Dry off this summer with a bright and fun beach towel. This towel is a necessity to have by the beach, pool, or lake this summer. It is made of cotton and has a solid panel that will make it easy to read your monogram.


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GDI, a beacon in the dark for humanity
By Xiao Qian | CHINA DAILY | Updated: 2022-12-03 09:01
This photo taken on Nov 12, 2022 shows an interior view of the media center for the 17th Group of 20 (G20) Summit in Bali, Indonesia. [Photo/Xinhua]
During the just-concluded G20 Summit and APEC Economic Leaders' Meeting, the Global Development Initiative previously proposed by Chinese President Xi Jinping offered a beacon of light to a world darkened by the COVID-19 pandemic, regional conflicts and geopolitical tensions.
Warmly welcomed, particularly by developing countries, the Global Development Initiative was first proposed by President Xi at the General Debate of the 76th Session of the United Nations General Assembly on Sept 21 last year. It calls on the world to stay committed to development as a priority and for countries to adopt a people-centered approach that is innovation-driven and results-oriented and seeks harmony between man and nature.
Aimed at building a global community of development, the Global Development Initiative seeks to expedite the implementation of the UN's 2030 Agenda for Sustainable Development. It affirms the UN system at the center of global governance and seeks collaboration with related international mechanisms.
UN Secretary-General Antonio Guterres has praised the Global Development Initiative as being of great importance to advancing the UN Sustainable Development Goals. He also stated in his meeting with Chinese State Councilor and Foreign Minister Wang Yi in September in New York that the two sides have been carrying out close and effective cooperation, and that the UN supports the initiative.
It is worth noting that China has recently called for a working mechanism to be set up at the UN level as soon as possible to advance the initiative and for related specialized UN agencies to synergize with the initiative's priority areas based on their portfolios.
China has always been a member of the big family of developing countries and has been making tremendous efforts to promote common development. The Global Development Initiative is a significant public good offered by China in this regard.
When China hosted the G20 summit in 2016, development was made a priority on the global macro policy agenda. Since then, Chinese leaders have repeatedly emphasized development in their speeches on many occasions.
During the G20 Summit in Bali, President Xi identified food and energy security as the most pressing challenges facing global development, and highlighted global trade, the digital economy, the green transition and fighting corruption as key factors driving global development. During the recent APEC Leaders' Meeting, he said it is "imperative" that development is placed at the center of the international agenda.
China is working hard to implement the Global Development Initiative, with many concrete measures announced. For example, it has upgraded the South-South Cooperation Assistance Fund to a Global Development and South-South Cooperation Fund and added $1 billion to the fund on top of the $3 billion already committed. So far this year, China has provided multiple batches of emergency food assistance to other developing countries in need, and recently pledged to provide additional emergency humanitarian assistance.
All these measures are expected to help developing countries tap their growth potential and bring new opportunities to them.
China has always believed that true development is development for all, and all countries should collaborate in, contribute to and benefit from global development.
Since the Global Development Initiative was proposed, it has been a public product that is open to the world. On separate occasions, China has expressed the hope that all countries will join the initiative. In less than one year, more than 60 countries have joined the Group of Friends of the Global Development Initiative.
China is also proactively aligning the initiative with the development strategies of relevant countries and regions. So far, the initiative has been aligned with the Pacific Roadmap for Sustainable Development and the 2050 Strategy for the Blue Pacific Continent by Pacific Island countries. African countries have also synergized the GDI with the African Union's Agenda 2063. Furthermore, the initiative is compatible with the ASEAN Community Vision 2025, according to Ong Tee Keat, chairman of the Center for New Inclusive Asia, Malaysia.
As mentioned by Chinese leaders on many occasions, China is willing to work with all parties to actively implement the Global Development Initiative to ensure that no country or individual is left behind.
With the entire world facing pronounced development problems, it is very timely for China to propose the Global Development Initiative and direct world attention to the pressing problems faced by the vast majority of people in developing countries. It is true that the initiative is in the process of being refined, having just been proposed. As State Councilor Wang Yi pointed out while speaking at the Ministerial Meeting of the Group of Friends of the Global Development Initiative in June this year, having sketched the outline, the fine details now need to be added.
The author is deputy director of the Center for International Security and Strategy, Tsinghua University. The views don't necessarily reflect those of China Daily.
The opinions expressed here are those of the writer and do not necessarily represent the views of China Daily and China Daily website.
If you have a specific expertise, or would like to share your thought about our stories, then send us your writings at opinion@chinadaily.com.cn, and comment@chinadaily.com.cn.
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Jonathan Ende
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Jonathan Ende focuses his practice on antitrust and competition, with particular strength in regulatory matters. He has a great deal of experience in internal investigations, and has a keen insight on the manufacturing and biomedical industries in particular. Read Jonathan Ende's full bio.
New Proposed CCPA Regulations Add Clarity to Process for Opting Out of Sale of Personal Information
By Jonathan Ende, Amy C. Pimentel and McDermott Will & Emery on Oct 21, 2020
Posted In Consumer Protection, Cybersecurity, Data Privacy, Data Transfers/Safe Harbor/Privacy Shield, Telehealth
On October 12, 2020, the California Department of Justice announced the release of a new, third set of proposed modifications to the California Consumer Privacy Act (CCPA) regulations. The proposed modifications amend a final set of regulations that were approved by the California Office of Administrative Law just two months earlier. The Third Set of...
Importance of CCPA Compliance Highlighted by First Round of Private Actions
By Jonathan Ende and McDermott Will & Emery on Jun 26, 2020
Posted In Consumer Protection, Data breach, Data Privacy
The first wave of California Consumer Privacy Act litigation has begun to roll in, and the complaints are already raising interesting questions about the scope of CCPA's private right of action. The actions assert a variety of claims under numerous theories and present a broad range of potential risks to businesses subject to CCPA. In...
CCPA Has Just Gone Into Effect, But Businesses May Need to Prepare for a New California Privacy Law
By Jonathan Ende on Jan 21, 2020
The California Consumer Privacy Act (CCPA) is not yet one month old, but movement has already started on a new California privacy law. In November 2019, the advocacy group Californians for Consumer Privacy, led by Alastair Mactaggart, the architect of CCPA, submitted a proposed California ballot initiative to the Office of the California Attorney General that would...
Though CCPA is Now Live, Questions About Its Constitutionality Linger
As businesses have scrambled to obtain compliance with the California Consumer Privacy Act (CCPA) in recent months, questions surrounding its constitutionality have arisen. As a broad, sometimes unclear state law that imposes significant obligations on businesses around the country, CCPA may be ripe for legal challenge. The strongest bases for such challenges appear to be:...
California Enacts a Groundbreaking New Privacy Law
By Amy C. Pimentel, Daniel F. Gottlieb, Jonathan Ende, Li Wang and Michael G. Morgan on Jun 29, 2018
Posted In Cybersecurity, Data Privacy
California's Senate and Assembly unanimously approved AB 375 (also known as the California Consumer Privacy Act of 2018), on June 28, 2018. This new consumer privacy bill will be the most progressive and comprehensive privacy law in the United States, reaching far beyond California's borders to give California consumers more visibility and control over their...


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Duff House ist ein im georgianischen Stil erbautes Schloss in der Nähe von Banff, Schottland.

Das Haus wurde von William Duff, Lord Braco, in Auftrag gegeben und unter der Leitung von William Adam errichtet. Der Grundstein dazu wurde am 11. Juni 1735 gelegt und das Hauptgebäude in seinen Grundzügen bis etwa 1740 erbaut. Jedoch zog sich der Innenausbau sowie die Fertigstellung des gesamten Ensembles bis in die 1760er Jahre hin, wobei der ursprüngliche Plan unvollendet blieb. Auch in diesem Zustand bildet Duff House den großartigsten Schlossneubau im frühen 18. Jahrhundert in Schottland.

Erbauung

Das Haus wurde für William Duff, Lord Braco und ab 1759 Earl of Fife, erbaut. Sein Vorhaben, in der Nähe der Stadt Banff/Aberdeenshire ein neues Schloss zu errichten, übergab er an William Adam, den seinerzeit führenden Architekten Schottlands. Auch wenn sich die ursprünglichen Plänen nicht erhalten haben, gibt ein Kupferstich für den Vitruvius Scoticus Aufschluss über den Entwurf, der jedoch nicht ausgeführt wurde. Dieser sah vor, den vierstöckigen Haupttrakt über zwei symmetrische Flügel mit Pavillons zu verbinden, ähnlich dem Entwurf für Hopetoun House. Das große Vermögen von Lord Braco erlaubte es ihm, einen völligen Neubau, anstatt wie ursprünglich geplant, nur den Umbau eines bestehenden Gebäudes zu veranlassen. Jedoch geriet er während des Baus in eine gerichtliche Auseinandersetzung mit seinem Architekten, die nicht nur den Bau verzögerten, sondern dazu führte, dass Lord Braco selbst nie in das Schloss einzog und als Rohbau stehen ließ.

Entwurf
Das Haus präsentiert sich in seinem Mittelblock als Quader mit einer aufwendig gestalteten, klassizistischen Fassade. Besonders auffällig sind die korinthischen Säulenreihen im Zentrum der Süd- und Nordfront. An allen vier Ecken wird das Haus von Kuppeltürmen begrenzt. Auf diese Weise erinnert es an Castle Howard und Houghton House. Der Innenausbau orientiert sich am französischen Rokoko und ist aufwändig mit geschnitzten Vertäfelungen, Pilastern, Deckenzier aus Pappmaché und Schablonenmalerei ausgestattet.

