2,3$.
It follows that our grading of rank $2$ is similar to $\GIII_{2}(G,\gamma,h)$, and, to ensure that $\dim V_e=2$, we must have $g_i\in G\setminus\langle h\rangle$
for all $i=1,2,3$. Now consider $\cC$ as the Cayley algebra, with product denoted by juxtaposition. Let $e_1$ and $e_2$ be the two nonzero orthogonal idempotents
of the quadratic algebra $\cC_e$, and consider the Peirce subspaces $\cU=\{x\in\cC\,|\, e_1x=x=xe_2\}$, $\cV=\{x\in\cC\,|\, e_2x=x=xe_1\}$,
so that $\cC=\cC_e\oplus\cU\oplus\cV$, and $\cU=\bigoplus_{i=1}^3\cU_{g_i}$, $\cV=\bigoplus_{i=1}^3\cV_{g_i^{-1}}$.
If we take, instead of $\varepsilon$, the para-unit $\varepsilon'=\omega e_1+\omega^2 e_2$ of $\cC_e$, then the corresponding para-Cayley subalgebra is
\[
\begin{split}
\cC'&=\{X\in V\,|\, X*\bigl(\varepsilon'\otimes 1\bigr)=
b_Q\bigl(\varepsilon'\otimes 1,X\bigr)\bigl(\varepsilon'\otimes 1\bigr) -X\}\\
&=(\cC_e\otimes 1) \oplus (\cU\otimes\xi) \oplus (\cV\otimes \xi^2).
\end{split}
\]
This shows that $\GIII_{2}(G,(g_1,g_2,g_3),h)$ is isomorphic to $\GIII_{2}(G,(hg_1,hg_2,hg_3),h)$, and also to $\GIII_{2}(G,(h^2g_1h^2g_2,h^2g_3),h)$.
As in the cases of rank $1$ and $4$, we see that $\GIII_{2}(G,\gamma,h)$ and $\GIII_{2}(G,\gamma,h^2)$ are similar, and the result follows.
\end{proof}
\section{Appendix: graded modules}\label{s:graded_mod}
In what follows, we assume the ground field $\FF$ to be {\em algebraically closed of characteristic $0$}. In \cite{EK14}, graded modules for the classical simple Lie algebras were studied. However, for the simple Lie algebra $\cL$ of type $D_4$, the computation of graded Brauer invariants of irreducible $\cL$-modules was restricted to the case when the $G$-grading on $\cL$ is a {\em matrix grading}, i.e., when $\wh{G}$ fixes the isomorphism class of the natural module. This covers the cases of Type I and II gradings. (As mentioned in the Introduction, Type II reduces to a matrix grading.) Our goal in this appendix is to complete the results of \cite{EK14} by considering Type III gradings on $\cL$.
We already showed in Subsection \ref{sse:lifting1} that, in this case, the graded Brauer invariants of the natural and half-spin modules are equal to the identity element of the $(G/H)$-graded Brauer group of $\FF$. But this is not sufficient to obtain the graded Brauer invariant of every irreducible module. In order to complete the calculation, we will need a few general remarks of independent interest.
\subsection{Background on algebraic groups}
Let $\cL$ be a finite-dimensional semisimple Lie algebra over $\FF$. The corresponding adjoint algebraic group $\bar\cG$ is the group of inner automorphisms $\inaut(\cL)$. Denote by $\tilde\cG$ the associated simply connected group. Once we fix a Borel subgroup $\wt{B}$ in $\tilde{\cG}$ and a maximal torus $\wt{T}$ in $\wt{B}$, we obtain the root system $\Phi$ of $\cL$ and a system of simple roots $\Pi$. The group of characters $\mathfrak{X}(\wt{T})$ is the group of integral weights $\Lambda$. Denote by $\Lambda^+$ the subset of dominant weights and by $\Lambda^{\mathrm{r}}$ the root lattice: $\Lambda^{\mathrm{r}}=\ZZ \Phi$.
For any (connected) semisimple algebraic group $\cG$ with the same root system, there are isogenies $\tilde\cG\xrightarrow{\tilde\pi}\cG\xrightarrow{\bar\pi}\bar\cG$. Moreover (see e.g. \cite[Theorem 8.17]{MT}), with $T=\tilde\pi(\wt{T})$, we have the root space decomposition:
\[
\cL=\Lie(\cG)=\Lie(T)\oplus\Bigl(\bigoplus_{\alpha\in\Phi}\cL_\alpha\Bigr),
\]
and for each root $\alpha\in\Phi$ there is a closed imbedding of algebraic groups $u_\alpha\colon\Gs_a\rightarrow \cG$ such that $tu_\alpha(c)t^{-1}=u_\alpha\bigl(\alpha(t)c\bigr)$ for all $t\in T$ and $c\in\FF$. As usual, we denote by $\Gs_a$ the additive group of $\FF$ and by $\Gs_m$ the one-dimensional torus $\FF^\times=\GL_1(\FF)$.
Once a Chevalley basis $\{h_i,x_\alpha\mid i=1,\ldots,\rank(\cL),\ \alpha\in\Phi\}$ is fixed, if $V$ is a faithful rational module for $\cG$ and we identify $\cG$ with a subgroup of $\GL(V)$, and hence $\cL$ with a subalgebra of $\frgl(V)$, then we may take $u_\alpha(c)=\exp(cx_\alpha)$ (see \cite{Steinberg}).
The group of characters $\mathfrak{X}(T)$ is a lattice with $\Lambda^{\mathrm{r}}=\mathfrak{X}(\wb{T})\leq \mathfrak{X}(T)\leq \Lambda=\mathfrak{X}(\wt{T})$, where $\wb{T}=\bar{\pi}(T)$. Given any symmetry of the Dynkin diagram $\tau\in\Aut(\mathrm{Dyn})$ such that $\tau$ preserves $\mathfrak{X}(T)$, there is a unique automorphism $\sigma_\tau\in\Aut(\cG)$ such that $\sigma_\tau\bigl(u_\alpha(c)\bigr)=u_{\tau(\alpha)}(c)$ for all $\alpha\in\Pi$ and $c\in\FF$ (see e.g. \cite[\S 23.7]{Chevalley} or \cite[p.~156]{Steinberg}). For the adjoint $\bar\cG$ or simply connected $\tilde \cG$, this allows the construction of the semidirect products $\bar\cG\rtimes \Aut(\mathrm{Dyn})$ and
$\tilde\cG\rtimes \Aut(\mathrm{Dyn})$. The first one is isomorphic to the automorphism group $\Aut(\cL)$, where to any $\tau\in\Aut(\mathrm{Dyn})$ as above we associate the unique automorphism of $\cL$, also denoted by $\sigma_\tau$, such that $\sigma_\tau(x_\alpha)=x_{\tau(\alpha)}$ for any $\alpha\in\Pi$ (see \cite[Chapter~IX]{Jacobson}).
Given a dominant weight $\lambda\in\Lambda^+$, consider its stabilizer $S_\lambda$ in $\Aut(\mathrm{Dyn})$. For $\tau\in S_\lambda$, let $\sigma=\sigma_\tau$ be the associated automorphism of $\cL$. Then $\sigma$ extends to an automorphism of the universal enveloping algebra $U(\cL)$ that preserves the left ideal $J(\lambda)$ in \cite[\S 21.4]{Humphreys}, and hence induces an element, also denoted by $\sigma$, in $\GL(V_\lambda)$, where $V_\lambda=U(\cL)/J(\lambda)$ is the irreducible module with highest weight $\lambda$.
By definition of $\sigma$, we have $\sigma(xv)=\sigma(x)\sigma(v)$ for all $x\in\cL$ and $v\in V_\lambda$.
The corresponding representation $\rho\colon\cL\rightarrow \frgl(V_\lambda)$ ``integrates'' to a representation $\tilde\rho\colon\tilde\cG\rightarrow \GL(V_\lambda)$ that extends to
\begin{equation}\label{eq:rhotilde}
\tilde\rho\colon\tilde\cG\rtimes S_\lambda\rightarrow \GL(V_\lambda).
\end{equation}
\subsection{Graded Brauer invariants of irreducible modules}
Let $G$ be an abelian group and let $\Gamma:\cL=\bigoplus_{g\in G}\cL_g$ be a $G$-grading on $\cL$. Recall that the grading is determined by a morphism of affine group schemes, $\eta_\Gamma\colon G^D\rightarrow \AAut(\cL)$. Because of our assumptions on the ground field $\FF$, it is sufficient to consider only the $\FF$-points, i.e., a morphism of algebraic groups $\wh{G}\to\Aut(\cL)$.
Strictly speaking, $\wh{G}$ is an algebraic group only if $G$ is finitely generated, while in general it is a pro-algebraic group. However, since we are dealing with gradings on finite-dimensional objects (algebras and modules), we may replace $G$ by a finitely generated subgroup in all arguments that deal with finitely many objects.
Following \cite{EK14}, we denote the image of a character $\chi\in\widehat{G}$ by $\alpha_\chi$, i.e., $\alpha_\chi(x)\bydef \chi(g)x$ for all $g\in G$ and $x\in\cL_g$. Thus any character $\chi\in\widehat{G}$ induces the automorphism $\alpha_\chi\in\Aut(\cL)$ and an associated diagram automorphism $\tau_\chi\in\Aut(\mathrm{Dyn})=\Aut(\cL)/\inaut(\cL)$. As in \cite{EK14}, given a dominant weight $\lambda\in\Lambda^+$, consider the \emph{inertia group}:
\[
K_\lambda\bydef \{ \chi\in\widehat{G} \mid \tau_\chi(\lambda)=\lambda\}.
\]
This is the inverse image of the subgroup $\bar{\cG}\rtimes S_\lambda\subset\bar\cG\rtimes\Aut(\mathrm{Dyn})\simeq\Aut(\cL)$ under $\eta_\Gamma$. We also write $H_\lambda\bydef K_\lambda^\perp\subset G$, where $\perp$ denotes the ``orthogonal complement'' in $\wh{G}$ of a subgroup of $G$, or vice versa.
Let $\pi\colon\tilde\cG\rtimes\Aut(\mathrm{Dyn})\rightarrow \bar\cG\rtimes\Aut(\mathrm{Dyn})$ be the natural projection, and consider preimages $\tilde\alpha_\chi$ in $\tilde\cG\rtimes\Aut(\mathrm{Dyn})$ for all $\chi\in\widehat{G}$. Recall that the kernel of $\pi$ equals the center of the simply connected group, $Z(\tilde\cG)$. Since $\widehat{G}$ is abelian, the commutator $[\alpha_{\chi_1},\alpha_{\chi_2}]=\alpha_{\chi_1}\alpha_{\chi_2}\alpha_{\chi_1}^{-1}\alpha_{\chi_2}^{-1}$ is trivial, and hence we obtain $[\tilde\alpha_{\chi_1},\tilde\alpha_{\chi_2}]\in Z(\tilde\cG)$ for all $\chi_1,\chi_2\in\widehat{G}$.
The action of the algebraic group $\Aut(\cL)=\bar\cG\rtimes\Aut(\mathrm{Dyn})$, or of $\tilde\cG\rtimes\Aut(\mathrm{Dyn})$, on $\cL=\Der(\cL)$ is the adjoint action. In particular, we have
\begin{equation}\label{eq:rhotildealpha}
\rho\bigl(\alpha_\chi(x)\bigr)=\tilde\rho(\tilde\alpha_\chi)\rho(x)\tilde\rho(\tilde\alpha_\chi)^{-1},
\end{equation}
for any $\chi\in K_\lambda$ and $x\in\cL$. On the other hand, the elements in the center $Z(\tilde\cG)$ act by scalars on $V_\lambda$, so there is an associated morphism
\[
\Psi_\lambda\colon Z(\tilde\cG)\rightarrow \Gs_m,
\]
defined by $\tilde\rho(z)=\Psi_\lambda(z)\id$ for all $z\in Z(\tilde\cG)$. In other words, $\Psi_\lambda$ is the restriction of $\lambda\in\mathfrak{X}(\wt{T})$ to $Z(\tilde\cG)$.
Recall from Section \ref{s:preliminaries} that the elements of the $G$-graded Brauer group $B_G(\FF)$ can be interpreted as alternating bicharacters $\hat{\beta}\colon\wh{G}\times \wh{G}\to\Gs_m$ (which factor through the homomorphism $\wh{G}\to\wh{G_0}$ where $G_0$ is the torsion subgroup of $G$).
There is a unique $(G/H_\lambda)$-grading on the associative algebra $\End(V_\lambda)$ such that $\rho\colon\cL\to\frgl(V_\lambda)$ is a homomorphism of $(G/H_\lambda)$-graded algebras. The \emph{Brauer invariant} $\Br(\lambda)$ (see \cite[Definition 4]{EK14}) is the class $[\End(V_\lambda)]$ in $B_{G/H_\lambda}(\FF)$. Together with the subgroup $H_\lambda$, it measures how far the irreducible module $V_\lambda$ is from admitting a $G$-grading compatible with the $G$-grading on $\cL$. To be precise, $V_\lambda$ admits such a grading if and only if $H_\lambda=1$ and $\Br(\lambda)$ is trivial. Moreover, knowing $H_\lambda$ and $\Br(\lambda)$ for all $\lambda\in\Lambda^+$ allows us to classify the simple objects in the category of finite-dimensional graded $\cL$-modules (see \cite[Theorem 8]{EK14}).
Using the above interpretation of graded Brauer groups, $\Br(\lambda)$ is identified with the commutation factor of the $(G/H_\lambda)$-graded matrix algebra $\End(V_\lambda)$, which is the alternating bicharacter $\hat{\beta}_\lambda\colon K_\lambda\times K_\lambda\rightarrow \Gs_m$ defined as follows. For every $\chi\in K_\lambda$, there exists an invertible element $u_\chi\in\End(V_\lambda)$, unique up to a scalar multiple, such that the action of $\chi$ on $\End(V_\lambda)$ (associated to the $(G/H_\lambda)$-grading) is the inner automorphism
$a\mapsto u_\chi a u^{-1}_\chi$, and then $\hat{\beta}_\lambda$ is defined by the equation $u_{\chi_1}u_{\chi_2}=\hat{\beta}_\lambda(\chi_1,\chi_2)u_{\chi_2}u_{\chi_1}$ for all $\chi_1,\chi_2\in K_\lambda$.
In view of Equation \eqref{eq:rhotildealpha}, we can take $u_\chi=\tilde{\rho}(\tilde{\alpha}_\chi)$, so we obtain a new interpretation of $\hat{\beta}_\lambda$, namely,
\begin{equation}\label{eq:new}
\hat{\beta}_\lambda(\chi_1,\chi_2)=\Psi_\lambda\bigl([\tilde\alpha_{\chi_1},\tilde\alpha_{\chi_2}]\bigr).
\end{equation}
This new point of view on Brauer invariants has the following consequences:
\begin{proposition}\label{pr:Br_trivial}
Let $\cL$ be a semisimple Lie algebra, with Dynkin diagram $\mathrm{Dyn}$, weight lattice $\Lambda$ and root lattice $\Lambda^{\mathrm{r}}$, endowed with a grading by an abelian group $G$. Let $\lambda\in\Lambda$ be a dominant weight.
\begin{enumerate}
\item If $\lambda\in\Lambda^{\mathrm{r}}$, then $\Br(\lambda)$ is trivial.
\item If $\inaut(\cL)$ is simply connected (i.e., $\Lambda=\Lambda^{\mathrm{r}}$) and $\Aut(\mathrm{Dyn})$ is trivial, then any $\cL$-module admits a compatible $G$-grading.
\end{enumerate}
\end{proposition}
\begin{proof}
We use the notation introduced in the previous subsection: $\cG$ is a connected algebraic group with $\Lie(\cG)=\cL$, etc. Recall that the isomorphism classes of irreducible representations of $\cG$ correspond bijectively to the weights in $\mathfrak{X}(T)\cap\Lambda^+$. In particular,
if $\lambda\in\Lambda^{\mathrm{r}}$, then the representation $\rho\colon\cL\rightarrow\frgl(V_\lambda)$ ``integrates'' to a representation $\bar\rho\colon\bar\cG\rtimes S_\lambda\rightarrow \GL(V_\lambda)$, so that $\tilde\rho=\bar\rho\circ \pi$, with $\tilde\rho$ as in \eqref{eq:rhotilde}.
Hence, $\tilde\rho\bigl(Z(\tilde\cG)\bigr)$ is trivial, so for any $\chi_1,\chi_2\in K_\lambda$, we have $\tilde\rho\bigl([\tilde\alpha_{\chi_1},\tilde\alpha_{\chi_2}]\bigr)=\id$ and thus $\hat\beta_\lambda(\chi_1,\chi_2)=\Psi_\lambda\bigl([\tilde\alpha_{\chi_1},\tilde\alpha_{\chi_2}]\bigr)=1$.
For the second part, if $\Aut(\mathrm{Dyn})$ is trivial, then $K_\lambda=\widehat{G}$ and $H_\lambda=1$ for any $\lambda\in\Lambda^+$. Moreover, $Z(\tilde\cG)=1$ here, and hence $\Br(\lambda)=1$.
\end{proof}
\begin{corollary}
Let $G$ be an abelian group and let $\cL$ be a simple Lie algebra of type $G_2$, $F_4$ or $E_8$, endowed with a $G$-grading.
Then any finite-dimensional module for $\cL$ admits a compatible $G$-grading.
\end{corollary}
In \cite{EK14}, this was remarked only for type $G_2$ (with a different argument).
\subsection{Brauer invariants for a Type III grading on the simple Lie algebra of type $D_4$}
Theorems 46 and 48 in \cite{EK14}, which compute Brauer invariants for Type I and II gradings on simple Lie algebras of series $D$, can now be completed with the next result, where $\omega_1,\omega_2,\omega_3,\omega_4$ denote the fundamental dominant weights of the simple Lie algebra of type $D_4$, with $\omega_1$, $\omega_3$ and $\omega_4$ corresponding to the natural and half-spin representations (the outer nodes of the Dynkin diagram), and $\omega_2$ to the adjoint representation (the central node of the diagram).
\begin{theorem}
Let $\cL$ be the simple Lie algebra of type $D_4$ over an algebraically closed field $\FF$ of characteristic $0$. Suppose $\cL$ is graded by an abelian group $G$ and that the grading is of Type III. Let $K=\langle h\rangle^\perp$, where $h\in G$ is the distinguished element (see Section \ref{s:lifting}). Then, for a dominant integral weight $\lambda=\sum_{i=1}^4 m_i\omega_i$, we have the following possibilities:
\begin{enumerate}
\item If $m_1=m_3=m_4$, then $H_\lambda=1$, $K_\lambda=\widehat{G}$ and $\Br(\lambda)=1$.
\item Otherwise $H_\lambda=\langle h\rangle$, $K_\lambda=K$ and $\Br(\lambda)=1$.
\end{enumerate}
\end{theorem}
\begin{proof}
If $m_1=m_3=m_4$, then the diagram automorphisms of order $3$ preserve $\lambda$, so $K_\lambda=\widehat{G}$ and $H_\lambda=1$. Moreover, both $\omega_2$ (the highest root) and $\omega_1+\omega_3+\omega_4$ are in the root lattice $\Lambda^{\mathrm{r}}$, so $\Br(\lambda)=1$ by Proposition \ref{pr:Br_trivial}.
Otherwise we get $K_\lambda=K$, $H_\lambda=\langle h\rangle$, and the associated grading by $\wb{G}\bydef G/\langle h\rangle$ is of Type I. As shown in Subsection \ref{sse:lifting1}, we have $\Br(\omega_1)=\Br(\omega_3)=\Br(\omega_4)=1$ in the $\wb{G}$-graded Brauer group. But, as we just observed, $\omega_2\in \Lambda^{\mathrm{r}}$, so $\Br(\omega_2)=1$, too. The result follows now from \cite[Proposition 10]{EK14}.
\end{proof}
\begin{corollary}
The simple $\cL$-module $V_\lambda$ admits a $G$-grading making it a graded $\cL$-module if and only if $m_1=m_3=m_4$.
\end{corollary}
\subsection{Analogy with Tits algebras}
Tits algebras (see e.g. \cite{Tits} or \cite[\S 27]{KMRT}) were introduced to study representations of semisimple algebraic groups over an arbitrary field.
In the present work, the ground field is assumed algebraically closed and of characteristic $0$, but we are interested in graded representations.
As before, let $G$ be a finitely generated abelian group and let $\cL$ be a semisimple Lie algebra endowed with a $G$-grading. Consider the associated morphism $\eta\colon\wh{G}\rightarrow \Aut(\cL)=\bar\cG\rtimes\Aut(\mathrm{Dyn})$, where $\bar \cG=\inaut(\cL)$. Any $\chi\in\wh{G}$ induces an automorphism $\alpha_\chi\in\Aut(\cL)$ and hence a diagram automorphism $\tau_\chi\in\Aut(\mathrm{Dyn})=\Aut(\cL)/\inaut(\cL)$. Thus $\wh{G}$ acts on $\Lambda$, $\Lambda^+$ and $\Lambda^{\mathrm{r}}$.
Let $H$ be the finite subgroup of $G$ such that $H^\perp$ is the inverse image of $\inaut(\cL)$ under $\eta$. Fix any subgroup $M$ of $G$ contained in $H$ and denote $K=M^\perp\subset\widehat{G}$ and $\wb{G}=G/M$. For any $\lambda\in\Lambda^+$ with $K\subset K_\lambda$, or, equivalently, $H_\lambda\subset M$, or $\lambda\in(\Lambda^+)^K$ (the set of dominant weights fixed under the action of $K$), we may consider the Brauer invariant of $\lambda$ in the $\wb{G}$-graded Brauer group $B_{\wb{G}}(\FF)$. Denote it by $\Br_{\wb{G}}(\lambda)$. In this way we get a map
\[
\begin{split}
(\Lambda^+)^K&\longrightarrow B_{\wb{G}}(\FF),\\
\lambda\quad &\mapsto\quad \Br_{\wb{G}}(\lambda),
\end{split}
\]
which is multiplicative by \cite[Proposition 10]{EK14} or directly from Equation \eqref{eq:new}.
Moreover, if $\lambda\in(\Lambda^{\mathrm{r}})^K$, then $\Br_{\wb{G}}(\lambda)=1$ by Proposition \ref{pr:Br_trivial}.
As in \cite[Corollary 3.5]{Tits} or \cite[Theorem 27.7]{KMRT}, this map ``extends'' to a group homomorphism
\[
\beta_{\wb{G}}\colon(\Lambda/\Lambda^{\mathrm{r}})^K\rightarrow B_{\wb{G}}(\FF),
\]
where for any $\lambda+\Lambda^{\mathrm{r}}\in(\Lambda/\Lambda^{\mathrm{r}})^K$ we consider the unique minimal weight $\hat\lambda\in \Lambda^+$ in the same class modulo $\Lambda^{\mathrm{r}}$ and define $\beta_{\wb{G}}(\lambda+\Lambda^{\mathrm{r}})\bydef\Br_{\wb{G}}(\hat\lambda)$.
\begin{remark}
The Brauer invariant $\Br(\lambda)$ of $\lambda\in \Lambda^+$ is precisely $\beta_{G/H_\lambda}(\lambda+\Lambda^{\mathrm{r}})$.
\end{remark}
In the setting of \cite{Tits}, a simply connected semisimple group over an arbitrary field $\FF$ is considered, and the group that acts on $\Lambda$, $\Lambda^+$ and $\Lambda^{\mathrm{r}}$ is the absolute Galois group $\Gamma$ of $\FF$, obtaining a group homomorphism $\beta\colon(\Lambda/\Lambda^{\mathrm{r}})^\Gamma\rightarrow B(\FF)$ (the classical Brauer group). Also, if $\lambda+\Lambda^{\mathrm{r}}\in \Lambda/\Lambda^{\mathrm{r}}$ is not $\Gamma$-invariant, then one has to consider the subgroup $\Gamma_\lambda\bydef\{\gamma\in\Gamma\;|\;\gamma(\lambda+\Lambda^{\mathrm{r}})=\lambda+\Lambda^{\mathrm{r}}\}$, the field $\FF_\lambda\bydef(\FF_{\mathrm{sep}})^{\Gamma_\lambda}$ and the homomorphism $(\Lambda/\Lambda^{\mathrm{r}})^{\Gamma_\lambda}\rightarrow B(\FF_\lambda)$. This is analogous to what we did restricting from $\wh{G}$ to $K_\lambda$.
\section*{Acknowledgments}
The second author would like to thank the Instituto Universitario de Matem\'aticas y Aplicaciones and Departamento de Matem\'aticas of the University of Zaragoza for support and hospitality during his visit in June 2014. Also, the authors are grateful to Kirill Zainoulline of the University of Ottawa for pointing out the analogy with Tits algebras.
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Why Every Law Firm Needs Market Research
If your law firm is one of many that will host its annual Partner retreat this spring it’s likely your agenda includes a set of challenges that are bedeviling the entire legal industry—from consolidation and commoditization to differentiation and the complexities of the modern digital marketplace.
These are complicated issues, and not every firm can, or should, address them in the same way. Which begs the question: how do you avoid falling back on generic advice and make the right marketing decisions for your firm in an ever-changing marketplace? Is there objective evidence you can bring to bear?
Download The Professional Services Guide to Research
Before I address these questions, I want to tell you the story of a remarkable law firm that you may not have heard of—one which is positioning itself for long-term success, whatever the market brings.
Nishith Desai Associates is a business-facing law firm based in Mumbai, India. Founded in 1989, the firm has grown steadily over the years and now has offices in eight cities around the world, from Silicon Valley to Singapore. But that’s not what makes this firm special. What is so remarkable is what they are doing today to future-proof their business.
They recognized that the future brings uncertainty, so they resolved to tackle the issue head-on. Nishith Desai Associates embraced the buzziest of buzzwords—“innovation”—but they committed themselves to do it right. To this end, here are just some of the amazing things they do:
- Conduct ongoing research into emerging areas of the law. They have published their findings in hundreds of articles and research papers, which they make freely available on their website. This research into bleeding-edge topics such as blockchain technology and self-driving vehicles positions the firm as thought leaders at the frontiers of social and technological change.
- Research their marketplace. Nishith Desai Associates knows that their clients don’t stand still, so they interview them on a regular basis to monitor their evolving needs.
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- Enhance the visibility of their attorneys. Many of their attorneys have acquired specialized expertise, so the firm is in the process of helping these individuals become more prominent in their field. This in turn, will take advantage of the halo effect and make the firm better known and more reputable.
- Represent many of the world’s most innovative companies. These associations strengthen the firm’s reputation as innovators, themselves.
- Build an institute devoted to innovation and the law. Early in 2018, Nishith Desai Associates will open the Blue Sky Innovation Centre, a multi-million dollar facility dedicated to exploring the legal issues of tomorrow.
The initial results are impressive. The firm’s annual revenue growth is several times the industry average, and The Financial Times has named them Asia-Pacific’s Most Innovative Law Firm, as well as India’s Most Innovative Law Firm four years in a row.
Most important, however, they are doing everything they can to sharpen their understanding of the changing marketplace so they can adjust course accordingly. The key take away from this story is not that your firm should try to emulate Nishith Desai Associates in every way but to recognize that as your clients change, your firm has to be prepared to change with them.
Which brings us back to the central point of this post. There is only one way to bring clarity to the marketplace and set your firm up for ongoing success: ask your clients and prospects directly. And the best way to do that is to conduct formal research on your market.
My colleague Liz Harr has written a terrific article explaining the basics of marketing research, so I don’t need to retread that territory here. Instead, I hope to convince you that going to the trouble and expense of conducting research is not only worth it, but absolutely essential if you want to thrive in the turbulent times ahead.
No doubt, research is a common—even vital—feature of your law practice. You rely on databases, media archives, witness interviews, expert opinions and countless other sources of primary and secondary research to shape and win your cases.
Marketing research should be no less influential in shaping the future of your firm. Imagine if you went to trial without an understanding of the case’s larger context. What if you had no idea what arguments your opponents might present? You would be at a significant (probably colossal) disadvantage.
Now imagine marketing your firm without understanding how today’s prospective clients seek out counsel, what characteristics they look for in a law firm, what marketing message instill them with confidence or how they choose one firm over another. These factors change over time. And firms that monitor these changes always have the advantage.
You don’t have to become another Nishith Desai Associates to prepare your firm for the future. You just need to be a few degrees more aware of what’s changing in the marketplace than your competitors. And research is the key. Start discovering what clients want and need today, so you can be ready to deliver those qualities in the year to come.
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And don’t try to conduct the interviews yourself. Hire a reputable third-party firm that is expert at coaxing insightful answers out of busy or unenthusiastic interviewees. People are also more inclined to provide honest feedback if the person asking the questions has no emotional stake in the answers.
To close, I’d like to instill a sense of urgency in you. If you are indeed planning a spring retreat (or even one later in the year), you need to start equipping yourself with the insights to make informed decision now. Research takes time—typically six to eight weeks—and you would be wise to hit the ground running in the winter months so you have time to turn your findings into solid strategic recommendations for your larger team.
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- Read more about this topic in our free Professional Services Guide to Research.
- Learn how to make your firm better known in the marketplace and leverage your expertise to build a powerful reputation. Check out our free Visible Firm® Guide today!
- Download our free book Inside the Buyer’s Brain to learn more about what sets growth leaders apart from runners-up.
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Don’t know where to start? Hinge can help your firm conduct the research you need to see your marketplace clearly. And when you are ready to take your firm’s marketing to the next level, our flagship Visible Firm® program is designed to drive unprecedented visibility and?
| 306,749
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TITLE: Inner product on a von Neumann algebra
QUESTION [3 upvotes]: Let $M$ be a $\sigma$-finite von Neumann algebra (one which admits a faithful normal state) acting on a Hilbert space $H$. Denote its faithful normal state by $\omega$.
We can define an inner product on $M$ by
$$\left< x, y \right> := \omega (y^*x).$$
Let $(\pi, K, \xi)$ be a cyclic GNS representation of $(M, \omega)$, then
$\omega (x) = \left<\xi, \pi(x)\xi\right>$. So we can see that the norm which comes from this inner product is
$$\|x\|_L^2= \left< x, x \right> = \omega(x^*x)= \left< \xi, \pi(x^*x) \xi \right> = \left\| \pi(x)\xi\right\|^2. $$
Since we can identify $M$ with $\pi(M)$ then we have that
$\|x\|_L = \|x\xi\|$.
We see from that the relation with the standard norm on $\mathcal{B}(H)$, denote it by $\| \cdot \|_{\infty}$, and $\| \cdot \|_L$ are the following $\|x\|_L \leq \|x\|_{\infty}$.
My questions are: Is $(M, \| \cdot \|_L)$ a Hilbert space? I know that if $M=\mathcal{B}(\mathbb{C}^n)$ that the answer is true and I doubt that it is true when $H$ is infinite dimensional.
If the answer is no, how the complition of $M$ with $\| \cdot \|_L$ looks?
If the answer is yes, are $\|\cdot \|_L$ and $\| \cdot \|_{\infty}$ equivalent?
REPLY [3 votes]: Is $(M, \Vert \cdot \Vert_L)$ a Hilbert space?
No, it is not in general a Hilbert space. This is clear from the commutative case.
If $M$ is of the form $L_\infty(\Omega, \mu)$ and the state $\omega$ is given by integrating against the finite measure $\mu$, then $(M, \Vert \cdot \Vert_L) = (L_\infty(\Omega,\mu) \cap L_2(\Omega, \mu), \Vert \cdot \Vert_2)$, which is usually not complete. Of course, one can complete $(M, \Vert \cdot \Vert_L)$ in the $\Vert \cdot \Vert_L$-norm to obtain a Hilbert space.
Are $\Vert \cdot \Vert_L$ and $\Vert \cdot \Vert_\infty$ equivalent?
No, they are usually not equivalent. The norms $\Vert \cdot \Vert_L$ and $\Vert \cdot \Vert_\infty$ are not usually equivalent as can be seen by considering elements $x_n$ in $L_\infty([0,1])$ with $\Vert x_n \Vert_2=1$ and
$\Vert x_n \Vert_\infty=n$.
What does the completion of $M$ with respect to $\Vert \cdot \Vert_L$ look like?
I'm not sure exactly what you're looking for here. Unless we have a fairly explicit description of $M$, it will be hard to give an explicit description of its completion with respect to $\Vert \cdot \Vert_L$. So I will give a very explicit example and then make some general remarks.
Let $\Gamma$ be a countable discrete group and let $L(\Gamma)$ be its group von Neumann algebra. I now recall how to construct $L(\Gamma)$. Let $\ell_2 (\Gamma)$ be the Hilbert space with orthonormal basis $\{\delta_g \mid g \in \Gamma\}$. Let $\lambda_g$ (for $g \in \Gamma$) be the bounded linear operator in $\ell_2(\Gamma)$ satisfying $\lambda_g \delta_h=\delta_{gh}$, for each $h \in \Gamma$. Then $L(\Gamma)$ is the von Neumann algebra (inside $B(\ell_2(\Gamma)$)
$$
\{ \lambda_g \mid g \in \Gamma \}'' = \overline{\textrm{span} \, \{\lambda_g \mid g\in \Gamma\}}^{wot},
$$
where $''$ denotes the double commutant, and $wot$ stands for the weak operator topology.
Let $\omega$ be the normal faithful state on $L(\Gamma)$ satisfying $$\omega(\lambda_g)=( \lambda_g \delta_e , \delta_e ),$$
where $e$ is the identity element of $\Gamma$. Then
the completion of $(M, \Vert \cdot \Vert_L)$ will be isometrically isomorphic to $\ell_2 (\Gamma)$ via the map sending $\lambda_g$ to $\delta_g$.
What can we say more generally?
Let's specialise for the moment to the case where the state $\omega$ is tracial. The completion of $(M,\Vert \cdot \Vert_L)$ is then denoted by $L_2(M)$ (and one can define $L_p(M)$ spaces as well by completing $M$ with respect to the norm $\Vert x \Vert_p= \omega( |x|^p)^{1/p}$). (In fact, if one works harder, one can define these noncommutative $L_p$-spaces when $\omega$ is not a trace.) These spaces can be considered as abstract completions of $M$ or they can be described spaces of (unbounded) operators satisfying certain conditions. But that approach gets quite technical.
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Chinese smartphone maker Oppo has overtaken the Apple for the first time to become the number two regarding sales in India behind the Samsung in August 2016. According to the reports of Germany-based Market Research GFK, Oppo has clocked the growth of 16 percent in comparison to the previous month.
Samsung which has been the leader of market sales for many years, and it has been retaining the top position with 46.7% value market share in the month.
As per the reports, Oppo constituted 8 percent of India’s market sales by value came top above the opposed Apple’s market sales to 6.3 percent. Reliance Lyf and Micromax have trailed upto 6.1% and 5.3% respectively.
Lenovo Z2 Plus Sale Comes with Exciting Offers on Amazon
“India is a priority market for us and we have caught the trend that users like taking photos and selfies by phones. Therefore, we will continue investing in photography technology, design and offering the best consumer experience,” said Sky Li, Oppo Global Vice President and President of Oppo India in a statement.
Oppo has recently launched the latest Oppo selfie expert Oppo F1s in India in the month of August. It came up with 16MP front camera, beauty function and fingerprint unlock. It also has good battery endurance. Oppo is also looking to invest Rs. 1000 crores to set up a manufacturing unit in Andhra Pradesh which is likely to create 20,000 jobs in the state.
Anonymous Report: Is Reliance Jio Really Selling the Customer Data for Revenue Generation to Compensate the Discount Rates Offered
The decrease in market share makes the launch of iPhone 7 and iPhone 7 Plus even more pivotal for Apple, which has been facing the tough competition from the Chinese mobile makers. In recent months the Chinese smartphones are gaining huge response for the users in terms of numbers and value.
In the Indian market, Oppo is expanding its offline presence by targeting 35,000 point of sales and 180 sales service centres to its customers for a better experience.
| 349,710
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If I married the guy I fell in love with in college, my kids wouldn't exist. I'd also probably be on my third marriage by now.
So I was slightly amused at the Valentine’s Day “Straight Talk” from Susan Patton in the Wall Street Journal last week, urging young women to wisely spend their time in college hunting for husbands.
“Despite all of the focus on professional advancement, for most of you the cornerstone of your future happiness will be the man you marry,” wrote Patton, who told ambitious 20-somethings to stop ordering sushi and swooning over "Downton Abbey" and start finding a suitor while surrounded by a campus of smart, interesting men.
The self-described “Princeton Mom” started finger wagging a year ago in The Daily Princetonian, where she encouraged co-eds to jump her youngest Princetonian son before he graduated and faded out of reach – dashing their hopes and dreams of ever having a happy, fulfilling life as a Princetonian wife.
Apparently that didn’t work. Now she has a new book – Marry Smart: Advice for Finding ‘The One’ – due out in March, and the wise editorial wizards at the WSJ thought their 70-to-death-year-old readership might find merit in her quaint ideas, cribbed from a Good Housekeeping advice column still smudged with Jell-O stains from 1952.
Women of Princeton, I beg of you, will somebody please sleep with this guy so his mom will shut up?
My daughters are four years away from college, but the thought that people like Patton still exist makes me realize I better pen my own advice. Here’s what I plan to tell them.
