text:
80s sex and protest art reassembled in Chicago
This Will Have Been looks back at the passionate, political art of Nan Goldin, Jenny Holzer, Richard Prince and co
Untitled (we will no longer be seen and not heard), (1985) by Barbara Kruger
"America is in the midst of a culture war," wrote professor James Davison Hunter in his book 1991 Culture Wars. This struggle, over the country's "most fundamental and cherished assumptions about how to order our lives", pitted the North American liberal and progressive movement against a Reaganite reactionary conservatism, which sought to put an end to the freer politics of the 1960s and 70s.
No Title (To Dust Cover…Shut) (1984) by Ray Pettibon
The conflict might not quite yet be over, but those early struggles are far enough behind us all for a little examination and reflection. This Will Have Been: Art, Love & Politics in the 1980s, which opens at the Museum of Contemporary Art Chicago on Feb 11, looks back at this time and features works by the likes of Jenny Holzer, Richard Prince, Barbara Kruger, Jeff Wall, Christian Marclay and Raymond Pettibon, all produced between 1979 and 1992, illustrating the struggle.
"During this era, the political sphere was dominated by the ideas of former US President Ronald Reagan and British Prime Minister Margaret Thatcher, the music scene was transformed by punk and the birth of hip-hop," the show's organizers write, "and our everyday lives were radically altered by a host of technological developments, from the Sony Walkman and the ATM to the appearance of MTV and the first personal computers."
Dream Girls (1989-90) by Deborah Bright
Anti-nuclear protests and the AIDS crisis coupled with a growing awareness of identity politics led many artists to produce forcefully didactic works, critical of both the American Right, and the establishment consensus more generally.
Kissing Doesn't Kill (1990) by Gran Fury
Interestingly, the show's curators also characterize the artists from this period as being the first generation to have grown up with a TV in their home. "They came of age in a culture saturated with images designed to promote desire - desire for objects, for lifestyles, for fame, for conformity, for anti-conformity. So too the majority of these artists lived through the heady days of the 1970s feminist movement and witnessed that broad-based social movement's demands for equality in all areas of life-work, family, and intimate relationships. It became the task of the 1980s to assimilate these powerful social forces - the rise of television and movements for social justice-as they converged."
Picture for Women (1979) by Jeff Wall
The show's organizers have kept to a strictly critical approach when organizing the exhibition, carefully avoiding any slips into nostalgia. The End is Near section looks at both the fear of a nuclear apocalypse and the malaise of postmodernism; Democracy examines the fresh promise of equality the Civil Rights movement brought about, as well as the inclusion of graffiti and posters into fine-art practice; Gender Trouble focuses on the artistic inheritances of feminism and gay liberation; meanwhile Desire and Longing looks at the commercial avarice of the era, as well as the abrupt (if perhaps, temporary) halt the AIDS virus brought to the sexual revolution.
Untitled (cowboys) 1987 by Richard Prince
The 80s has never been a hard age to classify. Yet, as every year passes, those ideas and historical forces seem to press in on our own time more forcefully. While the show's title suggests the struggle is over, the exhibits will perhaps make you question which side you think won.
For more on the show, go here. For more on how this period fits into the greater sweep of queer creativity, take a look at Art and Queer Culture; for more on television's influence, consider Art and Electronic Media; and of course don't forget our Jenny Holzer, Nan Goldin, Christian Marclay, Jeff Wall, Richard Prince and Ray Pettibon books. You can buy them all from the people who made them here.


Theodor Adorno: Is the man who deals with the absolute not necessarily claiming to be the thinking organ with the capacity to do so, and thus the absolute himself?
Blaise Pascal, Pensée #326: If he exalts himself, I humble him; if he humbles himself, I exalt him; and I always contradict him, until he understands that he is an incomprehensible monster.
Wittgenstein, Notebooks 1914-1916: Ethics does not treat of the world. Ethics must be a condition of the world, like logic. Ethics and Aesthetics are one.
Arthur Schopenhauer: The World as Will and Idea): Animals learn death first at the moment of death;…Man approaches death with the knowledge it is closer every hour, and this creates a feeling of uncertainty over his life, even for him who forgets in the business of life that annihilation is awaiting him. It is for this reason chiefly that we have philosophy and religion.
Georg Cantor: The fear of infinity is a form of myopia that destroys the possibility of seeing the actual infinite, even though in its highest form it has created and sustains us, and in its secondary transfinite forms occurs all around us and even inhabits our minds.
Augustine, Confessions: "He was not utterly unskilled in handling his own lack of training, and he refused to be rashly drawn into a controversy about those matters from which there would be no exit nor an easy way of retreat. This was an additional ground for my pleasure. For the controlled modesty of a mind that admits limitations is more beautiful than the things I was anxious to know about." .
Emmanuel Levinas: Infinity overflows the thought that thinks it.
Jacques Lacan: Desire, a function central to all human experience, is the desire for nothing nameable.
Jacques Lacan: We in fact address … those we do not know, true Others, true subjects. They are on the other side of the wall of language, there where in principle I never reach them. Fundamentally, it is them I'm aiming at every time I utter true speech,…I always aim at the true subject, and I have to be content with shadows. The subject is separated from the Others, the true ones, by the wall of language. —- Jacques Lacan, The Ego in Freud's Theory and in the Technique of Psychoanalysis.
Jacques Lacan, The Ethics of Psychoanalysis (1992): The deep dissatisfaction we find in every psychology—including the one we have founded thanks to psychoanalysis—derives from the fact that it is nothing more than a mask, and sometimes even an alibi, of the effort to focus on the problem of our own action—something that is the essence and very foundation of all ethical reflection.
Hegel, Phenomenology of the Spirit, On Sense-Certainty, Paragraph 95 -96: Sense-certainty itself has thus to be asked: What is the This? If we take it in the two-fold form of its existence, as the Now and as the Here, the dialectic it has in it will take a form as intelligible as the This itself. To the question, What is the Now? we reply, for example, the Now is night-time. To test the truth of this certainty of sense, a simple experiment is all we need: write that truth down. A truth cannot lose anything by being written down, and just as little by our preserving and keeping it. If we look again at the truth we have written down, look at it now, at this noon-time, we shall have to say it has turned stale and become out of date. The Now that is night is kept fixed, i.e. it is treated as what it is given out to be, as something which is; but it proves to be rather a something which is not. The Now itself no doubt maintains itself, but as what is not night; similarly, in its relation to the day which the Now is at present, it maintains itself as something that is also not day, or as altogether something negative. This self -maintaining Now is therefore not something immediate but something mediated; for, qua something that remains and preserves itself, it is determined through and by means of the fact that something else, namely day and night, is not. Thereby it is just as much as ever it was before, Now, and in being this simple fact, it is indifferent to what is still associated with it; just as little as night or day is its being, it is just as truly also day and night; it is not in the least affected by this otherness through which it is what it is. A simple entity of this sort, which is by and through negation, which is neither this nor that, which is a not-this, and with equal indifference this as well as that – a thing of this kind we call a Universal. The Universal is therefore in point of fact the truth of sense-certainty, the true content of sense-experience.
Here is Jung's reply to Hegel (1947): "A philosophy like Hegel's is a self-revelation of psychic background and, philosophically, a presumption. Psychologically, it amounts to an invasion by the unconscious. The peculiar high-flown language Hegel uses bears out this view: it is reminiscent of the megalomaniac language of schizophrenics, who use terrific spellbinding words to reduce the transcendent to subjective form, to give banalities the charm of novelty, or pass off commonplaces as searches wisdom. So bombastic a terminology is a symptom of weakness, ineptitude, and lack of substance. But that does not prevent the latest German philosopher from using the same crackpot power-words and pretending that it is not unintentional psychology.
Hagakure: Our bodies are given life from the midst of nothingness. Existing where there is nothing is the meaning of the phrase, 'Form is emptiness.' That all things are provided for by nothingness is the meaning of the phrase, 'Emptiness is form." One should not think that these are two separate things.