Geschichte
Nachdem Lord Braco nicht in Duff House einzog, scheint er es seinem ältesten Sohn, Lord MacDuff, übergeben zu haben, der es bis 1809 bewohnte. Die folgenden Earls of Fife lebten ebenso in Duff House, bis es der sechste Graf, der seit seiner Heirat mit Prinzessin Louise, der ältesten Tochter des späteren Edward VII., auch den Titel eines Duke of Fife trug, 1906 den Städten Banff und Macduff übergab. Seither diente es abwechselnd als Hotel, Sanatorium und Kriegsgefangenenlager. Nach dem Zweiten Weltkrieg wurde es schrittweise renoviert. Seit 1995 ist es Teil der National Galleries of Scotland. Dadurch sind im Haus wieder eine Gemäldesammlung mit Werken von El Greco, J. P. Cuyp, François Boucher und Allan Ramsays Mrs. Daniel Cunyngham sowie Möbel von Thomas Chippendale zu sehen.

Kriegsgefangenenlager und Bombardierung
Duff House diente im Zweiten Weltkrieg zeitweise als Kriegsgefangenenlager und führte die Bezeichnung "Prisoner of War Camp No5".

Am 22. Juli 1940 waren die Besatzungsangehörigen des in den westlichen Zugängen zum Englischen Kanal versenkten U-Boots U 26 praktisch einzige Insassen des Lagers. Ein Heinkel He 111 Bomber der Luftwaffe erkannte anhand der Stacheldrahtverhaue und sonstiger Sicherungsmaßnahmen eine militärische Anlage. Da sich die He 111 auf dem Rückflug von einem erfolglos verlaufenen Einsatz befand, entschloss sich der Flugzeugführer, seine vier noch an Bord befindlichen Bomben auf das Duff House zu werfen. 6 deutsche U-Boot Männer und 2 Soldaten der britischen Wachmannschaft fanden den Tod. Es handelte sich um den einzigen militärischen Zwischenfall in Großbritannien während des Zweiten Weltkriegs, bei dem Deutsche von Deutschen getötet wurden ("friendly fire").

Am 22. Juli 2018 wurde der Opfer der Bombardierung durch Enthüllung eines Gedenksteins im Park von Duff House gedacht. Im Rahmen der Andacht sprachen der stellvertretende Lord Lieutenant von Banffshire Andrew Simpson und der deutsche Generalkonsul aus Edinburgh Jens-Peter Voss.

Literatur
 Ian Gow, Timothy Clifford: Duff House. National Galleries of Scotland, Edinburgh 1995, ISBN 0903598558.
 Hugh Montgomery-Massingberd: Schlösser und Adelssitze in Schottland. Könemann, Köln 1998, ISBN 3-8290-0404-4.

Weblinks

 Duff House Website
The Friends of Duff House

Einzelnachweise

Schloss in Aberdeenshire
Erbaut in den 1740er Jahren
Georgianisches Bauwerk in Schottland
Kategorie-A-Bauwerk in Aberdeenshire
Schloss in Europa
Banff (Aberdeenshire)
Bauwerk von William Adam
Kriegsgefangenenlager (Vereinigtes Königreich)
Clan MacDuff

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ERICA BEEBE – UNIVERSITY OF MINNESOTA – Erica is on her way to start graduate school for her MFA in Documentary Filmmaking at the University of Edinburgh in London. While at the U of M, Erica was the sole videographer for the school's newspaper "the Minnesota Daily". Erica has also interned on a Documentary about Minnesota House Representative Ilhan Omar. She also worked as a freelance production assistant for "Women in Blue", a documentary women in the Minneapolis Police force.
During the 2016-2017 academic school year, I worked as the sole videographer for the University of Minnesota student newspaper, the Minnesota Daily. As videographer, I was responsible for pitching ideas that would be exciting to our readers, interviewing my subjects, filming, capturing sound, editing and sharing the finished product through the Daily's content management site. During my time at the Daily, I produced over two dozen videos that addressed a wide array of topics, from the burgeoning Minneapolis street art scene to the tumultuous political climate after President Trump's inauguration.
Over the past year, I have worked as a production intern for Time for Ilhan, a documentary about Minnesota State House Representative Ilhan Omar. As production intern, I conducted research on Ilhan's political journey, created graphics using InDesign and edited footage in Premiere Pro. I also operated a camera during sit-down interviews with Ilhan and her staff, as well as at public events such as the Women's March and Muslim Day at the Capitol.
Additionally, since last January, I have worked as a freelance production assistant for Women in Blue, a documentary series about women in the Minneapolis police force that is currently in production; a documentary about chronic alcoholism and the presence of wet-houses in Minneapolis, which is currently in the early stages of production; and finally, as a production assistant and second camera for Lady Parts Justice in the New World Order, a documentary about women's rights activist and comedian Lizz Winstead, which is currently in the final stages of production.


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We Will Make Nigerians Proud- Super Eagles Star Speaks Ahead of Match With Egypt
Kelechi Nwakali in action for Nigeria. Photo by Martin Rose Source: Getty Images
Impressive midfielder Kelechi Nwakali has confirmed his readiness to help the Super Eagles soar at the 2021 African Cup of Nations which started on Sunday, January 9, in Cameroon.
With 24 hours to the Super Eagles' first game against Egypt at the African Cup of Nations, the spirit of the Nigerian players in camp is high and ready to face the Pharaohs.
The game against Egypt is expected to be a tough encounter considering the pedigrees of these two nations and Kelechi Nwakali explained that the aim of the Super Eagles is to win.
According to the report on Complete Sports and GOAL, the former youth international stated clearly that he wants his invitation to the Super Eagles be a glorious one. Kelechi Nwakali's reaction about AFCON
"We are here to write our names in the history books, make our families and our country proud.
"If it happens that we win the competition, fine but the most important thing is to represent Nigeria in flying colours.
"That's the mindset of everybody, that's the first goal. It is football and we don't predict what will happen but we work as a team to win the competition."
Juventus come back from a two-goal deficit to defeat Jose Mourinho's Roma 4-3.
Full FA Cup Fourth-Round Fixture: Chelsea, Liverpool, and Man United Opponents Confirmed


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Reality, Health & Lifestyle
Season 13 Episode Guide
18 Episodes 2012 - 2012
Tue, Jan 3, 2012 120 mins
A new batch of couples arrive at the ranch and are shocked to learn they'll be split up to compete against each other in the Season 13 premiere. The new round has a "No Excuses" theme, and the trainers address the reasons the contestants are overweight.
Tue, Jan 10, 2012 120 mins
The teams bet on the percentage of weight they think they'll lose in order to gain an advantage at the weigh-in. Elsewhere, the contestants learn to shop for healthful foods; the teams compete in a puzzle-based challenge; and Dr. Huizenga performs sobering health assessments.
A Chinese-buffet temptation offers the winner a two-pound advantage at the weigh-in and an opportunity to match up players for a gym challenge. Later, the teams compete in a water-transfer challenge for the chance to video chat with loved ones; a player makes a shocking decision.
Tue, Jan 24, 2012 60 mins
The teams compete for exclusive use of the gym during a battering-ram challenge. Bob and Dolvett, meanwhile, try to get through to contestants they feel aren't putting in enough effort to shed pounds.
The players compete in a trampoline challenge to win the chance to exclude a contestant's weight from the weigh-in. Elsewhere, two eliminated players return for a chance to win a spot at the ranch, and personality clashes cause tension in the house.
Tue, Feb 7, 2012 120 mins
One team must work out for a week without the guidance of their trainer, who secretly monitors his players to track their progress. The losing team is determined by a dunk-tank challenge, in which Bob and Dolvett are suspended over water and must rely on the contestants to keep them dry.
Tue, Feb 14, 2012 120 mins
Two players visit home with their trainers in tow, and their weights alone count at the weigh-in. Elsewhere, the teams compete in an obstacle course in order to receive letters from loved ones for Valentine's Day; the players at the ranch must fend for themselves, though one contestant turns to nutritionist Rachel Beller for help preparing a special meal.
A temptation-challenge winner reorganizes the teams in an unpopular move that creates tension and suspicion among the players. Elsewhere, the teams race to assemble a large puzzle in order to win a special prize.
The contestants go home for 18 days and have a chance to earn immunity if they lose five percent of their body weight. While some struggle to stay on track, others benefit from the support of family and friends. Also: a home rowing challenge earns the winner $10,000.
Tue, Mar 6, 2012 120 mins
The teams switch trainers in a move designed to get them out of their comfort zones. Elsewhere, one trainer tries ballet; and the teams navigate a mud pit to locate weight plates in a messy challenge.
Tue, Mar 13, 2012 90 mins
The teams are dissolved, leaving the nine remaining players to compete on their own. Later, a tug-of-war contest finds the contestants competing for a shopping spree with Jessica Simpson; a trainer helps a player face her fear of heights; and a contestant turns the tables on a trainer.
Tue, Mar 20, 2012 120 mins
The contestants go to Hawaii to test their ability to lose weight during vacation. During the trip, the players face off in a trivia contest and a challenge involving tikis and leis. Later, surfer Bethany Hamilton gives the gang surf lessons.
Season 12 favorites Antone, Becky and Vinny return to judge a cooking contest and offer the players advice. Elsewhere, the contestants' exercise routines are spiced up with a Zumba class and mountain biking; and the players load food onto trucks during a pop challenge.
Tue, Apr 3, 2012 60 mins
The contestants receive makeovers before traveling to Washington, D.C., to meet Michelle Obama and debut their new looks to friends and family. Fashion pro Jeannie Mai and celebrity hairstylist Ken Paves assist with the transformations.
Tue, Apr 10, 2012 120 mins
Michelle Obama discusses health and fitness with the six remaining players and meets with their families. She also joins the gang for a workout at the White House. Elsewhere, Bob counsels a player who turns to food for comfort; and Dolvett leads a last-chance workout in front of the Ulysses S. Grant Memorial.
Tue, Apr 17, 2012 60 mins
Previous winners John Rhode, Olivia Ward and Patrick House return to take on a difficult obstacle course along with the players. Also: The contestant who loses the highest percentage of weight for the week wins an SUV; the trainers and players watch videos documenting their progress.
Eliminated players return to compete in a four-part challenge in order to win a spot in the final.
Tue, May 1, 2012 60 mins
The winner of the $250,000 prize is crowned in the Season 13 finale, and the eliminated players return to weigh in for the $100,000 at-home prize.