- College is a hunting ground for knowledge, not a husband. Be curious. Take classes that challenge and invigorate you.
- Intelligence is not a liability.
- College is the best time to seek new friendships and relationships. You will be surrounded by many different men, women and various beings in between. Break out of your comfort zone and get to know people different from you. You won’t regret it.
- Sleep with as many people as you can. Ha, just kidding. Be smart and selective, but don’t you dare believe that sex is a commodity or that somehow your value is tied to it. Contrary to Ms. Patton’s advice, you are not a cow to be bought.
- College is not the time to search for a sperm donor (or a sexually-transmitted disease). Condoms, please.
- Do not expect to find happiness in another human being. True happiness is knowing who you are and liking yourself, and that takes years to figure out.
- Don’t look for a partner to take care of you. Statistically, odds are you both will work, which makes it even more important to pursue a career that satisfies you.
- Feminism is not a bad word. It just means you want an equal shot at being human.
- Your education is far from over after graduation. Go on adventures. Climb a mountain. Backpack through a country. Learn a new language. You grow and change so much in your 20s. It is very important for you to know you can handle the world on your own.
- Don’t ever define yourself by your relationship status.
- We should all work at becoming the person we want in a mate. Finding a partner can be satisfying, but if you let that search consume you, life will be very dull.
- Eat lots of sushi and watch "Dowton Abbey" as much as possible.
| 171,663
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***
I'm working at two projects now. The first one is a cross-stitching picture, I'm using a set that I received from Ariadna some time ago. The second one will remain a secret, I just can tell you that it'll be something new in my artistic development. :)
Brak komentarzy:
Prześlij komentarz
Napisz, co o tym myślisz. :) Dziękuję. :)
| 132,954
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It's that time again: duck boat time! After securing a 5-1 World Series win against the Los Angeles Dodgers on Sunday, the Red Sox are coming home, and Boston's gearing up to fete them in style. Moments like Eduardo Nunez's three-run home run in Game 1 and Mitch Moreland's epic homer that clinched Game 4 will now make it into the history books, and somehow watching all 18 innings of Game 3 was worth it in the end. All that's left to do now is fire up the duck boats and take to the streets (again).
Here's everything you need to know about the Red Sox World Series victory parade.
Free, Awesome Activities You Can Do in Los Angeles for Every Day of the Week
Where and when will the parade start and end?
According to details announced by Boston Mayor Marty Walsh and the City of Boston, the parade kicks off bright and early at Fenway Park on Wednesday, October 31 at 11am. Mostly following Boylston St, the parade and its scores of duck boats will end at the corner of Cambridge St and New Sudbury St near Boston City Hall.
What is the parade route?
The route starts at Fenway Park on Landsdowne St and will hang a right on Ipswich St, continuing on Ipswich before taking a left on Boylston St. It will follow Boylston St through Back Bay and up to Boston Common opposite the flow of traffic before making a left onto Tremont St. From Tremont St it's a left onto Cambridge until you get to City Hall and New Sudbury St.
How do I get there?
Don't drive. Just don't. There is a long list of streets under a parking ban that goes into effect at 12am Tuesday and continues on additional streets through Wednesday. Even if you don't plan to park on them, parking will be hell, and you'll probably just make it harder for yourself and kill any time you saved by driving. "Do not try to drive to the parade," Walsh said at a press conference, encouraging folks to take public transit instead.
The MBTA and commuter rails will provide additional service for Sox fans on Wednesday morning. For more information on this, check out the city's website.
What will the weather be like?
The parade is on whether it rains, pours, hails, snows, or shines, but the weather forecast mostly looks good, leaning toward the "shine" end of the spectrum. As of mid-day Tuesday, the report for Wednesday is partly cloudy, with a high of around 56 degrees, low of 53 degrees, and a 10% chance of precipitation. Fingers crossed it stays that way.
What should I wear?
Obviously every shred and strip of Sox gear you own. And the weather doesn't spell doom for anyone out there, but my loving mom would probably advise me to bring a sweater, if not a coat, so I'm going to tell you to do the same.
Anything else I should know?
Have a blast! But also, for the love of Christ, stay safe. Don't be an asshole. Don't break stuff that doesn't belong to you. Don't break any bones that do belong to you! I won't tell you "don't drink too much," but definitely don't climb high things and jump off.
In his remarks at the press conference, the mayor emphasized safety to a city known for its rowdiness. .”
Sign up here for our daily Boston email and be the first to get all the food/drink/fun the Hub has to offer.
| 213,660
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This is the picture of class
by Jed Eckert on March 22, 2010 | Comments Off
Looking trashed out of her mind, Avril Lavigne exits a tattoo parlor with her husband Deryck Whibley. The couple reportedly received matching tattoos and you can see that they have their arms bandaged.
These two have been spending a lot of time together lately. Think they’re still getting divorced?
Anyone want to take a stab at what their tattoos say?
For the first word in gossip, friend us on Facebook and follow us on Twitter. You'll be glad you did!
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| 228,585
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E2 exclusive marketing and consulting partner of Fotmob since January 2017
After entering into a cooperation with E2 at the end of 2016, Fotmob, one of the world’s premier Livescore apps, has now benefited from its exclusive marketing and consulting services since the beginning of 2017.
In February 2017, the E2 sales team attended the Mobile World Congress in Barcelona, in order to improve their mobile expertise and pick up on the latest trends at an early stage. The goal is to be the world’s largest marketer of Livescore apps by 2018.
Team:
- Christian Czerko - Head of International Sales
Sebastian Paris - International Sales Manager
Benedikt Urban - International Sales Assistant
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Look. Trades and favours will be exchanged. You know.
Because JT, he’s an actor now. He’s jumping from project to project, many more projects than his girlfriend, who is supposed to be an actor too.
This week Justin Timberlake started work on yet another movie, I’m.Mortal. They may have changed the name. Am too lightheaded from Jakey and the Love & Other Drugs press screening to care or google. But the point is, he was at work on set today with Amanda Seyfried, and given that he tried to ride up on the situation with his previous attractive co-star, who’s to say that he won’t try with Amanda? It would be such a blow to poor Shelfy, what with Amanda getting first look at so many projects these days.
But he is an attractive motherf-cker, isn’t he? In a well cut suit, a perfect white shirt, his pants fit beautifully, a newly shaved head... yes, Justin is a goodlooking little punk, so full of his own sh-t.
I’m.Mortal is about eternal youth, a society that rewards it to the wealthy. In the future, we will buy Time the way we buy designer bags. Pip plays a poor boy who inherits Time but someone wants to take it away from him, so he kidnaps an heiress. It’s an intriguing plot, but he’s carrying the movie. So, yeah, there’s that.
Photos from Splashnewsonline.com
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Every grand and petit juror drawn and summoned to attend and serve at any term of a district court shall report to such court at the time and place designated in such summons. A failure to so report shall constitute contempt of court. On the first day of the term fixed for the attendance of either the grand or the petit jurors, or as soon thereafter as may be, the court shall ascertain whether the persons summoned to attend at such term as grand or petit jurors, as the case may be, have reported for duty as required by law; and, if it shall find a failure on the part of any person so summoned to report, it shall at once cause an attachment to issue against the juror, which shall be served by the sheriff or a deputy, and shall be forthwith arrested and brought before the court to be dealt with according to law. Nothing in this section contained shall render liable to jury duty any person who is exempt by law.
Official Publication of the State of Minnesota
Revisor of Statutes
| 232,757
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\begin{document}
\title{Cohomology of vertex algebras}
\author[Jos\'e I. Liberati]{Jos\'e I. Liberati$^*$}
\thanks {\textit{$^{*}$Ciem - CONICET (FAMAF), Medina Allende y
Haya de la Torre, Ciudad Universitaria, (5000) C\'ordoba,
Argentina.
e-mail: joseliberati@gmail.com}}
\address{{\textit{Ciem - CONICET (FAMAF), Medina Allende y
Haya de la Torre, Ciudad Universitaria, (5000) C\'ordoba -
Argentina. E-mail: joseliberati@gmail.com}}}
\date{March 25, 2016}
\maketitle
\begin{abstract}
Let $V$ be a vertex algebra and $M$ a $V$-module. We define the
first and second cohomology of $V$ with coefficients in $M$, and
we show that the second cohomology $H^{2}(V, M)$ corresponds
bijectively to the set of equivalence classes of square-zero
extensions of $V$ by $M$. In the case that $M=V$, we show that the
second cohomology $H^{2}(V, V)$ corresponds bijectively to the set
of equivalence classes of first order deformations of $V$.
\end{abstract}
\section{Introduction}\lbb{intro}
\
This is the second of a series of papers (see \cite{L}) trying to
extend certain restricted definitions and constructions developed
for vertex operator algebras to the general framework of vertex
algebras without assuming any grading condition neither on the
vertex algebra nor on the modules involved, and we make a strong
emphasis on the commutative associative algebra point of view
instead of the Lie theoretical point of view.
In this work we define the first and second cohomology of a vertex
algebra $V$ with coefficients in a $V$-module $M$, and we show
that the second cohomology $H^{2}(V, M)$ corresponds bijectively
to the set of equivalence classes of square-zero extensions of $V$
by $M$. In the case that $M=V$, we show that the second cohomology
$H^{2}(V, V)$ corresponds bijectively to the set of equivalence
classes of first order deformations of $V$. If we restrict it to a
vertex algebra given by an associative commutative algebra, then
we clearly obtain the Harrison's cohomology.
In \cite{H1,H2}, Huang developed the cohomology theory of graded
vertex algebras using analy- tical methods and complex variables. In
the present paper we develop the cohomology theory for vertex
algebras (without grading conditions) using algebraic methods and
formal variables, obtaining a very simplified, clear and nice
theory.
In the definition of $H^2(V,M)$, Huang used two complex
variables. In fact, in the proofs of the theorems that relate the
second cohomology with extensions and deformations \cite{H2},
Huang passed from two complex variables to one formal variable. We
directly use one formal variable (cf. \cite{BKV} versus
\cite{DK}). So, if we add grading conditions to our definitions
and constructions, then we obtain a simpler algebraic version of
the results in \cite{H2}.
We keep using the more comfortable notation introduced in
\cite{L}, where the map $Y(a,z)b$ is replaced by $a\z b$.
This paper is organized as follows. In section 2, we introduce the
basic definitions and notations. In section 3, we define the first
and second cohomology of a vertex algebra $V$ with coefficients in
a module $M$. In section 4, we show that the second cohomology
$H^{2}(V, M)$ corresponds bijectively to the set of equivalence
classes of square-zero extensions of $V$ by $M$. In section 5, in
the case that $M=V$, we show that the second cohomology $H^{2}(V,
V)$ corresponds bijectively to the set of equivalence classes of
first order deformations of $V$.
Unless otherwise specified, all vector spaces, linear maps and
tensor products are considered over an algebraically closed field
$\kk$ of characteristic 0.
\
\vskip .5cm
\section{Definitions and notation}\lbb{def }
\
In order to make a self-contained paper, in this section we
present the notion of vertex algebra and their modules. Our
presentation and notation differ from the usual one because we
want to emphasize the point of view that vertex algebras are
analog to commutative associative algebras with unit.
\
Throughout this work, we define $(x + y)^n$ for $n\in\ZZ$ (in
particular, for $n < 0$) to be the formal series
\begin{align*}
(x+y)^n=\sum_{k \in \mathbb{Z}_+} \binom{n}{k} x^{n-k}y^k,
\end{align*}
where $\binom{n}{k}=\frac{n(n-1)...(n-k+1)}{k!}$.
\
\definition \label{def VA}
A {\it vertex algebra} is a quadruple $\VA$ that consists of a
vector space $V$ equipped with a linear map
\begin{align}\label{zzzz}
\z \, :V &\otimes V \longrightarrow V((z))\\
a &\otimes b \, \, \, \longmapsto a\z b, \nonumber
\end{align}
a distinguished vector $\vac$ and $D\in\,$End$(V)$
satisfying the following axioms ($a,b,c \in V):$\\
\vskip.1cm \noindent$\bullet$ {\it Unit}:
\begin{align*}
\vac \z a= a \quad \textrm{ and }& \quad a \z
\vac=e^{zD}a;
\end{align*}
\noindent $\bullet$ {\it Translation - Derivation}: \vskip .1cm
\begin{align*}
(Da)\z b= \ddz (a\z b), \quad D(a\z b)=(Da)\z b + a\z (Db);
\end{align*}
\vskip .3cm
\noindent$\bullet$ {\it Commutativity}:
\begin{align*}
a \z b= e^{zD} (b\mz a);
\end{align*}
\vskip .3cm
\noindent$\bullet$ {\it Associativity}: For any $a,b,c\in V$,
there exist $l\in \NN$ such that
\begin{align} \label{associatt}
(z+w)^l\, \ (a \z b) \w c = (z+w)^l\, \ a\zw (b\w c) .
\end{align}
\vskip.3cm
\
Observe that the standard notation $Y(a,z)\, b$ for the
$z$-product in (\ref{zzzz}) has been changed. We adopted this
notation following the practical idea of the $\lambda$-bracket
in the notion of Lie conformal algebra (also called vertex Lie
algebra in the literature), see \cite{K}.
The commutativity axiom is known in the literature as
skew-symmetry (see \cite{LL, K}), but for us it really corresponds
to commutativity. We want to emphasize the point of view of a
vertex algebra as a generalization of an associative commutative
algebra with unit (and a derivation), as in \cite{BK, L1}, and
having in mind the usual (trivial) example of holomorphic vertex
algebra given by any associative commutative algebra $A$ with unit
$\one$ and a derivation $D$, with the $\z$-product defined by $a\z
b:=(e^{zD} a)\cdot b$. In the special case of $D=0$, we simply
have an associative commutative algebra with unit and $a\z b:= a
\cdot b$, and with the definition presented before, it is
immediate to see, without any computation, that this is a vertex
algebra, and this is the reason for us to call the usual
skew-symmetry axiom as commutativity.
An equivalent definition can be obtained by replacing the
associativity axiom by the {\it associator formula} (which is
equivalent to what is known in the literature as the iterate
formula (see \cite{LL}, p.54-55) or the $n$-product identity (see
\cite{BK})):
\begin{align} \label{associator}
(a \z b) \y c- a\zy (b\y c)= b\y (a\yz c - a\zy c),
\end{align}
for $a,b,c\in V$. Observe that in the last term of the associator
formula we can not use linearity to write it as a difference of $\
b\,_{\dot w}\big(\,a\,_{\dot {w+z}}\,c\big)$ and $\ b\,_{\dot
w}\big(\,a\,_{\dot {z+w}}\,c\big)$, because neither of these
expressions in general exists (see \cite{LL}, p.55). This
alternative definition of vertex algebra using the (iterate
formula or) the associator formula is essentially the original
definition given by Borcherds \cite{Bo1}, but in our case it is
written using the generating series in $z$ instead of the
$n$-products.
It is well known (see \cite{LL}) that in the two equivalent
definitions that we presented, some of the axioms can be obtained
from the others, but we prefer to make
emphasis on the properties of $D$ and the explicit formula for the multiplication
by the unit.
\
\begin{definition} A {\it module} over a vertex algebra $V$ is
a vector space $M$ equipped with an endomorphism $d$ of $M$ and a
linear map
\begin{align*}
&V\otimes M \longrightarrow M((z)) \\
&(a,u)\ \longmapsto \ a\, \Mz \, u
\end{align*}
satisfying the following axioms ($a \in V$ and $u\in M$):
\\
\vskip.1cm \noindent$\bullet$ {\it Unit}:
\begin{align*}
\label{1} \vac \Mz u= u;
\end{align*}
$\bullet$ {\it Translation - Derivation}:
\begin{align*}
(Da)\Mz u= \ddz (a\Mz u), \quad d(a\Mz u)=(Da)\Mz u + a\Mz (d\,u);
\end{align*}
\vskip.2cm \noindent $\bullet$ {\it Associativity}: For any
$a,b\in V$ and $u\in M$, there exist $l\in\NN$ such that
\begin{equation}\label{aaaa}
(z+w)^l\, \ (\,a\, _{\dot z} \, b\, )\Mw u\ =(z+w)^l\ \,a \Mzw (\, b\Mw u\,)\ .
\end{equation}
\end{definition}
\
\vskip .2cm
Sometimes, if everything is clear, we shall use $a\z u$ instead of
$a \Mz u$. Obviously, $V$ is a module over $V$. We follow
\cite{BK}, in the definition of module, because we need to work
with this $\kk[d]$-module structure
(similar to the situation of Lie conformal algebras \cite{K}).
Any $V$-module $M$ satisfies the {\it weak commutativity} or {\it
locality}: for all $a,b\in V$, there exist $k\in\NN$ such that
\begin{equation}\label{locality}
(x-y)^k \, a\x (b\y u)= (x-y)^k \,b\y (a\x u) \quad \hbox{for all }u\in M.
\end{equation}
Let $M$ and $N$ be $V$-modules, a {\it $V$-homomorphism} or a {\it
homomorphism of $V$-modules} from $M$ to $N$ is a linear map
$\varphi:M\to N$ such that for $a\in V$ and $u\in M$
$$
\varphi(a\z u)= a\z \varphi(u) \ \ \mathrm{ and }\ \ \varphi(d u)=
d \varphi(u) .
$$
A subspace $W$ of a vertex algebra $V$ is called an {\it ideal of}
$V$ if $a\z b\in W$ for all $a\in V$ and $b\in W$.
\
\vskip .5cm
\section{Definition of lower cohomologies}\lbb{first}
\
Let $V$ be a vertex algebra, and $M$ a (left) $V$-module. We
define the {\it right action} of $V$ on $M$ by
\begin{equation*}\label{}
m\z a=e^{zd}(a\mz m),
\end{equation*}
\vskip .3cm
\noi for $a\in V$ and $m\in M$. A linear map $f:V\rightarrow M$ is
called a {\it vertex derivation} if
\begin{equation*}
f(a\z b)=a\z f(b)+ f(a)\z b
\end{equation*}
\vskip .3cm
\noi for $a,b\in V$. We denote by VDer$(V,M)$ the space of all
such derivations.
Now, we consider
$$
0\xrightarrow{\quad\ \
\quad}C^0(V,M)\xrightarrow{\quad\de_0\quad}C^1(V,M)\xrightarrow{\quad\de_1\quad}C^2(V,M)
$$
\vskip .3cm
\noi with $C^0(V,M)=M$ and $\de_0\equiv 0$, therefore
$H^0(V,M)=$Ker $\de_0=M$. We define
\begin{equation*}
C^1(V,M)=\{ \ g\in {\rm Hom}_\kk (V,M) \ : \ g(da)=dg(a) \ \ {\rm
and }\ \ g(\vac)=0\ \}
\end{equation*}
\vskip .3cm
\noi and we take for $g\in C^1(V,M)$
\begin{equation*}
(\de_1 \, g)_z(a,b)=a\z g(b)-g(a\z b)+g(a)\z b
\end{equation*}
\vskip .4cm
\noi with $a,b\in V$. Hence, $H^1(V,M)=$Ker $\de_1=$VDer$(V,M)$.
We define $C^2(V,M)$ as the space of linear functions $f_z:V
\otimes V\to M((z))$ satisfying (for all $a,b\in V$)
\vskip.2cm
$\ast$
{\it Unit}:
\begin{align}\label{u}
f_z(a,\vac )=f_z(\vac , b)=0
\end{align}
$\ast$ {\it Translation - Derivation}:
\begin{align}\label{t}
\frac{d}{dz}
f_z(a,b)=f_z(da,b),\qquad d(f_z(a,b))=f_z(da,b)+f_z(a,db), \
\end{align}
$\ast$
{\it Symmetry}:
\begin{equation}\label{s}
f_z(a,b)=e^{zd}f_{-z}(b,a).
\end{equation}
\vskip .2cm
Now, let $Z^2(V,M)$ be the space of functions $f_z\in C^2(V,M)$
that satisfy that for all $a,b,c\in V$ there exists $n\in
\mathbb{N}$ such that
\begin{equation}\label{f-associa}
(x+z)^n \Big[f_z(a\x b,c)+f_x(a,b)\z c \Big]=(x+z)^n \Big[a\xz
f_z(b,c)+f_{x+z}(a,b\z c) \Big]
\end{equation}
and define $H^2(V,M)=Z^2(V,M)/$Im $\de_1$. If $V$ is an
associative commutative algebra, it is clear that we obtain the
Harrison's cohomology.
The associativity condition (\ref{associatt}) of a vertex algebra
produce the condition (\ref{f-associa}). But if we impose the
associator formula (\ref{associator}), we shall see after the
proof of Theorem \ref{h2-ext} that (\ref{f-associa}) could be
replaced by
\begin{align}\label{eee}
0= f_z(a\x b,c)+f_x(a,b)\z c & -a\xz
f_z(b,c)-f_{x+z}(a,b\z c)\\
& - b\z(f_{z+x}(a,c)
-f_{x+z}(a,c))-f_z(b,a\zx c- a\xz c).\nonumber
\end{align}
for all $a,b,c\in V$, and the RHS of (\ref{eee}) could be the
definition of $(\de_2 f)_{z,x}(a,b,c)$. In this case
$H^2(V,M)=$Ker $\de_2/$Im $\de_1$.
\begin{proposition}
$H^2(V,M)$ is well defined, that is Im $\de_1\subseteq Z^2(V,M)$.
\end{proposition}
\begin{proof}
Let $g:V\to M$ such that $dg(a)=g(da)$ and $g(\vac)=0$. We define
$f_z:V\otimes V\to M((z))$ by $f_z(a,b)=a\z g(b)-g(a\z b)+g(a)\z
b$. Now,
$$
f_z(\vac,b)=\vac\z g(b)-g(\vac\z b)+g(\vac)\z b=g(b)-g(b)=0,
$$
and
$$
f_z(a,\vac)=a\z g( \vac)-g(a\z \vac)+g(a)\z \vac=a\z
g(\vac)-g(e^{zd} a)+e^{zd} g(a)=0,
$$
therefore $f_z$ satisfies (\ref{u}). Now we prove that it
satisfies (\ref{t}):
\begin{align*}
\frac{d}{dz}f_z(a,b)=& da\z g(b)-g(da \z b)+d e^{zd}b\mz
g(a)-e^{zd}(db)\mz g(a)\\=& da\z g(b)-g(da \z b)+e^{zd}b\mz d
g(a)=f_z(da,b),
\end{align*}
and
\begin{align*}
d(f_z(a,b))=& da\z g(b)+ a\z dg(b)-g(da\z b)-g(a\z
db)+e^{zd}(db)\mz g(a)+e^{zd} b\mz dg(a) \\
=& f_z(da,b)+f_z(a,db).
\end{align*}
The symmetry (\ref{s}) follows by
\begin{align*}
e^{zd} f_{-z}(b,a)=& e^{zd} \, b\mz g(a)-e^{zd} g(b\mz
a)+e^{zd}\left(e^{-zd} a\z g(b)\right)\\
=& g(a)\z b-g(e^{zd} b\mz a)+a\z g(b)=f_z (a,b).
\end{align*}
Now, we should check (\ref{f-associa}):
\begin{align}\label{77}
f_z(a\x b,c)&+f_x(a,b)\z c - a\xz
f_z(b,c)-f_{x+z}(a,b\z c)=\qquad \qquad \qquad \nonumber\\
&=(a\x b)\z g(c)-a\xz(b\z
g(c))
-g((a\x b)\z c)+g(a\xz (b\z
c))\\
&\ \ \ \ -a\xz(g(b)\z c)
+(a\x g(b))\z c +(g(a)\x b)\z c-g(a)\xz (b\z c)\nonumber
\end{align}
using the associativity (\ref{aaaa}) of $M$, the first two terms
in the RHS is zero after the multiplication by $(x+z)^n$ for some
$n\in \mathbb{N}$. Similarly with the third and fourth terms, by
using the associativity (\ref{associatt}) of $V$. Now, consider
the fifth and sixth terms in (\ref{77}). Using that $e^{zd} a\x
w=a\xz (e^{zd} w)$, we have
\begin{align*}
(a\x g(b))\z c -a\xz(g(b)\z c)&=e^{zd} c\mz(a\x g(b))-a\xz
(e^{zd} c\mz g(b))\\ &=e^{zd}[c\mz(a\x g(b))- a\x(c\mz g(b))]
\end{align*}
which is zero after the multiplication by $(x+z)^n$ for some $n\in
\mathbb{N}$, due to locality (\ref{locality}) in the action of $V$
on $M$. Finally, consider the seventh and eighth terms in
(\ref{77}).
On one hand, we have
$$
(g(a)\x b)\z c=e^{zd}c\mz (g(a)\x b)=e^{zd}c\mz(e^{xd} b\mx
g(a))=e^{(z+x)d} c\mzmx (b\mx g(a)),
$$
and on the other hand, we have
$$
g(a)\xz(b\z c)=e^{xd}(b\z c)\mx (e^{zd} g(a))=e^{xd}(e^{zd} c
\mz b)\mx(e^{zd} g(a))=e^{(x+z)d} (c\mz b)\mx g(a),
$$
then using the associativity (\ref{aaaa}), both terms are equal after the
multiplication by $(x+z)^n$ for some $n\in \mathbb{N}$, finishing
the proof.
\end{proof}
\
\vskip .5cm
\section{Second cohomology and square-zero extensions}\lbb{ex}
\
\begin{definition}
(a) Let $V$ be a vertex algebra. A {\it square-zero ideal of $V$}
is an ideal $W$ of $V$ such that for any $a,b\in W$, $a \x b=0$.
(b) Let $V$ be a vertex algebra and $M$ a $V$-module. A {\it
square-zero extension $(\Lambda, f, g)$ of $V$ by $M$} is a vertex
algebra $\Lambda$ together with a surjective homomorphism $f:
\Lambda \to V$ of vertex algebras such that $\ker f$ is a
square-zero ideal of $\Lambda$ (so that $\ker f$ has the structure
of a $V$-module) and an injective homomorphism $g$ of $V$-modules
from $M$ to $\Lambda$ such that $g(M)=\ker f$.
(c) Two square-zero extensions $(\Lambda_{1}, f_{1}, g_{1})$ and
$(\Lambda_{2}, f_{2}, g_{2})$ of $V$ by $M$ are {\it equivalent}
if there exists an isomorphism of vertex algebras $h:
\Lambda_{1}\to \Lambda_{2}$ such that the diagram
$$\begin{CD}
0@>>> M @>>g_{1}> \Lambda_{1} @>>f_{1}>V @>>>0\\
@. @V1_{M}VV @VhVV @VV1_{V}V\\
0@>>> M @>>g_{2}> \Lambda_{2} @>>f_{2}>V @>>>0,
\end{CD}$$
is commutative.
\end{definition}
Now, we have the following result:
\begin{theorem}\label{h2-ext}
Let $V$ be a vertex algebra and $M$ a $V$-module. Then the set of
the equivalence classes of square-zero extensions of $V$ by $M$
corresponds bijectively to $H^{2}(V, M)$.
\end{theorem}
\begin{proof}
Let $(\Lambda, f, g)$ be a square-zero extension of $V$ by $M$.
Then there is an injective linear map $\Gamma: V\to \Lambda$ such
that the linear map $h: V\oplus M\to \Lambda$ given by $h(a,
u)=\Gamma(a)+g(u)$ is a linear isomorphism. By definition, the
restriction of $h$ to $M$ is the isomorphism $g$ from $M$ to $\ker
f$. Then the vertex algebra structure and the $V$-module structure
on $\Lambda$ give a vertex algebra structure and a $V$-module
structure on $V\oplus M$ such that the embedding $i_{2}: M\to
V\oplus M$ and the projection $p_{1}: V\oplus M\to V$ are
homomorphisms of vertex algebras. Moreover, $\ker p_{1}$ is a
square-zero ideal of $V\oplus M$, $i_{2}$ is an injective
homomorphism such that $i_{2}(M)=\ker p_{1}$ and the diagram
\begin{equation*}
\begin{CD}
0@>>> M @>i_{2}>> V\oplus M @>p_{1}>>V @>>>0\\
@. @V1_{M}VV @VhVV @VV1_{V}V\\
0@>>> M @>>g> \Lambda @>>f>V @>>>0
\end{CD}
\end{equation*}
of $V$-modules is commutative. So we obtain a square-zero
extension $(V\oplus M, p_{1}, i_{2})$ equivalent to $(\Lambda, f,
g)$. We need only consider square-zero extension of $V$ by $M$ of
the particular form $(V\oplus M, p_{1}, i_{2})$. Note that the
difference between two such square-zero extensions are in the
$\z$-product maps. So we use $(V\oplus M, \VMz , p_{1}, i_{2})$
to denote such a square-zero extension.
We now write down the $\z$-product map for $V\oplus M$ explicitly.
Since $(V\oplus M, \VMz, p_{1}, i_{2})$ is a square-zero extension
of $V$, there exists a linear map $\psi_z:V\otimes V\to M((z))$
such that
\begin{eqnarray}\label{Y_V+W}
(a,u)\VMz (b, v)=(a\z b, a\z v+ u\z b+\psi_z(a,b))
\end{eqnarray}
for $a,b\in V$ and $u,v\in M$.
Now, we shall prove that $V\oplus M$ with $\VMz$, the vacuum
vector $\vac_{V\oplus M}=(\vac,0)$ and $d_{V\oplus M}(a,u)=(d_V a,
d_M u)$, is a vertex algebra if and only if $\psi_z \in Z^2(V,M)$.
In order to simplify the proof, observe that in Proposition 4.8.1
in \cite{LL}, they showed that $V\oplus M$ with $\vac_{V\oplus
M}$, $d_{V\oplus M}$ and $\VMz$ corresponding to $\psi_z\equiv 0$,
is a vertex algebra. Therefore, when we check the axioms, we know
that all the terms without $\psi_z$ satisfy the corresponding
equation. So, in order to prove that $V\oplus M$ with $\VMz$ given
by (\ref{Y_V+W}) is a vertex algebra, we only need to see the
terms with $\psi_z$. For example, the element $(\one, 0)$
satisfies (for $a,b\in V$ and $u,v\in M$)
\begin{equation*}
(\vac ,0)\VMz (b,v)=(\vac\z b,\vac\z v +\psi_z(\vac,b))=(b,v)
\end{equation*}
and
\begin{equation*}
(a ,u)\VMz (\vac,0)=(a\z \vac,u\z \vac +\psi_z(a,\vac))=(e^{zd_V}
a, e^{zd_M} (\vac\mz u) + \psi_z(a,\vac))=e^{zd_{V\oplus M}}(a,u)
\end{equation*}
if and only if $\psi_z(\vac ,b)=0=\psi_z(a,\vac)$ for all $a,b\in
V$. From now on, we use $d=d_V=d_M$. A simple computation shows
that $V\oplus M$ satisfies the translation-derivation properties
if and only if for all $a,b\in V$
\begin{align*}
\frac{d}{dz}
\psi_z(a,b)=\psi_z(da,b),\qquad \hbox{ and }\ \
d(\psi_z(a,b))=\psi_z(da,b)+\psi_z(a,db).
\end{align*}
Now, consider the commutativity axiom, that is:
\begin{align*}
e^{zd_{V \oplus M}}(b,v)\VMmz (a,u)&=
e^{zd_{V \oplus M}}(b\mz a, b\mz u+e^{-zd} a\z
v+\psi_{-z}(b,a))\\
&=(e^{zd} b\mz a,e^{zd} b\mz
u+a\z v+e^{zd}\psi_{-z}(b,a))
\end{align*}
and $(a,u)\VMz (b,v)=(a\z b, a\z v+e^{zd}b\mz u+\psi_z(a,b))$.
Therefore, $\psi_z$ must satisfy $\psi_z(a,b)=e^{zd}
\psi_{-z}(b,a)$.
Similarly, expanding
$$
\Big((a,u)\VMx(b,v)\Big)\VMz (c,w) \qquad \hbox{ and } \qquad
(a,u)\VMxz \Big((b,v)\VMz(c,w)\Big),
$$
and taking the terms with $\psi_z$, it is easy to see that the
associativity axiom (\ref{associatt}) holds if and only if for all
$a,b,c\in V$ there exists $n\in \mathbb{N}$ such that
\begin{equation}\label{f-associal}
(x+z)^n \Big[\psi_z(a\x b,c)+\psi_x(a,b)\z c \Big]=(x+z)^n \Big[a\xz
\psi_z(b,c)+\psi_{x+z}(a,b\z c) \Big],
\end{equation}
proving that $V\oplus M$ is a vertex algebra if and only if
$\psi_z\in Z^2(V,M)$, and together with the projection
$p_1:V\oplus M\to V$ and the embedding $i_2:M\to V\oplus M$,
$V\oplus M$ is a square-zero extension of $V$ by $M$.
Next we prove that two elements of $Z^2(V,M)$ obtained in this way
differ by an element of $\delta_{1}C^{1}(V, M)$ if and only if the
corresponding square-zero extensions of $V$ by $M$ are equivalent.
Let $\psi, \phi \in Z^2(V,M)$ be two such elements obtained from
square-zero extensions $(V\oplus M, \zu, p_{1}, i_{2})$ and
$(V\oplus M, \zd , p_{1}, i_{2})$, respectively. Assume that
$\psi=\phi+\delta_{1}(g)$ where $g \in C^{1}(V, M)$.
We now define a linear map $h: V\oplus M\to V\oplus M$ by
$$h(a, u)=(a, u+g(a))$$
for $a\in V$ and $u\in M$. Then $h$ is a linear isomorphism and it
satisfies (for $a,b\in V$ and $u,v\in M$)
\begin{align}\label{hhh}
h\left((a,u)\zu (b,v)\right)&=(a\z b,a\z v+u\z b+\psi_z(a,b)+g(a\z b))
\nonumber\\
&=(a\z b,a\z v+u\z b+\phi_z(a,b)+a\z g(b)+g(a)\z b)\\
&=(a, u+g(a))\zd (b,v+g(b))\nonumber\\
&=h(a,u)\zd h(b,v).\nonumber
\end{align}
Thus $h$ is an isomorphism of vertex algebras from $(V\oplus M,
\zu, (\vac, 0))$ to $(V\oplus M, \zd , (\vac, 0))$ such that the
diagram
\begin{equation}\label{equivalence}
\begin{CD}
0@>>> M @>i_{2}>> V\oplus M @>p_{1}>>V @>>>0\\
@. @V1_{M}VV @VhVV @VV1_{V}V\\
0@>>> M @>i_{2}>> V\oplus M @>p_{1}>>V @>>>0\\
\end{CD}
\end{equation}
is commutative. Thus the two square-zero extensions of $V$ by $M$
are equivalent.
Conversely, let $(V\oplus M, \zu , p_{1}, i_{2})$ and $(V\oplus M,
\zd , p_{1}, i_{2})$ be two equivalent square-zero extensions of
$V$ by $M$. So there exists an isomorphism $h: V\oplus M\to
V\oplus M$ of vertex algebras such that (\ref{equivalence}) is
commutative. Let $h(a, u)=(f(a, u), g(a, u))$ for $a\in V$ and
$u\in M$. Then by (\ref{equivalence}), we have $f(a, u)=a$ and
$g(0, u)=u$. Since $h$ is linear, we have $g(a, u)=g(a, 0)+g(0,
u)=u+g(a, 0)$. So $h(a, u)=(a, u+g(a, 0))$. Taking $g(a)$ to be
$g(a, 0)$, we see that there exists a linear map $g: V\to M$ such
that $h(a, u)=(a, u+g(a))$. Using that $d\,h(a,0)=h(d(a,0))$ and
$h(\vac,0)=(\vac,0)$, it is clear that $g(da)=dg(a)$ and
$g(\vac)=0$. Thus, $g\in C^1(V,M)$.
Let $\psi$ and $\phi$ be elements of $Z^2(V,M)$ obtained from
$(V\oplus M, \zu , p_{1}, i_{2})$ and $(V\oplus M, \zd , p_{1},
i_{2})$, respectively. Then, since $h$ is a homomorphism of vertex
algebras, (\ref{hhh}) holds for $a,b\in V$ and $u,v \in M$. So
the two expressions in the middle of (\ref{hhh}) are equal. Thus,
we have $\psi=\phi+\de_1(g)$. Therefore, $\psi$ and $\phi$ differ
by an element of $\delta_{1}C^{1}(V, M)$.
\end{proof}
\
In the proof of Theorem \ref{h2-ext} (by taking the terms with
$\psi_z$), we saw in (\ref{f-associal}) that the associativity
axiom holds in $(V\oplus M, \VMz, (\vac,0))$ if and only if for
all $a,b,c\in V$ there exists $n\in \mathbb{N}$ such that
\begin{equation*}
(x+z)^n \Big[\psi_z(a\x b,c)+\psi_x(a,b)\z c \Big]=(x+z)^n \Big[a\xz
\psi_z(b,c)+\psi_{x+z}(a,b\z c) \Big].
