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TOURS.
The Malbork Castle consists of three separate castles: the High Castle, the Middle Castle and the Low Castle also called Przedzamcze. This object is worth seeing not only because of the interesting history and architecture. Since 1961, the Castle Museum has been thriving here, which has extensive documentation on the history of the castle, as well as unusual collections of amber, ceramics, textiles, paintings, engravings, sculptures, furniture, military items and many others.
ADDITIONAL INFORMATION:
- trip duration – 5h
- suggested departure time – 09:00 A.M.
- opening hours in the summer season from 09:00 A.M. to 07:00 P.M., in the winter season from 10:00 A.M. to 03:00 P.M.
- audio guides are available in Polish, English, German, Russian, French, Italian and Spanish, and for deaf people in sign language
- with prior reservation, it is possible to rent a guide, its price depends on the size of the group
PRICES:
- Group up to max. 3 people – Mercedes E Class w213 sedan – 145 Euro
- Group up to max. 7 people – Mercedes V Class minivan – 165 Euro
THE PRICE INCLUDES:
- private transport from a hotel or anywhere in Gdansk, Sopot or Gdynia, free time for visiting the castle in Malbork and surroundings and return transport
- a cultural, elegant and English-speaking driver with all licenses and permissions
- help with purchasing castle entrance tickets and organizing a guide
- water and Wi-Fi
- passenger insurance
- taxes, highway, parking lots, etc.
THE PRICE DOES NOT INCLUDE:
- entrance tickets to the castle
- guide fees
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Julie Andrews & Dike Van Dyke Signed Walt Disney's "Mary Poppins" 30.5x44.5 Custom Framed Display
- Lot number 4257006
- Total bids 0
- High bid $3,000.00
- Remaining time 4d 0h 8m
- ELITE NO RESERVE
Custom frame measures 30.5" x 44.5" in size, features an index card hand-signed by Julie Andrews, and has been hand-signed by Dick Van Dyke directly on the poster..
| 63,185
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WebMD Medical News
By
Denise Mann
Reviewed by
Louise Chang, MD.
.
SOURCES:Whitehouse, A. Pediatrics, 2011.Andrew Adesman, MD, chief, developmental & behavioral pediatrics, Steven & Alexandra Cohen Children's Medical Center of New York, New Hyde Park.Melissa Wexler Gurfein, speech-language pathologist, New York.Rahil Briggs, PsyD, infant-toddler psychologist/behavioral expert, Children' Hospital at Montefiore, New York.
Here are the most recent story comments.View All
The views expressed here do not necessarily represent those of WETM 18 Online
The Health News section does not provide medical advice, diagnosis or treatment. See additional information.
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TITLE: Solving limit without L'Hôpitale rule or infinity serieses
QUESTION [1 upvotes]: I've got this limit
$$
{
\lim_{x\to 0} \left( \frac{e\cos(x)-e}{\ln(1+\ln(2-e^{sin^2(x)}))} \right)
\
}$$
I tried to remove NaN situation by
$t=ln(2-e^{1-cos^2(x)}$).
But the problem was how can I know $1-cos(x)$ value .
So if there was any hint, please share it with me.
Thanks
REPLY [0 votes]: Let me first note that $ \lim\limits_{x \to 0} (\cos(x)+1) = 2$, or thus $ \lim\limits_{x \to 0} \frac{\cos(x)+1}{2} = 1$.
Assuming the limit you posted exists and is finite, we have:
$$\begin{align}
\lim_{x\to 0} \frac{e\cos(x)-e}{\ln(1+\ln(2-e^{\sin^2(x)}))}
&= e \left(\lim_{x\to 0} \frac{\cos(x)-1}{\ln(1+\ln(2-e^{\sin^2(x)}))} \right)\left(\lim_{x \to 0}\frac{\cos(x)+1}{2}\right)\\
&= \frac{e}{2} \lim_{x\to 0} \frac{\cos^2(x)-1}{\ln(1+\ln(2-e^{\sin^2(x)}))} \\
&= -\frac{e}{2} \lim_{x\to 0} \frac{\sin^2(x)}{\ln(1+\ln(2-e^{\sin^2(x)}))}\\
&= -\frac{e}{2} \lim_{t\to 0^+} \frac{t}{\ln(1+\ln(2-e^t))}
\end{align} \tag{1}
$$
where $t = \sin^2(x)$.
Now, the limit for $t$ you are left with can be solved in the way @ReneSchipperus showed in his answer:
$$\begin{align}\lim_{t\to 0^+} \frac{t}{\ln(1+\ln(2-e^t))} &= \lim_{t\to 0^+}\left( \frac{\ln(2-e^t)}{\ln(1+\ln(2-e^t))} \cdot \frac{1-e^t}{\ln(2-e^t)} \cdot \frac{t}{1-e^t}\right)\\
&= \left(\lim_{t\to 0^+} \frac{\ln(2-e^t)}{\ln(1+\ln(2-e^t))}\right) \cdot \left( \lim_{t\to 0^+}\frac{1-e^t}{\ln(2-e^t)}\right) \cdot \left(\lim_{t\to 0^+} \frac{t}{1-e^t}\right)\\
&= \left(\lim_{u\to 0^+} \frac{u}{\ln(1+u)}\right) \cdot \left( \lim_{v\to 0^+}\frac{v}{\ln(1+v)}\right) \cdot \left(-\lim_{w\to 0^+} \frac{\ln(1+w)}{w}\right)\\
\end{align}\tag{2}$$
I substituted $u = \ln(2-e^t)$, $v = 1-e^t$ and $w = 1-e^t$.
Now, the limit
$$\lim_{x\to 0} \frac{\ln(1+x)}{x} = 1 = \lim_{x\to 0} \frac{x}{\ln(1+x)}$$
is well-known, hence the limit in $(2)$ becomes $-1$ and thus the limit in $(1)$ becomes $\frac{e}{2}$.
| 176,290
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In a truly remarkable feat, Illinois has now gone almost 12 months without a budget. You read that right, 12 months. The budget impasse continues to drag on and House Democrat Leadership is sending signals they are not interested in real solutions as we near adjournment on May 31. On Friday, the House adjourned to take Saturday off despite the willingness of my colleagues and I to stay and work on a budget compromise.
Instead of working in a bipartisan manner towards a balanced budget, Chicago Democrats are instead playing games. On May 25, House Democrats rammed Senate Bill 2048 through the House, a $40 billion out of balance budget that would spend $7 billion more than we take in. This is a fantasy budget, it’s not real and we can’t afford it. This proposed budget is irresponsible and would only serve to plunge Illinois even further into the red. Your household has to live within its means, but Illinois continues to fail to do so. The simple truth is we cannot continue to spend money we just don’t have.
The lack of a budget is endangering our most vulnerable citizens with funding for programs such as autism, epilepsy, domestic violence and senior services jeopardized by the budget impasse. Many of our human services groups have been brought to the brink due to delays in payments that many are struggling to endure. We simply cannot allow our human service infrastructure to crumble due to the whims of the one party controlling the legislature.
It is important to note that there have been glimmers of hope in the budget making process. In April the legislature worked in a bipartisan manner to appropriate $600 million to fund higher education and MAP grants. This helped to avert significant university cutbacks and ensured that students would not lose their grant money, and on May 12 the legislature approved $700 million in stopgap funding to keep vital human services operating in the State of Illinois on a short-term basis until a complete budget is enacted.
The action on these two bipartisan bills are proof that the General Assembly can come together in a bipartisan manner to get work done and I hope that this bodes well for future budget negotiations.
However, time is short. If we reach June 30th without a balanced budget we will compound our problems by not having a budget for two consecutive fiscal years. Our state cannot continue in this manner.
Twelve months is far too long to wait, we need a balanced budget now.
Rep. Dwight Kay
112th Illinois House District
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Statistics is the science dedicated to the collection, organization, analysis, and interpretation of data. With the development of technology the collection of data has become ubiquitous, and it is vitally important that educated people be able to make sense of such large amounts of data.
If you have other suggestions for this page or any other pages on this website, please go here and enter your thoughts.
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ChoopyTrags (choopytrags) wrote in 4weekfilm, 2003-12-10 19:38:00 ChoopyTrags choopytrags 4weekfilm 2003-12-10 19:38:00 Previous Share Next Mood: mischievous The Turnip cries for Blood.... The hot beef injection was too much for the vegetarian to take. The syringe filled with boiled dead cows blood coarsed thru his veins as he screamed RUTABAYGA! RUTABAYGA! at the top of his lungs. It was amazing he was held down so long since he was dressed as a giant turnip. The Anti Vegan Society, a secret paramilitary black section of the government which took pride in turning these pansies back into men, had been scoping him out while he paraded himself in front of the Annual Carnivore Lovers parade, doing cartwheels and passing out turnip treats to the kids all week long. They had seen him arguing in front of butcher shops in loud tones until he was chased down the street by the butcher with a meat cleaver. They'd seen him kick men eating shishkebabs and slapping asians for eating dogkebobs. This menace had to be stopped, if the parade was to be sucessful and the eventual plan to bring blood lust to humankind a reality. They found him inside a supermarket dancing with another of his ilk, a woman dressed as a giant Carrot doing the twist to a crowd of awed spectators, who just wanted to get their lettuce and leave but these two whackos were in the way and so they had to watch them twist for three minutes while the muzak tweaked speakers blared out Chubby Checker inside the supermarktet.Suddenly they grabbed The Turnip and pulled him off to the back of the market in the butchers section and tied him down. "You will never eat a vegetable again freak" one of the agents dressed in black said, while pushing in his ear piece.Agent #2 got the cows blood ready,cooking it up under a ladle with a match and pulling it into the syringe. "No way," screamed the turnip," I love Ruttabaga,RUTTABAGA!". But it was all in vain and in his veins, the blood is delight no Mr. Renfield, The Turnips eyes turned red and his incisors grew, they released the harnesses holding him and he turned to one side and expelled all the vegetables he had in his system. He hunched over the best he could for a man dressed in a Turnip costume, and began breathing heavily, his nails grew long and he went into fits saying "I want some moooooo
| 352,664
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In addition to the top 13 teams, we will also acknowledge a "dubious achievement" for the college football weekend.
This week, because there is a huge game on a Monday night we chose to put the Boise State/Virginia Tech winner and loser in their respective slots rather than try to have each person pick the winner and make a mess of everything.
With that, here is the first week of rankings and dubious achievements.
*************************************************************
No. 1 --- Alabama
No. 2 --- Ohio State
No. 3 --- Boise St./Virginia Tech winner
No. 4 --- Nebraska
No. 5 --- Texas
No. 6 --- Oklahoma
No. 7 --- TCU
No. 8 --- Oregon
No. 9 --- Florida
No. 10 --- Iowa
No. 11 --- Wisconsin
No. 12 --- Miami
No. 13 --- Boise St./Virginia Tech loser
Dubious Achievement Awards:
Vettel: The Florida Gator offense. In 30-plus years of covering college football I don't think I've ever seen a "quality" team manage just 25 yards in three quarters of action. Actually, I can't remember a high school team that had just 25 yards that deep into a game.
Redman: USC's defense. In Lane and Monte Kiffin's debut as USC coaches, Hawaii tallied 588 yards and 36 points. The Warriors had 459 passing yards including 190 yards and two touchdowns from back up quarterbacks. The USC defense has about 20 five-star prospects compared to none for the Warriors. Any chance Dillon Baxter can play defense?
McKeeman: Southern Miss coach Larry Fedora. At least Jeff Bower could get his defense to hit a little harder. Fedora will be more of the same old sorry stuff for Southern Miss.
Jones: The yet-to-be-named North Carolina tutor. It was hard to watch the fourth quarter of the Tar Heels' game against LSU and wonder what might have been if they were at full strength.
FightinGators.com Top 13 -- Week 1
This season, the intrepid (and highly questionable) FightinGators.com Gainesville staff will be producing a weekly ranking of the top teams in college football. I know it's not any fun to yell at the national polls, but perhaps one closer to home will give you a good opportunity to vent.
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\begin{document}
\begin{abstract}
In the framework of reverse mathematics, we provide a lower bound on the statement that the order with three incomparable elements is a better quasi order. Specifically, we show that this statement implies the well foundedness of~$\varepsilon_0$ over~$\mathsf{RCA}_0$ and is thus unprovable in~$\mathsf{ACA}_0$.
\end{abstract}
\keywords{Better quasi order, reverse mathematics, unprovability, 3 is bqo}
\subjclass[2020]{06A07, 03B30, 03F15, 03F35}
\maketitle
\section{Introduction}
The notion of better quasi order ($\bqo$) is due to C.~Nash-Williams~\cite{nash-williams-bqo} and has proved extremely fruitful, e.\,g., in R.~Laver's~\cite{laver71} proof of R.~Fra\"iss\'e's conjecture that the scattered linear orders are well quasi ordered by embeddability. From the perspective of reverse mathematics, many questions about better quasi orders remain wide open, as one can see in a survey by A.~Marcone~\cite{marcone-survey-new} (to which we also refer for definitions and notation). In particular, we appear to have no lower bound on the statement that $3$ (the order with three incomparable elements) is $\bqo$, while the known upper bound is provided by~$\mathsf{ATR}_0$. The possibility that a strong theory may be needed to prove a property of such a small finite object seems quite intriguing. It gains additional relevance from A.~Montalb\'an's~\cite{montalban-fraisse} proof that (the countable case of) Laver's result can be established in~$\Pi^1_1\textsf{-CA}_0$, in which the (stronger but apparently related) statement that~$3$ is $\Delta^0_2$-$\bqo$ plays a crucial role.
Both Marcone and Montalb\'an have asked whether the statement that $3$ is $\bqo$ has a proof in any theory that is weaker than~$\mathsf{ATR}_0$ (see Questions~3.3 and~25 of \cite{marcone-survey-old} and~\cite{montalban-open-problems}, respectively). In the present paper, we narrow down the search for an answer, by showing that no proof in~$\aca_0$ can exist. More precisely, we show that the well foundedness of a standard notation system for~$\varepsilon_0$ (the proof theoretic ordinal of $\aca_0$) follows from the assumption that $3$ is $\bqo$, over the base theory~$\rca_0$. Let us note that our approach does not exclude any $\omega$-models, since well foundedness is a $\Pi^1_1$-property, while being $\bqo$ is $\Pi^1_2$-complete (as shown by Marcone~\cite{marcone-Pi12-complete}; cf.~also the discussion after Question~3.11 of~\cite{marcone-survey-new}).
Our argument employs the collection $H_f(Q)$ of all hereditarily finite sets over a given set~$Q$ of urelements. As we show in Section~\ref{sect:H_f}, the theory~$\rca_0$ proves that if $Q$ is $\bqo$ then so is~$H_f(Q)$ with a suitable order. On the one hand, this result is inspired by work of T.~Forster~\cite{forster-bqo}, in which $\bqo$s are characterized via transfinite hierarchies of powersets. On the other hand, it generalizes the result that $\rca_0$ proves closure of $\bqo$s under non-iterated powersets, which is due to Marcone (see Theorem~5.4 of~\cite{marcone-survey-old}). From our result on~$H_f(Q)$ we derive that if $3$ is $\bqo$ then so is any finite discrete order $n$, provably in~$\rca_0$. The point is that the quantification over~$n\in\mathbb N$ is internal. The corresponding result for arbitrary but externally fixed~$n$ is also due to Marcone (see the proof of Theorem~5.11 in~\cite{marcone-survey-old}). In Section~\ref{sect:eps_0} we show that any initial segment of $\varepsilon_0$ admits an order reflecting map into one of the~$H_f(n)$. This relates to another argument of Marcone (see the proof of Theorem~5.10 in~\cite{marcone-survey-old}) and its generalization by the present author~\cite{freund-higman-bqo}. Together with the previous results, it allows to derive the well foundedness of~$\varepsilon_0$ from the assumption that $3$ is $\bqo$.
\section{Hereditarily finite sets over $\bqo$s}\label{sect:H_f}
Working in $\rca_0$, we show that if $Q$ is $\bqo$ then so is the collection $H_f(Q)$ of hereditarily finite sets over~$Q$, with a suitable order relation. Over the same base theory, we conclude that if $3$ is $\bqo$ then so is any finite discrete order~$n$ (where the quantification over~$n\in\mathbb N$ is internal).
In the following definition, it is assumed that clauses~(i) and~(ii) produce distinct elements (as if we wrote $\langle 0,q\rangle$ and $\langle 1,\{a_0,\ldots,a_{n-1}\}\rangle$ for $q$ and $\{a_0,\ldots,a_{n-1}\}$).
\begin{definition}\label{def:H_f}
The collection $H_f(Q)$ of hereditarily finite sets over a given set~$Q$ is recursively generated as follows:
\begin{enumerate}[label=(\roman*)]
\item for each $q\in Q$ we have $q\in H_f(Q)$,
\item given any $a_0,\ldots,a_{n-1}\in H_f(Q)$, we get $\{a_0,\ldots,a_{n-1}\}\in H_f(Q)$.
\end{enumerate}
When $Q$ comes with a quasi order~$\leq$, the binary relation $\leq^\flat$ on $H_f(Q)$ is defined by the following recursive clauses:
\begin{align*}
p\leq q\quad&\Rightarrow\quad p\leq^\flat q,\\
\exists j<n:p\leq^\flat b_j\quad&\Rightarrow\quad p\leq^\flat\{b_0,\ldots,b_{n-1}\},\\
\forall i<m:a_i\leq^\flat q\quad&\Rightarrow\quad\{a_0,\ldots,a_{m-1}\}\leq^\flat q,\\
\forall i<m\exists j<n:a_i\leq^\flat b_j\quad&\Rightarrow\quad\{a_0,\ldots,a_{m-1}\}\leq^\flat\{b_0,\ldots,b_{n-1}\}.
\end{align*}
To each $a\in H_f(Q)$ we assign a finite support set $\supp(a)\subseteq Q$ that is recursively explained by $\supp(q):=\{q\}$ and $\supp(\{a_0,\ldots,a_{n-1}\}):=\bigcup_{i<n}\supp(a_i)$.
\end{definition}
The notation $\leq^\flat$ is traditional for the analogous relation on the non-iterated powerset, which is called Hoare order (see the paragraph before~\cite[Definition~6.4]{marcone-survey-new}).
\begin{lemma}\label{lem:H_f-basic}
For any quasi order~$Q$, the following holds:
\begin{enumerate}[label=(\alph*)]
\item The relation $\leq^\flat$ is a quasi order on~$H_f(Q)$.
\item For $a\in b\in H_f(Q)$ we have $a\leq^\flat b$.
\item If we have $a\leq^\flat b$, then any $p\in\supp(a)$ admits a $q\in\supp(b)$ with $p\leq q$.
\end{enumerate}
\end{lemma}
\begin{proof}
To give a precise formulation of inductive arguments, it can be convenient to consider the height function $\hth:H_f(Q)\to\mathbb N$ that is given by
\begin{equation*}
\hth(q):=0\quad\text{and}\quad\hth(\{a_0,\ldots,a_{n-1}\}):=1+\max(\{0\}\cup\{\hth(a_i)\,|\,i<n\}).
\end{equation*}
In part~(a), a straightforward induction over~$\hth(a)+\hth(b)+\hth(c)$ will confirm that $a\leq^\flat b$ and $b\leq^\flat c$ entail $a\leq^\flat c$. Another induction yields~$a\leq^\flat a$. Part~(b) can be established by induction on~$\hth(a)$. Indeed, assume that we have $b=\{b_0,\ldots,b_{n-1}\}$ and $a=b_j$ with $j<n$. When we have $a=\{a_0,\ldots,a_{m-1}\}$ and hence $a_i\in b_j$ for every $i<m$, the induction hypothesis yields $a_i\leq^\flat b_j$. We thus get $a\leq^\flat b$ by the fourth implication from Definition~\ref{def:H_f}. In case we have $a\in Q$, we can conclude by the second implication. Part~(c) can be checked by induction on~$\hth(a)+\hth(b)$.
\end{proof}
As mentioned in the introduction, the following can be seen as a special case of the transfinite powerset construction or, from a different perspective, as a generalization of the result for non-iterated powersets (see~\cite{forster-bqo} and~\cite[Theorem~5.4]{marcone-survey-old}). For notation that is not explained here, we refer to the aforementioned survey~\cite{marcone-survey-new}.
\begin{theorem}\label{thm:H_f}
If $(Q\leq)$ is $\bqo$ then so is $(H_f(Q),\leq^\flat)$, provably in~$\rca_0$.
\end{theorem}
\begin{proof}
To establish the contrapositive, we assume that $g:B\to H_f(Q)$ is a bad array for some barrier~$B$. Let us put
\begin{equation*}
B^\omega:=\textstyle\bigcup_{n\in\mathbb N}B^{n+1}\quad\text{with}\quad B^{n+1}:=\{s_0\cup\ldots\cup s_n\,|\,s_i\in B\text{ and }s_0\tl\ldots\tl s_n\}.
\end{equation*}
Note that $s\in B^\omega$ is decidable, since $s\in B^{n+1}$ can only hold when $n+1$ is bounded by the length of~$s$. In the rest of this proof, we implicitly assume $s_i\in B$ as well as $s_0\tl\ldots\tl s_n$ whenever we write $s_0\cup\ldots\cup s_n$. These assumptions entail that $n$ and the $s_i$ are uniquely determined, as $B$ is a barrier. To see this, note that removing the $i$~smallest elements of $s_0\cup\ldots\cup s_n$ yields $s_i\cup\ldots\cup s_n$, of which $s_i$ is an initial segment that is proper for~$i<n$. By the same reasoning we get
\begin{equation*}
s_0\cup\ldots\cup s_m\tl t_0\cup\ldots\cup t_n\quad\Rightarrow\quad s_{i+1}=t_i\text{ for }i\leq\min\{m-1,n\}.
\end{equation*}
For $s=s_0\cup\ldots\cup s_{n+1}$ we put $s^0:=s_0\cup\ldots\cup s_n$ and $s^1:=s_1\cup\ldots\cup s_{n+1}$. We note that this yields $s^0\tl s^1$. Furthermore, when $s$ and~$t$ lie in the same set $B^{n+2}$, the implication above ensures that $s\tl t$ implies $s^1=t^0$. We now construct a function $h:B^\omega\to H_f(Q)$ by recursion over sequences: For~$s\in B^1=B$ we put~$h(s):=g(s)$. Next, we consider $s\in B^{n+2}$. If we have $h(s^0)\in Q\subseteq H_f(Q)$ or $h(s^0)\leq^\flat h(s^1)$, then we set $h(s):=h(s^0)$. Otherwise, the definition of $\leq^\flat$ ensures that we can pick a value $h(s)\in h(s^0)$ with the following properties:
\begin{enumerate}[label=(\roman*)]
\item if $h(s^1)$ is an element of $Q\subseteq H_f(Q)$, then we have $h(s)\not\leq^\flat h(s^1)$,
\item otherwise we have $h(s)\not\leq^\flat b$ for all $b\in h(s^1)$.
\end{enumerate}
By induction on~$n$, we now show that $s\tl t$ entails $h(s)\not\leq^\flat h(t)$ when $s$ and $t$ lie in the same set~$B^{n+1}$. For $n=0$ it suffices to invoke the assumption that~$g$ is bad. In the induction step we first assume that we have $h(s^0)\in Q$ and hence $h(s)=h(s^0)$. By construction we have either $h(t)=h(t^0)$ or $h(t)\in h(t^0)$, so that the previous lemma yields $h(t)\leq^\flat h(t^0)$. So if we had $h(s)\leq^\flat h(t)$, we would get $h(s^0)\leq^\flat h(t^0)$. The latter contradicts the induction hypothesis, since we have $s^0\tl s^1=t^0$. Now assume that we have $h(s^0)\notin Q$. The induction hypothesis ensures $h(s^0)\not\leq^\flat h(s^1)$, so that we have $h(s)\in h(s^0)$ with properties~(i) and~(ii) from above. In case we have $h(s^1)=h(t^0)\in Q$, we get $h(s)\not\leq^\flat h(t^0)=h(t)$ by~(i). For $h(t^0)\notin Q$, we invoke the induction hypothesis to get $h(t^0)\not\leq^\flat h(t^1)$ and hence $h(t)\in h(t^0)=h(s^1)$. We~then obtain $h(s)\not\leq^\flat h(t)$ by property~(ii). For $t\in B$ we now set
\begin{equation*}
n(t):=\max\big\{\hth(h(s))\,|\,s\in B\text{ and }s\subseteq\{0,\ldots,\max(t)\}\big\},
\end{equation*}
where $\hth:H_f(Q)\to\mathbb N$ is the height function from the previous proof. Then
\begin{equation*}
B':=\{s_0\cup\ldots\cup s_n\,|\, n=n(s_0)\}
\end{equation*}
is a barrier with the same base as~$B$. Indeed, $s_0\cup\ldots\cup s_m\subset t_0\cup\ldots\cup t_n$ in $B'$ would yield $\max(s_0)\geq\max(t_0)$ and hence $m=n(s_0)\geq n(t_0)=n$. We could then use induction on $i\leq n$ to get $s_i\cup\ldots\cup s_m\subset t_i\cup\ldots\cup t_n$ and in particular $s_n\subset t_n$, against the assumption that $B$ is a barrier. By construction we have
\begin{alignat*}{3}
h(s_0\cup\ldots\cup s_{i+1})&=h(s_0\cup\ldots\cup s_i) \quad &&\text{when}\quad h(s_0\cup\ldots\cup s_i)\in Q,\\
h(s_0\cup\ldots\cup s_{i+1})&\in h(s_0\cup\ldots\cup s_i)\quad &&\text{otherwise}.
\end{alignat*}
Hence the height of $h(s_0\cup\ldots\cup s_i)$ decreases with~$i$ until an element of~$Q$ is reached. In view of $\hth(h(s_0))\leq n(s_0)$, it follows that $h$ restricts to a function from~$B'$ to~$Q$. To complete our proof of the contrapositive, it remains to show that this restriction is bad, i.\,e., that $s\tl t$ with $s,t\in B'$ entails $h(s)\not\leq h(t)$. We write $s=s_0\cup\ldots\cup s_m$ and $t=t_0\cup\ldots\cup t_n$. Note that we have $s_0\tl t_0$ and hence $\max(s_0)<\max(t_0)$, so that we get $m=n(s_0)\leq n(t_0)=n$. We can establish $h(s)\not\leq^\flat h(t_0\cup\ldots\cup t_i)=:b_i$ by induction on $i\in\{m,\ldots,n\}$. The case of~$i=m$ is covered by the above, since it concerns two elements from the same set $B^{m+1}$. The induction step reduces to $b_{i+1}\leq^\flat b_i$, which holds by the previous lemma, as we have $b_{i+1}=b_i$ or $b_{i+1}\in b_i$. For $i=n$ we obtain $h(s)\not\leq^\flat h(t)$ and hence $h(s)\not\leq h(t)$ in~$Q$.
\end{proof}
For $n\in\mathbb N$, the discrete order with underlying set $\{0,\ldots,n-1\}$ will itself be denoted by~$n$, as usual. To keep $\leq^\flat$ as notation for the induced order on~$H_f(n)$, we write~$\leq$ for the discrete order on~$n$, even though it coincides with equality.
\begin{proposition}\label{prop:H3-naturals}
There is an order embedding~$\mathbb N\ni n\mapsto\overline n\in H_f(3)$.
\end{proposition}
\begin{proof}
For the course of this proof, we declare that $a'\in H_f(3)$ for given $a\in H_f(3)$ denotes the element that is recursively defined by
\begin{equation*}
0':=1,\quad 1':=2,\quad 2':=0,\quad a':=\{a_0',\ldots,a_{n-1}'\}\,\,\text{for}\,\,a=\{a_0,\ldots,a_{n-1}\}.
\end{equation*}
A straightforward induction shows $a'''=a$. If $a,a',a''$ are pairwise incomparable, the same is clearly true for~$\{a,a'\},\{a,a''\},\{a',a''\}$. Now we recursively declare
\begin{equation*}
\overline 0:=\{0,1\}\quad\text{and}\quad\overline{n+1}:=\{\overline n,\overline n'\}.
\end{equation*}
Due to the above, we inductively learn that $\overline n,\overline n',\overline n''$ are pairwise incomparable. It follows that we have $\overline{n+1}\not\leq^\flat\overline n$, for otherwise we would get $\overline n'\leq^\flat a$ for some $a\in\overline n$. The latter entails $a\leq^\flat\overline n$ by Lemma~\ref{lem:H_f-basic}, so that we would obtain $\overline n'\leq^\flat\overline n$. In view of $\overline n\in\overline{n+1}$ we also get $\overline n\leq^\flat\overline{n+1}$. For $m<n$ we can now conclude $\overline m\leq^\flat\overline n$ as well as $\overline n\not\leq^\flat\overline m$ (otherwise $\overline{m+1}\leq^\flat\overline n\leq^\flat\overline m$).
\end{proof}
As explained above, the point of the following result is that the quantification over~$n$ is internal to the theory~$\rca_0$. The corresponding result for arbitrary but externally fixed~$n$ has been established by Marcone~\cite{marcone-survey-old}.
\begin{corollary}\label{cor:fin-bqo}
If $3$ is $\bqo$ then so is $n$ for every~$n\in\mathbb N$, provably in~$\rca_0$.
\end{corollary}
\begin{proof}
Given that $3$ is $\bqo$, we can infer that $4$ and $H_f(4)$ are $\bqo$ as well, by the aforementioned result of Marcone (see the proof of Theorem~5.11 in~\cite{marcone-survey-old}) and Theorem~\ref{thm:H_f} above. It remains to show that $H_f(4)$ contains antichains of arbitrary size~$n\in\mathbb N$. Since $H_f(3)$ is a suborder of $H_f(4)$, we can take the previous proposition to yield an embedding $\mathbb N\ni i\mapsto\overline i\in H_f(4)$ with $\emptyset\neq\supp(\overline i)\subseteq\{0,1,2\}$. The latter ensures that $3\in 4\subseteq H_f(4)$ is incomparable with any element of the form~$\overline i$, due to part~(c) of Lemma~\ref{lem:H_f-basic}. For $i<n$ we now put
\begin{equation*}
a_i:=\left\{\left\{\overline i,3\right\},\left\{\overline{2n-i}\right\}\right\}\in H_f(4).
\end{equation*}
It is straightforward to see that $a_0,\ldots,a_{n-1}$ is the desired antichain.
\end{proof}
\section{From hereditarily finite sets to well foundedness}\label{sect:eps_0}
In this section, we show that the well foundedness of~$\varepsilon_0$ can be derived from the assumption that $3$~is~$\bqo$, over $\rca_0$. Given that $\varepsilon_0$ is the proof theoretic ordinal of~$\aca_0$, we conclude that $3$ being $\bqo$ is unprovable in this theory.
To have a sound basis for the following arguments, we recall the definition of~$\varepsilon_0$ as a term system, which is of course standard.
\begin{definition}\label{def:eps_0}
We recursively define a set $\varepsilon_0$ of terms and, simultaneously, a binary relation~$\prec$ on this set:
\begin{enumerate}[label=(\roman*)]
\item Given $\alpha_0,\ldots,\alpha_{n-1}\in\varepsilon_0$ with $\alpha_{n-1}\preceq\ldots\preceq\alpha_0$, we add a term
\begin{equation*}
\langle\alpha_0,\ldots,\alpha_{n-1}\rangle\in\varepsilon_0.
\end{equation*}
\item For $\alpha=\langle\alpha_0,\ldots,\alpha_{m-1}\rangle$ and $\beta=\langle\beta_0,\ldots,\beta_{n-1}\rangle$ in $\varepsilon_0$, we declare
\begin{equation*}
\alpha\prec\beta\,\,\,\Leftrightarrow\,\,\,\begin{cases}
\text{$m<n$ and $\alpha_i=\beta_i$ for all~$i<m$},\\
\text{or there is $j<\min\{m,n\}$ with $\alpha_j\prec\beta_j$ and $\alpha_i=\beta_i$ for all~$i<j$}.
\end{cases}
\end{equation*}
\end{enumerate}
Note that we get $\langle\rangle\in\varepsilon_0$ by clause~(i) with $n=0$ and that $\alpha\preceq\beta$ abbreviates the disjunction of $\alpha\prec\beta$ and $\alpha=\beta$, where the latter denotes equality as terms.
\end{definition}
One may think of $\langle\alpha_0,\ldots,\alpha_{n-1}\rangle$ as the Cantor normal form $\omega^{\alpha_0}+\ldots+\omega^{\alpha_{n-1}}$. A straightforward induction shows that $\prec$ is a linear order on~$\varepsilon_0$.
\begin{definition}
The depth function $\dpt:\varepsilon_0\to\mathbb N$ is given by
\begin{equation*}
\dpt(\langle\alpha_0,\ldots,\alpha_{n-1}\rangle):=\max(\{0\}\cup\{\dpt(a_i)+1\,|\,i<n\}).
\end{equation*}
For $n\in\mathbb N$ we put $\omega_n:=\{\alpha\in\varepsilon_0\,|\,\dpt(\alpha)\leq n\}$.
\end{definition}
The following entails that $\varepsilon_0$ is well founded if the same holds for all~$\omega_n$.
\begin{lemma}\label{lem:omega_n-initial}
Each set $\omega_n$ is an initial segment of~$\varepsilon_0$.
\end{lemma}
\begin{proof}
Consider the length function $\len:\varepsilon_0\to\mathbb N$ with
\begin{equation*}
\len(\langle\alpha_0,\ldots,\alpha_{n-1}\rangle):=\len(\alpha_0)+\ldots+\len(\alpha_{n-1})+1.
\end{equation*}
An induction on~$\len(\alpha)+\len(\beta)$ shows that $\alpha\preceq\beta$ entails $\dpt(\alpha)\leq\dpt(\beta)$.
\end{proof}
In the previous section we have studied the collection~$H_f(m)$ of hereditarily finite sets over $m$ incomparable urelements. Let us now write $H_f^+(m)$ for the subcollection of sets that are hereditarily nonempty. This means that $H_f^+(m)$ is generated by clauses~(i) and~(ii) from Definition~\ref{def:H_f} with the condition~$n>0$ added in~(ii). The point is that we get $\supp(a)\neq\emptyset$ by induction on~$a\in H_f^+(m)$, which is relevant in connection with Lemma~\ref{lem:H_f-basic}(c). Our aim is to define functions $f_n:\omega_n\to H_f^+(2n+3)$ that are order reflecting. The following provides a convenient stepping stone.
\begin{definition}
Write $P_f(Q)$ for the collection of finite subsets of a given set~$Q$. When the latter comes with a quasi order~$\leq$, we declare
\begin{equation*}
\{p_0,\ldots,p_{m-1}\}\leq^\flat\{q_0,\ldots,q_{n-1}\}\quad:\Leftrightarrow\quad\forall i<m\exists j<n:p_i\leq q_j
\end{equation*}
to get a quasi order~$\leq^\flat$ on~$P_f(Q)$.
\end{definition}
An argument due to Marcone (see the proof of Theorem~5.10 in~\cite{marcone-survey-old}) suggests to define an order reflecting function from $\omega_{n+1}$ to $P_f(\mathbb N\times\omega_n)$. To construct~$f_{n+1}$ as above, we will use~$f_n$ to arrive in $P_f(\mathbb N\times H_f^+(2n+3))$. The latter can be embedded into~$H_f^+(2n+5)$, the intended codomain of~$f_{n+1}$, as we show in the following. Note that we always consider $P\times Q$ with the quasi order that is explained by
\begin{equation*}
\langle p,q\rangle\leq_{P\times Q}\langle p',q'\rangle\qquad\Leftrightarrow\qquad p\leq_P p'\quad\text{and}\quad q\leq_Q q'.
\end{equation*}
It is straightforward to see that the elements $\overline i\in H_f(3)$ from Proposition~\ref{prop:H3-naturals} are contained in~$H_f^+(3)$. For $m<n$ we view~$H_f^+(m)$ as a suborder of~$H_f^+(n)$.
\begin{definition}\label{def:g_n}
Let $g_n:\mathbb N\times H_f^+(2n+3)\to H_f^+(2n+5)$ be given by
\begin{equation*}
g_n(\langle i,a\rangle):=\big\{\big\{\overline i,2n+3\big\},\big\{a,2n+4\big\}\big\}.
\end{equation*}
To define $\hat g_n:P_f(\mathbb N\times H_f^+(2n+3))\to H_f^+(2n+5)$, we now set
\begin{equation*}
\hat g_n(\{p_0,\ldots,p_{m-1}\}):=\{g_n(p_0),\ldots,g_n(p_{m-1})\}\cup\{2n+3\}.
\end{equation*}
Here $2n+3$ is added to get a value in~$H_f^+(2n+5)$ even for~$m=0$.
\end{definition}
The following lemma confirms the desired property.
\begin{lemma}\label{lem:g_n}
The functions $g_n$ and $\hat g_n$ are order embeddings.
\end{lemma}
\begin{proof}
For any $i\in\mathbb N$ and $b\in H_f^+(2n+3)$ we have
\begin{equation*}
\emptyset\neq\supp(\overline i)\subseteq\{0,1,2\}\quad\text{and}\quad\emptyset\neq\supp(b)\subseteq\{0,\ldots,2n+2\}.
\end{equation*}
Thus $\{\overline i,2n+3\}$ and $\{b,2n+4\}$ are incomparable in~$H_f^+(2n+5)$, by Lemma~\ref{lem:H_f-basic}. One can conclude that $g_n(\langle i,a\rangle)\leq^\flat g_n(\langle j,b\rangle)$ is equivalent to the conjunction of
\begin{equation*}
\{\overline i,2n+3\}\leq^\flat\{\overline j,2n+3\}\quad\text{and}\quad\{a,2n+4\}\leq^\flat\{b,2n+4\}.
\end{equation*}
As $\overline i$ and $2n+3$ are incomparable, the first conjunct is equivalent to $\overline i\leq^\flat\overline j$ and hence to $i\leq j$, due to Proposition~\ref{prop:H3-naturals}. Similarly, we see that the second conjunct is equivalent to $a\leq^\flat b$ in $H_f^+(2n+3)\subseteq H_f^+(2n+5)$. Here it is crucial that we have $a\in H_f^+(2n+3)$ and hence $\supp(a)\neq\emptyset$, as we need to ensure $a\not\leq^\flat 2n+4$. We now know that $g_n$ is an embedding. To conclude that the same holds for~$\hat g_n$, it suffices to note that we have $2n+4\in\supp(g_n(\langle i,a\rangle))$ and hence $g_n(\langle i,a\rangle)\not\leq^\flat 2n+3$.
\end{proof}
Let us now complete the construction that was described above.
\begin{definition}
We recursively define $f_n:\omega_n\to H_f^+(2n+3)$ by setting
\begin{align*}
f_0(\langle\rangle)&:=0\in 3\subseteq H_f^+(3),\\
f_{n+1}(\langle\alpha_0,\ldots,\alpha_{m-1}\rangle)&:=\hat g_n\left(\big\{\langle 0,f_n(\alpha_0)\rangle,\ldots,\langle m-1,f_n(\alpha_{m-1})\rangle\big\}\right).
\end{align*}
In the context of~$\rca_0$, this should be read as a recursion over terms in~$\varepsilon_0$, where different $n$ are considered simultaneously.
\end{definition}
The following is crucial for our proof of well foundedness.
\begin{theorem}
The functions $f_n:\omega_n\to H_f^+(2n+3)$ are order reflecting.
\end{theorem}
\begin{proof}
We argue by induction on~$n$. For the step we assume $f_{n+1}(\alpha)\leq^\flat f_{n+1}(\beta)$ and derive $\alpha\preceq\beta$. Let us write $\alpha=\langle\alpha_0,\ldots,\alpha_{k-1}\rangle$ as well as $\beta=\langle\beta_0,\ldots,\beta_{m-1}\rangle$. Due to Lemma~\ref{lem:g_n} we obtain
\begin{equation*}
\big\{\left.\langle i,f_n(\alpha_i)\rangle\,\right|\,i<k\,\big\}\leq^\flat\big\{\left.\langle j,f_n(\beta_j)\rangle\,\right|\,j<m\,\big\}
\end{equation*}
in the order $P_f(\mathbb N\times H_f^+(2n+3))$. Given any~$i<k$ we thus find a $j<m$ such that we have $\langle i,f_n(\alpha_i)\rangle\leq\langle j,f_n(\beta_j)\rangle$ in $\mathbb N\times H_f^+(2n+3)$. This means that we have $i\leq j$ and $f_n(\alpha_i)\leq^\flat f_n(\beta_j)$, which yields $\alpha_i\preceq\beta_j$ by induction hypothesis. As clause~(i) of Definition~\ref{def:eps_0} ensures $\beta_{m-1}\preceq\ldots\preceq\beta_0$, we can conclude $\alpha_i\preceq\beta_i$. In particular, the given argument establishes~$k\leq m$. We get $\alpha\preceq\beta$ by Definition~\ref{def:eps_0}.
\end{proof}
With respect to the previous proof we note, again, that the implicit function from $\omega_{n+1}$ to $P_f(\mathbb N\times\omega_n)$ comes from an argument due to Marcone (see his proof of Theorem~5.10 in~\cite{marcone-survey-old}). Let us derive the desired conclusion.
\begin{corollary}
If $3$ is $\bqo$ then $\varepsilon_0$ is well founded, provably in~$\rca_0$.
\end{corollary}
\begin{proof}
Assume that $3$ is $\bqo$. Then so is $H_f^+(2n+3)\subseteq H_f(2n+3)$ for every~$n\in\mathbb N$, by Theorem~\ref{thm:H_f} and Corollary~\ref{cor:fin-bqo}. Now assume there was an infinite descending sequence $\alpha_0\succ\alpha_1\succ\ldots$ in~$\varepsilon_0$. We set $n:=\dpt(\alpha_0)$ to get~$\alpha_0\in\omega_n$. By Lemma~\ref{lem:omega_n-initial} we obtain $\alpha_i\in\omega_n$ for all~$i\in\mathbb N$. Given that $H_f^+(2n+3)$ is $\bqo$ and in particular a well quasi order, we find $i<j$ with $f_n(\alpha_i)\leq^\flat f_n(\alpha_j)$. Now the previous theorem yields $\alpha_i\preceq\alpha_j$, against the assumption that our sequence descends.
\end{proof}
As a consequence, we obtain the following independence result.
\begin{corollary}
The statement that $3$ is $\bqo$ cannot be proved in~$\aca_0$.
\end{corollary}
\begin{proof}
This can be inferred as $\varepsilon_0$ is the proof theoretic ordinal of $\aca_0$. The~following is one way to make this more explicit. If $\aca_0$ did prove that $3$ is~$\bqo$, it would also prove that $\varepsilon_0$ is well founded, by the previous corollary. But then~$\aca_0$ would prove the consistency of Peano arithmetic (see \cite{buchholz91} for a concise presentation of the relevant classical result). Since $\aca_0$ is a conservative extension of the latter (see, e.\,g.,~\cite[Theorem~III.1.16]{hajek91}), Peano arithmetic would thus prove its own consistency, against G\"odel's second incompleteness theorem.
\end{proof}
As in~\cite{marcone-survey-new}, we have worked with the definition of $\bqo$s in terms of barriers, which is prima facie more restrictive than the definition in terms of blocks. Clearly, the two previous corollaries remain valid when we switch to the less restrictive definition. In any case, the definitions are known to be equivalent in the presence of weak K{\H o}nig's lemma (see~\cite[Theorem~5.12]{cholak-RM-wpo}).
\bibliographystyle{amsplain}
\bibliography{3-bqo}
\end{document}
| 142,643
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This event is now sold out. There will be a stand-by line.
Spend all night at the Aero Theatre’s 13 2018. Oh, and did we mention free pizza during the second intermission and Monster drinks?
Films include:
JASON X, 2002, Warner Bros., 92 min. Dir. Jim Isaac. In the year 2455, a professor takes his students on a field trip to a long-abandoned Earth, where they discover the cryogenically frozen bodies of a young woman and the hockey-mask-wearing homicidal maniac Jason. Taken aboard the students' spaceship, Jason thaws out and resumes his murderous ways.
Only Known Print! 25th Anniversary! BODY MELT, 1993, 81 min. Dir. Philip Brophy. Made down under with a slew of Australian TV stars, this satirical gore-fest kicks in when residents of a Melbourne suburb receive free samples of a new vitamin, which produces such side effects as mutations, exploding body parts and liquefying flesh.
LINK, 1986, 103 min. Dir. Richard Franklin. Elisabeth Shue and Terence Stamp star in this U.K. horror thriller in which the title creature - a super-intelligent orangutan - turns on his masters with deadly results. Featuring a Saturn Award-nominated score by Jerry Goldsmith.
MAXIMUM OVERDRIVE, 1986, Rialto Pictures, 98 min. Dir. Stephen King..
30th Anniversary! ZOMBIE 3, 1988, 84 min. Dir. Lucio Fulci. Shortly before completing the sequel to his 1979 classic, Lucio Fulci became ill and left the Philippines-based production in the hands of writers Claudio Fragasso and Rosella Drudi (TROLL 2) and director Bruno Mattei (THE OTHER HELL), resulting in.
CURTAINS, 1983, 89 min. Dir. Richard Ciupka. In this cult favorite slasher film, a group of actresses converge on a director’s mansion to audition for a part; while the filmmaker has the casting couch on his mind, someone else has murder on theirs. With John Vernon, Samantha Eggar and Linda Thorson (from TV’s “The Avengers”). “The classiest, most chilling thriller to come along in quite a while … rich in surprises of a gripping, sensuous nature.” - The Hollywood Reporter
ANTROPOPHAGUS, 1980, 90 min. Dir. Joe D'Amato. This Italian-made “video nasty” sets sail with Tisa Farrow (Mia’s younger sister) and a group of tourists off to visit a Greek island, where they fall prey to insane, cannibalistic killer George Eastman.
Films in this Series at the Egyptian
Films in this Series at the Aero
| 347,643
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TITLE: On the number of Hall divisors of an integer
QUESTION [2 upvotes]: A Hall divisor of an integer $n$ is a divisor $d$ of $n$ such that $d$ and $n/d$ are coprime.
If $n$ is a positive integer, then $\varphi(n)$ is the number of integers $k$ in the range $1\leq k\leq n$ for which $\gcd(n, k) = 1$. We know that
$$\varphi(n)=n\prod_{p|n}(1-\dfrac{1}{p}).$$
Now how can we calculate the number of Hall divisors of an integer $n$?
REPLY [1 votes]: For any $n$, we can write $n$ as prime factorizations
$$
n=p_1^{e_1}p_2^{e_2}\dots p_k^{e_k}
$$
where $p_i$'s are prime numbers and $e_i\ge 1$ for all $i=1,\dots,k$.
Note that any hall divisor of $n$to be of the form $d=p_1^{f_1}p_2^{f_2}\dots p_k^{f_k}$ where $f_i=0 \textrm{ or } e_i$ since otherwise there will be common prime factor(s) between $d$ and $n/d$.
Since there are $k$ prime numbers for which each of them you have two choice in its power, i.e., $0$ or $e_i$ then there are $2^k$ hall divisors.
| 13,466
|
TITLE: Maximal subgroups of $\mathbb{Q}$
QUESTION [3 upvotes]: I have to prove that the group $(\mathbb{Q},+)$ hasn't got any maximal subgroup.
Let $H$ be a maximal subgroup of $G=\mathbb{Q}$. So $H$ is normal in $G$ and I can consider the quotient group $G/H$. My idea is the following: if I prove
1) this quotient hasn't got any maximal subgroup
2) this quotient has order $p$, where $p$ is a prime number
then I could consider $r \in G$ and obtain $pr \in H$. Therefore $p \frac{x}{p}=x \in H$ for any $x \in G$, so $H=G$, absurde!
But I can't prove 1) and 2)...
REPLY [4 votes]: $G/H$ has no proper non-trivial subgroup because any such subgroup would come from a subgroup of $G$ containing $H$, and $G$ and $H$ are the only possible choices because $H$ is maximal.
A non-trivial group $\Gamma$ with no proper non-trivial subgroup must be cyclic because any non-trivial element generates a subgroup. The group $\Gamma=\langle g \rangle$ cannot be infinite because $\langle g^2 \rangle$ would be a proper subgroup. Hence, $\Gamma$ is finite of order $n$. If $n=ab$ were composite, then $\langle g^a \rangle$ would be a proper subgroup. Thus, $\Gamma$ has prime order. Apply to $\Gamma=G/H$.
| 64,763
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TITLE: Use the definition of limit to prove that $\lim_{x \to 0} \frac{x}{x+1}=0$
QUESTION [4 upvotes]: Use the definition of limit to prove that $\lim_{x \to 0} \dfrac{x}{x+1}=0$
My attempt:
Let $\epsilon >0$ then I need to find $\delta>0$ such that if $0<|x|<\delta$ then $\left| \dfrac{x}{x+1} \right| < \epsilon$
What I was thinking is that, to get a $\delta$ that satisfies the definition, I should see what $\epsilon$ satisfies $|x|<\epsilon |x+1|$ and then I could define $\delta$ in terms of such an $\epsilon$.
If $\epsilon =1$ then: $0\leq|x|<|x+1|$ implies that $x^2 < x^2+2x+1$ and then $x>-\dfrac{1}{2}$. In particular, if $-\dfrac{1}{2}<x<\dfrac{1}{2} \Rightarrow |x|<\dfrac{1}{2} \Rightarrow \dfrac{1}{2} < x+1<\dfrac{3}{2}$ so that $|x+1|=x+1$ and then $\dfrac{1}{2}<|x+1| \Rightarrow \dfrac{1}{2|x+1|}<1$. Since $|x|<\dfrac{1}{2}$, then $\left| \dfrac{x}{x+1}\right| = \dfrac{|x|}{|x+1|} <\dfrac{1}{2|x+1|}<1$.
So $\delta=\dfrac{\epsilon}{2}$ stisfies the definition.
If $\epsilon \neq 1$ then:
$0 \leq |x|<\epsilon |x+1|$ Implies that: $x^2<{\epsilon}^2(x^2+2x+1) \Rightarrow (\epsilon ^2-1)x^2+2 \epsilon ^2x + \epsilon ^2 > 0$.
If $0< \epsilon <1 \Rightarrow \epsilon^2-1<0 \Rightarrow x^2 + \dfrac{2 \epsilon^2}{\epsilon^2-1}x+\dfrac{\epsilon^2}{\epsilon^2-1}<0 \Rightarrow \left(x+ \dfrac{\epsilon}{\epsilon-1} \right) \left(x+ \dfrac{\epsilon}{\epsilon+1} \right) < 0$
Let $ x_1 = -\dfrac{\epsilon}{\epsilon-1}$, $x_2=-\dfrac{\epsilon}{\epsilon+1}$. We see that $x_1 >0>x_2$. Then solving the inequality for $x$ we would have $x_2=-\dfrac{\epsilon}{\epsilon+1} < x < -\dfrac{\epsilon}{\epsilon-1} = x_1$.
I am expecting to get something like: $-\phi(\epsilon) < x < \phi(\epsilon) \Rightarrow |x|< \phi(\epsilon)$ on each case so that $\delta = \phi(\epsilon)$ satisfies the definition, but I don't know how to continue here. Could someone give me a hint? Maybe there is a simpler way to approach the exercise.
REPLY [0 votes]: Choosing a positive number $\delta < \frac{\epsilon}{1 + \epsilon}$ will make $\frac{\delta}{1 - \delta} < \epsilon$; if $0 < |x| < \delta$, then by the reverse triangle inequality $$\left|\frac{x}{x+1}\right| \le \frac{|x|}{1 - |x|} < \frac{\delta}{1 - \delta} < \epsilon$$
| 68,864
|
living in lilac7.7.10
Unknown - sorry
I am keeping cool thinking about these lovely shades of lilac. Paired with yellows and golds, or greens and greys - lilac is a lovely way to inject a fresh calm into your space.
I am reviving an old favorite for fans of fashion and interior design.. Runway to Room is where I take some of my favorite looks of the ...
| 220,920
|
TITLE: Show that if $a \equiv b\pmod{kn}$, then $a^k\equiv b^k\pmod{k^2n}$
QUESTION [2 upvotes]: Problem
Show that if $a \equiv b\pmod{2n}$, then $a^2\equiv b^2\pmod{4n}$. More generally, show that if $a \equiv b\pmod{kn}$, then $a^k\equiv b^k\pmod{k^2n}$. [Introduction to Higher Arithmetic, ex. 2.1]
Proof of the first statement
Note that $2n \mid a - b\implies 2n (a + b)\mid a^2 - b^2$ but the $\pmod{2n}$ congruence implies that $a$ and $b$ are of the same parity and so $2\mid (a + b)$ which implies that $2n \cdot 2\mid 2n (a + b)\mid a^2 - b^2$ or $a^2 = b^2\pmod{4n}$.
Attempt at proof of the general case
Multiplying $a \equiv b\pmod{kn}$ by itself $k$ times, we get $a^k \equiv b^k\pmod{kn}$, but it is still needed to prove that $k$ divides $a^k - b^k$. It maybe stands reasonable to consider the following expansion: $$(a - b) ^ k = a^k - {k \choose 1} a^{k-1}b^1 + {k \choose 2} a^{k-2}b^2 - \cdots \mp {k \choose 1} a^1b^{k-1} \pm b^k$$ Under $\pmod{kn}$, we can replace any $b$ with $a$ and vice versa, which might be useful. Maybe the trick is to sum binomial coefficients to a number (or two numbers) which are divisible by $k$. But sum over all of coefficients is $2^k$, which is not necessarily divisible by $k$.
So, I got stuck here.
REPLY [3 votes]: Note that $a\equiv b\pmod{kn}\implies a=b+kln$ for some integer $l$. Then $$a^k-b^k=(b+kln)^k-b^k=\binom k1b^{k-1}kln+\binom k2b^{k-2}(kln)^2+\cdots+(kln)^k$$ Every term has a factor of $k^2n$ from $(kln)^t$ for $1<t\le k$ but $\dbinom k1=k$ so $k^2n\mid (a^k-b^k)$.
| 6,913
|
TITLE: Find the direct summands of $\mathbb Z/ 36\mathbb Z$
QUESTION [4 upvotes]: I have this exercise:
Find the direct summands of the $\mathbb Z$-module $M = \mathbb Z/36 \mathbb Z$.
If $\bar T$ and $\bar N$ are direct summands of $M$ then $\bar T \cap \bar N = \{\bar 0\} = \{36 \mathbb Z\}$ (1). And $\bar T = T/36 \mathbb Z$ and $\bar N = N/36\mathbb Z$, where $T$ and $N$ are $\mathbb Z$-modules of $\mathbb Z$ so they are ideals of $\mathbb Z$, so they are $T = a\mathbb Z$ and $N = b\mathbb Z$ and $a,b \in \mathbb Z$ are divisors of $36$.
The condition (1) implies that for any $x \in \mathbb Z$ such that $a|x$ and $b|x$, we have $36 |x$. What does this mean? and how to proceed?
I am stuck and I need some help. Thanks.
REPLY [0 votes]: Continuing with your notation, let $T = a \mathbb {Z}$, $N = b \mathbb {Z}$, $\bar {T} = T/36 \mathbb Z$ and $\bar{ N} = N/36\mathbb Z$, where $a,b \in \mathbb Z$ are divisors of $36$.
The equality $\bar {T} + \bar{N} = \mathbb{Z} / 36 \mathbb {Z}$ implies that for all $d \in \mathbb {Z} $ there exists $x, y \in \mathbb {Z}$ such that $ ax + by + 36 \mathbb {Z} = d + 36 \mathbb {Z}$. By Bezout's identity $a$ and $b $ are be coprime.
But as you noted, for any $n \in \mathbb Z$ such that $a|n$ and $b|n$, we have $36 |n$. So the union of the divisors of a and b equals the divisors of $36$.
Hence $a=4$ and $b=9$.
| 9,748
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TITLE: How do i proof that that the map $ \varphi: Aut(F_n) \to GL_n(\mathbb{Z})$ is homomorpism?
QUESTION [2 upvotes]: I'm trying to proof that a map $ \varphi: Aut(F_n) \to GL_n(\mathbb{Z})$ is a homomorphism but i can't exactly define which is the function to show that.
The map $ \varphi $ is a map with for any $ \alpha \in Aut(F_n) $ the $(i,j)$-th entry of the matrix $ \varphi(a) $ is the sum of exponents of the letter $x_j$ in the $ a(x_i)$.
$F_n$ denotes the non-abelian free group generated by $n$ elements
REPLY [2 votes]: In general, if $\kappa \colon G \to G/K$ is natural projection where $K$ is characteristic in $G$, then the map $\kappa$ induces a homomorphism $\tilde{\kappa} \colon \mathop{Aut}(G) \to \mathop{Aut}(G/K)$ given by $$\tilde{\kappa}(\phi)(g K) = \phi(g)K$$ for every $g\in G$ and $\phi \in \mathop{Aut}(G)$.
By setting $G = F_n$ and $K = [F_n,F_n]$ you can check that the map you get exactly the one have described.
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Careers
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Individual will manage items from Receiving, Storage, Replenishment, Picking, and Shipping. Includes management of all Locations, Areas, Item Families, Labels and Footprints. Position reports to the Internal Logistics Mgr/Systems Administrator.
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Bachelor's degree in Electrical Computer Engineering
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Applicable work experience will also be considered. 2-5 years of related engineering experience in design and manufacture of electronic assemblies. Experience in electronics simulation, automobile or commercial vehicle industry would be beneficial.
Essential Functions:
- Support the development of prototype and production tools, fixtures, gauges, and equipment
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PHYSICAL REQUIREMENTS:
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So lets get to it and talk about Subscribe To Talk Space. membership to better help you can try it completely run the risk of complimentary attempted this out and my therapist stated that you can sort of separate those weekly sessions however you see fit – truly simple interacting with them they have a website that’s extremely simple to communicate with as well kind of provide you a concept about what it’s like utilizing it so it took me a while to find the best therapist if I’m being sincere so um I checked out a couple of therapists to begin with none really seemed like a great fit and that’s truly important I think even when discovering a therapist out in the real world with in-person interactions is that you need to really operate at finding one who’s going to be a truly excellent fit great connection with the therapist.
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By Nidal al-Mughrabi and Jeffrey Heller
GAZA/JERUSALEM (Reuters) -.
Israel charges the PA 40 million shekels ($11 million) a month for electricity, deducting that from the transfers of Palestinian tax revenues it collects on behalf of the Authority. Israel does not engage with Hamas, which it considers a terrorist group.
Last month, the Palestinian Authority informed Israel that it would cover only 70 percent of the monthly cost of electricity that the Israel Electric Corporation supplies to the Gaza Strip..
(Additional reporting by Ali Sawafta in Ramallah; Editing by Robin Pomeroy)
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TITLE: How to calculate aircraft range every second using initial range, bearing and elevation angle?
QUESTION [0 upvotes]: I have a problem where I have an initial range, bearing, and elevation angle of an aircraft. I want to calculate the $x$, $y$ and $z$ coordinates every second. Is there any formula to do so?
Initial course of the aircraft is 60 deg and speed is 245 m/s.
REPLY [0 votes]: As an example of how this could work, I'll assume that the $z$ axis points up, $y$ points north, and $x$ points east.
I'll also assume that course and bearing are both measured clockwise from north, that is, north is zero degrees and east is $90$ degrees,
and that a positive elevation angle is above horizontal.
I'll assume that range is measured in a straight line through space,
that is, it is not just the horizontal distance but includes the vertical component.
I'll also assume the aircraft is flying straight and level (not turning, climbing, or descending).
If any of those assumptions differ from the way your coordinates and angles are set up, the following formulas will have to be adjusted.
Let the initial range, bearing, and elevation of the aircraft
be $r_0,$ $\theta_0,$ and $\phi_0.$
Let $P_0$ be the initial position of the aircraft and $Q_0$ the point at the same $x,y$ coordinates in the $x,y$ plane.
Then the origin $O$ and the points $P_0$ and $Q_0$ are vertices of a right triangle with angle $\phi_0$ at $O$ and hypotenuse $r.$
The opposite leg $P_0Q_0$ gives the initial altitude $z_0,$
and the adjacent leg $OQ_0$ gives the horizontal distance to the initial point, which would equal the initial $y$ coordinate, $y_0,$
if $\theta_0 = 0$ but would be the initial $x$ coordinate, $x_0,$
if $\theta_0 = 90^\circ.$
\begin{align}
z_0 &= r_0 \sin\phi_0 \\
y_0 &= r_0 \cos\phi_0 \cos\theta_0 \\
x_0 &= r_0 \cos\phi_0 \sin\theta_0
\end{align}
Given a course angle $\psi$ and speed $v$ meters per second,
the velocity vector has components
\begin{align}
v_x &= v \sin\psi \\
v_y &= v \cos\psi \\
v_z &= 0
\end{align}
in the $x,$ $y,$ and $z$ directions, respectively.
The position after $t$ seconds is
\begin{align}
x(t) &= x_0 + v_x t \\
y(t) &= y_0 + v_y t \\
z(t) &= z_0 + v_z t.
\end{align}
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\begin{document}
\title[Solving a problem of angiogenesis]{Solving a problem of angiogenesis of degree three}
\author{Anastasios N. Zachos}
\dedicatory{Dedicated to Professors Dr. Alexander O. Ivanov and
Dr. Alexey A. Tuzhilin for their contributions on minimal trees}
\address{Greece}
\email{azachos@gmail.com}
\keywords{Universal absorbing Fermat-Torricelli set, Universal
Fermat-Torricelli minimum value, generalized Gauss problem,
weighted Fermat-Torricelli problem, weighted Fermat-Torricelli
point, absorbing Fermat-Torricelli tree, absorbing generalized
Gauss tree, evolutionary tree} \subjclass{51E12, 52A10, 52A55,
51E10}
\begin{abstract}
An absorbing weighted Fermat-Torricelli tree of degree four is a
weighted Fermat-Torricelli tree of degree four which is derived as
a limiting tree structure from a generalized Gauss tree of degree
three (weighted full Steiner tree) of the same boundary convex
quadrilateral in $\mathbb{R}^{2}.$ By letting the four variable
positive weights which correspond to the fixed vertices of the
quadrilateral and satisfy the dynamic plasticity equations of the
weighted quadrilateral, we obtain a family of limiting tree
structures of generalized Gauss trees which concentrate to the
same weighted Fermat-Torricelli tree of degree four (universal
absorbing Fermat-Torricelli tree). The values of the residual
absorbing rates for each derived weighted Fermat-Torricelli tree
of degree four of the universal Fermat-Torricelli tree form a
universal absorbing set. The minimum of the universal absorbing
Fermat-Torricelli set is responsible for the creation of a
generalized Gauss tree of degree three for a boundary convex
quadrilateral derived by a weighted Fermat-Torricelli tree of a
boundary triangle (Angiogenesis of degree three). Each value from
the universal absorbing set contains an evolutionary process of a
generalized Gauss tree of degree three.
\end{abstract}\maketitle
\section{Introduction}
We shall describe the structure of a generalized Gauss tree with
degree three and a weighted Fermat-Torricelli tree of degree four
with respect to a boundary convex quadrilateral
$A_{1}A_{2}A_{3}A_{4}$ in $\mathbb{R}^{2}.$
\begin{definition}{\cite[Section~2, pp.~2]{GilbertPollak:68}}\label{topology}
A tree topology is a connection matrix specifying which pairs of
points from the list
$A_{1},A_{2},A_{3},A_{4},A_{0},A_{0^{\prime}}$ have a connecting
linear segment (edge).
\end{definition}
\begin{definition}{\cite[Subsection~1.2, pp.~8]{Ci}\cite{GilbertPollak:68},\cite{IvanovTuzhilin:01}}\label{degreeSteinertree}
The degree of a vertex is the number of connections of the vertex
with linear segments.
\end{definition}
Let $A_{1}, A_{2}, A_{3}, A_{4}$ be four non-collinear points in
$\mathbb{R}^{2}$ and $B_{i}$ be a positive number (weight) which
corresponds to $A_{i}$ for $i=1,2,3,4.$
The weighted Fermat-Torricelli problem for four non-collinear
points (4wFT problem) in $\mathbb{R}^{2}$ states that:
\begin{problem}[4wFT problem]\label{4FTproblemR^2}
Find a point (weighted Fermat-Torricelli point) $A_{0}\in
\mathbb{R}^{2},$ which minimizes
\begin{equation}\label{tetrobjfunction}
f(A_{0})=\sum_{i=1}^{4}B_{i}\|A_{0}-A_{i}\|,
\end{equation}
where $\|\cdot\|$ denotes the Euclidean distance
\end{problem}
By letting $B_{1}=B_{2}=B_{3}=B_{4}$ in the 4wFT problem we obtain
the following two cases:
(i) If $A_{1}A_{2}A_{3}A_{4}$ is a convex quadrilateral, then
$A_{0}$ is the intersection point of the two diagonals
$A_{1}A_{3}$ and $A_{2}A_{4},$
(ii) If $A_{i}$ is an interior point of $\triangle
A_{j}A_{k}A_{l},$ then $A_{0}\equiv A_{i},$ for $i,j,k,l=1,2,3,4$
and $i\ne j\ne k\ne l.$
The characterization of the (unique) solution of the 4wFT problem
in $\mathbb{R}^{2}$ is given by the following result which has
been proved in \cite{BolMa/So:99} and \cite{Kup/Mar:97}:
\begin{theorem}{\cite[Theorem~18.37, p.~250]{BolMa/So:99},\cite{Kup/Mar:97}}\label{theor1}
Let $A_{0}$ be a weighted minimum point which minimizes
(\ref{tetrobjfunction}).
(a) Then, the 4wFT point $A_{0}$ uniquely exists.
(b) If for each point $A_{i}\in\{A_{1},A_{2},A_{3},A_{4}\}$
\begin{equation}\label{floatingcase}
\|\sum_{j=1, i\ne j}^{4}B_{j}\vec{u}_{ij}\|>B_i,
\end{equation}
for $i,j=1,2,3,4$ holds, then
($b_{1}$) $A_{0}$ does not belong to $\{A_{1},A_{2},A_{3},A_{4}\}$
and
($b_{2}$)
\begin{equation}\label{floatingequlcond}
\sum_{i=1}^{4}B_{i}\vec{u}_{0i}=\vec 0,
\end{equation}
where $\vec{u}_{kl}$ is the unit vector from $A_{k}$ to $A_{l},$
for $k,l\in\{0,1,2,3,4\}$
(Weighted Floating Case).\\
(c) If there is a point $A_{i}\in\{A_{1},A_{2},A_{3},A_{4}\}$
satisfying
\begin{equation}
\|{\sum_{j=1,i\ne j}^{4}B_{j}\vec{u}_{ij}}\|\le B_i,
\end{equation}
then $A_{0}\equiv A_{i}.$ (Weighted Absorbed Case).
\end{theorem}
The inverse weighted Fermat-Torricelli problem for four
non-collinear points (Inverse 4wFT problem) in $\mathbb{R}^{2}$
states that:
\begin{problem}{Inverse 4wFT problem}\label{inv4wFT}
Given a point $A_{0}$ which belongs to the convex hull of
$A_{1}A_{2}A_{3}A_{4}$ in $\mathbb{R}^{2}$, does there exist a
unique set of positive weights $B_{i},$ such that
\begin{displaymath}
B_{1}+B_{2}+B_{3}+B_{4} = c =const,
\end{displaymath}
for which $A_{0}$ minimizes
\begin{displaymath}
\sum_{i=1}^{4}B_{i}\|\|A_{0}-A_{i}\|.
\end{displaymath}
\end{problem}
By letting $B_{4}=0$ in the inverse 4wFT problem we derive the
inverse 3wfT problem which has been introduced and solved by S.
Gueron and R. Tessler in \cite[Section~4,p.~449]{Gue/Tes:02}.
In 2009, a negative answer with respect to the inverse 4wFT
problem is given in \cite[Proposition~4.4,p.~417]{Zachos/Zou:88}
by deriving a dependence between the four variable weights in
$\mathbb{R}^{2}.$ In 2014, we obtain the same dependence of
variable weights on some $C^{2}$ surfaces in $\mathbb{R}^{3}$ and
we call it the "dynamic plasticity of convex quadrilaterals"
(\cite[Problem~2, Definition~12,
Theorem~1,p.92,p.97-98]{Zachos:14}).
An important generalization of the Fermat-Torricelli problem is
the generalized Gauss problem (or full weighted Steiner tree
problem) for convex quadrilaterals in $\mathbb{R}^{2}$ which has
been studied on the K-plane (Sphere, Hyperbolic plane, Euclidean
plane) in \cite{Zachos:98}.
We mention the following theorem which provide a characterization
for the solutions of the (unweighted )Gauss problem in
$\mathbb{R}^{2}.$
\begin{theorem}{\cite[Theorem(*),pp.~328]{BolMa/So:99}}
Any solution of the Gauss problem is a Gauss tree (equally
weighted full Steiner tree) with at most two (equally weighted)
Fermat-Torricelli points (or Steiner points) where each
Fermat-Torricelli point has degree three, and the angle between
any two edges incident with a Fermat-Torricelli point is of
$120^{\circ}.$
\end{theorem}
We need to mention all the necessary definitions of the weighted
Fermat-Torricelli tree and weighted Gauss tree topologies, in
order to derive some important evolutionary structures of the
Fermat-Torricelli trees (Absorbing Fermat-Torricelli trees) and
Gauss trees (Absorbing Gauss trees) which have been introduced in
\cite[Definitions~1-7,p.~1070-1071]{Zachos:15}.
\begin{definition}\label{FTtopology3}
A weighted Fermat-Torricelli tree topology of degree three is a
tree topology with all boundary vertices of a triangle having
degree one and one interior vertex (weighted Fermat-Torricelli
point) having degree three.
\end{definition}
\begin{definition}\label{FTtopology4}
A weighted Fermat-Torricelli tree topology of degree four is a
tree topology with all boundary vertices of a convex quadrilateral
having degree one and one interior vertex (4wFT point) having
degree four.
\end{definition}
\begin{definition}{\cite[Subsection~3.7,pp.~6]{GilbertPollak:68}}\label{Steinertopology}
A weighted Gauss tree topology (or full Steiner tree topology) of
degree three is a tree topology with all boundary vertices of a
convex quadrilateral having degree one and two interior vertices
(weighted Fermat-Torricelli points) having degree three.
\end{definition}
\begin{definition}\label{Gaussrtree4}
A weighted Fermat-Torricelli tree of weighted minimum length with
a Fermat-Torricelli tree topology of degree four is called a
weighted Fermat-Torricelli tree of degree four.
\end{definition}
\begin{definition}{\cite[Subsection~3.7,pp.~6]{GilbertPollak:68},\cite{Zachos:15}}\label{Gaussrtree3}
A weighted Gauss tree of weighted minimum length with a Gauss tree
topology of degree three is called a generalized Gauss tree of
degree three or a full weighted minimal Steiner tree.
\end{definition}
In 2014, we study an important generalization of the weighted
Gauss (tree) problem that we call a generalized Gauss problem for
convex quadrilaterals in $\mathbb{R}^{2}$ by using a mechanical
construction which extends the mechanical construction of
Gueron-Tessler in the sense of P$\acute{o}$lya and Varigon
(\cite[Problem~1, Theorem~4, pp.1073-1075]{Zachos:15}).
We state a generalized Gauss problem for a weighted convex
quadrilateral $A_{1}A_{2}A_{3}A_{4}.$ in $\mathbb{R}^{2},$ such
that the weights $B_{i}$ which correspond to $A_{i}$ and
$B_{00^{\prime}}\equiv x_{G},$ satisfy the inequalities
\[|B_{i}-B_{j}|<B_{k}<B_{i}+B_{j},\]
and
\[|B_{t}-B_{m}|<B_{n}<B_{t}+B_{m}\]
where $x_{G}$ is the variable weight which corresponds to the
given distance $l\equiv \|A_{0}-A_{0^{\prime}}\|,$ for
$i,j,k\in\{1,4,00^{\prime}\},$ $t,m,n\in\{2,3,00^{\prime}\}$ and
$i\ne j\ne k,$ $t\ne m\ne n.$
\begin{problem}{\cite[Problem~1,p.~1073]{Zachos:15}}\label{genGaussproblem}
Given $l,$ $B_{1},$ $B_{2},$ $B_{3},$ $B_{4},$ find a generalized
Gauss tree of degree three with respect to $A_{1}A_{2}A_{3}A_{4}$
which minimizes
\begin{equation} \label{fundamentaleqxGauss}
B_{1}\|A_{1}-A_{0}\|+B_{2}\|A_{2}-A_{0}\|+B_{3}\|A_{3}-A_{0^{\prime}}\|+B_{4}\|A_{4}-A_{0^{\prime}}\|+x_{G}l.
\end{equation}
\end{problem}
For $l=0,$ we obtain a weighted Fermat-Torricelli tree of degree
four.
\begin{definition}{\cite[Definition~8, p.1076]{Zachos:15}}\label{varGauss} We call the variable $x_{G}$
which depend on $l,$ a generalized Gauss variable.
\end{definition}
\begin{definition}{\cite[Definition~9, p.1080]{Zachos:15}}\label{varGauss}\label{absorbingrate}
The residual absorbing rate of a generalized Gauss tree of degree
at most four with respect to a boundary convex quadrilateral is
\[\sum_{i=1}^{4}B_{i}-x_{G}.\]
\end{definition}
\begin{definition}{\cite[Definition~10, p.1080]{Zachos:15}}\label{AbsorbingGeneralizedtree}
An absorbing generalized Gauss tree of degree three is a
generalized Gauss tree of degree three with residual absorbing
rate
\[\sum_{i=1}^{4}B_{i}-x_{G}.\]
\end{definition}
In this paper, we prove that the weighted Fermat-Torricelli
problem for convex quadrilaterals cannot be solved analytically,
by extending the geometrical solution of E. Torricelli and
E.Engelbrecht for convex quadrilaterals in $\mathbb{R}^{2}$
(Section~2, Theorems~3,4)
In section~3, we derive a solution for the generalized Gauss
problem in $\mathbb{R}^{2},$ in the spirit of K. Menger which
depends only on five given positive weighta and five Euclidean
distances which determine a convex quadrilateral (Section~3,
Theorem~5).
In section~4, We give a new approach concerning the dynamic
plasticity of quadrilaterals by deriving a new system of equations
different from the dynamic plasticity equations which have been
deduced in \cite[Proposition~4.4,p.~417]{Zachos/Zou:88} and
\cite[Problem~2, Definition~12, Theorem~1,p.92,p.97-98]{Zachos:14}
(Section~4, Proposition~2). Furthermore, we obtain a surprising
connection of the dynamic plasticity of quadrilaterals with a
problem of Rene Descarted posed in 1638.
In section~5, we introduce an absorbing weighted Fermat-Torricelli
tree of degree four which is derived as a limiting tree structure
from a generalized Gauss tree of degree three of the same boundary
convex quadrilateral in $\mathbb{R}^{2}.$
Then, by assuming that the four variable positive weights which
correspond to the fixed vertices of the boundary quadrilateral and
satisfy the dynamic plasticity equations, we obtain a family of
limiting tree structures of generalized Gauss trees of degree
three which concentrate to the same weighted Fermat-Torricelli
tree of degree four in a geometric sense (Universal absorbing
Fermat-Torricelli tree).
Furthermore, we calculate the values of the universal rates of a
Universal absorbing tree regarding a fixed boundary quadrilaterals
(Section~5, Examples~2,3).
In section~6, we introduce a class of Euclidean minimal tree
structures that we call steady trees and evolutionary trees
(Section~6, Definitions~15,16). Thus, the minimum of the universal
absorbing Fermat-Torricelli set (Universal Fermat-Torricelli
minimum value) leads to the creation of a generalized Gauss tree
of degree three for the same boundary convex quadrilateral which
is derived by a weighted Fermat-Torricelli tree of degree four. A
universal absorbing Fermat-Torricelli minimum value corresponds to
the intersection point (4wFT point). This quantity is of
fundamental importance, because by attaining this value the
absorbing Fermat-Torricelli tree start to grow and will be able to
produce a generalized Gauss tree of degree three (Evolutionary
tree). Each specific value from the universal absorbing set gives
an evolutionary process of a generalized Gauss tree of degree
three regarding a fixed boundary quadrilateral by spending a
positive quantity from the storage of the universal
Fermat-Torricelli quantity which stimulates the evolution at the
4wFT point (Section~6, Example~4, Angiogenesis of degree three).
\section{Extending Torricelli-Engelbrecht's solution for convex quadrilaterals}\label{sec1}
The weighted Torricelli-Engelbrecht solution for a triangle
$\triangle A_{1}A_{2}A_{3}$ in the weighted floating case is given
by the following proposition:
\begin{lemma}{\cite{Gue/Tes:02},\cite{ENGELBRECHt:1877}}\label{trianglesolveFT}
If $A_{0}$ is an interior weighted Fermat-Torricelli point of
$\triangle A_{1}A_{2}A_{3},$ then
\begin{equation}
\angle
A_{i}A_{0}A_{j}\equiv\alpha_{i0j}=\arccos{\left(\frac{B_{k}^{2}-B_{i}^{2}-B_{j}^{2}}{2
B_{i}B_{j}}\right)}.
\end{equation}
\end{lemma}
Let $A_{1}A_{2}A_{3}A_{4}$ be a convex quadrilateral in
$\mathbb{R}^{2},$ $O$ be the intersection point of the two
diagonals and $B_{i}$ be a given weight which corresponds to the
vertex $A_{i},$ $A_{0}$ be the weighted Fermat-Torricelli point in
the weighted floating case (Theorem~) and $\vec{U}_{ij}$ be the
unit vector from $A_{i}$ to $A_{j},$ for $i,j=1,2,3,4.$
We mention the geometric plasticity principle of quadrilaterals in
$\mathbb{R}^{2},$
\begin{lemma}{\cite{Zachos/Zou:88},\cite[Definition~13,Theorem~3, Proposition~8, Corollary~4 p.~103-108]{Zachos:14}}\label{geomplasticityR2}
Suppose that the weighted floating case of the weighted
Fermat-Torricelli point $A_{0}$ point with respect to
$A_{1}A_{2}A_{3}A_{4}$ is satisfied:
\[\left\|
B_{i}\vec{U}_{ki}+B_{j}\vec{U}_{kj}+B_{m}\vec{U}_{km}\right\|>
B_{k},\] for each $i,j,k,m=1,2,3,4$ and $i\ne j\ne k\ne m.$ If
$A_{0}$ is connected with every vertex $A_{k}$ for $k=1,2,3,4$ and
we select a point $A_{k}^{\prime}$ with non-negative weight
$B_{k}$ which lies on the ray $A_{k}A_{0}$ and the quadrilateral
$A_{1}^{\prime}A_{2}^{\prime}A_{3}^{\prime}A_{4}^{\prime}$ is
constructed such that:
\[\left\|
B_{i}\vec{U}_{k^{\prime}i^{\prime}}+B_{j}\vec{U}_{k^{\prime}j^{\prime}}+B_{m}\vec{U}_{k^{\prime}m^{\prime}}\right\|>
B_{k},\] for each
$i^{\prime},j^{\prime},k^{\prime},m^{\prime}=1,2,3,4$ and
$i^{\prime}\ne j^{\prime}\ne k^{\prime}\ne m^{\prime}.$
Then the weighted Fermat-Torricelli point
$A_{0}^{\prime}$ is identical with $A_{0}.$
\end{lemma}
\begin{theorem}
The weighted Torricelli-Engelbrecht solution for
$A_{1}A_{2}A_{3}A_{4}$ is given by the following system of four
equations w.r. to the variables $\alpha_{102},$ $\alpha_{203},$
$\alpha_{304}$ and $\alpha_{401}:$
\begin{eqnarray}\label{first}
& \csc^{2}\alpha_{102} \csc^{2}\alpha _{304} \csc^{2}\alpha _{401}
\left(\cos\alpha_{102}-\sin\alpha_{102}\right) \left(\cos\alpha
_{304}-\sin\alpha
_{304}\right)\nonumber \\
{}& \left(\cos\left(\alpha_{102}-\alpha
_{304}\right)-\cos\left(\alpha_{102}+\alpha _{304}+2 \alpha
_{401}\right)-2 \sin\left(\alpha_{102}+\alpha
_{304}\right)\right)=0,
\end{eqnarray}
\begin{eqnarray}\label{second}
-B_1^2-2B_1 B_2\cos\alpha _{102}-B_2^2+B_3^2-2B_1 B_4 \cos\alpha
_{401} -2B_2 B_4 \cos\left(\alpha _{102}+\alpha _{401}\right)
-B_4^2=0,
\end{eqnarray}
\begin{eqnarray}\label{third}
\alpha_{304}=\arccos{\left(\frac{B_1^2+2B_1 B_2\cos\alpha
_{102}+B_2^2-B_3^2-B_4^2}{2 B_3 B_4}\right)}
\end{eqnarray}
and
\begin{eqnarray}\label{fourth}
\alpha_{203}=2\pi-\alpha_{102}-\alpha_{304}-\alpha_{401}.
\end{eqnarray}
\end{theorem}
\begin{proof}
Assume that we select $B_{1},$ $B_{2},$ $B_{3},$ $B_{4},$ such
that $A_{0}$ is an interior point of $\triangle A_{1}OA_{2}.$ By
applying the geometric plasticity principle of
Lemma~\ref{geomplasticityR2} we could choose a transformation of
$A_{1}A_{2}A_{3}A_{4}$ to the square
$A_{1}^{\prime}A_{2}^{\prime}A_{3}^{\prime}A_{4}^{\prime},$ where
$A_{0}^{\prime}=A_{0}$ and $A_{O}$ is an interior point of
$A_{1}^{\prime}O^{\prime}A_{2}^{\prime}$ where $O^{\prime}$ is the
intersection of $A_{1}^{\prime}A_{3}^{\prime}$ and
$A_{2}^{\prime}A_{4}^{\prime}.$
We consider the equations of the three circles which pass through
$A_{1},$ $A_{2},$ $A_{0},$ $A_{1},$ $A_{4},$ $A_{0}$ and $A_{3},$
$A_{4},$ $A_{0},$ respectively, which meet at $A_{0}:$
\begin{eqnarray}\label{circle1}
\left(x-\frac{a}{2}\right)^2+\left(y-\frac{1}{2} a \cot\alpha
_{102}\right)^2=\frac{1}{4} a^2 \csc^{2}\alpha _{102}
\end{eqnarray}
\begin{eqnarray}\label{circle2}
\left(y-\frac{a}{2}\right)^2+\left(x-\frac{1}{2} a \cot\alpha
_{401}\right)^2=\frac{1}{4} a^2 \csc^{2}\alpha _{401}
\end{eqnarray}
\begin{eqnarray}\label{circle3}
\left(x-\frac{a}{2}\right)^2+\left(y-(a-\frac{1}{2} a \cot\alpha
_{304})\right)^2=\frac{1}{4} a^2 \csc^{2}\alpha _{304}
\end{eqnarray}
By subtracting (\ref{circle2}) from (\ref{circle1}),
(\ref{circle3}) from (\ref{circle1}) and solving w.r. to $x, y$ we
get:
\begin{eqnarray}\label{countx}
x=\frac{a(-1+\cot\alpha_{102})(-1+\cot\alpha_{304})}{(-2+\cot\alpha_{102}+\cot\alpha_{304})
(-1+\cot\alpha_{401})}
\end{eqnarray}
and
\begin{eqnarray}\label{county}
y=-\frac{-a
\cot\alpha_{304}+a}{\cot\alpha_{102}+\cot\alpha_{304}-2}
\end{eqnarray}
By substituting (\ref{countx}) and (\ref{county}) in
(\ref{circle1}), we obtain (\ref{first}).
Taking into account the weighted floating equilibrium condition,
we get:
\begin{eqnarray}\label{vectorbalance1}
-B_{3} \vec{u}_{30}=B_{1} \vec{u}_{10}+B_{2} \vec{u}_{20}+B_{4}
\vec{u}_{40}
\end{eqnarray}
or
\begin{eqnarray}\label{vectorbalance2}
B_{1} \vec{u}_{10}+B_{2} \vec{u}_{20}=-B_{3}
\vec{u}_{30}-B_{4}\vec{u}_{40}.
\end{eqnarray}
By squaring both parts of (\ref{vectorbalance1}) we derive
(\ref{second}) and by squaring both parts of
(\ref{vectorbalance2}) we derive (\ref{third}).
\end{proof}
\begin{remark}\label{rem1}
A different approach was used in \cite[Solution~2.2,Example~2.4,
p.~413-414]{Zachos/Zou:88}, in order to derive a similar system of
equations w.r. to $\alpha_{401}$ and $\alpha_{102}.$
\end{remark}
\begin{example}\label{exam1}
By substituting $B_{1}=3.5,$ $B_{2}=2.5,$ $B_{3}=2,$ $B_{4}=1,$
$a=10$ in (\ref{first}) and (\ref{second}) and solving this system
of equations numerically by using for instance Newton method and
choosing as initial values $\alpha_{102}^{o}=2.7$ rad,
$\alpha_{401}^{o}=1.2$ rad, we obtain $\alpha_{102}=2.30886$ and
$\alpha_{401}=1.57801$ rad. By substituting $\alpha_{102}=2.30886$
and $\alpha_{401}=1.57801$ rad in (\ref{third}) we get
$\alpha_{304}=1.12492$ rad. From (\ref{fourth}), we get
$\alpha_{203}=1.2714$ rad. By substituting the angles
$\alpha_{i0j}$ in (\ref{countx}) and (\ref{county}), we derive
$x=4.0700893$ and $y=2.146831.$
\end{example}
\begin{theorem}\label{nonexist4FTsol}
There does not exist an analytical solution for the 4wFT problem
in $\mathbb{R}^{2}.$
\end{theorem}
\begin{proof}
The system of the two equations (\ref{first}) and (\ref{second})
taking into account (\ref{third}) cannot be solved explicitly w.r
to $\alpha_{102}$ and $\alpha_{401}.$ Therefore, by considering
(\ref{countx}) and (\ref{county}) we deduce that the location of
the 4wFT point $A_{0}$ cannot also be expressed explicitly via the
angles $\alpha_{i0j},$ for $i,j=1,2,3,4,$ for $i\ne j.$
\end{proof}
Thus, from Theorem~\ref{nonexist4FTsol} the position of a weighted
Fermat-Torricelli tree of degree four cannot be expressed
analytically and may be found by using numerical methods (see also
in \cite{Zachos/Zou:88}).
\section{An absorbing generalized Gauss-Menger tree in $\mathbb{R}^{2}$}
Let $A_{1}A_{2}A_{3}A_{4}$ be a boundary weighted convex
quadrilateral of a generalized Gauss tree of degree three in
$\mathbb{R}^{2}$ and $A_{0},$ $A_{0^{\prime}}$ are the two
weighted Fermat-Torricelli (3wFT) points of degree three which are
located at the convex hull of the boundary quadrilateral.
We denote by $l\equiv \|A_{0}-A_{0^{\prime}}\|,$ $a_{ij}\equiv
\|A_{i}-A_{j}\|,$ $\alpha_{ijk}\equiv \angle A_{i}A_{j}A_{k},$
$a_{10}\equiv a_{1},$ $a_{40}\equiv a_{4},$ $a_{20^{\prime}}\equiv
a_{2},$ $a_{30^{\prime}}\equiv a_{3}$ (Fig.~\ref{fig1n1}) and by
$B_{i}\equiv \frac{B_{i}^{\prime}}{\sum_{i=1}^{4}B_{i}^{\prime}},$
for $i, j, k\in\{0,0^{\prime},1,2,3,4\}$ and $i\ne j\ne k.$
\begin{figure}
\centering
\includegraphics[scale=0.75]{exampleuniv5}
\caption{A generalized Gauss Menger tree of degree three regarding
a boundary convex quadrilateral}\label{fig1n1}
\end{figure}
We proceed by giving the following two lemmas which have been
proved recently in \cite[Theorem~1]{Zachos:98} and
\cite{Zachos/Zou:13}.
\begin{lemma}{\cite[Theorem~1]{Zachos:98},\cite[Theorem~2.1,p.~485]{Zachos/Zou:13}}\label{FT K quadrilateral}
A generalized Gauss tree of degree three (full weighted Steiner
minimal tree) of $A_{1}A_{2}A_{3}A_{4}$ consists of two weighted
Fermat-Torricelli points $A_{0},$ $A_{0}^{\prime}$ which are
located at the interior convex domain with corresponding given
weights $B_{0},$ $B_{0^{\prime}}$ and minimizes the objective
function:
\begin{equation} \label{eq:B_1}
B_{1}a_{1}+B_{2}a_{2}+B_{3}a_{3}+B_{4}a_{4}+\frac{B_{0}+B_{0^{\prime}}}{2}l\to
min,
\end{equation}
such that:
\begin{equation}\label{ineq1}
|B_{i}-B_{j}|<B_{k}<B_{i}+B_{j},
\end{equation}
\begin{equation}\label{ineq2}
|B_{t}-B_{m}|<B_{n}<B_{t}+B_{m}
\end{equation}
where \[B_{00^{\prime}}\equiv\frac{B_{0}+B_{0^{\prime}}}{2},\]
for $i,j,k\in\{1,4,00^{\prime}\},$ $t,m,n\in\{2,3,00^{\prime}\}$
and $i\ne j\ne k,$ $t\ne m\ne n,$
\end{lemma}
Suppose that $B_{1},$ $B_{2},$ $B_{3},$ $B_{4},$ $B_{00^{\prime}}$
satisfy the inequalities (\ref{ineq1}) and (\ref{ineq2}).
\begin{lemma}{\cite[Theorem~2.2, p.~486]{Zachos/Zou:13}}\label{explicitsolution}
The location of $A_{0}$ and $A_{0^{\prime}}$ is given by the
relations:
\begin{equation}\label{base1}
\cot{\varphi}=\frac{B_{0}
a_{12}+B_{4}a_{14}\cos(\alpha_{214}-\alpha_{400^{\prime}})+B_{3}a_{23}\cos(\alpha_{123}-\alpha_{30^{\prime}0})}{B_{4}a_{14}\sin(\alpha_{214}-\alpha_{400^{\prime}})-B_{3}a_{23}\sin(\alpha_{123}-\alpha_{30^{\prime}0})},
\end{equation}
\begin{equation}\label{a1solvesystem}
a_{1}=\frac{a_{14}\sin(\alpha_{214}-\varphi-\alpha_{400^{\prime}})}{\sin(\alpha_{100^{\prime}}+\alpha_{400^{\prime}})}
\end{equation}
and
\begin{equation}\label{a2solvesystem}
a_{2}=\frac{a_{23}\sin(\alpha_{123}+\varphi-\alpha_{300^{\prime}})}{\sin(\alpha_{20^{\prime}0}+\alpha_{30^{\prime}0})}
\end{equation}
where $\varphi$ is the angle which is formed between the
line defined by $A_{1}$ and $A_{2}$ and the line which passes from
$A_{1}$ and it is parallel to the line defined
by $A_{0}$ and $A_{0}^{\prime}.$
\end{lemma}
\begin{definition}\label{GaussMengersolution} A generalized
Gauss-Menger tree is a solution of a generalized Gauss problem in
$\mathbb{R}^{2}$ for a boundary quadrilateral
$A_{1}A_{2}A_{3}A_{4}$ which depend on the Euclidean distances
$a_{ij}$ and the five given weights $B_{1},$ $B_{2},$ $B_{3},$
$B_{4}$ and $B_{00^{\prime}}.$
\end{definition}
By letting $B_{00^{\prime}}\equiv x_{G},$ we obtain an absorbing
generalized Gauss-Menger tree for a boundary quadrilateral.
\begin{theorem}\label{DependenceGaussmenger}
An absorbing generalized Gauss-Menger tree w.r. to a fixed convex
quadrilateral $A_{1}A_{2}A_{3}A_{4}$ depends only on the five
given weights $B_{i},$ $B_{2},$ $B_{3},$ $B_{4},$
$B_{00^{\prime}}\equiv x_{G}$ and the five given lengths $a_{12},$
$a_{23},$ $a_{34},$ $a_{41}$ and $a_{13}.$
\end{theorem}
\begin{proof}
Consider the Caley-Menger determinant which gives the volume of a
tetrahedron $A_{1}A_{2}A_{3}A_{4}$ in $\mathbb{R}^{3}.$
\begin{equation}\label{CaleyMengertetr}
288 V^{2} =\operatorname{det} \left(
\begin{array}{ccccc}
0 & a_{12}^{ 2} & a_{13}^{ 2} & a_{14}^{ 2} & 1 \\
a_{12}^{ 2} & 0 & a_{23}^{ 2} &a_{24}^{ 2} & 1 \\
a_{13}^{ 2} & a_{23}^{ 2} &0 &a_{34}^{ 2} & 1 \\
a_{14}^{ 2} & a_{24}^{ 2} &a_{34}^{ 2} & 0 & 1 \\
1 & 1 &1 & 1 & 0 \\
\end{array} \right).
\end{equation}
By letting $V=0$ in (\ref{CaleyMengertetr}), we obtain a
dependence of the six distances $a_{12},$ $a_{13},$ $a_{14},$
$a_{23},$ $a_{34}$ and $a_{24}.$ For instance, by solving a fourth
order degree equation w.r. to $a_{13}$ we derive that
$a_{13}=(a_{12},a_{14},a_{23},a_{34},a_{24}).$
By applying the cosine law in $\triangle A_{1}A_{2}A_{4}$ and
$\triangle A_{1}A_{2}A_{3}$ we get:
\begin{equation}\label{cosalapha214}
\alpha_{214}=\arccos \left(\frac{a_{12}^{ 2}+a_{14}^{ 2}-a_{24}^{
2}}{2 a_{12} a_{14} }\right),
\end{equation}
and
\begin{equation}\label{cosalapha123}
\alpha_{123}=\arccos \left(\frac{a_{12}^{ 2}+a_{23}^{ 2}-a_{13}^{
2}}{2 a_{12} a_{23} }\right).
\end{equation}
By Lemma~\ref{trianglesolveFT} and Lemma~\ref{FT K quadrilateral}
and considering that $A_{0}$ is the 3wFT point of $\triangle
A_{1}A_{4}A_{0^{\prime}}$ and $A_{0^{\prime}}$ is the 3wFT point
of $\triangle A_{2}A_{3}A_{0}$ we get:
\begin{equation}\label{ft01}
\alpha_{100^{\prime}}=\arccos\left(\frac{B_{4}^{2}-B_{1}^{2}-x_{G}^{2}}{2
B_{1}x_{G}}\right),
\end{equation}
\begin{equation}\label{ft02}
\alpha_{0^{\prime}04}=\arccos\left(\frac{B_{1}^{2}-B_{4}^{2}-x_{G}^{2}}{2
B_{4}x_{G}}\right),
\end{equation}
\begin{equation}\label{ft03}
\alpha_{104}=\arccos\left(\frac{x_{G}^{2}-B_{1}^{2}-B_{4}^{2}}{2
B_{1}B_{4}}\right),
\end{equation}
\begin{equation}\label{ft021}
\alpha_{00^{\prime}3}=\arccos\left(\frac{B_{2}^{2}-B_{3}^{2}-x_{G}^{2}}{2
B_{3}x_{G}}\right),
\end{equation}
\begin{equation}\label{ft022}
\alpha_{00^{\prime}2}=\arccos\left(\frac{B_{3}^{2}-x_{G}^{2}-B_{2}^{2}}{2
x_{G}B_{2}}\right),
\end{equation}
and
\begin{equation}\label{ft023}
\alpha_{20^{\prime}3}=\arccos\left(\frac{x_{G}^{2}-B_{2}^{2}-B_{3}^{2}}{2
B_{2}B_{3}}\right).
\end{equation}
Therefore, by replacing (\ref{ft01}), (\ref{ft02}), (\ref{ft022}),
(\ref{ft021}), (\ref{cosalapha214}), (\ref{cosalapha123}) in
(\ref{base1}), (\ref{a1solvesystem}), (\ref{a2solvesystem}) and
taking into account the dependence of the six distances $a_{ij},$
for $i,j=1,2,3,4,$ we derive that $\varphi,$ $a_{1}$ and $a_{2}$
depend only on $B_{1},$ $B_{2},$ $B_{3},$ $B_{4},$ $x_{G}$ and
$a_{12},$ $a_{13},$ $a_{23},$ $a_{34},$ $a_{24}.$
\end{proof}
\section{The dynamic plasticity of convex quadrilaterals}
In this section, we deal with the solution of the inverse 4wFT
problem in $\mathbb{R}^{2}$ which has been introduced in
\cite{Zachos/Zou:88} and developed in \cite{Zachos:14}, in order
to obtain a new system of equations of the dynamic plasticity of
weighted quadrilaterals w.r. to the four variable weights
$(B_{i})_{1234}$ for $i=1,2,3,4,$ which cover also the case
$(B_{1})_{1234}=(B_{3})_{1234}$ and
$(B_{2})_{1234}=(B_{4})_{1234}.$
First, we start by mentioning the solution of S. Gueron and R.
Tessler (\cite[Section~4,p.~449]{Gue/Tes:02}) of the inverse 3wFT
problem for three non-collinear points in $\mathbb{R}^{2}.$ By
letting $B_{i}=0$ in the inverse 4wFT problem for convex
quadrilaterals (Problem~\ref{inv4wFT}) we derive the inverse 3wFT
problem for s triangle.
Consider the inverse 3wFT problem for $\triangle A_{i}A_{j}A_{k}$
in $\mathbb{R}^{2}.$
\begin{lemma}{\cite[Section~4,p.~449]{Gue/Tes:02}}\label{inv3imp} The unique solution of the
inverse 3wFT problem for $\triangle A_{i}A_{j}A_{k}$ is given by
\begin{equation}\label{solinv3FT}
(\frac{B_{i}}{B_{j}})_{ijk}=\frac{\sin\alpha_{jik}}{\sin\alpha_{ijk}}.
\end{equation}
\end{lemma}
\begin{definition}{\cite{Zachos:14}} We call dynamic plasticity of a weighted Fermat-Torricelli tree of degree
four the set of solutions of the four variable weights with
respect to the inverse 4wFT problem in $\mathbb{R}^{2}$ for a
given constant value $c$ which correspond to a family of weighted
Fermat-Torricelli tree of degree four that preserve the same
Euclidean tree structure (the corresponding 4wFT point remains the
same for a fixed boundary convex quadrilateral), such that the
three variable weights depend on a fourth variable weight and the
value of $c.$
\end{definition}
By taking into account Lemma~\ref{inv3imp} for the triangles
$\triangle A_{1}A_{2}A_{3},$ $\triangle A_{1}A_{3}A_{4},$
$\triangle A_{1}A_{2}A_{4}$ and the weighted floating equilibrium
condition (\ref{floatingequlcond}) taken from Theorem~\ref{theor1}
(\cite{Zachos/Zou:88}, \cite{Zachos:14})
\begin{proposition}{\cite[Proposition~4.4,p.~417]{Zachos/Zou:88},\cite[Problem~2, Definition~12,
Theorem~1,p.92,p.97-98]{Zachos:14}}\label{dynamicplasticityR2}
Suppose that $A_{0}$ does not belong to the intersection of the
linear segments $A_{1}A_{3}$ and $A_{2}A_{4}.$ The dynamic
plasticity of the variable weighted Fermat-Torricelli tree of
degree four in $\mathbb{R}^{2}$ is given by the following three
equations:
\begin{eqnarray} \label{plastic1}
(\frac{B_2}{B_1})_{1234}=(\frac{B_2}{B_1})_{123}[1-(\frac{B_4}{B_1})_{1234}(\frac{B_1}{B_4})_{134}],\\
\label{plastic2}
(\frac{B_3}{B_1})_{1234}=(\frac{B_3}{B_1})_{123}[1-(\frac{B_4}{B_1})_{1234}(\frac{B_1}{B_4})_{124}],
\end{eqnarray}
and
\begin{equation}\label{invcond4}
(B_{1})_{1234}+(B_{2})_{1234}+(B_{3})_{1234}+(B_{4})_{1234}=c=const.
\end{equation}
\end{proposition}
It is worth mentioning that each quad of values
$\{(B_{1})_{1234},(B_{2})_{1234},(B_{3})_{1234},
(B_{4})_{1234}\},$ which satisfy simultaneously (\ref{plastic1}),
(\ref{plastic2}), (\ref{invcond4}) create a unique concentration
of different families of tetrafocal ellipses (Polyellipse or
Egglipse) to the same 4wFT point $A_{0}$ of
$A_{1}A_{2}A_{3}A_{4},$ A family of tetrafocal ellipses may be
constructed by selecting a decreasing sequence of real numbers
$c_{n}((B_{1})_{1234},(B_{2})_{1234},(B_{3})_{1234},(B_{4})_{1234}))$
\[c_{n}((B_{1})_{1234},(B_{2})_{1234},(B_{3})_{1234},(B_{4})_{1234});X)=\sum_{i=1}^{4}(B_{i})_{1234}\|A_{i}-X\|\]
which converge to the constant number
\[c((B_{1})_{1234},(B_{2})_{1234},(B_{3})_{1234},(B_{4})_{1234})\equiv
f(A_{0},(B_{1})_{1234},(B_{2})_{1234},(B_{3})_{1234},(B_{4})_{1234}).\]
or
\[c((B_{1})_{1234},(B_{2})_{1234},(B_{3})_{1234},(B_{4})_{1234})=\sum_{i=1}^{4}(B_{i})_{1234}\|A_{i}-A_{0}\|\]
or
\[c((B_{1})_{1234},(B_{2})_{1234},(B_{3})_{1234},(B_{4})_{1234})=\sum_{i=1}^{4}(B_{i})_{1234}a_{i}.\]
The concentration of different families of tetrafocal ellipses to
the same point provide a surprising connection with a problem
posed by R. Descartes in 1638. According to
\cite[Chapter~II,p.~235]{BolMa/So:99}, in a letter from August 23,
1638, R. Descartes invited P. de Fermat to investigate the
properties of tetrafocal ellipses in $\mathbb{R}^{2}.$ The dynamic
plasticity of weighted quadrilaterals solves the problem of
concentration of tetrafocal ellipse and offers a new property to
R.Descartes' problem.
We proceed by deriving a new system of dynamic plasticity
equations for a weighted Fermat-Torricelli tree of degree four
which also includes a class of weighted Fermat-Torricelli trees of
degree four which coincides with the two diagonals of the boundary
convex quadrilateral for $(B_{1})_{1234}=(B_{3})_{1234}$ and
$(B_{2})_{1234}=(B_{4})_{1234}.$
\begin{proposition}\label{dynamicplasticityR2n}
The dynamic plasticity of the variable weighted Fermat-Torricelli
tree of degree four in $\mathbb{R}^{2}$ is given by the following
three equations:
\begin{eqnarray} \label{plastic14n}
(B_{1})_{1234}^{2}+ (B_{2})_{1234}^{2}+2 (B_{1})_{1234}
(B_{2})_{1234}\cos\alpha_{102}= \nonumber\\
(B_{3})_{1234}^{2}+(B_{4})_{1234}^{2}+2 (B_{3})_{1234}
(B_{4})_{1234}\cos\alpha_{304},
\end{eqnarray}
\begin{eqnarray}\label{plastic24n}
(B_{1})_{1234}^{2}+ (B_{4})_{1234}^{2}+2 (B_{1})_{1234}
(B_{4})_{1234}\cos\alpha_{104}=\nonumber\\ (B_{2})_{1234}^{2}+
(B_{3})_{1234}^{2}+2 (B_{2})_{1234}
(B_{3})_{1234}\cos\alpha_{203},
\end{eqnarray}
and
\begin{equation}\label{invcond4n}
(B_{1})_{1234}+(B_{2})_{1234}+(B_{3})_{1234}+(B_{4})_{1234}=c const.
\end{equation}
\end{proposition}
\begin{proof}
Suppose that we select four positive weights $(B_{i})_{1234}(0)$
which correspond the vertex $A_{i}$ of the boundary convex
quadrilateral $A_{1}A_{2}A_{3}A_{4},$ such that the weighted
floating inequalities (\ref{floatingcase}) of Theorem~\ref{theor1}
hold, in order to locate the 4wFT point $A_{0}$ at the interior of
$A_{1}A_{2}A_{3}A_{4}.$
From the weighted floating (variable weighted) equilibrium
condition of the 4wFT point (\ref{floatingequlcond})we get
\begin{equation}\label{vect1}
(B_{1})_{1234}\vec{u}_{10}+(B_{2})_{1234}\vec{u}_{20}=-(B_{3})_{1234}\vec{u}_{30}-(B_{4})_{1234}\vec{u}_{40}
\end{equation}
and
\begin{equation}\label{vect2}
(B_{1})_{1234}\vec{u}_{10}+(B_{3})_{1234}\vec{u}_{30}=-(B_{2})_{1234}\vec{u}_{20}-(B_{4})_{1234}\vec{u}_{40},
\end{equation}
which yield (\ref{plastic14n}) and (\ref{plastic24n}),
respectively.
\end{proof}
\begin{corollary}\label{two equal}
If the variable weighted 4wFT point is the intersection point of
$A_{1}A_{3}$ and $A_{2}A_{4},$ then
\begin{equation}\label{2eq1}
(B_{1})_{1234}=(B_{3})_{1234}
\end{equation}
and
\begin{equation}\label{2eq2}
(B_{2})_{1234}=(B_{4})_{1234}.
\end{equation}
$(B_{1})_{1234}=(B_{3})_{1234}$ and
$(B_{2})_{1234}=(B_{4})_{1234}.$
\end{corollary}
\begin{proof}
By letting $\alpha_{102}=\alpha_{304}$ and
$\alpha_{104}=\alpha_{203},$ in (\ref{plastic14n}) and
(\ref{plastic24n}) we obtain (\ref{2eq1}) and (\ref{2eq2}),
respectively.
\end{proof}
\begin{remark}
The dynamic plasticity equations of
Theorem~\ref{dynamicplasticityR2} depend on the solutions of the
inverse 3FT problem for $\triangle A_{1}A_{2}A_{3},$ $\triangle
A_{1}A_{3}A_{4}$ and $\triangle A_{1}A_{2}A_{4}.$ Thus, the
corresponding 4wFT point $A_{0}$ remains at the interior of
$\triangle A_{1}A_{2}A_{3},$ in order to derive from the inverse
3wFT problem for $\triangle A_{1}A_{2}A_{3}$ the inverse wFT
problem for $A_{1}A_{2}A_{3}A_{4}.$ The dynamic plasticity
equations of Theorem~\ref{dynamicplasticityR2n} generalizes the
dynamic plasticity equations of Theorem~\ref{dynamicplasticityR2}
because $A_{0}$ could also lie on the side $A_{1}A_{3}$ of
$\triangle A_{1}A_{2}A_{3}$ and the side $A_{2}A_{4}$ of
$\triangle A_{1}A_{2}A_{4}.$ These are the cases where the inverse
3wFT problem for $\triangle A_{1}A_{2}A_{3}$ and $\triangle
A_{1}A_{2}A_{4}$ do not hold.
\end{remark}
\section{A universal Fermat-Torricelli minimal value of a family of absorbing generalized Gauss trees of degree three}
Suppose that an absorbing weighted Fermat-Torricelli tree of
degree four is derived as a limiting tree structure from an
absorbing generalized Gauss-Menger tree of degree three regarding
a fixed boundary quadrilateral $A_{1}A_{2}A_{3}A_{4}$ for a
specific value $c_{G}$ of the generalized Gauss variable.
We need to consider the following lemma which gives $\bar{B_{i}}$
as a linear function of $\bar{B_{4}}$ regarding a fixed variable
weighted Fermat-Torricelli tree of degree four in
$\mathbb{R}^{2}.$
\begin{lemma}{\cite[Corollary~4.5,p.~418]{Zachos/Zou:88}}\label{lemlinearB4}Let
$\sum_{1234}^{}\bar{B}:=(\bar{B_1})_{1234}(1+\frac{\bar{B_2}}{\bar{B_1}}+\frac{\bar{B_3}}{\bar{B_1}}+\frac{\bar{B_4}}{\bar{B_1}})_{1234}$.\\
If
$\sum_{1234}^{}\bar{B}=\sum_{123}^{}\bar{B}=\sum_{124}^{}\bar{B}=\sum_{134}^{}\bar{B}$,
then \[(\bar{B_i})_{1234}=x_i (\bar{B_4})_{1234}+ y_i, i=1,2,3:\]
\[(x_1,y_1)=(\frac{(\frac{\bar{B_1}}{\bar{B_4}})_{134}(\frac{\bar{B_2}}{\bar{B_1}})_{123}+(\frac{\bar{B_1}}{\bar{B_4}})_{124}(\frac{\bar{B_3}}{\bar{B_1}})_{123}-1
}{1+(\frac{\bar{B_2}}{\bar{B_1}})_{123}+(\frac{\bar{B_3}}{\bar{B_1}})_{123}},(\bar{B_1})_{123})\] \[(x_2,y_2)=(x_1(\frac{\bar{B_2}}{\bar{B_1}})_{123}-(\frac{\bar{B_1}}{\bar{B_4}})_{134}(\frac{\bar{B_2}}{\bar{B_1}})_{123}
,(\bar{B_2})_{123})\] \[(x_3,y_3)=(x_1(\frac{\bar{B_3}}{\bar{B_1}})_{123}-(\frac{\bar{B_1}}{\bar{B_4}})_{124}(\frac{\bar{B_3}}{\bar{B_1}})_{123},(\bar{B_3})_{123}).\]
\end{lemma}
\begin{theorem}\label{nounivconstant}
A universal constant does not exist for a unique fixed (variable)
weighted Fermat-Torricelli tree of degree four which is obtained
as a limiting tree structure from a family of variable weighted
Gauss-Menger trees (or full variable weighted Steiner trees) w.r.
to a fixed boundary convex quadrilateral $A_{1}A_{2}A_{3}A_{4},$
which depend on the variable weights $(\bar{B_{i}})_{1234}$ which
satisfy the dynamic plasticity equations
\begin{equation} \label{plastic1P4quadmod}
(\frac{\bar{B_2}}{\bar{B_1}})_{1234}=(\frac{\bar{B_2}}{\bar{B_1}})_{123}[1-(\frac{\bar{B_4}}{\bar{B_1}})_{1234}
(\frac{\bar{B_1}}{\bar{B_4}})_{134}],
\end{equation}
\begin{equation} \label{plastic2P5quadmod}
(\frac{\bar{B_3}}{\bar{B_1}})_{1234}=(\frac{\bar{B_3}}{\bar{B_1}})_{123}[1-(\frac{\bar{B_4}}{\bar{B_1}})_{1234}
(\frac{\bar{B_1}}{\bar{B_4}})_{124}],
\end{equation}
and
\begin{equation}\label{invcond4quadmod}
(\bar{B_{1}})_{1234}+(\bar{B_{2}})_{1234}+(\bar{B_{3}})_{1234}+(\bar{B_{4}})_{1234}=c constant.
\end{equation}
or
\begin{eqnarray} \label{plastic14nmod}
(\bar{B_{1}})_{1234}^{2}+ (\bar{B_{2}})_{1234}^{2}+2
(\bar{B_{1}})_{1234}
(\bar{B_{2}})_{1234}\cos\alpha_{102}= \nonumber\\
(\bar{B_{3}})_{1234}^{2}+(\bar{B_{4}})_{1234}^{2}+2
(\bar{B_{3}})_{1234} (\bar{B_{4}})_{1234}\cos\alpha_{304},
\end{eqnarray}
\begin{eqnarray}\label{plastic24nmod}
(\bar{B_{1}})_{1234}^{2}+ (\bar{B_{4}})_{1234}^{2}+2
(\bar{B_{1}})_{1234} (\bar{B_{4}})_{1234}\cos\alpha_{104}=\nonumber\\
(\bar{B_{2}})_{1234}^{2}+ (\bar{B_{3}})_{1234}^{2}+2
(\bar{B_{2}})_{1234} (\bar{B_{3}})_{1234}\cos\alpha_{203},
\end{eqnarray}
and
\begin{equation}\label{invcond4nmod}
(\bar{B_{1}})_{1234}+(\bar{B_{2}})_{1234}+(\bar{B_{3}})_{1234}+(\bar{B_{4}})_{1234}=c const.
\end{equation}
\end{theorem}
\begin{proof}
Suppose that we select four positive weights
$(\bar{B_{i}})_{1234}(0)$ which correspond the vertex $A_{i}$ of
the boundary convex quadrilateral $A_{1}A_{2}A_{3}A_{4},$ such
that the weighted floating inequalities (\ref{floatingcase}) of
Theorem~\ref{theor1} hold, in order to locate the 4wFT point
$A_{0}$ at the interior of $A_{1}A_{2}A_{3}A_{4}$ and particularly
at the interior of $\triangle A_{1}OA_{2},$ where $O$ is the
intersection of the diagonals $A_{1}A_{3}$ and $A_{2}A_{4},$
closer to the vertex $A_{1}.$
The location of the 3wFT points $A_{0}$ and $A_{0^{\prime}},$
respectively, is given by the relations
(Lemma~\ref{explicitsolution}):
\begin{equation}\label{base1nbar}
\cot{\varphi}=\frac{x_{G}
a_{12}+\bar{B_{4}}a_{14}\cos(\alpha_{214}-\alpha_{400^{\prime}})+\bar{B_{3}}a_{23}\cos(\alpha_{123}-\alpha_{30^{\prime}0})}{\bar{B_{4}}a_{14}\sin(\alpha_{214}-\alpha_{400^{\prime}})-\bar{B_{3}}a_{23}\sin(\alpha_{123}-\alpha_{30^{\prime}0})},
\end{equation}
\begin{equation}\label{a1solvesystemnbar}
a_{1}=\frac{a_{14}\sin(\alpha_{214}-\varphi-\alpha_{400^{\prime}})}{\sin(\alpha_{100^{\prime}}+\alpha_{400^{\prime}})}
\end{equation}
and
\begin{equation}\label{a2solvesystemnbar}
a_{2}=\frac{a_{23}\sin(\alpha_{123}+\varphi-\alpha_{300^{\prime}})}{\sin(\alpha_{20^{\prime}0}+\alpha_{30^{\prime}0})}
\end{equation}
where
\begin{equation}\label{ft01bar}
\alpha_{100^{\prime}}=\arccos\left(\frac{\bar{B_{4}}^{2}-\bar{B_{1}}^{2}-x_{G}^{2}}{2
\bar{B_{1}}x_{G}}\right),
\end{equation}
\begin{equation}\label{ft02bar}
\alpha_{0^{\prime}04}=\arccos\left(\frac{\bar{B_{1}}^{2}-\bar{B_{4}}^{2}-x_{G}^{2}}{2
\bar{B_{4}}x_{G}}\right),
\end{equation}
\begin{equation}\label{ft03bar}
\alpha_{104}=\arccos\left(\frac{x_{G}^{2}-\bar{B_{1}}^{2}-\bar{B_{4}}^{2}}{2
\bar{B_{1}}\bar{B_{4}}}\right),
\end{equation}
\begin{equation}\label{ft021bar}
\alpha_{00^{\prime}3}=\arccos\left(\frac{\bar{B_{2}}^{2}-\bar{B_{3}}^{2}-x_{G}^{2}}{2
\bar{B_{3}}x_{G}}\right),
\end{equation}
\begin{equation}\label{ft022bar}
\alpha_{00^{\prime}2}=\arccos\left(\frac{\bar{B_{3}}^{2}-x_{G}^{2}-\bar{B_{2}}^{2}}{2
x_{G}\bar{B_{2}}}\right),
\end{equation}
\begin{equation}\label{ft023bar}
\alpha_{20^{\prime}3}=\arccos\left(\frac{x_{G}^{2}-\bar{B_{2}}^{2}-\bar{B_{3}}^{2}}{2
\bar{B_{2}}\bar{B_{3}}}\right).
\end{equation}
and the variable weights $\bar{B_{i}}\equiv (\bar{B_{i}})_{1234}$
are taken from the system of equations (\ref{plastic1P4quadmod}),
(\ref{plastic2P5quadmod}), (\ref{invcond4quadmod}) or
(\ref{plastic14nmod}), (\ref{plastic24nmod}) and
(\ref{invcond4nmod}). Taking into account Lemma~\ref{lemlinearB4},
we express $(\bar{B_{i}})_{1234}$ as a function of
$\bar{B_{4}})_{1234}$
By taking into account the distance of $a_{12}$ from the line
defined by $A_{0}$ and $A_{0}^{\prime},$ and the distance of
$A_{2}$ from the line which passes through $A_{1}$ and is parallel
to the line defined by $A_{0}$ and $A_{0}^{\prime},$ we express
$l$ as a function w.r. to $\bar{B_{4}},$ $x_{G}$ and the five
Euclidean elements $a_{12},$ $a_{13},$ $a_{23},$ $a_{34},$
$a_{24},$ for $i=1,2,3,4.$
\begin{equation}\label{universalvaluesxG}
l=a_{1}\cos(\alpha_{100^{\prime}})+a_{2}\cos(\alpha_{20^{\prime}0})+a_{12}\cos(\varphi)
\end{equation}
By letting $l\equiv \epsilon$ a positive real number we derive a
nonlinear equation which depends only on the absorbing rate of the
generalized Gauss-Menger tree of degree three, where the 3wFT
point $A_{0}$ remains the same because the weights
$(\bar{B_{i}})_{1234}$ satisfy the dynamic plasticity equations of
the fixed variable weighted Fermat-Torricelli tree of degree four
concerning the same boundary quadrilateral $A_{1}A_{2}A_{3}A_{4}$
where the position of $A_{0}$ remains invariant in
$\mathbb{R}^{2}.$
By letting $l\equiv \epsilon_{i}$ a decreasing sequence which
converge to zero, $A_{0}^{\prime}$ will approach the fixed
position of $A_{0}.$
A universal constant $u_{c}$ may occur if $x_{G}(\epsilon_{k})\to
u_{c}$ for every positive real value of the variable weight
$(\bar{B_{4}})_{1234}.$ Thus, a weighted Fermat-Torricelli degree
four regarding the same boundary quadrilateral
$A_{1}A_{2}A_{3}A_{4}$ would exist with a constant absorbing rate
and could offer an analytical solution of the 4wFT problem in
$\mathbb{R}^{2}$ which is contradictory with the result of
Theorem~\ref{nonexist4FTsol}.
\end{proof}
The location of the 4wFT point for a convex quadrilateral or the
position of a weighted Fermat-Torricelli tree of degree four is
given by the following lemma which has been derived in
\cite{Zachos/Zou:88}
($(B_{1})_{1234}>(B_{2})_{1234}>(B_{3})_{1234}>(B_{4})_{1234}$):
\begin{lemma}{\cite[Formula~(5),(6),(7),(8),p.~413]{Zachos/Zou:88}}\label{location4ft}
The following system of equations allow us to compute the position
of the 4wFT point $A_{0}$ and provides a necessary condition to
locate it at the interior of the convex quadrilateral
$A_{1}A_{2}A_{3}A_{4}:$
\begin{equation} \label{eq:evquad3}
\cot(\alpha_{013})=\frac{\sin(\alpha_{213})-\cos(\alpha_{213})
\cot(\alpha_{102})- \frac{a_{31}}{a_{12}
}\cot(\alpha_{304}+\alpha_{401})}
{-\cos(\alpha_{213})-\sin(\alpha_{213}) \cot(a_{102})+
\frac{a_{31}}{a_{12} }},
\end{equation}
\begin{equation} \label{eq:evquad4}
\cot(\alpha_{013})=\frac{\sin(\alpha_{314})-\cos(\alpha_{314})
\cot(\alpha_{401})+ \frac{a_{31}}{a_{41}
}\cot(\alpha_{304}+\alpha_{401})}
{\cos(\alpha_{314})+\sin(\alpha_{314}) \cot(\alpha_{401})-
\frac{a_{31}}{a_{41} }},
\end{equation}
\begin{eqnarray} \label{eq:evquad2}
(B_3)_{1234}^2=(B_1)_{1234}^2+(B_2)_{1234}^2+(B_4)_{1234}^2+\nonumber\\
2(B_2)_{1234}(B_4)_{1234}\cos(\alpha_{401}+\alpha_{102})+2(B_1)_{1234}(B_2)_{1234}\cos(\alpha_{102})+\nonumber\\
+2(B_1)_{1234}(B_4)_{1234}\cos(\alpha_{401}),
\end{eqnarray}
\begin{equation} \label{eq:evquad1}
\cot(\alpha_{304}+\alpha_{401})=
\frac{(B_1)_{1234}+(B_2)_{1234}\cos(\alpha_{102})+(B_4)_{1234}\cos(\alpha_{401}))}{(B_4)_{1234}\sin(\alpha_{401})-(B_2)_{1234}\sin(\alpha_{102})},
\end{equation}
\begin{equation}\label{eq:evquad5}
\alpha_{203}=2\pi-\alpha_{102}-\alpha_{304}-\alpha_{401},
\end{equation}
\begin{equation}\label{a01quad4ft1}
a_{01}=a_{14}\frac{\sin(\alpha_{013}+\alpha_{314}+\alpha_{401})}{\sin\alpha_{401}}
\end{equation}
and
\begin{equation}\label{a04quad4ft2}
a_{04}=a_{14}\frac{\sin(\alpha_{013}+\alpha_{314})}{\sin\alpha_{401}}.
\end{equation}
\end{lemma}
\begin{example}\label{numuniversalminimumFTvalue2}
Given a rectangle $A_{1}A_{2}A_{3}A_{4}$ in $\mathbb{R}^{2}$ such
that: $A_{1}=(0,0),$ $A_{2}=(7,0),$ $A_{3}=(7,4),$ $A_{4}=(0,4),$
with side lengths $a_{12}=a_{34}=7$, $a_{23}=a_{41}=4$ and initial
positive weights $(B_{1})_{1234}=3,$ $(B_{2})_{1234}=2.5$
$(B_{3})_{1234}=1.7$ $(B_{4})=1.5$ which correspond to the
vertices $A_{1},$ $A_{2},$ $A_{3}$ and $A_{4},$ respectively. By
substituting these data in (\ref{eq:evquad3}), (\ref{eq:evquad4})
(\ref{eq:evquad2}), (\ref{eq:evquad1}), (\ref{eq:evquad5}) of
Lemma~\ref{location4ft} we obtain the location of the 4wFT point
$A_{0}=(2.8274502,1.2787811)$ which coincides with result obtained
by the Weiszfeld algorithm (\cite{Weis:37},\cite{Weis:09}) with 7
digit precision and we derive that:
\begin{eqnarray}\label{maindirections1}
\alpha_{102}=138.625^{\circ},
\alpha_{203}=50.1502^{\circ},\nonumber\\
\alpha_{304}=102.986^{\circ}, \alpha_{401}=68.2392^{\circ}.
\end{eqnarray}
\begin{figure}
\centering
\includegraphics[scale=0.75]{exampleuniv1}
\caption{A fixed weighted Fermat-Torricelli tree of degree four
having a universal Fermat-Torricelli minimum value $u_{FT}=\min
x_{G}=3.8088826$ with respect to a boundary weighted rectangle
taken from Example~2.}
\end{figure}
By substituting (\ref{maindirections1}) in
(\ref{plastic1P4quadmod}), (\ref{plastic2P5quadmod}) and
(\ref{invcond4quadmod}) or in (\ref{plastic14nmod}),
(\ref{plastic24nmod}) and (\ref{invcond4nmod}) we obtain the
dynamic plasticity equations of $A_{1}A_{2}A_{3}A_{4}:$
\begin{eqnarray}\label{dynamicplasticexam1}
(B_{1})_{1234}=4.2239621- 0.8159745 (B_4)_{1234},\nonumber\\
(B_{2})_{1234}=0.8393665+ 1.1070888 (B_4)_{1234},\nonumber\\
(B_{3})_{1234}=3.6366712- 1.2911143 (B_4)_{1234}.
\end{eqnarray}
Then, we replace (\ref{dynamicplasticexam1}) for
$(B_{4})_{1234}=1.5,$ $(B_{4})_{1234}=1.2,$ $(B_{4})_{1234}=1.7,$
$(B_{4})_{1234}=1.7728955,$ in the objective function
(\ref{eq:B_1}) of a generalized Gauss tree of degree three, taking
into account (\ref{base1}), (\ref{a2solvesystem}) and
(\ref{a1solvesystem}) from Lemma~\ref{explicitsolution} and by
maximizing (\ref{eq:B_1}) w.r. to $x_{G},$ we derive the following
table (Fig.~2).
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
$(B_{1})_{1234}$ & $(B_{2})_{1234}$ & $(B_{3})_{1234}$ & $(B_{4})_{1234}$ & $x_{G}$ & f \\
\hline
3& 2.5& 1.7& 1.5& 3.8192408 & 34.5746856\\
3.2447927& 2.1678731& 2.0873328 & 1.2& 3.8543169& 34.6371118\\
2.8368055& 2.7214176 & 1.4417756& 1.7 & 3.8096235& 34.5330567\\
2.7773246 & 2.8021194& 1.3476592& 1.7728955 & 3.8088826 & 34.5178864\\
\hline
\end{tabular}
By substituting $l=0.0000001$ and (\ref{dynamicplasticexam1}) in
(\ref{universalvaluesxG}) and by minimizing
(\ref{universalvaluesxG}) w.r. to the variables $x_{G}$ and
$(B_{4})_{1234},$ we obtain the universal minimum
Fermat-Torricelli value $u_{FT}=3.8088826$ for $(B_{4})_{1234}=
1.7728955$ (see also the above Table, Fig.~2).
The universal absorbing rate is
$\frac{u_{FT}}{\sum_{i=1}^{4}(B_{i})_{1234}}=\frac{3.8088826}{8.7}=0.4378025.$
\end{example}
\begin{example}\label{numuniversalminimumFTvalue3}
Given the same rectangle $A_{1}A_{2}A_{3}A_{4}$ in
$\mathbb{R}^{2}$ with the one considered in
Example~\ref{nounivconstant} and initial positive weights
$(B_{1})_{1234}=3.1,$ $(B_{2})_{1234}=2.3$ $(B_{3})_{1234}=1.7$
$(B_{4})=1.4$ which correspond to the vertices $A_{1},$ $A_{2},$
$A_{3}$ and $A_{4},$ respectively. By substituting these data in
(\ref{eq:evquad3}), (\ref{eq:evquad4}) (\ref{eq:evquad2}),
(\ref{eq:evquad1}), (\ref{eq:evquad5}) of Lemma~\ref{location4ft}
we obtain the location of the 4wFT point $A_{0}=(2.381487,
1.1855484)$ which coincides with result obtained by the Weiszfeld
algorithm (\cite{Weis:37},\cite{Weis:09}) with 7 digit precision
and we derive that:
\begin{eqnarray}\label{maindirections1new}
\alpha_{102}=139.138^{\circ},
\alpha_{203}=45.7542^{\circ},\nonumber\\
\alpha_{304}=98.8792^{\circ}, \alpha_{401}=76.2283^{\circ}.
\end{eqnarray}
\begin{figure}
\centering
\includegraphics[scale=0.75]{exampleuniv22}
\caption{A fixed weighted Fermat-Torricelli tree of degree four
having a universal Fermat-Torricelli minimum value $u_{FT}=\min
x_{G}=3.66326$ with respect to a boundary weighted rectangle taken
from Example~3.}
\end{figure}
By substituting (\ref{maindirections1new}) in
(\ref{plastic1P4quadmod}), (\ref{plastic2P5quadmod}) and
(\ref{invcond4quadmod}) or in (\ref{plastic14nmod}),
(\ref{plastic24nmod}) and (\ref{invcond4nmod}) we obtain the
dynamic plasticity equations of $A_{1}A_{2}A_{3}A_{4}:$
\begin{eqnarray}\label{dynamicplasticexam12}
(B_{1})_{1234}=4.1823652- 0.7731178 (B_4)_{1234},\nonumber\\
(B_{2})_{1234}=0.49794+ 1.2871855 (B_4)_{1234},\nonumber\\
(B_{3})_{1234}=3.8196947- 1.5140677(B_4)_{1234}.
\end{eqnarray}
By substituting $l=0.0000001$ and (\ref{dynamicplasticexam1}) in
(\ref{universalvaluesxG}) and by minimizing
(\ref{universalvaluesxG}) w.r. to the variables $x_{G}$ and
$(B_{4})_{1234},$ we obtain the universal minimum
Fermat-Torricelli value $u_{FT}=3.66326$ for $(B_{4})_{1234}=
1.8199325.$
The universal absorbing rate is
$\frac{u_{FT}}{\sum_{i=1}^{4}(B_{i})_{1234}}=\frac{3.66326}{8.5}=0.4309717$
(Fig.~3).
\end{example}
\begin{definition}\label{absorbingFermatTorricelli set}
A universal absorbing Fermat-Torricelli set with respect to a
fixed variable weighted Fermat-Torricelli tree of degree four is
the set of values of the Gauss variable $x_{G}$ which maximizes
the objective function of a generalized Gauss Menger tree of
degree three which is concentrated on the same weighted
Fermat-Torricelli tree of degree four, such that the variable
weights satisfy the dynamic plasticity equations of the weighted
boundary quadrilateral.
\end{definition}
\begin{definition}\label{Existenceuniversalminimumvalue}
We call a universal Fermat-Torricelli minimum value $u_{FT}$ the
minimum of the universal Fermat-Torricelli set regarding a fixed
variable weighted Fermat-Torricelli tree of degree four.
\end{definition}
\begin{remark}
The weighted Fermat-Torricelli tree of degree four taken from
example 2 has greater universal Fermat-Torricelli minimum value
$u_{FT}$ and universal absorbing rate than the weighted
Fermat-Torricelli tree of degree four w.r. to the same boundary
quadrilateral $A_{1}A_{2}A_{3}A_{4}$ with different weights
\end{remark}
\section{Steady trees and evolutionary trees for a boundary convex quadrilateral}\label{UniversalabsFTset}
Consider a universal Fermat-Torricelli set $U$ and minimum value
$u_{FT}$ which corresponds to a fixed variable weighted
Fermat-Torricelli tree of degree four regarding a boundary convex
quadrilateral, which is derived as a limiting tree structure from
a generalized Gauss tree of degree.
\begin{definition}\label{Steadytree}A steady tree of degree four is a
weighted Fermat-Torricelli tree of degree four having as storage
at the 4wFT point $A_{0}$ a positive quantity less than $u_{FT}.$
\end{definition}
\begin{definition}\label{Evolutiarytree}An evolutionary tree of degree three is
a generalized Gauss Menger tree of degree three which is derived
by a weighted Fermat-Torricelli tree of degree four having as
storage at the 4wFT point $A_{0}$ a positive quantity equal or
greater than $u_{FT}$ and then decreases by an absorbing rate
$0<a_{G}<u_{FT}.$
\end{definition}
\begin{example}
Consider the same weighted Fermar-Torricelli tree four that we
have obtained in Example~2 for the boundary rectangle
$A_{1}A_{2}A_{3}A_{4}.$ A steady tree is a weighted
Fermat-Torricelli tree of degree four with storage level at
$A_{0}$ less than $u_{FT}=3.8088826.$
When the storage quantity at $A_{0}$ reaches $u_{FT}=3.8088826,$
it stimulates the steady tree which starts to evolute
(Fig.~\ref{fig3}). Thus, the universal minimum Fermat-Torricelli
value unlocks the evolution of a generalized Gauss-Menger tree
which could be derived by a weighted Fermat-Torricelli tree of
degree four for the same boundary rectangle (or convex
quadrilateral).
For instance, if the storage level reaches
$u=3.8543169>u_{FT}=3.8088826$ and spends $a_{G}=0.5,$ then we
derive an evolutionary generalized Gauss tree having
$x_{G}=3.3543169$ (Fig.~\ref{fig4}).
\begin{figure}
\centering
\includegraphics[scale=0.75]{exampleuniv111}
\caption{An evolutionary tree of degree four having an initial
storage level $3.8543169$ and absorbing rate $a_{G}=0$ with
respect to a boundary weighted rectangle taken from Example~2 for
$(B_{1})_{1234}=3.2447927,$ $(B_{2})_{1234}=2.1678731,$
$(B_{3})_{1234}=2.0873328,$ $(B_{4})_{1234}=1.2.$}\label{fig3}
\end{figure}
\begin{figure}
\centering
\includegraphics[scale=0.75]{exampleuniv11}
\caption{An evolutionary tree of degree three having an initial
storage level $3.8543169$ and absorbing rate $a_{G}=0.5$ with
respect to a boundary weighted rectangle taken from Example~2 for
$(B_{1})_{1234}=3.2447927,$ $(B_{2})_{1234}=2.1678731,$
$(B_{3})_{1234}=2.0873328,$ $(B_{4})_{1234}=1.2.$}\label{fig4}
\end{figure}
The evolutionary Gauss-Menger tree is obtained by substituting
$x_{G}=3.3543169,$ $(B_{1})_{1234}=3.2447927,$
$(B_{2})_{1234}=2.1678731,$ $(B_{3})_{1234}=2.0873328,$
$(B_{4})_{1234}=1.2$ in (\ref{base1}), (\ref{a2solvesystem}) and
(\ref{a1solvesystem}) from Lemma~\ref{explicitsolution}. Thus, we
get:
\begin{eqnarray}
a_{1}=1.6642065, a_{2}= 2.7738702, a_{3}= 3.6321319,\nonumber\\
a_{4}=3.4873166, l=3.3543169.
\end{eqnarray}
Suppose that the storage level reaches $u=3.82>u_{FT}=3.8088826$
and spends $a_{G}=0.2,$ then we derive another evolutionary
weighted Fermat-Torricelli tree of degree four and a generalized
Gauss tree of degree three having $x_{G}=3.62$
(Figs.~\ref{fig5},~\ref{fig6},\ref{fig7}).
By substituting $u=x_{G}=3.82$ and the dynamic plasticity
equations (\ref{dynamicplasticexam1}) taken from Example~2 in
(\ref{eq:B_1}), the maximum of (\ref{eq:B_1}) w.r. to
$(B_{4})_{1234}$ yields $(B_{4})_{1234}=1.4901507$ or
$(B_{4})_{1234}=2.0556426.$ By letting $(B_{4})_{1234}=1.4901507$
in (\ref{dynamicplasticexam1}), we get:
$(B_{1})_{1234}=3.0080371,$ $(B_{2})_{1234}=2.4890958,$
$(B_{3})_{1234}=1.7127149,$
and by letting $(B_{4})_{1234}=2.0556426$ in
(\ref{dynamicplasticexam1}), we get: $(B_{1})_{1234}=2.5466101,$
$(B_{2})_{1234}=3.1151456,$ $(B_{3})_{1234}=0.9826002.$
\begin{figure}
\centering
\includegraphics[scale=0.75]{exampleuniv111}
\caption{An evolutionary tree of degree four having an initial
storage level $3.82$ and absorbing rate $a_{G}=0$ with respect to
a boundary weighted rectangle taken from Example~2 for
$(B_{1})_{1234}=3.0080371,$ $(B_{2})_{1234}=2.4890958,$
$(B_{3})_{1234}=1.7127149,$
$(B_{4})_{1234}=1.4901507.$}\label{fig5}
\end{figure}
\begin{figure}
\centering
\includegraphics[scale=0.75]{exampleuniv3}
\caption{An evolutionary tree of degree three having an initial
storage level $3.82$ and absorbing rate $a_{G}=0.2$ with respect
to a boundary weighted rectangle taken from Example~2
$(B_{1})_{1234}=3.0080371,$ $(B_{2})_{1234}=2.4890958,$
$(B_{3})_{1234}=1.7127149,$
$(B_{4})_{1234}=1.4901507.$}\label{fig6}
\end{figure}
The first evolutionary Gauss-Menger tree is obtained by
substituting $x_{G}=3.62,$ $(B_{1})_{1234}=3.0080371,$
$(B_{2})_{1234}=2.4890958,$ $(B_{3})_{1234}=1.7127149,$
$(B_{4})_{1234}=1.4901507$ in (\ref{base1}),
(\ref{a2solvesystem}) and (\ref{a1solvesystem}) from
Lemma~\ref{explicitsolution} (Fig.~\ref{fig6}). Thus, we get:
\begin{eqnarray}
a_{1}=2.5638686, a_{2}=3.4255328 , a_{3}=4.2080591 ,\nonumber\\
a_{4}=3.6397828, l=1.5309344.
\end{eqnarray}
The second evolutionary Gauss-Menger tree is obtained by
substituting $x_{G}=3.62,$ $(B_{1})_{1234}=2.5466101,$
$(B_{2})_{1234}=3.1151456,$ $(B_{3})_{1234}=0.9826002.$
$(B_{4})_{1234}=2.0556426.$ in (\ref{base1}),
(\ref{a2solvesystem}) and (\ref{a1solvesystem}) from
Lemma~\ref{explicitsolution} (Fig.~\ref{fig7}). Thus, we get:
\begin{eqnarray}
a_{1}=2.6836315, a_{2}=3.0204233 , a_{3}=4.0857226 ,\nonumber\\
a_{4}=3.6424502, l=1.8001622.
\end{eqnarray}
\begin{figure}
\centering
\includegraphics[scale=0.75]{exampleuniv4}
\caption{An evolutionary tree of degree three having an initial
storage level $3.82$ and absorbing rate $a_{G}=0.2$ with respect
to a boundary weighted rectangle taken from Example~2
$(B_{1})_{1234}=2.5466101,$ $(B_{2})_{1234}=3.1151456,$
$(B_{3})_{1234}=0.9826002.$
$(B_{4})_{1234}=2.0556426.$}\label{fig7}
\end{figure}
\end{example}
| 140,240
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Monday, June 9, 2014
Week of 06/09/2014
The GOP’s Hypocrisy Is Shameful
Here’s a little piece of advice for the GOP from one of their former card-carrying members:
If you’re going to criticize and castigate the President of the United States over a prisoner exchange, you better check your own record on the subject before you open your damned putrid mouths.
So President Barack Obama authorized a prisoner exchange with the Taliban in Afghanistan. They release one of ours, Bowe Bergdahl, and we release five people from our Guantanamo Base in Cuba.
And that had the GOP up in arms… as usual.
Apparently it’s in the script. Whatever it is that Obama does, the GOP is expected to scream bloody murder over it. They are expected to automatically condemn it, question its legality, accuse Obama of being a dictator, claim to know absolutely nothing about it, and do everything in their power to destroy, discredit, defame, or de-legitimize it.
And, of course, this GOP script gets regurgitated by conservative and neo-conservative talk radio personalities, Fox News, the conservatives on those other “liberal” cable news channels (that supposedly don’t exist), and of course the conservative and neo-conservative editorials and columnists. They all have to be on the same page, don’t you know?
So first there’s the idea that the exchange was “illegal”. Member of Congress weren’t supposedly “notified”.
Wait… this is the same group that refuses to talk to Obama unless he panders to whatever condition they know he won’t give in on? The same group that accuses of Obama of not giving in when he does, and of being a dictator when he shows strength, and being a wimp when he tries to get consensus? The same group that can’t even agree on the time of day even with an atomic clock? The same group that deliberately plays the political equivalent of Abbot and Costello’s “Who’s On First” skit with their rotating “critical issues”? The only “loop” they seem to be interested in when it comes to Obama is the kind that involves a rhetorical rope.
There is a good chance that Obama’s people may have reached out to the legislators and were turned away simply out of habit. Or Obama’s people only reached out to those within their own party. Or, as it was discovered, some of them left early because they just “couldn’t be bothered” with it. I have no doubt that whatever “bare minimum” standard that needed to be met was in this matter. It just was not the GOP’s “standard”, if they even have one, and that presumes too much nowadays.
Then there’s the idea that the United States would be coming to any kind of agreement with the Taliban, the same gang of thugs and criminals that we’ve been going after for over a decade. Oh, right, like we didn’t negotiate with the Vietnamese? Or the North Koreans? We negotiated with the Soviet Union to end the Cuban Missile Crisis. Not too long ago we negotiated with Russia to keep us from getting involved in Syria. This is the same White House that routinely brokers deals with career criminals we otherwise call “Too Big To Fail” so they would pay pocket change instead of jail time.
Besides, it’s not like Obama’s people are strangers to negotiating with idealistic extremists. After all, they’ve had to deal with the Tea Party crowd since 2010. And some of them have lately gotten into the habit of carrying around assault rifles in public places.
Then we have the questioning of the trade itself. According to the GOP script, these five prisoners in Gitmo are “dangerous terrorists”, otherwise why would they be in Gitmo in the first place?
Bear in mind that we’re not talking about a Super-Max correctional facility, where we have people that have been tried and convicted in a court of law so we know the seriousness of their crimes. These could be terrorist leaders, they could be foot soldiers, they could even be some sap who drove a cab or delivered groceries. We really won’t know and we only have the government’s “word” who they were before being snatched up and taken there.
We can’t try them in an actual tribunal – military or civilian – because some chickenhawk “genius” in the Bush Imperium decided it was okay to use “enhanced interrogation methods” on them and violated the Geneva Convention. So these were five people that we were going to have to let go at some point anyway.
Sorry conservatives and neo-conservatives, but you have only yourselves to blame for that one. If they weren’t hating America before they were captured, they sure as hell do now, thanks to you and your Koran-flushing “Our god is better than their god” buddies.
By the way, did you know we also held Haitian refugees there as well? Not just in the early 1990’s, but also after the 2010 earthquake. So we were keeping refugees in the same place as the supposedly “most dangerous terrorists”. Did that make those refugees equally “dangerous”?
Then, of course, the script demanded the GOP go after the serviceman himself. They attacked Bowe Bergdahl’s name, his character, even his family. According to the script, he’s “a deserter”, “a traitor”, even a supposed “sleeper agent”. They’ve even made comparisons to the fictional Showtime series “Homeland”.
Okay, first the “desertion” claims are being investigated. But here’s a possibility that the cons and neo-cons do not want to consider: that he was asked to do something “off the record” by our intelligence community and got captured for it. It certainly would explain the claims made by one of his fellow servicemen that he kept to himself and was studying other languages. But of course we will never really know, because the intelligence community won’t admit to it publicly.
Over fifty years ago, there was a little thing called the Bay of Pigs Invasion. Even today some people refuse to admit that there were American servicemen involved in that failed invasion, but I know there was because my father was one of them. So it is no stretch of the imagination to think that Bowe Bergdahl could have been caught up in something for our country as well.
But you know what? Even that isn’t the worst of it when it comes to the GOP. Character assassination and borderline slander are par for the course nowadays. We shouldn’t be surprised to see them use these tactics.
What makes this whole thing even worse, even more disgusting and despicable and disrespectful, is that prior to Obama actually going through with the trade-off, the GOP actually were demanding Bergdahl be released!
No, I am not exaggerating. I am not making this up. Members of the GOP were the ones that actually demanded the trade-off be made!
Cons and neo-cons were posting Twitter posts over and over and over again demanding that Bergdahl be released at all costs, and then immediately after the news reports came out that Obama actually did it, those same people were screaming bloody murder!
Before the release, Senator James Inhofe of Oklahoma (GOP) said "The mission to bring our missing soldiers home is one that will never end. It's important that we make every effort to bring this captured soldier home to his family."
After the release, he was complaining about “people who have killed Americans, people who are the brain power of Taliban.” Not exactly holding true to the “make every effort” part.
Then there’s Senator and failed Presidential Wannabe John McCain of Arizona (GOP), who actually said on CNN in February that he was open for such a trade-off. But after it actually happened, said that he "would not have made this deal." And this from someone who was a POW himself!
And true to the GOP script, any mention of online praise from the GOP on the release of Bergdahl were quickly erased in Orwellian fashion. “We’ve always been at war with Eastasia.”
Let’s get brutally honest here… the hypocritical actions of the GOP on this subject only further my disgust of the party that I once was a proud card-carrying member of!
You know, it is one thing to be critical of Obama’s actions after they happen. It is another thing entirely to be a blatant hypocrite about the subject, demanding action and then condemning it once it does happen.
The only thing that changed about the effort to get Bergdahl freed was that it was Barack Obama that did it. The GOP seemed to be quite content about screaming to the high heavens to get Bergdahl freed as long as it didn’t happen under Obama’s watch. They were raising Bergdahl up as a martyr for their unholy crusade. Now they’re treating him as a pariah. Why? Because of Obama!
All that talk about “support the troops” and not leaving anyone behind apparently means not a single damn thing to the GOP. “Make every effort”, they said! Or did they really mean “make every effort as long as that ‘that guy’ in the White house doesn’t actually do it and get all the credit for it”?
Keep in mind that I’m not the only one noticing this and is disgusted by the maliciously orchestrated actions of the GOP. Congressman Paul Labrador of Idaho (GOP) is also apparently disgusted by the blatant hypocrisy of his own party. Governor Butch Otter (GOP) is also apparently keeping a level head about this.
“I'm not going to speculate on anything beyond what I know,” Otter said, “and what I know is that after almost five years, a young man from Idaho no longer is in enemy hands. There are processes in place within the military and in Congress for whatever happens next. Those processes need to run their course.”
Sadly, though, I suspect these are the exceptions to the party, and not the rule.
The fact that the town of Hailey, Idaho, had to cancel their planned homecoming parade for Bergdahl due to the orchestrated hate calls coming in should speak volumes of how dangerously malicious this has become, and how the people behind this need to be reined in and shut down.
We’re not talking political disagreement or mild criticism anymore. We’re almost talking acts of domestic terrorism. And all in the name of partisan politics!
It was ten years ago almost to the day that the GOP attacked the credibility of then-Senator John Kerry, questioning his war record and picking apart his stances as a way to discredit him in the 2004 Presidential Election. It was ten years ago that then-Vice President Dick Cheney stood in front of the GOP at their national convention and went down Kerry’s political history, pointing out all of the times he reversed himself.
“Flip-flop, flip-flop, flip-flop,” he chanted.
Well now it’s the GOP’s turn to be picked apart and questioned and discredited for their orchestrated hypocrisy. It is high time the party answer to what they have wrought in the name of Obama-hate.
Flip-flop, flip-flop, flip-flop.
It is high time that Senator Inhofe and the rest of the GOP contingent that speak out of both sides of their mouths on this subject either apologize or else resign for being a disgrace to themselves and to the American people.
Flip-flop, flip-flip, flip-flop.
And for Senator McCain, himself a former prisoner of war and who should know better than anyone else in that party about what “supporting the troops” really means, his shameful hypocrisy should lead to his resignation. But, of course, it won’t. Because that would require something that the overbearing political party has sadly lacked these past few years: integrity.
Flip-flop, flip-flop, flip-flop.
As for the rest of the con and neo-cons out there who have flip-flopped on this subject, I say this: the next time you want a POW released from one of your wars, shut your hole, and rescue them yourselves!
2 comments:
Also, didn't their precious Saint Ronnie (secretly) negotiate with terrorists as well?
Yes, Reagan - or at least his people - negotiated with Iran during the "Iran-Contra" scandal.
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TITLE: How to show how primorials grow asymptotically?
QUESTION [9 upvotes]: The primorial $p_n\# $ is defined as the product of the first $n$ primes:
$$p_n\# = \prod_{k = 1}^n p_k.$$
Asymptotically, primorials grow like
$$p_n\# = e^{(1 + o(1))n\ln n)}.$$
How does one derive this asymptotic formula?
REPLY [11 votes]: Let $\pi(x)$ denote the prime counting function. Let $x\geq 1$. We will compute $\displaystyle\sum_{p\leq x\;prime} (\ln x - \ln p)$ where the sum is taken over all primes $p\leq x$. Note that
$\displaystyle\sum_{p\leq x\;prime} (\ln x - \ln p)=\displaystyle\sum_{p\leq x\;prime}\int^x_p 1/t dt=\int_1^x \frac{\pi(t)}{t}dt$.
Since there exists a constant $c$ such that $\pi(t)\leq \frac{ct}{\ln t}$, it follows that
$\displaystyle\sum_{p\leq x\;prime} (\ln x - \ln p)=O(\int_1^x \frac{1}{\ln t} dt)=o(x)$.
The prime number theorem gives
$\displaystyle\sum_{p\leq x\;prime}\ln p=\pi(x)\ln x - \displaystyle\sum_{p\leq x\;prime} (\ln x - \ln p)= \ln x (\frac{x}{ln x}+o(\frac{x}{ln x}))-o(x)=x+o(x)$.
Then $\displaystyle\sum_{k=1}^n \ln p_k = \displaystyle\sum_{p\leq p_n\;prime}\ln p=p_n+o(p_n)$. It is known that $p_n=n\ln n +o(n\ln n)$; see for example the Wikipedia article on the prime number theorem. Hence $\displaystyle\sum_{k=1}^n \ln p_k =n\ln n + o(n\ln n)$. Taking exponentials of both sides gives the result you asked for.
| 44,136
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TITLE: Gradient in terms of first fundamental form
QUESTION [3 upvotes]: In Do Carmo's Differential Geometry of Curves and Surfaces, I'm having a quite hard time trying to solve Excersise 14 on pages 101-102. He defines the gradient of a differentiable function $f:S\to \Bbb{R}$ as a differentiable map $\text{grad}f:S\to\Bbb{R}^3$ which assigns to each point $p\in S$ a vector $f(p)\in T_p(S)\subset \Bbb{R}^3$ such that
$$\qquad \langle \text{grad}f(p),v\rangle_p=df_p(v)\qquad\text{for all }v\in T_p(S)$$
The question is to express $\text{grad}f$ of $\phi(U)$ in terms of the coefficients $E,F,G$ of the first fundamental form in a parametrization $\phi:U\subset\Bbb{R}^2\to S$. The solution is
$$\qquad \text{grad}f=\frac{f_uG-f_vF}{EG-F^2}\phi_u\,+\,\frac{f_vE-f_uF}{EG-F^2}\phi_v$$
I'm not sure how to start. Sure that if $gradf$ sends points into vectors of the tangent plane, it must be a linear combination of $\phi_u$ and $\phi_v$, but I don't know how to use the information that $\langle \text{grad}f(p),v\rangle_p=df_p(v)$ $\,$for all $v\in T_p(S)$ to get the desired result. The $\dfrac{1}{EG-F^2}$ makes me think about the inverse matrix of the first fundamental form, but again, I don't see how can I relate one thing to the other. Any hints that can point me in the right direction would be appreciate. Thanks in advance!
REPLY [5 votes]: Let $X\colon U\subset \mathbb{R}^2\longrightarrow S$ be a parametrization of a regular surface $S$ and let $p=X(u,v)$. If $f\colon S\rightarrow \mathrm{R}$ is a differentiable function then $\mathrm{grad} f(p)\in T_pS$. Thus, $$\mathrm{grad} f (p)=\alpha X_u+\beta X_v,\ \ \ (\dagger)$$ where $\alpha,\beta $ are functions defined on $U$. From $(\dagger)$ you obtain the following two equations $$f_u=\alpha E+\beta F\ \ \&\ \ f_v=\alpha F+\beta G.$$
By solving the system and substituting into $(\dagger)$ you obtain your desired result.
| 210,879
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TITLE: Awkward behavior of $x^2 \sin\frac{1}{x}$ at $x=0$?
QUESTION [2 upvotes]: According to p. 107 of the book Advanced Calculus by Fitzpatrick,
Assuming that the periodicity and differentiability properties of the sine function are familiar, the following is an example of a differentiable function having a positive derivative at $x = 0$ but such that there is no neighborhood of $0$ on which it is monotonically increasing:
$$f(x) = \begin{cases}x^2\sin 1/x & \text{if}\ x\neq 0 \\
0 & \text{if}\ x = 0\end{cases}$$
The source of this counterintuitive behavior is that the derivative $f'$ is not continuous at $x = 0$.
There are two confusing things about it:
a. By the definition of derivative (at $x=0$), $$\lim_{x\to 0} \dfrac{f(x)-f(0)}{x-0} = \lim_{x\to 0} \dfrac{x^2 \sin \bigl(\frac{1}{x}\bigr)}{x} = 0,$$ since $\bigl\lvert\sin\bigl(\frac{1}{x}\bigr)\bigr\rvert \le 1$.
b. By use of rules for the derivative of products and quotients, $$f'(0) = \biggl[ 2x \sin \biggl(\frac{1}{x}\biggr) + x^2 \biggl(-\frac{1}{x^2} \cos \biggl(\frac{1}{x}\biggr)\biggr) \biggr]_{x=0} $$ which is not defined since $\cos\bigl(\frac{1}{0}\bigr)$ is not defined.
So why the text says that its derivative exists and its value is $>0$?
The function $f(x)$ is monotonically increasing because it is an odd function so for any neighborhood abound $x=0$, $f(x_1>0)>f(x_2<0)$. So why the text says otherwise?
REPLY [1 votes]: 1.- (a) is correct so $f'(0)=0$, but your conclusion on 1.- (b) is wrong. Notice that you are trying to compute the derivative of $f$ at $0$ by using the formula for $f$ on points different from zero. The formula
$$
f'(x) = 2x \sin \biggl(\frac{1}{x}\biggr) + x^2 \biggl(-\frac{1}{x^2} \cos \biggl(\frac{1}{x}\biggr)\biggr)
$$
holds for all $x\neq0$ but, as you know, this formula is undefined at $0$. Therefore, you can conlude that the derivative of $f$ is not continuous at $0$, since you cannot approximate $f'(0)$ by $f'(x)$ for $x$ near $0$. In other words,
$$
f'(0) \neq \lim_{x\rightarrow 0} f'(x).
$$
2.-
In general, odd functions are not monotonically increasing. For example,
$$
x^3 - x
$$
is odd, but it is not monotonic. Indeed, $x^3-x$ is increasing on the intervals $(-\infty,-\tfrac{1}{\sqrt{3}})$ and $(\tfrac{1}{\sqrt{3}},\infty)$, but is decreasing on $(-\tfrac{1}{\sqrt{3}},\tfrac{1}{\sqrt{3}})$.
Back to $f(x) = x^{2}\sin(\tfrac{1}{x})$, you can tell that it is not increasing because it is positive on the intervals $(\tfrac{1}{2k\pi},\tfrac{1}{(2k-1)\pi})$ but negative on $(\tfrac{1}{(2k+1)\pi},\tfrac{1}{2k\pi})$ for every $k\neq0$. Since every neighborhood of $0$ contains infinitely many intervals of those types, we conclude that $f(x)$ has infinitely many changes of sign on any neighborhood of $0$. Thus, it is not monotonic near $0$.
| 196,013
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Settle yourself into a comfy chair with your refreshing beverage of choice at the ready, because have I got quite a story for you. Hopefully one from which everybody can benefit.
I belong to a few screenwriting-oriented networking sites, and do what I can to engage with other members. I do what I can to be friendly, outgoing, and supportive with each connection.
Back in mid-July, I got an email from one such person. Their bio lists them as a “producer, screenwriter, and script consultant”. Would I be interested in a script swap? Despite having a few other reads already lined up, I’m always up for such a thing and agreed, telling them I’d try to get to it soon. Turns out they were in a similar situation.
They sent their script, and I sent mine. After a few days, I’d worked my way through the other projects and started in on their script.
Oh boy.
I won’t say it was awful, but I’d have to say in all honesty it simply wasn’t good. I’d also add that it made me seriously question their credentials.
Among the details:
-a passive protagonist I really didn’t care for, and who didn’t give me any reason to want to see them achieve their goal.
-a weak antagonist with a cartoonish goal
-underdeveloped story/bad structure, including several unresolved subplots and a big letdown of an ending
-unrealistic dialogue
-flat supporting characters
I pointed out what didn’t work for me and why, and offered suggestions of potential fixes. (I always make a point of never ever saying “this is how I’d do it”.) I’d estimate it was around 2 pages worth of notes, and they were free to use or ignore whatever they wanted.
I sent it out Friday afternoon.
Saturday morning, this was the email I got.
“Thanks, Paul.”
Seriously. That was it.
I came to two potential conclusions:
-I was an ignorant know-nothing boob to the nth degree with zero appreciation for their extraordinary skills (“How dare you not recognize my genius!”), and they were just saying “thanks” to be polite
-My notes were so cruel and inhuman, and if that was how we were going to play that game, then they’d be just as ruthless and grind my script into a bloody mess
Hyperbole on my part? Maybe, but check out their response again and think about what your reaction would be.
I figured it was one or the other, but all I could do now was wait (while working on other scripts, naturally).
Quick reminder – this was the end of July.
August passes. No response.
September. Still nothing. (but I did finish the outline of another script, so…yay)
Hmm. Several possibilities now.
-they still haven’t read it
-they read it, but haven’t gotten around to sending the notes
-they forgot. It happens.
-because of what I said about their script, they were deliberately not reading it OR sending the notes. To punish me, I guess?
September came to a close, and I figured I’d been patient enough.
I sent an email – “Know it’s been a while, and I’m sure you’ve been busy, but wanted to check in and see if you’ve had a chance to take a look at my script. Thanks.”
Five days later…
“Best script I ever read.”
Again, that was it.
I asked if they could elaborate. (note – this is my comedy)
Were there any parts you felt could use more work?
“Nope. Perfect.”
What did you think of the characters?
“Outstanding.”
Your thoughts on the jokes?
“I was rolling on the floor laughing.”
Anybody else find this just a tad suspicious, and, oh, total and utter bullshit?
No apology. No remorse. No attempt to make amends. Just a handful of “ain’t I hilarious?” bare minimum answers.
I really wanted to say something in response. Call them out for it. Tell them what an incredibly brazen dick move that was. I even came up with several scenarios to trap them in their sinister web of lies and deceit.
But in the end, I was getting all worked up for nothing. And this person is most definitely NOT worth it. All I’d lost was two hours of reading and writing notes, as well as severing our connection on that networking site. No skin off my nose.
I can only surmise they didn’t like what I had to say, so for whatever reason, decided to not read my script, and after being asked (reminded?) to uphold their end of the deal, took it one step further and opted to not even bother.
I don’t really mind that they didn’t read the script – especially after seeing their writing “skills” in action – but if you’re going to claim you’re a “professional”, then you damned well better act like it. No matter what.
Bet they wouldn’t have done this if I’d been a paying client. Thank goodness it never came to that.
Present yourself as someone who supposedly knows what they’re doing, but then show that’s not the case, and you’re just screwing yourself. Sometimes all you’ve got going for you is your reputation, and once that’s tarnished, you might never be able to restore it.
And let me also add that YOU CAME TO ME. You wanted MY help. And this is how you react because I didn’t like your script? Too fucking bad. Is this how you’re going to treat others who make similar comments? I may not be the most talented or analytical of writers, but at least I treat everybody with respect, even when they don’t deserve it.
When we read another writer’s script, we don’t want it to just be good. We want it to be so phenomenal we can’t believe we had the privilege of being able to read it.
Notes are about the script, not the writer. Of course you’re going to take criticism personally. But you can’t. I have no idea how much work you put into it, but are you more interested in making your script better, or getting a pat on the head and told “Good job”?
I hope this little incident doesn’t deter other writers from taking part in a script swap, including with me. Schedule permitting, I’m always happy to do so. Fortunately, most of my other script-swapping experiences have been of a significantly more positive nature. This was just one of those rare negative exceptions.
Hopefully you have a strong sense of what kind of writer/note-giver the other person is, and once those scripts are swapped, definitely make sure both of you hold up your respective ends of the bargain.
Because the last thing you want is to get on a writer’s bad side.
2 thoughts on “Cave scriptor, indeed*”
Hi Paul,
Thanks for sharing,
A few thoughts I want to share with you.
1- They could have not “appreciated” your notes as it should have been helpful to that person, but they thought it an insult. Also, they are not open as one should be to constructive criticism.
2.-Their way of a few words in a reply was just to shut you up and push you away. No respect at all in getting notes to you on your script. Maybe they felt slighted after reading your notes to them,
3-One may call themselves anything they wish or in their eyes that is what they see themselves as. I could call myself an accountant…after I balanced my checkbook. LOL (But I really can do accounting, baker, seamstress and a lot more.)
4-I see this as not supporting a fellow writer, but insulting them with a few words. EVEN IF the script was the BEST they have ever read, it would be nice to use examples to praise someone over. (example: I love how you gave the girlfriend the boot after she used the protagonist so much. HA HA Accidentially pushed her into the Alligator pit at the zoo)
5-I attended classes in how to give constructive criticism and that was when I was working in Law Enforcement. Go figure. Through experience, I understand this as there are some who have done similar to me. Or argued my notes that I didn’t charge them for. Then why did you ask me to give you notes?
Paul – you are very fair and giving person. Unfortunately, not everyone is like you.
Take care and today is another day – Lauren
I feel as though after reading what you wrote, you have worked out your disappointment. Sometimes it takes longer.
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PDP Chairmanship: Northern Leaders Pick Iyorchia Ayu As Consensus Candidate
The northern leaders of the Peoples Democratic Party (PDP) have picked Senator Iyorchia Ayu as their consensus candidate for the party’s chairmanship position zoned to the region.
This comes after the PDP leaders held a consensus meeting in Bauchi on Wednesday ahead of the party’s national convention.
It suffered a setback as some stakeholders in the party were reported to have shunned the meeting.
Chairman of the Convention Organising Committee and Adamawa State Governor, Umaru Fintiri, then adjourned the meeting to Thursday where the leaders chose Senator Iyorchia.
READ ALSO: Northern PDP Leaders Hold Consensus Meeting Ahead Of National Convention
The PDP leaders from the northern region had met to trim down the number of candidates contesting for the office of the national chairman of the PDP Professor Iyorchia.
