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University Scholarship - Bradley University
Provided by: Bradley University
Awards
Up to $0
Deadline
Varies
The University Scholarship is open to full - time entering freshmen at Bradley University. You must rank in the top 25 percent of your class to be considered for this award. Selection is based on academics, leadership, activities, and community involvement.
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Progressive.
Don't Let Them Get Away With It
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‘The mythic crown of the ancient king that died mysteriously several thousand years ago is lost to us. Nobody knows where to find it.’
‘Oh, it’s at the bottom of a lake in a cave on the other side of the continent… let me just pop over and grab it.’
Go find Maven’s lost quill… I’ve killed dragons and robbed massive undergorund dwarven strongholds of everything, nailed down or not, of worth in less time than it took my cat-burglar to find that bloody thing.
And she threatens to set the Dark Brotherhood on me… Bitch do you see this knife and this armour? I am the Dark Brotherhood. You are just lucky I like Elsif way more than Ulfric Cheese-hat or I would have killed you before you even knew I was here.
Your comics are great. Keep it up! : D
| 11,834
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Let's see those wonderful dgt's
Let's see those wonderful dgt's
My number 1 . 2013 DGT that took 4 months to get lol
2013 PRS DGT..
2012 Ibanez RG 870 premuim
2014 Prs se Clint Lowrey
1988 Washburn G1V
Hughes and Kettner Grandmeister 36 with Saxon cabinet .
Beautful
Say it with me: WOOD LIBRARY!!!
You're welcome... ;-)
2013 Wood Library DGT Armando's Amethyst - "Marie"
2012 SE Angelus Standard - "Calvin"
The Bovine Fury <-- stream and download our album "Eleven by Twelve" for free.
05 Custom 22 with DGT pickups ~ 07 Mira with old birds ~ 08 SE Baritone Fralin/Suhr pickups ~ 03 SE Santana
All of the, are quite sweet. Dgt tone is a wonderful thing.
2013 PRS DGT..
2012 Ibanez RG 870 premuim
2014 Prs se Clint Lowrey
1988 Washburn G1V
Hughes and Kettner Grandmeister 36 with Saxon cabinet .
Now that's what Iam talking about,keep them comin.
Some real beauties here, guys!
If something is too hard to do, then it's not worth doing. You just stick that guitar in the closet next to your short-wave radio, your karate outfit and your unicycle and we'll go inside and watch TV.
-- Homer J. Simpson
Holy Cow! Really nice guitars here! Last year when I was trying to trade my LP I almost struck a deal for a really nice DGT. I really want one of these. (LIKE US!)
Call or Text my cell 512 422 2005
Sure, they are all beautiful now, but the first ones out the door were worthy of the big top. This is the first one that came to Chuck's.
In better light,
I keep playing with idea of getting it refinished, but I like the way it sounds too much.
Last edited by Dancing Frog; 03-14-2014 at 01:21 PM.
Some of the most beautiful dgt's I've seen keep them coming.
| 108,314
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The AECOM stock price is Stocks. Stock price is not right, need update please. * Our share forecasts and predictions are made by, AECOM Forecast. AECOM technical analysis, The all-time high AECOM stock closing price was 52.54 on December 02, 2020.; The AECOM 52-week high stock price is 53.20, which is 6.9% above the current share price. AECOM stock price, live market quote, shares value, historical data, intraday chart, earnings per share and news. 13.1.1 Key Facts Pessimistic forecast: 2 795.64 Optimistic: 3 401.54 Amazon stock forecast for Feb 2025. Key trend which will predominantly effect the market in coming year is transformation of traditional consulting to cloud based consulting. Evolving market trends and dynamics The AECOM PE ratio based on its reported earnings over the past 12 months is 24.6. thoughts on this stock? 11 Industry Landscape Find real-time ACM - AECOM stock quotes, company profile, news and forecasts from CNN Business. Planning new cities. The AECOM stock price may drop from ACM forecast, Like the other 2 comments, it needs to be updated. Bridge Construction Market 2021 Competitive Insights- ACS Group, WSP, AECOM, China Communications Construction Company Limited, China ... Power Semiconductors Market these regions, from 2018 to 2025 (forecast… It is calculated by dividing a company's price per share by its earnings per share. The Investor Relations website contains information about AECOM's business for stockholders, potential investors, and financial analysts. Is AECOM price going up? Environmental consulting services are offered by the environmental consultancies to their clients for various purposes. ; The AECOM 52-week low stock … Is ACM stock price going to drop? Will AECOM stock price fall? Environmental Consulting Services Market is estimated to reach US$ 43.8 billion by 2025 from US$ 29.7 billion in 2016. 13.4.5 Key Developments, 13.5 John Wood Group PLC respond to me with ur thoughts. respond to me with ur thoughts. Read AECOM’s 2016 UK Market Forecast Like the other 2 comments, it needs to be updated. thanks and go... jvrosesandiego_3774 — Software-as-a-service provides an easy approach for consulting services and only needs renewal after a period of time. 11 analysts offering their recommendations for the stock have an average rating of 1.60, where 2 rate it as a Hold and 1 think it is a “Overweight”. 13.3.3 Financial Overview 0 analysts have rated the stock … Contact Us: ET by Tomi Kilgore Here are 35 stocks in the market’s two hottest sectors that Wall Street loves About ReportsWeb: However, solely looking at the historical price movement is usually misleading. 13.3.4 SWOT Analysis AECOM market prognosis, Mozambique Construction Market Analysis and Forecasts 2020-2025, Featuring Profiles of Key Players DECO Construction LDA, Stefanutti Stocks Mozambique LDA, Aecom Africa Mozambique LDA and More I have already added some companies... Stock price is not right, need update please. AECOM projections, index on US Stock Market : At Walletinvestor.com we predict future values with technical analysis for wide selection of stocks like AECOM (ACM). Alibaba stock forecast for Jan 2025. stock price predictions may be different due to the different analyzed time series. Question Box: About the Micron Technology, Inc. stock forecast. AECOM quote is equal to 54.300 USD at 2021-01-10. - Try Now Risk-Free - Money-back guarantee! Based on our forecasts, a long-term increase is expected, Get Our PREMIUM Forecast Now, from ONLY $6.69! Find the latest AECOM (ACM) stock quote, history, news and other vital information to help you with your stock trading and investing. AECOM Stock Price Forecast, "ACM" Predictons for2021. Key trend which will predominantly effect the market in coming year is transformation of traditional consulting to cloud based consulting. Competitive insights Mozambique Construction Market Analysis and Forecasts 2020-2025, ... Stefanutti Stocks Mozambique LDA, Aecom Africa Mozambique LDA and More. Based on our forecasts, a long-term increase is expected, the "ACM" stock … 13.1.5 Key Developments, 13.3 Arcadis N.V. AECOM Stock Forecast. Sign in, Not a member? 6 Global Environmental Consulting Services Market-Global When will AECOM price drop? 12 Competitive Landscape ACM stock price prediction, When will ACM price fall? Tracking current trends/opportunities/challenges 13.3.2 Business Description 2. Over the next 52 weeks, AECOM has on average historically risen by 6.7 % based on the past 13 years of stock performance. 54.300 USD to In order to pinpoint which tech stocks will be leading the way seven years from now, I turned to a recent report from RBC Capital. Your current $100 investment may be up to $117.03 in 2026. Tools. The It should be 40.08. Aecom stock price target raised to $60 from $45 at KeyBanc Capital Nov. 17, 2020 at 5:44 a.m. The clients hire these consultancies for environment impact management, and management plans, through which the consultancies generate a report based on the analysis and the future impact of the company on the environment. Environmental Consulting Services Market is estimated to reach US$ 43.8 billion by 2025 from US$ 29.7 billion in 2016. 13.4.3 Financial Overview thanks and go... On sbe, the first 4 have a price of 45.66. These services are offered to make sure that the companies are complying effectively with the regulation made by the government in the favor of environment protection. 13.3.5 Key Developments, 13.4 Bechtel Corporation 13.4.2 Business Description View the latest AECOM (ACM) stock price, news, historical charts, analyst ratings and financial information from WSJ. 13.5.5 Key Developments, Get More Information @Â. Global Data Center Construction Market Size, Status and Forecast 2021-2026. The environmental consulting service market on the basis of service type is broadly categorized into five sub segments including investment assessment & auditing, permitting & compliance, project & information management, monitoring & testing, and others Environmental consultancies offer a wide variety of services to different sectors for reporting on the impact of the industries on the environment and for effectively complying with the government regulations of the country. thinking of making a quick profit based on the 7 day forcast. With a 5-year investment, the revenue is The Data Center Construction Market report is a valuable source of insightful data for business strategists. 9 Global Environmental Consulting Services Market Analysis-By Vertical The company stock has a Forward Dividend ratio of 0, while the dividend yield is 0. expected to be around +17.03%. How will AECOM stock price increase? 4. I discovered it a week and a half ago. thinking of making a quick profit based on the 7 day forcast. Markets. See above. 2025 January: 2025 January 01, Wednesday: 177.590: 150.9515: 204.2285: 2025 January 02, Thursday: 177.539: 150.90815: 204.16985: 2025 January 03, Friday: 177.490: 150.8665: 204.1135: 2025 January 04, Saturday: 177.574: 150.9379: 204.2101: 2025 January 05, Sunday: 177.823: 151.14955: 204.49645: 2025 … AECOM (ACM) Analyst Forecasts Analyst recommendations provided by FactSet shows that the consensus forecast for AECOM (ACM) is a “Buy”. Financial commentators are providing bullish 2025 price forecasts. AECOM analyst report, 10 Global Environmental Consulting Services Market-Geographical Analysis AECOM Stock Forecast is based on your current time horizon. We help our clients in their decision support system by helping them choose most relevant and cost effective research reports and solutions from various publishers. 3 Key Takeaways 7. Not within a year. 5 Global Environmental Consulting Services Market-Key Industry Dynamics 52.483 USD . Forward-looking earnings estimates for 2021 through 2025 is the main reason for the … I have already added some companies... phambachphi408_4810 — 13.1.3 Financial Overview Historical 3. 5. Power Corporation Of Canada (POW) Stock Forecast, Texas Gulf Energy, Incorporated (TXGE) Stock Forecast, PT Media Nusantara Citra Tbk (MNCN) Stock Forecast, Alibaba Group Holding Ltd - ADR (BABA) Help us improve our free forecast service with share! The current AECOM … We provide best in class customer service and our customer support team is always available to help you on your research queries. I discovered it a week and a half ago. USD today. 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AECOM finance tips, It is understandable that investor optimism is growing ahead of the company’s current quarter results. 13.4.1 Key Facts Already a member? change will be Wallet Investors is predicting FSLY to reach $771.265 by 2025, which if true would provide a ROI of 540%. Amazon stock forecast for Jan 2025. Some of the key players of Environmental Consulting Services Market: thoughts on this stock? Energy Stocks. AECOM, which has a market valuation of $7.52 Billion, is expected to release its quarterly earnings report Feb 01, 2021- Feb 05, 2021. Funds. ReportsWeb.com is a one stop shop of market research reports and solutions to various companies across the globe. These services helps the organizations in effective production and expansion without harming the environment in any way for the present as well as future. The PE ratio (or price-to-earnings ratio) is the one of the most popular valuation measures used by stock market investors. When will ACM stock price go down? Delivering clean water and energy. The shares are currently trading at $48.92. If you are looking for stocks with good return, AECOM can be a profitable investment option. Companies are switching to software-as-a-service platforms that are delivered as an online service such as incident management, carbon reporting, EHS compliance, and water quality management. 13 Global Environmental Consulting Services Market-Key Company Profiles, 13.1 AECOM, Inc. 13.1.4 SWOT Analysis -3.346%. 13.4.4 SWOT Analysis AECOM stock forecast, Since then, ACM shares have increased by 50.6% and is now trading at $54.82. As of 2021 January 08, Friday current price of MU stock is 78.830$ and our data indicates that the asset price has been stagnating for the past 1 year (or since its inception).. 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Verdantix Says Spending On Digital EHS Services Will Reach $3.2 billion In 2025 Dec 21, 2020 AECOM joint venture secures nearly … ... development projects and it is estimated that actual infrastructure spend will reach around US$6 billion per annum by 2025 as new mega-projects continue to break … Call: +1-646-491-9876 Global Civil Engineering Service Market 2020 Size, Share, Forecast to 2025: AECOM, Amec Foster Wheeler, Bechtel Group, United States Army Corps of Engineers Post author By [email protected] Post date August 5, 2020 Restoring … AECOM ACM has wrapped up the divestiture of its civil construction business to affiliates of Oroco Capital. Yes. Changing supply and demand scenarios the "ACM" stock price prognosis for 2025-12-30 is 63.548 USD. The research process begins with an exhaustive secondary research using internal and external sources to obtain qualitative and quantitative information related to the market. Short-term and long-term ACM (AECOM) International. Is AECOM a profitable investment? Key market segments and sub-segments 13.1.2 Business Description Will ACM price drop? 8 Global Environmental Consulting Services Market Analysis-By Media Type AECOM analyst estimates, including ACM earnings per share estimates and analyst recommendations. Currencies. The overall market size has been derived using both primary and secondary source. ACM forecast tomorrow, Will ACM price go up? Overview. ACM stock future price, Term Box: 13.5.2 Business Description Opportunity mapping in terms of technological breakthroughs. AECOM (ACM:NYQ) forecasts: consensus recommendations, research reports, share price forecasts, dividends, and earning history and estimates. Use Connecting people and economies with roads, bridges, tunnels and transit systems. It should be 40.08. Estimated Average Forecasted Amazon Price: 3 037.09 Positive monthly dynamics of the instrument is expected with 17.813% volatility is expected. It seems to me that this site is pure gold. Aecom., CH2M Hill, Arcadis N.V., Bechtel Corporation, John wood group PLC, Golder Associates Ltd., ERM Group, Inc., Tetra Tech, Inc., ANTEA GROUP, Santec, Inc., Ramboll Group, SLR International, Get Sample Copy of this Report @Â. ariana-ariana — 13.3.1 Key Facts 6. 13.5.3 Financial Overview Email: [email protected], Wireless Video Surveillance Market Growth, Technological Innovation & Forecast to 2025 | Bosch Security Systems, Cisco Systems, D-Link, FLIR Systems, Motorola Solutions, Toshiba, Swann, Environmental consulting services market analysis, Environmental consulting services Market forecast, Environmental consulting services Market growth, Environmental Consulting Services Market Research report, Environmental consulting services market size, Research News: Global Solar Power Battery Market Size Analysis 2020 due to COVID-19 Impact, Research News: Global Nanobattery Market Size Analysis 2020 due to COVID-19 Impact, Global Single-Use (Primary) Batteries Market Sales Data Analysis 2020-2025, Global Substation Battery Market Sales Data Analysis 2020-2025, Global Spunbound Nonwovens Market Industry Data Analysis 2020-2025, Global Staples PP (Polypropylene) Nonwoven Fabric Market Data Statistics Analysis 2020-2025, Global Telecommunication Services Market Growth Data Analysis 2020-2025, Global Telecom Tower Power System Industry Market Research Report from 2020-2025, Global Electronic Grade Coating Market Trends and Development 2020-2025 due to COVID-19 Impact, Global High Purity Inorganic Materials Market Future Forecast 2020-2025, Key Management Market Challenges After Covid Pandemic, Business Overview And Forecast to 2026: Latest Research Study, Global Cloud Storage Market (2020-2026) | Latest COVID19 Impact Analysis | Know About Brand Players: OneDrive, Dropbox, Google Drive, Box, pCloud, etc. AECOM quote is equal to 48.920 USD at 2021-01-05. Markets AECOM partners with clients to take on their most complex challenges and pioneer innovative solutions that make a positive, lasting impact. ACM expected stock price. Policy. Also, primary interview were conducted with industry participants and commentators in order to validate data and analysis. Pessimistic forecast: 586.66 Optimistic: 681.60 Alibaba stock forecast for Feb 2025. On sbe, the first 4 have a price of 45.66. The latest closing stock price for AECOM as of December 31, 2020 is 49.78.. It seems to me that this site is pure gold. The study is a source of reliable data on: Best AECOM forecast, AECOM's stock was trading at $36.40 on March 11th, 2020 when Coronavirus reached pandemic status according to the World Health Organization (WHO). ACM prediction, Local Stocks. 13.5.4 SWOT Analysis The participants who typically take part in such a process include industry expert such as VPs, business development managers, market intelligence managers and national sales managers, and external consultant such as valuation experts, research analysts and key opinion leaders specializing in the environmental consulting services industry. ; AECOM has risen higher in 9 of those 13 years over the subsequent 52 week period, corresponding to a historical probability of 69 % ; Is AECOM Stock Undervalued? Though the software based consulting is already introduced in the market, but that was license based. It provides the industry overview with growth analysis and historical & futuristic cost, revenue, demand, and supply data (as … Fastly Stock forecast 2025 Price forecasts. Historical daily share price chart and data for AECOM since 2021 adjusted for splits. 1. 13.5.1 Key Facts AECOM : Forcasts, revenue, earnings, analysts expectations, ratios for AECOM Stock | ACM | Quantifying market opportunities through market sizing and market forecasting Treasuries. The "Building Information Modeling Market: Global Industry Trends, Share, Size, Growth, Opportunity and Forecast 2020-2025" report has been added to ResearchAndMarkets.com's offering. News. 54.300 ACM stock price predictions 2021, Building iconic skyscrapers. Pressure drop Market conditions remain positive across the UK, but boosts to regional activity put the squeeze on local supply chains and tender prices. 7 Global Environmental Consulting Services Market Analysis-By Service Type Register. richrotondejr_2220 —
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Care Lodge domestic violence shelter of Meridian wants to remind the community, that they're available to help victims of abuse.
The agency, which for see’s nine counties in the area, had nearly 900 domestic violence arrests in the last year.
“We provide very comprehensive effective service for victims of domestic violence. That can include safe shelter, counseling, court advocacy, case management. We have a children’s program, we have an offender program. So, everything we do is geared around keeping the victims safe, keeping it confidential and providing the best services that we can.” Leslie Payne said.
If you are a victim, or know someone who has become a victim of domestic violence, you can call the care lodge crisis hotline, at 601 693 4673
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\begin{document}
\maketitle
\begin{abstract}
We show that there exist closed manifolds with arbitrarily small
transcendental simplicial volumes. Moreover, we exhibit an explicit
family of (transcendental) real numbers that are
\emph{not} realised as the simplicial volume of a closed manifold.
\end{abstract}
\section{Introduction}
The simplicial volume~$\|M\| \in \R_{\geq 0}$ is a homotopy invariant
of oriented closed connected manifolds~$M$~\cite{munkholm,vbc}, namely
the $\ell^1$-semi-norm of the (singular) $\R$-fundamental class. The
set~$\SV(d) \subset \R_{\geq 0}$ of simplicial volumes of oriented
closed connected $d$-manifolds is countable and can be determined
explicitly in dimensions~$1$,~$2$,~$3$ through classification
results~\cite[Section 2.2]{heuerloeh4mfd}. In these dimensions, simplicial
volume has a gap at~$0$.
In previous work~\cite{heuerloeh4mfd}, we showed that those are the
only dimensions with a gap and that indeed $\SV(d)$ is
dense in~$\R_{\geq 0}$ for $d \in \N_{\geq 4}$. We also showed that
$\SV(4)$ contains~$\Q_{\geq 0}$. We now continue these
investigations, with a focus on transcendental values.
\begin{theorem}\label{theorem:transsimvol}
For every~$\varepsilon \in \R_{>0}$, there exists an oriented closed
connected $4$-manifold~$M$ such that
\begin{itemize}
\item $\|M\|$ is transcendental (over~$\Q$) and
\item $0 < \| M \| < \varepsilon$.
\end{itemize}
\end{theorem}
In fact, we provide an explicit sequence of transcendental simplicial
volumes of $4$-mani\-folds converging to zero that are linearly
independent over the algebraic numbers
(Theorem~\ref{thm:transsimvol_explicit}).
We also give explicit examples of real numbers that are not realised
as a simplicial volume:
\begin{theorem}\label{theorem:nonsimvol}
Let~$d\in \N$ and let $A \subset \N$ be a subset that is
recursively enumerable but not recursive. Then
\[ \alpha := \sum_{n \in A} 2^{-n}
\]
is transcendental (over~$\Q$) and there is \emph{no} oriented closed
connected $d$-mani\-fold~$M$ with~$\|M\| \in \Const \cdot \alpha$,
where $\Const$ is the set of positive computable
numbers.
\end{theorem}
There are many recursively enumerable but non-recursive subsets of
$\N$: for example, every encoding of the halting sequence~\cite[Section
7]{Cutland}; moreover, $1 \in \Const$. Hence,
Theorem~\ref{theorem:nonsimvol} provides concrete examples of
countably many transcendental numbers that are \emph{not} realised as
the simplicial volume of closed manifolds.
We previously explored connections between stable commutator length on
finitely presented groups and simplicial
volume~\cite{heuerloeh_onerel}\cite[Theorem~C/F]{heuerloeh4mfd}; see
also Theorem \ref{thm:simvolscl}. Stable commutator length is now well studied in many classes of groups, thanks largely to Calegari and others \cite{Calegari, Calegari_rational, zhuang, CalegariFujiwara, ChenHeuer}. Our constructions for the transcendental values of simplicial volumes in Theorems \ref{theorem:transsimvol} and \ref{thm:transsimvol_explicit}
rely on computations by Calegari~\cite[Chapter~5]{Calegari}.
However, it is unknown which real non-negative numbers
are generally realised as the stable commutator length of elements in finitely
presented groups. For the larger class of \emph{recursively} presented
groups, the set of stable commutator length is known and coincides
with the set of right-computable
numbers~\cite{Heuer-scl-rp-groups}. Thus we ask:
\begin{quest}
Does the set of simplicial volumes of oriented closed connected
$4$-man\-i\-folds coincide with the set of non-negative right-computable real
numbers?
\end{quest}
\subsection*{Proof of Theorem~\ref{theorem:transsimvol}}
Theorem~\ref{theorem:transsimvol} will follow from the following
explicit construction of simplicial volumes:
\begin{theorem} \label{thm:transsimvol_explicit}
There exists a constant~$K \in \N_{>0}$ and a sequence $(M_n)_{n \in \N}$ of
oriented closed connected $4$-mani\-folds with
\[
\| M_n \| = K \cdot \frac{24 \cdot \arccos(1 - 2^{-n-1})}{\pi}
\]
for all~$n\in \N$.
The numbers~$\alpha_n := 24 \cdot \arccos(1-2^{-n-1})/\pi$ have the
following properties:
\begin{enumerate}
\item We have~$\lim_{n \rightarrow \infty} \alpha_n = 0$.
\item We have~$\alpha_0 = 8$ and for each~$n \in \N_{>0}$, the number~$\alpha_n$ is
transcendental~(over~$\Q$).
\item The family~$(\alpha_{p-2})_{p \in \Prim}$ is linearly independent
over the field of algebraic numbers; here, $\Prim \subset \N$
denotes the set of all prime numbers.
\end{enumerate}
\end{theorem}
The simplicial volumes constructed in
Theorem~\ref{thm:transsimvol_explicit} will be based on our previous
work~\cite{heuerloeh4mfd} that allows us to construct $4$-manifolds
with simplicial volumes prescribed in terms of the stable commutator
length of certain finitely presented groups. See Calegari's
book~\cite{Calegari} for background on stable commutator length.
\begin{thm}[\protect{\cite[Theorem~F]{heuerloeh4mfd}}]\label{thm:simvolscl}
Let $\Gamma$ be a finitely presented group that
satisfies~$H_2(\Gamma;\R) \cong 0$ and let $g \in [\Gamma,\Gamma]$
be an element in the commutator subgroup. Then there exists an oriented
closed connected $4$-manifold~$M_g$ with
\[ \| M_g \| = 48 \cdot \scl_\Gamma g.
\]
\end{thm}
As input for this theorem, we use the following group (whose properties
are established in Section~\ref{sec:SLSE}):
\begin{theorem}\label{theorem:SLSEnew}
The central extension~$\widetilde \Gamma$ of~$\SLS$ corresponding
to the integral Euler class of~$\SLS$ is finitely presented. Moreover,
$H_1(\widetilde \Gamma;\Z)$ is finite and~$H_2(\widetilde \Gamma ;\R) \cong 0$.
\end{theorem}
It is known that the image of stable commutator length of
the central Euler class extension of~$\SLS$ contains arbitrarily small
transcendental numbers~\cite[Example~5.38]{Calegari}:
\begin{exmp}\label{exa:sclrot}
Let $\Gamma := \SLS$ and let $\widetilde \Gamma$ denote the central
extension of~$\SLS$ corresponding to the integral Euler class
of~$\SLS$. In other words, $\widetilde \Gamma$ is the pre-image
of~$\SLS$ under the canonical projection~$\widetilde\SL_2(\R)
\longrightarrow \SL_2(\R)$, where $\widetilde\SL_2(\R)$ denotes the
universal covering group of~$\SL_2(\R)$.
Then
\[ \scl_{\widetilde \Gamma} (\widetilde g) = \frac{\lvert\rot( \widetilde g)|}2
\]
for all~$\widetilde g \in \widetilde \Gamma$,
where $\rot \col \widetilde \Gamma \to \R_{\geq 0}$
denotes the rotation number~\cite[Example~5.38]{Calegari}.
Furthermore, for each~$g \in \Gamma$ with~$\lvert\tr(g)| \leq 2$,
there is a lift $\widetilde{g} \in \widetilde{\Gamma}$ of $g$ such
that~\cite[p.~145]{Calegari}
\[ \rot(\widetilde g) = \frac{\arccos (\tr g /2)}{\pi}.
\]
For~$n \in \N_{>0}$, we consider
\[ g_n :=
\begin{pmatrix}
2 & 1 + 2^{-n+1} \\
-1 & -2^{-n}
\end{pmatrix}
\in \Gamma
\]
and let $\widetilde g_n \in \widetilde \Gamma$ be the associated lift.
Then $\lim_{n \rightarrow \infty} \rot(\widetilde g_n) = 0$ and
\[ \scl_{\widetilde \Gamma}(\widetilde g_n)
= \frac{\lvert\rot(\widetilde g_n)|}2
= \frac{\arccos(\tr g_n / 2)}{2 \cdot \pi}
= \frac{\arccos(1 - 2^{-n-1})}{2 \cdot \pi}
= \frac{\alpha_n}{48}.
\]
However, a priori, it is not clear that $\widetilde g_n$ lies in the
commutator subgroup of~$\widetilde \Gamma$. Because $K :=
|H_1(\widetilde \Gamma;\Z)|$ is finite
(Theorem~\ref{theorem:SLSEnew}), we know that $h_n := \widetilde g_n{}^K
\in [\widetilde \Gamma,\widetilde \Gamma]$ for all~$n \in \N$. Moreover,
by construction,
\[ \scl_{\widetilde \Gamma} (h_n)
= K \cdot \scl_{\widetilde \Gamma} (\widetilde g_n)
= K \cdot \frac{\alpha_n}{48}.
\]
\end{exmp}
With these ingredients, we can complete the proof of
Theorem~\ref{thm:transsimvol_explicit} (and thus of
Theorem~\ref{theorem:transsimvol}):
\begin{proof}[Proof of Theorem~\ref{thm:transsimvol_explicit}/\ref{theorem:transsimvol}]
Let $\widetilde \Gamma$ be the
central Euler class extension of~$\SLS$ and let $(h_n)_{n\in \N}$ and~$K$ be as in
Example~\ref{exa:sclrot}. Applying
Theorem~\ref{thm:simvolscl} to~$h_n \in [\widetilde \Gamma,
\widetilde \Gamma]$ results in an oriented closed connected
$4$-manifold~$M_n$ with~$\|M_n\| = K \cdot \alpha_n$. Hence, $\lim_{n
\rightarrow \infty} \|M_n\| = K \cdot 24 \cdot \arccos(1) /\pi = 0$. If
$n>0$, then $\alpha_n$ is known to be transcendental
(Proposition~\ref{prop:antrans}). Moreover, Baker's theorem proves the
last part of Theorem~\ref{thm:transsimvol_explicit}
(Proposition~\ref{prop:anindep}).
\end{proof}
\subsection*{Proof of Theorem~\ref{theorem:nonsimvol}}
The proof of Theorem~\ref{theorem:nonsimvol} relies on the
following simple observation (proved in Section~\ref{sec:simvolrc},
where also the definition of right-computability is recalled):
\begin{theorem}\label{theorem:simvolrc}
Let $M$ be an oriented closed connected manifold. Then $\| M \|$
is a right-computable real number.
\end{theorem}
In contrast, the numbers~$\alpha$ in
Theorem~\ref{theorem:nonsimvol} are \emph{not}
right-computable~(Proposition \ref{prop:not right computable}) and thus, in particular, \emph{not}
algebraic, because every algebraic number is
computable~\cite[Section~6]{eisermann}. The product of a
computable number with a number that is not right-computable is also
not right-computable (Section~\ref{subsec:rc}). Therefore,
applying Theorem~\ref{theorem:simvolrc} proves
Theorem~\ref{theorem:nonsimvol}.
\subsection*{Organisation of this article}
In Section~\ref{sec:trans}, we prove the transcendence properties of
the $\arccos$-terms. In Section~\ref{sec:SLSE}, we solve the
group-theoretic problem for the proof of
Theorem~\ref{theorem:SLSEnew}. In Section~\ref{sec:simvolrc}, we prove
Theorem~\ref{theorem:simvolrc}.
\subsection*{Acknowledgements}
We would like to thank the anonymous referee for asking questions
about the Euler extension (which lead to
a simplification of the treatment of Theorem~\ref{theorem:SLSEnew}).
\section{Some transcendental numbers}\label{sec:trans}
In this section, for~$n \in \N_{\geq 0}$, we will investigate the
transcendence of the following real numbers
\[ \alpha_{n} := \frac{24 \cdot \arccos(1 - 2^{-n-1})}{\pi}.
\]
We will see that $\alpha_0 = 8$ and that $\alpha_n$ is transcendental (over the algebraic numbers) for every $n \geq 1$.
\subsection{Transcendence}
As a first step, we show that the~$\alpha_n$ are transcendental for $n \geq 1$, using
Niven's theorem.
\begin{thm}[Niven~\protect{\cite[Corollary~3.12]{niven}}]\label{thm:Niven}
Let $\trig \in \{\sin,\cos\}$ and let $x \in \Q$ with~$\trig(\pi \cdot x) \in \Q$.
Then $\trig(\pi \cdot x) \in \{0,\pm 1/2,\pm1\}$.
\end{thm}
\begin{prop}\label{prop:antrans}
For every $n \geq 1$, the number~$\alpha_n$ is transcendental over~$\Q$.
\end{prop}
\begin{proof}
A consequence of the Gelfond-Schneider theorem~\cite[Theorem~1]{lima} says that for any real algebraic number $x$, the expression $\arccos(x)/\pi$ is either rational or transcendental. Thus $\alpha_n$ is either rational or transcendental.
\emph{Assume} for a contradiction that $\alpha_n$ were rational. Then,
because $\cos(\pi/24 \cdot \alpha_n) = 1 - 2^{-n-1}$ is also rational, by
Niven's theorem (Theorem~\ref{thm:Niven}), we obtain
\[ 1 - \frac1{2^{n+1}} = \cos \Bigl(\frac\pi{24} \cdot \alpha_n\Bigr) \in \{0,\pm 1/2, \pm 1\}.
\]
However, this contradicts the hypothesis that $n \geq 1$.
Hence, $\alpha_n$ must be trans\-cendental.
\end{proof}
\subsection{Linear independence over the algebraic numbers}
We will now refine Proposition~\ref{prop:antrans}, using Baker's theorem.
\begin{thm}[Baker~\cite{baker}] \label{thm:baker}
Let $\Lambda \subset \{ \ln (\alpha) \in \mathbb{C} \mid \alpha \text{
algebraic over~$\Q$}\}$ be linearly independent over~$\Q$. Then
$\Lambda$ is linearly independent over the field of algebraic
numbers.
\end{thm}
\begin{prop}\label{prop:anindep}
Let $\Prim \subset \N$ be the set of prime numbers. Then the
sequence~$(\alpha_{p-2})_{p \in \Prim}$ is linearly independent over
the algebraic numbers.
\end{prop}
For the prime $p=2$ we compute that $\alpha_{p-2} = \alpha_0 = \frac{24
\arccos(1/2)}{\pi} = 8$, which is rational. Hence, Proposition
\ref{prop:anindep} includes a proof that $\alpha_{p-2}$ is
transcendental for every odd prime~$p$.
\begin{proof}
We will use Baker's Theorem \ref{thm:baker}. Rewriting~$\arccos$ as
$$
\arccos(z) = -i \cdot \ln\bigl(i \cdot z + \sqrt{1-z^2}\bigr),
$$
we see that
$$
\alpha_{p-2} = \frac{24 \cdot \arccos(1-2^{-p+1})}{\pi} = \frac{-24 \cdot i }{\pi} \cdot \ln(\gamma_p),
$$
where
$$
\gamma_p := i \cdot \frac{2^{p-1}-1}{2^{p-1}} + \frac{1}{2^{p-1}} \cdot \sqrt{2^p - 1}.
$$
We will show in Claim~\ref{claim:ln gamma lin indep} that for every
finite set~$\{ p_1, \ldots, p_k \}$ of distinct primes the family~$\{
\ln(\gamma_{p_j}) \}_{j \in \{1, \ldots, k\}}$ is linearly
independent over~$\Q$. As $\alpha_{p-2}$ is a uniform rescaling of~$\ln(\gamma_p)$,
this will imply by using Baker's Theorem that this family is
also linearly independent over the algebraic numbers.
We will show the linear independence of~$\{
\ln(\gamma_{p_j}) \}_{j \in \{1, \ldots, k\}}$
over~$\Q$ in several steps:
\begin{claim}
Let $(m_k)_{k \in \N}$ be a sequence of pairwise coprime positive
integers. Then, for every~$k \in \N_{\geq 2}$, we have that
$$
\sqrt{m_k} \not \in \Q[i, \sqrt{m_{1}}, \ldots, \sqrt{m_{k-1}}].
$$
\end{claim}
\begin{proof}
This follows from a classical result of Besicovitch~\cite{besicovitch}.
\end{proof}
\begin{claim} \label{claim: 2 power p not in field}
Let $\{ p_1, \ldots, p_k \}$ be a finite set of distinct primes. Then
$$
\sqrt{2^{p_k}-1} \not \in \Q[i, \sqrt{2^{p_{1}}-1}, \sqrt{2^{p_{2}}-1}, \ldots, \sqrt{2^{p_{k-1}} - 1}]
$$
\end{claim}
\begin{proof}
For all primes~$p, q \in \N$ with~$p \neq q$, the
Mersenne numbers $2^{p}-1$ and $2^q-1$ are coprime. We may conclude
using the previous claim.
\end{proof}
\begin{claim} \label{claim:powers lin indepd}
Let $\{ p_1, \ldots, p_{k} \}$ be a finite set of distinct primes and let~$n \in \N_{>0}$.
Then
$$
\gamma_{p_k}^n \not \in \Q[i, \sqrt{2^{p_{k-1}}-1}, \sqrt{2^{p_{k-2}}-1}, \ldots, \sqrt{2^{p_1} - 1}].
$$
\end{claim}
\begin{proof}
We compute that
\begin{align*}
\gamma_{p_k}^n
& = \Bigl( i \cdot \frac{2^{p_k-1}-1}{2^{p_k-1}} + \frac{1}{2^{p_k-1}} \cdot \sqrt{2^{p_k} - 1} \Bigr)^n
\\
& = \frac{1}{2^{n(p_k-1)}} \cdot
\sum_{j=0}^n {n \choose j} \cdot i^{n-j} \cdot (2^{p_k-1}-1)^{n-j}\cdot (2^{p_k}-1)^{\frac{j}{2}}.
\end{align*}
We see that the terms contributing to~$\sqrt{2^{p_k}-1}$ are the
terms where $j$ is odd and that there exist~$q_1,q_2 \in \Q$
with
$$
\gamma_{p_k}^n = i^n \cdot (q_1 + q_2 \cdot i \cdot \sqrt{2^{p_k}-1}).
$$
\emph{Assume} for a contradiction that $q_2$ were zero.
Then $\gamma_{p_k} \in \Q \cup i \cdot \Q$ and as
$|\gamma_{p_k}|=1$ we obtain~$\gamma_{p_k}^n \in \{\pm 1, \pm i \}$.
In particular, $\gamma_{p_k}$ is a root of unity. Therefore, there exists
an~$x \in \Q$ with
$$
\gamma_{p_k} = \cos (2 \pi \cdot x) + i \cdot \sin (2 \pi \cdot x).
$$
According to Niven's Theorem~\ref{thm:Niven}, by comparing with the
definition of~$\gamma_{p_k}$, we see that $\frac{2^{p_k}-1}{2^{p_k}}
\in \{0, \frac 12, 1 \}$. But if $p_k$ is a prime, then this never
happens. Hence, $q_2$ is non-zero, and so $\gamma_{p_k}^n \not \in
\Q[i, \sqrt{2^{p_1}-1}, \ldots, \sqrt{2^{p_{k-1}}-1}]$ by Claim~\ref{claim: 2 power p not in field}.
\end{proof}
\begin{claim} \label{claim:ln gamma lin indep}
Let $\{ p_1, \ldots, p_k \}$ be a finite set of distinct primes. Then the
corresponding family~~$\{
\ln(\gamma_{p_j}) \}_{j \in \{1, \ldots, k\}}$ is linearly independent over~$\Q$.
\end{claim}
\begin{proof}
\emph{Assume} for a contradiction that this family were linearly
dependent over~$\Q$, whence over~$\Z$. Thus, there
are integers $n_i \in \Z$, not all zero, such that
$$
\ln(\gamma_{p_1}^{n_1} \cdots \gamma_{p_k}^{n_k})
= n_1 \cdot \ln(\gamma_{p_1}) + \dots + n_k \cdot \ln(\gamma_{p_k}) = 0.
$$
Without loss of generality we may assume that $n_k > 0$.
Hence,
$$
\gamma_{p_1}^{n_1} \cdots \gamma_{p_k}^{n_k} \in \{1 + m \cdot 2 \pi i \mid m \in \Z \}.
$$
The left-hand side is algebraic over~$\Q$, but the right-hand side
is only algebraic if $m=0$. Thus, we conclude that
$\gamma_{p_1}^{n_1} \cdots \gamma_{p_k}^{n_k} = 1$; in other words,
$$\gamma_{p_k}^{n_k} = \gamma_{p_1}^{-n_1} \cdots \gamma_{p_{k-1}}^{-n_{k-1}}.
$$
Moreover, by construction, $\gamma_{p_1}^{-n_1} \cdots
\gamma_{p_{k-1}}^{-n_{k-1}} \in \Q[i, \sqrt{2^{p_1}-1}, \ldots,
\sqrt{2^{p_{k-1}}-1}]$. However, this contradicts
Claim~\ref{claim:powers lin indepd}. Thus, $\ln(\gamma_{p_1}),
\ldots, \ln(\gamma_{p_k})$ are linearly independent over $\Q$.
\end{proof}
This finishes the proof of Proposition~\ref{prop:anindep}.
\end{proof}
\section{Solving the group-theoretic problem}\label{sec:SLSE}
As the basic building block for our constructions we pick~$\SLS$ because
its low-degree (co)homology, its second bounded cohomology, and its
quasi-mor\-phisms are already known to basically have the right structure.
\subsection{Basic properties of~$\SLS$}
We collect basic properties of~$\SLS$
needed in the sequel; further information on the (bounded) Euler
class for circle actions can be found in the literature~\cite{BFH,Ghys_circle}.
\begin{prop}[low-degree (co)homology of~$\SLS$]\label{prop:lowdegSLS}
\hfil
\begin{enumerate}
\item The group~$\SLS$ is finitely presented.
\item The group~$H_1(\SLS;\Z)$ is finite (and non-trivial).
\item The group~$\SLS$ does \emph{not} admit any non-trivial
quasi-morphisms.
\item We have~$H^2_b (\SLS;\R) \cong \R$ and the bounded
Euler class~$\eu \R \SLS _b$ is a generator.
\item The evaluation map~$\langle \eu \Z \SLS, \args\rangle \colon
H_2(\SLS;\Z) \longrightarrow \Z$ has finite kernel and finite cokernel.
\end{enumerate}
\end{prop}
\begin{proof}
\emph{Ad~1.}
The group~$\SLS$ can be written as an amalgamated free
product of the form
\[ \SLS \cong \SL_2(\Z) *_{\Gamma_0(2)} \SL_2(\Z),
\]
where $\Gamma_0(2)$ is the subgroup of~$\SL_2(\Z)$ of those
matrices whose lower left entry is divisible by~$2$; this
leads to an explicit finite presentation~\cite[p.~81]{serretrees}.
\emph{Ad~2.}
In particular, one obtains that $H_1(\SLS;\Z) \cong \Z/3$ is
finite~\cite[Proposition~3.1]{ademnaffah}.
(Moreover, applying the Mayer-Vietoris sequence to the decomposition in
the proof of the first part allows to
compute the cohomology~$H^*(\SLS;\Z)$~\cite{ademnaffah}.)
\emph{Ad~3.}
This is one of many examples of groups acting on the circle
with this property~\cite[Example~5.38]{Calegari}.
\emph{Ad~4.}
This is a result of Burger and Monod: The
inclusion~$\SLS \longrightarrow \SL_2(\R)$ induces an
isomorphism~$H^2_{cb}(\SL_2(\R);\R) \longrightarrow
H^2_b(\SLS;\R)$~\cite[Corollary~24]{BurgerMonod}\cite[Corollary~4]{buchermonod}.
Moreover, $H^2_{cb}(\SL_2(\R);\R) \cong \R$, generated by the bounded
Euler class~\cite{burgermonodsl2}.
\emph{Ad~5.}
We abbreviate~$\Gamma := \SLS$. Because $\Gamma$ is finitely
presented, $H_2(\Gamma;\Z)$ is a finitely generated Abelian
group~\cite[II.5]{brown}. Moreover, it has been computed that
$H_2(\Gamma;\Q) \cong \Q$~\cite[Proposition~2.2]{moss}. Hence,
$H_2(\Gamma;\Z)$ is virtually~$\Z$ and it suffices to show that the
evaluation~$\langle \eu \Z \Gamma, \args\rangle \colon
H_2(\Gamma;\Z) \longrightarrow \Z$ is non-trivial.
As the space~$Q(\Gamma)$ of
quasi-morphisms (modulo trivial quasi-morphisms) is trivial, the
comparison map~$c_\Gamma \colon H_b^2(\Gamma;\R) \longrightarrow
H^2(\Gamma;\R)$ is injective~\cite[Theorem~2.50]{Calegari}. In particular, $\eu \R
\Gamma = c_\Gamma (\eu \R \Gamma _b)$ is non-trivial
in~$H^2(\Gamma;\R)$. Therefore, by the universal coefficient
theorem, also the evaluation $\langle \eu \Z \Gamma,\args\rangle
\colon H_2(\Gamma;\Z) \longrightarrow \Z$ associated with the
integral Euler class~$\eu \Z \Gamma \in H^2(\Gamma;\Z)$ is
non-trivial.
\end{proof}
\subsection{Imitating the universal central extension}
If $\Gamma$ is a perfect group, then its universal central extension~$E$
is a perfect group that satisfies~$H_2(E;\R)\cong 0$. The universal
central extension of~$\Gamma$ can be constructed as the central extension
corresponding to the cohomology class~$\varphi$ in~$H^2(\Gamma; H_2(\Gamma;\Z))$
whose evaluation map~$\langle \varphi,\args\rangle \colon H_2(\Gamma;\Z)
\longrightarrow H_2(\Gamma;\Z)$ is the identity map. Moreover, we may compute the
quasimorphisms on~$E$ from $H^2_b(\Gamma;\R)$, which in turn allows us to
compute the stable commutator length on~$E$ using Bavard's Duality
Theorem~\cite[Section 5]{heuerloeh4mfd}. The group~$\SLS$ is not
perfect, thus it does not have a universal central extension.
Instead, we will choose a central extension of~$\SLS$ that is able
to play the same role in our context.
\begin{prop}\label{prop:H_2}
Let $\Gamma$ be a finitely presented group with
finite~$H_1(\Gamma;\Z)$, let $A$ be a finitely generated
Abelian group, and let $E$ be a central
extension group of~$\Gamma$ that corresponds to
a class~$\varphi \in H^2(\Gamma;H)$ such that the
evaluation map~$\langle \varphi,\args\rangle \colon H_2(\Gamma;\Z)
\longrightarrow A$ has finite kernel
and finite cokernel.
Then:
\begin{enumerate}
\item The group~$E$ is finitely presented.
\item We have~$H_1(E;\R) \cong 0$ and $H_2(E;\R) \cong 0$.
\end{enumerate}
\end{prop}
\begin{proof}
The central extension group~$E$ fits into a short
exact sequence of the form
$\xymatrix{
1 \ar[r]
& A \ar[r]
& E \ar[r]
& \Gamma \ar[r]
& 1.}
$
\emph{Ad~1.}
Because $A$ is finitely generated, the central extension group~$E$
of~$\Gamma$ by~$A$ is also finitely presented.
\emph{Ad~2.}
Because the extension is central, we have the associated
exact sequence
\[ \makebox[0pt]{\xymatrix@=2em{
H_1(E;\Z) \otimes_\Z A \ar[r]
& H_2(E;\Z) \ar[r]
& H_2(\Gamma;\Z) \ar[r]^-\beta
& A \ar[r]
& H_1(E;\Z) \ar[r]
& H_1(\Gamma;\Z) \ar[r]
& 0
}}
\]
by Eckmann, Hilton, and Stammbach~\cite[(1.4) and Theorem~2.2]{eckmannhiltonstammbachI},
where
\begin{align*}
\beta \colon H_2(\Gamma;\Z) & \longrightarrow A \\
\alpha & \longmapsto \langle \varphi,\alpha\rangle.
\end{align*}
By assumption, $\beta$ has finite cokernel and $H_1(\Gamma;\Z)$ is finite.
Hence, $H_1(E;\Z)$ is finite and therefore also the left-most
term~$H_1(E;\Z) \otimes_\Z A$ is finite. As $\beta$ has
finite kernel, this implies that $H_2(E;\Z)$ is finite. Applying
the universal coefficient theorem, shows that $H_2(E;\R) \cong
H_2(E;\Z) \otimes_\Z \R \cong 0$.
\end{proof}
With these preparations, we can now give a proof of
Theorem~\ref{theorem:SLSEnew}:
\begin{proof}[Proof of Theorem~\ref{theorem:SLSEnew}]
We only need to combine Propositions~\ref{prop:lowdegSLS}
and~\ref{prop:H_2}. As $\widetilde\Gamma$ is finitely generated,
$H_1(\widetilde \Gamma;\R) \cong 0$ implies that $H_1(\widetilde
\Gamma;\Z)$ is finite.
\end{proof}
\subsection{More on almost universal extensions}
Let us mention that the same procedure as in the previous proofs also
works in other, similar, situations:
\begin{setup}\label{setup:scl}
Let $\Gamma$ be a group with a given orientation preserving continuous
action on~$S^1$ with the following properties:
\begin{itemize}
\item The group~$\Gamma$ is finitely presented.
\item The group~$H_1(\Gamma;\Z)$ is finite.
\item The group~$\Gamma$ does \emph{not} admit any non-trivial quasi-morphisms.
\item We have~$H^2_b(\Gamma;\R) \cong \R$ and the bounded Euler class~$\eu \R \Gamma_b$
is a generator.
\end{itemize}
In this situation, we denote the central extension group of~$\Gamma$ associated
with the Euler class~$\eu \Z \Gamma \in H^2(\Gamma;\Z)$ by~$\widetilde \Gamma$.
\end{setup}
We have already seen in the previous propositions that $\SLS$ fits
into this setup. Another prominent example is Thompson's group~$T$,
which is even perfect; the condition on~$H^2_b$ follows from explicit
cohomological computations~\cite[Proposition~5.6]{heuerloeh4mfd},
based on calculations by Ghys and Sergiescu~\cite{GS}.
\begin{prop}\label{prop:euler}
Let $\Gamma$ be as in Setup~\ref{setup:scl}. Then:
\begin{enumerate}
\item The evaluation map~$\langle \eu \Z \Gamma, \args\rangle \colon H_2(\Gamma;\Z)
\longrightarrow \Z$ is non-trivial.
\item Let $H := H_2(\Gamma;\Z)$, let $ m\in \N_{>0}$ be a generator
of~$\im \langle \eu \Z \Gamma,\args\rangle \subset \Z$ (which is
non-zero by the first part), and let $\varepsilon := 1/m \cdot
\langle \eu \Z \Gamma,\args\rangle \colon H \longrightarrow \Z$. Then
there exists a $\varphi \in H^2(\Gamma;\Z)$ with
\[ H^2(\id_\Gamma;\varepsilon) (\varphi) = \eu \Z \Gamma
\qand
\langle \varphi,\args\rangle = m \cdot \id_H.
\]
\item Let $E$ be the central extension group of~$\Gamma$ associated
with~$\varphi$. Then there exists an epimorphism~$\psi \colon E \longrightarrow \widetilde \Gamma$
with~$\psi|_H = \varepsilon \colon H \longrightarrow \Z$ and~$\ker \psi \subset H$.
\end{enumerate}
\end{prop}
\begin{proof}
\emph{Ad~1.}
This is the same universal coefficient theorem argument as
in the last part of (the proof of) Proposition~\ref{prop:lowdegSLS}.
\emph{Ad~2.}
By the naturality of the short exact sequence in the universal
coefficient theorem, we have the following commutative diagram
with exact rows:
\[ \xymatrix{
0 \ar[r]
& \Ext^1_\Z\bigl(H_1(\Gamma;\Z),H\bigr) \ar[r]
\ar[d]_{\Ext^1(\id;\varepsilon)}
& H^2(\Gamma;H) \ar[r]^-{\varphi \mapsto \langle \varphi,\args\rangle}
\ar[d]_{H^2(\id_\Gamma;\varepsilon)}
& \Hom_\Z(H,H) \ar[r]
\ar[d]^{f \mapsto \varepsilon \circ f}
& 0
\\
0 \ar[r]
& \Ext^1_\Z\bigl(H_1(\Gamma;\Z),\Z\bigr) \ar[r]
& H^2(\Gamma;\Z) \ar[r]_-{\varphi \mapsto \langle \varphi,\args\rangle}
& \Hom_\Z(H,\Z) \ar[r]
& 0
}
\]
The left vertical arrow is an epimorphism because $\varepsilon$ is
an epimorphism and the exactness properties of~$\Ext$ over the
principal ideal domain~$\Z$. Moreover, the right vertical arrow
maps~$m \cdot \id_H$ to~$m \cdot \varepsilon = \langle
\eu\Z\Gamma,\args\rangle$. A short diagram chase therefore proves the
existence of the desired class~$\varphi \in H^2(\Gamma;H)$ (e.g.,
using the four~lemma~\cite[Lemma~I.3.2]{maclanehomology}).
\emph{Ad~3.}
Because the extension classes are related
via~$H^2(\id_\Gamma;\varepsilon)(\varphi) = \eu \Z \Gamma$,
there exists a group homomorphism~$\psi \colon E \longrightarrow
\widetilde \Gamma$ with~$\psi|_H = \varepsilon$ that induces the
identity on~$\Gamma$:
\[ \xymatrix{
1 \ar[r]
& \Z \ar[r]
& \widetilde \Gamma \ar[r]
& \Gamma \ar[r]
& 1
\\
1 \ar[r]
& H \ar[r] \ar[u]^{\varepsilon}
& E \ar[r] \ar@{-->}[u]^{\psi}
& \Gamma \ar[r] \ar@{=}[u]
& 1
}
\]
As $\varepsilon \colon H \longrightarrow \Z$ is an epimorphism
also $\psi \colon E \longrightarrow \widetilde \Gamma$ is an
epimorphism. By construction, $\ker \psi \subset H$.
\end{proof}
\begin{corr}
Let $\Gamma$ be as in Setup~\ref{setup:scl}, let $H
:= H_2(\Gamma;\Z)$, and let~$E$ be the central extension group
of~$\Gamma$ associated with the class~$\varphi \in H^2(\Gamma;H)$ of
Proposition~\ref{prop:euler}. Then:
\begin{enumerate}
\item The group~$E$ is finitely presented and $H_2(E;\R) \cong 0$.
\item The epimorphism~$\psi \colon E \longrightarrow
\widetilde \Gamma$ of Proposition~\ref{prop:euler} induces an
isomorphism
\begin{align*}
Q(\psi) \colon Q(\widetilde \Gamma) & \longrightarrow Q(E)
\\
[f] & \longmapsto [f \circ \psi]
\end{align*}
and both spaces are one-dimensional.
Here, $Q$ denotes the space of quasi-morphisms modulo trivial
quasi-mor\-phisms.
\item In particular, $\scl_E([E,E]) = \scl_{\widetilde \Gamma}([\widetilde \Gamma,
\widetilde \Gamma])$
as subsets of~$\R$.
\end{enumerate}
\end{corr}
\begin{proof}
\emph{Ad~1.}
This follows directly from Proposition~\ref{prop:H_2}.
\emph{Ad~2.} We will use bounded cohomology in degree~$2$ to derive
the statement on quasi-morphisms;
we consider the commutative diagram
\[ \xymatrix{
0 \ar[r]
& Q(\widetilde \Gamma) \ar[r]^-\delta
\ar[d]_{Q(\psi)}
& H^2_b(\widetilde \Gamma;\R) \ar[r]^-{c^2_{\widetilde \Gamma}}
\ar[d]_{H^2_b(\psi;\R)}
& H^2(\widetilde \Gamma;\R)
\ar[d]^{H^2(\psi;\R)}
\\
0 \ar[r]
& Q(E) \ar[r]_-{\delta}
& H^2_b(E;\R) \ar[r]_-{c^2_{E}}
& H^2(E;\R)
}
\]
with exact rows.
By construction, the kernel of the epimorphism~$\psi \colon E
\longrightarrow \widetilde \Gamma$ lies in the Abelian group~$H$ and
thus is amenable. By the mapping theorem in bounded
cohomology~\cite[p.~40]{vbc}\cite[Theorem~4.3]{ivanov},
the induced map~$H^2_b(\psi;\R) \colon H^2_b(\widetilde \Gamma;\R)
\longrightarrow H^2_b(E;\R)$ is an isomorphism.
Because~$H_2(E;\R) \cong 0$, we also have~$H^2(E;\R) \cong 0$. Therefore,
$\delta \colon Q(E) \longrightarrow H^2_b(E;\R)$ is an isomorphism.
We now show that also $\delta \colon Q(\widetilde \Gamma)
\longrightarrow H^2_b(\widetilde \Gamma;\R)$ is an isomorphism: By
the mapping theorem in bounded cohomology, the extension
projection~$\widetilde \pi \colon \widetilde \Gamma \longrightarrow
\Gamma$ induces an isomorphism~$H^2_b(\widetilde \pi ;\R) \colon
H^2_b(\Gamma;\R) \longrightarrow H^2_b(\widetilde \Gamma;\R)$. As
$H^2_b(\Gamma;\R)$ is generated by the bounded Euler class, also
$H^2_b(\widetilde \Gamma;\R)$ is one-dimensional and generated by
\[ \widetilde \eurm := H^2_b(\widetilde \pi;\R)(\eu \R \Gamma_b).
\]
By naturality of the comparison map, we obtain that
\[ c^2_{\widetilde \Gamma} (\widetilde \eurm)
= H^2(\widetilde \pi;\R) (\eu \R \Gamma).
\]
By construction of the central Euler class extension~$\widetilde
\Gamma$, we have~$H^2(\widetilde\pi;\Z)(\eu \Z \Gamma) = 0 \in H^2(\widetilde
\Gamma;\Z)$. Therefore, $H^2(\widetilde\pi;\R) (\eu\R\Gamma) = 0$ and
so~$c^2_{\widetilde \Gamma}(\widetilde \eurm) = 0$. This shows that
$\delta \colon Q(\widetilde \Gamma) \longrightarrow H^2_b(\widetilde
\Gamma;\R)$ is an isomorphism.
Now commutativity of the left square in the diagram above shows
that $Q(\psi) \colon Q(\widetilde \Gamma) \longrightarrow Q(E)$
is an isomorphism.
\emph{Ad~3.} Let $[f] \in Q(\widetilde \Gamma) \cong \R$ be a
homogeneous generator, which exists by the second part; then $[f \circ \psi]$
is a homogeneous generator of~$Q(E)$. Bavard
duality~\cite{Bavard}\cite[Theorem~2.70]{Calegari} implies that
for all~$g \in [E,E]$, we have
\[ \scl_E (g)
= \frac{\bigl| f \circ \psi(g)\bigr|}{2 \cdot D_E(f \circ \psi)}
= \frac{\bigl| f(\psi(g))\bigr|}{2 \cdot D_{\widetilde \Gamma}(f)}
= \scl_{\widetilde \Gamma}\bigl(\psi(g)\bigr);
\]
the defects in the denominators are equal because $\psi$ is an
epimorphism. Again, because $\psi$ is an epimorphism, we conclude
that $\scl_E$ and $\scl_{\widetilde \Gamma}$ have the same image in~$\R$.
\end{proof}
\section{Right-computability of simplicial volumes}\label{sec:simvolrc}
We now turn to right-computability of the numbers occuring
as simplicial volumes. After recalling basic terminology in
Section~\ref{subsec:rc}, we will prove Theorem~\ref{theorem:simvolrc}
in Section~\ref{subsec:proofsimvolrc}.
\subsection{Right-computability}\label{subsec:rc}
We use the following version of (right-)computability of real numbers,
which is formulated in terms of Dedekind cuts. For basic notions of
(recursive) enumerability, we refer to the book of Cutland~\cite{Cutland}.
\begin{defn}[right-computable]
A real number~$\alpha$ is \emph{right-computable} if the set~$\{x
\in \Q \mid \alpha < x \}$ is recursively enumerable.
We say that $\alpha$ is \emph{computable} if both $\{x \in \Q \mid
\alpha < x \}$ and $\{x \in \Q \mid \alpha > x \}$ are recursively
enumerable.
\end{defn}
Further information on different notions of one-sided computability of
real numbers can be found in the work of Zheng and
Rettinger~\cite{zhengrettinger}.
There are only countably many recursively enumerable subsets of $\Q$
and thus the set of right computable and computable numbers is
countable.
We collect some easy properties:
\begin{lemma}
\hfil
\begin{enumerate}
\item
If $\alpha, \beta \in \R_{\geq 0}$ are right-computable and
non-negative, then so is $\alpha \cdot \beta \in \R$.
\item
If $\alpha \in \R_{\geq 0}$ is a real number and $c \in \R_{>0}$ a computable number such that
$c \cdot \alpha$ is right-computable, then $\alpha$ is right-computable.
\end{enumerate}
\end{lemma}
\begin{proof}
For the first part we observe that if $\alpha, \beta \geq 0$, then
$\{ x \in \Q \mid \alpha < x \}
\cdot \{ y \in \Q \mid \beta < y \} = \{ z \in \Q \mid \alpha \cdot
\beta < z \}$.
For the second part, let $\alpha \in \R_{\geq 0}$ be such that $c\cdot \alpha$ is
right-computable, where $c$ is computable. Since $c$ is computable
and positive, so is $c^{-1}$, thus $c^{-1}$ is in particular
right-computable. Hence $\alpha = c^{-1} \cdot (c \cdot \alpha)$ is the
product of non-negative right-computable numbers and thus right-computable.
\end{proof}
To a subset~$A \in \N$ we associate the number~$x_A := \sum_{n \in \N}
2^{-n}$. We relate the (right-)computability of~$x_A$ to the
computability of~$A$ as a subset of~$\N$, following Specker
\cite{specker}.
\begin{prop} \label{prop:not right computable}
Let $A \subset \N$ and let $x_A$ be defined as above. Then:
\begin{enumerate}
\item \label{item:left comp} If the set~$A$ is recursively enumerable, then $x_A$ is left-computable and $2-x_A = x_{\N \setminus A}$ is right-computable.
\item \label{item:comp} The set~$A$ is recursive if and only if $x_A$ is computable.
\item \label{item: not right comp number} If $A$ is recursively enumerable but not recursive, then $x_A$ is not right-computable.
\end{enumerate}
\end{prop}
\begin{proof}
The first two items are classical results of Specker~\cite{specker}. To see item~\ref{item: not right comp number}, let $A$ be recursively enumerable but not recursive. \emph{Assume} that $x_A$ is right-computable. By item~\ref{item:left comp}, $x_A$ is then also left-computable. Thus, $x_A$ is both left- and right-computable, whence computable. But by item~\ref{item:comp} this implies that $A$ is recursive, which contradicts our assumption.
\end{proof}
\begin{lemma}\label{lem:fractions}
Let $f \colon \N \longrightarrow \N$ be a function with the
following property: The set~$\{ (m,n) \in \N \times \N \mid f(m) \leq
n\} \subset \N \times \N$ is recursively enumerable. Then
\[ \inf_{m \in \N_{>0}} \frac{f(m)}m
\]
is right-computable.
\end{lemma}
\begin{proof}
Set~$S :=\{ (m,n) \in \N \times \N \mid f(m) \leq
n\}$ and observe that
$$
\inf_{m \in \N_{>0}} \frac{f(m)}{m} = \inf_{(m,n) \in S} \frac{n}{m}.
$$
There is a Turing machine that, as input, takes a rational number
and then enumerates all rational numbers above it. We may
diagonally use this Turing machine and the enumeration of~$S$ to
enumerate the set
$$
\Bigl\{ x \in \Q \Bigm| \exi{(m,n) \in S} \frac{n}{m} < x \Bigr\}
= \Bigl\{ x \in \Q \Bigm| \inf_{m \in \N_{>0}} \frac{f(m)}m < x \Bigr\}.
$$
Thus indeed $\inf_{m \in \N_{>0}} \frac{f(m)}m$ is right-computable.
\end{proof}
\subsection{Proof of Theorem~\ref{theorem:simvolrc}}\label{subsec:proofsimvolrc}
Let $M$ be an oriented closed connected manifold and $d := \dim
M$. Then $M$ is homotopy equivalent to a finite (simplicial)
complex~$T$~\cite{siebenmann,kirbysiebenmann}; let
$f \colon M \longrightarrow |T|$ be such a homotopy equivalence and for
a commutative ring~$R$ with unit, let
\[ [T]_R := H_d(f;R) \bigl([M]_R\bigr) \in H_d\bigl(|T|;R\bigr).
\]
If $R$ is a normed ring, then we write $\|\cdot\|_{1,R}$ for the
associated $\ell^1$-semi-norm on~$H_d(|T|;R)$.
Because $f$ is a homotopy equivalence, we have
\[ \| M\| = \bigl\| [M]_\R\bigr\|_{1,\R} = \bigl\| [T]_{\R} \bigr\|_{1,\R}.
\]
Moreover, the $\ell^1$-semi-norm with $\R$-coefficients can be
computed via rational coefficients~\cite[Lemma~2.9]{mschmidt}:
\[ \| M\| = \bigl\| [T]_{\R} \bigr\|_1
= \bigl\| [T]_{\Q} \bigr\|_{1,\Q}
= \inf_{m \in \N_{>0}} \frac{\bigl\| m \cdot [T]_{\Z}\|_{1,\Z}}{m}.
\]
The function~$m \longmapsto \| m \cdot [T]_{\Z}\|_{1,\Z}$ satisfies
the hypothesis of Lemma~\ref{lem:fractions}
(see Lemma~\ref{lem:isvenum} below). Applying Lemma~\ref{lem:fractions}
therefore shows that the number~$\|M\|$ is right-computable.
\begin{lemma}\label{lem:isvenum}
In this situation, the
subset
\[ \bigl\{ (m,n) \in \N \times \N \bigm| \| m \cdot [T]_{\Z} \|_{1,\Z} \leq n \bigr\} \subset \N \times \N
\]
is recursively enumerable.
\end{lemma}
\begin{proof}
We can use a straightforward enumeration of combinatorial
models of cycles~\cite[proof of Corollary~5.1]{loehodd}:
First, $H_d(|T|;\Z)$ is isomorphic to the simplicial
homology~$H_d(T;\Z)$ of~$T$. Therefore, we can (algorithmically)
determine a simplicial cycle~$z$ on~$T$ that represents the
class~$[T]_\Z$; this cycle can also be viewed as a singular cycle
on~$|T|$.
Inductive simplicial approximation of singular simplices shows that
for every singular cycle~$c \in C_d(|T|;\Z)$, there exists a
singular cycle~$c' \in C_d(|T|;\Z)$ with the following properties:
\begin{itemize}
\item The cycles $c$ and $c'$ represent the same homology class
in~$H_d(|T|;\Z)$.
\item The chain~$c'$ is a \emph{combinatorial singular chain}, i.e.,
all singular simplices in~$c'$ are simplicial maps from an
iterated barycentric subdivision of~$\Delta^d$
to an iterated barycentric subdivision of~$T$.
Here, each singular simplex in~$c'$ is the simplicial approximation
of a singular simplex in~$c$. In particular, in general, the image
of a singular simplex in~$c'$ might touch several simplices of~$T$
and might pass them several times.
\item We have~$|c'|_1 \leq |c|_1$.
\end{itemize}
This allows us to restrict attention to such combinatorial singular
chains. Moreover, the following operations can be performed by
Turing machines:
\begin{itemize}
\item Enumerate all iterated barycentric subdivisions of~$T$
and~$\Delta^d$.
\item Enumerate all simplicial maps between two finite simplicial
complexes.
\item Hence: Enumerate all combinatorial singular $\Z$-chains of~$T$.
\item Check, for given~$m \in \N$, whether a combinatorial singular
$\Z$-chain on~$T$ is a a cycle and represents the class~$m \cdot
[T]_\Z$ in~$H_d(|T|;\Z)$ (through comparison with the
corresponding iterated barycentric subdivision of~$z$ in
simplicial homology).
\item Compute the $1$-norm of a combinatorial singular $\Z$-chain.
\end{itemize}
In summary, we can enumerate the set~$\{(m,c) \mid m \in \N, c \in
C(m)\}$, where $C(m)$ is the set of all combinatorial $\Z$-cycles
of~$T$ that represent~$m \cdot [T]_\Z$ in~$H_d(|T|;\Z)$.
We now consider the following algorithm: Given~$m, n \in \N$,
we search for elements of $1$-norm at most~$n$ in~$C(m)$.
\begin{itemize}
\item If such an element is found (in finitely many steps), then the
algorithm terminates and declares that~$\|m \cdot [T]_\Z\|_{1,\Z}
\leq n$.
\item Otherwise the algorithm does not terminate.
\end{itemize}
From the previous discussion, it is clear that
this algorithm witnesses that the set~$\{(m,n) \in \N \times \N \mid
\|m \cdot [T]_\Z\|_{1,\Z} \leq n\}$ is recursively enumerable.
\end{proof}
This completes the proof of Theorem~\ref{theorem:simvolrc}.
\begin{rmk}
It should be noted that the argument above is constructive enough to
also give a slightly stronger statement (similar to the case of
integral simplicial volume~\cite[Remark~5.2]{loehodd}): The function
from the set of (finite) simplicial complexes (with vertices
in~$\N$) that triangulate oriented closed connected manifolds to the
set of subsets of~$\Q$ given by
\[ T \longmapsto \| \, |T|\,\|
\]
is semi-computable (and not only the resulting individual real
numbers) in the following sense: There is a Turing machine that
given such a triangulation~$T$ and $x \in \Q$ as input
\begin{itemize}
\item halts if $\| \,|T|\, \| < x$ and declares that $\| \,| T|\, \| < x$,
\item and does not terminate if~$ \| \,|T|\, \| \geq x$.
\end{itemize}
But it is known that this function is \emph{not}
computable~\cite[Theorem~2, p.~88]{weinberger}.
\end{rmk}
{\small
\bibliographystyle{alpha}
\bibliography{bib_l1}}
\vfill
\noindent
\emph{Nicolaus Heuer}\\[.5em]
{\small
\begin{tabular}{@{\qquad}l}
DPMMS, University of Cambridge \\
\textsf{nh441@cam.ac.uk},
\textsf{https://www.dpmms.cam.ac.uk/$\sim$nh441}
\end{tabular}}
\medskip
\noindent
\emph{Clara L\"oh}\\[.5em]
{\small
\begin{tabular}{@{\qquad}l}
Fakult\"at f\"ur Mathematik,
Universit\"at Regensburg,
93040 Regensburg\\
\textsf{clara.loeh@mathematik.uni-r.de},
\textsf{http://www.mathematik.uni-r.de/loeh}
\end{tabular}}
\end{document}
| 219,941
|
\begin{document}
\title{\textsc{Elliptical Tempered Stable Distribution and Fractional Calculus}}
\author{\textsc{Hassan A. Fallahgoul${}$}\thanks{Universit\'{e} libre de Bruxelles, Solvay Brussels School of Economics and Management, ECARES. 50, Av Roosevelt CP114, B1050 Brussels, Belgium. Tel: +32(0)26502792; Fax: +32(0)26504218; \texttt{hfallahg@ulb.ac.be}}
\hspace{0.05cm} and \textsc{Young S. Kim${}$}
\thanks{Stony Broke University, College of Business;
\texttt{Aaron.Kim@stonybrook.edu}
}}
\maketitle
\vspace{-0.8cm}
\begin{abstract}
A definition for elliptical tempered stable distribution, based on the characteristic function, have been explained which involve a unique spectral measure. This definition provides a framework for creating a connection between infinite divisible distribution, and particularly elliptical tempered stable distribution, with fractional calculus. Finally, some analytical approximations for the probability density function of tempered infinite divisible distribution, which elliptical tempered stable distributions are a subclass of them, are considered.
\\[2mm]
\noindent \textit{Keywords}: Tempered Stable Distribution, Elliptical Distribution, Fractional PDE.
\\[2mm]
\end{abstract}
\newpage
\section{Introduction}
Stable and tempered stable distributions, which are a subclass of Infinite Divisible Distributions (hereafter IDD), have been used for solving many practical problems (see, for example, \cite{Kim01}, \cite{Rachev01}, and \cite{Fallahgoul01} among others.). The application of them in finance back to work of Mandelbrot in \cite{Mandelbrot01, Mandelbrot02}. As a matter of fact, before his work the usual assumption was that the distribution of price changes in a speculative series is approximately normal distribution. In the other words, in the financial time series context the distribution of the innovations (white noise) in a model like autoregressive moving average of returns series was normal distribution. But, Mandelbort showed that the returns from financial assets have the properties like leptokurtosis and skewness which can not support by normality assumption. On the other hand, the central limit theorem property, which is used based on normality assumption, is a powerful instrument in financial time series. Therefore, Mandelbort suggested a model that have the central limit theorem property as well as be able to capture leptokurtosis and skewness of return series. He used the stable distribution, in name of stable paretion, to reach suitable model. Therefore, the stable and tempered stable distributions, as a heavy tailed distribution, have been became the most popular alternative to the normal distribution which has been rejected by numerous empirical studies (see, for example, \cite{Rachev01}, and \cite{Rachev02}, among others).
It is well known that stable distributions have infinite moments for some amounts of index parameter. In fact, they have infinite $ q-th $ moment for all $ q\geq\alpha $. This properties can be a disadvantage of the stable distributions in practical area. For example, in the context of time series a conclusion of literature is that the asset returns follow a model that the tails of it is heavier than normal and thinner than stable distribution (see, \cite{Rachev01}). In order to overcome this disadvantage, the L\'{e}vy measure of the stable distributions is multiplying by a tempering function. By this procedure a new class will be obtained, named tempered stable, such that they are not only IDD but also finite moments for all orders.\footnote{See, \cite{Rachev01}.}
The closed-form formula for the Probability Density Function (hereafter PDF) and Comulative Distribution Function (hereafter CDF) of the stable and tempered stable distribution, in the univarite and multivariate case, is not available. But, the characteristic function (hereafter CF) in the favorable form is accessible, which is used broadly instead of related PDF and CDF. But also, in the multivariate context the problem is still remain and working with related CF is not easy. For example, generating the multivariate stable and tempered stable distribution via the standard methods as fast fourier transform is so difficult even for lower dimensions.
Based on the computational facilities, the univariate stable and tempered stable distributions successfully have been used for practical problems (see, for example \cite{Kim01}, \cite{Rachev01}, \cite{Rachev02}, and \cite{Rachev03} among others.). But, regarding to the multivariate case just a little works are exist, albeit for multivariate stable distributions (see, for example \cite{Nolan01}, \cite{Press01}, \cite{Fallahgoul01} and \cite{Fallahgoul02}). This fact backs to the definitions of each of them. The definition of the both multivariate stable and tempered stable distributions is based on a complicated spectral measures, but since the CF of the earlier can be defined based on some projections,\footnote{For more details see \cite{taqqu}.} working with this distribution in practical area is more convenient than multivariate tempered stable distributions.
In this article we propose a suitable form of multivariate tempered stable distribution, named Elliptical Tempered Stable (hereafter ETS) distribution, and introduce some analytic approximations of the PDF of Tempered Infinite Divisible (hereafter TID), which ETS distribution is a subclass of it. More precisely, the contribution of the paper is threefold. First, a well-define definition for ETS distribution based on a unique spectral measure is introduced, and also some theoretical properties of it are considered. Second, this paper provides a framework for getting a connection between fractional calculus and TID. In fact, some fractional Partial Differential Equations (hereafter PDE) are introduced that the fundamental solution of them give the entire family of the PDF of TID. Last, the analytic approximations for the PDF of TID and ETS distribution are given.
After discussing the notation to be used in Section 2, the paper proceeds as follows. The definition of ETS distribution and the link between multivariate tempered stable distributions and fractional PDE whose solution gives nearly all the TID distributions is explained in Section 3. The analytic approximations for PDF of TID distribution and ETS distribution have been presented in Section 4.
\section{Notation and Basic Definitions}
Here we present the notation and some basic definitions used in this paper. First a word on notation. A random variable and vector will be shown in capital and capital bold character, respectively. And also, an element of a vector is shown by non-bold words but still it is capital.
Furthermore, the inner product and corresponding norm in $ \mathbb{R}^n $ are defined as $\langle\textbf{s},\textbf{t}\rangle=\displaystyle\sum_{i=1}^{n}s_it_i$ and $\|\textbf{s}\|=\langle\textbf{s}\rangle^{1/2}=(\displaystyle\sum_{i=1}^{n}s_i^2)^{1/2}$, respectively, where $\textbf{s}=(s_1,s_2,\cdots,s_n)$ and $\textbf{t}=(t_1,t_2,\cdots,t_n)$ are arbitrary elements in $\mathbb{R}^n$. Let $\mathbb{S}^n$ be the unite sphere in $\mathbb{R}^n$: $\mathbb{S}^n=\{X\in\mathbb{R}^n: |X|=1\}$. For real $x$, the function $sign(x)$ is defined as
$$
sign(x)=\left\{
\begin{array}{ll}
\frac{x}{|x|}, & \hbox{$x\neq0$;} \\
0, & \hbox{$x=0$.}
\end{array}
\right.
$$
Let $\textbf{x}=(x_1, x_2, \cdots, x_n)$ be a random vector in $\mathbb{R}^n$ and let
\begin{equation}\nonumber
\phi_{\textbf{x}}(\textbf{u})= E\exp(i\langle\textbf{u},\textbf{x}\rangle)= E\exp(i\displaystyle\sum_{i=1}^{n}u_ix_i),
\end{equation}
denote its Characteristic Function (hereafter CF). $\phi_{\textbf{x}}(\textbf{u})$ is also called the joint CF of the random variables $x_1, x_2, \cdots, x_n$. For any set as $ B $, $ \textbf{1}_{B} $ ia an indicator function, which $ \textbf{1}_{B}=1 $ for $ x\in B $ and $ \textbf{1}_{B}=0 $ for $x\neq B $. Furthermore, $ =^d $ means the both sides have the same distribution, and $ a\wedge b=\min\left(a,b \right) $.
In general, there are two different definitions for elliptical distributions that are equivalent:\footnote{See, [13].} (1) definition based on the stochastic representation, and; (2) definition based on the CF. The former is explained by four components: a vector of locations, a nonnegative random variable as ${\mathcal R}_{\alpha}$, a k-dimensional random vector uniformly distributed on ${{\mathbb S}}^{n{\rm -}{\rm 1}}$ as $U^{(k)}\ $ that stochastically independent of ${\mathcal R}_{\alpha}$, and a matrix as ${{\mathbf \Lambda }\in {\mathbb R}}^{n\times k}$.\footnote{The rank of matrix $ \boldsymbol{\Lambda} $ is equal $ k $.} In contrast, the definition based on CF has three components: a vector of locations, a dispersion matrix that reproduces the elliptical, and a density generator that controls the tail thickness. The connection between these two definitions backs to this fact that every affinely transformed spherical random vector is elliptically distributed.
Numerous distributions that are relevant for theoretical and practical works can be easily defined based on the stochastic representation definition of elliptical distributions: Gaussian, Laplace, Student-t, elliptical stable distribution (and hence Cauchy with $\alpha=1 $), and Kotz among others. For example, Gaussian $N({\boldsymbol\mu },{\rm \ }{\boldsymbol\Sigma })$ and Laplace ${\rm L(}{\boldsymbol\mu }{\rm ,\ }\lambda {\rm ,}{\mathbf \ }{\boldsymbol\Sigma }{\rm )}$ distributions are obtained if ${\mathcal R}_{\alpha}{\rm =}\sqrt{{\chi }^2_n}\ $ and ${\mathcal R}_{\alpha}{\rm =}\sqrt{{\chi }^2_n}\times \sqrt{\varepsilon (\lambda )}$, respectively where $\varepsilon (\lambda )$ is an exponential random variable with parameter $\lambda $ and stochastically independent of ${\chi }^2_n$.\footnote{It should be noted that the parameter $ \alpha $ in random variable $ {\mathcal R}_{\alpha} $ controls the tail thickness property.}
Before proceeding, recall that there is a L\'{e}vy measure related to the stable and tempered stable random variables, because the both of them are an IDD. Also, since tempered stable random variable has been obtained via tilting an IDD (stable distribution) and on the other hand tilting an IDD density leads to tilting of corresponding L\'{e}vy measure. So, the L\'{e}vy measure of the tempered stable distribution has been obtained via tilting the L\'{e}vy measure of a stable distribution. It should be noted that each IDD has been uniquely defined by the triple that named the \textit{L\'{e}vy triple}. Also, the \textit{L\'{e}vy-Khinchin} representation of the CF of the IDD which is based on the the L\'{e}vy triple, is very useful in study the theoretical properties of the stable random vectors and specially, more valuable, for different kinds of the tempered stable random vectors.\footnote{There are two major definitions for multivariate tempered stable distributions: (1) Rosinski's \cite{Rosinski01} multivariate tempered stable, and; (2) Bianchi et al. \cite{Kim02} multivariate tempered stable which known as Tempered Infinite Divisible (TID). }
In addition, we will see that the ETSD is a subclass of the symmetric Multivariate Normal Tempered Stable (MNTS) distribution.\footnote{MNTS distribution is obtained by a simple extension of normal tempered stable distribution. If the MNTS is symmetric, then the related random vector have the stochastic representation of elliptical tempered stable distribution. On the other hands, each symmetric MNTS distribution is a TID distribution, so we can reach the CF of symmetric MNTS distribution based on the spectral measure if we be able to reach the CF of symmetric TID distribution, see \cite{Kim01}.} And on the other hands, a symmetric MNTS is a subclass of the TID. So, we will continue this section with the structure of the L\'{e}vy measure of the TID, which plays important role in our paper. But first, recall the structure of the IDD.
The CF (\textit{L\'{e}vy-Khinchin} representation ) of an IDD $ \mu $ on $ \mathbb{R}^n $ can be written as $ \Phi(\textbf{u})=\exp\left( \Psi(\textbf{u})\right) $ where
\begin{equation}
\Psi(\textbf{u})=-\dfrac{1}{2}\left\langle \textbf{u},\textbf{A}\textbf{u}\right\rangle +i\left\langle \textbf{b}, \textbf{u}\right\rangle +\int_{\mathbb{R}^n}\left( e^{i\left\langle \textbf{u}, \textbf{x}\right\rangle }-1-i\dfrac{\left\langle \textbf{u}, \textbf{x}\right\rangle }{1+|\textbf{x}|^2} \right) M(dx),
\end{equation}
$ \textbf{A} $ is a symmetric nonnegative $ n\times n $ matrix, $ \textbf{b}\in\mathbb{R}^n $, and $ M $ satisfies
\begin{equation}\nonumber
\int_{\mathbb{R}^n}\left(\parallel x\parallel^2\wedge 1 \right)M(dx)<\infty, \qquad M({0})=0.
\end{equation}
The measure $ \mu $ uniquely defined by the L\'{e}vy triple $ (\textbf{A, \textbf{b, M}}) $, and we write $ \mu\sim IDD(\textbf{A, \textbf{b, M}} )$.
We will finish this section via the definition of the TID based on the Levy measure.
Based on the Rosinski's definition for multivariate tempered stable distribution ( Definition 2.1 of \cite{Rosinski01}), the L\'{e}vy measure of a multivariate tempered stable random vector without the Gaussian part in the polar coordinates is
\begin{equation}\label{eq03}
M(dr,du)=r^{-\alpha-1}q(r,u)dr\sigma(du),
\end{equation}
where $ q:(0,\infty)\times \mathbb{S}^{n-1}\longmapsto (0,\infty) $ is a Borel function such that $ q(\cdot,u) $ is completely monotone with $ q(\infty,u)=0 $ for each $ u\in\mathbb{S}^{n-1} $, and known as \textit{tempering function}. Likewise, $ \sigma $ is a finite measure on $ \mathbb{S}^{n-1} $ and $ 0<\alpha<2 $. On the other hands, the L\'{e}vy measure of a stable distribution on $ \mathbb{R}^n $ in polar coordinate is
\begin{equation}
M_0(dr,du)=r^{-\alpha-1}dr\sigma(du).
\end{equation}
A tempered stable distribution is characterized by the spectral measure as defined in Definition 2.4 of \cite{Rosinski01}, which based on Theorem 2.3 of \cite{Rosinski01} the L\'{e}vy measure of the tempered stable distributions can be written in the form
\begin{equation}\label{equ04}
M(A)=\int_{\mathbb{R}^n}\int_{0}^{\infty}I_A(tx)t^{-\alpha-1}e^{-t}dtR(dx),\quad A\in\textbf{B}(\mathbb{R}^n),
\end{equation}
where $ R $ is the spectral measure of the tempered stable distribution on $ \mathbb{R}^n $ such that
\begin{equation}\nonumber
\int_{\mathbb{R}^n}\left(\parallel x\parallel^2\wedge \parallel x\parallel^{\alpha} \right)R(dx)<\infty, \qquad R({0})=0.
\end{equation}
Now, let $ \left\lbrace Q(\cdot,u)\right\rbrace_{u\in S^{n-1}} $ be a measurable family of Borel measures on $ (0,\infty) $, then the tempering function $ q $ in (\ref{eq03}) can be represented as\footnote{Based on the structure of this function different kinds of distribution in the multivariate can be obtained. In fact, all of them (i.e., $ q(r,u) $ and $ q(r^2,u) $ are a subclass of Grabchak which the $ q $ is $ q(r^p,u)$ and $ p>0 $ (See, \cite{Grabchak}).}
\begin{equation}
q(r,u)=\int_{0}^{\infty}e^{-rs}Q(ds|u).
\end{equation}
Also, let measure $ Q $ be as
\begin{equation}
Q(A)=\int_{\mathbb{S}^{n-1}}\int_{0}^{\infty}I_A(ru)Q(dr|u)\sigma(du),\quad A\in\textbf{B}(\mathbb{R}^n),
\end{equation}
then $ Q({0})=0 $. Now, by defining measure $ R $ as
\begin{equation}\label{eq04}
R(A)=\int_{\mathbb{R}^{n}}I_A(\frac{x}{\parallel x\parallel^2})\parallel x\parallel^{\alpha}Q(dx),\quad A\in\textbf{B}(\mathbb{R}^n),
\end{equation}
we can see that measure $ R $ in equation (\ref{eq04}) satisfies in the conditions of the spectral measure of a tempered stable distribution.\footnote{See Theorem 2.3 \cite{Rosinski01}.}
Bianchi et al. \cite{Kim02} based on the Rosinski's approach in \cite{Rosinski01} define another class of multivariate tempered stable random vector, known as TID. In fact, they are modify the radial component of $ M_0 $ and obtain a probability distribution with lighter tails than stable ones.
\begin{definition}(TID measure, see \cite{Kim02})
Let $ \mu $ be an IDD measure on $ \mathbb{R}^n $ without Gaussian part (\textbf{A}=0). $ \mu $ is called TID if its L\'{e}vy measure $ M$ can be written in the polar coordinates as equation (\ref{eq03}), where $ \alpha $ is a real number, $ \alpha\in[0,2) $, $ \sigma $ is a finite measure on $ \mathbb{S}^{n-1} $, and $ q:(0,\infty)\times \mathbb{S}^{n-1}\longmapsto (0,\infty) $ is a Borel function as
\begin{equation}
q(r,u)=\int_{0}^{\infty}e^{-r^2s^2}Q(ds|u),
\end{equation}
with $ \left\lbrace Q(\cdot,u)\right\rbrace_{u\in S^{n-1}} $ be a measurable family of Borel measures on $ (0,\infty) $.
\end{definition}
\begin{remark}
The difference between the Rosinski and TID definition is just related to the tempering function ($ q(r,u) $).
\end{remark}
\begin{definition}(CF of TID, see \cite{Kim02})\label{TID:CF01}
Let $ \mu $ be a TID distribution with L\'{e}vy measure given by \ref{equ04}, $ \alpha\in[0,2) $, and $ \alpha\neq1 $. If the distribution has finite mean, i.e., $ \int_{\mathbb{R}^n}\parallel \textbf{x}\parallel \mu(dx)<\infty$, then
\begin{equation}\label{equ07}
\Phi_{\textbf{X}}(\textbf{u})=E\exp \left\lbrace i\left\langle \textbf{u},\textbf{X}\right\rangle \right\rbrace =\exp\left\lbrace \int_{\mathbb{R}^n}\psi_{\alpha}(\left\langle \textbf{u},\textbf{X}\right\rangle )R(dx) +i\left\langle \textbf{u},\textbf{m} \right\rangle \right\rbrace,
\end{equation}
where
\begin{eqnarray}
\nonumber\psi_{\alpha}(s)=2^{-\alpha/2-1}\left[ \Gamma\left( -\dfrac{\alpha}{2}\right) \left( _1F_1\left( -\dfrac{\alpha}{2},\dfrac{1}{2};\left( \dfrac{i\sqrt{2}s}{2}\right)^2 \right)-1 \right)\right.\\
\left.+i\sqrt{2}s\Gamma\left( \dfrac{1-\alpha}{2}\right) \left( _1F_1\left( \dfrac{1}{2}-\dfrac{\alpha}{2},\dfrac{3}{2};\left( \dfrac{i\sqrt{2}s}{2}\right)^2 \right)-1 \right) \right],
\end{eqnarray}
and $ \textbf{m}=\int_{\mathbb{R}^n} x\zeta(dx)$, and $ _1F_1 $ is the \textit{Kummer or confluent hypergeometric function of first kind}. Furthermore, if $ 0<\alpha<1 $, then the CF can be written in the alternative form
\begin{equation}
\Phi_{\textbf{X}}(\textbf{u})=E\exp \left\lbrace i\left\langle \textbf{u},\textbf{X}\right\rangle \right\rbrace =\exp\left\lbrace \int_{\mathbb{R}^n}\psi_{\alpha}^0(\left\langle \textbf{u},\textbf{X}\right\rangle )R(dx) +i\left\langle \textbf{u},\textbf{m}_0 \right\rangle \right\rbrace,
\end{equation}
where
\begin{eqnarray}
\nonumber\psi_{\alpha}(s)=2^{-\alpha/2-1}\left[ \Gamma\left( -\dfrac{\alpha}{2}\right) \left( _1F_1\left( -\dfrac{\alpha}{2},\dfrac{1}{2};\left( \dfrac{i\sqrt{2}s}{2}\right)^2 \right)-1 \right)\right.\\
\left.+i\sqrt{2}s\Gamma\left( \dfrac{1-\alpha}{2}\right) \left( _1F_1\left( \dfrac{1}{2}-\dfrac{\alpha}{2},\dfrac{3}{2};\left( \dfrac{i\sqrt{2}s}{2}\right)^2 \right) \right) \right].
\end{eqnarray}
\end{definition}
If random vector $ \textbf{X} $ be an TID, it will be given by $ \textbf{X}\sim TID_{\alpha}(R,\textbf{m}) $ and in the alternative form by $\textbf{X}\sim TID_{\alpha}^0(R,\textbf{m}_0) $ where $ \alpha\in(0,2) $ and $ \alpha\in(0,1) $, respectively. It should be noted that the spectral measure of the tempered stable distribution, $ R(dx) $, is unique.
\section{Elliptical Tempered Stable Distribution and Fractional PDE}
In this section, we provide the definition of the ETSD and then the link between fractional PDE and TID are presented. First, we provide the CF of the symmetric TID random vector.
\begin{lemma}\label{pro01}
$ \textbf{X}$ is symmetric TID random vector in $ \mathbb{R}^n $ with $ 0<\alpha<2 $ if and only if there exist a unique spectral finite measure such that
\begin{equation}\label{equ05}
E\exp \left\lbrace i\left\langle \textbf{u},\textbf{X}\right\rangle \right\rbrace =\exp\left\lbrace 2^{-\alpha/2-1}\Gamma(-\dfrac{\alpha}{2}) \int_{\mathbb{R}^n} \left( _1F_1\left( -\dfrac{\alpha}{2},\dfrac{1}{2};\dfrac{-1}{2} \left\langle \textbf{u},\textbf{X}\right\rangle^2 \right)-1 \right) R(dx) \right\rbrace.
\end{equation}
\end{lemma}
\begin{proof}
The proof is straightforward. It comes from this fact that a necessary and sufficient condition for an arbitrary random vector $ \textbf{X}$ to be symmetric is that its CF be real. And also, by defining a symmetric measure as $ \frac{1}{2}\left(R(dx)+R(-dx) \right) $ and replacing with measure $ R(dx) $ the appropriate result is obtained.
\end{proof}
\begin{remark}
In Lemma \ref{pro01}, random vector $ \textbf{X} $ is symmetric about zero. One can easily extend it to any point as \textbf{m} of $ \mathbb{R}^n $, which it is more convenience for the elliptical distributions.
\end{remark}
The following definition is a direct corollary of Lemma \ref{pro01}. Since, a random vector ${\mathbf X}$ is elliptically distributed if and only if there exist a vector $\textbf{m}$, a symmetric positive define matrix ${\mathbf \Sigma }$ and a function $\varphi {\rm :}{{\mathbb R}}^{{\rm +}}\mapsto {\mathbb R}$ such that the CF of ${\mathbf X}$ is of the form\footnote{See, \cite{Frahm}.}
\begin{equation}\label{equ01}
\Phi_\textbf{X}\left({\textbf{u}}\right){\rm =exp}\left(i\left\langle \textbf{m}, \textbf{u} \right\rangle\right){\rm \times }\varphi {\rm (}\textbf{u}^T{\mathbf \Sigma }\textbf{u}{\rm ).}
\end{equation}
So, if random vector $ \textbf{X} $ be a symmetric TID, then the CF of symmetric TID (equation (\ref{equ05})) can be shown as equation (\ref{equ01}), where the symmetric positive define matrix ${\mathbf \Sigma }$ can be extracted from the unique spectral measure $R{\rm (}dx{\rm )}$.
It should be noted that the extracting the symmetric positive define matrix ${\mathbf \Sigma }$, due to the complicated form of spectral measure $R{\rm (}dx{\rm )}$, is difficult. But, since each symmetric MNTS distribution is a subclass of the TID distribution, which it has the CF as Definition \ref{TID:CF01},\footnote{See \cite{Kim02}.} and also, every symmetric MNTS distribution is an elliptical distribution,\footnote{See \cite{Rachev03}.} the existence of this matrix in the mentioned form can be proved.
\begin{definition}\label{def05}
A random vector $ \textbf{X} $ is said an elliptical tempered stable random vector if there exist a unique measure such that the CF of it be as equation (\ref{equ05}).\footnote{It should be noted, due to the uniqueness of the spectral measure, Definition \ref{def05} is well defined.}
\end{definition}
Before explaining the connection between tempered stable random vectors and fractional PDE, some theoretical properties of ETSD are given. The symmetric MNTS distribution, which is a subclass of TID distribution, with the CF as Definition \ref{TID:CF01}, is come from a stochastic process. By setting the time variable equal 1, one can get this distribution (symmetric MNTS) from related stochastic process.\footnote{See, \cite{Kim01}.} Also, since the stable and tempered stable distributions don't have the closed form formula for PDF and CDF, the structure of their CF plays important role in the practical area. On the other hands, since the stable and tempered stable distribution are a subclass of IDD, and also based on the \textit{L\`{e}vy-Khinchine} formula the CF of the IDD can be obtained. Therefore, the L\`{e}vy-Khinchine formula provides a very useful tool for studying theoretical properties of these distributions. Also, because the uniqueness of the CF of a distribution, we check some properties of ETSD based the CF of symmetric MNTS which is exist in the suitable form.
The CF of the symmetric MNTS process can be obtained from related multivariate Lévy-Khinchine formula.\footnote{See, \cite{Kim01}} The CF of the symmetric MNTS process is given by
\begin{equation} \label{GrindEQ__43_}
{\Psi }_X{\rm (}u{\rm )=exp}\left[t{\rm \times }\left\{\frac{{\rm 2}\lambda }{\alpha }\left({\rm 1-}{\left({\rm 1-}\frac{{\rm 1}}{\lambda }\left({\rm -}\frac{{\rm 1}}{{\rm 2}}u^T\Sigma u\right)\right)}^{\frac{\alpha }{{\rm 2}}}\right){\rm +}i{\mu }^Tu\right\}\right],
\end{equation}
so, by replacing $t{\rm =1}$ in equation \eqref{GrindEQ__43_} we will get the CF of the ETSD as follow
\begin{equation} \label{equ06}
{\Psi }_X{\rm (}u{\rm )=exp}\left[\frac{{\rm 2}\lambda }{\alpha }\left({\rm 1-}{\left({\rm 1-}\frac{{\rm 1}}{\lambda }\left({\rm -}\frac{{\rm 1}}{{\rm 2}}u^T\Sigma u\right)\right)}^{\frac{\alpha }{{\rm 2}}}\right){\rm +}i{\mu }^Tu\right].
\end{equation}
\begin{remark}
As mentioned before, both definitions of elliptical distribution are equivalent. So, from equation (\ref{equ06}) we will be able to reach to the stochastid representation definition of the elliptical distribution, which In the next theorem, we will used this definition.
\end{remark}
Corollary \ref{The06} is the answer of this question that is there any connection between the PDF of an ETS random vector with related Gaussian underlying vector? If the answer is yes, the next question is that what is this connection? Corollary \ref{The06} explains more details in this regard. But, first the definition of a subordinator is need.
Univariate L\'{e}vy processes $ S_t, t\geq0 $ on $ \mathbb{R} $ with almost surely non-decreasing trajectories are called \textit{subordinators}. Now, let $ \alpha\in(0,2) $ and $ \theta>0 $. The purely non-Gaussian infinite divisible random variable $ T $ whose CF is given by
\begin{equation*}
\phi_N(u)=\exp\bigg( -\dfrac{2\theta^{1-\frac{\alpha}{2}}}{\alpha}((\theta-iu)^{\frac{\alpha}{2}}-\theta^{\frac{\alpha}{2}})\bigg),
\end{equation*}
is referred to as the tempered stable subordinator with parameter $ (\alpha,\theta) $.
\begin{corollary}\label{The06}
Let \textbf{X} be an ETSD as
\begin{equation}
\textbf{X}=^d\boldsymbol{\mu}+\textbf{C}\sqrt{T}\textbf{N},
\end{equation}
where ${\boldsymbol\mu }{\rm =(}{\mu }_{{\rm 1}},{\mu }_{{\rm 2}}{\rm ,}\cdots {\rm ,}{\mu }_n{\rm )}\in {{\mathbb R}}^n,{\mathbf C}{\rm =(}C_{{\rm 1}},C_{{\rm 2}}{\rm ,}\cdots {\rm ,}C_n{\rm )}\in {\rm (0,}\infty {{\rm )}}^n$ and $\textbf{N}=(N_1,N_2,\cdots,N_n)$ be a n-dimensional standard normal distributed with covariance matrix ${\left\{{\rho }_{r,s}\right\}}^n_{r,s{\rm =1}}$. And also, $T$ is the tempered stable subordinator and independent of ${\mathbf N}$. Then there is a one-to-one correspondence between the PDF of \textbf{N} and that of \textbf{X}.
\end{corollary}
\begin{proof}
Based on equation (\ref{equ06}), we will have
\begin{equation}\label{equ16}
\Psi_{X}(u) =\exp\left[ \dfrac{2\lambda}{\alpha}\left(1-\left(1-\dfrac{1}{\lambda}\left(-\dfrac{1}{2}\sum_{i=1}^{n}\sum_{j=1}^{n}u_iu_j\rho_{i,j} \right) \right)^{\frac{\alpha}{2}} \right) +i\mu^T u \right],
\end{equation}
where $ \rho_{i,j}=Cov(N_i,N_j) $.
Now, let $ \textbf{N}' $ be a standard normal vector on $ \mathbb{R}^n $ with covariance matrix $ \rho' $. We must show that if $ \textbf{X}=\boldsymbol\mu +C\sqrt{T}\textbf{N} $ has the same distribution as $ \textbf{X}'=\boldsymbol\mu +C\sqrt{T}\textbf{N}'$, then $ \textbf{N}' $ has the same distribution as $ \textbf{N} $. If two distributions be same, then the CF of them are same. So, from Equation (\ref{equ16}) we will have
\begin{equation}\nonumber
\nonumber\Psi_{\textbf{X}}(u) =\Psi_{\textbf{X}'}(u),
\end{equation}
or
\begin{eqnarray}\nonumber
\nonumber\exp\left[ \dfrac{2\lambda}{\alpha}\left(1-\left(1-\dfrac{1}{\lambda}\left(-\dfrac{1}{2}\sum_{i=1}^{n}\sum_{j=1}^{n}u_iu_j\rho_{i,j} \right) \right)^{\frac{\alpha}{2}} \right) +i\mu^T u \right]\\
\nonumber=\\
\nonumber\exp\left[ \dfrac{2\lambda}{\alpha}\left(1-\left(1-\dfrac{1}{\lambda}\left(-\dfrac{1}{2}\sum_{i=1}^{n}\sum_{j=1}^{n}u_iu_j\rho_{i,j}' \right) \right)^{\frac{\alpha}{2}} \right) +i\mu^T u \right],
\end{eqnarray}
therefore
\begin{equation}\nonumber
\sum_{i=1}^{n}\sum_{j=1}^{n}u_iu_j\rho_{i,j}=\sum_{i=1}^{n}\sum_{j=1}^{n}u_iu_j\rho_{i,j}',
\end{equation}
for any real number $ u_1,u_2,\cdots,u_n $. This identity between two polynomials leads to
\begin{equation}\nonumber
\rho_{i,j}=\rho_{i,j}'.
\end{equation}
Hence $ \textbf{N}=^d\textbf{N}' $.
\end{proof}
The next question is that what is the relation between two different ETS random vectors? Corollary \ref{The02} explains the relation between two different ETSD random vectors which the underlying vectors one of them is standard Gaussian and the other is any arbitrary Gaussian distribution.
\begin{corollary}\label{The02}
Let \textbf{X} be a centered elliptical tempered stable random vector in $ \mathbb{R}^n $ with standard normal underlying vector \textbf{N}.\footnote{The mean of centered random vector is that its mean is zero.} Then for any centered elliptical tempered stable random vector $ \textbf{X}' $ in $ \mathbb{R}^n $ there is a lower-triangular $n\times n $ matrix $ \Delta $ such that
\begin{equation}\nonumber
\textbf{X}'=^d\Delta\textbf{X}.
\end{equation}
\end{corollary}
\begin{proof}
Let $ S_{t} $ be a tempered stable subordinator.\footnote{See \cite{Kim01}} An \textit{elliptical tempered stable process} $ \textbf{X}_{t} $ is defined by subordinating a n-dimensional multivariate Brownian motion $ \textbf{B}_{s} $ and tempered stable subordinator $ S_{t} $. \footnote{Since the ETS distribution is a subclass (symmetric) of MNTS distribution, so the definition of the elliptical tempered stable process is well defined.} Therefore, the centered elliptical tempered stable process $ \textbf{X}_{t} $ is defined
as
\begin{equation}
\nonumber \textbf{X}_{t}=\textbf{B}_{S_{t}},
\end{equation}
where $ \textbf{B}_{s} $ denotes a centered Brownian motion with covariance matrix $ \boldsymbol{\Sigma} $. Since $ \textbf{B}_{s} $ can be degenerate as $ chol(\boldsymbol{\Sigma}) \textbf{W}_{s}$, we have
\begin{equation}
\nonumber\textbf{X}_{t}=chol(\boldsymbol{\Sigma}) \textbf{W}_{s},
\end{equation}
where $ \textbf{W}_{t} $ is a n-dimensional process with each dimension being an independent standard Wiener process, and $ chol(\boldsymbol{\Sigma}) $ is lower triangle Cholesky factor. On the other hands, the covariance matrix $ \boldsymbol{\Sigma} $ can be decomposed into its vector of standard deviations $ \sigma $ and correlation matrix \textbf{P} , i.e.,
\begin{equation}
\nonumber\boldsymbol{\Sigma}=diag(\sigma)\textbf{P}diag(\sigma),
\end{equation}
therefore,
\begin{equation}\label{equ03}
\textbf{X}_{t}=diag(\sigma)chol(\textbf{P}) \textbf{W}_{s},
\end{equation}
Now by setting $ t=1 $ in equation (\ref{equ03}), the favorite has been obtained.
\end{proof}
Now, we explain the link between fractional PDE, TID distribution, and particularly ETS distribution. In fact, we will introduce an initial value problem that the fundamental solution of it gives the PDF of TID distributions and particularly ETS distribution.
\begin{theorem}\label{The01}
Let $\textbf{X}_t$ be the position of a particle at time $t>0$, and $n-$dimentional Euclidean space $\mathbb{R}^n$. Let $P(\textbf{X},t)$ denote the density of $X_t$ where the vector $\textbf{X}=(X_1,X_2,\cdots,X_n)\in\mathbb{R}^n$. Then the fundamental solution of the fractional PDE as equation (\ref{equ30}) gives the PDF of the TID distribution
\begin{eqnarray}\label{equ30}
\frac{\partial P(\textbf{X},t)}{\partial t}=i\langle\textbf{m},\nabla P(\textbf{X},t)\rangle+\nabla_{R}^{\alpha}P(\textbf{X},t),
\end{eqnarray}
where $\textbf{m}\in\mathbb{R}^n$, $\nabla=(\frac{\partial}{\partial X_1},\frac{\partial}{\partial X_2},\cdots,\frac{\partial}{\partial X_n})$, and
\begin{equation}
\nabla_{R}^{\alpha}f(\textbf{X})=F^{-1}\left( \left[ \int_{\mathbb{R}^n}\psi_{\alpha}(\left\langle \textbf{u},\textbf{X}\right\rangle )R(dx)\right] \hat{f}(\textbf{u}) \right),
\end{equation}
where $ R $ and $ \psi $ are as the same as Definition \ref{TID:CF01}. And also, $ \hat{f}(\textbf{u}) $ and $ F^{-1} $ are the Fourier transform of $ f $ and inverse of Fourier transform, respectively. Furthermore, the initial condition of equation (\ref{equ30}) is $P(X_0=0)=1$.
\end{theorem}
\begin{proof}
Before we start the proof, let $ f(\textbf{X},t):\mathbb{R}^n\times\mathbb{R}^+\longrightarrow\mathbb{C} $ be a function of $ \textbf{X}\in\mathbb{R}^n $ and $ t\in\mathbb{R}^+ $ such that $ f\in L^1(\mathbb{R}^n) $. The Fourier and inverse of Fourier transform will be given as
\begin{eqnarray}
\hat{f}(\textbf{u})=F(f(\textbf{X}))=\int\exp(-i\left\langle\textbf{X},\textbf{u} \right\rangle )f(\textbf{X})dX,\\
f(\textbf{X})=F^{-1}(\hat{f}(\textbf{u}))=\dfrac{1}{(2\pi)^n}\int\exp(i\left\langle\textbf{X},\textbf{u} \right\rangle )f(\textbf{u})dX,
\end{eqnarray}
The strategy for proof is the same as procedure was explained by Fallahgoul \textit{et al.}\footnote{It should be noted that some other fractional PDEs for multivariate stable and geometric stable distributions have been introduced. The difference between all kinds of these fractional PDEs backs to the structure of the L\`{e}vy measure of each of them. More information regarding univarite and multivarite fractional PDE regradless stable and geometric stable distribution can be found in \cite{Fallahgoul01, Fallahgoul02}.}
Taking the Fourier transform of equation (\ref{equ30}) and Fourier transform properties, we obtain
\begin{eqnarray}\label{equ31}
\frac{\partial \hat{P}(\textbf{u},t)}{\partial t}=i\langle\textbf{m},\textbf{u} \rangle\hat{P}(\textbf{u},t)+ \left[ \int_{\mathbb{R}^n}\psi_{\alpha}(\left\langle \textbf{u},\textbf{X}\right\rangle )R(dx)\right] \hat{P}(\textbf{u},t) ,
\end{eqnarray}
where $ \hat{P}(\textbf{u},t) $ is the Fourier transform of $ P(\textbf{X},t) $ with respect to $ \textbf{X} $. It should be noted that the initial condition $P(\textbf{X},0)= \delta(\textbf{X}) $ also converts to $ \hat{P}(\textbf{u},t)=1$. Now, equation (\ref{equ31}) is an initial value problem which all coefficient of it is constant, so we will have
\begin{equation}\label{equ32}
\hat{P}(\textbf{u},t)=\exp\left(i\langle\textbf{m},\textbf{u}t \rangle+t\int_{\mathbb{R}^n}\psi_{\alpha}(\left\langle \textbf{u},\textbf{X}\right\rangle )R(dx) \right).
\end{equation}
Comparing equation (\ref{equ32}) and the CF of the TID distribution (equation (\ref{equ07})), we find that they are identical for a tempered stable distribution with $ \alpha\in(0,2) $. Therefore, the Green function solution of equation (\ref{equ30}) yields the entire PDF classes of the TID distribution.
\end{proof}
The following corollary are the direct results of Theorem \ref{The01}.
\begin{corollary}
Let the operator $ \nabla_{R}^{\alpha} $ be
\begin{equation}
\nabla_{R}^{\alpha}f(\textbf{X})=F^{-1}\left( \left[ \int_{\mathbb{R}^n}\psi_{\alpha}^0(\left\langle \textbf{u},\textbf{X}\right\rangle )R(dx)\right] \hat{f}(\textbf{u}) \right),
\end{equation}
in Theorem (\ref{The01}), and also $ \alpha\in(0,1) $. Then the Green function solution of equation (\ref{equ30}) yields the entire PDF classes of the TID distribution.
\end{corollary}
\begin{corollary}\label{Cor:ETSD}
Let the operator $ \nabla_{R}^{\alpha} $ be
\begin{equation}
\nabla_{R}^{\alpha}f(\textbf{X})=F^{-1}\left( \left[ 2^{-\alpha/2-1}\Gamma(-\dfrac{\alpha}{2}) \int_{\mathbb{R}^n} \left( _1F_1\left( -\dfrac{\alpha}{2},\dfrac{1}{2};\dfrac{-1}{2} \left\langle \textbf{u},\textbf{X}\right\rangle^2 \right)-1 \right) R(dx)\right] \hat{f}(\textbf{u}) \right),
\end{equation}
in Theorem (\ref{The01}), where $ \alpha\in(0,2)=\frac{p}{q} $. Then the fundamental solution $ \hat{P}(\textbf{X},t) $ of equation
\begin{eqnarray}\label{equ33}
\frac{\partial P(\textbf{X},t)}{\partial t}=\nabla_{R}^{\alpha}P(\textbf{X},t),
\end{eqnarray}
with the initial condition $ P(\textbf{X},0)=\delta(\textbf{X}) $ is the density of the ETSD.
\end{corollary}
\begin{remark}
In Corollary \ref{Cor:ETSD}, the location parameter is zero. But, based on the definition of symmetric TID about any real vector it can ba changed. In fact, in order to have the full form of stochastis representation definition of elliptical distribution, the location vector must be non-zero.
