text:
Campus Consortium
University of California Berkeley Graduate School of Journalism reporting fellow Sawsan Morrar, Columbia University Graduate School of Journalism reporting fellow Max Toomey, Washington University in St. Louis reporting fellow Tomek Cebrat, Davidson College reporting fellow Aman Madan, and University of California Berkeley Graduate School of Journalism reporting fellow Anna Clausen meet to discuss their work. Image by Kayla Sharpe. United States, 2017.
Reporting fellows spend two days in Washington D.C. presenting their reporting projects and networking with their peers. Image by Jin Ding. United States, 2017.
Welcome to the Pulitzer Center Campus Consortium Network
Learn More About Student Reporting Fellowships
The Campus Consortium is a network of partnerships between the Pulitzer Center and universities and colleges to engage with students and faculty on the critical global issues of our time.
At its core, the initiative aims to connect international reporting supported by the Pulitzer Center directly with communities across the U.S., expanding knowledge of the world, sparking conversations across disciplines and inspiring individuals to expand their horizons. We aim to accomplish these goals through campus visits by journalists and international reporting fellowship opportunities for students.
We are proud of our growing network of diverse partners: community colleges, state universities, liberal arts colleges and historically black colleges and universities. We coordinate with each campus on how best to launch our partnership and develop programming utilizing the expertise of Pulitzer Center staff and journalist grantees, and faculty and community leaders.
Campus Programs: The Center is eager for university departments to work together on programming to foster broader discussions and more nuanced analysis of concerns that span disciplines, from journalism to public health. This interdisciplinary approach benefits from the diverse topics Pulitzer Center grantees report on: resource extraction, climate change, maternal health, HIV/AIDS to name a few. Campus programming range from multi-day journalist visits to year-long for-credit seminars.
Reporting Fellowships: Students at Campus Consortium partners undertake international reporting projects they propose. Center staff and journalists act as mentors. Final projects are featured on the Pulitzer Center website, with publication possible via other news-media outlets. See reporting by fellows here.
The annual fellows weekend in Washington, D.C., further broadens the students' horizons, exposing them to the work of their peers and providing opportunities to network with professionals in journalism and other fields.
Costs: Average cost to a Campus Consortium partner is $10,000 to $15,000 per year. Cost for non-U.S. universities may range higher and modified Consortium partnerships are possible. At least $3,000 of the funding is reserved for the student reporting fellowship. The Center provides logistical and promotional support for Campus Consortium programming and mentorship for all reporting fellows via professional journalists and in-house staff from the onset of the projects through to completion.
To learn more about how your college or university can join the Pulitzer Center's growing Campus Consortium network, contact Ann Peters at 202-797-5267 or apeters[at]pulitzercenter.org or Holly Piepenburg at 202-734-3776 or hpiepenburg[at]pulitzercenter.org
April 13, 2012 | General news
Pulitzer Center's Jon Sawyer Participates in Narrative Arc Conference
Pulitzer Center Executive Director Jon Sawyer focuses on non-profit funding and new media innovation at Boston University's Narrative Arc Conference.
April 13, 2012 | Project news, Education news
Pulitzer Center-Wake Forest University Panel Draws Crowd, Inspires Discussion
Jennifer McDonald
Hundreds hear from Jon Sawyer and Cynthia Gorney at Wake Forest University community event focused on child marriage.
April 12, 2012 | Education news
Innovation Summit: GW Moving the Planet Forward
Campus Consortium Partner George Washington University convenes symposium on "Moving the Planet Forward."
April 03, 2012 | Project news
Radio Program Highlights Pulitzer Center Partnership With North Carolina Universities
Listen to Wake Forest Journalism Director Justin Catanoso discuss his school's partnership with the Pulitzer Center, Guilford College and High Point University.
Clark Atlanta University
Columbia University Graduate School of Journalism
Forsyth Technical Community College
Georgetown University Berkley Center for Religion, Peace, and World Affairs
Guttman Community College
Hampton University
McGill University Global Health Programs
Northwestern University Medill
Syracuse University S.I. Newhouse School of Public Communications
Texas Christian University Bob Schieffer College of Communication
The George Washington University Milken Institute School of Public Health
University of California Berkeley Graduate School of Journalism
University of Missouri School of Journalism
Westchester Community College
Yale Program on Climate Change Communication


Want to grow your Customer Experience competences?
It is very important to realise that Customer Experience and having a customer-centric mindset is often a cultural process within organizations. For that reason, a majority of leaders expressed that the existing culture within the company forms a barrier that has to be taken down before they can start managing daily operations with the customer in mind.
There are different approaches for looking at ways in which the customer can be at the centre of the organization, and two vivid examples really spring to mind. Steve Jobs always started with the customer experience and worked back towards the technology. Secondly, Richard Branson who, in contrast to the previous example, only looks from an 'employees-first' point of view. His view is that when you treat your employees well, they will love their jobs and bring that same good feeling to the customers who, in their turn, have a positive feeling about the organization and are willing to do more business with you.
Not every organization needs to have the exact same mindset as Richard Branson to achieve happy customers, but you need to have happy employees to really make a difference. HR has an important role in becoming customer-centric in all phases of the HR Cycle.
It starts before your onboarding process. During the recruitment process, recruiters need to focus more on talent or attitude and not so much on skills. Skills can (and should!) be learned and taught when your hire is settling into their job. Also, your recruiters need to recruit with purpose and really look for a good fit with the organization. And keep in mind: with every pair of hands you get a brain for free.
When you hire an employee, inspire them to pursue continuous development, participate in training sessions, courses and study. In that way, you give them the power to develop what they really want. You can also make this competitive or place development in a game-setting in order to make it more fun and an enjoyable challenge to keep on developing.
It is very important to drive cultural change within the organization as a whole. This is a continuous process that might be difficult to implement or keep alive, but it is worth it and employees will adopt some or all of the best practices. After all, they are designed to give employees a happier workplace.
In the masterclass, we had fun in an assignment to create a Museum of Cultural Change, where all participants shared Best practices.
Give the customer a chair in the meeting. Make a visual representation of your customer and assign him/her to a seat in the Boardrooms. In this way, the customer is always visible.
The first thing that may come to mind when thinking about rewarding employees for their efforts in customer excellence, may be to give them a bonus when their NPS has reached a certain level. This approach is brilliant, but be aware of possible gaming. You could also reward those who are positively named in customer surveys, give recognition to people behind the scenes and empower peers to celebrate each other's work. Or make it easier for your employees to balance their work/life, offer benefits that reduce stress levels and treat the people who deal with tough customer experiences with the utmost respect.
There are lots of ways to encourage your employees to work in a more customer-centric manner, you only have to be creative and persist in the execution! It won't be easy, since we all know Culture eats Strategy for breakfast. But this is where success is made and where the most fun can be had.
These are only some of the highlights Milou took away from attending the Customer Experience Masterclass. Would you like to know more? Find out more about our CX Masterclass options here.
This post is part of a wider series about all the 6 CX disciplines that represent the CXPA Framework around which the CCXP exam is structured and that we cover in the CX Masterclass.
Follow Rosaria Cirillo on Linkedin or complete the form below to be notified as soon as a new blog post is published.
Customer experience is the perception customers have of the interactions with your company along every touchpoint and communication. Great customer experiences don't happen by chance, but require the orchestration of 6 disciplines across different team. Would you like to learn more and grow your Customer Experience competences? Join our next CX Masterclass.


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Q: Stealing bits from a std::shared_ptr I am currently working on a class assignment where I need to implement a lock-free linked list.  The structure of each of my nodes is essentially:
class Node {
    std::shared_ptr<Node> next;
    long long             key;
}

I need to somehow embed an extra bit of data into the "next" pointer.  I cannot use an extra boolean field, because the "next" pointer needs to be updated atomically using compare_exchange_strong().  I also have to use a std::shared_ptr for its garbage collection abilities; performing memory reclamation on a lock-free data structure is beyond the scope of what we have studied in the course so far.
If I were using a plain old pointer, I could use bit manipulation operators to "twiddle" the high bit of the pointer, but clearly this won't work, because once I twiddle the bit, the pointer points to an invalid memory location, and when the assignment operator attempts to access the control block, it will cause a segmentation violation.
Can anyone offer any insights on how I can accomplish this?
Oh, and for those who are curious, I am using g++-4.8.2 on a Linux system.

A: Seems like you are trying to mix two different things here.
Meddling with the pointer in a shared_ptr does not strike me as a good idea. The shared pointer is thread safe, but you can accomplish that yourself if you are using atomic operations.
Anyway, a custom deleter as suggested before is done like this;
struct d { void operator()(foo* f){ delete f; } };

...and you create a shared_ptr w/this using;
shared_ptr<bar> b(new bar, d());

...but why not use atomic operations to start with?
Cheers!

redpajama_set_name: RedPajamaStackExchange


text:
Chester le Street Town vs Hartlepool United vs Chester le Street Town vs Hartlepool United
Durham County Women's Cup Sunday 20 January @ 1:30 PM Moor Park
After a very good away win (1-2) on Tuesday night at Norton, the Chester le Street Town Ladies were all ready to take on Hartlepool United Ladies in the quarter final of Durham FA's Ladies County Cup. A competition Chester have won 3 times and were runners up last season to Norton, in the last 4 seasons.
Hartlepool United Ladies play in the North East Regional Women's Football League (NERWFL) Premier Division and are currently 3rd in that league. The league is just 1 level below Chester so it was expected Hartlepool would put up a very good fight in trying to get through to the semi-finals.
The Cestrians were in a very relaxed mood as they turned up at Moor park, although it was soon apparent goalkeeper Robyn Parker would not be playing starting (although she did make an appearance near the end of the game) and Rachel Mellor would once again be in the number 1 shirt. New signing Kira Whittle was making her full home debut wearing the number 10 shirt.
Within the first 5 minutes Hartlepool showed they be no push overs and were up for the game. Chester tried to press Hartlepool's defence a couple of times but they came to nothing.
On 12 minutes the Cestrians continued to press forward, Sarah Burn took a throw on the right side to Anna Collin, who looked to put a cross into the box, but Hartlepool's defence were with her all the way and the ball went out for a goal kick.
Shortly after Hartlepool attacked the Chester goal but Rachel Mellor dealt with the high ball and was able to clear it.
Jade Clark was assigned to take a free kick about 25 yards out from the Hartlepool goal, unfortunately her effort sailed over the cross bar.
On 25 minutes Kira Whittle put a long throw in into the penalty area but it was thwarted by Hartlepool's defence. Chester thought they had a good shout for a penalty on 30 minutes, when Jessica Ellison looked to have been fouled inside the 18 yard box, but the referee waved it away.
As the first half progressed Hartlepool were forced to clear their lines quickly to keep Chester away from their 18 yard box and giving themselves a chance to re-group.
The first half was mostly a midfield battle as both teams struggled to put on a decent attack; until the 41st minute. A fantastic pass in behind the Hartlepool back 4, Kacie Elson raced through to beat the offside trap and smashed and early first time shot from 20 yards, which flew into the top right corner of the goal – 1 nil Chester.
Happy to go into half time 1 nil up, but everyone was hoping to see an improved performance in the second half.
Straight from the whistle in the second half Chester looked to be a completely different team; much better passing and build up play.
On 54 minutes Sarah Burn's went in for a 50/50 challenge but took a heavy knock. After a couple of minutes treatment and recovery on the side line, she was quickly back on the pitch.
65 minutes and some more good link up play between Katie Ellison and Kacie Elson but again dealt with by Hartlepool.
Lizzy Thompson-Clarke came off the bench and was soon in the thick of things, (having suffered from a hamstring injury in the last few games), she seemed to be running a lot easier. On 76 minutes she broke past one of Hartlepool's defenders and as the keeper came off her line, Lizzy took her chance and put the ball in the top left corner – 2 nil. A very composed finish. There was a sense of relief around the ground, but the players knew the game wasn't over.
Not long after and Lizzy was running at goal again, she took a knock on the back of her foot just inside the 18 yard box, as she went past the defender. She quickly turned her back to the keeper and set the ball back to Kacie Elson, who finished first time with another good goal, this one with her left foot.
Final score 3-0, there would be no scalps today.
Carl Chater (coach) had the following to say on both the game and the new recent signings for the team.
"We started slowly and made life difficult for ourselves in the first half but completely dominated the second half. We needed to be better in possession getting the ball forward with quality so we could exploit the space with our pace. We did this a number of times in the second half scoring 2 and maybe should have scored more. We're happy with how the last 2 games have gone and how the new additions have started. We now have competition for places all over the pitch which can only be a good thing. This will hopefully now push us on for the rest of the season."
I also asked Anna Collin (captain) a few questions after the game.
What did you make of today's overall team play?
AC; "It wasn't one of our better performances this season but thankfully we've come away with the win. We took a while to get going and never really passed the ball to the standard that we are capable of."
What can you take away from the game to improve on as a team?
AC; "It was a scrappy game today on a bobbly pitch which made it difficult to play good football. Some of our decision making is poor so lots to work on in training."
There have been some new players coming into the team how do you think they have done so far?
AC; "We have lots of new players for both teams which is great to see. They've all fit into the squad really well. I thought Jade Clark was great against Norton and Shannon Etherington played really well today."
Chester le Street Ladies 3 - 0 Chester le Street Town vs Hartlepool United vs Chester le Street Town vs Hartlepool United
Sunday 20 January 2019 @ 1:30 PM
Durham County Women's Cup
Chester United
Elizabeth Clarke	 1
Chester United	 2
20 Jan 19 1:30 PM	 Durham County Women's Cup Chester le Street Ladies vs Chester le Street Town vs Hartlepool United vs Chester le Street Town vs Hartlepool United	 3 - 0	 Match Centre
16 Jul 19 7:30 PM	 Pre-season Friendly Chester le Street Town vs Chester le Street Town vs Hartlepool United vs Chester le Street Town vs Hartlepool United	 Match Centre
Previous Match Chester le Street Town vs Durham City AFC
Next Match Chester le Street Town vs West Allotment Celtic