Source: Channels TV
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\begin{document}
\title{\bf
The $q$-Onsager algebra and the \\
positive part of $U_q({\widehat{\mathfrak{sl}}}_2)$
}
\author{
Paul Terwilliger}
\date{}
\maketitle
\begin{abstract}
The positive part $U^+_q$ of
$U_q({\widehat{\mathfrak{sl}}}_2)$ has a presentation
by two generators $X,Y$ that satisfy the $q$-Serre relations.
The $q$-Onsager algebra $\mathcal O_q$ has a presentation
by two generators
$A,B$ that satisfy the $q$-Dolan/Grady relations.
We give two results that describe how $U^+_q$ and $\mathcal O_q$ are related.
First,
we consider the filtration of $\mathcal O_q$ whose $n$th component
is spanned by the products of at most $n$ generators.
We show that the associated graded algebra
is isomorphic to $U^+_q$.
Second, we introduce an algebra
$\square_q$ and show how it is related to both $U^+_q$ and
$\mathcal O_q$. The algebra $\square_q$ is defined by
generators and relations. The generators are
$\lbrace x_i \rbrace_{i \in
\mathbb Z_4}$ where $\mathbb Z_4$ is the cyclic group of order 4.
For $i \in \mathbb Z_4$ the generators $x_i, x_{i+1}$ satisfy a $q$-Weyl relation,
and $x_i,x_{i+2}$ satisfy the $q$-Serre relations.
We show that $\square_q$ is related to
$U^+_q$ in the following way.
Let
$ \square^{\rm even}_q$
(resp.
$ \square^{\rm odd}_q$)
denote the subalgebra
of $ \square_q$ generated by
$x_0, x_2$ (resp.
$x_1, x_3$).
We show that
(i) there exists an algebra isomorphism
$U^+_q \to \square^{\rm even}_q$ that sends
$X\mapsto x_0$ and
$Y\mapsto x_2$;
(ii) there exists an algebra isomorphism
$U^+_q \to \square^{\rm odd}_q$ that sends
$X\mapsto x_1$ and
$Y\mapsto x_3$;
(iii) the multiplication map
$\square^{\rm even}_q
\otimes
\square^{\rm odd}_q
\to \square_q$,
$u \otimes v \mapsto uv$ is an isomorphism of
vector spaces.
We show that $\square_q$ is related to
$\mathcal O_q$ in the following way.
For nonzero scalars $a,b$
there exists an injective algebra
homomorphism
$ \mathcal O_q \to \square_q$ that sends
$A \mapsto a x_0+ a^{-1} x_1$ and
$B \mapsto
b x_2+ b^{-1} x_3$.
\bigskip
\noindent
{\bf Keywords}.
Quantum group,
grading, filtration, Onsager algebra.
\hfil\break
\noindent {\bf 2010 Mathematics Subject Classification}.
Primary: 17B37.
\end{abstract}
\section{Introduction}
There is a family of algebras called tridiagonal algebras
\cite[Definition~3.9]{TwoRel}
that come up in the theory of $Q$-polynomial distance-regular
graphs
\cite[Lemma~5.4]{tersub3}
and tridiagonal pairs
\cite[Theorem~10.1]{TD00},
\cite[Theorem~3.10]{TwoRel}.
A tridiagonal algebra has a presentation by
two generators
and two relations of a certain kind, called tridiagonal relations
\cite[Definition~3.9]{TwoRel}.
One example of a tridiagonal algebra is
the positive part
$U^+_q$ of
the quantum affine algebra
$U_q(\widehat {\mathfrak {sl}}_2)$
\cite[Corollary~3.2.6]{lusztig},
\cite[Lines~(18),~(19)]{LS99}.
The algebra $U^+_q$ has a presentation by generators $X,Y$ and
relations
\begin{eqnarray}
&&
X^3 Y - \lbrack 3 \rbrack_q X^2 Y X +
\lbrack 3 \rbrack_q X Y X^2 -Y X^3 = 0,
\label{eq:XXXYIntro}
\\
&&
Y^3 X - \lbrack 3 \rbrack_q Y^2 X Y +
\lbrack 3 \rbrack_q Y X Y^2 -X Y^3 = 0,
\label{eq:YYYXIntro}
\end{eqnarray}
where $\lbrack 3 \rbrack_q = (q^3-q^{-3})/(q-q^{-1})$.
The relations (\ref{eq:XXXYIntro}),
(\ref{eq:YYYXIntro}) are called
the
$q$-Serre relations
\cite{lusztig}.
Applications of $U^+_q$ to tridiagonal pairs can be found in
\cite[Example~1.7]{TD00},
\cite{shape},
\cite[Section~2]{nonnil},
\cite[Lemma~4.8]{TwoRel}.
The algebra
$U^+_q$ plays a prominant role in the theory of
$U_q(\widehat {\mathfrak {sl}}_2)$
\cite{beck3,
beck2,
beck,
damiani,
uqsl2hat,
qtet,
lusztig}.
Another example of a tridiagonal algebra is
the $q$-Onsager algebra $\mathcal O_q$
\cite[Section 2]{bas2},
\cite[Section 1]{bas1}.
This algebra
has a presentation by generators $A,B$
and relations
\begin{eqnarray}
&&
A^3 B - \lbrack 3 \rbrack_q A^2 B A +
\lbrack 3 \rbrack_q A B A^2 -B A^3 = (q^2-q^{-2})^2 (BA-AB),
\label{eq:AAABIntro}
\\
&&
B^3 A - \lbrack 3 \rbrack_q B^2 A B +
\lbrack 3 \rbrack_q B A B^2 -A B^3 = (q^2-q^{-2})^2 (AB-BA).
\label{eq:BBBAIntro}
\end{eqnarray}
The relations
(\ref{eq:AAABIntro}),
(\ref{eq:BBBAIntro})
are called the $q$-Dolan/Grady relations
\cite[Line~(5)]{bas1}.
Applications of
$\mathcal O_q$ to tridiagonal pairs
can be found in
\cite{TD00,
qRacahIT,
ITaug,
LS99,
TwoRel}.
The algebra $\mathcal O_q$ has applications to quantum integrable models
\cite{bas2,
bas1,
BK05,
bas4,
bas5,
bas6,
bas7,
bas8,
basnc,
basXXZ,
basKojima},
reflection equation algebras
\cite{basnc},
and
coideal subalgebras
\cite{bc, kolb}.
There are algebra homomorphisms
from $\mathcal O_q$ into
$U_q(\widehat {\mathfrak {sl}}_2)$
\cite[line (3.15)]{basXXZ},
\cite[line (3.18)]{basXXZ},
\cite[Example~7.6]{kolb},
the $q$-deformed loop algebra
$U_q(L({\mathfrak{sl}}_2))$
\cite[Prop.~2.2]{bas6},
\cite[Prop.~8.5]{qRacahIT},
\cite[Props.~1.1,~1.13]{ITaug},
and the
universal Askey-Wilson algebra $\Delta_q$
\cite[Sections~9,~10]{uaw}.
\medskip
\noindent Consider how
$U^+_q$
and
$\mathcal O_q$
are related.
These algebras have at least a superficial
resemblance, since
for
the
$q$-Serre relations
and
$q$-Dolan/Grady relations
their left-hand sides match.
In this paper our goal is to describe how
$U^+_q$ and
$\mathcal O_q$ are related on an
algebraic level.
Our first main result is
summarized as follows.
For notational convenience abbreviate $\mathcal O = \mathcal O_q$.
We consider the filtration
$\mathcal O_0 \subseteq \mathcal O_1 \subseteq \mathcal O_2 \subseteq \ldots $
of $\mathcal O$ such that for $n \in \mathbb N$
the subspace
$\mathcal O_n$ is spanned by the products
$g_1 g_2 \cdots g_r$ for which $0 \leq r \leq n$
and $g_i$ is among $A,B$ for $1 \leq i \leq r$.
We consider the associated graded algebra
$\overline{\mathcal O}$
in the sense of
\cite[p.~203]{carter}.
We show that
the algebras
$U^+_q$
and
$\overline{\mathcal O}$
are isomorphic.
For our second main result, we introduce
an algebra $\square_q$ and show how it is related
to both
$U^+_q$ and
$\mathcal O_q$. Let
$\mathbb Z_4 = \mathbb Z/4 \mathbb Z$ denote the cyclic
group of order 4.
The algebra $\square_q$ has a presentation by
generators $\lbrace x_i \rbrace_{i \in \mathbb Z_4}$ and relations
\begin{eqnarray*}
&&
\quad \qquad \qquad
\frac{q x_i x_{i+1}-q^{-1}x_{i+1}x_i}{q-q^{-1}} = 1,
\\
&&
x^3_i x_{i+2} - \lbrack 3 \rbrack_q x^2_i x_{i+2} x_i +
\lbrack 3 \rbrack_q x_i x_{i+2} x^2_i -x_{i+2} x^3_i = 0.
\end{eqnarray*}
We show that $\square_q$ is related to
$U^+_q$ in the following way.
Let
$ \square^{\rm even}_q$
(resp.
$ \square^{\rm odd}_q$)
denote the subalgebra
of $ \square_q$ generated by
$x_0, x_2$ (resp.
$x_1, x_3$).
We show that
(i) there exists an algebra isomorphism
$U^+_q \to \square^{\rm even}_q$ that sends
$X\mapsto x_0$ and
$Y\mapsto x_2$;
(ii) there exists an algebra isomorphism
$U^+_q \to \square^{\rm odd}_q$ that sends
$X\mapsto x_1$ and
$Y\mapsto x_3$;
(iii) the multiplication map
$\square^{\rm even}_q
\otimes
\square^{\rm odd}_q
\to \square_q$,
$u \otimes v \mapsto uv$ is an isomorphism of
vector spaces.
We show that $\square_q$ is related to
$\mathcal O_q$ in the following way.
For nonzero scalars $a,b$
there exists an injective algebra
homomorphism
$ \mathcal O_q \to \square_q$ that sends
\begin{eqnarray}
A \mapsto a x_0+ a^{-1} x_1, \qquad \qquad B \mapsto
b x_2+ b^{-1} x_3.
\label{eq:IntroMap}
\end{eqnarray}
Our two main results are obtained under the following
assumptions.
The underlying field is arbitrary.
The scalar $q$ is nonzero
and $q^2 \not=1$.
Our two main results are closely related, and
will be proved more or less simultaneously.
These proofs use only linear algebra, and
do not employ facts invoked from the literature.
Our proof strategy is to introduce several
algebras $\widetilde \square_q$, $\widehat \square_q$
that are related to $\square_q$ via surjective algebra homomorphisms
$\widetilde \square_q \to \widehat \square_q \to \square_q$.
The algebra $\widetilde \square_q$ is very general, and
an explicit basis will be given.
Using this basis
we will obtain some facts about
$\widetilde \square_q$, which give facts about
$\widehat \square_q$ and
$\square_q$
via the
homomorphisms
$\widetilde \square_q \to \widehat \square_q \to \square_q$.
These facts
yield
an algebra homomorphism $\mathcal O_q \to \widehat \square_q$
such that
the composition
$\mathcal O_q \to \widehat \square_q \to \square_q$ is
injective. This composition is the homomorphism
(\ref{eq:IntroMap}).
\medskip
\noindent Near the end of the paper we will discuss how
$\widehat \square_q$ and $\square_q$ are related to
$U_q(\widehat {\mathfrak {sl}}_2)$ and the $q$-tetrahedrom
algebra $\boxtimes_q$ from
\cite{qtet}. We will obtain a commuting diagram of
algebra homomorphisms
\begin{equation*}
\begin{CD}
\widehat \square_q @>>>
U_q(\widehat {\mathfrak {sl}}_2)
\\
@VVV @VVV \\
\square_q @>>>
\boxtimes_q
\end{CD}
\end{equation*}
In this diagram the homomorphism
$\square_q \to \boxtimes_q$ is injective.
Using the diagram we will explain
the homomorphisms $\mathcal O_q \to
U_q(\widehat {\mathfrak {sl}}_2) $
and
$\mathcal O_q \to
U_q(L({\mathfrak {sl}}_2)) $
that we mentioned earlier in this
section.
\section{Preliminaries}
\noindent We now begin our formal argument.
Throughout this paper the following notation
and assumptions are
in effect.
Recall the
natural numbers
$\mathbb N = \lbrace 0, 1, 2,\ldots \rbrace$ and
integers
$\mathbb Z = \lbrace 0, \pm 1, \pm 2,\ldots \rbrace$.
We will be discussing algebras.
An algebra is meant to be associative
and have a 1. A subalgebra has the same 1 as the parent
algebra. Let $\mathbb F$ denote a field and
let $\mathcal A$ denote an $\mathbb F$-algebra.
Let $H,K$ denote subspaces of
the $\mathbb F$-vector space $\mathcal A$.
Then $HK$
denotes the subspace
of $\mathcal A$ spanned by
$\lbrace hk | h\in H, k \in K\rbrace$.
By an {\it $\mathbb N$-grading} of $\mathcal A$
we mean a sequence $\lbrace \mathcal A_n\rbrace_{n \in \mathbb N}$
such that (i) each $\mathcal A_n$ is a subspace
of the $\mathbb F$-vector space $\mathcal A$;
(ii) $1 \in \mathcal A_0$;
(iii) the sum $\mathcal A = \sum_{n \in \mathbb N} \mathcal A_n$ is direct;
(iv) $\mathcal A_r \mathcal A_s \subseteq A_{r+s}$ for
$r,s\in \mathbb N$.
A $\mathbb Z$-grading of $\mathcal A$ is similarly defined.
\medskip
\noindent
We will be discussing algebras defined by generators and relations.
Let $T$ denote the $\mathbb F$-algebra with generators $x,y$ and
no relations; $T$ is often called a free algebra or tensor algebra.
The generators $x,y$ will be called {\it standard}.
For $n \in \mathbb N$, a {\it word of length $n$} in
$T$ is a product $g_1 g_2 \cdots g_n$ such that
$g_i$ is a standard generator
for $1 \leq i \leq n$.
We interpret the word of length zero to be the multiplicative
identity in $T$. The words in $T$ form a basis for the
$\mathbb F$-vector space $T$.
For $n \in \mathbb N$
the words of length $n$ in $T$ form a basis for
a subspace $T_n$ of $T$.
For example,
$1$ is a basis for
$T_0$ and $x,y$ is a basis for $T_1$.
By construction
the sum $T=\sum_{n\in \mathbb N} T_n$ is direct.
Also by construction
$T_r T_s = T_{r+s}$ for $r,s\in \mathbb N$.
By these comments
the sequence $\lbrace T_n\rbrace_{n \in \mathbb N}$ is
an $\mathbb N$-grading of $T$.
\medskip
\noindent
Fix $0 \not=q \in \mathbb F$ such that $q^2\not=1$.
Define
\begin{eqnarray*}
\lbrack n \rbrack_q = \frac{q^n-q^{-n}}{q-q^{-1}}
\qquad \quad n \in \mathbb Z.
\end{eqnarray*}
\noindent All unadorned tensor products are meant to be
over $\mathbb F$.
\section{The algebra $U^{+}_q$}
\noindent Later in the paper we will discuss the
quantum affine algebra
$U_q(\widehat {\mathfrak {sl}}_2)$. In the meantime
we consider
a certain subalgebra of
$U_q(\widehat {\mathfrak {sl}}_2)$, denoted $U^+_q$ and called the positive
part of
$U_q(\widehat {\mathfrak {sl}}_2)$.
\begin{definition}
\label{def:Aq}
\rm
(See \cite[Corollary~3.2.6]{lusztig}.)
Let $U^+=U^+_q$ denote the $\mathbb F$-algebra with
generators $X,Y$ and relations
\begin{eqnarray}
&&
X^3 Y - \lbrack 3 \rbrack_q X^2 Y X +
\lbrack 3 \rbrack_q X Y X^2 -Y X^3 = 0,
\label{eq:XXXY}
\\
&&
Y^3 X - \lbrack 3 \rbrack_q Y^2 X Y +
\lbrack 3 \rbrack_q Y X Y^2 -X Y^3 = 0.
\label{eq:YYYX}
\end{eqnarray}
The algebra $U^+$ is called the {\it positive part of
$U_q(\widehat {\mathfrak {sl}}_2)$}.
The relations (\ref{eq:XXXY}),
(\ref{eq:YYYX}) are called the
{\it $q$-Serre relations}.
\end{definition}
\begin{lemma}
\label{lem:xi}
Let $a,b$ denote nonzero scalars in $\mathbb F$.
Then there exists an automorphism of $U^+$ that sends
$X \mapsto a X$ and
$Y \mapsto b Y$.
\end{lemma}
\begin{proof} Use Definition
\ref{def:Aq}.
\end{proof}
\noindent
Recall the free algebra $T$ with standard generators
$x,y$.
There exists an $\mathbb F$-algebra homomorphism
$\mu: T \to U^+$ that sends $x\mapsto X$ and
$y\mapsto Y$.
The homomorphism $\mu$ is surjective.
We now describe the kernel of $\mu$.
Define elements $S_x, S_y$ in $T$ by
\begin{eqnarray}
\label{eq:SgenP}
&&S_x = x^3 y - \lbrack 3 \rbrack_q x^2 y x +
\lbrack 3 \rbrack_q x y x^2 -y x^3,
\\
&&
S_y = y^3 x - \lbrack 3 \rbrack_q y^2 x y +
\lbrack 3 \rbrack_q y x y^2 -x y^3.
\label{eq:SgenM}
\end{eqnarray}
Let
$S = T S_x T + T S_y T$
denote the 2-sided ideal of $T$ generated by
$S_x,S_y$.
Then $S$ is the kernel of $\mu$.
Recall the $\mathbb N$-grading $\lbrace T_n \rbrace_{n\in \mathbb N}$
of $T$.
For $n \in \mathbb N$ let
$U^+_n$ denote the image of
$T_n$ under $\mu$.
The $\mathbb F$-vector space $U^+_n$ is spanned
by the products $g_1g_2\cdots g_n$ such that
$g_i$ is among $X,Y$ for $1 \leq i \leq n$.
Let $\mu_n$ denote the
restriction of $\mu$ to $T_n$.
We view $\mu_n: T_n \to U^+_n$.
The kernel of $\mu_n $ is $S\cap T_n$.
The subspace $S \cap T_n$ is described as follows.
Note that $S_x,S_y \in T_4$.
So for $r,s\in \mathbb N$,
$T_r S_x T_s \subseteq T_{r+s+4}$
and
$T_r S_y T_s \subseteq T_{r+s+4}$.
Consequently for $n \in \mathbb N$,
\begin{eqnarray}
S\cap T_n =
\sum_{
\genfrac{}{}{0pt}{}{r,s \in \mathbb N}{r+s=n-4}
}
T_r S_x T_s+
\sum_{
\genfrac{}{}{0pt}{}{r,s \in \mathbb N}{r+s=n-4}
}
T_r S_y T_s.
\label{eq:Sn}
\end{eqnarray}
Assume for the moment that $n\leq 3$.
There does not exist $r,s \in \mathbb N$ such that
$r+s=n-4$. Therefore
$S \cap T_n=0$, so
$\mu_n: T_n \to U^+_n$ is an isomorphism.
Taking $n=0,1$ we see that
$1$ is a basis for $U^+_0$ and
$X,Y$ is a basis for $U^+_1$.
We show that the sequence
$\lbrace U^+_n \rbrace_{n \in \mathbb N}$ is an $\mathbb N$-grading
of $U^+$.
By
(\ref{eq:Sn}) and $T=\sum_{r\in \mathbb N} T_r $
we obtain
$S=\sum_{n\in \mathbb N} (S\cap T_n)$.
Therefore the sum
$U^+ = \sum_{n \in \mathbb N} U^+_n$ is direct.
Recall that $T_r T_s = T_{r+s}$ for $r,s \in \mathbb N$.
Applying $\mu$ we find that
$U^+_r U^+_s = U^+_{r+s}$ for
$r,s\in \mathbb N$.
By these comments the sequence
$\lbrace U^+_n \rbrace_{n \in \mathbb N}$ is an $\mathbb N$-grading
of $U^+$.
\section{The $q$-Onsager algebra}
In this section we recall the $q$-Onsager algebra
and discuss its basic properties.
\begin{definition}
\label{def:qOnsager}
\rm
(See
\cite[Section~2]{bas2},
\cite[Definition~3.9]{TwoRel}.)
Let $\mathcal O = \mathcal O_q$ denote the $\mathbb F$-algebra with
generators $A,B$ and relations
\begin{eqnarray}
&&
A^3 B - \lbrack 3 \rbrack_q A^2 B A +
\lbrack 3 \rbrack_q A B A^2 -B A^3 = (q^2-q^{-2})^2 (BA-AB),
\label{eq:AAAB}
\\
&&
B^3 A - \lbrack 3 \rbrack_q B^2 A B +
\lbrack 3 \rbrack_q B A B^2 -A B^3 = (q^2-q^{-2})^2 (AB-BA).
\label{eq:BBBA}
\end{eqnarray}
We call $\mathcal O$ the
{\it $q$-Onsager algebra}.
We call
(\ref{eq:AAAB}),
(\ref{eq:BBBA})
the {\it $q$-Dolan/Grady relations}.
\end{definition}
\noindent
Consider the elements
$\lbrace A^r B^s\rbrace_{r,s\in \mathbb N}$
in the $\mathbb F$-vector space $\mathcal O$.
We show that these elements are linearly independent.
Let $A^\vee, B^\vee$ denote commuting indeterminates.
Let $\mathbb F\lbrack A^\vee, B^\vee\rbrack$
denote the $\mathbb F$-algebra consisting of the polynomials
in
$ A^\vee, B^\vee$ that have all coefficients in
$\mathbb F$.
The elements $\lbrace (A^\vee)^r (B^\vee)^s\rbrace_{r,s\in \mathbb N}$
form a basis for
the $\mathbb F$-vector space $\mathbb F\lbrack A^\vee, B^\vee\rbrack$.
By the nature of the relations
(\ref{eq:AAAB}),
(\ref{eq:BBBA})
there exists an $\mathbb F$-algebra homomorphism
$\mathcal O \to
\mathbb F\lbrack A^\vee, B^\vee\rbrack$
that sends $A \mapsto A^\vee$ and
$B \mapsto B^\vee$.
By these comments
the elements $\lbrace A^r B^s\rbrace_{r,s\in \mathbb N}$
in $\mathcal O$
are linearly independent.
In particular the elements $1,A,B$ in $\mathcal O$
are linearly independent.
Recall the free algebra $T$ with standard generators
$x,y$. There exists an $\mathbb F$-algebra homomorphism
$\nu: T \to \mathcal O$ that sends $x \mapsto A$
and $y\mapsto B$. The homomorphism $\nu$ is surjective.
For $n \in \mathbb N$ let
$\mathcal O_n$ denote the image of $T_0+T_1+\cdots+T_n$ under $\nu$.
The $\mathbb F$-vector space $\mathcal O_n$ is
spanned by the products
$g_1g_2\cdots g_r$ such that $0 \leq r \leq n$ and
$g_i$ is among $A,B$ for $1 \leq i \leq r$.
For example
$\mathcal O_0 = \mathbb F 1$ and
$\mathcal O_1 = \mathbb F 1 + \mathbb F A + \mathbb F B$.
For notational convenience define $\mathcal O_{-1} = 0$.
For $n \in \mathbb N$ we have
$\mathcal O_{n-1} + \nu(T_n)=\mathcal O_n$
and in particular $\mathcal O_{n-1} \subseteq \mathcal O_n$.
Since $\nu$ is surjective we have
$\mathcal O = \cup_{n \in \mathbb N} \mathcal O_n$. By
construction $\mathcal O_r \mathcal O_s = \mathcal O_{r+s}$
for $r,s \in \mathbb N$. The sequence
$\lbrace \mathcal O_n \rbrace_{n \in \mathbb N}$
is a filtration of $\mathcal O$ in the sense of
\cite[p.~202]{carter}.
For $n \in \mathbb N$ consider the quotient $\mathbb F$-vector
space $\overline {\mathcal O}_n
= \mathcal O_n /\mathcal O_{n-1}$.
We view $\overline{\mathcal O}_0 =\mathcal O_0$.
The elements
$\overline{A}, \overline{B}$ form a basis for
$\overline {\mathcal O}_1$, where
$\overline{A}=A+\mathcal O_0$ and $\overline{B}= B+\mathcal O_0$.
Now consider
the formal direct sum
$\overline {\mathcal O} = \sum_{n \in \mathbb N}
\overline {\mathcal O}_n$.
We emphasize that for all elements
$u = \sum_{n \in \mathbb N} u_n$ in
$\overline {\mathcal O}$,
the summand $u_n$ is nonzero for finitely many $n \in \mathbb N$.
By construction
$\overline{\mathcal O}$
is an $\mathbb F$-vector space.
We next define a product
$\overline {\mathcal O} \times \overline {\mathcal O} \to
\overline {\mathcal O}$
that turns $\overline{\mathcal O}$ into an $\mathbb F$-algebra.
For $r,s\in \mathbb N$ the product sends
$\overline {\mathcal O}_r \times \overline {\mathcal O}_s
\to
\overline {\mathcal O}_{r+s}$ as follows.
For $u\in \mathcal O_r$ and
$v\in \mathcal O_s$ the product of
$u+\mathcal O_{r-1}$ and
$v+\mathcal O_{s-1}$ is
$uv+\mathcal O_{r+s-1}$.
We have turned
$\overline {\mathcal O}$
into an $\mathbb F$-algebra
\cite[p.~203]{carter}.
By construction
$
\overline {\mathcal O}_r
\overline {\mathcal O}_s =
\overline {\mathcal O}_{r+s}$ for $r,s \in \mathbb N$.
For $n\in \mathbb N$ the
$\mathbb F$-vector space $\overline{\mathcal O}_n$ is
spanned by the products $\overline{g}_1 \overline{g}_2 \cdots \overline{g}_n$
such that $g_i$ is among $A,B$ for $1 \leq i \leq n$.
The $\mathbb F$-algebra
$\overline{\mathcal O}$ is generated by
$\overline {A}, \overline {B}$.
The sequence $\lbrace
\overline {\mathcal O}_n \rbrace_{n \in \mathbb N}$ is
an $\mathbb N$-grading of $\overline {\mathcal O}$. The
$\mathbb F$-algebra
$\overline {\mathcal O}$ is called
the {\it graded algebra
associated with the filtration
$\lbrace \mathcal O_n \rbrace_{n \in \mathbb N}$}
\cite[p.~203]{carter}.
We now construct an $\mathbb F$-algebra homomorphism
$\overline {\nu}: T \to \overline{\mathcal O}$.
For $n \in \mathbb N$ let $\nu_n$ denote the
restriction of $\nu$ to $T_n$. We view
$\nu_n: T_n \to \mathcal O_n$. Consider the
composition
\begin{equation*}
\begin{CD}
\overline{\nu}_n: \quad T_n @>> \nu_n >
\mathcal O_n
@>> u\mapsto u+\mathcal O_{n-1} > \overline{\mathcal O}_n.
\end{CD}
\end{equation*}
The map $\overline{\nu}_n:T_n\to \overline{\mathcal O}_n$
is $\mathbb F$-linear and surjective.
Define an $\mathbb F$-linear map $\overline{\nu}:T \to
\overline{\mathcal O}$ that
acts on $T_n$ as $\overline{\nu}_n$ for
$n \in \mathbb N$.
By construction $\overline{\nu}(T_n)=\overline{\mathcal O}_n$
for $n \in \mathbb N$.
The map $\overline{\nu}$ sends
$x \mapsto \overline{A}$ and
$y \mapsto \overline{B}$.
\begin{lemma}
The map $\overline{\nu}:T\to \overline{\mathcal O}$
is an $\mathbb F$-algebra homomorphism.
\end{lemma}
\begin{proof} Pick $r,s \in \mathbb N$.
It suffices to show that
$ \overline{\nu}(uv)=\overline{\nu}(u) \overline{\nu}(v)$
for $u \in T_r$ and
$v \in T_s$.
This equation is routinely verified using the definition of
the map $\overline{\nu}$ and the algebra $\overline{\mathcal O}$.
\end{proof}
\noindent The algebras $U^+$ and $\overline{\mathcal O}$ are related
as follows.
Using
(\ref{eq:AAAB}),
(\ref{eq:BBBA})
we obtain
\begin{eqnarray}
&&
\overline {A}^3 \overline{B} - \lbrack 3 \rbrack_q
\overline {A}^2 \overline{B} \, \overline{A} +
\lbrack 3 \rbrack_q
\overline{A}\,\overline{ B}\, \overline{A}^2 -
\overline{B}\, \overline{A}^3 =0,
\label{eq:AAABgraded}
\\
&&
\overline {B}^3 \overline{A} - \lbrack 3 \rbrack_q
\overline {B}^2 \overline{A}\, \overline{B} +
\lbrack 3 \rbrack_q
\overline{B} \,\overline{ A} \,\overline{B}^2 -
\overline{A} \, \overline{B}^3 =0.
\label{eq:BBBAgraded}
\end{eqnarray}
By
(\ref{eq:AAABgraded}),
(\ref{eq:BBBAgraded})
there exists an $\mathbb F$-algebra
homomorphism
$\psi: U^+ \to \overline{\mathcal O}$ that sends
$X\mapsto \overline {A}$ and
$Y\mapsto \overline {B}$.
The homomorphism $\psi$ is surjective.
By construction
$\psi(U^+_n)=
\overline{\mathcal O}_n$ for $n \in \mathbb N$.
\begin{lemma}
\label{lem:munu}
The following diagram commutes:
\begin{equation*}
\begin{CD}
T @>I > >
T
\\
@V\mu VV @VV\overline{\nu} V \\
U^+ @>>\psi >
\overline{\mathcal O}
\end{CD}
\end{equation*}
\end{lemma}
\begin{proof} Each map in the diagram is an
$\mathbb F$-algebra homomorphism. To verify
that the diagram commutes, chase the standard generators
$x,y$ around the diagram.
\end{proof}
\begin{theorem}
\label{thm:main1}
The map $\psi : U^+\to \overline{\mathcal O}$ is
an isomorphism of $\mathbb F$-algebras.
\end{theorem}
\noindent The proof of Theorem
\ref{thm:main1} will be completed in
Proposition
\ref{prop:twoInj}.
\medskip
\noindent We mention one significance of
Lemma
\ref{lem:munu} and
Theorem
\ref{thm:main1}.
Consider the free algebra $T$ with $\mathbb N$-grading
$\lbrace T_n \rbrace_{n \in \mathbb N}$.
Pick $u \in T$. For $n \in \mathbb N$,
we say that $u$ is {\it $n$-homogeneous}
whenever $u \in T_n$. We say that $u$ is
{\it homogeneous} whenever there exists $n \in \mathbb N$
such that $u$ is $n$-homogeneous.
\begin{definition}\rm
A subset $\Omega \subseteq T$ is said to be {\it homogeneous}
whenever each element in $\Omega$ is homogeneous.
\end{definition}
\begin{proposition}
\label{prop:Omega}
Let $\Omega$ denote a homogeneous subset of $T$ such
that the vectors
$ \mu(z)$ $(z \in \Omega)$
form a basis for the $\mathbb F$-vector space $U^+$.
Then the vectors
$\nu(z)$ $(z \in \Omega)$ form
a basis for the $\mathbb F$-vector space
$\mathcal O$.
\end{proposition}
\begin{proof}
Recall the $\mathbb F$-algebra isomorphism
$\psi: U^+ \to \overline{\mathcal O}$ from
Theorem
\ref{thm:main1}.
For $n \in \mathbb N$ let
$\Omega_n$ denote the set of
$n$-homogeneous elements in
$\Omega$.
We have $\Omega_n \subseteq T_n$ and
$\mu(T_n)=U^+_n$,
so
$\mu(z) \in U^+_n$ for $z \in \Omega_n$.
By assumption $\Omega$ is homogeneous, so
$\Omega = \cup_{n\in \mathbb N}\Omega_n$.
Since the vectors
$\mu(z)$ $(z \in \Omega)$
form a basis for $U^+$
and the sum
$U^+=\sum_{n \in \mathbb N} U^+_n$ is direct, we see that
for $n \in \mathbb N$ the vectors
$\mu(z)$ $(z \in \Omega_n)$
form a basis for $U^+_n$.
We mentioned above Lemma
\ref{lem:munu}
that
$\psi(U^+_n) = \overline{\mathcal O}_n$,
so
the vectors
$\psi(\mu(z))$ $(z \in \Omega_n)$
form
a basis for $\overline{\mathcal O}_n$.
By
Lemma
\ref{lem:munu}
we have $\psi(\mu(z)) = \overline{\nu}(z)$ for $z \in \Omega_n$,
so the vectors
$\overline{\nu}(z)$ $(z \in \Omega_n)$
form
a basis for $\overline{\mathcal O}_n$.
Consequently the vectors
$\nu(z)$ $(z \in \Omega_n)$
form a basis for
a complement of
$\mathcal O_{n-1}$ in
$\mathcal O_n$.
Therefore the vectors
$\nu(z)$ $(z \in \Omega)$
form a basis for $\mathcal O$.
\end{proof}
\begin{note}\rm
Assume that $\mathbb F$ is algebraically
closed with characteristic zero, and $q$ is not
a root of unity.
In
\cite[Theorem 2.29]{shape}, T. Ito and the present
author display a homogeneous subset $\Omega$ of
$T$ such that the vectors
$\mu(z)$ $(z \in \Omega)$
form a basis for
the
$\mathbb F$-vector space $U^+$.
In \cite[Theorem~2.1]{ITaug}
the same authors show that the vectors
$\nu(z)$ $(z \in\Omega)$
form a basis for the $\mathbb F$-vector space
$\mathcal O$. We point out
that
\cite[Theorem~2.1]{ITaug} follows from
\cite[Theorem 2.29]{shape} and
Proposition \ref{prop:Omega}.
\end{note}
\section{The algebra $\square_q$}
\noindent We have been discussing the
algebra $U^+=U^+_q$ which is the positive part of
$U_q(\widehat {\mathfrak {sl}}_2)$, and the $q$-Onsager
algebra $ \mathcal O = \mathcal O_q$.
As we compare these algebras,
it is useful to bring in another algebra
$\square_q$.
In this section
we introduce $\square_q$, and describe how it
is related to
$U^+$ and $\mathcal O$.
\medskip
\noindent Let $\mathbb Z_4 = {\mathbb Z} /4 \mathbb Z$
denote the cyclic group of order $4$.
\begin{definition}
\rm
\label{def:boxqV1M}
Let $\square_q$ denote the $\mathbb F$-algebra with
generators $\lbrace x_i\rbrace_{i\in \mathbb Z_4}$
and relations
\begin{eqnarray}
&&
\quad \qquad \qquad
\frac{q x_i x_{i+1}-q^{-1}x_{i+1}x_i}{q-q^{-1}} = 1,
\label{eq:centralM}
\\
&&
x^3_i x_{i+2} - \lbrack 3 \rbrack_q x^2_i x_{i+2} x_i +
\lbrack 3 \rbrack_q x_i x_{i+2} x^2_i -x_{i+2} x^3_i = 0.
\label{eq:qSerreM}
\end{eqnarray}
\end{definition}
\noindent We have some comments.
\begin{lemma}
\label{lem:aut1M}
There exists an automorphism $\rho$ of
$\square_q$ that sends $x_i \mapsto x_{i+1}$ for
$i \in \mathbb Z_4$. Moreover $\rho^4=1$.
\end{lemma}
\begin{lemma}
\label{lem:aM}
For $0 \not=a \in \mathbb F$ there exists
an automorphism of $\square_q$ that sends
\begin{eqnarray*}
x_0 \mapsto a x_0, \qquad \quad
x_1 \mapsto a^{-1} x_1, \qquad \quad
x_2 \mapsto a x_2, \qquad \quad
x_3 \mapsto a^{-1} x_3.
\end{eqnarray*}
\end{lemma}
\noindent Our next goal is to describe how $\square_q$ is related to
$U^+$.
\begin{definition}
\label{def:squareTeToM}
\rm Define the subalgebras
$ \square^{\rm even}_q$,
$ \square^{\rm odd}_q$
of $ \square_q$ such that
\begin{enumerate}
\item[\rm (i)]
$ \square^{\rm even}_q$
is
generated by $x_0, x_2$;
\item[\rm (ii)]
$ \square^{\rm odd}_q$
is
generated by $x_1, x_3$.
\end{enumerate}
\end{definition}
\begin{proposition}
\label{thm:tensorDecPreM}
The following {\rm (i)--(iii)} hold:
\begin{enumerate}
\item[\rm (i)]
there exists an $\mathbb F$-algebra isomorphism
$U^+ \to \square^{\rm even}_q$ that sends
$X\mapsto x_0$ and
$Y\mapsto x_2$;
\item[\rm (ii)]
there exists an $\mathbb F$-algebra isomorphism
$U^+ \to \square^{\rm odd}_q$ that sends
$X\mapsto x_1$ and
$Y\mapsto x_3$;
\item[\rm (iii)]
the following is an isomorphism of
$\mathbb F$-vector spaces:
\begin{eqnarray*}
\square^{\rm even}_q
\otimes
\square^{\rm odd}_q
& \to & \square_q
\\
u \otimes v &\mapsto & uv
\end{eqnarray*}
\end{enumerate}
\end{proposition}
\noindent The proof of Proposition
\ref{thm:tensorDecPreM}
will be completed in Section 10.
\medskip
\noindent Next we describe how
$\square_q$ is related to $\mathcal O$.
\begin{proposition}
\label{prop:ABxyiLongerM}
Pick nonzero $a,b \in \mathbb F$.
Then there exists an $\mathbb F$-algebra
homomorphism
$ \sharp : \mathcal O \to \square_q$ that sends
\begin{eqnarray}
\label{eq:ABDefMapM}
A \mapsto a x_0+ a^{-1} x_1, \qquad \qquad B \mapsto
b x_2+ b^{-1} x_3.
\end{eqnarray}
\end{proposition}
\noindent The proof of Proposition
\ref{prop:ABxyiLongerM}
will be completed in Section 8.
In
Theorem \ref{thm:xiInj}
we will show that
the map $\sharp$ from
Proposition
\ref{prop:ABxyiLongerM} is injective.
\section{The algebra $\widetilde \square_q$}
\noindent In the previous section we introduced the
algebra $\square_q$. As we investigate $\square_q$,
it is useful to consider a certain homomorphic preimage
denoted $\widetilde \square_q$. In this section
we introduce
$\widetilde \square_q$ and describe its basic properties.
\begin{definition}
\label{def:boxqV2}
\rm
Let $\widetilde \square_q$ denote the $\mathbb F$-algebra with
generators $c^{\pm 1}_i, x_i$ $(i \in \mathbb Z_4)$
and relations
\begin{eqnarray}
&&c_i c^{-1}_i = c^{-1}_i c_i = 1,
\label{eq:cci}
\\
&&\mbox{$c^{\pm 1}_i$ is central in $\widetilde \square_q$},
\label{eq:cciCentral}
\\
&& \frac{q x_i x_{i+1}-q^{-1}x_{i+1}x_i}{q-q^{-1}} = c_i.
\label{eq:central2}
\end{eqnarray}
\end{definition}
\noindent We have some comments.
\begin{lemma}
\label{lem:aut2}
There exists an automorphism $\widetilde \rho$ of
$\widetilde \square_q$ that sends
$c_i \mapsto c_{i+1}$ and $x_i \mapsto x_{i+1}$ for
$i \in \mathbb Z_4$. Moreover $\widetilde \rho^4=1$.
\end{lemma}
\begin{lemma}
\label{lem:squareCanon}
There exists a unique $\mathbb F$-algebra homomorphism
$ \widetilde \square_q \to \square_q$ that
sends $c^{\pm 1}_i \mapsto 1$ and
$x_i \mapsto x_i$ for $i \in \mathbb Z_4$.
This homomorphism is surjective.
\end{lemma}
\begin{definition}
\label{def:Canon}
\rm The homomorphism
$ \widetilde \square_q \to \square_q$ from
Lemma
\ref{lem:squareCanon} will be called {\it canonical}.
\end{definition}
\begin{definition}
\label{def:TeTo}
\rm Define the subalgebras
$\widetilde \square^{\rm even}_q$,
$\widetilde \square^{\rm odd}_q$,
$\widetilde C$ of
$\widetilde \square_q$ such that
\begin{enumerate}
\item[\rm (i)]
$\widetilde \square^{\rm even}_q$
is
generated by $x_0, x_2$;
\item[\rm (ii)]
$\widetilde \square^{\rm odd}_q$
is
generated by $x_1, x_3$;
\item[\rm (iii)]
$\widetilde C$ is
generated by $\lbrace c^{\pm 1}_i\rbrace_{i \in \mathbb Z_4}$.
\end{enumerate}
\end{definition}
\begin{lemma}
\label{lem:xiadjT}
Let $\lbrace \alpha_i \rbrace_{i\in \mathbb Z_4}$
denote invertible elements in
$\widetilde C$.
Then there exists
an $\mathbb F$-algebra homomorphism
$\widetilde \square_q\to \widetilde \square_q $ that sends
$x_i \mapsto \alpha_i x_i $ and
$c_i \mapsto \alpha_i \alpha_{i+1} c_i$ for $i \in \mathbb Z_4$.
\end{lemma}
\begin{definition}
\label{def:alphaAutT}
\rm
The homomorphism in Lemma
\ref{lem:xiadjT} will be denoted by $\widetilde g(\alpha_0, \alpha_1, \alpha_2, \alpha_3)$.
\end{definition}
\begin{lemma}
\label{lem:alphaAutT}
Referring to Lemma
\ref{lem:xiadjT} and Definition
\ref{def:alphaAutT}, assume that $0 \not=\alpha_i \in \mathbb F$
for $i \in \mathbb Z_4$. Then
$\widetilde g(\alpha_0, \alpha_1, \alpha_2, \alpha_3)$ is an automorphism
of $\widetilde \square_q$. Its inverse is
$\widetilde g(\alpha^{-1}_0, \alpha^{-1}_1, \alpha^{-1}_2, \alpha^{-1}_3)$.
\end{lemma}
\begin{proof}
One checks that
$\widetilde g(\alpha_0, \alpha_1, \alpha_2, \alpha_3)$ and
$\widetilde g(\alpha^{-1}_0, \alpha^{-1}_1, \alpha^{-1}_2, \alpha^{-1}_3)$
are inverses.
Therefore
$\widetilde g(\alpha_0, \alpha_1, \alpha_2, \alpha_3)$ is invertible and
hence an automorphism of $\widetilde \square_q$.
\end{proof}
\noindent Our next goal is to obtain
an analog of
Proposition
\ref{thm:tensorDecPreM}
that applies to $\widetilde \square_q$.
\begin{lemma}
\label{lem:RR}
In $\widetilde \square_q$,
\begin{eqnarray*}
x_1 x_0 &=& q^2 x_0 x_1 + (1-q^2)c_0,
\\
x_1 x_2 &=& q^{-2} x_2 x_1 + (1-q^{-2})c_1,
\\
x_3 x_2 &=& q^2 x_2 x_3 + (1-q^2)c_2,
\\
x_3 x_0 &=& q^{-2} x_0 x_3 + (1-q^{-2})c_3.
\end{eqnarray*}
\end{lemma}
\begin{proof} These are reformulations of
(\ref{eq:central2}).
\end{proof}
\begin{definition}\rm
The four relations in Lemma
\ref{lem:RR} will be called {\it reduction rules} for
$\widetilde \square_q$.
\end{definition}
\noindent We now express the reduction rules in
a uniform way.
\begin{lemma}
\label{lem:RRuniform}
Referring to the algebra
$\widetilde \square_q$,
pick $u \in \lbrace x_0, x_2 \rbrace$ and
$v \in \lbrace x_1, x_3\rbrace $.
Then
\begin{eqnarray}
\label{eq:xycom}
vu = uv q^{\langle u,v\rangle} + \gamma(u,v)(1-q^{\langle u,v\rangle})
\end{eqnarray}
where
\bigskip
\centerline{
\begin{tabular}[t]{c|cc}
$\langle\,,\,\rangle$ & $x_1$ & $x_3$
\\ \hline
$x_0$ &
$2$ & $-2$
\\
$x_2$ &
$-2$ & $2$
\\
\end{tabular}
\qquad \qquad
\begin{tabular}[t]{c|cc}
$\gamma(\,,\,)$ & $x_1$ & $x_3$
\\ \hline
$x_0$ &
$c_0$ & $c_3$
\\
$x_2$ &
$c_1$ & $c_2$
\\
\end{tabular}}
\bigskip
\end{lemma}
\begin{proof} Use Lemma
\ref{lem:RR}.
\end{proof}
\begin{lemma}
\label{lem:uLong}
Fix $r \in \mathbb N$.
Referring to the algebra $\widetilde \square_q$,
pick
$u_i \in \lbrace x_0, x_2\rbrace $ for $1 \leq i \leq r$,
and also
$v \in \lbrace x_1, x_3 \rbrace$.
Then
\begin{eqnarray*}
&&
v u_1u_2\cdots u_r = u_1 u_2 \cdots u_r v
q^{
\langle u_1, v\rangle
+
\cdots
+
\langle u_r, v\rangle
}
\\
&& \qquad \qquad \qquad
+
\sum_{i=1}^r u_1\cdots u_{i-1} u_{i+1} \cdots u_r \gamma(u_i,v)
q^{\langle u_1,v\rangle + \cdots + \langle u_{i-1},v\rangle}
(1-q^{\langle u_i, v\rangle}).
\end{eqnarray*}
The functions $\langle \,,\,\rangle $ and
$\gamma (\,,\,)$ are from
Lemma
\ref{lem:RRuniform}.
\end{lemma}
\begin{proof} By Lemma
\ref{lem:RRuniform}
and induction on $r$.
\end{proof}
\begin{lemma}
\label{lem:vTe}
In the algebra $\widetilde \square_q$,
\begin{eqnarray*}
&&
x_1 \widetilde \square^{\rm even}_q
\subseteq
\widetilde \square^{\rm even}_q x_1 +
\widetilde \square^{\rm even}_q c_0 +
\widetilde \square^{\rm even}_q c_1,
\\
&&
x_3 \widetilde \square^{\rm even}_q
\subseteq
\widetilde \square^{\rm even}_q x_3 +
\widetilde \square^{\rm even}_q c_2 +
\widetilde \square^{\rm even}_q c_3.
\end{eqnarray*}
\end{lemma}
\begin{proof}
By Definition
\ref{def:TeTo} and
Lemma
\ref{lem:uLong}.
\end{proof}
\begin{lemma}
\label{lem:factor}
We have
\begin{eqnarray}
\label{eq:threeFactor}
\widetilde \square_q =
\widetilde \square^{\rm even}_q
\widetilde \square^{\rm odd}_q
\widetilde C.
\end{eqnarray}
\end{lemma}
\begin{proof}
By Definition
\ref{def:TeTo}
and
Lemma
\ref{lem:vTe} we obtain
$
\widetilde \square^{\rm odd}_q
\widetilde \square^{\rm even}_q \subseteq
\widetilde \square^{\rm even}_q
\widetilde \square^{\rm odd}_q
\widetilde C$.
By this and since $\widetilde C$ is central,
we see that
$
\widetilde \square^{\rm even}_q
\widetilde \square^{\rm odd}_q
\widetilde C
$
is a subalgebra of
$\widetilde \square_q$.
This subalgebra contains
$\widetilde \square^{\rm even}_q$,
$\widetilde \square^{\rm odd}_q$,
$\widetilde C$
and these together generate $\widetilde \square_q$.
The result follows.
\end{proof}
\begin{definition}
\label{def:Laurent}
\rm
Let $\lbrace \lambda_i \rbrace_{i\in \mathbb Z_4}$ denote
mutually commuting indeterminates. Let
$\mathbb F \lbrack
\lambda^{\pm 1}_0,
\lambda^{\pm 1}_1,
\lambda^{\pm 1}_2,
\lambda^{\pm 1}_3\rbrack$
denote the $\mathbb F$-algebra
consisting of the Laurent polynomials in
$\lbrace \lambda_i \rbrace_{i\in \mathbb Z_4}$ that have all coefficients
in $\mathbb F$.
We abbreviate
\begin{eqnarray}
\label{eq:Ldef}
L=
\mathbb F \lbrack
\lambda^{\pm 1}_0,
\lambda^{\pm 1}_1,
\lambda^{\pm 1}_2,
\lambda^{\pm 1}_3\rbrack.
\end{eqnarray}
\end{definition}
\noindent Recall the free algebra $T$ with standard generators
$x,y$.
\begin{lemma}
\label{lem:TTP}
The $\mathbb F$-vector space $T\otimes T\otimes L$
has a
unique
$\widetilde \square_q$-module structure
such that for all $u,v \in T$ and $w \in L$,
\begin{enumerate}
\item[\rm (i)]
$c^{\pm 1}_i$ sends
$u\otimes v \otimes w \mapsto
u\otimes v \otimes (\lambda^{\pm 1}_i w)$ for $i \in \mathbb Z_4$;
\item[\rm (ii)]
$x_0 $ sends
$u\otimes v \otimes w \mapsto
(xu)\otimes v \otimes w$;
\item[\rm (iii)]
$x_1 $ sends
$1\otimes v \otimes w \mapsto
1\otimes (xv) \otimes w$;
\item[\rm (iv)]
$x_2 $ sends
$u\otimes v \otimes w \mapsto
(yu)\otimes v \otimes w$;
\item[\rm (v)]
$x_3 $ sends
$1\otimes v \otimes w \mapsto
1\otimes (yv) \otimes w$.
\end{enumerate}
\end{lemma}
\begin{proof} We first show that
the $\widetilde \square_q$-module structure exists.
For notational convenience abbreviate $V = T \otimes T \otimes L$.
Let ${\rm End}(V)$ denote the $\mathbb F$-algebra
consisting of the $\mathbb F$-linear maps from
$V$ to $V$. We now define some elements in
${\rm End}(V)$. Momentarily abusing notation, we call
these elements
$\lbrace {c}^{\pm 1}_i \rbrace_{i \in \mathbb Z_4}$
and
$\lbrace {x}_i \rbrace_{i \in \mathbb Z_4}$.
For $i \in \mathbb Z_4$ there exist
${c}^{\pm 1}_i \in
{\rm End}(V)$
that satisfy (i).
There exists
$x_0 \in
{\rm End}(V)$
that satisfies (ii).
There exists
$x_2 \in
{\rm End}(V)$
that satisfies (iv).
We now define
$x_1 \in
{\rm End}(V)$
and
$x_3 \in
{\rm End}(V)$.
To do this, we specify how
$x_1$ and $x_3$
act on $u \otimes v \otimes w$.
Here we use Lemma
\ref{lem:uLong} as a guide.
Without loss of generality, we may assume that $u$ is a word in $T$.
Write $u$ as a product
$ u_1 u_2 \cdots u_r$ of standard generators.
The element $x_1$ sends
\begin{eqnarray*}
&&u\otimes v \otimes w \mapsto
u \otimes (xv) \otimes w
q^{
\langle u_1, x\rangle
+
\cdots
+
\langle u_r, x\rangle
}
\\
&& \qquad \quad
+
\sum_{i=1}^r u_1\cdots u_{i-1} u_{i+1} \cdots u_r
\otimes v \otimes w
\gamma(u_i,x)
q^{\langle u_1,x\rangle + \cdots + \langle u_{i-1},x\rangle}
(1-q^{\langle u_i, x\rangle}),
\end{eqnarray*}
and $x_3 $ sends
\begin{eqnarray*}
&&u\otimes v \otimes w \mapsto
u \otimes (yv) \otimes w
q^{
\langle u_1, y\rangle
+
\cdots
+
\langle u_r, y\rangle
}
\\
&& \qquad \quad
+
\sum_{i=1}^r u_1\cdots u_{i-1} u_{i+1} \cdots u_r
\otimes v \otimes w
\gamma(u_i,y)
q^{\langle u_1,y\rangle + \cdots + \langle u_{i-1},y\rangle}
(1-q^{\langle u_i, y\rangle}),
\end{eqnarray*}
where
\bigskip
\centerline{
\begin{tabular}[t]{c|cc}
$\langle\,,\,\rangle$ & $x$ & $y$
\\ \hline
$x$ &
$2$ & $-2$
\\
$y$ &
$-2$ & $2$
\\
\end{tabular}
\qquad \qquad
\begin{tabular}[t]{c|cc}
$\gamma(\,,\,)$ & $x$ & $y$
\\ \hline
$x$ &
$\lambda_0$ & $\lambda_3$
\\
$y$ &
$\lambda_1$ & $\lambda_2$
\\
\end{tabular}}
\bigskip
\noindent We just specified how $x_1$ and $x_3$ act on
$u \otimes v \otimes w$. For $u=1$
these actions become (iii) and (v) in the
lemma statement.
We have defined the elements
$\lbrace c^{\pm 1}_i\rbrace_{i \in \mathbb Z_4}$
and $\lbrace x_i\rbrace_{i \in \mathbb Z_4}$ in
${\rm End}(V)$.
One checks that these elements satisfy the defining relations
(\ref{eq:cci})--(\ref{eq:central2}) for
$\widetilde \square_q$.
This turns
$V$ into
a
$\widetilde \square_q$-module, and by construction
this
$\widetilde \square_q$-module
satisfies
the requirements in the lemma statement.
We have shown that
the $\widetilde \square_q$-module structure exists.
The
$\widetilde \square_q$-module structure is unique in
view of Lemma
\ref{lem:uLong}.
\end{proof}
\begin{proposition}
\label{prop:tensorDec}
The following {\rm (i)--(iv)} hold:
\begin{enumerate}
\item[\rm (i)]
there exists an $\mathbb F$-algebra isomorphism
$T\to \widetilde \square^{\rm even}_q$ that sends
$x\mapsto x_0$ and
$y\mapsto x_2$;
\item[\rm (ii)]
there exists an $\mathbb F$-algebra isomorphism
$T\to \widetilde \square^{\rm odd}_q$ that sends
$x\mapsto x_1$ and
$y\mapsto x_3$;
\item[\rm (iii)]
there exists an
$\mathbb F$-algebra isomorphism
$L\to \widetilde C$
that sends $\lambda^{\pm 1}_i \mapsto c^{\pm 1}_i$ for
$i \in \mathbb Z_4$;
\item[\rm (iv)]
the following is an isomorphism of
$\mathbb F$-vector spaces:
\begin{eqnarray*}
\widetilde \square^{\rm even}_q
\otimes
\widetilde \square^{\rm odd}_q
\otimes
\widetilde C
& \to & \widetilde \square_q
\\
u \otimes v \otimes c &\mapsto & uv c
\end{eqnarray*}
\end{enumerate}
\end{proposition}
\begin{proof}
There exists a surjective $\mathbb F$-algebra
homomorphism
$f_1:T\to \widetilde \square^{\rm even}_q$
that sends
$x \mapsto x_0$ and
$y \mapsto x_2$.
There exists a surjective $\mathbb F$-algebra
homomorphism
$f_2:T\to \widetilde \square^{\rm odd}_q$
that sends
$x \mapsto x_1$ and
$y \mapsto x_3$.
There exists a surjective $\mathbb F$-algebra
homomorphism
$f_3:L\to \widetilde C$
that sends
$\lambda^{\pm 1}_i \mapsto c^{\pm 1}_i$
for $i \in \mathbb Z_4$.
There exists an
$\mathbb F$-linear map $f: T\otimes T \otimes L \to
\widetilde \square_q$ that sends
$u\otimes v \otimes w \mapsto f_1(u)f_2(v)f_3(w)$ for all
$u,v \in T$ and $w \in L$. It suffices to show that $f$
is bijective.
The map $f$ is surjective by Lemma
\ref{lem:factor}.
To see that
$f$ is injective, view
$T\otimes T \otimes L$ as a
$\widetilde \square_q$-module
as in Lemma
\ref{lem:TTP}.
The map $f$ is injective because
the composition
\begin{equation*}
\begin{CD}
T\otimes T \otimes L @>> f >
\widetilde \square_q
@>> z \mapsto z(1\otimes 1 \otimes 1) > T \otimes T \otimes L
\end{CD}
\end{equation*}
is the identity map on $T\otimes T\otimes L$.
By these comments $f$ is bijective.
The result follows.
\end{proof}
\begin{proposition}
\label{prop:SquareBasis}
For the algebra $\widetilde \square_q$,
\begin{enumerate}
\item[\rm (i)] the following is a basis for the
$\mathbb F$-vector space $\widetilde \square^{\rm even}_q$:
\begin{eqnarray}
u_1 u_2 \cdots u_r \qquad \qquad r \in \mathbb N, \qquad
u_i \in \lbrace x_0, x_2\rbrace, \qquad 1 \leq i \leq r;
\label{eq:NevenBasis}
\end{eqnarray}
\item[\rm (ii)] the following is a basis for the
$\mathbb F$-vector space $\widetilde \square^{\rm odd}_q$:
\begin{eqnarray}
v_1 v_2 \cdots v_s \qquad \qquad s \in \mathbb N, \qquad
v_i \in \lbrace x_1, x_3\rbrace, \qquad 1 \leq i \leq s;
\label{eq:NoddBasis}
\end{eqnarray}
\item[\rm (iii)] the following is a basis for the
$\mathbb F$-vector space $\widetilde C$:
\begin{eqnarray}
c^{n_0}_0
c^{n_1}_1
c^{n_2}_2
c^{n_3}_3,
\qquad \qquad n_i \in \mathbb Z, \qquad i \in \mathbb Z_4;
\label{eq:Cbasis}
\end{eqnarray}
\item[\rm (iv)] the
$\mathbb F$-vector space $\widetilde \square_q$ has a basis consisting
of the elements
\begin{eqnarray}
u
v
c
\label{eq:basisWWC}
\end{eqnarray}
such that
$u$,
$v$,
$c$ are contained in the bases
$(\ref{eq:NevenBasis})$,
$(\ref{eq:NoddBasis})$,
$(\ref{eq:Cbasis})$, respectively.
\end{enumerate}
\end{proposition}
\begin{proof}
By Proposition
\ref{prop:tensorDec}.
\end{proof}
\section{Some calculations in $\widetilde \square_q$}
\noindent We continue to investigate the algebra $\widetilde \square_q$
from Definition
\ref{def:boxqV2}.
In this section, we obtain some results about
$\widetilde \square_q$
that will be used
in the proof of
Proposition
\ref{thm:tensorDecPreM}.
\begin{definition}
\label{def:Deltai}
\rm For the algebra $\widetilde \square_q$ define
\begin{eqnarray}
S_i = x^3_ix_{i+2} -
\lbrack 3 \rbrack_q x^2_i x_{i+2}x_i
+
\lbrack 3 \rbrack_q x_i x_{i+2}x^2_i
-
x_{i+2}x^3_i \qquad \qquad i \in \mathbb Z_4.
\label{eq:Delta}
\end{eqnarray}
\end{definition}
\noindent We note that the elements $S_i$ from
Definition
\ref{def:Deltai} are in the kernel of the canonical
homomorphism
$\widetilde \square_q \to \square_q$.
\begin{lemma}
We have
$S_0, S_2 \in \widetilde \square^{\rm even}_q$
and
$S_1, S_3 \in \widetilde \square^{\rm odd}_q$.
\end{lemma}
\begin{proof} By Definition
\ref{def:TeTo}(i),(ii) and
Definition
\ref{def:Deltai}.
\end{proof}
\begin{proposition}
\label{lem:deltaCom}
In the algebra $\widetilde \square_q$ the following
{\rm (i), (ii)} hold for
$i \in \mathbb Z_4$:
\begin{enumerate}
\item[\rm (i)]
$ x_{i+1} S_i =
q^4 S_i x_{i+1}$,
\item[\rm (ii)]
$x_{i-1} S_i =
q^{-4} S_i x_{i-1}$.
\end{enumerate}
\end{proposition}
\begin{proof}(i) Without loss of generality we may assume
that $i=0$. Our strategy is to
express
$
x_1 S_0-
q^4 S_0 x_1$
in the basis
for $\widetilde \square_q$ from Proposition
\ref{prop:SquareBasis}(iv).
By Definition
\ref{def:Deltai},
\begin{eqnarray}
x_1 S_0 = x_1 x^3_0 x_2 - \lbrack 3 \rbrack_q x_1x^2_0 x_2 x_0 +
\lbrack 3 \rbrack_q x_1x_0 x_2 x^2_0 - x_1 x_2 x^3_0,
\label{ex:x1Delta0}
\\
S_0 x_1 = x^3_0 x_2 x_1- \lbrack 3 \rbrack_q x^2_0 x_2 x_0 x_1+
\lbrack 3 \rbrack_q x_0 x_2 x^2_0 x_1 - x_2 x^3_0 x_1.
\label{ex:Delta0x1}
\end{eqnarray}
Consider the elements
\begin{eqnarray}
\label{ex:DeltaTerm}
x_1 x^3_0 x_2,
\qquad
x_1 x^2_0 x_2 x_0,
\qquad
x_1 x_0 x_2 x^2_0,
\qquad
x_1 x_2 x^3_0,
\qquad
S_0 x_1.
\end{eqnarray}
By
Lemma \ref{lem:uLong} and
(\ref{ex:Delta0x1}),
the elements
(\ref{ex:DeltaTerm}) are weighted sums involving
the following terms and coefficients:
\medskip
\centerline{
\begin{tabular}[t]{c|ccccc}
{\rm term} & {\rm $x_1x^3_0x_2$ coef.} & {\rm $x_1x^2_0x_2x_0$ coef.} &
{\rm $x_1x_0x_2x^2_0$ coef.} & {\rm $x_1x_2x^3_0$ coef.} &
{\rm $S_0 x_1$ coef.}
\\ \hline \hline
$x^3_0 x_2 x_1 $ & $q^4$ & $0$ & $0$ & $0$ & $1$
\\
$x^2_0 x_2 x_0x_1$ & $0$ & $q^4$ & $0$ & $0$ & $-\lbrack 3 \rbrack_q$
\\
$x_0 x_2 x^2_0x_1 $ & $0$ & $0$ & $q^4$ & $0$ & $\lbrack 3 \rbrack_q$
\\
$ x_2 x^3_0 x_1 $ &$0$ & $0$ &$0$ & $q^4$ & $-1$
\\
\hline
$x^2_0 x_2 c_0$ & $1-q^6$ & $q^2-q^4$ & $0$ & $0$ & $0$
\\
$x_0 x_2 x_0 c_0$ & $0$ & $1-q^4$ & $1-q^4$ & $0$ & $0$
\\
$ x_2 x^2_0 c_0$ & $0$ & $0$ & $1-q^2$ & $q^{-2}-q^4$ & $0$
\\
\hline
$ x^3_0 c_1$ & $q^6-q^4$ & $q^4-q^2$ & $q^2-1$ & $1-q^{-2}$ & $0$
\\
\end{tabular}}
\bigskip
\noindent
We can now easily write
$x_1S_0 - q^4 S_0 x_1$ in the basis for
$\widetilde \square_q$ from Proposition
\ref{prop:SquareBasis}(iv).
To do this,
expand
$x_1 S_0 - q^4 S_0 x_1$
using
(\ref{ex:x1Delta0})
and evaluate
the result using the
data in the above table.
After a routine cancellation we obtain
$x_1 S_0 - q^4 S_0 x_1=0$.
Therefore
$x_1 S_0 =q^4 S_0 x_1$.
\\
\noindent (ii) We proceed as in part (i) above.
Without loss of generality we may assume
that $i=0$. We express
$
x_3 S_0-
q^{-4} S_0 x_3$
in the basis
for $\widetilde \square_q$ from Proposition
\ref{prop:SquareBasis}(iv).
By Definition
\ref{def:Deltai},
\begin{eqnarray}
x_3 S_0 = x_3 x^3_0 x_2 - \lbrack 3 \rbrack_q x_3x^2_0 x_2 x_0 +
\lbrack 3 \rbrack_q x_3x_0 x_2 x^2_0 - x_3 x_2 x^3_0,
\label{ex:x3Delta0}
\\
S_0 x_3 = x^3_0 x_2 x_3- \lbrack 3 \rbrack_q x^2_0 x_2 x_0 x_3+
\lbrack 3 \rbrack_q x_0 x_2 x^2_0 x_3 - x_2 x^3_0 x_3.
\label{ex:Delta0x3}
\end{eqnarray}
Consider the elements
\begin{eqnarray}
\label{ex:DeltaTermx3}
x_3 x^3_0 x_2,
\qquad
x_3 x^2_0 x_2 x_0,
\qquad
x_3 x_0 x_2 x^2_0,
\qquad
x_3 x_2 x^3_0,
\qquad
S_0 x_3.
\end{eqnarray}
By
Lemma \ref{lem:uLong} and
(\ref{ex:Delta0x3}),
the elements
(\ref{ex:DeltaTermx3}) are weighted sums involving
the following terms and coefficients:
\medskip
\centerline{
\begin{tabular}[t]{c|ccccc}
{\rm term} & {\rm $x_3x^3_0x_2$ coef.} & {\rm $x_3x^2_0x_2x_0$ coef.} &
{\rm $x_3x_0x_2x^2_0$ coef.} & {\rm $x_3x_2x^3_0$ coef.} &
{\rm $S_0 x_3$ coef.}
\\ \hline \hline
$x^3_0 x_2 x_3 $ & $q^{-4}$ & $0$ & $0$ & $0$ & $1$
\\
$x^2_0 x_2 x_0x_3$ & $0$ & $q^{-4}$ & $0$ & $0$ & $-\lbrack 3 \rbrack_q$
\\
$x_0 x_2 x^2_0x_3 $ & $0$ & $0$ & $q^{-4}$ & $0$ & $\lbrack 3 \rbrack_q$
\\
$ x_2 x^3_0 x_3 $ &$0$ & $0$ &$0$ & $q^{-4}$ & $-1$
\\
\hline
$x^2_0 x_2 c_3$ & $1-q^{-6}$ & $q^{-2}-q^{-4}$ & $0$ & $0$ & $0$
\\
$x_0 x_2 x_0 c_3$ & $0$ & $1-q^{-4}$ & $1-q^{-4}$ & $0$ & $0$
\\
$ x_2 x^2_0 c_3$ & $0$ & $0$ & $1-q^{-2}$ & $q^{2}-q^{-4}$ & $0$
\\
\hline
$ x^3_0 c_2$ & $q^{-6}-q^{-4}$ & $q^{-4}-q^{-2}$ & $q^{-2}-1$ & $1-q^{2}$ & $0$
\\
\end{tabular}}
\bigskip
\noindent
Now to write
$x_3 S_0 - q^{-4} S_0 x_3$ in the basis for
$\widetilde \square_q$ from Proposition
\ref{prop:SquareBasis}(iv),
expand
$x_3 S_0 - q^{-4} S_0 x_3$
using
(\ref{ex:x3Delta0})
and evaluate
the result using the
data in the above table.
After a routine cancellation we obtain
$x_3 S_0 - q^{-4} S_0 x_3=0$.
Therefore
$x_3 S_0 =q^{-4} S_0 x_3$.
\end{proof}
\begin{corollary}
\label{cor:DeltaCom}
In the algebra $\widetilde \square_q$,
\begin{eqnarray*}
&&
S_0 \widetilde \square^{\rm odd}_q
=
\widetilde \square^{\rm odd}_q S_0 ,
\qquad \qquad
S_1 \widetilde \square^{\rm even}_q
=
\widetilde \square^{\rm even}_q S_1,
\\
&&
S_2 \widetilde \square^{\rm odd}_q
=
\widetilde \square^{\rm odd}_q S_2,
\qquad \qquad
S_3 \widetilde \square^{\rm even}_q
=
\widetilde \square^{\rm even}_q S_3.
\end{eqnarray*}
\end{corollary}
\begin{proof} By Definition
\ref{def:TeTo}(i),(ii)
and
Proposition
\ref{lem:deltaCom}.
\end{proof}
\section{More calculations in $\widetilde \square_q$
}
\noindent We continue to
investigate the algebra
$\widetilde \square_q$ from
Definition
\ref{def:boxqV2}.
In this section, we first obtain some results
about $\widetilde \square_q$. We then use these results
to prove
Proposition
\ref{prop:ABxyiLongerM}.
\medskip
\noindent
For the algebra $\widetilde \square_q$ define
\begin{eqnarray}
A = x_0+x_1, \qquad \qquad B = x_2+x_3.
\label{eq:AB}
\end{eqnarray}
Our next goal is to express
\begin{eqnarray}
&&
A^3 B - \lbrack 3 \rbrack_q A^2 B A +
\lbrack 3 \rbrack_q A B A^2 -B A^3 +
(q^2-q^{-2})^2 c_0(AB-BA)
\label{eq:terms}
\end{eqnarray}
in the basis
for $\widetilde \square_q$ from
Proposition
\ref{prop:SquareBasis}(iv).
\medskip
\noindent
For the rest of this section, the notation
(\ref{eq:AB})
is in effect.
\begin{lemma}
\label{lem:AAAB}
In the algebra $\widetilde \square_q$,
\begin{enumerate}
\item[\rm (i)]
the element $A^2$ is a weighted sum involving the following
terms and coefficients:
\medskip
\centerline{
\begin{tabular}[t]{c|cccc}
{\rm term} & $x^2_0$ & $x_0x_1$ & $x^2_1$ & $c_0$
\\ \hline
{\rm coefficient} & $1$ & $1+q^2$ & $1$ & $1-q^2$
\\
\end{tabular}}
\item[\rm (ii)]
the element $AB$ is a weighted sum involving the following
terms and coefficients:
\medskip
\centerline{
\begin{tabular}[t]{c|ccccc}
{\rm term} & $x_0x_2$ & $x_0x_3$ & $x_2x_1$ & $x_1x_3$ & $c_1$
\\ \hline
{\rm coefficient} & $1$ & $1$ & $q^{-2}$ & $1$ & $1-q^{-2}$
\\
\end{tabular}}
\item[\rm (iii)]
the element $BA$ is a weighted sum involving the following
terms and coefficients:
\medskip
\centerline{
\begin{tabular}[t]{c|ccccc}
{\rm term} & $x_2x_0$ & $x_0x_3$ & $x_2x_1$ & $x_3x_1$ & $c_3$
\\ \hline
{\rm coefficient} & $1$ & $q^{-2}$ & $1$ & $1$ & $1-q^{-2}$
\\
\end{tabular}}
\medskip
\end{enumerate}
\end{lemma}
\begin{proof} Use Lemma
\ref{lem:RR} and
(\ref{eq:AB}).
\end{proof}
\noindent Next we use Lemma
\ref{lem:AAAB} to
evaluate some terms in (\ref{eq:terms}).
Consider $A^3B=(A^2)(AB)$.
In this equation, evaluate the right-hand side
using
Lemma \ref{lem:AAAB}(i),(ii) and expand the result;
the details are
in the table below.
The table has two header columns that describe $A^2$,
and two header rows that describe $AB$.
The expressions inside parentheses
are ``out of order''
and will be subject to further reduction shortly.
\begin{eqnarray*} \qquad \qquad \qquad \qquad \qquad AB
\end{eqnarray*}
\centerline{
\begin{tabular}[t]{c}
\\
\\
\\
$A^2$
\\
\\
\end{tabular}
\qquad
\begin{tabular}[t]{cc|ccccc}
& & $1$ & $1$ & $q^{-2}$ & $1$ & $1-q^{-2}$
\\
& & $x_0x_2$ & $x_0x_3$ & $x_2x_1$ & $x_1x_3$ & $c_1$
\\ \hline
$1$ & $x^2_0$
& $x^3_0x_2$ & $x^3_0x_3$ & $x^2_0x_2x_1$ & $x^2_0x_1x_3$ & $x^2_0c_1$
\\
$1+q^2$ & $x_0x_1$ &
$x_0(x_1 x_0x_2)$ & $x_0(x_1 x_0)x_3$ & $x_0(x_1x_2)x_1$ & $x_0x^2_1x_3$ &
$x_0x_1c_1$
\\
$1$ & $x^2_1$ &
$(x^2_1x_0x_2)$ & $(x^2_1x_0)x_3$ & $(x^2_1x_2)x_1$ & $x^3_1x_3$ & $x^2_1c_1$
\\
$1-q^2$ & $c_0$ &
$x_0x_2c_0$ & $x_0x_3c_0$ & $x_2x_1c_0$ & $x_1x_3c_0$ & $c_0c_1$
\\
\end{tabular}}
\bigskip
\noindent Similarly for
$A^2BA=(A^2)(BA)$,
\begin{eqnarray*} \qquad \qquad \qquad \qquad \qquad BA
\end{eqnarray*}
\centerline{
\begin{tabular}[t]{c}
\\
\\
\\
$A^2$
\\
\\
\end{tabular}
\qquad
\begin{tabular}[t]{cc|ccccc}
& & $1$ & $q^{-2}$ & $1$ & $1$ & $1-q^{-2}$
\\
& & $x_2x_0$ & $x_0x_3$ & $x_2x_1$ & $x_3x_1$ & $c_3$
\\ \hline
$1$ & $x^2_0$
& $x^2_0x_2x_0$ & $x^3_0x_3$ & $x^2_0x_2x_1$ & $x^2_0 x_3x_1$ & $x^2_0c_3$
\\
$1+q^2$ & $x_0x_1$ &
$x_0(x_1 x_2x_0)$ & $x_0(x_1 x_0)x_3$ & $x_0(x_1x_2)x_1$ & $x_0x_1x_3x_1$ &
$x_0x_1c_3$
\\
$1$ & $x^2_1$ &
$(x^2_1x_2x_0)$ & $(x^2_1x_0)x_3$ & $(x^2_1x_2)x_1$ & $x^2_1x_3x_1$ & $x^2_1c_3$
\\
$1-q^2$ & $c_0$ &
$x_2x_0c_0$ & $x_0x_3c_0$ & $x_2x_1c_0$ & $x_3x_1c_0$ & $c_0c_3$
\\
\end{tabular}}
\bigskip
\noindent Similarly for
$ABA^2=(AB)(A^2)$,
\begin{eqnarray*} \qquad \qquad \qquad \qquad \qquad \qquad A^2
\end{eqnarray*}
\centerline{
\begin{tabular}[t]{c}
\\
\\
\\
\\
$AB$
\\
\\
\end{tabular}
\qquad
\begin{tabular}[t]{cc|cccc}
& & $1$ & $1+q^2$ & $1$ & $1-q^2$
\\
& & $x^2_0$ & $x_0x_1$ & $x^2_1$ & $c_0$
\\ \hline
$1$ & $x_0x_2$
& $x_0x_2x^2_0$ & $x_0x_2x_0x_1$ & $x_0x_2x^2_1$ & $x_0x_2c_0$
\\
$1$ & $x_0x_3$ &
$x_0(x_3 x^2_0)$ & $x_0(x_3 x_0)x_1$ & $x_0x_3x^2_1$ &
$x_0x_3c_0$
\\
$q^{-2}$ & $x_2x_1$ &
$x_2(x_1x^2_0)$ & $x_2(x_1x_0)x_1$ & $x_2x^3_1$ & $x_2x_1c_0$
\\
$1$ & $x_1x_3$ &
$(x_1x_3x^2_0)$ & $(x_1x_3x_0)x_1$ & $x_1x_3x^2_1$ & $x_1x_3c_0$
\\
$1-q^{-2}$ & $c_1$ &
$x^2_0c_1$ & $x_0x_1c_1$ & $x^2_1c_1$ & $c_0c_1$
\\
\end{tabular}}
\bigskip
\noindent Similarly for
$BA^3=(BA)(A^2)$,
\begin{eqnarray*} \qquad \qquad \qquad \qquad \qquad \qquad A^2
\end{eqnarray*}
\centerline{
\begin{tabular}[t]{c}
\\
\\
\\
\\
$BA$
\\
\\
\end{tabular}
\qquad
\begin{tabular}[t]{cc|cccc}
& & $1$ & $1+q^2$ & $1$ & $1-q^2$
\\
& & $x^2_0$ & $x_0x_1$ & $x^2_1$ & $c_0$
\\ \hline
$1$ & $x_2x_0$
& $x_2x^3_0$ & $x_2x^2_0x_1$ & $x_2x_0x^2_1$ & $x_2x_0c_0$
\\
$q^{-2}$ & $x_0x_3$ &
$x_0(x_3 x^2_0)$ & $x_0(x_3 x_0)x_1$ & $x_0x_3x^2_1$ &
$x_0x_3c_0$
\\
$1$ & $x_2x_1$ &
$x_2(x_1x^2_0)$ & $x_2(x_1x_0)x_1$ & $x_2x^3_1$ & $x_2x_1c_0$
\\
$1$ & $x_3x_1$ &
$(x_3x_1x^2_0)$ & $(x_3x_1x_0)x_1$ & $x_3x^3_1$ & $x_3x_1c_0$
\\
$1-q^{-2}$ & $c_3$ &
$x^2_0c_3$ & $x_0x_1c_3$ & $x^2_1c_3$ & $c_0c_3$
\\
\end{tabular}}
\bigskip
\noindent The above four tables
contain some
parenthetical expressions.
We will write these parenthetical expressions in the
basis for
$\widetilde \square_q$ from
Proposition
\ref{prop:SquareBasis}(iv).
For the parenthetical expressions of length two,
this is done using Lemma
\ref{lem:RR}.
For the remaining parenthetical expressions,
this will be done over the next three lemmas.
\begin{lemma}
\label{lem:NTL1}
In the algebra $\widetilde \square_q$,
\begin{enumerate}
\item[\rm (i)]
the element $x_1x_0x_2$ is a weighted sum involving the following
terms and coefficients:
\medskip
\centerline{
\begin{tabular}[t]{c|ccc}
{\rm term} & $x_0x_2x_1$ & $x_0c_1$ & $x_2c_0$
\\ \hline
{\rm coefficient} & $1$ & $q^2-1$ & $1-q^2$
\\
\end{tabular}}
\item[\rm (ii)]
the element $x^2_1x_0$ is a weighted sum involving the following
terms and coefficients:
\medskip
\centerline{
\begin{tabular}[t]{c|cc}
{\rm term} & $x_0x^2_1$ & $x_1c_0$
\\ \hline
{\rm coefficient} & $q^4$ & $1-q^4$
\\
\end{tabular}}
\item[\rm (iii)]
the element $x^2_1x_2$ is a weighted sum involving the following
terms and coefficients:
\medskip
\centerline{
\begin{tabular}[t]{c|cc}
{\rm term} & $x_2x^2_1$ & $x_1c_1$
\\ \hline
{\rm coefficient} & $q^{-4}$ & $1-q^{-4}$
\\
\end{tabular}}
\item[\rm (iv)]
the element $x_1x_2x_0$ is a weighted sum involving the following
terms and coefficients:
\medskip
\centerline{
\begin{tabular}[t]{c|ccc}
{\rm term} & $x_2x_0x_1$ & $x_2c_0$ & $x_0c_1$
\\ \hline
{\rm coefficient} & $1$ & $q^{-2}-1$ & $1-q^{-2}$
\\
\end{tabular}}
\medskip
\end{enumerate}
\end{lemma}
\begin{proof} Apply Lemma
\ref{lem:RR} repeatedly.
\end{proof}
\begin{lemma}
\label{lem:NTL2}
In the algebra $\widetilde \square_q$,
\begin{enumerate}
\item[\rm (i)]
the element $x_1x_3x_0$ is a weighted sum involving the following
terms and coefficients:
\medskip
\centerline{
\begin{tabular}[t]{c|ccc}
{\rm term} & $x_0x_1x_3$ & $x_1c_3$ & $x_3c_0$
\\ \hline
{\rm coefficient} & $1$ & $1-q^{-2}$ &
$q^{-2}-1$
\\
\end{tabular}}
\item[\rm (ii)]
the element $x_1x^2_0$ is a weighted sum involving the following
terms and coefficients:
\medskip
\centerline{
\begin{tabular}[t]{c|cc}
{\rm term} & $x^2_0x_1$ & $x_0c_0$
\\ \hline
{\rm coefficient} & $q^4$ & $1-q^4$
\\
\end{tabular}}
\item[\rm (iii)]
the element $x_3x^2_0$ is a weighted sum involving the following
terms and coefficients:
\medskip
\centerline{
\begin{tabular}[t]{c|cc}
{\rm term} & $x^2_0x_3$ & $x_0c_3$
\\ \hline
{\rm coefficient} & $q^{-4}$ & $ 1-q^{-4}$
\\
\end{tabular}}
\item[\rm (iv)]
the element $x_3x_1x_0$ is a weighted sum involving the following
terms and coefficients:
\medskip
\centerline{
\begin{tabular}[t]{c|ccc}
{\rm term} & $x_0x_3x_1$ & $x_1c_3$ & $x_3c_0$
\\ \hline
{\rm coefficient} & $1$ & $q^2-1$ & $1-q^2$
\\
\end{tabular}}
\medskip
\end{enumerate}
\end{lemma}
\begin{proof} Similar to the proof of
Lemma \ref{lem:NTL1}.
\end{proof}
\begin{lemma}
\label{lem:NTL3}
In the algebra $\widetilde \square_q$,
\begin{enumerate}
\item[\rm (i)]
the element $x^2_1x_0x_2$ is a weighted sum involving the following
terms and coefficients:
\medskip
\centerline{
\begin{tabular}[t]{c|cccc}
{\rm term} & $x_0x_2x^2_1$ & $x_0x_1c_1$ & $x_2x_1c_0$ & $c_0c_1$
\\ \hline
{\rm coefficient} &
$1$ & $q^4-1$ &
$q^{-2}-q^{2}$ &
$(1-q^2)(q^2-q^{-2})$
\\
\end{tabular}}
\item[\rm (ii)]
the element $x^2_1x_2x_0$ is a weighted sum involving the following
terms and coefficients:
\medskip
\centerline{
\begin{tabular}[t]{c|cccc}
{\rm term} & $x_2x_0 x^2_1$ & $x_0x_1c_1$ & $x_2x_1c_0$ & $c_0c_1$
\\ \hline
{\rm coefficient} & $1$ & $q^2-q^{-2}$ &
$q^{-4}-1$ & $(q^{-2}-1)(q^2-q^{-2})$
\\
\end{tabular}}
\item[\rm (iii)]
the element $x_1x_3x^2_0$ is a weighted sum involving the following
terms and coefficients:
\medskip
\centerline{
\begin{tabular}[t]{c|cccc}
{\rm term} & $x^2_0x_1x_3$ & $x_0x_1c_3$ & $x_0x_3c_0$ & $c_0c_3$
\\ \hline
{\rm coefficient} & $1$ & $q^2-q^{-2}$ &
$q^{-4}-1$ &
$(q^{-2}-1)(q^2-q^{-2})$
\\
\end{tabular}}
\item[\rm (iv)]
the element $x_3x_1x^2_0$ is a weighted sum involving the following
terms and coefficients:
\medskip
\centerline{
\begin{tabular}[t]{c|cccc}
{\rm term} & $x^2_0x_3x_1$ & $x_0x_1c_3$ & $x_0x_3c_0$ & $c_0c_3$
\\ \hline
{\rm coefficient} & $1$ & $q^4-1$ & $q^{-2}-q^{2}$
& $ (1-q^2)(q^2-q^{-2})$
\\
\end{tabular}}
\medskip
\end{enumerate}
\end{lemma}
\begin{proof} Similar to the proof of
Lemma \ref{lem:NTL1}.
\end{proof}
\noindent Referring to the algebra
$\widetilde \square_q$, we now write
the elements
\begin{eqnarray}
A^3B, \qquad A^2BA, \qquad ABA^2, \qquad BA^3
\label{eq:fiveTerms}
\end{eqnarray}
in the basis for $\widetilde \square_q$ from Proposition
\ref{prop:SquareBasis}(iv).
\begin{lemma}
\label{lem:expand}
In the algebra $\widetilde \square_q$, the elements
{\rm (\ref{eq:fiveTerms})} are weighted sums involving
the following terms and coefficients.
\medskip
\centerline{
\begin{tabular}[t]{c|ccccc}
{\rm term} & {\rm $A^3B$ coef.} & {\rm $A^2BA$ coef.} &
{\rm $ABA^2$ coef.} & {\rm $BA^3$ coef.}
\\ \hline \hline
$x^3_0x_2$ & $1$ & $0$ & $0$ & $0$
\\
$x^2_0x_2x_0$ & $0$ & $1$ & $0$ &$0$
\\
$x_0x_2x^2_0$ & $0$ & $0$ & $1$ & $0$
\\
$x_2x^3_0$ & $0$ & $0$ & $0$ & $1$
\\
\hline
$x^3_1x_3$ & $1$ & $0$ & $0$ & $0$
\\
$x^2_1x_3x_1$ & $0$ & $1$ & $0$ & $0$
\\
$x_1x_3x^2_1$ & $0$ & $0$ & $1$ & $0$
\\
$x_3x^3_1$ & $0$ & $0$ & $0$ & $1$
\\
\hline
$x^3_0x_3$ & $1$ & $q^{-2}$ & $q^{-4}$ & $q^{-6}$
\\
$x_2x^3_1$ & $q^{-6}$ & $q^{-4}$ & $q^{-2}$ & $1$
\\
\hline
$x_0x^2_1x_3$ & $q^{2}\lbrack 3\rbrack_q$ & $q^{2}$ & $0$ & $0$
\\
$x_0x_1x_3x_1$ & $0$ & $q^{2}+1$ & $q^{2}+1$ & $0$
\\
$x_0x_3x^2_1$ & $0$ & $0$ & $1$ & $\lbrack 3 \rbrack_q$
\\
\hline
$x^2_0x_2x_1$ & $\lbrack 3 \rbrack_q$ & $1$ & $0$ & $0$
\\
$x_0x_2x_0x_1$ & $0$ & $q^{2}+1$ & $q^{2}+1$ & $0$
\\
$x_2x^2_0x_1$ & $0$ & $0$ & $q^2$ & $q^2 \lbrack 3 \rbrack_q$
\\
\hline
$x^2_0x_1x_3$ & $q^{2}\lbrack 3\rbrack_q$ & $q^{2}+1$ & $1$ & $0$
\\
$x^2_0x_3x_1$ & $0$ & $1$ & $q^{-2}+1$ & $q^{-2}\lbrack 3 \rbrack_q$
\\
\hline
$x_0x_2x^2_1$ & $q^{-2}\lbrack 3\rbrack_q$ & $q^{-2}+1$ & $1$ & $0$
\\
$x_2x_0x^2_1$ & $0$ & $1$ & $q^{2}+1$ & $q^{2}\lbrack 3 \rbrack_q$
\\
\end{tabular}}
\bigskip
\centerline{
\begin{tabular}[t]{c|cccc}
{\rm term} & {\rm $A^3B$ coef.} & {\rm $A^2BA$ coef.} &
{\rm $ABA^2$ coef.} & {\rm $BA^3$ coef.}
\\ \hline \hline
$x^2_0 c_1$ & $(q^2-1)\lbrack 3 \rbrack_q$ & $q^2-q^{-2}$ & $1-q^{-2}$ & $0$
\\
$x^2_0 c_3$ & $0$ & $1-q^{-2}$ & $1-q^{-4}$ &
$q^{-2}(1-q^{-2})\lbrack 3 \rbrack_q $
\\
$x_0x_2 c_0$ & $(1-q^2)(2+q^2)$ & $q^{-2}-q^2$ & $1-q^2$ & $0$
\\
$x_2x_0 c_0$ & $0$ & $1-q^2$ & $q^{-2}-q^2$ & $(1-q^2)(2+q^2)$
\\
\hline
$x_0x_1 c_1$ & $(q^2-q^{-2})\lbrack 3 \rbrack_q$ & $2(q^2-q^{-2})$ &
$q^2-q^{-2}$ & $0$
\\
$x_0x_1 c_3$ & $0$ & $q^2-q^{-2}$ & $2(q^2-q^{-2})$
& $(q^2-q^{-2})\lbrack 3 \rbrack_q$
\\
$x_0x_3 c_0$ & $(1-q^2)(2+q^2)$ & $(q^{-2}-1)(q^2+2)$ &
$(q^{-2}-1)\lbrack 3 \rbrack_q$ & $(q^{-2}-1)(q^2+2)$
\\
$x_2x_1 c_0$ & $(q^{-2}-1)(2+q^2)$ & $(q^{-2}-1)\lbrack 3\rbrack_q$
& $(q^{-2}-1)(q^2+2)$ & $(1-q^2)(2+q^2)$
\\
\hline
$x^2_1 c_1$ & $q^{-2}(1-q^{-2})\lbrack 3 \rbrack_q$ & $1-q^{-4}$ &
$1-q^{-2}$& $0$
\\
$x^2_1 c_3$ & $0$ & $1-q^{-2}$ & $q^2-q^{-2}$ & $(q^2-1)\lbrack 3 \rbrack_q$
\\
$x_1x_3 c_0$ & $(1-q^2)(2+q^2)$ & $q^{-2}-q^2$ & $1-q^2$ & $0$
\\
$x_3x_1 c_0$ & $0$ & $1-q^{2}$ & $q^{-2}-q^2$ & $(1-q^2)(2+q^2)$
\\
\hline
$ c_0c_1$ & $-(q-q^{-1})^2(2+q^2)$ & $(q^{-2}-1)(q^2-q^{-2})$
& $-(q-q^{-1})^2$ & $0$
\\
$ c_0c_3$ & $0$ & $-(q-q^{-1})^2$ & $(q^{-2}-1)(q^2-q^{-2})$ & $-(q-q^{-1})^2(2+q^2)$
\\
\end{tabular}}
\bigskip
\end{lemma}
\begin{proof} In the four tables below
Lemma \ref{lem:AAAB}, evaluate
the parenthetical expressions using
Lemma \ref{lem:RR} along with Lemmas
\ref{lem:NTL1},
\ref{lem:NTL2},
\ref{lem:NTL3}.
\end{proof}
\noindent Recall the elements
$\lbrace S_i \rbrace_{i \in \mathbb Z_4}$
in $\widetilde \square_q$, from Definition
\ref{def:Deltai}.
\begin{proposition}
\label{prop:tildeMain}
For the algebra $\widetilde \square_q$ define
\begin{eqnarray*}
A = x_0+x_1, \qquad \qquad B = x_2+x_3.
\end{eqnarray*}
Then both
\begin{eqnarray}
&&
A^3 B - \lbrack 3 \rbrack_q A^2 B A +
\lbrack 3 \rbrack_q A B A^2 -B A^3 + (q^2-q^{-2})^2 c_0(AB-BA)
= S_0 + S_1, \qquad
\label{eq:1Assertion}
\\
&&
B^3 A - \lbrack 3 \rbrack_q B^2 A B +
\lbrack 3 \rbrack_q B A B^2 -A B^3 + (q^2-q^{-2})^2 c_2 (BA-AB)
=
S_2 + S_3.
\label{eq:2Assertion}
\end{eqnarray}
\end{proposition}
\begin{proof}
To obtain
(\ref{eq:1Assertion}),
write
(\ref{eq:terms}) in the basis for
$\widetilde \square_q$ from
Proposition
\ref{prop:SquareBasis}(iv).
To do this, evaluate the terms
(\ref{eq:fiveTerms})
using
Lemma
\ref{lem:expand},
and the term $c_0(AB-BA)$ using
Lemma
\ref{lem:AAAB}(ii),(iii).
Assertion
(\ref{eq:1Assertion})
follows after a routine computation.
Assertion
(\ref{eq:2Assertion})
is obtained from
(\ref{eq:1Assertion})
by applying the square of
the automorphism
$\widetilde \rho$ from Lemma
\ref{lem:aut2}.
\end{proof}
\noindent {\it Proof of Proposition
\ref{prop:ABxyiLongerM}}.
The composition
\begin{equation*}
\begin{CD}
\widetilde \square_q @>> \widetilde g(a,a^{-1},b,b^{-1}) >
\widetilde \square_q
@>> can > \square_q
\end{CD}
\end{equation*}
is an $\mathbb F$-algebra homomorphism.
Apply this homomorphism
to everything in Proposition
\ref{prop:tildeMain}.
$\hfill \square$
\section{The algebra $\widehat \square_q$}
\noindent We have been discussing the
algebras
$\square_q$ and
$\widetilde \square_q$.
As we compare these algebras,
it is useful to bring in
an ``intermediate'' algebra
$\widehat \square_q$
such that
the canonical homomorphism
$ \widetilde \square_q \to \square_q$
has a factorization
$ \widetilde \square_q \to \widehat \square_q \to \square_q $.
\begin{definition}
\rm
\label{def:boxqV1}
Let $\widehat \square_q$ denote the $\mathbb F$-algebra with
generators $c^{\pm 1}_i, x_i$ $(i \in \mathbb Z_4)$
and relations
\begin{eqnarray}
&& \quad \qquad \qquad c_ic^{-1}_i = c^{-1}_i c_i = 1,
\label{eq:Box1}
\\
&& \quad \qquad \qquad
\mbox{$c^{\pm 1}_i$ are central in $\widehat \square_q$},
\label{eq:Box2}
\\
&&
\quad \qquad \qquad
\frac{q x_i x_{i+1}-q^{-1}x_{i+1}x_i}{q-q^{-1}} = c_i,
\label{eq:central}
\\
&&
x^3_i x_{i+2} - \lbrack 3 \rbrack_q x^2_i x_{i+2} x_i +
\lbrack 3 \rbrack_q x_i x_{i+2} x^2_i -x_{i+2} x^3_i = 0.
\label{eq:Box4}
\end{eqnarray}
\end{definition}
\noindent We have some comments.
\begin{lemma}
\label{lem:aut1}
There exists an automorphism $\widehat \rho$ of
$\widehat \square_q$
that sends $c_i \mapsto c_{i+1}$ and $x_i \mapsto x_{i+1}$ for
$i \in \mathbb Z_4$. Moreover $\widehat \rho^4=1$.
\end{lemma}
\begin{lemma}
\label{lem:can01}
There exists an $\mathbb F$-algebra
homomorphism $\widetilde \square_q \to \widehat \square_q$
that sends $c^{\pm 1}_i \mapsto c^{\pm 1}_i$
and $x_i \mapsto x_i$ for
$i \in \mathbb Z_4$.
This homomorphism is surjective.
\end{lemma}
\begin{lemma}
\label{lem:can12}
There exists an $\mathbb F$-algebra
homomorphism
$\widehat \square_q \to \square_q$
that sends $c^{\pm 1}_i \mapsto 1$
and $x_i \mapsto x_i$ for
$i \in \mathbb Z_4$.
This homomorphism is surjective.
\end{lemma}
\begin{definition}
\label{def:cancan}
\rm The homomorphisms
$\widetilde \square_q \to \widehat \square_q$
from Lemma
\ref{lem:can01}
and
$\widehat \square_q \to \square_q$
from Lemma
\ref{lem:can12} will be called {\it canonical}.
\end{definition}
\begin{note}
\rm In Definition
\ref{def:Canon}
we defined the canonical
homomorphism
$\widetilde \square_q \to \square_q$,
and in
Definition
\ref{def:cancan} we defined the canonical homomorphisms
$\widetilde \square_q \to \widehat \square_q$ and
$\widehat \square_q \to \square_q$.
When we speak of the canonical homomorphism, it
should be clear from the context
which version we refer to.
\end{note}
\begin{lemma}
The following diagram commutes:
\begin{equation*}
\begin{CD}
\widetilde \square_q @>can > >
\widehat \square_q
\\
@VIVV @VVcan V \\
\widetilde \square_q @>>can >
\square_q
\end{CD}
\end{equation*}
\end{lemma}
\begin{proof} By Definitions
\ref{def:Canon},
\ref{def:cancan}.
\end{proof}
\begin{definition}
\label{def:squareTeTo}
\rm Define the subalgebras
$ \widehat \square^{\rm even}_q$,
$\widehat \square^{\rm odd}_q$,
$\widehat C$ of
$ \widehat \square_q$ such that
\begin{enumerate}
\item[\rm (i)]
$ \widehat \square^{\rm even}_q$
is
generated by $x_0, x_2$;
\item[\rm (ii)]
$ \widehat \square^{\rm odd}_q$
is
generated by $x_1, x_3$;
\item[\rm (iii)]
$\widehat C$ is
generated by $\lbrace c^{\pm 1}_i\rbrace_{i \in \mathbb Z_4}$.
\end{enumerate}
\end{definition}
\begin{lemma}
\label{lem:xiadj}
Let $\lbrace \alpha_i \rbrace_{i\in \mathbb Z_4}$
denote invertible elements in
$\widehat C$.
Then there exists
an $\mathbb F$-algebra homomorphism
$\widehat \square_q\to \widehat \square_q $ that sends
$x_i \mapsto \alpha_i x_i $ and
$c_i \mapsto \alpha_i \alpha_{i+1} c_i$ for $i \in \mathbb Z_4$.
\end{lemma}
\begin{definition}
\label{def:alphaAut}
\rm
The homomorphism in Lemma
\ref{lem:xiadj} will be denoted by $\widehat g(\alpha_0, \alpha_1, \alpha_2, \alpha_3)$.
\end{definition}
\begin{lemma}
\label{lem:alphaAut}
Referring to Lemma
\ref{lem:xiadj} and Definition
\ref{def:alphaAut}, assume that $0 \not=\alpha_i \in \mathbb F$
for $i \in \mathbb Z_4$. Then
$\widehat g(\alpha_0, \alpha_1, \alpha_2, \alpha_3)$ is an automorphism
of $\widehat \square_q$. Its inverse is
$\widehat g(\alpha^{-1}_0, \alpha^{-1}_1, \alpha^{-1}_2, \alpha^{-1}_3)$.
\end{lemma}
\begin{proof}
Similar to the proof of Lemma
\ref{lem:alphaAutT}.
\end{proof}
\noindent Our next goal is to obtain an analog of
Propositions \ref{thm:tensorDecPreM},
\ref{prop:tensorDec}
that applies to $\widehat \square_q$.
\begin{definition}
\label{def:J}
\rm
Let $J$ denote the 2-sided ideal of $\widetilde \square_q$
generated by $\lbrace S_i \rbrace_{i \in \mathbb Z_4}$.
Thus
\begin{eqnarray}
\label{eq:Jdef}
J = \sum_{i \in \mathbb Z_4}
\widetilde \square_q
S_i
\widetilde \square_q.
\end{eqnarray}
\end{definition}
\begin{lemma}
\label{lem:canKer}
The canonical homomorphism
$ \widetilde \square_q \to \widehat \square_q $ has
kernel $J$.
\end{lemma}
\begin{proof} Compare Definitions
\ref{def:boxqV2},
\ref{def:boxqV1}.
\end{proof}
\begin{definition}
\label{def:Jevenodd}
\rm
Let $J^{\rm even}$
(resp.
$J^{\rm odd}$)
denote the 2-sided ideal of
$\widetilde \square^{\rm even}_q$
(resp.
$\widetilde \square^{\rm odd}_q$)
generated by
$S_0, S_2$
(resp.
$S_1, S_3$).
Thus
\begin{eqnarray}
\label{eq:Jeve}
&&
J^{\rm even} =
\widetilde \square^{\rm even}_q
S_0
\widetilde \square^{\rm even}_q
+
\widetilde \square^{\rm even}_q
S_2
\widetilde \square^{\rm even}_q,
\\
&&
\label{eq:Jodd}
J^{\rm odd} =
\widetilde \square^{\rm odd}_q
S_1
\widetilde \square^{\rm odd}_q
+
\widetilde \square^{\rm odd}_q
S_3
\widetilde \square^{\rm odd}_q.
\end{eqnarray}
\end{definition}
\noindent Recall the free algebra $T$ with standard generators $x,y$.
Below
(\ref{eq:SgenM})
we defined the 2-sided ideal $S$ of $T$.
\begin{lemma}
\label{lem:imageIso}
The following {\rm (i), (ii)} hold:
\begin{enumerate}
\item[\rm (i)] $J^{\rm even}$ is the image of $S$
under the isomorphism
$T \to \widetilde \square^{\rm even}_q$
from Proposition \ref{prop:tensorDec}(i);
\item[\rm (ii)] $J^{\rm odd}$ is the image of $S$
under the isomorphism
$T \to \widetilde \square^{\rm odd}_q$
from
Proposition \ref{prop:tensorDec}(ii).
\end{enumerate}
\end{lemma}
\begin{proof}
Compare
Definition \ref{def:Jevenodd} with the definition of
$S$ below
(\ref{eq:SgenM}).
\end{proof}
\begin{lemma}
\label{lem:Jdesc}
Referring to the vector space isomorphism
from Proposition
\ref{prop:tensorDec}(iv), the preimage of
$J$ is
\begin{eqnarray}
\label{eq:Preim}
J^{\rm even} \otimes
\widetilde \square^{\rm odd}_q \otimes \widetilde C +
\widetilde \square^{\rm even}_q \otimes
J^{\rm odd} \otimes \widetilde C.
\end{eqnarray}
\end{lemma}
\begin{proof} Let $\widetilde m$ denote the isomorphism in question.
Let $J'$ denote the image of
(\ref{eq:Preim}) under $\widetilde m$.
We show that $J=J'$. We have
$J \supseteq J'$ by construction and since
each of $J^{\rm even},
J^{\rm odd}$
is contained in $J$.
To obtain
$J \subseteq J'$, by
(\ref{eq:Jdef})
it suffices to show
that $\widetilde \square_q S_i \widetilde \square_q \subseteq J'$
for $i \in \mathbb Z_4$. Let $i$ be given,
and first assume that $i$ is even.
By Lemma
\ref{lem:factor},
Corollary
\ref{cor:DeltaCom}, and since $\widetilde C$ is central in $\widetilde \square_q$,
\begin{eqnarray*}
\widetilde \square_q S_i \widetilde \square_q
=
\widetilde \square^{\rm even}_q
\widetilde \square^{\rm odd}_q
\widetilde C
S_i \widetilde \square_q
=
\widetilde \square^{\rm even}_q
S_i
\widetilde \square^{\rm odd}_q
\widetilde C
\widetilde \square_q
\subseteq
\widetilde \square^{\rm even}_q
S_i
\widetilde \square_q.
\end{eqnarray*}
By Lemma
\ref{lem:factor},
line (\ref{eq:Jeve}), and the definition of
$\widetilde m$,
\begin{eqnarray*}
\widetilde \square^{\rm even}_q
S_i
\widetilde \square_q =
\widetilde \square^{\rm even}_q
S_i
\widetilde \square^{\rm even}_q
\widetilde \square^{\rm odd}_q
\widetilde C
\subseteq
J^{\rm even}
\widetilde \square^{\rm odd}_q
\widetilde C
=
\widetilde m(
J^{\rm even}
\otimes \widetilde \square^{\rm odd}_q
\otimes \widetilde C)
\subseteq J'.
\end{eqnarray*}
We have shown
that $\widetilde \square_q S_i \widetilde \square_q \subseteq J'$
for $i$ even.
We similarly show that
$\widetilde \square_q S_i \widetilde \square_q \subseteq J'$ for $i$ odd.
Therefore $J\subseteq J'$. We have shown that $J=J'$,
and the result follows.
\end{proof}
\begin{lemma}
\label{lem:Jint}
In the algebra $\widetilde \square_q$,
\begin{enumerate}
\item[\rm (i)]
$
J \cap
\widetilde \square^{\rm even}_q
=
J^{\rm even}$;
\item[\rm (ii)]
$
J \cap \widetilde \square^{\rm odd}_q
=
J^{\rm odd}$;
\item[\rm (iii)]
$
J \cap
\widetilde C = 0$.
\end{enumerate}
\end{lemma}
\begin{proof}
By Lemma
\ref{lem:Jdesc} and since
neither of
$J^{\rm even},
J^{\rm odd}$ contains $1$.
\end{proof}
\noindent
Recall from
Definition
\ref{def:squareTeTo}
the subalgebras
$\widehat \square^{\rm even}_q$,
$\widehat \square^{\rm odd}_q$,
$\widehat C$ of $\widehat \square_q$.
\begin{lemma}
\label{lem:canonImage}
For the canonical homomorphism
$ \widetilde \square_q \to \widehat \square_q $,
the images of
$\widetilde \square^{\rm even}_q$,
$\widetilde \square^{\rm odd}_q$,
$\widetilde C$ are
$ \widehat \square^{\rm even}_q$
$ \widehat \square^{\rm odd}_q$,
$\widehat C$, respectively.
\end{lemma}
\begin{proof} By
Definitions
\ref{def:TeTo},
\ref{def:cancan},
\ref{def:squareTeTo}
and
Lemma
\ref{lem:can01}.
\end{proof}
\begin{definition}
\rm
For the canonical homomorphism
$ \widetilde \square_q \to \widehat \square_q$,
the restrictions to
$\widetilde \square^{\rm even}_q$,
$\widetilde \square^{\rm odd}_q$,
$\widetilde C$
induce surjective
$\mathbb F$-algebra homomorphisms
\begin{eqnarray}
\widetilde \square^{\rm even}_q \to \widehat \square^{\rm even}_q,
\qquad \qquad
\widetilde \square^{\rm odd}_q \to \widehat \square^{\rm odd}_q,
\qquad \qquad
\widetilde C \to \widehat C.
\label{eq:threeRes}
\end{eqnarray}
Each of the homomorphisms
(\ref{eq:threeRes})
will be called
{\it restricted canonical}.
\end{definition}
\begin{lemma}
\label{lem:KerRest}
The following {\rm (i)--(iii)} hold:
\begin{enumerate}
\item[\rm (i)] the restricted canonical homomorphism
$\widetilde \square^{\rm even}_q \to \widehat \square^{\rm even}_q$
has kernel $J^{\rm even}$;
\item[\rm (ii)] the restricted canonical homomorphism
$\widetilde \square^{\rm odd}_q \to \widehat \square^{\rm odd}_q$
has kernel $J^{\rm odd}$;
\item[\rm (iii)] the restricted canonical homomorphism
$\widetilde C \to \widehat C$
is a bijection.
\end{enumerate}
\end{lemma}
\begin{proof}
By Lemmas
\ref{lem:canKer},
\ref{lem:Jint}.
\end{proof}
\begin{proposition}
\label{thm:tensorDec}
The following {\rm (i)--(iv)} hold:
\begin{enumerate}
\item[\rm (i)]
there exists an $\mathbb F$-algebra isomorphism
$U^+ \to \widehat \square^{\rm even}_q$ that sends
$X\mapsto x_0$ and
$Y\mapsto x_2$;
\item[\rm (ii)]
there exists an $\mathbb F$-algebra isomorphism
$U^+ \to \widehat \square^{\rm odd}_q$ that sends
$X\mapsto x_1$ and
$Y\mapsto x_3$;
\item[\rm (iii)]
there exists an
$\mathbb F$-algebra isomorphism
$L\to \widehat C$
that sends $\lambda^{\pm 1}_i \mapsto c^{\pm 1}_i$ for
$i \in \mathbb Z_4$;
\item[\rm (iv)]
the following is an isomorphism of
$\mathbb F$-vector spaces:
\begin{eqnarray*}
\widehat \square^{\rm even}_q
\otimes
\widehat \square^{\rm odd}_q
\otimes
\widehat C
& \to & \widehat \square_q
\\
u \otimes v \otimes c &\mapsto & uvc
\end{eqnarray*}
\end{enumerate}
\end{proposition}
\begin{proof} (i)
Recall
the free algebra $T$ with
standard generators $x,y$.
By Proposition
\ref{prop:tensorDec}(i) there exists an
$\mathbb F$-algebra isomorphism
$T \to \widetilde \square^{\rm even}_q$
that sends $x \mapsto x_0$ and
$y \mapsto x_2$.
The inverse of this isomorphism will be denoted by
$\widetilde \theta$.
Consider the
$\mathbb F$-algebra homomorphism
$\mu: T \to U^+ $
that sends $x \mapsto X$ and
$y \mapsto Y$. By Lemma
\ref{lem:imageIso}(i),
the composition
\begin{equation*}
\begin{CD}
\widetilde \square^{\rm even}_q @>> \widetilde \theta >
T @>> \mu > U^+
\end{CD}
\end{equation*}
is surjective with kernel $J^{\rm even}$.
Therefore there exists an $\mathbb F$-algebra isomorphism
$
\theta:
\widetilde \square^{\rm even}_q /J^{\rm even}
\to
U^+$
that sends
$ x_0 + J^{\rm even} \mapsto X$ and
$ x_2 + J^{\rm even} \mapsto Y$.
The restricted canonical homomorphism
$\widetilde \square^{\rm even}_q \to \widehat \square^{\rm even}_q$
sends $x_0 \mapsto x_0$ and
$x_2 \mapsto x_2$. This map is surjective by
construction, and has kernel $J^{\rm even}$
by Lemma
\ref{lem:KerRest}(i).
Therefore there exists an $\mathbb F$-algebra isomorphism
$\vartheta :
\widetilde \square^{\rm even}_q/J^{\rm even} \to \widehat \square^{\rm even}_q$
that sends
$x_0+J^{\rm even} \mapsto x_0$ and
$x_2+J^{\rm even} \mapsto x_2$.
The composition
\begin{equation*}
\begin{CD}
U^+ @>>\theta^{-1} >
\widetilde \square^{\rm even}_q / J^{\rm even}
@>> \vartheta > \widehat \square^{\rm even}_q
\end{CD}
\end{equation*}
is the desired $\mathbb F$-algebra isomorphism.
\\
\noindent (ii), (iii). Similar to the proof of (i) above.
\\
\noindent (iv).
The multiplication map
$\widehat m:
\widehat \square^{\rm even}_q
\otimes
\widehat \square^{\rm odd}_q
\otimes
\widehat C
\to \widehat \square_q
$,
$u \otimes v \otimes c \mapsto uvc$
is $\mathbb F$-linear.
We show that $\widehat m$ is a bijection.
By Proposition
\ref{prop:tensorDec}(iv), the multiplication map
$\widetilde m:
\widetilde \square^{\rm even}_q
\otimes
\widetilde \square^{\rm odd}_q
\otimes
\widetilde C
\to \widetilde \square_q
$,
$u \otimes v \otimes c \mapsto uvc$
is an isomorphism of $\mathbb F$-vector spaces.
Recall the canonical homomorphism
$\widetilde \square_q \to \widehat \square_q $ from
Definition
\ref{def:cancan}.
By construction the following diagram commutes:
\begin{equation*}
\begin{CD}
\mbox{
$
\widetilde \square^{\rm even}_q \otimes
\widetilde \square^{\rm odd}_q \otimes
\widetilde C
$ } @>
u \otimes v \otimes c \mapsto
can(u)\otimes can(v) \otimes can(c)
>>
\mbox{ $
\widehat \square^{\rm even}_q \otimes
\widehat \square^{\rm odd}_q \otimes
\widehat C$
}
\\
@V\widetilde m VV @VV \widehat m V \\
\mbox{$\widetilde \square_q $} @>>{can}>
\mbox{$ \widehat \square_q $}
\end{CD}
\end{equation*}
The map $\widehat m$ is
surjective by these comments and
the last assertion of Lemma
\ref{lem:can01}.
The map $\widehat m$
is
injective
in view of
Lemma
\ref{lem:Jdesc}
along with Lemmas
\ref{lem:canKer},
\ref{lem:KerRest}.
Therefore $\widehat m$ is a bijection.
\end{proof}
\noindent For the rest of this section, we
describe how $\widehat \square_q$ is related to the $q$-Onsager algebra
$\mathcal O$.
\begin{proposition}
\label{prop:ABxy}
For the algebra $\widehat \square_q$, define
\begin{eqnarray*}
A = x_0+x_1, \qquad \qquad B = x_2+x_3.
\end{eqnarray*}
Then
\begin{eqnarray*}
&&
A^3 B - \lbrack 3 \rbrack_q A^2 B A +
\lbrack 3 \rbrack_q A B A^2 -B A^3 = (q^2-q^{-2})^2 c_0(BA-AB),
\\
&&
B^3 A - \lbrack 3 \rbrack_q B^2 A B +
\lbrack 3 \rbrack_q B A B^2 -A B^3 = (q^2-q^{-2})^2 c_2(AB-BA).
\end{eqnarray*}
\end{proposition}
\begin{proof}
Apply the canonical homomorphism $\widetilde \square_q \to \widehat \square_q$
to each side of
(\ref{eq:1Assertion}),
(\ref{eq:2Assertion}).
\end{proof}
\noindent The following more general
version of
Proposition
\ref{prop:ABxy} is obtained by applying
Lemma
\ref{lem:xiadj}.
\begin{corollary}
\label{prop:ABxyiLong}
Let
$\lbrace \alpha_i \rbrace_{i\in \mathbb Z_4}$ denote invertible
elements in
$\widehat C$.
For the algebra
$\widehat \square_q$,
define
\begin{eqnarray}
\label{eq:ABDef}
A = \alpha_0 x_0+ \alpha_1 x_1, \qquad \qquad B = \alpha_2 x_2+ \alpha_3 x_3.
\end{eqnarray}
Then
\begin{eqnarray}
&&
A^3 B - \lbrack 3 \rbrack_q A^2 B A +
\lbrack 3 \rbrack_q A B A^2 -B A^3 =
(q^2-q^{-2})^2 \alpha_0 \alpha_1 c_0(BA-AB),
\label{eq:qDGLong1}
\\
&&
B^3 A - \lbrack 3 \rbrack_q B^2 A B +
\lbrack 3 \rbrack_q B A B^2 -A B^3 =
(q^2-q^{-2})^2 \alpha_2 \alpha_3 c_2(AB-BA).
\label{eq:qDGLong2}
\end{eqnarray}
\end{corollary}
\begin{proof}
Apply the homomorphism
$\widehat g(\alpha_0,\alpha_1,\alpha_2,\alpha_3)$
from Definition
\ref{def:alphaAut}
to everything in
Proposition
\ref{prop:ABxy}.
\end{proof}
\begin{proposition}
\label{prop:ABxyiLonger}
Let
$\lbrace \alpha_i \rbrace_{i\in \mathbb Z_4}$ denote
elements in
$\widehat C$ such that
$\alpha_0 \alpha_1 c_0=1 $ and
$\alpha_2 \alpha_3 c_2=1$.
Then there exists an $\mathbb F$-algebra
homomorphism
$ \natural: \mathcal O \to \widehat \square_q$ that sends
\begin{eqnarray}
\label{eq:ABDefMap}
A \mapsto \alpha_0 x_0+ \alpha_1 x_1,
\qquad \qquad
B \mapsto \alpha_2 x_2+ \alpha_3 x_3.
\end{eqnarray}
\end{proposition}
\begin{proof}
By Definition
\ref{def:qOnsager}
and
Corollary
\ref{prop:ABxyiLong}.
\end{proof}
\noindent
In
Theorem \ref{thm:naturalINJ}
we will show that
the map $\natural$ from
Proposition
\ref{prop:ABxyiLonger} is injective.
\medskip
\noindent
Let the elements
$\lbrace \alpha_i \rbrace_{i\in \mathbb Z_4}$
and the map $\natural$ be as in
Proposition
\ref{prop:ABxyiLonger}. By construction
there exist nonzero $a,b \in \mathbb F$
such that
the canonical homomorphism $\widehat \square_q \to \square_q$
sends
\begin{eqnarray}
\alpha_0 \mapsto a, \qquad \quad
\alpha_1 \mapsto a^{-1}, \qquad \quad
\alpha_2 \mapsto b, \qquad \quad
\alpha_3 \mapsto b^{-1}.
\label{eq:canonsend}
\end{eqnarray}
Using $a,b$ we obtain the
$\mathbb F$-algebra homomorphism
$\sharp: \mathcal O \to \square_q$
as in
Proposition
\ref{prop:ABxyiLongerM}.
\begin{lemma}
\label{lem:naturalSharp}
With the above notation,
the following diagram commutes:
\begin{equation*}
\begin{CD}
\mathcal O @>\natural > >
\widehat \square_q
\\
@VIVV @VV can V \\
\mathcal O @>>\sharp>
\square_q
\end{CD}
\end{equation*}
\end{lemma}
\begin{proof}
Each map in the diagram is an $\mathbb F$-algebra
homomorphism.
The $\mathbb F$-algebra $\mathcal O$
is generated by $A,B$. To verify that the diagram commutes,
chase $A,B$ around the diagram
using
(\ref{eq:ABDefMapM}) and
(\ref{eq:ABDefMap}),
(\ref{eq:canonsend}).
\end{proof}
\section{The algebra $\square_q$, revisited}
\noindent
In Section 5 we defined the algebra $\square_q$, and
in Sections 6--9 we investigated its homomorphic
preimages
$\widetilde \square_q$,
$\widehat \square_q$.
In this section we return our attention to $\square_q$.
We will first prove Proposition
\ref{thm:tensorDecPreM}. We will then prove
Theorem
\ref{thm:main1}, and establish the
injectivity of the
maps $\sharp$ and $\natural$ from
Propositions
\ref{prop:ABxyiLongerM} and
\ref{prop:ABxyiLonger}, respectively.
\medskip
\noindent Recall the $\mathbb F$-algebra $L$ from
(\ref{eq:Ldef}). Let $L_0$ denote the ideal of $L$ generated by
$\lbrace \lambda_i -1 \rbrace_{i\in \mathbb Z_4}$.
Thus
\begin{eqnarray}
L_0 = \sum_{i \in \mathbb Z_4} L(\lambda_i-1).
\label{eq:L0}
\end{eqnarray}
The ideal $L_0$ is the kernel of the
$\mathbb F$-algebra
homomorphism $L\to \mathbb F$ that sends $\lambda_i \mapsto 1$
for $i \in \mathbb Z_4$.
The sum $L = L_0 + \mathbb F$ is direct.
\begin{definition}\rm
\label{def:K}
Let $K$ denote the 2-sided ideal of $\widehat \square_q$ generated by
$\lbrace c_i -1 \rbrace_{i \in \mathbb Z_4}$. Since the
$\lbrace c_i \rbrace_{i \in \mathbb Z_4}$ are central,
\begin{eqnarray*}
K = \sum_{i \in \mathbb Z_4} \widehat \square_q (c_i-1).
\end{eqnarray*}
\end{definition}
\begin{lemma}
\label{lem:Kmeaning}
The canonical homomorphism $\widehat \square_q \to \square_q$
has kernel $K$.
\end{lemma}
\begin{proof} Compare Definition
\ref{def:boxqV1M}
and Definition
\ref{def:boxqV1}.
\end{proof}
\noindent Recall from Definition
\ref{def:squareTeTo}
the subalgebras
$ \widehat \square^{\rm even}_q$,
$\widehat \square^{\rm odd}_q$,
$\widehat C$ of
$ \widehat \square_q$.
\begin{definition}\rm
\label{def:K0}
Let $K_0$ denote the ideal of $\widehat C$ generated by
$\lbrace c_i -1\rbrace_{i \in \mathbb Z_4}$. Thus
\begin{eqnarray}
\label{eq:K0}
K_0 = \sum_{i \in \mathbb Z_4} \widehat C (c_i-1).
\end{eqnarray}
\end{definition}
\begin{lemma}
The ideal $K_0 $ is the image of $L_0$ under the isomorphism
$L \to \widehat C$ from Proposition
\ref{thm:tensorDec}(iii).
\end{lemma}
\begin{proof} Compare
(\ref{eq:L0}),
(\ref{eq:K0}).
\end{proof}
\begin{lemma}
\label{lem:Kdesc}
Referring to the vector space isomorphism
from Proposition
\ref{thm:tensorDec}(iv), the preimage of $K$ is
\begin{eqnarray}
\label{eq:SSK}
\widehat \square^{\rm even}_q
\otimes
\widehat \square^{\rm odd}_q
\otimes
K_0.
\end{eqnarray}
\end{lemma}
\begin{proof} By
Definitions \ref{def:K},
\ref{def:K0}.
\end{proof}
\begin{lemma}
\label{lem:Kintersect}
In the algebra $\widehat \square_q$,
\begin{enumerate}
\item[\rm (i)] $K \cap \widehat \square^{\rm even}_q = 0$;
\item[\rm (ii)] $K \cap \widehat \square^{\rm odd}_q = 0$;
\item[\rm (iii)] $K \cap \widehat C= K_0$.
\end{enumerate}
\end{lemma}
\begin{proof} Use Lemma
\ref{lem:Kdesc}.
\end{proof}
\noindent
Recall from
Definition
\ref{def:squareTeToM}
the subalgebras
$\square^{\rm even}_q$,
$\square^{\rm odd}_q$
of
$\square_q$.
\begin{lemma}
\label{lem:canonImageA}
For the canonical homomorphism
$ \widehat \square_q \to \square_q $,
the images of
$\widehat \square^{\rm even}_q$,
$\widehat \square^{\rm odd}_q$,
$\widehat C$
are
$ \square^{\rm even}_q$
$ \square^{\rm odd}_q$,
$\mathbb F$,
respectively.
\end{lemma}
\begin{proof} By
Definitions
\ref{def:squareTeToM},
\ref{def:squareTeTo}
and Lemma
\ref{lem:can12}.
\end{proof}
\begin{definition}
\rm
For the canonical homomorphism
$ \widehat \square_q \to \square_q$,
the restrictions to
$\widehat \square^{\rm even}_q$,
$\widehat \square^{\rm odd}_q$,
$\widehat C$
induce surjective
$\mathbb F$-algebra homomorphisms
\begin{eqnarray}
\widehat \square^{\rm even}_q \to \square^{\rm even}_q,
\qquad \qquad
\widehat \square^{\rm odd}_q \to \square^{\rm odd}_q,
\qquad \qquad
\widehat C \to \mathbb F.
\label{eq:threeResA}
\end{eqnarray}
Each of the homomorphisms
(\ref{eq:threeResA})
will be called
{\it restricted canonical}.
\end{definition}
\begin{lemma}
\label{lem:KerRestA}
The following {\rm (i)--(iii)} hold:
\begin{enumerate}
\item[\rm (i)] the restricted canonical homomorphism
$\widehat \square^{\rm even}_q \to \square^{\rm even}_q$
is an isomorphism;
\item[\rm (ii)] the restricted canonical homomorphism
$\widehat \square^{\rm odd}_q \to \square^{\rm odd}_q$
is an isomorphism;
\item[\rm (iii)] the restricted canonical homomorphism
$\widehat C \to \mathbb F$
has kernel $K_0$.
\end{enumerate}
\end{lemma}
\begin{proof} By Lemmas
\ref{lem:Kmeaning},
\ref{lem:Kintersect}.
\end{proof}
\noindent {\it Proof of Proposition
\ref{thm:tensorDecPreM}}.
(i)
The desired isomorphism is the composition
of the isomorphism $U^+ \to \widehat \square^{\rm even}_q$
from Proposition
\ref{thm:tensorDec}(i)
and the isomorphism
$\widehat \square^{\rm even}_q \to \square^{\rm even}_q$
from
Lemma \ref{lem:KerRestA}(i).
\\
\noindent (ii)
The desired isomorphism is the composition
of the isomorphism $U^+ \to \widehat \square^{\rm odd}_q$
from Proposition
\ref{thm:tensorDec}(ii)
and the isomorphism
$\widehat \square^{\rm odd}_q \to \square^{\rm odd}_q$
from
Lemma \ref{lem:KerRestA}(ii).
\\
\noindent (iii) By Lemma
\ref{lem:Kdesc} and since $1 \not\in K_0$ we see that, for the
vector space isomorphism
in Proposition
\ref{thm:tensorDec}(iv),
the preimage of $K$ has zero intersection with
$\widehat \square^{\rm even}_q
\otimes
\widehat \square^{\rm odd}_q$.
The result follows.
$\hfill \square$
\medskip
\noindent We now turn our attention to
Theorem
\ref{thm:main1} and the maps
$\sharp$, $\natural$.
We comment on the notation.
In earlier sections we discussed the
algebras
$\square_q$,
$\square^{\rm even}_q$,
$\square^{\rm odd}_q$.
In our discussion going forward, in order to simplify
the notation
we will
drop the reference to $q$, and write
$\square=\square_q$,
$\square^{\rm even}=\square^{\rm even}_q$,
$\square^{\rm odd}=\square^{\rm odd}_q$.
\medskip
\noindent
Recall the $\mathbb F$-algebra $U^+=U^+_q$ and
its $\mathbb N$-grading
$\lbrace U^+_n\rbrace_{n \in \mathbb N}$.
\begin{definition}
\label{def:recall1}
\rm
Recall from Proposition
\ref{thm:tensorDecPreM}(i)
the $\mathbb F$-algebra isomorphism
$U^+ \to \square^{\rm even}$ that sends $X \mapsto x_0$
and $Y \mapsto x_2$.
Under this isomorphism,
for $n \in \mathbb N$ the image of
$U^+_n$ will be denoted
by $\square^{\rm even}_n$.
Recall from Proposition
\ref{thm:tensorDecPreM}(ii)
the $\mathbb F$-algebra isomorphism
$U^+ \to \square^{\rm odd}$ that sends $X \mapsto x_1$
and $Y \mapsto x_3$.
Under this isomorphism,
for $n \in \mathbb N$ the image of
$U^+_n$ will be denoted by
$\square^{\rm odd}_n$.
\end{definition}
\begin{lemma} The sequence
$\lbrace
\square^{\rm even}_n \rbrace_{n \in \mathbb N}$ is an $\mathbb N$-grading
of
$\square^{\rm even}$.
The sequence
$\lbrace
\square^{\rm odd}_n \rbrace_{n \in \mathbb N}$ is an $\mathbb N$-grading
of
$\square^{\rm odd}$.
\end{lemma}
\begin{proof}
By Definition
\ref{def:recall1} and since
$\lbrace
U^+_n \rbrace_{n \in \mathbb N}$ is an $\mathbb N$-grading
of $U^+$.
\end{proof}
\begin{definition}
\label{def:squareNPre}
\rm
Recall from Proposition
\ref{thm:tensorDecPreM}(iii)
the isomorphism of $\mathbb F$-vector spaces
$\square^{\rm even} \otimes
\square^{\rm odd} \to \square$.
Under this isomorphism,
for $r,s \in \mathbb N$ the image of
$
\square^{\rm even}_r
\otimes
\square^{\rm odd}_s
$
will be denoted by $\square_{r,s}$.
\end{definition}
\noindent Referring to Definition
\ref{def:squareNPre},
the sum $\square= \sum_{r,s \in \mathbb N} \square_{r,s}$ is
direct.
Moreover $1 \in \square_{0,0}$ and
$x_0,x_2 \in \square_{1,0}$ and
$x_1,x_3 \in \square_{0,1}$.
\begin{lemma}
\label{lem:rstcom}
The following
{\rm (i), (ii)} hold for $r,s,t \in \mathbb N$:
\begin{enumerate}
\item[\rm (i)]
$\square^{\rm even}_t \square_{r,s} \subseteq \square_{r+t,s}$;
\item[\rm (ii)]
$\square_{r,s}\square^{\rm odd}_t \subseteq \square_{r,s+t}$.
\end{enumerate}
\end{lemma}
\begin{proof} (i) By Definition
\ref{def:squareNPre} and since
$\square^{\rm even}_t
\square^{\rm even}_r \subseteq \square^{\rm even}_{r+t}$.
\\
\noindent (ii) By Definition
\ref{def:squareNPre} and since
$\square^{\rm odd}_s
\square^{\rm odd}_t \subseteq \square^{\rm odd}_{s+t}$.
\end{proof}
\begin{definition}
\label{def:squareN}
\rm
For $n \in \mathbb Z$
define
\begin{eqnarray*}
\square_n =
\sum_
{
\genfrac{}{}{0pt}{}{r,s \in \mathbb N}{r-s=n}
}
\square_{r,s}.
\end{eqnarray*}
\end{definition}
\noindent
Referring to Definition
\ref{def:squareN},
our next goal is to show that the sequence
$\lbrace \square_n\rbrace_{n \in \mathbb Z}$ is
a $\mathbb Z$-grading of $\square$.
\begin{lemma}
\label{lem:grade12a}
The sum
$\square= \sum_{n\in \mathbb Z} \square_n$ is direct.
Moreover
$ 1 \in \square_0$ and
$x_0,x_2 \in \square_1$ and
$x_1,x_3 \in \square_{-1}$.
\end{lemma}
\begin{proof} By Definition
\ref{def:squareN}
and the comments
above Lemma
\ref{lem:rstcom}.
\end{proof}
\begin{lemma}
\label{lem:grade12}
The following {\rm (i), (ii)}
hold for $n\in \mathbb Z$ and $t\in \mathbb N$:
\begin{enumerate}
\item[\rm (i)] $\square^{\rm even}_t \square_n \subseteq
\square_{n+t}$;
\item[\rm (ii)] $\square_n \square^{\rm odd}_t \subseteq
\square_{n-t}$.
\end{enumerate}
\end{lemma}
\begin{proof}
Use Lemma
\ref{lem:rstcom}
and Definition
\ref{def:squareN}.
\end{proof}
\begin{lemma}
\label{lem:grading1}
For $r,s\in \mathbb N$,
\begin{eqnarray}
\label{eq:outoforder}
\square^{\rm odd}_s \square^{\rm even}_r \subseteq
\sum_{\ell=0}^{{\rm {min}} (r,s)}
\square_{r-\ell,s-\ell}.
\end{eqnarray}
Moreover
$\square^{\rm odd}_s
\square^{\rm even}_r \subseteq \square_{r-s}$.
\end{lemma}
\begin{proof} To obtain
(\ref{eq:outoforder})
use Lemma
\ref{lem:uLong}
and induction on $s$.
The last assertion follows from
(\ref{eq:outoforder})
and Definition
\ref{def:squareN}.
\end{proof}
\begin{lemma}
\label{lem:grading2}
We have $\square_r \square_s \subseteq \square_{r+s}$
for
$r,s\in \mathbb Z$.
\end{lemma}
\begin{proof} By
Definition
\ref{def:squareN},
Lemma \ref{lem:grade12}, and the last assertion of
Lemma
\ref{lem:grading1}.
\end{proof}
\begin{proposition}
\label{prop:Grading}
The sequence $\lbrace \square_n\rbrace_{n \in \mathbb Z}$
is a $\mathbb Z$-grading of $\square$.
\end{proposition}
\begin{proof} By Lemmas
\ref{lem:grade12a},
\ref{lem:grading2}.
\end{proof}
\noindent
For the rest of this section, fix nonzero $a,b\in \mathbb F$.
Using $a,b$ we obtain the $\mathbb F$-algebra homomorphism
$\sharp: \mathcal O \to \square$ as in
Proposition
\ref{prop:ABxyiLongerM}.
Define
\begin{eqnarray}
A^+ = a x_0, \qquad
A^- = a^{-1} x_1, \qquad
B^+ = b x_2, \qquad
B^- = b^{-1} x_3
\label{def:ABPM}
\end{eqnarray}
so that
\begin{eqnarray}
\sharp(A) = A^+ + A^-, \qquad \qquad
\sharp(B) = B^+ + B^-.
\label{eq:ABxi}
\end{eqnarray}
\noindent By Lemma
\ref{lem:grade12a} and
(\ref{def:ABPM}),
\begin{eqnarray}
A^+ \in \square_1, \qquad
A^- \in \square_{-1}, \qquad
B^+ \in \square_{1}, \qquad
B^- \in \square_{-1}.
\label{eq:ABloc}
\end{eqnarray}
By
(\ref{eq:ABxi}),
(\ref{eq:ABloc}),
\begin{eqnarray}
\sharp (A) \in \square_1 + \square_{-1},
\qquad \qquad
\sharp (B) \in \square_1 + \square_{-1}.
\label{lem:xiAB}
\end{eqnarray}
\begin{lemma}
\label{lem:wordExpand}
Let $n \in \mathbb N$. For $1 \leq i \leq n$ pick
$g_i \in \lbrace A,B\rbrace$. Then
\begin{eqnarray}
\label{eq:xiExpand}
\sharp(g_1g_2\cdots g_n) =
\sum
g^{\epsilon_1}_1
g^{\epsilon_2}_2
\cdots
g^{\epsilon_n}_n,
\end{eqnarray}
where the sum is over all sequences
$\epsilon_1,
\epsilon_2,
\ldots,
\epsilon_n$
such that $\epsilon_i \in \lbrace +,-\rbrace$ for
$1 \leq i \leq n$.
\end{lemma}
\begin{proof} To verify
(\ref{eq:xiExpand}),
expand
the left-hand side
using
(\ref{eq:ABxi}) and the fact that $\sharp$
is an algebra homomorphism.
\end{proof}
\begin{lemma}
\label{lem:whereSummand}
Refer to Lemma
\ref{lem:wordExpand}.
In the sum on the right in
{\rm (\ref{eq:xiExpand})}, consider any summand
$g^{\epsilon_1}_1
g^{\epsilon_2}_2
\cdots
g^{\epsilon_n}_n$. This summand is contained in $\square_\ell$,
where
\begin{eqnarray*}
\ell
=
|\lbrace i|1 \leq i \leq n,\; \epsilon_i = +\rbrace |
-
|\lbrace i|1 \leq i \leq n, \,\epsilon_i = -\rbrace |.
\end{eqnarray*}
\end{lemma}
\begin{proof}
By Lemma
\ref{lem:grading2}
and
(\ref{eq:ABloc}).
\end{proof}
\begin{lemma}
\label{lem:spread}
For $n \in \mathbb N$,
\begin{eqnarray}
\sharp(\mathcal O_n) \subseteq \sum_{\ell=-n}^n \square_{\ell}.
\label{eq:spread}
\end{eqnarray}
\end{lemma}
\begin{proof}
We mentioned below Definition
\ref{def:qOnsager}
that
$\mathcal O_n$ is spanned by
the products
$g_1g_2\cdots g_r$ such that
$0 \leq r \leq n$ and
$g_i$ is among $A,B$ for $1 \leq i \leq r$.
The result follows from this and
Lemmas
\ref{lem:wordExpand},
\ref{lem:whereSummand}.
\end{proof}
\begin{definition}
\label{def:projGrad}
\rm
For $r\in \mathbb Z$ define an $\mathbb F$-linear
map $\pi_r: \square \to \square$ such that
$(\pi_r-I) \square_r = 0$ and
$\pi_r \square_s = 0$ for $s \in \mathbb Z$, $s\not=r$.
Thus $\pi_r$ is the projection from $\square$ onto
$\square_r$.
\end{definition}
\begin{lemma}
\label{lem:comp}
For $n \in \mathbb N$ and
$r \in \mathbb Z$,
the composition
\begin{equation*}
\begin{CD}
\mathcal O_n @>> \sharp >
\square @>> \pi_r > \square
\end{CD}
\end{equation*}
is zero unless $-n \leq r \leq n$.
\end{lemma}
\begin{proof}
By Lemma
\ref{lem:spread}
and Definition
\ref{def:projGrad}.
\end{proof}
\begin{definition}
\label{def:comp}
Pick $n \in \mathbb N$ and recall the $\mathbb F$-vector
space
$\overline{\mathcal O}_n = \mathcal O_n /\mathcal O_{n-1}$
from Section 4.
Define the $\mathbb F$-linear map
$\varphi_n : \mathcal O_n \to \square$ to be
the composition
\begin{equation}
\begin{CD}
\varphi_n: \mathcal O_n @>> \sharp >
\square @>> \pi_n > \square.
\end{CD}
\label{eq:compVarphi}
\end{equation}
By Lemma
\ref{lem:comp}
we have $\varphi_n(\mathcal O_{n-1})=0$.
Therefore
$\varphi_n$ induces an $\mathbb F$-linear
map $\overline{\varphi}_n:
\overline{\mathcal O}_n \to \square$ that sends
$u+\mathcal O_{n-1} \mapsto \varphi_n(u)$ for all
$u \in \mathcal O_n$.
\end{definition}
\begin{lemma}
\label{lem:basisAction}
For $n \in \mathbb N$
the map
$\overline{\varphi}_n: \overline {\mathcal O}_n \to \square$
is described as follows.
For $1 \leq i \leq n$ pick $g_i \in \lbrace A,B\rbrace$.
Then $\overline{\varphi}_n$ sends
$\overline{g}_1 \overline{g}_2 \cdots \overline{g}_n \mapsto
g^+_1g^+_2 \cdots g^+_n$, where we recall $A^+= a x_0$
and $B^+ = b x_2$.
\end{lemma}
\begin{proof}
By construction
$
\overline{g}_1 \overline{g}_2 \cdots \overline{g}_n
=
g_1 g_2 \cdots g_n+\mathcal O_{n-1} \in \overline{\mathcal O}_n$.
By this and Definition
\ref{def:comp},
the map
$\overline{\varphi}_n$ sends
$\overline{g}_1 \overline{g}_2 \cdots \overline{g}_n$
to $\varphi_n(
g_1 g_2 \cdots g_n)$, which is
equal to
$\pi_n(\sharp(g_1 g_2 \cdots g_n))$.
To compute this last quantity, apply
$\pi_n$ to each side of
(\ref{eq:xiExpand}), and
evaluate the result using
Lemma
\ref{lem:whereSummand} and
Definition
\ref{def:projGrad}.
For the sum
on the right in
(\ref{eq:xiExpand}),
$\pi_n$ fixes
the summand
$g^+_1g^+_2 \cdots g^+_n$ and sends the remaining
summands to zero.
The result follows.
\end{proof}
\begin{lemma}
\label{lem:rsvarphi}
Let $r,s \in \mathbb N$. Then
$\overline{\varphi}_r(u)
\overline{\varphi}_s(v)
=
\overline{\varphi}_{r+s}(uv)$
for
all $u\in \overline{\mathcal O}_r$ and
$v\in \overline{\mathcal O}_s$.
\end{lemma}
\begin{proof} Use Lemma
\ref{lem:basisAction}.
\end{proof}
\begin{definition}
\label{def:varphi}
Define an $\mathbb F$-linear map
$\overline{\varphi} :\overline{\mathcal O} \to
\square$ that acts on
$\overline{\mathcal O}_n$ as $\overline{\varphi}_n$
for $n \in \mathbb N$.
\end{definition}
\begin{lemma} The map
$\overline{\varphi} :\overline{\mathcal O} \to
\square$ from Definition
\ref{def:varphi} is an $\mathbb F$-algebra homomorphism.
\end{lemma}
\begin{proof}
It suffices to check that
$\overline{\varphi}(u)
\overline{\varphi}(v)=
\overline{\varphi}(uv)$
for
all $u,v\in \overline{\mathcal O}$.
This checking is routine
using Lemma
\ref{lem:rsvarphi}, Definition
\ref{def:varphi}, and
the definition of the $\mathbb F$-algebra
$\overline{\mathcal O}$ in Section 4.
\end{proof}
\begin{lemma}
\label{lem:barphiAction}
The map
$\overline{\varphi} :\overline{\mathcal O} \to
\square$ from Definition
\ref{def:varphi} sends
$\overline{A} \mapsto ax_0$ and
$\overline{B} \mapsto bx_2$.
\end{lemma}
\begin{proof} We have
$\overline{A}, \overline{B} \in \overline{\mathcal O}_1$.
By
Lemma
\ref{lem:basisAction} and
Definition
\ref{def:varphi},
\begin{eqnarray*}
\overline{\varphi}(\overline{A}) =
\overline{\varphi}_1(\overline{A}) = A^+ = ax_0,
\qquad \qquad
\overline{\varphi}(\overline{B}) =
\overline{\varphi}_1(\overline{B}) = B^+ = bx_2.
\end{eqnarray*}
\end{proof}
\noindent
By
Lemma
\ref{lem:xi} there exists an automorphism of
$U^+$ that sends $X \mapsto aX$ and
$Y \mapsto bY$. Denote this automorphism by $\phi$.
Recall the surjective $\mathbb F$-algebra homomorphism
$\psi: U^+ \to \overline{\mathcal O}$ from
above
Lemma
\ref{lem:munu}.
By Proposition
\ref{thm:tensorDecPreM}(i)
there exists an injective $\mathbb F$-algebra
homomorphism $U^+ \to \square$ that sends
$X \mapsto x_0$ and
$Y \mapsto x_2$.
Call this homomorphism $\flat $.
\begin{lemma}
\label{lem:bigCom}
With the above notation, the following diagram commutes:
\begin{equation*}
\begin{CD}
U^+ @>\psi > >
\overline{\mathcal O}
\\
@V \phi VV @VV\overline{\varphi} V \\
U^+ @>>\flat>
\square
\end{CD}
\end{equation*}
\end{lemma}
\begin{proof} Each map in the diagram
is an $\mathbb F$-algebra homomorphism.
The $\mathbb F$-algebra $U^+$ is generated by
$X,Y$.
To verify that the diagram commutes,
chase $X,Y$ around
the diagram using
Lemma
\ref{lem:barphiAction}.
\end{proof}
\begin{proposition}
\label{prop:twoInj}
Theorem
\ref{thm:main1}
holds. Moreover
the map
$\overline{\varphi}$ is injective.
\end{proposition}
\begin{proof}
Consider the commuting diagram in Lemma
\ref{lem:bigCom}.
By construction $\phi $ is an isomorphism,
$\flat$ is injective,
and $\psi$ is surjective.
Now since the diagram commutes,
$\psi$ and $\overline{\varphi}$ are injective.
The map $\psi$ is an isomorphism since
it is both injective and surjective.
Therefore Theorem
\ref{thm:main1}
holds.
\end{proof}
\begin{theorem}
\label{thm:xiInj}
The map $\sharp $ from
Proposition
\ref{prop:ABxyiLongerM} is injective.
\end{theorem}
\begin{proof}
Pick a nonzero $u \in \mathcal O$.
We show that
$\sharp (u)\not=0$.
There exists a unique $n \in \mathbb N$ such that $
u \in \mathcal O_n$ and
$ u \not\in \mathcal O_{n-1}$.
The map $\overline{\varphi}_n$ is injective,
by Definition
\ref{def:varphi} and since $\overline{\varphi}$ is injective by
Proposition
\ref{prop:twoInj}. By this and
Definition
\ref{def:comp}, the kernel of $\varphi_n$
is equal to
$\mathcal O_{n-1}$. This kernel does not
contain $u$, so
$\varphi_n(u) \not=0$. Now $\sharp (u)\not=0$ in view of
(\ref{eq:compVarphi}).
The result follows.
\end{proof}
\begin{theorem}
\label{thm:naturalINJ}
The map $\natural$ from Proposition
\ref{prop:ABxyiLonger}
is injective.
\end{theorem}
\begin{proof} By Lemma
\ref{lem:naturalSharp}
and Theorem
\ref{thm:xiInj}.
\end{proof}
\section{The $q$-tetrahedron algebra $\boxtimes_q$ }
\noindent The
$q$-tetrahedron algebra $\boxtimes_q$ was introduced in
\cite{qtet} and investigated further in
\cite{irt},
\cite{miki}.
In this section, we consider how
$\mathcal O$ and $\square_q$ are related to $\boxtimes_q$.
We first
display an injective $\mathbb F$-algebra
homomorphism $\square_q \to
\boxtimes_q$. Next, we compose this map with
the map
$\sharp: \mathcal O \to \square_q$ from
Proposition
\ref{prop:ABxyiLongerM},
to get an injective $\mathbb F$-algebra
homomorphism
$\mathcal O \to \boxtimes_q$.
\begin{definition}
\label{def:tet}
\rm
(See \cite[Definition~6.1]{qtet}.)
Let $\boxtimes_q$ denote the $\F$-algebra defined by generators
\begin{eqnarray}
\lbrace x_{ij}\;|\;i,j \in
\mathbb Z_4,
\;\;j-i=1 \;\mbox{\rm {or}}\;
j-i=2\rbrace
\label{eq:gen}
\end{eqnarray}
and the following relations:
\begin{enumerate}
\item[\rm (i)] For $i,j \in
\Z_4$ such that $j-i=2$,
\begin{eqnarray}
x_{ij}x_{ji} =1.
\label{eq:tet1}
\end{eqnarray}
\item[\rm (ii)] For $i,j,k \in
\Z_4$ such that $(j-i,k-j)$ is one of $(1,1)$, $(1,2)$,
$(2,1)$,
\begin{eqnarray}
\label{eq:tet2}
\frac{qx_{ij}x_{jk} - q^{-1} x_{jk}x_{ij}}{q-q^{-1}}=1.
\end{eqnarray}
\item[\rm (iii)] For $i,j,k,\ell \in
\Z_4$ such that $j-i=k-j=\ell-k=1$,
\begin{eqnarray}
\label{eq:tet3}
x^3_{ij}x_{k\ell}
-
\lbrack 3 \rbrack_q
x^2_{ij}x_{k\ell} x_{ij}
+
\lbrack 3 \rbrack_q
x_{ij}x_{k\ell} x^2_{ij}
-
x_{k\ell} x^3_{ij} = 0.
\end{eqnarray}
\end{enumerate}
We call
$\boxtimes_q$ the {\it $q$-tetrahedron algebra}.
\end{definition}
\begin{lemma}
\label{lem:aut1Tet}
There exists an automorphism $\varrho$ of
$\boxtimes_q$ that sends each generator
$x_{ij} \mapsto x_{i+1,j+1}$.
Moreover $\varrho^4=1$.
\end{lemma}
\begin{lemma}
\label{lem:mapn}
There exists an $\mathbb F$-algebra homomorphism
$\square_q \to \boxtimes_q$ that sends
$x_i \mapsto x_{i-1,i}$ for $i \in \mathbb Z_4$.
\end{lemma}
\begin{proof} Compare Definitions
\ref{def:boxqV1M},
\ref{def:tet}.
\end{proof}
\begin{definition}
\label{def:canonBoxTet}
\rm
The homomorphism
$\square_q \to \boxtimes_q$ from
Lemma \ref{lem:mapn}
will be called {\it canonical}.
\end{definition}
\noindent Our next goal is to show that the canonical homomorphism
$\square_q \to \boxtimes_q$ is injective.
\begin{definition}
\label{def:tetsub}
\rm
(See \cite[Section~4.1]{miki}.)
Define the subalgebras
$ \boxtimes^{\rm even}_q$,
$ \boxtimes^{\rm odd}_q$,
$ \boxtimes^{\times }_q$
of $ \boxtimes_q$ such that
\begin{enumerate}
\item[\rm (i)]
$ \boxtimes^{\rm even}_q$
is
generated by $x_{30}, x_{12}$;
\item[\rm (ii)]
$ \boxtimes^{\rm odd}_q$
is
generated by $x_{01}, x_{23}$;
\item[\rm (iii)]
$ \boxtimes^{\times }_q$
is
generated by
$
x_{02}, x_{20},
x_{13}, x_{31}
$.
\end{enumerate}
\end{definition}
\begin{proposition}
\label{prop:miki}
{\rm (See
Miki \cite[Prop.~4.1]{miki}.)}
The following {\rm (i)--(iv)} hold:
\begin{enumerate}
\item[\rm (i)]
there exists an $\mathbb F$-algebra isomorphism
$U^+ \to \boxtimes^{\rm even}_q$ that sends
$X\mapsto x_{30}$ and
$Y\mapsto x_{12}$;
\item[\rm (ii)]
there exists an $\mathbb F$-algebra isomorphism
$U^+ \to \boxtimes^{\rm odd}_q$ that sends
$X\mapsto x_{01}$ and
$Y\mapsto x_{23}$;
\item[\rm (iii)]
the $\mathbb F$-algebra
$\boxtimes^{\times }_q$ has a presentation by
generators
$x_{02},
x_{20},
x_{13},
x_{31}$ and relations
\begin{eqnarray*}
x_{02}x_{20} =
x_{20}x_{02} = 1, \qquad \qquad
x_{13}x_{31} =
x_{31}x_{13} = 1;
\end{eqnarray*}
\item[\rm (iv)]
the following is an isomorphism of
$\mathbb F$-vector spaces:
\begin{eqnarray*}
\boxtimes^{\rm even}_q
\otimes
\boxtimes^{\times }_q
\otimes
\boxtimes^{\rm odd}_q
& \to & \boxtimes_q
\\
u \otimes v \otimes w &\mapsto & uvw
\end{eqnarray*}
\end{enumerate}
\end{proposition}
\begin{note}
\rm
In \cite[Prop.~4.1]{miki} Miki
assumes that $\mathbb F = \mathbb C$ and $q$ is
not a root of unity. We emphasize that
Proposition
\ref{prop:miki} holds without this assumption.
There is a proof of
Proposition
\ref{prop:miki} that is
analogous to our proof of
Proposition
\ref{thm:tensorDecPreM}.
\end{note}
\begin{proposition}
\label{prop:sBI}
The canonical homomorphism
$\square_q \to \boxtimes_q$ is injective.
\end{proposition}
\begin{proof} Compare
Proposition
\ref{thm:tensorDecPreM} and
Proposition
\ref{prop:miki}.
\end{proof}
\begin{proposition}
\label{cor:tetqmain}
Pick nonzero $a,b \in \mathbb F$.
Then there exists an $\mathbb F$-algebra homomorphism
$\mathcal O \to
\boxtimes_q$ that sends
\begin{eqnarray*}
A \mapsto a x_{30}+ a^{-1} x_{01},
\qquad \qquad
B \mapsto b x_{12}+ b^{-1} x_{23}.
\end{eqnarray*}
This homomorphism is injective.
\end{proposition}
\begin{proof}
The desired homomorphism is the composition
of the homomorphism
$\sharp: \mathcal O \to \square_q$ from
Proposition
\ref{prop:ABxyiLongerM}
and the canonical homomorphism
$\square_q \to \boxtimes_q$.
The last assertion follows from
Theorem
\ref{thm:xiInj}
and Proposition
\ref{prop:sBI}.
\end{proof}
\section{The quantum affine algebra
$U_q(\widehat {\mathfrak {sl}}_2)$ }
\noindent In the literature on the
$q$-Onsager algebra $\mathcal O$,
there are algebra homomorphisms from $\mathcal O$ into
the quantum algebra $U_q(\widehat {\mathfrak {sl}}_2)$ due to
P.~Baseilhac and S.~Belliard
\cite[line (3.15)]{basXXZ},
\cite[line (3.18)]{basXXZ}
and S.~Kolb
\cite[Example~7.6]{kolb}.
Also, there are algebra homomorphisms from $\mathcal O$ into
the $q$-deformed loop algebra
$U_q(L({\mathfrak{sl}}_2))$ due to
P.~Baseilhac
\cite[Prop.~2.2]{bas6},
and also
T.~Ito and the present author
\cite[Prop.~8.5]{qRacahIT},
\cite[Props.~1.1,~1.13]{ITaug}.
In the case of \cite[Example~7.6]{kolb} and
\cite[Props.~1.1,~1.13]{ITaug}
the homomorphism was shown to be injective. In fact,
all of the above homomorphisms are injective,
and this can be established using the results from the previous
sections of the present paper.
In each case the proof is similar.
In this section we illustrate what is going on with a single example
\cite[Example~7.6]{kolb}.
This example involves
$U_q(\widehat {\mathfrak {sl}}_2)$, which we now define.
\begin{definition}
\label{def:Chevalley}
\rm
(See \cite[p.~262]{cp3}.) Let
$U_q(\widehat {\mathfrak {sl}}_2)$ denote the $\mathbb F$-algebra
with generators $e^{\pm}_i$, $k^{\pm 1}_i$, $i \in \lbrace 0,1\rbrace$
and the following relations:
\begin{eqnarray*}
&& k_i k^{-1}_i = k^{-1}_i k_i = 1,
\\
&& k_0 k_1 = k_1 k_0,
\\
&& k_i e^{\pm}_i k^{-1}_i = q^{\pm 2}e^{\pm}_i,
\\
&& k_i e^{\pm}_j k^{-1}_i = q^{\mp 2}e^{\pm}_j, \qquad i\not=j,
\\
&&
\lbrack e^+_i, e^-_i\rbrack = \frac{k_i - k^{-1}_i}{q-q^{-1}},
\\
&&
\lbrack e^{\pm}_0, e^{\mp}_1\rbrack = 0,
\\
&&
(e^{\pm}_i)^3 e^{\pm}_j
-
\lbrack 3 \rbrack_q
(e^{\pm}_i)^2 e^{\pm}_j e^{\pm}_i
+
\lbrack 3 \rbrack_q
e^{\pm}_i e^{\pm}_j (e^{\pm}_i)^2
-
e^{\pm}_j (e^{\pm}_i)^3 = 0, \qquad i \not=j.
\end{eqnarray*}
\noindent We call
$e^{\pm}_i$, $k^{\pm 1}_i$, $i \in \lbrace 0,1\rbrace$
the {\it Chevalley generators} for
$U_q(\widehat {\mathfrak {sl}}_2)$.
\end{definition}
\noindent In the following three lemmas we describe some automorphisms of
$U_q(\widehat {\mathfrak {sl}}_2)$; the proofs are routine
and omitted.
\begin{lemma}
\label{lem:omegaAut}
There exists an automorphism of
$U_q(\widehat {\mathfrak {sl}}_2)$ that sends
\begin{eqnarray*}
e^+_i \mapsto e^-_i, \qquad \qquad e^-_i \mapsto e^+_i,
\qquad \qquad k^{\pm 1}_i \mapsto k^{\mp 1}_i,
\qquad \qquad i \in \lbrace 0,1\rbrace.
\end{eqnarray*}
\end{lemma}
\begin{lemma}
\label{lem:autZeta}
There exists
an automorphism of
$U_q(\widehat {\mathfrak {sl}}_2)$ that sends
\begin{eqnarray*}
e^+_i \mapsto e^+_i k_i,
\qquad \qquad
e^-_i \mapsto k^{-1}_i e^-_i,
\qquad \qquad
k^{\pm 1}_i \mapsto k^{\pm 1}_i,
\qquad \qquad i \in \lbrace 0,1\rbrace.
\end{eqnarray*}
\end{lemma}
\begin{lemma}
\label{lem:uqsl2Aut}
Let $\varepsilon_0,\varepsilon_1$ denote nonzero scalars in $\mathbb F$.
Then there exists an automorphism of
$U_q(\widehat {\mathfrak {sl}}_2)$ that sends
\begin{eqnarray*}
e^+_i \mapsto \varepsilon_i e^+_i,
\qquad \qquad
e^-_i \mapsto \varepsilon^{-1}_i e^-_i,
\qquad \qquad
k^{\pm 1}_i \mapsto k^{\pm 1}_i,
\qquad \qquad i \in \lbrace 0,1\rbrace.
\end{eqnarray*}
\end{lemma}
\begin{definition}
\label{def:xi}
\rm
The automorphism of
$U_q(\widehat {\mathfrak {sl}}_2)$
from
Lemma \ref{lem:uqsl2Aut}
will be denoted by $\xi(\varepsilon_0,\varepsilon_1)$.
\end{definition}
\noindent We now recall the
equitable presentation of
$U_q(\widehat {\mathfrak {sl}}_2)$.
\begin{lemma}
\label{thm:equitUq}
{\rm (See \cite[Theorem~2.1]{uqsl2hat}.)}
The $\mathbb F$-algebra $U_q(\widehat {\mathfrak {sl}}_2)$
has a presentation
with generators $y^{\pm }_i$, $k^{\pm 1}_i$, $i \in \lbrace 0,1\rbrace$
and relations
\begin{eqnarray*}
&&k_i k^{-1}_i = k^{-1}_i k_i = 1,
\\
&&
\mbox{\rm $k_0 k_1$ is central},
\\
&& \frac{q y^+_i k_i - q^{-1} k_i y^+_i}{q-q^{-1}} = 1,
\\
&& \frac{q k_i y^-_i - q^{-1} y^-_i k_i}{q-q^{-1}} = 1,
\\
&& \frac{q y^{-}_i y^+_i - q^{-1} y^+_i y^{-}_i}{q-q^{-1}} = 1,
\\
&& \frac{q y^{+}_i y^-_j - q^{-1} y^-_j y^{+}_i}{q-q^{-1}} = k^{-1}_0 k^{-1}_1,
\qquad \qquad i \not=j,
\\
&&
(y^{\pm}_i)^3 y^{\pm}_j -
\lbrack 3 \rbrack_q
(y^{\pm}_i)^2 y^{\pm}_j y^{\pm}_i
+
\lbrack 3 \rbrack_q
y^{\pm}_i y^{\pm}_j (y^{\pm}_i)^2
-
y^{\pm}_j (y^{\pm}_i)^3 = 0, \qquad \qquad i\not=j.
\end{eqnarray*}
An isomorphism with the presentation in Definition
\ref{def:Chevalley} is given by:
\begin{eqnarray}
k^{\pm}_i &\mapsto& k^{\pm}_i,
\label{lem:kmove}
\\
y^-_i &\mapsto& k^{-1}_i + e^-_i(q-q^{-1}),
\label{lem:ymmove}
\\
y^+_i &\mapsto & k^{-1}_i -
k^{-1}_i e^+_i
q(q-q^{-1}).
\label{lem:ypmove}
\end{eqnarray}
\noindent The inverse of this isomorphism is given by:
\begin{eqnarray*}
k^{\pm}_i &\mapsto& k^{\pm}_i,
\\
e^-_i &\mapsto & (y^-_i - k^{-1}_i)(q-q^{-1})^{-1},
\\
e^+_i &\mapsto & (1-k_i y^+_i)q^{-1}(q-q^{-1})^{-1}.
\end{eqnarray*}
\end{lemma}
\begin{definition}
\label{def:tau}
\rm Referring to Lemma
\ref{thm:equitUq}, we call
$y^{\pm}_i$, $k^{\pm 1}_i$, $i \in \lbrace 0,1\rbrace$
the {\it equitable generators} for
$U_q(\widehat {\mathfrak {sl}}_2)$.
The isomorphism described in
(\ref{lem:kmove})--(\ref{lem:ypmove}) will be denoted by
$\tau$.
\end{definition}
\noindent
We now describe an $\mathbb F$-algebra
homomorphism $\widehat \square_q \to
U_q(\widehat {\mathfrak {sl}}_2)$ and
an
$\mathbb F$-algebra
homomorphism
$U_q(\widehat {\mathfrak {sl}}_2) \to
\boxtimes_q$.
In this description we use
the equitable presentation of
$U_q(\widehat {\mathfrak {sl}}_2)$.
\begin{lemma}
\label{lem:sigmai}
There exists an $\mathbb F$-algebra
homomorphism $\widehat \sigma: \widehat \square_q \to
U_q(\widehat {\mathfrak {sl}}_2)$ such that
\bigskip
\centerline{
\begin{tabular}[t]{c||cccc|cccc}
$u$
& $x_0$ & $x_1$ & $x_2$
& $x_3$ &
$c_0$ & $c_1$ & $c_2$ & $c_3$
\\
\hline
$\widehat \sigma(u)$ &
$y^-_0$ & $y^+_0$ & $y^-_1$ & $y^+_1$ &
$1$ & $k^{-1}_0 k^{-1}_1$ & $1$ & $k^{-1}_0 k^{-1}_1$
\end{tabular}}
\medskip
\end{lemma}
\begin{proof} Compare the defining relations
(\ref{eq:Box1})--(\ref{eq:Box4})
for $\widehat \square_q$, with the defining relations
for
$U_q(\widehat {\mathfrak {sl}}_2)$ given in Lemma
\ref{thm:equitUq}.
\end{proof}
\begin{lemma}
\label{lem:sigmaiT}
There exists an $\mathbb F$-algebra
homomorphism $ \sigma:
U_q(\widehat {\mathfrak {sl}}_2) \to
\boxtimes_q$
such that
\bigskip
\centerline{
\begin{tabular}[t]{c||cccc|cccc}
$u$ & $y^-_0$ & $y^+_0$ & $y^-_1$
& $y^+_1$ & $k_0$ & $k^{-1}_0$ & $k_1$ & $k^{-1}_1$
\\ \hline
$\sigma(u)$
& $x_{30}$ & $x_{01}$ & $x_{12}$ & $x_{23}$
& $x_{13}$ & $x_{31}$ & $x_{31}$ & $x_{13}$
\\
\end{tabular}}
\medskip
\end{lemma}
\begin{proof}
Compare the defining relations for
$U_q(\widehat {\mathfrak {sl}}_2)$ given in
Lemma
\ref{thm:equitUq},
with the defining relations for
$\boxtimes_q$ given in
Definition
\ref{def:tet}.
\end{proof}
\begin{lemma}
\label{lem:backtoCan}
The following diagram commutes:
\begin{equation*}
\begin{CD}
\widehat \square_q @>\widehat \sigma > >
U_q(\widehat {\mathfrak {sl}}_2)
\\
@V can VV @VV \sigma V \\
\square_q @>>can >
\boxtimes_q
\end{CD}
\end{equation*}
\end{lemma}
\begin{proof}
Each map in the diagram is an $\mathbb F$-algebra homomorphism.
To verify that the diagram commutes, chase the
$\widehat \square_q $
generators $x_i, c^{\pm 1}_i$ $(i \in \mathbb Z_4)$ around the diagram using
Definitions
\ref{def:cancan},
\ref{def:canonBoxTet}
and
Lemmas \ref{lem:sigmai},
\ref{lem:sigmaiT}.
\end{proof}
\noindent We now obtain
the homomorphism $\mathcal O \to
U_q(\widehat {\mathfrak {sl}}_2)$ due to Kolb
\cite[Example~7.6]{kolb}.
\begin{proposition}
\label{prop:Kolb}
{\rm (See \cite[Example~7.6]{kolb}.)}
For
$i \in \lbrace 0,1\rbrace$ pick
$s_i\in \mathbb F$ and $0 \not= \gamma_i \in \mathbb F$.
