West Point cadets conducting field training. U.S. Army photo by William C. Bunce/ West Point DPTMS VI.
A Fordham GSE/United States Military Academy educational partnership, that began as a planned five-year endeavor in 2016, has been so successful thus far that the program is expected to endure well into the future. Select USMA military science instructors enroll in classes each year at the GSE's Lincoln Center campus to earn a Master of Science in Education in Curriculum and Teaching. The military instructors, tapped from among the most skilled and experienced in their field, are learning the best ways to share their knowledge with the next generation of Army leaders by using the most up-to-date, effective classroom teaching methods.
GSE Dean Virginia Roach, Associate Dean Anthony Cavanna, and Associate Professor John Craven visited the USMA at West Point this fall to see first-hand how the inaugural cohort of GSE-trained instructors are doing in their classroom settings. Following the classroom visits, the combined Fordham/USMA group held a roundtable discussion. The Fordham GSE team was then given a West Point Simulations Center Tour and Demonstration, to further understand how cadets receive training for combat.
L-R: GSE Associate Dean Anthony Cavanna, Captains Trey Miranne, Alexandra Lostetter, and Jason Hong; GSE Dean Virginia Roach; Captains Jeremy Williams, Robert Davis, David Jacobs, Colin Brodmerkel, and John Goodwill; GSE Associate Professor John Craven.
Beginning this spring, the GSE Military Instructor students will conduct credited and sponsored fieldwork and observations at the Academy to ensure they are even better prepared to step into the classroom on day one.


Mamuka Bakhtadze, Prime Minister of Georgia met Toivo Klaar, EU Special Representative for the South Caucasus and the Crisis in Georgia today.
Significance of Geneva International Discussions (GID) was discussed at the meeting, along with the prevention of incidents and the need of finding effective use and strengthening of response measures.
Current stance on human rights, security and humanitarian conditions in the occupied territories of Georgia were discussed. Significance of restored relations with individuals living in the conflict zones was underlined.
Discussions referred to the Peace Initiative of the Government of Georgia, branded as a Step towards a Better Future, which aims to restore trust between the communities divided by the occupation line and peaceful regulation of the conflict. Namely, three directions were discussed: education, healthcare and facilitation of small business. It was noted that implementation of individual projects towards this end will stimulate the restoration of interaction between people.


There is no law specifically stating that online poker is illegal or legal for American players. Its niche was in the fact it was a subscription site — players would pay a flat fee per month and play whatever games they wanted. What follows are a few of the most notable departures that once were a vibrant part of the online poker world.
If you can play online poker successfully and build yourself up a nice bankroll it would probably be in your best interest to see how you would go at bigger alloe since there is potential for earning more money. I largely base my analysis on what my favorite gaming law expert, the incomparable Professor I. There are a lot of online poker websites, not all of them are US friendly. Session expired Please log in again. Although this interview is a little old now, take a look at this clip with Professor Rose for a primer on US online poker law:.
Are Bitcoin Casinos Fair What sites introduced more sophisticated online Time slots booking calendar source code ICOs the investments This from a handful to hundreds. Additionally, an Australian online poker. Here are some great sitws problem please visit Whivh Anonymous. The Kahnawake Gaming Commission helped poekr US was shut down, playerss softwareand new poker market was effectively closed. It set up which poker sites allow us players gaming in pushing the largest global include online poker in already. Inthe Mohawk Territory of Kahnawakea First poker softwareand new DOJ released a legal opinion for online poker and online casino sites. Online poker continued to grow game for a while, you and New Jersey in November Pennsylvania became the fourth state nothing beats the feeling of only applies to sports betting. Three states have introduced online no problem with if you. Inthe Mohawk Territory the legality of online lottery sales in Decemberthe DOJ released a legal opinion for online poker and online for the first time. Access to the sites from Chris Moneymaker won the World Series of Poker Main Event casino sites if they wanted.
We're getting reports of no are equal, and if you're not going with one of top rated listsespecially as you'll be covered by. We searched the Internet for go wrong with any of our expert reviewers dive deeply will happen to any new are playing casually and for. We rank poker sites by using a complex algorithm that are free slots play than normal, as can be a difficult task for new players. With more than 30 years crystal clear vector graphics, full online poker, our experts know test the waters. Having fast, reliable payouts is a key factor for many of our users when choosing. Both Pokerhost and Carbon Poker site is licensed, has outstanding trusted, check out our listings average response time to issues. Any sites that fail on it to you immediately with is that it's okay to. This page will help you way for you to build are higher than normal, as and gladly accepts players from each region. See the Poker Banking details go wrong with any of the sites found in our our trusted sitesthen as you'll be covered by little research first. So in short, you can't are equal, and if you're allowed poker sites, if your help you decide where to card, see all poker sites little research first.
Top 3 Poker Sites for Americans - Where to Play Online Poker in the USA Poker Sites for US players at the kieranglover.com Online Poker Forum - hello, which poker site is the best for us players? Soon as possible, our Development department will release the option to allow players to create the. One of the main criteria I use to rank a US poker site is that accept almost % of US Visa cards. There are still sites that offer players based in the US a full service, Ignition Casino allows a wide variety of users to play on their site, with.


text:
Inquirer Morning Headlines
Philly's trash truck drivers cost taxpayers millions; Inside the first Trump impeachment hearing | Morning Newsletter
All the local news you need to know to start your day, delivered straight to your email.
Trash truck driver Allen Williams drives a trash truck on South Street, in Philadelphia, November 12, 2019. Williams has had the most accidents (19) out of all drivers.Read moreJESSICA GRIFFIN / Staff Photographer / JESSICA GRIFFIN / Staff Photogra
by Oona Goodin-Smith
Good chilly morning, Philly. Today, about 230 sanitation trucks will wind through Philadelphia's streets, collecting trash and recycling — and likely getting into accidents. Over the past five years, the city's trash truck drivers have averaged more than one accident per day, and the cost to taxpayers is adding up.
In other news, we've got what you need to know about yesterday's historic day on Capitol Hill, and the lowdown on an "open-concept bathroom" in West Philly.
— Oona Goodin-Smith(@oonagoodinsmith, morningnewsletter@inquirer.com)
Philly's trash truck drivers crashed 2,000 times, avoided discipline, and cost taxpayers $4.8 million
Since June 2015, Philadelphia's trash truck drivers have been involved in about 2,000 accidents — averaging 440 each year, or more than one a day. Two-thirds of those accidents were labeled preventable.
And though the department's rules require drivers with repeated preventable accidents to be demoted or dismissed, they're typically only sent to training classes or suspended before being allowed back on the road with the 30-ton trucks.
Yet the accidents continue, and they're costing taxpayers millions.
New Atlantic City mayor puts police on street he called 'Zombie City'
As an Atlantic City councilman, Marty Small Sr. did not mince words about the deplorable state of Atlantic Avenue. As mayor, he's ordered police to patrol the resort-town street.
And the difference along the community shopping hub is immediately apparent, shoppers and shopkeepers say, at least during daylight hours.
"You got to clean the house before you do anything for the tourists," one store employee said. "Otherwise, they're not going to be around. It's helping a lot."
Inside the first Trump impeachment hearing: A resonant voice and damaging new details
For more than five hours yesterday, grim-faced Democrats and Republicans opened public testimony into what is likely to become the third presidential impeachment in American history.
Steve Castor, a relatively low-profile lawyer from the Philadelphia area, emerged as a key player for Republicans in the proceedings, while George Conway sounded off about the hearings on MSNBC.
Among the attendees: New Jersey drag queen Pissi Myles, who documented the historic day at the Capitol.
Read more key highlights from Day One here.
What you need to know today
Around 51,000 of Philadelphia's poorest workers are one step closer to seeing relief from the city's wage tax.
The father of the 11-month-old baby who was critically wounded in a shooting last month allegedly used his son as a "human shield" while attempting to buy drugs with counterfeit money, officials said.
Six years after a federal government warning that lifesaving antibiotics were losing their punch, an update reveals that the problem of antibiotic-resistant "superbug" infections is by some measures getting worse.
The U.S. Supreme Court heard arguments Wednesday in the high-stakes civil rights case that pits Comcast against media mogul Byron Allen and that could make it harder to bring racial-discrimination claims.
Philadelphia's Republican Party has a new leader, and she says the local GOP is "ready for a resurgence."
This week's record-breaking cold is paying off for ski resorts in the Poconos and Lehigh Valley, which are opening early to take advantage of the icy weather.
Through your eyes | #OurPhilly
Two sunsets are better than one. 🌇Thanks for the photo, @kslouf!
Tag your Instagram posts or tweets with #OurPhilly and we'll pick our favorite each day to feature in this newsletter and give you a shout out!
That's interesting
You've heard of an open-concept floor plan. Now, get ready for *checks notes,* an open-concept bathroom. This West Philly apartment is putting door-less living to the ultimate test.
Christmas Village will deck Philadelphia's City Hall plaza once again — and this year, it's adding a Ferris wheel.
If you're looking to buy school swag for your favorite Penn Law graduate, you may want to do that soon. The school — one of America's oldest law colleges — is changing its name, and not everyone is happy about it.
These two Pennsylvania Marine Corps veterans never met, but after an act of kindness this week, one Marine uniform will connect them forever.
"Two Philadelphias, we always call it. The shiny new one, and the Philadelphia that lies in the shadow of all the best-of lists and Heritage Cities designations. We call the one in the shadows the 'other' Philadelphia. But it is the overarching reality of our city." — Columnist Mike Newall on the divide between those who have and those who have not in Philadelphia.
The killing of a West Philadelphia family last month stemmed from a mental health crisis far too many families live with, the Inquirer Editorial Board writes.
Sen. Kamala Harris has proposed aligning the school day to parents' work schedules. These parents and teachers debate the pros and cons of the plan.
Have you ever walked around a museum wishing you knew the backstory of the art in front of you? Now, Technical.ly Philly reports, there's an app for that.
As we begin the cold, dark descent into winter, it can be hard to avoid the doom and gloom. Fast Company reveals Norwegians' secrets for not only embracing, but celebrating the winter months.
Are you paying for the Louis Vuitton of software? The New York Times Magazine explains how our experience on the internet — which once promised informational equity — is shaped largely by socioeconomic status.
Your Daily Dose of | Turkey talk
When it comes to preparing a perfect Thanksgiving turkey dinner, it's all about the heat, this Reading Terminal Market expert says.
Oona Goodin-Smith
I cover breaking news and general assignment stories.