text:
Daniel Dobbeni is Executive in Residence at Vlerick Business School. In his role he is chairman of the advisory board of the Centre for Energy Management.
Daniel Dobbeni started his career in nuclear engineering. He moved to chief technology officer for Research & Development focused on power generation. Since 1996 he is involved in the electricity transmission business where he became Chief Operation Officer of Elia System Operator SA in charge of the Belgian customer services, control centres, European regulation and grid codes. From 2003 to 2012, he was CEO and chairman of the Executive Committee of the Elia Group. Following the acquisition of 50Hertz (Germany) in 2010 he was chairman of the board of Eurogrid International SA (till 2013) and became Chairman of the board of 50Hertz. With his experience he helps the school in building the Energy plaform and since a couple of years he has been a sparring partner for professor Meeus in building a credible platform and network in Energy.
Job Title : Chairman Vlerick Energy Centre
Expert in
Energy Markets
Daniel Dobbeni Chairman Vlerick Energy Centre

redpajama_set_name: RedPajamaCommonCrawl


A compressed version of the same picture or pictures replaces the original high-resolution picture or pictures. Reduce the size of your publication by using linked pictures Each time you insert a picture in your publication, the publication grows in size.... Those people can use Microsoft inbuilt photo editing tool Paint. It is very easy to use and you can also merge two or more images using Microsoft Paint. It is very easy to use and you can also merge two or more images using Microsoft Paint.
A compressed version of the same picture or pictures replaces the original high-resolution picture or pictures. Reduce the size of your publication by using linked pictures Each time you insert a picture in your publication, the publication grows in size.... 15/06/2011�� Send one picture to paint first > in the paint window, use the cursor and " stretch " the side towards the right side to make room for the next picture > go to the next picture folder > right click at the 2nd pic > Copy > return to Paint window > top left, click Paste > move the 2nd picture to the open slot > now you have 2 pic in one.
29/08/2017�� To open Paint, type paint in the search box on the taskbar, and then select Paint from the list of results. With the Windows 10 Creators Update, try creating in three dimensions with Paint 3D. In addition to the new 3D capabilities, many of the classic 2D features from Microsoft Paint are �... As part of the creator's update to Windows 10, Microsoft has introduced Paint 3D, which is a modern Windows app. As opposed to the old classic Paint "program". The classic Paint is now deprecated and may be removed in future updates of Windows 10.


text:
// Copyright 2019 Google LLC
//
// 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
//
//     https://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.google.sps.servlets;

import com.google.appengine.api.datastore.DatastoreService;
import com.google.appengine.api.datastore.DatastoreServiceFactory;
import com.google.appengine.api.datastore.Entity;
import com.google.appengine.api.datastore.Key;
import com.google.appengine.api.datastore.KeyFactory;
import com.google.appengine.api.datastore.PreparedQuery;
import com.google.appengine.api.datastore.Query;
import com.google.appengine.api.datastore.Query.FilterOperator;
import com.google.appengine.api.datastore.Query.FilterPredicate;
import com.google.appengine.api.users.UserService;
import com.google.appengine.api.users.UserServiceFactory;
import com.google.common.flogger.FluentLogger;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Enumeration;
import java.util.List;
import javax.servlet.annotation.WebServlet;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

/**
 * Servlet that processes users responses to exam questions and stores them in datastore.
 *
 * @author Aidan Molloy
 * @author Róisín O'Farrell
 * @author Klaudia Obieglo
 */
@WebServlet("/examResponse")
public class ExamResponseServlet extends HttpServlet {
  private static final FluentLogger logger = FluentLogger.forEnclosingClass();

  /**
   * doPost process the information from the exam form response and send it to the datastore.
   *
   * @param request  provides request information from the HTTP servlet
   * @param response response object where servlet will write information to
   */
  @Override
  public void doPost(HttpServletRequest request, HttpServletResponse response) throws IOException {
    // Only logged in users should access this page.
    UserService userService = UserServiceFactory.getUserService();
    if (!userService.isUserLoggedIn()
        || !userService.getCurrentUser().getEmail().contains("@google.com")) {
      logger.atWarning().log("User is not logged in.");
      response.sendError(HttpServletResponse.SC_UNAUTHORIZED);
      return;
    }
    logger.atInfo().log("user=%s", userService.getCurrentUser());
    PrintWriter out = response.getWriter();
    response.setContentType("text/html");
    String examID = request.getParameter("examID");

    Enumeration<String> parameterNames = request.getParameterNames();
    String possibleMarks = null;
    String expected = "none";
    String type = null;
    try {
      String email = userService.getCurrentUser().getEmail();
      DatastoreService datastore = DatastoreServiceFactory.getDatastoreService();
      while (parameterNames.hasMoreElements()) {
        String questionID = parameterNames.nextElement();
        if (questionID.equals("examID")) {
          continue;
        }
        String[] questionAnswer = request.getParameterValues(questionID);
        // Correct mcq questions with pre-defined answers
        try {
          Long questionIDl = Long.parseLong(questionID);
          // Get questionType, mcqAnswer and possibleMarks using Key as Question has a known ID/Name
          Key questionKey = KeyFactory.createKey("Question", questionIDl);
          Entity qs = datastore.get(questionKey);
          type = (String) qs.getProperty("type");
          if (type.equals("MCQ")) {
            String mcqAnswer = (String) qs.getProperty("mcqAnswer");
            int questionNum = Integer.parseInt(mcqAnswer);
            possibleMarks = (String) qs.getProperty("marks");
            List<String> answerList = (List<String>) qs.getProperty("mcqPossibleAnswers");
            expected = (String) answerList.get(questionNum - 1);
          }
        } catch (Exception e) {
          System.out.println("Cannot Find Question");
          logger.atInfo().log("Cannot find question: %s", e);
        }
        Entity examResponseEntity = new Entity(questionID, email);
        examResponseEntity.setProperty("email", email);
        examResponseEntity.setProperty("answer", questionAnswer[0]);
        if (expected.equals(questionAnswer[0]) && type.equals("MCQ")) {
          examResponseEntity.setProperty("marks", possibleMarks);
        } else if (!expected.equals(questionAnswer[0]) && type.equals("MCQ")) {
          examResponseEntity.setProperty("marks", "0");
        } else {
          examResponseEntity.setProperty("marks", null);
        }
        datastore.put(examResponseEntity);
      }
      if (email != null && examID != null) {
        examTaken(email, Long.parseLong(examID));
      }
    } catch (Exception e) {
      logger.atSevere().log("There was an error: %s", e);
      response.sendError(HttpServletResponse.SC_BAD_REQUEST);
      return;
    }
    out.println("
Responses Saved.
");
    out.println("Return to dashboard");
  }