Then
for $ U_q(\widehat {\mathfrak {sl}}_2)$
the elements
\begin{eqnarray}
B_i = e^-_i -\gamma_i e^+_i k^{-1}_i + s_i k^{-1}_i
\qquad \qquad i \in \lbrace 0,1\rbrace
\label{eq:BiKolb}
\end{eqnarray}
satisfy
\begin{eqnarray}
&&
B_0^3 B_1 - \lbrack 3 \rbrack_q B_0^2 B_1 B_0 +
\lbrack 3 \rbrack_q B_0 B_1 B_0^2 -B_1 B_0^3 = q(q+q^{-1})^2 \gamma_0
(B_1B_0-B_0B_1),
\label{eq:AAABKolb}
\\
&&
B_1^3 B_0 - \lbrack 3 \rbrack_q B_1^2 B_0 B_1 +
\lbrack 3 \rbrack_q B_1 B_0 B_1^2 -B_0 B_1^3 =
q(q+q^{-1})^2 \gamma_1(B_0B_1-B_1B_0).
\label{eq:BBBAKolb}
\end{eqnarray}
\end{proposition}
\begin{proof}
Let $\lambda$ denote an indeterminate.
Replacing $\mathbb F$ by its algebraic closure if necessary, we
may assume without loss that
$\mathbb F$ is algebraically closed.
There exist scalars
$\lbrace \alpha_i \rbrace_{i \in \mathbb Z_4}$ in $\mathbb F$
such that
$\alpha_0, \alpha_1$ are the roots of the
polynomial
\begin{eqnarray*}
\lambda^2 - s_0 \lambda + \gamma_0 q (q-q^{-1})^{-2}
\end{eqnarray*}
and $\alpha_2, \alpha_3$ are the roots of the polynomial
\begin{eqnarray*}
\lambda^2 - s_1 \lambda + \gamma_1 q (q-q^{-1})^{-2}.
\end{eqnarray*}
By construction
\begin{eqnarray*}
&&
\alpha_0 + \alpha_1
=
s_0,
\qquad \qquad
\alpha_0 \alpha_1 = \gamma_0 q (q-q^{-1})^{-2},
\\
&&
\alpha_2 + \alpha_3
=
s_1,
\qquad \qquad
\alpha_2 \alpha_3 = \gamma_1 q (q-q^{-1})^{-2}.
\end{eqnarray*}
Note that $\alpha_i \not=0$ for $i \in \mathbb Z_4$.
Define
\begin{eqnarray*}
\varepsilon_0 = \alpha_0 (q-q^{-1}),
\qquad \qquad
\varepsilon_1 = \alpha_2 (q-q^{-1}).
\end{eqnarray*}
Note that $\varepsilon_0 \not=0$ and $\varepsilon_1 \not=0$.
In the algebra $\widehat \square_q$ define
\begin{eqnarray*}
A = \alpha_0 x_0 + \alpha_1 x_1, \qquad \qquad
B = \alpha_2 x_2 + \alpha_3 x_3.
\end{eqnarray*}
Then
$A,B$ satisfy
(\ref{eq:qDGLong1}),
(\ref{eq:qDGLong2}) by
Corollary
\ref{prop:ABxyiLong}.
Consider the composition
\begin{eqnarray}
\label{eq:cdlong}
\begin{CD}
\zeta: \quad \widehat \square_q @>> \widehat \sigma >
U_q(\widehat {\mathfrak {sl}}_2)
@>> \tau >
U_q(\widehat {\mathfrak {sl}}_2)
@>>
\xi(\varepsilon_0,\varepsilon_1)
>
U_q(\widehat {\mathfrak {sl}}_2),
\end{CD}
\end{eqnarray}
where $\widehat \sigma$ is from
Lemma
\ref{lem:sigmai},
$\tau$ is from
Definition
\ref{def:tau},
and
$\xi(\varepsilon_0,\varepsilon_1)$ is from
Definition \ref{def:xi}.
By construction $\zeta:\widehat \square_q \to
U_q(\widehat {\mathfrak {sl}}_2)$ is an $\mathbb F$-algebra homomorphism.
Using (\ref{eq:cdlong}) and
$k_0 e^+_0 = q^{2} e^+_0 k_0$ we find that
$\zeta$ sends $A\mapsto B_0$. Similarly $\zeta$ sends
$B \mapsto B_1$.
By
Lemma
\ref{lem:sigmai},
$\widehat \sigma$ sends $c_0\mapsto 1$ and $c_2 \mapsto 1$. Therefore
$\zeta$ sends
$c_0\mapsto 1$ and $c_2 \mapsto 1$.
Applying $\zeta$ to each side of
(\ref{eq:qDGLong1}),
(\ref{eq:qDGLong2})
we obtain
(\ref{eq:AAABKolb}),
(\ref{eq:BBBAKolb}).
\end{proof}
\begin{proposition}
\label{prop:KolbB}
{\rm (See \cite[Example~7.6]{kolb}.)}
Referring to Proposition
\ref{prop:Kolb}, assume that
\begin{eqnarray}
\label{eq:gammaChoice}
\gamma_0 = q^{-1}(q-q^{-1})^2, \qquad \qquad
\gamma_1 = q^{-1}(q-q^{-1})^2.
\end{eqnarray}
Then there exists an $\mathbb F$-algebra homomorphism
$\mathcal O \to
U_q(\widehat {\mathfrak {sl}}_2)$ that sends
$A\mapsto B_0$ and
$B\mapsto B_1$.
This homomorphism is injective.
\end{proposition}
\begin{proof} The desired $\mathbb F$-algebra
homomorphism
$\mathcal O \to U_q(\widehat {\mathfrak {sl}}_2)$
exists, since under the assumption
(\ref{eq:gammaChoice}) the relations
(\ref{eq:AAABKolb}),
(\ref{eq:BBBAKolb})
become the $q$-Dolan/Grady relations.
Call the above homomorphism $\partial$.
We show that $\partial$ is injective.
Consider the composition
\begin{eqnarray*}
\begin{CD}
\eta:
\quad
\mathcal O @>> \partial >
U_q(\widehat {\mathfrak {sl}}_2)
@>> \xi(\varepsilon^{-1}_0, \varepsilon^{-1}_1) >
U_q(\widehat {\mathfrak {sl}}_2)
@>>
\tau^{-1}
>
U_q(\widehat {\mathfrak {sl}}_2)
@>> \sigma > \boxtimes_q,
\end{CD}
\end{eqnarray*}
where
$\xi(\varepsilon^{-1}_0,\varepsilon^{-1}_1)$ is from
Definition \ref{def:xi},
$\tau$ is from
Definition
\ref{def:tau}, and
$\sigma$ is from
Lemma
\ref{lem:sigmaiT}.
By construction
$\eta:\mathcal O
\to
\boxtimes_q$
is an $\mathbb F$-algebra homomorphism.
One checks that $\eta$
coincides with the $\mathbb F$-algebra homomorphism
$\mathcal O \to \boxtimes_q$ from Proposition
\ref{cor:tetqmain}, where $a=\alpha_0$ and $b=\alpha_2$.
The map from Proposition
\ref{cor:tetqmain} is injective,
so $\partial$ is injective.
\end{proof}
\begin{note}\rm
In
\cite[Example~7.6]{kolb} Kolb assumes
that $\mathbb F$ has
characteristic zero and $q$ is not a root of unity.
We emphasize that
Propositions
\ref{prop:Kolb} and
\ref{prop:KolbB} hold without this assumption.
\end{note}
\section{Directions for future research}
\noindent In this section we give some suggestions for future research.
\begin{problem}\rm
For the $\mathbb F$-algebras
$\widetilde \square_q$,
$\widehat \square_q$, $\square_q$ find their
automorphism group.
\end{problem}
\begin{problem}\rm
The Lusztig automorphisms of $U_q(\widehat {\mathfrak {sl}}_2)$ are
described in
\cite[p.~294]{damiani}. Find analogous automorphisms for
$\widetilde \square_q$,
$\widehat \square_q$, $\square_q$.
\end{problem}
\begin{problem}\rm
\label{prob:N}
Referring to the algebra $\square_q$, for
$i \in \mathbb Z_4$ define
\begin{eqnarray*}
N_i = \frac{q(1-x_ix_{i+1})}{q-q^{-1}} =
\frac{q^{-1}(1-x_{i+1}x_i)}{q-q^{-1}}.
\end{eqnarray*}
Note that $N_i x_i = q^{2}x_i N_i$ and
$N_i x_{i+1} = q^{-2}x_{i+1} N_i$.
Determine how $N_i$ is related to $x_{i+2}$ and $x_{i+3}$.
\end{problem}
\begin{problem}\rm
Referring to Problem \ref{prob:N},
determine how $N_i$, $N_j$ are
related for $i,j \in \mathbb Z_4$.
\end{problem}
\begin{problem}\rm
Referring to Problem \ref{prob:N},
consider the $q$-exponential
$E_i={\rm exp}_q (N_i)$.
For $u \in \square_q$ compute
$E_i u E_i^{-1}$ and determine if the result is contained in
$\square_q$. If it always is, then conjugation by $E_i$ gives an
automorphism of $\square_q$. In this case, describe the subgroup of
${\rm Aut}(\square_q)$ generated by
$\lbrace E^{\pm 1}_i\rbrace_{i \in \mathbb Z_4}$.
\end{problem}
\begin{conjecture}\rm
Let $V$ denote a finite-dimensional irreducible
$\widetilde \square_q$-module. Then
$V$ becomes a $\boxtimes_q$-module
such that for $i \in \mathbb Z_4$
the action of $x_i$ on $V$ is a scalar multiple of
the action of $x_{i-1,i}$ on $V$. The
$\boxtimes_q$-module $V$ is irreducible.
\end{conjecture}
\begin{conjecture}\rm Let $A,B$ denote a tridiagonal pair over $\mathbb F$
that has $q$-Racah type in the sense of
\cite[p.~259]{bockting}.
Then the underlying vector space $V$ becomes a $\square_q$-module
on which $A$ (resp. $B$) is a linear combination of $x_0, x_1$
(resp.
$x_2, x_3$).
The $\square_q$-module $V$ is irreducible.
\end{conjecture}
\begin{problem}\rm
For the $\mathbb F$-algebras
$\widetilde \square_q$,
$\widehat \square_q$, $\square_q$ find their center.
\end{problem}
\section{Acknowledgments}
The author thanks
Pascal Baseilhac, Stefan Kolb,
and Kazumasa Nomura
for giving this paper a close reading and offering valuable
suggestions.
| 112,755
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“Before Her” is a short memoir of Jacqueline Woodson's romantic history. It's warm and moving in places. I didn't love it, but it was really enjoyable - I'm giving my first read of The One collection 3 stars.
Tag: Short Stories!
Book Review: The Goddess of Macau by Graeme Hall
"The Goddess of Macau" is a wonderful collection of stories that I devoured over the past couple of weeks. I didn't love them all, but that's only natural in a collection. One story in particular stood out for me and will probably stay with me for some time. I'd definitely recommend this for short stories fans and those interested in learning a little about the culture of Macau. It’s 4 stars from me!!
Blog Tour: Two Lives by A Yi!
| 362,717
|
\begin{document}
\title{On extensions of the Penrose inequality with angular momentum }
\author{Wojciech Kulczycki}
\author{Edward Malec}
\affiliation{Instytut Fizyki im.~Mariana Smoluchowskiego, Uniwersytet Jagiello\'nski, {\L}ojasiewicza 11, 30-348 Krak\'{o}w, Poland}
\begin{abstract}
We numerically investigate the validity of recent modifications of the Penrose inequality that include angular momentum. Formulations expressed in terms of asymptotic mass and asymptotic angular momentum are contradicted. We analyzed numerical solutions describing polytropic stationary toroids around spinning black holes.
\end{abstract}
\pacs{04.20.-q, 04.25.Nx, 04.40.Nr, 95.30.Sf}
\maketitle
\section{Introduction}
The Cosmic Censorship Hypothesis, originally formulated by Roger Penrose more than half century ago \cite{Penrose1969}, can be understood as a statement that classical general relativity is self-contained, when describing regions exterior to black holes. Penrose has
argued that the Cosmic Censorship Hypothesis cannot be true if a body collapsing to a black hole fails to satisfy
the inequality
\begin{equation}M_\mathrm{ADM} \ge \sqrt{\frac{S}{16\pi }}, \label{1}
\end{equation}
where $S$ is the area of the outermost apparent horizon that surrounds the body and $M_\mathrm{ADM}$ is the asymptotic mass of the spacetime \cite{Penrose1973}. (Herein and in what follows we always assume asymptotic flatness of a spacetime.) Failure in satisfying of (\ref{1}) would imply the existence of a ``naked singularity'' and a loss of predictability in a collapsing system.
There is probably no exaggeration in saying that this idea has shaped the development of classical and mathematical general relativity in the last five decades.
The Penrose inequality has been proven or checked numerically in a number of special cases --- conformally flat systems with matter \cite{KMS}, Brill gravitational waves in Weyl geometries \cite{Koc} and various foliations of spherically symmetric systems \cite{MNOM, Hayward}. Most remarkably, it was proven in the important so-called Riemannian case, when the apparent horizon coincides with a minimal surface \cite{Huisken, Huisken1, Bray}. There exist scenarios for a general proof \cite{Frauendiener,MMS}, but there are no easy prospects for their implementation. For more information see specialized reviews, for instance \cite{KM, Mars}.
Christodoulou introduced the concept of an irreducible mass $M_\mathrm{irr}=\sqrt{S/(16\pi )}$ \cite{Christodoulu}, where $S$ is the area of the event horizon. It appears that for the Kerr spacetime endowed with the asymptotic mass $M_\mathrm{ADM}$ and the angular momentum $J_\mathrm{ADM}$ one has the relation $M^2_\mathrm{ADM}=M^2_\mathrm{irr}+\frac{ J^2_\mathrm{ADM}}{4M^2_\mathrm{irr}} =\frac{S}{16\pi }+\frac{4\pi J^2_\mathrm{ADM}}{S}$. An analog of this formula might be used in order to define the quasilocal mass of a black hole (assuming that the linear momentum of the black hole vanishes) in terms of its area and quasilocal angular momentum $J_\mathrm{BH}$:
$M_\mathrm{BH} =\sqrt{M^2_\mathrm{irr}+\frac{ J_\mathrm{BH}^2}{4M^2_\mathrm{irr}}} =\sqrt{\frac{S}{16\pi }+\frac{4\pi J_\mathrm{BH}^2}{S}}$.
This concept of the quasilocal mass of a black hole is commonly used in the literature.
There exist formulations of the Penrose inequality that involve the asymptotic mass and quasilocal angular momentum \cite{Mars,Dain_Gabach},
\begin{equation}M_\mathrm{ADM} \ge \left(\frac{S}{16\pi }+\frac{4\pi J_\mathrm{BH}^2}{S}\right)^{1/2}; \label{2}
\end{equation}
here $S$ and $J_\mathrm{BH}$ are the area and quasilocal angular momentum of the outermost apparent horizon. Anglada \cite{Anglada} and Khuri \cite{Khuri} have proved other versions of (\ref{2}) under the assumption of axial symmetry,
\begin{equation}
M_\mathrm{ADM} \ge \left( \frac{S}{16\pi }+\frac{ J_\mathrm{BH}^2}{\tilde R^2(S)}\right)^{1/2};
\label{3}
\end{equation}
where $\tilde R(S) $ is some linear measure of the boundary of the black hole.
In what follows we shall study numerically the validity of (\ref{2}) and related inequalities in a class of stationary configurations consisting of a black hole and a torus. The order
is following. Next Section contains a short description of equations and relevant quantities. Section 3 gives a concise summary of the numerical procedure. The main results are reported in Section 4. We conclude the paper with a summary.
\section{Equations}
We assume a \textit{stationary} metric of the form
\begin{align}
ds^{2} & =-\alpha^{2}{dt}^{2}+r^{2}\sin^{2}\theta\psi^{4}\left(d\varphi+\beta dt\right)^{2}\nonumber \\
& \qquad+\psi^{4}e^{2q}\left(dr^{2}+r^{2}d\theta^{2}\right). \label{metric}
\end{align}
Here $t$ is the time coordinate, and $r$, $\theta$, $\varphi$
are spherical coordinates. In this paper the gravitational constant $G=1$ and the
speed of light $c=1$. We assume axial symmetry and employ the stress-momentum
tensor
\[
T^{\alpha\beta}=\rho hu^{\alpha}u^{\beta}+pg^{\alpha\beta},
\]
where $\rho$ is the baryonic rest-mass density, $h$ is the specific
enthalpy, and $p$ is the pressure. Metric functions $\alpha $,
$\psi $, $q $ and $\beta $ in \eqref{metric}
depend on $r$ and $\theta$ only.
The forthcoming Einstein equations have been found in \cite{MSH} and checked by authors of \cite{prd2018a, prd2018b}; the present formulation follows closely the description of \cite{prd2018a, prd2018b}.
Below $K_{ij}$ denotes the extrinsic curvature of the $t = \mathrm{const}$ hypersurface. The conformal extrinsic curvature $\hat K_{ij}$ is defined as $\hat K_{ij} = \psi^2 K_{ij}$. The only nonzero component $\beta$ of the shift vector is separated into two parts, $\beta = \beta_\mathrm{K} + \beta_\mathrm{T}$. Here $\beta_\mathrm{T}$ is a part of the shift vector related to the rotating torus \cite{MSH}. Functions $\beta_\mathrm{K}$ and $\beta_\mathrm{T}$ are determined as follows. The nonzero components of $\hat K_{ij}$ can be written in the form
\[ \hat K_{r \varphi} = \frac{H_\mathrm{E} \sin^2 \theta}{r^2} + \frac{\psi^6}{2 \alpha} r^2 \sin^2 \theta \partial_r \beta_\mathrm{T}, \]
\[ \hat K_{\theta \varphi} = \frac{H_\mathrm{F} \sin \theta}{r} + \frac{\psi^6}{2 \alpha} r^2 \sin^2 \theta \partial_\theta \beta_\mathrm{T}. \]
As in \cite{MSH}, we choose the functions $H_\mathrm{E}$ and $H_\mathrm{F}$ to be expressed by the formulae obtained for the Kerr metric of mass $m$ and the spin parameter $a$, written in the form (\ref{metric}). In explicit terms they read \cite{BrandtSeidel1995}
\[ H_\mathrm{E} = \frac{m a \left[ (r_\mathrm{K}^2 - a^2) \Sigma_\mathrm{K} + 2 r_\mathrm{K}^2 (r_\mathrm{K}^2 + a^2) \right]}{\Sigma_\mathrm{K}^2}, \]
\[ H_\mathrm{F} = - \frac{2 m a^3 r_\mathrm{K} \sqrt{r_\mathrm{K}^2 - 2 m r_\mathrm{K} + a^2} \cos \theta \sin^2 \theta}{\Sigma_\mathrm{K}^2}, \]
where
\[ r_\mathrm{K} = r \left( 1 + \frac{m}{r} + \frac{m^2 - a^2}{4 r^2} \right), \]
and
\[ \Sigma_\mathrm{K} = r_\mathrm{K}^2 + a^2 \cos^2 \theta. \]
It appears that for the Kerr metric one has
\[ \hat K_{r \varphi} = \frac{H_\mathrm{E} \sin^2 \theta}{r^2} \]
and
\[ \hat K_{\theta \varphi} = \frac{H_\mathrm{F} \sin \theta}{r}. \]
The function $\beta_\mathrm{K}$ has to be computed from the relation \cite{MSH}
\begin{equation}
\frac{\partial \beta_\mathrm{K}}{\partial r} = \frac{2 H_\mathrm{E} \alpha}{r^4 \psi^6}.
\label{eqbetak}
\end{equation}
The function $\beta_\mathrm{T}$, with suitable boundary conditions (see next Section) is found from the differential equation (\ref{46}).
In what follows we apply the puncture method as implemented in \cite{MSH}. Let $\Phi = \alpha \psi$ and assume the puncture at $r = 0$. Define $r_\mathrm{s} = \frac{1}{2}\sqrt{m^2 - a^2}$, and
\[ \psi = \left(1 + \frac{r_\mathrm{s}}{r} \right) e^\phi, \quad \Phi = \left(1 - \frac{r_\mathrm{s}}{r} \right)e^{-\phi} B. \]
Then the surface $r = r_\mathrm{s}$ is an apparent horizon.
Einstein equations read
\begin{widetext}
\begin{subequations}
\label{main_sys}
\begin{eqnarray}
\left[ \partial_{rr} + \frac{1}{r } \partial_r + \frac{1}{r^2} \partial_{\theta \theta} \right] q & = & S_q, \label{47}\\
\left[ \partial_{rr} + \frac{2 r }{r^2 - r_\mathrm{s}^2} \partial_r + \frac{1}{r^2} \partial_{\theta \theta} + \frac{ \cot{\theta}}{r^2} \partial_\theta \right] \phi & = & S_\phi, \label{44} \\
\left[ \partial_{rr} + \frac{3 r^2 + r_\mathrm{s}^2}{r(r^2 - r_\mathrm{s}^2)} \partial_r + \frac{1}{r^2} \partial_{\theta \theta} + \frac{2 \cot{\theta}}{r^2} \partial_\theta \right] B & = & S_B, \label{45} \\
\left[ \partial_{rr} + \frac{4 r^2 - 8 r_\mathrm{s} r + 2 r_\mathrm{s}^2}{r(r^2 - r_\mathrm{s}^2)} \partial_r + \frac{1}{r^2} \partial_{\theta \theta} + \frac{3 \cot{\theta}}{r^2} \partial_\theta \right] \beta_\mathrm{T} & = & S_{\beta_\mathrm{T}}, \label{46}
\end{eqnarray}
\end{subequations}
where source terms $S_\phi, S_B, S_{\beta_\mathrm{T}}, S_q $ are:
\begin{subequations}
\begin{flalign}
&S_q = -8 \pi e^{2q} \left( \psi^4 p - \frac{\rho h u_\varphi^2}{r^2 \sin^2 \theta} \right) + \frac{3 A^2}{\psi^8} + 2 \left[ \frac{r - r_\mathrm{s}}{r(r + r_\mathrm{s})} \partial_r + \frac{\cot \theta}{r^2} \partial_\theta \right] b + \left[ \frac{8 r_\mathrm{s}}{r^2 - r_\mathrm{s}^2} + 4 \partial_r (b - \phi) \right] \partial_r \phi \nonumber\\
&+ \frac{4}{r^2} \partial_\theta \phi \partial_\theta (b - \phi), \label{5001}\\
&
S_\phi =
- 2 \pi e^{2q} \psi^4 \left[ \rho_\mathrm{H} - p + \frac{\rho h u_\varphi^2}{\psi^4 r^2 \sin^2 \theta} \right] - \frac{A^2}{\psi^8} - \partial_r\phi \partial_r b - \frac{1}{r^2} \partial_\theta \phi \partial_\theta b - \frac{1}{2} \left[ \frac{r - r_\mathrm{s}}{r (r + r_\mathrm{s})} \partial_r b + \frac{\cot \theta}{r^2} \partial_\theta b \right],\label{5002} \\
&
S_B = 16 \pi B e^{2q} \psi^4 p, \label{5003}\\
&
S_{\beta_\mathrm{T}} = \frac{16 \pi \alpha e^{2q} \tilde J}{r^2 \sin^2 \theta} - 8 \partial_r \phi \partial_r \beta_\mathrm{T} + \partial_r b \partial_r \beta_\mathrm{T} - 8 \frac{\partial_\theta \phi \partial_\theta \beta_\mathrm{T}}{r^2} +\frac{\partial_\theta b \partial_\theta \beta_\mathrm{T}}{r^2}\label{5004}
\end{flalign}
\end{subequations}
\end{widetext}
and
\[ A^2 = \frac{\hat K^2_{r \varphi}}{r^2 \sin^2 \theta} + \frac{\hat K^2_{\theta \varphi}}{r^4 \sin^2 \theta}, \]
\[ \rho_\mathrm{H} = \rho h (\alpha u^t)^2 - p, \]
\[ \tilde J = \rho h \alpha u^t u_\varphi, \]
\[ B = e^b. \]
In the rest of this paper we will deal with polytropes $p(\rho)=K\rho^{\gamma }$.
Then one has the specific enthalpy
\[
h(\rho)=1+ \frac{\gamma p}{(\gamma -1) \rho}.
\]
The 4-velocity $(u^{\alpha})=(u^{t},0,0,u^{\varphi})$ is normalized,
$g_{\alpha\beta}u^{\alpha}u^{\beta}=-1$. The coordinate angular velocity
reads
\begin{equation}
\Omega=\frac{u^{\varphi}}{u^{t}}.\label{Omega_def}
\end{equation}
We define the angular momentum per unit
inertial mass $\rho h$ \cite{FM}
\begin{equation}
j\equiv u_{\varphi}u^{t}.\label{j_def}
\end{equation}
It is known since early 1970's that general-relativistic Euler equations are solvable under the condition that $j\equiv j(\Omega)$ \cite{Butterworth_Ipser,Bardeen_1970}.
Within the fluid region, the Euler equations $\nabla_{\mu}T^{\mu\nu}=0$
can be integrated,
\begin{equation}
\int j(\Omega) d\Omega+\ln\left(\frac{h}{u^{t}}\right)=C,\label{uf}
\end{equation}
where $C$ is a constant.
We shall use in this paper the rotation laws obtained in \cite{KM2020}
\begin{eqnarray}
\label{momentum}
j(\Omega ) &\equiv &-\frac{1}{1-3\delta}\frac{d}{d\Omega} \ln \left( 1- ( a\Omega )^2\right. \nonumber\\
&& \left.- \kappa w^{1- \delta }\Omega^{1+\delta }(1- a\Omega )^{1-\delta} \right) .
\end{eqnarray}
Here $J_\mathrm{BH}$ and $a=J_\mathrm{BH}/m$ are the angular momentum and the spin parameter of the central black hole, respectively. $\delta $ is a free parameter and $\kappa =\frac{1-3\delta }{1+\delta }$. Let us remark, that (\ref{momentum}) supplements former rotation recipes that have been formulated in \cite{MM, APP2015} and (in the case of the Keplerian rotation around spinning black holes) \cite{prd2018a, prd2018b}.
The Keplerian rotation corresponds to the parameter
$\delta =-1/3$ and $\kappa =3$.
The rotation curves --- angular velocities as functions of spatial coordinates $\Omega(r,\theta )$ --- can be recovered from Eq.\ (\ref{j_def}),
\begin{equation}
j(\Omega)=\frac{V^{2}}{\left(\Omega+\beta\right)\left(1-V^{2}\right)}.\label{rotation_law}
\end{equation}
Here the squared linear velocity is given by
\[
V^{2}=r^{2}\sin^{2}\theta\left(\Omega+\beta\right)^{2}\frac{\psi^{4}}{\alpha^{2}}.
\]
The central black hole is defined by the puncture method \cite{BrandtSeidel1995}. The black hole is surrounded by a minimal two-surface $S $ (the horizon) embedded in a fixed hypersurface of constant time, and
located at $r=r_{\mathrm{s}} = \sqrt{m^2 - a^2}/2$, where $m$ is a mass parameter. Its area is denoted as $S$ and its angular momentum $J_{\mathrm{BH}}$ follows from the Komar expression
\begin{equation}
J_{\mathrm{BH}}=\frac{1}{4}\int_{0}^{\pi/2}\frac{r^{4}\psi^{6}}{\alpha}\partial_{r}\beta\sin^{3}\theta d\theta.
\end{equation}
We would like to point that the angular momentum is given rigidly on the event horizon $S $, in terms of
data taken from the Kerr solution and
independently of the content of mass in a torus, $J_\mathrm{BH} = m a$ \cite{MSH}. The mass of the black
hole is then defined in terms of its area and the angular momentum,
\begin{equation}
M_{\mathrm{BH}}=\sqrt{\frac{S}{16\pi }+\frac{4\pi J_{\mathrm{BH}}^{2}}{S}}.
\label{PI}
\end{equation}
Asymptotic (total) mass $M_\mathrm{ADM}$ and angular momentum $J_\mathrm{ADM}$ can be defined as apropriate Arnowitt-Deser-Misner charges, and they can be computed by means of corresponding volume integrals \cite{MSH}.
Thus we have
\begin{eqnarray}
\label{j1}
M_\mathrm{ADM}&=&\sqrt{m^2-a^2}-2\int_{r_\mathrm{s}}^\infty \left(r^2-r^2_\mathrm{s}\right)dr\int_0^{\pi/2}\sin \theta d\theta S_\phi, \nonumber\\
J_\mathrm{T} &=& 4 \pi \int_{r_\mathrm{s}}^\infty r^2 dr \int_0^{\pi/2} \sin \theta d \theta \rho \alpha u^t \psi^6 e^{2 q} h u_\varphi, \nonumber\\
J_\mathrm{ADM}&=&J_\mathrm{BH}+J_\mathrm{T}.
\end{eqnarray}
Here $J_\mathrm{T}$ is the angular momentum deposited within the torus.
A circumferential radius corresponding to the coordinate $r=\mathrm{const}$ on the symmetry plane $\theta =\pi /2$
is given by
\begin{equation}
r_\mathrm{C}(r)=r\psi^2(r,\theta =\pi/2).
\label{rad}
\end{equation}
One can define an alternative mass of the apparent horizon in terms of $r_\mathrm{C}(r)$,
\begin{equation}
M_\mathrm{C}\equiv \frac{r_\mathrm{C}(r_\mathrm{s})} {2}.
\label{mc}
\end{equation}
In the Kerr spacetime one has exactly $M_\mathrm{C}=M_\mathrm{BH}=m$. It is known that in numerically obtained spacetimes with gaseous toroids the first equality holds with a good accuracy, albeit depending on spin \cite{MSH, prd2018a,prd2018b}. The second equality is true only approximately for relatively light disks, and it is not true for heavy tori.
\section{Description of numerics}
The numerical method bases on \cite{MSH} and it has been presented in more details in \cite{prd2018b}. Here we use a more general rotation law (\ref{momentum}) and different linear algebra routines --- the PARDISO library \cite{pardiso} instead of LAPACK \cite{lapack}. In what follows we briefly summarize the main points.
The solutions are found iteratively. In each iteration one solves the Einstein equations (\ref{main_sys}) and (\ref{5001})-(\ref{5004}), Eq. (\ref{eqbetak}) and two hydrodynamic equations: (\ref{uf}) and (\ref{rotation_law}). Eqs. (\ref{main_sys}) (with their source terms (\ref{5001})-(\ref{5004})) are solved using a fixed-point method (we use the PARDISO library \cite{pardiso}) with respect to functions $\phi$, $B$, $\beta_{\mathrm{T}}$ and $q$. The function $\beta_\mathrm{K}$ is computed by integration of Eq.\ (\ref{eqbetak}). Eqs. (\ref{uf}) and (\ref{rotation_law}) are used to calculate the specific enthalpy $h$ and the angular velocity $\Omega$, respectively. Constants ($C$ and $w$) that appear in these two equations are computed by solving them at boundary points $(r_1,\pi/2)$ and $(r_2,\pi/2)$, using the Newton-Raphson method. Here $r_1$ and $r_2$ are the values of the inner and outer radii of the torus, respectively; they are given a priori.
The free hydrodynamic data consist of the maximal baryonic density $\rho_\mathrm{max}$ and the polytropic index $\gamma $. We assume from now on $\gamma =4/3$. The baryonic density $\rho$ is calculated from the specific enthalpy $h$ using the polytropic formula
\[
\rho=\left[\frac{h-1}{4K } \right]^{\frac{1}{1/3}}.
\]
This yields (in each iteration) the constant $K$ as a function of the maximal value of the specific enthalpy $h_\mathrm{max}$ and $\rho_\mathrm{max}$
\[
K= \frac{h_\mathrm{max}-1}{4\rho_\mathrm{max}^{1/3}}.
\]
We have assumed axial and equatorial symmetry and the puncture method with the puncture at $r=r_\mathrm{s}\equiv\sqrt{m^2-a^2}/2$ . We should add that the mass $m$ and the spin $a$ (which appears also in (\ref{momentum}))
are given apriori. Thus it suffices that the numerical grid covers the region $r_\mathrm{s}\le r< \infty$ and $0\le\theta\le\pi/2$ with suitable boundary conditions \cite{MSH}. We have at the equator ($\theta=\pi/2$): $\partial_{\theta}\phi=\partial_{\theta}B=\partial_{\theta}\beta_{\mathrm{T}}=\partial_{\theta}q=0$. The regularity conditions along the axis ($\theta=0$) read: $\partial_{\theta}\phi=\partial_{\theta}B=\partial_{\theta}\beta_{\mathrm{T}}=0$. It is required that $q(\theta=0)=0$; this is due to the local flatness of the metric. The puncture formalism implies the boundary conditions at the horizon $r=r_\mathrm{s}$: $\partial_{r}\phi=\partial_{r}B=\partial_{r}\beta_{\mathrm{T}}=\partial_{r}q=0$ and $ \partial_{rr}\beta_{\mathrm{T}}=\partial_{rrr}\beta_{\mathrm{T}}=0$. These last two conditions on $\beta_\mathrm{T}$ follow from a careful analysis of Eq. (\ref{46}) that yields stringent conditions at $r=r_{\mathrm{s}}$ \cite{MSH}.
At the outer boundary of the numerical domain the boundary conditions are obtained from the multipole expansion and the conditions of asymptotic flatness. Thus we have for $r\to\infty$:
\begin{eqnarray}
\phi & \sim & \frac{M_{1}}{2r}, \ \ \ \ \ B \sim 1-\frac{B_{1}}{r^{2}}, \nonumber \\
\beta_{\mathrm{T}} & \sim & -\frac{2J_{1}}{r^{3}}, \ \ \ \ \ q \sim \frac{q_{1}\sin^{2}\theta}{r^{2}}.\label{boundinf}
\end{eqnarray}
Herein the constants $M_{1}$, $B_{1}$, $J_{1}$ and $q_{1}$ are given by
\begin{equation}
M_{1}=-2\int_{r_{\mathrm{s}}}^{\infty}(r^{2}-r_{\mathrm{s}}^{2})dr\int_{0}^{\pi/2}\sin\theta d\theta S_{\phi},\label{m1}
\end{equation}
\begin{equation}
B_{1}=\frac{2}{\pi}\int_{r_{\mathrm{s}}}^{\infty}dr\frac{(r^{2}-r_{\mathrm{s}}^{2})^{2}}{r}\int_{0}^{\pi/2}d\theta\sin^{2}\theta S_{B},\label{b1}
\end{equation}
\begin{equation}
J_{1}=4\pi\int_{r_{\mathrm{s}}}^{\infty}r^{2}dr\int_{0}^{\pi/2}\sin\theta d\theta\rho\alpha u^{t}\psi^{6}e^{2q}hu_{\varphi},\label{j1}
\end{equation}
\begin{eqnarray}
q_{1} & = & \frac{2}{\pi}\int_{r_{\mathrm{s}}}^{\infty}drr^{3}\int_{0}^{\pi/2}d\theta\cos(2\theta)S_{q}\nonumber \\
& & -\frac{4}{\pi}r_{\mathrm{s}}^{2}\int_{0}^{\pi/2}d\theta\cos(2\theta)q(r_{\mathrm{s}},\theta).\label{q1}
\end{eqnarray}
Finally, we add that the Kerr solution emerges in our method as a vacuum limit $\rho_\mathrm{max}\rightarrow 0$.
\section{The Penrose inequality in stationary black hole-torus systems}
In the rest of the paper we always assume that $\Omega>0$ and the mass parameter $m=1$. Corotating disks have $a>0$, while counterrotating disks have negative spins: $a<0$.
The disk's boundaries are numerically defined by the condition that the specific enthalpy is
$h=1 $.
The results of numerical calculations are provided in the forthcoming Table. We shall describe its content in more detail in the second part of this Section, when referring to new
proposals of Penrose-type inequalities. In the first part we will refer to canonical versions.
\subsection{On inequalities (\ref{1}) and (\ref{2})}
The mass $M_\mathrm{BH}$ of the apparent horizon is defined in terms of the area and the quasilocal (Komar-type) angular momentum; see (\ref{PI}). For such a choice one has --- as discussed above --- the relation $M_\mathrm{C} \approx M_\mathrm{BH}$. This is a kind of a virial relation; we shall treat its fullfilment as a test for the selfconsistency and correctness of our numerical description. Let us remark, that there exists another --- exact --- virial relation, discussed in \cite{MSH}, that can be used to check the numerical self-consistency.
Our polytropic matter within a torus satisfies the dominant energy condition. There is no analytic proof, but there exists numerical evidence \cite{MSH,prd2018a,prd2018b} that the asymptotic mass $M_\mathrm{ADM}$ is not smaller than the quasilocal mass $M_\mathrm{BH}=\sqrt{\frac{S}{16\pi }+\frac{4\pi J^2_\mathrm{BH}}{S}}$. Thus (\ref{2}) should hold,
\begin{equation}
M_\mathrm{ADM} \ge \sqrt{ \frac{S}{16\pi }+\frac{4\pi J_\mathrm{BH}^2}{S}}.
\label{3c}
\end{equation}
Obviously, the original Penrose inequality (\ref{1}) also holds true.
The above statements of this subsection should be true, whenever there exist numerical solutions. The inspection of the Table confirms this expectation.
\subsection{On new proposals}
Inequality (\ref{3}) uses the size measure $\tilde R(S)$ of the apparent horizon \cite{Anglada}. This quantity is difficult to calculate, but we can use a bound, that was shown in \cite{Anglada}, that $\tilde R(S)$ is not larger than $\sqrt{10}M_\mathrm{C}$, which in turn is approximated by $\sqrt{10}M_\mathrm{BH}$. Thus the necessary condition for the validity of (\ref{3}) reads
\begin{equation}
M_\mathrm{ADM}^2 \ge \frac{S}{16\pi }+\frac{ J_\mathrm{BH}^2}{10 M^2_\mathrm{BH}}
\label{3a}
\end{equation}
Inequality (\ref{3a}) is valid in all our numerical examples reported in the forthcoming Table --- compare relevant values in the column denoted as $M_\mathrm{ADM}$ with suitable entries in the last column denoted as $I_3$. This does not mean, however, that (\ref{3}) is confirmed, since (\ref{3a}) constitutes only the necessary condition.
The inequality
\begin{equation}
M_\mathrm{BH}^2 \ge \frac{S}{16\pi }+\frac{ J_\mathrm{BH}^2}{4 M^2_\mathrm{BH}}
\label{3b}
\end{equation}
follows directly from the definition of $M_\mathrm{BH}$, since $M_\mathrm{BH}^2 \ge \frac{S}{16\pi }$. The equality occurs for $a=J_\mathrm{BH}/m=0$. The Table confirms that --- compare relevant values in the column denoted as $M_\mathrm{BH}$ with suitable entries in the last column denoted as $I_3$.
The quasilocal inequalities (\ref{2}) and (\ref{3}) are awkward in a sense, since they require the use of quasilocal measures of the angular momentum. There exists a conserved quantity related with Killing vectors, in stationary and axially symmetric quantities, that gives rise to a distinguished (Komar-type) quasilocal measure of the angular momentum. We used this fact in Section 2. Unfortunately, there is no such a quasilocal measure
in general spacetimes. The question arises, whether one can replace $J_\mathrm{BH}$ by its global counterpart $J_\mathrm{ADM}$, that is whether
\begin{equation}
M_\mathrm{ADM}^2 \ge \frac{S}{16\pi }+\frac{4\pi J^2_\mathrm{ADM}}{S},
\label{4}
\end{equation}
or (in a weaker formulation)
\begin{equation}
M_\mathrm{ADM}^2 \ge \frac{S}{16\pi }+\frac{ J^2_\mathrm{ADM}}{4M_\mathrm{ADM}^2},
\label{5}
\end{equation}
at least for stationary and asymptotically flat spacetimes with compact material systems.
This restriction to compact material systems is necessary, since it is easy to envisage a classical mechanical system, with an arbitrarily large angular momentum, so that both inequalities (\ref{4}) and (\ref{5}) are broken. In all examples considered below the circumferential radii of the outermost part of tori are smaller than 39 $M_\mathrm{ADM}$.
Let us mention here the recent work of Kopi\'nski and Tafel \cite{Tafel}, in which they consider spacetimes arising from a class of perturbations of the spinless Schwarzschild geometry. These perturbations carry an angular momentum, that yields the asymptotic value $J_{\mathrm{ADM}} $. Kopi\'nski and Tafel prove that (\ref{5}) is valid for such spacetimes.
When using the Table, in order to test the inequality (\ref{4}), one should compare entries of the column designated as $M_{\mathrm{ADM}} $ with relevant terms in the column denoted as $I_1$.
One can see, that (\ref{4}) is not valid for systems with heavy toroids, irrespective of the spin of the central black holes --- see the cases H1 -- H{15}, L3MR and L5M1 -- L5M5. Let us remark, that a similar conclusion can be drawn from Table II of \cite{prd2016}, but that numerical analysis have used a perturbative approach, and therefore it is not convincing.
For lighter disks, the inequality (\ref{4}) is valid for spins of the black hole that are not too big, but it is broken if $a$ is large enough --- see just a few examples corresponding to $a=0.9$: L5, L10, L15 and L20.
We have found only one counterexample to the inequality (\ref{5}) --- see the case L3MR and compare relevant elements in the columns denoted as $M_\mathrm{ADM}$ and $I_2$.
\section{Conclusions}
We numerically investigate the validity of various versions of the Penrose inequality, in particular those that include angular momentum. This is done by analyzing a stationary, axially symmetric system consisting of a black hole and a rotating polytropic torus. The original version formulated by Penrose \cite{Penrose1973} is always true. Formulations, that bound the mass $M_\mathrm{ADM}$ by quasilocal quantities --- the area of the black hole $S$, its mass $M_\mathrm{BH}$ and angular momentum $J_\mathrm{BH}$ --- are also satisfied in our numerical solutions. We have found, however, counterexamples to those versions of the inequality, that are expressed in terms of the asymptotic angular momentum $J_\mathrm{ADM}$.
\begin{table*}
\caption{\label{tab_sols}
Table. Black hole --- tori solutions. Subsequent columns contain
(from the left to the right): the solution number, the rotation law
parameter $\delta$, the black-hole spin parameter
$a$, the inner radius of the torus $r_{1}$, the outer radius of
the torus $r_{2}$, the total asymptotic mass $M_\mathrm{ADM}$,
the black hole mass $M_\mathrm{BH}$, the black hole surface $S$, the toroid angular
momentum $J_\mathrm{T}$, the total angular momentum $J_\mathrm{ADM}$ and variants of terms in various Penrose inequalities: $I_1=\sqrt{S/(16\pi)+4\pi J^2_\mathrm{ADM}/S}$, $I_2=\sqrt{S/(16\pi)+J^2_\mathrm{ADM}/(4M_\mathrm{ADM}^2)}$, $I_3=\sqrt{S/(16\pi)+J_\mathrm{BH}^2/(4M_\mathrm{BH}^2)}$.
The solutions were obtained assuming $m=1$, $\kappa=(1-3\delta)/(1+\delta)$
and $\gamma=4/3$.}
\begin{ruledtabular}
\begin{tabular}{c c c c c c c c c c c c c}
No. & $\delta$ & $a$ & $r_1$ & $r_2$ & $M_\mathrm{ADM}$ & $M_\mathrm{BH}$ & $S$ & $J_\mathrm{T}$ & $J_{\mathrm{ADM}} $ & $I_1$ & $I_2$ & $I_3$\\
\hline
L1 & $-1/3$ & $-0.9$ & $6$ & $41$ & $1.100$ & $1.001$ & $36.19$ & $0.5204$ & $-0.3796$ & $0.8775$ & $0.8659$ & $0.9604$\\
L2 & $-1/3$ & $-0.5$ & $6$ & $41$ & $1.100$ & $1.003$ & $47.26$ & $0.5008$ & $8.431\times 10^{-4}$ & $0.9697$ & $0.9697$ & $1.001$\\
L3 & $-1/3$ & $0$ & $6$ & $41$ & $1.100$ & $1.005$ & $50.79$ & $0.4846$ & $0.4846$ & $1.034$ & $1.029$ & $1.005$\\
L4 & $-1/3$ & $0.5$ & $6$ & $41$ & $1.100$ & $1.003$ & $47.23$ & $0.4883$ & $0.9883$ & $1.095$ & $1.068$ & $1.001$\\
L5 & $-1/3$ & $0.9$ & $6$ & $41$ & $1.100$ & $1.000$ & $36.17$ & $0.4947$ & $1.395$ & $1.181$ & $1.059$. & $0.9601$\\
\hline
L6 & $-1/7$ & $-0.9$ & $6$ & $41$ & $1.100$ & $1.001$ & $36.21$ & $0.4758$ & $-0.4242$ & $0.8848$ & $0.8704$ & $0.9605$\\
L7 & $-1/7$ & $-0.5$ & $6$ & $41$ & $1.100$ & $1.004$ & $47.32$ & $0.4528$ & $-4.716\times 10^{-2}$ & $0.9705$ & $0.9704$ & $1.002$\\
L8 & $-1/7$ & $0$ & $6$ & $41$ & $1.100$ & $1.006$ & $50.87$ & $0.4326$ & $0.4326$ & $1.029$ & $1.025$ & $1.006$\\
L9 & $-1/7$ & $0.5$ & $6$ & $41$ & $1.100$ & $1.004$ & $47.28$ & $0.4331$ & $0.9331$ & $1.083$ & $1.059$ & $1.001$\\
L10 & $-1/7$ & $0.9$ & $6$ & $41$ & $1.100$ & $1.001$ & $36.18$ & $0.4371$ & $1.337$ & $1.158$ & $1.0436$ & $0.9602$\\
\hline
L11 & $-1/10$ & $-0.9$ & $6$ & $41$ & $1.100$ & $1.001$ & $36.21$ & $0.4676$ & $-0.4324$ & $0.8862$ & $0.8712$ & $0.9605$\\
L12 & $-1/10$ & $-0.5$ & $6$ & $41$ & $1.100$ & $1.004$ & $47.32$ & $0.4446$ & $-5.543\times 10^{-2}$ & $0.9707$ & $0.9706$ & $1.002$\\
L13 & $-1/10$ & $0$ & $6$ & $41$ & $1.100$ & $1.006$ & $50.87$ & $0.4242$ & $0.4242$ & $1.028$ & $1.024$ & $1.006$\\
L14 & $-1/10$ & $0.5$ & $6$ & $41$ & $1.100$ & $1.004$ & $47.28$ & $0.4242$ & $0.9242$ & $1.081$ & $1.057$ & $1.001$\\
L15 & $-1/10$ & $0.9$ & $6$ & $41$ & $1.100$ & $1.001$ & $36.18$ & $0.4277$ & $1.328$ & $1.154$ & $1.041$ & $0.9602$\\
\hline
L16 & $0$ & $-0.9$ & $6$ & $41$ & $1.100$ & $1.001$ & $36.21$ & $0.4477$ & $-0.4525$ & $0.8897$ & $0.8733$ & $0.9605$\\
L17 & $0$ & $-0.5$ & $6$ & $41$ & $1.100$ & $1.004$ & $47.32$ & $0.4246$ & $-7.535\times 10^{-2}$ & $0.9710$ & $0.9709$ & $1.002$\\
L18 & $0$ & $0$ & $6$ & $41$ & $1.100$ & $1.006$ & $50.88$ & $0.4041$ & $0.4041$ & $1.026$ & $1.023$ & $1.006$\\
L19 & $0$ & $0.5$ & $6$ & $41$ & $1.100$ & $1.004$ & $47.29$ & $0.4030$ & $0.9030$ & $1.076$ & $1.053$ & $1.001$\\
L20 & $0$ & $0.9$ & $6$ & $41$ & $1.100$ & $1.001$ & $36.18$ & $0.4052$ & $1.305$ & $1.145$ & $1.035$ & $0.9602$\\
\hline
H1 & $0$ & $-0.9$ & $8$ & $20$ & $2.000$ & $1.011$ & $37.88$ & $5.307$ & $4.407$ & $2.683$ & $1.403$ & $0.9756$\\
H2 & $0$ & $0$ & $8$ & $20$ & $2.000$ & $1.079$ & $58.49$ & $4.879$ & $4.879$ & $2.506$ & $1.628$ & $1.079$\\
H3 & $0$ & $0.9$ & $8$ & $20$ & $2.000$ & $1.007$ & $37.29$ & $4.942$ & $5.842$ & $3.499$ & $1.695$ & $0.9702$\\
\hline
H4 & $-0.2$ & $-0.9$ & $8$ & $20$ & $2.000$ & $1.011$ & $37.88$ & $5.362$ & $4.462$ & $2.713$ & $1.414$ & $0.9755$\\
H5 & $-0.2$ & $0$ & $8$ & $20$ & $2.000$ & $1.078$ & $58.46$ & $4.938$ & $4.938$ & $2.531$ & $1.639$ & $1.078$\\
H6 & $-0.2$ & $0.9$ & $8$ & $20$ & $2.000$ & $1.007$ & $37.28$ & $5.013$ & $5.913$ & $3.539$ & $1.711$ & $0.9702$\\
\hline
H7 & $-0.4$ & $-0.9$ & $8$ & $20$ & $2.000$ & $1.011$ & $37.87$ & $5.406$ & $4.506$ & $2.737$ & $1.422$ & $0.9755$\\
H8 & $-0.4$ & $0$ & $8$ & $20$ & $2.000$ & $1.078$ & $58.41$ & $4.988$ & $4.988$ & $2.552$ & $1.648$ & $1.078$\\
H9 & $-0.4$ & $0.9$ & $8$ & $20$ & $2.000$ & $1.007$ & $37.27$ & $5.075$ & $5.975$ & $3.574$ & $1.724$ & $0.9701$\\
\hline
H10 & $-0.6$ & $-0.9$ & $8$ & $20$ & $2.000$ & $1.011$ & $37.87$ & $5.436$ & $4.536$ & $2.753$ & $1.428$ & $0.9754$\\
H11 & $-0.6$ & $0$ & $8$ & $20$ & $2.000$ & $1.077$ & $58.35$ & $5.032$ & $5.032$ & $2.572$ & $1.656$ & $1.077$\\
H12 & $-0.6$ & $0.9$ & $8$ & $20$ & $2.000$ & $1.007$ & $37.26$ & $5.134$ & $6.034$ & $3.608$ & $1.737$ & $0.9700$\\
\hline
H13 & $-0.8$ & $-0.9$ & $8$ & $20$ & $2.000$ & $1.011$ & $37.88$ & $5.437$ & $4.537$ & $2.753$ & $1.428$ & $0.9756$\\
H14 & $-0.8$ & $0$ & $8$ & $20$ & $2.000$ & $1.077$ & $58.28$ & $5.074$ & $5.074$ & $2.591$ & $1.664$ & $1.077$\\
H15 & $-0.8$ & $0.9$ & $8$ & $20$ & $2.000$ & $1.007$ & $37.24$ & $5.219$ & $6.119$ & $3.657$ & $1.755$ & $0.9698$\\
\hline
L3MR & $-1/3$ & $0$ & $50$ & $75$ & $2.000$ & $1.016$ & $51.90$ & $9.984$ & $9.984$ & $5.016$ & $2.695$ & $1.016$\\
L5M1 & $-1/3$ & $0.9$ & $6$ & $41$ & $1.650$ & $1.003$ & $36.66$ & $3.379$ & $4.279$ & $2.647$ & $1.553$ & $0.9646$\\
L5M2 & $-1/3$ & $0.9$ & $6$ & $41$ & $2.475$ & $1.008$ & $37.46$ & $8.382$ & $9.282$ & $5.445$ & $2.064$ & $0.9717$\\
L5M3 & $-1/3$ & $0.9$ & $6$ & $41$ & $3.713$ & $1.017$ & $38.79$ & $17.57$ & $18.47$ & $10.55$ & $2.637$ & $0.9836$\\
L5M4 & $-1/3$ & $0.9$ & $6$ & $41$ & $5.569$ & $1.033$ & $41.17$ & $35.47$ & $36.37$ & $20.11$ & $3.389$ & $1.004$\\
L5M5 & $-1/3$ & $0.9$ & $6$ & $41$ & $8.353$ & $1.066$ & $46.01$ & $72.76$ & $73.66$ & $38.51$ & $4.512$ & $1.046$\\
\end{tabular}
\end{ruledtabular}
\end{table*}
| 67,112
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TITLE: Map from $\phi_N: \{1, ..., N\} \rightarrow \{1, ..., r\}$, subsequence of $(\phi_N)_{N=1}^\infty$ that stabilises for each $k \in \mathbb{N}$.
QUESTION [1 upvotes]: In these lecture notes on MIT Opencourseware, on top of page 14 (which is page 2 really), a rough proof is provided for the equivalence of theorems 1.1 and 1.2. These are Schur's theorems but you really don't need to know any of that to understand my problem. You don't even need to read the theorems. I must be missing something really basic here. Theorem 1.2 clearly implies theorem 1.1, but in showing the converse (that theorem 1.1 implies theorem 1.2), the lecturer uses proof by contradiction and negates theorem 1.2. In it, he argues, suppose that given $r \in \mathbb{N}$ (the number of colours), for any $N \in \mathbb{N}$ there exists a colouring $\phi_N: \{1, ..., N\} \rightarrow \{1, ..., r\}$ of these $N$ natural numbers such that there is no monochrome $x, y, z$ (that is, all of the same colour) such that $x + y = z$. Then he considers the infinite sequence $(\phi_N)_{N=1}^\infty$, and in particular an infinite subsequence of it, such that for every $k \in \mathbb{N}$, the value of $\phi_N(k)$ stabilises as $N$ increases. That's the sentence I have a problem with.
How do we know that for every $k \in \mathbb{N}$, we can find a subsequence of $(\phi_N)_{N=1}^\infty$ such that $\phi_N(k)$ stabilises? I mean, let's take $k=1$ for example, and suppose $\exists N \in \mathbb{N}$ such that $\phi_N(1) = s$ for some $s \in \{1, ..., r\}$. Then how do we know that there even exists $M > N$ such that $\phi_M(1) = s$? In plain words, how do we know that some $\phi_N$ will map $1$ onto $s$ ever again (after it's done so once for some $N$)? From what I can see, it's entirely possible that the $\phi_N$ may never map $1$ onto $s$ ever again, while maintaining the condition that the mappings (colourings) $\phi_N$ never permits a monochrome solution $x, y, z \in \mathbb{N}$ to $x + y = z$. What am I missing (other than brain cells)?
Please let me know if I haven't been clear.
REPLY [0 votes]: For some $s_1$, there are infinitely many $N$ such that $\phi_N(1)=s_1$. Let $A_1$ be the set of such $N$, and let $N_1$ be its first element. Now since $A_1$ is infinite, there is some $s_2$ such that there are infinitely many $N\in A_1$ such that $\phi_N(2)=s_2$. Let $A_2$ be the set of such $N$, and let $N_2$ be the first element of $A_2$ that is greater than $N_1$. Now since $A_2$ is infinite, there is some $s_3$ such that there are infinitely many $N\in A_2$ such that $\phi_N(3)=s_3$. Let $A_3$ be the set of such $N$, and let $N_3$ be the first element of $A_3$ that is greater than $N_2$.
Continuing in this fashion, we get an increasing sequence $(N_k)$ and a sequence of elements $s_k\in\{1,\dots,r\}$ such that $\phi_{N_\ell}(k)=s_k$ for all $\ell\geq k$. In particular, $\phi_{N_\ell}(k)$ stabilizes for each $k$.
Here's a more abstract perspective. We may extend each $\phi_N$ in some arbitrary way to assume they defined on all of $\mathbb{N}$. So, $(\phi_N)$ is then a sequence in the space $\{1,\dots,r\}^{\mathbb{N}}$, which is compact and metrizable in the product topology. Thus, $(\phi_N)$ has a convergent subsequence. But convergence in the product topology is just pointwise convergence, and convergence in $\{1,\dots,r\}$ is just stabilization, so this means there is a subsequence on which $\phi_N(k)$ stabilizes for each $k$.
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TITLE: Existence of a supporting hyperplane
QUESTION [1 upvotes]: In $\mathbb R^n$, let $C$ be a closed convex set not equal to $\mathbb R^n$ itself.
I'd like to prove that the boundary of C: $\delta C$ is the set of all supporting points of $C$.
For the first inclusion, let $z\in \delta C$ .
So as to prove that $z$ belongs to the set of all supporting points of $C$, I think I have to prove the existence of $a \in \mathbb R^n \setminus C$ such that the projection of $a$ on $C$ is $z$.
How to build such an $a$ ?
I know that a sequence $x_n :\mathbb N \rightarrow\mathbb R^n \setminus C$ such that $x_n \rightarrow z$ exists, but it doesn't help much...
Thanks for your help.
REPLY [1 votes]: It does help that you can find the sequence $x_n \to z$, $x_n \not \in C$. :)
Take such sequence. Let $z_n$ be the projection of $x_n$ on $C$. Now, let $y_n \not \in C$ be the point on the line connecting $x_n$ and $z_n$, distance $1$ apart from $C$. You can prove that $y_n$ converges to some point $y$, which is distance $1$ from $C$. What is more, since the projection of each $y_n$ in $z_n$, and $z_n \to z$, you can conclude that $y$ is the sought point!
Note that the above still needs some work to be done. Treat this as an exercise.
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726 Comments
- SheilaNoya, on 11/03/2007, -18/+256People are less afraid now to admit that they don't believe in God. When I was growing up in the 1950's, admitting that you were an atheist would get you fired from your job, accused of being a communist, and shunned by 99.9% of the public. Fortunately, society has progressed a lot since then and we no longer let the churches scare us into fake submission.
- insllvn, on 11/07/2007, -24/+163Please, God, deliver us from your followers!
- lucidguru, on 11/02/2007, -25/+138The rich, the educated and a growing online youth no longer have any reasons to believe in imaginary deities. Life is too good, and we have too much time to think about the meaning of life and come up with our own conclusions; why would we believe some fictional story?
Hopefully enough people will come around and realize that religion is nothing more than a business that has unfair tax advantages. THEY DON'T PAY ANY! Churches, synagogues and other "religious" places should not be exempt. They should be taxed like all businesses regardless of their beliefs.
- inactive, on 10/31/2007, -13/+114Your idea of "heaven" was crowded? THAT"s funny!
- Incosian, on 11/02/2007, -16/+114Religion began when the first scoundrel met the first fool.
- CannedMango, on 10/31/2007, -13/+105You may as well be bragging about getting to be first in line to see Santa this Christmas
- MJG2007, on 11/02/2007, -11/+97
How many drive by shootings were there in the 1950s? - more than most people really imagine. Street gangs isn't exactly a new concept and violence both with and without firearms isn't something that started after 1950 (ever hear of the gangsters in Chicago or New York during prohibition?).
How many babies were born out of wedlock in the 1950s? 13% of teen pregnancies in 1957 (the highest year recorded for teen pregnancy) were to unwed mothers. Of course back then, more women tended to get married in their teens (my mother and two sisters were married in the 50's and each of them were 15 years old at the age of marriage) I'm pretty sure that you aren't suggesting we return to the days when half of all girls married before their 18th birthday (and one wonders how many of those were the results of "accidents" involving a failed rabbit test).
How many school shootings were there in the 1950s? - That we can agree there were not many. Most "hoodlums" just quit going to school which sort of cut down on the incidents of violence IN the schools.
How many high school dropouts were there in the 1950s? In 1950, around HALF of all students never completed high school. Are you trying to suggest that half of all students never graduate high school in the United States today? (Hint: the dropout rate is around 13 percent)
What percentage of the population was on welfare in the 1950s? The percentage of people living in abject poverty was much higher too.
How much violent crime was there in the 1950s? More than most people would think. Don't confuse reporting of violent crime with the actual violent crime rate (think about the difference between domestic violence as it was treated in 1950 and how it is treated today).
We like to romanticize the past, but the 1950's while in some respects was better than today, in other respects it was much much worse.
- bugsy187, on 11/02/2007, -8/+89It's kind of strange how some christians get pleasure telling people they'll go to hell.
- inactive, on 10/30/2007, -10/+72Theres no hell, either.
- Wrathernaut, on 11/02/2007, -13/+74Aw come on, heaven's great... Why wouldn't you want to spend eternity worshiping a narcissistic sadist?
- WiseWeasel, on 10/31/2007, -15/+71That's fine; as a neo-athiest, I'll be going to iHeaven xTreme... Enjoy your stay in regular heaven with all the other schlubs, sucker!
- inactive, on 11/02/2007, -1/+54As an atheist, I personally follow most of the code of morals laid down by Jesus in the bible. "Do unto others", "turn the other cheek", etc. It's not difficult.
But I don't believe that God exists or that Jesus was the son of God.
What's the big problem?
- floridiot2, on 11/07/2007, -5/+58Ohhh how I wish for a nation where no one cares what religion you are.. What a world that might be..
- WiseWeasel, on 10/30/2007, -7/+52Correlation does not imply causation. You fail at logic. Here are a couple more selected correlations to counter your "argument":
How many cases of polio were there in the 1950s?
What was the average life expectancy in the 1950s?
What was the average level of education in the 1950s?
What was the average inflation-adjusted income in the 1950s?
How many black presidential candidates were there in the 1950s?
What were your job prospects as a minority background citizen in the 1950s?
I doubt most people would want to be living in the 1950s rather than today, and your arguments otherwise are incredibly weak.
- SquigglyP, on 10/30/2007, -6/+50The removal of the word 'God' from everything is a reflection of the fact that:
1) The people of this country are starting to realize that there are more than just christians living here
2) There is supposed to be a separation of church and state (tho try telling that to the ***** in command)
You are free to believe what you want to believe. So am I. The state's job isn't to give your religion free advertising.
- ajimmykid, on 10/31/2007, -5/+45if you can't bring yourself to believe in the notion that there may not BE a God. I feel sorry for you.
- FuzzyWhisper, on 10/30/2007, -5/+45Right, except a healthy marriage doesn't usually entail one spouse threatening the other with eternal torture in a pit of fire if he or she doesn't "choose" to be loving and respectful.
- bitcloud, on 11/02/2007, -9/+48"Secular America" means "Tolerant America"
people who aren't so arrogant as to assume knowledge of EVERYTHING are a little easier to get along with generally... (they can also share their knowledge and allow you to improve upon that knowledge.. it's quite nice really)
- zenoizen, on 10/30/2007, -8/+47Actually, people are turning away from the word of god because it's entirely false.
- inactive, on 10/31/2007, -10/+49How many people have actually picked up the Bible and read it. The book is full of hatred, murder, incest, rape, man’s inhumanity to man, anti-women’s rights. And that is only the Bible’s good points.
The laws and society we have today was built by men who had the guts to tell the Bishop of Rome and his henchmen that they are wrong. Christianity is a religion of war and violence not of peace. The reformation of Christianity was started by Islamic ideals. Blasphemy definitely not. It is ironic that most people get their religious knowledge from a Parson on a pulpit and not from the source. Do your own historical research from verifiable facts. Islamic Moorish Spain was a centre of learning. Those concepts found their way throughout Europe into the Golden Age of science and logic.
The Middle Eastern Islam which is show on TV, is not Islam but a bastardised ideology by extremist control freaks operating under the disguise of Islam. Worst still is the extremist Christian fundamentalist ideology promoted in the USA under the disguise of democracy.
If you want an insight into humanity, read the collected wisdom of Buddhist monks. Pure Buddhism is not a religion but a philosophical sub-set of the Hindu religion with the religious beliefs removed. Pure Buddhism deals with ethical and moral dilemmas and how to live harmoniously in a society.
Where every other religion has conquered the world by stealth violence; the fundamental humanitarian teachings contained within Buddhism have been quietly integrated within modern societies.
There is no proof there is a GOD; only an ideology professing to a man made concept. No-one knows what happens after death; it is pure speculation. What is important is what you do with your life when you are alive. An ***** by any other name is still an *****. There are plenty of religious wolves masquerading as puritanical lambs.
Mankind will be psychologically free when the last church stone falls on and kills the last living preacher.
The moral of the story is: ***** will always baffle BRAINS.
- JackHorner, on 10/31/2007, -11/+49organized religion is something for humanity to overcome in order to be at peace with each other. We must let go of these foolish beliefs before its too late.
- inactive, on 11/07/2007, -0/+38Clearly you're not married.
- vuke69, on 10/31/2007, -2/+40You say strange, I say sick.
- JigoroKano, on 11/02/2007, -8/+45By not taxing Churches, the government is already favoring them over other scams.
- inactive, on 11/02/2007, -10/+44"To have faith is to deny reason. To claim both is to have neither."
- pintomp3, on 10/31/2007, -7/+41if only. the number of evangelical christians has been rising and is estimated at as high as 100 million. many of our elected officials incorrectly believe this country was founded on christian principles. we are getting closer to banning abortions, to banning gay marriage, and teaching creationism in schools. we have bogus law schools and universities. there is an entire paraprofessional system set up to compete with the secular one. america was always secular, the rise is the evangelical america.
- richIsBored, on 10/31/2007, -5/+35If I'm to believe in your God I have to make one of two possible concessions. Either he's mute or a total douche.
I mean here you are believing in him for fear of going to hell. And if your notion of God turns out to be true kudos to you for making the right choice.
But seriously, what's it like deciding to believe in something simply because your fate in the hereafter depends on it? I mean, you can reveal whatever life experiences you've had to substantiate your beliefs but in the end your god is threatening you and you're just doing what is necessary to save your ass aren't you?
And if we put whatever spiritual experiences you've allegedly had aside, what's left? A book that's supposedly his word? But who told you that? Did God? And if not him then how do you know the men who speak for him are any authority on the subject? How do you know that what you've been taught isn't just something your father taught you and his father before him and so on until we get to one guy who made the whole thing up?
I mean in order to accept this notion you have to either believe that God is incapable of speaking to anyone except for a select few prophets who he had play secretary for him by writing the Bible, or that he's just a douche and refuses to speak to you directly.
I dunno. Maybe you actually can hear him. That must be nice. As for me, all he has to do is pipe up. Until then I think the whole thing is a crock of *****.
- spyrochaete, on 11/02/2007, -3/+33The ratio of good atheists is the same as the ratio of good Christians. How can you respect a religion that eternally damns good people because they're good for the wrong reason? Sounds like an elitist xenophobic country club to me.
- aroq, on 11/07/2007, -15/+44"I have faith" is synonymous with "I believe something for no reason."
Enjoy your cop-out, ye faithful.
- inactive, on 10/30/2007, -4/+33This article does not account for the "blind-faith" or the "lip-service" generations. Belief in God is either disappearing completely or reaching a "fundamental" level. In maybe 30 years or so you will have the religious zealous vs the anti-religious.
- unearth, on 11/07/2007, -1/+30-Wait, before you go, what's heaven like?
-It's fine... There's a shortage of chairs...
-Oh.
-...Yeah..
- scoot2006, on 10/31/2007, -1/+29If there is a god and I end up in front of it when I die, I'm giving it the finger for being such a complete ***** to so many people.
Oh, and keep your ***** pitty.
- Tippis, on 10/31/2007, -2/+30...especially since doing so means they'll go there themselves ;D
- KraftDinner101, on 10/30/2007, -7/+35I refuse to think that I "walk" with a figment of someone else's imagination.
- unearth, on 10/30/2007, -1/+28It already exists, and it's called Sweden. Those people have it made.
- Dumbledorito, on 10/30/2007, -2/+29As was noted by those who collected them, why do you think there were all those school films trying to curb behaviors if those behaviors weren't a problem?
- BbIT, on 10/30/2007, -7/+33Spot on. Spot on.
- Marthinus, on 11/03/2007, -7/+32As a very serious Christian I am very happy about the current situation where no one is forced to be a 'Christian' just because of group pressure. I am of the opinion that a lot of the problems the Christians have today that relates to hypocrisy etc. was caused by pressure 'Christians', so now at least if you do not want to do what the Bible says you can fully reject it and carry on with your life instead of being critical and unhappy in Church.
- thomas, on 10/30/2007, -3/+28God is leaving us to fend for ourselves? At what point in history did god have our backs? If you payed attention in history class you would know that things are getting better not worse. And you can ask for gods blessings all you want all you are doing is wasting your life.
- wdcampbell3, on 10/30/2007, -7/+31No doubt helped by people such as Richard Dawkins, Christopher HItchens:
And of course Marcus Brigstocke :)
I do agree that blind faith is a dangerous and often maliciously leveraged force. Is it the age of the Brights?
- petewiz, on 10/30/2007, -1/+25Even if there is a Heaven, ***** it. I don't want to spend an eternity with someone ("God") who makes the purpose of human existence to praise him; especially someone who would tolerate disgusting human atrocities (slavery, genocide, war) as well as (according to their own mythology) allowing evil incarnate ("the Devil") set up shop and torture people for all eternity, just for things as petty as being gay or not going to his weekly fan-club meetings. Sounds like a sadistic, *****-up egomaniac to me.
- otakushark, on 10/30/2007, -3/+25Probably 50% of the world's population asks the god of their choice daily for "blessings" and it doesn't seem to have any effect on the way the world is going.
- vuke69, on 11/01/2007, -7/+29"...and the world began it's ascent into happiness, peace and prosperity when god was abandoned"
- SquigglyP, on 10/31/2007, -0/+22damn, all you need is a beat and that will go straight to #1 on the R&B charts.
- TheSpook, on 10/30/2007, -4/+25> "As a Christian we are taught to respect and tolerate the other religions and beliefs that exist in our world."
... as long as you're not gay or anything.
- Arcesius, on 10/31/2007, -3/+24god that's so ***** up... I'm amazed christians think that because someone's an atheist means that they don't have any morals... I'd propose that anyone who gets their "morals" from "God" is on less firm footing than someone, like myself, whose moral code is founded upon the notion that you get what you give, you reap what you sow (yes, yes, I know that's the christian concept of it... albeit largely ignored), karma's a bitch, for every action there's a reaction, and stuff like that...
- derkmerkin, on 10/30/2007, -2/+23well *****, if you think thats hard, try being a PS3 supporter on digg.
- SquigglyP, on 10/30/2007, -4/+25you lost me at "we believe in an Omnipotent God"
- Orat, on 10/30/2007, -2/+23This is the fire and brimstone speech of the 21st century- disgust and resentment thinly veiled with fake pity, which they believe is translated as genuine concern.
- zenoizen, on 10/31/2007, -14/+34Yeah, churches should start paying taxes in these United States. Especially Christian churches, as Christianity is and always has been as much a political movement as a religion.
- Show 51 - 100 of 717 discussions
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Last Updated: Aug. 15, 2014
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French Corner Cellars at the Adirondack Wine & Food Festival
French Corner Cellars at the Adirondack Wine & Food Festival
Mark your calendars for June 23rd & 24th for the 2018 Adirondack Wine & Food Festival in Lake George, NY, where French Corner Cellars will be doing a live demonstration of our wine cabinets. The event will showcase the best the Adirondacks and NY have to offer including distilleries, wineries, cideries, breweries, meaderies, artisanal food, food trucks, and more.
French Corner Cellars will be teaming up with Marcella’s appliance to bring you demonstrations for the Big Green Egg grill, our own Avintage Revolution 195, as well as an Iron Chef-styled cook off featuring Marcella’s own Mike Morrill – a CIA graduate – and Ny State’s Chef William Cornelius along with other special guests. French Corner Cellars will be offering presentations and education on the undervalued importance of wine serving temperature, and give the opportunity to view our highest performing Avintage wine cabinet!
With more than eighty tents and kiosks to sample (and purchase) from, there is more than enough for you and your family to take in over the two-day period, so make sure to bring your appetite, your thirst, and your sunglasses. We’ll see you there!
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How To Write Letters Of Explanations To Mortgage Underwriters
This BLOG On How To Write Letters Of Explanations To Mortgage Underwriters Was UPDATED On July 16, 2017
Mortgage Borrowers who are either buying a home or homeowners who are refinancing their home loan go through the mortgage process by completing and submitting a mortgage application called the 1003. Completing a mortgage application officially starts the mortgage process.
- The mortgage application, also known as the 1003, asks information about income, debts, and assets.
- It also asks disclosures such as if the borrower filed bankruptcy, had a foreclosure, had a judgment, had tax liens, or short sale.
- It is of utmost importance that mortgage applicants answer these questions as accurately and truthfully as possible.
- All information stated on mortgage application will be verified.
- Letters of explanations to mortgage underwriters will need to submitted on matters mortgage underwriters need clarification.
Documents Required To Process Mortgage
Prior to processing and underwriting your mortgage loan application, the mortgage loan officer will ask borrowers to provide financial documents such as the following:
- Two years tax returns
- 2 months bank statements
- W-2’s for the past two years
- 30 days most recent paycheck stubs
- Bankruptcy papers if it applies
- Foreclosure papers if it applies
- Divorce decree
- Other documents such as judgment release and tax release paperwork.
Mortgage Processing
Once these documents are submitted, the loan officer will then start the mortgage process.
- The mortgage loan application will be assigned a mortgage processor and prepared for underwriting.
- Once the mortgage processor makes sure that all documents are complete with no missing pages, the loan will be submitted to the lender’s underwriting department and assigned a mortgage underwriter.
Mortgage Underwriting
- It is the job of the underwriter to approve mortgage loan.
- Underwriters issue conditional mortgage loan approvals.
- Mortgage Underwriters issue clear to close.
- Mortgage application and the document submitted along with credit reports is reviewed by the mortgage loan underwriter.
- The mortgage loan underwriter fully understand credit and financial profile and any questions, letters of explanations to mortgage underwriters need to be submitted.
How Underwriters Underwrite Mortgage Files
Once mortgage loan application is assigned to an underwriter, the underwriter will got through mortgage application, credit reports, and all of the documents submitted.
- The mortgage loan underwriter will review 2 years tax returns.
- The tax returns will need to be verified with the Internal Revenue Service through 4506T.
- The underwriter will review W-2’s and will want to see a verification of employment letter from current employer.
- The underwriter will also check bank statements and see if there are any overdrafts in checking or savings account in the past 12 months.
- The mortgage underwriter will also review bankruptcy papers for borrowers who filed a prior bankruptcy.
- Underwriters will also review derogatory items on credit report and see the patterns of credit history.
- The mortgage underwriter will assess risk level and if he or she believes whether borrowers will pay future mortgage payments.
- The mortgage underwriter will then issue you a conditional mortgage loan approval.
Types of letters of explanations to mortgage underwriters
Prior to issuing a conditional approval, you may be required to write letters of explanations to mortgage underwriters for questionable items.
- Letters of explanations to mortgage underwriters need to be brief and to the point and a wrong statement can blow the deal.
- Mortgage broker will help write the letter of explanation and review it prior to submitting it.
Typical Requests For LOX
Some examples of letters of explanations to mortgage underwriters are the following:
- Borrowers who have unsatisfied collection accounts, mortgage underwriters will want to explain why the account went into arrears and the debt has not been paid.
- Some valid reasons can be loss of job, divorce, medical issues, identity theft, or other unforeseen circumstances.
- Borrowers who had a bankruptcy and/or foreclosure, underwriter will want to know a letter of explanation on why they filed bankruptcy or went through foreclosure.
- Some reasons can be due to loss of job, closure of business, medical issues, or divorce.
- Too many credit inquiries.
- Every credit inquiry within the past 90 days will need letters of explanations to mortgage underwriters.
- Some reasons can be shopping for a mortgage, shopping for a lower interest rate credit card, or some other reasonable explanation where it will prove to the mortgage loan underwriter that the borrower were not desperately seeking new credit due to financial troubles.
- Self employed individuals might need to prove that their business is most likely to continue via letters of explanation to mortgage underwriters.
- Job gap
- Letters of explanations to mortgage underwriters will need to be submitted if borrowers had job gaps in the past two years or periods of unemployment.
Clear To Close
- Once mortgage loan applicants have supplied letters of explanations to mortgage underwriters, they will be issued a conditional approval.
- There will be conditions that should easily be satisfied.
- Such conditions may be insurance information, the most recent bank statements, the most recent pay check stubs, and pulling a current credit check on you and/or co-borrowers.
- Once these terms have been satisfied, the mortgage loan underwriter will issue a clear to close which is the clear go ahead to schedule closing.
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TITLE: Proving $2\cos(k\theta) = \sum_{r=0}^{\lfloor k/2\rfloor} c(k,r) (2\cos \theta)^{k-2r}$ for $c(k,r)$ as defined
QUESTION [1 upvotes]: Let $k$ be a positive integer. Define for $n\ge 1, 0\leq r\leq \lfloor n/2\rfloor$, the integers $c(n,r)$ so that $c(1,0) = 1, c(2,0) = 1, c(2,1) = -2$ and for $n\ge 3$,
$$\begin{cases}
c(n,0)\phantom{/2} = 1,\\
c(n,r)\phantom{/2} = c(n-1, r) - c(n-2, r-1), &1\leq r\leq (n-1)/2,\\
c(n,n/2) = (-1)^{n/2} 2, &\text{ if $n$ is even}\end{cases}$$
Prove that $2\cos(k\theta) = \sum_{r=0}^{\lfloor k/2\rfloor} c(k,r) (2\cos \theta)^{k-2r}$ for any real number $\theta$.
I was thinking of doing a proof by induction, but it seems very complicated to do.
If $k=1, 2$, then
$$2\cos \theta = 2\cos\theta \quad\text{and}\quad 2\cos (2\theta) = (2\cos\theta)^2 + (-2)(2\cos \theta)^{2-2}$$
so the claim holds. So assume the claim holds for all $1\leq n < m$ and $m\ge 3$.
Suppose $m$ is odd. We want to show that
$$2\cos m\theta = \sum_{r=0}^{(m-1)/2} c(m,r) (2\cos \theta)^{m-2r}$$ We know by induction that
$$2\cos(m-1)\theta = \sum_{r=0}^{(m-1)/2} c(m-1, r)(2\cos \theta)^{m-1-2r}$$ and
$$2\cos(m-2)\theta = \sum_{r=0}^{(m-3)/2} c(m-2, r)(2\cos\theta)^{m-2-2r}$$
But then if I use the cosine addition formula on
$$2\cos m\theta = 2\cos ((m-1)\theta + (m-1)\theta - (m-2)\theta)$$ then I get an expression involving sines for which I don't have an explicit formula. Also, if I were to simplify the resulting expression, it would be very complicated.
Edit: Here's my attempt for even m, based on the first answer for odd m. There's likely a shorter approach though (see Mindlack's comment).
If m is even, then $2\cos ((m-1)\theta) = \sum_{r=0}^{(m-2)/2} c(m-1, r) (2\cos \theta)^{m-1-2r}$ and $2\cos ((m-2)\theta) = \sum_{r=0}^{(m-2)/2} c(m-2, r) (2\cos\theta)^{m-2-2r}$. We have
$\begin{align}
2\cos m\theta &= 2\cos \theta \cdot 2\cos ((m-1)\theta) - 2\cos((m-2)\theta)\\
&= \sum_{r=0}^{(m-2)/2} c(m-1, r) (2\cos \theta)^{m-2r} - \sum_{r=0}^{(m-2)/2}c(m-2, r)(2\cos \theta)^{(m-2) - 2r}\\
&= c(m-1, 0)(2\cos \theta)^m +\sum_{r=0}^{(m-4)/2} c(m-1, r + 1)(2\cos \theta)^{m-2 - 2r} -\sum_{r=0}^{(m-2)/2}c(m-2, r)(2\cos \theta)^{m-2-2r}\\
&= (2\cos \theta)^m - c(m-2, \frac{m-2}2) + \sum_{r=0}^{(m-4)/2}c(m, r+1)(2\cos \theta)^{m-2-2r}\\
&= \sum_{r=1}^{(m-2)/2}c(m, r)(2\cos \theta)^{m-2r} + c(m, m/2) +(2\cos \theta)^m\\
&= \sum_{r=0}^{m/2} c(m,r)(2\cos \theta)^{m-2r}\end{align}$
REPLY [0 votes]: Remarks: The proof for even $m$ is similar. So I omit it.
The proof for odd $m$:
By the inductive hypothesis, we have
$$2\cos (m - 1)\theta = \sum_{r=0}^{(m - 1)/2} c(m - 1,r)(2\cos \theta)^{m - 1 - 2r}$$
and
$$2\cos (m - 2)\theta = \sum_{r=0}^{(m - 3)/2} c(m - 2,r)(2\cos \theta)^{m - 2 - 2r}.$$
Using the identity $\cos m \theta + \cos (m - 2)\theta = 2\cos \theta \cos (m-1)\theta$, we have
\begin{align*}
2\cos m x &= 2\cos\theta \cdot 2 \cos (m - 1)\theta - 2\cos (m - 2)\theta\\
&= \sum_{r=0}^{(m - 1)/2} c(m - 1,r)(2\cos \theta)^{m - 2r} - \sum_{r=0}^{(m - 3)/2} c(m - 2,r)(2\cos \theta)^{m - 2 - 2r}\\
&= c(m - 1, 0)(2\cos \theta)^m + \sum_{r=1}^{(m - 1)/2} c(m - 1,r)(2\cos \theta)^{m - 2r} \\
&\qquad - \sum_{r=0}^{(m - 3)/2} c(m - 2,r)(2\cos \theta)^{m - 2 - 2r}\\
&= (2\cos \theta)^m + \sum_{r=0}^{(m - 1)/2 - 1} c(m - 1, 1 + r)(2\cos \theta)^{m - 2 - 2r} \\
&\qquad - \sum_{r=0}^{(m - 3)/2} c(m - 2,r)(2\cos \theta)^{m - 2 - 2r}\\
&= (2\cos \theta)^m + \sum_{r=0}^{(m - 3)/2} [c(m - 1, 1 + r) - c(m - 2, r)](2\cos \theta)^{m - 2 - 2r}\\
&= (2\cos \theta)^m + \sum_{r=0}^{(m - 3)/2} c(m, 1 + r)(2\cos \theta)^{m - 2 - 2r}\\
&= (2\cos \theta)^m + \sum_{r=1}^{(m - 3)/2 + 1} c(m, r)(2\cos \theta)^{m - 2r}\\
&= (2\cos \theta)^m - c(m, 0)(2\cos \theta)^m + \sum_{r=0}^{(m - 1)/2} c(m, r)(2\cos \theta)^{m - 2r}\\
&= \sum_{r=0}^{(m - 1)/2} c(m, r)(2\cos \theta)^{m - 2r}
\end{align*}
where we have used $c(m - 1, 0) = c(m, 0) = 1$.
The desired result follows.
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Want some tips to revive droopy eyelids? Follow these guidelines to avoid having tired and aging eyelids and banish the droopy eyes look.
Similarly to eyelashes, eyelids also play an essential role when it comes of creating a dazzling look. Indeed as the most important canvas for the eye makeup its condition can influence the success of our image. This area is one of the most sensitive spots on our face.
Consequently it is worth devoting some extra time and attention to its care. Droopy eyelids can make us look tired and older than we really are. That's why it is useful to experiment with the following tips to revive droopy eyelids.
Prevention is just as essential as treatment, why not spare ourselves of facing droopy or puffy eyelids too soon. Moreover in cases some might face serious surgery to eliminate this problem. These are some DIY homemade remedies to treat this unaesthetic issue.
- As mentioned earlier this area of of our face is ultra-sensitive to any damages. The best way to avoid any unpleasant consequences, as puffiness or dark circles is to quit rubbing our eyes. This is one of the worst habits, people often get used to. However if they could see the long-term result they would certainly stop.
- Rubbing is a harsh and damaging factor, by hurting the soft and fine tissues in this spot, we might generate some serious trouble. Stop increasing the chance for the appearance of redness, dark circles and droopy eyelids.
- The proper water intake can prevent droopy eyelids. Hydration plays a crucial role in maintaining the health of both eyes and the area around these.
- If you deprive your organism from the proper amount of water it switches to a water retention mode. This means that both body in general and the eyelids too will retain the liquids turning into real water bags.
- This will finally result in droopy eyelids. Drink at least 8 glasses of water per day and not a drop of it 2 hours before going to bed, in order to avoid the swelling.
- Cucumbers are magical remedies for droopy eyelids. Apply fresh slices of cucumber on the critical area. Leave them on for 15-20 minutes and lean back. This homemade method will help you in maintaining your eyes and eyelids refreshed and moisturized. Potato facials and slices work just as well, however take care of eliminating all the dirt from these. The colder the better, avoid warming them, it would only enhance the swelling process.
- Chamomile tea bags have also a great effect on eyelids. After you drunk your daily, tea, put the teabags in the refrigerator until these become cool and apply them on your eyes. Herbal teas are widely used in the beauty care, however in this case you should avoid mint tea bags. These might prove to be too harsh with this sensitive area. Leave them for other homemade skin care rituals.
- Simple spoons were also used by our grannies to reduce the puffiness of their eyes. This trick still works nowadays, as a quick remedy you can place them into the refrigerator until these reach the desired temperature. Again cold ones are the perfect solution for your tired eyelids.
elf1219 says:
Posted on 16 Aug 2009
i am so going to try some of these
cherryontop says:
Posted on 18 Aug 2009
I am totally going to try these!
Anonymous says:
Posted on 07 Aug 2013
Lame
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| 45,567
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\begin{document}
\title{Congruence Boolean Lifting Property}
\author{George GEORGESCU and Claudia MURE\c SAN\thanks{Corresponding author.}\\ \footnotesize University of Bucharest\\ \footnotesize Faculty of Mathematics and Computer Science\\ \footnotesize Academiei 14, RO 010014, Bucharest, Romania\\ \footnotesize Emails: georgescu.capreni@yahoo.com; c.muresan@yahoo.com, cmuresan@fmi.unibuc.ro}
\date{\today }
\maketitle
\begin{abstract} We introduce and study the Congruence Boolean Lifting Property (CBLP) for congruence--distributive universal algebras, as well as a property related to CBLP, which we have called $(\star )$. CBLP extends the so--called Boolean Lifting Properties (BLP) from MV--algebras, BL--algebras and residuated lattices, but differs from the BLP when particularized to bounded distributive lattices. Important classes of universal algebras, such as discriminator varieties, fulfill the CBLP. The main results of the present paper include a characterization theorem for congruence--distributive algebras with CBLP and a structure theorem for semilocal arithmetical algebras with CBLP. When we particularize the CBLP to the class of residuated lattices and to that of bounded distributive lattices and we study its relation to other Boolean Lifting Properties for these algebras, interesting results concerning the image of the reticulation functor between these classes are revealed.\\ {\em 2010 Mathematics Subject Classification:} Primary: 08B10; secondary: 03C05, 06F35, 03G25, 08B05.\\ {\em Keywords:} Boolean Lifting Property; Boolean center; lattice; residuated lattice; reticulation; (congruence--distributive, congruence--permutable, arithmetical) algebra; discriminator variety.\end{abstract}
\section{Introduction}
A unital ring has the {\em Idempotent Lifting Property} ({\em ILP} or {\em LIP}) iff its idempotents can be lifted modulo every left ideal. The ILP is closely related to important classes of rings such as clean rings, exchange rings, Gelfand rings, maximal rings etc.. Several algebraic and topological characterizations of commutative unital rings with ILP are collected in \cite[Theorem $1.7$]{mcgov}.
In studying the ILP for commutative unital rings, it is essential that the set of idempotents of a commutative unital ring $R$ is a Boolean algebra (called the {\em Boolean center of $R$}). There are many algebraic structure to which one can associate a ``Boolean center``: bounded distributive lattices, $lu$--groups, MV--algebras, BL--algebras, residuated lattices etc.. For all of these algebras, a lifting condition for the elements of the Boolean center, similar to the ILP for rings, can be defined. In \cite{blpiasi}, \cite{blpdacs}, \cite{ggcm}, \cite{dcggcm}, we have defined and studied the Boolean Lifting Properties for residuated lattices and bounded distributive lattices, we have provided algebraic and topological characterizations for them, and established links between them, by means of the reticulation functor (\cite{eu3}, \cite{eu1}, \cite{eu}, \cite{eu2}, \cite{eu4}, \cite{eu5}, \cite{eu7}). We have also studied a type of generalization of the lifting properties for universal algebras in \cite{dcggcm} and \cite{eudacs}.
The issue of defining a condition type Boolean Lifting Property in the context of universal algebras naturally arises. Such a condition needs to extend the Boolean Lifting Properties in the particular cases of the structures mentioned above, thus, first of all, it needs to be defined for a class of universal algebras which includes these particular kinds of structures. This idea has started the research in the present paper. The Congruence Boolean Lifting Property (CBLP), which we study in this article, is a lattice--theoretic condition: a congruence--distributive algebra ${\cal A}$ has {\em CBLP} iff its lattice of congruences, ${\rm Con}({\cal A})$, is such that the Boolean elements of each of its principal filters are joins between the generator of that filter and Boolean elements of ${\rm Con}({\cal A})$. CBLP is the transcription in the language of universal algebra of property $(3)$ from \cite[Lemma $4$]{banasch}. In the particular cases of MV--algebras, BL--algebras and residuated lattices, CBLP is equivalent to the Boolean Lifting Property; however, this does not hold in the case of bounded distributive lattices, which all have CBLP, but not all have the Boolean Lifting Property. Consequently, the CBLP is only a partial solution to the issue mentioned above, but its study is motivated by the properties of the universal algebras with CBLP which we prove in this paper. Significant results on the Boolean Lifting Property for MV--algebras, BL--algebras and residuated lattices can be generalized to CBLP in important classes of universal algebras.
Section \ref{preliminaries} of this paper contains some well--known notions and previously known results from universal algebra that we need in the rest of the paper. The results in the following sections are new, excepting only the results cited from other papers.
Section \ref{directproducts} is a collection of results concerning the form of congruences, and that of Boolean, finitely generated, prime and maximal congruences, in finite, as well as arbitrary direct products of congruence--distributive algebras from an equational class. These results serve as preparatives for the properties we obtain in the following sections.
In Section \ref{thecblp}, we introduce the property whose study is the aim of this paper: Congruence Boolean Lifting Property (abbreviated CBLP). We define and study the CBLP in equational classes of algebras whose lattice of congruences is distributive and has the property that its last element is compact. We prove a characterization theorem for the CBLP through algebraic and topological conditions, and identify important classes of algebras with CBLP, which include the class of bounded distributive lattices and discriminator varieties. We also study the behaviour of CBLP with respect to quotients and direct products, and its relation to a property we have called $(\star )$. Both CBLP and $(\star )$ extend properties which we have studied for residuated lattices in \cite{ggcm}, \cite{dcggcm}; many of the results in this section are inspired by results in \cite{ggcm}, \cite{dcggcm}. The preservation of the CBLP by quotients gives it an interesting behaviour in non--distributive lattices; in order to illustrate the related properties, we provide some examples.
Section \ref{cblpversusblp} is concerned with the particularizations of CBLP to the class of residuated lattices and that of bounded distributive lattices. For these algebras we have studied a property called the Boolean Lifting Property (BLP): for residuated lattices, in a restricted form in \cite{eu3}, \cite{eu4}, and, in the form which also appears in the present article, in \cite{ggcm} and \cite{dcggcm}; for bounded distributive lattices, in \cite{blpiasi}, \cite{blpdacs} and \cite{dcggcm}. It turns out that the CBLP and BLP coincide in the case of residuated lattices, but not in the case of bounded distributive lattices, where, as shown in Section \ref{thecblp}, the CBLP always holds, unlike the BLP. The reticulation functor from the category of residuated lattices to that of bounded distributive lattices preserves the BLP for filters (\cite{dcggcm}); clearly, the situation is different for the CBLP. These considerations make it easy to notice some properties concerning the image through the reticulation functor of certain classes of residuated lattices, including MV--algebras and BL--algebras, more precisely to exclude certain classes of bounded distributive lattices from these images.
In Section \ref{semilocal}, we return to the setting of universal algebra, and study the CBLP in semilocal arithmetical algebras; we find that, out of these algebras, the ones which fulfill the CBLP are exactly the finite direct products of local arithmetical algebras.
\section{Preliminaries}
\label{preliminaries}
We refer the reader to \cite{bal}, \cite{blyth}, \cite{bur}, \cite{gralgu} for a further study of the notions we recall in this section.
We shall denote by $\N $ the set of the natural numbers and by $\N ^*=\N \setminus \{0\}$. Throughout this paper, any direct product of algebras of the same type shall be considerred with the operations of that type of algebras defined componentwise. Also, throughout this paper, whenever there is no danger of confusion, for any non--empty family $(M_i)_{i\in I}$ of sets, by $\displaystyle (x_i)_{i\in I}\in \prod _{i\in I}M_i$ we shall mean: $x_i\in M_i$ for all $i\in I$. Finally, throughout this paper, a non--empty algebra will be
called a {\em trivial algebra} iff it has only one element, and a {\em non--trivial algebra} iff it has at least two distinct elements.
Throughout this section, $\tau $ will be a universal algebra signature and ${\cal A}$ will be a $\tau $--algebra, with support set $A$. The {\em congruences} of ${\cal A}$ are the equivalences on $A$ which are compatible to the operations of ${\cal A}$. Let ${\rm Con}({\cal A})$ be the set of the congruences of ${\cal A}$, which is a complete lattice with respect to set inclusion. Notice that, for every $\phi ,\theta \in {\rm Con}({\cal A})$, $\phi \subseteq \theta $ iff $\phi $ is a refinement of $\theta $, that is all the congruence classes of $\theta $ are unions of congruence classes of $\phi $. In the complete lattice $({\rm Con}({\cal A}),\subseteq )$, the meet of any family of congruences of ${\cal A}$ is the intersection of those congruences. The join, however, does not coincide to the union; we shall use the common notation, $\vee $, for the join in ${\rm Con}({\cal A})$. Also, clearly, $\Delta _{\cal A}$ is the first element of the lattice ${\rm Con}({\cal A})$, and $\nabla _{\cal A}$ is the last element of the lattice ${\rm Con}({\cal A})$, where we have denoted by $\Delta _{\cal A}=id_A=\{(a,a)\ |\ a\in A\}$ and $\nabla _{\cal A}=A^2$. Any $\phi \in {\rm Con}({\cal A})$ such that $\phi \neq \nabla _{\cal A}$ is called a {\em proper congruence of ${\cal A}$.}
Since the intersection of any family of congruences of ${\cal A}$ is a congruence of ${\cal A}$, it follows that, for every subset $X$ of $A^2$, there exists the smallest congruence of ${\cal A}$ which includes $X$; this congruence is denoted by $Cg(X)$ and called {\em the congruence of ${\cal A}$ generated by $X$.} Whenever the algebra ${\cal A}$ needs to be specified, we shall denote $Cg(X)$ by $Cg_{\cal A}(X)$. It is obvious that the join in the lattice $({\rm Con}({\cal A}),\subseteq )$ is given by: for all $\phi ,\psi \in {\rm Con}({\cal A})$, $\phi \vee \psi =Cg(\phi \cup \psi)$. The congruences of ${\cal A}$ generated by finite subsets of $A^2$ are called {\em finitely generated congruences.} For every $a,b\in A$, $Cg(\{(a,b)\})$ is denoted, simply, by $Cg(a,b)$, and called the {\em principal congruence generated by $(a,b)$} in ${\cal A}$. Clearly, finitely generated congruences are exactly the finite joins of principal congruences, because, for any $X\subseteq A^2$, $\displaystyle Cg(X)=\bigvee _{(a,b)\in X}Cg(a,b)$ and, since every congruence is reflexive, $Cg(\emptyset )=\Delta _{\cal A}=Cg(a,a)$ for all $a\in A$. The set of the finitely generated congruences of ${\cal A}$ shall be denoted by ${\cal K}({\cal A})$. It is well known (\cite{bur}) that ${\rm Con}({\cal A})$ is an {\em algebraic lattice,} that is a complete lattice such that each of its elements is a supremum of compact elements, and that the compact elements of ${\rm Con}({\cal A})$ are exactly the finitely generated congruences of ${\cal A}$: ${\cal K}({\cal A})$ coincides to the set of the compact elements of ${\rm Con}({\cal A})$.
The $\tau $--algebra ${\cal A}$ is said to be:
\begin{itemize}
\item {\em congruence--distributive} iff the lattice ${\rm Con}({\cal A})$ is distributive;
\item {\em congruence--permutable} iff each two congruences of ${\cal A}$ permute (with respect to the composition of binary relations);
\item {\em arithmetical} iff it is both congruence--distributive and congruence--permutable.\end{itemize}
Throughout the rest of this section, the $\tau $--algebra ${\cal A}$ shall be considerred non--empty and congruence--distributive.
For any $n\in \N ^*$ and any non--empty $\tau $--algebras ${\cal A}_1$, ${\cal A}_2$, $\ldots $, ${\cal A}_n$, with support sets $A_1$, $\ldots $, $A_n$, respectively, if $\displaystyle {\cal A}=\prod _{i=1}^n{\cal A}_i$, then:
\begin{itemize}
\item if $\theta _i\in {\rm Con}({\cal A}_i)$ for all $i\in \overline{1,n}$, then we denote: $\theta _1\times \ldots \times \theta _n=\{((x_1,\ldots ,x_n),(y_1,\ldots ,y_n))\in A^2\ |\ (\forall \, i\in \overline{1,n})\, ((x_i,y_i)\in \theta _i)\}$;
\item if $\theta \in {\rm Con}({\cal A})$, then, for all $i\in \overline{1,n}$, we denote by: $\displaystyle pr_i(\theta )=\{(x,y)\in A_i^2\ |\ (\exists \, (x_1,\ldots ,x_n),(y_1,\ldots ,y_n)\in A)\, (((x_1,\ldots ,x_n),(y_1,\ldots ,y_n))\in \theta ,x_i=x,y_i=y)\}$.\end{itemize}
More generally, for any non--empty family $({\cal A}_i)_{i\in I}$ of non--empty $\tau $--algebras, if $A_i$ is the support set of ${\cal A}_i$ for each $i\in I$ and $\displaystyle {\cal A}=\prod _{i\in I}{\cal A}_i$, then:
\begin{itemize}
\item if $\theta _i\in {\rm Con}({\cal A}_i)$ for all $i\in I$, then we denote: $\displaystyle \prod _{i\in I}\theta _i=\{((x_i)_{i\in I},(y_i)_{i\in I})\in A^2\ |\ (\forall \, i\in I)\, ((x_i,y_i)\in \theta _i)\}$;
\item if $\theta \in {\rm Con}({\cal A})$, then, for all $i\in I$, we denote by: $\displaystyle pr_i(\theta )=\{(x,y)\in A_i^2\ |\ (\exists \, (x_t)_{t\in I},(y_t)_{t\in I}\in A)\, (((x_t)_{t\in I},$\linebreak $(y_t)_{t\in I})\in \theta ,x_i=x,y_i=y)\}$.\end{itemize}
\begin{lemma}{\rm \cite{bj}} Let $n\in \N ^*$, ${\cal A}_1$, ${\cal A}_2$, $\ldots $, ${\cal A}_n$ be non--empty congruence--distributive $\tau $--algebras, and assume that $\displaystyle {\cal A}=\prod _{i=1}^n{\cal A}_i$. Then:
\begin{enumerate}
\item\label{totdinbj1} given any $\theta _i\in {\rm Con}({\cal A}_i)$ for every $i\in \overline{1,n}$, it follows that $\theta _1\times \ldots \times \theta _n\in {\rm Con}({\cal A})$ and, for each $i\in \overline{1,n}$, $pr_i(\theta _1\times \ldots \times \theta _n)=\theta _i$;
\item\label{totdinbj0} given any $\theta \in {\rm Con}({\cal A})$, it follows that: for each $i\in \overline{1,n}$, $pr_i(\theta )\in {\rm Con}({\cal A}_i)$, and $\theta =pr_1(\theta )\times \ldots \times pr_n(\theta )$;
\item\label{totdinbj2} the function $\displaystyle f:\prod _{i=1}^n{\rm Con}({\cal A}_i)\rightarrow {\rm Con}({\cal A})$, defined by $f(\theta _1,\ldots ,\theta _n)=\theta _1\times \ldots \times \theta _n$ for all $\theta _1\in {\rm Con}({\cal A}_1)$, $\ldots $, $\theta _n\in {\rm Con}({\cal A}_n)$, is a bounded lattice isomorphism, whose inverse is defined by: $f^{-1}(\theta )=(pr_1(\theta ),\ldots ,pr_n(\theta ))$ for all $\theta \in {\rm Con}({\cal A})$.\end{enumerate}\label{totdinbj}\end{lemma}
A proper congruence $\phi $ of ${\cal A}$ is called a {\em prime congruence} iff, given any $\theta _1,\theta _2\in {\rm Con}({\cal A})$, if $\theta _1\cap \theta _2\subseteq \phi $, then $\theta _1\subseteq \phi $ or $\theta _2\subseteq \phi $. The maximal elements of the set of proper congruences of ${\cal A}$ are called {\em maximal congruences.} We shall denote by ${\rm Spec}({\cal A})$ the set of the prime congruences of ${\cal A}$, by ${\rm Max}({\cal A})$ the set of the maximal congruences of ${\cal A}$ and by $\displaystyle {\rm Rad}({\cal A})=\bigcap _{\theta \in {\rm Max}({\cal A})}\theta $.
Let us consider the following hypothesis:
\begin{flushleft}
\hspace*{20pt} (H)$\quad $ $\nabla _{\cal A}\in {\cal K}({\cal A})$ (that is: $\nabla _{\cal A}$ is a compact element of ${\rm Con}({\cal A})$; equivalently: $\nabla _{\cal A}$ is a finitely generated congruence of ${\cal A}$).\end{flushleft}
Clearly, ${\cal K}({\cal A})$ is a sublattice of ${\rm Con}({\cal A})$, because, for all $X\subseteq A^2$ and $Y\subseteq A^2$, $Cg(X)\vee Cg(Y)=Cg(X\cap Y)$ and $Cg(X)\cap Cg(Y)=Cg(X\cup Y)$. Furthermore, $\Delta _{\cal A}=Cg(\emptyset )\in {\cal K}({\cal A})$, hence, if (H) is satisfied, then ${\cal K}({\cal A})$ is a bounded sublattice of ${\rm Con}({\cal A})$.
The following lemma is well known and straightforward.
\begin{lemma}
If ${\cal A}$ fulfills (H), then:
\begin{enumerate}
\item\label{folclor1} any proper congruence of ${\cal A}$ is included in a maximal congruence;
\item\label{folclor2} ${\rm Max}({\cal A})\subseteq {\rm Spec}({\cal A})$;
\item\label{folclor3} any proper congruence of ${\cal A}$ equals the intersection of the prime congruences that include it.\end{enumerate}\label{folclor}\end{lemma}
Clearly, if ${\cal A}$ fulfills (H), then: ${\cal A}$ is non--trivial iff $\Delta _{\cal A}\neq \nabla _{\cal A}$ iff $\Delta _{\cal A}$ is a proper congruence of ${\cal A}$ iff ${\cal A}$ has proper congruences iff ${\cal A}$ has maximal congruences, where the last equivalence follows from Lemma \ref{folclor}, (\ref{folclor1}).
We say that the {\em Chinese Remainder Theorem} ({\em CRT}, for short) holds in ${\cal A}$ iff: for all $n\in \N ^*$, any $\theta _1,\ldots ,\theta _n\in {\rm Con}({\cal A})$ and any $a_1,\ldots ,a_n\in A$, if $(a_i,a_j)\in \theta _i\vee \theta _j$ for all $i,j\in \overline{1,n}$, then there exists an $a\in A$ such that $(a,a_i)\in \theta _i$ for all $i\in \overline{1,n}$.
\begin{proposition}{\rm \cite{bj}} CRT holds in ${\cal A}$ iff ${\cal A}$ is arithmetical.\label{propozitie5.6}\end{proposition}
For every congruence $\theta $ of ${\cal A}$, we shall denote:
\begin{itemize}
\item by $[\theta )$ the principal filter of the lattice ${\rm Con}({\cal A})$ generated by $\theta $, that is $[\theta )=\{\phi \in {\rm Con}({\cal A})\ |\ \theta \subseteq \phi \}$; as is the case for any lattice filter, $[\theta )$ is a sublattice (not bounded sublattice) of ${\rm Con}({\cal A})$, and $h_{\textstyle \theta }:{\rm Con}({\cal A})\rightarrow [\theta )$, $h_{\textstyle \theta }(\phi )=\phi \vee \theta $ for all $\phi \in {\rm Con}({\cal A})$, is a bounded lattice morphism;
\item for any $a\in A$, by $a/\theta $ the equivalence class of $a$ with respect to $\theta $, and by $A/\theta $ the quotient set of $A$ with respect to $\theta $; in what follows, we shall assume that $A/\theta $ becomes an algebra of the same kind as ${\cal A}$, with the operations defined canonically, and we shall denote by ${\cal A}/\theta $ the algebraic structure of $A/\theta $;
\item by $p_{\textstyle \theta }:A\rightarrow A/\theta $ the canonical surjection with respect to $\theta $;
\item for any $X\subseteq A^2$ and any $Y\subseteq A$, by $X/\theta =\{(p_{\textstyle \theta }(a),p_{\textstyle \theta }(b))\ |\ (a,b)\in X\}=\{(a/\theta ,b/\theta )\ |\ (a,b)\in X\}$ and by $Y/\theta =p_{\textstyle \theta }(Y)=\{a/\theta \ |\ a\in Y\}$;
\item by $s_{\textstyle \theta }:{\rm Con}({\cal A}/\theta )\rightarrow [\theta )$ the function defined by: for any $\alpha \in {\rm Con}({\cal A}/\theta )$, $s_{\textstyle \theta }(\alpha )=\{(a,b)\in A^2\ |\ (p_{\textstyle \theta }(a),p_{\textstyle \theta }(b))$\linebreak $\in \alpha \}$; as shown in \cite{bur}, $s_{\textstyle \theta }$ is a bounded lattice isomorphism, whose inverse is defined by $s_{\textstyle \theta }^{-1}(\phi )=\phi /\theta =\{(p_{\textstyle \theta }(a),p_{\textstyle \theta }(b))\ |\ (a,b)\in \phi \}=\{(a/\theta ,b/\theta )\ |\ (a,b)\in \phi \}$ for every $\phi \in [\theta )$; consequently, if ${\cal A}$ and ${\cal A}/\theta $ fulfill (H), then ${\rm Max}({\cal A}/\theta)=s_{\textstyle \theta }^{-1}({\rm Max}({\cal A})\cap [\theta ))=\{\phi /\theta \ | \phi \in {\rm Max}({\cal A}),\theta \subseteq \phi \}$; thus, if ${\cal A}$ and ${\cal A}/\theta $ fulfill the hypothesis (H), then ${\rm Max}({\cal A}/{\rm Rad}({\cal A}))=s_{\textstyle {\rm Rad}({\cal A})}^{-1}({\rm Max}({\cal A}))$, which is isomorphic to ${\rm Max}({\cal A})$, and ${\rm Rad}({\cal A}/{\rm Rad}({\cal A}))={\rm Rad}({\cal A})/{\rm Rad}({\cal A})=\{\nabla _{\cal A}/{\rm Rad}({\cal A})\}$.\end{itemize}
\begin{lemma}{\rm \cite[Theorem $2.3$, (iii)]{bj}}
For every $\theta \in {\rm Con}({\cal A})$ and any $X\subseteq A^2$, $Cg_{{\cal A}/\theta }(X/\theta )=(Cg_{\cal A}(X)\vee \theta )/\theta $.\label{superlema}\end{lemma}
For every bounded distributive lattice $L$, we shall denote by ${\cal B}(L)$ the {\em Boolean center} of $L$, that is the set of the complemented elements of $L$. Then ${\cal B}(L)$ is a Boolean algebra, and, given any bounded distributive lattice $M$ and any bounded lattice morphism $f:L\rightarrow M$, the image of the restriction ${\cal B}(f)$ of $f$ to ${\cal B}(L)$ is included in ${\cal B}(M)$. Thus ${\cal B}$ becomes a covariant functor from the category of bounded distributive lattices to the category of Boolean algebras.
Clearly, if $(L_i)_{i\in I}$ is an arbitrary family of bounded distributive lattices, then $\displaystyle {\cal B}(\prod _{i\in I}L_i)=\prod _{i\in I}{\cal B}(L_i)$.
A congruence $\theta $ of ${\cal A}$ is called a {\em factor congruence} iff there exists $\theta ^*\in {\rm Con}({\cal A})$ such that $\theta \vee \theta ^*=\nabla _{\cal A}$, $\theta \cap \theta ^*=\Delta _{\cal A}$ and $\theta \circ \theta ^*=\theta ^*\circ \theta $. In other words, the factor congruences of ${\cal A}$ are the elements of ${\cal B}({\rm Con}({\cal A}))$ that permute with their complement. If the algebra ${\cal A}$ is arithmetical, then it is congruence--permutable, thus the set of its factor congruences coincides to ${\cal B}({\rm Con}({\cal A}))$.
All finite direct products of algebras in this paper are considerred non--empty.
\begin{lemma}{\rm \cite{bj}} Let $n\in \N ^*$ and consider $n$ arithmetical algebras, ${\cal A}_1$, ${\cal A}_2$, $\ldots $, ${\cal A}_n$. Then the following are equivalent:
\begin{enumerate}
\item\label{dinbj1} ${\cal A}$ is isomorphic to the direct product $\displaystyle \prod _{i=1}^n{\cal A}_i$;
\item\label{dinbj2} there exist $\alpha _1,\ldots ,\alpha _n\in {\cal B}({\rm Con}({\cal A}))$ such that $\alpha _i\vee \alpha _j=\nabla _{\cal A}$ for all $i,j\in \overline{1,n}$ such that $i\neq j$, $\displaystyle \bigcap _{i=1}^n\alpha _i=\Delta _{\cal A}$ and ${\cal A}_i$ is isomorphic to ${\cal A}/_{\textstyle \alpha _i}$ for each $i\in \overline{1,n}$.\end{enumerate}\label{dinbj}\end{lemma}
For every bounded distributive lattice $L$, we denote by ${\rm Id}(L)$ the lattice of ideals of $L$, by ${\rm Spec}_{\rm Id}(L)$ the set of the prime ideals of $L$, by ${\rm Max}_{\rm Id}(L)$ the set of the maximal ideals of $L$ and by ${\rm Rad}_{\rm Id}(L)$ the intersection of all maximal ideals of $L$. $L$ is said to be {\em ${\rm Id}$--local} iff it has exactly one maximal ideal.
We recall that a bounded distributive lattice $L$ is called:
\begin{itemize}
\item a {\em normal} lattice iff, for every $x,y\in L$ such that $x\vee y=1$, there exist $e,f\in L$ such that $e\wedge f=0$ and $x\vee e=y\vee f=1$;
\item a {\em B--normal} lattice iff, for every $x,y\in L$ such that $x\vee y=1$, there exist $e,f\in {\cal B}(L)$ such that $e\wedge f=0$ and $x\vee e=y\vee f=1$;
\item a {\em conormal} lattice iff its dual is normal;
\item a {\em B--conormal} lattice iff its dual is B--normal.\end{itemize}
Throughout the rest of this section, $L$ will be a bounded distributive lattice. Throughout the rest of this paper, by lattice we shall mean bounded distributive lattice.
Clearly, any B--normal lattice is normal, and any B--conormal lattice is conormal. Trivially, any Boolean algebra is B--normal and B--conormal.
\begin{lemma}{\rm \cite{brdi}} The following are equivalent:
\begin{itemize}
\item $L$ is ${\rm Id}$--local;
\item for all $x,y\in L$, $x\vee y=1$ implies $x=1$ or $y=1$.\end{itemize}\label{lema2.5}\end{lemma}
\begin{lemma}{\rm \cite{bur}} ${\rm Rad}_{\rm Id}(L)=\{a\in L\ |\ (\forall \, x\in L)\, (a\vee x=1\Rightarrow x=1)\}$.\label{lema2.7}\end{lemma}
\section{Direct Products of Algebras}
\label{directproducts}
In this section, we obtain a series of results concerning the congruences of finite and those of arbitrary direct products of congruence--distributive algebras from an equational class, results which we need in the sequel.
Throughout the rest of this paper, $\tau $ will be a universal algebra signature, ${\cal C}$ shall be an equational class of congruence--distributive $\tau $--algebras and, unless mentioned otherwise, ${\cal A}$ will be a non--empty algebra from ${\cal C}$, with support set $A$. See the notations in Section \ref{preliminaries} for what follows.
\begin{lemma} Let $n\in \N ^*$, ${\cal A}_1,\ldots ,{\cal A}_n$ be algebras from ${\cal C}$, with support sets $A_1,\ldots ,A_n$, respectively, and assume that $\displaystyle {\cal A}=\prod _{i=1}^n{\cal A}_i$.
\begin{enumerate}
\item\label{lemaI1} For all $i\in \overline{1,n}$, let $X_i\subseteq A_i^2$, and let $X=\{((a_1,\ldots ,a_n),(b_1,\ldots ,b_n))\ |\ (\forall \, i\in \overline{1,n})\, ((a_i,b_i)\in X_i)\}$. Then $Cg_{\cal A}(X)=Cg_{{\cal A}_1}(X_1)\times \ldots \times Cg_{{\cal A}_n}(X_n)$.
\item\label{lemaI2} Let $X\subseteq A^2$ and, for all $i\in \overline{1,n}$, $X_i=\{(x,y)\in A_i^2\ |\ (\exists \, ((a_1,\ldots ,a_n),(b_1,\ldots ,b_n))\in X)\, (a_i=x,b_i=y)\}$. Then $Cg_{\cal A}(X)=Cg_{{\cal A}_1}(X_1)\times \ldots \times Cg_{{\cal A}_n}(X_n)$.\end{enumerate}\label{lemaI}\end{lemma}
\begin{proof} (\ref{lemaI1}) Clearly, $Cg_{{\cal A}_1}(X_1)\times \ldots \times Cg_{{\cal A}_n}(X_n)\supseteq X$, thus $Cg_{{\cal A}_1}(X_1)\times \ldots \times Cg_{{\cal A}_n}(X_n)\supseteq Cg_{\cal A}(X)$; also, for all $i\in \overline{1,n}$, $pr_i(Cg_{\cal A}(X))\supseteq X_i$, thus $pr_i(Cg_{\cal A}(X))\supseteq Cg_{{\cal A}_i}(X_i)$, hence $Cg_{\cal A}(X)\supseteq Cg_{{\cal A}_1}(X_1)\times \ldots \times Cg_{{\cal A}_n}(X_n)$ by Lemma \ref{totdinbj}. Therefore $Cg_{\cal A}(X)=Cg_{{\cal A}_1}(X_1)\times \ldots \times Cg_{{\cal A}_n}(X_n)$.
\noindent (\ref{lemaI2}) Denote $Y=\{((a_1,\ldots ,a_n),(b_1,\ldots ,b_n))\ |\ (\forall \, i\in \overline{1,n})\, ((a_i,b_i)\in X_i)\}\subseteq A^2$. Clearly, $X\subseteq Y$, thus $Cg_{\cal A}(X)\subseteq Cg_{\cal A}(Y)=Cg_{{\cal A}_1}(X_1)\times \ldots \times Cg_{{\cal A}_n}(X_n)$, by (\ref{lemaI1}). Also, clearly, for all $i\in \overline{1,n}$, $pr_i(Cg_{\cal A}(X))\supseteq X_i$, thus $pr_i(Cg_{\cal A}(X))\supseteq Cg_{{\cal A}_i}(X_i)$, hence $Cg_{\cal A}(X)\supseteq Cg_{{\cal A}_1}(X_1)\times \ldots \times Cg_{{\cal A}_n}(X_n)$ by Lemma \ref{totdinbj}. Therefore $Cg_{\cal A}(X)=Cg_{{\cal A}_1}(X_1)\times \ldots \times Cg_{{\cal A}_n}(X_n)$.\end{proof}
\begin{proposition} Let $n\in \N ^*$, ${\cal A}_1,\ldots ,{\cal A}_n$ be algebras from ${\cal C}$, and assume that $\displaystyle {\cal A}=\prod _{i=1}^n{\cal A}_i$. Then:
\begin{enumerate}
\item\label{propO1} the function $\displaystyle g:\prod _{i=1}^n{\cal B}({\rm Con}({\cal A}_i))\rightarrow {\cal B}({\rm Con}({\cal A}))$, defined by $g(\theta _1,\ldots ,\theta _n)=\theta _1\times \ldots \times \theta _n$ for all $\theta _1\in {\cal B}({\rm Con}({\cal A}_1)),\ldots ,\theta _n\in {\cal B}({\rm Con}({\cal A}_n))$, is a Boolean isomorphism;
\item\label{propO2} the function $\displaystyle h:\prod _{i=1}^n{\cal K}({\cal A}_i)\rightarrow {\cal K}({\cal A})$, defined by $h(\theta _1,\ldots ,\theta _n)=\theta _1\times \ldots \times \theta _n$ for all $\theta _1\in {\cal K}({\cal A}_1),\ldots ,\theta _n\in {\cal K}({\cal A}_n)$, is a bijection.\end{enumerate}\label{propO}\end{proposition}
\begin{proof} (\ref{propO1}) Since $\displaystyle {\cal B}(\prod _{i=1}^n{\rm Con}({\cal A}_i))=\prod _{i=1}^n{\cal B}({\rm Con}({\cal A}_i))$, it follows that $g={\cal B}(f)$, the image through the functor ${\cal B}$ of the bounded lattice isomorphism $f$ from Lemma \ref{totdinbj}, (\ref{totdinbj2}), hence $g$ is a Boolean isomorphism.
\noindent (\ref{propO2}) With the notations in Lemma \ref{lemaI}, (\ref{lemaI1}), if $X_1,\ldots ,X_n$ are finite, then $X$ is finite, hence, if $\theta _i\in {\cal K}({\cal A}_i)$ for all $i\in \overline{1,n}$, then $\theta _1\times \ldots \times \theta _n\in {\cal K}({\cal A})$, thus $h$ is well defined. With the notations in Lemma \ref{lemaI}, (\ref{lemaI2}), if $X$ is finite, then $X_1,\ldots ,X_n$ are finite, thus $h$ is surjective. Finally, since $h$ is the restriction to $\displaystyle \prod _{i=1}^n{\cal K}({\cal A}_i)$ of the function $f$ from Lemma \ref{totdinbj}, (\ref{totdinbj2}), and $f$ is injective, it follows that $h$ is injective. Therefore $h$ is bijective.\end{proof}
\begin{lemma}
Let $({\cal A}_i)_{i\in I}$ be a non--empty family of non--empty algebras from ${\cal C}$, assume that $\displaystyle {\cal A}=\prod _{i\in I}{\cal A}_i$, let $J$ be an arbitrary non--empty set and, for every $i\in I$, let $(\theta _{i,j})_{j\in J}\subseteq {\rm Con}({\cal A}_i)$. Then $\displaystyle \bigcap _{j\in J}(\prod _{i\in I}\theta _{i,j})=\prod _{i\in I}(\bigcap _{j\in J}\theta _{i,j})$.\label{interscong}\end{lemma}
\begin{proof} For all $i\in I$, let $A_i$ be the support set of ${\cal A}_i$ and $a_i,b_i\in A_i$. Then: $\displaystyle ((a_i)_{i\in I},(b_i)_{i\in I})\in \bigcap _{j\in J}(\prod _{i\in I}\theta _{i,j})$ iff, for all $i\in I$ and all $j\in J$, $(a_i,b_i)\in \theta _{i,j}$, iff $\displaystyle ((a_i)_{i\in I},(b_i)_{i\in I})\in \prod _{i\in I}(\bigcap _{j\in J}\theta _{i,j})$, hence the equality in the enunciation.\end{proof}
\begin{lemma} Let $({\cal A}_i)_{i\in I}$ be a non--empty family of non--empty algebras from ${\cal C}$, and assume that $\displaystyle {\cal A}=\prod _{i\in I}{\cal A}_i$. Then:
\begin{enumerate}
\item\label{lemal1} for any $\displaystyle (\theta _i)_{i\in I}\in \prod _{i\in I}{\rm Con}({\cal A}_i)$, it follows that $\displaystyle \prod _{i\in I}\theta _i\in {\rm Con}({\cal A})$ and, for each $j\in \overline{1,n}$, $\displaystyle pr_j(\prod _{i\in I}\theta _i)=\theta _j$;
\item\label{lemal0} for any $\theta \in {\rm Con}({\cal A})$, it follows that: for each $i\in I$, $pr_i(\theta )\in {\rm Con}({\cal A}_i)$, and $\displaystyle \theta \subseteq \prod _{i\in I}pr_i(\theta )$;
\item\label{lemal2} the function $\displaystyle f:\prod _{i\in I}{\rm Con}({\cal A}_i)\rightarrow {\rm Con}({\cal A})$, defined by $\displaystyle f((\theta _i)_{i\in I})=\prod _{i\in I}\theta _i$ for all $\displaystyle (\theta _i)_{i\in I}\in \prod _{i\in I}{\rm Con}({\cal A}_i)$, is an injective bounded lattice morphism.\end{enumerate}\label{lemal}\end{lemma}
\begin{proof} For every $i\in I$, let $A_i$ be the support set of ${\cal A}_i$.
\noindent (\ref{lemal1}) Let $\displaystyle (\theta _i)_{i\in I}\in \prod _{i\in I}{\rm Con}({\cal A}_i)$, and let $\omega $ be an operation symbol from $\tau $, of arity $k\in \N ^*$. For each $j\in \overline{1,k}$, let $\displaystyle (a_{j,i})_{i\in I},(b_{j,i})_{i\in I}\in \prod _{i\in I}A_i$ and, for every $i\in I$, let $\omega ^{\textstyle {\cal A}_i}(a_{1,i},\ldots ,a_{k,i})=a_i\in A_i$ and $\omega ^{\textstyle {\cal A}_i}(b_{1,i},\ldots ,b_{k,i})=b_i\in A_i$. Assume that, for all $j\in \overline{1,k}$, $\displaystyle ((a_{j,i})_{i\in I},(b_{j,i})_{i\in I})\in \prod _{i\in I}\theta _i$, that is, for all $i\in I$ and all $j\in \overline{1,k}$, $(a_{j,i},b_{j,i})\in \theta _i$. For all $i\in I$, since $\theta _i\in {\rm Con}({\cal A}_i)$, it follows that $(a_i,b_i)\in \theta _i$; thus $\displaystyle ((a_i)_{i\in I},(b_i)_{i\in I})\in \prod _{i\in I}{\rm Con}({\cal A}_i)$. Since $\displaystyle {\cal A}=\prod _{i\in I}{\cal A}_i$, the following hold: $\omega ^{\textstyle {\cal A}}((a_{1,i})_{i\in I},\ldots ,(a_{k,i})_{i\in I})=(\omega ^{\textstyle {\cal A}_i}(a_{1,i},\ldots ,a_{k,i}))_{i\in I}=(a_i)_{i\in I}$ and $\omega ^{\textstyle {\cal A}}((b_{1,i})_{i\in I},\ldots ,(b_{k,i})_{i\in I})=(\omega ^{\textstyle {\cal A}_i}(b_{1,i},\ldots ,b_{k,i}))_{i\in I}=(b_i)_{i\in I}$. Therefore $\displaystyle (\omega ^{\textstyle {\cal A}}((a_{1,i})_{i\in I},\ldots ,(a_{k,i})_{i\in I}),$\linebreak $\displaystyle \omega ^{\textstyle {\cal A}}((b_{1,i})_{i\in I},\ldots ,(b_{k,i})_{i\in I}))\in \prod _{i\in I}\theta _i$. Hence $\displaystyle \prod _{i\in I}\theta _i\in {\rm Con}({\cal A})$. It is immediate that, for all $j\in I$, $\displaystyle pr_j(\prod _{i\in I}\theta _i)=\theta _j$.
\noindent (\ref{lemal0}) Straightforward.
\noindent (\ref{lemal2}) (\ref{lemal1}) ensures us that the image of $f$ is, indeed, included in ${\rm Con}({\cal A})$. Bounded distributive lattices form an equational class, thus $\displaystyle \prod _{i\in I}{\rm Con}({\cal A}_i)$ is a bounded distributive lattice, with the operations defined componentwise; from this it is straightforward that $f$ is a bounded lattice morphism. The injectivity of $f$ follows from the second statement in (\ref{lemal1}).\end{proof}
Throughout the rest of this paper, we shall assume that all non--empty algebras from ${\cal C}$ fulfill the hypothesis (H) (see Section \ref{preliminaries}).
\begin{proposition} Let $n\in \N ^*$, ${\cal A}_1$, ${\cal A}_2$, $\ldots $, ${\cal A}_n$ be non--empty algebras from ${\cal C}$, and assume that $\displaystyle {\cal A}=\prod _{i=1}^n{\cal A}_i$. Then:
\begin{enumerate}
\item\label{specprod1} $\displaystyle {\rm Max}({\cal A})=\bigcup _{i=1}^n\{\nabla _{{\cal A}_{\scriptstyle 1}}\times \ldots \times \nabla _{{\cal A}_{\scriptstyle i-1}}\times \theta \times \nabla _{{\cal A}_{\scriptstyle i+1}}\times \ldots \times \nabla _{{\cal A}_{\scriptstyle n}}\ |\ \theta \in {\rm Max}({\cal A}_i)\}$; consequently, $\displaystyle |{\rm Max}({\cal A})|=\sum _{i=1}^n|{\rm Max}({\cal A}_i)|$ and ${\rm Rad}({\cal A})={\rm Rad}({\cal A}_1)\times \ldots \times {\rm Rad}({\cal A}_n)$;
\item\label{specprod2} $\displaystyle {\rm Spec}({\cal A})=\bigcup _{i=1}^n\{\nabla _{{\cal A}_{\scriptstyle 1}}\times \ldots \times \nabla _{{\cal A}_{\scriptstyle i-1}}\times \theta \times \nabla _{{\cal A}_{\scriptstyle i+1}}\times \ldots \times \nabla _{{\cal A}_{\scriptstyle n}}\ |\ \theta \in {\rm Spec}({\cal A}_i)\}$; consequently, $\displaystyle |{\rm Spec}({\cal A})|=\sum _{i=1}^n|{\rm Spec}({\cal A}_i)|$.\end{enumerate}\label{specprod}\end{proposition}
\begin{proof} (\ref{specprod1}) It is immediate that, for every $i\in \overline{1,n}$ and every $\theta \in {\rm Max}({\cal A}_i)$, $\nabla _{{\cal A}_{\scriptstyle 1}}\times \ldots \times \nabla _{{\cal A}_{\scriptstyle i-1}}\times \theta \times \nabla _{{\cal A}_{\scriptstyle i+1}}\times \ldots \times \nabla _{{\cal A}_{\scriptstyle n}}\in {\rm Max}({\cal A})$.
Now, given any $\phi \in {\rm Max}({\cal A})$, it follows that $\phi $ is a proper congruence, thus there exist $a,b\in A$ such that $(a,b)\notin \phi $, that is, for some $i\in \overline{1,n}$, $(a_i,b_i)\notin \phi _i$, where we have denoted $a=(a_1,\ldots ,a_n)$, $b=(b_1,\ldots ,b_n)$ and $\phi =\phi _1\times \ldots \times \phi _n$, with $a_k,b_k\in A_k$ (the support set of ${\cal A}_k$) and $\phi _k\in {\rm Con}({\cal A}_k)$ for all $k\in \overline{1,n}$. So $\phi _i\neq \nabla _{{\cal A}_{\scriptstyle i}}$. Assume by absurdum that there exists a $j\in \overline{1,n}$ such that $j\neq i$ and $\phi _j\neq \nabla _{{\cal A}_{\scriptstyle j}}$. Then $\phi \subsetneq \nabla _{{\cal A}_{\scriptstyle 1}}\times \ldots \times \nabla _{{\cal A}_{\scriptstyle i-1}}\times \theta \times \nabla _{{\cal A}_{\scriptstyle i+1}}\times \ldots \times \nabla _{{\cal A}_{\scriptstyle n}}\subsetneq \nabla _{\cal A}$, which is a contradiction to the maximality of $\phi $. Hence $\phi =\nabla _{{\cal A}_{\scriptstyle 1}}\times \ldots \times \nabla _{{\cal A}_{\scriptstyle i-1}}\times \phi _i\times \nabla _{{\cal A}_{\scriptstyle i+1}}\times \ldots \times \nabla _{{\cal A}_{\scriptstyle n}}$. It is clear that the proper congruence $\phi _i\in {\rm Max}({\cal A}_i)$, because otherwise we would get another contradiction to the maximality of $\phi $.
Therefore ${\rm Max}({\cal A})$ has the form in the enunciation, otherwise written $\displaystyle {\rm Max}({\cal A})=\bigcup _{i=1}^n(\{\nabla _{{\cal A}_{\scriptstyle 1}}\}\times \ldots \times \{\nabla _{{\cal A}_{\scriptstyle i-1}}\}\times {\rm Max}({\cal A}_i)\times \{\nabla _{{\cal A}_{\scriptstyle i+1}}\}\times \ldots \times \{\nabla _{{\cal A}_{\scriptstyle n}}\})$, from which the expression of its cardinality follows by noticing that, since $\nabla _{{\cal A}_{\scriptstyle i}}\notin {\rm Max}({\cal A}_i)$ for any $i\in \overline{1,n}$, the sets $\{\nabla _{{\cal A}_{\scriptstyle 1}}\}\times \ldots \times \{\nabla _{{\cal A}_{\scriptstyle i-1}}\}\times {\rm Max}({\cal A}_i)\times \{\nabla _{{\cal A}_{\scriptstyle i+1}}\}\times \ldots \times \{\nabla _{{\cal A}_{\scriptstyle n}}\}$, with $i\in \overline{1,n}$, are mutually disjoint, and that they are in bijection to the sets ${\rm Max}({\cal A}_i)$, with $i\in \overline{1,n}$, respectively. The formula of ${\rm Rad}({\cal A})$ now follows by Lemma \ref{interscong}.
\noindent (\ref{specprod2}) Let $i\in \overline{1,n}$, $\theta \in {\rm Spec}({\cal A}_i)$ and denote $\phi =\nabla _{{\cal A}_{\scriptstyle 1}}\times \ldots \times \nabla _{{\cal A}_{\scriptstyle i-1}}\times \theta \times \nabla _{{\cal A}_{\scriptstyle i+1}}\times \ldots \times \nabla _{{\cal A}_{\scriptstyle n}}$. Let $\alpha ,\beta \in {\rm Con}({\cal A})$ such that $\alpha \cap \beta \subseteq \theta $. Then, if $\alpha =\alpha _1\times \ldots \times \alpha _n$ and $\beta =\beta _1\times \ldots \times \beta _n$, with $\alpha _k,\beta _k\in {\rm Con}({\cal A}_k)$ for all $k\in \overline{1,n}$, it follows that $\alpha _i\cap \beta _i\subseteq \theta \in {\rm Spec}({\cal A}_i)$, thus $\alpha _i\subseteq \theta $ or $\beta _i\subseteq \theta $, thus $\alpha \subseteq \phi $ or $\beta \subseteq \phi $. Therefore $\phi \in {\rm Spec}({\cal A})$.
Now let $\phi \in {\rm Spec}({\cal A})$, so $\phi $ is a proper congruence of ${\cal A}$, from which, just as above for maximal congruences, we get that, if $\phi =\phi _1\times \ldots \times \phi _n$, with $\phi _k\in {\rm Con}({\cal A}_k)$ for all $k\in \overline{1,n}$, then there exists an $i\in \overline{1,n}$ such that $\phi _i$ is a proper congruence of ${\cal A}_i$. If there also exists a $j\in \overline{1,n}$ with $j\neq i$ and $\phi _j$ a proper congruence of ${\cal A}_j$, then, by denoting $\alpha =\phi _1\times \ldots \times \phi _{i-1}\times \nabla _{{\cal A}_{\scriptstyle i}}\times \phi _{i+1}\times \ldots \times \phi _n$ and $\beta =\phi _1\times \ldots \times \phi _{j-1}\times \nabla _{{\cal A}_{\scriptstyle j}}\times \phi _{j+1}\times \ldots \times \phi _n$, we get that $\alpha \cap \beta =\phi $, but $\alpha \nsubseteq \phi $ and $\beta \nsubseteq \phi $, which is a contradiction to the primality of $\phi $. Hence $\phi =\nabla _{{\cal A}_{\scriptstyle 1}}\times \ldots \times \nabla _{{\cal A}_{\scriptstyle i-1}}\times \phi _i\times \nabla _{{\cal A}_{\scriptstyle i+1}}\times \ldots \times \nabla _{{\cal A}_{\scriptstyle n}}$, with $\phi _i\neq \nabla _{{\cal A}_{\scriptstyle i}}$. Assume by absurdum that $\phi _i\notin {\rm Spec}({\cal A}_i)$, that is there exist $\alpha _i,\beta _i\in {\rm Con}({\cal A}_i)$ such that $\alpha _i\cap \beta _i\subseteq \phi _i$, but $\alpha _i\nsubseteq \phi _i$ and $\beta _i\nsubseteq \phi _i$. Denote $\alpha =\nabla _{{\cal A}_{\scriptstyle 1}}\times \ldots \times \nabla _{{\cal A}_{\scriptstyle i-1}}\times \alpha _i\times \nabla _{{\cal A}_{\scriptstyle i+1}}\times \ldots \times \nabla _{{\cal A}_{\scriptstyle n}}$ and $\beta =\nabla _{{\cal A}_{\scriptstyle 1}}\times \ldots \times \nabla _{{\cal A}_{\scriptstyle i-1}}\times \beta _i\times \nabla _{{\cal A}_{\scriptstyle i+1}}\times \ldots \times \nabla _{{\cal A}_{\scriptstyle n}}$. Then $\alpha \cap \beta =\nabla _{{\cal A}_{\scriptstyle 1}}\times \ldots \times \nabla _{{\cal A}_{\scriptstyle i-1}}\times (\alpha _i\cap \beta _i)\times \nabla _{{\cal A}_{\scriptstyle i+1}}\times \ldots \times \nabla _{{\cal A}_{\scriptstyle n}}\subseteq \nabla _{{\cal A}_{\scriptstyle 1}}\times \ldots \times \nabla _{{\cal A}_{\scriptstyle i-1}}\times \phi _i\times \nabla _{{\cal A}_{\scriptstyle i+1}}\times \ldots \times \nabla _{{\cal A}_{\scriptstyle n}}=\phi $, but $\alpha \nsubseteq \phi $ and $\beta \nsubseteq \phi $, which is a contradiction to the primality of $\phi $. Hence $\phi _i\in {\rm Spec}({\cal A}_i)$.
Therefore ${\rm Spec}({\cal A})$ has the form in the enunciation, otherwise written $\displaystyle {\rm Spec}({\cal A})=\bigcup _{i=1}^n(\{\nabla _{{\cal A}_{\scriptstyle 1}}\}\times \ldots \times \{\nabla _{{\cal A}_{\scriptstyle i-1}}\}\times {\rm Spec}({\cal A}_i)\times \{\nabla _{{\cal A}_{\scriptstyle i+1}}\}\times \ldots \times \{\nabla _{{\cal A}_{\scriptstyle n}}\})$, from which the expression of its cardinality follows by noticing that, since $\nabla _{{\cal A}_{\scriptstyle i}}\notin {\rm Spec}({\cal A}_i)$ for any $i\in \overline{1,n}$, the sets $\{\nabla _{{\cal A}_{\scriptstyle 1}}\}\times \ldots \times \{\nabla _{{\cal A}_{\scriptstyle i-1}}\}\times {\rm Spec}({\cal A}_i)\times \{\nabla _{{\cal A}_{\scriptstyle i+1}}\}\times \ldots \times \{\nabla _{{\cal A}_{\scriptstyle n}}\}$, with $i\in \overline{1,n}$, are mutually disjoint, and that they are in bijection to the sets ${\rm Spec}({\cal A}_i)$, with $i\in \overline{1,n}$, respectively.\end{proof}
\begin{proposition}
Let $({\cal A}_i)_{i\in I}$ be a non--empty family of non--empty algebras from ${\cal C}$, and assume that $\displaystyle {\cal A}=\prod _{i\in I}{\cal A}_i$. Then:
\begin{enumerate}
\item\label{specprodarb1} $\displaystyle {\rm Max}({\cal A})\supseteq \bigcup _{j\in I}\{\prod _{i\in I}\phi _i\ |\ \phi _j\in {\rm Max}({\cal A}_j),(\forall \, i\in I\setminus \{j\})\, (\phi _i=\nabla _{{\cal A}_{\scriptstyle i}})\}$; consequently, $\displaystyle |{\rm Max}({\cal A})|\geq \sum _{j\in I}|{\rm Max}({\cal A}_j)|$ and $\displaystyle {\rm Rad}({\cal A})\subseteq \prod _{j\in I}{\rm Rad}({\cal A}_j)$;
\item\label{specprodarb2} $\displaystyle {\rm Spec}({\cal A})\supseteq \bigcup _{j\in I}\{\prod _{i\in I}\phi _i\ |\ \phi _j\in {\rm Spec}({\cal A}_j),(\forall \, i\in I\setminus \{j\})\, (\phi _i=\nabla _{{\cal A}_{\scriptstyle i}})\}$; consequently, $\displaystyle |{\rm Spec}({\cal A})|\geq \sum _{j\in I}|{\rm Spec}({\cal A}_j)|$.\end{enumerate}\label{specprodarb}\end{proposition}
\begin{proof} (\ref{specprodarb1}) For every $j\in I$, the following hold: $\displaystyle \prod _{i\in I}{\cal A}_i={\cal A}_j\times \prod _{i\in I\setminus \{j\}}{\cal A}_i$, hence, according to Proposition \ref{specprod}, (\ref{specprod1}), applied for $n=2$, $\displaystyle {\rm Max}({\cal A})={\rm Max}(\prod _{i\in I}{\cal A}_i)={\rm Max}({\cal A}_j\times \prod _{i\in I\setminus \{j\}}{\cal A}_i)=\{\theta _j\times \nabla _{\prod _{i\in I\setminus \{j\}}{\cal A}_i}\ |\ \theta _j\in {\rm Max}({\cal A}_j)\}\cup \{\nabla _{{\cal A}_j}\times \theta \ |\ \theta \in {\rm Max}(\prod _{i\in I\setminus \{j\}}{\cal A}_i)\}\supseteq \{\theta _j\times \nabla _{\prod _{i\in I\setminus \{j\}}{\cal A}_i}\ |\ \theta _j\in {\rm Max}({\cal A}_j)\}=\{\theta _j\times \prod _{i\in I\setminus \{j\}}\nabla _{{\cal A}_i}\ |\ \theta _j\in {\rm Max}({\cal A}_j)\}$, and the latter set is isomorphic to ${\rm Max}({\cal A}_j)$, so its cardinality coincides to $|{\rm Max}({\cal A}_j)|$. Therefore, $\displaystyle {\rm Max}({\cal A})\supseteq \bigcup _{j\in I}\{\theta _j\times \prod _{i\in I\setminus \{j\}}\nabla _{{\cal A}_i}\ |\ \theta _j\in {\rm Max}({\cal A}_j)\}$, and the sets in this union are, obviously, mutually disjoint, hence $\displaystyle |{\rm Max}({\cal A})|\geq \sum _{j\in I}|{\rm Max}({\cal A}_j)|$. Also, the previous inclusion shows that $\displaystyle {\rm Rad}({\cal A})\subseteq \bigcap _{j\in I}\bigcap _{\theta _j\in {\rm Max}({\cal A}_j)}(\theta _j\times \prod _{i\in I\setminus \{j\}}\nabla _{{\cal A}_i})=\prod _{j\in I}\bigcap _{\theta _j\in {\rm Max}({\cal A}_j)}\theta _j=\prod _{j\in I}{\rm Rad}({\cal A}_j)$, by Lemma \ref{interscong}.
\noindent (\ref{specprodarb2}) Analogously to (\ref{specprodarb1}), but applying Proposition \ref{specprod}, (\ref{specprod2}), instead of Proposition \ref{specprod}, (\ref{specprod1}).\end{proof}
\section{Introducing the Congruence Boolean Lifting Property}
\label{thecblp}
In this section we introduce the property we call CBLP, which constitutes the subject of this paper, identify important classes of congruences and classes of algebras which fulfill CBLP, prove a structure theorem for algebras with CBLP, and study CBLP in quotient algebras, in direct products of algebras and in relation to other significant properties concerning congruence--distributive algebras.
Until mentioned otherwise, $\theta $ shall be an arbitrary but fixed congruence of ${\cal A}$. Let us consider the functions $u_{\textstyle \theta }:{\rm Con}({\cal A})\rightarrow {\rm Con}({\cal A}/\theta )$ and $v_{\textstyle \theta }:{\rm Con}({\cal A})\rightarrow [\theta )$, defined by: for all $\alpha \in {\rm Con}({\cal A})$, $u_{\textstyle \theta }(\alpha )=(\alpha \vee \theta )/\theta $ and $v_{\textstyle \theta }(\alpha )=\alpha \vee \theta $.
\begin{lemma}
\begin{enumerate}
\item\label{uvstheta1} $u_{\textstyle \theta }$ and $v_{\textstyle \theta }$ are bounded lattice morphisms;
\item\label{uvstheta2} the first diagram below (in the category of bounded distributive lattices) is commutative, and hence the second diagram below (in the category of Boolean algebras) is commutative; since $s_{\textstyle \theta }$ is a bounded lattice isomorphism (see Section \ref{preliminaries}), it follows that ${\cal B}(s_{\textstyle \theta })$ is a Boolean isomorphism:\end{enumerate}
\vspace*{-10pt}
\begin{center}
\begin{tabular}{cc}
\begin{picture}(120,53)(0,0)
\put(0,30){${\rm Con}({\cal A})$}
\put(35,33){\vector (1,0){40}}
\put(15,27){\vector (3,-2){33}}
\put(94,26){\vector (-3,-2){33}}
\put(50,38){$u_{\textstyle \theta }$}
\put(23,11){$v_{\textstyle \theta }$}
\put(80,10){$s_{\textstyle \theta }$}
\put(76,30){${\rm Con}({\cal A}/\theta )$}
\put(50,0){$[\theta )$}
\end{picture}
&\hspace*{20pt}
\begin{picture}(160,53)(0,0)
\put(0,30){${\cal B}({\rm Con}({\cal A}))$}
\put(49,33){\vector (1,0){40}}
\put(15,27){\vector (3,-2){33}}
\put(110,27){\vector (-3,-2){34}}
\put(56,38){${\cal B}(u_{\textstyle \theta })$}
\put(7,10){${\cal B}(v_{\textstyle \theta })$}
\put(93,8){${\cal B}(s_{\textstyle \theta })$}
\put(91,30){${\cal B}({\rm Con}({\cal A}/\theta ))$}
\put(50,0){${\cal B}([\theta ))$}
\end{picture}
\end{tabular}\end{center}\label{uvstheta}\end{lemma}
\vspace*{-7pt}
\begin{proof} Straightforward.\end{proof}
\begin{lemma}
The following are equivalent:
\begin{enumerate}
\item\label{camcalalr1} $\theta \subseteq {\rm Rad}({\cal A})$ and $\theta \in {\cal B}({\rm Con}({\cal A}))$;
\item\label{camcalalr2} $\theta=\Delta _{\cal A}$.\end{enumerate}\label{camcalalr}\end{lemma}
\begin{proof} (\ref{camcalalr2})$\Rightarrow $(\ref{camcalalr1}): Obvious.
\noindent (\ref{camcalalr1})$\Rightarrow $(\ref{camcalalr2}): Assume that $\theta \subseteq {\rm Rad}({\cal A})$ and $\theta \in {\cal B}({\rm Con}({\cal A}))$. The fact that $\theta \in {\cal B}({\rm Con}({\cal A}))$ means that there exists $\phi \in {\rm Con}({\cal A})$ such that $\theta \cap \phi =\Delta _{\cal A}$ and $\theta \vee \phi =\nabla _{\cal A}$. Now the fact that $\theta \subseteq {\rm Rad}({\cal A})$, that is $\theta \subseteq \mu $ for all $\mu\in {\rm Max}({\rm Con}({\cal A}))$, shows that $\phi \vee \mu =\nabla _{\cal A}$, for all $\mu \in {\rm Max}({\rm Con}({\cal A}))$. Assume by absurdum that $\phi \neq \nabla _{\cal A}$. Then $\phi \subseteq \mu $ for some $\mu\in {\rm Max}({\rm Con}({\cal A}))$, according to Lemma \ref{folclor}. We get that $\nabla _{\cal A}=\theta \vee \phi \subseteq \mu \vee \mu =\mu $, a contradiction to $\mu\in {\rm Max}({\rm Con}({\cal A}))\subseteq {\rm Con}({\cal A})\setminus \{\nabla _{\cal A}\}$. Thus $\phi =\nabla _{\cal A}$, hence $\theta =\theta \cap \nabla _{\cal A}=\theta \cap \phi =\Delta _{\cal A}$.\end{proof}
\begin{corollary}
If $\theta \subseteq {\rm Rad}({\cal A})$, then ${\cal B}(u_{\textstyle \theta })$ and ${\cal B}(v_{\textstyle \theta })$ are injective.\label{candsrad}\end{corollary}
\begin{proof} Since $s_{\textstyle \theta }$ is a bounded lattice isomorphism, it follows that ${\cal B}(s_{\textstyle \theta })$ is a Boolean isomorphism. Now Lemma \ref{uvstheta}, (\ref{uvstheta2}), shows that ${\cal B}(v_{\textstyle \theta })$ is injective iff ${\cal B}(u_{\textstyle \theta })$ is injective. Let $\alpha \in {\cal B}({\rm Con}({\cal A}))$ such that ${\cal B}(v_{\textstyle \theta })(\alpha )=\theta $, that is $\alpha \vee \theta =\theta $, so that $\alpha \subseteq \theta \subseteq {\rm Rad}({\cal A})$. Thus $\alpha \in {\cal B}({\rm Con}({\cal A}))$ and $\alpha \subseteq {\rm Rad}({\cal A})$, hence $\alpha =\Delta _{\cal A}$ by Lemma \ref{camcalalr}. The fact that ${\cal B}(v_{\textstyle \theta })$ is a Boolean morphism now shows that ${\cal B}(v_{\textstyle \theta })$ is injective.\end{proof}
\begin{definition}
We say that $\theta $ has the {\em Congruence Boolean Lifting Property} (abbreviated {\em CBLP}) iff ${\cal B}(u_{\textstyle \theta })$ is surjective.\end{definition}
\begin{remark}
As shown by Lemma \ref{uvstheta}, (\ref{uvstheta2}), since ${\cal B}(s_{\textstyle \theta })$ is bijective, we have: $\theta $ has CBLP iff ${\cal B}(v_{\textstyle \theta })$ is surjective.
Furthermore, according to Corollary \ref{candsrad}, if $\theta \subseteq {\rm Rad}({\cal A})$, then we have: $\theta $ has CBLP iff ${\cal B}(u_{\textstyle \theta })$ is bijective iff ${\cal B}(v_{\textstyle \theta })$ is bijective.\label{usiv}\end{remark}
\begin{remark}
Obviously, $v_{\theta }$ is surjective, because, for any $\psi \in [\theta )\subseteq {\rm Con}({\cal A})$, $v_{\theta }(\psi )=\psi \vee \theta =\psi $.\label{vesurj}\end{remark}
\begin{definition}
Let $\Omega \subseteq {\rm Con}({\cal A})$. We say that ${\cal A}$ has the {\em $\Omega $--Congruence Boolean Lifting Property} (abbreviated {\em $\Omega $--CBLP}) iff every $\omega \in \Omega $ has CBLP. We say that ${\cal A}$ has the {\em Congruence Boolean Lifting Property (CBLP)} iff ${\cal A}$ has ${\rm Con}({\cal A})$--CBLP.\end{definition}
The definition of CBLP is inspired by a property in \cite[Lemma $4$]{banasch}.
\begin{proposition}\begin{enumerate}
\item\label{toatebool1} If $[\theta )\subseteq {\cal B}({\rm Con}({\cal A}))$, then each $\phi \in [\theta )$ has CBLP and fulfills ${\cal B}({\rm Con}({\cal A}/\phi ))={\rm Con}({\cal A}/\phi )$.
\item\label{toatebool0} If ${\cal B}({\rm Con}({\cal A}))={\rm Con}({\cal A})$, then ${\cal A}$ has CBLP and, for each $\phi \in {\rm Con}({\cal A})$, ${\cal B}({\rm Con}({\cal A}/\phi ))={\rm Con}({\cal A}/\phi )$.\end{enumerate}\label{toatebool}\end{proposition}
\begin{proof} (\ref{toatebool1}) Assume that $[\theta )\subseteq {\cal B}({\rm Con}({\cal A}))$, and let $\phi \in [\theta )$, so that $[\phi )\subseteq [\theta )\subseteq {\cal B}({\rm Con}({\cal A}))$. Let $\gamma \in {\cal B}({\rm Con}({\cal A}/\phi ))$. Then $\gamma \in {\rm Con}({\cal A}/\phi )$, so $\gamma =\alpha /\phi $, for some $\alpha \in [\phi )\subseteq {\cal B}({\rm Con}({\cal A}))$, thus $\gamma =\alpha /\phi =(\alpha \vee \phi )/\phi =u_{\phi }(\alpha )={\cal B}(u_{\phi })(\alpha )$. Therefore ${\cal B}({\rm Con}({\cal A}/\phi ))\supseteq {\cal B}(u_{\phi })({\cal B}({\rm Con}({\cal A})))\supseteq {\rm Con}({\cal A}/\phi )\supseteq {\cal B}({\rm Con}({\cal A}/\phi ))$, hence ${\cal B}(u_{\phi })({\cal B}({\rm Con}({\cal A})))={\cal B}({\rm Con}({\cal A}/\phi ))={\rm Con}({\cal A}/\phi )$, thus $\phi $ has CBLP.
\noindent (\ref{toatebool0}) This statement can be derived from Remarks \ref{vesurj} and \ref{usiv}, but it also follows from (\ref{toatebool1}), by the fact that ${\rm Con}({\cal A})=[\Delta _{\cal A})$.\end{proof}
\begin{remark} $\Delta _{\cal A}$ and $\nabla _{\cal A}$ have CBLP. For $\Delta _{\cal A}$, we can apply Remark \ref{usiv} and the fact that $\Delta _{\cal A}\subseteq {\rm Rad}({\cal A})$, or we can notice that $[\Delta _{\cal A})={\rm Con}({\cal A})$ and $v_{\Delta _{\cal A}}$ is the identity of ${\rm Con}({\cal A})$, thus it is a bounded lattice isomorphism, hence ${\cal B}(v_{\Delta _{\cal A}})$ is a Boolean isomorphism, so it is surjective, thus $\Delta _{\cal A}$ has CBLP. For $\nabla _{\cal A}$ we can apply Proposition \ref{toatebool}, (\ref{toatebool1}), or simply notice that $[\nabla _{\cal A})=\{\nabla _{\cal A}\}$, thus ${\cal B}([\nabla _{\cal A}))=\{\nabla _{\cal A}\}$, hence ${\cal B}(v_{\nabla _{\cal A}})$ is surjective, thus $\nabla _{\cal A}$ has CBLP.\label{deltanabla}\end{remark}
In what follows, the complementation in the Boolean algebra ${\cal B}({\rm Con}({\cal A}))$ shall be denoted by $\neg \, $, and, for every $\phi \in {\rm Con}({\cal A})$, the complementation in the Boolean algebra ${\cal B}([\phi ))$ shall be denoted by $\neg _{\phi }$. Notice that $\theta /\theta =\Delta _{{\cal A}/\theta }$ and $\nabla _{\cal A}/\theta =\nabla _{{\cal A}/\theta }$.
\begin{remark} If $\theta \in {\rm Max}({\cal A})$, then the following hold:
\begin{itemize}
\item $[\theta )=\{\alpha \in {\rm Con}({\cal A})\ |\ \theta \subseteq \alpha \}=\{\theta ,\nabla _{\cal A}\}$, with $\theta \neq \nabla _{\cal A}$, thus $[\theta )$ is the two--element chain, which is a Boolean algebra, so ${\cal B}([\theta ))=[\theta )=\{\theta ,\nabla _{\cal A}\}$;
\item ${\rm Con}({\cal A}/\theta )=\{\alpha /\theta \ |\ \alpha \in [\theta )\}=\{\theta /\theta ,\nabla _{\cal A}/\theta \}$, with $\theta /\theta \neq \nabla _{\cal A}/\theta $, because $s_{\textstyle \theta }$ is injective (see Section \ref{preliminaries}); thus ${\rm Con}({\cal A}/\theta )$ is the two--element chain, which is a Boolean algebra, thus ${\cal B}({\rm Con}({\cal A}/\theta ))={\rm Con}({\cal A}/\theta )=\{\theta /\theta ,\nabla _{\cal A}/\theta \}$.\end{itemize}\label{R1}\end{remark}
\begin{remark} ${\cal B}([\theta ))=\{\theta ,\nabla _{\cal A}\}$ iff ${\cal B}({\rm Con}({\cal A}/\theta ))=\{\theta /\theta ,\nabla _{\cal A}/\theta \}$. Indeed, since $s_{\textstyle \theta }$ is a bounded lattice isomorphism (see Section \ref{preliminaries}), it follows that ${\cal B}(s_{\textstyle \theta }):{\cal B}({\rm Con}({\cal A}/\theta ))\rightarrow {\cal B}( [\theta ))$ is a Boolean isomorphism, whose inverse is ${\cal B}(s_{\textstyle \theta }^{-1})$. Therefore ${\cal B}([\theta ))={\cal B}(s_{\textstyle \theta })({\cal B}({\rm Con}({\cal A}/\theta )))=\{s_{\textstyle \theta }(\alpha )\ |\ \alpha \in {\cal B}({\rm Con}({\cal A}/\theta ))\}$ and ${\cal B}({\rm Con}({\cal A}/\theta ))={\cal B}(s_{\textstyle \theta }^{-1})({\cal B}([\theta )))=\{s_{\textstyle \theta }^{-1}(\alpha )\ |\ \alpha \in {\cal B}([\theta ))\}$, hence the equivalence above.\label{R2}\end{remark}
\begin{lemma}\begin{enumerate}
\item\label{L2(1)} If $\theta \in {\rm Spec}({\cal A})$, then ${\cal B}([\theta ))=\{\theta ,\nabla _{\cal A}\}$ and ${\cal B}({\rm Con}({\cal A}/\theta ))=\{\theta /\theta ,\nabla _{\cal A}/\theta \}$.
\item\label{L2(2)} If $\theta \in {\rm Max}({\cal A})$, then ${\cal B}([\theta ))=\{\theta ,\nabla _{\cal A}\}$ and ${\cal B}({\rm Con}({\cal A}/\theta ))=\{\theta /\theta ,\nabla _{\cal A}/\theta \}$.\end{enumerate}\label{L2}\end{lemma}
\begin{proof} (\ref{L2(1)}) Assume that $\theta \in {\rm Spec}({\cal A})$. Let $\alpha \in {\cal B}([\theta ))$ and denote $\beta=\neg _{\theta }\alpha $. Then $\alpha \cap \beta =\theta $, thus $\alpha \cap \beta \subseteq \theta $, hence $\alpha \subseteq \theta $ or $\beta \subseteq \theta $ since $\theta $ is a prime congruence. But $\alpha ,\beta \in [\theta )$, that is $\theta \subseteq \alpha $ and $\theta \subseteq \beta $. Therefore $\alpha =\theta $ or $\beta =\theta $. If $\beta =\theta $, then $\alpha =\neg _{\theta }\beta =\neg _{\theta }\theta =\nabla _{\cal A}$. Hence ${\cal B}([\theta ))\subseteq \{\theta ,\nabla _{\cal A}\}$. But, clearly, $\{\theta ,\nabla _{\cal A}\subseteq {\cal B}([\theta ))\}$. Therefore ${\cal B}([\theta ))=\{\theta ,\nabla _{\cal A}\}$. Hence ${\cal B}({\rm Con}({\cal A}/\theta ))=\{\theta /\theta ,\nabla _{\cal A}/\theta \}$ by Remark \ref{R2}.
\noindent (\ref{L2(2)}) This is part of Remark \ref{R1}, but also follows from (\ref{L2(1)}) and Lemma \ref{folclor}, (\ref{folclor1}).\end{proof}
\begin{lemma}\begin{enumerate}
\item\label{L1(1)} If ${\cal B}([\theta ))=\{\theta ,\nabla _{\cal A}\}$, then $\theta $ has CBLP.
\item\label{L1(2)} If ${\cal B}({\rm Con}({\cal A}/\theta ))=\{\theta /\theta ,\nabla _{\cal A}/\theta \}$, then $\theta $ has CBLP.\end{enumerate}\label{L1}\end{lemma}
\begin{proof} (\ref{L1(1)}) Assume that ${\cal B}([\theta ))=\{\theta ,\nabla _{\cal A}\}$. Since, clearly, $\Delta _{\cal A},\nabla _{\cal A}\in {\cal B}({\rm Con}({\cal A}))$, it follows that ${\cal B}(v_{\textstyle \theta })({\cal B}({\rm Con}({\cal A})))\supseteq {\cal B}(v_{\textstyle \theta })(\{\Delta _{\cal A},\nabla _{\cal A}\})=\{v_{\textstyle \theta }(\Delta _{\cal A}),v_{\textstyle \theta }(\nabla _{\cal A})\}=\{\Delta _{\cal A}\vee \theta ,\nabla _{\cal A}\vee \theta \}=\{\theta ,\nabla _{\cal A}\}={\cal B}([\theta ))$, thus ${\cal B}(v_{\textstyle \theta })({\cal B}({\rm Con}({\cal A})))={\cal B}([\theta ))$, that is ${\cal B}(v_{\textstyle \theta })$ is surjective, so $\theta $ has CBLP.
\noindent (\ref{L1(2)}) By (\ref{L1(1)}) and Remark \ref{R2}.\end{proof}
\begin{proposition}\begin{enumerate}
\item\label{primecblp1} Any prime congruence of ${\cal A}$ has CBLP.
\item\label{primecblp2} Any maximal congruence of ${\cal A}$ has CBLP.\end{enumerate}\label{primecblp}\end{proposition}
\begin{proof} (\ref{primecblp1}) By Lemma \ref{L2}, (\ref{L2(1)}), and Lemma \ref{L1}.
\noindent (\ref{primecblp2}) By (\ref{primecblp1}) and Lemma \ref{folclor}, (\ref{folclor2}).\end{proof}
\begin{lemma} ${\cal B}({\rm Con}({\cal A}))\subseteq {\cal K}({\cal A})$.\label{boolfg}\end{lemma}
\begin{proof} Let $\phi \in {\cal B}({\rm Con}({\cal A}))$. Then $\displaystyle \phi =Cg(\phi )=\bigvee _{(a,b)\in \phi }Cg(a,b)$, $\phi \cap \neg \, \phi =\Delta _{\cal A}$ and $\phi \vee \neg \, \phi =\nabla _{\cal A}$, so, by the hypothesis (H), there exists a finite set $X\subseteq \phi $ such that $Cg(X)\vee \neg \, \phi =\nabla _{\cal A}$. But then $Cg(X)\subseteq \phi $, thus $Cg(X)\cap \neg \, \phi \subseteq \phi \cap \neg \, \phi =\Delta _{\cal A}$, hence we also have $Cg(X)\cap \neg \, \phi =\Delta _{\cal A}$. Therefore $Cg(X)=\neg \, \neg \, \phi =\phi $, so $\phi =Cg(X)\in {\cal K}({\cal A})$, hence ${\cal B}({\rm Con}({\cal A}))\subseteq {\cal K}({\cal A})$.\end{proof}
\begin{corollary}
$u_{\textstyle \theta }({\cal K}({\cal A}))={\cal K}({\cal A}/\theta )$.\label{corsuper}\end{corollary}
\begin{proof} By Lemma \ref{superlema}, both inclusions hold.\end{proof}
\begin{proposition}\begin{enumerate}
\item\label{prop1(1)} If ${\cal B}({\rm Con}({\cal A}))={\cal K}({\cal A})$ and ${\cal B}({\rm Con}({\cal A}/\theta ))={\cal K}({\cal A}/\theta )$, then $\theta $ has CBLP.
\item\label{prop1(2)}
If every non--empty algebra ${\cal N}$ from ${\cal C}$ has ${\cal B}({\rm Con}({\cal N}))={\cal K}({\cal N})$, then every non--empty algebra from ${\cal C}$ has CBLP.\end{enumerate}\label{prop1}\end{proposition}
\begin{proof} (\ref{prop1(1)}) If ${\cal A}$ and ${\cal A}/\theta $ are such that their Boolean congruences coincide to their compact congruences, then, by Corollary \ref{corsuper}, ${\cal B}({\rm Con}({\cal A}/\theta ))={\cal K}({\cal A}/\theta )=u_{\textstyle \theta }({\cal K}({\cal A}))=u_{\textstyle \theta }({\cal B}({\rm Con}({\cal A}/\theta )))={\cal B}(u_{\textstyle \theta })({\cal B}({\rm Con}({\cal A}/\theta )))$, thus $\theta $ has CBLP.