\end{remark}
\section{Analytic Approximation of Solution of Fractional PDE}
In this section, we derive the PDF approximation of multivariate tempered stable distributions by using the Homotopy Perturbation Method (hereafter HPM), Adomian Decomposition Method (hereafter ADM), and Variational Iteration Method (hereafter VIM).\footnote{See, for example \cite{He01}, \cite{Abdelrazec}, and \cite{He02}, among others.} It should be noted that the same procedure for getting the approximation of the PDF of fractional PDE regardless the ETS distribution can be applied.\footnote{It should be noted the details of computations have been omitted, and the reader can refer to \cite{Fallahgoul02} and \cite{Fallahgoul01} for more details. }
\subsection{PDF Approximation of the TID Distribution Using the HPM}
To solve equation \ref{equ30} with initial condition $P{\rm (}X{\rm (0)=0)=1}$ using the HPM, we construct the following homotopy
\begin{equation} \label{GrindEQ__5_1_}
H{\rm (}p,V{\rm )=(1-}p{\rm )}\left(\frac{\partial V}{\partial t}{\rm -}\frac{\partial P_0}{\partial t}\right){\rm +}p\left(\frac{\partial V}{\partial t}{\rm +}<\textbf{m}{\rm ,}\nabla V>{\rm -}{\nabla }^{\alpha }_R\right){\rm =0.}
\end{equation}
Suppose the solution of equation \eqref{equ30} has the form
\begin{equation} \label{GrindEQ__5_2_}
V{\rm (}\textbf{X},t{\rm )=}p^0V_0{\rm +}p^{{\rm 1}}V_{{\rm 1}}{\rm +}p^{{\rm 2}}V_{{\rm 2}}{\rm +}\cdots {\rm +}p^{n{\rm -}{\rm 1}}V_{n{\rm -}{\rm 1}}{\rm +}\cdots {\rm .}
\end{equation}
Substituting equation \eqref{GrindEQ__5_2_} into equation \eqref{GrindEQ__5_1_}, and comparing coefficients of terms with identical powers of $p$, leads to
\[p^0{\rm :\ \ \ \ \ \ \ \ }\frac{\partial V_0}{\partial t}{\rm -}\frac{\partial P_0}{\partial t}{\rm =0,}\]
\[p^{{\rm 1}}{\rm :\ \ \ \ \ \ \ \ }\frac{\partial V_{{\rm 1}}}{\partial t}{\rm =}<\textbf{m}{\rm ,}\nabla V_0>{\rm +}{\nabla }^{\alpha }_RV_0,\]
\[\vdots \]
\[p^{n{\rm +1}}{\rm :\ \ \ \ \ \ \ \ }\frac{\partial V_{n{\rm +1}}}{\partial t}{\rm =}<\textbf{m}{\rm ,}\nabla V_n>{\rm +}{\nabla }^{\alpha }_RV_n.\]
For simplicity, we take $V_0{\rm (}\textbf{X},t{\rm )=}P_0{\rm (}\textbf{X},t{\rm )}$. According to compared coefficients, we derive the following recurrent relation
\[V_0{\rm (}\textbf{X},t{\rm )=}P_0{\rm (}\textbf{X},t{\rm ),}\]
\[V_{{\rm 1}}{\rm (}\textbf{X},t{\rm )=}\int^t_0{}\left({\rm }<\textbf{m}{\rm ,}\nabla V_0>{\rm +}{\nabla }^{\alpha }_RV_0\right)dt{\rm ,\ \ \ \ \ \ \ \ }V_{{\rm 1}}{\rm (}\textbf{X}{\rm ,0)=0,}\]
\[\vdots \]
\[V_{n{\rm +1}}{\rm (}\textbf{X},t{\rm )=}\int^t_0{}\left({\rm }<\textbf{m}{\rm ,}\nabla V_n>{\rm +}{\nabla }^{\alpha }_RV_n\right)dt{\rm ,\ \ \ \ \ \ \ \ }V_{n{\rm +1}}{\rm (}\textbf{X}{\rm ,0)=0.}\]
The solution is
\[P\left(\textbf{X},t\right){\rm =}\mathop{{\rm lim}}_{p\to {\rm 1}}V\left(\textbf{X},t\right)\]
\[{\rm =}V_0{\rm (}\textbf{X},t{\rm )+}V_{{\rm 1}}{\rm (}\textbf{X},t{\rm )+}V_{{\rm 2}}{\rm (}\textbf{X},t{\rm )+}\cdots {\rm +}V_{n{\rm +1}}{\rm (}\textbf{X},t{\rm )+}\cdots {\rm ,}\]
\[{\rm =}\boldsymbol{\delta}(\textbf{X}) {\rm +}\int^t_0{}\left({\rm }<\textbf{m}{\rm ,}\nabla V_0>{\rm +}{\nabla }^{\alpha }_RV_0\right)dt{\rm +}\int^t_0{}\left({\rm }<\textbf{m}{\rm ,}\nabla V_{{\rm 1}}>{\rm +}{\nabla }^{\alpha }_RV_{{\rm 1}}\right)dt{\rm +}\cdots \]
\[{\rm +}\int^t_0{}\left({\rm }<\textbf{m}{\rm ,}\nabla V_n>{\rm +}{\nabla }^{\alpha }_RV_n\right)dt{\rm +}\cdots {\rm .}\]
Therefore,
\[P{\rm (}\textbf{X},t{\rm )=}\boldsymbol{\delta}(\textbf{X}){\rm +}\sum^{\infty }_{k{\rm =0}}{}\left[\int^t_0{}\left({\rm }<\textbf{m}{\rm ,}\nabla V_k>{\rm +}{\nabla }^{\alpha }_RV_k\right)dt\right].\]
\subsection{PDF Approximation of TID Distribution Using the ADM}
We will solve equation (\ref{equ30}) with initial condition $P{\rm (}\textbf{X}{\rm (0)=0)=1}$ using the ADM. To do so, we construct the following recurrence relation
\[V_0{\rm (}\textbf{X},t{\rm )=}P_0{\rm (}\textbf{X},t{\rm )=}\boldsymbol{\delta}(\textbf{X}),\]
\[V_{k{\rm +1}}{\rm (}\textbf{X},t{\rm )=}\int^t_0{}\left({\rm }<v{\rm ,}\nabla V_k>{\rm +}{\nabla }^{\alpha }_RV_k\right)dt{\rm ,\ \ \ \ \ \ \ \ }k\ge {\rm 0.}\]
So, the solution is obtained as
\[V_{{\rm 1}}{\rm (}\textbf{X},t{\rm )=}\int^t_0{}\left({\rm -}<v{\rm ,}\nabla V_0>{\rm +}{\nabla }^{\alpha }_RV_0\right)dt,\]
\[\vdots \]
\[V_{n{\rm +1}}\left(\textbf{X},t\right){\rm =}\int^t_0{}\left({\rm }\left\langle v{\rm ,}\nabla V_n\right\rangle {\rm +}{\nabla }^{\alpha }_RV_n\right)dt.\]
\[\vdots \]
Therefore,
\begin{equation}\nonumber
P{\rm (}\textbf{X},t{\rm )=}\sum^{\infty }_{k{\rm =0}}{}V_k{\rm (}\textbf{X},t{\rm )=}\boldsymbol\delta {\rm (}\textbf{X}{\rm )+}\sum^{\infty }_{k{\rm =0}}{}\left[\int^t_0{}\left({\rm }<v{\rm ,}\nabla V_k>{\rm +}{\nabla }^{\alpha }_RV_k\right)dt\right].
\end{equation}
\subsection{PDF Approximation of TID Distribution Using the VIM}
The analytic approximation of the PDF of TID via VIM is considered. In fact, the fundamental solution of equation (\ref{equ30}) with initial condition $P{\rm (}\textbf{X}{\rm (0)=0)=1}$ via VIM is obtained. We set the following recurrence relation
\begin{equation}\label{equ08}
V_{n+1}(\textbf{X},t)=V_{n}(\textbf{X},t)+\lambda\int_{0}^{t}\left( \dfrac{\partial V_n(\textbf{X},s)}{\partial s}+(\left\langle \textbf{m},\nabla V_n(\textbf{X},s)\right\rangle -\nabla_R^{\alpha}V_n(\textbf{X},s)\right) ds,\qquad n=0,1,2,\cdots
\end{equation}
so
\begin{equation*}
\delta V_{n+1}(\textbf{X},t)=\delta V_{n}(\textbf{X},t)+\delta\int_{0}^{t}\lambda\left( \dfrac{\partial P_n(\textbf{X},s)}{\partial s}+(\left\langle \textbf{m},\nabla P_n(\textbf{X},s)\right\rangle -\nabla_R^{\alpha}P_n(\textbf{X},s)\right) ds=0.
\end{equation*}
Based on the stationary condition of equation (\ref{equ08}), i.e. $ \lambda+1=0, \lambda'=0 $, the Lagrange multiplier turns out to be $ \lambda=-1 $.
Now, by substituting $ \lambda=-1 $ in equation (\ref{equ08}), we get the following variational iteration formula
\begin{equation*}
V_{n+1}(\textbf{X},t)=V_{n}(\textbf{X},t)-\int_{0}^{t}\left( \dfrac{\partial V_n(\textbf{X},s)}{\partial s}+(\left\langle \textbf{m},\nabla V_n(\textbf{X},s)\right\rangle -\nabla_R^{\alpha}V_n(\textbf{X},s)\right) ds,\qquad n=0,1,2,\cdots
\end{equation*}
where $ V_0(\textbf{X},t)=P_0(\textbf{X},t)= \delta(\textbf{X})$.
Therefore
\begin{equation}
\begin{tabular}{lllll}
$P(\textbf{X},t)$&$=$&$\lim\limits_{n\longrightarrow\infty}V_n(\textbf{X},t)$\\&
$=$&$V_0(\textbf{X},t)+\sum_{k=0}^{\infty}\left[ \int_{0}^{t}(-\left\langle \textbf{m},\nabla V_k\right\rangle +\nabla_R^{\alpha}V_k)dt\right] ,$\\&
$=$&$\delta(\textbf{X})+\sum_{k=0}^{\infty}\left[ \int_{0}^{t}(-\left\langle \textbf{m},\nabla V_k\right\rangle +\nabla_R^{\alpha}V_k)dt\right].$
\end{tabular}
\end{equation}
In this manner, the rest of the components of the VIM can be obtained. If we compute more terms, then we can show that $ P(\textbf{X}, t) $ is the TID's PDF with respect to $ \textbf{X} $, as $ \textbf{X}\sim TID_{\alpha}(R,t\textbf{m}) $.
\section{Acknowledgment}
Hassan A. Fallahgoul acknowledges financial support from a Belgian Federal Science Policy Office (BELSPO) grant. He is a beneficiary of a mobility grant from the BELSPO co-funded by the Marie Curie Actions from the European Commission. Also, the comments and support of Professor David Veredas and Davy Paindaveine are gratefully acknowledged.
\bibliography{mybibfile}{}
\end{document}
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\begin{document}
\begin{frontmatter}
\title{Term Models of Horn Clauses over Rational Pavelka Predicate Logic}
\runningtitle{Term Models of Horn Clauses over Rational Pavelka Predicate Logic}
\runningauthor{Costa}
\author[A,B]{\fnms{Vicent} \snm{Costa}},
\author[A,B,C]{\fnms{Pilar} \snm{Dellunde}}
\address[A]{Universitat Aut\`onoma de Barcelona\\
}
\vspace{-0.4cm}
\address[B]{Artificial Intelligence Research Institute (IIIA - CSIC)\\
Campus UAB, 08193 Bellaterra, Catalonia \\
}
\vspace{-0.4cm}
\address[C]{Barcelona Graduate School of Mathematics\\
\email{vicent@iiia.csic.es}\\
\email{pilar.dellunde@uab.cat}\\}
\begin{abstract}
This paper is a contribution to the study of the universal Horn fragment of predicate fuzzy logics, focusing on the proof of the existence of free models of theories of Horn clauses over Rational Pavelka predicate logic. We define the notion of a term structure associated to every consistent theory T over Rational Pavelka predicate logic and we prove that the term models of T are free on the class of all models of T. Finally, it is shown that if T is a set of Horn clauses, the term structure associated to T is a model of T.
\end{abstract}
\begin{keyword} Horn clause \sep term model \sep free model \sep Rational Pavelka predicate logic.
\end{keyword}
\end{frontmatter}
\maketitle
\section{Introduction}
\noindent Free models and Horn clauses have a relevant role in classical logic and logic programming. On the one hand, free models, which appeared first in category theory (see for instance \cite[Def. 4.7.17]{BaWe98}), are crucial in universal algebra and, thereby, in model theory. In the context of logic programming, free structures, introduced in \cite{GoThWaWr75} and also named \emph{initial} (as for instance in \cite[Def. 2.1 (i)]{Mak87}), are important in logic programming, since they allow a procedural interpretation of the programs and admitting free structures makes reasonable the \emph{negation as failure} (see for instance \cite{Mak87}). In the context of abstract data types, Tarlecki \cite{Tar85} characterizes abstract algebraic institutions which admit free constructions. On the other hand, the significant importance of Horn clauses in classical logic was detailed in \cite{Hod93}, while it is well-known that Horn clauses are used both as a specification and as a programming language in Prolog, the most common language in logic programming. \smallskip
In the context of fuzzy logics, several definitions of Horn clause have been proposed in the literature, but there is not a canonical one yet. An extensive and important work in predicate fuzzy logics has been done by B\v{e}lohl\'avek and Vychodil (see \cite{Be02,Be03,BeVy05,BeVic06,BeVic06b,Vy15}). Even if the work of these authors also adopts Pavelka-style, it differs from our approach: we do not restrict Horn clauses to fuzzy equalities and we work in the general semantics of \cite{Ha98}.
Another approach is shown in \cite{DuPra96}, where Dubois and Prade discuss different possibilities of defining \emph{fuzzy rules} and they show how these different semantics can be captured in the framework of fuzzy set theory and possibility theory. We find also that, in the context of fuzzy logic programming, there is a rich battery of proposals of Horn clauses which differ depending on the programming approach chosen. Some reference here are \cite{Voj01, Ebra01}. \smallskip
With the goal of developing a systematic study of the universal Horn fragment of predicate fuzzy logics from a model-theoretic point of view, we took in \cite{CoDe16} the syntactical definition of Horn clause of classical logic. Starting by this general and basic definition we studied the existence of free models of theories of Horn clauses in MTL$\forall$. As a generalisation of a group-theoretic construction, Mal'tsev showed in classical logic that any theory of Horn clauses has a free model. In the present paper, a definition of Horn clause in RPL$\forall$ using evaluated formulas is introduced. Consequently, we prove the existence of free models of theories of RPL$\forall$- Horn clauses showing in RPL$\forall$ an analogous result to Mal'tsev's one. The advantage of using these RPL$\forall$-Horn clauses instead of the ones of \cite{CoDe16} lies in the fact that the former can be better settled in the context of fuzzy logic programming. For instance, from a syntactical point of view, basic RPL$\forall$-Horn clauses are a particular case of the clauses used in \cite{Cao04}\smallskip
The paper is organized as follows. Section 2 contains the preliminaries on RPL$\forall$. In Section 3 we introduce the definition of a term structure associated to a consistent theory and prove that when this structure is a model of the associated theory, the term structure is free on the class of all models of the theory. In Section 4 we define the notion of RPL$\forall$-Horn clause and it is shown that whenever the associated theory is a set of RPL$\forall$-Horn clauses, the term structure is a model of this theory.
\section{Preliminaries}
\noindent In this section we introduce the basic notions and results of RPL$\forall$, the first-order extension of Rational Pavelka Logic. For an extensive presentation of RPL$\forall$ see \cite[Ch.3.3 and Ch.5.4]{Ha98} and \cite[Ch.VIII]{CiHaNo11}.
\begin{defi} \textbf{Rational Pavelka Predicate Logic \cite[Ch.VIII]{CiHaNo11}} \label{RPL} Rational Pavelka Predicate Logic RPL$\forall$ is the expansion of \L$\forall$ by adding a truth constant for each rational number $r$ in $[0,1]$ and by adding the axioms RPL1 and RPL2. The following is an axiomatic sytem for RPL$\forall$: \begin{description}
\item[($\mathbf{\text{\L}} 1$)] $\varphi\rightarrow(\psi\rightarrow\varphi)$
\item[($\mathbf{\text{\L}} 2$)] $(\varphi\rightarrow\psi)\rightarrow((\psi\rightarrow\xi)\rightarrow(\varphi\rightarrow\xi))$
\item[($\mathbf{\text{\L}} 3$)] $(\neg\psi\rightarrow\neg\varphi)\rightarrow(\psi\rightarrow\varphi)$
\item[($\mathbf{\text{\L}} 4$)] $((\varphi\rightarrow\psi)\rightarrow\psi)\rightarrow((\psi\rightarrow\varphi)\rightarrow\varphi)$
\item[($\text{RPL}1$)] $(\overline{r}\rightarrow \overline{s})\leftrightarrow\overline{r\rightarrow s}$
\item[($\text{RPL}2$)] $(\overline{r}\&\overline{s})\leftrightarrow\overline{ r\& s}$
\item[($\forall 1$)] $(\forall x)\varphi(x)\rightarrow\varphi(t)$, where the term $t$ is substitutable for $x$ in $\varphi$.
\item[($\forall 2$)] $(\forall x)(\xi\rightarrow\varphi)\rightarrow(\xi\rightarrow(\forall x)\varphi(x))$, where $x$ is not free in $\xi$.
\end{description}
The rules are Modus Ponens and Generalization, that is, from $\varphi$ infer $(\forall x)\varphi$.
\end{defi}
\noindent A \emph{theory} $\Phi$ is a set of sentences. We denote by $\Phi\vdash_{\text{RPL}\forall}\varphi$ the fact that $\varphi$ is provable in RPL$\forall$ from the set of formulas $\Phi$. From now on, when it is clear from the context, we will write $\vdash$ to refer to $\vdash_{\text{RPL}\forall}$. We say that a theory $\Phi$ is \emph{consistent} if $\Phi\not\vdash\overline{0}$.
\begin{defi}
An \emph{evaluated formula} $(\varphi,r)$ in a language of RPL$\forall$ is a formula of the form $\overline{r}\rightarrow\varphi$, where $r\in[0,1]$ is a rational number and $\varphi$ is a formula without truth constants apart from $\overline{0}$ and $\overline{1}$. We say that an evaluated formula $(\varphi,r)$ is \emph{atomic} whenever $\varphi$ is atomic.
\end{defi}
Now we introduce the semantics of the predicate languages. Let $[0,1]_{\text{RPL}}$ be the standard RPL-algebra \cite[Def.2.2.5, Ch.II]{CiHaNo11}, a \emph{structure} for a predicate language $\mathcal{P}$ of the logic RPL$\forall$ has the form $\langle[0,1]_{\text{RPL}}, \textbf{M}\rangle$, where $ \textbf{M}=\langle M, (P_M)_{P\in Pred}, (F_M)_{F\in Func} \rangle$, $M$ is a non-empty domain; for each $n$-ary predicate symbol $P\in Pred$, $P_{\mathrm{\mathbf{M}}}$ is an $n$-ary fuzzy relation $M$, i.e., a function $M^n\rightarrow[0,1]_{\text{RPL}}$ (identified with an element of $[0,1]_{\text{RPL}}$ if $n=0$); for each $n$-ary function symbol $F\in Func$, $F_{\mathrm{\mathbf{M}}}$ is a function $M^n\rightarrow M$ (identified with an element of $M$ if $n=0$).
An \textbf{M}-\emph{evaluation} of the object variables is a mapping $v$ which assigns an element from $M$ to each object variable. Let $v$ be an \textbf{M}-evaluation, $x$ a variable, and $a\in M$. Then by $v[x\mapsto a]$ we denote the \textbf{M}-evaluation such that $v[x\mapsto a](x)=a$ and $v[x\mapsto a](y)=v(y)$ for each object variable $y$ different from $x$. We define the \emph{values} of terms and the \emph{truth values} of formulas in the structure $\langle[0,1]_{\text{RPL}}, \textbf{M}\rangle$\space for an evaluation $v$ recursively as follows: given $F\in Func, P\in Pred$ and $c$ a connective of RPL:
\begin{itemize}
\item $||x||^{\small{[0,1]_{\text{RPL}}}}_{\mathrm{\mathbf{M}},v}=v(x)$
\item $||F(t_1,\ldots,t_n)||^{\small{[0,1]_{\text{RPL}}}}_{\mathrm{\mathbf{M}},v}=F_{\mathrm{\mathbf{M}}}(||t_1||^{\small{[0,1]_{\text{RPL}}}}_{\mathrm{\mathbf{M}},v},\ldots,||t_n||^{\small{[0,1]_{\text{RPL}}}}_{\mathrm{\mathbf{M}},v})$
\item $||P(t_1,\ldots,t_n)||^{\small{[0,1]_{\text{RPL}}}}_{\mathrm{\mathbf{M}},v}=P_{\mathrm{\mathbf{M}}}(||t_1||^{\small{[0,1]_{\text{RPL}}}}_{\mathrm{\mathbf{M}},v},\ldots,||t_n||^{\small{[0,1]_{\text{RPL}}}}_{\mathrm{\mathbf{M}},v})$
\item $||c(\varphi_1,\ldots,\varphi_n)||^{\small{[0,1]_{\text{RPL}}}}_{\mathrm{\mathbf{M}},v}=c_{[0,1]_{\text{RPL}}}(||\varphi_1||^{\small{[0,1]_{\text{RPL}}}}_{\mathrm{\mathbf{M}},v},\ldots,||\varphi_n||^{\small{[0,1]_{\text{RPL}}}}_{\mathrm{\mathbf{M}},v})$
\item $||(\forall x)\varphi||^{\small{[0,1]_{\text{RPL}}}}_{\mathrm{\mathbf{M}},v}=inf\{||\varphi||^{\small{[0,1]_{\text{RPL}}}}_{\mathrm{\mathbf{M}},v[x\rightarrow a]}\mid a\in M\}$
\item $||(\exists x)\varphi||^{\small{[0,1]_{\text{RPL}}}}_{\mathrm{\mathbf{M}},v}=sup\{||\varphi||^{\small{[0,1]_{\text{RPL}}}}_{\mathrm{\mathbf{M}},v[x\rightarrow a]}\mid a\in M\}.$
\end{itemize}
Observe that, since the universe of the standard RPL-algebra is the interval of real numbers $[0,1]$, which is complete, all the infima and suprema in the definition of the semantics of the quantifiers exist.
\bigskip
\noindent For every formula $\varphi$, possibly with variables, we write $||\varphi||^{[0,1]_{\text{RPL}}}_{\textbf{M}}=$ $$inf\{||\varphi||^{[0,1]_{\text{RPL}}}_{\textbf{M},v}\mid \text{ for every } \textbf{M}\text{ -evaluation } v\},$$ \noindent we say that $\langle[0,1]_{\text{RPL}}, \textbf{M}\rangle$ is a \emph{model of a sentence} $\varphi$ if $||\varphi||^{[0,1]_{\text{RPL}}}_{\textbf{M}}=1$; and that $\langle[0,1]_{\text{RPL}}, \textbf{M}\rangle$ is a \emph{model of a theory} $\Phi$ if $||\varphi||^{[0,1]_{\text{RPL}}}_{\textbf{M}}=1$ for every $\varphi\in\Phi$. \bigskip
In particular, given a structure $\langle[0,1]_{\text{RPL}}, \textbf{M}\rangle$ and two formulas $\varphi$ and $\psi$:
$$||\varphi\&\psi||^{[0,1]_{\text{RPL}}}_{\textbf{M}}=max\{||\varphi||^{[0,1]_{\text{RPL}}}_{\textbf{M}}+||\psi||^{[0,1]_{\text{RPL}}}_{\textbf{M}}-1,0\}$$
$$||\varphi\rightarrow\psi||^{[0,1]_{\text{RPL}}}_{\textbf{M}}=min\{1-||\varphi||^{[0,1]_{\text{RPL}}}_{\textbf{M}}+||\psi||^{[0,1]_{\text{RPL}}}_{\textbf{M}},1\}.$$
\begin{defi}
\label{def:mapping structures} Let $\langle[0,1]_{\text{RPL}},\mathrm{\mathbf{M}}\rangle$ and $\langle[0,1]_{\text{RPL}},\mathrm{\mathbf{N}}\rangle$ be structures, and $g$ be a mapping from $M$ to $N$. We say that $g$ is a \emph{homomorphism from} $\langle[0,1]_{\text{RPL}},\mathrm{\mathbf{M}}\rangle$ \emph{to} $\langle[0,1]_{\text{RPL}},\mathrm{\mathbf{N}}\rangle$ if for every $n$-ary function symbol $F$, any $n$-ary predicate symbol $P$ and $d_1,\ldots,d_n\in M$, \begin{itemize}
\item[(1)] $\space g(F_{\mathrm{\mathbf{M}}}(d_1,\ldots,d_n))=F_{\mathrm{\mathbf{N}}}(g(d_1),\ldots,g(d_n))$, and\newline
\item[(2)] $ P_{\mathrm{\mathbf{M}}}(d_1,\ldots,d_n)=1\Rightarrow P_{\mathrm{\mathbf{N}}}(g(d_1),\ldots,g(d_n))=1.$
\end{itemize}
\end{defi}
Throughout the paper we assume that all our languages have a binary predicate symbol $\approx$ and we extend the axiomatic system of RPL$\forall$ in \cite[Ch.VIII]{CiHaNo11} with the following axioms of similarity and congruence.
\smallskip
\begin{defi} $\textbf{ \cite[Definitions 5.6.1, 5.6.5]{Ha98}}$ \label{def similarity}
\begin{itemize}
\item[S1.] $(\forall x)x\approx x$
\item[S2.] $(\forall x)(\forall y)(x\approx y\rightarrow y\approx x$)
\item[S3.] $(\forall x)(\forall y)(\forall z)(x\approx y \& y\approx z\rightarrow x\approx z)$ \end{itemize}
\begin{itemize}
\item[C1.] For each $n$-ary function symbol $F$, \end{itemize}
$(\forall x_1)\dotsb(\forall x_n)(\forall y_1)\dotsb(\forall y_n)(x_1\approx y_1\&\dotsb \& x_n\approx y_n\rightarrow F(x_1,\ldots,x_n)\approx F(y_1,\ldots,y_n))$
\begin{itemize}
\item[C2.] For each $n$-ary predicate symbol $P$, \end{itemize}
$(\forall x_1)\dotsb(\forall x_n)(\forall y_1)\dotsb(\forall y_n)(x_1\approx y_1\&\dotsb \& x_n\approx y_n\rightarrow (P(x_1, \ldots, x_n)\leftrightarrow P(y_1,\ldots, y_n))$
\end{defi}
\begin{defi} \label{provability}
Let $\Phi$ be a theory over RPL$\forall$, $\varphi$ a formula in a language of RPL$\forall$ and $r\in[0,1]$ a rational number. \newline \newline
(i) The \emph{truth degree} of $\varphi$ over $\Phi$ is $||\varphi||_{\Phi}=$
$$inf\{ ||\varphi||^{[0,1]_{\emph{RPL}}}_{\emph{\textbf{M}}}\mid \langle[0,1]_{\emph{RPL}},\emph{\textbf{M}}\rangle \text{ is a model of } \Phi\}.$$
(ii) The \emph{provability degree} of $\varphi$ over $\Phi$ is$$|\varphi|_{\Phi}=sup\{r\mid \Phi\vdash\overline{r}\rightarrow\varphi\}.$$
\end{defi}
\begin{theorem} \textbf{Pavelka-style completeness \cite[Th.5.4.10]{Ha98}} \label{pavelka}
Let $\Phi$ be a theory over RPL$\forall$ and $\varphi$ a formula in a language of RPL$\forall$. Then, $|\varphi|_{\Phi}=||\varphi||_{\Phi}$.
\end{theorem}
\section{Term structures}
\noindent In this section we introduce the notion of term structure associated to a consistent theory $\Phi$ over RPL$\forall$, and prove that whenever the term structure is a model of $\Phi$, the structure is free on the class of models of $\Phi$. Term structures have been used extensively in classical logic, for instance, to prove the satisfiability of a set of consistent sentences (see for example \cite[Ch.V]{EbiFlu94}).
\begin{defi}\label{relacio}
Let $\Phi$ be a consistent theory, we define a binary relation on the set of terms, denoted by $\sim$, in the following way: For every terms $t_1,t_2$,
\begin{center}
$t_1\sim t_2$ if and only if $|t_1\approx t_2|_{\Phi}=1$.
\end{center}
\end{defi}
Using Axioms $\forall1$, S1, S2 and S3, it can be proven that $\sim$ is an equivalence relation. Next lemma, which states that the equivalence relation $\sim$ is compatible with the symbols of the language, is proved using Axioms $\forall1$, C1, C2 and \cite[Remark 3.18]{Ha98}.
\begin{lemma} \label{compatible}
For any consistent theory $\Phi$, the following holds: If $t_i\sim t'_i$ for every $1\leq i\leq n$, then \begin{itemize}
\item[(i)] For any $n$-ary function symbol $F$, $F(t_1,\ldots,t_n)\sim F(t'_1,\ldots,t'_n)$.
\item[(ii)] For any $n$-ary predicate symbol $P$ and rational number $r\in[0,1]$, \begin{center}
$|(\overline{r}\rightarrow P(t_1,...,t_n))\leftrightarrow (\overline{r}\rightarrow P(t'_1,...,t'_n))|_{\Phi}=1$
\end{center}
\end{itemize}
\end{lemma}
From now on, for any term $t$ we denote by $\overline{t}$ the $\sim$-class of $t$.
\begin{defi}\textbf{Term Structure} \label{term structure}
Let $\Phi$ be a consistent theory. We define the following structure $\langle[0,1]_{\text{RPL}},\mathrm{\mathbf{T}}^{\Phi}\rangle$, where $T^{\Phi}$ is the set of all equivalence classes of the relation $\sim$ and
\begin{itemize}
\item For any $n$-ary function symbol $F$ and terms $t_1,\ldots,t_n$,
$$F_{\mathrm{\mathbf{T}}^{\Phi}}(\overline{t_1},\ldots,\overline{t_n})=\overline{F(t_1,\ldots,t_n)}$$
\item For any $n$-ary predicate symbol $P$ and terms $t_1,\ldots,t_n$,
$$ P_{\mathrm{\mathbf{T}}^{\Phi}}(\overline{t_1},\ldots,\overline{t_n})=|P(t_1,\ldots,t_n)|_{\Phi} $$
\end{itemize}
We call $\langle[0,1]_{\text{RPL}},\mathrm{\mathbf{T}}^{\Phi}\rangle$ the \emph{term structure associated to $\Phi$}.
\end{defi}
Notice that for $0$-ary functions, that is, for individual constants, $c_{\mathrm{\mathbf{T}}^{\Phi}}=\overline{c}$. Given a consistent theory $\Phi$, let $e^{\Phi}$ be the following $\mathrm{\mathbf{T}}^{\Phi}$-evaluation: $e^{\Phi}(x)=\overline{x}$ for every variable $x$. We call $e^{\Phi}$ the \emph{canonical evaluation of} $\langle[0,1]_{\text{RPL}},\mathrm{\mathbf{T}}^{\Phi}\rangle$.
\begin{lemma} \label{terms}
Let $\Phi$ be a consistent theory, the following holds: \begin{itemize}
\item[(i)] For any term $t$, $||t||^{[0,1]_{\emph{RPL}}}_{\mathrm{\mathbf{T}}^{\Phi},e^{\Phi}}=\overline{t}$.
\item[(ii)] For any atomic formula $\varphi$, $||\varphi||^{[0,1]_{\emph{RPL}}}_{\mathrm{\mathbf{T}}^{\Phi},e^{\Phi}}=1$ if and only if $|\varphi|_{\Phi}=1$.
\item[(iii)] For any evaluated atomic formula $(\varphi,s)$, $||(\varphi,s)||^{[0,1]_{\emph{RPL}}}_{\mathrm{\mathbf{T}}^{\Phi},e^{\Phi}}=1$ if and only if $|(\varphi,s)|_{\Phi}=1$.
\end{itemize}
\end{lemma}
\begin{proof}
The proofs of (i) and (ii) are straightforward. Regarding (iii), let $(\varphi,s)=(P(t_1\ldots,t_n),s)$, we have:
\medskip$\begin{array}{lr} ||(P(t_1\ldots,t_n),s)||^{[0,1]_{\text{RPL}}}_{\mathrm{\mathbf{T}}^{\Phi},e^{\Phi}}=1 & \text{iff}
\\[1ex] s\leq||P(t_1\ldots,t_n)||^{[0,1]_{\text{RPL}}}_{\mathrm{\mathbf{T}}^{\Phi},e^{\Phi}} & \text{iff}
\\[1ex] s\leq P_{\textbf{T}^{\Phi}}(\overline{t_1}\ldots,\overline{t_n})& \text{iff}
\\[1ex] s\leq |P(t_1\ldots,t_n)|_{\Phi} \text{ iff } |\overline{s}\rightarrow P(t_1,\ldots, t_n)|_{\Phi}=1. & \end{array}$
\smallskip
\noindent The last equivalence is proved from \cite[Remark 3.18]{Ha98}. \end{proof}
\bigskip
Since the simplest well-formed formulas are atomic formulas, Lemma \ref{terms} (ii) can be read as saying that term structures are minimal with respect to atomic formulas. By Theorem \ref{pavelka}, $|\varphi|_{\Phi}=||\varphi||_{\Phi}$ and, by Lemma \ref{terms} (ii), the term structure $\langle[0,1]_{\text{RPL}},\mathrm{\mathbf{T}}^{\Phi}\rangle$ only assigns the truth value $1$ to those atomic formulas that have $1$ as their truth value in every model $\langle[0,1]_{\text{RPL}},\textbf{M}\rangle$ of $\Phi$. By a similar argument, Lemma \ref{terms} (iii) states that the term structure $\langle[0,1]_{\text{RPL}},\mathrm{\mathbf{T}}^{\Phi}\rangle$ is minimal with respect to evaluated atomic formulas.
From an algebraic point of view, the minimality of the term structure is revealed by the fact that the structure is \emph{free}. The following theorem proves that in case that the term structure associated to a theory is a model of that theory, the term structure is free.
Working in predicate fuzzy logics (and, in particular, in RPL$\forall$) allows to define the term structure associated to a theory using similarities instead of crisp identities. This leads us to a notion of free structure restricted to the class of reduced models of that theory. Remember that \emph{reduced structures} are those whose Leibniz congruence is the identity. By \cite[Lemma 20]{De12}, a structure $\langle[0,1]_{\text{RPL}},\mathrm{\mathbf{M}}\rangle$ is reduced iff it has the \emph{equality property} (EQP) (that is, for any $d,e\in M$, $ || d\approx e||^{[0,1]_{\text{RPL}}}_{\mathrm{\mathbf{M}}}=1$ iff $d=e$). Observe that, by using Definitions \ref{relacio} and \ref{term structure} and the fact that $\sim$ is an equivalence relation, it can be proven that $\langle[0,1]_{\text{RPL}},\mathrm{\mathbf{T}}^{\Phi}\rangle$ is a reduced structure.
\begin{theorem} \label{free thm}
Let $\Phi$ be a consistent theory such that $\langle[0,1]_{\emph{RPL}},\mathrm{\mathbf{T}}^{\Phi}\rangle$ is a model of $\Phi$. Then $\langle[0,1]_{\emph{RPL}},\mathrm{\mathbf{T}}^{\Phi}\rangle$ is free on the class of all the reduced models $\langle[0,1]_{\emph{RPL}},\mathrm{\mathbf{N}}\rangle$ of $\Phi$. That is, for every reduced model of $\Phi$ $\langle[0,1]_{\emph{RPL}},\mathrm{\mathbf{N}}\rangle$ and every $\mathrm{\mathbf{N}}$-evaluation $v$, there is a unique homomorphism $g$ from $\langle[0,1]_{\emph{RPL}},\mathrm{\mathbf{T}}^{\Phi}\rangle$ to $\langle[0,1]_{\emph{RPL}},\mathrm{\mathbf{N}}\rangle$ such that for every variable $x$, $g(\overline{x})=v(x)$.
\end{theorem}
\begin{proof} Let $\langle[0,1]_{\text{RPL}},\mathrm{\mathbf{N}}\rangle$ be a reduced model of $\Phi$ and $v$ an $\textbf{N}$-evaluation.
We define $g$ by: $g(\overline{t})=|| t ||^{[0,1]_{\text{RPL}}}_{\mathrm{\mathbf{N}},v}$ for every term $t$. We show that $g$ is the claimed homomorphism. \newline
Let us first check that $g$ is well-defined. Let $t_1,t_2$ be terms with $\overline{t_1}=\overline{t_2}$, i.e., $t_1\sim t_2$, that is, $|t_1\approx t_2|_{\Phi}=1$. From Theorem \ref{pavelka} we have $||t_1\approx t_2||_{\Phi}=1$. Since $||\Phi||^{[0,1]_{\text{RPL}}}_{\mathrm{\mathbf{N}}}=1$, it follows that $||t_1\approx t_2||^{[0,1]_{\text{RPL}}}_{\mathrm{\mathbf{N}}}=1$ and, in particular, $||t_1\approx t_2||^{[0,1]_{\text{RPL}}}_{\mathrm{\mathbf{N}},v}=1$. From this and the fact that $\langle[0,1]_{\text{RPL}},\mathrm{\mathbf{N}}\rangle$ is reduced, we deduce, by \cite[Lemma 20]{De12}, that $||t_1||^{[0,1]_{\text{RPL}}}_{\mathrm{\mathbf{N}},v}=||t_2||^{[0,1]_{\text{RPL}}}_{\mathrm{\mathbf{N}},v}$, i.e., $g(\overline{t_1})=g(\overline{t_2})$. \smallskip
The task is now to see that $g$ satisfies the conditions (1) and (2) of Definiton \ref{def:mapping structures}. For any $0$-function symbol $c$, $ c_{\mathrm{\mathbf{T}}^{\Phi}}=\overline{c}=c_{\textbf{N}}$ by Definition \ref{term structure}. Let $\overline{t_1},\ldots,\overline{t_n}\in T^{\Phi}$ and $F$ be an $n$-ary function symbol, $F_{\mathrm{\mathbf{T}}^{\Phi}}(\overline{t_1},\ldots,\overline{t_n})=\overline{F(t_1,\ldots,t_n)}$ by Definition \ref{term structure}. Then, by the definition of $g$,
\begin{center} $g(F_{\mathrm{\mathbf{T}}^{\Phi}}(\overline{t_1},\ldots,\overline{t_n}))=g(\overline{F(t_1,\ldots,t_n)})=$\end{center}
\begin{center} $F_{\textbf{N}}(|| t_1||^{[0,1]_{\text{RPL}}}_{\mathrm{\mathbf{N}},v},\ldots ,|| t_n ||^{[0,1]_{\text{RPL}}}_{\mathrm{\mathbf{N}},v})=F_{\textbf{N}}(g(\overline{t_1}),\ldots,g(\overline{t_n}))$. \end{center}
\smallskip
Let $P$ be an $n$-ary predicate symbol such that $P_{\mathrm{\mathbf{T}}^{\Phi}}(\overline{t_1},\ldots,\overline{t_n})=1$. By Definition \ref{term structure} and Theorem \ref{pavelka}, $1=P_{\mathrm{\mathbf{T}}^{\Phi}}(\overline{t_1},\ldots,\overline{t_n})=|P(t_1,\ldots,t_n)|_{\Phi}=||P(t_1,\ldots,t_n)||_{\Phi}$.
\noindent Consequently, $||P(t_1,\ldots,t_n)||^{[0,1]_{\text{RPL}}}_{\mathrm{\mathbf{N}}}=1$, because $||\Phi||^{[0,1]_{\text{RPL}}}_{\mathrm{\mathbf{N}}}=1$. Thus \newline
$|| P(t_1,\ldots,t_n)||^{[0,1]_{\text{RPL}}}_{\mathrm{\mathbf{N}},v}=1$. Therefore $P_{\mathrm{\mathbf{N}}}(|| t_1||^{[0,1]_{\text{RPL}}}_{\mathrm{\mathbf{N}},v},\ldots ,|| t_n ||^{[0,1]_{\text{RPL}}}_{\mathrm{\mathbf{N}},v})=1$, that is, $P_{\mathrm{\mathbf{N}}}(g(\overline{t_1}),\ldots,g(\overline{t_n}))=1$.
\smallskip
Finally, since the set $\{\overline{x}\mid x\text{ is a variable}\}$ generates the universe $T^{\Phi}$ of the term structure associated to $\Phi$, $g$ is the unique homomorphism such that for every variable $x$, $g(\overline{x})=v(x)$.
\end{proof}
\bigskip
Observe that in languages in which the similarity symbol is interpreted by the crisp identity, by using an analogous argument to the one in Theorem \ref{free thm}, we obtain that the term structure is free in the class of all the models $\langle[0,1]_{\text{RPL}},\mathrm{\mathbf{M}}\rangle$ of the theory and not only in the class of the reduced ones.
\section{RPL$\forall$-Horn Clauses}
\noindent In the previous section we have seen that if the term structure associated to a theory $\Phi$ is a model of $\Phi$, then the structure is free in the class of all models of $\Phi$. In this section, we show in Theorem \ref{model} that whenever $\Phi$ is a theory of RPL$\forall$-Horn clauses, $\langle[0,1]_{\text{RPL}},\textbf{T}^{\Phi}\rangle$ is a model of $\Phi$. Theorem \ref{model} gains in interest if we realize that it proves (using Theorem \ref{free thm}) the existence of free models of theories of RPL$\forall$-Horn clauses. Let us first define the notion of RPL$\forall$-Horn clauses. \smallskip
In predicate classical logic, a \emph{basic Horn formula} is a formula of the form $ \alpha_{1}\wedge\dotsb \wedge\alpha_{n}\rightarrow\beta$, where $n$ is a natural number and $\alpha_1,\ldots,\alpha_n,\beta$ are atomic formulas. Notice that there is not a unique way to extend this definition in fuzzy logics, where we have different conjunctions and implications. In this section we present one way to define Horn clauses over RPL$\forall$ extending the classical definition.
\smallskip
\begin{defi}\textbf{Basic RPL$\forall$-Horn Formula}\label{strong basic} A \emph{basic \emph{RPL}$\forall$-Horn formula} is a formula of the form
$$(\alpha_1,r_1)\&\dotsb\&(\alpha_n,r_n)\rightarrow(\beta,s)$$
where $(\alpha_1,r_1)\ldots,(\alpha_n,r_n),(\beta,s)$ are evaluated atomic formulas and $n$ is a natural number. Observe that $n$ can be $0$. In that case the basic RPL$\forall$-Horn formula is an evaluated atomic formula.
\end{defi}
\begin{defi} \textbf{Quantifier-free RPL$\forall$-Horn Formula} \label{qf Horn} A \emph{quantifier-free \emph{RPL}$\forall$-Horn formula} is a formula of the form $\phi_1\&\dotsb\&\phi_m$, where $m$ is a natural number and $\phi_i$ is a basic RPL$\forall$-Horn formula for every $1\leq i\leq m$.
\end{defi}
\begin{defi} \textbf{RPL$\forall$-Horn Clause} \label{Horn}
A \emph{\emph{RPL}$\forall$-Horn clause} is a formula of the form $Q\gamma$, where $Q$ is a (possibly empty) string of universal quantifiers $(\forall x)$ and $\gamma$ is a quantifier-free RPL$\forall$-Horn formula. \end{defi}
\begin{ex} \label{example}
Let $\mathcal{P}$ be a predicate language with a unary predicate symbol $P$, a binary predicate symbol $R$ and $a$ an individual constant. The following formulas are examples of RPL$\forall$-Horn clauses: \begin{itemize}
\item[(1)] $(P(a),0.5)$,
\item[(2)] $(P(a),0.6)\&(R(a,x),0.3)$,
\item[(3)] $(P(a),0.5)\rightarrow(R(a,a),0.1)$,
\item[(4)] $(P(a),0.6)\&(R(a,x),0.3)\rightarrow(P(x),0.8)$,
\item[(5)] $(\forall x)((P(x),0.6)\&(R(a,x),0.3))$,
\item[(6)] $(\forall x)((P(x),0.6)\&(R(a,x),0.3)\rightarrow(P(a),0.9))$.
\end{itemize}
\end{ex}
Observe that, in general, RPL$\forall$-Horn clauses are not evaluated, only the atomic RPL$\forall$-Horn clauses are evaluated formulas.
A weak version of RPL$\forall$-Horn clauses can be introduced by substituting each strong conjunction $\&$ appearing in the formula by the weak conjunction $\wedge$. Although in this paper we do not present this weak version, all the results we prove are also true for weak RPL$\forall$-Horn clauses. In classical logic, the set of all Horn clauses is recursively defined, because the formula $(\forall x)(\varphi\wedge\psi)$ is logically equivalent to $(\forall x)\varphi\wedge(\forall x)\psi$. In RPL$\forall$ these two formulas are also logically equivalent, so the set of the weak version of fuzzy RPL$\forall$-Horn clauses is recursively definable. However, this is not the case for fuzzy RPL$\forall$-Horn clauses. Indeed, let $P$ and $R$ be unary predicate symbols, consider the structure $\langle[0,1]_{\text{RPL}},\textbf{M}\rangle$ such that $M=\{a,b\}$, $ P_{\textbf{M}}(a)=R_{\textbf{M}}(b)=0.4$ and $P_{\textbf{M}}(b)=R_{\textbf{M}}(a)=0.7$. Then, $||(\forall x)((P(x),1)\&(R(x),1))||^{[0,1]_{\text{RPL}}}_{\textbf{M}}=0.1$, but $||(\forall x)((P(x),1))\&(\forall x)((R(x),1))||^{[0,1]_{\text{RPL}}}_{\textbf{M}}=0$.
We now see that for any consistent theory of RPL$\forall$-Horn clauses $\Phi$, the term structure associated to $\Phi$ is a model of $\Phi$. To show that, we need the following lemmas and the notion of rank of a formula. Our definition of rank is a variant of the notion of \emph{syntactic degree of a formula} of \cite[Def. 5.6.7]{Ha98}). Let $\varphi$ be a formula, the \emph{rank} of $\varphi$, denoted by $rk(\varphi)$ is defined by:
\begin{itemize}
\item $rk(\varphi)=0$ if $\varphi$ is atomic;
\item $rk(\neg\varphi)=rk((\exists x)\varphi)=rk((\forall x)\varphi)=rk(\varphi)+1$;
\item $rk(\varphi\circ\psi)=rk(\varphi)+rk(\psi)$ for every binary propositional connective $\circ$.
\end{itemize}
\noindent Note that since the set of RPL$\forall$-Horn clauses is not recursively definable, induction on the complexity of the clause cannot be applied. Hence it is applied on the rank of the clauses. Such induction can be used to prove next lemma.
\begin{lemma} \label{Horn substitucio}
Let $\varphi$ be an RPL$\forall$-Horn clause where $x_1,\ldots,x_m$ are pairwise distinct free variables. Then, for every terms $t_1,\ldots,t_m$, the substitution $$\varphi (t_1,\ldots,t_m/x_1,\ldots,x_m)$$ is an RPL$\forall$-Horn clause.
\end{lemma}
\begin{lemma} \label{lemma th}
For any consistent theory $\Phi$ and any evaluated atomic formula $(\varphi,s)$,
$$||(\varphi,s)||^{[0,1]_{\emph{RPL}}}_{\emph{\textbf{T}}^{\Phi}}=||(\varphi,s)||_{\Phi}.$$
\end{lemma}
\begin{proof}
It is enough to show that for any rational number $t\in[0,1]$, $||(\varphi,s)||^{[0,1]_{\text{RPL}}}_{\textbf{T}^{\Phi}}\geq t \text{ iff } ||(\varphi,s)||_{\Phi}\geq t.$ Let $t\in[0,1]$ be a rational number, we have: $$||(\varphi,s)||^{[0,1]_{\text{RPL}}}_{\textbf{T}^{\Phi}}\geq t \text{ iff } ||\overline{t}\rightarrow(\overline{s}\rightarrow\varphi)||^{[0,1]_{\text{RPL}}}_{\textbf{T}^{\Phi}}=1 \text{ iff }$$
$$||\overline{t}\&\overline{s}\rightarrow\varphi||^{[0,1]_{\text{RPL}}}_{\textbf{T}^{\Phi}}=1 \text{ iff } ||\varphi||^{[0,1]_{\text{RPL}}}_{\textbf{T}^{\Phi}}\geq t*_{\text{\L}}s\text{ iff}$$
$$||\varphi||^{[0,1]_{\text{RPL}}}_{\textbf{M}}\geq t*_{\text{\L}}s\text{ for every model } \langle[0,1]_{\text{RPL}},\textbf{M}\rangle \text{ of } \Phi\text{ iff}$$
$$\text{ for any model } \langle[0,1]_{\text{RPL}},\textbf{M}\rangle \text{ of } \Phi,$$
$$||\overline{t}\rightarrow(\overline{s}\rightarrow\varphi)||^{[0,1]_{\text{RPL}}}_{\textbf{M}}=1.$$
\noindent The second and latter equivalence are proved by using \cite[Def.2.2.4 (Axioms 5a and 5b)]{Ha98}. The latter expression is equivalent to $||(\varphi,s)||^{[0,1]_{\text{RPL}}}_{\textbf{M}}\geq t$ for every model $\langle[0,1]_{\text{RPL}},\textbf{M}\rangle$ of $\Phi$, i.e., $||(\varphi,s)||_{\Phi}\geq t.$ \end{proof}
\bigskip
\begin{lemma} \label{desigualtat}
For any consistent theory $\Phi$ and any evaluated atomic sentences \newline {\footnotesize $(\varphi_1,s_1),\dots,(\varphi_n,s_n)$}, {\footnotesize $$||(\varphi_1,s_1)\&\dotsb\&(\varphi_n,s_n)||^{[0,1]_{\emph{RPL}}}_{\emph{\textbf{T}}^{\Phi}}\leq||(\varphi_1,s_1)\&\dotsb\&(\varphi_n,s_n)||_{\Phi}.$$}
\end{lemma}
\begin{proof}
By Lemma \ref{lemma th}, it is clear for $n=1$. For the sake of clarity, we present the proof for the case $n=2$, but the argument is analogous for the cases with $n> 2$. First, by Lemma \ref{lemma th} we have:\newline\newline
{\footnotesize $||(\varphi_1,s_1)\&(\varphi_2,s_2)||^{[0,1]_{\text{RPL}}}_{\textbf{T}^{\Phi}}$$=||(\varphi_1,s_1)||^{[0,1]_{\text{RPL}}}_{\textbf{T}^{\Phi}}*_{\text{\L}}||(\varphi_2,s_2)||^{[0,1]_{\text{RPL}}}_{\textbf{T}^{\Phi}}=$
$||(\varphi_1,s_1)||_{\Phi}*_{\text{\L}}||(\varphi_2,s_2)||_{\Phi}.$
}
\bigskip
\noindent Since for any model {\footnotesize$\langle[0,1]_{\text{RPL}},\textbf{M}\rangle$} of $\Phi$, {\footnotesize $||(\varphi_1,s_1)||_{\Phi}\leq||(\varphi_1,s_1)||^{[0,1]_{\text{RPL}}}_{\textbf{M}}$} and {\footnotesize$||(\varphi_2,s_2)||_{\Phi}\leq||(\varphi_2,s_2)||^{[0,1]_{\text{RPL}}}_{\textbf{M}}$}, we have that for any model {\footnotesize $\langle[0,1]_{\text{RPL}},\textbf{M}\rangle$} of $\Phi$,
{\footnotesize
$$||(\varphi_1,s_1)||_{\Phi}*_{\text{\L}}||(\varphi_2,s_2)||_{\Phi}\leq ||(\varphi_1,s_1)||^{[0,1]_{\text{RPL}}}_{\textbf{M}}*_{\text{\L}}||(\varphi_2,s_2)||^{[0,1]_{\text{RPL}}}_{\textbf{M}}=$$
$ ||(\varphi_1,s_1)\&(\varphi_2,s_2)||^{[0,1]_{\text{RPL}}}_{\textbf{M}}.$
}
\newline
\noindent Therefore, since $||(\varphi_1,s_1)\&(\varphi_2,s_2)||_{\Phi}$ is the infimum, we have \newline\newline {\footnotesize $||(\varphi_1,s_1)||_{\Phi}*_{\L}||(\varphi_2,s_2)||_{\Phi}\leq ||(\varphi_1,s_1)\&(\varphi_2,s_2)||_{\Phi}$.
}
\smallskip
\smallskip
\noindent Consequently, \newline\newline {\footnotesize $||(\varphi_1,s_1)\&(\varphi_2,s_2)||^{[0,1]_{\text{RPL}}}_{\textbf{T}^{\Phi}}\leq ||(\varphi_1,s_1)\&(\varphi_2,s_2)||_{\Phi}$.}
\end{proof}
\smallskip
\begin{theorem} \label{model}
Let $\Phi$ be a consistent theory. For every RPL$\forall$-Horn clause $\varphi$ without free variables, $$\text{If } |\varphi|_{\Phi}=1\text{, then } ||\varphi||^{[0,1]_{\emph{RPL}}}_{\emph{\textbf{T}}^{\Phi}}=1.$$ \end{theorem}
\begin{proof}
Let $\varphi$ be an RPL$\forall$-Horn clause without free variables. We proceed by induction on $rk(\varphi)$. \smallskip
\smallskip
\underline{$rk(\varphi)=0$.} We can distinguish three subcases:
\smallskip
1) If $\varphi=(\psi,s)$ is an evaluated atomic formula, the statement holds by Lemma \ref{lemma th} (iii). \smallskip
2) Let $\varphi=(\psi_1,s_1)\&\dotsb\&(\psi_n,s_n)\rightarrow(\psi,s)$ be a basic RPL$\forall$-Horn formula, where $(\psi_1,s_1),\ldots,(\psi_n,s_n),(\psi,s)$ are evaluated atomic formulas. By hypothesis and Theorem \ref{pavelka} we have:
\begin{center}$1=|(\psi_1,s_1)\&\dotsb\&(\psi_n,s_n)\rightarrow(\psi,s)|_{\Phi}=||(\psi_1,s_1)\&\dotsb\&(\psi_n,s_n)\rightarrow(\psi,s)||_{\Phi}$.\end{center}
\noindent Therefore, $||(\psi_1,s_1)\&\dotsb\&(\psi_n,s_n)||_{\Phi}\leq ||(\psi,s)||_{\Phi}$.
\smallskip
\noindent By Lemmas \ref{lemma th} and \ref{desigualtat}, $|| (\psi,s)||^{[0,1]_{\text{RPL}}}_{\textbf{T}^{\Phi}}=||(\psi,s)||_{\Phi}$ and
$$||(\psi_1,s_1)\&\dotsb\&(\psi_n,s_n)||^{[0,1]_{\text{RPL}}}_{\textbf{T}^{\Phi}}\leq ||(\psi_1,s_1)\&\dotsb\&(\psi_n,s_n)||_{\Phi}.$$
\noindent Therefore $||(\psi_1,s_1)\&\dotsb\&(\psi_n,s_n)||^{[0,1]_{\text{RPL}}}_{\textbf{T}^{\Phi}}\leq || (\psi,s)||^{[0,1]_{\text{RPL}}}_{\textbf{T}^{\Phi}}.$ That is, $$ ||(\psi_1,s_1)\&\dotsb\&(\psi_n,s_n)\rightarrow(\psi,s)||^{[0,1]_{\text{RPL}}}_{\textbf{T}^{\Phi}}=1.$$
\smallskip
3) If $\varphi=\phi_1\&\dotsb\&\phi_m$ is a conjunction of basic RPL$\forall$-Horn formulas, \begin{center}
$||\phi_1\&\dotsb\&\phi_m||^{[0,1]_{\text{RPL}}}_{\mathrm{\mathbf{T}}^{\Phi}}=1$ iff \end{center}
\begin{center}
$||\phi_i||^{[0,1]_{\text{RPL}}}_{\mathrm{\mathbf{T}}^{\Phi}}=1$ for every $1 \leq i \leq m$.
\end{center}
From 1) and 2), $|\phi_i|_{\Phi}=1$ for every $1\leq i\leq m$ and thus $|\phi_1\&\dotsb\&\phi_m|_{\Phi}=1$. \newline
\underline{$rk(\varphi)=n+1$.} Let $\varphi=(\forall x)\psi$ be such that $\psi$ is an RPL$\forall$-Horn clause of rank $n$. Assume inductively that for any RPL$\forall$-Horn clause without free variables $\xi$ of rank less or equal than $n$ and such that $|\xi|_{\Phi}=1$, $||\xi||^{[0,1]_{\text{RPL}}}_{\mathrm{\mathbf{T}}^{\Phi}}=1$. By assumption and Axiom $\forall 1$, \begin{center}
$\Phi\vdash(\forall x)\psi\rightarrow\psi(t/x)$ for every term $t$.
\end{center}
From Axiom \L 2, $sup\{r\mid\Phi\vdash\overline{r}\rightarrow\varphi\}=1$ implies that $sup\{r\mid\Phi\vdash\overline{r}\rightarrow\psi(t/x)\}=1$ for any term $t$. That is, $|\psi(t/x)|_{\Phi}=1$ for every term $t$. \smallskip
Since $\psi$ has rank $n$ and is an RPL$\forall$-Horn clause by Lemma \ref{Horn substitucio}, we can apply the inductive hypothesis and conclude that $||\psi(t/x)||^{[0,1]_{\text{RPL}}}_{\mathrm{\mathbf{T}}^{\Phi}}=1$ for any term $t$. So, by Lemma \ref{terms} (i), $||\psi(x)||^{[0,1]_{\text{RPL}}}_{\mathrm{\mathbf{T}}^{\Phi},v[x\mapsto\overline{t}]}=1$ for every element $\overline{t}$ of the domain, and thus we get $||(\forall x)\psi||^{[0,1]_{\text{RPL}}}_{\mathrm{\mathbf{T}}^{\Phi}}=1$.
\end{proof}
\section{Conclusions and Future Work}
\noindent The present paper is another step towards a systematic study of theories of Horn clauses over predicate fuzzy logics from a model-theoretic point of view, a study that we started in \cite{CoDe16} and which is still in progress. In particular, here we have proved the existence of free models of theories of Horn clauses in RPL$\forall$. \smallskip
Future work will be devoted to study the broad approach taken in \cite[Ch.8]{CiHaNo11} to fuzzy logics with enriched languages. We shall see if RPL$\forall$-Horn clauses introduced here can be generalized to that logics with enriched languages. Later, since one of our next goals is to solve the open problem (formulated by Cintula and H\'ajek in \cite{CiHa10}) about the characterization of theories of fuzzy Horn clauses in terms of quasivarieties, we will analyze quasivarieties and try to define them in the context of fuzzy logics using recent results on products over fuzzy logics like \cite{De12}. \smallskip
Herbrand structures have been important in model theory and in the foundations of logic programming. Therefore, as a continuation of the present work, we would like to characterize the free Herbrand model in the class of the Herbrand models of theories of RPL$\forall$-Horn clauses without equality. Finally, we will focus on a generalization of Herbrand structure, \emph{fully named models}, in order to show that two types of minimality for these models (specifically free models and $A$-generic models) are equivalent.
\section*{Acknowledgments}
\noindent We would like to thank the referees for their useful comments. This project has received funding from the European Union's Horizon 2020 research and innovation program under the Marie Sklodowska-Curie grant agreement No 689176 (SYSMICS project). Pilar Dellunde is also partially supported by the project RASO TIN2015-71799-C2-1-P (MINECO/FEDER) and the grant 2014SGR-118 from the Generalitat de Catalunya.
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HYDERABAD: Who wouldn’t jump at a wedding invitation? But here is a guy who more often than not turns down an invitation unless it meets up to his principles. Aleem Khan Falaki gets good number of invitations, but he makes it to only those weddings where no dowry or dinner is imposed on the bride’s family. “Dowry and reception dinners are the root of all evils”, he remarks.
He is out to alter the course of history. And how? Through vigorous campaign he is trying to create awareness about the scourge of dowry which has made marriage a big problem in the Muslim community. A one-man army against the social menace, Falaki pushes his agenda with passion, compassion, humour, satire and style. He reaches out to the community through books, talks, songs and short films. Over the years he has stayed focussed on this task. For him life is a mission and not an intermission.
Under the aegis of the Socio Reforms Society (SRS), he has authored four books and made six short movies – all targeting the dowry system. “It is a crime against Islam, human rights and women’s rights. The only solution is to boycott such marriages. This is the biggest jihad of the times,” he says.
His latest movie, Khuddar, got 12,500 views on YouTube within three days of its release. Next month, Falaki is coming out with his fifth book – Nikah Ya Vivah. The message is pushed through in simple language covering all aspects of the issue.
Falaki accepts no excuses, only results. And he has a band of ten members committed to the cause. They go about delivering the message to youngsters in colleges and the community at large through the Friday sermons. But it is the short movies which are succeeding in pricking the conscience of the people.
In Khuddar, a drunk person is pushed out of the mosque and everyone beats him for his audacity to enter the Masjid in an intoxicated state. But the man raises some pertinent questions which leave the congregation red-faced. “If alcohol is prohibited in Islam so is usury, cheating, beggary and corruption.
Are not those who demand dowry corrupt. Why don’t you push all of them out of the masjid”, he asks.
In one of his books – Mard Bhi Bikte Hain .... Jahez Ke Liye (Men too are sold for dowry), Falaki tells how dowry is illegal according to the Quran and Hadith. Falaki is filled with a burning desire to eradicate dowry system. And his dedication is showing results. So far more than 2,500 persons have signed a pledge to boycott marriages where dowry or dinner is taken from the bride’s side whether by consent or demand. His plea: Say no to such marriages.
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Strong, with milk and sugar. Anything to stay away from those energy drinks.
Strong and black. Sometimes espresso from my maker or instant.
black with 1 splenda
Originally Posted by howarmat black with 1 splenda Saccharineligeous!!
Black, like my Berry
I've been having my as a triple or quad shot Americano.
If I drink coffee straight with no additives, it's either from Peet's or a local brew that knows how to make a good roast.
Starbucks on the other hand... I add pure sugar, milk, and cream.
Originally Posted by JR A If I drink coffee straight with no additives, it's either from Peet's or a local brew that knows how to make a good roast.
Starbucks on the other hand... I add pure sugar, milk, and cream. If it's a local brew I'm going with beer.
Today I had two cups of double espressos shots with Kahlua.
mmm....guess you could describe it as black with just enough low fat milk to cool it down so I can drink it right away...not a huge fan of waiting
and yes the Q10 wait is killing me !!!!!
Black with 1 teaspoon sugar in the raw, like a blackberry dark and slightly sweet.
Originally Posted by cl8baller Are you a morning drinker? Yes. Sometimes I would drink in the evening, only when I have evening classes though.
Do you drink coffee religiously or on one-of-those days? A religious drinker.
How do you have your coffee? It depends on where I go. If I go to Starbucks, I usually get a Grande white chocolate mocha with whip cream.
If I go to McDonald's or Tim Hortons, then I would get a large coffee with two cream and four sugar.
What methods do you use? The method of transaction, where I have other people make it for me and I give them my money in return.
Like i like my women, white as snow and sweet as ****.
I've almost moved beyond brewing it. In another year or so, I'll be using the grounds like chew.
Originally Posted by the_sandman_454 I've almost moved beyond brewing it. In another year or so, I'll be using the grounds like chew. Now there's a great idea.
I eat chocolate covered espresso beans. amazing.
My grandfather got me into habit of drinking my coffee "like a good dutch man", cream and sugar (although i don't use too much, i have to use SOME of both in honor of my grandpa)
Blue, Jamaican blue mountain, damn(ed) fine coffee! A friend I'm gonna do some work for just laid a pound of the stuff on me tonight, I haven't had this stuff in over ten years, what a treat!
Nice hot cup of coffee with sugar and Bailey's
I had a dirty chai today. Dirty chai latte (makes 2) 12 oz. whole milk1 cinnamon stick2 cardamom pods5 peppercorns, pink or black1 1/2 inch chunk of fresh ginger, peeled, and partially smashed with the flat of a large knife2 Tbs. honey2!
With milk.
Posted using CrackBerry App on BB10
Fresh from Tim Horton's right now
Sent from my BlackBerry 9930 using Tapatalk
With my blackberry of course - seriously I was amazed to find a subforumn for this. LoL
put a scoop of chocolate whey protein powder in today and it tasted pretty damn good
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My next sip of coffee will be my first. (I'm 40+)
I like my coffee like I like my women: all over my crotch mere seconds after being picked up at the Starbuck's drive-thru.
Posted via CB.
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Holy Rosary Catholic School truly honors the call to develop socially and morally responsible individuals able to lead and succeed in a secular world. Children can only feel secure and fulfilled after having achieved the self-respect that results from discipline and mastery over his or her actions.
Discipline, Social Skills and Leadership Development in Our School
The true goal of any discipline program is to assist the students program that all faculty and staff are trained in and use is the model “The Well Managed Classroom” from Boys Town.
Teachers present lessons each week to their students, and parents have the opportunity to review and reinforce the skills at home with their children. Students benefit from the consistency of this powerful program that teaches both social skills and life skills.
The Holy Rosary.
Students have many opportunities to develop leadership skills. Our active Student Council allows students in fourth through eighth grades to participate in a democratic system while directing many service projects that benefit the school and entire community. The peer helping system plays an important role in the school, whereby older students model appropriate behavior through example and by acting as a “pal” to younger students. Many academic competitions throughout the year, open to all students beginning with first grade, help our children to experience the joy of achieving their potential while fostering self-disciple and perseverance.
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Nov 25, 2021
The holiday season can feel overwhelming with the need to do
everything that comes flying at us. Yikes! --- There go the goals
out the window till the New Year’s?! The crazy pull of everything
at this time of year can feel like:
-Your routines come off the tracks during the holidays
-You feel like hitting your...
Nov 22, 2021
I'm BORED!
Have you ever said that about your goals?
You start off super excited, make a great plan, and then bleh - you're over it.
It happens to all of us and it can be a real momentum buster.
Today on the podcast, I'm chatting about what you can do when you just aren't into your goals anymore. How do you know...
Nov 18, 2021
Why do we struggle with routines?! Maybe you are thinking … “I
keep trying routines and nothing seems to stick.”
In today’s episode Cara is sharing in her interview that failing routines are happening because of one of these two reasons:
No. 1 You are trying to do too much too fast.
No.2 You don't have an...
Nov 15, 2021
Did you know that not every goal should be broken down the
same?
Often when we look at our goals, we try to break things down identically and we can get stuck. I teach 3 different types of goals - numerical, project-based and routine-based. Today on the podcast, I'm breaking down numerical goals and how you can...
Nov 11, 2021
Are you feeling aligned and happy about your home and stuff?
As we gear up and get ready for the upcoming holiday season we are really excited to chat about our home and how we can simplify our stuff before MORE stuff comes in.
Listen in to today’s episode to hear great tips about:
- how we can have conversations...
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Abstract
This study reports a novel green chemistry approach to assemble copper-nanowires/reduced-graphene-oxide hybrid coatings onto inorganic and organic supports. Such films are robust and combine sheet resistances (<30 Ω sq−1).
Keywords
- copper nanowires
- electrochromic device
- flexible electronics
- graphene
- transparent conductor
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Serving as the community relations office of the university, the Office of External Relations exists to integrate Indiana University's Kokomo campus with the 11-county region that it serves in north central Indiana. In addition to being the only public university offering four-year and graduate degrees in the area, IU Kokomo is vital to the continued economic growth of the region, both as an employer and as a provider of highly qualified, well-educated employees.
Through its Office of Development, external relations represents the Indiana University Foundation and solicits donations, endowments, planned gifts, and non-research related grants for the campus.
The Office of Alumni Relations provides opportunities for alumni and students to be involved in planning the future and preserving the heritage of IU Kokomo. Additionally, the office plans and implements campus ceremonies, including Commencement. The Office of Public Affairs serves as the government relations office on campus and works closely with Hoosiers for Higher Education.
The Office of Communications and Marketing publicizes campus activities and programs; coordinates internal and external communications; places advertising; manages Web site design and continuity; coordinates all media relations and press releases; and designs campus publications.
2300 S. Washington St., Kokomo, IN 46904-9003 |
Ph: (765) 453-2000
Last updated:
19 July 2005 |
2005, The Trustees of
Indiana University |
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one of the groups who are more likely to die of influenza are the very young. Often those most at risk for die.
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The Dynamics of Labor Migration from Central Asia to Russia. Kyrgyz Republic in the EEU. dynamic materials and the of Western Societies. Biosciences Biotechnology Research Asia. Indo-European
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Ohjelma: One Quart Radio
The phenomenon of the playlist!
No longer the exclusive domain of radio programmers, the playlist has become many people’s entry point for listening to and finding new music. While ’playlist’ status becomes increasingly important for the music industry as well as public, has the playlist stunted the critical consensus and reduced the influence of radio? Or does the success of the playlist reflect a growing trend in self curation and an obsession with data? As technology advances what role does the playlist serve in delivering our entertainment thrills in the future?
Joining Nick in the studio are Tommi Muhli and Oskari Onninen.
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Entry Level Private Elementary Education Position, GimpoSmall World
Entry Level Private Elementary Education Position, Gimpo
Small World 어학원 Is an English academy owned by an American. We are looking for a positive and fun teacher to start May 18th, 2015. This is a full time entry/intermediate level position located in Gimpo, South Korea. Some experience in teaching/public speaking is required. Certification preferred. Korean language not required. We will sponsor your E2 Visa. Teachers must procure all documentation required for teaching in Korea on time.
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Martinez: We need to be ready against Barnsley
Everton manager Roberto Martinez says he side must not underestimate their hosts tonight as they take on Barnsley in the league cup.
The Catalan also praised the South Yorkshire sides academy and says they have brought through a number of fantastic youngsters over the last few years who all get a good football education.
“You need to be ready – you can’t go into any round thinking your league status gives you an advantage on the pitch,”
“The focus will be there and the understanding of who we are facing will be there.
“Talking about winning a competition when you first go into it is a bit stupid but we need to make sure we are as prepared and as good as we can in order to progress.”
The manager praised the Barnsley academy for producing brilliant youngsters including John Stones and Mason Holgate who now play for the blues.
“They do a terrific job with youngsters and give a really good education, so it’s no surprise why they have got such a young team,” added the manager.
“We are expecting a very talented Barnsley team that looks after the ball well. It’s going to be a really good contest in that respect.
“I’m looking forward to the game and the ex-players [Stones and Holgate] will be looking forward to re-living successful memories they had at that ground.”
What will the score be tonight?
Everton Fan? Register for FREE on our Everton messageboard –
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\section{Polynomial invariants from CSM cycles}\label{Sec:polynomial}
In this section we show how CSM cycles of matroids behave under deletions and contractions, and we use this to express their degrees in terms of the coefficients
of the characteristic polynomial. We also provide a conjectural presentation of Speyer's $g$-polynomial of a matroid in terms of CSM cycles.
\subsection{Deletion and contraction of CSM cycles}
Recall that $\{\bfe_0, \bfe_1, \dotsc, \bfe_n\}$ denotes the standard basis of the lattice $\ZZ^{n+1} \subseteq \RR^{n+1}$.
Fix $i \in \{0, \dots, n\}$, and let $\pi_i \colon \R^{n+1} \to \R^n$ denote the linear projection that forgets the $i$-th coordinate.
With this projection in mind, we label the elements of the standard basis of $\Z^n \subseteq \R^n$ by
$\bfe_k$ for $k \neq i$. We will also denote by $\pi_i$ the induced map $\pi_i : \RR^{n+1}/\mathbf 1 \to \RR^{n}/\mathbf 1$.
Let $M$ be a loopless matroid in $\Mat_{n+1}$.
The flats of the deletion $M \backslash i$ and the contraction $M/i$ of $i$ from $M$ are
\[
\L(M\setminus i) = \{ F \backslash i \mid F \in \L(M) \} \quad \text{ and } \quad \L(M/i) = \{ F \backslash i \mid F \in \L(M) \text{ and } i \in F\};
\]
see, for example, \cite[Section 7]{White1}.
The map $\pi_i$ sends the cone $\sigma_ \F$ of the Bergman fan $\B(M)$ corresponding to a flag of flats
$\F: =\{ \emptyset \subsetneq F_1 \subsetneq \dots \subsetneq F_{k} \subsetneq \{0,\dotsc,n\}\}$ in $M$ to
the cone $\sigma_{\F'}$ where $\F'$ is the flag
$\F' \coloneqq\{ \emptyset \subseteq F_1\setminus i \subseteq \dots \subseteq F_{k}\setminus i \subseteq \{0,\dotsc,n\} \setminus i\}$.
It follows that the image of $\B(M)$ under $\pi_i$ is the Bergman fan $\B(M \backslash i)$.
Let $\delta$ denote the restriction of $\pi_i$ to $\B(M)$. The surjective map $\delta \colon \B(M) \to \B(M\backslash i)$ is called the {\bf deletion map} with respect to the element $i$.
The next proposition states that when $i$ is not a coloop of $M$ this deletion map is an {\bf open tropical modification} along a {\bf tropical rational function} $f \colon \R^{n} / \mathbf{1} \to \R$. We refer the reader to \cite{ShawInt} and \cite{BriefIntro} for an introduction to tropical modifications and tropical rational functions.
\begin{prop}\cite[Proposition 2.25]{ShawInt}\label{prop:deletionMod}
Let $M \in \Mat_{n+1}$ be a loopless matroid and assume $i \in \{0, \dots, n\}$ is not a coloop of $M$.
Then the deletion map
$\delta \colon \B(M) \to \B(M\backslash i)$ is an open tropical modification along a tropical rational function $f: \R^{n} / \mathbf{1} \to \R$
such that
$\divis_{\B(M\backslash i)}(f) = \B(M/i)$.
\end{prop}
Proposition \ref{prop:deletionMod} is expressing the following fact.
If $i$ is not a coloop of $M$ then $M$ and $M \backslash i$ are matroids of the same rank, and thus their Bergman fans are of the same dimension.
The map $\delta$ is one to one except above a codimension-$1$ subset of $\B(M \backslash i)$, which is exactly the Bergman fan $\B(M / i)$.
The pre-image of $\delta$ over any point in $\B(M / i)$ is a half-line in direction ${\bf e}_i$.
The Bergman fan $\B(M)$ can be obtained from the graph of $f$ restricted to $\B(M \backslash i)$ by adding
cones in the direction ${\bf e}_i$ over the
image of $\B(M/i)$.
\begin{exa}
Consider the uniform matroid $M=U_{3,4}$ on the set $\{0, 1, 2, 3\}$.
Then $M\backslash 3$ is the uniform matroid $U_{3,3}$ and $M / 3$ is the uniform matroid $U_{2,3}$.
As we have seen in Example \ref{ex:coarse}, the Bergman fan $\B(M)$ is the union of the cones in $\R^4 / \mathbf{1}$ of the form $\cone\{\mathbf e_i, \mathbf e_j\}$ for all distinct $i,j\in\{0,1,2,3\}$. The Bergman fan $\B(M\setminus 3)$ is all of $\R^3/ \mathbf{1}$, and $\B(M/3)$ is the union of the three rays in $\R^3/ \mathbf{1}$ in the directions $\mathbf e_i$ for $i\in\{0,1,2\}$.
Let $\pi_3\colon \R^{4} \to \R^{3}$ be the linear projection with kernel generated by ${\bf e}_3$.
This map induces the deletion map $\delta \colon \B(U_{3,4}) \to \B(U_{3,3})$, depicted in Figure \ref{fig:mod}.
The tropical rational function $f: \R^{3}/ \mathbf{1} \to \R$ from Proposition \ref{prop:deletionMod} is in this case $f(\bfx_0, \bfx_1,\bfx_2)=\min\{\bfx_0, \bfx_1,\bfx_2\}$.
\begin{figure}[ht]
\begin{center}
\includegraphics[scale=1.8]{modification}
\caption{The deletion map $\delta \colon \B(U_{3,4}) \to \B(U_{3,3}).$}
\label{fig:mod}
\end{center}
\end{figure}
\end{exa}
A deletion map between Bergman fans induces pushforward and pullback maps on tropical cycles.
\begin{defi}\cite[Definition 2.16]{ShawInt} \label{pushModCon}
Let $\delta \colon \B(M) \to \B(M \backslash i)$ be the deletion map with respect to a non-coloop element $i$ of the loopless matroid $M$.
For any $k$, the {\bf pushforward} and {\bf pullback maps} on tropical cycles are maps
$$\delta_{\ast} \colon \mathcal{Z}_k(\B(M)) \to \mathcal{Z}_k(\B(M\backslash i)) \qquad \text{and} \qquad
\delta^{\ast} \colon \mathcal{Z}_k(\B(M\backslash i)) \to \mathcal{Z}_k(\B(M)).$$
The pushforward of a tropical cycle $Z \in \mathcal{Z}_k(\B(M))$ is supported on the polyhedral complex $\delta(Z)$, and has weights described in \cite[Definition 2.16(1)]{ShawInt}.
The pullback of a cycle $Z \in \mathcal{Z}_k(\B(M \backslash i))$ is
the modification of $Z$ along the tropical polynomial function $f \colon \R^{n} / \mathbf{1} \to\R$
associated to $\delta$ by Proposition \ref{prop:deletionMod}.
\end{defi}
Both the pushforward and pullback maps induced by a deletion map $\delta \colon \B(M) \to \B(M \backslash i)$ are group homomorphisms. Moreover, the composition $\delta_{\ast}\delta^{\ast}$ is the identity in $\mathcal{Z}_k(\B(M\backslash i))$ \cite[Proposition 2.23]{ShawInt}.
We now use the pushforward and pullback homomorphisms to relate the CSM cycles of a matroid with the CSM cycles of its deletion and contraction with respect to a non-coloop element $i$.
\begin{prop}\label{prop:pullback}
Let $\delta \colon \B(M) \to \B(M \backslash i)$ be the deletion map with respect to a non-coloop element $i$ of the loopless matroid $M$. Then
\begin{equation}\label{eq:pullback}
\csm_k(M) = \delta^*\csm_k(M\backslash i) - \delta^*\csm_{k}(M/i)
\end{equation}
and
\begin{equation}\label{eq:pushforward}
\delta_{*}\csm_k(M) = \csm_k(M\backslash i) - \csm_{k}(M/i).
\end{equation}
\end{prop}
For the proof of Proposition \ref{prop:pullback} we need the following matroidal result, which we record as a separate lemma.
\begin{lemma}\label{lem:loopscoloops}
Let $S \subseteq T$ be subsets of the ground set of a matroid $M$, and suppose $i \in T \setminus S$.
\begin{enumerate}[label=\it{\alph*}\normalfont{)}]
\item \label{lem:coloop} If $T \setminus i$ is a flat of $M$ then $i$ is a coloop of $M|T/S$.
\item \label{lem:loop} If $S \cup i$ is a flat of $M$ but $S$ is not a flat of $M$ then $i$ is a loop of $M|T/S$.
\item \label{lem:notloopcoloop} If $S, T$ are flats of $M$ but $T \setminus i$ is not a flat of $M$ then $i$ is neither a loop nor a coloop of $M|T/S$.
\end{enumerate}
\end{lemma}
\begin{proof}
Recall that the circuits of the minor $M|T/S$ are the minimal nonempty subsets of the form $C \setminus S$,
where $C$ is a circuit of $M$ contained in $T$ \cite[Section 7]{White1}. This description implies that
$i$ is a coloop of $M|T/S$ if and only if in $M$ the element $i$ is not in the closure of $T\setminus i$. Similarly, $i$ is a loop of
$M|T/S$ if and only if in $M$ the element $i$ is in the closure of $S$. The three assertions in the lemma follow directly from these facts.
\end{proof}
\begin{proof}[Proof of Proposition \ref{prop:pullback}]
The second equation follows directly from the first one by applying $\delta_*$.
To prove \eqref{eq:pullback}, suppose $\sigma_ \F$ is a $k$-dimensional cone of $\B(M)$ corresponding to the flag of flats
$\F: =\{ \emptyset = F_0 \subsetneq F_1 \subsetneq \dots \subsetneq F_{k} \subsetneq F_{k+1} = \{0,\dotsc,n\}\}$ in $M$.
The cone $\delta(\sigma_ \F)$ is the cone $\sigma_{\F'}$ where
$\F' \coloneqq\{ F'_0 \subseteq F'_1 \subseteq \dots \subseteq F'_{k} \subseteq F'_{k+1}\}$
is the chain of flats of $M\setminus i$ defined by $F'_l \coloneqq F_l \setminus i$ for all $l$.
Assume first that $\sigma_{\F}$ is contained in the graph of the function $f \colon \R^{n} / \mathbf{1} \to \R$ restricted to $\B(M \backslash i)$, where $f$ is the tropical rational function of the modification $\delta$.
In this case $\sigma_{\F'}$ has the same dimension as $\sigma_ \F$, and so the chain $\F'$ has also length $k+1$.
By the pullback formula for tropical cycles, the weight of the cone $\sigma_ \F$ in $\delta^*(\csm_k(M\backslash i) - \csm_k(M/i))$ is equal to the weight of
the cone $\sigma_{\F'}$ in the cycle $\csm_k(M\backslash i) - \csm_k(M/i)$.
To show that $\sigma_\F$ has the same weight in both cycles, we thus need to show that
\begin{equation}\label{eq:equalweights}
\prod_{l=0}^k \beta(M|F_{l+1}/F_l) = \prod_{l=0}^k \beta(( M \backslash i) |F'_{l+1}/F'_l) + \prod_{l= 0}^k \beta(( M/ i) |F'_{l+1}/F'_l).
\end{equation}
Let $m$ be such that $i \notin F_m$ and $i \in F_{m+1}$.
For any $l < m$, by Lemma \ref{lem:loopscoloops} \ref{lem:coloop} the element
$i$ is a coloop in $M|(F_{l+1}\cup i)/F_l$, and thus its deletion is the same as its contraction, i.e.,
$M|F_{l+1}/F_l = ( M \backslash i) |F'_{l+1}/F'_l = (M/ i) |F'_{l+1}/F'_l$.
Moreover, since $\sigma_\F$ is in the graph of the function $f$,
for any $l \geq m$ we have that $F_{l+1} \setminus i$ is not a flat of $M$,
otherwise the cone of $\B(M)$ corresponding to the chain of flats
$\{ F'_0 \subseteq F'_1 \subseteq \dots \subseteq F'_{l+1} \subseteq F_{l+2} \subseteq \dotsb \subseteq F_{k+1}\}$ would be below the graph of $f$, contradicting Proposition \ref{prop:deletionMod}.
Therefore, by Lemma \ref{lem:loopscoloops} \ref{lem:loop}, for any $l > m$ we have that
$i$ is a loop in $M|F_{l+1}/(F_l \setminus i)$, and thus again $M|F_{l+1}/F_l = (M / i) |F'_{l+1}/F'_l = (M \setminus i) |F'_{l+1}/F'_l$.
When $l=m$, Lemma \ref{lem:loopscoloops} \ref{lem:notloopcoloop} shows that $i$ is neither a loop nor a coloop of $M|F_{m+1}/F_m$, and so we have
$$\beta(M|F_{m+1}/F_m) = \beta(( M \backslash i) |F'_{m+1}/F'_l) + \beta(( M/ i) |F'_{m+1}/F'_m).$$
Multiplying all these equations proves Equation \eqref{eq:equalweights}. This shows that the cycles $\csm_k(M)$ and $\delta^*(\csm_k(M \backslash i) - \csm_k(M/i))$ agree in the graph $\Gamma_f$ of the function $f$.
By the pullback formula for tropical cycles, any cone of the cycle $\delta^*(\csm_k(M \backslash i) - \csm_k(M/i))$ is either contained in $\Gamma_f$ or it contains the direction $\mathbf{e}_i$. Moreover, the weights of the cones contained in $\Gamma_f$, together with the balancing condition, determine the pullback cycle completely.
Similarly, each $k$-dimensional cone of the coarse subdivision of $|\B(M)|$ is either in $\Gamma_f$ or it contains the $\mathbf{e}_i$ direction. Since the support of the cycle $\csm_k(M)$ is the $k$-skeleton of this coarse subdivision (Proposition \ref{prop:nonzerocones}), the weights in $\csm_k(M)$ of the cones in the $\mathbf{e}_i$ direction are also determined by the weights of the cones in $\Gamma_f$ together with the balancing condition. This shows that the cycles $\csm_k(M)$ and $\delta^*(\csm_k(M \backslash i) - \csm_k(M/i))$ must be the same.
\end{proof}
\subsection{Degrees of CSM cycles and the characteristic polynomial}
We now relate the degrees of the CSM cycles of a matroid to the coefficients of its characteristic polynomial.
If $Z$ and $Z'$ are two tropical cycles
in $\R^{n+1} / \mathbf{1}$, we denote by $Z \cdot Z'$ their stable intersection, and by $Z^k$ the stable intersection of $k$ copies of $Z$; see \cite[Section 3.6]{MaclaganSturmfels}.
\begin{defi}\label{def:degree}
The degree of a $0$-dimensional tropical cycle $Z$ in $\R^{n+1} / \mathbf{1}$ is $\deg(Z) \coloneqq \sum_{z \in Z} w_Z(z)$.
The degree of a $k$-dimensional tropical cycle $Z$ in $\R^{n+1} / \mathbf{1}$ is
$$\deg(Z) \coloneqq \deg( Z \cdot \B(U_{n, n+1})^k).$$
\end{defi}
\begin{exa}\label{ex:degcsmUniform}
Consider the uniform matroid $U_{d+1, n+1}$.
By Example \ref{ex:csmcycleUniform} we have $$\csm_k(U_{d+1, n+1}) = (-1)^{d-k}\binom{n-k-1}{d-k} \B(U_{k+1, n+1})$$ for all $0\leq k\leq d$.
The degree of $\B(U_{k+1, n+1})$ is 1, and so $\deg(\csm_k(U_{d+1, n+1})) = (-1)^{d-k}\binom{n-k-1}{d-k}$.
\end{exa}
The following result generalizes \cite[Theorem 3.5]{Huh:MaxLikely} and \cite[Theorem 1.2]{Aluffi:Grothendieck} to all matroids, not necessarily representable in characteristic 0. Recall that $\overline{\chi}_M$ denotes the reduced characteristic polynomial of the matroid $M$.
\begin{thm}\label{thm:hvector}
If $M \in \Mat_{n+1}$ is a rank $d+1$ matroid then
$$\sum_{k=0}^d \deg(\csm_{k}(M))\,t^k = \overline{\chi}_M(1+t).$$
\end{thm}
\begin{exa}\label{ex:degcsm0d}
The $0$-dimensional CSM cycle of a rank $d+1$ matroid $M$ has degree equal to $(-1)^{d}\beta(M)$, which is equal to the constant coefficient $\overline{\chi}_M(1)$ of the polynomial $\overline{\chi}_M(1+t)$.
The $d$-dimensional CSM cycle of $M$ is equal to the tropical cycle $\B(M)$, which has degree $1$ if $M$ is loopless and $0$ otherwise.
This is the leading coefficient of $\overline{\chi}_M(1+t)$.
\end{exa}
We require the next proposition to prove Theorem \ref{thm:hvector}.
\begin{prop}\label{prop:degpullback}
Let $\delta : \B(M) \to \B(M\backslash i)$ be the deletion map with respect to a non-coloop element $i$ of $M$.
For any $k$-dimensional tropical cycle $Z \in \mathcal{Z}_k(\B(M\backslash i))$,
we have $$\deg(Z) = \deg(\delta^*Z).$$
\end{prop}
\begin{proof}
To aid with notation we assume that $i = n$.
The tropical cycle $\B(U_{n-1, n}) \in \mathcal{Z}_{n-2}(\R^{n}/ \mathbf{1})$ is the tropical hypersurface of the tropical polynomial
$h(\bfx_0, \dots, \bfx_{n-1}) = \min\{ \bfx_0, \bfx_1, \dots, \bfx_{n-1}\}$ on $\R^{n}/\mathbf{1}$.
Let $C_n$ denote the matroid consisting of a single coloop $n$. Then $\B(U_{n-1, n} \oplus C_n) \in \mathcal{Z}_{n-1}(\R^{n+1} / \mathbf{1})$ is also a tropical hypersurface defined by the polynomial $\tilde{h}(\bfx_0, \dots, \bfx_n) = h(\bfx_0 ,\dots, \bfx_{n-1})$.
Let $\pi: \RR^{n+1} / \mathbf{1} \to \RR^{n}/ \mathbf{1}$ be the map induced by the linear projection $\RR^{n+1} \to \RR^{n}$ which forgets the $n$-th coordinate.
Then $\pi^{\ast} \divis(h) = \divis(\tilde{h})$, which implies that $\pi^{\ast} \B(U_{n-1, n}) = \B(U_{n-1, n} \oplus C_n)$.
We have that $\pi_{\ast} \delta^{\ast}Z = \delta_{\ast} \delta^{\ast}Z = Z$.
Applying the projection formula in \cite[Proposition 4.8]{AllermannRau}
yields
\begin{align*}
\B(U_{n-1, n})^k \cdot Z & = \B(U_{n-1, n})^{k-1} \cdot (\divis(h) \cdot \pi_{\ast} \delta^{\ast}Z) \\
& = \B(U_{n-1, n})^{k-1} \cdot ( \pi_{\ast} (\pi^{\ast} \divis(h) \cdot \delta^{\ast}Z) ) \\
& = \B(U_{n-1, n})^{k-1} \cdot \pi_{\ast} (\B(U_{n-1, n} \oplus C_n) \cdot \delta^{\ast} Z) .
\end{align*}
Repeatedly applying this argument $k$ times we obtain
$\B(U_{n-1, n})^k \cdot Z= \pi_{\ast}( \B(U_{n-1, n} \oplus C_n)^k \cdot \delta^{\ast}Z)$.
The degree of a zero cycle is preserved under the pushforward map, and so we have
$\deg( Z) = \deg( \B(U_{n-1, n} \oplus C_n)^k \cdot \delta^{\ast}Z)$.
We will now show that $\deg( \delta^{\ast} Z) = \deg ( \B(U_{n-1, n} \oplus C_n)^k \cdot \delta^{\ast}Z )$.
Let $X \coloneqq \B(U_{n, n+1}) - \B(U_{n-1, n} \oplus C_n).$
Since $n$ is not a coloop of $M$, the support of the tropical cycle $X$ is contained in the closed connected component of $\R^{n+1} / \mathbf{1}$ defined by
$$\Gamma_f(\B(M\backslash n))^- \coloneqq \{ \mathbf x \in \R^{n+1} / \mathbf{1} \mid \mathbf x \cdot \mathbf e_n \leq f (\pi(\mathbf{x})) \}.$$
To compute the stable tropical intersection $\delta^{\ast}Z \cdot X$, denote by $X_{\epsilon}$ the translate of $X$ by $ \epsilon \mathbf e_n$ for $\epsilon < 0$. Then $X_{\epsilon} \cap \delta^{\ast} Z = \emptyset$, and so
$\delta^{\ast}Z \cdot X = 0$. Moreover, we have
$$\delta^{\ast}Z \cdot \big[\B(U_{n, n+1})^k - \B(U_{n-1, n} \oplus C_n)^k\big]
= \delta^{\ast}Z \cdot X \cdot \left[ \sum_{j = 0}^{k-1}
\B(U_{n, n+1})^{k-1-j} \cdot \B(U_{n-1, n} \oplus C_n)^j\right],$$
which is equal to zero by the associativity of the intersection product.
This shows the equality of degrees $\deg( Z) = \deg(\delta^{\ast}Z)$ and proves the lemma.
\end{proof}
\begin{proof}[Proof of Theorem \ref{thm:hvector}]
Both the reduced characteristic polynomial and the CSM cycles satisfy a recursion
via deletions and contractions. More precisely, if $M$ is a loopless matroid and $i$ is not a coloop of $M$, we have
$$\overline{\chi}_M(\lambda) = \overline{\chi}_{M \backslash i}(\lambda) - \overline{\chi}_{M / i}(\lambda) \qquad \text{ and } \qquad
\csm_k(M) = \delta^*(\csm_k(M \backslash i) - \csm_{k}(M/i)),$$
where the equality on the right-hand side follows from Proposition \ref{prop:pullback}.
Since degree is preserved under pullbacks by Proposition \ref{prop:degpullback}, in this case we have
\begin{equation}\label{eq:degreerecursion}
\deg(\csm_k(M)) = \deg(\csm_k(M \backslash i)) - \deg(\csm_{k}(M/i)).
\end{equation}
If $M$ has any loops then
$$\overline{\chi}_M(\lambda) = 0 \qquad \text{ and } \qquad \csm_k(M) = \emptyset.$$
It thus suffices to check the statement for matroids $M$ with no loops and where all the elements are coloops, i.e., $M = U_{d+1, d+1}$.
In this case, the tropical cycle $\B(M)$ is the same as $\R^{n+1} / \mathbf{1}$ equipped with weight $1$ everywhere. The only non-trivial CSM cycle is $\csm_d(\B(M)) = \B(M)$, which is of degree $1$. Therefore
$$\sum_{k=0}^n (-1)^k \text{deg}(\csm_{k}(M))t^k = t^d,$$
whereas by Example \ref{ex:betauniform},
$$\overline{\chi}_M(t+1) =\sum_{k = 0}^d (-1)^{d-k}\binom{d}{k}(t+1)^k= t^d,$$
confirming the desired result.
\end{proof}
\subsection{Conjecture: The $g$-polynomial as intersection numbers}
In this section we give a conjectured presentation of Speyer's $g$-polynomial of a matroid using CSM cycles.
For a general rank $d$ matroid on $n$ elements, the $g$-polynomial of $M$ is defined by way of the $K$-theory of the Grassmannian $\text{Gr}(d,n)$ \cite{FinkSpeyer}. This polynomial is a valuative matroid invariant in the sense of Section \ref{Sec:val} \cite[Section 4]{FinkSpeyer}.
Conjecture \ref{conj:gpoly} describes the coefficients of the $g$-polynomial as intersection numbers in the Bergman fan of $M$ between CSM cycles and certain tropical cycles defined recursively from them.
This formula would offer a Chow theoretic description of this matroid invariant from $K$-theory.
There is an intersection product for tropical cycles contained in Bergman fans of matroids
\cite{ShawInt, FrancoisRau}.
If $M \in \Mat_{n+1}$ is a loopless rank $d+1$ matroid and $\mathcal Z_k(\B(M))$ denotes the group of $k$-dimensional tropical cycles whose support is contained in $\B(M)$, this intersection product gives rise to a
bilinear pairing
$$ \mathcal{Z}_{d-k}(\B(M)) \times \mathcal{Z}_{d-l}(\B(M)) \to \mathcal{Z}_{d-k-l}(\B(M)) $$
for any $k,l$ such that $k+l \leq d$.
In particular, for any $Z \in \mathcal{Z}_k(\B(M))$, the intersection product $\B(M) \cdot Z$ in the matroidal cycle $\B(M)$ is simply $Z$.
Using this product we define a
collection of new tropical cycles $n_k(M) \in \mathcal{Z}_k(\B(M))$ for $k = 0, \dots , d$. Firstly, we set
$$n_d(M) \coloneqq \csm_d(M) = \B(M).$$
Let $A$ be the tropical cycle in $\mathcal{Z}_{d-1}(\B(M))$ obtained by taking the tropical stable intersection in $\R^{n+1} / \mathbf{1}$ of $\B(M)$ with the standard tropical hyperplane $\B(U_{n, n+1})$.
For $k <d$ we define $n_k(M)$ recursively by the formula
\begin{equation}\label{formula:Ncycles}
n_{d-k}(M) \coloneqq (-1)^{k} A^{k} - \Bigg [\sum_{i= 0}^{k-1} \csm_{d-k+i}(M) \cdot n_{d-i}(M) \Bigg ],
\end{equation}
where the intersection products above are now in $\B(M)$.
\begin{conj}\label{conj:gpoly}
The $g$-polynomial of a loopless rank $d+1$ matroid $M \in \Mat_{n+1}$ is equal to
\begin{equation}\label{eqn:conj}
g_M(t) = \sum_{k = 0}^d (-1)^{d-k}\,\deg(\csm_{k}(M)\cdot n_{d-k}(M))\,t^{k+1},
\end{equation}
where the intersection products occur in the matroidal cycle $\B(M)$ of $M$.
\end{conj}
\begin{exa} For a loopless matroid $M$ of rank $d+1$,
Formula \eqref{formula:Ncycles} gives
\begin{align*}
n_{d-1}(M) &= -A - \csm_{d-1}(M),\\
n_{d-2}(M) &= A^2 + A \cdot \csm_{d-1}(M) + \csm_{d-1}^2(M) - \csm_{d-2}(M).
\end{align*}
The linear, quadratic, and cubic coefficients of the
polynomial on the right hand side of Equation \eqref{eqn:conj} are
up to sign
\begin{align*}
\csm_0(M) \cdot n_{d}(M) &= \csm_0(M) = (-1)^d \beta(M), \\
- \csm_1(M) \cdot n_{d-1}(M) &= \deg (\csm_1(M)) + \csm_1(M)\cdot\csm_{d-1}(M),\\
\csm_2(M) \cdot n_{d-2}(M) &= \deg (\csm_2(M)) + \deg (\csm_2(M) \cdot \csm_{d-1}(M)) \\
&\quad + \csm_2(M) \cdot \csm_{d-1}^2(M) - \csm_2(M) \cdot \csm_{d-2}(M).
\end{align*}
Consider the case $d=2$, so $M \in \Mat_{n+1}$
is a matroid of rank $3$ and $\B(M)$ is a $2$-dimensional tropical cycle.
The intersection products above are
\begin{align*}
\csm_0(M) \cdot n_2(M) &= \beta(M), \\
- \csm_1(M) \cdot n_{1}(M) &= \deg (\csm_1(M)) + \csm_1^2(M),\\
\csm_2(M) \cdot n_{0}(M) &= 1 + \deg (\csm_{1}(M)) + \csm_{1}^2(M) - \beta(M).
\end{align*}
For simplicity, let us assume that $M$ has no double points. By repeatedly applying Equation \eqref{eq:degreerecursion}, we find that $\deg(\csm_1(M)) = -(n-2)$.
Moreover,
the formula for intersection products of tropical cycles in $2$-dimensional Bergman fans in \cite[Definition 3.6]{BrugalleShaw} gives us
$$\csm_1^2(M) = (n-2)^2 - \sum_{\substack{ F \in \mathcal{L}(M) \\ r(F) = 2}}(|F| -2)^2.$$
It can be verified that these formulae produce the coefficients of the $g$-polynomials in the examples of rank $3$ matroids presented in \cite[Section 10]{Speyer:Ktheory}.
\end{exa}
\begin{exa}
Suppose $M$ is the uniform matroid $M=U_{d+1, n+1}$.
In this case we have $A = \B(U_{d, n+1}) \in \mathcal{Z}_{d-1}(\B(M))$ and
$A^k = \B(U_{d-k+1, n+1}) \in \mathcal{Z}_{d-k}(\B(M))$.
By Example \ref{ex:csmcycleUniform}, we have $\csm_k(M) = (-1)^{d-k} \binom{n- k-1}{d-k} A^{d-k}$.
We claim that $n_{d-k}(M) = \binom{n-d-1}{k}A^k$ for all $0 \leq k \leq d$. This formula
is true when $k = 0$, so assume that it
holds for all $l < k$ and proceed by induction.
By Formula (\ref{formula:Ncycles}) we have
$$n_{d-k}(M) = \left[(-1)^{k} - \sum_{i= 0}^{k-1} (-1)^{k-i} \binom{n-d+k-i-1}{k-i} \binom{n-d-1}{i} \right]A^k.$$
Then the fact that $n_{d-k}(M) = \binom{n-d-1}{k}A^k$ follows from the binomial identity
$$(-1)^k = \sum_{i = 0}^k (-1)^{k-i} \binom{m+k-i}{k-i}\binom{m}{i}$$
when $m = n-d-1$.
From these expressions we conclude that
$$\deg \left[ ( -1)^{d-k}\csm_{k}(M) \cdot n_{d-k}(M) \right] = \binom{n- k-1}{d-k} \binom{n-d-1}{k}.$$
This coincides with the formula for the coefficients of the $g$-polynomial for uniform matroids \cite[Proposition 10.1]{Speyer:Ktheory}.
\end{exa}
| 184,309
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TITLE: Find the locus of the point of intersection of 2 normals to a parabola
QUESTION [1 upvotes]: This is a question I am struggling with as I work through a pure maths book as a hobby:
Prove that the normal to the parabola $y^2=4ax$ at the point $at^2,2at)$ has the equation $y+tx=2at+at^3$. The normals at the points $P(ap^2, 2ap)$ and $Q(aq^2,2aq)$ intersect at the point R. Find the coordinates of R in terms of $(p+q)$ and $pq$. if O is the vertex of the parabola and P and Q are variable points such that $\angle POQ$ is a right angle, find the locus of R; verify that it is a parabola and find the coordinates of the vertex.
For the first part:
$y=2at \rightarrow \frac{dy}{dt}=2a\\
x=at^2 \rightarrow \frac{dx}{dt}=2at\\
\frac{dy}{dx}=\frac{2a}{2at}=\frac{1}{t}\\$
$\rightarrow$ Gradient of normal = $-t \rightarrow$ equation of normal at $(at^2,2at)$ is:
$\rightarrow y-2at={-t}(x-at^2)\\
\rightarrow y+tx=2at+at^3$
So we then find the point of intersection of the 2 normals at P and Q at point R:
$2ap+ap^3-px=2aq+aq^3-qx$
This eventually gives coordinates of
$\\x=a\{2+(p+q)^2 -pq\}\\ y=-apq(p+q)$
Now I need to use this to find the locus of R. I can find no way get rid of the p's and q's to express y in terms of x. I feel the fact that POQ is a right angle is relevant but cannot see how, unless it is merely to indicate that P and Q are on opposite sides of the x-axis. The book says the answer is $y^2=16a(x-6a)$
REPLY [2 votes]: Slope of $OP$ is $$\frac{2ap}{ap^2}=\frac2p$$
Similarly, slope of $OQ$ is $2/q$. If they are perpendicular, it means
$$\frac2p\frac2q=-1$$or $pq=-4$. Plug this into your equation for $y$, and write $(p+q)$ in terms of $y$. Then use this expression, and plug it into $x$.
| 112,228
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1621 GLENEAGLES $81,500
Kokomo, Indiana 46902
United States
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- Beds
- 3
- Baths
- 1.50
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- Gas,Forced Air
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- Year Built
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- One Story
- Garage Type
- Attached
- Garage Size
- 240 01-16-2019.
Listing provided courtesy of .
| 224,358
|
The links below are provided as a service to American Breast Care retailers. Please visit the links below on a frequent, regular basis to keep you to up to date. Medicare Policies change often.
Please direct all questions regarding billing and policy to your DME MAC as ABC cannot interpret Medicare policy.
Click here to visit the Medicare Pricing, Data Analysis and Coding (PDAC) website for Healthcare Common Procedure Coding System (HCPCS) coding verification. Visit this site to identify the proper billing code for a DMEPOS item.
Your DME MAC is a valuable tool and resource. Please visit your DME MAC’s website for Medical policies (LCDs), policy articles, fee schedules, Medicare Supplier Manual and forms. Be sure to review and download the “Local Coverage Determination (LCD) for External Breast Prostheses” and the “Local Coverage Article for External Breast Prostheses”. Remember to sign up for your DME MAC’s listserv/newsletter to keep you informed.
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| 327,909
|
Hello Adcock Eagles! I am so happy to be part of this staff and community! I moved to Las Vegas in 1998 from Washington state after obtaining my Bachelor’s in Education from Eastern Washington University. I taught first and second grade for seven years, and during that time I attended UNLV where I got a Master’s Degree in Library Science. I was then lucky enough to open a new school as the Librarian and I remained at that school for the next twelve years. I have endorsements in STEM, TESL and GATE, and am now using my GATE endorsement to teach my wonderful students here at Adcock. Being a GATE Specialist is something I am very blessed and honored to be doing. I love waking up each morning knowing that my students will be walking into my room full of curiosity and a love of learning. We strive to “think outside the box” in GATE and have so much fun with our project based learning units!
Vicki Korzeniewski
GATE
smithv
Contact Specialist
| 260,901
|
TITLE: Is there a mistake in the problem? Continuity of a two-variable function.
QUESTION [2 upvotes]: Let $\phi : \mathbb{R} \rightarrow \mathbb{R}$ be a $C^2$ function, such that $\phi (0) = 0$ and $\phi''(0) \neq 0$. If
$$
f(x,y) =
\begin{cases}
\frac{x\phi(y) - y\phi(x)}{x^2+y^2}, &\text{if } (x,y) \neq (0,0) \\
0, &\text{if } (x,y) = (0,0)
\end{cases},
$$
then show that $f$ is continuous, but not $C^1$.
I have tried solving it. The problem is, that if $\phi(x)=\sqrt{x}$, then $f$ would not be continuous (taking the path $y=2x$, and letting $x \rightarrow 0$ makes the limit $-\infty \neq 0$). Did I not understand the question properly? Is taking $\phi(x) = \sqrt{x}$ as a counter example a mistake? Is there some mistake with the problem? And if not, could I get some sort of hint on solving it, as I have tried a few methods, but I can't seem to solve it.
Edit: I may have solved it, but I am not sure if it is correct.
Let $\epsilon > 0$. We have
\begin{align}
|f(x,y) - f(0,0)| &= \frac{|x\phi(y)-y\phi(x)|}{x^2+y^2} \\
&\leq \frac{|x||\phi(y)|+|y||\phi(x)|}{x^2+y^2} \\
&\leq \frac{\sqrt{x^2+y^2}(|\phi(y)| + |\phi(x)|)}{x^2+y^2}, \quad \text{since }|x|,|y|\leq\sqrt{x^2+y^2} \\
&= \frac{|\phi(x)| + |\phi(y)|}{\sqrt{x^2+y^2}}.
\end{align}
For any of those $x,y \in \mathbb{R}^*$, because $\phi$ is continuous at $0$ with $\phi(0) = 0$, by having $\epsilon_1 = y^2 > 0$ and $\epsilon_2 = x^2 > 0$, there must be some $\delta_1, \delta_2 > 0$, such that if $|x|\leq\delta_1$ and $|y|\leq\delta_2$, then $|\phi(x)|\leq\epsilon_1=y^2$ and $|\phi(y)|\leq\epsilon_2=x^2$. From this, we pick $\delta_m = \min\{\delta_1, \delta_2\}$ and have $|x|,|y|\leq\delta_m$.
Continuing
\begin{align}
\frac{|\phi(x)| + |\phi(y)|}{\sqrt{x^2+y^2}} &\leq \frac{x^2+y^2}{\sqrt{x^2+y^2}} \\
&= \sqrt{x^2+y^2} \\
&\leq |x| + |y| \\
&\leq 2\delta_m
\end{align}
We pick $\delta = \min\{\epsilon, 2\delta_m\}$, making $f$ continuous at $(0,0)$.
Could someone tell me if taking $\epsilon_1 = y^2$ and $\epsilon_2 = x^2$ was a mistake?
REPLY [2 votes]: You cannot let $\epsilon_1=x^2$ and then apply the epsilon-delta definition of continuity. The $\epsilon_1$ you use must be a fixed number, not function of $x$.
As soon as you applied the inequality
$$
|f(x,y) - f(0,0)|
\leq \frac{|x||\phi(y)|+|y||\phi(x)|}{x^2+y^2},
$$
you were doomed to failure. The problem is that there that $\phi'(x)$ may very well be nonzero, so that $\phi(x)$ and $\phi(y)$ are approximately linear in $x,y$ near zero, meaning the numerator and the denominator are both quadratic in $x$ and $y$. Thus, as $x,y\to0$, you approach a constant, but it may not be zero.
Before applying the upper bound, you had the numerator was $x\phi(y)-y\phi(x)$. There, the linear parts of $\phi$ cancel out, making the numerator go to zero at a faster order and allowing you to show the limit is zero.
How to make all this rigorous? By Taylor's theorem, given $\epsilon>0$, you can choose $\delta>0$ so $|x|<\delta$ implies $|\phi(x)-\phi(0)-\phi'(0)x|=|\phi(x)-\phi'(0)x|\le \epsilon |x|$. Then when $|x|+|y|<\delta$,
\begin{align}
|f(x,y) - f(0,0)| &= \frac{|x\phi(y)-y\phi(x)|}{x^2+y^2} \\
&\leq \frac{|x\phi'(0)y-y\phi'(0)x|+|x(\phi(y)-\phi'(0)y)|+|y(\phi(x)-\phi'(0)x)|}{x^2+y^2} \\
&\leq \frac{0+\epsilon(|x||y|+|y||x|)}{x^2+y^2}.
\end{align}
and I think you can take it from here.
| 60,399
|
TITLE: Why aren't jet engines used in cars?
QUESTION [14 upvotes]: From my "research", I found that there have been efforts to make cars with jet engines in them (mainly race cars), but all of them relied on the thrust produced by the jet engine.
Why can't a jet engine be used to turn a driveshaft the same way an ICE turns a driveshaft?
This would mean not making use of any thrust generated by the jet engine and only leveraging the rotational force produced by the jet engine.
REPLY [4 votes]: One thing that hasn't been mentioned yet is that although gas turbine engines have an excellent power to weight ratio, they are less efficient than piston engines of similar power output, especially in small sizes.
Gas turbines typically use only a third of the oxygen in the inlet air, or to put it another way, they ingest three times the air required to burn the fuel. This is because the hot parts of the gas turbine (especially the turbine blades) are subject to constant high temperature and would melt if the minimum air required for combustion was used. On the other hand, gasoline engines use the minimum air required for combustion (as do diesel engines when operating at full throttle.) The combustion temperature in a piston engine is therefore higher than in a gas turbine, but it does not damage the cylinder walls because they see only the average temperature of the full cycle.
The low flame temperature in a gas turbine limits the theoretical Carnot efficency. Because the gas turbine uses three times the mass of air required for combustion, a lot of energy is carried away in the exhaust. A more detailed analysis will show that these two statements are more or less equivalent.
Thus a truck diesel engine can achieve 50% efficiency (with the help of a turbocharger) which is unheard of in a gas turbine. Combined cycle power stations achieve better than 50% efficiency by raising steam from the exhaust gas of gas turbines to extract extra energy. A typical mass balance is: 100MW fuel, 40MW shaft power from the gas turbine, 40MW steam raised from the exhaust, 20MW lost to the chimney. The steam raised can then be used to generate electricity at 40% efficiency, yielding another 16MW. Total 56MW electric from 100MW fuel.
Installing a combined cycle plant in a car (or even a truck) is prohibitively complex. Therefore, piston engines offer better efficiency.
This is especially the case in small sizes. Better efficiency is achieved by high combustion chamber pressure. But the pressure that can be developed / recovered by a single row of blades is proportional to the square of the blade velocity, which means that rotors in small gas turbines have to spin very fast or have many rows of blades on them in order to achieve decent combustion chamber pressure.
| 5,550
|
TITLE: Rayleigh Quotient on Legendre´s Differential Equation
QUESTION [0 upvotes]: We know that Legendre's DE
$$ \frac{d}{d\phi}\bigg(sin\phi\frac{dg}{d\phi}\bigg) + \big(\mu - \frac{m^2}{sin\phi}\big)g = 0 $$ can be transformed by letting $x = cos\phi$ into
$$ \frac{d}{dx}\bigg((1-x^2)\frac{dg}{dx}\bigg) + \big(\mu - \frac{m^2}{1-x^2}\big)g = 0 $$
I want to prove, by using Rayleigh's quotient, that $\mu \geq 0$. Can $\mu = 0$ be an eigenvalue? If not, are there extra conditions that asure so?
For a general Sturm-Liouville problem given by
$$ \frac{d}{dx}\bigg(p(x)\frac{dg}{dx}\bigg) + \big(\mu\sigma(x) + q(x)\big)g = 0 $$
Rayleigh's Quotient is given by:
$$\mu = \frac{-p(x)g(x)g'(x)|_{-1}^1+\int_{-1}^1[p(x)g_x^2(x)-q(x)g^2(x)]dx}{\int_{-1}^{1}g^2(x)\sigma(x)dx} $$
where $g_x(x) = g'(x)$. In this particular case we have that
$$ p(x) = 1-x^2; \ \sigma(x) = 1; \ q(x) = -\frac{m^2}{1-x^2}$$
Therefore from $p(x)$ it's clear that our expression
$$-p(x)g(x)g'(x)|_{-1}^1 = -p(1)g(1)g'(1) + p(-1)g(-1)g'(-1) = 0$$
and our denominator is positive, so it remains the other integral is positive.
$$ \int_{-1}^1[p(x)g_x^2(x)-q(x)g^2(x)]dx = \int_{-1}^1[(1-x^2)g_x^2(x)+\frac{m^2}{1-x^2}g^2(x)]dx$$
This is where I got stuck, how can I prove this?
That said integral is positive.
Also, there's conditions on $g(\pm 1) < \infty$. I was thinking, are there conditions that give $\mu = 0$ as an eigenvalue?
REPLY [1 votes]: Consider
$$
g(x)=\frac{1}{2}\ln\frac{1+x}{1-x}.
$$
For this,
$$
\frac{dg}{dx}=\frac{1}{2}\left(\frac{1}{1+x}+\frac{1}{1-x}\right)=\frac{1}{1-x^2} \\
\frac{d}{dx}\left((1-x^2)\frac{dg}{dx}\right)=0.
$$
So $0$ is an eigenvalue, when $m=0$. The function $g$ is square integrable, and $Lg=0$ where
$$
Ly=-\frac{d}{dx}\left((1-x^2)\frac{dy}{dx}\right).
$$
This is in addition to the constant function $1$ also being a solution of $Lg=0$. Both $1$ and $\ln\frac{1+x}{1-x}$ are legitimate solutions of $Lg=0$. The ordinary Legendre problem has $0$ for an eigenvalue, with eigenfunction $g=1$.
You can also show that
$$
g_m(x) = (1-x^2)^{m/2}g^{(m)}(x)
$$
is a solution of
$$
-\frac{d}{dx}\left((1-x)^2\frac{dg_m}{dx}\right)+\frac{m^2}{1-x^2}g_m = 0.
$$
This holds for $m=0,1,2,3,\cdots$.
| 99,260
|
A red and black leather jacket worn by Michael Jackson in the video to 'Thriller' has sold for $1.8m (£1.1m) at auction a day after the second anniversary of his death.