redpajama_set_name: RedPajamaCommonCrawl


text:
Chinese electric car brand NIO says 2 killed in vehicle fall
The logo for Chinese automaker NIO is seen at the Shanghai Auto Show in Shanghai on April 19, 2021. Two people involved in testing for the electric car brand NIO died when one of its vehicles fell three stories from a Shanghai parking structure on Thursday, the company said Friday, June 24, 2022. (AP Photo/Ng Han Guan)[ASSOCIATED PRESS/Ng Han Guan]
BEIJING (AP) — Two people involved in testing for the electric car brand NIO died when one of its vehicles fell three stories from a Shanghai parking structure, the company said Friday.
The crash Thursday was under investigation but appeared to be an accident and "not caused by the vehicle," the company said in a statement. It said the employees who died were "digital cabin testers," one from NIO and the other from a partner.
The vehicle fell from the third floor of a parking structure adjacent to the company's Shanghai Innovation Port building, the company said.
Photos in Chinese media showed the vehicle lying on its side with its roof caved in, surrounded by broken glass and emergency workers.
NIO Inc., founded in 2014, is a Chinese-Western hybrid with bases in Shanghai, London and Silicon Valley and the most prominent premium-priced Chinese brand in a crowded electric vehicle industry with dozens of competitors. Its early investors included Chinese tech giants Tencent, Baidu and Lenovo.
The company has a service that allows owners to swap drained batteries instead of recharging them. It has set up 143 battery-swap stations in cities including Beijing and Shanghai.
NIO Inc.: www.nio.com


Somehow Little Miss is already six months old. I know. Crazy talk.
That means it's time for her six month appointment!
She's always so happy in the waiting room!
Once we got back into the room, it was time to weigh the big girl!
She weighed in at fifteen pounds twelve ounces. Little peanut. Sitting in the forty sixth percentile!
Then they measured her. Twenty-seven inches! The ninetieth percentile.
Olivia loved that the little table they have her sit on is covered in paper. It's her favorite toy.
She pretty much destroyed it.
Our pediatrician came in and we chatted about what she's about to and any concerns I had. There weren't a whole lot. He commented on how great she was sitting up for her age. Which was really reassuring since Preston was crawling at this point. I told our pediatrician that and he told me that Preston was definitely not the norm and Olivia was sitting better than most babies her age! Way to go Livs!
Then came my least favorite part. Shots.
Olivia did so awesome. She only cried for a couple of seconds. Then was over it. The nurse was trying to sneak out since she said she knew Liv must not be happy with her. But Liv was irritated that the nurse wouldn't pay attention to her.
We get to go back in a month for her second flu shot and then at nine months but she doesn't need shots again until she's one I think! Hurray!


text:
Śląsk Song and Dance Ensemble (full name: Polish National Song and Dance Ensemble "Śląsk" in memory of Stanisław Hadyna, in Polish: Zespół Pieśni i Tańca "Śląsk" im. Stanisława Hadyny) is one of the largest Polish folk ensembles. It was founded on 1 July 1953 by Stanisław Hadyna and is named after the Silesia ("Śląsk") region. It is based in Koszęcin.

Stanislaw Hadyna, the son of a Silesian teacher, is well known for his collection of folk songs. The world of legends and fantasies, ever-present in the tunes of Silesia, fascinated him from his earliest years and made him decide to form an ensemble which could render the richness and originality of Silesian folklore. To achieve his aim he needed talented and sensitive young people who were as yet unmannered by the stage. The performers were chosen from among 12,000 candidates and the castle in Koszecin became their home base. The first performance by "Slask" in the Autumn of 1954 in Warsaw was something of a revelation for the whole country.
The aim of the group is to show Silesia's age-old folklore which retains its own very specific identity in a large and much varied area. (excerpts from notes on LP "Slask, The Polish Song and Dance Ensemble, vol. 3". Muza. Polskie Nagrania SX 348)

Śląsk originally focused on the folk traditions of Upper Silesia, Cieszyn Silesia, and the Beskids. It was later expanded to include all Polish regions. The ensemble has performed more than 6,000 shows for over 20 million people worldwide. It has performed in the United States, Canada, Australia, France, Germany, Italy, Belgium, Sweden, the Netherlands, China, Mexico, and many other countries, as well as the Vatican.

The Śląsk ensemble has worked with such notable Polish composers as Wojciech Kilar.

Some of its most notable songs are Szła dzieweczka, Helokanie, Ondraszek, Głęboka studzienka, Karolinka.

Awards
 Music and Art Award of Silesia (Silesian Music Prize in 1958, Artistic Prize in 1965)
 Honorary award from the President of the United States
 Honorary award from the ministry of culture of the Soviet Union
 Award from the Silesian Voivodeship
 Gold Medal from The World Peace Council in Stockholm in 1959

References

External links

Official site

Silesian culture
Polish dances
Polish music
Polish folk groups
1953 establishments in Poland

redpajama_set_name: RedPajamaWikipedia


branches 1 13 years Sam Hocevar * Creating repository.
tags 49 11 years Sam Hocevar * Imported original public domain release, for future reference.
trunk 48 11 years Sam Hocevar * Add autom4te.cache to the svn:ignore list.


text:
Downtown Revitalization and Historic Preservation in one Local Community
with Paul Kanitra, Mayor, Borough of Point Pleasant Beach
https://comcastnewsmakers.com:443/Videos/2021/10/12/Downtown-Revitalization-and-Historic-Preservation-in-one-Local-Community
Point Pleasant Beach Mayor Paul Kanitra discusses downtown revitalization and historic preservation and how his community has found a way to blend the two concepts in their local community. Recorded October 8, 2021.
Hosted by: Jill Horner Produced by: Freedom Newsmakers Team
#New Jersey #Economy and Business #Community #General Interest #Government #Local
Other videos hosted by Jill Horner
Increasing Mental Health Services for County Residents
Diane Ellis-Marseglia, Vice Chair of the Bucks County Commissioners, discusses mental health services available through the county. Recorded January 14, 2022.
The Potential of Expanding Passenger Rail Options Including Between Philadelphia...
The Potential of Expanding Passenger Rail Options Including Between Philadelphia and Reading
Pennsylvania State Representative Joe Ciresi discusses what bringing back a passenger rail route between Philadelphia and Reading could mean for the region. Recorded January 14, 2022.
Camden County is Focused on Helping Workers Obtain the Skills they Need for Futu...
Camden County is Focused on Helping Workers Obtain the Skills they Need for Future Careers
Camden County Commissioner Jonathan Young discusses The Camden County Workforce Development Board and what it will mean for the region. Recorded January 14, 2022.
Bringing Black History and Culture Beyond Museum Walls
Ralph E. Hunter, Sr., Founder and President of the African American Heritage Museum of Southern New Jersey, discusses the museum's pop up, traveling exhibits, including one on Jackie Robinson, and how they bring Black history and culture beyond museum walls. Recorded January 14, 2022.
Providing Children with Everyday Essentials including Clothing
Michal Smith, Executive Director of Cradles to Crayons Philadelphia, discusses the organization's efforts to provide children with everyday essentials including clothing, diapers, and toiletries. Recorded January 14, 2022.
Digital Skills for Young Adults
Dan Rhoton, Executive Director of Hopeworks, discusses how his organization works with 17 to 26 year olds to help them achieve their highest potential by teaching them the digital skills they need to succeed. Recorded January 14, 2022.
Collaboraction Theatre for Social Justice and Equity
All Collaboraction Theatre works are original. Collaboraction incites change through original devised theatre and deep community collaboration.
Greater Washington Urban League
For more than 80 years the Greater Washington Urban League has been empowering communities and changing lives, their President and CEO George Lambert sits down with Elena Russo to discuss how they are fighting for excluded and endangered populations.
Food-Insecurity in South Jersey
Jill Horner speaks with Joe Njoroge, Interim President and CEO from the Food Bank of South Jersey, about Food-Insecurity in South Jersey. Interview recorded 9/19/2018.
Don't Wait for a Tragedy to be Prepared. How to Do Business in the Future.
Don't Wait for a Tragedy to be Prepared. How to Do Business...
Founded in 1986, the mission of the WBDC is to support and accelerate business development and growth, targeting women and serving other underserved communities, in order to strengthen their participation in, and impact on, the economy.


text:
package org.apache.beam.runners.spark.translation.streaming;

import org.apache.beam.runners.spark.translation.EvaluationContext;
import org.apache.beam.runners.spark.translation.SparkRuntimeContext;
import org.apache.beam.sdk.Pipeline;
import org.apache.beam.sdk.coders.Coder;
import org.apache.beam.sdk.transforms.AppliedPTransform;
import org.apache.beam.sdk.transforms.PTransform;
import org.apache.beam.sdk.util.WindowedValue;
import org.apache.beam.sdk.values.PInput;
import org.apache.beam.sdk.values.POutput;
import org.apache.beam.sdk.values.PValue;

import org.apache.spark.api.java.JavaRDD;
import org.apache.spark.api.java.JavaRDDLike;
import org.apache.spark.api.java.JavaSparkContext;
import org.apache.spark.api.java.function.VoidFunction;
import org.apache.spark.streaming.api.java.JavaDStream;
import org.apache.spark.streaming.api.java.JavaDStreamLike;
import org.apache.spark.streaming.api.java.JavaStreamingContext;

import java.util.LinkedHashMap;
import java.util.LinkedHashSet;
import java.util.Map;
import java.util.Queue;
import java.util.Set;
import java.util.concurrent.LinkedBlockingQueue;