\noindent (\ref{prop1(2)}) If every non--empty algebra ${\cal N}$ from ${\cal C}$ has ${\cal B}({\rm Con}({\cal N}))={\cal K}({\cal N})$, then ${\cal B}({\rm Con}({\cal A}))={\cal K}({\cal A})$ and, for each $\phi \in {\rm Con}({\cal A})$, ${\cal B}({\rm Con}({\cal A}/\phi ))={\cal K}({\cal A}/\phi )$, hence, according to (\ref{prop1(1)}), each $\phi \in {\rm Con}({\cal A})$ has CBLP, that is ${\cal A}$ has CBLP. Since ${\cal A}$ is an arbitrary non--empty algebra from ${\cal C}$, it follows that every non--empty algebra from ${\cal C}$ has CBLP.\end{proof}
\begin{corollary}
Any bounded distributive lattice has CBLP.\label{d01cblp}\end{corollary}
\begin{proof} By Proposition \ref{prop1} and the fact that, if ${\cal A}$ is a bounded distributive lattice, then, according to \cite[p. $127$]{blyth}, $\displaystyle {\cal B}({\rm Con}({\cal A}))=\{\bigvee _{i=1}^nCg(a_i,b_i)\ |\ n\in \N ^*,(\forall \, i\in \overline{1,n})\, (a_i,b_i\in A)\}={\cal K}({\cal A})$.\end{proof}
\begin{remark} In bounded non--distributive lattices, the CBLP is neither always present, nor always absent. Indeed, let ${\cal D}$ be the diamond and ${\cal P}$ be the pentagon, with the elements denoted as in the following Hasse diagrams:\vspace*{-15pt}
\begin{center}
\begin{tabular}{ccccc}
\begin{picture}(40,100)(0,0)
\put(30,25){\line(0,1){40}}
\put(30,25){\line(1,1){20}}
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\put(30,65){\line(1,-1){20}}
\put(30,65){\line(-1,-1){20}}
\put(30,25){\circle*{3}}
\put(10,45){\circle*{3}}
\put(3,42){$a$}
\put(30,45){\circle*{3}}
\put(33,42){$b$}
\put(50,45){\circle*{3}}
\put(53,42){$c$}
\put(30,65){\circle*{3}}
\put(28,15){$0$}
\put(28,68){$1$}
\put(25,0){${\cal D}$}
\end{picture}
&\hspace*{10pt}
\begin{picture}(40,100)(0,0)
\put(30,25){\line(1,1){10}}
\put(30,25){\line(-1,1){20}}
\put(30,65){\line(1,-1){10}}
\put(30,65){\line(-1,-1){20}}
\put(40,35){\line(0,1){20}}
\put(30,25){\circle*{3}}
\put(10,45){\circle*{3}}
\put(3,43){$x$}
\put(40,35){\circle*{3}}
\put(43,33){$y$}
\put(30,65){\circle*{3}}
\put(40,55){\circle*{3}}
\put(43,53){$z$}
\put(25,0){${\cal P}$}
\put(28,15){$0$}
\put(28,68){$1$}
\end{picture}
&\hspace*{30pt}
\begin{picture}(40,100)(0,0)
\put(20,30){\line(0,1){15}}
\put(20,30){\circle*{3}}
\put(20,45){\circle*{3}}
\put(16,20){$\Delta _{\cal D}$}
\put(16,48){$\nabla _{\cal D}$}
\put(5,0){${\rm Con}({\cal D})$}
\end{picture}
&\hspace*{10pt}
\begin{picture}(40,100)(0,0)
\put(20,45){\line(1,1){10}}
\put(20,45){\line(-1,1){10}}
\put(20,65){\line(1,-1){10}}
\put(20,65){\line(-1,-1){10}}
\put(20,45){\line(0,-1){15}}
\put(16,20){$\Delta _{\cal P}$}
\put(16,68){$\nabla _{\cal P}$}
\put(20,30){\circle*{3}}
\put(23,40){$\gamma $}
\put(2,53){$\alpha $}
\put(32,52){$\beta $}
\put(20,45){\circle*{3}}
\put(10,55){\circle*{3}}
\put(30,55){\circle*{3}}
\put(20,65){\circle*{3}}
\put(5,0){${\rm Con}({\cal P})$}
\end{picture}
&\hspace*{30pt}
\begin{picture}(40,100)(0,0)
\put(20,25){\line(1,1){10}}
\put(20,25){\line(-1,1){10}}
\put(20,45){\line(1,-1){10}}
\put(20,45){\line(-1,-1){10}}
\put(20,25){\circle*{3}}
\put(-9,33){$x/\gamma $}
\put(33,33){$y/\gamma =z/\gamma $}
\put(10,35){\circle*{3}}
\put(30,35){\circle*{3}}
\put(20,45){\circle*{3}}
\put(13,15){$0/\gamma $}
\put(13,48){$1/\gamma $}
\put(12,0){${\cal P}/\gamma $}
\end{picture}
\end{tabular}
\end{center}\vspace*{-5pt}
Let us denote, for any set $M$ and any partition $\pi $ of $M$, by $eq(\pi )$ the equivalence on $M$ which corresponds to $\pi $; also, if $\pi =\{M_1,\ldots ,M_n\}$ for some $n\in \N ^*$, then we shall denote by $eq(M_1,\ldots ,M_n)=eq(\pi )$.
The well--known fact that the classes of a congruence of a lattice $L$ are convex sublattices of $L$ make it easy to prove that ${\rm Con}({\cal D})=\{\Delta _{\cal D},\nabla _{\cal D}\}$, which is isomorphic to the two--element Boolean algebra, ${\cal L}_2$, and ${\rm Con}({\cal P})=\{\Delta _{\cal P},\alpha ,\beta ,\gamma ,\nabla _{\cal P}\}$, where $\alpha =eq(\{0,y,z\},\{x,1\})$, $\beta =eq(\{0,x\},\{y,z,1\})$ and $\gamma =eq(\{0\},\{x\},\{y,z\},\{1\})$, with the lattice structure represented above.
Thus ${\cal B}({\rm Con}({\cal D}))={\rm Con}({\cal D})=\{\Delta _{\cal D},\nabla _{\cal D}\}$, hence ${\cal D}$ has CBLP by Remark \ref{deltanabla}. The lattice structure of ${\rm Con}({\cal P})$ is the one represented above. By Remark \ref{deltanabla}, $\Delta _{\cal P}$ and $\nabla _{\cal P}$ have CBLP. $[\alpha )$ and $[\beta )$ are isomorphic to the standard Boolean algebra, ${\cal L}_2$: $[\alpha )=\{\alpha ,\nabla _{\cal P}\}$ and $[\beta )=\{\beta ,\nabla _{\cal P}\}$, thus ${\cal B}([\alpha ))=[\alpha )=\{\alpha ,\nabla _{\cal P}\}$ and ${\cal B}([\beta ))=[\beta )=\{\beta ,\nabla _{\cal P}\}$, hence $\alpha $ and $\beta $ have CBLP by Lemma \ref{L1}, (\ref{L1(1)}). But ${\cal P}/\gamma $ is isomorphic to the four--element Boolean algebra, ${\cal L}_2^2$, which, being a finite Boolean algebra, is isomorphic to its congruence lattice, so ${\rm Con}({\cal P}/\gamma )$ is isomorphic to ${\cal L}_2^2$, thus ${\cal B}({\rm Con}({\cal P}/\gamma ))={\rm Con}({\cal P}/\gamma )$ is isomorphic to ${\cal L}_2^2$, while the lattice structure of ${\rm Con}({\cal P})$ shows that ${\cal B}({\rm Con}({\cal P}))=\{\Delta _{\cal P},\nabla _{\cal P}\}$, which is isomorphic to ${\cal L}_2$, thus ${\cal B}(u_{\gamma }):{\cal B}({\rm Con}({\cal P}))\rightarrow {\cal B}({\rm Con}({\cal P}/\gamma ))$ can not be surjective, which means that $\gamma $ does not have CBLP. Therefore ${\cal P}$ does not have CBLP.\label{exdsip}\end{remark}
Now let us recall some definitions and results from \cite[Chapter $4$]{bj} and \cite[Chapter IV, Section $9$]{bur} concerning discriminator varieties. The {\em discriminator function} on a set $A$ is the mapping $t:A^3\rightarrow A$ defined by: for all $a,b,c\in A$,$$t(a,b,c)=\begin{cases}a, & \mbox{if }a\neq b,\\ c, & \mbox{if }a=b.\end{cases}$$A {\em discriminator term} on the algebra ${\cal A}$ is a term from the first order language associated to $\tau $ with the property that $t^{\cal A}$ is the discriminator function on $A$. The algebra ${\cal A}$ is called a {\em discriminator algebra} iff there exists a discriminator term on ${\cal A}$. An equational class ${\cal D}$ is called a {\em discriminator equational class} iff it is generated by a class of algebras which have a common discriminator term (equivalently, iff the subdirectly irreducible algebras from ${\cal D}$ have a common discriminator term).
\begin{proposition}{\rm \cite{bj}} Let ${\cal D}$ be a discriminator equational class and ${\cal A}$ be an algebra from ${\cal D}$. Then:\begin{itemize}
\item\label{discriminator1} ${\cal A}$ is an arithmetical algebra;
\item\label{discriminator2} any compact congruence of ${\cal A}$ is principal;
\item\label{discriminator3} any principal congruence of ${\cal A}$ is a factor congruence.\end{itemize}\label{discriminator}\end{proposition}
\begin{corollary}
All non--empty algebras from a discriminator equational class which satisfy (H) have CBLP.\label{cordiscrim}\end{corollary}
\begin{proof} Let ${\cal D}$ be a discriminator equational class and ${\cal A}$ be an algebra from ${\cal D}$ which satisfies (H). Then, by Proposition \ref{discriminator}, ${\cal A}$ is an arithmetical algebra, thus its set of factor congruences coincides to ${\cal B}({\rm Con}({\cal A}))$, hence ${\cal K}({\cal A})\subseteq {\cal B}({\rm Con}({\cal A}))$. By Lemma \ref{boolfg}, the converse inclusion holds, as well. Therefore ${\cal B}({\rm Con}({\cal A}))={\cal K}({\cal A})$, so ${\cal A}$ has CBLP by Proposition \ref{prop1}.\end{proof}
\begin{remark}
Among the discriminator equational classes with all members satisfying (H), there are important classes of algebras of logic such as: Boolean algebras, Post algebras, $n$--valued MV--algebras, monadic algebras, cylindric algebras etc.. Recently, in \cite{lpt}, it has been proven that ${\rm G\ddot{o}del}$ residuated lattices form a discriminator equational class. By Corollary \ref{cordiscrim}, it follows that all the algebras in these classes have CBLP.\label{remdiscrim}\end{remark}
From here until the end of this section, $\theta $ shall no longer be fixed.
For any $\theta \in {\rm Con}({\cal A})$, we shall denote by $V(\theta )=\{\pi \in {\rm Spec}({\cal A})\ |\ \theta \subseteq \pi \}$ and by $D(\theta )={\rm Spec}({\cal A})\setminus V(\theta )$.
\begin{lemma}
Let $\sigma ,\tau \in {\rm Con}({\cal A})$, $I$ be a non--empty set and $(\theta _i)_{i\in I}\subseteq {\rm Con}({\cal A})$. Then:\begin{enumerate}
\item\label{ltop1} $V(\sigma )=\emptyset $ iff $D(\sigma )={\rm Spec}({\cal A})$ iff $\sigma =\nabla _{\cal A}$;
\item\label{ltop2} $V(\sigma )={\rm Spec}({\cal A})$ iff $D(\sigma )=\emptyset $ iff $\sigma =\Delta _{\cal A}$;
\item\label{ltop3} $V(\sigma \cap \tau )=V(\sigma )\cup V(\tau )$ and $D(\sigma \cap \tau )=D(\sigma )\cap D(\tau )$;
\item\label{ltop4} $\displaystyle V(\bigvee _{i\in I}\theta _i)=\bigcap _{i\in I}V(\theta _i)$ and $\displaystyle D(\bigvee _{i\in I}\theta _i)=\bigcup _{i\in I}D(\theta _i)$;
\item\label{ltop5} $V(\sigma )\subseteq V(\tau )$ iff $D(\sigma )\supseteq D(\tau )$ iff $\sigma \supseteq \tau $;
\item\label{ltop6} $V(\sigma )=V(\tau )$ iff $D(\sigma )=D(\tau )$ iff $\sigma =\tau $;
\item\label{ltop7} $\{D(\theta )\ |\ \theta \in {\rm Con}({\cal A})\}$ is a topology on ${\rm Spec}({\cal A})$.\end{enumerate}\label{ltop}\end{lemma}
\begin{proof} (\ref{ltop1}) $D(\sigma )={\rm Spec}({\cal A})$ iff $V(\sigma )=\emptyset $ iff $\sigma =\nabla _{\cal A}$, according to Lemma \ref{folclor}, (\ref{folclor1}) and (\ref{folclor2}).
\noindent (\ref{ltop2}) By Lemma \ref{folclor}, (\ref{folclor3}), $\displaystyle \bigcap _{\pi \in {\rm Spec}({\cal A})}\pi =\Delta _{\cal A}$. $D(\sigma )=\emptyset $ iff $V(\sigma )={\rm Spec}({\cal A})$ iff $\sigma \subseteq \pi $ for all $\pi \in {\rm Spec}({\cal A})$ iff $\displaystyle \sigma \subseteq \bigcap _{\pi \in {\rm Spec}({\cal A})}\pi =\Delta _{\cal A}$ iff $\sigma =\Delta _{\cal A}$.
\noindent (\ref{ltop3}) Every $\pi \in {\rm Spec}({\cal A})$ satisfies: $\pi \in V(\sigma )\cup V(\tau )$ iff $\sigma \subseteq \pi $ or $\tau \subseteq \pi $ iff $\sigma \cap \tau \subseteq \pi $ iff $\pi \in V(\sigma \cap \tau )$. Thus $V(\sigma \cap \tau )=V(\sigma )\cup V(\tau )$, hence $D(\sigma \cap \tau )=D(\sigma )\cap D(\tau )$.
\noindent (\ref{ltop4}) Every $\phi \in {\rm Con}({\cal A})$ satisfies: $\displaystyle \phi \in \bigcap _{i\in I}V(\theta _i)$ iff, for all $i\in I$, $\theta _i\subseteq \phi $ iff $\displaystyle \bigvee _{i\in I}\theta _i\subseteq \phi $ iff $\displaystyle \phi \in V(\bigvee _{i\in I}\theta _i)$. Thus $\displaystyle V(\bigvee _{i\in I}\theta _i)=\bigcap _{i\in I}V(\theta _i)$, hence $\displaystyle D(\bigvee _{i\in I}\theta _i)=\bigcup _{i\in I}D(\theta _i)$.
\noindent (\ref{ltop5}) $D(\sigma )\supseteq D(\tau )$ iff $V(\sigma )\subseteq V(\tau )$. $\sigma \subseteq \tau $ clearly implies $V(\sigma )\subseteq V(\tau )$. $\tau \in V(\tau )$, therefore $V(\sigma )\subseteq V(\tau )$ implies $\tau \in V(\sigma )$, that is $\sigma \subseteq \tau $.
\noindent (\ref{ltop6}) By (\ref{ltop5}).
\noindent (\ref{ltop7}) By (\ref{ltop1}), (\ref{ltop2}), (\ref{ltop3}) and (\ref{ltop4}).\end{proof}
\begin{lemma}
Let $\sigma ,\tau \in {\rm Con}({\cal A})$. Then: $D(\sigma )=V(\tau )$ iff $\sigma ,\tau \in {\cal B}({\rm Con}({\cal A}))$ and $\tau =\neg \, \sigma $.\label{lprep}\end{lemma}
\begin{proof} By Lemma \ref{ltop}, (\ref{ltop1}), (\ref{ltop2}), (\ref{ltop3}), (\ref{ltop4}) and (\ref{ltop6}), the following hold: $D(\sigma )=V(\tau )$ iff $D(\sigma )={\rm Spec}({\cal A})\setminus D(\tau )$ iff $D(\sigma )\cup D(\tau )={\rm Spec}({\cal A})$ and $D(\sigma )\cap D(\tau )=\emptyset $ iff $D(\sigma \vee \tau )=D(\nabla _{\cal A})$ and $D(\sigma \cap \tau )=D(\Delta _{\cal A})$ iff $\sigma \vee \tau =\nabla _{\cal A}$ and $\sigma \cap \tau =\Delta _{\cal A}$ iff $\sigma ,\tau \in {\cal B}({\rm Con}({\cal A}))$ and $\tau =\neg \, \sigma $.\end{proof}
\begin{lemma}
The set of the clopen sets of the topological space $({\rm Spec}({\cal A}),\{D(\theta )\ |\ \theta \in {\rm Con}({\cal A})\})$ is $\{V(\alpha )\ |\ \alpha \in {\cal B}({\rm Con}({\cal A})\}))$.\label{lclp}\end{lemma}
\begin{proof} The set of the closed sets of the topological space $({\rm Spec}({\cal A}),\{D(\theta )\ |\ \theta \in {\rm Con}({\cal A})\})$ is $\{V(\theta )\ |\ \theta \in {\rm Con}({\cal A})\})$. Hence a subset $S\subseteq {\rm Spec}({\cal A})$ is clopen in this topological space iff $S=D(\sigma )=V(\tau )$ for some $\sigma ,\tau \in {\rm Con}({\cal A})$, which is equivalent to $\sigma ,\tau \in {\cal B}({\rm Con}({\cal A})$ and $\tau =\neg \, \sigma $ according to Lemma \ref{lprep}. Hence, by Lemma \ref{lprep}, $S$ is clopen iff $S=V(\tau )$ for some $\tau \in {\cal B}({\rm Con}({\cal A})$.\end{proof}
We recall that a topological space $(X,{\cal T})$ is said to be {\em strongly zero--dimensional} iff, for every $U,V\in {\cal T}$ such that $X=U\cup V$, there exist two clopen sets $C$ and $D$ of $(X,{\cal T})$ such that $C\subseteq U$, $D\subseteq V$, $C\cap D=\emptyset $ and $C\cup D=X$.
\begin{note}
The equivalence between statements (\ref{blpbnorm1}) and (\ref{blpbnorm2}) in the next proposition is implied by \cite[Lemma $4$]{banasch} in the particular case when the intersection in ${\rm Con}({\cal A})$ is completely distributive with respect to the join, but, for the sake of completeness, we shall provide a proof for it in our setting.\end{note}
\begin{proposition}
The following are equivalent:
\begin{enumerate}
\item\label{blpbnorm1} ${\cal A}$ has CBLP;
\item\label{blpbnorm2} the lattice ${\rm Con}({\cal A})$ is B--normal;
\item\label{blpbnorm0} for any $n\in \N ^*$ and every $\phi _1,\ldots ,\phi _n\in {\rm Con}({\cal A})$ such that $\phi _1\vee \ldots \vee \phi _n=\nabla _{\cal A}$, there exist $\alpha _1,\ldots ,\alpha _n\in {\cal B}({\rm Con}({\cal A}))$ such that $\alpha _1\cap \ldots \cap \alpha _n=\Delta _{\cal A}$ and $\phi _1\vee \alpha _1=\ldots =\phi _n\vee \alpha _n=\nabla _{\cal A}$;
\item\label{blpbnorm3} for every $\phi, \psi \in {\cal K}({\cal A})$ such that $\phi \vee \psi =\nabla _{\cal A}$, there exist $\alpha ,\beta \in {\cal B}({\rm Con}({\cal A}))$ such that $\alpha \cap \beta =\Delta _{\cal A}$ and $\phi \vee \alpha =\psi \vee \beta =\nabla _{\cal A}$;
\item\label{blpbnorm5} for any $n\in \N ^*$ and every $\phi _1,\ldots ,\phi _n\in {\cal K}({\cal A})$ such that $\phi _1\vee \ldots \vee \phi _n=\nabla _{\cal A}$, there exist $\alpha _1,\ldots ,\alpha _n\in {\cal B}({\rm Con}({\cal A}))$ such that $\alpha _1\cap \ldots \cap \alpha _n=\Delta _{\cal A}$ and $\phi _1\vee \alpha _1=\ldots =\phi _n\vee \alpha _n=\nabla _{\cal A}$;
\item\label{blpbnorm6} the topological space $({\rm Spec}({\cal A}),\{D(\theta )\ |\ \theta \in {\rm Con}({\cal A})\})$ is strongly zero--dimensional.\end{enumerate}\label{blpbnorm}\end{proposition}
\begin{proof} (\ref{blpbnorm1})$\Rightarrow $(\ref{blpbnorm2}): Let $\phi ,\psi \in {\rm Con}({\cal A})$ such that $\phi \cap \psi =\nabla _{\cal A}$. Denote $v={\cal B}(v_{(\phi \cap \psi )}):{\cal B}({\rm Con}({\cal A}))\rightarrow {\cal B}([\phi \vee \psi ))$. By Remark \ref{usiv}, the Boolean morphism $v$ is surjective. $\varphi ,\psi \in {\cal B}([\phi \vee \psi ))$, because $\psi $ is the complement of $\phi $ in the lattice $[\phi \vee \psi )$, that is $\psi =\neg _{(\phi \vee \psi )}\, \phi $. Thus there exists $\alpha \in {\cal B}({\cal A})$ such that $\alpha \vee (\psi \cap \phi )=v(\alpha )=\phi =\neg _{(\phi \vee \psi )}\, \psi $, hence $\psi =\neg _{(\phi \vee \psi )}\, \phi =\neg _{(\phi \vee \psi )}\, v(\alpha )=v(\neg \, \alpha )=\neg \, \alpha\vee (\phi \cap \psi )$. Therefore $\phi \vee \neg \, \alpha =\alpha \vee (\psi \cap \phi )\vee \neg \, \alpha =(\psi \cap \phi )\vee \alpha \vee \neg \, \alpha =(\psi \cap \phi )\vee \nabla _{\cal A}=\nabla _{\cal A}$, $\psi \vee \alpha =\neg \, \alpha\vee (\phi \cap \psi )\vee \alpha =(\phi \cap \psi )\vee \alpha \vee \neg \, \alpha =(\phi \cap \psi )\vee \nabla _{\cal A}=\nabla _{\cal A}$ and, of course, $\alpha \cap \neg \, \alpha =\Delta _{\cal A}$. So ${\rm Con}({\cal A})$ is B--normal.
\noindent (\ref{blpbnorm2})$\Rightarrow $(\ref{blpbnorm1}): Let $\theta \in {\rm Con}({\cal A})$ and let us denote by $v={\cal B}(v_{\theta }):{\cal B}({\rm Con}({\cal A}))\rightarrow {\cal B}([\theta ))$. Let $\phi \in {\cal B}([\theta ))$, so that there exists $\psi \in {\rm Con}({\cal A})$ such that $\phi \vee \psi =\nabla _{\cal A}$ and $\phi \wedge \psi =\theta $, that is $\psi =\neg _{\theta }\, \phi $. Since $\phi \vee \psi =\nabla _{\cal A}$ and ${\rm Con}({\cal A})$ is B--normal, it follows that there exist $\alpha ,\beta \in {\cal B}({\rm Con}({\cal A}))$ such that $\alpha \cap \beta =\Delta _{\cal A}$ and $\alpha \vee \phi =\beta \vee \psi =\nabla _{\cal A}$. Then $\theta \vee \alpha =(\phi \cap \psi )\vee \alpha =(\phi \vee \alpha )\cap (\psi \vee \alpha )=\nabla _{\cal A}\cap (\psi \vee \alpha )=\psi \vee \alpha $. $\alpha \cap \beta =\Delta _{\cal A}$ in the Boolean algebra ${\cal B}({\rm Con}({\cal A}))$, thus $\beta \leq \neg \, \alpha $, so, since $\beta \vee \psi =\nabla _{\cal A}$, it follows that $\nabla _{\cal A}=\beta \vee \psi \leq \neg \, \alpha \vee \psi $, hence $\neg \, \alpha \vee \psi =\nabla _{\cal A}$. Therefore $\alpha =\alpha \cap (\neg \, \alpha \vee \psi)=(\alpha \cap \neg \, \alpha )\vee (\alpha \cap \psi )=\Delta _{\cal A}\vee (\alpha \cap \psi )=\alpha \cap \psi $, thus $\alpha \leq \psi $, so $\psi \vee \alpha =\psi $. Hence $\theta \vee \alpha =\psi \vee \alpha =\psi $, so that $v(\alpha )=\theta \vee \alpha =\psi $, thus $v(\neg \, \alpha )=\neg _{\theta }\, v(\alpha )=\neg _{\theta }\, \psi =\phi $. Hence $v$ is surjective, that is $\theta $ has CBLP. Therefore ${\cal A}$ has CBLP.
\noindent (\ref{blpbnorm2})$\Rightarrow $(\ref{blpbnorm3}): Trivial.
\noindent (\ref{blpbnorm3})$\Rightarrow $(\ref{blpbnorm2}): Let $\phi, \psi \in {\rm Con}({\cal A})$ such that $\phi \vee \psi =\nabla _{\cal A}$. Since the lattice ${\rm Con}({\cal A})$ is algebraic, it follows that there exist $(\phi _i)_{i\in I}\subseteq {\cal K}({\cal A})$ and $(\psi _j)_{j\in J}\subseteq {\cal K}({\cal A})$ such that $\displaystyle \phi =\bigvee _{i\in I}\phi _i$ and $\displaystyle \psi =\bigvee _{j\in J}\psi _j$, thus $\displaystyle \bigvee _{i\in I}\phi _i\vee \bigvee _{j\in J}\psi _j=\nabla _{\cal A}$. Since $\nabla _{\cal A}$ is compact, it follows that there exist $I_0\subseteq I$ and $J_0\subseteq J$ such that $I_0$ and $J_0$ are finite and $\displaystyle \bigvee _{i\in I_0}\phi _i\vee \bigvee _{j\in J_0}\psi _j=\nabla _{\cal A}$. Let $\displaystyle \phi _0=\bigvee _{i\in I_0}\phi _i$ and $\displaystyle \psi _0=\bigvee _{j\in J_0}\psi _j$. Then $\phi _0\vee \psi _0=\nabla _{\cal A}$ and, clearly, $\phi _0,\psi _0\in {\cal K}({\cal A})$. It follows that there exist $\alpha ,\beta \in {\cal B}({\rm Con}({\cal A}))$ such that $\alpha \cap \beta =\Delta _{\cal A}$ and $\phi _0\vee \alpha =\psi _0\vee \beta =\nabla _{\cal A}$. Since, obviously, $\phi _0\subseteq \phi$ and $\psi _0\subseteq \psi$, we obtain $\phi \vee \alpha =\psi \vee \beta =\nabla _{\cal A}$.
\noindent (\ref{blpbnorm0})$\Rightarrow $(\ref{blpbnorm2}): Trivial: just take $n=2$.
\noindent (\ref{blpbnorm2})$\Rightarrow $(\ref{blpbnorm0}): Assume that the lattice ${\rm Con}({\cal A})$ is B--normal, and let $n\in \N ^*$ and $\phi _1,\ldots ,\phi _n\in {\rm Con}({\cal A})$ such that $\phi _1\vee \ldots \vee \phi _n=\nabla _{\cal A}$. Then, by \cite[Proposition $12$]{blpiasi}, it follows that there exist $\beta _1,\ldots ,\beta _n\in {\cal B}({\rm Con}({\cal A}))$ such that $\beta _1\vee \ldots \vee \beta _n=\nabla _{\cal A}$, $\beta _i\cap \beta _j=\Delta _{\cal A}$ for all $i,j\in \overline{1,n}$ with $i\neq j$ and $\phi _i\supseteq \beta _i$ for all $i\in \overline{1,n}$. For all $i\in \overline{1,n}$, let $\alpha _i=\neg \, \beta _i\in {\cal B}({\rm Con}({\cal A}))$. Then $\alpha _1\cap \ldots \cap \alpha _n=\neg \, (\beta _1\vee \ldots \vee \beta _n)=\neg \, \nabla _{\cal A}=\Delta _{\cal A}$ and, for all $i\in \overline{1,n}$, $\phi _i\vee \alpha _i\geq \beta _i\vee \alpha _i=\beta _i\vee \neg \, \beta _i=\nabla _{\cal A}$, thus $\phi _i\vee \alpha _i=\nabla _{\cal A}$.
\noindent (\ref{blpbnorm5})$\Rightarrow $(\ref{blpbnorm3}): Trivial: just take $n=2$.
\noindent (\ref{blpbnorm3})$\Rightarrow $(\ref{blpbnorm5}): We apply induction on $n\in \N ^*$. For $n=1$, we have $\phi _1=\nabla _{\cal A}$. We may take $\alpha _1=\Delta _{\cal A}\in {\cal B}({\rm Con}({\cal A}))$, and we get $\phi _1\vee \alpha _1=\nabla _{\cal A}\vee \Delta _{\cal A}=\nabla _{\cal A}$.
Now assume that the statement in (\ref{blpbnorm5}) is valid for some $n\in \N ^*$, and let $\phi _1,\ldots ,\phi _{n+1}\in {\cal K}({\cal A})$ such that $\phi _1\vee \ldots \vee \phi _{n+1}=\nabla _{\cal A}$. Then $\phi _1\vee \ldots \vee \phi _n\in {\cal K}({\cal A})$ and $(\phi _1\vee \ldots \vee \phi _n)\vee \phi _{n+1}=\nabla _{\cal A}$, so, by the hypothesis (\ref{blpbnorm3}), it follows that there exist $\alpha ,\beta \in {\cal B}({\rm Con}({\cal A}))$ such that $\alpha \cap \beta =\Delta _{\cal A}$, $\phi _{n+1}\vee \beta =\nabla _{\cal A}$ and $(\phi _1\vee \alpha )\vee \ldots \vee (\phi _n\vee \alpha )=\phi _1\vee \ldots \vee \phi _n\vee \alpha =\nabla _{\cal A}$. Then $\alpha \in {\cal K}({\cal A})$ by Lemma \ref{boolfg}, hence $\phi _1\vee \alpha ,\ldots ,\phi _n\vee \alpha \in {\cal K}({\cal A})$, thus, by the induction hypothesis, it follows that there exist $\gamma _1,\ldots ,\gamma _n\in {\cal B}({\rm Con}({\cal A}))$ such that $\gamma _1\cap \ldots \cap \gamma _n=\Delta _{\cal A}$ and $\phi _1\vee \alpha \vee \gamma _1=\ldots =\phi _n\vee \alpha \vee \gamma _n=\nabla _{\cal A}$. Let $\alpha _{n+1}=\beta \in {\cal B}({\rm Con}({\cal A}))$ and, for all $i\in \overline{1,n}$, $\alpha _i=\alpha \vee \gamma _i\in {\cal B}({\rm Con}({\cal A}))$. Then, for all $i\in \overline{1,n+1}$, $\phi _i\vee \alpha _i=\nabla _{\cal A}$, and $\alpha _1\cap \ldots \cap \alpha _{n+1}=(\alpha \vee \gamma _1)\cap (\alpha \vee \gamma _n)\cap \beta =[\alpha \vee (\gamma _1\cap \ldots \cap \gamma _n)]\cap \beta =(\alpha \vee \Delta _{\cal A})\cap \beta =\alpha \cap \beta =\Delta _{\cal A}$.
\noindent (\ref{blpbnorm2})$\Rightarrow $(\ref{blpbnorm6}): Let $U,V\in \{D(\theta )\ |\ \theta \in {\rm Con}({\cal A})\}$ such that $U\cup V={\rm Spec}({\cal A})$, that is $U=D(\sigma )$ and $V=D(\tau )$ for some $\sigma ,\tau \in {\rm Con}({\cal A})$ such that $D(\nabla _{\cal A})={\rm Spec}({\cal A})=D(\sigma )\cup D(\tau )=D(\sigma \vee \tau )$, which means that $\sigma \vee \tau =\nabla _{\cal A}$, according to Lemma \ref{ltop}, (\ref{ltop1}), (\ref{ltop4}) and (\ref{ltop6}). Since the lattice ${\rm Con}({\cal A})$ is B--normal, it follows that there exist $\alpha ,\beta \in {\cal B}({\rm Con}({\cal A}))$ such that $\alpha \cap \beta =\Delta _{\cal A}$ and $\alpha \vee \tau =\beta \vee \sigma =\nabla _{\cal A}$, hence $\beta \subseteq \neg \, \alpha $ and thus $\neg \, \alpha \vee \sigma =\nabla _{\cal A}$. We have obtained $\alpha \vee \tau =\neg \, \alpha \vee \sigma =\nabla _{\cal A}$, thus $\neg \, \alpha \subseteq \tau $ and $\alpha =\neg \, \neg \, \alpha \subseteq \sigma $, therefore $D(\neg \, \alpha )\subseteq D(\tau )=V$ and $D(\alpha )\subseteq D(\sigma )=U$, by Lemma \ref{ltop}, (\ref{ltop5}). By Lemma \ref{ltop}, (\ref{ltop1}), (\ref{ltop2}), (\ref{ltop3}) and (\ref{ltop4}), $D(\alpha )\cap D(\neg \, \alpha )=D(\alpha \cap \neg \, \alpha )=D(\Delta _{\cal A})=\emptyset $ and $D(\alpha )\cup D(\neg \, \alpha )=D(\alpha \vee \neg \, \alpha )=D(\nabla _{\cal A})={\rm Spec}({\cal A})$. By Lemma \ref{lclp}, $D(\alpha )$ and $D(\neg \, \alpha )$ are clopen sets of the topological space $({\rm Spec}({\cal A}),\{D(\theta )\ |\ \theta \in {\rm Con}({\cal A})\})$. Therefore this topological space is strongly zero--dimensional.
\noindent (\ref{blpbnorm6})$\Rightarrow $(\ref{blpbnorm2}): Let $\sigma ,\tau \in {\rm Con}({\cal A})$ such that $\sigma \vee \tau =\nabla _{\cal A}$. Then $D(\sigma )\cup D(\tau )=D(\sigma \vee \tau )=D(\nabla _{\cal A})={\rm Spec}({\cal A})$, according to Lemma \ref{ltop}, (\ref{ltop4}) and (\ref{ltop1}). By the hypothesis of this implication and Lemma \ref{lclp}, it follows that there exist $\alpha ,\beta \in {\cal B}({\rm Con}({\cal A}))$ such that $D(\alpha )\subseteq D(\sigma )$, $D(\beta )\subseteq D(\tau )$, $D(\alpha )\cap D(\beta )=\emptyset $ and $D(\alpha )\cup D(\beta )={\rm Spec}({\cal A})$. Then $D(\alpha )\cup D(\tau )=D(\beta )\cup D(\sigma )={\rm Spec}({\cal A})$. By Lemma \ref{ltop}, (\ref{ltop1}), (\ref{ltop2}), (\ref{ltop3}), (\ref{ltop4}) and (\ref{ltop6}), it follows that $D(\alpha \cap \beta )=D(\Delta _{\cal A})$ and $D(\alpha \vee \tau )=D(\beta \vee \sigma )=D(\nabla _{\cal A})$, thus $\alpha \cap \beta =\Delta _{\cal A}$ and $\alpha \vee \tau =\beta \vee \sigma =\nabla _{\cal A}$, hence the lattice ${\rm Con}({\cal A})$ is B--normal.\end{proof}
\begin{corollary} ${\cal A}$ has CBLP iff, for all $\theta \in {\rm Con}({\cal A})$, ${\cal A}/\theta $ has CBLP.\label{corolar4.7}\end{corollary}
\begin{proof} Assume that ${\cal A}$ has CBLP, which means that ${\cal A}$ is B--normal, according to Proposition \ref{blpbnorm}. Let $\theta \in {\rm Con}({\cal A})$. Let us prove that the lattice $[\theta )$ is B--normal. So let $\phi ,\psi \in [\theta )$ such that $\phi \vee \psi =\nabla _{\cal A}$. Then, since ${\cal A}$ is B--normal, it follows that there exist $\alpha ,\beta \in {\cal B}({\rm Con}({\cal A}))$ such that $\alpha \cap \beta =\Delta _{\cal A}$ and $\phi \vee \alpha =\psi \vee \beta =\nabla _{\cal A}$. Let us consider the Boolean morphism ${\cal B}(v_{\textstyle \theta }):{\cal B}({\rm Con}({\cal A}))\rightarrow {\cal B}([\theta ))$. We have: ${\cal B}(v_{\textstyle \theta })(\alpha ),{\cal B}(v_{\textstyle \theta })(\beta )\in {\cal B}([\theta ))$, ${\cal B}(v_{\textstyle \theta })(\alpha )\cap {\cal B}(v_{\textstyle \theta })(\beta )={\cal B}(v_{\textstyle \theta })(\alpha \cap \beta )={\cal B}(v_{\textstyle \theta })(\Delta _{\cal A})=\theta $, ${\cal B}(v_{\textstyle \theta })(\phi )\vee {\cal B}(v_{\textstyle \theta })(\alpha )={\cal B}(v_{\textstyle \theta })(\phi \vee \alpha )={\cal B}(v_{\textstyle \theta })(\nabla _{\cal A})=\nabla _{\cal A}$ and ${\cal B}(v_{\textstyle \theta })(\psi )\vee {\cal B}(v_{\textstyle \theta })(\beta )={\cal B}(v_{\textstyle \theta })(\psi \vee \beta )={\cal B}(v_{\textstyle \theta })(\nabla _{\cal A})=\nabla _{\cal A}$, hence the lattice $[\theta )$ is B--normal. Since the lattices ${\rm Con}({\cal A}/\theta )$ and $[\theta )$ are isomorphic, it follows that ${\rm Con}({\cal A}/\theta )$ is B--normal, hence ${\cal A}/\theta $ has CBLP, according to Proposition \ref{blpbnorm}. For the converse implication, just take $\theta =\Delta _{\cal A}$, so that ${\cal A}/\theta ={\cal A}/\Delta _{\cal A}$ is isomorphic to ${\cal A}$.\end{proof}
\begin{corollary}
Let $n\in \N ^*$, ${\cal A}_1,\ldots ,{\cal A}_n$ be algebras and $\displaystyle {\cal A}=\prod _{i=1}^n{\cal A}_i$. Then: ${\cal A}$ has CBLP iff, for all $i\in \overline{1,n}$, ${\cal A}_i$ has CBLP.\label{corolar4.8}\end{corollary}
\begin{proof} By Lemma \ref{totdinbj}, (\ref{totdinbj2}), ${\rm Con}({\cal A})$ is isomorphic to $\displaystyle \prod _{i=1}^n{\rm Con}({\cal A}_i)$. By \cite[Propositions $12$ and $13$]{blpiasi}, $\displaystyle \prod _{i=1}^n{\rm Con}({\cal A}_i)$ is B--normal iff, for all $i\in \overline{1,n}$, ${\rm Con}({\cal A}_i)$ is B--normal. By Proposition \ref{blpbnorm}, we obtain that: ${\cal A}$ has CBLP iff ${\rm Con}({\cal A})$ is B--normal iff, for all $i\in \overline{1,n}$, ${\rm Con}({\cal A}_i)$ is B--normal, iff, for all $i\in \overline{1,n}$, ${\cal A}_i$ has CBLP.\end{proof}
In the following results, we shall designate most lattices by their underlying sets.
\begin{remark} It is well known and straightforward that, if $L$ is a lattice, $S$ is a sublattice of $L$ and $\theta \in {\rm Con}(L)$, then $\theta \cap S^2\in {\rm Con}(S)$.\label{congrsl}\end{remark}
\begin{remark} Let $L$ be a lattice with $1$ and $M$ be a lattice with $0$. We shall denote by $L\dotplus M$ the ordinal sum between $L$ and $M$. Let $c$ be the common element of $L$ and $M$ in the lattice $L\dotplus M$. Using a notation from Remark \ref{exdsip}, for any $\phi \in {\rm Con}(L)$ and any $\psi \in {\rm Con}(M)$, we shall denote by $\phi \dotplus \psi =eq((L/\phi \setminus c/\phi )\cup \{c/\phi \cup c/\psi \}\cup (M/\psi \setminus c/\psi ))$.
\begin{enumerate}
\item\label{ordsum1} Then ${\rm Con}(L\dotplus M)=\{\phi \dotplus \psi \ |\ \phi \in {\rm Con}(L),\psi \in {\rm Con}(M)\}$.
Indeed, it is straightforward that, for any $\phi \in {\rm Con}(L)$ and any $\psi \in {\rm Con}(M)$, we have $\phi \dotplus \psi \in {\rm Con}(L\dotplus M)$, and the fact that $L\dotplus M$ has no other congruences follows from Remark \ref{congrsl}.
\item\label{ordsum2} It is easy to notice that $(L\dotplus M)/(\Delta _L\dotplus \nabla _M)\cong L$ and $(L\dotplus M)/(\nabla _L\dotplus \Delta _M)\cong M$.
\item\label{ordsum3} It is immediate that, for all $\alpha ,\phi \in {\rm Con}(L)$ and all $\beta ,\psi \in {\rm Con}(M)$, $\alpha \dotplus \beta \subseteq \phi \dotplus \phi $ iff $\alpha \subseteq \phi $ and $\beta \subseteq \psi $ iff $\alpha \times \beta \subseteq \phi \times \psi $, which, together with the form of ${\rm Con}(L\dotplus M)$ established above, shows that the mapping $\phi \dotplus \psi \mapsto \phi \times \psi $ is an order isomorphism, and thus a bounded lattice isomorphism between ${\rm Con}(L\dotplus M)$ and ${\rm Con}(L\times M)$, which, in turn, is isomorphic to ${\rm Con}(L)\times {\rm Con}(M)$ by Lemma \ref{totdinbj}, (\ref{totdinbj2}). From this and Proposition \ref{propO}, (\ref{propO1}), we deduce that the Boolean algebras ${\cal B}({\rm Con}(L\dotplus M))$, ${\cal B}({\rm Con}(L\times M))$ and ${\cal B}({\rm Con}(L))\times {\cal B}({\rm Con}(M))$ are isomorphic.\end{enumerate}
In what follows, we shall keep the notations from this remark.\label{ordsum}\end{remark}
\begin{corollary}\begin{enumerate}
\item\label{ordscudsip1} The lattice of congruences of any ordinal sum of finite lattices which are either distributive or isomorphic to the diamond is a Boolean algebra, hence any such ordinal sum has CBLP (regardless of whether it is finite). The lattice of congruences of any direct product of finite lattices which are either distributive or isomorphic to the diamond is a Boolean algebra, hence any such direct product has CBLP (regardless of whether it is finite).
\item\label{ordscudsip2} If $L$ is a lattice with $1$, $M$ is a lattice with $0$, and $L\dotplus M$ has the CBLP, then both $L$ and $M$ have the CBLP.
\item\label{ordscudsip3} Any ordinal sum of bounded lattices which contains the pentagon does not have the CBLP.\end{enumerate}\label{ordscudsip}\end{corollary}
\begin{proof} (\ref{ordscudsip1}) As we have seen in Remark \ref{exdsip}, ${\rm Con}({\cal D})$ is isomorphic to the two--element Boolean algebra. According to a result in \cite{blyth}, the lattice of congruences of any finite distributive lattice is a Boolean algebra. By Remark \ref{ordsum}, (\ref{ordsum3}), it follows that, if $L$ is an ordinal sum of finite lattices which are either distributive or isomorphic to ${\cal D}$, then ${\rm Con}(L)$ is a Boolean algebra, that is ${\cal B}({\rm Con}(L))={\rm Con}(L)$, hence $L$ has CBLP by Proposition \ref{toatebool}, (\ref{toatebool0}), and the same holds if if $L$ is a direct product of finite lattices which are either distributive or isomorphic to ${\cal D}$.
\noindent (\ref{ordscudsip2}) By Corollary \ref{corolar4.7} and Remark \ref{ordsum}, \ref{ordsum2}.
\noindent (\ref{ordscudsip3}) By (\ref{ordscudsip2}) and Remark \ref{exdsip}, in which we have shown that ${\cal P}$ does not have CBLP.\end{proof}
\begin{remark} If a lattice has CBLP, then its sublattices do not necessarily have CBLP. To illustrate this property, we provide an example of a non--modular lattice with CBLP. Let $E$ be the following bounded non--distributive lattice, in which ${\cal P}$ is embedded; we know from Remark \ref{exdsip} that ${\cal P}$ does not have the CBLP.\vspace*{-15pt}
\begin{center}
\begin{tabular}{cc}
\begin{picture}(40,100)(0,0)
\put(40,20){\line(0,1){60}}
\put(40,20){\line(1,1){30}}
\put(40,20){\line(-1,1){30}}
\put(40,80){\line(1,-1){30}}
\put(40,80){\line(-1,-1){30}}
\put(40,20){\circle*{3}}
\put(40,40){\circle*{3}}
\put(40,60){\circle*{3}}
\put(40,80){\circle*{3}}
\put(10,50){\circle*{3}}
\put(70,50){\circle*{3}}
\put(3,48){$a$}
\put(33,38){$b$}
\put(43,57){$d$}
\put(73,48){$c$}
\put(38,10){$0$}
\put(38,84){$1$}
\put(37,-2){$E$}\end{picture}
&\hspace*{45pt}
\begin{picture}(40,100)(0,0)
\put(20,25){\line(0,1){40}}
\put(20,25){\circle*{3}}
\put(20,45){\circle*{3}}
\put(20,65){\circle*{3}}
\put(16,15){$\Delta _E$}
\put(23,43){$\varepsilon $}
\put(16,68){$\nabla _E$}
\put(6,-2){${\rm Con}(E)$}
\end{picture}\end{tabular}\end{center}\vspace*{-5pt}
By using Remark \ref{congrsl} and the calculations in Remark \ref{exdsip}, it is easy to obtain that, if we denote by $\varepsilon =eq(\{0\},\{a\},\{b,d\},\{c\},\{1\})$, then ${\rm Con}(E)=\{\Delta _E,\varepsilon ,\nabla _E\}$, which is isomorphic to the three--element chain (hence ${\cal B}({\rm Con}(E))=\{\Delta _E,\nabla _E\}$ is isomorphic to the two--element Boolean algebra, but we do not even need its form). By Remark \ref{deltanabla}, $\Delta _E$ and $\nabla _E$ have CBLP. $E/\varepsilon $ is isomorphic to ${\cal D}$, hence, by Remark \ref{exdsip}, ${\rm Con}(E/\varepsilon )=\{\Delta _{ E/\varepsilon },\nabla _{E/\varepsilon }\}$, which is isomorphic to the two--element Boolean algebra, thus ${\cal B}({\rm Con}(E/\varepsilon ))={\rm Con}(E/\varepsilon )=\{\Delta _{E/\varepsilon },\nabla _{E/\varepsilon }\}$, hence $\varepsilon $ has CBLP by Lemma \ref{L2}, (\ref{L2(2)}).\end{remark}
\begin{corollary} If an algebra has CBLP, then its subalgebras do not necessarily have CBLP.\end{corollary}
\begin{remark} Let $\theta \in {\rm Con}({\cal A})$. Then the inequality of cardinalities $|{\cal B}({\rm Con}({\cal A}/\theta ))|\leq |{\cal B}({\rm Con}({\cal A}))|$ does not imply the surjectivity of ${\cal B}(u_{\theta })$, that is it does not imply that $\theta $ has CBLP.
Indeed, let $Z$ be the ordinal sum between ${\cal P}$ and the two--element chain, ${\cal L}_2$, which, as Corollary \ref{ordscudsip}, (\ref{ordscudsip3}), ensures us, does not have CBLP:\vspace*{-8pt}
\begin{center}
\begin{tabular}{ccccc}
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\put(3,38){$x$}
\put(43,27){$y$}
\put(43,47){$z$}
\put(28,10){$0$}
\put(34,57){$u$}
\put(28,83){$1$}
\put(0,-17){$Z={\cal P}\dotplus {\cal L}_2$}
\end{picture}
&\hspace*{9pt}
\begin{picture}(60,100)(0,0)
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\put(0,63){$\zeta _1$}
\put(34,63){$\zeta _7$}
\put(54,63){$\zeta _8$}
\put(0,43){$\zeta _5$}
\put(34,43){$\zeta _6$}
\put(54,43){$\zeta _4$}
\put(28,15){$\zeta _3$}
\put(74,23){$\zeta _2$}
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\put(26,89){$\nabla _Z$}
\end{picture}
&\hspace*{14pt}
\begin{picture}(60,100)(0,0)
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\put(26,15){$\Delta _Z$}
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\put(6,-17){${\cal B}({\rm Con}(Z))$}\end{picture}
&\hspace*{9pt}
\begin{picture}(60,100)(0,0)
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\put(23,15){$0/\zeta _4$}
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\put(-10,42){$x/\zeta _4$}
\put(53,42){$y/\zeta _4=z/\zeta _4$}
\put(21,-17){$Z/\zeta _4$}\end{picture}
&\hspace*{44pt}
\begin{picture}(60,100)(0,0)
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\put(26,15){$\Delta _{Z/\zeta _4}$}
\put(26,68){$\nabla _{Z/\zeta _4}$}
\put(2,42){$\xi $}
\put(53,42){$\zeta $}
\put(-28,-17){${\rm Con}(Z/\zeta _4)={\cal B}({\rm Con}(Z/\zeta _4))$}\end{picture}\end{tabular}\end{center}\vspace*{12pt}
The congruences of $Z$ are easy to calculate by using Remark \ref{ordsum}, (\ref{ordsum1}), the calculations in Remark \ref{exdsip} and the fact that the finite Boolean algebra ${\cal L}_2$ is isomorphic to its lattice of congruences: ${\rm Con}({\cal L}_2)=\{\Delta _{{\cal L}_2},\nabla _{{\cal L}_2}\}$: ${\rm Con}(Z)=\{\Delta _Z,\zeta _1,\zeta _2,\zeta _3,\zeta _4,\zeta _5,\zeta _6,\zeta _7,\zeta _8,\nabla _Z \}$, where $\zeta _1=eq(\{0,x,y,z,u\},\{1\})$, $\zeta _2=eq(\{0\},\{x\},\{y\},\{z\},$\linebreak $\{u,1\})$, $\zeta _3=eq(\{0\},\{x\},\{y,z\},\{u\},\{1\})$, $\zeta _4=eq(\{0\},\{x\},\{y,z\},\{u,1\})$, $\zeta _5=eq(\{0,y,z\},\{x,u\},\{1\})$, $\zeta _6=eq(\{0,x\},\{y,z,u\},\{1\})$, $\zeta _7=eq(\{0,y,z\},\{x,u,1\})$ and $\zeta _8=eq(\{0,x\},\{y,z,u,1\})$, with the lattice structure represented above. Therefore ${\cal B}({\rm Con}(Z))=\{\Delta _Z,\zeta _1,\zeta _2,\nabla _Z\}$, which is isomorphic to the four--element Boolean algebra, ${\cal L}_2^2$. Now let us look at the congruence $\zeta _4$. $Z/\zeta _4$ is isomorphic to ${\cal L}_2^2$, hence it is isomorphic to its lattice of congruences: ${\rm Con}(Z/\zeta _4)={\cal B}({\rm Con}(Z/\zeta _4))=\{\Delta _{Z/\zeta _4},\xi ,\zeta ,\nabla _{Z/\zeta _4}\}$, where $\xi =eq(\{0/\zeta _4,x/\zeta _4\},\{y/\zeta _4,1/\zeta _4\})$ and $\zeta =eq(\{0/\zeta _4,y/\zeta _4\},\{x/\zeta _4,1/\zeta _4\})$. Thus ${\cal B}({\rm Con}(Z/\zeta _4))$ is isomorphic to ${\cal B}({\rm Con}(Z))$. Let us calculate ${\cal B}(u_{\zeta _4})$ in each element of ${\cal B}({\rm Con}(Z))$: ${\cal B}(u_{\zeta _4})(\Delta _Z)=u_{\zeta _4}(\Delta _Z)=\Delta _{Z/\zeta _4}$, ${\cal B}(u_{\zeta _4})(\nabla _Z)=u_{\zeta _4}(\nabla _Z)=\nabla _{Z/\zeta _4}$, ${\cal B}(u_{\zeta _4})(\zeta _1)=u_{\zeta _4}(\zeta _1)=(\zeta _1\vee \zeta _4)/\zeta _4=\nabla _Z/\zeta _4=\nabla _{Z/\zeta _4}$ and ${\cal B}(u_{\zeta _4})(\zeta _2)=u_{\zeta _4}(\zeta _2)=(\zeta _2\vee \zeta _4)/\zeta _4=\zeta _4/\zeta _4=\Delta _{Z/\zeta _4}$, hence ${\cal B}(u_{\zeta _4})({\cal B}({\rm Con}(Z)))=\{\Delta _{Z/\zeta _4},\nabla _{Z/\zeta _4}\}\subsetneq {\cal B}({\rm Con}(Z/\zeta _4))$, so ${\cal B}(u_{\zeta _4})$ is not surjective, which means that $\zeta _4$ does not have CBLP.\end{remark}
We say that ${\cal A}$ is {\em local} iff it has exactly one maximal congruence.
\begin{corollary}
If the algebra ${\cal A}$ is local, then the lattice ${\rm Con}({\cal A})$ is ${\rm Id}$--local.\label{corolar2.6}\end{corollary}
\begin{proof} Assume that ${\cal A}$ is local and let $\theta $ be the unique maximal congruence of ${\cal A}$. Let $\alpha ,\beta \in {\rm Con}({\cal A})$ such that $\alpha \vee \beta =\nabla _{\cal A}$. Assume by absurdum that $\alpha \neq \nabla _{\cal A}$ and $\beta \neq \nabla _{\cal A}$. Then, according to Lemma \ref{folclor}, (\ref{folclor1}), it follows that $\alpha \subseteq \theta $ and $\beta \subseteq \theta $, thus $\alpha \vee \beta \subseteq \theta $, which is a contradiction to the choice of $\alpha $ and $\beta $. Hence $\alpha =\nabla _{\cal A}$ or $\beta =\nabla _{\cal A}$.\end{proof}
\begin{lemma}
Any {\rm Id}--local lattice is B--normal.\label{lema2.8}\end{lemma}
\begin{proof} By Lemma \ref{lema2.5}.\end{proof}
\begin{corollary}
Any local algebra has CBLP.\label{corolar4.9}\end{corollary}
\begin{proof} Assume that ${\cal A}$ is a local algebra. Then, by Corollary \ref{corolar2.6}, Lemma \ref{lema2.8} and Proposition \ref{blpbnorm}, it follows that ${\rm Con}({\cal A})$ is {\rm Id}--local, thus it is B--normal, hence ${\cal A}$ has CBLP.\end{proof}
\begin{corollary}
Any finite direct product of local algebras has CBLP.\label{finprodloc}\end{corollary}
We recall that a {\em normal algebra} is an algebra whose lattice of congruences is normal.
\begin{remark}\begin{itemize}
\item Any algebra with CBLP is a normal algebra.
\item Any local algebra is a normal algebra.
\item Any bounded distributive lattice is a normal algebra.\end{itemize}
The first statement follows from Proposition \ref{blpbnorm}, while the second follows from the first and Corollary \ref{corolar4.9}, and the third follows from the first and Corollary \ref{d01cblp}.\label{remarca4.10}\end{remark}
\begin{lemma}
Let $\theta \in {\rm Con}({\cal A})$. If $\theta \vee {\rm Rad}({\cal A})=\nabla _{\cal A}$, then $\theta =\nabla _{\cal A}$.\label{lema4.11}\end{lemma}
\begin{proof} Let $\theta \in {\rm Con}({\cal A})$ such that $\theta \vee {\rm Rad}({\cal A})=\nabla _{\cal A}$. Since ${\rm Rad}({\cal A})\subseteq \phi $ for all $\phi \in {\rm Max}({\cal A})$, it follows that $\theta \vee \phi =\nabla _{\cal A}$ for all $\phi \in {\rm Max}({\cal A})$. Assume by absurdum that $\theta $ is a proper congruence of ${\cal A}$, so that $\theta \subseteq \phi _0$ for some $\phi _0\in {\rm Max}({\cal A})$, by Lemma \ref{folclor}, (\ref{folclor1}). But then it follows that $\phi _0=\theta \vee \phi _0=\nabla _{\cal A}$, which is a contradiction to the fact that $\phi _0$ is a maximal, and thus a proper congruence of ${\cal A}$. Therefore $\theta =\nabla _{\cal A}$.\end{proof}
\begin{proposition}
If ${\cal A}$ is a normal algebra, then ${\rm Rad}({\cal A})$ has CBLP.\label{propozitie4.12}\end{proposition}
\begin{proof} We shall consider the Boolean morphism ${\cal B}(v_{\textstyle {\rm Rad}({\cal A})}):{\cal B}({\rm Con}({\cal A}))\rightarrow {\cal B}([{\rm Rad}({\cal A})))$. Let $\phi \in {\cal B}([{\rm Rad}({\cal A})))$, so that $\phi \vee \psi =\nabla _{\cal A}$ and $\phi \cap \psi ={\rm Rad}({\cal A})$ for some $\psi \in {\rm Con}({\cal A})$. But the algebra ${\cal A}$ is normal, thus the lattice ${\rm Con}({\cal A})$ is normal, hence there exist $\alpha ,\beta \in {\rm Con}({\cal A})$ such that $\alpha \cap \beta =\Delta _{\cal A}$ and $\phi \vee \alpha =\psi \vee \beta =\nabla _{\cal A}$, thus $\alpha \vee \beta \vee \phi =\alpha \vee \beta \vee \psi =\nabla _{\cal A}$, so $\alpha \vee \beta \vee {\rm Rad}({\cal A})=\alpha \vee \beta \vee (\phi \wedge \psi )=(\alpha \vee \beta \vee \phi )\wedge (\alpha \vee \beta \vee \psi )=\nabla _{\cal A}\wedge \nabla _{\cal A}=\nabla _{\cal A}$, hence $\alpha \vee \beta =\nabla _{\cal A}$, by Lemma \ref{lema4.11}. We have obtained that $\alpha \vee \beta =\nabla _{\cal A}$ and $\alpha \wedge \beta =\Delta _{\cal A}$, thus $\alpha ,\beta \in {\cal B}({\rm Con}({\cal A}))$. $\phi =\phi \vee \Delta _{\cal A}=\phi \vee (\alpha \wedge \beta)=(\phi \vee \alpha )\wedge (\phi \vee \beta )=\nabla _{\cal A}\wedge (\phi \vee \beta)=\phi \vee \beta $. We obtain: ${\cal B}(v_{\textstyle {\rm Rad}({\cal A})})(\beta )=\beta \vee {\rm Rad}({\cal A})=\beta \vee (\phi \cap \psi )=(\beta \vee \phi )\cap (\beta \vee \psi )=\phi \cap \nabla _{\cal A}=\phi $. Therefore ${\cal B}(v_{\textstyle {\rm Rad}({\cal A})})$ is surjective, hence ${\rm Rad}({\cal A})$ has CBLP by Remark \ref{usiv}.\end{proof}
The following property generalizes property $(\star )$ for residuated lattices from \cite{ggcm}, \cite{dcggcm}.
\begin{definition}
We say that ${\cal A}$ satisfies {\em the property $(\star )$} iff: for all $\theta \in {\rm Con}({\cal A})$, there exist $\alpha \in {\cal K}({\cal A})$ and $\beta \in {\cal B}({\rm Con}({\cal A}))$ such that $\alpha \subseteq {\rm Rad}({\cal A})$ and $\theta =\alpha \vee \beta $.\end{definition}
\begin{proposition}
If ${\cal A}$ satisfies $(\star )$, then ${\cal A}$ has CBLP.\label{propozitie4.14}\end{proposition}
\begin{proof} Assume that ${\cal A}$ satisfies $(\star )$, and let $\phi ,\psi \in {\cal K}({\cal A})$ such that $\phi \vee \psi =\nabla _{\cal A}$. Condition $(\star )$ ensures us that $\phi =\alpha \vee \beta $ and $\psi =\gamma \vee \delta $ for some $\alpha ,\gamma \in {\cal K}({\cal A})$ such that $\alpha \subseteq {\rm Rad}({\cal A})$ and $\gamma \subseteq {\rm Rad}({\cal A})$, and some $\beta ,\delta \in {\cal B}({\rm Con}({\cal A}))$. Then $\beta \vee \delta \in {\cal B}({\rm Con}({\cal A}))$ and $\beta \vee \delta \vee {\rm Rad}({\cal A})\supseteq \beta \vee \delta \vee \alpha \vee \gamma =\alpha \vee \beta \vee \gamma \vee \delta =\phi \vee \psi =\nabla _{\cal A}$, so $\beta \vee \delta \vee {\rm Rad}({\cal A})=\nabla _{\cal A}$, hence $\beta \vee \delta =\nabla _{\cal A}$ by Lemma \ref{lema4.11}. Then $\neg \, \beta ,\neg \, \delta \in {\cal B}({\rm Con}({\cal A}))$, $\neg \, \beta \cap \neg \, \delta=\neg \, (\beta \vee \delta )=\neg \, \nabla _{\cal A}=\Delta _{\cal A}$, $\phi \vee \neg \, \beta =\alpha \vee \beta \vee \neg \, \beta =\alpha \vee \nabla _{\cal A}=\nabla _{\cal A}$ and $\psi \vee \neg \, \delta =\gamma \vee \delta \vee \neg \, \delta =\gamma \vee \nabla _{\cal A}=\nabla _{\cal A}$. By Proposition \ref{blpbnorm}, it follows that ${\cal A}$ has CBLP.\end{proof}
\begin{proposition}
${\cal A}$ satisfies $(\star )$ iff, for all $\theta \in {\rm Con}({\cal A})$, ${\cal A}/\theta $ satisfies $(\star )$.\label{starcaturi}\end{proposition}
\begin{proof} Assume that ${\cal A}$ satisfies $(\star )$, and let $\theta \in {\rm Con}({\cal A})$. Let $\phi \in {\rm Con}({\cal A})$ such that $\theta \subseteq \phi $. Then there exist $\alpha \in {\cal K}({\cal A})$ and $\beta \in {\cal B}({\rm Con}({\cal A}))$ such that $\alpha \subseteq {\rm Rad}({\cal A})$ and $\phi =\alpha \vee \beta $. We obtain: $\phi =\phi \vee \theta =\phi \vee \theta \vee \theta $, thus $\phi /\theta =(\phi \vee \theta \vee \theta)/\theta =(\alpha \vee \theta \vee \beta \vee \theta)/\theta =(\alpha \vee \theta )/\theta \vee (\beta \vee \theta )/\theta $. Since $\beta \in {\cal B}({\rm Con}({\cal A}))$, we have $(\beta \vee \theta )/\theta =u_{\textstyle \theta }(\beta )={\cal B}(u_{\textstyle \theta })(\beta )\in {\cal B}({\rm Con}({\cal A}/\theta ))$, while $(\alpha \vee \theta )/\theta \in {\cal K}({\cal A}/\theta )$ by Corollary \ref{corsuper}. Finally, $\displaystyle (\alpha \vee \theta )/\theta \subseteq ({\rm Rad}({\cal A})\vee \theta )/\theta =((\bigcap _{\mu \in {\rm Max}({\cal A})}\mu )\vee \theta )/\theta \subseteq ((\bigcap _{\stackrel{\scriptstyle \mu \in {\rm Max}({\cal A})}{\mu \supseteq \theta }}\mu )\vee \theta )/\theta =(\bigcap _{\stackrel{\scriptstyle \mu \in {\rm Max}({\cal A})}{\mu \supseteq \theta }}\mu )/\theta =\bigcap _{\stackrel{\scriptstyle \mu \in {\rm Max}({\cal A})}{\mu \supseteq \theta }}\mu /\theta ={\rm Rad}({\cal A}/\theta )$. Therefore ${\cal A}/\theta $ satisfies $(\star )$.
For the converse implication, just take $\theta =\Delta _{\cal A}$, so that ${\cal A}/\theta ={\cal A}/\Delta _{\cal A}$ is isomorphic to ${\cal A}$.\end{proof}
\begin{proposition}
Let $n\in \N ^*$ and ${\cal A}_1,\ldots ,{\cal A}_n$ be algebras such that $\displaystyle {\cal A}=\prod _{i=1}^n{\cal A}_i$. Then: ${\cal A}$ satisfies $(\star )$ iff, for all $i\in \overline{1,n}$, ${\cal A}_i$ satisfies $(\star )$.\end{proposition}
\begin{proof} Assume that, for all $i\in \overline{1,n}$, ${\cal A}_i$ satisfies $(\star )$, and let $\theta \in {\rm Con}({\cal A})$. For all $i\in \overline{1,n}$, let $\theta _i=pr_i(\theta )\in {\rm Con}({\cal A}_i)$ by Lemma \ref{totdinbj}. Then, for each $i\in \overline{1,n}$, there exist $\alpha _i\in {\cal K}({\cal A}_i)$ and $\beta _i\in {\cal B}({\rm Con}({\cal A}_i))$ such that $\alpha _i\subseteq {\rm Rad}({\cal A}_i)$ and $\theta _i=\alpha _i\vee \beta _i$. Let $\alpha =\alpha _1\times \ldots \times \alpha _n,\beta =\beta _1\times \ldots \times \beta _n$. By Proposition \ref{propO}, (\ref{propO2}) and (\ref{propO1}), we have $\alpha \in {\cal K}({\cal A})$ and $\beta \in {\cal B}({\rm Con}({\cal A}))$. Also, $\alpha =\alpha _1\times \ldots \times \alpha _n\subseteq {\rm Rad}({\cal A}_1)\times \ldots \times {\rm Rad}({\cal A}_n)={\rm Rad}({\cal A})$ by Lemma \ref{totdinbj} and Proposition \ref{specprod}, (\ref{specprod1}), and $\theta =\theta _1\times \ldots \times \theta _n=(\alpha _1\vee \beta _1)\times \ldots \times (\alpha _n\vee \beta _n)=(\alpha _1\times \ldots \times \alpha _n)\vee (\beta _1\times \ldots \times \beta _n)=\alpha \vee \beta $, by Lemma \ref{totdinbj}, (\ref{totdinbj2}). Therefore ${\cal A}$ satisfies $(\star )$.
Now assume that ${\cal A}$ satisfies $(\star )$, and, for all $i\in \overline{1,n}$, let $\theta _i\in {\rm Con}({\cal A}_i)$. Denote $\theta =\theta _1\times \ldots \times \theta _n\in {\rm Con}({\cal A})$, by Lemma \ref{totdinbj}. Then there exist $\alpha \in {\cal K}({\cal A})$ and $\beta \in {\cal B}({\rm Con}({\cal A}))$ such that $\alpha \subseteq {\rm Rad}({\cal A})$ and $\theta =\alpha \vee \beta $. For each $i\in \overline{1,n}$, let $\alpha _i=pr_i(\alpha )$ and $\beta _i=pr_i(\beta )$. By Lemma \ref{totdinbj}, Proposition \ref{propO}, (\ref{propO2}) and (\ref{propO1}), and Proposition \ref{specprod}, (\ref{specprod1}), it follows that, for all $i\in \overline{1,n}$, $\theta _i=pr_i(\theta )=pr_i(\alpha \vee \beta )=pr_i(\alpha )\vee pr_i(\beta )=\alpha _i\vee \beta _i$, $\alpha _i\in {\cal K}({\cal A}_i)$, $\beta _i\in {\cal B}({\rm Con}({\cal A}_i))$ and $\alpha _i=pr_i(\alpha )\subseteq pr_i({\rm Rad}({\cal A}))={\rm Rad}({\cal A}_i)$. Thus, for all $i\in \overline{1,n}$, ${\cal A}_i$ satisfies $(\star )$.\end{proof}
\section{CBLP Versus BLP in Residuated Lattices and Bounded Distributive Lattices}
\label{cblpversusblp}
In this section, we recall some results on the Boolean Lifting Property (BLP) for residuated lattices and bounded distributive lattices, as well as the reticulation functor between these categories of algebras, and obtain new results, concerning the relationships between CBLP and BLP in these categories, and the behaviour of the reticulation functor with respect to CBLP. From these results it is easy to derive notable properties concerning the image of the reticulation functor.
We refer the reader to \cite{bal}, \cite{gal}, \cite{haj}, \cite{ior}, \cite{kow}, \cite{pic}, \cite{tur} for a further study of the results on residuated lattices that we use in this section. For the results on bounded distributive lattices, we refer the reader to \cite{bal}, \cite{blyth}, \cite{bur}.
Throughout this section, all algebras will be designated by their underlying sets.
We recall that a {\em residuated lattice} is an algebra $(A,\vee ,\wedge ,\odot ,\rightarrow ,0,1)$ of type $(2,2,2,2,0,0)$ such that $(A,\vee ,\wedge ,0,1)$ is a bounded lattice, $(A,\odot ,1)$ is a commutative monoid and every $a,b,c\in A$ satisfy {\em the law of residuation}: $a\leq b\rightarrow c$ iff $a\odot b\leq c$, where $\leq $ is the order of $(A,\vee ,\wedge )$. The operation $\odot $ is called {\em product} or {\em multiplication}, and the operation $\rightarrow $ is called {\em implication} or {\em residuum}. It is well known that residuated lattices form an equational class. A {\em $G\ddot{o}del$ algebra} is a residuated lattice in which $\odot =\wedge $.
Throughout this section, unless mentioned otherwise, $(A,\vee ,\wedge ,\odot ,\rightarrow ,0,1)$ shall be an arbitrary residuated lattice. We recall the definitions of the derivative operations $\neg \, $ (the {\em negation}) and $\leftrightarrow $ (the {\em equivalence} or the {\em biresiduum}) on the elements of $A$: for all $a,b\in A$, $\neg \, a=a\rightarrow 0$ and $a\leftrightarrow b=(a\rightarrow b)\wedge (b\rightarrow a)$. We also recall that, for all $a\in A$ and any $n\in \N $, we denote: $a^0=1$ and $a^{n+1}=a^n\odot a$. Next we shall recall some things about the arithmetic of a residuated lattice, its Boolean center, its filters and congruences, as well as the Boolean Lifting Property in a residuated lattice, and we shall prove several new results regarding these notions.
\begin{lemma}{\rm \cite{bal}, \cite{gal}, \cite{haj}, \cite{ior}, \cite{kow}, \cite{pic}, \cite{tur}} For any $a,b\in A$, the following hold:
\begin{enumerate}
\item\label{aritmlr1} $a\rightarrow b=1$ iff $a\leq b$; $a\leftrightarrow b=1$ iff $a=b$;
\item\label{aritmlr2} $a\odot (a\rightarrow b)\leq b$.
\end{enumerate}\label{aritmlr}\end{lemma}
A {\em filter} of $A$ is a non--empty subset $F$ of $A$ such that, for all $x,y\in A$:
\begin{itemize}
\item if $x,y\in F$, then $x\odot y\in F$;
\item if $x\in F$ and $x\leq y$, then $y\in F$.\end{itemize}
The set of the filters of $A$ is denoted by ${\rm Filt}(A)$. $({\rm Filt}(A),\subseteq )$ is a bounded poset, with first element $\{1\}$ and last element $A$. Clearly, a filter equals $A$ iff it contains $0$.
The intersection of any family of filters of $A$ is a filter of $A$, hence, for any $X\subseteq A$, there exists a smallest filter of $A$ which includes $X$; this filter is denoted by $[X)$ and called the {\em filter generated by $X$}. For every $x\in A$, $[\{x\})$ is denoted, simply, by $[x)$, and called the {\em principal filter generated by $x$}. Clearly, $[\emptyset )=\{1\}=[1)$, while, for any $\emptyset \neq X\subseteq A$, $[X)=\{a\in A\ |\ (\exists \, n\in \N ^*)\, (\exists \, x_1,\ldots ,x_n\in X)\, (x_1\odot \ldots \odot x_n\leq a)\}=\{a\in A\ |\ (\exists \, n\in \N )\, (\exists \, x_1,\ldots ,x_n\in X)\, (x_1\odot \ldots \odot x_n\leq a)\}$, where we make the convention that the product of the empty family is $1$. Thus, for any $x\in A$, $[x)=\{a\in A\ |\ (\exists \, n\in \N ^*)\, (x^n\leq a)\}=\{a\in A\ |\ (\exists \, n\in \N )\, (x^n\leq a)\}$. We denote by ${\rm PFilt}(A)$ the set of the principal filters of $A$.
For every $F,G\in {\rm Filt}(A)$, we denote by $F\vee G=[F\cup G)$. Moreover, for any $(F_i)_{i\in I}\subseteq {\rm Filt}(A)$, we denote by $\displaystyle \bigvee _{i\in I}F_i=[\bigcup _{i\in I}F_i)$. $({\rm Filt}(A),\vee ,\cap ,\{1\},A)$ is a complete bounded distributive lattice, orderred by set inclusion. ${\rm PFilt}(A)$ is a bounded sublattice of ${\rm Filt}(A)$, because $\{1\}=[1)$, $A=[0)$ and, for all $x,y\in A$, $[x)\vee [y)=[x\odot y)$ and $[x)\cap [y)=[x\vee y)$.
To every filter $F$ of $A$, one can associate a congruence $\sim _F$ of $A$, defined by: for all $x,y\in A$, $x\sim _Fy$ iff $x\leftrightarrow y\in F$. Let $F$ be a filter of $A$. The congruence class of any $x\in A$ with respect to $\sim _F$ is denoted by $x/F$, and the quotient set of $A$ with respect to $\sim _F$ is denoted by $A/F$. Residuated lattices form an equational class, thus $A/F$ becomes a residuated lattice, with the operations defined canonically. We shall denote by $p_F:A\rightarrow A/F$ the canonical surjective morphism. Notice that $1/F=F$. For any $x,y\in A$, $x\leq y$ implies $x/F\leq y/F$, and $x/F\leq y/F$ iff $x\rightarrow y\in F$. For any $X\subseteq A$, we denote by $X/F=p_F(X)=\{x/F\ |\ x\in X\}$. We have: ${\rm Filt}(A/F)=\{G/F\ |\ G\in {\rm Filt}(A),F\subseteq G\}$. It is well known and straightforward that the function $h_A:{\rm Filt}(A)\rightarrow {\rm Con}(A)$, for all $F\in {\rm Filt}(A)$, $h_A(F)=\sim _F$, is a bounded lattice isomorphism; hence the bounded lattice ${\rm Con}(A)$ is distributive.
By ${\cal B}(A)$ we denote the set of the complemented elements of the underlying bounded lattice of $A$, which, although not necessarily distributive, is uniquely complemented, and has ${\cal B}(A)$ as a bounded sublattice. Moreover, ${\cal B}(A)$ is a Boolean algebra. ${\cal B}(A)$ is called the {\em Boolean center of $A$}.
If $B$ is a residuated lattice and $f:A\rightarrow B$ is a residuated lattice morphism, then $f({\cal B}(A))\subseteq {\cal B}(B)$, thus, just as in the case of bounded distributive lattices, we can define ${\cal B}(f):{\cal B}(A)\rightarrow {\cal B}(B)$ by: ${\cal B}(f)(x)=f(x)$ for all $x\in {\cal B}(A)$. Then ${\cal B}(f)$ is a Boolean morphism. Hence ${\cal B}$ becomes a covariant functor from the category of residuated lattices to the category of Boolean algebras. We believe that there is no danger of confusion between this functor and the functor ${\cal B}$ from the category of bounded distributive lattices to the category of Boolean algebras.
\begin{proposition}{\rm \cite{kow}} Any residuated lattice is an arithmetical algebra and satisfies (H).\end{proposition}
\begin{definition}{\rm \cite{ggcm}}
For any filter $F$ of $A$, we say that $F$ has the {\em Boolean Lifting Property} (abbreviated {\em BLP}) iff the Boolean morphism ${\cal B}(p_F):{\cal B}(A)\rightarrow {\cal B}(A/F)$ is surjective; also, we say that $\sim _F$ has the {\em Boolean Lifting Property (BLP)} iff $F$ has BLP.
We say that $A$ has the {\em Boolean Lifting Property (BLP)} iff each filter of $A$ has the BLP (equivalently, iff each congruence of $A$ has the BLP).\end{definition}
\begin{remark}{\rm \cite{ggcm}}
For any filter $F$ of $A$, ${\cal B}(A)/F\subseteq {\cal B}(A/F)$ and the image of ${\cal B}(p_F)$ is ${\cal B}(A)/F$, hence: $F$ has BLP iff ${\cal B}(A)/F={\cal B}(A/F)$ iff ${\cal B}(A)/F\supseteq {\cal B}(A/F)$.\end{remark}
\begin{lemma}{\rm \cite{ggcm}}
${\cal B}({\rm Filt}(A))=\{[e)\ |\ e\in {\cal B}(A)\}$.\label{boolfilt}\end{lemma}
Let us define $i_A:A\rightarrow {\rm Filt}(A)$, for all $x\in A$, $i_A(x)=[x)$. Clearly, $i_A$ is an injective bounded lattice anti--morphism between the underlying bounded lattice of $A$ and ${\rm Filt}(A)$. Now let us define ${\cal B}(i_A):{\cal B}(A)\rightarrow {\cal B}({\rm Filt}(A))$, for all $e\in {\cal B}(A)$, ${\cal B}(i_A)(e)=[e)$.
\begin{lemma}
${\cal B}(i_A)$ is well defined and it is a Boolean anti--isomorphism.\label{atentie}\end{lemma}
\begin{proof} By Lemma \ref{boolfilt}, ${\cal B}(i_A)$ is well defined and surjective. Since $i_A$ is injective, it follows that ${\cal B}(i_A)$ is injective. Since $i_A$ is a bounded lattice anti--morphism, ${\cal B}(A)$ is a bounded sublattice of $A$ and ${\cal B}({\rm Filt}(A))$ is a bounded sublattice of ${\rm Filt}(A)$, it follows that ${\cal B}(i_A)$ is a bounded lattice anti--morphism between two Boolean algebras, thus ${\cal B}(i_A)$ is a Boolean anti--morphism. Hence ${\cal B}(i_A)$ is a Boolean anti--isomorphism.\end{proof}
Now let $(F_i)_{i\in I}$ be a non--empty family of filters of $A$ such that $\displaystyle A\subseteq \bigvee _{i\in I}F_i$, that is $\displaystyle A=\bigvee _{i\in I}F_i$, that is $\displaystyle 0\in \bigvee _{i\in I}F_i=[\bigcup _{i\in I}F_i)$, which means that there exist $n\in \N ^*$ and $\displaystyle x_1,\ldots ,x_n\in \bigcup _{i\in I}F_i$ such that $x_1\odot \ldots \odot x_n\leq 0$, that is $x_1\odot \ldots \odot x_n=0$. Then $x_1\in F_{\textstyle i_1},\ldots ,x_n\in F_{\textstyle i_n}$ for some $i_1,\ldots ,i_n\in I$. So $0=x_1\odot \ldots \odot x_n\in [F_{\textstyle i_1}\cup \ldots \cup F_{\textstyle i_n})=F_{\textstyle i_1}\vee \ldots \vee F_{\textstyle i_n}$, thus $A=F_{\textstyle i_1}\vee \ldots \vee F_{\textstyle i_n}$, hence $A\subseteq F_{\textstyle i_1}\vee \ldots \vee F_{\textstyle i_n}$. Therefore $A$ is a compact element of the bounded distributive lattice ${\rm Filt}(A)$. Since ${\rm Filt}(A)$ is isomorphic to ${\rm Con}(A)$, it follows that $\nabla _A$ is a compact element of the bounded distributive lattice ${\rm Con}(A)$, which means that $A$ fulfills the hypothesis (H).
Until mentioned otherwise, $F$ will be a filter of $A$, arbitrary but fixed. We shall denote by $\delta _F:{\rm Filt}(A)\rightarrow {\rm Filt}(A/F)$ the function defined by: for all $G\in {\rm Filt}(A)$, $\delta _F(G)=(G\vee F)/F$.
\begin{lemma}\begin{enumerate}
\item\label{lamunca1} For all $a\in A$, $([a)\vee F)/F=[a/F)$.
\item\label{lamunca2} For all $J,K\in {\rm Filt}(A)$ such that $F\subseteq J$ and $F\subseteq K$, $(J\vee K)/F=J/F\vee K/F$.
\item\label{lamunca3} $\delta _F$ is well defined and it is a bounded lattice morphism.
\item\label{lamunca4} The following diagram is commutative:
\vspace*{-30pt}
\begin{center}\begin{picture}(140,80)(0,0)
\put(30,50){$A$}
\put(24,5){$A/F$}
\put(100,50){${\rm Filt}(A)$}
\put(93,5){${\rm Filt}(A/F)$}
\put(21,30){$p_F$}
\put(33,48){\vector(0,-1){34}}
\put(113,30){$\delta _F$}
\put(110,48){\vector(0,-1){34}}
\put(61,57){$i_A$}
\put(38,53){\vector(1,0){60}}
\put(57,12){$i_{A/F}$}
\put(45,8){\vector(1,0){45}}\end{picture}\end{center}
\vspace*{-20pt}
\item\label{lamunca5} $F$ has BLP iff ${\cal B}(\delta _F)$ is surjective.\end{enumerate}\label{lamunca}\end{lemma}
\begin{proof} (\ref{lamunca1}) Let $a\in A$. Then $[a/F)=\{b/F\ |\ b\in A,(\exists \, n\in \N )((a/F)^n\leq b/F)\}=\{b/F\ |\ b\in A,(\exists \, n\in \N )\, (a^n/F\leq b/F)\}=\{b/F\ |\ b\in A,(\exists \, n\in \N )\, (a^n\rightarrow b\in F)\}$ and $([a)\vee F)/F=[[a)\cup F)/F=[\{a\}\cup F)/F=\{b/F\ |\ b\in [\{a\}\cup F)\}=\{b/F\ |\ (\exists \, n,k\in \N )\, (\exists \, x_1,\ldots ,x_k\in F)\, (a^n\odot x_1\odot \ldots \odot x_n\leq b)\}=\{b/F\ |\ (\exists \, n\in \N )\, (\exists \, x\in F)\, (a^n\odot x\leq b)\}$, since any product of a finite family of elements of $F$ belongs to $F$, and the converse is trivial; we shall be using this property repeatedly in what follows. Let $b\in A$ such that $b/F\in [a/F)$. Then $a^n\rightarrow b\in F$ for some $n\in \N $, thus there exists an $x\in F$ such that $a^n\rightarrow b=x$, so $x\rightarrow a^n\rightarrow b$, hence $a^n\odot x\leq b$ by the law of residuation, therefore $b/F\in ([a)\vee F)/F$. In what follows, we shall be using the law of residuation without mentioning it. Now let $b\in A$ such that $b/F\in ([a)\vee F)/F$. Then there exist $n\in \N $ and $x\in F$ such that $a^n\odot x\leq b$, thus $x\leq a^n\rightarrow b$, hence $a^n\rightarrow b\in F$, therefore $b/F\in [a/F)$. Therefore $([a)\vee F)/F=[a/F)$.
\noindent (\ref{lamunca2}) Let $J$ and $K$ be as in the enunciation. Then $(J\vee K)/F=[J\cup K)/F=\{a/F\ |\ a\in [J\cup K)\}=\{a/F\ |\ a\in A,(\exists \, n,k\in \N )\, (\exists \, x_1,\ldots ,x_n\in J)\, (\exists \, y_1,\ldots ,y_k\in K)\, (x_1\odot \ldots \odot x_n\odot y_1\odot \ldots \odot y_k\leq a)\}=\{a/F\ |\ a\in A,(\exists \, x\in J)\, (\exists \, y\in K)\, (x\odot y\leq a)$. And $J/F\vee K/F=[J/F\cup K/F)=\{a/F\ |\ a\in A,(\exists \, n,k\in \N )\, (\exists \, x_1,\ldots ,x_n\in J)\, (\exists \, y_1,\ldots ,y_k\in K)\, (x_1/F\odot \ldots \odot x_n/F\odot y_1/F\odot \ldots \odot y_k/F\leq a/F)\}=\{a/F\ |\ a\in A,(\exists \, n,k\in \N )\, (\exists \, x_1,\ldots ,x_n\in J)\, (\exists \, y_1,\ldots ,y_k\in K)\, ((x_1\odot \ldots \odot x_n)/F\odot (y_1\odot \ldots \odot y_k)/F\leq a/F)\}=\{a/F\ |\ a\in A,(\exists \, x\in J)\, (\exists \, y\in K)\, (x/F\odot y/F\leq a/F)\}=\{a/F\ |\ a\in A,(\exists \, x\in J)\, (\exists \, y\in K)\, ((x\odot y)/F\leq a/F)\}$. If $x,y,a\in A$ such that $x\odot y\leq a$, then $(x\odot y)/F\leq a/F$, thus $(J\vee K)/F\subseteq J/F\vee K/F$. Now let $a\in A$ such that $a/F\in J/F\vee K/F$, thus there exist $x\in J$ and $y\in K$ such that $(x\odot y)/F\leq a/F$, that is $(x\odot y)\rightarrow a\in F$, so $(x\odot y)\rightarrow a=z$ for some $z\in F$, thus $z\leq (x\odot y)\rightarrow a$, that is $x\odot y\odot z\leq a$. We have: $x\in J$, $y\in K$ and $z\in F\subseteq K$, thus $y\odot z\in K$. So $a/F\in (J\vee K)/F$. Therefore $(J\vee K)/F=J/F\vee K/F$.
\noindent (\ref{lamunca3}) For all $G\in {\rm Filt}(A)$, $G\vee F\supseteq F$, thus $(G\vee F)/F\in {\rm Filt}(A/F)$, so $\delta _F$ is well defined. $\delta _F(\{1\})=(\{1\}\vee F)/F=F/F=\{1/F\}$; $\delta _F(A)=(A\vee F)/F=A/F$. Now let $G,H\in {\rm Filt}(A)$. By (\ref{lamunca2}), we have: $\delta _F(G\vee H)=(G\vee H\vee F)/F=(G\vee F\vee H\vee F)/F=(G\vee F)/F\vee (H\vee F)/F=\delta _F(G)\vee \delta _F(H)$. By the distributivity of the lattice of filters of a residuated lattice, we have: $\delta _F(G\cap H)=((G\cap H)\vee F)/F=((G\vee F)\cap (H\vee F))/F=(G\vee F)/F\cap (H\vee F)/F=\delta _F(G)\cap \delta _F(H)$. Therefore $\delta _F$ is a bounded lattice morphism.
\noindent (\ref{lamunca4}) Let $a\in A$. By (\ref{lamunca1}), $\delta _F(i_A(a))=\delta _F([a))=([a)\vee F)/F=[a/F)=i_{A/F}(a/F)=i_{A/F}(p_F(a))$. Therefore $\delta _F\circ i_A=i_{A/F}\circ p_F$.
\noindent (\ref{lamunca5}) By taking the restrictions to the Boolean centers in the commutative diagram in (\ref{lamunca4}), we get the following commutative diagram, where we have denoted by ${\cal B}(i_A)$ the restriction of $i_A$ to ${\cal B}(A)$, and the same goes for ${\cal B}(i_{A/F})$:
\vspace*{-23pt}
\begin{center}\begin{picture}(140,80)(0,0)
\put(22,50){${\cal B}(A)$}
\put(16,5){${\cal B}(A/F)$}
\put(100,50){${\cal B}({\rm Filt}(A))$}
\put(93,5){${\cal B}({\rm Filt}(A/F))$}
\put(7,30){${\cal B}(p_F)$}
\put(33,48){\vector(0,-1){34}}
\put(123,30){${\cal B}(\delta _F)$}
\put(121,48){\vector(0,-1){34}}
\put(58,57){${\cal B}(i_A)$}
\put(44,53){\vector(1,0){54}}
\put(53,12){${\cal B}(i_{A/F})$}
\put(51,8){\vector(1,0){39}}\end{picture}\end{center}
\vspace*{-3pt}
Thus ${\cal B}(\delta _F)\circ {\cal B}(i_A)={\cal B}(i_{A/F})\circ {\cal B}(p_F)$. By Lemma \ref{atentie}, ${\cal B}(i_A)$ and ${\cal B}(i_{A/F})$ are Boolean anti--isomorphisms. Hence: $F$ has BLP iff ${\cal B}(p_F)$ is surjective iff ${\cal B}(\delta _F)$ is surjective.\end{proof}
Now let us consider the congruence $h_A(F)=\sim _F$ associated to $F$ and the bounded lattice morphism $u_{\textstyle \sim _F}:{\rm Con}(A)\rightarrow {\rm Con}(A)/\sim _F$, for all $\theta \in {\rm Con}(A)$, $u_{\textstyle \sim _F}(\theta )=(\theta \vee \sim _F)/\sim _F$ (see Section \ref{thecblp}).
\begin{lemma}
The following diagram is commutative:\vspace*{-24pt}
\begin{center}\begin{picture}(140,80)(0,0)
\put(22,50){${\rm Filt}(A)$}
\put(16,5){${\rm Filt}(A/F)$}
\put(103,50){${\rm Con}(A)$}
\put(96,5){${\rm Con}(A/F)$}
\put(26,30){$\delta _F$}
\put(37,48){\vector(0,-1){34}}
\put(123,30){$u_{\textstyle \sim _F}$}
\put(121,48){\vector(0,-1){34}}
\put(70,57){$h_A$}
\put(53,53){\vector(1,0){48}}
\put(65,12){$h_{A/F}$}
\put(60,8){\vector(1,0){34}}\end{picture}\end{center}\vspace*{-10pt}\label{capace}\end{lemma}
\begin{proof} Let $G\in {\rm Filt}(A)$. Then $h_{A/F}(\delta _F(G))=h_{A/F}((G\vee F)/F)=\sim _{(G\vee F)/F}=\{(x/F,y/F)\ |\ x,y\in A,x/F\leftrightarrow y/F\in (G\vee F)/F\}=\{(x/F,y/F)\ |\ x,y\in A,(x\leftrightarrow y)/F\in (G\vee F)/F\}$ and $u_{\textstyle \sim _F}(h_A(G))=u_{\textstyle \sim _F}(\sim _G)=(\sim _G\vee \sim _F)/\sim _F=(h_A(G)\vee h_A(F))/\sim _F=h_A(G\vee F)/\sim _F=\sim _{G\vee F}/\sim _F=\{(x/\sim _F,y/\sim _F)\ |\ x,y\in A, (x,y)\in /\sim _{G\vee F}\}=\{(x/F,y/F)\ |\ x,y\in A, x\leftrightarrow y\in G\vee F\}$. For any $x,y\in A$, if $x\leftrightarrow y\in G\vee F$, then $(x\leftrightarrow y)/F\in (G\vee F)/F$; conversely, if $(x\leftrightarrow y)/F\in (G\vee F)/F$, then $(x\leftrightarrow y)/F=z/F$ for some $z\in G\vee F$, thus $(x\leftrightarrow y)\leftrightarrow z\in F$, so, since $(x\leftrightarrow y)\leftrightarrow z\leq z\rightarrow (x\leftrightarrow y)$, it follows that $z\rightarrow (x\leftrightarrow y)\in F$, that is $z\rightarrow (x\leftrightarrow y)=t$ for some $t\in F$, hence $t\leq z\rightarrow (x\leftrightarrow y)$, thus $t\odot z\leq x\leftrightarrow y$, with $t\in F\subseteq G\vee F$ and $z\in G\vee F$, hence $t\odot z\in G\vee F$, thus $x\leftrightarrow y\in G\vee F$. Hence $h_{A/F}(\delta _F(G))=u_{\textstyle \sim _F}(h_A(G))$. Therefore $h_{A/F}\circ \delta _F=u_{\textstyle \sim _F}\circ h_A$.\end{proof}
\begin{proposition}\begin{enumerate}
\item\label{happy1} For every filter $F$ of $A$: $F$ has BLP iff $\sim _F$ has CBLP.
\item\label{happy2} $A$ has BLP iff $A$ has CBLP.\end{enumerate}\label{happy}\end{proposition}
\begin{proof} (\ref{happy1}) By applying the functor ${\cal B}$ from the category of bounded distributive lattices to the category of Boolean algebras to the commutative diagram in Lemma \ref{capace}, we get the following commutative diagram in the category of Boolean algebras:\vspace*{-23pt}
\begin{center}\begin{picture}(140,80)(0,0)
\put(22,50){${\cal B}({\rm Filt}(A))$}
\put(13,5){${\cal B}({\rm Filt}(A/F))$}
\put(111,50){${\cal B}({\rm Con}(A))$}
\put(106,5){${\cal B}({\rm Con}(A/F))$}
\put(18,30){${\cal B}(\delta _F)$}
\put(43,48){\vector(0,-1){34}}
\put(136,30){${\cal B}(u_{\textstyle \sim _F})$}
\put(134,48){\vector(0,-1){34}}
\put(75,57){${\cal B}(h_A)$}
\put(67,53){\vector(1,0){43}}
\put(70,12){${\cal B}(h_{A/F})$}
\put(71,8){\vector(1,0){33}}\end{picture}\end{center}
\vspace*{-4pt}
This means that: ${\cal B}(h_{A/F})\circ {\cal B}(\delta _F)={\cal B}(u_{\textstyle \sim _F})\circ {\cal B}(h_A)$. $h_A$ and $h_{A/F}$ are bounded lattice isomorphisms, hence ${\cal B}(h_A)$ and ${\cal B}(h_{A/F})$ are Boolean isomorphisms. By Lemma \ref{lamunca}, (\ref{lamunca5}), we get that: $F$ has BLP iff ${\cal B}(\delta _F)$ is surjective iff ${\cal B}(u_{\textstyle \sim _F})$ is surjective iff $\sim _F$ has CBLP.
\noindent (\ref{happy2}) By (\ref{happy1}) and the fact that $h_A:{\rm Filt}(A)\rightarrow {\rm Con}(A)$, for all $F\in {\rm Filt}(A)$, $h_A(F)=\sim _F$, is a bijection.\end{proof}
\begin{definition}{\rm \cite{dcggcm}} $A$ is a {\em Gelfand residuated lattice} iff any prime filter of $A$ is included in a unique maximal filter of $A$.\end{definition}
\begin{proposition}{\rm \cite{dcggcm}} $A$ is Gelfand iff the lattice ${\rm Filt}(A)$ is normal.\label{lrgelfand}\end{proposition}
\begin{corollary}
$A$ is Gelfand iff the lattice ${\rm Con}(A)$ is normal.\label{gelfand}\end{corollary}
\begin{proof} By Proposition \ref{lrgelfand} and the fact that the bounded distributive lattices ${\rm Filt}(A)$ and ${\rm Con}(A)$ are isomorphic.\end{proof}
The following corollary is part of \cite[Theorem $6.20$]{dcggcm}, but here we provide a different proof for it, by using the equivalence between CBLP and BLP in residuated lattices.
\begin{corollary}
Any residuated lattice with BLP is Gelfand.\label{cordinp2}\end{corollary}
\begin{proof} By Proposition \ref{happy}, (\ref{happy2}), Proposition \ref{blpbnorm}, Corollary \ref{gelfand} and the trivial fact that any B--normal lattice is normal.\end{proof}
An element of $a\in A$ is said to be {\em idempotent} iff $a^2=a$. The set of the idempotents of $A$ is denoted by ${\cal I}(A)$. An element of $a\in A$ is said to be {\em regular} iff $\neg \, \neg \, a=a$. The set of the regular elements of $A$ is denoted by ${\rm Reg}(A)$.
\begin{definition}{\rm \cite{dcggcm}} Let $F$ be an arbitrary filter of $A$. We say that $F$ has the {\em Idempotent Lifting Property} (abbreviated {\em ILP}) iff ${\cal I}(A/F)={\cal I}(A)/F$.
We say that $A$ has the {\em Idempotent Lifting Property (ILP)} iff all of its filters have the ILP.\end{definition}
\begin{proposition}{\cite{dcggcm}}
Neither of the properties BLP and ILP in residuated lattices implies the other.\end{proposition}
\begin{proposition}{\rm \cite{dcggcm}} For any filter $F$ of $A$, ${\rm Reg}(A/F)={\rm Reg}(A)/F$.\label{regtriv}\end{proposition}
MV--algebras form a subclass of the class of BL--algebras, which, in turn, form a subclass of the class of residuated lattices. If $A$ is a BL--algebra, then so is $A/F$ for any $F\in {\rm Filt}(A)$; the same goes for MV--algebras.
\begin{proposition}{\rm \cite{dcggcm}} Any BL--algebra is a Gelfand residuated lattice.\label{blgelfand}\end{proposition}
\begin{proposition}{\rm \cite{bal}, \cite{gal}, \cite{haj}, \cite{ior}, \cite{kow}, \cite{pic}, \cite{tur}} If $A$ is an MV--algebra, then ${\cal B}(A)={\cal I}(A)$.\label{bvsi}\end{proposition}
\begin{corollary}
If $A$ is an MV--algebra, then:\begin{itemize}
\item for any filter $F$ of $A$: $\sim _F$ has CBLP iff $F$ has BLP iff $F$ has ILP;
\item $A$ has CBLP iff $A$ has BLP iff $A$ has ILP.\end{itemize}\end{corollary}
\begin{proof} By Propositions \ref{happy} and \ref{bvsi}.\end{proof}
The equivalences not involving CBLP in the previous corollary were proven in \cite{dcggcm} by using Proposition \ref{bvsi}. Now let us investigate them for BL--algebras. Of course, by Proposition \ref{happy}, a BL--algebra has BLP iff it has CBLP, and, furthermore, any filter of it has BLP iff its associated congruence has CBLP.
\begin{proposition}{\rm \cite{bal}, \cite{gal}, \cite{haj}, \cite{ior}, \cite{kow}, \cite{pic}, \cite{tur}} If $A$ is a BL--algebra, then ${\cal B}(A)={\cal I}(A)\cap {\rm Reg}(A)$.\label{bvsir}\end{proposition}
\begin{corollary}
If $A$ is a BL--algebra, then:\begin{enumerate}
\item\label{lpsbl1} for any filter $F$ of $A$: if $F$ has ILP, then $F$ has BLP;
\item\label{lpsbl2} if $A$ has ILP, then $A$ has BLP.\end{enumerate}\label{lpsbl}\end{corollary}
\begin{proof} (\ref{lpsbl1}) If $F$ has ILP, then ${\cal I}(A/F)={\cal I}(A)/F$, so, by Propositions \ref{bvsir} and \ref{regtriv}, ${\cal B}(A/F)={\cal I}(A/F)\cap {\rm Reg}(A/F)={\cal I}(A)/F\cap {\rm Reg}(A)/F={\cal B}(A)/F$, thus $F$ has BLP.
\noindent (\ref{lpsbl2}) By (\ref{lpsbl1}).\end{proof}
Throughout the rest of this section, unless mentioned otherwise, $(L,\vee ,\wedge ,0,1)$ shall be an arbitrary bounded distributive lattice. We shall denote the set of the filters of $L$ by ${\rm Filt}(L)$, and the set of the ideals of $L$ by ${\rm Id}(L)$. To each filter $F$ of $L$, one can associate a congruence $\equiv _F$ of $L$, defined by: for any $x,y\in L$, $x\equiv _Fy$ iff $x\wedge a=y\wedge a$ for some $a\in F$; the mapping $F\mapsto \equiv _F$ is an embedding of the bounded distributive lattice ${\rm Filt}(L)$ into ${\rm Con}(L)$. For every $F\in {\rm Filt}(L)$, any $x\in L$ and any $X\subseteq L$, we shall denote by $x/F=x/\equiv _F$ and $X/F=X/\equiv _F$. Dually, to each ideal $I$ of $L$, one can associate a congruence $\approx _I$ of $L$, defined by: for any $x,y\in L$, $x\approx _Iy$ iff $x\vee a=y\vee a$ for some $a\in I$; the mapping $I\mapsto \approx _I$ is a bounded lattice embedding of ${\rm Id}(L)$ into ${\rm Con}(L)$.
\begin{definition}{\rm \cite{blpiasi},\cite{blpdacs},\cite{dcggcm}} We say that a congruence $\sim $ of $L$ has the {\em Boolean Lifting Property (BLP)} iff ${\cal B}(L/\sim )={\cal B}(L)/\sim $. We say that $L$ has the {\em Boolean Lifting Property (BLP)} iff all of its congruences have the BLP.
We say that a filter $F$ of $L$ has the {\em Boolean Lifting Property (BLP)} iff $\equiv _F$ has the BLP, that is iff ${\cal B}(L/F)={\cal B}(L)/F$. We say that $L$ has the {\em Boolean Lifting Property for filters ({\rm Filt}--BLP)} iff all of its filters have the BLP.
Similarly, we say that an ideal $I$ of $L$ has the {\em Boolean Lifting Property (BLP)} iff $\approx _I$ has the BLP, and we say that $L$ has the {\em Boolean Lifting Property for ideals ({\rm Id}--BLP)} iff all of its ideals have the BLP.\end{definition}
Clearly, in any bounded distributive lattice $L$, the BLP implies the Filt--BLP and Id--BLP, and the BLP is self--dual, while the Filt--BLP and Id--BLP are duals of each other.
See in \cite{eu3}, \cite{eu1}, \cite{eu}, \cite{eu2}, \cite{eu4}, \cite{eu5}, \cite{dcggcm} the definition of {\em the reticulation functor ${\cal L}$} from the category of residuated lattices to the category of bounded distributive lattices, which takes every residuated lattice $A$ to the unique (up to a bounded lattice isomorphism) bounded distributive lattice ${\cal L}(A)$ whose prime spectrum is homeomorphic to that of $A$, where the prime spectra are the sets of the prime filters of ${\cal L}(A)$, respectively $A$, endowed with the Stone topologies. ${\cal L}(A)$ is called the {\em reticulation of $A$}. The bounded distributive lattice ${\cal L}(A)$ is isomorphic to the dual of ${\rm PFilt}(A)$ (\cite{eu1}, \cite{eu}).
\begin{proposition}{\rm \cite[Proposition $5.19$]{dcggcm}} $A$ has BLP iff ${\cal L}(A)$ has Filt--BLP.\label{propQ}\end{proposition}
\begin{remark} In bounded distributive lattices, CBLP always holds, as proven in Corollary \ref{d01cblp}. But the next remark contains an example of a bounded distributive lattice without Filt--BLP, thus without BLP. See, in what follows, whole classes of bounded distributive lattices without BLP.\end{remark}
\begin{remark} Trivially, the functor ${\cal L}$ preserves the CBLP, by Corollary \ref{d01cblp}. But ${\cal L}$ does not reflect the CBLP. Indeed, let us consider the following example of residuated lattice from \cite{ior}: $A=\{0,a,b,c,d,1\}$, with the following Hasse diagram, with $\odot =\wedge $ and $\rightarrow $ defined by the following table:\vspace*{-5pt}
\begin{center}\begin{tabular}{cc}
\begin{picture}(50,70)(0,0)
\put(25,10){\line(1,1){10}}
\put(25,10){\line(-1,1){10}}
\put(25,30){\line(1,-1){10}}
\put(25,30){\line(-1,-1){10}}
\put(25,30){\line(0,1){15}}
\put(25,30){\circle*{3}}
\put(25,10){\circle*{3}}
\put(25,45){\circle*{3}}
\put(15,20){\circle*{3}}
\put(35,20){\circle*{3}}
\put(24,1){$0$}
\put(23,48){$1$}
\put(8,17){$a$}
\put(37,17){$b$}
\put(28,29){$c$}
\end{picture} & \begin{picture}(120,70)(0,0)
\put(30,30){\begin{tabular}{c|ccccc}
$\rightarrow $ & $0$ & $a$ & $b$ & $c$ & $1$ \\ \hline
$0$ & $1$ & $1$ & $1$ & $1$ & $1$ \\
$a$ & $b$ & $1$ & $b$ & $1$ & $1$ \\
$b$ & $a$ & $a$ & $1$ & $1$ & $1$ \\
$c$ & $0$ & $a$ & $b$ & $1$ & $1$ \\
$1$ & $0$ & $a$ & $b$ & $c$ & $1$\end{tabular}}\end{picture}\end{tabular}\end{center}
According to \cite[Example $3.5$]{ggcm}, the residuated lattice $A$ does not have BLP, because its filter $[c)$ does not have the BLP, hence, by Proposition \ref{happy}, (\ref{happy2}), $A$ does not have CBLP. But, according to Corollary \ref{d01cblp}, ${\cal L}(A)$ has CBLP, as does every bounded distributive lattice.
Concerning the structure of ${\cal L}(A)$, we may notice that ${\cal L}(A)$ has the same Hasse diagram as $A$, because $\odot =\wedge $ in $A$ ($A$ is a ${\rm G\ddot{o}del}$ algebra) and hence, according to a result in \cite{eu2}, ${\cal L}(A)$ is isomorphic to the underlying bounded lattice of $A$. In \cite[Example $1$]{blpiasi}, we have proven that this bounded distributive lattice does not have Filt--BLP, since its filter $[c)$ does not have BLP. Of course, this and Proposition \ref{propQ} provide another proof for the fact that $A$ does not have the BLP, but this issue is, actually, trivial here, because the underlying bounded lattice of a ${\rm G\ddot{o}del}$ algebra is distributive, since, in any residuated lattice, $\odot $ is distributive with respect to $\vee $, and, obviously, the filters of a ${\rm G\ddot{o}del}$ algebra coincide with the filters of its bounded lattice reduct, and so do the congruences associated to these filters, hence the BLP in a ${\rm G\ddot{o}del}$ algebra coincides with the Filt--BLP in its bounded lattice reduct.\label{rcuex}\end{remark}
Here is an extended version of one of the results recalled above:
\begin{proposition}{\rm \cite{dcggcm}} The following are equivalent:\begin{enumerate}
\item\label{rezulttare0} $A$ is a Gelfand residuated lattice;
\item\label{rezulttare1} ${\rm Filt}(A)$ is a normal lattice;
\item\label{rezulttare2} ${\rm PFilt}(A)$ is a normal lattice;
\item\label{rezulttare3} ${\cal L}(A)$ is a conormal lattice;
\item\label{rezulttare4} any prime filter of $A$ is included in a unique maximal filter of $A$;
\item\label{rezulttare5} any prime filter of ${\cal L}(A)$ is included in a unique maximal filter of ${\cal L}(A)$.\end{enumerate}\label{rezulttare}\end{proposition}
\begin{remark} In \cite{blpiasi}, by noticing that, if $L$ is a bounded distributive lattice which is not local and in which $\{1\}$ is a prime filter, then $L$ does not satisfy condition (\ref{rezulttare5}) from Proposition \ref{rezulttare}, and thus $L$ is not conormal, we have pointed out that, for instance, an ordinal sum between a bounded distributive lattice which is not local (for example, a direct product of at least two non--trivial chains) and a non--trivial chain is not a conormal bounded distributive lattice. Such a lattice is ${\cal L}(A)$ from Remark \ref{rcuex}, which is the ordinal sum between ${\cal L}_2^2$ and ${\cal L}_2$, where ${\cal L}_2$ is the two--element chain.\label{tarecool}\end{remark}
\begin{remark} Concerning the class of the B--normal lattices, notice that it includes all congruence lattices of bounded distributive lattices, according to Corollary \ref{d01cblp} and Proposition \ref{blpbnorm}, and it also includes all congruence lattices of algebras satisfying (H) from any discriminator equational class, according to Corollary \ref{cordiscrim} and Proposition \ref{blpbnorm} (see examples of classes of such algebras in Remark \ref{remdiscrim}).\label{totcool}\end{remark}
\begin{corollary}\begin{enumerate}
\item\label{sohappy1} The image of the class of Gelfand residuated lattices through the reticulation functor is included in the class of conormal bounded distributive lattices, thus it is not the whole class of the bounded distributive lattices.
\item\label{sohappy0} The image of the class of the residuated lattices with BLP through the reticulation functor is included in the class of conormal bounded distributive lattices, thus it is not the whole class of the bounded distributive lattices.
\item\label{sohappy4} The image of the class of the residuated lattices with CBLP through the reticulation functor is included in the class of conormal bounded distributive lattices, thus it is not the whole class of the bounded distributive lattices.
\item\label{sohappy2} The image of the class of BL--algebras through the reticulation functor is included in the class of conormal bounded distributive lattices, thus it is not the whole class of the bounded distributive lattices.
\item\label{sohappy3} The image of the class of MV--algebras through the reticulation functor is included in the class of conormal bounded distributive lattices, thus it is not the whole class of the bounded distributive lattices.\end{enumerate}\label{sohappy}\end{corollary}
\begin{proof} (\ref{sohappy1}) By Proposition \ref{rezulttare} and Remark \ref{tarecool}, which actually provides quite a productive method for obtaining bounded distributive lattices which are outside of the image through the reticulation functor of the class of Gelfand residuated lattices, and thus of any of the classes mentioned in (\ref{sohappy0}), (\ref{sohappy4}), (\ref{sohappy2}), (\ref{sohappy3}) (see just below).
\noindent (\ref{sohappy0}) By (\ref{sohappy1}) and Corollary \ref{cordinp2}.
\noindent (\ref{sohappy4}) By (\ref{sohappy0}) and Proposition \ref{happy}, which shows that the class of the residuated lattices with BLP coincides to the class of the residuated lattices with CBLP.
\noindent (\ref{sohappy2}) By (\ref{sohappy1}) and Proposition \ref{blgelfand}.
\noindent (\ref{sohappy3}) By (\ref{sohappy2}) and the fact that the class of MV--algebras is included in the class of BL--algebras.\end{proof}
Now let us see if the statements in Proposition \ref{rezulttare} which refer to normality or conormality remain equivalent if we replace these properties by B--normality and B--conormality, respectively.
\begin{corollary}
${\cal B}({\rm PFilt}(A))={\cal B}({\rm Filt}(A))$.\label{corboolfilt}\end{corollary}
\begin{proof} Since ${\rm PFilt}(A)$ is a bounded sublattice of the bounded distributive lattice ${\rm Filt}(A)$, it follows that\linebreak ${\cal B}({\rm PFilt}(A))\subseteq {\cal B}({\rm Filt}(A))$. But, according to Lemma \ref{boolfilt}, ${\cal B}({\rm Filt}(A))=\{[e)\ |\ e\in {\cal B}(A)\}\subseteq {\rm PFilt}(A)$, hence ${\cal B}({\rm Filt}(A))=\{[e)\ |\ e\in {\cal B}(A)\}\subseteq {\cal B}({\rm PFilt}(A))$ since ${\cal B}({\rm Filt}(A))$ is a Boolean algebra. Therefore ${\cal B}({\rm PFilt}(A))={\cal B}({\rm Filt}(A))$.\end{proof}
\begin{proposition} The following are equivalent:\begin{enumerate}
\item\label{blpretic0} $A$ has CBLP;
\item\label{blpretic1} $A$ has BLP;
\item\label{blpretic2} ${\rm Con}(A)$ is a B--normal lattice;
\item\label{blpretic3} ${\rm Filt}(A)$ is a B--normal lattice;
\item\label{blpretic4} ${\rm PFilt}(A)$ is a B--normal lattice;
\item\label{blpretic5} ${\cal L}(A)$ is a B--conormal lattice;
\item\label{blpretic6} ${\cal L}(A)$ has Filt--BLP;
\item\label{blpretic7} ${\rm PFilt}(A)$ has Id--BLP.\end{enumerate}\label{blpretic}\end{proposition}
\begin{proof} (\ref{blpretic0})$\Leftrightarrow $(\ref{blpretic1}): By Proposition \ref{happy}, (\ref{happy2}).
\noindent (\ref{blpretic0})$\Leftrightarrow $(\ref{blpretic2}): By Proposition \ref{blpbnorm}.
\noindent (\ref{blpretic3})$\Rightarrow $(\ref{blpretic4}): By Corollary \ref{corboolfilt} and the fact that ${\rm PFilt}(A)$ is a bounded sublattice of ${\rm Filt}(A)$.
\noindent (\ref{blpretic4})$\Rightarrow $(\ref{blpretic3}): Let $F,G\in {\rm Filt}(A)$ such that $F\vee G=A$. Then $x\odot y=0$ for some $x\in F$ and $y\in G$. So $[x)\vee [y)=[x\odot y)=[0)=A$, hence, according to Corollary \ref{corboolfilt}, there exist $H,K\in {\cal B}({\rm PFilt}(A))={\cal B}({\rm Filt}(A))$ such that $H\cap K=\{1\}$ and $[x)\vee H=[y)\vee K=A$. But $x\in F$ and $y\in G$, thus $[x)\subseteq F$ and $[y)\subseteq G$, hence $A=[x)\vee H\subseteq F\vee H$ and $A=[y)\vee K\subseteq G\vee K$, thus $F\vee H=G\vee K=A$. Therefore ${\rm Filt}(A)$ is B--normal.
\noindent (\ref{blpretic1})$\Leftrightarrow $(\ref{blpretic6}): By Proposition \ref{propQ}.
\noindent (\ref{blpretic4})$\Leftrightarrow $(\ref{blpretic5}) and (\ref{blpretic6})$\Leftrightarrow $(\ref{blpretic7}): By the fact that ${\cal L}(A)$ is isomorphic to the dual of ${\rm PFilt}(A)$.\end{proof}
\begin{corollary} Any bounded distributive lattice which is not B--conormal does not belong to the image of the class of the residuated lattices with BLP (equivalently, with CBLP) through the reticulation functor. Consequently, any bounded distributive lattice which is not conormal does not belong to the image of the class of the residuated lattices with BLP (equivalently, with CBLP) through the reticulation functor.\end{corollary}
\begin{remark} The previous corollary shows that, for instance, the bounded distributive lattices constructed as in Remark \ref{tarecool} do not belong to the image of the class of residuated lattices with BLP (equivalently, with CBLP) through the reticulation functor.\end{remark}
\begin{corollary}
The following are equivalent:
\begin{itemize}
\item the lattice ${\rm PFilt}(A)$ is normal and it is not B--normal;
\item the lattice ${\rm PFilt}(A)$ is normal and it does not have Id--BLP;
\item the lattice ${\cal L}(A)$ is conormal and it is not B--conormal;
\item the lattice ${\cal L}(A)$ is conormal and it does not have Filt--BLP;
\item $A$ is Gelfand and it does not have BLP;
\item $A$ is Gelfand and it does not have CBLP.\end{itemize}\end{corollary}
\begin{proof} By Propositions \ref{rezulttare} and \ref{blpretic}.\end{proof}
\begin{example} Let $A$ be the residuated lattice in Remark \ref{rcuex}, which does not have BLP, and whose underlying bounded lattice is isomorphic to ${\cal L}(A)$. The prime filters of ${\cal L}(A)$ are $[a)$ and $[b)$, which coincide to its maximal filters, hence $A$ is Gelfand by Proposition \ref{rezulttare}. So $A$ is a Gelfand residuated lattice without BLP (equivalently, without CBLP).
Furthermore, since $A$ has $\odot =\wedge $, $A$ is a ${\rm G\ddot{o}del}$ algebra, so, by \cite[Corollary $4.5$]{dcggcm}, it follows that $A$ has ILP, hence, by Corollary \ref{lpsbl}, (\ref{lpsbl2}), we get that $A$ is not a BL--algebra. So this is an example of a Gelfand residuated lattice which is not a BL--algebra, that is a counter--example for the converse of Proposition \ref{blgelfand}.\end{example}
\section{CBLP in Semilocal Algebras}
\label{semilocal}
In this section, we study the CBLP in semilocal arithmetical algebras.
Throughout this section, we shall assume that all the algebras from ${\cal C}$ are arithmetical, and that the algebra ${\cal A}$ is non--trivial. Consequently, ${\rm Max}({\cal A})$ is non--empty.
We say that ${\cal A}$ is {\em semilocal} iff ${\rm Max}({\cal A})$ is finite.
The results in this section generalize the results on semilocal residuated lattices from \cite[Section $6$]{ggcm}.
\begin{proposition}
The following are equivalent:
\begin{enumerate}
\item\label{propozitie5.2(0)} ${\cal A}$ is semilocal and satisfies $(\star )$;
\item\label{propozitie5.2(1)} ${\cal A}$ is semilocal and has CBLP;
\item\label{propozitie5.2(2)} ${\cal A}$ is semilocal and ${\rm Rad}({\cal A})$ has CBLP;
\item\label{propozitie5.2(3)} there exist $n\in \N ^*$ and $\alpha _1,\ldots ,\alpha _n\in {\cal B}({\rm Con}({\cal A}))$ such that $\displaystyle \bigcap _{i=1}^n\alpha _i=\Delta _A$, $\alpha _i\vee \alpha _j=\nabla _A$ for all $i,j\in \overline{1,n}$ with $i\neq j$ and ${\cal A}/\alpha _i$ is local for all $i\in \overline{1,n}$;
\item\label{propozitie5.2(4)} ${\cal A}$ is isomorphic to a finite direct product of local algebras.\end{enumerate}\label{propozitie5.2}\end{proposition}
\begin{proof} (\ref{propozitie5.2(0)})$\Rightarrow $(\ref{propozitie5.2(1)}): By Proposition \ref{propozitie4.14}.
\noindent (\ref{propozitie5.2(1)})$\Rightarrow $(\ref{propozitie5.2(2)}): Trivial.
\noindent (\ref{propozitie5.2(2)})$\Rightarrow $(\ref{propozitie5.2(3)}): Let $n\in \N ^*$ be the cardinality of ${\rm Max}({\cal A})$ and ${\rm Max}({\cal A})=\{\phi _1,\ldots ,\phi _n\}$. Then, according to \cite[Lemma 2]{afgg}, ${\cal A}/{\rm Rad}({\cal A})$ is isomorphic to $\displaystyle \prod _{i=1}^n{\cal A}/\phi _i$, hence, by Lemma \ref{dinbj}, it follows that there exist $\theta _1,\ldots ,\theta _n\in {\rm Con}({\cal A})$ such that ${\rm Rad}({\cal A})\subseteq \theta _i$ for all $i\in \overline{1,n}$ and the following hold:
\begin{flushleft}\begin{tabular}{cl}
$(a)$ & ${\rm Rad}({\cal A})\subseteq \theta _i$ for all $i\in \overline{1,n}$;\\
$(b)$ & $\theta _i/{\rm Rad}({\cal A})\in {\cal B}({\rm Con}({\cal A}/{\rm Rad}({\cal A})))$ for all $i\in \overline{1,n}$;\\
$(c)$ & $\displaystyle \bigcap _{i=1}^n\theta _i/{\rm Rad}({\cal A})=\Delta _{\textstyle {\cal A}/{\rm Rad}({\cal A})}$;\\
$(d)$ & $\theta _i/{\rm Rad}({\cal A})\vee \theta _j/{\rm Rad}({\cal A})=\nabla _{\textstyle {\cal A}/{\rm Rad}({\cal A})}$ for all $i,j\in \overline{1,n}$ such that $i\neq j$;\\
$(e)$ & for all $i\in \overline{1,n}$, ${\cal A}/\phi _i$ is isomorphic to $({\cal A}/{\rm Rad}({\cal A}))/_{\textstyle (\theta _i/{\rm Rad}({\cal A}))}$, which in turn is isomorphic to ${\cal A}/\theta _i$ by\\
& the Second Isomorphism Theorem, hence ${\cal A}/\phi _i$ is isomorphic to ${\cal A}/\theta _i$.\end{tabular}\end{flushleft}
Since ${\rm Rad}({\cal A})\subseteq \theta _i$ for all $i\in \overline{1,n}$, it follows that $\displaystyle {\rm Rad}({\cal A})\subseteq \bigcap _{i=1}^n\theta _i$. From $(c)$ we obtain: $\displaystyle (\bigcap _{i=1}^n\theta _i)/{\rm Rad}({\cal A})=\Delta _{\textstyle {\cal A}/{\rm Rad}({\cal A})}$, thus, for all $a,b\in A$ such that $\displaystyle (a,b)\in \bigcap _{i=1}^n\theta _i$, it follows that $(a/{\rm Rad}({\cal A}),b/{\rm Rad}({\cal A}))\in \Delta _{\textstyle {\cal A}/{\rm Rad}({\cal A})}$, that is $a/{\rm Rad}({\cal A})=b/{\rm Rad}({\cal A})$, which means that $(a,b)\in {\rm Rad}({\cal A})$; hence $\displaystyle \bigcap _{i=1}^n\theta _i\subseteq {\rm Rad}({\cal A})$. Therefore:
\begin{flushleft}\begin{tabular}{cl}
$(f)$ & $\displaystyle \bigcap _{i=1}^n\theta _i={\rm Rad}({\cal A})$.\end{tabular}\end{flushleft}
From $(d)$ we get that: for all $i,j\in \overline{1,n}$ such that $i\neq j$, $(\theta _i\vee \theta _j)/{\rm Rad}({\cal A})=\nabla _{\textstyle {\cal A}/{\rm Rad}({\cal A})}=\nabla _{\cal A}/{\rm Rad}({\cal A})$, so $s_{\textstyle {\rm Rad}({\cal A})}(\theta _i\vee \theta _j)=s_{\textstyle {\rm Rad}({\cal A})}(\nabla _{\cal A})$, hence, by the injectivity of $s_{\textstyle {\rm Rad}({\cal A})}$:
\begin{flushleft}\begin{tabular}{cl}
$(g)$ & $\theta _i\vee \theta _j=\nabla _{\cal A}$ for all $i,j\in \overline{1,n}$ such that $i\neq j$.\end{tabular}\end{flushleft}
From $(b)$ and the fact that ${\rm Rad}({\cal A})$ has CBLP, it follows that there exist $\alpha _1,\ldots ,\alpha _n\in {\cal B}({\rm Con}({\cal A}))$ such that, for all $i\in \overline{1,n}$, $\theta _i/{\rm Rad}({\cal A})=(\alpha _i\vee {\rm Rad}({\cal A}))/{\rm Rad}({\cal A})$, that is $s_{\textstyle {\rm Rad}({\cal A})}(\theta _i)=s_{\textstyle {\rm Rad}({\cal A})}(\alpha _i\vee {\rm Rad}({\cal A}))$, thus $\theta _i=\alpha _i\vee {\rm Rad}({\cal A})$ by the injectivity of $s_{\textstyle {\rm Rad}({\cal A})}$. We obtain that $\displaystyle (\bigcap _{i=1}^n\alpha _i)\vee {\rm Rad}({\cal A})=\bigcap _{i=1}^n(\alpha _i\vee {\rm Rad}({\cal A}))=\bigcap _{i=1}^n\theta _i={\rm Rad}({\cal A})$ according to $(f)$, thus $\displaystyle \bigcap _{i=1}^n\alpha _i\subseteq {\rm Rad}({\cal A})$. But $\displaystyle \bigcap _{i=1}^n\alpha _i\in {\cal B}({\rm Con}({\cal A}))$, hence $\displaystyle \bigcap _{i=1}^n\alpha _i=\Delta_{\cal A}$ by Lemma \ref{camcalalr}. From $(g)$ we obtain: for all $i,j\in \overline{1,n}$ such that $i\neq j$, $\alpha _i\vee \alpha _j\vee {\rm Rad}({\cal A})=\theta _i\vee \theta _j={\rm Rad}({\cal A})$, hence $\alpha _i\vee \alpha _j=\nabla _{\cal A}$ by Lemma \ref{lema4.11}.
For every $i\in \overline{1,n}$, since $\phi _i\in {\rm Max}({\cal A})$, it follows that $\phi _i\neq \nabla _{\cal A}$, thus ${\cal A}/\phi _i$ is non--trivial, hence, by $(e)$, ${\cal A}/\theta _i$ is non--trivial, thus $\theta _i\neq \nabla _{\cal A}$; but $\theta _i=\alpha _i\vee {\rm Rad}({\cal A})$, so $\alpha _i\leq \theta _i<\nabla _{\cal A}$, thus $\alpha _i\neq \nabla _{\cal A}$, so ${\cal A}/\alpha _i$ is non--trivial, thus $\Delta _{\cal A}$ is a proper congruence of ${\cal A}$, hence ${\rm Max}({\cal A}/\alpha _i)$ is non--empty by Lemma \ref{folclor}, (\ref{folclor1}). According to Proposition \ref{specprod}, (\ref{specprod1}), $\displaystyle \sum _{i=1}^n|{\rm Max}({\cal A}/\alpha _i)|=|{\rm Max}(\prod _{i=1}^n{\cal A}/\alpha _i)|=|{\rm Max}({\cal A}/{\rm Rad}({\cal A}))|=n$. It follows that, for all $i\in \overline{1,n}$, $|{\rm Max}({\cal A}/\alpha _i)|=1$, that is ${\cal A}/\alpha _i$ is local.
\noindent (\ref{propozitie5.2(3)})$\Leftrightarrow $(\ref{propozitie5.2(4)}): By Lemma \ref{dinbj}.
\noindent (\ref{propozitie5.2(4)})$\Rightarrow $(\ref{propozitie5.2(1)}): By Corollaries \ref{corolar4.8} and \ref{corolar4.9}.
\noindent (\ref{propozitie5.2(4)})$\Rightarrow $(\ref{propozitie5.2(0)}): By Corollary \ref{finprodloc}.\end{proof}
\begin{corollary}
If ${\cal A}$ is semilocal, then: ${\cal A}$ satisfies $(\star )$ iff ${\cal A}$ has CBLP iff ${\rm Rad}({\cal A})$ has CBLP.\label{corolar5.2,5}\end{corollary}
\begin{corollary}
If ${\cal A}$ is finite, then: ${\cal A}$ satisfies $(\star )$ iff ${\cal A}$ has CBLP iff ${\rm Rad}({\cal A})$ has CBLP.\label{corolar5.3}\end{corollary}
\begin{openproblem}
Find an arithmetical algebra fulfilling (H) with CBLP that does not satisfy $(\star )$.\end{openproblem}
\begin{remark}
Concerning the open problem above, note that, according to Corollary \ref{corolar5.2,5}, an arithmetical algebra fulfilling (H) with CBLP that does not satisfy $(\star )$ is not semilocal. Finding an algebra which is not semilocal is easy: for instance, according to Proposition \ref{specprodarb}, (\ref{specprodarb1}), a direct product of an infinite family of non--trivial algebras is not semilocal. In some particular cases, such as that of ${\rm G\ddot{o}del}$ algebras, CBLP is preserved by arbitrary direct products, according to Proposition \ref{happy} and \cite[Proposition $5.2$]{ggcm}; unfortunately, in this particular case, so is $(\star )$. The open problem above may prove difficult.\end{remark}
\begin{corollary}
If ${\rm Rad}({\cal A})$ has CBLP, then: ${\cal A}$ is semilocal iff it is isomorphic to a finite direct product of local algebras.\label{corolar5.4}\end{corollary}
\begin{corollary}
If ${\cal A}$ is normal, then: ${\cal A}$ is semilocal iff it is isomorphic to a finite direct product of local algebras.\label{corolar5.5}\end{corollary}
\begin{proof} By Corollary \ref{corolar5.4} and Proposition \ref{propozitie4.12}.\end{proof}
\begin{definition}{\rm \cite{afgg}} The algebra ${\cal A}$ is said to be {\em maximal} iff, given any index set $I$, any family $(a_i)_{i\in I}\subseteq A$ and any family $(\theta _i)_{i\in i}\subseteq {\rm Con}({\cal A})$ with the property that, for any finite subset $J$ of $I$, there exists an $x_J\in A$ such that $(x_J,a_i)\in \theta _i$ for all $i\in J$, it follows that there exists an $x\in A$ such that $(x_i,a_i)\in \theta _i$ for all $i\in I$.\end{definition}
\begin{lemma}{\rm \cite{afgg}} Any maximal algebra is semilocal.\label{lema5.8}\end{lemma}
\begin{lemma}{\rm \cite{afgg}} Let $n\in \N ^*$, ${\cal A}_1,\ldots ,{\cal A}_n$ be algebras and $\displaystyle {\cal A}=\prod _{i=1}^n{\cal A}_i$. Then: ${\cal A}$ is maximal iff ${\cal A}_i$ is maximal for every $i\in \overline{1,n}$.\label{lema5.9}\end{lemma}
\begin{proposition} The following are equivalent:
\begin{enumerate}
\item\label{propozitie5.10(0)} ${\cal A}$ is maximal and satisfies $(\star )$;
\item\label{propozitie5.10(1)} ${\cal A}$ is maximal and has CBLP;
\item\label{propozitie5.10(2)} ${\cal A}$ is maximal and ${\rm Rad}({\cal A})$ has CBLP;
\item\label{propozitie5.10(3)} ${\cal A}$ is isomorphic to a finite direct product of maximal local algebras.\end{enumerate}\label{propozitie5.10}\end{proposition}
\begin{proof} (\ref{propozitie5.10(0)})$\Leftrightarrow $(\ref{propozitie5.10(1)}): By Proposition \ref{propozitie4.14}.
\noindent (\ref{propozitie5.10(1)})$\Leftrightarrow $(\ref{propozitie5.10(2)}): By Lemma \ref{lema5.8} and Proposition \ref{propozitie5.2}.
\noindent (\ref{propozitie5.10(1)})$\Rightarrow $(\ref{propozitie5.10(3)}): By Lemma \ref{lema5.8}, Proposition \ref{propozitie5.2} and Lemma \ref{lema5.9}.
\noindent (\ref{propozitie5.10(3)})$\Rightarrow $(\ref{propozitie5.10(1)}): Assume that ${\cal A}$ is isomorphic to a finite direct product of maximal local algebras. Then ${\cal A}$ is maximal by Lemma \ref{lema5.9}, and ${\cal A}$ has CBLP by Corollaries \ref{corolar4.8} and \ref{corolar4.9}.
\noindent (\ref{propozitie5.10(3)})$\Rightarrow $(\ref{propozitie5.10(0)}): By Lemma \ref{lema5.9} and Corollary \ref{finprodloc}.\end{proof}
\begin{corollary} If ${\rm Rad}({\cal A})$ has CBLP, then: ${\cal A}$ is maximal iff it is isomorphic to a finite direct product of maximal local algebras.\label{corolar5.11}\end{corollary}
\begin{corollary} If ${\cal A}$ is normal, then: ${\cal A}$ is maximal iff it is isomorphic to a finite direct product of maximal local algebras.\label{corolar5.12}\end{corollary}
\begin{proof} By Proposition \ref{propozitie4.12} and Corollary \ref{corolar5.11}.\end{proof}
| 133,550
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ASBURY PARK OKTOBERFEST SET FOR OCTOBER 17-19
The Anchor’s Bend and Asbury Festhalle & Biergarten bar have teamed up to bring Asbury Park Oktoberfest to the Anchor’s Bend inside the iconic Asbury Park Convention Hall the weekend of Friday, Oktober 17, Saturday Oktober 18 and Sunday Oktober 19.
New Downtown Asbury Park business, Asbury Festhalle & Biergarten, is set to open its doors in November, but with the support of The Anchor’s Bend a traditional Oktoberfest will be held thanks to this community effort.
“In our commitment to fostering the Asbury Park community as a whole, The Anchor’s Bend is more than happy to welcome the Biergarten to town. We are excited for what the Biergarten team will bring to Asbury Park, as I’m sure they’re only going to add new energy to the ever-growing eclectic nature that makes this city so very special,” said James Douglas.
“It was an awesome welcoming gesture and great idea for us to team up with The Anchor’s Bend. Asbury Festhalle & Biergarten has received such a warm welcome -- we cannot wait to open. In the meantime, come get your polka on and celebrate Oktoberfest at Anchor’s Bend!”said Jennifer Lampert of the Asbury Festhalle & Biergarten.
Asbury Park Oktoberfest features six draft beers from Radeberger, HofbrauHaus, Hacker-Pschoor Kostritzer, Brooklyn and Paulener breweries and a specialty keg from Weltenburger. Bottled beers include selections from Schöfferhofer, Augesteiner, Flying Fish and Königbreweries. A special, limited quantity of Paulener Stern Weiss bottles will be featured and beer cocktails will also be available.
Festivities include Masskrugstemmen [stein-holding] competitions and live music from Polka Floyd, Alex Meixner and Bavarski.
Schedule of Events
Friday 6pm-2pm
8pm: Specialty Keg tapping Weltenburger
9pm: Masskrugstemmen Competition (Stein holding competition) Great prizes!
10pm-1am: Live-Musik with Polka Floyd
Saturday Noon-6pm
1pm: Specialty Keg tapping Weltenburger
2pm: Masskrugstemmen Competition (Stein holding competition) Great prizes!
2pm-5pm Live-Musik with Alex Meixner:
Grammy Nominated Accordionist plays traditional polkas to rock and pop!
Sunday Noon-6pm
1pm: Specialty Keg tapping Weltenburger 1pm-4pm Live-Musik with Bavarski:
2pm: Masskrugstemmen Competition (Stein holding competition) Great prizes! Get your Polka on Brooklyn style!
FOR FURTHER INFORMATION-
The Anchor's Bend Bar is a casual bar and grille featuring select beers and fine spirits. It is located at the northeast corner of the Grand Arcade within Asbury Park's historic Convention Hall at 1300 Ocean Avenue. For more information, find The Anchor's Bend on Facebook at.
Asbury Festhalle & Biergarten, located at 527 Lake Avenue, aims to be the first truly authentic Austro-Hungarian Biergarten on the Jersey Shore. It is currently slated for a late November opening. For more information, visit or find them on Facebook at.
| 99,192
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New Delhi: In the Joint conference of Chief Ministers and Chief Justices of High Courts held at Delhi, the Present Prime Minister of the Country, Narendra Modi addressed the audience in the inauguration function.
The idea behind such forum, as was stated by PM, Modi, is for increasing transparency and strengthening the nation, by strengthening the concerned organs thereof.
There have be raised, many time, a concern about anti-corruption, however, no one found able to give proper and effective solution, as per PM, Modi.
He appreciate the work of persons engaged in ‘law and order industry’ as they are doing divine work, which is the work God requires one to do. Further, he said, these people having many responsibilities as they are the one whom others look up to. He shown our luckiness to have such manpower in the law and order wing of the nation.
In future, the quality manpower in the law and order filed will be seen, as per PM, Modi. The Judiciary must be powerful and perfect one and If it is the national requirement to have perfect or powerful judiciary, we need to ask, he said.
PM, Modi, pointed in his speech that, bringing technology as quickly as possible in the filed of judiciary will lead to easier qualitative change in it.
Further, by pointing the availability of major bulk of law, he state that now there is trend adopted by law making agencies to enact new laws, even though ‘we have enough laws’. He said, ministers are elected to go to the Parliament for providing effective laws, however, the working time at the Parliament is how being wasted, he pointed by saying “you all know what happens in the Parliament”.
Moreover, in his speech, PM Modi, specifically stated that there is a committee set by him, for recommending the laws to be repealed.
Prime Minster, Narendra Modi, seen talking about qualitative development in law and justice organ of the State. He asked the new generations to come forward and take part in it and provide the work of quality.
By Faim Khalilkhan Pathan
| 112,415
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Profile Information
Mike Wilson is a passionate content writer and a part time blogger by profession. He love writing and marketing technical as well as non-technical articles and blogs. He writes unbiased and informative articles and reviews on verity of categories like: eCommerce| Data Entry & Data Management | ePublishing | Advanced Application Development & Cloud Solutions | Digital Marketing.
Setting Up Magento for the Search EnginesBlog Post: September 23, 2014
It's a great read!! Contains all the useful insights for setting up Magento for search engines. Thanks a lot @Lewis for sharing this stuff. I think, you should see this too: SEO Best Practices for your eCommerce Store – Part II. Contains some great insights for improving an eCommerce site’s performance with SEO best practices at all times.
Will be waiting for your response on this :)
Thanks!
| 397,142
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MANY of us think that the gym is the best place to burn fat and calories – but that’s not the only case.
There are now specially-designed garments that can help you burn calories no matter where you are or what you’re doing.
So if you’re struggling to slim down, yet unwilling to go the extra mile at the gym, then it’s time to update your closet by investing in Be-fit Firming Pants.
Be-fit Firming Pants is essentially compression apparel driven by the success of the Kodenshi fibre that is made using state-of-the-art Ginrei knitting technology from Japan.
It works by integrating ultra-fine ceramic particles into the fibre. When worn, the fabric absorbs body heat to release far infra-red rays (FIR) back into our skin to improve blood circulation, resulting in higher body metabolism which helps to burn calories faster.
The patented Kodenshi fibre absorbs temperatures of between 34°C and 37°C emitted by the human body.
Since the FIR consists only of electromagnetic waves, which get molecules to generate heat, the wave itself is not heat and does not feel hot.
With this proprietary technology, Be-fit Firming Pants feel lighter and breathable, ensuring maximum comfort.
It is also designed to help reduce tummy fats and regain a better body shape, while for athletes, it also helps to improve sports performance and endurance.Only available at CNI. Kindly logon to for further information.
| 164,969
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How much does it cost to ship a SUV from Seattle to Utah?
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| 285,851
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\begin{document}
\title{A method to find generators of a semi-simple Lie group via the
topology of its flag manifolds}
\author{Ariane Luzia dos Santos\thanks{
Address: FCL - Unesp, Departamento de Ciências da Educação. Rodovia
Araraquara-Jaú, km 1, 14.800-901 Araraquara, São Paulo, Brasil. e-mail:
ariane@fclar.unesp.br} \and Luiz A. B. San Martin\thanks{
Supported by CNPq grant n$^{\mathrm{o}}$ 303755/2009-1 and FAPESP grant n$^{
\mathrm{o}}$ 07/06896-5}\thanks{
Address: Imecc - Unicamp, Departamento de Matemática. Rua Sérgio Buarque de
Holanda, 651, Cidade Universitária Zeferino Vaz, 13083-859 Campinas, São
Paulo, Brasil. e-mail: smartin@ime.unicamp.br}}
\date{}
\maketitle
\begin{abstract}
In this paper we continue to develop the topological method started in
Santos-San Martin \cite{ariasm} to get semigroup generators of semi-simple
Lie groups. Consider a subset $\Gamma \subset G$ that contains a semi-simple
subgroup $G_{1}$ of $G$. Then $\Gamma $ generates $G$ if $\mathrm{Ad}\left(
\Gamma \right) $ generates a Zariski dense subgroup of the algebraic group $
\mathrm{Ad}\left( G\right) $. The proof is reduced to check that some
specific closed orbits of $G_{1}$ in the flag manifolds of $G$ are not
trivial in the sense of algebraic topology. Here, we consider three
different cases of semi-simple Lie groups $G$ and subgroups $G_{1}\subset G$.
\end{abstract}
\noindent
\textit{AMS 2010 subject classification:} $20M05, 22E46, 22F30$
\noindent
\textit{Key words and phrases:} Semigroup generators of groups, semi-simple
Lie groups, flag manifolds.
\section{Introduction}
In this paper we continue to develop the topological method started in
Santos-San Martin \cite{ariasm} to get semigroup generators of semi-simple
Lie groups.
The method is based on the notion of flag type of a semigroup that arouse
from the results of \cite{smroconex}, \cite{sminv}, \cite{smsurv}, \cite{SM}
, \cite{SMT}, \cite{smord}, \cite{smmax} and \cite{smsanhom}. By these
results if $G$ is a connected noncompact semi-simple Lie group with finite
center and $S\subset G$ is a proper semigroup with nonempty interior, then
the flag type of $S$ allows to select a flag manifold $\mathbb{F}_{\Theta }$
of $G$ that contains a subset $C_{\Theta }$ (the so called invariant control
set) which is invariant by $S$ and is contained in an open Bruhat cell $
\sigma _{\Theta }$ of $\mathbb{F}_{\Theta }$. Since $\sigma _{\Theta }$ is
diffeomorphic to an Euclidean space $\mathbb{R}^{N}$ it follows that $
C_{\Theta }$ is contractible in $\mathbb{F}_{\Theta }$.
Hence if one can show that a subset $\Gamma \subset G$ does not leave
invariant a contractible subset on any flag manifold of $G$, then it is
possible to conclude that $\Gamma $ generates $G$ if the semigroup generated
by $\Gamma $ has nonempty interior. Actually thanks to a result by Abels
\cite{abel} this last condition can be changed by asking that $\mathrm{Ad}
\left( \Gamma \right) $ generates a Zariski dense subgroup of the algebraic
group $\mathrm{Ad}\left( G\right) $.
The problem of semigroup generation of groups has several motivations. One
of them comes from control theory where the controllability problem is
translated into the semigroup generation problem. Controllability results on
semi-simple Lie groups where obtained in Jurdjevic-Kupka \cite{JK, JK1},
Gauthier-Kupka-Sallet \cite{GKS} and El Assoudi-Gauthier-Kupka \cite{EGK}.
In another direction we mention the 1.5 generation problem studied in Abels
\cite{abel}, Abels-Vinberg \cite{abelsV} and references therein, which
consists in finding pairs of generators starting from one element of the
pair.
Here, as in our previous paper \cite{ariasm}, we take as generator a subset $
\Gamma \subset G$ that contains a semi-simple subgroup $G_{1}$ of $G$. In
this setting the problem is reduced to check that some specific closed
orbits of $G_{1}$ in the flag manifolds of $G$ are not trivial in the sense
of algebraic topology (see Proposition \ref{propSemiSimpleGeral} below).
In \cite{ariasm} we pursued this approach with an eye in the controllability
results mentioned above. Thus in \cite{ariasm} the group $G$ is a connected
complex simple Lie group and $G_{1}$ is a subgroup $G\left( \alpha \right) $
with Lie algebra isomorphic to $\mathfrak{sl}\left( 2,\mathbb{C}\right) $
generated by the root spaces $\mathfrak{g}_{\pm \alpha }$ associated to the
roots $\pm \alpha $ of the complex Lie algebra $\mathfrak{g}$ of $G$. In
that case the method was successfully applied because the relevant orbits of
$G_{1}=G\left( \alpha \right) $ are flag manifolds of $\mathfrak{sl}\left( 2,
\mathbb{C}\right) $ so that diffeomorphic to $S^{2}$. We used De Rham
cohomology $H^{2}\left( \mathbb{F}_{\Theta },\mathbb{R}\right) $ to prove
that these orbits are not homotopic to a point for any flag manifold $
\mathbb{F}_{\Theta }$.
In this paper we consider three different cases of semi-simple Lie groups $G$
and subgroups $G_{1}\subset G$.
In the first one we take $G$ whose Lie algebra $\mathfrak{g}$ is the split
real form of a complex simple Lie algebra and $G_{1}=G\left( \alpha \right) $
is a subgroup generated by root spaces analogous to the complex case. In the
real case we are led to check whether some closed curves are null homotopic
and hence work with the fundamental groups of the flag manifolds. Contrary
to the complex Lie algebras there are only a few cases where the relevant
orbits of $G_{1}=G\left( \alpha \right) $ are null homotopic. Namely when $
\mathfrak{g}=\mathfrak{sl}\left( n,\mathbb{R}\right) $, $\mathfrak{g}=
\mathfrak{sp}\left( n,\mathbb{R}\right) $ and $\alpha $ is a long root and
when $\mathfrak{g}$ is the split real form of $G_{2}$ and $\alpha $ is a
short root. In these cases the subgroups \ $G\left( \alpha \right) $ are not
contained in proper semigroups with interior points and a subset $\Gamma $
generates $G$ if $G\left( \alpha \right) \subset \Gamma $ (any root $\alpha $
) and the group generated by $\Gamma $ is Zariski dense (see Theorem \ref
{teoreal} below). In the remaining cases our method breaks down. We give an
example of a proper semigroup with nonempty interior in $\mathrm{Sp}\left( n,
\mathbb{R}\right) $ that contains $G\left( \alpha \right) $ for several
short roots $\alpha $, showing that the result is indeed not true in this
case.
In another direction we take an irreducible finite dimensional
representation $\rho _{n}:\mathfrak{sl}\left( 2,\mathbb{C}\right)
\rightarrow \mathfrak{sl}\left( n,\mathbb{C}\right) $, $n\geq 2$, and get a
\linebreak subgroup $G_{1}=\langle \exp \rho _{n}\left( \mathfrak{sl}\left(
2,\mathbb{C}\right) \right) \rangle \subset \mathrm{Sl}\left( n,\mathbb{C}
\right) $. We prove that $G_{1}$ is not contained in any proper semigroup
with nonempty interior of $\mathrm{Sl}\left( n,\mathbb{C}\right) $. It
follows that a subset $\Gamma $ is a semigroup generator of $\mathrm{Sl}
\left( n,\mathbb{C}\right) $ if it contains $\langle \exp \rho _{n}\left(
\mathfrak{sl}\left( 2,\mathbb{C}\right) \right) \rangle $ and the group
generated by $\Gamma $ is Zariski dense (see Theorem \ref{teoParaRepreSl2}).
The algebraic topological fact that permits the proof of this result is
Proposition \ref{propfiblinhas2}. It shows that the projective orbit $
G_{1}\cdot \left[ v_{0}\right] $ of the highest weight space is not
contractible in the projective space of the vector space of the
representation. This noncontractibility follows from the fact that the
tautological bundle of the projective space restricts to a nontrivial bundle
on the orbit
In our third case we take a complex Lie group $G$ and a complex subgroup $
G_{1}$ such that the Lie algebra $\mathfrak{g}_{1}$ of $G_{1}$ contains a
regular real element of the Lie algebra $\mathfrak{g}$ of $G$. Examples of
this case are the inclusion in $\mathfrak{sl}\left( n,\mathbb{C}\right) $ of
the classical Lie algebras $B_{l}=\mathfrak{so}\left( 2l+1,\mathbb{C}\right)
$, $C_{l}=\mathfrak{sp}\left( 2l,\mathbb{C}\right) $ and $D_{l}=\mathfrak{so}
\left( 2l,\mathbb{C}\right) $. In this case we prove that $G_{1}$ is not
contained in a proper semigroup with nonempty interior of $G$ and hence get
generators of $G$ in the same spirit of the other cases (see Theorem \ref
{teoParaSemiComplex}). Here we exploit the same technique provided by
Proposition \ref{propfiblinhas2} by realizing the flag manifolds of $G$ as
projective orbits in spaces of representations. The proof is not much
different from the case of representations of $\mathfrak{sl}\left( 2,\mathbb{
C}\right) $.
\section{Notation and background}
Let $G$ be a connected real semi-simple Lie group with finite center and Lie
algebra $\mathfrak{g}$. For $G$ and $\mathfrak{g}$ we use the following
notation.
\begin{itemize}
\item Let $\theta $ be a Cartan involution of $\mathfrak{g}$ and $\mathfrak{g
}=\mathfrak{k}\oplus \mathfrak{s}$ the corresponding Cartan decomposition.
\item If $\mathfrak{a}\subset \mathfrak{s}$ is a maximal abelian subalgebra
its set of roots is denoted by $\Pi $. Let $\Pi ^{+}$ be a set of positive
roots with
\begin{equation*}
\Sigma =\{\alpha _{1},\ldots ,\alpha _{l}\}\subset \Pi ^{+}
\end{equation*}
corresponding simple system of roots. We have $\Pi =\Pi ^{+}\dot{\cup}(-\Pi
^{+})$ and any $\alpha \in \Pi ^{+}$ is a linear combination $\alpha
=n_{1}\alpha _{1}+\cdots +n_{l}\alpha _{l}$ and $n_{i}\geq 0$, $i=1,\ldots
,l $ are integers. The support of $\alpha $, $\mathrm{supp}\alpha $, is the
subset of $\Sigma $ where $n_{i}>0$.
\item The Cartan-Killing form of $\mathfrak{g}$ is denoted by $\langle \cdot
,\cdot \rangle $. If $\alpha \in \mathfrak{a}^{\ast }$ then $H_{\alpha }\in
\mathfrak{a}$ is defined by $\alpha (\cdot )=\langle H_{\alpha },\cdot
\rangle $, and $\langle \alpha ,\beta \rangle =\langle H_{\alpha },H_{\beta
}\rangle $.
\item We write
\begin{equation*}
\mathfrak{a}^{+}=\{H\in \mathfrak{a}:\forall \alpha \in \Pi ^{+},\,\alpha
(H)>0\}
\end{equation*}
for the Weyl chamber defined by $\Pi ^{+}$.
\item The root space of a root $\alpha $ is
\begin{equation*}
\mathfrak{g}_{\alpha }=\{X\in \mathfrak{g}:\forall H\in \mathfrak{a}
,\,[H,X]=\alpha (H)X\}.
\end{equation*}
If $\mathfrak{g}$ is a split real form of a complex simple Lie algebra then $
\dim _{\mathbb{R}}\mathfrak{g}_{\alpha }=1$.
\item For a root $\alpha $, $\mathfrak{g}(\alpha )$ is the subalgebra
generated by $\mathfrak{g}_{\alpha }$ and $\mathfrak{g}_{-\alpha }$. We have
\begin{equation*}
\mathfrak{g}(\alpha )=\mathrm{span}_{\mathbb{R}}\{H_{\alpha }\}\oplus
\mathfrak{g}_{\alpha }\oplus \mathfrak{g}_{-\alpha }\approx \mathfrak{sl}(2,
\mathbb{R}).
\end{equation*}
We let $G(\alpha )$ be the connected Lie subgroup of $G$ with Lie algebra $
\mathfrak{g}(\alpha )$.
\item Let $K=\langle \exp \mathfrak{k}\rangle $ be the maximal compact
subgroup of $G$ with Lie algebra $\mathfrak{k}$.
\item $\mathcal{W}$ is the Weyl group. Either $\mathcal{W}$ is the group
generated by the reflections $r_{\alpha }$, $\alpha \in \Pi $, $r_{\alpha
}(\beta )=\beta -\frac{2\langle \alpha ,\beta \rangle }{\langle \alpha
,\alpha \rangle }\alpha $, or $\mathcal{W}=M^{\ast }/M$ where $M^{\ast }=
\mathrm{Norm}_{K}(\mathfrak{a})$ is the normalizer of $\mathfrak{a}$ in $K$
and $M=\{g\in K:\mathrm{Ad}(g)H=H\,\mbox{for all}\,H\in \mathfrak{h}\}$ is
the centralizer of $\mathfrak{a}$ in $K$.
\item $\mathfrak{n}^{+}=\sum_{\alpha \in \Pi ^{+}}\mathfrak{g}_{\alpha }$
and $\mathfrak{n}^{-}=\sum_{\alpha \in \Pi ^{+}}\mathfrak{g}_{-\alpha }$.
\item Given the data $\mathfrak{a}$ and $\Pi ^{+}$ (or $\Sigma $) there
exists the minimal parabolic subalgebra $\mathfrak{p}=\mathfrak{m}\oplus
\mathfrak{a}\oplus \mathfrak{n}^{+}$. A subset $\Theta \subset \Sigma $
defines the standard parabolic subalgebra by
\begin{equation*}
\mathfrak{p}_{\Theta }=\mathfrak{p}+\sum_{\alpha \in \langle \Theta \rangle }
\mathfrak{g}_{\alpha }
\end{equation*}
where $\langle \Theta \rangle =\{\alpha \in \Pi :$ $\mathrm{supp}\alpha
\subset \Theta $ or $\mathrm{supp}(-\alpha )\subset \Theta \}$ is the set of
roots spanned by $\Theta $. When $\Theta =\emptyset $ we have $\mathfrak{p}
_{\emptyset }=\mathfrak{p}$.
\item For $\Theta \subset \Sigma $, $P_{\Theta }$ is the parabolic subgroup
with Lie algebra $\mathfrak{p}_{\Theta }$:
\begin{equation*}
P_{\Theta }=\mathrm{Norm}_{G}(\mathfrak{p}_{\Theta })=\{g\in G:\mathrm{Ad}(g)
\mathfrak{p}_{\Theta }\subset \mathfrak{p}_{\Theta }\}.
\end{equation*}
\item The flag manifold $\mathbb{F}_{\Theta }=G/P_{\Theta }$ does not depend
on the specific group $G$ with Lie algebra $\mathfrak{g}$. The origin of $
G/P_{\Theta }$, the coset $1\cdot P_{\Theta }$, is denoted by $b_{\Theta }$.
\end{itemize}
Now let $S\subset G$ be a subsemigroup with $\mathrm{int}S\neq \emptyset $.
We recall some results of \cite{sminv, smsurv, SMT} that are on the basis of
our topological approach to get generators of $G$.
We let $S$ act on a flag manifold $\mathbb{F}_{\Theta }$ by restricting the
action of $G$. An invariant control set for $S$ in $\mathbb{F}_{\Theta }$ is
a subset $C\subset \mathbb{F}_{\Theta }$ such that $\mathrm{cl}(Sx)=C$ for
every $x\in C$, where $Sx=\{gx\in \mathbb{F}_{\Theta }:g\in S\}$. Since $
\mathrm{int}S\neq \emptyset $ such a set is closed, has nonempty interior
and is in fact invariant, that is, $gx\in C$ if $g\in S$ and $x\in C$.
\begin{lema}
\textrm{{(\cite[Theorem 3.1]{sminv})}} \label{lemuniqueics}In any flag
manifold $F_{\Theta }$ there is a unique invariant control set for $S$,
denoted by $C_{\Theta }$.
\end{lema}
To state the geometric property of $C_{\Theta }$ to be used later we discuss
the dynamics of the vector fields $\widetilde{H}$ on a flag manifold $
\mathbb{F}_{\Theta }$ whose flow is $e^{tH}$, with $H$ in the closure $
\mathrm{cl}\mathfrak{a}^{+}$ of the Weyl chamber $\mathfrak{a}^{+}$. It is
known that $\widetilde{H}$ is a gradient vector field with respect to some
Riemmannian metric on $\mathbb{F}_{\Theta }$; cf. \cite[Proposition 3.3 (ii)]
{dkv} and \cite{fepase}.
Hence the orbits of $\widetilde{H}$ are either fixed points of $e^{tH}$ or
trajectories flowing between fixed points of $H$. Moreover, $\widetilde{H}$
has a unique attractor fixed point set, say $\mathrm{att}_{\Theta }(H)$,
that has an open and dense stable manifold $\sigma _{\Theta }(H)$; cf. \cite
{dkv, fepase}. This means that if $x\in \sigma _{\Theta }(H)$ then its $
\omega $-limit set $\omega (x)$ is contained in $\mathrm{att}_{\Theta }(H)$.
This attractor has the following algebraic expressions
\begin{equation*}
\mathrm{att}_{\Theta }(H)=Z_{H}\cdot b_{\Theta }=K_{H}\cdot b_{\Theta },
\end{equation*}
cf. \cite[Corollary 3.5]{dkv} and \cite{fepase}. Here $Z_{H}=\{g\in G:
\mathrm{Ad}(g)H=H\}$ is the centralizer of $H$ in $G$ and $K_{H}=Z_{H}\cap K$
is the centralizer in $K$. We note that $\mathrm{att}_{\Theta }(H)$ is a
connected manifold because $Z_{H}=M(Z_{H})_{0}=(Z_{H})_{0}M$ where $
(Z_{H})_{0}$ is the identity component of $Z_{H}$ and $M$ is the centralizer
of $\mathfrak{a}$ in $K$; see \cite[Lemma 1.2.4.5]{W}. Hence $Z_{H}\cdot
b_{\Theta }=(Z_{H})_{0}\cdot b_{\Theta }$, since $M\cdot b_{\Theta
}=b_{\Theta }$; cf. \cite[Theorem 7.101]{K}.
The stable set $\sigma _{\Theta }(H)$ is also described algebraically by
\begin{equation*}
\sigma _{\Theta }(H)=N_{H}^{-}Z_{H}\cdot b_{\Theta }
\end{equation*}
where
\begin{equation*}
N_{H}^{-}=\exp \mathfrak{n}_{H}^{-}\quad \mathrm{and}\quad \mathfrak{n}
_{H}^{-}=\sum_{\gamma (H)<0}\mathfrak{g}_{\gamma },
\end{equation*}
cf. \cite[Corollary 3.5]{dkv}. In particular if $H$ is regular, that is, $
H\in \mathfrak{a}$ and $\alpha (H)>0$ for $\alpha \in \Pi ^{+}$, then $
Z_{H}=MA$, which fixes $b_{\Theta }$. Hence
\begin{equation*}
\mathrm{att}_{\Theta }(H)=Z_{H}\cdot b_{\Theta }=\{b_{\Theta }\}\qquad H\in
\mathfrak{a}.
\end{equation*}
Actually, in the regular case the fixed points are isolated because $
\widetilde{H}$ is the gradient of a Morse function; cf. \cite{dkv, fepase}.
Also, $\mathfrak{n}_{H}^{-}=\mathfrak{n}^{-}$ (notation as above) and the
stable set is $N^{-}\cdot b_{\Theta }$ the open Bruhat cell.
The following statement is a well known result from the Bruhat decomposition
of the flag manifolds; cf. \cite{dkv, K, W}.
\begin{proposicao}
In any flag manifold $\mathbb{F}_{\Theta }$ the open Bruhat cell $N^{-}\cdot
b_{\Theta }$ is diffeomorphic to an Euclidean space $\mathbb{R}^{d}$. The
diffeomorphism is $X\in \mathfrak{n}_{\Theta }^{-}\mapsto e^{X}\cdot
b_{\Theta }$, where $\mathfrak{n}_{\Theta }^{-}=\sum \{\mathfrak{g}_{\alpha
}:\alpha <0$ and $\alpha \notin \langle \Theta \rangle \}$.
\end{proposicao}
Set $h=e^{H}$, $H\in \mathfrak{a}^{+}$. It follows from the gradient
property of $\widetilde{H}$ that $\lim_{n\rightarrow +\infty
}h^{n}x=b_{\Theta }$ for any $x\in N^{-}\cdot b_{\Theta }$.
Now, we say that $g\in G$ is regular real if it is a conjugate $g=aha^{-1}$
of $h=\exp H$, $H\in \mathfrak{a}^{+}$ with $a\in G$. Then we write $\sigma
_{\Theta }( g) =g\cdot \sigma _{\Theta}( H) $ and we call this the stable
set of $g$ in $\mathbb{F}_{\Theta }$. The reason for this name is clear: $
g^{n}=(aha^{-1}) ^{n}=ah^{n}a^{-1}$ and hence $g^{n}x\rightarrow gb_{\Theta
} $ if $x\in \sigma _{\Theta }( g) $.
The following lemma was used in \cite{sminv} to prove the above Lemma \ref
{lemuniqueics}.
\begin{lema}
\textrm{{(\cite[Lemma 3.2]{sminv})}} There exists a regular real $g\in
\mathrm{int}S$.
\end{lema}
Now we can state the following result of \cite{SMT} which is basic to our
approach.
\begin{teorema}
\label{teoregre}Suppose that $S\neq G$. Then there exists a flag manifold $
\mathbb{F}_{\Theta }$ such that the invariant control set $C_{\Theta
}\subset \sigma _{\Theta }(g)$ for every regular real $g\in \mathrm{int}S$.
\end{teorema}
\begin{corolario}
\label{coreuclid}If $S\neq G$ then there exists a flag manifold $\mathbb{F}
_{\Theta }$ such that for every flag manifold $\mathbb{F}_{\Theta _{1}}$
such that $\Theta \subset \Theta _{1}$ the invariant control set $C_{\Theta
_{1}}$ in $\mathbb{F}_{\Theta _{1}}$ is contained in a subset $\mathcal{E}
_{\Theta _{1}}$ diffeomorphic to an Euclidean space.
\end{corolario}
\begin{profe}
If $\Theta _{1}\subset \Theta $ then the canonical projection $\pi :\mathbb{F
}_{\Theta }\rightarrow \mathbb{F}_{\Theta _{1}}$ is equivariant under the
actions of $G$. This implies that the open Bruhat cells are projected onto
open cells\ and $\pi \left( C_{\Theta }\right) =C_{\Theta _{1}}$. Hence $
C_{\Theta _{1}}$ is contained in an open cell $\mathcal{E}_{\Theta _{1}}$ if
this happens to $C_{\Theta }$.
\end{profe}
In particular if $\Theta _{1}=\Sigma \setminus \{\alpha \}$ contains $\Theta
$ if $\alpha \in \Sigma \setminus \Theta $ so in the minimal flag manifold $
\mathbb{F}_{\Sigma \setminus \{\alpha \}}$ the invariant control set is
contained in open cells.
\begin{corolario}
\label{coreuclidminimal}If $S\neq G$ then there exists a minimal flag
manifold $\mathbb{F}_{\Theta }$ such that $C_{\Theta }$ is contained in a
subset $\mathcal{E}_{\Theta }$ diffeomorphic to an Euclidean space.
\end{corolario}
\vspace{12pt}
\noindent \textbf{Remark:} It can be proved that there exists a minimal $
\Theta \left( S\right) $ satisfying the condition of Theorem \ref{teoregre}.
This $\Theta \left( S\right) $ (or rather the flag manifold $\mathbb{F}
_{\Theta \left( S\right) }$) is called the flag type or parabolic type of $S$
. Several properties of $S$ are derived from this flag type, e.g. the
homotopy type of $S$ as in \cite{smsanhom} or the connected components of $S$
as in \cite{smroconex}.
\section{Semi-simple subgroups: Set up\label{secSemiGeral}}
Let $\mathfrak{g}$ be a noncompact semi-simple Lie algebra and $G$ be a
connected Lie group with Lie algebra $\mathfrak{g}$ and finite center. In
this section we take a semi-simple subalgebra $\mathfrak{g}_{1}\subset
\mathfrak{g}$ and the corresponding connected subgroup $G_{1}=\langle \exp
\mathfrak{g}_{1}\rangle $. We ask whether there is a proper semigroup $
S\subset G$ with $\mathrm{int}S\neq \emptyset $ such that $G_{1}\subset S$.