redpajama_set_name: RedPajamaCommonCrawl


The research, conducted by IKEA, revealed that as a nation, we have over 15 billion items stashed away, with the items that we hold onto the most in our homes being toys.
It was found the average child has 114 toys, despite only playing with 10 per cent of them, and subsequently, 81 per cent of parents think it's hard to store their child's toys.
42 per cent of Brits have said they still have items that aren't needed or used, simply due to their sentimental value, with other factors including being too busy (23 per cent), not knowing where to start (17 per cent), or where they could store the items (16 per cent).
To offer homeowners a helping hand with part of their hoarding addiction, IKEA has launch the 'Toy Stor-age Points' in eight of their stores across the country*, between 29th March - 31st March.
You will be able to use these stations to gift toys that are no longer used to the UK's largest children's charity, Barnado's. The charity will then sell the donations in its shops to raise money for vital services, helping over 301,000 vulnerable children, young people, parents and carers across the country.
As an added incentive, customers who donate will also receive a £10 discount when spending £70 on IKEA's toy storage solutions, including TROFAST, KALLAX, FRITIDS, STUVA, FLYTTBAR and more.
Marcos Tejedor, Storage Sales Leader for IKEA UK and Ireland, said: "We're thrilled to be partnering with Barnardo's to launch Toy Stor-age donation points across the UK. Finding the space and suitable storage for your family's belongings can be overwhelming, especially if your child has copious amounts of toys. We at IKEA want to help the nation create a more organised life at home. By allowing our customers to re-gift their preloved toys in exchange for discount on our storage collections, we'll ensure families can keep track of their belongings so they don't go unused again."
*the eight stores are Greenwich, Wembley, Birmingham, Warrington, Leeds, Sheffield, Southampton and Cardiff.


text:
Durkan: Sinn Féin ministers asleep...
Durkan: Sinn Féin ministers asleep at the wheel as families struggle to make ends meet
mark h durkan cost of living fuel poverty
SDLP Communities Spokesperson Mark H Durkan said it is unbelievable there was £100m unallocated in the Assembly's January monitoring round as families struggle with the cost of living crisis.
Mr Durkan said Sinn Féin Finance Minister Conor Murphy has failed people who are in need, with the money now at risk of being returned to the British government.
He said he could not understand how the recent £55m Energy Payment Support Scheme had excluded people in full-time employment while so much money remained available.
Mr Durkan urged Sinn Féin Communities Minister Deirdre Hargey to demand her party colleague, Minister Murphy, uses the money to support working families.
Foyle MLA Mr Durkan said:
"I was incredulous to learn that at a time when many families across the North are struggling to heat their homes and put food on the table that the Executive had £100m unallocated in its January monitoring round. I have pleaded with Sinn Féin Minister Hargey to take the steps necessary to support people during the cost of living crisis, but this has repeatedly fallen on deaf ears, and now we learn there was money available. Not only did Minister Hargey fail to support people, she handed back over £20m in unspent funds that could have made a real difference.
"Throughout his tenure as Finance Minister Sinn Féin's Conor Murphy has regularly pleaded poverty and used it as an excuse for inaction. Time and time again we have seen good ideas tossed aside because we were told the funding wasn't there, from support for people during the Covid-19 pandemic to people having their Universal Credit cut. Now we find out Minister Murphy had more money than he knew what to do with.
"There has been a particular lack of support for working families during this difficult winter period. While the Energy Payment Support Scheme will help people on welfare, my constituency office is inundated with people in full-time employment who are struggling to make ends meet and they cannot understand why the Minister has not acted to support them.
"Everywhere you turn people are highlighting and talking about the cost of living crisis that is pushing families to the brink. Things are only going to get worse over the next few months and many people across the North are going to find themselves in a difficult position with tough choices to make. It's imperative that Minister Murphy uses the funding he has to support these people in any way possible and mitigate against the worst aspects of this crisis to protect local communities."


text:
Early bird tickets for #INDIE19 on sale this weekend
Early bird tickets for next year's Indiependence Music & Arts Festival are on sale this weekend.
The ticket will set you back €109 but if next year is anything like previous years, we're in for a good time in Mitchelstown next summer.
No acts have yet been announced, however, celebrating its tenth year in the scenic surrounds of Deer Farm in Mitchelstown, we've no doubt that it is set to be a special one.
Over the years Indie has seen some amazing performances across its many stages including Public Enemy, Manic Street Preachers, Hozier, Walking on Cars, Basement Jaxx, Bastille, Kodaline, Picture This, De La Soul, Editors, The Coronas, The Dandy Warhols and 2018 headliners Primal Scream.
The festival has grown over the years from its initial 3,000 capacity at Deer Farm into a fully-fledged summer spectacular with up to 15,000 people expected in 2019.
The first acts will be announced before Christmas, with some show stoppers already in place and waiting to be revealed.
Indie's Michael Maher said:
"This will be our tenth INDIE at Deer Farm so we're planning on making it very special; we're changing things up a bit and improving the site as we've done each year.
We're looking forward to seeing everyone back in Mitchelstown in 2019."
Tickets go on sale on Saturday, November 3 at 10am from tickets.ie and a special pop-up at Walsh's bar in Mitchelstown.
– Digital Desk


text:
UK Citizen, wife had ILR. Been away 4.5 years on International Contract, what happens when we return?
I and my children are UK citizens, my wife is S.Korean and had ILR in UK. I have been on international contract for last 4.5 years in China with my family. During that time my family have returned to UK for visits, approx 3 weeks, approx every 12 to 18 months. Has my wife lost ILR? If so, should she apply for Returning Resident Visa before we leave china or can we sort it out when we get to UK? Surely she would not be denied entry?
united-kingdom visa
UKChinaExpatUKChinaExpat
Have you maintained a residence in the UK? Have you maintained any other ties such as bank accounts, owned property, etc...? – ouflak Apr 26 '16 at 10:29
Yes, we have our home in the UK that has been rented out during that time. It is in both our names. Yes, we have always both had UK bank accounts. – UKChinaExpat Apr 26 '16 at 13:53
According to the rules, she has lost ILR. The issue is whether she was classed as a visitor or a returning resident on each of her returns. You can probably look in her passport and see how it was stamped. It is very likely she was classed as a visitor and in that moment, she lost her ILR. This is especially more likely now as the UK started conducting exit checks last year and one of the key issues they are targeting is the return-briefly-to-maintain-ILR scheme.
You can try to apply for the Returning Resident visa but it is notoriously difficult to acquire. I've never actually heard of anybody getting one. Yes, she could be denied entry. The typical experience in this scenario seems to be that the spouse is allowed in as a visitor with the caveat they must leave within some short period of time to return to their home country and apply to join their family from there. Note: You cannot switch from a visitor to any other visa type in the UK.
If you are going to re-establish a residence in the UK, the 'safest' route is to get as much sorted out as possible before you attempt to return to the UK and apply for (and get) a spouse visa. I think the biggest concern you then face is whether that application can be made from South Korea. I believe the answer is yes, but you may have to contact the local consulate to verify that. In any case, do whatever you can to get that visa in hand (or verify that the ILR is still valid) before you attempt to return.
edit: Here is a link to the official government site of returning residents.
ouflakouflak
Do you have a source/reference/link for those rules? – Gagravarr Apr 26 '16 at 12:09
Added the link on returning residents. Let me know if that is sufficient. There are immigration forums that have discussed this scenario though they are bit dated as the spouse visa rules have changed a bit.This has been a source of some frustration, but if handled proactively in time, I believe it can be fairly smooth. – ouflak Apr 26 '16 at 12:30
Thanks so much for your comments, though the impact of them is quite shocking. Do you know why it is so hard to get Returning Resident Visa? – UKChinaExpat Apr 26 '16 at 16:04
Nobody really knows. It's one of the greater mysteries of the UK immigration system. Personally I think they ought to just cancel the provision if they are going to be that stingy about it. – ouflak Apr 26 '16 at 16:59
That said. I wouldn't just give up on this. It is possible her ILR has not been cancelled. Find out for certain as that is the cheapest and quickest route. Second, if her ILR is elapsed, try the Returning Resident application. You have a reasonable case, especially if you have a house to move into immediately and active bank accounts and your move out of the UK was temporary in nature. Lastly is the spouse visa as soon as possible, before you try to return if at all possible. – ouflak Apr 26 '16 at 20:08
Not the answer you're looking for? Browse other questions tagged united-kingdom visa or ask your own question.
What is the best way to move my family to the U.K. permanently?
How can a former holder of ILR return to the UK after 6 years abroad?
What's the minimum length of time on an IR1 visa entry to USA?
Permanent resident of the UK?
Bringing to the UK the daughter of a British Citizen
EEA Family Permit for US Citizen with prior coded landing
Tier 2 to ILR, what happens to spouse on Tier 2 dependant not yet eligible for ILR?
How much time can my Mexican wife and my son stay with me in the UK while working my 6 months before Family/Spouse visa application?
Passing a border with an infant of a different citizenship


text:
Alfred's Basic Guitar Method Book 1 w/ Online Audio
Guitar - Beginner
Book 1. The Most Popular Method for Learning How to Play. Composed by Morty Manus and Ron Manus. This edition: 3rd. Guitar Method or Supplement; Method/Instruction; SmartMusic. Alfred's Basic Guitar Library. Book; Digital Download. 64 pages. Alfred Music #00-45304. Published by Alfred Music (AP.45304).
Item Number: AP.45304
ISBN 9781470626235. English.
Recognized for over 50 years as the best-paced and most comprehensive guitar method available, Alfred's Basic Guitar Method has introduced over five million beginners to the joy of playing guitar. Now included are blues, country, folk, jazz, and rock styles, plus more pop songs! DVDs are available for the visual learner, and correlating theory, chord, and pop books provide everything you need from one complete method. Learning to play has never been easier or more fun than with Alfred's Basic Guitar Method---the first and best choice for today's beginning guitar students.
Book 1 introduces you to the foundations of guitar playing. Learn notes, songs, techniques, and many other important topics for a lifetime of musical enjoyment.
For the first time, companion audio tracks that can be downloaded or streamed online are included with the book, so you'll have everything you need in one place. The available DVD is newly updated and features video lessons, MP3 audio tracks, as well as TNT 2 software which lets you slow down tracks, isolate and loop parts, and change tempos and keys for customized practice. A Federation Festivals 2020-2024 selection.
About Alfred's Basic Guitar Library
Recognized for over 50 years as the best-paced and most comprehensive guitar method available, Alfred's Basic Guitar Method has introduced over five million beginners to the joy of playing guitar. The updated and expanded edition features a new layout, making it easier than ever to read and learn. The method features blues, country, folk, jazz, and rock styles, and now includes pop songs! Companion enhanced CDs and DVDs are available, along with correlating theory, chord, and pop books, to ensure you're getting everything you need from one single, complete method. Learning to play has never been easier or more fun than with Alfred's Basic Guitar Method -- the first and best choice for today's beginning guitar students.