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HVDS CASE STUDY
by admin	| Posted on July 3, 2019
So, we require the economical way to provide the electrical energy by State Electricity Boards to various consumers at minimum voltage drop and reduce the regulation of voltage. The installed capacity of electricity in the State increased from MW to to Isha Bansal et al in [10] presented a paper, in which it was presented to implement a distribution system having negligible losses, high quality of supply and reliability as the substantial amount of generated power is being wasted as losses. Total load losses per annum units units 5. The present paper emphasis on the re-designing for the existing distribution network for fulfilment of the above objectives of distribution feeder. Log In Sign Up.
Now voltage drop can be estimated at various power factors and also at various temperatures of conductors. The subdivision is working on the installation of HVDS system in this feeder and calculations made above is an attempt for success of the above work made by State Electricity Board. Thus, the distribution sectors require economical system to provide electrical energy at a suitable prize and at a minimum voltage drop to reduce the voltage regulation. Installed capacity generation in Punjab IV. The total installed capacity in India is Sub-transmission and distribution systems constitute the link between electricity utilities and consumers. For this purpose, two stage methodology is used.
In 1st stage, the power losses including the theft and losses in the line and transformer losses for both LT and HT systems are determined and then in 2 nd stage, determination of annual savings and payback period is carried out alongwith the complete comparison of LVDS and HVDS system of 11kV feeder.
Estimation of Current at different power studdy On the basis of the current estimation at reference power factor of 0.
Total iron losses per annum units units 4. The installation of HVDS system in considerable area of the sub-division, is the main technique which is also applied in the present work to evaluate the proposed re-designing of existing distribution network and its future planning.
Thus required calculation of HT transformer losses are as under in table no. The results are satisfactorily obtained in the above case study, theoretically. The installed capacity of electricity in the State increased from MW to to The subdivision is working on the installation of HVDS system in this feeder and calculations made above is an attempt for success of the above work made by State Electricity Board.
HOMEWORK 4.4 MONGODB
Case Study: High Voltage Distribution System (HVDS) Implementation in BESCOM and MGVCL
Thus, the net power loss per stusy can be calculated as considering working for hours of a day with days per annum. From the result, it is also realised that the causes of voltage drop on the feeder was mainly due to high impedance level as compared to the permissible value and this high impedance is caused by poor jointing and terminations, use of undersized conductors and different types of conductor materials etc.
Thus, this sector especially, the distribution sectors require economical system to provide electrical energy at a suitable prize and at a minimum voltage drop to reduce the voltage regulation. The present paper emphasis on the re-designing for the existing distribution network for fulfilment of the above objectives of distribution feeder.
Now voltage drop can be estimated at various power factors and also at uvds temperatures of conductors. The performance of the feeder may be improved by the HVDS system installation, which results in saving of total losses per annum and an annual savings of Calculations of power losses of studt Ganga Nagar Road Feeder The size of conductor used in the 11kV feeder, which is 30 and 20mm 2 and the resistance for this particular conductor is 0.
So, we require the economical way to provide the electrical energy by State Electricity Boards to various consumers at minimum voltage drop and reduce the regulation of voltage.
LT Transformer losses As per the information derived from the 33kV State Electricity board of subdivision, on LT side, a large number of transformer of capacity of kVA is used to supply the power to the consumers at the end point of each section.
Sub- transmission and distribution systems constitute the link between electricity utilities and consumers. This results in improving the stability as well as energy handling capacity of the system at minimum cost. Total power losses per annum Help Center Find new research papers in: Click here to sign up. This is due to, in most cases, voltage drops is a major concern in low voltage distribution systems and not very particular about voltage drop in the high voltage sides leaving it unattended.
TERM PAPER INTRODUCTION TAGALOG
Calculative Analysis of 11kV Distribution Feeder. Soloman Nunoo et al in [4] presented a paper analysing the causes and effects of voltage drop on hvss 11KV GMC sub- transmission feeder in Tarkwa, Ghana. Estimation of Current in feeder lines On the basis of the data of feeder, the calculation of current flowing through the feeder lines can be calculated as under in table 4.
So, there will be voltage drop, line losses and system reliability comes into the act. For this purpose, two stage methodology is used. Thus, the distribution sectors require economical system to provide electrical energy at a suitable prize and at a minimum voltage drop to reduce the voltage regulation. Sarang Pande et al hvfs [9] presented a paper, in which a method for energy losses calculation is presented.
The total installed capacity in India is Studdy paper demonstrates the capability of load factor and load loss factor to calculate the power losses of the network. Installed capacity generation in Punjab IV.
Devi Lal State Institute of Engg. At power factor Unity and at various temperatures On the basis of the above parameters, the voltage drop calculations had been estimated in table no.
As the distribution network is located far away from the sources of power generation and the other infrastructure of electrical power system. Thus core and copper losses occurred in the transformer also contributed to the total power losses per annum in LT system So, for kVA rating transformer, the fixed value of no-load losses and full load losses are W and W respectively.
HIGH PSRR LDO THESIS
EDEXCEL RESISTIVITY CASE STUDY
DELL VWORKSPACE CASE STUDY
PTDF RESEARCH PROPOSAL
UPFC MODELLING THESIS
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Series Editors: Beller, M., Dixneuf, P.H., Dupont, J., Fürstner, A., Glorius, F., Gooßen, L.J., Nolan, S.P., Okuda, J., Oro, L.A., Willis, M., Zhou, Q.-L.
The individual volumes of Topics in Organometallic Chemistry are thematic. Review articles are generally invited by the volume editors. All chapters from Topics in Organometallic Chemistry are published OnlineFirst with an individual DOI. In references, Topics in Organometallic Chemistry is abbreviated as Top Organomet Chem and cited as a journal.
Get the table of contents of every new volume published in Topics in Organometallic Chemistry.


Find the Land Rover Discovery Sport for sale at Land Rover Chandler. This SUV provides your drive with a plethora of performance features that will keep you safe on and off the beaten path. There's also an abundance of safety features to give you peace of mind on the Gilbert roads. Our sales team is standing by to get you behind the wheel of the model that best suits your budget and driving needs, and we always have rotating vehicle specials to make the car-buying process that much less stressful.
When you're ready to take your Gilbert commute to the next level, stop by Land Rover Chandler to pursue the Land Rover models for sale in Chandler, AZ! Contact us today to see how the Discovery Sport stands up to competition like the Acura MDX and BMW X5.


The end objectives in working with Mindy include In the Now Moment learning (some say AH-HA's) and growth to move toward creating and fulfilling one's life purpose, dreams, and goals.
Mindy's individual sessions are a combination of her gifts as consultant and coach – the true Edge Runner that she is! Although possibilities are limitless, Mindy works with her clients exactly where they are, pointing them inward.
Mindy shares human experiences and concepts to assist the client in moving from where they are to where they desire to be. Mindy's work is In The Now Moment, and unlike a traditional therapist or counselor she does not encourage dragging up old baggage and skeletons in the subconscious mind. Mindy literally coaches her clients to create the life he/she desires. She coaches In The Now and helps the client to shift old thought patterns, taking them to their Edge of discovering new ways of living and being in a higher reality. She is highly skilled in seeing what others may not, and she capitalizes on these gifts in helping those she serves.
Some people experience profound changes in perception that immediately get them on track with only one session, while others prefer to work individually with Mindy over time. Only the client can determine what works best for them.
Contact Mindy for a free 15 minute consultation. Then YOU decide if she is a good fit for working with YOU!
All too often life gets in the way, and we need an opportunity to unplug from work, routines, Facebook, email, texts, family members, and the world news to take a look at a picture much bigger than we realize we are a part of. During the individual intensive you will learn about some discoveries that can change everything about your life. This is a learning, personal growth and development 'time-out' to give you a firm grasp of the philosophies and principles Mindy has learned through her experiences; shares and teaches to others – – which can be life altering.
During the minimum three days you will have one-on-one sessions with Mindy twice each day. The sessions average 2 hours each usually one in the morning and one in afternoon. In Mindy's In The Now Moment approach, if the situation presents to proceed through all 4 hours in one setting, then with the client's agreement, all 4 hours are completed, and the remainder of client's day is free for contemplation and directed exploration of the day's learnings. There will be breaks during the sessions for reflection, review of supporting materials or exercises for practice of new teachings. The sessions consist of discussion and focused learning that is specifically oriented to the client's life and current situation.
Contact Mindy to discuss customizing additional number of days or number of people involved.
Client is responsible for all travel, accommodation, and meal arrangements in Charlotte.
Mindy has clients across the United States and around the globe. With her professional background, cross cultural experience and gifts, under certain circumstances, Mindy will come to you – domestically or internationally. She views the world as her home and the countries of the globe as rooms in her house. The guidelines and teachings are exactly the same as if the client visited Charlotte, NC. Internationally, depending on location, language translation requirements, among cost of living indices Mindy's minimum intensive charges vary.
For serious inquiries contact Mindy to discuss possible arrangements, fees, number of persons, and requirements for this custom designed service.
Please DO NOT pay for a consultation until it is confirmed with Mindy or her office.
Contact Mindy by email for information on Intensives. Intensives are designed specifically for Client and his/her needs and can be held in Charlotte or another location.
Additional customized packages are available. Please contact Mindy to discuss.
Client is responsible for accommodation, meals, and travel to Charlotte, NC for Intensives held there.