The signed garment far exceeded its lower estimate of $200,000 (£125,600) when sold this weekend at Julien's in Beverly Hills.
Auction administrator Darren Julien told the press that the lot was purchased by Texan entrepreneur Milton Verret, who plans to take it 'on tour' to raise money for children's hospitals.
Jackson wore two jackets in the video and the second garment is on display at the Rock and Roll Hall of Fame in Cleveland. Julien had previously described the jacket as "the most recognised and significant piece of pop culture".
A portion of the sale proceeds will be donated to the Shambala wildlife reserve, which is home to two Bengal tigers once owned by Jackson named Thriller and Sabu.
Marilyn Monroe's 'subway' dress from The Seven Year Itch also exceeded its estimate when sold at auction earlier this month. The actress wore the outfit in a scene where air from a passing train blows the dress up around her head.
It had been predicted to sell for between $1m (£618,000) and $2m (£1.2 million), but was ultimately purchased for $4.6m (£2.8m).
Watch the full 13-minute video to 'Thriller' by Michael Jackson below:
| 123,263
|
Batman (1966)
1 audio problem in The Great Escape (1)
Directed by: Leslie H. Martinson
Starring: Adam West, Burt Ward, Cesar Romero, Lee Meriwether
Genres: Action, Adventure, Comedy, Crime, Family
The Great Escape (1) - S3-E21
Audio problem: When the camera shows the flames of the exhaust, the accompanying sound is wrong. The turbine starting up was played when the flames were seen, instead of the roaring sound effect.
Add timeMovie Nut
| 288,177
|
Jamaica to welcome Courtyard by Marriot Hotel in third quarter of 2015
Marriott International has announced that it will open the first Courtyard by Marriott hotel in Kingston in Jamaica in the third quarter of 2015. The hotel group will have 129 guest rooms and suites, and features green building design, construction, and operations as it seeks a LEED Silver Certification.
Marriott International will be opening a brand new hotel in the Caribbean this autumn in a project that is being developed by Caribe Hospitality Jamaica.
The hotel group will debut its Courtyard by Marriott brand in Kingston, Jamaica with a new hotel in the third quarter of 2015.
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Facilities at the Courtyard by Marriott hotel include the brand's signature fresh business lobby, a lobby bar, a casual restaurant, flexible meeting space, a swimming pool and a modern fitness centre.
Laurent de Kousemaeker, the Chief Development Officer, for Caribbean and Latin America at Marriott International, commented:
"With the success of the brand across the region, the Courtyard by Marriott brand has become a favourite amongst regional travellers.
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Jamaica is an island nation located in the Caribbean Sea and one of the world's leading tourist destinations. With large white sand beaches fringed with warm turquoise ocean waters, soaring mountains, cascading waterfalls, verdant undulating green hills, and dense forests, holidaymakers will have plenty of opportunities to admire the country's stunning natural scenery and sites.
Popular tourist attractions include the Bob Marley Museum, Kingston Harbour, and Rose Hall House, a former sugar...
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\begin{document}
\title{Computation of generalized Killing spinors on reductive homogeneous spaces}
\author[Andree Lischewski]{Andree Lischewski}
\address[Andree Lischewski]{Humboldt-Universit\"at zu Berlin, Institut f\"ur Mathematik\\
Rudower Chaussee 25, Room 1.310, D12489 Berlin, Germany}
\email{lischews@mathematik.hu-berlin.de}
\begin{abstract}
We determine the holonomy of generalized Killing spinor covariant derivatives of the form $D= \nabla + \Omega$ on pseudo-Riemannian reductive homogeneous spaces in a purely algebraic and algorithmic way, where $\Omega : TM \rightarrow \Lambda^*(TM)$ is a left-invariant homomorphism. This is essentially an application of the theory of invariant principal bundle connections defined over homogeneous principal bundles. Moreover, the algorithm allows a computation of the associated Killing superalgebra in certain cases. The procedure is demonstrated by determining the super\texttt{}symmetries of certain homogeneous M2 duals, which arise in M-theory.\\
\newline
\smallskip
\noindent \textbf{\keywordsname} \textit{Killing spinors, homogeneity, holonomy, supergravity}
\end{abstract}
\maketitle
\tableofcontents
\section{Introduction}
The holonomy algebra of the Levi Civita connection of a symmetric space $(H/K,g)$ can be computed in purely algebraic terms once the Lie algebraic structure of $\mathfrak{h}$ and $\mathfrak{k}$ is known. In particular, it is easy to determine the space of parallel tensors or spinors on a given symmetric space. More generally, \cite{fu,brm} presents an algorithmic procedure which turns the Killing transport equation on a reductive homogeneous space $H/K$ with $H$-invariant metric $g$ and reductive split $\mathfrak{h} = \mathfrak{k} \oplus \mathfrak{n}$ into something algebraic, making it possible to determine the Lie algebra of Killing vector fields, which in general need not to coincide with $\mathfrak{h}$.\\
Turning to the spinorial analogue, we ask whether also the space of geometric Killing spinors, i.e. solutions $\ph \in \Gamma(S^g)$ of $\nabla^{S^g}_X \ph = \lambda \cdot X \cdot \ph$ for some $\lambda \in \R \cup i \R$ on $(H/K,g)$ can be determined using purely algebraic methods. Killing spinors on pseudo-Riemannian reductive homogeneous spaces have been studied intensively in \cite{bfkg,kath}. The latter reference shows how to construct pseudo-Riemannian reductive homogeneous spaces admitting Killing spinors as so called T-dual spaces of Riemannian homogeneous spaces with Killing spinors, which was one of the first construction methods of Killing spinors in arbitrary signature.\\
More generally than Killing spinors, we consider in this article a pseudo-Riemannian reductive homogeneous space $(M=H/K,g)$ and ask for the existence of spinor fields $\ph \in \Gamma(S^g)$ which are parallel wrt. the connection
\begin{align}
D = \nabla^{S^g} + \Omega \cdot , \label{tre}
\end{align}
where $\nabla^{S^g}$ is the Levi-Civita spin connection, $\Omega: TM \rightarrow Cl(TM,g)$ is $H-$invariant and $\Omega(X)$ acts on spinor fields by Clifford multiplication $\cdot$ for fixed $X \in TM$. Let us elaborate on two examples in which the study of (\ref{tre}) appears naturally.\\
First, generalized Killing spinor equations on Lorentzian manifolds of the form (\ref{tre}) arise in physics in supergravity theories when setting the gravitino variation to zero, i.e. $D-$parallel sections give supersymmetries, cf. \cite{jose2,of12}. In this case $\Omega$ is made up of other bosonic fields of the theory. Often the spinors are additionally subject to further algebraic equations. In this situation one is not only interested in knowing whether there are $D-$parallel spinors but also the dimension of the space of $D-$parallel spinors, i.e. the number of unbroken supersymmetries, is important. For instance, the homogeneity theorem states that backgrounds in certain supergravity theories which preserve a sufficiently large fraction of supersymmetry are automatically homogeneous, cf. \cite{jhom}. Moreover, symmetric backgrounds of 11- and 10-dimensional supergravity have been classified recently in \cite{jose,jn} under the additional assumption that all the other bosonic data are also invariant. However, it is yet unclear which of these solutions to the bosonic field equations are also supersymmetric, i.e. one has to go through the classification list and distinguish those backgrounds which admit solutions to (\ref{tre}). This is work in progress. Group-theoretical methods have earlier been used in \cite{ksh} to construct the Killing spinors of special classes of homogeneous supergravity backgrounds.\\
Second, examples of the generalized Killing spinor equation (\ref{tre}) also appear in a more mathematical context, cf. \cite{am}: Let $(\widehat{N},\widehat{g})$ be a Riemannian spin manifold with oriented hypersurface $(N,g)$. If $\widehat{\ph}$ is a parallel spinor on $\widehat{N}$, then its restriction $\ph$ to $N$ satisfies (\ref{tre}) with $\Omega$ being $\frac{1}{2} \cdot$ the Weingarten tensor of the embedding. Conversely, if $W \in \Gamma(End(TN))$ is any symmetric tensor field and $\ph \in \Gamma(N,S^g)$ is a spinor satisfying (\ref{tre}) with $\Omega = \frac{1}{2} \cdot W$ and all data are real analytic, then there is an ambient space $\widehat{N}= N \times (- \epsilon, \epsilon)$ in which $N$ embeds with Weingarten tensor $W$ and $\ph$ extends to a parallel spinor on $\widehat{N}$.\\
Thus, given a reductive homogeneous spin manifold $(M^n=H/K,g)$ and some $H-$invariant $\Omega \in Sym(TM)$, one can use (\ref{tre}) to decide whether $(M^n,g)$ can be embedded as oriented hypersurface $M^n \subset \widehat{M}$ with Weingarten tensor $\Omega$ into a space $\widehat{M}$ admitting a parallel spinor.\\
\newline
Motivated by these examples, we ask for an algebraic algorithm which solves (\ref{tre}), i.e. determines the dimension of the space of $D-$parallel spinors on a given pseudo-Riemannian reductive homogeneous space $(H/K,g)$ with reductive split $\mathfrak{h}= \mathfrak{k} \oplus \mathfrak{n}$. Under further generic assumptions, namely that the spin structure is homogeneous, as to be made precise in section \ref{se1}, we find in section \ref{se3} that such an algorithm does indeed exist. The only data which enter are the Lie algebraic structure of $\mathfrak{k} \oplus \mathfrak{n}$ and the $Ad_K$-invariant inner product on $\mathfrak{n}$ which corresponds to $g$. The algorithm is essentially an application of Wang's theorem and the theory of invariant connections on principal bundles as studied in detail in \cite{kn1,kn2}. Thus, the theory which is underlying the algorithm is not new but here it is presented in a way such that it is directly accessible for concrete computations. It has earlier been used in \cite{hammerl1} to compute the conformal holonomy group of the product of two spheres. In section \ref{se5} we show based on our previous results how the Killing superalgebra of a supergravity background defined over a pseudo-Riemannian reductive homogeneous space can be computed purely algebraically. \\
Section \ref{se6} applies these results to an interesting class of 11-dimensional $M-$theory backgrounds studied in \cite{fu}: Motivated by the search for new gravity duals to M2 branes with $N>4$ supersymmetry, equivalently characterized as M-theory backgrounds with Killing superalgebra $\mathfrak{osp}(N|4)$ for $N>4$, one classifies homogeneous M-theory backgrounds with symmetry Lie algebra $\mathfrak{so}(n) \oplus \mathfrak{so}(3,2)$. One finds a number of new backgrounds for $n=5$ of the form $S^4 \times X^7$, where the Lorentzian factor $X^7$ is reductive homogeneous under the action of $SO(3,2)$. However, it remains unclear how much supersymmetries these backgrounds preserve. We study two examples of such backgrounds in section \ref{se6} and determine the space of $D-$parallel spinors using the algorithm developed.\\
The final section \ref{coa} extends the algorithm to conformal geometry and oulines how to solve the twistor equation on a reductive homogeneous space.\\
\newline
\textbf{Acknowledgment} The author gladly acknowledges support from the DFG (SFB 647 - Space Time Matter at Humboldt University Berlin) and the DAAD (Deutscher Akademischer Austauschdienst / German Academic Exchange Service). Furthermore, the author would like to thank José Figueroa-O'Farrill and Noel Hustler for many helpful discussions.
\section{Facts about reductive homogeneous spaces} \label{se1}
Our notation for (reductive) homogeneous spaces follows \cite{aa,hammerl1}: Let $M=H/K$ be a connected homogeneous space for some Lie group $H$ and closed subgroup $K$. We shall in addition assume that $H/K$ is \textit{reductive}, i.e. there exists a -from now on fixed- subspace $\mathfrak{n}$ of $\mathfrak{h}$ such that
\begin{align*}
\mathfrak{h} = \mathfrak{k} \oplus \mathfrak{n} \text{ and } \left[\mathfrak{k},\mathfrak{n} \right] \subset \mathfrak{n}.
\end{align*}
This allows a natural identification $T_{eK}M \cong \mathfrak{n}$, where $e \in H$ is the neutral element. $H-$invariant tensor fields on $H/K$ correspond to $Ad_{K}:K \rightarrow GL(\mathfrak{n})$- invariant tensors of the same type on $\mathfrak{n}$, where the correspondence is given by evaluating the tensor field at the origin $eK \in H/K$.
Let $g$ be a $H-$\textit{invariant} signature $(p,q)$-metric on $H/K$, i.e. for each $h \in H$ left multiplication $l_h$ with $h$ is an isometry. $g$ corresponds to an $Ad_{K}$-invariant scalar product $\langle \cdot, \cdot \rangle_{\mathfrak{n}}$ of the same signature on $\mathfrak{n}$, i.e. $Ad_{K}$ takes values in $O(\mathfrak{n}, \langle \cdot, \cdot \rangle_{\mathfrak{n}})$.\\
\newline
We briefly describe spin structures on oriented pseudo-Riemannian reductive homogeneous spaces. Consider the $SO(p,q)$-bundle $\mathcal{P}^g \rightarrow M$ of oriented orthonormal frames of $(M,g)$. We have
\begin{align*}
\mathcal{P}^g \cong H \times_{Ad_{K}} SO(\mathfrak{n}),
\end{align*}
and
\begin{align*}
\mathcal{P}^g \times_{SO(p,q)}\R^n \cong TM \cong H\times_{Ad_{K}} \mathfrak{n},
\end{align*}
where the latter isomorphism is given by $dl_h d\pi_e X \mapsto [h,X]$ for $X \in T_{eK} \left(H/K \right)$.\\
Any lift of the isotropy representation $Ad_{K}$ to the spin group $Spin(\mathfrak{n})$, i.e. any map $\widetilde{Ad}_{K}:K \rightarrow Spin(\mathfrak{n})$ such that the diagram
\begin{align*}
\begin{xy}
\xymatrix{
Spin(\mathfrak{n}) \ar[rd]^{\lambda} & \\
K \ar@{->}[r]^{Ad_{K}} \ar[u]^{\widetilde{Ad}_{K}} & SO(\mathfrak{n}) \\
}
\end{xy}
\end{align*}
commutes, allows us to fix a \textit{homogeneous spin structure} $(\mathcal{Q}^g=H \times_{\widetilde{Ad}_{K}}Spin(\mathfrak{n}),f^g)$ of $(M,g)$ (cf. \cite{bfkg}), where $f^g:\mathcal{Q}^g \rightarrow \mathcal{P}^g$ is simply the double covering $\lambda:Spin(p,q) \rightarrow SO(p,q)$ in the second factor. From now on we shall always assume that $(M,g)$ admits a homogeneous spin structure and think of this structure as being fixed. For algebraic properties of Clifford algebras $Cl$, their Clifford groups $Cl^*$ and spinor modules $\Delta$, we refer to \cite{ba81,lm,har}. The real or complex spinor bundle is the associated bundle $S^g := \mathcal{Q}^g \times_{Spin(p,q)}\Delta_{p,q}$.
\section{The algorithm} \label{se3}
Let $(M,g)$ be a connected and oriented $n-$dimensional pseudo-Riemannian reductive homogeneous spin manifold of signature $(p,q)$ with fixed decomposition $M=H/K$, $\mathfrak{h} = \mathfrak{k} \oplus \mathfrak{n}$ and corresponding $Ad_{K}$-invariant inner product $\langle \cdot, \cdot \rangle_{\mathfrak{n}}$ on $\mathfrak{n}$. Let $\nabla=\nabla^{S^g} :\Gamma(S^g) \rightarrow \Gamma(T^*M \otimes S^g)$ denote the spinor covariant derivative induced by the (lift of the) Levi Civita connection.
We want to compute the number of linearly independent spinor fields which are parallel wrt. the \textit{modified covariant derivative}
\begin{align*}
D &= \nabla + \Omega: \Gamma(S^g) \rightarrow \Gamma(T^*M \otimes S^g),\\
D_X \ph &= \nabla_X \ph + \Omega(X) \cdot \ph \text{ for }X \in \mathfrak{X}(M), \ph \in \Gamma(S^g),
\end{align*}
where $\Omega: TM \rightarrow Cl(TM,g) \cong \Lambda^*(TM)$ is a vector bundle homomorphism which is \textit{left-invariant}, i.e. $l_h^* \Omega = \Omega$ for $h \in H$, or in more detail
\begin{align}
l_{h^{-1}}^*\left(\Omega \left(dl_h(X)\right) \right) \stackrel{!}{=} \Omega(X) \text{ for all }X \in TM, h \in H. \label{dufff}
\end{align}
To this end, we introduce the homogeneous $Cl^*(\mathfrak{n},\langle \cdot, \cdot \rangle_{\mathfrak{n}})$-bundle $\overline{\mathcal{Q}}:= H \times_K Cl^*(\mathfrak{n},\langle \cdot, \cdot \rangle_{\mathfrak{n}})$, where $K \rightarrow Cl^*(\mathfrak{n},\langle \cdot, \cdot \rangle_{\mathfrak{n}})$ acts by trivial extension of $\widetilde{Ad}^{H/K}:K \rightarrow Spin(\mathfrak{n}) \subset Cl^*(\mathfrak{n},\langle \cdot, \cdot \rangle_{\mathfrak{n}})$. Obviously, there is a natural $H-$left action $L$ on $\overline{\mathcal{Q}}$ and we call a connection $A \in \Omega^1(\overline{\mathcal{Q}},Cl(\mathfrak{n},\langle \cdot, \cdot \rangle_{\mathfrak{n}}))$ $H-$invariant iff $L_h^* A = A$ for $h \in H$. In the usual way, the covariant derivative $D$ on $S^g \cong \overline{\mathcal{Q}} \times_{Cl^*(\mathfrak{n},\langle \cdot, \cdot \rangle_{\mathfrak{n}})} \Delta_{p,q}$ is induced by a connection $A$ on $\overline{\mathcal{Q}}$.\\
One easily verifies that $D$ is even induced by a $H-$\textit{invariant} connection. Checking this is essentially just a reformulation of the isometry-invariance of the Levi Civita connection and the assumption (\ref{dufff}). However, this observation enables us to use further results from \cite{kn1,kn2} which allow the algebraic computation of the holonomy algebra and curvature of invariant connections defined over homogeneous principal bundles.\\
\newline
Carrying these steps out, results with the mentioned theoretical background in a purely algebraic algorithm: To this end, let
\begin{align*}
T_1,...,T_n \text{ be an oriented orthonormal basis of }\mathfrak{n},\\
L_1,...,L_m \text{ be a basis of }\mathfrak{k}.
\end{align*}
There are constants $d^{ki}_v$ such that $\left[ L_k,T_i\right] = \sum_v d^{ki}_v T_v$.\\
\newline
\textit{Step 1}:\\
The connection $D$ is equivalently encoded (in the sense of \cite{kn1,kn2}) in a linear map
\begin{align*}
\alpha=\alpha_{g} + \alpha_{\Omega} : \mathfrak{h} \rightarrow Cl(p,q),
\end{align*}
taking values in\footnote{At this point, one has to make a choice: Either, one fixes an orthonormal basis of $\mathfrak{n}$ and works with the Clifford algebra $Cl(p,q)$ of $\R^{p,q}$ only, what we will do, or one works more abstractly in the Clifford algebra $Cl(\mathfrak{n},\langle \cdot, \cdot \rangle_{\mathfrak{n}})$.} the Clifford algebra $Cl(p,q)=Cl(\R^n,\langle \cdot, \cdot \rangle_{p,q})$ of $\R^{p,q}$ with standard pseudo-orthonormal basis $(e_1,...,e_n)$, and it splits into parts
$\alpha_{g}$ and $\alpha_{\Omega}$, describing $\nabla$ and $\Omega$, respectively. Here,
$\alpha_{g}: \mathfrak{h} \rightarrow \mathfrak{spin}(p,q) \subset Cl(p,q)$ is given by
\begin{equation} \label{ntensor}
\begin{aligned}
\alpha_g(L_k) &= \frac{1}{2} \cdot \sum_{i<j} d^{ki}_j \cdot e_i \cdot e_j \in \mathfrak{spin}(p,q), \\
\alpha_g(T_i) &= \frac{1}{2} \cdot \sum_{i<j} N^i_{ab} e_a \cdot e_b \in \mathfrak{spin}(p,q),
\end{aligned}
\end{equation}
where (cf. \cite{hammerl1,hammerl2}) $N^i_{ab}=N^i_{[ab]}=\frac{1}{2} \cdot \left(\langle [T_i,T_a]_{|\mathfrak{n}},T_b \rangle_{\mathfrak{n}}- \langle [T_i,T_b]_{|\mathfrak{n}},T_a \rangle_{\mathfrak{n}} - \langle [T_a,T_b]_{|\mathfrak{n}},T_i \rangle_{\mathfrak{n}}\right)$ and for any $A \in \mathfrak{h}$ we let $A_{|\mathfrak{n}}$ denote its projection to $\mathfrak{n}$.
\begin{bemerkung}
More invariantly, the map $\alpha_{g}: \mathfrak{h} \rightarrow \mathfrak{spin}(\mathfrak{n},\langle \cdot, \cdot \rangle_{\mathfrak{n}})$ is given as $\lambda_*^{-1} \circ \widetilde{\alpha}_g$, for $\widetilde{\alpha}_g: \mathfrak{h} \rightarrow \mathfrak{so}(\mathfrak{n},\langle \cdot, \cdot \rangle_{\mathfrak{n}})$ uniquely determined by $\langle \alpha(X)Y,Z \rangle = \frac{1}{2} \left(\langle [X,Y]_{|\mathfrak{n}},Z \rangle_{\mathfrak{n}}- \langle [X,Z]_{|\mathfrak{n}},Y \rangle_{\mathfrak{n}} - \langle [Y,Z]_{|\mathfrak{n}},X \rangle_{\mathfrak{n}} \right)$, where $X \in \mathfrak{h}$ and $Y,Z \in \mathfrak{n}$.
\end{bemerkung}
The $\Omega-$part $\alpha_{\Omega} : \mathfrak{n} \rightarrow Cl(p,q)$, which lives only on $\mathfrak{n}$, is the evaluation of $\Omega$ at $eK \in M$. More precisely, for fixed $i \in \{1,...,n\}$, we have that
\[ \Omega(T_i) = \sum_{I} \Omega^i_{I} T_I \in Cl(\mathfrak{n},\langle \cdot, \cdot \rangle_{\mathfrak{n}}). \]
The $\Omega^i_I$ are constants, the sum runs over all multi-indicees $(i_1 < i_2 <...i_k)$ for $k \leq n$ and $T_I:=T_{i_1} \cdot...\cdot T_{i_k}$. In this notation,
\begin{align*}
\alpha_{\Omega}(T_i)=\sum_{I} \Omega^i_{I} e_I \in Cl(p,q).
\end{align*}
\textit{Step 2}:\\
One introduces the curvature map $\kappa:\Lambda^2 \mathfrak{n} \rightarrow Cl(p,q)$, which measures the failure of $\alpha$ being a Lie algebra homomorphism. Concretely, one computes for $i<j$
\begin{align*}
\kappa(T_i,T_j):= [\alpha(T_i),\alpha(T_j)]_{Cl(p,q)} - \alpha([T_i,T_j]_{\mathfrak{h}}),
\end{align*}
and determines the space $\widehat{Im}(\kappa):= \text{span}\{ \kappa(T_i,T_j) \mid i<j\} \subset Cl(p,q)$ and its dimension.\\
\newline
\textit{Step 3}:\\
The holonomy algebra $\mathfrak{hol}(D) \subset \mathfrak{gl}(\Delta_{p,q}) \cong Cl(p,q)$ (or $Cl(p,q) \cong \mathfrak{gl}(\Delta_{p,q}) \oplus \mathfrak{gl}(\Delta_{p,q})$), where $\Delta_{p,q}$ denotes the appropriate spinor module in signature $(p,q)$, can be determined as follows\footnote{We use the notation from \cite{hammerl1,hammerl2} where this construction is reviewed. Moreover, this reference presents some examples and shows how the procedure can be applied to certain Cartan geometries which allows the determination of the conformal holonomy algebra of conformal structures over homogeneous spaces.}:
$\mathfrak{hol}(D)$ is the $\mathfrak{h}$-module generated by $\widehat{Im}(\kappa)$, i.e.
\begin{align} \label{sumy}
\mathfrak{hol}(D) = \widehat{Im}(\kappa) + \left[\alpha(\mathfrak{h}),\widehat{Im}(\kappa)\right]_{Cl(p,q)}+\left[\alpha(\mathfrak{h}),\left[\alpha(\mathfrak{h}),\widehat{Im}(\kappa)\right]\right]_{Cl(p,q)}+...,
\end{align}
that is one starts with $\widehat{Im}(\kappa) \subset \mathfrak{hol}(D)$ which has been computed before. One adds all elements of type $[\alpha(T_i \text{ or }L_j),\kappa(T_a,T_b)]$. If dim $\widehat{Im}(\kappa) + \left[\alpha(\mathfrak{h}),\widehat{Im}(\kappa)\right]_{Cl(p,q)}= $dim $\widehat{Im}(\kappa)$, we are already done. Otherwise, one adds elements of $\left[\alpha(\mathfrak{h}),\left[\alpha(\mathfrak{h}),\widehat{Im}(\kappa)\right]\right]_{Cl(p,q)}$ until the dimension of the sum in (\ref{sumy}) becomes stable, which will happen
after at most $(\text{dim }\Delta_{p,q})^2$ steps.\\
\newline
\textit{Step 4}:\\
Once $\mathfrak{hol}(D)$ is known, one computes its natural action on spinors (obtained by restriction of an irreducible representation of $Cl(p,q)$ on $\Delta_{p,q}$ to $\mathfrak{hol}(D)$. In particular, $Ann(\mathfrak{hol}(D)) := \{ v \in \Delta_{p,q} \mid \mathfrak{hol}(D) \cdot v = 0 \}$ can be computed, which by the holonomy principle is isomorphic to the space of parallel spinors wrt. $D$ (for $M$ being simply-connected).
\begin{bemerkung}
If actually $(M=H/K,g)$ is a symmetric space, the algorithm simplifies: First, the map $\alpha_g$ is simply the trivial extension of $\widetilde{ad}_{K}: \mathfrak{k} \rightarrow \mathfrak{spin}(\mathfrak{n})$, i.e. the tensor $N$ in (\ref{ntensor}) vanishes. Moreover, (\ref{sumy}) simplifies to
\begin{align}
\mathfrak{hol}(D) = \widehat{Im}(\kappa) + \left[\alpha(\mathfrak{n}),\widehat{Im}(\kappa)\right]+\left[\alpha(\mathfrak{n}),\left[\alpha(\mathfrak{n}),\widehat{Im}(\kappa)\right]\right]+... \label{holfo}
\end{align}
Note that (\ref{holfo}) generalizes a well-known formula for the holonomy of the Levi Civita connection on symmetric spaces, i.e. where $\Omega=0$ and thus also $\alpha(\mathfrak{n}) = 0$.
\end{bemerkung}
\section{The associated Killing superalgebra} \label{se5}
Let us now in addition assume that $(M,g)$ is space-and time oriented, which allows a global symmetric squaring of spinor fields to vector fields, $(\ph_1,\ph_2) \rightarrow V_{\ph_1,\ph_2}$, given by $g(V_{\ph_1,\ph_2},X) = \langle \ph_1, X \cdot \ph_2 \rangle_{S^g}$ for a $Spin^+(p,q)$ invariant inner product on the spinor module\footnote{The definition of $V_{\ph_1,\ph_2}$ might also involve taking the real or imaginary part, depending on $(p,q)$} . Assume moreover that for $\ph_i$ being parallel wrt. $D$, the associated vector is Killing (as true for geometric Killing spinors in certain signatures, cf. \cite{boh}, or Killing spinors in 11-dimensional supergravity).\\
Furthermore we assume that the simply-connected reductive homogeneous space is of the form $(M,g)=(H/K,g)$ such that\footnote{Every $X \in \mathfrak{h}$ generates a Killing vector field $X^*$ on $M$ by setting $X^*(hK):= \frac{d}{dt}_{|t=0} \text{exp}(tX) hK$. We assume that these are all Killing vector fields.} $\mathfrak{h} \cong Kill(M,g)$, the space of all Killing vector fields. Under these assumptions, the determination of the associated \textit{Killing superalgebra} (cf. \cite{jose2}) $\mathfrak{g}=\mathfrak{g}_0 \oplus \mathfrak{g}_1$ is purely algebraic:\\
By definition, the odd part is the space of $D-$parallel spinors, i.e. $\mathfrak{g}_1 \stackrel{\ph \mapsto \ph(eK)}{\cong} \{ v \in \Delta_{p,q} \mid \mathfrak{hol} (D) \cdot v =0 \}$ and the even part is given by Killing vector fields, $\mathfrak{g}_0 = \mathfrak{h} = \mathfrak{k} \oplus \mathfrak{n}$, where we use the isomorphism to $Kill(M,g)$ given by
\begin{align}
Kill(M,g) \ni X \mapsto \left(\left(\nabla X \right)_{eK}, X(eK) \right).\label{i}
\end{align}
In particular, we identify $\mathfrak{k}$ with a subspace of $\mathfrak{so}(\mathfrak{n}) \cong \mathfrak{spin}(\mathfrak{n})$. \\
The brackets in $\mathfrak{g}=\mathfrak{g}_0 \oplus \mathfrak{g}_1$ can now be computed as follows: The even-even bracket, which is classically the usual Lie bracket on vector fields, is simply (minus) the bracket in $\mathfrak{h}$. The odd-even bracket is classically given as the spinorial Lie derivative
\begin{align*}
L_{X} \ph := \nabla_X \ph + \frac{1}{2} \underbrace{\left(\nabla(X) \right)}_{\in \mathfrak{so}(TM) \cong \Lambda^2(TM)} \cdot \ph \stackrel{D \ph = 0}{=} - \Omega(X) \cdot \ph + \frac{1}{2} {\left(\nabla(X) \right)} \cdot \ph
\end{align*}
for $X \in Kill(M,g)$ and $D \ph = 0$. Thus, in the algebraic picture, for $(\beta,t) \in \mathfrak{k} \oplus \mathfrak{n}=\mathfrak{g}_0$, this corresponds to
\begin{align*}
\g_0 \otimes \g_1 \ni (\beta,t) \otimes v \mapsto -\Omega_{eK}(t) \cdot v + \frac{1}{2} \beta \cdot v \in \g_1,
\end{align*}
In order to express the odd-odd-bracket, which squares a $D-$parallel spinor $\ph$ to its Dirac current $V_{\ph,\ph} \in Kill(M,g)$, we differentiate $V{_{\ph}}:=V_{\ph,\ph}$ to obtain,
\begin{align} \label{frt}
g(\nabla_X V_{\ph},Y) = \langle (\epsilon_1 \cdot Y \cdot \Omega(X) + \epsilon_2 \cdot \Omega(X)^T \cdot Y)\cdot \ph, \ph \rangle_{S^g},
\end{align}
where $\epsilon_i$ are $(p,q)$-dependent signs and $\Omega(X)^T$ denotes the transpose of $\Omega(X)$ considered as endomorphism acting on spinors. Thus the bracket is by polarization under the isomorphism (\ref{i}) uniquely determined by
\begin{align*}
S^2 \mathfrak{g}_1 \ni v \circ v \mapsto (\alpha_{v},t_{v}) \in \g_0,
\end{align*}
where $t_v$ is the vector in $\mathfrak{n}$ algebraically associated to $v (\leftrightarrow \ph(eK)) \in \mathfrak{g}_1$ and $\alpha_v \in \mathfrak{so}(\mathfrak{n})$ denotes the skew symmetric endomorphism $\nabla V_{\ph}$ given by (\ref{frt}) evaluated at $eK$.\\
\newline
That is, under the assumptions made the structure of $\mathfrak{g}$ can be calculated in a purely algebraic way and $\mathfrak{g}$ can then be analyzed further via its Levi decomposition.
\section{Application to a class of homogeneous M2-duals} \label{se6}
In general, let $(M,g,F)$ be a classical $M-$theory background, i.e. $(M,g)$ is a 11-dimensional connected Lorentzian spin manifold with mostly $+$metric and Clifford algebra convention $x \cdot x = + ||x||^2$ for $x \in \R^{1,10}$, $F$ is a closed $4-$form and we demand the triple to satisfy the bosonic field equations
\begin{equation} \label{fwedge}
\begin{aligned}
d \ast F &= \frac{1}{2} F \wedge F, \\
Ric(X,Y) &= \frac{1}{2} \langle X \invneg F, Y \invneg F \rangle - \frac{1}{6} g(X,Y) |F|^2.
\end{aligned}
\end{equation}
Setting the gravitino variation to zero in a purely bosonic background yields the Killing spinor equation
\begin{align}
D_X \ph = \nabla_X \ph + \underbrace{\frac{1}{6}(X \invneg F) \cdot \ph + \frac{1}{12} \left(X^{\flat} \wedge F\right) \cdot \ph}_{=:\Omega(X) \cdot \ph}. \label{sugra}
\end{align}
A background of 11-dimensional supergravity is called supersymmetric iff it admits nontrivial solutions to (\ref{sugra}). Motivated by the search for new homogeneous M2 duals, \cite{fu} obtains new families of solutions to (\ref{fwedge}):
The geometry in this case is locally isometric to a product $M= H/K \cong SO(5)/SO(4) \times SO(3,2)/SO(3)$. Let $L_{ab}, L_{a5}$ denote the standard generators of $\mathfrak{so}(5)$, where $a,b=1,2,3,4$, and let $J_{ij},J_{iA},J_{45}$ denote the standard generators of $\mathfrak{so}(3,2)$, where $i,j=1,2,3$ and $A=4,5$. Then
\begin{align*}
\mathfrak{k}&=\mathfrak{k}_1 \oplus \mathfrak{k}_2 = \text{span} \{L_{ab} \} \oplus \text{span} \{J_{ij} \},\\
\mathfrak{n}&=\mathfrak{n}_1 \oplus \mathfrak{n}_2 = \text{span} \{L_{a5} \} \oplus \text{span} \{J_{iA},J_{45} \} \cong \R^4 \oplus \R^7.
\end{align*}
Given the ordered basis $X_{\mu}=(J_{45},L_{15},...,L_{45},J_{14},...,J_{34},...,J_{15},...,J_{35})$, let $\theta^{\mu}$ denote the canonical dual basis for $\mathfrak{n}^*$. Then an $H-$invariant inner product is given by
\begin{align}
g = -(\theta^0)^2+\gamma_1 ((\theta^1)^2 + ...+(\theta^4)^2) + \gamma_2 ((\theta^5)^2 + ...+(\theta^7)^2)+ \gamma_3 ((\theta^8)^2 + ...+(\theta^{10})^2),\label{metrik}
\end{align}
for real parameters $\gamma_{1,2,3} > 0$. Let $(a_0,...,a_{10})$ denote the $g-$pseudo orthonormal basis of $\mathfrak{n}$ obtained by rescaling elements of $X_{\mu}$ with appropriate positive constants.\\
Special choices of the $\gamma_i$ and specifying certain closed $H-$invariant 4-forms on $M=S^4 \times X^7$ yield $M-$theory backgrounds, i.e. solutions to (\ref{fwedge}). Let us consider preserved supersymmetry of two of them in more detail. To this end, we observe that $H-$invariance of $F$ implies $H-$invariance of $\Omega$ and thus makes the algorithm developed before applicable:\\
\newline
\textbf{A supersymmetric Freund-Rubin background. } In this geometry, $\gamma_1= \frac{4}{9}, \gamma_2 = \gamma_3 = \frac{2}{3}$. The Lorentzian factor $X^7$ admits an invariant Lorentzian Sasaki-Einstein structure. The triple $(M,g,F:= \frac{9}{2} \cdot vol_{S^4})$ is a solution to (\ref{fwedge}) and $F$ is obviously $H-$invariant. We turn to (complex) spinor fields on $(M,g)$:
Let $(e_0,...,e_{10})$ denote the standard basis of $\R^{1,10}$. By $e_i$ we also label the Clifford action of the vector $e_i$ on the complex spinor module $\Delta_{1,10}^{\C}$. We work with the identification
\[ \Delta_{1,10}^{\C} \cong \Delta_4^{\C} \otimes \Delta_{1,6}^{\C}, \]
(cf. also \cite{ldr}) under which Clifford action becomes
\begin{align}
e_{i=1,...,4} \rightarrow e_i \otimes Id\text{ and }e_{i=0,5,...,10} \rightarrow vol_{S^4} \otimes e_i.
\end{align}
In particular, $vol_{S^4}$ acts as the identity on $\Delta_{1,6}^{\C}$. We fix the isometry $\eta:\mathfrak{n} \rightarrow \R^{1,10}$ mapping $a_i$ to $e_i$. It follows directly from the various definitions that the map $\alpha$ describing $D$ splits into $\alpha= \alpha_1 + \alpha_2$, where $\alpha_i : \mathfrak{h}_i \rightarrow Cl(\mathfrak{n}_i)$. As moreover $\alpha_1(X_1)$ and $\alpha_2(X_2)$ commute when acting on spinors, also $\kappa$ splits. The $S^4$-factor is equipped with a multiple of the round standard metric and therefore symmetric. Whence, $\alpha_1(X) = \frac{1}{6} \cdot \frac{2}{3} \cdot \eta(X) \cdot \eta_*F$ for $X \in \mathfrak{n}_1$. Moreover, using that the adjoint action of $\mathfrak{k}_1$ on $\mathfrak{n}_1$ is identified with the identity map, one obtains for $i \neq j$
\begin{align*}
\kappa_1(L_{i5},L_{j5}) &= [\alpha_1(L_{i5}),\alpha_1(L_{j5})] - \alpha_1([L_{i5},L_{j5}])\\
&=\frac{2}{6^2} \cdot \left( \left(\frac{2}{3}\right)^2 \cdot e_i \cdot e_j \cdot \left(\frac{9}{2} \cdot e_1 \cdot e_2 \cdot e_3 \cdot e_4 \right)^2 \right) - \frac{1}{2} \cdot e_i \cdot e_j =0,
\end{align*}
that is $\kappa_1 \equiv 0$. In particular, $\mathfrak{hol}(\alpha_1) = \{0\}$, which reflects the fact that $S^4$ admits a full space of geometric Killing spinors.
Consequently,
\begin{align}
\mathfrak{hol}(D) = \mathfrak{hol}(\alpha_2) \subset Cl_{1,6} \subset Cl_{1,10}, \label{stre}
\end{align}
and $\alpha_2$ precisely encodes (in the sense of \cite{kn1,kn2}) the connection $\nabla_X + \frac{3}{8} X \cdot$ on $X^7$. (\ref{stre}) and the holonomy-principle direcly reveal that there is a basis of $D-$parallel spinors of the form $\ph_1 \otimes \ph_2$, where $\ph_1$ is a (combination of) Killing spinors on $S^4$ and $\ph_2$ is a geometric Killing spinor to the Killing number $-\frac{3}{8}$ on $X^7$. On the other hand, we have collected all the algebraic ingredients to compute with the algorithm that the following elements lie in $\mathfrak{hol}(\alpha_2)$:
\begin{equation} \label{eq1}
\begin{aligned}
\kappa_2(a_5,a_{10}) &=e_5 \cdot e_{10} + e_7 \cdot e_8,\\
\kappa_2(a_6,a_{10}) &=e_6 \cdot e_{10} + e_7 \cdot e_9,\\
\kappa_2(a_5,a_9) &=e_5 \cdot e_9 + e_6 \cdot e_8,\\
\kappa_2(a_0,a_{10}) &=e_7 + e_0 \cdot e_{10},\\
\kappa_2(a_0,a_{9}) &=e_6 + e_0 \cdot e_{9}.\\
\end{aligned}
\end{equation}
It is known from the general theory (cf. \cite{boh,kath}) that every Lorentzian Sasaki-Einstein manifold admits 2 geometric Killing spinors, which in our case also follows from running the algorithm from section \ref{se3}. On the other hand, it is easy to verify from the definitions that $\mathfrak{hol}(\alpha_2) \subset \mathfrak{spin}(1,6) \oplus \R^{1,6} \cong \mathfrak{spin}(2,6)$. As $X$ is Lorentzian Einstein Sasaki, we must in fact have, cf. \cite{baer,boh,kath} that $\mathfrak{hol}(\alpha_2) \subset \mathfrak{su}(1,3)$. Moreover, $\lambda_* (\mathfrak{hol}(\alpha_2)) \subset \mathfrak{so}(2,6)$ acts irreducible on $\R^{2,6}$ as follows easily from an inspection of the elements (\ref{eq1})\footnote{An element $h \in \mathfrak{spin}(2,6)$ acts via $\lambda_*$ on $\R^{2,6}$ as $x \mapsto [h,x]=h \cdot x - x \cdot h \in \R^{2,6}$. From this it follows easily that there is no proper subspace of $\R^{2,6}$ preserved by alle elements (\ref{eq1}).}. However, there is no proper subgroup of $SU(1,3)$ which acts irreducile on $\R^{2,6}$ as shown in \cite{scal}. This already implies that
\[\mathfrak{hol}(\alpha_2)=\mathfrak{su}(1,3), \]
which can also be derived by using the algorithm only. Thus,
\begin{align*}
\mathfrak{hol}(D) = \mathfrak{su}(1,3) \subset \mathfrak{spin}(2,6) = \mathfrak{spin}(1,6) \oplus \R^{1,6} \subset Cl_{1,6} \subset Cl_{1,10}.
\end{align*}
Thus, $X^7$ is a generic Lorentzian Einstein Sasaki manifold and all $D-$parallel spinors are given by tensor products of geometric Killing spinors on $S^4$ and the 2 linearly independent geometric Killing spinors on $X^7$ which define its Sasaki structure. Real $D-$parallel spinors are obtained by imposing additional Majorana conditions (cf. \cite{har,lm,br}).\\
\newline
\textbf{A circle of backgrounds. } This family $(M,g,F)$ of $M-$theory backgrounds is specified from (\ref{metrik}) by the choice $\gamma_1 = \gamma_2 = \gamma_3 = \frac{4}{9}$ and the $\alpha \in \R$-depended $H-$invariant 4-form
\[ F = F_1 + F_2 -\frac{1}{3} \theta^{1234} + \frac{1}{\sqrt{3}} \theta^0 \wedge \text{Re} \left(e^{i \alpha} (\theta^5+ i \theta^8) \wedge (\theta^6+ i \theta^9) \wedge (\theta^7+ i \theta^{10}) \right). \]
Note that $\alpha$ (and henceforth also $D$) does not split into connections on the factors for this geometry. It is very easy to deduce from the algorithm that there are no $D-$parallel spinors in this situation: In fact, we have with the same notation as in the previous case for $i \neq j$:
\begin{equation} \label{ho1}
\begin{aligned}
\kappa_1(L_{i5},L_{j5}) &= [\alpha(L_{i5}),\alpha(L_{j5})] - \alpha([L_{i5},L_{j5}]) \\
&= \left(\left( -\frac{1}{18} \cdot \frac{4}{9} \cdot \left(\frac{1}{3} \cdot \left( \frac{3}{2}\right)^4 \right) + \frac{2}{{12}^2} \cdot \frac{4}{9} \cdot (\eta_*(F_2))^2 \right) - \frac{1}{2}\right) \cdot e_i \cdot e_j.\\
& \in \mathfrak{hol} (\alpha)
\end{aligned}
\end{equation}
A $D-$parallel spinor requires that $\kappa_1(L_{i5},L_{j5})$ considered as endomorphism acting on spinors has a kernel. As $e_i \cdot e_j$ acts invertible on spinors, the expression in brackets in (\ref{ho1}) must be singular. However, the endomorphism $(\eta_*(F_2))^2$ has eigenvalues $0$ and $\approx \pm \left(\frac{3}{2}\right)^4 \cdot 60,75$ only. Whence (\ref{ho1}) does not annihilate any nonzero spinor, i.e. there are no $D-$parallel spinor fields on $M$. \\
More intuitively, but less rigorous, one sees this by plugging in the ansatz $\ph = \ph_1 \otimes \ph_2$ for a $D-$parallel spinor. $\ph_1$ then has to satisfy a geometric Killing spinor equation on $S^4$ but with the wrong Killing constant. However, note that for the algorithm we did not have to make a particular ansatz for the spinor.
\section{A conformal analogue} \label{coa}
If we turn our attention to conformal (spin) geometry, i.e. metrics which differ by multiplication with a positive function, or conformal supergravity and are interested in the study of first order conformally covariant differential equations on spinors, the geometric Killing spinor equation is replaced by the conformally covariant twistor equation (cf. \cite{bfkg,lei,leihabil,bl})
\begin{align*}
\nabla_X \ph + \frac{1}{n} X \cdot D^g \ph = 0 \text{ for } X \in TM,
\end{align*}
where $D^g$ is the $Spin$ Dirac operator (cf. \cite{ba81,fr}). It's solutions are called twistor spinors. Consider the covariant derivative $\widetilde{D}$ on $S^{g,2}:=S^g \oplus S^g$ given by
\begin{align*}
\widetilde{D}_X \begin{pmatrix} \ph \\ \phi \end{pmatrix} = \begin{pmatrix} \nabla_X^{S^g} & -X \cdot \\ \frac{1}{2}K^g(X) \cdot & \nabla^{S^g}_X \end{pmatrix} \begin{pmatrix} \ph \\ \phi \end{pmatrix}=: \nabla_X \begin{pmatrix} \ph \\ \phi \end{pmatrix} + \widetilde{\Omega}(X) \left(\begin{pmatrix} \ph \\ \phi \end{pmatrix} \right),
\end{align*}
where $K^g := \frac{1}{n-2} \cdot \left( \frac{scal^g}{2(n-1)} \cdot g - Ric^g \right)$ is the Schouten tensor and $\widetilde{\Omega}:TM \rightarrow End(S^{g,2})$. If $\ph$ is a twistor spinor, then $\begin{pmatrix} \ph \\ -\frac{1}{n} \cdot D^g \ph \end{pmatrix}$ is $\widetilde{D}$-parallel (cf. \cite{bfkg}), and conversely, the first slot of a $\widetilde{D}$-parallel spinor is a twistor spinor.\\
\newline
Suppose now that $(M=H/K,g)$ is a reductive homogeneous pseudo-Riemannian space with $\mathfrak{h}= \mathfrak{k}\oplus \mathfrak{n}$. We identify $\mathfrak{n} \cong \R^{p,q}$ by means of some fixed orthonormal basis. There is a natural homogeneous bundle over $M$ admitting an invariant connection whose holonomy coincices with that of $\widetilde{D}$. To this end, we enlarge $\R^{p,q}$ to $\R^{p+1,q+1}$ by introducing new lightlike directions $e_{\pm}$ such that $\langle e_+, e_- \rangle = 1$. Clearly, $\R^{p+1,q+1} = \R e_- \oplus \R^{p,q}\oplus \R e_+$ as $O(p,q)-$modules. We define the annihilation spaces $Ann(e_{\pm}):=\{ v \in \Delta_{p+1,q+1} \mid e_{\pm}\cdot v = 0 \}$. It follows that for every $v \in \Delta_{p+1,q+1}$ there is a unique $w \in \Delta_{p+1,q+1}$ such that $v=e_+ w + e_- w$, leading to a decomposition
\begin{align*}
\Delta_{p+1,q+1} = Ann(e_+) \oplus Ann(e_-).
\end{align*}
$Ann(e_{\pm})$ is acted on by $Spin(p,q) \hookrightarrow Spin(p+1,q+1)$ and there is an isomorphism $\chi: Ann(e_-) \rightarrow \Delta_{p,q}$ of $Spin(p,q)$-modules leading to the identification
\begin{equation}
\begin{aligned} \label{deco}
\Pi: {\Delta_{p+1,q+1}}_{|Spin(p,q)} & \rightarrow \Delta_{p,q} \oplus \Delta_{p,q}, \\
v=e_+w+e_-w & \mapsto (\chi(e_-e_+w),\chi(e_-w)).
\end{aligned}
\end{equation}
We identify $\Delta_{p+1,q+1}$ and $\Delta_{p,q} \oplus \Delta_{p,q}$ by means of $\Pi$. Now we introduce the bundle
\[\widehat{Q}:=H \times_{\widetilde{Ad}_K} Spin(p+1,q+1), \]
where $\widetilde{Ad}_K : K \rightarrow Spin(p,q) \hookrightarrow Spin(p+1,q+1)$. Obviously, $S^{g,2} \cong \widehat{Q} \times_{Spin(p+1,q+1)} \Delta_{p+1,q+1}$ and under this identification we obtain with (\ref{deco}) that
\begin{align*}
\widetilde{D}_X \psi = \nabla \psi + \underbrace{(X \cdot s_+ + \frac{1}{2}K^g(X) \cdot s_-) \cdot \psi}_{=\widetilde{\Omega}(X) \cdot \psi},
\end{align*}
where $\nabla$ is induced by the Levi Civita connection, $s_{\pm}$ are the obvious global lightlike sections in $H \times_{Ad_K} \R^{p+1,q+1}$ and $\psi \in \Gamma(S^{g,2})$. From this description it becomes immediate that $\widetilde{D}$ is induced by a connection $A \in \Omega^1(\widehat{Q},\mathfrak{spin}(p+1,q+1))$ in the usual way. As the curvature tensor is isometry-invariant, $\widetilde{\Omega}$ is $H-$invariant and it follows as for the Killing spinor equation that $A$ is $H-$invariant. That is, $A$ is again equivalently characterized (in the sense of \cite{kn1,kn2}) by a linear map
\[ \alpha = \alpha_g + \alpha_{\widetilde{\Omega}} : \mathfrak{h} \rightarrow \mathfrak{spin}(p+1,q+1), \]
where $\alpha_g:\mathfrak{h} \rightarrow \mathfrak{spin}(p,q) \hookrightarrow \mathfrak{spin}(p+1,q+1)$ describes the Levi Civita connection and is given by (\ref{ntensor}) and $\alpha_{\widetilde{\Omega}}$ lives only on $\mathfrak{n}$ and is given by
\[ \alpha_{\widetilde{\Omega}}(t) = \widetilde{\Omega}_{ek}(t) = t \cdot e_+ + \frac{1}{2}K^g(t) \cdot e_- \in \mathfrak{spin}(p+1,q+1).\]
The curvature tensor of a reductive homogneous space, and thus also $K^g$, can be computed purely algebraically (cf. \cite{aa}). Again formula (\ref{sumy}) applies and yields $\mathfrak{hol}(\widetilde{D}) \cong \mathfrak{hol}(\alpha) \subset \mathfrak{spin}(p+1,q+1)$. One then computes it's natural action on $\Delta_{p+1,q+1}$ and twistor spinors for $(H/K,g)$ are in bijective correspondence to the trivial subrepresentations of $\mathfrak{hol}(\alpha)$. This yields a purely algebraic procedure to solve the twistor equation on any reductive homogeneous spin manifold.
\small
\bibliographystyle{amsalpha}
\bibliography{literatur}
\end{document}
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\begin{document}
\begin{center}
{\Large \textbf{On a class of distributions generated by
stochastic mixture of the extreme order statistics of a sample of
size two}}
\end{center}
\begin{center}
{\large \textbf{S. M. Mirhoseini$^{\mathrm{a}}$, Ali
Dolati$^{\mathrm{b}}$ and M. Amini$^{\mathrm{c}}$}}
\end{center}
\begin{center}
\textit{$^{\mathrm{a,c}}$Department of Statistics, Faculty of
Mathematical Science, Ferdowsi University,P.O. Box 91775-1159,
Mashhad, Iran \\ \textrm{mmirhoseini@yazduni.ac.ir}
\\ $^{\mathrm{b}}$Department of Statistics, Faculty of
Mathematics, Yazd University, Yazd, 89195-741, Iran\\[0pt]\textrm{adolati@yazduni.ac.ir}
\\[0pt]}
\end{center}
\begin{center}
\textbf{Abstract}
\end{center}
\noindent This paper considers a family of distributions
constructed by a stochastic mixture of the order statistics of a
sample of size two. Various properties of the proposed model are
studied. We apply the model to extend the exponential and
symmetric Laplace distributions. An extension to the bivariate
case is considered.
\bigskip
\noindent AMS(2000) Subject classification: 62F15.
\medskip
\noindent \textit{Keywords}: Aging characteristics, copula, hazard
rate function, mixture, stochastic orders; transformation.
\section{Introduction}
Different methods may be used to introduce a new parameter to a
family of distributions to increase flexibility for modeling
purposes. Marshall and Olkin \cite{Marshal97} introduced a method
for adding a parameter to a family of distributions and applied it
to the exponential and Weibull models. Jones \cite{Jones} used the
distribution of order statistics to provide new families of
distributions with extra parameters. The well--known
Farlie--Gumbel-Morgenstern (FGM, for short) family of bivariate
distributions with the given univariate marginal distributions
$F_1$ and $F_2$, is defined by
\begin{equation}
H(x,y)=F_{1}(x)F_{2}(y)\{ 1+\lambda
\bar{F_{1}}(x)\bar{F_{2}}(y)\}, \label{FGM}
\end{equation}
where $\lambda \in \lbrack -1,1]$; see, Drouet-Mari and Kotz
(\cite{Kotz2001}, Chapter 5) for a good review. For a given
univariate cumulative distribution function $F$, the univariate
version of (\ref{FGM}) may be considered as
\begin{eqnarray}
G_{\lambda}[F](x)= F(x)\{1+\lambda \bar{F}(x)\}, \label{m2}
\end{eqnarray}
for all $x$ and $-1\leq \lambda \leq 1$. The family of
distributions defined by (\ref{m2}) is comparable with the
Marshall-Olkin \cite{Marshal97} family of distributions, which
also called the proportional odds model
\cite{Kirmani2001,Marshal97}, given by
\begin{equation}
H(x)=\frac{F(x)}{1-(1-\alpha)\bar{F}(x)}, \quad\ -\infty<x<\infty,
\quad\ \alpha>0.
\end{equation}
Note that $H$ with $0<\alpha<1$ could be written as
\begin{equation}
H(x)=F(x)\sum_{k=0}^{\infty}\{(1-\alpha)\bar{F}(x)\}^k,
\end{equation}
and hence for $0<\lambda<1$, (\ref{m2}) is a first-order
approximation to the proportional odds model.
The aim of the present paper is to investigate different
properties of (\ref{m2}). We first provide a physical
interpretation for this model in Section 3. Some preservation
results of stochastic orderings and aging properties are given in
Section 4. A generalization of the ordinary exponential
distribution which exhibits both increasing and decreasing hazard
rate functions and a skew extension of the symmetric Laplace
distribution are given in Section 5. Bivariate case is discussed
in Section 5. In Section 1 we recall some notions that will be
used in the sequel.
\section{Preliminaries}
Let us recall some notions of stochastic orderings and aging
concepts that will be useful in this paper. Let $X$ be a
continuous random variable with the cdf $F$, the survival function
$\bar{F}=1-F$, the probability density function (pdf) $f$, the
residual life survival function $\bar{F}_t(x)=P(X>x+t|X>t)$ and
the hazard rate function $h_F(x)=f(x)/\bar{F}(x)$. Then $F$ is
said to have: (i) increasing (decreasing) hazard rate IHR (DHR) if
$h_F(x)$ is increasing (decreasing) in $x$; (ii) increasing
(decreasing) hazard rate average IHRA (DHRA) if
$\int_{0}^{t}h_F(x)dx/t$ is increasing (decreasing) in $t$; (iii)
new better (worse) than used NBU (NWU) property if
$\bar{F}_t(x)\leq (\geq) \bar{F}(x)$, for all $x\geq 0$ and
$t\geq0$. The implications
$$
\text{\rm IHR} \Longrightarrow \text{\rm IHRA} \Longrightarrow
\text{\rm NBU} \quad {\rm and} \quad \text{\rm DHR}
\Longrightarrow \text{\rm DHRA} \Longrightarrow \text{\rm NWU},
$$
are well known. See \cite{bar1975} for more detail. The following
definitions will be used for various stochastic comparisons. Let
$F_1$ and $F_2$ be two cdfs with the corresponding pdfs $f_1$ and
$f_2$, the hazard rate functions $h_{F_{1}}$, $h_{F_{2}}$, and the
quantile functions $F^{-1}_1$ and $F^{-1}_2$, respectively, where
$F^{-1}_i=\text{sup}\{x|F_i(x)\leq u\}$, for $0\leq u\leq1$. The
cdf $F_1$ is said to be smaller than $F_2$ in (i) stochastic order
$(F_1\prec_{\rm st}F_2)$ if $F_1(x)\geq F_2(x)$ for all $x$; (ii)
hazard rate order $(F_1\prec_{\rm hr}F_2)$ if $h_{F_{1}}(x)\geq
h_{F_{2}}(x)$ for all $x$; (iii) likelihood ratio order
$(F_1\prec_{\rm lr}F_2)$ if $f_2(x)/f_1(x)$ is non-decreasing in
$x$; (iv) convex transform order $(F_1\prec_{\rm c}F_2)$ if
$F^{-1}_{2}F_1(x)$ is convex in $x$ on the support of $F_1$; (v)
star order $(F_1\prec_{*}F_2)$ if $F^{-1}_{2}F_1(x)/x$ is
increasing in $x\geq 0$; (vi) superadditive order $(F_1\prec_{\rm
su}F_2)$ if $F^{-1}_{2}F_1(x+y)\geq
F^{-1}_{2}F_1(x)+F^{-1}_{2}F_1(y)$; (vii) dispersive order
$(F_1\prec_{\rm disp}F_2)$ if $F^{-1}_{2}F_1(x)-x$ increases in
$x$. The implications $F_1\prec_{\rm lr}F_2\Longrightarrow
F_1\prec_{\rm hr}F_2\Longrightarrow F_1\prec_{\rm st}F_2$ are well
known. See \cite{Shaked} for an extensive study of these notions.
\section{Genesis of family (2)}
Let $X_{1}$ and $X_{2}$ be two independent and identically
distributed random variables having the survival function
$\bar{F}=1-F.$ For $-1\leq \lambda \leq 1,$ let $Z$ be a Bernulli
random variable, independent of $X_i$s, with
$P(Z=1)=\frac{1+\lambda}{2}$ and $P(Z=0)=\frac{1-\lambda}{2}$.
Consider the stochastic mixture
\begin{equation}
U=ZX_{(1)}+(1-Z)X_{(2)},\label{mrep}
\end{equation}
where $X_{(1)}=\min (X_{1},X_{2})$ and $X_{(2)}=\max
(X_{1},X_{2})$ are the corresponding order statistics of $X_{1}$
and $X_{2}$. Since the distribution functions of $X_{(2)}$ and
$X_{(1)}$ are given by $F_{(2)}(x)=F^{2}(x)$ and
$F_{(1)}(x)=2F(x)-F^{2}(x)$, respectively, then the cdf of $U$,
denoted by $G_{\lambda}[F]$, is given by
\begin{eqnarray}
G_{\lambda}[F](x)&=&\frac{1+\lambda}{2}F_{(1)}(x)+\frac{1-\lambda}{2}F_{(2)}(x)\nonumber\\&=&
F(x)\{1+\lambda \bar{F}(x)\},
\end{eqnarray}
for all $x$ and $-1\leq \lambda \leq 1$. Clearly $G_{0}[F]=F$,
$G_{-1}[F]=F_{(2)}$, and $G_{1}[F]=F_{(1)}$. Since
$G_{\lambda}[F](.)$ is increasing in $\lambda$, we have the
inequality
$$
F_{(2)}(x)\leq G_{\lambda}[F](x)\leq F_{(1)}(x),
$$
for all $x$ and $-1\leq \lambda \leq 1$.
In the following result we show that the transformation (\ref{m2})
is ``unique'', in the sense that given a distribution $F$, this
generates a unique distribution or a family of distributions.
\begin{proposition} Let $F_1$ and $F_2$ be two distribution
functions such that $G_\lambda[F_1]=G_\lambda[F_2]$ for every
$\lambda\in[-1,1]$. Then $F_1=F_2$.
\end{proposition}
\noindent \textbf{Proof.} Suppose that $\lambda> 0$ (the case
$\lambda=0$ is trivial and the case $\lambda<0$ the result could
be proved similar). Then, $G_\lambda[F_1]=G_\lambda[F_2]$, is
equivalent to
\begin{equation}
[F_1(x)-F_2(x)][1-\lambda(F_1(x)+F_2(x)-1)]=0\label{m3},
\end{equation}
for each $x$. Suppose there exist a point $x_0\in R$ such that
---without loss of generality--- $F_1(x_0)<F_2(x_0)$. Then the
equality (\ref{m3}) is equivalent to
$F_1(x_0)+F_2(x_0)=\frac{1}{\lambda}+1$. Since $1\leq
\frac{1}{\lambda}$ and $F_1(x_0)<F_2(x_0)<1$, we must have
$F_2(x_0)>1$. This absurd, so that we conclude that $F_1=F_2$.
\section{Properties}
The survival function, the probability density function and the hazard rate
function corresponding to (\ref{m2}) are given by
\begin{equation}
\bar{G}_{\lambda}[F](x)=\bar{F}(x)\{1-\lambda F(x)\},\label{Gbar}
\end{equation}
\begin{equation}
g_{\lambda}[F](x)=f(x)\{1+\lambda(1-2F(x))\}\label{m5}
\end{equation}
and
\begin{equation}
h_{G}(x;\lambda)=\frac{g_{\lambda}[F](x)}{\bar{G}_{\lambda}[F](x)}=h_{F}(x)\left(1+\frac{\lambda
\bar{F}(x)}{1-\lambda F(x)}\right)\label{hr},
\end{equation}
respectively, where, $h_{F}(x)$ is the hazard rate function of
$F$.
It follows from (\ref{hr}) that
\begin{equation*}
\lim_{x\rightarrow -\infty}
h_{G}(x;\lambda)=(1+\lambda)\lim_{x\rightarrow -\infty}h_{F}(x),
\quad\ \lim_{x\rightarrow
\infty}h_{G}(x;\lambda)=\lim_{x\rightarrow \infty} h_{F}(x),
\end{equation*}
\begin{equation*}
h_{F}(x)\leq h_{G}(x;\lambda)\leq (1+\lambda)h_{F}(x), \quad\
(-\infty<x<\infty, 0\leq \lambda\leq 1),
\end{equation*}
\begin{equation*}
\quad\ (1+\lambda)h_{F}(x)\leq h_{G}(x;\lambda)\leq h_{F}(x),
\quad\ (-\infty<x<\infty, -1\leq \lambda\leq 0).
\end{equation*}
Let $\bar{F}_t(x)=\frac{\bar{F}(x+t)}{\bar{F}(t)}$ be the residual
life survival function corresponding to cdf $F$. Then from
(\ref{Gbar}), the residual life survival function of the generated
distribution $G_{\lambda}[F]$, denoted by
$\bar{G}_{\lambda,t}[F](x)$, is given by
\begin{eqnarray}
\bar{G}_{\lambda,t}[F](x)&=&\nonumber\frac{\bar{G}_{\lambda}[F](x+t)}{\bar{G}_{\lambda}[F](t)}
\\\nonumber&=&\bar{F}_t(x)\left(\frac{1-\lambda F(x+t)}{1-\lambda F(t)}\right)\\\nonumber
&=&\bar{F}_t(x)\left(1-\frac{\lambda \bar{F}(t)}{1-\lambda
F(t)}\frac{F(x+t)-F(t)}{\bar{F}(t)}\right)\\\nonumber&=&\bar{F}_{t}(x)\{1-\beta
F_{t}(x)\}\\&=&\bar{G}_{\beta}[F_t](x), \label{rlife}
\end{eqnarray}
where $\beta=\beta(t)=\frac{\lambda \bar{F}(t)}{1-\lambda F(t)}$
and $F_{t}(x)=1-\bar{F}_{t}(x)$. Thus the residual life survival
function of $G_{\lambda}[F]$ is the transformed version of the
residual life survival function of $F$ under (\ref{m2}), with a
new parameter.
By solving the equation
$F(x)\{1+\lambda(1-F(x)\}=G_{\lambda}[F](x)$, with respect to $F$,
one gets
\begin{equation*}
F(x)=\frac{1+\lambda-\sqrt{(1+\lambda)^2-4\lambda
G_{\lambda}[F](x)}}{2\lambda},\label{inv}
\end{equation*}
which gives the the quantile function of $G_{\lambda}[F]$ as
\begin{equation}
G^{-1}_{\lambda}[F](q)=F^{-1}\left( \frac{1+\lambda
-\sqrt{(1+\lambda)^{2}-4\lambda q}}{2\lambda}\right), \quad 0\leq
q\leq1. \quad \label{quantil}
\end{equation}
Note that
$\lim_{\lambda\rightarrow0}G^{-1}_{\lambda}[F](q)=F^{-1}(q)$. In
particular, the median of $G_{\lambda}[F]$ is given by
\begin{equation*}
G^{-1}_{\lambda}[F](0.5)=F^{-1}\left(\frac{1+\lambda
-\sqrt{1+\lambda^{2}}}{2\lambda}\right).
\end{equation*}
\subsection{Stochastic comparisons}
In this section we provide some results for stochastic orderings and aging
properties of a given cdf under the transformation (\ref{m2}).
\begin{proposition} For a given cdf $F$, we have
\noindent a) \ \ (i) If $F$ is {\rm IHR} ({\rm IHRA, NBU}) and
$-1\leq \lambda\leq 0$, then $G_{\lambda}[F]$ is {\rm IHR (IHRA,
NBU)}.
\indent (ii) If $F$ is {\rm DHR (DHRA, NWU)} and $0\leq
\lambda\leq 1$, then $G_{\lambda}[F]$ is {\rm DHR (DHRA, NWU)}.
\noindent b) $F\prec_{lr} G_{\lambda}[F]$ (consequently,
$F\prec_{\rm hr} G_{\lambda}[F]$ and $F\prec_{\rm st}
G_{\lambda}[F]$), if $-1\leq\lambda\leq 0$ and
$G_{\lambda}[F]\prec_{\rm lr}F$ (consequently, $F\prec_{\rm hr}
G_{\lambda}[F]$ and $F\prec_{\rm st} G_{\lambda}[F]$) if
$0\leq\lambda\leq1$.
\noindent c)the parametric family $\{G_{\lambda}[F]\}$ of
distributions is decreasing in $\lambda$ in the likelihood ratio
order. Consequently, $G_{\lambda}[F]$ is decreasing in the hazard
rate and stochastic orders.
\end{proposition}
\begin{proposition} Suppose that $F_1$ and $F_2$
be two given CDFs such that $F_1\prec_{\rm st} F_2$. Then
$G_{\lambda}[F_1]\prec_{\rm st} G_{\lambda}[F_2]$ for every
$\lambda\in [-1,1]$.
\end{proposition}
\noindent \textbf{Proof.} Since $F_1\prec_{\rm st}F_2$ implies
that $F_1(x)\geq F_2(x)$ and $\bar{F}_1(x)\leq \bar{F}_2(x)$, for
all $x$; we have
$G_{\lambda}[F_1](x)=F_1(x)\{1+\lambda\bar{F}_1(x)\}\geq
F_2(x)\{1+\lambda\bar{F}_2(x)\}=G_{\lambda}[F_2](x)$ for $\lambda<
0$ and $\bar{G}_{\lambda}[F_1](x)=\bar{F}_1(x)\{1-\lambda
\bar{F}_1(x)\}\leq \bar{F}_2(x)\{1-\lambda
F_2(x)\}=\bar{G}_{\lambda}[F_2](x)$ for $\lambda
>0$, which completes the proof.
\begin{proposition} Let $F_1$ and $F_2$ be two given
cdfs and let $G_{\lambda}[F_1]$ and $G_{\lambda}[F_2]$ be their
transformed versions using (\ref{m2}). Then
$$
F_1\prec_{\rm order}F_2\Rightarrow G_{\lambda}[F_1]\prec_{\rm
order}G_{\lambda}[F_2],
$$
where $\prec_{\rm order}$ is any one of the orders $\prec_{\rm
c}$, $\prec_{*}$ $\prec_{\rm su}$ and $\prec_{\rm disp}$.
\end{proposition}
\noindent \textbf{Proof.} From (\ref{quantil}) it is easy to see
that
\begin{equation*}
G^{-1}_{\lambda}[F_2]\left(G_{\lambda}[F_1](x)\right)=F^{-1}_2\left(F^{-1}_1(x)\right),
\end{equation*}
for all $x$, which gives the required result.
\subsection{A symmetry property}
The transformation map (\ref{m2}) can be applied to any symmetric
or asymmetric distribution. The following result shows the effect
of this transformation on the symmetry property of the parent
distribution.
\begin{proposition} Let $X$ with the cdf $F$, be a symmetric
random variable about zero (i.e., $X$ and $-X$ have the same
distribution) and let $Y_{\lambda}$ be a random variable
distributed according to $G_{\lambda}[F]$, $-1\leq\lambda\leq 1$.
Then $Y_{-\lambda}$ and $-Y_{\lambda}$ have the same distribution.
\end{proposition}
\noindent \textbf{Proof.} Since $X$ is symmetric about zero, then
$F(x)=1-F(-x)=\bar{F}(-x)$ for all $x$. From (\ref{m2}) and
(\ref{Gbar}) we have
\begin{eqnarray*}
P(-Y_{\lambda}\leq y)&=&\nonumber\bar{G}_{\lambda}[F](-y)\\
&=&\nonumber\bar{F}(-y)\{1-\lambda F(-y)\}\\&=&\nonumber
F(y)\{1-\lambda
\bar{F}(y)\}\\&=&\nonumber G_{-\lambda}[F](y)\\
&=&\nonumber P(Y_{-\lambda}\leq y),
\end{eqnarray*}
which completes the proof.
\section{Examples}
\subsection{The transformed exponential distribution}
In particular case that $F$ is an exponential
distribution with the parameter $\theta$, the two--parameter
distribution generated using (\ref{m2}) has the cdf
\begin{equation}
G(x;\lambda,\theta)=(1-e^{-\theta x})(1+\lambda e^{-\theta x}),
\quad\ x,\theta>0, -1\leq \lambda\leq1, \label{ME}
\end{equation}
and the corresponding density function
\begin{equation}
g(x;\lambda,\theta)=\theta e^{-\theta x}\{1+\lambda (2e^{-\theta
x}-1)\}.
\end{equation}
For the density function $g$, we have that $\log g(x;\lambda
,\theta )$, is concave for $-1\leq \lambda \leq 0$ and convex for
$0\leq \lambda \leq 1.$ As a result for $0\leq \lambda \leq 1$,
$g(x;\lambda,\theta)$ is decreasing, and for $-1\leq \lambda < 0$,
$g(x;\lambda,\theta)$ is unimodal. By solving the equation $d\log
g(x;\lambda ,\theta )/dx=0$, it is readily verified that the
density function $g(x;\lambda ,\theta )$ has the mode equal to
zero for $\lambda
>-\frac{1}{3}$ and $-\frac{1}{\theta }\ln ( \frac{\lambda-1
}{4\lambda})$ for $\lambda<-\frac{1}{3}$.
From (\ref{hr}), the hazard rate function of this distribution is
given by
\begin{equation*}
h(x;\lambda ,\theta )=\frac{\theta \{ 1+\lambda (2e^{-\theta
x}-1)\} }{1+\lambda (e^{-\theta x}-1)}.
\end{equation*}
It may be noticed that while the exponential distribution has a
constant hazard rate function, the generated cdf $G$, has
increasing hazard rate for $-1\leq\lambda < 0,$ and decreasing
hazard rate for $0< \lambda \leq1$, which follows using the
log-convexity and the log-concavity of the density function.
From (\ref{rlife}) the residual life survival function
corresponding to (\ref{ME}), is given by
\begin{equation}
\bar{G_{t}}(x;\lambda ,\theta )=e^{-\theta x}\{ 1+\beta
(e^{-\theta x}-1)\}, \label{rlifeME}
\end{equation}
where $\beta=\beta(t)=\lambda e^{-\theta t}\{ 1+\lambda
(e^{-\theta t}-1)\}^{-1}$. The limit distribution as
$t\rightarrow\infty$ is an ordinary exponential distribution
because the limit of $\beta(t)$ is 0.
From (\ref{rlifeME}) the mean residual life function of a random
variable $X$ having cdf (\ref{ME}), could be obtained as
\begin{eqnarray*}
m(t;\lambda ,\theta ) &=&\nonumber E(X-t|X>t) \\
&=&\nonumber\int_{0}^{\infty}\bar{G_{t}}(x;\lambda ,\theta )dx\\
&=&\frac{1+\lambda (\frac{1}{2}e^{-\theta
t}-1)}{\theta[1+\lambda(e^{-\theta t}-1)]},
\end{eqnarray*}
which is increasing in $t$ for $0\leq\lambda\leq 1$ and decreasing
for $-1\leq\lambda\leq 0$, with $\lim_{t \rightarrow \infty
}m(t;\lambda ,\theta )=1/\theta=E(X;0,\theta)$ and $
\lim_{t\rightarrow 0}m(t;\lambda ,\theta
)=(2-\lambda)/2\theta=E(X;\lambda,\theta)$; and hence
$$
\frac{1}{\theta} \leq m(t;\lambda ,\theta )\leq
\frac{2-\lambda}{2\theta } \text{ \ \ \ } (-1\leq \lambda \leq 0),
$$
and
$$
\text{ \ \ \ } \frac{2-\lambda}{2\theta }\leq m(t;\lambda ,\theta )\leq\frac{1}{\theta} \text{ \ \ }
(0\leq \lambda \leq 1).
$$
The moment generating function of this distribution is given by
\begin{equation*}
M(t)=E(e^{tX})=\frac{\theta \{2\theta -(1+\lambda )t\}}{(\theta
-t)(2\theta -t)}.
\end{equation*}
By straightforward integration the raw moments are found to be
\begin{equation*}
E(X^{r})=\frac{(1+\lambda(2^{-r}-1))r!}{\theta ^{r}},
\end{equation*}
for $r\in N.$
Since for the exponential distribution with the parameter $\theta$
we have $F^{-1}(q)=-\frac{1}{\theta}\text{ln}(1-q)$, $0<q<1$, then
from (\ref{quantil}) the quantile function of the generated
distribution is given by
\begin{equation*}
G^{-1}(q)=-\frac{1}{\theta }\ln \left( \frac{\lambda
-1+\sqrt{(1+\lambda)^{2}-4\lambda q}}{2\lambda }\right).
\end{equation*}
Note that if $\lambda \rightarrow 0,$ then $G^{-1}(q)\rightarrow
-\frac{1}{\theta }\ln (1-q).$
It may be noticed that for the generated distribution,
median($X$), mode($X$) and $E(X)$ are all decreasing in $\lambda$,
$\theta$ and $\text{mod}(X)\leq \text{median}(X)\leq E(X)$.
\subsection{A class of skew--Laplace distributions}
The classical symmetric Laplace distribution has the pdf
\begin{equation*}
f(x;\theta)=\frac{1}{2\theta}e^{-\frac{|x|}{\theta}},
\end{equation*}
and cdf
\begin{equation*}
F(x;\theta)=\left\{
\begin{array}{ll}
\frac{1}{2}e^{\frac{x}{\theta}}, & \mbox{{\rm} $x\leq 0$}, \\
\noalign{\smallskip}1-\frac{1}{2}e^{\frac{-x}{\theta}},&
\mbox{{\rm } $x\geq 0,$}
\end{array}
\right.\label{SLD}
\end{equation*}
where $-\infty<x<\infty$ and $\theta>0$. The symmetric Laplace
distribution has been used as an alternative to the normal
distribution for modeling heavy tails data. Different forms of the
skewed Laplace distributions have been introduced and studied by
various authors. Recently, Kozubowski and Nadarajah
\cite{Kozub2008} identified over sixteen variations of the Laplace
distribution. In the following we propose a new version of the
skewed Laplace distribution using (\ref{m2}). The cdf and pdf of
the generated model are given by
\begin{equation*}
G_{\lambda}(x;\theta)=\left\{
\begin{array}{ll}
\frac{1}{2}e^{\frac{x}{\theta}}\{1+\lambda(1-\frac{1}{2}e^{\frac{x}{\theta}})\}, & \mbox{{\rm} $x\leq 0$}, \\
\noalign{\smallskip}1-\frac{1}{2}e^{\frac{-x}{\theta}}\{1-\lambda(1-\frac{1}{2}e^{\frac{-x}{\theta}})\},&
\mbox{{\rm } $x\geq 0,$}
\end{array}
\right.\label{SLG}
\end{equation*}
and
\begin{equation*}
g_{\lambda}(x;\theta)=\left\{
\begin{array}{ll}
\frac{1}{2\theta}e^{\frac{x}{\theta}}\{1+\lambda(1-e^{\frac{x}{\theta}})\}, & \mbox{{\rm} $x\leq 0$}, \\
\noalign{\smallskip}\frac{1}{2\theta}e^{\frac{-x}{\theta}}\{1-\lambda(1-e^{\frac{-x}{\theta}})\},&
\mbox{{\rm } $x\geq 0,$}
\end{array}
\right.\label{SLg}
\end{equation*}
respectively. The moment generating function of $G_{\lambda}$, is
given by
\begin{equation*}
M(t)=\frac{1-\lambda \theta t}{1-(\theta t)^2}+\frac{\lambda
\theta t}{4-(\theta t)^2},
\end{equation*}
and the raw moments are found to be
\begin{equation*}
E(X^r)=\left\{
\begin{array}{llll}
\frac{r!\lambda\theta^{r}(1-2^{r+1})}{2^{r+1}}, & \mbox{{\rm if $r$ is odd, }}\\
\noalign{\smallskip}r!\theta^{r},& \mbox{{\rm if $r$ is even. }}
\end{array}
\right.\label{Mom}
\end{equation*}
The expectation, variance, skewness and the kurtosis are given by
\begin{eqnarray*}
\text{E}(X)&=&-\frac{3}{4}\lambda\theta , \\
\text{Var}(X)&=&\theta^{2}(1-\frac{9}{16}\lambda^{2}), \\
\text{Skewness}(X)&=&\frac{18\lambda(4+3\lambda^{2})}{(9\lambda^{2}-32)\sqrt{32-9\lambda^{2}}}, \\
\text{Kurtosis}(X)&=&\frac{6144-243\lambda^{4}-2592\lambda^{2}}{(32-9\lambda^{2})^{2}}.
\end{eqnarray*}
It may be noticed that the skewness of $G_{\lambda}$ is decreasing
in $\lambda$, and then $-1.1423\leq \text{Skewness}(X)\leq
1.1423$. It is positive for $-1\leq\lambda\leq 0$, and negative
for $0\leq\lambda\leq 1$.
\section{Bivariate case}
\subsection{Construction}
A large number of bivariate distributions
have been proposed in literature. A very wide survey on bivariate
distributions are given in \cite{Bala2009} and \cite{Kotz2000}.
The method used to construct the family of distributions given by
(\ref{m2}) also lends itself well to the construction of bivariate
distributions whose univariate marginal cdf are of the form
(\ref{m2}).
\begin{proposition} Let $F$ be a bivariate cdf with
the univariate
marginal cdfs $F_1$, $F_2$ and the associated survival function $\overline{F}(x,y)=1-F_1(x)-F_2(y)+F(x,y)$.
Then, for every $-1\leq\lambda\leq1$, the
function $G_{\lambda}:\text{R}^2\rightarrow[0,1]$, defined by
\begin{equation}
G_{\lambda}(x,y)=(1+\lambda)\left(F_1(x)F_2(y)+F(x,y)\overline{F}(x,y)\right)-\lambda
F^{2}(x,y), \label{Bivariate}
\end{equation}
is a bivariate cdf with the univariate marginal distributions
\begin{equation}
G_1(x)=F_1(x)\{1+\lambda \overline{F}_1(x)\} \quad \text{and}
\quad G_2(y)=F_2(y)\{1+\lambda \overline{F}_2(y)\}. \label{uni}
\end{equation}
\end{proposition}
\noindent \textbf{Proof.} To prove this, let $(X_1,Y_1)$ and
$(X_2,Y_2)$ be two independent random vector having common
bivariate cdf $F$ and the univariate
marginal cdfs $F_1$ (of $X_i$) and $F_2$ (of $Y_i$), $i=1,2$.
Let $X_{(1)},X_{(2)}$ and $Y_{(1)},Y_{(2)}$ be their corresponding
order statistics. For $-1\leq\lambda\leq1$, consider the random
pair $(V_1,V_2)=(X_{(1)},Y_{(1)})$ with probability
$\frac{1+\lambda}{2}$ and $(V_1,V_2)=(X_{(2)},Y_{(2)})$ with
probability $\frac{1-\lambda}{2}$. Then, it is straightforward to
verify that $(V_1,V_2)$ have the joint cdf (\ref{Bivariate}) with
$G_\lambda(x,\infty)=G_1(x)$ and $G_\lambda(\infty,y)=G_1(y)$.
Note that the special case $F(x,y)=F_1(x)F_2(y)$, the cdf
(\ref{Bivariate}) reduces to
\begin{equation*}
G_{\lambda}(x,y)=F_1(x)F_2(y)\left\{F_1(x)F_2(y)+(1+\lambda)(\overline{F}_1(x)+\overline{F}_2(y))\right\},\label{FGM2}
\end{equation*}
which may serve as a competitor to the FGM family of distributions
with the univariate margins of the form (\ref{uni}).
\subsection{Underlying copula}
A bivariate distribution $F$ can be written in the form $F(x, y)
=C\{F_1(x),F_2(y)\}$, where $C$ is the copula associated with $F$;
see Nelsen \cite{Nelsen2006} for more detail. The function
$\hat{C}$ defined by
$\hat{C}(u,v)=u+v-1+C(1-u,1-v)=\overline{C}(1-u,1-v)$, is the
survival copula associated with $C$ and, moreover
$\overline{F}(x,y)=\hat{C}\{\overline{F}_1(x),\overline{F}_2(y)\}=\overline{C}\{F_1(x),F_2(y)\}$.
The following result shows the relationship between the copula
associated with the baseline cdf $F$ and the copula of generated
cdf $G_{\lambda}$.
\begin{proposition} Let $C_{\lambda}$ be the copula
of the cdf $G_{\lambda}$ defined by (\ref{Bivariate}) and let $D$
be the copula of the baseline cdf $F$. Then
\begin{eqnarray}
C_{\lambda}(\psi(u),\psi(v))&=&(1+\lambda)\left\{uv+D(u,v)\overline{D}(u,v)\right\}-\lambda
D^2(u,v), \label{cop}
\end{eqnarray}
for all $0<u,v<1$, and $-1\leq\lambda\leq1$, where
$$
\psi(t)=t+\lambda t(1-t), \quad 0<t<1.
$$
\end{proposition}
\noindent \textbf{Proof.} Notice that using the definition of a
copula, the bivariate cdf ({\ref{Bivariate}) can be rewritten as
\begin{eqnarray}
C_{\lambda}\{G_1(x),G_2(y)\}&=&(1+\lambda)\left(F_1(x)F_2(y)+D\{F_1(x),F_2(y)\}\overline{D}\{F_1(x),F_2(y)\}\right)\nonumber\\&-&\lambda
D^{2}\{F_1(x),F_2(y)\}, \label{cop2}
\end{eqnarray}
where $G_1$ and $G_2$ are given by (\ref{uni}). By applying the
transformations $u=F_1(x)$ and $v=F_2(y)$ on both sides of
(\ref{cop2}) we readily obtain the required result.
\begin{remark} Note that if the baseline copula $D$ is symmetric, i.e., $D(u,v)=D(v,u)$, for all $u,v\in (0,1)$, then the generated copula $C_{\lambda}$
defined in (\ref{cop}) is symmetric.
\end{remark}
\begin{proposition} The family of copulas $\{C_{\lambda}\}$ defined in (\ref{cop}) is positively ordered for all $-1<\lambda\leq 1$
and any baseline copula $D$; i.e., $C_{\lambda_{1}}(u,v)\geq
C_{\lambda_{2}}(u,v)$ for all $u,v\in(0,1)$ whenever $\lambda
_{1}\geq \lambda_{2}.$
\end{proposition}
\noindent \textbf{Proof.} For any two
copulas $C_{\lambda_1}$ and $C_{\lambda_2}$ of the form (\ref{cop}), one has
$$C_{\lambda_1}\{\psi_{\lambda_1}(u),\psi_{\lambda_1}(v)\}-C_{\lambda_2}\{\psi_{\lambda_2}(u),\psi_{\lambda_2}(v)\}=(\lambda_1-\lambda_2)\{D(u,v)(1-u)(1-v)+uv(1-D(u,v))\},$$
where $\psi_{\lambda}(t)=t+\lambda t(1-t)$, $0<t<1$. Since
$D(u,v)(1-u)(1-v)+uv(1-D(u,v))\geq 0$ for all $u,v\in(0,1)$ and
the function $\psi_{\lambda}(t)=t+\lambda t(1-t)$ is increasing in
$\lambda$ for all $t\in(0,1)$, it is easy to see that
$$C_{\lambda_1}(u,v)-C_{\lambda_2}(u,v)\geq 0,$$ for all $u,v\in
(0,1)$ and $\lambda_1\geq\lambda_2$, which completes the proof.
\bigskip
We now consider some special cases.
\begin{example}
For the special case that $D(u,v)=uv$, i.e.,
$F(x,y)=F_1(x)F_2(y)$, we have
\begin{equation*}
C_{\lambda}\{\psi(u),\psi(v)\}=uv\{uv+(1+\lambda)(2-u-v)\}.
\end{equation*}
\end{example}
\begin{example}
Suppose that $D=M$, where $M(u,v)={\rm min}(u,v)$, is the
Fr\'{e}chet--Hoeffding upper bound copula (see \cite{Nelsen2006});
which means that the baseline cdf $F$ is the cdf of two perfect
positive dependent random variable $X$ and $Y$. Since
$\overline{M}(u,v)=M(1-u,1-v)$ for every $u,v\in(0,1)$, it is easy
to verify that $M(u,v)\overline{M}(u,v)=M(u,v)-uv$. By applying
(\ref{cop}) to $M$, from the fact that for non--decreasing
function $\psi$,
$\text{min}\{\psi(u),\psi(v)\}=\psi(\text{min}(u,v))$ we obtain
\begin{eqnarray*}
C_{\lambda}(\psi(u),\psi(v))&=&M(u,v)\{1+\lambda(1-M(u,v)\}\nonumber\\&=&\psi\{M(u,v)\}\nonumber\\&=&M(\psi(u),\psi(v)),
\end{eqnarray*}
that is for all $\lambda \in (-1,1)$,
$$
C_{\lambda}(u,v)=M(u,v).
$$
Thus the functional transformation (\ref{Bivariate}) preserves the
perfect dependence of the parent distribution.
\end{example}
\begin{example}
Suppose that $D=W$, where $W(u,v)={\rm max}(u+v-1,0)$, is the
Fr\'{e}chet--Hoeffding lower bound copula (see \cite{Nelsen2006});
which means that the baseline cdf $F$ is the cdf of two perfect
negative dependent random variable $X$ and $Y$. It is easy to
verify that $\overline{W}(u,v)W(u,v)=0$, for every $u,v\in(0,1)$.
Thus (\ref{cop}) gives
$$
C_{\lambda}(\psi(u),\psi(v))=uv+\lambda\{uv-W^2(u,v)\}.
$$
\end{example}
\section{Discussion}
We have introduced a method for constructing a new family of
distributions from any given one. We deliberately restricted our
attention to the study of some general properties of the proposed
model in univariate as well as the bivariate case. The attentive
reader will agree that the construction presented here leaves room
for more studies beyond what accomplished in this work. In our
next investigation we aim to make deeper contributions to the
distribution theory connected to the bivariate case.
\bigskip
\noindent \textbf{\large Acknowledgements} \smallskip
The authors would like to thank an anonymous referee for his/her
valuable suggestions on an earlier version of this paper.
\bigskip
| 23,777
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| 13,215
|
TITLE: Computing the Jacobian $J_F$ with $F = h \circ f$
QUESTION [1 upvotes]: Let $f : \mathbb{R}^l \rightarrow{} \mathbb{R}^m$ and $h : \mathbb{R}^m \rightarrow{} \mathbb{R}^o$
and let $F = h \circ f$ with $F : \mathbb{R}^l \rightarrow{} \mathbb{R}^o$
I want to compute the Jacobian using Forward mode accumulation in one path.
I do understand the automatic differentiation way of working in the forward mode for simple cases.
So I think based on my knowledge if I want to compute $J_F$ using as many paths as I want I could do $l*m$ paths. Since I am constrained with only one path, I start to get confused. I know I can do something with the initialization but it's very confused in my mind. Could you please help me to understand how to implement the forward mode just in one path using the Jacobian of $f$ and $h$ along the way?
Thanks
REPLY [2 votes]: It's just the matrix multiplication of the two Jacobians. In fact, this can be seen as the reason why matrix multiplication is defined like that. If you see a $m\times n$ matrix as a linear function from $\mathbb R^n \to \mathbb R^m$, then the Jacobian of the matrix is the matrix itself. The matrix multiplication is just the composition of the two linear functions, so the Jacobian of the composed function is just the matrix multiplication of the two Jacobians.
Locally (that is, in a infinitesimal neighborhood), all differentiable functions can be seen as a linear function of the derivatives of the inputs, therefore you can use matrices to describe the local behavior of all multivariable differentiable functions.
Let's write the components of the function $f$ as $f^\mu$ where $\mu$ is an index that goes from $1$ to $m$. Similarly, you can write $g$ as $g^\mu$. In this way, can write a vector as a symbol with indices but treat them as numbers, (as in, $f^1$ is the first element of the result of $f(x)$, which is in $\mathbb R$.
Then you can write the partial derivative $\frac{\partial f^\mu}{\partial x^\nu}$ as $f^\mu{}_{,\nu}$, mind the comma. It is easy to see that the $f^\mu{}_{,\nu}$ is exactly the Jacobian of $f$, as in $f^1{}_{,2} \equiv \partial f^1/\partial x_2$, is the $(1,2)$-th element of the Jacobian. These notations are called the Ricci Calculus.
The chain rule says:
$$
\left(\frac{\partial}{\partial x^\nu}g(f(x))\right)^\mu=(g\circ f)^\mu{}_{,\nu}=\sum_\gamma g^\mu{}_{,\gamma}f^\gamma{}_{,\nu}\equiv g^\mu{}_{,\gamma}f^\gamma{}_{,\nu}
$$
The last one is to simplify the notation of the summation using Einstein notation, invented by Einstein for manipulating tensors, which is a part of Ricci Calculus as well.
You can see that $\mu$ goes from $1$ to $o$, $\gamma$ goes from $1$ to $m$, and $\nu$ goes from $1$ to $l$. This immediately gives the matrix multiplication of the two Jacobians.
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| 212,103
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TITLE: How to write this inequality in terms of Schur Complement?
QUESTION [1 upvotes]: I know the basis about Schur-Complement. Anyway, while looking at this inequality to apply it in order to solve for $\lambda$ such that the the matrix is definite positive, I got a little bit confused because the lack of non inverse terms.
${ X }^{ T }\left( PA+{ A }^{ T }P \right) X<-\lambda { X }^{ T }AX$
May I... ?
$\begin{bmatrix} 0 & { X } \\ { X }^{ T } & { \left( PA+{ A }^{ T }P+\lambda A \right) }^{ -1 } \end{bmatrix}$
Or,
$\begin{bmatrix} { X }^{ T }\left( PA+{ A }^{ T }P+\lambda A \right) X & { P } \\ P & 0 \end{bmatrix}$
Or even,
$\begin{bmatrix} { X }^{ T }\left( PA+{ A }^{ T }P \right) X & X \\ { X }^{ T } & -\lambda { A }^{ -1 } \end{bmatrix}$
I'm sorry if this is a silly or even stupid question, but i'm lost :(
Any references, books, or examples that You could recommend me?
REPLY [1 votes]: I don't see the point of using Schur's complements here.
In fact, looking for $\lambda$ satisfying inequality
$$X^T\left( PA+A^TP \right) X<-\lambda X^TAX\tag{1}$$
is the same as looking for the bounds of
$$\dfrac{X^TBX}{X^TAX} \ \ \text{where} \ \ B:=PA+A^TP \tag{2}$$
which is a classical issue :
If $A$ is positive-definite, consider a Cholesky decomposition $A=C^TC$, set $Y=CX$, transforming (2) into the Rayleigh quotient :
$$\dfrac{YC^{-T}BC^{-1}Y}{Y^TY} $$
which is known to take all values in interval $[\lambda_{min},\lambda_{max}]$
where $\lambda_{min},\lambda_{max}$ are the extreme eigenvalues of matrix $D:=C^{-T}BC^{-1}$.
Connected :
Proof: Ratio of matrix traces and difference of traces
Ratio of two quadratic vector forms
| 29,280
|
\begin{document}
\begin{abstract}
To every $n$-dimensional lens space $L$, we associate a congruence lattice $\mathcal L$ in $\Z^m$, with $n=2m-1$ and
we prove a formula relating the multiplicities of Hodge-Laplace eigenvalues on $L$ with the number of lattice elements of a given $\norma{\cdot}$-length in $\mathcal L$.
As a consequence, we show that two lens spaces are isospectral on functions (resp.\ isospectral on $p$-forms for every $p$) if and only if the associated congruence lattices are $\norma{\cdot}$-isospectral (resp.\ $\norma{\cdot}$-isospectral plus a geometric condition).
Using this fact, we give, for every dimension $n\ge 5$, infinitely many examples of Riemannian manifolds that are isospectral on every level $p$ and are not strongly isospectral.
\end{abstract}
\maketitle
\section{Introduction}
Two compact Riemannian manifolds $M$ and $M'$ are said to be $p$-isospectral if the spectra of their Hodge-Laplace operator $\Delta_p$, acting on $p$-forms, are the same.
Many examples of non-isometric isospectral manifolds have been constructed showing connections between the spectra and the geometry of a Riemannian manifold (e.g.\ \cite{Mil}, \cite{Vigneras}, \cite{Ik80}, \cite{Go01}, \cite{Schueth}). In \cite{Su} Sunada gave a general method that allowed constructing many examples; however, the resulting manifolds are always strongly isospectral, that is, they are isospectral for every natural, strongly elliptic operator acting on sections of a natural vector bundle over $M$; in particular, they are $p$-isospectral for all $p$ (see \cite{Ik83}, \cite{Gi}, \cite{Wo2} for applications in the case of spherical space forms).
The converse question is a problem that has been present for some time, i.e.\ whether $p$-isospectrality for all $p$ implies strong isospectrality (see for instance J.~A.~Wolf~\cite[p.~323]{Wo2}).
Manifolds that are $p$-isospectral for some values of $p$ only have been investigated by several authors.
C.~Gordon \cite{Go} gave the first example of this type in the context of nilmanifolds and A.~Ikeda \cite{Ik88} showed, for each $p_0>0$, lens spaces that are $p$-isospectral for all $p<p_0$ and are not $p_0$-isospectral. For more examples see \cite{Gornet}, \cite{MRp}, \cite{MRl}, \cite{GM}.
In this paper, we will show a connection between the spectra of lens spaces and the one-norm spectra of their associated congruence lattices, and, as a consequence, we will give many pairs of $p$-isospectral lens spaces for every $p$ that are far form being strongly isospectral.
These examples are the first of this kind to our best knowledge.
The main approach is as follows.
To each $2m-1$-dimensional lens space $L=L(q;s_1, \dots, s_m)$, we associate the congruence lattice in $\Z^m$ defined by
\begin{equation*}
{\mathcal L=\mathcal L(q;s_1,\dots,s_m) = \{(a_1,\dots,a_m) \in \Z^m: a_1s_1+\dots +a_ms_m\equiv0\pmod q\}.}
\end{equation*}
In the first main result, Theorem~\ref{thm3:dim V_k,p^Gamma}, we give a formula for the multiplicity of each eigenvalue of $\Delta_p$ on a lens space, in terms of the multiplicities of the weights of certain representations of $\SO(2m)$ and the number of elements with a given $\norma{\cdot}$-length in the associated congruence lattice (see \eqref{eq3:dim V_k,p^Gamma}).
In particular, the multiplicity of the eigenvalue $k(k+n-1)$ of the Laplace-Beltrami operator $\Delta_0$ simplifies to \eqref{eq3:dim V_k,1^Gamma}
\begin{equation*}
\sum_{r=0}^{\lfloor k/2\rfloor}\binom{r+m-2}{m-2} N_{\mathcal L}(k-2r),
\end{equation*}
where $N_{\mathcal L}(h)$ denotes the number of elements in $\mathcal L$ with $\norma{\cdot}$-length $h$.
As a consequence, we prove that two lens spaces $L$ and $L'$ are $0$-isospectral if and only if the associated lattices $\mathcal L$ and $\mathcal L'$ are isospectral with respect to $\norma{\cdot}$ (Theorem~\ref{thm3:characterization}).
Remarkably, it turns out that two lens spaces are $p$-isospectral for every $p$, if and only if the corresponding lattices are $\norma{\cdot}$-isospectral and satisfy an additional geometric condition: for each $k\in\N$ and $0\leq \zz\leq m$ there are the same number of elements $\mu$ in each lattice having $\norma{\mu} = k$ and exactly $\zz$ coordinates equal to zero.
We call such lattices $\norma{\cdot}^*$-isospectral.
In Section~\ref{sec:finiteness} we define, for any congruence lattice ${\mathcal L}$, a finite set of numbers $N_{\mathcal L}^{\mathrm{red}}(k,\zz)$ that
count the number of lattice points of a fixed norm $k$ in a small cube, having exactly $\zz$ zero coordinates.
We show that these numbers determine the $p$-spectrum of the associated lens space.
In particular, one can decide with finitely many computations, whether two lens spaces are $p$-isospectral for all $p$.
We implemented an algorithm in Sage~\cite{Sage} obtaining all examples in dimensions $n=5$ and $7$ for values of $q$ up to $300$ and $150$ respectively (see Tables~\ref{table:m=3} and \ref{table:m=4} in Section~\ref{sec:examples}). We also include some open questions on the nature of the existing examples. We point out that in \cite{DD} the authors make a nice improvement, answering one of our questions.
As a next step we exhibit many pairs of $\norma{\cdot}^*$-isospectral congruence lattices.
We do this in Section~\ref{sec:families} proving that, for any $r\geq7$ and $t$ positive integers such that $r$ is coprime to $3$, the congruence lattices
$$
\mathcal L(r^2t;\;1,\;1+rt,\;1+3rt)
\quad\text{and}\quad
\mathcal L(r^2t;\;1,\;1-rt,\;1-3rt)
$$
are $\norma{\cdot}^*$-isospectral.
In order to prove the equality of $N_{\mathcal L}(k,\zz)$ and $N_{\mathcal L'}(k,\zz)$ for every $k$ and $\zz$ we develop a procedure to compute these numbers.
By using the results in the previous sections, we obtain the corresponding results for lens spaces in Section~\ref{sec:all-p-iso}.
The pairs of $\norma{\cdot}^*$-isospectral congruence lattices from Section~\ref{sec:families} produce an infinite family of pairs of $5$-dimensional lens spaces that are $p$-isospectral for all $p$.
This allows to obtain lens spaces that are $p$-isospectral for all $p$ in arbitrarily high dimensions by using a result of Ikeda (Theorem~\ref{thm7:high-dim}).
We point out that the resulting lens spaces are homotopically equivalent but cannot be homeomorphic to each other (see Lemma~\ref{lem:homotequiv}).
It is well known that two non-isometric lens spaces cannot be strongly isospectral (see Proposition~\ref{prop7:lens-non-strongly}).
In particular, isospectral non-isometric lens spaces cannot be constructed by the Sunada method.
In the case of the simplest pair of $5$-dimensional lens spaces $L=L(49;1,6,15)$ and $L'=L(49;1,6,20)$ that are $p$-isospectral for all $p$, we give many representations $\tau$ of $K=\SO(5)$ for which the associated natural strongly elliptic operators do not have the same spectrum (Section~\ref{sec:tau-isospectrality}). Actually, these lens spaces are very far from being strongly isospectral.
The method based on representation theory to characterize $0$-spectrum for a lens space could also be applied to many other spaces, for example, orbifold lens spaces (see Remark~\ref{rem7:orbifolds}), arbitrary spherical space forms and more general locally symmetric spaces of compact type.
\section{Preliminaries}\label{sec:prelim}
Let $G$ be a compact Lie group and let $K$ be a compact subgroup, and let $X=G/K$ endowed with a $G$-invariant Riemannian metric induced by a $G$-biinvariant metric on $G$.
We shall assume that $G$ is semisimple.
Let $\Gamma$ be a discrete subgroup of $G$ that acts freely on $X$, thus the manifold $\Gamma\ba X$ inherits a locally $G$-invariant Riemannian structure.
\subsection{Homogeneous vector bundles}\label{subs:vector-bundles}
For each finite dimensional unitary representation $(\tau, W_\tau)$ of $K$, we consider the \emph{homogeneous vector bundle}
$$
E_\tau= G \times_\tau W_\tau \longrightarrow X=G/K
$$
(see for instance \cite[\S 2.1]{LMRrepequiv}).
We recall that the space $\Gamma^{\infty}(E_\tau )$ of smooth sections of $E_\tau$ is isomorphic to the space $C^\infty(G/K;\tau)$ of smooth functions $C^\infty(G/K;\tau):=\{ f: G \rightarrow W_\tau$ such that $f(xk)= \tau(k^{-1})f(x)\}$.
We form the vector bundle $\Gamma\ba E_\tau$ over the manifold $\Gamma\ba G/K$ and denote it by $L^2(\Gamma\ba E_\tau)$ the closure of $C^\infty (\Gamma\ba G/K;\tau)$ with respect to the inner product
$
(f_1,f_2)= \int_{\Gamma\ba X} \langle f_1(x),f_2(x)\rangle \;\mathrm{d}x$, where $\langle\,\,\rangle$ is a $\tau$-invariant inner product on $W_\tau$.
The complexification $\mathfrak g$ of the Lie algebra $\mathfrak g_0$ of $G$ and the universal enveloping algebra $U(\mathfrak g)$ act on $C^\infty(G/K;\tau)$ by left invariant differential operators in the usual way.
We shall denote by $C=\sum X_i^2$ the \emph{Casimir} element of $\mathfrak g$, where $X_1,\dots,X_n$ is any orthonormal basis of $\mathfrak g$;
$C$ lies in the center of $U(\mathfrak g)$ and defines second order elliptic \emph{differential operators}
$\Delta_\tau$ on $C^\infty(G/K;\tau)$ and $\Delta_{\tau,\Gamma}$ on $\Gamma\ba E_\tau$. The Casimir element $C$ acts on an irreducible representation $V_\pi$ of $G$ by a scalar $\lambda(C,\pi)$.
Consider the left regular representation of $G$ on $L^2( E_\tau) \simeq L^2(G/K;\tau)$. By Frobenius reciprocity, the multiplicity of an irreducible representation $\pi$ of $G$ equals $[\tau:\pi_{|K}] : = \dim Hom_K(V_\tau, V_\pi)$. We thus have
\begin{equation}\label{eq2:L^2(G/K, tau)}
L^2(G/K;\tau)=\sum_{\pi\in\widehat G} [\tau:\pi_{|K}]\, V_\pi.
\end{equation}
Thus, by taking $\Gamma$-invariants,
\begin{equation}\label{eq2:L^2(G/K, tau)}
L^2(\Gamma\ba G/K;\tau)=\sum_{\pi\in\widehat G} [\tau:\pi_{|K}]\, V_\pi^\Gamma,
\end{equation}
where $V_\pi^\Gamma$ is the space of $\Gamma$-invariant vectors in $V_\pi$. We set $d_\pi^\Gamma=\dim V_\pi^\Gamma$.
Similarly we may consider the right regular representation of $G$ on
$$L^2(\Gamma \ba G) = \sum_{\pi \in\widehat G} n_{\pi}(\Gamma) \pi\,.$$
By Frobenius reciprocity, we get in this case that $n_{\pi}(\Gamma)= d_\pi^\Gamma=\dim V_\pi^\Gamma$.
Hence we have the decomposition
\begin{equation}\label{eq2:L^2(GammaG)}
L^2(\Gamma\ba G)=\sum_{\pi\in\widehat G} d_\pi^\Gamma\, V_\pi.
\end{equation}
Hence, taking into account \eqref{eq2:L^2(G/K, tau)} and \eqref{eq2:L^2(GammaG)} we obtain:
\begin{prop}\label{prop2:tau-equiv=>tau-iso}
Let $(G,K)$ be a symmetric pair of compact type and let $\Gamma$ be a discrete cocompact subgroup of $G$ that acts freely on $X=G/K$.
Let $\Delta_{\tau,\Gamma}$ be the Laplace operator acting on the sections of the homogeneous vector bundle $\Gamma\ba E_\tau$ of the manifold $\Gamma\ba X$.
If $\lambda\in\R$, the multiplicity $d_{\lambda}(\tau,\Gamma)$ of the eigenvalue $\lambda$ of $\Delta_{\tau,\Gamma}$ is given by
\begin{equation}\label{eq2:mult_lambda}
d_{\lambda}(\tau, \G)=
\sum_{\pi\in \widehat G :\, \lambda(C,\pi)=\ld}
d_\pi^\Gamma \;[\tau:\pi_{|K}] .
\end{equation}
\end{prop}
In the case when $\Gamma$ is a finite abelian group inside a maximal torus $T$ of $G$ one can further write the dimension $d_\pi^\Gamma$
of the space of $\Gamma$-invariants in $V_\pi$ in a simple way in terms of weight multiplicities (see Lemma \ref{lem3:L_Gamma}).
\subsection{Spherical space forms}
In this subsection we will recall the description of the $p$-spectrum of the Hodge-Laplace operator on spherical space forms.
We will restrict our attention to odd dimensions, namely, spaces $\Gamma\ba S^{2m-1}$ where $\Gamma$ is a finite subgroup of $\SO(2m)$ that acts freely on $S^{2m-1}$. We first recall some general facts on the representation theory of compact Lie groups.
We note that if a discrete (finite) subgroup $\Gamma\subset\Ot(n+1)$ acts freely on $S^n$, then it must necessarily be included in $\SO(n+1)$, thus $\Gamma\ba S^n$ is an orientable manifold.
We set $G=\SO(2m)$.
We fix the standard maximal torus in $G$,
\begin{equation*}
T=\left\{t=
\diag\left(
\left[\begin{smallmatrix}\cos(2\pi\theta_1)&-\sin(2\pi\theta_1) \\ \sin(2\pi\theta_1)&\cos(2\pi\theta_1)
\end{smallmatrix}\right]
,\dots,
\left[\begin{smallmatrix}\cos(2\pi\theta_m)&-\sin(2\pi\theta_m) \\ \sin(2\pi\theta_m)&\cos(2\pi\theta_m)
\end{smallmatrix}\right]
\right)
:\theta\in\R^m
\right\}.
\end{equation*}
The Lie algebra of $T$ is given by
\begin{equation}\label{eq2:h_0}
\mathfrak h_0=\left\{ H=
\diag\left(
\left[\begin{smallmatrix}0&-2\pi \theta_1\\ 2\pi \theta_1&0\end{smallmatrix}\right]
, \dots,
\left[\begin{smallmatrix}0&-2\pi \theta_m\\ 2\pi \theta_m&0\end{smallmatrix}\right]
\right)
:\theta\in\R^m
\right\}.
\end{equation}
Note that $t=\exp(H)$ if $t\in T$ and $H\in \mathfrak h_0$ as above.
The Cartan subalgebra $\mathfrak h:=\mathfrak h_0\otimes_\R \C$ is given as in \eqref{eq2:h_0} with $\theta_1,\dots,\theta_m\in\C$, and in this case we let $\varepsilon_j\in\mathfrak h^*$ be given by $\varepsilon_j(H)=2\pi i\theta_j$ for any $1\leq j\leq m$.
The weight lattice for $G=\SO(2m)$ is $P(G)=\bigoplus_{j=1}^m\Z\varepsilon_j$.
We fix the standard system of positive roots $\Delta^+(\mathfrak g,\mathfrak h)=\{\varepsilon_i\pm\varepsilon_j: 1\leq i<j\leq m\}$, with system of simple roots $\{\varepsilon_j-\varepsilon_{j+1}: 1\leq j\leq m-1\}\cup\{\varepsilon_{m-1}+\varepsilon_m\}$ and dominant weights of the form $\sum_{j=1}^m a_j\varepsilon_j\in P(G)$ such that $a_1\geq\dots\geq a_{m-1}\geq |a_m|$.
We denote by $\{e_1,\dots,e_{2m}\}$ the standard basis of $\R^{2m}$.
If $K= \{k \in \SO(2m): k e_{2m} = e_{2m}\} \simeq\SO(2m-1)$, then we take the maximal torus $T\cap K$, thus the Cartan subalgebra associated $\mathfrak h_K$ can be seen as included in $\mathfrak h$.
Under this convention, the positive roots are $\{\varepsilon_i\pm\varepsilon_j:1\leq i<j\leq m-1\}\cup\{\varepsilon_i:1\leq i\leq m-1\}$, the simple roots are $\{\varepsilon_j-\varepsilon_{j
+1}: 1\leq j\leq m-2\}\cup \{\varepsilon_{m-1}\}$, the weight lattice of $K$ is $P(K)=\bigoplus_{j=1}^{m-1} \Z \varepsilon_j$ and $\mu=\sum_{j=1}^{m-1}a_j\varepsilon_j\in P(K)$ is dominant if and only if $a_1\geq\dots \geq a_m\geq0$.
We consider on $\mathfrak g=\so(2m,\C)$ the inner product given by $\langle X,Y\rangle=-(2n-2)^{-1}B(X,\theta Y)$, where $B$ is the Killing form and $\theta$ is the Cartan involution.
One can check that $\langle X,Y\rangle=\operatorname{Trace}(XY)$ for $X,Y\in\mathfrak g$, and this inner product induces on $G/K=S^{2m-1}$ the Riemannian metric of constant sectional curvature $1$.
Furthermore, $\{\varepsilon_1,\dots,\varepsilon_m\}$ is an orthonormal basis of $\mathfrak h^*$.
If $\Gamma$ is a finite subgroup of $G$ acting freely on $S^{2m-1}$, denote by $\Delta_{p,\Gamma}$ the Hodge-Laplace operator on $p$-forms on the spherical space form $\Gamma\ba S^{2m-1}$.
That is, $\Delta_{p,\Gamma}=dd^*+d^*d:\bigwedge^p T^* M\to \bigwedge^p T^* M$, where $d$ is the differential, $d^*$ the codifferential and $\bigwedge^p T^* M$ is the $p$-exterior cotangent bundle of $M$.
As usual, the \emph{$p$-spectrum} of $\Gamma\ba S^{2m-1}$ stands for the spectrum of $\Delta_{p,\Gamma}$ and we say that two manifolds are \emph{$p$-isospectral} if their $p$-spectra coincide.
Spherical space forms are always orientable, thus their $p$-spectra coincide with the $(2m-1-p)$-spectra for all $0\leq p\leq 2m-1$.
We next describe the $p$-spectrum of any odd-dimensional spherical space form $\Gamma\ba S^{2m-1}$ in terms of $\Gamma$-invariants.
We first introduce some more notation.
Let $\mathcal E_0=\{0\}$ and
$$
\mathcal E_p=\{\lambda_{k,p} := k^2 + k(2m-2) + (p-1)(2m-1-p): k\in\N\}
$$
for $1\leq p\leq m$.
A known and useful fact is that $\mathcal E_p$ and $\mathcal E_{p+1}$ are disjoint for every $0\leq p\leq m-1$ (see for instance \cite[Rmk. after Thm.~4.2]{IkTa} \cite[Rmk.~1.14]{Ik88} and \cite[Thm.~1.1]{LMRrepequiv}).