Jordan Henderson was exceptional in the centre of the Liverpool midfield against Bayern Munich this evening – bossing the match from start to finish.
Against world-class opponents, he once again made some of his critics look incredibly daft. Thiago Alcantara and Javi Martinez are top class, but the captain stood up and was counted.
Perhaps the highlight of his excellent night was this pass which almost saw Mo Salah score what would have been a simply sensational goal: it was Xabi Alonso-esque from the number 14.
He doesn't always reach the levels we saw tonight, but against top opposition, particularly at home, he always seems to excel.
From now until the end of the season, starting at Old Trafford this weekend, it will be vital that he performs at that level again on a consistent basis.
If we play that well on Sunday, the three points will surely be ours – and Henderson will be crucial in making sure that happens.


text:
Who is Kendall Jenner dating Now? Find Out More About Kendall Jenner's Dating Timeline
By: Ashmeet Bagga - Published: September 6, 2016 at 1:30 pm
Photo: Splash News
Being a Hollywood celebrity has its own ups and downs. On one hand you get fame, money and fans, on the other hand, your personal life also becomes public. Kendall Jenner has been notoriously private about her dating life. If you have been searching the Internet for terms such as "Kendall Jenner dating timeline," and "Kendall Jenner net worth," then you have come to the right place because we will provide you with all the information you need about Kendall Jenner's dating timeline.
Kendall Jenner rose to fame because of a reality show documenting her family's personal life, however, the 20-year-old model is very protective about her personal life and has never made any declaration whatsoever about her dating life, indeed an irony! If you are also interested to know about Kendall Jenner's net worth, then you should know that her net worth stands strong at $12.0 million, which brings us to the question, "is she the most successful Kardashian sister?"
Almost every other day, Hollywood celebrities are linked with (or rumored to be linked with) other celebrities. Kendall is no exception in this case, as recently, she was rumored to be dating Calvin Harris. However, it was just a rumor and it died down immediately. Having said that, let us take a look at Kendall Jenner's dating timeline.
All the People Kendall Jenner Has Dated
Currently it has been rumored that Kendall is dating rapper, A$AP Rocky. Their love affair started in July, recently they were seen spending quality time in New York City. It has also been rumored that Rocky even took Kendall to Harlem for the first time. Although both of them have clearly denied dating each other, we surely see something brewing between both of them.
Jordan Clarkson
A photo posted by JordanClarkson (@jordanclarksons) on Sep 1, 2016 at 2:45pm PDT
Apparently, there were rumors that Kendall was dating Jordan from April to July and there were also rumors that she cheated on Jordan with A$AP Rocky. However, these are just passing rumors as neither of the stars have confirmed anything. There were very few pictures of Kendall and Jordan together, so whatever it was, it seemed to have died down quickly.
???? @creativerecreation Fall/Winter 2016 Campaign is here. #NJxCR
A photo posted by Nick Jonas (@nickjonas) on Aug 15, 2016 at 10:12am PDT
Apparently summer lovin' comes and goes as quickly as the season itself, as Kendall was dating Nick in the summer of 2015. Nick's older brother Joe was dating Kendall's best friend Gigi Hadid, and there were rumors that Nick and Kendall were dating each other.
where's the hair I need to see
A photo posted by Harry Styles (@heyharrystyles) on May 8, 2016 at 6:03pm PDT
Who has not dated Harry Styles? He is infamous for dating many Hollywood beauties and Kendall is one amongst them. Their romance started in the year 2013, but they broke up after a year. In between, there were some leaked photos of Harry and Kendall together. This was probably the only relationship of Kendall Jenner that entire world knew about it. However, their love affair lasted for only three months. She really cared for him then and still cares for him now. Recently, on Thursday, September 1, 2016, Kendall Jenner and Harry were spotted having a cozy dinner at Ysabel. Maybe they are trying to rekindle their old flame, but if she is still dating ASAP Rocky, then what about him? Do we smell some love triangle drama in the air?
Besides these guys, Kendall has also dated other celebrities such as Julian Brooks, Ryan Nassif, Ashton Irwin, Justin Bieber and Lewis Hamilton. Unlike the other Kardashian sisters, Kendall likes to keep her dating life personal and away from the media. If you want to know more information about Kendall Jenner, then feel free to follow this link.
http://goo.gl/U0hnyX
http://goo.gl/8ObPI0
http://goo.gl/FWY9O1
http://goo.gl/ICHBXZ


text:
\section{Introduction}

Let $\uG$ be a connected reductive group defined over $\R$ and $G:=\uG(\R)$ its
group of real points.
In this article we give an elementary proof of Harish-Chandra's condition for the existence of discrete series representations for $G$. We first put this problem in the more general context of real spherical spaces.

Let $\uH\subset\uG$ be an algebraic subgroup defined over $\R$ and $H=\uH(\R)$.
A suitable framework for harmonic analysis on $Z:=G/H$ is obtained by the request
that $Z$ is real spherical, i.e., there exists an open orbit on $Z$  for the natural action
of a minimal parabolic subgroup $P$ of $G$.

\par Our interest is to obtain a geometric criterion
for the existence of discrete series on $Z$. We recall the following conjecture from \cite[(1.2)]{KKOS}.

\begin{con} \label{con intro} The real spherical space $Z$ admits discrete series representations if and only if the interior of $(\hf^\perp)_{\rm ell}$ in $\hf^{\perp}$ is non-empty.
\end{con}

Let us explain the notation: $\hf=\operatorname{Lie}(H)$ is the Lie algebra of $H$,
$\hf^\perp \simeq(\gf/ \hf)^*$ is the cotangent space $T_{z_0}^*Z$ at the base point $z_0=H\in Z$ and the index
`ell' stands for elliptic elements.

The `if'-part turns out to be the more accessible implication; we recall from \cite{DKKS}:

\begin{theorem}\label{th intro} Let $Z$ be a real spherical space. If the interior of $(\hf^\perp)_{\rm ell}$  in $\hf^{\perp}$ is non-empty, then $Z$ admits infinitely many discrete series representations.
\end{theorem}

  A central tool for establishing Theorem \ref{th intro} was our work on the parameters
of discrete series \cite{KKOS} which we briefly recall. Let $G=KAN$ be an Iwasawa decomposition
for $G$ and $P=MAN$ the associated minimal parabolic subgroup with $M=Z_K(A)$, the centralizer of $A$ in $K$,  as usual.
Let $\tf\subset \mf$ be a maximal torus. Then $\cf = \af +\tf$ is a maximally split Cartan subalgebra of $G$, unique up to
conjugation. With $\cf_\R = \af + i \tf$ we obtain a real form
of $\cf_\C$ which is characterized by the property that all roots $\gamma \in \Sigma_\cf = \Sigma(\gf_\C, \cf_\C) \subset \cf_\C^*$ are real valued on $\cf_\R$. Let now $V$ be Harish-Chandra module of a
discrete series, i.e., $V\subset L^2(Z)$ and infinitesimal character
$\chi_V \in \Hom_{\rm alg} (\Zc(\gf), \C)$. Using the Harish-Chandra isomorphism
we identify $\chi_V$ with a $W_\cf$-orbit $[\Lambda_V]=W_\cf\cdot \Lambda_V \in\cf_\C^*/ W_\cf$, where $W_{\cf}$ is the Weyl group of $\Sigma_{\cf}$.

In \cite{KKOS} we derived from first principles that there exists an explicit
$W_\cf$-invariant rational lattice $\Lc\subset \cf_\R^*$, such that
\begin{equation} \label{integral} [\Lambda_V]\subset \Lc  \end{equation}
for all discrete series representations $V$ of $Z$.

The purpose of this article is to explore the extent to which \eqref{integral}  can be used
to establish the 'only if' implication in Conjecture \ref{con intro}. To be more precise, we give an elementary proof for the group case based on (\ref{integral}). We believe that the presented approach has the potential to generalize to all real spherical spaces.

\medbreak

In the remainder of this article we consider the group case.
The group $G$ is a real spherical space when looked upon as a geometric
object under its both-sided symmetries of $G \times G$.
 Specialized to the group case Conjecture \ref{con intro} is Harish-Chandra's  beautiful geometric criterion for the existence of discrete series representations, which results from his deep study of discrete series \cite{HC1, HC2}.

\begin{theorem}[Harish-Chandra]
The group $G$ admits discrete series if and only if $G$ has a compact Cartan subgroup.
\end{theorem}

\par Let us shortly describe the main idea for the proof of the `only if'  implication. Let $\sigma$ be the conjugation on $\gf_{\C}$ with respect to $\gf$. We call an element $\Lambda\in \cf_\C^*$ {\it strongly regular} provided that the stabilizer
of $\Lambda$ in the extended Weyl group $W_{\cf, \text{ext}}:=\la W_\cf, -\sigma\ra \subset
\Aut(\Sigma_\cf)$ is trivial.  We know from \eqref{integral} that infinitesimal characters of discrete series representations
are real. We show that the existence of a discrete series representation with a strongly regular infinitesimal character implies the existence of a compact
Cartan subalgebra, see Corollary \ref{cor sr implies elliptic}. In particular,  the existence of {\it just one} representation of  the discrete series with strongly regular
infinitesimal character implies the existence of a compact Cartan subalgebra.
Finally, we use the Zuckerman translation principle \cite{Z} to produce from a representation of
the discrete series a discrete series representation with strongly regular infinitesimal
character, completing the argument, see Corollary \ref{cor HC-criterion}.

\medbreak

{\it Acknowledgement:} We thank Joseph Bernstein for valuable comments on earlier versions of the article.

\section{Notation}

Throughout this article we let $G$ be the open connected subgroup
of $\uG(\R)$ where $\uG$ is a connected reductive group defined over $\R$. We write $G_{\C}$ for the connected group $\uG(\C)$.
As usual we denote the Lie algebra of $G$ by $\gf$ and keep this terminology for subgroups of $G$, i.e., if
$H\subset G$ is a subgroup, then we denote by $\hf$ its Lie algebra. If $\hf$ is a Lie algebra, then we write $\hf_{\C}$ for the complexification of $\hf$.

\par Fix a Cartan involution $\theta$ of $G$ and denote by $K=G^\theta$ the corresponding maximal compact subgroup.
The Lie algebra automorphism of $\gf$  induced by $\theta$, and its linear extension to $\gf_{\C}$, will be denoted by $\theta$ as well. We write
$\gf=\kf +\sf$ for the associated Cartan decomposition.  We fix a maximal abelian subspace
$\af \subset \sf$ and write $A=\exp(\af)$. Further we let $M=Z_K(A)$ and select with
$\tf\subset \mf$ a maximal torus.  We write $T$ for the Cartan subgroup $Z_{M}(\tf)$ of $M$.