\end{equation*}
Recall that we can replace the associativity axiom
(\ref{associatt}) in the definition of vertex algebra, by the
associator formula (\ref{associator}). By taking the terms with
$\psi_z$, it is possible to prove that the associator formula
holds in $(V\oplus M, \VMz, (\vac,0))$ if and only if for all
$a,b,c\in V$ we have
\begin{align*}
0= \psi_z(a\x b,c)+\psi_x(a,b)\z c & -a\xz
\psi_z(b,c)-\psi_{x+z}(a,b\z c)\\
& - b\z(\psi_{z+x}(a,c)
-\psi_{x+z}(a,c))-\psi_z(b,a\zx c- a\xz c).\nonumber
\end{align*}
We avoid replacing the associativity or associator formula by the
Jacobi identity because we want to make emphasis that a vertex
algebra is a generalization of an associative commutative algebra,
and that the cohomology must be a generalization of Harrison
cohomology.
\vskip .3cm
\
\section{Second cohomology and first order deformations}
\
\begin{definition}
(a) Let $t$ be a formal variable and let $(V,\z , \one, d)$ be a
vertex algebra. A {\it first order deformation of $V$} is
a family of $z$-products of the
form
\begin{equation*}
a\sz b= a\z b + t \, f_z(a,b)
\end{equation*}
with $a,b\in V$, where $f_z:V\otimes V \to V((z))$ is a linear
map (independent of $t$), such that $(V,\szz , \vac,d)$ is a family
of vertex algebras up to the first order in $t$ (ie. modulo
$t^2$). More precisely, the quadruple $(V, \szz, \one, d)$
satisfies the following conditions:
\vskip.1cm
$\ast$ {\it Unit}:
\begin{equation}\label{uu}
\vac \sz a= a \quad \textrm{ and } \quad a \sz \vac=e^{zd}a;
\end{equation}
$\ast$ {\it Translation - Derivation}:
\begin{align}\label{dd}
(da)\sz b= \ddz (a\sz b), \quad d(a\sz b)=(da)\sz b + a\sz (db);
\end{align}
$\ast$ {\it Commutativity}:
\begin{align} \label{ssss}
a \sz b= e^{zd} (b\smz a);
\end{align}
$\ast$ {\it Associativity up to the first order in $t$}: For any
$a,b,c\in V$, there exist $l\in \NN$ such that
\begin{align} \label{aa}
(z+w)^l\, \ (a \sz b) \sw c = (z+w)^l\, \ a\, \szw \, (b\sw c) \quad
\hbox{ mod } t^2.
\end{align}
\vskip.3cm
(b) Two first order deformations $\szu$ and $\szd$ of $(V, \z ,
\one,d)$ are {\it equivalent} if there exists a family $\phi_{t}:
V \to V[t]$, of linear maps of the form $\phi_{t}=1_{V}+tg$ where
$g: V\to V$ is a linear map such that
\begin{equation*}
\phi_{t}(a\szu b)=\phi_{t}(a)\szd \phi_{t}(b) \quad \hbox{ mod }
t^2
\end{equation*}
for $a,b \in V$.
\end{definition}
We have:
\begin{theorem}\label{deform}
Let $V$ be a vertex algebra. Then the set of the equivalence
classes of first order deformations of $V$ corresponds
bijectively to $H^{2}(V, V)$.
\end{theorem}
\begin{proof}
Let $\szz$ be a first order deformation of $V$. By definition,
there exists a linear map $f_z:V\otimes V\to V((z))$ such that
\begin{equation}\label{ss}
a\sz b=a\z b+ t \, f_z(a,b)
\end{equation}
for $a,b\in V$, and $(V,\szz ,\vac, d)$ is a family of vertex
algebras up to the first order in $t$.
The unit properties (\ref{uu}) for $(V,\szz ,\vac, d)$ gives
$$
\vac \sz a=\vac\z a+t\, f_z(\vac,a)=a
$$
and
$$
a\sz \vac=a\z \vac+t\, f_z(a,\vac)=e^{zd} a
$$
for $a\in V$. So, they are equivalent to
\begin{equation}\label{11}
f_z(a,\vac )=0=f_z(\vac, a) \quad\hbox{ for all } a\in V.
\end{equation}
Similarly, the coefficient of $t^{0}$ of the
Translation-Derivation properties (\ref{dd}) corresponds exactly
to the Translation-Derivation properties of $(V,\z ,\vac, d)$, and
the coefficient of $t^{1}$ in (\ref{dd}) corresponds exactly to
the following properties on $f_z$:
\begin{equation}\label{22}
\ddz f_z(a,b)=f_z(d a,b), \quad \hbox{ and }\quad
d(f_z(a,b))=f_z(da,b)+f_z(a,db).
\end{equation}
Now, the coefficient of $t^{0}$ of the Commutativity property
(\ref{ssss}) corresponds exactly to the Commu- tativity property of
$(V,\z ,\vac, d)$, and the coefficient of $t^{1}$ in (\ref{ssss})
corresponds exactly to the following property on $f_z$:
\begin{equation}\label{4444}
f_z(a,b)=e^{zd} f_{-z}(b,a).
\end{equation}
In the same way, using (\ref{ss}), we take the expansions modulo
$t^2$ of the expressions
$$
(a \sz b) \sw c \quad \hbox{ and }\quad \ a\, \szw \, (b\sw c)
$$
and we consider the coefficients of $t^{0}$ and $t^{1}$ of them.
By a direct computation, we can see that the coefficient of $t^0$
of the associativity property (\ref{aa}) corresponds exactly to
the associativity property of $\z$, and the coefficient of $t^1$
corresponds exactly to the following property: for all $a,b,c\in
V$ there exists $n\in\mathbb{N}$ such that
\begin{equation}\label{33}
(w+z)^n \Big[f_w(a\z b,c)+f_z(a,b)\w c \Big]=(w+z)^n \Big[a\zw
f_w(b,c)+f_{z+w}(a,b\w c) \Big].
\end{equation}
Therefore, using (\ref{11}), (\ref{22}),(\ref{4444}) and
(\ref{33}), we have seen that
$$
a\sz b=a\z b+t\, f_z(a,b)
$$
is a first order deformation of $V$
if and only if $f_z\in Z^2(V,V)$.
Now, we prove that two first order deformations of $V$ are
equivalent if and only if the difference between the corresponding
elements in $Z^2(V,V)$ is in Im $\de_1$.
Consider two first order deformations of $V$ given by
$$
a\szu b=a\z b+t\,\psi_z(a,b)\quad \hbox{ and }\quad a\szd b=a\z
b+\phi_z(a,b),
$$
where $\psi_z$ and $\phi_z$ are in $Z^2(V,V)$. They are equivalent
if and only if there exists $f_t=1_V+t\, g$ where $g:V\to V$ is a
linear map such that
\begin{equation}\label{44}
f_{t}(a\szu b)=f_{t}(a)\szd f_{t}(b) \quad \hbox{ mod } t^2
\end{equation}
for $a,b \in V$. Now, since
$$
f_t(a\szu b)=f_t(a\z b+t\,\psi_z(a,b))=a\z b+t\,\psi_z(a,b)+t\,
g(a\z b) \quad \hbox{ mod } t^2
$$
and
$$
f_t(a)\szd f_t (b)=(a+t\,g(a))\szd (b+t\,g(b))= a\z b +t\, a\z
g(b)+t\,g(a)\z b+t\, \phi_z(a,b) \quad \hbox{ mod } t^2
$$
then (\ref{44}) is equivalent to
$$
\psi_z(a,b)-\phi_z(a,b)=a\z g(b)-g(a\z b)+g(a)\z b
$$
for all $a,b\in V$. Therefore, it is equivalent to
$\psi_z-\phi_z=(\de_1 g)_z$, finishing the proof.
\end{proof}
\vskip .5cm
\vskip .5cm
\subsection*{Acknowledgements}
The author would like to thank M. V. Postiguillo and C. Bortni for
their constant help and support throughout this work. Special
thanks to C. Boyallian.
\bibliographystyle{amsalpha}
| 93,652
|
by Fiona Spreadborough
A Cardo Mariano Planta tea would be perfect, very strong, bitter, yet full of healing :)
I use the park lavvies when I wanna pee, wanna wash, wanna change my sannie, fill up my water bottle, look at myself in its half cracked mirror. Half cracked mirror reflects half cracked life. I don’t have a hairbrush, I hope one day someone leaves one behind at the sink, even a comb would be dope.
I’m kinda special, I’m a statistic, woohoo, one in every 122 people, who, around the World are now refugees, displaced or abandoned. Look on the bright side I tell myself, at least I don’t need to worry about traipsing round Ikea for furniture.
I collect peoples leftover sandwiches, biscuits and papers from the bins and watch people’s lives as they walk through the park or sit on a bench, talk of love, hate, bills, kids, affairs, life and death. I know lots of people but they don’t know me, they never see me, cause my clothes are ragged, my hair is matte, my shoes are holey, my eyes are hollow but I’ve got some banging hoop earrings and my heart is good! And I always share the leftovers with the tiny sparrows in my bush, my chirpy housemates, sometimes to bloody chirpy!!
I’m getting on a bit now but I can read and I see all the young ones and lost babies and oroughwar-torn lives that roam zombified in this glorious, leafy green, urban jungle, from the children’s swings, to rose garden, to duck pond, to prostitute corner, it once was the pensioners bowling green but council cutbacks! Nowhere for bui doi to go, the dusts of life, endlessly trembling from looking into the mouth of hell, smelling the breath of Satan.
My only crime was being a naughty catholic strumpet and getting preggas out of wedlock; it was irrelevant I was raped by my brother-in-law. “Be Jesus” me Mammie wailed as she closed our shiny, gloss black front door with brass letterbox and number 69 glowing above the spy hole. She always done the sign of the cross when she stepped over that threshold and she refused to give our address out over the telephone, even if it was the Lord himself on the other end! That door was closed, not with me behind it, no, slammed hard in my face, so hard I think I still have a chipping of gloss black paint blocking up my right nostril, years of pain, I mean years of paint trapped somewhere deep inside me.
So where do you go when no one wants you? I got my bush in the park, but I’ve been everywhere and nowhere, I sometimes sleep near the local gym in winter as they have big warm air vents that send out invisible hugs and cuddles for me to fall asleep with and pretend that Brad Pitt is really my boyfriend and my life has just been a film and I’m about to win an Oscar. I do tend to dream big even though I am homeless, well it cost nothing.
My comrades of the “unclean, dusts of life” are very international and no they aren’t fucking drug dealers and whores, theirs many children, half the Worlds refuges are children. Syria, Afghanistan, Somalia, Rakhine, Colombia, Venezuela, Sudan, Congo, Mongolia, Central African Republic, it goes on and on, the places are endless, the faces nameless and blameless and me from North London via Dublin well that’s almost tame. I have a lot of time in my earthy burrow under my bush to read the endless print I pull from the bins.
I met a young girl recently in the park at Prostitute corner, pretty but scared, squatting down with a small bowl begging. Her fingers around the bowl were so bony, I thought she was going to break and turn to dust in front of my eyes.
She spoke little but I understood.
“I’m from Rakhine, from Chibok, from Iraq, from Mongolia, from hell,” she said
“I have 50 million starving babies pushing out from my pelvis, fighting to latch onto an impoverished breast, hanging between my legs because they don’t have the strength to pull themselves up and I don’t have the will or courage to lift them. I don’t want to see their faces or their eyes because their dying. Born only to be raped, tortured, abused, slaughtered, heads torn from their limbs used as footballs, suicide vests strapped to their skeletal weak frames. No food, no clothes, no home, no money, no hope,” she whispered.
“The umbilical cord isn’t a lifeline it’s a hangman’s noose."
I wrapped my arms around her, no words said, just heartbeats that drummed it, just breath that released it, our fear deep inside, scared to hope, what are we to BECOME?
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It is one I my new year resolutions for 2012 to try out making my own custom polishes and frankenpolish, I have been really excited about trying it so yesterday I decided I wanted make a Holographic franken polish out of a mixture of OPI, Nicole By OPI, Orly & Color Club polishes. Half way through making the frankenpolish I decided it might look nice to add some INM Silver Hologram top coat to the mixture not realising it was a quick dry formula and it made my frankenpolish go all thick and gluggy I tried adding a lot of polish thinner but it wasn't helping so I gave up.
There must have been some sort of new years miracle last night and this afternoon when I tried swatching the color again the formula had magically thinned out enough for me to paint my nails with it, I am sorry I have included so many pics in this post but I was really had to get good photos of the holographic effect.
I don't have a name for this polish yet please let me know if you have any good ideas?
It's a gorgeous mix! Congrats!
Beautiful!
Underwater Radiance maybe? So pretty though!
Oh pretty! I Like it!
Holobuloo :0)
Wow...it looks as good as store bought!..How about "Sacre Blue"??...luvin it!
| 50,792
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Effect of exogenously added rhamnolipids on citric acid production yield
AbstractThe influence of a biosurfactant (rhamnolipids) on the effectiveness of citric acid production by Yarrowia lipolytica from sunflower oil was studied. The surfactant-mediated solubilization of the hydrophobic substrate was assessed by particle size distribution characteristics with and without the presence of sunflower oil hydrolization products. The presence of rhamnolipids contributed to a decrease of the oil droplet size, most notably for samples containing sunflower oil and its hydrolization products. The citric acid yield for cultures not supplemented with rhamnolipids was at 82.9 g/l, with a 1:0.04 citric acid to isocitric acid ratio (CA:ICA). The addition of rhamnolipids at 1 g/l resulted in a 5% increased citric acid yield (87.1 g/l), however a decrease (79.0 g/l) was observed for samples containing 5 g/l of rhamnolipids. The rhamnolipids-induced emulsification of sunflower oil did not seem to influence the citric acid production efficiency. Additional research revealed that the biosurfactant was degraded by yeast cells during the bioconversion process. The possible explanations of this phenomenon include the utilization of rhamnolipids as an alternative carbon source or microbial destabilization of micelles formed by this biosurfactant due to potential bioavailability issues.
Keywords: Yarrowia lipolytica, citric acid, rhamnolipids, sunflower oil
African Journal of Biotechnology Vol. 12(21), pp. 3313-3320
Published
2016-03-14
Issue
Section
Articles
Copyright for articles published in this journal is retained by the journal.
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By Michael McConnell
By Michael McConnell
The notion of coin collecting appeals to people of all ages; but for novice collectors, it can be difficult to go about identifying
- Take the coins pictured in this article, for instance: can you tell which coin is worth over $1,000.00? The answer: they all are -- except for the oldest coin, which is a 1,800-year-old Roman coin. Even though this is the oldest coin pictured by far it only sells for $80.00. Surprised? Most people are.
Coin values 101: basic tips for new collectors
Coin values 101: basic tips for new collectors
Here are some basic things to understand about coin values, using the examples pictured as a reference guide:
Age doesn’t matter. Coins are very durable, and that means unless they have been purposely destroyed, most coins are still around somewhere. In other words, a coin’s value is never determined based on age alone. The first coin pictured is a 1,800-year-old ancient coin – and while it is fascinating from a historical context, it is much more common than you might expect. Furthermore, collectors don’t focus on ancient coins as much as they do on other types of collectible coins. Then again, there are some very rare and expensive ancient coins on the market as well.
Popularity is key. The Lincoln Cent, pictured, is a very common coin in general. However, it also ranks among the most popular collectors’ series – and because the 1909-S VDB has the lowest mintage of that series with just 484,000 minted pieces, it is therefore a highly valued coin. Mintage, like age, is not a sole determining factor, however. There are some coins with very low mintage, but no active collector base. If there are only 1,000 minted pieces of something but only 900 collectors who want one, then that coin will not have great value on the market.
“Mint condition” matters. The next coin you see is an 1890-CC (Carson City) Silver Dollar in gem un-circulated condition. This coin has a mintage of over two million pieces, making it a relatively common piece. However, only a small number of coins of this date have maintained such a high-grade condition. This makes them very valuable despite their high mintage.
Precious metals add value: Next you will see a one-ounce Gold Eagle. This piece is also very common; but despite being a mass-produced bullion coin, it is very valuable because of the precious metals value of gold. Compare this Gold Eagle with the 100-year-old $20 gold piece pictured hereafter. If commonly circulated, this coin might also bear value purely based on its gold content. Unlike the Gold Eagle, however, it could also have value that greatly exceeds the gold value if it is a rare date within the series or in very high-grade condition.
I hope this column has helped you gain a basic understanding of how rare coins derive their value. Additional research can make the process easier and infinitely fascinating for enthusiastic collectors; but it is always best to have a professional evaluate your coin collection to ensure both accurate information and lasting enjoyment. To learn more, request information about a coin or start your own collection, visit our experienced
at The Coin Shop in La Jolla today or contact us online:
.
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\begin{document}
\title{Third Mac Lane cohomology}
\author{H.-J. Baues}
\address{
Max-Planck-Institut f\"ur Mathematik\\
Vivatsgasse 7\\
Bonn 53111\\
Germany} \email{baues@mpim-bonn.mpg.de}
\author{M. Jibladze}
\address{
A. Razmadze Mathematical Institute\\
M. Alexidze st. 1\\
Tbilisi 0193\\
Georgia} \email{jib@rmi.acnet.ge}
\author{T. Pirashvili}
\address{
A. Razmadze Mathematical Institute\\
M. Alexidze st. 1\\
Tbilisi 0193\\
Georgia} \email{pira@rmi.acnet.ge}
\begin{abstract}MacLane cohomology is an algebraic version of the topological
Hochschild cohomology.
Based on the computation of the third author (see Appendix below) we
obtain an interpretation of the third Mac~Lane cohomology
of rings
using certain kind of crossed extensions of rings in the quadratic
world. Actually we obtain two such interpretations corresponding to
the two monoidal structures on the category of square groups.
\end{abstract}
\maketitle
\section{Introduction}\label{intro}
Let $R$ be a ring and $M$ a bimodule over $R$. Then there are three
essential cohomology theories associated to the pair $(R,M)$ due to
Hochschild, Shukla, and Mac~Lane, see \cite{hh, sh, macl}. These theories are connected by
natural maps (\cite{shukla})
$$
\H^n(R;M)\to\Sh^n(R;M)\xto{\tau^n}\HML^n(R;M).
$$
It is known that Mac~Lane cohomology coincides with topological
Hochschild cohomology (\cite{fw}) and coincides also with
Baues-Wirsching cohomology of the category $\modr R$ of finitely
generated free $R$-modules (\cite{JP}). We study the cohomologies in
dimension $n=3$. In this case the elements in $\H^3(R;M)$ are
represented by split crossed extensions and elements in $\Sh^3(R;M)$
are represented by all crossed extensions of $R$ by $M$ in the
monoidal category $(\Ab,\ox)$ of abelian groups. See \cite{jll},
\cite{baumin} and \cite{shukla}. Here a crossed extension of $R$ by
$M$ is an exact sequence in $\Ab$,
$$
0\to M\xto\iota C_1\xto\d C_0\xto q R\to0,
$$
where $C_0$ is a ring, $C_1$ is a bimodule over it, $q$ is a ring
homomorphism and $\iota$ and $\d$ are $C_0$-biequivariant maps
satisfying $b\d(c)=\d(b)c$ for $b,c\in C_1$.
A similar result for group cohomology in dimension 3 is due to
Mac~Lane and J.~H.~C.~Whitehead \cite{MW}.
The main goal of this paper is the construction of appropriate
crossed extensions of $R$ by $M$ which represent classes in the
third Mac~Lane cohomology $\HML^3(R;M)$.
To this end we recall that for any small category $\c$ with a
natural system $D$ on it, the Baues-Wirsching cohomology group
$H^3(\c;D)$ can be represented by linear track extensions of $\c$ by
$D$, see \cite{P1,P2,B}. Hence by the isomorphism
$$
\HML^3(R;M)\cong H^3(\modr R;\Hom_R(-,-\ox_RM))
$$
elements of $\HML^3(R;M)$ are represented by linear track extensions
of $\modr R$. Such a description, however, is available for any
category $\c$ and does not restrict to the specific nature of
Mac~Lane cohomology of a ring.
In order to find specific crossed extensions for $\HML^3(R;M)$ we
have to proceed from linear algebra to quadratic algebra. Here
``linear algebra'' is the algebra of rings and modules. A ring is a monoid
in the monoidal category $(\Ab,\ox)$ and a module is an object in
$\Ab$ together with an action of such a monoid.
In ``quadratic algebra'' abelian groups are replaced by square
groups. In fact, if one considers endofunctors of the category of
groups which preserve filtered colimits and reflexive coequalizers,
then abelian groups can be identified with linear endofunctors and
square groups can be identified with quadratic endofunctors
(\cite{square}). The category $\SG$ of square groups contains the
category $\Ab$ as a full subcategory since a linear endofunctor is
also quadratic. Composition of functors leads to monoidal structures
$\t$ and $\square$ in such a way that $(\Ab,\t)$ is a monoidal
subcategory of $(\SG,\square)$. There is also another monoidal structure
$\tl$ on $\SG$ such that the identity of $\SG$ is a lax monoidal
functor $(\SG,\tl)\to(\SG,\square)$. Here $\tl$ is symmetric while $\square$
is highly nonsymmetric. Compare \cite{bjp}.
Crossed extensions in the monoidal categories $(\SG,\square)$ or
$(\SG,\tl)$ are defined similarly to the case $(\Ab,\t)$ above, see
section \ref{crex}. As a main result we prove in this paper the
following theorem, compare the more detailed version \ref{xext}
below.
\subsection{Main theorem}
\begin{The}\label{mainth}
Elements in the third Mac~Lane cohomology group $\HML^3(R;M)$ are in
1-1 correspondence with equivalence classes of linearly generated
crossed extensions of $R$ by $M$ in the monoidal category
$(\SG,\tl)$, or in the monoidal category $(\SG,\square)$.
\end{The}
Such an interpretation of the group $\HML^3$ was missing for many
years; in terms of obstruction theory the problem first arose in the
classical paper of Mac~Lane \cite{maclo}. The theorem is based on
the quadratic theory developed in \cite{square,bjp} and emphasizes
importance of the quadratic algebra of square groups. A crucial step
in the proof of the theorem relies on the vanishing result achieved
by the third named author in the Appendix.
In dimension three, the map $\tau^3$ fits in the exact sequence (see
\cite{JPmoambe}, \cite{shukla})
$$
0\to\Sh^3(R;M)\xto{\tau^3}\HML^3(R;M)\xto\nu\H^0(R;{}_2M)\to
\Sh^4(R;M)\xto{\tau^4}\HML^4(R;M)
$$
where $_2M=\{m\in M\mid 2m=0\}$.
As an application of the theorem we describe the connecting
homomorphism $\nu$ in terms of crossed extensions in $\SG$, see
Section \ref{nu}.
It follows from the relationship between $\Sh^3(R,M)$ and crossed
extensions of rings that $\Sh^3(R,M)$ describes homotopy types of
those chain algebras $C_*$ with $H_0(C_*)=R$, $H_1(C_*)=M$, and
$H_i(C_*)=0$ for $i\ne0,1$. On the other hand, it follows from the
relationship between Mac~Lane and topological Hochschild cohomology
that $\HML^3(R,M)$ describes homotopy types of ring spectra
$\Lambda$ with $\pi_i(\Lambda)=0$, $i\ne0,1$, $\pi_0(\Lambda)=R$ and
$\pi_1(\Lambda)=M$ \cite{laza}. Thus our result shows that crossed
extensions of $R$ by $M$ in $\SG$ are algebraic models of such ring
spectra. It follows that the homomorphism $\nu$ is an obstruction
for such a ring spectrum to be representable by a chain algebra.
In Section \ref{obs2cat}
we give an application of our results to
the theory of 2-categories.
\section{Crossed extensions}\label{crex}
We shall apply the following general notion of crossed extension to
the monoidal categories $(\Ab,\t)$, $(\SG,\square)$ and $(\SG,\tl)$
where the category $\SG$ of square groups is defined in \ref{sqgr}.
\subsection{Crossed extensions}\label{crexte}
Let $(\bf V,\boxtimes)$ be a monoidal category and let $L$ be a
monoid in $\bf V$. Recall that a \emph{$L$-biobject} is a tuple
$(A,l,r)$, where $A$ is an object in $\bf V$ and $l:L\boxtimes A\to
A$ and $r:A\boxtimes L\to A$ are respectively left and right actions
of $L$ on $A$ which are compatible in a natural way. We let
$_{L}{\bf V}_{L}$ be the category of $L$-biobjects in $\bf V$. In
particular the monoid structure on $L$ defines also a structure of a
$L$-biobject on $L$. In what follows we always consider $L$ as a
biobject with this particular structure.
A \emph{crossed $L$-biobject} is a diagram $C=(\d:B\to L)$ in the
category $_{L}{\bf V}_{L}$ such that the following diagram commutes:
$$\xymatrix{B\boxtimes B\ar[r]^{\Id\boxtimes \d}\ar[d]_{\d\boxtimes \Id}&B\boxtimes
L\ar[d]^r\\ L\boxtimes B\ar[r]_l&L}$$
Let $R$ be a monoid in $(\V,\boxtimes)$, let $M$ be an $R$-biobject
and assume that exact sequences are defined in $\V$. Then a
\emph{crossed extension} of $R$ by $M$ in $(\V,\boxtimes)$ is an
exact sequence
\begin{equation}\label{crext}
0\to M\xto\iota C_1\xto\d C_0\xto q R\to0
\end{equation}
where $\d$ is a crossed $C_0$-biobject as above, $q$ is a morphism
of monoids and $\iota$ is a morphism in $_{C_0}\V_{C_0}$. A morphism
between crossed extensions of $R$ by $M$ is a commutative diagram
$$
\xymatrix{
0\ar[r]&M\ar[r]\ar[d]_{\Id}&C_1\ar[r]^\d\ar[d]^{f_1}&C_0\ar[r]\ar[d]^{f_0}&R\ar[r]\ar[d]^\Id&0\\
0\ar[r]&M\ar[r]&C_1'\ar[r]^{\d'}&C_0\ar[r]&R\ar[r]&0 }
$$
where $f_0$ is a morphism of monoids and $f_1$ is
$f_0$-biequivariant. Let
\begin{equation}
\xext(R;M)^{\V,\boxtimes}
\end{equation}
be the category of such crossed extensions and morphisms and let
\begin{equation}
\Xext(R;M)^{\V,\boxtimes}
\end{equation}
be the set of connected components of this category.
One readily checks that crossed extensions in $(\Ab,\ox)$ are the
extensions defined in Section \ref{intro}. Hence if $R$ is a ring
and $M$ is an $R$-bimodule then one has canonical bijections (see
\cite{shukla}, \cite[page 42]{jll}, \cite{baumin})
\begin{equation}\label{shuxt}
\Xext(R;M)^{\Ab,\ox}\approx\Sh^3(R;M)
\end{equation}
and
\begin{equation}
\Xext_\Z(R;M)^{\Ab,\ox}\approx\H^3(R;M).
\end{equation}
Here $\Xext_\Z(R;M)^{\Ab,\ox}$ is the set of connected components of
the following subcategory $\xext_\Z(R;M)^{\Ab,\ox}$ of
$\xext(R;M)^{\Ab,\ox}$: its objects are $\Z$-split crossed
extensions $(\iota,\d,q)$ in $(\Ab,\ox)$, that is, with arrows
$\iota$, $\d$ and $q$ admitting a $\Z$-splitting; morphisms in
$\xext_\Z(R;M)^{\Ab,\ox}$ are morphisms $(f_0,f_1)$ in
$\xext(R;M)^{\Ab,\ox}$ such that both $f_0$ and $f_1$ are
$\Z$-split.
\subsection{Square groups}\label{sqgr}
A \emph{square group} is a diagram
$$
A = (\xymatrix{A_e \ar[r]^H & A_{ee} \ar[r]^P & A_e})
$$
where $A_{ee}$ is an abelian group and $A_e$ is a group. Both
groups are written additively. Moreover $P$ is a homomorphism and
$H$ is a quadratic map, meaning that the \emph{cross effect}
$$(x \mid y)_H = H(x+y)-H(y) -H(x)$$
is linear in $x,y \in A_e$. In addition the following identities are
satisfied
\begin{align*}
(Pa \mid y)_H&=0,\\
P(x\mid y)_H&=[x,y],\\
PHP(a)&=P(a)+P(a).
\end{align*}
Here $[x,y]=-y-x+y+x$, $a,b \in A_{ee}$ and $x,y \in A_e$. It
follows from the first two identities that $P$ maps to the center of
$A_e$. The second identity shows also that $$A^{\mathrm{ad}}:={\sf
Coker}(P)$$ is abelian. Hence $A_e$ is a group of nilpotence class
2. It follows from the axioms that the function $T=HP-\Id_{A_{ee}}$
is an automorphism of $A_{ee}$ and $T^2=\Id_{A_{ee}}$. Moreover, the
function $\Delta:A_e\to A_{ee}$ is linear, where
$$\Delta (x) = HPH(x) + H(x+x) -4H(x)$$
and furthermore one has the induced homomorphisms $$(-,-)_H:A\ad\t A\ad\to
A_{ee}$$ and $$\Delta:A\ad\to A_{ee}.$$
We refer to \cite{square} and \cite{bjp} for more information on
square groups. We denote by $\SG$ the category of square groups. In
what follows we identify abelian groups and square groups with
$A_{ee}=0$. In this way we obtain a full embedding of categories
$$\Ab\subset\SG
$$
This inclusion corresponds to the fact that any linear functor is
quadratic. The inclusion $\Ab\subset\SG$ has a left adjoint given by
$A\mapsto A\ad$.
The category $\SG$ has two monoidal structures $\square:\SG\x \SG\to
\SG$ \cite{square} and $\tl:\SG\x \SG\to \SG$ \cite{bjp}, which are
related via a binatural transformation $$\sigma_{X,Y}:X\square Y \to
X\tl Y$$ such that the identity functor together with $\sigma$
defines a lax monoidal functor $\Id:(\SG,\tl)\to (\SG,\square)$
\cite{bjp}. The monoidal category structure $\square$ is highly
nonsymmetric, while
the monoidal category structure $\tl$ is symmetric. For the definitions of the products
$\square$ and $\tl$ on $\SG$ we refer the reader to \cite{square} and
\cite{bjp} respectively. Below we shall, however, describe
explicitly the notion of crossed extension in $(\SG,\square)$ and in
$(\SG,\tl)$. Since $(\Ab,\t)$ is a monoidal subcategory both in
$(\SG,\square)$ and in $(\SG,\tl)$, we see that a ring $R$, i.~e. a
monoid in $(\Ab,\t)$, is also a monoid in $(\SG,\square)$ and in
$(\SG,\tl)$. Let $M$ be an $R$-bimodule. Then crossed extensions
$$
0\to M\xto\iota C_{(1)}\xto\d C_{(0)}\xto q R\to0
$$
are defined in $(\SG,\square)$ and in $(\SG,\tl)$ by \eqref{crext}. Such
an extension is \emph{linearly generated} if $R$ as an additive group is
generated by the image of the linear
elements of ${C_{(0)}}_e$ in $R$. Here an element $x\in {C_{(0)}}_e$ is
\emph{linear} provided $H(x)=0$.
As a main result we prove the quadratic analogue of \eqref{shuxt}.
\begin{The}\label{xext}
Let $R$ be a ring and let $M$ be an $R$-bimodule. Then there are
natural bijections
$$
\Xext_L(R;M)^{\SG,\tl}\approx\Xext_L(R;M)^{\SG,\square}\approx\HML^3(R;M)
$$
where the index $L$ indicates the full subcategories of linearly
generated crossed extensions. The first bijection is induced by the
lax monoidal functor $(\SG,\square)\to(\SG,\tl)$.
\end{The}
The proof of this result is given in Section \ref{proof261}.
{\bf Remark}. For the second bijection in the theorem we use the isomorphism
$$
\HML^3(R;M)\cong H^3(\modr R;D_M)
$$
where $\modr R$ is the category of finitely generated free right $R$-modules
and $D_M=\Hom_R(-,-\ox_RM)$. Here we use the following
interpretation of crossed biobjects from \ref{crexte}.
Let $(\V,\boxtimes)$ be a monoidal category and assume that finite
colimits exist in $\V$. Let $\pair(\V)$ be the category of pairs in
$\V$, objects are morphisms $V=\left(V_1\xto\d V_0\right)$ in $\V$
and morphisms $V\to W$ are pairs $\alpha=(\alpha_1:V_1\to
W_1,\alpha_0:V_0\to W_0)$ in $\V$ with $\d\alpha_1=\alpha_0\d$. Then
$(\pair(\V),\ubx)$ is a monoidal category with $\ubx$ defined by the
diagram with the inner square pushout
$$
\xymatrix{ V_1\boxtimes W_1\ar@{}[ddrr]|{\mathrm{push}}
\ar[rr]^{1 \boxtimes\d}\ar[dd]_{\d \boxtimes 1}
&&V_1\boxtimes W_0\ar[dd]\ar@/^/[dddr]^{\d \boxtimes 1}\\
\\
V_0\boxtimes W_1\ar[rr]\ar@/_/[drrr]_{1 \boxtimes\d}&&(V\ubx W)_1\ar[dr]^\d\\
&&&**[r]V_0 \boxtimes W_0:=(V\ubx W)_0.}
$$
One readily checks that a crossed $L$-biobject $C=(\d:B\to L)$ in
\ref{crexte} is the same as a monoid in $(\pair(\V),\ubx)$. Hence
the action of the monoid $C$ on an object $X$ in $\pair(\V)$ is
defined, compare section 5.1 in \cite{Ba}. In this case we call $X$
a \emph{$C$-module}.
{\bf Addendum}. For a crossed extension $C$ of $R$ by $M$ in either
$(\SG,\square)$ or $(\SG,\tl)$ let $\modr C$ be the category of
finitely generated free left $C$-modules. Then $\modr C$ is a linear
track extension which represents an element
$$
\brk{\modr C}\in H^3(\modr R;D_M)
$$
and the bijections from \ref{xext} carry the component of the
crossed extension $C$ in $(\SG,\tl)$, resp. in $(\SG,\square)$, to the
class $\brk{\modr C}$.
Since a crossed extension $C$ is also a monoid in the category of
pairs, we will also call $C$ a ``pair algebra''. The addendum makes use
of modules over such pair algebras.
\subsection{Square rings and quadratic rings}
A monoid in the
monoidal category $(\SG,\square)$ is termed a \emph{square ring},
while a monoid in the monoidal category $(\SG,\tl)$ is termed a
\emph{quadratic ring}.
More explicitly (see \cite{BHP},
\cite{square}, \cite{iwa}), to provide a square group $Q$ with a
\emph{square ring structure} is the same as to give additionally a
multiplicative monoid structure on $Q_e$. The multiplicative unit of
$Q_e$ is denoted by $1$. One requires that this monoid structure
induces a ring structure on the abelian group $Q\ad$ through the
canonical projection
$$
Q_e\to Q\ad, \ \ \ a\mapsto \bar{a}.
$$
Moreover the abelian group $Q_{ee}$ must be a $Q\ad\t Q\ad\t
(Q\ad)\op$-module with action denoted by $(\bar{x}\t \bar{y})\cdot
a\cdot\bar{z}\in Q_{ee}$ for $\bar{x}, \bar{y}, \bar{z}\in Q\ad$,
$a\in Q_{ee}$. In addition the following conditions must be
satisfied where $H(2)=H(1+1)$
\begin{enumerate}
\item[(i)]
$x(y+z)=xy+xz$
\item[(ii)]
$(x+y)z=xz+yz+P((\bar{x}\t \bar{y})\cdot H(z))$
\item[(iii)]
$(x\mid y )_H=(\bar{y}\t \bar{x})\cdot H(2)$
\item [(iv)]
$T((\bar{x}\t \bar{y})\cdot a\cdot\bar{z})=(\bar{y}\t \bar{x})\cdot
T(a)\cdot\bar{z}$
\item[(v)]
$P(a\cdot x)=P(a)x$
\item[(vi)]
$P((\bar{x}\t \bar{x})\cdot a) =xP(a)$
\item[(vii)]
$H(xy)=(\bar{x}\t \bar{x})\cdot H(\bar{y})+H(x)\cdot \bar{y}$
\end{enumerate}
Under the equivalence ${\sf Quad}(\Gr)\cong\SG$ square rings
correspond to monads on the category of groups, whose underlying
functors lie in ${\sf Quad}(\Gr)$.
A \emph{quadratic ring structure} on a square group $C$ is given by a
multiplicative monoid structure on $C_e$ and a ring structure on
$C_{ee}$. The multiplicative unit of $C_e$ is denoted by $1$. One
requires that these structures satisfy the following additional
conditions.
\begin{enumerate}
\item[(i)] $x(y+z)=xy+xz$,
\item[(ii)] $(x+y)z=xz+yz+P(\cref HyxH(z))$.
\end{enumerate}
Thus $C\ad$ is a ring. Moreover the maps
\begin{align*}
-T&:C_{ee}\to C_{ee},\\
(-\mid -)_H&:C\ad\t C\ad\to C_{ee}
\end{align*}
are ring homomorphisms, in other words one has
\begin{enumerate}
\item[(iii)] $(x\mid y)_H (u\mid v)_H=(xu\mid yv)_H$,
\item[(iv)] $T(ab)+T(a)T(b)=0$.