Let
\begin{equation}\label{eq2:Lambda_p}
\Lambda_p = \begin{cases}
0 \quad&\text{if }p=0,\\
\varepsilon_1+ \varepsilon_2+\dots+\varepsilon_{p}
\quad&\text{if }1 \leq p\leq m.
\end{cases}
\end{equation}
Let $\pi_{k,p}$ denote the irreducible representation of $\SO(2m)$ with highest weight $k\varepsilon_1+\Lambda_p$ for $0\leq p<m$, and let $\pi_{k,m}$ denote the sum of the irreducible representations with highest weights $k\varepsilon_1+\Lambda_m$ and $k\varepsilon_1+\overline \Lambda_m$, where $\overline \Lambda_m= \varepsilon_1 + \dots +\varepsilon_{m-1} -\varepsilon_m$.
We will usually write $\pi_{k\varepsilon_1}$ and $\pi_{\Lambda_p}$ in place of $\pi_{k,0}$ and $\pi_{0,p}$ respectively.
\begin{prop}\label{prop2:p-spectrum}
Let $\Gamma\ba S^{2m-1}$ be a spherical space form and let $p$ be such that $0\leq p\leq m-1$.
If $\lambda$ is an eigenvalue of $\Delta_{p,\Gamma}$ then $\lambda\in\mathcal E_p\cup\mathcal E_{p+1}$. Its multiplicity is given by
\begin{align*}
d_\lambda (0,\Gamma) &=
\dim V_{\pi_{k\varepsilon_1}}^\Gamma = n_\Gamma(\pi_{k\varepsilon_1})
\quad \text{if } \lambda= k^2 + k(2m-2)\in \mathcal E_0\cup\mathcal E_1,
\quad \textrm{ for } p=0, \\
d_\lambda (p,\Gamma) & =
\begin{cases}
\dim V_{\pi_{k,p}}^\Gamma = n_\Gamma(\pi_{k,p})
&\text{if } \lambda=\lambda_{k,p}\in\mathcal E_p,\\
\dim V_{\pi_{k,p+1}}^\Gamma = n_\Gamma(\pi_{k,p+1}) &\text{if } \lambda=\lambda_{k,p+1}\in\mathcal E_{p+1},
\end{cases}
\quad \textrm{ for } 1\le p \le m-1.
\end{align*}
\end{prop}
When $\Gamma={1}$ this description appears in \cite{IkTa}, the case for general $\Gamma$ involves only minor modifications (see \cite[Thm.~1.1]{LMRrepequiv}).
The following proposition follows from \cite[Thm.~4.2]{IkTa} (see also \cite[Prop.~2.1]{Ik88}).
It will be a useful tool to prove one of the main results in the next section.
We include a proof for completeness.
\begin{prop}\label{prop2:charact}
Let $\Gamma\ba S^{2m-1}$ and $\Gamma'\ba S^{2m-1}$ be spherical space forms.
Then
\begin{enumerate}\item[(i)]
$\Gamma\ba S^{2m-1}$ and $\Gamma'\ba S^{2m-1}$ are $0$-isospectral if and only if, for every $k\in \N$,
\begin{equation*}
\dim V_{\pi_{k\varepsilon_1}}^{\Gamma} = \dim V_{\pi_{k\varepsilon_1}}^{\Gamma'}.
\end{equation*}
Generally, for any $0\le p\le m-1$, $\Gamma\ba S^{2m-1}$ and $\Gamma'\ba S^{2m-1}$ are $p$-isospectral if and only if, for every $k\in \N$,
\begin{equation*}
\dim V_{\pi_{k,p}}^{\Gamma} = \dim V_{\pi_{k,p}}^{\Gamma'}
\quad\text{and}\quad
\dim V_{\pi_{k,p+1}}^{\Gamma} = \dim V_{\pi_{k,p+1}}^{\Gamma'}.
\end{equation*}
\item[(ii)]
$\Gamma\ba S^{2m-1}$ and $\Gamma'\ba S^{2m-1}$ are $p$-isospectral for every $p$, if and only if
\begin{equation*}
\dim V_{\pi_{k,p}}^{\Gamma} = \dim V_{\pi_{k,p}}^{\Gamma'}
\end{equation*}
for every $1\leq p\leq m$ and every $k\in\N$.
\end{enumerate}
\end{prop}
\begin{proof}
By Proposition~\ref{prop2:p-spectrum}, if $\lambda\in\R$ is an eigenvalue of $\Delta_{p,\Gamma}$ then $\lambda\in \mathcal E_p\cup \mathcal E_{p+1}$ for some $k\in\N$ and its multiplicity is $\dim V_{\pi_{k,p}}^{\Gamma}$ or $\dim V_{\pi_{k,p+1}}^{\Gamma}$ depending on whether $\lambda$ is in $\mathcal E_p$ or in $\mathcal E_{p+1}$.
Since $\mathcal E_p\cap \mathcal E_{p+1}$ is empty, then (i) follows.
Note that $\pi_{k,0}$ and $\pi_{k,1}$ are the irreducible representations of $\SO(2m)$ with highest weight $k\varepsilon_1$ and $(k+1)\varepsilon_1$ respectively.
Item (ii) follows from (i) since $p$-isospectrality for $0\leq p\leq m-1$ implies $p$-isospectrality for every $p$.
\end{proof}
\section{Isospectrality conditions for lens spaces}
This section contains the first main result in this paper that gives a characterization of pairs of lens spaces that are either $0$-isospectral or $p$-isospectral for every $p$ (see Theorem~\ref{thm3:characterization}) in terms of geometric properties of their associated lattices.
Odd dimensional lens spaces can be described as follows: for each $q\in\N$ and $s_1,\dots,s_m\in\Z$ coprime to $q$, denote
\begin{equation}\label{eq3:L(q;s)}
L(q;s_1,\dots,s_m) = \langle\gamma\rangle \ba S^{2m-1}
\end{equation}
where
\begin{equation}\label{eq3:gamma}
\gamma=
\diag\left(
\left[\begin{smallmatrix}\cos(2\pi{s_1}/q)&-\sin(2\pi{s_1}/q) \\ \sin(2\pi{s_1}/q)&\cos(2\pi{s_1}/q)
\end{smallmatrix}\right]
,\dots,
\left[\begin{smallmatrix}\cos(2\pi{s_m}/q)&-\sin(2\pi{s_m}/q) \\ \sin(2\pi{s_m}/q)&\cos(2\pi{s_m}/q)
\end{smallmatrix}\right]
\right)
\end{equation}
The element $\gamma$ generates a cyclic group of order $q$ in $\SO(2m)$ that acts freely on $S^{2m-1}$.
Sometimes we shall abbreviate $L(q;\bs)$ in place of $L(q; s_1,\dots,s_m)$, where $\bs$ stands for the vector $\bs=(s_1,\dots,s_m)\in\Z^m$.
The following fact is well known (see \cite[Ch.~V]{Co} or \cite[\S12]{Mil2}).
\begin{prop}\label{prop3:lens-isom}
Let $L=L(q;\bs)$ and $L'=L(q;\bs')$ be lens spaces.
Then the following assertions are equivalent.
\begin{enumerate}
\item $L$ is isometric to $L'$.
\item $L$ is diffeomorphic to $L'$.
\item $L$ is homeomorphic to $L'$.
\item There exist $\sigma$ a permutation of $\{1,\dots,m\}$, $\epsilon_1,\dots,\epsilon_m\in\{\pm1\}$ and $t\in\Z$ coprime to $q$ such that
$$
s_{\sigma(j)}'\equiv t\epsilon_js_j\pmod{q}
$$
for all $1\leq j\leq m$.
\end{enumerate}
\end{prop}
The next definition will play a main role in the rest of this paper.
\begin{defi}
Let $q\in\N$ and $\bs=(s_1, \ldots, s_m) \in\Z^m$ such that each entry $s_j$ is coprime to $q$.
We associate to the lens space $L(q;\bs)$ the \emph{congruence lattice }
\begin{equation}\label{eq3:Lambda(q;s)}
\mathcal L(q;s_1,\dots,s_m) =\{(a_1,\dots,a_m)\in\Z^m: a_1s_1+\dots+a_ms_m\equiv 0\pmod q\}.
\end{equation}
\end{defi}
For $\mu=(a_1,\dots,a_m)\in\Z^m$, we set $\norma{\mu}=\sum_{j=1}^m |a_j|$.
\begin{prop}\label{prop3:isometrias}
Let $L(q;\bs)$, $L(q;\bs')$ be lens spaces with $\mathcal L(q,\bs)$ and $\mathcal L(q,\bs')$ the associated lattices.
Then, $L(q;\bs)$ and $L(q;\bs')$ are isometric if and only if $\mathcal L(q;\bs)$ and $\mathcal L(q;\bs')$ are $\norma{\cdot}$-isometric.
\end{prop}
\begin{proof}
By Proposition~\ref{prop3:lens-isom}, $L$ and $L'$ are isometric if and only if there exist $t$ coprime to $q$ and $\varphi$, a composition of permutations and changes of signs, such that $\varphi(t\bs)=\varphi(t s_1,\dots, t s_m)= (s_1', \ldots, s'_m)=\bs'$.
Hence $\mathcal L(q,\bs')=\mathcal L(q,\varphi (\bs))=\varphi (\mathcal L(q,\bs))$ with $\varphi$ a $\norma{\cdot}$-isometry.
In order to prove the converse assertion, we first show that every $\norma\cdot$-linear isometry of $\R^n$ is a composition of permutations and changes of signs.
If $T$ is a $\norma{\cdot}$-linear isometry of $\R^n$, then for each $1\le k \le n$, $T(\varepsilon_k)= \sum_{j=1}^n c_{k,j} \varepsilon_j$ with $\sum |c_{k,j}| =1$. We claim that $c_{k,j}\ne 0$ for at most one value of $j$. Otherwise, there are $h,k,\ell$ such that $c_{k,\ell}c_{h,\ell}\ne 0$. Hence $|c_{k,\ell} +\delta c_{h,\ell}|< |c_{k,\ell}| + |c_{h,\ell}|$, for $\delta =1$ or $\delta =-1$.
Thus, for this choice of $\delta$ we have $2 = \norma{T(\varepsilon_k) +\delta T(\varepsilon_h)} = \sum_{j=1}^n |c_{k,j} + \delta c_{k,j}|<
\sum_{j=1}^n |c_{k,j}| + |c_{k,j}|=2$, a contradiction.
Now, suppose conversely that $\varphi$ is a $\norma{\cdot}$-isometry between $\mathcal L(q,s)$ and $\mathcal L(q,s')$.
The previous paragraph ensures that $\varphi$ is given by
$$
\varphi(a_1,\dots,a_m)=(\epsilon_{\sigma(1)} a_{\sigma(1)},\dots, \epsilon_{\sigma(m)} a_{\sigma(m)})
$$
with $\sigma$ a permutation of $\{1,\dots,m\}$ and $\epsilon_j=\pm1$ for all $j$, and satisfies $\mathcal L(q;\bs')= \varphi(\mathcal L(q;\bs))$, thus $\mathcal L(q;\bs')= \mathcal L(q;\varphi(\bs))$.
For each $2\leq j\leq m$, the vector $$(-s_j',0,\dots,0,s_1',0,\dots,0)$$ lies in $\mathcal L(q;\bs')$, thus $-s_j' \epsilon_{\sigma(1)} s_{\sigma(1)} + s_1' \epsilon_{\sigma(j)} s_{\sigma(j)}\equiv 0 \pmod q$ since it is also in $\mathcal L(q;\varphi(\bs))$.
Then, if $t\in\Z$ is such that $ t\epsilon_{\sigma(1)} s_{\sigma(1)}\equiv s_1'\pmod q$, one has that $s_j' \equiv t \epsilon_{\sigma(j)} s_{\sigma(j)} \pmod q$ for every $j$.
Hence $L$ and $L'$ are isometric to each other.
This completes the proof.
\end{proof}
The goal of this section is to write the $p$-spectrum of a lens space in terms of the $\norma{\cdot}$-length spectrum of the associated congruence lattice.
To do this we will express the numbers $\dim V_{\pi_{k,p}}^\Gamma$ in terms of weight multiplicities of representations of $G=\SO(2m)$.
We will identify the weight lattice $P(G) = \bigoplus_{j=1}^m \Z \varepsilon_j$ with $\Z^m$ via the correspondence $\sum_{j=1}^m a_j\varepsilon_j$ $\mapsto$ $(a_1,\dots,a_m)$.
\begin{lemma}\label{lem3:L_Gamma}
Let $\Gamma =\langle \gamma \rangle$ where $\gamma$ is as in \eqref{eq3:gamma}. Let $L = L(q;s_1,\dots,s_m)$ be the corresponding lens space and let $\mathcal L=\mathcal L(q;s_1,\dots,s_m)$ be the associated lattice.
If $(\pi,V_\pi)$ is a finite dimensional representation of $\SO(2m)$, then
\begin{equation}\label{eq:Gammainvar}
\dim V_\pi^\Gamma=
\sum_{\mu\in \mathcal L}\, m_\pi(\mu),
\end{equation}
where $m_\pi(\mu)$ denotes the multiplicity of the weight $\mu$ in $\pi$.
\end{lemma}
\begin{proof}
One has that $V_\pi=\oplus_{\mu\in P(G)} V_\pi(\mu)$, where $V_\pi(\mu)$ is the $\mu$-weight space, i.e.\ the space of vectors $v$ such that $\pi(h)v=h^\mu v$ for every $h\in T$.
Here, $h^\mu=e^{\mu(X_h)}$ where $X_h$ is any element in $\mathfrak h_0$ satisfying $\exp(X_h)=h$.
Thus, $V_\pi^\Gamma= \oplus_{\mu\in P(G)} V_\pi(\mu)^\Gamma$.
Now, $v\in V_\pi(\mu)$, $v\ne 0$, is $\Gamma$-invariant if and only if $\gamma^\mu=1$, hence
$
\dim V_\pi^\Gamma =\sum_{\mu:\gamma^\mu=1} \dim V_{\pi}(\mu)=\sum_{\mu:\gamma^\mu=1} m_\pi(\mu).
$
We let
$$
H_\gamma=\diag\left( \left(\begin{smallmatrix}0&-2\pi s_1/q\\ 2\pi s_1/q&0\end{smallmatrix}\right) ,\dots, \left(\begin{smallmatrix}0&-2\pi s_m /q\\ 2\pi s_m/q&0\end{smallmatrix}\right) \right),
$$
thus $\exp (H_\gamma)=\gamma$.
If $\mu=\sum_{j=1}^{m} a_j\varepsilon_j\in P(\SO(2m))$ then
$$
\gamma^\mu = e^{\mu(H_\gamma)} = e^{-2\pi i\left(\frac{a_1s_1+\dots+a_ms_m}{q}\right)}=1
$$
if and only if $a_1s_1+\dots+a_ms_m\equiv 0\pmod q$, that is, $\mu \in \mathcal L$.
\end{proof}
Let $\mathcal L$ be an arbitrary sublattice of $\Z^m$. For $\mu\in\Z^m$ we set $Z(\mu)=\#\{j:1\leq j\leq m,\, a_j=0\}$.
We denote, for any $0\leq \zz \leq m$ and any $k\in\N_0:=\N\cup\{0\}$,
\begin{align}
N_\mathcal L(k) &= \#\left\{\mu\in \mathcal L : \norma{\mu}=k\right\},\label{eq:Nk}\\
N_{\mathcal L}(k,\zz ) &= \#\left\{\mu\in \mathcal L : \norma{\mu}=k,\; Z(\mu)=\zz \right\}.\label{eq:Nkz}
\end{align}
\begin{defi}\label{def3:isosp-latt}
Let $\mathcal L$ and $\mathcal L'$ be sublattices of $\Z^m$.
\begin{enumerate}
\item[(i)] $\mathcal L$ and $\mathcal L'$ are said to be \emph{$\norma{\cdot}$-isospectral} if $N_{\mathcal L}(k) = N_{\mathcal L'}(k)$ for every $k\in\N$.
\item[(ii)]
$\mathcal L$ and $\mathcal L'$ are said to be \emph{$\norma{\cdot}^*$-isospectral} if $N_{\mathcal L} (k,\zz ) = N_{\mathcal L'}(k,\zz )$ for every $k\in\N$ and every $0\leq \zz \leq m$.
\end{enumerate}
\end{defi}
We will need two useful lemmas on weight multiplicities.
The first one follows from well known facts, but we could not find it stated in the form below, so we include a short proof here.
Recall that $\Lambda_p$ is given by \eqref{eq2:Lambda_p} and $ \pi_{0,p} = \pi_{\Lambda_p}$ is the exterior representation of $\SO(2m)$ on $\bigwedge^p \C^{2m}$ for $0 \le p \le m$.
\begin{lemma}\label{lem3:mult-k=1-p=1}
Let $k\in\N$ and $0\leq p\leq m$.
If $\mu=\sum_{j=1}^m a_j\varepsilon_m\in\Z^m$ we have
\begin{align}
m_{\pi_{k\varepsilon_1}}(\mu) =&
\begin{cases}
\binom{r+m-2}{m-2} & \text{ if }\, \norma{\mu}=k-2r \;\text{ with } r\in \N_0,\\
0 & \text{ otherwise,}
\end{cases}
\label{eq3:mult-pi_k,1} \\
m_{\pi_{\Lambda_{p}}}(\mu) =&
\begin{cases}
\binom{m-p+2r}{r} & \text{if }\,\norma{\mu}=p-2r \;\text{ with } r\in \N_0, \text{ and } |a_j|\leq1\;\forall\,j,\\
0&\text{otherwise.}
\end{cases}
\label{eq3:mult-pi_1,p}
\end{align}
\end{lemma}
\begin{proof}
It is well known that the representation $\pi_{k\varepsilon_1}$ can be realized in the space of harmonic homogeneous polynomials $\mathcal H_k$ of degree $k$ in $m$ variables.
Moreover, $\mathcal P_k\simeq\mathcal H_k\oplus\mathcal P_{k-2}$ where $\mathcal P_k$ denotes the space of homogeneous polynomials of degree $k$, thus
\begin{equation}\label{eq3:H_k=P_k+r^2P_k-2}
m_{\pi_{k\varepsilon_1}}(\mu) = m_{{\mathcal P}_k}(\mu) - m_{{\mathcal P}_{k-2}}(\mu).
\end{equation}
In order to find the weights of $\mathcal P_k$, we set $f_j(x)=x_{2j-1}+ix_{2j},\, f_{j+m}=x_{2j-1}-ix_{2j}\in\mathcal P_1$ for each $1\leq j\leq m$.
It can be easily seen that the polynomials $f_1^{l_1}\dots f_{2m}^{l_{2m}}$ with $\sum_{j=1}^{2m} l_j=k$ form a basis of $\mathcal P_k$ given by weight vectors.
Indeed, $h\in T$ acts on $f_1^{l_1}\dots f_{2m}^{l_{2m}}$ by multiplication by $h^\mu$ where $\mu=\sum_{j=1}^m (l_j-l_{j+m})\varepsilon_j$.
It follows that $\mu=\sum_{j=1}^m a_j\varepsilon_j\in\Z^m$ is a weight of $\mathcal P_k$ if and only if there are $l_1,\dots,l_{2m}\in\N_0$ such that $a_j=l_j-l_{j+m}$ and $\sum_{j=1}^{2m} l_j=k$.
Furthermore, one checks that the last condition is equivalent to $k-\norma{\mu}=2r$ with $r\in \N_0$. Hence, $m_{\mathcal P_k}(\mu)$ equals the number of different ways one can write $r$ as an ordered sum of $m$ different nonnegative integers, which equals $\binom{r+m-1}{m-1}$.
This implies that
\begin{equation*}\label{eq3:mult-P_k}
m_{\mathcal P_k}(\mu) =
\begin{cases}
\binom{r+m-1}{m-1} &\text{ if }r=\frac12(k-\norma{\mu})\in\N_0,\\
0 & \text{ otherwise.}
\end{cases}
\end{equation*}
This formula and \eqref{eq3:H_k=P_k+r^2P_k-2} imply \eqref{eq3:mult-pi_k,1}.
We now prove the second assertion.
The representation $\pi_{\Lambda_p}$ can be realized as the complexified $p$-exterior representation $\bigwedge^p(\C^{2m})$ with the canonical action of $\SO(2m)$.
Let $\{e_1,\dots,e_{2m}\}$ denote the canonical basis of $\C^{2m}$.
For $1\leq j\leq m$, we set $v_j=e_{2j-1}-i e_{2j}$ and $v_{j+m}=e_{2j-1}+i e_{2j}$.
Hence $\{v_1,\dots,v_{2m}\}$ is also a basis of $\C^{2m}$ and
\begin{equation}\label{eq5:base_p-forms}
\left\{v_{i_1}\wedge\dots\wedge v_{i_p}: 1\leq i_1<i_2<\dots<i_p\leq 2m\right\}
\end{equation}
is a basis of $\bigwedge^p(\C^{2m})$.
For $I=\{1\leq i_1<i_2<\dots<i_p\leq 2m\}$ we write $\omega_I=v_{i_1}\wedge\dots\wedge v_{i_p}$.
One can check that $h\in T$ acts on $\omega_I$ by multiplication by $h^\mu$ where
$\mu=\sum_{j=1}^m a_j\varepsilon_j$ is given by
$$
a_j=\begin{cases}
1&\quad\text{if $j\in I$ and $j+m\notin I$,}\\
-1&\quad\text{if $j\notin I$ and $j+m\in I$,}\\
0&\quad\text{if both $j,j+m\in I$, or $j,j+m\notin I$.}
\end{cases}
$$
Thus, an arbitrary element $\mu=\sum_j a_j\varepsilon_j\in\Z^m$ is a weight of $\bigwedge^p(\C^{2m})$ if and only if $|a_j|\leq 1$ for all $j$ and $p-\norma{\mu}\in2\N_0$.
Let $\mu=\sum_{j=1}^m a_j\varepsilon_j\in\Z^m$ be such that $|a_j|\leq 1$ for all $j$ and $r=\frac12(p-\norma{\mu})\in\N_0$.
Let $I_\mu=\{i:1\leq i\leq m, \, a_i=1\}\cup\{i:m+1\leq i\leq 2m,\, a_{i-m}=-1\}$.
Thus $I_\mu$ has $p-2r$ elements.
It is a simple matter to check that $\omega_I$ is a weight vector with weight $\mu$ if and only if $I$ has $p$ elements, $I_\mu\subset I$ and $I$ has the property that $j\in I\smallsetminus I_\mu \iff j+m\in I\smallsetminus I_\mu$ for $1\leq j\leq m$.
One can check that there are $\binom{m-p+2r}{r}$ choices for $I$, hence the claim follows.
\end{proof}
The second lemma is crucial in the proof of Theorem~\ref{thm3:characterization}~(ii).
We recall that $\pi_{k,p}$ is the irreducible representation of $\SO(2m)$ with highest weight $k\varepsilon_1+\Lambda_p$ if $p<m$ and, when $p=m$, the sum of the irreducible representations with highest weights $k\varepsilon_1+\Lambda_m$ and $k\varepsilon_1+\overline\Lambda_m$.
\begin{lemma}\label{lem3:mult-k-p}
Let $\mu,\mu'\in P(\SO(2m))\simeq \Z^m$.
If $\norma{\mu}=\norma{\mu'}$ and $Z(\mu)=Z(\mu')$ then $m_{\pi_{k,p}}(\mu) = m_{\pi_{k,p}}(\mu')$ for every $k\in\N$ and every $1\leq p\leq m$.
\end{lemma}
\begin{proof}
We say that a finite dimensional representation $\sigma$ of $\SO(2m)$ \emph{satisfies condition $(\star)$} if $m_\sigma(\mu)=m_\sigma(\mu')$ for every $\mu$ and $\mu'$ such that $\norma{\mu}=\norma{\mu'}$ and $Z(\mu)=Z(\mu')$.
We see, by Lemma~\ref{lem3:mult-k=1-p=1}, that $\pi_{k\varepsilon_1}$ and $\pi_{\Lambda_p}$ satisfy $(\star)$ for every $k$ and $p$.
Next we show that $\sigma := \pi_{k\varepsilon_1} \otimes \pi_{\Lambda_p}$ also satisfies $(\star)$.
Let $\mu=\sum_{i=1}^m a_i \varepsilon_i$ and $\mu'=\sum_{i=1}^m a_i' \varepsilon_i$ in $\Z^m$ be such that $\norma{\mu}=\norma{\mu'}$ and $Z(\mu)=Z(\mu')$.
We fix a bijection $\varrho:[1,m]\to [1,m]$ so that $a_i'\neq0$ if and only if $a_{\varrho(i)}\neq0$.
We have that
\begin{equation}\label{eq3:m_sigma}
m_\sigma(\mu)
=\sum_{\eta\in \Z^m} m_{\pi_{\Lambda_p}}(\eta) \;m_{\pi_{k\varepsilon_1}}(\mu-\eta)
\end{equation}
and a similar expression for $m_\sigma(\mu')$ (see for instance \cite[Ex.~V.14]{Knapp}).
Both sums are already over the weights of $\pi_{\Lambda_p}$, that is, over the weights $\eta=\sum_{i=1}^m b_i \varepsilon_i$ such that $|b_i|\leq 1$ for all $i$ and $\norma{\eta}=p-2r$ for some $r\in\N$, by Lemma~\ref{lem3:mult-k=1-p=1}.
To each such $\eta$ we associate $\eta'=\sum_{i=1}^m b_i' \varepsilon_i$ defined by $b_i'=b_{\varrho(i)}$ for every $i$ such that $a_i'=0$ and $b_i'= \textrm{sg}(a_{\varrho(i)})\, \textrm{sg}({a_i'}) \, b_{\varrho(i)}$ for every $i$ such that $a_i'\neq 0$.
One can check that $\norma{\eta}=\norma{\eta'}$, $Z(\eta) =Z(\eta')$ and furthermore $\norma{\mu-\eta}=\norma{\mu'-\eta'}$, thus $m_{\pi_{\Lambda_p}}(\eta) \, m_{\pi_{k\varepsilon_1}}(\mu-\eta) = m_{\pi_{\Lambda_p}}(\eta') \, m_{\pi_{k\varepsilon_1}}(\mu'-\eta')$.
By \eqref{eq3:m_sigma} we have that $m_\sigma(\mu) = m_\sigma(\mu')$ as asserted.
By Steinberg's formula (see for instance \cite[Ex.~17-Ch.~ IX]{Knapp}), the representation $\sigma$ decomposes as
\begin{equation}\label{eq3:descomp-sigma}
\chi_\sigma
= \sum_\mu m_{\pi_{\Lambda_p}}(\mu) \, \mathrm{sgn}(\mu+k\varepsilon_1+\rho) \, \chi_{(\mu+k\varepsilon_1+\rho)^\vee-\rho},
\end{equation}
where $\chi_\sigma$ denotes the character of the representation $\sigma$, $\rho=\sum_{j=1}^m (m-j)\varepsilon_j$, half the sum of positive roots, $\eta^\vee$ denotes the only dominant weight in the same Weyl orbit as $\eta$, and
$$
\mathrm{sgn}(\mu) =
\begin{cases}
0 &\text{ if $\omega \mu=\mu$ for some nontrivial $\omega\in W$,}\\
\mathrm{sgn}(\omega) &\text{ otherwise, where $\omega\mu$ is dominant}.
\end{cases}
$$
Note that the sum in \eqref{eq3:descomp-sigma} is over the weights of $\pi_{\Lambda_p}$, described in \eqref{eq3:mult-pi_1,p}.
Moreover, the character of the representation $\pi_{k,p}$ appears in the sum on the right-hand side in \eqref{eq3:descomp-sigma} and this is the only time it does, hence $\pi_{k,p}$ appears exactly once in the decomposition of $\sigma$.
Now the proof of the lemma is completed by an inductive argument in $k$ and $p$ by checking that any other irreducible representation $\pi_{k',p'}$ that appears in \eqref{eq3:descomp-sigma} satisfies $k'<k$, or else $k'=k$ and $p'<p$, thus $\pi_{k',p'}$ satisfies $(\star)$ by the strong inductive hypothesis.
Finally, since $\sigma$ also satisfies $(\star)$ then $\pi_{k,p}$ also does.
\end{proof}
The next theorem gives an explicit formula for $\dim V_{\pi_{k,p}}^\Gamma$ in terms of weight multiplicities $m_{\pi_{k,p}}(\mu)$ and of the numbers $N_{\mathcal L} (k,\zz )$, when $L=\Gamma\ba S^{2m-1}$ is a lens space with congruence lattice $\mathcal L$.
\begin{thm}\label{thm3:dim V_k,p^Gamma}
Let $L=\Gamma\ba S^{2m-1}$ be a lens space with associated lattice $\mathcal L$ and let $k\in\N$ and $0\leq p\leq m$.
Then
\begin{equation}\label{eq3:dim V_k,p^Gamma}
\dim V_{\pi_{k,p}}^{\Gamma}=
\sum_{r=0}^{\lfloor(k+p)/2\rfloor} \;\sum_{\zz =0}^m \; m_{\pi_{k,p}}(\mu_{r,\zz })\; N_{\mathcal L}(k+p-2r,\zz ),
\end{equation}
where $\mu_{r,\zz }$ is any weight such that $Z(\mu_{r,\zz })=\zz $ and $\norma{\mu_{r,\zz }} = k+p-2r$.
In the particular case when $p=0$ we have that
\begin{equation}\label{eq3:dim V_k,1^Gamma}
\dim V_{\pi_{k\varepsilon_1}}^{\Gamma}=
\sum_{r=0}^{\lfloor k/2\rfloor}\binom{r+m-2}{m-2} N_{\mathcal L}(k-2r).
\end{equation}
\end{thm}
\begin{proof}
By Lemma~\ref{lem3:L_Gamma} we have that
\begin{equation*}
\dim V_{\pi_{k,p}}^{\Gamma} = \sum_{\mu\in \mathcal L} m_{\pi_{k,p}}(\mu).
\end{equation*}
The sum is finite since it is a sum over the weights $\mu$ of $\pi_{k,p}$.
These weights are of the form $k\varepsilon_1+\Lambda_{p}-\nu$ with $\nu$ a sum of positive roots, if $p<m$, and of the form $k\varepsilon_1+\Lambda_{m}-\nu$ or $k\varepsilon_1+ \overline \Lambda_{m}-\nu$, if $p=m$.
Since $\norma{\alpha}=\norma{\varepsilon_i\pm\varepsilon_j}=2$ for every positive root $\alpha$ of $\,\mathfrak{so}(2m,\C)$ (see Section~\ref{sec:prelim}), then $m_{\pi_{k,p}}(\mu)=0$ unless $\norma{k\varepsilon_1+\Lambda_{p}}-\norma{\mu}=k+p-\norma{\mu}\in2\N_0$.
Hence
\begin{align*}
\dim V_{\pi_{k,p}}^{\Gamma}
&=
\sum_{r=0}^{\lfloor (k+p)/2\rfloor} \;\sum_{\zz =0}^m \;\sum_{\mu\in \mathcal L:\, Z(\mu)=\zz , \atop \norma{\mu}=k+p-2r} m_{\pi_{k,p}}(\mu).
\end{align*}
Since, by Lemma~\ref{lem3:mult-k-p}, the value of $m_{\pi_{k,p}}(\mu)$ depends only on $\norma{\mu}$ and $Z(\mu)$, the last sum equals the number of weights $\mu$ such that $\norma{\mu}=k+p-2r$ and $Z(\mu)=\zz $, times the multiplicity of any such weight.
This proves \eqref{eq3:dim V_k,p^Gamma}.
In the case when $p=0$, the multiplicity $m_{\pi_{k\varepsilon_1}}(\mu)$ is as given in \eqref{eq3:mult-pi_k,1}.
Thus
\begin{align*}
\dim V_{\pi_{k\varepsilon_1}}^{\Gamma}
&= \sum_{r=0}^{\lfloor k/2\rfloor} \sum_{\mu\in \mathcal L:\atop \norma{\mu}=k-2r} \binom{r+m-2}{m-2}
= \sum_{r=0}^{\lfloor k/2\rfloor}\binom{r+m-2}{m-2} N_{\mathcal L}(k-2r). \notag
\end{align*}
This completes the proof.
\end{proof}
We now state the first main result in this paper.
\begin{thm}\label{thm3:characterization}
Let $L=\Gamma\ba S^{2m-1}$ and $L'=\Gamma'\ba S^{2m-1}$ be lens spaces with associated congruence lattices $\mathcal L$ and $\mathcal L'$ respectively.
Then
\begin{enumerate}
\item[(i)] $L$ and $L'$ are $0$-isospectral if and only if $\mathcal L$ and $\mathcal L'$ are $\norma{\cdot}$-isospectral.
\item[(ii)] $L$ and $L'$ are $p$-isospectral for all $p$ if and only if $\mathcal L$ and $\mathcal L'$ are $\norma{\cdot}^*$-isospectral.
\end{enumerate}
\end{thm}
\begin{proof}
Proposition~\ref{prop2:charact}~(i) (resp.\ (ii)) says that $L$ and $L'$ are $0$-isospectral (resp.\ $p$-isospectral for all $p$) if and only if, for every $k\in\N$, $\dim V_{\pi_{k\varepsilon_1}}^{\Gamma} = \dim V_{\pi_{k\varepsilon_1}}^{\Gamma'}$ (resp. $\dim V_{\pi_{k,p}}^{\Gamma} = \dim V_{\pi_{k,p}}^{\Gamma'}$ for every $k\in\N$ and every $1\leq p\leq m$).
Hence, in the converse direction, (i) and (ii) follow immediately from \eqref{eq3:dim V_k,1^Gamma} and \eqref{eq3:dim V_k,p^Gamma} respectively.
We now assume that $L$ and $L'$ are $0$-isospectral.
We shall prove by induction that
\begin{equation}\label{eq3:N(k)=N'(k)}
N_{\mathcal L}(k) = N_{\mathcal L'}(k)
\end{equation}
for every $k \in\N$.
The case $k=0$ is clear, since both sides are equal to one.
Suppose that \eqref{eq3:N(k)=N'(k)} holds for every $k<k_0$.
By \eqref{eq3:dim V_k,1^Gamma} we have that
$$
\sum_{r\geq0} \binom{r+m-2}{m-2} N_{\mathcal L}(k_0-2r)
=
\sum_{r\geq0} \binom{r+m-2}{m-2} N_{\mathcal L'}(k_0-2r).
$$
All the terms with $r>0$ on both sides are equal by assumption, hence this equality implies that also $N_{\mathcal L}(k_0)=N_{\mathcal L'}(k_0)$. This proves (i).
We next prove (ii).
Assume that $L$ and $L'$ are $p$-isospectral for all $p$.
We shall prove that
\begin{equation}\label{eq3:N(k,z)=N'(k,z)}
N_{\mathcal L}(k,\zz ) = N_{\mathcal L'}(k,\zz )
\qquad\forall\zz:
0\leq\zz \leq m,
\end{equation}
for every $k\in\N$.
We use an inductive argument on $k$.
The case $k=0$ is again clear.
We suppose that \eqref{eq3:N(k,z)=N'(k,z)} holds for every $k<k_0$.
For each $1\leq p\leq m$, if we let $k=k_0-p$, then, by \eqref{eq3:dim V_k,p^Gamma}, since $L$ and $L'$ are $p$-isospectral, we have that
$$
\sum_{r\geq0} \sum_{\zz =0}^m \; m_{\pi_{k,p}}(\mu_{r,\zz })\; N_{\mathcal L}(k_0-2r,\zz )
=
\sum_{r\geq0} \sum_{\zz =0}^m \; m_{\pi_{k,p}}(\mu_{r,\zz })\; N_{\mathcal L'}(k_0-2r,\zz ),
$$
where $\mu_{r,\zz }$ is any weight satisfying $\norma{\mu_{r,\zz }}=k_0-2r$ and $Z(\mu_{r,\zz })=\zz $.
By assumption, all terms in both sides with $r>0$ coincide. Thus
\begin{equation*}
\sum_{\zz =0}^{m-1} \; m_{\pi_{k,p}}(\mu_{0,\zz })\; N_{\mathcal L}(k_0,\zz )
=
\sum_{\zz =0}^{m-1} \; m_{\pi_{k,p}}(\mu_{0,\zz })\; N_{\mathcal L'}(k_0,\zz ).
\end{equation*}
Note that the terms $\zz =m$ in both sides have been deleted since they are both equal to zero.
To prove our claim it suffices to show that the $m\times m$-matrix $(m_{\pi_{k,p}}(\mu_{0,\zz }) )_{p,\zz}$ with $p=1,\dots,m$ and $\zz =0,\dots, m-1$ is invertible. We claim that this matrix has $1$'s on the anti-diagonal and it is `upper-triangular' with respect to the anti-diagonal, hence it has determinant $\pm 1$.
Now, the element $\mu_{0,\zz }$ is any weight in $\Z^m$ such that $\norma{\mu_{0,\zz }}=k_0$ and $Z(\mu_{r,\zz })=\zz $, thus we may pick
\begin{equation*}
\mu_{0,\zz } = (k_0-m+\zz +1)\varepsilon_1 + \varepsilon_2 + \dots + \varepsilon_{m-\zz }.
\end{equation*}
If $m-\zz=p$ (i.e.\ $(p,\zz)$ is on the antidiagonal), then $\mu_{0,\zz}=k\varepsilon_1+\Lambda_p$.
If $p<m$, then $\pi_{k,p}$ has highest weight $k\varepsilon_1+\Lambda_p$, hence $m_{\pi_{k,p}}(\mu_{0,\zz })=1$.
On the other hand, if $m-\zz<p$ then $\mu_{0,\zz } $ cannot be a weight since
$k\varepsilon_1 +\Lambda_p - \mu_{0,\zz }$ is not a sum of positive roots given that the coefficient of $\varepsilon_1$ equals $m-\zz-p<0$.
The case $p=m$ is very similar and its verification is left to the reader.
\end{proof}
\begin{rem}\label{rem3:non-strongly}
Two spherical space forms $\Gamma\ba S^{n}$ and $\Gamma'\ba S^{n}$ are said to be \emph{strongly isospectral} if for any strongly elliptic natural operator $D$ acting on sections of a natural bundle $E$ over $S^n$, the associated operators $D_\Gamma$ and $D_{\Gamma'}$ acting on sections of the bundles $\Gamma\ba E$ and $\Gamma'\ba E$ have the same spectrum.
Isospectral manifolds constructed by Sunada's method are strongly isospectral.
It is a well known fact that non-isometric lens spaces cannot be strongly isospectral (see Proposition~\ref{prop7:lens-non-strongly}).
\end{rem}
\begin{rem}\label{rem3:ej-Ikeda}
Ikeda in \cite{Ik80} gave many pairs of non-isometric lens spaces that are $0$-isospectral.
The simplest such pair is $L(11;1,2,3)$ and $L(11;1,2,4)$ in dimension $5$.
In light of Theorem~\ref{thm3:characterization}~(i), the associated congruence $3$-dimensional lattices $\mathcal L=\mathcal L(11;1,2,3)$ and $\mathcal L'=\mathcal L(11;1,2,4)$ must be $\norma{\cdot}$-isospectral.
However, it is a simple matter to check that $\mathcal L$ and $\mathcal L'$ are not $\norma{\cdot}^*$-isospectral.
In fact, it is easy to see that $\pm(2,-1,0)$ and $\pm(1,1-1)$ are the only vectors in $\mathcal L$ with $1$-norm equal to $3$, while $\pm(2,-1,0)$ and $\pm(0,2,-1)$ are those with $1$-norm equal to $3$ lying in $\mathcal L'$.
This implies that $N_{\mathcal L}(3,0)=2 \neq N_{\mathcal L'}(3,0)=0$ and $N_{\mathcal L}(3,1)=2 \neq N_{\mathcal L'}(3,1)=4$, proving the assertion.
As we shall see in Section~\ref{sec:families}, there exist infinitely many pairs of congruence lattices that are $\norma{\cdot}^*$-isospectral in dimension $m=3$.
Such examples do not exist for $m=2$, since Ikeda and Yamamoto showed that two $0$-isospectral $3$-dimensional lens spaces are isometric (\cite{IY}, \cite{Y}).
Also, in the relevant paper \cite{Ik88}, Ikeda produced for each given $p_0$ pairs of lens spaces that are $p$-isospectral for every $0\le p \le p_0$ but are not $p_0+1$ isospectral.
\end{rem}
\section{Finiteness conditions}\label{sec:finiteness}
In this section we give a necessary and sufficient condition for two $m$-dimensional $q$-congruence lattices to be $\norma{\cdot}^*$-isospectral, by comparison, for the two lattices, of a finite set of numbers of cardinality at most $\binom{m+1}{2}q$.
Thus, in light of the connection with lens spaces in Theorem~\ref{thm3:characterization}~(ii), one can check with finitely many computations whether two lens spaces are $p$-isospectral for all $p$.
In Section~\ref{sec:examples} we will show many examples of $\norma{\cdot}^*$-isospectral lattices found with a computer.
We first need to introduce some notions and notations.
For $q\in\N$ we set
$$
C(q)=\left\{\textstyle\sum_j a_j\varepsilon_j\in\Z^m : |a_j|<q\;,\,\forall\,j\right\}.
$$
An element in $C(q)$ will be called \emph{$q$-reduced}.
We define an equivalence relation in $\Z^m$ as follows: if $\mu=\sum_j a_j\varepsilon_j$, $\mu'=\sum_j a_j'\varepsilon_j\in \Z^m$ then $\mu\sim \mu'$ if and only if $\mu-\mu'\in (q\Z)^m$ and $a_j a_j'\geq 0$ for every $j$ such that $ a_j\not \equiv 0 \pmod q$.
This relation induces an equivalence relation in $\Z^m$ and also in any $q$-congruence lattice $\mathcal L$ since $q\Z^m\subset\mathcal L$.
Furthermore, $C(q)$ and $C(q)\cap\mathcal L$ give a complete set of representatives of $\sim$ on $\Z^m$ and $\mathcal L$ respectively.
We now consider the number of $q$-reduced elements $\mathcal L$ with a fixed norm and a fixed number of zeros.
\begin{defi}
Let $\mathcal L$ be a $q$-congruence lattice as in \eqref{eq3:Lambda(q;s)}.
For any $k\in\N_0$ and $0\leq \zz \leq m$, we set
\begin{equation*}
N_{\mathcal L}^{\mathrm{red}}(k,\zz )
= \#\{\mu\in C(q)\cap \mathcal L: \norma{\mu} = k,\; Z(\mu)=\zz \}.
\end{equation*}
\end{defi}
We note that any element $\mu\in \Z^m$ lying in the regular tetrahedron $\norma{\mu}<q$ is $q$-reduced, thus $N_{\mathcal L}(k,\zz ) = N_{\mathcal L}^{\mathrm{red}}(k,\zz )$ for every $k<q$. Also, for each of the $m-\zz$ nonzero coordinates $a_i$ of a $q$-reduced element one has $|a_i|\le q-1$, thus
$N_{\mathcal L}^{\mathrm{red}}(k,\zz )=0$ for every $k>(m-\zz)(q-1)$. Hence, the total number of possibly nonzero numbers $N_{\mathcal L}^{\mathrm{red}}(k,\zz )$ is at most $\binom{m+1}{2}q$.
We have mentioned above that every element in a $q$-congruence lattice $\mathcal L$ is equivalent to one and only one $q$-reduced element in $\mathcal L$.
As one should expect, the finite set of $N_{\mathcal L}^{\mathrm{red}}(k,\zz)$'s determines the numbers $N_{\mathcal L}(k,\zz)$, for every $k, \zz$. That is, if $N_{\mathcal L}^{\mathrm{red}}(k,\zz)=N_{\mathcal L'}^{\mathrm{red}}(k,\zz)$ for every $k$ and $\zz$, then $\mathcal L$ and $\mathcal L'$ are $\norma{\cdot}^*$-isospectral.
The next theorem shows this fact by giving an explicit formula for $N_{\mathcal L}(k,\zz)$ in terms of the $N_{\mathcal L}^{\mathrm{red}}(k,\zz)$. This formula will also allow us to prove that the numbers $N_{\mathcal L}(k,\zz)$ determine the numbers $N_{\mathcal L}^{\mathrm{red}}(k,\zz)$.
\begin{thm}\label{thm4:finitud}
Let $\mathcal L$ and $\mathcal L'$ be two $q$-congruence lattices as in \eqref{eq3:Lambda(q;s)}.
\begin{enumerate}
\item[(i)]
If $k=\alpha q+r\in \N$ with $0\leq r < q$, then
\begin{equation}\label{eq4:N^*-N^red}
N_{\mathcal L}(k,\zz )= \sum_{s=0}^{m-\zz } 2^s\binom{\zz +s}{s} \sum_{t=s}^{\alpha}
\binom{t-s+m-\zz -1}{m-\zz -1} \,
N_{\mathcal L}^{\mathrm{red}}(k-tq,\zz +s).
\end{equation}
\item[(ii)]
$N_{\mathcal L}(k,\zz ) = N_{\mathcal L'}(k,\zz )$ for every $k$ and $\zz $ if and only if $N_{\mathcal L}^{\mathrm{red}}(k,\zz ) = N_{\mathcal L'}^{\mathrm{red}}(k,\zz )$ for every $k$ and $\zz $.
\end{enumerate}
\end{thm}
\begin{proof}
We begin by proving (i).
We fix $0\leq r<q$ and we write $k=\alpha q+r$ for some $\alpha \in\N_0$.
When $\alpha=0$ \eqref{eq4:N^*-N^red} is reduced to the identity $N_{\mathcal L}(r,\zz ) = N_{\mathcal L}^{\mathrm{red}}(r,\zz )$, which is valid.
For convenience, in the rest of this proof, we say that $\mu$ is \emph{of type} $(k,\zz )$ if $\norma{\mu}=k$ and $Z(\mu)=\zz $.
Now assume that $\alpha=1$.
In this case \eqref{eq4:N^*-N^red} is reduced to
$$
N_{\mathcal L}(q+r,\zz ) =
N_{\mathcal L}^{\mathrm{red}}(q+r,\zz ) +
(m-\zz ) N_{\mathcal L}^{\mathrm{red}}(r,\zz ) +
2(\zz +1) N_{\mathcal L}^{\mathrm{red}}(r,\zz +1).
$$
There are three terms in the right hand side.
Also, if $\mu$ is an element of type $(q+r,\zz )$ and $\mu_0$ is the only element in $C(q)$ such that $\mu\sim\mu_0$, then there are three possible different types for $\mu_0$, namely $(q+r,\zz )$, $(r,\zz )$ and $(r,\zz +1)$.
Next, we check the correspondence between the three terms and the three types, in the same order that are given.
The first term corresponds to the elements in $\mathcal L$ of type $(q+r,\zz )$ which are already reduced.
The second term corresponds to the elements in $\mathcal L$ that are equivalent to a reduced element of type $(r,\zz )$.
Indeed, if $\mu=\sum_i a_i\varepsilon_i \in\mathcal L\cap C(q)$ is of type $(r,\zz )$, then for each nonzero coordinate $i$ of $\mu$ (there are $m-\zz $ of them), the element $\mu+\frac{a_i}{|a_i|}q\varepsilon_i$ has type $(q+r,\zz )$ and lies in the lattice, since $\pm q\varepsilon_i\in q\Z^m \subset \mathcal L$.
Regarding the third term, for each $\mu\in\mathcal L\cap C(q)$ of type $(r,\zz +1)$ and each zero coordinate $i$ of $\mu$ (there are $\zz+1$ of them), the element $\mu\pm q\varepsilon_i$ has type $(q+r,\zz )$.
The detailed description done in the particular case $\alpha=1$ will help to understand the general case.
Let $\mu\in\mathcal L$ of type $(k,\zz )$ and denote by $\mu_0$ the only element in $C(q)\cap\mathcal L$ such that $\mu\sim\mu_0$. One can check that $\mu_0$ is of type $(k-tq,\zz +s)$ for some $0\leq s\leq m-\zz $ and some $s\leq t\leq \alpha$.
Assume that $\mu_0$ is of type $(k-tq,\zz +s)$.
For each choice of $s$ zero coordinates, $i_1,\dots,i_s$, of $\mu_0$, the element $\mu_1:=\mu_0 \pm q \varepsilon_{i_1}\pm\dots\pm q \varepsilon_{i_s}$ has type $(k-tq+sq,\zz )$.
There are $2^s\binom{\zz +s}{s}$ different ways to choose $\mu_1$ from $\mu_0$.
Now, it remains to add $\pm q$ (depending on the sign of the coordinate) $(t-s)$-times in the $m-\zz $ nonzero coordinates.
This can be done in as many ways as the number of ordered partitions of $t-s$ into $m-\zz$ parts, that is, the number of ways of writing $t-s\in\N_0$ as a sum of $m-\zz $ non-negative integers. This equals $\binom{t-s+m-\zz -1}{m-\zz -1}$ and establishes formula \eqref{eq4:N^*-N^red}.
We next prove (ii). In one direction the assertion follows from \eqref{eq4:N^*-N^red}.
We now assume that $N_{\mathcal L}(k,\zz ) = N_{\mathcal L'}(k,\zz )$ for every $k$ and $\zz $.
We write $k=\alpha q+r$ with $0\leq r<q$.
We argue by induction on $\alpha$.
When $\alpha =0$, $N_{\mathcal L}(k,\zz ) = N_{\mathcal L}^{\mathrm{red}}(k,\zz )$ and similarly for $\mathcal L'$, thus $N_{\mathcal L}^{\mathrm{red}}(k,\zz ) = N_{\mathcal L'}^{\mathrm{red}}(k,\zz )$ for every $k<q$.
We assume that $N_{\mathcal L}^{\mathrm{red}}(k,\zz ) = N_{\mathcal L'}^{\mathrm{red}}(k,\zz )$ holds for every $k=\alpha q+r$ with $\alpha <\alpha_0\in\N$.
Clearly, $N_{\mathcal L}^{\mathrm{red}}(\alpha_0 q+r,m) = N_{\mathcal L'}^{\mathrm{red}}(\alpha_0 q+r,m)=0$.
We proceed by induction on $\zz$, decreasing from $m$ to $0$.
Suppose that $N_{\mathcal L}^{\mathrm{red}}(\alpha_0 q+r,\zz ) = N_{\mathcal L'}^{\mathrm{red}}(\alpha_0 q+r,\zz )$ for every $\zz>\zz_0$.
By \eqref{eq4:N^*-N^red}, $N_{\mathcal L}(\alpha_0 q+r,\zz_0)$ can be written as a linear combination of the $N_{\mathcal L}^{\mathrm{red}}(\alpha q+r,\zz)$ for $\alpha\leq \alpha_0$ and $\zz \geq \zz_0$, and similarly for $N_{\mathcal L'}(\alpha_0 q+r,\zz_0)$.
Thus, by the inductive hypothesis, we obtain that $N_{\mathcal L}(\alpha_0 q+r,\zz_0) = N_{\mathcal L'}(\alpha_0 q+r,\zz_0)$ as asserted.
\end{proof}
\section{Computations and questions}\label{sec:examples}
In this section we shall use the finiteness theorem of Section~\ref{sec:finiteness} to produce, with the help of a computer, many examples of pairs of non-isometric congruence lattices that are $\norma{\cdot}^*$-isospectral.
In light of Theorem~\ref{thm3:characterization}~(ii), each such pair gives rise to a pair of non-isometric lens spaces that are $p$-isospectral for all $p$.
We next explain the computational procedure to find $\norma{\cdot}^*$-isospectral lattices.
For each $m$ and $q$, one finds first, by using Propositions~\ref{prop3:lens-isom} and \ref{prop3:isometrias}, a complete list of non-isometric $q$-congruence lattices in $\Z^m$.
Then, for each lattice $\mathcal L$ in the list, one computes the (finitely many) numbers $N_{\mathcal L}^{\mathrm{red}}(k,\zz )$ for $0\leq \zz \leq m$ and $0\leq k\leq (m-\zz)(q-1)$.
Next, for each pair of lattices, one compares their associated sets of numbers.
Finally, the program puts together the lattices for which these numbers coincide.
By Theorem~\ref{thm4:finitud}, such lattices are mutually $\norma{\cdot}^*$-isospectral.
By the procedure above, using the computer program Sage~\cite{Sage}, we found all $\norma{\cdot}^*$-isospectral $m$-dimensional $q$-congruence lattices for $m=3$, $q\leq 300$ and $m=4$, $q\leq 150$ (see Tables~\ref{table:m=3} and \ref{table:m=4}).
We point out that all such lattices come in pairs for these values of $q$ and $m$ (see Question~\ref{que6:families}).
In the tables, the parameters $[s_1,\dots,s_m]$ and $[s_1',\dots,s_m']$ in a row indicate the corresponding $\norma{\cdot}^*$-isospectral lattices
$\mathcal L(q;s_1,\dots, s_m)$ and $\mathcal L(q;s_1',\dots, s_m')$ as in \eqref{eq3:Lambda(q;s)}.
\begin{table}
\caption{Pairs of $\norma{\cdot}^*$-isospectral $q$-congruence lattices of dimension $m=3$ for $q\leq 300$.}\label{table:m=3}
{
\begin{tabular}[t]{c@{\;\;[}r@{,\,}r@{,\,}r@{\,]\quad[}r@{,\,}r@{,\,}r@{\,]\;\;}c}
$q$& $s_1$&$s_2$&$s_3$& $s_1'$&$s_2'$&$s_3'$\\ \hline
49& 1& 6& 15& 1& 6& 20 & *\\
64& 1& 7& 17& 1& 7& 23 & *\\
98& 1& 13& 29& 1& 13& 41 & *\\
100& 1& 9& 21& 1& 9& 29 & *\\
100& 1& 9& 31& 1& 9& 39\\
121& 1& 10& 23& 1& 10& 32 & *\\
121& 1& 10& 34& 1& 10& 43\\
121& 1& 10& 45& 1& 10& 54\\
121& 1& 21& 34& 1& 21& 54\\
121& 1& 21& 45& 1& 21& 56\\
128& 1& 15& 33& 1& 15& 47 & *\\
147& 1& 20& 43& 1& 20& 62 & *\\
169& 1& 12& 27& 1& 12& 38 & *\\
169& 1& 12& 53& 1& 12& 64\\
169& 1& 12& 66& 1& 12& 77\\
169& 1& 25& 40& 1& 25& 64\\
169& 1& 25& 53& 1& 25& 77\\
169& 1& 38& 53& 1& 38& 79\\
169& 1& 12& 40& 1& 12& 51\\
169& 1& 25& 66& 1& 25& 79\\
192& 1& 23& 49& 1& 23& 71 & *\\
196& 1& 13& 29& 1& 13& 41 & *\\
196& 1& 13& 57& 1& 13& 69\\
196& 1& 41& 71& 1& 41& 85\\
196& 1& 13& 43& 1& 13& 55\\
196& 1& 13& 71& 1& 13& 83\\
196& 1& 27& 43& 1& 27& 69\\
196& 1& 27& 57& 1& 27& 83 & *\\
200& 1& 19& 41& 1& 19& 59 & *\\
200& 1& 19& 61& 1& 19& 79\\
242& 1& 21& 45& 1& 21& 65 & *
\end{tabular}
\qquad
\begin{tabular}[t]{c@{\;\;[}r@{,\,}r@{,\,}r@{\,]\quad[}r@{,\,}r@{,\,}r@{\,]\;\;}c}
$q$& $s_1$&$s_2$&$s_3$& $s_1'$&$s_2'$&$s_3'$\\ \hline
242& 1& 21& 67& 1& 21& 87 \\
242& 1& 21& 89& 1& 21&109 \\
242& 1& 43& 67& 1& 43&109 \\
242& 1& 43& 89& 1& 43&111 \\
245& 1& 34& 71& 1& 34&104 & *\\
256& 1& 15& 33& 1& 15& 47 & *\\
256& 1& 15& 81& 1& 15& 95 \\
256& 1& 31& 81& 1& 31&111 \\
256& 1& 47& 97& 1& 47&113 \\
256& 1& 15& 97& 1& 15&111 \\
256& 1& 31& 49& 1& 31& 79 \\
256& 1& 31& 65& 1& 31& 95 & *\\
289& 1& 16& 35& 1& 16& 50 & *\\
289& 1& 16& 86& 1& 16&101 \\
289& 1& 16&120& 1& 16&135 \\
289& 1& 33& 69& 1& 33&101 \\
289& 1& 33& 86& 1& 33&118 \\
289& 1& 50& 69& 1& 50&118 \\
289& 1& 50&103& 1& 50&137 \\
289& 1& 67& 86& 1& 67&137 \\
289& 1& 16& 52& 1& 16& 67 \\
289& 1& 16& 69& 1& 16& 84 \\
289& 1& 16&103& 1& 16&118 \\
289& 1& 33& 52& 1& 33& 84 \\
289& 1& 67&103& 1& 67&120 \\
289& 1& 33&103& 1& 33&135 \\
289& 1& 50& 86& 1& 50&135 \\
289& 1& 33&120& 1& 33&137 \\
294& 1& 41& 85& 1& 41&125 & *\\
300& 1& 29& 61& 1& 29& 89 & *\\
300& 1& 29& 91& 1& 29&119
\end{tabular}} \\[3mm]
Pairs marked with $*$ belong to the family to be given in Section~\ref{sec:families}.
\end{table}
\begin{table}
\caption{Pairs of $\norma{\cdot}^*$-isospectral $q$-congruence lattices of dimension $m=4$ for $q\leq 150$.}\label{table:m=4}
{
\begin{tabular}[t]{c@{\;\;[}r@{,\,}r@{,\,}r@{,\,}r@{\,]\quad[}r@{,\,}r@{,\,}r@{,\,}r@{\,]}}
$q$& $s_1$&$s_2$&$s_3$& $s_4$& $s_1'$&$s_2'$&$s_3'$&$s_4'$\\ \hline
49& 1& 6& 8& 20& 1& 6& 8& 22 \\
81& 1& 8& 10& 26& 1& 8& 10& 28 \\
81& 1& 8& 10& 35& 1& 8& 10& 37 \\
81& 1& 8& 19& 37& 1& 8& 26& 37 \\
98& 1& 13& 15& 41& 1& 13& 15& 43 \\
100& 1& 9& 11& 29& 1& 9& 11& 31 \\
100& 1& 9& 21& 39& 1& 9& 29& 31 \\
121& 1& 10& 12& 32& 1& 10& 12& 34 \\
121& 1& 10& 12& 54& 1& 10& 12& 56 \\
121& 1& 10& 23& 56& 1& 10& 32& 56
\end{tabular}
\qquad
\begin{tabular}[t]{c@{\;\;[}r@{,\,}r@{,\,}r@{,\,}r@{\,]\quad[}r@{,\,}r@{,\,}r@{,\,}r@{\,]}}
$q$& $s_1$&$s_2$&$s_3$& $s_4$& $s_1'$&$s_2'$&$s_3'$&$s_4'$\\ \hline
121& 1& 10& 34& 54& 1& 10& 43& 45 \\
121& 1& 21& 23& 54& 1& 21& 23& 56 \\
121& 1& 10& 12& 43& 1& 10& 12& 45 \\
121& 1& 10& 23& 43& 1& 10& 32& 34 \\
121& 1& 10& 23& 45& 1& 10& 32& 54 \\
121& 1& 10& 23& 54& 1& 10& 32& 45 \\
121& 1& 10& 34& 56& 1& 10& 43& 56 \\
144& 1& 11& 13& 47& 1& 11& 13& 49 \\
144& 1& 11& 25& 59& 1& 11& 35& 49 \\
147& 1& 20& 22& 62& 1& 20& 22& 64
\end{tabular}}
\end{table}
Next we will attempt to explain in a unified manner the examples appearing in the tables.
Let $r$ and $t$ be positive integers and set $q=r^2t$, $r>1$.
We let $\ww= 1+rt$, considered as an element of $(\Z/q\Z)^\times$, the group of units of $\Z/q\Z$.
Then, the inverse of $\ww$ modulo $q$ is $\ww^{-1}:=1-rt$.
Clearly, for every $k \in \Z$,
\begin{equation*}
\ww^k \equiv 1+krt\pmod{q}.
\end{equation*}
In particular, $\ww$ has order $r$ in $(\Z/q\Z)^\times$.
For example, the pairs considered in Section~\ref{sec:families} can be written in the form
\begin{equation}\label{eq:r2t-inverses}
\mathcal L=\mathcal L(q;\ww^{0},\ww^{1}, \ww^{3})\qquad\text{and}\qquad
\mathcal L'=\mathcal L(q;\ww^{0},\ww^{-1}, \ww^{-3}).
\end{equation}
We note that all pairs in the tables have a description in terms of suitable powers of $\ww$ for some choices of $r$ and $t$ such that $q=r^2t$.
For instance, the simplest example in Table~\ref{table:m=3}, if we take $r=7$ and $t=1$ can be written as
\begin{equation}\label{eq6:ex-m=3-q=49}
\begin{array}{l}
\mathcal L(49;1,6,15)
=
\mathcal L(q;\ww^{0},-\ww^{-1}, \ww^{2})
\cong_1
\mathcal L(q;\ww^0, \ww^{1}, \ww^{3}),\\
\mathcal L(49;1,6,20)
=
\mathcal L(q;\ww^{0},-\ww^{-1}, -\ww^{-3})
\cong_1
\mathcal L(q;\ww^{0},\ww^{-1}, \ww^{-3}),
\end{array}
\end{equation}
where $\cong_1$ denotes isometric in $\norma{\cdot}$.
Indeed, in both cases we multiplied by an appropriate power of $\theta$ and then we reordered the terms.
Furthermore, the first pair in Table~\ref{table:m=4}, if $r=7$ and $t=1$ becomes
\begin{equation}\label{eq6:ex-m=4-q=49}
\begin{array}{l}
\mathcal L(49;1,6,8,20)
=
\mathcal L(q;\ww^0, -\ww^{-1}, \ww^{1}, -\ww^{-3})
\cong_1
\mathcal L(q;\ww^0, \ww^{2}, \ww^{3}, \ww^{4}),\\
\mathcal L(49;1,6,8,22)
=
\mathcal L(q;\ww^0, -\ww^{-1}, \ww^{1}, \ww^{3})
\cong_1
\mathcal L(q;\ww^0, \ww^{-2}, \ww^{-3}, \ww^{-4}).
\end{array}
\end{equation}
We point out that all examples shown in Tables~\ref{table:m=3} and \ref{table:m=4} respond to the following description:
\begin{equation}\label{eq6:inversos}
\mathcal L(q;\ww^{d_0},\ww^{d_1},\dots,\ww^{d_{m-1}})
\quad\text{and}\quad
\mathcal L(q;\ww^{-d_0},\ww^{-d_1},\dots,\ww^{-d_{m-1}}),
\end{equation}
where $q=r^2t$, $r>1$, $\ww=1+rt$ and $0=d_0<d_1<\dots<d_{m-1}<r$.
However, note that for some choices of $m$, $r$ and $t$, there are sequences $0=d_0<d_1<\dots<d_{m-1}<r$ such that the lattices defined as in \eqref{eq6:inversos} are not $\norma{\cdot}^*$-isospectral.
For example, this is the case when
$m=3$, $r=8$, $t=1$ and
$[d_0,d_1,d_2]=[0,1,4]$.
The following questions come up naturally.
\begin{question}\label{que6:conditions}
Give conditions on the sequence $0=d_0<d_1<\dots<d_{m-1}<r$ for lattices as in \eqref{eq6:inversos} to be $\norma{\cdot}^*$-isospectral.
\end{question}
\begin{question}
Are there examples of $\norma{\cdot}^*$-isospectral lattices that are not of the type in \eqref{eq6:inversos} for some choice of $\theta$?
\end{question}
\begin{question}\label{que6:families}
Are there families of $\norma{\cdot}^*$-isospectral lattices having more than two elements?
\end{question}
We have carried out computations for small values of $m$ and $q$ and in this search
we have not found any such family yet.
Next, we give a particular sequence as in \eqref{eq6:inversos} that is very likely to give $\norma{\cdot}^*$-isospectral pairs in all dimensions under rather general conditions on $r$, for instance, if $r$ is prime. This motivation makes it worth showing that this sequence always gives non-isometric lattices.
\begin{prop}\label{lem6:non-isom}
Let $m\geq3$, $r\geq m+3$, $t\in\N$ and set $q=r^2t$ and $\ww=1+rt$, then the $q$-congruence lattices
\begin{equation}\label{eq6:inv-non-isom}
\mathcal L(q;\ww^0, \ww^2, \ww^{3}, \dots, \ww^{m})
\quad\text{and}\quad
\mathcal L(q;\ww^0, \ww^{-2}, \ww^{-3}, \dots, \ww^{-m})
\end{equation}
are not $\norma{\cdot}$-isometric.
\end{prop}
\begin{proof}
In general, if $0=d_0<d_1<\dots<d_{m-1}<r$, we associate to $\mathcal L(q;\ww^{d_0},\ww^{d_1},\dots,\ww^{d_{m-1}})$ the following ordered partition of $r$:
$$
r=(d_1-d_0)+\dots+(d_{m-1}-d_{m-2})+(r-d_{m-1}).
$$
By using Propositions~\ref{prop3:lens-isom} and \ref{prop3:isometrias}, one can check that two lens spaces with partitions $r=a_1+\dots+a_m$ and $r=b_1+\dots+b_m$ are $\norma{\cdot}$-isometric if and only if there is $l \in \Z$ such that $a_j=b_{j+l}$ for every $j$, where the index $j+l$ is taken in the interval $[0,m-1]$$ \pmod m$.
In our case, the ordered partitions for the lattices $\mathcal L(q;\ww^0, \ww^2, \ww^{3}, \dots, \ww^{m})$ and $\mathcal L(q;\ww^0, \ww^{-2}, \ww^{-3}, \dots, \ww^{-m}) {\cong}_1 \mathcal L(q;\ww^0, \ww^{1}, \dots, \ww^{m-2},\ww^{m})$ are
\begin{align*}
r=2+1+\dots+1+(r-m),\\
r=1+\dots+1+2+(r-m).
\end{align*}
The assertion now follows since $r-m\geq3$.
\end{proof}
\section{Families of $\norma{\cdot}^*$-isospectral lattices} \label{sec:families}
The goal of this section is to construct an infinite two-parameter family of pairs of $\norma{\cdot}^*$-isospectral lattices in $\Z^m$ for $m=3$.
Together with Theorem~\ref{thm3:characterization}~(ii), this will produce infinitely many pairs of non-isometric $5$-dimensional lens spaces, isospectral on $p$-forms for every $p$.
Although our construction does not give all of the examples for $m=3$, the list given in Section~\ref{sec:examples} shows that most of the examples can be obtained by a slight variation of the method used in this section.
Throughout this section, we fix $r,t\in\N$, $r>1$. We single out (see \eqref{eq3:Lambda(q;s)}) the congruence lattices
\begin{equation}
\begin{array}{rcl}
\mathcal L &=& \mathcal L(\;r^2t;\;1,\;1+rt,\;1+3rt),\\
\mathcal L' &=& \mathcal L(\;r^2t;\;1,\;1-rt,\;1-3rt).
\end{array}
\end{equation}
In other words, $\mathcal L$ and $\mathcal L'$ are defined by the equations
\begin{equation}\label{eq5:L-L'}
\begin{array}{rc@{\;}c@{\;}c@{\;}c@{\;}c@{\;\;}c@{\;\;}l}
\mathcal L:&\quad
a&+&(1+rt)b&+&(1+3rt)c&\equiv &0 \pmod{r^2t},\\[1mm]
\mathcal L':&\quad
a&+&(1-rt)b&+&(1-3rt)c&\equiv &0 \pmod{r^2t},
\end{array}
\end{equation}
or equivalently by
\begin{equation}\label{eq5:L-L'2}
\begin{array}{rc@{\;}c@{\;}c@{\;}c@{\;}c@{\;\;}c@{\;\;}l}
\mathcal L:&\quad
a+b+c&+rt(b+3c)&\equiv &0 \pmod{r^2t},\\[1mm]
\mathcal L':&\quad
a+b+c&-rt(b+3c)&\equiv &0 \pmod{r^2t}.
\end{array}
\end{equation}
Our goal is to prove that $\mathcal L$ and $\mathcal L'$ are $\norma{\cdot}^*$-isospectral for every $r$ not divisible by $3$.
By Proposition~\ref{lem6:non-isom}, $\mathcal L$ and $\mathcal L'$ are non-isometric for $r\geq6$.
The first pair in this family is for $r=7$ and $t=1$, namely $\mathcal L=\mathcal L(49;1,8,22)$ and $\mathcal L'=\mathcal L(49;1,-6,-20)$.
We point out that this pair is isometric to the simplest pair in Table~\ref{table:m=3} by \eqref{eq6:ex-m=3-q=49}.
We recall that $\mathcal L$ and $\mathcal L'$ are said to be $\norma{\cdot}^*$-isospectral if $N_{\mathcal L}(k,\zz ) = N_{\mathcal L'}(k,\zz )$ for every $k\in\N$ and every $0\leq \zz \leq m=3$ (where $N_{\mathcal L}(k,\zz )$ is as in \eqref{eq:Nkz}).
We shall first prove that this equality holds easily for $\zz =1,2,3$.
\begin{lemma}\label{lem:z=1,2,3}
For any $\zz =1,2,3$ and any $k \in \N$, one has that $N_\mathcal L (k,\zz )= N_{\mathcal L'} (k,\zz )$.
\end{lemma}
\begin{proof}
The assertion is clear for $\zz =3$. Also, it is easy to check for $\zz =2$ since the elements $(sr^2t,0,0),(0,sr^2t,0),(0,0,sr^2t)$ for $s\in\Z$, $s\neq0$ are the only ones in both lattices having exactly two coordinates equal to zero.
For $\zz =1$, it is not hard to give a $\norma{\cdot}$-preserving bijection between the sets $\{\eta\in\mathcal L:Z(\eta)=1\}$ and $\{\eta'\in\mathcal L':Z(\eta')=1\}$. Namely one has
\begin{align*}
(a,b,0)\in\mathcal L\quad&\Longleftrightarrow\quad (b,a,0)\in\mathcal L', \\
(a,0,c)\in\mathcal L\quad&\Longleftrightarrow\quad (c,0,a)\in\mathcal L', \\
(0,b,c)\in\mathcal L\quad&\Longleftrightarrow\quad (0,c,b)\in\mathcal L',
\end{align*}
for every nonzero integers $a$, $b$ and $c$.
For example, $(a,b,0)\in\mathcal L$ $\iff$ $a+(1+rt)b\equiv0\pmod{r^2t}$ $\iff$ $(1-rt)a+b\equiv0\pmod{r^2t}$ $\iff$ $(b,a,0)\in\mathcal L$.
The second and the third rows follow in a similar way, multiplying by $1-3rt$ and $1-4rt$ respectively.
\end{proof}
\begin{rem}
It now remains to prove that $N_\mathcal L (k,0) = N_{\mathcal L'}(k,0)$ for every $k$, which,
by Lemma~\ref{lem:z=1,2,3}, is equivalent to show that $\mathcal L$ and $\mathcal L'$ are $\norma{\cdot}$-isospectral, since $N_{\mathcal L} (k) = \sum _{\zz =0}^3 N_\mathcal L (k,\zz )$.
We see that, remarkably, in light of Theorem~\ref{thm3:characterization}~(i), the previous lemma allows us to reduce the verification of $p$-isospectrality for all $p$ of the associated lens spaces, to prove that they are just $0$-isospectral.
\end{rem}
\begin{thm}\label{thm4:isolattices}
For any $r$ and $t$ positive integers with $r\not\equiv0 \pmod3$, the lattices $\mathcal L$ and $\mathcal L'$ in \eqref{eq5:L-L'} are $\norma{\cdot}^*$-isospectral.
\end{thm}
\begin{proof}
By Lemma~\ref{lem:z=1,2,3}, it remains to prove that $N_\mathcal L (k,0) = N_{\mathcal L'}(k,0)$ for every $k$.
This is clearly true for $k=0$, hence we will assume that $k>0$.
The proof consists in showing that the number of elements with a fixed one-norm in each octant is the same for both lattices.
Since lattices have central symmetry, we have
\begin{align*}
\tfrac12 \,N_{\mathcal L}(k,0) = N_{\mathcal L}^{{+}{+}{+}}(k) + N_{\mathcal L}^{{+}{+}{-}}(k) + N_{\mathcal L}^{{+}{-}{+}}(k) + N_{\mathcal L}^{{+}{-}{-}}(k),
\end{align*}
where the signs in the supra-indexes indicate the signs of the coordinates.
That is, $N_{\mathcal L}^{{+}{+}{-}}(k)$ is the number of $\eta=(a,b,c)\in\mathcal L$ such that $\norma{\eta}=k$, $a>0$, $b>0$ and $c<0$.
We will show that $N_{\mathcal L}^{{+}{+}{+}}(k)=N_{\mathcal L'}^{{+}{+}{+}}(k)$, $N_{\mathcal L}^{{+}{+}{-}}(k)=N_{\mathcal L'}^{{+}{+}{-}}(k)$ and so on.
We first examine the octant ${+}{+}{+}$.
Any vector here has the form
\begin{equation}\label{eq5:eta+++}
\eta=(k-x,x-y,y)
\qquad\text{with} \quad 0<y<x<k.
\end{equation}
By \eqref{eq5:L-L'2}, we have that $\eta\in\mathcal L$ (resp.\ $\eta\in\mathcal L'$) if and only if $k+rt(x+2y)\equiv 0\pmod{r^2t}$ (resp.\ $k-rt(x+2y)\equiv 0\pmod{r^2t}$).
Thus $N_{\mathcal L}^{{+}{+}{+}}(k)=N_{\mathcal L'}^{{+}{+}{+}}(k)=0$ unless $k$ is divisible by $rt$.
We write $k=\omega rt$ for some positive integer $\omega$.
Then
\begin{equation}\label{eq5:+++}
\begin{array}{lcr@{\;}l}
\eta\in\mathcal L&
\Longleftrightarrow &
&x+2y\equiv -\omega\pmod{r},
\\[1mm]
\eta\in\mathcal L'&
\Longleftrightarrow &
&x+2y\equiv \;\;\,\omega\pmod{r}.
\end{array}
\end{equation}
In order to count the number of solutions in \eqref{eq5:+++}, we split the set of integer points $(x,y)$ satisfying $0<y<x<k$ into squares and triangles as follows.
We take the squares $\{(x,y): \alpha r\leq x<(\alpha+1)r, \; \beta r<y\leq (\beta+1)r\}$ for $1\leq\beta<\alpha\leq \omega t-1$, together with the triangles $\{(x,y): \alpha r<x<(\alpha+1),\; \alpha r<y<x\}$ for $0\leq\alpha\leq wt-1$.
We note that the points which are the upper-left corner of the squares near the diagonal (when $\alpha=\beta+1$) are not in the orginal set; this will be taken into account in the computations.
There are $\binom{\omega t}{2}$ squares and $\omega t$ triangles.
Set
\begin{equation*}
A(r,\xi)=\#\{(x,y)\in\Z^2: 0<y<x<r, \; x+2y\equiv\xi\pmod r\}.
\end{equation*}
Since we are working modulo $r$, the number of elements of $\mathcal L$ (resp.\ $\mathcal L'$) in any triangle is always the same and is equal to $A(r,-\omega)$ (resp.\ $A(r,\omega)$).
Thus we have $\omega t A(r,-\omega)$ (resp.\ $\omega t A(r,\omega)$) elements in $\mathcal L$ (resp.\ $\mathcal L'$) in the union of all triangles.
On the other hand, if $\omega\not\equiv 0\pmod r$, there are exactly $r$ elements in $\mathcal L$ (or in $\mathcal L'$) in each square, thus we have $\binom{\omega t}{2} r$ elements in $\mathcal L$ (or in $\mathcal L'$) and in the union of all the squares.
When $\omega\equiv 0\pmod r$, one has the same quantity minus $\omega t-1$ elements, since, as noticed above, we have to exclude the vertices $(\alpha \omega t,(\alpha+1) \omega t)$ for $1\leq \alpha\leq \omega t-1$ which lie
in the squares next to the diagonal $x=y$.
Summing up, we get
\begin{equation*}
N_{\mathcal L}^{{+}{+}{+}}(\omega rt) =
\begin{cases}
\omega t\, A(r,-\omega) + \binom{\omega t}{2} r
&\quad\text{if $\omega\not\equiv 0\pmod r$},\\
\omega t\, A(r,-\omega) + \binom{\omega t}{2} r-\omega t +1
&\quad\text{if $\omega\equiv 0\pmod r$},
\end{cases}
\end{equation*}
and the same for $N_{\mathcal L'}^{{+}{+}{+}}(\omega rt)$ replacing $A(r,-\omega)$ by $A(r,\omega)$.
The next lemma gives a formula for $A(r,\xi)$ showing that $A(r,\omega)=A(r,-\omega)$, hence $N_{\mathcal L}^{{+}{+}{+}}(\omega rt)=N_{\mathcal L'}^{{+}{+}{+}}(\omega rt)$.
\begin{lemma}\label{lem5:A(r,omega)}
Let $r$ and $\xi$ be integers such that $r\not\equiv0\pmod3$.
If $r$ is odd, then
\begin{equation*}
A(r,\xi) =
\begin{cases}
\tfrac{r-3}{2}
\quad&\text{if $\xi\not\equiv0\pmod r$,} \\
\tfrac{r-1}{2}
\quad&\text{if $\xi\equiv0\pmod r$.}
\end{cases}
\end{equation*}
If $r$ is even, then
\begin{equation*}
A(r,\xi) =
\begin{cases}
\tfrac{r}2-1
\quad&\text{if $\xi\not\equiv0\pmod r$ and $\xi$ is odd,}\\
\tfrac{r}2-2
\quad&\text{if $\xi\not\equiv0\pmod r$ and $\xi$ is even,}\\
\tfrac{r}2-1
\quad&\text{if $\xi\equiv0\pmod r$.}
\end{cases}
\end{equation*}
\end{lemma}
We will often use the standard notation $\lfloor u\rfloor=\max\{d\in\Z: d\leq u\}$ and $\lceil u\rceil=\min\{d\in\Z: d\geq u\}$ for the floor and ceiling of a real number respectively.
\begin{proof}
We may assume that $0\leq \xi<r$.
Suppose that $x+2y\equiv \xi\pmod r$; thus, $x=\gamma r+\xi-2y$ for some $\gamma\in\Z$.
One can check that $1\leq\gamma\leq 2$ if $\xi=0$ and $0\leq \gamma\leq 2$ if $\xi>0$, since $0<y<x<r$.
Furthermore, the restrictions $y<x$ and $x<r$ are equivalent to
\begin{equation}\label{eq5:intervalo-beta}
\begin{array}{rcl@{\hspace{10mm}}rcl}
y+1&\leq& \gamma r+\xi-2y, &
\gamma r+\xi-2y&\leq& r-1\\[1mm]
y&\leq &\tfrac{\gamma r+\xi-1}{3}, &
\tfrac{(\gamma-1) r+\xi+1}{2}&\leq& y\\[1mm]
y&\leq &\lfloor\tfrac{\gamma r+\xi-1}{3}\rfloor, &
\lceil\tfrac{(\gamma-1) r+\xi+1}{2}\rceil&\leq& y.
\end{array}
\end{equation}
If $\xi=0$, then $\gamma=1$ implies $1\leq y\leq \lfloor\tfrac{r-1}{3}\rfloor$ and $\gamma=2$ implies $\lceil\tfrac{r+1}{2}\rceil\leq y\leq \lfloor\tfrac{2r-1}{3}\rfloor$, thus
$$
A(r,0) = \lfloor\tfrac{r-1}{3}\rfloor + \lfloor\tfrac{2r-1}{3}\rfloor+1-\lceil\tfrac{r+1}{2}\rceil = r-\lceil\tfrac{r+1}{2}\rceil,
$$
which is our assertion for $\xi\equiv0\pmod r$.
Similarly, if $\xi>0$, then $\gamma=0$ implies $1\leq y\leq \lfloor\tfrac{\xi-1}{3}\rfloor$, $\gamma=1$ implies $\lceil\tfrac{\xi+1}{2}\rceil\leq y\leq \lfloor\tfrac{r+\xi-1}{3}\rfloor$ and $\gamma=2$ implies $\lceil\tfrac{r+\xi+1}{2}\rceil\leq y\leq \lfloor\tfrac{2r+\xi-1}{3}\rfloor$, thus
$$
A(r,\xi) = \lfloor\tfrac{\xi-1}{3}\rfloor+ \lfloor\tfrac{r+\xi-1}{3}\rfloor + \lfloor\tfrac{2r+\xi-1}{3}\rfloor
+2-\lceil\tfrac{\xi+1}{2}\rceil -\lceil\tfrac{r+\xi+1}{2}\rceil
=r+\xi-\big( \lceil\tfrac{\xi+1}{2}\rceil +\lceil\tfrac{r+\xi+1}{2}\rceil\big).
$$
The rest of the proof is straightforward.
\end{proof}
We continue with the proof of Theorem~\ref{thm4:isolattices}, now considering the octant ${+}{-}{-}$.
Any vector in this octant can be written as
\begin{equation}\label{eq5:eta+--}
\eta=(k-x,y-x,-y)
\qquad\text{with} \quad 0<y<x<k,
\end{equation}
then, by \eqref{eq5:L-L'2}, we have that
\begin{equation}\label{eq5:+--1}
\begin{array}{lcr@{\;}l}
\eta\in\mathcal L&
\Longleftrightarrow &
&\;\;\,rt(x+2y)\equiv k-2x\pmod{r^2t},
\\[1mm]
\eta\in\mathcal L'&
\Longleftrightarrow &
&-rt(x+2y)\equiv k-2x\pmod{r^2t}.
\end{array}
\end{equation}
In both cases we have $2x\equiv k\pmod{rt}$.
We fix $k=\omega rt+k_0$ with $0\leq k_0<rt$, thus $x$ must satisfy $2x\equiv k_0\pmod{rt}$.
We first assume that $rt$ is odd.
Then there exists only one $x_0$ satisfying $2x_0\equiv k_0\pmod{rt}$ and $0\leq x_0<rt$.
We write any other solution as $x=\alpha rt+x_0$ for some $\alpha$.
The restriction $0<x<k$ is equivalent to
\begin{equation}\label{eq5:intervalo-alpha}
\begin{array}{rcccl}
1&\leq& \alpha rt+x_0 &\leq &\omega rt+k_0-1, \\[1mm]
-\lfloor\tfrac{x_0-1}{rt}\rfloor=\lceil\tfrac{1-x_0}{rt}\rceil &\leq& \alpha &\leq& \omega+ \lfloor\tfrac{k_0-1-x_0}{rt}\rfloor.
\end{array}
\end{equation}
On the other hand, by \eqref{eq5:+--1}, we have that
\begin{equation}\label{eq5:+--2}
\begin{array}{lcr@{\;}l}
\eta\in\mathcal L&
\Longleftrightarrow &
2y\equiv &-x_0+\left(\omega-2\alpha + \tfrac{k_0-2x_0}{rt}\right) \pmod{r},
\\[1mm]
\eta\in\mathcal L'&
\Longleftrightarrow &
2y\equiv &-x_0-\left(\omega-2\alpha + \tfrac{k_0-2x_0}{rt}\right) \pmod{r}.
\end{array}
\end{equation}
Since $r$ is odd, these equations always have a solution $y$, which is unique modulo $r$.
For each $\alpha$ satisfying \eqref{eq5:intervalo-alpha}, denote respectively by $y_\alpha$ and $y_\alpha'$ the solutions of \eqref{eq5:+--2} for $\mathcal L$ and $\mathcal L'$ such that $0\leq y_\alpha,y_\alpha'<r$.
We write the solutions as $y=\beta r+y_\alpha$ and $y'=\beta' r+y_{\alpha'}$.
Now, the restriction $0<y<x$ is equivalent to
\begin{equation}\label{eq5:intervalo-beta}
\begin{array}{rcccl}
1&\leq& \beta r+y_\alpha &\leq &\alpha rt+x_0-1, \\[1mm]
-\lfloor\tfrac{y_\alpha-1}{r}\rfloor=\lceil\tfrac{1-y_\alpha}{r}\rceil &\leq& \beta &\leq& \alpha t+ \lfloor\tfrac{x_0-1-y_\alpha}{r}\rfloor.
\end{array}
\end{equation}
Hence
\begin{equation}\label{eq5:N_L}
N_{\mathcal L}^{{+}{-}{-}}(k)
= \sum_{\alpha=-\lfloor\tfrac{x_0-1}{rt}\rfloor}^{\omega+ \lfloor\tfrac{k_0-1-x_0}{rt}\rfloor}
\left(\alpha t+1+ \lfloor \tfrac{x_0-1-y_\alpha}{r} \rfloor+\lfloor \tfrac{y_\alpha-1}r\rfloor\right).
\end{equation}
The same formula holds for $N_{\mathcal L'}^{{+}{-}{-}}(k)$ replacing $y_\alpha$ by $y_\alpha'$.
Then $N_{\mathcal L}^{{+}{-}{-}}(k)-N_{\mathcal L'}^{{+}{-}{-}}(k)$ is equal to
\begin{equation}\label{eq5:N_L-N_L'}
C:=\sum_{\alpha=-\lfloor\tfrac{x_0-1}{rt}\rfloor}^{\omega+ \lfloor\tfrac{k_0-1-x_0}{rt}\rfloor}
\left(\lfloor \tfrac{x_0-1-y_\alpha}{r} \rfloor+\lfloor \tfrac{y_\alpha-1}r\rfloor-\lfloor \tfrac{x_0-1-y_\alpha'}{r} \rfloor-\lfloor \tfrac{y_\alpha'-1}r\rfloor\right)
\end{equation}
The proof in the case when $rt$ is odd will be completed by showing that $C=0$.
We first suppose that $k_0$ is even and nonzero, thus $k_0=2x_0$ with $0<x_0<rt/2$ and $x_0<k_0$.
Then, \eqref{eq5:N_L-N_L'} implies that
\begin{equation*}
C=\sum_{\alpha=0}^{\omega}
\left(\lfloor \tfrac{x_0-1-y_\alpha}{r} \rfloor
-\lfloor \tfrac{x_0-1-y_\alpha'}{r} \rfloor
+\lfloor \tfrac{y_\alpha-1}r\rfloor
-\lfloor \tfrac{y_\alpha'-1}r\rfloor\right).
\end{equation*}
But a careful look at \eqref{eq5:+--2} shows that the solutions of both equations are related by the equation $y_\alpha'=y_{\omega-\alpha}$ for every $0\leq \alpha\leq\omega$; hence, $C=0$.
If $k_0=0$, then $x_0=0$ and the sum in \eqref{eq5:N_L-N_L'} runs through the interval $1\leq\alpha\leq \omega-1$.
Hence, $C=0$ since $y_\alpha'=y_{\omega-\alpha}$ for every $1\leq \alpha\leq\omega-1$ by \eqref{eq5:+--2}.
Now suppose that $k_0$ is odd, then $k_0=2x_0-rt$ with $rt/2\leq x_0<rt$ and $k_0<x_0$.
In this case the sum in \eqref{eq5:N_L-N_L'} runs through the interval $0\leq\alpha\leq\omega-1$ and $y_\alpha'=y_{\omega-1-\alpha}$ for every $1\leq\alpha\leq \omega-1$ by \eqref{eq5:+--2}, hence $C=0$.
We now assume that $rt$ is even.
We recall that $x$ must satisfy $2x\equiv k_0\pmod{rt}$.
Clearly, when $k$ is odd, $N_{\mathcal L}^{{+}{-}{-}}(k) = N_{\mathcal L'}^{{+}{-}{-}}(k)=0$; thus, we assume that $k$ is even.
Let $x_0$ be the only integer such that $2x_0\equiv k_0\pmod{rt}$ and $0\leq x_0<\frac{rt}{2}$.
Thus $x_0=k_0/2\leq k_0$ and a general solution has the form $x=\alpha\frac{rt}{2}+x_0$.
Similarly, as in \eqref{eq5:intervalo-alpha}, one can check that the restriction $0<x<k$ is equivalent to $-\lfloor2\tfrac{x_0-1}{rt}\rfloor \leq \alpha\leq 2\omega + \lfloor2\tfrac{k_0-x_0-1}{rt}\rfloor$, or more precisely, $1\leq \alpha\leq 2\omega-1$ if $x_0=0$ and, $0\leq \alpha \leq 2\omega$ if $x_0>0$.
Thus in this case, from \eqref{eq5:+--1} we have that
\begin{equation}\label{eq5:+--3}
\begin{array}{lcr@{\;}l}
\eta\in\mathcal L&
\Longleftrightarrow &
2y\equiv &-x_0-\alpha\tfrac{rt}{2}+(\omega-\alpha) \pmod{r},
\\[1mm]
\eta\in\mathcal L'&
\Longleftrightarrow &
2y\equiv &-x_0-\alpha\tfrac{rt}{2}-(\omega-\alpha) \pmod{r}.
\end{array}
\end{equation}
If $r$ is odd, then both equations always have a solution $y$, which is unique modulo $r$.
When $r$ is even, we assume that $-x_0-\alpha\tfrac{rt}{2}+\omega-\alpha$ is even since both equations do not have any solution otherwise.
Thus, equations in \eqref{eq5:+--3} have unique solutions modulo $\tfrac r2$.
Let $y_\alpha$ and $y_\alpha'$ be the smallest non-negative solutions of \eqref{eq5:+--3} for $\mathcal L$ and $\mathcal L'$ respectively.
A similar argument as in \eqref{eq5:intervalo-beta} implies that $N_{\mathcal L}^{{+}{-}{-}}(k)$ is equal to the sum over $-\lfloor2\tfrac{x_0-1}{rt}\rfloor \leq \alpha\leq 2\omega + \lfloor2\tfrac{k_0-x_0-1}{rt}\rfloor$ of the terms
\begin{equation}
\begin{cases}
\alpha\tfrac{t}{2} +1 +\lfloor\tfrac{x_0-y_\alpha-1}{r}\rfloor+ \lfloor\tfrac{y_\alpha-1}{r}\rfloor
&\text{ if $r$ is odd},\\
\alpha t +1 +\lfloor2\tfrac{x_0-y_\alpha-1}{r}\rfloor+ \lfloor2\tfrac{y_\alpha-1}{r}\rfloor
&\text{ if $r$ is even},
\end{cases}
\end{equation}
and the same formula holds for $N_{\mathcal L'}^{{+}{-}{-}}(k)$ replacing $y_\alpha$ by $y_\alpha'$.
But, for arbitrary $r$, \eqref{eq5:+--3} implies that $y_\alpha' = y_{2\omega-\alpha}$ for every $0\leq \alpha\leq 2\omega$; then, $N_{\mathcal L}^{{+}{-}{-}}(k)=N_{\mathcal L'}^{{+}{-}{-}}(k)$.
This concludes the proof for the octant ${+}{-}{-}$.
Entirely similar arguments apply to the octants ${+}{+}{-}$ and ${+}{-}{+}$, by considering the elements written as $(k-x,x-y,-y)$ and $(k-x,-y,x-y)$ for $0<y<x<k$ respectively.
\end{proof}
\begin{rem}
The previous proof gives an explicit formula for $N_{\mathcal L}^{{+}{+}{+}}(k)$, $N_{\mathcal L}^{{+}{-}{-}}(k)$, $N_{\mathcal L}^{{+}{+}{-}}(k)$ and $N_{\mathcal L}^{{+}{-}{+}}(k)$ for every $k$; thus, also for $N_{\mathcal L}(k,0)$.
Actually, we have checked with the computer that the formulas hold for $k\leq 1000$.
A formula for $N_{\mathcal L}^{{+}{+}{+}}(k)$ was included before Lemma~\ref{lem5:A(r,omega)}.
An explicit expression for $N_{\mathcal L}^{{+}{-}{-}}(k)$ could also be given but the formula must be divided into many cases, namely, $rt$ odd, $rt$ even and $r$ odd, $rt$ even and $r$ odd, and (following the notation inside the proof) with each of these subdivided into $k_0$ odd, $k_0>0$ even, $k_0=0$ (subdivided again by $y_\alpha=0$, $y_\alpha>0$).
Similar complications occur for the octants ${+}{+}{-}$ and ${+}{-}{+}$.
Any of these expressions mentioned above contains a main term and a residual term written as a sum of floors of rational numbers.
For example, when $rt$ is odd and $k$ is even and not divisible by $rt$, \eqref{eq5:+--1} implies that
\begin{equation*}
N_{\mathcal L}^{{+}{-}{-}}(k) = t\binom{\omega+1}{2}+\omega+1+\sum_{\alpha=0}^{\omega}
\left(\lfloor\tfrac{x_0-1-y_\alpha}{r}\rfloor+\lfloor\tfrac{y_\alpha-1}{r}\rfloor\right),
\end{equation*}
where $\omega=\lfloor k/{rt}\rfloor$, $x_0$ is the only integer such that $2x_0\equiv k\pmod {rt}$ and $0 \leq x_0<rt$ and $y_\alpha$ is the only solution of $2y_\alpha\equiv -x_0-\omega+2\alpha\pmod r$ satisfying $0\leq y_\alpha<r$.
It is easy to give an expression for $N_{\mathcal L}(k,\zz)$ for $\zz$ equal to $2$ and $3$.
It is also possible for $\zz=1$ in a similar ---and simpler--- way as in the previous proof.
This implies that we can compute explicitly every $p$-spectrum of the lens space $L(r^2t; 1, 1+rt, 1+3rt)$ by using the formula in Theorem~\ref{thm3:dim V_k,p^Gamma}.
\end{rem}
\begin{rem}
In a previous version \cite{LMRhodgeiso_old} of this article, we proved Theorem~\ref{thm4:isolattices} (for $rt$ odd) with a completely different method which was more involved but gave useful additional geometric information on the lattices.
\end{rem}
\section{Lens spaces $p$-isospectral for every $p$}\label{sec:all-p-iso}
In this section we summarize the spectral properties of lens spaces that can be obtained from the results on congruence lattices in the previous three sections, in light of the characterization in Theorem~\ref{thm3:characterization}.
It also contains information on the geometric, topological and spectral properties of the examples.
\begin{thm}\label{thm7:all-p-iso-lens}
For any $r$ and $t$ positive integers with $r\geq7$ and $r\not\equiv0 \pmod3$, the lens spaces
\begin{equation*}
L(r^2t; 1,1+rt,1+3rt)
\quad\text{and}\quad
L(r^2t; 1,1-rt,1-3rt)
\end{equation*}
are $p$-isospectral for all $p$ but not strongly isospectral.
\end{thm}
Tables~\ref{table:m=3} and \ref{table:m=4} give more such pairs in dimensions $5$ and $7$ respectively.
The proof of $p$-isospectrality for all $p$ follows immediately from Theorems~\ref{thm3:characterization} and \ref{thm4:isolattices}.
The non-isometry comes from Proposition~\ref{lem6:non-isom}.
They are not strongly isospectral by the following general fact, which follows from well known results.
We include a proof for completeness.
\begin{prop}\label{prop7:lens-non-strongly}
If $L$ and $L'$ are strongly isospectral lens spaces, then they are isometric.
\end{prop}
\begin{proof}
We first assume that $\Gamma\ba S^{2m-1}$ and $\Gamma'\ba S^{2m-1}$ are strongly isospectral spherical spaces forms, where $\Gamma$ and $\Gamma'$ are arbitrary finite subgroups of $\Ot(2m)$ acting freely on $S^{2m-1}$.
By Proposition~1 in \cite{Pe1}, the subgroups $\Gamma$ and $\Gamma'$ are representation equivalent, i.e.\ $L^2(\Gamma\ba \Ot(2m))$ and $L^2(\Gamma'\ba \Ot(2m))$ are equivalent representations of $\Ot(2m)$.
Hence, $\Gamma$ and $\Gamma'$ are almost conjugate in $\Ot(2m)$ (see Lemma~2.12 in \cite{Wo2}).
In our case, $L=\Gamma\ba S^{2m-1}$ and $L'=\Gamma'\ba S^{2m-1}$ are lens spaces with $\Gamma$ and $\Gamma'$ cyclic subgroups of $\SO(2m)$.
Since almost conjugate cyclic subgroups are necessarily conjugate, then $L$ and $L'$ are isometric.
\end{proof}
We observe that the examples in Theorem~\ref{thm7:all-p-iso-lens} allow to obtain pairs of Riemannian manifolds in every dimension $n\ge 5$ that are $p$-isospectral for all $p$ and are not strongly isospectral.
Indeed, for this purpose, we may just take $M=L\times S^k$ and $M'=L'\times S^k$, for any $k\in \N_0$, where $L$, $L'$ is any pair of non-isometric lens spaces in dimension $5$ satisfying $p$-isospectrality for every $p$.
In relation to lens spaces of higher dimensions we have the following result.
\begin{thm}\label{thm7:high-dim}
For any $n_0 \ge 5$, there are pairs of non-isometric lens spaces of dimension $n$, with $n>n_0$, which are $p$-isospectral for all $p$.
\end{thm}
\begin{proof}
We will apply Theorem~\ref{thm4:isolattices}, together with an extension of a duality result of Ikeda. For each $q\in\N$ and $n=2m-1$ odd, denote by $\mathfrak L_0(q,m)$ the classes of non-isometric $n$-dimensional lens spaces $L(q;s_1,\dots,s_m)$ such that $s_i\not\equiv \pm s_j\pmod q$ for all $i\neq j$.
Set $h=(\phi(q)-2m)/2$, where $\phi$ is the Euler function.
To each lens space $L=L(q;s_1,\dots,s_m)$ in $\mathfrak L_0(q,m)$, one associates the lens space $\overline L=L(q;\bar s_1,\dots,\bar s_h)$, where the parameters $\bar s_1,\dots,\bar s_h$ are chosen so that the set $\{\pm s_1,\dots,\pm s_m,\pm \bar s_1,\dots,\pm \bar s_h\}$ exhausts the coprime classes module $q$.
We thus obtain a new lens space $\overline L$ of dimension $2h-1 = \phi(q)-2m-1$.
By \cite[Thm.~3.6]{Ik88}, if $q$ is prime, $L$ and $L'$ in $\mathfrak L_0(q,m)$ are $p$-isospectral for all $p$ if and only if $\overline L$ and $\overline{ L'}$ are $p$-isospectral for all $p$.
Now, for each $q=r^2$, $r$ an odd prime, $r\not\equiv0\pmod3$, in Theorem~\ref{thm4:isolattices} we have obtained lens spaces $L$ and $L'$ in $\mathfrak L_0(r^2,3)$ that are $p$-isospectral for all $p$. Now, by an extension of Ikeda's argument in \cite[Thm 3.6]{Ik88} for $q$ prime ---to be sketched below--- one can show that the associated lens spaces $\overline L$ and $\overline {L'}$ are $p$-isospectral for all $p$. These lens spaces have dimension $2h-1=\phi(r^2)-7 = r^2-r-7$, a quantity that tends to infinity when $r$ does, thus the assertion in the theorem immediately follows.
We now explain why Ikeda's argument also works in the case $q=r^2$, $r$ prime. One has that $L,L'$ are isospectral for every $p$ if and only if they have the same generating functions (see \cite[Thm 2.5]{Ik88}). Thus, one needs to show that the analogous sums for $\overline L$ and $\overline{L'}$ are equal to each other.
The generating function for $L$ is given as a sum over the elements in the cyclic group generated by $g$ (see \cite[Thm 2.5]{Ik88}), which can be split into a subsum over $g^k$ with $(k,q)=1$ plus a subsum over $g^k$ with $(k,q)=r$ plus a term corresponding to the identity element (i.e.\ $k=0$) and similarly for the generating function for $L'$, with $g'$ in place of $g$. As asserted, the total sums are equal to each other for $L$ and $L'$.
It turns out that to prove the assertion for $\overline L$ and $\overline{L'}$ it suffices to show that the subsums just mentioned are equal to each other for $L$ and $L'$ (the contribution for $k=0$ is the same in both cases). But it is not hard to show that this is true for the second subsums (hence also for the first ones) for the lens spaces corresponding to the lattices in Theorem~\ref{thm4:isolattices}, by taking into account that both lattices are of the form $\mathcal L(q;s_1,\ldots, s_n)$ with $s_i\equiv\pm 1 \pmod q$. This concludes the proof.
\end{proof}
\begin{rem}
In Section~\ref{sec:finiteness} we have seen that the finite set of $N_{\mathcal L}^{\mathrm{red}}(k,\zz )$ determines whether two $q$-congruence lattices are $\norma{\cdot}^*$-isospectral.
Moreover, we point out that these numbers also determine explicitly each individual $p$-spectrum of a lens space $L=\Gamma\ba S^{2m-1}$ for $0\le p \le n=2m-1$.
Indeed, by Proposition~\ref{prop2:p-spectrum}, the multiplicities in the $p$-spectrum of $L$ depend only on the numbers $\dim V_{\pi_{k,p}}^\Gamma$ and $\dim V_{\pi_{k,p+1}}^\Gamma$ which, by expression \eqref{eq3:dim V_k,p^Gamma}, are determined by the $N_{\mathcal L}(k,\zz )$ which, in turn, can be computed by using equation \eqref{eq4:N^*-N^red} if we know the numbers $N_{\mathcal L}^{\mathrm{red}}(k,\zz )$.
\end{rem}
\begin{rem}\label{rem7:orbifolds}
If the discrete subgroup $\Gamma$ of $\SO(n+1)$ acts possibly with fixed points on $S^n$, then $\Gamma\ba S^n$ is a good orbifold.
For instance, in our case, if we take $L(q;s_1,\dots,s_m)$ as in \eqref{eq3:L(q;s)} with $s_1,\dots,s_m$ not necessarily coprime to $q$ and $\gcd(q,s_1,\dots,s_m)=1$, we obtain an \emph{orbifold lens space}.
See \cite{Shams} for an extension of Ikeda's result to orbifold lens spaces.
Most of the results in this paper also work for orbifold lens spaces.
For instance, the determination of the $p$-spectrum in Theorem~\ref{thm3:dim V_k,p^Gamma} via Proposition~\ref{prop2:p-spectrum} and the characterizations in Theorem~\ref{thm3:characterization} between lens spaces and congruence lattices.
Furthermore, Section~\ref{sec:finiteness} also works for congruence lattices $\mathcal L(q;s_1, \dots,s_m)$ without the assumption that the $s_j$ are coprime to $q$.
Proposition~\ref{prop7:lens-non-strongly} is also valid in this context; that is, strongly isospectral orbifold lens spaces are necessarily isometric.
\end{rem}
We now show that the lens spaces constructed in Section~\ref{sec:families} are homotopically equivalent to each other.
We note that they cannot be simply homotopically equivalent (see \cite[\S31]{Co}) since in this case they would be homeomorphic.
\begin{lemma} \label{lem:homotequiv}
The lens spaces $L(r^2t;1, 1+rt, 1+3rt)$, $L(r^2t;1, 1-rt, 1-3rt)$, $r\not\equiv 0\pmod 3$, associated to the congruence lattices in Theorem~\ref{thm4:isolattices} are homotopically equivalent to each other.
\end{lemma}
\begin{proof}
We have seen that $L=L(q; \theta^0, \theta^1, \theta^{3})$ and $L'=L(q; \theta^0, \theta^{-1}, \theta^{-3})$, where $\theta =1+rt$.
The condition for homotopy equivalence of $L$ and $L'$ (see \cite[(29.6)]{Co}) is that $\pm \theta^{8}\equiv d^3 \pmod {r^2t}$, for some $d \in\Z$.
We claim that $r$ divides $\phi(r^2t)$. Indeed, we can write $q= \prod_j p_j^{2v_{p_j}(r)+ v_{p_j}(t)}$ a product over primes $p_j$. We have
$$ \phi(q)= r \prod_j p_j^{v_{p_j}(r)+ v_{p_j}(t)-1}(p_j-1).$$
Furthermore, this implies that if $r$ is odd then $2r$ divides $\phi(r^2t)$.
We first assume that $rt$ is odd. Then $H:=\Z_{r^2t}^\times$ is a cyclic group of order $\phi(r^2t)$.
Thus, if $\omega$ is a generator of this group, then $\omega ^{\phi(r^2t)/2r}$ has order $2r$. Hence, since $H$ is cyclic, $\theta = \omega^{\pm h}$ for some $h=j\phi(r^2t)/2r$ with $(j,2r)=1$.
Now, if $3$ divides $\phi(r^2t)$, since $(3,2r)=1$, then $\theta = \left(\omega^\frac{\phi(r^2t)j}{2r3}\right)^3$, as asserted. If $3$ does not divide $\phi(r^2t)$, then the map $x\mapsto x^3$ is surjective, hence $\theta$ is again in the image.
This proves the assertion for $rt$ odd.
In case $rt$ is even, then $\Z_{r^2t}^\times$ is a cyclic group $H$ times an abelian 2-group $K$.
Again the map $x\mapsto x^3$ in $K$ is surjective.
By a similar argument as before we show that $\theta$ is in the image of $x\mapsto x^3$ in $H$.
\end{proof}
\begin{rem}
In \cite{DR}, P.~Doyle and the third named author showed examples of disconnected flat orbifolds in dimension two that are $p$-isospectral for every $p$ but are not strongly isospectral.
\end{rem}
\section{$\tau$-isospectrality}\label{sec:tau-isospectrality}
In this section we give complementary spectral information, showing in a direct way the non-strong isospectrality of the pairs in Theorem~\ref{thm7:all-p-iso-lens}.
To make the computations easier, we consider in the sequel only the simplest pair
\begin{align*}
L&=L(49;1,6,15),\\
L'&=L(49;1,6,20)
\end{align*}
of non-isometric lens spaces $p$-isospectral for all $p$.
This pair is associated with the first row in Table~\ref{table:m=3} and it is isometric to the first pair in the family in Theorem~\ref{thm7:all-p-iso-lens} (see \eqref{eq6:ex-m=3-q=49}).
We denote by $\Gamma$ and $\Gamma'$ the finite cyclic subgroups of the torus $T\subset SO(6)$ of order $q=49$ such that $L=\Gamma\ba S^{5}$ and $L'=\Gamma'\ba S^{5}$.
We write $G=\SO(2m)$ and $K=\SO(2m-1)$ as in Section~\ref{sec:prelim}.
Any representation $\tau$ of $K$ induces a strongly elliptic natural operator $\Delta_{\tau,\Gamma}$ on the smooth sections of a natural bundle on a spherical space form $\Gamma\ba S^{2m-1}$.
By using representation theory, we will exhibit many choices of representations $\tau$ of $K$ such that $L$ and $L'$ are not $\tau$-isospectral.
\begin{lemma}\label{prop7:no_tau-iso}
The lens spaces $L=L(49;1,6,15)$ and $L'=L(49;1,6,20)$ are not $\tau$-isospectral for every irreducible representation $\tau$ of $\SO(5)$ with highest weight of the form $b_1\varepsilon_1 + b_2\varepsilon_2$ where
\begin{equation}\label{eq7:tau-weights}
4\geq b_1\geq 3\geq b_2\geq0.
\end{equation}
\end{lemma}
\begin{proof}
We choose $\Lambda_0=4\varepsilon_1+3\varepsilon_2$ and we let $\pi_{\Lambda_0}$ be the irreducible representation of $G$ with highest weight $\Lambda_0$.
The Casimir element $C$ acts on $\pi_{\Lambda_0}$ by $\lambda_0=\lambda(C,\pi_{\Lambda_0} ) = \langle \Lambda_0+\rho,\Lambda_0+\rho\rangle-\langle \rho,\rho\rangle = (6^2+4^2)-(2^2+1^2)=47$.
By \eqref{eq2:mult_lambda}, the multiplicity $d_{\lambda_0}(\tau,\Gamma)$ of the eigenvalue $\lambda_0$ of $\Delta_{\tau,\Gamma}$ is
\begin{equation}\label{eq:dlambda}
d_{\lambda_0}(\tau,\Gamma)=\sum_\pi \dim V_\pi^{\Gamma} [\tau:\pi],
\end{equation}
where the sum is over the irreducible representations $\pi$ of $\SO(6)$ such that $\lambda(C,\pi)=\lambda_0=47$.
A similar expression is valid for $d_{\lambda_0}(\tau,\Gamma')$.
Now if $\pi$ has highest weight $\Lambda=a_1\varepsilon_1+a_2\varepsilon_2+a_3\varepsilon_3$ ($a_i\in\Z$ and $a_1\geq a_2\geq|a_3|$) and $\lambda(C,\pi)=\lambda_0$, then we have $(a_1+2)^2+(a_2+1)^2+a_3^2=\langle\Lambda+\rho, \Lambda+\rho\rangle = \langle\Lambda_0+\rho, \Lambda_0+\rho\rangle =52.$
By taking congruence modulo $4$, we see that the numbers $a_1+2>a_2+1>a_3$ are even.
Hence
$
\left(\frac{a_1+2}{2}\right)^2 + \left(\frac{a_2+1}{2}\right)^2 + \left(\frac{a_3}{2}\right)^2=13.
$
It is easy to check that this implies that $a_1=4$, $a_2=3$ and $a_3=0$, therefore $\Lambda=\Lambda_0$ and hence, there is only one irreducible representation $\pi$ of $G$ with $\lambda(C,\pi)=47$, namely $\pi =\pi_{\Lambda_0}$.
On the other hand, by Lemma~\ref{lem3:L_Gamma} we have that $\dim V_{\pi_0}^{\Gamma} = \sum_{\mu\in \mathcal L} m_{\pi_0}(\mu)$, where $\mathcal L$ is the associated congruence lattice given by \eqref{eq3:Lambda(q;s)} and similarly for $\mathcal L'$.
We compute by using Sage~\cite{Sage} the weights of $\pi_0$ (i.e.\ $\mu\in\Z^m$ such that $m_{\pi_0}(\mu)>0$) and their respective multiplicities:
$$
\begin{array}{ll@{\quad\qquad}ll@{\qquad}ll}
4\varepsilon_1 + 3\varepsilon_2 & 1 &
4\varepsilon_1 + 2\varepsilon_2 \pm \varepsilon_3 & 1 &
3\varepsilon_1 + 3\varepsilon_2 \pm \varepsilon_3 & 2 \\
3\varepsilon_1 + 2\varepsilon_2 \pm 2\varepsilon_3 & 2 &
4\varepsilon_1 + \varepsilon_2 & 1 &
3\varepsilon_1 + 2\varepsilon_2 & 4 \\
3\varepsilon_1 + \varepsilon_2 \pm \varepsilon_3 & 4 &
2\varepsilon_1 + 2\varepsilon_2 \pm \varepsilon_3 & 5 &
3\varepsilon_1 & 4 \\
2\varepsilon_1 + \varepsilon_2 & 9 &
\varepsilon_1 + \varepsilon_2 \pm \varepsilon_3 &12 &
\varepsilon_1 &16
\end{array}
$$
(only the dominant weights are shown, since weights in the same Weyl group orbit have the same multiplicity).
A weight $\mu=\sum_i a_i\varepsilon_i$ is in $\mathcal L$ (resp.\ $\mathcal L'$) if and only if
$$
a_1+6a_2+15a_3\equiv0\pmod{49}\quad (\text{resp.\ } a_1+6a_2+20a_3\equiv0\pmod{49}).
$$
With computer aid one checks that the weights of $\pi_{\Lambda_0}$ that satisfy these congruences (i.e lying in $\mathcal L$ (resp.\ $\mathcal L'$) are $\pm(4\varepsilon_1+3\varepsilon_3)$, $\pm(-\varepsilon_1+2\varepsilon_2-4\varepsilon_3)$ and $\pm(3\varepsilon_1-3\varepsilon_2+\varepsilon_3)$ (resp.\ $\pm(-3\varepsilon_2-4\varepsilon_3)$ and $\pm(-3\varepsilon_1+2\varepsilon_2+2\varepsilon_3)$).