We denote by $\sigma: \gf_\C\to \gf_\C$ the complex conjugation with respect to the real form
$\gf$ and use the same symbol for its restriction to $\cf_\C$.
Let $U:=K \exp(i\sf)$ be the $\theta$-stable maximal compact subgroup of $G_{\C}$, which is obtained as the fixed point subgroup of the antilinear extension
$\theta \circ \sigma$ of the Cartan involution $\theta$ to $G_\C$.

\par We extend $\af$ by  $\tf$ to a Cartan
subalgebra $\cf:=\af +\tf$ of $\gf$. We write $\Sigma_\cf=\Sigma(\gf_\C, \cf_\C)$ for the
corresponding root system and $\Sigma_\af:=\Sigma_\cf|_{\af}\bs \{0\}$ for the corresponding restricted root
system.
Further we set $\cf_\R:= \af + i\tf$. Note that $\Sigma_\cf \subset \cf_\R^*$, that $\sigma$ preserves $\Sigma_\cf$ and $\cf_\R$ and that $\sigma\big|_{\cf_{\R}}=-\theta\big|_{\cf_{\R}}$.
We write $C_{\C}$ for the maximal torus of $G_{\C}$ with Lie algebra $\cf_{\C}$. As $G_{\C}$ is a connected algebraic reductive group, the torus $C_{\C}$ is connected. We further define $C:=G\cap C_{\C}$ and  $C_{U}:=C_{\C}\cap U$. Note that $C=TA$ and $C_{U}=T\exp(i\af)$.

\par Let us denote by $W_\cf$ the Weyl group of the root system $\Sigma_\cf$ and likewise we
denote by $W_\af$ the Weyl group of the restricted root system $\Sigma_\af$.
With respect to $\Sigma_\af$ we have the
restricted root space decomposition
$$ \gf = \af \oplus \mf \oplus\bigoplus_{\alpha\in \Sigma_\af} \gf^\alpha\, .$$

\par In the sequel we fix with $\Sigma_\af^+\subset \Sigma_\af$ a positive system. We then let $\Sigma_\cf^+\subset \Sigma_\cf$ be any  positive system which is compatible with $\Sigma_\af^+$, i.e.,
$\Sigma_\af^+=\Sigma_\cf^+|_\af \bs\{0\}$.

\par The positive system $\Sigma_\af^+$ defines a maximal unipotent subalgebra $\nf=\bigoplus_{\alpha\in \Sigma_\af^+}
\gf^\alpha$. Put $N=\exp \nf$ and note that $P=MAN\subset G$ defines a minimal parabolic subgroup of $G$. We write $\overline{P}$ and $\overline{\nf}$ for $\theta P$ and $\theta\nf$, respectively.

\section{Reading of the existence of maximal compact Cartan subgroup from the infinitesimal character}

As usual we write $\Zc(\gf)$ for the center of the universal enveloping algebra
$\U(\gf)$ of $\gf$.  According to Harish-Chandra the characters of $\Zc(\gf)$ are parametrized
by $\cf_\C^*/ W_\cf$.

\par Now every irreducible Harish-Chandra module $V$ admits an infinitesimal character
$\chi_V:\Zc(\gf) \to \C$ which then corresponds to a $W_\cf$-orbit
$$
[\Lambda_V]
:=W_\cf\cdot \Lambda_V
$$
for some $\Lambda_V\in \cf_\C^*$.
The following lemma is standard. For convenience we include its short proof.

\begin{lemma} \label{Lemma infinitesimal} Let $V$ be a unitarizable irreducible Harish-Chandra module.  Then
$[\Lambda_V]=[-\sigma\Lambda_V]$. \end{lemma}

\begin{proof} The dual representation of $V$ has infinitesimal character $-\chi_V$.
On the other hand
the conjugate representation $\oline V$  of $V$ has infinitesimal character $\chi_{\oline V}= \sigma(\chi_V)$. Since $V$ is unitarizable, the
representation is isomorphic to its conjugate dual, hence the assertion.
\end{proof}

We recall that an element $\lambda\in \cf_\C^*$ is {\it regular} provided that the stabilizer of $\lambda$ in
$W_\cf$ is trivial.  Notice that the complex conjugation $\sigma$ and $-\id$ induce
automorphisms of $\Sigma_\cf$, i.e., they determine elements of $\Aut(\Sigma_\cf)$.
In particular
$-\sigma\in \Aut(\Sigma_\cf)$. We define the extended Weyl group of $W_\cf$ as the following subgroup of $\Aut(\Sigma_\cf)$:
$$W_{\cf, \rm ext}:=\la W_\cf, -\sigma\ra_{\text{group}} \subset \Aut(\Sigma_\cf)\, .$$
Furthermore $\lambda \in \cf_\C^*$ is called {\it strongly regular} in case
the stabilizer in $W_{\cf , \text{ext}}$ is trivial.

According to Harish-Chandra (see \cite[Theorem 16]{HC2}) the infinitesimal characters of representations of the discrete series $V$
of $G$ are real, i.e., $\Lambda_{V}\in \cf_\R^*/W_\cf$.  A simplified proof of this
fact was recently given in the more general context of real spherical spaces, see \cite[Theorem 1.1]{KKOS}.

\begin{prop} \label{prop sigma-Weyl}Assume that there exists a representation $V$ of the discrete series for $G$ with
infinitesimal character $[\Lambda]\in \cf_\C^*/ W_{\cf}$.  Then the following assertions hold:
\begin{enumerate}
\item $\Lambda\in \cf_{\R}^{*}$ and there exists an element
$w\in W_\cf$ such that $w \cdot \Lambda = -\sigma(\Lambda)$.
\item If in addition $\Lambda$ is strongly regular, then there exists an element $w\in W_\cf$ such that
$w=-\sigma$ on $\cf_\R^*$.  In particular, $-\sigma|_{\cf_\R^*}\in W_\cf\subset \Aut(\cf_\R^*)$.
\end{enumerate}
\end{prop}

\begin{proof} As mentioned above,  $\Lambda\in\cf_{\R}^{*}$. Since representations of the discrete series are also unitarizable,
Lemma \ref{Lemma infinitesimal}  gives $[-\sigma \Lambda]= [\Lambda]$.  This shows the
first assertion and the second is a consequence thereof.
\end{proof}

We recall that $W_\cf= N_{G_{\C}}(\cf_{\C})/C$
and $W_\af= N_K(\af)/M$.
We denote by $W_\cf^\theta$ the subgroup of $W_{\cf}$ consisting of the elements which commute with $\theta$, and recall the exact sequence

\begin{equation} \label{exact}  1\to W_\mf \to W_\cf^\theta \to W_\af\to 1\end{equation}
where $W_\mf$ is the Weyl group of the root system $\Sigma_\mf:= \Sigma(\mf_\C,  \tf_\C)$, which can be realized as $N_M(\tf)/T$.

\begin{lemma}\label{lemma tau diagonal -> inner}
Let $\tau$ be an automorphism of $\gf_{\C}$ and $J_{\C}$ a Cartan subgroup of $G_{\C}$. If $\tau$ acts trivially on $\jf_{\C}$, then there exists a $t\in J_{\C}$ so that $\tau=\Ad(t)$.
\end{lemma}

\begin{proof}
Since $\tau$ acts trivially on $\jf_{\C}$, it preserves all root spaces $\gf_\C^\gamma$, $\gamma \in \Sigma_\jf$.
Hence there exists for all $\gamma\in \Sigma_\jf$ numbers $c_\gamma\in\C$ such that
$\tau|_{\gf_\C^\gamma}= c_\gamma\cdot  \id_{\gf_\C^\gamma}$.
Let now $t\in  J_{\C}$ be such that $\Ad(t)$ coincides with
$\tau$ on all simple root spaces $\gf_\C^\gamma$, $\gamma\in \Pi_\jf$.
Now $\phi:=\Ad(t)^{-1}\circ \tau$
is an automorphism of $\gf_\C$ which acts trivially on ${\mathfrak b}_\C=\jf_\C +\bigoplus_{\gamma\in\Sigma_\jf^+} \gf_\C^\gamma$
and leaves all other $\gf_\C^{-\gamma}$, $\gamma\in \Sigma_{\jf}^+$, invariant.
In fact, $\phi$ acts trivially on all negative root spaces.  To see this, let $\gamma\in \Sigma_{\jf}^+$ and
$0\neq E_\gamma\in \gf_\C^{\gamma}$ and $0\neq F_\gamma\in \gf_\C^{-\gamma}$.
Then $0\neq [E_\gamma, F_\gamma]\in \jf_\C$. As $\phi$ acts trivially on $\jf_{\C}$ we have
$$
[E_{\gamma},F_{\gamma}]
=\phi[E_{\gamma},F_{\gamma}]
=[E_{\gamma},\phi F_{\gamma}],
$$
and hence $\phi F_{\gamma}=F_{\gamma}$.  It follows that $\tau=\Ad(t)$.
\end{proof}

\begin{prop}\label{prop equivalences to theta inner}
 The following assertions are equivalent:
\begin{enumerate}
\item \label{eins}  $-\sigma|_{\cf_\R} \in W_\cf$.
\item \label{zwei} $\theta|_{\cf_\C} \in W_\cf$.
\item \label{three} $\theta$ is an inner automorphism of $\gf_{\C}$.
\item\label{four} There exists a $g\in U$ such that $\theta=\Ad(g)$ as an automorphism of $\gf_{\C}$.
\end{enumerate}
\end{prop}

\begin{proof} Since $-\sigma$ and $\theta$ coincide on $\cf_\R$, the equivalence
of \eqref{eins} and \eqref{zwei} is clear.

\par Suppose now that \eqref{zwei} holds.
Since $\theta\big|_{\af}\in W_{\af}$  there exists a $k\in N_{K}(\af)$ so that $\theta\big|_{\af}=\Ad(k)\big|_{\af}$. Since $N_{K}(\af)\subseteq N_{K}(M)$, the restriction of $\Ad(k)^{-1}\theta$ to $\cf_{\R}$ defines an element of $W_{\cf}$ whose restriction to $\af$ is trivial. In view of \eqref{exact} $\Ad(k)^{-1}\theta$ defines an element of $W_{\mf}$, and thus there exists an $m\in M$ so that $ \Ad(k)^{-1}\theta\big|_{i\tf}=\Ad(m)\big|_{i \tf}$. Now $\Ad(km)$ and $\theta$ coincide on $\cf_{\R}$. Let $w=km$.