\end{enumerate}
Let us observe that $T(abc)=T(a)T(b)T(c)$. Furthermore the following equations hold
\begin{enumerate}
\item[(v)] $ P(a\Delta(x))=P(a)x,$
\item[(vi)] $P((x\mid x)_Ha)=xP(a),$
\item[(vii)] $H(xy)=(x\mid x)_HH(y)+H(x)\Delta(y).$
\end{enumerate}
It follows from the axioms that $\Delta:C\ad\to C_{ee}$ is a ring
homomorphism \cite{bjp}.
Let $\QR$ (resp. $\SR$) denote the category of quadratic (resp.
square) rings. One has the full embedding of categories ${\sf
Rings}\subset\QR $ (resp. ${\sf Rings}\subset\SR $) which identifies
rings with quadratic (resp. square) rings $C$ satisfying $C_{ee}=0$.
This inclusion has a left adjoint given by $R\mapsto R\ad$.
There is also a functor
$$
U:\QR\to \SR
$$
which assigns to a quadratic ring $C$ a square ring, whose
underline square group is the same, while the $C\ad\t C\ad\t
(C\ad)\op$-module structure on $C_{ee}$ is given by
$$ (\bar{x}\t \bar{y})a\bar{z}=(y\mid x)_Ha\Delta(z).$$
The initial object in the category of quadratic rings (resp. square
rings) is $\Znil$, which is given by $$(\Znil)_e=\Z=\ (\Znil)_{ee},
\ \ P=0, \ \ H(x)=\frac{x(x-1)}{2}.$$
We now extend the monoid ring construction to quadratic rings and
square rings. For a monoid $S$ one puts
$$
\Znil[S]_{ee}=\Z[S]\t\Z[S],
$$
where $\Z[S]$ is the free abelian group generated by $S$. We take
$\Znil[S]_e$ to be the free nil$_2$-group generated by $S$. The
homomorphism $P$ is given by $P(s\t t)=[t,s]$, $s,t\in S$, while the
quadratic map $H$ is uniquely defined by
$$
H(s)=0, \ \ (s\mid t)_H=t\t s\ \ s,t\in S.
$$
One has
$$
\Znil[S]\ad=\Z[S].
$$
There is a unique quadratic (resp. square) ring structure on
$\Znil[S]$ for which the multiplication on $\Znil[S]_e$ extends the
multiplication on the monoid $S$ and such that the ring structure
(resp. $\Z[S]\t \Z[S]\t\Z[S]\op$-module structure) on
$\Znil[S]_{ee}=\Z[S]\t\Z[S]$ is the obvious one (resp. is given by
$(x\t y)(s\t t) z=xsz\t ytz$). In this case $Q\ad=\Z[S]$ is the
usual monoid ring of $S$. The functor $\Znil[-]:{\sf Monoids}\to
\QR$ (resp. $\Znil[-]:{\sf Monoids}\to \SR$) is left adjoint to the
functor
$$
{\sf L}:\QR\to \sf Monoids \ \ \ ({\rm resp}. \ \ {\sf L}:\SR\to \sf
Monoids),
$$
where ${\sf L}(Q)$ consists of \emph{linear elements} of $Q$, that
is
$$
{\sf L}(Q)=\{x\in Q_e\mid H(x)=0\}.
$$
The equality $H(xy)=(\bar{x}\t\bar{x})H(y)+H(x)\bar{y}$ shows that
linear elements indeed form a multiplicative submonoid of $Q_e$.
\subsection{Quadratic pair algebras and crossed
square rings}
Now we consider crossed biobjects in the monoidal
categories $(\SG,\square)$ and $(\SG,\tl)$. Actually we restrict
ourselves to considering only those crossed biobjects $\d:C_1\to
C_0$ which induce isomorphism on ee-level. This is the condition
which implies that $\Cok(\d)$ is a usual ring and $\Ker(\d)$ is a
usual bimodule. This forces us to introduce the following
definition.
A \emph{quadratic pair module} (qpm for short) is
a morphism of square groups $\d:C_1\to C_0$
such that the homomorphism $\d_{ee}:{C_1}_{ee}\to{C_0}_{ee}$ is an identity map.
Thus explicitly a quadratic pair module $C$ is
given by a diagram
$$
\xymatrix{&C_{ee}\ar[dl]_P\\
C_{1}\ar[r]_\d&C_{0}\ar[u]_H}
$$
where $C_{1}$ and $C_{0}$ are groups, $C_{ee}$ is an abelian group,
$P$ and $\d$
are group homomorphisms and $H$ is a quadratic map,
and moreover the following
identities are satisfied for any $a\in C_{ee}$, $r,s\in C_{1}$ and
$x,y\in C_{0}$:
\begin{align*}
PH\d P(a)&=2P(a);\\
H(x+\d P(a))&=H(x)+H\d P(a);\\
PH(\d(r)+\d(s))&=PH\d(r)+PH\d(s)+[r,s];\\
\d PH(x+y)&=\d PH(x)+\d PH(y)+[x,y].
\end{align*}
The category of qpm's is denoted by $\QPM$. If $C$ is a qpm, then
$\Im(\d)$ is a normal subgroup of $C_0$ containing the commutator
subgroup of $C_0$. Thus
$$h_0(C):=\Cok(\d)$$
is an abelian group. Moreover
$$h_1(C):=\Ker(\d)$$
is a central subgroup of $C_{1}$.
We have an exact sequence of square groups
$$0\to h_1(C)\to C_{(1)}\xto{(\d,\Id)} C_{(0)}\to h_0(C)\to 0.$$
Here $(C_{(1)})_{ee}=(C_{(0)})_{ee}=C_{ee}$, $(C_{(1)})_{e}=C_{1}$ and
$(C_{(0)})_e=C_{0}.$ The
structural maps are given by
$$ P^{{C}_{(0)}}=\d P, \ \ \ P^{{C}_{(1)}}=P,$$
$$ H^{{C}_{(0)}}=H, \ \ \ H^{{C_{(1)}}}=H\d.$$
A qpm together with crossed biobject structure in the monoidal
category $(\SG,\tl)$ is called a \emph{quadratic pair algebra}
(shortly qpa). Thus a qpa is a qpm $C$ together with a ring
structure on $C_{ee}$ and a quadratic ring structure on $C_{(0)}$.
Additionally a two-sided action of $C_{0}$ on $C_{1}$ is given, which is
associative and unital and
the following identities are satisfied for all $x,y\in C_{0}$,
$r,s\in C_{1}$, $a,b\in C_{ee}$:
\begin{enumerate}
\item[(i)] $ P((x|x)_Ha\Delta(y))=xP(a)y$
\item[(ii)] $\d(x r y)=x\d (r) y$
\item[(iii)] $ \d(r)s=r\d(s)$
\item[(iv)] $x(r+s)=xr+xs $
\item[(v)] $r(x+y)=rx+ry$
\item[(vi)] $(x+y)r=xr+yr+ P ((y|x)_HH\d(r))$
\item[(vii)] $(r+s)x=rx+sx+P( (s|r)_{H\d}H(x)).$
\end{enumerate}
If $C$ is a qpa, then $C_{(0)}$ is a quadratic ring and the
multiplication on $C_{0}$ yields the multiplication on $h_0(C)$
which equips $h_0(C)$ with a structure of a ring. Moreover
$h_1(C)$ is a bimodule over $h_0(C)$.
A qpm together with a crossed biobject structure in the monoidal
category $(\SG,\square)$ is called a \emph{crossed square ring}
(shortly csr). Thus a csr is a qpm $C$ together with a square ring
structure on $C_{(0)}$ and a two-sided action of $C_{0}$ on $C_{1}$,
which is associative and unital and
such that
the following identities hold for all $x,y,z,t\in C_{0}$,
$r,s\in C_{1}$, $a,b\in C_{ee}$:
\begin{enumerate}
\item[(i)] $P ((\bar{x}\t \bar{x})\cdot a\cdot y)=x\cdot P(a)\cdot y$
\item[(ii)] $\d(x\cdot r\cdot y)=x\cdot \d (r)\cdot y$
\item[(iii)] $ \d(r)s=r\d(s)$
\item[(iv)] $x(r+s)=xr+xs $
\item[(v)] $r(x+y)=rx+ry$
\item[(vi)] $(x+y) r=xr+yr+ P ((\bar{x}\t \bar{y})\cdot H\d r)$
\item[(vii)] $(r+s)x=rx+sx+P((\bar{\d r}\t \bar{\d s})\cdot Hx).$
\end{enumerate}
In a crossed square ring the quotient $R=\Cok(\d)$ has a ring
structure and $\Ker(\d)$ is a bimodule over $R$.
We denote by
$\CS$
the category of crossed square rings.
Let $R$ be a ring and $M$ be a bimodule over $R$. A \emph{quadratic
ring extension} (resp. \emph{crossed square ring extension}) of
$R$ by $M$ is an exact sequence
$$0\to M\xto{i} C_{(1)}\xto{(\d,\Id)} C_{(0)}\xto{p} R\to 0$$
where $(\d,\Id):C_{(1)}\to C_{(0)}$ is a qpa (resp. csr), the induced
homomorphisms $p:\Cok(\d)\to R$ is an isomorphism of rings and the
induced homomorphism $i:M\to \Ker(\d)$ is an isomorphism of
bimodules over $\Cok(p)$. Here $M$ is considered as a bimodule over
$\Cok(\d)$ via the isomorphism $p:\Cok(\d)\to R$.
Using the definition of the products $\square$ and $\tl$ from
\cite{square}, resp. \cite{bjp}, one readily checks:
\begin{Le}
A crossed extension of $R$ by $M$ in $(\SG,\square)$ is isomorphic
to a crossed square ring extension. A crossed extension of $R$ by
$M$ in $(\SG,\tl)$ is isomorphic to a quadratic ring extension.
\end{Le}
Hence we have explicitly described the objects in the category of
the main theorem \ref{xext}.
\begin{Exm}\label{ztildanil} Let $Q$ be a square ring. One can
consider the quotient $Q_{ee}/(\Id-T)$, where as usual $T=HP-\Id$.
Let $\tP:Q_{ee}\to Q_{ee}/(\Id-T) $ be the canonical projection. It
is clear that the homomorphism $P:Q_{ee}\to Q_e$ factors through
$Q_{ee}/(\Id-T)$. We denote by $\d:Q_{ee}/(\Id-T)\to Q_e$ the
quotient map. Then $$\xymatrix{& Q_{ee}\ar[dl]_{\tP}\\
Q_{ee}/(\Id-T)\ar[r]_{\ \ \d} &Q_e\ar[u]_H}$$ is a crossed square
ring. Thus Theorem \ref{xext} assigns to any square ring $Q$ an
element in $\HML^3(Q\ad, Q^{re})$, where $Q^{re}
=\Ker(\d:Q/(\Id-T)\to Q_e)$. In particular, for the square ring
$\Znil$ one obtains the following crossed square ring
$$\xymatrix{& \Z\ar[dl]_{1}\\
\Z/2\Z\ar[r]_{0}&\Z\ar[u]_H}$$ where $H(x)=\frac{x^2-x}{2}$ which
defines an element of $\HML^3(\Z,\Z/2\Z)=\Z/2\Z$, which is actually
the generator.
\end{Exm}
\subsection{The homomorphism $\nu:\HML^3(R,M)\to
\H^0(R,\,_2M)$}\label{nu}
Let $R$ be a ring and let $M$ be a bimodule over $R$. Take a crossed
square ring extension $(\d)$ of $R$ by $M$
$$\xymatrix{
0\ar[r]& M\ar[r]^i& C_{(1)}\ar[r] ^{(\d,\Id)} &C_{(0)} \ar[r]^p&R\ar[r]&0}$$ and set
$$\upsilon(w):=P H(2)$$
Since $H(1)=0$ it follows that $\d P H(2)=\d P(1\mid 1)_H=0$. On the
other hand $2P H(2)=P H\d P H(2)=0$. Thus $\upsilon(w)\in \, _2\,M$.
Actually $$\upsilon(w)\in \, \H^0(R,\, _2\,M)$$ and $\nu$ yields a
well-defined map
$$
\nu:\Xext_L(R;M)^{\SG,\square}\to\H^0(R;{}_2M).
$$
\begin{Le}
Kernel of $\nu$ coincides with the image of
$$
\Xext(R;M)^{\Ab,\t}\to\Xext_L(R;M)^{\SG,\square}.
$$
\end{Le}
In fact we obtain the lemma directly by the exact sequence in
section 1 and the bijections (\ref{shuxt}) and \ref{xext}. We snow show
the lemma more directly in terms of crossed extensions.
\begin{proof}
If $\d$ is a crossed ring extension then $C_{ee}=0$ and a fortiori
$H=0$, thus $\nu(\d)=0$. Conversely, assume $(\d)$ is a crossed
square ring extension with $\nu(\d)=0$. Without loss of generality
one can assume that $C_{(0)}$ is a monoid square ring
$C_{(0)}=\Znil[L]$ (see Section \ref{proof261} below). In this case
$C_{ee}=\Z[L]\t \Z[L]$ and $H(2)=(1\mid 1)_H=1\t 1.$ The equality
$P((\bar{x}\t \bar{x}\cdot m\cdot y)=x\cdot P(m)\cdot y$ shows that
$P$ factors through $\Lambda^2(\Z[L])$. Thus one gets the following
diagram
$$\xymatrix{&&& 0\ar[d]&&\\
&& \Lambda ^2(\Z[L])\ar[r]^\id\ar[d]^{\tP}&\Lambda ^2(\Z[L])\ar[d]^{[-,-]}&&\\
0\ar[r]&M\ar[r]\ar[d]_\Id&C_1\ar[r]^\d\ar[d]&C_{0}\ar[r]\ar[d]&R\ar[r]
\ar[d]_\Id&0\\
0\ar[r]&M\ar[r]&\Cok{\tP}\ar[r]\ar[d]&\Z[L]\ar[r]\ar[d]&R\ar[r]&0\\
&&0&0&& }$$ Since $C_{0}$ is a free nil$_2$-group on $L$ the
commutator map $[-,-]$ is a monomorphism. It follows that $\tP$ is
also a monomorphism, the bottom row is exact and $(\d)$ is
equivalent to
$$0\to M\to \Cok{\tP}\to \Z[L]\to R\to 0$$
which is a crossed ring extension in $(\Ab,\t)$.
\end{proof}
\subsection{Application to ring spectra} Since Mac~Lane
cohomology and topological Hochschild cohomology are isomorphic for
discrete rings it follows that for any ring spectrum $\Lambda$ with
$\pi_i(\Lambda)=0$ for $i\ne0,1$ there is a well-defined element
$k(\Lambda)\in \HML^3(\pi_0(\Lambda),\pi_1(\Lambda))$ known as the
first Postnikov invariant (see \cite{laza}) and any element in this
group comes in this way. Thus linearly generated crossed square
rings and quadratic pair algebras can be used to model such ring
spectra. The explicit functor from the category of crossed square
rings to the category of ring spectra can be constructed as follows.
By Corollary \ref{lodjan} below one can associate to any crossed
square ring an internal groupoid in the category of square rings and
hence an internal groupoid in the category of algebraic theories
(see \ref{sqringtheories} below). Now using the nerve construction one
obtains a simplicial object in the category of algebraic theories.
Then one can use the well-known construction of Schwede
\cite{schwede} to obtain a ring spectrum in a functorial way.
\section{Recollections}
\subsection{Preliminaries on double categories and internal categories}
Let $\A$ be a category with finite limits. An \emph{internal
category} $C$ in $\A$ consists of the following data: objects $C_0$
(\emph{object of objects}), $C_1$ (\emph{object of morphisms}) and
morphisms $s,t:C_1\to C_0$ (\emph{source} and \emph{target}),
$i:\C_0\to \C_1$ (\emph{identity}), $m:C_2\to C_1$
(\emph{composition}) satisfying associativity and unitality
conditions. Here $C_2$ is defined by the pullback diagram
$$
\xymatrix{
G_2\ar[r]^{p_1}\ar[d]_{p_2}& G_1\ar[d]^s\\
G_1\ar[r]_t&G_0
}
$$
We denote by $\Cat(\A)$ the category of internal categories in $\A$.
Let us also recall that an internal category $C$ is called an
\emph{internal groupoid} provided the diagram
$$
\xymatrix{
G_2\ar[r]^{m}\ar[d]_{p_2}& G_1\ar[d]^s\\
G_1\ar[r]_s&G_0
}
$$
is a pullback diagram. We denote by $\Gpd(\A)$ the category of
internal groupoids in $\A$.
An internal category in the category of sets $\Sets$ is nothing but
a small category, while a groupoid object in the category of sets
$\Sets$ is a groupoid. We write $\Cat$ and $\Gpd$ instead of
$\Cat(\Sets)$ and $\Gpd(\Sets)$.
Let $A$ be an object of $\A$, then we can consider the internal
groupoid $A^{dis}$ with $(A^{dis})_0=A=(A^{dis})_1$ and $s=t=\Id_A$.
An internal category is called \emph{discrete} if it is isomorphic
to $A^{dis}$ for some $A$. We will need also an internal groupoid
$A^{adis}$ with $(A^{adis})_0=A$, $(A^{adis})_1=A\x A$, where $s$
and $t$ are the projections. An internal category is called
\emph{antidiscrete} if it is isomorphic to $A^{adis}$ for some $A$.
Let $\B$ be a category with finite limits and let $F:\A\to\B$ be a
functor which preserves finite limits. Then obviously $F$ yields
functors $\Cat(\A)\to\Cat(\B)$ and $\Gpd(\A)\to\Gpd(\B)$ which will
be also denoted by $F$.
Let us recall that a \emph{double category} is an internal category
in the category $\Cat$ of small categories. Let $\D$ be a double
category with the object category $\D_0$ and morphism category
$\D_1$.
We have a functor ${\sf Ob}:\Cat\to\Sets$, which assigns to a
category $\C$ the set of objects of $\C$. Since ${\sf Ob}$ preserves
inverse limits, for any double category $\D$ we obtain a category
$O(\D)$, whose morphisms are objects of $\D_1$ and objects are
objects of $\D_0$. A double category $\D$ is a \emph{2-category} if
$O(\D)$ is a discrete category. Equivalently a 2-category is a
category enriched in the category $\Cat$. Let us recall how one gets
such an enrichment.
Let $\D$ is a 2-category. Then objects of the category $\D_0$ are
called simply objects of $\D$, while morphisms of the category
$\D_0$ are called simply morphisms of $\D$. Let $f,g:A\to B$ be
morphisms of $\D$. Then $A$ and $B$ are also objects in $\D_1$ and
we can consider the set of all morphisms $\alpha:A\to B$ in $\D_1$
such that $s(\alpha)=f$ and $t(\alpha)=g$. Such an $\alpha$ is
called a 2-morphism from $f$ to $g$. Thus for objects $A$ and $B$ we
have a category $\D(A,B)$ with objects morphisms from $A$ to $B$ in
the category $\D_0$ and morphisms from $f:A\to B$ to $g:A\to B$
being all 2-morphisms from $f$ to $g$.
Conversely, if $\B$ is a a
category enriched in the category $\Cat$, then one can consider the
following categories $\B_0$ and $\B_1$. The category $\B_0$ has the
same objects as $\B$, while morphisms in $\B_0$ are 1-arrows of
$\B$. The category $\B_1$ has the same objects as $\B_0$. The
morphisms $A\to B$ in $\B_1$ are 2-arrows $\alpha:f\then f_1$ where
$f,f_1:A\to B$ are 1-arrows in $\B$. Composition in $\B_1$ is given
by $(\bb:x\then x_1)(\aa:f\then f_1):=(\bb\aa:xf\then x_1f_1)$,
where
$$
\bb\aa=\bb f_1+x\aa=x_1\aa+\bb f.
$$
One furthermore has the source and target functors
$$
\xymatrix{
\B_1\ar@<1ex>[r]^s\ar@<-1ex>[r]_t&\B_0
},
$$
with $s(\alpha:f\then f_1)=f$, $t(\alpha:f\then f_1)=f_1$, and the
``identity'' functor $i:\ta_0\to\ta_1$ assigning to an 1-arrow $f$
the identity 2-arrow $0_f:f\then f$. One easily sees that in this
way we obtain a double category such that after applying the functor
${\sf Ob}:\Cat\to\Sets$ one gets a discrete category.
\subsection{Preliminaries on Baues-Wirsching cohomology of small categories
}\label{relcoh}
For a small category $\c$ we denote by $\f\c$ the \emph{category of
factorizations} of $\c$ \cite{BW}. Objects of $\f\c$ are morphisms
of $\c$, and a morphism from $\alpha:x\to y$ to $\beta:u\to v$ is a
pair $(\nu:u\to x,\psi:y\to v)$ of morphisms in $\c$ such that
$\beta = \psi \alpha \nu$, that is, one has a commutative diagram
$$
\xymatrix{ x\ar[rr]^{\alpha} & & y \ar[d]^{\psi}\\
u \ar[u]^{\nu}\ar[rr]^{\beta} & & v }
$$
Composition in $\f\c$ is defined by $(\nu,\psi)(\nu',\psi')= (\nu'
\nu, \psi \psi')$. A \emph{natural system} on $\c$ is a covariant
functor $D:\f\c\to\Ab$. Now, following \cite{BW}, one defines the
cohomology $\H^*(\c,D)$ as the cohomology of the cochain complex
$\F^*(\c,D)$ given by
$$
\F^n(\c,D)= \prod_{ c_0 \stackrel{\alpha_1}{\leftarrow} \cdots
\stackrel{\alpha_n}{\leftarrow} c_n} D_{ \alpha_1 \cdots \alpha_n}
$$
with the coboundary map
$$
d : \F^n(\c,D) \to \F^{n+1}(\c,\D)
$$
given by
$$
\begin{array}{ll}
(df)(\alpha_1, \cdots, \alpha_{n+1}) &= (\alpha_1)_* f(\alpha_2,
\cdots, \alpha_{n+1}) \\ \\
& + \sum_{i=1}^n (-1)^i f( \alpha_1, \cdots, \alpha_i
\alpha_{i+1}, \cdots, \alpha_{n+1}) \\ \\
& + (-1)^{n+1} (\alpha_{n+1})^* f (\alpha_1, \cdots, \alpha_{n}).
\end{array}
$$
Here, and in the rest of the paper, we use the following convention.
For a diagram $u\xto{\beta}x\xto{\alpha}y\xto{\gamma}v$ and elements
$a \in D_\beta$, $ b \in D_{\gamma}$, we write $\alpha_*a$ and
$\alpha^*b$ for the image of the elements $a$ and $b$ under the
homomorphisms $D(id_u, \alpha): D_\beta \to D_{\alpha\beta}$ and
$D(\alpha, id_v): D_\gamma \to D_{\gamma\alpha}$ respectively.
We also need the relative cohomologies of small categories. Let
$p:\ka\to\c$ be a functor which is identity on objects and
surjective on morphisms. Let $D:\f\c\to\Ab$ be a natural system on
$\c$. We have an induced natural system $p^*D$ on $\ka$ given by
$g\mapsto D_{pg}$, which we will, abusing notation, still denote by
$D$. Then $p$ yields a monomorphism of cochain complexes
$\F^*({\c},D)\to \F^*(\ka,D)$. We let $\F^*(\c,\ka;D)$ be the
cokernel of this homomorphism. The {\it $n$-th dimensional relative
cohomology} $\H^n(\c,\ka;D)$ is defined as the $(n-1)$-th homology
of the cochain complex $\F^*(\c,\ka;D)$. Then one has an exact
sequence
$$0\to \H^0({\c},D)\to \H^0(\ka,D)\to \H^1(\c, \ka;D)\to \cdots \to$$
$$\to \H^n({\c },D)\to \H^n(\ka,D)\to \H^{n+1}(\c,\ka;D)\to \cdots.$$
We have a functor $\f\c \to \c\op \times \c$ which sends an arrow
$\alpha:c \to d$ to the pair $(c,d)$. This functor allows us to
conclude that any bifunctor gives rise to a natural system. Thus for
any bifunctor $D:\c\op \times \c\to \Ab$ we have well-defined
cohomology groups $H^*(\c,D)$.
Among many equivalent definitions of the Mac~Lane cohomology
\cite{JP} for our purposes the most convenient is via the
Baues-Wirsching cohomology of small categories \cite{BW}. Let $R$ be
a ring. Let $\Mod R$ be the category of right $R$-modules and let
$\modr R$ be the full subcategory of finitely generated free right
$R$-modules. To avoid set-theoretic complications we will assume
that objects of $\modr R$ are natural numbers and morphisms from $y$
to $x$, $x,y\in \N$ are $(x\x y)$-matrices with entries in $R$. We
write $f=(f_i^k)$ for a morphism $y\to x$, where $f_i^k\in R, 1\le
i\le x, 1\le k\le y$.
For an $R$-$R$-bimodule $M$, we denote by $D_M:(\Modr R)\op\x \Modr
R\to \Ab$ the bifunctor given by
$$
D_M(X,Y):=\Hom_R(X,Y\t_R M), \ \
X, Y \in \Modr R.
$$
Now one defines the \emph{Mac~Lane cohomology of $R$ with
coefficients in $M$} by
$$
\HML^*(R,M):=\H^*(\modr R,D_M).
$$
We refer to \cite{JP} and Chapter 13 of \cite{jll} for relationship
between different definitions of Mac~Lane cohomology. We use this
definition of $\HML^*$ also if $R$ is a square ring or a quadratic
ring.
\subsection{Third Baues-Wirsching cohomology and linear track extensions}
We recall the relationship between third Baues-Wirsching cohomology
and linear track extensions. We start with recalling the definition of track
categories.
A \emph{track category} is a groupoid enriched category, i.~e. a
2-category such that all of its 2-morphisms are invertible.
Equivalently a track category $\ta$ is an internal groupoid in the
category $\Cat$ such that $O(\ta)$ is a discrete category. We will
use the following notation for track categories. Composition of
morphisms will be denoted by juxtaposition; for 2-arrows we will use
additive notation, so composition is + and identity 2-arrows are
denoted by 0. The hom-category for objects $A$, $B$ of a track
category will be denoted by $\hog{A,B}$. If there is a 2-arrow
$\alpha:f\then g$ between maps $f,g\in\Ob(\hog{A,B})$, we will say
that $f$ and $g$ are homotopic and write $f\ho g$. We have the
\emph{homotopy category} $\ta_\ho=\ta_0/\ho$. Objects of $\ta_\ho$
are objects in $\Ob(\ta)$, while morphisms of $\ta_\ho$ are
homotopy classes of morphisms in $\ta_0$. A map $f$ in $\ta$ is
called a \emph{homotopy equivalence} if the class of $f$ in
$\ta_\ho$ is an isomorphism.
Two track categories $\ta$, $\ta'$ are called \emph{weakly
equivalent} if there is an enriched functor $F:\ta\to\ta'$ which
induces equivalences of hom-groupoids
$\hog{X,Y}_\ta\to\hog{FX,FY}_{\ta'}$ and is \emph{essentially
surjective}, i.~e. any object of $\ta'$ is homotopy equivalent to
one of the form $FX$.
Let $\c$ be a small category and let
$D$ be a natural system on $\c$. A \emph{linear track extension of $\c$ by $D$} denoted by
$$
0\to D\to \ta_1 \rightrightarrows \ta_0\to \c \to 0
$$
is a pair $(\ta, \tau)$. Here $\ta$ is a track category equipped
with a functor $q:\ta_0\to\c$ which is full and identity on objects.
In addition for maps $f,g$ in $\ta_0$ we have $q(f)=q(g)$ iff $f\ho
g$. In other words the functor $q$ identifies $\c$ with $\ta_\ho$.
Furthermore, for each map $f:A\to B$ in $\ta_0$ there is given an
isomorphism of groups $\tau_f:D_{qf}\to {\ta}(f,f)$, such that for
any $\xi:f\then g$ and $a\in D_{qf}=D_{qg}$ one has
$$
\xi+\tau_f(a)=\tau_g(a)+\xi.
$$
Furthermore for any diagram $\buildrel e \over \longrightarrow
\buildrel f \over \longrightarrow \buildrel h \over \longrightarrow$
additionally one has
$$h_*\tau_f(a)=\tau_{hf}(h_*a),$$
$$ e^*\tau_f(a)=\tau_{fe}(e^*a).$$
For a category $\c$ and a natural system $D:\f\c\to \Ab$ we denote
by ${\Track}(\c,D)$ the category of all linear track extensions of
$\c$ by $D$, where the morphisms are the obvious ones.
Linear track extensions of categories were first described in the
preprint of \cite{B} and the following theorem in a slightly
different terminology first was
proved in \cite{P1} (see also \cite{P2}) and was proved
by different methods in \cite{BD}.
\begin{The}\label{h3}\cite{P1} For a small category $\c$ and a natural system
$D:\f\c\to \Ab$ there exists a natural bijection between the set of
connected components of the category $\Track(\c,D)$ and third
cohomology:
$$
\pi_0(\Track(\c,D))\cong \H^3(\c,D).
$$
\end{The}
The proof of Theorem \ref{h3} given in \cite{P1} and \cite{P2} is
based on the following Theorem \ref{fardobiti}, which is going to be
crucial in this paper as well.
Let $p:\ka\to \c$ be a functor which is identity on objects and
surjective on morphism. Let $D:\f\c\to \Ab$ be a natural system on
$\c$. We denote by $\Track(\c,\ka;D)$ the subcategory of
$\Track(\c,D)$ whose objects are track categories $\ta$ satisfying
$\ta_0=\ka$,
$$
0\to D\to \ta_1 \rightrightarrows \ka \buildrel {q}\over \to \c \to
0,
$$
whereas morphisms are those morphisms in $\Track(\c,D)$ which are
identity on $\ka$.
\begin{The}\label{fardobiti}\cite{P1},\cite{P2}
For a small category $\c$, a bifunctor $D:\c\op\x \c\to \Ab$ and a
functor $p:\ka\to \c$ which is identity on objects and surjective on
morphisms, the category $\Track(\c,\ka;D)$ is a groupoid and there
exists a natural bijection
$$
\pi_0(\Track(\c,\ka;D))\cong \H^3(\c,\ka;D).
$$
\end{The}
\subsection{Relative track extensions of algebraic theories}
An \emph{algebraic theory} is a category with finite coproducts. A
\emph{morphism of algebraic theories} is a functor preserving finite
coproducts. We denote the coproduct by $\vee$. Let $\c$ be an
algebraic theory. A natural system $D:\f\c\to \Ab$ is called
\emph{cartesian} if for any arrow $f:c=c_1\vee \cdots \vee c_n\to d$
the natural map
$$
D_f\to D_{f_1}\x \cdots \x D_{f_n},
$$
given by $x\mapsto ((i_1)^*x,\cdots ,(i_n)^*x)$, is an isomorphism.
Here $i_k:c_k\to c$ is the standard inclusion and $f_k=i_k\circ f:c_k\to
d$. For example if $D:\c\x\c\op\to\Ab$ is a bifunctor such that for
all $c,d$ and $x$ from $\c$ one has an isomorphism
$$
D(c\vee d,x)\cong D(c,x)\x D(d,x)
$$
natural in $c,d$ and $x$, then the natural system corresponding to $D$ is
cartesian.
Let $\c$ be an algebraic theory and let $D:\f\c\to\Ab$ be a
cartesian natural system. A track extension
$$
0\to D\to \ta_1 \rightrightarrows \ta_0\xto{p} \c \to 0
$$
is called a \emph{track extension of algebraic theories} if $\ta_0$
is an algebraic theory and the functor $p$ is a morphism of
algebraic theories.
\begin{Le}\label{trth} Let $\c$ be an algebraic theory and let $D:\f\c\to\Ab$ be a cartesian
natural system. Let $0\to D\to \ta_1 \rightrightarrows \ta_0\xto{p} \c \to 0$ be a
track extension of algebraic theories. Then $\ta_1$ is also an algebraic theory
and $s,t:\ta_1\to \ta_0$ are morphisms of algebraic theories.
\end{Le}
\begin{proof}
Let $\aa:f\then g$ and $\aa':f'\then g'$ be tracks, where $f,g:A\to
B$ and $f',g': A'\to B$ are 1-morphisms. We have to show that there
is a unique track $(\aa, \aa'):(f,g)\then (f', g')$ such that
$i_A^*(\aa,\aa')= \aa$ and $ i_{A'}^*(\aa, \aa')= \aa'$, where
$(f,g):A\vee A'\to B$ is the unique 1-morphism with $i_A(f,g)=f$ and
$i_{A'}(f,g)=g$. Here $i_A:A\to A\vee B$ is the canonical inclusion
and similarly for $i_{A'}$. First we show the existence of such a
track. By assumption $f \ho g$ and $f'\ho g'$. Since $p$ preserves
finite coproducts, it follows that $(f, f')\ho (g, g')$. Hence there
exists a track $\eta: (f, f')\then (g, g')$. Since
$i_A^*(\eta):f\then g$, there exists a unique element $x\in D_{pf}$
such that $i_A^*(\eta)=\aa+ \sigma_{f}(x)$. Similarly there exists a
unique element $x'\in D_{pf'}$ such that $i_{A'}^*(\eta)=\aa'+
\sigma_{f'}(x')$. By our assumptions there is a unique element $y\in
D_{(pf,pf'}$ such that $i_A(y)=x$ and $i_{A'}(y)=x'$. Then the track
$\xi=\eta-\sigma_{(f, f')}(y)$ satisfies the condition required. To
prove uniqueness one observes that if $\xi$ and $\eta$ both satisfy
the condition, then they will differ by an element $z\in
D_{(pf,pf')}$, whose restrictions to $D_{pf}$ and $D_{pf'}$ are
zero, hence it is itself zero and the lemma follows.
\end{proof}
\begin{Le}\label{trcogr}
Let $0\to D\to \ta_1 \rightrightarrows \ta_0\xto{p} \c \to 0$ be a
track extension of algebraic theories and let $\nu:X\to X\vee X$ be
an internal cogroup in $\ta_0$. Then $X$ is also a cogroup in
$\ta_1$, where the cogroup structure is given by the morphism
$0:\nu\then\nu$.
\end{Le}
\begin{proof}
By Lemma \ref{trth} the ``identity functor'' $\ta_0\to\ta_1$
respects finite coproducts and therefore carries a cogroups to
cogroups.
\end{proof}
\subsection{Quadratic functors, quadratic categories and square objects}
We now recall the relationship between square groups and quadratic
functors. We consider endofunctors $F:\Gr\to\Gr$ of the category of
groups with $F(0)=0$. Additionally we assume that $F$ preserves
filtered colimits and reflexive coequalizers. The last condition
means that for any simplicial group $G_*$ the canonical homomorphism
$\pi_0(F(G_*))\to F(\pi_0(G_*))$ is an isomorphism. Such a functor
$F$ is completely determined by the restriction of $F$ to the
subcategory of finitely generated free groups.
The second cross-effect $F(X|Y)$ of $F$ is a bifunctor defined via
the short exact sequence
$$
0\to F(X|Y)\to F(X\vee Y)\to F(X)\x F(Y)\to 0.
$$
Here $\vee$ denotes the coproduct in the category of groups and the
last map is induced by the canonical projections:
$r_1=(\Id_X,0):X\vee Y\to X$ and $r_2=(0,\Id_Y):X\vee Y\to Y$. A
functor $F$ is called \emph{linear} if the second cross effect
vanishes. Moreover, $F$ is called \emph{quadratic} if $F(X|Y)$ is
linear in $X$ and $Y$. Let ${\sf lin}(\Gr)$ (resp. ${\sf
Quad}(\Gr)$) be the category of such linear (resp. quadratic)
endofunctors. Any endofunctor in ${\sf lin}(\Gr)$ is isomorphic to a
functor $T$ of the form $ T(X)=A\t X\ab$ where $A$ is an abelian
group. Therefore there is an equivalence of categories
$$
{\sf lin}(\Gr)\simeq \Ab.
$$
Let $F:\Gr\to \Gr$ be a quadratic functor. We associate with $F$ a
square group ${\sf cro}(F)$ as follows. We put
$$
{\sf cro}(F)_e=F(\Z), \ \ {\sf cro}(F)_{ee} =F(\Z\mid\Z).
$$
The homomorphism $P$ of the square group ${\sf cro}(F)$ is the
restriction of the homomorphism $(\Id,\Id)_*: F(\Z \vee \Z)\to
F(\Z)$. We denote by $e_1$ and $e_2$ the canonical free generators
of $\Z\vee\Z$. The map $H$ is given by
$$
H(x)=\mu_*(x)-p_2(\mu_*x)-p_1(\mu_*x)
$$
Here $\mu:\Z\to \Z \vee\Z$ is the unique homomorphism which sends $1$ to
$e_1+e_2$, while $p_1$ and $p_2$ are endomorphisms of $\Z\vee\Z\to \Z\vee\Z$
such that $p_i(e_i)=e_i$, $i=1,2$ and $p_i(e_j)=0$, if $i\not =j$.
The main result of \cite{square} claims that the functor
$$
{\sf cro}: {\sf Quad}(\Gr)\to \SG
$$
is an equivalence of categories. Under this equivalence square rings
corresponds to monads on the category of groups, whose underlying
functors lie in ${\sf Quad}(\Gr)$.