/**
 * Streaming evaluation context helps to handle streaming.
 */
public class StreamingEvaluationContext extends EvaluationContext {

  private final JavaStreamingContext jssc;
  private final long timeout;
  private final Map<PValue, DStreamHolder<?>> pstreams = new LinkedHashMap<>();
  private final Set<DStreamHolder<?>> leafStreams = new LinkedHashSet<>();

  public StreamingEvaluationContext(JavaSparkContext jsc, Pipeline pipeline,
      JavaStreamingContext jssc, long timeout) {
    super(jsc, pipeline);
    this.jssc = jssc;
    this.timeout = timeout;
  }

  /**
   * DStream holder Can also crate a DStream from a supplied queue of values, but mainly for
   * testing.
   */
  private class DStreamHolder<T> {

    private Iterable<Iterable<T>> values;
    private Coder<T> coder;
    private JavaDStream<WindowedValue<T>> dStream;

    DStreamHolder(Iterable<Iterable<T>> values, Coder<T> coder) {
      this.values = values;
      this.coder = coder;
    }

    DStreamHolder(JavaDStream<WindowedValue<T>> dStream) {
      this.dStream = dStream;
    }

    @SuppressWarnings("unchecked")
    JavaDStream<WindowedValue<T>> getDStream() {
      if (dStream == null) {
        // create the DStream from values
        Queue<JavaRDD<WindowedValue<T>>> rddQueue = new LinkedBlockingQueue<>();
        for (Iterable<T> v : values) {
          setOutputRDDFromValues(currentTransform.getTransform(), v, coder);
          rddQueue.offer((JavaRDD<WindowedValue<T>>) getOutputRDD(currentTransform.getTransform()));
        }
        // create dstream from queue, one at a time, no defaults
        // mainly for unit test so no reason to have this configurable
        dStream = jssc.queueStream(rddQueue, true);
      }
      return dStream;
    }
  }

  <T> void setDStreamFromQueue(
      PTransform<?, ?> transform, Iterable<Iterable<T>> values, Coder<T> coder) {
    pstreams.put((PValue) getOutput(transform), new DStreamHolder<>(values, coder));
  }

  <T> void setStream(PTransform<?, ?> transform, JavaDStream<WindowedValue<T>> dStream) {
    PValue pvalue = (PValue) getOutput(transform);
    DStreamHolder<T> dStreamHolder = new DStreamHolder<>(dStream);
    pstreams.put(pvalue, dStreamHolder);
    leafStreams.add(dStreamHolder);
  }

  boolean hasStream(PTransform<?, ?> transform) {
    PValue pvalue = (PValue) getInput(transform);
    return pstreams.containsKey(pvalue);
  }

  JavaDStreamLike<?, ?, ?> getStream(PTransform<?, ?> transform) {
    return getStream((PValue) getInput(transform));
  }

  JavaDStreamLike<?, ?, ?> getStream(PValue pvalue) {
    DStreamHolder<?> dStreamHolder = pstreams.get(pvalue);
    JavaDStreamLike<?, ?, ?> dStream = dStreamHolder.getDStream();
    leafStreams.remove(dStreamHolder);
    return dStream;
  }

  // used to set the RDD from the DStream in the RDDHolder for transformation
  <T> void setInputRDD(
      PTransform<? extends PInput, ?> transform, JavaRDDLike<WindowedValue<T>, ?> rdd) {
    setRDD((PValue) getInput(transform), rdd);
  }

  // used to get the RDD transformation output and use it as the DStream transformation output
  JavaRDDLike<?, ?> getOutputRDD(PTransform<?, ?> transform) {
    return getRDD((PValue) getOutput(transform));
  }

  public JavaStreamingContext getStreamingContext() {
    return jssc;
  }

  @Override
  public void computeOutputs() {
    for (DStreamHolder<?> streamHolder : leafStreams) {
      computeOutput(streamHolder);
    }
  }

  private static <T> void computeOutput(DStreamHolder<T> streamHolder) {
    streamHolder.getDStream().foreachRDD(new VoidFunction<JavaRDD<WindowedValue<T>>>() {
      @Override
      public void call(JavaRDD<WindowedValue<T>> rdd) throws Exception {
        rdd.rdd().cache();
        rdd.count();
      }
    }); // force a DStream action
  }

  @Override
  public void close() {
    if (timeout > 0) {
      jssc.awaitTerminationOrTimeout(timeout);
    } else {
      jssc.awaitTermination();
    }
    //TODO: stop gracefully ?
    jssc.stop(false, false);
    state = State.DONE;
    super.close();
  }

  private State state = State.RUNNING;

  @Override
  public State getState() {
    return state;
  }

  //---------------- override in order to expose in package
  @Override
  protected <InputT extends PInput> InputT getInput(PTransform<InputT, ?> transform) {
    return super.getInput(transform);
  }
  @Override
  protected <OutputT extends POutput> OutputT getOutput(PTransform<?, OutputT> transform) {
    return super.getOutput(transform);
  }

  @Override
  protected JavaSparkContext getSparkContext() {
    return super.getSparkContext();
  }

  @Override
  protected SparkRuntimeContext getRuntimeContext() {
    return super.getRuntimeContext();
  }

  @Override
  protected void setCurrentTransform(AppliedPTransform<?, ?, ?> transform) {
    super.setCurrentTransform(transform);
  }

  @Override
  protected AppliedPTransform<?, ?, ?> getCurrentTransform() {
    return super.getCurrentTransform();
  }

  @Override
  protected <T> void setOutputRDD(PTransform<?, ?> transform,
      JavaRDDLike<WindowedValue<T>, ?> rdd) {
    super.setOutputRDD(transform, rdd);
  }

  @Override
  protected <T> void setOutputRDDFromValues(PTransform<?, ?> transform, Iterable<T> values,
      Coder<T> coder) {
    super.setOutputRDDFromValues(transform, values, coder);
  }

  @Override
  protected boolean hasOutputRDD(PTransform<? extends PInput, ?> transform) {
    return super.hasOutputRDD(transform);
  }
}

redpajama_set_name: RedPajamaGithub


Warsaw - Lanka's Jataka Buddhist Tales written for the mankind to gain knowledge and morality originally n in prakrit Sri Lanka have been released in Polish on November 24, In Asia and Pacific Museum in Warsaw.
"We are proud of their achievements. It is our privilege to publish this work in Poland said Janusz Krzyzowski, head of IPCC, to the audience which included several well-known polish intellectuals from Polish Academy of Science, Colegium Civitas, Ibn Khaldun Institute in Warsaw and others.
Janusz Krzyżowski studied medicine in the Medical Academy in Łódź and philosophy at the Warsaw University. He received a MA Degree at the university and in 1964 and Doctor Degree in Charile at Humboldt University, Berlin. In 1971 he acquired himseif with prospects of application of yoga techniques in contemporary medicnine during his stay in Santa Cruz Institute, Bombay and Lonavla. In 1986 he was decorated with the Golden Cross of academic didactic activity. Dr Krzyżowski is a member of the Polish Society of Earth Science. As a Doctor he took part in scientific expeditions organised by the Polish Society of Earth Science: in l986 Tikicaca expedition, in 1987 dramatical expedition in Tibet, in 1990 the mountains of Thailand and others. He is a passionate traveler, he practices tramping. His takes the particular active interest in the countries of oriental culture.


text:
A-class submarine may refer to one of these classes of submarine:

 , a class of 13 submarines of the Royal Navy that served in World War I
 , a class of six built for the Royal Danish Navy, 1911–1914
 , a class built for the Royal Norwegian Navy in 1913
 , also later known as Soviet A-class submarine, submarines launched in 1916
 , a class of three Spanish submarines launched in 1917
 , a class of seven United States Navy submarines built before World War I; also known as the "A class"
  (also called the Archeron class), a class of sixteen submarines of the Royal Navy that served in World War II; also known as the "A class"
 The Japanese Type A submarine, three classes submarines:
 I-9-class submarine (Type A1 submarine)
 I-12-class submarine (Type A2 submarine)
 I-13-class submarine (Type AM submarine)

Submarine classes


text:
\section{Introduction}

The Lambda algebra $\Lambda$ was constructed in the ``six author
paper'' \cite{6-author}, and numerous authors have studied it over the
past 40 years.  In \cite{wang.j*;hopf}, Wang studied an algebra
endomorphism $\theta$ of $\Lambda$, and this endomorphism is the main
focus of this paper.

$\Lambda$ is the bigraded differential $\mathbb{F}_{2}$--algebra with generators
$\lambda_{n} \in \Lambda^{1,n+1}$ for $n \geq 0$, with relations
and differential given as follows:
\begin{gather*}
\sum_{i+j=n} \tbinom{i+j}{i} \lambda_{i-1+m} \lambda_{j-1+2m} \quad
\quad \textup{for} \ m \geq 1, n \geq 0, \\
d(\lambda_{n-1}) = \sum_{i+j=n} \tbinom{i+j}{j} \lambda_{i-1}
\lambda_{j-1}.
\end{gather*}
Let $A$ be the mod 2 Steenrod algebra.  The following result and its
unstable analogue are the main reasons for studying the Lambda algebra.

\begin{thm}[Bousfield et al.~\cite{6-author}]
\label{thm-6-author}
$H^{s,t}(\Lambda) \cong \ext{A}{s,t}$.  Indeed, one
can view $\Lambda$ as the $E_{1}$--term of the classical Adams spectral
sequence converging to the 2--component of the stable homotopy groups
of spheres.
\end{thm}

Given the free algebra $F = \textup{alg}_{\mathbb{F}_{2}}(\lambda_{0},
\lambda_{1}, \dotsc)$ of which $\Lambda$ is a quotient, define a map
$\wtilde{\theta} \co F \rightarrow F$ by $\wtilde{\theta}
(\lambda_{n}) = \lambda_{2n+1}$.

\begin{prop}[Wang \cite{wang.j*;hopf}]
\label{prop-wang-lambda}
$\wtilde{\theta}$ induces a map $\theta \co \Lambda \rightarrow
\Lambda$ of differential graded algebras, and this map is one-to-one.
\end{prop}

\begin{cor}
The subalgebra of $\Lambda$ generated by $\{ \lambda_{2n+1} :
n \geq 0 \}$ is a sub-differential graded algebra, which is isomorphic
to $\Lambda$.
\end{cor}

Note that the isomorphism $\theta \co \Lambda \smash{\xrightarrow{\cong}}
\im (\theta)$ doubles internal degrees.  There is an endomorphism, called
$\SSq^{0}$, of $H^{*}\Lambda = \ext{A}{*}$ which also doubles internal
degrees.  One way to describe this map is as follows: the graded dual
of $A$ is a commutative Hopf algebra over $\mathbb{F}_{2}$, and the Frobenius
map is a Hopf algebra map.  $\SSq^{0}$ is the induced map on $\ext{A}{*}$.
See May \cite[Proposition 11.10]{may;steenrod} for this result, and see
Palmieri
\cite{palmieri;quotient} for more discussion of $\SSq^{0}$ acting on
Ext.

\begin{prop}\label{prop-theta}
$\theta \co \Lambda \rightarrow \Lambda$ induces the map
\[
\SSq^{0} \co \ext{A}{*} \rightarrow \ext{A}{*}.
\]
\end{prop}

Bob Bruner told me how to prove this.

\begin{proof}
According to Priddy \cite{priddy;koszul}, $\lambda_{n}$ is
represented in the cobar construction by $[\xi_{1}^{n+1}]$.  According
to May \cite[Proposition 11.10]{may;steenrod}, $\SSq^{0}$ is induced by
the dual to the Frobenius, so $\SSq^{0}$ takes $[\xi_{1}^{n+1}]$ to
$[\xi_{1}^{2n+2}]$; thus it agrees with the map induced by $\theta$,
which sends $\lambda_{n}$ to $\lambda_{2n+1}$.
\end{proof}

The goal of this paper is to use the DGA endomorphism $\theta \co
\Lambda \rightarrow \Lambda$ to study the action of $\SSq^{0}$ on Ext over
the Steenrod algebra.  The main result is the following; the result
when $s \leq 3$ was first proved by Wang.

\begin{thm}\label{thm-main}
Consider $\SSq^{0} \co \ext{A}{s,*} \rightarrow
\ext{A}{s,*}$.  When $s \leq 3$, this map is
injective.  When $s=4$, the kernel is one-dimensional:
\[
\ker (\SSq^{0}|\Ext_{A}^{4}) = \Span (h_{0}^{4}).
\]
\end{thm}

This theorem is proved in \fullref{sec-bss} using a Bockstein
spectral sequence argument.  The Bockstein spectral sequence is also
of interest in its own right.  We also invert $\theta$ to construct
the ``cocomplete Lambda algebra'' in \fullref{sec-cocomplete}.