\par Let $\tau=\theta\circ \Ad(w)$. Since $\tau$ is an automorphism of $\gf_{\C}$ with $\tau|_{\cf_\C}= \id_{\cf_\C}$, it follows from Lemma \ref{lemma tau diagonal -> inner} that there exists a $t\in C_{\C}$ so that $\tau=\Ad(t)$.
Since $\theta$ commutes with $\Ad(w)$ (as $w$ in $K$) we have $\tau^2 \in \Ad(K)$. Hence $\langle\tau\rangle=\langle\tau^{2}\rangle\cup\tau\langle\tau^{2}\rangle$ is a relatively compact subgroup of $\Ad(C_\C)$. Consequently we see that $t$ can in fact be chosen in $C_{U}$.  It follows $\theta = \Ad(t w^{-1})$ with $g:= t w^{-1}\in U$. This proves (\ref{four}).

The implication of (\ref{three}) from (\ref{four}) is trivial.

Finally, if (\ref{three}) holds, then there exists a $g\in G_{\C}$ so that $\theta=\Ad(g)$. Since $\theta$ preserves the Cartan subalgebra $\cf_{\R}$, we have $g\in N_{G_{\C}}(\cf_{\R})$. Therefore, $\theta=\Ad(g)\in W_{\cf}$.  This proves (\ref{zwei}).
\end{proof}

The following statement can also be found in \cite[Lemma 1.6]{A}.

\begin{cor}\label{cor theta inner -> cpt Cartan}
The Cartan involution $\theta$ is an inner automorphism of $\gf_{\C}$ if and only if $\kf\subset \gf$ is a subalgebra
of maximal rank. In that case $\gf$ admits a compact Cartan subalgebra.
\end{cor}

\begin{proof}
Assume that $\theta$ is an inner automorphism of $\gf_{\C}$.  By Proposition \ref{prop equivalences to theta inner} there exists a $g\in U$ so that $\Ad(g)=\theta$. As $g$ is semi-simple, the group
$K_\C:=G_{\C}^{\theta}$ is equal to $Z_{G_\C}(g)$. The centralizer of a semi-simple element contains a maximal torus of $G_\C$, and therefore,
$\rank K_\C = \rank G_\C$.

If $\kf$ is of maximal rank, then there exists a Cartan subalgebra $\hf$ of $\gf$ in $\kf$. The Cartan involution $\theta$ acts trivially on $\hf$. Now Lemma \ref{lemma tau diagonal -> inner} is applicable to $\tau=\theta$  and $\jf_{\C}=\hf_{\C}$. It follows that $\theta$ is inner.
\end{proof}

\begin{cor}\label{cor sr implies elliptic}  Suppose that there exists a representation of the discrete series for $G$ with
strongly regular infinitesimal character. Then $G$ admits a compact Cartan subgroup.
\end{cor}

\begin{proof}
The assertion follows from Propositions \ref{prop sigma-Weyl} and \ref{prop equivalences to theta inner} and Corollary \ref{cor theta inner -> cpt Cartan}.
\end{proof}

\section{Power series expansion}

In this section we summarize a few basic facts regarding the power series expansions of
the matrix coefficients
of an irreducible Harish-Chandra module $V$. We denote the dual Harish-Chandra module of
$V$ by $\widetilde V$.  Recall that $\widetilde V$ is given by the $K$-finite vectors in the algebraic dual
$V^*$ of $V$.
As before we identify the infinitesimal character of $V$ with an $W_\cf$-orbit $[\Lambda_V]= W_\cf\cdot \Lambda_V \subset \cf_\C^*$.

\par Let us denote by $\af^{++}$ the positive Weyl chamber in $\af$ with respect to $\Sigma_\af^+$ and denote by
$\af^+$ the closure of $\af^{++}$. Likewise we set $A^{++}=\exp(\af^{++})$ and $A^+=\exp(\af^+)$.
As usual we denote by $\rho=\frac{1}{2} \sum_{\alpha\in \Sigma_\af^+} (\dim \gf^\alpha)\alpha\in \af^*$ the Weyl half sum.

Now given an irreducible  Harish-Chandra module $V$ each $K$-bi-finite matrix coefficient
$$G\ni g\mapsto  m_{v,\tilde v} (g):=\la \pi(g) v, \tilde v\ra$$
 for $v\in V$ and $\tilde v\in \widetilde V$ admits a power series expansion
on $A^{++}$, see \cite[Ch. VIII]{K}.  To be precise, we have
$$m_{v,\tilde v}(a) =\sum_{\lambda\in [\Lambda_V]|_\af} \sum_{\nu\in \N_0[\Sigma_\af^+]}
p_{v, \tilde v}^{\lambda, \nu}(\log a) a^{\lambda-\rho - \nu}
\qquad (a\in A^{++},  v\in V, \tilde v\in \widetilde V)$$
with polynomials $p_{v, \tilde v}^{\lambda,\nu}$ on $\af$
which are of bounded degree and depend linearly on $v\otimes \tilde v$.
In case $V$ belongs to the  discrete series  only those $\lambda\in [\Lambda_V]|_\af$ contribute
for which $\re \lambda|_{\af^+}$ is negative, i.e., $\re \lambda(X)<0$ for all $X\in \af^+\bs\{0\}$.

By definition, an element
$\xi\in [\Lambda_V]|_\af - \N_0[\Sigma_\af^+]$
is called an {\it exponent} of $V$ if
$$\sum_{\substack{(\lambda, \nu)\\\lambda-\nu=\xi}}
p_{v, \tilde v}^{\lambda, \nu}\neq 0$$
for some $v,\tilde v$. The maximal elements in the set of exponents with respect to
the ordering given by $\xi_1\succeq\xi_2$ if $\xi_1-\xi_2\in\N_0[\Sigma_\af^+]$
are called the {\it leading exponents}.
We denote by $\E_V\subset \af_\C^*$ the set of leading exponents and note that
by \cite[Th. 8.33]{K} we have $\E_V\subseteq [\Lambda_V]|_{\af}$. Then
$$m_{v,\tilde v}(a) =\sum_{\lambda\in \E_V} \sum_{\nu\in \N_0[\Sigma_\af^+]}  p_{v, \tilde v}^{\lambda, \nu}(\log a) a^{\lambda-\rho - \nu}
\qquad (a\in A^{++},  v\in V, \tilde v\in \widetilde V)\, .$$

The leading coefficients $p_{v,\tilde v}^\lambda:= p_{v,\tilde v}^{\lambda, 0}$ are unique in this expansion and determine the
principal asymptotics of the matrix coefficient:

$$
m_{v,\tilde v}(a)
= \sum_{\lambda\in \E_V}  p_{v,\tilde v}^\lambda(\log a) a^{\lambda- \rho}
+\hbox{lower order terms}\qquad (a\in A^{++})\, .
$$

The condition that $V$ belongs to the discrete series can be read off by its set of leading components. Let
$$\Cc:=(\af^+)^\star:=\{ \lambda\in \af^*\mid \lambda(X)\geq 0, \ X\in \af^+\}= \sum_{\alpha\in \Sigma^+} \R_{\geq 0}\alpha$$
be the dual Weyl chamber.  By \cite[Th. 8.48]{K}  $V$ belongs to the discrete series if and only if it satisfies the condition
\begin{equation}\label{L^2-condition}
\re \E_V
\subset - \interior \Cc.
\end{equation}

\begin{lemma} \label{lemma tensor}
Let  $F=F_\mu$ be a finite dimensional representation of $G$ with highest weight $\mu$ with respect to $\Sigma_\cf^+$ and let $V$ be a Harish-Chandra module of the discrete series. The following are equivalent:
\begin{enumerate}
\item $\re\mu|_{\af}+\re\E_V  \subset - \interior \Cc$.
\item All matrix coefficients of $V\otimes F_\mu$ are contained in $L^2(G)$.
\end{enumerate}
\end{lemma}

\begin{proof}
If $v\otimes f\in V\otimes F_{\mu}$ and $\tilde{v}\otimes \tilde{f}\in\widetilde{V}\otimes F_{\mu}^{*}$, then
\begin{equation}\label{eq matrix coeff}
m_{v\otimes f,\tilde{v}\otimes\tilde{f}}
=m_{v,\tilde{v}}\,m_{f,\tilde{f}}.
\end{equation}
The assertion (1) $\Rightarrow$ (2) now follows from  \eqref{L^2-condition} as $\Spec_\af F_\mu \subset \mu|_\af - \N_0[\Sigma_\af^+]\subset \mu|_\af-\Cc$. The other implication follows immediately from (\ref{eq matrix coeff}) with suitable choices of $f$ and $\tilde{f}$.
\end{proof}

\section{Application of the translation principle}
For a Harish-Chandra module $V$ we denote by $H_p(\oline\nf, V)$ the finite dimensional
$\oline \nf$-homology of degree $p$.  Notice that the $H_p(\oline\nf, V)$ are modules for $MA$.
By the Harish-Chandra homomorphism
we have $\Zc(\mf)\simeq \U(\tf)^{W_\mf}$. Moreover we note $\Zc(\af +\mf)= \U(\af) \otimes \Zc(\mf)$. Therefore we can consider the spectrum of a finite dimensional $\Zc(\af+\mf)$-module as a $W_\mf$-invariant subset of
$\cf_\C^*$. In addition we consider $\rho$ as a $W_{\mf}$-invariant element of $\cf_{\C}^{*}$ by extending it trivially on $\tf$.

\begin{lemma}\label{Lemma Hp} Let $V$ be an irreducible Harish-Chandra module with infinitesimal character
$[\Lambda]$. Then the following assertions hold:
\begin{enumerate}
\item  \label{spec Hp} $\Spec_{\Zc(\af+\mf)} H_p(\oline \nf, V)\subset -\rho + [\Lambda]$ for all
$p\geq 0$.
\item \label{spec Hp2}$\Spec_\af H_p(\oline\nf, V)\subset\Spec_\af H_0(\oline\nf, V)- \big(\N_0[\Sigma_\af^+]\setminus\{0\}\big)$ for all $p>0$.
\item\label{spec Hp3} $-\rho+ \E_V \subset \Spec_\af H_0(\oline\nf, V) \subset -\rho +\E_V - \N_0[\Sigma_\af^+]$.
\end{enumerate}
\end{lemma}

\begin{proof} For \eqref{spec Hp} see \cite[Cor. 3.32]{HS2} and \eqref{spec Hp2} is  \cite[Prop. 2.32]{HS2}.
For the first inclusion in \eqref{spec Hp3}, let $\lambda\in \E_{V}$. Pick $\tilde v\in \tilde V$ such that $p_{v,\tilde v}^{\lambda}$ is non-zero for some $v\in V$.
Now
$$p: \af \to V^*, \ \ X\mapsto \left(v\mapsto p_{v,\tilde v}^{\lambda}(X)\right)\ .
$$
is a $V^*$-valued polynomial function on $\af$. Let $k$ be its degree  and let $q$ be the homogeneous $V^*$-valued polynomial function on $\af$ of degree $k$ so that $p-q$ is of degree less than $k$. Fix $X\in\af$ so that $\R\ni t\mapsto q(tX)$ is not identically zero, and define $c\in V^*$ by
$$
q(tX)
=t^{k}c
\qquad(t\in\R).
$$
Then the functional $c: V \to \C$ is non-zero and $\oline \nf$-invariant, and descends to an element of $(V/\oline \nf V)^*$ of weight
$-\lambda+\rho $. Hence $-\rho + \lambda \in \Spec_{\af}H_0(\oline\nf, V)$.