Let $\C$ be an algebraic theory with zero object $0$. We will say
that $\C$ is equipped with a structure of \emph{quadratic theory} if
each object $C$ in $\C$ is equipped with a cogroup structure
$\nu_C:C\to C\vee C$ and the functor $\C(C,-):\C\to \Gr$ is
quadratic. Thus for all $X$ and $Y$ in $\C$ one has the following
short exact sequence of groups
$$
0\to \C(C;X\mid Y)\to \C(C,X\vee Y)\to \C(C,X)\x \C(C,Y)\to 0
$$
and $\C(C;X\mid Y)$ is linear in $X$ and $Y$. This definition is
equivalent but not identical to the one given in \cite{BHP}.
\begin{Le}\label{lec} Let $\c$ be an additive category,
$D:\c\op\x \c\to \Ab$ be a biadditive bifunctor and
$$0\to D\to \ta_1 \rightrightarrows \ta_0\xto{p} \c \to 0$$
be a linear track extension. If $\ta_0$ is a quadratic theory and
$p$ preserves finite coproducts, then $\ta_1$ is also a quadratic
theory.
\end{Le}
\begin{proof} By Lemma \ref{trth} the category $\ta_1$ is an algebraic theory.
It is quite easy to show that the zero object in $\ta_0$ remains
also a zero object in $\ta_1$. By Lemma \ref{trcogr} any object in
$\ta_1$ has a canonical cogroup structure. We claim that
$$\ta_1(X, Y\mid Z)\cong \ta_0(X, Y\mid Z)\x \ta_0(X, Y\mid Z),$$
which implies that $\ta_1(X, -)$ is a quadratic functor and hence
Lemma. To prove the claim we put
$$
r_1=(\Id,0):Y\vee Z\to Y \ \ \mathrm{ and} \ \ r_2=(0,\id):Y\vee Z\to Z.
$$
By definition of the cross-effect $\ta_0(X, Y\mid Z)$ consists of
1-morphisms $f:C\to Y\vee Z$ such that $r_1f=0=r_2f$. On the other
hand $\ta_1(X, Y\mid Z)$ consists of tracks $\aa:f\then g$ such
that $r_1f=0=r_2f$, $r_1g=0=r_2g$ and $r_{1*}\aa=0=r_{2*}\aa$. Thus
our claim is equivalent to the following: suppose $f,g:X\to Y\vee Z$
are 1-morphisms such that $r_1f=0=r_2f$, $r_1g=0=r_2g$. Then there
exists a unique track $\aa:f\then g$ such that
$r_{1*}\aa=0=r_{2*}\aa$. If $\aa$ and $\bb$ both satisfy the
assertion, then $\bb=\aa =\sigma_f(x)$ with $x\in D(X,Y\vee
Z)=D(X,Y)\oplus D(X,Z)$. The conditions $r_{1*}\aa=0=r_{2*}\aa=
r_{1*}\bb=r_{2*}\bb$ show that $x=0$. Hence we proved uniqueness.
Now we prove the existence. Since $\c$ is an additive category, $p$
respects coproducts and $r_1f=0=r_2f$ it follows that $p(f)=0$ in
$\c$. Similarly $p(g)=0$. In particular $f\ho g$. Thus there exist a
track $\xi:f\then g$. Then $r_{1*}(\xi):0\then 0$. Hence there
exists a unique $y\in D(X,Y)$ such that $r_{1*}(\xi)=\sigma_0(y)$.
Similarly $r_{2*}(\xi)=\sigma_0(z)$ for uniquely defined $z\in
D(X,Z)$. Since $D$ is biadditive, we have $(y,z)\in D(X,Y\vee Z)$.
It is clear that $\aa=\xi-\sigma_f(y,z)$ satisfies the conditions of
the claim.
\end{proof}
\subsection{Square rings and single sorted quadratic
theories}\label{sqringtheories} We recall the relationship between
square rings and quadratic categories \cite{BHP}. A quadratic theory
is a \emph{single sorted quadratic theory} if the objects of $\C$
are natural numbers and the coproduct on objects corresponds to the
addition of natural numbers. Thus each object ${\bf n}$ in $\C$ is
an $n$-fold coproduct of $\bf 1$. We additionally require that the
cogroup structure on ${\bf n}$ is the $n$-fold coproduct of the
cogroup structure on $\bf 1$.
Assume $\C$ is a single sorted quadratic theory. Then one has the
square ring ${\sf cro}(\C)$ with
$$
{\sf cro}(\C)_e={\sf cro}(\C({\bf 1},-))_e=\C({\bf 1},{\bf 1})
$$
and
$${\sf cro}(\C)_{ee}={\sf cro}(\C({\bf 1},-))_{ee}=\C({\bf 1};{\bf 1}\mid {\bf 1}).$$
The main result of \cite{BHP} shows that the functor ${\sf cro}$
from the category of single sorted quadratic theories to the
category of square rings is an equivalence of categories. The
inverse functor is given by $Q\mapsto\modr Q$. Here the objects of
the category $\modr Q$ are natural numbers, while morphisms from
$y$ to $x$, $x,y\in \N$ are defined by product sets
$$
{\sf Mor}(y,x):=(\prod_{k=1}^y\prod_{i=1}^x
Q_e)\x(\prod_{k=1}^y\prod_{1\le i<j\le x}Q_{ee}).
$$
For a morphism $f:y\to x$ we write $f=(f_i^k,f^k_{ij})$.
If $g=(g^s_k,g^s_{kl})$ is a morphism $z\to y$, then the composite
$fg=((fg)_i^s,(fg)^s_{ij})$ is given by
$$(fg)^s_i=f^1_i\circ g_1^s+\cdots +f_i^y\circ
g^s_y+\sum_{k<l}P((\bar{f}_i^k\t \bar{f}_i^l)\cdot g^s_{kl})$$
$$(fg)^s_{ij}=\sum_k(f^k_{ij}\cdot \bar{g}^s_k+
\sum_{i<l}((\bar{f}_i^k\t \bar{f}_i^l)\cdot g^s_{kl}+ (\bar{f}_i^l
\t \bar{f}_j^{k} )\cdot T g^s_{kl}+\overline{(f^l_i\cdot
g^s_l)}\t\overline{(f^k_j\cdot g^s_k)}\cdot H(2))$$ Actually $\modr
Q$ is a single sorted quadratic theory, the group structure on hom's
is defined by the formula:
$$(f_i^k,f_{ij}^k)+({f'}_i^k,{f'}_{ij}^k)=(f_i^k+{f'}_i^k,f_{ij}^k+{f'}^k_{ij}+
e_{ij}^k)$$ where $$e_{ij}^k=(\bar{f}_i^k\t \bar{f'}_j^k)\cdot
H(2).$$ To get more hints on the category $\modr Q$, we recall that
a \emph{right $Q$-module} \cite{BHP} is nothing but a right
$Q$-object in the monoidal category $(\SG,\square)$. More
explicitly, a right $Q$-module is a group $M$ together with maps
$M\x Q_{e}\to M$, $(m,x)\to mx$ and $M\x M\x Q_{ee}\to M$,
$(m,n,a)\mapsto [m,n]_a$ satisfying the following identities.
\begin{align*}
m1&=m,\\
(mx)y&=m(xy),\\
m(x+y)&=mx+my,\\
(m+n)x&=mx+nx+[m,n]_{H(a)},\\
mP(a)&=[m,m]_a,\\
[m,n]_{Ta}&=[n,m]_a,\\
[mx,ny]_a&=[m,n]_{(x\otimes y)a},\\
[[m,n]_a,z]_b&=0.
\end{align*}
Moreover $[m,n]_a$ is linear in $m,n$ and $a$ and lies in the center
of $M$. We denote by $\Mod Q$ the category of all right $Q$-modules.
It is a standard fact of universal algebra that the forgetful
functor $\Mod Q\to \sf Sets$ has the left adjoint, whose values on a
set $X$ is called the free right $Q$-module generated by the set
$X$. Now one checks directly \cite{BHP} that the category $\modr Q$
is equivalent to the category of finitely generated free right
$Q$-modules.
Let us observe that for $Q=\Znil$, the category of right
$\Znil$-modules is nothing but the category $\Nil$ of groups of
nilpotence class two. More generally, if $S$ is a monoid then the
category of right modules over the square ring $Q=\Znil[S]$ is
isomorphic to the category of pairs $(G,\alpha)$, where $G$ is a
group of nilpotence class two and $\alpha:S\to \Hom(G,G)$ is an
action of $S$ on $G$ via group homomorphisms.
\subsection{Internal groupoids and crossed objects}
We describe now internal groupoids in the category of square groups.
Actually results obtained in this section are very particular case
of much more general results of Gogi Janelidze \cite{gogia}.
Let $A$ and $G$ be square groups. An \emph{action of $G$ on $A$} is
a homomorphism of abelian groups $\xi:A\ad\t G\ad\to A_{ee}$.
In particular we have the action of $A$ on $A$ given by $(-,-)_H$,
which is called the \emph{adjoint action of $A$ on itself}.
Let $\xi$ be an action of $G$ on $A$. The \emph{semi-direct} product
of $G$ and $A$ denoted $G\rtimes A$ is a square group defined as
follows. As a set $(G\rtimes A)_e$ is the cartesian product $G_e\x
A_e$ while the group structure is given by
$$(g,x)+(h,y)=(g+h,x+y+P\xi(g,h)).$$
Moreover one puts
$$(G\rtimes A)_{ee}=G_{ee}\oplus A_{e},$$
$$P(u,a)=(Pu,Pa),$$
$$H(g,x)=(H(g),H(x)-\xi(x,g)).$$
One easily sees that
$$[(g,x),(h,y)]=([g,h],[x,y]+P\xi(x,h)-P\xi(y,g))$$
and
$$((g,x)\mid (h,y))_H=((g\mid)_H,(x\mid y)_H+HP\xi(x,h)-P\xi(x,h)-P\xi(y,g)).$$
Based on these identities one easily checks that $(G\rtimes A)_e$ is
really a square group and one has the following split short exact
sequence of square groups
$$0\to A\to A\rtimes G\to G\to 0$$
with obvious maps. Conversely, let
$$0\to A\xto{i} B\xto{p} G\to 0$$
be a short exact sequence. Assume $j:G\to B$ is a morphism of
square groups with $pj=\Id_G$. Then
$$\xi(x,g):=(i_e(x)\mid j_e(g))_H$$
defines an action of $G$ on $A$ and the maps
$f_e(g,x)=j_e(g)+i_e(x)$ and $f_{ee}(g,x)=j_{ee}(g)+i_{ee}(x)$
define an isomorphism $f=(f_e,f_{ee}):G\rtimes A\to B$ of square
groups.
A \emph{crossed square group} is a morphism of square groups
$\d:A\to G$ together with an action of $G$ on $A$ such that $\d$ is
compatible with the action of $G$, where $G$ acts on itself via the
adjoint action and the action of $A$ on $A$ given via $\d$ coincides
with the adjoint action of $A$ on itself. In other words a
homomorphism $\xi:A\ad\t G\ad\to A_{ee}$ of abelian groups is given
and the following identity holds
$$\d_{ee}\xi(x,g)=(\d_e(x),g)_H,$$
$$\xi(x,\d_e(y))=(x,y)_H.$$
We denote by $\CSG$ the category of crossed square groups. The
following is a specialization of the main result of \cite{gogia}.
\begin{Le}
Any internal category in the category of square groups is an
internal groupoid. Thus $\Cat(\SG)=\Gpd(\SG)$ and there is an
equivalence of categories
$$\Gpd(\SG)\cong \CSG.$$
\end{Le}
\begin{proof} The first fact is a general property of so
called Maltsev categories \cite{gogia}. The second part can be
proved by modifying the argument of Loday in \cite{jllcat} based on
our description of split short exact sequences. Alternatively one
can check directly that the above definition is a specialization of
the general notion of Janelidze and use the main result of
\cite{gogia} on relationship between internal groupoids and crossed
objects in so called semi-abelian categories. The checking is an
easy exercise because of the explicit description of the coproduct
in the category $\SG$ of square groups given in \cite{square}.
\end{proof}
Since the functors $(-)_e:\SG\to \Gr$ and $(-)_{ee}:\SG\to \Ab$
preserve limits any internal groupoid $X$ in $\SG$ gives rise to two
internal groupoids $X_e$ and $X_{ee}$ in the category of groups and
abelian groups respectively.
An internal groupoid $X\in\Gpd(\SG)$ is called ee-antidiscrete
provided $X_{ee}$ is antidiscrete. We denote by $\Gpd_{adee}(\SG)$
the category of ee-antidiscrete internal groupoids in the category
of square groups.
\begin{Le}\label{lodjan}
The equivalence $\Gpd(\SG)\cong {\sf cross}(\SG)$ restricts to an
equivalence of categories $$\Gpd_{adee}(\SG)\cong \QPM.$$
\end{Le}
\begin{proof}
We have to show that qpms are exactly crossed square groups
$\d:A\to G$ for which $\d_{ee}:A_{ee}\to G_{ee}$ is the identity
map. But this is clear, because after identification of $A_{ee}$ and
$G_{ee}$ via $\d_e$, the action $\xi$ of $G$ on $A$ becomes
redundant, $\xi(x,g)=(\d_e(x)\mid x)_H$.
\end{proof}
Let $\Gpd_{adee}(\SR)$ denote the category of ee-antidiscrete
groupoid objects in the category of square rings. Lemma
\ref{lodjan} implies the following result.
\begin{Le}\label{lodgog} There is an equivalence of categories
$$\Gpd_{adee}(\SR)\cong \CS.$$
\end{Le}
\section{Proof of the main result}
\subsection{Relative Mac~Lane cohomology}
Let $R$ be a ring and let $M$ be a bimodule over $R$. Assume also that a
surjective morphism $p:Q\to R$ is given from a square
ring $Q$ to $R$. We denote by $\xext(R,Q;M)^{\SG,\square}$ the
subcategory of the category $\xext(R,M)^{\SG,\square}$ whose objects
are crossed square ring extensions of the form
$$\xymatrix{
0\ar[r]& M\ar[r]& C_{(1)}\ar[r] ^{(\d,\Id)} &
C_{(0)}\ar[r]^p&R\ar[r]&0}$$ with
$C_{(0)}=Q$. Morphism are such morphisms of crossed square ring
extensions which are identity on $Q$
$$\xymatrix{
0\ar[r]& M\ar[r]\ar[d]_\Id& C_{(1)}\ar[d]^f\ar[r] ^{(\d,\Id)} &Q
\ar[d]^\Id\ar[r]^p&R\ar[d]^\id
\ar[r]&0\\
0\ar[r]& M\ar[r]& C_{(1)}'\ar[r] ^{(\d',\Id)} &Q
\ar[r]^p&R\ar[r]&0.}$$ Then the category
$\xext(R,Q;M)^{\SG,\square}$ is a groupoid.
Quite similarly, for a given surjective morphism $p:Q\to R$ from a
quadratic ring $Q$ to $R$, we denote by $\xext(R,Q;M)^{\SG,\tl}$ the
subcategory of the category $\xext(R,M)^{\SG,\tl}$ whose objects are
crossed square ring extensions of the form
$$
\xymatrix{
0\ar[r]& M\ar[r]& C_{(1)}\ar[r] ^{(\d,\Id)} &C_{(0)} \ar[r]^p&R\ar[r]&0
}
$$
with $C_{(0)}=Q$. Then the category $\xext(R,Q;M)^{\SG,\tl}$ is a
groupoid.
\begin{Le}\label{monreliso} Let $R$ be a ring and let $L$ be a monoid
and let $p:\Znil[L]\to R$ be a surjective morphism of quadratic
rings (and hence also a surjective morphism of square rings). Then
for any $R$-bimodule $M$ the functor $\xext(R,M)^{\SG,\tl}\to
\xext(R,M)^{\SG,\square}$ yields an equivalence of categories
$$\xext(R,\Znil[L];M)^{\SG,\tl}\xto\simeq\xext(R,\Znil[L]
;M)^{\SG,\square}.
$$
\end{Le}
\begin{proof} It is straightforward to check that
the conditions posed on $C_{1}$ and $\d$ in the definition of
quadratic pair algebra and in the definition of crossed square ring
are the same provided $C_{(0)}=\Znil[L]$.
\end{proof}
Let us turn back to an epimorphism $Q\to R$ for a square ring $R$.
The set of connected components of $\xext(R,Q;M)^{\SG,\square}$ has
the following cohomological description. In order to give the
precise statement we first extend the definition of the Mac~Lane
cohomology to square rings and then we introduce the relative
cohomology groups.
Let $Q$ be a square ring, then $Q\ad$ is a ring, which we denote by
$R$. There is an obvious functor $$q:\modr Q\to \modr R$$ which is
identity on objects and on morphisms it is given by
$$q((f_i^k,f_{ij}^k)):=(\bar{f}_i^k)$$ For any bimodule $M$ over the
ring $R$ we let $D_M$ be the bifunctor on $\modr R$ given by
$$(X,Y)\mapsto \Hom_R(X,Y\t _R M)$$
By abuse of notation we will denote by $D_M$ also the induced bifunctor on
$\modr Q$. Then we put
$$
\HML^*(Q,M):=H^*(\modr Q,D_M)
$$
Thanks to Section \ref{relcoh} we recover for usual rings the classical Mac
Lane cohomology. Using the relative cohomology of small categories defined in
Section \ref{relcoh} one can also define the relative Mac Lane cohomology groups
$\HML^*(R,Q;M)$ to be $H^*(\modr R,\modr Q;D_M).$ Thus one has the following
long exact sequence
\begin{multline*}
0\to \HML^0(R;M)\to \HML^0(Q;M)\to \HML^1(R,Q;M)\to\cdots\\
\to \HML^n(R;M)\to \HML^n(Q;M)\to \HML^{n+1}(R,Q;M)\to\cdots.
\end{multline*}
The proof of the isomorphisms
in Theorem \ref{xext} is based on a computation given in
Appendix and on the following result.
\begin{The}\label{fardobitisq}
Let $p:Q\to R$ be a surjective morphism from a square ring $Q$ to a ring
$R$.
Then
$$\pi_0(\xext(R,Q;M)^{\SG,\square})\approx \HML^3(R,Q;M).
$$
\end{The}
\begin{proof}
Let ${\Tracks}(\modr R, \modr Q;D_M)$ denote the category of such
abelian track categories $\ta$ that the corresponding homotopy
category $\ta_\ho$ is $\modr R$, underlying category $\ta_0$ is
$\modr Q$ and the corresponding natural system is given by the
bifunctor $D_M$. We now construct the functor
$$
\chi:\xext(R,Q;M)^{\SG,\square}\to {\Tracks}(\modr R, \modr Q;D_M)
$$
as follows. Let
$$
\xymatrix{
0\ar[r]& M\ar[r]& \tilde{Q}\ar[r] ^{(w,\Id)} &Q \ar[r]^p&R\ar[r]&0,
}
$$
be a crossed square ring extension. The underlying category of the
track category $\chi(\omega)$ is $\modr Q$. If ${\bf
f}=(f^k_i,f^k_{ij})$ and ${\bf g}=(g^k_i,g^k_{ij})$ are morphisms
$y\to x$, $x,y\in \N$ in $\modr Q$, then a track $\bf f\then g$ is a
collection $(h^k_i)$ of elements in $\tilde{Q}_e$ such that
$\d(h^k_i)=f^k_i-g^k_i$ for all $1\le i\le x$ and $1\le k\le y$. Now
the result follows from the fact that $\chi$ is an isomorphism of
categories. The inverse of $\chi$ is given as follows. Let $\ta$ be
a track category such that $\ta_\ho=\modr R$ and $\ta_0=\modr Q$. By
Lemma \ref{lec} $\ta$ is an internal groupoid in the category of
quadratic theories. By applying the functor $\sf cro$ one obtains an
internal groupoid in the category of square rings. Moreover, the
proof of Lemma \ref{lec} shows that this groupoid is
ee-antidiscrete, therefore by Lemma \ref{lodgog} it defines an
object in $\xext(R,Q;M)^{\SG,\square}$, which is the value of the
inverse of $\chi$.
\end{proof}
\subsection{A pullback construction}\label{pbc}
We now give a construction in the category of crossed square ring
extensions which is needed in the proof of Theorem \ref{xext}.
Let
$$\xymatrix{
0\ar[r]& M\ar[r]& C_{(1)}\ar[r] ^{(\d,\Id)} &C_{(0)}\ar[r]^p&R\ar[r]&0,}
$$
be a crossed square ring extension and let $f:Q_{(0)}\to C_{(0)}$ be a
morphism of square rings, such that $p\circ f_e:Q_{0}\to R$ is
surjective. Based on this data we construct the following crossed
square ring
$$
\xymatrix{& Q_{ee}\ar[dl]_{P^Q}\\
Q_{1}\ar[r]_{\d^Q} &Q_{0}\ar[u]_{H^{Q_0}}}
$$
where the group $Q_{1}$ is defined by the pullback diagram
$$\xymatrix{Q_{1}\ar[r]^{\d^Q}\ar[d]_{g_e}&Q_{0}\ar[d]^{f_e}\\
C_{1}\ar[r]_\d& C_{0}}$$ and $P^Q=(P^{C}\circ f_{ee},
P^{Q_0}):Q_{ee}\to Q_{1}$. Then one has the following crossed square
ring extension
$$\xymatrix{
0\ar[r]& M\ar[r]& Q_{(1)}\ar[r] ^{\d^Q} &Q_{(0)} \ar[r]&R\ar[r]&0.}$$
One easily sees that
$$\xymatrix{
0\ar[r]& M\ar[r]\ar[d]_\Id& Q_{(1)}\ar[d]^g\ar[r] ^{\d^Q} &Q_{(0)}
\ar[d]^f\ar[r]^{pf}&R\ar[d]^\id
\ar[r]&0\\
0\ar[r]& M\ar[r]& C_{(1)}\ar[r] ^{\d} &C_{(0)} \ar[r]^p&R\ar[r]&0}$$ is a morphism
of crossed square ring extensions.
We call this construction the pullback construction and write
$f^*\d$ instead of $(\d^Q)$. Assume now that $(\d)$ is linearly
generated and the composite ${\sf L}(Q_{(0)})\to {\sf L}(C_{(0)})\to R$ is
surjective, then one easily sees that $(f^*\d)$ is also linearly
generated.
Of course a similar constructions works for quadratic pair algebras.
\subsection{Proof of Theorem \ref{xext}}\label{proof261} Let
$$
\xymatrix{
0\ar[r]& M\ar[r]& \tilde{Q}\ar[r] ^{(w,\Id)} &Q \ar[r]^q&R\ar[r]&0,}
$$
be an object of $\xext_L(R,M)^{\SG,\square}$. For simplicity we
denote this object by $(w)$. Then it can be also considered as an
object of $\xext(R,Q;M)^{\SG,\square}$ and therefore $(w)$ defines
an element in $\HML^3(R,Q;M)$ thanks to Theorem \ref{fardobitisq}.
Then the boundary homomorphism gives an element in $\HML^3(R,M)$. In
this way we get a map
$$
\zeta: \pi_0(\xext_L(R,M)^{\SG,\square})\to \HML^3(R;M).
$$
Composing it with $\pi_0(\xext_L(R,M)^{\SG,\tl}) \to
\pi_0(\xext_L(R,M)^{\SG,\square})$ we obtain the map
$$
\zeta': \pi_0(\xext_L(R,M)^{\SG,\tl}) \to \HML^3(R;M).
$$
We have to show that these maps are bijections. Take an $a\in
\HML^3(R;M)$. Take any surjective homomorphism $L\to R$ from a free
monoid $L$ to the multiplicative monoid of the ring $R$. It yields a
surjective morphism $r:\Znil[L]\to R$. Here $\Znil[L]$ can be
considered as a square ring as well as a quadratic ring. Since
$\HML^i(\Znil[L];D_M)=0$ for $i=2,3$ (see Theorem \ref{nilnul} in
Appendix), we have an isomorphism
$$
\partial:\HML^3(R,\Znil[L];M)\cong\HML^3(R;M).
$$
Let $b=\partial\1(a)\in\HML^3(R,\Znil[L];M)$ be the element
corresponding to $a$. Thanks to Theorem \ref{fardobitisq} the
element $b$ defines a crossed square ring extension of $R$ by $M$
$$
\xymatrix{
0\ar[r]& M\ar[r]& \tilde{Q}\ar[r]^-{(v,\Id)}&\Znil[L]
\ar[r]&R\ar[r]&0}
$$
which is also linearly generated by construction and therefore is an
object of $\xext_L(R,M)^{\SG,\square}$. By Lemma \ref{monreliso} it
can be considered also as a quadratic pair algebra extension. Hence
$\zeta$ and $\zeta'$ are surjections. It remains to show that
$\zeta$ and $\zeta'$ are injections as well. Suppose
$\zeta(w)=\zeta(w')$ (resp. $\zeta'(w)=\zeta'(w')$). We have to show
that $(w)$ and $(w')$ are in the same connected component. Let ${\sf
L}(Q)$ be the monoid of linear elements in $Q$. Via $q$ it maps to
the multiplicative submonoid $q({\sf L}(Q))$ of $R$. Take any
surjective homomorphism of monoids $F\to q({\sf L}(Q))$ with $F$ a
free monoid. It has a lifting to a monoid homomorphism $F\to {\sf
L}(Q)$, which yields a square (resp. quadratic) ring homomorphism
$t:\Znil[F]\to Q$. The homomorphism $t$ satisfies all conditions on
$f$ in Section \ref{pbc} and hence yields a morphism of crossed
square ring extensions (resp. quadratic pair algebra extensions)
$t^*(w)\to w$. Thus without loss of generality we can assume that
$(w)$ and $(w')$ are chosen in such a way that $Q=\Znil[F]$ and
$Q'=\Znil[F']$. Let $L$ and $r$ be the same as above (see the proof
of surjectivity of $\zeta$). Since $L\to R$ is surjective,
$q(F)\subset R$ and $F$ is free, there exists a morphism of monoids
$F\to L$ such that for the induced morphism $k:Q=\Znil[F]\to
\Znil[L]$ one has $q=r\circ k$. Thus one has the following
commutative diagram
$$\xymatrix{\HML^3(R,\Znil[L];M)\ar[r]\ar[d]^{k^*}& \HML^3(R,M)\\
\HML^3(R,Q;M)\ar[ur]}$$
Since both morphisms in the diagram with target $HML^3(R,M)$ are
isomorphisms, it follows that $k^*:\HML^3(R,\Znil[L];M)\to
\HML^3(R,Q;M)$ is also an isomorphism. Considering an extension
corresponding to ${k^*}^{-1}(w)$ one sees that there exists a
morphism of square ring extensions (resp. quadratic pair algebra
extensions)
$$\xymatrix{0\ar[r]& M\ar[r]\ar[d]^{\Id}& \tilde{Q}\ar[d]\ar[r]^-{(w,\Id)}
&Q\ar[d]^k
\ar[r]^q&R\ar[r]\ar[d]^\Id&0\\
0\ar[r]& M\ar[r]& \bar{Q}\ar[r]^-{(\bar{w},\Id)} &\Znil[L]
\ar[r]&R\ar[r]&0}$$ In a similar manner we find a morphism of square
ring extensions (resp. quadratic pair algebra extensions)
$$\xymatrix{0\ar[r]& M\ar[r]\ar[d]^{\Id}& \tilde{Q'}\ar[d]\ar[r]^-{(w',\Id)}
&Q\ar[d]^k
\ar[r]^{q'}&R\ar[r]\ar[d]^\Id&0\\
0\ar[r]& M\ar[r]& \bar{Q'}\ar[r]^-{(\bar{w'},\Id)} &\Znil[L]
\ar[r]&R\ar[r]&0}$$ Since the square ring extensions (resp.
quadratic pair algebra extensions) $(\bar{w})$ and $(\bar{w'})$ lie
in the same groupoid $\xext(R,\Znil[L] ;M)^{\SG,\square}$ and their
classes in $ \HML^3(R,\Znil[L];M)$ are the same, it follows that
they are isomorphic in the groupoid $\xext(R,\Znil[L]
;M)^{\SG,\square}$. Therefore we have the following diagram in $
\xext(R,M)^{\SG,\square}$ (resp. $\xext(R,M)^{\SG,\tl} $):
$$
(w')\ot (\bar{w}')\ \cong (\bar{w})\to (w),
$$
hence the result.
\section{Application to theory of
2-categories}\label{obs2cat}
In this section we introduce the notion of 2-additive track
category, which is the 2-categorical analogue of
additive category and we prove a strengthening theorem for such
2-additive track categories.
\subsection{Abelian track categories}
A track category is \emph{abelian} if for any map $f:X\to Y$, the
group $\Aut(f)$ of tracks from $f$ to itself is abelian. Any track
category which fits in a linear track extension is abelian. Converse
is also true: any abelian track category defines a natural system
$D=D_{\ta}$ on $\ta_\ho$ and a linear track extension
$$
0\to D_\ta\to \ta_1 \rightrightarrows \ta_0\to \ta_\ho \to 0.
$$
The natural system $D_\ta$ and the linear track extension are unique
up to isomorphism (see \cite{BJ}).
\subsection{Track theories}
A \emph{coproduct} $A\vee B$ in a track category $\ta$ is an object
$A\vee B$ equipped with 1-morphisms $i_1:A\to A\vee B$, $i_2:B\to
A\vee B$ such that the induced functor
$$
(i_1^*,i_2^*):\hog{A\vee B,X}\to \hog{A,X}\x \hog{B,X}
$$
is an equivalence of groupoids for all objects $X\in\ta$. The
coproduct is \emph{strong} if the functor $(i_1^*,i_2^*)$ is an
isomorphism of groupoids. By duality we have also notion of
\emph{product} and \emph{strong product}. A \emph{zero object} in
a track category $\ta$ is an object $0$ such that the categories
$\hog{0,X}$ and $\hog{X,0}$ are equivalent to the trivial groupoid
for all $X\in\ta$.
A \emph{strong zero object} in a track category $\ta$ is an object $0$
such that all categories $\hog{0,X}$ and $\hog{X,0}$ are trivial
groupoids.
A \emph{track theory} (resp. \emph{strong track theory}) is a small track
category $\ta$ possessing finite coproducts (resp. strong coproducts).
Morphisms of track theories are enriched functors which are compatible with
coproducts. An equivalence of track theories is a track theory morphism which is
a weak equivalence and two track theories are called \emph{equivalent} if they
are made so by the smallest equivalence relation generated by these.
The following is a particular case of a general result of Power
\cite{Pow}. For a cohomological proof we refer to \cite{streng}.
\begin{The}\label{power}
Any abelian track theory is equivalent to a strong one.
\end{The}
If $\ta$ is an abelian track theory, then the corresponding category
$\ta_\ho$ is an algebraic theory and the natural system $D_\ta$ is
cartesian. Conversely, if $\ta$ is an abelian track category such
that $\ta_\ho$ is an algebraic theory and $D_\ta$ is cartesian, then
$\ta$ is an abelian track theory.
Moreover an abelian track theory
is strong if and only if $\ta_0$ is an algebraic theory and the
canonical functor $\ta_0\to\ta_\ho$ is a morphism of algebraic
theories.
\subsection{2-Additive track categories}
Now we introduce an analogue of additive categories in the 2-world.
Let $\ta$ be a track theory with zero object. Then for any objects
$A$ and $B$ of $\ta$, there is a map $p_1: A\vee B\to A$ and tracks
$p_1i_1\then \id_A$, $p_1i_2\then 0$. Similarly for $p_2:A\vee B\to B$.
A \emph{2-additive track category} is an abelian track
theory with strong zero object, such that the following conditions hold
i) for any two objects $A$
and $B$ the coproduct $A\vee B$ is also a product via $p_1:A\vee B\to A$
and $p_2:A\vee B\to B$
ii) for any morphism $f:A\to B$ there exists a morphism $g:A\to B$ and
a track $hd\then 0$, where $d:A\to A\vee A$ and $h:A\vee A\to B$ are
morphisms with tracks $hi_1\then f$, $hi_2\then g$, $p_1d\then \id_A$,
$p_2d\then \id_A$.
It is clear that the homotopy category
$\ta_\ho$ of a 2-additive track theory is an additive
category. The following is a direct consequence
of \cite{BJ} and \cite{beitrage}.
\begin{Le}
Let $\ta$ be an abelian track category. Then $\ta$ is a 2-additive
track category iff $\ta_\ho$ is an additive category and the
corresponding natural system $D_\ta$ is a biadditive bifunctor.
\end{Le}
It follows that a 2-additive track category determines a triple
$(\ta_\ho,D_\ta,\Ch(\ta)\in H^3(\ta_\ho;D_\ta))$. Conversely for an
additive category $\C$, a biadditive bifunctor $D:\C\op\x\C\to\Ab$
and an element $a\in H^3(\C;D)$ there exists a 2-additive track
category unique up to equivalence such that $\ta_\ho=\C$, $D_\ta=D$
and $\Ch(\ta)=a$.
\subsection{Strongly and very strongly 2-additive track theories}
As we said Theorem \ref{power} asserts that any track theory is
equivalent to one with strong coproducts. In particular, any
2-additive track category is equivalent to one which possesses
strong products. Since the dual of an additive track category is
still a track theory, we see that it is also equivalent to one which
possesses strong coproducts. Can we always get strong products and
coproducts simultaneously? In other words, is every 2-additive track
category $\ta$ equivalent to a very strongly 2-additive track
theory? Here a 2-additive track category is called \emph{very strongly
2-additive} if it admits a strong zero object $0$, strong finite
coproducts and for any two object $A$ and $B$ the strong coproduct
$A\vee B$ is also a strong coproduct via $p_1:A\vee B\to A$ and
$p_2:A\vee B\to B$. The answer is given by the following result, which also
shows that the number 2 plays an important r\^ole
in the theory of 2-categories.
\begin{The} \label{verystongstrengthening}
Let $\ta$ be a small 2-additive track category with homotopy
category $\C=\ta_\ho$ and canonical bifunctor $D=D_{\ta}$. Let $_2D$
be the two-torsion part of $D$. Then there is a well-defined element
$\nu(\ta)\in H^0(\C;{}_2D)$, which is nontrivial in general and such
that $\nu(\ta)=0$ iff $\ta$ is equivalent to a very strongly
2-additive track theory. The class $\nu(\ta)$ for example is zero
provided homs of the additive category $\C$ are modules either over
$\Z[\frac12]$ or over $\F_2$ (the field with two elements).
\end{The}
\begin{proof} First one observes that a 2-additive track category
$\ta$ is very strongly 2-additive iff the category $\ta_0$ is
additive. For simplicity we restrict ourself to the case of single
sorted theries. Then $\ta_\ho=\modr R$ for a ring $R$. In this case
one has an isomorphism $D(X,Y)\cong\Hom(X,M\t_R Y)$ natural in $X$
and $Y$, where $M=D(R,R)$ is a bimodule over $R$. We claim that up
to equivalence single sorted very strongly 2-additive track
categories $\ta$ with fixed $\ta_\ho$ and $D_\ta$ are in bijection
with $\Sh^3(R;M)$. Indeed, if
$$
0\to M\to C_1\xto\d C_0\to R\to 0
$$
is an object of ${\sf Cross}(R,M)$, then we have the following very
strongly 2-additive track category $\ta$. Objects of $\ta$ are the
same as the objects of $\modr R$, i.~e. natural numbers. For any
natural numbers $n$ and $m$ the maps from $n$ to $m$ (which is the
same as objects of the groupoid $\ta(n,m)$) are $m\times
n$-matrices with coefficients in $C_0$. For $f,g\in Mat_{m\times
n}(S)$ the set of tracks $f\to g$ is given by
$$
\Hom_{\ta(n,m)}(f,g)=\left\{h \in \mathrm{Mat}_{m\times n}(C_1)\
\mid\ \d (h)=f-g\right\}.
$$
Composition of 1-arrows is given by the usual multiplication of
matrices, while composition of tracks is given by the addition of
matrices. One easily checks that in this way one really obtains a
very strongly 2-additive track theory $\ta(\d)$. It is clear that
$\ta_\ho=\modr R$, where $R={\sf Coker}(\d)$ and the bifunctor
corresponding to $\ta$ is $D=\Hom(-,M\t_R-)$. Conversely, assume
$\ta$ is a single sorted very strongly 2-additive track category
with $\ta_\ho=\modr R$ and $D_\ta=\Hom(-,M\t_R-)$. Then $\ta_0$ is a
single sorted additive category and therefore it is equivalent to
$\modr S$, where $S={\sf End}_{\ta_0}(1)$. Restriction of the
quotient functor $\ta\to \ta_0$ yields a homomorphism of rings $S\to
R$. One defines $X$ to be the set of pairs $(h,x)$, where $x\in
\Hom_{\ta_0}(1.1)$ and $h:x\then 0$ is a track in the groupoid
$\ta(1,1)$. Moreover we put $\partial=\partial_{\ta}(h,x)=x$. Then
$X$ carries a structure of a bimodule over $S$, and
$$
0\to M\to X\buildrel\partial\over\to S\to R\to 0
$$
is a crossed extension and the claim follows from isomorphism
(\ref{shuxt}). As we said, up to equivalence single sorted 2-additive track
categories $\ta$ with fixed $\ta_\ho$ and $D_\ta$ are in bijection
with $\HML^3(R,M)$. Therefore the exact sequence $0\to \Sh^3(R,M)\to
\HML^3(R,M)\xto{\nu} H^0(R, \,_2M)$ together with Proposition 9.1.1 of
\cite{shukla} implies the result.