Taking into account their multiplicities we obtain that $\dim V_{\pi_{\Lambda_0}}^{\Gamma} = 2+2+4=8$ and $\dim V_{\pi_{\Lambda_0}}^{\Gamma'} = 2+4=6$.
By applying the branching law from $\SO(6)$ to $\SO(5)$ (see for instance \cite[Thm.~8.1.4]{GW}) , for every irreducible representation $\tau$ of $\SO(5)$ with highest weight of the form $b_1\varepsilon_1 + b_2\varepsilon_2$ with $ 4\geq b_1\geq 3\geq b_2\geq 0$ one has that $[\tau,{\pi_{\Lambda_0}}_{|K}] =1$.
Thus, we obtain from \eqref{eq:dlambda} that $$d_{\lambda_0}(\tau,\Gamma)=\dim V_{\pi_0}^{\Gamma}[\tau:\pi_0]=8, \qquad d_{\lambda_0}(\tau,\Gamma')=6. $$
Thus, $L$ and $L'$ cannot be $\tau$-isospectral for any $\tau$ as in the statement.
\end{proof}
\begin{rem}\label{rem7:strongly_tau-iso}
We note that the assertion in Lemma~\ref{prop7:no_tau-iso} followed by comparing the multiplicities of $\lambda=\lambda(C,\pi)$ for only one choice of $\pi\in\widehat G$ satisfying $\dim V_{\pi}^\Gamma \neq \dim V_{\pi}^{\Gamma'}$.
By computer methods using Sage~\cite{Sage}, we have checked that there are many different choices of $\pi$ that allow to find many other $K$-types $\tau$ such that the lens spaces $L$ and $L'$ are not $\tau$-isospectral.
\end{rem}
We believe that there are only finitely many irreducible representations $\tau$ of $\SO(5)$ such that $L$ and $L'$ are $\tau$-isospectral.
\bigskip
\noindent\emph{Acknowledgement.}
The authors wish to thank Peter G.~Doyle for very stimulating conversations on the subject of this paper.
The first named author wishes to thank the support of the Oberwolfach Leibniz Fellows programme (Germany) in May--July 2013 and in August--November 2014.
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\section{Some Formulae of Taubes}\label{AppE}
In this appendix, we summarize some formulae of the Seiberg-Witten equations from Taubes' paper \cite[Section 2]{Taubes96}. Although the primary applications of \cite{Taubes96} focus on symplectic 4-manifolds, it is well-known that some of them generalize to any Riemannian 4-manifolds. This observation forms the basis of the finiteness result in Section \ref{Sec25}. For the sake of completeness, we record their statements and prove Lemma \ref{L25.3}.
\smallskip
Given an oriented Riemannian 4-manifold $X$, consider the Seiberg-Witten equations on $X$ perturbed by a self-dual 2-form $\omega^+\in \Gamma(X, i\Lambda^+ X)$:
\begin{align}
\half \rho_4(F_{A^t}^+)-(\Phi\Phi^*)_0&=\rho_4(\omega^+),\label{EE.1}\\
D_A^+\Phi&=0. \label{EE.2}
\end{align}
The 2-form $\omega^+$ is not assumed to be closed, and $X$ is not necessarily compact. Set
\[
F\colonequals \half F_{A^t}\in \Omega^2(X, i\R).
\]
Then the curvature tensor $F_{A}|_{S^+}\in \Gamma(X, i\Lambda^2X\otimes\End(S^+))$ can be written as
\begin{equation}\label{EE.3}
F_A|_{S^+}=F\otimes\Id_{S^+}+\SU,
\end{equation}
where $\SU$ is the traceless part of $F_A|_{S^+}$ and is independent of the \spinc connection $A$. By the Weitzenb\"{o}ck formula \cite[(4.14)]{Bible}, if $(A,\Phi)$ solves \eqref{EE.1}\eqref{EE.2}, then
\begin{equation}\label{EE.4}
\half \Delta_A\Phi+\half |\Phi|^2\Phi=-\rho_4(\omega^+)\Phi-\frac{s}{4}\Phi.
\end{equation}
Our goal is to find explicit formulae for $
d^*F,\ \Delta |F|^2 \text{ and } \Delta |\nabla_A \Phi|^2. $
\begin{lemma}\label{LE.1} For any solution $(A,\Phi)$ to the perturbed Seiberg-Witten equations \eqref{EE.1}\eqref{EE.2}, we have \[
d^*F=2d^*F^+=2d^*F^-=i\im \langle \Phi,\nabla_A\Phi\rangle+2d^*\omega^+.
\]
\end{lemma}
\begin{proof}
Since $F$ is a closed 2-form, $
dF^-=-dF^+$ and $d^*F^-=d^*F^+.
$
To compute $d^*F^+$, we pick a local orthonormal framing $\{e_i\}_{1\leq i\leq 4}$ such that $\nabla_{e_i}e_k=0$ at $x\in X$. Moreover, we exploit the formula from \cite[Lemma 5.13]{Spin}:
\[
d^*F^+=-(\nabla_{e_i} F^+)(e_i,\cdot).
\]
By the curvature equation \eqref{EE.1}, we have
\begin{align*}
(F^+-\omega^+)(e_i, e_k)&=-\frac{1}{4}\tr(\rho_4(F^+-\omega^+)\rho_4(e_i)\rho_4(e_k))=-\frac{1}{4}\tr((\Phi\Phi^*)_0\rho_4(e_i)\rho_4(e_k)).
\end{align*}
Now we use the Dirac equation \eqref{EE.2} to compute:
\begin{align*}
I\colonequals &-\nabla_{e_i}(F^+-\omega^+)(e_i, e_k)\\
=&\frac{1}{4}\tr([(\nabla_{e_i}^A\Phi)\Phi^*+\Phi(\nabla_{e_i}^A\Phi)^*-\re\langle \Phi, \nabla_{e_i}^A\Phi\rangle \otimes\Id_{S^+}]\rho_4(e_i)\rho_4(e_k)))\\
=&\frac{1}{4}\tr(\rho_4(e_i)\rho_4(e_k)(\nabla_{e_i}^A\Phi)\Phi^*)+\frac{1}{4}\tr(\Phi\underbrace{(\nabla_{e_i}^A\Phi)^*\rho_4(e_i)}_{=0}\rho_4(e_k))+\half \re\langle \Phi,\nabla_{e_k}^A\Phi\rangle. \\
\end{align*}
For the first term, we commute $\rho_4(e_i)$ and $\rho_4(e_k)$ to derive:
\begin{align*}
I=&-\frac{1}{4}\tr(\rho_4(e_k)\underbrace{\rho_4(e_k)(\nabla_{e_i}^A\Phi)}_{=0}\Phi^*)+\half\tr(\rho_4(e_k)\rho_4(e_k)(\nabla_{e_i}^A\Phi)\Phi^*)+ \half\re\langle \Phi,\nabla_{e_k}^A\Phi\rangle\\
=&-\half \langle \nabla_{e_k}^A\Phi, \Phi\rangle+\half \re\langle \Phi,\nabla_{e_k}^A\Phi\rangle=\half\cdot i\im \langle \Phi, \nabla_{e_k}^A\Phi\rangle.
\end{align*}
We conclude that $2d^*F^+=2d^*\omega^++i\im \langle \Phi,\nabla_{e_k}^A\Phi\rangle\cdot \omega_k$, where $\{\omega_i\}_{1\leq i\leq 4}$ are co-vectors dual to $\{e_i\}$.
\end{proof}
Now we are read to compute the Hodge Laplacian of the curvature 2-form $F$.
\begin{proposition}
For any solution $(A,\Phi)$ to the perturbed Seiberg-Witten equations \eqref{EE.1}\eqref{EE.2}, we have
\[
(d+d^*)^2 F+|\Phi|^2 F=\langle \nabla_A\Phi\wedge \nabla_A\Phi\rangle+2dd^*\omega^++I(\Phi,\Phi),
\]
where $\langle \nabla_A\Phi\wedge \nabla_A\Phi\rangle$ denotes the imaginary valued 2-form
\[
\sum_{i,j} \omega_i\wedge \omega_j\cdot \langle \nabla_{e_i}^A\Phi, \nabla_{e_j}^A\Phi\rangle=2i\sum_{i<j}\omega_i\wedge \omega_j \cdot \im \langle \nabla_{e_i}^A\Phi, \nabla_{e_j}^A\Phi\rangle,
\]
and $I(\Phi,\Phi)=\sum_{i<k}\omega_i\wedge \omega_k\cdot i\im \langle \Phi, \SU(e_i, e_k)\Phi\rangle$ is a symmetric bilinear form.
\end{proposition}
\begin{proof}Since $dF=0$, it suffices to compute $dd^*F$. We exploit the formula from \cite[Lemma 5.13]{Spin}:
\[
d\nu=\omega_i\wedge \nabla_{e_i}\nu
\]
for any $\nu\in \Omega^*(X,i\R)$.
In particular, set $\nu=i\im\langle \Phi,\nabla_A\Phi\rangle$:
\begin{align*}
d\nu&=\omega_i\wedge \nabla_{e_i}(\omega_k\otimes i\im\langle \Phi,\nabla_{e_k}\Phi\rangle)=\omega_i\wedge \omega_k\cdot (i\im\langle \nabla_{e_i}^A\Phi,\nabla_{e_k}^A\Phi\rangle+i\im\langle \Phi,\nabla_{e_i}^A\nabla_{e_k}^A\Phi\rangle)\\
&=\sum_{i<k}\omega_i\wedge \omega_k\cdot (2i\im\langle \nabla_{e_i}^A\Phi,\nabla_{e_k}^A\Phi\rangle+i\im\langle \Phi, F_A(e_i,e_k)\Phi\rangle)\\
&=\langle \nabla_A\Phi\wedge \nabla_A\Phi\rangle-|\Phi|^2 F+I(\Phi, \Phi).
\end{align*}
At the last step, we used the decomposition
$
F_A|_{S^+}=\half F_{A^t}\otimes \Id_{S^+}+\SU
$ from \eqref{EE.3}.
\end{proof}
Finally, we address the Laplacian of $|\nabla_A\Phi|^2$. Note that
\[
\half \Delta|\nabla_A\Phi|^2+|\Hess_A\Phi|^2=\re\langle (\nabla_A^*\nabla_A)\nabla_A\Phi,\nabla_A\Phi\rangle.
\]
We start with an explicit formula for the commutator $\Delta_A\nabla_A-\nabla_A\Delta_A$ where $\Delta_A\colonequals \nabla_A^*\nabla_A$.
\begin{lemma}\label{LE.3} For any \spinc connection $A$ and any spinor $\Phi$, we have an identity:
\begin{align*}
\langle \Delta_A\nabla_A\Phi,\nabla_A\Phi\rangle&=\langle \nabla_A(\Delta_A\Phi),\nabla_A\Phi\rangle-\Ric(e_i, e_j)\re\langle \nabla_{e_i}\Phi,\nabla_{e_j}\Phi\rangle\\
&\qquad+\re\langle (d^*_AF_A)\Phi, \nabla_A\Phi\rangle-2\re\langle F_A(e_i,e_j)\nabla_{e_i}\Phi,\nabla_{e_j}\Phi\rangle.
\end{align*}
\end{lemma}
\begin{remark}\label{RE.4} The last two terms can be recognized as follows:
\begin{align*}
d_A^*F_A&=d^*F\otimes\Id_{S^+}+d^*_{A_0}\SU,\\
\re\langle F_A(e_i,e_j)\nabla_{e_i}\Phi,\nabla_{e_j}\Phi\rangle&=\re\langle \SU(e_i,e_j)\nabla_{e_i}\Phi,\nabla_{e_j}\Phi\rangle+2\sum_{i<j} F(e_i, e_j)\cdot i\im\langle \nabla_{e_i}\Phi,\nabla_{e_j}\Phi\rangle\\
&=\re\langle \SU(e_i,e_j)\nabla_{e_i}\Phi,\nabla_{e_j}\Phi\rangle-\langle F, \langle\nabla_A\Phi\wedge\nabla_A\Phi\rangle\rangle.
\end{align*}
Note that $d^*_{A_0}\SU$ is independent of the reference \spinc connection $A_0$.
\end{remark}
\begin{proof}[Proof of Lemma \ref{LE.3}] Let $\{e_i=\partial_i\}$ in a normal coordinate at $x\in X$ and $\{\omega_i\}_{1\leq i\leq 4}$ be the co-vectors dual to $\{e_i\}$. Then
\begin{align*}
\nabla_A\Phi&=\omega_i\otimes\nabla_{e_i}\Phi,\\
\Delta_A\nabla_A\Phi&=-\omega_i\otimes \nabla_{e_j}\nabla_{e_j}\nabla_{e_i}\Phi-\nabla_{e_j}\nabla_{e_j}\omega_i\otimes\nabla_{e_i}\Phi\\
&=-\omega_i\otimes \nabla_{e_i}\nabla_{e_j}\nabla_{e_j}\Phi-2\omega_i\otimes F_A(e_j, e_i)\nabla_{e_j}\Phi\\
&\qquad+\omega_i\otimes (d^*_AF_A)(e_i)\Phi-\nabla_{e_j}\nabla_{e_j}\omega_i\otimes\nabla_{e_i}\Phi,\\
\Delta_A\Phi&=-\nabla_{e_j}\nabla_{e_j}\Phi-\langle \nabla_{e_j}\omega_i,\omega_j\rangle \nabla_{e_i}\Phi,\\
\nabla_A \Delta_A\Phi&=-\omega_i\otimes \nabla_{e_i}\nabla_{e_j}\nabla_{e_j}\Phi-\omega_k\otimes \langle\nabla_{e_k} \nabla_{e_j}\omega_i,\omega_j\rangle \nabla_{e_i}\Phi,
\end{align*}
Now take inner products with $\nabla_A\Phi$. To find the Ricci curvature, use relations:
\begin{align*}
\langle \nabla_{e_j}\nabla_{e_j}\omega_i, \omega_k\rangle&=-\langle\nabla_{e_j}\nabla_{e_j} e_k, e_i\rangle,& \langle \nabla_{e_k}\nabla_{e_j}\omega_i, \omega_j\rangle&=-\langle\nabla_{e_k}\nabla_{e_j} e_j, e_i\rangle,
\end{align*}
and $\nabla_{e_j}e_k=\nabla_{e_k}e_j$ in a normal neighborhood.
\end{proof}
\begin{proposition}\label{PB.5} For any solution $(A,\Phi)$ to the perturbed Seiberg-Witten equations, we have
\begin{align*}
&\half \Delta |\nabla_A\Phi|^2+|\Hess_A\Phi|^2+\half |\Phi|^2|\nabla_A\Phi|^2+|\langle \nabla_A\Phi, \Phi\rangle|^2+\frac{s}{4}|\nabla_A\Phi|^2\\
=&2\langle F,\langle\nabla_A\Phi\wedge \nabla_A\Phi\rangle\rangle+J_1+J_2
\end{align*}
where
\begin{align*}
J_1&=-2\langle \SU(e_i, e_j)\nabla_{e_i}\Phi,\nabla_{e_j}\Phi\rangle-\Ric(e_i, e_j)\re\langle \nabla_{e_i}\Phi,\nabla_{e_j}\Phi\rangle-\re\langle \rho_4(\omega^+)\nabla_A\Phi,\nabla_A\Phi\rangle,\\
J_2&=\re\langle (d_{A_0}^*\SU)\Phi,\nabla_A\Phi\rangle+\re\langle 2d^*\omega^+\otimes \Phi, \nabla_A\Phi\rangle\\
&\hspace{1.527in}-\re\langle \rho_4(\nabla\omega^+)\Phi,\nabla_A\Phi\rangle-\frac{1}{4}\re\langle ds\otimes \Phi, \nabla_A\Phi\rangle.
\end{align*}
In particular, $|J_1|\leq C|\nabla_A\Phi|^2$ and $|J_2|\leq C|\nabla_A\Phi||\Phi|$ for some function $C:X\to \R_{>0}$ depending only on $(g_X,\omega^+)$.
\end{proposition}
\begin{proof} By Lemma \ref{LE.3} and Remark \ref{RE.4}, it suffices to compute
\[
\re\langle (d^*F)\Phi,\nabla_A\Phi\rangle \text{ and }\re\langle \nabla_A\Delta_A\Phi, \nabla_A\Phi\rangle.
\]
For the first term, we apply Lemma \ref{LE.1}:
\begin{align*}
\re\langle (d^*F)\Phi,\nabla_A\Phi\rangle&=-|\im \langle\Phi,\nabla_A\Phi\rangle|^2+\re\langle 2d^*\omega^+\otimes \Phi, \nabla_A\Phi\rangle.
\end{align*}
For the second term, we apply $\Delta_A$ to \eqref{EE.4} to compute
\begin{align*}
\nabla_A\Delta_A\Phi+\half |\Phi|^2\nabla_A\Phi+\re\langle\Phi,\nabla_A\Phi\rangle \Phi=\rho_4(\omega^+)\nabla_A\Phi-\rho_4(\nabla\omega^+)\Phi-\frac{s}{4}\nabla_A\Phi-\frac{1}{4}ds\otimes\Phi.
\end{align*}
To conclude, take the inner product with $\nabla_A\Phi$:
\begin{align*}
&\re\langle \nabla_A\Delta_A\Phi, \nabla_A\Phi\rangle+\half |\Phi|^2|\nabla_A\Phi|^2+|\re\langle\Phi,\nabla_A\Phi\rangle|^2\\
=&-\re\langle \rho_4(\omega^+)\nabla_A\Phi,\nabla_A\Phi\rangle-\re\langle \rho_4(\nabla\omega^+)\Phi,\nabla_A\Phi\rangle-\frac{s}{4}|\nabla_A\Phi|^2-\frac{1}{4}\re\langle ds\otimes \Phi, \nabla_A\Phi\rangle.\qedhere
\end{align*}
\end{proof}
Finally, let us state the corresponding results for 3-manifolds from which one can easily deduce Lemma \ref{L25.3}.
\begin{proposition}\label{PE.6} Let $(Y,g_Y)$ be any Riemannian 3-manifold, $\Rm_Y$ be the curvature tensor and $\omega\in \Omega^2(Y, i\R)$ be a closed 2-form. For any solution $(B,\Psi)$ to the 3-dimensional Seiberg-Witten equations \eqref{3DSWEQ}, write
\[
F=\half F_{B^t}\in \Omega^2(Y, i\R).
\]
Then we have
\[
d^*F=i\im\langle \Psi,\nabla_B\Psi\rangle+d^*\omega,
\]
and
\begin{align*}
\half \Delta |\nabla_B\Psi|^2+&|\Hess_B\Psi|^2+\half |\Phi|^2|\nabla_B\Psi|^2+|\langle \nabla_B\Psi, \Psi\rangle|^2\\
=&2\langle F,\langle\nabla_B\Psi\wedge \nabla_B\Psi\rangle\rangle+J_1(\nabla_B\Psi,\nabla_B\Psi)+J_2(\Psi,\nabla_B\Psi).
\end{align*}
where $J_1$ and $J_2$ are certain bilinear maps depending only on $R_Y$, $\omega$ and their first derivatives. In particular, if
\[
\|\Rm_Y\|_{L^\infty_1}, \|\omega\|_{L^\infty_1}<\infty,
\]
then
$$|J_1(\nabla_B\Psi,\nabla_B\Psi)|\leq C|\nabla_B\Psi|^2\text{ and }|J_2(\nabla_B\Psi,\Psi)|\leq C|\nabla_B\Psi||\Psi|,$$
for some constant $C>0$.
\end{proposition}
Proposition \ref{PE.6} follows from its 4-dimensional analogue: Lemma \ref{LE.1} and Proposition \ref{PB.5}.
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\begin{document}
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{\Large \bf
\begin{center}
Generalized Auslander-Reiten conjecture and derived equivalences
\end{center}}
\medskip
\centerline{{\bf Shengyong Pan}}
\begin{center} Department of Mathematics,\\
Beijing Jiaotong University, Beijing 100044,\\
People's Republic of China\\ E-mail:shypan@bjtu.edu.cn \\
\end{center}
\bigskip
\renewcommand{\thefootnote}{\alph{footnote}}
\setcounter{footnote}{-1} \footnote{2000 Mathematics Subject
Classification: 18E30,16G10;16S10,18G15.}
\renewcommand{\thefootnote}{\alph{footnote}}
\setcounter{footnote}{-1} \footnote{Keywords: generalized
Auslander-Reiten conjecture, derived equivalences.}
\begin{abstract}
In this note, we prove that the generalized Auslander-Reiten
conjecture is preserved under derived equivalences between Artin
algebras.
\end{abstract}
\section{Introduction}
In the representation theory of Artin algebras, one of the most
important open problems is the Nakayama conjecture which predicts
that an Artin algebra $A$ is self-injective provided that all terms
in a minimal injective resolution of $A$ are projective. Mueller
\cite{M} in the late sixties proved that the Nakayama conjecture
holds for every Artin algebra if and only if for any Artin algebra
$A$ any finitely generated generator-cogenerator $M$, the vanishing
$\Ext^{n}_A(M,M)=0$, for $n\geq1$, implies that $M$ is projective.
In this connection and based on Mueller's result, Auslander-Reiten
proposed several stronger conjectures, and in particular the
following:
(ARC) \quad Let $M$ be a finitely generated module over an Artin
algebra $A$ such that $\Ext_A^i(M,M)=0=\Ext_A^i(M,A), $ for
$i\geq1$. Then $M$ is projective.
The above conjecture, widely known as {\it the Auslander-Reiten
Conjecture}, implies the Nakayama conjecture, and, as
Auslander-Reiten proved, it holds in all cases where the Nakayama
conjecture is known to be true. It should be noted that the above
conjectures, which are trivial consequences of the finitistic
dimension conjecture, are still open. Auslander-Reiten conjecture
has been verified for some special classes of Artin algebras and
commutative Noetherian rings \cite{CH,W1,W3}.
In this note, we consider the following generalization version of
Auslander and Reiten conjecture which can be stated as follows:
(GARC) \quad Let $A$ be an Artin algebra. Let $X$ be a finitely
generated $A$-module and $r$ a non-negative integer. If
$$
\Ext_A^i(M,M)=0=\Ext_A^i(M,A),
$$
for $i> r$, then $\pd(X)\leq r$, where $\pd(X)$ is the projective
dimension of $X$.
In case $r=0$, (GARC) is (ARC). In \cite {W1}, the generalized
Auslander-Reiten conjecture holds for Artin algebras for which any
finitely generated module has an ultimately closed projective
resolution. It also holds for all algebras which satisfy the
Auslander-Reiten conjecture. In \cite{W2}, it was proved that the
generalized Auslander-Reiten conjecture is stable under tilting
equivalences.
The aim of this note is to show that the generalized Auslander
Reiten conjecture is preserved by derived equivalences, as it was
done for the finiteness of finitistic dimension conjecture
\cite{PX}. Then we generalize the main result of \cite{W2},
answering in the affirmative a question of Wei.
Our main result reads as follows:
\begin{Theo}
Suppose that $A$ and $B$ are Artin algebras. Assume that $A$ and $B$
are derived equivalent. Then $A$ satisfies the generalized
Auslander-Reiten conjecture if and only if so does $B$. \label{thm1}
\end{Theo}
In view of the importance of the Nakayama conjecture and the
(Generalized) Auslander-Reiten conjecture, it is highly desirable to
have as much as possible information about classes of algebras
satisfying the conjectures. The main result indicates that the
validity of the generalized Auslander-Reiten conjecture for an Artin
algebra depends on its derived equivalence class and in this way one
produces further classes of algebras satisfying the conjecture.
The contents of this paper are organized as follows. In Section
\ref{pre}, we recall some definitions and notations on derived
categories and derived equivalences. In Section \ref{i}, we prove
our main result, Theorem \ref{thm1}.\\
\noindent{\bf Acknowledgements.} The author would like to
acknowledge the Fundamental Research Funds for the Center University
(2011JBM131) and postdoctoral granted financial support from China
Postdoctoral Science Foundation (20100480188), during which this
work was carried out. The author also would like to thank the
referee for his/her helpful comments that improve the paper.
\section{Preliminaries}\label{pre}
In this section, we shall recall some definitions and notations on
derived categories and derived equivalences, and basic results which
are needed in the proofs of our main results.
Let $\mathscr{A}$ be an abelian category. For two morphisms $\alpha:
X\ra Y$ and $\beta: Y\ra Z$, their composition is denoted by
$\alpha\beta$. An object $X\in\mathscr{A}$ is called a additive
generator for $\mathscr{A}$ if $\add(X)=\mathscr{A}$, where
$\add(X)$ is the additive subcategory of $\mathscr{A}$ consisting of
all direct summands of finite direct sums of the copies of $X$. A
complex $\cpx{X}=(X^i,d_{X}^i)$ over $\mathscr{A}$ is a sequence of
objects $X^i$ and morphisms $d_{X}^i$ in $\mathscr{A}$ of the form:
$\cdots \ra X^i\stackrel{d^i}\ra X^{i+1}\stackrel{d^{i+1}}\ra
X^{i+2}\ra\cdots$, such that $d^id^{i+1}=0$ for all
$i\in\mathbb{Z}$. If $\cpx{X}=(X^i,d_{X}^i)$ and
$\cpx{Y}=(Y^i,d_{Y}^i)$ are two complexes, then a morphism $\cpx{f}:
\cpx{X}\ra\cpx{Y}$ is a sequence of morphisms $f^i: X^i\ra Y^i$ of
$\mathscr{A}$ such that $d^i_{X}f^{i+1}=f^id^i_{Y}$ for all
$i\in\mathbb{Z}$. The map $\cpx{f}$ is called a chain map between
$\cpx{X}$ and $\cpx{Y}$. The category of complexes over
$\mathscr{A}$ with chain maps is denoted by $\C{\mathscr{A}}$. The
homotopy category of complexes over $\mathscr{A}$ is denoted by
$\K{\mathscr{A}}$ and the derived category of complexes is denoted
by $\D{\mathscr{A}}$.
Let $R$ be a commutative Artin ring, and let $A$ be an Artin
$R$-algebra. We denote by $A$-Mod and $A$-mod the categories of left
$A$-modules and finitely generated left $A$-modules, respectively.
The full subcategories of $A$-Mod and $A$-mod consisting of
projective modules and finitely generated projective modules are
denoted by $A$-Proj and $_A\mathcal {P}$, respectively. Denote by
$_{A}\mathcal {X}^{>r}$ the category of $A$-modules $X$ satisfied
$\Ext_{A}^{i}(X,A)=0$ for $i> r$, where $r$ is a non-negative
integer. Particularly, if an $A$-module $X$ satisfies
$\Ext_{A}^{i}(X,A)=0$ for $i>0$, then $X$ is said to be a
Cohen-Macaulay $A$-module.
Recall that a homomorphism $f: X\ra Y$ of $A$-modules is called a
radical map provided that for any $A$-module $Z$ and homomorphisms
$g: Y\ra Z$ and $h: Z\ra X$, the composition $hfg$ is not an
isomorphism. A complex of $A$-modules is called a radical complex if
its differential maps are radical maps. Let $\Kf{A}$ and $\Kb{A}$
denote the homotopy category of bounded above and bounded complexes
of $A$-modules, respectively. We denote by $\Df{A}$ and $\Db{A}$ the
derived category of bounded above and bounded complexes of
$A$-modules, respectively.
The fundamental theory on derived equivalences has been established
by Rickard \cite{Ri1}.
\begin{Theo}$\rm \cite[Therem\;6.4]{Ri1}$\label{2.1}
Let $A$ and $B$ be rings. The following conditions are equivalent.
$(i)$ $\Db{A\Modcat}$ and $\Db{B\Modcat}$ are equivalent as
triangulated categories.
$(ii)$ $\Kf{\Pmodcat{A}}$ and $\Kf{\Pmodcat{B}}$ are equivalent as
triangulated categories.
$(iii)$ $\Kb{\Pmodcat{A}}$ and $\Kb{\Pmodcat{B}}$ are equivalent as
triangulated categories.
$(iv)$ $\Kb{_{A}\mathcal {P}}$ and $\Kb{_{B}\mathcal {P}}$ are
equivalent as triangulated categories.
$(v)$ $B$ is isomorphic to $\End_{\Db{A}}(\cpx{T})$ for some complex
$\cpx{T}$ in $\Kb{_{A}\mathcal {P}}$ satisfying
\qquad $(1)$ $\Hom_{\Db{A}}(\cpx{T},\cpx{T}[n])=0$
for all $n\neq 0$.
\qquad $(2)$ $\add(\cpx{T})$, the category of direct summands of
finite direct sums of copies of $\cpx{T}$, generates
$\Kb{_{A}\mathcal {P}}$ as a triangulated category.
\end{Theo}
\noindent{\bf Remarks.} (1) The rings $A$ and $B$ are said to be
derived equivalent if $A$ and $B$ satisfy the conditions of the
above theorem. The complex $\cpx{T}$ in Theorem 2.1 is called a
tilting complex for $A$.
(2) By \cite[Corollary 8.3]{Ri1}, two Artin $R$-algebras $A$ and $B$
are said to be derived equivalent if their derived categories
$\Db{A}$ and $\Db{B}$ are equivalent as triangulated categories. By
Theorem 2.1, Artin algebras $A$ and $B$ are derived equivalent if
and only if $B$ is isomorphic to the endomorphism algebra of a
tilting complex $\cpx{T}$. If $\cpx{T}$ is a tilting complex for
$A$, then there is an equivalence $F: \Db{A}\ra\Db{B}$ that sends
$\cpx{T}$ to $B$. On the other hand, for each derived equivalence
$F: \Db{A}\ra\Db{B}$, there is an associated tilting complex
$\cpx{T}$ for $A$ such that $F(\cpx{T})$ is isomorphic to $B$ in
$\Db{B}$.
\section{Generalized Auslander-Reiten conjecture is invariant under derived equivalences}\label{i}
In this section, we shall prove Theorem \ref{thm1}.
We shall give the proof as a sequence of lemmas.
Now we suppose that $A$ and $B$ are Artin algebras. Let $F:
\Db{A}\lra \Db{B}$ be a derived equivalence and let $\cpx{P}$ be the
tilting complex associated to $F$. Without loss of generality, we
assume that $\cpx{P}$ is a radical complex of the following form
$$
0\ra P^{-n}\ra P^{-n+1} \ra \cdots\ra P^{-1}\ra P^{0}\ra 0.
$$
Then we have the following fact.
\begin{Lem} $\rm\cite[lemma\, 2.1]{hx}$
Let $F: \Db{A}\lra \Db{B}$ be a derived equivalence between Artin
algebras $A$ and $B$. Then there is a tilting complex
$\bar{P}^{\bullet}$ for $B$ associated to the quasi-inverse of $F$
of the form
$$
0\ra \bar{P}^{0}\ra \bar{P}^{1} \ra \cdots\ra \bar{P}^{n-1}\ra
\bar{P}^{n}\ra 0,
$$ with the differential being radical maps.
\end{Lem}
Suppose that $\cpx{X}$ is a complex of $A$-modules. We define the
following truncations:
$\tau_{\geq 1}(\cpx{X}): \cdots\ra0\ra0\ra X^{1}\ra X^{2}\ra\cdots$,
$\tau_{\leq 0}(\cpx{X}): \cdots\ra X^{-1}\ra X^{0}\ra 0\ra0\cdots$.
The following lemma, proved in \cite[Lemma\,2.1]{PX}, will be used
frequently in our proofs below.
\begin{Lem}\label{PX} Let $m,t,d \in \mathbb{N}$, $X^{\bullet}, Y^{\bullet} \in
\Kb{A}$. Assume that $X^{i}=0$ for $i<m$, $Y^{j}=0$ for $j>t$, and
$\Ext^{l}(X^{i},Y^{j})=0$ for all $i,j\in \mathbb{N}$ and $l\geq d$.
Then $\Hom_{\Db{A}}(X^{\bullet},Y^{\bullet}[l])=0$ for $l\geq
d+t-m$.
\end{Lem}
The following lemma is inspired by \cite[lemma 3.3]{P}, we have a
variation.
\begin{Lem} Let $F:
\Db{A}\lra \Db{B}$ be a derived equivalence between Artin algebras
$A$ and $B$, and let $G$ be the quasi-inverse of $F$. Suppose that
$\cpx{P}$ and $\bar{P}^{\bullet}$ are the tilting complexes
associated to $F$ and $G$, respectively. Let $r$ be a non-negative
integer. Then
$(i)$ For $X\in _A\mathcal {X}^{>r}$, the complex $F(X)$ is
isomorphic in $\Db{B}$ to a radical complex $\bar{P}^{\bullet}_{X}$
of the form
$$
0\ra \bar{P}_{X}^{0}\ra \bar{P}_{X}^{1} \ra \cdots\ra
\bar{P}_{X}^{n-1}\ra \bar{P}_{X}^{n}\ra 0
$$
with $\bar{P}_{X}^{0}\in_B\mathcal {X}^{>r}$ and $\bar{P}_{X}^{i}$
projective $B$-modules for $1\leq i\leq n$.
$(ii)$ For $Y\in_B\mathcal {X}^{>r}$, the complex $G(Y)$ is
isomorphic in $\Db{A}$ to a radical complex $P^{\bullet}_{Y}$ of the
form
$$
0\ra P_{Y}^{-n}\ra P_{Y}^{-n+1} \ra \cdots\ra P_{Y}^{-1}\ra
P_{Y}^{0}\ra 0
$$
with $P_{Y}^{-n}\in_A\mathcal {X}^{>r}$ and $P_{Y}^{i}$ projective
$A$-modules for $-n+1\leq i\leq 0$.\label{2.3}
\end{Lem}
{\bf Proof}. We only to show the first case. The proof of ($ii$) is
similar to that of ($i$).
($i$) For an $A$-module $X$ with $\Ext_A^i(X,A)=0$ for $i>r$, by
\cite[Lemma 3.1]{hx}, we see that the complex $F(X)$ is isomorphic
in $\Db{B}$ to a complex $\bar{P}^{\bullet}_{X}$ of the form
$$
0\ra \bar{P}_{X}^{0}\ra \bar{P}_{X}^{1} \ra \cdots\ra
\bar{P}_{X}^{n-1}\ra \bar{P}_{X}^{n}\ra 0,
$$
with $\bar{P}_{X}^{i}$ projective $B$-modules for $i>0$. We only
need to show that $\Ext_B^{i>r}(\bar{P}_{X}^{0},B)=0$, that is,
$\Ext^{i}_{B}(\bar{P}_{X}^{0},B)=0$ for $i> r$, where $r$ is a
non-zero integer. Indeed, there exists a distinguished triangle
$$\bar{P}^{+}_{X}\ra\bar{P}^{\bullet}_{X}\ra\bar{P}^{0}_{X}\ra
\bar{P}^{+}_{X}[1]$$ in $\Kb{B}$, where $\bar{P}^{+}_{X}$ denotes
the complex $\tau_{\geq 1}(\cpx{\bar{P}_{X}})$. For each
$i\in\mathbb{Z}$, applying the functor $\Hom_{\Db{B}}(-,B[i])$ to
the above distinguished triangle, we get an exact sequence
\begin{eqnarray*}
\cdots\ra\Hom_{\Db{B}}(\bar{P}^{+}_{X}[1],B[i])\ra
\Hom_{\Db{B}}(\bar{P}^{0}_{X},B[i])\ra
\Hom_{\Db{B}}(\bar{P}^{\bullet}_{X},B[i])\\\ra
\Hom_{\Db{B}}(\bar{P}^{+}_{X},B[i])\ra\cdots. \end{eqnarray*} On the
other hand, $\Hom_{\Db{B}}(\bar{P}^{+}_{X},B[i])\simeq
\Hom_{\Kb{B}}(\bar{P}^{+}_{X},B[i])=0$ for $i> r$. By Lemma \ref{PX}
and $\End^{i}_{A}(X,A)=0$ for $i>r$, we get
$\Hom_{\Db{B}}(\bar{P}^{\bullet}_{X},B[i])\simeq\Hom_{\Db{A}}(X,P^{\bullet}[i])=0$
for all $i> r$. Consequently, we get
$\Hom_{\Db{B}}(\bar{P}^{0}_{X},B[i])=0$ for all $i> r$ by the above
exact sequence. Therefore,
$$
\End^{i}_{B}(\bar{P}_{X}^{0},B)\simeq
\Hom_{\Db{B}}(\bar{P}^{0}_{X},B[i])=0, \;\;\;\text{for}\;\;\; i>
r.$$ This implies that $\bar{P}_{X}^{0}\in_B\mathcal {X}^{>r}$.
$\square$
Choose an $A$-module $X\in_{A}\mathcal {X}^{>r}$, by Lemma
\ref{2.3}, we know that $F(X)$ is isomorphic in $\Db{B}$ to a
radical complex of the form
$$
0\ra \bar{P}_{X}^{0}\ra \bar{P}_{X}^{1} \ra \cdots\ra
\bar{P}_{X}^{n-1}\ra \bar{P}_{X}^{n}\ra 0
$$
such that $\bar{P}_{X}^{0}\in _{B}\mathcal {X}^{>r}$ and
$\bar{P}_{X}^{i}$ are projective $B$-modules for $1\leq i\leq n$. In
the following, we try to define an additive functor $\underline{F}:
\underline{_{A}\mathcal {X}^{>r}}\ra \underline{_{B}\mathcal
{X}^{>r}}$, where $\underline{_{A}\mathcal {X}^{>r}}$ denotes the
stable category of $_{A}\mathcal {X}^{>r}$ , in which objects are
the same as the objects of $_{A}\mathcal {X}^{>r}$ and, for two
objects $X,Y$ in $\underline{_{A}\mathcal {X}^{>r}}$, their morphism
set is the quotient of $\Hom_{\mathcal {X}_{A}^{>r}}(X,Y)$ modulo
the homomorphisms that factors through projective modules.
\begin{Lem} Let $F:\Db{A}\lra \Db{B}$
be a derived equivalence. Then there is an additive functor
$\underline{F}: \underline{_{A}\mathcal
{X}^{>r}}\ra\underline{_{B}\mathcal {X}^{>r}}$ sending $X$ to
$\bar{P}_{X}^{0}$, such that the following diagram
$$\xymatrix{
\underline{_{A}\mathcal {X}^{>r}}\ar[r]^(.35){{\rm can}}\ar[d]_{\underline{F}} &
\Db{A}/\Kb{\pmodcat{A}}\ar[d]^{{F}}\\
\underline{_{B}\mathcal {X}^{>r}}\ar[r]^(.35){{\rm can}} & \Db{B}/\Kb{\pmodcat{B}}
}$$
is commutative up to natural isomorphism.
\end{Lem}
{\bf Proof.} By composing of the embedding functor $_{A}\mathcal
{X}^{>r}\hookrightarrow \Db{A}$ with the localization functor
$\Db{A}\ra \Db{A}/\Kb{\pmodcat{A}}$, we obtain a natural functor
$_{A}\mathcal {X}^{>r}\ra \Db{A}/\Kb{\pmodcat{A}}$. Since the
projective $A$-module is sending to zero in
$\Db{A}/\Kb{\pmodcat{A}}$, we get a canonical functor
$\underline{_{A}\mathcal {X}^{>r}}\ra \Db{A}/\Kb{\pmodcat{A}}$.
There is a functor between $\Db{A}$ and $\Db{B}/\Kb{\pmodcat{B}}$
which is the composition of $F:\Db{A}\lra \Db{B}$ and the
localization functor $\Db{B}\lra\Db{B}/\Kb{\pmodcat{B}}$. Since $F$
is an equivalence, we see that
$F(\Kb{\pmodcat{A}})=\Kb{\pmodcat{B}}$ by Theorem \ref{2.1}. Thus,
there is a functor between $\Db{A}/\Kb{\pmodcat{A}}$and
$\Db{B}/\Kb{\pmodcat{B}}$ induced by $F$, which we also denoted by
$F$. In the following, we will show that the above diagram is
commutative up to natural isomorphism.
For each $f:X\ra Y$ in $_{A}\mathcal {X}^{>r}$, we denote by
$\underline{f}$ the image of $f$ in $\underline{_{A}\mathcal
{X}^{>r}}$. By Lemma \ref{2.3}, we have a distinguished triangle
$$\bar{P}^{+}_{X}\stackrel{i_X}\ra
F(X)\stackrel{j_X}\ra\bar{P}^{0}_{X}\stackrel{m_X}\ra\bar{P}^{+}_{X}[1]\quad
\text{in}\quad \Db{B}.$$ Moreover, for each $f:X\ra Y$ in
$_{A}\mathcal {X}^{>r}$, there is a commutative diagram in $\Db{B}$
$$\xymatrix{
\bar{P}^{+}_{X}\ar^{i_{X}}[r]\ar^{\alpha_{f}}[d]
&F(X)\ar^{j_{X}}[r]\ar^{F(f)}[d]
& \bar{P}^{0}_{X}\ar^{m_{X}}[r]\ar^{\beta_{f}}[d]&\bar{P}^{+}_{X}[1]
\ar^{\alpha_{f[1]}}[d]\\
\bar{P}^{+}_{Y}\ar^{i_{Y}}[r] & F(Y)\ar^{j_{Y}}[r]
& \bar{P}^{0}_{Y}\ar^{m_{Y}}[r]& \bar{P}^{+}_{Y}[1] .}$$
Since $\Hom_{\Db{B}}(\bar{P}^{+}_{X},\bar{P}^{0}_{Y})\simeq
\Hom_{\Kb{B}}(\bar{P}^{+}_{X},\bar{P}^{0}_{Y})=0$, it follows that
$i_{X}F(f)j_{Y}=0$. Then there exists a homomorphism $\alpha_{f}:
\bar{P}^{+}_{X}\ra \bar{P}^{+}_{Y}$. Note that $B$-mod is fully
embedding into $\Db{B}$, hence $\beta_{f}$ is a morphism of
$B$-modules which is in $_{B}\mathcal {X}^{>r}$. If there is another
morphism $\beta'_{f}$ such that $j_{X}\beta'_{f}=F(f)j_{Y}$, then
$j_{X}(\beta_{f}-\beta'_{f})=0$. Thus $\beta_{f}-\beta'_{f}$ factors
through $\bar{P}^{+}_{X}[1]$. There is a distinguished triangle
$$\bar{P}^{1}_{X}[-1]\ra\tau_{\geq
1}(\bar{P}^{+}_{X}[1])\stackrel{a}\ra\bar{P}^{+}_{X}[1]\stackrel{b}\ra\bar{P}^{1}_{X}\quad\text{in}\quad
\Db{A}.$$ Suppose that $\beta_{f}-\beta'_{f}=gh$, where $g: X\ra
\bar{P}^{+}_{X}[1]$ and $h:\bar{P}^{+}_{X}[1]\ra Y$. Since
$\Hom_{\Db{A}}(\tau_{\geq
1}(\bar{P}^{+}_{X}[1]),Y)\simeq\Hom_{\Kb{A}}(\tau_{\geq
1}(\bar{P}^{+}_{X}[1]),Y)=0$, it follows that $ah=0$. Then there is
a map $x: \bar{P}^{1}_{X}\ra Y$, such that $h=bx$. Thus, we get
$\beta_{f}-\beta'_{f}=gbx$, which implies that
$\beta_{f}-\beta'_{f}$ factors through a projective $B$-module.
Therefore, the morphism $\underline{\beta_{f}}$ in
$\Hom_{\underline{\mathcal
{X}_{B}^{>r}}}(\bar{P}^{0}_{X},\bar{P}^{0}_{Y})$is uniquely
determined by $f$.
Let $f: X\ra Y$ and $g: Y\ra Z$ be morphisms in $_{A}\mathcal
{X}^{>r}$. Then there are commutative diagrams as follows:
$$\xymatrix{
\bar{P}^{+}_{X}\ar^{i_{X}}[r]\ar^{\alpha_{fg}}[d]
&F(X)\ar^{j_{X}}[r]\ar^{F(fg)}[d]
& \bar{P}^{0}_{X}\ar^{m_{X}}[r]\ar^{\beta_{fg}}[d]&\bar{P}^{+}_{X}[1]
\ar^{\alpha_{fg[1]}}[d]\\
\bar{P}^{+}_{Z}\ar^{i_{Z}}[r] & F(Z)\ar^{j_{Z}}[r]
& \bar{P}^{0}_{Z}\ar^{m_{Z}}[r]& \bar{P}^{+}_{Z}[1] }$$
and $$\xymatrix{
\bar{P}^{+}_{X}\ar^{i_{X}}[r]\ar^{\alpha_{f}}[d]
&F(X)\ar^{j_{X}}[r]\ar^{F(f)}[d]
& \bar{P}^{0}_{X}\ar^{m_{X}}[r]\ar^{\beta_{f}}[d]&\bar{P}^{+}_{X}[1]
\ar^{\alpha_{f[1]}}[d]\\
\bar{P}^{+}_{Y}\ar^{i_{Y}}[r]\ar^{\alpha_{g}}[d] &
F(Z)\ar^{j_{Y}}[r]\ar^{F(g)}[d]
& \bar{P}^{0}_{Z}\ar^{m_{Y}}[r]\ar^{\beta_{f}}[d]& \bar{P}^{+}_{Z}[1]\ar^{\alpha_{g[1]}}[d]
\\\bar{P}^{+}_{Z}\ar^{i_{Z}}[r] & F(Z)\ar^{j_{Z}}[r]
& \bar{P}^{0}_{Z}\ar^{m_{Y}}[r]& \bar{P}^{+}_{Z}[1] .}$$
Then we have $F(fg)j_{Z}=j_{X}\beta_{fg}$ and
$F(f)F(g)j_{Z}=F(fg)j_{Z}=j_{X}\beta_{f}\beta_{g}$. Therefore,
$j_{X}(\beta_{fg}-\beta_{f}\beta_{g})=0$. By the uniqueness of
$\underline{\beta_{fg}}$, we have
$\underline{\beta_{fg}}=\underline{\beta_{f}}$
$\underline{\beta_{g}}$. Moreover, if $X$ is a projective
$A$-module, then by Lemma \ref{2.3}, we know that $F(X)$ is
isomorphic in $\Db{B}$ to a radical complex of the form
$$
0\ra \bar{P}_{X}^{0}\ra \bar{P}_{X}^{1} \ra \cdots\ra
\bar{P}_{X}^{n-1}\ra \bar{P}_{X}^{n}\ra 0
$$
such that $\bar{P}_{X}^{i}$ are projective $B$-modules for $0\leq
i\leq n$. Thus, if $f$ factors through a projective $A$-module, then
we see that $\beta_{f}$ also factors through a projective
$B$-module.
For each $X\in_{A}\mathcal {X}^{>r}$, we define
$\underline{F}(X):=\bar{P}_{X}^{0}$. Set
$\underline{F}(\underline{f})=\underline{\beta_{f}}$, for each
$\underline{f}\in\Hom_{\underline{\mathcal {X}_{A}}^{>r}}(X,Y)$.
Then $\underline{F}$ is well-defined and an additive functor. The
last statement is discussed in \cite[Proposition 3.5]{P}, we omit it
here. $\square$
The next lemma is useful in our proof of the main result.
\begin{Lem}For $X\in _{A}\mathcal{X}^{>r}$, we have:
For each positive integer $k>r$, there is an isomorphism
$$
\beta_k:
\Hom_{\Db{A}}(X,X[k])\ra\Hom_{\Db{B}}(\underline{F}(X),\underline{F}(X)[k])
$$
Here we denote the image of $g$ under $\beta_k$ by
$\beta_k(g)$.\label{2.4}
\end{Lem}
{\bf Proof.} For $X\in _{A}\mathcal{X}^{>r}$, by Lemma \ref{2.3},
$F(X)=\bar{P}^{\bullet}_{X}$ is isomorphic in $\Db{B}$ to a complex
of the form
$$
0\ra \bar{P}_{X}^{0}\ra \bar{P}_{X}^{1} \ra \cdots\ra
\bar{P}_{X}^{n-1}\ra \bar{P}_{X}^{n}\ra 0
$$
with $\bar{P}_{X}^{0}\in_B\mathcal {X}^{>r}$. Consequently, there is
a distinguished triangle in $\Db{B}$
$$
\bar{P}^{+}_{X}\ra\bar{P}^{\bullet}_{X}\ra\bar{P}^{0}_{X}\ra
\bar{P}^{+}_{X}[1],
$$
where $\bar{P}^{+}_{X}$ is the complex $0\ra\bar{P}_{X}^{1} \ra
\cdots\ra \bar{P}_{X}^{n-1}\ra \bar{P}_{X}^{n}\ra 0$.
For a morphism $f:X\ra X[k]$, and it is easy to see that
$i_{X}F(f)j_{X}[k]\in\Hom_{\Db{B}}(\bar{P}^{+}_{X},\bar{P}^{0}_{X}[k])\simeq
\Hom_{\Kb{B}}(\bar{P}^{+}_{X},\bar{P}^{0}_{X}[k])=0$. Then there is
a map $b_f: \bar{P}^{0}_{X}\ra \bar{P}^{0}_{X}[k]$, we can form the
following commutative diagram
$$
\xymatrix{ \bar{P}^{+}_{X}\ar^{i_{X}}[r]\ar^{a_{f}}[d]
&F(X)\ar^{j_{X}}[r]\ar^{F(f)}[d]
& \bar{P}^{0}_{X}\ar^{m_{X}}[r]\ar@{.>}^{b_f}[d]&\bar{P}^{+}_{X}[1]
\ar^{a_{f[1]}}[d]\\
\bar{P}^{+}_{X}[k]\ar^{i_{X}[k]}[r] & F(X)[k]\ar^{j_{X}[k]}[r]
& \bar{P}^{0}_{X}[k]\ar^{m_{X}}[r]& \bar{P}^{+}_{X}[k+1] .}
$$
We claim that the morphism $b_f$ is uniquely determined by the above
commutative diagram. In fact, if there is another map $b'_f$ such
that $j_{X}b'_f=F(f)j_{X}[k]$, then we get $j_{X}(b_f-b'_f)=0$.
Therefore, $b_f-b'_f$ factors through $\bar{P}^{+}_{X}[1]$. Since
$\Hom_{\Db{B}}(\bar{P}^{+}_{X}[1],\bar{P}^{0}_{X}[k])\simeq
\Hom_{\Kb{B}}(\bar{P}^{+}_{X},\bar{P}^{0}_{X}[k-1])=0$, we have
$b_f-b'_f=0$. Hence, $b_f=b'_f$. Thus, we define a morphism
$$
\beta_k:
\Hom_{\Db{A}}(X,X[k])\ra\Hom_{\Db{B}}(\underline{F}(X),\underline{F}(X)[k]),
$$
by sending $f$ to $b_f$. Next, we will show that $\beta_k$ is an
isomorphism.
Firstly, it is injective. Assume that $\beta_k(f)=b_f=0$. Then
$F(f)j_{X}[k]=0$, and consequently, $F(f)$ factors through
$\bar{P}^{+}_{X}[k]$. It follows that $GF(f)$ factors through
$G(\bar{P}^{+}_{X})[k]$, that is, the map $f: X\ra X[k]$ factors
through $G(\bar{P}^{+}_{X})[k]$, say $f=xy$, for some $x: X\ra
G(\bar{P}^{+}_{X})[k] $ and $y: G(\bar{P}^{+}_{X})[k]\ra X[k]$.
Since $\Hom_{\Db{A}}(\cpx{Q_X},X)\simeq\Hom_{\Kb{A}}(\cpx{Q_X},X)$,
we deduce that $y$ can be chosen to be a chain map. Set
$G(\bar{P}^{+}_{X})=Q^{\bullet}_{X}$. Then $G(\bar{P}^{+}_{X})$ is a
radical projective bounded complex $Q^{\bullet}_{X}$ of the form
$$
0\ra Q^{-m}\cdots\ra Q^{-1}_{X}\ra Q^{0}_{X}\ra Q^{1}_{X}\ra 0,
$$
where $m$ is a positive integer. Indeed, by the distinguished
triangle
$\bar{P}^{+}_{X}\ra\bar{P}^{\bullet}_{X}\ra\bar{P}^{0}_{X}\ra
\bar{P}^{+}_{X}[1],$ we get $H^i(G(\bar{P}^{+}_{X}))=0$ for $i>1$,
where $H^i(G(\bar{P}^{+}_{X}))$ is $i$-th cohomology group of
$G(\bar{P}^{+}_{X})$. We claim that
$$
\Hom_{\Db{A}}(X,Q^{\bullet}_{X}[k])\simeq\Hom_{\Kb{A}}(X,Q^{\bullet}_{X}[k]).$$
So, it suffices to show that for the complex $Q^{\bullet}_{X}$ of
the form $ 0\ra Q_{X}^{-1}\ra Q_{X}^{0}\ra0$, we get the result.
There is a distinguished triangle $$(\ast)\quad\quad
Q_{X}^{-1}[k]\ra Q_{X}^{0}[k]\ra Q_{X}^{\bullet}[k]\ra
Q_{X}^{-1}[k+1]\quad \text{in} \quad \Kb{A}.$$ Applying the functors
$\Hom_{\Kb{A}}(X,-)$, $\Hom_{\Db{A}}(X,-)$ to ($\ast$), we obtain
the following commutative diagram
$$\xymatrix{
\Hom_{\Kb{A}}(X,Q_{X}^{-1}[k])\ar[r]\ar^{\simeq}[d] &
\Hom_{\Kb{A}}(X,Q_{X}^{0}[k])\ar[r]\ar^{\simeq}[d]
& \Hom_{\Kb{A}}(X,Q_{X}^{\bullet}[k])\ar[r]\ar[d]& \Hom_{\Kb{A}}(X,Q_{X}^{-1}[k+1]) \ar^{\simeq}[d]\\
\Hom_{\Db{A}}(X,Q_{X}^{-1}[k])\ar[r] &
\Hom_{\Db{A}}(X,Q_{X}^{0}[k])\ar[r]
& \Hom_{\Db{A}}(X,Q_{X}^{\bullet}[k])\ar[r]& \Hom_{\Db{A}}(X,Q_{X}^{-1}[k+1]).}$$ Since
$\End^{i}_{A}(X,A)=0$ for $i>r$, it follows that
$\Hom_{\Db{A}}(X,Q_{X}^{-1}[k+1])=0$ for $k>r$. Moreover,
$\Hom_{\Kb{A}}(X,Q_{X}^{-1}[k+1])=0$. We thus get
$\Hom_{\Db{A}}(X,Q^{\bullet}_{X}[k])\simeq\Hom_{\Kb{A}}(X,Q^{\bullet}_{X}[k])$.
Therefore, $x$ is chosen to be a chain map. Consequently, we see
that $f=xy=0$. This shows that $\beta_k$ is injective.
Next, we can prove that $\beta_k$ is surjective. For a map $b:
\bar{P}^{0}_{X}\ra\bar{P}^{0}_{X}[k]$, we have
$j_Xbm_X[k]\in\Hom_{\Db{B}}(F(X),\bar{P}^{+}_{X}[1])$. It follows
from Lemma \ref{PX} that
$$\Hom_{\Db{B}}(F(X),\bar{P}^{+}_{X}[k+1])\simeq\Hom_{\Db{A}}(X,G(\bar{P}^{+}_{X})[k+1])=0 \quad \mbox{for}\quad k>r.$$
Then there is a map $c: F(X)\ra F(X)[k]$ such that
$cj_{X}[k]=j_{X}b$. Since $F$ is an equivalence, it follows that
$c=F(f)$ for some $f: X\ra X[k]$. Hence, $b=\beta_k(f)$. Therefore,
$\beta_k$ is surjective. $\square$
We now have all the ingredients to complete the proof of our main
theorem.
\medskip
{\bf Proof of Theorem \ref{thm1}.} We assume that the generalized
Auslander-Reiten conjecture is true for $B$. If $X$ is an $A$-module
which satisfies $\Ext_A^i(X,X)=0=\Ext_A^i(X,A)$ for $i> r$, then it
follows from Lemma \ref{2.3} that,
$\underline{F}(X)=\bar{P}_{X}^{0}$ satisfies
$\Ext_B^{i}(\bar{P}_{X}^{0},B)=0$ for $i>r$. By Lemma \ref{2.4}, it
is easy to see that, for $i>r$
$$\Ext_B^i(\bar{P}_{X}^{0},\bar{P}_{X}^{0})
\simeq\Hom_{\Db{B}}(\bar{P}_{X}^{0},\bar{P}_{X}^{0}[i])\simeq\Hom_{\Db{A}}(X,X[i])=0.$$
Since we assume that $B$ satisfies the generalized Auslander-Reiten
conjecture, we see that the $\pd(\bar{P}_{X}^{0})\leq r$. We can
take a projective resolution $P^\bullet_{\bar{P}_{X}^{0}}$ of
$\bar{P}_{X}^{0}$. Therefore, by the distinguished triangle
$\bar{P}^{0}_{X}[-1]\ra\bar{P}^{+}_{X}\ra\bar{P}^{\bullet}_{X}\ra\bar{P}^{0}_{X},$
we can take a projective resolution $P^{\bullet}_{\bar{\cpx{P}_X}}$
of $\bar{\cpx{P}_X}$ by the mapping cone of
$P^\bullet_{\bar{P}_{X}^{0}}[-1]$ and $\bar{P}^{+}_{X}$. Thus, we
get $P^{\bullet}_{\bar{\cpx{P}_X}}\in\Kb{\pmodcat{B}}$. It follows
that $X\simeq G(\bar{\cpx{P}_X})\simeq
G(P^{\bullet}_{\bar{\cpx{P}_X}})$. Then it is easy to see that
$\pd_A(X)<\infty$. Let $0\ra P_s\ra\cdots \ra P_r\ra \cdots\ra
P_0\ra X\ra 0$ be a projective resolution of $X$. Since
$\Ext_A^i(X,A)=0$ for $i> r$, we conclude that
$$0\ra \Hom_A(\Omega^r(X),A)\ra\cdots\ra \Hom_A(P_s,A) \ra 0$$
is a split exact sequence. Then, we get $\Hom_A(\Omega^r(X),A)$ is a
projective $A^{op}$-module and consequently, $\Omega^r(X)$ is a
projective $A$-module. It follows that $\pd_A(X)\leq r$.
Similarly, we can prove that the converse is also true. $\square$
As a corollary of Theorem \ref{thm1}, we re-obtain the following
result of Wei \cite[Theorem 3.7]{W2}.
\begin{Koro} $\rm\cite[Theorem\,3.7]{W2}$ Let $A$ be an Artin algebra and $T$ be a tilting
$A$-module with $\End_A(T)=B$. Then $A$ satisfies the generalized
Auslander-Reiten conjecture if and only if so does $B$.
\end{Koro}
\bigskip
{\footnotesize
| 171,159
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TITLE: If $A$ is normal and upper triangular then it is diagonal
QUESTION [14 upvotes]: Let $A$ be a normal matrix in Mat$_{n\times n}(\mathbb C)$, if $A$ is upper triangular then it is diagonal
(Normal means $AA^*=A^*A$, where $A^*$ is the conjugate transpose of $A$)
If I consider the diagonal of $AA^*$, let denote $(a_{ij})=A$ and $(â_{ij})_{i,j}=AA^*$ then, since $AA^*=A^*A$
$â_{ii}=\sum\limits_{k=1}^na_{ik}\overline{a}_{ik}=\sum\limits_{k=1}^n\overline{a_{ki}}{a}_{ki}$
$\implies\sum\limits_{k=1}^n|a_{ik}|^2=\sum\limits_{k=1}^n|a_{ki}|^2$.
If I take $i=n$ then it follows that $a_{in}=0, \forall 1\le i\le n-1$ and continuing in this manner the upper diagonal entries are zero, Is this correct ?
Can I show it in another way, because in a previous exercise I had to show that ''If A is normal and nilpotent then $A=0$'' so using this can I decompose $A$ into diagonal and nilpotent matrix, then show that the nilpotent part is zero ?
REPLY [14 votes]: Let's prove this by induction. Suppose $A$ is an $n\times n$ upper-triangular, normal matrix.
If $n=1$, this is trivial.
If $n=2$, we have
$$ A = \begin{pmatrix}
a & b\\
0 & c
\end{pmatrix}$$
So
\begin{align*}
0 = AA^* - A^*A &= \begin{pmatrix}
a & b\\
0 & c
\end{pmatrix}\begin{pmatrix}
\overline{a} & 0\\
\overline{b} & \overline{c}
\end{pmatrix}- \begin{pmatrix}
\overline{a} & 0\\
\overline{b} & \overline{c}
\end{pmatrix}\begin{pmatrix}
a & b\\
0 & c
\end{pmatrix} \newline
&= \begin{pmatrix}
|a|^2 + |b|^2 & b\overline{c}\\
\overline{b}c & |c|^2
\end{pmatrix} - \begin{pmatrix}
|a|^2 & \overline{a}b \\
a\overline{b} & |b|^2+|c|^2
\end{pmatrix} \newline
&= \begin{pmatrix}
|b|^2 & b\overline{c}-\overline{a}b\\
\overline{b}c -a\overline{b} & |b|^2
\end{pmatrix}_.
\end{align*}
Therefore $|b|^2 = 0$. Thus $b = 0$ and $A$ is diagonal.
Now assume the result holds for $n-1$.
Then $$A = \begin{pmatrix}
a & B\\
0 & C
\end{pmatrix}$$
where $B$ is an $1\times (n-1)$ matrix and $C$ is an $(n-1)\times(n-1)$ upper-triangular matrix.
Since $A$ is normal,
\begin{align*}
0 = A^*A-AA^* &= \begin{pmatrix}
\overline{a} & 0\\
B^* & C^*
\end{pmatrix}\begin{pmatrix}
a & B\\
0 & C
\end{pmatrix} - \begin{pmatrix}
a & B\\
0 & C
\end{pmatrix}\begin{pmatrix}
\overline{a} & 0\\
B^* & C^*
\end{pmatrix} \newline
&= \begin{pmatrix}
|a|^2 & \overline{a}B\\
aB^* & B^*B + C^*C
\end{pmatrix} - \begin{pmatrix}
|a|^2 + BB^* & BC^*\\
CB^* & CC^*
\end{pmatrix} \newline
&= \begin{pmatrix}
-|B|^2 & \overline{a}B-BC^*\\
aB^*-CB^* & B^*B + C^*C-CC^*
\end{pmatrix}_.
\end{align*}
Thus $|B| = 0$ and so $B=0$. Hence $C^*C-CC^* = 0$, i.e. $C$ is normal. By the hypothesis of induction, $C$ is diagonal. Finally,
$$ A = \begin{pmatrix}
a & 0\\
0 & C
\end{pmatrix}_.$$
So $A$ is diagonal.
| 158,841
|
Here’s how much Warren Buffett spends on his haircuts
Warren Buffett, the 84-year-old head of Berkshire Hathaway, may be a billionaire. But he doesn’t spend like one.
He famously still lives in the Omaha, Nebraska, house he bought in 1958 for $31,500.
And, according to a new story by Market Watch, he’s a long-time patron of Omaha barber Stan Docekal—who charges him $18 (tip not included) for a hair trimming every two or three weeks.
Docekal, who has cut Buffett’s hair for about 23 years, is also in his early 80s. Desperate to pick up tidbits about the Oracle of Omaha, journalists have interviewed the barber many times over the years. Apparently Buffett’s activities during haircuts include listening to oldies music, watching CNBC, and reading the his mail, a newspaper, or an annual report.
Is $18 a good deal?
Technically it’s higher than the national average of about $14 for a men’s cut, according to a recent study. But it’s pretty darn cheap for the third richest person in the world.
Read more: This Is How Much It Costs to Live Next Door to Warren Buffett
| 418,680
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TITLE: Double Factorial
QUESTION [2 upvotes]: I am having trouble proving/understanding this question.
Let $n=2k$ be even, and $X$ a set of $n$ elements. Define a factor to be a partition of
$X$ into $k$ sets of size $2$. Show that the number of factors is equal to $1 \dot\ 3 \dot\ 5 \cdots\ (2k-1)$.
Lets suppose our set $X=\{x_1,x_2,...,x_n\}$ contains $4$ elements, i.e $X=\{x_1,x_2,x_3,x_4\}$ then $k=2$ implies $k = (x_1,x_2), \ (x_3,x_4)$ but the numbers of factors equals $3$. What am I doing wrong trying to understand this?
REPLY [3 votes]: Let the $2k$ elements of $X$ be people. We want to count the number of ways to divide these $2k$ people into $k$ teams of $2$ each.
Line up the people from left to right in order of age, or weight, or student number.
The leftmost person can choose her team mate in $2k-1$ ways.
For every such way, the leftmost person not yet chosen can choose her team mate in $2k-3$ ways.
For every choice made so far, the leftmost person not yet chosen can choose her team mate in $2k-5$ ways.
And so on.
If the "and so on" is not viewed as rigorous enough, we rewrite the proof as a proof by induction. We assume that the result holds for $n=2k$, and prove that the result holds for $n=2(k+1)$. Alicia has $2k+1$ choices of team mate. For every choice, there are $2k$ people left, and by the induction assumption they can be divided into teams in $(2k-1)(2k-3)\cdot (3)(1)$ ways. Thus the $2(k+1)$ people can be divided into teams in $(2k+1)(2k-1)\cdots(3)(1)$ ways. This completes the induction step.
Remark: In your example, we can list the divisions into teams explicitly. Let the people be $A, B, C, D$. Then $A$ can be paired with $B$, $C$, or $D$. Once this is done the other pairing is determined.
| 76,150
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TITLE: Discrete Mathematics Power of 3
QUESTION [5 upvotes]: If m is a power of $3$, n is a power of $3$, prove that $m+n$ is never a power of $3$.
This is the question that I was given, unfortunately, my teacher doesn't like to teach and I' left reading a $1000$ page book by myself answering these random questions that aren't covered in our book.
Edit: I think I really need help defining what is meant by a power of $3$. When they say $m$ is a power of $3$ does that mean $m=3\enspace\text{or}\enspace A^m = \enspace\text{or}\enspace A^3$ ?
if so $A^m + B^n$ with $m$ & $n$ being powers of $3 == A^3 + B^3$ and therefore $m+n = 3+3 = 6$ and a power of $6$ is not a power of $3$ ?
REPLY [4 votes]: Powers of 3 and cubes are different things. Given an exponent $\alpha$ that is a positive integer, $3^\alpha$ is a power of 3. If you flip that, however, $\alpha^3$, you have a cube, and the only way that's also a power of 3 is if $\alpha = 1$ or 3.
The first few powers of 3 are: 1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907. These are listed in Sloane's OEIS A000244.
The first few cubes are 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, 2197, 2744, 3375, 4096, 4913, 5832, 6859, 8000, 9261, 10648, 12167, 13824, 15625, 17576, 19683, 21952, 24389. These are listed in Sloane's A000578.
The only numbers in common to both lists are 1 and 27.
So if $m$ is a power of 3 and $n$ is also a power of 3, meaning that $m = 3^\alpha$ and $n = 3^\beta$ (it doesn't matter if $\alpha = \beta$, as long as they're both positive integers), then $m + n = 3^\alpha + 3^\beta$.
Not much help there, until you notice, like the other answerer already mentioned, that powers of 3 are odd, but adding up two of them gives an even number.
Now, to prove that two nonzero cubes can never add up to a cube, well, Fermat claimed to have a wonderful proof of that which was just a wee bit too long for the margin.
One more thing. If a number is a power of 3 other than 1, its representation in the ternary numeral system consists of a single digit 1 followed by one or more digits 0. If a number is the sum of two powers of 3, then its representation in ternary is either:
A single digit 2 followed by one more digits 0, or
A digit 1, followed by one or more digits 0, followed by a 1 and then one or more digits 0, or
Two digits 1 with one or more digits 0 in between them.
These numbers are also listed in Sloane's OEIS. The first few are 6, 10, 12, 18, 28, 30, 36, 54, 82, 84, 90.
| 185,355
|
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| 15,340
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Tuesday, July 11, 2017
Doses of History
Hello everyone!
I know it's been. . . awhile, but I'm here to tell you I've started a new blog! It's called Doses of History and, as opposed to this one, focuses on all history, not just The Gilded Age Era (though that is included, of course). History, clearly, fascinates me, so I've created this new blog as a place to share the pieces of it that I find interesting, and also to, hopefully, have discussions about it with some of you guys!
Since this is a blog about the Gilded Age. . . I did write a post about the Vanderbilt chateau in NYC, pictured up above, on the new blog, so CLICK HERE to view that post. I've thought about transferring and editing some of the posts on this blog and posting them on the new blog, since I won't be posting on this one very much anymore, and haven't for quite awhile anyways.
Anyways, hope to see you over at my new blog!
Sunday, January 11, 2015
Oscar G. Mayer Mansion!
Friday, November 28, 2014
Bellefontaine.
The'
Giraud Foster and His Beloved Bellefontaine
Two Great Country Estates ~ Lenox, MA.
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Friday, June 13, 2014
Villa Philbrook and The Phillips
.
| 190,762
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Tax Director
PwC Singapore
Rising Star (Financial Times-RBS Coutts Women in Asia awards 2010)
Young Accountant of the Year
A trailblazer, Kexin graduated as Singapore Management University’s (SMU) first female Valedictorian in 2005 on dual scholarship. Over her last 12 years with PwC, Kexin has advised top tier multinational, institutional and business owner clients in Singapore, London and Jakarta in her dynamic international tax career.
Volunteering actively since 17, Kexin has served in a wide range of hands-on to leadership roles across community, government and politics. She has also represented Singapore internationally in various premier emerging leader programmes in US, China, Taiwan, Malaysia and Switzerland.
A Chartered Accountant (Singapore) and Accredited Tax Practitioner, Kexin is the youngest governing Council Member in the Institute of Singapore Chartered Accountants’ 54-year history. Concurrently, she serves as Council Member of the Singapore Chamber of Commerce in Indonesia and chairs SMU’s Accountancy Alumni Advisory Board.
Widely profiled in the media, a frequent public speaker and multiple award winner, Kexin was named ‘Young Accountant of the Year’ at the inaugural Singapore Accountancy Awards 2014 and by Prestige magazine as one of Singapore’s ’40 under 40’ to watch 2015. Kexin believes in “me” time and is an (amateur) home cook.
| 283,965
|
TITLE: Among any $11$ integers, sum of $6$ of them is divisible by $6$
QUESTION [3 upvotes]: Using pigeon hole principle show that among any $11$ integers, sum of $6$ of them is divisible by $6$.
I thought we can be sure that we have $6$ odd or $6$ even numbers among $11$ integers.Suppose that we have $6$ odd integers. Obviously sum of these $6$ odd numbers is divisible by $2$, so it remains to show that this sum is divisible by $3$ as well. But how?
REPLY [4 votes]: We first show that among any 5 integers, sum of 3 of them is divisible by 3. The residue classes modulo 3 are $[0], [1], [2]$. By Pigeonhole principle, we have two cases:
each of these residue classes must have atleast one of the five integers, or
one residue class must have 3 of these 5 integers belonging to it.
In the former case, let $x_0 \equiv 0 \mod 3$, $x_1 \equiv 1 \mod 3$ and $x_2 \equiv 2 \mod3$. Then summing $x_1, x_2, x_3$, we get, $x_1 + x_2 + x_3 \equiv 0 \mod 3$.
In the latter case, we have 3 integers among the 5, say $x_1, x_2, x_3$ such that, $x_1 \equiv x_2 \equiv x_3 \equiv k \mod 3$, again summing these three we get $x_1 + x_2 + x_3 \equiv 3k \equiv 0 \mod 3$.
This proves that among any 5 integers, sum of some 3 of them is divisible by 3.
Now, we have 11 integers. By the previous result, we can choose 3 of them such that there sum is divisible by 3. Denote this sum by $s_1$. Now, we are left with 8 integers, again, choose 3 of them such that there sum is divisible by 3. Denote this by $s_2$. Now, we are left with 5 integers. Choose $s_3$ similarly.
Thus we have 3 sums: $s_1, s_2, s_3$ (each of which are sums of 3 integers). These sums are divisible by 3. So, each of these sums are congruent to either 0 or 3 modulo 6.
Now, since there are 3 sums, and two residue clases ($[0], [3]$), by Pigeonhole principle, one residue class must have two sums belonging to it. Let $s_i$ and $s_j$ be those sums. Either, $s_i \equiv s_j \equiv 0 \mod 6$ or $s_i \equiv s_j \equiv 3 \mod 6$. In both the cases, $s_i + s_j \equiv 0 \mod 6$.
Since, $s_i$ and $s_j$ are both sum of 3 integers, $s_i + s_j$ is a sum of 6 integers (which is divisible by 6). This completes the proof.
| 131,375
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Section 401, acts Aug. 27, 1940, ch. 689, §1, 54 Stat. 858; Aug. 18, 1941, ch. 362, §11, 55 Stat. 628, authorized for the period ending the later of June 30, 1942, or 6 months after the termination of the authority under section 352 of this Appendix the President to order reserve and retired personnel to active service. See section 471 of this Appendix.
Section 402, act Aug. 27, 1940, ch. 689, §2, 54 Stat. 859, related to laws and regulations governing personnel called to active service.
Section 403, acts Aug. 27, 1940, ch. 689, §3, 54 Stat. 859; Sept. 16, 1940, ch. 720, §8(d), (f), 54 Stat. 891; July 28, 1942, ch. 529, §1, 56 Stat. 723; Dec. 8, 1944, ch. 548, §2, 58 Stat. 799, related to service and health certificates and reemployment rights. See section 459 of this Appendix.
Section 404, act Aug. 27, 1940, ch. 689, §4, 54 Stat. 860, made applicable the Soldiers’ and Sailors’ Civil Relief Act and section 101 et seq. of this Appendix. See section 501 et seq. of this Appendix.
Section 405, act Aug. 27, 1940, ch. 689, §5, 54 Stat. 860, suspended all laws in conflict with sections 401 to 405 of this Appendix.
Act Feb. 6, 1942, ch. 42, 56 Stat. 50, related to pay of persons inducted in erroneous rank or grade under sections 401 to 405 of this Appendix.
| 316,242
|
\begin{document}
\maketitle
\begin{abstract}
Let $K$ be a number field, and $g \geq 2$ a positive integer. We define $c_K(g)$ as the smallest integer $n$ such that there exist infinitely many $\overline{K}$-isomorphism classes of genus $g$ hyperelliptic curves $C/K$ with all Weierstrass points in $K$ having potentially good reduction outside $n$ primes in $K$. We show that $c_K(g) > \pi_{K, \textrm{odd}}(2g) + 1$, where $\pi_{K, \textrm{odd}}(n)$ denotes the number of odd primes in $K$ with norm no greater than $n$, as well as present a summary of various conditional and unconditional results on upper bounds for $c_K(g)$.
\end{abstract}
{\renewcommand{\thefootnote}{}
\footnotetext{2020 \textit{Mathematics Subject Classification}. 11G30.}
}
\section{Introduction}
For a given number field $K$, the problem of studying the reduction behaviour of hyperelliptic curves $C/K$ has attracted significant interest over the last few decades. Indeed, there exist well-known effective algorithms for determining all hyperelliptic curves $C/K$ of a given genus $g$ with potentially good reduction outside a given finite set of primes $S$, with Merriman--Smart \cite{merrimansmart, smartbinaryforms} effectively carrying out this computation to list all genus 2 curves $C/\mathbb{Q}$ with good reduction outside 2. \\
For our purposes, we shall be interested in the reduction of hyperelliptic curves $C/K$ with all of its Weierstrass points lying in $K$. We first recall that a prime $\mathfrak{p}$ in $K$ is considered \textrm{odd} if it lies above an odd rational prime (or equivalently has odd norm), and define $\pi_{K, \textrm{odd}}(n)$ as the number of odd primes in $K$ with norm no greater than $n$. We also define $\mathcal{B}_\textrm{odd}(C/K)$ as the set of odd primes $\mathfrak{p}$ in $K$ for which $C/K$ does not have potential good reduction at $\mathfrak{p}$. Our main result is the following: \\
\refstepcounter{mainnum} \label{thm:maintheorem}
\textbf{Theorem \themainnum.} (\textit{Theorem \ref{thm:norm}, Theorem \ref{thm:mainlowerboundtheorem}}) Let $K$ be a number field. Then for all genus $g$ hyperelliptic curves $C/K$ with all Weierstrass points in $K$, we have $\mathfrak{p} \in \mathcal{B}_\textrm{odd}(C/K)$ for all odd primes $\mathfrak{p}$ satisfying $N_{K/\mathbb{Q}}(\mathfrak{p}) < 2g$.
Furthermore, there are only finitely many $\overline{K}$-isomorphism classes of genus $g$ hyperelliptic curves $C/K$ with all Weierstrass points in $K$ satisfying $\# \mathcal{B}_\textrm{odd}(C/K) \leq \pi_{K, \textrm{odd}}(2g) + 1$. \\
This gives us the lower bound $c_K(g) > \pi_{K, \textrm{odd}}(2g) + 1$. Applying Theorem \ref{thm:maintheorem} to $K = \mathbb{Q}$ gives the following corollaries: \\
\refstepcounter{mainnum}
\textbf{Corollary \themainnum.} (\textit{Corollary \ref{thm:rationalWeier}, Corollary \ref{thm:box}}) Let $C/\mathbb{Q}$ be a hyperelliptic curve with rational Weierstrass points. Then $C$ cannot have potentially good reduction at any odd prime $p \leq 2g$. Moreover, there is no genus 2 curve $C/\mathbb{Q}$ with rational Weierstrass points having potentially good reduction outside one prime. \\
Finally, in Section 4, we also prove the following various conditional and unconditional upper bounds for $c_K(g)$. \\
\refstepcounter{mainnum}
\label{thm:mainupperboundtheorem}
\textbf{Theorem \themainnum.} (\textit{Theorem \ref{thm:upperDickson}, Theorem \ref{thm:upper}, Theorem \ref{thm:uppergenDickson}}) Let $K$ be a number field of degree $n$. Then $c_K(g) \leq (\frac{2}{\log{2}} + o(1)) n g \log{g}$. Furthermore, under the assumption of the Hardy-Littlewood prime $k$-tuples conjecture for $K$, we have $c_K(g) \leq 2g - 1 + n \pi(2g)$, and under the assumption of Schinzel’s hypothesis H for $K$, we have moreover that
\begin{equation*}
c_K(g) \leq \sum_{\substack{1 \leq d < g, \textrm{ or} \\ d < 2g, \, d \textrm{ even}}} \frac{n}{[K(\zeta_d) : \mathbb{Q}(\zeta_d)]} + 1 + n \pi(2g) .
\end{equation*}
Precise statements of the Hardy--Littlewood prime $k$-tuples conjecture and the Schinzel hypothesis H are provided in Section 4. This hence gives rise to the following corollaries: \\
\refstepcounter{mainnum}
\textbf{Corollary \themainnum.} (\textit{Corollary \ref{thm:upperCor1}, Corollary \ref{thm:upperCor2}}) Let $K$ be a number field of degree $n$ with no non-trivial abelian subfields, and suppose that Schinzel's hypothesis H holds for $K$. Then if $K$ is abelian (and hence of prime degree) with conductor $\mathfrak{f}_K$, then
\begin{equation*}
c_K(g) \leq \begin{cases}
\frac{3}{2} g \big(1 + \frac{n-1}{\mathfrak{f}_K} \big) + 1 + n \pi(2g) & \textrm{ if $\mathfrak{f}_K$ odd, } \\[2mm]
\frac{3}{2} g \big( 1 + \frac{4(n-1)}{3 \mathfrak{f}_K} \big) + 1 + n \pi(2g) & \textrm{ if $\mathfrak{f}_K$ even, }
\end{cases}
\end{equation*}
otherwise $c_K(g) \leq \frac{3}{2} g + n \pi (2g)$ if $K$ is non-abelian. \\
Let $C/K$ be a hyperelliptic curve of genus $g$ with all Weierstrass points in $K$. We first recall that $C$ is isomorphic to a curve of the form
\begin{equation*}
y^2 = cx(x-1)(x-a_1)(x-a_2) \cdots (x-a_{2g-1})
\end{equation*}
for some $c, a_i \in K$, $a_i \not = 0, 1$, as all Weierstrass points of $C$ lie in $K$. This is called the \textit{Rosenhain normal form} of $C$. \\
We shall now apply the machinery of cluster pictures to study these curves, introduced by Dokchitser, Dokchitser, Maistret, Morgan \cite{dokchitser}.
\section{Cluster pictures}
We first quickly recall the notion of cluster pictures. Indeed, let $C/K$ be a hyperelliptic curve, given by $C : y^2 = f(x)$, and let $\mathcal{R}$ denote the roots of $f$, with $\mathcal{P}(\mathcal{R})$ being the power set of $\mathcal{R}$. Let $\mathfrak{p}$ be an odd prime in $K$, and let $v_\mathfrak{p}$ denote the discrete normalised $p$-adic valuation induced by $\mathfrak{p}$. The \textbf{cluster picture} $\Sigma_{\mathfrak{p}} \subset \mathcal{P}(\mathcal{R})$ associated to $C$ (with respect to $\mathfrak{p}$) is given by the following set:
\begin{equation*}
\Sigma_{\mathfrak{p}} := \big\{ \mathfrak{s} \in \mathcal{P}(\mathcal{R}) \;|\; \forall x \in \mathfrak{s}, \, v_{\mathfrak{p}}(x - z) \geq d \textrm{ for some } z \in \overline{K}, d \in \mathbb{Q} \big\} .
\end{equation*}
We say that $\Sigma_{\mathfrak{p}}$ is \textit{trivial} if it only contains the singleton elements and $\mathcal{R}$ itself. \\
Using these cluster pictures, we shall make use of the following theorem of Dokchitser, Dokchitser, Maistret, Morgan, which allows us to directly read off whether a hyperelliptic curve $C/K$ has potentially good reduction at $\mathfrak{p}$. \\
\refstepcounter{mainnum} \label{thm:potgoodredn}
\textbf{Theorem \themainnum.} \cite[p.~4]{dokchitser2} Let $C/K$ be a hyperelliptic curve of genus $g$, and let $\mathfrak{p}$ be an odd prime in $K$. Then $C$ has potentially good reduction at $\mathfrak{p}$ if and only if $\Sigma_{\mathfrak{p}}$ has no proper clusters of size $< 2g + 1$. \\
From this theorem, we can prove the following proposition. \\
\refstepcounter{mainnum} \label{thm:lambdas}
\textbf{Proposition \themainnum.} Let $C/K$ be a hyperelliptic curve with Weierstrass points in $K$, given in Rosenhain normal form
\begin{equation*}
y^2 = c x(x-1)(x-\lambda_1) \dots (x-\lambda_{2g-1}) , \qquad c, \lambda_i \in K .
\end{equation*}
Let $\mathfrak{p}$ be an odd prime of $K$. Then $C$ has potentially good reduction at $\mathfrak{p}$ if and only if we have $v_\mathfrak{p}(\lambda_i) = v_\mathfrak{p}(\lambda_i - 1) = 0$ for all $i \in 1, \dots, 2g-1$, and $v_\mathfrak{p}(\lambda_i - \lambda_j) = 0$ for all distinct $i, j \in \{1, \dots, 2g-1\}$. (i.e. the values $\lambda_i, \lambda_i - 1, \lambda_i - \lambda_j$ are all $\mathfrak{p}$-units) \\
\textit{Proof.} Let $C/K$ be given in the above form, and let $\mathcal{R}$ denote the Weierstrass points, i.e. $\mathcal{R} := \{0, 1, \lambda_1, \dots, \lambda_{2g-1} \} $. Then by Theorem \ref{thm:potgoodredn}, since $|\mathcal{R}| = 2g+1$, we have that $C$ has potentially good reduction at $\mathfrak{p}$ if and only if $\Sigma_\mathfrak{p}$ is trivial. \\
Note that $\Sigma_\mathfrak{p}$ is trivial if and only if $v_\mathfrak{p}(r_i - r_j)$ is constant over all distinct pairs $r_i, r_j \in \mathcal{R}$. However, since $v_\mathfrak{p}(1 - 0) = 0$, this implies that $v_\mathfrak{p}(\lambda_i) = v_\mathfrak{p}(\lambda_i - 1) = 0$ for all $i$, and that $v_\mathfrak{p}(\lambda_i - \lambda_j) = 0 $ for all $i, j$, which yields the result. \qed \\
This immediately implies the following corollary: \\
\refstepcounter{mainnum} \label{cor:Cpgr}
\textbf{Corollary \themainnum.} Let $K$ be a number field, and let $S$ be a finite set of primes of $K$, and assume that $S$ consists of all even primes of $K$. Let $\mathcal{O}_S^\times$ denote the set of $S$-units in $K$. Then for a given hyperelliptic curve $C/K$ of the above form, $C$ has potentially good reduction outside $S$ if and only if $\lambda_i$ and $\lambda_i - 1$ are in $\mathcal{O}_S^\times$, and if $\lambda_i - \lambda_j$ are in $\mathcal{O}_S^\times$. \\
We can now prove our main theorem. \\
\refstepcounter{mainnum} \label{thm:norm}
\textbf{Theorem \themainnum.} Let $C/K$ be a hyperelliptic curve with Weierstrass points in $K$. Then $C$ cannot have potentially good reduction at any odd prime $\mathfrak{p}$ such that $N_{K/\mathbb{Q}}(\mathfrak{p}) \leq 2g$. \\
\textit{Proof.} Let $C$ be given by its Rosenhain normal form:
\begin{equation*}
y^2 = cx(x-1)(x-\lambda_1) \cdots (x-\lambda_{2g-1})
\end{equation*}
and assume for contradiction that $C$ has potentially good reduction at $\mathfrak{p}$, where $\mathfrak{p}$ is a prime ideal of $K$ such that $N(\mathfrak{p}) \leq 2g$. \\
We have by Theorem \ref{thm:lambdas} that $\lambda_1, \dots, \lambda_{2g-1}$ must all be $\mathfrak{p}$-units. Furthermore, note that each of the roots $0, 1, \lambda_1, \dots, \lambda_{2g-1}$ must yield distinct values under the reduction map $\mathcal{O}_K \to \mathcal{O}_K / \mathfrak{p}$. However, this is a contradiction if $2g + 1 > N(\mathfrak{p})$, noting that $\# \mathcal{O}_K / \mathfrak{p} = N(\mathfrak{p})$. \qed \\
We remark that the above inequality is tight, since given any prime $\mathfrak{p}$ with $N(\mathfrak{p}) > 2g$, we can simply let $0, 1, \lambda_1, \dots, \lambda_{2g-1}$ be some distinct representative elements in the residue field to yield an example of a curve $C$ with good reduction at $\mathfrak{p}$. \\
This result immediately implies that there are only finitely many $\overline{K}$-isomorphism classes of genus $g$ hyperelliptic curves $C/K$ with Weierstrass points in $K$ having potentially good reduction outside at most $\pi_{K, \textrm{odd}}(2g)$ odd primes. However, remarkably we can go one step further: \\
\refstepcounter{mainnum} \label{thm:mainlowerboundtheorem}
\textbf{Theorem \themainnum.} There are only finitely many $\overline{K}$-isomorphism classes of genus $g$ hyperelliptic curves $C/K$ with rational Weierstrass points in $K$ having potentially good reduction outside at most $\pi_{K, \textrm{odd}}(2g) + 1$ primes. \\
In order to prove the above theorem, we shall make use of the following elementary (albeit technical) lemma: \\
\refstepcounter{mainnum} \label{thm:mainlemma}
\textbf{Lemma \themainnum.} Let $K$ be a number field, and $S$ a fixed finite set of primes of $K$. Then there exist only finitely many odd primes $\mathfrak{p}$ such that there exist distinct $T$-units $x, y, z \in \mathcal{O}_T^\times$ where $T = S \cup \{\mathfrak{p} \}$ such that $x-y$, $x-z$, and $y-z$ are all $T$-units, and assuming $v_\mathfrak{p}(x), v_\mathfrak{p}(y), v_\mathfrak{p}(z), v_\mathfrak{p}(x-y), v_\mathfrak{p}(x-z), v_\mathfrak{p}(y-z)$ are not all equal. \\
The proof of this lemma proceeds by analysing various three-term $S$-unit equations, and thus we shall make essential use of the following finiteness result: \\
\refstepcounter{mainnum} \label{thm:sunit3finite}
\textbf{Theorem \themainnum.} \cite[p.~131]{sunit3} Let $K$ be a number field and $S$ a fixed finite set of primes of $K$. Then the equation $u+v+w = 1$ has only finitely many solutions in $u, v, w \in \mathcal{O}_S^*$, such that $u, v, w \not = 1$ (i.e. only finitely many non-degenerate solutions). \\
With the above theorem under our belt, we can now prove Lemma \ref{thm:mainlemma}. \\
\textit{Proof of Lemma \ref{thm:mainlemma}.} We first fix some odd prime $\mathfrak{p}$ of $K$, and shall aim to derive a set of equations which can only be satisfied for finitely many $\mathfrak{p}$. With this in mind, we can first assume without loss of generality that $v_\mathfrak{p}(x) \geq v_\mathfrak{p}(y) \geq v_\mathfrak{p}(z)$. For brevity we denote $s := \frac{x}{z}$ and $t := \frac{y}{z}$ and define $a := v_\mathfrak{p}(s)$ and $b := v_\mathfrak{p}(t)$, noting that $a, b$ are non-negative integers where $a \geq b$.
We also define $c := v_\mathfrak{p}(s - 1)$, $d := v_\mathfrak{p}(t-1)$ and $e := v_\mathfrak{p}(s-t)$, noting that $s, t$ and $s-t$ are all $T$-units. \\
The proof now proceeds by considering the various cases for $a$ and $b$. In each case, the main idea is to obtain a three term $S$-unit equation from which we will obtain only finitely many solutions.
\begin{itemize}
\item \textbf{Case 1: } $a, b > 0$ and $a > b$. This implies $c = d = 0$ and $e = b$, and thus we have
\begin{equation*}
s - 1 = u, \quad \textrm{and} \quad
t - 1 = v,
\end{equation*}
for some $u, v \in \mathcal{O}_S^\times$. As $v_\mathfrak{p}(u-v) = v_\mathfrak{p}(t)$, we have that $t/(u-v)$ is an $S$-unit, and thus by rearranging, we obtain the three term $S$-unit equation:
\begin{equation} \label{eq:case1}
\frac{t}{u-v} u - \frac{t}{u-v} v - v = 1 .
\end{equation}
At this stage, we would like to apply Theorem \ref{thm:sunit3finite} in order to conclude that there are only finitely many solutions to the above equation. We must therefore check that we do not obtain (or only obtain finitely many) degenerate solutions where one of the above terms equals 1:
\begin{itemize}
\item[(i)] If the first term of (\ref{eq:case1}) is 1, then $t = v-u$ which implies $x = 0$, contradiction.
\item[(ii)] If the second term of (\ref{eq:case1}) is 1, then $t v = v-u$, which implies $s = t (1-v)$ and hence $1-v$ has positive $\mathfrak{p}$-adic valuation. But $v-1 = t - 2$ which yields a contradiction, since $\mathfrak{p}$ is odd.
\item[(iii)] If the third term of (\ref{eq:case1}) is 1, then $t = 0$, contradiction.
\end{itemize}
Therefore, we have a three-term $S$-unit equation with no degenerate solutions. Thus, there are only finitely many solutions to (\ref{eq:case1}), and thus only finitely many $v$, and thus clearly only finitely many $\mathfrak{p}$, noting that $b$ is positive.
\item \textbf{Case 2: } $a, b > 0$ and $a = b$. As before, we have
\begin{equation*}
s - 1 = u, \quad \textrm{and} \quad
t - 1 = v
\end{equation*}
for some $u, v \in \mathcal{O}_S^\times$. Noting that $v_\mathfrak{p}(t/s) = 0$, by rearranging, we obtain the three term $S$-unit equation:
\begin{equation} \label{eq:case2}
\frac{t}{s} u + \frac{t}{s} - v = 1
\end{equation}
Once again, we check the three cases:
\begin{itemize}
\item[(i)] If the first term of (\ref{eq:case2}) is 1, then $tu = s$ and $t = vs$ which implies $uv = 1$. Now by multiplying the first two equations we get
\begin{equation*}
st - (s+t) + 1 = (s-1)(t-1) = uv = 1
\end{equation*}
which implies $v_\mathfrak{p}(s+t) = v_\mathfrak{p}(st) = 2a > a$, and thus $v_\mathfrak{p}(s-t) = v_\mathfrak{p}(s+t - 2t) = a$ as $\mathfrak{p}$ odd. This thus yields the following two-term $S$-unit equation:
\begin{equation*}
\frac{s}{t} - \frac{s-t}{t} = 1
\end{equation*}
which implies finitely many values for $\frac{s}{t}$ and thus for $u$, and so only finitely many values for $\mathfrak{p}$.
\item[(ii)] If the second term of (\ref{eq:case2}) is 1, then $x = y$, contradiction.
\item[(iii)] If the third term of (\ref{eq:case2}) is 1, then $u = 1$ and thus $x/z = 2$, contradiction.
\end{itemize}
Therefore, as before, only finitely many solutions.
\item \textbf{Case 3: } $a > 0$ and $b = 0$. We therefore have $c = 0$ and $e = 0$. This yields
\begin{equation*}
s - 1 = u, \quad
s - t = w
\end{equation*}
for some $u, w \in \mathcal{O}_S^\times$, which yields the three term $S$-unit equation:
\begin{equation} \label{eq:case3}
w + t - u = 1
\end{equation}
\begin{itemize}
\item[(i)] If $w = 1$, then $v_\mathfrak{p}(t-1) = v_\mathfrak{p}(s - 2) = 0$, as $\mathfrak{p}$ odd. Therefore $t - 1 = v$ for some $v \in \mathcal{O}_S^\times$. This yields a 2-term $S$-unit equation, of which there are only finitely many solutions for $t, v$, and thus for $u$, hence only finitely many for $p$.
\item[(ii)] If $t = 1$, then $y = z$, contradiction.
\item[(iii)] If $u = -1$. then $x = 0$, contradiction.
\end{itemize}
\item \textbf{Case 4: } $a = b = 0$ and $c > d$. This implies $e = d$ which yields the three term $S$-unit equation:
\begin{equation} \label{eq:case4}
\frac{t-1}{s-t} t - \frac{t-1}{s-t} s + t = 1
\end{equation}
Firstly, if $d = 0$, then $v_\mathfrak{p}(t-1) = 0$ which yields a two term $S$-unit equation of which there are only finitely many solutions. Thus, we may assume $d > 0$.
\begin{itemize}
\item[(i)] If the first of (\ref{eq:case4}) is 1, then $s - t = t(t-1)$ which implies $s = t^2$. This yields
\begin{equation*}
s - 1 = (t-1)(t+1)
\end{equation*}
which implies $t+1$ has positive $\mathfrak{p}$-adic valuation. But $t + 1 = (t-1) + 2$ which yields a contradiction as $\mathfrak{p}$ odd.
\item[(ii)] If the second term of (\ref{eq:case4}) is 1, then $t-1 = t - s$ and so $s = 1$, contradiction.
\item[(iii)] If the third term of (\ref{eq:case4}) is 1, then $y = z$, contradiction.
\end{itemize}
\item \textbf{Case 5: } $a = b = 0$ and $c = d$. Again, note that if $c = d = 0$, then $s$ and $t$ satisfy two-term $S$-unit equations, of which there are only finitely many solutions. This thus implies only finitely many $p$, since we'd then have $v_\mathfrak{p}(s-t) > 0$ by assumption.
Now assume $c, d \not = 0$, and note that $c, e$ must necessarily be positive. We obtian the three term $S$-unit equation:
\begin{equation} \label{eq:case5}
\frac{s-1}{t-1} - \frac{s-1}{t-1}t + s = 1
\end{equation}
\begin{itemize}
\item[(i)] If the first term of (\ref{eq:case5}) is 1, then $s = t$, contradiction.
\item[(ii)] If the second term of (\ref{eq:case5}) is 1, then $(s-1)t = - (t-1)$ and $(s-1) = -s(t-1)$ which implies $st = 1$. By a dual argument to the above case 4(i), we have $v_\mathfrak{p}(t+1) = 0$. This implies
\begin{equation*}
e = v_\mathfrak{p}(s - t) = v_\mathfrak{p}(1-t^2) = v_\mathfrak{p}(1-t) = c
\end{equation*}
This implies that we have the following two term $S$-unit equation:
\begin{equation*}
\frac{s-1}{t-1} - \frac{s-t}{t-1} = 1
\end{equation*}
which implies only finitely many solutions for $\frac{s-1}{t-1}$. Therefore, this gives finitely many $t$, and thus finitely many $\mathfrak{p}$.
\item[(iii)] If $s = 1$, then $x = z$, contradiction.
\end{itemize}
\item \textbf{Case 6: } $a = b = 0$ and $c < d$. Done analogously to case 4.
\end{itemize}
Therefore, in each case, we obtain only finitely many valid primes $\mathfrak{p}$, which concludes the proof. \qed \\
We note that effectively obtaining a list of all possible primes $\mathfrak{p}$ depends entirely on the effectiveness of solving the above three term $S$-unit equations. From the results of \cite{sunit3}, no finite algorithm has been found to determine all possible solutions, however one can obtain an explicit bound on the number of possible $\mathfrak{p}$, which for a fixed number of terms, is exponential in $|S|$ \cite[p.~132]{sunit3}. \\
We are now finally ready to prove Theorem \ref{thm:mainlowerboundtheorem}: \\
\textit{Proof of Theorem \ref{thm:mainlowerboundtheorem}.} Let $C/K$ be a hyperelliptic curve of genus $g$ given in Rosenhain normal form $C : y^2 = x(x-1)(x-\lambda_1) \cdots (x - \lambda_{2g-1})$
with Weierstrass points in $K$. By Theorem \ref{thm:norm}, $C$ cannot have potentially good reduction at any odd primes with norm less than $2g$. Now assume $C$ has potentially good reduction outside exactly $\pi_{K,\textrm{odd}}(2g)+1$ odd primes $S$. \\
Thus, $S$ must consist of all $\pi_{K,\textrm{odd}}(2g)$ primes with norm below $2g$, plus one additional prime $\mathfrak{p}$. Now by Corollary \ref{cor:Cpgr}, we must have that $\lambda_1, \lambda_2, \lambda_1 - 1, \lambda_2 - 1$ and $\lambda_1 - \lambda_2$ are all $S$-units. Therefore, by Lemma \ref{thm:mainlemma}, there are only finitely many possible primes $\mathfrak{p}$, and thus by either applying a theorem of Faltings \cite[p.~25]{cornellsilverman} or by finiteness of solutions to $S$-unit equations, we obtain only finitely many $\overline{K}$-isomorphism classes of hyperelliptic curves with potentially good reduction outside $S$. \qed \\
We do remark that only the Weierstrass points $0, 1, \lambda_1,$ and $\lambda_2$ were used in the proof above, whilst Corollary \ref{cor:Cpgr} does include constraints on all the Weierstrass points $\lambda_i$. Indeed, if we were to use all $\lambda_i$, we would expect to prove a significantly stronger lower bound for $c_K(g)$.
A heuristic argument suggests that if we generalise Lemma \ref{thm:mainlemma} where we adjoin an additional $k$ primes $T := S \cup \{ \mathfrak{p}_1, \dots, \mathfrak{p}_k \}$, then assuming we don't encounter any degenerate solutions, this could yield a potential linear lower bound of $g + \pi_{K, \textrm{odd}}(2g)$ for $c_K(g)$. However, besides extending the case bash analysis for small values of $k$, we do not know at this stage how to produce such a proof for arbitrary $k$. \\
We shall now restrict our attention to the specific case where $C$ is a hyperelliptic curve over $\mathbb{Q}$, with all of its Weierstrass points in $\mathbb{Q}$.
\section{Hyperelliptic curves with rational Weierstrass points}
Firstly, it's worth stating the application of Theorem \ref{thm:norm} to the rational case: \\
\refstepcounter{mainnum} \label{thm:rationalWeier}
\textbf{Corollary \themainnum.} Let $C/\mathbb{Q}$ be a hyperelliptic curve with rational Weierstrass points. Then $C$ cannot have potentially good reduction at any odd prime $p \leq 2g$. \\
Note that this clearly implies that no genus 2 hyperelliptic curve with rational Weierstrass points has potentially good reduction at 3. This corollary can be applied to give a short proof of the following result from Box and Le Fourn \cite{box}, which was originally proven using a two-dimensional analogue of Baker's and Runge's method applied to the Siegel variety $A_2(2)$ (i.e. the moduli space of principally polarised abelian surfaces with full $2$-torsion).\\
\refstepcounter{mainnum} \label{thm:box}
\textbf{Corollary \themainnum.} \cite[p.~3]{box} There is no genus 2 hyperelliptic curve $C$ over $\mathbb{Q}$ such that all Weierstrass points of $C$ are rational and $C$ has potentially good reduction at all but one of the primes. \\
\textit{Proof.} As shown above, such a curve $C$ cannot have potentially good reduction at $3$. Now assume for contradiction such a curve has potentially good reduction outside $3$. By applying Corollary \ref{cor:Cpgr}, we can now effectively compute all genus 2 curves $C/\mathbb{Q}$ with rational Weierstrass points having potentially good reduction outside $S = \{2, 3\}$. \\
By Corollary \ref{cor:Cpgr}, we proceed by solving the $S$-unit equation $x+y = 1$, where $x, y \in \mathcal{O}_S^\times$.
These solutions can be computed using existing algorithms, such as those described by von K\"{a}nel and Matschke \cite{vonkanel}. Using their Sage \cite{sage} implementation, we obtained 21 solutions, and can conclude that any such curve must be isomorphic to one of the following curves:
\begin{align*}
C_1 : y^2 &= cx(x-1)(x-2)(x-3)(x-4), \quad \textrm{with } \Delta_\textrm{min} = 2^{18} 3^4 \\
C_2 : y^2 &= cx(x-2)(x-3)(x-4)(x-6), \quad \textrm{with } \Delta_\textrm{min} = 2^{14} 3^6
\end{align*}
where $c \in \mathbb{Z}$ is some squarefree integer. \\
After verifying that each of the curves above do not have potentially good reduction at 2, this gives us our contradiction, and thus the result holds. \qed \\
\section{Upper bounds}
It's worth mentioning some of the results we can obtain regarding upper bounds for $c_K(g)$. Most of the more interesting results are conditional on various conjectures concerning the distribution of primes. \\
For a given number field $K$, we recall that a $k$-tuple $(h_1, \dots, h_k)$ of distinct elements in $\mathcal{O}_K$ is \textit{admissible} if the set $\{h_1, \dots, h_k\}$ does not consist of all residues mod $\mathfrak{p}$, for every prime $\mathfrak{p}$ in $K$. We also say that an element $x \in \mathcal{O}_K$ is \textit{prime} if the principal ideal generated by $x$ is prime. We now first recall the Hardy-Littlewood prime $k$-tuples conjecture for $K$: \\
\textbf{Conjecture.} (\textit{Hardy-Littlewood prime $k$-tuples conjecture for number fields}) Let $K$ be a number field and $(h_1, \dots, h_k)$ an admissible $k$-tuple in $\mathcal{O}_K$. Then there exist infinitely many $x \in \mathcal{O}_K$ such that each of $x+h_1, \dots, x+h_k$ is prime. \\
Notably, one can therefore prove the following result, conditional on the assumption of the above conjecture. \\
\refstepcounter{mainnum} \label{thm:upperDickson}
\textbf{Theorem \themainnum.} Let $K$ be a number field of degree $n$. Under the assumption of the Hardy-Littlewood prime $k$-tuples conjecture for $K$, then $c_K(g) \leq 2g - 1 + n \pi(2g)$. \\
\textit{Proof.} For a given genus $g \geq 2$, we consider the following admissible prime $2g-1$ tuple $(h_1, \dots, h_{2g-1})$:
\begin{equation} \label{eq:admissibletuple}
(0, \; (2g)!, \; 2 \cdot (2g)!, \; 3 \cdot (2g)!, \; \dots, (2g-2) \cdot (2g)! )
\end{equation}
Now by the prime $k$-tuples conjecture, there exist infinitely many primes $p$ in $K$ such that $p+h_1, \dots, p+h_{2g-1}$ are all prime. Thus, for each such prime $p$, we can construct the genus $g$ hyperelliptic curve $C_p/K$ as
\begin{equation*}
C_p : y^2 = x(x-p-h_1)(x-p-h_2) \cdots (x-p-h_{2g-1}) (x - 2p - 2g \cdot (2g)! ) .
\end{equation*}
As the only possible primes of bad reduction are those which divide the differences between Weierstrass points, it's clear that the only possible primes of bad reduction are either the primes $p+h_1, \dots, p+h_{2g-1}$ or the primes dividing $(2g)!$, of which there are at most $n \pi(2g)$. \\
This therefore yields the existence of infinitely many genus $g$ hyperelliptic curves $C/K$ satisfying $\# \mathcal{B}_\textrm{odd} (C/K) \leq 2g-1 + n \pi(2g)$. Furthermore, as this yields infinitely many different sets of bad primes $\mathcal{B}_\textrm{odd} (C/K)$, this gives rise to infinitely many $\overline{K}$-isomorphism classes of such curves. This therefore yields the conditional bound $c_K(g) \leq 2g-1 + n \pi(2g)$. \qed \\
Whilst the above result gives a conditional linear upper bound for $c_K(g)$, it's worth noting that we can also give an unconditional linearithmic bound for $c_K(g)$. \\
\refstepcounter{mainnum} \label{thm:upper}
\textbf{Theorem \themainnum.} Let $K$ be a number field of degree $n$. We have $c_K(g) \leq (\frac{2}{\log{2}} + o(1)) n g \log{g}$. \\
\textit{Proof.} Let $(h_1, \dots, h_{2g-1})$ be the same admissible prime tuple as given in (\ref{eq:admissibletuple}). We shall apply the result of Murty and Vatwani \cite[p.~183]{murtyvatwani}, which asserts the existence of infinitely many integers $k$ such that $(k + h_1) \cdots (k + h_{2g-1})$ has at most $(\frac{2}{\log{2}} + o(1)) g \log{g}$ prime divisors in $\mathbb{Q}$. By therefore considering the set of genus $g$ hyperelliptic curves
\begin{equation*}
C_k : y^2 = x(x-k-h_1) \cdots (x-k-h_{2g}) (x - 2k - 2g \cdot (2g)!)
\end{equation*}
this yields the desired upper bound. \qed \\
It's tempting to ask how far we can push our conditional upper bounds. Whilst a sublinear bound is almost certainly out of reach, we can sharpen the above theorem if we furthermore assume the following generalisation to the Hardy-Littlewood prime $k$-tuples conjecture. This goes by various different names, often called Schinzel's hypothesis H, generalised Dickson's conjecture, or the generalised Bunyakovsky conjecture. \\
\textbf{Conjecture.} (\textit{Schinzel's hypothesis H for number fields}) Let $K$ be a number field and $(f_1, \dots, f_k)$ a collection of $k$ distinct nonconstant irreducible polynomials in $\mathcal{O}_K[x]$, such that
for all primes $\mathfrak{p}$ in $K$, there exists an $n \in \mathcal{O}_K$ where $v_\mathfrak{p}(f_1(n) f_2(n) \cdots f_k(n)) = 0$.
There there exist infinitely many $x \in \mathcal{O}_K$ such that each of $f_1(x), \dots, f_k(x)$ is prime. \\
Under the assumption of the above conjecture, we can prove the following sharpened upper bound for $c_K(g)$. \\
\refstepcounter{mainnum} \label{thm:uppergenDickson}
\textbf{Theorem \themainnum.} Let $K$ be a number field of degree $n$. Assuming Schinzel's hypothesis H for $K$, we have that
\begin{equation} \label{eq:uppermaingenbound}
c_K(g) \leq \sum_{\substack{1 \leq d < g, \textrm{ or} \\ d < 2g, \, d \textrm{ even}}} \frac{n}{[K(\zeta_d) : \mathbb{Q}(\zeta_d)]} + 1 + n \pi(2g)
\end{equation}
\textit{Proof.} For brevity, we shall denote $\alpha := (2g)!$. The idea is to consider, for infinitely many $k$, genus $g$ hyperelliptic curves of the form
\begin{equation*}
C_k : y^2 = x(x-1)(x+1)(x-\alpha k)(x+\alpha k)(x-(\alpha k)^2)(x+(\alpha k)^2) \cdots (x- (\alpha k)^{g-1} ) (x + (\alpha k)^{g-1} )
\end{equation*}
We note that the only possible primes of bad reduction are those which divide $2 \alpha k$, $(\alpha k)^d - 1$, or $(\alpha k)^d + 1$ for some $d < g$. Under the assumption of Schinzel's hypothesis H, it thus suffices to count the number of irreducible factors of $(\alpha x)^d \pm 1$ over $K$.