  /**
   * Updates UserExams entity to reflect the user has taken the exam.
   * @param email Users Email
   * @param examID ID of the Exam the user has taken
   */
  public void examTaken(String email, Long examID) {
    /*Marks what exam a user has taken by storing that exam id in their
    * UserInfo.
    */
    Query queryUser = new Query("UserExams").setFilter(new FilterPredicate(
          "email", FilterOperator.EQUAL, email));
    DatastoreService datastore = DatastoreServiceFactory.getDatastoreService();
    PreparedQuery pq = datastore.prepare(queryUser);
    Entity user = pq.asSingleEntity();
    if (user != null) {
      //add to examsTaken list
      if (user.getProperty("taken") == null) {
        List<Long> examsTakenList = new ArrayList<>();
        examsTakenList.add(examID);
        user.setProperty("taken", examsTakenList);
      } else {
        List<Long> examsTakenList = (List<Long>) user.getProperty("taken");
        examsTakenList.add(examID);
        System.out.println("taken list has curr exam");
        user.setProperty("taken", examsTakenList);
      }
      //remove examID from exams To Do list as exam has been taken
      if (user.getProperty("available") != null) {
        List<Long> availableExams = (List<Long>) user.getProperty("available");
        availableExams.remove(Long.valueOf(examID));
      }
      datastore.put(user);
    } else {
      Entity userExamEntity = new Entity("UserExams", email);
      userExamEntity.setProperty("available", null);
      userExamEntity.setProperty("created", new ArrayList<>());
      userExamEntity.setProperty("email", email);
      List<Long> examsTakenList = new ArrayList<>();
      examsTakenList.add(examID);
      userExamEntity.setProperty("taken", examsTakenList);
      datastore.put(userExamEntity);
    }
  }
}

redpajama_set_name: RedPajamaGithub


Posted in Farm News	| Comments Off on Capestone Chickens Sunbathing…!
Posted in Farm News	| Comments Off on Introducing….Capestone Office Dog!
Capestone starring in Wright Food Emporium.
Have you sampled our Pembrokeshire Poussin yet?
Posted in Uncategorized	| Comments Off on Have you sampled our Pembrokeshire Poussin yet?


text:
Welcome to OCTANORM® Germany
Sign in | Contact | DE
THE FUTURE IS NOW!
When Benjamin Bruder joined OCTANORM two years ago in 2019, no one knew what the following years would hold for the exhibition industry. In the beginning, Benjamin Bruder worked his way through all departments of our company and then focused on sales, marketing and our OSPI network. Thanks to his studies in International Business Management and his subsequent professional experience, he quickly gained a foothold and continued to expand his skills and areas of responsibility. With the successful appearance at EuroShop 2020 at the latest, he was ready to start full speed together with the company. As we all know, this plan was put on hold by the pandemic. But Benjamin's optimism persisted through all that time:
"The solidarity and commitment of our team during these past months, the trusting cooperation with both our longtime and our new business partners and the strong believe that there is a successful future for OCTANORM have encouraged me in my willingness and decision to take on even more responsibility."
We are therefore more than pleased that Benjamin Bruder will be joining his father Hans Bruder in the management of the company as of 1st July 2021. He succeeds Jörg Staeger, who is retiring from management but will continue to support the company as a representative of the main shareholder in a consultative function.
"My thanks go to my father, to Jörg Staeger, and to the entire staff for all the trust and support I have received over the past two years. With this amazing team behind me, I am very much looking forward to this new challenge and working together with you as a strong partner of our company."
In the future, our Cleanroom and Project fields are to be specifically promoted and put on a par with exhibition sector. Time was already used for this during the pandemic and new business areas were established and expanded.
To put it in the words of Benjamin Bruder:
"The future is now, so let's get started!"
Office Mannheim
Pick up address
OCTANORM® Germany
current: en-de


Now you don't have to utilize another system to check eligibility, all the information you need is in one place saving you time and reducing errors.
The Links system checks your customer's eligibility status for commercial payers, Medicaid, and Medicare every night at midnight so you'll never deliver equipment to patients without the coverage to pay for the equipment.
Any time you open a patient order or profile, we automatically check the customer's eligibility. You get real-time data so that if their coverage has changed between midnight and the time you review the order, you always have up to date information.


text:
Q: Get audioSessionId of song running in background In a Project.,when i equalize like
equalizer = new Equalizer(0, mediaPlayer.getAudioSessionId());

Equalizer is working fine. but when i don't have sessionId. Is there any global sessionId(0 is according to me) . because when i initialize equalizer(0,0). it is not working. why?
I set audioSessionId to 0 because i don't have access to a song which is run by any other app. Or how could i get audioSession id of song.
Or is there any other way to equalize a song run by another app.

redpajama_set_name: RedPajamaStackExchange


text:
'Morning Joe'; Jim Lehrer
"Morning Joe" co- hosts Joe Scarborough and Mika Brzezinski talk about their show and how it started. Jim Lehrer describes his 19th novel, "Oh, Johnny," whose protagonist is a young marine during WWII, and reflects on his career as an anchor for PBS NewsHour.
Jim Lehrer Mika Brzezinski Joe Scarborough
Entertainment Books Media
Jim Lehrer talks about his 19th novel, "Oh, Johnny" and his career as an anchor for PBS NewsHour. 17:43
'Morning Joe'
Entertainment, Media
"Morning Joe" co- hosts Joe Scarborough and Mika Brzezinski talk about their show and how it started. 34:29
Mika Brzezinski; Gabourey Sidibe
Journalist Mika Brzezinski shares her new memoir, "All Things at Once." Actor Gabourey Sidibe on the surprise hit film, "Precious." 53:17
Walter Cronkite reflects on how network news has moved away from hard news reporting. 20:39
Tony Award-winning playwright Michael Frayn reflects on the inspiration for his new play, "Democracy." 23:28