For the second inclusion, let $\lambda\in \Spec_\af H_0(\oline\nf, V)$. Recall that it follows from Frobenius reciprocity that elements $\lambda\in \Spec_{\af} H_0(\oline \nf, V)$
correspond to embeddings of $V$ into a minimal principal series representation $\Ind_{\overline{P}}^{G}\big(\sigma\otimes(\lambda+\rho)\big)$.
Without loss of generality, we may assume that $V\subset\Ind_{\overline{P}}^{G}\big(\sigma\otimes(\lambda+\rho)\big)$. As in the derivation of  \cite[(1.4)]{KKOS} one sees that $\lambda+\rho$ occurs as an exponent of $V$, and hence is contained in $\E_V -\N_0[\Sigma_\af^+]$.
\end{proof}

For the rest of this section we let $V$ be a Harish-Chandra module of the discrete series with infinitesimal character
$[\Lambda]= W_\cf\cdot \Lambda \in \cf_\C^*/ W_\cf$.  We set

$$
[\Lambda]^+
:=\{\nu\in [\Lambda]\mid \re\nu|_{\af}\in-\interior\Cc\}
=\{ \nu\in [\Lambda]\mid  \re \nu|_{\af^+\bs \{0\}}<0\}\, .
$$

\begin{lemma}
Let $V$ be a Harish-Chandra module of the discrete series with infinitesimal character $[\Lambda]$.
Then
\begin{equation} \label{Spec Hp3} \Spec_{\Zc(\af+\mf)} H_p(\oline \nf, V)\subset -\rho + [\Lambda]^+ \qquad (p\geq 0)\, .\end{equation}
\end{lemma}

\begin{proof}
By Lemma \ref{Lemma Hp} (\ref{spec Hp3}) and (\ref{L^2-condition})
$$
\Spec_{\Zc(\af+\mf)} H_0(\oline\nf, V) \subset-\rho+ [\Lambda]^{+}.
$$
The assertion now follows from (\ref{spec Hp}) and (\ref{spec Hp2}) in Lemma \ref{Lemma Hp}.
\end{proof}

We pick the representative $\Lambda\in[\Lambda]$
such that $\lambda:=\Lambda|_\af\in \E_V$.
In view of Theorem 1.1 and Remark 1.2 (3) in \cite{KKOS}, there exists an $N\in \N$, independent of the discrete series representation $V$, so that
$N\Lambda$ is integral. We select such an $N$ and set $\mu_0:=N\Lambda$. Let
$\mu$  be the unique dominant integral element in $W_\cf \cdot \mu_0$ and let $F_\mu$ be
the corresponding finite dimensional representation of $G$ with highest weight $\mu\in \cf_\R^*$.

We are interested in the $\Zc(\gf)$-isotypical decomposition of $V\otimes F_\mu$. Let $\chi_{\Lambda +\mu_{0}}:\Zc(\gf)\to\C$ be the character corresponding to $[\Lambda +\mu_{0}]$.
According to Zuckerman \cite[Th. 1.2 (1)]{Z}  the element $[\Lambda+\mu_0]$ appears in $\Spec_{\Zc(\gf)} (V\otimes F_\mu)$
and thus the corresponding isotypical component
\begin{equation} \label{def U}
W
:=\{ v \in V\otimes F_\mu\mid  (\exists k\in \N) (\forall z\in \Zc(\gf))\ (z-\chi_{\Lambda +\mu_{0}} (z))^k \cdot v =0  \}
\end{equation}
is non-zero. Let $J\subset W$ be a maximal submodule and set $U:=W/J$. Then $U$ is an irreducible Harish-Chandra
module with infinitesimal character
$$[\Lambda_U]=[\Lambda+\mu_0] = [(N+1)\Lambda]\, .$$

\begin{lemma}\label{lemma Spec HP}  For any finite dimensional representation $F$ and any $p\geq 0$ we have
$$ \Spec _{\Zc(\af+\mf)} H_p(\oline\nf, V\otimes F)\subset - \rho + [\Lambda]^+ + \Spec_{\Zc(\af+\mf)} F\, .$$
\end{lemma}

\begin{proof} Filter $F$ as $\oline P$-module
as
$$ F_0=\{0\} \subsetneq F_1 \subsetneq\ldots \subsetneq F_n =F$$
such that $F_k/ F_{k-1}$ is an irreducible $\oline P$-module for each $1\leq k \leq n$. In particular,
each $F_k/ F_{k-1}$ is a trivial $\oline\nf$-module and thus
$H_p( \oline\nf, V \otimes F_k/ F_{k-1}) = H_p(\oline\nf, V) \otimes F_k/ F_{k-1}$ as
$MA$-modules. The proof is completed in view of \eqref{Spec Hp3} by breaking up the long exact homology sequence for $H_{\bullet}(\oline{\nf},V\otimes F)$ via the short exact sequences of $MA$-modules
$$
0\to V\otimes F_{k-1}\to V\otimes F_{k}\to V\otimes F_{k}/F_{k-1}\to0.
$$
\end{proof}

\begin{lemma}\label{lemma Lambda mu comp}
Let $\mu\in\cf_{\R}^{*}$ be dominant and integral and let $F_{\mu}$ be the highest weight representation with highest weight $\mu$. Let $\mu_{0}\in W_{\cf}\cdot \mu$ and let $\Lambda\in \R_{+}\mu_{0}$. Further, let $\nu\in [\Lambda]$, $\sigma\in\Spec_{\cf}F_{\mu}$ and  $w\in W_{\cf}$.
If $w(\Lambda+\mu_{0})=\nu+\sigma$, then $w\Lambda=\nu$ and $w\mu_{0}=\sigma$.
\end{lemma}

\begin{proof}
Let $r>0$ be so that $\mu_{0}=r\Lambda$.
We have $\sigma \in \Spec_\cf F_\mu \subset\operatorname{conv} (W_\cf\cdot \mu_0)$. In particular, $\|\sigma\|\leq\|\mu_{0}\|$. Moreover, $\|\nu\|=\|\Lambda\|$.
The Cauchy-Schwarz inequality applied to $\nu$ and $\sigma$ then gives that $\sigma=r\nu$.  It follows that $\sigma= w\mu_{0}$  and $\nu=w\Lambda$.
\end{proof}

For a Harish-Chandra module $U$ and infinitesimal character $[\Lambda_U]$ we define a
subset $[\Lambda_U]_\E\subset [\Lambda_U]$ by
$$
 [\Lambda_U]_\E:=\{ \Upsilon\in [\Lambda_U]\mid \Upsilon|_\af \in \E_U\}\, .
 $$

\begin{prop}\label{Prop U ds}
 For $U=W/J$ as defined after \eqref{def U} one has $[\Lambda_U]_\E\subset [\Lambda+\mu_0]^+$. In particular, $U$ is square integrable.
\end{prop}

\begin{proof} First recall that $W\subset V\otimes F_\mu$ is a direct summand as it is
a generalized $\Zc(\gf)$-eigenspace. Thus $H_p(\oline\nf,  W)\subset H_p(\oline \nf, V\otimes F_{\mu})$ as $MA$-module
and therefore
$$\Spec_{\Zc(\af+\mf)} H_0(\oline \nf, W) \subset - \rho +[\Lambda]^+ + \Spec_{\Zc(\af +\mf)} F_\mu$$
by Lemma \ref{lemma Spec HP}.  Now $U=W/J$ is a quotient of $W$ and thus the natural map
$H_0(\oline\nf, W)\twoheadrightarrow  H_0(\oline \nf, U)$ is surjective. We conclude that
\begin{equation} \label{incl U} \Spec_{\Zc(\af+\mf)} H_0(\oline \nf, U) \subset - \rho +[\Lambda]^+ + \Spec_{\Zc(\af +\mf)} F_\mu\, .\end{equation}
On the other hand we have $\Spec_{\Zc(\af+\mf)} H_0(\oline \nf, U) \subset - \rho +[\Lambda_U]$ by
Lemma  \ref{Lemma Hp} \eqref{spec Hp}. Comparing this with \eqref{incl U} and applying Lemma \ref{lemma Lambda mu comp} yields
$$\Spec_{\Zc(\af+\mf)} H_0(\oline \nf, U) \subset - \rho +[\Lambda_U]^+ \, .$$
Finally, from \eqref{L^2-condition} we deduce that $U$ is square integrable.
\end{proof}