This paper studies some of the same issues as
\cite{palmieri;quotient}, but uses a different approach.  As with that
paper, this work raises more questions than it answers; here are
some.

\begin{question}
\begin{enumerate}
\item Is $(\SSq^{0})^{-1} \ext{A}{*}$ finite dimensional in each bidegree?
\item Even better, for fixed $s$, is there a uniform bound on
\[
\dim (\SSq^{0})^{-1} \ext{A}{s,t}?
\]
\item H\uhorn{}ng has asked the following: for fixed $s$, is there a bound
on the ``$\SSq^{0}$--nilpotence height'' of elements in
$\ext{A}{s,*}$?  That is, is there a number $N=N(s)$ so that, for
$z \in \ext{A}{s,*}$, either $(\SSq^{0})^{N}(z)=0$ or
$(Sq^{0})^{i}(z)\neq 0$ for all $i$?  See Hu'ng
\cite{hung;cohomology-steenrod} for more details.
\end{enumerate}
\end{question}

The Bockstein spectral sequence of \fullref{sec-bss} may be
helpful in answering the last question, and perhaps the cocomplete
Lambda algebra (\fullref{sec-cocomplete}) could be helpful for the
first two.

\section{Recollections on the Lambda algebra}\label{sec-lambda}

In this section, we recall some facts about the Lambda algebra.
$\Lambda$ was first constructed in the ``six author paper''
\cite{6-author}.  Wang \cite{wang.j*;hopf} gave alternate forms for
the relations and the differential, described the endomorphism
$\theta$, and used Lambda algebra calculations to prove the Hopf
invariant one problem.  See also Ravenel's green book \cite[Section
3.3]{ravenel;green}, which summarizes the basic facts about $\Lambda$,
and also describes the Curtis algorithm; Priddy's Koszul resolutions
paper \cite{priddy;koszul} gives a purely algebraic construction of
$\Lambda$; and Richter has given elementary proofs of some of the
properties of $\Lambda$ in \cite{richter;lambda}.

$\Lambda$ is a bigraded differential $\mathbb{F}_{2}$--algebra; it is defined to
be
\[
\Lambda = \textup{alg}_{\mathbb{F}_{2}}(\lambda_{0}, \lambda_{1}, \lambda_{2},
\dotsc) / (\text{relations}),
\]
graded by putting $\lambda_{n}$ in bidegree $(1,n+1)$.  There are two
forms for the relations and the differential, the ``symmetric'' form
\begin{gather}\label{reln-symm}
\sum_{i+j=n} \tbinom{i+j}{i} \lambda_{i-1+m} \lambda_{j-1+2m} \quad
\quad \textup{for} \quad m \geq 1, n \geq 0, \\
d(\lambda_{n-1}) = \sum_{i+j=n} \tbinom{i+j}{j} \lambda_{i-1}
\lambda_{j-1},
\end{gather}
and the ``admissible'' form
\begin{gather}\label{reln-admissible}
\lambda_{i} \lambda_{2i+1+n} = \sum_{j \geq 0} \tbinom{n-j-1}{j}
\lambda_{i+n-j} \lambda_{2i+1+j} \quad \quad \textup{for} \quad i, n
\geq 0, \\
d(\lambda_{n}) = \sum_{j \geq 1} \tbinom{n-j}{j}
\lambda_{n-j}\lambda_{j-1}.
\end{gather}
Wang \cite{wang.j*;hopf} showed that these two sets of relations
generate the same ideal in the free algebra on the $\lambda_{n}$'s.
One can show that $d$ behaves formally like taking the commutator with
$\lambda_{-1}$; Bruner noted this in \cite{bruner;augmented}.

We refer to the first grading in $\Lambda$ as the
\emph{homological} grading, and the second as the \emph{internal}
grading.  Note that some authors define $\lambda_{n}$ to have degree
$(1,n)$ rather than $(1,n+1)$, in which case the second grading is the
\emph{stem} grading, the difference between the internal and
homological gradings.

\begin{definition}
A monomial $\lambda_{i_{1}} \dotsb \lambda_{i_{s}}$ is
\emph{admissible} if $2i_{r} \geq i_{r+1}$ for $1 \leq r \leq s-1$.
\end{definition}

\begin{prop}[Bousfield et al.~\cite{6-author}]
\label{prop-admissible}
The admissible monomials form a basis for $\Lambda$.
\end{prop}

Note that, while it is easy to show from the relations above that the
admissible monomials span $\Lambda$, it is not trivial to show that
they are linearly independent.  Similarly, it is not obvious that the
differential $d$ respects the relations.  See \cite{richter;lambda}
for elementary proofs.

As remarked in \fullref{thm-6-author}, the cohomology
$H^{*}(\Lambda,d)$ is $\Lambda$'s raison d'\^etre.

\section{A Bockstein spectral sequence}\label{sec-bss}

Let
\[
\Lambda' = \Lambda / \theta \Lambda,
\]
and consider the short exact sequence of chain complexes
\begin{equation}\label{eqn-ses}
0 \rightarrow \Lambda \xrightarrow{\theta} \Lambda \rightarrow
\Lambda' \rightarrow 0.
\end{equation}
After taking cohomology, this gives a Bockstein spectral sequence
with
\begin{gather}\label{eqn-bss}
E_{1}^{s,t,u} =
\begin{cases}
H^{s+t,u}(\Lambda'), & \text{when} \ \ s \geq 0, \\
0 & \text{otherwise},
\end{cases} \\
d_{r} \co E_{r}^{s,t,u} \rightarrow E_{r}^{s+r,t-r+1,u/2^{r}},
\end{gather}
converging to $H^{s+t,u} (\Lambda) = \ext{A}{s+t,u}$.  The phrase
``Bockstein spectral sequence'' is perhaps ambiguous, so some details
may be helpful.  One way to construct it is to filter $\Lambda$ by
setting $F^{s}\Lambda = \im \theta^{s}$.  Because $\theta$ is
injective, $F^{s}\Lambda$ is isomorphic to $\Lambda$; more precisely,
$F^{s}\Lambda$ in bidegree $(i,2^{s}j)$ is isomorphic to $\Lambda$ in
bidegree $(i,j)$.  Up to a similar doubling of internal degrees,
$F^{s}\Lambda / F^{s+1}\Lambda$ is isomorphic to $\Lambda'$.  The
Bockstein spectral sequence is the one which arises from taking
cohomology of this filtered chain complex.  Thus for fixed $s \geq 0$
and $t$ we have $E_{1}^{s,t} \cong E_{1}^{s+i,t-i}$ for all $i$
with $i \geq 0$.  Given a class $x \in E_{1}^{s,0}$, we write
$\wbar{\theta}^{-i} x$ for the image of $x$ under this
isomorphism.  The differentials are ``$\wbar{\theta}$--periodic''
and determined by their effect on $E_{r}^{*,0}$. For any $x \in
E_{r}^{s,0}$, there is a differential
\[
d_{r} \co x \longmapsto \wbar{\theta}^{r} y,
\]
if and only if for any $i \geq 0$, there is a differential
\[
d_{r} \co \wbar{\theta}^{i} x \longmapsto \wbar{\theta}^{i+r} y.
\]
Furthermore, as the notation suggests, the differentials determine
extensions in that they reflect the action of the map induced by
$\theta$, namely $\SSq^{0}$.  That is, if $x \in H^{n}(\Lambda')$ is an
infinite cycle and not a boundary in the spectral sequence, then it
survives to a ``$\SSq^{0}$--periodic'' element $\wbar{x}$ in
$H^{n}(\Lambda)$. More precisely, in this situation
$\wbar{\theta}^{i} x$ survives to $E_{\infty}^{n+i,-i}$ for each
$i \geq 0$, and this infinite family corresponds in the abutment to
the family
\[
\{(\SSq^{0})^{i}\wbar{x} \in H^{2^{i} n}(\Lambda) : i \geq 0 \}.
\]
The presence of a differential $d_{r} \co x \longmapsto y$ from a class
$x \in H^{n,u}(\Lambda')$ to a class $y \in H^{n+1,u/2^{r}}(\Lambda')$ means
that $r$ ``copies'' of $y$ survive to $E_{\infty}$: the elements
$\wbar{\theta}^{i}y$ survive for $0 \leq i \leq r-1$.  These
elements correspond in the abutment to a finite $\SSq^{0}$--family
\[
\{\wbar{y}, \SSq^{0} \wbar{y}, (\SSq^{0})^{2} \wbar{y}, \dotsc,
(\SSq^{0})^{r-1} \wbar{y} \},
\]
where $\wbar{y} \in H^{n+1,u/2^{r}}(\Lambda)$.  Thus differentials
coming from the $n$--line in $H^{*}(\Lambda')$ give information about
the $\SSq^{0}$--action on $H^{n+1}(\Lambda)$.

To compute the $r$th differential on an element $x \in E_{r}$, lift
$x$ to a class in $H^{*}(\Lambda')$, represent it by a class in
$\Lambda$, and take the coboundary.  Since $x$ has survived to
the $E_{r}$--term, this coboundary is in the image of $\theta^{r}$, so
apply $(\theta^{r})^{-1}$.  Project back to a class in
$H^{*}(\Lambda')$.  For example, $\lambda_{2} \lambda_{2}
\lambda_{1}$ represents a class at the $E_{1}$--term.  Lift it back to
the class with the same name in $\Lambda$ and take its coboundary.  The
result is $\lambda_{1}^{4}$.  This is in the image of $\theta$, and
applying $\theta^{-1}$ gives $\lambda_{0}^{4}$.  The class
$\lambda_{0}^{4}$ projects to a nonzero class in $H^{*}(\Lambda')$.
Thus there is a differential $d_{1}(\lambda_{2} \lambda_{2}
\lambda_{1}) = \lambda_{0}^{4}$.  More generally, there is a
differential
\begin{equation}\label{eqn-d1}
d_{1} (\lambda_{2} \lambda_{2} \lambda_{1}^{n}) = \lambda_{0}^{n+3}
\end{equation}
for $n \geq 1$, reflecting the fact that $\lambda_{0}^{n+3}$
represents a nonzero class in $H^{*}(\Lambda)$, while
$\lambda_{1}^{n+3}=\theta (\lambda_{0}^{n+3})$ is a boundary in
$\Lambda$, and thus $\SSq^{0} [\lambda_{0}^{n+3}]=0$ for all $n \geq 1$.