text:
AntiAhmadiyya Movement in Islam
Al-Fatwa International 7
Founder & President: Syed Abdul Hafeez Shah Editor: Dr. Syed Rashid Ali
No.7 / Rabi-ul-Awwal 1419 / June 1999
Dear Readers!
Syedna Muhammad was the Last Prophet in the long line of Prophets of Allah who brought Holy Quran as the Final Message that the Creator has sent for the guidance of humanity for all times to come. We, the Ummah of Muhammad (May Love and Peace of Allah eternally shower upon Him), have been exclusively honoured with this Divine Gift, and were given the glad tidings of this bounty in the following words by Allah:
"This day I have perfected your religion for you, completed My Favor upon you, and have chosen for you Islam as a way of life for you." (Quran 5.4)
This belief in the Finality of Message and the Messenger and the Perfection of their Religion is what distinguishes Muslim nation from all other religions, and stands as a barrier against all heretical movements and disintegrating forces which arose from time to time in the vast world of Islam. Quran, books of Hadith and Seerah, books of history, and various scriptures of other religions, have all testified to Syedna Muhammad being the Last and Final Messenger of Allah and is synonymous with the belief of Perfection of Religion and Quran being the Final Revealed Message from Allah for the guidance of his creations. Allah says in Quran:
"Muhammad is not the father of any man among you, but he is the Messenger of Allah and the Seal of the Prophets; and Allah is Aware of all things." (Quran 33.40)
Holy Prophet Muhammad announced:
"Verily, Risalat and Nabuwwat are terminated. So there shall be neither a Rasool (Messenger) nor a Nabi (Prophet) after me." (Tirmidhi vol.2 p.51; Masnad Ahmed)
All Prophets had given the glad tidings of the coming of a new Prophet but Holy Prophet Muhammad warned us that:
"In my Ummah there shall be born thirty liars, each of them claiming that he is a prophet BUT I AM THE KHATAM-UN-NABIYEEN, THERE SHALL BE NO PROPHET AFTER ME." (Abu Dawood, Tirmidhi)
Jamaat Ahmadiyya and Revolt against Syedna Muhammad's Prophethood
Among the different anti-Islamic movements that have arisen from time to time during the last 1400-year history of Islam, Qadianism/Ahmadiyya Movement is unique. Mirza Ghulam Ahmad Qadiani, the founder of Ahmadiyya Movement, has attacked both, the Message (Quran) and the Messenger, revolting against the Prophethood of Syedna Muhammad as well as attempting to alter and distort the message of Quran. Disguising themselves as the champions of Islam, they are deceitfully spreading their false doctrine, taking advantage of the ignorance of unsuspecting Muslims in various parts of the globe. Muslims, by and large, know the basic tenets of Islam, but they are ignorant of the pious fraud of Ahmadiyya Movement in the name of Islam. Few people know that this man, who started as a preacher and defender of Islam, was programmed by his colonial masters – The British Raj – to undermine the Love of Muhammad Rasoolullah in the hearts of Muslims and undermine the spirit of Jihad in the hearts of Muslims by distorting/altering the Quranic teachings, thus attempting to divide the Ummah.
Mirza Ghulam claimed to be a prophet:
"Dead are those who did not accept one Holy Messenger. Blessed is the one who had recognized me. Amongst ALL THE WAYS of Allah I am the last Way and from all His Lights, I am the Last Light. Unfortunate is the one leaves me, since without me there is all darkness." (Kishtee-e-Nooh, Roohani Khazain vol. 19 p.61)
Later he claimed to be the Second Advent of Holy Prophet Muhammad .
"'Muhammadur Rasoolullah wal lazeena m'ahoo, ashiddao 'ala alkuffare rohamao bainahum' in this revelation God has named me Muhammad as well as Messenger." (Roohani Khazain vol. 18 p.207)
"When I am the Holy Prophet re-incarnate and when all the accomplishments of Muhammad including the prophethood are reflected in the mirror of my shadiness, then who is the separate man who has claimed prophethood?" (Ek Ghalti Ka Izala, p.8; Roohani Khazain, vol.18, p.212)
"I am the only one selected in this Ummah to get the name of 'Prophet'. No one else deserved this name.... and it was bound to happen... so that as mentioned in the Sahih Hadith that 'there will be only one such person', that prophecy is fulfilled." (Haqeeqatul Wahi, Roohani Khazain vol. 22 p.407)
Thus Mirza stole the title of Khatam-un-Nabiyeen – the Final Prophet – from Holy Prophet Muhammad and applied it on himself. Later on Mirza lays the foundation for a parallel ummah, by labeling the entire Muslim nation as Kafirs and Bastards:
"Except for the CHILDREN OF PROSTITUTES, whose hearts have been sealed by God, everyone else believes in me and has accepted my prophethood." (Aina-e-Kamalat-e-Islam, Roohani Khazain vol.5 p.547)
"God has revealed to me that anyone to whom my message has reached and he has not accepted me, he is not a Muslim." (Letter Mirza to Dr. Abdul Hakeem)
Mirza's son, Mirza Bashir Ahmad declared:
"Any person who believes in Moses but does not believe in Christ, or believes in Christ but does not believe in Muhammad, or believes in Muhammad but does not believe in the Promised Messiah, is not only a Kafir, but a 'PUKKA' (confirmed) Kafir, and is out of the fold of Islam." (Kalimatul Fasl, page 110, by Mirza Bashir Ahmad)
Alteration in Holy Quran and Ahmadiyya Jamaat
Ahmadiyya Jamaat boasts of printing Holy Qurans in several languages. What people do not know is that Mirza Ghulam and his Jamaat have systematically attempted to distort the Message of Quran. The alteration in Quran can be done in following ways:
Alteration in text of Quran Alteration in translation and interpretation of text Misapplication of Quranic verses Abrogation of Quranic laws
Mirza Gulam A. Qadiani is guilty on all four counts. While the Arabic text of Holy Qurans has not been tempered with but he has helped himself freely with the distortion in translations, interpretations and misapplications of various verses.
Alteration in text of Quran:
Despite the fact that the Arabic text in printed Quran's of Jamaat Ahmadiyya have not been altered, but Mirza Gulam have altered the Arabic text of various verses while quoting them in his books. Few examples are attached with this leaflet.
Visit http://alhafeez.org/rashid/alter/alter3.htm for further details:
Alteration in Translation & Interpretation:
Examples of alteration in the meanings of Holy Quran are plenty, the most famous being the verse 33:40: Muhammad is not the father of any of your men, but (he is) the Messenger of Allah, and the Seal of the Prophets: and Allah has full knowledge of all things. In the light of this verse and Hadiths of Holy Prophet about His Finality as a Prophet and Messenger of Allah, all through the ages Muslims have unanimously believed Rasoolullah Muhammad to be the Last and Final Messenger of Allah – the Khatam-un-Nabiyeen. But ask any Qadiani/Ahmadi follower of Mirza Ghulam and they will play the same record: No No! This Seal (Khatam) is not the seal of Finality, it is the seal of approval for future prophets. But then, after sneaking through this back door of prophethood, he closes it again by saying: "Blessed is he who has recognized me. Of all the paths to God I am the Last Path, and of all his Lights, I am the last Light. Unfortunate is he who forsakes me, because without me all is darkness". (Roohani Khazain, vol.19, p.61)
But they have no reply to this question: What do you call a man, who steals the title Khatam-un-Nabiyeen from the Holy Prophet?
They are dumbfounded!
Misapplication of Quranic Verses:
Holy Quran is full of verses where Allah is praising His Most Beloved Prophet Muhammad . For example: al-Quran 9:33, 17:1, 3:31, 48:1-2, 36:1-3, 108:1, 21:107 etc. are all famous verses, which were exclusively revealed to Glorify the Prophet of Islam.
However open the books of Mirza Gulam, especially Haqeeqatul Wahi, vol.22 Roohani Khazain, and you will find that from pages 77 to 111, a shameless plagiarization of the Quranic text, which Mirza mentions as samples of his revelations/inspirations, revelaed again to praise Mirza! For instance:
"He it is Who hath sent His Messenger with the guidance and the Religion of Truth, that He may make it triumph over all religions." (As-Saff 61:9) (Tazkirah pp.387-388, 4th Edition)
"Say (O Muhammad, to mankind): If ye love Allah, follow me; Allah will love you." (Aal-Imran 3:31) (Haqiqat-ul-Wahi p.82)
"Nor doth he speak of (his own) desire. It is but a Revelation revealed." (An-Najm 53.3-4) (Tazkirah p.378 )
"Say (O Muhammad): O mankind! Lo! I am the Messenger of Allah to you all." (Al-A'raf 7:158) (Tazkirah p.352, 4th Edition)
"Lo! Those who swear allegiance unto thee (Muhammad) swear allegiance only unto Allah. The Hand of Allah is over their hands."(Al-Fath 48:10)(Haqiqat-ul-Wahi, 80)
"Say: I am only a human being like yourselves. My Lord reveals unto me that your God is only One God." (Al-Kahf l8:110) (Haqiqat-ul-Wahi, p.81)
"Verily We have given thee (O Muhammad) a manifest victory, that Allah may forgive thee of thy faults of the past and the future."(Al-Fath48:1)(Haqiqat-ul-Wahi p.94)
"Verily We have sent unto you (O men!) a Messenger as a witness over you as we sent a Messenger unto Pharaoh." (Al-Muzammil 73:15) (Haqiqat-ul-Wahi, p.101)
"To thee have We granted the Fountain (of Abundance)". (Al-Kauthar 108:1) (Haqiqat-ul-Wahi, p.102)
"Yaseen. By the Quran, full of Wisdom, thou art, indeed one of the Messengers sent on a straight path." (Ya-Sin 36: 1-3) (Haqiqat-ul-Wahz, p.107, Tazkirah, p.479)
"And thou (Muhammad) threwest not (a handful of dust) when thou did St throw, but Allah threw." (Al-Anfal 8:17) (Haqiqat-ul-Wahi p.70)
"Verse of Holy Quran (17:1) 'Subhan alLazee asra be'abdehee lailan min al-Masjid alHaram ila alMasjid alAqsa.....' Glory be to Him who took His Slave in the night from Masjid Haram to Masjid Aqsa...., its literal and true application is that Mosque which has been built by Mirza's father and whose extension has been done by Mirza." (Collection of Advertisement vol.3 p.286)
Imagine! For centuries Muslims were stupid to fight over the possession of the Holy City of Jerusalem. According to Mirza Ghulam Masjid Aqsa is in Qadian and Holy Prophet went to Qadian and not Jerusalem on Night of Ascension. What a service for the Jews!!
Abrogation of Quranic Law:
The concept of Jihad in Islam is a legitimate war with well defined aims and purposes: to suppress tyranny, insure the right of a man to his home and freedom within his nation, prevent persecution in religion, and guarantee freedom of belief to all people. Allah says:
"And fight them until persecution is no more, and religion is for Allah. But if they desist, then let there be no hostility except against wrongdoers (Zalimeen)." (Quran 2.193)
"And slay them wherever you find them, and drive them out of the places whence they drove you out.." (Quran 2.191)
"And wage war on all idolaters as they are waging war on all of you. And you know that Allah is with those who keep their duty (unto Him)." (Quran 9.36)
Despite these clear injunctions from Allah, at the behest of his masters, Mirza Ghulam abrogated the concept of Jihad. He wrote:
"From my early age till now when I am 65 years of age, I have been engaged , with my pen and tongue, in an important task to turn the hearts of Muslims towards the true love & Goodwill & sympathy for the British Government and to obliterate the idea of Jihad from the hearts of stupid (Muslims). " (Kitab-ul-Bariyah, Roohani Khazain vol. 13 p.350)
"For the sake of British Government, I have published & distributed 50,000 leaflets in this country and other Islamic countries (against Jihad)..the result is that hundreds of thousands of people have given up their filthy ideas about Jihad." (Roohani Khazain vol. 15 p.114)
It is all these insults against the Holy Prophet of Islam and attempts to distort the message of Quran, which led the entire Muslim Ummah to declare Mirza Ghulam and his Ahmadiyya Jamaat as non-Muslim and out of the fold of Islam. But they continue to portray themselves as the only champions of Islam and thus luring the ignorant and innocent Muslim masses, especially in Africa, from the Islam of Holy Prophet Muhammad to the deceitful doctrine of Mirza Ghulam's 'Ahmadi Islam'. May Allah protect all Muslims from this fraud in the name of Islam and show the ignorant Ahmadi followers true face of Mirza Ghulam Qadiani, for they have no access to his original writings. Ameen.
Wassalam to those who follow the hidayah
A.A.M.I. PO Box 11560
Dibba Fujairah U.A.E.
rasyed@emirates.net.ae
http://alhafeez.org/rashid/