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PlayStation 4Xbox One
Red Dead Online Beta Rollout Begins On November 27, 2018
By Jenni Lada November 26, 2018
People looking to experience the Red Dead Online beta may want to save time this week. Rockstar has announced a staggered rollout that will give certain people priority access, depending on what version they own and when they last played. The earliest someone can play is November 27, 2018, with general access opening on November 30, 2018.
To make things easier to understand, here is a list going over when people will be able to play Red Dead Online.
November 27, 2018: Red Dead Redemption Ultimate Edition owners can begin playing through the online mode.
November 28, 2018: People who played Red Dead Redemption 2 on October 26, 2018, launch day, can begin going online.
November 29, 2018: Anyone who played Red Dead Redemption 2 between October 26-29, 2018 will have access to Red Dead Online.
November 30, 2018: Everyone who owns Red Dead Redemption 2 can play Red Dead Online.
In Red Dead Online, people create a custom character and play online alone or with friends, exploring areas they have seen in Red Dead Redemption 2. Up to seven people can form a group together and undertake activities.
Red Dead Redemption 2 is available for the PlayStation 4 and Xbox One.


If you had tried to access this blog over the last few days, you would have experienced very slow loading time or website being inaccessible. That was because, my trusted web-host of 9 years suddenly seems to have made some major changes on their servers and all my websites were loading very slow. One was totally down.
I tried fixing the issue with their tech support, but they were not able to help me. Previously, for the most part, either I didn't have to contact the support or my issue was solved quickly over chat. It never escalated beyond that. But this time, even after a couple of days, there was no solution and my websites were losing traffic quickly.
I did some research on alternative hosting platforms, and I finally zeroed-in on Amazon Web Services Lightsail Cloud VPS hosting. I have taken three separate VPS instances on their cloud – one for each site. As this service is being offered by the cloud computing pioneer – Amazon, and they had very good reviews all over the net, I decided to shift my websites to AWS.
But the process of shifting was not easy – it took my 3.5 days to shift my three sites, including the one that was down. Somehow I managed to finish it today. A few more tasks are pending, though.
I am glad that I decided to go through this difficult technical process (esp. for a non-techie like me) – I was able to learn many new things about servers and hosting.
BTW, if you see any issues from your end – like slow loading, images not being displayed, commenting difficulties, etc. let me know. I have not tested this site much in the new environment. Also tell me if you find this website loading faster now.


text:
\section{Introduction}
\label{secINT}

There have been a number of discrete analogs of the curvature of smooth surfaces and manifolds.  The oldest such analog is the angle defect (also known as the angle deficiency) at a vertex $v$ of a triangulated polyhedral surface $M$, which is given by $\kdk vM = 2\pi - \sum_{\alpha \ni v} \alpha$, where the $\alpha$ are the angles at $v$ of the triangles containing $v$.  This curvature function goes back at least as far as Descartes (see \cite{FE}).  The angle defect satisfies various properties one would expect a curvature function to satisfy, notably an analog of the Gauss-Bonnet Theorem, which is that $\sum_{v \in M} \kdk vM = 2\pi \chi(M)$, where the summation is over all the vertices of $M$, and where $\chi(M)$ is the Euler characteristic of $M$.

The angle defect, and related constructs involving sums of angles in polyhedra, have been widely studied in dimension $2$ and higher, both for convex polytopes, for example in \cite{SH2} and \cite{GR2}, and more generally, for example in \cite{BA1}, \cite{C-M-S}, \cite{G-S2} and \cite{BL4}.

The above discrete analogs of smooth curvature are geometric in nature, making use of interior or exterior angles in simplices and polyhedral cells.  It would be interesting to know if there is a purely combinatorial definition of curvature, making use of only the face relations of the cells in the complex, that satisfies properties analogous to smooth curvature.  We consider here the combinatorial analog of curvature, called combinatorial Ricci curvature, which is defined in \cite{FORM6}.  (There are other definitions of discrete curvature that make use of the term ``Ricci'' in their names, such as the ``simplicial Ricci tensor'' of \cite{A-M-M} and the ``discrete Ricci curvature'' of \cite{FRIT}, but these approaches are very different from \cite{FORM6}).  An earlier approach that is more comparable to combinatorial Ricci curvature\ is the combinatorial analog of curvature given in \cite{STON1} and \cite{STON2}, and we discuss the latter briefly in Section~\ref{secDSTO}; this same approach, though with a factor of $\frac 12$, was used many years later in the context of graphs by \cite{HIGU}, and by a number of papers that refer to that one, perhaps unaware that the formula had appeared in \cite{STON1}.

Some, though unfortunately not all, of the properties of smooth curvature have analogs for combinatorial Ricci curvature.  An analog of Myers' Theorem for combinatorial Ricci curvature\ is given in Theorem~6.1 of \cite{FORM6}, which says that if $K$ is an appropriately nicely behaved CW complex, and if the combinatorial Ricci curvature\ is positive at every edge in $K$, then $\pi_1 (K)$ is finite; see \cite{MYER} for the original smooth version of this theorem.  Theorem~7.2 of \cite{FORM6} shows that for any simplicial complex $K$ of dimension at least $2$ that is a combinatorial manifold, there is a subdivision $M$ of $K$ such that $\ricc e < 0$ for every edge $e$ of $M$; see \cite{GAO1}, \cite{GA-YA}, \cite{BROO1}, \cite{LOHK1} and \cite{LOHK2} for the smooth analog of that result for Ricci curvature of Riemannian manifolds in dimensions $3$ or higher.

The analogy between combinatorial Ricci curvature\ and smooth curvature breaks down in dimension $2$, because there can be no smooth analog of Theorem~7.2 of \cite{FORM6} in dimension $2$, due to the Gauss-Bonnet Theorem for smooth surfaces, which implies that a smooth surface with non-negative Euler characteristic cannot have curvature that is everywhere negative.  Combinatorial Ricci curvature, therefore, does not satisfy an analog of the Gauss-Bonnet Theorem, and that again shows that combinatorial Ricci curvature\ is not entirely analogous to smooth curvature.

The purpose of this note is to offer a way to resolve this anomaly of combinatorial Ricci curvature\ in  dimension $2$.  Specifically, we describe a way to extend combinatorial Ricci curvature\ to all cells of a \nd{2}dimensional polyhedral complex, and we  prove an analog of the Gauss-Bonnet Theorem for this extended curvature.  We also provide an answer to the question of everywhere negative curvature in the orientable case.

Combinatorial Ricci curvature\ is the \nd{1}dimensional case of a more general definition of curvature in \cite{FORM6} that applies to cells in all dimensions, and which is defined as follows.  Let $\alpha$ and $\eta$ be \ncells p\ in a CW complex.  The \ncells p\ $\alpha$ and $\eta$ are called parallel neighbors if they are either both the faces of a common \ncell {(p+1)}, or they both have a common \nface {(p-1)}, but not both.  The curvature at $\alpha$ is then defined, in the notation of \cite{FORM6}, by
\begin{equation}\label{eqBBC}
\#\{\text{\ncells {(p+1)} $\beta > \alpha$}\} + \#\{\text{\ncells {(p-1)} $\gamma < \alpha$}\} - \#\{\text{parallel neighbors of $\alpha$}\},
\end{equation}
where $<$ denotes the relation of being a face.  (In \cite{FORM6} the above definition is also given with weights on the cells, though we do not do that here.)  Combinatorial Ricci curvature\ is the special case of this curvature when $p = 1$, and is denoted $\ricc e$ for every edge $e$ of the CW complex.

We will restrict our attention to \nd{2}dimensional cell complexes.  From the point of view of combinatorial Ricci curvature\ and the fundamental group, restricting to dimension $2$ is no loss, because both are computed entirely in the \nd{2}skeleton of a cell complex.

Because Equation~(\ref{eqBBC}) uses only the face relations of the cells of a CW complex, it is  more clear, and slightly more general, to formulate our discussion in the context of posets, which we do in Section~\ref{secDRC}; in Section~\ref{secSUR} we will return to  \nd{2}dimensional polyhedral complexes.

Although the definition of curvature in \cite{FORM6} carries over directly to ranked posets, as long as every element covers, and is covered by, finitely many elements, the approach we take here uses a slight variant of that definition, where we replace the number of parallel neighbors with the more convenient set of all neighbors (which means that non-parallel neighbors are double counted).

\section{Discrete Curvature on Ranked Posets}
\label{secDRC}

We assume that the reader is familiar with basic properties of posets.  See \cite{STAN}*{Chapter~3} for details.  Let $P$ be a poset.  We let $<$ denote the partial order relation on $P$, and we write $a \prec b$ if $b$ covers $a$.  A function $\func {\rho}{P}{\{0, 1, \ldots, r\}}$ for some $r \in \nn \cup \{0\}$ is a rank function\ for $P$ if it satisfies the following conditions: for $a, b \in P$, if $a$ is a minimal element then $\rho (a) = 0$, and if $a \prec b$ then $\rho (a) + 1 = \rho (b)$.    A poset is ranked\ if it has a rank function.  If a poset has a rank function, the rank function\ is unique.
The rank of such a poset is the smallest possible $r$.  If $P$ is ranked\ and has rank $r$, and if $i \in \{0, 1, \ldots, r\}$, let $\rkiset Pi = \{x \in P \mid \rho (x) = i\}$ and $\rkinum Pi = \setcar {\rkiset Pi}$.