This may not be a good way to compute the cohomology of the Lambda
algebra, but it is a way to study $\SSq^{0}$ acting on it.  A little
analysis leads one to believe that most of the cohomology of $\Lambda$
is ``$\SSq^{0}$--periodic''.  The word ``most'' in the previous sentence
doesn't have any meaning really, but consider this: given a monomial
$M=\lambda_{n_{1}} \dotsb \lambda_{n_{k}}$ in $\Lambda$ which maps to
a nonzero class in $\Lambda'$, then at least one $n_{i}$ is even.
Recall that the differential is a derivation, and note that
$d(\lambda_{2m})$ is a sum of terms $\lambda_{i} \lambda_{j}$ with one
of $i$ and $j$ even, the other odd.  Thus applying $d$ to $M$ (in
$\Lambda$) yields a sum of monomials which, before applying the
Adem relations, have the same number of even $\lambda$s as $M$ does.

Furthermore, the Adem relations preserve the parity of the number of
even $\lambda$s.  So there is really only one way to get cohomology
classes in $\Lambda'$ which are not images of cohomology classes in
$\Lambda$: all terms in the differential must contain an even number
of even $\lambda$s, and the Adem relations on those terms must convert
them into terms with no even $\lambda$s.  The resulting boundary is
zero in the quotient $\Lambda'$, and hence one gets a cocycle in
$H(\Lambda')$ which supports a boundary in the spectral sequence.
Thus such cocycles are precisely those classes in $H(\Lambda')$ which
are cocycles but which are not in the image of any cocycle in
$H(\Lambda)$.

Let's look for such things.  First, here is a summary of the previous
paragraphs.

\begin{lemma}\label{lemma-parity}
If a class $x \in H^{*}(\Lambda')$ supports a nonzero
differential in the Bockstein spectral sequence~\eqref{eqn-bss}, then
$x$ lifts to a sum of monomials in $\Lambda'$ which
each contain a positive and even number of even $\lambda$s.
\end{lemma}

Here are two other useful observations in our search.

\begin{lemma}[Wang {\cite[Proposition 1.9]{wang.j*;hopf}}]
\label{lemma-key-fact}
Given $x \in \Lambda'$ in positive stem degree, if
$d(x) = 0$ and $x$ is not a boundary, then $x$ is homologous to an
element $y$ with ``odd ending integers'' -- that is, every term in the
admissible expression for $y$ ends in an odd lambda.
\end{lemma}

(Wang's proof works $\Lambda'$.)

\begin{lemma}[Wang {\cite[Proposition 1.8.3]{wang.j*;hopf}}]
\label{wang-1.8.3}
Let $\lambda_{n_{1}} \dotsb \lambda_{n_{r}} \in \Lambda$ be an
admissible sequence.  Then $(d \lambda_{n_{1}}) \lambda_{n_{2}} \dotsb
\lambda_{n_{r}}$ is a sum of admissible terms with leading integers at
most $n_{1}-1$.
\end{lemma}

\begin{note}\label{note-splitting}
There is an obvious vector space splitting of the short exact sequence
\[
0 \rightarrow \Lambda \xrightarrow{\theta} \Lambda \rightarrow
\Lambda' \rightarrow 0,
\]
in which one views $\Lambda'$ as being spanned by the admissible
monomials containing at least one even lambda.  We will often take
this point of view.
\end{note}

The following proposition and theorem are the main results in this
section.

\begin{prop}
Consider the Bockstein spectral sequence~\eqref{eqn-bss}.
\begin{enumerate}
\item There are no nonzero differentials coming from
$H^{1}(\Lambda')$, and thus $\SSq^{0}$ is injective on $H^{2}\Lambda =
\ext{A}{2}$.
\item There are no nonzero differentials coming from
$H^{2}(\Lambda')$, and thus $\SSq^{0}$ is injective on $H^{3}\Lambda =
\ext{A}{2}$.
\end{enumerate}
\end{prop}

Wang proved this result in \cite[Proposition 2.3]{wang.j*;hopf}.

\begin{proof}
(a) \fullref{lemma-parity} implies that any class which supports a
nonzero differential must be in homological degree at least 2.

(b) By \fullref{lemma-parity}, if a class supports a differential,
then it is a sum of terms of the form $\lambda_{2m} \lambda_{2n}$, and
hence is in even stem degree.  Also, \fullref{lemma-key-fact} says
that any cohomology class is cohomologous to a sum of odd-ending
monomials.  Therefore parity considerations imply that any class in an
even stem on the 2--line is cohomologous to a sum of terms of the form
$\lambda_{2i+1} \lambda_{2j+1}$, and hence is zero in $\Lambda'$.
\end{proof}

\begin{thm}\label{thm-degree-three}
In the Bockstein spectral sequence~\eqref{eqn-bss}, the only
differential emanating from $H^{3}(\Lambda')$ is $d_{1} \co \lambda_{2}
\lambda_{2} \lambda_{1} \longmapsto \lambda_{0}^{4}$.  Hence the
kernel of $\SSq^{0}$ on $\Ext^{4}$ is spanned by $h_{0}^{4}$.
\end{thm}

\begin{proof}
Suppose that $y \in H^{3}(\Lambda')$ supports a differential.  By
\fullref{lemma-key-fact}, we may assume that $y$ is a sum of
admissible monomials each of which ends in an odd lambda, and by
\fullref{lemma-parity}, each term in $y$ has two even lambdas, and
thus $y$ is a sum of terms of the form $\lambda_{\text{even}}
\lambda_{\text{even}} \lambda_{\text{odd}}$.

Following Wang, we write $y$ in the form
\[
y=\lambda_{2n} y_{1} + y',
\]
where $y'$ is a sum of admissible terms with leading term less than
$\lambda_{2n}$ (and hence leading term no larger than
$\lambda_{2n-2}$), and $y_{1}$ is a polynomial in homological degree
2; note that the stem degree of $y_{1}$ is odd.  We also assume that
$y$ is chosen from its cohomology class so that its leading term, in
the lexicographic ordering, is as small as possible.

View $y$ as being an element of $\Lambda$, via the splitting mentioned
in \fullref{note-splitting}.  Since $y$ is a cocycle in $\Lambda'$, the
boundary of $y$ in $\Lambda$ must consist of all odd terms.  So by
\fullref{wang-1.8.3}, $y_{1}$ must be a cocycle in $\Lambda$.  If
$y_{1}$ were a coboundary, then $y$ would be cohomologous to a class
with smaller leading term, so this can't happen by the minimality
assumption.  Now we appeal to Wang's computation of $H^{2}(\Lambda)$.
By \cite[Proposition 2.4]{wang.j*;hopf}, the only odd ending cocycles
in odd stems are $\lambda_{0} \lambda_{2^{m}-1}$, which in admissible
form is
\[
\sum_{j=1}^{m-1} \lambda_{2^{m}-2^{j}} \lambda_{2^{j}-1}
= \lambda_{2^{m}-2} \lambda_{1} + \lambda_{2^{m}-4} \lambda_{3} +
\text{other terms}.
\]
Thus $y$ is of the form
\[
y=(2n, 2^{m}-2, 1) + (2n, 2^{m}-4, 3) + \text{(smaller terms)},
\]
where ``smaller'' is with respect to the lexicographic ordering.
The coboundary of $y$ is
\[
d (y) = (2n-1, 2^{m}-3, 1, 1) + (2n-1, 2^{m}-4, 2, 1) +
\text{(smaller terms)}.
\]
The first term is all odd, so is zero in $\Lambda'$, but the second
term cannot be canceled by any of the smaller terms.  Thus it must
not be present, which means that $m$ must be 2: $y$ is of the form
\[
y=(2n, 2, 1) + \text{(smaller terms)}.
\]
The leading term of each ``smaller term'' is at most $\lambda_{2n-2}$.
We will show that if $n$ is bigger than 1, then the coboundary of $y$
cannot be zero in $\Lambda'$.  To do this, we will examine the
coboundary of $\lambda_{2n} \lambda_{2} \lambda_{1}$, and find terms
in it which cannot be canceled.  There are three cases, depending on
the congruence class of $n$ mod 4.

\textbf{Case 1} ($n \equiv 0 \mod 4$)\qua
Modulo terms with leading term less than $\lambda_{2n-4}$, here are
the coboundaries of all monomials in the appropriate bidegree:
\begin{align*}
d(\lambda_{2n} \lambda_{2} \lambda_{1})
&=
\lambda_{2n-1} \lambda_{1} \lambda_{1} \lambda_{1} +
\lambda_{2n-2} \lambda_{1} \lambda_{2} \lambda_{1} +
\lambda_{2n-4} \lambda_{3} \lambda_{2} \lambda_{1}, \\
d(\lambda_{2n-2} \lambda_{2} \lambda_{3})
&=
\lambda_{2n-2} \lambda_{1} \lambda_{2} \lambda_{1} +
\lambda_{2n-5} \lambda_{2} \lambda_{2} \lambda_{3}, \\
d(\lambda_{2n-2} \lambda_{4} \lambda_{1})
&=
\lambda_{2n-2} \lambda_{2} \lambda_{1} \lambda_{1} +
\lambda_{2n-3} \lambda_{3} \lambda_{1} \lambda_{1} +
\lambda_{2n-3} \lambda_{2} \lambda_{2} \lambda_{1} +
\lambda_{2n-5} \lambda_{2} \lambda_{4} \lambda_{1}, \\
d(\lambda_{2n-4} \lambda_{6} \lambda_{1})
&=
\lambda_{2n-4} \lambda_{3} \lambda_{2} \lambda_{1} +
\lambda_{2n-5} \lambda_{5} \lambda_{1} \lambda_{1} +
\lambda_{2n-5} \lambda_{4} \lambda_{2} \lambda_{1} +
\lambda_{2n-5} \lambda_{3} \lambda_{3} \lambda_{1} \\
& \quad \quad +  \lambda_{2n-6} \lambda_{4} \lambda_{3} \lambda_{1} +
\lambda_{2n-6} \lambda_{3} \lambda_{4} \lambda_{1}, \\
d(\lambda_{2n-4} \lambda_{4} \lambda_{3})
&=
\lambda_{2n-5} \lambda_{2} \lambda_{2} \lambda_{3} +
\lambda_{2n-6} \lambda_{2} \lambda_{3} \lambda_{3}.
\end{align*}
Any terms which are all odd go to zero in $\Lambda'$, so we may
ignore them.  Given the monomial $\lambda_{2n}\lambda_{2}
\lambda_{1}$, the only way to cancel $\lambda_{2n-4} \lambda_{3}
\lambda_{2} \lambda_{1}$ in its coboundary is to add $\lambda_{2n-4}
\lambda_{6} \lambda_{1}$, but then there is no way to cancel the term
$\lambda_{2n-5} \lambda_{4} \lambda_{2} \lambda_{1}$.

\textbf{Case 2} ($n \equiv 2 \mod 4$)\qua  If $n=2$, then the leading term of $y$
is $\lambda_{4} \lambda_{2} \lambda_{1}$.  This leads to a permanent
cycle in the spectral sequence: the element
\[
\lambda_{5} \lambda_{1} \lambda_{1} +
\lambda_{4} \lambda_{2} \lambda_{1} +
\lambda_{2} \lambda_{2} \lambda_{3}
\]
is a cocycle in $\Lambda$, and maps to $\lambda_{4} \lambda_{2}
\lambda_{1} + \lambda_{2} \lambda_{2} \lambda_{3}$ in $\Lambda'$.