It is well known that there exist compatible Cartan decompositions $
\mathfrak{g}_{1}=\mathfrak{k}_{1}\oplus \mathfrak{s}_{1}$ and $\mathfrak{g}=
\mathfrak{k}\oplus \mathfrak{s}$ such that $\mathfrak{k}_{1}\subset
\mathfrak{k}$ and $\mathfrak{s}_{1}\subset \mathfrak{s}$ (see Warner \cite[
Lemma 1.1.5.5]{W}). If $\mathfrak{a}_{1}\subset \mathfrak{s}_{1}$ is maximal
abelian then there exists a maximal abelian $\mathfrak{a}\subset \mathfrak{s}
$ with $\mathfrak{a}_{1}\subset \mathfrak{a}$. Denote by $\Pi _{1}$ the
roots of $\left( \mathfrak{g}_{1},\mathfrak{a}_{1}\right) $ and by $\Pi $
the roots of $\left( \mathfrak{g},\mathfrak{a}\right) $. Any $\alpha _{1}\in
\Pi _{1}$ is the restriction to $\mathfrak{a}_{1}$ of some $\alpha \in \Pi $
. Take $H_{1}\in \mathfrak{a}_{1}$ regular (in $\mathfrak{g}_{1}$). Then
there exists a Weyl chamber $\mathfrak{a}^{+}\subset \mathfrak{a}$ with $
H_{1}\in \mathrm{cl}\mathfrak{a}^{+}$. If $\alpha \in \Pi $ is positive
w.r.t $\mathfrak{a}^{+}$ then $\alpha \left( H_{1}\right) \geq 0$ so that we
get compatible Iwasawa decompositions $\mathfrak{g}_{1}=\mathfrak{k}
_{1}\oplus \mathfrak{a}_{1}\oplus \mathfrak{n}_{1}$ and $\mathfrak{g}=
\mathfrak{k}\oplus \mathfrak{a}\oplus \mathfrak{n}$ with $\mathfrak{k}
_{1}\subset \mathfrak{k}$, $\mathfrak{a}_{1}\subset \mathfrak{a}$ and $
\mathfrak{n}_{1}\subset \mathfrak{n}$ where
\begin{equation*}
\mathfrak{n}_{1}=\sum_{\alpha _{1}\in \Pi _{1}^{+}}\left( \mathfrak{g}
_{1}\right) _{\alpha _{1}}\, ,\quad \mathfrak{n}=\sum_{\alpha \in \Pi ^{+}}
\mathfrak{g}_{\alpha }
\end{equation*}
and $\Pi _{1}^{+}$ is the set of roots of $\left( \mathfrak{g}_{1},\mathfrak{
a}_{1}\right) $ that are positive on $H_{1}$ and $\Pi ^{+}$ the roots
positive on $\mathfrak{a}^{+}$.
In the sequel we keep fixed these compatible Iwasawa decompositions. Let $
\Sigma $ be the simple system of roots in $\Pi ^{+}$. Then the standard
parabolic subalgebras $\mathfrak{p}_{\Theta }\subset \mathfrak{g}$ and
subgroups $P_{\Theta }\subset G$ are built from subsets $\Theta
\subset\Sigma $. For the corresponding flag manifolds $\mathbb{F}
_{\Theta}=G/P_{\Theta }$ we write $b_{\Theta }=1\cdot P_{\Theta }$ for their
origins.
By the construction of the compatible Iwasawa decomposition from the choice
of $H_{1}\in \mathfrak{a}_{1}\cap \mathrm{cl}\mathfrak{a}^{+}$ we have that $
b_{\Theta }$ belongs to the attractor fixed point set $\mathrm{att}
_{\Theta}(H_{1})=Z_{H_{1}}\cdot b_{\Theta }=K_{H_{1}}\cdot b_{\Theta }$ of
the one-parameter semigroup $e^{tH_{1}}$, $t\geq 0$. The corresponding
stable set $\sigma _{\Theta }(H_{1})$ is open and dense in $\mathbb{F}
_{\Theta }$.
Now let $S$ be a semigroup with $\mathrm{int}S\neq \emptyset $ such that $
G_{1}\subset S$. Denote by $C_{\Theta }$ the unique $S$-invariant control
set in $\mathbb{F}_{\Theta }$. The set $C_{\Theta }$ is compact and has
nonempty interior. Hence $C_{\Theta }\cap \sigma _{\Theta }(H_{\alpha
})\neq\emptyset $. If $x\in C_{\Theta }\cap \sigma _{\Theta }(H_{\alpha })$
then $y=\lim_{t\rightarrow +\infty }e^{tH_{\alpha }}\cdot x$ belongs to $
C_{\Theta}\cap \mathrm{att}_{\Theta }(H_{\alpha })$, because $C_{\Theta }$
is closed. Hence we get the following lemma.
\begin{lema}
\label{lemintericsatrac}If $G_{1}\subset S$ then $C_{\Theta }\cap \mathrm{att
}_{\Theta }(H_{1})\neq \emptyset $.
\end{lema}
Now the idea is to look at the topology of the orbits $G_{1}\cdot y$ with $
y\in \mathrm{att}_{\Theta }(H_{1})$. Clearly if $y\in C_{\Theta }$ and $
G_{1}\subset S$ then $G_{1}\cdot y\subset C_{\Theta }$. On the other hand,
by Theorem \ref{teoregre} and its corollaries, there are flag manifolds
where $C_{\Theta }$ is contained in a contractible Euclidean subset $
\mathcal{E}_{\Theta }$ if $S$ is proper. Hence if we achieve to prove that
none of the orbits $G_{1}\cdot y$ with $y\in \mathrm{att}_{\Theta }(H_{1})$
are contractible then we can conclude that $G_{1}$ is not contained in a
proper semigroup with nonempty interior. In principle this
noncontractibility property must by checked on every flag manifold but by
Corollary \ref{coreuclidminimal} it is enough to look at the minimal ones.
These arguments can be used to get semigrouop generators of $G$. In fact, if
$\Gamma $ is a subset that contains $G_{1}$ and generates a semigroup $S$
with nonempty interior then $S=G$ provided we have noncontractibility of the
orbits $G_{1}\cdot y$ through the attractor fixed point set.
Actually, thanks to a result by Abels \cite{abel} it is enough to assume
that the group generated by $\Gamma $ is Zariski dense in the following
sense: The group $\mathrm{Ad}\left( G\right) $ is algebraic and hence
endowed with the Zariski topology. We say that $B\subset G$ is Zariski dense
in case $\mathrm{Ad}\left( B\right) $ is dense in $\mathrm{Ad}\left(
G\right) $ with respect to the Zariski topology. With this terminology it is
proved in \cite{abel}, as a consequence of Corollary 5, that the semigroup $
S $ generated by $\Gamma $ has non empty interior provided i) the group
generated by $\Gamma $ is Zariski dense and ii) $S$ contains a non-constant
smooth curve.
These comments yield the following fact that reduces the problem of finding
semigroup generators to the topology of orbits of $G_{1}$.
\begin{proposicao}
\label{prop:zariski} \label{propSemiSimpleGeral}Let $\mathfrak{g}_{1}\subset
\mathfrak{g}$ be a semi-simple Lie subalgebra. Choose compatible Iwasawa
decompositions $\mathfrak{g}_{1}=\mathfrak{k}_{1}\oplus \mathfrak{a}
_{1}\oplus \mathfrak{n}_{1}\subset \mathfrak{g}=\mathfrak{k}\oplus \mathfrak{
a}\oplus \mathfrak{n}$, so that $\mathrm{cl}\mathfrak{a}^{+}$ contains a
regular element $H_{1}\in \mathfrak{a}_{1}$. Let $G_{1}=\langle \exp
\mathfrak{g}_{1}\rangle $ and suppose that for every (minimal) flag manifold
$\mathbb{F}_{\Theta }$ the orbits $G_{1}\cdot y$ through the attractor fixed
point set $\mathrm{att}_{\Theta }(H_{1})$ are not contractible in $\mathbb{F}
_{\Theta }$. Then a subset $\Gamma \subset G$ generates $G$ as a semigroup
if $G_{1}\subset \Gamma $ and the subgroup generated by $\Gamma $ is Zariski
dense.
\end{proposicao}
In the special case when $H_{1}\in \mathfrak{a}^{+}$, that is, $\mathfrak{g}
_{1}$ contains a regular real element of $\mathfrak{g}$, the attractor fixed
point $\mathrm{att}_{\Theta }(H_{1})$ reduces to $b_{\Theta }$. In this case
we need to check contractibility only of the orbits $G_{1}\cdot b_{\Theta }$
through the origin. For later reference we record this fact.
\begin{corolario}
\label{corSemiSimpleGeral}With the notation of Proposition \ref{prop:zariski}
suppose that $H_{1}\in \mathfrak{a}^{+}$. Then the same result holds with
the assumption that $G_{1}\cdot b_{\Theta }$ is not contractible in any
(minimal) flag manifold.
\end{corolario}
\section{Split real forms and subgroups $G\left( \protect\alpha \right) $}
We assume in this section that $\mathfrak{g}$ is a split real form of a
complex simple Lie algebra and $G$ is a connected Lie group with Lie algebra
$G$ and having finite center. Take a Cartan decomposition $\mathfrak{g}=
\mathfrak{k}\oplus \mathfrak{s}$ and a maximal abelian $\mathfrak{a}\subset
\mathfrak{s}$. If $\alpha $ is a root of the pair $\left( \mathfrak{g},
\mathfrak{a}\right) $ then the subalgebra $\mathfrak{g}\left( \alpha \right)
$ generated by the root spaces $\mathfrak{g}_{\pm \alpha }$ is isomorphic to
$\mathfrak{sl}\left( 2,\mathbb{R}\right) $. Precisely, we choose $X_{\alpha
}\in \mathfrak{g}_{\alpha }$, $X_{-\alpha }=-\theta X_{\alpha }\in \mathfrak{
g}_{-\alpha }$ such that $\langle X_{\alpha },X_{-\alpha }\rangle =1$. Then
the isomorphism is given by
\begin{equation*}
X_{\alpha }\leftrightarrow \left(
\begin{array}{cc}
0 & 1 \\
0 & 0
\end{array}
\right) ,\qquad H_{\alpha }\leftrightarrow \left(
\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}
\right) ,\qquad X_{-\alpha }\leftrightarrow \left(
\begin{array}{cc}
0 & 0 \\
1 & 0
\end{array}
\right) .
\end{equation*}
In order to apply the ideas of Section \ref{secSemiGeral} we fix a Weyl
chamber $\mathfrak{a}^{+}$ such that $H_{\alpha }\in \mathrm{cl}\mathfrak{a}
^{+}$. Then $\alpha $ is positive w.r.t. $\mathfrak{a}^{+}$ and we get
compatible Iwasawa decompositions $\mathfrak{g}\left( \alpha \right) =
\mathfrak{k}_{H_{\alpha }}\oplus \langle H_{\alpha }\rangle \oplus \mathfrak{
g}_{\alpha }\subset \mathfrak{g}=\mathfrak{k}\oplus \mathfrak{a}\oplus
\mathfrak{n}$ where $\mathfrak{k}_{H_{\alpha }}$ is spanned by $
A_{\alpha}=X_{\alpha }-X_{-\alpha }$. We put $G\left( \alpha \right)
=\langle \exp \mathfrak{g}\left( \alpha \right) \rangle $.
Our objective is to check whether the conditions of Proposition \ref
{propSemiSimpleGeral} are satisfied by $G\left( \alpha \right) $. The
following result reduces the question to the orbit through the origin $
b_{\Theta }$ of a flag manifold $\mathbb{F}_{\Theta }$. (Where the origins
are given by the standard parabolic subalgebras defined from the Weyl
chamber $\mathfrak{a}^{+}$.)
\begin{teorema}
\label{teoOrbitsAtt}In a flag manifold $\mathbb{F}_{\Theta }$ take $y\in
\mathrm{att}_{\Theta }(H_{\alpha })$. Then the orbit $G(\alpha )\cdot y$ is
a circle $S^{1}$ homotopic to the orbit $G(\alpha )\cdot b_{\Theta }$.
\end{teorema}
The proof this theorem requires some lemmas. We start by looking at the
orbit $G(\alpha )\cdot b_{\Theta }$. By compatibility of the Iwasawa
decompositions it follows that the parabolic subalgebra $\mathfrak{p}
_{\alpha }=\langle H_{\alpha }\rangle \oplus \mathfrak{g}_{\alpha }$ of $
\mathfrak{g}\left( \alpha \right) $ is contained in the isotropy subalgebra
at $b_{\Theta }$ for any flag manifold $\mathbb{F}_{\Theta }$. To get the
inclusion of the parabolic subgroup of $G\left( \alpha \right) $ as well we
perform a somewhat standard computation on $\mathfrak{sl}(2,\mathbb{R})$ and
$\mathfrak{sl}\left( 2,\mathbb{C}\right) $ (cf. \cite{K}, Chapter VII.5).
\begin{proposicao}
\label{propcentroreal}Let $G(\alpha )\subset G$ be the connected subgroup
with Lie algebra $\mathfrak{g}(\alpha )\approx \mathfrak{sl}(2,\mathbb{R})$
and $M$ the centralizer of $\mathfrak{a}$ in $K$. Then the center $Z\left(
G(\alpha )\right) $ of $G(\alpha )$ is contained in $M$.
\end{proposicao}
\begin{profe}
Let $\widetilde{G}=\widetilde{\mathrm{Sl}(2,\mathbb{R})}$ be the universal
covering and denote by $\widetilde{\exp }:\mathfrak{sl}(2,\mathbb{R}
)\rightarrow \widetilde{\mathrm{Sl}(2,\mathbb{R})}$ the exponential map.
Take the basis
\begin{equation*}
A=\left(
\begin{array}{cc}
0 & -1 \\
1 & 0
\end{array}
\right) ,\qquad H=\left(
\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}
\right) ,\qquad S=\left(
\begin{array}{cc}
0 & 1 \\
1 & 0
\end{array}
\right)
\end{equation*}
of $\mathfrak{sl}(2,\mathbb{R})$. The center $Z\left( \widetilde{\mathrm{Sl}
(2,\mathbb{R})}\right) $ of $\widetilde{\mathrm{Sl}(2\mathbb{R})}$ is the
kernel of its adjoint representation $\widetilde{\mathrm{Ad}}$ which is
explicitly given by
\begin{equation*}
Z\left( \widetilde{\mathrm{Sl}(2,\mathbb{R})}\right) =\{\widetilde{\exp }
(k\pi A):k\in \mathbb{Z}\}\approx \mathbb{Z},
\end{equation*}
since the center is contained in a one-parameter group $\widetilde{\exp }
(tA) $.
Now $\mathfrak{g}(\alpha )\approx \mathfrak{sl}(2,\mathbb{R})$ with the
isomorphism given by $H\leftrightarrow H_{\alpha }^{\vee }=\frac{2H_{\alpha }
}{\langle \alpha ,\alpha \rangle }$, $A\leftrightarrow A_{\alpha }=X_{\alpha
}-X_{-\alpha }$ and $S\leftrightarrow S_{\alpha }=X_{\alpha }+X_{-\alpha }$,
where $X_{\pm \alpha }\in \mathfrak{g}_{\pm \alpha }$ and $\langle X_{\alpha
},X_{-\alpha }\rangle =1$.
Suppose first that $G=\mathrm{Aut}_{0}\mathfrak{g}$ is the adjoint group.
Then $G(\alpha )=\widetilde{\mathrm{Sl}(2,\mathbb{R})}/D$ with $D\subset
Z\left( \widetilde{\mathrm{Sl}(2,\mathbb{R})}\right) $ given by
\begin{equation*}
D=\{\widetilde{\exp }(kn_{0}\pi A):k\in \mathbb{Z}\}\approx n_{0}\mathbb{Z}
\end{equation*}
where
\begin{equation*}
n_{0}=\min \{n>0:e^{n\pi \mathrm{ad}\left( A_{\alpha }\right) }=\mathrm{id}
\}.
\end{equation*}
It follows that $Z\left( G(\alpha )\right) =Z\left( \widetilde{\mathrm{Sl}(2,
\mathbb{R})}\right) /D\approx \mathbb{Z}_{n_{0}}$ is generated by $e^{\pi
\mathrm{ad}\left( A_{\alpha }\right) }$. Complexifying and doing
computations in $\mathrm{Sl}(2,\mathbb{C})$, it turns out that $e^{\pi
\mathrm{ad}(A_{\alpha })}=e^{\pi \mathrm{ad}(iH_{\alpha }^{\vee })}$. This
last term belongs to $M$ showing that $Z\left( G(\alpha )\right) \subset M$
when $G=\mathrm{Aut}_{0}\mathfrak{g}$. For a general $G$ the same result is
obtained by taking adjoints.
\end{profe}
The next lemma will ensure that the isotropy subalgebra at $b_{\Theta }$ for
the action of $G\left( \alpha \right) $ is exactly $\mathfrak{p}_{\alpha
}=\langle H_{\alpha }\rangle \oplus \mathfrak{g}_{\alpha }$.
\begin{lema}
\label{lemsuport}If $\alpha $ is a root with $H_{\alpha }\in \mathrm{cl}
\mathfrak{a}^{+}$, then $\mathrm{supp}\alpha =\Sigma $.
\end{lema}
\begin{profe}
See \cite[Proposition 3.3]{ariasm}.
\end{profe}
Now we can describe the orbits through the origins of the flag manifolds.
\begin{lema}
\label{lemesse1}Let $b_{\Theta }$ be the origin of a flag manifold $\mathbb{F
}_{\Theta }$ and $\beta $ a positive root. Then $G(\beta )\cdot b_{\Theta }$
is either a circle $S^{1}$ or it reduces to a point. If $\beta \notin
\langle \Theta \rangle $ then $\dim \,G(\beta )\cdot b_{\Theta }=1$. In
particular, $\dim G(\alpha )\cdot b_{\Theta }=1$ if $H_{\alpha }\in \mathrm{
cl}\mathfrak{a}^{+}$.
\end{lema}
\begin{profe}
Let $\mathfrak{g}\left( \beta \right) _{b_{\Theta }}$ and $G\left( \beta
\right) _{b_{\Theta }}$ be the isotropy subalgebra and subgroup,
respectively, at $b_{\Theta }$ for the action of $G\left( \beta \right) $.
The subalgebra $\mathfrak{g}\left( \beta \right) _{b_{\Theta }}$ contains
the parabolic subalgebra of $\mathfrak{g}\left( \beta \right) $ given by $
\mathfrak{p}_{\beta }=\mathrm{span}\{H_{\beta }\}\oplus \mathfrak{g}_{\beta
}\subset \mathfrak{p}_{\Theta }$. This implies that $G\left( \beta \right)
_{b_{\Theta }}$ contains the identity component $\left( P_{\beta }\right)
_{0}$ of the parabolic subgroup $P_{\beta }=\mathrm{Norm}_{G\left( \beta
\right) }\mathfrak{p}_{\beta }\subset G\left( \beta \right) $. If $\mathfrak{
g}\left( \beta \right) _{b_{\Theta }}=\mathfrak{p}_{\beta }$ then $G\left(
\beta \right) _{b_{\Theta }}$ is a union of connected components of $
P_{\beta }$. But $P_{\beta }=Z\left( G(\alpha )\right) (P_{\beta })_{0}$ and
$Z\left( G(\alpha )\right) \subset M\subset P_{\Theta }$ by Proposition \ref
{propcentroreal}. Hence, in this case $G\left( \beta \right) _{b_{\Theta
}}=P_{\beta }$ and the orbit $G\left( \beta \right) \cdot b_{\Theta
}=G\left( \beta \right) /P_{\beta }\approx S^{1}$. On the other hand if $
\mathfrak{p}_{\beta }\neq \mathfrak{g}\left( \beta \right) _{b_{\Theta }}$
then $\mathfrak{g}\left( \beta \right) _{b_{\Theta }}=\mathfrak{g}\left(
\beta \right) $, so that $G\left( \beta \right) _{b_{\Theta }}=G\left( \beta
\right) $ and the orbit $G\left( \beta \right) \cdot b_{\Theta }$ reduces to
a point.
Now, if $\beta \notin \langle \Theta \rangle $ then $\mathfrak{g}_{-\beta }$
has empty intersection with the isotropy subalgebra $\mathfrak{p}_{\Theta }$
which implies that $\mathfrak{g}\left( \beta \right) _{b_{\Theta }}=
\mathfrak{p}_{\beta }$. Hence $G\left( \beta \right) \cdot b_{\Theta }$ is a
circle $S^{1}$ if $\beta \notin \langle \Theta \rangle $. The last statement
is a consequence of Lemma \ref{lemsuport}.
\end{profe}
We proceed now to look at the orbits of $G(\alpha )$ through $y\in \mathrm{
att}_{\Theta }(H_{\alpha })$. We recall that $\mathrm{att}_{\Theta
}(H_{\alpha })=\left( K_{H_{\alpha }}\right) _{0}\cdot b_{\Theta }$, where $
\left( K_{H_{\alpha }}\right) _{0}$ is the identity component of the compact
part of the centralizer $Z_{H_{\alpha }}$ of $H_{\alpha }$. The Lie algebra $
\mathfrak{k}_{H_{\alpha }}$ of $K_{H_{\alpha }}$ is spanned by $A_{\alpha
}=X_{\alpha }-X_{-\alpha }$.
Let $y=u\cdot b_{\Theta }\in \mathrm{att}_{\Theta }(H_{\alpha })$ with $u\in
\left( K_{H_{\alpha }}\right) _{0}$. Then
\begin{equation*}
G(\alpha )\cdot y=u\left( u^{-1}G(\alpha )u\right) \cdot b_{\Theta }.
\end{equation*}
The group $u^{-1}G(\alpha )u$ is isomorphic to $G(\alpha )$ and its Lie
algebra $\mathrm{Ad}(u^{-1})\mathfrak{g}(\alpha )$ is isomorphic to $
\mathfrak{g}(\alpha )$ and hence to $\mathfrak{sl}(2,\mathbb{R})$. Since $
\mathrm{Ad}\left( u\right) H_{\alpha }=H_{\alpha }$, the decomposition of $
\mathrm{Ad}(u^{-1})\mathfrak{g}(\alpha )$ into root spaces is given by
\begin{equation*}
\mathrm{Ad}\left( u^{-1}\right) \mathfrak{g}(\alpha )=\langle H_{\alpha
}\rangle \oplus \mathrm{Ad}\left( u^{-1}\right) \mathfrak{g}_{\alpha }\oplus
\mathrm{Ad}\left( u^{-1}\right) \mathfrak{g}_{-\alpha }.
\end{equation*}
The subspace $\mathfrak{p}_{u}=\langle H_{\alpha }\rangle \oplus \mathrm{Ad}
(u^{-1})\mathfrak{g}_{\alpha }$ is a parabolic subalgebra of $\mathrm{Ad}
(u^{-1})\mathfrak{g}(\alpha )$. Denote by $P_{u}$ the corresponding
parabolic subgroup.
\begin{lema}
\label{isotropia}The subgroup $P_{u}$ is contained in the isotropy subgroup
at $b_{\Theta }$ of the action of $u^{-1}G(\alpha )u$.
\end{lema}
\begin{profe}
We have that $\langle H_{\alpha }\rangle $ is contained in the isotropy
algebra, because $b_{\Theta }$ is a fixed point of $e^{tH_{\alpha }}$. To
see that the same occurs with $\mathrm{Ad}(u^{-1})(\mathfrak{g}_{\alpha })$,
note that if $0\neq X\in \mathrm{Ad}(u^{-1})(\mathfrak{g}_{\alpha })$ then $
\mathrm{ad}(H_{\alpha })(X)=\alpha (H_{\alpha })X$ because $u$ centralizes $
H_{\alpha }$. Hence $X$ is an eigenvector of $\mathrm{ad}(H_{\alpha })$
associated to the eigenvalue $\alpha (H_{\alpha })=\langle \alpha ,\alpha
\rangle >0$. Since $H_{\alpha }\in \mathrm{cl}\mathfrak{a}^{+}$ we have that
the eigenspaces of $\mathrm{ad}(H_{\alpha })$ associated to positive
eigenvalues are contained in $\mathfrak{n}^{+}$. Therefore $X\in \mathfrak{n}
^{+}$ so that $\mathrm{Ad}(u^{-1})(\mathfrak{g}_{\alpha })\subset \mathfrak{n
}^{+}$, implying that $\mathrm{Ad}(u^{-1})(\mathfrak{g}_{\alpha })$ is
contained in the isotropy subalgebra at $b_{\Theta }$.
Now, the proof follows as in Lemma \ref{lemesse1} by checking that any
connected component of $P_{u}$ is contained in the isotropy subgroup at $
b_{\Theta }$.
\end{profe}
As a complement to this lemma we have the following homotopy property. From
it the proof of Theorem \ref{teoOrbitsAtt} follows quickly.
\begin{lema}
\label{homotopia}The orbit $\left( u^{-1}G(\alpha )u\right) \cdot b_{\Theta
} $ is a circle $S^{1}$ \ homotopic to the orbit $G(\alpha )\cdot b_{\Theta
} $, by a homotopy that fixes $b_{\Theta }$.
\end{lema}
\begin{profe}
Let $u_{t}\in \left( K_{H_{\alpha }}\right) _{0}$, $t\in \lbrack 0,1]$, be a
continuous curve with $u_{0}=1$ e $u_{1}=u$. Define the continuous map $\psi
:[0,1]\times G(\alpha )\rightarrow \mathbb{F}_{\Theta }$ by
\begin{equation*}
\psi (t,g)=u_{t}^{-1}gu_{t}\cdot b_{\Theta }.
\end{equation*}
This map has the factorization
\begin{equation*}
\begin{array}{ccc}
\lbrack 0,1]\times G(\alpha ) & \longrightarrow & \mathbb{F}_{\Theta } \\
\downarrow & \nearrow & \\
\left[ 0,1\right] \times \left( G\left( \alpha \right) /P_{\alpha }\right) &
&
\end{array}
\end{equation*}
that defines a continuous map $\phi :[0,1]\times \left( G(\alpha )/P_{\alpha
}\right) \rightarrow \mathbb{F}_{\Theta }$ by $\phi (t,gP_{\alpha })=\psi
(t,g)$. Indeed, $\phi $ is well defined because if $h\in P_{\alpha }$ then $
u_{t}^{-1}hu_{t}\in P_{u_{t}}$ and, by the previous lemma, $
(u_{t}^{-1}hu_{t})\cdot b_{\Theta }=b_{\Theta }$. Hence
\begin{equation*}
\psi (gh)=u_{t}^{-1}gu_{t}(u_{t}^{-1}hu_{t})\cdot b_{\Theta
}=u_{t}^{-1}gu_{t}\cdot b_{\Theta }=\phi \left( g\right)
\end{equation*}
and so $\phi $ is well defined. The function $\phi $ is continuous and if $
b_{\alpha }$ denotes the origin of $G(\alpha )/P_{\alpha }$ then, again by
the previous lemma, we have that $\phi (t,b_{\alpha })=b_{\Theta }$ for all $
t\in \lbrack 0,1]$.
Therefore, looking at $G(\alpha )/P_{\alpha }$ as a circle $S^{1}$ with
distinguished point $b_{\alpha }=1\cdot P_{\alpha }$ we see that $\phi $ is
a homotopy between $\phi (0,\cdot )$ whose image is $G(\alpha )\cdot
b_{\Theta }$ and $\phi (1,\cdot )$ whose image is $\left( u^{-1}G(\alpha
)u\right) \cdot b_{\Theta }$. This homotopy fixes $b_{\Theta }$.
\end{profe}
Now we can finish the proof of Theorem \ref{teoOrbitsAtt}: Let $u\in \left(
K_{H_{\alpha }}\right) _{0}$ be such that $y=ub_{\Theta }$. Then $G(\alpha
)\cdot y=u\left( u^{-1}G(\alpha )u\right) \cdot b_{\Theta }$, so that $
G(\alpha )\cdot y$ and $\left( u^{-1}G(\alpha )u\right) \cdot b_{\Theta }$
are homotopic by a homotopy defined by a curve $t\in \left[ 0,1\right]
\mapsto u_{t}\in \left( K_{H_{\alpha }}\right) _{0}$ with $u_{0}=1$ and $
u_{1}=u$. By the previous lemma it follows that $G(\alpha )\cdot y$ and $
G(\alpha )\cdot b_{\Theta }$ are homotopic.
Combining Theorem \ref{teoOrbitsAtt} with Proposition \ref
{propSemiSimpleGeral} we arrive at once at the following result.
\begin{teorema}
\label{teohomotopyorbits}Let $\alpha $ be a root and $\mathfrak{a}^{+}$ a
Weyl chamber with $H_{\alpha }\in \mathrm{cl}\mathfrak{a}^{+}$. Define the
standard parabolic subgroups $P_{\Theta }$ with respect the positive roots
associated to $\mathfrak{a}^{+}$ and denote by $b_{\Theta }$ the origin of $
\mathbb{F}_{\Theta }=G/P_{\Theta }$. Take a subset $\Gamma \subset G$ with $
G\left( \alpha \right) \subset \Gamma $. Assume that
\begin{enumerate}
\item for every minimal flag manifold $\mathbb{F}_{\Theta }$ the orbit $
G\left( \alpha \right) \cdot b_{\Theta }$ (which is a closed curve) is not
homotopic to a point and
\item the subgroup generated by $\Gamma $ is Zariski dense.
\end{enumerate}
Then $\Gamma $ generates $G$ as semigroup.
\end{teorema}
As will be seen below condition (1) of this theorem holds in three cases
namely when $\mathfrak{g}=\mathfrak{sl}\left( n,\mathbb{R}\right) $, $\alpha
$ is a long root of $\mathfrak{sp}\left( n,\mathbb{R}\right) $ and $\alpha $
is a short root of the $G_{2}$ digram.
\subsection{The fundamental group of minimal flag manifolds}
We look here at the fundamental groups of the minimal flag manifolds and the
homotopy classes of the orbits $G\left( \alpha \right) \cdot b_{\Theta }$
appearing in Theorem \ref{teohomotopyorbits}.
Fix a simple system of roots $\Sigma $. For a root $\alpha $ we choose $
X_{\alpha }\in \mathfrak{g}_{\alpha }$, $X_{-\alpha }=-\theta X_{\alpha }\in
\mathfrak{g}_{-\alpha }$ such that $\langle X_{\alpha },X_{-\alpha }\rangle
=1$, and write $A_{\alpha }=X_{\alpha }-X_{-\alpha }\in \mathfrak{k}_{\alpha
}$ and $S_{\alpha }=X_{\alpha }+X_{-\alpha }$. If $\mathbb{F}_{\Theta }$, $
\Theta \subset \Sigma $, is a flag manifold we write $b_{\Theta }$ for its
origin defined by $\Sigma $. Then for a root $\alpha $ the orbit $G\left(
\alpha \right) \cdot b_{\Theta }$ is the closed curve $e^{tA_{\alpha }}\cdot
b_{\Theta }$. We denote by $c_{\alpha }^{\Theta }$ (or simply $c_{\alpha }$)
the homotopy class of this curve in the fundamental group of $\mathbb{F}
_{\Theta }$.
The fundamental groups of real flag manifolds were described in Wiggerman
\cite{wig} by generators and relations. We recall the result of \cite{wig}
the case of a split real form $\mathfrak{g}$.
\begin{teorema}
\label{teofundwig}The fundamental group $\pi _{1}\left( \mathbb{F}_{\Theta
}\right) $ of $\mathbb{F}_{\Theta }$ is generated by $c_{\alpha }$ with $
\alpha \in \Sigma $, subjected to the relations
\begin{enumerate}
\item $c_{\alpha }=1$ if $\alpha \in \Theta $ and
\item $c_{\alpha }c_{\beta }c_{\alpha }^{-1}c_{\beta }^{-\varepsilon \left(
\alpha ,\beta \right) }=1$ where $\varepsilon \left( \alpha ,\beta \right)
=\left( -1\right) ^{\langle \alpha ^{\vee },\beta \rangle }$ and $\langle
\alpha ^{\vee },\beta \rangle =\frac{2\langle \alpha ^{\vee },\beta \rangle
}{\langle \alpha ,\alpha \rangle }$ is the Killing number.
\end{enumerate}
\end{teorema}
The first relation says that $\pi _{1}\left( \mathbb{F}_{\Theta }\right) $
is in fact generated by $c_{\alpha }$ with $\alpha \in \Sigma \setminus
\Theta $. However it is convenient to include in the statement the
generators $c_{\alpha }$, $\alpha \in \Theta $, because they enter in the
second set of relations. Namely, $c_{\beta }c_{\beta }^{-\varepsilon \left(
\alpha ,\beta \right) }=c_{\alpha }c_{\beta }c_{\alpha }^{-1}c_{\beta
}^{-\varepsilon \left( \alpha ,\beta \right) }=1$ if $\alpha \in \Theta $
and $\beta \in \Sigma \setminus \Theta $. Notice that this relation implies
that $c_{\beta }^{2}=1$, $\beta \in \Sigma \setminus \Theta $, if there
exists a root $\alpha \in \Theta $ such that the Killing number $\langle
\alpha ^{\vee },\beta \rangle $ is odd.
From these generators and relations it is easy to get the fundamental groups
of the minimal flag manifolds. Given a root $\beta \in \Sigma $ we write $
\Theta _{\beta }=\Sigma \setminus \{\beta \}$ and take the corresponding
minimal flag manifold $\mathbb{F}_{\Theta _{\beta }}$. These exaust the
minimal flag manifolds, except for the diagram $A_{1}$ when the only flag
manifold is given by $\Theta =\emptyset $.
\begin{proposicao}
\label{propze2}The fundamental group of a minimal flag manifold $\mathbb{F}
_{\Theta _{\beta }}$ is $\pi _{1}\left( \mathbb{F}_{\Theta _{\beta }}\right)
=\mathbb{Z}_{2}$ except in for $A_{1}$ or when $\beta $ is the long root in
the $C_{l}$ diagram. In the exceptions $\pi _{1}\left( \mathbb{F}_{\Theta
_{\beta }}\right) =\mathbb{Z}$.
\end{proposicao}
\begin{profe}
By Theorem \ref{teofundwig} the fundamental group of $\mathbb{F}_{\Theta
_{\beta }}$ is cyclic and generated by $c_{\beta }$. If the diagram is not $
A_{1}$ then $\beta $ is linked to another root $\alpha $. If the link is a
simple edge then $\langle \alpha ^{\vee },\beta \rangle =-1$ and hence $
c_{\beta }^{2}=1$ so that $\pi _{1}\left( \mathbb{F}_{\Theta _{\beta
}}\right) =\mathbb{Z}_{2}$. The Killing number $\langle \alpha ^{\vee
},\beta \rangle $ is also odd in the $G_{2}$ diagram ($-1$ or $-3$) or in
case $\alpha $ is a long root and $\beta $ a short root. A glance at the
Dynkin diagrams shows that every root $\beta $ has such link if $\beta $ is
not the long root in the $C_{l}$ diagram. If $\beta $ is this long root then
there are no relations involving $c_{\beta }$ and hence $\pi _{1}\left(
\mathbb{F}_{\Theta _{\beta }}\right) $ is cyclic infinity.
Finally, the flag manifold of $A_{1}$ is the circle $S^{1}$ and hence has $
\mathbb{Z}$ as fundamental group.
\end{profe}
We proceed now to look at the homotopy condition of Theorems \ref
{teoOrbitsAtt} and \ref{teohomotopyorbits}. Here we change slightly the
point of view. In those theorems we started with a root $\alpha $ and
choosed a Weyl chamber containing $H_{\alpha }$ in its closure. Here we fix
a Weyl chamber $\mathfrak{a}^{+}$ (and hence an origin $b_{\Theta }$ in a
flag manifold $\mathbb{F}_{\Theta }$) and take a root $\mu $ such that $
H_{\mu }\in \mathrm{cl}\mathfrak{a}^{+}$. Since the Weyl group $\mathcal{W}$
acts transitively on the set of Weyl chambers, there is no loss of
generality in fixing $\mathfrak{a}^{+}$ in advance. In fact, by $wH_{\alpha
}=H_{w\alpha }$ it follows that for any root $\alpha $ there exists a root $
\mu =w\alpha $ such that $H_{\mu }\in \mathrm{cl}\mathfrak{a}^{+}$, that is,
$H_{\alpha }\in \mathrm{cl}\left( w^{-1}\mathfrak{a}^{+}\right) $ . If $
\overline{w}\in M^{\ast }$ is a representative of $w$ then $G\left( \alpha
\right) =\overline{w}^{-1}G\left( \mu \right) \overline{w}$ and $G\left(
\alpha \right) \cdot \overline{w}^{-1}b_{\Theta }=\overline{w}^{-1}G\left(
\mu \right) \overline{w}\cdot \overline{w}^{-1}b_{\Theta }=\overline{w}
^{-1}\left( G\left( \mu \right) \cdot b_{\Theta }\right) $. The point $
\overline{w}^{-1}b_{\Theta }$ is the origin corresponding to $w^{-1}
\mathfrak{a}^{+}$ and a curve $t\in \left[ 0,1\right] \mapsto g_{t}\in G$
with $g_{0}=1$ and $g_{1}=\overline{w}$ realizes a homotopy between the two
orbits $G\left( \alpha \right) \cdot \overline{w}^{-1}b_{\Theta }$ and $
G\left( \mu \right) \cdot b_{\Theta }$. Therefore we can restrict our
analysis to roots $\mu $ with $H_{\mu }\in \mathrm{cl}\mathfrak{a}^{+}$ and
the Weyl chamber $\mathfrak{a}^{+}$ previously fixed.
The action of $\mathcal{W}$ on the set of roots is either transitive (for
the simply laced diagrams $A_{l}$, $D_{l}$, $E_{6}$, $E_{7}$ and $E_{8}$) or
has two orbits the long and short roots (for the other diagrams $B_{l}$, $
C_{l}$, $F_{4}$ and $G_{2}$). This implies that for the simply laced
diagrams there is just one root (the highest positive root) $\mu $ with $
H_{\mu }\in \mathrm{cl}\mathfrak{a}^{+}$ while in the other cases there is
the highest root and a short root as well in the closure of the Weyl chamber.
To look at the classes $c_{\mu }^{\Theta }$ with $H_{\mu }\in \mathrm{cl}
\mathfrak{a}^{+}$ in the fundamental groups of the minimal flag manifolds we
realize them as orbits of projective representations.
\subsection{Projective and spherical orbits}
To look at the fundamental group of the flag manifolds we shall exploit
their realizations as projective orbits of representations. Given the simple
system of roots $\Sigma =\{\alpha _{1},\ldots ,\alpha _{l}\}$ let $\{\omega
_{1},\ldots ,\omega _{l}\}$ be the set of basic weights defined by
\begin{equation*}
\langle \alpha _{i}^{\vee },\omega _{j}\rangle =\frac{2\langle \alpha
_{i},\omega _{j}\rangle }{\langle \alpha _{i},\alpha _{j}\rangle }=\delta
_{ij}.
\end{equation*}
Any $\omega =p_{1}\omega _{1}+\cdots +p_{l}\omega _{l}$, $p_{i}\in \mathbb{Z}
$, $p_{i}\geq 0$, is the highest weight of a representation $\rho _{\omega }$
of $\mathfrak{g}$ in a vector space $V_{\omega }$. Write $G=\langle \exp
\rho _{\omega }\left( \mathfrak{g}\right) \rangle $ for the group with Lie
algebra $\mathfrak{g}$ that integrates the representation.
Let $V\left( \omega \right) \subset V_{\omega }$ be the one-dimensional
weight space of $\omega $. For $v\in V\left( \omega \right) $, $v\neq 0$,
put $V^{+}\left( \omega \right) =\mathbb{R}^{+}v$ for the ray spanned by $v$.
Consider the projective orbit $G\cdot V\left( \omega \right) =\{g\cdot
V\left( \omega \right) :g\in G\}$ and the spherical orbit $G\cdot
V^{+}\left( \omega \right) =\{g\cdot V^{+}\left( \omega \right) :g\in G\}$.
The isotropy subalgebra at both $V\left( \omega \right) $ and $V^{+}\left(
\omega \right) $ is the parabolic subalgebra $\mathfrak{p}_{\Theta _{\omega
}}$ where
\begin{equation*}
\Theta _{\omega }=\{\alpha _{i}\in \Sigma :p_{i}=0\}=\{\alpha \in \Sigma
:\alpha \left( \omega \right) =0\}.
\end{equation*}
Hence the identity component of the parabolic subgroup $P_{\Theta _{\omega
}} $ is contained in the isotropy subgroups of $V\left( \omega \right) $ and
$V^{+}\left( \omega \right) $. It turns out that the isotropy subgroup at $
V\left( \omega \right) $ is the whole $P_{\Theta _{\omega }}$, since $m\cdot
v=\pm v$ if $m\in M$. Therefore $G\cdot V\left( \omega \right) \approx
\mathbb{F}_{\Theta _{\omega }}$ with the origin $b_{\Theta _{\omega }}\in
\mathbb{F}_{\Theta _{\omega }}$ being identified to $V\left( \omega \right)
\in G\cdot V\left( \omega \right) $. Also the the map $G\cdot V^{+}\left(
\omega \right) \rightarrow G\cdot V\left( \omega \right) $, $g\cdot
V^{+}\left( \omega \right) \mapsto g\cdot V\left( \omega \right) $ is a
covering because the isotropy subgroup at $V^{+}\left( \omega \right) $ is
an open subgroup of $P_{\Theta _{\omega }}$. This covering has one or two
leaves, depending if $-V^{+}\left( \omega \right) $ belongs or not to the
orbit $G\cdot V^{+}\left( \omega \right) $.
To look closely at the covering $G\cdot V^{+}\left( \omega \right)
\rightarrow G\cdot V\left( \omega \right) $ we take a root $\alpha $. The
orbit $G\left( \alpha \right) \cdot b_{\Theta }$ is the curve $\left( \exp
tA_{\alpha }\right) \cdot b_{\Theta }$, $t\in \left[ 0,\pi \right] $, where $
A_{\alpha }=X_{\alpha }-X_{-\alpha }$. This means that $m_{\alpha }v=\pm v$
if $v\in V\left( \omega \right) $, where $m_{\alpha }=\exp \pi \rho _{\omega
}\left( A_{\alpha }\right) $. The following lemma gives the sign in this
equality.
\begin{lema}
\label{lemmpesomax}$m_{\alpha }v=\left( -1\right) ^{\omega \left( H_{\alpha
}^{\vee }\right) }v$.
\end{lema}
\begin{profe}
The Lie algebra $\rho _{\omega }\left( \mathfrak{g}\right) $ as well as the
Lie group $G=\langle \exp \rho _{\omega }\left( \mathfrak{g}\right) \rangle $
can be complexified. In the complexification we have $m_{\alpha }=\exp \pi
\rho _{\omega }\left( A_{\alpha }\right) =\exp i\pi \rho _{\omega }\left(
H_{\alpha }^{\vee }\right) $ (cf. Proposition \ref{propcentroreal}). Hence $
m_{\alpha }v=e^{i\pi \omega \left( H_{\alpha }^{\vee }\right) }v=\left(
-1\right) ^{\omega \left( H_{\alpha }^{\vee }\right) }v$ if $v\in V\left(
\omega \right) $.
\end{profe}
Now let $\omega =\omega _{j}$ be a basic weight, so that $\Theta _{\omega
_{j}}=\Sigma \setminus \{\alpha _{j}\}$, and $\mathbb{F}_{\Theta _{\omega
_{j}}}$ is a minimal flag manifold. By the very definition of the basic
weights we have $\omega _{j}\left( H_{\alpha _{j}}^{\vee }\right) =1$, so
that $m_{\alpha _{j}}v=-v$ if $v\in V\left( \omega _{j}\right) $. This
implies that the spherical orbit $G\cdot V^{+}\left( \omega _{j}\right) $ is
a double covering of $\mathbb{F}_{\Theta _{\omega _{j}}}$.
On the other hand we have, by Proposition \ref{propze2}, that except in two
cases the fundamental group of a minimal flag manifold is $\mathbb{Z}_{2}$.
Combining these facts we get the universal covering space of the minimal
flag manifolds.
\begin{proposicao}
\label{propunivercover}Let $\omega _{j}$ be a basic weight in a diagram
different from $A_{1}$ and such that $\alpha _{j}$ is not the long root of
the $C_{l}$ diagram. Let $\mathbb{F}_{\Theta _{\omega _{j}}}$ be the
corresponding minimal flag manifold. Then the spherical orbit $G\cdot
V^{+}\left( \omega _{j}\right) $ is the simply connected universal cover of $
\mathbb{F}_{\Theta _{\omega _{j}}}$.
\end{proposicao}
\begin{profe}
In fact, $\mathbb{F}_{\Theta _{\omega _{j}}}$ has only two coverings since
its fundamental group is $\mathbb{Z}_{2}$. Since $G\cdot V^{+}\left( \omega
_{j}\right) \rightarrow \mathbb{F}_{\Theta _{\omega _{j}}}$ is a double
covering it follows that $G\cdot V^{+}\left( \omega _{j}\right) $ is indeed
the simply connected cover.
\end{profe}
As a consequence we have that a closed curve in $\mathbb{F}_{\Theta _{\omega
_{j}}}$ is null homotopic if and only if its lifting to $G\cdot V^{+}\left(
\omega _{j}\right) $ is a closed curve. This yields the following criterion
to decide if an orbit $G\left( \alpha \right) \cdot b_{\Theta }$ is null
homotopic.
\begin{proposicao}
Let $\mathbb{F}_{\Theta _{\omega _{j}}}$ be a minimal flag manifold with $
\omega _{j}$ as in Proposition \textrm{{\ref{propunivercover}}}. If $\alpha $
is a root then $G\left( \alpha \right) \cdot b_{\Theta _{\omega _{j}}}$ is
null homotopic if and only if the Killing number $\omega _{j}\left(H_{\alpha
}^{\vee }\right) $ is even.
\end{proposicao}
\begin{profe}
The orbit $G\left( \alpha \right) \cdot b_{\Theta _{\omega _{j}}}$ is the
closed curve $\left( \exp tA_{\alpha }\right) \cdot b_{\Theta _{\omega
_{j}}} $, $t\in \left[ 0,\pi \right] $. In terms of the projective orbit of
the representation this curve is given by $\exp t\rho _{\omega _{j}}\left(
A_{\alpha }\right) \cdot V\left( \omega _{j}\right) $, $t\in \left[ 0,\pi
\right] $. The lifting of this curve to $G\cdot V^{+}\left( \omega
_{j}\right) $ starting at $V^{+}\left( \omega _{j}\right) $ is $\exp t\rho
_{\omega _{j}}\left( A_{\alpha }\right) \cdot V^{+}\left( \omega _{j}\right)
$, $t\in \left[ 0,\pi \right] $. This lifting is a closed curve if and only
if $m_{\alpha }\cdot v=\exp \pi \rho _{\omega _{j}}\left( A_{\alpha }\right)
\cdot v=v$ for $v\in V\left( \omega _{j}\right) $. By Lemma \ref{lemmpesomax}
we have $m_{\alpha }v=\left( -1\right) ^{\omega _{j}\left( H_{\alpha }^{\vee
}\right) }v$, so that $G\left( \alpha \right) \cdot b_{\Theta _{\omega
_{j}}} $ is null homotopic if and only if $\omega _{j}\left( H_{\alpha
}^{\vee }\right) $ is even.
\end{profe}
If $\alpha =n_{1}\alpha _{1}+\cdots +n_{l}\alpha _{l}$ where $\Sigma
=\{\alpha _{1},\ldots ,\alpha _{l}\}$ is the simple system of roots then we
can compute $\omega _{j}\left( H_{\alpha }^{\vee }\right) $ explicitly from
the coefficients $n_{j}$. In fact, since $\langle \alpha _{i},\omega
_{j}\rangle =0$ if $i\neq j$ and $2\langle \alpha _{j},\omega _{j}\rangle
=\langle \alpha _{j},\alpha _{j}\rangle $, we get
\begin{equation}
\omega _{j}\left( H_{\alpha }^{\vee }\right) =\frac{2\langle \alpha ,\omega
_{j}\rangle }{\langle \alpha ,\alpha \rangle }=n_{j}\frac{2\langle \alpha
_{j},\omega _{j}\rangle }{\langle \alpha ,\alpha \rangle }=n_{j}\frac{
\langle \alpha _{j},\alpha _{j}\rangle }{\langle \alpha ,\alpha \rangle }.
\label{fornumkillingpesoraiz}
\end{equation}
By this formula we see that $G\left( \alpha \right) \cdot b_{\Theta _{\omega
_{j}}}$ is null homotopic if $n_{j}$ is even and the roots have the same
length. We observe the following possibilities when the roots $\alpha _{j}$
and $\alpha $ have different length.
\begin{enumerate}
\item $\alpha _{j}$ is long and $\alpha $ is short with $\langle \alpha
_{j},\alpha _{j}\rangle =2\langle \alpha ,\alpha \rangle $. Then $G\left(
\alpha \right) \cdot b_{\Theta _{\omega _{j}}}$ is null homotopic regardless
the coefficient $n_{j}$.
\item $\alpha $ is long and $\alpha _{j}$ is short with $\langle \alpha
,\alpha \rangle =2\langle \alpha _{j},\alpha _{j}\rangle $. Then $G\left(
\alpha \right) \cdot b_{\Theta _{\omega _{j}}}$ is not null homotopic if $
n_{j}\neq 4p$.
\end{enumerate}
\subsection{Roots in the closure of the Weyl chamber}
\label{sectionclosure}
Now we can look at the homotopy class of the orbits $G\left( \alpha \right)
\cdot b_{\Theta }$ in a minimal flag manifold, for a root $\alpha $ with $
H_{\alpha }\in \mathrm{cl}\mathfrak{a}^{+}$. Concerning the highest roots we
write their coefficients above the simple roots in the Dynkin diagrams:
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\put(139,12){$\alpha_l$}
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\put(43,30){\line(-1,2){5}}
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}
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\put(127,30){\line(1,0){20}}
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\put(147,38){$2$}
\put(98,33){\line(0,1){20}}
\put(98,56){\circle{6}}
\put(95,64){$2$}
\put(16,20){${\alpha}_1$}
\put(42,20){${\alpha}_2$}
\put(68,20){${\alpha}_3$}
\put(94,20){${\alpha}_4$}
\put(120,20){${\alpha}_5$}
\put(146,20){${\alpha}_6$}
\put(102,53){${\alpha}_7$}
\end{picture}
}
\put(220,0){
\begin{picture}(200,70)(0,0)
\put(0,27){$E_8$}
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\put(68,20){${\alpha}_3$}
\put(94,20){${\alpha}_4$}
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\put(146,20){${\alpha}_6$}
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\end{picture}
}
\end{picture}
Consider first the highest root $\mu $. By inspecting the coefficients of $
\mu $ we can apply formula (\ref{fornumkillingpesoraiz}) to get the
following to cases where $G\left( \mu \right) \cdot b_{\Theta }$ is not null
homotopic in whatsoever minimal flag manifold:
\begin{enumerate}
\item $A_{l}$, because the coefficients are $n_{j}=1$ and the roots have the
same length.
\item $C_{l}$, in this case the highest root $\mu $ has coefficients $
n_{j}=2 $ only with respect to the short roots $\alpha _{j}$, $j<l$. So that
$\omega _{j}\left( H_{\mu }^{\vee }\right) =1$ for every $j$.
\end{enumerate}
In the diagrams $D_{l}, E_{6}, E_{7}$ and $E_{8} $, the roots have the same
length and the coefficients $n_{j}$ are even for several $j $. This means
that for these diagrams $G\left( \mu \right) \cdot b_{\Theta }$ is null
homotopic on several minimal flag manifolds. We have also that in the
diagrams $B_{l}, G_{2}$ and $F_{4}$, the Killing number $\omega _{j}\left(
H_{\mu }^{\vee }\right) $ is even if $\alpha _{j}$ is a short simple root
for several $j $. Hence for these diagrams $G\left( \mu \right) \cdot
b_{\Theta }$ is null homotopic on several minimal flag manifolds.
Now let $\overline{\mu }$ be the short root with $H_{\overline{\mu }}\in
\mathrm{cl}\mathfrak{a}^{+}$ in the diagrams with multiple edges. For $B_{l}$
, $C_{l}$ and $F_{4}$, the Killing number $\omega _{j}\left( H_{\overline{
\mu }}^{\vee }\right) $ is even if $\alpha _{j}$ is a long simple root.
Hence for these diagrams $G\left( \overline{\mu }\right) \cdot b_{\Theta }$
is null homotopic on several minimal flag manifolds.
On the other hand, in $G_{2}$ we have $\overline{\mu }=\alpha _{1}+3\alpha
_{2}$. Hence by formula (\ref{fornumkillingpesoraiz}), $\omega _{j}\left( H_{
\overline{\mu }}^{\vee }\right) $ is odd for $j=1,2$. This means that for $
G_{2}$ the orbits $G\left( \overline{\mu }\right) \cdot b_{\Theta }$ in the
minimal flag manifolds are not null homotopic.
Having these facts about the fundamental groups Theorem \ref
{teohomotopyorbits} yields immediately following result.
\begin{teorema}
\label{teoreal}Let $\Gamma \subset G$ be a subset such that $G\left( \alpha
\right) \subset \Gamma $ and the group generated by $\Gamma $ is Zariski
dense. Then $\Gamma $ generates $G$ in the following cases:
\begin{enumerate}
\item $\mathfrak{g}=\mathfrak{sl}\left( l+1,\mathbb{R}\right) $.
\item $\mathfrak{g}=\mathfrak{sp}\left( l,\mathbb{R}\right) $ and $\alpha $
is a long root.
\item $\mathfrak{g}$ is the split real form associated to $G_{2}$ and $
\alpha $ is a short root.
\end{enumerate}
\end{teorema}
\subsection{\protect\medskip Example: Short roots in $\mathfrak{sp}\left( n,
\mathbb{R}\right) $}
\label{secexampl}
We present an\textbf{\ }example showing that the result of Theorem \ref
{teoreal} does not hold if $\alpha $ is a short root of the symplectic Lie
algebra $\mathfrak{sp}\left( l,\mathbb{R}\right) $. Let $Q$ be the quadratic
form in $\mathbb{R}^{2l}$ whose matrix with respect to the standard basis $
\{e_{1},\ldots ,e_{l},f_{1},\ldots ,f_{l}\}$ is
\begin{equation*}
\lbrack Q]=\left(
\begin{array}{ll}
0 & \mathrm{id}_{l\times l} \\
\mathrm{id}_{l\times l} & 0
\end{array}
\right).
\end{equation*}
Define the subset of the projective space $C\subset \mathbb{P}^{2l-1}$ by
\begin{equation*}
C=\{[v]\in \mathbb{P}^{2l-1}:Q\left( v\right) \geq 0\}
\end{equation*}
where $[v]=\mathrm{span}\{v\}$, $0\neq v\in \mathbb{R}^{2l}$. It is easily
seen that $C$ is compact and has interior $\mathrm{int}C=\{[v]\in \mathbb{P}
^{2l-1}:Q\left( v\right) >0\}$.
Now define $S\subset \mathrm{Sp}{\left( l,\mathbb{R}\right) }$ to be the
compression semigroup
\begin{equation*}
S=\{g\in \mathrm{Sp}\left( l,\mathbb{R}\right) {:gC\subset C\}.}
\end{equation*}
We claim that $\mathrm{int}S\neq \emptyset $. In fact, $X=[Q]$ is itself a
symplectic matrix and for any $0\neq v\in \mathbb{R}^{2l}$ we have
\begin{equation*}
\frac{d}{dt}Q\left( e^{tX}v\right) =\left( e^{tX}v\right) ^{T}\left(
X^{T}[Q]+[Q]X\right) e^{tX}v=2\left( e^{tX}v\right) ^{T}e^{tX}v>0,
\end{equation*}
so that $t\mapsto Q\left( e^{tX}v\right) $ is strictly increasing. This
implies that for $t>0$, $e^{tX}C\subset \mathrm{int}C\subset C$ hence $
e^{tX}\in S$ if $t\geq 0$. Actually, $e^{tX}\in \mathrm{int}S$ if $t>0$ by
continuity with respect to the compact-open topology.
Now, take a symplectic matrix of the form
\begin{equation*}
Y=\left(
\begin{array}{ll}
A & 0 \\
0 & -A^{T}
\end{array}
\right) .
\end{equation*}
Then $Y^{T}[Q]+[Q]Y=0$, so that $e^{tY}$, $t\in \mathbb{R}$, is an isometry
of $Q$, that is, $Q\left( e^{tY}v\right) =Q\left( v\right) $ for $v\in
\mathbb{R}^{2l-1}$. Hence $e^{tY}\in S$, $t\in \mathbb{R}$, which in turn
implies that the group
\begin{equation*}
G=\{\left(
\begin{array}{ll}
g & 0 \\
0 & \left( g^{-1}\right) ^{T}
\end{array}
\right) \in \mathrm{Sp}\left( l,\mathbb{R}\right) :g\in \mathrm{Gl}
^{+}\left( l,\mathbb{R}\right) \}
\end{equation*}
is contained in $S$. But for any short root $\alpha =\lambda _{i}-\lambda
_{j}$, $i\neq j$, we have $G\left( \alpha \right) \subset G$, concluding our
example.
\section{Representations of $\mathfrak{sl}\left( 2,\mathbb{C}\right) $}
In this section we take an irreducible representation of $\mathfrak{sl}
\left( 2,\mathbb{C}\right) $ on $\mathbb{C}^{n+1}$, that is, a homomorphism $
\rho _{n}:\mathfrak{sl}\left( 2,\mathbb{C}\right) \rightarrow \mathfrak{sl}
\left( n+1,\mathbb{C}\right) $ and look at the subgroup $G_{1}=\langle \exp
\rho _{n}\left( \mathfrak{sl}\left( 2,\mathbb{C}\right) \right) \rangle
\subset \mathrm{Sl}\left( n+1,\mathbb{C}\right) $. As an application of the
previous results will show the following result.
\begin{teorema}
\label{teoParaRepreSl2}Let $\rho _{n}:\mathfrak{sl}\left( 2,\mathbb{C}
\right) \rightarrow \mathfrak{sl}\left( n+1,\mathbb{C}\right) $ be the
irreducible $\left( n+1\right) $-dimensional representation of $\mathfrak{sl}
\left( 2,\mathbb{C}\right) $ and suppose that $\Gamma \subset \mathrm{Sl}
\left( n+1,\mathbb{C}\right) $ is a subset containing $G_{1}=\langle \exp
\rho _{n}\left( \mathfrak{sl}\left( 2,\mathbb{C}\right) \right) \rangle $
such that the group generated by $\Gamma $ is Zariski dense. Then $\Gamma $
generates $\mathrm{Sl}\left( n+1,\mathbb{C}\right) $ as a semigroup.
\end{teorema}
Given the basis
\begin{equation*}
X=\left(
\begin{array}{cc}
0 & 1 \\
0 & 0
\end{array}
\right) ,\qquad H=\left(
\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}
\right) ,\qquad Y=\left(
\begin{array}{cc}
0 & 0 \\
1 & 0
\end{array}
\right)
\end{equation*}
of $\mathfrak{sl}\left( 2,\mathbb{C}\right) $ we choose a basis $
\{v_{0},v_{1},\ldots ,v_{n}\}$ of $\mathbb{C}^{n+1}$ such that
\begin{equation*}
\rho _{n}\left( H\right) =\mathrm{diag}\{n,n-2,\ldots ,-n+2,-n\},
\end{equation*}
$\rho _{n}\left( Y\right) v_{j}=v_{j+1}$ and $\rho _{n}\left( X\right)
v_{j}=j\left( n-j+1\right) v_{j-1}$. The diagonal matrix $\rho _{n}\left(
H\right) $ is a regular real element of $\mathrm{Sl}\left( n+1,\mathbb{C}
\right) $ and is contained in the standard Weyl chamber $\mathfrak{a}^{+}$
formed by real diagonal matrices with strict decreasing eigenvalues. Hence $
G_{1}=\langle \exp \rho _{n}\left( \mathfrak{sl}\left( 2,\mathbb{C}\right)
\right) \rangle $ is in the case covered by Corollary \ref
{corSemiSimpleGeral}. To apply it we must look at the homotopy properties of
the orbits $G_{1}\cdot b_{\Theta }$ on the minimal flag manifolds of $
\mathrm{Sl}\left( n+1,\mathbb{C}\right) $, that is on the Grassmannians $
\mathrm{Gr}_{k}\left( n+1\right) $ of $k$-dimensional subspaces of $\mathbb{C
}^{n+1}$.
By the choice of the standard Weyl chamber $\mathfrak{a}^{+}$ the origin of $
\mathrm{Gr}_{k}\left( n+1\right) $ is the subspace $b_{k}$ spanned by $
\{v_{0},v_{1},\ldots ,v_{k-1}\}$. An easy computation shows that the
parabolic subalgebra $\mathfrak{p}=\mathrm{span}_{\mathbb{C}}\{\rho
_{n}\left( H\right) ,\rho _{n}\left( X\right) \}$ is contained in the
isotropy subalgebra at any $b_{k}$. Since the corresponding parabolic
subgroup $P\subset G_{1}$ is connected ($G_{1}$ is a complex Lie group) it
follows that the isotropy subgroups at $b_{k}$ contain $P$. On the other $Y$
does not belong to the isotropy subalgebras of $b_{k}$. This implies that
the isotropy subgroup at $b_{k}$ of the action of $G_{1}$ is the parabolic
subgroup $P$. Hence all the orbits $G_{1}\cdot b_{k}$ are $G_{1}/P\approx
S^{2}$.
In the sequel it will be proved that the orbits $G_{1}\cdot b_{k}$ are not
contractible in $\mathrm{Gr}_{k}\left( n+1\right) $. This will be done by
showing that a canonical line bundle over $\mathrm{Gr}_{k}\left( n+1\right) $
is not trivial when restricted (pull backed) to $S^{2}\approx G_{1}/P$. This
approach is based on the following proposition about the restriction of the
tautological line bundle $\mathbb{C}^{n}\setminus \{0\}\rightarrow \mathbb{P}
^{n}$ to the projective orbit $S^{2}\approx \langle \exp \rho _{n}\left(
\mathfrak{sl}\left( 2,\mathbb{C}\right) \right) \rangle \left[ v_{0}\right] $
in the irreducible representation.
\begin{proposicao}
\label{propfiblinhas2}Let $\rho _{n}:\mathfrak{sl}\left( 2,\mathbb{C}\right)
\rightarrow \mathfrak{sl}\left( n+1,\mathbb{C}\right) $ be the irreducible
representation of $\mathfrak{sl}\left( 2,\mathbb{C}\right) $ on $\mathbb{C}
^{n+1}$, $n\geq 1$. With the notation as above let $G_{1}\cdot \left[ v_{0}
\right] \subset \mathbb{P}^{n}$ be the projective orbit of the highest
weight space $\left[ v_{0}\right] $. Then $G_{1}\cdot \left[ v_{0}\right] $
is diffeomorphic to $S^{2}$.
Let $\pi :\mathbb{C}^{n+1}\setminus \{0\}\rightarrow \mathbb{P}^{n}$ the
tautological line bundle over $\mathbb{P}^{n}$ and denote by $\tau _{n}$ its
restriction (pull back) to $S^{2}=G_{1}\cdot \left[ v_{0}\right] $. Then $
\tau _{n}$ is not a trivial line bundle. In fact $\tau _{n}$ is represented
by the homotopy class $n\in \mathbb{Z}=\pi _{1}\left( \mathbb{C}_{\times
}\right) $.
\end{proposicao}
\begin{profe}
For the proof we view $\tau _{n}$ as a bundle $\xi _{1}\cup _{a}\xi _{2}$
obtained by clutching along the equator $S^{1}$ trivial bundles $\xi _{1}$
and $\xi _{2}$ on the north and south hemispheres. Such a bundle is trivial
if and only if the clutching function $a:S^{1}\rightarrow \mathbb{C}_{\times
}$ is homotopic to a point (see Husemoller \cite{hus}, Chapter 9.7).
The set $\sigma =\{[e^{z\rho _{n}\left( Y\right) }v_{0}]\in S^{2}:z\in
\mathbb{C}\}$ is an open Bruhat cell in the flag manifold $S^{2}=G_{1}\cdot
\left[ v_{0}\right] $ of $\mathfrak{sl}\left( 2,\mathbb{C}\right) $ since $
\mathfrak{n}^{-}=\mathrm{span}_{\mathbb{C}}\{Y\}$ is the nilpotent component
of an Iwasawa decomposition. Hence $\sigma $ is open and dense. Actually, $
\sigma =S^{2}\setminus \{[v_{n}]\}$ because $[v_{n}]$ is the space of lowest
weight. Furthermore the map
\begin{equation*}
\phi :z\in \mathbb{C}\longmapsto \lbrack e^{z\rho _{n}\left( Y\right)
}v_{0}]\in S^{2}\setminus \{[v_{n}]\}
\end{equation*}
is a chart. This map yields the following section over $S^{2}\setminus
\{[v_{n}]\}$ of the tautological bundle:
\begin{equation*}
\chi _{1}:[e^{z\rho _{n}\left( Y\right) }v_{0}]\in S^{2}\setminus
\{[v_{n}]\}\longmapsto e^{z\rho _{n}\left( Y\right) }v_{0}.
\end{equation*}
The same way
\begin{equation*}
\psi :w\in \mathbb{C}\longmapsto \lbrack e^{w\rho _{n}\left( X\right)
}v_{n}]\in S^{2}\setminus \{[v_{0}]\}
\end{equation*}
is a chart and
\begin{equation*}
\chi _{2}:[e^{w\rho _{n}\left( X\right) }v_{n}]\in S^{2}\setminus
\{[v_{0}]\}\longmapsto e^{w\rho _{n}\left( X\right) }v_{n}
\end{equation*}
is a section over $S^{2}\setminus \{[v_{0}]\}$.
The change of coordinates $\psi ^{-1}\circ \phi :\mathbb{C}\setminus
\{0\}\rightarrow \mathbb{C}\setminus \{0\}$ between the charts is
\begin{equation*}
w=\psi ^{-1}\circ \phi \left( z\right) =\dfrac{1}{z}.
\end{equation*}
In fact, we have
\begin{equation*}
e^{z\rho _{n}\left( Y\right) }v_{0}=\left( 1,z,\frac{z^{2}}{2!},\ldots ,
\frac{z^{n}}{n!}\right) \qquad e^{w\rho _{n}\left( X\right) }v_{n}=\left(
p_{n}\left( w\right) ,\ldots ,p_{2}\left( w\right) ,p_{1}\left( w\right)
,1\right)
\end{equation*}
where $p_{j}\left( w\right) $ is a polynomial with positive coefficients and
$p_{1}\left( w\right) =nw$. If $z\neq 0$ then $\left( 1,z,\frac{z^{2}}{2!}
,\ldots ,\frac{z^{n}}{n!}\right) $ and
\begin{equation*}
\left( \frac{n!}{z^{n}},\frac{n!}{z^{n-1}},\frac{n!}{2!z^{n-2}},\ldots ,
\frac{n}{z},1\right)
\end{equation*}
span the same line. Hence, if $z,w\neq 0$ then $\left[ e^{z\rho _{n}\left(
Y\right) }v_{0}\right] =\left[ e^{w\rho _{n}\left( X\right) }v_{n}\right] $
if and only if $n/z=nw$, which means that $w=1/z$ as claimed. It follows
that $|\psi ^{-1}\circ \phi \left( z\right) |=1$ if $|z|=1$ so that $
S^{1}=\{z:\left\vert z\right\vert =1\}$ is the same equator of $S^{2}$ in
both charts.
Now the clutching function $a:S^{1}\rightarrow \mathbb{C}_{\times }$ is
given by $\chi _{2}\left( x\right) =a\left( x\right) \chi _{1}\left(
x\right) $ with $x$ in the equator $S^{1}$. To get it take $w=\psi
^{-1}\circ \phi \left( z\right) $. Then
\begin{eqnarray*}
\chi _{2}\left( w\right) &=&e^{w\rho _{n}\left( X\right) }v_{0}=\left( \frac{
n!}{z^{n}},\ldots ,\frac{n}{z},1\right) \\
&=&\frac{n!}{z^{n}}\chi _{1}\left( z\right) =n!w^{n}\chi _{1}\left( z\right)
.
\end{eqnarray*}
Hence, for $x\in S^{1}$, $a\left( x\right) =n!x^{n}$ which is not homotopic
to a point, showing that the bundle is not trivial.
\end{profe}
This proposition implies that the projective orbit $G_{1}\cdot \left[ v_{0}
\right] \approx S^{2}$ ($G_{1}=\langle \exp \rho _{n}\left( \mathfrak{sl}
\left( 2,\mathbb{C}\right) \right) \rangle $) in $\mathbb{P}^{n}$ of the
highest weight space is not contractible and hence not contained in an open
Bruhat cell. In fact, if the orbit were contractible then the restriction of
the tautological bundle on it would be trivial.
The same approach yields also the noncontractibility of the orbits $
G_{1}\cdot b_{k}$ on the Grassmannians $\mathrm{Gr}_{k}\left( n+1\right) $, $
1\leq k\leq n-1$, where \linebreak $b_{k}=\mathrm{span}\{v_{0},\ldots
,v_{k-1}\}$. To this end we view the Grassmannian $\mathrm{Gr}_{k}\left(
n+1\right) $ as a subset of the projective space $\mathbb{P}\left( \wedge
^{k}\mathbb{C} ^{n+1}\right) $ of the $k$-fold exterior power of $\mathbb{C}
^{n+1}$. Namely $\mathrm{Gr}_{k}\left( n+1\right) $ is identified to the
projective $\mathrm{\ Sl}\left( n+1,\mathbb{C}\right) $-orbit of the highest
weight space $\left[ \xi _{k}\right] \in \mathbb{P}\left( \wedge ^{k}\mathbb{
C}^{n+1}\right) $ where $\xi _{k}=v_{0}\wedge \cdots \wedge v_{k-1}$. Via
this identification $G_{1}\cdot b_{k}$ is identified to the projective orbit
of $G_{1}$ through $\left[ \xi _{k}\right] $.
Consider the representation of $\mathfrak{sl}\left( 2,\mathbb{C}\right) $ on
$\wedge ^{k}\mathbb{C}^{n+1}$ obtained by composing the representation of $
\mathrm{Sl}\left( n+1,\mathbb{C}\right) $ with $\rho _{n}$. Denote it by $
\rho _{n}$ as well. Let
\begin{equation*}
V_{k}=\mathrm{span}\{\rho _{n}^{j}\left( Y\right) \xi _{k}:j\geq 0\}
\end{equation*}
be the $\mathfrak{sl}\left( 2,\mathbb{C}\right) $-irreducible subspace of $
\wedge ^{k}\mathbb{C}^{n+1}$ that contains $\xi _{k}$. Since \linebreak $
\rho _{n}\left( X\right) \xi _{k}=0$ and $\xi _{k}$ is an eigenvector of $
\rho _{n}\left( H\right) $ with eigenvalue $\lambda $ we have $\dim
V_{k}=\lambda +1$. Now, $\lambda $ is the sum of the $k$-largest eigenvalues
of the matrix $\rho _{n}\left( H\right) $, that is,
\begin{equation*}
\lambda =\sum_{j=0}^{k-1}\left( n-2j\right) =k\left( n-\left( k-1\right)
\right) >0.
\end{equation*}
Hence $\dim V_{k}>0$ and the $\mathfrak{sl}\left( 2,\mathbb{C}\right) $
representation on $V_{k}$ is the non trivial irreducible representation $
\rho _{m}$ with $m=k\left( n-k+1\right) $.
Now it is clear that the projective orbit $S^{2}=G_{1}\cdot \left[ \xi _{k}
\right] $ is contained in the projective space $\mathbb{P}\left(
V_{k}\right) $. By Proposition \ref{propfiblinhas2} we have that the
restriction to $G_{1}\cdot \left[ \xi _{k}\right] $ of the tautological
bundle $V_{k}\setminus \{0\}\rightarrow \mathbb{P}\left( V_{k}\right) $ is
not trivial. But the tautological bundle of $\mathbb{P}\left( V_{k}\right) $
is the restriction to it of the tautological bundle of $\mathbb{P}\left(
\wedge ^{k}\mathbb{C}^{n+1}\right) $. We conclude that the restriction to $
G_{1}\cdot \left[ \xi _{k}\right] $ of the tautological bundle of $\mathbb{P}
\left( \wedge ^{k}\mathbb{C}^{n+1}\right) $ is a nontrivial bundle. Hence
the orbit $G_{1}\cdot \left[ \xi _{k}\right] \approx G_{1}\cdot b_{k}\approx
S^{2}$ is not contractible in $\mathbb{P}\left( \wedge ^{k}\mathbb{C}
^{n+1}\right) $. A fortiori it is not contractible in $\mathrm{Gr}_{k}\left(
n+1\right) $.
Hence the group $G_{1}=\langle \exp \rho _{n}\left( \mathfrak{sl}\left( 2,
\mathbb{C}\right) \right) \rangle \subset \mathrm{Sl}\left( n+1,\mathbb{C}
\right) $ falls in the conditions of Proposition \ref{propSemiSimpleGeral}
and Corollary \ref{corSemiSimpleGeral}, concluding the proof of Theorem \ref
{teoParaRepreSl2}.
\section{Semi-simple subgroups containing regular elements}
We consider here two complex semi-simple Lie algebras $\mathfrak{g}
_{1}\subset \mathfrak{g}$ such that $\mathfrak{g}_{1}$ contains a regular
real element of $\mathfrak{g}$. That is, if $\mathfrak{g}_{1}=\mathfrak{k}
_{1}\oplus \mathfrak{a}_{1}\oplus \mathfrak{n}_{1}\subset \mathfrak{g}=
\mathfrak{k}\oplus \mathfrak{a}\oplus \mathfrak{n}$ are compatible Iwasawa
decompositions then there exists a Weyl chamber $\mathfrak{a}^{+}\subset
\mathfrak{a}$ such that $\mathfrak{a}_{1}\cap \mathfrak{a}^{+}\neq \emptyset
$. For this case we can apply essentially the same proof of Theorem \ref
{teoParaRepreSl2} to get analogous semigroup generators of $G$ by subsets
containing $G_{1}=\langle \exp \mathfrak{g}\rangle $.
\begin{teorema}
\label{teoParaSemiComplex}Let $\mathfrak{g}_{1}\subset \mathfrak{g}$ be
complex semi-simple Lie algebras and suppose that $\mathfrak{g}_{1}$
contains a regular real element of $\mathfrak{g}$. Let $G$ be a connected
Lie group with Lie algebra $\mathfrak{g}$ and put $G_{1}=\langle \exp
\mathfrak{g}\rangle $. Then a subset $\Gamma \subset G$ generates $G$ as a
semigroup provided the group generated by $\Gamma $ is Zariski dense and $
G_{1}\subset \Gamma $.
\end{teorema}
Before proving the theorem we note that $G$ has finite center because $
\mathfrak{g}$ is complex, hence the results on semigroups can be applied to $
G$. Also we can prove the theorem only for $G$ simply connected. For
otherwise we have the simply connected cover $\pi :\widetilde{G}\rightarrow
G $ and $\Gamma $ generates $G$ if and only if $\pi ^{-1}\left( \Gamma
\right) $ generates $\widetilde{G}$. Hence we assume from now on that $G$ is
simply connected.
The proof of the theorem is based on Proposition \ref{propfiblinhas2}. Fix a
Weyl chamber $\mathfrak{a}^{+}\subset \mathfrak{a}$ such that $\mathfrak{a}
_{1}\cap \mathfrak{a}^{+}\neq \emptyset $. By Corollary \ref
{corSemiSimpleGeral} we are required to prove that in any flag manifold $
\mathbb{F}_{\Theta }$ of $G$ the orbit $G_{1}\cdot b_{\Theta }$ is not
contractible where $b_{\Theta }$ is the origin of $\mathbb{F}_{\Theta }$
defined by means of $\mathfrak{a}^{+}$.
To this end we realize the flag manifolds of $G$ as projective orbits in
irreducible representations of $\mathfrak{g}$. If $\omega $ \ is a dominant
weight of $\mathfrak{g}$ (w.r.t $\mathfrak{a}^{+}$) let $\rho _{\omega }$ be
the irreducible representation of $\mathfrak{g}$ with highest weight $\omega
$. The space of the representation is denoted by $V_{\omega }$ and the
highest weight space is spanned by $v_{\omega }\in V_{\omega }$.
Since $G$ is simply connected it represents on $V_{\omega }$.
It is known that the orbit $G\cdot \left[ v_{\omega }\right] $ of the
highest weight space in the projective space $\mathbb{P}\left( V_{\omega
}\right) $ is a flag manifold of $G$. Precisely, if $\Sigma $ is the simple
system of roots $\Theta =\{\alpha \in \Sigma :\langle \alpha ,\omega \rangle
=0\}$ then $G\cdot \left[ v_{\omega }\right] $ equals $\mathbb{F}_{\Theta }$
as homogeneous spaces.
Our objective is to prove that $G_{1}\cdot \left[ v_{\omega }\right] $ is
not contractible in $G\cdot \left[ v_{\omega }\right] $. Clearly it is
enough to check that $G_{1}\cdot \left[ v_{\omega }\right] $ is not
contractible in $\mathbb{P}\left( V_{\omega }\right) $. We prove this by
applying Proposition \ref{propfiblinhas2} to a representation of a copy of $
\mathfrak{sl}\left( 2,\mathbb{C}\right) $ inside $\mathfrak{g}_{1}$.
Let $\Pi _{1}$ be the set of roots of $\left( \mathfrak{g}_{1},\mathfrak{a}
_{1}\right) $. As before if $\alpha _{1}\in \Pi _{1}$ \ then $H_{\alpha
_{1}}\in \mathfrak{a}_{1}$ is defined by $\alpha _{1}\left( \cdot \right)
=\langle H_{\alpha _{1}},\cdot \rangle $. The subspace $\mathfrak{a}_{1}$ is
spanned by $H_{\alpha _{1}}$, $\alpha _{1}\in \Pi _{1}$. Since $\mathfrak{a}
_{1}\cap \mathfrak{a}^{+}\neq \emptyset $ and $\omega $ is strictly positive
on $\mathfrak{a}^{+}$ it follows that there exists $\alpha _{1}\in \Pi _{1}$
such that $\omega \left( H_{\alpha _{1}}\right) \neq 0$.
We choose $\alpha _{1}$ with $\omega \left( H_{\alpha _{1}}\right) \neq 0$
such that $\alpha _{1}$ is positive for the chosen compatible Iwasawa
decomposition, that is, $\left( \mathfrak{g}_{1}\right) _{\alpha
_{1}}\subset \mathfrak{n}^{+}$. We denote by $\mathfrak{g}_{1}\left( \alpha
_{1}\right) $ the subalgebra of $\mathfrak{g}_{1}$ spanned by the root
spaces $\left( \mathfrak{g}_{1}\right) _{\pm \alpha _{1}}$, which is
isomorphic to $\mathfrak{sl}\left( 2,\mathbb{C}\right) $, and put $
G_{1}\left( \alpha _{1}\right) =\langle \exp \mathfrak{g}_{1}\left( \alpha
_{1}\right) \rangle \subset G_{1}$.
To get a representation of $\mathfrak{g}_{1}\left( \alpha _{1}\right) $ take
a generator $Y$ of $\left( \mathfrak{g}_{1}\right) _{-\alpha _{1}}$ and let $
V_{\omega }\left( \alpha _{1}\right) $ be the subspace spanned by $\rho
_{\omega }\left( Y\right) ^{j}v_{\omega }$, $j\geq 0$. Since $\left(
\mathfrak{g}_{1}\right) _{\alpha _{1}}\subset \mathfrak{n}^{+}$ and $
v_{\omega }$ is a highest weight vector we have $\rho _{\omega }\left(
\left( \mathfrak{g}_{1}\right) _{\alpha _{1}}\right) v_{\omega }=0$. Also $
v_{\omega }$ is an eigenvector of $\rho _{\omega }\left( H_{\alpha
_{1}}\right) $ with eigenvalue $\omega \left( H_{\alpha _{1}}\right) \neq 0$
. Hence by the usual construction of the irreducible representations of $
\mathfrak{sl}\left( 2,\mathbb{C}\right) $ the subspace $V_{\omega }\left(
\alpha _{1}\right) $ is invariant and irreducible by $\rho _{\omega }\left(
\mathfrak{g}_{1}\left( \alpha _{1}\right) \right) $. Since the eigenvector $
\omega \left( H_{\alpha _{1}}\right) \neq 0$ we have $\dim V_{\omega }\left(
\alpha _{1}\right) \geq 2$ and we get a nontrivial representation of $
\mathfrak{sl}\left( 2,\mathbb{C}\right) $ on $V_{\omega }\left( \alpha
_{1}\right) $. In this representation $v_{\omega }$ is a highest weight
vector because $\rho _{\omega }\left( \left( \mathfrak{g}_{1}\right)
_{\alpha _{1}}\right) v_{\omega }=0$. A posteriori $\omega \left( H_{\alpha
_{1}}\right) =\dim V_{\omega }\left( \alpha _{1}\right) -1>0$.
Now we can apply Proposition \ref{propfiblinhas2} to the representation of $
\mathfrak{sl}\left( 2,\mathbb{C}\right) $ on $V_{\omega }\left( \alpha
_{1}\right) $ to conclude that the restriction to $G_{1}\left( \alpha
_{1}\right) \cdot \left[ v_{\omega }\right] $ of the tautological bundle $
V_{\omega }\left( \alpha _{1}\right) \setminus \{0\}\rightarrow \mathbb{P}
\left( V_{\omega }\left( \alpha _{1}\right) \right) $ is not trivial. This
implies, by restricting twice, that the tautological bundle $V_{\omega
}\setminus \{0\}\rightarrow \mathbb{P}\left( V_{\omega }\right) $ restricts
to a nontrivial bundle on the orbit $G_{1}\left( \alpha _{1}\right) \cdot
\left[ v_{\omega }\right] $. Hence the orbit $G_{1}\left( \alpha _{1}\right)
\cdot \left[ v_{\omega }\right] $ is not contractible on $\mathbb{P}\left(
V_{\omega }\right) $ so that it is not contractible on $\mathbb{F}_{\Theta
}=G\cdot \left[ v_{\omega }\right] $. Finally, the orbit $G_{1}\cdot \left[
v_{\omega }\right] $ is not contractible as well since it contains $
G_{1}\left( \alpha _{1}\right) \cdot \left[ v_{\omega }\right] $. By
Corollary \ref{corSemiSimpleGeral} this finishes the proof of Theorem \ref
{teoParaSemiComplex}.
\vspace{12pt}
\noindent
\textbf{Example:} The standard realizations of the classical complex simple
Lie algebras of types $B_{l}$, $C_{l}$ and $D_{l}$ as subalgebras of
matrices satisfy the condition of Theorem \ref{teoParaSemiComplex} as
subalgebras of the appropriate $\mathfrak{sl}\left( n,\mathbb{C}\right) $.
In fact, for $C_{l}$ and $D_{l}$ one has Cartan subalgebras of diagonal $
2l\times 2l$ matrices $\mathrm{diag}\{\Lambda ,-\Lambda \}$ with $\Lambda $
diagonal $l\times l$, while for $B_{l}$ the Cartan subalgebra is given by $
\mathrm{diag}\{0,\Lambda ,-\Lambda \}$. The $\mathfrak{a}_{1}$ are given by
such matrices with real entries. A quick glance to these diagonal matrices
shows that in any case $\mathfrak{a}_{1}$ contains diagonal matrices with
distinct eigenvalues, and hence regular real elements of $\mathfrak{sl}
\left( n,\mathbb{C}\right) $, $n=2l$ or $2l+1$.
| 201,843
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Hello hello! I know I've been such a lousy blogger. Again. And believe me, I'm not happy about it but life goes on anyway. And I do sincerely hope you still read (and perhaps sometimes even wait for...) my rare writings. Because at least one fresh winter outfit from January 24 is currently waiting for being published, plus an elder ensemble where I'm presenting my cute thrifted shirt dress (Mili, I practically promise I will certainly share that look!).
Oh and last Sunday I photographed quite a few new clothes and two pairs of shoes, pretty much all bar one thrifted. And if there are gardening fans among my readers - I will also show the seeds I've so far bought this year. And I'm so excited for the incoming gardening season and already can't wait to start sowing everything! Who's with me, guys??
Therefore, this is what's waiting for you in my next publications but today is a completely different day because...
On January 14 I paid a visit to the Parliament of Estonia in our capital town Tallinn with some of my colleagues. First of all the most persistent of us climbed 215 excruciatingly narrow and uncomfortable stone steps that lead us to the top of a watch tower called Tall Hermann. At the top of Tall Hermann, 95 metres above sea level our beautiful blue, black and white Estonian flag was hoisted for the very first time on December 12, 1918. Then, after decades of Soviet occupation it was hoisted again on February 24, 1989. That's our Independence Day. But to transmit at least fractions of the amazing panorama from the top of the tower I've added photos 1-4 here in this post. My favourite is the one above where my hair is totally gone with the wind!
Anyway, the climb was followed by a truly interesting and educational excursion inside the building of the Parliament which is one part of the Toompea Castle. By the way, that building is unique among all the parliament buildings of the world since it is the only expressionistic one out there. And to see some of the cool geometric, mainly triangular shapes and other details of the lofty assembly hall explore my first collage below.
Now, last but not least we could discover the Stenbock House - the Seat of Government and the Government Office of Estonia. And my little collage below shows you a glimpse of the Green Hall on the right. It is named after the colour of its walls and Empire style furniture upholstery and the Prime Minister uses this room for smaller consultations and for meetings with ambassadors.
Those groovy green bushes above, my favourite detail there, derive from the Government Session Hall, the largest room in the house. The Government of Estonia holds its weekly sessions there on Thursdays. We also visited the Office of the Prime Minister where I sat down on his statuesque chair that has served all of Estonia's heads of government. It sure felt special, hence my questionable look I guess. Oh and the very last photograph presents the spectacular view from the balcony of the Stenbock House.
So, I hope you enjoyed today's unusual photos and didn't mind studying fragments about the Estonian political buildings. As always, let me know what you think, and see you already in my next, probably more conventional post. Take care and stay warm and cosy!
Therefore, this is what's waiting for you in my next publications but today is a completely different day because...
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| 279,653
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TITLE: In infinite-dimensional space, does linear mapping always have a fixed point?
QUESTION [2 upvotes]: In infinite-dimensional space, does linear mapping that satisfy the strictly nonexpansive always have a fixed point?
I believe the answer is no and we can construct a counterexample to show that such as in $X=c_0$ or $X=l^\infty$ but I am still working on constructing my example.
Any help would be appreciated
REPLY [1 votes]: Again, there is no linear mapping that is without a fixed point since trivially the zero vector will always satisfy $T(0) = 0$ for any linear mapping $T:V\to V$. It may be that a linear map has only this one fixed point, e.g. the zero map that sends every $x\in V$ to zero necessarily has but that one fixed point. Indeed any strictly contractive linear map, $||T(x-y)|| \lt ||x-y||$ when $x\neq y$, has no fixed point other than $T(0)=0$.
However an interesting question is whether there can be a choice of initial point $x_0$ such that the sequence defined by $x_{n+1} = T(x_n)$ has no convergent subsequence under the restriction that $||T(x-y)|| \le ||x-y||$.
A standard example of this is the shift operator on $\ell^\infty$. That is, define $T(x)$ to be the result of shifting all the entries of $x$ to the right (and introducing a new zero entry to fill the first component). For any choice of nonzero $x$, the sequence $T^n(x)$ will neither converge nor have any convergent subsequence using the $\infty$-norm.
The shift operator on $\ell^\infty$ satisfies $||T(x-y)|| = ||x-y||$.
| 139,048
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We are sorry! The publisher (or author) gave us the instruction to take down this book from our catalog. But please don't worry, you still have more than 1 million other books to choose from, so you can read without limits!
Data Information Literacy
Jake Carlson
Publisher: Purdue University Press
Summary "e;data information literacy"e; "e;DIL Toolkit,"e;.
| 149,647
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Hello apathetic.
Let's see - is there anything good going on? Just school. I go back for my second class in two weeks. I'm really excited although it's going to be harder than the first one. But I know I can do it.
I've got a lot of other stresses happening in my life. I won't comment on my marriage, because it never fails that something that I post here turns out to bite me in the ass.
Shit - I thought I had a lot to say but everything I try to write I wind up erasing it. Nothing seems to be coming out right and I'm feeling physically like shit right now. All I want to do is cry.
I know y'all want to offer comfort and advice - in fact that seems to be all you do because my life is just that pathetic. But I'm going to leave this up - just because I guess. Maybe a reminder to myself that I'm still alive somewhere inside. But I'm turning off comments.
I'm just grateful for any of y'all's company.
Stay sane inside insanity ~ and never forget your towel.
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TITLE: Permutation and combination question. Whats wrong in my attempt.
QUESTION [0 upvotes]: My Question:
6 different books are to be distributed among three peoples {R/S/G} . Then number of ways of distribution of books such that each person gets atleast one book.
My attempt:
$$\binom{6}{3} 3! 3^3 = 3240$$
because we can choose first three books with $\binom{6}{3}$ and give them to three people with $3!$. Then we can distribute remaining three books with $27$ different ways.
REPLY [2 votes]: This question can be considered as the following:
Let $A$ and $B$ be sets such that $|A| = 6$ and $|B| = 3$. How many
surjections $A \rightarrow B$ are there?
When number of surjections is asked, you should use "Stirling numbers of the second kind" in order to solve the problem. Then answer of this question is $$3!S(6,3) = 6*90 = 540$$
Here, the number $S(6,3)$ is the number of ways of distributing $6$ different books to $3$ identical people (I don't know how can something like this be possible but you can consider it this way). Since the people are different, you can change the order with $3!$ so the result follows.
Now the problem with your solution is assume you have books $1,2,3,4,5,6$. Then in the first choose, suppose you choose $1,2,3$ and give it them to $R,S,G$ respectively. Then you are distributing the remaining books, suppose $4,5,6$ to $R,S,G$, respectively, similar to the first distribution. But this is same as choosing the first three books as $4,5,6$ and distributing the remaining three as $1,2,3$, giving each people two books. So you are over-counting the possibilities. I also suggest you to learn the Stirling numbers of the second kind and its relation with number of surjections (It is about Inclusion-Exclusion Principle).
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This is going to make things even weider between Betsy Morgan and her old HuffPo pals.
The former HuffPo CEO is going to work for Glenn Beck's news site The Blaze, according to Media Decoder.
Morgan left HuffPo in 2009. The Blaze, which is just a four-month old site, is off to a blistering start.
This is more surprising than Mara Laisson from NPR appearing on Fox News, but hey, it's all business and Morgan has seen a site go from small to enormous.
Great hire.
For more background on The Blaze, see here >
| 134,624
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TITLE: Varieties with infinitely many etale covers and rational points
QUESTION [9 upvotes]: Let $X$ be a (smooth projective geometrically connected) variety over a field $k$.
Consider the set Et$(X,k)$ of finite etale covers $Y\to X$ over $k$, with $Y$ geometrically connected over $k$.
Assume Et$(X,k)$ is infinite. Consider the following question:
Does $X$ have a $k$-rational point?
The answer should be negative in general. In fact, I think one can construct a surface with infinitely many etale covers but no rational points by taking the product of two curves $C$ and $D$ over $k$, where $C$ has infinitely many etale covers and a rational point, but $D$ doesn't have any rational points. Then $C\times D$ has no rational points, but infinitely many covers.
What if $X$ is a curve? Is $X(k)$ non-empty?
Note that the converse is true if we consider curves of positive genus. That is, if $X$ is a curve of positive genus over $k$ with a $k$-rational point, then it has infinitely many etale covers.
I'm mainly interested in the characteristic zero case, but comments on the situation in positive characteristic would also be interesting.
REPLY [7 votes]: Maybe I misunderstand something, but don't all curves have etale covers? Embed $X$ in $J^1$ (divisors of degree $1$ modulo linear equivalence). Then $J^1$ is a torsor for the Jacobian $J$ and since $J$ has etale covers, e.g. coming from multiplication by an arbitrary $n$, $J^1$ does too. Certainly, for those curves with a rational divisor of degree one, they have covers, as $J^1$ is isomorphic to $J$.
EDIT: Upon further reflection, I guess it's not true that $J^1$ always has covers, as it may not be in the divisible part of the Weil-Chatelet group of $J$. But there definitely exist curves with no points having divisors of degree one, and therefore covers of arbitrarily large degree.
However, you question is a good one and you might be heading in the direction of Grothendieck's section conjecture: For finitely generated fields $k$, $X(k)$ is non-empty if and only if there is a section $G_k \to \pi_1(X)$ of the canonical projection $\pi_1(X) \to G_k$, where $G_k$ is the absolute Galois group of $k$ and $\pi_1$ is the etale fundamental group.
| 155,902
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Figures.
Citation: Freeman ZN, Dorus S, Waterfield NR (2013) The KdpD/KdpE Two-Component System: Integrating K+ Homeostasis and Virulence. PLoS Pathog 9(3): e1003201. doi:10.1371/journal.ppat.1003201
Editor: Chetan E. Chitnis, International Centre for Genetic Engineering and Biotechnology, India
Published: March 28, 2013
Copyright: © 2013 Freeman et al. This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.
Funding: This work was supported by a BBSRC funded PhD studentship for Zoe Freeman (Bath University, UK). The funders had no role in study design, data collection and analysis, decision to publish, or preparation of the manuscript.
Competing interests: The authors have declared that no competing interests exist.
Introduction
Two-component systems (TCSs) are widespread regulatory systems that enable microbes to control their cellular functions and respond appropriately to a diverse range of stimuli such as pH, osmolarity, quorum signals, or nutrient availability. The two components are a histidine kinase (HK), which senses an environmental signal, and a response regulator (RR), which mediates a cellular response, typically by altering expression of target genes. The genes kdpD and kdpE together encode the KdpD/KdpE TCS, which is well-studied for its regulation of the Kdp-ATPase potassium ion (K+) pump operon kdpFABC. Although best studied in Escherichia coli, highly homologous genes are found in most other bacteria, and are assumed to function similarly in K+ homeostasis. However, the signalling networks regulating basic bacterial physiology and bacterial virulence are sometimes intricately linked [1], and TCSs are no exception. A classic example of a TCS with pleiotropic roles in homeostasis and virulence is PhoP/PhoQ, whose regulon both mediates responses to Mg2+-limited environments and governs virulence and intra-macrophage survival in a range of Gram negative species [2], [3]. More recently, the GraS/GraR TCS of Staphylococcus aureus has also been shown to possess functional diversity—it was best known for controlling resistance to antimicrobial peptides, but further investigation has revealed links to quorum sensing, stress response, cell wall metabolism, and regulation of haemolytic and fibrinogen-binding proteins [4]. Here, we review evidence from diverse and clinically relevant bacterial species, indicating that the KdpD/KdpE TCS responds to pathogenesis-related signals, directly regulates virulence genes, and mediates stress resistance.
KdpD/KdpE and the Regulation of K+ Homeostasis during Pathogenesis
Regulation of K+ Homeostasis by KdpD/KdpE
K+ uptake in bacteria occurs via different combinations of nonspecific channels and specialised transport systems [5]. Constitutively expressed systems, such as Trk, satisfy general K+ requirements, but the Kdp-ATPase is a specialised, inducible higher-affinity K+ pump [6]. The Kdp-ATPase itself is encoded by the structural operon kdpFABC. The Kdp-ATPase is widespread throughout the prokaryotes and the KdpD/KdpE TCS is found in over 1,082 bacterial and archaeal species [7]. Expression of the operon is triggered by three stimuli perceived and integrated by the HK KdpD: K+ concentration, osmolarity, and ATP concentration [8]. Turgor pressure has also been suggested as a stimulus for KdpD/KdpE activation, although this has been a subject of debate [8]. Mutation analysis suggests that an activating stimulus causes the inhibition of the phospho-KdpE-specific phosphatase activity of KdpD, leading to an accumulation of phospho-KdpE. This in turn binds to an operator sequence in the promoter DNA to activate transcription of kdpFABC [9]. Its regulation by a TCS makes Kdp the only known bacterial K+ transport system whose expression is strongly controlled at the level of transcription [5].
Regulation of K+ Homeostasis during Pathogenesis
K+ is the single most abundant ion in the intracellular environment and its regulation is crucial for maintenance of cell turgor and for diverse processes contributing to normal homeostasis [5]. Many studies have shown that K+ regulation is critical for bacterial virulence [10]–[13]. Similarly, it is an intricate aspect of host response to pathogens; in neutrophils, the active transport of K+ across the phagosomal membrane releases antimicrobial peptides, activating proteases and enabling Reactive Oxygen Species (ROS)–mediated killing of engulfed bacteria [14]. Thus, K+ sequestration from a common limited supply may be critical to the competition between the bacterium and host. Considering the importance of K+ regulation in virulence, it may not be surprising that genes involved in K+ regulation and transport are critical to the survival of pathogenic bacteria. Transcriptomic evidence reveals that transcription of the kdpD gene is repressed at least 3-fold and in some cases up to 20-fold in Staphylococcus aureus in response to phagocytosis by human neutrophils (after 3 h of interaction) [15], and up-regulated in Mycobacterium avium during early growth in human macrophages (at 48 h after infection) [16].
Role of KdpD/KdpE for Virulence and Pathogen Survival
KdpD/KdpE Aids Bacterial Survival by Increasing K+ Uptake
Although the precise mechanisms of action are yet to be determined, infection models suggest that KdpD/KdpE increases the ability of some bacteria to cause disease or to survive within a host cell or animal (Table 1). For example, increased transcription of kdpD/kdpE was concurrent with increased survival of S. aureus in macrophages and deletion of the TCS resulted in attenuated survival [17], [18]. In human neutrophils, a Y. pestis kdpD/kdpE mutant was more readily killed compared to the wild-type strain [3]. A study that aimed to identify genes involved in persistent infection of the Caenorhabditis elegans intestine highlighted the impact of Salmonella typhimurium kdpD [19]. Further characterisation revealed that kdpD was in fact required for S. typhimurium infection of the nematode and also for survival in macrophage cell lines. In the nematode feeding assays, worms that had grazed on the kdpD mutant strain lived significantly longer than those fed on the wild-type bacteria, and in macrophages, intracellular growth and cytotoxicity were reduced [19]. These observations from intracellular models are generally consistent with the importance of K+ scavenging in K+ limited environments, as would be found in a phagosomal vacuole.
Other KdpD/KdpE Mechanisms Influence Bacterial Virulence and Survival
There are, however, several observations that do not fully fit with the K+ requirement models. Evidence from S. aureus suggests that K+ availability has little to do with KdpD/KdpE activity. Despite the fact that human blood is K+ rich, S. aureus kdpD/kdpE is still up-regulated and its deletion results in attenuated survival in that environment [18]. Additionally, S. aureus kdpD/kdpE transcription is altered not just during phagocytosis (when it is repressed); it is enhanced during biofilm formation and in response to microbicidal neutrophil extracts [20], [21], situations in which K+ is not necessarily limited. Although, it is important to remember that it is the KdpD/KdpE phosphorylation status that is ultimately important for activity rather than the transcription level of the genes themselves. Lastly, it has been suggested that the S. aureus Kdp-ATPase does not even function as a predominant K+ transporter [17]. It should be noted, however, that this suggestion was based on results from the deletion of only one KdpD/KdpE system when in fact some S. aureus strains possess two systems, with one located on the Type II SCCmec mobile genetic element [22]. Therefore, clarification of whether the strain used by Xue et al. (NCTC 8325) carries one or more functional Kdp operons would be pertinent.
KdpE Is a Regulator of Diverse Virulence Loci
As well as a regulator of the Kdp-ATPase, recent evidence supports the role of KdpE as a direct regulator of S. aureus virulence factors [17], [18]. KdpE has been shown to bind directly to the DNA promoters of a range of virulence genes and it has been demonstrated that deletion of kdpD/kdpE altered the level of transcription of over 100 genes, including (i) the surface protein gene spa (Staphylococcal protein A), (ii) the capsule synthesis gene cap, (iii) the alpha-toxin gene hla, (iv) the metalloproteinase aureolysin (aur), (v) the lipase gene geh, and (vi) the gamma-haemolysin gene hlgB [18]. Figure 1B depicts virulence-related regulatory targets of KdpD/KdpE in S. aureus. A positive regulatory effect was elicited upon the colonisation genes (spa and cap), and a negative regulatory effect was elicited upon the local invasion enzymes and toxin genes (hla, aur, geh, and hlgB) [17]. S. aureus kdpD is most highly transcribed at low external K+. This means that colonisation is enabled in response to low K+ (in the external environment, for example), and once in higher K+ conditions (in blood or cytoplasm), local pathogenesis is enabled. The spa gene has previously been shown to enhance virulence in mouse and macrophage models due to its antiphagocytic role [23].
(A) The conventional model is that KdpD/KdpE stimulates transcription of the Kdp-ATPase in response to cytoplasmic ionic and ATP concentrations and possibly also turgor pressure [9]. (B) In S. aureus KdpD is affected directly or indirectly by QS systems, and KdpE regulates many downstream genes including virulence factors by directly binding to their promoters [17], [18]. (C) In EHEC, KdpE can also be activated by the QseC histidine kinase, which senses host adrenergic signals as well as bacterial quorum sensing (QS) signals [24]. In vitro its regulatory targets include the ler gene, which controls the “locus of enterocyte effacement" (LEE) genes. Under gluconeogenic conditions, KdpE interacts with Cra to optimally regulate ler; both proteins bind to the promoter, perhaps through bending of the DNA [25]. The downstream regulatory cascade is integral to lesion formation in the host gut [24], [25]. (D) Recently identified accessory components in nonpathogenic E. coli link the pathway to additional input stimuli or modulate KdpD activity [34], [37], [50].
Activation of virulence genes via KdpE has also been reported for Enterohaemorrhagic E. coli (EHEC) strain O157:H7 [24]. The adrenergic receptor QseC enables EHEC to sense host adrenaline and noradrenaline, as well as sensing bacterial signalling molecules (Figure 1C). This means EHEC can exploit host signals to regulate its metabolic, virulence, and stress response genes via a complex signalling cascade. In addition to its cognate RR, the QseC HK activates two additional RRs, one of which is KdpE [24]. In an in vitro system employing liposomes and purified proteins, over 80 QseC-regulated genes were found to be activated via KdpE. Amongst its targets were the type 3 secretion system (T3SS) and its effectors (via regulation of ler), which are necessary genes for “attachment and effacement" lesions of the intestinal epithelia (see Figure 1) [24]. These results show that KdpE has at least the potential to regulate critical virulence genes during infection. Further in vitro research has shown that KdpE directly interacts with a second transcription factor, the catabolite repressor/activator protein Cra, to regulate ler in a glucose-dependent manner [25]. Visualisation of infected HeLa cells revealed that deletion of kdpE resulted in slightly fewer lesion pedestals than wild-type, though the effect was more pronounced for the equivalent cra mutation. Importantly, however, in a double mutant, full complementation was achieved only when both proteins were expressed, indicating that both transcription factors are co-involved in regulating that aspect of EHEC virulence. Additional in vivo studies using a noncomplemented kdpE knockout strain and whole organism infection models would further help to dissect to what extent KdpE plays a direct role in EHEC virulence.
Signal Integration via the Sensor Kinase KdpD and the Relevance for Host Infection
Rather than representing two distinct concepts, bacterial virulence and survival may be considered “two sides of the same coin." Genes that promote the survival of a pathogen in its host—whether via stress response or by direct regulation of virulence genes—may be classed as virulence factors. While EHEC and S. aureus provide examples for the direct regulation of virulence genes by KdpE, there remains a more general implication of the importance of KdpD/KdpE in virulence mediated by stress resistance.
Stress Sensing and Resistance
There is a significant body of evidence supporting a link between KdpD/KdpE and the bacterial stress response, which in itself has clear relevance to survival within a host. With K+ concentration being an activating stimulus for Kdp-ATPase expression, it is understandable that osmotic stress (salt shock) can be linked to activity of KdpD/KdpE. However, studies in several bacteria also link kdpD/kdpE to resistance to oxidative and antimicrobial stresses specifically. Importantly, these bacteria are all either obligate or facultative intracellular pathogens. All have been reported in some instance to be able to resist host killing [26]–[28], and the majority are known to spend at least part of their lifecycle in the harsh phagosomal compartment—an acidic environment containing a barrage of hydrolytic enzymes, microbicidal agents, and reactive oxygen species (ROS) including hydrogen peroxide (H2O2). If KdpD/KdpE does mediate the sensing of such stresses encountered in the host, then the effect of knocking them out in those infection models would be explained. Table 2 summarises the experimental evidence where KdpD/KdpE has been implicated in stress resistance.
Further evidence that KdpD can sense host-derived signals—including evidence for a role in stress response—was seen in Bacillus cereus, in which exposure to chitosan led to up-regulation of the kdpABC and kdpD genes [29]. Chitosan is an antibacterial polysaccharide, thought to induce loss of ions and cellular turgor by forming pores in bacterial membranes. This might suggest that the Kdp-ATPase was up-regulated in order to recover lost K+. Yet two observations suggest otherwise. One is that a deletion mutant of the Kdp-ATPase structural genes actually showed no impairment of growth in conditions of either K+-limitation or salt-stress, bringing into question the importance of Kdp for K+ regulation in B. cereus. The other is that exposure to benzalkonium chloride (a disinfectant that generates leakage of intracellular ions in much the same way as chitosan) did not have the same effect of up-regulating the B. cereus kdp genes [30]. Taken together this suggests that ion loss alone was not being sensed; rather, some other form of “antimicrobial stress" may have been. Indeed there is precedence for a TCS being directly activated by host-elicited antimicrobial peptides (AMPs) in the case of PhoP/PhoQ [31]. It is therefore tempting to speculate that the biologically derived antimicrobial chitosan may be sensed by KdpD in a manner akin to the sensing of small innate immune molecules by PhoP.
Quorum Sensing
As discussed earlier, in various bacteria KdpD can integrate multiple stimuli including K+ concentration, osmolarity, possibly turgor pressure, and even host adrenergic signals. Importantly, in S. aureus KdpD/KdpE can also respond to population density. The two studies that showed direct regulation of S. aureus virulence factors also reported that KdpD/KdpE can integrate signals from two quorum sensing (QS) systems—the luxS/AI-2 system [18] and the Agr system [17]. Therefore, KdpE-regulated S. aureus virulence genes are expressed appropriately according to both K+ concentration (an indication of whether the bacterium is extra- or intracellular), and population density (indicating whether there are enough bacteria present to make a particular response appropriate, e.g. toxin production) [17]. As such, S. aureus uses KdpD/KdpE in an adaptive manner, utilizing its “traditional" sensor activity in an integrated fashion with input from QS systems.
Accessory Components of KdpD
Accessory proteins (also known as auxiliary factors) are known to affect TCSs by (i) fine-tuning the ratio of the HK's kinase- to phosphatase-activity, (ii) linking the system to other regulatory networks, or (iii) enabling the sensing of additional stimuli [32], [33]. Table 3 summarises accessory proteins that have been identified for KdpD, and E. coli accessory factors are also shown in Figure 1D. It is not clear if there is any one specific region of the KdpD protein structure that is responsible for fine-tuning sensor kinase activity through accessory protein interactions. For example, it has been suggested that the IIANtr protein (see below) interacts with the C-terminal transmitter domain of KdpD [34]. On the other hand, the N-terminus includes a conserved “KdpD" subdomain with a regulatory ATP-binding site [35], [36], and a universal stress protein (KdpD-Usp) subdomain, which is highly variable between species. The E. coli universal stress protein UspC is known to interact with the N-terminal Usp-domain thus altering its activity [37], although not through an influence on KdpD kinase activity.
Is the Usp Domain of KdpD Involved in Stress Sensing?
A fundamental feature of Usp proteins is that they accumulate under stress conditions, and indeed UspC binds to the KdpD-Usp domain in response to osmotic stress, stimulating kdpFABC expression by acting as a scaffold between the active KdpD/phospho-KdpE/DNA complex [37]. This enables Kdp-ATPase expression at K+ levels at which it would normally be repressed. It has previously been suggested that the KdpD-Usp domain might be pivotal in the sensing of additional stimuli [19], [38]. UspA proteins protect E. coli against superoxide stress, and the UspA proteins of Photorhabdus species and Yersinia enterocolitica are thought to play a role in stress sensing when invading insect hosts [37], [39], [40]. Photorhabdus asymbiotica is a facultative intracellular pathogen of insects and humans. Interestingly, when P. asymbiotica kdpD/kdpE was expressed in a nonpathogenic E. coli strain, the presence of the transgenic TCS conferred an ability to the E. coli strain to persist within insect phagocytic cells [38]. Although further molecular experiments would be required in order to identify the precise mechanism behind the phenomenon, this observation does show that there are differences in the functioning of the KdpD/KdpE TCSs between the two species. The greatest evolutionary divergence of the kdpD and kdpE genes from P. asymbiotica, E. coli, and other bacterial species does indeed reside within the KdpD-Usp domain. Interestingly, domain swapping experiments revealed that this domain is also crucial for activation of KdpD. While the Usp-domain has a net-positive charged surface, integration of a net-negative charged surface was shown to block the activity of the protein. Therefore, this domain can not only act as an interaction surface for other proteins, but also appears important for internal (de)activation of KdpD [41]. It has been suggested that binding of other Usp proteins might either alter the secondary structure of the protein or perhaps facilitate activation by shaping a more positive surface at this position. The ability of KdpD/KdpE to sense and respond to stress may thus be mediated via this Usp-binding domain, and the variability of that domain between species might contribute to differences in bacterial virulence and lifestyle.
Overview
A study of the TCSs of M. tuberculosis in 2003 found that deletion of kdpD/kdpE unexpectedly resulted in hypervirulence in mice [42]. Although no targets of KdpD/KdpE other than the Kdp-ATPase itself had been identified at the time, the result prompted the suggestion that KdpD/KdpE deletion had de-repressed virulence genes. It is possible that several mycobacterial lipoproteins identified in a study by Steyn et al. [43] might be targets of KdpD/KdpE regulation. Since then, a diverse set of observations in a range of species supports the role of KdpD/KdpE in promoting survival and virulence through various mechanisms distinct from K+ regulation. In most cases it is not yet clear exactly how KdpD/KdpE impacts on virulence—it is likely that the link to stress response aids bacterial persistence within host cells. KdpD/KdpE's role in virulence is likely to be highly variable across species and may ultimately be an indicator of species-specific adaptation.
In both Gram positive and Gram negative pathogens, KdpD/KdpE functions as a previously unrecognised adaptive TCS of stimuli during host infection, integrating cues from the host (phagocytosis, host-derived antimicrobials, and adrenergic hormones) in addition to bacterial quorum-sensing signals and the outside ionic concentrations. A recent review of “moonlighting" bacterial proteins focused upon glycolytic enzymes and molecular chaperones in a range of pathogens. Just as the evidence suggests for KdpD/KdpE, this review emphasized that key metabolic proteins and molecular chaperones, essential for dealing with the bacterium's response to stress, also have unexpected functions in virulence [44]. Like key metabolic enzymes and molecular chaperones, the KdpD/KdpE TCS usually plays a conserved role in maintaining normal cellular function. It seems that the conserved TCS KdpD/KdpE provides a further example of bacterial protein “moonlighting." It is suggested that proteins can display multifunctionality due to “biochemical" or “geographical" conditions: either through possessing additional biochemical reactions or by being located in a different cellular location (such as the pole of an asymmetric cell), or both [44]. There appear to be no reports of KdpE or KdpD being located anywhere other than the cytoplasm and cytoplasmic membrane, but evidence does suggest that the Usp domain is a prime candidate as an additional “biochemical" site. Data regarding the effects of KdpD/KdpE in pathogenesis are somewhat limited to date but come from a range of bacterial genera and experimental approaches. The evidence reveals a spectrum of Kdp function from a clearly homeostasis-based role to a highly pleiotropic one. Given its function in K+ homeostasis, the KdpD/KdpE TCS is extremely important for bacteria in challenging environments. We argue that it is therefore likely to be a focal point for selection in relation to survival during infection. It should be noted that no additional functions have yet been ascribed to KdpD/KdpE in nonpathogenic E. coli; accessory components have been identified that interact with E. coli KdpD, but these can be reasonably linked to K+ homeostasis. Examples include thioredoxins, which are integral in stress response pathways involving osmotic stress, the H-NS nucleoid protein, which is a transcriptional repressor that can respond to K+ concentration [45]; and the IIANtr enzyme. This enzyme has been shown to connect K+ homeostasis with carbon starvation through direct interaction of nonphosphorylated IIANtr with KdpD activating the kinase activity [46]. Nonpathogenic E. coli are not typically capable of surviving phagocytosis, and yet when expressing the KdpD/KdpE TCS from P. asymbiotica, E. coli cells did persist intracellularly [38] implying a fundamental difference between the abilities of the two systems to sense or respond to the phagocytic environment. The evidence from F. novicida can be adequately explained by a role in K+ regulation; mutants in the kdpFABC operon (as well as those in regulatory genes) were attenuated in the fruit fly model, which indicates that the Kdp-ATPase itself is important in this species [47]. Yet in F. tularensis, only the kdpD gene mutant was identified as attenuated in the mouse model; kdpFABC operon genes were apparently less critical than within the gene encoding the sensor kinase [48]. In EHEC, KdpE regulates K+ uptake and osmolarity alongside important virulence genes; the EHEC QseC HK, which activates KdpE, has even been investigated as a potential target for novel “anti-virulence" drugs [24], [49]. KdpE interacts with Cra, which itself coordinates the E. coli response to sugar availability, thereby linking metabolism to pathogenesis. It has been suggested that during its evolution EHEC has co-opted “established mechanisms for sensing the metabolites and stress cues in the environment, to induce virulence factors in a temporal and energy-efficient manner, culminating in disease" [25]. The clearest K+ independent role in virulence so far is in S. aureus, in which KdpD responds to population density and KdpE actually regulates over 100 genes.
Concluding Remarks and Future Directions
In these cases where KdpD/KdpE has been shown to regulate additional virulence genes, it is not yet clear how this has evolved. For example, did the promiscuous regulation by KdpE evolve through changes in the KdpE DNA-binding domain, or has there been widespread selection for the evolution of KdpE operator sequence sites across a range of virulence loci? Furthermore, how taxonomically widespread is the expanded regulatory capacity of KdpE, and has the same mechanism occurred in each case? If specific virulence genes have evolved KdpE-responsive promoters, then we may expect to see similar virulence genes under the control of KdpD/KdpE in different pathogens by virtue of horizontal gene transfer. EHEC's T3SS and lesion-causing genes were acquired on a mobile genetic element, and it has been suggested that Cra and KdpE (originally regulators of nonpathogenic functions) have been “co-opted by a pathogen to regulate virulence factors encoded within a horizontally acquired [pathogenicity island]" [25].
KdpD/KdpE is positioned at a molecular hub that is fundamentally relevant to homeostasis during infection. This central position within regulatory networks controlling homeostasis, in conjunction with the radical stress associated with pathogenic host interactions, is likely to explain how KdpD/KdpE has evolved a role in pathogenicity in some taxa. As such, KdpD/KdpE is likely subjected to strong selection to acquire novel functions in mediating virulence pathways. We might expect that the selection pressures, or the novel functions that result from them, would be variable across different bacteria with different hosts and infection parameters. This taxa-specific evolutionary novelty is reflected in the phylogenetically diverse yet discontinuous nature of the data available so far. Figure 2 depicts the phylogenetic relationship between bacterial taxa where evidence supports a pleiotropic role of KdpD/KdpE. The majority of cases come from the well-studied gamma-Proteobacteria clade, but there is also evidence from the more distantly related Firmicutes and Actinobacteria. The common association of KdpD/KdpE within an operon containing the structural genes KdpABC suggests that a role in K+ homeostasis is ancestral and that KdpD/KdpE's adaptation for virulence or survival has arisen more recently and possibly multiple times. Whether K+ regulation by KdpD/KdpE pre-dates more specific host-responsive activities or whether these evolved in parallel is a key question with broad implications on the evolution of virulence. Ultimately, a detailed understanding (across a range of pathogenic and nonpathogenic bacteria) of KdpD/KdpE's connectivity within homeostatic response, stress response, and virulence pathways will be needed to address this question.
Evidence supporting the role of KdpD/KdpE in virulence (V) or survival (S) is indicated across diverse bacterial species, all of which are capable of intracellular replication to some extent. The relevant references are also indicated. Phylogenetic relationships are as suggested by Battistuzzi et al. (2004) (not drawn to scale) [51].
The perception that Kdp serves a general homeostatic function alone means that the Kdp regulatory network has not been specifically targeted in virulence studies. The studies discussed in this review, however, point to a clear need for more detailed experiments, with a systematic approach across multiple taxa. Specifically, useful tools would include (i) genetic knock-outs (plus/minus complementation) of KdpD and KdpE (individually and together) and the structural genes kdpFABC or (ii) combinations of transgenic and/or native TCSs and Kdp-ATPases from species observed to show different effects of KdpD/KdpE. These genetic tools could be used for transcriptomic analyses and in vivo studies. Transcriptomic analyses could identify networks of genes co-regulated in response to different challenges (K+ limitation, AMPs, oxidative or osmotic stress) or during infection of cells and whole organisms. In vivo studies in host organisms could examine the effects on acute virulence (measuring LD50, time to death, or host weight gain) or chronic persistence (how long bacteria can be retrieved). Ideally, it would be informative to re-investigate the species discussed in this review—and new species—using a coordinated and comparable experimental approach. To contrast with the data available so far, it would be interesting to investigate species that cannot replicate intracellularly such as Pseudomonas spp. (gamma-Proteobacteria) or B. cereus (Firmicutes). Studies involving closely related species with diverse lifestylessuch as the Burkholderia, which range from environmental organisms to obligate and opportunistic pathogens with varied host ranges—might be useful for dissecting the effects of host and lifestyle on KdpD/KdpE function.
We predict that KdpD/KdpE will continue to be demonstrated to function in K+ homeostasis and virulence (directly or via stress response etc.) to different extents in different species. These differences are likely to correlate with whether the species in question leads a predominantly pathogenic or more environmentally based lifestyle. In the former case, the type of host response a species encounters is also likely to be important. Finally, studies of TCSs such as PhoP/PhoQ, Cra, and KdpD/KdpE suggest that other central homeostasis bacterial regulators will also prove to have various adaptive roles in virulence. The application of sophisticated systems-level studies across diverse bacterial pathogens may help to clarify this phenomenon and take us closer to understanding the regulatory breadth and complexity of pathogenesis.
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Anne McClelland
In late March I called a meeting of the XaaS Channel Optimization Advisory Board to find out how these TSIA members were coping with the lockdown and their new way of doing business with employees, partners, and customers. These technology vendors were, by and large, making significant investments in time, effort, and energy in their partner communities, making them feel cared for, invested in, and encouraged.
However, when surveying the TSIA membership in July, 50% of the participants in that survey believed that vendors had not provided enough financial support to their partners during the COVID-19 crisis.
It is clear that vendors attempted to provide a lot of support to their partners. But with such unknowns surrounding the future and how this economic “black swan” will play out, vendors may have held back financial support until they could see clearly the path to the future.
At that Advisory Board meeting, the technology vendors discussed the following partner care initiatives to invest in their partner communities:
As the Advisory Board members discussed these things they began to talk of the future. What will the future hold? How do they plan for the future? Can they plan? What will happen to the channel partners? How many will remain standing when the dust clears? If their company is “paying it forward” now, will that reap revenue and customer returns after the crisis is over?
During a webinar that I held on April 1st, I asked the audience about their opinion of how COVID-19 would impact the partner channel, and it was clear that they were concerned. Nearly 60% of the audience said that partners who were not profitable prior to the crisis would go out of business.
Nearly 50% of the participants predicted a swell of acquisitions in the partner channel.and over 70% predicted that partners who were not already delivering “as a Service” customer value prior to the crisis would likely become irrelevant. And, over 68% of them believed that customers will only continue to buy from partners that deliver solutions which customers consider mission critical.
The final question was around how their company was helping others to cope. From the data, one can see that over 75% of the companies were encouraging innovation and leveraging the company’s strengths to help others (customers, partners, employees). Additional ways to help included driving online engagement opportunities, focusing even more on customer, and partner success.
In the TSIA XaaS Partner Trends Ecosystem survey which concluded in late April, we asked the question, “How are you using your channel partner differently in this time of COVID-19?” (see Figure 3)
You can see from the data that the majority of companies (75%) were engaging in communications with their partners to both brainstorm with them and find out how the vendors could help the partners during the crisis. It is clear that driving more 2-way communication between partners and technology vendors has been one of the key focus areas of most vendor companies during this time.
In another Advisory Board meeting held in early May, when asked what the technology vendors were most concerned about at that time, the answers were fascinating. The discussions did not center around the “lock-down” or around the work-from-home transition as they had before, but were instead focused on new challenges that were visible on the horizon, such as:
From a recent TSIA Rapid Research Response survey focused on the partner ecosystem, only 30% of the respondents said that they knew of partner companies that are either folding or being acquired. Additionally, the respondents were evenly split between whether or not technology vendors had provided enough financial support to partners during this COVID crisis.
The more interesting data centered around perspectives of the future.
When asked the question, “Moving forward, do you believe technology vendors will pull more revenue into their direct channel?”, 40% of the respondents said “yes.” Also, when asked if they believed that customers will be more interested in Cloud and subscription-based offerings in the future, 90% said “yes,” which correlates to other TSIA surveys which include similar questions.
The final question surrounded the disruption factor of Cloud and subscription business models to the relationship between technology providers and their partners. You can see the responses below in Figure 4.
Of the respondents, 90% feel that these business models will either be somewhat disruptive or very disruptive to the relationships between technology providers and their partners. This is consistent with my anecdotal findings from meeting with technology providers around the world.
The data that we collect at TSIA clearly shows that the COVID-19 crisis and subsequent lockdowns have caused a dramatic acceleration of use and investment in “as-a-Service” offerings by companies large and small. I have spoken to those serving the Financial Services industry recently and I am told that even the banking industry, who are typically insistent upon having only “on premises” technology solutions are now forced to the Cloud due to the fact that their offices are closed as a result of the pandemic.
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Apartment mh30984
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TITLE: Analysis and categories
QUESTION [3 upvotes]: Can the concept of topology be generalized to deal with categories instead of sets such that one can define continuous mappings between categories possibly using categories with countably or uncountably infinite set of objects and morphisms ?
REPLY [3 votes]: It is not very clear what you mean so I hope the following answer, according to two possible interpretations of what you mean, is along what you have in mind. The notion of metric space (which is related to topology) naturally generalized to categories. The terminology is 'weighted category'. Basically, a weighted category is a category in which every arrow is assigned a non-negative real number (possibly $\infty $) in such a way that identities are assigned $0$, and the assignment in sub-additive with respect to composition. Grandis has some papers with things related to weighted categories, in particular in relation to the fundamental category of directed spaces.
A different answer is provided by what are known as Grothendieck topologies and sites. This is a bit more technical so I'll refer you to this link on the nLab for more information.
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Stay up to date with the most current information.
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\begin{document}
\maketitle
\dedicatory
\begin{abstract}
In this paper, we show that the infinitesimal Torelli theorem implies the existence of deformations of automorphisms. In the first part, we use Hodge theory and deformation theory to study the deformations of automorphisms of complex projective manifolds. In the second part, we use crystalline cohomology to explore the $p$-adic analogues of the first part, which generalizes a result of Berthelot and Ogus. The study of the deformations of automorphisms also provides criterions characterizing when the action of the automorphism group of a variety on its cohomology is faithful.
\end{abstract}
\tableofcontents
\section {Introduction}
The Hodge conjecture characterizes when a Betti cohomology class of a complex projective manifold can be represented by an algebraic cycle. Let $X_0$ be a complex projective manifold with a Betti cohomology class $Z_0$ of even degree, say $Z_0\in \Ho^{2d}(X_0,\Q)$. The Betti cohomology class $Z_0$ is called a class $of$ $Hodge$ $type$ if $(Z_0)\otimes {\C}$ is in $ \Ho^{2d}(X_0,\C)\cap \Ho^d(X_0,\Omega_{X_0/\C}^d)$. One expects that $Z_0$ is representable by an algebraic cycle of dimension $d$ if $Z_0$ is of Hodge type.
Let $\pi:X\rightarrow S$ be a smooth projective morphism over a complex quasi-projective variety $S$, and let $0$ be a point of $S$. Deligne \cite{Deligne} shows that, for a family of horizontal Betti cohomology classes $\{Z_t\}_{t\in S}$ of degree $2d$ on $X/S$, all the classes are of Hodge type if $Z_0$ is. Therefore, the Hodge conjecture predicts that $\{Z_t\}_{t\in S}$ are representable by algebraic cycles if $Z_0$ is. This prediction is what we call the variational Hodge conjecture, see \cite[Conjecture 9.6]{MP}.
Suppose that $S$ is smooth and connected. Assume that the Betti cohomology class $Z_0$ is representable by a local complete intersection $Z$ in $X_0$ of codimension $d$. Bloch~\cite{Bloch} defines a semi-regularity map $$s:\Ho^1(Z,N_{Z/X_0})\rightarrow \Ho^{d+1}(X_0,\Omega_{X_0/\C}^{d-1})$$ and shows that if $s$ is injective, then the Betti cohomology classes $\{Z_t\}_{t\in S}$ are representable by algebraic cycles.
In this paper, we show a similar result for automorphisms. Let \[\{h_t:\Ho^m(X_t,\Q)\rightarrow \Ho^m(X_t,\Q)\}_{t\in S}\] be a continuous family of maps. Suppose that $h_0$ is $\Ho^m(g_0,\Q)$ for some automorphism $g_0$ of $X_0$ and $X_0$ satisfies the infinitesimal Torelli theorem of degree $m$, i.e., the cup product
\begin{equation}\label{introcup}
\Ho^1(X_0,T_{X_0})\rightarrow \bigoplus\limits_{p+q=m} \mathrm{Hom}(\Ho^p(X_0,\Omega_{X_0/\mathbb{C}}^q), \Ho^{p+1}(X_0,\Omega_{X_0/\mathbb{C}}^{q-1}))
\end{equation}
is injective. We show that there is a family of automorphisms $\{g_t:X_t \rightarrow X_t\}_{t\in U}$ over an open neighborhood $U$ of $0\in S$ such that $h_t=\Ho^m(g_t,\Q)$. This motivates the following definition.
\begin{defi}\label{condition1}
Let $\pi:X\rightarrow S$ be a smooth projective morphism over a complex quasi-projective variety $S$, and let $0$ be a point of $S$. For a non-negative integer $m$, an automorphism $g_0$ of $X_0$ is $of~m$-$Hodge~type~$on $X/S$ if $\Ho^m(g_0,\mathbb{Q})$ is the stalk \[h_0: ( \Ri^m\pi_*\mathbb{Q})_0\xrightarrow{} (\Ri^m\pi_*\mathbb{Q})_0\] of an endomorphism $h$ of the local system $\Ri^m\pi_*\mathbb{Q}$ at the point $0$ where $\Ri^m\pi_*$ is the $m$-th derived functor of $\pi_*$.
\end{defi}
\begin{remark} \label{rmkhdg}Let $\Omega^{\bullet}_{X/S}$ be the de Rham complex of $X/S$, and let $\Omega^{\bullet\geq p}_{X/S}$ be the complex obtained from $\Omega^{\bullet}_{X/S}$ by replacing terms in degrees less than $p$ by $0$. The $Hoge$ $filtration$ $\F^p( \R^m \pi_*\Omega^{\bullet}_{X/S})$ of $\R^m\pi_*\Omega^{\bullet}_{X/S}$ is given by $$\mathrm{Im}\left(\R^m \pi_*\Omega^{\bullet\geq p}_{X/S}\rightarrow \R^m\pi_*\Omega^{\bullet}_{X/S} \right),$$ cf. \cite[Definition 4.7]{MP}. By the results of Deligne \cite{Deligne}, the map $h$ has the following remarkable property. Under the natural comparison $$\Ri^m\pi_*\C\otimes_{\C} \OO_S \cong \R^m\pi_*\Omega^{\bullet}_{X/S},$$
the map $h\otimes\OO_S$ preserves the Hodge filtration of $ \R^m\pi_*( \Omega^{\bullet}_{X/S})$, namely,
\begin{equation} \label{hodge}
h\otimes\OO_S\left( \F^p \R^m\pi_*( \Omega^{\bullet}_{X/S})\right) \subseteq \F^p \R^m\pi_*( \Omega^{\bullet}_{X/S}) \text{~for~all~p}.
\end{equation}
\end{remark}
\begin{example}
For an integer $m$, if $g_0$ is an automorphism of $X_0$ with $\Ho^m(g_0,\Q)=\Id_{\Ho^m(X_0,\Q)}$, then $g_0$ is of $m$-Hodge type on $X/S$ by taking \[ h=\Id:\Ri^m\pi_*(\Q)\rightarrow \Ri^m\pi_*(\Q).\]
\end{example}
It is obvious that $g_0$ is of $m$-Hodge type on $X_{U}/U$ for all $m$ if $g_0$ has a deformation $g:X|_{U}\rightarrow X|_{U}$ over an open neighborhood $U$ of $0$ in $S$ by taking $h=\Ri^m\pi_*(g)$. It is easy to see that $h\otimes\OO_S$ preserves the Hodge filtration of $ \R^m\pi_*( \Omega^{\bullet}_{X_U/U})$. Conversely, we have one of the main theorems of this paper.
\begin{theorem}\label{thmdeform}
Let $\pi: X\rightarrow S$ be a smooth projective morphism from a complex variety $X$ to a smooth curve $S$, and let $g_0$ be an automorphism of the fiber $X_0$ of $\pi$ over a point $0$ of $S$. Suppose that $X_0$ satisfies the infinitesimal Torelli theorem of degree $m$ for some integer $m$. If the automorphism $g_0$ is of $m$-Hodge type on $X/S$,
then $g_0$ has a deformation \[g:X|_{U}\rightarrow X|_{U}\] over an open neighborhood $U$ of $0\in S$.
\end{theorem}
One interesting application of Theorem \ref{thmdeform} is Corollary \ref{faithful} characterizing when the action of the automorphism group of a variety on its cohomology is faithful. Historically speaking, the faithfulness question of the action is first explored for varieties of low dimensions. It is well known that the faithfulness is confirmed for algebraic curves of genus at least 2 by Klein. The case of $K3$ surfaces is confirmed by Burns, Rapoport, Shafarevich, Ogus, among others. However, few higher dimensional varieties are known to have a positive answer to this question. Using Corollary \ref{faithful}, one can show that the automorphism groups of higher dimensional varieties (e.g. complete intersections, see \cite{PAN2} for details) act on their cohomology faithfully as long as they satisfy the infinitesimal Torelli theorem and the general members of these varieties have no non-trivial automorphisms.
The second part of this paper is to show the $p$-adic analogue of Theorem \ref{thmdeform}.
Recall from \cite[Definition 4.7]{MP} that the $q$-th de Rham cohomology $\Ho^q_{\DR}(\mathcal{X}/B)$, for a smooth proper morphism $g:\mathcal{X}\rightarrow B$ of noetherian schemes is defined as the coherent $\OO_B$-module $\R^qg_*\Omega^{\bullet}_{\mathcal{X}/B}$. The $Hodge~filtration$ of $\Ho^q_{\DR}(\mathcal{X}/B)$ is given by
\[ \F^p( \R^q g_*\Omega^{\bullet}_{\mathcal{X}/B}):=\mathrm{Im}\left(\R^qg_*\Omega^{\bullet\geq p}_{\mathcal{X}/B}\rightarrow \R^qg_*\Omega^{\bullet}_{\mathcal{X}/B} \right).\]
With these notions, the $p$-adic analogue of the variational Hodge conjecture can be formulated as follows. Let $k$ be an algebraically closed field of positive characteristic, and let $W(k)$ be the Witt ring of $k$ with the fraction field $K$. Suppose that $\pi': X\rightarrow W(k)$ is a smooth proper morphism with the closed fiber $X_0$. Let $\cris^{*}(X_0/W(k))$ be the crystalline cohomology of $X_0$. Based on the connection between crystalline cohomology and Hodge theory, ~Bloch, Esnault, and~Kerz \cite[Conjecture 1.2]{BEK},~Maulik and~Poonen \cite[Conjecture 9.2]{MP}, Fontaine and~Messing propose the following conjecture.\\% \\
\noindent\textsc{Conjecture.} The rational crystalline cycle class of an algebraic cycle $Z_0\in \cris^{2r}(X_0/W(k))_K$, expressed as a de Rham class on $X_K/K$ under the natural comparison $\cris^{2r}(X_0/W(k))_K\cong \Ho^{2r}_{\DR}(X_{K}/K)$, is the cycle class of an algebraic cycle on $X/K$, if and only if $Z_0$ is in the right level of the Hodge filtration $\F^r\Ho^{2r}_{\DR}(X_K/K)$. \\
Inspired by the conjecture above, it is natural to ask for the $p$-adic analogue of Theorem \ref{thmdeform}. The natural way to formulate the $p$-adic analogue is to replace the morphism $\pi:X\rightarrow S$ in Theorem \ref{thmdeform} by the morphism $\pi ': X\rightarrow W(k)$ over $W(k)$. In this case, $X_0$ should be the closed fiber of $\pi'$. Furthermore, by Hodge theory, a notable feature of $X/S$ is that its Hodge-de Rham spectral sequence degenerates at $E_1$ with locally free terms. Therefore, we introduce the following definition.
\begin{defi}
Let $k$ be an algebraically closed field. A $W(k)$-scheme $X$ over the Witt ring $W(k)$ of $k$ is $of~Hodge~type$ if the structure morphism $\pi': X\rightarrow W(k)$ is smooth and projective such that the Hodge-de Rham spectral sequence of $\pi'$ degenerates at $E_1$ page and the terms are locally free.
\end{defi}
Motivated by (\ref{hodge}) in Remark \ref{rmkhdg}, we make the following analogous definition.
\begin{defi}\label{hodgetype}
Let $k$ be an algebraically closed field, and let $\pi ': X\rightarrow \Spec W(k)$ be a smooth proper morphism over the Witt ring $W(k)$. For a non-negative integer $m$, an automorphism $g_0$ of the closed fiber $X_0$ of $\pi'$ is $of$ $m$-$Hodge$ $type$ on $X/W(k)$ if the map \[\cris^m(g_0): \cris^m(X_0/W)\rightarrow \cris^m(X_0/W)\] preserves the Hodge filtrations under the natural identification $\cris^m(X_0/W)\cong \Ho^m_{\DR}(X/W)$.
\end{defi}
Now, we are able to state the $p$-adic analogue of Theorem \ref{thmdeform}.
\begin{theorem}\label{corolifting}
Let $k$ be an algebraically closed field, and let $X$ be a $W(k)$-scheme of Hodge type with the closed fiber $X_0$. Suppose that $X_0$ satisfies the infinitesimal Torelli theorem of degree $m$ for some integer $m$, i.e., the cup product\begin{equation}\label{cupproduct}
\Ho^1(X_0,T_{X_0})\rightarrow
\bigoplus\limits_{p+q=m} \mathrm{Hom}(\Ho^q(X_0,\Omega^p_{X_0/k}),\Ho^{q+1}(X_0,\Omega^{p-1}_{X_0/k}))
\end{equation}is injective. If $g_0$ is an automorphism of $X_0$ of $m$-Hodge type on $X/W(k)$, then one can lift $g_0$ to an automorphism $g:X\rightarrow X$ over $W(k)$.
\end{theorem}
Theorem \ref{corolifting} is predicted by the $p$-adic variational Hodge conjecture, see \cite[Remark 6.5]{EO}. Ogus \cite[Corollary 2.5]{Ogus} proves a special case of this theorem for automorphisms of K3 surfaces. Later on, Berthelot and Ogus \cite[Theorem 3.15]{BO} prove a special case of the theorem for abelian varieties.
To use Theorem \ref{corolifting}, we have to verify that $X_0$ satisfies the infinitesimal Torelli theorem. In fact, it is verified for complete intersections and cyclic coverings in \cite{PAN2} and \cite{PAN3}. Therefore, it follows from Theorem \ref{corolifting} that, for many smooth projective varieties over an arbitrary field, their automorphism groups act faithfully on their \'etale cohomology, see Corollaries \ref{faithful} and \ref{padicfaithful}. For the applications of the faithfulness of the automorphism group action on the cohomology to arithmetic and moduli problems, we refer to \cite{PAN1}, \cite{PAN2}, \cite{PAN3}, \cite{DL1}, and \cite{DL2}.
To end this introduction, we outline the main ingredients of the proofs of Theorem \ref{thmdeform} and Theorem \ref{corolifting}. Theorem \ref{thmdeform} involves some techniques of homological algebra, deformation theory and the theory of de Rham cohomology. The key step of the proof is to use the Riemann-Hilbert correspondence to translate a morphism between local systems into a morphism between vector bundles with integrable connections, see Section \ref{deforms}. To prove Theorem \ref{corolifting}, this kind of translation is replaced by the rigidity of crystalline cohomology, see Sections \ref{cryobs} and \ref{bloch}.
\textbf{Acknowledgments.} The author is very grateful to~S.~Bloch for some suggestions on this project during his visiting in Washington University in St.~Louis. The author is also very grateful to~H.~Esnault for pointing out some references and typos of the manuscript. The author also thanks ~L.~Illusie and~M.~Kerr for their interest in this project, and~J.~de Jong for giving lectures on crystalline cohomology when the author was a graduate student in Columbia University. One part of the paper was written in Morningside Center of Mathematics in Beijing. The author thanks~Y.~Tian and~W.~Zheng for their invitation and warm hospitality.
\section{Homological Algebra and de Rham Cohomology} \label{homoalg}
In this section, we prove some results of homological algebra. We also cite some results of the theory of de Rham cohomology. They will be used in Section \ref{deforms} to prove Theorem \ref{thmdeform}.
\begin{lemm}\label{lemcom1} Suppose that we have the following exact sequences and diagram
\[\xymatrix{ &0 \ar[r] &M \ar@{->}[ddl] \ar@{=}[dl] \ar[r] &C \ar@{->}[ddl] \ar[r] &C_0\ar[r] \ar[dl] \ar@{->}[ddl] &0\\
0 \ar[r] &M\ar@{=}[d] \ar[r]|-{\hole} &B \ar[d] \ar@{}[dr] \ar[r]|-{\hole} &B_0 \ar[d]\ar[r]|-{\hole} &0\\
0 \ar[r] &M \ar[r] &A\ar[r] &A_0\ar[r] &0}\]
in an abelian category.
There is a morphism $C\rightarrow B$ filling in the diagram such that the whole diagram commutes.
\end{lemm}
\begin{proof}
Note that the exact sequence $0\rightarrow M \rightarrow B \rightarrow B_0\rightarrow 0$
is the pull-back of
\[0\rightarrow M \rightarrow A \rightarrow A_0\rightarrow 0\] via the morphism $B_0\rightarrow A_0$ and the exact sequence \[0\rightarrow M \rightarrow C \rightarrow C_0\rightarrow 0\] is the pull-back of
$0\rightarrow M \rightarrow A \rightarrow A_0\rightarrow 0$ via the morphism $C_0\rightarrow A_0.$ The lemma follows from this remark.
\end{proof}
\begin{lemm}\label{lemcom2} Suppose that we have exact sequences and an diagram (in an abelian category) as follows.
\begin{equation}\label{commu}
\xymatrix{ 0 \ar[r] &M_0 \ar@/_1pc/@{->}[ddrr] \ar@{>->}[d] \ar[r] &B_0 \ar@{>->}[d] \ar[r] &A_0 \ar@{=}[d] \ar[r] \ar@/^0.9pc/@{->}[ddrr] &0\\
0 \ar[r] &M_0' \ar@/_1pc/@{->}[ddrr] \ar[r] &B_0' \ar@/^0.9pc/@{->}[ddrr]\ar[r] &A_0\ar[r]\ar@/^0.9pc/@{->}[ddrr] &0\\
& &0 \ar[r] &M_1 \ar@{>->}[d] \ar[r] &B_1 \ar@{>->}[d] \ar[r] &A_1 \ar@{=}[d] \ar[r] &0\\
& &0 \ar[r] &M_1' \ar[r] &B_1' \ar[r] &A_1\ar[r] &0\\
}
\end{equation}
Assume that the morphisms $M_0\rightarrow M'_0$ and $M_1\rightarrow M'_1$ are monic. There is a morphism $B_0\rightarrow B_1$ filling in the diagram such that the whole diagram commutes.
\end{lemm}
\begin{proof}
From the commutative diagram (\ref{commu}), we have a commutative diagram as follows
\[\xymatrix{ M_0 \ar@{->}[drr] \ar@{>->}[d] \ar[r] & B_0 \ar@{>->}[d]\\
M_0' \ar@{->}[drr] \ar@{->>}[d] \ar[r] & B_0' \ar@{->}[drr] \ar@{->>}[d] &M_1 \ar@{>->}[d] \ar[r] & B_1 \ar@{>->}[d]\\
\cok_0 \ar@{->}[drr] \ar@{=}[r] & \cok_0\ar@{->}[drr] & M_0' \ar@{->>}[d] \ar[r] & B_0' \ar@{->>}[d] \\
&& \cok_1\ar@{=}[r] &\cok_1 }.\]
It induces a morphism $B_0\rightarrow B_1$ filling into the diagram (\ref{commu}) such that the whole diagram commutes.
\end{proof}
Let $A$ be a $\C$-algebra with an ideal $I$, and let $\pi:X\rightarrow S$ be a smooth projective morphism where $S=\Spec (A)$. Denote by $S_0$ the affine scheme $\Spec (A_0)$ where $A_0=A/I$. Suppose that $X_0$ is the pull back of $X/S$ along $S_0\hookrightarrow S$. Denote the natural immersion $X_0\hookrightarrow X$ by $i$.
\[\xymatrix{ X_0 \ar[d]_{\pi_0} \ar@{}[rd]|-{\Box} \ar@{^{(}->}[r]^i &X \ar[d]^{\pi} \\
S_0 \ar@{^{(}->}[r] &S }\]
Let $K_{X/S/\C}\in \Ext^1_X(\Omega^1_{X/S},\Omega^1_{S/\C}\otimes \OO_X)$
be the Kodaira-Spencer class of $X/S$, i.e., the class of the extension
\[0\rightarrow \Omega^1_{S/\C}\otimes \OO_X\rightarrow \Omega^1_{X/\C}\rightarrow \Omega^1_{X/S}\rightarrow 0.\] Let $\beta\in \Ext^1_{X_0}(\Omega^1_{X_0/S_0},\Omega^1_{S/\C}\otimes \OO_{X_0})$ be the class of the extension $K_{X/S/\C}\otimes \OO_{S_0}$
\begin{equation} \label{beta}
0\rightarrow \Omega^1_{S/\C}\otimes \OO_{X_0}\rightarrow \Omega^1_{X/\C}|_{X_0}\rightarrow \Omega^1_{X_0/S_0}\rightarrow 0.
\end{equation}
Recall that $\Ho^q _{\DR}(X/S)$ is the $q$-th de Rham cohomology given by $\R^q\pi_*(\Omega^{\bullet}_{X/S})$ where $\Omega^{\bullet}_{X/S}$ is the de Rham complex of $X/S$ and $\Ri^q\pi_*$ is the $q$-th derived functor of $\pi_*$. We recall the following propositions.
\begin{prop}$($\cite[Theorem 3.2]{Bloch} and \cite{Del}$)$\label{prop0} Let S be a scheme over $\Spec (\mathbb{Q})$, and let $\pi:X\rightarrow S$ be a proper and smooth morphism. Then
\begin{enumerate} \label{thmlf}
\item The sheaves $\Ri^q \pi_*(\Omega^p_{X/S})$ are locally free of finite type and commute with base change.
\item The spectral sequence
\[E^{p,q}_1=\Ri^q \pi_*(\Omega^p_{X/S})\Longrightarrow \Ho^{p+q}_{\DR}(X/S)\]
degenerates at $E_1$.
\item The sheaves $\Ho^*_{\DR}$ are locally free of finite type and commute with base change.
\end{enumerate}
Let $S$ be a smooth $\mathbb{C}$-scheme. There is a canonical integrable connection, namely, the Gauss-Manin connection
\[\nabla :\Ho^q_{\DR}(X/S)\rightarrow \Ho^q_{\DR}(X/S)\otimes_{\OO_S}\Omega^1_{S/\C}.\]
The spectral sequence in Proposition \ref{thmlf} (2) induces a (Hodge) filtration
\[0\subseteq \F^q \subseteq \F^{q-1}\cdots \subseteq \F^1\subseteq \F^0=H^q_{\DR}(X/S)\]
such that \begin{itemize}
\item $\F^0,\F^1,\ldots,\F^q$ are locally free $\OO_S$-module,
\item and the Griffiths's Transversality $\nabla(\F^p)\subseteq \F^{p-1}\otimes \Omega^1_{S/\C}$.
\end{itemize}
\end{prop}
\begin{prop} $($\cite[Proposition 3.6]{Bloch} and \cite{Griff}\label{prop1}$)$
With the same notations as above, the Gauss-Manin connection $\nabla$ is related to the Kodaira-Spencer class $K_{X/S/\C}$ by the following commutative diagram
\[\xymatrix{\F^p/\F^{p+1}\ar@{=}[d]\ar[rr]^{\nabla} &&\F^{p-1}/\F^{p}\otimes \Omega^1_{S/\C} \ar@{=}[d]\\
\Ri^{q-p}\pi_*(\Omega^p_{X/S}) \ar[rr]^<<<<<<<<<<{\cup K_{X/S/\C}} &&\Ri^{q-p+1}\pi_*(\Omega^{p-1}_{X/S})\otimes \Omega^1_{S/\C}}.\]
\end{prop}
Note that we have the following proposition about integrable connections and stratifications.
\begin{prop}$($\cite[Proposition 3.7, Proposition 3.8, Remark 3.9]{Bloch}$)$ \label{prop2}
\begin{enumerate}
\item Let $k$ be a field of characteristic $0$, $M$ a finite $k[[t_1,\ldots,t_r]]$-module with integrable connection $\nabla$. Let $M^{\nabla}=\Ker(\nabla)$. Then $M=M^{\nabla}\otimes_k k[[t_1,\ldots,t_r]]$.
\item Let $A$ be a complete, local, augmented $\C$-algebra (e.g. $A$ artinian), $S=\Spec(A)$, and $X_0\subseteq X$ be the closed fiber. Then $$\Ho^*_{\DR}(X/S)\cong \Ho^*(X_0,\C)\otimes_{\C}A.$$ It gives a stratification on $\Ho^*_{\DR}(X/S).$ Cohomology classes of the form $$c\otimes 1, c\in \Ho^*(X_0,\C)$$ are said to be horizontal.
\end{enumerate}
\end{prop}
\section{Deformations of Automorphisms for Algebraic Manifolds} \label{deforms}
In this section, we use the notations as in Section \ref{homoalg}. Let $g_0$ be an automorphism of $X_0$ over $S_0=\Spec(A_0)$.
\begin{lemm} \label{lemmlift}
Let $A$ be a $\C$-algebra with square zero ideal $\II$, and let $A_0$ be $A/\II$. Denote by $S$ (resp. $S_0$) $\Spec(A)$ (resp. $\Spec(A_0)$). Suppose that the natural map $d:\II \rightarrow \Omega^1_{S/\C}\otimes \OO_{S_0}$ is injective and $g_0^*\beta=\beta$, see (\ref{beta}) for the definition of $\beta$. Then the automorphism $g_0$ is unobstructed, i.e., one can extend $g_0$ to an automorphism \[g:X/A\rightarrow X/A\] over $\Spec (A)$.
\end{lemm}
\begin{proof}
To extend $g_0$ to a morphism $g$ over $\Spec (A)$, it suffices to find a morphism $h: \OO_X \rightarrow (g_0^{-1})_*(\OO_X)$ of sheaves of rings such that it fills into the following diagram
\begin{equation} \label{ex0}
\xymatrix{0 \ar[r] &\II \OO_{X} \ar[r] \ar[d] &\OO_X \ar[d]^{h} \ar[r] &i_*\OO_{X_0}\ar[r] \ar[d]^{(g_0^{-1})^*} &0\\
0 \ar[r] &\II (g^{-1}_{0})_*\OO_{X} \ar[r] &(g^{-1}_{0})_*\OO_{X}\ar[r] &(g^{-1}_{0})_*\OO_{X_0}\ar[r] &0}
\end{equation}
where we abuse $(g_0^{-1})_*$ to represent $i_*\circ (g_0^{-1})_*$ and $i$ is the natural inclusion $X_0 \hookrightarrow X$.
In fact, we have a commutative diagram
\[\xymatrix{0 \ar[r] &\II \OO_X (=\II\otimes \OO_{X_0})\ar[d]^{d\otimes 1} \ar[r]&\OO_{X} \ar[d]|-{\pi\circ d} \ar@{}[dr] \ar[r] &i_*\OO_{X_0} \ar[d]|-{i_*(d)}\ar[r] &0\\
0 \ar[r] &i_*(\Omega^1_{S/\mathbb{C}}\otimes \OO_{X_0}) \ar[r] &i_*(\Omega^1_{X/\mathbb{C}}|_{X_0})\ar[r] &i_*\Omega^1_{X_0/S_0}\ar[r] &0}\]where $\pi$ is the quotient $\Omega^1_{X/\C}\rightarrow \Omega^1_{X/\C}|_{X_0}$. We denote the first exact sequence in the previous diagram by $\alpha$. Recall that $\beta$ (see Section \ref{homoalg}) is the class of the extension
\[0\rightarrow \Omega^1_{S/\C}\otimes \OO_{X_0}\rightarrow \Omega^1_{X/\C}|_{X_0}\rightarrow \Omega^1_{X_0/S_0}\rightarrow 0.\]We have that \[(i_*(d))^*(i_*(\beta))= (d\otimes 1)_* (\alpha).\]We hope it will cause no confusion if we also denote $\beta$ by $i_*(\beta)$. It gives rise to the following diagram
\begin{equation}\label{ex1}
\xymatrix{ &0 \ar[r] &\II \OO_{X} \ar@{->}[ddl] \ar[dl]_{d\otimes 1} \ar[r] &\OO_X \ar[dl] \ar@{->}[ddl]|->>>>>>>{\pi \circ d} \ar[r] &i_*\OO_{X_0}\ar[r] \ar@{=}[dl] \ar@{->}[ddl] &0\\
0 \ar[r] &i_*(\Omega^1_{S/\mathbb{C}}\otimes \OO_{X_0})\ar@{=}[d] \ar[r]|-{\hole} &B \ar[d] \ar@{}[dr] \ar[r]|-{\hole} &i_*\OO_{X_0} \ar[d]|-{i_*(d)}\ar[r]|-{\hole} &0\\
0 \ar[r] &i_*(\Omega^1_{S/\mathbb{C}}\otimes \OO_{X_0}) \ar[r] &i_*(\Omega^1_{X/\mathbb{C}}|_{X_0})\ar[r] &i_*\Omega^1_{X_0/S_0}\ar[r] &0}
\end{equation}
where the middle short exact sequence is represented by $(i_*(d))^*(\beta)$. Note that $g_{0*}=(g_0^{-1})^*$, $g_0^*=(g_0^{-1})_*$ and the bottom of (\ref{ex1}) consists of $\OO_{X_0}$-modules.
We can pull back the diagram above via $g_0^*$. It gives rise to a commutative diagramm as (\ref{ex1}),
\begin{equation}\label{ex2}
\xymatrix{ 0 \ar[r] &\II (g^{-1}_{0})_*\OO_{X} \ar@{->}@/^5.5pc/[dd] \ar[d]_{d\otimes 1} \ar[r] &(g^{-1}_{0})_*\OO_{X} \ar[d] \ar@{->}@/^2.8pc/[dd]|->>>>>>>{\pi \circ d} \ar[r] &(g^{-1}_{0})_*\OO_{X_0}\ar[r] \ar@{=}[d] \ar@{->}@/^3.2pc/[dd] &0\\
0 \ar[r] &g_0^*(\Omega^1_{S/\mathbb{C}}\otimes \OO_{X_0})\ar@{=}[d] \ar[r]|-{\hole} &D \ar[d] \ar@{}[dr] \ar[r]|-{\hole} &(g^{-1}_{0})_*\OO_{X_0} \ar[d]|-{ (g^{-1}_{0})_*(d)}\ar[r]|-{\hole} &0\\
0 \ar[r] &g_0^*(\Omega^1_{S/\mathbb{C}}\otimes \OO_{X_0})\ar@{=}[d] \ar[r] &(i\circ g_0)_*(\Omega^1_{X/\mathbb{C}}|_{X_0^{g_0}})\ar[r]\ar@{=}[d] &g_0^*\Omega^1_{X_0/S_0}\ar[r] \ar@{=}[d] &0\\
0\ar[r]& i_*(\Omega^1_{S/\mathbb{C}}\otimes \OO_{X_0})\ar[r] & C \ar[r]&i_*(\Omega^1_{X_0/S_0})\ar[r] &0 }
\end{equation}
where $X_0^{g_0}$ is $\xymatrix{X_0 \ar@/^1pc/[rr]^{i\circ g_0} \ar[r]_{g_0} &X_0 \ar[r]_i &X}$ and we abuse $g_0^*$ to represent $i_*\circ g^*_0$.
Note that the pull-back $g_0^*$ is given by
\[\xymatrix{\Ext^1(\Omega^1_{X_0/S_0},\Omega^1_{S/\C}\otimes \OO_{X_0}) \ar[r] \ar[dr]_{g_0^*}
&\Ext^1(g_0^*\Omega^1_{X_0/S_0},g_0^*(\Omega^1_{S/\C}\otimes \OO_{X_0}))\ar@{=}[d]\\
&\Ext^1(\Omega^1_{X_0/S_0},\Omega^1_{S/\C}\otimes \OO_{X_0})
}\]
where the vertical identification follows from the differential map $$dg_0: g_0^*\Omega^1_{X_0/S_0}\xrightarrow{\cong} \Omega^1_{X_0/S_0}.$$
The assumption $g_0^*\beta =\beta$ follows that the exact sequence at the bottom of (\ref{ex2}) is the exact sequence at the bottom of (\ref{ex1}). It follows from Lemma \ref{lemcom1} that there is a map $u:B\rightarrow D$ filling into the digram
\begin{equation} \label{ex3}
\xymatrix{
0 \ar[r] &i_*(\Omega^1_{S/\mathbb{C}}\otimes \OO_{X_0})\ar[d] \ar[r] &B \ar[d]^u \ar[r] &i_*\OO_{X_0} \ar[d]^{(g_0^{-1})_*} \ar[r] &0\\
0 \ar[r] &g_0^*(\Omega^1_{S/\mathbb{C}}\otimes \OO_{X_0})\ar[r] &D \ar[r] &(g^{-1}_{0})_*\OO_{X_0} \ar[r]&0
}
\end{equation}
such that the whole diagram commutes.
Recall that $d:\II \rightarrow \Omega^1_{S/\C}\otimes \OO_{S_0}$ is injective and $X\rightarrow S$ is smooth. It implies that the maps $d\otimes 1$ in (\ref{ex1}) and (\ref{ex2}) are injective. It follows from Lemma \ref{lemcom2} and (\ref{ex3}) that there is a map $h: \OO_X \rightarrow (g_0^{-1})_*(\OO_X)$ filling into the diagram (\ref{ex0}) above such that the whole diagram commutes. We have proved the lemma.
\end{proof}
Now we are able to prove Theorem \ref{thmdeform}.
\begin{proof}
We translate the assumption of the theorem in terms of the Gauss-Manin connection as follows, cf. Proposition \ref{prop0}, Proposition \ref{prop1} and Proposition \ref{prop2}. By the Riemann-Hilbert correspondence, the horizontal map $h$ in Definition \ref{condition1} (cf. \ref{hodge}) induces the following commutative diagram
\begin{equation}\label{horizontal}
\xymatrix{\Ri^{m-p}\pi_*(\Omega _{X/S}^p)\ar[d]^{h_{ \C}} \ar[rr]^>>>>>>>>>>{\nabla} & & \Ri^{m-p+1}\pi_*(\Omega _{X/S}^{p-1})\otimes \Omega^1_{S/\C}\ar[d]^{h_{\C}\otimes \Id_{\Omega^1_{S/\C}}} \\
\Ri^{m-p}\pi_*(\Omega _{X/S}^p) \ar[rr]^>>>>>>>>>>{\nabla} & & \Ri^{m-p+1}\pi_*(\Omega _{X/S}^{p-1})\otimes \Omega^1_{S/\C} } .
\end{equation}
where $\nabla$ is given by the cup product of $K_{X/S/\C}$. We first show the theorem when $S=\Spec (\C[[t]])$.
Let $S_N$ be $\Spec (\C[[t]]/(t)^{N+1})$. Note that the maps
\[d:(t)^N/(t)^{N+1}\rightarrow \Omega^1_{S_N/\C}\otimes \OO_{S_{N-1}}\]are injective, cf. \cite[Theorem 7.1]{Bloch}. Denote by $X_N$ the pull-back $X\times_S S_N$. Assume that $g_0$ can extend to an automorphism $g_N$ over $S_N$. We base change to $S_N$ via $S_N\subseteq S$. We have the Kodaira-Spencer class $K_{X_{N+1}/S_{N+1}/\C}$ and
\begin{center}
$\beta=K_{X_{N+1}/S_{N+1}/\C}\otimes \OO_{S_N}$ with $\nabla(-)=(-)\cup \beta$,
\end{center}
cf. \cite[(4.1)]{Bloch}, Proposition \ref{prop1} and Proposition \ref{prop2}. Note that $$\Omega^1_{S/\C}=\C[t]/(t)^{N+1} dt=S_N dt=\Omega^1_{S/\C}\otimes_{\OO_S} S_N$$ and $h_{\C}\otimes S_N$ is given by $g_N^*:\Ho^{m-p}(X_N,\Omega ^{p}_{X_N/S_N})\rightarrow \Ho^{m-p}(X_N,\Omega ^{p}_{X_N/S_N})$. Therefore, the commutativity of the diagram (\ref{horizontal}) ($h_{\C}\nabla =\nabla h_{\C}$) gives that
\begin{center}
$g_N^*(-)\cup \beta= g_N^{*}(-\cup \beta)$ in $\Ho^{m-p+1}(X_N,\Omega ^{p-1}_{X_N/S_N})\otimes \Omega^1_{S_{N+1/\C}}$,
\end{center}
cf. \cite[Proposition 4.2]{Bloch}. It follows that $g_N^*(-)\cup \beta=g_N^{*}(-)\cup g_N^*(\beta)$.
We claim that the element\[g_N^{*}(\beta)-\beta \in \Ext^1(\Omega^1_{X_N/S_N,}\Omega^1_{S_{N+1}/\C}\otimes \OO_{X_N})\] is zero. If the claim holds, then it follows from Lemma \ref{lemmlift} that $g_N$ is liftable to an automorphism $g_{N+1}:X_{N+1}\rightarrow X_{N+1}$ over $S_{N+1}$. Therefore, by the Grothendieck existence theorem, we have an automorphism $\hat{g}:X\rightarrow X$ over $S$ such that $\hat{g}|_{X_0}=g_0$.
Note that $\Omega^1_{S_{N+1}/\C}=\C[t]/(t)^{N+1} dt=S_N dt$. Therefore, we have $$\Omega^1_{S_{N+1}/\C}\otimes \OO_{X_N}=\OO_{X_N}.$$ To show the claim, it suffices to show the cup product
\begin{equation}\label{injv}
\xymatrix{\Ext^1(\Omega^1_{X_N/S_N},\Omega^1_{S_{N+1}/\C}\otimes \OO_{X_N})\ar[d] \\
\bigoplus\limits_{p+q=m} \mathrm{Hom}(\Ho^p(X_N,\Omega_{X_N/S_N}^q), \Ho^{p+1}(X_N,\Omega_{X_N/S_N}^{q-1}\otimes\Omega^1_{S_{N+1}/\C}\otimes \OO_{X_N} ))\ar@{=}[d]\\
\bigoplus\limits_{p+q=m} \mathrm{Hom}(\Ho^p(X_N,\Omega_{X_N/S_N}^q), \Ho^{p+1}(X_N,\Omega_{X_N/S_N}^{q-1})\otimes\Omega^1_{S_{N+1}/\C})}
\end{equation}
is injective. We show the injectivity of the cup product by the induction on the length of $S_N$.