We note that a ranked\ poset need not be graded, using the terminology of \cite{STAN}*{Chapter~3}; the term ``ranked'' is used in \cite{ROS00}*{Section~11.1.3}, though we use the definition of a rank function\ given in \cite{STAN}*{Chapter~3}.

Let $P$ be a poset.  The order complex of $P$, denoted $\ordcom P$, is the simplicial complex with a vertex for each element of $P$, and a simplex for each non-empty chain of elements of $P$.  It is a standard fact that this construction yields a simplicial complex.
Suppose that $P$ is finite.  The Euler characteristic of $P$, denoted $\eulp P$ is defined by $\eulp P = \chi(\ordcom P)$.

The reason to define $\eulp P$ as $\chi(\ordcom P)$ is that a finite poset in general has no natural rank function, in contrast to the simplicial complex $\ordcom P$, which is naturally ranked by the dimensions of the simplices, and this natural rank function is what allows the Euler characteristic of simplicial complexes to be defined.

Suppose, however, that $P$ is a finite ranked\ poset.  Then there is a more direct approach to defining the Euler characteristic of $P$, as given in the following definition.

\begin{definition}\label{defnADQ}
Let $P$ be a finite ranked poset of rank $r$.  The \textbf{ranked Euler characteristic} of $P$ is the number $\eulgr P = \sum_{i=0}^r (-1)^i \rkinum Pi$.
\qef\end{definition}

If $P$ is the face poset of a finite regular CW complex, or in particular a polyhedral complex, or simplicial complex, then $\eulgr P = \eulp P$.  In general, however, it is not the case that $\eulgr P$ and $\eulp P$ are equal.

We also need the following definition regarding posets.

\begin{definition}\label{defnBCA}
Let $P$ be a poset.  The poset $P$ is \textbf{covering-finite} if for any $a \in P$, there are finitely many $b \in P$ such that $a \prec b$, and there are finitely many $c \in P$ such that $c \prec a$.
\qef\end{definition}

Throughout this section, let $P$ be a covering-finite\ ranked poset of rank $r$.  Let $i \in \{0, 1, \ldots, r\}$, and $x \in \rkiset Pi$.
Let
\begin{gather*}
A_i(x) = \setcar {\{y \in \rkiset P{i+1} \mid x \prec y\}}, \quad B_i(x) = \setcar {\{z \in \rkiset P{i-1} \mid z \prec x\}}\\
U_i(x) = \sum_{y \succ x} B_{i+1}(y), \quad D_i(x) = \sum_{z \prec x} A_{i-1}(z),
\end{gather*}
and
\begin{align*}
N_i(x) = \osetcar \{\{w \in \rkiset Pi &\mid x \prec v \text{ and } w \prec v \text{ for some } v \in \rkiset P{i+1}\}\\
 &\mathrel{\bigtriangleup} \{w \in \rkiset Pi \mid u \prec x \text{ and } u \prec w \text{ for some } u \in \rkiset P{i-1}\}\csetcar,
\end{align*}
where $\bigtriangleup$ denotes symmetric difference, and summation over the empty set is taken to be zero.

The reader can verify the following equalities:
\begin{align}
\sum_{x \in \rkiset Pi} A_i(x) &= \sum_{y \in \rkiset P{i+1}} B_{i+1}(y)\label{eqBAB1}\\
\sum_{x \in \rkiset Pi} U_i(x)
 &= \sum_{y \in \rkiset P{i+1}} [B_{i+1}(y)]^2\label{eqBAB2}\\
\sum_{x \in \rkiset Pi} D_i(x)
 &= \sum_{z \in \rkiset P{i-1}} [A_{i-1}(z)]^2.\label{eqBAB3}
\end{align}

Equation~(\ref{eqBBC}) in the case $p = 1$, which defines combinatorial Ricci curvature, can be rewritten as
\begin{equation}\label{eqRIC}
\ricc e = A_1(e) + B_1(e) - N_1(e)
\end{equation}
for all $e \in \rkiset P1$.

We now define our discrete curvature\ functions on covering-finite\ ranked posets of rank $2$.

\begin{definition}\label{defnBAY}
let $P$ be a covering-finite\ ranked poset of rank $2$.  For each $i \in \{0, 1, 2\}$, let $\func {R_i}{\rkiset Pi}{\mathbb{R}}$ be defined by
\begin{align*}
R_0(v) &= 1 + \frac 32A_0(v) - [A_0(v)]^2,
\\
R_1(e) &= 1 + 6A_1(e) + \frac 32B_1(e) - U_1(e) - D_1(e),
\\
R_2(\sigma) &= 1 + 6B_2(\sigma) - [B_2(\sigma)]^2,
\end{align*}
for all $v \in \rkiset P0$ and $e \in \rkiset P1$ and $\sigma \in \rkiset P2$.
\qef\end{definition}

We will see in Lemma~\ref{lemBEBX} that for certain posets, including the face posets of all \nd{2}dimensional polyhedral complexes, the function $R_1$ equals $\ricci$.

The choice of coefficients in Definition~\ref{defnBAY}, particularly $6$ and $\frac 32$, may seem unmotivated.  They were chosen simply because they relate properly to combinatorial Ricci curvature.  A slight variation in the coefficients can also be used, but the above choice appears to be the simplest possible.

The following analog of the Gauss-Bonnet Theorem is a trivial consequence of Definition~\ref{defnBAY} and Equations~(\ref{eqBAB1})--(\ref{eqBAB3}); the details are left to the reader.

\begin{theorem}\label{thmBAK}
Let $P$ be a finite ranked poset of rank $2$.  Then
\[
\sum_{v \in \rkiset P0} R_0(v) - \sum_{e \in \rkiset P1} R_1(e) + \sum_{\sigma \in \rkiset P2} R_2 (\sigma) = \eulgr P.
\]
\end{theorem}

We now turn to a less trivial result, which says something about the nature of the poset $P$ in the case that the $R_1$ is everywhere positive, somewhat analogously to Theorem~6.1 of \cite{FORM6}.  Our result is weaker than that theorem, due to the fact that not all posets are as nicely behaved as the face posets of CW complexes.

In the case of Gaussian curvature of compact smooth surfaces, the classical Gauss-Bonnet Theorem implies that if the curvature is everywhere positive, then the Euler characteristic of the surface is positive.  A similar result holds for the polyhedral curvature defined in \cite{BA1}, and for the combinatorial approach of \cite{STON1} and \cite{STON2} (see Section~\ref{secDSTO} for a brief discussion of that approach).  Unfortunately, because of the negative coefficient for the $R_1$ terms in Theorem~\ref{thmBAK}, it is not possible to deduce from this version of the Gauss-Bonnet Theorem that if each of $R_0$, $R_1$ and $R_2$ are everywhere positive, then the Euler characteristic is positive.  It turns out, as we now see, that it is nonetheless true that positive $R_1$ implies that the ranked Euler characteristic\ is positive.  In fact, all that is required is that the average value of $R_1$ is positive (in contrast to Theorem~6.1 of \cite{FORM6}, which requires $\ricci$ to be positive everywhere).

We start with a definition.

\begin{definition}\label{defnBAV}
Let $P$ be a finite ranked poset of rank $2$.  Let
\[
\bari R1 = \frac 1{\rkinum P1} \sum_{e \in \rkiset P1} R_1(e), \qquad \bari A1 = \frac 1{\rkinum P1} \sum_{e \in \rkiset P1} A_1(e), \qquad \bari B1 = \frac 1{\rkinum P1} \sum_{e \in \rkiset P1} B_i(e).
\]
The poset $P$ is \textbf{sufficiently covered} if
\begin{equation}\label{eqBEE}
[\bari A1 + \bari B1]^2 - 6\bari A1 - \frac 32\bari B1 - 1 \ge 0.\tag*{\qefsymbol}
\end{equation}
\end{definition}

\begin{remark}\label{remkCCC}
It is straightforward to verify that if $P$ is a finite ranked poset of rank $2$ and if $\bari B1 = 2$, then $P$ is sufficiently covered\ if and only if $\bari A1 \ge 2$;
in particular, that would hold for the face poset of any simplicial surface, and more generally for any polyhedral map of a surface (as defined in Section 21.1 of \cite{BR-SC}).  If $\bari B1 \ge \frac {20}9$, then $P$ is sufficiently covered\ regardless of the value of $\bari A1$.
\qex\end{remark}

In the following theorem, our analog of Theorem~6.1 of \cite{FORM6}, we restrict our attention to sufficiently covered\ posets.  After the theorem, we will give a simple example that shows the necessity of some such restriction.  A relation between similar average values and the topology of \nd{3}manifolds is discussed in \cite{L-S}, so it is not surprising that such averages are used here as well; it is not clear whether the results of \cite{L-S} are related to our approach.  In \cite{L-S} the averages take place in simplicial complexes, so $\bari B1 = 2$, and therefore only $\bari A1$ is considered.  Moreover, while there is a very simple formula for $\bari A1$ in a simplicial complex, as used in \cite{L-S}, that is not the case for posets that are not the face posets of simplicial complexes.