Now assume that $n\geq 6$.  Modulo terms with leading term less than
$\lambda_{2n-5}$, here are the coboundaries of all monomials in the
appropriate bidegree:
\begin{align*}
d(\lambda_{2n} \lambda_{2} \lambda_{1})
&=
\lambda_{2n-1} \lambda_{1} \lambda_{1} \lambda_{1} +
\lambda_{2n-2} \lambda_{1} \lambda_{2} \lambda_{1} +
\lambda_{2n-5} \lambda_{4} \lambda_{2} \lambda_{1}, \\
d(\lambda_{2n-2} \lambda_{2} \lambda_{3})
&=
\lambda_{2n-2} \lambda_{1} \lambda_{2} \lambda_{1} +
\lambda_{2n-5} \lambda_{2} \lambda_{2} \lambda_{3}, \\
d(\lambda_{2n-2} \lambda_{4} \lambda_{1})
&=
\lambda_{2n-2} \lambda_{2} \lambda_{1} \lambda_{1} +
\lambda_{2n-3} \lambda_{3} \lambda_{1} \lambda_{1} +
\lambda_{2n-3} \lambda_{2} \lambda_{2} \lambda_{1} +
\lambda_{2n-5} \lambda_{2} \lambda_{4} \lambda_{1}, \\
d(\lambda_{2n-4} \lambda_{6} \lambda_{1})
&=
\lambda_{2n-4} \lambda_{3} \lambda_{2} \lambda_{1} +
\lambda_{2n-5} \lambda_{5} \lambda_{1} \lambda_{1} +
\lambda_{2n-5} \lambda_{4} \lambda_{2} \lambda_{1} +
\lambda_{2n-5} \lambda_{3} \lambda_{3} \lambda_{1}, \\
d(\lambda_{2n-4} \lambda_{4} \lambda_{3})
&=
\lambda_{2n-5} \lambda_{2} \lambda_{2} \lambda_{3}.
\end{align*}
The only way to cancel the term $\lambda_{2n-5} \lambda_{4}
\lambda_{2} \lambda_{1}$ in the coboundary of $\lambda_{2n}\lambda_{2}
\lambda_{1}$ is to add $\lambda_{2n-4} \lambda_{6} \lambda_{1}$ to
it, but the coboundary of the resulting sum has a term $\lambda_{2n-4}
\lambda_{3} \lambda_{2} \lambda_{1}$, which cannot be canceled.

\textbf{Case 3} ($n$ odd)\qua  Assume that $2n > 2$.  Then the
first ``other term'' in the coboundary of $y$ is $\lambda_{2n-3}
\lambda_{2} \lambda_{2} \lambda_{1}$.  To cancel this, the only
appropriate smaller terms are of the form $\lambda_{2n-2} \lambda_{2j}
\lambda_{2k+1}$, where $2j+2k+1 = 5$.  So the terms are
$\lambda_{2n-2} \lambda_{2} \lambda_{3}$ and $\lambda_{2n-2}
\lambda_{4} \lambda_{1}$.  Modulo terms with leading term less than
$\lambda_{2n-3}$, their coboundaries are as follows:
$\lambda_{2n-3}$:
\begin{align*}
d(\lambda_{2n} \lambda_{2} \lambda_{1}) &= \lambda_{2n-1}\lambda_{1}
\lambda_{1} \lambda_{1} + \lambda_{2n-3} \lambda_{2} \lambda_{2}
\lambda_{1}, \\
d(\lambda_{2n-2} \lambda_{2} \lambda_{3}) &=
\lambda_{2n-2} \lambda_{1} \lambda_{2} \lambda_{1}, \\
d(\lambda_{2n-2} \lambda_{4} \lambda_{1})
 &= \lambda_{2n-2} \lambda_{2} \lambda_{1} \lambda_{1} +
\lambda_{2n-3} \lambda_{3} \lambda_{1} \lambda_{1} +
\lambda_{2n-3} \lambda_{2} \lambda_{2} \lambda_{1}.
\end{align*}
While we can cancel the term $\lambda_{2n-3} \lambda_{2} \lambda_{2}
\lambda_{1}$, we cannot cancel $\lambda_{2n-2} \lambda_{2}
\lambda_{1} \lambda_{1}$.

So if $y$ supports a differential in the Bockstein spectral sequence,
then $y$ has leading term $\lambda_{2} \lambda_{2} \lambda_{1}$.
There are no smaller terms in the same bidegree, so the potential
cocycle equals its leading term: $y=\lambda_{2} \lambda_{2}
\lambda_{1}$.
\end{proof}

This proof is not ideal.  The technical aspects are cumbersome, and
the proof will not generalize well to higher dimensions.

\subsection[Comments on Sq on the 5--line]{Comments on $\SSq^{0}$ on the 5--line}

What about the next degree?  All of the classes in the ideal generated
by $\lambda_{0}^{4}$ must be hit by differentials
\[
d_{1} \co \lambda_{2^{n}-1} \lambda_{2} \lambda_{2} \lambda_{1}
\longmapsto \lambda_{2^{n-1}-1} \lambda_{0}^{4}, \quad \text{for $n \geq
5$}.
\]
Why $n \geq 5$?  Note that the class $\lambda_{2^{n-1}-1} \lambda_{0}^{4}$
is
a coboundary in $\Lambda'$ when $n=2$, since $\lambda_{1}
\lambda_{0}$ is the coboundary of $\lambda_{2}$.  Thus $\lambda_{1}
\lambda_{0}^{4}$ cannot be the target of a differential in the
spectral sequence.  It is a coboundary when
$n=3$, since $\lambda_{3} \lambda_{0} \lambda_{0}$ is cohomologous to
$\lambda_{1}^{3}$, and $\lambda_{1}^{3} \lambda_{0}$ is the coboundary
of $\lambda_{1} \lambda_{1} \lambda_{2}$.  It is a coboundary when
$n=4$: a calculation shows that it is the coboundary of
\[
\begin{split}
\lambda_{8} \lambda_{0} \lambda_{0} \lambda_{0} +
\lambda_{6} \lambda_{2} \lambda_{0} \lambda_{0} +
\lambda_{5} \lambda_{1} \lambda_{2} \lambda_{0} +
\lambda_{4} \lambda_{4} \lambda_{0} \lambda_{0} +
\lambda_{4} \lambda_{2} \lambda_{2} \lambda_{0} +
\lambda_{4} \lambda_{1} \lambda_{1} \lambda_{2} \\
+
\lambda_{3} \lambda_{3} \lambda_{2} \lambda_{0} +
\lambda_{2} \lambda_{4} \lambda_{1} \lambda_{1} +
\lambda_{2} \lambda_{2} \lambda_{4} \lambda_{0} +
\lambda_{2} \lambda_{2} \lambda_{2} \lambda_{2} +
\lambda_{1} \lambda_{2} \lambda_{4} \lambda_{1} +
\lambda_{1} \lambda_{1} \lambda_{2} \lambda_{4}.
\end{split}
\]

As above, suppose that $y=\lambda_{n} y_{1} + y'$ is a cycle in
$\Lambda'$ with odd-ending integers, in which the leading integers in
the polynomial $y_{1}$ are at most $2n$, and the leading integers in
$y'$ are at most $n-1$.  We may also assume that among classes in
$\Lambda'$ with odd-ending integers which are homologous to $y$, $y$
has the smallest leading integer.

(We will also view $y$ as an element of $\Lambda$, using the obvious
splitting of the quotient map \fullref{note-splitting}.)

If $n$ is odd, then $y_{1}$ must be a cycle in $\Lambda'$.  In this
case the boundary of $y$ is
\[
d y = \lambda_{n} (dy_{1}) + (d\lambda_{n}) y_{1} + dy'.
\]
The first term goes to zero in $\Lambda'$.  We may assume that
$y_{1}$ is nonzero in $\Lambda'$, and so is not all odd.  We may also
assume that $y_{1}$ is not a boundary.  So either $y_{1}$ is a cycle
in $\Lambda$, or is a non-cycle but has an all-odd boundary.  Suppose
that $y_{1}$ is a cycle in $\Lambda$.  By Wang's results, we know all
of the cycles in $\Lambda^{3}$, and all of the even-dimensional ones
are homologous to an all-odd monomial, and hence to something zero in
$\Lambda'$.  Therefore we can ignore this case, and we may assume
that $y_{1}$ is a cycle in $\Lambda'$ which is not the image of a
cycle in $\Lambda$.  By our previous calculation, this means that
$y_{1}=\lambda_{2} \lambda_{2} \lambda_{1}$:
\[
y = \lambda_{2m+1} \lambda_{2} \lambda_{2} \lambda_{1} + y'.
\]
By an analysis similar to the proof of \fullref{thm-degree-three},
we can show that if $2m+1 \equiv 1 \pmod{4}$, then $y$ cannot be the
leading term of a cocycle.  If $2m+1 \equiv 3 \pmod{4}$, then
$\lambda_{2m+1} \lambda_{2} \lambda_{2} \lambda_{1}$ is the leading
term in the coboundary of $\lambda_{2n+4} \lambda_{2} \lambda_{1}$,
and thus
\[
\lambda_{2m+1} \lambda_{2} \lambda_{2} \lambda_{1} + y' +
d(\lambda_{2n+4} \lambda_{2} \lambda_{1})
\]
is homologous to $y$ but has smaller leading term.

As a consequence, we may assume that the leading integer $n$ is even.
In this case, then $y_{1}$ must be a cycle in $\Lambda$, and as in the
proof of \fullref{thm-degree-three}, the minimality assumption on
the leading integer of $y$ means that $y_{1}$ cannot be a boundary.  A
result of Wang \cite[Proposition 2.10]{wang.j*;hopf} lists
representatives of all of the cohomology classes of length 3 in an
even stem: they are
\[
\lambda_{2} \lambda_{3} \lambda_{3}, \ \lambda_{0} \lambda_{2^{i}-1}
\lambda_{2^{j}-1}.
\]
Thus we could write down all of the possibilities for the term
$\lambda_{2m} y_{1}$.  This is somewhat complicated, though, because
these are not the admissible forms for the degree 3 cycles, and even
if we had them, left multiplication by $\lambda_{2m}$ could make some
of the resulting terms inadmissible.

Computer calculations in internal degrees up to 82 have found two
phenomena: in this range, there is only one differential other than
those coming from the ideal generated by $\lambda_{2} \lambda_{2}
\lambda_{1}$:
\[
d_{1}(\lambda_{14} \lambda_{6} \lambda_{5} \lambda_{3} +
\text{(smaller terms)}) = \lambda_{6} \lambda_{2} \lambda_{1}
\lambda_{1} \lambda_{1} + \text{(smaller terms)}.
\]
(This reflects the fact that $Ph_{2} \in \ext{A}{5,16}$ is in the
kernel of $\SSq^{0}$.)  Also, the source of every differential in this
range except for $\lambda_{2} \lambda_{2} \lambda_{1} \lambda_{1}$ is
cohomologous to a class of the form
\[
\lambda_{2m} \lambda_{6} \lambda_{5} \lambda_{3} + \text{(smaller
terms)}.
\]
This is a relatively small range of dimensions, but one wonders if
these patterns hold for the rest of the 4--line.  Since we don't have
enough evidence to make a good conjecture, we pose questions.

\begin{question}
\begin{enumerate}
\item Is the kernel of $\SSq^{0}$ on $\Ext^{5}$ is spanned by the ``obvious''
classes $h_{i}h_{0}^{4}$ with $i \geq 5$, plus the class $Ph_{2}$ in
the 11--stem?
\item Also, consider an element of the form $y=\lambda_{2k} y_{1} + y'
\in \Lambda'$ which supports a differential in the spectral sequence.
Assume that $y$ is not cohomologous to $\lambda_{2} \lambda_{2}
\lambda_{1} \lambda_{1}$.  Does the lexicographically smallest element
in the cohomology class of $y$ have leading term $\lambda_{2m}
\lambda_{6} \lambda_{5} \lambda_{3}$?
\end{enumerate}
\end{question}

These may be interesting questions, but the methods used in proving
\fullref{thm-main} and \fullref{thm-degree-three} are only going to
get harder in higher filtrations, so some other ideas are needed.