\end{proof}
{\bf Remark.} One can describe the function $\nu$ in Theorem
\ref{verystongstrengthening} as follows. Let $\ta$ be a 2-additive
track theory. Let $\vee$ denote the weak coproduct in $\ta$ and let
$0$ be the weak zero object. For objects $X,Y$ one has therefore
``inclusions'' $i_1:X \to X\vee Y$ and $i_2:Y \to X\vee Y$. Since
$X\vee Y$ is also a weak product of $X$ and $Y$ in $\ta$ it follows
that one has also projection maps $p_1:X\vee Y\to X$ and $p_2:X\vee
Y\to Y$.
For each $X$ we choose maps $i_X:X\to X\vee X$ and $t_X:X\vee Y\to
Y\vee X$ in such a way that classes of $i_X$ and $t_X$ in $\ta_\ho$
are the codiagonal and twisting maps in the additive category
$\ta_\ho$. It follows that there is a unique track
$$
\aa_X:i_X\then t\circ i_X
$$
such that $p_{i*}(\aa_X)=0$ for $i=1,2$. Now, let $(1,1):X\vee X\to
X$ be a map which lifts the codiagonal map in $\ta_\ho$. Then
$(1,1)_*\aa_X$ is a track $\id_X\to\id_X$ and therefore it differs
from the trivial track by an element $\nu(X)\in D(X,X)$. One can
prove that the assignment $X\mapsto\nu(X)$ is the expected one.
\subsection{Strongly additive track categories}
A 2-additive track category $\ta$ is called \emph{strongly
2-additive} if $\ta_0$ is quadratic.
\begin{The}\label{stronglyadditive}
Any 2-additive track theory is equivalent to a strongly 2-additive one.
\end{The}
\begin{proof}
We continue to restrict ourselves to the single sorted case. In this
case $\ta_\ho$ is the category $\modr R$ for a ring $R$ and
$D=\Hom_R(-,(-)\t_R M)$ for an $R$-bimodule $M$. Thus ${\sf
Ch}((\ta))\in \HML^3(R;M)$ and therefore it belongs to
$\HML^3(R,Q;M)$ for a square ring $Q$ thanks to the proof of Theorem
\ref{xext} given in Section \ref{proof261}. Thus the element ${\sf
Ch}((\ta))$ has a realization via track category $\ta'$ such that
$(\ta')_0=\modr Q$ and we are done.
\end{proof}
\appendix
\section{Cohomology of free monoid square rings}
\centerline{\sc T. Pirashvili}
\
\
Here we prove the following result.
\begin{The} \label{nilnul}
Let $L$ be a free monoid and let $Q=\Znil[L]$ be the corresponding
monoid square ring. Then for any $R$-$R$-bimodule $B$ one has
$$
\HML^2( Q,B)=0= \HML^3( Q,B).
$$
\end{The}
Proof of Theorem \ref{nilnul} is given in Section \ref{proofnil}.
The argument is a modification of the one given in \cite{Nil}.
\subsection{Auxiliary results}
For a ring $R$ we denote by ${\bf F}(R)$ or simply by $\bf F$ the
category of all covariant functors from the category $\modf R$ of
finitely generated free right $R$-modules to the category $\Mod R$
of all right $R$-modules. It is well known \cite{JP} that
$$
\HML^*(R,B)\cong \Ext^*_{\bf F}(\Id, (-)\t_RB)
$$
We need the following result, which is an easy consequence of
Theorem 9.2.1 \cite{shukla} and the fact that ${\sf SH}^i(R,-)=0$
for all $i\ge 2$, provided $R$ is a free ring.
\begin{Le}\label{freering} Let $R$ be a free ring and let $B$ be an
$R$-$R$-bimodule. Then one has $\HML^2(R, B)=0$
and $\HML^3(R, B)\cong {\sf H}^0 (R,\, _2
B)$, where ${\sf H}^*(R,-)$ denotes the Hochschild cohomology of $R$.
\end{Le}
We also need the following vanishing result.
\begin{Le}\cite{add}\label{vanishing}
Let $R$ be a ring and let $$T:(\modf R)\x (\modf R)\to \Mod R$$ be a
bifunctor, which is covariant in both variables and
$T(0,X)=0=T(X,0)$ for all $X\in \modf R$. Then for any additive
functor $F:\modf R\to \Mod R$ one has
$$
\Ext^*_{\bf F}(F, T^d)=0=\Ext^*_{\bf F}(T^d,F),
$$
where $T^d(X)=T(X,X)$.
\end{Le}
In the following we need the simplicial derived functors of the
functor $(-)\ad:Q$-$Mod\to R$-$Mod$, which are denoted by
$$
\Tor^Q_*(-,R):\Mod Q\to \Mod R.
$$
We recall the definition of these functors. According to
\cite{quillen} the category of simplicial objects in the category of
right $Q$-modules has a closed model category structure where a
morphism $f:X_*\to Y_*$ of simplicial objects is a weak equivalence
(resp. fibration) when it is so in the category of simplicial sets.
Let $M$ be a right $Q$-module and let $X_*$ be a cofibrant
replacement of $M$. By \cite{quillen} one can assume that each
$X_n$, $n\ge 0$ is a free right $Q$-module. We also have
$\pi_iX_*=0$ for $i>0$ and $\pi_0X_*=M$. Now one puts
$$\Tor^Q_*(M,R):=\pi_*(X_*\ad).$$
It is well known that these are well-defined functors. Since $\Mod
R\subset \Mod Q$, one can consider also the restriction of
$\Tor^Q_*(-,R)$ to $\Mod(R)$ (see Proposition \ref{ssnil} below).
\begin{Pro}\label{ssnil}
For any square ring $Q$ and for any $R$-$R$-bimodule $B$, one has the following
spectral sequence
$$E^2_{pq}=\Ext^p_{\bf F}(\Tor_q^Q(-,R),F)\Longrightarrow \HML^{p+q}( Q,B), $$
where $R=Q\ad$, $F(-)=(-)\tp _R B$.
\end{Pro}
{\it Proof}. Proposition follows immediately from the spectral sequence
(8.2.2) and Lemma 8.3.1 of \cite{Nil}.
\subsection{Computation of $\Tor^Q$} In this section we give a computation of
$\Tor$-groups involved in Proposition \ref{ssnil}. It is based on
Lemma \ref{qucfo} below, which is the specialization of the exact
sequence (4.1) of \cite{qucf}. Let us recall that Eilenberg and
Mac~Lane \cite{EM} defined the quadratic functor
$$\Omega:\Ab\to \Ab$$ such that it commutes with filtered colimits,
$$\Omega(A\oplus B)=\Omega(A)\oplus \Omega(B)\oplus \Tor(A,B)$$
and moreover $$\Omega(\Z)=0, \ \ \Omega(\Z/n\Z)=\Z/n\Z.$$
\begin{Le}\cite{qucf} \label{qucfo} Let $X_*$ be a simplicial abelian group, which is degreewise free and
has homotopy groups $\pi_i=\pi_i(X_*)$. Then one has
$$\pi_0(\La ^2X_*)=\La ^2 (\pi_0)$$
$$0\to \pi_1\t \pi_0\to \pi_1(\La^2(X_*))\to \Omega(\La ^2\pi_0)\to 0$$
$$0\to \pi_2\t \pi_0 \oplus \Gamma \pi_1
\to \pi_2(\La ^2 (X_*))\to \Tor( \pi_1,\pi_0)\to 0$$
\end{Le}
Let us recall that if $G$ is a free class two nilpotent group, then
one has the following short exact sequence
$$0\to \La^2(G\ab)\to G\to G\ab\to 0,$$
where the first nontrivial map is induced by $(x,y)\mapsto
-x-y+x+y$. Assume now that $L$ is a monoid and $Q=\Znil[L]$ is the
corresponding monoid square ring. As we already mentioned a right
$Q$-module is the same as a nilpotent group of class two together
with an action of $L$ via group homomorphisms. It follows that if
$X$ is a free right $Q$-module, then $X$ is also free as a
nilpotent group of class two. Furthermore, $X\ad$ in this case is
simply $X\ab$, thus we have the following Lemma.
\begin{Le}\label{sesla} Let $L$ be a monoid and let $Q=\Znil[L]$ be
the monoid square ring. Then, for any free right $Q$-module $X$, one has the
following short exact sequence
$$0\to \La^2(X\ad)\to X\to X\ad\to 0,$$
in the category of modules over the ring $R=Q\ad=\Z[L]$, where the
first nontrivial map is induced by $(x,y)\mapsto -x-y+x+y$, and
$\La^2(X\ad)$ is an $R$-module via the diagonal action of $L$.
\end{Le}
We would like to use these results in the following situation.
\begin{Pro}\label{qtor} Let $L$ be a monoid and let $Q=\Znil[L]$ be
the monoid square ring. Then, for any free right module $M$ over the
ring $R=Q\ad$, one has the following natural isomorphisms
$$\Tor_0^Q(M,R)\cong M$$
$$\Tor_1^Q(M,R)\cong \La^2(M)$$
$$\Tor_2^Q(M,R)\cong M\tp \La^2(M)$$
$$\Tor_3^Q(M,R)\cong (M\tp M\tp \La^2(M))\oplus (\Gamma (\La ^2(M)))$$
\end{Pro}
{\it Proof}. Let $M$ be a free $R$-module. Let us take a free simplicial
resolution $Y_*$ of $M$ in the category of $Q$-modules. Thanks to Lemma
\ref{sesla} one has an exact sequence
$$0\to \La ^2X_*\to Y_*\to X_*\to 0,$$
where $X_*=Y_*\ad$. Since $\pi_iY_*=0$ for $i>0$ and
$\pi_0Y_*=M$ we have $\pi_0X_*=M$ and
$\pi_{i+1}X_*= \pi_i\La ^2(X_*)$. Since $M$ is a free abelian
group, one can use Lemma \ref{qucfo} to get
$$\pi_1(X_*)\cong \La ^2(M), \ \ \ \pi_2 (X_*)= M\t \La ^2 (M)$$
$$\pi_3(X_*)\cong \La ^2(M)\t M^{\t 2}\oplus \Gamma (\La ^2M)$$
Comparing with definition of simplicial derived functors we obtain the
expected result.
\subsection{Universal quadratic functors}
Let $A$ be an abelian group. We set
$$P(A)=I(A)/I^3(A),$$
where $I(A)$ is the augmentation ideal of the group algebra of $A$.
Let $p:A\to P(A)$ be the map given by $p(a)= (a-1)({\sf mod} \
I^3(A))$. Then $p$ is a \emph{quadratic map}, meaning that the
\emph{cross-effect} $$(a\mid b)_p:=p(a+b)-p(a)-p(b)$$ is linear in
$a$ and $b$. Actually $p$ is a universal quadratic function $p:A\to
P(A)$ (see \cite{approximation}). A quadratic map $f:A\to B$ of
abelian groups is called \emph{homogeneous} if $f(-a)=f(a)$. It is
well known that for any abelian group $A$ there exists a universal
homogeneous quadratic function $\gamma:A\to \Gamma(A)$. If $A$ is a
module over a monoid ring $R=\Z[L]$, then $P(A)$, $\Gamma(A)$, $A\tp
A$ are also $R$-modules, where the action of $x\in L$ is given by
$$p(a)x=p(ax), \ \ (\gamma(a))x=\gamma(ax), \ \ (a\tp b)x=ax\tp bx.$$
\begin{Le}\label{galatoad} If $F\in \bf F$ is an additive functor, then
$$\Hom_{\bf F}(\Gamma\circ \La^2,F)=0=\Hom_{\bf F}(\La^2,F)$$
\end{Le}
{\it Proof}. Let us recall that if $T\in \bf F$ is a functor with $T(0)=0$,
then the second cross-effect of $T$ fits in the decomposition
$$
T(A\oplus B)\cong T(A)\oplus T(B) \oplus T(A\mid B).
$$
Putting $B=A$ and using the codiagonal morphism $(\Id,\Id):A\oplus
A\to A$ one obtains a natural transformation $\eta_A:T(A\mid A)\to
T(A)$. It is clear that any natural transformation from $T$ to an
additive functor factors through ${\sf Coker}(\eta)$. We first take
$T= \Gamma(\La ^2)$. Since the second cross-effect of $\Gamma\circ
\La^2$ contains as a direct summand the term $\Gamma(A\t B)$ and for
$A=B$ it maps via $\eta$ surjectively to $\Gamma(\La^2 A)$, we
conclude that there is no nontrivial map from $\Gamma\circ \La^2$ to
any additive functor. Similarly for $\Hom_{\bf F}(\La^2,F)$.
\begin{Le}\label{pesappr} Let $L$ be a free monoid and let $R=\Z[L]$ be the
corresponding monoid ring. Then
$$\Ext^p_{\bf F}(P,F)=0$$
provided $F$ is additive and $2\le p\le 4$.
\end{Le}
{\it Proof}. Since $R=\Z[L]$ is torsion free as an abelian group and
$F$ is an additive functor the main result of \cite{approximation}
shows that one has an isomorphism
$$\Ext^p_{\bf F}(P,F)\cong \Ext^p_{\bf Q}(P,F), \ \ $$
provided $p\le 4$. Here $\bf Q$ is the abelian category of quadratic
functors from $\modf R$ to $\Mod R$. For the functor $P\tp R$, which
is given by $X\mapsto P(X)\tp R$, one has an isomorphism (see
\cite{approximation})
$$\Hom_{\bf Q}(P\tp R,T)\cong T(R), \ \ T\in \bf Q$$ It follows that $P\tp R$
is a projective object in $\bf Q$. Thus one can use the bar-resolution
$$0\leftarrow P\leftarrow P\tp R\leftarrow P\tp R\tp R\leftarrow \cdots$$
to get a projective resolution of $P$ in the category $\bf Q$. In particular
one has an isomorphism
$$\Ext^*_{\bf Q}(P,F)\cong H^*(R,F(R))$$
and the result follows from the fact that the Hochschild cohomology
vanishes for free rings in dimensions $\ge 2$.
\subsection{Proof of Theorem \ref{nilnul}} \label{proofnil} We put
$F=(-)\tp_RB\in \bf
F$. Thanks to Proposition \ref{ssnil} one has the following spectral sequence
$$E^2_{pq}=\Ext^p_{\bf F}(\Tor_q^Q(-,R),F)\Longrightarrow \HML^{p+q}(
Q,B).
$$
By Proposition \ref{qtor} restriction of the functor $\Tor_*^Q(-,R)$
to the category $\modf R$ is given by
$$\Tor_0^Q(-,R)=\Id, $$
$$ \Tor_1^Q(-,R)= \La^2,$$
$$ \Tor_2^Q(-,R)= \Id\t \La^2,$$
$$\Tor_3^Q(-,R)= (\La ^2\t \Id^{\t 2})\oplus (\Gamma\circ \La^2),$$
Since $F$ is additive, Lemma \ref{vanishing} shows that
$$E_2^{p0}= \Ext^p_{{\bf F}}(\Id,F),$$
$$E_2^{p1}= \Ext^p_{{\bf F}}(\La^2,F),$$
$$E_2^{p2}= 0$$
$$E_2^{p3}= \Ext^p_{{\bf F}}(\Gamma\circ \La^2,F).$$
We also have $$E^{03}_2=0=E^{01}_2$$ thanks to Lemma \ref{galatoad}. Moreover
$E_2^{20}=0$ by Lemma \ref{freering}. Thus it suffices to
show that the following differentials of the spectral sequence
$$d_2:E^{11}_2=\Ext^1_{{\bf F}}(\La^2, F)\to E^{30}_2= \Ext^{3}_{{\bf F}}(\Id,F)$$
and
$$d_2:E^{12}_2=\Ext^2_{{\bf F}}(\La^2,F)\to E^{40}_2=\Ext^{4}_{{\bf F}}(\Id,F)$$
are isomorphisms. Let us observe that in general the differential
$$d_2:\Ext^p_{{\bf F}}(\La ^2, F)\to \Ext^{p+2}_{{\bf F}}(\Id,F)$$
is given by the cup product with $e\in \Ext^2_{{\bf F}}(\Id,\La^2)$
corresponding to the extension
$$0\leftarrow \Id\leftarrow P^2\leftarrow \Id^{\t 2}\leftarrow \La ^2 \leftarrow 0$$
We have $e=e_1\cup e_2$, where $e_1$ corresponds to the extension
$$0\to \La^2 \to \Id^2\to {\sf Sym}^2\to 0$$
while $e_2$ corresponds to the extension
$$
0\to {\sf Sym}^2\to P\to \Id\to 0,
$$
where ${\sf Sym}^2$ is the second symmetric power and the first
nontrivial map is induced by the assignment $a\tp b\mapsto (a\mid
b)_p$, while the second map is given by $p(a)\mapsto a.$
It follows from Lemma \ref{vanishing} that the cup product with $e_1$ yields an
isomorphism
$$\Ext^p_{{\bf F}}(\La ^2, F)\to \Ext^{p+1}_{{\bf F}}({\sf Sym ^2},F), \ p\ge -1.$$
Similarly Lemma \ref{pesappr} shows that the map
$$\Ext^p_{{\bf F}}({\sf Sym} ^2, F)\to \Ext^{p+1}_{{\bf F}}(\Id,F)$$
induced by the cup product with $e_2$ is an
isomorphism if $2\le p\le 3$ and we are done.
| 199,023
|
\begin{document}
\title[q-Calculus Revisited]{\emph{q}-Calculus Revisited}
\author{Si Hyung Joo}
\address{Department of Industrial Engineering, Chonnam National University, 77, Yongbong-ro, Buk-gu, Gwangju, 61186, Republic of Korea}
\ead{innovation@jnu.ac.kr}
\vspace{10pt}
\begin{indented}
\item[]June 2021
\end{indented}
\begin{abstract}
In this study, a new representation is obtained for \emph{q}-calculus, as proposed by Borges [Phyica A 340 (2004) 95], and a new dual \emph{q}-integral is suggested.
\vspace{2pc}
\noindent{\it Keywords}: \emph{q}-calculus, \emph{q}-exponential, \emph{q}-logarithm
\end{abstract}
\section{Introduction}
With a generalization of the Boltzmann-Gibbs entropy \cite{Tsallis88}, the \emph{q}-logarithm and \emph{q}-exponential functions were first proposed by Tsallis \cite{Tsallis94}.
\begin{eqnarray}
\lnq x &\equiv \frac{x^{1-q}-1}{1-\emph{q}}
\qquad \qquad \qquad & (x>0) \\
\expq x &\equiv [1+(1-q)x]_{+}^{1/(1-q)}
&(x,q \in \mathbb{R}) \\
& \mbox{where } [A]_+ \equiv \max \{ A,0 \} \nonumber
\end{eqnarray}
A \emph{q}-calculus associated with non-extensive statistical mechanics and thermodynamics was developed by Borges in 2004 \cite{Borges04}. He developed a primal \emph{q}-derivative operator, $D_{(q)}$, for which the \emph{q}-exponential function is an eigenfunction, as the ordinary exponential function is the eigenfunction of the ordinary derivative operator.
\begin{equation}
D_{(q)} \Big( \expq x \Big) = \expq x \label{eq1}
\end{equation}
A primal \emph{q}-integral operator, $I_{(q)}$, which is the inverse operator of the primal \emph{q}-derivative operator, was also developed. The primal \emph{q}-integral of a \emph{q}-exponential function is a \emph{q}-exponential function.
\begin{equation}
I_{(q)} \Big( \expq x \Big) = \expq x + c \label{eq2}
\end{equation}
In general, the following relationships hold for the primal \emph{q}-derivative and \emph{q}-integral.
\begin{eqnarray}
D_{(q)} \Big( {I_{(q)}}_a^x \Big( f(t) \Big) \Big) = f(x) \label{eq3} \\
{I_{(q)}}_a^x \Big( D_{(q)} \Big( F(t) \Big) \Big) = F(x) +c \nonumber \\
\mbox{where } {I_{(q)}}_a^x \Big( f(t) \Big) = F(x) - F(a), F(t) ={I_{(q)}} \Big( f(t) \Big) - c \nonumber
\end{eqnarray}
For the \emph{q}-logarithm function, which is the inverse function of the \emph{q}-exponential function, he developed a dual \emph{q}-derivative operator, $D^{(q)}$. The dual \emph{q}-derivative of the \emph{q}-logarithm function is $1/x$, analogous to the fact that the ordinary derivative operator on the logarithm function gives $1/x$.
\begin{equation}
D^{(q)} \Big( \lnq x \Big) = \frac{1}{x} \label{eq4}
\end{equation}
Finally, he suggested a dual \emph{q}-integral operator, $I^{(q)}$, which is the inverse operator of the dual \emph{q}-derivative. However, the dual \emph{q}-integral of $1/x$ is found to be $\ln (x) - \frac{1-q}{x} + c$, and the following relationship does not hold:
\begin{equation}
I^{(q)} \Big( \frac{1}{x} \Big) = \lnq x + c \label{eq5}
\end{equation}
In general, the following relationship does not hold for the dual \emph{q}-derivative and \emph{q}-integral, and it is a significant weakness of the dual \emph{q}-calculus suggested by Borges.
\begin{eqnarray}
D^{(q)} \Big( {I^{(q)}}_a^x \Big( f(t) \Big) \Big) = f(x) \label{eq6} \\
{I^{(q)}}_a^x \Big( D^{(q)} \Big( F(t) \Big) \Big) = F(x) +c \nonumber \\
\mbox{where } {I^{(q)}}_a^x \Big( f(t) \Big) = F(x) - F(a), F(t)={I^{(q)}} \Big( f(t) \Big) - c \nonumber
\end{eqnarray}
To address this issue, a new representation of \emph{q}-calculus with a new dual \emph{q}-integral operator is proposed herein, which satisfies equation (\ref{eq1}) $\sim$ (\ref{eq6}) with a modification of ordinary addition to \emph{q}-difference ($\qminus$) or \emph{q}-addition ($\qplus$) in equations (\ref{eq5}) and (\ref{eq6}):
\begin{eqnarray}
D_{[q]} \Big( F(x) \Big) & \equiv \lim_{y \rightarrow x} \frac{F(x) - F(y)}{\ln[E_{q}(x)] - \ln[E_{q}(y)]} \\
\stackrel[{[q]}] {\vphantom{q}} {I} ^{x}_{x_0} \Big( f(t) \Big)
& \equiv \int^{x}_{t=x_0} f(t) \ \mathrm{d} u(t) \\
& \quad \mbox{where } u(t) = \ln \Big[ \vert 1+(1-q)t \vert ^\frac{1}{1-q} \Big] \nonumber \\
D^{[q]}F(x) & \equiv \lim_{y \rightarrow x} \frac{ \ln( \expq {F(x)}) - \ln( \expq {F(y)})}{x - y} \\
\stackrel[\vphantom{q}] {{[q]}} {I} ^{x}_{x_0} \Big( f(t) \Big)
& \equiv \ln_{q} \left[ \exp \left( \int^{x}_{x_0} f(t) \mathrm{d} t \right) \right]
\end{eqnarray}
The new representation of \emph{q}-calculus is based on the concept of primal and dual \emph{q}-tangent lines, analogous to ordinary tangent lines. The primal and dual \emph{q}-derivatives are defined as the slope of the primal and dual \emph{q}-tangent lines at each point on the curve $y=f(x)$; this is analogous to the fact that the ordinary derivative of a function is the slope of the tangent line at each point on the curve $y=f(x)$. The primal and dual \emph{q}-integrals are defined as the signed primal and dual \emph{q}-area between the curve $y=f(x)$ and the horizontal axis, as the ordinary integral of a function is the signed area between the curve $y=f(x)$ and the horizontal axis.
The remainder of this paper is organized as follows. Section 2 provides some background on \emph{q}-algebra, \emph{q}-calculus of Borges, and ordinary calculus. Section 3 involves the derivation of a new representation of \emph{q}-calculus and the new dual \emph{q}-integral operator. Section 4 presents the relationship between the primal and dual \emph{q}-derivatives and integrals.
\section{Review of \emph{q}-Algebra, \emph{q}-Calculus and Ordinary Calculus}
\subsection{q-Algebra}
\emph{q}-algebra was first proposed by Borges \cite{Borges04}.
\begin{eqnarray}
x \qplus y &\equiv x + y + (1-q)xy \\
x \qminus y &\equiv \frac{x - y}{1 + (1-q)y} \\
x \qtimes y &\equiv [x^{1-q}+y^{1-q}-1]_{+}^{1/(1-q)}
\qquad (x, y >0) \\
x \qdiv y &\equiv [x^{1-q}-y^{1-q}+1]_{+}^{1/(1-q)}
\qquad (x, y >0) \\
x^{\otimes_{q}^{n}} &\equiv \underbrace{ x \qtimes x \qtimes x \qtimes \cdots \qtimes x}_{n \ times} \
= [nx^{1-q}-(n-1)]_{+}^{1/(1-q)}\\
n \qproduct x &\equiv \underbrace{ x \qplus x \qplus x \qplus \cdots \qplus x}_{n \ times} \
= \frac{1}{1-q}\{[1+(1-q)x]^{n}-1\}
\end{eqnarray}
The properties of the \emph{q}-logarithm and the \emph{q}-exponential can be expressed as follows.
\begin{eqnarray}
\lnq {xy} &= \lnq x \qplus \lnq y, \qquad
\expq x \expq y &= \expq {x \qplus y} \\
\lnq {x \qtimes y} &= \lnq x + \lnq y, \qquad
\expq x \qtimes \expq y &= \expq {x + y}\\
\lnq {x/y} &= \lnq x \qminus \lnq y, \qquad
\expq x / \expq y &= \expq {x \qminus y} \\
\lnq {x \qdiv y} &= \lnq x - \lnq y, \qquad
\expq x \qdiv \expq y &= \expq {x - y}
\end{eqnarray}
\subsection{q-Calculus}
Borges \cite{Borges04} defined primal and dual \emph{q}-derivatives and \emph{q}-integrals as follows. \\
\begin{eqnarray}
D_{(q)} \Big( f(x) \Big) &\equiv \lim_{y \rightarrow x} \frac{f(x)-f(y)}{x \qminus y}=[1+(1-q)x] \frac{\mathrm{d}f(x)}{\mathrm{d}x} \label{eqp1} \\
I_{(q)} \Big( f(x) \Big) &\equiv \int \frac{f(x)}{1+(1-q)x}{\mathrm{d}x} \label{eqp3}\\
D^{(q)} \Big( f(x) \Big) &\equiv \lim_{y \rightarrow x} \frac{f(x) \qminus f(y)}{x - y} \
=\frac{1}{1+(1-q)f(x)} \frac{\mathrm{d}f(x)}{\mathrm{d}x} \label{eqp2}\\
I^{(q)} \Big( f(x) \Big) &\equiv \int [1+(1-q)f(x)]f(x){\mathrm{d}x}
\end{eqnarray}
\subsection{Ordinary Calculus}
\subsubsection{Derivative}
Let $f(x)$ be the ordinary derivative function of $F(x)$.
\begin{equation}
\frac{\mathrm{d}}{\mathrm{d}x}F(x) \equiv
\lim_{t \rightarrow x} \frac{F(x)-F(t)}{x - t} =f(x) \\
\end{equation}
A derivative operation is a function that takes a function $F(x)$ as an argument and produces another function $f(x)$ as an output.
In ordinary calculus, $f(x_0)$, the value of $f(x)$ evaluated at $x_0$, represents the slope of the tangent line $\mathcal{T}(C, P): y=T(x;F(x), x_0)=k_{(C,P)}x+c_{(C,P)}$ at the point $P=(x, y)=(x_0, F(x_0))$ on the curve $C: y=F(x)$.\\
\begin{figure}[h]
\centering
\begin{tikzpicture}[
declare function={f(\x)=0.2*exp(\x/2)+1;}
]
\newcommand\XTickA{2}
\newcommand\XTickB{6}
\begin{axis}[
axis lines=center,
xmin=-1.2,xmax=10.2,
ymin=-1,ymax=9.2,
width=12cm,
xtick=data,
ytick=data,
typeset ticklabels with strut,
yticklabel style = {yshift=0.25cm},
xticklabels={$x_0$,$t$},
yticklabels={$F(x_0)=y_0$,$F(t)=y_t$},
xlabel=$x$,
ylabel=$y$,
domain=-0.5:7,
smooth
]
\addplot [ycomb,
mark=*,
mark options={black,mark size=2pt},
gray, dashed,
samples at={\XTickA,\XTickB}] {f(x)}
;
\addplot [mark=none] coordinates {(\XTickA, {f(\XTickA)})} node[above left] {P};
\addplot [mark=none] coordinates {(\XTickB, {f(\XTickB)})} node[above left] {Q};
\addplot [mark=none] coordinates {(0,0)} node[above left] {0};
\addplot [xcomb, gray, dashed, samples at={\XTickA,\XTickB}] {f(x)};
\addplot [name path = fcurve] {f(x)} node[above right] {$y=F(x)$};
\addplot [red, domain=-0.5:7] {(0.1*exp(\XTickA/2))*(x-\XTickA)+f(\XTickA)} node[above right] {$y=T(x;C,\mathrm{P})$};
\addplot [blue, domain=-0.5:7] {(f(\XTickB)-f(\XTickA))/(\XTickB-\XTickA)*(x-\XTickA)+f(\XTickA)} node[above right] {$y=L(x,C,\mathrm{P},\mathrm{Q})$};
\end{axis}
\end{tikzpicture}
\caption{Derivative as the slope of the tangent line} \label{fig:F1}
\end{figure}
Let the line passing through two points $P_i=(x_i,F(x_i))$ and $P_j=(x_j,F(x_j))$ on the curve $C$ be $\mathcal{L}(C, P_i, P_j): y=L(x; F(x), x_i, x_j)$.\\
\begin{equation}
L(x; F(x), x_i, x_j)= \frac{F(x_i)-F(x_j)}{x_i-x_j}(x-x_i)+F(x_i)
\end{equation}
The line $\mathcal{L}(C, P, Q)$ passing through $P=(x_0, F(x_0))$ and another point $Q=(t, F(t))$ on the curve $C$ tends to $\mathcal{T}(C, P)$, as $Q$ approaches $P$, and the slope of $\mathcal{L}(C, P, Q)$ converges to that of $ \mathcal{T}(C, P)$.
\begin{equation}
f(x_0) = \lim_{t \rightarrow x_0} \frac{L(x_0;F(x), x_0, t)-L(t;F(x), x_0, t)}{x_0 - t}=k_{(C,P)}
\end{equation}
Note that
\begin{equation}
\frac{\mathrm{d}}{\mathrm{d}x} H(x) = \frac{\mathrm{d}}{\mathrm{d}x} F(x) \quad \Longleftrightarrow \quad H(x)=F(x)+c
\end{equation}
If $C$ is a line, that is, $C: F(x) = kx+c$, the curve (line) $C$ itself is regarded as the tangent line at each point on it, and the slope of the tangent line is constant, $k$, for all points on $C$.
\subsubsection{Integral}
A definite integral operation is a function that takes a function $f(x)$ and a pair of values $(x_L, x_H)$ as arguments and outputs the difference between the values of primitive function, $F(x)$, evaluated at $x_H$ and $x_L$.
\begin{equation}
\int_{x_L}^{x_H} f(x) \mathrm{d}x \equiv F(x_H) -F(x_L).
\end{equation}
Let the signed area between the curve $C':y=f(x)$ and the horizontal axis ($y=0$) from $x_L$ to $x_H$ be $A(f(x), x_L, x_H)$. The value of $A(f(x), x_L, x_H)$ is equal to the difference between the two values of the primitive function $F(x)$ evaluated at $x_H$ and $x_L$, \\
\begin{equation}
\int_{x_L}^{x_H} f(x) \mathrm{d}x \equiv F(x_H) -F(x_L) = A(f(x), x_L, x_H).
\end{equation}
The relationship is clear when the primitive function is $F(x)=kx+c$, and its derivative is a constant; that is, $\frac{\mathrm{d}}{\mathrm{d}x}F(x)=k$ (see Figure 2).
\begin{figure}[h]
\centering
\begin{tikzpicture}[
declare function={f(\x)=\x+11;},
declare function={g(\x)= 3;},
declare function={w(\x)= 10;}
]
\newcommand\XTickA{1.5}
\newcommand\XTickB{4}
\begin{axis}[
axis lines=center,
xmin=-1.2,xmax=8.2,
ymin=-1.2,ymax=18.2,
width=12cm,
xtick=data,
ytick=data,
typeset ticklabels with strut,
yticklabel style = {yshift=0.25cm},
xticklabels={$x_L$,$x_H$},
yticklabels={$y_L=kx_L+c$,$y_H=kx_H+c$},
xlabel=$x$,
y label style={anchor=east},
ylabel=$y$,
domain=-0.5:6,
smooth
]
\addplot [ycomb,
mark=*,
mark options={black,mark size=2pt},
gray, dashed,
samples at={\XTickA,\XTickB}] {f(x)};
\addplot [xcomb, gray, dashed, samples at={\XTickA,\XTickB}] {f(x)};
\addplot [name path = fcurve, red] {f(x)} node[above] {$y=F(x)=kx+c$};
\addplot [mark=none] coordinates {(0,11)} node[above left] {\emph{c}};
\addplot [mark=none] coordinates {(0,10)} node[below left] {0};
\addplot [name path = gcurve, blue] {g(x)} node[above] {$y=f(x)=k$};
\addplot [mark=none] coordinates {(0,3)} node[above left] {\emph{k}};
\addplot [mark=none] coordinates {(0,0)} node[below left] {0};
\addplot [name path = g,domain=\XTickA:\XTickB] {g(x)};
\addplot[name path = h,domain=\XTickA:\XTickB] {0};
\addplot [mark=none] coordinates {(0,7)} node[left] {\emph{y}};
\addplot [mark=none, white, very thick] coordinates {(0, 8) (0, 8.5)};
\addplot [mark=none, -stealth, black] coordinates {(0, 7.5) (0, 8)};
\addplot [gray!20] fill between[of=g and h];
\addplot [name path = gcurve, -stealth, domain=-1.2:8.2] {w(x)} node[above left] {\emph{x}};
\addplot [mark=none, -stealth, gray] coordinates {(0.5, 12.5) (0.5, 15)};
\addplot [mark=none] coordinates {(0.5,14)} node[right] {$k(x_H-x_L)$};
\addplot [mark=none] coordinates {(1.7,2)} node[right] {$A(k,x_L,x_H)$};
\addplot [mark=none] coordinates {(1.7,1)} node[right] {$=k(x_H-x_L)$};
\end{axis}
\end{tikzpicture}
\caption{Definite integral and the signed area when $\frac{\mathrm{d}}{\mathrm{d}x}F(x)=k$} \label{fig:F2}
\end{figure}
\vspace{20pt}
In general, $A(f(x), x_L, x_H)$ can be evaluated as follows:
\begin{figure}[h]
\centering
\begin{tikzpicture}[
declare function={f(\x)=0.05*exp(\x/2)+0.2;}
]
\newcommand\XTickA{2}
\newcommand\XTickB{6}
\newcommand\NoP{8}
\begin{axis}[
axis lines=center,
xmin=-0.7,xmax=7.2,
ymin=-0.2,ymax=1.8,
width=11cm,
xtick=data,
ytick=\empty,
xticklabels={$x_L$,$x_H$},
xlabel=$x$,
ylabel=$y$,
domain=-0.3:6.3,
smooth
]
\addplot [ycomb,
mark=none,
mark options={black,mark size=2pt},
gray,
samples at={\XTickA,\XTickB}] {f(x)};
\
\addplot [xcomb, gray, samples at={\XTickA,\XTickB}] {f(x)};
\addplot [mark=none] coordinates {(0,0)} node[below right] {0};
\addplot [name path = fcurve] {f(x)} node[left] {$C':y=f(x)$};
\addplot [name path = f,domain=\XTickA:\XTickB] {f(x)};
\addplot[name path = g,domain=\XTickA:\XTickB] {0};
\foreach \t in {1,...,\NoP} {
\addplot [name path = seg_f, domain=\XTickA+(\t-1)*((\XTickB-\XTickA)/\NoP):\XTickA+\t*((\XTickB-\XTickA)/\NoP)] {f(\XTickA+\t*((\XTickB-\XTickA)/\NoP))};
\addplot [name path = seg_g, draw=none, domain=\XTickA+(\t-1)*((\XTickB-\XTickA)/\NoP):\XTickA+\t*((\XTickB-\XTickA)/\NoP)] {0};
\addplot [gray!10] fill between [of = seg_f and seg_g];
}
\addplot[dashed] coordinates{(\XTickA, {f(\XTickA)}) (\XTickA, {f(\XTickA+((\XTickB-\XTickA)/\NoP))})};
\addplot [gray!50] fill between[of=f and g];
\foreach \t in {2,...,\NoP} {
\addplot[dashed] coordinates{(\XTickA+(\t-1)*((\XTickB-\XTickA)/\NoP), 0) (\XTickA+(\t-1)*((\XTickB-\XTickA)/\NoP), {f(\XTickA+\t*((\XTickB-\XTickA)/\NoP))})};
}
\addplot [mark=none] coordinates {({\XTickA-0.15},-0.1)} node[below] [font=\fontsize{7}{0}] {$(=x_0)$};
\addplot [mark=none] coordinates {({\XTickA+(1)*((\XTickB-\XTickA)/\NoP)},-0.05)} node[below] [font=\fontsize{8}{0}] {$x_1$};
\addplot [mark=none] coordinates {({\XTickA+(2)*((\XTickB-\XTickA)/\NoP)},-0.05)} node[below] [font=\fontsize{8}{0}] {$x_2$};
\addplot [mark=none] coordinates {({\XTickA+((floor(\NoP/2)+1)-1)*((\XTickB-\XTickA)/\NoP)},-0.05)} node[below] [font=\fontsize{8}{0}] {$\cdots$};
\addplot [mark=none] coordinates {({\XTickA+(\NoP-1)*((\XTickB-\XTickA)/\NoP)},-0.05)} node[below] [font=\fontsize{8}{0}] {$x_{n-1}$};
\addplot [mark=none] coordinates {({\XTickB+0.15},-0.1)} node[below] [font=\fontsize{7}{0}] {$(=x_n)$};
\addplot [mark=none] coordinates {({\XTickA+(1)*((\XTickB-\XTickA)/\NoP)},0.05)} node[above left] [font=\fontsize{8}{0}] {$A_1$};
\addplot [mark=none] coordinates {({\XTickA+(2)*((\XTickB-\XTickA)/\NoP)},0.05)} node[above left] [font=\fontsize{8}{0}] {$A_2$};
\addplot [mark=none] coordinates {({\XTickA+((floor(\NoP/2)+1)-1)*((\XTickB-\XTickA)/\NoP)},0.05)} node[above left] [font=\fontsize{8}{0}] {$\cdots$};
\addplot [mark=none] coordinates {({\XTickA+(\NoP-1)*((\XTickB-\XTickA)/\NoP)},0.05)} node[above left] [font=\fontsize{8}{0}] {$A_{n-1}$};
\addplot [mark=none] coordinates {({\XTickB},0.05)} node[above left] [font=\fontsize{8}{0}] {$A_n$};
\end{axis}
\end{tikzpicture}
\caption{General evaluation of signed area} \label{fig:F3}
\end{figure}
Let $T=\{x_0, x_1, \cdots, x_{n-1}, x_n\}$ be the $n$-partitions of $[x_L, x_H]$, $\Delta t_i = x_i-x_{i-1}$, and the length of the longest sub-interval is the norm of the partition, $\|T\|$. The signed area $A_i=A((f(x), x_{i-1}, x_{i})$ between the curve $C':y=f(x)$ and the horizontal axis ($y=0$) over the $i$-th partition $[x_{i-1}, x_i]$ can be approximated by the signed area of a rectangle with height $f(x_i)$ and base $\Delta t_i$; that is, $A_i \approx f(x_i)\Delta t_i$. Note that $A_i = f(x_i)\Delta t_i$ if $f(x)$ is constant over the partition.