\begin{theorem}\label{thmBAL}
Let $P$ be a finite sufficiently covered\ ranked poset of rank $2$.  If $\bari R1 > 0$, then $\eulgr P > 0$.
\end{theorem}

\begin{proof}
Suppose that $\bari R1 > 0$.  By Equations~(\ref{eqBAB2}) and (\ref{eqBAB3}), we see that
\begin{align*}
\frac 12 \bari R1\rkinum P1 < \bari R1\rkinum P1 &= \sum_{e \in \rkiset P1} R_1(e) = \sum_{e \in \rkiset P1} \bigl[1 + 6A_1(e) + \frac 32B_1(e) - U_1(e) - D_1(e)\bigr]\\
 &=  \rkinum P1 + 6\rkinum P1 \bari A1 + \frac 32\rkinum P1\bari B1 - \sum_{\sigma \in \rkiset P2} [B_2(\sigma)]^2 - \sum_{v \in \rkiset P0} [A_0(v)]^2,
\end{align*}
and hence
\[
\sum_{v \in \rkiset P0} [A_0(v)]^2 + \sum_{\sigma \in \rkiset P2} [B_2(\sigma)]^2 < \rkinum P1\bigl[1 - \frac 12 \bari R1 + 6\bari A1 + \frac 32\bari B1\bigr].
\]
Let $d = 1 - \frac 12 \bari R1 + 6\bari A1 + \frac 32\bari B1$.  The inequality
\[
\frac 1n \left[\sum_{i=1}^n a_i\right]^2 \le \sum_{i=1}^n \left(a_i\right)^2
\]
implies that
\[
\frac 1{\rkinum P0}\left[\sum_{v \in \rkiset P0} A_0(v)\right]^2 + \frac 1{\rkinum P2}\left[\sum_{\sigma \in \rkiset P2} B_2(\sigma)\right]^2 < \rkinum P1d.
\]
By Equation~(\ref{eqBAB1}) we deduce that
\[
\frac 1{\rkinum P0}\left[\sum_{e \in \rkiset P1} B_1(e)\right]^2 + \frac 1{\rkinum P2}\left[\sum_{e \in \rkiset P1} A_1(e)\right]^2 < \rkinum P1d.
\]
Hence
\[
\frac 1{\rkinum P0}[\rkinum P1 \bari B1]^2 + \frac 1{\rkinum P2}[\rkinum P1 \bari A1]^2 < \rkinum P1d,
\]
and therefore
\[
[\bari B1]^2\frac {\rkinum P1}{\rkinum P0} + [\bari A1]^2\frac {\rkinum P1}{\rkinum P2} < d.
\]
Let $a = [\bari A1]^2$ and $b = [\bari B1]^2$.  Then
\begin{equation}\label{eqBAC}
b\frac {\rkinum P1}{\rkinum P0} + a\frac {\rkinum P1}{\rkinum P2} < d.
\end{equation}

Because $P$ is a ranked poset of rank $2$, it follows that $\bari A1 \ne 0$ and $\bari B1 \ne 0$.  Hence $a, b > 0$.  Because $b\frac {\rkinum P1}{\rkinum P0} + a\frac {\rkinum P1}{\rkinum P2} > 0$, then $d > 0$.

By hypothesis we know that $P$ is sufficiently covered\ and $\frac 12 \bari Ri > 0$, and hence
\[
\left[\bari B1 + \bari A1\right]^2 \ge 1 + 6\bari A1 + \frac 32\bari B1 > 1 - \frac 12 \bari R1 + 6\bari A1 + \frac 32\bari B1,
\]
which is the same as $(\sqrt a + \sqrt b)^2 > d$.

Let
\[
U = \{(x, y) \in \rrr{2} \mid x > 0 \text{ and } y > 0 \text{ and } ax + by < d\},
\]
and let $\func f{\rrr{2}}{\mathbb{R}}$ be defined by $f(x, y) = (x - 1)(y - 1)$ for all $(x, y) \in \rrr{2}$.  The level curve for $f$ with value $c = 1$ is the hyperbola $y = \frac 1{x - 1} + 1$,
and the function $f$ has value less than $1$ between the branches of this hyperbola.
Using the condition $(\sqrt a + \sqrt b)^2 > d$,
the reader can verify that the set $U$ is between the two branches of the hyperbola (find the point on the upper branch of the hyperbola at which the tangent line is parallel to the line $ax + by = d$).  It follows that $f(x, y) < 1$ for all $(x, y) \in U$.

Combining the fact that $\frac {\rkinum P1}{\rkinum P0} > 0$ and $\frac {\rkinum P1}{\rkinum P2} > 0$ with Equation~(\ref{eqBAC}), we know that $(\frac {\rkinum P1}{\rkinum P0}, \frac {\rkinum P1}{\rkinum P2}) \in U$, and therefore $(\frac {\rkinum P1}{\rkinum P0} - 1)(\frac {\rkinum P1}{\rkinum P2} - 1) < 1$.  On the other hand, we see that
\begin{align*}
\left(\frac {\rkinum P1}{\rkinum P0} - 1\right) \left(\frac {\rkinum P1}{\rkinum P2} - 1\right) &= \frac {\rkinum P1 - \rkinum P0}{\rkinum P0} \cdot \frac {\rkinum P1 - \rkinum P2}{\rkinum P2}\\
 &= \frac {\rkinum P2 - \eulgr P}{\rkinum P0} \cdot \frac {\rkinum P0 - \eulgr P}{\rkinum P2}\\
 %
 &= \frac 1{\rkinum P0\rkinum P2}[\eulgr P]^2 - \frac {\rkinum P0 + \rkinum P2}{\rkinum P0\rkinum P2}\eulgr P + 1.
\end{align*}
It now follows that
\[
\frac 1{\rkinum P0\rkinum P2}[\eulgr P]^2 - \frac {\rkinum P0 + \rkinum P2}{\rkinum P0\rkinum P2}\eulgr P + 1 < 1,
\]
and hence
\[
[\eulgr P]^2 - (\rkinum P0 + \rkinum P2)\eulgr P < 0.
\]
Because $x^2 - (\rkinum P0 + \rkinum P2)x < 0$ if and only if $0 < x < \rkinum P0 + \rkinum P2$, we conclude that $\eulgr P > 0$.
\end{proof}

The hypothesis in Theorem~\ref{thmBAL} that $P$ is sufficiently covered\ cannot be dropped.  Let $P$ be the poset shown in Figure~\ref{figDIRIr}.  Then $R_1(e) = \frac 52$, and so $\bari R1 = \frac 52$, and yet $\eulgr P = -1$.  Hence, some hypothesis on $P$ is required for the theorem to hold.

\begin{figure}[ht]
\centering\includegraphics{figDIRIr.pdf}
\caption{}\label{figDIRIr}
\end{figure}

The analog of Myers' Theorem in Theorem~6.1 of \cite{FORM6} has a much stronger conclusion than our Theorem~\ref{thmBAL}.  Specifically, the main part of the proof of Theorem~6.1 of \cite{FORM6} consists of proving that if an appropriately nicely behaved connected CW complex has everywhere positive combinatorial Ricci curvature\ that is bounded away from zero, then the CW complex is bounded, in the sense that there is an upper bound on the lengths of paths between vertices.  If the CW complex is regular and covering-finite, that would imply that the CW complex is finite.  Unfortunately, the analogous result does not hold when $R_1$ is everywhere positive and bounded away from zero on a covering-finite\ ranked poset of rank $2$, or even when all three of $R_0$, $R_1$ and $R_2$ are everywhere positive and bounded away from zero, as seen in the following example.  Let $Q$ be the poset shown in Figure~\ref{figDIRIm}, where the pattern repeats infinitely.  Let $v, e, x \in P$ be the elements shown in the figure.  Then $R_0(v) = \frac 32$, and $R_1(e) = 4$ and $R_2(x) = 9$, and yet $Q$ is infinite.  We note that $\ricci$ works no better for this poset, because $\ricc e = 2$.

\begin{figure}[ht]
\centering\includegraphics{figDIRIm.pdf}
\caption{}\label{figDIRIm}
\end{figure}

Finally, we note that whereas in the special case of compact smooth surfaces, the Gauss-Bonnet Theorem immediately implies that if the average curvature is positive then so is the Euler characteristic, it is not so simple in our present context of finite ranked posets of rank $2$, because our analog of the Gauss-Bonnet Theorem, Theorem~\ref{thmBAK}, makes use of $R_0$, $R_1$ and $R_2$, whereas Theorem~\ref{thmBAL} uses only $R_1$, and hence the latter theorem does not follow from the former.

\section{$2$-Dimensional Polyhedral Complexes}
\label{secSUR}

We now relate our curvature to the problem with combinatorial Ricci curvature\ that was mentioned at the start of Section~\ref{secINT}.

First, we note that for an arbitrary covering-finite\ ranked poset of rank $2$, it is not necessarily the case that $R_1$ equals $\ricci$.  For example, let $P$ be the poset shown in Figure~\ref{figDIRIr}.  Then $R_1(e) = \frac 52$, but $\ricc e = 2$.