\section{The cocomplete Lambda algebra}\label{sec-cocomplete}

Let $\theta^{-1} \Lambda$ be the direct limit of the diagram
\[
\Lambda \xrightarrow{\ \theta\ } \Lambda \xrightarrow{\ \theta\ } \Lambda
\xrightarrow{\ \theta\ } \dotsb.
\]
Call $\theta^{-1} \Lambda$ the \emph{cocomplete Lambda algebra}.  For any real
number $r$, write ${\textstyle \Z[\frac{1}{2}]}_{>r}$ for the set of elements of ${\textstyle \Z[\frac{1}{2}]}$
which are greater than $r$, and similarly for ${\textstyle \Z[\frac{1}{2}]}_{\geq r}$.

\begin{prop}
In this proposition, all indices $i$, $j$, $m$, $n$, and $i_{r}$ are
assumed to be in ${\textstyle \Z[\frac{1}{2}]}$.
\begin{enumerate}
\item $\theta^{-1} \Lambda$ is a $\Z \times {\textstyle \Z[\frac{1}{2}]}_{>0}$--graded
$\mathbb{F}_{2}$--algebra with generators $\lambda_{n}$ in bidegree $(1,n+1)$
for all $n > -1$.
\item The relations in $\theta^{-1} \Lambda$ are generated by
\[
\sum_{i+j=n} \tbinom{i+j}{i} \lambda_{i-1+m} \lambda_{j-1+2m} \quad
\quad \text{for} \ m > 0, \ n \geq 0.
\]
This is the symmetric form of the relations.
\item Alternatively, the relations are generated by
\[
\lambda_{i} \lambda_{2i+1+n} = \sum_{j \geq 0}
\tbinom{n-j-2^{-N(j,n)}}{j} \lambda_{i+n-j} \lambda_{2i+1+j}
\quad \quad \text{for} \ i > -1, \ n \geq 0,
\]
where (after \cite[1.7]{llerena-hung}) $N=N(j,n)$ is the least integer
so that $2^{N}j$ and $2^{N}n$ are integers.  This is the admissible
form of the relations.
\item The differential in $\theta^{-1} \Lambda$ is given by
\[
d(\lambda_{n-1}) = \sum_{\substack{i+j=n \\ i, j > 0}}
\tbinom{i+j}{j} \lambda_{i-1}\lambda_{j-1} \quad \quad
\text{(symmetric form)}.
\]
\item Alternatively, the differential is given by
\[
d(\lambda_{n}) = \sum_{j > 0}
\tbinom{n-j}{j} \lambda_{n-j}\lambda_{j-1} \quad \quad
\text{(admissible form)}.
\]
\item The admissible monomials form a basis for $\theta^{-1} \Lambda$.  (As in
$\Lambda$, a monomial $\lambda_{i_{1}} \dotsb \lambda_{i_{s}}$ in
$\theta^{-1} \Lambda$ is \emph{admissible} if $2i_{r} \geq i_{r+1}$ for $1
\leq r \leq s-1$.)
\item The cohomology of $\theta^{-1} \Lambda$ is equal to
$(\SSq^{0})^{-1} \ext{A}{*}$, which in turn is equal to
$\ext{\widehat{A}}{*}$, where $\widehat{A}$ is the ``complete Steenrod
algebra,'' as studied in \cite{arnon;dickson} and
\cite{llerena-hung}.
\end{enumerate}
\end{prop}

Note that for integers $a$ and $b$, $\tbinom{a}{b} \equiv
\tbinom{2a}{2b} \pmod{2}$, and this allows one to define mod 2 binomial
coefficients for elements of ${\textstyle \Z[\frac{1}{2}]}$.  As a consequence, all of the
sums here are finite.  This is essentially because $\theta^{-1} \Lambda$ is
constructed as a direct limit; in contrast, the complete Steenrod
algebra $\widehat{A}$ is constructed as an inverse limit, and the Adem
relations there are infinite sums -- see \cite[1.7]{llerena-hung}.
Similarly, the admissible monomials in $\widehat{A}$ do not span, while
the admissible monomials in $\theta^{-1} \Lambda$ do.  The relation with
$\widehat{A}$ also explains the terminology ``cocomplete Lambda
algebra.''

\begin{proof}
In general, all of this follows from the colimit description of
$\theta^{-1} \Lambda$.  In more detail: for part (b), applying $\theta$ to
the symmetric Adem relation~\eqref{reln-symm} indexed by $m$ and $n$
yields
\begin{align*}
(\text{symmetric Adem relation})_{m,n} &=
\sum_{i+j=n} \tbinom{i+j}{i} \lambda_{i-1+m} \lambda_{j-1+2m} \\
 & \stackrel{\theta}{\longmapsto} \sum_{i+j=n} \tbinom{i+j}{i}
 \lambda_{2i-1+2m} \lambda_{2j-1+4m} \\
 &= \sum_{2i+2j=2n} \tbinom{2i+2j}{2i} \lambda_{2i-1+2m} \lambda_{2j-1+4m} \\
 &= \sum_{i+j=2n} \tbinom{i+j}{i} \lambda_{i-1+2m} \lambda_{j-1+4m} \\
 &= (\text{symmetric Adem relation})_{2m,2n}.
\end{align*}
In $\Lambda$, one has symmetric Adem relations for all integers $m
\geq 1$ and $n \geq 0$; thus after inverting $\theta$, one needs
relations for all $m, n \in {\textstyle \Z[\frac{1}{2}]}$ with $m > 0$ and $n \geq 0$.

Similarly, for part (c), applying $\theta$ to the admissible Adem
relation \eqref{reln-admissible} indexed by $i$ and $n$ yields
\begin{align*}
(\text{admissible Adem } &\text{relation})_{i,n} =
\lambda_{i} \lambda_{2i+1+n} + \sum_{j \geq 0} \tbinom{n-j-1}{j}
\lambda_{i+n-j} \lambda_{2i+1+j} \\
 & \stackrel{\theta}{\longmapsto}
\lambda_{2i+1} \lambda_{4i+3+2n} + \sum_{j \geq 0} \tbinom{n-j-1}{j}
\lambda_{2i+1+2n-2j} \lambda_{4i+3+2j} \\
 &= \lambda_{2i+1} \lambda_{4i+3+2n} + \sum_{j \geq 0} \tbinom{2n-2j-2}{2j}
\lambda_{2i+1+2n-2j} \lambda_{4i+3+2j} \\
 &= \lambda_{2i+1} \lambda_{4i+3+2n} + \sum_{j \geq 0} \tbinom{2n-j-2}{j}
\lambda_{2i+1+2n-j} \lambda_{4i+3+j} \\
 &= (\text{admissible Adem relation})_{2i+1,2n}
\end{align*}
In $\Lambda$, one has admissible Adem relations for all non-negative
integers $i$ and $n$, so in $\theta^{-1} \Lambda$, one gets admissible Adem
relations for all $i \in {\textstyle \Z[\frac{1}{2}]}_{>-1}$ and $n \in {\textstyle \Z[\frac{1}{2}]}_{\geq 0}$.
The change of the binomial coefficient from $\tbinom{n-j-1}{j}$ to
$\tbinom{2n-j-2}{j}$ explains the presence of the integer $N(j,n)$ in
the formula.

The two forms of the differentials in $\theta^{-1} \Lambda$ are derived
similarly.

Since the admissible monomials form a basis for $\Lambda$, and since
$\theta$ is injective on basis elements, part (f) follows.

Since homology commutes with colimits, part (g) follows.  See also
\cite[5.3]{palmieri;quotient} for the isomorphism between
$(\SSq^{0})^{-1} \ext{A}{*}$ and $\ext{\widehat{A}}{*}$.
\end{proof}

The action of $\theta$ on $\theta^{-1} \Lambda$ yields an action of an
infinite cyclic group; this action is free in each positive degree in
$\theta^{-1} \Lambda$, and trivial in degree zero.  Under this action, the
$\lambda_{n}$'s are partitioned into orbits, and each orbit contains a
unique $\lambda_{n}$ with $n$ an even integer.  Among these,
$\lambda_{0}$ is the only cocycle.  Hence $H^{1}(\theta^{-1} \Lambda)$ is
spanned by $\{\theta^{k} (\lambda_{0}) : k \in \Z \}$.

One would hope that computing $H^{*}(\theta^{-1} \Lambda)$ in higher
dimensions would be simpler than computing $H^{*}(\Lambda)$, because
of this symmetry.  We have been unable to take advantage of this so
far, unfortunately.

We remark that $\Lambda$ is a Koszul algebra.  Let $A_{\mathsf{Lie}}$ be the
``Steenrod algebra for simplicial Lie algebras,'' which is generated
by elements $\SSq^{n}$ and satisfies the usual Adem relations, but has
$\SSq^{0}=0$.  Priddy provided a criterion in \cite[5.3]{priddy;koszul}
to check whether an algebra is a Koszul algebra, and he showed in
\cite[8.3--4]{priddy;simplicial-lie} that both $A_{\mathsf{Lie}}$ and $\Lambda$
satisfy this condition.  Furthermore, he showed in
\cite[9.1]{priddy;simplicial-lie} that $\ext{A_{\mathsf{Lie}}}{*}$ is isomorphic to
the opposite algebra to $\Lambda$, and pointed out in
\cite[9.4]{priddy;koszul} that
\[
\ext{\ext{A_{\mathsf{Lie}}}{*}}{*} \cong A_{\mathsf{Lie}}.
\]
In other words, $\Lambda$ is the ``Koszul dual'' of $A_{\mathsf{Lie}}$.

As a consequence, $\theta^{-1} \Lambda$ is a Koszul algebra, in a slightly
unconventional sense (since it is $\Z \times {\textstyle \Z[\frac{1}{2}]}$--graded, rather
than $\Z \times \Z$--graded).  So one should be able to compute
$\ext{\theta^{-1} \Lambda}{*}$ pretty easily; the result should be the
``complete Steenrod algebra for simplicial Lie algebras'' (which is
built from $A_{\mathsf{Lie}}$ using an inverse limit, just as $\widehat{A}$ is built
from $A$).  The details are left for the interested reader.

redpajama_set_name: RedPajamaArXiv


Platamonas is located on the South-East part of Pieria prefecture and it is 36 kilometers off Katerini, which is capital of the prefecture. It belongs to East Olympus Municipality (which consists of the Municipal Departments of Leptokaria, Nei Pori, Panteleimonas, Platamonas, and Scotina), it is 8422 km2 in extent and it has about 2.200 resident population. The access is easy via the National Speedway E75 (Athens - Thessaloniki) or by rail.
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text:
/*
 * Licensed under the Apache License, Version 2.0 (the "License");
 * you may not use this file except in compliance with the License.
 * You may obtain a copy of the License at
 *
 *     http://www.apache.org/licenses/LICENSE-2.0
 *
 * Unless required by applicable law or agreed to in writing, software
 * distributed under the License is distributed on an "AS IS" BASIS,
 * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
 * See the License for the specific language governing permissions and
 * limitations under the License.
 */
package com.facebook.presto.operator.scalar;

import com.facebook.presto.operator.Description;
import com.facebook.presto.spi.type.StandardTypes;
import com.facebook.presto.type.SqlType;
import com.google.common.base.Ascii;
import com.google.common.base.Charsets;
import io.airlift.slice.Slice;
import io.airlift.slice.Slices;

import javax.annotation.Nullable;

import java.nio.ByteBuffer;
import java.nio.CharBuffer;

import static com.google.common.base.Preconditions.checkArgument;

public final class StringFunctions
{
    private StringFunctions() {}

    @Description("convert unicode code point to a string")
    @ScalarFunction
    @SqlType(StandardTypes.VARCHAR)
    public static Slice chr(@SqlType(StandardTypes.BIGINT) long codepoint)
    {
        char[] utf16 = Character.toChars((int) codepoint);
        ByteBuffer utf8 = Charsets.UTF_8.encode(CharBuffer.wrap(utf16));
        return Slices.wrappedBuffer(utf8.array(), 0, utf8.limit());
    }