Firstly, we define two functors as follows:
\[T(M)=\Ext^1_{\OO_{X_N}}(\Omega^1_{X_N/S_N},M)\]
and \[S(M)=\bigoplus\limits_{p+q=m} \mathrm{Hom}(\Ho^p(X_N,\Omega_{X_N/S_N}^q), \Ho^{p+1}(X_N,\Omega_{X_N/S_N}^{q-1}\otimes M )) \]
from the category of $\OO_{X_N}$-modules to the category of $S_N$-modules. The cup product is a natural transformation between these two functors
\[\cup_M: T(M)\rightarrow S(M).\]The injectivity of the vertical arrow (\ref{injv}) is the same as the injectivity of \[\cup_{\OO_{X_N}}:T(\OO_{X_N})\rightarrow S(\OO_{X_N}).\] We consider the following exact sequence
\[0\rightarrow (t)^{N}\otimes \OO_{X_N}(= (t^N)\otimes_{\mathbb{C}} \OO_{X_0})\rightarrow \OO_{X_N}\rightarrow \OO_{X_{N-1}}\rightarrow 0.\]By the base change theorem \cite[Chapter III, Section 12]{H}, we have the following commutative diagram of exact sequences
\[\xymatrix{ &(t^N)\otimes_{\mathbb{C}} T(\OO_{X_0}) \ar[d]^{(t^N)\otimes_{\mathbb{C}} \cup} \ar[r] &T(\OO_{X_N})\ar[r] \ar[d]^{\cup} & T(\OO_{X_{N-1}})\ar[d]^{\cup}\\
0\ar[r] &(t^N)\otimes_{\mathbb{C}} S(\OO_{X_0}) \ar[r] &S(\OO_{X_N})\ar[r] & S(\OO_{X_{N-1}}) \ar[r] &0.
}\]The first vertical arrow $(t^N)\otimes_{\mathbb{C}} \cup$ is injective by the assumption of the theorem that $X_0$ satisfies the infinitesimal Torelli theorem of degree $m$. So the cup product $\cup_{\OO_{X_N}}$ is injective by the induction. It implies the theorem in the case $S=\Spec (\C[[t]])$.
The general case follows from the case when $S=\Spec (\C[[t]])$ by base change along $\Spec (\widehat{\OO_{S,0}})=\Spec (\C[[t]])\hookrightarrow S$.
\end{proof}
Apply Theorem \ref{thmdeform} with $h=\Id_{\mathrm{R}^m\pi_*\mathbb{Q}}$ in Definition \ref{condition1}. We obtain the following corollary.
\begin{cor}\label{faithful}
Suppose that a smooth projective $X_0$ is a fiber of a smooth family of projective variety $\pi: X\rightarrow S$. Assume that the cup product \[\Ho^1(X_0,T_{X_0})\rightarrow \bigoplus\limits_{p+q=m} \mathrm{Hom}(\Ho^p(X_0,\Omega_{X_0}^q), \Ho^{p+1}(X_0,\Omega_{X_0}^{q-1}))\]is injective for some $m$. Then the kernel of $\Aut(X_s)\rightarrow \Aut(\mathrm{H}^m(X_s,\Q))$ is trivial for all $s\in S$ if the kernels are trivial over an open dense subset of $S$.
\end{cor}
\begin{proof}
Let $f_0$ be an automorphism of $X_0$ in $$\mathrm{Ker_0}:( \Aut(X_0)\rightarrow \Aut(\mathrm{H}^m(X_0,\Q)).$$ Definition \ref{condition1} can be verified by taking $h=\Id_{\Ri^m\pi_*(\mathbb{Q})}$. Therefore, by Theorem \ref{thmdeform}, $f_0$ can be deformed to an automorphism $f_s\in \Ker_s$ of $X_s$ where $s\in S$ is a general point. By the assumption of the corollary, we conclude that $f_0$ is a specialization of identities, therefore, $f_0=\Id_{X_0}$.
\end{proof}
\begin{remark}
One can use Corollary \ref{faithful} to show that the automorphism groups of higher dimensional varieties (e.g. complete intersections) act on their cohomology faithfully as long as they satisfy the infinitesimal Torelli theorem and the general members of these varieties have no non-trivial automorphisms.
\end{remark}
\section{Crystalline Cohomology and Obstructions} \label{cryobs}
In this section, we use the notations following
\cite{Ber} and \cite{BO}. We will carry out a proof of Theorem \ref{corolifting} through the rest of the paper. Suppose that $S\rightarrow T$ is a closed immersion of affine schemes with the square zero ideal sheaf $\II$ and the prime $p$ is nilpotent on $T$. The pair $(T,\II)$ has a natural P.D structure such that $\II^{[a]}=0$ if $a\geq 2$. Suppose that $X$ and $Y$ are smooth projective schemes over $T$ with reductions $X_0$ and $Y_0$ over $S$. We have two inclusions
\begin{center}
$i:Y_0\hookrightarrow Y$ and $j:X_0\hookrightarrow X$.
\end{center}Let $f_0:X_0\rightarrow Y_0$ be an isomorphism between $X_0$ and $Y_0$. It gives rise to a map \begin{equation}\label{fcris}
\cris^k(f_0/T):\cris^k(Y_0/T)\rightarrow \cris^k (X_0/T).
\end{equation}
By the comparison theorem of crystalline cohomology and de Rham cohomology, we can view this map as $\cris^k(f_0):\Ho^{k}_{\DR}(Y/T)\rightarrow \Ho^{k}_{\DR}(X/T)$.
If we restrict $S$ in Theorem \ref{thmdeform} to a small disk $\Delta$ with center $0$, then the local system $\Ri^m\pi_* \Q|_{\Delta}$ is trivial and the the map $\Ho^m(g_0)$ induces the horizontal map \[h_{\Delta}:\Ri^m\pi_* \Q|_{\Delta}\rightarrow \Ri^m\pi_* \Q|_{\Delta}\] (cf. Definition \ref{condition1}) with the stalk $h_0=\Ho^m(g_0)$. In the case of mixed characteristic, the map $\cris^k(f_0)$ replaces the role of the horizontal morphism $h_{\Delta}$.
\begin{Assumption}\label{assumption}
Throughout the rest of the paper, we fix the notations and the assumptions as follows.
Let $\T^{\bullet}$ be a complex. We denote by $\T^{\bullet}[m]$ the complex \[(\T^{\bullet}[m])_n=\T_{m+n}\text{~ with differentials~} d_{(\T^{\bullet}[m])_n}=d_{\T^{\bullet}_{m+n}}:(\T^{\bullet}[m])_n\rightarrow (\T^{\bullet}[m])_{n+1}.\]
Let $T$ be an affine scheme with the square zero ideal sheaf $\II$. Suppose that a prime $p$ is nilpotent on $T$. Let $S$ be the closed subscheme of $T$ defined by $\II$. Let $X$ (resp. $Y$) be a smooth projective scheme over $T$ with reductions $X_0$ (resp. $Y_0$) over $S$.
Suppose that the Hodge-de Rham spectral sequences of $X/T$ and $Y/T$ degenerate at $E_1$ and the terms are locally free, so that the Hodge and de Rham cohomology sheaves commute with base change.\\
\end{Assumption}
We denote by $\F_X$ (resp. $\F_Y$) the Hodge filtration ($\F_{Hdg}$) for $X/T$ (resp. $Y/T$). Since $\cris^k(f_0)\otimes \Id_S=\Ho^k_{\DR}(f_0)$ preserves the Hodge filtrations, the map $$\F^p_{Y}\Ho^k_{\DR}(Y/T)\rightarrow gr_{\F_{X}}^{p-1}\Ho^k_{\DR}(X/T)\otimes\OO_S$$is zero. The map $\cris^k(f_0)$ induces a map:
\[
\begin{aligned}
\F^p_{Y}\Ho^k_{\DR}(Y/T)\rightarrow gr_{\F_X}^{p-1}\Ho^k_{\DR}(X/T)\otimes \II&=\Ho^{k-p+1}(X,\Omega_{X/T}^{p-1})\otimes \II \\
&=\Ho^{k-p+1}(X_0,\Omega_{X_0/S}^{p-1})\otimes \II
\end{aligned}
\]
which factors through
\begin{equation} \label{defrohf}
\rho(f_0)_p:\F^p_{Y_0}\Ho^k_{\DR}(Y_0/S)\rightarrow \Ho^{k-p+1}(X_0,\Omega_{X_0/S}^{p-1})\otimes \II .
\end{equation}
We denote $\rho(f_0)_p$ by $\rho(f_0)$ for simplicity. On the other hand, the obstruction $ob(f_0)$ of extending $f_0$ to a $T$-morphism $X\rightarrow Y$ is an element of $\Ext^1_{X_0}(f^*_0\Omega^1_{Y_0/S},\OO_{X_0}\otimes \II)$. The following proposition relates $\rho(f_0)$ and $ob(f_0)$.
\begin{prop} \label{generalremark}
With the notations and assumptions in \ref{assumption}, let $f_0:X_0\rightarrow Y_0$ be an isomorphism between $X_0$ and $Y_0$. We have the following commtative diagram
\begin{equation}
\xymatrix{\F_{Y_0}^p\Ho^k_{\DR}(Y_0/S) \ar[d]^{\proj} \ar[rr]^{\rho(f_0)} &&gr_{\F_{X_0}}^{k-p+1}\Ho^k_{\DR}(X_0/S)\otimes\II \ar@{=}[d]\\
\Ho^{k-p}(Y_0,\Omega^p_{Y_0/S})\ar[rr]^{\pm ob(f_0)\cup} &&\Ho^{k-p+1}(X_0,\Omega^{p-1}_{X_0/S})\otimes \II .}
\end{equation}
\end{prop}
For $p=1$, Proposition \ref{generalremark} is \cite[Proposition 3.20]{derham}. The idea of the proof of Proposition \ref{generalremark} is similar to the proof of \cite[Proposition 3.20]{derham}, but the arugment for the case of higher degrees $p\geq 2$ is much more complicated. The proof of Proposition \ref{generalremark} depends on the concrete descriptions of the maps of $\rho({f_0})$ and $ob(f_0)\cup$. We give these concrete descriptions in the rest of this section and present a proof of Proposition \ref{generalremark} in Section \ref{bloch}.
Let us cite some results from \cite{BO}. Recall that there are short exact sequences as follows (\cite[Chapter 5, 5.2 (3)]{BO}):
\[0\rightarrow \jj_{X_0/T} \rightarrow \OO_{X_0/T} \rightarrow i_{X_0/T*}(\OO_{X_0})\rightarrow 0 \]and
\[0\rightarrow \jj_{X/T} \rightarrow \OO_{X/T} \rightarrow i_{X/T*}(\OO_{X})\rightarrow 0. \]
There is a natural morphism of topoi:
\[u_{X/T}:(X/T)_{cris}\rightarrow X_{zar},\]
given by \[\left(u_{X/T*}(\mathcal{F}))(U)=\Gamma((U/T)_{cris},j^*_{cris} \mathcal{F}\right)\]where $\mathcal{F}\in (X/T)_{cris}$ and $j:U\rightarrow X$ is an open immersion, cf. \cite[Proposition 5.18]{BO}.
\begin{lemm}\label{quo}\label{lemm41}
Let $\jj$ be the quotient of $j_{cris*}(\jtp_{X_0/T}) $ by $ \jtp_{X/T}$. We have a short exact sequence \[0\rightarrow \jtp_{X/T}\rightarrow j_{cris*}(\jtp_{X_0/T})\xrightarrow{Q} \jj \rightarrow 0.\]Then (in the derived category)
\[R\ux(\jj)= \II \otimes \Ox^{p-1}[-p+1]. \]
\end{lemm}
\begin{proof}
By \cite[Theorem 2.1]{Trans}, \cite[Proposition 3.4.1]{Ber}, $(X_0)_{zar}=X_{zar}$ and $\II^2=0$, we have that (in the derived category)
\begin{equation} \label{lemma41com}
\xymatrix{ & Ru_{X/T*}(\jtp_{X/T})\ar[r] \ar[d]_{\cong} &Ru_{X/T*}\left(j_{cris*}(\jtp_{X_0/T})\right)\ar[r]\ar[d]_{\cong} & Ru_{X/T*}\jj \ar[r]^>>>{+1} \ar@{-->}[d]_{h}& \\
0 \ar[r]& \F^p_{X}\Omega_{X/T}^{\bullet} \ar[r] & \F^p_{X_0}\Omega_{X/T}^{\bullet}\ar[r]& \II\otimes \Omega^{p-1}_{X/T}[-p+1]\ar[r]&0}
\end{equation}
where \[\F^p_X\Omega _{X/T}^{\bullet}=\left(\ldots\rightarrow 0\rightarrow \Omega^p_{X/T}\rightarrow \Omega ^{p+1}_{X/T}\rightarrow \ldots \right)\]
and
\begin{equation} \label{filtration}
\F^p_{X_0}\Omega _{X/T}^{\bullet}= \sum \limits_{a+b=p} \II^{[a]}\F_X^b\Omega _{X/T}^{\bullet}=\left(\ldots\rightarrow 0\rightarrow \II\Omega_{X/T}^{p-1}\rightarrow \Omega^p_{X/T}\rightarrow \Omega_{X/T}^{p+1}\rightarrow \ldots\right),
\end{equation}
see \cite[Theorem 2.1]{Trans} and \cite[Theorem 7.2]{BO}. Since a derived category is a triangulated category, the induced isomorphism $h$ between the distinguished triangles is an isomorphism. We have proved the lemma.
\end{proof}
\begin{lemm}\label{lemm42}With the notations as in Lemma \ref{lemm41}, we have
\[\cris^k(X,\jtp_{X/T}) \cong \Ho^k(X,\Ox^{{\bullet}\geq p}) \cong \F_X^p\Ho^k_{\DR}(X/T)\]
\end{lemm}
\begin{proof}
The lemma follows from \cite[7.2.1]{BO} and our assumption \ref{assumption}.
\end{proof}
\subsection*{Description of $\rho(f_0)$}
In this subsection, we define a map \[\theta:Ru_{Y/T*}\jtp_{Y/T} \rightarrow Rf_{0*}(\II\otimes \Omega^{p-1}_{X/T}[-p+1])\]which induces the map $\rho(f_0)$.
First of all, we make a map $\rho$ as follows. Recall that we have a map $f_{0cris}^{-1} \jtp_{Y_0/T}\rightarrow \jtp_{X_0/T}$ (see \cite[Chapter III]{Ber}). The map $\rho$ is given as follows:
\[\xymatrix{R\uy\jtp_{Y/T}\ar[r] \ar[d]^{\rho}& R\uy i_{cris*}\jtp_{Y_0/T} \ar@{=}[d] \\
i_{zar*}Rf_{0*}Ru_{X_0/T*}\jtp_{X_0/T} & i_{zar*}Ru_{Y_0/T}^*\jtp_{Y_0/T}\ar[l]^{\psi}}\]
where the map $\psi$ is induced by
\[Ru_{Y_0/T*}\jtp_{Y_0/T}\rightarrow Ru_{Y_0/T*}Rf_{0cris*}(\jtp_{X_0/T})\cong Rf_{0zar*}Ru_{X_0/T*}(\jtp_{X_0/T}).\]
The quotient $Q$ in Lemma \ref{lemm41} gives rise to a map
\begin{equation}\label{hatrho}
\widehat{\rho}=i_*Rf_{0*}\left(Ru_{X/T*}(Q)\right): i_*Rf_{0*}[Ru_{X/T*}(j_*\jtp_{X_0/T})] \rightarrow i_*Rf_{0*}(Ru_{X/T*}(\jj)).
\end{equation}
We define $\theta:=\hat{\rho}\circ \rho$. In other words, the map\[\theta : Ru_{Y/T}\jtp_{Y/T} \rightarrow Rf_{0*}(\II \otimes \Omega^{p-1}_{X/T}[-p+1]) \]is defined as
\[ \xymatrix{ Ru_{Y/T*}\jtp_{Y/T} \ar[r]^<<<<{\rho} & i_*Rf_{0*}Ru_{X_0/T*}(\jtp_{X_0/T}) \ar@{=}[r] & i_*Rf_{0*}j_*Ru_{X_0/T*} (\jtp_{X_0/T})\ar@{=}[dll] \\
i_*Rf_{0*}[Ru_{X/T*}(j_*\jtp_{X_0/T})] \ar[r]_{\widehat{\rho}}& i_*Rf_{0*}(Ru_{X/T*}(\jj)) \ar@{=}[r] &Rf_{0*}(\II \otimes \Omega^{p-1}_{X/T}[-p+1]) } \]
where the identity in the first row follows from $(X_0)_{zar}=X_{zar}$ and the last identity in the second row follows from Lemma \ref{quo} and $(Y_0)_{zar}=Y_{zar}.$ Furthermore, the following diagram is commutative, cf. \cite[the proof of Proposition 3.20, page 184] {derham}:
\begin{equation}\label{thetadiag}
\xymatrix{\Ho^k(Y/T,\jtp_{Y/T}) \ar[d] \ar@{=}[r] &\F_{Y}^p \Ho^{k}_{\DR}(Y/T) \ar[rr]^>>>>>>>>>{\rho(f_0)} & & gr_{\F_{X_0}}^{k-p+1}\Ho^k_{\DR}(X_0/S)\otimes\II \ar@{=}[d] \\
\Ho^k(Y/T,i_*\jtp_{Y_0/T}) \ar[d] \ar[r] & \Ho^k (X/T,j_*\jtp_{X_0/T}) \ar[rr]^{G_0} \ar[d]_{G_1} & & \Ho^{k-p+1}(X,\Omega_{X/T}^{p-1}\otimes \II )\ar[d]\\
\Ho^k(Y/T,\OO_{Y/T})\ar[r]^{\Ho^k_{cris}(f_0)} & \Ho^k(X/T,\OO_{X/T})\ar[rr] & &\Ho^{k-p+1}(X,\Omega_{X/T}^{p-1}) }
\end{equation}
where the identity in the first row is due to Lemma \ref{lemm42}, the map $G_0$ is induced by the map $Q$ in Lemma ~\ref{lemm41} and the vertical map $G_1$ is induced by \[j_{cris*}(\jtp_{X_0/T}) \hookrightarrow j_{cris*}\OO_{X_0/T}\cong \OO_{X/T} .\]
The diagram (\ref{thetadiag}) makes it clear that $\rho(f_0)$ is induced by the local map $\theta$.
On the other hand, we have the following diagram:
\begin{equation}\label{dbstar}
\xymatrix{Ru_{Y/T*}\jtp_{Y/T} \ar@/^2pc/[rr]^{\theta}\ar[r]^<<<<<<{\rho} \ar[d]^{\proj} & Rf_{0*} Ru_{X_0/T*}(\jtp_{X_0/T}) \ar[r]^>>>>{\widehat{\rho}} \ar[d] & Rf_{0*}(\II\otimes \Omega^{p-1}_{X/T}[-p+1]) \ar[d]\\
Ru_{Y/T*}\left(\jtp_{Y/T}/\jthp_{Y/T}\right) \ar[r]^<<<{\Psi} & Rf_{0*}Ru_{X_0/T*}\left(\jtp_{X_0/T}/\jthp_{X_0/T}\right) \ar[r] & Rf_{0*}(\II/\II^{[2]}\otimes \Ox^{p-1}[-p+1] ).}
\end{equation}
The second row of the digram gives rise to a map
\begin{equation}\label{star}
Ru_{Y/T*}\left(\jtp_{Y/T}/\jthp_{Y/T}\right) \rightarrow Rf_{0*}(\II/\II^{[2]}\otimes \Ox^{p-1}[-p+1] ).
\end{equation}
Let us describe the map $\Psi$ more precisely. The map $\Psi$ is given by
\[\xymatrix{ Ru_{Y/T*}\jtp_{Y/T}/\jthp_{Y/T} \ar[dr]^{\Psi} \ar[r] & Ru_{Y/T*}i_{cris*}(\jtp_{Y_0/T}/\jthp_{Y_0/T})\ar@{=}[r] &i_{zar*}Ru_{Y_0/T*}(\jtp_{Y_0/T}/\jthp_{Y_0/T})\ar[d]^{\phi}\\
&Rf_{0*}Ru_{X_0/T}(\jtp_{X_0/T}/\jthp_{X_0/T}) \ar@{=}[r] & i_{*}Ru_{Y_0/T*}Rf_{0*}(\jtp_{X_0/T}/\jthp_{X_0/T}) }\]where the existence of the first arrow in the first row follows from the fact that $i_{cris*}$ is exact (\cite[Proposition 6.2]{BO}), the identities in the first and the second rows follow from \cite[Proposition 3.4.1]{Ber}, the arrow $\phi$ is induced by the morphism
\[\jj^{[k]}_{Y_0/T}\rightarrow f_{0cris *}\jj^{[k]}_{X_0/T}.\]
\begin{lemm}\label{dif}\label{lemm43}With the notations as above, we have
\[Ru_{Y/T*}\left(\jtp_{Y/T}/\jthp_{Y/T}\right)\cong \Oy^p[-p].\]
\end{lemm}
\begin{proof}
By the proof of the filtered Poincar\'e lemma \cite[6.13 and 7.2]{BO}, we have ( in the derived category)
\[\xymatrix{&Ru_{Y/T*}\jthp_{Y/T}\ar[r] \ar[d]_{\cong} &Ru_{Y/T*}\jtp_{Y/T} \ar[r]\ar[d]_{\cong} &Ru_{Y/T*}(\jtp_{Y/T}/\jthp_{Y/T})\ar[r]^>>>>{+1} \ar@{-->}[d]_{h}& \\
0\ar[r]& \F_Y^{p+1}\Omega_{Y/T}^{\bullet}\ar[r] &\F_Y^p\Omega^{\bullet}_{Y/T}\ar[r] &\Omega^p_{Y/T}[-p]\ar[r]&0}\]
where the map $h$ is an isomorphism. We have proved the lemma.
\end{proof}
By Lemma \ref{lemm43}, the bottom arrow (\ref{star}) of the diagram (\ref{dbstar}) induces a morphism
\begin{equation} \label{map1}
\widetilde{\rho(f_0)}: \Omega^p_{Y/T}[-p] \cong Ru_{Y/T*}\left(\jj^{[p]}_{Y/T}/\jj^{[p+1]}_{Y/T}\right) \rightarrow Rf_{0*}(\II/\II^{[2]} \otimes \Omega^{p-1}_{X/T}[-p+1]).
\end{equation}
The morphism (\ref{map1}) gives rise to an "adjoint" morphism (in the derived category) and we denote it by $\rho(f_0)$ as well:
\begin{equation} \label{rhof0}
\rho(f_0):f_0^* \Omega_{Y_0/S}^p[-p]\rightarrow \Omega^{p-1}_{X_0/S}[-p+1]\otimes \II/\II^{[2]}=\Omega^p_{X_0/S}[-p]\otimes \II.
\end{equation}
This "is" an element of
\[\Ho^0\mathbb{R}Hom(f_0^* \Omega_{Y_0/S}^p[-p], \Omega^{p-1}_{X_0/S}[-p+1]\otimes \II/\II^{[2]})=\mathrm{Ext}^1_{\OO_{X_0}}\left( f_0^* \Omega_{Y_0/S}^p, \Omega^{p-1}_{X_0/S}\otimes \II/\II^{[2]}\right).\]
\subsection{de Rham description of $\rho(f_0)$}
In this subsection, we describe the map $\rho(f_0)$ in (\ref{rhof0}) in terms of de Rham theory. Using the graph of $f_0$, we have an immersion \[X_0\xrightarrow{(j, i\circ f_0)} X\times_T Y .\] Let $D$ be the PD envelope of $X_0$ in $X\times_T Y$ and denote by $\pi_D: D \rightarrow X\times_T Y$.
Let $\ojj$ be the ideal of $X_0$ in $D$. It follows from \cite[Theorem 7.2]{BO} that
\[
Ru_{X_0/T*}\jj^{[k]}_{X_0/T}\cong \F_{X_0}^k\Omega^{\bullet}_{D/T} \cong \left(\ojj^{[k]}\rightarrow \ojj^{[k-1]}\Omega^1_{D/T}\rightarrow\ojj^{[k-2]}\Omega^2_{D/T}\rightarrow \ldots \right)
\]
where $\Omega_{D/T}^{\bullet}$ is the de Rham complex of $D/T$ induced by the natural connection
\begin{equation}\label{na}
\nabla : \OO_D\rightarrow \OO_D\otimes \Omega_{X\times_TY}^1=\Omega_{D/T}^1
\end{equation} of $\OO_D$, cf. \cite[ Exercise 6.4 and Theorem 7.1]{BO}. The $s$-th term of $\Omega_{D/T}^{\bullet}$ is given by $\OO_D\otimes \Omega_{X\times_T Y}^s $.
\begin{lemma} \label{lemmacom}With the same notations as above, we have that
\[\xymatrix{
gr_{\F_{X_0}}^p(\Omega ^{\bullet}_{D/T})= ( \ojtp / \ojthp \ar[r]^>>>d & \ojj^{[p-1]} / \ojtp \otimes \Omega^1_{X\times Y/T} \ar[r]^<<<d&\ldots \OO_{X_0}\otimes \Omega _{X\times Y/T}^p )}\]
\[\xymatrix{
gr_{\F_{X_0}}^p(\Omega ^{\bullet}_{X/T}) = ( \II^{[p]} / \II^{[p+1]} \ar[r]^>>>d & \II^{[p-1]} / \II^{[p]} \otimes \Omega^1_{X_0/S} \ar[r]^>>>d & \cdots\ar[r]^<<<d & \OO_{X_0}\otimes \Omega _{X_0/S}^p)
} \]
\[\xymatrix{ & =(0\ar[r] &0 \ar[r] &\ldots \ar[r] &0 \ar[r]& \II/\II^{[2]}\otimes \Omega^{p-1}_{X_0/S}\ar[r] & \Omega _{X_0/S}^p )}\]
where the first terms of both complexes are of degree zero and the differentials follow the rule \cite[Page 238 (1.3.6)]{Ber} $($\cite[ Exercise 6.4]{BO}$)$.
\end{lemma}
\begin{proof}
It follows from the proof of the filtered Poincar\'e Lemma \cite[(6.13) and (7.2)]{BO} or \cite[Chapter V, 2]{Ber}. See (\ref{filtration}).
\end{proof}
\begin{remark}\label{reduced}
Let $J$ be the ideal sheaf of $X_0$ in $X\times_T Y$.It follows from \cite[Remark 3.20 3)]{BO} that $J$ is maped into $\ojj$. Therefore, we have $J \ojj^{s}\subseteq \ojj^{s+1}$. It follows that all the terms of $gr_{\F_{X_0}}^p(\Omega ^{\bullet}_{D/T})$ in Lemma \ref{lemmacom} are $\OO_{X_0}$-module in a natural way, i.e., the $s$-th component is
$$\left(gr_{\F_{X_0}}^p(\Omega ^{\bullet}_{D/T})\right)_s=\ojj^{[p-s]} / \ojj^{[p-s+1]} \Omega^s_{X\times Y/T} =\ojj^{[p-s]} / \ojj^{[p-s+1]} \Omega^s_{X_0\times Y_0/S}|_{X_0}. $$
\end{remark}
Furthermore, we have a diagram as follows
\[ \xymatrix{ X_0\ar[r] \ar[d]_{f_0} &D \ar[dr]^{\pi_D} \ar[r]^{\pi_X} \ar[d]_{\pi_Y} &X \\
Y_0\ar[r]_j & Y & X\times_T Y \ar[l]^{\Pr_Y} \ar[u]_{\Pr_X} }\]
where $\Pr_X$ and $\Pr_Y$ are the natural projections.
It follows from the construction of the natural connection $\nabla$ (\ref{na}) (\cite[Corollary 6.3 and Exercise 6.4]{BO}) that there is a natural commutative diagram
\[\xymatrix{ \OO_{X\times_T Y}\ar[r]^{d} \ar[d]_{\pi_{D*}}&\Omega^1_{X\times_T Y/T} \ar[d]\\
\OO_{D} \ar[r]^<<<<<<<{\nabla} & \OO_D\otimes \Omega_{X\times_T Y/T}^1} \]
which induces a map $\pi_D^*: \Omega_{X\times_T Y/T}^{\bullet} \rightarrow\Omega_{D/T}^{\bullet} $. Therefore, we have a commutative diagram
\[\xymatrix{ \Pr_Y^*\Omega^{1}_{Y/T} \ar@/_1pc/[dr]_{\pi_Y^*} \ar[r]^{\Pr_Y^*} & \Omega_{X\times_T Y/T}^{1} \ar[d]_{\pi_D^*} & \Pr_X^*\Omega^{1}_{X/T} \ar[l]_{\Pr_X^*}\ar@/^1pc/[dl]^{\pi_X^*}\\
&\Omega_{D/T}^{1}
}.\]The map (\ref{fcris}) \[\cris^k(f_0/T):\cris^k(Y_0/T)\rightarrow \cris^k (X_0/T)\]can be considered as ``topological extension" of $\cris^k(f_0/S)$. The map $f_0$ does not necessarily have a ``real" extension $f:X \rightarrow Y$. However, we can use $\pi_Y$ to calculate $\cris^k(f_0/T)$ by \cite[Remark 7.5]{BO} and \cite[Chapter V, Lemma 2.3.3, Corollary 2.3.4]{Ber}. Namely, the map $\cris^k(f_0/T)$ is given by
\[\cris^k(Y_0/T)=\Ho^k(Y,\Omega_{Y/T}^{\bullet}) \xrightarrow{\pi_Y^*} \Ho^k(X\times_T Y,\Omega_{D/T}^{\bullet}) =\cris^k (X_0/T)\]
Moreover, the map $\pi_X^*$ induces a filtered quasi-isomorphism$$(\Omega^{\bullet}_{X/T},\F_{X_0}) \xrightarrow{\cong} (\Omega^{\bullet}_{D/T}, \F_{X_0}),$$ which is also denoted by $\pi_X^*$, see \cite[Theorem 7.2.2 and Remark 7.5]{BO} and \cite[Chapter V, Corollary 2.3.5]{Ber}. Recall from the last subsection that the map $\rho(f_0)$ is induced by the map $\theta$. Therefore, by Remark \ref{reduced} and the diagrams (\ref{thetadiag}), (\ref{dbstar}) and (\ref{lemma41com}), the morphism $\rho(f_0)$ (\ref{rhof0}) is given by (in the derived category):
\begin{equation} \label{rhof}
\xymatrix{ f_0^*\Omega_{Y_0/S}^p[-p]\ar[dr]_{\rho(f_0)} \ar@{=}[r] &f_0^*gr^p_{\F_{Y_0}}\Omega ^{\bullet}_{Y_0/S} \ar[r]^{\pi_Y^*} & gr^p_{\F_{X_0}}\Omega^{\bullet}_{D/T} &gr^p_{\F_{X_0}}\Omega^{\bullet}_{X/T} \ar[l]_{\pi^*_X}^{\cong} \ar@/^1pc/[dll]^{\proj}\\
&\Omega^{p-1}_{X_0/S}\otimes \II/\II^{[2]} [-p+1]}
\end{equation}
where $\proj$ is the natural map following from Lemma \ref{lemmacom}.
\begin{remark} \label{moreex}
To let the maps in (\ref{rhof}) be more explicit, we make the following remarks.
\begin{itemize}
\item The map $\proj$ is given by
\[gr_{\F_{X_0}}^p(\Omega ^{\bullet}_{X/T})=(\ldots \rightarrow \II/\II^{[2]}\otimes \Omega^{p-1}_{X_0/S} \rightarrow \ldots) \rightarrow \II/\II^{[2]}\otimes \Omega^{p-1}_{X_0/S},\]
cf. Lemma \ref{lemmacom}. By the diagram in the proof of Lemma \ref{lemm41}, the map $\proj$ is induced by $Ru_{X/T*}(Q) $ where $Q$ is the quotient in Lemma \ref{lemm41}.
\item By Lemma \ref{lemmacom}, it is clear that $\pi_Y^*:f_0^*\Omega_{Y_0/S}^p[-p]\rightarrow gr^p_{\F_{X_0}}\Omega^{\bullet}_{D/T}$ maps $u\in f^*_0\Omega^p_{Y_0/S}$ to the element $(0,0,0,u,0)$ in
\[\begin{aligned}
&\Omega^p_{X_0/S}\oplus \left(\Omega^{p-1}_{X_0/S}\otimes f_0^*\Omega^1_{Y_0/S}\right)\oplus \left( \Omega^1_{X_0/S}\otimes f^*_0\Omega^{p-1}_{Y_0/S}\right) \oplus f^*_0\Omega^p_{Y_0/S}\oplus (\ldots)\\
& =\Omega^p_{X_0\times Y_0/S}|_{X_0}=\left(gr^p_{\F_{X_0}}\Omega^{\bullet}_{D/T}\right) _p.
\end{aligned}\]
\end{itemize}
\end{remark}
\subsection{Description of cup product $ob(f_0)\cup-$}
Note that we have a natural injection \begin{equation}\label{eq:inclusion}
incl:\Omega^p_{Y_0/S}\hookrightarrow \Omega^1_{Y_0/S}\otimes \Omega^{p-1}_{Y_0/S}
\end{equation}
associating to $dx_1\wedge\ldots\wedge dx_p$ the element
\[\sum\limits_{i=1}^p (-1)^i dx_i\otimes dx_1\wedge\ldots\wedge \widehat{dx_i}\wedge\ldots\wedge dx_p.\]
Given an element in \[\Ext^1_{\OO_{X_0}}(f^*_0\Omega^1_{Y_0/S},\OO_{X_0}\otimes \II/\II^{[2]}),\]say the obstruction element $ob(f_0)$ of the map $f_0$ with respect to $S\hookrightarrow T$, the cup product of this element induces a morphism (in the derived category) from $f^*_0\Omega ^p_{Y_0/S}$ to $\Omega^{p-1}_{X_0/S}\otimes \II/\II^{[2]}[1]$ via the inclusion (\ref{eq:inclusion}). Denote the cup product (in the derived category) by
\begin{equation} \label{cupwithobs}
ob(f_0)\cup : f^*_0\Omega ^p_{Y_0/S}\rightarrow \Omega^{p-1}_{X_0/S}\otimes \II/\II^{[2]}[1]=\Omega^{p-1}_{X_0/S}\otimes \II[1].
\end{equation}
In the following, we describe the map $ob(f_0)\cup\text{ } $.
Let $\II_0$ be the ideal of $X_0$ in $X\times_T Y$, and let $\II_1$ be the ideal of $X_0$ in the $X_0\times_S Y_0$ (via the graph of $f_0$). There is an exact sequence of $\OO_{X_0}$-modules:
\[0\rightarrow \II\OO_{X_0/S}\rightarrow \II_0/\II_0^2\rightarrow \II_1/\II_1^2\rightarrow 0,\]see \cite[(2.7)]{Trans}.
This extension corresponds to the obstruction element \[ob(f_0)\in \Ext^1_{\OO_{X_0}}(\II_1/\II_1^2,\II \OO_{X}).\]We identify $\II_1/\II_1^2$ with $f_0^*\Omega^1_{Y_0/S}$, which gives the exact sequence \cite[Page 186]{derham}
\[0\rightarrow \OO_{X_0}\otimes \II/\II^{[2]} \rightarrow \ojj/\ojt \rightarrow f_0^*\Omega^1_{Y_0/S} \rightarrow 0.\]Let $A^{\bullet}$ be the two terms complex:
\begin{equation} \label{ttA}
\ojj/\ojt\rightarrow \II_1/\II_1^2\left(=f_0^*\Omega^1_{Y_0/S}\right)\end{equation}
where the first term is of degree zero.
In particular, there is a quasi-isomorphism as follows
\begin{equation}\label{qisiso}
w: \OO_{X_0}\otimes \II/\II^{[2]} \xrightarrow{\makebox[2cm]{qis}} A^{\bullet}.
\end{equation}It gives rise to an element \[\overline{ob(f_0)}\in \mathrm{Ext}^1(f_0^*\Omega^1_{Y_0/S},\II/\II^{[2]} \otimes \OO_{X_0})\] as follows.
\begin{equation}\label{obs1}
\xymatrix{\OO_{X_0}\otimes \II/\II^{[2]}[1]\ar[r]^>>>>{w}_>>>>{\cong} & A^{\bullet}\\
&f_0^*\Omega^1_{Y_0/S}\ar@{_{(}->}[u] \ar@{->}[ul]^{\overline{ob(f_0)}}}
\end{equation}
We also have a natural map \[gr^1_{\F_{X_0}}\Omega^{\bullet}_{D/T}\rightarrow A^{\bullet}\] between the complexes,
\[\xymatrix{ gr^1_{\F_{X_0}}\Omega^{\bullet}_{D/T}\ar[d] \ar@{=}[r] &(\ojj/\ojt \ar@{=}[d]\ar[r] & \Omega^1_{D/T}/\ojj \Omega^1_{D/T})\ar[d]^{\phi}\\
A^{\bullet} \ar@{=}[r]& (\ojj/\ojt\ar[r] &f_0^*\Omega^1_{Y_0/S}) }\]
where $\Omega^1_{D/T}$ is $\OO_D\otimes \Omega_{X\times Y/T}^1$ by definition (see \cite[Chapter 7]{BO}) and $\phi$ is the natural projection
\[ \Omega^1_{D/T}/\ojj \Omega^1_{D/T}\left(=\Omega^1_{X_0/S}\oplus f^*_0\Omega_{Y_0/S}^1\right) \rightarrow f_0^*\Omega^1_{Y_0/S} .\]
Recall that there is a natural quasi-isomorphism: $\xymatrix{gr^1_{\F_{X_0}}\Omega^{\bullet}_{D/T} \ar[r]^{\cong} &gr^1_{\F_{X_0}}\Omega^{\bullet}_{X/T}} $. It gives rise to a commutative diagram as follows.
\begin{equation} \label{obs2}
\xymatrix{f^*_0\Omega^1_{Y_0/S}[-1]\ar[r]\ar[dr]\ar@{->}@/_2pc/[ddr]_{\overline{ob(f_0)}} &gr^1_{\F_{X_0}}\Omega^{\bullet}_{D/T}\ar[d] & gr^1_{\F_{X_0}}\Omega^{\bullet}_{X/T} \ar@/^2pc/[ddl] \ar[l]^{\cong}\\
& A^{\bullet} \\
& \OO_{X_0}\otimes \II/\II^{[2]}\ar[u]^{\cong}}
\end{equation}
It gives rise to the morphism $ob(f_0)\cup$ (\ref{cupwithobs}) as the composition of the following maps:
\begin{equation} \label{cup}
\xymatrix{f_0^*\Omega^p_{Y_0/S}[-1]\ar[r] & gr^1_{\F_{X_0}}\Omega^{\bullet}_{D/T}\otimes f_0^*\Omega_{Y_0/S}^{p-1} \ar[r] & A^{\bullet}\otimes f_0^*\Omega_{Y_0/S}^{p-1} \\
\II/\II^{[2]} \otimes f_0^*\Omega^{p-1}_{Y_0/S}\ar[urr]^{qis}_{\cong}\ar[r]_{\Id\otimes f^*_0}& \II/\II^{[2]} \otimes \Omega_{X_0/S}^{p-1} }.
\end{equation}
where $f_0^*:f^*_0\Omega_{Y_0/S}^{p-1}\rightarrow \Omega^{p-1}_{X_0/S}$ is the natural differential induced by \[df_0: f^*_0\Omega_{Y_0/S}^{1}\rightarrow \Omega^{1}_{X_0/S}.\] The first arrow $f_0^*\Omega^p_{Y_0/S}[-1]\rightarrow gr^1_{\F_{X_0}}\Omega^{\bullet}_{D/T}\otimes f_0^*\Omega_{Y_0/S}^{p-1} $ is given by
\[\xymatrix{ f^*_0\Omega^p_{Y_0/S}\ar@{^{(}->}[r]^{f_0^*(incl)} \ar@{->}[dr] &f_0^*\Omega^1_{Y_0/S}\otimes f_0^*\Omega^{p-1}_{Y_0/S} \ar[d]^h \\
&\Omega^1_{X_0/S}\otimes f_0^*\Omega^{p-1}_{Y_0/S}\oplus f_0^*\Omega^1_{Y_0/S}\otimes f_0^*\Omega^{p-1}_{Y_0/S}}\]
where the term at the bottom is the degree-one term of the complex \[gr^1_{\F_{X_0}}\Omega^{\bullet}_{D/T}\otimes f_0^*\Omega_{Y_0/S}^{p-1}\]and \begin{equation}\label{maph}
h=(f_0^*\otimes \Id_{f_0^*\Omega^{p-1}_{Y_0/S}}, \Id_{f^*_0\Omega^1_{Y_0/S}\otimes f_0^* \Omega^{p-1}_{Y_0/S}}).
\end{equation}
The map (\ref{cup}) induces the cup product map
\[ob(f_0)\cup-:\Ho^{k-p}(Y_0,\Omega_{Y_0/S}^p)=\Ho^{k-p}(X_0,f_0^*\Omega_{Y_0/S}^p)\rightarrow \Ho^{k-p+1}(X_0,\Omega_{X_0/S}^{p-1})\otimes \II.\]
\section{$p$-adic Deformations of Automorphisms}\label{bloch}
In this section, we provide a general criterion lifting automorphisms of smooth projective varieties from positive characteristic to characteristic zero, see Theorem \ref{corolifting}. The key point to show this criterion is Proposition \ref{generalremark}. In the following, we will show Proposition \ref{generalremark} in a cautious way. We follow the notations and the assumptions \ref{assumption} in Section \ref{cryobs}. For $p=1$, Proposition \ref{generalremark} is \cite[Proposition 3.20]{derham}. In the following, we assume $p\geq 2$.
\\
\noindent \textbf{Comparison of the maps $\rho(f_0)$ and $ob(f_0)\cup$.}\\
To show Proposition \ref{generalremark}, it suffices to prove
\begin{equation}\label{toproveidentity}
\rho(f_0)=-[ob(f_0)\cup-]
\end{equation}for $p\geq 2$. Recall that $D$ is the PD envelope of $X_0$ in $X\times Y$ and $A^{\bullet}$ is the two term complex (\ref{ttA}). Note the concrete descriptions of $\rho(f_0)$ (\ref{rhof}) and $ob(f_0)\cup$ (\ref{cupwithobs}). To prove the identity \ref{toproveidentity}, it suffices to show the existence of the following diagrams:
\[\xymatrix{ f^*_0\Omega^p_{Y_0/S} [-p]\ar@{}[dr]|-{III} \ar[r]\ar[d] & gr^p_{\F_{X_0}}\Omega ^{\bullet}_{D/T}\ar@{}[dr]|-{II} \ar@{-->}[d]^{F} &gr^p_{\F_{X_0}}\Omega^{\bullet}_{X/T}\ar[l]_{\pi_X}^{\cong} \ar@{-->}[d]^{H}\\
gr^1_{\F_{X_0}}\Omega^{\bullet}_{D/T}\otimes f^*_0\Omega^{p-1}_{Y_0/S}[-p+1]\ar[r] &A^{\bullet}\otimes f^*_0\Omega^1_{Y_0/S}[-p+1] &\II/\II^{[2]} \otimes f^*_0\Omega^{p-1}_{Y_0/S}[-p+1] \ar[l]^{qis}_{Q} }\]
and
\[\xymatrix{gr^p_{\F_{X_0}}\Omega^{\bullet}_{X/T} \ar[rr]^{-\proj}\ar@{-->}[d]^{H}\ar@{}[drr]|-{I} & & \II/\II^{[2]}\otimes \Omega^{p-1}_{X_0/S}[-p+1]\ar@{=}[d]\\
\II/\II^{[2]} \otimes f^*_0\Omega^{p-1}_{Y_0/S}[-p+1] \ar[rr]_{\Id \otimes f^*_0 } & &\II/\II^{[2]}\otimes \Omega^{p-1}_{X_0/S}[-p+1]
}\]
where the map $\proj$ is the natural map (cf. Lemma \ref{lemmacom}) \[gr^p_{\F_{X_0}}\Omega^{\bullet}_{X/T}=(\ldots\rightarrow \II/\II^{[2]}\otimes \Omega^{p-1}_{X_0/S}\rightarrow \ldots ) \rightarrow \II/\II^{[2]}\otimes \Omega^{p-1}_{X_0/S}[-p+1].\]
In the following, we construct the maps $F$ and $H$ so that the diagrams commute.
The map $H$ is given by
\[\xymatrix{gr^p_{\F_{X_0}}\Omega^{\bullet}_{X/T} \ar@{=}[r]& (\II/\II^{[2]}\otimes \Omega_{X_0/S}^{p-1}\ar[r]^d \ar@/^2pc/[rr]^{\Id_{\II/\II^{[2]}}\otimes [-(f_0^*)^{-1}]} &\OO_{X_0}\otimes \Omega^p_{X_0/S}) & \II/\II^{[2]}\otimes f^*_0\Omega^{p-1}_{Y_0/S}[-p+1] },\]
see Lemma \ref{lemmacom}.
A direct diagram chasing can verify the commutativity of diagram $I$. We will only show the commutativity of diagrams $II$ and $III$.
First of all, we define the map $F:gr^p_{\F_{X_0}}\Omega^{\bullet}_{D/T} \rightarrow A^{\bullet}\otimes f^*_0\Omega^1_{Y_0/S}[-p+1]$.
Define the map $F_{p-1}$ as follows $(p\geq 2)$:
\[\xymatrix{\ojj/\ojj^{[2]}\otimes \Omega^{p-1}_{X_0\times Y_0/S}|_{X_0} \ar[d]_{F_{p-1}} \ar@{=}[r] & \ojj/\ojj^{[2]}\otimes(\Omega^{p-1}_{X_0/S}\oplus \ldots \oplus f^*_0\Omega ^{p-1}_{Y_0/S})\ar[dl]^{h_1}\\
\ojj/\ojj^{[2]}\otimes f_0^*\Omega^{p-1}_{Y_0/S} }\]
where $h_1=\Id_{\ojj/\ojj^{[2]}}\otimes (-(f_0^*)^{-1},0,\Id_{f_0^*\Omega ^{p-1}_{Y_0/S}})$.
Define the map $F_p$ as follows:
\[\xymatrix{ \Omega^p_{X_0\times Y_0/S}|_{X_0} \ar@{=}[r] \ar[d]_{F_p} & \Omega^p_{X_0/S}\oplus \Omega^{p-1}_{X_0/S}\otimes f_0^*\Omega^1_{Y_0/S}\oplus \Omega^1_{X_0/S}\otimes f^*_0\Omega^{p-1}_{Y_0/S} \oplus f^*_0\Omega^p_{Y_0/S}\oplus (\ldots) \ar[dl]^{h_2}\\
f_0^*\Omega^1_{Y_0/S} \otimes f_0^*\Omega^{p-1}_{Y_0/S}}\]where $h_2=(0,-(f_0^*)^{-1}\otimes \Id_{f^*_0\Omega ^1_{Y_0/S}},0,f^*_{0}(incl),0)$.
The map $F:$ (cf. Lemma \ref{lemmacom})
\[(\ldots\rightarrow \ojj/\ojj^{[2]}\otimes \Omega^{p-1}_{X_0\times Y_0/S}|_{X_0} \rightarrow \Omega^p_{X_0\times Y_0/S}|_{X_0}) \rightarrow (\ojj/\ojj^{[2]}\otimes f_0^*\Omega^{p-1}_{Y_0/S}\rightarrow f_0^*\Omega^1_{Y_0/S} \otimes f_0^*\Omega^{p-1}_{Y_0/S} )\]
is given by the diagram (see Lemma \ref{lem1})
\begin{equation}\label{diagramcom}
\xymatrix{ \Omega^p_{X_0\times Y_0/S}|_{X_0} \ar[rr]^{F_p} & & f_0^*\Omega^1_{Y_0/S} \otimes f_0^*\Omega^{p-1}_{Y_0/S} \\
\ojj/\ojj^{[2]}\otimes \Omega^{p-1}_{X_0\times Y_0/S}|_{X_0} \ar[rr]^{F_{p-1}}\ar[u]^d & &\ojj/\ojj^{[2]}\otimes f_0^*\Omega^{p-1}_{Y_0/S} \ar[u]^{d} .}
\end{equation}
\begin{lemma}\label{lem1}
The map $F$ defined above is a morphism between complexes, i.e. the diagram (\ref{diagramcom}) is commutative.
\end{lemma}
\begin{proof}
We show the commutativity of digram (\ref{diagramcom}). In fact, let $B$ be an element $u\otimes (a,\ldots,b)$ of $$ \ojj/\ojj^{[2]}\otimes(\Omega^{p-1}_{X_0/S}\oplus \ldots \oplus f^*_0\Omega ^{p-1}_{Y_0/S})= \ojj/\ojj^{[2]}\otimes \Omega^{p-1}_{X_0\times Y_0/S}|_{X_0}.$$ Then we have \[d(B)=u(da+db)+(du)\cdot (a,\ldots,b)\]
\[=0+(du)\cdot (a,\ldots,b)=(du)\cdot (a,\ldots,b) \in \Omega^p_{X_0\times Y_0/S}|_{X_0}\]where $"\cdot"$ is the wedge product, $\Omega^1_{X_0\times Y_0/S}|_{X_0}=\Omega^1_{X_0/S}\oplus f_0^*\Omega^1_{Y_0/S}$ with projections $\Pr_1$, $\Pr_2$ and \[d(u)=(\Pro_1(d(u)),\Pro_2(d(u))).\]Therefore, we have
\begin{equation}\label{eq1}
\begin{aligned}
F_p(d(B))&=F_p((du)\cdot (a,\ldots,b))\\
&=\Pro_2(d(u))\cdot\left(-(f_0^*)^{-1}(a)\right)+\Pro_2(d(u))\cdot b\\
&=\Pro_2(du)\cdot(b-(f^*_0)^{-1}(a)).
\end{aligned}
\end{equation}
On the other hand, we have
\begin{equation}\label{eq2}
\begin{aligned}
d\left(F_{p-1}(B)\right) &=d\left( \Id_{\ojj/\ojj^{[2]}}\otimes (-(f_0^*)^{-1},0,\Id_{f_0^*\Omega ^{p-1}_{Y_0/S}})(u\otimes (a,\ldots,b) )\right)\\
& =d\left(u\otimes (b-(f_0^*)^{-1}(a))\right)\\
&=\Pro_2(du)\cdot (b-(f_0^*)^{-1}(a))
\end{aligned}
\end{equation}
where the last equality follows from the definition of the complex $A^{\bullet}$, see (\ref{ttA}), namely, we have $d_{A^{\bullet}}=\Pro_2\circ d$
\[\xymatrix{\ojj/\ojj^{[2]}\ar[d]^d \ar[r]^{d_{A^{\bullet}}}& f_0^*\Omega^1_{Y_0/S}\\
\Omega^1_{D/T}/\ojj \Omega^1_{D/T} \ar@{=}[r] &\Omega^1_{X_0/S}\oplus f_0^*\Omega^1_{Y_0/S} \ar[u]^{\Pro_2}
.}\]By the equalities (\ref{eq1}) and (\ref{eq2}), we have proved the lemma.
\end{proof}
Let $(\Id,0)$ be the natural inclusion
\[\Omega_{X_0/S}^{p-1}\hookrightarrow \Omega^{p-1}_{X_0\times Y_0/S}|_{X_0}=\Omega^{p-1}_{X_0/S}\oplus \ldots\] and similar for \[\Omega_{X_0/S}^{p}\hookrightarrow \Omega^{p}_{X_0\times Y_0/S}|_{X_0}=\Omega^{p}_{X_0/S}\oplus \ldots\]Denote by "$in$" the natural map
\[in: \II\OO_{D} \rightarrow \ojj.\]
\begin{lemma} \label{lem2}
The diagram $II$ commutes.
\end{lemma}
\begin{proof}
To prove the commutativity of the diagram $II$, it suffices to prove that the following diagrams (\ref{diag1}) and (\ref{diag2}) commute.
Recall that $w: \OO_{X_0}\otimes \II/\II^{[2]}\rightarrow A^{\bullet}$ is a quasi-isomorphism, see (\ref{qisiso})).
It is induced by the map $"in"$. We claim there is a commutative diagram as follows:
\begin{equation}\label{diag1}
\xymatrix{ \II/\II^{[2]}\otimes \Omega^{p-1}_{X_0/S}\ar[rrr]^{H=\Id_{\II/\II^{[2]}}\otimes [-(f_0^*)^{-1}]}\ar[d]^{\overline{in}\otimes (\Id,0)} & & &\II/\II^{[2]}\otimes f^*_0\Omega^{p-1}_{Y_0/S} \ar[d]^{w\otimes \Id_{\Omega^{p-1}_{Y_0/S}}}\\
\ojj/\ojj^{[2]}\otimes \Omega^{p-1}_{X_0\times Y_0/S}|_{X_0} \ar@{=}[r] & \ojj/\ojj^{[2]}\otimes(\Omega^{p-1}_{X_0/S}\oplus\ldots)\ar[rr]_{F_{p-1}}& &\ojj/\ojj^{[2]}\otimes f^*_0\Omega^{p-1}_{Y_0/S} .}
\end{equation}
In fact, we have
\[
\begin{aligned}
w\otimes \Id_{\Omega^{p-1}_{Y_0/S}}(H(u\otimes a))&=w\otimes \Id_{\Omega^{p-1}_{Y_0/S}}(u\otimes [-(f_0^*)^{-1}(a)])\\
&=\overline{in(u)}\otimes [-(f_0^*)^{-1}(a)]
\end{aligned}
\]for $u\otimes a\in \II/\II^{[2]}\otimes \Omega^{p-1}_{X_0/S}$. On the other hand, we have
\[\begin{aligned}
F_{p-1}(\overline{in}\otimes (\Id,0)(u\otimes a))&=F_{p-1}(\overline{in(u)}\otimes (a,0))\\
&=\overline{in(u)}\otimes [-(f_0^*)^{-1}(a)].
\end{aligned}\]
We have proved that the diagram (\ref{diag1}) commutes. To show the lemma, it remains to verify the commutativity of the following diagram
\begin{equation}\label{diag2}
\xymatrix{\Omega^{p}_{X_0/S} \ar[rrr] \ar[d]^{(\Id,0)} & & &0 \ar[d] \\
\Omega^p_{X_0\times Y_0/S}|_{X_0} \ar@{=}[r] &\Omega^p_{X_0/S}\oplus (\ldots) \ar[rr]^{F_{p}}& & f_0^*\Omega^1_{Y_0/S}\otimes f^*_0\Omega^{p-1}_{Y_0/S}
}.
\end{equation}
In fact, it is clear that \[F_p((\Id,0)(a))=F_p((a,0,\ldots,0))=0\] for $a\in \Omega^{p}_{X_0/S}$. We have proved that the diagram commutes.
In summary, we show the diagram $II$ commutes.
\end{proof}
\begin{lemma}\label{lem3}
The diagram $III$ commutes.
\end{lemma}
\begin{proof}
It suffices to show the following diagram commutes.
\[\xymatrix{f^*_0\Omega^p_{Y_0/S}\ar@{^{(}->}[d]^{f^*_0(incl)}\ar[rr]^{\pi_Y^*} &&\Omega^p_{X\times Y}|_{X_0} \ar[dd]^{F_p}\\
f_0^*\Omega^1_{Y_0/S}\otimes f_0^*\Omega^{p-1}_{Y_0/S}\ar[d]^h & \\
\Omega^1_{X_0/S}\otimes f_0^*\Omega^{p-1}_{Y_0/S}\oplus f_0^*\Omega^{1}_{Y_0/S}\otimes f_0^*\Omega^{p-1}_{Y_0/S} \ar[rr]^{Pr_2} &&f_0^*\Omega^{1}_{Y_0/S}\otimes f_0^*\Omega^{p-1}_{Y_0/S}
}\]
where $h$ is the map (\ref{maph}) and the term $f_0^*\Omega^{1}_{Y_0/S}\otimes f_0^*\Omega^{p-1}_{Y_0/S}$ at the right corner is the p-th term of the complex $A^{\bullet}\otimes f_0^*\Omega^{p-1}_{Y_0/S}[-p+1]$.
In fact, we have \[\pi_Y^*(u)=(0,0,0,u,0)\in
\Omega^p_{X_0/S}\oplus \Omega^{p-1}_{X_0/S}\otimes f_0^*\Omega^1_{Y_0/S}\oplus \Omega^1_{X_0/S}\otimes f^*_0\Omega^{p-1}_{Y_0/S} \oplus f^*_0\Omega^p_{Y_0/S}\oplus (\ldots)\] for $u\in f^*_0\Omega^p_{Y_0/S}$, see Remark \ref{moreex}. Therefore, we conclude that
\[F_p(\pi_Y^*(u))=F_p((0,0,0,u,0))=f_0^*(incl(u)).\] On the other hand, we have
\[\begin{aligned}
\Pro_2(h(f_0^*(incl(u))) &= \Pro_2\left( (f_0^*\otimes \Id_{f_0^*\Omega^{p-1}_{Y_0/S}}, \Id_{f^*_0\Omega^1_{Y_0/S}\otimes f_0^* \Omega^{p-1}_{Y_0/S}}) (incl(u))\right)\\
&=f_0^*(incl(u))
\end{aligned}\]
Comparing the identities above, we have proved the commutativity of the diagram.
\end{proof}
Now, we can finish the proof of Proposition \ref{generalremark}.
\begin{proof}
Proposition \ref{generalremark} follows from Lemma \ref{lem1}, Lemma \ref{lem2} and Lemma \ref{lem3}.
\end{proof}
We are able to prove Theorem \ref{corolifting}.
\begin{proof}
Let $h:X
\rightarrow W(k)$ be the structure map of $X$ over the Witt ring $W(k)$. Suppose that $\pi$ is the
uniformizer of $W(k)$. We have smooth morphisms
\[
h_n:X_n\rightarrow W_n
\]
where $h_n=h|_{W_n}$ is the restriction of $h$ to $W_n=W(k)/(\pi^{n+1})$. Note that $(\pi^{n+1})$ is square-zero ideal of $W_{n+1}$. Therefore, the $W_{n+1}$-module $(\pi^{n+1})$ is a $W_n$-module as well. For each $h_n$, we
have the natural cup product $\Psi_n$ as follows
\[\xymatrix{
\Ri^1h_{n*}(T_{X_n/W_n})\otimes_{W_n} (\pi^{n+1}) \ar[d]\\
\bigoplus\limits_{p+q=m}\mathrm{Hom}(\Ri^qh_{n*}(\Omega^p_{X_n/W_n}),\Ri^{q+1}h_{n*}(\Omega^{p-1}_{X_n/W_n})\otimes (\pi^{n+1})).}
\]
It follows from our assumptions that the Hodge-de Rham spectral sequence of $X_n/W_n$ degenerates at
$E_1$ and their terms are locally free so that the Hodge and de Rham cohomology sheaves
commute with base change. Therefore, the cup product $\Psi_n$ is $\Psi_0\otimes_k(\pi^{n+1})$. Since $\Psi_0$ is the cup product (\ref{cupproduct}) which is injective, the cup product $\Psi_n$ is injective.
Let $g_n:X_n\rightarrow X_n$ be a lifting of $g_0$ over $W_n$. Note that $g_n$ is an automorphism of $X_n$ over $W_n$. The map $\Psi_n$ induces an injection $\widehat{\Psi}_n$
\[
\xymatrix{\Ri^1h_{n*}g^*_nT_{X_n/W_n}\otimes (\pi^{n+1}) = \Ri^1h_{0*}(g_0^*T_{X_0/k})\otimes_k (\pi^{n+1})\ar@/^15pc/[dd]^{\widehat{\Psi}_n} \ar@{^{(}->}[d]\\
\bigoplus\limits_{p+q=m} \mathrm{Hom}(\Ri^q h_{n*}(g_n^*\Omega^p_{X_n/W_n}),\Ri^{q+1}h_{n*}(g_n^*\Omega^{p-1}_{X_n/W_n})\otimes (\pi^{n+1}))\ar@{=}[d]\\
\bigoplus\limits_{p+q=m}\mathrm{Hom}(\Ri^q h_{0*}(\Omega^p_{X_0/k}),\Ri^{q+1}h_{0*}(\Omega^{p-1}_{X_0/k})\otimes_k (\pi^{n+1})).}
\]
On the other hand, the map $g_n:X_n/W_{n+1}\rightarrow X_n/W_{n+1}$ induces a map $\cris^{p+q}(g_n)$ as follows
\[
\xymatrix@C+24pt{\cris^{p+q}(X_n/W_{n+1})\ar@{=}[d]\ar[r] &\cris^{p+q}(X_n/W_{n+1})\ar@{=}[d]\\
\Ho^{p+q}_{\DR}(X_{n+1}/W_{n+1})\ar[r]^-{\Ho^{p+q}_{cris}(g_n)}&
\Ho^{p+q}_{\DR}(X_{n+1}/W_{n+1}).}
\]
The map $\cris^{p+q}(g_n)\otimes {W_n}$ can be identified with
$\Ho_{\DR}^{p+q}(g_n)$ and hence it preserves the Hodge filtrations. Therefore, as (\ref{defrohf}), the map $\cris^{p+q}(g_n)\otimes {W_n}$
induces a diagram
\[
\xymatrix{\F_{Hdg}^p \Ho_{\DR}^{p+q}(X_{n+1}/W_{n+1})\ar[r] \ar[dr]& gr_{\F}^{p-1} \Ho_{\DR}^{p+q}(X_{n+1}/W_{n+1})\bigotimes_{W_{n+1}} (\pi^{n+1}) \ar@{=}[d]\\
& \Ho^{q+1}(X_n,\Omega^{p-1}_{X_n/W_n})\bigotimes_{W_n} (\pi^{n+1}) }
\]
by the fact that $(\pi^{n+1})$ is a square zero ideal in $W_{n+1}$. In particular, we have that
\[
\xymatrix{\F_{Hdg}^p \Ho_{\DR}^{p+q}(X_{n}/W_{n})\ar[r] \ar@{->>}[d]_{\proj}& \Ho^{q+1}(X_n,\Omega^{p-1}_{X_n/W_n})\bigotimes_{W_n} (\pi^{n+1})\\
\Ho^{p}(X_n,\Omega^q_{X_n/W_n}) \ar[ru]_{\rho(g_n)_q},}
\]
see (\ref{defrohf}) for the definition of $\rho(g_n)_q$.
We apply Proposition $\ref{generalremark}$ to the case $f_0=g_n$, $S=\Spec W_n$ and $T=\Spec(W_{n+1})$. It follows that
\[
\bigoplus\limits_{p+q=m}\rho(g_n)_q=\pm \widehat{\Psi_n}(ob(g_n))
\]
where $ob(g_n)$ is the obstruction element in
$$\Ho^1(X_n,g^*_n T_{X_n/W_n}\bigotimes (\pi^{n+1}))=\Ho^1(X_0,g^*_0 T_{X_0/k})\bigotimes_k (\pi^{n+1}).$$ Since $\cris^m(g_0)$
preserves the Hodge filtration of $\cris^m(X_0/W)$, we conclude that $\rho(g_n)_q$ are zero by the construction of $\rho(g_n)_q$, cf. (\ref{defrohf}) .
It follows from the
injectivity of $\widehat{\Psi_n}$ that $ob(g_n)$ is zero.
Hence, we have a formal automorphism $\lim\limits_{\leftarrow} g_n$ on the
formal scheme $\lim\limits_{\leftarrow} X_n$. By the Grothendieck's existence
theorem, the formal automorphism comes from an automorphism
$g:X/W\rightarrow X/W$. In other words, we can lift
$g_0$ over $k$ to $g$ over $W(k)$.
\end{proof}
\begin{cor} \label{padicfaithful}
With the notations and assumptions as in Theorem \ref{corolifting}, we suppose that $g_0$ is an automorphism of $X_0$ over $k$ such that the order of $g_0$ is finite. If $\et^m(f_0,{\mathbb{Q}_l})=\Id$ where $l\neq char(k)$, then one can lift $g_0$ to an automorphism over $W(k)$\[g:X/W(k)\rightarrow X/W(k).\] In particular, if the automorphism group $\Aut(X_0)$ is finite and $\Aut(X_{K})$ acts on $\et^m(X_{K},\mathbb{Q}_l)$ faithfully where $K$ is the fraction field of $W(k)$, then $\Aut(X_0)$ acts on $\et^m(X_{0},\mathbb{Q}_l)$ faithfully.
\end{cor}
\begin{proof}
Note that \[\det(\Id-g_0^*t,\cris^m(X_0/W)_{K})=\det(\Id-g_0^*t,\et^m(X_0,\Ql)),\]
see \cite[Theorem 2]{KM} and \cite[3.7.3 and 3.10]{Ill1}. The finiteness of the order $ord(g_0)$ implies that $\et^m(g_0,{\mathbb{Q}_l})=\Id$ if and only if $\cris^m(g_0)_{K}=\Id$ since both $\et^m(g_0,{\mathbb{Q}_l})$ and $\cris^m(g_0)_{K}$ can be diagonalizable. The corollary follows from Theorem \ref{corolifting}.
\end{proof}
\bibliographystyle{acm}
| 207,818
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The Lowlands from Yeti is more than a blanket: It’s a ground cover, a cozy accessory, and a luxury item that’s made to last for years. The special fabric repels dirt, burrs, and pet hair. The waterproof outer layer protects the padded and insulated interior, so you stay dry and comfortable.
Planning to work remotely and need a way to charge a phone or even a laptop? The Yeti 200X Portable Power Station weighs only five pounds and is the size of a lunchbox, but it can charge a laptop up to four times (or a smartphone up to 20 times) on one single charge of its own. It’s equipped with a high-speed power delivery port capable of powering up USB-C laptops, tablets, mirrorless cameras, and phones in a fraction of the time of other portable chargers.
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| 157,267
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Megastar Inflatable Rainbow & Cloudy Bouncer Water Pool Combo Kids Slide 8.40 x 5.80 x 2.55 mtr
Dhs. 3,899
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- The tall mesh walls around the This Megastar Inflatable Rainbow and cloudy Bouncer Water Pool keeps your children safe while allowing them free to bounce, and you can watch their movements at any time.
- The bouncer stakes in each corner ensure the stability and firmness of the entire bouncer castle.
- Equipped with a.
- An inflatable slide is connected to the jumping area. Your child will be keen to have fun in the jumping area and jump over and over again, then slide down Splishing into the Water pool.
- Built-in basketball rim provides extra entertainment.
- Big entrance for convenient access. Build a happy childhood for your child.
Product Features :
- Activities : 3 slides , 1 jumping area , ball/ Water pool , Climber
- Age - 3+ years.
- Usage : wet & Dry Air Blower: 750W
- Material : Jumping part is PVC tarpaulin,else is 420D fabric,materials comply with ASTM and EN71 toy safety standards
- Sewing
- Double-stitched reinforced seams and jump surface
- Commercial grade sewing style
- Inflate and deflate
- Easy to set up and take down within 2-3minutes
- Netting and mesh
- High quality netting or mesh with small holes for safety
- Blower
- Powerful blower with CE certificate and correct plug
- Opening
- Velcro opening for safe and easy to go or out
- Air tube
- Long fill tube for placing blower away from the bouncer
- Product size: 840 x 580 x 255 cm
- Package Size: 49 x 48 x 72 cm
- Product weight: 33.2 kg
- shipping weight : 45 kg
| 356,225
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TITLE: Proving existence of basis such that the matrix of $f$ is given by $\begin{pmatrix} A&0\\0&0 \end{pmatrix}$
QUESTION [1 upvotes]: Consider a sesquilinear form $f$ over a finite dimensional vectorspace
$V$. Show that the is a basis of $V$ such that the matrix of $f$ is
given by $$A_{f} = \begin{pmatrix} A_g &0\\0&0 \end{pmatrix}$$ with
$g=f_{|W}$ and $W\le V$ such that $V = W \oplus \operatorname{rad}(f)$. Moreover, $\operatorname{rad}(g)$ is trivial.
My attempt:
This seems like a rather trivial theorem. If $V = W \oplus \operatorname{rad}(f)$, then $f = f_{|W} \oplus_{\perp} 0_{|\operatorname{rad}(f)}$. So it's clear that we only have to consider the matrix of $f_{|W}=: g$. This is $A_g$, and the matrix of the zero map is of course the zero matrix. This explains the given matrix representation of $f$.
How do I prove this rigorously?
REPLY [1 votes]: So You take an orthonormal basis of $rad(f)$ and complete it to an orthonormal basis of $V$, the basis vectors not in $rad(f)$ forming an orthonormal basis of $W$ and rearrange the vectors. Then the matrix representation of $f$ with respect to this basis clearly has the desired form. The only thing that deserves a proof is that $rad(g)=\{0\}$. For this assume $w_1\in rad(g)$ i.e.
$$g(w_1,w)=0\text{ for all }w\in W$$
$$\implies f(w_1,w)=0\text{ for all }w\in W$$
$$\implies f(w_1,v)=0\text{ for all }v\in V$$
since any $v\in V$ can be written in the form $v=w+u$ where $w\in W,u\in rad (f)$ and thus
$$w_1\in W\cap rad(f)=\{0\}\implies w_1=0.$$
| 213,554
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Description
Prussia – Sword – Officer – Lion Head – Garde Regiment
This is an interesting variation of a Prussian Lion Head officer sword. It has two small red eyes which serve as the eyes of the Lion. The head itself is highly detailed. The grip of the sword is the traditional shark skin, which shows some minor wear. The grip is single wire wrapped.
Below the tang we see a fine Garde Star. This makes this sword attributed to one of the elite Prussian Garde-Regiments. The sword is housed in a well made silver toned scabbard with one hanger. There is a small rust spot near the tip of the scabbard and there are a couple of dents as well. The blade is quite simple and has a single fuller and is in excellent condition. On the edge is hallmarked for the well known firm of quality swords M Neumann Hoflieferant Berlin. This firm was a military effects firm that specialized in selling uniforms, caps, swords, etc. I don’t believe that they actually produced swords but rather offered them under license.
An overall lovely example of an M. Neumann sword with a Garde-Regiment attribution.
| 234,468
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TITLE: Probability of earning money on game slot.
QUESTION [1 upvotes]: Let's say you are playing a game on the game slot where probability of getting
a number $k$ is $p_k=\frac{2^{k-1}}{3^k}$, if number you got happens to have a reminder of one when dividing by three you earn 10 dollars, in case you got a number that is divisible by three, you earn nothing, but you don't lose anything either (we could say that you earn zero dollars in that case), if it happens that you got the number that has reminder two when dividing by three you lose ten dollars. Find the probability that amount of money player is going to earn is 50 to 100 dollars after 1000 attepmts.
This is my attempt:
probability of earning 10 dollars in one attempt is:
$P(X_j=10)=\sum_{k=0}^{\infty} \frac{2^{3k}}{3^{3k+1}}=\frac{9}{19}$
probability of "earning" zero dollars in one attempt is:
$P(X_j)=\sum_{k=0}^{\infty} \frac{2^{3k-1}}{3^{3k}}=\frac{9}{38}$
probability of losing 10 dollars in one attempt is:
$P(X_j)=\sum_{k=0}^{\infty} \frac{2^{3k+1}}{3^{3k+2}}=\frac{11}{38}$
Now i am supposed to analyze what happens when i have thousand attempts, money earned then is sum of all money player earned in every single attempt so we have $$Y_{1000}=\sum_{k=1}^{1000}X_k$$
Now i could simply solve this by implementing central limit theorem , however, i need to find $E(Y_{1000})$ and $Var(Y_{1000})$
As far as i know, expectation should be
$E(Y_{1000})=\sum_{k=1}^{1000}X_kp_k$
but i don't know how to determine this, so i cannot progress any further, any help appreciated!
REPLY [1 votes]: Your probabilities for the individual outcomes are wrong. The probability for $X_j=10$ is right. The probability for $X_j=-10$ should be $\frac23$ of that; here you wrote the right sum but got the wrong result. The probability for $X_j=0$ should again be $\frac23$ of that; here you wrote the wrong sum, as $k$ should start at $1$. (The $p_k$ are only normalized if we let $k$ start at $1$.)
Thus
\begin{eqnarray*}
\mathsf P(X_j=10)&=&\frac9{19}\;,\\
\mathsf P(X_j=0)&=&\frac4{19}\;,\\
\mathsf P(X_j=-10)&=&\frac6{19}\;.\\
\end{eqnarray*}
The expectation of a sum of random variables is the sum of the individual expectations. (Here no independence is required.) The variance of a sum of independent random variables is the variance of the individual variances. (Here independence is required.)
The expected gain from one round is
$$
\mathsf E[X_j]=\frac9{19}\cdot10+\frac6{19}\cdot(-10)=\frac{30}{19}\;.
$$
The expected squared gain from one round is
$$
\mathsf E[X_j^2]=\frac9{19}\cdot10^2+\frac6{19}\cdot(-10)^2=\frac{1500}{19}\;.
$$
Thus the variance of the gain from one round is
$$
\mathsf E[X_j^2]-\mathsf E[X_j]^2=\frac{1500}{19}-\left(\frac{30}{19}\right)^2=\frac{27600}{361}\;.
$$
The expectation and variance for $1000$ rounds are therefore $\frac{30000}{19}$ and $\frac{27600000}{361}\approx\left(\frac{5254}{19}\right)^2$, respectively. The range $[50,100]$ is almost six standard deviations away from the mean, so the desired probability is very low.
| 113,387
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FINA Artistic Swimming Technical Committee holds innovation meeting in Paris
Artistic swimming is a traditional sport with a long-lasting history but the discipline continuously looks at improving and modernise.
As an important Olympic sport in the Summer Olympics programme since Los Angeles 1984, artistic swimming needs to stay attractive and evolve with its time to maintain the sports fans’ interest and to attract new generations of athletes.
In parallel to the first stage of the FINA Artistic Swimming World Series 2019 in Paris, France, from March 1-3, some of the world’s biggest experts in the discipline are holding major discussions.
The latest key development the Technical Committee, led by Chairwoman Lisa Schott, is looking at is building a designed computer software that would analyse movements and standardise the way difficulties are being judged. The execution and synchronised impression will both remain subject to the judges’ expertise.
The aim of this innovation is to improve and transform the current judging system which is based on human assessment at the moment. It can sometimes be quite controversial and provoke conflict of interest within the artistic swimming community.
The new computer software would be able to analyse the imposed elements performed in a routine and attribute marks independently from a human being action.
Swimmers, female and male, are judged in three areas:
1. Execution — how well the skill was carried out.
2. Synchronisation — how in-time with one another the swimmers are.
3. Difficulty — how difficult the attempted skills were.
With this new system, providing graphics and a comprehensive and pre-determined fair calculation system, scores would be awarded more partially and the level of excellence demonstrated would be considered more objectively.
| 297,607
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| 271,781
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U3A to host annual open day at Beaudesert RSL Services Club next week
BEAUDESERT University of the Third Age is hosting its annual open day on January 24 with a free morning tea.
U3A said it would have tutors and facilitators on hand to demonstrate many of the more than 20 classes and activities available this year at at the RSL club function room in William Street from 10 am to 12.30 pm.
Groups range from the mysteries of certain card games (500 and Bolivia) to Nordic walking, croquet and mosaics.
Members will be available to talk to all comers about their own particular activities, including the newly formed Travel and Adventure Group, which meets monthly to share ideas and photographs and has already organized a week-long trip to Norfolk Island and several day outings.
U3A president Andy Fechner said people sometimes wondered about the meaning behind "University of the Third Age."
Mr Fechner said the Third Age meant the retirement years, when people could enjoy the stimulus of mental and physical activities in a social environment, after the earlier ages of their lives as children and working adults.
"Basically, a university is a community of students and that's what we all are, although not faced with exams and awards, and not necessarily in academic fields," he said.
"Our tutors are volunteers with experience and expertise.
"We offer retirees the chance to re-discover an old interest, or to learn something new."
Visitors are welcome to simply turn up at the open day, or get more information at 0408 801 780.
Details of courses are available at beaudesertu3a.com
| 370,148
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\begin{document}
\title[Gromov--Witten theory of stacks]{Gromov--Witten theory\\of Deligne--Mumford stacks}
\author[D. Abramovich]{Dan Abramovich}
\thanks{Research of D.A. partially supported by NSF grant DMS-0335501}
\address{Department of Mathematics, Box 1917, Brown University,
Providence, RI, 02912, U.S.A}
\email{abrmovic@math.brown.edu}
\author[T. Graber]{Tom Graber}
\thanks{Research of T.G. partially supported by NSF grant DMS-0301179
and an Afred P. Sloan research fellowship.}
\address{Department of Mathematics\\
California Institute of Technology\\
Pasadena, CA \\
U.S.A.} \email{graber@caltech.edu}
\author[A. Vistoli]{Angelo Vistoli}
\thanks{Research of A.V. partially supported by the PRIN Project ``Geometria
sulle variet\`a algebriche'', financed by MIUR}
\address{Scuola Normale Superiore\\ Piazza dei Cavalieri 7\\
56126 Pisa\\ Italy}
\email{angelo.vistoli@sns.it}
\maketitle
\setcounter{tocdepth}{1}
\tableofcontents
\section{Introduction}
Gromov\ddash Witten theory of orbifolds was introduced in the symplectic
setting in \cite{Chen-Ruan}. In \cite{AGV} we adapted the theory to
algebraic geometry, using Chow rings and the language of stacks. The
latter work is, to a large extent, a research announcement, as
detailed proofs were not given. The main purpose of this paper is
to complete that work and lay the algebro-geometric theory on a
sounder footing.
It appears from the emerging literature that the language of stacks
-- whether algebraic or differential -- is imperative in making
serious computations in orbifold Gromov\ddash Witten theory (see, e.g.
\cite{Cadman2}, \cite{Tseng}). It can therefore be hoped that some of
the material here should be useful in the symplectic setting as
well.
The fact that a few years have passed since the release of our paper
\cite{AGV} may be one cause for a regretful slow development of
applications. On the other hand, we believe recent developments
have enabled us to set the foundations in a much better way than was
possible at the time of the paper \cite{AGV}. Most important among
these are Olsson's papers \cite{Olsson-Hom} and
\cite{Olsson-twisted}.
\subsection{Twisted stable maps}
Algebro-geometric treatments of Gromov\ddash Witten theory of a smooth
projective variety $X$ rely on Kontsevich's moduli stacks
$\ocM_{g,n}(X, \beta)$, parametrizing $n$-pointed stable maps from
curves of genus $g$ to $X$ with image class $\beta\in \H_2(X, \ZZ)$,
see \cite{BM}. When one replaces the manifold $X$ with an
orbifold, one needs to replace the curves in the stable maps by
orbifold curves - this is a phenomenon discussed in detail elsewhere
(see \cite{AV-families}). The stack of stable maps is thus replaced
by the stack of \emph{twisted stable maps}, denoted $\cK_{g,n}(\cX,
\beta)$ in \cite{AV}, where it was constructed.
The proof of the main theorem in \cite{AV} is not ideal
as it relies on ad-hoc arguments and requires verifying Artin's
axioms one by one. An alternative approach which is much more
conceptual is given in Olsson's papers \cite{Olsson-Hom} and
\cite{Olsson-twisted}. Artin's axioms still need to be verified, but
in a more general and cleaner situation. In Olsson's paper
\cite{Olsson-twisted} one also finds a direct construction of the
stack of twisted pre-stable curves, which is an important tool in
the theory developed here. See Section \ref{Sec:twisted-curves} (and
Appendix \ref{roots}) for a quick review.
An analogous space of \emph{orbifold stable maps} was constructed by
W. Chen and Y. Ruan in \cite{Chen-Ruan} using very different
methods. In spirit the two constructions describe the same thing, and one expects that the resulting Gromov--Witten numbers are identical.
For quotient stacks by a finite group $\cX= [V/G]$, Jarvis, Kaufmann
and Kimura considered in \cite{JKK} maps of pointed admissible
$G$-covers to $V$. This was revisited in \cite{AGOT}. Lev
Borisov recently discovered in his mail archives two letters from
Kontsevich, dated from July 1996, where Gromov--Witten theory of a
quotient stack by a finite group is outlined precisely using
admissible $G$-covers. See \cite{cimenotes} for a reproduced text.
When the target stack $\cX$ is smooth, the stack $\cK_{g,n}(\cX,
\beta)$ admits a perfect obstruction theory in the sense of
\cite{BF} and so one gets a virtual fundamental class
$[\cK_{g,n}(\cX, \beta)]^\vir$, which then gives rise to a
Gromov-Witten theory.
\subsection{Gromov--Witten classes - manifold case}
We follow a formalism for Gromov--Witten theory suited for Chow
rings developed in \cite{GP} (a paper longer
overdue than this one). A similar formalism was given in \cite{EK}, Section~3. Consider classes $\gamma_1,\ldots,\gamma_n
\in A^*(X)$. Define the associated Gromov--Witten classes to be
$$\langle \gamma_1,\ldots,\gamma_n ,*\rangle^X_{g,\beta} \ \ = \\ e_{n+1\ *}(e_1^*\gamma_1 \cup\ldots \cup e_n^*\gamma_n\cap[\ocM_{g,n+1}(X,\beta)]^\vir)\in A^*(X).$$ An important part of Gromov--Witten theory is concerned with relations these classes satisfy. In particular, in genus 0, they satisfy the Witten--Dijkgraaf--Verlinde--Verlinde (WDVV) equation.
These classes are convenient for defining the quantum product on
either the Chow ring or cohomology ring. The associativity of that
product is a consequence of the WDVV equation.
A fundamental ingredient of Gromov--Witten theory is a description
of the ``boundary" of the moduli stack of maps. The locus of nodes
$\Sigma$ on the universal curve of the stack of stable maps
$\ocM_{0,n}(X,\beta)$ is a partial normalization of this boundary,
and has the beautiful description
$$\Sigma\ \ =\ \ \coprod_{\substack{A\sqcup B = \{1,\ldots, n\}\\\beta_1+\beta_2\ \ =\ \ \beta}} \ocM_{0,A \sqcup \bigstar}(X,\beta_1)\ \mathop\times\limits_X \ \ocM_{0,B \sqcup \bullet}(X,\beta_2). $$ Here the fibered product is taken with respect to the \emph{evaluation maps} $e_{\bigstar}: \ocM_{0,A \sqcup\bigstar}(X,\beta_1)\to X$ and $e_{\bullet}: \ocM_{0,B \sqcup\bullet}(X,\beta)\to X$.
At the bottom of this is the fact that if $C = C_1 \sqcup_p C_2$ is
a nodal curve with node $p$ separating it into two subcurves $C_1$
and $C_2$, then $C$ is a coproduct of $C_1$ and $C_2$ over $p$. It
follows from the universal property of a coproduct that
$$\Hom(C, X) \ \ =\ \ \Hom(C_1,X) \mathop\times\limits_{\Hom(p,X)} \Hom(C_2,X).$$ Since $ \Hom(p,X) = X$ we get the familiar formula
$$\Hom(C, X) \ \ = \ \ \Hom(C_1,X) \ \mathop\times\limits_{X}\ \Hom(C_2,X).$$ The decomposition of the boundary is immediate from this.
\subsection{Boundary of moduli - orbifold case}
When analyzing the orbifold case something new happens. The source
curve is a twisted curve $\cC = \cC_1 \sqcup_\cG \cC_2$ with node
$\cG$ which is a gerbe banded by $\bmu_r$ for some $r$. We show in Proposition \ref{Prop:gluing} that
the coproduct $\cC_1\sqcup^\cG \cC_2$ of $\cC_1$ and $\cC_2$ over $\cG$ exists, and it is immediate that $\cC_1\sqcup^\cG \cC_2\to \cC$ is an isomorphism. It follows
again that
$$\Hom(\cC, \cX) \ \ =\ \ \Hom(\cC_1,\cX)
\mathop\times\limits_{\Hom(\cG,\cX)} \Hom(\cC_2,\cX),$$ see Appendix
\ref{Sec:gluing}. Now the data of
a gerbe with a map to $\cX$ is not a point of $\cX$, but of a
fascinating gadget $\riX$ we call \emph{the rigidified cyclotomic
inertia stack}, see Section \ref{Sec:riX}. (A different notation $\bar {\mathcal{X}}_1$
was used in \cite{AGV}, but we were convinced that the
present notation, as used by
Cadman\cite{Cadman1,Cadman2}, is more
appropriate). We get the formula
$$\Hom(\cC, \cX) \ \ = \ \ \Hom(\cC_1,\cX) \mathop\times\limits_{\riX}
\Hom(\cC_2,\cX).$$ The fibered product uses a natural morphism
$\Hom(\cC_2, \cX) \to \riX$ corresponding to $\cG \to \cX$, and a {\em
twisted} map $\Hom (\cC_1,\cX) \to \riX$ corresponding to $\cG \to
\cX$ \emph{with the band inverted}. This is necessary since the glued
curve $\cC$ is \emph{balanced}.
The resulting map of moduli stacks
$$ \coprod_{\substack{ A\sqcup B = \{1,\ldots, n\}\\ \beta_1+\beta_2 =
\beta}} \cK_{0,A
\sqcup\checkbullet}(\cX,\beta_1)\mathop\times\limits_{\riX} \cK_{0,B
\sqcup \bullet}(\cX,\beta_2) \ \ \lrar \ \ \cK_{0,A \sqcup
B}(\cX,\beta)$$ is crucial for Gromov--Witten theory of stacks. Here
the fibered product is taken using an evaluation map $$e_\bullet:
\cK_{0,B \sqcup \bullet}(\cX,\beta_2) \ \ \lrar \ \ \riX$$ and a
\emph{twisted evaluation map}
$$\check e_\checkbullet: \cK_{0,A \sqcup \checkbullet}(\cX,\beta_1) \
\ \lrar \ \ \riX$$ necessary to make the glued curves balanced,
see Section \ref{Sec:evaluation}.
\subsection{Gromov--Witten classes - orbifold case}
We can now define the orbifold Gromov--Witten classes by integrating
along evaluation maps. We use the formalism of Chow rings, though
any cohomology theory satisfying some reasonable assumptions works.
Since evaluation maps land naturally in $\riX$, \emph{the correct
cohomological theory to use is not that of $\cX$ but rather of
$\riX$}. The fact that something like $A^*(\riX)_\QQ$ or
$\H^*(\riX,\QQ)$ is of interest was recognized by physicists (see
e.g. \cite{DHVW}, \cite{Zaslow}) where
``twisted
sectors" and ``orbifold Euler characteristics" were considered. Some
mathematical reasons were discussed in \cite{AGV}. Also, in
Kontsevich's remarkable messages to Borisov the same phenomenon
occurs. But all these are reasoned by analogy, ``delicious
reciprocal reflections, furtive caresses, inexplicable quarrels"
\cite{Weil}. Alas, in orbifold Gromov--Witten theory all this becomes
mundane once one understands that it
all follows from the fact that
$\cC$ is a coproduct.
Now, given $\gamma_1,\ldots,\gamma_n \in A^*(\riX)_\QQ$ we can
consider the class $$\check e_{n+1\ *}(e_1^*\gamma_1 \cup\ldots \cup
e_n^*\gamma_n\cap [\cK_{g,n+1}(\cX,\beta)]^\vir)\in A^*(\riX)_\QQ,$$
where $e_i :\cK_{g,n+1}(\cX , \beta) \to \riX$ are the evaluation
maps of Section \ref{Sec:evaluation}. It turns
out that in the orbifold theory we need to multiply
this by the function $r:\riX \to \ZZ$ describing the index of the
gerbe (but see the second part of Proposition
\ref{Prop:lifted-evaluation} for a way out
of that annoyance). We thus define the Gromov--Witten
classes to be
$$\langle \gamma_1,\ldots,\gamma_n ,*\rangle_{g,\beta} \ \ = \ \ r
\cdot \check e_{n+1\ *}(e_1^*\gamma_1 \cup\ldots \cup
e_n^*\gamma_n\cap [\cK_{g,n+1}(\cX,\beta)]^\vir).$$ With this
definition Gromov--Witten theory goes through almost as in the
manifold case, though the proofs require some interesting changes. The
main result is Theorem \ref{Thm:WDVV}, in which the WDVV equation is
proven.
\subsection{Acknowledgements}The authors are grateful to L.
Borisov, B. Fantechi, T. Kimura, R. Kaufmann, M. Kontsevich, M. Olsson, and H. Tseng for
helpful conversations. The two referees checked the paper very thoroughly, which is much appreciated by the authors.
\section{Chow rings, cohomology and homology of stacks}\label{classical}
\subsection{Intersection theory on Deligne--Mumford stacks}
Throughout the paper we work over a fixed base field $k$ of
characteristic $0$. Let $\cX$ be a separated Deligne--Mumford stack
of finite type over $k$, $\pi: \cX \to X$ its moduli space.
The rational Chow group $A_{*}(\cX)_{\QQ}$ is defined as in
\cite{Mumford}, \cite{Gillet} and \cite{Vistoli}. One defines the
group of cycles on $\cX$ as the free abelian group on closed
integral substacks of $\cX$, then divides by rational equivalence. There is also an integral version of the theory, developed in \cite{EG} for quotient stacks and in \cite{Kresch} in general, but we will not need it.
These groups are covariant for proper morphisms of Deligne--Mumford
stacks. The pushforward $\pi_{*}: A_{*}(\cX)_{\QQ} \to
A_{*}(X)_{\QQ}$ is given by the following formula. Let $\cV$ be a
closed integral substack of $\cX$, and call $V$ its moduli space;
this is an integral scheme of finite type over $k$. Because of the
hypothesis on the characteristic, the natural morphism $V \to X$ is
a closed embedding, and we have
\[
\pi_{*}[\cV] = \frac{1}{r}[V],
\]
where $r$ is the order of the stabilizer of a generic geometric point of $\cV$.
The homomorphism $\pi_{*}$ is an isomorphism. In what follows we
will always identify $A_{*}(\cX)_{\QQ}$ and $A_{*}(X)_{\QQ}$ via
$\pi_{*}$.
If $\cX$ is proper, we denote by
\[
\int_{\cX}: A_{*}(\cX)_{\QQ} \to \QQ
\]
the pushforward $A_{*}(\cX)_{\QQ} \to A_{*}(\Spec k)_{\QQ} = \QQ$
If $f: \cX \to \cY$ is an {l.c.i.}~morphism of constant codimension
$k$, then for any morphism $\cY' \to \cY$ we have a \emph{Gysin
homomorphism} $f^{!}:A_{*}(\cY')_{\QQ} \to A_{*}(\cX\times_{\cY}\cY')_{\QQ}$, of degree $-k$. These commute with proper pushfowards, and among
themselves (\cite{Vistoli}).
As in \cite{Fulton}, Chapter~17, with every such stack $\cX$ we can associate a
bivariant ring $A^{*}(\cX)_{\QQ}$, that gives a contravariant
functor from the 2-category of algebraic stacks of
finite type over $k$ to the category of commutative rings. The
product of two classes $\alpha$ and $\beta$ in $A^{*}(\cX)_{\QQ}$
will be denoted by $\alpha\beta$, or by $\alpha\cup \beta$. By
definition, $A_{*}(\cX)_{\QQ}$ is a module over $A^{*}(\cX)_{\QQ}$;
we indicate the result of the action of a bivariant class $\alpha
\in A^{i}(\cX)_{\QQ}$ on a class of cycles $\xi\in A_{j}(\cX)_{\QQ}$
as a cap product $\alpha\cap \xi\in A_{j-i}(X)_{\QQ}$. The pullback
ring homomorphism $A^{*}(X)_{\QQ} \to A^{*}(\cX)_{\QQ}$ is also an
isomorphism. The projection formula holds: if $f : \cX \to \cY$ is a
proper homomorphism, $\beta \in A^{*}(\cY)_{\QQ}$ and $\xi \in
A_{*}(\cX)_{\QQ}$, then
\[
f_{*}(f^{*}\beta \cap \xi) = \beta\cap f_{*}\xi.
\]
If $\cE$ is a vector bundle on $\cX$, then there are Chern classes
$c_{i}(\cE) \in A^{i}(\cX)$, satisfying the usual formal properties.
Suppose that $\cX$ is smooth: then the homomorphism
$A^{*}(\cX)_{\QQ} \to A_{*}(\cX)_{\QQ}$ defined by $\alpha \mapsto
\alpha \cap [\cX]$ is an isomorphism. In this case we can identify
$A^{*}(\cX)_{\QQ}$ with $A_{*}(\cX)_{\QQ}$.
The moduli space $X$ will not be smooth. However, assuming for the
moment that $\cX$ is connected, hence irreducible, and we call $r$
the order of the automorphism group of a generic geometric point of
$\cX$, then by the projection formula we have
\[
\pi_{*}(\pi^{*}\alpha \cap [\cX]) = \alpha\cap \pi_{*}[\cX] =
\frac{1}{r}\alpha \cap [X]
\]
for any $\alpha\in A^{*}(X)_{\QQ}$; hence the homomorphism
$A^{*}(X)_{\QQ} \to A_{*}(X)_{\QQ}$ defined by $\alpha \mapsto
\alpha \cap [X]$ is also an isomorphism. The same holds without
assuming that $\cX$ is connected, because if $\cX_{i}$ are the
connected components of $\cX$ and $X_{i}$ is the moduli space of
$\cX_{i}$, then $X = \coprod_{i} X_{i}$. Hence $A_{*}(X)_{\QQ}$
inherits a ring structure from that of $A^{*}(X)_{\QQ}$, even though
$X$ is in general singular.
However, one should be careful: the isomorphism $\pi_{*}:
A_{*}(\cX)_{\QQ} \to A_{*}(X)_{\QQ}$ is not a homomorphism of rings,
unless $\cX$ is generically a scheme, because in general the
identity $[\cX]$ of $A_{*}(\cX)$ is not carried into the identity
$[X]$ of $A_{*}(X)$.
Suppose that $\cX$ is smooth and proper: then $X$ is a complete
variety with quotient singularities. We say that an element of the
group $A_{1}(X)_{\QQ}$ of 1-dimensional cycles is \emph{numerically
equivalent to $0$} if $\int_{X}\alpha\cup \xi = 0$ for all $\alpha
\in A^{1}(X)_{\QQ}$. The elements of $A_{1}(X)$ whose images in
$A_{1}(X)_{\QQ}$ are numerically equivalent to $0$ form a subgroup;
we denote by $N(X)$ the quotient group. This is finitely generated.
Furthermore we denote by $N^{+}(X)$ the submonoid of
$N(X)$ consisting of effective cycles.
Let $\cE$ be a vector bundle on $\cX$. If $\xi\in A_{1}(X)$ is an
integral 1-dimensional class, we denote
\[
c_{1}(\cE)\cdot \xi := \int_{\cX}c_1(\cE)\cap\xi',
\]
where $\xi'$ is the class in $A_{1}(\cX)_{\QQ}$ such that
$\pi_{*}\xi'$ equals the image of $\xi$ in $A_{1}(X)_{\QQ}$.
Notice the following fact. If $\alpha$ is the class in
$A^{1}(X)_{\QQ}$ such that $\pi^{*}\alpha = c_{1}(\cE)$, then
\begin{align*}
c_{1}(\cE)\cdot \xi &= \int_{\cX}c_1(\cE)\cap\xi'\\
&= \int_{X}\pi_{*}(c_1(\cE)\cap\xi')\\
& = \int_{X}\alpha\cap\xi;
\end{align*}
hence $c_{1}(\cE)\cdot \xi$ only depends on the class of $\xi$ in
$N(X)$. This allows us define the rational number $c_{1}(\cE)\cdot
\beta$ for a class $\beta\in N(X)$. It is easy to see that the denominators
in $c_{1}(\cE)\cdot \beta$ are uniformly bounded, but the following
proposition gives a natural bound:
\begin{proposition}\label{Prop:bound-denominators}
Assume that $\cX$ is proper. For each geometric point $p: \Spec
\overline{k} \to \cX$ denote by $e_{p}$ the exponent of the
automorphism group of $p$, and call $e$ the least common multiple of
the $e_{p}$ for all geometric points of $\cX$. Then
\[
c_{1}(\cE)\cdot \beta \in \frac{1}{e}\ZZ
\]
for any vector bundle $\cE$ on $\cX$ and any $\beta\in N(X)$.
\end{proposition}
\begin{proof}
First note that $c_{1}(\cE)\cdot \beta \in \ZZ$ when $\cE = \pi^*\cM$ for a bundle $\cM$ on $X$. Indeed, in this case
\begin{align*}c_{1}(\cE)\cdot \beta &= \int_\cX \pi^*c_{1}(\cM)\cap \beta' \\
&= \int_X c_{1}(\cM)\cap \beta\in \ZZ.
\end{align*}
We may substitute $\cE$ with its determinant, and assume that $\cE$
is a line bundle.
The following is a standard fact; we include a proof below for lack of a suitable reference.
\begin{lemma}
Let $\cL$ be a line bundle on $\cX$. The line bundle $\cL^{\otimes
e}$ on $\cX$ is the pullback of a line bundle $\cM$ on $X$.
\end{lemma}
To conclude the proof of the proposition,
\begin{align*}
c_{1}(\cE)\cdot \beta &= \frac{1}{e}c_{1}(\cL^{\otimes e})\cdot \beta\\
&= \frac{1}{e}\pi^* c_{1}(\cM)\cdot \beta \in \frac{1}{e}\ZZ
\end{align*}
as required. \qed
\begin{proof}[Proof of the lemma.] Observe that this is equivalent to the statement
that $\pi_{*}\cL^{\otimes e}$ is a line bundle on $X$, and the
adjunction homomorphism $\pi^{*}\pi_{*}\cL^{\otimes e} \to
\cL^{\otimes e}$ is an isomorphism. This is a local statement in
the \'etale topology.
Let $p:\Spec \overline{k} \to \cX$ be a geometric point, $G_{p}$ its
automorphism group. By definition, the exponent of
$G_{p}$ divides $e$. The action of $G_{p}$ on the fiber $\cL_{p}$ of
$\cL$ at $p$ is given by a 1-dimensional character $\chi: G_{p} \to
\overline{k}^{*}$, and therefore $\chi^{e}$ is the trivial character.
This implies that the action of $G_{p}$ on the fiber of
$\cL^{\otimes e}$ is trivial.
There is an \'etale neighborhood $\Spec\overline{k} \to U \to X$ of the image of $p$ in $X$, such that the pullback $U \times_{X}\cX$ is isomorphic to the quotient $[V/G_{p}]$, where $V$ is a scheme on which $G_{p}$ acts with a fixed geometric point $q: \Spec \overline{k} \to V$, with an invariant morphism $V \to U$ mapping $q$ to $p$; then the pullback $\cL_{[V/G_{p}]}$ corresponds to a $G_{p}$-equivariant locally free sheaf $\cL_{V}$ on $V$. We may assume that $U$ is affine. Then $V$ is also affine; since the characteristic of the base field is $0$, we can take an invariant non-zero element the in fiber of $\cL^{\otimes e}_{V}$, and extend it to an invariant section of $\cL^{\otimes e}_{V}$. By restricting $V$ we may also assume that this section does not vanish anywhere: and then $\cL^{\otimes e}_{V}$ is trivial as a $G_{p}$-equivariant line bundle. This implies that the restriction $\cL^{\otimes e}_{[V/G_{p}]}$ is trivial, and so $\cL^{\otimes e}_{[V/G_{p}]}$ is the pullback of the trivial line bundle on $U$. This concludes the proof of the Lemma and of the Proposition.
\end{proof}\noqed
\end{proof}
\subsection{Homology and cohomology of smooth Deligne--Mum\-ford stacks}
In this section the base field will be the field $\CC$ of complex
numbers. If $\cX$ is an algebraic stack of finite type over $\CC$,
then we can define the classical homology and cohomology of $\cX$ as
the homology and cohomology of the simplicial scheme associated with
a smooth presentation of $\cX$. More precisely, if $X_{1} \double X_{0}$ is a smooth presentation of $\cX$, we can obtain from it a simplicial scheme $X_{\bullet}$ in the usual fashion. To this we associate a simplicial space by taking the classical topology on each $X_{i}$. The homotopy type of the realization of this simplicial space is by definition the homotopy type of $\cX$, and its homology and cohomology are the homology and cohomology of $\cX$.
Alternatively, one can define the
classical site of $\cX$, and define cohomology as the
sheaf-theoretic cohomology of a constant sheaf on this site.
Homology can be defined by duality, as usual.
In this paper we are only going to need the cohomology and homology
with \emph{rational} coefficients of a \emph{proper} Deligne--Mumford
stack (mostly in the smooth case)
$\cX$. In this case a much more elementary approach is available.
Let $\cX$ be a separated Deligne--Mumford stack of finite type over
$\CC$, and let $\pi: \cX \to X$ be its moduli space. We define the homology
and cohomology $\H_{*}(\cX,\QQ)$ and $\H^{*}(\cX,\QQ)$ as
$\H_{*}(X,\QQ)$ and $\H^{*}(X,\QQ)$, respectively.
If $f: \cX \to \cY$ is a morphism of separated Deligne--Mumford
stacks of finite type over $\CC$ with moduli spaces $\pi: \cX \to X$
and $\rho: \cY \to Y$, this induces a morphism $g: X \to Y$ of
algebraic varieties over $\CC$. We define the pushforward $f_{*}:
\H_{*}(\cX,\QQ) \to \H_{*}(\cY,\QQ)$ and the pullback $f^{*}:
\H^{*}(\cX,\QQ) \to \H^{*}(\cY,\QQ)$ as
\[
g_{*}: \H_{*}(X,\QQ) \to \H_{*}(Y,\QQ)
\quad\text{and}\quad
g^{*}: \H^{*}(X,\QQ) \to \H^{*}(Y,\QQ)
\]
respectively.
In particular, the pushfoward
\[
\pi_{*}: \H_{*}(\cX,\QQ) \to \H_{*}(X,\QQ),
\]
is the identity function $\H_{*}(\cX,\QQ) \to \H_{*}(X,\QQ)$, and the pullback
\[
\pi^{*}: \H^{*}(X,\QQ) \to \H^{*}(\cX,\QQ)
\]
is the identity $\H^{*}(X,\QQ) \to \H^{*}(\cX,\QQ)$.
With these definitions, $\H^{*}$ becomes a covariant functor from the
2-category of separated stacks of finite type over $\CC$ to the
category of graded abelian $\QQ$-vector spaces. Similarly, $\H^{*}$
becomes a contravariant functor from the same 2-category to the
category of graded-commutative $\QQ$-algebras.
Also, the cap product $\cap: \H^{*}(X, \QQ) \otimes \H_{*}(X,\QQ) \to
\H_{*}(X,\QQ)$ can be interpreted as a cap product $\cap: \H^{*}(\cX,
\QQ) \otimes \H_{*}(\cX,\QQ) \to \H_{*}(\cX,\QQ)$. If $f: \cX \to
\cY$ is a morphism of stacks, the projection formula
\[
f_{*}(f^{*}\alpha \cap \xi) = \alpha\cap f_{*}\xi
\]
for any $\alpha\in \H^{*}(\cY,\QQ)$ and any $\xi\in \H_{*}(\cY, \QQ)$, holds.
Now assume that $\cX$ is proper. We define the cycle homomorphism
\[
\cyc_{\cX}: A_{*}(\cX)_{\QQ} \to \H_{*}(\cX,\QQ)
\]
as the composition
\[
A_{*}(\cX)_{\QQ} \overset{\pi_{*}}{\longrightarrow}
A_{*}(X)_{\QQ} \overset{\cyc_{X}}{\longrightarrow}
\H_{*}(X,\QQ).
\]
It is easy to see that the cycle homomorphism gives a natural
transformation of functors from the 2-category of proper
Deligne--Mumford stacks over $\CC$ to the category of graded
$\QQ$-vector spaces. If $\cV$ is a closed substack of $\cX$, this
has a fundamental class $[\cV]$ in $A_{*}(\cX)_{\QQ}$; we
denote its image in $\H_{*}(\cX,\QQ)$ also by $[\cV]$, and call it
the \emph{homology fundamental class} of $\cV$.
Now assume that $\cX$ is smooth and proper. Then $X$ is a variety
with quotient singularities, hence it is a rational homology
manifold; thus we have Poincar\'e duality, that is, the homomorphism
\[
\PD_{X} := -\cap[X] : \H^{*}(X,\QQ) \longrightarrow \H_{*}(X, \QQ)
\]
is an isomorphism. From this we get Poincar\'e duality on $\cX$;
however, one should be a little careful here, because the
fundamental class $[\cX]$, that we want to use to define Poincar\'e
duality on $\cX$, does not coincide with the fundamental class
$[X]$. In fact, if $\cX_{i}$ are the connected components of $\cX$,
then their moduli spaces $X_{i}$ are the connected components of $X$.
We have that $\pi_{*}[\cX_{i}] = \frac{1}{r_{i}}[X_{i}]$, if $r_{i}$
is the order of the automorphism of a generic geometric point of
$\cX_{i}$; hence we get the formula
\[
[\cX] = \sum_{i}\frac{1}{r_{i}}[X_{i}]
\]
in $A_{*}(\cX)_{\QQ}$, and hence also in $H_{*}(\cX,\QQ)$.
We define the Poincar\'e duality homomorphism on $\cX$ as
\[
\PD_{\cX} := -\cap[\cX] : \H^{*}(\cX,\QQ) \longrightarrow \H_{*}(\cX, \QQ);
\]
this is an isomorphism, because it is an isomorphism for every
connected component of $\cX$.
\section{The cyclotomic inertia stack and its rigidification} \label{Sec:riX}
\subsection{Cyclotomic inertia}
Let $\cX$ be a finite type Deligne--Mumford stack over $k$.
\begin{definition} We define a category $\iX[r]$, fibered over the
category of schemes, as follows:
\begin{enumerate}
\item Objects $\iX[r](T)$ consist of pairs $(\xi,\alpha)$ where $\xi$ is an object of $\cX$ over $T$, and
$$\alpha: (\bmu_r)_T \to \cAut_T(\xi)$$ is an injective morphism of
group-schemes. Here $(\bmu_r)_T$ is $\bmu_r \times T$.
\item An arrow from $(\xi,\alpha)$ over $T$ to $(\xi',\alpha')$ over $T'$ is an arrow $F: \xi \to \xi'$ making the following diagram commutative:
$$\xymatrix{
(\bmu_r)_T \ar[r]\ar_\alpha[d] & (\bmu_r)_{T'} \ar^{\alpha'}[d]\\
\cAut_T(\xi) \ar[r]\ar[d] & \cAut_{T'}(\xi') \ar[d]\\
T \ar[r] & T'
}$$
where $(\bmu_r)_T \to (\bmu_r)_{T'} $ is the projection and $ \cAut_T(\xi) \to \cAut_{T'}(\xi') $ is the map induced by $F$.
\end{enumerate}
\end{definition}
It is evident that this category is fibered in groupoids over the category of schemes. There is an obvious morphism $\iX[r]\to \cX$ which sends $(\xi,\alpha)$ to $\xi$.
\begin{proposition}\label{Prop:inertia-finite}\label{Prop:X1-rep+finite}
The category $\iX[r]$ is a Deligne--Mumford stack, and the functor $\iX[r]\to \cX$ is representable and finite.
\end{proposition}
\begin{proof}
We verify that $\iX[r]\to \cX$ is representable and finite, which implies that $\iX[r]$ is a Deligne--Mumford stack. Consider a morphism $T \to \cX$ corresponding to an object $\xi$. The fibered product $\iX[r] \times_\cX T$ is an open and closed subscheme of the finite $T$-scheme
$\Hom_T( (\bmu_r)_T, \cAut_T(\xi)).$
\end{proof}
\begin{proposition}
Given an isomorphism $\bmu_r \stackrel{\phi}{\lrar} \ZZ/r\ZZ$ there is an induced isomorphism $\iX[r]\simeq \cI(\cX,r)$, where $ \cI(\cX,r) \hookrightarrow \cI(\cX)$ is the open and closed substack of the inertia stack of $\cX$, consisting of pairs $(\xi,g)$ with $g\in \cAut(\xi)$ of order $r$ at each point.
\end{proposition}
\begin{proof}
The data of an element $g\in \cAut(\xi)$ over $T$ of order $r$ at each point is equivalent to an injective group-scheme homomorphism $(\ZZ/r\ZZ)_T \to \cAut_T(\xi)$. Composing with $\bmu_r \stackrel{\phi}{\lrar} \ZZ/r\ZZ$ we get the result.
\end{proof}
\begin{corollary}\label{Cor:X1-smooth}
When $\cX$ is smooth, $\iX[r]$ is smooth as well.
\end{corollary}
\begin{proof}
It suffices to check this after extension of base field, so we may assume there exists an isomorphism $\bmu_r \stackrel{\phi}{\lrar} \ZZ/r\ZZ$, and by the Proposition it suffices to check the result for $\cI(\cX)$, which is well known.
\end{proof}
\begin{definition}
We define $\iX\ \ =\ \ \sqcup_r\ \iX[r],$ and we name it \emph{the cyclotomic inertia stack} of $\cX$.
\end{definition}
Note that, since $\cX$ is of finite type, $\iX[r]$ is empty except
for finitely many $r$, so $\iX$ is also of finite type, and in fact
finite over $\cX$ by Proposition \ref{Prop:inertia-finite}.
\subsection{Alternative description of cyclotomic inertia}
There is another less evident description of $\iX[r]$, which we give in the following definition.
\begin{definition}
We define a category $\iX[r]'$ over the category of schemes as follows.
\begin{enumerate}
\item Objects over a scheme $T$ consist of representable morphisms $\phi:(\cB\bmu_r)_T \to \cX$.
\item An arrow $\phi \to \phi'$ over $f: T \to T'$ is a 2-morphism $\rho: \phi\to \phi'\circ f_*$ making the following diagram commutative:
$$\xymatrix{ (\cB\bmu_r)_T \ar^{f_*}[rr]\ar[rd]_\phi &&(\cB\bmu_r)_{T'} \ar[ld]^{\phi'} \\
& \cX
}$$
\end{enumerate}
\end{definition}
It is clear that this category is fibered in groupoids over the category of schemes.
\begin{definition}
We define a morphism of fibered categories
$$\iX[r]' \to \iX[r]$$ as follows:
\begin{enumerate}
\item Given an object $$\xymatrix{
\cB(\bmu_r)_T \ar^\phi[r]\ar[d] & \cX \\ T
}$$ we obtain a pair $(\xi, \alpha)$ as follows: $\xi$ is obtained by composing $\phi$ with the section $T\to \cB(\bmu_r)_T$ associated with the trivial $\bmu_r$-torsor $\bone_{\bmu_r,T}$. The homomorphism $\alpha$ is the associated map of automorphisms
$$ (\bmu_r)_T = \cAut_T(\bone_{\bmu_r,T}) \lrar \cAut_T(\xi),$$ which is injective since $\phi$ is representable.
\item Given an arrow $\rho$ as above, we obtain $F: \phi(\bone_{\bmu_r,T}) \to \phi'(\bone_{\bmu_r,T'})$ by completing the following diagram:
$$\xymatrix{
\phi(\bone_{\bmu_r,T}) \ar[d]_\rho\ar@{..>}[rrdd]^F \\
(\phi'\circ f_*)(\bone_{\bmu_r,T})\ar@{=}[d]\\
\phi'\left((\bone_{\bmu_r,T'})_T\right)\ar[rr] &&\phi'(\bone_{\bmu_r,T'})
}$$
\end{enumerate}
\end{definition}
\begin{proposition}
The morphism $\iX[r]' \to \iX[r]$ is an equivalence of fibered categories.
\end{proposition}
\begin{proof}
It is enough to show that for each scheme $T$, the induced functor on the fiber $\iX'(r)(T) \to \iX[r](T)$ is an equivalence. In the following proof we will write the action of a group on a torsor on the left, not on the right, as is more customary.
{\sc Step 1: the functor is faithful.} Given two objects $\phi, \phi':\cB(\bmu_r)_T \to \cX$, suppose that $\alpha: \phi \to \phi'$ is a 2-arrow. We need to show that for any $T$-scheme $f: U \to T$ and any $\bmu_{r}$-torsor $P \to U$, the arrow $\alpha_{P\to U}: \phi(P\to U) \to \phi'(P\to U)$ is uniquely determined by $\alpha_{\bone_{\bmu_{r},T}}: \phi(\bone_{\bmu_{r},T}) \to \phi'(\bone_{\bmu_{r},T})$. Let $\{U_{i} \to U\}$ be an \'etale covering, such that the pullbacks $P_{i} \to U_{i}$ are all trivial. Then the pullbacks of $\phi(P\to U)$ and $\phi'(P\to U)$ to $\cX(U_{i})$ are $\phi(P_{i}\to U_{i})$ and $\phi'(P_{i}\to U_{i})$, respectively, and, since $\cX$ is a stack, the restrictions of a morphism $\phi(P\to U) \to \phi'(P\to U)$ is determined by its restriction to $\phi(P_{i}\to U_{i}) \to \phi'(P_{i}\to U_{i})$. So we may assume that $P\to U$ is trivial. But then there is a cartesian arrow $P \to \bone_{\bmu_{r},T}$, and an induced diagram
\[
\xymatrix@C+15pt{
\phi(P\to U) \ar[r]^{\alpha_{P\to U}}\ar[d]
& \phi'(P\to U) \ar[d]\\
\phi(\bone_{\bmu_{r},T})\ar[r]^{\alpha_{\bone_{\bmu_{r},T}}}
& \phi'(\bone_{\bmu_{r},T})
}
\]
that proves what we want.
{\sc Step 2: the functor is fully faithful.} Assume that $\beta:
\phi(\bone_{\bmu_{r},T}) \to \phi'(\bone_{\bmu_{r},T})$ is an arrow
in $\cX(T)$, commuting with the actions of $\bmu_{r}$. First
consider the case of a trivial $\bmu_{r}$-torsor $P\to U$ over a
$T$-scheme. Choose a trivialization that induces a cartesian arrow
$P \to \bone_{\bmu_{r},T}$. Then, by definition of cartesian arrow,
there is a unique dotted arrow in $\cX(U_{i})$ that we can insert in
the diagram
\[
\xymatrix@C+15pt{
\phi(P\to U) \ar@{.>}[r]^{\alpha_{P\to U}}\ar[d]
& \phi'(P\to U) \ar[d]\\
\phi(\bone_{\bmu_{r},T})\ar[r]^{\beta}
& \phi'(\bone_{\bmu_{r},T})
}
\]
making it commutative. This arrow $\alpha_{P\to U}$ is independent of the chosen
trivialization, because $\beta$ commutes with the actions of $\mu_{r}$, and two
trivializations differ by a morphism $U \to \bmu_{r}$.
If $P \to U$ is not necessarily trivial, choose a covering $\{U_{i}
\to U\}$ such that the pullbacks $P_{i} \to U_{i}$ are trivial. We
have arrows $\alpha_{P_{i} \to U_{i}}: \phi(P_{i} \to U_{i})\to
\phi'(P_{i} \to U_{i})$ in $\cX(U_{i})$, and their pullbacks to
$U_{i}\times_{U}U_{j}$ coincide; hence they glue together to given
an arrow $\alpha_{P \to U}: \phi(P \to U)\to \phi'(P \to U)$. It is
easy to see that $\alpha_{P \to U}$ does not depend on the covering,
and defines a 2-arrow $\phi\to \phi'$ whose image in $\iX'(r)$
coincides with $\beta$.
{\sc Step 3: the functor is essentially surjective.} Let there be given an object $(\xi,\alpha)$ of $\iX[r](T)$, and let us construct a morphism $\phi: \cB(\bmu_r)_T \to \cX$ of fibered categories, whose image in $\iX'(r)(T)$ is isomorphic to $(\xi,\alpha)$.
Let $P \to U$ be a $\bmu_{r}$-torsor, where $U$ is a $T$-scheme: we will define an object $\eta$ of $\cX(U)$, that is a twisted version of the pullback $\xi_{U}$ to $U$, by descent theory, using the action of $\bmu_{r}$ on $\xi$. The facts that we are going to use are all in \cite{descent}, Sections 3.8 and 4.4.
Consider the pullback $\xi_{P}$ of $\xi$ to $P$. The morphism $\bmu_{r}\times P \to P\times_{U}P$ defined as a natural transformation via Yoneda's Lemma by the usual rule $(\zeta, p) \mapsto (\zeta p,p)$ is an isomorphism. The pullbacks of $\xi$ to $P\times_{U}P = \bmu_{r}\times P$ along the first and second projection coincide with the pullback $\xi_{\bmu_{r}\times P}$. On the other end, the group scheme $\bmu_{r}$ acts on $\xi$, so the projection $\bmu_{r}\times P \to \bmu_{r}$ induces an automorphism of $\xi_{\bmu_{r}\times P}$, that gives descent data for $\xi_{P}$ along the \'etale covering $P \to U$. These descent data are effective, and define an object $\eta$ of $\cX(U)$. So we have assigned to every object of $\cB(\bmu_{r})_{T}(U)$ an object of $\cX(U)$; this is easily seen to extend to a morphism of fibered categories $\phi: \cB(\bmu_{r})_{T} \to \cX$.
Let $P = (\bmu_{r})_{T} \to T$ be the trivial torsor. We claim that the object $\eta \eqdef \phi(P \to T)$ of $\cX(T)$ is isomorphic to $\xi$. In fact, the object with descent data defining $\xi$ is $\xi_{P}$, with the descent data given by the identity on $\xi_{P\times_{T}P}$. Then the projection $P \to \bmu_{r}$ defines an automorphism of $\xi_{P}$ in $\cX(P)$, that is easily seen to descend to an isomorphism $\xi \simeq \eta$ in $\cX(T)$. This isomorphism is $\bmu_{r}$-equivariant, because $\bmu_{r}$ is commutative, hence the image of $\phi$ in $\iX(T)$ is isomorphic to $(\xi,\alpha)$, as we wanted.
\end{proof}
\subsection{The stack of gerbes in $\cX$}
We introduce a stack $\riX$ closely related to $\iX$, which will
play an important role below. It will be defined in terms of
morphisms of gerbes. Recall that a gerbe over
a scheme $X$ is an fppf stack $F$ over
$X$ such that
\begin{itemize}
\item there exists an fppf covering $\{X_i \to X\}$ such that
$F(X_i)$ is not empty for any $i$, and
\item given two objects $a$ and $b$ of $F(T)$, where $T$ is an
$X$-scheme, there exists a covering $\{T_i \to T\}$ such that the
pullbacks $a_{T_i}$ and $b_{T_i}$ are isomorphic in $F(T_i)$.
\end{itemize}
Slightly more generally, a stack $F$ over a {\em stack} $\cX$ is a gerbe if for any morphism $V \to \cX$ with $V$ a scheme, $F_V \to V$ is a gerbe.
If $G$ is a sheaf of abelian groups over $X$, we say that $F$ is
banded by $G$ if for each object $a$ of $F(T)$ we have an
isomorphism of sheaves of groups $Aut_T(a)$ with $G_T$. This should
be functorial, in the obvious sense.
If $G$ is not abelian, one can still define a gerbe banded by $G$,
but the definition is more subtle. If $G$ is a group scheme of
finite type, it is not hard to see that every gerbe banded by $G$ is
an algebraic stack.
We first need the following definition.
\begin{definition}
Define a 2-category $\tworiX[r]$ with a functor to the category of schemes as
follows:
\begin{enumerate}
\item An object over a scheme $T$ is a pair $(\cG,\phi)$, where $\cG \to T$ is a gerbe banded by $\bmu_{r}$, and $\cG \stackrel{\phi}{\to} \cX$ is a representable morphism.
\item A morphism $(F,\rho): (\cG,\phi) \to (\cG',\phi')$ consists of a morphism $F : \cG \to \cG'$ over some $f: T \to T'$, compatible with the bands,
and a 2-morphism
$\rho: \phi\to \phi'\circ F$ making the following diagram commutative:
$$\xymatrix{ \cG \ar^{F}[rr]\ar[rd]_\phi && \cG' \ar[ld]^{\phi'} \\
& \cX
}$$
\item A 2-arrow $(F,\rho) \to (F_{1},\rho_{1})$ is a usual 2-arrow $\sigma: F \to F_{1}$ compatible with $\rho$ and $\rho_{1}$ in the sense that the following diagram is commutative:
\[
\xymatrix{ & \phi \ar[dl]_{{\rho}} \ar[dr]^{\rho_{1}}\\
\phi'\circ F \ar[rr]^{\phi'(\sigma)} && \phi'\circ F_{1}
}
\]
\end{enumerate}
\end{definition}
\begin{lemma}
The 2-category $\tworiX[r]$ is equivalent to a category.
\end{lemma}
\begin{proof}
Since all 2-arrows are isomorphisms, it suffices to show that the automorphism group of any 1-arrow is trivial. This is the content of the following general lemma.
\end{proof}
\begin{lemma} \label{lem:representable->1-category}
Suppose given a diagram
$$\xymatrix{ \cG \ar^{F}[rr]\ar[rd]_\phi && \cG' \ar[ld]^{\phi'} \\
& \cX }$$ where $\cG$, $\cG'$ and $\cX$ are categories fibered in
groupoids over a base category, with a 2-arrow $\rho: \phi \to
\phi'\circ F$ making the diagram commutative. Assume $\phi'$ is
faithful. Then an automorphism of $F$ compatible with $\rho$ is the
identity.
\end{lemma}
\begin{proof}
Take an object $\xi$ of $\cG$ over some $T$ in the base category. We get a diagram
\[
\xymatrix{ & \phi(\xi) \ar[dl]_{{\rho_{\xi}}} \ar[dr]^{\rho_{\xi}}\\
\phi'( F(\xi)) \ar[rr]^{\phi'(\sigma_{\xi})} && \phi'( F(\xi))
}
\]
But $\phi'(\sigma_{\xi})$ lies over the identity $T \to T$. Since $\cX$ is fibered in groupoids, it follows that $ \phi'(\sigma_{\xi})$ is the identity. Since $\phi'$ is faithful, $\sigma_{\xi}$ is the identity, which is what we wanted.
\end{proof}
\begin{definition}\label{definition-riX}
We define $\riX[r]$ to be the category associated with the 2-category $\tworiX[r]$,
where arrows in $\riX[r]$ are 2-isomorphism classes of 1-arrows in
$\tworiX[r]$.
\end{definition}
We note that $\riX[r]$ is a category fibered in groupoids over the
category of schemes.
\begin{remark}\label{rigidification-remark}
There is a tautological morphism $\iX[r] \to \riX[r]$, defined as follows. An object of $\iX[r](T)$ corresponds to a representable morphism $(\cB\bmu_{r})_{T} \to \cX$. Since $(\cB\bmu_{r})_{T}$ is a gerbe over $T$ banded by $\bmu_{r}$, this gives an object of $\riX$. An arrow in $\iX[r]$ gives an arrow in $\riX[r]$ in the obvious way.
We can also define a fibered category, an object of
which is an object of $\riX[r]$, together with a section of the
gerbe. There is an obvious forgetful functor into $\riX[r]$, which
exhibits it as the universal gerbe over $\riX$. Since a gerbe over
$T$ banded by $\bmu_{r}$ with a section has a canonical isomorphism
with $(\cB\bmu_{r})_{T}$, this universal gerbe is evidently
isomorphic to $\iX[r] \to \riX$.
\end{remark}
In the next section we give an alternative description of $\riX[r]$,
which in particular shows that it is a Deligne--Mumford stack.
\begin{definition}
We define $\riX\ \ =\ \ \sqcup_r\ \riX[r],$ and we name it \emph{the
stack of cyclotomic gerbes} in $\cX$.
\end{definition}
\subsection{The rigidified cyclotomic inertia stack}
Consider the stack $\iX[r]$. By definition, for every object
$(\xi,\alpha)$ of $\iX[r]\bigl(T\bigr)$ we have a canonical {\em
central} embedding $(\bmu_{r})_{T}
\stackrel{\iota_{(\xi,\alpha)}}{\hookrightarrow} \cAut_{T}
(\xi,\alpha)$. By \cite{ACV}, Theorem 5.1.5 (see also Appendix~\ref{Sec:rigidification}), there
exists a {\rm rigidification} denoted there $\iX[r] \to
\iX[r]^{\bmu_{r}}$. We are adopting the better notation proposed by
Romagny in \cite{Romagny}, and denote this by
$\iX[r]\thickslash\bmu_{r}$.
\begin{proposition}
We have an equivalence of fibered categories
$$\iX[r]\thickslash\bmu_{r} \to \riX[r]$$ so that the following
diagram is commutative:
\[
\xymatrix{
& \iX[r]\ar[dl]\ar[dr]\\ \iX[r]\thickslash\bmu_{r}\ar[rr]&&\riX[r]
}
\]
The right diagonal arrow is described in Remark~\ref{rigidification-remark}.
\end{proposition}
\begin{proof}
We will use the modular interpretation
$\iX[r]\slashprime \bmu_{r}$ of the rigidified stack
$\iX[r]\thickslash \bmu_{r}$ given in
Proposition~\ref{prop:equivalence-rigidification}. There is an
obvious morphism of fibered categories $\iX[r]\slashprime \bmu_{r}
\to \riX[r]$, defined as follows. An object of $\iX[r]\slashprime
\bmu_{r}(T)$ consists of a gerbe $\cG \to T$ banded by $\bmu_{r}$,
and a $\bmu_{r}$-2-equivariant morphism $\cG \to \iX[r]$, in the sense of Section~\ref{subsection-moduli-interpretation}: this can
be composed with the projection $\iX[r] \to \cX$ to get a
representable morphism $\cG \to \cX$. This function on objects can
be extended to a function on arrows, and it defines the desired
functor.
We claim that this an equivalence: let us construct an inverse $\riX
\to \iX\slashprime \bmu_{r}$. Consider an object of $\riX$,
consisting of a gerbe $\cG \to T$ banded by $\bmu_{r}$, and a
representable morphism $\phi: \cG \to \cX$. Given an object $\xi$ of
$\cG(U)$, where $U$ is a $T$-scheme, the morphism $\phi$ induces a
homomorphism of group-schemes $\alpha: H_{U} = \cAut_{U,\cG}(\xi)\to \cAut_{U,\cX}(\phi\xi)$; and this homomorphism is an
embedding, because a representable morphism is also faithful, so
$(\xi,\alpha)$ is an object of $\iX$. This function from objects of
$\cG$ to objects of $\iX$ extends naturally to a morphism $\cG \to
\iX$; and this morphism is $\bmu_{r}$-2-equivariant, by definition.
We leave it to the reader to extend this
to a functor $\riX \to \iX\slashprime \bmu_{r}$, and show that it
gives an inverse to the functor $\iX[r]\slashprime \bmu_{r} \to
\riX[r]$ above. The commutativity of the diagram is straighforward.
\end{proof}
\begin{corollary}
If $\cX$ is smooth, $\riX$ is smooth as well. If $\cX$ is proper, $\riX$ is proper as well.
\end{corollary}
\begin{proof}
This follows from Proposition \ref{Prop:X1-rep+finite} and Corollary \ref{Cor:X1-smooth}, since the morphism $\iX \to \riX$ is proper, \'etale and surjective.
\end{proof}
\subsection{Changing the band by a group automorphism}\label{Sec:changing-band}
There is an involution $\iota:\riX\to \riX$ defined as follows:
given a gerbe $\cG \to T$ banded by a group-scheme $G$ and an
automorphism $\tau: G \to G$, we can change the banding of the
gerbe through the automorphism $\tau$. Applying this procedure to
the gerbe $\cG$ of an
object $\cG \to \cX$ of $\riX[r](T)$, we get another object
$\sideset{^\tau}{}{\mathop{\cG}}
\to \cX$
of $\riX[r](T)$. When $\tau:\bmu_{r}\to \bmu_{r}$ is the inversion
automorphism $\zeta \mapsto \zeta^{-1}$, this induces an involution
of $\riX[r]$. Applying this to each piece of $\riX$ separately, we
obtain the desired involution of $\iota:\riX\to \riX$.
\subsection{The tangent bundle lemma}
\begin{lemma}\label{tangent} Let $S$ be a scheme, and let $f:S \to \riX$ be
a morphism. Let
$$\xymatrix{\cG \ar[r]^F \ar[d]_\pi & \cX\\
S}$$
be the associated diagram. Then there is a canonical isomorphism between
$\pi_*(F^*(T_\cX))$ and $f^*(T_{\riX})$.
\end{lemma}
\begin{proof}
Consider the universal gerbe $\iX \to \riX$ and the diagram of smooth stacks
$$\xymatrix{\iX \ar^{F_1}[r]\ar^\varpi[d] & \cX \\ \riX . }
$$
Given a morphism $f:S \to \riX$ as in the lemma, we have a fiber diagram
$$\xymatrix{\cG \ar^g[r]\ar@/^1.5pc/[rr]^{F} \ar_\pi[d] &\iX \ar^{F_1}[r]\ar^\varpi[d] & \cX \\ S \ar^f[r] & \riX . }
$$
Since $\varpi$ is flat and $\varpi_*$ is exact on coherent sheaves, for any locally free sheaf $\cH$ on $\iX$ we have
$f^* \varpi_* \cH = \pi_*g^*\cH$. Therefore it suffices to check that
$\varpi_*(F_1^*(T_\cX))$ and $T_{\riX}$ are canonically isomorphic.
We have a natural morphism of sheaves $T_{\iX} \to F_1^* T_\cX$, giving a morphism
$\varpi_*T_{\iX} \to \varpi_* F_1^* T_\cX$. Since $\iX \to \riX$ is an \'etale gerbe, we have $\varpi_*T_{\iX} \cong T_{\riX}$, giving a morphism
$T_{\riX} \to \varpi_* F_1^* T_\cX$. We can check that this is an isomorphism by pulling back to geometric points.
Over a geometric point $y$ of $\riX$, we can identify the
fiber of $\varpi$ with $\cB\bmu_r$. This gives a lift of $y$ to $\iX$, and
so the point $y$ maps to a geometric
point $x$ of $\cX$, with stabilizer $G$. We can locally describe
$\cX$ around $x$ as $[U/G]$. The pullback $T$ of the tangent space of $\cX$
to $y$ has a natural action of $\bmu_r$, and the fiber of
$\varpi_*(F_1^*T_\cX)$ at $y$ is naturally the space of invariants
$T^{\bmu_r}$. Given a local chart of $\cX$ of the form $[U/G]$, the stack $\iX$ has a local chart
given by $[U^{\bmu_r}/C(\bmu_r)]$. Since $T_{U^{\bmu_r}}=T_U^{\bmu_r}$, we obtain a natural isomorphism $T_{\riX,y} \cong T^{\bmu_r}$, which is what we need.
\end{proof}
\section{Twisted curves and their maps} \label{Sec:twisted-curves}
The foundation of the theory of stable maps to an orbifold rests on the notion of a \emph{twisted curve}. Over an algebraically closed field, a twisted curve is a connected, one-dimensional Deligne--Mumford stack which is \'etale locally a nodal curve, and which is a scheme outside the marked points and the singular locus. Moreover, we will always include the condition that the nodes be balanced, that is, formally locally near a node, the stack is isomorphic to $$\big[\Spec (k[x,y]/(xy))\, \big/\, \bmu_r\big]$$ where the action of $\mu_r$ is given by $\zeta (x,y) = (\zeta \cdot x, \zeta^{-1} \cdot y)$.
In particular, the coarse moduli space of a twisted curve is always a nodal curve.
The notion of a family of twisted curves is straightforward, but involves one novelty. Although the relative coarse moduli scheme of a family of twisted pointed curves is a family of prestable curves -- and hence comes with sections corresponding to the marked points -- a family of twisted curves need not have sections. The marked point on each fiber instead gives rise to a \emph{gerbe} banded by $\bmu_r$ where $r$ is the order of the inertia group at the twisted point.
\subsection{The stack of twisted curves}\label{Sec:stack-twisted-curves}
If we define a groupoid $\fM^{\tw}_{g,n}$ whose $S$ points are given
by families of twisted curves over $S$, that is morphisms $\pi: \cC
\to S$ which are flat, together with a collection of $n$ disjoint
gerbes over $S$ embedded in $\cC$, such that the geometric fibers of
$\pi$ are $n$ pointed twisted curves of genus $g$, then the
following results are proven in \cite{Olsson-twisted} .
\begin{enumerate}
\item $\fM^{\tw}_{g,n}$ is a smooth algebraic stack, locally of finite type.
\item If we bound the topological type, including the twisting at marked points and nodes, we get a stack of finite type.
\item A formal deformation space $\Delta^{\tw}$ of a twisted curve
$\cC$ with nodes $q_{1},\ldots, q_{s}$ of indices
$r_{1},\ldots,r_{s}$ can be obtained from a given formal deformation
space $\Delta = \Spf R$ of the coarse curve $C$ as follows: Let
$D_{i}$ be the divisor in $\Delta$ corresponding to $q_{i}$, with
defining equations $f_{i}$. Then $$\Delta^{\tw} = \Spf R\llbracket x_{1},\ldots,x_{s}\rrbracket/\left(x_{1}^{r_{1}} - f_{1},\ldots x_{s}^{r_{s}} - f_{s}\right)$$ is a formal deformation space of $\cC$.
\end{enumerate}
\subsection{The smooth locus of a twisted pointed curve}
The theory in this section is due to the authors and to C.~Cadman
independently (\cite{Cadman1}, Section~2). The reader is advised to read Appendix \ref{roots} for the notion of root stacks.
Suppose that $C \to S$ is an $n$-pointed nodal curve; call $s_i
\colon S \to C$ the sections, $S_i \subseteq C$ their images,
$\sigma_i$ the canonical section of $\cO(S_i)$. Given positive
integers $d_1$, \dots,~$d_n$, we define a stack over $S$ as the
fibered product
\[
\cC[d_1, \dots, d_n]\ \ =\ \ \radice{d_1}{(\cO(S_1),
\sigma_1)/C}\ \mathop\times\limits_C\ \cdots\ \mathop\times\limits_C\
\radice{d_n}{(\cO(S_n), \sigma_n)/C}.
\]
See Appendix \ref{roots} for an explanation of the notation.
The locus where the projection
$$\radice{d_i}{(\cO(S_i), \sigma_i)/C} \to C$$ is not an isomorphism
coincides with $S_i$ when $d_{i} > 1$; if $d_{i} = 1$ the projection
is an isomorphism. Outside of the locus $S_i$
the stack $\cC[d_1, \dots, d_n]$ is isomorphic to $C$; over $S_i$ we
have an embedding $$\radice{d_i}{\cO_C(S_i)}\mid_{S_i} \into
\radice{d_i} {(\cO(S_i),\sigma_i)/C};$$ since we have that for $j
\neq i$ the morphism $\radice{d_j}{(\cO(S_j), \sigma_j)/C} \to C$ is
an isomorphism in a neighborhood of $S_i$, we also have an embedding
$$
\radice{d_i} {\cO_C(S_i)}\mid_{S_i} \into \cC[d_1, \dots, d_n].
$$
It
is easily checked that, after taking the
$\radice{d_i}{\cO_C(S_i)}\mid_{S_i}$ as markings, the stack
$\cC[d_1, \dots, d_n]$ is a twisted curve over $S$ with moduli space
equal to $C$.
The following theorem shows that the smooth part of a twisted curve
is uniquely characterized by its moduli space and its indices along
the markings. If $\cC \to S$ is a twisted curve, we will denote its
smooth part by $\cC_\mathrm{sm}$.
\begin{theorem}\label{thm:description-twistedcurves}
Let $\cC \to S$ be an $n$-pointed twisted curve with moduli space
$\pi: \cC \to C$. Assume that its index at the $i\th$-section is
constant for all $i$, and call it $d_i$. Denote by $\Sigma_i$ the
$i\th$ marking, by $s_i \colon S \to C$ the section corresponding to
$\Sigma_i$, and by $N_i$ the normal bundle to $S$ along $s_i$.
\begin{enumerate}
\item There is a canonical isomorphism of twisted curves
\[
\cC_\mathrm{sm} \simeq \cC[d_1, \dots, d_n]_\mathrm{sm}
\]
inducing the identity on $C_\mathrm{sm}$.
\item There is a canonical isomorphism of gerbes over $S$
\[
\Sigma_i\simeq \radice{d_i}{N_i/S}.
\]
In particular, $\Sigma_{i}$ is canonically banded by $\bmu_{d_{i}}$.
\end{enumerate}
\end{theorem}
\begin{proof}
For part~(1), notice that we have an equality of a Cartier divisor $\pi^{*}S_{i} = d_{i}\Sigma_{i}$ on $\cC$; this induces a morphism $\cC \to \cC[d_1, \dots, d_n]$. This is an isomorphism outside of the marked points and the nodes, and is easily seen to be representable outside of the nodes. To check that the restriction $\cC_\mathrm{sm} \to \cC[d_1, \dots, d_n]_\mathrm{sm}$ is an isomorphism it is enough to restrict to the geometric fibers of $S$, because $\cC$ and $\cC[d_1, \dots, d_n]$ are flat over $S$; but the morphism $\cC_\mathrm{sm} \to \cC[d_1, \dots, d_n]_\mathrm{sm}$ restricted to a geometric fiber is representable, finite and birational, and the stacks appearing are smooth, hence it is an isomorphism.
Part~(2) follows immediately from part~(1).
\end{proof}
\subsection{Twisted stable maps}
Let $\cX$ be a Deligne--Mumford stack, $g,n$ non-negative integers
and $\beta$ a curve class on the coarse moduli space $X$. Associated
with
this data we have the stack $\cK_{g,n}(\cX,\beta)$ of $n$-pointed
twisted stable maps into $\cX$ of genus $g$ and class $\beta$. This
classifies stable maps from $n$-pointed twisted genus $g$ curves to
$\cX$ of degree $\beta$. Precisely, an object of this stack over a
scheme $T$ consists of the data of a family of twisted curves $\cC
\to T$, $n$ gerbes $\Sigma_i \subset \cC$, and a representable
morphism $\cC \to \cX$, such that the induced maps of underlying
coarse moduli spaces give a family of $n$ pointed genus $g$ stable
maps to $X$. We refer the reader to \cite{AV} for a construction
and a more detailed discussion of this stack.
Let $\cC \to T$ be a twisted $n$-pointed curve with $T$ connected,
and let $1\leq i \leq n$. Assume that the index of the $i$-th
marking is the integer $r$. Note that, by Theorem
\ref{thm:description-twistedcurves}, part (2) we have that
$\Sigma_{i}^{\cC}$ is canonically banded by $\bmu_{r}$.
\subsection{Evaluation maps}\label{Sec:evaluation}\hfil
\begin{definition}\hfil
\begin{enumerate}
\item Assume given a twisted stable map $f: \cC \to \cX$ over a base $T$. We define
$$e_{i}(f)\in \riX[r]\bigl(T\bigr)$$
to be the object associated with the diagram
\[
\xymatrix{
\Sigma_{i}^{\cC} \ar[r]^{f\mid_{ \Sigma_{i}^{\cC}}} \ar[d] & \cX \\
T
}
\]
By Definition~\ref{definition-riX}, this defines a morphism
$$e_{i}: \cK_{g,n}(\cX,\beta) \to \riX,$$
which we call \emph{the $i$-th evaluation map}. \item The morphism
$\check e_{i} := \iota\circ e_{i}$, where $\iota:\riX\to \riX$ is
the involution defined in section \ref{Sec:changing-band}, is called
\emph{the $i$-th twisted evaluation map}.
\end{enumerate}
\end{definition}
\subsection{The virtual fundamental class}
The key technical point in developing Gromov-Witten theory for $\cX$ is the
construction of the \emph{virtual fundamental class} $[\cK_{g,n}(\cX,\beta)]^\vir$ in $A_*(\cK_{g,n}(\cX,\beta))$. By \cite{BF}
and \cite{LT}, what is needed to construct this class is a perfect obstruction theory on this moduli stack. Following the methods of \cite{BF}, we will mean by this a morphism in the derived category
$$\phi:E \to \bL_{\cK_{g,n}(\cX,\beta)/\fM^\tw_{g,n}}$$
such that
\begin{itemize}
\item $E$ is locally equivalent to a two term complex of locally free sheaves, and
\item $\H^0(\phi)$ is an isomorphism and $\H^{-1}(\phi)$ is surjective.
\end{itemize}
As in the case of ordinary stable maps, there is a natural perfect
obstruction theory with $E = \bR\pi_*(f^*T\cX)^\vee$. The proof of
this is identical to the proof for
ordinary stable maps since what is needed are formal properties of
the cotangent complex and Illusie's results \cite{Illusie} relating
these to the deformation theory of morphisms. Since the theory of
the cotangent complex for Artin stacks has been developed in \cite{LMB} and corrected
in \cite{Olsson-sheaves}, and since
Illusie explicitly works in the general setting of ringed topoi, all
the necessary generalizations have already been established.
Specifically, in the
discussion of \cite{Behrend} page 604, immediately after Proposition
4, one relies on the claim that $$\phi:\bR\pi_*(f^*T\cX)^\vee \to
\bL_{\cK_{g,n}(\cX,\beta)/\fM^\tw_{g,n}}$$ is a perfect {\em
relative} obstruction theory. This relative case, discussed in
section 7 of \cite{BF} (page 84 onward), reworks the absolute case
discussed earlier in that paper. The crucial result in \cite{BF} is
Proposition 6.3, where $\cC$ is assumed to be a Gorenstein and {\em
projective} curve. As explained above, projectivity is not necessary
for deformation theory (i.e. \cite{BF} Theorem 4.5) - it works just
as well for a proper Deligne--Mumford stack. Both in \cite{BF}
Proposition 6.3 and in \cite{BF} Lemma 6.1 on which it relies, one
also needs relative duality, which is ``well known" for proper
Gorenstein Deligne--Mumford stacks; for a twisted curve $\cC$ with a
projective coarse moduli space it can be shown using a finite flat
Galois covering $D \to \cC$, ramified over auxiliary sections, which
can be constructed locally over the base.
We remark that one additional
feature of $E$ that is required in \cite{BF} is that $E$ admit a
global resolution (see discussion before \cite{BF}, Proposition 5.2).
Kresch's work on intersection theory for Artin stacks \cite{Kresch}
(see specifically section 5.2 there)
has removed the need for this hypothesis.
\section{The boundary of moduli}
\subsection{Boundary of the stack of twisted curves}
We will need to study the geometry of the moduli stack of pre-stable
twisted curves, $\fM^\tw_{g,n}$, as described in
\cite{Olsson-twisted}. In particular we are interested in the
structure of the boundary. We consider the following category
$$\twD(g_1;A\,|\,g_2;B)$$ fibered in groupoids over the category of
schemes. Informally, this category
parametrizes nodal twisted curves, with a distinguished node
separating the curve in two connected components, one of genus $g_1$
containing the markings in a subset $A \subset \{1,\ldots n\}$, the
other, of genus $g_2$, containing the markings in the complementary
set $B$. More formally the objects over a scheme $S$ consist of
commutative diagrams
$$\xi\ \ \ =\ \ \ \left(
\vcenter{\xymatrix{*+<12pt,12pt>{\cG_1}\ar^{\alpha}[rr] \ar@{^{(}->}[d]& &*+<12pt,12pt>{\cG_2} \ar@{^{(}->}[d] \\ \cC_1 \ar[dr] && \cC_2\ar[dl] \\ &S}}\right)$$
where
\begin{enumerate}
\item $\cC_1 \to S$ is a pre-stable twisted curve of genus $g_1$ with marking in $A\sqcup \checkbullet$,
\item $\cC_2 \to S$ is a pre-stable twisted curve of genus $g_2$ with marking in $B\sqcup \bullet$,
\item $\cG_1$ and $\cG_2$ are the markings on $\cC_1$ and $\cC_2$ corresponding to $\checkbullet$ and $\bullet$, respectively,
and
\item $\alpha$ is an isomorphism inverting the band.
\end{enumerate}
An arrow of $\twD(g_1;A\,|\,g_2;B)$ is a fiber diagram
$$
\vcenter{\xymatrix{
*+<12pt,12pt>{\cG_1}\ar^{\alpha}[rr]\ar[rrrd] \ar@{^{(}->}[d]& &*+<12pt,12pt>{\cG_2} \ar@{^{(}->}[d]\ar[rrrd] \\
\cC_1 \ar[dr]\ar[rrrd] && \cC_2\ar[dl]\ar[rrrd] &
*+<12pt,12pt>{\cG'_1}\ar^{\alpha'}[rr] \ar@{^{(}->}[d]& &*+<12pt,12pt>{\cG'_2} \ar@{^{(}->}[d]
\\
&S\ar[rrrd] & &\cC'_1 \ar[dr] && \cC'_2\ar[dl] \\
&&&&S'
}}$$
This, in particular, includes the data of a 2-isomorphism in the following square:
$$\xymatrix{
\cG_1\ar^{\alpha}[r] \ar[d]& \cG_2 \ar[d] \ar@{=>}[dl]\\
\cG'_1\ar^{\alpha'}[r] & \cG'_2
}$$
\begin{remark}
In any reasonable framework of 2-stacks, $\twD(g_1;A\,|\,g_2;B)$ should be the fibered product
$$\fM^\tw_{g_1,A\sqcup\checkbullet}\ \ \mathop\times\limits_{\mathop\sqcup\limits_r\fB\cB\bmu_r}\ \ \fM^\tw_{g_2,B\sqcup\bullet}.$$
Here $\fB\cB\bmu_r$ is the classifying 2-stack of $\cB\bmu_r$, which parametrizes gerbes banded by $\bmu_r$, and the the first morphism underlying the product has the band inverted. Since this 2-stack occurs only as the basis of the fibered product, the result is still a 1-category.
There is a similar construction for non-separating nodes, which we will not describe explicitly here.
\end{remark}
\begin{proposition}
The category $\twD(g_1;A\,|\,g_2;B)$ is a smooth algebraic stack, locally of finite type over the base field $k$.
\end{proposition}
\begin{proof}
Consider the universal gerbes $\cG_1$ and $\cG_2$
corresponding to the markings $\checkbullet$ and $\bullet$ over the
product
$\fM^\tw_{g_1,A\sqcup\checkbullet}\times\fM^\tw_{g_2,B\sqcup\bullet}$.
By Theorem 1.1 of \cite{Olsson-Hom}, there exists an algebraic stack
$$ \Isom_{\fM^\tw_{g_1,A\sqcup\checkbullet}\times\fM^\tw_{g_2,B\sqcup\bullet}}(\cG_1, \cG_2)$$
parametrizing isomorphisms between $\cG_1$ and $\cG_2$. The ``change
of band" isomorphism $\bmu_r\to \bmu_r$ induced by such an
isomorphism $\cG_1 \to \cG_2$ is locally constant, and
$\twD(g_1;A\,|\,g_2;B)$ is the locus where it is the inversion
isomorphism. Since the $\cG_i$ are \'etale gerbes, the smoothness of
the Isom stack follows immediately from that of $\fM^\tw_{g,n}$.
\end{proof}
\begin{proposition}
We have a natural representable morphism
$$gl: \twD(g_1;A\,|\,g_2;B) \to \fM^\tw_{g_1+g_2, A\sqcup B}$$ induced by gluing the
two families of curves over $\twD$ into a family of reducible curves
with a distinguished node.
\end{proposition}
{\bf Proof.} Fix an object of $\twD(g_1;A\,|\,g_2;B)$ over $S$. By
Proposition \ref{Prop:gluing} applied to the diagram $$\xymatrix
{*+<12pt,12pt>{\cG_1} \ar@{^{(}->}[r]\ar@{^{(}->}[d] & \cC_2\\\cC_1},$$ we have an
associated family of nodal curves $\cC:= \cC_1 \cup_{\cG_1} \cC_2.$
Representability follows from a straightforward comparison of
isotropy groups. \qed
\begin{definition} \label{Def:gothic-r}
We define the locally constant function
\[
\fr : \twD(g_1;A\,|\,g_2;B) \to \ZZ
\]
(this is a Gothic ``r'') which takes a nodal twisted curve to the index of the node.
\end{definition}
\subsection{Boundary of the stack of twisted stable maps}
There is an analogous gluing map on the spaces of morphisms.
\begin{proposition}\hfill
\begin{enumerate}
\item
Consider the evaluation morphisms
$$ \check e_{ \checkbullet}: \cK_{g_1,A\sqcup \checkbullet}(\cX,\beta_1) \to \riX$$
and
$$ e_{ \bullet}: \cK_{g_2,B\sqcup \bullet}(\cX,\beta_2) \to \riX$$
There exists a natural representable morphism
$$\cK_{g_1,A\sqcup \checkbullet}(\cX,\beta_1) \times_{\riX}\cK_{g_2,B\sqcup \bullet}(\cX,\beta_2) \ \ \lrar \ \ \cK_{g_1+g_2,A\sqcup B}(\cX,\beta_1+\beta_2).$$
\item
Consider the evaluation morphisms
$$ \check e_{\checkbullet}\times e_{\bullet}: \cK_{g-1,A\sqcup\{\checkbullet,\bullet\}}(\cX,\beta) \to \riXs$$
There exists a natural representable morphism
$$\cK_{g-1,A\sqcup\{\checkbullet,\bullet\}}(\cX,\beta) \times_{\riXs}\riX \ \ \lrar \ \ \cK_{g,A}(\cX,\beta).$$
\end{enumerate}
\end{proposition}
\begin{proof}
We prove the first statement, the second being similar, replacing Proposition \ref{Prop:gluing} with Corollary \ref{Cor:limits}.
We give the morphism on the level of $S$-valued points. We have an identification of objects
\begin{equation}\label{Diag:esuoh}
\cK_{g_1,A\sqcup \checkbullet}(\cX,\beta_1) \mathop\times\limits_{\ \riX}\cK_{g_2,B\sqcup \bullet}(\cX,\beta_2)\ \big(\ S \ \big) \ \ = \ \ \left\{
\vcenter{\xymatrix{*+<12pt,12pt>{\cG_1}\ar^{\alpha}[rr] \ar@{^{(}->}[d]& &*+<12pt,12pt>{\cG_2} \ar@{^{(}->}[d] \\ \cC_1 \ar[dr]\ar[r] &\cX& \cC_2\ar[l]\ar[dl] \\ &S}}\right\}
\end{equation}
where the diagram is a 1-commutative diagram of stacks, $\cG_i \subset \cC_i$ are the markings corresponding to $\checkbullet$ and $\bullet$, respectively, and $\alpha: \cG_1 \to \cG_2$ is the isomorphism \emph{inverting the band} induced by $ \check e_{\checkbullet}$ and $ e_{\bullet}$.
By Proposition \ref{Prop:gluing} applied to the diagram $$\xymatrix {*+<12pt,12pt>{\cG_1} \ar@{^{(}->}[r]\ar@{^{(}->}[d] & \cC_2\\\cC_1},$$ we have an associated
family of nodal curves $\cC:= \cC_1 \cup_{\cG_1} \cC_2.$ Since Diagram
(\ref{Diag:esuoh}) is commutative, we are given a 2-isomorphism between the two resulting maps $\cG_1 \to \cX$. Therefore, by the universal property of $\cC$, we have a morphism $\cC \to \cX$, which is clearly a twisted stable map over $S$.
\end{proof}
\begin{proposition}\label{cartesian}
We have a cartesian diagram
$$\xymatrix{{\mathop\coprod\limits_{\beta_1+\beta_2 = \beta} \cK_{g_1,A\sqcup \checkbullet}(\cX,\beta_1) \times_{\riX}\cK_{g_2,B\sqcup \bullet}(\cX,\beta_2) \ar[r]\ar[d]} &
\cK_{g_1+g_2,A\sqcup B}(\cX,\beta)\ar[d] \\
\twD(g_1;A\,|\,g_2;B)\ar^{gl}[r] &\fM^\tw_{g_1+g_2, A\sqcup B}
}$$
\end{proposition}
\begin{proof}
For convenience of notation we will use the shorthand
$$\xymatrix{{ \mathop\coprod\limits_{\beta_1+\beta_2 = \beta}\cK_1(\beta_1)\times_{\riX}\cK_2(\beta_2) \ar[r]\ar[d]} &
\cK(\beta)\ar[d] \\
\twD\ar[r] &\fM^\tw
}$$
The diagram gives a morphism $$\mathop\coprod\limits_{\beta_1+\beta_2 = \beta} \cK_1(\beta_1) \times_{\riX}\cK_2(\beta_2) \to \cK(\beta) \times_{\fM^\tw} \twD. $$ We construct a morphism in the reverse direction as follows. It suffices to restrict attention to points over a connected base scheme $S$. In this case an object on the right hand side is a triple
$$ \left( f:\cC \to \cX,\ \xi, \ \phi\right) ,$$
where $\xi$ is an object of $\twD(S)$ and $\phi$ is an isomorphism between the resulting objects in $\twM$, namely between $\cC$ and $\cC_1\cup_{\cG_1}\cC_2$. Since $f:\cC \to \cX$ is stable, so are the resulting morphisms $\cC_i\to \cX$, with the additional markings taken into account. By the connectedness of the base, these maps have constant image classes $\beta_1,\beta_2$. With this, the diagram describing $\xi$ is completed to a diagram as in the description \ref{Diag:esuoh} of an object on the left hand side, which is what we needed.
\end{proof}
\subsection{Gluing and virtual fundamental classes}
For Gromov-Witten theory, one of the key points is that a similar
statement holds for the virtual fundamental class. First, a crucial
fact is that fundamental classes exist for the stacks of twisted
curves we need. This is because, as mentioned in Section
\ref{Sec:stack-twisted-curves}, away from a closed substack of
arbitrarily high codimension, the stack of twisted curves of bounded
indices is of finite type. We will show below (Lemma
\ref{Lem:gluing-finite}) that $gl:\twD \to \twM$ is a finite
unramified morphism. By \cite{Kresch}, Section 4.1 it induces a
pull-back homomorphism on Chow groups
$$gl^! : A_*(\cK(\beta)) \lrar \mathop\oplus\limits_{\beta_1+\beta_2=\beta}
A_*(\cK_1(\beta_1)\times_{\riX} \cK_2(\beta_2)).$$ Pulling back the
virtual class on $\cK(\beta)$ gives us a candidate for the virtual
fundamental class of the boundary. There is another natural Chow
class living on the fibered product spaces coming from the pull back
by the diagonal morphism $\Delta: \riX \to \riXs$. The splitting
axiom in Gromov-Witten theory identifies these two classes.
\begin{proposition}\label{Prop:virtual}
$$gl^![\cK_{g,A\cup B}(\cX,\beta)]^\vir = \sum_{\beta_1+\beta_2=\beta}
\Delta^!([\cK_{g_1,A\sqcup \checkbullet}(\cX,\beta_1)]^\vir \times
[\cK_{g_2,B\sqcup \bullet}(\cX,\beta_2)]^\vir).$$
\end{proposition}
\begin{proof}
This proof is very similar to the proof of the splitting axiom
for Gromov-Witten theory of schemes. First, we observe that the
left hand side of our equation is the
pullback of a relative virtual fundamental class under a change of base.
It follows by Proposition 7.2 of \cite{BF}
that this left hand side is the relative virtual fundamental class
of $\cK_1 \times_{\riX} \cK_2$ over $\twD $ with respect to the
relative perfect obstruction theory $R\pi_*(f^*T\cX)$.
We need to compare this to the right hand side. Here we use the
basic compatibility result for virtual fundamental classes. The
class $[\cK_1]^\vir \times [\cK_2]^\vir$ is the virtual fundamental
class associated with the relative perfect obstruction theory on
$\cK_1 \times \cK_2$ given by $R\pi_{1*}(f_1^*T\cX)\oplus
R\pi_{2*}(f_2^*T\cX)$. This is an immediate consequence of
Proposition 5.7 of \cite{BF}. By considering the normalization
sequence for the family of nodal curves with a distinguished node
over $\cK_1 \times_{\riX} \cK_2$, we get the following distinguished
triangle:
$$ R\pi_*(f^*T\cX) \to R\pi_{1*}(f_1^*T\cX)\oplus R\pi_{2*}(f_2^*T\cX)
\to R\pi_{\Sigma *} (f_\Sigma^*T\cX) $$ where $f_\Sigma$ denotes the
restriction of $f$ to the gerbe which is the intersection of $\cC_1$
with $\cC_2$. By Proposition 5.10 of \cite{BF}, in order to prove
the equality we want, we just need to identify $R\pi_{\Sigma
*}(f_\Sigma^* T\cX)$ with the normal bundle of the map $\Delta$. The
normal bundle of $\Delta$ is obviously $T\riX$. Applying Lemma
\ref{tangent} to $S=\cK_1\times_{\riX}\cK_2$ with the morphism
$F=f_\Sigma$ gives our result.
\end{proof}
An identical argument yields the analogous splitting axiom for a
non-separating node.
\begin{proposition}
$$gl^![\cK_{g,A}(\cX,\beta)]^\vir =
\Delta^![\cK_{g-1,A\sqcup \{ \bullet, \checkbullet\}
}(\cX,\beta_1)]^\vir.$$
\end{proposition}
\section{Gromov--Witten classes}
\subsection{Algebraic Gromov--Witten classes}
\begin{definition}
We define a locally constant function
$r: \riX \to \ZZ$ by evaluating on geometric points:
$(x,\cG) \mapsto r$, where $\cG$ is a gerbe banded by $\bmu_r$. We view $r$ as an
element in $A^0(\riX)$.
\end{definition}
We now define \emph{Gromov--Witten Chow classes:}
\begin{definition}\label{Def:gromov-witten-classes} Fix integers $g,n$, Chow classes
$\gamma_i\in A^*(\riX)_\QQ$, and a curve class $\beta$. We define a
class in $A_*(\riX)_\QQ$ by the formula
$$\langle \gamma_1,\ldots,\gamma_n, * \rangle_{g,\beta}^\cX \ \ = \ \
r\ \cdot\ \check e_{n+1\,*}\left( \left(\prod_{i=1}^n e_i^*\gamma_i\right)\cap\left[\cK_{g,n+1}(\cX,\beta)\right]^\vir\right).$$
\end{definition}
We will suppress the superscript $\cX$ when the target is clear, and the genus $g$ when $g=0$.
We will write $\langle \gamma_1, \gamma_2,\delta_A, * \rangle_{g,\beta}^\cX$
for an expression of the type $\langle \gamma_1,
\gamma_2,\delta_{i_1},\ldots,\delta_{i_m}, * \rangle_{g,\beta}^\cX$
with $A = \{i_1,\ldots i_m\}$ subject to the convention $i_1<\cdots<
i_m$. The factor $r$ comes very
naturally in the proof, though see Proposition
\ref{Prop:lifted-evaluation} below for a way to avoid this
factor.
\subsubsection{Alternative formalism} \label{Sec:formalism}
There is at least one other way to define Gromov\ddash Witten classes
introduced in \cite{AGV}, and it is necessary to compare
them. In
that paper we defined $$\ocM_{g,n}(\cX,\beta) = \Sigma_1^\cC
\mathop\times\limits_{\cK_{g,n}(\cX,\beta)} \cdots
\mathop\times\limits_{\cK_{g,n}(\cX,\beta)} \Sigma_n^\cC,$$ the
moduli stack of twisted stable maps with sections at the markings. It
has degree $(\prod r_i)^{-1}$ over $\cK_{g,n}(\cX,\beta)$. There are
clearly natural evaluation maps
$$
{e}^\cM_i:\ocM_{g,n}(\cX,\beta) \to \iX
$$
to the \emph{non-rigidified}
cyclotomic inertia stack.
Given $n$ Chow
classes $\widetilde\gamma_i\in A^*(\iX)_\QQ$ on the non-rigidified
cyclotomic inertia stack, we defined classes
$$\langle \widetilde\gamma_1,\ldots,\widetilde\gamma_n, *
\rangle_{g,\beta}^\cX \ \ = \ \
{\check{{e}}}^\cM_{n+1\,*}\left( (\prod r_i) \left(\prod_{i=1}^n
\tilde{e}^{\cM\,*}_i\widetilde\gamma_i\right)\cap
\left[\ocM_{g,n+1}(\cX,\beta)\right]^\vir\right).$$ This formalism is
used in the work of Tseng \cite{Tseng}.
There is another way to write down the same classes without need to
introduce the stack
$\ocM_{g,n}(\cX,\beta)$. Even though liftings $\tilde{e}_i:
\cK_{g,n}(\cX,\beta) \to \iX$ do not necessarily exist, $$\xymatrix{ &
\iX\ar[d]^{\varpi}\\
\cK_{g,n}(\cX,\beta)\ar[r]_{e_i}\ar@{.>}[ru]|{\not\exists\, \tilde e_i} & \riX
}$$
the
isomorphism between the rational Chow groups (or cohomology groups) of
$\iX$ and $\riX$ enables us to move from one to the other on the
intersection theory level.
A lifting $\tilde e_{i\,*}$ of $e_{i\,*}$ is obtained by composing
with the non-multiplicative isomorphism
$$(\varpi_*)^{-1}: A^*(\riX)_\QQ \to A^*(\iX)_\QQ.$$ Note
$(\varpi_*)^{-1} = r \cdot \varpi^*$, so $$\tilde e_{i\,*} =
(\varpi_*)^{-1} \circ e_{i\,*} = r \varpi^*\circ e_{i\,*}.$$
Similarly we define $$\tilde e_i^* = e_i^*\circ (\varpi^*)^{-1}.$$
Since $(\varpi^*)^{-1}= r\cdot \varpi_*$ we can also write $\tilde
e_i^* = r\cdot e_i^*\circ \varpi_*.$ (We remark that the
corresponding formula in \cite{AGV} has $r$ mistakenly replaced by
$r^{-1}$.)