We know that $(\alpha x)^d \pm 1$ factorises over $\mathbb{Q}$ as
\begin{equation*}
(\alpha x)^d - 1 = \prod_{i | d} \Phi_i(\alpha x), \quad \textrm{and} \quad (\alpha x)^d + 1 = \prod_{\substack{i | 2d \\ i \not | d}} \Phi_i(\alpha x),
\end{equation*}
where $\Phi_i(x)$ denotes the $i$-th cyclotomic polynomial. Furthermore, the factorisation of $\Phi_i(x)$ over $K$ can be given as $\Phi_i(\alpha x) = f_{i,1}(\alpha x) \cdots f_{i, \ell_i}(\alpha x)$ where each $f_{i,j}(x)$ has degree $[K(\zeta_i) : K]$, and hence $\ell_i = \frac{\varphi(i)}{[K(\zeta_i) : K]} = \frac{n}{[K(\zeta_d) : \mathbb{Q}(\zeta_d)]}$ by tower law. \\
Now Schinzel's hypothesis H states that we can find infinitely many primes $p$ in $K$ such that $f_{i,j}(\alpha p)$ are all prime, noting that the factor of $\alpha$ ensures we have no local obstructions to primality. By thus counting the primes dividing $(\alpha p)^d \pm 1$, the prime $p$, and the primes dividing $2 \alpha$, this therefore yields the conditional bound
\begin{equation*}
c_K(g) \leq \sum_{\substack{1 \leq d < g, \textrm{ or} \\ d < 2g, \, d \textrm{ even}}}
\frac{n}{[K(\zeta_d) : \mathbb{Q}(\zeta_d)]}
+ 1 + n \pi(2g)
\end{equation*}
which yields our result. \qed \\
Whilst the above construction does not necessarily improve upon the result given in Theorem \ref{thm:upperDickson} for all fields $K$, one can obtain the following two corollaries: \\
\refstepcounter{mainnum} \label{thm:upperCor1}
\textbf{Corollary \themainnum.} Let $K$ be a primitive abelian number field of (necessarily prime) degree $n$ and conductor $\mathfrak{f}_K$. Then assuming Schinzel's hypothesis H for $K$, we have
\begin{equation*}
c_K(g) \leq \begin{cases}
\frac{3}{2} g \big(1 + \frac{n-1}{\mathfrak{f}_K} \big) + 1 + n \pi(2g) & \textrm{ if $\mathfrak{f}_K$ odd, } \\[2mm]
\frac{3}{2} g \big( 1 + \frac{4(n-1)}{3 \mathfrak{f}_K} \big) + 1 + n \pi(2g) & \textrm{ if $\mathfrak{f}_K$ even. }
\end{cases}
\end{equation*}
\textit{Proof.} Since $K$ has conductor $\mathfrak{f}_K$, this implies $[K(\zeta_d) : \mathbb{Q}(\zeta_d)] = 1$ if $\mathfrak{f}_K$ divides $d$, and $[K(\zeta_d) : \mathbb{Q}(\zeta_d)] = n$ otherwise, noting that $K$ is primitive. We can therefore easily evaluate the bound given in (\ref{eq:uppermaingenbound}) as
\begin{equation*}
c_K(g) \leq
\sum_{\substack{1 \leq d < g, \textrm{ or} \\ d < 2g, \, d \textrm{ even} \\ \mathfrak{f}_K | d}} \! n \; +
\sum_{\substack{1 \leq d < g, \textrm{ or} \\ d < 2g, \, d \textrm{ even} \\ \mathfrak{f}_K \not | d}} \! 1
\quad
+ 1 + n \pi(2g)
\end{equation*}
If $\mathfrak{f}_K$ is odd, this evaluates to
\begin{equation*}
c_K(g) \leq \frac{3}{2} \frac{gn}{\mathfrak{f}_K} + \big( \frac{3}{2} g - \frac{3}{2} \frac{g}{\mathfrak{f}_K} \big) + 1 + n \pi(2g)
\end{equation*}
whilst if $\mathfrak{f}_K$ is even, this yields
\begin{equation*}
c_K(g) \leq
\frac{2gn}{\mathfrak{f}_K}
+ \big( \frac{3}{2} g - \frac{2g}{\mathfrak{f}_K} \big) + 1 + n \pi(2g)
\end{equation*}
which proves the result. \qed \\
\refstepcounter{mainnum} \label{thm:upperCor2}
\textbf{Corollary \themainnum.} Let $K$ be a number field such that its maximal abelian subfield is $\mathbb{Q}$. Then assuming Schinzel’s hypothesis H for $K$, we have $c_K(g) \leq \frac{3}{2} g + n \pi (2g)$. \\
\textit{Proof.} The above condition implies that $K \cap \mathbb{Q}(\zeta_d) = \mathbb{Q}$ for all $d$, and thus the bound given in (\ref{eq:uppermaingenbound}) implies our result. \qed \\
Finally, it's worth mentioning that the bound given in (\ref{eq:uppermaingenbound}) does not necessarily represent the optimal conditional bound for all genera $g$, even over $\mathbb{Q}$. For example, under the assumption of Schinzel's hypothesis H, there exist infinitely many integers $k$ such that $k$, $\alpha k - 1$, $\alpha k + 1$, $(\alpha k)^2 + 1$, $(\alpha k)^2 - 2(\alpha k) - 1$, and $(\alpha k)^2 - 2(\alpha k) + 1$ are all prime, where $\alpha := 7!$. \\
From this, one can therefore conditionally construct infinitely many genus $5$ curves $C/\mathbb{Q}$ of the form:
\begin{align*}
C_k : y^2 &= x \cdot
\big(x - (\alpha k)^2(\alpha k-1)(\alpha k+1)\big) \cdot \big(x + (\alpha k)^2(\alpha k-1)(\alpha k+1) \big) \\
&\qquad \cdot
\big(x - \alpha k(\alpha k-1)(\alpha k+1)\big) \cdot \big(x + \alpha k(\alpha k-1)(\alpha k+1) \big) \\
&\qquad \cdot \big(x - (\alpha k-1)(\alpha k+1)\big) \cdot\big(x + (\alpha k-1)(\alpha k+1) \big) \\
&\qquad \cdot
\big(x - \alpha k(\alpha k-1)^2\big) \cdot \big(x + \alpha k(\alpha k-1)^2 \big) \\
&\qquad \cdot
\big(x - \alpha k(\alpha k+1)^2\big) \cdot\big(x + \alpha k(\alpha k+1)^2 \big)
\end{align*}
which yields a conditional bound of $c_\mathbb{Q}(5) \leq 10$, and thus one better than the bound of 11 given by (\ref{eq:uppermaingenbound}). \\
Besides the above example, we should mention however that we haven't found any better examples for higher genera over $\mathbb{Q}$, noting that a naive computational search quickly becomes unmanageable for large genus hyperelliptic curves. \\
Finally, I would like to give my sincere thanks to my supervisor, Samir Siksek, for his amazing support and many insightful comments.
\input{bib}
\smallskip \footnotesize
\textsc{Mathematics Institute, University of Warwick, CV4 7AL, United Kingdom} \par \nopagebreak
\textit{E-mail address}: \texttt{robin.visser@warwick.ac.uk}
\end{document}
| 165,588
|
\begin{document}
\begin{Large}
\centerline{\bf Proof of W.M.Schmidt's conjecture }
\centerline{\bf concerning successive minima of a lattice}
\vskip+1.5cm \centerline{\bf Moshchevitin N.G. \footnote{ The
research was supported by the grant RFBR 09-01-00371a
}
}
\end{Large}
\vskip+1.5cm
\centerline{\bf Abstract} We prove W.M.Schmidt's conjecture about
a one-parameter family of lattices related to simultaneous
Diophantine approximations.
AMS 2000 subject classification: 11H06, 11J13. \vskip+1.5cm
\section{Introduction}
Consider real numbers $\xi_j \in[0,1), 1\le j\le n$. For a real $x$ we denote by $|x|$ the absolute value of $x$.
We know since Dirichlet that for any real $N>1$ the inequalities
$$
|x|\le N,\,\,\, \max_{1\le j\le n}|x\xi_j -y_j|\le N^{-1/n}
$$
have a solution in integers $x\neq 0, y_1,...,y_n$. Similarly for
any real $N>1$ the inequalities
$$
|x|\le N,\,\,\, \left(\sum_{j=1}^n|x\xi_j -y_j|^2\right)^{1/2}\le
2w_n^{-1/n} N^{-1/n}
$$
have a solution in integers $x\neq 0, y_1,...,y_n$ (here $w_n$
stands for the volume of the unit ball in the $n$-dimensional
Euclidean space).
In this paper we work with Euclidean space $\mathbb{R}^{n+1}$ with
coordinates $(x,y_1,...,y_n)$ and with Euclidean space
$\mathbb{R}^{n}$ with coordinates $(y_1,...,y_n)$.
Consider an $(n+1)$-dimensional vector $ \xi = (1,\xi_1,...,\xi_n)
\in \mathbb{R}^{n+1}$.
For a vector $y =(y_1,...,y_n) \in \mathbb{R}^n$ we define $|y | $
to be the Euclidean norm of $y$. So, $|y|
=\sqrt{y_1^2+...+y_n^2}.$ We also use the notation $|y|_s =
\max_{1\le j\le n} |y_j|$ for the sup-norm of a vector $y
=(y_1,...,y_n) \in \mathbb{R}^n$.
For a
real $N\ge 1$ and a vector $\xi $ we define a matrix
$$
{\cal A} (\xi, N) = \left(
\begin{array}{ccccc}
N^{-1} & 0& 0& \cdots &0 \cr N^{\frac{1}{n}} \xi_1 &
-N^{\frac{1}{n}} & 0&\cdots & 0 \cr N^{\frac{1}{n}} \xi_2 &0&
-N^{\frac{1}{n}} & \cdots & 0 \cr \cdots &\cdots &\cdots &\cdots
\cr N^{\frac{1}{n}} \xi_n &0&0&\cdots &- N^{\frac{1}{n}}
\end{array}\right)
$$
and a lattice
$$
\Lambda (\xi, N) = {\cal A} (\xi, N)\mathbb{Z}^{n+1}.
$$
Consider the $ (n+1)$-dimensional unit cube
$${\cal U} =
\{ z= (x,y_1,...,y_n)\in \mathbb{R}^{n+1}:\,\,\, \max ( |x|, |y|_s
)\le 1 \}
$$
and a convex 0-symmetric body
$${\cal W} =
\{ z= (x,y_1,...,y_n)\in \mathbb{R}^{n+1}:\,\,\, \max ( |x|, |y|
)\le 1 \}
$$
For a natural $l ,\, 1\le l \le n+1$ let $\lambda_l (\xi , N)$ be
the $l$-th successive minimum of ${\cal U}$ with respect to $
\Lambda (\xi, N)$ and let $\mu_l (\xi , N)$ be the $l$-th
successive minimum of ${\cal W}$ with respect to $ \Lambda
(\xi, N)$. By the second Minkowski theorem for convex body (see
\cite{Cassil}, Ch. VIII or \cite{SCH}, Ch.IV) we have
$$
\frac{1}{(n+1)!} \le \prod_{l=1}^{n+1} \lambda_l (\xi , N)
\le 1 ,\,\,\,
\frac{2^{n}}{w_n(n+1)!} \le \prod_{l=1}^{n+1} \mu_l (\xi , N)
\le \frac{2^{ n}}{w_n}.
$$
In the case $n=1$ we have $\mu_l (\xi , N)=\lambda _l (\xi , N), l
= 1,2$ .
Suppose that $\xi_1/\xi_2\not\in \mathbb{Q}$. Since there are arbitrary large values of $N$ with
$\mu_1(\xi , N)=\mu_2 (\xi , N)$, it may never happen that $
\mu_1(\xi , N)\to 0,\, N\to +\infty$.
Consider the general case. Suppose that numbers $\xi_1,...,\xi_n$
are linearly independent over $\mathbb{Z}$ together with $1$. Then
for $1\le k \le n$ there exist arbitrary large values of $N$ such
that $\mu_k(\xi , N)=\mu_{k+1} (\xi , N)$. But in the case $n>1$
it may happen that $ \mu_1(\xi , N)\to 0,\, N\to +\infty$.
Moreover from A.Khinthcine's result \cite{HINS} (see also
\cite{DS}) it follows that it may happen that $ \mu_{n-1}(\xi ,
N)\to 0,\, N\to +\infty$.
In this paper we prove the following theorem.
{\bf Theorem 1.}\,\,\,{\it Let $ 1\le k \le n-1$. Then there exist
real numbers $\xi_j \in[0,1), 1\le j\le n$, such that
$\bullet$\,\, $ 1,\xi_1,...,\xi_n$ are linearly independent over
$\mathbb{Z}$;
$\bullet$\,\, $\mu_k (\xi, N) \to 0$ as $ N\to \infty$;
$\bullet$\,\, $\mu_{k+2} (\xi, N) \to \infty$ as $ N\to \infty$. }
We make two remarks.
{\bf Remark 1.} The analogous result for $\lambda_l (\xi , N)$
was conjectured by W.M. Schmidt in \cite{S}. In this paper we
consider the Euclidean norm only for simplicity reasons. We must
note that the main result is valid not only for the Euclidean norm
but also for the sup-norm $|\cdot |_s$ (as it was conjectured in
W.M. Schmidt's paper \cite{S}).
{\bf Remark 2.} It is shown in Section 3 that Theorem 1 becomes
trivial without the condition on $ 1,\xi_1,...,\xi_n$ to be
linearly independent over $\mathbb{Z}$.
The construction in the proof of Theorem 1 in the case $k=1$ is
very simple. It
is close to the construction from \cite{MUMN}, where the
author gives a counterexample to J. Lagarias' conjecture
concerning the behavior of consecutive best simultaneous
Diophantine approximations (see \cite{Lag2}). We give a complete
proof of Theorem 1 in the case $ k=1$ in Section 2.
In the case $k>1$ the construction in the proof of Theorem 1 is a
little bit more difficult. It is close to procedures from
\cite{AM},\cite{MMZ} (See the author's review \cite{ME} for
related topics). We give a complete proof of Theorem 1 in the case
$ k >1$ in Sections 3-5.
In the proof we shall need the following notation.
By
$\mu_l ({\cal C};L )$ we denote the $l$-th successive minimum of a
convex 0-symmetric set ${\cal C}$ with respect to a lattice $L$.
Let $w_l$ denote the volume of the unit ball in the
$l$-dimensional Euclidean space.
For a set ${\cal M} \subset \mathbb{R}^{n+1}$ we denote by
$\overline{\cal M} $ the smallest closed set containing ${\cal
M}$. We also denote the smallest linear and affine subspaces of
$\mathbb{R}^{n+1}$ containing
${\cal M}$, by
${\rm span}\,{\cal M} $ and ${\rm
aff}\,{\cal M} $, respectively.
Consider a sublattice $\Lambda \subset \mathbb{Z}^{n+1}$. By
${\rm dim}\,\Lambda$ we denote the dimension of the linear
subspace ${\rm span}\, \Lambda$. A sublattice $\Lambda\subset
\mathbb{Z}^{n+1}$ is defined to be {\it complete} if
$$
\Lambda = ({\rm span}\,\Lambda )\cap \mathbb{Z}^{n+1},
$$
that is in the linear subspace ${\rm span}\, \Lambda$ there is no
integer points different from points of $\Lambda$.
For every positive $Q$ and $\sigma$ we define a cylinder ${\cal
C}_\xi (Q, \sigma) \subset \mathbb{R}^{n+1}$ as follows:
$$
{\cal C}_\xi (Q, \sigma) = \left\{ z=(x,y) \,\, x\in
\mathbb{R},\,\, y =(y_1,...,y_n) \in \mathbb{R}^n :\,\,\, |x|<Q,
\left( \sum_{j=1}^n|y_j-\xi_j x|^2\right)^{1/2}<\sigma \right\}.
$$
The quantities $ \mu_l(\xi , N)$ coincide with the successive
minima of $\mathbb{Z}^{N+1}$ with respect to $ {\cal C}_\xi (N,
N^{-1/n} )$, that is
$$
\mu_l(\xi , N) = \mu_l ({\cal C}_\xi (N, N^{-1/n}
);\mathbb{Z}^{n+1}).
$$
\section{ Proof of Theorem 1: case $ k=1$}
We need two auxiliary
results - Lemmas A and B.
{\bf Lemma A.}\,\,\,{\it Let $\xi = \left( 1, \frac{a_1}{q}, ... ,
\frac{a_n}{q}\right)\in [0,1]^{n+1}$ be a rational vector. Suppose
that for integers $q,a_1,...,a_n \in \mathbb{Z}$ we have
$$ q\ge 1, (q,a_1,...,a_n) = 1.
$$
Then for any positive $U >0$ and any natural $i$ there exists a
positive real number
$$
\eta = \eta (\xi, i, U) >0
$$
such that for every real vector $\xi' = (1, \xi_1',...,\xi_n')$
under condition $ |\xi' - \xi| < \eta $ the inequalities
$$
\mu_1 (\xi', N) \le i^{-1},\,\,\,\, \mu_2(\xi', N) \ge i$$ are
valid for all $N$ in the interval
$$
(2qi\sqrt{n+1})^n \le N \le U.
$$}
Proof. First of all we note that for $N \ge q$ we obviously have
$$
\mu_1 (\xi, N) \le qN^{-1}.
$$
Besides that, the Euclidean distance between the one-dimensional
subspace $ {\rm span} \, \xi $
and the set $ \mathbb{Z}^{n+1}\setminus {\rm span} \, \xi $ is
not less than $ (q\sqrt{n+1})^{-1}$. Thus, in order to catch an integer point,
independent with $\xi$,
in the
cylinder $t{\cal C}_\xi (N, N^{-1/n})$, we should take $t$ to be not less than $ N^{1/n}(q\sqrt{n+1})^{-1}$. Hence
$$ \mu_2(\xi , N ) \ge N^{\frac{1}{n}}q^{-1}.
$$
From the hypothesis of Lemma A we deduce that the inequalities
$$
\mu_1 (\xi, N) \le (2i)^{-1},\,\,\,\, \mu_2(\xi, N) \ge 2i$$ hold
for all $N \ge (2qi\sqrt{n+1})^n$. Now Lemma A follows from the
observation that
for any $l$ the function $ \mu_l (\xi , N)$ is a continuous function in $\xi$ and $N$.
Lemma A is proved.
{\bf Lemma B.}\,\,\,{\it Let $\Gamma $ be a two-dimensional
complete sublattice of $\mathbb{Z}^{n+1}$. Let $R$ be the
two-dimensional fundamental volume of $\Gamma$ and let $\rho
=\rho(\Gamma ) >0$ be the Euclidean distance between $ {\rm span
} \,\Gamma$ and $\mathbb{Z}^{n+1}\setminus \Gamma$.
Suppose
that $ \xi = (1,\xi_1,...,\xi_n)\in {\rm span } \,\Gamma$. Then
for any positive $N$ we have the following estimates:
$$
\mu_1(\xi ,N ) \le N^{\frac{1-n}{2n}} R^{\frac{1}{2}},\,\,\, \mu_3
(\xi, N) \ge N^{\frac{1}{n}} \rho .
$$}
Proof. First of all we prove the upper bound for $\mu_1(\xi ,N
)$.
The intersection of the cylinder
${\cal C}_\xi (N, N^{-1/n})$ with $ {\rm span } \,\Gamma$ is an
$0$-symmetric parallelogram,whose two-dimensional volume greater
than or equal to $4N^{\frac{n-1}{n}}$. Suppose that $
4t^2N^{\frac{n-1}{n}} > 4R$ for some $t >0$. Then, by the
Minkowski convex body theorem there is a nonzero point of
$\Gamma$ inside the parallelogram $t{\cal C}_\xi (N,
N^{-1/n})\cap {\rm span } \,\Gamma$. So, for any $ t
> N^{\frac{1-n}{2n}} R^{\frac{1}{2}}$ the cylinder $t{\cal C}_\xi (N, N^{-1/n})$ contains a nonzero integer point and the upper bound for
$\mu_1(\xi ,N )$ is proved.
To prove the lower bound for
$\mu_3
(\xi, N) $ we need to take into account that if
the cylinder $t{\cal C}_\xi
(N, N^{-1/n})$ contains more than two linearly independent integer
points, then one of these points does not belong to $ \Gamma$
and
$ tN^{-1/n}\ge \rho$.
Lemma B is proved.
Now we describe the inductive procedure which gives the proof of
Theorem 1 in the case $k=1$.
The set of all $n$-dimensional sublattices of $\mathbb{Z}^{n+1}$
is countable. We fix an enumeration of this set and let
$$ L_1, L_2,..., L_i , ...
$$
be the sequence of all $n$-dimensional complete sublattices of
$\mathbb{Z}^{n+1}$.
Set $\pi_i = {\rm span} \,L_i$. Suppose that $$\pi_1 = \{z=(x,y_1,...,y_n)\in\mathbb{R}^{n+1}:\,\ x=0\}.$$
We construct a sequence of rational vectors
$$
\xi_i = \left( 1, \frac{a_{1,i}}{q_i}, ... ,
\frac{a_{n,i}}{q_i}\right),\,\,\,\, q_i,a_{1,i},...,a_{n,i} \in
\mathbb{Z},\,\,\, q_i\ge 1, (q_i,a_{1,i},...,a_{n,i}) = 1,\,\,\, i
= 1,2,3,... $$ with $ q_i \to \infty$ as $ i \to \infty$,
a sequence of
two-dimensional complete sublattices $\Gamma_{i+1} \subset
\mathbb{Z}^{n+1},\,\, i=1,2,3,...$,
and a sequence of positive real numbers
$$
\eta_1,\eta_2,...,\eta_i, ...
$$
satisfying the following conditions (i) - (iv).
(i) For every $ i \ge 1$ we have
$$
\xi_i, \xi_{i+1} \in {\rm span}\, \Gamma_{i+1}.
$$
(ii) The closed ball $\overline{\cal B}_i$ of radius $\eta_i$
centered at $\xi_i$ and has no common points with the subspace
$\pi_i$:
$$
\overline{\cal B}_i\bigcap \pi_i = \varnothing.
$$
(iii) The balls $\overline{\cal B}_i$ form a nested sequence:
$$
\overline{\cal B}_1 \supset \overline{\cal B}_2 \supset ...
\supset \overline{\cal B}_i .
$$
(iv) Let $H_0=1$ and
$$
H_i = (4q_i (i+1)\sqrt{n+1})^n,\,\, i = 1,2,3,... \,\, .
$$
Then for any $ i \ge 1$, for any $\xi \in\overline{\cal B}_i$
and for any $N$, such that
$$ H_{i-1}\le N <H_i,
$$
the following inequalities holds:
$$
\mu_1(\xi, N) \le i^{-1},\,\,\, \mu_3 (\xi, N) \ge i.
$$
Suppose all the objects are already constructed. Then from (ii),
(iii) one can easily see that $\lim_{i\to \infty} \eta_i = 0$.
Then the intersection $\cap_{i\in \mathbb{N}} \overline{\cal B}_i$
consists of the only one point. Note that for every $i$ the center
$\xi_i$ of the ball $\overline{\cal B}_i$ has its first
coordinate equal to one. So the unique point from the intersection
$\cap_{i\in \mathbb{N}} \overline{\cal B}_i$ is of the form
$$
\xi
=(1,\xi_1,...,\xi_n) $$ (as $\xi_i \to \xi,\,\, i \to \infty$).
Then
$$
\mu_1(\xi, N) \le i^{-1},\,\,\, \mu_3 (\xi, N) \ge i,\,\,\,
H_{i-1}\le N< H_i.
$$
Hence
$$
\mu_1(\xi, N) \to 0,\,\,\, \mu_3 (\xi, N) \to \infty,\,\,\, N\to
\infty,
$$
and it follows from the conditions (ii) and (iii) that
$1,\xi_1,...,\xi_n$ are linearly independent over $\mathbb{Z}$.
This proves Theorem 1 in the case $k=1$.
We start our inductive procedure with the vector
$$
\xi_1 = (1, \underbrace {0,...,0}_{n\,\,\text{times}}) .$$ Then $
H_1 = (8\sqrt{n+1})^n$. The sublattice $\Gamma_1$ is not defined
yet and we do not care about the condition (i) at this stage.
The condition (ii) is obviously satisfied for any choice of $
\eta_1 < 1$. The conditions (iii) is empty. Recall that $H_0 =
1$. To satisfy the condition (iv) we take $\eta_1$ to be small
enough.
Now we pass to the inductive step. Suppose that all the objects
$\xi_i, \Gamma_i, \eta_i $, $i \le t$ satisfying conditions (i)
-- (iv) are already constructed. We describe how to construct the
$(t+1)$-th set of objects.
First of all we can take a two-dimensional complete sublattice
$\Gamma_{t+1}$ satisfying the conditions
$$
q_t\xi_t = (q_t, a_{1,t},...,a_{n,t} ) \in \Gamma_{t+1},\,\,\,
\Gamma_{t+1} \not\subset \pi_{t+1} .
$$
Let $R_{t}$ be the two-dimensional fundamental volume of
$\Gamma_{t+1}$ and let $\rho_{t}$ be the Euclidean distance
between $
{\rm span } \,\Gamma_{t+1}$ and
$\mathbb{Z}^{n+1}\setminus \Gamma_{t+1}$.
Set
$$
U_{t} = \max\left( (2(t+1)\rho_{t}^{-1})^n,\,
(2(t+1)R_t^{1/2})^{\frac{2n}{n-1}} \right).
$$
Now we apply Lemma A with $\xi = \xi_t, i = 2(t+1)$ and $ U =
U_t$. We get a positive $\eta_t'$, such that for every $\xi '$
under the condition $|\xi ' - \xi_t|< \eta_t'$ one has
$$
\mu_1 (\xi', N) \le (2t+2)^{-1},\,\,\,\, \mu_2(\xi', N) \ge 2t+2
$$
for every $N$ in the interval
$$
H_t = (4q_t(t+1)\sqrt{n+1})^n \le N \le U_t. $$
Obviously, there is an integer point
$$
(q_{t+1},a_{1,t+1},...,a_{n,t+1}) \in \Gamma_{t+1}\setminus
\pi_{t+1}, \,\,\, q_{t+1} \ge q_t,\,\,\,
(q_{t+1},a_{1,t+1},...,a_{n,t+1}) = 1,
$$ such that for
$$
\xi_{t+1} = \left( 1, \frac{a_{1,t+1}}{q_{t+1}}, ... ,
\frac{a_{n,t+1}}{q_{t+1}}\right)
$$
we have
$$
|\xi_{t+1} - \xi_t | < \frac{\min ( \eta_t, \eta_t ')}{2}.
$$
Since $\xi_{t+1} \in \Gamma_{t+1}$, we can apply Lemma B with $\xi
= \xi_{t+1}, \Gamma =\Gamma_{t+1}$. This gives that for any $N$
under the condition
\begin{equation}
N \ge U_t \label{mai}
\end{equation}
one has
\begin{equation}
\mu_1 (\xi_{t+1}, N) \le (2(t+1))^{-1},\,\,\,\, \mu_3(\xi_{t+1},
N) \ge 2(t+1). \label{maine}
\end{equation}
But $|\xi_{t+1} - \xi_t|<\eta_t'$ and $\mu_3(\xi_{t+1}, N)
\ge\mu_2(\xi_{t+1}, N) $. So, the inequalities (\ref{maine}) are
valid not only for $N$ in the interval (\ref{mai}) but also for
$N$ in the interval $ N \ge H_t$. Having constructed $
\xi_{t+1}$, we define $H_{t+1} $ from the condition (iv) of the
$(t+1)$-th step of the inductive process.
Now we
take into account that
for any $l$ the function $ \mu_l (\xi , N)$ is a continuous function in $\xi$ and $N$.
This means that we can find a number $\eta_{t+1} < {\min ( \eta_t,
\eta_t ')}/{2}, $ such that
$$
\mu_1 (\xi , N) \le (t+1)^{-1},\,\,\,\, \mu_2(\xi, N) \ge t+1
$$
for all $\xi$ under the condition
$$
|\xi - \xi_{t+1}| \le \eta_{t+1}
$$
and all $N$ in the interval
$$
H_t \le N < H_{t+1}.
$$
Moreover, since $\xi_{t+1}\not\in \pi_{t+1}$, we can take
$\eta_{t+1}$ to be small enough, so that the ball $\overline{\cal
B}_{t+1}$ of radius $ \eta_{t+1}$ centered at $ \xi_{t+1}$ and
has no common points with $\pi_{t+1}$. The $(t+1)$-th step of the
inductive procedure is described completely and hence Theorem 1 in
the case $ k = 1$ is proved.
\section{Lemmas concerning successive minima and badly
approximable numbers}
In this section we start the proof of Theorem 1 in the general
case.
Let $l$ be an integer and $ 1\le l \le n$. First of all we should
say that everywhere in the sequel we consider $l$-dimensional
lattices $L$ such that for some real vector $\xi =
(1,\xi_1,...,\xi_n) \in \mathbb{R}^{n+1}$ one has
$$
\xi \in {\rm span }\, L.
$$
This condition leads to the following corollary concerning linear
subspace ${\rm span }\, L $. Consider affine subspace
$${\cal P} =\{
z=(x,y_1,...,y_n)\in \mathbb{R}^{n+1}:\,\, x = 1\} \subset
\mathbb{R}^{n+1}.
$$
Then the intersection
$$
{\rm span }\, L\cap {\cal P}
$$
is an affine subspace of dimension $ l-1$.
We
prove some auxiliary results on successive minima and badly
approximable numbers.
For positive integer $l$ and positive $\sigma \in (0,1)$ define
\begin{equation}\label{starr}
Q_1= Q_1(\sigma, l,s) = 2^{l-1}w_{l-1}^{-1} \sigma^{1-l}s,\,\,\,
Q_2 =Q_2(\sigma, l,s) = 2^{l-1}Q_1 = 2^{2(l-1)}w_{l-1}^{-1}
\sigma^{1-l}s.
\end{equation}
{\bf Lemma 1.}\,\,\,{\it Consider an integer $l$, such that $
2\le l\le n+1$. Consider a complete sublattice $ L \subseteq
\mathbb{Z}^{n+1}$ and suppose that $ {\rm dim}\, L = l$ and $
\xi \in {\rm span}\, L$. Suppose that the fundamental
$l$-dimensional volume of the lattice $L$ is equal to $s$.
Consider a cylinder ${\cal C} = {\cal C}_\xi (Q, \sigma)$. Suppose
also that for some positive $Q$ and $\sigma $ we have $ {\cal C}
\bigcap L = \{ 0\}. $ Then the following upper bounds are valid:
$$
\mu_1 ({\cal C} )\le Q_1Q^{-1}, \,\,\,\, \mu_m ({\cal C} )\le Q_2
Q^{-1} ,\,\,\, 2\le m \le l.
$$}
Proof. Consider the cylinder
$$
{\cal C}^{(1)}= {\cal C}_\xi (Q_1,\sigma)\bigcap {\rm span }\,L.
$$
As $ \xi \in {\rm span }\,L$ we see that each section
$$
{\cal C}^{(1)}\cap \{z=(x,y_1,...,y_n) \in \mathbb{R}^{n+1}:\,\, x
= x_0\}, \,\,\, |x_0| \le Q_1
$$
is a $(l-1)$-dimensional ball of the volume $w_{l-1}\sigma^{l-1}$.
Let $ H_1$ be the distance between $(l-1)$-dimensional facets of
the cylinder ${\cal C}^{(1)}$.
Then
the $l$-volume
of $ {\cal C}^{(1)}$ is equal to
$$
w_{l-1} H_1 \sigma^{l-1}.$$ But $H_1 \ge 2Q_1$. So the $l$-volume
of $ {\cal C}^{(1)}$ is
$$
\ge 2w_{l-1} Q_1 \sigma^{l-1} = 2^ls.
$$
(in the equality here we take into account the definition of $Q_1$
from (\ref{starr})).
By the
Minkowski's theorem for convex body, there is a nonzero integer
points $\zeta^{(1)} \in {\cal C}^{(1)}\cap L$.
So, $\mu_1 ({\cal C} )\le Q_1Q^{-1}$
and the bound for the first successive minimum is proved.
Here we should note that as $\sigma <1$ the first coordinate of
$\zeta^{(1)}$ is not equal to zero.
Now we describe an inductive process of constructing
linearly independent integer point $ \zeta^{(1)} ,...,
\zeta^{(l)}$ with non-zero first coordinates which ensure the
upper bound for the successive minima under consideration.
Suppose that linearly independent integer points $ \zeta^{(1)} ,..., \zeta^{(\nu)}\in {\rm span }\, L$ with $1\le \nu
\le l-1$ are already constructed. Set $ \pi = {\rm span }
(\zeta^{(1)} ,..., \zeta^{(\nu)})\subset {\rm span } \, L$. As
all the points $ \zeta^{(1)} ,..., \zeta^{(\nu)}$ are linearly
independent we see that ${\rm dim }\,\pi = \nu < l$. Note that the
dimension of the affine subspace $ \pi' = \pi\cap
\{z=(x,y_1,...,y_n) \in \mathbb{R}^{n+1}:\,\, x = Q\}$ is equal to
$ {\rm dim }\,\pi-1< l-1 $.
Consider the facet $ {\cal B} = \overline{\cal
C}\cap \{ x = Q\}$. This facet is an $n$-dimensional ball of
radius $\sigma$ centered at $Q \xi$.
Note that as $ \xi \in {\rm span}\, L$ we see that the
intersection ${\rm span}\,L\cap\{ x = Q\}$ is a
$(l-1)$-dimensional affine subspace. Consider the intersection $
{\cal B}\cap {\rm span}\, L$. As $Q \xi\in {\rm span}\, L $ we see
that this intersection is a $(l-1)$-dimensional ball centered at
$Q \xi$.
We have the following situation. In the $(l-1)$-dimensional affine
subspace ${\rm span}\,L\cap\{ x = Q\}$ there are the ball $ {\cal
B}\cap {\rm span}\, L$ of dimension $l-1$ and the affine subspace
$ \pi' $ of dimension ${\rm dim }\, \pi' < l-1$. So there
exists a $n$-dimensional ball ${\cal B}'\subset {\cal B} \subset
\{ x = Q\}$ of radius $\sigma/2$ centered at a certain point
$\Xi \in {\rm span}\,L\cap\{ x = Q\}$ and such its $(l-1)$-dimensional section $ {\cal B}'\cap {\rm span}\, L$ does not intersect with
$\pi'$:
$$ {\cal B}'\cap {\rm span}\,L \cap \pi =\varnothing.$$
In fact as $ \pi' \subset {\rm span}\,L \cap\{ x = Q\}$, it means
that
$$ {\cal B}'\cap \pi =\varnothing.$$
Put $ \xi^{(\nu+1)} = \frac{\Xi}{Q}$ and consider the cylinder
$$
{\cal C}^{(\nu+1)}= {\cal C}_{\xi^{(\nu+1)}} (Q_2,\sigma /2
)\bigcap {\rm span}\, L .
$$
As $\xi^{(\nu+1)}\in {\rm span}\, L $ from (\ref{starr}) we see
that the
$l$-volume
of $ {\cal C}^{(\nu+1)}$ is equal to
$$
w_{l-1} H_2\left(\frac{\sigma}{2}\right)^{l-1} ,$$ where $
H_2\ge 2Q_2$ is the distance between $(l-1)$-dimensional facets of
the cylinder ${\cal C}^{(\nu+1)}$. So the
$l$-volume
of $ {\cal C}^{(\nu+1)}$ is
$$
\ge 2w_{l-1} Q_2\left(\frac{\sigma}{2}\right)^{l-1} =
2^{l}s.
$$
Applying again the Minkowski theorem for convex body we get a nonzero integer point $\zeta^{(\nu+1)} \in
{\cal C}^{(\nu+1)}\cap L$.
As $\sigma <1$ we see that the first coordinate of
$\zeta^{(\nu+1)}$ is not equal to zero. Moreover the first
coordinate of the point $\zeta^{(\nu+1)}$ is greater than $Q$ as
there is no nonzero integer points in ${\cal C} \bigcap {\rm
span}\, L$ and
$$
{\cal C}^{(\nu+1)} \cap \{z=(x,y_1,...,y_n) \in
\mathbb{R}^{n+1}:\,\, |x| \le Q\} \subset {\cal C} = {\cal
C}_\xi (Q, \sigma).$$ So
$$
\zeta^{(\nu+1)}\not \in {\cal C}.
$$
But
$$
\pi \cap {\cal C}^{(\nu+1)} \subset {\cal C}.
$$
So $\zeta^{(\nu+1)}\not \in \pi$. It means that $\zeta^{(\nu+1)}$
is independent of $ \zeta^{(1)} ,..., \zeta^{(\nu)}$.
To conclude the proof we make two following observations:
1. each point $z = (x,y_1,...,y_n)\in {\cal C}^{(\nu+1)}$
satisfies $ |x| \le Q_2$;
2. for the section $ {\cal C}^{(\nu+1)}\cap \{ x = Q\}$ one has
$$
{\cal C}^{(\nu+1)}\cap \{ x = Q\} \subset {\cal C} \cap \{ x = Q\}
,
$$
and the section ${\cal C} \cap \{ x = Q\}$ is a ball of radius
$\sigma$ centered at $Q\xi$.
So
$$
{\cal C}^{(\nu+1)}\,\, \subset \,\, {\cal C}_{\xi^{(\nu+1)}}
(Q_2,Q_2Q^{-1}\sigma).$$ Now $ \mu_m ({\cal C} )\le Q_2Q^{-1} $
for $ 2\le m \le l. $ Lemma 1 is proved.
{\bf Lemma 2.}\,\,\,{\it Let $ 2\le l\le n+1$. Consider a
complete sublattice $ L \subseteq \mathbb{Z}^{n+1}$ and suppose
that $ {\rm dim}\, L = l$ and $ \xi \in {\rm span}\, L$.
Suppose that the fundamental $l$-dimensional volume of $L$ is
equal to $s$. Suppose also that for some $Q ,\sigma >0 $ we have
$$
{\cal C}_\xi (Q, \sigma)\bigcap L = \{ 0\}.
$$
Then for any $M , \delta >0$ the following upper bound is valid:
$$
\mu_{l}({\cal C}_\xi (M,\delta) )\le Q_2 Q^{-1} \max( QM^{-1},
\sigma \delta^{-1}).
$$
}
{\bf Corollary.}\,\,\,{\it Suppose that the conditions of Lemma 2
are satisfied. Then for the cylinder ${\cal C}_\xi (N,N^{ -1/n})$
we have
$$
\mu_{l}(\xi , N)\le Q_2 Q^{-1} \max( QN^{-1}, \sigma N^{ 1/n}).
$$}
Proof of Lemma 2. Put $ t = \max( QM^{-1}, \sigma \delta^{-1})$.
Then
$$
{\cal C}_\xi (Q, \sigma)\subset t {\cal C}_\xi (M,\delta),
$$
and applying Lemma 1 we see that
$$
\mu_{l}({\cal C}_\xi (M,\delta) ) = t \mu_{l }(t{\cal C}_\xi
(M,\delta) )\le t\mu_{l}({\cal C}_\xi (Q, \sigma))\le Q_2 Q^{-1}
t.
$$
Lemma 2 is proved.
Put $ \xi_0 = 1$. For a real vector $\xi = (\xi_0,\xi_1,...,\xi_n
)=(1,\xi_1,...,\xi_n ) \in\mathbb{R}^{n+1}$ we define ${\rm
dim}_\mathbb{Q}\xi $ to be the maximal integer $t$, such that the
components $ \xi_{j_1},...,\xi_{j_t}, 0\le j_1,...,j_t \le n+1$
are linearly independent over $\mathbb{Q}$. For example, the
equality ${\rm dim}_\mathbb{Q}\xi = 1$ occurs only if $\xi \in
\mathbb{Q}^{n+1}\setminus \{ 0\}$ and the equality ${\rm
dim}_\mathbb{Q}\xi = n+1$ occurs only if all the components
$1,\xi_1,...,\xi_n$ are linearly independent over $\mathbb{Q}$.
Obviously, if ${\rm dim}_\mathbb{Q}\xi = l, \,\,\, 1\le l \le
n+1$, then there is a complete sublattice $ L \subseteq
\mathbb{Z}^{n+1}$, such that
$
{\rm dim}\, L = l $ and $ \xi \in {\rm span}\, L$. Moreover,
$$
{\rm dim}_\mathbb{Q}\xi =\min\{ l\in \mathbb{N}:\,\, \text{there
exists a sublattice} \, L \subseteq \mathbb{Z}^{n+1}, \,\,\text{
such that}\,\,
{\rm dim}\, L = l\,\,\text{ and}\,\,
\xi \in {\rm span}\, L \}.$$
Let us now we consider a complete sublattice $ L \subseteq
\mathbb{Z}^{n+1}$, such that
$
{\rm dim}\, L = l\ge 2 $ and let us consider a vector
$ \xi =(1,\xi_1,...,\xi_n )\in {\rm span}\, L$ (then $ {\rm
dim}_\mathbb{Q}\xi\le l$). We shall say that
$\xi $ is {\it $\gamma$-badly approximable with
respect to } $L$ (briefly $(L,\gamma)$-BAD) if for any nonzero
integer point $ \zeta =(q,a) = (q,a_1, ...,a_n) \in L $ with
$q\neq 0$ one has
\begin{equation}
|q\xi - \zeta| \ge \gamma |q|^{-1/(l-1)}. \label{BAD}
\end{equation} We should note that for any $(L,\gamma)$-BAD vector
$\xi$ and any $Q\ge 1$ the cylinder
\begin{equation}
{\cal C}_\xi (Q, \sigma_Q)\bigcap {\rm span}\, L,\,\,\, \sigma_Q
= \gamma Q^{-1/(l-1)} \label{CY} \end{equation} contains no
nonzero integer points inside. A vector $\xi \in {\rm span}\, L$
is defined to be {\it badly approximable with respect to } $L$
(briefly $ L$-BAD) if (\ref{BAD}) holds with some positive
$\gamma$. It is easy to see that if vector $\xi$ is badly
approximable with respect to $L$ and ${\rm dim}\, L = l$ then
$\xi \in {\rm span}\,\Lambda$ and $ {\rm dim}_\mathbb{Q}\xi = l$.
Let $W\ge 1$. It is necessary for us to consider vectors $\xi \in
{\rm span} \, L$, such that the cylinder (\ref{CY}) contains no
nonzero integer point inside only for $ Q\ge W$. We define such
vectors to be {\it $(\gamma, W)$-badly approximable with respect
to } $L$ (briefly $(L,\gamma, W)$-BAD). A vector $\xi \in {\rm
span} L$ is $(\gamma, W)$-badly approximable with respect to
$L$ iff (\ref{BAD}) holds for all $\zeta $ with $|q|\ge W$ and for
all $q$ under the condition $1\le |q|\le W$ the following
inequality holds instead of (\ref{BAD}):
$$
|q\xi - \zeta| \ge \gamma W^{-1/(l-1)}.
$$
It is obvious that a vector $\xi \in {\rm span} \, L$ is
$(\Lambda,\gamma)$-BAD iff it is $(\Lambda,\gamma, 1)$-BAD.
{\bf Example 1.}\,\,\, Consider the space $\mathbb{R}^{n+1} $
related to coordinates $x,y_1,...,y_n$. Consider the case when
real algebraic integers $1,\alpha_1,...,\alpha_{l-1}$ form a basis
of a real algebraic field ${\cal K}$ of degree $l\ge 2$. Then
there exists a constant $\gamma = \gamma ({\cal K}) $, such that
for all natural $q$ we have
$$
\left( \sum_{j=1}^{l-1}||q\alpha_j||^2\right)^{1/2} \ge \gamma
q^{-1/(l-1)}
$$
(see \cite{Cas}, Chapter V, \S 3) and hence the
$(n+1)$-dimensional vector
$$
(1,\alpha_1,...,\alpha_{l-1},\underbrace
{0,...,0}_{n+1-l\,\,\text{times}})
$$
is $(L, \gamma ({ \cal K}))$-BAD where $L = \mathbb{Z}^{n+1}\cap
\{y_l = ...= y_n = 0\}$.
Define
$$
G_1= G_1(l,s,\gamma)=2^{2(l-1)}w^{-1}_{l-1}s\gamma^{1-l},\,\,\,
G_2= G_2(l,s,\gamma)=
2^{2(l-1)}w^{-1}_{l-1}s\gamma^{-\frac{(l-1)^2}{l}}.
$$
{\bf Lemma 3.}\,\,\,{\it Let $ L \subseteq \mathbb{Z}^{n+1}$ be
a complete sublattice, such that
$
{\rm dim}\, L = l \ge 2 $ and let
$ \xi =(1,\xi_1,...,\xi_n )\in {\rm span} \, L$ be an $(L,\gamma , W)$-BAD
vector. Let $s$ be the $l$-dimensional fundamental volume of $L $.
Consider positive $M,\delta $ and the cylinder
${\cal C}= {\cal C}_\xi (M,\delta)$. Then the following statements
hold.
1) If
\begin{equation}
\left(M\gamma \delta^{-1}\right)^{\frac{l-1}{l}}\le W
\label{EN0B}
\end{equation}
then
\begin{equation}
\mu_l ({\cal C}) \le G_1 WM^{-1}.
\label{C1C0B}
\end{equation}
2) If
\begin{equation}
\left(M\gamma \delta^{-1}\right)^{\frac{l-1}{l}}\ge W \label{ENB}
\end{equation}
then
\begin{equation}
\mu_l ( {\cal C}) \le G_2
M^{-\frac{1}{l}} \delta^{\frac{1-l}{l}}
.
\label{C1CB}
\end{equation}
}
{\bf Remark.}\,\, We actually construct in the proof
$l$ nonzero linearly independent integer points $\zeta_j \in L$ lying in
the cylinder $\mu_l ({\cal C}) \cdot \overline {\cal C}$.
It is seen from the construction that in the
case 2) of Lemma 3 each ray $[0,\zeta_j)$, $ 1\le j \le l$
intersects the facet $\{ x = M\}$ of the cylinder ${\cal C}= {\cal C}_\xi (M,\delta)$.
Proof of Lemma 3.
For $Q\ge W$ the cylinder (\ref{CY}) has no nonzero integer
points. By Lemma 2 for any $ Q\ge W$ we have
$$
\mu_{l}( {\cal C})\le G_1 \max( QM^{-1}, \gamma Q^{-1/(l-1)}
\delta^{-1}).
$$
Consider
$$
m(M,\delta , W) =\min_{Q\ge W} \max( QM^{-1}, \gamma Q^{-1/(l-1)}
\delta^{-1}) .
$$
If (\ref{EN0B}) holds we have
$$
m(M,\delta , W)= WM^{-1}.
$$
If (\ref{ENB}) holds we see that
$$
m(M ,\delta, W)= \gamma^{\frac{l-1}{l}}\delta^{\frac{1-l}{l}}
M^{-\frac{1}{l}} .
$$
Lemma 3 follows.
Lemma 3 applied to the cylinder ${\cal C}_\xi (N, N^{-1/n})$ gives
the following
{\bf Corollary 1.}\,\,\,{\it Let $\xi =(1,\xi_1,...,\xi_n )
\in \mathbb{R}^{n+1}$. Let $ L \subseteq \mathbb{Z}^{n+1}$ be a
complete sublattice such that
$
{\rm dim}\, L = l \ge 2 $ and let
$ \xi \in {\rm span}\, L$ be a $(L,\gamma , W)$-BAD
vector. Let $s$ be the $l$-dimensional fundamental volume of $L $.
Then
1) for any positive $N$ under the condition
\begin{equation}
N \le \gamma^{-\frac{n}{n+1}} W^{\frac{ln}{(n+1)(l-1)}}
\label{EN0}
\end{equation}
one
has
\begin{equation}
\mu_l (\xi , N) \le G_1 WN^{-1}.
\label{C1C0}
\end{equation}
2) for any $N$ under the condition
\begin{equation}
N \ge \gamma^{-\frac{n}{n+1}} W^{\frac{ln}{(n+1)(l-1)}}
\label{EN}
\end{equation}
one
has
\begin{equation}
\mu_l (\xi , N) \le G_2N^{\frac{l-n-1}{nl}}.
\label{C1C}
\end{equation}
}
{\bf Corollary 2.}\,\,\,{\it Let $ 2\le l \le n$. Let $\xi \in
\mathbb{R}^{n+1}$. Let $ L \subseteq \mathbb{Z}^{n+1}$ be a
complete sublattice such that
$
{\rm dim}\, L = l $ and let
$ \xi \in {\rm span}\, L$ be a $L$-BAD
vector.Then
$$
\mu_l (\xi , N) \to 0,\,\,\,\mu_{l+1} (\xi , N) \to +\infty,\,\,\, N \to \infty .
$$}
{\bf Remark 1.}\,\,\, Obviously, for $l = 1$ in the case $ {\rm
dim}_\mathbb{Q}\xi = 1$ we have $$
\mu_1 (\xi , N) \to 0,\,\,\,\mu_{2} (\xi , N) \to +\infty,\,\,\, N \to \infty .
$$
{\bf Remark 2.}\,\,\ Of course, the assertion of Corollary 2
enforces the components $1,\xi_1,...,\xi_n$ to be linearly
dependent over $\mathbb{Q}$.
Proof of Corollary 2. The statement about $\mu_l (\xi , N)$ follows immediately
from Corollary 1 of
Lemma 3
as the exponent in the right hand side of (\ref{C1C}) is negative. We
prove the statement about $\mu_{l+1} (\xi , N)$.
Suppose that $f_1, ...,f_l\in L $ form a basis of $L$. Then it
can be completed to a basis $f_1,...,f_l, g_{l+1},...,g_{n+1}$ of
the entire integer lattice $\mathbb{Z}^{n+1}$. Let $L '$ be the
sublattice generated by $g_{l+1},...,g_{n+1}$. Then $
\mathbb{Z}^{n+1}= L \oplus L '$, \,\,$ {\rm dim }\, L ' = n+1-l$
and $ {\rm span }\,L \cap {\rm span }\,L ' =\{ 0\}.$ Consider the
$n+1-l$ dimensional linear subspace $\pi \subset \mathbb{R}^{n+1}
$, orthogonal to ${\rm span }\,L $. Then $\pi \oplus {\rm span
}\,L= \mathbb{R}^{n+1}$ and any two vectors $ u\in \pi,v \in {\rm
span }\,L$ are orthogonal. Hence the orthogonal projection of $L'$
onto $\pi$ is a lattice $L ''$, such that ${\rm span}\, L '' =
\pi$. Let $\omega=\omega (L )>0 $ be the length of the shortest
nonzero vector in $L ''$. Then for any integer point $\zeta \in
\mathbb{Z}^{n+1} \setminus L$ the Euclidean distance from $\zeta $
to $ {\rm span }\,L$ is not less than $ \omega$. Suppose that
$ \xi \in {\rm span }\,L$ and that the cylinder $ {\cal C}_\xi
(tN, tN^{-1/n} )$ contains $ l+1$ linearly independent integer
points. Then at least one of these points belongs to
$\mathbb{Z}^{n+1} \setminus L$. Hence $ tN^{-1/n}
> \omega (L )$ and $\mu_{l+1} (\xi , N) \ge \omega (L
)N^{1/n}\to +\infty,\,\, N\to \infty$. The Corollary is proved.
{\bf Lemma 4.}\,\,\,{\it Let $ 2\le l \le n+1$. Let $s$ be the $l
$-dimensional fundamental volume of a lattice $L$, ${\rm dim}\ L =
l .$ Suppose that a vector $\xi = (1,\xi_1,...,\xi_n) \in {\rm
span }\,L $ and positive numbers $\gamma $ and $ T \ge 1$ satisfy
the equality
\begin{equation}
{\cal C}_{\xi } (T,\gamma T^{-1/(l-1)}) \bigcap L = \{ 0\}.
\label{EMP}
\end{equation}
Let
\begin{equation} \gamma ^*= \gamma^*(\gamma,L) = \min \left( 3^{-2}\gamma ,
3^{-l-2} (2 w_{l-1}l !)^{-\frac{1}{l-1}} s^{\frac{1}{l-1}}
\right) .
\label{prime}
\end{equation}
Then there exists an $(L,\gamma^*, T)$-BAD vector $\xi^*=
(1,\xi_1^*,...,\xi_n^*) \in {\rm span }\, L$, such that
\begin{equation}
|\xi^*-\xi |< \gamma T^{-\frac{l }{l-1}} \label{prim}.
\end{equation} }
Proof. Put $T_\nu = 3^{ (l-1)\nu }T,\,\, \nu = 0,1,2,... $. To
prove Lemma 4 it suffices to construct a sequence of cylinders
$$
{\cal C}^{(\nu )} = {\cal C}_{\xi^{(\nu ) }} (T_\nu,9\gamma^*
T_\nu^{-1/(l-1)}),\,\,\,\,\, \xi^{(\nu ) }\in {\rm span }\,L
$$
such that
(i) for every $\nu$ we have ${\cal C}^{(\nu )} \cap L =\{ 0\}$;
(ii) the section $ \{ x = T_{\nu}\}\cap {\cal C}^{(\nu +1)} $ of
the cylinder ${\cal C}^{(\nu + 1)}$ lies inside the facet $ \{ x =
T_{\nu}\}\cap \overline {{\cal C}^{(\nu )}} $ of the preceding
cylinder ${\cal C}^{(\nu )} $; moreover, the distance between the
centers of ${\cal B}$ and ${\cal B}'$ does not exceed $
3\gamma^* T_\nu^{-1/(l-1)} $.
If such cylinders ${\cal C}^{(\nu )} $ are constructed and ${\cal
C}^{(0 )} = {\cal C}_{\xi } (T,\gamma T^{-1/(l-1)})$ then the
vector $ \xi ^* =\lim_{\nu \to + \infty} \xi^{(\nu )}$ satisfies
(\ref{prim}). Moreover, it follows from (ii) that $\xi ^*$ is an
$(L,\gamma^*, T)$-BAD vector. Indeed, for an integer point $\zeta
=(q,a_1,...,a_n)$ with $ T_\nu \le q\le T_{\nu+1} = 3^{l-1}T_\nu$
we have
$$
|q\xi^*-\zeta|\ge 3\gamma^*T_{\nu+1}^{-1/(l-1)}\ge
\gamma^*q^{-1/(l-1)}.
$$
We now describe the inductive process, which constructs the sequence of
cylinders ${\cal C}^{(\nu )}$. Suppose that ${\cal C}^{(\nu )}$
is already constructed. Consider the cylinder
$${\cal C}'=
{\cal C}_{\xi^{(\nu ) }} (T_{\nu+1},3^{l+1 }\gamma^*
T_\nu^{-1/(l-1)}).
$$
We prove that there exists a linear subspace ${\cal L} \subset
{\rm span} \, L $ of dimension ${\rm dim }\, {\cal L} = l-1$
containing all the integer points $\zeta \in L \bigcap {\cal C} '
$. Suppose that there are $l $ linearly independent integer points
$$
\zeta^{(1)},..., \zeta^{(l)} \in \Gamma\bigcap{\cal C} ' .
$$
Then the $l$-dimensional volume $V$ of the convex hull ${\rm
conv}\,(0, \zeta^{(1)},..., \zeta^{(l)})$ is bounded from below by
the fundamental volume of $L$:
\begin{equation}
V \ge s ( l!)^{-1}. \label{contra}\end{equation} On another hand,
the volume of ${\rm conv}\,(0, \zeta^{(1)},..., \zeta^{(l)})$
admits an upper bound based on the relation ${\rm conv}\,(0,
\zeta^{(1)},..., \zeta^{(l)} ) \subset {\cal C} ' . $ Taking into
account (\ref{prime}) we see that
\begin{equation}
V \le T_{\nu+1}\left( 3^{l+1}\gamma^*
T_\nu^{-1/(l-1)}\right)^{l-1} \le s (2 l !)^{-1}
\label{contrb}\end{equation}
Relations (\ref{contra},\ref{contrb}) contradict each other, which means that all the integer points from the cylinder under
consideration lie in a subspace ${\cal L}$.
Let ${\cal B}$ be the $(l-1)$-dimensional
facet $\{ x = T_\nu \}$ of ${\cal C}^{(\nu )}\cap {\rm span}\, L$.
In fact ${\cal B}$ is an $(l-1)$-dimensional open ball of radius
$3\gamma^* T_\nu^{-1/(l-1)}$ centered at
$T_\nu \xi^{(\nu )}\in {\rm span}\, L$. There is an $(l-1)$-dimensional open ball ${\cal B}'\subset {\cal B}$ of radius
$\gamma ^* T_\nu^{-1/(l-1)} = 3\gamma ^* T_{\nu+1}^{-1/(l-1)}$ and centered at a certain point
of the affine subspace $\{ x = T_\nu \}\cap {\rm span}\, L$, such that $ {\cal B}'\cap {\cal L} = \varnothing$ and the point
$T_\nu \xi^{(\nu )}$ lies on the boundary of ${\cal B}'$. Let
$(T_\nu,\Xi_1,...,\Xi_n)$ be the center of ${\cal B}'$. Put
$$
\xi^{(\nu+1)} = \left( 1, \frac{\Xi_1}{T_\nu},...,
\frac{\Xi_n}{T_\nu}\right)\in {\rm span}\, L.$$ As ${\cal C}^{(\nu
+1)} \subset {\cal C}'$, we see that
there are no nonzero points of $L$
in ${\cal C}^{(\nu+1)}$
and (i) is valid with $\nu$ replaced by $\nu+1$.
From the construction we see that (ii) is also valid for $\nu+1$. Lemma 4 is proved.
{\bf Remark 1.}\,\,\, Lemma 4 is obtained by well-known arguments
(see \cite{SCH}, Chapter 3, \S 2). The constant $3^{-l-2}
(2w_{l-1}l !)^{-\frac{1}{l-1}} s^{\frac{1}{l-1}}$ in (\ref{prime})
may be slightly improved but this is of no importance for the
proof of our main result.
\section{Lemmas concerning two sublattices}
{\bf Lemma 5.}\,\,\,{\it Let $\Gamma \subset \mathbb{Z}^{n+1}$ be
a complete lattice, such that $ {\rm dim } \, \Gamma = k+1\ge
3. $
Let $R$ be
the $(k+1)$-dimensional fundamental volume of $\Gamma $. Let
vector $\xi = (1,\xi_1,...,\xi_n) \in {\rm span }\,\Gamma $,
$\xi_j \in (0,1)$, be
$(\Gamma,\gamma ,W)$-BAD. Consider a positive number $\kappa $, such that
\begin{equation}
\kappa \le \gamma W^{-\frac{k+1}{k}}.
\label{kappasmall}
\end{equation}
Then there is a sublattice $\Lambda\subset \Gamma$, ${\rm dim }\,
\Lambda = k$ satisfying the following two conditions:
1)
the Euclidean distance from $\xi \in \mathbb{R}^{n+1}$
to ${\rm span }\, \Lambda \cap \{ x= 1\} $ does not exceed $\kappa$;
2) the $k$-dimensional fundamental volume $r$ of $\Lambda$
admits the following upper bound:
$$
r\le G (\gamma, \Gamma )\kappa^{-\frac{1}{k+1}},
$$
where \begin{equation}
G(\gamma,\Gamma )=
2^{2k^2 } w_k^{-k}w_{k-1} k!\gamma^{-\frac{k^3}{k+1}} R^k
.
\label{je}
\end{equation}
}
Proof. We take
$$
M =(\gamma \kappa^{-1})^{\frac{k}{k+1}},\,\,\, \delta = M\kappa.
$$
By (\ref{kappasmall}) we have
$$
(M\gamma \delta^{-1})^{\frac{k}{k+1}} =(\gamma
\kappa^{-1})^{\frac{k}{k+1}} \ge W.
$$
We now can apply the statement 2) of Lemma 3 for $ l = k+1, L =
\Gamma$. Then (\ref{C1CB}) gives the inequality
\begin{equation}
\mu = \mu_k ({\cal C}_\xi (M, \delta))\le \mu_{k+1} ({\cal
C}_\xi (M, \delta))\le 2^{2k}w_k^{-1} R\gamma^{-k}. \label{u}
\end{equation}
We see now that the cylinder $\mu\overline{\cal C}_\xi (M,
\delta)$ has $k$ linearly independent integer points
$\zeta_1,...,\zeta_k$. Define $\Lambda = {\rm span}
(\zeta_1,...,\zeta_k) \cap
\mathbb{Z}^{n+1}$. The remark after Lemma 3 shows that the
condition 1) is satisfied.
Let us obtain the needed upper bound for the
fundamental volume $r$ of $\Lambda $. As
$$
{\rm conv}(0,\zeta_1,..,\zeta_k) \subset {\rm span}\Lambda \bigcap
\mu\overline{\cal C}_\xi (M,
\delta)
$$
(here ${\rm conv}(0,\zeta_1,..,\zeta_k)$ stands for the convex
hull of the points $0,\zeta_1,..,\zeta_k\in \mathbb{R}^{n+1}$)
we see that
\begin{equation}
r\le k! \mu^k w_{k-1} \delta^{k-1}M, \label{v}
\end{equation}
and the
required upper bound follows from (\ref{u},\ref{v}). Lemma is
proved.
Let $\Gamma $ be a
sublattice as in Lemma 5. For each lattice $\Gamma$ and for
every$\gamma$ small enough there exist $ (\Gamma, \gamma, W)$-BAD
vectors (see Example 1). Consider a $ (\Gamma, \gamma, W)$-BAD
vector $ \xi = (1,\xi_1,...,\xi _n) \in {\rm span } \,\Gamma$.
Then for any $ T\ge W$ the cylinder $ {\cal C}_\xi (T, \gamma
T^{-1/k}) $ contains no nonzero points of $\Gamma$. Let ${\cal B}$
be the facet $ \{ x = T\}$ of the cylinder $ {\cal C}_\xi (T,
\gamma T^{-1/k})\cap {\rm span } \,\Gamma$. This facet is a $k$
dimensional ball of radius $\gamma T^{-1/k}$
centered at $T\xi$.
Lemma 5 with
$$
\kappa = \frac{\gamma}{4n} \, T^{-\frac{k+1}{k}}
$$
implies that there is a $k$-dimensional complete sublattice
$\Lambda \subset \Gamma$ with $k$-dimensional fundamental volume
$r$ satisfying the condition
$$
r\le G (\gamma, \Gamma )(4n\gamma^{-1})^{\frac{1}{k+1}} T^{\frac{1}{k}},
$$
and such that the intersection of $ {\rm span } \,\Lambda$ with ${\cal B}$ is a $(k-1)$-dimensional ball ${\cal B} '\subset {\cal B}$ with the
center $\Xi'$ and radius $\ge\gamma T^{-1/k}/2$. Take another
$k-1$-dimensional ball ${\cal B} ''\subset {\cal B}'$ of radius
$2^{-3}\gamma T^{-1/k}$ centered at $\Xi '$. Then the distance
from ${\cal B} ''$ to the boundary of ${\cal B}$ is greater than
$2^{-3}\gamma T^{-1/k}$. Put $ \xi'=\Xi'/T$.
Then the cylinder
\begin{equation}\label{lllw} {\cal C}_{\xi'} (T, 2^{-3}\gamma T^{-1/k} )\cap {\rm span }
\,\Lambda={\cal C}_{\xi'} (T, \gamma ' T^{-1/(k-1)} )\cap {\rm
span } \,\Lambda,\,\,\, \gamma' = 2^{-3}\gamma
T^{\frac{1}{k(k-1)}}
\end{equation}
contains no nonzero points of $\Lambda$ and
we can apply Lemma 2 with $ l = k $ and $L=\Lambda$. Thus we
obtain a $(\Lambda,\hat{\gamma }, T)$-BAD vector $\hat{\xi }$ with
\begin{equation}
\hat{\gamma } = \hat{\gamma } ( \gamma , T, \Lambda)
= \gamma^* (2^{-3}\gamma
T^{\frac{1}{k(k-1)}},\Lambda ). \label{hat}
\end{equation}
As the cylinder (\ref{lllw}) has no nonzero points of $\Lambda$ we
see from the Minkowski theorem on convex bodies that
$$
2T\cdot w_{k-1} \left(2^{-3} \gamma T^{-\frac{1}{k}}\right)^{k-1}
< 2^k r
$$
or
\begin{equation}
r > 2^{-2(k-1)} T^{\frac{1}{k} }w_{k-1} \gamma^{k-1}. \label{vo}
\end{equation}
Note that from (\ref{hat},\ref{prime},\ref{vo}) we see that
\begin{equation}
\hat{\gamma } \ge \underline{C}(k)\gamma
T^{\frac{1}{k(k-1)}},\label{hatmin}
\end{equation}
where
\begin{equation}
\underline{C}(k)= \min\left(3^{-5} , 2^{-1-\frac{1}{k-1}} 3^{-k-1}
(k!)^{-\frac{1}{k-1}}\right).
\label{ce}
\end{equation}
Now we put
\begin{equation}
Z_1 (\gamma, k ) = (\underline{C}(k)\gamma )^{-\frac{n}{n+1}},
\label{zet1}
\end{equation}
\begin{equation}
Z_2 (i,\gamma, \Gamma ) = \left( i\cdot 2^{2(k-1)}w_{k-1}^{-1}
G(\gamma,\Gamma ) (4n\gamma^{-1})^{\frac{1}{k+1}}
(\underline{C}(k)\gamma )^{-\frac{(k-1)^2}{k}}
\right)^{\frac{nk}{n+1-k}}.
\label{zet2}
\end{equation}
{\bf Lemma 6.}\,\,\,{\it For the vector $\hat{\xi}$ defined above
and for $N $ under the condition
\begin{equation}\label{starr}
N \ge Z(i,\gamma,\Gamma, T) = \max ( Z_1 (\gamma, k ) T^{\frac{n(k+1)}{(n+1)k}},
Z_2 (i,\gamma, \Gamma ) T^{\frac{n }{ k(n+1-k)}})
\end{equation}
we have
$$
\mu_k (\hat{\xi}, N) \le i^{-1}.
$$}
Proof. As $ N \ge Z_1 (\gamma, k ) T^{\frac{n(k+1)}{(n+1)k}}$, we
see from (\ref{zet1},\ref{ce},\ref{hatmin}) that the condition
(\ref{EN}) of the case 2) of Corollary 1 to Lemma 3 is satisfied.
Then due to (\ref{C1C}) we have
$$
\mu_k (\hat{\xi}, N) \le 2^{2(k-1)} w_{k-1}^{-1} r \hat{\gamma
}^{-\frac{(k-1)^2}{k}} N^{\frac{k-n-1}{nk}}.
$$
It remains to make use of the inequality $ N \ge
Z_2 (i,\gamma, \Gamma ) T^{\frac{n }{ k(n+1-k)}}
$ and of the formulas (\ref{zet2},\ref{vo},\ref{hatmin},\ref{ce}).
Lemma 6 follows.
We shall need the following notation related to a pair of
sublattices.
Let $\Lambda \subset \Gamma \in \mathbb{Z}^{n+1}$ be complete
sublattices such that
$$
{\rm dim}\, \Gamma
>{\rm dim}\,\Lambda .
$$
Then $\Gamma$ can be partitioned into classes $ \pmod{\Lambda}$:
$$
\Gamma = \bigcup_{\alpha \in \mathbb{Z}^v} \Gamma_\alpha,\,\,\,
\Gamma_0 =\Lambda,\,\,\, v = {\rm dim}\,(\,{\rm span }\,\Gamma )
-{\rm dim}\,(\,{\rm span }\,\Lambda ),
$$
so that the affine subspaces ${\rm aff}\, \Gamma_\alpha$ are
parallel ${\rm span }\,\Lambda ={\rm aff}\, \Gamma_0$. Here by
${\rm aff}\, \Omega$ we mean the smallest affine subspace of
$\mathbb{R}^{n+1}$ containing $\Omega$.
Denote by $R = R (\Lambda,\Gamma ) >0$ the minimal distance
between points $ z^{(1)},z^{(2)}$, where $ z^{(1)}\in
\Gamma\setminus \Lambda $ and $z^{(2)} \in {\rm span }\,\Lambda$.
For our purpose we need not the $R (\Lambda, \Gamma )$ itself but
a little bit different distance $\rho = \rho (\Lambda,\Gamma )$
which we define now. Recall that as it was pointed out in the very
beginning of Section 3 all $k$-dimensional lattices $\Lambda$
under the consideration admit the property
$$
{\rm dim}\,( \Lambda \cap {\cal P} ) = k-1
$$
(here ${\cal P} =\{ z=(x,y_1,...,y_n)\in \mathbb{R}^{n+1}:\,\, x =
1\}$). For such a lattice $\Lambda$ and for a lattice $\Gamma
\supset \Lambda$ of dimension $ {\rm dim}\, \Gamma > {\rm dim}\,
\Lambda$ we consider the following objects.
Put ${\cal L} = {\rm span
}\,\Lambda \cap {\cal P}$. Let ${\cal G}$ be the parallel
projection of $\Gamma \setminus \Lambda$ along $ {\rm span
}\,\Lambda $ onto ${\cal P}$. Then we define $\rho = \rho
(\Lambda,\Gamma )$ to be the minimal distance between points $
z^{(1)},z^{(2)}$, where $ z^{(1)}\in {\cal G}$ and $z^{(2)} \in
{\cal L} = {\rm span }\,\Lambda \cap {\cal P}$. We see that $\rho
= \rho (\Lambda,\Gamma ) >0$. In fact, $ \rho (\Lambda,\Gamma
)\ge \rho (\Lambda , \mathbb{Z}^{n+1} )
>0$.
Now we give two more lemmas.
{\bf Lemma 7.}\,\,\,{\it Let $\Lambda \subset \Gamma\subset
\mathbb{Z}^{n+1}$ be complete sublattices, such that
$$
{\rm dim} \, \Gamma = k+1 = {\rm dim}\, \Lambda
+1 ,$$ and
$\rho = \rho (\Lambda,\Gamma ).
$
Let $\xi = (1,\xi_1,...,\xi_n) \in {\rm span }\,\Lambda$ be a
$(\Lambda ,\gamma ,W )$-BAD vector with some positive $\gamma$ and
$W\ge 1$. Put
\begin{equation}
\gamma ' =\gamma ' (\gamma , \Lambda , \Gamma )=
\gamma^{\frac{k-1}{k}} 2^{-\frac{1}{k}} \rho^{+\frac{1}{k}}
\label{gammaprime}
\end{equation}
and
\begin{equation}
A_1 = A_1(\gamma , \Lambda , \Gamma )
= \max\left( (\rho (2\gamma ')^{-1})^{\frac{k}{k-1}} , (2\gamma
' \rho^{-1})^k\right). \label{AAA}
\end{equation}
Suppose that
\begin{equation}
T \ge A_1W^{\frac{k}{k-1}}
\ge\max\left( (\rho (2\gamma ')^{-1}W)^{\frac{k}{k-1}} , (2\gamma
' \rho^{-1})^k\right).
\label{te}
\end{equation}
Let $\xi' = (1,\xi_1',...,\xi_n') \in {\rm span }\,\Gamma $
satisfy the following two conditions:
1) the orthogonal projection of vector $\xi '$ on the subspace
${\rm span }\,\Lambda$ is of the form $\lambda \xi $ with some
positive $\lambda$;
2) for the Euclidean norm we have $| \xi ' - \xi | =
(2T)^{-1}\rho$.
Then
$$
{\cal C}_{\xi '} (T,\gamma' T^{-1/k}) \bigcap \Gamma = \{ 0\}.
$$}
Proof.
The $(k+1$-dimensional linear subspace ${\rm span}\, \Gamma$
contains parallel $k$-dimensional affine subspaces $ {\rm aff}\,
\Gamma_i, i\in \mathbb{Z}$. Each such subspace $ {\rm aff}\,
\Gamma_i$ divides the subspace ${\rm span}\, \Gamma$ into two
"half-subspaces" with the common boundary $ {\rm aff} \,\Gamma_i$.
The situation with the $k$-dimensional affine subspace ${\rm
span}\, \Gamma \cap \{ x = T\}$ and $(k-1)$-dimensional affine
subspaces $ {\rm aff}\, \Gamma_i \cap \{ x = T\}, i\in \mathbb{Z}$
is quite similar. We should note that the Euclidean distance
between neighboring subspaces
$ {\rm aff}\, \Gamma_i \cap \{ x = T\}$ and $ {\rm aff}\, \Gamma_{i+1} \cap \{ x =
T\}$ is exactly $\rho$.
Without loss of generality we may suppose that the point
$T\xi'\in {\rm span }\,\Gamma \cap \{ x = T\}$ lies
in the same "half-subspace" (with respect to ${\rm aff }\,\Gamma_0
\cap \{ x = T\} = {\rm span }\,\Lambda \cap \{ x = T\}$) as the
set ${\rm span }\,\Gamma_1 \cap \{ x = T\}$. From the conditions
1), 2) we see that the distance from the point $T\xi'$ to the
subspace ${\rm aff }\,\Gamma_0 \cap \{ x = T\}$ is equal to the
distance from $T\xi'$ to the subspace ${\rm aff }\,\Gamma_1 \cap
\{ x = T\}$ and is equal to $\rho/2$.
Define $ H = 2\gamma'\rho^{-1} T^{\frac{k-1}{k}} $. Then
by definition of $H$ and (\ref{te}), we have $H\le T$. Note that
the distance from each point of the form $t\xi, H\le t\le T$ to
the corresponding affine subspace ${\rm span }\,\Lambda \cap \{
x=t\}$ is greater than $\gamma' T^{-1/k}$. So
$$
{\cal C}_{\xi '} (T,\gamma' T^{-1/k})\bigcap \{ z=(x,y_1,...,y_n)
: |x| \ge H\} \bigcap {\rm span}\, \Lambda = \varnothing.
$$
It means that the cylinder ${\cal C}_{\xi '} (T,\gamma' T^{-1/k})$
intersected with the domain $\{ H\le x\le T\}$ has no points of
the lattice $\Lambda$. But the distance from each point of the
form $t\xi, H\le t\le T$ to the corresponding affine subspace
${\rm aff }\,\Gamma_1 \cap \{ x=t\}$ is greater than the distance
from $t\xi $ to ${\rm span }\,\Lambda \cap \{ x=t\}$. So the
distance from each point of the form $t\xi, H\le t\le T$ to the
any affine subspace ${\rm aff }\,\Gamma_i \cap \{ x=t\}, i \neq
0$
is greater than $\gamma' T^{-1/k}$ also. So
$$
{\cal C}_{\xi '} (T,\gamma' T^{-1/k})\bigcap \{ z=(x,y_1,...,y_n)
: |x| \ge H\} \bigcap \Gamma = \varnothing.
$$
From the other hand if $ 0\le t\le H$ then the distance from
$t\xi$ to any ${\rm aff }\,\Gamma_i \cap \{ x=t\}, i \neq
0$ is again greater than $\gamma' T^{-1/k}$. Hence
$$
{\cal C}_{\xi '} (T,\gamma' T^{-1/k})\bigcap \Gamma = {\cal
C}_{\xi '} (H,\gamma' T^{-1/k})\bigcap \Lambda.
$$
But (\ref{gammaprime}) implies that $ \gamma ' T^{-1/k} =\gamma
H^{-1/(k-1)}$. As $\xi$ is a $(\Lambda ,\gamma, W )$-BAD vector
we see that
$$
{\cal C}_{\xi '} (H,\gamma' T^{-1/k})\bigcap \Lambda\subseteq
{\cal C}_{\xi } (H,\gamma H^{-1/(k-1)})\bigcap \Lambda = \{ 0\}.
$$
(Note that from (\ref{te}) it follows that $ H\ge W$.)
Lemma 7 is proved.
{\bf Lemma 8.}\,\,\,{\it In the notation of Lemma 5, let $r$ be
the $k$-dimensional fundamental volume of the lattice $\Lambda $,
vector $\xi '$ be defined in Lemma 7 and let $\xi''=
(1,\xi_1'',...,\xi_n'') \in {\rm span }\,\Gamma $ be a vector
satisfying
\begin{equation}
|\xi''-\xi ' |< \rho (4T)^{-1}. \label{ex}
\end{equation}
Set
\begin{equation}
A_2 =A_2 (\gamma, \Lambda ,\Gamma ) = 3\rho \gamma^{-1}/4,
\label{aaa}
\end{equation}
\begin{equation}
B_1 = B_1 (\gamma )=
(\sqrt{2}\gamma )^{-\frac{n}{n+1}},\,\,
B_2 = B_2(\Lambda ,\Gamma )=
\left(\frac{2\sqrt{2}}{3\rho}\right)^{\frac{n}{n+1}}, \label{BBBB}
\end{equation}
\begin{equation}
C_1 = C_1(\gamma, \Lambda ) = 2^{2k-2}w_{k-1}^{-1}
r\gamma^{1-k},\,\,
C_2= C_2 (\gamma, \Lambda )
=2^{\frac{4k^2-3k-1}{2k}}w_{k-1}^{-1}r\gamma^{-\frac{(k-1)^2}{k}},
\label{ceodin}
\end{equation}
$$
C_3= C_3 (\gamma, \Lambda ,\Gamma )
=2^{\frac{k^2-3k-1}{2k}} 3^ {\frac{1}{k}}
w_{k-1}^{-1}r\gamma^{-\frac{(k-1)^2}{k}}\rho^{\frac{1}{k}}.
$$
Suppose that
\begin{equation}
T \ge A_2W^{\frac{k}{k-1}}.
\label{te0}
\end{equation}
Then the following statements are valid:
1) for $N$ in the interval
\begin{equation}
N \le B_1 W^{\frac{kn}{(k-1)(n+1)}}
\label{en00}
\end{equation}
we have
\begin{equation}
\mu_{k} (N, \xi '') \le C_1 W N^{-1};
\label{LL00} \end{equation}
2) for $N$ in the interval
\begin{equation}
B_1 W^{\frac{kn}{(k-1)(n+1)}}\le
N\le B_2T^{\frac{n}{n+1}} \label{en0}
\end{equation}
we have
\begin{equation}
\mu_{k} (N, \xi '') \le C_2 N^{\frac{k-n-1}{nk}} ;
\label{LL0} \end{equation}
3) for $N$ in the interval
\begin{equation}
N\ge B_2T^{\frac{n}{n+1}} \label{en1}
\end{equation}
we have
\begin{equation}
\mu_{k} (N, \xi '') \le C_3 T^{-\frac{1}{k}} N^{\frac{1}{n}}.
\label{LL} \end{equation}
}
{\bf Corollary.}\,\,\,{\it Under the conditions of Lemma 8, for
$N$ in the interval
\begin{equation}
H(i,\gamma , \Lambda, W) = \max\left( (C_1 (\gamma, \Lambda ) i
W) , (C_2 (\gamma , \Lambda ) i)^{\frac{nk}{n+1-k}}\right) \le
N\le (i C_3 (\gamma, \Lambda ,\Gamma ))^{-n} T^{\frac{n}{k}}
\label{inte}
\end{equation}
we have the following inequality:
\begin{equation}
\mu_{k} (N, \xi '') \le i^{-1}.
\label{mumain} \end{equation}}
Proof of Lemma 8. First of all, let us consider the case 3).
Set
$$
M = B_2^{\frac{n+1}{n}}TN^{-\frac{1}{n}}\le N,\,\,\,\delta
=N^{-\frac{1}{n}}/\sqrt{2}.
$$
It follows from the definition of $\xi '$ and (\ref{ex}) that
\begin{equation}
{\cal C}_{\xi ''} (N, N^{-1/n}) \supset {\cal C}_{\xi } (M,
\delta)\cap {\rm span }\,\Lambda . \label{false}
\end{equation}
Hence
$$
\mu_{k} (N, \xi '')\le \mu_{k}({\cal C}_{\xi } (M, \delta)),
$$
so it is suffices to obtain the corresponding upper bound for the
latter successive minimum. We observe that by (\ref{te0}) we have
$$
(M\delta^{-1}\gamma)^{\frac{k-1}{k}} = \left(\frac{4}{3}
T\rho^{-1}\gamma \right)^{\frac{k-1}{k}}\ge W.
$$
Applying the statement 2) of Lemma 3 we obtain (\ref{C1CB}) with $
l=k, s=r$. Now (\ref{LL}) follows from (\ref{C1CB}).
Consider the case 2). Since $ M \ge N$, the relation (\ref{false})
may be false, so we have
\begin{equation}
{\cal C}_{\xi ''} (N, N^{-1/n}) \supset {\cal C}_{\xi } (N,
\delta)\cap {\rm span }\,\Lambda . \label{falseno}
\end{equation}
Hence
$$
\mu_{k} (N, \xi '')\le \mu_{k}({\cal C}_{\xi } (N, \delta)),
$$
It follows from (\ref{en0}) that
$$
(N\delta^{-1}\gamma)^{\frac{k-1}{k}} \ge W.
$$
Let us apply the statement 2) of Lemma 3 for the cylinder ${\cal
C}_{\xi } (N, \delta)$ from (\ref{falseno}). Then the conclusion
(\ref{C1CB}) of Lemma 3 with our parameters leads to (\ref{LL0}).
Finally, we consider the case 1). Again, we have $M\ge N$. So we
must use the relation (\ref{falseno}). But (\ref{en00}) implies
that
$$
(N\delta^{-1}\gamma)^{\frac{k-1}{k}} \le W.
$$
Applying the statement 1) of Lemma 3 we get from (\ref{C1C0B})
the desired inequality (\ref{LL00}).
Lemma 8 is proved.
\section{Proof of Theorem 1: general case}
Now we are able to give a proof of Theorem 1 in the case $ k\ge 2$. We begin with the same consideration of the countable set
of all the $n$-dimensional complete sublattices of the integer lattice
$\mathbb{Z}^{n+1}$. We fix an enumeration of this set and let
$$ L_1, L_2,..., L_i , ...
$$
be all these lattices. Set $\pi_i = {\rm span} \,L_i$. Suppose
that $$\pi_1 = \{z=(x,y_1,...,y_n)\in\mathbb{R}^{n+1}:\,\ x=0\}.$$
Let $2\le k \le n-1$. We construct a sequence of real numbers
$$
\eta_1> \eta_2>...>\eta_i >...
$$
decreasing to zero, a sequence of positive real numbers
$$
\gamma_1,\gamma_2,...,\gamma_i,... ,
$$
two sequences of real numbers
$$
W_1,W_2,...,W_i,...,
$$
$$
H_1,H_2,...,H_i, ... ,$$
$$W_i \ge 1,\,\,\,
W_i, H_i \to +\infty,\,\,\, i \to +\infty,
$$
two sequences
of complete sublattices
$$
\Lambda_1,\Lambda_2,...,\Lambda_{i-1},\Lambda_i , ... ,
$$
$$
\Gamma_2,\Gamma_3,...,\Gamma_{i},\Gamma_{i+1},... ,
$$
and a sequence of vectors
$$
\xi_i = (1,\xi_{i,1},..., \xi_{i,n})\in \mathbb{R}^{n+1}
$$
satisfying the following conditions (i) -- (vii). Further, suppose
$r_i$ be the $k$-dimensional fundamental volume of $\Lambda_i$
and let $R_i$ be the $(k+1)$-dimensional fundamental volume of
$\Gamma_i$.
(i) For every $i\in \mathbb{N}$ we have
$$
\Lambda_i \subset \mathbb{Z}^{n+1}, \,\,\, {\dim }\, \Lambda_i =
k;
$$
$$
\Gamma_{i+1} \subset \mathbb{Z}^{n+1},\,\,\, {\dim }\,
\Gamma_{i+1} = k+1;
$$
$$
\Lambda_{i}, \Lambda_{i+1} \subset \Gamma_{i+1}.
$$
(ii) For every $i\in \mathbb{N}$ the vector $\xi_i$ is $(\Lambda_i
,\gamma_i, W_i)$-BAD.
(iii) The $n$-dimensional closed ball $\overline{\cal B}_i\subset
\{ z = (x,y_1,..., y_n) \in \mathbb{R}^{n+1}:\,\, x= 1\}$ of
radius $\eta_i$ is centered at $\xi_i$ and has no common points
with $\pi_{i}$.
(iv) The balls defined in (iii) form a nested sequence
$$
\overline{\cal B}_1 \supset \overline{\cal B}_2 \supset ...
\supset \overline{\cal B}_i .
$$
(v) For every $i\ge 2$ the following inequality holds:
\begin{equation}
H_i \ge
\max \left( H(2(i+1),\gamma_i,\Lambda_i, W_i), \frac{4(i+1)}{\rho
(\Lambda_i, \mathbb{Z}^{n+1} )}\right)
\label{AH10}
\end{equation}
(here the value of $\rho (\cdot, \cdot)$ for two lattices is
defined in Section 4 before Lemma 7 and $H(\cdot,\cdot,\cdot, \cdot)$
is defined in (\ref{inte})).
(vi) For every $i\ge 2$, every $\xi \in
\overline{\cal B}_i$ and for every real $N$ in the interval
$H_{i-1} \le N< H_i $
one
has
$$
\mu_k (\xi, N) \le i ^{-1} .
$$
(vii) For every $i\ge 2$, every $\xi \in \overline{\cal B}_i$ and
every real $N$ in the interval $H_{i-1}^n \le N< H_i^n $
one
has
$$
\mu_{k+2} (\xi, N) \ge i .
$$
Suppose that
all these objects are already constructed. Then we have Theorem
1 proved in the case $ k \ge 2$. Indeed, if we consider the
unique vector $\xi =(1,\xi_1,...,\xi_n)$ from the intersection $
\cap_{i\in \mathbb{N}} \overline{\cal B}_i$, then the components
$1,\xi_1,...,\xi_n $ are linearly
independent over $\mathbb{Z}$ due to (iii), and
$$\lim_{N \to +\infty }
\mu_k (\xi, N) =0,\,\,\,\,\lim_{N \to +\infty } \mu_{k+2} (\xi, N)
=+\infty
$$
due to (vi) and (vii).
We now describe an inductive process, which constructs all the
objects mentioned.
First of all, put $ W_1=H_1=1$,
$$
\Lambda_1 = \mathbb{Z}^{n+1}\bigcap
\{z=(x,y_1,...,y_n)\in\mathbb{R}^{n+1}:\,\,\, y_k = ...=y_n = 0\},
$$
Take $\xi$ to be a $(\Lambda,\gamma_1,1)$-BAD vector with some positive $\gamma_1$
(we can take such a vector from Example 1). We do not define
$\Gamma_1$.
Obviously, $\rho (\Lambda_1, \mathbb{Z}^{n+1}) = 1 $.
The conditions (i) -- (vii) for $ i = 1$ are satisfied (note that the conditions (v) -- (vii) are empty).
Assume that all the objects $ \eta_i , \gamma_i, W_i, H_i
,\Lambda_i , \Gamma_{i}, \xi_i $
for every natural $i$ up to $t$
are constructed
to satisfy the conditions (i) -- (vii). Let us describe the
construction for $ i = t+1$.
Consider the integer vector $q\in \mathbb{Z}^n$ orthogonal to the
subspace $\pi_{t+1}$. For any small positive $\varepsilon$ there
exists
an integer vector $q'$ such that the angle between $q$ and $q'$ is less than $\varepsilon$ and
$q' \not\in \Lambda_t$. Put
$$
\Gamma_{t+1} = \mathbb{Z}^{n+1} \cap {\rm span}\, (\Lambda_t\cup
q').
$$
Then $ \Gamma_{t+1}$ is a complete sublattice of dimension ${\rm
dim}\, \Gamma_{t+1} = k+1$, $\Gamma_{t+1} \supset \Lambda_t$, and
$$ {\rm span} \,\Gamma_{t+1} \not\subset \pi_{t+1}.
$$
Moreover for the vector $ q' \in \Gamma_{t+1}$ and any nonzero
vector $p\in \pi_{t+1}$ the angle between $ p$ and $q'$ is greater
than $\frac{\pi}{2}-\varepsilon$.
Let $R_{t+1}$ be the $(k+1)$-dimensional fundamental volume of
$\Gamma_{t+1}$. Set
$$
\rho_t^{(1)} = \rho (\Lambda_t, \mathbb{Z}^{n+1}),\,\,\,
\rho_t^{(2)} = \rho (\Lambda_t, \Gamma_{t+1}),\,\,\, \rho_t^{(3)}
= \rho (\Gamma_{t+1}, \mathbb{Z}^{n+1}).
$$
Set
$$
E_j(t) = A_j(\gamma_t ,
\Lambda_t,\Gamma_{t+1})W_t^{\frac{k}{k-1}},\,\, j =1,2,
$$
where the right hand sides are defined by (\ref{AAA},\ref{aaa}),
and set
$$
E_3(t) = \frac{3\rho^{(2)}_t}{4\eta_t},\,\,\, E_4(t) =
\frac{2^{n+3}(t+1)^{n+1} \rho^{(2)}_t}{\rho^{(1)}_t
(\rho^{(3)}_t)^n}.
$$
We also need one more quantity $E_5(t)$ defined as follows.
First, we put
$$
Z_1(t) = Z_1 (\gamma^* (\gamma_t^{\frac{k-1}{k}} 2^{-\frac{1}{k}}
(\rho_t^{(2)})^{\frac{1}{k}}, \Gamma_{t+1}),k),
$$
$$
Z_2(t) = Z_2 (2(t+1), \gamma^* (\gamma_t^{\frac{k-1}{k}}
2^{-\frac{1}{k}} (\rho_t^{(2)})^{\frac{1}{k}},
\Gamma_{t+1}),\Gamma_{t+1}),
$$
where $Z_1(\cdot,\cdot ),Z_2(\cdot,\cdot,\cdot )$ are defined by
(\ref{zet1},\ref{zet2}) and $\gamma^*(\cdot,\cdot )$ is defined
by (\ref{prime}). Then we put
$$
E_5(t)= \max \left( (Z_1(t))^{\frac{(n+1)k}{n(n-k)}} (2(t+1) C_3
(\gamma_t,\Lambda_t,\Gamma_{t+1}))^{\frac{(n+1)k}{n-k}} ,
(Z_2(t))^{\frac{(n+1-k)k}{n(n-k)}} (2(t+1) C_3
(\gamma_t,\Lambda_t,\Gamma_{t+1}))^{\frac{(n+1-k)k}{n-k}}\right),
$$
where $C_3(\cdot,\cdot,\cdot )$ is defined by (\ref{ceodin}). Note
that as $k\le n-1$ all the exponents are positive (particulary, $
n-k \ge 1$ and all the denominators in the exponents are nonzero).
Put
$$
T_t = \max_{1\le j \le 5} E_j (t).
$$
Since $ T_t \ge E_1(t), E_2 (t)$, we can apply Lemmas 7,8 to the
lattices $\Lambda = \Lambda_t, \Gamma = \Gamma_{t+1}$. Denote by
$\xi_t '$ the real $(n+1)$-dimensional vector satisfying the
conditions 1), 2) of Lemma 7. Consider the ball
$$
{\cal B}_t' = \{ \xi = (1,\xi_1,...,\xi_n) :\,\, |\xi - \xi_t' | <
\rho^{(2)}_t (4T_t)^{-1}\}.
$$
Note that for $\xi \in {\cal B}_t' $ we have $|\xi - \xi_t| <
\frac{3\rho^{(2)}_t}{4T_t}.$
Since $ T_t \ge E_3(t)$, we have
$$
{\cal B}_t' \subset {\cal B}_t .
$$
Note that
$$
H_t \ge H(2(t+1),\gamma_t,\Lambda_t,W_t)
$$
by the inductive conjecture (v). So by Corollary of Lemma 8 we see that for every $N$ in the interval
$ H_t \le N \le H_t',\,\,\, $ where
\begin{equation} H_t' = \left(2 (t+1) C_3 (\gamma_t, \Lambda_t, \Gamma_{t+1})\right)^{-n }T_t^{\frac{n}{k}},
\label{ahtprim}
\end{equation}
and every $\xi \in {\cal B}_t'\cap {\rm span}\,\Gamma_{t+1}$ we
have
$$
\mu_k (\xi, N) \le (2(t+1))^{-1}.
$$
Let us prove that for any $N\ge H_t^n$ and for any $\xi \in {\cal
B}_t'\cap {\rm span} \,\Gamma_{t+1} $ we have
\begin{equation}
\mu_{k+2} (\xi , N) \ge 2(t+1). \label{dd}
\end{equation}
To do this let us put
$$
U = T_t \cdot \frac{\rho^{(1)}_t}{3(t+1)\rho^{(2)}_t} .
$$
Recall that
$$
|\xi_t - \xi_t'| =\frac{\rho_t^{(2)}}{2T_t}
$$
and $\xi_t \in {\rm span }\,\Lambda_t$. So for every $N$ in the interval $ H_t^n \le N \le U$ we have
$$
|N\xi - N\xi_t| \le U \cdot \frac{3\rho^{(2)}_t}{4T_t}\le
\frac{\rho^{(1)}_t}{4(t+1)} ,
$$
and so the distance between $N\xi $ and $ {\rm span } \Lambda $
does not exceed $ \frac{\rho^{(1)}_t}{4(t+1)} $. But it follows
from the condition (v) of the $t$-th step that for the considered
values of $N$ we have
$$
N^{-\frac{1}{n}} \le H_t^{-1} \le \frac{\rho^{(1)}_t}{4(t+1)}.
$$
We see that the maximal distance between a point of the cylinder
$ C_\xi (N, N^{-1/n})$ and the subspace ${\rm span}\, \Lambda_t$
is $\le \frac{\rho^{(1)}_t}{2(t+1)}$.
Recall that ${\rm dim}\, \Lambda_t = k$.
Hence the cylinder $ 2(t+1)\cdot C_\xi (N, N^{-1/n})$ cannot
contain $ k+1$ linearly independent integer points inside for $
H_t^n \le N \le U$, so in this case we have the inequality
$$
\mu_{k+1} (\xi , N) \ge 2(t+1)
$$
(and thus, the inequality (\ref{dd})).
Suppose that $ N \ge U$. Then we deduce from the inequality $
T_t \ge E_4(t)$ that
$$
N^{-\frac{1}{n}} \le U^{-\frac{1}{n}} \le
\frac{\rho^{(3)}_t}{2(t+1)}.
$$
So the distance between a point of
$ C_\xi (N, N^{-1/n})$ and the linear subspace ${\rm span}\,
\Gamma_{t+1}$ is
$\le \frac{\rho^{(3)}_t}{2(t+1)}$.
Hence the cylinder
$ 2(t+1)\cdot C_\xi (N, N^{-1/n})$ cannot
contain $ k+2$ linearly independent integer points inside. This
implies (\ref{dd}) in the case $ N \ge U$.
We have proved the following statement: for any $ \xi \in {\cal B}_t'\cap {\rm span}\, \Gamma_{t+1}$ we have
\begin{equation}
\mu_{k+2} (\xi , N) \ge 2(t+1),\,\,\, N \ge H_t^n,
\label{kaplusdva}
\end{equation}
\begin{equation}
\mu_{k} (\xi , N) \le (2(t+1))^{-1},\,\,\, H_t \le N \le H_t ',
\label{ka}
\end{equation}
where $ H_t'$ is defined by (\ref{ahtprim}). Moreover, if
$$ \gamma '_t = \gamma '(\gamma_t , \Lambda_t , \Gamma_{t+1} ),
$$
where $\gamma ' (\cdot , \cdot , \cdot )$ is defined by
(\ref{gammaprime}) then it follows from Lemma 7
that the cylinder
$$
{\cal C}_{\xi_t '} (T_t, \gamma_t' T_t^{-1/k} )
$$
contains no nonzero points of $\Gamma_{t+1}$. Recall that we
constructed $\Gamma_{t+1}$ to satisfy the condition $ {\rm span}
\,\Gamma_{t+1} \not\subset \pi_{t+1}. $ Moreover for the vector $
q' \in \Gamma_{t+1}$ and any nonzero vector $p\in \pi_{t+1}$ the
angle between $ p$ and $q'$ is greater than
$\frac{\pi}{2}-\varepsilon$.
We can find a
$k$-dimensional ball ${\cal }B'' $ of radius $\gamma_t'
T_t^{-1/k}/2$ inside the facet $\{ x = T_t\}\cap {\rm span}\,
\Gamma_{t+1}$ of the cylinder $ {\cal C}_{\xi_t '} (T_t, \gamma_t'
T_t^{-1/k} ) \cap {\rm span}\, \Gamma_{t+1} $ (in fact, this facet
is a $k$-dimensional ball of radius $\gamma_t' T_t^{-1/k}$) such
that ${\cal }B''\cap ({\rm span}\, \Gamma_{t+1}\cap \pi_{t+1}\cap
\{ x = T_t\})=\varnothing$. Let $\Xi''$ be the center of ${\cal
}B'' $. Put $\xi_t'' = \Xi'' / T_t$. Then $ \xi_t''\in {\rm
span}\, \Gamma_{t+1}$. From the construction we see that
$n$-dimensional ball with the center at point $\Xi''$ and radius
$\gamma_t' T_t^{-1/k}/4$ has no common points with the subspace
$\pi_{t+1}$.
We get a cylinder
$$
{\cal C}_{\xi_t ''} (T_t, \gamma_t' T_t^{-1/k}/4 )
$$
with
no nonzero points of $\Gamma_{t+1}$
inside it and with $\xi_t'' \in {\rm span}\, \Gamma_{t+1}$.
By Lemma 4 we construct a $(\Gamma_{t+1},\gamma^*_t,
T_t)$-BAD vector $\xi^*_t \in {\rm span}\, \Gamma_{t+1}$ with
$$
\gamma_t^* = \gamma^* (\gamma_t'/4, \Gamma_{t+1} )
$$
( $\gamma ^* ( \cdot , \cdot )$ defined by (\ref{prime})), such
that the facet $\{ x = T_t\} $ of
$$
{\cal C}_{\xi_t^*} (T_t, \gamma_t^* T_t^{-1/k} )
$$
lies inside the facet $\{ x = T_t\} $ of $ {\cal C}_{\xi_t '}
(T_t, \gamma_t' T_t^{1/k} ) $ and
does not intersect $\pi_{t+1}$. Hence the ball
$$
{\cal B}^*_t = \{ \xi = (1,\xi_1,...,\xi_n) \in \mathbb{R}^{n+1}:\,\,\,
|\xi - \xi^*_t|< \gamma_t^* T_t^{-(k+1)/k}\}
$$
enjoys the following properties:
\begin{equation}
{\cal B}^*_t\subset {\cal B}_t',\,\,\, {\cal B}^*_t\bigcap
\pi_{t+1} = \varnothing . \label{ball}
\end{equation}
Recall that $\xi^*_t \in {\rm span}\, \Gamma_{t+1}$ is a $
(\Gamma_{t+1},\gamma^*_t, T_t)$-BAD vector.
Applying Lemma 5
to the lattice $ \Gamma =\Gamma_{t+1}$, $
(\Gamma_{t+1},\gamma^*_t, T_t)$-BAD vector $\xi^*_t$ and $\kappa =
\gamma_t^* T_t^{-(k+1)/k}/(4n)$
we get a complete $k$-dimensional lattice $\Lambda_{t+1}$ with fundamental volume
\begin{equation}
r_{t+1} \le G (\gamma_t^*, \Gamma_{t+1} )
(\gamma_t^*/4n)^{-\frac{1}{k+1}} (T_t)^{\frac{1}{k}} \label{voll}
\end{equation}
(here $G(\cdot ,\cdot )$ is defined by (\ref{je})),
such that
$ \Lambda_{t+1} \subset \Gamma_{t+1} , $
and
the Euclidean distance between $\xi^*_t $
and ${\rm span }\, \Lambda_{t+1} \cap \{ x= 1\} $ does not
exceed $\gamma_t^* (T_t)^{-(k+1)/k}/4n$.
Next, we apply the construction described in Section 4 after Lemma
5 and
obtain a
$(\Lambda_{t+1},\gamma_{t+1}, T_{t+1})$-BAD vector
$$
\xi_{t+1} \in {\rm span}\,\Lambda_{t+1}.
$$
We set
$$W_{t+1} = T_t.
$$
In the notation of Section 4 we have
$$
\xi_{t+1} =\hat {\xi^*_t},
$$
$$
\gamma_{t+1} = \hat{\gamma } ( \gamma_t^*, T_t, \Lambda_{t+1})=
\gamma^* (2^{-3} \gamma_t^*T_t^{\frac{1}{k(k-1)}}, \Lambda_{t+1})
$$ (here $
\hat{\gamma } ( \cdot, \cdot,\cdot )$ is defined by (\ref{hat})
and $\gamma ^* ( \cdot,\cdot )$ is defined by (\ref{prime})).
Now we set
\begin{equation}
H_{t+1} = \max \left( Z(2(t+2),\gamma_t^*,\Gamma_{t+1}, T_t),
H(2(t+1),\gamma_{t+1},\Lambda_{t+1}, T_{t+1}),
\frac{4(t+2)}{\rho (\Lambda_{t+1}, \mathbb{Z}^{n+1} )}
\right),
\label{newah}
\end{equation}
where $Z(\cdot ,\cdot ,\cdot, \cdot)$ is defined in
(\ref{starr}) and $H (\cdot ,\cdot ,\cdot, \cdot)$ is defined in (\ref{inte}).
Note that due to (\ref{ka}) we have
$$
\mu_{k} (\xi_{t+1} , N) \le (2(t+1))^{-1},\,\,\, H_t \le N \le H_t '
$$
$H_{t+1}$ may be greater than $H_t'$. But it
follows from the inequality $ T_t \ge E_5 (t) $ and the
definition (\ref{ahtprim}) of $H_t'$ that
$$
Z(2(t+1),\gamma_t^*,\Gamma_{t+1}, T_t)
\le H_t'
. $$ Lemma 6 implies that for
$$
N \ge Z(2(t+1),\gamma_t^*,\Gamma_{t+1}, T_t)
$$
we have
$$
\mu_{k} (\xi_{t+1} , N) \le (2(t+1))^{-1}.
$$
Hence
$$
\mu_{k} (\xi_{t+1} , N) \le (2(t+1))^{-1},\,\,\, H_t \le N \le H_{t+1}.
$$
On the other hand, it follows from (\ref{kaplusdva}) that
$$ \mu_{k+2} (\xi_{t+1} , N) \ge 2(t+1),\,\,\, N \ge H_t^n.
$$
For every $l$ the function $ \mu_l (\xi , N)$ is a continuous
function in $\xi$ and $N$. So, there exists $ \eta_{t+1}
>0
$,
such that
\begin{equation}
\mu_{k} (\xi , N) \le (t+1)^{-1},\,\,\,
\forall \xi : \, |\xi - \xi_{t+1} |\le \eta_{t+1}, \,\,\,\forall
N:\,
H_t \le N \le H_{t+1},
\label{kaa}
\end{equation}
\begin{equation}
\mu_{k+2} (\xi , N) \ge t+1,\,\,\, \forall \xi : \, |\xi -
\xi_{t+1} |\le \eta_{t+1}, \,\,\,\forall N:\, H_t^n \le N \le
H_{t+1}^n, \label{kaplusdva1}
\end{equation}
and
\begin{equation}
{\cal B}_{t+1} = \{ \xi = (1,\xi_1,...,\xi_n )\in \mathbb{R}^{n+1}
:\, |\xi - \xi_{t+1} |\le \eta_{t+1}\} \subset {\cal B}^*_t
\subset {\cal B}_{t} \label{ball1}
\end{equation}
Now for
the objects $ \eta_i , \gamma_i, W_i, H_i ,\Lambda_i ,
\Gamma_{i}, \xi_i $ with
$i = t+1$ we have the following statements.
The condition
(i) is satisfied by the construction.
The condition
(ii) is satisfied
since $ \xi_{t+1}$ is a $(\Lambda_{t+1},\gamma_{t+1},
W_{t+1})$-BAD vector.
The condition
(iii)
follows from (\ref{ball}).
The condition
(iv) follows from (\ref{ball1}).
The condition
(v) follows from the definition (\ref{newah}) of $H_{t+1}$.
The condition
(vi) follows from (\ref{ka}).
The condition
(vii) follows from (\ref{kaplusdva1}).
The inductive procedure is described completely and Theorem 1 for
$ k \ge 2 $ is proved.
\vskip+1.0cm
The author thanks the anonymous referee for important suggestions.
| 215,964
|
White Cloud
Created by the design team
of Andy Cao and Xavier Perrot,
‘Bai Yun’, White Cloud, is a sculpted cumulus cloud hovering over an undulating surface of compacted granite, crushed glass and oyster shell. The piece evolved from Cao Perrot Studio’s former installation ‘Lullaby Garden’, constructed in 2005. ‘Bai Yun’ is crafted with swirls of wire mesh supported by a series of slender posts. The Cloud’s shimmering presence is enhanced by thousands of clear, cut crystals catching the light from morning to moonlight.
| 60,981
|
Apr 06,2009
Charges dropped against ANC's Zuma
by UPI
PRETORIA, South Africa -- South Africa's chief prosecutor said Monday he dropped corruption charges against African National Congress leader Jacob Zuma.
Mokotedi Mpshe said it was "neither possible nor desirable" to pursue a case against Zuma, who is expected to become president after elections later in April, the BBC reported.
Mpshe said his decision was one of the most difficult he has ever made.
Zuma was charged in October 2005 but in September 2006 the prosecutors' case against the South African politician collapsed. A year later, a judge ordered that the corruption case against Zuma could not proceed but prosecutors won an appeal in January that allowed them to refile charges.
Mpshe said his decision was related to possible illegal manipulation of the country's prosecution system, based on secretly recorded conversations with the former head of the National Prosecution Authority, the BBC said.
"I have come to the difficult conclusion that it is neither possible nor desirable for the NPA to continue with the prosecution of Mr. Zuma," Mphse said. "It is a difficult decision because the NPA has expended considerable resources on this matter, and it has been conducted by a committed and dedicated team of prosecutors and investigators who have handled a difficult case with utmost professionalism and who have not been implicated in any misconduct."
Zuma has said the allegations, which involve a multimillion-dollar arms deal, are part of a political plot against him, the British broadcaster said.
| 221,830
|
TITLE: If $[R:Q]$ is cardinality equal to $R$?
QUESTION [0 upvotes]: We know real number set $\mathbb R$ as a linear space over rational numbers $\mathbb Q$ is infinite dimensional, if $[\mathbb R:\mathbb Q]$ is cardinality equal to $\mathbb R$?
We can do field extension for $\mathbb Q$ to get $\mathbb R$, I think do countable extension canont get $\mathbb R$ ,so the cardinal of $[\mathbb R:\mathbb Q]$ is strictly greater than the cardinality of $\mathbb Q$.
I have another guess: maybe the problem is like "Continuum hypothesis" is undecidable?
REPLY [0 votes]: The answer depends on whether or not we accept the axiom of choice.
With choice, Ned’s comment is the solution. Take $B$ to be a basis for $\mathbb{R}$ as a $\mathbb{Q}$-vector space. Every point in $\mathbb{R}$ can be written as a $\mathbb{Q}$-linear combinations of points of $B$. The number of such combinations is given by $\sum_{n \in \mathbb{N}}|(\mathbb{Q} \times B)^n|$. Since we have choice, for all $n \in \mathbb{N}$ we have that $| (\mathbb{Q} \times B)^n| = |\mathbb{Q} \times B| = \max\{|\mathbb{Q}|, |B|\}$. It follows that $|\mathbb{R}| = \sum_{n \in \mathbb{N}} \max\{|\mathbb{Q}|, |B|\} = \max\{|\mathbb{Q}|, |B|\}$. Thus we may conclude $|\mathbb{R}| = |B|$, which is equivalent to saying that $[\mathbb{R}:\mathbb{Q}] = |\mathbb{R}|$. This cardinal arithmetic argument relies on choice though.
The rules we used to make this argument are:
For infinite sets $A,B$, we have $|A \times B| = |A + B| = \max\{A \times B\}$, where $+$ is disjoint union.
$\sum_{n \in \mathbb{N}} |A| = |\mathbb{N}| \times |A|$
Without choice, there is no guarantee that such a basis $B$ exists. Even if we assume there is a basis, I think there are models of set theory in which the cardinality of $B$ cannot be pinned down as $|\mathbb{R}|$.
| 73,485
|
\begin{document}
\title[Folding maps on a cross-cap]
{Folding maps on a cross-cap\\
}
\author[M.~Barajas~S.]{Mart\'{i}n Barajas S.}
\address[M.~Barajas~S.]{Proyecto Curricular de Matem\'aticas, Facultad de Ciencias y Educaci\'on,
Universidad Distrital F.J.C., Bogot\'a, Colombia.}
\email{mbarajass@udistrital.edu.co}
\subjclass[2010]{Primary 53A05; secondary 57R45}
\keywords{Crosscaps, folding maps, singularities}
\date{\today}
\begin{abstract}
We study the reflectional symmetry of a surface in the Euclidean $3$-dimensional space with a cross-cap singularity with respect to planes.
This symmetry is picked up by the singularities of folding maps on the cross-cap.
We give a list of the generic singularities that appear in the members of the family of folding maps on a cross-cap
and characterise them geometrically.
\end{abstract}
\maketitle
\setlength{\baselineskip}{14pt}
\section{Introduction}\label{section1}
Given a plane $W$ in the Euclidean 3-space $\mathbb{R}^3$ with normal vector $\eta$, the \textit{folding map} with respect to $W$ is the map $f_W:\mathbb{R}^3\rightarrow\mathbb{R}^3$ given by
\begin{equation}\label{eqfold}
f_W(p)=q+\lambda^2 \eta,
\end{equation}
where $q$ is the projection of $p$ into $W$ along $\eta$ and $\lambda$ is the distance between $p$ and $W$. Thus, $p$ and its reflection with respect to $W$ have the same image under $f_W$.
Varying the plane $W$ gives the \textit{family of folding maps}.
Bruce and Wilkinson studied in \cite{Bruce1991,Wilkinson1991} the family of folding maps restricted to a smooth surface $M$ in $\mathbb{R}^3$. A member of the family is locally a map-germ $\mathbb R^2,0\to\mathbb R^3,0$. They showed that the singularities types of the folding map, when allowing smooth changes of coordinates in the source and target, capture some aspect of the extrinsic geometry of the surface $M$.
For instance, the folding map is singular at $p\in M$ if and only if $W$ is a member of the pencil of planes containing the normal line of $M$ at $p$. The singularity is more degenerate than a cross-cap if and only if $W$ is orthogonal to a principal direction of $M$ at $p$. A certain type of singularities of the folding maps ($S_2$) occurs along a curve on the surface called {\it the sub-parabolic curve}. It turns out that the sub-parabolic curve is the locus of points where a principal curvature is extremal along the lines of the other principal curvature.
Another singularity type ($B_2$) occurs on the {\it ridge curve} which
is the locus of points where a principal curvature is extremal along its own lines of principal curvature.
Bruce and Wilkinson also proved that the bifurcation set of the family if folding maps is the dual of the union of the focal and symmetry sets of $M$. This captures, for instance, the parabolic set on the focal surface of $M$. An analogous work was carried out in \cite{Izumiya2010} for surfaces in the hyperbolic or the Sitter 3-dimensional spaces.
The cross-cap singularity occurs stably on parametrized surfaces in $\mathbb R^3$.
The extrinsic differential geometry of the cross-cap was initiated by Bruce and West in \cite{BruceWest1998, West1995}. Since then, several work was carried out on the subject; see for example \cite{Barajas2017,BarajasKabata2016,Dias2016,Fukui2012,Garcia2000,Tari2007}. This paper is part of this ongoing work on the geometry of the cross-cap.
We consider here the reflectional symmetry of a cross-cap with respect to planes and consider folding maps on the cross-cap.
We obtain the list of generic singularities of these folding maps
(Theorem \ref{th1}) and describe their associated geometry (Theorem \ref{th2}). We give some preliminaries in Section \ref{section1} and \ref{section2}
\section{Preliminaries}\label{section2}
\subsection{Geometric cross-cap}
Whitney showed that maps $\mathbb{R}^2\rightarrow\mathbb{R}^3$ can have a stable local singularity under smooth changes of coordinates in the source and target. A model of this singularity is given by $f(x,y)=(x,xy,y^2)$.
A map germ $g:\mathbb{R}^2,0\rightarrow\mathbb{R}^3,0$ is said to have a cross-cap singularity if it is $\mathcal{A}$-equivalent to $f$ (we write $f\sim_{\mathcal{A}}g$), that is, there exist germs of diffeomorphisms $h:\mathbb{R}^2,0\rightarrow\mathbb{R}^2,0$ and $k:\mathbb{R}^3,0\rightarrow\mathbb{R}^3,0$, such that $k\circ f=g\circ h$.
The image of the map-germ $g$ is a germ of a singular surface called cross-cap.
West showed in \cite{West1995} that a parametrization of a cross-cap can be taken, by a suitable choice of a coordinate system in the source and isometries in the target, in the form
\begin{equation}\label{Par_Cros}
\phi(x,y)=(x,xy+p(y),ax^2+bxy+y^2+q(x,y)),
\end{equation}
where $a$ and $b$ are constant real numbers. We write
$$
\begin{array}{rcl}
p(y)&=&p_3y^3+p_4y^4+O(5),\\
q(x,y)&=&\sum^{3}_{k=0}q_{3k}x^{3-k}y^k+\sum^{4}_{k=0}q_{4k}x^{4-k}y^k+O(5),
\end{array}
$$
where $O(l)$ denotes a reminder in $(x,y)$ of order $l$.
A cross-cap with a parametrization as in \eqref{Par_Cros} is called \textit{geometric cross-cap} (in contrast to the $\mathcal A$-model $f$ above where the geometry is destroyed by the diffeomorphisms in the target).
A key feature of the cross-cap is its double point curve. It is a regular curve
in the source given by $x=-p_3y^2+\text{h.o.t.}$ for a cross-cap parametrized as in \eqref{Par_Cros} (see \cite{West1995}). This curve is mapped by $\phi$ to a segment of a curve ending at the cross-cap point, see Figure \ref{fig3}.
The tangent space to the cross-cap at the cross-cap point is one dimensional. We called this space the \textit{tangential line}. The orthogonal complement to the tangential line is called the \textit{normal plane}. The tangent cone at the cross-cap point is also an important object for the study of the differential geometry of the cross-cap. For a cross-cap with a parametrization as in \eqref{Par_Cros}
the tangential line is parallel to the $x$-axis, the tangent cone coincides with the $xz$-plane and the normal plane is the $yz$-plane (see Figure \ref{fig3}).
\begin{figure}[h!!!]
\includegraphics[width=7.5cm,clip]{Cone_Outros.pdf}\\
\caption{The tangential line, tangent cone and normal plane to an elliptic cross-cap.}
\label{fig3}
\end{figure}
The parabolic curve on a geometric cross-cap is studied in \cite{BruceWest1998,West1995}. When $a\ne 0$, the parabolic set
in the source has a Morse singularity. If $a<0$, the parabolic set is an isolated point (the cross-cap point) and every regular point on the surface is hyperbolic. The cross-cap is called a hyperbolic cross-cap (Figure \ref{fig2}, right).