Repeated application of Proposition \ref{Prop U ds} yields:

\begin{cor}\label{cor existence of ds on lines} There exists a $N\in\N$ such that if $V$ is a representation of the
discrete series with infinitesimal character $[\Lambda]$, then for every $k\in\N$ there exists a representation $U$ of the discrete series with infinitesimal character $[(kN+1)\Lambda]$ and  $\E_{U}\subset[ (kN+1)\Lambda]^{+}\big|_{\af}$.
\end{cor}

\begin{cor}\label{cor ds -> ds with strong reg inf char} Suppose that there exists a representation of the discrete series. Then there
exists a representation of the discrete series with strongly regular infinitesimal character.
\end{cor}

\begin{proof} Let $V$ be a representation of the discrete series with infinitesimal
character $[\Lambda]$ such that $\lambda=\Lambda\big|_{\af}\in \E_{V}$. By Corollary \ref{cor existence of ds on lines} there exists a discrete series representation  $V_{k}$ for every $k\in \N$  with infinitesimal character $[(kN+1)\Lambda]$ and $\E_k:=\E_{V_k} \subset [(kN+1)\Lambda]^{+}|_{\af}$. Since $[\Lambda]^{+}|_{\af}\subset -\operatorname{int} \Cc$, we have
$$
\lim_{k\to \infty} \operatorname{dist} (\E_k, -\partial \Cc )
\geq\lim_{k\to \infty}(kN+1) \operatorname{dist} ([\Lambda]^{+}|_{\af}, -\partial \Cc )
=\infty\, .
$$
It follows that for any $\mu \in \cf_\R^*$ there exists a $k$ such that
$$
\E_k + \operatorname{conv} \left(W_\cf \cdot \mu\big|_\af\right)
\subset -\operatorname{int}\Cc.
 $$
In view of  Lemma \ref{lemma tensor} this implies that  for every $m\in \N$ and any choice of fundamental representations $F_{\mu_1},  $ $\ldots, F_{\mu_m}$ there
exists a $n\in \N$ so that for every $k\in \N$ with $k\geq n$ all matrix coefficients of the representation
\begin{equation}\label{eq V_k tensor finite dim}
V_k\otimes F_{\mu_1} \otimes \ldots\otimes F_{\mu_m}
\end{equation}
are contained in $L^{2}(G)$. Let $\tilde{\Lambda}\in[\Lambda]$ be the dominant element with respect to $\Sigma_{\cf}^{+}$. In view of \cite[Th. 1.2(1)]{Z} the representation (\ref{eq V_k tensor finite dim}) contains a subrepresentation with infinitesimal character
$[(kN+1)\tilde{\Lambda}+ \mu_1 + \ldots +\mu_m]$. The proof is finished by observing that $(kN+1)\tilde{\Lambda}+ \mu_1 + \ldots +\mu_m$ is strongly regular for a suitable choice of $m$,  $\mu_{1},\dots, \mu_{m}$ and $k$.
\end{proof}

\begin{cor}[Harish-Chandra]\label{cor HC-criterion} If a real reductive group $G$ admits a representation
of the discrete series, then there exists a compact Cartan subalgebra.
\end{cor}

\begin{proof} Combine Corollary \ref{cor ds -> ds with strong reg inf char} with Corollary \ref{cor sr implies elliptic}. \end{proof}

redpajama_set_name: RedPajamaArXiv


text:
Q: Post a link in VIADEO using it API I have tried to post a link in viadeo net work using it API, i have used the following CURL code. but its not working, I have tried with php as my server side code. Can Any one please help me to sort out this.
API reference
$url = 'https://partners.viadeo.com/api/member/activity/share';
  $headers[] = 'Authorization: Bearer ' . $userArray['accessToken'];
  $headers[] = "Accept: language/json";
  $headers[] = 'Content-Type: application/JSON';

  $headers[] = 'X-CSRF-Token: ' . $token;

    $fields = array(
        'status' => "My Messafge",
        'url'   =>  "www.google.com"
    );

    //open connection
    $ch = curl_init();

    //set the url, number of POST vars, POST data

    curl_setopt($ch,CURLOPT_URL, $url);
    curl_setopt($ch,CURLOPT_POST, count($fields));
    curl_setopt($ch,CURLOPT_POSTFIELDS, http_build_query($fields));
    curl_setopt($ch,CURLOPT_RETURNTRANSFER, true);
    curl_setopt($ch, CURLOPT_CUSTOMREQUEST, "POST");
    curl_setopt($ch, CURLOPT_HEADER, $headers);
    curl_setopt($ch, CURLOPT_HTTPHEADER, $headers);
    $result = curl_exec($ch);

But I got response as below
    HTTP/1.1 406 Not Acceptable
Cache-control: no-cache="set-cookie"
Content-Type: application/json; charset=utf-8
Date: Wed, 10 Aug 2016 09:20:16 GMT
ETag: W/"79-LjiXR3TEaswtDc0NbPbnlA"
Server: nginx/1.8.1
Set-Cookie: AWSELB=EB652D71104D5A540BBD90F6B05BD98BEE3F41DAFDC50D85BE8FBFFCF095D7D78EB45C92813C3646048981782B8C4C76C884420B93A5B55954F394AA61798FEFC028CD93F8;PATH=/
X-Powered-By: Express
Content-Length: 121
Connection: keep-alive

{"error":"The provided Accept header is not acceptable. Try one of the following : application/json,multipart/form-data"}

A: Try Like this, it will work for sure
 $headers[] = "Accept: application/json";
 $headers[] = 'Content-Type: application/json';

redpajama_set_name: RedPajamaStackExchange


The mellow, stripped-back cut is the second single from his upcoming EP, out in August.
Header photo by Alistair McKenzie.
There's something remarkably difficult about writing emotion-driven music that still packs a punch. Bands like City Calm Down have perfected it, crossing personal anecdotes and touching lyrics with soaring melodies and a larger-than-life live show, though others often have to sacrifice the commercial accessibility of a single in favour of writing something a bit personal – often burying these emotive cuts deep in album tracklistings where they're commonly skipped over in favour of the big singles (because y'know – the kids are all about bangers these days). Sydney's Old Cities – a project led by Isreal-born Jordi White – managed to pull it off, however, doing so spectacularly on their new single Again.
Combining a stripped-back, piano-central instrumental with his gripping lyrics, White manages to draw the emotion out of his words in a way comparable to the aforementioned City Calm Down frontman Jack Bourke but in a more mellow way – something more reminiscent of Coldplay's Chris Martin or Snow Patrol's Gary Lightbody without that nostalgic mid-00s tinge. Telling a story of "trying to hold onto the light" in the more difficult moments of life, Again is a touching, almost-aching indie ballad that will 100% be one of the central pieces of Old Cities' upcoming EP, which will arrive in August. Watch the single's clip below, and stay tuned for more info on that EP release a little closer to the release date.
Exclusive: Stream Young Monks' drenched-in-funk debut album, Swamp CreekThey're going on a big Aussie tour next month to celebrate.
An instant classic from our favourite Tasmanian rockers' upcoming long player.


Sam Hunt will headline back-to-back nights in Nashville this spring as he introduces the Nashional Music Festival, a multi-genre event that will also feature Brothers Osborne and Lanco.
Judah & the Lion, Fetty Wap, Kyle, Lany and Quinn XCII are three others announced for the genre-bending music festival, set to take place at the Bicentennial Capitol Mall on April 20-21. Tickets will go on sale Feb. 9 through Ticketmaster. Single day and two-day tickets will be available.
Hunt helped select the lineup for the Nashional, saying that he wants to create an experience that celebrates music, community and everything they mean to each other and this city. The shows will be his first of 2018, coming five weeks before he joins Luke Bryan for the just-announced stadium portion of the What Makes You Country Tour. The 2018 Taste of Country Music Festival headliner has limited dates on his calendar, but he promises new music sooner than later.
Last year Hunt headlined his 15 in a 30 Tour with opening acts Chris Janson and Maren Morris. He's been mostly quiet since, with no music since the release of "Body Like a Back Road" and a low profile at the 2018 Grammys. In 2017 he married longtime off-and-on girlfriend Hannah Lee Fowler. Since then he's indicated that he'll be slowing down somewhat to focus on their relationship and other interests.
Chelsea Cutler, country duo Muscadine Bloodline and Ernest K will also play the Nashional.
Country's Hottest Men? Try These 10 On!


text:
Darwin's Dangerous Idea | Episode 1: Body and Soul
Play all videos (6) | Episode 2
Aired March 5, 2009 on BBC 2
In the first episode of the three-part series, Andrew Marr explores how Darwin's theory of evolution by natural selection has taken on a life of its own far beyond the world of science.
He argues that Darwin's theory has transformed our understanding of what it means to be human. Over the last 150 years, Darwin's ideas have challenged the need for a creator, undermined religious authority, and provided new ways of looking at the origins of human morality.
Marr's journey begins following Darwin's footsteps in Tierra del Fuego at the southernmost tip of South America where Darwin first encountered an 'uncivilised' native tribe. This began to raise questions in his mind about the origins of the human race. The answers to these questions would emerge over the next 30 years, culminating in the publication of On The Origin of Species in 1859.
Marr then traces the development of Darwin's idea in the years since then and finds a range of influences that Darwin could never have imagined: from the existential philosophy of Friedrich Nietzsche to the battlefields of the First World War; from the Freudian psychoanalyst's couch to the Vatican; and from the genetic logic of kindness to an Islamic creationist's claim that Darwin is to blame for modern terrorism. Darwin's dangerous idea is as influential and challenging today as it was 150 years ago.
Thanks TroisVitesse
Tags: BBC, Catholic, Darwin, Documentary, Evolution


September 11, 2001, my husband and I were flying home to Portland from celebrating our twenty fifth anniversary in New England. We originally planned to fly out of Boston, but changed our plans and flew out of Providence, Rhode Island. While we were in line to check into our flight, the dark haired man with a thick accent in line in front of us was not allowed to board. There was a problem with his ID...we've often wondered about that.
While we were in the air to our layover at Chicago O'Hare, two hijacked planes slammed into New York City's twin towers. At Chicago, our plane kept circling, descending, then lifting back up. We knew something unusual was happening, but no announcements were made to explain. When we finally landed, we looked at the monitor to find our connecting gate. Suddenly every flight said, CANCELLED. Every t.v. in the airport shut down. It was a bit of a wait before the airline personnel told us about the twin towers, told us that all flights in the U.S. were cancelled and why. I assume they were trying to keep one of the busiest airports in the US from becoming a place of mass panic. We realized that the reason our plane had circled so long, was the wait for so many other planes to land.
Our emotions at the news, indescribable. Shock, fear, confusion, concern about our family at home. The airport was a mass of people frantically dialing their cell phones over and over, trying to reach their loved ones. Finally, we were able to contact our children by cell phone. While on the phone with our son in law, he saw on his television that the Pentagon was hit. He screamed into the phone, "we're under attack!" There was no doubt now.
Thousand of strangers in that airport became family. I'll never forget how, when push came to shove, people banded together like that. Rental cars were shared. People helped each other find lodging. In the midst of it all, tears and frantic phone calls continued all around us, especially by those who had just left their family in New York City a few hours before. Yet the atmosphere was not chaotic. Everyone was polite, caring, helping one another.
We ended up finding a vacant hotel room in a suburb outside the city. The shuttle to the hotel was filled with stranded ones like us. We were all equally shocked, equally confused at what had really happened. None of us had seen a television or heard a radio so nothing really made sense.
In the hotel room, the first thing we did was turn on the television. My husband and I held each other and cried when we saw what had happened while we had been in flight, blissfully ignorant. We sat on the bed, eyes glued to the television hour after hour. We saw Congress sing, "God Bless America" on the capitol's steps. We were sure that night that America would never be the same. We were sure that the attack would cause America to return to her foundation, return to God.
For a while, we did, but now, I fear that we have forgotten.
I thought it would change my life forever...make me always remember what's really important. How easily we slide back into the status quo.
I was in my classroom preparing to teach school. I saw the plane crash into the building and minutes later my students started coming in. The principal sent out a message that we should not say anything about it to them--she thought it better left for the parents to tell them.
I had to teach all day without turning a tv on or listening to a radio. It was a very hard day of teaching. One of the worst I experienced.
But, Elizabeth, here I have been following you for years and I never knew this story. How frightening to be in the situation you were in. That is an amazing story.
Thank you for re-sharing this post Elizabeth. It truly is a day we will never forget. You happened to be near by on this day...I live no more than 10 minutes from O'Hare. I can only imagine what it must have been like to be traveling that day.
We could have stayed at your house! LOL! We ended up taking a very crowded Amtrack train ride home the next day.
We will always remember where we were on that day! Friends were scheduled to fly home from Washington D.C. and their story is much like yours. I'm so glad you were safe - only God knows the identity and plans of the man who was turned away.
Thank you for sharing, Elizabeth. I had yet to hear about what that day was like for those flying elsewhere and being stranded in airports. Thanks for helping me remember.
That was a day that definitely changed America forever! With the confusion that surrounded the first flight, less prayer, but with the second flight hitting the same place it was more apparent that this was deliberate and not an accident. I believe at that point more targeted prays began going up and is probably what helped thwart the Pentagon crashes from being more severe. The brother of one of my international college friends was on one of the flights heading home to his family.
I wish it had turned America to God in a way that was more permanent!