The basic comparison result is
\begin{proposition}\label {Prop:lifted-evaluation}
\begin{enumerate}
\item For Chow classes $\gamma_i\in A^*(\riX)_\QQ$, we have
$$\varpi^*\langle \gamma_1,\ldots,\gamma_n, *
\rangle_{g,\beta}^\cX \ \ = \ \ \langle
\varpi^*\gamma_1,\ldots,\varpi^*\gamma_n, *
\rangle_{g,\beta}^\cX $$
\item For Chow classes $\widetilde\gamma_i\in A^*(\iX)_\QQ$, we have
$$\langle \widetilde\gamma_1,\ldots,\widetilde\gamma_n, *
\rangle_{g,\beta}^\cX \ \ = \ \
{{(\iota\circ \tilde{e}_{n+1})}}_{*}\left( \left(\prod_{i=1}^n
\tilde{e}_i^*\widetilde\gamma_i\right)\cap
\left[\cK_{g,n+1}(\cX,\beta)\right]^\vir\right).$$
\end{enumerate}
\end{proposition}
The first part shows that, if one identifies $A^*(\riX)_\QQ$ and
$A^*(\iX)_\QQ$ \emph{using the multiplicative homomorphism $\varpi^*$},
the Gromov--Witten classes are unchanged. The second part says that,
if one is willing to pretend a lifting $\tilde
e_i:\cK_{g,n}(\cX,\beta) \to \iX$ exists, all the factors of $r$ are
removed from the formalism. While the rigidified inertia stack and evaluation map $
e_i:\cK_{g,n}(\cX,\beta) \to \riX$ is what arises naturally, the
formalism using $\tilde
e_i$ and $\iX$ is probably the most convenient
one to work with, and has been used in the work of Cadman
\cite{Cadman1},\cite{Cadman2}. We will use this formalism in our
example in section \ref{Sec:example}. A direct comparison was carried
out in the example in \cite{cimenotes}.
\begin{proof}
This is immediate using the non-cartesian commutative diagram
\[
\xymatrix{
\save+<0pt,+10pt>*+{\cM_{g,n+1}(\cX,\beta)}\ar_\rho[d]\ar[r]^(0.6){\tilde e^\cM_i}\restore
&\iX\ar^\varpi[d]\\ \cK_{g,n+1}(\cX,\beta)\ar[r]^(0.6){e_i} \ar@{.>}[ru]^{\tilde e_i}
& \riX,
}
\]
where $\deg \rho = (\prod r_i)^{-1}$ and $\deg \varpi = 1/r_i$.
\end{proof}
\subsection{The WDVV equation}
In genus 0 we have the \emph{Witten--Dijkgraaf--Verlinde--Verlinde} (WDVV) equation:
\begin{theorem}\label{Thm:WDVV}
\begin{eqnarray*}
\lefteqn{\sum_{\beta_1+\beta_2 = \beta}\ \ \sum_{A \sqcup B = \{1,\ldots ,n\}}
\left\langle \left\langle \gamma_1, \gamma_2,\delta_A,*\right\rangle_{\beta_1},\gamma_3, \delta_B,*\right\rangle_{\beta_2} = } \\
& &\sum_{\beta_1+\beta_2 = \beta}\ \ \sum_{ A \sqcup B = \{1,\ldots ,n\}}
\left\langle \left\langle \gamma_1, \gamma_3,\delta_A,*\right\rangle_{\beta_1},\gamma_2, \delta_B,*\right\rangle_{\beta_2}
\end{eqnarray*}
\end{theorem}
Note that the two lines differ precisely in the positions of $\gamma_2$ and $\gamma_3$.
\begin{proof}
Consider the stabilization morphism
$$st: \cK_{0,n+4}(\cX,\beta) \to \ocM_{0,4}$$ corresponding to forgetting the map to $\cX$, passing to coarse curves, and forgetting all the marking except the last four.
We show the equality by showing that both sides equal
$$\Psi \ = \ r \ \cdot \ \check e_{n+4\ *}\left( \left (st^*[pt] \cup \prod_{i=1}^n e_i^*\delta_i \cup \prod_{j=1}^3 e_{n+j}^*\gamma_j\right)\cap\left[\cK_{0,n+4}(\cX,\beta)\right]^\vir\right).$$
Write $\hat A = A \sqcup\{n+1,n+2\}$ and $\hat B = B \sqcup\{n+3,n+4\}$
Consider the following diagram:
$$\xymatrix{\coprod \cK_{0,\hat A\sqcup \checkbullet}(\cX, \beta_1) \times_{\riX}\cK_{0,\hat B \sqcup \bullet}(\cX, \beta_2) \ar^(.8)l[r] \ar[d]
& \cK\ar[d]\ar@/^2pc/[dd]^{st}\\
\coprod \twD(\hat A|\hat B)\ar^{gl}[r]& \fM^\tw_{0,n+4}\ar[d]
\\
& \ocM_{0,4}
}$$
\begin{proposition}\label{Prop:divisor}
$$st^*[pt] \cap [\cK]^\vir = l_*\left( e^*_\bullet r \ \cdot \ gl^![\cK]^\vir\right).$$
\end{proposition}
\begin{proof}
We expand the diagram into the following cartesian diagram:
$$\xymatrix{
{}\save+<0pt,-0pt>
*+{\coprod^{\vphantom{\riX}} \cK_{0,\hat A \sqcup \checkbullet}(\cX, \beta_1) \mathop\times\limits_{\riX}\cK_{0,\hat B \sqcup \bullet}(\cX, \beta_2)}
\ar^l[rrr] \ar_\phi[d]
\restore
& & & \cK\ar[d]\ar@/^2pc/[dddd]^{st}\\
\coprod \twD(\hat A|\hat B)\ar^{gl}[rrr]\ar[d] &&& \fM^\tw_{0,n+4}\ar[d]
\\
\twD(12|34)\ar^{i_\fP}[r] &\fP\ar[d]\ar[rr] && \fM^\tw_{0,4}\ar[d]
\\
& \fM_{0,3}\times \fM_{0,3} \ar^(.7){i_\fQ}[r] & \fQ \ar[r]\ar[d] & \fM_{0,4} \ar[d]
\\
&& \{pt\}\ar[r] & \ocM_{0,4}
}$$
\begin{lemma} All the morphisms in
$$\fM^\tw_{0,n+4} \lrar \fM^\tw_{0,4}\lrar \fM_{0,4} \lrar \ocM_{0,4}
$$
are flat.
\end{lemma}
\begin{proof}
The first arrow just forgets the first $n$ markings which is smooth. The second is locally a finite morphism of smooth stacks. The third is a dominant morphism from a smooth stack to a smooth curve.
\end{proof}
\begin{lemma}\label{Lem:gluing-finite}
All horizontal arrows in this diagram are finite and unramified.
\end{lemma}
\begin{proof}
We first show the result for the arrow
$\fM_{0,3}\times \fM_{0,3} \to\fM_{0,4}$. Consider the universal
curve $f_{0,4}:\fC_{0,4} \to \fM_{0,4}$, which is clearly a proper
and representable morphism. There is a closed substack
$\Sing(f_{0,4}) \subset \fC_{0,4}$ consisting of the nodes of the
universal curve, schematically defined by the first Fitting ideal of
$\Omega^{1}_{\fC_{0,4}/\fM_{0,4}}$. Since $\Sing(f_{0,4}) \to
\fM_{0,4}$ is representable, quasi-finite and proper, it is a finite
morphism.
Inside $\Sing(f_{0,4})$ we have an open and closed substack
$\Sigma\subset \Sing(f_{0,4})$ consisting of nodes separating the
marking numbered $1,2$ from the markings numbered $3,4$. We claim
that there is an isomorphism $\Sigma\simeq \fM_{0,3}\times
\fM_{0,3}$ over $\fM_{0,4}$, which proves the required property.
We construct a morphism $\Sigma \to \fM_{0,3}\times \fM_{0,3}$ as
follows. The pullback $\fC_{\Sigma} \to \Sigma$ of the universal
pre-stable curve has a canonical section landing at the appropriate
node. The normalization of $\fC_{\Sigma}$ is the disjoint union of
two families of $3$-pointed curves, obtained by separating the node
marked by the section above. This gives the required morphism.
A morphism in the other direction can be constructed as follows. We
have two universal families $\fC_{1} = \fC_{0,3} \times \fM_{0,3}$
and $\fC_{2} = \fM_{0,3} \times
\fC_{0,3}$. We have a non-cartesian diagram
\[
\xymatrix{
\save+<0pt,+10pt>*+{\fC_{1}\sqcup \fC_{2}}\ar[d]\ar[r]\restore
&\fC_{0,4}\ar[d]\\
\fM_{0,3}\times \fM_{0,3}\ar[r] & \fM_{0,4}
}
\]
where the diagonal arrow is the gluing map of the third marking of $\fC_{1}$ with the first marking of $\fC_{2}$. Composing the third section $ \fM_{0,3}\times \fM_{0,3} \to \fC_{1}$ with this diagonal arrow (or, for that matter, using the first section of $\fM_{0,3}\times \fM_{0,3} \to \fC_{2}$), we get a morphism $ \fM_{0,3}\times \fM_{0,3} \to \fC_{0,4}$, which obviously lands in $\Sigma$. It is a simple local computation to check that the two arrows are inverses of each other.
The arrow $i_\fP$ is the embedding of the reduced substack in $\fP$. All other arrows
arise by base change.
\end{proof}
As a consequence, we can make sense of the following lemma.
\begin{lemma}
$$i_{\fQ\, *} \left[ \fM_{0,3}\times \fM_{0,3}\right] \ \ = \ \ [\fQ]$$
and, using Definition \ref{Def:gothic-r},
$$i_{\fP\, *}\left( \fr \ \cdot \left[ \twD(12|34)\right]\right) \ \ = \ \ [\fP]$$
\end{lemma}
The first statement is well-known---see \cite{Behrend}, Proposition 8. The second follows from the deformation theory of twisted curves, as mentioned in section \ref{Sec:stack-twisted-curves}. Proposition \ref{Prop:divisor} now follows from the lemmas using the projection formula and the observation that $e^*_\bullet r = e^*_{\checkbullet} r = \phi^*\fr$.
\end{proof}
\begin{corollary}
\begin{eqnarray*}
\Psi &=& \ r \ \cdot \ \check e_{n+4\ *}\left( \prod_{i=1}^n e_i^*\delta_i \cup \prod_{j=1}^3 e_{n+j}^*\gamma_j\right)\cap l_*\left( e^*_\bullet r \ \cdot \ gl^![\cK]^\vir\right)\\
&=&\ r \ \cdot \ (\check e_{n+4}\circ l)_*\ \left( e^*_\bullet r \ \cdot \prod_{i=1}^n l^*e_i^*\delta_i \cup \prod_{j=1}^3 l^*e_{n+j}^*\gamma_j\ \cap \ gl^![\cK]^\vir\right)
\end{eqnarray*}
\end{corollary}
To simplify notation, we fix $$\cK_1= \cK_{0,\hat A \sqcup \checkbullet}(\cX, \beta_1) , \ \cK_2= \cK_{0,\hat B \sqcup \bullet}(\cX, \beta_2) ,
\text{ and } \ \cK= \cK_{0,n+4}(\cX, \beta) $$
and $$\eta_1= e_{n+1}^*\gamma_1\cup e_{n+2}^*\gamma_2\cup \prod_{i\in A}e_i^*\delta_i \ \ \text{ and } \ \
\eta_2 = e_{n+3}^*\gamma_3\cup \prod_{i\in B} e_i^*\delta_i .$$
Consider the following diagram.
$$\xymatrix{\cK_1 \times_{\riX} \cK_2 \ar[r]^(.6){p_2}\ar[d]_{p_1}& \cK_2 \ar[r]^{\check e_{n+4}}\ar[d]^{e_\bullet} & \riX\\
\cK_1\ar[r]^{\check e_{\checkbullet}} & \riX}$$
The expression $\left\langle \left\langle \gamma_1, \gamma_2,\delta_A,*\right\rangle_{\beta_1},\gamma_3, \delta_B,*\right\rangle_{\beta_2}$ is by definition
$$r\ \cdot \ \check e_{n+4\, *}\left(\eta_2\cup e_\bullet ^*\left(r\cdot \check e_{\checkbullet\, *}\left(\eta_1\cap[\cK_1]^\vir\right)\right)\cap [\cK_2]^\vir\right)
.$$
By the following lemma, this expression equals
$$r \cdot \check e_{n+4 \,*}(e_\bullet^*r \ \cdot\ p_{2\,*}(p_2^*\eta_2 \cup p_1^*\eta_1 \cap \Delta^! ([\cK_1]^\vir\times[\cK_2]^\vir)).$$
Applying Proposition \ref{Prop:virtual} and summing over $A,B$ and $\beta_1,\beta_2$ gives the Theorem.
\end{proof}
\begin{lemma}\label{Lem:virtual-gysin}
\begin{align*}
\eta_2\cup & e_\bullet ^*\left(\check e_{\checkbullet\, *}\left(\eta_1\cap[\cK_1]^\vir\right)\right)\cap [\cK_2]^\vir\\
&= p_{2\,*}(p_2^*\eta_2 \cup p_1^*\eta_1 \cap \Delta^! ([\cK_1]^\vir\times[\cK_2]^\vir).
\end{align*}
\end{lemma}
\begin{proof}
Set $\xi_{i} = \eta_{i}\cap [\cK_{i}]^{\vir}$; then the left hand side of the equality is
\[
e_\bullet ^*\left(\check e_{\checkbullet\, *}\left(\xi_{1}\right)\right)\cap \xi_{2},
\]
while the right hand side is
\[
p_{2\,*}\Delta^! (\xi_{1}\times\xi_{2}).
\]
Consider the following cartesian diagram
\[
\xymatrix{
\cK_{1}\times_{\riX}\cK_{2} \ar[r]^{\qquad p_{2}}\ar[d] & \cK_{2} \ar[r]^{e_{\bullet}} \ar[d]^{\Gamma_{e_{\bullet}}} & \riX \ar[d]^{\Delta}\\
\cK_{1}\times\cK_{2} \ar[r]^{\check e_{\checkbullet}\times \mathrm{id}\quad} \ar[d]_{\pi_{\cK_{1}}} & \riX\times \cK_{2} \ar[r]^{\mathrm{id} \times e_{\bullet}\quad}\ar[d]^{\pi_{\riX}} & \riX\times \riX \\
\cK_{1}\ar[r]^{\check e_{\checkbullet}} &\riX
}
\]
We have
\begin{align*}
e_\bullet ^*\left(\check e_{\checkbullet\, *}\left(\xi_{1}\right)\right)\cap \xi_{2}
&= \Gamma_{e_\bullet}^* \left( \pi_{\riX}^{*}\check e_{\checkbullet\, *}\xi_{1}\right)\ \cap \ \xi_{2} \\
&= \Gamma_{e_\bullet}^* \left(\check e_{\checkbullet\, *}\xi_{1}\ \times \ \xi_{2}\right) \\
&=\Gamma_{e_\bullet}^* \left( (\check e_{\checkbullet}\times \mathrm{id})_{*}(\xi_{1}\ \times \ \xi_{2})\right) \\
&=p_{2\,*}\Gamma_{e_\bullet}^! \left(\xi_{1}\ \times \ \xi_{2}\right) \\
&=p_{2\,*}\Delta^{!} \left(\xi_{1}\ \times \ \xi_{2}\right).\qedhere
\end{align*}
\end{proof}
\subsection{Topological Gromov--Witten classes}
In this section the base field will be $k = \CC$.
The definition of a Gromov--Witten cohomology class is the same as in Definition~\ref{Def:gromov-witten-classes}:
\begin{definition} Fix integers $g,n$, classes $\gamma_i\in \H^*(\riX)_\QQ$, and a curve class $\beta\in N^{+}(X)$. We define
$$\langle \gamma_1,\ldots,\gamma_n, * \rangle_{g,\beta}^\cX \ \ = \ \
r\ \cdot\ \check e_{n+1\,*}\left( \left(\prod_{i=1}^n e_i^*\gamma_i\right)\cap\left[\cK_{g,n+1}(\cX,\beta)\right]^\vir\right)$$
where by $\left[\cK_{g,n+1}(\cX,\beta)\right]^\vir$ we mean the homology class in $\H_{*}\bigl(K_{g,n+1}(\cX,\beta),\QQ\bigr)$ corresponding to the virtual fundamental class
\[
\left[\cK_{g,n+1}(\cX,\beta)\right]^\vir \in
A_{*}\bigl(K_{g,n+1}(\cX,\beta)\bigr)_\QQ.
\]
\end{definition}
In genus 0 we have again the WDVV equation. To state it precisely we
need a sign convention.
We restrict the discussion to cohomology classes
$\gamma_i$ and $\delta_i$ which are homogeneous with respect to the
usual grading in $\H^*(\riX)$ - in fact homogeneous parity
suffices; the formulas extend to the inhomogeneous case, but are less clean.
For $A\sqcup B = \{1,\ldots ,n\}$ we can write $\delta_A =
\delta_{i_1}\wedge \cdots \wedge \delta_{i_m}$, where
$A = \{i_1,\ldots i_m\}$ subject to the ordering convention $i_1<\cdots<
i_m$, and similarly for $\delta_B$. We define signs
$(-1)^{\epsilon_1(A)}$ and $(-1)^{\epsilon_2(A)}$ so that
$$
\left( \gamma_1 \wedge \gamma_2\wedge
\gamma_3\right)\ \wedge\ \left(\delta_1 \wedge \cdots\wedge
\delta_n\right)\ \ =\ \ (-1)^{\epsilon_1(A)} \left( \gamma_1 \wedge
\gamma_2 \wedge \delta_A\right)\ \wedge\
\left( \gamma_3 \wedge \delta_B\right)
$$
and
$$
\left( \gamma_1 \wedge \gamma_2\wedge
\gamma_3\right)\ \wedge\ \left( \delta_1 \wedge \cdots\wedge
\delta_n\right)\ \ =\ \ (-1)^{\epsilon_2(A)} \left( \gamma_1 \wedge
\gamma_3 \wedge \delta_A\right)\ \wedge\
\left( \gamma_2 \wedge \delta_B\right),
$$
Of course the products could vanish, but the signs are formally well defined depending only on the parity of the classes $\gamma_i$ and $\delta_i$.
With these conventions we have:
\begin{theorem}\label{Thm:WDVV-homology}
\begin{eqnarray*}
\lefteqn{\sum_{\beta_1+\beta_2 = \beta}\ \ \sum_{A \sqcup B = \{1,\ldots ,n\}}
(-1)^{\epsilon_1(A)} \left\langle \left\langle \gamma_1, \gamma_2,\delta_A,*\right\rangle_{\beta_1},\gamma_3, \delta_B,*\right\rangle_{\beta_2} = } \\
& &\sum_{\beta_1+\beta_2 = \beta}\ \ \sum_{A \sqcup B = \{1,\ldots
,n\}} (-1)^{\epsilon_2(A)} \left\langle \left\langle \gamma_1,
\gamma_3,\delta_A,*\right\rangle_{\beta_1},\gamma_2,
\delta_B,*\right\rangle_{\beta_2}.
\end{eqnarray*}
\end{theorem}
The proof is identical to that of WDVV in the algebraic case.
We stress that the degrees used in determining the signs above are the
standard degrees in cohomology, which may be different from those in
the age grading defined in the next section.
\subsection{Gromov--Witten numbers}
The usual definition of Gromov--Witten numbers works without changes in our context.
\begin{definition}
Fix integers $g,n$, classes $\gamma_i\in \H^*(\riX)_\QQ$, and a curve class $\beta\in N^{+}(X)$. We define
$$\langle \gamma_1,\ldots,\gamma_n \rangle_{g,\beta}^\cX \ \ = \ \
\int_{K_{g,n}(\cX,\beta)}\left( \left(\prod_{i=1}^n e_i^*\gamma_i\right)\cap\left[\cK_{g,n}(\cX,\beta)\right]^\vir\right)$$
\end{definition}
The two definitions are connected by the following Proposition.
First, some notation: let $\alpha_{1},\ldots,\alpha_{M}$ be a basis
for the cohomology of $\riX$. We write $g_{ij} = \int_{\riX}
\alpha_{i}\cap \iota^{*}\alpha_{j}$ and denote by $g^{ij}$ the inverse
matrix.
\begin{proposition}\hfil
\begin{enumerate}
\item
$\displaystyle
\langle \gamma_1,\ldots,\gamma_n \rangle_{g,\beta}^\cX =
\int_{\riX}\frac{1}{r}
\langle\gamma_1,\ldots,\gamma_{n-1}, * \rangle_{g,\beta}^\cX
\cap \iota^{*}(\gamma_{n}).
$
\item
$\displaystyle
\langle\gamma_1,\ldots,\gamma_{n-1}, * \rangle_{g,\beta}^\cX\ =\ r\cdot \sum_{i,j=1}^{M} \langle \gamma_1,\ldots,\gamma_{n-1},\alpha_{i} \rangle_{g,\beta}^\cX\ g^{{ij}}\ \alpha_{j}
$
\end{enumerate}
\end{proposition}
\begin{proof}
This is immediate from the projection formula, noting
that $\check e_{n} = \iota \circ e_{n}$.
\end{proof}
As in Section \ref{Sec:formalism} and Proposition
\ref{Prop:lifted-evaluation} we can again make these formulas
appear even more analogous
to the usual manifold case (i.e. we can remove the factors of $r$)
by multiplying the intersection form on $\H^*(\riX[r])$ by $1/r$. We
will denote this modified intersection form by $\tilde g_{ij}$, and correspondingly define $\tilde g^{ij}=r\cdot g^{ij}$. This
is equivalent to pulling back the classes to $\iX$ (or considering
directly classes on $\iX$) and doing the
intersection there.
We can now state the WDVV equation in its
classical form:
\begin{theorem}\label{bigwdvv}
\begin{eqnarray*}
\lefteqn{\sum_{\substack{\beta_1+\beta_2 = \beta \\ A \sqcup B = \{1,\ldots ,n\}}}\sum_{i,j=1}^{M}
(-1)^{\epsilon_1(A)} \left\langle \delta_A,\gamma_1, \gamma_2,\alpha_{i}\right\rangle_{\beta_1} \ \tilde g^{ij}\ \left\langle\alpha_j,\delta_B,\gamma_3, \gamma_{4}\right\rangle_{\beta_2} = } \\
& &\sum_{\substack{\beta_1+\beta_2 = \beta\\ A \sqcup B = \{1,\ldots ,n\}}}\sum_{i,j=1}^{M}
(-1)^{\epsilon_2(A)} \left\langle \delta_A,\gamma_1, \gamma_3,\alpha_{i}\right\rangle_{\beta_1}\ \tilde g^{ij}\ \left\langle\alpha_j, \delta_B,\gamma_2, \gamma_{4}\right\rangle_{\beta_2}.
\end{eqnarray*}
\end{theorem}
\section{The age grading}
\subsection{The age of a sheaf}
Consider the group-scheme $\bmu_{r}$ over a field, with its
representation ring $\mathrm{R}\bmu_{r}$. Each character
$\lambda\colon \bmu_{r} \to \bG_{\mathrm{m}}$ is of the form $t
\mapsto t^{k}$ for a unique integer $k$ with $0 \leq k \leq r-1$;
following M. Reid (see e.g. \cite{Ito-Reid}), we define the \emph{age} of
$\lambda$ as $k/r$. Since these characters form a basis for the
representation ring of $\bmu_{r}$, this extends to a unique additive
homomorphism $\age \colon \mathrm{R} \bmu_{r} \to \QQ$.
Now let $\cG \to T$ be a gerbe banded by $\bmu_{r}$, and let $\cE$ be a locally free sheaf on $\cG$. There is an \'etale covering $\{T_{i}\to T\}$ with sections $T_{i} \to \cG$, inducing an isomorphism $\cG_{T_{i}} \simeq \cB(\bmu_{r})_{T_i}$. Then the pullback of $\cE$ to $\cG_{T_{i}}$ becomes a locally free sheaf $\cE_{T_{i}}$ on $T_{i}$ with an action of $\bmu_{r}$; and the age of each fiber is a locally constant invariant. Furthermore, this invariant is independent of the section, and the age of $\cE$ is a locally constant function on $T$.
Consider a connected scheme $T$, and an object of $\riX(T)$, consisting of a gerbe $\cG \to T$ and a representable morphism $f:\cG \to \cX$. Then the \emph{age} of the object is a rational number defined to be the age of the locally free sheaf $f^{*}\mathrm{T}_{\cX}$. This number only depends on the connected component of $\riX$ containing the image of $T$.
\begin{definition}
The age of a connected component $\Omega$ of $\riX$ is the age of any object of $\Omega(T)$, where $T$ is a connected scheme.
\end{definition}
The age is called the \emph{degree-shifting number} in \cite{Chen-Ruan}.
\subsection{Riemann-Roch for twisted curves}
Let $\cC$ be a balanced twisted curve over an algebraically closed field, $\cE$ a coherent sheaf on $\cC$, that is locally free at the nodes of $\cE$ (for the applications needed in this paper, the case of a locally free sheaf is sufficient: however, the added generality helps with the proof). Call $\pi \colon \cC \to C$ the coarse curve, $p_{1}$, \dots,~$p_{n}$ the marked points on $C$. For each $i$ call $r_{i}$ the index of $\cC$ at $p_{i}$. For each $i$ choose a section $p_{i} \to \Sigma_{i}$ of the marking gerbe $\Sigma_{i} \to p_{i}$, inducing an isomorphism $\Sigma_i \simeq \cB\bmu_{r_i}$. If $\cE$ is locally free at $p_{i}$, we have defined above the age of $\cE$ at $p_{i}$ as $ \age_{p_{i}}(\cE) = \age(\cE\mid_{\Sigma_{i}})$.
In the general case, when $\cE$ has torsion, this is not the correct definition. Consider the embedding
\[
\iota_{i} \colon \cB\bmu_{r_{i}} \cong \Sigma_{i} \into \cC;
\]
it induces a pullback in the K-theory of coherent sheaves of finite projective dimension
\[
\iota_{i}^{*} \colon \mathrm{K}_{0}(\cC) \longrightarrow
\mathrm{K}_{0}(\cB\bmu_{r_{i}}) = \mathrm{R}\bmu_{r_{i}}
\]
via the usual formula
\[
\iota_i^{*}\cE = [\cE \otimes_{\cO_{\cC}}\cO_{\Sigma_{i}}]
- [\tor_{1}^{\cO_{\cC}}(\cE, \cO_{\Sigma_{i}})];
\]
the correct general definition is
\[
\age_{p_{i}}(\cE) = \age(\iota_{i}^{*}\cE).
\]
This gives an additive homomorphism
\[
\age_{p_{i}} \colon \mathrm{K}_{0}(\cC) \longrightarrow \QQ.
\]
By $\chi(\cE)$ we denote as usual the Euler characteristic of $\cE$ on $\cC$. For each $i$ we have $\H^{i}(\cC, \cE) = \H^{i}(C, \pi_{*}\cE)$, so the Euler characteristic is finite. In particular, since $\pi_{*}\cO_{\cC} = \cO_{C}$, we have
\[
\chi(\cO_{\cC}) = 1 - g
\]
where $g$ is the arithmetic genus of $C$.
We define the \emph{degree} of $\cE$ as follows. First assume that $\cC$ is smooth and irreducible. Take an ordinary connected smooth curve $D$ with a finite morphism $\phi \colon D \to \cC$ (it is not hard to see that this exists). Call $d$ the degree of $\phi$. Then we set
\[
\deg_{\cC} \cE = \frac{1}{d}\deg_{D}\phi^{*}\cE.
\]
It is easy to see that the degree is independent of the choice of $\phi$: if $\phi' \colon D' \to \cC$ is another choice, call $D''$ a component of the normalization of the fibered product $D \times_{\cC} D'$. This dominates both $D$ and $D'$, and then one uses standard properties of the degree.
If $\cC$ is not irreducible, take the normalization $\nu \colon
\overline{\cC} \to \cC$, pull back $\cE$, and sum the degrees over
the irreducible components of $\overline{\cC}$.
The degree is a rational number. If $\cE$ is locally free, then it
follows from the definition that the degree of $\cE$ is also the
degree of $\det \cE$, as usual; and we also have the formula
\[
\deg_{\cC} \cE = c_{1}(\cE) \cdot [\cC] = \int_{\cC}c_{1}(\cE) \cdot [\cC].
\]
\begin{theorem} We have
\[
\chi(\cE) = \rk(\cE)\chi(\cO_{\cC}) + \deg \cE
- \sum_{i=1}^{n} \age_{p_{i}}(\cE).
\]
\end{theorem}
It is worth noticing that the stack structure at the nodes does not
intervene in the formula. This is due to the fact that the curve is
balanced.
This theorem can be deduced from To\"en's Riemann-Roch
theorem for stacks, but as it is not too much harder, we give a
direct argument.
\begin{proof}
In the following we will use the fact that Euler characteristic,
rank, degree and age are all additive invariants in
$\mathrm{K}_{0}(\cC)$.
Assume that $\cC$ is smooth.
First of all, assume that there exists a coherent sheaf $\cF$ on $C$
such that $\pi^{*}\cF = \cE$. Then the adjunction homomorphism $\cF
\to \pi_{*}\pi^{*}\cF$ is an isomorphism, because $\pi$ is flat.
Then $\chi(\cE) = \chi(\cF)$, $\deg_{\cC} \cE = \deg_{C}\cF$, and
$\age_{p_{i}}\cE = 0$ for all $i$; hence the formula follows from
ordinary Riemann--Roch applied to $\cF$.
The kernel and cokernel of the adjunction homomorphism
\[
\pi^{*}\pi_{*}\cE \longrightarrow \cE
\]
are torsion, supported at the marked points of $\cC$; hence by additivity it is enough to prove the formula for torsion sheaves supported in the stack locus. Each such sheaf is an extension of sheaves of the form $\iota_{i*}L_{k}$, where $0 \leq k \leq r_{i}-1$, $\iota_{i}\colon \cB\bmu_{r_{i}}$ is the inclusion, and $L_{k}$ is the 1-dimensional representation of $\bmu_{r_{i}}$ defined by the character $\bmu_{r_{i}} \to \GG_{\mathrm{m}}$, $t \mapsto t^{k}$. So it suffices to prove the formula for the sheaves $\iota_{i*}L_{k}$.
It is easy to see that
\[
\deg \iota_{i*}L_{k} = \frac{1}{r_{i}}
\]
for any $k$. Also we see that
\[
\chi(\iota_{i*}L_{k}) =
\begin{cases}
1 & \text{if } k = 0\\
0 & \text{if } k \neq 0.
\end{cases}
\]
To complete the calculation let us compute $\age_{p_{i}} \iota_{i*}L_{k}$. Let $\cI_{\Sigma_{i}}$ be the sheaf of ideals of $\Sigma_{i}$ in $\cC$; by with tensoring $\iota_{i*}L_{k}$ the sequence
\[
0 \longrightarrow \cI_{\Sigma_{i}} \longrightarrow \cO_{\cC}
\longrightarrow \cO_{\Sigma_{i}} \longrightarrow 0
\]
we obtain that
\[
\iota_{i}^{*}\iota_{i*}L_{k} = [L_{k}]
- [\mathrm{T}^{\vee}_{\Sigma_{i}} \otimes L_{k}]
\]
where $\mathrm{T}^{\vee}_{\Sigma_{i}}$ is the cotangent space to $\Sigma_{i}$ in $\cC$ (this is a very particular case of the self-intersection formula in K-theory). But by definition of the isomorphism $\Sigma_{i} \cong \cB\bmu_{r_{i}}$, the tangent space to $\Sigma_{i}$ is $L_{1}$, hence $\mathrm{T}^{\vee}_{\Sigma_{i}}$ is $L_{r_{i}-1}$. From this we obtain the remarkable formula
\[
\age_{p_{i}} \iota_{i*}L_{k} =
\begin{cases}\displaystyle
-\frac{r_{i}-1}{r_{i}} & \text{if } k=0\\\displaystyle
\frac{1}{r_{i}} & \text{if } k\neq 0.
\end{cases}
\]
Now we plug in the values of the invariants and check the formula for the sheaves $\iota_{i*}L_{k}$. This completes the proof when $\cC$ is smooth.
In the general case, let $q_{1}$, \dots,~$q_{s}$ be the nodes of $C$. Call $l_{i}$ the index of $\cC$ at the node $q_{i}$; call $\Theta_{i}$ the residual gerbe of $\cC$ at the unique point living over $q_{i}$. Consider the normalization $\nu\colon \overline{\cC} \to \cC$; the moduli space $\overline{C}$ of $\overline{\cC}$ is the normalization of $C$. Call $q'_{i}$ and $q''_{i}$ the two inverse images of $q_{i}$ in $\overline{C}$, and call $\Theta'_{i}$ and $\Theta''_{i}$ the residual gerbe of $\cC$ over the unique point over $q'_{i}$ and $q''_{i}$ respectively. The natural morphisms $\Theta_{i} \to \Theta'_{i}$ and $\Theta_{i} \to \Theta''_{i}$ are isomorphisms of gerbes. Furthermore, $\Theta'_{i}$ and $\Theta''_{i}$ are isomorphic to $\cB\bmu_{l_{i}}$, where the isomorphisms are chosen so that the action of $\bmu_{l_{i}}$ on the tangent space to $\cC$ at the gerbe is given by the embedding $\bmu_{l_{i}} \subseteq \GG_{\mathrm{m}}$. We give $\overline{\cC}$ the structure of a twisted curve by using these gerbes as a marking.
The fact that $\cC$ is balanced has the following important consequence: the two bandings by $\bmu_{l_{i}}$ on $\Theta_{i}$ that we obtain via the two isomorphisms $\Theta_{i} \cong \Theta'_{i}$ and $\Theta_{i} \cong \Theta''_{i}$ are opposite. Consider the pullback $\nu^{*}\cE$; its restrictions to $\Theta'_{i}$ and $\Theta''_{i}$ are isomorphic to the restriction of $\cE$ to $\Theta_{i}$, and they give dual representations of $\bmu_{l_{i}}$. This implies that
\[
\age_{q'_{i}} \nu^{*}\cE + \age_{q''_{i}} \nu^{*}\cE = c_{i},
\]
where $c_{i}$ is the codimension of the space of invariants of the restriction of $\nu^{*}\cE$ to $\Theta'_{i}$ or $\Theta''_{i}$.
We have an exact sequence
\[
0 \longrightarrow \cO_{\cC} \longrightarrow \nu_{*}\cO_{\overline{\cC}}
\longrightarrow \bigoplus_{i=1}^{s}\cO_{\Theta_{i}} \longrightarrow 0
\]
from which we deduce that
\[
\chi(\cO_{\overline{\cC}}) = \chi(\nu_{*}\cO_{\overline{\cC}})
= \chi(\cO_{\cC}) + s.
\]
By tensoring with $\cE$, and keeping in mind that $\cE$ is locally free at the nodes of $\cC$, we get a sequence
\[
0 \longrightarrow \cE \longrightarrow \nu_{*}\nu^{*}\cE
\longrightarrow \bigoplus_{i=1}^{s}(\cE \otimes_{\cO_{\cC}} \cO_{\Theta_{i}})
\longrightarrow 0.
\]
By taking invariants, and then using Riemann--Roch on the smooth twisted curve $\overline{\cC}$ together with the formulas above, we get
\begin{align*}
\chi(\cE) &= \chi(\nu_{*}\nu^{*}\cE)
- \sum_{i=1}^{s} \dim_{k}\H^{0}(\cE \otimes_{\cO_{\cC}} \cO_{\Theta_{i}})\\
&= \chi(\nu^{*}\cE)
- \sum_{i=1}^{s}(\rk\cE - c_{i})\\
&= \rk(\cE)\chi(\cO_{\overline{\cC}}) + \deg(\nu^{*}\cE)
- \sum_{i=1}^{n}\age_{p_{i}}(\cE)
- \sum_{i=1}^{s}(\age_{q'_{i}} \nu^{*}\cE + \age_{q''_{i}} \nu^{*}\cE)\\
&\qquad- \sum_{i=1}^{s}(\rk\cE -
\age_{q'_{i}} \nu^{*}\cE - \age_{q''_{i}} \nu^{*}\cE)\\
&= \rk(\cE)\bigl(\chi(\cO_{\cC}) + s\bigr) + \deg(\cE)
- \sum_{i=1}^{n}\age_{p_{i}}(\cE) - s \rk \cE\\
&= \rk(\cE)\chi(\cO_{\cC}) + \deg(\cE) - \sum_{i=1}^{n}\age_{p_{i}}(\cE).
\end{align*}
This concludes the proof.\end{proof}
\subsection{The \stringy Chow group and its grading}
Let $\cX$ be a smooth Deligne--Mumford stack as in the introduction. We define the rational \stringy Chow group
of $\cX$ to be $$\stChow{\cX}: = A^*(\riX)_\QQ.$$ We make this into a graded group using the following rule (see \cite{Chen-Ruan}, \cite{DHVW}, \cite{Zaslow}):
$$\stChow[a]{\cX} = \oplus_\Omega A^{a-\age(\Omega)}(\Omega)_\QQ, $$
where the sum is taken over all connected components $\Omega$ of
$\riX$.
\subsection{The small quantum Chow ring}
The small quantum Chow ring is an algebra over the completed monoid-algebra $\QQ\llbracket N^+(X) \rrbracket$, where we denote the monomial corresponding to a class $\beta$ by $q^\beta$. As a group, the quantum Chow ring is $$\QA^* (\cX) : = \stChow{\cX} \llbracket N^+(X)\rrbracket.$$
We define a product on $\QA^* (\cX)$ by specifying the product of monomials, as follows:
$$ \gamma_1 * \gamma_2 = \sum_{\beta\in N^+(X)} \left\langle \gamma_1,\gamma_2,*\right\rangle_{0,\beta}\, q^\beta.$$
We define the \emph{degree} of $q^{\beta}$ to be
$\beta \cdot c_{1}(\cT_{\cX})$.
\begin{theorem}
The product defined above makes $\QA^* (\cX)$ into a commutative, associative ``pro-$\QQ$-graded'' ring, in the sense that the product of two homogeneous elements of degrees $a,b$ is homogeneous of degree $a+b$.
\end{theorem}
\begin{proof}
Commutativity is immediate. Associativity is, as usual, a consequence of Theorem \ref{Thm:WDVV}. It remains to check the claim about grading.
Consider the summand $ \left\langle \gamma_1,\gamma_2,*\right\rangle_{0,\beta}\, q^\beta$ in the formula above, where we assume that $\gamma_{1}$ and $\gamma_{2}$ are each supported on a single component $\Omega_{1},$ respectively $\Omega_{2}\subset\riX$, of corresponding ages $a_{1}, a_{2}$. We need to show that it has degree $$\deg \gamma_{1}+\deg \gamma_{2} = (\codim_{\Omega_{1}} \gamma_{1} +{a_{1}}) + (\codim_{\Omega_{2}} \gamma_{2} +{a_{2}}).$$ Similarly, it is enough to calculate the degree of a single term
\[
\left\langle \gamma_1,\gamma_2,*\right\rangle_{0,\beta,\Omega_{3}}\, q^\beta
\]
of $ \left\langle \gamma_1,\gamma_2,*\right\rangle_{0,\beta}\, q^\beta$ lying in the Chow group of a component $\Omega_{3}\subset \riX$ having age $\check a_{3}$.
Given a stable map $f: \cC \to \cX$ corresponding to a geometric point of a component $\cK\subset \cK_{0,3}(\cX,\beta)$ with evaluations $e_{1}:\cK \to \Omega_{1}$, $e_{2}:\cK \to \Omega_{2}$ and $\check e_{3}:\cK \to \Omega_{3}$, the bundle $f^{*}\cT_{\cX}$ has ages $a_1, a_2$ and $a_{3}$ at the three markings, with $$a_{3} + \check a_{3} = \dim \cX - \dim \Omega_{3}.$$ This is explained in \cite{Chen-Ruan}, the point being that $\dim_{\xi,\alpha} \Omega_{3} = \mathrm{rank} (\cT_{\cX}^{\mu_{r(\Omega_{3})}}) = \dim \cX - (a_{3}+ \check a_{3})$, as noted in the last section. We denote the class of $f_{*}\cC$ by $\beta'$, and clearly we have $\pi_{*}\beta' = \beta$.
We can now calculate dimensions at $f:\cC \to \cX$ using Riemann-Roch: first, the dimension of $[\cK]^{\vir}$ is given by
$$ \chi (f^{*}\cT_{\cX}) = c_{1}(\cT_{\cX})\cdot \beta' + \dim \cX - a_{1}-a_{2}-a_{3}.$$
Therefore we have
\begin{align*}
\dim((e_{1}^{*} \gamma_{1}\cup e_{2}^{*} \gamma_{2})\cap [\cK]^{\vir})
&= \deg q^{\beta} + \dim \cX -a_{1}-a_{2}-a_{3}\\
&\phantom{= \deg q^{\beta} + \dim \cX}- \codim_{\Omega_{1}} \gamma_{1} - \codim_{\Omega_{2}} \gamma_{2} \\
&= \deg q^{\beta} + \dim \cX -\deg \gamma_{1}- \deg \gamma_{2} - a_{3}.
\end{align*}
Pushing forward by $\check e_{3}$ we obtain
\begin{align*}
\codim &\left(\check e_{3\,*} ((e_{1}^{*} \gamma_{1}\cup e_{2}^{*}) \gamma_{2})\cap [\cK]^{\vir})\right) \\
&= \dim\Omega_{3} - (\deg q^{\beta} + \dim \cX -\deg \gamma_{1}- \deg \gamma_{2} - a_{3} ) \\
&= \deg \gamma_{1}+ \deg \gamma_{2} + (a_{3} +\dim \Omega_{3} - \dim \cX) - \deg q^{\beta} \\
&= \deg \gamma_{1}+ \deg \gamma_{2} - \check a_{3}- \deg q^{\beta}
\end{align*}
and therefore
\begin{align*}
\deg \left(\check e_{3\,*} ((e_{1}^{*} \gamma_{1}\cup e_{2}^{*}) \gamma_{2})\cap [\cK]^{\vir})\right) = \deg \gamma_{1}+ \deg \gamma_{2} - \deg q^{\beta}.
\end{align*}
It follows that
\begin{align*}
\deg \left(\check e_{3\,*} ((e_{1}^{*} \gamma_{1}\cup e_{2}^{*}) \gamma_{2})\cap [\cK]^{\vir})q^{\beta}\right) = \deg \gamma_{1}+ \deg \gamma_{2},
\end{align*}
as required.
\end{proof}
We note that, while the grading has rational degrees, the
denominators which appear are bounded. The fact that the $q^{\beta}$
have degrees with bounded denominators follows from the fact that the
ring is finitely generated, and more explicitly through
Proposition~\ref{Prop:bound-denominators}; and the denominators in
the ages of a connected component are bounded by the exponent of the
automorphism group of a geometric point of $\cX$.
An identical construction using the topological Gromov-Witten classes, gives us a definition
of the small quantum cohomology ring $\QH^*(\cX)$. This is an interesting ring
structure on $\H^*(\riX)\llbracket N(X)^+\rrbracket$. Its associativity follows from Theorem \ref{Thm:WDVV-homology}.
\subsection{Stringy Chow ring}
By setting the $q$'s to zero in $\QA^*(\cX)$, we get an interesting product structure on $\stChow{\cX}$ which we
refer to as the stringy Chow ring. This is the Chow analogue of the
\emph{orbifold cohomology} or \emph{Chen-Ruan cohomology ring}, which arises by doing exactly the same thing to the small quantum cohomology ring of $\cX$. In \cite{AGV} we show that it is possible to define these products with integral coefficients, provided one works with $A^*(\iX)$ instead of $A^*(\riX)$.
\subsection{The big quantum cohomology ring}
As in the Gromov-Witten theory of a manifold, the full topological WDVV equation,
Theorem \ref{bigwdvv}, is
essentially equivalent to the associativity of a big quantum cohomology ring
whose multiplication is defined in terms of the basis $\{
\alpha_i\}$ of $\H^*(\riX)$ as
$$\mu * \nu\ \ =\ \ \sum_{ij} \sum_n \ \frac{1}{n!}\ \big\langle \mu, \nu, T^n, \alpha_i \big\rangle_\beta\ q^\beta\
\tilde g^{ij}\ \alpha_j .$$ Here $T=\sum \alpha_i x_i$ should be
expanded formally, so that the resulting product is a power series
in the $x_i$ and $q_i$ whose coefficients record every genus zero
Gromov-Witten invariant including the classes $\mu$ and $\nu$. The same arguments as in the previous subsection show
that this gives rise to a pro-$\QQ$-graded, associative, commutative ring structure on
$\H^*(\riX)\llbracket N^+(X)\rrbracket\llbracket x_1,\ldots , x_M\rrbracket$.
\section{A few useful facts}
We collect here some useful facts analogous to standard properties of Gromov-Witten
invariants for manifolds. While this will not constitute an exhaustive list, we hope that readers will see how standard facts and techniques from Gromov-Witten theory carry over to this context. Most sources of difference come from the need to systematically replace $\cX$ with $\riX$ at various points in formulating the theory, or from a difference in the relationship between the spaces of twisted stable maps and their universal curves, which we explain now.
\subsection{Universal curve}
A critical fact in ordinary Gromov-Witten theory is that $\ocM_{g,n+1}(X,\beta)$ is the
universal curve over $\ocM_{g,n}(X,\beta)$. While this is not true for the orbifold theory,
we have a similar result.
\begin{proposition}\label{universalcurve} The
universal curve over $\cK_{g,n}(\cX,\beta)$ is naturally identified
with the open and closed substack $\cU \subseteq \cK_{g,n+1}(\cX,\beta)$ for
which the $(n+1)$st marked point is untwisted. Moreover, if we
consider the flat morphism $\pi: \cU \to \cK_{g,n}(\cX,\beta)$, then
we have an equality of virtual fundamental classes $[\cU]^\vir =
\pi^*[\cK_{g,n}(\cX,\beta)]^\vir$, where $[\cU]^\vir$ denotes the
restriction of $[\cK_{g,n+1}(\cX,\beta)]^\vir$ to $\cU$.
\end{proposition}
Another way to say this, which is useful in practice, is that the universal curve $\cU$
fits in the cartesian square:
$$\xymatrix{\cU\ar[r]\ar[d]& \cX\ar[d]^i\\
\cK_{g,n+1}(\cX,\beta)\ar[r] & \riX}
$$
where $i$ denotes the inclusion of $\cX\cong \riX[1]$ into the inertia stack.
\begin{proof}
The identification of the universal curve is essentially Corollary 9.1.3 in \cite{AV}. The identification of the virtual classes follows by the same reasoning as the analogous statement in \cite{Behrend}.
\end{proof}
\subsection{Degree zero invariants}
The identification of the $(n+1)$-pointed space with the universal curve in ordinary Gromov-Witten theory implies that the degree zero invariants carry very little information. In particular, it implies that degree zero, genus zero invariants all vanish with the exception of the three point invariants, which simply compute the classical cohomology ring. This is not at all true for the orbifold theory, which is precisely why the ring structure on orbifold cohomology is interesting.
However, one consequence of this fact does continue to hold. If we let $1$ denote the fundamental class of the identity component of $\riX$, then we have the following fact.
\begin{proposition} The Gromov-Witten number $\langle 1, \delta_1, \ldots , \delta_n \rangle_{g,\beta}$
vanishes unless $g=0$ and $n=2$, in which case we have
$$\langle 1, \delta_1, \delta_2 \rangle_{0,0}= \int_{\riX} \frac{1}{r} \delta_1 \cup \iota^*\delta_2.$$
\end{proposition}
\begin{proof}
Given Proposition \ref{universalcurve}, the vanishing portion follows from the same arguments as in ordinary Gromov-Witten theory (see \cite{KM}). The precise formula comes from the easy fact that $e_1^{-1}(\riX[1])$ in $\cK_{0,3}(\cX,0)$ is naturally isomorphic to $\iX$ and we can identify $e_2$ with the standard rigidification map. Using this identification, $e_3$ is then identified with the composition of the rigidification map with $\iota$ and the result follows.
\end{proof}
This has the immediate consequence that $1$ is the identity element in both the small and big quantum cohomology rings. (A completely analogous lemma shows that 1 is the identity in the quantum Chow ring.)
\subsection{Gravitational descendants}
In studying higher genus Gromov\ddash Witten theory it is important to include the descendant classes. These are analogues
of the Mumford--Morita--Miller classes on $\overline{M}_{g,n}$ involving the Chern
classes of the normal bundles of the $n$ sections.
One can define these classes in the orbifold theory in much the
same way as in the ordinary theory. On $\cK_{g,n}(\cX,\beta)$, one defines
$n$ tautological line bundles, $\cL_i$. There is more than one
way to define these, but the most straightforward is to take $\cL_i$ to
be the bundle whose fiber at a point is the cotangent space to the
corresponding \emph{coarse} curve at the $i$th marked point, in other words, the pullback via the $i$-th section on the universal \emph{coarse} curve of the sheaf of relative differentials. We
remark that while one might like to use the cotangent space to the
twisted curve, these will not form a line bundle on
$\cK_{g,n}(\cX,\beta)$, but only on the gerbe corresponding to the
$i$th marking. However, one could decide to push down the Chern
class of that ``twisted" bundle, which would simply be $1/r$ times the
Chern class of the $\cL_i$ as we are defining them here. We will not
use that convention.
Let $\psi_i = c_1(\cL_i)$. Then given classes $\delta_1, \ldots
\delta_n$ in $\H^*(\riX)$, we define the following invariants:
$$\langle \, \tau_{a_1}(\delta_1) \cdots
\tau_{a_n}(\delta_n)\, \rangle_{g,\beta} =
\int_{\cK_{g,n}(\cX,\beta)]^\vir} e_1^*(\delta_1) \cup \psi_1^{a_1}
\cup \cdots \cup e_n^*(\delta_n)\cup \psi_n^{a_n}.$$
For simplicity of statements we define any class involving $\tau_{-1}$ in the formulas below to be 0.
With these definitions, the standard equations among descendants for manifolds (cf. \cite[Section 1.2]{Pandharipande}) hold in the orbifold setting with no changes (keeping in mind that 1 means the fundamental class of the identity component of the inertia stack).
\begin{theorem}
Assume $(\beta,g,n)$ is not any of $\beta=0, g=0,
n<3$ or $\beta=0,g=1,n=0$. Then
\begin{enumerate}
\item
(Puncture or String Equation)
\begin{align*}
\langle\, \tau_{a_1}(\gamma_1)& \cdots \tau_{a_n}(\gamma_n)\tau_0(1)\,\rangle_{g,\beta} \\&=
\sum_{i=1}^n \langle\, \tau_1(\gamma_1) \cdots \tau_{a_{i-1}}(\gamma_
{i-1})\tau_{a_i-1}(\gamma_i)
\tau_{a_{i+1}}(\gamma_{i+1}) \cdots \tau_{a_n}(\gamma_n)\,\rangle_{g,\beta}
\end{align*}
\medskip
\item (Dilaton Equation)
$$
\langle\, \tau_{a_1}(\gamma_1) \cdots \tau_{a_n}(\gamma_n)\tau_1(1)\,\rangle_{g,\beta} =
(2g-2+n)\langle\,\tau_{a_1}(\gamma_1) \cdots \tau_{a_n}(\gamma_n)\,\rangle_{g,\beta}
$$
\medskip
\item (Divisor Equation) For $\gamma$ in $\H^2(X) \subset \H^2_{\mathrm{orb}}(X)$ (but not for an
arbitrary element of $\H^2_{\mathrm{orb}}(X)$), we have
\begin{align*}
\langle\, \tau_{a_1}(\gamma_1)&\cdots
\tau_{a_n}(\gamma_n)\tau_0(\gamma)\,{\rangle}_{g,\beta}
\\=&
\left(\int_\beta \gamma\right) \cdot \langle \, \tau_{a_1}(\gamma_1) \cdots \tau_
{a_n}(\gamma_n)\,\rangle_{g,\beta}
\\&+ \sum_{i=1}^n \langle\, \tau_1(\gamma_1) \cdots
\tau_{a_{i-1}}(\gamma_{i-1})\tau_{a_i-1}(\gamma_i\cup \gamma)
\tau_{a_{i+1}}(\gamma_{i+1}) \cdots \tau_{a_n}(\gamma_n)\,\rangle_{g,\beta}
\end{align*}
\end{enumerate}
\end{theorem}
\begin{proof}
We reduce to the untwisted case. In all these equations
the intersection happens on the open and closed substack $\cU \subset
\cK_{g,n+1}(\cX, \beta)$ of Proposition \ref{universalcurve}. The key
commutative diagram is the following:
$$\xymatrix{
\cU \ar[r]^{\mu_{n+1}} \ar[d]_{\pi_\cK}
& \ocM_{g,n+1}(X, \beta)\ar[d]^{\pi_\cM} \\
\cK_{g,n}(\cX, \beta)\ar[r]^{\mu_{n}}
& \ocM_{g,n}(X, \beta)
}$$
This is not quite a fiber diagram, but $\cU$ has the same moduli space
as the fibered product, so the projection formula still holds. If
$\alpha$ is a cohomology class on $\cK_{g,n}(\cX, \beta)$ and $\psi$
is a cohomology class on $\ocM_{g,n+1}(X, \beta)$, we have by
proposition \ref{universalcurve}
$$ \int_{[\cU]^\vir} \pi_\cK^*\alpha\cup\mu_{n+1}^*\psi \ \ = \ \
\int_{[\cK_{g,n}(\cX, \beta)]^\vir} \alpha \cup
\mu_n^*\pi_{\cM\,*}\psi.$$ So any identity which holds for
$\pi_{\cM\,*}\psi$ (keeping in mind that $X$ may be singular) can be
used here.
For all the equations use $\alpha = \prod_{i=1}^n e_i^*\gamma_i$. For
the Puncture Equation use $\psi = \prod_{i=1}^{n} \psi_i^{a_i}$ and
the equation
$\pi_{\cM\,*}\psi = \sum_{i=1}^n(\prod_{i=1}^{n}
\psi_i^{a_i})/\psi_i$. For the Dilaton Equation use $\psi =
(\prod_{i=1}^{n} \psi_i^{a_i})\psi_{n+1}$ with $\pi_{\cM\,*}\psi =
(2g-2+n) \prod_{i=1}^{n} \psi_i^{a_i}$. For the Divisor equation
use
$\psi = (\prod_{i=1}^{n} \psi_i^{a_i}) \cup
e_{\cM_{n+1},n+1}^*\gamma$, with the equation $\pi_{\cM\,*}\psi =
\int_\beta \gamma \cdot \prod_{i=1}^{n} \psi_i^{a_i} + \sum_{i=1}^n
(\prod_{i=1}^{n}\psi_i^{a_i})/\psi_i \cap e_{\cM_{n},i}^*\gamma,$
noticing that $e_i^*\gamma_i \cup u_n^* e_{\cM, i}^* \gamma =
e_i^*(\gamma_i\cup \gamma)$.
\end{proof}
\begin{remark} There is also a Topological Recursion Relation valid in this context, treated in \cite{Tseng}, section 2.5.5.
\end{remark}
\section{An example: the weighted projective line}\label{Sec:example}
We conclude by giving a nontrivial
calculation of the quantum Chow
ring of a stack.
We consider one of the simplest possible classes of examples -- the
weighted projective lines.
We fix two positive integers, $a$ and $b$, and consider the one
dimensional weighted projective space $\cX=\proj(a,b)$ which is the
stack quotient of a punctured two dimensional affine space by the
action of $\GG_m$ with weights $a$ and $b$. The coarse moduli space
of $\proj(a,b)$ is always $\proj^1$. If $a$ and $b$ are relatively
prime, then $\proj(a,b)$ is a twisted curve. Otherwise this stack
has a generic stabilizer. We will denote by $0$, the point with
stabilizer $\mu_a$ and by $\infty$ the point with stabilizer
$\mu_b$.
For the convenience of the reader, we collect the basic facts about
this stack here. A morphism from a scheme $Z$ to $\proj(a,b)$ is
given by choosing a line bundle $L$ on $Z$ together with sections
$s_1 \in \Gamma(Z,L^{\otimes a})$ and $s_2 \in \Gamma(Z,L^{\otimes
b})$ with no common zeroes. 2-morphisms are given by morphisms
between line bundles which take the sections to the sections. Note
that if $a=b=1$ we get the usual description of $\proj^1$ and there
are no nontrivial 2-automorphisms. By descent, we get the same
description of maps from a stack to $\proj(a,b)$. We let $\oh(1)$ denote the
line bundle on $\proj(a,b)$ corresponding to the identity
morphism. Then $\Pic
(\proj(a,b)) = \ZZ \oh(1)$ and there are sections of $\oh(a)$ and
$\oh(b)$ vanishing at $\infty$ and 0 respectively. The degree of
$\oh(1)$ is $\frac{1}{ab}$. Finally, $T\proj(a,b) \cong \oh(a+b)$.
The inertia stack of this stack is straightforward to describe. Note
that since $\GG_m$ is an abelian group, the quotient presentation of
$\cX$ endows the inertia group of each point of $\proj(a,b)$ with an
embedding in $\GG_m$. (This character of the isotropy group is also
its action on the fiber of $\oh(1)$.) Because of this, each
irreducible component of $I_{\mu_r}\cX$ is canonically associated with
the unit in $\ZZ/r\ZZ$ which relates that fixed embedding to the one
obtained by composition with the embedding of $\mu_r$ in $\GG_m$.
Below we will use the obvious convention which identifies the set of
elements of $\ZZ/d\ZZ$ with the set of units in $\ZZ/k\ZZ$ for all
positive integers $k$ dividing $d$.
Let $d=gcd(a,b)$. For each element of $\ZZ/d$, there is a one
dimensional component of $I_\mu\cX$ which is isomorphic to
$\proj(a,b)$. For each element of $\ZZ/a\ZZ$ which is not divisible
by $a/d$, there is a zero dimensional component of $I_\mu(\cX)$
lying over the point $0$. Each of these components is isomorphic to
$\cB\mu_a$. Similarly, for each element of $\ZZ/b\ZZ$ not divisible by
$b/d$ there is a component lying over the point $\infty$ which is
isomorphic to $\cB\mu_b$. We hope that the following picture of the
inertia stack of $\proj(4,6) \cong \ocM_{1,1}$ will make this labeling system clear:
\begin{center}
\setlength{\unitlength}{1mm}
\begin{picture}(80,55)(0,0)
\put( 5,14){1}
\put(10,15){\blacken\ellipse{1.5}{1.5}}
\put(15,14){(age 6/12)}
\put( 7,17){``$x$''}
\put( 5,44){3}
\put(10,45){\blacken\ellipse{1.5}{1.5}}
\put(15,44){(age 6/12)}
\put(50, 9){(age 8/12)}
\put(70,10){\blacken\ellipse{1.5}{1.5}}
\put(75, 9){1}
\put(50,19){(age 4/12)}
\put(70,20){\blacken\ellipse{1.5}{1.5}}
\put(75,19){2}
\put(50,39){(age 8/12)}
\put(70,40){\blacken\ellipse{1.5}{1.5}}
\put(75,39){4}
\put(50,49){(age 4/12)}
\put(70,50){\blacken\ellipse{1.5}{1.5}}
\put(75,49){5}
\put(67,52){``$y$''}
\path(5,0)(75,0)
\put(10,0){\blacken\ellipse{1.5}{1.5}}
\put(70,0){\blacken\ellipse{1.5}{1.5}}
\put(0,-1){0}
\path(5,30)(75,30)
\put(10,30){\blacken\ellipse{1.5}{1.5}}
\put(70,30){\blacken\ellipse{1.5}{1.5}}
\put(0,29){1}
\put(21,32){``$\zeta$''}
\end{picture}
\end{center}
To give a presentation of the quantum Chow ring we need to choose
generators. A convenient way to
make this choice is as follows. Choose integers $m$ and $n$ such
that $ma+nb=d$. Set $A=a/d$ and $B=b/d$. We take $\zeta$ to be the
fundamental class of the one dimensional component of $\riX$ corresponding to
$1\in \ZZ/d\ZZ$ (if $d=1$, take $\zeta=1$), we let $x$ be the fundamental class of the
component lying over $0$ which corresponds to $n\in \ZZ/a\ZZ$, and
we let $y$ be the fundamental class of the component lying over
$\infty$ corresponding to $m\in \ZZ/b\ZZ$. In the example of
$\PP(4,6)$ above, we chose $n=1, m=-1$ and indicated the components where the resulting
$x,y$ as well as $\zeta$ serve as fundamental classes. (If $d=a$ or $d=b$ some
of these zero dimensional components don't exist. We will ignore
this case in what follows, but the results all hold with essentially
identical proofs if we take $x$ or $y$ to be the fundamental class
of the appropriate zero dimensional substack of the inertia stack
associated with $m$ or $n$ in $\ZZ/d\ZZ$.) One consequence of
choosing $x$ and $y$ in this manner is that they have minimal age.
Under this convention, we find that $\deg(x)=1/A$ and $\deg(y)=1/B$,
while $\deg(\zeta)=0$.
Following reasoning similar to that at the end of \cite{AGV} it is
easy to calculate that the stringy Chow ring of $\proj(a,b)$ is
$\QQ[\zeta,x,y]\, /\, \langle xy, Ax^A-By^B\zeta^{n-m}, \zeta^d-1\rangle$. (Note that the factor $\zeta^{n-m}$ in the second relation is missing in \cite{AV-families}.)
The N\'eron-Severi group of $\proj(a,b)$ has rank one, so to compute
the quantum Chow ring, we need to introduce one further generator,
$q$. We normalize this by selecting our generator of
$N(\proj(a,b))$ to correspond to the minimal positive degree map
from a twisted curve to $\proj(a,b)$. If $gcd(a,b)=1$, then this
minimal degree map can be taken to be the identity map, if not, then
the presence of a generic stabilizer of order $d$ forces any map
from a twisted curve to have degree divisible by $d$, we will see
that there does exist a map of degree exactly $d$. In other words,
we take the generator of $N(\proj(a,b))$ to be $d$ times the
fundamental class.
Since we know that $\QH^*(\cX)$ is free as a $\QQ\llbracket q\rrbracket$ module, it
follows that there is a presentation for $\QH^*(\cX)$ of the form
$$\QQ\llbracket q\rrbracket[\zeta,x,y]\ \big/\ \langle R_1, R_2, R_3 \rangle$$
where we have
$$R_1 \equiv xy \mod q$$
$$R_2 \equiv Ax^A-By^B\zeta^{n-m} \mod q$$
$$R_3 \equiv \zeta^d-1 \mod q.$$
Since the degree of $q$ is $1/A + 1/B$, no monomial in the
generators containing a $q$ can possibly have degree equal to the
degrees of $R_2$ or $R_3$. Hence all that remains to compute is the
quantum product of $x$ with $y$. By degree considerations, the only
possibility for the form of $R_1$ is then
$$xy=q(c_0+c_1\zeta + \cdots +c_{d-1}\zeta^{d-1})$$ where the $c_i$
are rational numbers. We will establish that $c_0=1$ and that the
other $c_i$ all vanish. As $c_0$ is the coefficient of the
fundamental class in $x*y$, it is determined by considering the
moduli space of maps from a 3 pointed stacky $\proj^1$ which
actually has only two stacky points of indices $a$ and $b$. We let
$C_{a,b}$ denote this curve. $\Pic(C_{a,b})$ is generated by line
bundles $L_0$ and $L_\infty$ of degrees $\frac 1a$ and $\frac 1b$
satisfying the single relation $L_0^{\otimes a} \cong
L_\infty^{\otimes b}$. In particular, when $a$ and $b$ are not
relatively prime, so that $C_{a,b} \not\cong \proj(a,b)$ we find
that there is torsion in the Picard group. Also, the restriction map
$r_0: \Pic (C_{a,b}) \to \Pic(\cB\mu_a)$ takes $L_0$ to the standard
generator. (A possibly confusing point here is that even when $a$
and $b$ are relatively prime so that $C_{a,b}\cong \proj(a,b)$, our
identification of the isotropy groups of the substacks
supported at 0 with $\mu_a$ are different, since in $\proj(a,b)$ we
are using the action of $\mu_a$ on the fiber of $\oh(1)$ as the
standard representation, whereas on $C_{a,b}$ we are using the
action on the tangent space.)
We are looking for a map of degree $d$ from $C_{a,b}$ to
$\proj(a,b)$. Since we can compute the degree of a morphism $f:
C_{a,b} \to \proj(a,b)$ by comparing the degree of $\oh(1)$ to the
degree of $f^*\oh(1)$ we see that we need to find a line bundle $L$
of degree $\frac{d}{ab}$ such that $L^{\otimes a}$ has a section
vanishing at $\infty$ and $L^{\otimes b}$ has a section vanishing at
$0$. If we denote this line bundle $L= L_0^{\otimes z_0} \otimes
L_\infty^{\otimes z_\infty}$ then in order for the first point to
evaluate to the correct component of the inertia stack, we need that
$z_0 \equiv n \mod a$ and for the second point we have the analogous
condition that $z_\infty \equiv m \mod b$. It follows that we must
have $L\cong L_0^m\otimes L_\infty^n$. Thus the relevant space of
morphisms from $C_{a,b}$ to $\proj(a,b)$ can be identified with the
space of pairs $s_1 \in \Gamma(L^{\otimes a}), s_2 \in
\Gamma(L^{\otimes b})$ modulo the action of $\CC^*$ acting by scalar
multiplication on $L$. Since we are assuming
that $d\neq a$ and $d \neq b$, there is a unique section of both
$L^{\otimes a}$ and $L^{\otimes b}$ by degree considerations. We conclude that the space of 3 pointed maps with irreducible source curve is the quotient of $\GG_m \times \GG_m$ by the linear action of $\GG_m$ with
weights $a$ and $b$. This is simply $\GG_m \times \mu_d$.
There are two additional points where the source is reducible with a component collapsed over either zero or infinity. The reader may verify that the full moduli space is isomorphic to $C_{a,b} \times \cB\mu_d$, but to compute the relevant pushforward, the exact structure of the compactification is irrelevant.
Since this space has the expected
dimension, the virtual fundamental class is the ordinary fundamental
class, which pushes forward to the fundamental class of $\proj(a,b)$
and we conclude that $c_0=1$. To see that the other $c_i$
vanish we just need to verify that there are no representable
morphisms of minimal degree from a three pointed twisted genus zero
curve where the first two points are as before, but the third
twisted point has nontrivial stacky structure. To find such a map,
we would need to find an integer $D|d$, and a line bundle of degree
$\frac{d}{ab}$ on $C_{a,b,D}$ satisfying all of the conditions as
before as well as being nontrivial when restricted to $\cB\mu_D$. This
is obviously impossible.
We conclude that
$$\QA^*(\proj(a,b))\ \ =\ \ \QQ\llbracket q\rrbracket[\zeta,x,y]\ \big/\ \langle
xy-q,\, Ax^A-By^B\zeta^{n-m},\, \zeta^d-1\rangle.$$
\appendix
\section{Gluing of algebraic stacks along closed substacks} \label{Sec:Gluing}
\subsection{} We introduce a gluing construction for
Artin stacks.
\begin{proposition}\label{Prop:gluing}
Let $Z,X_1,X_2$ be algebraic stacks, and assume given
$$ \xymatrix{ *+<12pt,12pt>{Z} \ar@{^{(}->}[r]^{i_1} \ar@{^{(}->}[d]_{i_2} & X_1 \\
X_2 } $$
where $i_1,i_2$ are closed embeddings. Then there exists
an algebraic stack $X$ and a diagram
$$ \xymatrix{ *+<12pt,12pt>{Z} \ar@{^{(}->}[r]^{i_1} \ar@{^{(}->}[d]_{i_2} & X_1\ar[d] \\
X_2 \ar[r] & X} $$
such that the diagram is co-cartesian, namely, for any algebraic
stack $\cM$, the natural functor
$$\Hom(X,\cM) \to \Hom(X_1,\cM)\ \ \mathop{\times}\limits_{\Hom(Z,\cM)}\
\ \Hom(X_2,\cM)$$
is an equivalence of categories. Such $X$ is unique up to a unique isomorphism.
\end{proposition}
\begin{corollary}\label{Cor:limits}
Let $Z, Y$ be algebraic stacks and $i_1, i_2: Z \hookrightarrow Y$ closed embeddings with disjoint images. Then there exists
\begin{enumerate}
\item an algebraic stack $X$,
\item a morphism $\pi: Y \to X$, and
\item a 2-isomorphism $\alpha: \pi\circ i_1 \to\pi\circ i_2$
\end{enumerate}
such that $(X,\pi,\alpha) =
\displaystyle\varinjlim( Z \double Y)$.
\end{corollary}
In other words, we can glue together the two copies of $Z$ in $Y$, obtaining $X$
\subsection{Gluing of schemes and algebraic spaces}\label{Sec:gluing}
Given a scheme $Z$ together with a pair of closed embeddings
of $Z$ in schemes $X_1$ and $X_2$, we can define a new scheme $X_1\cup_Z X_2$.
It is determined by the universal property that a morphism
from $X_1 \cup_Z X_2$ to a scheme $W$ is given by a morphism from $X_1$
to $W$ and a morphism from $X_2$ to $W$ whose restrictions to $Z$
agree. To construct this scheme, we can just do the construction
for affines, where this amounts to taking a fibered product of rings.
In the following we use the fact that a similar construction exists in
the category of algebraic spaces.
One way to prove
that it exists is to use our proof below, replacing ``algebraic
space'' by ``scheme'' and ``algebraic
stack'' by ``algebraic space''.
Our first lemma shows that the universal property of the gluing of
algebraic spaces is preserved in the 2-category of stacks.
\begin{lemma}\label{Lem:gluing-is-coproduct} Let $\cM$ be an
algebraic stack, and let
$$ \xymatrix{ *+<12pt,12pt>{Z} \ar@{^{(}->}[r]^{i_1} \ar@{^{(}->}[d]_{i_2} & X_1\ar[d] \\
X_2 \ar[r] & X} $$
be a co-cartesian diagram of algebraic spaces, where $i_1,i_2$ are
closed embeddings.
Then the natural functor
$$\cM(X) \to \cM(X_1) \times_{\cM(Z)} \cM(X_2)$$
is an equivalence, where the fibered product is taken in the sense of categories.
\end{lemma}
\emph{Proof.} We construct a functor inverse to the given one. Let $R
\double U $ be a presentation of $\cM$. Assume given an object of the fibered product on the right hand side, namely
\begin{enumerate}
\item objects $f_i\in \cM(X_i) $, and
\item an isomorphism $\alpha: i_1^* f_1 \to i_2^* f_2$.
\end{enumerate}
Explicitly in terms of the presentation, we are given $U_i =
f_i^*U$ and $R_i = f_i^*R$ and morphisms of groupoids
$$ \xymatrix{ R_i \doubledown \ar[r] &R \doubledown \\
U_i \ar[r] & U. }
$$
Moreover, $\alpha$ gives an isomorphism
$$i_1^* ( R_1 \double U_1) \to i_2^* ( R_2 \double U_2).$$
Since $R_i$ and $U_i$ are algebraic spaces, we can form a groupoid in
algebraic spaces $R_X \double U_X$ presenting $X$, by gluing $U_1,U_2$
along the
morphism $\alpha:i^*U_1 \to i^*U_2$, and similarly for $R_X$. We
obtain a morphism of groupoids
$$ \xymatrix{ R_X \doubledown \ar[r] &R \doubledown \\
U_X \ar[r] & U }
$$
giving a morphism $X \to \cM$.
The construction of the functor on the level of arrows, and the fact
that the functors are inverses, is left to the reader. \qed
\subsection{Extension of atlases}
\begin{lemma} Let $Z\subset X$ be a closed embedding of
schemes. Assume $U_Z \to Z$ is a smooth morphism. Then there exists a
Zariski open covering $U_Z' \to U_Z$ and a smooth morphism $U_X' \to
X$ with $U_X' \times_X Z \cong U_Z'$.
\end{lemma}
\begin{proof}
We may assume $U_Z,Z$ and $X$ are affine, and embed $U_Z
\subset \bbA^n_Z$ for some $n$. The subscheme $U_Z$ is locally a complete
intersection in $\bbA^n_Z$, so for every point $u\in U_Z$ we can find
elements $(g_1,\ldots g_N)\subset \cO_Z[x_1,\ldots,x_n]$ and a
Zariski neighborhood $V \subset \bbA^n_Z$ such that $U_Z^u :=
U_Z \cap V
\subset V$ is the complete intersection of the zero schemes of
$g_i$. Choose $\tilde g_i\subset \cO_X[x_1,\ldots,x_n]$ lifting
$g_i$, and let $W$ be the zero scheme of $(\tilde g_1,\ldots,\tilde
g_N)$ in $\bbA^n_X$. Clearly $W\cap V = U_Z^u$, and so $W \to X$ is
smooth along points of $U_Z^u$. There exists a neighborhood $U_X^u
\subset W$ of $u$ containing $U_Z^u$ which is smooth over $X$. We
can take
\[
U_Z' = \bigcup_u U_Z^u, \quad U_X' = \bigcup_u U_X^u.\qedhere
\]
\end{proof}
\begin{lemma}
Let $i_j:Z \to X_j$ be closed embeddings of algebraic stacks,
$j=1,2$. There exist schemes $U_j$ and smooth surjective morphisms
$U_j \to
X_j$ such that
$$ Z\times_{X_1}U_1 \cong Z\times_{X_2}U_2. $$
\end{lemma}
\emph{Proof.} Let $V_j \to X_j$ be smooth and surjective. We have a pullback diagram
$$\xymatrix{ i_j^* V_j \ar[r]\ar[d]& V_j\ar[d]\\
Z\ar^{i_j}[r] & X_j.
}$$
Consider the fibered product $\tilde V = i_1^* V_1 \times_Z i_2^* V_2$. Applying the previous lemma with $Z\subset X$ replaced by the closed embedding $i_1^*V_1 \subset V_1$, and $U_Z \to Z$ replaced by $\tilde V \to i_1^*V_1$, there is a Zariski-open covering $ \tilde V_1 \to \tilde V$ and a smooth $\tilde U_1\to X_1$ with $i_1^*\tilde U_1 \cong \tilde V_1$. Applying the same procedure for $i_1^*V_1 \subset V_1$, we obtain a Zariski-open covering $ \tilde V_2 \to \tilde V$ and a smooth $\tilde U_2\to X_2$ with $i_2^*\tilde U_2 \cong \tilde V_2$. Replacing $\tilde V_1$ and $\tilde V_2$ by a common Zariski-refinement $\tilde V_{12}$, and replacing $\tilde U_j$ by a suitable Zariski-refinements $U_j$ lifting $\tilde V_{12}$, the lemma is proven. \qed
\subsection{The construction}\hfill
\begin{proof}[Proof of \ref{Prop:gluing}]
By the previous lemma, there is a choice of schemes $U_i$
and smooth
surjective morphisms $U_i \to X_i$ with $i_1^*U_1 \cong i_2^*U_2= U_Z$. Set
$R_i = U_i \times_{X_i} U_i$, so that $R_i \double U_i$ is a
presentation of $X_i$, and $i_1^*R_1 \cong i_2^*R_2= R_Z \double U_Z$
is a presentation of $Z$. Set $U = U_1 \cup_{U_Z} U_2$ and $R =
R_1\cup_{R_Z} R_2$. We have a groupoid $R \double U$, with a diagram
of groupoids
\begin{equation}\label{cubediagram}
\xymatrix{
& R_Z\ar[ld]\doubledown\ar[rd] &\\
R_1\doubledown\ar[rd]& U_Z\ar[ld]\ar[rd] & R_2\doubledown\ar[ld]\\
U_1\ar[rd] & R \doubledown & U_2\ar[ld] \\
& U &}
\end{equation}
Let $X = [R\double U]$ be the quotient. We claim this is the desired stack.
Let $\cM$ be an algebraic stack. By definition, we have
$$\Hom(X,\cM)\ \ =\ \
\varprojlim\left(\cM(U)\ \double \ \cM(R) \right).$$
By Lemma \ref{Lem:gluing-is-coproduct}, we have
$$\cM(U)\ \ =\ \
\cM(U_1)\ \ \mathop{\times}\limits_{\cM(U_Z)}\ \ \cM(U_2)
\ \ = \ \ \varprojlim\left(\cM(U_1\sqcup U_2)\ \double \ \cM(U_Z) \right)
$$
and
$$\cM(R)\ \ =\ \
\cM(R_1)\ \ \mathop{\times}\limits_{\cM(R_Z)}\ \ \cM(R_2)
\ \ = \ \ \varprojlim\left(\cM(R_1\sqcup R_2)\ \double \ \cM(R_Z) \right).$$
Taking $\cM$-valued points in diagram (\ref{cubediagram}), we get
\begin{equation}\label{reversediagram}
\xymatrix{
& \cM(R_Z)&\\
\cM(R_1)\ar[ur]& \cM(U_Z) \doubleup & \cM(R_2)\ar[ul]\\
\cM(U_1)\doubleup\ar[ur] & & \cM(U_2)\doubleup\ar[ul] }
\end{equation}
We thus have
\begin{align*}
&\Hom(X,\cM) = \\
&= \varprojlim
\Big( \varprojlim\left(\cM(U_1\sqcup U_2)\double\cM(U_Z) \right) \double
\varprojlim
\left(\cM(R_1\sqcup R_2)\double\cM(R_Z) \right)\Big) \\
&= \varprojlim\, (\text{ diagram (\ref{reversediagram})\ })\\
&= \varprojlim\left(
\cM(U_1) \double \cM(R_1)
\right)
\mathop{\times}\limits_{\varprojlim
\left(\cM(U_Z) \rightrightarrows \cM(R_Z)\right)}
\left(\varprojlim
\cM(U_2) \double \cM(R_2)
\right)\\
&= \Hom(X_1,\cM) \ \mathop{\times}\limits_{\Hom(Z,\cM)}\ \Hom(X_2,\cM).
\qedhere\end{align*}
\end{proof}
\begin{proof}[Proof of corollary \ref{Cor:limits}]
Consider the morphisms
$$j_1 := i_1\sqcup i_2: Z \sqcup Z \to Y \ \ \text{and} \ \ j_2 :=
i_2\sqcup i_1: Z \sqcup Z \to Y.$$ These morphisms are closed
embeddings by the empty intersection hypothesis. Let $$\tilde X = Y
\mathop\cup\limits_{Z \sqcup Z}Y.$$ There is a canonical free $G =
\ZZ/2\ZZ$-action on $\tilde X$ arising from the action on the gluing
diagram. We claim that $$X = \left[\tilde X \ \big/ \ G \right]$$ is
the desired colimit. Indeed, $X$ is the colimit of the following
diagram:
\[
\xymatrix{
G \times (Z \sqcup Z) \doubleright \doubledown & Z \sqcup Z \doubledown \\
G \times (Y \sqcup Y) \doubleright & Y \sqcup Y.
}
\]\end{proof}
\section{Taking roots of line bundles}
The following constrution is due independently to the authors and to C.~Cadman (\cite{Cadman1}, Section~2).
\subsection{The root of a line bundle}
If $L$ is a line bundle on a scheme $S$ and $d$ is a positive
integer, we will denote by $\radice{d}{L/S}$ the stack over $S$ of
$d\th$ roots of $L$; an object of $\radice{d}{L/S}$ over $T \to S$
is a line bundle $M$ over $T$, together with an isomorphism of
$M^{\otimes d}$ with the pullback $L_T$ of $L$ to $T$. The arrows
are defined in the obvious way. This stack $\radice{d}{L/S}$ is a
gerbe over $S$ banded by $\mu_d$; its cohomology class in the flat
cohomology group $\H^2(S, \mu_d)$ is obtained from the class $[L]
\in \H^1(S, \gm)$ via the boundary homomorphism $\partial\colon
\H^1(S, \gm) \to \H^2(S, \mu_d)$ obtained from the Kummer exact
sequence
\[
0 \longrightarrow \mu_{d,S}
\longrightarrow\mathbb{G}_{\mathrm{m},S}
\xrightarrow{(-)^d} \mathbb{G}_{\mathrm{m},S}
\longrightarrow 0.
\]
It is clear that if $T \to S$ is a morphism of schemes, then
$\radice{d}{L_T/T} = \radice{d}{L/S} \times_S T$.
The stack $\radice{d}{L/S}$ can be described directly as the quotient stack
$[L^0/\gm]$. Here $L^0$ is the total space of the $\gm$-bundle associated
with $L$ (or, in more down to earth terms, $L$ minus the zero section),
and the action of $\gm$ over $L^0$ is defined by the formula $\rho(\alpha)
x = \alpha^{-d}x$.
Equivalently, $$\radice{d}{L/S}\ =\ S \times_{\cB\GG_m}{\cB\GG_m},$$ where the fibered product is taken with respect to the classifying morphism $[L]: S \to {\cB\GG_m}$ on the left and the $d$-th power map ${\cB\GG_m}\to {\cB\GG_m}$ on the right. There is a universal line bundle $\mathcal{L}$ over
$\radice{d}{L/S}$, which is the quotient $[\mathbb{A}^1 \times L^0/ \gm]$
by the action defined by $\alpha(u, x) = (\alpha u, \rho(\alpha)x)$.
\subsection{The root of a line bundle with a section}\label{roots}
Now, given a section $\sigma\colon S \to L$ we can define a variant of
this, which we denote by $\radice{d}{(L, \sigma)/S}$, in which the objects
over a scheme $T \to S$ consist of triples
$$(M,\phi,\tau)$$
where
\begin{enumerate}
\item $M$ is a line bundle over $T$,
\item $\phi:M^{\otimes d} \simeq L_T$ is an isomorphism, and
\item $\tau$ is a section of $M$ such that $\phi(\tau^m) = \sigma$.
\end{enumerate}
If $Y$ is the
scheme-theoretic zero locus of $\sigma$, then the restriction of the
stack $\radice{d}{(L, \sigma)/S}$ to $S \setminus Y$ is equal to $S
\setminus Y$. Its restriction to $Y$ is more interesting; it does
not coincide with $\radice{d}{L_{Y}/Y}$, because an object of the stack
$\radice{d}{(L, \sigma)/S}\mid_Y$ over $T \to Y$ consists of a
$d\th$ root $M$ of $L_T$, plus a section of $M$ whose $d\th$ power
is 0; but the section itself is not necessarily 0. However, the
morphism $\radice{d}{L_{Y}/Y} \to \radice{d}{(L, \sigma)/S}\mid_Y$
defined by sending a $d\th$ root $M$ of $L_T$ on a scheme $T$ over
$Y$ to the same $M$ together with the zero section is a closed
embedding, defined by a nilpotent sheaf of ideals on $\radice{d}{(L,
\sigma)/S}\mid_Y$. Thus $\radice{d}{(L, \sigma)/S}\mid_Y$ contains a
canonical gerbe banded by $\mu_{d}$, supported over the zero scheme
of $\sigma$. The forgetful map $\radice{d}{(L, \sigma)/S}\mid_Y\to
\radice{d}{L_{Y}/Y}$ identifies $\radice{d}{(L, \sigma)/S}\mid_Y$ as
the $d$-th infinitesimal neighborhood of $\radice{d}{L_{Y}/Y}$ in
its universal line bundle.
The stack $\radice{d}{(L, \sigma)/S}$ can also be decribed as a quotient
stack. Consider the universal line bundle $\mathcal{L} = [\mathbb{A}^1
\times L^0/ \gm]$ described above, and the morphism $\Phi\colon \mathcal{L}
\to L$ induced by the $\gm$-invariant morphism $\mathbb{A}^1 \times L^0 \to
L$ defined by $(u, x) \mapsto u^d x$; then
\[
\radice{d}{(L, \sigma)/S} =
\Phi^{-1} \sigma(S).
\]
In other words: if we call
$V_\sigma\subseteq
\mathbb{A}^1 \times L^0$ the inverse image in $\mathbb{A}^1 \times L^0$ of
the embedding $\sigma \colon S \into L$; then
\[
\radice{d}{(L, \sigma)/S} = [V_\sigma/\gm].
\]
In particular, assume that $L = \cO$, so that $\sigma$ is a
regular function on $S$. In this case $L^0 = S \times \gm$, and if we
denote by $W_\sigma = \mathbb{A}^1 \times S$ the subscheme defined by the
equation $t^d - \sigma(x) = 0$, where $t$ is a coordinate on
$\mathbb{A}^1$, then there is an isomorphism $W_\sigma \times \gm \simeq
V_\sigma$.
An equivalent description exists here too: Let $\cU = [\AA^1/\GG_m]$, the classifying stack for line bundles with section. Then $$\radice{d}{(L, \sigma)/S} = S \times_\cU \cU,$$ where the map on the left is the classifying map and the map $\cU \to \cU$ on the right is the $d$-th power map.
\section{Rigidification}\label{Sec:rigidification}
\subsection{The setup}
We recall the concept of \emph{rigidification} of an algebraic
stack, as presented in \cite{ACV}, see also the related treatment in
\cite{Romagny}.
Let $H$ be a flat finitely presented separated group scheme over a
base scheme $\bbS$, ${\cX}$ an algebraic stack over $\bbS$. We say that $\cX$ has \emph{an $H$-2-structure}
if for each object $\xi \in {\cX}(T)$ there is an embedding
$$
\iota_\xi\colon H(T)\into \Aut_T (\xi),
$$
which is compatible with
pullback, in the following sense: given two objects $\xi\in
{\cX}(T)$ and $\eta\in{\cX}(T)$, and an arrow $\phi\colon \xi \to
\eta$ in ${\cX}$ over a morphism of schemes $f\colon S \to T$, the
natural pullback homomorphisms
$$\phi^*\colon \Aut_T (\eta)\to \Aut_S (\xi)$$ and $$f^* \colon H(T) \to H(S)$$
commute with the embedding, that is, $\iota_\xi f^* = \phi^*\iota_\eta$.
This condition can also be expressed as follows. Let $\phi\colon
\xi \to \eta$ be an arrow in ${\cX}$ over a morphism of schemes
$f\colon S \to
T$, and
$g \in H(T)$. Then
the diagram
\[
\xymatrix{
\xi\ar[r]^\phi\ar[d]^{f^*g} & \eta\ar[d]^g \\
\xi\ar[r]^\phi & \eta
}
\]
commutes. In particular, by taking $\xi = \eta$ and $\phi$ to be in
$\Aut_S(\xi)$, we see that $H(S)$ must be in the center of
$\Aut_S(\xi)$; in particular, $H$ is an abelian group
scheme.
The simplest example of such a situation is when $\cX \to T$ is a
gerbe banded by $H$; in this case the embedding $H(S)\into \Aut_S
(\xi)$ is an isomorphism of group schemes.
Then we have the following result.
\begin{theorem}(\cite{ACV}, Theorem~5.1.5)
There is a smooth surjective finitely presented morphism of algebraic stacks $\cX \to \cX\thickslash H$ satisfying the following properties:
\begin{enumerate}
\item For any object $\xi\in \cX(T)$ with image $\eta\in \cX\thickslash H(T)$, we have that $H(T)$ lies in the kernel of $\Aut_T( \xi) \to \Aut_T(\eta)$.
\item The morphism $\cX \to \cX\thickslash H$ is universal for morphisms of stacks $\cX \to \cY$ satisfying (1) above.
\item If $T$ is the spectrum of an algebraically closed field, then in (1) above, we have
\[
\Aut_T(\eta) = \Aut_T( \xi)/H(T).
\]
\item A moduli space for $\cX$ is also a moduli space for $\cX\thickslash H$.
\end{enumerate}
Furthermore, if $\cX$ is a Deligne--Mumford stack, then $\cX\thickslash H$ is
also a Deligne--Mumford stack and the morphism $\cX \to \cX\thickslash H$ is
\'etale.
\end{theorem}
The notation in \cite{ACV} is $\cX^{H}$; here we adopt the better
notation $\cX\thickslash H$ proposed by Romagny in \cite{Romagny}.
This stack $\cX\thickslash H$ is called the \emph{$H$-rigidification
of $\cX$}. For example, if $\cG \to T$ is a gerbe banded by $H$,
then $\cG\thickslash H$ is isomorphic to $T$.
The stack $\cX\thickslash H$ is obtained as the fppf stackification
of a prestack $\cX^{H}_{\mathrm{pre}}$, that has the same objects as
$\cX$; this has the property that for any object $\xi$ of $\cX$ over
an $\bbS$-scheme $T$, the sheaf of automorphisms of
$\cAut_{T,\cX^{H}_{\mathrm{pre}}}(\xi)$ in $\cX^{H}_{\mathrm{pre}}$
is the quotient sheaf of $\cAut_{T,\cX}(\xi)$ by the normal subgroup
sheaf $H_{T}$.
\subsection{Moduli interpretation of the stack $\cX\thickslash H$}\label{subsection-moduli-interpretation}
The construction of rigidification is functorial, in the
sense described below. First let $\phi\colon \cX \to \cY$ be a morphism of algebraic
stacks endowed with $H$-2-structures. We say that $\phi$ is \emph{$H$-2-equivariant} if for each
$\bbS$-scheme $T$ and object $\xi$ of $\cX(T)$, the homomorphism of
group-schemes
\[
H(T) \into \Aut_{T,\cX}(\xi) \longrightarrow \Aut_{T,\cY}(\phi(\xi))
\]
defined by $\phi$ coincides with the given embedding $H(T)
\hookrightarrow \Aut_{T,\cY}(\phi\xi)$.
Define a 2-category $\cX\slashsecond H$ over the category of schemes over $\bbS$, as follows.
\begin{enumerate}
\item An object over a scheme $T$ is a pair $(\cG,\phi)$, where $\cG \to T$ is a gerbe banded by $H$, and $\cG \stackrel{\phi}{\to} \cX$ is an $H$-2-equivariant morphism of fibered categories.
\item A morphism $(F,\rho): (\cG,\phi) \to (\cG',\phi')$ consists of a morphism $F : \cG \to \cG'$ over some $f: T \to T'$, compatible with the bands, and a 2-morphism $\rho: \phi\to \phi'\circ F$ making the following diagram commutative:
$$\xymatrix{ \cG \ar^{F}[rr]\ar[rd]_\phi && \cG' \ar[ld]^{\phi'} \\
& \cX
}$$
\item A 2-arrow $(F,\rho) \to (F_{1},\rho_{1})$ is a usual 2-arrow $\sigma: F \to F_{1}$ compatible with $\rho$ and $\rho_{1}$ in the sense that the following diagram is commutative:
\[
\xymatrix{ & \phi \ar[dl]_{{\rho}} \ar[dr]^{\rho_{1}}\\
\phi'\circ F \ar[rr]^{\phi'(\sigma)} && \phi'\circ F_{1}
}
\]
\end{enumerate}
From Lemma~\ref{lem:representable->1-category} we see that $\cX\slashsecond H$ is equivalent to a 1-category, that we denote by $\cX\slashprime H$. This is easily checked to be a category fibered in groupoids over the category of schemes over $\bbS$.
\begin{proposition}\label{prop:equivalence-rigidification}
There is an equivalence of fibered categories between $\cX\thickslash H$ and $\cX\slashprime H$.
\end{proposition}
\begin{proof}
Given an $H$-2-equivariant morphism $\cG \to \cX$, where $\cG \to T$ is a gerbe banded by $H$, there is an induced morphism $T = \cG\thickslash H \to \cX\thickslash H$. This gives a function from the objects of $\cX\slashprime H$ to the objects of $\cX\thickslash H$, that extends to a functor in the obvious way.
In the other direction, given an object $\xi$ of $\cX\thickslash H(T)$, consider the fibered product $\cG := T \times_{\cX\thickslash H} \cX \to T$. If $U$ is a scheme over $T$, an object of $\cG(U)$ is a pair $(\zeta,\alpha)$, where $\zeta$ is an object of $\cX(U)$, and $\alpha$ is an isomorphism between $\zeta$ and the pullback $\xi_{U}$ in $\cX\thickslash H(U)$. We claim that $\cG$ is a gerbe over $T$, and there is a unique banding of $\cG$ by $H$ making the projection $\cG \to \cX$ $H$-2-equivariant.
Both statements are fppf local on $T$, so we may assume that the given object $\xi$ of $\cX\thickslash H (T)$ comes from an object $\widetilde{\xi}$ of $\cX(T)$. Then the pair $(\widetilde{\xi}_{U},\mathrm{id})$ is an object of $\cG(U)$, showing the existence of local sections. Two objects $(\zeta_{1}, \alpha_{1})$ and $(\zeta_{2}, \alpha_{2})$ in $\cG(U)$ are locally isomorphic, because the morphism of sheaves of sets $\cHom_{U,\cX}(\zeta_{1}, \zeta_{2}) \to \cHom_{U,\cX\thickslash H}(\zeta_{1}, \zeta_{2})$ is an $H$-torsor.
Also, given an object $(\zeta,\alpha)$ of $\cG(U)$, its automorphism group in $\cG(U)$ is the kernel of the homomorphism $\Aut_{U,\cX}(\zeta) \to \Aut_{U,\cX\thickslash H}(\zeta)$, that is exactly $H(U) \subseteq \Aut_{U,\cX}(\zeta)$.
So $\cG$ is a gerbe banded by $H_{T}$. This function from the objects of $\cX$ to the objects of $\cX\slashprime H$ extends to a functor in the obvious way.
It is immediate to see that the composition $\cX\thickslash H \to \cX\slashprime H \to \cX\thickslash H$ is isomorphic to the identity. Let us show that the composition $\cX\slashprime H \to \cX\thickslash H \to \cX\slashprime H$ is also isomorphic to the identity. Given a gerbe $\cG \to T$ and an $H$-2-equivariant morphism $\cG \to \cX$, the induced morphism $\cG \to T \times_{\cX\thickslash H} \cX$ is a morphism of gerbes banded by $H$; and any such morphism is an isomorphism.
\end{proof}
\subsection{The rigidification as a quotient} Here is another way to think of the rigidification. Given the collection of data $\iota_\xi : H(T) \hookrightarrow \Aut_T(\xi)$ for all objects $\xi$ of $\cX$ required for forming the rigidification, we have an associated action $$\cB H \times \cX \to \cX,$$ defined as follows. First, given a scheme $T$, an object of $(\cB H \times \cX)\ (T)$ consists of a principal $H$-bundle $P \to T$ together with an object $\xi\in \cX(T)$. We need to form a new object $P\star \xi \in \cX(T)$. We do this as follows: the pullback $\xi_P$ of $\xi$ to $P$ admits a left diagonal action of $H$ coming from the two actions on $P$ (inverted, as to make it into a left action) and on $\xi$. By the descent axiom for $\cX$ there is a quotient object on $T$ which we call $P\star \xi \in \cX(T)$. The assumptions on $\iota_\xi$ guarantee that the formation of $P\star \xi$ is functorial, giving the required morphism $\cB H \times \cX \to \cX$. The particular case $\cX = \cB H$ shows that $\cB H$ is indeed a group stack, and in general one shows the morphism $\cB H \times \cX \to \cX$ is an action. The morphism $\cX \to \cX \thickslash H$ is easily seen to be invariant. From the fact that $\iota_\xi$ is injective one obtaines that $$\cB H \times \cX\ \ \lrar\ \ \cX\mathop\times\limits_{\cX \thickslash H}\cX $$
is an isomorphism, so this action is free and, whatever the 2-categorical quotient should mean, up to equivalence we obtain
$$ \cX / \cB H \ \ \simeq \ \ \cX \thickslash H.$$
This is certainly in agreement with our previous moduli interpretation of $\cX \thickslash H$: a principal $\cB H$-bundle $\cG \to T$ is simply a gerbe banded by $H$, and saying that $\cG \to \cX$ is equivariant translates to our requirement on the map of automorphisms to coincide with the homomorphisms $\iota_\xi$.
The case of non-central actions is more complicated and is worked
out in the appendix to \cite{AOV}.
| 37,370
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Juneau, AK (KINY) - The Alaska House passed House Bill 316, which was introduced by Rep. Harriet Drummond (D-Anchorage) that would protect an Alaskan's ability to work despite past convictions for marijuana possession.
This bill would also restrict public access to record related to simple marijuana possession in the wake of the legalization of marijuana in November of 2014. The Alaska CourtView system would also be wiped clean of all marijuana convictions classified as a VIA misdemeanor.
From a press release, Rep. Drummond said, .”
House Bill 316 passed in the House by a 30-10 vote and will head to the Senate for consideration.
| 162,247
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\begin{document}
\title{\bf Influence of conformal symmetry on the amplitude ratios for O($N$) models}
\author{H. A. S. Costa}
\email{hascosta@ufpi.edu.br}
\affiliation{\it Departamento de F\'\i sica, Universidade Federal do Piau\'\i, 64049-550, Teresina, PI, Brazil}
\author{P. R. S. Carvalho}
\email{prscarvalho@ufpi.edu.br}
\affiliation{\it Departamento de F\'\i sica, Universidade Federal do Piau\'\i, 64049-550, Teresina, PI, Brazil}
\begin{abstract}
In this Letter we compute analytically the effect of conformal symmetry on the radiative corrections to the amplitude ratios for O($N$) $\lambda\phi^{4}$ massless scalar field theories in curved spacetime for probing the two-scale-factor universality hypothesis. For that we employ three distinct and independent field-theoretic renormalization group methods. The amplitude ratios values obtained are identical when computed through the three distinct methods, thus showing their universal character. Furthermore, they are the same as that obtained in flat spacetime, then satisfying the two-scale-factor universality hypothesis.
\end{abstract}
\maketitle
\section{Introduction}\label{Introduction}
\par Symmetry is one of the most important properties to consider if we want to describe the behavior of physical systems. For example in the high energy physics scenario, the standard model (SM) of elementary particles and fields which describes three of the four elementary interactions of nature, namely electromagnetic, weak and strong interactions, is based on gauge symmetries \cite{Itzykson,Peskin}. The remaining elementary interaction, which is not part of the SM, \emph{i. e.} gravitation \cite{Misner.Thorne.Wheeler,Wald} is also a gauge theory. On the other hand, in the low energy physics realm the importance of symmetry considerations is not quite different as well. In fact, the universal critical scaling behavior based on the scaling hypothesis \cite{doi10106311696618,PhysicsPhysiqueFizika.2.263} of systems undergoing a continuous phase transition is characterized by the determination of a set of universal critical exponents. These universal critical exponents do not depend on nonuniversal properties of the system as the form of the lattice and critical temperature but on universal ones as the dimension $d$, $N$ and symmetry of some $N$-component order parameter and if the interactions among their constituents are of short- or long-range type. Obviously the $d$ \cite{PhysRevB.86.155112,PhysRevE.71.046112} and $N$ \cite{PhysRevLett.110.141601,Butti2005527,PhysRevB.54.7177} parameters are more evident to study as opposed to the symmetry one \cite{CARVALHO2017290,Carvalho2017}. Surprisingly, the critical behavior of many distinct systems as a fluid and a ferromagnet is characterized by the same set of critical exponents. When this happens we say that the different systems belong to the same universality class. The universality class approached here will be the O($N$) one which generalizes and encompass some particular models, namely the Ising ($N=1$), XY ($N=2$), Heisenberg ($N=3$), self-avoiding random walk ($N=0$), spherical ($N \rightarrow \infty$) \cite{Pelissetto2002549} ones. The critical exponents are not the only universal quantities in describing the critical behavior of systems undergoing a phase transition. This role is similarly played by others universal quantities, although being harder to compute than the former ones, called amplitude ratios \cite{V.PrivmanP.C.Hohenberg}. Then emerges the concept of two-scale-factor universality \cite{PhysRevLett.29.345}, where now there are eleven independent universal amplitude ratios since we have chosen two independent length scales as the order parameter and conjugate field scales for example. While the critical exponents are universal quantities, the amplitudes themselves are not since they depend on nonuniversal properties of the system. The universal quantities are, in fact, some ratios involving some non universal amplitudes, where the nonuniversal properties, explicitly expressed in the amplitudes themselves, cancel out in the middle of computations and thus turn out the amplitude ratios to be universal quantities.
\par In this Letter we have to investigate the influence of conformal symmetry on the values of the amplitude ratios for massless O($N$) $\lambda\phi^{4}$ scalar field theory in curved spacetime. When we try to renormalize a massless theory in curved spacetime following the conventional program, some divergences yet persist \cite{0305-4470-13-3-023}. The fixed value of $\xi = 1/6$ defines the theory as being invariant under conformal transformations. The amplitude ratios are a result of the fluctuating properties of a fluctuating scalar quantum field $\phi$ whose mean value we can identify to the order parameter, the magnetization of the system below the critical temperature $T_{c}$ for example. Since the theory involves properties of the system at the low temperature phase, as the magnetization, we have to describe a theory with spontaneous symmetry breaking since some amplitude ratios involve a few critical amplitudes computed below the critical temperature. Another feature showed through the lower temperature phase is that due to the presence of Goldstone modes, we have that some amplitudes and thus some amplitude ratios are not defined for every $N$ but only for Ising-like systems for which $N = 1$. These are the cases of both $C^{-}$ and $\xi_{0}^{-}$ amplitudes. The effect of quantum field fluctuations is evaluated as coming from the radiative quantum corrections to the renormalized effective potential with spontaneous symmetry breaking. If we do not take these loop corrections into account, we are limited to obtain the amplitude ratios values associated to the mean field or Landau approximation \cite{ZinnJustin}. We compute the amplitude ratios up to one-loop order. The effective potential with spontaneous symmetry breaking is renormalized in the normalization conditions method \cite{Amit}, minimal subtraction scheme \cite{Amit} and massless Bogoliubov-Parasyuk-Hepp-Zimmermann (BPHZ) method \cite{BogoliubovParasyuk,Hepp,Zimmermann}, where the external momenta of Feynman diagrams are held at fixed values in the first method and arbitrary in the last two ones. Although the application of a single method in computing the amplitude ratios is enough, the application of two another ones, besides to be useful as a check of the final results, must furnish the same amplitude ratios values as the renormalization group program demands since these physical quantities are universal. The same task was approached in a flat spacetime with Lorentz symmetry breaking mechanism \cite{Neto2017}. We then follow the steps of Ref. \cite{Neto2017} taken originally in Ref. \cite{BrezinLeGuillouZinnJustin}. In this Letter, the fluctuating quantum field is embedded on a curved spacetime and considering its non-minimal interaction with the curved background of the form $\xi R\phi^{2}$, where $\xi$ and $R$ are the non-minimal interaction coupling constant and the scalar curvature $R = g^{\mu\nu}R_{\mu\nu}$, respectively \cite{PhysRevD.20.2499,0305-4470-13-3-022,0305-4470-13-2-023,Bunch1981,Buchbinder1989,Class.QuantumGrav.}. We have to expand the free propagator of the theory and to make our calculations up to linear powers in $R$ and $R_{\mu\nu}$. Also up to linear powers in $R$ and $R_{\mu\nu}$, the present authors performed a computation of the critical exponents in an early work \cite{Costa_2019}. Now we proceed to obtain the corresponding amplitude ratios.
\section{Normalization conditions method}\label{Normalization conditions method}
\par The critical amplitudes \cite{Neto2017}
\begin{eqnarray}\label{uhdfuhfuh}
\begin{array}{lcr}
\mbox{\textrm{Critical isochore: $T > T_{c}$, $H = 0$}} & & \\
\mbox{} \xi = \xi_{0}^{+}t^{-\nu}, \chi = C^{+}t^{-\gamma}, C_{s} = \frac{A^{+}}{\alpha_{+}}t^{-\alpha} & \\ [10pt]
\mbox{\textrm{Critical isochore: $T < T_{c}$, $H = 0$}} & & \\
\mbox{} \xi = \xi_{0}^{-}t^{-\nu}, \chi = C^{-}t^{-\gamma}, C_{s} = \frac{A^{-}}{\alpha_{-}}t^{-\alpha}, M = B(-t)^{\beta} & \\ [10pt]
\mbox{\textrm{Critical isotherm: $T = T_{c}$, $H \neq 0$}} & & \\
\mbox{} \xi = \xi_{0}^{c}|H|^{-\nu_{c}}, \chi = C^{c}|H|^{-\gamma_{c}}, C_{s} = \frac{A^{c}}{\alpha_{c}}|H|^{-\alpha_{c}}, H = DM^{\delta} & \\ [10pt]
\mbox{\textrm{Critical point: $T = T_{c}$, $H = 0$}} & & \\
\mbox{} \chi(p) = \widehat{D}p^{\eta-2} & \\ \end{array} \nonumber
\end{eqnarray}
are obtained through the renormalized effective potential with spontaneous symmetry breaking, equation of state and the longitudinal and transverse correlation functions, respectively \cite{Neto2017}
\begin{eqnarray}\label{fhdldglkjdflk}
&& \mathcal{F}(t,M,g^{\ast}) = \frac{1}{2}tM^{2} + \frac{1}{4!}g^{\ast}M^{4} + \frac{1}{4}\left[ Nt^{2} + \frac{N+2}{3}tg^{\ast}M^{2} + \frac{N+8}{36}(g^{\ast}M^{2})^{2} \right]\parbox{10mm}{\includegraphics[scale=1.0]{fig10.eps}}_{SP} + \nonumber \\ && \frac{1}{2}\int d^{d}q\Bigg\{ \ln\Bigg[ 1 + \Bigg(t + \frac{g^{\ast}M^{2}}{2}\Bigg)G_{0}(q) \Bigg] + (N-1)\ln\Bigg[ 1 + \Bigg(t + \frac{g^{\ast}M^{2}}{6}\Bigg)G_{0}(q) \Bigg] - \nonumber \\ && \frac{N+2}{6}tg^{\ast}M^{2}G_{0}(q) \Bigg\},
\end{eqnarray}
\begin{eqnarray}\label{jglkfjkflj}
&& H/M = t + \frac{1}{6}g^{\ast}M^{2} + \frac{1}{2}g^{\ast}\Biggl\{ \Biggl[ \parbox{12mm}{\includegraphics[scale=1.0]{fig1.eps}}_{(1)} + (t + g^{\ast}M^{2}/2)\parbox{10mm}{\includegraphics[scale=1.0]{fig10.eps}}_{SP} \Biggl] + \nonumber \\ && \frac{N - 1}{3}\Biggl[ \parbox{12mm}{\includegraphics[scale=1.0]{fig1.eps}}_{(N - 1)} + (t + g^{\ast}M^{2}/6)\parbox{10mm}{\includegraphics[scale=1.0]{fig10.eps}}_{SP} \Biggl] \Biggl\},\nonumber \\
\end{eqnarray}
\begin{eqnarray}\label{uhduhuhg}
&& \Gamma_{L}(P^{2},t,M) = P^{2} + \frac{1}{6}R_{\mu\nu}\frac{\partial}{\partial P_{\mu}}\frac{P^{\nu}}{(P^{2} + t + g^{\ast}M^{2}/2)^{2}} + t + \frac{1}{2}g^{\ast}M^{2} + \nonumber \\ && \frac{1}{2}g^{\ast}\Biggl\{ \Biggl[ \parbox{12mm}{\includegraphics[scale=1.0]{fig1.eps}}_{(1)} + (t + g^{\ast}M^{2}/2)\parbox{10mm}{\includegraphics[scale=1.0]{fig10.eps}}_{SP} \Biggl] + \frac{N - 1}{3}\Biggl[ \parbox{12mm}{\includegraphics[scale=1.0]{fig1.eps}}_{(N - 1)} + (t + g^{\ast}M^{2}/6)\parbox{10mm}{\includegraphics[scale=1.0]{fig10.eps}}_{SP} \Biggl] + \nonumber \\ && g^{\ast}M^{2}\Biggl[ \Biggl(\parbox{10mm}{\includegraphics[scale=1.0]{fig10.eps}}_{SP} - \parbox{10mm}{\includegraphics[scale=1.0]{fig10.eps}}_{(1)} \Biggl) + \frac{N - 1}{9}\Biggl(\parbox{10mm}{\includegraphics[scale=1.0]{fig10.eps}}_{SP} - \parbox{10mm}{\includegraphics[scale=1.0]{fig10.eps}}_{(N - 1)} \Biggl) \Biggl] \Biggl\},
\end{eqnarray}
\begin{eqnarray}\label{uhdsuhdfuhdfu}
&& \Gamma_{T}(P^{2},t,M) = P^{2} + \frac{1}{6}R_{\mu\nu}\frac{\partial}{\partial P_{\mu}}\frac{P^{\nu}}{(P^{2} + t + g^{\ast}M^{2}/6)^{2}} + t + \frac{1}{6}g^{\ast}M^{2} + \nonumber \\ && \frac{1}{6}g^{\ast}\Biggl\{ \Biggl[ \parbox{12mm}{\includegraphics[scale=1.0]{fig1.eps}}_{(1)} + (t + g^{\ast}M^{2}/2)\parbox{10mm}{\includegraphics[scale=1.0]{fig10.eps}}_{SP} \Biggl] + (N + 1)\Biggl[ \parbox{12mm}{\includegraphics[scale=1.0]{fig1.eps}}_{(N - 1)} + (t + g^{\ast}M^{2}/6)\parbox{10mm}{\includegraphics[scale=1.0]{fig10.eps}}_{SP} \Biggl] + \nonumber \\ && \frac{2}{3}g^{\ast}M^{2}\Biggl[ \Biggl(\parbox{10mm}{\includegraphics[scale=1.0]{fig10.eps}}_{SP} - \parbox{10mm}{\includegraphics[scale=1.0]{fig10.eps}}_{(1,(N - 1))} \Biggl) \Biggl] \Biggl\},
\end{eqnarray}
where $M =\,<\phi>$, $g$ and $t$ are the renormalized magnetization (as a mean value of the renormalized field), dimensionless coupling constant and composite field coupling constant. The dimensionless and dimensionful coupling constants are related through $\lambda = g\kappa^{\epsilon/2}$, where $\kappa$ is some arbitrary momentum scale. The quantity $g^{\ast}$ is the dimensionless coupling constant evaluated at its nontrivial fixed point value. The nontrivial fixed point is computed from the nontrivial solution of the equation $\beta(g^{\ast}) = 0$ \cite{Amit}, where $\beta(g)$ is the $\beta$-function giving the flow of the renormalized dimensionless coupling constant flowing from some arbitrary value to the renormalized one when the renormalized theory is attained. This nontrivial fixed point is responsible for the computation of the loop quantum corrections to the amplitude ratios beyond the Landau approximation. The renormalized massless free propagator $G_{0}(q) \equiv \parbox{12mm}{\includegraphics[scale=1.0]{fig9.eps}}$ is given by
\begin{eqnarray}
G_{0}(q) = \frac{1}{q^{2}} + \frac{(1/3 - \xi)R}{(q^{2})^{2}} - \frac{2R_{\mu\nu}q^{\mu}q^{\nu}}{3(q^{2})^{3}},
\end{eqnarray}
where we have to set $\xi \rightarrow \xi(d)$ at the middle of Feynman diagrams calculations to get rid the infrared divergences that yet would persist in the theory if we would maintain $\xi$ arbitrary and $\xi(d) = [(d - 2)/4(d - 1)]$ at arbitrary dimensions less than four \cite{HATHRELL1982136,PhysRevD.25.1019,Buchbinder.Odintsov.Shapiro,Parker.Toms,0264-9381-25-10-103001}, where $d = 4 - \epsilon$. . The subscript ``$SP$'' in the ``fish'' diagram $\parbox{8mm}{\includegraphics[scale=.8]{fig10.eps}}_{SP}$ means that this diagram is to be evaluated in the so called symmetry point, where the external momenta $P$ are held at fixed values according to $P_{i}\cdot P_{j} = (\kappa^{2}/4)(4\delta_{ij}-1)$, implying that $(P_{i} + P_{j})^{2} \equiv P^{2} = \kappa^{2}$ for $i\neq j$ \cite{Amit}. The nontrivial fixed point can be computed after we have evaluated the $\beta$-function displayed just below
\begin{eqnarray}
\beta(g) = -\epsilon g + \frac{N + 8}{6}\Bigg[ 1 + \frac{1}{2}\epsilon + \Bigg(\frac{R}{6\mu^{2}} - \frac{R_{\mu\nu}\widehat{P}^{\mu}\widehat{P}^{\nu}}{3\mu^{2}}\Bigg)\epsilon \Bigg]g^{2} - \frac{3N + 14}{12}g^{3},
\end{eqnarray}
where we have used the following evaluated Feynman diagrams
\begin{eqnarray}\label{khjhuh}
\parbox{10mm}{\includegraphics[scale=1.0]{fig10.eps}}_{SP} = \frac{1}{\epsilon} \Biggr(1 + \frac{1}{2}\epsilon + \frac{R}{6\mu^{2}}\epsilon - \frac{R_{\mu\nu}\widehat{P}^{\mu}\widehat{P}^{\nu}}{3\mu^{2}}\epsilon \Biggr),
\end{eqnarray}
\begin{eqnarray}\label{gfjfgujfguj}
\parbox{12mm}{\includegraphics[scale=0.8]{fig21.eps}}_{SP} = \frac{1}{2\epsilon^{2}} \Biggr(1 + \frac{3}{2}\epsilon + \frac{R}{3\mu^{2}}\epsilon - 2\frac{R_{\mu\nu}P^{\mu}P^{\nu}}{3\mu^{4}}\epsilon \Biggr).
\end{eqnarray}
We emphasize that in the present method, the $\beta$-function depends on the nonuniversal curved spacetime parameters $R$ and $R_{\mu\nu}$ as well as the corresponding nontrivial fixed point
\begin{eqnarray}\label{klkoilkj}
g^{\ast} = \frac{6\epsilon}{(N + 8)}\Bigg\{ 1 + \epsilon\left[ \frac{(9N + 42)}{(N + 8)^{2}} -\frac{1}{2} - \Bigg(\frac{R}{6\mu^{2}} - \frac{R_{\mu\nu}\widehat{P}^{\mu}\widehat{P}^{\nu}}{3\mu^{2}}\Bigg) \right]\Bigg\}.
\end{eqnarray}
Thus, following the notation of Ref. \cite{Neto2017} we obtain the critical amplitudes
\begin{eqnarray}\label{hduhsduhsdu}
&& A^{+} = \frac{N}{4}\Bigg[ 1 + \Bigg( \frac{4}{4 - N} + A_{N} \Bigg)\epsilon \Bigg],
\end{eqnarray}
\begin{eqnarray}\label{jdoijdio}
A^{-} = 1 + \Bigg( \frac{N}{4 - N} - \frac{4 - N}{2(N + 8)}\ln 2 + A_{N} \Bigg)\epsilon ,
\end{eqnarray}
where
\begin{eqnarray}\label{hduhsduhsduu}
&& A_{N} = \frac{1}{2} - \frac{9N + 42}{(N + 8)^{2}} - \frac{4 - N}{(N + 8)} - \frac{(N + 2)(N^{2} + 30N + 56)}{2(4 - N)(N + 8)^{2}} + \nonumber \\ && \frac{2(N + 2)}{N + 8}\Bigg(\frac{R}{6\mu^{2}} - \frac{R_{\mu\nu}\widehat{P}^{\mu}\widehat{P}^{\nu}}{3\mu^{2}}\Bigg),
\end{eqnarray}
\begin{eqnarray}\label{jhdshudfhuifh}
C^{+} = 1 - \frac{N + 2}{2(N + 8)} \Bigg[ 1 + 2\Bigg(\frac{R}{6\mu^{2}} - \frac{R_{\mu\nu}\widehat{P}^{\mu}\widehat{P}^{\nu}}{3\mu^{2}}\Bigg) \Bigg]\epsilon ,
\end{eqnarray}
\begin{eqnarray}\label{ygdsuygsduygds}
C^{-} = \frac{1}{2}\Bigg\{1 - \frac{1}{6}\Bigg[ 4 + \ln 2 + 2\Bigg(\frac{R}{6\mu^{2}} - \frac{R_{\mu\nu}\widehat{P}^{\mu}\widehat{P}^{\nu}}{3\mu^{2}}\Bigg) \Bigg]\epsilon\Bigg\},
\end{eqnarray}
\begin{eqnarray}\label{fhduhdfu}
D = \frac{1}{6}g^{\ast(\delta - 1)/2}\Bigg\{1 + \Bigg[ 1 - \ln 2 - \frac{N - 1}{N + 8}\ln 3 + 2\Bigg(\frac{R}{6\mu^{2}} - \frac{R_{\mu\nu}\widehat{P}^{\mu}\widehat{P}^{\nu}}{3\mu^{2}}\Bigg) \Bigg]\epsilon\Bigg\},
\end{eqnarray}
\begin{eqnarray}\label{huhfuhfuhf}
&& B = \Bigg(\frac{N + 8}{\epsilon}\Bigg\{1 - \frac{3}{N + 8}\Bigg[ 1 + \ln 2 + 2\Bigg(\frac{R}{6\mu^{2}} - \frac{R_{\mu\nu}\widehat{P}^{\mu}\widehat{P}^{\nu}}{3\mu^{2}}\Bigg) \Bigg]\epsilon - \nonumber \\ && \Bigg[\frac{9N + 42}{(N + 8)^{2}} -\frac{1}{2} - \Bigg(\frac{R}{6\mu^{2}} - \frac{R_{\mu\nu}\widehat{P}^{\mu}\widehat{P}^{\nu}}{3\mu^{2}}\Bigg) \Bigg]\epsilon \Bigg\}\Bigg)^{1/2},
\end{eqnarray}
\begin{eqnarray}\label{hjfdhjfhjf}
\xi_{0}^{+} = 1 - \frac{N + 2}{4(N + 8)} \Bigg[ 1 + 2\Bigg(\frac{R}{6\mu^{2}} - \frac{R_{\mu\nu}\widehat{P}^{\mu}\widehat{P}^{\nu}}{3\mu^{2}}\Bigg) \Bigg]\epsilon ,
\end{eqnarray}
\begin{eqnarray}\label{kjlkjhlkh}
\xi_{0}^{-} = 2^{-1/2}\Bigg\{1 - \frac{1}{12}\Bigg[ \frac{7}{2} +\ln 2 + 2\Bigg(\frac{R}{6\mu^{2}} - \frac{R_{\mu\nu}\widehat{P}^{\mu}\widehat{P}^{\nu}}{3\mu^{2}}\Bigg) \Bigg]\epsilon\Bigg\},
\end{eqnarray}
\begin{eqnarray}\label{fdgfdsadfs}
&& \xi_{0}^{T} = \Biggl( \frac{\epsilon}{N + 8}\Bigg\{1 + \frac{3}{N + 8}\Bigg[ \frac{5}{6} + \ln 2 + 2\Bigg(\frac{R}{6\mu^{2}} - \frac{R_{\mu\nu}\widehat{P}^{\mu}\widehat{P}^{\nu}}{3\mu^{2}}\Bigg) \Bigg]\epsilon + \nonumber \\ && \Bigg[\frac{9N + 42}{(N + 8)^{2}} -\frac{1}{2} - \Bigg(\frac{R}{6\mu^{2}} - \frac{R_{\mu\nu}\widehat{P}^{\mu}\widehat{P}^{\nu}}{3\mu^{2}}\Bigg)\Bigg]\epsilon \Bigg\}\Biggl)^{1/(d - 2)},\nonumber \\
\end{eqnarray}
\begin{eqnarray}\label{ouypuopuo}
&& C^{c} = \frac{2D^{1/\delta}}{g^{\ast 1/2\beta}} \Bigg( 1 - \frac{9}{2(N + 8)}\Bigg\{ \Biggl[ 1 - \ln 2 + 2\Bigg(\frac{R}{6\mu^{2}} - \frac{R_{\mu\nu}\widehat{P}^{\mu}\widehat{P}^{\nu}}{3\mu^{2}}\Bigg) \Bigg] + \nonumber \\ && \frac{N - 1}{9}\Biggl[ 1 - \ln 6 + 2\Bigg(\frac{R}{6\mu^{2}} - \frac{R_{\mu\nu}\widehat{P}^{\mu}\widehat{P}^{\nu}}{3\mu^{2}}\Bigg) \Bigg] + \frac{2(N + 8)}{27} \Bigg\}\epsilon\Bigg),
\end{eqnarray}
\begin{eqnarray}\label{bxvbcvbxvc}
&& \xi_{0}^{c} = \frac{2^{1/2}D^{1/2\delta}}{g^{\ast 1/4\beta}} \Bigg( 1 - \frac{9}{4(N + 8)}\Bigg\{ \Biggl[ 1 - \ln 2 + 2\Bigg(\frac{R}{6\mu^{2}} - \frac{R_{\mu\nu}\widehat{P}^{\mu}\widehat{P}^{\nu}}{3\mu^{2}}\Bigg) \Bigg] + \nonumber \\ && \frac{N - 1}{9}\Biggl[ 1 - \ln 6 + 2\Bigg(\frac{R}{6\mu^{2}} - \frac{R_{\mu\nu}\widehat{P}^{\mu}\widehat{P}^{\nu}}{3\mu^{2}}\Bigg) \Bigg] + \frac{N + 14}{27} \Bigg\}\epsilon\Bigg)^{1/2},
\end{eqnarray}
\begin{eqnarray}\label{hjdghssf}
\widehat{D} = 1,
\end{eqnarray}
where the remaining evaluated Feynman diagrams needed to obtain the critical amplitudes aforementioned are the ones shown in the Eq. (\ref{khjhuh}) and in the ones (\ref{fldsjlkfdjlk})-(\ref{tyrttffyty})
\begin{eqnarray}\label{fldsjlkfdjlk}
\parbox{12mm}{\includegraphics[scale=1.0]{fig1.eps}}_{(1)} = -\frac{t + g^{\ast}M^{2}/2}{\epsilon}\Biggl[ 1 - \frac{1}{2}\ln \Biggl( t + \frac{g^{\ast}M^{2}}{2} \Biggl)\epsilon \Biggl],
\end{eqnarray}
\begin{eqnarray}\label{fldsjlkfdjl}
&& \parbox{11mm}{\includegraphics[scale=1.0]{fig1.eps}}_{(N - 1)} = -\frac{t + g^{\ast}M^{2}/6}{\epsilon}\Biggl[ 1 - \frac{1}{2}\ln \Biggl( t + \frac{g^{\ast}M^{2}}{6} \Biggl)\epsilon \Biggl],
\end{eqnarray}
\begin{eqnarray}
\parbox{10mm}{\includegraphics[scale=1.0]{fig10.eps}}_{(1)} = \frac{1}{\epsilon} \Biggr[1 - \frac{1}{2}\epsilon - \frac{1}{2}\epsilon L_{1}(P^{2}) + \frac{R}{6\mu^{2}}\epsilon L_{R,1}(P^{2}) - \frac{R_{\mu\nu}P^{\mu}P^{\nu}}{3\mu^{4}}\epsilon L_{R_{\mu\nu},1}(P^{2}) \Biggr],
\end{eqnarray}
\begin{eqnarray}
&& \parbox{10mm}{\includegraphics[scale=1.0]{fig10.eps}}_{(N - 1)} = \nonumber \\ && \frac{1}{\epsilon} \Biggr[1 - \frac{1}{2}\epsilon - \frac{1}{2}\epsilon L_{(N-1)}(P^{2}) + \frac{R}{6\mu^{2}}\epsilon L_{R,(N - 1)}(P^{2}) - \frac{R_{\mu\nu}P^{\mu}P^{\nu}}{3\mu^{4}}\epsilon L_{R_{\mu\nu},(N - 1)}(P^{2}) \Biggr],
\end{eqnarray}
\begin{eqnarray}\label{assdxe}
&& \parbox{10mm}{\includegraphics[scale=1.0]{fig10.eps}}_{1,(N - 1)} = \nonumber \\ && \frac{1}{\epsilon} \Biggr[1 - \frac{1}{2}\epsilon - \frac{1}{2}\epsilon L_{1,(N-1)}(P^{2}) + \frac{R}{6\mu^{2}}\epsilon L_{R,[1,(N - 1)]}(P^{2}) - \frac{R_{\mu\nu}P^{\mu}P^{\nu}}{3\mu^{4}}\epsilon L_{R_{\mu\nu},[1,(N - 1)]}(P^{2}) \Biggr],
\end{eqnarray}
\begin{eqnarray}\label{et5rtgdgff}
L_{1}(P^{2}) = \int_{0}^{1}dx \ln \left[\frac{x(1-x)P^{2} + t + \frac{g^{\ast}M^{2}}{2}}{\mu^{2}}\right],
\end{eqnarray}
\begin{eqnarray}\label{jhkjoji}
L_{(N-1)}(P^{2}) = \int_{0}^{1}dx \ln \left[\frac{x(1-x)P^{2} + t + \frac{g^{\ast}M^{2}}{6}}{\mu^{2}}\right],
\end{eqnarray}
\begin{eqnarray}\label{gjfghdfyr}
L_{1,(N-1)}(P^{2}) = \int_{0}^{1}dx \ln \left[\frac{x(1-x)P^{2} + x(t + \frac{g^{\ast}M^{2}}{2}) + (1-x)(t + \frac{g^{\ast}M^{2}}{6})}{\mu^{2}}\right],\nonumber \\
\end{eqnarray}
\begin{eqnarray}\label{zcxvcgfd}
L_{R,1}(P^{2}) = \int_{0}^{1}d x\frac{x(1 - x)}{\frac{x(1 - x)P^{2}}{\mu^{2}} + \frac{t + \frac{g^{\ast}M^{2}}{2}}{\mu^{2}}},
\end{eqnarray}
\begin{eqnarray}\label{jlkjiyiuj}
L_{R_{\mu\nu},1}(P^{2}) = \int_{0}^{1}d x\frac{x^{2}(1 - x)^{2}}{\left[\frac{x(1 - x)P^{2}}{\mu^{2}} + \frac{t + \frac{g^{\ast}M^{2}}{2}}{\mu^{2}}\right]^{2}},
\end{eqnarray}
\begin{eqnarray}\label{bmbnbmv}
L_{R,(N - 1)}(P^{2}) = \int_{0}^{1}d x\frac{x(1 - x)}{\frac{x(1 - x)P^{2}}{\mu^{2}} + \frac{t + \frac{g^{\ast}M^{2}}{6}}{\mu^{2}}},
\end{eqnarray}
\begin{eqnarray}\label{wereretjjgdhg}
L_{R_{\mu\nu},(N - 1)}(P^{2}) = \int_{0}^{1}d x\frac{x^{2}(1 - x)^{2}}{\left[\frac{x(1 - x)P^{2}}{\mu^{2}} + \frac{t + \frac{g^{\ast}M^{2}}{6}}{\mu^{2}}\right]^{2}},
\end{eqnarray}
\begin{eqnarray}\label{ipkokljgdhg}
L_{R,[1,(N - 1)]}(P^{2}) = \int_{0}^{1}d x\frac{x(1 - x)}{\frac{x(1 - x)P^{2}}{\mu^{2}} + \frac{x(t + \frac{g^{\ast}M^{2}}{2}) + (1-x)(t + \frac{g^{\ast}M^{2}}{6})}{\mu^{2}}},
\end{eqnarray}
\begin{eqnarray}\label{tyrttffyty}
L_{R_{\mu\nu},[1,(N - 1)]}(P^{2}) = \int_{0}^{1}d x\frac{x^{2}(1 - x)^{2}}{\left[\frac{x(1 - x)P^{2}}{\mu^{2}} + \frac{x(t + \frac{g^{\ast}M^{2}}{2}) + (1-x)(t + \frac{g^{\ast}M^{2}}{6})}{\mu^{2}}\right]^{2}},\quad\quad
\end{eqnarray}
where $\widehat{P}^{\mu}$ is dimensionless and unitary such that $P^{\mu} = \kappa \widehat{P}^{\mu}$ \cite{EurophysLett10821001}. Thus as for the $\beta$-function and nontrivial fixed point, the amplitudes themselves depend on the nonuniversal curved spacetime parameters $R$ and $R_{\mu\nu}$. But now, if we apply the expressions above for the amplitudes, we have the cancelling of the curved spacetime parameters $R$ and $R_{\mu\nu}$ in the amplitude ratios computation and obtain that the curved spacetime amplitude ratios are identical to their flat spacetime counterparts \cite{V.PrivmanP.C.Hohenberg}. These results show that the amplitude ratios values are insensible to the conformal symmetry thus maintaining the two-scale-factor universality hypothesis validity intact. In fact, as the symmetry under consideration is one present in the space where the quantum field is embedded and not in its internal one, the curved spacetime amplitude ratios values must be the same as that of flat sapetime ones. Just a change in the internal symmetry of the field could modify the amplitude ratios values.
\section{Minimal subtraction scheme}\label{Minimal subtraction scheme}
\par This present method is much more general and elegant than the earlier \cite{Amit}, since in the normalization conditions method, the external momenta are fixed at some value, namely the symmetry point while in the present one they are kept at general arbitrary values. In the present method we consider only the divergent part of the diagrams and not both divergent and finite ones as in the earlier method. Thus, for example, instead of considering the $\parbox{8mm}{\includegraphics[scale=.8]{fig10.eps}}_{SP}$ diagram in the Eqs. (\ref{fhdldglkjdflk})-(\ref{uhdsuhdfuhdfu}) we have to consider the $[\parbox{8mm}{\includegraphics[scale=.8]{fig10.eps}}]_{S}$ one where $[\parbox{8mm}{\includegraphics[scale=.8]{fig10.eps}}]$ is that of Eq. (\ref{klijouio}) and $[~~~]_{S}$ means that the only part of the diagram we have to consider is its singular and one so on for the remaining diagrams. Thus, in this method, we obtain amplitudes which are different from that obtained through the earlier method. Their values in this method are given by
\begin{eqnarray}\label{terreteu}
&& A^{+} = \frac{N}{4}\Bigg[ 1 + \Bigg( \frac{4}{4 - N} + A_{N}^{\prime} \Bigg)\epsilon \Bigg],
\end{eqnarray}
\begin{eqnarray}\label{eytwertyer}
A^{-} = \Bigg[1 + \Bigg( \frac{N}{4 - N} - \frac{4 - N}{2(N + 8)}\ln 2 + A_{N}^{\prime} \Bigg)\epsilon\Bigg],
\end{eqnarray}
where
\begin{eqnarray}\label{reytreytrwe}
A_{N}^{\prime} = - \frac{9N + 42}{(N + 8)^{2}} - \frac{4 - N}{(N + 8)} - \frac{(N + 2)(N^{2} + 30N + 56)}{2(4 - N)(N + 8)^{2}},
\end{eqnarray}
\begin{eqnarray}\label{uiyiupoiu}
&& C^{+} = 1,
\end{eqnarray}
\begin{eqnarray}\label{iuoipiyo}
&& C^{-} = \frac{1}{2}\Bigg[1 - \frac{1}{6}( 3 + \ln 2 )\epsilon\Bigg],
\end{eqnarray}
\begin{eqnarray}\label{tretrety}
D = \frac{1}{6}g^{\ast(\delta - 1)/2}\Bigg[1 - \frac{1}{2}\Bigg( \ln 2 + \frac{N - 1}{N + 8}\ln 3 \Bigg)\epsilon\Bigg],
\end{eqnarray}
\begin{eqnarray}\label{tuytyut}
B = \Bigg\{\frac{N + 8}{\epsilon}\Bigg[1 - \frac{3\ln 2}{N + 8}\epsilon - \frac{9N + 42}{(N + 8)^{2}}\epsilon \Bigg]\Bigg\}^{1/2},
\end{eqnarray}
\begin{eqnarray}\label{xcvbcxzvcxzv}
\xi_{0}^{+} = 1,
\end{eqnarray}
\begin{eqnarray}\label{xvnbxvznbx}
\xi_{0}^{-} = 2^{-1/2}\Bigg[1 - \frac{1}{12}\Bigg( \frac{5}{2} +\ln 2 \Bigg)\epsilon\Bigg],
\end{eqnarray}
\begin{eqnarray}\label{tryteyt}
\xi_{0}^{T} = \Biggl\{ \frac{\epsilon}{(N + 8)}\Bigg[1 + \frac{3}{N + 8}\Bigg( -\frac{1}{6} + \ln 2 \Bigg)\epsilon + \frac{9N + 42}{(N + 8)^{2}}\epsilon \Bigg]\Biggl\}^{1/(d - 2)},
\end{eqnarray}
\begin{eqnarray}\label{bcnbvcbc}
C^{c} = \frac{2D^{1/\delta}}{g^{\ast 1/2\beta}} \Bigg\{ 1 - \frac{9}{2(N + 8)}\Bigg[ - \ln 2 - \frac{N - 1}{9}\ln 6 + \frac{2(N + 8)}{27} \Bigg]\epsilon\Bigg\},
\end{eqnarray}
\begin{eqnarray}\label{bvnbcvb}
\xi_{0}^{c} = \frac{2^{1/2}D^{1/2\delta}}{g^{\ast 1/4\beta}} \Bigg\{ 1 - \frac{9}{4(N + 8)}\Bigg[ - \ln 2 - \frac{N - 1}{9}\ln 6 + \frac{N + 14}{27} \Bigg]\epsilon\Bigg\}^{1/2},
\end{eqnarray}
\begin{eqnarray}\label{eetetee}
\widehat{D} = 1,
\end{eqnarray}
where the nontrivial fixed point is obtained through the corresponding $\beta$-function
\begin{eqnarray}
\beta(g) = -\epsilon g + \frac{N + 8}{6}g^{2} - \frac{3N + 14}{12}g^{3},
\end{eqnarray}
computed from the evaluated Feynman diagrams (\ref{klijouio})-(\ref{bnnmbj}) just displayed below
\begin{eqnarray}\label{klijouio}
\parbox{10mm}{\includegraphics[scale=1.0]{fig10.eps}} = \frac{1}{\epsilon} \Biggr[1 - \frac{1}{2}\epsilon - \frac{1}{2}\epsilon J(P^{2}) + \frac{R}{6\mu^{2}}\epsilon J_{R}(P^{2}) - \frac{R_{\mu\nu}P^{\mu}P^{\nu}}{3\mu^{4}}\epsilon J_{R_{\mu\nu}}(P^{2}) \Biggr],
\end{eqnarray}
\begin{eqnarray}\label{bnnmbj}
\parbox{12mm}{\includegraphics[scale=0.8]{fig21.eps}} = \frac{1}{2\epsilon^{2}} \Biggr[1 - \frac{1}{2}\epsilon - \epsilon J(P^{2}) + \frac{R}{3\mu^{2}}\epsilon J_{R}(P^{2}) - 2\frac{R_{\mu\nu}P^{\mu}P^{\nu}}{3\mu^{4}}\epsilon J_{R_{\mu\nu}}(P^{2})\Biggr],
\end{eqnarray}
where
\begin{eqnarray}\label{ugujdfjgdhg}
J_{R}(P^{2}) = \int_{0}^{1}d x\frac{x(1 - x)}{x(1 - x)P^{2}},
\end{eqnarray}
\begin{eqnarray}\label{ufgfghujjgdhg}
J_{R_{\mu\nu}}(P^{2}) = \int_{0}^{1}d x\frac{x^{2}(1 - x)^{2}}{\left[x(1 - x)P^{2}\right]^{2}}.
\end{eqnarray}
The nontrivial fixed point s given by
\begin{eqnarray}\label{klkoilkj}
g^{\ast} = \frac{6\epsilon}{(N + 8)}\Bigg\{ 1 + \epsilon\left[ \frac{(9N + 42)}{(N + 8)^{2}} \right]\Bigg\}.
\end{eqnarray}
As this method is more general and elegant, the $R$ and $R_{\mu\nu}$ terms dependent on the curved spacetime cancel out already in the amplitude ratios, $\beta$-function and nontrivial fixed point computation, such that the corresponding curved spacetime quantities themselves are the same as that of flat spacetime. Thus, the resulting curved spacetime amplitude ratios are automatically the same as their flat spacetime counterparts.
\section{Massless BPHZ method}\label{Massless BPHZ method}
\par The massless BPHZ method \cite{BogoliubovParasyuk,Hepp,Zimmermann} is different from the two earlier ones. While in the two former ones the divergences of the theory are absorbed by a procedure starting from the bare theory, in the former one the divergences are absorbed starting from the renormalized theory by introducing counterterms diagrams. Then only the divergent part of the resulting diagrams are to be considered and the operator which represents this operation is given by $\mathcal{K}(~~~)$ \cite{Neto2017} and acts in the corresponding diagrams. Thus, although the present method is distinct from that of earlier section, at least at the loop level considered here, they lead to the same results. Now we have to present our conclusions.
\section{Conclusions and perspectives}
\par In this Letter we have probed the effect of conformal symmetry on the curved spacetime amplitude ratios for massless O($N$) $\lambda\phi^{4}$ scalar field theories in curved spacetime. By applying three distinct and independent field-theoretic renormalization group methods, the amplitude ratios have been evaluated in the normalization conditions, minimal subtraction scheme and massless BPHZ methods, respectively, where the external momenta of Feynman diagrams have been held at fixed values in the former method and at arbitrary values in the later ones. We have found, at least at the loop level considered, identical curved spacetime amplitude ratios values when obtained through the three methods. We emphasize the importance of employing more than one method for computing the amplitude ratios since we can check the final results when obtained through that methods. Furthermore, the curved spacetime amplitude ratios obtained having been the same as their flat spacetime counterparts. This fact has confirmed the two-scale-factor universality hypothesis validity since the symmetry probed here has been one defined in the spacetime where the field is embedded and not in its internal one. We believe that the influence of conformal symmetry on the critical properties of systems undergoing continuous phase transitions can be analogously probed in future works through the evaluation of finite-size scaling effects, corrections to scaling etc.
\section*{Acknowledgements}
\par HASC and PRSC would like to thank CAPES (brazilian funding agency) for financial support. Furthermore, PRSC would like to thank CNPq (brazilian funding agency) for financial support through Grant number: 431727/2018.
\bibliography{apstemplate}
\end{document}
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A few weeks back I gave a Chickens 101 workshop in Concord, NH for people who are planning to establish chicken flocks in their backyards.
It was an introductory class on how to get chicks and then, once you have them, how to take care of the birds until they reach egg laying age.
The second workshop (cleverly named Chickens 102) scheduled for April 30th will cover some of the problems you need to be aware of with chickens and basic chicken first aid. (It’s also going to cover “coning” but we won’t get into that right now.)
Because chickens are new to Concord – it’s a probationary program where they are letting people have 5 hens on their property – there are going to be a lot of people who will need chicken support, at least initially. For this reason, not only am I holding the workshops, but I’ve also written an article for the Concord Monitor which covers some of the chicken basics.
The Monitor sent down a photographer (Brad) to get “action shots” for the story and he and I had a lovely chat about chickens in general. Brad was intrigued with the diversity in our flock and asked all kinds of chicken questions. (See, even during a photo shoot, our birds are fowl ambassadors, sharing the chicken love.)
I told Brad about Charlie and so he also grabbed a few photos of her (she happened to be posing perfectly on a couch.)
Brad took some pictures of Charlie our house chicken, – I wrote to the Editor, feeling that I needed to tell her why there was a photo of a chicken on a couch in the set. I gave her a quick version of Charlie’s history and explained that she was our adopted, rescued chicken who now lives indoors with a human flock and a dog.
Her response?
“What a crazy thing!”
But knowing a good story when she hears one, the editor asked me to write a short piece about Charlie that will run alongside the Introduction to Chickens article on April 22.
So much like our Good-Egg literary chickens, our video-character chicken, and our painting for a playground hen, we now have a new little celebrity-bird in our midst.
Our little chick; Charlie is brilliantly showing others that despite the fact that life may not have dealt you the best hand, with the help of others, and by being a member of a flock that cares and looks out for you, you can overcome anything and succeed.
3 responses to “Lesson 532 – Our newest celebrity chick”
Coning? Never heard of it. Charlie is looking cute as usual.
Coning – a made up word (on my part) that means : Using a killing cone to humanely kill a chicken.
Not a pleasant thought but something I teach in my classes to those who make the decision to have chickens.
Wendy
Oh, I see. Yes, I’ve heard of a killing cone before but never used one. When the necessity arises we use a board, a brick and a wire loop to keep the chicken calm. We get them calm and in position for a once-off no-miss chop. A quick exit is the goal. Not a nice thing, but we all need to be ready for the day it has to be done if we breed chickens. I have to admit that in the case of a couple of indoor ‘pet’ chickens we have been sooky and taken them to the vet. They just felt like one of the dogs or cats, not like one of the chickens being grown for meat.
| 138,253
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When you may need me
For your change to be successful you need the willing cooperation of people outside your authority. You’ve organised meetings and workshops, spoken energetically at every opportunity and written until blue in the pen. But stuff isn’t happening.
You feel that, despite doing the right things, it’s hard to make change stick. The unexpected too often causes delay. Traditional, programmatic approaches seem too hard. Business cases look unrealistically ambitious and depend too much on hand-wavy assumptions about “culture change”. Things are just too unclear, too complex.
You may be using agile methods but progress is slow because your wider organisation isn’t very agile. Maybe there is too much emphasis on technology. You sense a lack of connection between those creating new capabilities and those who use or support them. Perhaps there is conflict.
What I do
I support organisational change programmes. I work on business change, technology reform and digital transformation; which I also write and speak about. Typical engagements range from a few days on something specific to longer periods providing more general support.
My skills are in...
Change strategy - I work with management teams and individuals to develop ideas, approaches and behaviours that bring about genuine change in complex, uncertain situations.
Communication - I write things, create presentations and make videos that explain complicated stuff simply and capture imaginations.
Facilitation - I run events, workshops and conferences in inventive ways that connect and inspire people. (Without yoga, eye-closing or hand-holding.)
What you get
What you get depends on the situation. I cook from fresh.
This case study is typical...
Here are some other examples in each of my areas of specialism...
Change Strategy
Four Ex Model
A government organisation I worked with was struggling to get innovations into operation effectively. I found that the problem was the model of delivery; so I developed a new one – which I called the Four Ex Model. They liked it. And it fixed things.
Communication
Gubbins of Government
There was confusion about the future architecture of government services. I made a video called the Gubbins of Government. It went viral. People were less confused. And many were delighted. I am often asked to speak about it.
Facilitation
Digital Women Unconference
Digital Leaders were seeking an alternative approach to a significant meeting - their Digital Women Unconference. I had persuaded the team that an unconference style would be appropriate and ran the event in a format known as "World Cafe". It turned in to a great big buzz of a meeting. See this I wrote afterwards: We should talk more with each other
What I get
What I get is the satisfaction of using my skills and everything else that I am to enable others to make progress in tough situations. I like helping people and the acknowledgement that comes from it. With luck, I get to make a living as well.
Ultimately...
Change is very much easier when the people involved understand and believe in what is happening. Get this right… and useful stuff just starts to happen.
| 39,517
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TITLE: Bhargava's work on the BSD conjecture
QUESTION [19 upvotes]: How much would Bhargava's results on BSD improve if finiteness of the Tate-Shafarevich group, or at least its $\ell$-primary torsion for every $\ell$, was known? Would they improve to the point of showing $100$% of elliptic curves over $\mathbf{Q}$ satisfy the BSD conjecture? (which, of course, would still not prove the BSD conjecture)
I've just attended a very nice seminar talk about the topic, and I'm curious to get some expert info.
REPLY [8 votes]: This was meant to be a comment, but it won't fit, so here we go.
Just a few naive observations.
For completeness:
Weak BSD For any global field $K$ and any abelian variety $A$ defined over $K$, we have:
$$\text{ord}_{s=1}L(A/K,s) = \text{rk}_{\mathbf{Z}}A(K).$$
BSD For any global field $K$ and any abelian variety $A$ defined over $K$, we have:
(1) $\text{Sha}(A/K)$ is a finite abelian group.
(2) $\rho :=\text{ord}_{s=1}L(A/K,s) = \text{rk}_{\mathbf{Z}}A(K).$
(3) $\lim_{s\to 1}L(A/K,s)(s-1)^{-\rho} = \frac{R_A\cdot\Omega_A\cdot\#\text{Sha}(A/K)}{|\Delta_K|^{1/2}\cdot \# A(K)_{\rm tor}\cdot\# A^{\vee}(K)_{\rm tor}}.$
Super-weak BSD The proportion of abelian varieties defined over $K$ that satisfy Weak BSD above, is 100%.
If $K$ is the global function field of a smooth projective geometrically irreducible curve over $\mathbf{F}_q$, then it is known by work of Tate, Milne, Bauer, Schneider, Kato, Trihan, that Weak BSD is equivalent to BSD, and BSD is in turn equivalent to finiteness of the $\ell$-primary torsion in the Tate-Shafarevich group for some prime $\ell$ (allowed to be the characteristic of $K$).
If $K$ is a number field, the above stream of equivalences is still expected to be true, but not known yet, likely because there's still no robust cohomological method to put us in the position to mimic the flat cohomology/crystalline-syntomic cohomology methods employed in the positive characteristic case, which Iwasawa theory for abelian varieties was partially meant for. Let us grant for a moment the following:
Expectation Let $K$ be a number field. For any abelian variety $A$ defined over $K$, the following are equivalent:
(1) $\text{Sha}(A/K)$ is finite.
(2) For all primes $\ell$, the $\ell$-primary torsion subgroup in $\text{Sha}(A/K)$ is finite.
(3) For some prime $\ell$, the $\ell$-primary torsion subgroup in $\text{Sha}(A/K)$ is finite.
(4) Weak BSD is true for $A$.
(5) Strong BSD is true for $A$.
This expectation, a Theorem if $K$ were a positive characteristic global function field, reveals the problem is really about showing finiteness of $\text{Sha}(A/K)(\ell)$.
Over number fields, it is not even known that $\text{Sha}(A/K)(\ell)$ vanishes for almost all primes $\ell$.
The takeaway from the Bhargava-Skinner-Zhang paper, and from the answers and comments here, is that knowledge of finiteness $\text{Sha}(A/K)(\ell)$ does not actually help much or at all to improve their progress on Super-weak BSD ($\text{Sha}$-finiteness enters through the parity conjecture, a theorem unconditionally, and definitely a much weaker statement than finiteness of $\text{Sha}(A/K)(\ell)$), which to me just means such methods fail to get to the point of the problem itself, and will not solve it.
In other words, I don't see any trace of the ability to produce interesting algebraic cycles on abelian varieties, into them, and according to the Expectation above, true in char $p$ though open in char $0$ so far, this should be the whole point.
Regardless, clearly Super-weak BSD, which is what the discussion in the question, comments, and around the Bhargava-Skinner-Zhang paper, were about, is not equivalent to weak BSD under any circumstances, no matter that $K = \mathbf{Q}$ and $A$ is $1$-dimensional.
Showing 100% of elliptic curves over $\mathbf{Q}$ satisfy the BSD conjecture does not show weak BSD (as the OP took care to make clear in his/her question). Such result, if ever available, should be regarded as being motivational only.
EDIT: I should also add that several of the averages (both unconditional and conjectural) that are key to the Bhargava-Skinner-Zhang methods, fail in char $p$, while I'd regard a method towards BSD to be "promising", if it were able to make progress or settle its char $p$ counterpart, first.
| 158,694
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President
lisa@multicultural.com | Phone 212.242.3351
Lisa Skriloff founded Multicultural Marketing Resources, Inc. (MMR) to bring visibility to the nation’s top ethnic and niche marketing resources available to corporations and to journalists. Established in 1994, MMR is a public relations and marketing firm representing the leading experts in marketing to Hispanics, Asian Americans, African Americans, LGBT consumers and other ethnic and lifestyle groups. MMR is also the publisher of The Source Book of Multicultural Experts, MMRNews and a Speakers Showcase.
Prior to starting the company business, Lisa had a 10-year career at The New York Times, where she held a number of director level positions in teaches “Marketing to the New Majority: How to Reach the Multicultural Consumer,” at New York University’s (NYU) SCPS and has thought, contributing to Travel Agent Magazine, The NY Daily News, Jaxfax and Fodors Guides, and is the founding editor of Multicultural Travel News (MTN) and Dance Travel News. She specializes in multicultural travel, Spanish language destinations, dance and cruise travel.
She is co-author of the book “Men Are From Cyberspace: The Single Woman’s Guide to Finding Love Online” published by St. Martin’s Press in December 1997.
Watch video of Lisa Skriloff, President, Multicultural Marketing Resources interviewed (during PRSA’s 2010 Sliver Anvil Judging) about diversity, Census 2010 and the growth of multicultural communications.
| 185,879
|
Brazil Cerrado - Brazil Cerrado
$11.75
Brazil Cerrado
******* Back in Stock********
screen size 17/18
Cerrado Region - Minas Gerais Brazil
classic sweet clean coffee from one of Brazils best growing regions
Cup Profile
full body smooth sweet clean cup tasting of chocolate and caramel notes - nice as a base for espresso blends or roast for filter or plunger to take advantage of the sweetness
| 415,129
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\section{Safety Lemmas: Proofs of Lemmas \ref{lem:causal1}, \ref{lem:causal2}}
\label{app:safety}
\subsection{Preliminary Lemmas}
\begin{lemma}\label{lem:vc_after_apply}
Consider a node $s$ and a point $P$ at which the $\texttt{Apply\_Inqueue}$ action is performed such that the if statement on line \ref{line:applycondition}returns true. Suppose that just before point $P$, $InQueue.Head = (j,X,v,t).$ Then, at any point $Q$ which is identical to or after point $P$, we have $vc^{s,P} \geq t.ts$.
\end{lemma}
\begin{proof}
When a node $s$ performs an $\texttt{Apply}\_\texttt{InQueue}$ internal action, at the beginning of this action, $(j,X,v,t)$ is the head of $InQueue$, and $t.ts[p]\leq vc[p]$ for all $p\neq j$, and $t.ts[j]=vc[j]+1$.
On performing this $\texttt{Apply}\_\texttt{InQueue}$ action, the $j$th component of local vector clock is incremented to $vc[j]=t.ts[j]$.
So after this $\texttt{Apply}\_\texttt{InQueue}$ action is complete, $ vc[p]\geq t.ts[p]$ for all $p\neq j$ and $vc[j]=t.ts[j]$, so $vc\geq t.ts$.
\end{proof}
\begin{lemma}\label{lem:listtag}
Consider a point $P$ of an execution $\beta$ of $CausalEC$ such that at server $s$, the list $L_s^{P}[X]$ is non-empty. At any point $Q$ of $\beta$ that comes after $P$, at least one of the following statements is true:
\begin{itemize}
\item $L_s^{Q}[X]$ is non-empty and $ L_s^{Q}[X].Highesttagged.tag \geq L_s^{P}[X].Highesttagged.tag$, or
\item $M_s^{Q}.tagvec[X] \geq L_s^{P}[X].Highesttagged.tag$
\end{itemize}
\end{lemma}
\begin{proof}
Let $t=L_s^{P}[X].Highesttagged.tag.$ Since $L_s^{P}[X]$ is non-empty, there exists a tuple $(t,v)$ in $L_s^{P}[X].$ If $(t,v)$ exists in $L_s^{Q}[X]$ then $L_s^{Q}[X].Highesttagged.tag \geq t = L_s^{P}[X].Highesttagged.tag$, and therefore the lemma statement holds. We consider these case where $(t,v)$ does not exist in $L_s^{Q}[X]$. Let $Q'$ be the first point after $P$ at which tuple $(t,v)$ does not exist in $L_s^{Q}[X];$ note that $Q'$ is no later than $Q$.
From the protocol, a $\texttt{Garbage\_Collection}$ action took place at $Q'.$ In the execution of the action, it executed either line \ref{line:GC1} or line \ref{line:GC2} in Algorithm \ref{alg:internal_actions}. From the code in these lines, we have $t \leq tmax[X].$ From the definition of $tmax[X]$ in line \ref{line:tmax}, we conclude that there is a tag $t' \geq t$ such that server $s$ has an element of $(t',s)$ in $DelL_s^{Q'}[X].$ From Lemma \ref{lem:tag-delete-ordering}, we infer that $t' \leq M_s^{Q'}.tagvec[X].$ We thus infer that $L_s^{P}[X].Highesttagged.tag = t \leq t' \leq M_s^{Q'}.tagvec[X] \stackrel{(a)}{\leq} M_s^{Q}.tagvec[X],$ where $(a)$ follows from Lemma \ref{lemma:tagsalwaysincrease}. This completes the proof.
\end{proof}
\begin{lemma}\label{lem:vc}
Consider an execution $\beta$ of $CausalEC$.
At any point $P$ be $\beta,$ for any server $s$, we have:
\begin{enumerate}[(a)]
\item If list $L[X]$ is non-empty, then, for any entry $(t,v)$ in $L_s^{P}[X],$ we have $vc_s^P \geq t.ts$
\item $vc_s^P \geq M_s^{P}.tagvec[X].ts$ for any object $X$, and
\end{enumerate}
\end{lemma}
\begin{proof}
We begin the proof with the following claim.
\begin{claim}
Let $t=M_s^{P}[X].tag.$
There, either $t = \vec{0}$ or there is a point $Q$ before point $P$ such that there exists a tuple of the form $(t,v)$ in $L_s^{Q}[X].$
\label{claim:mtagslist}
\end{claim}
\begin{proof}[Proof of Claim]
If $t \neq \vec{0},$ then the $M_s[X].tag$ was updated in an $\texttt{Encoding}$ action either in lines \ref{line:updatetag1} or \ref{line:updatetag2} at some point $Q'$ that is not later than $P$. Based on the conditions to execute these lines in lines \ref{line:conditionforupdate} and \ref{line:Forloopupdatetag2}, there exists a tuple of the form $(t,v)$ in $L_s^{Q}[X],$ at point $Q'.$ Furthermore, since an $\texttt{Encoding}$ action does not add elements to the list $L_s[X]$, this element must exist in the list at some point $Q$ before $Q'.$
\end{proof}
The claim above implies that if $(a) \Rightarrow (b).$ To see this, suppose as a contradiction, $(b)$ is violated, that is, then $t=M_s^{P}{[X]}.tag.ts > vc_s^{P}$ for some point $P$. Then the claim shows that there is a point $Q$ before $P$ such that an element $(t,v)$ is in $L_s[X]^{Q}$ at some point $Q$ before $P$. Therefore $L_s^{Q}[X].Highesttagged.ts \geq t \geq vc_s^{P} \stackrel{(1)}{\geq} vc_s^{Q},$ where in $(1)$ we have used Lemma \ref{lem:vconlyincreases}. This is a contradiction of $(a)$.
So to complete the proof, it suffices to show $(a).$
\underline{Proof of (a)}
Let $P$ be the first point of $\beta$ where $(a)$ is violated.
At point $P$ where $L_s^{P}[X]$ is non-empty, and there is an element $(t,v) \in L_s^{P}[X]$ such that $t.ts \leq vc_{s}^{P}.$
Since $vc_s$ cannot decrease (Lemma \ref{lem:vconlyincreases}, we infer that an element $(t,v)$ is added at point $P$ to $L_s^{P}[X]$ by executing:
\begin{enumerate}[(i)]
\item lines
\ref{line:writeaddtolist} in Algorithm \ref{alg:inputactions_clients}, or
\item line \ref{line:list_applyaction} in Algorithm \ref{alg:internal_actions}, or
\item line \ref{line:readaddtolist} in Algorithm \ref{alg:input_actions}, or
\item line \ref{line:valrespaddtolist} in Algorithm \ref{alg:input_actions}.
\end{enumerate}
We show a contradiction in each case.
\underline{(i):}From the protocol, $L_s^{P}[X].Highesttagged.ts = vs_s^{P},$ which does not violate $(a),$ a contradiction.
\underline{(ii):} From the protocol, we know that before point $P$, the element at $\texttt{Inqueue}.Head$ had tag $t$ and value $v$. Furthermore, the fact that Line \ref{line:list_applyaction} implies that the condition in Line \ref{line:list_applyaction} returned true. From Theorem \ref{lem:vc_after_apply}, we have $vc_s^{P} \geq t.ts$. Therefore $(a)$ is not violated at point $P$.
\underline{(iii) and (iv)}
In these cases, because of conditions in lines \ref{line:valrespencoded_readcheck}, \ref{line:valresp_handle_condition}, a tuple with tag-vector $tvec$ exists in $ReadL_s^{P}$ with $tvec[X] = t$ at the point of receipt of the $\texttt{val\_resp}$ or $\texttt{val\_resp\_encoded}$ message.
Such an element is added at a point $Q$ before point $P$ either at line \ref{line:addtoreadl} in Algorithm \ref{alg:inputactions_clients}, or line \ref{line:readlentryencoding} in Algorithm \ref{alg:internal_actions}. In either case, we have $tvec= M_s^{Q}.tagvec$. Because we assume that $(a)$ is contradicted at point $P$, we have $t.ts = tagvec[X].ts > vc_s^{P}.$ From Lemma \ref{lem:vconlyincreases}, we have $tagvec[X].ts > vc_s^{Q},$ which readily implies that $M_s^{Q}.tagvec[X].ts > vc_s^{Q}.$ Therefore, point $Q$ violates $(b)$. Claim \ref{claim:mtagslist} implies that there is a point $R$ before $Q$ which violates $(a)$. Because $Q$ is no later than $P$, this violates the hypothesis that $P$ is the first point at which $(a)$ is violated. This completes the proof.
\end{proof}
\begin{lemma}\label{lem:readlist}
Consider an execution $\beta$ of $CausalEC$.
At any point $P$ be $\beta,$ for any server $s$, suppose there exists an entry $(clientid, opid, X, tagvec, w_1, w_2, \ldots, w_N)$ in $ReadL_{s}^{P}$. Then, for all objects $\overline{X} \in \mathcal{X}_{s},$ we have:
\begin{itemize}
\item $ M_s^P.tagvec[\overline{X}] \geq tagvec[\overline{X}],$ and
\item $vc_s^P \geq tagvec[\overline{X}].ts$
\end{itemize}
\label{lem:readLtagsmallerthanvc}
\end{lemma}
\begin{proof}
Suppose the entry $(clientid, opid, X, tvec, w_1, w_2, \ldots, w_N)$ is added to $ReadL_s$ at point $Q$, which is before $P$.
From the algorithm, at point $Q$, the node executed line \ref{line:addtoreadl} in Algorithm \ref{alg:inputactions_clients} or Line \ref{line:readlentryencoding} in Algorithm \ref{alg:internal_actions}. In either case, we have $tvec=M_s^{Q}.tagvec.$ Consider any object $\overline{X} \in \mathcal{X}_{s}$. From Lemma \ref{lemma:tagsalwaysincrease}, we have $tvec[\overline{X}] = M_s^{Q}.tagvec[\overline{X}] \leq M_s^{P}.tagvec[\overline{X}]$ as desired. Further, this readily implies that $tvec[\overline{X}].ts \leq M_s^{P}.tagvec[\overline{X}].ts$. From Lemma \ref{lem:vc}, we have $M_s^{P}.tagvec[\overline{X}].ts \leq vc_s^{P}.$ Therefore, we have $tvec[\overline{X}].ts \leq vc_s^{P}.$ This completes the proof.
\end{proof}
\begin{lemma}
Consider any execution $\beta$ of $CausalEC.$
\begin{enumerate}[(i)]
\item
If an item $(t,v)$ is in list $L[X],$ then $v$ is the value of the unique write with tag $t$
\item at any point $P$ in the execution,
$M_s^{P}.val = \Phi_{s}(w_1, w_2, \ldots, w_N)$ where $w_i$ is the value of the unique write with tag $M_s^{P}.tag[O_i]$ if $M_s^{P}.tag[O_i] \neq 0$, and $w_{i} = \vec{0}$ if $M_s^{P}.tag[O_i] = \vec{0}.$
\item When node $s$ sends a $\langle \texttt{val\_resp\_encoded}, \bar{M}, clientid, opid, X, tagvec \rangle $, then
$$M.val = \Phi_{s}(w_1, w_2, \ldots, w_N)$$ where $w_i$ is the value of the unique write with tag $\bar{M}.tagvec[O_i]$ if $\bar{M}.tag[O_i] \neq 0$, and $w_{i} = \vec{0}$ if $\bar{M}.tagvec[O_i] = \vec{0}.$
\item
Suppose there exists an entry $(clientid, opid, X, tagvec, w_1, w_2, \ldots, w_N)$ in $ReadL_{s}^{P}$. Then $w_{i}$ is either equal to $\bot$, or $\Phi_{i}(v_1, v_2, \ldots, v_K)$ where $v_i$ is the value of the unique write with tag $tagvec[O_i]$.
\end{enumerate}
\label{lem:valuesarelegitimate}
\end{lemma}
\begin{proof}[Proof Sketch]
The lemma holds because of the following claims.
\begin{claim}
Consider an execution $\beta$ where at point $P$, a $\texttt{val\_resp\_encoded}$ message is received. Suppose in execution $\beta$, at every point before point $P$, all the invariants stated by the lemma holds. Then, the lemma holds at point $P$ as well.