text:
\section{Introduction}
 It plays an important role in global differential
geometry to investigate geometrical and topological structures of
manifolds. After the pioneering rigidity theorem for closed minimal
submanifolds in a sphere due to Simons \cite{Simons}, several
striking rigidity results for minimal submanifolds were proved by
Chern, do Carmo, Kobayashi, Lawson, Yau and others (see \cite{Chern,
Ejiri, Lawson, Li, Yau}, etc.). In 1979, Ejiri \cite{Ejiri} obtained
the following rigidity theorem for compact minimal submanifold in a unit sphere.\\\\
\textbf{Theorem A.} \emph{Let $M$ be an $n(\geq4)$-dimensional
  oriented compact simply connected minimal submanifold in $S^{n+p}$.
If the Ricci curvature of $M$ satisfies $Ric_M \geq n-2,$ then $M$
is either the totally geodesic submanifold $S^n$, the Clifford torus
$S^{m}\big(\sqrt{\frac{1}{2}}\big)\times
S^{m}\big(\sqrt{\frac{1}{2}}\big)$ in $S^{n+1}$ with $n=2m$, or
$\mathbb{C}P^{2}(\frac{4}{3})$ in $S^7$. Here
$\mathbb{C}P^{2}(\frac{4}{3})$ denotes the $2$-dimensional complex
projective space minimally immersed into $S^7$ with constant
holomorphic sectional curvature $\frac{4}{3}$.}\\

The pinching constant above is the best possible in even dimensions.
In 1987, Sun \cite{Sun} studied the rigidity of compact oriented
submanifolds of dimension $n\geq 4$ with parallel mean curvature and
pinched Ricci curvature in a sphere, and partially generalized
Ejiri's rigidity theorem. In 1990s, Shen \cite{ShYB} and Li \cite{Li2} extended Ejiri's
rigidity theorem to the case of 3-dimensional compact minimal
submanifolds in a sphere. Moverover,  Li
\cite {Li2} discussed $n(\geq5)$-dimensional
compact minimal submanifolds in $S^{n+p}$, and obtained that if $n$ is odd, and
$Ric_{M}>n-2-\frac{1}{n-1}$, then $M$ is the totally geodesic. In 2011, Xu and Tian \cite{XT} obtained a
refined version of the Ejiri rigidity theorem without the assumption
that $M$ is simply connected.

 Set $UM=\bigcup_{x\in M}U_xM$, where $U_xM=\{u\in T_xM:\|u\|=1\}$, and $T_xM$ is the tangent space at $x\in M$. In 1986, Gauchman \cite{Gauchman} proved the following rigidity theorem for compact minimal submanifolds in a unit sphere.\\\\
 \textbf{Theorem B.} \emph{Let $M$ be an n-dimensional compact minimal submanifold  in $S^{n+p}$. If $$||h(u,u)||^{2}\leq C(n,p)$$ for all $u\in UM$, where $h$ is the second fundamental form of $M$, then either $||h(u,u)||^{2}=0$, and $M$
is totally geodesic, or $n$ is even, and $||h(u,u)||^{2}=C(n,p)$.
 Here $$C(n,p)=\left\{\begin{array}{ll} 1,&\mbox{\ for\ } p=1,\mbox{\ and\ } n\mbox{\  is even},\\
\frac{n}{n-1}, &\mbox{\ for\ } p=1,\mbox{\ and\ } n\mbox{\  is odd},\\
 \frac{1}{3}, &\mbox{\ for\ }  p\geq2, \mbox{\ and\ } n\mbox{\  is even},\\
 \frac{n}{3n-2}, &\mbox{\ for\ }  p\geq2, \mbox{\ and\ } n\mbox{\  is odd}.
 \end{array}\right.$$}

When $n$ is even, the above rigidity theorem is optimal. In 1991, Leung \cite{Leung}  proved that if  $M^n$ is an odd-dimensional compact minimal submanifold  in $S^{n+p}$ with flat normal connection, and if
$||h(u,u)||^{2}\leq \frac{n}{n-1}$ for all $u\in UM$, then $M$ is totally geodesic. Let $M=S^{m}\big(\sqrt{\frac{m}{n}}\big)\times
S^{m+1}\big(\sqrt{\frac{m+1}{n}}\big)$ be a Clifford minimal hypersurface
in $S^{n+1}$ with $n=2m+1$, then $\frac{n-1}{n+1}\leq||h(u,u)||^{2}\leq \frac{n+1}{n-1}$. Based on these facts, Leung proposed the following conjecture.\\\\
\textbf{Conjecture C.} \emph{Let  $M^n$ be an odd-dimensional compact minimal submanifold  in $S^{n+p}$. \\
(i) If $||h(u,u)||^{2}\leq \frac{n}{n-1}$ for all $u\in UM$, then $M$ is homeomorphic to a sphere.\\
(ii) If $||h(u,u)||^{2}<\frac{n+1}{n-1}$ for all $u\in UM$, then $M$ is homeomorphic to a sphere.}\\

In 2001, Hasanis and Vlachos \cite{HV} proved if $M^n(n\geq5)$ is an odd-dimensional oriented compact minimal submanifold in $S^{n+p}$, and if $Ric_M>n-2-\frac{2}{n-1}$, then $M$ is homeomorphic to a sphere. Moreover, they showed that the condition $Ric_M>n-2-\frac{2}{n-1}$ is equivalent to $||h(u,u)||^{2}<\frac{n+1}{n-1}$ when $M$ is a compact minimal submanifold in $S^{n+p}$ with flat normal connection. In 2016, Qian and Tang \cite{QT} gave some counter examples  to Conjecture C, which are focal submanifolds of isoparametric hypersurfaces in unit spheres with $g=4$.