However, as we now show, it is the case that $R_1$ equals $\ricci$ for the following class of posets.

\begin{definition}\label{defnBAW}
Let $P$ be a ranked poset of rank $2$.  The poset $P$ is \textbf{almost polyhedral} if it is  covering-finite, and if the following conditions hold.  Let $w \in \rkiset P0$, and $a, b \in \rkiset P1$ and $\tau \in \rkiset P2$.  Suppose $a \ne b$.
\begin{enumerate}
\item
$B_1(a) = 2$.
\item
There is at most one $v \in \rkiset P0$ such that $v \prec a, b$.
\item
There is at most one $\sigma \in \rkiset P2$ such that $a, b \prec \sigma$.
\item
If $w < \tau$, then $[w, \tau]$ has cardinality four.
\qef\end{enumerate}
\end{definition}

The face poset of every \nd{2}dimensional polyhedral complex, and in particular every simplicial complex, is an almost polyhedral\ poset.  The face poset of a polyhedral map on a compact surface is also an almost polyhedral\ poset.  However, the set of almost polyhedral\ posets neither contains, nor is contained in, the set of face posets of all \nd{2}dimensional regular CW complexes.  The poset in Figure~\ref{figMCOM7} is almost polyhedral\ but not the face poset of a regular CW complex, because if it were, then the boundary of $m$ would be disconnected.  On the other hand, The face poset of the CW complex with two vertices, two edges and one disk, seen in Figure~\ref{figMCOM8}, is not almost polyhedral.  We note that Condition~(4) in Definition~\ref{defnBAW} is found in a number of places, such as Proposition~2.2 of \cite{BJOR2} and Definition~3.3 of \cite{FORM6}.

\begin{figure}[ht]
\centering\includegraphics{figMCOM7.pdf}
\caption{}\label{figMCOM7}
\end{figure}

\begin{figure}[ht]
\centering\includegraphics{figMCOM8.pdf}
\caption{}\label{figMCOM8}
\end{figure}

\begin{lemma}\label{lemBEBX}
Let $P$ be an almost polyhedral\ ranked poset of rank $2$.  Then $R_1(e) = \ricc e$ for all $e \in \rkiset P1$.
\end{lemma}

\begin{proof}
Let $e \in \rkiset P1$.  Let
\begin{align*}
U &= \{a \in \rkiset P1 \mid e \prec \tau \text{ and } a \prec \tau \text{ for some } \tau \in \rkiset P2\}\\
\intertext{and}
V &= \{a \in \rkiset P1 \mid v \prec e \text{ and } v \prec a \text{ for some } v \in \rkiset P0\}.
\end{align*}
Then $N_1(e) = \setcar {U \mathrel{\bigtriangleup} V}$.

If $\sigma \in \rkiset P2$ and $e \prec \sigma$, let
\[
C_\sigma = \{a \in \rkiset P1 \mid a \prec \sigma \text{ and } a \notin V\}
\]
and if $v \in \rkiset P0$ and $v \prec e$, let
\[
D_v = \{a \in \rkiset P1 \mid v \prec a \text{ and } a \notin U\}.
\]
Then $U - V = \bigcup_{\sigma \succ e} C_\sigma$ and $V - U = \bigcup_{v \prec e} D_v$.  Observe that $e \in U \cap V$.  It follows that $e \notin C_\sigma$ for all $\sigma \in \rkiset P2$ such that $e \prec \sigma$, and $e \notin D_v$ for all $v \in \rkiset P0$ such that $v \prec e$.

Let $u, w \in \rkiset P0$.  Suppose that $u, w \prec e$ and $u \ne w$.  We claim that $D_u \cap D_w= \emptyset$ and that $\setcar {D_u} = A_0(u) - A_1(e) - 1$.

Let $d \in D_u \cap D_w$.  Then $u, w \prec d$, and because $d \ne e$, we have a contradiction to Condition~(2) of Definition~\ref{defnBAW}.  Hence $D_u \cap D_w= \emptyset$.

Let $S = \{a \in \rkiset P1 \mid u \prec a\}$.  Then
\[
S = \{a \in \rkiset P1 \mid u \prec a \text{ and } a \notin U\} \cup \{a \in \rkiset P1 \mid u \prec a \text{ and } a \in U - \{e\}\} \cup \{e\},
\]
where this union is disjoint.

Let
\[
\func f{\{a \in \rkiset P1 \mid u \prec a \text{ and } a \in U - \{e\}\}}{\{\sigma \in \rkiset P2 \mid e \prec \sigma\}}
\]
be defined as follows.  Let $m \in \{a \in \rkiset P1 \mid u \prec a \text{ and } a \in U - \{e\}\}$.  By the definition of $U$ there is some $\tau \in \rkiset P2$ such that $m, e \prec \tau$.  Then $\tau$ is unique by Condition~(3) of Definition~\ref{defnBAW}, and let $f(m) = \tau$.  Let $n \in \{a \in \rkiset P1 \mid u \prec a \text{ and } a \in U - \{e\}\}$, and suppose that $f(m) = f(n)$.  Then $u \prec m, e, n$ and $m, e, n \prec f(m) = f(n)$, which contradicts Condition~(4) of Definition~\ref{defnBAW}, unless $m = n$.  Hence $f$ is injective.  Let $\eta \in \{\sigma \in \rkiset P2 \mid e \prec \sigma\}$.  Then $u \prec e \prec \eta$, and by Condition~(4) of Definition~\ref{defnBAW} there is a unique $c \in \rkiset P1$ such that $c \ne e$ and $u \prec c \prec \eta$.  Then $c \in {\{a \in \rkiset P1 \mid u \prec a \text{ and } a \in U - \{e\}\}}$ and $f(c) = \eta$.  Hence $f$ is surjective.

Because $f$ is bijective, we see that
\[
\setcar S = \setcar {\{a \in \rkiset P1 \mid u \prec a \text{ and } a \notin U\}} + \setcar {\{\sigma \in \rkiset P2 \mid e \prec \sigma\}} + 1,
\]
which implies that $A_0(u) = \setcar {D_u} + A_1(e) + 1$, and hence $\setcar {D_u} = A_0(u) - A_1(e) - 1$.

Using a similar argument as above, together with the fact that $B_1(e) = 2$, which holds by Condition~(1) of Definition~\ref{defnBAW}, it is seen that if $\eta, \tau \in \rkiset P2$, and if $e \prec \eta, \tau$ and $\eta \ne \tau$, then $C_\eta \cap C_\tau = \emptyset$ and $\setcar {C_\eta} = B_2(\eta) - 3$.

It follows that
\begin{align*}
N_1(e) &= \setcar {U - V} + \setcar {V - U} = \sum_{\sigma \succ e} \setcar {C_\sigma} + \sum_{v \prec e} \setcar {D_v}\\
 &= \sum_{\sigma \succ e} \left[B_2(\sigma) - 3\right] +\sum_{v \prec e} \left[A_0(v) - A_1(e) - 1\right].
\end{align*}

Equation~(\ref{eqRIC}) then yields
\begin{align*}
\ricc e &= A_1(e) + B_1(e) - N_1(e)\\
 &= A_1(e) + 2 - \sum_{\sigma \succ e} \left[B_2(\sigma) - 3\right] - \sum_{v \prec e} \left[A_0(v) - A_1(e) - 1\right]\\
 &= A_1(e) + 2 - \sum_{\sigma \succ e} B_2(\sigma) + 3A_1(e) - \sum_{v \prec e} A_0(v) + 2A_1(e) + 2\\
 &= 4 + 6A_1(e) - U_1(e) - D_1(e)\\
 & = 1 + \frac 32B_1(e) + 6A_1(e) - U_1(e) - D_1(e) = R_1(e).\tag*{\qedsymbol}
\end{align*}
\renewcommand{\qedsymbol}{}\end{proof}
\vspace{-\baselineskip}

Combining Theorem~\ref{thmBAK}, Lemma~\ref{lemBEBX} and Remark~\ref{remkCCC}, we now see that by having $R_0$ and $R_2$ available to us, there is a Gauss-Bonnet Theorem for almost polyhedral\ ranked posets of rank $2$ that incorporates $\ricci$.

\begin{corollary}\label{coroBBA}
Let $P$ be a finite almost polyhedral\ ranked poset of rank $2$.  Then
\[
\sum_{v \in \rkiset P0} R_0(v) - \sum_{e \in \rkiset P1} \ricc e + \sum_{\sigma \in \rkiset P2} R_2 (\sigma) = \eulgr P.
\]
\end{corollary}

Restricting our attention to polyhedral complexes allows us to make use of the standard Euler characteristic.  If $K$ is a polyhedral complex, we let $\ksim Ki$ denote the collection of all \ncells i\ of $K$, for each $i \in \{0, 1, 2\}$.

\begin{corollary}\label{coroBBY}
Let $K$ be a \nd{2}dimensional polyhedral complex.  Then
\[
\sum_{v \in \ksim K0} R_0(v) - \sum_{e \in \ksim K1} \ricc e + \sum_{\sigma \in \ksim K2} R_2 (\sigma) = \chi(P).
\]
\end{corollary}

The following corollary is deduced immediately from from Theorem~\ref{thmBAL}, Lemma~\ref{lemBEBX} and Remark~\ref{remkCCC}, together with the fact that $\bari B1 = 2$ in an almost polyhedral\ ranked poset of rank $2$.

\begin{corollary}\label{coroBAX}
Let $P$ be a finite almost polyhedral\ ranked poset of rank $2$.  Suppose that $\bari A1 \ge 2$.  If the average value of $\ricci$ is positive, then $\eulgr P > 0$.
\end{corollary}

In a polyhedral surface we have $\bari A1 = 2$, and hence the following holds.

\begin{corollary}\label{coroBAJA}
Let $K$ be a compact connected polyhedral surface.  If the average value of $\ricci$ is positive, then $\chi(K) > 0$, and hence $\pi_1 (K)$ is finite.
\end{corollary}

The observation in Corollary~\ref{coroBAJA} that $\pi_1 (K)$ is finite is trivial, because the classification of compact connected surfaces implies that if a compact connected polyhedral surface has positive Euler characteristic, then it is either $S^2$ or $P^2$, and in both cases it has finite fundamental group.  We stated this trivial conclusion in the corollary only for comparison with Theorem~6.1 of \cite{FORM6}.  The latter is a much stronger result, because it allows for complexes that are not surfaces, though of course the proof in \cite{FORM6} is much more substantial.