The signed area between the curve $C'$ and the x axis over $[x_L, x_H]$ can be approximated by the sum of areas of rectangles, that is, $A(f(x), x_L, x_H) \approx \sum_{i=1}^{n} f(x_i)\Delta t_i$. Note that $A = \sum_{i=1}^{n} f(x_i)\Delta t_i$ if $f(x)$ is constant over $[x_L, x_H]$.
As the norm of the partition approaches zero, the sum of areas of rectangles converges to the signed area between the curve $C$ and the horizontal axis ($y=0$) over $[x_L, x_H]$, and the definite integral is defined as the limit.
\begin{equation}
\int_{x_L}^{x_H} f(x) \mathrm{d}x = A(f(x), x_L, x_H)\equiv \lim_{\|T\| \rightarrow 0} \sum_{n=1}^{n} f(x_i)\Delta t_i
\end{equation}
In contrast, an indefinite integral is a function that takes a function $f(x)$ as an argument and produces a family of functions $\mathcal{F}= \left\{ F(x)+c, c \in \mathbb{R} \right\}$. The codomain of an indefinite integral operation is a set of families of functions, and the image of an indefinite integral is a translation family of functions along the y axis.
\begin{equation}
\int f(x)dx = F(x) + c = \mathcal{F}
\end{equation}
We can recover the exact function $F(x)$ only if we know a point $P=(x_0, F(x_0))$ on the curve $C$ with $f(x)$.
\begin{equation}
F(x) = \int_{x_0}^x f(t)dt + F(x_0) = A(f(x), x, x_0)+F(x_0)
\end{equation}
\newpage
\section{New Representation of \emph{q}-Calculus and New Dual \emph{q}-integral}
\subsection{Primal}
\subsubsection{Primal q-derivative}
Let a family of curves $\mathcal{L}_{q (k_q, \cdot)} = \left\{ y=L_{q}(x; k_q, c) \right\}$ have a constant primal \emph{q}-derivative (defined by equation (\ref{eqp1})), $k_q$, at every point in their domain.
$L_{q}(x; k_q, c)$ must satisfy the condition $D_{(q)} \Big( L_{q}(x; k_q, c) \Big) = k_q$.
\begin{equation}
D_{(q)} \Big( L_{q}(x; k_q, c) \Big)
= \left\{ 1+(1-q)x \right\} \frac{\mathrm{d}}{\mathrm{d}x} L_{q}(x; k_q, c)
= k_q
\end{equation}
Therefore,
\begin{equation}
\eqalign{
L_{q}(x; k_q, c) & = \int \frac{k_q}{1+(1-q)x} \mathrm{d}x \cr
& = \frac{k_q}{1-q} \cdot \ln(\vert 1+(1-q)x \vert ) + c \cr
& = k_q \cdot \ln( \vert 1+(1-q)x \vert ^\frac{1}{1-q}]+c \cr
& = k_q \cdot \ln[ E_{q}(x) ] +c \\
}
\end{equation}
where
\begin{equation}
E_{q}(x) \equiv \vert 1+(1-q)x \vert ^\frac{1}{1-q}
\end{equation}
Note that $E_{q}(x)= \expq x$, where $1+(1-q)x >0$.
Let us call the curve $\mathcal{L}_q (k_q, c):y=L_{q}(x; k_q, c)=k_q \cdot ln(E_{q}(x))+c$ a primal $\emph{q}$-line with primal $\emph{q}$-slope $k_q$ and $y$-intercept $c$.
Note that $\mathcal{L}_q (0, c):y=L_{q}(x; 0, c)$ is a horizontal line, $y=c$.
The line $L_{q}(x; k_q, c)$ has an interesting property:
\begin{equation}
\fl \frac{L_{q}(x_i; k_q, c)-L_{q}(x_j; k_q, c)}{\ln(E_{q}(x_i))-\ln(E_{q}(x_j))}=k_q, \forall x_i, x_j \in \mathbb{R}, x_i \neq x_j, x_i, x_j \neq \frac{1}{q-1} \label{eqpa1}
\end{equation}
From this finding, we define a new primal \emph{q}-derivative operator $D_{[q]}$,
\begin{equation}
D_{[q]} \Big( F(x) \Big) \equiv \lim_{y \rightarrow x} \frac{F(x) - F(y)}{\ln[E_{q}(x)] - \ln[E_{q}(y)]}
\end{equation}
Note that
\begin{equation}
D_{[q]} \Big( H(x) \Big) = D_{[q]} \Big( F(x) \Big) \quad \Longleftrightarrow \quad H(x)=F(x)+c.
\end{equation}
$\expq x$ is the eigenfunction of $D_{[q]}$.
\begin{equation}
\fl D_{[q]} \Big( \expq x \Big)
= \lim_{y \rightarrow x} \frac{\expq x - \expq y}{\ln[E_{q}(x)] - \ln[E_{q}(y)]}
= \expq x, \ \mbox{if} \ \ 1+(1-q)x >0
\end{equation}
In general, $D_{[q]}$ is equivalent to $D_{(q)}$; that is, $D_{[q]} \Big( F(x) \Big) = D_{(q)} \Big( F(x) \Big)$, in general, because
\begin{equation}
\lim_{y \rightarrow x} \frac{\ln[E_{q}(x)] - \ln[E_{q}(y)]}{x \qminus y} = 1
\end{equation}
\begin{equation}
\eqalign{
D_{[q]} \Big( F(x) \Big) & \equiv \lim_{y \rightarrow x} \frac{F(x) - F(y)}{\ln[E_{q}(x)] - \ln[E_{q}(y)]} \cr
& = \lim_{y \rightarrow x} \frac{F(x) - F(y)}{\ln[E_{q}(x)] - \ln[E_{q}(y)]} \frac{\ln[E_{q}(x)] - \ln[E_{q}(y)]}{x \qminus y} \cr
& = \lim_{y \rightarrow x} \frac{F(x) - F(y)}{x \qminus y} = D_{(q)} \Big( F(x) \Big) \cr
}
\end{equation}
Let $\mathcal{L}_q(C, P_i, P_j) : y = L_q(x; F(x), x_i, x_j)$ be a primal $\emph{q}$-line that passes through the two points $P_i=(x_i, y_i)=(x_i, F(x_i))$ and $P_j=(x_j, y_j)=(x_i, F(x_i))$ on the curve $C: y=F(x)$.
$L_q(x; F(x), x_i, x_j))$ turns out to be
\begin{eqnarray}
L_q(x; F(x), x_i, x_j)) = k_q \cdot \ln(E_{q}(x))+c \\
k_q = \frac{F(x_i)-F(x_j)}{\ln(E_{q}(x_i))-\ln(E_{q}(x_j))} \\
c = y_i-k \cdot \ln(E_{q}(x_i))
\end{eqnarray}
$L_q(x; F(x), x_i, x_j))$ can also be expressed as follows:
\begin{eqnarray}
L_q(x; F(x), x_i, x_j)) = k_q \cdot \ln(E_{q}(x \qminus x_i))+ F(x_i) \\
L_q(x; F(x), x_i, x_j)) = k_q \cdot \ln(E_{q}(x \qminus x_j))+ F(x_j)
\end{eqnarray}
\begin{figure}[h]
\centering
\begin{tikzpicture}[
declare function={f(\x)= (max(1 - 0.1*x, 0))^(-10);},
declare function={g(\x)= (1 + 0.1)^(-10)*(ln((max(1 - 0.1*(x+1), 0))^(-10))+1);},
declare function={h(\x)=
((1 + (-0.1))^(-10) - (1 + 0.1)^(-10))/(ln((1 + (-0.1))^(-10)) - ln((1 + 0.1)^(-10)))
*ln((max(1 - 0.1*x, 0))^(-10))
+(1 + 0.1)^(-10)
-((1 + (-0.1))^(-10) - (1 + 0.1)^(-10))/(ln((1 + (-0.1))^(-10)) - ln((1 + 0.1)^(-10)))
*ln((1 + 0.1)^(-10))
;}
]
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\begin{axis}[
axis lines=center,
xmin=-5,xmax=9,
ymin=-1,ymax=7,
width=10cm,
xtick=data,
ytick=\empty,
yticklabel style = {above right},
xticklabels={$x_0$,$t$},
yticklabels={$F(x_0)$,$F(t)$},
xlabel=$x$,
ylabel=$y$,
smooth
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mark=*,
mark options={black,mark size=2pt},
gray, dashed,
samples at={\XTickA,\XTickB}] {f(x)};
\addplot [mark=none] coordinates {(0,0)} node[below right] {0};
\addplot [name path = fcurve, domain=-4:1.7] {f(x)} node[right] {$C:y=F(x)$};
\addplot [red, name path = gcurve, domain=-4:6] {g(x)} node[below right]{};
\addplot [blue, name path = hcurve, domain=-4:2.5] {h(x)} node[above right] {$\mathcal{L}_q(C, P, Q)$};
\addplot [mark=none] coordinates {(\XTickA, {f(\XTickA)})} node[above left] {P};
\addplot [mark=none] coordinates {(\XTickB, {f(\XTickB)})} node[right] {Q};
\addplot [mark=none] coordinates {(3,2.5)} node[red, below right] {$\mathcal{T}_q(C, P)$};
\end{axis}
\end{tikzpicture}
\caption{Primal \emph{q}-derivative as the slope of the primal \emph{q}-tangent line} \label{fig:F4}
\end{figure}
We call $\mathcal{T}_q(C, P): y=T_q(x; F(x), x_0)=k_{q(C,P)} \cdot \ln(E_q(x))+c_{q(C,P)}$ as the primal $\emph{q}$-tangent line of the curve $C: y=F(x)$ at the point $P=(x_0, F(x_0))$ when $F(x_0)=T_{q}(x_0)$ and $D_{[q]} \Big( F(x) \Big) \Big\vert_{x=x_0} = k_{q(C,P)}$.
$D_{[q]} \Big( F(x) \Big) \Big\vert_{x=x_0}$ is the primal $\emph{q}$-slope of the primal $\emph{q}$-tangent line at point $P$ on the curve $C$.
The primal $\emph{q}$-line $\mathcal{L}_q(C, P, Q)$ passing through $P=(x_0, F(x_0))$ and another point $Q=(t, F(t))$ on the curve $C$ tends to $\mathcal{T}_q(C, P)$ as $Q$ approaches $P$, and the primal $\emph{q}$-slope of $\mathcal{L}_q(C, P, Q)$ converges to that of $\mathcal{T}_q(C, P)$.
\begin{eqnarray}
D_{[q]} \Big( F(x) \Big) \Big\vert_{x=x_0} & = \lim_{t \rightarrow x_0} \frac{L_q(x_0; F(x), x_0, t)-L_q(t; F(x), x_0, t)}{x_0 - t} \\
& =k_{q(C,P)} \nonumber
\end{eqnarray}
\subsubsection{Primal q-integral} If $f(x)= D_{[q]} \Big( F(x) \Big)$, the definite primal \emph{q}-integral of $f(x)$ from $x_L$ to $x_h$, denoted as $\stackrel[{[q]}] {\vphantom{q}} {I} ^{x_H}_{x_L} \Big( f(x) \Big)$, is equal to $F(x_H) - F(x_L)$.
\begin{equation}
\stackrel[{[q]}] {\vphantom{q}} {I} ^{x_H}_{x_L} \Big( f(x) \Big) = F(x_H) - F(x_L)
\end{equation}
Equation (\ref{eqpa1}) gives us a hint on how a primal \emph{q}-integral can be related to a (deformed) signed area.
Consider the most immediate example, the case of primal \emph{q}-lines.
Let $F(x)=k_q \cdot ln(E_{q}(x))+c$ and $f(x)= D_{[q]} \Big( F(x) \Big) = k_q$. Let us consider the transformation $(x, y)=(x, f(x)) \rightarrow (u, v)= (\ln(E_{q}(x)), f(x))$.
\begin{figure}[h]
\centering
\begin{tikzpicture}[
declare function={f(\x)=(-1.5)*ln(abs(1-0.2*x))+11;},
declare function={g(\x)= 3;},
declare function={w(\x)= 10;}
]
\newcommand\XTickA{1}
\newcommand\XTickB{3.5}
\newcommand\XTickC{6.5}
\newcommand\XTickD{9}
\begin{axis}[
axis lines=center,
xmin=-1.2,xmax=10.2,
ymin=-1.2,ymax=18.2,
width=12cm,
xtick=data,
ytick=data,
typeset ticklabels with strut,
xticklabel style = {yshift=0.25cm, scale=0.65},
yticklabel style = {yshift=0.3cm, scale=0.85},
xticklabels={$u_{L_1}$,$u_{H_1}$, $u_{H_2}$,$u_{L_2}$},
yticklabels={${y_{L_{1}}, y_{H_{2}}}$,${y_{H_{1}}, y_{L_{2}}}$, ,},
x label style={anchor=north, scale=0.8},
xlabel=$u$,
y label style={anchor=east, scale=0.85},
ylabel=$y$,
domain=-0.5:9,
smooth
]
\addplot [ycomb,
mark=*,
mark options={black,mark size=1pt},
gray, dashed,
samples at={\XTickA,\XTickB,\XTickC,\XTickD}] {f(x)};
\addplot [mark=none, white, very thick] coordinates {(\XTickA, 7) (\XTickA, 10)};
\addplot [mark=none, white, very thick] coordinates {(\XTickB, 7) (\XTickB, 10)};
\addplot [mark=none, white, very thick] coordinates {(\XTickC, 7) (\XTickC, 10)};
\addplot [mark=none, white, very thick] coordinates {(\XTickD, 7) (\XTickD, 10)};
\addplot [xcomb, gray, dashed, samples at={\XTickC,\XTickD}] {f(x)};
\addplot [name path = fcurve, red, domain=-0.5:4.8] {f(x)} node[above, scale=0.65] {$y=L_{q}(x; k_q,c)=k_{q} \cdot ln(E_{q}(x))+c$};
\addplot [name path = fcurve, red, domain=5.2:10] {f(x)} node[above] {};
\addplot [name path = hcurve, -stealth, domain=-1.2:10.2] {w(x)} node[below left, scale=0.8] {\emph{x}};
\addplot [mark=none] coordinates {(0,10)} node[below left, scale=0.85] {0};
\addplot [mark=none] coordinates {(0,11)} node[below left, scale=0.85] {\emph{c}};
\addplot [mark=none, purple, dashed] coordinates {(5, 9.5) (5, 16)};
\addplot [mark=none] coordinates {(5,9)} node[purple, scale=0.65] {$x=-\frac{1}{1-q}$};
\addplot [mark=none] coordinates {(\XTickA,9.3)} node[scale=0.8] {$x_{L_1}$};
\addplot [mark=none] coordinates {(\XTickB,9.3)} node[scale=0.8] {$x_{H_1}$};
\addplot [mark=none] coordinates {(\XTickC,9.3)} node[scale=0.8] {$x_{L_2}$};
\addplot [mark=none] coordinates {(\XTickD,9.3)} node[scale=0.8] {$x_{H_2}$};
\addplot [mark=none, dashed, gray] coordinates {(6.5, {(-1.5)*ln(abs(1-0.2*6.5))+11}) (7.3, {(-1.5)*ln(abs(1-0.2*6.5))+11})};
\addplot [mark=none, -stealth, gray] coordinates {(7, {(-1.5)*ln(abs(1-0.2*9))+11}) (7, {(-1.5)*ln(abs(1-0.2*6.5))+11})};
\addplot [mark=none] coordinates {(7.3,{(-1.5)*ln(abs(1-0.2*3.8))+11})} node[right, scale=0.75] {$y_{H_{1}}-y_{L_{1}}$};
\addplot [mark=none] coordinates {(7.7,{(-1.5)*ln(abs(1-0.2*3))+11})} node[right, scale=0.75] {$= - (y_{H_{2}}-y_{L_{2}}) $};
\addplot [mark=none, white, very thick] coordinates {(0, 8) (0, 8.8)};
\addplot [mark=none, -stealth, black] coordinates {(0, 7.5) (0, 8)};
\addplot [mark=none] coordinates {(0,7)} node[left, scale=0.85] {\emph{v}};
\addplot [name path = gleft, blue, domain= 0:4.5] {g(x)} node[above, scale=0.65] {};
\addplot [mark=none] coordinates {(0,3)} node[above left, scale=0.65] {$k_{q}$};
\addplot [mark=none] coordinates {(0,0)} node[below left, scale=0.65] {0};
\addplot [mark=none] coordinates {(4.7,0)} node[below left, scale=0.85] {\emph{u}};
\addplot [mark=none, white, very thick] coordinates {(4.2, 0) (4.8, 0)};
\addplot [mark=none, -stealth, black] coordinates {(4,0) (4.5,0)};
\addplot [mark=none] coordinates {(\XTickA,-0.8)} node[scale=0.5] {$=ln(E_q(x_{L_1}))$};
\addplot [mark=none] coordinates {(\XTickB,-0.8)} node[scale=0.5] {$=ln(E_q(x_{H_1}))$};
\addplot [mark=none] coordinates {(5,7)} node[left, scale=0.85] {\emph{v}};
\addplot [name path = gright, blue, domain=5:9.5] {g(x)} node[above, scale=0.65] {};
\addplot [mark=none] coordinates {(5,3)} node[above left, scale=0.65] {$k_{q}$};
\addplot [mark=none] coordinates {(5,0)} node[below left, scale=0.65] {0};
\addplot [mark=none, -stealth, black] coordinates {(5, -1.2) (5, 8)};
\addplot [mark=none] coordinates {(\XTickC,-0.8)} node[scale=0.5] {$=ln(E_q(x_{H_2}))$};
\addplot [mark=none] coordinates {(\XTickD,-0.8)} node[scale=0.5] {$=ln(E_q(x_{L_2}))$};
\addplot [name path = g1,domain=\XTickA:\XTickB] {g(x)};
\addplot [name path = h1,domain=\XTickA:\XTickB] {0};
\addplot [name path = g2,domain=\XTickC:\XTickD] {g(x)};
\addplot [name path = h2,domain=\XTickC:\XTickD] {0};
\addplot [gray!10] fill between[of=g1 and h1];
\addplot [gray!40] fill between[of=g2 and h2];
\addplot [mark=none] coordinates {(1.2,1.5)} node[right, scale=0.65] {$A_q(k_q, u_{L_1}, u_{H_1})$};
\addplot [mark=none] coordinates {(6.7,1.5)} node[right, scale=0.65] {$A_q(k_q, u_{L_2}, u_{H_2})$};
\end{axis}
\end{tikzpicture}
\caption{Relationship between the primal \emph{q}-integral and signed primal \emph{q}-area} \label{fig:F5}
\end{figure}
Figure 5 shows the graph of $y=F(x)$ at the top and that of $u$ and $v$ at the bottom.
The bottom left graph shows the case $x_{L_1},x_{H_1}<-\frac{1}{1-q}$, and the bottom right graph shows the case $x_{L_2},x_{H_2}>-\frac{1}{1-q}$. $x_{L_2}$ and $x_{H_2}$ are set as $x_{L_2} = -\frac{2}{1-q} - x_{H_1}$ and $x_{H_2} = -\frac{2}{1-q} - x_{L_1}$.
${y_{L_{1}} = y_{H_{2}}}$ and ${y_{H_{1}} = y_{L_{2}}}$ because
\begin{eqnarray}
E_{q}(x) & = \vert 1+(1-q)x \vert ^\frac{1}{1-q} \\
& = \vert -1-(1-q)x \vert ^\frac{1}{1-q} = \left \vert 1+(1-q) \left( - \frac{2}{1-q}-x \right) \right \vert ^\frac{1}{1-q} \nonumber \\
& = E_{q}( - \frac{2}{1-q}-x ) \nonumber
\end{eqnarray}
Let $A_q(k_q, u_{L}, u_{H})$ be the signed rectangular area formed by $v=k_q$ and $v=0$ over the range $[u_L, u_H]$; this is called the primal \emph{q}-area.
\begin{equation}
A_q(k_q, u_L, u_H) = k_q \cdot ( u_H - u_L )
\end{equation}
With equation (\ref{eqpa1}), we find that the signed area of each shaded rectangle in the $u-v$ graph is equal to the corresponding definite integral.
\begin{eqnarray}
A_q(k_q, u_{L_1}, u_{H_1}) & = k_q \cdot ( u_{H_1} - u_{L_1} ) \\
& = k_q \cdot \left\{ \ln(E_{q}(x_{H_1}))-\ln(E_{q}(x_{L_1})) \right\} \nonumber \\
& = L_{q}(x_{H_1}; k_q, c) - L_{q}(x_{L_1}; k_q, c) \nonumber \\
& = \stackrel[{[q]}] {\vphantom{q}} {I} ^{x_{H_1}}_{x_{L_1}} \Big( L_{q}(x; k_q, c) \Big) \nonumber \\
A_q(k_q, u_{L_2}, u_{H_2}) & = k_q \cdot ( u_{H_2} - u_{L_2} ) \\
& = k_q \cdot \left\{ \ln(E_{q}(x_{H_2}))-\ln(E_{q}(x_{L_2})) \right\} \nonumber \\
& = L_{q}(x_{H_2}; k_q, c) - L_{q}(x_{L_2}; k_q, c) \nonumber \\
& = \stackrel[{[q]}] {\vphantom{q}} {I} ^{x_{H_2}}_{x_{L_2}} \Big( L_{q}(x; k_q, c) \Big) \nonumber
\end{eqnarray}
$y_{H_1} - y_{L_1} = -( y_{H_2} - y_{L_2} ), $ because ${y_{L_{1}} = y_{H_{2}}}$ and ${y_{H_{1}} = y_{L_{2}}}$.
From this, we find that
\begin{equation}
\stackrel[{[q]}] {\vphantom{q}} {I} ^{x_H}_{x_L} \Big( L_{q}(x; k_q, c) \Big)
= - \stackrel[{[q]}] {\vphantom{q}} {I} ^{-\frac{2}{1-q} - x_L}_{-\frac{2}{1-q} - x_H} \Big( L_{q}(x; k_q, c) \Big)
\end{equation}
From this, we get
\begin{eqnarray}
\fl \stackrel[{[q]}] {\vphantom{q}} {I} ^{x_H}_{x_L} \Big( L_{q}(x; k_q, c) \Big)
& = \stackrel[{[q]}] {\vphantom{q}} {I} ^{-\frac{1}{1-q}}_{x_L} \Big( L_{q}(x; k_q, c) \Big)
+\stackrel[{[q]}] {\vphantom{q}} {I} ^{x_H}_{-\frac{1}{1-q}} \Big( L_{q}(x; k_q, c) \Big) \\
& = \stackrel[{[q]}] {\vphantom{q}} {I} ^{-\frac{1}{1-q}}_{x_L} \Big( L_{q}(x; k_q, c) \Big)
- \stackrel[{[q]}] {\vphantom{q}} {I} ^{-\frac{1}{1-q}}_{-\frac{2}{1-q} - x_H} \Big( L_{q}(x; k_q, c) \Big) \nonumber \\
& = \stackrel[{[q]}] {\vphantom{q}} {I} ^{-\frac{2}{1-q} - x_H}_{x_L} \Big( L_{q}(x; k_q, c) \Big), \nonumber \\
& \quad \mbox{when} \ x_L < -\frac{1}{1-q} < x_H \nonumber
\end{eqnarray}
Based on the above findings, we define the definite primal \emph{q}-integral of $f(x)$ from $x_L$ to $x_h$, $\stackrel[{[q]}] {\vphantom{q}} {I} ^{x_H}_{x_L} \Big( f(x) \Big)$, as follows:
\begin{eqnarray}
\stackrel[{[q]}] {\vphantom{q}} {I} ^{x_H}_{x_L} \Big( f(x) \Big) = F(x_H) - F(x_L) \equiv A_q(f(x), u_L, u_H)
\end{eqnarray}
We derive the primal \emph{q}-integral of the \emph{q}-exponential function, $\stackrel[{[q]}] {\vphantom{q}} {I} ^{x_H}_{x_L} \Big( \expq x \Big)$, with $A_q(k_q, u_{L}, u_{H})$.
We can use $A_q(k_q, u_{L}, u_{H}) = k_q \cdot \left\{ \ln(\expq {x_{H}})-\ln(\expq {x_{L}}) \right\}$ instead of $A_q(k_q, u_{L}, u_{H}) = k_q \cdot \left\{ \ln(E_{q}(x_{H}))-\ln(E_{q}(x_{L})) \right\}$ because $k_q = \expq x =0$ when $1+(1-q)x \leq 0$.
Split $[x_L, x_H]$ into $n$-partitions, $T_q=\{x_L=x_0, x_1, x_2, \cdots, x_{n-1}, x_n = x_H\}$, where $x_i = x_L \qplus (i \qproduct t)$,
\begin{equation}
t= \frac{1}{1-q} \left[ \left\{ 1 + (1-q) \cdot (x_H \qminus x_L) \right\}^{\frac{1}{n}}-1 \right]
\end{equation}
\begin{figure}[h]
\centering
\begin{tikzpicture}[
declare function={f(\x)=0.2*exp(\x/2);},
declare function={g(\x)= 0.1*(max(1 - 0.1*x, 0))^(-10)+6.2;},
declare function={w(\x)= 6;}
]
\newcommand\XTickA{2}
\newcommand\XTickB{6}
\newcommand\NoP{8}
\begin{axis}[
axis lines=center,
xmin=-0.7,xmax=8.2,
ymin=-0.5,ymax=9.2,
width=13cm,
xtick=data,
ytick=\empty,
xtick=\empty,
x label style={anchor=north, scale=0.8},
xlabel=$u$,
y label style={anchor=east, scale=0.85},
ylabel=$f(x)$,
domain=-0.5:7,
smooth
]
\addplot [ycomb,
mark=none,
mark options={black,mark size=2pt},
gray,
samples at={\XTickA,\XTickB}] {f(x)};
\
\addplot [xcomb, gray, samples at={\XTickA,\XTickB}] {f(x)};
\addplot [mark=none] coordinates {(0,0)} node[below right] {0};
\addplot [name path = ocurve, red, domain=-0.5:2.7] {g(x)} node[right] {$f(x)=e_q (x)$};
\addplot [name path = hcurve, -stealth, domain=-0.7:8.2] {w(x)} node[below left, scale=0.8] {\emph{x}};
\addplot [mark=none] coordinates {(0,6)} node[below left, scale=0.85] {0};
\addplot [mark=none] coordinates {(0,6.3)} node[above left, scale=0.85] {1};
\addplot [mark=none, white, very thick] coordinates {(0, 5) (0, 5.5)};
\addplot [mark=none, -stealth, black] coordinates {(0, 4.5) (0, 5)};
\addplot [mark=none] coordinates {(0,5)} node[left, scale=0.85] {\emph{v}};
\addplot [mark=none, dashed, gray] coordinates {(1.5, 6) (1.5, {0.1*(max(1 - 0.1*1.5, 0))^(-10)+6.2})};
\addplot [mark=none, dashed, gray] coordinates {(2, 6) (2, {0.1*(max(1 - 0.1*2, 0))^(-10)+6.2})};
\addplot [mark=none] coordinates {(1.5,6)} node[below, scale=0.8] {$x_{L}$};
\addplot [mark=none] coordinates {(2,6)} node[below, scale=0.8] {$x_{H}$};
\addplot [name path = fcurve, blue, domain=-0.5:6.5] {f(x)} node[below right] {$v=e(u)$};
\addplot [name path = f, blue, domain=\XTickA:\XTickB] {f(x)};
\addplot[name path = g,domain=\XTickA:\XTickB] {0};
\foreach \t in {1,...,\NoP} {
\addplot [name path = seg_f, domain=\XTickA+(\t-1)*((\XTickB-\XTickA)/\NoP):\XTickA+\t*((\XTickB-\XTickA)/\NoP)] {f(\XTickA+\t*((\XTickB-\XTickA)/\NoP))};
\addplot [name path = seg_g, draw=none, domain=\XTickA+(\t-1)*((\XTickB-\XTickA)/\NoP):\XTickA+\t*((\XTickB-\XTickA)/\NoP)] {0};
\addplot [gray!30] fill between [of = seg_f and seg_g];
}
\addplot[dashed] coordinates{(\XTickA, {f(\XTickA)}) (\XTickA, {f(\XTickA+((\XTickB-\XTickA)/\NoP))})};
\foreach \t in {2,...,\NoP} {
\addplot[dashed] coordinates{(\XTickA+(\t-1)*((\XTickB-\XTickA)/\NoP), 0) (\XTickA+(\t-1)*((\XTickB-\XTickA)/\NoP), {f(\XTickA+\t*((\XTickB-\XTickA)/\NoP))})};
}
\addplot [mark=none] coordinates {({\XTickA-0.3},-0.05)} node[below] [font=\fontsize{8}{0}] {$u_L=u_0$};
\addplot [mark=none] coordinates {({\XTickA+(1)*((\XTickB-\XTickA)/\NoP)},-0.05)} node[below] [font=\fontsize{8}{0}] {$u_1$};
\addplot [mark=none] coordinates {({\XTickA+(2)*((\XTickB-\XTickA)/\NoP)},-0.05)} node[below] [font=\fontsize{8}{0}] {$u_2$};
\addplot [mark=none] coordinates {({\XTickA+((floor(\NoP/2))-1)*((\XTickB-\XTickA)/\NoP)},-0.05)} node[below] [font=\fontsize{8}{0}] {$\cdots$};
\addplot [mark=none] coordinates {({\XTickA+((floor(\NoP/2)+1)-1)*((\XTickB-\XTickA)/\NoP)},-0.05)} node[below] [font=\fontsize{8}{0}] {$u_{i-1}$};
\addplot [mark=none] coordinates {({\XTickA+(\NoP-3)*((\XTickB-\XTickA)/\NoP)},-0.05)} node[below][font=\fontsize{8}{0}] {$u_{i}$};
\addplot [mark=none] coordinates {({\XTickA+(\NoP-1)*((\XTickB-\XTickA)/\NoP)},-0.05)} node[below] [font=\fontsize{8}{0}] {$u_{n-1}$};
\addplot [mark=none] coordinates {({\XTickB+0.35},-0.05)} node[below] [font=\fontsize{8}{0}] {$u_n=u_H$};
\addplot [mark=none] coordinates {({\XTickA+(1)*((\XTickB-\XTickA)/\NoP)+0.05},0.05)} node[above left] [font=\fontsize{8}{0}] {$A_{q1}$};
\addplot [mark=none] coordinates {({\XTickA+(2)*((\XTickB-\XTickA)/\NoP)+0.05},0.05)} node[above left] [font=\fontsize{8}{0}] {$A_{q2}$};
\addplot [mark=none] coordinates {({\XTickA+((floor(\NoP/2)+1)-1)*((\XTickB-\XTickA)/\NoP)},0.05)} node[above left] [font=\fontsize{8}{0}] {$\cdots$};
\addplot [mark=none] coordinates {({\XTickA+(5)*((\XTickB-\XTickA)/\NoP)+0.05},0.05)} node[above left] [font=\fontsize{8}{0}] {$A_{qi}$};
\addplot [mark=none] coordinates {({\XTickB+0.1},0.05)} node[above left] [font=\fontsize{8}{0}] {$A_{qn}$};
\addplot[dashed] coordinates{(0, {f(\XTickA+5*((\XTickB-\XTickA)/\NoP))}) ({\XTickA+(5)*((\XTickB-\XTickA)/\NoP)}, {f(\XTickA+5*((\XTickB-\XTickA)/\NoP))})};
\addplot[mark=none] coordinates {(0, {f(\XTickA+5*((\XTickB-\XTickA)/\NoP))})} node[left] [font=\fontsize{8}{0}] {$f(x_i)$};
\end{axis}
\end{tikzpicture}
\caption{Definite primal \emph{q}-integral of a \emph{q}-exponential function} \label{fig:F6}
\end{figure}
$T_q$ transforms into $U_q = \left\{ \ln [ \expq {x_L} ] = u_0, u_1, u_2, \cdot, u_{n-1}, u_n = \ln [ \expq {x_H} ] \right\}$, where $u_i = \ln[ \expq {x _i} ]$, with a transformation $(x,y) = (x, f(x))=(x, \expq x) \rightarrow (u, v) = (\ln[ \expq x ],f(x)) = (\ln[ \expq x ], \expq x )$.
\begin{equation}
u_i = \ln [ \expq {x_i} ] = \ln [ \expq {x_L} ] + i \cdot \ln \left[ \left\{ \expq t \right\} \right]
\end{equation}
\begin{equation}
u_i - u_{i-1} = \ln \left[ \left\{ \expq t \right\} \right]
\end{equation}
\begin{eqnarray}
\expq t & = \left\{1 + (1-q) \cdot (x_H \qminus x_L) \right\}^{\frac{1}{n(1-q)}} \\
& = \left\{ \expq {x_H \qminus x_L} \right\}^{\frac{1}{n}} \nonumber \\
\ln \Big( \expq t \Big) & = \frac{1}{n(1-q)} \ln \left[ 1 + (1-q) \cdot (x_H \qminus x_L) \right] \label{eqpa2}\\
& = \frac{1}{n} \ln \left[ \expq {x_H \qminus x_L} \right] \nonumber
\end{eqnarray}
Equation (\ref{eqpa2}) shows that the norm of the partition $U_q$, $\|U_q\|$, approaches zero when the number of partitions approaches infinity.