 Since 1970s, the sphere theorems for general submanifolds have been
investigated by several authors \cite{Lawson2,Shiohama1,Shiohama2,Vlachos,Xia,Xin,XHZ,XZ}. Recently, the authors \cite{XG2} generalized the Ejiri rigidity theorem for
minimal submanifolds in a sphere due to Ejiri \cite{Ejiri} and Xu-Tian \cite{XT} to compact
submanifolds with parallel mean curvature in space forms, and proved the following topological sphere theorem for compact submanifolds in a space form with nonnegative constant curvature.\\\\
\textbf{ Theorem D.} \emph{Let $M$ be an $n(\geq4)$-dimensional compact submanifold in
$F^{n+p}(c)$ with  $c\geq 0$,
 if  $Ric_M>(n-2)(c+H^2),$
then $M$ is homeomorphic to a sphere.} \\

Moreover, Xu-Leng-Gu \cite{XG3}  investigated the compact submanifolds with odd dimension in
  space forms, and proved the following topological
   sphere theorem. \\\\
\textbf{Theorem E.} \emph{Let $M$ be an $n(\geq5)$-dimensional
compact submanifold in $F^{n+p}(c)$ with  $c\geq0$. If $n$ is odd,
and
$$Ric_M>(n-2-\epsilon_n)(c+H^2),$$  then M
is homeomorphic to a sphere. Here $\epsilon_n =\frac{4}{n^3-2n^2-n-2}.$}\\

In this paper, we study the compact submanifolds in space forms, and prove the following topological sphere theorems.\\\\
\textbf{Theorem 1.1.} \emph{Let $M$ be an $n(\geq4)$-dimensional compact submanifold in
$F^{n+p}(c)$ with  $c\geq0$. If the scalar curvature of M satisfies  $R>n(n-2)(c+H^2)$, and $$Ric_M\geq (n-2-\frac{2\sigma_n}{2n-\sigma_n})(c+H^2),$$
then $M$ is homeomorphic to a sphere. Here $\sigma_n=sgn(n-4)((-1)^n+3)$, and $sgn(\cdot)$ is the standard sign function.}\\\\
\textbf{Remark.} If $Ric_M>(n-2)(c+H^2)$, then $R>n(n-2)(c+H^2)$. Therefore, Theorem 1.1 is a generalization of Theorem D.\\\\
  \textbf{Theorem 1.2.} \emph{Let $M$ be an $n(\geq4)$-dimensional compact oriented submanifold in
$F^{n+p}(c)$ with  $c\geq0$. If  $R>n(n-2)(c+H^2)$, and  $$Ric_M\leq (n-2+\frac{2\tau_n}{2n+\tau_n})(c+H^2),$$
then $M$ is homeomorphic to a sphere. Here $\tau_n=(-1)^n+3$.}\\

The following examples show that the pinching conditions in Theorems
1.1 and 1.2 are the best possible.\\\\
\textbf{Example 1.3.} Let
$S^{n+1}(\frac{1}{\sqrt{c+H^2}})$ be the totally umbilic sphere in
$F^{n+p}(c)$. Here $H$ is a nonnegative constant.

(i) Let
$M=S^{m}\big(\frac{1}{\sqrt{2(c+H^2)}}\big)\times
S^{m}\big(\frac{1}{\sqrt{2(c+H^2)}}\big)$ be a Clifford hypersurface
in $S^{n+1}(\frac{1}{\sqrt{c+H^2}})$ with $n=2m$ and $c+H^2>0$.  Then $M$ is a
compact submanifold in $F^{n+p}(c)$ with constant mean curvature $H$
and constant Ricci curvature $Ric_M\equiv (n-2)(c+H^2)$. Moreover, $R=n(n-2)(c+H^2).$

(ii) Let
$M=S^{m-1}\big(\sqrt{\frac{m-1}{n(c+H^2)}}\big)\times
S^{m+1}\big(\sqrt{\frac{m+1}{n(c+H^2)}}\big)$ be a Clifford hypersurface
in $S^{n+1}(\frac{1}{\sqrt{c+H^2}})$ with  $n=2m$ and  $c+H^2>0$.  Then $M$ is a
compact submanifold in $F^{n+p}(c)$ with constant mean curvature $H$,  $Ric(e_i)\equiv (n-2-\frac{4}{n-2})(c+H^2)$ for $1\leq i\leq m-1$, and $Ric(e_j)\equiv (n-2+\frac{4}{n+2})(c+H^2)$ for $m\leq j\leq n$. Moreover, $R=n(n-2)(c+H^2).$

(iii) When $n=2m+1$, let
$M=S^{m}\big(\sqrt{\frac{m}{n(c+H^2)}}\big)\times
S^{m+1}\big(\sqrt{\frac{m+1}{n(c+H^2)}}\big)$ be a Clifford hypersurface
in $S^{n+1}(\frac{1}{\sqrt{c+H^2}})$ with  $c+H^2>0$. Then $M$ is a
compact submanifold in $F^{n+p}(c)$ with constant mean curvature $H$,  $Ric(e_i)\equiv (n-2-\frac{2}{n-1})(c+H^2)$ for $1\leq i\leq m$, and $Ric(e_j)\equiv (n-2+\frac{2}{n+1})(c+H^2)$ for $m+1\leq j\leq n$. Moreover, $R=n(n-2)(c+H^2).$\\