Next, we turn to the question of everywhere negative curvature on polyhedral surfaces.  For flexibility, we use polyhedral maps on surfaces, observing that any triangulation of a compact connected surface can be thought of as a polyhedral map.  If $K$ is a polyhedral map on a surface, we let $\ksim Ki$ denote the collection of all \ncells i\ of $K$, for each $i \in \{0, 1, 2\}$.

\begin{theorem}\label{thmBAJ}
Let $K$ be a polyhedral map on a compact connected surface.
\begin{enumerate}
\item\label{thmBAJ3}
$R_0$, $\ricci$ and $R_2$ have negative values at all cells of $K$ if and only if $B_2(\eta) \ge 7$ for all $\eta \in \ksim K2$.
\item\label{thmBAJ4}
If $\chi (K) \ge 0$, then not all three of $R_0$, $\ricci$ and $R_2$ can have negative values at all cells of $K$.
\item\label{thmBAJ5}
If $\chi (K) < 0$ and $K$ is orientable, there is polyhedral map on the underlying space of $K$ such that $R_0$, $\ricci$ and $R_2$ have negative values at all cells of the polyhedral map.
\end{enumerate}
\end{theorem}

\begin{proof}
By Lemma~\ref{lemBEBX}, we will replace $\ricci$ with $R_1$.  We ask when each of $R_0$, $R_1$ and $R_2$ has negative values.

Let $v \in \ksim K0$.  Then $R_0(v) < 0$ means $1 + \frac 32A_0(v) - [A_0(v)]^2 < 0$, and it is straightforward to verify that this condition holds if and only if $A_0(v) \ge 2$, which is always true for a polyhedral map of a surface.  Hence $R_0(v) < 0$ is always true.

Let $e \in \ksim K1$.  Because $K$ is a polyhedral map of a surface then $A_1(e) = 2$ and $B_1(e) = 2$.  Hence $R_1(e) < 0$ if and only if $16 - U_1(e) - D_1(e) < 0$, which is equivalent to $\sum_{v \prec e} A_0(v) + \sum_{\sigma \succ e} B_2(\sigma) > 16$.  The edge $e$ has two vertices, and $A_0(v) \ge 2$ for all vertices $v \in \ksim K0$, and therefore if $\sum_{\sigma \succ e} B_2(\sigma) > 12$ then $R_1(e) < 0$; this condition on $\sum_{\sigma \succ e} B_2(\sigma)$ is not necessary for obtaining $R_1(e) < 0$.  In particular, because $e$ is contained in two faces, if $B_2(\sigma) \ge 7$ for all $\sigma \in \ksim K2$ then $R_1(e) < 0$.

Let $\sigma \in \ksim K2$.  Then $R_2(\sigma) < 0$ means $1 + 6B_2(\sigma) - [B_2(\sigma)]^2 < 0$, and it is straightforward to verify that this condition holds if and only if $B_2(\sigma) \ge 7$.

Putting the above considerations together immediately implies Part~(\ref{thmBAJ3}) of this theorem.

Suppose $\chi (K) \ge 0$.  It can be verified that $K$ must have a face with no more than $6$ edges; details are left to the reader.  Part~(\ref{thmBAJ4}) of this theorem then follows immediately from Part~(\ref{thmBAJ3}).

Now suppose that $K$ is orientable and $\chi (K) < 0$.  Then the genus of $K$ is greater than or equal to $2$.  By Theorem~2(a) of \cite{M-S-W1} there is a polyhedral surface $L$ with the same genus as $K$, and with each face having $3$ edges, and each vertex contained in $7$ edges; this surface can be rectilinearly embedded in $\rrr{3}$.  (Such a polyhedral surface is called equivelar.)  Let $M$ be the dual map of $L$.  Then each face of $M$ has $7$ edges.  It follows immediately from Part~(\ref{thmBAJ3}) of this theorem that $R_0$, $R_1$ and $R_2$ have negative values at all cells of $M$, which is Part~(\ref{thmBAJ5}) of the theorem.
\end{proof}

The glaring omission in Theorem~\ref{thmBAJ} is that Part~(\ref{thmBAJ5}) of the theorem treats only the orientable case.  It does not appear to be known whether for every non-orientable compact connected surface $K$ with $\chi (K) < 0$, there is a polyhedral map $M$ of the underlying space of $K$ such that every face of $M$ has at least $7$ edges.  It would be interesting to know if that holds.

\section{Comparison with D.\ Stone's Approach}
\label{secDSTO}

The approach taken in \cite{FORM6} is not the only combinatorial approach to curvature of CW complexes.  Another (in fact earlier) approach was taken in \cite{STON1} and \cite{STON2}, where an analog of Myers' Theorem was proved in dimension $2$, using ideas similar to those in \cite{FORM6}; the results of the latter are stronger than the former.  We now offer a brief comparison of our approach in the case of polyhedral surfaces with the approach of \cite{STON1} and \cite{STON2}.

In \cite{FORM6}, the combinatorial Ricci curvature\ is located at the edges; in our approach the discrete curvature\ for polyhedral surfaces is located at all cells, though from the point of view of the analog of Myers' Theorem, the curvature of interest is located at the edges.  By contrast, the curvature defined in \cite{STON1} and \cite{STON2} is located at the vertices, and is defined as follows.  Let $K$ be a polyhedral surface, and let $v \in \ksim K0$.  Using our notation, the curvature at $v$ is given by the formula
\begin{equation}
\label{eqSTO}
R^*(v) =  2 - \sum_{\sigma > v} \left(1 - \frac 2{B_2(\sigma)}\right),
\end{equation}
where the summation is over all \ncells 2\ of $K$ that contain $v$.

It is simple to verify that an analog of the Gauss Bonnet Theorem, which is $\sum_{v \in \ksim K0} R^*(v) = 2 \chi (K)$, holds for this definition of curvature for all polyhedral surfaces.  However, in contrast to the analog of the Gauss-Bonnet Theorem in Theorem~\ref{thmBAK}, which holds for all finite ranked posets of rank $2$, the analog of the Gauss Bonnet Theorem for $R^*$ does not hold for all \nd{2}dimensional simplicial complexes, as examples show; we omit the details.

When $K$ is a polyhedral surface, the number of \ncells 2\ that contain $v$ equals the number of edges that contain $v$, which is $A_0(v)$.  Hence we can trivially rewrite Equation~\ref{eqSTO} as
\begin{equation}
\label{eqSTY}
R^*(v)= 2 - A_0(v) + \sum_{\sigma > v}\frac 2{B_2(\sigma)}.
\end{equation}
In contrast to the formula for $R^*$ given in Equation~\ref{eqSTO}, which is precisely as given in \cite{STON1} and \cite{STON2}, the formula given in Equation~\ref{eqSTY} does satisfy the analog of the Gauss Bonnet Theorem for all \nd{2}dimensional simplicial complexes.  However, while the formula in Equation~\ref{eqSTY} can, similarly to combinatorial Ricci curvature, be extended unchanged to the context of covering-finite\ ranked posets, the analog of the Gauss-Bonnet Theorem does not hold for this definition of $R^*$ for all finite ranked poset of rank $2$, as examples show; again, we omit the details.

We now return to the case where $K$ is a polyhedral surface.  The analog of the Gauss-Bonnet Theorem for $R^*$, which in contrast to Corollary~\ref{coroBBA} does not have $-1$ coefficients for any terms, implies that if $\chi (K) \ge 0$ then it cannot be the case that $R^*(w) < 0$ for all $w \in \ksim K0$, which is analogous to Theorem~\ref{thmBAJ} (\ref{thmBAJ4}).

Similarly to one direction of Theorem~\ref{thmBAJ} (\ref{thmBAJ3}), if $B_2(\eta) \ge 7$ for all $\eta \in \ksim K2$, then
\[
R^*(v)  \le 2 - A_0(v) + \sum_{\sigma > v}\frac 27 = 2 - A_0(v) + \frac 27 A_0(v)
\le 2 - \frac 57 \cdot 3 < 0
\]
for all $v \in \ksim K0$, because $A_0(v) \ge 3$.   (We note, however, that whereas this condition is also a necessary condition in Theorem~\ref{thmBAJ} (\ref{thmBAJ3}), it is not necessary for $R^*(v)$ for all $v \in \ksim K0$; for example, by \cite{M-S-W1}*{Theorem~2(c)} there is a polyhedral surface $M$ such that $\chi (M) = -8$, that each face has $5$ edges, and each vertex is contained in $4$ edges; it is seen that $R^*(v) < 0$ for all $v \in M$.)

The proof of Theorem~\ref{thmBAJ} (\ref{thmBAJ5}) also shows that if $K$ is orientable and $\chi (K) < 0$, there is polyhedral surface $M$ with the same underlying space as $K$ such that $R^*(w) < 0$ for all $w \in \ksim K0$.

Finally, we note that $R^*$ does not work any better than $\ricci$, or $R_0$, $R_1$ and $R_2$ in relation to a possible analog of Myers' Theorem for posets.  Let $P$ be the poset shown in Figure~\ref{figDIRIm}, and $v \in P$ be the element shown in the figure.  The reader can verify that $R^*(v) = 3$.  It is therefore not the case that $R^*(w)$ is positive and bounded away from zero for all $w \in \rkiset P0$ guarantees that the poset is finite.

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\end{document}

redpajama_set_name: RedPajamaArXiv