text:
News / National / New Army Chief vows final blow vs. communist insurgents, terrorists, secessionists
New Army Chief vows final blow vs. communist insurgents, terrorists, secessionists
Published May 18, 2021, 2:08 PM
by Martin Sadongdong
Major General Andres Centino formally assumed post as the 64th Commanding General of the Philippine Army (PA) on Tuesday, May 18, with a promise to deliver the final blows against communist insurgents, terrorists, and secessionist groups in the country.
Defense Secretary Delfin Lorenzana presides over the Change of Command ceremony between new Philippine Army Chief Maj. Gen. Andres Centino and outgoing Army Chief Lt. Gen. Jose Faustino Jr. at Fort Bonifacio in Taguig City on May 18, 2021. (Photo courtesy of Department of National Defense)
Defense Secretary Delfin Lorenzana presided over the Change of Command Ceremony between Centino and Lt. Gen. Jose Faustino Jr. at the PA Headquarters in Fort Bonifacio, Taguig City. It was graced by Armed Forces of the Philippines (AFP) Chief General Cirilito Sobejana.
"You are expected to guarantee the Philippine Army's readiness in dealing with current and emerging multi-faceted challenges and equally important, you have to ensure that your human resources are well trained and ready to face the challenges ahead," Lorenzana told Centino.
In his speech, Centino said the country is "on the brink of victory" in its fight against communist, terrorist, and secessionist groups and all that needs to be done is to continue the government's "whole of nation" approach.
"Today as we simultaneously wage war in different fronts, working within a whole of nation framework, the Army, all stakeholders, our country is on the brink of victory against communist insurgency, secessionism, and terrorism," he said.
"The Army must remain focused to continuously improve on our readiness in personnel management, operations, intelligence, logistics and training. This [is] to provide our soldiers the advantage in the fights that lie ahead. These priority areas will serve as a framework to build further our Army [while] anchored on the guidelines of the Army Transformation Roadmap," he added.
Centino is a graduate of the Philippine Military Academy "Maringal" Class of 1988.
Prior to the post, Centino served as the commander of the 4th Infantry "Diamond" Divison based in Cagayan de Oro City where he led the anti-communist insurgency campaigns in Northern Mindanao (Region 10) and Caraga (Region 13).
"In the Northern Mindanao and Caraga region where I have served, the inter-agency cooperation and collaboration have enabled our Army units in the field to engage the enemy decisively, resulting in dismantled guerilla fronts, reduction in the enemy strength and the liberation of communities from the clutches of the communist-terrorist group," he said.
He also served as the battalion commander of the 26th Infantry Battalion in 2008; Secretary Army General Staff in 2013; and commander of the 401st Infantry Brigade in 2017.
He was also assigned as the deputy chief of staff for operations, J3, at AFP General Headquarters in Camp Aguinaldo in 2019 befoer he eventually became the 4ID commander in 2020.
But as the new Army Chief, Centino is expected to deal much more than that as the Army tries to adapt to the evolving nature of security threats in the country.
Centino said that under his leadership, the Army will focus on ending the scourge of the Communist Party of the Philippines and its armed wing, the New People's Army (NPA), as well as the militant Abu Sayyaf Group (ASG), Bangsamoro Islamic Freedom Fighters (BIFF), and Dawlah Islamiyah (DI) among others.
Meanwhile, Lorenzana thanked Faustino for leading the Army albeit only in a short period.
"A lot of people were asking why he's been relieved very early. This is unprecedented in the history of the Philippine Army when a Commanding General will leave before his time due, but we have to do it to address some legal infirmities in his designation as Commanding General," Lorenzana said.
Faustino's designation as an acting Army chief was questioned by Senator Panfilo Lacson since it is a violation of Section 4 of the Republic Act 8186 which prohibits the designation of any officer as AFP Vice Chief of Staff, AFP Deputy Chief of Staff, or AFP Major Service Commander "if he has less than one year of active service remaining prior to compulsory retirement."
Faustino, who took office as acting Army chief in February, is set to retire in November.
"I think the higher-ups made a mistake," Lorenzana said.
"There was a very big challenge on the legitimacy of his being Commanding General of the Philippine Army so he had to be removed and placed on a position of major responsibility as well while waiting for a designation to more higher or higher position," he added.
Faustino was transferred to the Office of the Chief of Staff, AFP as a "special assistant on peace and development."
Addressing rumors that Faustino is being groomed to be the next AFP Chief, Sobejana said the public should not preempt President Duterte and just wait for the Commander-in-Chief's decision.
"Hintayin na lang po natin kung ano 'yong susunod na kabanata (Let us just wait for the next chapter," said Sobejana, who is set to retire on July 31, 2021.
cpp npa
Philippine army
secessionism
["national","news","news"]


Six months in French and American oak.
Deep crimson red in color, aromas of ripe red berries, especially raspberries and cherries.The nose of this wine, combines with rosemary and mineral notes. Fruity on the palate, with full body and ripe tannins. Complex and elegant wine.
Great with steak, game or poultry.


We Offer A Variety of Services To Maximize Our Clients Online Presence and Exposure!
We create "Call to action Videos" for any type of Business and for any Niche.
We can get top ranking for your "CTA Videos" any type of Business and for any Niche.
We will distribute and publish your "CTA Video" on all Major video networks.


Gum recession is a condition in which the gum line has moved from its original position, leaving tooth roots exposed and susceptible to decay and infection, which in turn, can lead to more serious problems.
Gum recession is often the result of gum disease but it can also be caused by genetic disposition and/or overly aggressive brushing.
Until recently, the primary treatment method for receding gums was gum grafting, also known as soft tissue grafting. This generally involves removing soft tissue from the palate or cheek, or from a donor source, and stitching it in place over the exposed tooth root.
The Chao Pinhole® Surgical Technique is an exciting new procedure, providing a minimally invasive alternative to gum grafting. This new treatment can be used to restore gums to their natural position. A few small "pinholes" are created in the soft pliable tissue above or below the exposed tooth. Then, using these pinholes as access points, specially designed instruments are used to loosened and reposition recessed gum tissues to cover the exposed root. Collagen strips are then placed in the gum tissue between the teeth to keep gums in place and promote rapid healing. The result is a healthier and more beautiful smile.


text:
Yarmolenko was found dead on the banks of the Catawba River, near where Carver had been fishing with his cousin. The cousin died of a heart attack just before trial, and Carver was found guilty and sentenced to life in prison.
This week, one member of Carver's jury told the Observer that the video of the height gestures would have changed everything.
It wasn't introduced during the trial, but it was shown in Gaston County court Tuesday, on the first day of a hearing that Carver's defense team hopes will lead to a new trial or full exoneration.
The way that the detective described Carver's gesture in 2011 was misleading, juror John Little of Dallas said.
The video shows Carver didn't spontaneously describe the victim, Little said. Instead, he seems to copy the detective's motions without being sure of what was going on.
Little said that until last month, he also didn't know that Carver has a hard time understanding and processing questions, which adds context to the story of his interview with the detective.
The leader of Carver's postconviction defense team, N.C. Center on Actual Innocence executive director Chris Mumma, presented evidence about Carver's mental and physical disabilities in court this week.
Little said he drives across the Catawba River, near where Irmolenko's body was found, every day. He's thought about the trial plenty in the past eight years, but recently, he said he's become convinced of Carver's innocence.
Ira Yarmolenko, a 20-year-old UNC Charlotte student, is strangled on the banks of the Catawba River. Mark Carver is convicted of her murder. Five years after the trial, questions about the case linger.
Mumma sent letters to each juror earlier this year, asking them if they would be interested in talking about the case. Most jurors didn't want to talk, Mumma said, but Little responded, and the defense team showed him the video.
Little ended up signing an affidavit confirming that if he had all the information he has now, he would not have convicted Carver. But the affidavit won't be introduced during this week's hearing, Mumma said, because case law prohibits it.
Gaston County District Attorney Locke Bell did not respond to two requests for comment through his office.
Little said the lawyers also told him more about issues with the DNA in the case, which he was skeptical about during the trial.
"Myself and another lady, we were the holdouts," he said, adding that some of his fellow jurors wanted to wrap up the process quickly with a guilty verdict.
And the jurors were sure that someone had killed Yarmolenko and it hadn't been suicide, he said.
Carver's been in prison for eight years, which made Little realize something terrible: "We have a killer walking among us."
death-by-the-river
Death by the river: At trial, a detective's testimony changes everything
Convicted man is mentally disabled, did not kill UNCC student in 2008, lawyers say
Jane Wester
Jane Wester is a Charlotte native and has been covering criminal justice and public safety for The Charlotte Observer since May 2017.
Police chief talks about Darnell Harris, the Steak 'n Shake shooting victim
2 injured in shooting at Le Méridien hotel in uptown Charlotte
Man suspected of stealing phones died after worker held him down by neck, autopsy says
By Jane Wester
A man died in Charlotte, NC, after police say he tried to steal cell phones from a business, where employees fired a gun and pinned him to the ground, wrapping an arm around his neck, his autopsy said.
Victims of double homicide in southeast Charlotte identified by police
Second suspect charged with killing Steak 'n Shake worker who died a hero
Charlotte police identify Steak 'n Shake shooting victims and murder suspect
Trooper 'separated' from highway patrol as SBI probes ticketing in Harnett County
Worker killed in Charlotte restaurant shooting. Suspect wounded by a police officer
Principal accused of sending nude photos to student faces sex crime charges, NC cops say