\section{Notation and lemmas}
 Throughout this paper,
let $M$ be an $n$-dimensional compact Riemannian manifold
isometrically immersed into an $(n+p)$-dimensional complete and
simply connected space form $F^{n+p}(c)$. We shall make use of the
following convention on the range of indices:
$$ 1\leq A,B,C,\ldots\leq n+p;\ 1\leq i,j,k,\ldots\leq n;\ n+1\leq
\alpha,\beta,\gamma,\ldots\leq n+p.$$ We let \{$e_{A}$\} be local
orthonormal frames in $F^{n+p}(c)$ such that, restricted to $M$, the
$e_{i}$'s are tangent to \emph{M}. Let \{$\omega _{A}$\} and
\{$\omega _{AB}$\} be the dual frame field and the connection
1-forms of $F^{n+p}(c)$ respectively. Restricting these forms to
\emph{M}, we have
\begin{eqnarray}&&\omega_{\alpha i}=\sum_{j} h^{\alpha}_{ij}\omega_{j},  \, \,
h^{\alpha}_{ij}=h^{\alpha}_{ji},\nonumber\\
&&h=\sum_{\alpha,i,j}h^{\alpha}_{ij}\omega_{i}\otimes\omega_{j}\otimes
e_{\alpha},\,\,\xi=\frac{1}{n}\sum_{\alpha,i}h^{\alpha}_{ii}e_{\alpha},\nonumber
\\
&&R_{ijkl}=c(\delta_{ik}\delta_{jl}-\delta_{il}\delta_{jk})+\sum_{\alpha}(h^{\alpha}_{ik}h^{\alpha}_{jl}-h^{\alpha}_{il}h^{\alpha}_{jk}), \label{Gauss}\\
&&R_{\alpha\beta kl}=\sum_{i}(h^{\alpha}_{ik}h^{\beta}_{il}-h^{\alpha}_{il}h^{\beta}_{ik}),\label{Codazzi}
\end{eqnarray}
where $h, \xi, R_{ijkl},$ and $R_{\alpha\beta kl}$ are the second
fundamental form, the mean curvature vector, the curvature tensor
and the normal curvature tensor of $M$, respectively.  We define
$$S=|h|^{2}, \ H=|\xi|, \ H_{\alpha}=(h^{\alpha}_{ij})_{n\times
n}.$$
Denote by $Ric(u)$ the Ricci curvature of $M$ in direction of $u\in
UM$. From the Gauss equation, we have
\begin{equation}Ric(e_i)=(n-1)c
+\sum_{\alpha,j}\big[h_{ii}^{\alpha}h_{jj}^{\alpha}
                          -(h_{ij}^{\alpha})^2\big].\label{Ricci}\end{equation}
Set $Ric_{\min}(x)=\min_{u\in U_{x}M}Ric(u)$.
The scalar curvature $R$ of $M$ is given by
\begin{equation}R=n(n-1)c+n^{2}H^{2}-S.\label{Gauss2}\end{equation}
Choose an
orthonormal basis $\{e_{i}\}$ in $T_{x}M$ such that $e_{j_0}=u,\,\,
span\{e_{j_1},\ldots,e_{j_k}\}=V_{x}^k$, where the indices $1\le
j_0,j_1,\ldots,j_k\le n$ are distinct with each other.
 Using the notations of \cite{GX}, we set
\begin{eqnarray}&& Ric^{[s]}([e_{j_0},\ldots,e_{j_{n-1}}])  = \sum_ {p=0}^{s-1}\sum_
 {q=0}^{n-1}R_{j_pj_{q}j_pj_{q}},\nonumber\\
 &&Ric^{[s]}_{\min}(x)=\min_{
 \{e_{j_0},\ldots,e_{j_{n-1}}\}\subset T_{x}M} Ric^{[s]}([e_{j_0},\ldots,e_{j_{n-1}}]),\nonumber\\
 &&Ric^{[s]}_{\max}(x)=\max_{
 \{e_{j_0},\ldots,e_{j_{n-1}}\}\subset T_{x}M}
 Ric^{[s]}([e_{j_0},\ldots,e_{j_{n-1}}]).\label{2-5}\end{eqnarray}
  \textbf{Definition 2.1(\cite{GX}).} \emph{We call  $Ric^{[s]}([e_{j_0},\ldots,e_{j_{n-1}}])$ the $s$-th weak Ricci curvature of $M$, respectively.}\\

The nonexistence theorem for stable currents
in a compact Riemannian manifold $M$ isometrically immersed into
$F^{n+p}(c)$ is employed to eliminate the homology groups
$H_{q}(M;\mathbb{Z})$ for $0<q<n$, which was initiated by
Lawson-Simons \cite{Lawson2} and extended by Xin \cite{Xin}.\\\\
\textbf{Theorem 2.1.} \emph{Let $M^{n}$ be a compact submanifold in
$F^{n+p}(c)$ with $c\geq0$. Assume that$$
\sum_{k=q+1}^{n}\sum_{i=1}^{q}[2|h(e_{i},e_{k})|^{2}-\langle
h(e_{i},e_{i}),h(e_{k},e_{k})\rangle]<q(n-q)c$$ holds for any
orthonormal basis $\{e_{i}\}$ of $T_xM$ at any point $x\in M$, where
q is an integer satisfying $0<q<n$. Then there do not exist any
stable q-currents. Moreover,
$H_{q}(M;\mathbb{Z})=H_{n-q}(M;\mathbb{Z})=0,$ and $\pi_{1}(M)=0$
when $q=1$. Here $H_{i}(M;\mathbb{Z})$ is the $i$-th homology group
of M with integer
coefficients.}\\

To prove the sphere theorems for submanifolds, we need
to eliminate the fundamental group
$\pi_1(M)$ under the s-th weak Ricci curvature and the scalar curvature pinching condition, and get the following lemma.\\\\
  \textbf{Lemma 2.2.} \emph{Let $M$ be an $n(\geq4)$-dimensional
 compact submanifold in $F^{n+p}(c)$ with  $c\geq0$.
 Assume that one of the following  conditions holds: \\
  $(i)$ $R+2Ric_{\min}>n(n-1)(c+H^2),$\\
  $(ii)$  $3R-2Ric^{[n-1]}_{\max}>n(n-1)(c+H^2),$\\then $H_{1}(M;\mathbb{Z})=H_{n-1}(M;\mathbb{Z})=0,$ and $\pi_1(M)=0$.}\\\\
\textbf{Proof.} From (\ref{Gauss2}), we have
$$S-nH^2=n(n-1)(c+H^2)-R.$$ This together with (\ref{Ricci})
implies that
\begin{eqnarray} &&\sum_{k=2}^{n}[2|h(e_{1},e_{k})|^{2}-\langle
h(e_{1},e_{1}),h(e_{k},e_{k})\rangle]\nonumber\\
&=&2\sum_{\alpha}\sum_{k=2}^{n}(h^{\alpha}_{1k})^{2}-
\sum_{\alpha}\sum_{k=2}^{n}h^{\alpha}_{11}h^{\alpha}_{kk}\nonumber\\
&=&\sum_{\alpha}\sum_{k=2}^{n}(h^{\alpha}_{1k})^{2}-Ric(e_1)+(n-1)c\nonumber\\
&\leq&\frac{1}{2}(S-nH^2)-Ric(e_1)+(n-1)c\nonumber\\
&=&\frac{n(n-1)}{2}(c+H^2)-\frac{R}{2}-Ric(e_1)+(n-1)c.
\label{2-6} \end{eqnarray}
Since $$2Ric_{\min}+R>n(n-1)(c+H^2),$$ we have $$\sum_{k=2}^{n}[2|h(e_{1},e_{k})|^{2}-\langle
h(e_{1},e_{1}),h(e_{k},e_{k})\rangle]<(n-1)c.$$
On the other hand, we also get from (\ref{2-6}) that
\begin{eqnarray} &&\sum_{k=2}^{n}[2|h(e_{1},e_{k})|^{2}-\langle
h(e_{1},e_{1}),h(e_{k},e_{k})\rangle]\nonumber\\
&\leq&\frac{n(n-1)}{2}(c+H^2)-\frac{3R}{2}+\sum_{k=2}^nRic(e_k)+(n-1)c.
\end{eqnarray}
 Therefore, if $$3R-2Ric^{[n-1]}_{\max}>n(n-1)(c+H^2),$$  then
$$\sum_{k=2}^{n}[2|h(e_{1},e_{k})|^{2}-\langle
h(e_{1},e_{1}),h(e_{k},e_{k})\rangle]<(n-1)c.$$
Hence, it follows from Theorem 2.1 that $H_{1}(M;\mathbb{Z})=H_{n-1}(M;\mathbb{Z})=0,$
 and  $\pi_1(M)=0$. This proves Lemma 2.2.\qed \\

 If $$Ric_M>\frac{n(n-1)}{n+2}(c+H^2),$$  then we can directly obtain
$$2Ric_{\min}+R>n(n-1)(c+H^2).$$ Therefore, we have the following corollary  from Lemma 2.2.\\\\
 \textbf{Corollary 2.3 (\cite{XG2}).} \emph{Let $M$ be an $n(\geq4)$-dimensional
 compact submanifold in $F^{n+p}(c)$ with  $c\geq0$.
 If the Ricci curvature of M satisfies
  $$Ric_M>\frac{n(n-1)}{n+2}(c+H^2),$$ then $H_{1}(M;\mathbb{Z})=H_{n-1}(M;\mathbb{Z})=0,$ and $\pi_1(M)=0$.}\\