If $a>0$, the parabolic set in the source is the union of two transverse curves and the cross-cap is called an elliptic cross-cap (Figure \ref{fig2}, left). When $a=0$, the cross-cap is called a parabolic cross-cap (Figure \ref{fig2}, center). For a generic parabolic cross-cap, the parabolic set in the source has a cusp singularity.
We say that a geometric cross-cap is generic if $a\neq0$.
\begin{figure}[h!!!]
\includegraphics[width=12.5cm,clip]{Class_cross_west.pdf}\\
\caption{Geometric cross-caps (\cite{BruceWest1998,West1995}).}
\label{fig2}
\end{figure}
Let $p$ be a regular point on the cross-cap surface, and suppose that it is not an umbilic point. Let $v_i$ be a principal direction and $\kappa_i$ its associated principal curvature at $p$, for $i=1,2$. Denote by $v_i(\kappa_j)$ the directional derivative of $\kappa_j$ along $v_i$, $i,j=1,2$. The point $p$ is a sub-parabolic point relative to $v_i$ if $v_i(\kappa_j)(p)=0$, $i\neq j$. Similarly, $p$ is a ridge point relative to $v_i$ if $v_i(\kappa_i)(p)=0$.\\
\begin{prop}[\cite{West1995}]
Suppose that $C$ is a regular curve in the source that passes through the origin, parametrised by
$
\gamma(t)=\left(\alpha t+O(2),\beta t+O(2)\right).
$
Then as we approach the cross-cap point along the curve parametrised by $\phi\circ\gamma$ (for $\phi$ as in \eqref{Par_Cros}) one principal curvature tends to
\begin{equation}\label{k1cross-cap}
\frac{2\left(a\alpha^2-\beta^2\right)}{\alpha\left(\alpha^2+\left(2\beta+\alpha b\right)^2\right)^{\frac{1}{2}}}
\end{equation}
and the other tends to infinity.
\end{prop}
We denote by $\kappa_1$ the principal curvature that tends to \eqref{k1cross-cap} and by $\kappa_2$ the other one. The limiting principal vectors $\nu_i$ associate to $\kappa_i$ are defined in \cite{Fukui2012} by a natural limiting process. The ridge curve on a cross-cap is studied in \cite{Fukui2012} and \cite{West1995} and the sub-parabolic curve in \cite{Fukui2012}.\\
It is shown in \cite{Fukui2012} that there is at least one and at most three regular sub-parabolic curves relative to $\nu_2$ and there are no sub-parabolic points relative to $\nu_1$ at the cross-cap. Denote by $(w_1,w_2)$ the tangent directions of these curves in the source, then
\begin{equation}\label{eq_subparab}
abw_1^3+(b^2+2a+1)w_1^2w_2+3bw_1w_2^2+2w_2^3=0,
\end{equation}
where $a,b$ are the coefficients in the parametrisation \eqref{Par_Cros}.
There are two ridge curves relative to $\nu_2$ at the cross-cap and their tangent directions $(w_1,w_2)$ in the source satisfy
\begin{equation}\label{eq_ridge1}
\left(bw_1-2w_2\right)w_1=0.
\end{equation}
In \cite{Tari2007} the author studied the lines of principal curvatures on a cross-cap and showed that there is generically one single topological model. In particular, there are three separatrices of the foliations, one is given by $y=-\frac{1}{2}q_{31}x^2+\text{h.o.t.}$ and the other two are given by $x=\lambda_i y^2+\text{h.o.t.}$, $i=1,2$, with
\begin{equation}\label{eq_sep_lc}
\lambda_i^2+3p_3\lambda_i-2=0.
\end{equation}
\section{The singularities of the folding maps on a cross-cap}\label{section3}
Consider a plane in $\mathbb R^3$ whose points $p$ satisfy
\begin{equation*}
W_{(\eta,\delta)}: \left\langle p,\eta\right\rangle=\delta,
\end{equation*}
with $\eta\in\mathbb{S}^2$ and $\delta\in\mathbb{R}$.
Varying $\eta$ and $\delta$ gives all the planes in $\mathbb R^3$.
For a fixed geometric cross-cap $\phi$ as in \eqref{Par_Cros}, the family $F:\mathbb{R}^2\times\mathbb{S}^2\times\mathbb{R}\rightarrow\mathbb{R}^3$
of folding maps on the cross-cap
is given by
\begin{eqnarray*}
F(x,y,\eta,\delta) & = & \phi(x,y)+(\left\langle \eta,\phi(x,y) \right\rangle-\delta)(\left\langle \eta,\phi(x,y) \right\rangle-\delta-1)\eta.
\end{eqnarray*}
We write $f_{(\eta,\delta)}$ for the restriction of $F$ to the plane $W_{(\eta,\delta)}$.
This is locally a map-germ $\mathbb{R}^2,0\rightarrow\mathbb{R}^3,0$.
The singularities of such map-germs under the action of the group $\mathcal A$ of smooth changes of coordinates in the source and target are classified by Mond in \cite{Mond1982,Mond1985}. We determine here the $\mathcal A$-singularities of the map-germs $f_{(\eta,\delta)}$.
Before that, we need some notation.\\
Let $\mathcal{E}_n$ denote the local ring of germs of functions $\mathbb{R}^n,0\rightarrow\mathbb{R}$ and $\mathcal{M}_n$ its maximal ideal. Denote by $\mathcal{E}(2,3)$ the 3-tuples of elements in $\mathcal{E}_2$.
The tangent space to the $\mathcal{A}$-orbit of $f: \mathbb{R}^2,0\rightarrow\mathbb{R}^3,0$ at the germ $f$ is given by
\begin{equation*}
T\mathcal{A}\cdot f=\mathcal{M}_2\cdot\left\{f_{x},f_{y}\right\}+f^*\left(\mathcal{M}_3\right)\cdot\left\{e_1,e_2,e_3\right\},
\end{equation*}
where $f_{x}$ and $f_y$ are the partial derivatives of $f$, $\left\{e_1,e_2,e_3\right\}$ denotes the standard basis vectors of $\mathbb{R}^3$ considered as elements of $\mathcal{E}(2,3)$ and $f^*\left(\mathcal{M}_3\right)$ is the pull-back of the maximal ideal in $\mathcal{E}_3$. \\
The extended tangent space is defined as
\begin{equation*}
T_e\mathcal{A}\cdot f=\mathcal{E}_2\cdot\left\{f_{x},f_{y}\right\}+f^*\left(\mathcal{E}_3\right)\cdot\left\{e_1,e_2,e_3\right\},
\end{equation*}
and the $\mathcal{A}_e$-codimension of the germ $f$ is
\begin{equation*}
\mathcal{A}_e\text{-cod}\left(f\right)=\dim_{\mathbb{R}}\frac{\mathcal{E}(2,3)}{T_e\mathcal{A}\cdot f}.
\end{equation*}
Let $f\in\mathcal{M}_n\cdot\mathcal{E}(n,p)$. A $q$-parameter unfolding $(q,F)$ of $f$ is a map-germ
\begin{equation*}
F:\mathbb{R}^n\times\mathbb{R}^q,(0,0)\rightarrow\mathbb{R}^p\times\mathbb{R}^q,(0,0)
\end{equation*}
in the form $F(x,u)=\left(\bar{f}(x,u),u\right)$, with $\bar{f}(x,0)=f(x)$. The family
\begin{equation*}
\bar{f}:\mathbb{R}^n\times\mathbb{R}^q,(0,0)\rightarrow\mathbb{R}^p,0\end{equation*}
is called a $q$-parameter deformation of $f$. Let $I$ be the identity element in $\mathcal{A}$. A morphism between two unfoldings $(q_1,F)$ and $(q_2,G)$ is a pair $\left(\alpha,\psi\right):(q_1,F)\rightarrow(q_2,G)$ with $\alpha:\mathbb{R}^{q_1},0\rightarrow\mathcal{A},I$, $\psi:\mathbb{R}^{q_1},0\rightarrow\mathbb{R}^{q_2},0$, such that $\bar{f}_u=\alpha(u)\cdot\bar{g}_{\psi(u)}$. The unfolding $(q_1,F)$ is then said to be induced from $(q_2,G)$ by $\left(\alpha,\psi\right)$. An unfolding $(q_1,F)$ of a map-germ $f$ is said to be $\mathcal{A}_e$-versal if any unfolding $(q_2,G)$ of $f$ can be induced from $(q_1,F)$.\\
Now we state a fundamental theorem on unfoldings. Given an unfolding $F(x,u)=\left(f(x,u),u\right)$, the initial speeds, $\dot{F}_i\in\mathcal{E}(n,p)$, of $F$ are defined by
\begin{equation*}
\dot{F}_i(x)=\frac{\partial f}{\partial u_i}(x,0),\quad\text{for}\quad i=1,\ldots,q.
\end{equation*}
\begin{thm}[\cite{Martinet1982}]\label{thm_versal-unfolding}
An unfolding $(q,F)$ of $f\in\mathcal{M}_n\cdot\mathcal{E}(n,p)$ is $\mathcal{A}_e$-versal if and only if
\begin{equation*}
T_e\mathcal{A}\cdot f+\mathbb{R}\cdot\left\{\dot{F}_1,\ldots,\dot{F}_q\right\}=\mathcal{E}(n,p).
\end{equation*}
\end{thm}
Our goal in is to classify germs of folding maps on a cross-cap and to analise the deformations in the members of the family of folding
maps. We start with the following result.
\begin{prop}\label{prop3.1}
For any $(\eta,\delta)\in\mathbb{S}^2\times\mathbb{R}$,
the folding map $f_{(\eta,\delta)}$ on a cross-cap as in \eqref{Par_Cros}
is singular at the origin. The
singularity is more degenerate than a cross-cap if, and only if, the plane
$W_{(\eta,\delta)}$ contains the origin, that is, $\delta=0$.
\end{prop}
\begin{proof}
The first part follows by observing that $\frac{\partial f_{(\eta,\delta)}}{\partial y}(0,0)=(0,0,0)$ for all $\left(\eta,\delta\right)\in\mathbb{S}^2\times\mathbb{R}$,
thus \mbox{rank $df_{(\eta,\delta)}(0,0)\leq 1$}. For the second part we use Whitney's criteria for recognition of the cross-cap singularity (see \cite{Whitney1943}).
\end{proof}
We list below the non-stable $\mathcal A$-singularities that occur at the origin in the members of the family of folding maps on a cross-cap. By Proposition \ref{prop3.1} above, $\delta=0$. We denote $f_{(\eta,0)}$ by $f_\eta$ and write $\eta=(\alpha,\beta,\gamma)$.
\begin{thm}\label{th1}
For a generic cross-cap $\phi$ as in \eqref{Par_Cros}, the folding map $f_\eta$ has a singularity $\mathcal{A}$-equivalent to one in Table \ref{table1} when $\eta$ is transverse to the tangent cone, and to one in Table \ref{table2} when $\eta$ is in the tangent cone but is not parallel to the tangential direction.\\
When $\eta$ is parallel to the tangential direction,
$f_\eta$ has corank 2 and is $\mathcal A$-equivalent to
\begin{equation}\label{eq_corank2}
(x^2,xy+y^3,y^2+Ax^3+Bx^2y+Cxy^2+y^3),
\end{equation}
with $\Theta(A,B,C)\neq0$, where
\begin{eqnarray*}
\Theta(A,B,C) & = & -10240A^3+\left(-2560C^2-3840B-1440C-135\right)A^2+\\
& + & \left(-160C^4+2880B^2C+800BC^2-20C^3+60B^2+\right.\\
& + & \left. 90BC\right)A+40B^2C^3-540B^4-180B^3C+5B^2C^2-20B^3.
\end{eqnarray*}
\newpage
\begin{table}[ht!!!]
\caption{Singularities of the folding map $f_{\eta}$ when $\eta$ transverse to the tangent cone.}
\begin{center}
{\begin{tabular}{llcc}
\hline
Type & Normal form & $\mathcal{A}_e$-cod.& Conditions\\
\hline
$B^{\pm}_{2}$ & $(x,y^2,x^2y\pm y^5)$ & $2$ & \mbox{$\alpha\neq0$, $\gamma\neq p_3\alpha$} \\
$B^{\pm}_{3}$ & $(x,y^2,x^2y\pm y^7)$ & $3$ & \mbox{$\alpha\neq0$, $\gamma=p_3\alpha$}, $(*)$\\
$B^{\pm}_{4}$ & $(x,y^2,x^2y\pm y^9)$ & $4$ & \mbox{$\alpha\neq0$, $\gamma=p_3\alpha$, $(**)$}\\
$C^{\pm}_{3}$ & $(x,y^2,xy^3\pm x^3y)$ & $3$ & \mbox{$\alpha=0$, $\Phi(\beta,\gamma)\neq0$} \\
$C^{\pm}_{4}$ & $(x,y^2,xy^3\pm x^4y)$ & $4$ & \mbox{$\alpha=0$, $\Phi(\beta,\gamma)=0$} \\
$F_{1,0}$ & $(x,y^2,x^3y+A_1xy^5+B_1y^7)$ & $4$ & \mbox{$\beta=1$, $4A_1^3+27B_1^2\neq0$}
\\
\hline
\end{tabular}
}
\end{center}
\label{table1}
\end{table}
where $\Phi(\beta,\gamma)=-2b\beta^3+(4a-b^2+2)\beta^2\gamma+\gamma^3$ and $F_{1,0}$ is a unimodal singularity (see \cite[page 29]{Mond1982}).\\
\noindent {\footnotesize
$(*)$ The singularity $B^{\pm}_{3}$ occurs for generic $\eta$ on the plane curve $\gamma=p_3\alpha$ in $\mathbb{S}^2$.\\
$(**)$ The singularity $B^{\pm}_{4}$ occurs for special values of $\eta$ on the curve $\gamma=p_3\alpha$
whose expression too lengthy to reproduce here, see \cite{Barajas2017} for details}.\\
\begin{small}
\begin{table}[ht!!!]
\caption{Singularities of the folding map $f_{\eta}$ when $\eta$ is in the tangent cone but is not parallel to the tangential direction.}
\begin{center}
{
\begin{tabular}{llcc}
\hline
Type & Normal form & $\mathcal{A}_e$-cod.& Conditions\\ \hline
$P_{3}$ & $(x,xy+y^3,xy^2+ky^4)$ & $3$ & \mbox{$\alpha\neq -p_3\gamma$, $k\neq\frac{1}{2},1,\frac{3}{2}$} \\
$P_{4}(\frac{1}{2})$ & $(x,xy+y^3,xy^2+\frac{1}{2}y^4)$ & $4$ & \mbox{$\alpha\neq -p_3\gamma$, $\Psi(\frac{1}{2})=0$} \\
$P_{4}(\frac{3}{2})$ & $(x,xy+y^3,xy^2+\frac{3}{2}y^4)$ & $4$ & \mbox{$\alpha\neq -p_3\gamma$, $\Psi(\frac{3}{2})=0$} \\
$P_{4}(1)$ & $(x,xy+y^3,xy^2+y^4)$ & $4$ & \mbox{$\alpha\neq -p_3\gamma$, $\Psi(1)=0$}\\
$R_{4}$ & $(x,xy+y^6+A_2y^7,xy^2+y^4+B_2y^6)$ & $4$ & \mbox{\hspace{0.7cm}$\alpha=-p_3\gamma$} \\
$T_{4}$ & $(x,xy+y^3,y^4)$ & $4$ & $\gamma=1$ \\
\hline
\end{tabular}
}
\end{center}
\label{table2}
\end{table}
\end{small}
\noindent where $\Psi(k)=\Psi(k,\alpha,\gamma)=-2k\alpha(\alpha+p_3\gamma)+1$.
\end{thm}
\begin{proof}
The proof follows by making successive changes of coordinates on the jet level and using the
conditions for a map-germ to be $\mathcal A$-equivalent to one in Mond's list. When $\eta$ is parallel to the tangential direction, the condition $\Theta(A,B,C)\ne 0$ is for the germ \eqref{eq_corank2} to be 3-$\mathcal A_1$-determined. Details of the calculations can be found in \cite{Barajas2017}. The conditions for the folding map $f_{\eta}$ to have the different types of singularities listed are given on the vector $\eta$, therefore the result is true for an open set in the space of geometric cross-caps.
\end{proof}
\begin{prop}
The non-stable singularities of $f_{\eta}$ are not $\mathcal{A}_e$-versally unfolded by the family $F$.
\end{prop}
\begin{proof}
Fix $\eta_0=\left(\alpha_0,\beta_0,\gamma_0\right)\in\mathbb{S}^2$ and consider o map germ $f_{(\eta_0,0)}$. Let $\xi=\left(u,v,w\right)\in\mathbb{R}^3$ and consider a real number $\delta\neq0$, $\left|\delta\right|<\varepsilon$ ($\varepsilon$ small). Thus, $f_{\left(\bar{\eta},\delta\right)}$ with $\bar{\eta}=\frac{\eta_0+\xi}{\left\|\eta_0+\xi\right\|}\in\mathbb{S}^2$ is a deformation of $(\eta_0,0)$. Let $F(x,y,u,v,w,\delta)=\left(f(x,y,u,v,w,\delta),u,v,w,\delta\right)$, where $f(x,y,u,v,w,\delta)=f_{\left(\bar{\eta},\delta\right)}(x,y)$ and
$$f(x,y,0,0,0,0)=f_{(\eta_0,0)}(x,y).$$
It follows that $F$ is a 4-parameter unfolding of the map germ $f_{(\eta_0,0)}$. By Proposition \ref{prop3.1}, the map germ $f_{(\bar{\eta},\delta)}$ has singularity type cross-cap at origin. By changes of coordinates in the source and target, the map germ $j^2f_{\left(\eta_0,0\right)}$ is $\mathcal{A}$-equivalent to
\begin{equation}\label{2jetfeta0}
j^2f_{\left(\eta_0,0\right)}\sim_{\mathcal{A}} \left(x,-\gamma_0\left(\gamma_0-b\beta_0\right)xy+\beta_0\gamma_0y^2,\beta_0\left(\gamma_0-b\beta_0\right)xy-\beta_0^2y^2\right).
\end{equation}
We consider the 2-jet of $f_{\left(\eta_0,0\right)}$ since
the map germ $f_{\left(\bar{\eta},\delta\right)}$, with $\delta\neq0$, has a cross-cap singularity. We take, without loss of generality,
$f_{\left(\eta_0,0\right)}$ as \eqref{2jetfeta0}. Thus, the initial speeds of $F$ are given by
\begin{align*}
\dot{F}_u(x,y) =& \left(0, 2\alpha_0xy, 2b\alpha_0xy+2\alpha_0y^2\right), \\
\dot{F}_v(x,y) =&\left(0, (b\gamma_0+2\beta_0)xy+\gamma_0y^2, \gamma_0xy\right), \\
\dot{F}_w(x,y) =& \left(0, b\beta_0xy+\beta_0y^2, (2b\gamma_0+\beta_0)xy+2\gamma_0y^2\right), \\
\dot{F}_\delta(x,y) =& \left(0,(2b\beta_0\gamma_0+2\left(1-\gamma_0^2\right))xy+2\gamma_0\beta_0y^2,\right. \\
& \left.2\left(\beta_0\gamma_0+b\left(1-\beta_0^2\right)\right)xy+2\left(1-\beta_0^2\right)y^2\right).
\end{align*}
Note that the Jacobian ideal of $j^2f_{\left(\eta_0,0\right)}$ is generated by
\begin{align*}
\left(j^2f_{\left(\eta_0,0\right)}\right)_x =& \left(1,-\gamma_0\left(\gamma_0-b\beta_0\right)y,\beta_0\left(\gamma_0-b\beta_0\right)y\right),\\
\left(j^2f_{\left(\eta_0,0\right)}\right)_y =& \left(0,-\gamma_0\left(\gamma_0-b\beta_0\right)x+2\beta_0\gamma_0y,\beta_0\left(\gamma_0-b\beta_0\right)x-2\beta_0^2y\right).
\end{align*}
It follows that
\begin{align*}
\left(j^2f_{\left(\eta_0,0\right)}\right)_x &- \left(j^2f_{\left(\eta_0,0\right)}\right)^*\left(1\right)\cdot e_1= \left(0,-\gamma_0\left(\gamma_0-b\beta_0\right)y,\beta_0\left(\gamma_0-b\beta_0\right)y\right),\\
\left(j^2f_{\left(\eta_0,0\right)}\right)_y &+ \left(\gamma_0-b\beta_0\right)\left(j^2f_{\left(\eta_0,0\right)}\right)^*\left(x\right)\cdot\left(\gamma_0 e_2-\beta_0 e_3\right) = \left(0,2\beta_0\gamma_0y,-2\beta_0^2y\right).
\end{align*}
It follows that it is not possible to simultaneously obtain the vectors $\left(0,y,0\right)$ and $\left(0,0,y\right)$ in the extended tangent space $T_e\mathcal{A}\left(j^2f_{\left(\eta_0,0\right)}\right)$. Therefore $T_e\mathcal{A}\left(j^2f_{\left(\eta_0,0\right)}\right)+\mathbb{R}\left\{\dot{F}_u,\dot{F}_v,\dot{F}_w,\dot{F}_\delta\right\}\neq\mathcal{E}(2,3)$, which concludes the proof.
\end{proof}
\section{The geometry of the folding maps on a cross-cap}\label{section4}
In this section we characterize geometrically some singularities that occur in the members of the family of folding maps on a cross-cap.
We use the concepts in \S \ref{section2} and the maps which we define below.
Let $\theta(t)=(t,\alpha(t))$ be a germ of a regular curve with $\alpha'(0)=a_1\ne 0$
and consider its image $\mu(t)=\phi\circ\theta(t)$ on the cross-cap. Note that $\mu$ is a regular curve and $\mu'(0)$ is parallel to the tangential line. When $\alpha'(0)=0$ the image on the cross-cap is a singular curve. Curves with this condition will be considered later.
Let $N(p)$ denote the unit normal vector to the cross-cap surface away from the cross-cap point. At the cross-cap point, the surface has no well defined normal vector, that is, $N(p)$ does not extend to the cross-cap point. However, a simple calculation shows that
\begin{equation*}
\lim_{t\rightarrow0}N\left(\mu(t)\right)=\left(0,\frac{-\left(b+2a_1\right)}{\left(\left(b+2a_1\right)^2+1\right)^{\frac{1}{2}}},\frac{1}{\left(\left(b+2a_1\right)^2+1\right)^{\frac{1}{2}}}\right).
\end{equation*}
Therefore, the normal vector has a limiting direction along the curve $\mu$ at the cross-cap point, and this limiting direction is parallel to $(0,-(b+2a_1),1)$. Observe that the limiting direction depends only on $\theta'(0)$ and not on the curve $\mu$. Thus, we can define the following map.
\begin{definition}
Let $X$ be a cross-cap as in \eqref{Par_Cros} and let $N_0X$ denote its normal plane at the origin. We call the {\rm limiting normal map} of $X$ the map $LN:\mathbb{R}^2\rightarrow N_0X$ given by
\begin{equation*}
LN(v_1,v_2)=\left(0,-(bv_1+2v_2),v_1\right).
\end{equation*}
\end{definition}
The limiting normal map induces a bijection between $\mathbb{S}^1\subset\mathbb{R}^2$ and $\mathbb{S}^1\subset N_{(0,0,0)}X$.
Consider now the family of curves $\varpi_{\lambda}(t)=\left(\beta_{\lambda}(t),t\right)$, with
$\beta_{\lambda}(0)=\beta_{\lambda}'(0)=0$ and $\beta_{\lambda}''(0)=\lambda$. We have $\varpi'_{\lambda}(0)=(0,1)$ and
$
LN(0,1)=\left(0,-2,0\right).
$
(The direction $LN(0,1)$ is orthogonal to the tangent cone of the cross-cap.) Thus, the limiting normal direction to the cross-cap is the same for all the members of
the family of curves $\varpi_{\lambda}$. The image of $\varpi_{\lambda}(t)$ by $\phi$ is a singular curve on the cross-cap
and its limiting tangent direction belongs to the tangent cone $TC_0X$ of the cross-cap.
More precisely,
$$
\lim_{t\rightarrow0}\left(\phi\circ\varpi_{\lambda}\right)'(t)=\left(2\lambda,0,2\right).
$$
\begin{definition}
We call {\rm the limiting tangent map} of the cross-cap as in \eqref{Par_Cros} the map
$LT:\mathbb{R}^2\rightarrow TC_0X$ given by
$$
LT(\lambda,1)=\frac{1}{2}\lim_{t\rightarrow0}\left(\phi\circ\varpi_{\lambda}\right)'(t)=
\left(\lambda,0,1 \right).
$$
\end{definition}
We use the limiting tangent and normal maps of the cross-cap to characterise the singularities of the folding maps.
\begin{thm}\label{th2}
Let $X$ be a cross-cap parametrised as in \eqref{Par_Cros}.
We have the following characterization of the singularities of the folding maps $f_{\eta}$ on $X$.
\begin{enumerate}
\item[(i)] The $C_4$ singularity occurs when folding with respect to a plane generated
by the tangential direction of the cross-cap and $LN(w)$, where $w$ is a tangent
direction at the origin to a sub-parabolic curve.
\item[(ii)] The $F_{1,0}$ and $T_4$ singularities occur when folding with respect to
the plane generated by the tangential direction of the cross-cap and
$LN(w)$, where the $w$ is a tangent
direction at the origin to a ridge curve relative to $\nu_2$.
In particular $f_\eta$ has a singularity type $F_{1,0}$ when $\eta$ is orthogonal to the tangent cone.
\item[(iii)] The $P_4(\frac{1}{2})$ singularity occurs when folding with respect
to the plane generated by $LN(0,1)$ and the limiting tangent direction to the double point curve.
\item[(iv)] The $P_4(\frac{3}{2})$ singularity occurs when folding with respect to the plane
generated by $LN(0,1)$ and the limiting tangent direction to the separatrix of principal foliations.
\item[(v)] When $\eta$ is parallel to the limiting tangent direction of the double point curve, the singularity of $f_{\eta}$ is of type $R_4$.
\item[(vi)] The corank 2 map singularity of $f_{\eta}$
occur when $\eta$ is parallel to the tangential direction, i.e., the folding plane is the normal plane.
\end{enumerate}
\end{thm}
\begin{proof}
We will characterize, in each case, the normal direction $\eta$ to the folding plane.
(i) Consider the map germ $f_\eta$ such that $\eta=(0,\beta,\gamma)$ satisfying $\Phi(\beta,\gamma)=0$ as in the Theorem \ref{th1}. By direct calculations we have that
\begin{equation*}
(LN)^{-1}((0,-\gamma,\beta))=(w_1,w_2),
\end{equation*}
where $\lambda,\mu$ satisfies
\begin{equation*}
abw_1^3+(b^2+2a+1)w_1^2w_2+3bw_1w_2^2+2w_2^3=0,
\end{equation*}
which is the same condition in the equation \eqref{eq_subparab}.
(ii) Take the map germ $f_\eta$ such that $\eta=(0,1,0)$, thus by the Theorem \ref{th1}, $f_\eta$ has a $F_{1,0}$ singularity at the origin. Follows that
\begin{equation*}
(LN)^{-1}(0,0,1)=\left(1,-\frac{b}{2}\right)=(w_1,w_2)\quad\text{satisfying}\quad bw_1+2w_2=0,
\end{equation*}
which satisfies the equation \eqref{eq_ridge1}. Now consider the map germ $f_\eta$ such that $\eta=(0,0,1)$ and again by the Theorem \ref{th1}, $f_\eta$ has a $T_{4}$ singularity at the origin. Using the same idea
\begin{equation*}
(LN)^{-1}(0,1,0)=\left(0,-\frac{1}{2}\right)=(w_1,w_2)\quad\text{satisfying}\quad w_1=0,
\end{equation*}
which is the other one condition that satisfy the equation \eqref{eq_ridge1}.
(iii) Consider the map germ $f_\eta$ such that $\eta=(\alpha,0,\gamma)$ satisfying $\Psi\left(\frac{1}{2}\right)=0$ and $\alpha\neq-p_3\gamma$. Note that
\begin{equation}\label{eq_psi12}
\Psi\left(\frac{1}{2}\right)=\Psi\left(\frac{1}{2},\alpha,\gamma\right)=-\alpha\gamma p_3+(1-\alpha^2)=\gamma(\gamma-p_3\alpha).
\end{equation}
Again we consider the orthogonal a $\eta$ in the tangent cone and applying to the equation \eqref{eq_psi12} and equaling to zero we obtain
\begin{equation*}
\Psi\left(\frac{1}{2},-\gamma,\alpha\right)=\alpha(\alpha+p_3\gamma)=0.
\end{equation*}
As $\alpha\neq0$ follows that $\alpha+p_3\gamma=0$. Calculating $(LT)^{-1}(-p_3,0,1)=(-p_3,1)$, which correspond with the 2-jet of curve of double points of the cross-cap $x=-p_3y^2$.
(iv) Analogous as (iii) take a map germ $f_\eta$ such that $\eta=(\alpha,0,\gamma)$ satisfying $\Psi\left(\frac{3}{2}\right)=0$ and $\alpha\neq-p_3\gamma$. It follows that
\begin{equation}\label{eq_psi32}
\Psi\left(\frac{3}{2}\right)=\Psi\left(\frac{3}{2},\alpha,\gamma\right)=-3\alpha\gamma p_3-3\alpha^2+1=-3\alpha\gamma p_3-2\alpha^2+\gamma^2.
\end{equation}
As before we consider the orthogonal a $\eta$ in the tangent cone and applying to the equation \eqref{eq_psi32} we obtain
\begin{equation}\label{eq_psi32lambda}
\Psi\left(\frac{3}{2},-\gamma,\alpha\right)=3\alpha\gamma p_3+\alpha^2-2\gamma^2=\lambda^2+3p_3\lambda-2=0,
\end{equation}
where $\lambda=\frac{\alpha}{\gamma}$. Via $(LT)^{-1}$, the vectors $(-\gamma,0,\alpha)$ in the tangent cone satisfying the equation \eqref{eq_psi32lambda} correspond with the curves $x=\lambda y^2$ where $\lambda$ satisfies the same equation above. (Compare with the equation \eqref{eq_sep_lc}).
(v) As $\eta$ is tangent to $(-p_3,0,1)$, in particular satisfies $\alpha=-p_3\gamma$, which is exactly the condition in the Theorem \ref{th1} for the germ $f_\eta$ to have a singularity at the origin of type $R_4$.
(vi) In this case $\eta=(1,0,0)$ and by definition, $f_\eta$ is the folding map respect to the normal plane. The rest follows from Theorem \ref{th1}.
\end{proof}
\textit{Acknowledgements:} The results in this paper are a part of the author's Ph.D. thesis. He would like to thank
Farid Tari for his help, advice and friendship. The author was supported by a CAPES doctoral grant.
\nocite{*}
| 23,449
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39.
Yep.
I have the gray hair to prove it.
One more year til the big 4-0.
I told my mom today that I remembered when SHE turned 40. I was 17 and we threw her a surprise party. And now, I'm one year away from that same milestone. How CRAZY is that?
As it is, I had a wonderful birthday weekend (oh yeah...at this age, it becomes a two-day celebration!). Last night I partied like a vampire at our annual "Twilight" viewing party. It was Breaking Dawn this time - and we wore our wedding veils. And drank blood red drinks out of Stryofoam cups. (And those last two sentences made no sense to anyone who hasn't seen the movie).
Anyway...back to turning 39 today.
I've been thinking back on the last 10 years ... you know, when I was 29 and on the cusp of my 30s.
10 years ago I was planning my wedding, working full time, and had no gray hair or stretch marks. I was 40 pounds lighter (at least). I wasn't blogging. It seems like a million years ago. And the thought of turning 30 didn't phase me at all.
Fast forward to today. I'll be celebrating my 10 year anniversary in October. I have two beautiful children who are my world. I'm still trying to find the 'right' job for me - looking for my happy place, I suppose. Oh and I have to get my hair colored every eight weeks because I have so much gray hair. What a difference 10 years makes.
As for turning 40 in another year?
Bring. It. On.
I have a few things planned for the next 365 days (or is it 366?). First, I'm participating in my first 5K race in two weeks. Then, when the weather get warmer (and my motivation comes out of hibernation), I am going to start "training" with the couch-to-5K program (again). And my end goal? I want to run a half marathon before I turn 40.
There. I said it.
It's "on paper" now....so I sort of have to do it, right? (lol)
I want to run a half marathon (that's 13.1 miles!) before I turn 40.
Preferably a Disney half marathon, since if I'm going to torture myself it might as well be at my happy place.
A plus to this insanity? Maybe I'll actually lose those extra 40 pounds before I turn 40.
Oh and I'm also going to find an amazing job very soon.
I just know it.
2.12.2012
39 and holding
Posted by Traci at Sunday, February 12, 2012
Labels:
birthdays,
Disney,
NaBloPoMo
| 222,550
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Guangzhou Gaaye Electronics Co., Ltd.
High boron glass can transmit ultraviolet rays, about 60% of the ultraviolet rays can be transmitted, and the production process of high boron glass is not difficult, and it is the same as our household energy-saving lamps, so the ultraviolet lamp tube of high boron glass is also carried out through the production line produced. Because of the production line, the output of this lamp is very sufficient. Sufficient output and relatively simple production process represent low prices. The price of the high boron glass ultraviolet lamp is between a few dollars to tens of dollars, and because of the material, its performance effect is not so good; the high boron glass ultraviolet lamp has a very good light attenuation. Strong, especially as the use time increases, the degree of attenuation will be higher and higher. Not only the transmittance is better, the UV lamp of quartz material has a low light attenuation degree. After four to five thousand hours of use, the light is still 70%-80% of the original, and the light is not extinguished, and the light attenuation will be low. lower and lower. Another feature of the quartz lamp tube is that it can emit ozone. The ultraviolet lamp tube itself has a sterilization and disinfection effect. Coupled with the blessing of the ozone effect, the sterilization effect is even higher.
helios quartz uv lamp,quartz tube uv lamp,quartz glass uv lamp,quartz uv light, quartz for uv lamp,quartz tube for uv lamp,quartz lamp uv lamp,quartz uv
| 261,733
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TITLE: A permutation of $1,2,3, \ldots, n$ is chosen at random . The probability that $1$ and $2$ are neighbors is
QUESTION [1 upvotes]: A permutation of $1,2,3, \ldots, n $ is chosen at random . The probability that $1$ and $2$ are neighbors is -
$(1) \frac {1}{n} $
$(2) \frac {2}{n} $
$(3) \frac {1}{n-1} $
$(4) \frac {1}{n-2} $
I think the correct answer is $(2) $ $ \frac{2}{n}$.
Reasoning : Total number of permutation $=n!$ .
Now consider $1,2$ as single object permute $(1,2),3, \ldots, n$. They are $(n-1)!$ in numbers. For each of these permutations $(12)$ and $(21)$ are counted differently, so there are $2(n-1)!$ permutation such that $1,2$ are neighbors . .
So the probability $=\frac {2(n-1)!}{n!}=\frac{2}{n}$
Is my solution correct? If it is not, then supply a proof and possible explanation why my solution is incorrect. Thank you. Your efforts are highly appreciated.
REPLY [1 votes]: If "$1$ and $2$ are neighbour" means that "$1$ and $2$ are neighboring," the solution you have provided appears to be correct. It is $$\frac{2 \times (n-1)!}{n!}=\frac{2}{n}$$
As you have solved.
| 117,490
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Joris Teepe: Going Dutch
Yet this is far from a blowing session. In addition to being an impeccable bassist with a precise but muscular attack, Teepe's a gifted composer, and his tunes are often memorable to the point of becoming stuck for hours in the listener's head. "The Princess and the Monster is based on an infectious, easy-paced piano vamp. With a harmonized two-horn head that's pure Blue Note and tremendous solos from Brecker, Braden, and ColliganTeepe's bass bobbing and jabbing between their notesit's irresistible.
Yet it could be better. Teepe's got a tendency towards overwriting, and the twelve minutes of "The Princess and the Monster are simply overstuffed with parts that were, to paraphrase the old anti-jazz cliché, probably more fun to play than hear. There's a recurring section where the horns or piano slip into a static phrase while Jackson or Teepe solos; Jackson's undeniably explosive here, but, really, this part just hamstrings the song's momentum. The title cut's slightly overcooked as well, with some very musicianly and impressivebut ultimately superfluousstops and starts built into its structure.
There's nothing at all wrong with the bluesy, sly "Nanananana. It's got more hard boppish harmonized horns in its theme, but the real essence of the tune is its phenomenal, thrilling call and response between Brecker and Braden that leads almost immediately to them simultaneous soloing over some very sympathetic accompaniment from the rhythm section. The quintet sounds completely integrated here, although each player is to some extent playing autonomously.
"Inventions in Maine is a slightly eccentric, almost Monkish blues played by the trio of Teepe, Colligan, and Jackson alone. It's got some great walking bass from the leader and some of Jackson's finest playing on the CD. His playing alongside Colligan's long, thematically impeccable solo is positively polyrhythmic at times; it's a little busy, but it works. Wayne Shorter's "Footprints is also played by the trio alone, and while it's beginning to seem that every jazz musician on earth has now recorded "Footprints (Shorter wrote some other tunes, you know), the version here's pretty seductive. Teepe's arrangement tweaks the song metrically and his almost contrapuntal bass linedoubled at times by Colligan's left handadds an ominous minor quality. Colligan's especially good here, but then, he's good everywhere on this disc. No contemporary pianist's solos combine swing, melody, and unity of theme more effectively.
This is, I think, Teepe's best band overall on record. Teepe's always memorable writing and arranging could at times do with a bit more austerity, but the shortcomings in those areas are offset by the band's pristine cohesion.
Visit Joris Teepe on the web.
Track Listing
The Princess and the Monster; Footprints; Nanananana; The Healing; Inventions in Maine; Going Dutch; Syeeda's Song Flute; Win Win Situation.
Personnel
Joris Teepe: bass; George Colligan: piano, Fender Rhodes; Gene Jackson: drums; Randy Brecker: trumper and flugelhorn; Don Braden: tenor and soprano sax, flute
Album information
Title: Going Dutch | Year Released: 2005 | Record Label: Twinz Records
| 395,712
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April 12, 2014
By Kasandra Martinson
kmartinson@csufresno.edu
GoBulldogs.com
By Kasandra Martinson
FRESNO, Calif. --- The Fresno State men's tennis team defeated UNLV 4-3 in a Mountain West dual at Spalding G. Wathen Tennis Center on Saturday afternoon. This was the Bulldogs fourth straight home victory in the past seven days.
"Our guys did a great job battling from start to finish today in the heat against a tough UNLV team," said Fresno State head coach Evan Austin.
After splitting the first two doubles matches with the Rebels, senior Sam MacNeil and freshman Eric Komati rallied for the 'Dogs on court two to defeat Jakob Amilon and Dimitar Petrov of UNLV (8-5) to clinch the important doubles point for Fresno State.
With an early lead over the Rebels, Komati then won a quick singles match on court five (6-2,6-2) to extend the Bulldogs lead to 2-0. Freshman John Darmstaedter took his point shortly after at the four slate over Dimitar Petrov 6-4,6-2.
UNLV took the next two points on the No. 1 and 6 slates to close the gap at 3-2. The Rebels then pushed play into three sets on courts two and three to make for an exciting finish.
With a one point lead, MacNeil stepped up for the team and sealed the match for the Bulldogs defeating Tamas Batyi of UNLV, 7-6 (8-6), 2-6, 6-1.
Winning eight of their last nine matches played, the Bulldogs improve to 11-8, 3-3 MW this season.
"We need to rest up and be ready to go again tomorrow against Sacramento State," added Austin.
Fresno State is back in action tomorrow at noon as they host Sacramento State for Senior Day at Spalding G. Wathen Tennis Center.
Fresno State 4, UNLV 3
Singles
1. Ace Matias (UNLV) def. Sai Kartik Nakireddi (FS) 7-6 (7-3), 6-2
2. Denys Pume (UNLV) def. Nikolas Papic (FS) 6-7 (1-7), 7-5, 5-1, WD
3. Sam MacNeil (FS) def. Tamas Batyi (UNLV) 7-6 (8-6), 2-6, 6-1
4. John Darmstaedter (FS) def. Dimitar Petrov (UNLV) 6-4, 6-2
5. Eric Komati (FS) def. Willie Sublette (UNLV) 6-2, 6-2
6. Jakob Amilon (UNLV) def. Alexandros Georgios (FS) 6-2, 4-6, 6-2
Doubles
1. Ace Matias/Denys Pume (UNLV) def. Sai Kartik Nakireddi/John Darmstaedter (FS) 8-3
2. Eric Komati/Sam MacNeil (FS) def. Jakob Amilon/Dimitar Petrov (UNLV) 8-5
3. Nikolas Papic/Alexandros Georgios (FS) def. Willie Sublette/Tamas Batyi (UNLV) 8-3
Order of finish: Doubles (1,3,2); Singles (5,4,1,6,3,2)
"Catch the Wave"--Stay connected with the Bulldogs on Facebook (fresnostateathletics), Instagram (fresnostateathletics) and Twitter (@FSAthletics)
| 130,887
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TITLE: Find integer $n$ where $1+\frac{n}{10} < \alpha < 1+\frac{n+1}{10}$
QUESTION [1 upvotes]: On first look, doesn't any number suffice? For example, if I let $n=1$ then $$1+\frac{1}{10} < \alpha < 1+\frac{2}{10}$$ is valid right? Tho, I think the question is asking in the context of the rest of the question? Am at part iv
I don't really get the reasoning behind the answer tho
How does the pieces fall into place? Why the expression for $g(x)$. How do I get the seemingly arbitrary numbers 1.7 and 1.8?
REPLY [1 votes]: The expression for $g(x)$ comes from (iii). Apparently $\alpha \ln \alpha - \ln(1+\alpha) = 0$, so $g(\alpha) = 0$. As for how to get the numbers 1.7 and 1.8: The question sort of indicates that you'll probably find something fairly near 1, so I guess just trying values $g(1+\frac{n}{10})$ for $n$ from $0$ to $10$ until you get a change of sign could be expected to work. There's probably a better way.
As for your first question, yes, of course it's in the context of the rest of the question. Otherwise what is $\alpha$? How could you claim that any $n$ works, without knowing what $\alpha$ is?
| 65,808
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TITLE: Possible polygons in uniform polyhedra
QUESTION [1 upvotes]: Summary: What are the possible regular polygons that can be created by the orbit of a 3D point under $O_h$ or $I_h$?
I've been reading about uniform polyhedra for a while now. Something that has stricken me as odd is that, excluding prisms and antiprisms, there are only certain regular polygons that can appear on a uniform polyhedron. When the symmetry group of the figure is a subgroup of $O_h$, the possible number of sides are 3, 4, 6 and 8. Likewise, when the symmetry group is a subgroup of $I_h$, the corresponding numbers of sides are 3, 4, 5, 6 and 10. And moreover, polyhedra with more than 4 sides are always invariant under some rotation of the group. One could prove this by looking at the classification of uniform polyhedra, but that feels like cheating. Instead, I'd like to prove the more general fact that the orbit of a point under one of these groups can only create the corresponding regular polygons.
I've tried looking at the individual symmetries of a regular $n$-gon creating a counterexample, trying to prove they must be contained in the greater symmetry group, to no avail. I've also tried more combinatorial approaches, as well as stuff with vectors, but those just seem to lead to humongous casework. I'm completely stuck at the moment. Any help?
REPLY [1 votes]: $O_h = [3,4]$ has 2-gonal, 3-gonal and 4-gonal rotational symmetry axes. So clearly you could generate a regular triangle and a square ($n$-gon). But when placing the edges somewhat more apart you also can generate a hexagon and an octagon ($2n$-gon).
Same for $I_h = [3,5]$, it has 2-gonal, 3-gonal and 5-gonal rotational symmetry axes. So clearly you could generate a regular triangle and a regular pentagon ($n$-gon). But when placing the edges somewhat more apart you also can generate a square, a hexagon and a decagon ($2n$-gon).
--- rk
| 62,635
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TITLE: Permutation Problem
QUESTION [0 upvotes]: There are n children that are randomly seated. In how many ways will the oldest child always be seated to the right of the youngest child? (It does not have to be directly to the right).
REPLY [1 votes]: HINT: In half of the possible seatings the oldest child is seated to the right of the youngest, and in the other half the oldest child is seated to the left of the youngest. Why? And how many seatings are there altogether?
Added:
If we consider only those arrangements in which the youngest child is not in the leftmost position, how many are there with the oldest child to the right of the youngest?
An equivalent question that’s a little easier to talk about:
Of those permutations of $[n]=\{1,\dots,n\}$ whose first element is not $1$, how many have $n$ to the right of $1$?
For each $k\in[n]$ there are $(n-1)!$ permutations of $[n]$ with $k$ in the first position. We’re given that $k\ne 1$. If $1<k<n$, half of these permutations have $1$ to the left of $n$, and half have $1$ to the right of $n$. If $k=n$, clearly $n$ cannot be to the right of $1$. Thus, there are $$\frac{(n-2)(n-1)!}2=\frac{n!-2(n-1)!}2=\frac{n!}2-(n-1)!\tag{1}$$
permutations that do not have $1$ in the first position and do have $n$ to the right of $1$.
Note that the final expression in $(1)$ implicitly shows another way of arriving at the result. There are altogether $\frac{n!}2$ permutations of $[n]$ that have $n$ to the right of $1$, so we need only remove those that have $1$ in the first position. Every permutation with $1$ in the first position has $n$ to the right of $1$, so we must remove all of them. Since there are $(n-1)!$ such permutations, we must subtract $(n-1)!$.
| 20,106
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Reviews // 2 Articles
//Reviews
Labyrinth (1986)
The Dark Crystal (1982)
Labyrinth (1986)
29 Feb 2004 // 7:00 PM
Sarah is the origin and end of the fiction, its cause and effect simultaneously.
The Dark Crystal (1982)
8 Dec 2003 // 7:00 PM
It's ability to inspire terror is indicative of its creators' amazing craftsmanship.
| 240,670
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Japan
Sleek accommodation on this laid-back, modern-art loving island, with views over the tranquil Sea of Japan, as well as of Kusama Yayoi's giant spotted pumpkin on the seashore. Desert island bliss and a change of pace guaranteed.
Gotanji, Naoshima, Kagawa 7613110 Japan
+81 (0)87 892 3223
Google map: bit.ly/AAfBjY
Send your feedback or queries to been.there@guardian.co.uk
Search Been there
| 193,629
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Elite Camp
Elite Camp
Mission:
The Bulldog Elite Camp is designed to give campers entering 9th grade and higher who want to play at the collegiate level an opportunity to compete against other campers who want the same experience while also being observed by college recruiters. We provide professional instruction, personal attention, and an atmosphere conducive to learning.
IMPORTANT
Once you see the message that your registration has been added, click the cart located to the top right to check out. If you are registering more than one camper, please repeat the process on this page twice before checking out. This ensures we get the registration form for each player you are registering for camp.
Please see additional information below.
Additional Information
At A Glance
Who: Open to all males entering 9th grade and higher.
Where: On the campus of Yale University, Payne Whitney Gymnasium and the Lanman Center. Campers must enter through Payne Whitney Gymnasium, which is located at the end of Grove Street on Tower Parkway in New Haven, CT. Overnight campers will be housed at Albertus Magnus College located at 700 Prospect Street in New Haven, CT.
When: June 1–3, 2018 or August 3–5, 2018 Camp will conclude on Sunday at noon.
Time: Check-in begins at 4 p.m. at Albertus Magnus College for overnight campers and at 6 p.m. at the Yale Lanman Center for commuters.
Cost: Commuter $350; Overnight $525.
Phone: 203.432.1484
Highlights
- Excellent Instruction from the Yale Staff
- Personal Attention
- Player/Coach Ratio of 8:1
- Skill Development
- Games vs. Great Competition
- College Admissions and Financial Aid Information
- Great Accommodations
- Active college recruiting by more than 60 colleges
33 states were represented at last year’s camp and 62 colleges attended the camp for the purpose of recruiting, including some of the best academic schools in the northeast. In addition, 14 players on Yale’s current roster were Bulldog Elite Campers..
Commuter drop-off and pickup will be at Payne Whitney Gym; overnight housing will be at nearby Albertus Magnus College, Dominican Hall..
LOCATIONS
Payne Whitney Gym at Yale University
70 Tower Pkwy, New Haven, CT 06511
Albertus Magnus College - Dominican Hall
700 Prospect Street, New Haven, CT 06511
| 321,995
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ISLAMABAD: Army chief General Raheel Sharif rushed to Kabul today to deliver a warning to Afghan authorities to take decisive action against sanctuaries of the Tehreek-e-Taliban Pakistan (TTP), or else Pakistan would go for ‘hot pursuit.
General Raheel is traveling to the Afghan capital after security agencies found evidence that the Peshawar attack, which killed 141 people, including 132 children, was planned inside Afghanistan by the Mullah Fazlullah group.
The army chief will meet his Afghan counterpart and Afghan President Ashraf Ghani, as well as the ISAF commander to present evidence of the Peshawar massacre’s linkage with TTP sanctuaries in Kunar and Nuristan province. A security source told the channel.
RAW, complete support in eliminating terrorist in his area of responsibility.
| 352,961
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Presbyterians Week Headlines
[1] New Christian Observer Articles and Features
[2] "John Calvin: Myth and Reality" is Theme of 2009 Calvin Studies Society Conference at Calvin College
[3] Greenville Presbyterian Theological Seminary Hosts “John Calvin: 500 Years in Retrospect” Conference 10-12 March 2009
[4] Hampton Road United Reformed Church of Southport, England, Closes after 119 Years
[5] History of the Hungarian Reformed Church by Imre Revesz and George A.F. Knight Listed by AntiQbook
[6] PCUSA Mission Yearbook for Prayer and Study Now Available in Electronic Formats
[7] HRFA Senior Care Facility in Ligonier, Pennsylvania Seeks Refund of Real Estate and School Taxes Paid 2006-2007
[8] Death of Church of Scotland Minister the Rev. Douglas Clyne
[9] Answers Research Journal Editor Tells of Mockery, Derision, and Bogus Paper Submission from Darwinian Scientists
[10] AINA Reports Turkey Nearer to Admitting WWI Genocide of 1 Million in Armenia
[1] New Christian Observer Articles and Features
The online Christian Observer has several new articles and features for 2009 including:
John Knox’s Theology of Prayer – by Brian Golez Najapfour – explores John Knox and his theology of prayer as being an indication of true faith, and “an earnest and familiar talking with God.”
The Christian Observer in 1845 – a 1845 letter to the editor of the Philadelphia, Pennsylvania-based Christian Observer by Congregationalist minister and abolitionist Leonard Bacon, which sheds light on political and ecclesial controversies and the tone of public discourse surrounding these issues in the 1845 United States of America.
Christian Observer - Our Renewed Vision And Expanding Mission - 2009 – Associate Editor Chuck Huckaby shares the editorial vision of the Christian Observer as the format of the former monthly print magazine is changing to meet the capabilities of Internet technologies, explains the current ways of accessing and receiving Christian Observer resources, and solicits your help in recruiting new writers that love the Lord and are seeking to faithfully uphold His Word in order to carry the Christian Observer far into the future, D.V.
Additionally, links to Princeton Theological Seminary’s 2009 daily readings of Calvin’s Institutes, Dr. Robert LaMay’s Sabbath School lessons for January 2009, weekly] "John Calvin: Myth and Reality" is Theme of 2009 Calvin Studies Society Conference at Calvin College
The Meeter Center for Calvin Studies at Calvin College and Calvin Theological Seminary will host the 2009 Calvin Studies Society conference “John Calvin: Myth and Reality” 16-18 April 2009.
Information is available at the Meeter Center website.
+ Calvin College 3201 Burton Southeast, Grand Rapids, Michigan, 49546, 616-526-6000
[3] Greenville Presbyterian Theological Seminary Hosts “John Calvin: 500 Years in Retrospect” Conference 10-12 March 2009
Greenville Presbyterian Theological Seminary (GPTS) is hosting their Spring Theology Conference on 10-12 March 2009 at the Woodruff Road Presbyterian Church in Simpsonville, South Carolina.
The conference theme is “John Calvin: 500 Years in Retrospect – A 21st Century Assessment.” Speakers include Dr. Joel Beeke, president of Puritan Reformed Theological Seminary, and Dr. Joseph Pipa, president of GPTS.
Information is available at the GPTS website.
+ Greenville Presbyterian Theological Seminary, 200 East Main Street, Post Office Box 690, Taylors, South Carolina, 29687, 864-322-2717, Fax: 864/322-2719, info@gpts.edu
[4] Hampton Road United Reformed Church of Southport, England, Closes after 119 Years
Hampton Road United Reformed Church (URC United Kingdom) of Southport, England, closed permanently after a service of thanksgiving held 4 January 2008.
The congregation that once numbered sixty had dwindled to fifteen people, and the costs of maintaining the church building had become too burdensome.
+ Trinity Mirror North West & North Wales Ltd., Post Office Box 48, Old Hall Street, Liverpool L69 3EB, England, 01704-536655, visiternews@southportvisiter.co.uk
+ United Reformed Church, Church House, 86 Tavistock Place, London WC1H 9RT, England, 020-7916-2020, Fax: 020-7916-2021, urc@urc.org.uk
[5] History of the Hungarian Reformed Church by Imre Revesz and George A.F. Knight Listed by AntiQbook
AntiQbook lists a 163-page hard cover edition in good condition of History of the Hungarian Reformed Church by Imre Revesz and George A.F. Knight published in 1956 by Hungarica Americana of Washington DC, USA, for EUR€22.50 or approximately US$31.23.
+ AntiQbook, Havenstraat 17, 3441 BH Woerden, Nederland, antiquariaat@hoeksteenboekhandel.nl
[6] PCUSA Mission Yearbook for Prayer and Study Now Available in Electronic Formats
The Presbyterian Church (PCUSA) Mission Yearbook for Prayer and Study, which dates from 1892, is now available via daily email delivery, podcast, and RSS feed, as well as in the traditional paper format orderable by calling 800-524-2612 or through the PCUSA website.
+ Presbyterian Church (PCUSA), 100 Witherspoon Street, Louisville, Kentucky 40202, 888-728-7228, Fax: 502-569-8005
[7] HRFA Senior Care Facility in Ligonier, Pennsylvania Seeks Refund of Real Estate and School Taxes Paid 2006-2007
The Bethlen Home in Ligonier, Pennsylvania, which is operated by the Hungarian Reformed Federation of America (HRFA), has filed suit against Westmoreland County, Pennsylvania seeking the return of more than US$34 thousand in real estate taxes it paid 2006-2007. It also wants the Ligonier Valley School District to refund more than US$146 thousand in school taxes.
A 2007 Pennsylvania Supreme Court ruling that nonprofit organizations passing a five-point test can be exempt from real estate taxes has triggered tax appeals by senior-care facilities throughout the state.
Bethlen operates a 96-bed skilled-nursing center and a 10-bed unit that cares for patients suffering from dementia.
+ Hungarian Reformed Federation of America, 2001 Massachusetts Avenue Northwest, Washington DC 20036-1011, 202-328-2630, 202- 328-7984, hrfa@hrfa.orgThis e-mail address is being protected from spam bots, you need JavaScript enabled to view it
[8] Death of Church of Scotland Minister the Rev. Douglas Clyne
Church of Scotland minister the Rev. Douglas Clyne, 67, died from a heart attack in Elgin, Scotland on 28 December 2008.
Clyne was the minister at Fraserburgh Old Parish Church in Fraserburgh, Scotland for thirty-one years, before becoming pastoral assistant at Old High Saint Stephen’s Church in Inverness, Scotland.
The funeral was held at Old High Saint Stephen’s Church on 7 January 2009.
+
[9] Answers Research Journal Editor Tells of Mockery, Derision, and Bogus Paper Submission from Darwinian Scientists
Approaching the one-year anniversary of Answers in Genesis’ peer-reviewed Answers Research Journal on 9 January 2009, editor Dr. Andrew Snelling looks back on the first year of publication as exceeding expectations both quantitatively and qualitatively, and looks optimistically ahead to 2009 due to fourteen papers being already submitted or promised before the new year began, and cautiously with “a new journal [having] to first demonstrate its credibility and acceptability in order to attract and encourage the best creationist researchers to submit their papers to it.”
Dr. Snelling additionally described the mockery, derision, and even the submission of a bogus paper—by Darwinian scientists that want their view of creation to be the only one allowed in the public square and to be taught in the schools. Snelling describes how a Darwinian scientist formerly affiliated with Queens University Belfast, using an assumed name, submitted a paper purporting to be based on legitimate scientific research, but in actuality was full of factual errors and what Snelling characterized as “not serious scientific research of the calibre of that in [the pseudonymed scientist’s list of prior] publications.”
+ Answers in Genesis, 2800 Bullittsburg Church Road, Petersburg, Kentucky 41080, 859-727-2222
[10] AINA Reports Turkey Nearer to Admitting WWI Genocide of 1 Million in Armenia
Abdulmesih BarAbrahem of the Assyrian International News Agency reports that Turkey is moving closer to admitting to and apologizing for the genocide of 1 million Armenians during WWI, 750 thousand of whom were Assyrians.
+ Assyrian International News Agency,The Banbridge Leader, 25 Bridge Street, Banbridge, County Down BT32 3JL, Northern Ireland, 028406-62745
Wednesday, January 7, 2009
Presbyterians Week Headlines
| 230,698
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The current state of feature films may be as bleak as it has ever been. But the art of the documentary movie remains vibrant. Here are three I recently enjoyed.
White Horse As the nuclear plant meltdowns in Japan were in the headlines, I came across this modest black & white 18-minute documentary on HBO that made its human emotional consequences palpable. It follows a refugee from the 1986 Chernobyl nuclear plant disaster as he returns to the home he left as a child. The bleak landscape left behind feels chilling as the filmmakers return with Russian Maxym Surkov to the Soviet era apartment block he left behind as a child. They climb the stairs to Surkovs childhood home, the decaying building makes an apt metaphor for the devastation the meltdown wrought even beyond the effects of radiation. His pain is wrenchingly affecting as he inspects the looted apartment of his youth and finds small remaining remnants of his youth. A poster of a white horse on the wall of his boyhood bedroom provides the films title and serves as a symbol for the lost innocence and childhood of the former Chernobyl resident. Though short and simple, the film is redolent with drama and emotion that make the potential dangers of nuclear power compelling by the tragedys effects on just one human soul.
Joan Rivers A Piece Of Work Much as I have found Rivers funny over the years since her days as a housewife comedian appearing on The Tonight Show, something about her was increasingly irritating to me. Its a testament to both her and the filmmakers that this account of a year in her life at age 77 warmed me to Rivers in an indelible fashion. It documents the still-working star on the far side of her peak of fame as she deals with her increasing irrelevance and loneliness, showing the person behind the persona with touching grace.
The brash, bold, irreverent and wacky humorist becomes a genuine person who feels and suffers like the rest of us behind the comedic guise, making this a moving portrayal of a pop culture icon in decline.
Dolley Madison I am not a big fan of documentaries that attempt to dramatically recreate history with actors, being more a just the facts, maam kind of guy. But this PBS American Experience film about the wife of President James Madsion who all but birthed the role of First Lady in our culture and political landscape is an exception to the rule. In part it succeeds by only using her own words for the dramatic dialog, and through the strength of actress Eve Best who portrays her. Plus its a story rife with early tragedy and social ostracism that grows rich with warmth and eventual triumph as Dolley develops a full and dynamic partnership with Madison, her second husband who was 17 years her senior, and defines a wifes role in the White House in a critical time in our nations history. One might not be able to call Dolley Madison the first American feminist, but she certainly proved how women could rise to dynamism and be a loving wife and mother many decades before the term was coined.
From The Progressive Populist, June 15, 2011
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Water Polo Splits Two Matches on Saturday
Water Polo Splits Two Matches on Saturday
).
October 31, 2018
Filed under Sports, Water Polo
Hang on for a minute...we're trying to find some more stories you might like.
By Maria Trivelpiece
Fordham Water Polo lost to Bucknell and beat James Madison on Saturday (Courtesy of Fordham Athletics).
The start to their season was anything but ideal, but the Fordham University men’s water polo team is not discouraged. Through hard-fought battles and many one-goal games, the squad has been making a comeback in the hopes of ending its season on a much higher note than it started. After a successful week of four victories, the Rams headed into this weekend with ample confidence.
They began their Saturday off strong, as they defeated Johns Hopkins University 13-8. Freshman Demitris Koukias was stellar, scoring three goals. Joining him in leading the team, as always, was junior Jake Miller-Tolt, who performed well with two goals and two assists. The Rams’ offense was on fire, and they hoped to carry the momentum into their nightcap, where they faced off against No. 17 Bucknell University.
The Rams were not as successful in match two, as the Bison defeated Fordham 14-8. There were still some highlights in the loss, as senior Ian Watson netted two goals, and the six other players shared the scoring. The team effort was there offensively, but the defense struggled to silence the dynamic Bucknell offense.
The weekend was not yet over as the Rams had one more match the following day against the LaSalle Explorers. Looking to avenge its Saturday night loss, Fordham jumped on their opponent early. Miller-Tolt had six goals and senior Patrick Coffey had seven saves in goal. Senior Magnus Sims and junior Tristin Knoflick each two scores of their own leading the Rams to a 15-7 victory. With the win, Fordham improves to 4-5 in conference play and 12-13 overall.
The Rams have four matches before they head into postseason play. Next week, they will face off against Johns Hopkins, Navy, George Washington and Iona. After that, they will be back in the Bronx at Francis B. Messmore Aquatic Center for the MAWPC tournament. The Fordham University men’s water polo team seems to be gaining momentum. They are peaking at a time that could be just right, and their postseason hopes may be achievable.
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PikoPixel 1.0 Beta 4
An extremely easy to use image editor that provides a collection of basic tools that can be used to create beautiful pixel art projects.
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By default, the PikoPixel canvas comes with a grid and with a background pattern that can be easily customized via the appropriate settings panels (navigate to the Canvas menu to access them).
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To sum up, PikoPixel provides basic drawing tools that will help you create pixel art. Since the app comes with a grid, background patterns and the possibility to work with multiple layers, you will be able to generate fore complex effects: your imagination is the limit.
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CLECs Gain Ground with SMBs
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Pittsburgh, PA (Sports Network) - The Pittsburgh Pirates placed second baseman Neil Walker on the 15-day disabled list Friday with a right hand laceration.
The move is retroactive to April 27.
Walker is batting .253 with eight RBI through 23 games this season.
To fill Walker's roster spot, the Pirates recalled infielder Jordy Mercer from Triple-A Indianapolis.
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\begin{document}
\maketitle
\begin{abstract}
We study degenerations of Bethe subalgebras $B(C)$ in the Yangian $Y(\mathfrak{gl}_n)$, where $C$ is a regular diagonal matrix. We show that closure of the parameter space of the family of Bethe subalgebras, which parametrizes all possible degenerations, is the Deligne-Mumford moduli space of stable rational curves $\overline{M_{0,n+2}}$. All subalgebras corresponding to the points of $\overline{M_{0,n+2}}$ are free and maximal commutative. We describe explicitly the ``simplest'' degenerations and show that every degeneration is the composition of the simplest ones. The Deligne-Mumford space $\overline{M_{0,n+2}}$ generalizes to other root systems as some De Concini-Procesi resolution of some toric variety. We state a conjecture generalizing our results to Bethe subalgebras in the Yangian of arbitrary simple Lie algebra in terms of this De Concini-Procesi resolution.
\end{abstract}
\makeatletter
\@setabstract
\makeatother
\section{Introduction}
\subsection{Bethe subalgebras in the Yangian.} Yangian for $\fgl_n$ is the associative algebra, historically one of the first examples of {\em quantum groups}. The Yangian $Y(\fgl_n)$ is a Hopf algebra
deforming the enveloping algebra $U(\fgl_n[t])$, where
$\fgl_n[t]$ is the (infinite dimensional) Lie algebra of $\fgl_n$-valued polynomials.
This algebra was considered in the works of L. Fadeev and St.-Petersburg school in the relation with the inverse scattering method, see e.g. \cite{TF,T}.
There is a family $B(C)$ of commutative subalgebras in $Y(\fgl_n)$ parameterized by complex matrices $C\in {\rm Mat}_n$ called Bethe subalgebras. This family originates from the integrable models in statistical mechanics and algebraic Bethe ansatz. For details and links on Yangians, we refer the reader to the survey \cite{molev} and to the book \cite{molev2} by A.~Molev.
Denote by $T$ the maximal torus in $GL_n$ i.e. the subgroup of diagonal matrices in $GL_n$. In the present paper we restrict ourselves to Bethe subalgebras with $C\in T$. Let $T^{reg}$ be the set of regular elements of the torus, i.e. the set of matrices from $T$ with pairwise distinct eigenvalues. We will frequently use the embedding $GL_n\subset \fgl_n={\rm Mat}_n$ and regard $C$ as an element of the Cartan subalgebra $\fh\subset\fgl_n$.
In \cite{nazol} Nazarov and Olshanski showed that $B(C)$ is a free polynomial algebra and that it is a maximal commutative subalgebra in $Y(\fgl_n)$ for all $C \in T^{reg}$.
For non-regular $C\in T\backslash T^{reg}$, the subalgebra $B(C)$ becomes smaller. But there is a natural way to assign a commutative subalgebra of the same size as for $C \in T^{reg}$ to any $C_0\in \fh\backslash T^{reg}$ by taking some {\em limit} of $B(C)$ as $C\to C_0$. For example, one can get the Gelfand-Tsetlin subalgebra of the Yangian as the $t\to0$ limit of some $1$-parametric family of Bethe subalgebras with $C(t)\in T^{reg}$ for $t\ne0$ and $C(0)=E_{11}$. In general, such limit subalgebra $\lim\limits_{C\to C_0}B(C)$ is not unique since it depends on the $1$-parametric family $C(t)$ such that $C(0)=C_0$. Our goal is to describe all possible limit subalgebras.
The images of Bethe subalgebras in the universal enveloping algebra $U(\fgl_n)$ under the evaluation homomorphism are known as ``shift of argument subalgebras''. The problem of describing all limits for the shift of argument subalgebras was posed by Vinberg in late 1990-s. The answer was given by V.Shuvalov in \cite{shuvalov} and later in more algebro-geometric terms by L.~Aguirre, G.~Felder and A.~Veselov in \cite{AFV}. Their description is roughly as follows. Shift of argument subalgebras themselves are parametrized by regular diagonal $n\times n$-matrices up to proportionality and up to adding a scalar matrix. The latter can be regarded as the space of configurations of $n$ pairwise distinct points on the complex line. It turns out that the limit shift of argument subalgebras are parametrized by the Deligne-Mumford closure of this space \cite{AFV,shuvalov}, all limit subalgebras are free \cite{shuvalov} and maximal commutative \cite{taras}, and moreover there is an explicit inductive procedure generating the limit subalgebras from smaller shift of argument subalgebras assigned to smaller $n$. It is natural to expect a similar description for limit Bethe subalgebras in the Yangian.
\subsection{Limits of Bethe subalgebras.} The limit subalgebras can be defined in purely algebro-geometric terms (we will do this in Section~2). Roughly, the construction of Bethe algebras can be regarded as a regular map from $T^{reg}$ to the ``Grassmannian'' of subspaces in $Y(\fgl_n)$ of the same ``dimension'' as $B(C)$. The space $T^{reg}$ is noncompact while the Grassmannian is (in appropriate sense) compact. So we can take the closure of the image of this map and obtain new subalgebras with the same Poincare series. We call such subalgebras {\em limit subalgebras.}
Since the subalgebras $B(C)$ do not change under dilations of $C$, the parameter space for the family $B(C)$ is the quotient $T^{reg}/\mathbb{C}^*$ of the set of regular elements of the torus by the subgroup of scalar matrices. Following \cite{AFV} we can regard the space $T^{reg}/\mathbb{C}^*$ as the moduli space $M_{0,n+2}$ of rational curves with $n+2$ marked points (we can assign to a matrix $C$ with the eigenvalues $z_1,\ldots,z_n$ the curve $\mathbb{P}^1$ with the marked points $0,z_1,\ldots,z_n,\infty$). Therefore the limit subalgebras of the family $B(C)$ are parametrized by some compactification of the space $M_{0,n+2}$. The main results of the present paper can be summarized as follows:
\medskip
\noindent {\bf Main Theorem.} \emph{The closure of $T^{reg}/\mathbb{C}^*$ which parametrizes the limit subalgebras is isomorphic to the Deligne-Mumford compactification $\overline{M_{0,n+2}}$. All limit subalgebras are free polynomial algebras and maximal commutative subalgebras in $Y(\fgl_n)$.}
\medskip
In fact we will describe all limit subalgebras as products of smaller Bethe subalgebras and some shift of argument subalgebras (which are the images of Bethe subalgebras in the universal enveloping algebra $U(\fgl_n)$ under the evaluation map).
It is natural to expect this result from the perspective of bispectral duality of Mukhin, Tarasov and Varchenko \cite{MTV} which relates the image of Bethe algebra in a tensor product of evaluation representations with the algebra of higher Hamiltonians of the trigonometric Gaudin model for $\fg=\fgl_k$. \footnote{We thank Evgeny Mukhin for pointing this out.} The latter is the image of a Gaudin subalgebra in the tensor product of $n+1$ copies of the universal enveloping algebra of $\fgl_k$. On the other hand in \cite{ryb2} we prove that the closure of the parameter space for this family of Gaudin algebras is $\overline{M_{0,n+2}}$. The problem in this approach is that we have to deal with the images of Bethe subalgebras in some specific representations of the Yangian, so the closure of the parameter space could be different from that of the subalgebras in the Yangian itself. We use different approach based on shift of argument subalgebras and the Olshanski centralizer construction of the Yangian.
The main idea of our proof is to reduce the problem of describing the closure to the similar one for the family of shift of argument subalgebras in $U(\fgl_N)$ where $N$ is big enough. For this we use on the one hand the Olshanski centralizer construction \cite{olsh} which approximates the Yangian $Y(\fgl_n)$ by the centralizer subalgebras of the form $U(\fgl_{n+k})^{\fgl_k}$, and, on the other hand, the results of Shuvalov \cite{shuvalov} and Tarasov \cite{taras} which describe the limit subalgebras for the family of shift of argument subalgebras.
\subsection{Shift of argument subalgebras.}
Let $\hat F(C)\subset U(\fgl_n)$ be the image of $B(C)\subset Y(\fgl_n)$ under the evaluation homomorphism $Y(\fgl_n)\to U(\fgl_n)$. The subalgebra $\hat F(C)$ does not change after adding a scalar matrix to $C$. The associated graded of $\hat F(C)$ is a Poisson commutative subalgebra $F(C)\subset S(\fgl_n)$ known as \emph{shift of argument subalgebra} since it is generated by the adjoint invariants from $S(\fgl_n)$ shifted by $tC$ for all $t\in\BC$, see \cite{nazol}.
Suppose $C$ is a diagonal matrix with pairwise distinct eigenvalues $z_1,\ldots,z_n$. Then the algebra $F(C)$ contains the quadratic elements $H_i:=\sum\limits_{j\ne i}\frac{e_{ij}e_{ji}}{z_i-z_j}$ which are the coefficients of (an appropriate version of) the $KZ$ connection. Moreover both $F(C)$ and $\hat F(C)$ are uniquely determined by the subspace $Q_C\subset S(\fgl_n)$ which is the linear span of the $H_i$'s. Note that $H_i$ do not change under simultaneous affine transformations of the $z$'s, hence the space of parameters of the corresponding shift of argument subalgebras is naturally the configuration space of $n$ pairwise distinct points on the affine line or, equivalently, the configuration space $M_{0,n+1}$ of $n+1$ pairwise distinct points on the projective line. From the results of Aguirre, Felder and Veselov \cite{AFV} it follows that the closure of the family of subspaces $Q_C\subset S(\fgl_n)$ is the Deligne-Mumford compactification $\overline{M_{0,n+1}}$, which is crucial for our proof.
\subsection{Centralizer construction.} Let $A_0=\BC[\mathcal{E}_1,\mathcal{E}_2, \ldots, ]$ be the filtered polynomial algebra of infinitely many generators such that $\deg \mathcal{E}_i=i$. The Olshanski centralizer construction \cite{olsh} is the collection of surjective homomorphisms of filtered algebras $Y(\fgl_n)\otimes A_0\to U(\fgl_{n+k})^{\fgl_k}$ generalizing the evaluation map. The intersection of kernels of such homomorphisms is known to be zero, so this collection of homomorphisms can be regarded as an {\em an asymptotic isomorphism} i.e. for each filtered component of $Y(\fgl_n)\otimes A_0$ there is $K\in\BZ$ such that for any $k>K$ the restriction of the above homomorphism to this filtered component is an isomorphism. The idea of our proof is to analyze the parameter spaces of the images of $B(C)$ under the centralizer construction maps. Since centralizer construction is an asymptotic isomorphism the closure of the parameter space stabilizes for $k>>0$.
\subsection{Plan of the proof.} We prove that the image of $B(C)\otimes A_0$ in the centralizer algebra $U(\fgl_{n+k})^{\fgl_k}$ is contained in some non-regular shift of argument subalgebra $\hat F(C^{(k)})$ (Proposition~\ref{incl}) and moreover $\hat F(C^{(k)})$ is asymptotically isomorphic to $B(C)\otimes A_0$. We show that the closure of the parameter space for the subalgebras $\hat F(C^{(k)})$ is $\overline{M_{0,n+2}}$ (realized as a subvariety in $\overline{M_{0,n+k+1}}$ which parametrizes all limit shift of argument subalgebras in $U(\fgl_{n+k})$). Since the closure of the parameter space for $\hat F(C^{(k)})$ does not depend on $k$, the closure of the parameter space for Bethe subalgebras in $Y(\fgl_n)$ is $\overline{M_{0,n+2}}$ as well.
Next, we deduce from the results of Shuvalov and Tarasov that any limit of the algebras $\hat F(C^{(k)})$ is a free polynomial algebra and a maximal commutative subalgebra in $U(\fgl_{n+k})^{\fgl_k}$. Since $B(C)\otimes A_0$ is asymptotically isomorphic to $\hat F(C^{(k)})$, the same is true for limit Bethe subalgebras corresponding to points of $\overline{M_{0,n+2}}$ (Theorem \ref{result}). Using Shuvalov's description of limit shift of argument subalgebras we explicitly describe the simplest limits corresponding to generic points of codimension $1$ strata in $\overline{M_{0,n+2}}$ in terms of Bethe subalgebras for smaller Yangians and shift of argument subalgebras for smaller Lie algebras (Theorem \ref{result2}). Iterating this procedure we obtain the explicit description of all limit Bethe subalgebras.
\subsection{Generalization to Yangians of other types.}
To any semisimple Lie algebra $\fg$ (and even more generally, to any Kac-Moody algebra $\fg$) one can assign the Yangian $Y(\fg)$ , the quantum group generated by the rational R-matrix of $\fg$. To any element $C$ of the Cartan torus $T$ of the corresponding Lie group $G$ one can assign a Bethe subalgebra generated by traces of the products of $C$ with the R-matrix in all integrable representations. This is a commutative subalgebra which is expected to be maximal for regular $C$. The variety parameterizing all possible limits of such commutative subalgebras is a compactification of the set $T^{reg}$ of regular elements of the torus (well-defined as a pro-algebraic scheme). Our Theorem~\ref{result} states that in the case $\fg=\fgl_n$ it is $\overline{M_{0,n+2}}$. The natural generalization of this statement to Lie algebras of other types is the De Concini -- Procesi closure \cite{DCP} of the complement to the following arrangement of subvarieties in a toric variety. Consider the toric variety $X$ (acted on by the maximal torus $T\subset G$) which corresponds to the fan determined by the root hyperplanes. Equivalently, $X$ is the closure of a generic $T$-orbit in the flag variety $G/B$. We can regard $T^{reg}$ as a complement of an arrangement of hypersurfaces in $X$. Following De Concini and Procesi \cite{DCP}, one can construct a compactification $M_{\fg}$ of $T^{reg}$ by blowing up all indecomposable intersections of the hypersurfaces in $X$.
\begin{conjj} $M_{\fg}$ is the parameter space for limit subalgebras of the family of Bethe subalgebras in the Yangian $Y(\fg)$.
\end{conjj}
\begin{rem}
Note that $\overline{M_{0,n+2}}$ is the De Concini--Procesi closure of $T^{reg}$ for $\fg=\fgl_n$.
\end{rem}
\begin{rem}
For infinite root systems (say, for affine Kac-Moody) $M_\fg$ is not well-defined as an algebraic scheme but is still well-defined as a pro-algebraic scheme. On the other hand, in this case the Bethe subalgebras themselves live not in the Yangian but rather in its completion which is an inverse limit of some quotients of the Yangian. So we can generalize our conjecture to Yangians of infinite-dimensional Lie algebras by stating that two inverse limits are isomorphic.
\end{rem}
\subsection{Application to crystals.} By analogy with the shift of argument algebras and Gaudin algebras, we expect that for real values of the parameter the corresponding (limit) Bethe algebra has simple spectrum in any irreducible finite-dimensional (or integrable) representation of the Yangian. Following \cite{ryb2}, one can assign a covering of $M_\fg(\BR)$ to any irreducible representation of $Y(\fg)$. The fiber of such covering is just the set of joint eigenlines for the elements of a Bethe algebra in this representation. For Kirillov-Reshetikhin modules, we expect a natural bijection of the fiber of this covering with the corresponding Kirillov-Reshetikhin crystal hence obtaining an action of the fundamental group $\pi_1(M_\fg(\BR))$ on the crystal. We also expect that this action can be described in purely combinatorial terms.
\subsection{The paper is organized as follows.} In section~2 we recall some known facts about Yangians and Bethe subalgebras. In section~3 we recall the definition of the moduli space $\overline{M_{0,n+1}}$. In section 4 we discuss some facts about shift of argument subalgebras. In Section 5 we formulate the main theorems. In section 6 we prove the main theorems.
\subsection{Acknowledgements}
We thank Michael Finkelberg and Alexander Molev for helpful remarks and references. We thank the referee for careful reading of the first version of the text. This work has been funded by the Russian Academic Excellence Project '5-100'. The work was finished during A.I.'s internship at MIT supported by NRU HSE. A.I. is deeply indebted to MIT and especially to R.~Bezrukavnikov and P.~Etingof, for providing warm hospitality and excellent working conditions. The work of A.I. and L.R. was supported in part by the Simons Foundation. The work of L.R. has been supported by the Russian Science Foundation under grant 16-11-10160.
\section{Yangians and Bethe subalgebras.}
\subsection{Definitions.} We follow the notations and conventions of \cite{molev2}.
\begin{defn}
{\itshape Yangian for $\fgl_n$} is a complex unital associative algebra with countably many generators $t_{ij}^{(1)}, t_{ij}^{(2)}, \ldots $ where $1 \leqslant i,j \leqslant n$, and the defining relations
$$ [t_{ij}^{(r+1)},t_{kl}^{(s)}] - [t_{ij}^{(r)},t_{kl}^{(s+1)}] = t_{kj}^{(r)}t_{il}^{(s)} - t_{kj}^{(s)}t_{il}^{(r)}, $$
where $r,s \geqslant 0$ and $t_{ij}^{(0)} = \delta_{ij}$. This algebra is denoted by $Y(\fgl_n)$.
\end{defn}
It is convenient to consider the formal series
$$t_{ij}(u) = \delta_{ij} + t_{ij}^{(1)} u^{-1} + t_{ij}^{(2)} u^{-2} + \ldots \in Y(\fgl_n)[[u^{-1}]].$$
We denote by $T(u)$ the matrix whose $ij$-entry is $t_{ij}(u)$.
We regard this matrix as the following element of $Y(\fgl_n)[[u^{-1}]] \otimes \rm{End} \ \mathbb{C}^n$:
$$T(u) = \sum_{i,j=1}^n t_{ij}(u) \otimes e_{ij},$$
where $e_{ij}$ stands for the standard matrix units.
Consider the algebra
$$Y(\fgl_n)[[u^{-1}]] \otimes (\rm{End} \ \mathbb{C}^n)^{\otimes n}.$$
For any $a \in \{1, \ldots ,n\}$ there is an embedding $$
i_a: Y(\fgl_n)[[u^{-1}]] \otimes \rm{End} \ \mathbb{C}^n\to Y(\fgl_n)[[u^{-1}]] \otimes (\rm{End} \ \mathbb{C}^n)^{\otimes n}
$$
which is an identity on $Y(\fgl_n)[[u^{-1}]]$ and embeds $\rm{End} \ \mathbb{C}^n$ as the $a$-th tensor factor in $(\rm{End} \ \mathbb{C}^n)^{\otimes n}$. Denote by $T_a(u)$ the image of $T(u)$ under this embedding.
The symmetric group $S_n$ acts on $Y(\fgl_n)[[u^{-1}]] \otimes (\rm{End} \ \mathbb{C}^n)^{\otimes n}$ by permuting the tensor factors. This action factors through the embedding $S_n\hookrightarrow (\rm{End} \ \mathbb{C}^n)^{\otimes n}$ hence the group algebra $\BC[S_n]$ is a subalgebra of $Y(\fgl_n)[[u^{-1}]] \otimes (\rm{End} \ \mathbb{C}^n)^{\otimes n}$. Let $S_m$ be the subgroup of $S_n$ permuting the first $m$ tensor factors. Denote by $A_m$ the antisymmetrizer $$\sum_{\sigma \in S_m} (-1)^{\sigma}\sigma \in \mathbb{C}[S_m]\subset Y(\fgl_n)[[u^{-1}]] \otimes (\rm{End} \ \mathbb{C}^n)^{\otimes n}.$$
Let $C$ be a $\mathbb{C}$-valued $n \times n$ matrix. For any $a \in \{1, \ldots ,n\}$ denote by $C_a$ the element $i_a(1\otimes C)\in Y(\fgl_n)[[u^{-1}]] \otimes (\rm{End} \ \mathbb{C}^n)^{\otimes n}$.
\begin{defn}
For any $1 \leqslant k \leqslant n$ introduce the series with coefficients in $Y(\fgl_n)$ by
$$\tau_k(u,C) = \frac{1}{k!} \tr A_k C_1 \ldots C_k T_1(u) \ldots T_k(u-k+1),$$
where we take the trace over all $k$ copies of $ \rm{End} \ \mathbb{C}^n$.
\end{defn}
\begin{rem}
$\tau_n(u,C)=\det C\ \tau_n(u,E)$, and hence, does not depend on $C$ up to proportionality. The series $n! \cdot \tau_n(u,E)$ is called quantum determinant and usually denoted by ${\rm qdet} \, T(u)$.
\end{rem}
The following result is classical (see e.g. \cite{molev2,nazol}):
\begin{prop} The coefficients of $\tau_k(u,C)$, $k=1,\ldots,n$ pairwise commute. The coefficients of ${\rm qdet} \, T(u)$ generate the center of $Y(\fgl_n)$.
\end{prop}
\begin{defn}
The commutative subalgebra generated by the coefficients of $\tau_k(u,C)$ is called {\itshape Bethe
subalgebra}. We denote it by $B(C)$.
\end{defn}
In \cite{nazol} Nazarov and Olshanski proved the following theorem:
\begin{thm}\label{MaxFreeTheorem}
Suppose that $C \in \End \mathbb{C}^n$ has pairwise distinct nonzero eigenvalues. Then the subalgebra $B(C)$ in $Y(\fgl_n)$ is maximal commutative. The coefficients of $\tau_1(u,C), \ldots, \tau_n(u,C)$ are free generators of $B(C)$.
\end{thm}
From now on we suppose that $C$ is diagonal and $C_{aa} = \lambda_a,\ a=1,\ldots,n$. Let us write $\tau_k(u,C)$ explicitly in the generators $t_{ij}(u)$. Straightforward computation gives the following well known Lemma:
\begin{lem}
\label{vid}
We have
$$\tau_k(u,C) = \sum_{1 \leqslant a_1< \ldots < a_k \leqslant n} \lambda_{a_1} \ldots \lambda_{a_k} t_{a_1 \ldots a_k}^{a_1\ldots a_k}(u),$$
where $$t^{a_1 \ldots a_k}_{b_1 \ldots b_k} = \sum_{\sigma \in S_k} (-1)^\sigma \, \cdot \, t_{a_{\sigma(1)}b_1}(u) \ldots t_{a_{\sigma(k)}b_k}(u-k+1) = \sum_{\sigma \in S_k} (-1)^\sigma \, \cdot \, t_{a_1 b_{\sigma(1)}}(u-k+1) \ldots t_{a_k b_{\sigma(k)}}(u) $$ are the quantum minors.
\end{lem}
\begin{comment}\begin{proof}
Note that $\tau_k (u,C) = {\rm tr} \, A_k \cdot M_k$, where $M_k = C_1 T_1(u) \otimes C_2 T_2(u-1) \otimes \ldots \otimes C^{(k)} T_k(u-k+1).$
It is clear that in the basis $e_{i_1} \otimes \ldots \otimes e_{i_k}$ with the standard order we have $$M_{i_1 \ldots i_k; j_1 \ldots j_k} = \lambda_{i_1} \ldots \lambda_{i_k} t_{i_1 j_1}(u) \cdot t_{i_2 j_2}(u-1) \cdot \ldots \cdot t_{i_k j_k}(u-k+1).$$
Then in the same basis we have $$A_{i_1 \ldots i_k; j_1 \ldots j_k} = \sum_{\sigma \in S_k | j_p = i_{\sigma(p)}} (-1)^{\sigma}. $$
Note that if there exists $i_x = i_y$ then $(A_k \cdot M_k)_{i_1 \ldots i_k; j_1 \ldots j_k} = 0$. Suppose that $i_x \ne i_y$ for $x \ne y$.
Then
\begin{multline*} (A_k \cdot M_k)_{i_1 \ldots i_k; j_1 \ldots j_k} = \sum_{l_1, \ldots, l_k} A_{i_1 \ldots i_k; l_1 \ldots l_k} M_{l_1 \ldots l_k; j_1 \ldots j_k} = \sum_{\sigma \in S_k } (-1)^{\sigma} M_{i_{\sigma(1)} \ldots i_{\sigma(k)}; j_1 \ldots j_k} \end{multline*}
This implies that the trace is equal to
\begin{multline*}
\sum_{i_1 \ldots i_k} (A_k \cdot M_k)_{i_1 \ldots i_k; i_1 \ldots i_k} = \sum_{i_1 \ldots i_k} \sum_{\sigma \in S_k} (-1)^{\sigma} \lambda_{i_{\sigma(1)}} \ldots \lambda_{i_{\sigma(k)}} t_{i_{\sigma(1)} i_1}(u) \cdot \ldots \cdot t_{i_{\sigma(k)} i_k}(u-k+1) = \\ = k! \sum_{i_1 < \ldots < i_k} \sum_{\sigma \in S_k} (-1)^{\sigma} \lambda_{i_1} \ldots \lambda_{i_k} t_{i_1 i_{\sigma(1)}}(u) \cdot \ldots \cdot t_{i_k i_{\sigma(k)}}(u-k+1)
\end{multline*}
The last equation follows from skew-symmetricity of quantum minors.
\end{proof}
\end{comment}
\subsection{Filtration on Yangian.}
We define an ascending filtration on $Y(\fgl_n)$ by setting the degree of the generators as
\begin{equation*}
\label{Degree}\deg t_{ij}^{(r)} = r.
\end{equation*}
Denote by $Y_r(\fgl_n)$ the vector subspace consisting of all elements with degrees not greater than $r$.
Consider the associated graded algebra
\label{filtration}
\begin{equation}
CY(\fgl_n) = \bigoplus_{r \geqslant 0} Y_r(\fgl_n) / Y_{r-1}(\fgl_n).
\end{equation}
There is the following analog of PBW-theorem for $Y(\fgl_n)$ (see \cite{molev}):
\begin{thm}
\label{PBW}$CY(\fgl_n)$ is the algebra of polynomials in the variables $\bar t_{ij}^{(r)}$, where $\bar t_{ij}^{(r)}$ is the image of $t_{ij}^{(r)}$ in $Y_r(\fgl_n) / Y_{r-1}(\fgl_n)$.
\end{thm}
The polynomial algebra $CY(\fgl_n)$ has a natural Poisson structure. For any two elements $x,y \in Y(\fgl_n)$ of degrees $p,q$ respectively the Poisson bracket of their images $\bar x,\bar y$ in $CY(\fgl_n)$ is
$$\{\bar x,\bar y\} = [x,y] \mod Y_{p+q-2}(\fgl_n).$$
\subsection{Some homomorphisms between the Yangians.}
Let us define two different embedding of $Y(\fgl_n)$ to $Y(\fgl_{n+k})$:
$$i_k: Y(\fgl_n) \to Y(\fgl_{n+k}); \ t^{(r)}_{ij} \mapsto t^{(r)}_{ij}$$
$$\varphi_k: Y(\fgl_n) \to Y(\fgl_{n+k}); \ t^{(r)}_{ij} \mapsto t^{(r)}_{k+i,k+j}$$
According to PBW-theorem these maps are injective.
Define a homomorphism
$$\pi_n: Y(\fgl_n) \to U(\fgl_n); \ t_{ij}(u) \mapsto \delta_{ij} + E_{ij}u^{-1}.$$ Here $E_{ij}$ are the standard generators of $\mathfrak{gl_n}$.
$\pi_n$ is a surjective homomorphism from $Y(\fgl_n)$ to $U(\fgl_n)$ known as \emph{evaluation} homomorphism.
By definition, put
$$\omega_n: Y(\fgl_n) \to Y(\fgl_n); \ T(u) \mapsto (T(-u-n))^{-1}.$$
It is well-known that $\omega_n$ is an involutive automorphism of $Y(\fgl_n)$.
We define a homomorphism
$$\psi_k = \omega_{n+k} \circ \varphi_k \circ \omega_n: Y(\fgl_n) \to Y(\fgl_{n+k}).$$
Note that $\psi_k$ is injective.
\begin{prop}\label{ipsi}\cite[Proposition 1.13.1]{molev2} $i_{k}(Y(\fgl_{n}))$ is centralized by $\psi_{n}(Y(\fgl_k))$ in $Y(\fgl_{n+k})$.
\end{prop}
We will need a bit more precise statement:
\begin{lem}
\label{commuteyang} The homomorphisms $i_{k}$ and $\psi_{n}$ define an embedding
$i_{k}\otimes\psi_{n}: Y(\fgl_{n}) \otimes Y(\fgl_k) \hookrightarrow Y(\fgl_{n+k})$.
\end{lem}
\begin{proof}
It suffices to show that the associated graded $\gr (i_{k}\otimes\psi_{n}): CY(\fgl_{n}) \otimes CY(\fgl_k) \to CY(\fgl_{n+k})$ is injective. But the latter is sends $t_{ij}^{(r)}\otimes1$ to $t_{ij}^{(r)}$ and $1\otimes t_{ij}^{(r)}$ to $t_{n+i,n+j}^{(r)}+L$, where $L$ is an expression of $t_{ij}^{(s)}$ with $s<r$. Hence it is clearly injective.
\end{proof}
\subsection{Centralizer construction.}
Consider the map
\begin{equation}
\Phi_k: Y(\fgl_n) \to U(\fgl_{n+k}) \ \text{given by} \ \Phi_k = \pi_{n+k} \circ \omega_{n+k} \circ i_k.
\end{equation}
From ~\cite[Proposition 8.4.2]{molev2} it follows that ${\rm Im} \, \Phi_k \subset U(\fgl_{n+k})^{\fgl_k}$. Here we use an embedding $$\mathfrak{gl}_k \to \mathfrak{gl}_{n+k}, \ E_{ij} \to E_{i+n, j+n}.$$
Let $A_0 = \mathbb{C}[\mathcal{E}_1, \mathcal{E}_2, \ldots]$ be the polynomial algebra of infinite many variables. Define a grading on $A_0$ by setting $\deg \mathcal{E}_i = i$. For any $k$ we have a surjective homomorphism
$$z_k: A_0 \to Z(U(\fgl_{n+k}));\ \mathcal{E}_i \to \mathcal{E}_i^{(n+k)}.$$
where $\mathcal{E}_i^{(n+k)}$, $i=1,2,3,\ldots$ are the following generators of $Z(U(\fgl_{n+k}))$ of degree $i$, see \cite[Section 8.2]{molev2}:
$$ 1+\sum\limits_{i=1}^{n+k}\mathcal{E}_iu^{-i}=\pi_{n+k}({\rm qdet } T(U)).
$$
Consider the algebra $Y(\fgl_n) \otimes A_0$. This algebra has a well-defined ascending filtration given by
$$\deg a \otimes b = \deg a + \deg b.$$
For any $k \geqslant 0$ we define homomorphisms of filtered algebras
$$\eta_k: Y(\fgl_n) \otimes A_0 \to U(\fgl_{n+k})^{\fgl_k}; \ a \otimes b \to \Phi_k(a) \cdot z_k(b)$$
Denote by $(Y(\fgl_n) \otimes A_0)_N$ the $N$-th filtered component, i.e. the vector space of the elements of degree not greater than $N$. From \cite[Theorem 8.4.3]{molev2} we have:
\begin{thm}
\label{qasym}
The sequence $\{\eta_k\}$ is an asymptotic isomorphism. This means that for any $N$ there exists $K$ such that for any $k>K$ the restriction of $\eta_k$ to the $N$-th filtered component $(Y(\fgl_n) \otimes A_0)_N$ is an isomorphism of vector spaces $(Y(\fgl_n) \otimes A_0)_N \simeq U(\fgl_{n+k})^{\fgl_k}_N$.
\end{thm}
According to Lemma~\ref{commuteyang}, there is a tensor product of two commuting Yangians $i_{m}(Y(\fgl_{n-m}))$ and $\psi_{n-m}(Y(\fgl_{m}))$ inside $Y(\fgl_n)$. Let us see what happens with these two commuting Yangians under the centralizer construction map $\eta_{k}:Y(\fgl_n)\otimes A_0\to U(\fgl_{n+k})^{\fgl_k}$.
The algebra $U(\fgl_{n+k})$ is generated by $E_{ij}, 1 \leqslant i,j \leqslant n+k$; $\fgl_k$ is generated by $E_{ij}, n+1 \leqslant i,j \leqslant n+k$. The subalgebra $U(\fgl_{m+k})\subset U(\fgl_{n+k})$ is generated by $E_{ij}, n-m+1 \leqslant i,j \leqslant n+k$. Hence $U(\fgl_{m+k})^{\fgl_k}$ and $U(\fgl_{n+k})^{\fgl_{m+k}}$ are naturally subalgebras in $U(\fgl_{n+k})^{\fgl_k}$.
\begin{lem}
The restriction of $\eta_k$ to $i_m(Y(\fgl_{n-m}))$ is $\eta_{k+m}:Y(\fgl_{n-m})\to U(\fgl_{n+k})^{\fgl_{k+m}}$.
The restriction of $\eta_k$ to $\psi_{n-m}(Y(\fgl_m))$ is $\eta_{k}:Y(\fgl_m)\to U(\fgl_{m+k})^{\fgl_k}$.
\end{lem}
\begin{proof}
The first statement is immediate from the following commutative diagram:
\begin{center}
\[ \begin{diagram}
\node{Y(\fgl_{n-m})} \arrow[2]{e,t}{i_{m}}
\arrow{ese,b}{i_{k+m}}
\node[2]{Y(\fgl_n)} \arrow{s,r}{i_k}\\
\node[3]{Y(\fgl_{n+k})}
\end{diagram}\]
\end{center}
To prove second statement, consider another diagram:
\begin{center}
\[ \begin{diagram}
\node{Y(\fgl_m)}
\arrow{e,t}{\psi_{n-m}}
\arrow{s,r}{i_k}
\node{Y(\fgl_n)}
\arrow{s,r}{i_k} \\
\node{Y(\fgl_{m+k})} \arrow{e,t,..}{\psi_{n-m}}
\arrow{s,r}{\omega_{m+k}}
\node{Y(\fgl_{n+k})} \arrow{s,r}{\omega_{n+k}}\\
\node{Y(\fgl_{m+k})} \arrow{e,t}{\varphi_{n-m}}
\node{Y(\fgl_{n+k})}
\end{diagram}\]
\end{center}
The lower square is commutative by definition so it is enough to prove that upper square is commutative as well. But it follows from \cite[Lemma 1.11.2]{molev2}.
\end{proof}
\subsection{Definition of limit subalgebras.}\label{def-lim}
It is possible to construct new commutative subalgebras as limits of Bethe subalgebras.
We describe here what the "limit" means.
Let $T^{reg}$ be the set of regular elements of the maximal torus $T$ of $GL_n$. Let $C$ be an element of $T^{reg}$. Consider $B_r(C):= Y_r(\fgl_n) \cap B(C)$. In \cite{nazol} it was proved that the images of the coefficients of $\tau_1(u,C), \ldots, \tau_n(u,C)$ freely generate the subalgebra $\bar B(C) = \gr B(C)\subset CY(\fgl_n)$. Hence the dimension $d(r)$ of $B_r(C)$ does not depend on $C$.
Therefore for any $r \geqslant 1$ we have a map $\theta_r$ from $T^{reg}$ to $\bigtimes_{i=1}^r {\rm Gr}(d(i),\dim Y_{i}(\fgl_n))$ such that $C \to (B_1(C), \ldots, B_r(C))$.
Denote the closure of $\theta_r(T^{reg})$ (with respect to Zariski topology) by $Z_r$. There are well-defined projections $\rho_r: Z_r \to Z_{r-1}$ for all $r \geqslant 1$. The inverse limit $Z = \varprojlim Z_r$ is well-defined as a pro-algebraic scheme and is naturally a parameter space for some family of commutative subalgebras which extends the family of Bethe subalgebras.
Indeed, any point $X\in Z$ is a sequence $\{x_r\}_{r \in \mathbb{N}}$ where $x_r \in Z_r$ such that $\rho_r(x_r) = x_{r-1}$. Every $x_r$ is a point in $\bigtimes_{i=1}^r {\rm Gr}(d(i),\dim Y_{i}(\fgl_n))$ i.e. a collection of subspaces $B_{r,i}(X)\subset Y_{i}(\fgl_n)$ such that $B_{r,i}(X)\subset B_{r,i+1}(X)$ for all $i<r$. Since $\rho_r(x_r) = x_{r-1}$ we have $B_{r,i}(X)= B_{r-1,i}(X)$ for all $i<r$. Let us define the subalgebra corresponding to $X\in Z$ as $B(X):= \bigcup_{r=1}^{\infty} B_{r,r}(X)$. We claim that this subalgebra is commutative and we call it {\em limit subalgebra}. Indeed, $B(X)$ is a commutative subalgebra because being a commutative subalgebra is a Zariski-closed condition: we have $B_r(C)\cdot B_s(C)\subset B_{r+s}(C)$ for all $C\in T^{reg}$ for all $r,s$ and this product is commutative, hence we get the same for the product $B_{r,r}(X)\cdot B_{s,s}(X)=B_{r+s,r}(X)\cdot B_{r+s,s}(X)\subset B_{r+s,r+s}(X)$. Note that this subalgebra has the same Poincare series as $B(C)$.
\begin{rem}
In \cite{shuvalov} the limits shift of argument subalgebras are defined in the analytic topology, just as limits of $1$-parametric families of generic shift of argument subalgebras. But it is well-known that the closure of an affine algebraic variety in a complex projective space with respect to Zariski topology coincides with its closure with respect to the analytic topology. So the closure of the parameter space is the same for both definitions of the limit. In our work we use both approaches.
\end{rem}
\section{Moduli spaces of stable rational curves}
\subsection{The space $\overline{M_{0,n+1}}$.} Let $\overline{M_{0,n+1}}$ denote the Deligne-Mumford space of stable rational curves with $n+1$ marked points. The points of $\overline{M_{0,n+1}}$ are isomorphism classes of curves of genus $0$, with $n+1$ ordered marked points and possibly with nodes, such that each component has at least $3$ distinguished points (either marked points or nodes). One can represent the combinatorial type of such a curve as a tree with $n+1$ leaves with inner vertices representing irreducible components of the corresponding curve, inner edges corresponding to the nodes and the leaves corresponding to the marked points. Informally, the topology of $\overline{M_{0,n+1}}$ is determined by the following rule: when some $k$ of the distinguished points (marked or nodes) of the same component collide, they form a new component with $k+1$ distinguished points (the new one is the intersection point with the old component). In particular, the tree describing the combinatorial type of the less degenerate curve is obtained from the tree corresponding to the more degenerate one by contracting an edge.
The space $\overline{M_{0,n+1}}$ is a smooth algebraic variety. It can be regarded as a compactification of the configuration space $M_{0,n+1}$ of ordered $(n+1)$-tuples $(z_1,z_2,\ldots,z_{n+1})$ of pairwise distinct points on $\BC\BP^1$ modulo the automorphism group $PGL_2(\BC)$. Since the group $PGL_2(\BC)$ acts transitively on triples of distinct points, we can fix the $(n+1)$-th point to be $\infty\in\BC\BP^1$ and fix the sum of coordinates of other points to be zero. Then the space $M_{0,n+1}$ gets identified with the quotient ${\rm Conf}_n / \BC^*$ where ${\rm Conf}_n:=\{(z_1,\ldots,z_n)\in\BC^n\ |\ z_i\ne z_j,\ \sum\limits_{i=1}^n z_i=0\}$, and the group $\BC^*$ acts by dilations. Under this identification of $M_{0,n+1}$, the space $\overline{M_{0,n+1}}$ is just the GIT quotient of the iterated blow-up of the subspaces of the form $\{z_{i_1}=z_{i_2}=\ldots=z_{i_k}\}$ in $\BC^{n-1}$ by the natural $\BC^*$ action by dilations.
\subsection{Stratification and operad structure on $\overline{M_{0,n+1}}$.}
The space $\overline{M_{0,n+1}}$ is stratified as follows. The strata are indexed by the combinatorial types of stable rational curves, i.e. by rooted trees with $n$ leaves colored by the marked points $z_1,\ldots,z_n$ (the root is colored by $z_{n+1}=\infty$). The stratum corresponding to a tree $T$ lies in the closure of the one corresponding to a tree $T'$ if and only if $T'$ is obtained from $T$ by contracting some edges.
The spaces $\ol{M_{0,n+1}}$ form a topological operad. This means that one can regard each point of the space $\ol{M_{0,n+1}}$ as an $n$-ary operation with the inputs at marked points $z_1,\ldots,z_n$ and the output at $\infty$. Then one can substitute any operation of this form to each of the inputs. More precisely, for any partition of the set $\{1,\ldots,n\}$ into the disjoint union of subsets $M_1,\ldots,M_k$ with $|M_i|=m_i\ge1$ there is a natural substitution map $\gamma_{k;M_1,\ldots,M_k}:\ol{M_{0,k+1}}\times\prod\limits_{i=1}^k\ol{M_{0,m_i+1}}\to\ol{M_{0,n+1}}$ which attaches the $i$-th curve $C_i\in\ol{M_{0,m_i+1}}$ to the $i$-th marked point of the curve $C_0\in\ol{M_{0,k+1}}$ by gluing the $m_{i}+1$-th marked point of each $C_i$ with the $i$-th marked point of $C_0$. In fact all substitution maps $\gamma_{k;M_1,\ldots,M_k}$ are compositions of the elementary ones with $m_1=\ldots=m_{k-1}=1$.
The compositions of the substitution maps are indexed by rooted trees describing the combinatorial type of the (generic) resulting curves. In particular, each stratum of $\ol{M_{0,n+1}}$ is just the image of the product of the open strata of appropriate $\prod\ol{M_{0,m+1}}$ under some composition of substitution maps. In particular, strata of codimension $1$ are just the images of the open strata of $\ol{M_{0,k+1}}\times\ol{M_{0,n-k+1}}$ under the elementary substitution maps and can be obtained as the limit set when the points $z_1,\ldots,z_k$ collide and other points stay isolated. Since each substitution map is a composition of the elementary ones, generic point of each stratum can be obtained iterating this limiting procedure.
\section{Shift of argument subalgebras.}
\subsection{The algebras $F(C)$.} Let $\fg$ be a reductive Lie algebra. To any $C\in\fg^*$ one can assign a Poisson-commutative subalgebra in $S(\fg)$ with respect to the standard Poisson bracket (coming from the universal enveloping algebra $U(\fg)$ by the PBW theorem).
Let $ZS(\fg)=S(\fg)^{\fg}$ be the center of $S(\fg)$ with
respect to the Poisson bracket. The algebra $F(C)\subset S(\fg)$
generated by the elements $\partial_{C}^n\Phi$, where $\Phi\in
ZS(\fg)$, (or, equivalently, generated by central elements of
$S(\fg)=\BC[\fg^*]$ shifted by $tC$ for all $t\in\BC$) is
Poisson-commutative and has maximal possible transcendence degree. More precisely, we have the following
\begin{thm}\label{mf} ~\cite{mf} For regular semisimple $C\in\fg$ the algebra $F(C)$
is a free commutative subalgebra in $S(\fg)$ with
$\frac{1}{2}(\dim\fg+\rk\fg)$ generators (this means that
$F(C)$ is a commutative subalgebra of maximal possible
transcendence degree). One can take the elements
$\partial_{C}^n\Phi_k$, $k=1,\dots,\rk\fg$,
$n=0,1,\dots,\deg\Phi_k$, where $\Phi_k$ are basic
$\fg$-invariants in $S(\fg)$, as free generators of $F(C)$.
\end{thm}
\begin{rem} In particular for $\fg=\fgl_n$, the algebra $F(C)$ has $n-k$ generators of degree $k$ for each $k=1,\ldots,n$.
\end{rem}
Let $\fh\subset\fg$ be a Cartan subalgebra of the Lie algebra
$\fg$. Denote by $\Delta_+$ the set of positive roots of
$\fg$. We assume that $C$ is a regular semisimple
element of the Cartan subalgebra $\fh\subset\fg=\fg^*$. The
linear and quadratic part of the subalgebras $F(C)$
can be described as follows (see \cite{vinberg}):
\begin{gather*}F(C)\cap\fg=\fh,\\ F(C)\cap S^2(\fg)=S^2(\fh)\oplus
Q_C,\ \text{where}\
Q_C=\{\sum\limits_{\alpha\in\Delta_+}\frac{\langle\alpha,h\rangle}{\langle\alpha,C\rangle}e_{\alpha}e_{-\alpha}|h\in\fh\}.\end{gather*}
The results of Vinberg \cite{vinberg} and Shuvalov \cite{shuvalov} imply that the limit subalgebra of the family $F(C)$ is uniquely determined by its quadratic component. Hence the variety parameterizing limit shift of argument subalgebras in $S(\fg)$ is just the closure of the family $Q_C$ in the Grassmannian $Gr(\rk\fg, \dim S^2(\fg))$. In \cite{shuvalov} Shuvalov described the closure of the family of
subalgebras $F(C)\subset S(\fg)$ under the condition
$C\in \fh^{reg}$ (i.e., for regular $C$ in the fixed Cartan
subalgebra). In particular, the following holds:
\begin{thm}\label{shuvalov}\cite{shuvalov} Suppose that $C(t)=C_0+tC_1+t^2C_2+\dots\in \fh^{reg}$
for generic $t$. Let $\fz_{\fg}(C_i)$ be the centralizer of $C_i$ in $\fg$. Set $\fz_k=\bigcap\limits_{i=0}^k\fz_{\fg}(C_i)$,
$\fz_{-1}=\fg$. Then we have
\begin{enumerate}
\item the subalgebra $\lim\limits_{t\to0}F(C(t))\subset S(\fg)$
is generated by elements of $S(\fz_k)^{\fz_k}$ and their
derivatives of any order along $C_{k+1}$ for all $k=-1,0,\ldots$. \item
$\lim\limits_{t\to0}F(C(t))$ is a free commutative algebra.
\end{enumerate}
\end{thm}
Moreover, according to the results of A.~Tarasov \cite{taras} the subalgebras dicussed above (both $F(C)$ and the limit ones) are maximal Poisson-commutative subalgebras in $S(\fg)$ (i.e. coincide with their Poisson centralizers).
\subsection{Lifting to $U(\fg)$.} In \cite{ryb4} the subalgebras $F(C)$ were quantized, i.e. the existence of such commutative subalgebras $\hat F(C)\subset U(\fg)$ that $\gr \hat F(C) = F(C)$ was proved. In \cite{ryb} it is proved that $F(C)$ is the Poisson centralizer of the space $Q_C$ for generic $C$ hence the lifting of $F(C)$ to $U(\fg)$ is unique for generic $C$. In the case $\fg=\fgl_n$ we have a stronger statement due to A.~Tarasov \cite{taras2}: any subalgebra from this family (including the limit ones) can be uniquely lifted to the universal enveloping algebra and, moreover, there is a particular choice of the generators such that this lifting is just the symmetrization map on the generators. In particular, for $\fg=\fgl_n$, the varieties parameterizing limit subalgebras of the families $F(C)$ and $\hat F(C)$ are the same. In fact, we expect that the latter is true for arbitrary $\fg$.
In the case $\fg=\fgl_n$ the subalgebra $\hat F(C)\subset U(\fgl_n)$ is in fact the image of $B(C)\subset Y(\fgl_n)$ under the evaluation homomorphism $Y(\fgl_n)\to U(\fgl_n)$ (we can assume that $C$ is nondegenerate since $\hat F(C)$ does not change after adding a scalar matrix to $C$). This fact plays crucial role in what follows. Unfortunately this does not generalize to arbitrary $\fg$ since there is no evaluation homomorphism in general.
\subsection{$\ol{M_{0,n+1}}$ and shift of argument subalgebras.}
In \cite{shuvalov} the explicit description of limit subalgebras is given together with a set-theoretical description of the parameter space. The results of Shuvalov in the $\fgl_n$ case can be reformulated in the following way.
The shift of argument subalgebras in $S(\fgl_n)$ depend on a parameter $C\in T^{reg}$ and do not change under the transformations $C\mapsto aC+bE$. So the parameter space for the subalgebras $F(C)\subset S(\fgl_n)$ can be regarded as $M_{0,n+1}$.
\begin{prop}\label{prop-fclosure} The closure of the parameter space for the shift of argument subalgebras in $S(\fgl_n)$ is $\ol{M_{0,n+1}}$. The same is true for the parameter space for the family $\hat F(C)$.
\end{prop}
\begin{proof} For $\fg=\fgl_n$ the regular Cartan element $C$ is a diagonal matrix with pairwise distinct eigenvalues $z_1,\ldots,z_n$. From the results of Shuvalov \cite{shuvalov} and Vinberg \cite{vinberg} it follows that any limit subalgebra is uniquely determined by its quadratic graded component. According to the results of Tarasov \cite{taras2}, the lifting of any limit shift of argument subalgebra is unique hence uniquely determined by the quadratic component as well. Note that in this case $Q_C$ is the linear span of the quadratic elements $H_i:=\sum\limits_{j\ne i}\frac{e_{ij}e_{ji}}{z_i-z_j}$ which are the coefficients of (an appropriate version of) the $KZ$ connection. In particular the space $Q_C$ does not change under simultaneous affine transformations of the $z$'s, hence the space of parameters is naturally the configuration space of $n$ pairwise distinct points on the affine line or, equivalently, the configuration space $M_{0,n+1}$ of $n+1$ pairwise distinct points on the projective line. In \cite{AFV} Aguirre, Felder and Veselov considered the same subspaces $Q_C$ universally (i.e. as subspaces in the Drinfeld-Kohno holonomy Lie algebra) and showed that the closure of the family $Q_C$ is the Deligne-Mumford compactification $\overline{M_{0,n+1}}$. This means that the parameter space for limit shift of argument subalgebras (both classical and quantum) for $\fg=\fgl_n$ is $\overline{M_{0,n+1}}$.