    @Description("concatenates given strings")
    @ScalarFunction
    @SqlType(StandardTypes.VARCHAR)
    public static Slice concat(@SqlType(StandardTypes.VARCHAR) Slice str1, @SqlType(StandardTypes.VARCHAR) Slice str2)
    {
        Slice concat = Slices.allocate(str1.length() + str2.length());
        concat.setBytes(0, str1);
        concat.setBytes(str1.length(), str2);
        return concat;
    }

    @Description("length of the given string")
    @ScalarFunction
    @SqlType(StandardTypes.BIGINT)
    public static long length(@SqlType(StandardTypes.VARCHAR) Slice slice)
    {
        return slice.length();
    }

    @Description("greedily removes occurrences of a pattern in a string")
    @ScalarFunction
    @SqlType(StandardTypes.VARCHAR)
    public static Slice replace(@SqlType(StandardTypes.VARCHAR) Slice str, @SqlType(StandardTypes.VARCHAR) Slice search)
    {
        return replace(str, search, Slices.EMPTY_SLICE);
    }

    @Description("greedily replaces occurrences of a pattern with a string")
    @ScalarFunction
    @SqlType(StandardTypes.VARCHAR)
    public static Slice replace(@SqlType(StandardTypes.VARCHAR) Slice str, @SqlType(StandardTypes.VARCHAR) Slice search, @SqlType(StandardTypes.VARCHAR) Slice replace)
    {
        String replaced = str.toString(Charsets.UTF_8).replace(
                search.toString(Charsets.UTF_8),
                replace.toString(Charsets.UTF_8));
        return Slices.copiedBuffer(replaced, Charsets.UTF_8);
    }

    @Description("reverses the given string")
    @ScalarFunction
    @SqlType(StandardTypes.VARCHAR)
    public static Slice reverse(@SqlType(StandardTypes.VARCHAR) Slice slice)
    {
        Slice reverse = Slices.allocate(slice.length());
        for (int i = 0, j = slice.length() - 1; i < slice.length(); i++, j--) {
            reverse.setByte(j, slice.getByte(i));
        }
        return reverse;
    }

    @Description("returns index of first occurrence of a substring (or 0 if not found)")
    @ScalarFunction("strpos")
    @SqlType(StandardTypes.BIGINT)
    public static long stringPosition(@SqlType(StandardTypes.VARCHAR) Slice string, @SqlType(StandardTypes.VARCHAR) Slice substring)
    {
        if (substring.length() > string.length()) {
            return 0;
        }

        for (int i = 0; i <= (string.length() - substring.length()); i++) {
            if (string.equals(i, substring.length(), substring, 0, substring.length())) {
                return i + 1;
            }
        }

        return 0;
    }

    @Description("suffix starting at given index")
    @ScalarFunction
    @SqlType(StandardTypes.VARCHAR)
    public static Slice substr(@SqlType(StandardTypes.VARCHAR) Slice slice, @SqlType(StandardTypes.BIGINT) long start)
    {
        return substr(slice, start, slice.length());
    }

    @Description("substring of given length starting at an index")
    @ScalarFunction
    @SqlType(StandardTypes.VARCHAR)
    public static Slice substr(@SqlType(StandardTypes.VARCHAR) Slice slice, @SqlType(StandardTypes.BIGINT) long start, @SqlType(StandardTypes.BIGINT) long length)
    {
        if ((start == 0) || (length <= 0)) {
            return Slices.EMPTY_SLICE;
        }

        if (start > 0) {
            // make start zero-based
            start--;
        }
        else {
            // negative start is relative to end of string
            start += slice.length();
            if (start < 0) {
                return Slices.EMPTY_SLICE;
            }
        }

        if ((start + length) > slice.length()) {
            length = slice.length() - start;
        }

        if (start >= slice.length()) {
            return Slices.EMPTY_SLICE;
        }

        return slice.slice((int) start, (int) length);
    }

    // TODO: Implement a more efficient string search
    @Nullable
    @Description("splits a string by a delimiter and returns the specified field (counting from one)")
    @ScalarFunction
    @SqlType(StandardTypes.VARCHAR)
    public static Slice splitPart(@SqlType(StandardTypes.VARCHAR) Slice string, @SqlType(StandardTypes.VARCHAR) Slice delimiter, @SqlType(StandardTypes.BIGINT) long index)
    {
        checkArgument(index > 0, "Index must be greater than zero");

        if (delimiter.length() == 0) {
            if (index > string.length()) {
                // index is too big, null is returned
                return null;
            }
            return string.slice((int) (index - 1), 1);
        }

        int previousIndex = 0;
        int matchCount = 0;

        for (int i = 0; i <= (string.length() - delimiter.length()); i++) {
            if (string.equals(i, delimiter.length(), delimiter, 0, delimiter.length())) {
                matchCount++;
                if (matchCount == index) {
                    return string.slice(previousIndex, i - previousIndex);
                }
                // noinspection AssignmentToForLoopParameter
                i += (delimiter.length() - 1);
                previousIndex = i + 1;
            }
        }

        if (matchCount == index - 1) {
            // returns last section of the split
            return string.slice(previousIndex, string.length() - previousIndex);
        }

        // index is too big, null is returned
        return null;
    }

    @Description("removes spaces from the beginning of a string")
    @ScalarFunction("ltrim")
    @SqlType(StandardTypes.VARCHAR)
    public static Slice leftTrim(@SqlType(StandardTypes.VARCHAR) Slice slice)
    {
        int start = firstNonSpace(slice);
        return slice.slice(start, slice.length() - start);
    }

    @Description("removes spaces from the end of a string")
    @ScalarFunction("rtrim")
    @SqlType(StandardTypes.VARCHAR)
    public static Slice rightTrim(@SqlType(StandardTypes.VARCHAR) Slice slice)
    {
        int end = lastNonSpace(slice);
        return slice.slice(0, end + 1);
    }

    @Description("removes spaces from the beginning and end of a string")
    @ScalarFunction
    @SqlType(StandardTypes.VARCHAR)
    public static Slice trim(@SqlType(StandardTypes.VARCHAR) Slice slice)
    {
        int start = firstNonSpace(slice);
        if (start == slice.length()) {
            return Slices.EMPTY_SLICE;
        }

        int end = lastNonSpace(slice);
        assert (end >= 0) && (end >= start);

        return slice.slice(start, (end - start) + 1);
    }

    private static int firstNonSpace(Slice slice)
    {
        for (int i = 0; i < slice.length(); i++) {
            if (slice.getByte(i) != ' ') {
                return i;
            }
        }
        return slice.length();
    }

    private static int lastNonSpace(Slice slice)
    {
        for (int i = slice.length() - 1; i >= 0; i--) {
            if (slice.getByte(i) != ' ') {
                return i;
            }
        }
        return -1;
    }

    @Description("converts the alphabets in a string to lower case")
    @ScalarFunction
    @SqlType(StandardTypes.VARCHAR)
    public static Slice lower(@SqlType(StandardTypes.VARCHAR) Slice slice)
    {
        Slice upper = Slices.allocate(slice.length());
        for (int i = 0; i < slice.length(); i++) {
            upper.setByte(i, Ascii.toLowerCase((char) slice.getByte(i)));
        }
        return upper;
    }

    @Description("converts all the alphabets in the string to upper case")
    @ScalarFunction
    @SqlType(StandardTypes.VARCHAR)
    public static Slice upper(@SqlType(StandardTypes.VARCHAR) Slice slice)
    {
        Slice upper = Slices.allocate(slice.length());
        for (int i = 0; i < slice.length(); i++) {
            upper.setByte(i, Ascii.toUpperCase((char) slice.getByte(i)));
        }
        return upper;
    }

    @Description("get the largest of the given values")
    @ScalarFunction
    @SqlType(StandardTypes.VARCHAR)
    public static Slice greatest(@SqlType(StandardTypes.VARCHAR) Slice value1, @SqlType(StandardTypes.VARCHAR) Slice value2)
    {
        return value1.compareTo(value2) > 0 ? value1 : value2;
    }

    @Description("get the smallest of the given values")
    @ScalarFunction
    @SqlType(StandardTypes.VARCHAR)
    public static Slice least(@SqlType(StandardTypes.VARCHAR) Slice value1, @SqlType(StandardTypes.VARCHAR) Slice value2)
    {
        return value1.compareTo(value2) < 0 ? value1 : value2;
    }
}


Teint 006 provides a dark form unified to sensitive and prone to enrrojecerse. Thanks to its delicate texture but Compact, is fixed to the base extending, of this form, its duration.
These powders provide a matte finish and an anti-inflammatory and soothing action, thanks to the formulated combination of plants with active principles of cornmeal, rosa canina, vitamin and grape seed oil and.
* Product with organic ingredients.
Put color to your cheeks in an easy way and with natural ingredients!
Take a little product and apply with the scum that is included in the container. Insist on the zone T: contour of eyes, nose and forehead.
MICA, ZEA MAYS STARCH (ZEA MAYS (CORN) STARCH) *, ZINC STEARATE, SQUALANE, ROSA CANINA FRUIT OIL *, CITRUS AURANTIUM DULCIS PEEL WAX (CITRUS AURANTIUM DULCIS (ORANGE) PEEL WAX), HYDROGENATED VEGETABLE OIL, VITIS VINIFERA SEED OIL (VITIS VINIFERA (GRAPE) SEED OIL) *, PRUNUS AMYGDALUS DULCIS OIL (PRUNUS AMYGDALUS DULCIS (SWEET ALMOND) OIL), CITRUS AURANTIUM DULCIS PEEL OIL (CITRUS AURANTIUM DULCIS (ORANGE) PEEL OIL), GLYCERYL CAPRYLATE CI 77820 (SILVER), PARFUM (FRAGRANCE), CITRONELLOL, GERANIOL, LIMONENE [+ / ¬ MAY CONTAIN: CI] [77891 (TITANIUM DIOXIDE), CI 77491 (IRON OXIDES), CI 77492 (IRON OXIDES), CI 77499 (IRON OXIDES)].


text:
Rock remove filter
Minnesota Historical Society remove filter
Rock (7) remove filter
Rock County Community Library (Plum Creek Library System)
Bridge No. L-4646: Pre-Development Construction Plans and Specifications
City of Beaver Creek
To hire a qualified consultant to develop architectural drawings for Bridge No. L-4646 (Spring Brook Bridge), listed in the National Register of Historic Places.
Luverne Downtown Historic District National Register Evaluation
City of Luverne
To acquire professional services that will comprehensively evaluate Luverne's commercial district for possible inclusion in the National Register of Historic Places
National Guard Armory National Register Evaluation
Nobles County Historical Society
To hire a qualified historian to complete an evaluation to determine eligibility for listing in the National Register of Historic Places for the Worthington National Guard Armory.
Palace Theater: Palace Theatre Restoration Phase III (HVAC)
To study and prepare preliminary design work for replacement of the existing heating and cooling system in the 1915 Historic Palace Theatre, listed on the National Registrar of Historic Places
The Women of Southwest Minnesota and the Great War: Phase III (Publication)
Society for the Study of Local and Regional History
To hire qualified professionals to publish an essay on women in Southwest Minnesota during World War I.
The Women of Southwest Minnesota and the Great War
To hire a qualified historian to conduct primary source research on the history of women's activities in southwestern Minnesota during World War I.