Let $A_{qi}$ be the primal \emph{q}-area formed by $v = \expq {x_i}$ and $v=0$ over the $i$-th partition $[u_{i-1}, u_i]$.
\begin{equation}
\fl \expq {x_i} = \expq {x_L \qplus (i \qproduct t)} = \expq {x_L} \cdot \expq {i \qproduct t} = \expq {x_L} \cdot \left\{ \expq t \right\} ^i
\end{equation}
\begin{eqnarray}
A_{qi} & = A_q(\expq {x_i}, u_{i-1}, u_i) = \expq {x_i} \cdot (u_i - u_{i-1}) \\
& = \left[ \expq {x_L} \cdot \left\{ \expq t \right\} ^i \right] \ln \left[ \expq t \right] \nonumber \\
& = \expq {x_L} \cdot \left\{ 1 + (1-q) \cdot (x_H \qminus x_L) \right\}^{\frac{i}{n(1-q)}} \cdot \frac{1}{n(1-q)} \ln \left[ 1 + (1-q) \cdot (x_H \qminus x_L) \right] \nonumber \\
& = \expq {x_L} \cdot \frac{\ln(z)}{1-q} \cdot \frac{1}{n} {\left( z^{\frac{1}{n(1-q)}} \right) }^i \nonumber \\
& \quad \mbox{where} \ z = 1 + (1-q) \cdot (x_H \qminus x_L) \nonumber
\end{eqnarray}
$A_q(k_q, u_{L}, u_{H})$ can be evaluated as the sum of $A_{qi}$ when the norm of the partition $U_q$, $\|U_q\|$, approaches zero,
\begin{eqnarray}
\sum_{i=1}^{n} A_{qi} & = \sum_{i=1}^{n} \left\{ \expq {x_L} \cdot \frac{\ln(z)}{1-q} \cdot \frac{1}{n} {\left( z^{\frac{1}{n(1-q)}} \right) }^i \right\} \\
& = \expq {x_L} \cdot \frac{\ln(z)}{1-q} \cdot \frac{1}{n} \cdot \frac{z^{\frac{1}{n(1-q)}} \left(z^{\frac{n}{n(1-q)}}-1 \right) }{z^{\frac{1}{(1-q)}}-1} \nonumber \\
& = \expq {x_L} \cdot \frac{\ln(z)}{1-q} \cdot \left(z^{\frac{1}{1-q}}-1 \right) \cdot \frac{1}{n} \cdot \frac{z^{\frac{1}{n(1-q)}}}{z^{\frac{1}{n(1-q)}}-1} \nonumber
\end{eqnarray}
\begin{eqnarray}
A_q(k_q, u_{L}, u_{H}) &= \lim_{\|U_q\| \rightarrow 0} \sum_{i=1}^{n} A_{(q)i} \\
& = \expq {x_L} \cdot \frac{\ln(z)}{1-q} \cdot \left(z^{\frac{1}{1-q}}-1 \right) \cdot \lim_{n \rightarrow \infty } \Big( \frac{1}{n} \frac{z^{\frac{1}{n(1-q)}}}{z^{\frac{1}{n(1-q)}}-1} \Big) \nonumber \\
& = \expq {x_L} \cdot \frac{\ln(z)}{1-q} \cdot \left(z^{\frac{1}{1-q}}-1 \right) \cdot \frac{1-q}{\ln (z)} \nonumber \\
& = \expq {x_L} \cdot \left( z^{\frac{1}{1-q}}-1 \right) \nonumber \\
& = \expq {x_L} \cdot \left\{ \expq {x_H \qminus x_L} - 1 \right\} \nonumber \\
& = \expq {x_L} \cdot \left\{ \frac{\expq {x_H}}{\expq {x_L}} -1 \right\} \nonumber \\
& = \expq {x_H} - \expq {x_L} \nonumber
\end{eqnarray}
Therefore,
\begin{equation}
\stackrel[{[q]}] {\vphantom{q}} {I} ^{x_H}_{x_L} \Big( \expq x \Big) = F(x_H) -F(x_L) = \expq {x_H} - \expq {x_L}
\end{equation}
If $F(x) = \expq x$, we know that $D_{[q]} \Big( F(x) \Big) = \expq x$ and $F(0) = 1$, for all \emph{q}.
\begin{eqnarray}
F(x) = 1 + \stackrel[{[q]}] {\vphantom{q}} {I} ^{x}_{0} \Big( \expq t \Big) = 1 + \left\{ \expq {x} - \expq {0} \right\} = \expq {x}
\end{eqnarray}
Indefinite primal \emph{q}-integral of the \emph{q}-exponential function can be expressed as follows:
\begin{eqnarray}
F(x) = \stackrel[{[q]}] {\vphantom{q}} {I} \Big( \expq x \Big) = \stackrel[{[q]}] {\vphantom{q}} {I} ^{x}_{x_0} \Big( \expq t \Big) = \expq {x} - \expq {x_0} = \expq {x} + c
\end{eqnarray}
In general, the signed area between the trajectory formed by $(u, v)$ over the range $[x_L, x_H]$ and $v=0$ can be evaluated using the Riemann–Stieltjes integral, $\int^{x_H}_{x=x_L} f(x) \ \mathrm{d} u(x)$ \cite{Lang93}, where $u(x) = \ln(E_{q}(x))$.
Therefore, when $f(x)= D_{[q]} \Big( F(x) \Big)$,
\begin{eqnarray}
\stackrel[{[q]}] {\vphantom{q}} {I} ^{x_H}_{x_L} \Big( f(x) \Big) & = F(x_H) - F(x_L) \label{eqpa3} \\
& = \int^{x_H}_{x=x_L} f(x) \ \mathrm{d} u(x) \nonumber \\
& = \int^{x_H}_{x=x_L} \frac{f(x)}{1+(1-q)x} \ \mathrm{d} x \nonumber
\end{eqnarray}
Equation (\ref{eqpa3}) shows that the new primal \emph{q}-integral, $\stackrel[{[q]}] {\vphantom{q}} {I}$, is equivalent to the primal \emph{q}-integral $I_{(q)}$ in equation (\ref{eqp3}).
Equation (\ref{eq5}) holds for $D_{[q]}$ and $\stackrel[{[q]}] {\vphantom{q}} {I}$ because they are equivalent to $D_{(q)}$ and $I_{(q)}$, respectively.
When $f(x)= D_{[q]} \Big( F(x) \Big)$ and $F(x_0) = -c$, the new primal indefinite \emph{q}-integral can be expressed as follows:
\begin{equation}
\stackrel[{[q]}] {\vphantom{q}} {I} ^{x}_{x_0} \Big( f(t) \Big) = F(x) + c
\end{equation}
\newpage
\subsection{Dual}
\subsubsection{Dual q-derivative}
Let a family of curves $\mathcal{L}^{q}_{(k^q, \cdot)} = \left\{ y=L^{q}(x; k^q, c) \right\}$ have a constant dual \emph{q}-derivative (as defined by equation (\ref{eqp2})), $k^q$, at every point on their domain.
$L^{q}(x; k^q, c)$ must satisfy the condition $D^{(q)} L^{q}(x; k^q, c) = k^q$,
\begin{equation}
D^{(q)} L^{q}(x; k^q, c)
= \frac{1}{1+(1-q)L^{q}(x; k^q, c)} \frac{\mathrm{d}}{\mathrm{d}x} L^{q}(x; k^q, c)
= k^q
\end{equation}
Therefore,
\begin{equation}
\eqalign{
L^{q} (x; k^q, c) & = c' \cdot \exp((1-q)k^q x) - \frac{1}{1-q} \cr
& = \frac{c}{1-q} \exp((1-q)k^q x)- \frac{1}{1-q} \cr
& = \left\{ \frac{1}{1-q} \exp((1-q)k^q x) + \frac{1}{1-q} \right\} + \frac{c-1}{1-q} \cr
& \quad + (1-q) \frac{c-1}{1-q} \left\{ \frac{1}{1-q} \exp((1-q)k^q x) + \frac{c-1}{1-q} \right\} \cr
& = \left\{ \frac{1}{1-q} \exp((1-q)k^q x) + \frac{1}{1-q} \right\} \qplus \frac{c-1}{1-q} \cr
& = \lnq {\exp(k^q x)} \qplus \frac{c-1}{1-q}
}
\end{equation}
Note that ${\mathcal{L}^q}_{(k^q, \cdot)}$ is a $\qplus$ translation family of a curve along the y axis.
Consider the curve $\mathcal{L}^q(k^q, c):y=L^{q}(x; k^q, c)= \lnq { \exp(k^q x)} \qplus \frac{c-1}{1-q}$ as the dual $\emph{q}$-line with dual $\emph{q}$-slope $k^q$ and $y$-intercept $\frac{c-1}{1-q}$.
$L^{q}(x; k^q, c)$ also has the property,
\begin{equation}
\fl \frac{ \ln( \expq {L^{q}(x_i; k^q, c)}) - \ln( \expq {L^{q}(x_j; k^q, c)})}{x_i-x_j}=k^q, \forall x_i, x_j \in \mathbb{R}, x_i \neq x_j .\label{eqda4}
\end{equation}
From this finding, we define a new dual \emph{q}-derivative operator $D^{[q]}$ as
\begin{equation}
D^{[q]} \Big( F(x) \Big) \equiv \lim_{y \rightarrow x} \frac{ \ln( \expq {F(x)}) - \ln( \expq {F(y)})}{x - y}
\end{equation}
Note that
\begin{equation}
D^{[q]} \Big( H(x) \Big) = D^{[q]} \Big( F(x) \Big) \quad \Longleftrightarrow \quad H(x)=F(x) \qplus c \label{eqda1}
\end{equation}
It follows that $D^{[q]} \lnq x = \frac{1}{x}$.
\begin{equation}
\eqalign{
D^{[q]} \lnq x
& = \lim_{y \rightarrow x} \frac{ \ln( \expq {\lnq x}) - \ln( \expq {\lnq y})}{x - y} \cr
& = \lim_{y \rightarrow x} \frac{ \ln(x) - \ln(y)}{x - y} = \frac{1}{x}
}
\end{equation}
$D^{[q]}$ turns out to be equivalent to $D^{(q)}$; that is, $D^{[q]}F(x) = D^{(q)}F(x)$, in general, because
\begin{equation}
\lim_{y \rightarrow x} \frac{ \ln(\expq {F(x)}) - \ln(\expq {F(y)})}{F(x) \qminus F(y)} = 1
\end{equation}
\begin{equation}
\eqalign{
D^{[q]} \Big( F(x) \Big) & \equiv
\lim_{y \rightarrow x} \frac{ \ln( \expq {F(x)}) - \ln( \expq {F(y)})}{x - y} \cr
& = \lim_{y \rightarrow x} \frac{ \ln( \expq {F(x)}) - \ln( \expq {F(y)})}{F(x) \qminus F(y)}
\frac{F(x) \qminus F(y)}{x - y} \cr
& = \lim_{y \rightarrow x} \frac{F(x) \qminus F(y)}{x - y} = D^{(q)}\Big( F(x) \Big) \cr
}
\end{equation}
\begin{figure}[h]
\centering
\begin{tikzpicture}[
declare function={f(\x)= (x^(-0.3)-1)/(-0.3);},
declare function={g(\x)= (((0.5)/exp(1)*exp(2*x))^(-0.3)-1)/(-0.3);},
declare function={h(\x)=
(((((0.5)/exp((0.5)*(1/(3.5)*ln((1+(-0.3)*(((4*2)^(-0.3)-1)/(-0.3)))^(1/(-0.3))))))^(-0.3))^(1/(-0.3))
*exp((1/(3.5)*ln((1+(-0.3)*(((4*2)^(-0.3)-1)/(-0.3)))^(1/(-0.3))))*x))^(-0.3)-1)/(-0.3)
;}
]
\newcommand\XTickA{0.5}
\newcommand\XTickB{4}
\begin{axis}[
axis lines=center,
xmin=-1,xmax=8,
ymin=-3,ymax=4,
width=10cm,
xtick=data,
ytick=\empty,
xticklabel style = {xshift=-0.2cm, above},
yticklabel style = {above right},
xticklabels={$x_0$,$t$},
yticklabels={$F(x_0)$,$F(t)$},
xlabel=$x$,
ylabel=$y$,
smooth
]
\addplot [ycomb,
mark=*,
mark options={black,mark size=2pt},
gray, dashed,
samples at={\XTickA,\XTickB}] {f(x)};
\addplot [mark=none] coordinates {(0,0)} node[below left] {0};
\addplot [name path = fcurve, domain=0.1:5.5] {f(x)} node[right] {$y=F(x)$};
\addplot [red, name path = gcurve, domain=-0.5:6] {g(x)} node[above] {$\mathcal{T}^{q}(C, P)$};
\addplot [blue, name path = hcurve, domain=-0.5:6] {h(x)} node[above] {$\mathcal{L}^{q}(C, P, Q)$};
\addplot [mark=none] coordinates {(\XTickA, {f(\XTickA)})} node[below right] {P};
\addplot [mark=none] coordinates {(\XTickB, {f(\XTickB)})} node[above] {Q};
\end{axis}
\end{tikzpicture}
\caption{Dual \emph{q}-derivative as the slope of the dual \emph{q}-tangent line} \label{fig:F7}
\end{figure}
Let $\mathcal{L}^q(C, P_i, P_j): y = L^q(x; F(x), x_i, x_j)$ be the dual $\emph{q}$-line that passes through the two points $P_i=(x_i, y_i)=(x_i, F(x_i))$ and $P_j=(x_j, y_j)=(x_j, F(x_j))$ on the curve $C: y=F(x)$.
$L^{q}(x; F(x), P_i, P_j))$ turns out to be
\begin{equation}
L^{q}(x; F(x), x_i, x_j))
= \lnq { \exp(k^q x)} \qplus \frac{c-1}{1-q}
\end{equation}
\begin{eqnarray}
k^q = \frac{ \ln( \expq {F(x_i)}) - \ln( \expq {F(x_j)})}{x_i-x_j} \\
\frac{c-1}{1-q} = y_i \qminus \lnq { \exp(k^q x_i)}
\end{eqnarray}
$L^{q}(x; F(x), x_i, x_j))$ can also be expressed as follows:
\begin{eqnarray}
L^{q}(x; F(x), x_i, x_j)) = \lnq {\exp(k^q (x-x_i)} \qplus F(x_i) \\
L^{q}(x; F(x), x_i, x_j)) = \lnq {\exp(k^q (x-x_j)} \qplus F(x_j)
\end{eqnarray}
We call $\mathcal{T}^{q}(C, P): y=T^{q}(x; F(x), x_0)=\lnq {\exp({k^q}_{(C, P)} x)} \qplus \frac{c^q_{(C,P)}-1}{1-q}$ as the dual $\emph{q}$-tangent line of the curve $C: y=F(x)$ at the point $P=(x_0, F(x_0))$ when $F(x_0)=T^{q}(x_0)$ and $D^{[q]} \Big( F(x) \Big) \Big\vert_{x=x_0} = {k^q}_{(C,P)}$.
$D^{[q]} \Big( F(x) \Big) \Big\vert_{x=x_0}$ is the dual $\emph{q}$-slope of the dual $\emph{q}$-tangent line at the point $P$ on the curve $C$.
The dual $\emph{q}$-line $L^{q}(x; C, P, Q)$ passing through $P=(x_0, F(x_0))$ and another point $Q=(t, F(t))$ on the curve $C$ tends to $\mathcal{T}^{q}(C, P)$ as $Q$ approaches $P$, and the dual $\emph{q}$-slope of $L^{q}(x; C, P, Q)$ converges to that of $\mathcal{T}^{q}(C, P)$.
\begin{eqnarray}
\fl D^{[q]} \Big( F(x) \Big) \Big\vert_{x=x_0} & = \lim_{t \rightarrow x_0} \frac{ \ln(\expq {L^{q}(x_0; F(x), x_0, t)}) - \ln(\expq {L^{q}(t; F(x), x_0, t)})}{x_0 - t} \\
& = {k^q}_{(C,P)} \nonumber
\end{eqnarray}
\subsubsection{New dual q-integral}
Let the indefinite dual \emph{q}-integral operator, $\stackrel[\vphantom{q}] {{[q]}} {I}$, be the inverse operator of the dual \emph{q}-derivative operator, $D^{[q]}$.
Because the $\qplus$-translation family of a function has the same \emph{q}-derivative (see equation (\ref{eqda1})), the indefinite dual \emph{q}-integral operator should produce a $\qplus$-translation family of a function.
\begin{equation}
f(x)= D^{[q]} \Big( F(x) \Big) \quad \Longrightarrow \quad \stackrel[\vphantom{q}] {{[q]}} {I} \Big( f(x) \Big) = F(x) \qplus c.
\end{equation}
Let us denote the definite dual \emph{q}-integral of $f(x)$ from $x_L$ to $x_h$ as $\stackrel[\vphantom{q}] {{[q]}} {I} ^{x_H}_{x_L} \Big( f(x) \Big)$.
It is desirable and natural that a definite dual \emph{q}-integral vanishes when the integrating range is zero.
\begin{equation}
\stackrel[\vphantom{q}] {{[q]}} {I} ^{c}_{c} \Big( f(x) \Big) = 0,\qquad \forall c \in \mathbb{R} . \label{eqda2}
\end{equation}
From equation (\ref{eqda2}), the function $F(x)$, with $D^{[q]} \Big( F(x) \Big) = f(x)$ and $F(x_0)=0$, can be represented as follows:
\begin{equation}
F(x) = \stackrel[\vphantom{q}] {{[q]}} {I} ^{x}_{x_0} \Big( f(t) \Big).
\end{equation}
If $H(x)= F(x) \qplus c$, $D^{[q]} \Big( H(x) \Big) = f(x)$ and $H(x_0)= F(x_0) \qplus c = c$. $H(x)$ can be represented as
\begin{eqnarray}
H(x) & = F(x) \qplus c = \stackrel[\vphantom{q}] {{[q]}} {I} ^{x}_{x_0} \Big( f(t) \Big) \qplus c \\
& = \stackrel[\vphantom{q}] {{[q]}} {I} ^{x}_{x_0} \Big( f(t) \Big) \qplus H(x_0) \nonumber
\end{eqnarray}
Therefore,
\begin{eqnarray}
\stackrel[\vphantom{q}] {{[q]}} {I} ^{x}_{x_0} \Big( f(t) \Big) & = H(x) \qminus H(x_0) = \Big( F(x) \qplus c \Big) \qminus \Big( F(x_0) \qplus c \Big) \\
& = F(x) \qminus F(x_0) \nonumber
\end{eqnarray}
Therefore,
\begin{eqnarray}
\stackrel[\vphantom{q}] {{[q]}} {I} ^{x_H}_{x_L} \Big( f(x) \Big) & = F(x_H) \qminus F(x_L) = \lnq { \frac{\expq {F(x_H)}}{\expq {F(x_L)}} } \label{eqda3}
\end{eqnarray}
From equation (\ref{eqda3}), we find that the following relationship holds for the definite dual \emph{q}-integral.
\begin{equation}
\stackrel[\vphantom{q}] {{[q]}} {I} ^{x_H}_{x_L} \Big( f(x) \Big)
= \stackrel[\vphantom{q}] {{[q]}} {I} ^{c}_{x_L} \Big( f(x) \Big)
\qplus \stackrel[\vphantom{q}] {{[q]}} {I} ^{x_H}_{c} \Big( f(x) \Big),\qquad \forall c \in \mathbb{R}
\end{equation}
Equation (\ref{eqda4}) gives us a hint on how a definite dual \emph{q}-integral can be related to a signed area.
Consider the immediate example, the case of dual \emph{q}-lines.
Let $F(x)=L^{q}(x; k^q, c)=\lnq {\exp(k^q x)} \qplus \frac{c-1}{1-q}$ and $f(x)= D^{[q]} \Big( F(x) \Big) = k^q$.
\begin{eqnarray}
\stackrel[\vphantom{q}] {{[q]}} {I} ^{x_H}_{x_L} \Big( k^q \Big)
& = L^{q}(x_H; k^q, c) \qminus L^{q}(x_L; k^q, c) \\
& = \lnq { \frac{\expq {L^{q}(x_H; k^q, c)}}{\expq {L^{q}(x_L; k^q, c)}} } \nonumber
\end{eqnarray}
Let us consider the transformation $(x,Y)=(x, F(x))=(x, L^{q}(x; k^q, c)) \rightarrow (x, w)= (x, \ln ( \expq Y )) = (x, \ln ( \expq {L^{q}(x; k^q, c)} )) $.
\begin{figure}[h]
\centering
\begin{tikzpicture}[
declare function={f(\x)=(-4)*exp(((-0.2)*2.5*\x))+5+21;},
declare function={h(\x)=\x+11;},
declare function={g(\x)= 3;}
]
\newcommand\XTickA{1}
\newcommand\XTickB{5}
\begin{axis}[
axis lines=center,
xmin=-1.2,xmax=10.2,
ymin=-1.2,ymax=28.2,
width=12cm,
xtick=data,
ytick=data,
typeset ticklabels with strut,
yticklabel style = {yshift=0.3cm, scale=0.65},
xticklabels={$x_L$,$x_H$},
yticklabels={$Y_L$,$Y_H$},
xlabel=$x$,
y label style={anchor=east, scale=0.85},
ylabel=$Y$,
domain=-0.5:9,
smooth
]
\addplot [ycomb,
mark=*,
mark options={black,mark size=1pt},
gray, dashed,
samples at={\XTickA,\XTickB}] {f(x)};
\addplot [xcomb, gray, dashed, samples at={\XTickA,\XTickB}] {f(x)};
\addplot [name path = fcurve, red, domain=-0.15:7] {f(x)} node[above, scale=0.65] {$Y=L^{q}(x; k^q,c)= \lnq {exp(kx)} \qplus \frac{c-1}{1-q}$};
\addplot [name path = hcurve1, -stealth, domain=-1.2:10.2] {20} node[above left] {\emph{x}};
\addplot [mark=none] coordinates {(0,20)} node[below left, scale=0.65] {0};
\addplot [mark=none] coordinates {(0,22)} node[left] {$\frac{c-1}{1-q} $};
\addplot [name path = hcurve1, -stealth, domain=-1.2:10.2] {10} node[above left] {\emph{x}};
\addplot [mark=none] coordinates {(0,10)} node[below left, scale=0.65] {0};
\addplot [mark=none] coordinates {(0,17)} node[right, scale=0.65] {$w= \ln (\emph{e}_q(Y))$};
\addplot [mark=none] coordinates {(7.3,15)} node[right, blue, scale=0.65] {$=k^qx + \ln[ \expq {\frac{c-1}{1-q}} ]$};
\addplot [mark=none, white, very thick] coordinates {(0, 18) (0, 18.8)};
\addplot [mark=none, -stealth, black] coordinates {(0, 17.5) (0, 18)};
\addplot [name path = hcurve, blue, domain=-0.5:7] {h(x)} node[below right, scale=0.65] {$w= \ln \left[ \expq {L^{q}(x; k^q, c)} \right]$};
\addplot [mark=none, gray, dashed] coordinates {(0, 12) (1, 12)};
\addplot [mark=none, gray, dashed] coordinates {(0, 16) (5, 16)};
\addplot [mark=*, mark options={black,mark size=1pt}] coordinates {(1, 12)};
\addplot [mark=*, mark options={black,mark size=1pt}] coordinates {(5, 16)};
\addplot [mark=none] coordinates {(0, 12)} node[left, scale=0.65] {$w_L$};
\addplot [mark=none] coordinates {(0, 16)} node[left, scale=0.65] {$w_H$};
\addplot [mark=none, -stealth, gray] coordinates {(0.5, 12) (0.5, 16)};
\addplot [mark=none] coordinates {(0.5,14)} node[right, scale=0.75] {$w_{H}-w_{L}$};
\addplot [mark=none] coordinates {(0,7)} node[left, scale=0.65] {\emph{y}};
\addplot [mark=none, white, very thick] coordinates {(0, 8) (0, 8.8)};
\addplot [mark=none, -stealth, black] coordinates {(0, 7.5) (0, 8)};
\addplot [mark=none] coordinates {(0,0)} node[below left, scale=0.65] {0};
\addplot [name path = gcurve, blue, domain=-0.5:7] {g(x)} node[above, scale=0.65] {$y=D^{[q]} L^{q}(x; k^q, c)$};
\addplot [mark=none] coordinates {(0,3)} node[above left, scale=0.65] {$k^{q}$};
\addplot [name path = g,domain=\XTickA:\XTickB] {g(x)};
\addplot [name path = h,domain=\XTickA:\XTickB] {0};
\addplot [gray!20] fill between[of=g and h];
\addplot [mark=none] coordinates {(1.5,1.5)} node[right, scale=1] {$A^q(k^q, x_{L}, x_{H})$};
\end{axis}
\end{tikzpicture}
\caption{Relationship between the dual \emph{q}-integral and signed dual \emph{q}-area} \label{fig:F8}
\end{figure}
Figure 8 shows the graph of $Y=F(x)$ at the top, the graph of $x$ and $w$ in the middle, and the graph of $y=f(x)$ at the bottom.
With equation (\ref{eqda4}), we find that the signed area of the shaded rectangle at the bottom graph is equal to $w_H-w_L$ in the middle graph.
Let $A(k^q, x_{L}, x_{H})$ be the ordinary signed rectangular area formed by $y=k^q$ and $y=0$ over the range $[x_L, x_H]$,
\begin{eqnarray}
A(k^q, x_L, x_H) & = k^q \cdot ( x_H - x_L ) \\
& = w_H-w_L \nonumber \\
& = \ln \left[ \expq {L^{q}(x_H; k^q, c)} \right] -\ln \left[ \expq {L^{q}(x_L; k^q, c)} \right] \nonumber \\
& = \ln \Big[ \frac{\expq {L^{q}(x_H; k^q, c)}}{\expq {L^{q}(x_L; k^q, c)}} \Big] \nonumber \\
& = \ln \Big[ \expq {\lnq {\frac{\expq {L^{q}(x_H; k^q, c)}}{\expq {L^{q}(x_L; k^q, c)}}}} \Big] \nonumber \\
& = \ln \left[ \expq {L^{q}(x_H; k^q, c) \qminus L^{q}(x_L; k^q, c)} \right] \nonumber \\
& = \ln \left[ \expq {\stackrel[\vphantom{q}] {{[q]}} {I} ^{x_H}_{x_L} \Big( k^q \Big)} \right] \nonumber
\end{eqnarray}
Let $A^q(k^q, x_L, x_H) \equiv \lnq {\exp \left[ A(k^q, x_L, x_H) \right]}=\lnq {\exp \left[ k^q \cdot ( x_H - x_L ) \right]}$ be a dual \emph{q}-area,
\begin{eqnarray}
\stackrel[\vphantom{q}] {{[q]}} {I} ^{x_H}_{x_L} \Big( k^q \Big)
& = \lnq {\exp \left[ A(k^q, x_L, x_H) \right]} \\
& = A^q(k^q, x_L, x_H) \nonumber
\end{eqnarray}
For $f(x)=D^{[q]} \Big( F(x) \Big)$, in general, the ordinary signed area, $A(f(x), x_{L}, x_{H})$, formed by $y=f(x)$ and $y=0$ over the range $[x_L, x_H]$ is equal to the ordinary definite integral $\int^{x_H}_{x_L} f(x) \mathrm{d}x$. Therefore, the following relationship holds true:
\begin{eqnarray}
\stackrel[\vphantom{q}] {{[q]}} {I} ^{x_H}_{x_L} \Big( f(x) \Big)
& = A^q(f(x), x_L, x_H) \label{eqda5}\\
& = \lnq {\exp \left[ A(f(x), x_L, x_H) \right]} \nonumber \\
& = \ln_q \Big( \exp \left[ \int^{x_H}_{x_L} f(x) \mathrm{d}x \right] \Big) \nonumber
\end{eqnarray}
Using equation (\ref{eqda5}), $\stackrel[\vphantom{q}] {{[q]}} {I} ^{x}_{x_0} \Big( \frac{1}{t} \Big)$ is found to be a \emph{q}-logarithmic function,
\begin{eqnarray}
\stackrel[\vphantom{q}] {{[q]}} {I} ^{x}_{x_0} \Big( \frac{1}{t} \Big)
& = \ln_q \Big( \exp \left[ \int^{x}_{x_0} \frac{1}{t} \mathrm{d}t \right] \Big) \\
& = \lnq {\exp \left[ \ln (x) - \ln (x_0) \right]} \nonumber \\
& = \lnq {\frac{x}{x_0} } = \lnq {x} \qminus \lnq {x_0} \nonumber
\end{eqnarray}
If $F(x) = \lnq x$, we know that $D^{[q]} \Big( \lnq x \Big) = \frac{1}{x}$ and $F(1) = 0$ for all \emph{q},
\begin{eqnarray}
F(x) = \stackrel[\vphantom{q}] {{[q]}} {I} ^{x}_{1} \Big( \frac{1}{t} \Big) = \lnq {x} \qminus \lnq {1} = \lnq {x}
\end{eqnarray}
The indefinite dual \emph{q}-integral of $\frac{1}{x}$ can be expressed as follows, and equation (\ref{eq5}) holds with a modification of ordinary addition to \emph{q}-difference ($\qminus$).
\begin{eqnarray}
F(x) = \stackrel[\vphantom{q}] {{[q]}} {I} \Big( \frac{1}{t} \Big)
\stackrel[\vphantom{q}] {{[q]}} {I} ^{x}_{x_0} \Big( \frac{1}{t} \Big)
= \lnq {x} \qminus \lnq {x_0} = \lnq {x} \qminus c.
\end{eqnarray}
Equation (\ref{eq6}) also holds with a modification of ordinary addition to \emph{q}-difference ($\qminus$).
\begin{eqnarray}
D^{[q]} \Big( \stackrel[\vphantom{q}] {{[q]}} {I} ^{x}_{a} \Big( f(x) \Big) \Big) & = D_{[q]} \Big( F(x) \qminus F(a) \Big) = f(x) \\
\stackrel[\vphantom{q}] {{[q]}} {I} ^{x}_{a} \Big( D^{[q]} \Big( F(x) \Big) \Big) & = \stackrel[\vphantom{q}] {{[q]}} {I} ^{x}_{a} \Big( f(x) \Big) = F(x) \qminus c
\end{eqnarray}
\section{Relationship Between the Primal and Dual}
\subsection{Relationship Between Primal and Dual q-derivatives}
Let $G(y)$ be the inverse function of $F(x)$, that is, $y=F(x)$ and $x=G(y)$, and $P=(x_0, y_0)$, $Q=(x_1, y_1)$ be points on the curve $C: y=F(x)$. The curve $C$ can also be represented by $C: x=G(y)$.
Let $\mathcal{L}_q(C, P, Q)$ be a primal \emph{q}-line that passes through $P$ and $Q$.
\begin{equation}
\mathcal{L}_q(C, P, Q): y = k_q \cdot \ln(E_{q}(x \qminus x_1))+ y_1 \label{eqpp1}
\end{equation}
$\mathcal{L}_q(C, P, Q)$ can also be represented as
\begin{equation}
\mathcal{L}_q(C, P, Q): x = \lnq {\exp(\frac{1}{k_q} (y-y_1)} \qplus x_1 \label{eqpp2}
\end{equation}
Equations (\ref{eqpp1}) and (\ref{eqpp2}) show that the primal \emph{q}-line that passes through $P$ and $Q$ is the dual \emph{q}-line passing through $P$ and $Q$, $\mathcal{L}^q(C, P, Q)$, and $k^q$; the dual \emph{q}-slope of $\mathcal{L}^q(C, P, Q)$ is $\frac{1}{k_q}$.
Let $\mathcal{T}_q(C, P)$ and $\mathcal{T}^q(C, P)$ be the primal and dual \emph{q}-tangent lines at $P$ on $C$, respectively, and let ${k_q}^*$ and $k^{q*}$ be the primal and dual \emph{q}-slopes of the corresponding \emph{q}-tangent line.
As $Q$ approaches $P$, $k_q$ converges to ${k_q}^*$, and $k^q$ converges to $k^{q*}$.
Therefore, the primal \emph{q}-derivative and the dual \emph{q}-derivative are inversely related.
\begin{equation}
k^{q*} = \frac{1}{{k_q}^*}.
\end{equation}
\subsection{Relationship Between Primal and Dual q-integrals}
Let $f(x)=D_{[q]} \Big( F(x) \Big)$ and $g(x)=D^{[q]} \Big( G(x) \Big)$,
\begin{eqnarray}
\stackrel[{[q]}] {\vphantom{q}} {I} ^{x_1}_{x_0} \Big( f(x) \Big) = y_1 - y_0 \qquad
\stackrel[\vphantom{q}] {{[q]}} {I} ^{y_1}_{y_0} \Big( g(y) \Big) = x_1 \qminus x_0 \\
\frac{\stackrel[{[q]}] {\vphantom{q}} {I} ^{x_1}_{x_0} \Big( f(x) \Big)}{\stackrel[\vphantom{q}] {{[q]}} {I} ^{y_1}_{y_0} \Big( g(y) \Big)} = \frac{y_1 - y_0}{x_1 \qminus x_0} = k_q = \frac{1}{k^q} \\
\qquad \mbox{where } k_q = \frac{y_1 - y_0}{\ln \left[ E_{q}(x_1)\right]-\ln \left[ E_{q}(x_0) \right]} \nonumber
\end{eqnarray}
Therefore, the definite primal and dual \emph{q}-integrals are proportionally related to the \emph{q}-slope of the \emph{q}-line passing through two points $P$ and $Q$.
\ack
The author would like to thank Prof. Borges for his helpful advice and insights on the issue examined in this study.
\section*{Reference}
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Guest Editorial: Happy birthday, Mr. President; we can still learn from you
One year ago, our state celebrated the 100th birthday of President Ronald Reagan, a true Illinoisan, to recognize his accomplishments and celebrate the prosperity he brought to America.
As we now reflect on what would have been President Reagan’s 101st birthday this past Monday, it’s an ideal time to highlight his signature economic philosophies that our state could emulate as we attempt to navigate our way out of our own financial calamity. Ronald Reagan demonstrated the courage it takes to be an effective public servant.
I first met candidate Reagan soon out of college at Illinois State University, when he asked me to lead his Illinois campaign. My passion for politics and my understanding of his message of lower taxes and less government spending made my answer an easy “yes.” As the Illinois state director for then-Governor Reagan’s first presidential campaign in 1980, his wife Nancy and I traveled to many parts of Illinois in a station wagon with a California State Trooper to campaign.
America’s financial turnaround overseen by President Reagan after some very dismal years in the 1970s can help illuminate a way forward for our state leaders to avoid further financial problems in Illinois. Ronald Reagan governed from the perspective that economies flourish and jobs are created when taxes are lowered, regulations on job creators are lessened and government spending is reduced. He often left room for compromise to avoid gridlock, but he always stayed true to his fiscal beliefs. For the state to recover from its financial situation, leaders in Illinois should follow Reagan’s lead and adopt these same principles.
President Reagan was the only U.S. president born and raised in Illinois. He was born in Tampico, spent his formative years in Dixon and attended college in Eureka. The character he developed in Illinois would eventually position him as the most influential leader of the free world.
Happy birthday, Mr. President. Thank you for the life lessons. We in Illinois can learn from you.
Dan Rutherford
Illinois State Treasurer
| 221,502
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TITLE: A topological property that holds by countable union/intersection but not by uncountable union/intersection?
QUESTION [2 upvotes]: I was wondering about a problem related to regions in differentiable manifolds and I wanted to use an induction. Firstly I didn't notice that in an open set only could be countable connected components, so I was thinking to use an induction in a union with uncountable index.
My question is: Can you give me an example of a topological property that holds by countable union/intersection but not by uncountable union/intersection?
Thank you.
REPLY [3 votes]: A set is meager if it is the countable union of nowhere dense sets. While meager-ness is preserved under countable unions (since countable $\times$ countable = countable), it is not preserved under arbitrary uncountable unions. For example, in $\mathbb{R}$ the whole space is not meager$^*$, but $$\mathbb{R}=\bigcup_{x\in\mathbb{R}}\{x\}$$ and every singleton is meager in $\mathbb{R}$.
$^*$This is not obvious! The Baire category theorem states that $\mathbb{R}$ with the usual topology, and many other spaces, is not meager. The Baire category theorem has many applications; I don't think I can write a good summary here, but if you google around (and search on this site) you'll find many uses of it.
Relatedly, there is a rich hierarchy of "types of set" beyond just open and closed. For example, we inductively define the classes $\Sigma_\alpha^0$ and $\Pi^0_\alpha$ for $\alpha$ a countable ordinal) as:
$\Sigma^0_1$= open, $\Pi^0_1$= closed.
$\Sigma^0_\alpha$ is the set of countable unions of sets, each of which is $\Pi^0_\beta$ for some $\beta<\alpha$; and $\Pi^0_\alpha$ is the set of countable intersections of sets, each of which is $\Sigma^0_\alpha$ for some $\beta<\alpha$.
(Note that $X$ is $\Sigma^0_\alpha$ iff the complement of $X$ is $\Pi^0_\alpha$. A set is $\Delta^0_\alpha$ if it is both $\Sigma^0_\alpha$ and $\Pi^0_\alpha$.)
This hierarchy - the Borel hierarchy - makes sense for arbitrary spaces. For $\mathbb{R}$ with the usual topology, and related spaces, it is particularly nice: for example, in $\mathbb{R}$ with the usual topology we have that every $\Sigma^0_\alpha$ (resp. $\Pi^0_\alpha$) set is $\Sigma^0_\gamma$ (resp. $\Pi^0_\gamma$) whenever $\alpha<\gamma$, and this hierarchy is strict ($\Sigma^0_\gamma$ properly contains $\Sigma^0_\alpha$ for $\gamma>\alpha$). We can of course go beyond the Borel hierarchy (for example, via the projective hierarchy) but already the Borel hierarchy contains basically every set of reals you can easily explicitly describe.
Each level of the Borel hierarchy has some amount of closure: e.g. $\Sigma^0_\alpha$ is closed under countable unions but not intersections and $\Pi^0_\alpha$ is closed under countable intersections but not unions. It's a good exercise to show that "countable" is necessary here in general (think about $\alpha=1$). So these provide additional examples.
Incidentally, you may find topics in descriptive set theory of particular interest; Kechris' book is an excellent introduction.
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