\end{proof}
One can define the subalgebra $F(X)\subset S(\fgl_n)$ corresponding to a degenerate curve $X\in \ol{M_{0,n+1}}$ recursively as follows. Let $X_\infty$ be the irreducible component of $X$ containing the marked point $\infty$. To any distinguished point $\lambda\in X_\infty$ we assign the number $k_\lambda$ of marked points on the (reducible) curve $X_\lambda$ attached to $X_\infty$ at $\lambda$. Let $C$ be the diagonal matrix with the eigenvalues $\lambda$ of multiplicity $k_\lambda$ for all distinguished points $\lambda\in X_\infty$. Then the corresponding shift of argument subalgebra $F(C)\subset S(\fgl_n)$ is centralized by the Lie subalgebra $\bigoplus\limits_{\lambda}\fgl_{k_\lambda}$ in $\fgl_n$ and contains the Poisson center $S(\bigoplus\limits_\lambda \fgl_{k_\lambda})^{\bigoplus\limits_\lambda \fgl_{k_\lambda}}$. The subalgebra corresponding to the curve $X$ is just the product of $F(C)\subset S(\fgl_n)$ and the subalgebras corresponding to $X_\lambda$ in $S(\fgl_{k_\lambda})\subset S(\fgl_n)$ for all distinguished points $\lambda\in X_\infty$ (for this we need to define the point $\infty$ on each $X_\lambda$ -- it is just the intersection with $X_\infty$).
Having a unique lifting of each $F(C)$ to the universal enveloping algebra $U(\fgl_n)$ we can do the same and attach a commutative subalgebra in $U(\fgl_n)$ to any stable rational curve $X\in\ol{M_{0,n+1}}$. Since every limit shift of argument subalgebra has a unique lifting to $U(\fgl_n)$, every limit subalgebra of the family $\hat F(C)\subset U(\fgl_n)$ is of this form. Now we can make a precise statement:
\begin{prop}
\label{tensorprod}
The subalgebra $F(X)$ corresponding to a degenerate curve $X$ is the tensor product $F(C)\otimes_{S(\bigoplus\limits_\lambda \fgl_{k_\lambda})^{\bigoplus\limits_\lambda \fgl_{k_\lambda}}}\bigotimes\limits_\lambda F(X_\lambda)$. The same is true for the lifting $\hat F(X)\subset U(\fgl_n)$.
\end{prop}
\begin{proof}
In order to prove this proposition we need the following Lemma:
\begin{lem}
\label{tensor} Suppose that $\fg$ is a reductive Lie algebra, $\fg_0$ -- reductive subalgebra of $\fg$. Then the subalgebras $U(\fg)^{\fg_0}$ and $U(\fg_0)$ in $U(\fg)$ are both free $U(\fg_0)^{\fg_0}$-modules. Moreover, the product of these subalgebras in $U(\fg)$ is:
$$U(\fg)^{\fg_0} \cdot U(\fg_0) \simeq U(\fg)^{\fg_0} \otimes_{U(\fg_0)^{\fg_0}} U(\fg_0).$$
The same holds for the associated graded algebras. Namely, $S(\fg)^{\fg_0}$ and $S(\fg_0)$ are free $S(\fg_0)^{\fg_0}$-modules, and
$$S(\fg)^{\fg_0} \cdot S(\fg_0) \simeq S(\fg)^{\fg_0} \otimes_{S(\fg_0)^{\fg_0}} S(\fg_0).$$
\end{lem}
\begin{proof} We can assume without loss of generality that $\fg_0$ does not contain nontrivial ideals of $\fg$. Then this is a particular case of Knop's theorem on Harish-Chandra map for reductive group actions (see \cite{Kn}, Theorem 10.1 and items (d) and (e) of the Main Theorem). Indeed, the Main Theorem of \cite{Kn} states that for any reductive $H$ and any smooth affine $H$-variety $X$ the algebra $D(X)^H$ of $H$-invariant differential operators and its commutant $U(X)$ in the algebra $D(X)$ are both free modules over the center of $D(X)^H$. Moreover, the product $D(X)^H\cdot U(X)\subset D(X)$ is the tensor product $D(X)^H\otimes_{ZD(X)^H} U(X)$ of $D(X)^H$ and $U(X)$ over the center of $D(X)^H$. To get the desired statement we just apply this to the $H=G\times G_0$-action on $X=G$, where $G$ acts from the right and $G_0$ acts on from the left. For this case we have $D(X)^{G\times G_0}$ is $U(\fg)^{\fg_0}$ and its commutant in $D(X)$ is $U(X)=U(\fg_0)\otimes_\BC U(\fg)$ (generated by momenta of the left $G_0$-action and the right $G$-action). By Theorem 10.1 of \cite{Kn}, the center of $U(\fg)^{\fg_0}$ is $U(\fg_0)^{\fg_0}\otimes_\BC U(\fg)^{\fg}$ and the algebra $U(X)$ contains $U(\fg_0)\otimes_\BC U(\fg)^\fg$ as the subalgebra of (right) $G$-invariants. Hence by the Main Theorem of \cite{Kn} we have the desired assertion.
Theorems 9.4 and 9.8 of \cite{Kn} imply the same for the associated graded algebras (in fact the Main Theorem of \cite{Kn} is a consequence of the same fact for the associated graded algebras).
\end{proof}
\begin{comment}
\begin{cor}
\label{corinter}
Let $A=\lim\limits_{t\to0} F(C(t))$ where $C(t)=C_0+t \cdot C_1\in \fh_{reg}$. Then we have $A=F(C_0)\otimes_{ZU(\fz_\fg(C_0))}F(C_1)$ where $F(C_1)$ is the shift of argument subalgebra in $U(\fz_\fg(C_0))$. In particular $A\cap U(\fg)^{\fz_\fg(C_0)} = F(C_0)$ and $A\cap U(\fz_\fg(C_0)) = F(C_1)\subset U(\fz_\fg(C_0))$. The same is true for the associated graded subalgebras in $S(\fg)$.
\end{cor}
\end{comment}
Now let us describe the simplest limit subalgebras corresponding to the case when all the curves $X_\lambda$ are irreducible.
\begin{lem}
\label{size}
Let $C_0=\diag(\underbrace{\lambda_1, \ldots, \lambda_1}_{k_1}, \ldots, \underbrace{\lambda_l, \ldots, \lambda_l}_{k_l})$ and $C_i=\diag(\underbrace{\mu_{i,1}, \ldots, \mu_{i,k_i}}_{k_i})$ for $i=1,\ldots,l$ such that $\lambda_r\ne\lambda_s$ and $\mu_{i,r}\ne\mu_{i,s}$ for $r\ne s$. Then the element
\begin{eqnarray*}C(t) := C_0 + t \cdot \diag(C_1, \ldots, C_l)\end{eqnarray*}
is regular (as an element of the \emph{Lie algebra} $\fgl_n$) for small $t$.
The subalgebra $\lim_{t \to 0} F(C(t))$ is the tensor product $$F(C_0) \otimes_{Z(S(\fgl_{k_1}\oplus \ldots \oplus \fgl_{k_l}))} (F(C_1)\otimes\ldots\otimes F(C_l)).$$
Here $F(C_i)$ is a subalgebra in $S(\fgl_{k_i})\subset S(\fgl_{k_1}\oplus \ldots \oplus \fgl_{k_l})$.
Moreover, $$\lim_{t \to 0} F(C(t)) \cap S(\fgl_{n})^{\fgl_{k_1}\oplus \ldots \oplus \fgl_{k_l}} = F(C_0).$$
Here $n=k_1+ \ldots +k_l$.
The same holds for quantum shift of argument subalgebras $\hat F(C)$.
\end{lem}
\begin{proof}
From \cite{shuvalov}
it follows that $\lim_{t \to 0} = F(C_0) \cdot (F(C_1)\otimes\ldots\otimes F(C_l))$.
We know that $F(C_0) \subset S(\fgl_n)^{\fgl_{k_1} \oplus \ldots \oplus \fgl_{k_l}}$ (because $[\partial_C, {\rm ad} \, x] = \partial_{[C,x]}$) and $(F(C_1)\otimes\ldots\otimes F(C_l)) \subset S(\fgl_{k_1} \oplus \ldots \oplus \fgl_{k_l})$.
Moreover, from proof of \cite[Lemma 1]{shuvalov} it follows that $F(C_0) \cap (F(C_1)\otimes\ldots\otimes F(C_l)) = Z(S(\fgl_{k_1}\oplus \ldots \oplus \fgl_{k_l}))$. Using Lemma \ref{tensor} we obtain the result.
\end{proof}
Since the closure of the parameter space for subalgebras $F(C)$ is $\overline{M_{0,n+1}}$, any limit subalgebra can be obtained by iterating the above degeneration. Iterating limits from Lemma above we obtain Proposition ~\ref{tensorprod}.
\end{proof}
Lemma~\ref{size} implies the following formula for the Poincare series of $F(C)$ for any diagonal $C$. Denote by $Z_k(x)$ the formal power series $\prod\limits_{i=1}^k(1-x^i)^{-1}$. We set $P_k(x):=\prod\limits_{i=1}^k Z_i(x)$. Note that $P_n(x)$ is the Poincare series of $F(C)$ for regular $C$.
\begin{lem}\label{fsize} Let $C=\diag(\underbrace{\lambda_1, \ldots, \lambda_1}_{k_1}, \ldots, \underbrace{\lambda_l, \ldots, \lambda_l}_{k_l})$ such that $\lambda_r\ne\lambda_s$. Then $F(C)$ is a free polynomial algebra with the Poincare series $$ P_n(x)\prod\limits_{i=1}^{l}\frac{Z_{k_i}(x)}{P_{k_i}(x)}.
$$
\end{lem}
\begin{proof} Straightforward from Lemma~\ref{size}.
\end{proof}
\begin{lem}
\label{l36}
Let $\mathfrak{A}$ be a family of subalgebras of the form $F(\diag(C,\underbrace{0, \ldots, 0}_{k})), C \in T^{reg} \subset GL_n$. Then \\
1) Every limit subalgebra of the family $\mathfrak{A}$ is a maximal commutative subalgebra of $S(\fgl_{n+k})^{\fgl_k}$. \\
2) Every limit subalgebra of this family is a free polynomial algebra with the following Poincare series:
$$P(x) = \prod \dfrac{1}{(1-x)^{n+1}} \cdot \dfrac{1}{(1-x^2)^{n+1}} \cdot \ldots \cdot \dfrac{1}{(1-x^k)^{n+1}} \cdot \dfrac{1}{(1-x^{k+1})^{n}} \cdot \dfrac{1}{(1-x^{k+2})^{n-1}} \cdot \ldots \cdot \dfrac{1}{(1-x^{n+k})}.$$
3) The closure of the parameter space for the family $\mathfrak{A}$ is $\ol{M_{0,n+2}}$.
\end{lem}
\begin{proof}
Let $a(t)$ belongs to $T^{reg}$ for small $t$ and consider
$b(t) = \diag(a(t),\underbrace{0, \ldots, 0}_{k})$.
Let $D = (\underbrace{0, \ldots, 0}_n, \mu_1 \ldots \mu_k)$ and degree $N$ such that $N > \deg a(t)$ and $c(t) = b(t) + t^N D$ is regular for small $t$.
Then
\begin{equation}
\label{f1}
\lim_{t \to 0} F(c(t)) = \lim_{t \to 0} F(b(t)) \otimes_{Z(S(\fgl_k))} F(D)
\end{equation} is free and maximal commutative.
Suppose that $\lim_{t \to 0} F(b(t))$ is not maximal. This implies that there exists an element $x \in S(\fgl_{n+k})^{\fgl_k}$ such that $\{x,\lim_{t \to 0}F(b(t))\} = 0$.
This means that $x$ commutes with every element from the limit subalgebra (\ref{f1}). But the limit subalgebra is maximal therefore $x \in \lim_{t \to 0} F(c(t))\cap S(\fgl_{n+k})^{\fgl_k}$ which is $\lim_{t \to 0} F(b(t))$ by Lemma ~\ref{size}. Thus $\lim_{t \to 0} F(b(t))$ is maximal.
\begin{comment}
Further, from~\cite{shuvalov} follows that $\lim_{t \to 0} F(a(t))$ has the following system of free generators:
some free generators of $\lim_{t \to 0}F(C_1(t))$, some free generators of $F(C_2)$ and generators of $S(\fgl_k)^{\fgl_k}$. So, first and third set are free generators of $\lim_{t \to 0} F(C_1(t))$.
\end{comment}
The second assertion follows from the fact that Poincare series of limit subalgebra \\ $\lim_{t \to 0} F(b(t))$ is the same as for algebra $F(b(t))$ for generic $t$ and from Lemma~\ref{fsize}.
To prove third assertion let fix $C_1 \in \fgl_k$ and consider a family of subalgebras of the form $$F(C_0) \otimes_{Z(S(\fgl_k))} F(C_1),$$ where $C_0 = \diag(C,\underbrace{0, \ldots, 0}_{k}), C \in \fh^{reg}$. This family is parameterized by \\ $\gamma_{n+1;\{1\},\{2\},\ldots,\{n\},\{n+1,\ldots,n+k\}}(M_{0,n+2} \times \{pt\}) \subset \gamma_{n+1;\{1\},\{2\},\ldots,\{n\},\{n+1,\ldots,n+k\}}(M_{0,n+2} \times M_{0,k+1}) \subset M_{0, n+k+1}$. But the closure of $M_{0, n+2} \times \{pt\}$ in $\ol{M_{0,n+k+1}}$ is $\ol{M_{0,n+2}} \times \{pt\}$ therefore the closure of the parameter space for the family $\mathfrak{A}$ is $\ol{M_{0,n+2}}$.
\end{proof}
\begin{cor} The same results (Lemmas \ref{size}, \ref{fsize} and \ref{l36}) are true for the liftings of shift of argument subalgebras to the universal enveloping algebra $U(\fgl_n)$.
\end{cor}
\begin{proof}
Follows from the fact that shift of argument subalgebras admits unique lifting to $U(\fgl_n)$ \\ ~\cite{taras2}.
\end{proof}
\section{The results.}
The subalgebra $B(C)$ does not change under dilations of $C$, so the space of parameters for the family of Bethe subalgebras in the Yangian $Y(\fgl_n)$ is $T^{reg}/\mathbb{C}^*$. Let $Z$ be the pro-algebraic scheme defined in Section~\ref{def-lim} (note that $Z$ is naturally a closure of $T^{reg}/\mathbb{C}^*$). By analogy with the shift of argument subalgebras, we can regard $T^{reg}/\mathbb{C}^*$ as the moduli space $M_{0,n+2}$, i.e. the space of rational curves with $n+2$ marked points. More precisely, to any matrix $C$ with the eigenvalues $z_1,\ldots,z_n$, we can assign a rational curve $\mathbb{P}^1$ with the marked points $0,z_1,\ldots,z_n,\infty$. Our main result is the following
\begin{thm}
\label{result}
\begin{enumerate}
\item $Z$ is a smooth algebraic variety isomorphic to $\overline{M_{0,n+2}}$.
\item For any point $X\in\overline{M_{0,n+2}}$, the corresponding commutative subalgebra $B(X)$ in $Y(\fgl_n)$ is free and maximal commutative.
\end{enumerate}
\end{thm}
Next, the operad structure on the spaces $\ol{M_{0,n+1}}$ leads to a recursive description of the limit subalgebras analogous to that of the limit shift of argument subalgebras. Let $X_\infty$ be the irreducible component of $X\in\ol{M_{0,n+2}}$ containing the marked point $\infty$. We identify $X_\infty$ with the standard $\mathbb{CP}^1$ in such a way that the marked point $\infty$ is $\infty$ and the point where the curve containing the marked point $0$ touches $X_\infty$ is $0$. To any distinguished point $\lambda\in X_\infty$ we assign the number $k_\lambda$ of marked points on the maximal (possibly reducible) curve $X_\lambda$ attached to $X_\infty$ at $\lambda$ (we set $k_\lambda=1$ if $X_\lambda$ is a (automatically, marked) point). Let $C$ be the diagonal $(n-k_0)\times(n-k_0)$-matrix with the eigenvalues $\lambda$ of multiplicity $k_\lambda$ for all distinguished points $0\ne\lambda\in X_\infty$. Then the subalgebra $i_{k_0}(B(C))$ centralized by the Lie subalgebra $\bigoplus\limits_{\lambda\ne0}\fgl_{k_\lambda}$ in $\fgl_{n-k_0}\subset i_{k_0}(Y(\fgl_{n-k_0}))\subset Y(\fgl_n)$ and by the complement sub-Yangian $\psi_{n-k_0}(Y(\fgl_{k_0}))$.
The subalgebras corresponding to boundary points of $\ol{M_{0,n+2}}$ can be inductively described as follows:
\begin{thm}\label{result1} \begin{enumerate} \item The limit Bethe subalgebra corresponding to the curve $X\in\ol{M_{0,n+2}}$ is the product of the following $3$ commuting subalgebras: first, $i_{k_0}(B(C))\subset i_{k_0}(Y(\fgl_{n-k_0}))\subset Y(\fgl_n)$, second, the subalgebra corresponding to $X_0$ in the complement sub-Yangian $\psi_{n-k_0}(Y(\fgl_{k_0}))\subset Y(\fgl_n)$ and third, the limit shift of argument subalgebras $\hat F(X_\lambda)$ in $U(\fgl_{k_\lambda})\subset i_{k_0}(Y(\fgl_{n-k_0}))\subset Y(\fgl_n)$ for all distinguished points $\lambda\ne0$ (again we define the point $\infty$ on each $X_\lambda$ just as the intersection with $X_\infty$).
\item $i_{k_0}(B(C))$ contains the center of every $U(\fgl_{k_\lambda})\subset i_{k_0}(Y(\fgl_{n-k_0}))$. The above product is in fact the tensor product $$\psi_{n-k_0}(B(X_0))\otimes_{\BC}i_{k_0}(B(C))\otimes_{ZU(\bigoplus_{\lambda\ne0} \fgl_{k_\lambda})}\bigotimes_{\lambda\ne0}\hat F(X_\lambda).$$
\end{enumerate}
\end{thm}
\begin{comment}
\begin{rem}
$\hat F(X_\lambda)$ is the limit quantum shift of argument subalgebra defined in Proposition ~\ref{tensorprod}.
\end{rem}
\end{comment}
\begin{thm}
\label{result2}
1) Let $C_0=\diag(\lambda_1, \ldots, \lambda_{n-k})$ and $C_1=\diag(\mu_1, \ldots, \mu_k)$. Suppose that
$C(t) = \diag(C_0, \underbrace {0, \ldots, 0}_k)
+ t\cdot\diag(\underbrace{0, \ldots, 0}_{n-k}, C_1) \in T^{reg}$ (i.e. both $C_0$ and $C_1$ are regular and non-degenerate). Then
$$\lim_{t \to 0} B(C(t)) = i_{k}(B(C_0)) \otimes \psi_{n-k}(B(C_1)).$$ \\
2) Let $C_0=\diag(\underbrace{\lambda_1, \ldots, \lambda_1}_{k_1}, \ldots, \underbrace{\lambda_l, \ldots, \lambda_l}_{k_l})$ be a non-degenerate matrix and $C_i=\diag(\underbrace{\mu_{i,1}, \ldots, \mu_{i,k_i}}_{k_i})$ for $i=1,\ldots,l$ such that $\lambda_r\ne\lambda_s$ and $\mu_{i,r}\ne\mu_{i,s}$ for $r\ne s$. Let
\begin{eqnarray*}C(t) := C_0 + t \cdot \diag(C_1, \ldots, C_l).\end{eqnarray*}
Then
$$\lim_{t \to 0} B(C(t)) = B(C_0) \otimes_{\bigotimes\limits_{i=1}^lZ(U(\fgl_{k_i}))} \bigotimes\limits_{i=1}^l \hat F(C_i),$$
where $\hat F(C_i)$ is the shift of argument subalgebra in $U(\fgl_{k_i}) \subset U(\fgl_n) \subset Y(\fgl_n)$ (the copy of $U(\fgl_n)$ in the Yangian $Y(\fgl_n)$ is generated by $t_{ij}^{(1)}$).
\end{thm}
\begin{rem} {\em
Theorems ~\ref{result1} and ~\ref{result2} are equivalent. Indeed, Theorem~\ref{result2} is a particular case of Theorem~\ref{result1}. On the other hand, any degenerate curve $X$ can be obtained from a non-degenerate one by, first, taking a limit from part 1 of Theorem~\ref{result2} ($m$ times where $m$ is the number of irreducible components of $X$ on the way between the marked points $0$ and $\infty$) and, second, taking a limit from part 2 of Theorem~\ref{result2} for each of that $m$ components. So Theorem~\ref{result1} follows from Theorem~\ref{result2}.}
\end{rem}
\begin{rem} {\em For example, Theorem~\ref{result2} allows to describe explicitly the subalgebra corresponding to the degenerate curve which has exactly $n$ components between the points $0$ and $\infty$ with a unique marked point on each component (i.e. the so-called ``caterpillar curve''). This subalgebra is the same as $\lim\limits_{t_i\to0}B(\diag(1,t_1,t_1t_2,\ldots,t_1\cdot\ldots\cdot t_{n-1}))$. According to Theorem~\ref{result2} this limit subalgebra is the Gelfand-Tsetlin subalgebra of $Y(\fgl_n)$.
The opposite example is the degenerate $2$-component curve such that one component contains $0$ and $\infty$ while the other component contains all other marked points. The subalgebras corresponding to such curve have the form $\lim\limits_{t\to0}B(E+\diag(t\lambda_1,\ldots,t\lambda_n))$ which is according to Theorem~\ref{result2} the subalgebra generated by $B(E)$ and $\hat{F}(\diag(\lambda_1,\ldots,\lambda_n))$.}
\end{rem}
\section{Proof of the main Theorems.}
Let $C = \diag(\underbrace{\lambda_1, \ldots, \lambda_1}_{k_1}, \underbrace{\lambda_2, \ldots, \lambda_2}_{k_2}, \ldots, \underbrace{\lambda_l, \ldots, \lambda_l}_{k_l})\in T$ be a non-regular element from the maximal torus. Let $d_i(C)$ be the number of homogeneous degree $i$ generators of $F(C)$ (i.e. the multiplicity of the factor $(1-x^i)$ in the Poincare series of $F(C)$).
\begin{prop}\label{bethesize}(Lower bound for the size of Bethe subalgebra)
There is a set of algebraically independent elements of $B(C)$ which consists of $\min(d_i(C)+i-1,n)$ elements of degree $i$ for all $i\in\BZ_{>0}$.
\end{prop}
\begin{proof}
It is enough to check that in $\bar B(C) = \gr B(C) \subset CY(\fgl_n)$ we have enough algebraically independent generators (see $\ref{filtration}$).
The images of $\tau_l(u,C)$ are
\begin{equation}
\bar\tau_l(u,C) = \sum_{i_1 < \ldots <i_l} \lambda_{i_1} \ldots \lambda_{i_l} \left(\sum_{\sigma \in S_l} \bar{t}_{i_1 i_{\sigma(1)}}(u) \ldots \bar{t}_{i_l i_{\sigma(l)}}(u)\right).
\end{equation}
Now we can change generators $\bar{t}_{ij}^{(p)} \to \tilde t_{ij}^{(p)} = \lambda_i \bar t_{ij}^{(p)}$.
Note that there is a filtration in $CY(\fgl_n)$ given by
$$\deg \tilde t_{ij}^{(p)} = 1.$$
According to this filtration the leading term of the coefficient of $\bar\tau_l(u,C)$ at $u^{-p}$ for $p \geqslant l$ is $$\sum_{i_1 <\ldots <i_l} \lambda_{i_1} \ldots \lambda_{i_l} \sum_{p_1 + \ldots + p_l = p; \, p_1, \ldots, p_l> 0} \Delta_{i_1 \ldots i_l}^{(p_1 \ldots p_l)}=\sum_{i_1 <\ldots <i_l} \sum_{p_1 + \ldots + p_l = p; \, p_1, \ldots, p_l> 0} \tilde \Delta_{i_1 \ldots i_l}^{(p_1 \ldots p_l)}.$$
Here we set
\begin{equation}
\Delta_{i_1 \ldots i_m}^{(p_1 \ldots p_m)} = \det \begin{pmatrix}
\bar t_{i_1 i_1}^{(p_1)} & \bar t_{i_1 i_2}^{(p_1)} & \dots & \bar t_{i_1 i_m}^{(p_1)} \\
\bar t_{i_2 i_1}^{(p_2)} & \bar t_{i_2 i_2}^{(p_2)} & \dots & \bar t_{i_2 i_m}^{(p_2)} \\
\vdots & \vdots & \vdots & \vdots \\
\bar t_{i_m i_1}^{(p_m)} & \bar t_{i_m i_2}^{(p_m)} & \dots & \bar t_{i_m i_m}^{(p_m)}
\end{pmatrix};\ \tilde \Delta_{i_1 \ldots i_m}^{(p_1 \ldots p_m)} = \det \begin{pmatrix}
\tilde t_{i_1 i_1}^{(p_1)} & \tilde t_{i_1 i_2}^{(p_1)} & \dots & \tilde t_{i_1 i_m}^{(p_1)} \\
\tilde t_{i_2 i_1}^{(p_2)} & \tilde t_{i_2 i_2}^{(p_2)} & \dots & \tilde t_{i_2 i_m}^{(p_2)} \\
\vdots & \vdots & \vdots & \vdots \\
\tilde t_{i_m i_1}^{(p_m)} & \tilde t_{i_m i_2}^{(p_m)} & \dots & \tilde t_{i_m i_m}^{(p_m)}
\end{pmatrix}
\end{equation}
We see that, for $p\ge l$ the leading terms of such coefficients do not depend on $C$ if we change generators to $\tilde t_{ij}^{(p)}$. Moreover, we see that the leading terms of the coefficients of $\bar \tau_l(u,C)$ at $u^{-l}$ consists of only $\tilde t_{ij}^{(1)}$, coefficients of $u^{-l-1}$ of $\bar\tau_l(u,C)$ contain $\tilde t_{ij}^{(2)}$ and do not contain $ \tilde t_{ij}^{(p)}, p >2$ and so on. Generally, we have the following
\begin{lem}
The leading term of the coefficient of $\bar \tau_l(u,C)$ at $u^{-l-k}$ does contain $\tilde t_{ij}^{(k+1)}$ and does not contain $\tilde t_{ij}^{(p)}, p > k+1$.
\end{lem}
The leading terms of the coefficients of $u^{-p}, p \leqslant l$ of $\tau_l(u,C)$ are polynomials in $\tilde t_{ij}^{(1)}$ hence can be regarded as elements of $S(\fgl_n)\subset CY(\fgl_n)$ generated by $\tilde t_{ij}^{(1)}$.
\begin{lem} The leading terms of the coefficients of $u^{-p}, p \leqslant l$ of $\tau_l(u,C)$, $l=1,\ldots,n$ generate the shift of argument subalgebra $F(C)$ in $S(\fgl_n)$.
\end{lem}
\begin{proof} Set $\bar T^{(1)}:=\sum\limits_{i,j}\bar t_{ij}^{(1)}\otimes e_{ij}\in S(\fgl_n)\otimes Mat_n$. Then the generating function for the leading terms is ${\rm Tr} \, \Lambda^lC(E+T^{(1)}u^{-1})= {\rm Tr} \, \Lambda^l (CE+\tilde T^{(1)}u^{-1})$ which is the generating function for the derivatives of the $l$-th coefficient of the characteristic polynomial.
\end{proof}
To see that the set of coefficients of $u^{-l-k}$ of $\bar\tau_l(u,C)$ are algebraically independent let us compute differential at point $\tilde t_{ij}^{(1)} = \delta_{i, j-1}$, $\tilde t_{ij}^{(p)} = 0$. It is straightforward computation that
$$\diff \tilde \Delta_{i_1 \ldots i_l}^{(p_1 \ldots p_l)} =
\begin{cases}
(-1)^{l-1} \cdot \diff \tilde t_{i_1, i_1 + l - 1}^{(p-l+1)}, & p_l = p-(l-1), p_{l-1} = 1, \ldots, p_1 = 1, \\
& i_2 = i_1 + 1, \ldots i_l = i_{l-1} + 1\\
0, & {\text else.}
\end{cases}$$
Hence the differential of the coefficient of $\bar\tau_l(u,C)$ at $u^{-l-k}$ is a non-zero linear combination of $\diff \tilde t_{ij}^{(k+1)}$ with $j-i = l-1$. Hence differentials of all coefficients with $p \geqslant l$ are linearly independent. Therefore if we have algebraic dependence between some coefficients, then it can only consist of coefficients $u^{-p}$ with $p \leqslant l$.
The following tableaux shows the greater number $r$, such that there is some $t_{ij}^{(r)}$ at this coefficient:
\renewcommand{\arraystretch}{1.4}
\begin{center}
\begin{tabular}{|m{1.5cm} | m{1.2cm} | m{1.2cm} | m{1.2cm} | m{1.2cm} | m{1.2cm}| m{1.2cm}| m{1.2cm}| m{1.2cm}|}
\hline
& $u^{-1}$ & $u^{-2}$ & $u^{-3}$ & \ldots & $u^{-(n-1)}$ & $u^{-n}$ & $u^{-(n+1)}$ &\ldots \\
\hline
$\tau_1(u,C)$ & 1 & 2 & 3 & \ldots & $n-1$ & $n$ & $n+1$ & \ldots \\
\hline
$\tau_2(u,C)$ & 1 & 1 & 2 & \ldots & $n-2$ & $n-1$ & $n$ & \ldots \\
\hline
$\tau_3(u,C)$ & 1 & 1 & 1 & \ldots & $n-3$ & $n-2$ & $n-1$ & \ldots \\
\hline
$\ldots$ & & & & & & & & \\
\hline
$\tau_n(u,C)$ & 1 & 1 & 1 & \ldots & 1 & 1 & 2 & \ldots\\
\hline
\end{tabular}
\end{center}
On the other hand, the coefficients of $u^{-p}, p \leqslant l$ of $\tau_l(u,C)$, $l=1,\ldots,n$ generate the shift of argument subalgebra $F(C)$ in $S(\fgl_n)$ hence we have the desired number of algebraically independent generators.
\end{proof}
Recall the following fact from \cite{molev2}:
\begin{lem}
\label{togd}
\cite{molev2}[Theorem 1.10.7]
${\rm qdet} \, T(u) \cdot \omega_n(t^{j_{m+1}\ldots j_n}_{i_{m+1}\ldots i_n}(-u+n-1)) = \sgn p \cdot \sgn q \cdot t^{i_1 \ldots i_m}_{j_1\ldots j_m}(u)$, where $p = \left(\begin{smallmatrix}
1 & 2 & \ldots & n \\
i_1 & i_2 & \ldots & i_n \end{smallmatrix}\right), q = \left(\begin{smallmatrix}
1 & 2 & \ldots & n \\
j_1 & j_2 & \ldots & j_n \end{smallmatrix}\right) \in S_n$.
\end{lem}
This implies the following
\begin{lem}
\label{betheImage}
Suppose that $C \in T$ is non-degenerate. Then
$\omega_n(B(C)) = B(C^{-1})$.
\end{lem}
\begin{proof}
Note that the series ${\rm qdet} \, T(u)$ is invertible so multiplying by ${\rm qdet} \, T(u)$ is a bijection on a set of generators of $B(C)$. Moreover $T(u) \to T(u-c)$ is invertible automorphism of $Y(\fgl_n)$ so we can consider $\tau_k(-u+n-1)$ instead of $\tau_k(u,C)$ as a generators of $B(C)$.
By Lemma \ref{togd}
\begin{multline*}
(\det C)^{-1} \cdot \qdet T(u) \cdot \omega_n(\tau_k(-u+n-1,C)) = \\ =(\det C)^{-1} \cdot \sum_{1 \leqslant a_1< \ldots < a_k \leqslant n} \lambda_{a_1} \ldots \lambda_{a_k} \qdet T(u) \cdot \omega_n(t_{a_1, \ldots, a_k}^{a_1,\ldots,a_k}(-u+n-1)) = \\ = (\det C)^{-1} \sum_{1 \leqslant a_1< \ldots < a_k \leqslant n} \lambda_{a_1} \ldots \lambda_{a_k} t_{a_{k+1}, \ldots, a_n}^{a_{k+1},\ldots,a_n}(u) =\\= \sum_{1 \leqslant a_1< \ldots < a_k \leqslant n} \lambda_{a_{k+1}}^{-1} \ldots \lambda_{a_n}^{-1} t_{a_{k+1}, \ldots, a_n}^{a_{k+1},\ldots,a_n}(u) = \tau_{n-k}(u, C^{-1}).
\end{multline*}
\end{proof}
\subsection{Proof of Theorem ~\ref{result2}, part 1.}
\label{pf1} We have $i_{k}(B(C_0))$ in the limit subalgebra. To its generators one can just take the standard generators of the Bethe subalgebra
$\tau_k(u,C_0 + t \cdot C_1), 1 \leqslant k \leqslant n,$ and put $t = 0$.
On the other hand, from Lemma \ref{betheImage} we have the following:
\begin{multline*}
\lim_{t \to 0} B(a(t)) = \lim_{t \to 0} \omega_n(B(a(t)^{-1})) = \omega_n(\lim_{t \to 0} B(a(t)^{-1})) \supset \\ \supset (\omega_n \circ \varphi_{n-k}) (B(C_1^{-1})) = (\omega_n \circ \varphi_{n-k} \circ \omega_{k}) (B(C_1)) = \psi_{n-k}(B(C_1)).
\end{multline*}
So the limit subalgebra contains $i_k(B(C_0)) \cdot \psi_{n-k}(B(C_1))$. The latter is $i_k(B(C_0)) \otimes \psi_{n-k}(B(C_1))$ due to Lemma \ref{commuteyang}.
Since $C_0$ and $C_1$ are regular elements of $gl_k$ and $gl_{n-k}$ respectively, the tensor product has $i_k(B(C_0)) \otimes \psi_{n-k}(B(C_1))$ has the same Poincare series as generic Bethe algebra, therefore the limit is in fact equal to $i_k(B(C_0)) \otimes \psi_{n-k}(B(C_1))$. So Theorem~\ref{result2} (1) is proved.
\begin{prop}
\label{incl}
Let $C\in T$ be any non-degenerate diagonal matrix. Set $C^{(k)}:=\diag(C,\underbrace{0,\ldots,0}_k)$. Then $\eta_k(B(C) \otimes A_0) \subset \hat F(C^{(k)})$ for any non-degenerate $C \in \fh$ and $k \in \mathbb{N}$.
Moreover, the restrictions of $\{\eta_k\}$ to $B(C)$ give an asymptotic isomorphism of $B(C)$ and $\hat{F}(C^{(k)})$, $k\to\infty$.
The same is true for all limit algebras of the family $B(C)$, i.e. we have $\eta_k(\lim\limits_{t\to0}B(C(t)) \otimes A_0) \subset \lim\limits_{t\to0}\hat F(C(t)^{(k)})$.
\end{prop}
\begin{proof}
We have $\eta_k = \pi_{n+k} \circ \omega_{n+k} \circ i_k$. Consider $a(t) = \diag(C,\underbrace{0, \ldots, 0}_k)+ t\cdot\diag(0, \ldots , 0, C^{\prime})$ and subalgebra $B(a(t))$ in $Y(\fgl_{n+k})$. Then by Theorem~\ref{result2} (1) the limit subalgebra $\lim\limits_{t\to0}B(a(t))$ is equal to $i_n(B(C^{(k)})) \otimes \psi_{n-k}(B(C^{\prime}))$. Note that $B(C^{(k)}) = i_k(B(C))$. We have
\begin{multline*}
\pi_{n+k} \circ \omega_{n+k}(\lim_{t \to 0} B(a(t)) \otimes A_0) \subset \lim_{t \to 0}(\pi_{n+k} \circ \omega_{n+k}(B(a(t)) \otimes A_0) = \\ = \lim_{t \to 0} \hat F(a(t))) = \hat F(C^{(k)}) \otimes_{Z(U(\fgl_k))} \hat F(C^{\prime}).
\end{multline*}
We use here the fact that $\pi_{n+k}(B(a(t)) = \hat F(a(t)^{-1})$, see ~\cite{taras2}.
But $\eta_k(B(C)) \subset U(\fgl_{n+k})^{\fgl_k}$ therefore $\eta_k(B(C) \otimes A_0) \subset \hat F(C^{(k)}) \otimes_{Z(U(\fgl_k))} \hat F(C^{\prime})\cap U(\fgl_{n+k})^{\fgl_k} = \hat F(C^{(k)})$.
The fact that $\{\eta_k\}$ is asymptotic isomorphism follows from Lemma~\ref{bethesize}.
For limit algebras it follows from the fact that $\eta_k(\lim_{t \to 0}(B(C(t))\otimes A_0)) \subset \lim_{t \to 0} F(C(t)^{(k)},0, \ldots,0)$ and the fact that limit algebras have the same Poincare series.
\end{proof}
\subsection{Proof of Theorem ~\ref{result2}, part 2.}
Analogously to \ref{pf1}, we have $B(C_0)$ in the limit subalgebra. Also we know that
$$\eta_k(\lim_{t \to 0} (B(C(t))\otimes A_0) \subset \lim_{t \to 0} \hat{F}(C(t)^{(k)}) = \hat F(C_0^{(k)}) \otimes_{ZU(\mathfrak{z}(C_0))
} \hat F(C_1).$$
We see that $B(C_0) \otimes_{Z(U(\mathfrak{z}({C_0})))} \hat F(C_1) \otimes A_0$ maps to $\hat F(C_0^{(k)}) \otimes_{ZU(\mathfrak{z}(C_0))} \hat F(C_1)$.
Moreover,$\eta_k$ is asymptotic isomorphism and restriction of $\eta_k$ to $U(\mathfrak{z}(C_0)) \subset Y(\fgl_{n})$ is identity, so we have $\hat F(C_1)$ in the limit subalgebra as well. But using Proposition~\ref{bethesize} we see that the Poincare series of $B(C_0) \otimes_{Z(Y(\fgl_k))} \otimes \hat F(C_1)$ coincides with that of $B(C(t))$ for generic $t$.
\subsection{Proof of Theorem ~\ref{result}}
From Proposition ~\ref{incl} we know that family \{$\eta_k$\} is asymptotic isomorphism between $B(C) \otimes A_0$ and $\hat F(\diag(C,\underbrace{0,\ldots,0}_k))$. According to Lemma ~\ref{l36} all algebras $\hat F(\diag(C,\underbrace{0,\ldots,0}_k))$ form the family $\mathfrak{A}$. Limit subalgebras of this family are parameterized by $\ol{M_{0,n+2}}$ for any $k > 0$. Therefore limit Bethe subalgebras are parameterized by $\ol{M_{0,n+2}}$ as well.
Additionally, all limit algebras of the family $\mathfrak{A}$ are free and maximal commutative by Lemma ~\ref{l36}. Therefore all limit Bethe subalgebras in the Yangian are free and maximal commutative as well.
\subsection{Proof of Theorem ~\ref{result1}} We have already proved the first part of the Theorem.
We only need to see that $i_{k_0}(B(C))$ contains the center of every $U(\fgl_{k_\lambda})\subset i_{k_0}(Y(\fgl_{n-k_0}))$.
We know that the image $\eta_k(B(C) \otimes A_0)$ contains the center of every $U(\fgl_{k_\lambda})$ for any $k$.
Also we know that $A_0$ maps to the center of $U(gl_{n+k})$ by $\eta_k$. But from ~\cite[Lemma 10.1]{Kn} it follows that the generators of the center of $U(gl_{n+k})$ and of the centers of all $U(\fgl_{k_\lambda})$ are algebraically independent.
Using last time the fact that $\eta_k$ is asymptotic isomorphism we obtain the result.
\begin{comment}
\section{Appendix.}
Argument shift method is a particular case of the famous
Magri--Lenard construction \cite{Ma}. Let $R$ be a commutative
algebra equipped with two compatible Poisson brackets,
$\{\cdot,\cdot\}_1$ and $\{\cdot,\cdot\}_2$, (i.e., any linear
combination of $\{\cdot,\cdot\}_1$ and $\{\cdot,\cdot\}_2$ is a
Poisson bracket). Let $Z_t$ be the Poisson center of $R$ with
respect to $\{\cdot,\cdot\}_1+t\{\cdot,\cdot\}_2$. Let $A$ be the
subalgebra of $R$ generated by all $Z_t$ for generic $t$.
\begin{fact} The subalgebra $A\subset R$ is commutative with respect to any Poisson bracket
$\{\cdot,\cdot\}_1+t\{\cdot,\cdot\}_2$.
\end{fact}
\begin{proof}
Suppose $a\in Z_{t_1},\ b\in Z_{t_2}$ with $t_1\ne t_2$. The
expression $\{a,b\}_1+t\{a,b\}_2$ is linear in $t$, and, on the
other hand, it vanishes at two distinct points, $t_1$ and $t_2$.
This means that $\{a,b\}_1+t\{a,b\}_2=0$ for all $t$.
Now suppose $a,b\in Z_{t_0}$. Since $t_0$ is generic, there exists
a continuous function $a(s)$ such that $a(t_0)=a$, and for $s$ in
a certain neighborhood of $t_0$ we have $a(s)\in Z_s$. For any $s$
in a punctured neighborhood of $t_0$ we have
$\{a(s),b\}_1+t\{a(s),b\}_2=0$, and therefore
$\{a,b\}_1+t\{a,b\}_2=0$.
\end{proof}
\begin{cor} Suppose that
$ZS(\fg)=S(\fg)^{\fg}$ is the center of $S(\fg)$ with respect to
the Poisson bracket, and let $\mu\in\fg^*$. Then the algebra
$F(C)\subset S(\fg)$ generated by the elements
$\partial_{C}^n\Phi$, where $\Phi\in ZS(\fg)$, (or,
equivalently, generated by central elements of $S(\fg)=\BC[\fg^*]$
shifted by $tC$ for all $t\in\BC$) is commutative with respect
to the Poisson bracket.
\end{cor}
\begin{proof}
Take the Poisson--Lie bracket as $\{\cdot,\cdot\}_1$, and the
"frozen argument" bracket as $\{\cdot,\cdot\}_2$; this means that
for the generators we have
$$
\{x,y\}_2=\mu([x,y])\quad x,y\in\fg.
$$
Then the algebra $Z_t$ is generated by central elements of
$\BC[\fg^*]=S(\fg)$ shifted by $tC$.
\end{proof}
Since the Lie algebra $\fg$ is reductive we can identify $\fg$
with $\fg^*$ and write $C\in\fg$.
\begin{fact}\label{mf} ~\cite{mf} For regular semisimple $C\in\fg$ the algebra $F(C)$
is a free commutative subalgebra in $S(\fg)$ with
$\frac{1}{2}(\dim\fg+\rk\fg)$ generators (this means that
$F(C)$ is a commutative subalgebra of maximal possible
transcendence degree). One can take the elements
$\partial_{C}^n\Phi_k$, $k=1,\dots,\rk\fg$,
$n=0,1,\dots,\deg\Phi_k$, where $\Phi_k$ are basic
$\fg$-invariants in $S(\fg)$, as free generators of $F(C)$.
\end{fact}
\begin{rem} For $\fg=\fgl_N$ we have: \\
$N$ generators of degree 1; \\
$N-1$ of degree $2$; \\
\ldots \\
1 of degree $N$.
\end{rem}
In \cite{shuvalov} Shuvalov described the closure of the family of
subalgebras $F(C)\subset S(\fg)$ under the condition
$C\in T^{reg}$ (i.e., for regular $C$ in the fixed Cartan
subalgebra). In particular, the following assertion is proved
in~\cite{shuvalov}.
\begin{fact}\label{shuvalov} Suppose that $C(t)=C_0+tC_1+t^2C_2+\dots\in T^{reg}$
for generic $t$. Set $\fz_k=\bigcap\limits_{i=0}^k\fz_{\fg}(C_i)$
(where $\fz_{\fg}(C_i)$ is the centralizer of $C_i$ in $\fg$),
$\fz_{-1}=\fg$. Then we have
\begin{enumerate}
\item the subalgebra $\lim\limits_{t\to0}F(C(t))\subset S(\fg)$
is generated by all elements of $S(\fz_k)^{\fz_k}$ and their
derivatives (of any order) along $C_{k+1}$ for all $k$. \item
$\lim\limits_{t\to0}F(C(t))$ is a free commutative algebra.
\end{enumerate}
\end{fact}
The following results were obtained by Tarasov.
\begin{fact}\label{commutant1}\cite{taras} The subalgebras
$F(C)$ and the limit subalgebras of the type
$\lim\limits_{t\to0}F(C(t))$ are maximal commutative
subalgebras, i.e., they coincide with their Poisson centralizers
in $S(\fg)$.
\end{fact}
The {\em symmetrization map} $\sigma:S(\fg)\to U(\fg)$ is defined
by the following property:
$$
\sigma(x^k)=x^k \quad \forall\ x\in\fg,\ k=0,1,2,\dots
$$
\begin{fact}\label{symmetrization}\cite{taras2} For $\fg = gl_n$, a certain system of generators of
$F(C)$ and of the limit subalgebras of the type
$\lim\limits_{t\to 0} F(C(t))$ can be lifted to commuting
elements of $U(\fg)$ by the symmetrization map. This gives rise to
a unique lifting of $F(C)$ and the limit subalgebras to the universal enveloping algebra.
\end{fact}
\begin{comment}
\begin{rem} The system of generators of $A_{\mu}$ of the limit subalgebras of the type
$\lim\limits_{t\to0}A_{\mu(t)}$ to be lifted by the symmetrization
is chosen explicitly in \cite{taras2}. It is, up to proportionality,
the system of the elements $\partial_{\mu}^n\Phi_k$,
$k=1,\dots,r-1$, $n=0,1,\dots,\deg\Phi_k$, (where $\Phi_k\in
S(gl_n)^{gl_n}$ are the coefficients of the characteristic
polynomial as functions on $gl_n$) and their limits, respectively.
We shall only use that this system of generators up to
proportionality is continuous in the parameter~$\mu$.
\end{rem}
\begin{thm}
\label{shiftdeg}
Let $\fg$ be a reductive Lie algebra, $\fh$ -- Cartan subalgebra, $T^{reg}$ -- regular elements of $\fh$. Suppose that $C \in T^{reg}$. Then
any shift of argument subalgebra $F(C)$ is free. Particulary, derivatives of basic invariants along $C$ are free generators of $F(C)$.
The set of the degrees of these generators can be described as the set of the heights, each of which is increased by 1,
of all positive roots of the algebra and $\rk \fg$ additional 1's.
\end{thm}
\end{comment}
\newpage
| 9,460
|
TITLE: Self-intersection number of a loop on a surface
QUESTION [1 upvotes]: Let $M$ be an orientable surface$($without boundary$)$, possibly non-compact. Let $\alpha\in \pi_1(M)$ be a primitive element i.e. $\alpha$ is not a proper power of some other element. Let $f:\Bbb S^1\to M$ be any loop representing $\alpha$. Consider the lifting problem
Here, $M_\alpha$ be the cover corresponding to the subgroup $\langle\alpha\rangle$ of $\pi_1(M)$ and $\widetilde M$ be the universal cover of $M$. Also, all unleveled maps are covering maps.
Note that $\pi_1(M)$ acts on $\widetilde M$ via covering transformations, and $M_\alpha$ be an open annulus$($any open connected surface with a finitely generated fundamental group is homeomorphic to the interior of a closed surface$)$.
$\textbf{Problem 1:}$ Consider two sets $$\mathscr A_f:=\big\{g\in
\pi_1(M):\text{ the map } \Bbb R\xrightarrow{g\cdot \ell}\widetilde
M\to M_\alpha\text{ runs from one end to other of the open annulus
}M_\alpha\big\},$$ $$\mathscr B_f:=\big\{(z,w)\in \Bbb S^1\times\Bbb
S^1:z\not= w\text{ and }f(z)=f(w)\big\}$$ Is the cardinality of
$\mathscr B$ two-times the cardinality of $\mathscr A$, i.e. $|\mathscr A_f|=\frac{1}{2}|\mathscr B_f|$?
$\textbf{Problem 2:}$ Give a Riemannian metric on $M$ such that
$\alpha$ can be represented by a shortest loop $f^\#:\Bbb S^1\to M$,
i.e. $f^\#$ has the minimum length in its free homotopy class. Is
$|\mathscr A_f|\geq |\mathscr A_{f^\#}|$?
REPLY [1 votes]: For problem one, the answer is "no" as there may be "bigons". The pair of intersections forming the bigon contribute to $B_f$ but not to $A_f$.
For problem two, the answer is "yes" as the shortest representative has no bigons.
A version of this is discussed in the "Primer on mapping class groups" by Farb and Margalit under the name "the bigon criterion".
| 170,760
|
Severe Spring Floods Forecast for Upper Midwest, South and East
WASHINGTON, DC, March 16, 2010 (ENS) – An unusually wet and snowy winter will increase the potential for spring flood events, the National Weather Service warned today. Seasonal flooding already has begun and is forecast to continue through spring in parts of the Midwest, said the meteorologists.
Forecasters said the South and East are more susceptible to flooding than usual this year as an El Nino influenced winter left the area soggier than usual.
Overall, more than a third of the area in the lower 48 states has an above average flood risk after one of the snowiest winter seasons in many years. December precipitation was up to four times above average.
The highest threat is in the Dakotas, Minnesota and Iowa, including along the Red River Valley where crests could approach the record levels set last year.
“It’s a terrible case of deja vu, but this time the flooding will likely be more widespread,” said NOAA Administrator Dr. Jane Lubchenco. “As the spring thaw melts the snowpack, saturated and frozen ground in the Midwest will exacerbate the flooding of the flat terrain and feed rising rivers and streams.
“We will continue to refine forecasts to account for additional precipitation and rising temperatures, which affect the rate and severity of flooding,” she said.
North Dakota Governor John Hoeven said Sunday that President Barack Obama has approved his request for a federal declaration for the state under the Disaster Relief and Emergency Assistance Act due to weather conditions and forecasts indicating a strong potential for severe spring flooding. Disaster relief was approved for North Dakota 18 counties and the Spirit Lake Nation reservation.
The state and many communities and counties are already allocating resources and preparing for widespread flooding.
Today, Governor Hoeven joined a group of Fargo South High School students to help pass sandbags from pallets into backyards of houses along the Red River in south Fargo. The sandbags are being used to create flood barriers in an effort to protect homes from the rising water of the Red River.
In South Dakota, Governor Mike Rounds says 12 counties and two American Indian reservations will get federal financial assistance to help recover from the Christmas blizzard of 2009 through a presidential disaster declaration signed last week by President Obama.
“I’m pleased that federal aid has been promptly approved by the President,” Governor Rounds said. “Unfortunately, the area struck by the Christmas-period blizzard continues to be plagued by an excess of moisture. We are now monitoring rivers and streams that are at or near flood stage.”
Saturated soil conditions, high water content in the current snowpack, full ponds and wetlands that result in minimal available surface storage, and the prospects of additional moisture and ice jams through the spring snowmelt are all contributing to the flood forecast.
Until early March, consistently cold temperatures limited snow melt and runoff, and the ground is frozen to a depth as much as three feet below the surface, meteorologists said.
D, director of the National Weather Service.
“Though El Nino is forecast to continue at least through spring, its influence on day-to-day weather should lessen considerably,” said Hayes.
Without a strong El Nino influence, climate forecasting for the spring months April through June is more challenging, but NOAA’s Climate Prediction Center says odds currently favor wetter-than-average conditions in coastal sections of the Southeast.
The center predicts warmer-than-average temperatures across the western third of the nation and Alaska, and below-average temperatures in the extreme north-central and south-central United States.
This is Flood Safety Awareness Week, says Hayes, who is reminding all drivers across the country “Turn Around, Don’t Drown.
“It floods somewhere in the United States or its territories nearly every day of the year, killing approximately 100 people on average and causing nearly $7 billion in damages,” said Hayes. .
| 22,758
|
Common Threads: The Humanities, Arts, and Social Sciences at MIT
Part of the "Common Threads" video, produced for the MIT 150th celebration, this clip features the vital role of humanities, arts, and social sciences at MIT. The roots of MIT's strengths in the humanities, arts, and social sciences go back to its founding days, and by mid 20th century, the Institute had created a dedicated School for these disciplines. Today, MIT SHASS is home to research that has a global impact, and to superb graduate programs, all recognized as among the finest in the world. The School's mission also includes a core part of an MIT education: the School's faculty help all undergraduates gain a range of critical skills and cultural/historical perspectives that enable MIT graduates to serve as leaders—at home and around the globe.
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“The expansion of the ice was only about half as large in the summer than decades ago and only about half as thick as 130 years ago. The disappearance of summer sea ice in the Arctic is one of the first landmines in this minefield, one of the tipping points that we set off first when we push warming too far,” said Rex.
Arctic Ice Melt Means Less Rain For Our Fibershed Communities in the Western U.S.
How does Arctic sea ice melting impact our Fibershed communities in California and throughout the Western United States? According to research initially conducted in 2004 by Jacob Sewall and Lisa Sloan (1), Arctic sea ice disappearance has a “substantial impact on water resources in western North America.”
“Where the sea ice is reduced, heat transfer from the ocean warms the atmosphere, resulting in a rising column of relatively warm air,” Sewall said. This warming air blocks the movement of precipitation to our land area and sends storm tracks northward.
2017 image illustrates what was known as a ‘ridiculously resilient ridge’ a high-pressure system that blocked storm tracks from entering the Western United States for months during the only time of year when precipitation would naturally fall.
Climate Change Impacts Are Here Now, And Innovation is Needed
In a recently issued State of the Planet report by the Swedish Academy of Sciences (2), the authors noted that “Climate change impacts are hitting people harder and sooner than envisioned a decade ago. This is especially true for extreme events, like heat waves, droughts, wildfires, extreme precipitation, floods, storms, and variations in their frequency, magnitude, and duration.” (3)
The analysis identifies a particularly damaging and exacerbating force of climate change: “Greater inequality leads to more rapid environmental degradation.” (4) Pointing to a needed transformation of the economic status quo, the authors’ synthesis informs all aspects of our society, from policy to individual lifestyle choices: “Biosphere stewardship involves caring for, looking after, and cultivating a sense of belonging in the biosphere, ranging from people and environments locally to the planet as a whole. Such stewardship is not a top-down approach forced on people, nor solely a bottom-up approach. It is a learning-based process with a clear direction, a clear vision, engaging people to collaborate and innovate across levels and scales as integral parts of the systems they govern.” (5)
The accelerating pace of climate change presents an enormous but not impossible challenge for producers in our local fibershed. Local growers face mounting pressures from changing weather conditions and natural disasters, with two of the most prominent in California and the Western United States being wildfires and drought. Impacts of climate change and natural disasters extend beyond the environmental consequences. There is also a real economic and social toll on communities. Our regional fibersheds are particularly vulnerable in an economic system that favors globalized industrial supply chains over small-scale, regional production. However, our regional producers are poised to be more resilient and dynamic to meet the realities of climate change.
Farmers in Fibershed Communities are Stepping Up to Adapt to the Challenge
The climate circumstances make our producers’ work all that more important and urgent. Fibershed is working directly with farmers and ranchers that are mitigating the effects of climate change through carbon farming practices. Carbon farming practices enhance soil health and increase soil carbon through various land management practices. These techniques work to sequester carbon and reduce greenhouse gas emissions.
Hedgerow planting at True Grass Farms. Image courtesy of True Grass Farms.
When it comes to drying conditions and drought, in particular, our producers are working to improve soil health, increase soil water holding capacity, and reduce landscape demand for water. This work is necessary in order to restore hydrological cycles and regenerate soil.
We know that poor land management practices, such as intensive tillage, can further diminish soil quality and reduce the land’s ability to absorb and retain water. As a result, soil dries out faster, and less groundwater is replenished. There is an immediate need to scale carbon farming practices on par with the rate at which we see climate change accelerating. Together with our partners, Fibershed is working to support small-scale growers to develop viable and resilient regional production. Three examples of this work in practice include:
- At PT Ranch, 525 acres sit at the base of the Sierra mountains, where melted snowcap hydrates the land. Historically, with the abundance of moisture in the soil, grass has grown plentifully. But with the increasing droughts, the Taylor family needs to be more strategic. For the most part, they have been able to raise about 150 animals, year-round, by managed grazing on the grass alone, along with some feed such as pellets of grain for the poultry. Continue reading how the Taylor family discovered the interconnected benefits of mitigating climate change, producing healthy food, and providing high-quality employment opportunities.
- At Integrity Alpacas, three acres of the property have been designated for work within Fibershed’s Climate Beneficial™ program. These efforts include planting fruitless mulberry trees (which provide shelter, shade, and a food source for the alpacas, along with general carbon sequestration) and bringing the property’s Delta clay soil back to life. Continue reading about Integrity Alpacas plans to ensure the land is better off than where it started.
- At Fortunate Farm, compost application is among several other carbon farming practices that make up the farm’s active Carbon Farm Plan. According to Gowan Batist, farmer and co-owner, partnership with a local brewing company provides the farm with spent barley that is incredible for compost as it is pathogen-free and has a good ratio of carbon to nitrogen. The spent grains are mixed with hardwood chips from local tree trimming companies, then formed into windrows to compost, and finally spread onto crop fields, pasture, and invasive gorse patches (cut down and reseeded with grass seeds). Gowan notes that after covering old invasive gorse patches with compost and reseeding, she has seen many native plant species return voluntarily, such as thimbleberry, salmonberry, elderberry, cascara, and wax myrtle. Continue reading about Fortune Farm’s extensive Carbon Farm Plan.
The benefits of carbon farm practices go beyond adapting to changing conditions. Additional co-benefits of adopting better land management strategies include higher productivity, increased water retention, hydrological function, biodiversity, and resilience. At Fibershed, we are reimaging how we think about the connection between soil, water, and climate change. These practices are essential for the reduction of future climate impacts.
Sources:
- Sewall, J. O., & Sloan, L. C. (2004). Disappearing Arctic Sea Ice Reduces Available Water in the American West. Geophysical Research Letters, Vol. 31, L06209, 2004..
- Folke, C., Polasky, S., Rockström, J. et al. Our future in the Anthropocene biosphere. Ambio 50, 834–869(2021).
- Downing, A.S., A. Bhowmik, D. Collste, S.E. Cornell, J. Donges, I. Fetzer, T. Häyhä, J. Hinton, et al. 2019. Matching scope, purpose, and uses of planetary boundaries science. Environmental Research Letters 14: 073005.
- Enqvist, J.P., S. West, V.A. Masterson, L.J. Haider, U. Svedin, and M. Tengö. 2018. Stewardship as a boundary object for sustainability research: Linking care, knowledge and agency. Landscape and Urban Planning 179: 17–37., Chaplin-Kramer, R., R.P. Sharp, C. Weil, E.M. Bennett, U. Pascual, K.K. Arkema, K.A. Brauman, B.P. Bryant, et al. 2019. Global modelling of nature’s contributions to people. Science 366: 255–258., Plummer, R., J. Baird, S. Farhad, and S. Witkowski. 2020. How do biosphere stewards actively shape trajectories of social-ecological change? Journal of Environmental Management 261: 110139.
-., Clark, W.C., L. van Kerkhoff, L. Lebel, and G. Gallopi. 2016. Crafting usable knowledge for sustainable development. Proceedings of the National Academy of Sciences, USA 113: 4570–4578., Nyström, M., J.-B. Jouffray, A. Norström, P.S. Jørgensen, V. Galaz, B.E. Crona, S.R. Carpenter, and C. Folke. 2019. Anatomy and resilience of the global production ecosystem. Nature 575: 98–108.
All images by Paige Green unless otherwise noted.
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SOUNDS GOOD E-Book on all Amazon sites
SOUL, FUNK and JAZZ FUSION On CD - Exception Remasters
Just Click Below To Purchase for £3.95
Thousands of E-Pages - All Details and In-depth Reviews From Discs
"...Fire And Water..." fifth and final '2on1' set issued 26 November 2016 in the UK (see full list below) deals with his ninth and tenth studio LPs from 1970 and 1971 and comes bolstered up with a huge seventeen Bonus Tracks on 2CDs.) – "In Philadelphia/Don't Knock My Love" by WILSON PICKETT on Edsel EDSK 7122 (Barcode 740155711238) offers 2LPs Remastered onto 2CDs plus Seventeen Bonus Tracks and plays out as follows:
Disc 1 (63:52 minutes):
1. Run Joey Run
2. Help The Needy
3. Come Right Here
4. Bumble Bee (Sting Me)
5. Don't Let The Green Grass Fool You
6. Get Me Back On Time, Engine Number Nine [Side 2]
7. Days Go By
8. International Playboy
9. Ain’t No Doubt About It
Tracks 1 to 9 are his 9th studio album "Wilson Pickett In Philadelphia" - released September 1970 in the USA on Atlantic SD 8270 (Stereo Only) and April 1971 in the UK (with the same tracks) as "Engine No. 9" on Atlantic 2400 026 (Stereo Only).
BONUS TRACKS:
10. International Playboy (Remix)
Track 10 is the A-side of Atlantic 45-2961 – a US 7" single from 1973
11. Pray For The Rain
12. Let It Come Naturally
Tracks 11 and 12 are from "Funky Midnight Mover: The Atlantic Studio Recordings (1962-1978)" released 2009 in the USA on Rhino Handmade RHM2 07753
13. Funky Broadway (Live In Ghana)
14. Land Of 1000 Dances (Live in Ghana)
Tracks 13 and 14 are from the Various Artists LP "Soul To Soul – Music from The Original Soundtrack Recorded Live In Ghana, West Africa" on Atlantic SD 7207
15. Un' Avventura
Track 15 is the A-side of a 1969 Italian-only 7" single on Atlantic ATL NP 03097
16. Heaven
17. Can't Stop A Man In Love
Tracks 16 and 17 are from "Funky Midnight Mover: The Atlantic Studio Recordings (1962-1978)" released 2009 in the USA on Rhino Handmade RHM2 07753
18. One Step Away
19. Funk Factory
Tracks 18 and 19 are the A&B-sides of a 1972 US 7” single on Atlantic 45-2878
Disc 2 (68:44 minutes):
1. Fire And Water
2. (Your Love Has Brought Me) A Mighty Long Way
3. Covering The Same Old Ground
4. Don't Knock My Love (Part 1)
5. Don't Knock My Love (Part 2)
6. Call My Name, I’ll Be There
7. Hot Love
8. Not Enough Love To Satisfy
9. You Can't Judge A Book By Its Cover
10. Pledging My Love
11. Mama Told Me Not To Come
12. Woman Let Me Be Down Home
Tracks 1 to 12 are his 10th studio album "Don't Knock My Love" - released December 1971 in the USA on Atlantic SD 8300 and January 1972 in the UK on Atlantic K 40319.
BONUS TRACKS:
13. Don't Knock My Love (Part 2) Single Version - 1972 US 7" single on Atlantic 45-2797, B-side
14. If You Need Me (Remake)
15. Don't Forget The Bridge
16. Rock Of Ages
17. Many Roads To Travel
18. Hope She'll Be Happier
19. Believe I'll Shout
Tracks 14 to 19 are from "Funky Midnight Mover: The Atlantic Studio Recordings (1962-1978)" released 2009 in the USA on Rhino Handmade RHM2 07753 - Track 15 mistaken credited as Atlantic 45-2909
Each of these five card digipaks comes in a gatefold with Volume 5 being the only double-disc issue. Like the four others it sports a comprehensive 16-page booklet in the left flap with new liner notes from legendary Soul writer TONY ROUNCE – a man whose name has graced literally hundreds of quality CD reissues. He goes into all the chart statistics for 1970 and 1971 - his Gamble & Huff project "In Philadelphia" with the second LP embracing Rock artists like Free and Randy Newman - the concert at Accra in Ghana that resulted in the "Soul To Soul" Soundtrack LP in 1972 (Atlantic SD 7207) – his progression to RCA Records and recording at Muscle Shoals and finally to Little Richard attending his funeral in 2006 paying homage in his sermon to one of the Soul greats. Mastered by PHIL KINRADE – the tracks are licensed from Warners and are therefore the US 1996 Rhino versions of old and those 2009 Rhino Handmade rarities – full and punchy Bill Inglot and Dan Hersch CD Remasters from original tapes.
The "Wilson Pickett In Philadelphia" LP (called "Engine No. 9" in the UK after the song became a hit) is dominated by two song writing forces – Kenneth Gamble and Leon Huff and a team of four other staff writers which included Bunny Sigler. Gamble & Huff threw three tunes into the ring – the opener "Run Joey Run", the two-parter single and hooky-as-Hell Bobby Eli guitar-fuzz of "Get Me Back On Time, Engine Number Nine" (a No. 3 R&B hit in October 1970 for Pickett on Atlantic 45-2765) and the final LP cut "Ain’t No Doubt About It" – a mid-tempo stroller with choppy keyboards and sweet brass accompaniment (could even have been another 45). Bunny Sigler had his hands in the melodrama of "Days Go By", the don’t worry baby pluck of "Come Right Here" (used as B-side to "International Playboy" in May 1973 on Atlantic 45-2961) and one the album’s genuine gems - "Don’t Let The Green Grass Fool You" – co-written with John Bellmon, Reginald Turner and Jerry Akines who also contributed "Bumble Bee (Sting Me)". The LP peaked at a respectable No. 12 on the US R&B charts with Atlantic in fact taking advantage of the hits on it and pumping out "The Best Of Wilson Pickett, Vol. II" in May 1971 – rewarded with an even higher No. 8 placing (the last time Pickett would break the Top 10).
As you can imagine - the Bonus Tracks are a mixed bag of killer vs. throwaway. It’s not surprising that Atlantic returned to the catchy legend-in-my-own-time "International Playboy" song in 1973 with a remix where Wilson lists ladies in Rome and New Orleans all too willing to cook more than egg fu yung for the Wicked Alabama lad (one of the Bonus Tracks on CD1). It’s very cool to hear the band count-in "Pray For The Rain" (one of the 6CD Rhino Handmade outtakes) but the vocal never quite takes off – better is the six-minutes of "Heaven" – all strings and ‘reach out and touch me’ lyrics. Sung half in Italian and the rest in English "Un Avventura" is clearly dubbed off a disc and is a weird hybrid that’s interesting but not much else. "Can’t Stop A Man In Love" dates from 23 Feb 1972 and is excellent. But best is the non-album track "Funk Factory" that was paired with the ballad "One Step Away" on Atlantic 45-2878 – a No. 11 R&B hit in June 1972. "Funk Factory" is a truly great Pickett groove co-written by him with Brad Shapiro and Luther Dixon – ending Disc 1 on a bit of a high.
You would think with three singles off the "Don’t Knock My Love" LP – one of which was the huge title track and another US No. 1 R&B hit – that the album would have charted higher than No. 23 - but it didn’t - a sign of things to come. Like many of the smarter players of the day Wilson Pickett was quick to take on board Rock acts whose music had a swing and swagger suited to Soul - here he tackles Free by doing "Fire and Water" from their 1970 Island Records LP of the same name and Randy Newman's "Mama Told Me (Not To Come)" which had become a No. 1 rock hit for Three Dog Night. He then mixed these with Brad Shapiro songs like "Call My Name, I'll Be There", "Hot Love" and the huge title track - a co-write on Part 1 with WP. There's even a hint of Area Code 615 and Paul Butterfield in the harmonica-driven "(Your Love Has Brought Me) A Mighty Long Way" - a fantastic brassy groover that follows perfectly after "Fire And Water" (a Jack Avery, Earl Simms and Carlton McWilliams song itself covered by Bonnie Bramlett in 1975 on her "It’s Time" LP on Capricorn Records). This sexy little beast then segues into a set of strings that introduce the ballad "Covering The Same Old Ground" - a George Jackson song Atlantic used as the B-side to the November 1972 US single for "Mama Told Me (Not To Come)". The audio on this is gorgeous - full of warmth and even though the strings and girls threaten to drown the whole thing on occasion - Wilson howls his heart out and saves the day.
Edsel's decision to not separate the segue continues with the two-part title track which applies also to the lesser and ever so slightly hammy "Call My Name, I'll Be There". Another of the album's cool cuts is the fuzzed-up guitar of "You Can't Judge A Book By Its Cover" - a song with a Stevie Wonder writing credit. The Johnny Ace standard "Pledging My Love" was the B-side to "Fire And Water" which made R&B No. 2 in January 1972 on Atlantic 45-2852. It ends on another B-side with the funky potential of being an A - "Woman Let Me Be Down Home" which Atlantic used as the flipside to "Call My Name, I’ll Be There" on Atlantic 45-2824 in August 1971. Wilson lists off his country-boy complaints – his woman constantly telling him what to do, say and wear (oh dear).
Collectors are going to love the sheer in-yer-face fuzzed-up funkiness of "Don’t Knock My Love, Pt. 2" in Mono (one of the Bonus Tracks) coming on like a mash-up between a James Brown/Chi-Lites/Sly Stone instrumental – what a Funky Winner! Of the 1971 and 1972 unreleased recordings – and even though the vocal is not quite right - I like the mellow cover of Bill Withers track "Hope She’ll Be Happier" where the highly-polished acoustic guitars and piano make it feel more Carole King "Tapestry" than "Don’t Knock My Love". And the Wilson-penned Jesus song "Rock Of Ages" (another Bonus) is akin to Aretha's passion on the landmark "Amazing Grace" double.
The original Rhino CDs have been deleted for years and the Rhino Handmade set hard to find and liable to make your bang manager nervous – so this pairing of Pickett’s lesser-heard career is a very welcome reissue indeed. From here he would go on to limited success with RCA...
"...Start It off!" - Pickett roars to the band as they kick in with the righteous-groove of "Believe I'll Shout" – a holy-roller funk work out with a driving organ backbeat.
Not everything on these two CDs is unmitigated genius by any stretch – but there are so many great moments like the above that I'm digging it big time. I suggest you dig in too and enjoy the illicit fruit of the wicked man's labours...)-->
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\begin{document}
\title{Non-differentiable Bohmian trajectories}
\author{Gebhard Gr\"{u}bl and Markus Penz \\
Theoretical Physics Institute, Universit\"{a}t Innsbruck,\\
Technikerstr 25, A-6020 Innsbruck Austria}
\maketitle
\begin{abstract}
A solution $\psi $ to Schr\"{o}dinger's equation needs some degree of
regularity in order to allow the construction of a Bohmian mechanics from
the integral curves of the velocity field $\hbar \Im \left( \bigtriangledown
\psi /m\psi \right) .$ In the case of one specific non-differentiable weak
solution $\Psi $ we show how Bohmian trajectories can be obtained for $\Psi $
from the trajectories of a sequence $\Psi _{n}\rightarrow \Psi .$ (For any
real $t$ the sequence $\Psi _{n}\left( t,\cdot \right) $ converges
strongly.) The limiting trajectories no longer need to be differentiable.
This suggests a way how Bohmian mechanics might work for arbitrary initial
vectors $\Psi $ in the Hilbert space on which the Schr\"{o}dinger evolution $
\Psi \mapsto e^{-iht}\Psi $ acts.
\end{abstract}
\section{Introduction}
Quantum mechanis often is praised as a theory which unifies classical
mechanics and classical wave theory. Quanta are said to behave either as
particles or waves, depending on the type of experiment they are subjected.
But where in the standard formalism can the particles of the interpretive
talk be found? Perhaps only to some degree in the reduction postulate
applied to position measurements. In reaction to this unsatisfactory state
of affairs, Bohmian mechanics introduces a mathematically precise particle
concept into quantum mechanical theory. The fuzzy wave functions are
supplemented by sharp particle world lines. Through this additional
structure some quantum phenoma like the double slit experiment have lost
their mystery.
Clearly the additional structure of particle world lines brings along its
own mathematical problems. Ordinary differential equations are generated
from solutions of partial differential equations. A mathematically
convincing general treatment so far has been given for a certain type of
wave functions which do not exhaust all possible quantum mechanical
situations. Exactly this fact has led some workers to doubt that a Bohmian
mechanics exists for all initial states $\Psi _{0}$ of a Schr\"{o}dinger
evolution $t\mapsto e^{-iht}\Psi _{0}.$ We shall show on one specific case
of a counter example $\Psi _{0}$ how the problem might be resolved in
general. We approximate the state $\Psi _{0},$ for which the Bohmian
velocity field does not exist, by states which do have one. Their integral
curves turn out to converge to limit curves which can be taken to constitute
the Bohmian mechanics of the state unamenable to Bohmian mechanics on first
sight.
\section{Bohmian evolution for $\protect\psi \in C^{2}$}
Let $\psi :\mathbb{R}\times \mathbb{R}^{s}\rightarrow \mathbb{C}$ be twice
continuously differentiable, i.e. $\psi \in C^{2}\left( \mathbb{R}\times
\mathbb{R}^{s}\right) ,$ and let $\psi $ obey Schr\"{o}dinger's partial
differential equation
\begin{equation}
i\hbar \partial _{t}\psi \left( t,x\right) =-\frac{\hbar ^{2}}{2m}\Delta
\psi \left( t,x\right) +V\left( x\right) \psi \left( t,x\right)
\label{Schroed}
\end{equation}
with $V:\mathbb{R}^{s}\rightarrow \mathbb{R}$ being smooth, i.e., $V\in
C^{\infty }\left( \mathbb{R}^{s}\right) .$ From $\psi ,$ which is called a
classical solution of Schr\"{o}dinger's equation, a deterministic time
evolution $x\mapsto \gamma _{x}\left( t\right) $ of certain points $x\in
\mathbb{R}^{s}$ can be derived: If there exists a unique maximal solution $
\gamma _{x}:I_{x}\rightarrow \mathbb{R}^{s}$ to the implicit first order
system of ordinary differential equations
\begin{equation}
\rho _{\psi }\left( t,\gamma \left( t\right) \right) \dot{\gamma}\left(
t\right) =j_{\psi }\left( t,\gamma \left( t\right) \right) \label{BMmotion}
\end{equation}
with the initial condition $\gamma \left( 0\right) =x,$ one takes $\gamma
_{x}$ as the evolution of $x.$ Here $\rho _{\psi }:\mathbb{R}\times \mathbb{R
}^{s}\rightarrow \mathbb{R}$ and $j_{\psi }:\mathbb{R}\times \mathbb{R}
^{s}\rightarrow \mathbb{R}^{s}$ with
\begin{equation}
\rho _{\psi }\left( t,x\right) =\left\vert \psi \left( t,x\right)
\right\vert ^{2}\text{ and }j_{\psi }\left( t,x\right) =\frac{\hbar }{m}\Im
\left[ \overline{\psi \left( t,x\right) }\nabla _{x}\psi \left( t,x\right)
\right] \label{Current}
\end{equation}
obey the continuity equation $\partial _{t}\rho _{\psi }\left( t,x\right) =-$
div$j_{\psi }\left( t,x\right) $ for all $\left( t,x\right) \in \mathbb{R}
\times \mathbb{R}^{s}.$ From now on we shall drop the index $\psi $ from $
\rho _{\psi }$ and $j_{\psi }.$
For certain solutions\footnote{
The simplest explicitly solvable example is provided by the plane wave
solution $\psi \left( t,x\right) =e^{-i\left\vert k\right\vert ^{2}t+ik\cdot
x}.$ Its Bohmian evolution $\Phi $ obeys $\Phi \left( t,x\right) =tk.$
Another explicitly solveable case is given by a Gaussian free wave packet.} $
\psi $ the curves $\gamma _{x}$ can be shown to exist on a maximal domain $
I_{x}=\mathbb{R}$ for all $x\in \mathbb{R}^{s}:$ \emph{If} $\psi $ has no
zeros, then the velocity field $v=j/\rho $ is a $C^{1}$-vector field. $v$
then obeys a local Lipschitz condition such that the maximal solutions are
unique. \emph{If} in addition there exist continuous nonnegative real
functions $\alpha ,\beta $ with $\left\vert v\left( t,x\right) \right\vert
\leq \alpha \left( t\right) \left\vert x\right\vert +\beta \left( t\right) $
then all maximal solutions equation (\ref{BMmotion}) are defined on $\mathbb{
R}$ and the general solution
\[
\Phi :\bigcup\nolimits_{x\in \mathbb{R}^{s}}I_{x}\times \left\{ x\right\}
\rightarrow \mathbb{R}^{s}\text{ with }\Phi \left( t,x\right) =\gamma
_{x}\left( t\right)
\]
extends to all of $\mathbb{R}\times \mathbb{R}^{s}.$ (Thm 2.5.6, ref. \cite
{Aul}) Due to the uniqueness of maximal solutions the map $\Phi \left(
t,\cdot \right) :\mathbb{R}^{s}\rightarrow \mathbb{R}^{s}$ is a bijection
for all $t\in \mathbb{R}.$ It obeys\footnote{
Here $\Phi \left( t,\Omega \right) =\left\{ \Phi \left( t,x\right)
\left\vert x\in \Omega \right. \right\} .$}
\begin{equation}
\int_{\Phi \left( t,\Omega \right) }\rho \left( t,x\right)
d^{s}x=\int_{\Omega }\rho \left( 0,x\right) d^{s}x \label{Tranport}
\end{equation}
for all $t\in \mathbb{R}$ and for all open subsets $\Omega \subset \mathbb{R}
^{s}$ with sufficiently smooth boundary such that the integral theorem of
Gauss can be applied to the space time vector field $\left( \rho ,j\right) $
on the domain $\bigcup\nolimits_{t^{\prime }\in \left( 0,t\right) }\Phi
\left( t^{\prime },\Omega \right) .$ \cite{DTe}
These undisputed mathematical facts have instigated Bohm's amendment of
equation (\ref{Schroed}) in order to explain the fact that \emph{macroscopic
bodies usually are localized much stricter than their wave functions suggest.
}
In Bohm's completion of nonrelativistic quantum mechanics it is assumed that
any closed system has at any time, in addition to its wave function, a
position in its configuration space and that this position evolves according
to the general solution $\Phi $ induced by the wave function. One says that
the position is guided by $\psi $ since $\Phi $ is completely determined by $
\psi $ (and no other forces than the ones induced by $\psi $ are allowed to
act on the position).$\ $More specifically, $\gamma _{x}$ is assumed to give
the position evolution for an isolated particle with wave function $\psi
\left( 0,\cdot \right) $ and position $x$ -- both at time $t=0.$
As is common in standard quatum mechanics, $\psi \left( 0,\cdot \right) $ is
supposed to obey
\[
\int_{\mathbb{R}^{s}}\left\vert \psi \left( 0,x\right) \right\vert
^{2}d^{s}x=1.
\]
The nonnegative density $\rho \left( 0,\cdot \right) $ is interpreted as the
probability density of the position which the particle has at time $t=0.$
Since an initial position $x$ is assumed to evolve into $\gamma _{x}\left(
t\right) ,$ the position probability density at time $t$ is then, due to
equation (\ref{Tranport}), given by $\rho \left( t,\cdot \right) .$ In
particular, Bohm's completion gives the position probabilities among all the
other spectral probability measures a fundamental status, since the
empirical meaning of the other ones, as for instance momentum probabilities,
all are deduced from position probabilities.
There are classical solutions of Schr\"{o}dinger's equation, whose general
solution $\Phi $ \emph{does not extend} to all of $\mathbb{R}\times \mathbb{R
}^{s}.$ An obstruction to do so can be posed by the zeros of $\psi .$ In the
neigbourhood of such zero the velocity field $v=j/\rho $ may be unbounded
and $v$ then lacks a continuous extension into the zero. As an example
consider a time $0$ wave function $\psi \left( 0,\cdot \right) :\mathbb{R}
^{2}\rightarrow \mathbb{C},$ for which $\psi \left( 0,x,y\right)
=x^{2}+iy^{2}$ within a neighbourhood $U$ of its zero $\left( x,y\right)
=\left( 0,0\right) .$ Within $U$ for the velocity field follows
\[
\frac{m}{\hbar }v^{1}\left( 0,x,y\right) =\Im \frac{\partial _{x}\psi \left(
0,x,y\right) }{\psi \left( 0,x,y\right) }=-\frac{2xy^{2}}{x^{4}+y^{4}}.
\]
Hence for $0<\left\vert \phi \right\vert <\pi /2$ we have $v^{1}\left( r\cos
\phi ,r\sin \phi \right) \rightarrow -\infty $ for $r\rightarrow 0$ with $
\phi $ fixed. Thus the implicit Bohmian evolution equation (\ref{BMmotion})
is singular in a zero of the wave function whenever the velocity field does
not have a continuous extension into it. As a consequence the evolution $
\gamma _{x}$ of such a zero $x$ is not defined by equation (\ref{BMmotion}).
As a related phenomenon there are solutions to equation (\ref{BMmotion})
which begin or end at a finite time because they terminate at a zero of $
\psi .$ A nice example \cite{Bndl} for this to happen provide the zeros of
the harmonic oscillator wave function $\psi :\mathbb{R}^{2}\rightarrow
\mathbb{C}$ with
\[
\psi \left( t,x\right) =e^{-\frac{x^{2}}{2}}\left( 1+e^{-2it}\left(
1-2x^{2}\right) \right) .
\]
E.g., the points $\left\vert x\right\vert =1$ are zeros of $\psi \left(
t,\cdot \right) $ at the times $t\in \pi \mathbb{Z}.$ They are singularities
of $v$ since
\[
\lim_{t\rightarrow 0}t\Im \frac{\partial _{x}\psi \left( t,\pm 1\right) }{
\psi \left( t,\pm 1\right) }=\pm 2.
\]
Note however that $\Im \frac{\partial _{x}\psi \left( 0,x\right) }{\psi
\left( 0,x\right) }=0$ for $x\neq \pm 1.$
There are more challenges to Bohmian mechanics. The notion of distributional
solutions to a partial differential equation like (\ref{Schroed}) raises the
question whether these solutions support a kind of Bohmian particle motion
like the classical solutions do. After all quantum mechanics employs such
distributional solutions.
\section{Bohmian evolution for $\Psi _{t}\in C_{h}^{\infty }$}
In standard quantum mechanics the classical solutions, i.e. the $C^{2}$
-solutions of equation (\ref{Schroed}), do not represent all physically
possible situations. Rather a more general quantum mechanical evolution is
abstracted from equation (\ref{Schroed}). It is given by the socalled weak
solutions
\[
\Psi _{0}\mapsto \Psi _{t}=e^{-iht}\Psi _{0}\text{ for all }\Psi _{0}\in
L^{2}\left( \mathbb{R}^{s}\right)
\]
with $\hbar h$ being a self-adjoint, ususally unbounded hamiltonian
corresponding to equation (\ref{Schroed}). The domain $D_{h}$ of $h$ does
not comprise all of $L^{2}\left( \mathbb{R}^{s}\right) ,$ yet it is dense in
$L^{2}\left( \mathbb{R}^{s}\right) .$ Since $h$ is self-adjoint, the
exponential $e^{-iht}$ has a unique continuous extension to $L^{2}\left(
\mathbb{R}^{s}\right) .$ This unitary evolution operator $e^{-iht}$
stabilizes the domain of $h$ as a dense subspace of $L^{2}\left( \mathbb{R}
^{s}\right) .$ Thus if and only if an initial vector $\Psi _{0}$ belongs to $
D_{h},$ equation (\ref{Schroed}) generalizes to
\begin{equation}
\lim_{\varepsilon \rightarrow 0}\left\Vert i\frac{\Psi _{t+\varepsilon
}-\Psi _{t}}{\varepsilon }-h\Psi _{t}\right\Vert =0 \label{SchroedQM}
\end{equation}
for all $t\in \mathbb{R}.$ For $\Psi _{0}\notin D_{h}$ equation (\ref
{SchroedQM}) does not hold for any time.
Yet the construction of Bohmian trajectories needs much more than the
evolution $\Psi _{0}\mapsto \Psi _{t}$ within $L^{2}\left( \mathbb{R}
^{s}\right) ,$ since the elements of $L^{2}\left( \mathbb{R}^{s}\right) $
are equivalence classes $\left[ f\right] $ of functions $f\in \mathcal{L}
^{2}\left( \mathbb{R}^{s}\right) .$ It rather needs a trajectory of
functions instead of a trajectory of equivalence classes of
square-integrable functions. \emph{If} there exists a function $\psi \in
C^{1}\left( \mathbb{R}\times \mathbb{R}^{s}\right) $ such that $\Psi \left(
t,\cdot \right) =\left[ \psi \left( t,\cdot \right) \right] $ holds for all $
t\in \mathbb{R},$ then $\psi $ is unique and the Bohmian equation of motion (
\ref{BMmotion}) can be derived from the evolution $\Psi _{0}\mapsto \Psi
_{t} $ through $\psi .$ When does there exist such $\psi ?$
Due to Kato's theorem (e.g. Thm X.15 of ref. \cite{RS}) the Schr\"{o}dinger
hamiltonians $h,$ corresponding to potentials $V$ from a much wider class
than just $C^{\infty }\left( \mathbb{R}^{s}:\mathbb{R}\right) ,$ have the
same domain as the free hamiltonian $-\Delta ,$ namely the Sobolev space $
W^{2}\left( \mathbb{R}^{s}\right) .$ This is the space of all those $\Psi
\in L^{2}\left( \mathbb{R}^{s}\right) $ which have all of their
distributional partial derivatives up to second order being regular
distributions belonging to $L^{2}\left( \mathbb{R}^{s}\right) .$ Since $
D_{h} $ is stabilized by the evolution $e^{-iht},$ for any $\Psi _{0}\in
D_{h}$ there exists for any $t\in \mathbb{R}$ a function $\psi \left(
t,\cdot \right) \in \mathcal{W}^{2}\left( \mathbb{R}^{s}\right) $ such that
\begin{equation}
e^{-iht}\Psi _{0}=\left[ \psi \left( t,\cdot \right) \right] . \label{Repr}
\end{equation}
However, for this family of time parametrized functions $\psi \left( t,\cdot
\right) $ the Bohmian equation of motion in general does not make sense
since $\psi \left( t,\cdot \right) $ need not be differentiable in the
classical sense.\footnote{
Only for $s=1$ Sobolev's lemma (Thm IX.24 in Vol 2 of ref. \cite{RS}) says
that $\left[ \psi \left( t,\cdot \right) \right] $ has a $C^{1}$
representative within $\mathcal{L}^{2}\left( \mathbb{R}\right) .$ From such
a $C^{1}$ representative $\psi \left( t,\cdot \right) $ the current $j$
follows as a continuous vector field and a continuous velocity field $v$ can
be derived outside the zeros of $\psi .$ However, $v$ does not need to obey
the local Lipschitz condition implying the local uniquenss of its integral
curves.}
Therefore some stronger restriction of initial data than $\Psi _{0}\in D_{h}$
is needed in order to supply the state evolution $\Psi _{0}\mapsto
e^{-iht}\Psi _{0}$ with Bohm's amendment. For a restricted set of initial
states $\left( x,\Psi _{0}\right) $ and for a fairly large class of static
potentials a$\ $Bohmian evolution has indeed been constructed in \cite{BDG}
and \cite{TeTu}. There it is shown that for any $\Psi _{0}\in
\bigcap\nolimits_{n\in \mathbb{N}}D_{h^{n}}=:C_{h}^{\infty }$ there exists
\begin{itemize}
\item a (time$\ $independent) subset $\Omega \subset \mathbb{R}^{s}$
\item for any $t$ a square-integrable function $\psi \left( t,\cdot \right) $
\end{itemize}
such that the restriction of $\psi \left( t,\cdot \right) $ to $\Omega $
belongs to $C^{\infty }\left( \Omega \right) $ and equation (\ref{Repr})
holds. The set $\Omega $ is obtained by removing from $\mathbb{R}^{s}$ first
the points where the potential function $V$ is not $C^{\infty },$ second the
zeros of $\psi \left( 0,\cdot \right) ,$ and third those points $x$ for
which the maximal solution $\gamma _{x}$ does not have all of $\mathbb{R}$
as its domain. Surprizingly, $\Omega $ is still sufficiently large, since
\[
\int_{\Omega }\left\vert \psi \left( 0,x\right) \right\vert ^{2}d^{s}x=1.
\]
On this reduced set $\Omega $ of initial conditions a Bohmian evolution $
\Phi :\mathbb{R}\times \Omega \rightarrow \mathbb{R}^{s}$ can be
constructed. Thus if $\Psi _{0}\in C_{h}^{\infty }$ and if the initial
position $x$ is distributed within $\mathbb{R}^{s}$ with probability density
$\left\vert \psi \left( 0,\cdot \right) \right\vert ^{2}$ then the global
Bohmian evolution $\gamma _{x}$ of $x$ exists with probability $1.$
\section{Bohmian evolution for $\Psi _{t}\in L^{2}\smallsetminus
C_{h}^{\infty }$}
How about initial conditions $\Psi _{0}\in L^{2}\left( \mathbb{R}^{s}\right)
\smallsetminus C_{h}^{\infty }?$ Can the equation of motion (\ref{BMmotion})
still be associated with $\Psi _{0}?$ Hall has devised a specific
counterexample $\Psi _{0}\notin D_{h}$ which leads to a wave function $\psi $
which at certain times is nowhere differentiable with respect to $x$ and
thus renders impossible the formation of the velocity field $v.$ Therefore
it has been brought forward that the Bohmian amendment of standard quantum
mechanics is \textquotedblleft formally incomplete\textquotedblright\ and it
has been claimed that the problem is unlikely to be resolved. \cite{Hal}
A promising way to tackle the problem is to succesively approximate the
initial condition $\Psi _{0}\notin D_{h}$ by a strongly convergent sequence
of vectors $\left( \Psi _{0}^{n}\right) \in C_{h}^{\infty }.$ For each of
the vectors $\Psi _{0}^{n}$ a Bohmian evolution $\Phi _{n}$ exists. We do
not know whether it has actually been either disproven or proven that the
sequence of evolutions does converge to a limit $\Phi $ and that the limit
depends on the chosen sequence $\Psi _{0}^{n}\rightarrow \Psi _{0}.$
Here we shall explore this question within the simplified setting of a
spatially one dimensional example. We will make use of an equation for $
\gamma _{x}$ which has already been pointed out in \cite{BDG} and which does
not rely on the differentiability of $j.$ In this case equation (\ref
{Tranport}) can be generalized in order to determine a nondifferentiable
Bohmian trajectory $\gamma _{x}$ by choosing $\Omega =\left( -\infty
,x\right) $ in (\ref{Tranport}) as follows.
Consider first the case of a $C^{2}$-solution of equation (\ref{Schroed})
generating a general solution $\Phi :\mathbb{R}\times \mathbb{R}\rightarrow
\mathbb{R}$\ of the Bohmian equation of motion (\ref{BMmotion}). Since
because of their uniqueness the maximal solutions do not intersect, we have $
\Phi \left( t,\left( -\infty ,x\right) \right) =\left( -\infty ,\Phi \left(
t,x\right) \right) =\left( -\infty ,\gamma _{x}\left( t\right) \right) .$
From this it follows by means of equation (\ref{Tranport}) that
\begin{equation}
\int_{-\infty }^{\gamma _{x}\left( t\right) }\rho \left( t,y\right)
dy=\int_{-\infty }^{x}\rho \left( 0,y\right) dy. \label{Tranport_1}
\end{equation}
As a check we may take the derivative of equation (\ref{Tranport_1}) with
respect to $t.$ This yields
\[
\rho \left( t,\gamma _{x}\left( t\right) \right) \dot{\gamma}_{x}\left(
t\right) +\int_{-\infty }^{\gamma _{x}\left( t\right) }\partial _{t}\rho
\left( t,y\right) dy=0.
\]
Making use of local probability conservation $\partial _{t}\rho =-\partial
_{x}j$ we recover by partial integration equation (\ref{BMmotion}).
Now observe that the equation (\ref{Tranport_1}) for $\gamma _{x}\left(
t\right) $ is meaningful not only when $\psi $ is a square integrable $C^{2}$
-solution of equation (\ref{Schroed}) but also if $\psi \left( t,\cdot
\right) $ is an arbitrary representative of $\Psi _{t}$ with arbitrary $\Psi
_{0}\in L^{2}\left( \mathbb{R}\right) .$ In order to make this explicit let $
E_{x}:L^{2}\left( \mathbb{R}\right) \rightarrow L^{2}\left( \mathbb{R}
\right) $ with $x\in \mathbb{R}$ denote the spectral family of the position
operator. For the orthogonal projection $E_{x}$ holds
\[
\left( E_{x}\varphi \right) \left( y\right) =\left\{
\begin{array}{cc}
\varphi \left( y\right) & \text{for }y<x \\
0 & \text{otherwise}
\end{array}
\right. .
\]
The expectation value $\left\langle \Psi ,E_{x}\Psi \right\rangle $ of $
E_{x} $ with a unit vector $\Psi \in L^{2}\left( \mathbb{R}\right) $ thus
yields the cumulative distribution function of the position probability
given by $\Psi .$ If we define $F:\mathbb{R}^{2}\rightarrow \left[ 0,1\right]
$ through $F\left( t,x\right) =\left\langle \Psi _{t},E_{x}\Psi
_{t}\right\rangle ,$ then equation (\ref{Tranport_1}) is equivalent to
\begin{equation}
F\left( t,\gamma _{x}\left( t\right) \right) =F\left( 0,x\right) .
\label{Tranport_2}
\end{equation}
Thus, the graph $\left\{ \left( t,\gamma _{x}\left( t\right) \right)
\left\vert t\in \mathbb{R}\right. \right\} $ of a trajectory is a subset of
the level set of $F$ which contains the point $\left( 0,x\right) .$ If $\Psi
_{n}$ is a sequence in $L^{2}\left( \mathbb{R}^{s}\right) $ which converges
to $\Psi $ then
\begin{eqnarray*}
\lim_{n\rightarrow \infty }F_{n}\left( t,x\right) &=&\lim_{n\rightarrow
\infty }\left\langle \Psi _{n},e^{iht}E_{x}e^{-iht}\Psi _{n}\right\rangle
=\lim_{n\rightarrow \infty }\left\Vert E_{x}e^{-iht}\Psi _{n}\right\Vert ^{2}
\\
&=&\left\Vert E_{x}e^{-iht}\Psi \right\Vert ^{2}=F\left( t,x\right)
\end{eqnarray*}
because $e^{-iht},E_{x},$ and $\left\Vert \cdot \right\Vert ^{2}$ are
continuous mappings.
Note that for any $t\in \mathbb{R}\ $the function $F\left( t,\cdot \right) :
\mathbb{R}\rightarrow \left[ 0,1\right] $ is continuous and monotonically
increasing. Furthermore $\lim_{x\rightarrow -\infty }F\left( t,x\right) =0$
and $\lim_{x\rightarrow \infty }F\left( t,x\right) =1.$ The monotonicity is
a strict one if $\psi \left( t,\cdot \right) $ does not vanish on any
interval. Thus for any $\left( t,x\right) \in \mathbb{R}^{2}$ there exists
at least one $\gamma _{x}\left( t\right) \in \mathbb{R}$ such that equation (
\ref{Tranport_2}) holds. (For those values $t$ for which $F\left( t,\cdot
\right) $ is strictly increasing, there exists exactly one $\gamma
_{x}\left( t\right) \in \mathbb{R}$ such that equation (\ref{Tranport_2})
holds.) The function $F$ cannot be constant in an open neighbourhood of some
point $\left( t,x\right) $ if the hamiltonian is bounded from below. Thus
for any $x\in \mathbb{R},$ for which there does not exist a neighbourhood on
which $F\left( 0,\cdot \right) $ is constant, we now \emph{define} $\gamma
_{x}:\mathbb{R}\rightarrow \mathbb{R}$ to be the unique \emph{continuous}
mapping for which
\[
F\left( t,\gamma _{x}\left( t\right) \right) =F\left( 0,x\right) .
\]
Note that $\gamma _{x}:\mathbb{R}\rightarrow \mathbb{R}$ is continuous, yet
need not be differentiable.
\section{Hall's counter example}
Let us now illustrate this construction of not necessarily differentiable
Bohm\-ian trajectories by means of a solution $t\mapsto \Psi _{t}\notin
D_{h} $ of the Schr\"{o}dinger equation (\ref{SchroedQM}) describing a
particle confined to a finite interval on which the potential $V$ vanishes.
This solution has been used by Hall \cite{Hal} as a counter example to
Bohmian mechanics. Similar ones have been used in order to illustrate an
\textquotedblleft irregular\textquotedblright\ decay law $t\mapsto
\left\vert \left\langle \Psi _{0},\Psi _{t}\right\rangle \right\vert ^{2}.$
\cite{EF} Both works have made extensive use of Berry's earlier results
concerning this type of wave functions. \cite{Ber}
The (reduced) classical Schr\"{o}dinger equation corresponding to the
quantum dynamics is
\begin{equation}
i\partial _{t}\psi \left( t,x\right) =-\partial _{x}^{2}\psi \left(
t,x\right) \label{Schroed1}
\end{equation}
for all $\left( t,x\right) \in \mathbb{R}\times \left[ 0,\pi \right] $
together with the homogeneous Dirichlet boundary condition $\psi \left(
t,0\right) =\psi \left( t,\pi \right) =0$ for all $t\in \mathbb{R}.$ The
corresponding hamiltonian's domain $D_{h}$ is the set of all those $\Psi \in
L^{2}\left( 0,\pi \right) $ which have an absolutely continuous
representative $\psi $ vanishing at $0$ and $\pi $ and whose distributional
derivatives up to second order belong to $L^{2}\left( 0,\pi \right) .$ As an
initial condition we choose the equivalence class of the function
\[
\psi \left( 0,x\right) =1/\sqrt{\pi }\text{ for all }x\in \left[ 0,\pi
\right] .
\]
Since within the class $\Psi _{0}=\left[ \psi \left( 0,\cdot \right) \right]
$ there does not exist an absolutely continuous function vanishing at $0$
and $\pi $ the equivalence class $\Psi _{0}$ does not belong to $D_{h}.$ As
a consequence for any $t$ the vector $\Psi _{t}=e^{-iht}\Psi _{0}$ does not
belong to $D_{h}.$ This in turn implies that $\Psi _{t}$ does not have a
representative within the class of $C^{2}$-functions on $\left[ 0,\pi \right]
$ with vanishing boundary values.
The hamiltonian $h$ is self-adjoint. An orthonormal basis formed by
eigenvectors of $h$ is represented by the functions $u_{k}$ with
\[
u_{k}\left( x\right) =\sqrt{\frac{2}{\pi }}\sin \left( kx\right) \text{ for }
0\leq x\leq \pi \text{ and }k\in \mathbb{N}.
\]
For $n\in \mathbb{N}$ the $C^{\infty }$-function $\psi _{n}:\mathbb{R}
^{2}\rightarrow \mathbb{C}$ with
\[
\psi _{n}\left( t,x\right) =\frac{4}{\pi \sqrt{\pi }}\sum_{k=0}^{n}\frac{
e^{-i\left( 2k+1\right) ^{2}t}}{2k+1}\sin \left[ \left( 2k+1\right) x\right]
\]
is a classical solution to the Schr\"{o}dinger equation (\ref{Schroed1}) on $
\mathbb{R}^{2}$ and fulfills homogeneous Dirichlet boundary conditions at $
x=0$ and $x=\pi .$ Furthermore $\psi _{n}$ is periodic not only in $x$ but
also in $t$ with period $2\pi .$ More precisely $\psi \left( t,\cdot \right)
$ is an odd trigonometric polynomial of degree $2n+1$ for any $t\in \mathbb{R
}.$ In addition $\psi _{n}\left( t,\cdot \right) $ also is even with respect
to reflection at $\pi /2,$ i.e., it holds
\[
\psi _{n}\left( t,\frac{\pi }{2}-x\right) =\psi _{n}\left( t,\frac{\pi }{2}
+x\right)
\]
for all $x\in \mathbb{R}.$ The functions $\psi _{n}\left( \cdot ,x\right) $
are trigonometric polynomials of degree $\left( 2n+1\right) ^{2}.$
As is well known, the sequence $\left( \psi _{n}\left( 0,\cdot \right)
\right) _{n\in \mathbb{N}}$ converges pointwise on $\mathbb{R}.$ Its limit
is the odd, piecewise constant $2\pi $-periodic function $\sigma \left(
0,\cdot \right) $ with
\[
\lim_{n\rightarrow \infty }\sqrt{\pi }\psi _{n}\left( 0,x\right) =\sqrt{\pi }
\sigma \left( 0,x\right) =\left\{
\begin{array}{cc}
1 & \text{for }0<x<\pi \\
0 & \text{for }x\in \left\{ 0,\pi \right\}
\end{array}
\right. .
\]
$\sigma \left( 0,\cdot \right) $ is discontinuous at $x\in \pi \cdot \mathbb{
Z}.$ For any $t\in \mathbb{R}$ the sequence $\left( \psi _{n}\left( t,\cdot
\right) \right) _{n\in \mathbb{N}}$ converges pointwise on $\mathbb{R}$ to a
function $\psi \left( t,\cdot \right) .$ For rational $t/\pi $ this function
is piecewise constant. \cite{Ber} However for irrational $t/\pi $ the real-
and imaginary parts of $\psi \left( t,\cdot \right) $ restricted to any open
real interval have a graph with noninteger dimension. \cite{Ber} Thus for
irrational $t/\pi $ the function $\psi \left( t,\cdot \right) $ is
non-differentiable on any real interval. As an illustration we give in
Figure \ref{psix} the graph of
\[
x\mapsto \Re \sqrt{\pi }\psi _{500}\left( \frac{\pi }{\sqrt{12}},\pi
x\right)
\]
for $0<x<1/2$ together with the partial sum over $k\in \left\{ 501,\ldots
750\right\} $ visible as the small noisy signal along the abscissa
\[
x\mapsto \Re \sqrt{\pi }\left( \psi _{750}\left( \pi /\sqrt{12},\pi x\right)
-\psi _{500}\left( \pi /\sqrt{12},\pi x\right) \right) .
\]
\begin{figure}[h!]
\begin{center}
\includegraphics[scale=0.5]{plotx.png}
\caption{Real part of $\protect\psi_{500}$ at a fixed time}
\label{psix}
\end{center}
\end{figure}
Similarly, for given $x\in \left( 0,\pi \right) $ the mapping
$t\mapsto \psi
\left( t,x\right) $ does not belong to the set of piecewise $C^{1}$
-functions on $\left[ 0,2\pi \right] .$ This can be seen as follows. First
note that for given $x$ the $2\pi $-periodic function $\psi \left( \cdot
,x\right) $ has the Fourier expansion
\begin{eqnarray*}
\psi \left( t,x\right) &=&\sum_{k=1}^{\infty }c_{n}e^{-int}\text{ where} \\
c_{n} &=&\left\{
\begin{array}{ll}
\frac{4}{\pi \sqrt{\pi }}\frac{1}{2k+1}\sin \left[ \left( 2k+1\right) x
\right] & \text{for }n=\left( 2k+1\right) ^{2}\text{ with }k\in \mathbb{N}
\\
0 & \text{otherwise}
\end{array}
\right. .
\end{eqnarray*}
Assume now that $\psi \left( \cdot ,x\right) $ is piecewise $C^{1}.$ Then,
according to a well known property of Fourier coefficients, there exists a
positive real constant $C$ such that $n\left\vert c_{n}\right\vert <C$ for
all $n\in \mathbb{N}.$ This implies that
\begin{equation}
\left( 2k+1\right) \left\vert \sin \left[ \left( 2k+1\right) x\right]
\right\vert <C^{\prime }\text{ for all }k\in \mathbb{N}
\label{decayestimate}
\end{equation}
with the positive constant $C^{\prime }=\pi \sqrt{\pi }C/4.$ However, for $
x\notin \pi \cdot \mathbb{Z}$ there exists some real constant $\varepsilon
>0 $ such that the set $\left\{ k\in \mathbb{N}:\left\vert \sin \left[
\left( 2k+1\right) x\right] \right\vert >\varepsilon \right\} $ contains
infinitely many elements. Thus for $x\notin \pi \cdot \mathbb{Z}$ the
estimate (\ref{decayestimate}) cannot hold and therefore the function $
t\mapsto \psi \left( t,x\right) $ cannot be piecewise $C^{1}$ on $\left[
0,\pi \right] .$
In Figure \ref{PsiT} we plot the time dependence
\[
t\mapsto \Re \sqrt{\pi }\psi _{15}\left( \pi t,\pi /2\right) =\frac{4}{\pi }
\sum_{k=0}^{15}\frac{\left( -1\right) ^{k}}{2k+1}\cos \left( \left(
2k+1\right) ^{2}\pi t\right)
\]
for $0<t<1/2$ together with the partial sum
\[
t\mapsto \Re \sqrt{\pi }\left( \psi _{20}\left( \pi t,\pi /2\right) -\psi
_{15}\left( \pi t,\pi /2\right) \right)
\]
(noisy signal along abscissa).
\begin{figure}[h!]
\begin{center}
\includegraphics[scale=0.5]{plott.png}
\caption{Real part of $\protect\psi_{15}$ at $x=\protect\pi /2$}
\label{PsiT}
\end{center}
\end{figure}
The restriction of the limit $\sigma \left( 0,\cdot \right) $ to
$\left[ 0,\pi \right] $ represents the same $L^{2}$ element as $\psi
\left( 0,\cdot \right) $ does. Thus $\lim_{n\rightarrow \infty
}\left\Vert \left[ \widetilde{\psi _{n}}\left( 0,\cdot \right)
\right] -\Psi _{0}\right\Vert =0, $ when $\widetilde{\psi _{n}}$
denotes the restriction of $\psi _{n}$ to $\mathbb{R}\times \left[
0,\pi \right] .$ Correspondingly $\rho _{n}\left( 0,\cdot \right)
=\left\vert \psi _{n}\left( 0,\cdot \right) \right\vert ^{2}$
converges towards the density of the equipartition on $\left[ 0,\pi
\right] . $ Additionally, due to the continuity of the evolution
operator $e^{-iht},$ the sequence of equivalence classes $\left[
\widetilde{\psi _{n}}\left( t,\cdot \right) \right] \in
C_{h}^{\infty }$ approximates the $L^{2}$ vector
$\Psi _{t}=e^{-iht}\Psi _{0},$ i.e., for all $t$ holds
\[
\lim_{n\rightarrow \infty }\left\Vert \left[ \widetilde{\psi _{n}}\left(
t,\cdot \right) \right] -\Psi _{t}\right\Vert =0.
\]
Since also $E_{x}:L^{2}\left( \mathbb{R}\right) \rightarrow L^{2}\left(
\mathbb{R}\right) $ is continuous, the time dependent cumulative position
distribution function $F:\mathbb{R}\times \left[ 0,\pi \right] \rightarrow
\left[ 0,1\right] $ with $F\left( t,x\right) :=\left\langle \Psi
_{t},E_{x}\Psi _{t}\right\rangle $ obeys
\[
F\left( t,x\right) =\lim_{n\rightarrow \infty }\left\langle \left[
\widetilde{\psi _{n}}\left( t,\cdot \right) \right] ,E_{x}\left[ \widetilde{
\psi _{n}}\left( t,\cdot \right) \right] \right\rangle =\lim_{n\rightarrow
\infty }\int_{0}^{x}\left\vert \psi _{n}\left( t,y\right) \right\vert
^{2}dy.
\]
The level lines of the functions $F_{n}:\mathbb{R}\times \left[ 0,\pi \right]
\rightarrow \left[ 0,1\right] $ with $F_{n}\left( t,x\right)
:=\int_{0}^{x}\left\vert \psi _{n}\left( t,y\right) \right\vert ^{2}dy$ thus
converge to the continuous level lines of $F.$
Figure \ref{traj_app} shows some level lines of $F_{n}$ for $n=5,10,20$
starting off at equal positions at $t=0.$ The level lines inherit the period
$\pi /4$ of $F\left( \cdot ,x\right) ,$ which has this periode since the
frequencies appearing in the even function $\left\vert \psi _{n}\left( \cdot
,x\right) \right\vert ^{2}$ are $0,8,16,\ldots $
\begin{figure}[h!]
\begin{center}
\includegraphics[scale=0.5]{traj_mehrere_n5,10,20.png}
\caption{Level lines of $F_{n}$ for $n=5,10,20$}
\label{traj_app}
\end{center}
\end{figure}
Figure \ref{Traj1000} shows the case $n=1000.$ Increasing $n$ from $20$ to $
1000$ hardly alters the level lines.
\begin{figure}[h!]
\begin{center}
\includegraphics[scale=0.6]{traj_mehrere_n1000.png}
\caption{Level lines of $F_{1000}$} \label{Traj1000}
\end{center}
\end{figure}
\newpage
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TITLE: Inductive proof that $n(n-1)(n+1)$ is divisible by $6$
QUESTION [2 upvotes]: I am trying to prove that $n(n-1)(n+1)$ is divisible by $6$ for all $n$ in $\mathbb{N}$.
My attempt:
The result certainly holds for $n=0$. Suppose now that $n > 0$. Assume that $P(k)$ is true for all $k<n$. In particular $P(n-1)$ is true. Thus $(n-1)(n-2)n$ is divisible by 6. But
$n(n-1)(n+1) = (n-1)(n-2)n + 3n(n-1)$
Now I don't know how to proceed from here. It is not immediately apparent to me that the right hand side of the expression is divisible by 6.
REPLY [0 votes]: Since the OP already has an answer, here is one more approach for "fun" (in the spirit of killing a fly with a sledgehammer).
The product of the first two terms is $n^2 - n$ which is divisible by $2$, and the product of all three terms is $n^3 - n$ which is divisible by $3$, as follows in each case from Fermat's Little Theorem.
Since $\gcd(2,3) = 1$, divisibility by each implies the expression is divisible by $2 \cdot 3 = 6$. QED.
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Home » Solid State (SSD, flash key, DNA, etc.) » Lexar Multi-Card 2-in-1 USB 3.1 Reader for SD and microSD Cards
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Description
TECH TIP: You cannot use any conventional round wire 75 series recoil springs in the P-07. Conventional springs will coil bind, rendering the firearm inoperable. The factory recoil spring weight is 20#’s and only works with “hot” standard pressure 9mm ammunition. For the ideal recoil spring using “target” ammo, see our 15RS.
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Product Description
Smarty is a beautiful dolls pram in a modern color and it is suitable for dolls up to 46cm. Conveniently, the shoulder bag and shopping basket can store everything, what is needed for a trip by the doll's mum and her darling. Delivery in a beautiful showbox. Handle height (cm): 56, Product Measurements (cm): 58 x 38,5 x 55,5
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TITLE: Hilbert-irreducible Banach space
QUESTION [2 upvotes]: A Banach space $X$ is called Hilbert-irreducible if it satisfies the following condition:
If a subspace $Y\subset X$ satisfies the parallelogram equality, then $Y$ is necessarilly a one dimensional space.
Does $M_{n}(\mathbb{R})$ with operator norm satisfy this property? What is an example of an infinite dimensional Banach space with this property?
REPLY [7 votes]: In other words, a (real) Banach space $X$ is Hilbert irreducible iff it has no $2$-dimensional subspace isometric to $\mathbb R^2$ with the Euclidean norm.
In $M_n(\mathbb R)$, the subspace $Y$ consisting of matrices whose entries below the first row are $0$ satisfies the parallelogram law.
The space $\mathbb c$ of real sequences converging to $0$ with supremum norm is Hilbert irreducible. To prove this, consider two linearly independent members $x$ and $y$ of $c$. It is easy to show that there is $\epsilon > 0$ such that $\|x + t y\|$ is
an affine function of $t$ for $0 < t < \epsilon$. On the other hand, in $\mathbb R^2$ with Euclidean norm $\|(1,t)\|$ is strictly
convex.
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