\label{lem:val_resp_encoded_integrity1}
\end{claim}
\begin{proof}
At point $P$, a $\langle\texttt{val\_resp\_encoded},\bar{M},clientid, opid,requestedtags, \rangle$ message is received by server $s$ from server $s'$. If the condition in \ref{line:valrespencoded_readcheck} in Algorithm \ref{alg:input_actions} is not satisfied, then the server state does not change, the lemma continues to hold. Henceforth, we assume that the condition in \ref{line:valrespencoded_readcheck}, that is, there is a tuple $(clientid, opid, \overline{X}, requestedtags, \overline{w})$ in $ReadL_s^{P}.$
We show that if $Error1[X] =Error2[X]=0$ for all $X \in \mathcal{X}_{s'},$ then by the time line \ref{line:errorcondition} is executed, $Modified\_Codeword$ contains a codeword that is equal to $\Phi_{s'}(v_1, v_2, \ldots, v_K)$ where $v_\ell, \ell=1,2,\ldots,K$ represents the value of the unique write whose tag is equal to $requestedtags[X_\ell].$
Before entering the for loop, because theitem (iii) of the lemma statement holds before point $P$, and $Modified\_Codeword$ is initialized to $\overline{M}.val$ in line \ref{line:modifiedcodeword_init}, we know that
$Modified\_Codeword = \Phi_{s'}(w_1, w_2, \ldots, w_K)$
where $w_{i}$ is the value of the unique write with tag $\overline{M}.tag[X_i].$ Based on Definition \ref{def:ecobjectstored}, it follows that we can write $$Modified\_Codeword = \Phi_{s'}(\overline{w}_1, \overline{w}_2, \ldots, \overline{w}_K)$$ where
$$ \overline{w}_{k} =\left\{\begin{array}{cc} \textrm{the value of the write with tag}requestedtags[X_k] & \textrm{if }X_k \notin \mathcal{X}_{s'} \\ \textrm{the value of the write with tag}\overline{M}.tagvec[X_k]& \textrm{if }X_k \in \mathcal{X}_{s'} \end{array}\right.$$
We examine the for loop in Line \ref{line:valrespencoded_objectforloop} iteratively over the object in $\mathcal{X}_{s'}$, and show that the loop ensures that for each $X \in \mathcal{X}_{k},$ the codeword in $Modified\_Codeword$ is modified to reflect the encoding of the value of the write with tag $requestedtags[X].$
For an object $X \in \mathcal{X}_{s'}$, if $requestedtags[X] = \overline{M}.tagvec[X],$ then no modification is required and the For loop \ref{line:valrespencoded_objectforloop} proceeds to the next object. If $requestedtags[X] = \overline{M}.tagvec[X],$ since the hypothesis of the claim states that $Error1[X] \neq 1$, we infer from condition in line \ref{line:metaerrorcondition1} that either $\overline{M}.tagvec[X] = \mathbf{0}$ or line \ref{line:Modifiedcodeword1} is executed. In either case, by the time line \ref{line:metaerrorcondition2} is checked, $Modified\_Codeword$ stores a codeword that corresponds to value $\mathbf{0}$ for object $X$. Because $Error2[X] \neq 1,$ line \ref{line:Modifiedcodeword2} is executed. At the end of this line, $Modified\_Codeword$ stores a codeword that corresponds to value of the write $requestedtags[X]$ for object $X$.
Therefore,at the point where line \ref{line:errorcondition} is executed,
$$Modified\_Codeword = \Phi_{s'}(\overline{v}_1, \overline{v}_2, \ldots, \overline{v}_K)$$ where
$$ \overline{v}_{k} = \textrm{the value of the write with tag}requestedtags[O_k]$$
This implies that if $ReadL_s^{P}$ or if $L_s^{P}[X]$ are modified in line \ref{line:valrespencoded_addtuple} or line \ref{line:readaddtolist} are consistent with the (i),(iv) stated in the lemma.
\end{proof}
\begin{claim}
Consider an execution $\beta$ where at point $P$, a $\texttt{val\_resp\_encoded}$ is sent by server $s$. Suppose in execution $\beta$, at every point before $P$, all the invariants stated by the lemma holds. Then, the lemma holds at point $P$ as well.
\label{lem:val_resp_encoded_integrity2}
\end{claim}
\begin{proof}
Note that point $P$ is the point of receipt of a $\texttt{val\_inq}$ message. From Line \ref{line:Responsetovalinqinit} in Algorithm \ref{alg:input_actions}, we know that $ResponsetoValInq$ is initialized to $M.val$. Because invariant (ii) of the lemma is true by point $P$, we know that after executing line \ref{line:Responsetovalinqinit}, we have
$ResponsetoValInq.val = \Phi_{s'}({w}_1, w_2, \ldots, w_K)$ where
$w_{k}, k=1,2,\ldots,K$ is the value of the write with tag $M_s^{P}.tagvec[X_k]$. Also note that $ResponsetoValInq.tag$ is equal to $M_s[X].tag$. Since invariant (i) stated by the lemma is true by point $P$, and the action at point $P$ does not modify $L_s^{P}[X]$, the invariant is true at point $P$ as well. Specifically, any value $(t,v)$ in $L_s^P[X]$ has the property that $v$ is the value of the unique write with tag $t$. Therefore, after executing \ref{line:ResponsetoValInqvalupdate1}, \ref{line:ResponsetoValInqtagupdate1}, or after executing \ref{line:ResponsetoValInqvalupdate2}, \ref{line:ResponsetoValInqtagupdate2}, the following invariant is true:
$ResponsetoValInq.val = \Phi_{s'}(\overline{w}_1, \overline{w}_2, \ldots, \overline{w}_K)$
where $\overline{w}_{k}, k=1,2,\ldots,K$ is the value of the write with tag $ResponsetoValInq.tagvec[X_k].$
Therefore $(iii)$ is true at point $P.$ Furthermore, the action at point $P$ does not modify $M_s^{P}$ or $ReadL_{s}^{P}.$ Since these variables satisfy Lemma invariants (ii),(iv) before point $P$ they satisfy (ii) and (iv) at point $P$ as well.
\end{proof}
The lemma can then be proved by induction on the sequence of states in $\beta.$ The lemma is clearly true at the initial point of $\beta.$ Assume that the lemma is true point $P$ Of $\beta.$ We can show that the lemma is true at point $P'$ which immediately succeeds $P$. We omit the mechanical details for the sake of brevity, and only provide a sketch here. Informally, at point $P'$, $L[X]$ satisfies (i) because it is either updated by writes, or $\texttt{Apply\_Inqueue}$ actions, or receipt of $\texttt{val\_resp}$ or $\texttt{val\_resp\_encoded}$ messages. Because $(i),(ii),(iii),(iv)$ are satisfied at point $P'$, the only non-trivial case is to show (i) after the receipt of $\texttt{val\_resp\_encoded}$ message, which is shown in Claim \ref{lem:val_resp_encoded_integrity1}.
Because $(i), (ii)$ is satisfied at $P'$, and because $M.tagvec$ and $M.val$ are only updated as a part of $\texttt{Encoding}$ action, the result of the actions satisfy $(ii)$ at point $P$ as well. Claim \ref{lem:val_resp_encoded_integrity2} shows that (iii) is satisfied. $ReadL$ is updated only on receipt of a read, or receipt of $\texttt{val\_resp\_encoded}$ message, or on behalf of an encoding operation. Because of Because $(ii), (iii)$ are satisfied at $P'$, and because of Claim \ref{lem:val_resp_encoded_integrity1}, $ReadL$ satisfies $(iv).$ as well.
\subsection{Proof of Lemma \ref{lem:causal1}}
\begin{proof}
To show that $\leadsto$ is a partial order, it suffices to consider any execution $\beta$ and show that:
\begin{enumerate}[(I)]
\item For any two distinct operations $\pi_1,\pi_2$ in $\beta$, $\pi_1\leadsto \pi_2 \Rightarrow \pi_2 \not \leadsto \pi_1$.
\item $\pi_1 \leadsto \pi_2, \pi_2 \leadsto \pi_3 \Rightarrow \pi_1 \leadsto \pi_3$.
\end{enumerate}
\underline{Proof of {(I)}}
Suppose for a contradiction that $\pi_1 \leadsto \pi_2$, and $\pi_2 \leadsto \pi_1$.
By Definition \ref{def:ourcausalordering}, $\pi_1,\pi_2$ both acquire timestamps, and $ts(\pi_1) \leq ts(\pi_2)$, and $ts(\pi_2) \leq ts(\pi_1)$. This implies that $ts(\pi_1)=ts(\pi_2)$.
\begin{claim}
$\pi_1,\pi_2$ are both read operations.
\end{claim}
\begin{proof}[Proof of Claim]
Because of Lemma \ref{lem:uniquetag1} which states that distinct write operations have different timestamps, $\pi_1$ and $\pi_2$ cannot both be write operations. We show that even one of them cannot be a write operation. Suppose as a contradiction, $\pi_1$ is a write and $\pi_2$ is a read. From the definition of $\leadsto$ in Definition \ref{def:ourcausalordering}, we cannot have $\pi_2 \leadsto \pi_1.$ Therefore $\pi_1$ is not a write. By a symmetric argument, $\pi_2$ is also not a write.
\end{proof}
Because $ts(\pi_1)=ts(\pi_2)$ and $\pi_1 \leadsto \pi_2,$ Definition \ref{def:ourcausalordering} implies that $\pi_1$ and $\pi_2$ happen at a same client and $\pi_1 \leadsto \pi_2$. Because $ts(\pi_1)=ts(\pi_2)$ and $\pi_2 \leadsto \pi_1,$ we have $\pi_2 \leadsto \pi_1.$ However, this is a contradiction to our earlier conclusion that $\pi_1 \leadsto \pi_2;$ this completes the proof.
\underline{Proof of {(II)}}
Consider three operations $\pi_1, \pi_2, \pi_3$ such that $\pi_1 \leadsto \pi_2, \pi_2 \leadsto \pi_3.$ This implies that $\pi_1,\pi_2$ acquire timestamps in $\beta$. If $\pi_3$ does not acquire a timestamp in $\beta$, then Definition \ref{def:ourcausalordering} implies that $\pi_1 \leadsto \pi_3$ to complete the proof.
We now assume that $\pi_3$ acquires a timestamp in $\beta.$
From Definition \ref{def:ourcausalordering}, we have $ts(\pi_1) \leq ts(\pi_2) \leq ts(\pi_3).$ If at least one of these inequalities is strict, we readily have $ts(\pi_1)< ts(\pi_3) \Rightarrow \pi_1 \leadsto \pi_3,$ which completes the proof.
We now consider the case where both these inequalities are not strict, that is, $ts(\pi_1) = ts(\pi_2)=ts(\pi_3).$ The fact that $\pi_1 \leadsto \pi_3 \leadsto \pi_3$ implies that $\pi_2, \pi_3$ are both read operations issued at the same client with $\pi_2 \rightarrow \pi_3$. The fact that $\pi_1 \leadsto \pi_2$ with $ts(\pi_1) = ts(\pi_2)$ implies that, $\pi_1$ is a write operation or $\pi_1 \rightarrow \pi_2;$ in either case, we show that $\pi_1 \leadsto \pi_3$ to complete the proof.
Specifically, if $\pi_1$ is a write operation, then combined with the fact that $ts(\pi_1) = ts(\pi_3),$ Definition \ref{def:ourcausalordering} implies that $\pi_1 \leadsto \pi_3$. On the other hand, if $\pi_1 \rightarrow \pi_2,$ because we have already shown that $\pi_2 \rightarrow \pi_3,$ we have $\pi_1 \rightarrow \pi_3.$ Definition \ref{def:ourcausalordering} implies that $\pi_1 \leadsto \pi_3$ to complete the proof.
\end{proof}
\subsection{Proof of Lemma \ref{lem:causal2}}
\begin{proof}
Let $\pi$ be an operation that terminates in $\beta$. Without loss of generality, assume that $\pi$ is issued by client $c$ in $\mathcal{C}_{s}$ for some server $s.$
Based on the protocol $\pi$ returns can return because at a point $P$ in $\beta$, server $s$ sends a $\langle \texttt{read-return}, opid, v\rangle$ to the client. This message is sent on executing one of the following lines:
\begin{enumerate}
\item Line \ref{line:sendtoread1} in Algorithm \ref{alg:inputactions_clients}.
\item Line \ref{line:readimmediatereturn} in Algorithm \ref{alg:inputactions_clients}, or
\item Line \ref{line:read-send-val_resp_encoded} in Algorithm \ref{alg:input_actions}, or
\item Line \ref{line:valresp_handle} in Algorithm \ref{alg:input_actions}, or
\item Line \ref{line:readreturn_apply} in Algorithm \ref{alg:internal_actions}.
\end{enumerate}
We show that the lemma holds for each of these cases. Note that $ts(\pi) = vc_s^{P}.$
\underline{1. Line \ref{line:sendtoread1} in Algorithm \ref{alg:inputactions_clients}.}
In this case point $P$ is the point of receipt of a $\langle \texttt{write}, opid, X, v\rangle$ message from a write operation $\phi$. By Definition \ref{def:ts_tag}, we have $ts(\phi) = vs_{s}^{P}.$
Therefore $\pi$ returns the value $v$ of a write operation $\phi$ with $ts(\phi) = ts(\pi).$
\underline{2. Line \ref{line:readimmediatereturn} in Algorithm \ref{alg:inputactions_clients}.}
In this case, the value $v$ is one that forms a tuple $(t,v)$ in list $L[X].$ The value $v$ is the value of the unique write operation $\phi$ with timestamp $ts(\phi) = t.ts.$ Furthermore, from Lemma \ref{lem:listtag}, we have $t.ts \leq vs_s^{P}.$ Therefore $\pi$ returns the value of a write operation $\phi$ with $ts(\phi) \leq ts(\pi).$
\underline{3. Line \ref{line:read-send-val_resp_encoded} in Algorithm \ref{alg:input_actions}}
Note that at the point of receipt of the $\texttt{val\_resp\_encoded}$ with operation id $opid$ and tag vector $requestedtags,$ the fact that Line \ref{line:read-send-val_resp_encoded} was executed implies there exists an entry in $ReadL$ with operation id $opid$ and tag vector $requestedtags.$
From Lemma \ref{lem:valuesarelegitimate}, the input to the function $\chi_{T_O}$ on Line \ref{line:valresp_handle_decoding} is a set whose elements are of the form $(\overline{i},{v}_{\overline{i}})$ where ${v}_{i} = \Phi_i(w_1, w_2, \ldots, w_K)$ where, for $\ell \in \{1,2,\ldots,K\},$ we have $w_\ell$ is the value of the unique write with tag $requestedtags[X_\ell].$ Therefore, the output of the decoding function, which is returned to the read, is the value $w$ of the value of the unique write $\phi$ with tag $requestedtags[X].$ Since, at this point, there exists a tuple in $ReadL$ with tag vector $requestedtags,$ Lemma \ref{lem:readLtagsmallerthanvc} implies that $vc_{s}^{P} \geq requestedtags[X].ts.$ Therefore, the read returns the value of a write $\phi$ with $ts(\phi) \leq ts(\pi)$.
\underline{4. Line \ref{line:valresp_handle} in Algorithm \ref{alg:input_actions}}
This is the point of a $\langle \texttt{val\_resp}, X, v, clientid, opid, tvec\rangle.$ From the condition on Line \ref{line:valresp_handle_condition}, we know that there exists ane entry in $ReadL$ with operation id $opid$ and tag vector $tvec$ at point $P$. Furthermore, from Lemma \ref{lem:valuesarelegitimate}, we infer that $v$ is the value of the unique write $\phi$ with tag $tvec[X].$ From Lemma \ref{lem:readLtagsmallerthanvc}, we conclude that $tvec[X].ts \leq vc_s^{P} = ts(\pi).$ Therefore, the read returns the value of a write $\phi$ with $ts(\phi) \leq ts(\pi)$.
\underline{5. Line \ref{line:readreturn_apply} in Algorithm \ref{alg:internal_actions}}
The proof is similar to Case $2.$ The value $v$ is one that forms a tuple $(t,v)$ in list $L[X].$ This is because the tuple is added to list $L[X]$ in the same action in Line \ref{line:list_applyaction}. The value $v$ is the value of the unique write operation $\phi$ with timestamp $ts(\phi) = t.ts.$ Furthermore, from Lemma \ref{lem:listtag}, we have $t.ts \leq vs_s^{P}.$ Therefore $\pi$ returns the value of a write operation $\phi$ with $ts(\phi) \leq ts(\pi).$
\end{proof}
\remove{
\underline{Proof of $(i)$}
$(i)$ holds at point $P$. For a server $s$, if no new element is added to the list at point $P'$, then $(i)$ holds at $P'$ as well. We consider the case where an element of the form $(t,v)$ is added to a list $L_s[X]$ for an object $X$ at point $P'$.
From the protocol, an item is added to list $L[X]$ on four instances: (a) a write to object $X$ in Line \ref{line:writeaddtolist} in Algorithm \ref{alg:inputactions_clients}, (b) an $\texttt{Apply\_Inqueue}$ action to object $X$ in Line \ref{line:list_applyaction} in Algorithm \ref{alg:internal_actions}, (c), an decoding action in Line \ref{line:readaddtolist} in Alg. \ref{alg:input_actions}, or (d) on receiving a $\texttt{val\_resp}$ message in Line \ref{line:valrespaddtolist}. The lemma is true if line (a) is executed because $t$ is the tag of the unique write that sent the $\texttt{write}$ message, and $v$ is the value of the write. The lemma is true if line (b) is executed on receipt of an $\langle \texttt{app},X,v,t\rangle$ from node $s'$, and $(t,v)$ is added to $L_s[X]$. This is because the $\langle \texttt{app},X,v,t\rangle$ is sent by $s'$ on executing Line \ref{line:writesendapply}, and the code implies that $v$ is the value of the write with tag $t$. The lemma is also true in case $(d),$ because the $\langle \texttt{val\_resp}, clientid, opid, \overline{X}, v, \overline{tvec}angle$ message sent by server $s'$ in Line \ref{line:val_resp_send} contains value $v$ where $(v,\overline{tvec}[X]) \in L_{s'}^Q[X],$ at some point $Q$ before $P$, and therefore, $v$ is the value of the write operation with tag $\overline{tvec}[X].$ Thus the property holds at point $P$ as well.
We now consider case $(c),$ which is the receipt of $\langle \texttt{val\_resp\_encoded}, \overline{M},clientid, opid, \overline{X}, tvec \langle$ message at point $P'$. Note that by point $P$, statement $(iv)$ of the lemma is true. Therefore, for, for $\overline{i} \neq s,$ a tuple $(\overline{i},v)$ that is input to $\psi_{T_O}$ on line \ref{line:valresp_handle_decode}, has $v = \Phi_{\overline{i}}(w_1, w_2, \ldots, w_K)$ where $w_{\ell}$ is the value of the write with tag $tvec[X_\ell].$ We argue that a tuple $(\overline{s},v)$ that is input to \ref{line:valresp_handle_decode} also has the property that $v = \Phi_{\overline{i}}(w_1, w_2, \ldots, w_K)$ where $w_{\ell}$ is the value of the write with tag $tvec[X_\ell].$
Therefore, the decoding in
\ref{line:valresp_handle_condition} outputs $v$, where $v$ is the value of the unique writes of tag $requestedtags[X],$ where $X$ is the object of operation $opid.$ In particular, the value $v$ is added to an internal read in the
{\color{red} TBD}}
\end{proof}
\section{Liveness Lemmas: Proofs of Lemmas \ref{lem:error1} and \ref{lem:error2}}
\label{app:liveness}
The main goal of this section is to prove Lemmas \ref{lem:error1} and \ref{lem:error2}.
\subsection{Preliminary Lemmas}
\begin{lemma}
Let $P,Q$ be two points in an execution $\beta$ of $CausalEC$ such that $P$ comes after $Q.$ Then for any node $s,$ for any object $X$, $M_s^{P}.tagvec[X]\geq M_s^{Q}.tagvec[X].$
\label{lemma:tagsalwaysincrease}
\end{lemma}
\begin{proof}
From the protocol code in Algorithms \ref{alg:inputactions_clients} and \ref{alg:internal_actions}, we note that at any server $s$, the only actions that can change $M_s.tagvec[X]$ are $\texttt{Encoding}$ actions. To show the lemma, it therefore suffices to show the following claim:
\begin{claim}
Consider two consecutive points $P,Q$ of the execution, where $P$ is a point preceding an $\texttt{Encoding}$ action, and $Q$ is the point of the $\texttt{Encoding}$ action. Then $M_s^{Q}.tagvec[X] \geq M_s^{P}.tagvec[X].$
\end{claim}
If $X \in \mathcal{X}_{s},$ then line \ref{line:updatetag1} of Algorithm \ref{alg:internal_actions} is the only assignment that can sets $M_s^{P}.tagvec[X] = L_s^{Q}[X].Highesttagged.tag.$ To execute line \ref{line:updatetag1}, we note that the condition in the For loop in line \ref{line:Forloopupdatetag1} needs to be satisfied, and this condition dictates that $L_s^{Q}[X].Highesttagged.tag >M_s^{Q}.tagvec[X]$. Therefore, $M_s^{P}.tagvec[X] >M_s^{P}.tagvec[X]$.
If $X \notin \mathcal{X}_{s},$ then the argument is similar to the case of $X \in \mathcal{X}_{s},$ with the For loop condition in \ref{line:Forloopupdatetag2} coupled with the construction of set $\overline{U}$ in line \ref{line:Delete2_Ubarset} ensuring that
the assignment of line \ref{line:updatetag2} results in $M_s^{P}.tagvec[X] >M_s^{Q}.tagvec[X]$.
\end{proof}
\begin{lemma}
Let $P$ be a point in an execution $\beta$ of $CausalEC$. For any node $s$, either $M_s^{P}.tagvec[X] = \vec{0},$ or there is some point $P'$ which is no later than (and possibly equal to) $P$ in $\beta$ such that a tuple of the form $(M_s^{P}.tagvec[X],val)$ in $L[X]$ at point $P',$ that is, in $L_s^{P'}[X].$
\label{lem:pointsexec}
\end{lemma}
\begin{proof}
Let $P'$ be the first point where $M_s^{P'}.tagvec[X] = M_s^{P}.tagvec[X].$ Note that $P'$ is no later than $P$ in $\beta.$ Since, $M_s.tagvec[X]$ can only be changed by an $\texttt{Encoding}$ internal action, we infer that an $\texttt{Encoding}$ was performed at $P.$ Furthermore, as a part of the action, line \ref{line:updatetag1} or line \ref{line:updatetag2} in Algorithm \ref{alg:internal_actions} was executed at $P'.$ If $X \in \mathcal{X},$ from the protocol in line \ref{line:updatetag1}, we conclude that $L_s^{P'}[X].Highesttagged.tag = M_s^{P'}.tagvec[X].$ If $X \notin \mathcal{X}$, from the construction of set $\overline{U}$ in line \ref{line:Delete2_Ubarset}, we conclude that line \ref{line:updatetag2} updates the tag only if there exists $val'$ that satisfies $(M_s^{P'}.tagvec[X], val') \in L_{s}^{P'}[X]$.
Furthermore, from the protocol, every element in $L[X]$ is of the form $(tag,val)$ with $val$ taking the value of the unique write whose tag is $tag$. Therefore, there exists a tuple of the form $(M_s^{P}.tagvec[X],val)$ at point $P'$ in $L[X]$ at server $s.$
\end{proof}
\begin{lemma}
Let $P$ be a point in an execution $\beta$ of $CausalEC$. For any node $s$, for an object $X$, let $t$ be the highest tag such that node $s$ has sent a message of the form $\langle \texttt{del},X,t\rangle$ or added a tuple of the form $(t,s)$ to $DelL_s[X]$ before point $P$. Then $t \leq M_s^{P}.tagvec[X]$.
\label{lem:tag-delete-ordering}
\end{lemma}
\begin{proof}
We consider two cases: $X \in \mathcal{X}_{s}$ and $X \notin \mathcal{X}_{s}$
\underline{Case (I): $X \in \mathcal{X}_{s}$}
Based on the protocol, node $s$ has sends $\langle \texttt{del},X,t\rangle$ message or adds a tuple of the form $(t,s)$ in either Line \ref{line:Delete_send1} or Line \ref{line:Delete_send3} of Algorithm \ref{alg:internal_actions}. Let $Q$ denote a point of the execution where one of these internal actions is performed. To show the lemma, it suffices to show that $M_s^P.tagvec[X] \geq M_s^Q.tagvec[X]$ for at any point $P$ that comes after $Q$.
\underline{Case (IA) - Line \ref{line:Delete_send1}:} This line is performed as a part of the \texttt{Encoding} internal action. As a part of the state changes in this internal action, in line \ref{line:updatetag1}, the node sets $M_s^Q.tagvec[X]$ to be $t$. From Lemma \ref{lemma:tagsalwaysincrease}, we conclude that at any point $P$ that comes after $Q$, we have $M_s^P.tagvec[X] \geq M_s^Q.tagvec[X] = t.$ This completes the proof.
\underline{Case (IB) - Line \ref{line:Delete_send3}:} Notice that $X \in \mathcal{X}_{s},$ which implies that $s \in R_{s}^{Q}.$
Based on the condition imposed in Line \ref{line:Delete_cond3}, the fact that line (\ref{line:Delete_send3}) was executed implies that $(t,v) \in DelL^{s}_Q[X]$. Specifically, the tuple $(t,v)$ was added at some point $Q'$ before $Q,$ and this was added as a part of line \ref{line:Delete_send1} in an $\texttt{Encoding}$ action. From the result of Case (IA), and because $P$ comes after $Q'$, we conclude that a have $M_s^{P}.tagvec[X] \geq M_s^{Q'}.tagvec[X] = t.$ This completes the proof.
\underline{Case (II): $X \notin \mathcal{X}_{s}$}
Based on the protocol, node $s$ has sends $\langle \texttt{del},X,t\rangle$ message or adds a tuple of the form $(t,s)$ to $DelL[X]$ in Line \ref{line:Delete_send2} of Algorithm \ref{alg:internal_actions} as a part of the $\texttt{Encoding}$ internal action. Let $Q$ denote the point of the execution where this internal action is performed. As a part of the state changes in this internal action, in line \ref{line:updatetag2}, the node sets $M_s^{Q}.tagvec[X]$ to be $t$. From Lemma \ref{lemma:tagsalwaysincrease}, we conclude that at any point $P$ that comes after $Q$, we have $M_s^{P}.tagvec[X] \geq M_s^{Q}.tagvec[X] = t.$ This completes the proof.
\end{proof}
\begin{lemma}
Let $P$ be a point in an execution $\beta$ of $CausalEC$. For any node $s$ and any object $X$, suppose no tuple of the form $(M_s^{P}.tagvec[X],val)$ exists in $L_s^{P}[X].$ Then, node $s$ received a tuple of the from $\langle \texttt{del},X,M_s^{P}.tagvec[X],val\rangle$ from every node in the system by some point before $P$.
\label{lem:currenttag-delete}
\end{lemma}
\begin{proof}
From Lemma \ref{lem:pointsexec}, we know that there is a point $P'$ at or before $P$ such that tuple of the form $(M_s^{P}.tagvec[X],val)$ exists in $L_s^{P'}[X].$ Let $Q$ be the first point after $P'$ such that this tuple does not exist in $L_s^{P'}[X].$ By the hypothesis of the lemma $Q$ can be no later than $P$. From the protocol, since the only action that removes objects from $L_s^{P}[X]$ is a $\texttt{Garbage\_Collection}$ action, $\texttt{Garbage\_Collection}$ action took place at $Q$, and the tuple was removed from $L_s[X]$ as a part of line \ref{line:GC1} or \ref{line:GC3} or \ref{line:GC2} of Algorithm \ref{alg:internal_actions}. From Lemma \ref{lem:tag-delete-ordering}, we infer that for any tag $t > M_s^{Q}.tagvec[X],$ the element $\{(t,s)\}$ does not belong to $Del_s^{Q}[X].$ Therefore, $tmax_s^{Q}[X] \stackrel{(a)}{\leq} M_s^Q.tagvec[X].$ Furthermore, $M_s^Q.tagvec[X] \stackrel{(b)}{\leq} M_s^P.tagvec[X]$ because $Q$ is no later than $P$. Since lines \ref{line:GC1}, \ref{line:GC3}, \ref{line:GC2} can only remove elements from $L_s^{Q}[X]$ with tags no bigger than $tmax_s^{Q}[X]$, we can conclude that:
\begin{itemize}
\item Inequalities $(a),(b)$ are in fact met with equality, which further implies that $t{max}_s^{Q}[X] = M_s^P.tagvec[X]$, and
\item Line \ref{line:GC1} is executed at point Q.
\end{itemize}
Since $t{max}_s^{Q}[X] = M_s^{P}.tagvec[X]$ and line \ref{line:GC1} was executed we conclude that $M_s^{P}.tagvec[X] \in \overline{S}_{s}^{Q},$ where $\overline{S}_{s}^{Q}$ is formed in line \ref{line:s1set}. The definition of line \ref{line:s1set} implies that at $Q,$ $\{i:(M_s^{P}.tagvec[X],i)
\in DelL[X]$ is equal to $\mathcal{N}.$ Since tuples are only added to $DelL[X]$ on receipt of $\langle \texttt{del},X,t,val\rangle$ messages in Line \ref{line:delmessage} in Algorithm \ref{alg:input_actions}, we conclude that node $s$ received a tuple of the from $\langle \texttt{del},X,M_s^{P}.tagvec[X],val\rangle$ from every node in the system by some point before $Q$, and hence before $P$.
\end{proof}
\begin{lemma}
Let $P$ be a point of an execution $\beta$ such that for a node $s$ at point $P$, there exists $val$ such that there is tuple $(M_s^{P}.tagvec[X], val)$ in $L_s^{P}[X]$ for some object $X$. Let $Q$ be a point in $\beta$ at or after $P$ such that a tuple $(clientid,opid, X,tvec,i,\overline{w})$ exists in $ReadL_s^{Q}$ where $tvec[X]=M_s^{P}.tagsvec[X].$ If $M_s^{Q}.tagvec[X] >M_s^{P}.tagvec[X],$ then there exists a tuple $(M_s^{P}.tagvec[X], val)$ in $L_s^{Q}[X].$
\label{lem:listsavespendingread}
\end{lemma}
\begin{proof}
Let $P'$ be the first point after $P$ such that $M_s^{P'}.tagvec[X] > M_s^{P}.tagvec[X]$. Notice that $P'$ is no later than $Q.$ The proof involves three claims: Claim \ref{claim:l1}, \ref{claim:l2} and \ref{claim:l3}.
\begin{claim}
There exists a tuple $(M_s^{P}.tagvec[X], val)$ in $L_{s}^{P'}[X].$
\label{claim:l1}
\end{claim}
\begin{proof}[Proof of Claim \ref{claim:l1}]
At the point $P''$ which is immediately before $P',$ we have $M_s^{P''}.tagvec[X] = M_s^{P}.tagvec[X]$, because $P'$ be the first point after $P$ such that $M_s^{P'}.tagvec[X] > M_s^{P}.tagvec[X]$.
From the protocol, it readily follows that an $\texttt{Encoding}$ action took place at $P',$ and line \ref{line:updatetag1} or line \ref{line:updatetag2} of Algorithm \ref{alg:internal_actions} was executed. Since the condition in line \ref{line:conditionforupdate} or \ref{line:Delete2_condition} returns true for the execution of line \ref{line:updatetag1} or line \ref{line:updatetag2}, we conclude that a tuple of the form $(M_s^{P}.tagvec[X], val)$ exists in $L_s^{P'}[X].$
\end{proof}
\begin{claim}
The tuple $(clientid,opid, X,tvec,i,\overline{w})$ in $ReadL_s^{Q}$ with $tvec[X]=M_s^{P}.tagvec[X]$ is added to $ReadL_s$ at some point before $P'.$
\label{claim:l2}
\end{claim}
\begin{proof}[Proof of Claim \ref{claim:l2}]
Consider any tuple $(clientid,opid, X,tvec,i,\overline{w})$ in $ReadL_s^{Q}$ with $tvec[X]=M_s^{P}.tagvec[X].$
At $P',$ we have $M_s^{P'}.tagvec[X] > M_s^{P}.tagvec[X].$ Because of Lemma \ref{lemma:tagsalwaysincrease}, at any point $R$ after $P',$ $M_s^{R}.tagvec[X] > M_s^{P}.tagvec[X].$ From the protocol, the tuple is added to $ReadL$ during execution of line \ref{line:addtoreadl} in Algorithm \ref{alg:inputactions_clients} or \ref{line:readlentryencoding} in Algorithm \ref{alg:internal_actions}. If the tuple is added at point $Q'$, by examining both lines, we have $tvec[X]=M_s^{Q'}.tagvec[X].$ However, $M_s^{Q'}.tagvec[X] = M_s^{P}.tagvec[X].$ Because for any point $R$ at or after $P'$, we have $M_s^{R}.tagvec[X] > M_s^{P}.tagvec[X],$ we conclude that $Q'$ must be before $P'.$
\end{proof}
\begin{claim}
If a \texttt{Garbage\_Collection} action that takes place between $P'$ and $Q,$ the action does not remove the tuple $(M_s^{P}.tagvec[X], val)$ from $L[X].$
\label{claim:l3}
\end{claim}
\begin{proof}[Proof of Claim \ref{claim:l3}]
Suppose a \texttt{Garbage\_Collection} is performed at point $Q''$ that is between $P'$ and $Q.$ Because $Q''$ comes after $P'$, we have $M_s^{Q''}.tagvec[X] > M_s^{P}.tagvec[X]$. Because of Claim \ref{claim:l2}, the tuple of the form $(clientid,opid X,tvec,i,v)$ with $tvec[X] = M_s^{P}.tagvec[X]$ in $ReadL_s^{Q}$ that is noted in the hypothesis of the lemma is in $ReadL_{s}$ at every point between $P'$ and $Q$, specifically, it is in $ReadL_{s}^{Q''}.$ The \texttt{Garbage\_Collection} action removes elements from $L[X]$ for $X \in \mathcal{X}_{s}$ only if it executes line \ref{line:GC1}, line \ref{line:GC3} or line \ref{line:GC2}. However these lines do not remove elements from $T$ which is found in Line \ref{line:pendingreads}. Because $ReadL_s^{Q''}$ has a tuple of the form $(clientid,opid, X,tvec,i,v)$ with $tvec[X] = M_s^{P}.tagvec[X]$ and because $M_s^{P}.tagvec[X] < M_s^{Q''}.tagvec[X],$ we conclude $tagvec[X]$ is in the set $T$. Therefore, the \texttt{Garbage\_collection} action at point $Q''$ does not remove the tuple $(M_{s}^{P}.tagvec[X], val)$ from $L[X].$
\end{proof}
\end{proof}
\begin{lemma}
Consider a node $s$ and an object $X \notin \mathcal{X}_{s}.$ Let $P$ be a point of an execution $\beta$ such that $M_s^{P}.tagvec[X] > \mathbf{0}.$ Then, for any server $s'$ such that $X \in \mathcal{X}_{s'},$ node $s$ has received a message $\langle \texttt{del}, X,t)\rangle$ from server $s'$ before point $P$ for some $t \geq M_{s}^{P}.tagvec[X]$
\label{lem:handlenonpresentobjects}
\end{lemma}
\begin{proof}
For $X \notin \mathcal{X}_{s},$ $M_s.tagvec[X]$ is updated in Line \ref{line:updatetag2} in Algorithm \ref{alg:internal_actions}. From the condition to execute Line \ref{line:updatetag2} in Line \ref{line:Delete2_condition}, we infer that node $s$ received a $\langle \texttt{del},X,t\rangle$ with $t \geq M_{s}^{P}.tagvec[X]$ no later than point $P$ from all nodes in $\{i: X \in \mathcal{X}_{i}\}$. Specifically it received the message from node $s'$ no later than $P.$
\end{proof}
\begin{lemma}
Consider nodes $s,s'$ and an object $X$ such that $X \notin \mathcal{X}_{s}, X \in \mathcal{X}_{s'}.$ At any point $P$ be a point of an execution $\beta,$ we have $M_{s}^{P}.tagvec[X] \leq M_{s'}^{P}.tagvec[X].$
\label{lem:nonpresentobjectshavesmallertags}
\end{lemma}
\begin{proof} If $M_s^P.tagvec[X]=\mathbf{0},$ the lemma is trivially true. Otherwise,
the hypothesis of Lemma \ref{lem:handlenonpresentobjects} holds, and from the lemma, we infer that node $s'$ sent a $\langle \texttt{del},X,t\rangle$ before $P$ with $t \geq M_{s}^{P}.tagvec[X]$. Lemma \ref{lem:tag-delete-ordering} therefore implies that $M_{s'}^{P}.tagvec[X] \geq M_s^{P}.tagvec[X].$
\end{proof}
\begin{lemma}
Let $m$ be a message sent from $s$ and delivered to node $s'$ in an execution $\beta$. Let $P$ be the point of sending of message $m$ from $s$ and $P'$ be the point of receipt at $s'$. For an object $X \in \mathcal{X}_{s'},$ if $M_s^{P}.tagvec[X] < M_{s'}^{P'}.tagvec[X]$ and if there is a point $Q$ before $P'$ such that node $s'$ sends a $\langle \texttt{del}, X, M_s^{P}.tagvec[X] \rangle$ message to any node in the system at point $Q$, then there there exists a tuple of the form $(M_s^{P}.tagvec[X],val)$ in list $L_{s'}^{P'}[X]$,
\label{lem:objectpresentfordecoding1}
\end{lemma}
\begin{proof}
Based on the hypothesis of the lemma, node $s'$ sent a message of the form $\langle \texttt{del},X,v,M_s^{P}.tagvec[X]\rangle$ at some point $Q$ that is before $P'$. Based on the protocol, at point $Q$, we have $M_{s'}^{Q}.tagvec[X] = M_s^P.tagvec[X].$ Furthermore, at point $P'$ which is after $Q,$ $M_{s'}^{P'}.tagvec[X] > M_{s'}^{Q}.tagvec[X].$ Let $Q'$ be the first point after $Q$ at which $M_{s'}^{Q'}.tagvec[X] > M_{s}^{P}.tagvec[X].$
Based on the protocol, because $X \in \mathcal{X}_{s'}$, we infer that an $\texttt{Encode}$ action takes place at $Q'$. Based on the $\texttt{Encode}$ action steps, at $Q'$, a tuple of the form $({M}_{s}^{P}.tagvec[X],val)$ exists in $L_{s'}^{Q'}[X].$ We claim that node $s'$ does not delete this tuple from $L_{s'}[X]$ between $Q'$ and $P'.$ To show the claim, observe that elements are deleted from $L[X]$ at $s'$ only because of the $\texttt{Garbage\_Collection}$ internal action at node $s'$. We argue that no $\texttt{Garbage\_Collection}$ performed by node $s'$ between $Q'$ and $P'$ deletes the tuple.
Consider a $\texttt{Garbage\_Collection}$ action performed by node $s'$ at point $Q''$ that is between $Q'$ and $P'$.
Because $Q''$ is after $Q'$, from Lemma \ref{lemma:tagsalwaysincrease}, we note that $M_{s'}^{Q''}.tagvec[X] \geq M_{s'}^{Q'}.tagvec[X] > {M}_{s}^{P}.tagvec[X]$. Because of Lemma \ref{lem:tag-delete-ordering}, node $s$ does not send a $\langle\texttt{Del},X,tag\rangle$ for some $tag$ larger than $M_{s}^{P}.tagvec[X]$ from node $s$ by point $P$. Because of the FIFO nature of the channel, node $s'$ does not receive a $\langle\texttt{Del},X,tag\rangle$ for some $tag$ larger than $M_{s}^{P}.tagvec[X]$ from node $s$ by point $P'$. Therefore, as a part of the $\texttt{Garbage\_Collection}$ internal action performed by node $s$ at t $Q''$, we have $t_{max}^{Q''}[X] \leq M_s^{P}.tagvec[X].$ Because $M_{s'}^{Q''}.tagvec[X] > M_{s}^{P}.tagvec[X],$ we have $t_{max,s'}^{Q''}[X] < M_{s'}^{Q''}.tagvec[X].$ Because $X \in \mathcal{X}_{s'}$ and $t_{max,s'}^{Q''}[X] < M_{s'}^{Q''}.tagvec[X]$, lines \ref{line:GCcondition1} and \ref{line:GCcondition3} are not true for object $X$. the $\texttt{Garbage\_Collection}$ action performed at $Q''$ executes line \ref{line:GC2}, which only deletes tags that are strictly smaller than $t{max}[X]$ in Line \ref{line:GC2} in Algorithm \ref{alg:internal_actions}. As $tmax_{s}^{Q''}[X] \leq M_s^{P}.tagvec[X],$ the tuple of the form $(M_s^{P}.tagvec[X], val)$ is not deleted by the $\texttt{Garbage\_Collection}$ action at $Q''.$
\end{proof}
\begin{lemma}
Consider two nodes $s,s'$, and consider some message $m$ sent from $s$ to $s'$ in an execution $\beta$. Let $P$ be the point of sending of message $m$ from $s$ and $P'$ be the point of receipt. If $M_s^{P}.tagvec[X] \neq M_{s'}^{P'}.tagvec[X],$ then at least one of the following statements is true:
\begin{itemize}
\item There exists a tuple of the form $(M_{s}^{P}.tagvec[X],val)$ in list $L_{s}^{P}[X]$, or
\item $M_s^{P}.tagvec[X] < M_{s'}^{P'}.tagvec[X]$ and $X \notin \mathcal{X}_{s'},$ or
\item $M_s^{P}.tagvec[X] < M_{s'}^{P'}.tagvec[X]$ and there exists a tuple of the form $(M_{s}^{P}.tagvec[X],val)$ in list $L_{s'}^{P'}[X]$.
\end{itemize}
\label{lem:objectpresentfordecoding}
\end{lemma}
\begin{proof}
Consider the hypothesis of the lemma.
Because $M_{s}^{P}.tagvec[X] \neq M_{s'}^{P'}.tagvec[X],$ and because tags are comparable, we infer that either (I): $M_{s}^{P}.tagvec[X] > M_{s'}^{P'}.tagvec[X],$ or
(II) $M_{s}^{P}.tagvec[X] < M_{s'}^{P'}.tagvec[X].$ We consider cases (I) and (II) separately and prove that the lemma statement holds for both cases.
\underline{Case (I): :$M_{s}^{P}.tagvec[X] > M_{s'}^{P'}.tagvec[X].$} From Lemma \ref{lem:pointsexec}, node $s$ consists of a tuple of the form $(M_{s}^{P}.tagvec[X],val)$ in its list $L^{s}[X]$ for some $val \in \mathcal{V}$ at some point before point $P$.
We argue that this tuple exists in the list at point $P$ as well.
Lemma \ref{lem:tag-delete-ordering} implies that every message sent by node $s'$ before $P'$ of the form $\langle \texttt{del},X,v,t\rangle$ has $t \leq M_{s'}^{P'}.tagvec[X].$ Because $M_{s'}^{P'}.tagvec[X]^{P'} < M_{s}^{P}.tagvec[X]$ and because $P$ comes before $P'$, we conclude that node $s'$ does not send a message of the form $\langle \texttt{del},X,v,t\rangle$ where $t \geq M_{s}^{P}.tagvec[X]$ before point $P$ to any node. Therefore, node $s$ does not receive a $\langle \texttt{del},X,v,t\rangle$ message $t \geq M_{s}^{P}.tagvec[X]$ from node $s'$ before point $P$. Therefore, for any point $Q$ that is not later than point $P,$ we have $t_{max,s}^{Q}[X] < M_{s}^{P}.tagvec[X]$.
We conclude that $s$ contains an item of the form $(M.tagvec[X],val)$ in $L[X]$ at point $P$. This is because (i) a tuple of the form $(M_{s}^{P}.tagvec[X],val)$ is exists in $L_{s}[X]$ at node $s$ at some point before $P$, (ii) based on the protocol, this tuple is not deleted by node $s$ before a point $R$ such that $t_{max,s}^{R}[X] \geq M_s^P.tagvec[X]$ and (iii) at every point $Q$ that is no later than $P$, we have $t_{max,s}^Q[X] < M_s^P.tagvec[X]$.
\underline{Case (II):~$M_{s}^{P}.tagvec[X] < M_{s'}^{P'}.tagvec[X].$} If a tuple of the form $(M.tagvec_s^P[X],val)$ exists in $L_s^P[X]$ or if $X \notin \mathcal{X}_{s'}$ then the statement of the lemma is true. So it suffices to prove the lemma statement for the scenario where a tuple of the form $(M_s^P.tagvec[X],val)$ does not exist in $L_s^P[X]$ for $X \in \mathcal{X}_{s'}.$ In this case, the hypothesis of Lemma \ref{lem:currenttag-delete} holds and therefore, we infer that node $s'$ sent a message of the form $\langle \texttt{del},X,v,M_s^P.tagvec[X]\rangle$ at some point $Q$ that is before $P$. Because $M_s^P.tagvec[X] < M_{s'}^{P'}.tagvec[X]$ and $X \in \mathcal{X}_{s'}$ the hypothesis of Lemma \ref{lem:objectpresentfordecoding1} holds. Therefore, we conclude that there exists a tuple of the form $(M_s^P.tagvec[X], val)$ in $L_{s'}^{P'}[X]$.
\end{proof}
\begin{lemma}
Consider two nodes $s,s'$, and consider some message $m$ sent from $s$ to $s'$ in an execution $\beta$. Let $P$ be the point of sending of message $m$ from $s$ and $P'$ be the point of receipt. Consider an object $X \in \mathcal{X}_{s'}.$ If $M_{s}^{P}.tagvec[X] < M_{s'}^{P'}.tagvec[X],$ then there exists a tuple of the form $(M_{s'}^{P'}.tagvec[X],val)$ in list $L_{s'}^{P'}[X].$
\label{lem:objectpresentfordecoding2}
\end{lemma}
\begin{proof}
From Lemma \ref{lem:pointsexec}, we infer that there exists a point $Q$ before $P'$ such that a tuple of the form $(M_{s'}^{P'}.tagvec[X],val)$ exists in list $L_{s'}^{P'}[X].$ Since $\texttt{Garbage\_Collection}$ internal actions are the only actions that remove entries from list $L_s[X],$ it suffices to show that no $\texttt{Garbage\_Collection}$ action that occurs between $Q$ and $P'$ removes the element $(M_{s'}^{P'}.tagvec[X],val)$ from the list $L_s[X].$ We show this next.
From Lemma \ref{lem:tag-delete-ordering}, we infer that node $s$ does not send a $\langle \texttt{Del},X,t,val \rangle$ message before point $P$ for any $t > M_s^P.tagvec[X].$ Because of the FIFO nature of the channel, node $s'$ does not receive a $\langle \texttt{Del},X,t,val \rangle$ from node $s'$ for any $t > M_{s}^{P}.tagvec[X]$ before point $P'$. Thus, for any $\texttt{Garbage\_Collection}$ that takes place before $P',$ the variable $tmax_{s}[X]$ in line \ref{line:tmax} is no bigger than $M_{s}^{P}.tagvec[X].$ Therefore, the state changes that take place in lines \ref{line:GC1} and \ref{line:GC2} of Algorithm \ref{alg:internal_actions} does not remove the element $(M_{s'}^{P'}.tagvec[X],val)$ from the list $L_s[X].$ \end{proof}
\begin{lemma}
Consider two nodes $s,s'$, and consider some message $m$ sent from $s$ to $s'$ in an execution $\beta$. Let $P$ be the point of sending of message $m$ from $s$ and $P'$ be the point of receipt. Consider an object $X$ such that $X \notin \mathcal{X}_{s}, X \in \mathcal{X}_{s'}.$ If $M_s^P.tagvec[X] \neq M_{s'}^{P'}.tagvec[X],$ then there exists a tuple of the form $(M_{s'}^{P'}.tagvec[X],val)$ in list $L_{s'}^{P'}[X].$
\label{lem:objectpresentfordecoding3}
\end{lemma}
\begin{proof}
Because $X \in \mathcal{X}_{s'}, X \notin \mathcal{X}_{s},$ from Lemma \ref{lem:nonpresentobjectshavesmallertags}, we have $M_{s}^{P}.tagvec[X] \stackrel{(a)}{\leq} M_{s'}^{P}.tagvec[X].$ Because $P'$ comes after $P$, from Lemma \ref{lemma:tagsalwaysincrease}, we have $M_{s'}^{P}.tagvec[X] \stackrel{(b)}{\leq} M_{s'}^{P'}.tagvec[X].$ The lemma hypothesis that $M_{s}^{P}.tagvec[X] \neq M_{s'}^{P'}.tagvec[X]$ implies that at least one of the inequalities $(a),(b)$ is strict and $M_{s}^{P}.tagvec[X] < M_{s'}^{P'}.tagvec[X].$ Thus, the hypothesis of Lemma \ref{lem:objectpresentfordecoding2} is satisfied and its conclusion implies that there exists a tuple of the form $(M_{s'}^{P'}.tagvec[X],val)$ in list $L_{s'}^{P'}[X].$
\end{proof}
\subsection{Proof of Lemma \ref{lem:error1}}
\begin{proof}
Since the only action that modifies $Error1$ variable is the receipt of a $\texttt{val\_resp\_encoded}$ message, it suffices to consider such a point. Every $\texttt{val\_resp\_encoded}$ message is sent in response to a $\texttt{val\_inquiry}$ message. Let point $P$ denote the point of sending of the $\langle \texttt{val\_inq},clientid, opid, \overline{X},tvec\rangle$ from node $s$, let $Q$ be the point of receipt of this message at node $s'$ and let $R$ denote the point of receipt of $\langle \texttt{val\_resp\_encoded},\overline{M},clientid,opid,\overline{X},reqtvec\rangle$ message at node $s.$ From the protocol, we conclude that the point of sending of the $\texttt{val\_resp\_encoded}$ from node $s'$ is is also $Q$. Also note that $tvec = M_s^{P}.tagvec$.
It suffices to consider every object $X \in \mathcal{X}_{s'}$ and argue that on receipt of $\texttt{val\_resp\_encoded}$ message, the action performed by node $s$ does not set $Error1[X] \leftarrow 1.$ Note that $Error1[X]$ is not set to $1$ if $\overline{M}.tagvec[X] = {M}_{s}^P.tagvec[X].$ So it suffices to assume that $\overline{M}.tagvec[X] \neq {M}_{s}^{P}.tagvec[X]$ for the remainder of the proof.
\begin{claim}
If there exists an element $(M_{s'}^{Q}.tagvec[X],val)$ in $L_{s'}^{Q'}[X],$ then we have $\overline{M}.tagvec[X] \in \{\mathbf{0}, M_{s}^{P}.tagvec[X]\}$. Furthermore, if an element $M_{s}^{P}.tagvec[X]$ also exists in $L_{s'}^{Q'}[X],$ then $\overline{M}.tagvec[X] = M_{s}^{P}.tagvec[X]$.
\label{claim:error1vals}
\end{claim}
\begin{proof}
If there exists an element $(M_{s'}^{Q}.tagvec[X],val)$, then at point $Q$ node $s'$ the condition in \ref{line:responsetovalinqcondition1} in Algorithm \ref{alg:input_actions} returns true. Therefore the node execute line \ref{line:ResponsetoValInqtagupdate1} which sets $\overline{M}.tagvec[X] = \mathbf{0}$. Further, if $M_{s}^{P}.tagvec[X]$ also exists in $L_{s'}^{Q'}[X],$ then the condition in \ref{line:responsetovalinqcondition2} returns true, and line \ref{line:ResponsetoValInqtagupdate2} is executed. This line sets $\overline{M}.tagvec[X] = M_{s}^{P}.tagvec[X]$
\end{proof}
\begin{claim} There exists $val \in \mathcal{V}$ such that\\ (I) $(M_{s'}^{Q'}.tagvec[X],val) \in L_{s'}^{Q}[X]$ or\\ (II) $(M_{s'}^{Q'}.tagvec[X],val) \in L_{s}^{R}[X].$
\end{claim}
\begin{proof}[Proof of Claim:] If $X \in \mathcal{X}_{s},$ then the $\texttt{val\_resp\_encoded}$ satisfies the hypothesis of Lemma \ref{lem:objectpresentfordecoding}, whose statement implies either (I) $(M_{s'}^{Q}.tagvec[X],val) \in L_{s'}^{Q'}[X]$ or (II) $(M_{s'}^{Q}.tagvec[X],val) \in L_{s}^{R}[X] .$ If, on the other hand, $X \notin \mathcal{X}_{s},$ then the $\texttt{val\_inq}$ message from $s$ to $s'$ satisfies the hypothesis of Lemma \ref{lem:objectpresentfordecoding3}, which implies that (I) $(M_{s'}^{Q'}.tagvec[X],val)\in L_{s'}^{Q'}[X].$
\end{proof}
We consider cases (I) and (II) separately.
\underline{Case (I): $(M_{s'}^{Q}.tagvec[X],val) \in L_{s'}^{Q}[X]$}
From, Lemma \ref{claim:error1vals} we know that $\bar{M}.tagvec[X] \stackrel{(a)}{=} \mathbf{0},$ or $\bar{M}.tagvec[X] \stackrel{(b)}{=} M_s^{P}.tagvec[X].$
In both cases (a) and (b), based on the protocol, we conclude that in line \ref{line:seterror1} of Algorithm \ref{alg:input_actions} does not return true and $Error1[X]$ is not set to $1$ .
\underline{Case (II): $(M_{s'}^{Q}.tagvec[X],val) \notin L_{s'}^{Q}[X], (M_{s'}^{Q}.tagvec[X],val) \in L_{s}^{R}[X]$}
From the protocol, in this case,$\bar{M}.tagvec[X] = M_{s'}^{Q}.tagvec[X].$ On receipt of this message at server $s$ at point $R$, because $(M_{s'}^{Q}.tagvec[X],val) \in L[X]^{s,R},$ we note that that the condition in line \ref{line:metaerrorcondition1} holds. Therefore $Error1[X]
$ is not set to $1$ at point $R$.
\end{proof}
\subsection{Proof of Lemma \ref{lem:error2}}
Since the only action that modifies $Error2$ variable is the receipt of a $\texttt{val\_resp\_encoded}$ message, it suffices to consider the point of receipt an arbitrary $\texttt{val\_resp\_encoded}$ message and show the the corresponding action does not set $Error2$ to $1$. Every $\texttt{val\_resp\_encoded}$ message is sent in response to a $\texttt{val\_inquiry}$ message. Let point $P$ denote the point of sending of the $\langle \texttt{val\_inq},\overline{M},clientid, opid,\overline{X},tvec\rangle$ message from node $s$, let $Q$ be the point of receipt of this message at node $s'$ and let $R$ denote the point of receipt
of the $\rangle \texttt{val\_resp\_encoded},\overline{M},clientid,opid,\overline{X},reqtvec\rangle$ message at node $s.$ From the protocol, we conclude that the point of sending of the $\texttt{val\_resp\_encoded}$ from node $s'$ is is also $Q$. Also note that $tvec = M_s^{P}.tagvec$.
It suffices to take every object $X \in \mathcal{X}_{s'}$ such that $M_{s}^{P}.tagvec[X] \neq {M}_{s'}^{Q}.tagvec[X]$ and show that the for loop in line \ref{line:seterror2} in Algorithm \ref{alg:input_actions} does not set $Error2_s^{L}[X] \leftarrow 1.$ Consider object $X \in \mathcal{X}_{s'}.$ Because, $M_{s}^{P}.tagvec[X] \neq M_{s'}^{Q}.tagvec[X],$ Cases (I), (II) below are exhaustive.
\underline{Case (I): $ M_{s}^{P}.tagvec[X]< M_{s'}^{Q}.tagvec[X].$} Because a message $m = \texttt{val\_inquiry}$ is sent at point $P$ from node $s$ and arrives at point $Q$ at node $s'$ and because $X \in \mathcal{X}_{s'}$, from Lemma \ref{lem:objectpresentfordecoding2}, we note that there exists an element of the form $(val2,M_{s'}^{Q}.tagvec[X]) \in L_{s'}^{Q}[X].$ Therefore, from Claim \ref{claim:error1vals} in Lemma \ref{lem:error1}'s proof, we have $\overline{M}.tagvec[X] \in \{\mathbf{0},M_{s}^{P}.tagvec[X]\}.$ From Lemma \ref{lem:objectpresentfordecoding} applied to the $\texttt{val\_inq}$ message, we know that there exists $val1$ such that either (I-A) $(M_{s}^{P}.tagvec[X],val1) \in L_{s'}^{Q}[X]$ or (I-B) $(M_{s}^{P}.tagvec[X],val1) \in L_{s}^{P}[X] .$ We consider cases (I) and (I-B) separately.
\underline{Case (I-A): $(M_{s}^{P}.tagvec[X],val1) \in L_{s'}^{Q}[X]$}
In this case, because there exists an element of the form $(val2,M_{s}^{Q}.tagvec[X]) \in L_{s}^{Q}[X],$ we conclude that $\overline{M}.tagvec[X] = M_s^{P}.tagvec[X].$ From the protocol, the condition in line \ref{line:metaerrorcondition} in Algorithm \ref{alg:input_actions} is not satisfied; therefore $Error2[X]$ is not set to 1 at point $R$.
\underline{Case (I-B): $(M_{s}^{P}.tagvec[X],val1) \in L_{s}^{P}[X], (M_{s}^{P}.tagvec[X],val1) \notin L_{s'}^{Q}[X]$}
At point $R$, there are two possibilities (i) there exists a tuple of the form $$(clientid,opid,X,tvec,M_s^{P}.tagvec,\overline{v})$$ exists in $ReadL_{s}^{R}$ with $tvec[X] = M_{s}^{P}.tagvec[X],$ or (ii) there exists no such tuple. Based on the protocol, in case $(ii)$ , $Error2[X]$ is not set to $1$ at point $R.$ So we only consider case $(i)$ here. The point $R$ satisfies the hypothesis of Lemma \ref{lem:listsavespendingread}. From the lemma statement, there are only two possibilities: $M_{s}^{P}.tagvec[X] \in L_{s}^{R}[X]$ or $M_s^{P}.tagvec[X] = M_s^{R}.tagvec[X].$ If
$M_{s}^{P}.tagvec[X] \in L_{s}^{R}[X]$, coupled with Lemma \ref{lem:error1}, we conclude that the line condition in line \ref{line:metaerrorcondition2} returns true; therefore condition $Error2[X]$ is not set to $1$. We consider the case where $M_{s}^{P}.tagvec[X] = M_{s}^{R}.tagvec[X].$
Because we are considering case (I)-B, $(M.tagvec_s^{P}[X],val)$ does not belong to $L_{s'}^{Q}[X]$. In this case, we note that node $s'$ does not send a $\langle del, X, M_s^{P}.tagvec[X]\rangle$ at any point before $Q.$ To see this consider the contradictory hypothesis that the node sent such a message. Then combined with the fact that $M_{s}^{P}.tagvec[X] < M_{s}^{Q}.tagvec[X], X \in \mathcal{X}_{s'}$ we infer from Lemma \ref{lem:objectpresentfordecoding1} that $(M_{s}^{P}.tagvec[X],val)$ belongs to $L_{s'}^{Q}[X]$, which violates the hypothesis of Case (I-B). By the FIFO nature of the channel, we infer that a message $\langle \texttt{del}, X, M_{s}^{P}.tagvec[X]\rangle$ is not received by node $s$ from node $s'$ by point $R$. From the contrapositive of Lemma \ref{lem:currenttag-delete}, and because $M_{s}^{P}.tagvec[X] = M_{s}^{R}.tagvec[X],$ we infer that a tuple of the form $(M_{s}^{P}.tagvec[X],val)$ exists in $L_{s}^{R}[X].$
Thus, we have shown that a tuple of the form $(M_{s}^{P}.tagvec[X],val)$ belongs to $L_{s}^{R}[X]$ for every object $X \in \mathcal{X}_{s'}.$ Combined with Lemma \ref{lem:error1}, we infer that the condition in line \ref{line:metaerrorcondition2} is true, and therefore $Error2[X]$ is not set to $0.$
\underline{Case (II): $M_{s}^{P}.tagvec[X] > M_{s'}^{Q}.tagvec[X].$}
From Lemma \ref{lem:pointsexec}, we observe that there is a point $P'$ before $P$ such that a tuple of the form $(M_s^P.tagvec[X],val)$ belongs to $L_{s}^{P'}[X].$ We show that this tuple is not removed from $L[X]$ before point $R.$ To show this, it suffices to show that for any $\texttt{Garbage\_Collection}$ action performed by node $s$ at point $R'$ which is between $P'$ and $R$, the element $(M_{s}^{P}.tagvec[X],val)$ is not removed from the list $L_{s}^{R'}[X].$ From Lemma \ref{lem:tag-delete-ordering}, we infer that the highest tag $t$ such that node $s'$ has sent a message of the form $\langle \texttt{del},X, t\rangle$ satisfies $t \leq M_{s'}^{Q}.tagvec[X] < M_{s}^{P}.tagvec[X].$ Therefore, node $s$ does not receive a tuple of the form $\langle \texttt{del},X, t\rangle$ with $t \geq M_{s}^{P}.tagvec[X]$ from node $s$ before point $R.$ Therefore, $R$, $tmax_{s}^{R'}[X] < M_{s}^{P}.tagvec[X]$. Therefore, from the protocol lines \ref{line:GC1}, \ref{line:GC3}, \ref{line:GC2} in Algorithm \ref{alg:internal_actions}, we infer that the element $(M_{s}^{P}.tagvec[X],val)$ is not removed from the list $L_s^{R'}[X].$ This completes the proof.
\section{Eventual Consistency - Proofs of Lemmas \ref{lem:eventually_apply} and \ref{lem:returntagofread}}
\label{app:eventual_proof}
\subsection{Proof of Lemma \ref{lem:eventually_apply}}
\begin{proof}
According to the algorithm, tuple $(s',X,v,t)$ is sent to every node other than $s$, eventually this tuple will be in every other node $s$'s $InQueue$. For node $s$, because $InQueue$ is a priority queue, and because there are finite number of timestamps that are smaller than $t.ts$, the number of tuples that are ever placed ahead of ahead of $(s,X,v,t)$ is finite.
We begin with the following claim.
\begin{claim}
Consider any any tuple $(\overline{s},\overline{X},\overline{v},\overline{t})$ in $Inqueue.Head.$ If we index all the writes by clients in $\mathcal{C}_{\overline{s}}$ as $\pi_{1},\pi_{2}, \ldots, $ in the order of the arrival of the corresponding $\texttt{write}$ messages at server $\overline{s}$, then:
$\pi_{\overline{t}.ts[\overline{s}]}$ is a write to object $\overline{X}$ with value $v$. Further $\overline{t}.ts=ts(pi_{\overline{t}.ts[\overline{s}]}).$
\label{claim:writeordering}
\end{claim}
\begin{proof}
The claim follows from the protocol at server $\overline{s}.$ Specifically, the server increments $vc_{\overline{s}}[\overline{s}]$ by $1$ on receiving a $\texttt{write}$ message, and never decrements it. Therefore, $\pi_{\overline{t}.ts[\overline{s}}$ is a write to object $\overline{X}$ with value $v$. Since any $\langle \texttt{app},\overline{s},\overline{X},\overline{v},\overline{t})\rangle$ message has the $\overline{t}$ being the timestamp, $\overline{X}$ being the object and $\overline{v}$ being the value of the write operation, so the claim follows.
\end{proof}
\begin{claim}
For any servers $s,\overline{s}$ for any point $Q$ of $\beta$, let $\pi$ denote the $vc_s^{Q}[\overline{s}]$-th write operation at server $s$ as per the ordering of Claim \ref{claim:writeordering}. Then $vc_s^{Q} \geq ts(\pi).$
\label{claim:writeordering2}
\end{claim}
\begin{proof}
From the protocol, server $s$ only increments $vc_{s}[\overline{s}]$ on an $\texttt{Apply\_Inqueue}$ action, where a tuple from $\overline{s}$ at $Inqueue.Head$ is processed and line \ref{line:applycondition} returns true. Consider the tuple $(\overline{s},\overline{X},\overline{v},\overline{t})$ which corresponds to $\pi$. This tuple is processed by point $Q$ and on processing this tuple, \ref{line:applycondition} is true, and line \ref{line:vcincrement_apply} is executed. Therefore, after the $\texttt{Apply\_Inqueue}$ that processes this tuple, $vc_{s} \geq \overline{t}.ts = ts(\pi).$ The claim follows for point $Q$ applying Lemma \ref{lem:vconlyincreases}.
\end{proof}
We prove the lemma via induction on $t.$
\underline{Base case:}
Consider a tag $(s_1,O_1,v_1,t_1)$ such that server $s$ receives adds no tuple to $Inqueue$ with a tag smaller than $t_1$ in $\beta.$
\begin{claim}
$t_1.ts$ is a $N$ dimensional vector such that the $s_1$th component of $t_1.ts$ is equal to $1$, and all other components are $0.$
\end{claim}
\begin{proof}
BLet $\pi_1$ be the write operation with the smallest tag in $\beta$, that is $tag(\pi_1) = t_1.$ From the assumption, it follows that $\pi_1$ is issued by a client in $\mathcal{C}_{s_1}.$ Note that $t_1.ts = ts(\pi_1).$ Because there is no tuple with tag smaller than $t_1$ in any servers $Inqueue,$ $\pi_1$ is the first write at $s_1$. From Claim \ref{claim:writeordering}, $t_1.ts[s_1]=1.$ Suppose there is a server $s' \neq s_1$ such that $t_1.ts[s'] \neq 0.$ Let $\pi_2$ be the $t_1.ts[s']$th write at server $s'.$ From Claim \ref{claim:writeordering2}, we have $ts(\pi_2) < ts(\pi_1),$ which violates the assumption that $\pi_1$ is the write with the smallest tag - a contradiction.
\end{proof}
Because $t_1$ has the smallest tag, the corresponding tuple $(s_1,O_1,v_1,t_1)$ is in $Inqueue.Head$ for any server $s$. Note that that before this tuple is processed at server $s$, $vc_{s}[s_1]=0.$ From the above claim, we note that the first $Apply\_Inqueue$ action performed after the addition of the tuple $(s_1,O_1,v_1,t_1)$ returns true of line \ref{line:applycondition}. Therefore, line \ref{line:inqueue_remove} is executed and this tuple is removed from $Inqueue_s.$
\underline{Inductive step:}
We make the inductive hypothesis that for any tuple $(\overline{s}, \overline{X}, \overline{v}, \overline{t})$ with $\overline{t} < t,$ there is a point $\overline{Q}$ such that, at any point after $\overline{Q}$, the tuple is not in $Inqueue_s$. There are two possibilities: (i) the tuple $(s,X,v,t)$ is not in $Inqueue_s$ at any p
oint after $\overline{Q}$, or (ii) there is a point $P'$ after $\overline{Q}$ such that the tuple $(s,X,v,t)$ is in $Inqueue_s^{P'}$. In case $(i)$ the lemma readily holds. So we consider case $(ii).$ By the inductive hypothesis, we recognize that at point $P'$, and at all points following $P'$ where $(s,X,v,t) \in \textrm{Inqueue},$ the tuple is actually $Inqueue.Head.$
\begin{claim}
For any server $\overline{s} \neq s',$ $vc_{s}^{P'}[\overline{s}] \geq t.ts[\overline{s}]$
\label{claim:vcts1}
\end{claim}
\begin{proof}
Let $\pi$ be the $t.ts[\overline{s}]$th write performed at node $\overline{s}$ as per the ordering in Claim \ref{claim:writeordering}. From Claim \ref{claim:writeordering2}, $vc_{s}^{P'} \stackrel{(a)}{\geq} ts(\pi)$. From Lemma \ref{claim:writeordering}, we have $ts(\pi)[\overline{s}] \stackrel{(b)}{=} t.ts[\overline{s}]$. From $(a)$ and $(b),$ the clam follows.
\end{proof}
\begin{claim}
For any server, $vc_{s}^{P'}[s'] + 1 = t.ts[{s}']$
\label{claim:vcts2}
\end{claim}
\begin{proof}
Let $\phi,\pi$ respectively be the $t.ts[s']-1$th and $ t.ts[s']$ th writes performed at node $\overline{s}$ as per the ordering in Claim \ref{claim:writeordering}.
From Theorem \ref{thm:causallyconsistent}, we know that $ts(\phi) < ts(\pi).$ By the inductive hypothesis, we know that the $\texttt{app}$ message sent on behalf of $\phi$ has been emptied from the Inqueue by server $s$. From the server protocol, we have $vc_{s}^{P'} \stackrel{(a)}{\geq} t.ts[s'] - 1.$ Further, from the protocol, any $\texttt{app}$.
From the protocol, server $s$ only increments $vc_{s}[s']$ on an $\texttt{Apply\_Inqueue}$ action, where a tuple from $s'$ at $Inqueue.Head$ is processed and line \ref{line:applycondition} returns true. Since the server has not processed tuple $(s,X,v,t)$ by point $P'$, we know that $vc_{s}^{P'} \stackrel{(b)}{<} t.ts[s'].$ From $(a)$ and $(b)$, the lemma follows.
\end{proof}
Consider the first $\texttt{Apply\_Inqueue}$ action after $P'$. Claims \ref{claim:vcts1}, \ref{claim:vcts2} imply that line \ref{line:applycondition} is true for this line. Therefore, line \ref{line:inqueue_remove} is executed. This completes the proof.
\end{proof}
\subsection{Proof of Lemma \ref{lem:returntagofread}}
We mimic the proof of Lemma \ref{lem:readLtagsmallerthanvc}, with appropriate modifications.
\begin{proof}
Let $\pi$ be an operation that terminates in $\beta$. Without loss of generality, assume that $\pi$ is issued by client $c$ in $\mathcal{C}_{s}$ for some server $s.$
Based on the protocol $\pi$ returns because at a point $Q$ in $\beta$, server $s$ sends a $\langle \texttt{read-return}, opid, v\rangle$ to the client. This message is sent on executing one of the following lines:
\begin{enumerate}
\item Line \ref{line:sendtoread1} in Algorithm \ref{alg:inputactions_clients}.
\item Line \ref{line:readimmediatereturn} in Algorithm \ref{alg:inputactions_clients}, or
\item Line \ref{line:read-send-val_resp_encoded} in Algorithm \ref{alg:input_actions}, or
\item Line \ref{line:valresp_handle} in Algorithm \ref{alg:input_actions}, or
\item Line \ref{line:readreturn_apply} in Algorithm \ref{alg:internal_actions}.
\end{enumerate}
We show that the lemma holds for each of these cases. Note that $ts(\pi) = vc_s^{Q}.$
\underline{1. Line \ref{line:sendtoread1} in Algorithm \ref{alg:inputactions_clients}.}
In this case point $Q$ is the point of receipt of a $\langle \texttt{write}, opid, X, v\rangle$ message from a write operation $\phi$, and the read $\pi$ with operation id $opid$ returns the value of $\phi$. By Definition \ref{def:ts_tag}, we have $ts(\phi) = vc_s^{Q}$. Because $Q$ comes after $P$, from Lemma \ref{lem:vc_after_apply}, we have $ts(\phi) = vc_s^{Q} > vc_{s}^{P}.$ From Lemma \ref{lem:vc}, we have $vc_s^{P} > M_s^{P}[X].tag.ts,$ and $vc_{s}^{P} \geq L_s^{P}[X].Highesttagged.tag$ if $L_s^{P}[X]$ is non-empty. Therefore, $ts(\phi) > M_s^{P}[X].tag,$ and if $L_s^{P}[X]$ is non-empty, $tag(\phi) \geq L_s^{P}[X].Highesttagged.tag.$
\underline{2. Line \ref{line:readimmediatereturn} in Algorithm \ref{alg:inputactions_clients}.}
Let $Q$ be the point of receipt of the $\langle \texttt{read},opid,X\rangle$ message from the read $\pi.$ Note that $Q$ is after $P$.
The value $v$ returned by $\pi$ is one that forms a tuple $(t,v)$ in list $L[X],$ where $t = L_s^{Q}[X].Highesttagged.tag$. By Lemma \ref{lem:valuesarelegitimate}, $v$ is the value of the unique write $\phi$ with $tag(\phi)=t$. Because Line \ref{line:readimmediatereturn} is executed only if \ref{line:readimmediatereturncheck1} is satisfied, we infer that $t \geq M_s^Q.tagvec[X]$. Since Lemma \ref{lemma:tagsalwaysincrease} implies that $M_s^Q.tagvec[X] \geq M_s^P.tagvec[X]$, we have $t \geq M_s^P.tagvec[X]$ as desired.
If $L_s^{P}[X]$ is non-empty, then Lemma \ref{lem:listtag} combined with the fact that $t = L_s^Q[X].Highesttagged.tag \geq M_s^Q.tagvec[X]$ implies that $t \geq L_s^P[X].Highesttagged.tag$ as desired.
\underline{3. Line \ref{line:read-send-val_resp_encoded} in Algorithm \ref{alg:input_actions}:}
Note that at the point $R$ of receipt of the $\texttt{val\_resp\_encoded}$ with operation id $opid$ and tag vector $requestedtags,$ the fact that Line \ref{line:read-send-val_resp_encoded} was executed implies there exists an entry in $ReadL_s^R$ with operation id $opid$ and tag vector $requestedtags.$
From Lemma \ref{lem:valuesarelegitimate}, the input to the function $\chi_{T_O}$ on Line \ref{line:valresp_handle_decoding} is a set whose elements are of the form $(\overline{i},{v}_{\overline{i}})$ where ${v}_{i} = \Phi_i(w_1, w_2, \ldots, w_K)$ where, for $\ell \in \{1,2,\ldots,K\},$ we have $w_\ell$ is the value of the unique write with tag $requestedtags[X_\ell].$
Therefore, the output of the decoding function, which is returned to the read, is the value $w$ of the value of the unique write $\phi$ with tag $requestedtags[X].$ It remains to show that $requestedtags[X] \geq M_s^P.tagvec[X]$
Suppose the entry operation id $opid$, and tag vector $requestedtags$ is added to $ReadL_s$ at point $Q$.
From the algorithm, at point $Q$, the node executed line \ref{line:addtoreadl} in Algorithm \ref{alg:inputactions_clients} or Line \ref{line:readlentryencoding} in Algorithm \ref{alg:internal_actions}. In either case, we have $requestedtags=M_s^{Q}.tagvec.$ Because $Q$ comes after $P$, we have from Lemma \ref{lemma:tagsalwaysincrease}, $requestedtags[X] \geq M_s^P.tagvec[X]$ as desired.
\underline{4. Line \ref{line:valresp_handle} in Algorithm \ref{alg:input_actions}}
Let $R$ be the point of a $\langle \texttt{val\_resp}, X, v, clientid, opid, tvec\rangle.$ From the condition on Line \ref{line:valresp_handle_condition}, we know that there exists an entry in $ReadL_s^{R}$ with operation id $opid$ and tag vector $tvec$. Let $Q$ be the point before $R$ where this entry was added to $ReadL_s.$ From Lemma \ref{lem:valuesarelegitimate}, we infer that $v$ is the value of the unique write $\phi$ with tag $tvec[X].$ It remains to show that $tvec[X] \geq M_s^P.tagvec[X]$. The proof is similar to Case 3.
Suppose the entry operation id $opid$, and tag vector $tvec$ is added to $ReadL_s$ at point $Q$.
From the algorithm, at point $Q$, the node executed line \ref{line:addtoreadl} in Algorithm \ref{alg:inputactions_clients} or Line \ref{line:readlentryencoding} in Algorithm \ref{alg:internal_actions}. In either case, we have $tvec=M_s^{Q}.tvec.$ Because $Q$ comes after $P$, we have from Lemma \ref{lemma:tagsalwaysincrease}, $tvec[X] \geq M_s^P.tagvec[X]$ as desired.
\underline{5. Line \ref{line:readreturn_apply} in Algorithm \ref{alg:internal_actions}}
The proof is similar to Cases $3$ and $4$. Line \ref{line:readreturn_apply} is executed in an $\texttt{Apply\_Inqueue}$ internal action. The value $v$ returned is the value of the unique write operation $\phi$ with tag $t.$ The fact that Line \ref{line:readreturn_apply} was executed implies that the condition in \ref{line:readreturn_apply_condition} is satisfied. This condition implies that an entry with operation $opid$ and tag vector $tvec$ exists in $ReadL$ at the point of execution of the $\texttt{Apply\_Inqueue}$ action. The condition also implies that $t \geq tvec[X]$. It remains to show that $tvec[X] \geq M_s^{P}.tvec[X].$ The argument is similar to cases $3$ and $4$.
Suppose the entry operation id $opid$, and tag vector $tvec$ is added to $ReadL_s$ at point $Q$.
From the algorithm, at point $Q$, the node executed line \ref{line:addtoreadl} in Algorithm \ref{alg:inputactions_clients} or Line \ref{line:readlentryencoding} in Algorithm \ref{alg:internal_actions}. In either case, we have $tvec=M_s^{Q}.tagvec.$ Because $Q$ comes after $P$, we have from Lemma \ref{lemma:tagsalwaysincrease}, $tvec[X] \geq M_s^P.tagvec[X]$ as desired.
\end{proof}
\section{Proof of Lemma \ref{lem:storagecostkey}}
\label{app:storage}
\subsection{Preliminary Lemmas}
\begin{lemma}
Consider a fair execution $\beta$ of $CausalEC$ where every server is non-halting. Consider a point $P$ and an entry $(t,val) \in L_s[X]$ for some server $s$ and some object $X$. Then,
\begin{enumerate}[(a)]
\item there is a point $Q$ in $\beta$ such that, for every point $Q'$ after $Q$ in $\beta,$ we have $M_s^{Q'}.tagvec[X] \geq t$.
\item there is a point $\overline{Q}$ in $\beta$ such that, by $\overline{Q}$ every node $s$ has at least one entry $(t_{s'},s')$ in $DelL_s[X]$ with $t_{s'} \geq t,$ for every node $s' \in \mathcal{N}.$
\end{enumerate}
\label{lem:Mexceedslisteventually}
\end{lemma}
\begin{proof}
We show $(a)$ first.
\underline{Proof of $(a)$}
Assume the hypothesis of the lemma, that is $(t,val) \in L_s^{P}[X]$ in execution $\beta$. Suppose that at some point $Q$ in $\beta$, $M_s^{Q}.tagvec[X] \geq t$. Then, because of Lemma \ref{lemma:tagsalwaysincrease}, we know that at any point $Q'$ after $Q$, $M_s^{Q'}.tagvec[X] \geq M_s^{Q}.tagvec[X] \geq t.$ Therefore the statement of the lemma holds. If possible, suppose there exists no such point $Q$. Then there is a point $P'$ such that, at any point $Q'$ after $P',$ $M_s^{Q'}.tagvec[X] = M_s^{P'}.tagvec[X] < t$. We will show a contradiction. We consider two cases: $X \in \mathcal{X}_{s}$ and $X \notin \mathcal{X}_{s}.$
\underline{Case 1: $X \in \mathcal{X}_{s}$}
\begin{claim}
At any point of the execution after $P$, the tuple $(t,val)$ is in $L_s[X].$
\label{claim:rr0}
\end{claim}
\begin{proof}
From Lemma \ref{lem:tag-delete-ordering} and the hypothesis that $t > M_s^{P'}.tagvec[X]$ implies that for any entry $(s,t')$ in $DelL_s[X]$ at any point of $\beta$, we have $t' < t$. Items are removed from $L_s[X]$ only as a part of $\texttt{Garbage\_Collection}$ actions in line \ref{line:GC1} or \ref{line:GC2}. By examining lines \ref{line:GC1}, \ref{line:GC2}, $\texttt{Garbage\_Collection}$ action removes the element $(t,val)$ from $L_s[X]$. Therefore, at any point of the execution after $P$, the tuple $(t,val)$ is in $L_s[X].$
\end{proof}
In particular, consider a point $Q'$ after $P'$ where an $\texttt{Encoding}$ action is performed. At this point, because $(t,v)$ is in $L_s^{Q'}[X],$ we have $L_s^{Q'}[X].Highesttagged > t.$ Therefore, the condition for the for loop in line \ref{line:Forloopupdatetag1} in Algorithm \ref{alg:internal_actions} returns true for object $X$. If the condition in line \ref{line:conditionforupdate} is satisfied, then line \ref{line:updatetag1} is executed and $M_s^{Q'}.tagvec[X]$ gets updated to a tag that is at least $t$, which is a contradiction to our earlier assumption that $M_s.tagvec[X]$ is always smaller than $t$ for any point after $P'$.
If \ref{line:conditionforupdate}
returns false, then line \ref{line:readlentryencoding} is executed. By Theorem \ref{lem:read_terminates_f=0}, the read eventually responds and there is a point after $P''$ such that $(M_s^{P'}.tagvec[X], v)$ belongs to $L_s[X].$ Let $Q'$ be the first point after $P''$ where $(M_s^{P'}.tagvec[X], v)$ lies in $L_s[X].$ Let $Q''$ be the point of the
the first $\texttt{Encoding}$ action performed after $Q''$ in $\beta.$ The result follows from the following claims:
We make the following claim:
\begin{claim}
A $\texttt{Garbage\_Collection}$ action performed between $Q'$ and $Q''$ does not remove the element $(M_s^{P'}.tagvec[X], v)$.
\label{claim:rr1}
\end{claim}
\begin{claim}
In the $\texttt{Encoding}$ action performed at $Q'',$ line \ref{line:conditionforupdate} is executed and returns true.
\label{claim:rr2}
\end{claim}
Once we show the above claims, the lemma statement readily follows, line \ref{line:updatetag1} is executed as a part of the $\texttt{Encoding}$ action at $Q'$, and $M_s^{P'}.tagvec[X]$ gets updated to a tag that is at least $t$, which is a contradiction.
\begin{proof}[Proof of Claim \ref{claim:rr1}]
Consider a $\texttt{Garbage\_Collection}$ action performed at point $U$ between $Q'$ and $Q''.$ Because of Lemma \ref{lem:tag-delete-ordering} and because our assumption implies that $M_s^{P'}.tagvec[X] = M_s^{U}.tagvec[X]$, we infer that $tmax_s^{U}[X] \leq M_s^{P'}.tagvec[X].$ If the inequality is strict, then from lines \ref{line:GC1}, \ref{line:GC2}, we infer that the element $(M_s^{P'}.tagvec[X], v)$ is not deleted from $L_s^{U}[X].$ If the inequality is not strict, we have $tmax_s^{U}[X] = M_s^{P'}.tagvec[X] = M_s^{U}.tagvec[X].$ Note that point $Q'$, $(t,val)$ is in $L_s^{U}[X]$, so $L_s^{U}[X].Highesttagged.tag > M_s^{U}.tagvec[X]$. Therefore, the condition in line \ref{line:GCcondition1} returns false. Therefore line \ref{line:GC2} is executed (and not line \ref{line:GC1}), and from the line, the element $(M_s^{P'}.tagvec[X], v)$ is not deleted from $L_s[X]$.
\end{proof}
\begin{proof}[Proof of Claim \ref{claim:rr2}]
From Claim \ref{claim:rr0}, $(t,val) \in L_s^{U}[X] $ and $t > M_s^{P'}$. Therefore, for object $X$, line \ref{line:Forloopupdatetag1} returns true. Therefore, line \ref{line:conditionforupdate} in Algorithm \ref{alg:internal_actions} is executed as a part of the $\texttt{Encoding}$ action.
Because at point $Q',$ we have $(M_s^{P'}.tagvec[X], v), \in L_s^{Q'}[X]$, and because of Claim \ref{claim:rr1}, we have $(M_s^{P'}.tagvec[X], v), \in L_s^{U}[X].$ Therefore, line \ref{line:conditionforupdate} returns true.
\end{proof}
\underline{Case 2: $X \notin \mathcal{X}_{s}$}
Let $R=\{i \in \mathcal{N}: X \in \mathcal{X}_{i}\}.$ Note that $s \notin R.$
\begin{claim}
In $\beta$, eventually, every node $s' \in R$ sends a $\langle \texttt{del},X,t_{s'}\rangle$ message to all the other nodes for some $t_{s'}\geq t$. In addition, node $s$ eventually adds a $(\overline{t}_{s'},s)$ element to $DelL_{s}[X]$ eventually with $\overline{t}_{s'} \geq t$.
\label{claim:rr3}
\end{claim}
\begin{proof}
Consider any node $s'$ such that $X \in \mathcal{X}_{s'}$, that is $s' \in R$. Because the lemma has been shown for Case $1,$ we know that there is a point $Q'$ such that $M_{s'}^{Q'}.tagvec[X] \geq t.$ In particular, at the point where $M_{s'}.tagvec[X]$ becomes at least $t$, it executes line \ref{line:Delete_send1}. Therefore, it adds an element $(\overline{t}_{s'}, s')$ element to $DelL_{s'}[X]$ and sends a $\langle \texttt{del},X,\overline{t}_{s'}\rangle$ message to every other server $s'' \in R$ with tag $\overline{t}_{s'} \geq t.$ By a similar argument, by some point $Q'',$ it receives $\langle \texttt{del},X,\overline{t}_{s''}\rangle$ message from every other node $s'' \in R$ with $\overline{t}_{s''} \geq t.$
Consider the first $\texttt{Garbage\_Collection}$ action after $Q^{''}$ performed by server $s'.$
Note that by $Q'',$ $s'$ has received a $\texttt{del}$ message from server in $R$ for object $X$ with a tag at least $t$. Therefore $t$ belongs to the set $U$ executed in Line \ref{line:Delete3_Uset}, which implies that $t_{s'} = \max(U) \geq t.$ Therefore, in line \ref{line:Delete_send3}, it sends a $\texttt{del}$ message to all nodes with a tag $t_{s'}$ which is at least $t$.
\end{proof}
From the above claim, we conclude that in $\beta$, node $s$ receives at least one $\langle \texttt{del},X,t_{s'}\rangle$ from every node $s'$ in $R$ with $t_{s'} \geq t$. Let $\overline{P}$ be the first point no sooner than $P$ at which the node $s$ has received such messages.
Consider any $\texttt{Encoding}$ action performed by node $s$ after $\overline{P}$. At point $\overline{P}$, from our earlier assumption $L_s^{\overline{P}}[X].Highesttagged.tag \geq t > M_{s}^{\overline{P}}.tagvec[X].$ Because the $\texttt{Encoding}$ is performed after $\overline{P}$, we have $t \in U,$ where $U$ is found on executing line \ref{line:Delete2_Uset}. Further, because $(M_{s}^{P'}.tagvec[X],v) \in L_{s}^{Q''}[X]$, we have $t \in \overline{U},$ where $\overline{U}$ is found in line \ref{line:Delete2_Ubarset}. Thus, $\max(U \cap \overline{U}) \geq t > M_{s}^{\overline{P}}.tagvec[X],$ which implies that line \ref{line:Delete2_condition} returns true and line \ref{line:updatetag2} is executed. Thus, after the $\texttt{Encoding}$ action, we have $M_s.tagvec[X] = \max(U \cap \overline{U}) \geq t,$ which completes the proof.
\underline{Proof of (b)}
It suffices to show that (i) every node $s'$ sends a $\texttt{del}$ message for object $X$ with tag at least $t$ eventually to every other node $s$, and (ii) that every node $s$ adds an element $(t_s,s)$ to $DelL_{s}[X]$ with $t_{s} \geq t$ eventually.
From Claim \ref{claim:rr3}, we know that every node $s$ with $X \in \mathcal{X}_{s}$ eventually sends $\texttt{del}$ message with tag $t_{s} \geq t.$ Therefore, the lemma holds if $X \in \mathcal{X}_{s}$.
For a node $s$ such that $X \notin \mathcal{X}_{s}$, because we have shown property (a) of the lemma, line \ref{line:updatetag2} is executed with updating $M_s.tagvec[X]$ with a tag at least $t$. We note that an update of the $M_{s}.tagvec[X]$ variable is also associated with the sending of a $\texttt{del}$ message in line \ref{line:Delete_send2}, and the adding of a $(t_s,s)$ element to $DelL_s[X].$ Therefore, the lemma holds.
\end{proof}
\begin{lemma}
Consider a fair execution $\beta$ where every server is non-halting. Let $\pi$ be a write to object $X$ that completes in $\beta$ with value $v$. Then,
\begin{enumerate}[(a)]
\item there is a point $P$ in $\beta$ after which, for any server $s,$ for any point $Q$ which is after $P$, $M_s^{Q}.tagvec[X] \geq tag(\pi).$
\item there is a point $Q$ in $\beta$ by which every server $s$ has, for every server $s' \in \mathcal{N}$, at least one tuple $\langle t_{s'}, s'\rangle$ in $Del_{s}^{Q}[X]$ with $t_{s'} \geq tag(\pi)$
\end{enumerate}
\label{lem:Meventuallyexceedseverywrite}
\end{lemma}
\begin{proof}
We show (a) first.
Suppose $\pi$ is issued by a client in $\mathcal{C}_{s'},$ for some server $s'$.
Based on the protocol, on receiving the value write $\pi,$ server $s$ adds element $(tag(\pi),v)$ to $L_s[X]$ and sends an $\langle \texttt{app},X,v,tag(\pi)\rangle$ message to all other servers. From Lemma \ref{lem:eventually_apply}, and from line \ref{line:list_applyaction} in Algorithm \ref{alg:internal_actions}, we infer that in $\beta,$ for any server $s$, the element $(tag(\pi),v)$ is eventually added to $L_s[X]$ in $\beta.$ Combined with Lemma \ref{lem:Mexceedslisteventually} (a), we conclude that there is a point $P'$ such that, for every point after $P_s,$ $M_s.tagvec[X] > tag(\pi).$ Therefore, property (a) of the lemma is satisfied by choosing point $P$ to be the latest of $P_1, P_2, \ldots, P_N$.
Statement (b) of Lemma \ref{lem:Meventuallyexceedseverywrite} similarly follows from statement (b) of Lemma \ref{lem:Mexceedslisteventually}.
\end{proof}
\begin{lemma}
Consider any infinite execution $\beta$ of $CausalEC$, and consider point $P$ of $\beta$, server $s$ and object $X$. At least one of the following statements is true:
\begin{enumerate}
\item An element $(M_s^{P}.tagvec[X],v)$ is not added to $L_s[X]$ at any point of the execution after $P$, for any value $v$
\item There exists a point $Q$ of $\beta$ after $P$ such that $M_s^{Q}.tagvec[X] > M_s^{P}.tagvec[X]$
\end{enumerate}
\label{lem:listisnotaddedbeyond}
\end{lemma}
\begin{proof}
From Lemma \ref{lem:pointsexec}, we know that there is a point before $P$ at which an element is added to $L_{s}[X]$ with tag $M_s^{P}.tagvec[X]$. From the protocol, we note that the first such point of addition is either on receiving a write in line \ref{line:writeaddtolist} in Algorithm \ref{alg:inputactions_clients}, or an apply action in line \ref{line:applycondition} in Algorithm \ref{alg:internal_actions}.
For any tag, each node receives either an apply message, or a message from a write with that tag, and such a message is received only once. Therefore, we note that an element with tag $M_s^{P}.tagvec[X]$ is never added after point $P$ to $L_s[X]$ via \ref{line:writeaddtolist} in Algorithm \ref{alg:inputactions_clients}, or an $\texttt{Apply\_Inqueue}$ action in line \ref{line:applycondition} in Algorithm \ref{alg:internal_actions}.
To complete the proof, we consider the case where for any point $\overline{Q}$ of a fair execution $\beta$ that is after point $P$, $M_s^{\overline{Q}}.tagvec[X] = M_s^{P}.tagvec[X]$. We aim to show that that an element with tag $M_s^{P}.tagvec[X]$ is not added after point $P$ to $L_s[X]$ on behalf of lines \ref{line:valrespaddtolist}, or line \ref{line:valrespencoded_addtuple} in Algorithm \ref{alg:input_actions}. Note that these lines are executed at server $s$ on receipt of a $\texttt{val\_resp}$ or a $\texttt{val\_resp\_encoded}$ message from some server $s',$ with parameters $clientid=\textrm{localhost},$ object $X$, and a tag vector $tvec$ with $tvec[X] = M_s^{P}.tagvec[X].$ Note that server $s'$ sent such a message on receipt of a $\texttt{val\_inq}$ message from server $s$. Based on the code, the sending of such a $\texttt{val\_inq}$ from server $s$ with $clientid=\textrm{localhost}$ occurs on line \ref{line:valinq_send2}. However, this line is executed if there exists an element $(t,v)$ in $L_s[X]$ with $t > M_s^{P}.tagvec[X].$ Lemma \ref{lem:Mexceedslisteventually}, there is a point $Q$ after $P$ where $M_s^{Q}.tagvec[X] \geq t > M_s^{P}.tagvec[X]$. This is a contradiction. Therefore, that an element with tag $M_s^{P}.tagvec[X]$ is not added after point $P$ to $L_s[X]$ in $\beta$.
\end{proof}
\begin{lemma}
Consider an infinite execution $\beta$ of $CausalEC$, and consider point $P$ of $\beta$, server $s$ and object $X$.
For a tag $t < M_s^{P}.tagvec[X]$, then eventually, there is a point $Q$ such that, after $Q$ no element $(t,v)$ is added to $L_s[X]$ for any value $v$.
\label{lem:listisnotaddedbeyond2}
\end{lemma}
\begin{proof}
For tag $t$, at node $s$, an element $(t,v)$ is added to $L_s[X]$ at most one time in $\beta$ via line \ref{line:writeaddtolist} in Algorithm \ref{alg:inputactions_clients}, or line \ref{line:applycondition} in Algorithm \ref{alg:internal_actions}. We aim to show that that an element with tag $t$ is not added eventually, after a point $Q$, to $L_s[X]$ on behalf of lines \ref{line:valrespaddtolist}, or line \ref{line:valrespencoded_addtuple} in Algorithm \ref{alg:input_actions}. From the code, any element $\langle clientid,opid,X, tvec, \overline{w}\rangle$ that is added to $ReadL_s$ at point $Q$ has $tvec=M_s^Q.tagvec.$ If $Q$ is after $P$, then from Lemma \ref{lemma:tagsalwaysincrease}, we have $tagvec[X] = M_s^Q.tagvec[X] \geq M_s^P.tagvec[X] > t$. Therefore, an element $\langle \texttt{localhost},opid,X, tvec, \overline{w}\rangle$ with $tvec[X] = t$ is added at most a finite number of times to $ReadL_s$ in $\beta$ in line \ref{line:readlentryencoding}. Correspondingly, a finite number of $\texttt{val\_inq}$ messages are sent on line \ref{line:valinq_send2} with a parameter tag vector $tvec$ and object $X$ satisfying $tvec[X] = t.$ Since lines \ref{line:valrespaddtolist}, or line \ref{line:valrespencoded_addtuple} in Algorithm \ref{alg:input_actions} are executed on $\texttt{val\_resp}$ and $\texttt{val\_resp\_encoded}$ messages that are responses to $\texttt{val\_inq}$ messages, these lines are executed a finite number of times with parameters $tvec$ and object $X$ satisfying $tvec[X]=t.$ Therefore, eventually, there is a point $Q$ after which an element with tag $t$ is not added to $L_s[X]$ on behalf of lines \ref{line:valrespaddtolist}, or line \ref{line:valrespencoded_addtuple} in Algorithm \ref{alg:input_actions}. This completes the proof.
\end{proof}
\remove{
\begin{lemma}
Consider a fair execution $\beta$ where every server is non-halting. Let $\pi$ be a write to object $X$ that completes in $\beta$ with value $v$. Then, there is a point $P$ in $\beta$ after which, for any server $s,$ there is no entry of the form $(tag(\pi),val)$ in $L_s[X]$ for any value $val.$
\label{lem:finalstorage}
\end{lemma}
}
\subsection{Proof of Lemma \ref{lem:storagecostkey}}
\begin{proof}
We consider two cases (I) $\pi$ is the highest tagged write to object $X$ in $\beta,$ and (II) there is a write in $\beta$ with a tag larger than $tag(\pi)$ in $\beta$.
\underline{Case (I) $\pi$ is the highest tagged write to object $X$ in $\beta.$}
Consider any server $s$. From Lemma \ref{lem:Meventuallyexceedseverywrite}, we know that there is a point $Q$ such that:
\begin{itemize}
\item server $s$ has, for every server $s' \in \mathcal{N}$, an element $(t_{s'},s')$ in $DelL_{s}^{Q}[X]$ with $t_{s'} \geq tag(\pi).$
\item $M_s^{Q}.tagvec[X] \geq tag(\pi).$
\end{itemize}
Furthermore, since there is no write to object $X$ with a tag larger than $tag(\pi)$, so we have $t_{s'} = tag(\pi), \forall s'$ at $Q$, and $M_s^{Q}.tagvec[X] = tag(\pi).$ In fact, because of Lemma \ref{lemma:tagsalwaysincrease}, at every point after $Q$, we have $M_s.tagvec[X] = tag(\pi).$
Consider the first $\texttt{Garbage\_Collection}$ action after ${Q}$. We claim that line \ref{line:GCcondition1} returns true. Since there is an element $(tag(\pi),s')$ for every $s' \in \mathcal{N}$ in $DelL_s[X],$ the set $S$ in line \ref{line:sset} and the set $\overline{S}$ in line \ref{line:s1set} both contain $tag(\pi).$ Therefore, $t_{max}[X] = tag(\pi)$ in line \ref{line:tmax}. Further, if $L_s[X]$ is non-empty, since there is no tag larger than $tag(\pi)$ in $\beta$ for a write to object $X$, $L_s[X].Highesttagged.tag \leq t_{max}[X] = M_s.tagvec[X].$
Therefore, line \ref{line:GCcondition1} returns true and \ref{line:GC1} is executed. Further, because $tag(\pi) = M_s.tagvec[X],$ it does not belong to set ${T}$ identified in line \ref{line:pendingreads}. Therefore, line \ref{line:GC1} removes any element with tag $tag(\pi)$ from $L_s[X].$ Because $tag(\pi)$ is the largest tag in $\beta$ for objrct $X$, note that statement $(2)$ of Lemma \ref{lem:listisnotaddedbeyond} is not satisfied after the $\texttt{Garbage\_Collection}$. From the lemma statement, we conclude that statement (1) of Lemma \ref{lem:listisnotaddedbeyond} holds. That is, we know that any element with tag $tag(\pi)$ is not added again to $L_s[X]$ after the point of $\texttt{Garbage\_Collection}$ action. Therefore, the lemma holds with point $P$ being the point of the $\texttt{Garbage\_Collection}$ action.
\underline{Case (II) $\pi$ is not the highest tagged write to object $X$ in $\beta.$}
Let $\phi$ be a write to object with tag larger than $tag(\pi).$ Then from Lemma \ref{lem:Meventuallyexceedseverywrite}, we know that
we know that there is a point $Q'$ such that:
\begin{itemize}
\item server $s$ has, for every server $s' \in \mathcal{N}$, an element $(\overline{t}_{s'},s')$ in $DelL_{s}^{Q}[X]$ with $\overline{t}_{s'} \geq tag(\phi) > tag(\pi).$
\item $M_s^{Q}.tagvec[X] \geq tag(\phi) > tag(\pi).$
\end{itemize}
Since the hypothesis of Lemma \ref{lem:listisnotaddedbeyond2} is satisfied with tag $t = tag(\pi)$, we conclude that there is eventually a point $Q''$ such that, after $Q'',$ no element is added with tag $t=tag(\pi)$ to $L_s[X].$ Since entries added into $ReadL_s$ at a point $\overline{P}$ have parameter tag vector $M_s^{\overline{P}}.tagvec,$ no entry is added with parameter tag vector $tvec$ satisfying $tvec[X]=tag(\pi)$ after $Q$ to $ReadL.$
Therefore, entries with parameter tag vector $tvec$ satisfying $tvec[X]=tag(\pi)$ are added at most a finite number of times to $ReadL_s$ in $\beta.$ From Theorem \ref{lem:read_terminates_f=0}, we note that every pending read in $ReadL_s$ is eventually returned and the entry is cleared. Let $Q'''$ be the point where every read with tag vector $tvec$ satisfying $tvec[X]=tag(\pi)$ is cleared from $ReadL_s.$
Let $\overline{Q}$ be the latest of $Q', Q'', Q'''.$
Consider the first $\texttt{Garbage\_Collection}$ action after $\overline{Q}$, suppose this is performed at point ${P}$. Since $P$ is no earlier than $Q'$ there is an element $(t_{s'},s')$ for every $s' \in \mathcal{N}$ with $t_{s'} > tag(\phi)$ in $DelL_s^{P}[X],$ the set $S$ in line \ref{line:sset} contains $tag(\pi).$ Therefore, $tmax_{s}^{P}[X] > tag(\pi)$ in line \ref{line:tmax}. Further, since the action is performed no sooner than $Q''',$ $tag(\pi) \notin T,$ where $T$ is determined in line \ref{line:pendingreads}. Therefore, the $\texttt{Garbage\_Collection}$ action executes \ref{line:GC1}, \ref{line:GC3} or \ref{line:GC2}, which all remove any element with tag $tag(\pi)$ from $L_s^{{P}}[X].$
Further, since ${P}$ is after $Q'',$ an element with tag $tag(\pi)$ is not added after ${P}$ to $L_s[X].$ Therefore, the lemma holds, that is, after $P$ there is no entry with tag $tag(\pi)$ in $L_s^{Q}[X].$
\end{proof}
\remove{
\subsection{Proof of Theorem \ref{thm:storagecost}}
\underline{Proof of (a):}
Consider any execution $\beta$ satisfying the hypothesis of the theorem. By Lemma \ref{lem:finalstorage}, for every write $\pi$ in $\beta$, there is a point $P_{\pi}$ after which there is no entry of the form $(tag(\pi),val)$ in $\cup_{X \in \mathcal{X}}\cup_{s \in \mathcal{N}}L_s[X].$ Consider the point $P$ which is the latest of $\{P_{\pi}: \pi \textrm{is a write in}\beta\};$ because $\beta$ has a finite number of writes, point $P$ exists. For any point $Q$ after $P$, we have $\cup_{s\in \mathcal{N},X \in \mathcal{X}}L_s^{Q}[X]=\{\}.$
\underline{Proof of (b):}
Because $\beta$ has a finite number of writes, for any server $s$, there are only a finite number of elements added to $Inqueue_{s}$ in $\beta$. Statement (b) readily follows from Lemma \ref{lem:eventually_apply}.
\underline{Proof of (c):}
For any server $s$, an entry $(clientid, opid, tvec,X,\overline{w})$ is added to $ReadL_s$ on receiving a read in line \ref{line:addtoreadl}, or on performing an encoding action in line \ref{line:readlentryencoding}. Since there are only a finite number of read and write operations, these lines are executed only a finite number of times - at most once for every read operation received by server $s$, and at most once for every write operation in $\beta.$ Therefore, to prove the theorem, it suffices to show that every entry in $ReadL$ is eventually removed in $\beta.$
For any entry made to $ReadL_s$ in line \ref{line:addtoreadl} in Algorithm \ref{alg:inputactions_clients} or line \ref{line:readlentryencoding} in Algorithm, \ref{alg:internal_actions} Theorem \ref{lem:read_terminates_f=0} ensures that a response is eventually received. Every action that receives the response eventually removes the read from $ReadL$ in line \ref{line:read-remove1}in Algorithm \ref{alg:inputactions_clients}, or in lines \ref{line:read-remove2}, \ref{line:read-remove3} in Algorithm \ref{alg:input_actions}, or in line \ref{line:read-remove4}, \ref{line:read-remove5} in Algorithm \ref{alg:internal_actions}.
}
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Last week, I thankfully got an invite to use Measure Map. I have been using it non-stop, making tests, jotting down my likes and my suggestions, etc., and emailed Measure Map asking for permission to write about it. So, for those of you that have not received a copy, hopefully this look at Measure Map will clear a few things up for you. There has been quite some hype about Measure Map and I feel that there definitely should. It is very organized, writes out statistics in readable sentences rather just numbers in a table, integrates Flash and Ajax very well and uses both technologies to compliment each other for certain actions, and a whole lot more. Ok, lets get started. We’ll start at the beginning, installing Measure Map.
I was very curious as to how Measure Map’s setup and installation process was going to be like. This is because it has features such as viewing stats on your site itself, posts, comments of each individual post, and a way to track clicks going out of your site. The process was actually quite simple and seeing how to do it and what they needed explained how they manage to grab the specific blog stats. When you get your invite and go to the installation page, it will ask for your site name, site address, and time zone, and importantly what blogging software you use. They give you a choice between five different software and you can also choose other. Why does it ask for your software? Because when you install Measure Map, it will provide you with the exact code that you need to place into your software (note: you can get it for any software later on as well). For example, WordPress has different code to place in your site then TypePad. WordPress has three steps and that is to include a line of JavaScript in the Footer Template, some code in the Post Template, and lastly code in the Comments Template. For each step, Measure Map provides easy to follow instructions and even screenshots of the blog software so you can very easily set it up. Now, remember, that not all software is the same for installation. TypePad had two steps, Blogger had one step, and Other is fairly similar to WordPress’s installation. Once you include the code on your site, visit it just to make sure everything is running smoothly and then log into Measure Map and everything should be ready to go.
Note: Clicking on the screenshots will open a larger image
When you first login, you will see the overview page of your blogs statistics. The interface for Measure Map uses an excellent blend of colors, great typography that is easy to read, and graphics/icons that show exactly what you are looking it, not to mention it all is Web Standard too and looks great in text-only format as well. On the top of the overview page, you will see four boxes: Visitors, Links, Comments, and Posts. Below the icons in the boxes are the figures for each box, such as how many visitors came to your site today or how many comments were made today. What I also like is that they give a little description under each box saying something like, “came to your blog. That’s 100 fewer than an average day,” which compares the statistics for the day to your overall statistics. Now, looking below the boxes are two columns. The first column gives an overview of popular posts for the day. It shows how many visits you have received to your posts. This is great because now I can still see what posts people are viewing easily, for example, I see that my Photoblog Webware Roundup and Visualizing Del.icio.us posts are still attracting visitors. On the right column is a soon to be feature, “What’s Happening on your Blog.” I have been told that they are still working hard on this to get it complete. This is still in alpha release, so I am not expecting everything to be perfect nor complete. But looking at this screenshot from Jeffrey Veen, it shows the column with information, such as traffic being lower, possible comment spam, etc.. I also see that there are RSS feeds on both columns, while on the alpha release there are not. I am not sure if these features will be in the final release or not, but they sure do look great.
One feature that you must be aware of is the really neat date range selection at the top of every Measure Map statistics page. At the top right, you will see a gray box with the date and an icon of a calendar. Click on this and you will see a graph drop down. It is a bar graph of visitors for every day since you started using Measure Map. When you hover over a bar, you can see how many visitors you have received that day. You can click on a days bar and it will then bring up those results for that specific day throughout Measure Map. I feel this is a very simple and great way to view past statistics. Now, what if you want a bigger range, for example, viewing last weeks stats. Measure Map has covered this as well. You will see two gray arrows on the left and right of the selected bar. Click and drag one and you will see that it starts to select other days. Let go and the Flash graph will then get Ajax to kick in and update the statistics below with that date range. That is not all that this top date selection does. It also allows you to see all the posts that you have made. On the bottom of the graph, you will see icons that look like paper. Putting your mouse over one will show you the name of the post and the date the post was made. When you click on it, it will bring you to the posts detail page.
Measure Map puts your statistics in a very understandable format. Instead of just numbers, it explains your statistics to you with sentences. For example, viewing the visitors page, you will see:
849 visitors came to your blog
That’s 220 more than an average day.
15% of those visitors had been to your blog before
That’s 127 visitors, 43 more than an average day.
Seeing statistics written out like this makes it much easier to give yourself a visual on how your sites performance is. I can easily see if my site is having a good day and how many regular visitors have come back for a visit. The visitors page doesn’t show much else except a bar graph on the bottom showing the last thirty days and how many visitors you received each day. But how about we go look at a the Links page (screenshot above). I find links page very helpful for tracking my sites statistics and view this more often then any other page. I can easily see how many links are referring to my site, how many are new referrals that my site hasn’t received traffic from before, and what I love is that you can even see statistics for links out of your site, meaning links that your visitors leave your site with. When you enter the page, it will write out your stats again in clear sentences saying how many new links (referrals) you have received for that day and how many visitors they have attracted and compare the stats to past statistics to give you an average. You can also see how many links in total you have received. I really like how Measure Map decided to split new links and total links. It helps gives a better visual on the performance of your site and how active it is. On the links page, you don’t only see statistics for incoming links, but outgoing links as well. For example, when I publish this post, I will be able to see how many times my visitors have clicked on links going to Measure Map and what links they were. This helps me see my visitors interests. This is not as extensive as MyBlogLog.com, but it gets the job done rather well and is enough to please me.
On the Links page that I have talked about above, you will also see a link that says something around, “102 search terms used.” Clicking this will bring you to the search engine overview page. This page definitely opened my eyes about the power of search engines because I seriously did not realize how many refers I get from search engines a day. I was very easily able to see how many visitors came from Google and I was also able to see how much MSN and Yahoo dislike my site. You can also view each individual term that the search engines are sending you. Clicking on a search engine will make a table of each term and how many times it was used to get to your site (right now, “Fruitcast” is the top search term). Thanks for opening my eyes Measure Map. I know that other statistic software that I use shows information like this as well, but Measure Map has shown it a bit clearer and just made myself more aware of it by showing me results for individual days and posts. One thing that I am wondering though is what search engines are included, does it just show the major search engines?
As I said above, Measure Map uses a great blend of Flash and Ajax throughout the application. The countries section uses Flash to show a map and statistics for each country and shows the advantage of using Macromedia Flash for applications. It shows the map of the world with the countries being different shades of red, depending on the amount of visitors, giving a heat map type effect. I can see that I received a good portion of visitors from United States and Spain because they are the brightest shade of red. Not only that, but when placing my cursor over a country, it tells me a percentage of visitors from the overall map and how many visitors I had from the country. You can also see a small table on the bottom left of the map that shows the top five countries. On the top left you can also see a method to zoom in and out of the map to focus on small areas and you can also click and drag on the map to move around.
Something that I found eye catching was the browser statistics page. It presents the statistics again using Flash and in an easy to read bar graph. Instead of using simple bars, they went ahead and used Browser logos instead. Even though that is a very small adjustment to the normal bar graph, it actually makes it easier to take in. I just look at the graph and don’t have to look at any text to understand what I am seeing.
I have gone over some of what Measure Map can accomplish with you, now let me go over some issues that I came across and some of my opinions/suggestions about Measure Map. Believe me, there aren’t many issues that I have come across. The team has really done an excellent job. The first thing is minor compatibility issues in Firefox/Flock. When viewing statistics on pages such as countries where Flash is in use and then clicking on the top date selection to bring down the Flash bar graph, it overlaps ontop of the other Flash document, in this case the world map. However, I noticed that it does as it should in Safari, which is shifting the map down so the bar graph that you open has its own space. Next! I noticed that when viewing the Links page and seeing links to some referrals and outgoing links that if they are long, they are not shortened or word wrapped, but instead continue going through the column warping the table a bit. I would be happy with word wrapping these long links or maybe just cutting them short with a “…” at the end of them to show they are long addresses. Now, in the same links page, when viewing incoming links, I got annoyed with one small thing, but may not be the case with everyone. It deals with Bloglines. I get quite a bit of referrals from Bloglines from my readers. This is great that it picks this up, but what I am seeing is that every incoming link from Bloglines is different because of the unique user IDs and whatnot in their addresses. So, when viewing my referrals, I saw a whole page of only Bloglines. What I am thinking is some kind of way to group links from services such as this. Search for part of the URL from Bloglines and group them all together as if it is one link and that would make it a lot cleaner and easier to see how many visitors came from Bloglines. That’s all I’ve got! I didn’t really see any other issues with the service yet.
Overall, I feel that Measure Map is excellent and definitely deserves the hype that it has been receiving. If you have received an invitation for Measure Map and have any suggestions, send an email over to the Measure Map team. They seem to be very open to what their users have to say. I have suggested some sort of RSS feed tracking or even FeedBurner implementation. They have replied saying that they are actually working on their first iteration of RSS tracking and should be out shortly and said they have also had given FeedBurner some thought and may be included at some point. Great work to the team of Measure Map. I absolutely love it and cannot wait for its official release and sure others are excited as well.
View Measure Map - Get to know your Blog.
Deadliner says:
Solution Watch - solutions for people slacking off on their writing duties
I really am busted now, but I wanted to hook up to this site in that it has so many resources that might make deadlining easier. This is a really extensive site and deserves more than this pithy soundbite, so
November 8th, 2005 at 8:59 pm
I was really looking forward in adding Measure Map to my website. However, I’m gonna have a major problem. I run my own CMS, so it’ll probably not work, which is a shame. I might still have use for it on another site though.
November 10th, 2005 at 6:55 am
Trovster, you may be right. Measure Map is not for every kind of site (which I should have mentioned), but strongly for blogs and news related sites that have multiple entries. If your CMS has a news like system in it, you can most likely hook up Measure Map to it. But if you just have single content pages, Measure Map may not be the right service for your site.
Thanks for the comment!
November 10th, 2005 at 12:23 pm
My website is a blog kind of site, so maybe they’ll be use for it yet on that one. I hope so. Might be a pain to integrate it though…
November 11th, 2005 at 6:15 am
» Blog Archive » Google Groups : Troubleshooting says:
[...] s “XX vistors came to your blog. That’s N more than an average day” (see SolutionWatch and about a dozen other blogs for gushing reviews. Speaking of which, wha [...]
November 15th, 2005 at 10:06 am
Razvan Antonescu » Web Metrics says:
[...] not hapilly coexist on the web metrics market. Additional readings: Solution Watch - A look at Measure Map Techcrunch - (Just a bit) More on Measure Map Jeffrey Veen - Welc [...]
November 16th, 2005 at 5:02 am
Just a Memo says:
Measure Map - Alpha Test, Invite Received
What a nice way start a Wednesday, aka. “Hump Day”. Other than the FREE Wednesday lunch meals we get here at MA, I finally received an invite to give Measure Map a test-drive early this morning. Thank you MM/AP-family!…
November 16th, 2005 at 7:04 pm
Blog Analytics with Blogbeat » Solution Watch says:
[...] ics, but leans more toward blog statistics with simplistic monitoring, as does MeasureMap (Review), which is still in private beta. Blogbeat provides all of the basic statistics in [...]
January 25th, 2006 at 3:23 am
Razvan Antonescu » MeasureMap aquisition email says:
[...] Razvan Antonescu | posted in General, Google Trackback URL | Comment RSS Feed Tag at del.icio.us | Incoming links yactions.buildButton( ’save’, ‘My_Web’ ); yactions.buildButton( ‘blog’, ‘360′); [...]
February 15th, 2006 at 1:03 am
Google Acquires Adaptive Path’s Measure Map at BenBishop.me.uk says:
[...] Adaptive Path’s Measure Map has been snapped up by Google according to Google’s Blog. Measuremap, though still in private beta, came to prominence as a free blog stats service that has been attracting great reviews. Using a great combination of Ajax and Flash, Measuremap gives you your information put into digestible chunks that is in layman’s terms. A product clearly with much potential there has been reports of performance problems recently though on larger sites which is ironic given the recent issues that Analytics had when first launched. The integration into Analytics is sure to follow and it will be interesting to see the results and how competitors like Mint respond. Their homepage has already been updated to reflect this takeover; with a link to Google’s blog at the top and a ‘Measure Map is a production of Google’. [...]
February 15th, 2006 at 3:06 pm
futuria.hr » Blog Archive » Measure Map u rukama Googlea says:
[...] Google je prije nekoliko dana kupio još jednu firmu koja radi software za web analitiku. Radi se o Measure Mapu, koji je za razliku od robusnog Google analyticsa namjenjen prvenstveno manjim sajtovima i blogovima. Ako vas zanima kako Measure Map funkcionira u praksi i želite ga instalirati na svoj blog ili sajt pogledajte recenziju i screenshotove na Solutionwatchu. [...]
February 17th, 2006 at 6:45 am
» Lowdown on Performancing Metrics Online Marketing Blog says:
[...]. [...]
March 13th, 2006 at 3:33
Brightlamp Technology says:
Google Announces Analytics for Blogs….
Google have announced that they have purchased Adaptive Path’s Measure Map. This is an analytics package specifically tailored for blogs….
April 18th, 2006 at 3:54 pm
Hello, I’m looking for a measure map invite.
Do you have one?
Thanks
Wayne
June 8th, 2006 at 12:09 am
| 33,202
|
TITLE: Solving the simplest coupled nonlinear ODES for chemical kinetics
QUESTION [1 upvotes]: I am just trying to get the integrated form for the kinetics of the reaction
$A + B \rightarrow C + D$ characterized by:
$$
-\dfrac{d[A]}{dt} = -\dfrac{d[B]}{dt} = k[A][B] \; .
$$
As you note, this is a system of two nonlinear coupled ordinary differential equations, I will put that in the familiar notation for mathematicians or physicists:
$$
\dfrac{dx(t)}{dt} = -k x(t) y(t)
$$
$$
\dfrac{dy(t)}{dt} = -k x(t) y(t)
$$
Can anyone helping me with some reference or ideas to solve it?
REPLY [4 votes]: Since the rate of change of $ x $ is the same as the rate of change of $y $ you really only a single equation of with one variable. We write,
\begin{equation}
x = y + c
\end{equation}
where the constant $ c $ is determined by your initial conditions,
\begin{equation}
c = x (0) - y (0)
\end{equation}
(in your case it is the difference between the concentrations at the start). Inserting in this relation we have,
\begin{equation}
x' = -k ( x ^2 - x c )
\end{equation}
This is now a simple equation to solve,
\begin{equation}
\int _{x _0 } ^{ x} \frac{ d x }{ - k ( x ^2 - x c ) } = \int _0 ^t d t
\end{equation}
where $ x _0 \equiv x ( 0 ) $. Resorting to Mathematica (though you could use partial fractions) gives,
\begin{equation}
\log \left[ \frac{ x }{ x _0 } \frac{ x _0 - c }{ x - c } \right] = kt
\end{equation}
Isolating for $ x $ I get,
\begin{equation}
x = - \frac{ c e ^{ kt} }{ \alpha - e ^{ kt} }
\end{equation}
where $ \alpha \equiv ( x _0 - c ) / x _0 $.
Plotting $x$[red] and $y$[blue] for different $x0$ and setting $c=1$ we have,
$\hspace{2cm}$
REPLY [3 votes]: Hints:
Conclude that $y-x=c$ is a constant.
Use separation of variables $-k\int \!\mathrm{d}t= \int \!\frac{\mathrm{d}x}{x(x+c)}$.
REPLY [3 votes]: The other answers address how to solve this analytically, but I like numerical solutions to things so here goes:
$$\frac{d}{dt} \begin{bmatrix}x \\ y \end{bmatrix} = -\begin{bmatrix}kxy\\kxy\end{bmatrix}$$
which can be solved using any number of numerical methods. For simplicity, we can take the second-order Runge-Kutta method where $i$ is the time index.
Step 1:
$$\begin{aligned}x^{i+1/2} &= x^i - \Delta t k x^i y^i\\y^{i+1/2} &= y^i - \Delta t k x^i y^i\end{aligned}$$
Step 2:
$$\begin{aligned}x^{i+1} &= \frac{1}{2}\left(x^i + x^{i+1/2} - \Delta t k x^{i+1/2} y^{i+1/2}\right)\\y^{i+1} &= \frac{1}{2}\left(y^i + y^{i+1/2}- \Delta t k x^{i+1/2} y^{i+1/2}\right)\end{aligned}$$
This method is then marched until you reach your physical time of interest or until you reach steady state (which is $t \rightarrow \infty$ technically but may be truncated by measuring the residuals and stopping when they approach zero.
Some care must be taken when choosing your time step, $\Delta t$. Too large and the integration will be unstable. Too small and you're wasting time.
Analytical methods are awesome and all but the numerical approach will work for much more complicated mechanisms and with way more coupled equations. So it's a good tool to have in the box.
| 50,919
|
TITLE: Is there a theorem or axiom stating that integers added to integers always yields integers?
QUESTION [5 upvotes]: I have finished a small proof for school but I realize it relies on the statement that integers added to integers yields integers. I assumed this statement was pretty much just accepted, but considering how central it is to my proof I wanted to cite the theorem or law or something. So far no internet search has turned up anything useful.
REPLY [2 votes]: Notation: write $\mathbb{Z}$ for the set of integers and $\mathbb{R}$ for the set of real numbers.
Case 0. Perhaps we're viewing $\mathbb{Z}$ as a standalone object, independent of its status as a subset of the real line.
Then by definition, it is a set $\mathbb{Z}$ equipped with functions $$+:\mathbb{Z}, \mathbb{Z} \rightarrow \mathbb{Z}, \qquad -:\mathbb{Z} \rightarrow \mathbb{Z}, \qquad 0:\mathbb{Z},\qquad \cdot : \mathbb{Z}, \mathbb{Z} \rightarrow \mathbb{Z}$$
satisfying certain conditions.
(The problem then remains of showing that this mathematical structure actually exists.)
Anyway, in this case, there is no need to prove that the sum of two integers is an integer; its just a consequence what functions are (and/or how they work).
Case 1. On the other hand, perhaps we're viewing $\mathbb{Z}$ as a subset of $\mathbb{R}$. In this case, in principle we need to prove that the sum of two integers is an integers. This is essentially immediate from the defintion, however.
Definition. $\mathbb{Z}$ is the least subset of $\mathbb{R}$ such that:
$1 \in \mathbb{Z}$
$-1 \in \mathbb{Z}$
$x+y \in \mathbb{Z}$ whenever $x,y \in \mathbb{Z}$
Notice that the sum of two integers is itself always an integer, essentially by definition (again!)
It remains to show that this thing actually exists, in other words that $\mathbb{R}$ actually has a least subset satisfying these three conditions. We can prove this from accepted mathematical principles as follows.
Let $\mathbb{Z}$ denote the intersection of all subsets of $\mathbb{R}$ satisfying the above three condtions.
Show that $\mathbb{Z}$ must itself satisfy these conditions.
Show that if a subset $J$ of $\mathbb{R}$ satisfies these conditions, then $\mathbb{Z} \subseteq J$.
| 12,167
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Reconstruction consists of complete curb and gutter replacement, roadway reconstruction, water main replacement, storm sewer repairs and upgrades, and sanitary sewer repairs.
Links
* You can also click on the map below for additional information
The project is substantially complete. The contractor is finishing up a few cleanup items. Final paving is scheduled for spring 2020. Please contact Cody Pietz - Project Inspector, at 952-356-6839 if you have any questions or concerns.
The contractor’s 30 day maintenance period on the sod is ending soon. Please pay attention for a mailer on sod care arriving soon.
| 137,204
|
Pearls
“A famous pearl – it even had a name – that was once owned by Elizabeth Taylor sold for eleven point eight mill yesterday,” Admin says, tossing the paper to Spence.
Spence reads. “Hey! I like this part where they lose the pearl in Vegas and find her dog chewing on it.”
“Thought you would,” says Admin. “Remind you of anyone?”
And then for some reason they’re both looking at me.
And our Pearl:
Tags: Elizabeth Taylor, Pearl
This entry was posted on Wednesday, December 14th, 2011 at 7:32 am and is filed under Chet The Dog. You can follow any responses to this entry through the RSS 2.0 feed. Both comments and pings are currently closed.
8:13 am on December 14th, 2011
If I had my choice, I’d take the four legged Pearl over the round one that Liz wore.
Hey, Folks, we have only today and two more to put AJ on top. Sign on and vote before you do anything else. It’s really important.
8:17 am on December 14th, 2011
Spence’s Pearl is a cutie.
The sale of the “famous pearl” for 11.8 mil proves how screwed our value system is.
Good morning all.
8:19 am on December 14th, 2011
Good Morning.
I always liked Liz Taylor. I hope that just a small portion of the proceeds goes to dogs. The pearl went for almost 5 times the estimated selling price!
I’m off to vote.
8:36 am on December 14th, 2011
Today is my last day of teaching. This time next week (after final grades are in) I’ll be retired.
8:38 am on December 14th, 2011
B.Stover: Congratulations to you!
8:50 am on December 14th, 2011
Thank you Dawson.
8:51 am on December 14th, 2011
Professor Stover: Congrats! Are you still up for your retirement bash this weekend?
Dawson: I hope you’re feeling better.
Tub: You are a great reminder each day. Is your test today? Good luck!
8:52 am on December 14th, 2011
Grrrr….iPhone. It should be Tyb not tub. Sheesh.
9:09 am on December 14th, 2011
Snowhook Bella: Thank you! Yes to the party.
TYB: Good luck.
9:13 am on December 14th, 2011
What a day, Bee. Wonderful, but a bet you get a little teary too.
Totally agree with TYB, would much rather have Spence’s little Pearl than Liz’s pearl. So much more fun, and emotionally gratifying.
Party this weekend. There is some goat aging underground right now. What else is on the menu?
9:39 am on December 14th, 2011
B., Happy Last Day of Teaching.
9:41 am on December 14th, 2011
Last day for B! Hurray!
Can’t see enough of little Pearl.
Dawson: Feeling better?
A bit iffy on Goat – no matter how it is prepared.
Hi SnowBella.
Off to vote.(TYB)
9:46 am on December 14th, 2011
Sickening!….All that money spent by the filthy rich on jewelry when there is so hunger and suffering in the world right now… sniff! ….Makes me sick to my stomach!…puke!
Card!… Well my friend woke up this morning and looked at what she made last night and decided she didn’t like it very much. Especially after her mother said, “Why didn’t you make Sunflowers?” Really? …..grunt!…. So she took the little card to the busy-place with her and gave it to the gal who wrangled her into making it…sniff!….That gal says she really likes it and is going to have it framed. My friend told her not to spend a lot of money on it….heh! …The gal said if my friend didn’t like it after it was framed then she wouldn’t give it to their boss….deal!
Stover!…Congrats!…snort!
9:51 am on December 14th, 2011
B&M: This wouldn’t happen to be the same goat that was, er, shall we say tenderized by Siber-H? I’m planning on bringing the fixings for root beer floats. Maybe I’ll whip up a batch of Bev’s almond bark popcor, too.
Off to work. I’ve got a long day ahead of me kicked off with a bad commute. Ugh.
9:53 am on December 14th, 2011
Second!….snort!
9:53 am on December 14th, 2011
Rio, your friend is an artiste and her cow-orkers (tee hee, orking a cow) need to know that her beautiful work takes time.
10:02 am on December 14th, 2011
Rio, Liked the card, also noticed it said shellabrate. Very cute. As far as our cards go, have been trying to decide on the favorites. They are all so pretty! Think we are leaning towards the candles and the large one with the dark red flowers. Not sure what they are – they look sort of like Gerber daises. It is hard to decide.
Yes Rebecca, that is the same goat. It will be delish by this weekend. Prime beef usually ages about 16 days. heh
10:06 am on December 14th, 2011
Goat!…I’m not eating any ole aged goat at Stover’s party…grunt!
10:22 am on December 14th, 2011
Party! Party! Party! Party!
I wonder if Audrey chews on Pearl?
Today is the first day I have felt there is some hope that I will get well. AND Zoe’s sick. Jeez Louise!
The counter guys are back (Thank Dog).
Have to think about what we’re going to bring. The Pack wants to help dig up the goat. (yuck)
Bella, (what the heck are you doing up so early?) I am putting an order in for the Root Bear float.
Riö, I would suggest that next time a co-worker of your friend wants a card in one day that she hand them a card, some paper and glue and say, “Here you go!”
10:27 am on December 14th, 2011
Card!…snort!…Get this, my friend tried to get the little card away from the wrangler so she could take more time and do something much nicer, but the wrangler showed it to another person and they both said the boss would love it….grunt!… So my friend has to let go and quit stressing over it…sniff! …Oh and now both of those other guys want little cards…heh!
10:50 am on December 14th, 2011
Dear Rio – please tell your friend “An artist’s harshest critic is always their own soul”. We are 100% positive that the card she made is lovely, and the fact the cow-orkers(Thanks Sam) now want cards is testament enough that the card is perfect.
Besides, she needs to save her best work for her upcoming art shows! (hint, hint)
Bee! Congratulations and we hope the day is SHORT! But we are going to miss the party this weekend…waaaahhhhh!! The show goes to 10 at night (!) and after 12 hours of showtime each day, we expect Mama will be crawling into bed instead of putting her dancing shows on and digging into goat fondue. Maybe the Pack can bring us a doggie bag – wait…what are we thinking…it’s the PACK. No doggie bag would ever survive the trip.
10:52 am on December 14th, 2011
That was supposed to be “dancing s h o e s”, but thank goodness we mistyped, or the big bad moderator would be hunched over his delete button, maniacal chortling echoing thru the halls.
11:02 am on December 14th, 2011
Dog gone it! I thought The Royal Mum was going to be on Dancing With the Stars!
11:03 am on December 14th, 2011
I voted!
Today between 11 and 1 we are supposed to get two new windows in the kitchen. I will let you know if it happens or if we have a “fun” window adventure like Melanie’s fun remodeling adventures.
BBL
11:03 am on December 14th, 2011
Oh no! Now Zoe is sick. Please somehow let her take it easy.
I think I had the flu (despite flu shot) I went to the doctor (because I have to have a doctor’s note) and he looked at my throat and listened to my chest (for show) then wrote the usual prescription for antibiotics. I wasn’t going to fill the prescription but the man in this house insisted saying he had never seen me so sick. I’m not telling whether or not I’ve actually been taking those pills but I feel almost human today and will go back to work tomorrow.
Rio: Maybe you should quit your day job and go a’quilling?
B.Stover: I will be working the 12 hour shifts this weekend. Sob, sob. I will miss your party.
11:04 am on December 14th, 2011
From now on whenever we need to write that word that starts with sh and ends with oe that we say ‘shows’ instead and then the moderator can go to hail.
11:19 am on December 14th, 2011
So nice to have a another photo of Spence’s Pearl instead of a dumb piece of jewelry.
Hurrah for Stover! I know you’ll have a successful last day of teaching.
I am sorry Zoe is sick. She has to look after herself right now and concentrate on getting well.
11:34 am on December 14th, 2011
Gah!…Goat Fondue!…choke!
11:41 am on December 14th, 2011
Don’t worry Rio. I am not digging up that goat.
12:14 pm on December 14th, 2011
Melanie – can you imagine Mama on Dancing with the Stars. No doubt she would be paired up with Mad Max, and the Rhumba would turn into the Rumble! They wouldn’t be able to show clips from the practice sessions due to the “violence on tv” guidelines.
Rio – how about “goat-tage and green”?
12:22 pm on December 14th, 2011
Wrath!….All I know is SiberH has now brought the Wrath of Goats upon all of us…snort! … We will be lucky to get our Christmas cards sent after this….wuffle!
12:27 pm on December 14th, 2011
Siber-H: Thank Dog!
12:29 pm on December 14th, 2011
Dear Plunderers,
We will be at the party.
There better not be any “lamb” served.
Signed,
The Marauding Goats
12:30 pm on December 14th, 2011
OH NOOOO!!!
12:31 pm on December 14th, 2011
Rio: Oh, we’ll get them sent all right. The question is – will they be received? Beware of the TMGs on the rampage!
12:36 pm on December 14th, 2011
They’ll come to the party in disguises for sure. Let’s just hope it’s not as “corgis”. Where are those suits anyway? Haven’t they been reported as missing?
12:50 pm on December 14th, 2011
Party!….snort!…. The Goat Rising will build momentum and they will try to crash Stover’s big –Big-BIG party!….grunt! …. We will have to organize Goat-Guard-Duty (GGD) the night of the party…sniff! ….SiberH will take the first GGD watch from 6:00pm our time till 7:00pm…grunt!…. Alice will take the next 3 or 4 hours after that….chuffle!
12:57 pm on December 14th, 2011
Siber, You certainly stirred things up with that goat business.
We like the idea of dancing shows.
If Zoe is sick, does that mean she can’t go visit Dan? Is he having a connip?
Very foggy day here, glad we are not traveling afar. Dangerous out there, even if drivers are not texting!
1:00 pm on December 14th, 2011
Rio, What exactly do you mean by Goat Rising? Do you mean an uprising, or do you mean the goat is going to be rising up out of the burial plot? That could be a little S. Kingish!
1:05 pm on December 14th, 2011
This weather is dreadful! It’s 37 degrees with rain on the way. Skiing started yesterday with only 3 trails open, but many opening days are limited like this, just not so late into the season. After all that beautiful snow that they made, it has turned into ‘mashed potatoes.’ That’s a skiing term that aptly describes the ski trail surface right now.
1:07 pm on December 14th, 2011
Barb!… It was a “little” play on words…sniff!…. Goat Rising (or Uprising) at the same time meaning the dead aged Goat coming up out of the ground ….heh!
1:14 pm on December 14th, 2011
Skiing!…. Mashed Potatoes! That is the kind of snow condition that makes Orthopedic Surgeons Rich!Rich!Rich!….grunt! ….Stay off your skis MollysMom! ,,,,snort!….We don’t want you laid up during Christmas too!…sniffle!
1:28 pm on December 14th, 2011
OK Rio, thanks, we are breathing a little easier now. Maybe we jumped to conclusions, since we are currently reading King. 400 pages so far. Good, but I bet that book weighs 5 pounds!
1:44 pm on December 14th, 2011
TMGs…All Sibers will be doing MG guard duty at the party. Be Afraid, Be Very Afraid
Mollypop’s Mom, I think that warm system that went through here is what is headed your way. It ruined our snow at the lower elevations, but snowed in the mountains and ski resorts.
1:47 pm on December 14th, 2011
Rio – I’m not going to ski yet, but since I’m on Safety Patrol, I can’t avoid it for long. Since we’re only able to open 3 trails (out of 52), they don’t need me right now.
The Ski Patrol is the one that has to be there because they do all the first aid. Safety Patrol assists them if needed, and directs traffic around the accidents. I’m a Certified Outdoor First Responder and can cut jackets open to stop bleeding, but thankfully I have never needed to do that. A lot of what we do is catch the troublemakers who make it so dangerous for all the families who come here to have fun. I can even kick those guys out for using bad language.
1:49 pm on December 14th, 2011
1 hour for Mr.Husky and 3-4 hours for Alice? Hmmm That means you really really really trust me!
Doesn’t it?
It couldn’t mean anything else could it? Rio??????
1:52 pm on December 14th, 2011
I’m glad we have an experienced goat dispatcher (Siber H) in case those odoriferous creatures try crashing B.Stover’s party.
1:58 pm on December 14th, 2011
OOOO Mollypop’s Mom I am impressed. Can I get you to call the window people?
So I put all the stuff in the way into boxes. I took down the shelf across the sink window that is just perfect for plants which involved emptying half of the cupboard next to the sink so I could get in there with a screwdriver. All that stuff is out on the counter. The shelf is in the dining room on the table. I took the before pictures and we are ready and waiting.
11 to 1.
Finally at 12:30 a phone call.
Nope, not coming. They are cold and wet and need to go home and warm up. Someone else will call and reschedule.
I (the only sighted one) will not be here Thursday or Friday so the next possible day is Monday. They don’t do weekends. The house is a mess and we can’t find basic things that were on the shelf (or hanging below it). Things like measuring cups and spoons and whisks and cheese cutters are all piled in boxes.
The counter is full of spices from the cupboard so these is very little room to do prep work. And it is blocking the toaster.
Not happy.
Not happy AT ALL!
2:16 pm on December 14th, 2011
Dear Staff,
Heh heh heh.
Our revenge has just begun.
Signed,
The Marauding Goats
2:30 pm on December 14th, 2011
GRRRRRR darn goats….
I did call and expess my displeasure. They will call me back (out to lunch) sniff.
I am going to take a nap. That means they will call just as soon as I fall asleep.
grouch.
2:31 pm on December 14th, 2011
GRRRRRR darn goats….
I did call and express my displeasure. They will call me back (out to lunch) sniff.
I am going to take a nap. That means they will call just as soon as I fall asleep.
grouch.
2:54 pm on December 14th, 2011
Staff: Those window guys are terrible. I would be sorely tempted to fire them and hire someone else. Another case of lousy customer service. Unlike Stormy Kromer. Did you vote?
I got 3 calls from work yesterday: 1. Was the potluck still happening on Friday night in the Long Term Care? Yes, I said and will you tell the boss I have a doctor’s note and am calling in sick for tomorrow’s shift? Yes, she would. 2. There is a nursing review meeting tonight at blah blah blah. I won’t be coming as I am sick. 3. The nursing review meeting is cancelled until blah blah blah. Oy veh! Got a call early this morning. “Are you coming to work?” “No…I called in sick yesterday.” The person I gave the message to didn’t pass it on. ((sigh))
2:54 pm on December 14th, 2011
B.
Congratulations on your retirement and best wishes for a very long and happy one.
3:02 pm on December 14th, 2011
Targets!….Uh oh!…snort!…It looks like the Goats are targeting Staff and Dawson today!…gruffle! ….Those dern Goats always pick on the
haplesshelpless ones….sniff!
3:07 pm on December 14th, 2011
Hello all! Tyb, thanks for the daily reminders… we voted!
Bee, Congratulations on your retirement! Our mom says she will live vicariously thru you, so be sure to post your daily happenings.
What happened with YSAS’ audition yesterday? We are hoping …
3:13 pm on December 14th, 2011
Oh! and we definitely agree that Spencer’s Pearl is way better to look at and makes us happier than any old 16th century $11.8 million little stupid pearl. We’d like to eat that stupid pearl. We would only lick Spencer’s Pearl.
3:32 pm on December 14th, 2011
B.Stover: I started reading To the Lighthouse while I was in Minneapolis (a beautiful lavender colored boxed book from Staff’s collection) but it was quickly deemed too educational for vacation reading and discarded in favor of Tony Hillerman novels. Now I have started it again. Thank dog I do not live in the time of Mrs. Ramsey, as to which fact the man who lives in this house can surely attest.
He has gone out just now…to look for a job so he says. He says I am the cause of that grey patch of hair he has right in the middle of the top of his head. I say NO WAY! If I was the cause of the grey it would not be little patches here and there but a solid color throughout. And by the way…get a haircut!
3:55 pm on December 14th, 2011
Hair!….Heh!heh!…I can tell that Dawson is feeling better…snort!
3:56 pm on December 14th, 2011
Repairs!…Hey Dawson! When are you going to repair The Fan?…sniff!…You can’t leave it the way it is, you know….grunt!
4:07 pm on December 14th, 2011
Snowhook can assist with goat duty if needed.
4:14 pm on December 14th, 2011
Staff: so sorry about your windows. We feel your pain. 🙂
The counter guys have left for the day. They had this idea that the Pack does nut live in the house. So when I said when will it be safe to let them in just incase they wanted to go counter surfing, the guys were flabbergasted! The Pack had to wait two hours to go into the kitchen.
After that Zoe who is very sick decided to get the mail and left the door open. Is lammed it in Katies’s face so only four got out. I started the Element and they piled in for a ride. Right into the puppy play yard where I shut the gate got them into the Stockade, opened the gate again and drove out.
And that was a restful day. Think I will now go downstairs because Zoe is now sweeping the rugs.
4:14 pm on December 14th, 2011
Hi Howlers – Haven’t heard from the kats lately. What’s up Buddy and T.C.?
Dawson’s feeling better (I can tell) But, get well Zoe!
Looks like the MGs are branching out into other ventures. Is there nothing and nowhere that is safe?
4:22 pm on December 14th, 2011
Melanie: Never a dull moment at your house!
4:27 pm on December 14th, 2011
Oh yeah…The Fan!
4:38 pm on December 14th, 2011
Four years ago when I retired as department chair we had a party. I told everyone that I didn’t want another party now that I’m retiring from teaching. Today near the end of class, the department members marched down the hall with a noise clanker and the chair was carrying a huge cake. They came in my classroom. The students knew about it. (And the cake was delish!)
4:46 pm on December 14th, 2011
Hip! Hip! Hoorah for B.Stover! The Easy Chair calleth!
4:46 pm on December 14th, 2011
Melanie: Can you had one of the Pack sit on Zoe? It is clear that she won’t sit still and rest otherwise.
4:49 pm on December 14th, 2011
Dawson,
I have hydrogen peroxide in my ear and it is bubbling away!
How long do I leave it in?
4:51 pm on December 14th, 2011
ZoE!!!
REST!!!!
fer cryin’ out loud
4:56 pm on December 14th, 2011
Staff!…heh!snort!heh!…Do I dare ask why you have hydrogen peroxide in your ears?….chorkel! …(hapless seems to fit afterall)….heh!
5:01 pm on December 14th, 2011
Hey!
Dawson told me she does it.
5:03 pm on December 14th, 2011
Love the music with this video….
5:07 pm on December 14th, 2011
Peroxide is very good for clearing out the ears.
Thank you all. The easy chair is where I’m sitting right now!
Dawson: I hope you enjoy Lighthouse. Let me know.
I agree that Staff should go elsewhere for the windows.
Hope Zoe starts feeling better.
5:09 pm on December 14th, 2011
Staff, I know what you mean. I have had two calls on my land line today, both between 1-2:00 when I was trying to take a nap! How do they know?
5:17 pm on December 14th, 2011
Uh Rio: Tell your friend the job just got double difficult. There are <b.twostones loose. I did a practice run at putting the #1 stone in place and it went right where it should first time. Heh. Then I tried to get it out so I could put the glue on and it fell into all those intricately curled pieces of paper. You can’t just put your big paws in the middle of all that. Came up with an recipe holder alligator clip that retrieved the stone nicely. I cannot figure out where the second stone came from.
Um Staff: I think that’s long enough. Heh, heh.
5:29 pm on December 14th, 2011
Dawson, How do you like Hillerman? Siber and Wookie have read all of it I think, and we have read almost all.
5:50 pm on December 14th, 2011
Love Hillerman. I’d read a book here and there but Staff has them in order so I think I am up to book 5 now. I’ll have to check our library once life slows down.
5:52 pm on December 14th, 2011
Oops!
I screwed up the bold up there.
6:01 pm on December 14th, 2011
AbbietheKitty,
I think that music is the theme to Star Wars, isn’t it?
6:05 pm on December 14th, 2011
B. – Congratulations, you did it!
6:16 pm on December 14th, 2011
Wolfie here. I finally got the ipaw away from mom. Grandma Jean is doing much better. At some point in time (August, we think) she had a massive heart attack. The doctors can’t make her any better but she is hanging in there.
I voted!
I am going to the party and those darn goats better keep their cotton picking hands off of our toffees!
6:31 pm on December 14th, 2011
Thanks Mollypop’s Mom.
Oh….toffee at the party!
6:42 pm on December 14th, 2011
Toffee AND Root Beer Floats at the party we can’t go to???
waaaahhhhhhh!!!!!
we’re connipping and saying “CARPOLA!”
7:04 pm on December 14th, 2011
Dawson, Correction, we did not read all the Hillermans, just all the Chee/Leaphorn ones. Really liked them alot.
7:18 pm on December 14th, 2011
T&G: I can’t say the same for the toffee but I will set aside some root beer floats fixings for you. I will also set aside the fixings for boot rear floats. 😉
7:24 pm on December 14th, 2011
Oh, and congratulations, Bee!
School’s out…….FOREVER!!!
7:27 pm on December 14th, 2011
LOL. Thank you Wolfie.
Snowhook Bella: I love root beer floats.
7:34 pm on December 14th, 2011
Boot rear floats! That’s a good one.
7:55 pm on December 14th, 2011
Well I want Root Beer Floats. Not those others. And toffee!!! Can I have some with White Chocolate on it. Dawson made some for me when she was here and it was AWSOME.
Congratulations on the retirement B. Quite an achievement. I hope you had fun when they surprised you!
I am going to training for my busy place tomorrow and Friday and I told the parents that I would make them a sack lunch. Then I told them it would be a Tuna sandwiche, a bag of Frito’s, a chocolate bar and an apple. That is what Mom used to make us years ago for school.
Unfortunately they didn’t know I was joking and so now I have to go shopping.
BBL
8:14 pm on December 14th, 2011
Congratulations to B and to all the students lucky enough to have been in her classes over the years!
Busy day here in Chevy Chase. Mom’s quilt guild had their annual Christmas party at our place today. 53 quilters — wow, what a group. Actually, I was only around at the very beginning and the end, as dad, grandpa and I had a “guys” afternoon out (mom was a bit concerned that meeting and greeting 52 quilters might tire me out too much). Anyways, us guys had a grand time out and about the town, and some of my favoritemost quilting ladies were stlll around when we got home. No goats though, thank dog. Wookie
8:34 pm on December 14th, 2011
Thank you Staff and Wookie. That does sound like a busy day Wookie.
I’m watching White Christmas which I haven’t seen in years.
8:41 pm on December 14th, 2011
Just a quick check in before we head off to the LAST SHOW OF THE SEASON…WOOO HOOOO!!!
As some of you know, CHarris actually made an even bigger donation to the MMCS than the wonderful birch bark vase of flowers. There were 10 more flowers in the package she sent to Melanie. However, the postage rates kind of put a damper on selling the flowers individually, so we came up with a solution that we hope will satisfy everyone.
We are taking the other 10 flowers to the Cultural Survival show. We will have them out for sale at the suggested price of $10 each and will have an explanation sign of how they were made, and that proceeds go to Snowhook Kennel.
We have been meaning to email CHarris about this – but time has kind of slipped away from us. Someone just mentioned that Christmas is next week!!! YIKES
Talk to you all later – and thanks for saving us some Root Beer Floats!
8:50 pm on December 14th, 2011
White!…..Stover we are watching White Christmas too!…snort!….So far this week we have seen the begging, the end and now we are seeing the middle where they are on the train to Vermont singing about Snow…heh!…Course she is also trying to make little paper objects for her mothers gift…sniff!….And that damned Beau is on the Easy!Down!Boy! chair with her again…grunt!…He is making up for lost time last night when she chased us both upstairs….wuffle!
8:50 pm on December 14th, 2011
Best of luck at the show this weekend! We are keeping all paws crossed that people are buying this last weekend before Christmas.
9:10 pm on December 14th, 2011
T&G: that sounds like a great solution.
B. Stover: Congrats. I can’t imagine what it will feel like when I leave the school for the last time in May.
Well…I made 4 doz cookies to take to school. I put them in a plastic baking dish and rubber-banded the lid in place. This morning I couldn’t find them. I thought I was losing my mind. Before I left for work, I checked the dog pen. There was the pan with 6 cookies in it. So the dogs waited until I went to bed, the tall one got the pan off the counter (I thought I heard a “thunk”) and drug it down the stairs, through the garage and out the arctic dog door, took off the rubber bands and proceed to feast. AND both dogs were hungry for breakfast when I got up this morning and fed them.
9:29 pm on December 14th, 2011
Wow! This may be the grand counter surfing achievement reported here on the blog.
Wait till Trigger hears this one.
9:39 pm on December 14th, 2011
OMD!
C.Harris: Now that is talent. Trigger’s valedictorian!
9:39 pm on December 14th, 2011
Siber-H: I had a dog years ago that didn’t even reach my knees. She stole something off the middle of my dining room table. I swear she had to nose a chair up to he table to climb up and get it. I had another dog that loved bread and I had to keep the bread in the microwave for protection. When I took a trip, I always had to clue in the house sitter not to leave bread out under any circumstances.
9:44 pm on December 14th, 2011
Dear CHarris,
Your dogs had help.
Our revenge continues
Signed,
The Marauding Goats
9:51 pm on December 14th, 2011
Fur Flowers!
I NEED those fur flowers! I got distracted last time but not now! I will send a hundred dollar donation to Snowhook (and by the way…every donation to Cultural Survival is matched right now) and send the money for postage to T&G or whosoever.
There is no need to take them to the show.
Rio: I got one stone in…I have no clue where the other one goes. Did your friend throw an extra one in just in case?
9:53 pm on December 14th, 2011
TMGs, Accomplished counter surfing canines do not need the help of stinky goats.
10:00 pm on December 14th, 2011
Dawson!…. She says if you are really sure you cannot see where the loose stone belongs then she thinks it must have been one that she lost while trying to glue it on….heh!…You can see how that would happen. It must have finally come out of its hiding place during its journey across the US and Canada…snort!
10:04 pm on December 14th, 2011
Charris!….I wonder why those guys of yours left the 6 cookies!….snort!…If I had been there there would not be any left-over cookies….sniff! ….I love cookies! Love, love love Cookies!…wheeze!….Someone here has not baked cookies in ages. She always used to do a huge bunch with Grandma right about this time of year….wuffle!….Her mother makes that kind of activity so dreadful that I bet she doesn’t even attempt to bake one lousy cookie….grunt!
10:12 pm on December 14th, 2011
You really are Awesome Dawson! That is wonderful.
Bee, Glad you had a great last day. Enjoy your retirement. Are you going to do any part time jobs?
T & G’s Mum, Good luck with the last show of the season. Sell out the house!
Going to hit the hay. Work tomorrow. Yuck!
10:13 pm on December 14th, 2011
Rio: Tell your friend I am even more impressed with her patience and ability after the gyrations I went through to get that one stone glued back on to it’s rightful place. I am grateful only the one came loose.
10:19 pm on December 14th, 2011
Bling!….Dawson, if you ever get your little cards in the mail…snort!….you will notice that not a lot of bling was used on very many of them….heh!
10:26 pm on December 14th, 2011
Vest!…..Hey, Dawson….sniff!… Since I found a buyer for your fishing vest will you be sending in an extra gift in Agent Molly’s name?….snort!…. I can’t remember if the Vest was one of the Items that had that clause in it….wuffle!
10:29 pm on December 14th, 2011
Work!
They just called to make sure I was coming in tomorrow. Yes, I am. They gave me the scoop on the mix-up this morning. The one I asked to pass the message on says she did. My boss says she did not. Hmmmmm.
10:33 pm on December 14th, 2011
Rio: I had to send a check to Snowhook earlier so I made that matching gift for the vest at the same time in the hopes that it would sell. Thanks for
getting ridselling that vest for me.
10:49 pm on December 14th, 2011
It is the Hour of the Wolf
Night All
11:09 pm on December 14th, 2011
Heh, I remember Dawson moaning about missing the flowers when she was here. We just had a strange incident with AbbietheKitty.
I was sitting upstairs reading a Midnight Louie book and I heard this strange dragging sound from the bedroom. It seemed to be coming towards the door so I waited. Finally I saw Abbie’s hind quarters stagger slowly into view. She was moving very quietly but still I could hear something being dragged over the rug. I wondered what she could be dragging. She struggled into view with a cloth shopping bag wrapped tightly around her chest. It was wound so tightly that I couldn’t tell which way to unwind it. After I unwound it about 10 times it came off.
She claims some goats got into the bedroom and pushed her into the bag and when she tried to get out of it they swung her around and tightened it up.
Hmmmmm.
I wonder…..
Good thing I sent that vest off first thing yesterday.
12:57 am on December 15th, 2011
Congrats, B! We are certainly jealous here and wish Mom could retire. Instead she and Dad have to get this new house paid off ASAP so work, work, work is in the future (unless the lottery gods decide to bless us with their winnings!)
TMG have been very busy lately with their dastardly deeds. It’s a good thing we have quite a few Sibes on guard duty around the blog. What is they got to Book#5????
LOVE White Christmas and most of the Christmas movies you guys have been mentioning. Have to check out the ones we’ve never heard of before.
Pearl is so beautiful!!
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Asia Supermarket Birdseye View
Posted on: 29-09-2017 @ 14:42
QMAC have recently been awarded the following projects:
All projects are on site and primarily in the early stages of development. The total value of this recently appointed work exceeds £12m.
Posted on: 29-09-2017 @ 14:42
Posted on: 18-08-2017 @ 11:21
Posted on: 26-05-2017 @ 11:33
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TITLE: Problem of Liouville's theorem
QUESTION [6 upvotes]: Here is my problem:
Let $f(z)$ be an entire function such that $|f '(z)| < |f(z)|$ for all $z \in \mathbb{C}$,
Show that there exists a constant A such that $|f(z)| < A*e^{|z|}$ for all $z \in \mathbb{C}$.
I am trying to use Liouville's theorem to prove and try to set $g(z)=|f '(z)|/|f(z)| < 1$ and then g(z) is constant. I am not sure if my thinking is right and how to prove this problem?
Thanks
REPLY [6 votes]: Consider $g(z) = \frac{f'(z)}{f(z)}$. Since $|f'(z)| \lt |f(z)|$ we see that $f$ has no zeros, so $g$ is entire. Thus, $g$ is an entire function satisfying $|g(z)| \lt 1$, so by Liouville it is constant. This means that $f'(z) = C f(z)$ for some $C \lt 1$.
Since $f$ is entire we can write $f(z) = \sum_{n=0}^{\infty} a_n z^n$ for all $z \in \mathbb{C}$ and the equation $f'(z) = C f(z)$ leads to
$$f'(z) = \sum_{n=0}^{\infty} (n +1) a_{n+1} z^{n} = C \sum_{n=0}^{\infty} a_n z^n.$$
By comparing coefficients we see that $a_1 = C a_0$, $2 a_2 = C a_1$, $3a_3 = Ca_2$, etc, so that $a_n = \frac{C^{n}}{n!} a_0$. In other words, $f(z) = \sum_{n=0}^{\infty} \frac{a_{0}}{n!} (Cz)^n = a_{0} e^{Cz}$.
Now let us prove the estimate. Since $C \lt 1$ we have for all $z \in \mathbb{C}$ that $$|f(z)| \leq \sum_{n=0}^{\infty} \frac{|a_0|}{n!}\,|Cz|^n \leq \sum_{n=0}^{\infty} \frac{|a_0|}{n!} |z|^n = |a_0| e^{|z|},$$
almost as we wanted. By choosing $A \gt |a_0|$ we get the desired strict inequality $|f(z)| \lt A e^{|z|}$.
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Photochemical Sciences Ph.D. Dissertations
Title
Probing the Photochemistry of Rhodopsin Through Population Dynamics Simulations
Date of Award
2019
Document Type
Dissertation
Degree Name
Doctor of Philosophy (Ph.D.)
Department
Photochemical Sciences
First Advisor
Massimo Olivucci (Advisor)
Second Advisor
Andrew Gregory (Other)
Third Advisor
Hong Lu (Committee Member)
Fourth Advisor
Alexey Zayak (Committee Member)
Abstract
The primary event in vision is induced by the ultrafast photoisomerization of rhodopsin, the dim-light visual pigment of vertebrates. While spectroscopic and theoretical studies have identified certain vibrationally coherent atomic motions to promote the rhodopsin photoisomerization, how exactly and to what degree such coherence is biologically related with its isomerizing efficiency (i.e. the photoisomerization quantum yield) remains unknown. In fact, in the past, the computational cost limited the simulation of the rhodopsin photoisomerization dynamics, which could be carried out only for a single molecule or a small set of molecules, therefore lacking the necessary statistical description of a molecular population motion.
In this Dissertation I apply a hybrid quantum mechanics/molecular mechanics (QM/MM) models of bovine rhodopsin, the verterbrate visual pigment, to tackle the basic issues mentioned above. Accordingly, my work has been developing along three different lines comprising the development, testing and application of new tools for population dynamics simulation: (I) Development of a suitable protocol to investigate the excited state population dynamics of rhodopsins at room temperature. (II) A correlation between the phase of a hydrogen-out-of-plane (HOOP) motion at the decay point and the outcome of the rhodopsin photoisomerization. (III) A population “splitting” mechanism adopted by the protein to maximize its quantum yield and, therefore, light sensitivity.
In conclusion, my Dissertation reports, for the first time, a connection between the initial coherent motion of a population of rhodopsin molecules and the quantum efficiency of their isomerization. The photoisomerization efficiency is ultimately determined by the way in which the degree of coherence of the excited state population motion is modulated by the protein sequence and conformation.
Recommended Citation
Yang, Xuchun, "Probing the Photochemistry of Rhodopsin Through Population Dynamics Simulations" (2019). Photochemical Sciences Ph.D. Dissertations. 113.
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January 1: I got a lot of things on my mind
I’m looking at my body through a new spy satellite
I try to lift a finger but I don’t think I can make a call
So tell me if I move, ’cause I don’t feel anything at
… it won’t work
Dead Man (Carry Me), Jars of Clay
Archive for January, 2010
Posted in SHM on Friday 1st January, 2010| 1 Comment »
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The National Oceanic and Atmospheric Administration and U.S. Fish and Wildlife Service issued a final rule last week and FWS share jurisdiction for loggerhead sea turtles listed under the ESA.
"The change will help increase protection of the loggerhead and frees up resources to be used where its needed most," explained Sanibel-Captiva Conservation Foundation biologist Amanda Bryant. "It will allow researchers and conservationists to target certain areas where loggerheads are in most danger."
Special to the BREEZE / courtesy of SCCF volunteer Julie ReedLoggerhead sea turtle. (These photos were taken from a safe distance)
On March 6, 2010, the two agencies proposed to list seven distinct population segments of loggerheads, also known as DPSs, as endangered and two as threatened. In the final rule issued Sept. 16, five were listed as endangered and four as threatened.
"We fall into a population of loggerheads that will remain listed as threatened," said Bryant, who followed the ruling to see if it would effect conservation on the islands..
Loggerhead Sea Turtle (Caretta caretta)
Although they are Florida's most commonly observed sea turtle, loggerheads are rare throughout most of their global range. They are found in marine waters from warm-temperate seas through the sub-tropics. Loggerheads are named for their overly proportioned head.
Nesting/hatching season: May through October
Adult shell length: 31 to 43 inches
Adult weight: 155 to 375 pounds
Age at maturity: 30 to 35 years
Local status: threatened
Diet: Loggerheads eat a wide variety of animals and are one of the few predators of large hard-shelled invertebrates. Examples of food items include jelly animals, copepods, sea slugs, hydroids for post hatchling; mollusks, crabs and sea pens for juveniles to adults.
In addition, some of the fisheries bycatch effects appear to have been resolved through requirement of turtle excluder devices in shrimp trawlers, and longline fishery effort has declined due to fish stock decreases and economic reasons.
"Both agencies agreed that loggerhead sea turtle conservation benefits from an approach that recognizes regionally varying threats," said Cindy Dohner, FWS southeast regional director. "Today's listing of separate distinct population segments will help us better assess, monitor and address threats, and evaluate conservation successes on a regional scale."iana Ocean, North Pacific Ocean and South Pacific Ocean - and two others were listed as threatened - South Atlantic Ocean and Southwest Indian Ocean.
"I think they made a sound decision," Bryant said about the status change. "They looked at a lot of data and I hope it helps."."
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TITLE: Find error in abstract algebra proof
QUESTION [0 upvotes]: I suspect that the proof below is flawed. I did not use the hypothesis "$\ker(h) \subseteq \ker(k)$" when proving sufficiency.
Lemma. $ $ Let $G$, $H$, $K$ be groups, let $h : G \to H$ and $k : G \to K$ be homomorphisms, and suppose that $h$ is surjective. In order for there to exist a homomorphism $f : H \to K$ such that $f \circ h = k$, it is necessary and sufficient that $\ker(h) \subseteq \ker(k)$. The homomorphism $f$ is then unique.
Necessity.$\ $ If there is a homomorphism $f$ such that $f \circ h = k$, and if $h(x) = e$, then $$k(x) = f(h(x)) = f(e) = e.$$
Sufficiency.$\ $ There exists a section $s : H \to G$ of $h$, so that $h \circ s = I_H$. Then $h \mid s(H) = s^{-1}$, so $s$ is a homomorphism. Choose $f = k \circ s$.
Uniqueness.$\ $ If $f, g : H \to K$ are homomorphisms such that $f \circ h = k = g \circ h$, then $f = k \circ s = g$.
REPLY [2 votes]: There is an error when you say that $h$ has a section. Consider the surjection $\Bbb{Z} \to \Bbb{Z}/2$. It doesn't have a section since there are no nontrivial maps $\Bbb{Z}/2 \to \Bbb{Z}$.
Now, if you want to prove it, just realise that $H\cong G/ker(h)$ and $h$ is the quotient map. Then use the universal property of the quotient group.
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TITLE: Balls and Bins (Find no of balls)
QUESTION [1 upvotes]: How many balls do you need to throw randomly into n bins so that the probability that none of the bins is empty is at least 1/100?
I am trying to solve this question by first finding the expected no of balls to be thrown so that every bin has at least one ball which comes out to be the following (similar to coupon-collector problem)
$$
n*\sum_{i = 1}^{n} \frac{1}{n} = n * H_{n}
$$
Now, I am trying to use Markov's inequality :
$$
Pr[X\geqslant a] \leqslant \frac{E[X]}{a}
$$
But, I am stuck here as I am unable to connect it with the question.
Please help.
REPLY [1 votes]: Lemma:
Let $X$ be the number of empty bins when $m$ balls are thrown uniformly and independently at random in $n$ bins. Then $$\mathbb{E}[X] = n\left( 1 - \frac{1}{n}\right)^m \leq n e^{-m/n}.$$
Proof:
Let $X_i$ be the random variable denoting whether bin $i$ is empty or not (i.e., $X_i = 1$ if bin $i$ is empty and $X_i = 0$ if bin $i$ contains at least one ball).
Observe that $X = \sum_{i=1}^n X_i$, and that all $X_i$'s are independent. The probability that bin $i$ receives a ball is $\frac{1}{n}$, hence the probability that it remains empty after tossing one ball is $Pr(X_i = 1) = 1 - \frac{1}{n}$. Since the throws are supposed independent, after $m$ throws, we have $$Pr(X_i = 1) = \left( 1 - \frac{1}{n}\right)^m.$$
Since $X$ is a sum of independent Bernouilli random variables, we directly obtain $$\mathbb{E}[X] = n \left( 1 - \frac{1}{n}\right)^m \leq n e^{-m/n}.$$
Now, letting $m \geq n \ln(n \cdot 100)$, we have $$\mathbb{E}[X] \leq n e^{-n \ln(100n)/n} = \frac{n}{100n} = \frac{1}{100}.$$
Concluding with Markov's inequality that tells us $Pr(X \geq 1) \leq \mathbb{E}[X]$, we obtain $$Pr(X = 0) \geq 1 - \frac{1}{100}.$$
In plain text, this means that the probability to have no empty bin is greater that $1 - \frac{1}{100}$, or that the probability to have at least an empty bin is smaller that $\frac{1}{100}$.
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