 \section{Proof of the Main Theorem}
In this section, we will give the proof of our Main Theorem. we first need
to eliminate the homology groups
$H_{q}(M;\mathbb{Z})$ for $1<q<n-1$ under the s-th weak Ricci curvature and scalar curvature pinching condition.\\\\
\textbf{Lemma 3.1.} \emph{Let $M$ be an $n(\geq4)$-dimensional
compact submanifold in $F^{n+p}(c)$ with  $c\geq0$.
 Assume that one of the following  conditions holds: \\
  $(i)$ $(n-q)(q-1)R+(n-2q)Ric^{[q]}_{\min}
> n^2(n-q-1)(q-1)(c+H^2),$ \\
$(ii)$  $q(n-q-1)R-(n-2q)Ric^{[n-q]}_{\max}
> n^2(n-q-1)(q-1)(c+H^2),$\\
where q is some integer in $[2, \frac{n}{2}]$. Then $H_{q}(M;Z)=H_{n-q}(M;Z)=0$.}\\\\
\textbf{Proof.} Assume that $ 2\leq q\leq \frac{n}{2}$. Setting
$$T_{\alpha}:=\frac{trH_{\alpha}}{n},$$
  we have $\sum_{\alpha}T_{\alpha}^{2}=H^2.$  Then we get from (\ref{Ricci})
  \begin{equation}Ric(e_i)=(n-1)c+\sum_{\alpha}\Big[nT_{\alpha}h^{\alpha}_{ii}-(h^{\alpha}_{ii})^2
  -\sum_{j\neq i}(h^{\alpha}_{ij})^2\Big].\label{3.1}
  \end{equation}
Then we get\begin{eqnarray}
&&\sum_{k=q+1}^{n}\sum_{i=1}^{q}[2|h(e_{i},e_{k})|^{2}-\langle
h(e_{i},e_{i}),h(e_{k},e_{k})\rangle]\nonumber\\
&=&2\sum_{\alpha}\sum_{k=q+1}^{n}\sum_{i=1}^{q}(h^{\alpha}_{ik})^{2}-
\sum_{\alpha}\sum_{k=q+1}^{n}\sum_{i=1}^{q}h^{\alpha}_{ii}h^{\alpha}_{kk}\nonumber\\
&=&\sum_{\alpha}\Big[2\sum_{k=q+1}^{n}\sum_{i=1}^{q}(h^{\alpha}_{ik})^{2}-
a\Big(\sum_{i=1}^{q}h^{\alpha}_{ii}\Big)\Big(nT_{\alpha}-\sum_{i=1}^{q}h^{\alpha}_{ii}\Big)\nonumber\\
&&-(1-a)\Big(\sum_{k=q+1}^{n}h^{\alpha}_{kk}\Big)\Big(nT_{\alpha}-\sum_{k=q+1}^{n}h^{\alpha}_{kk}\Big)\Big]\label{3.2}
\end{eqnarray}
for some $a\in(0,1)$.
Then it follows from (\ref{3.1}) and the Cauchy-Schwarz inequality that
\begin{eqnarray}&&\sum_{k=q+1}^{n}\sum_{i=1}^{q}[2|h(e_{i},e_{k})|^{2}-\langle
h(e_{i},e_{i}),h(e_{k},e_{k})\rangle]\nonumber\\
&\leq&\sum_{\alpha}\Big[2\sum_{k=q+1}^{n}\sum_{i=1}^{q}(h^{\alpha}_{ik})^{2}-
anT_{\alpha}\sum_{i=1}^{q}h^{\alpha}_{ii}+tq\sum_{i=1}^{q}(h^{\alpha}_{ii})^2\nonumber\\
&&-(1-a)nT_{\alpha}\sum_{k=q+1}^{n}h^{\alpha}_{kk}+(1-a)(n-q)\sum_{k=q+1}^{n}(h^{\alpha}_{kk})^2\Big]\nonumber\\
&\leq&\sum_{\alpha}\Big[2\sum_{k=q+1}^{n}\sum_{i=1}^{q}(h^{\alpha}_{ik})^{2}-aq\sum_{k\neq i}\sum_{i=1}^{q}(h^{\alpha}_{ik})^{2}-
(1-a)(n-q)\sum_{i\neq k}\sum_{k=q+1}^{n}(h^{\alpha}_{ik})^{2}\Big]\nonumber\\
&&+ aq\sum_{i=1}^{q}[(n-1)c-Ric(e_i)]+(1-a)(n-q)\sum_{k=q+1}^{n}[(n-1)c-Ric(e_k)]
\nonumber\\
&&+an(q-1)\sum_{\alpha}\sum_{i=1}^{q}T_{\alpha}h^{\alpha}_{ii}
+(1-a)n(n-q-1)\sum_{\alpha}\sum_{k=q+1}^{n}T_{\alpha}h^{\alpha}_{kk}.\label{3.3}
\end{eqnarray}
Since $q\geq2$, and $n-q\geq2$, we get
\begin{eqnarray}&&\sum_{\alpha}\Big[aq\sum_{k\neq i}\sum_{i=1}^{q}(h^{\alpha}_{ik})^{2}+
(1-a)(n-q)\sum_{i\neq k}\sum_{k=q+1}^{n}(h^{\alpha}_{ik})^{2}\Big]\nonumber\\
&\geq&\sum_{\alpha}\Big[2a\sum_{k\neq i}\sum_{i=1}^{q}(h^{\alpha}_{ik})^{2}+
2(1-a)\sum_{i\neq k}\sum_{k=q+1}^{n}(h^{\alpha}_{ik})^{2}\Big]\nonumber\\
&\geq&\sum_{\alpha}2\sum_{k=q+1}^{n}\sum_{i=1}^{q}(h^{\alpha}_{ik})^{2}.\label{3.4}
\end{eqnarray}
Let $a=\frac{n-q-1}{n-2}$, then we have $a(q-1)=(1-a)(n-q-1).$
This together with (\ref{3.3}), (\ref{3.4}) implies
\begin{eqnarray}&&\sum_{k=q+1}^{n}\sum_{i=1}^{q}[2|h(e_{i},e_{k})|^{2}-\langle
h(e_{i},e_{i}),h(e_{k},e_{k})\rangle]-q(n-q)c\nonumber\\
&\leq& -(2a-1)\sum_{i=1}^{q}Ric(e_i)-(1-a)(n-q)R
\nonumber\\
&&+[aq^2+(1-a)(n-q)^2](n-1)c\nonumber\\
&&+an^2(q-1)\sum_{\alpha}T^2_{\alpha}-q(n-q)c\nonumber\\
&=&-\frac{n-2q}{n-2}\sum_{i=1}^{q}Ric(e_i)-\frac{(n-q)(q-1)}{n-2}R
\nonumber\\
&& +\frac{n^2(n-q-1)(q-1)}{n-2}(c+H^2)\label{3.5}
\end{eqnarray}
It follows from the condition (i) and (\ref{3.5}) that
\begin{eqnarray}&&\sum_{k=q+1}^{n}\sum_{i=1}^{q}[2|h(e_{i},e_{k})|^{2}-\langle
h(e_{i},e_{i}),h(e_{k},e_{k})\rangle]-q(n-q)c\nonumber\\
&\leq&-\frac{n-2q}{n-2}Ric^{[q]}_{\min}-\frac{(n-q)(q-1)}{n-2}R
\nonumber\\&& +\frac{n^2(n-q-1)(q-1)}{n-2}(c+H^2)\nonumber\\
&<&0.\label{3.6}\end{eqnarray}
We also get from (\ref{3.5}) and condition (ii) that
\begin{eqnarray}&&\sum_{k=q+1}^{n}\sum_{i=1}^{q}[2|h(e_{i},e_{k})|^{2}-\langle
h(e_{i},e_{i}),h(e_{k},e_{k})\rangle]-q(n-q)c\nonumber\\
&\leq&\frac{n-2q}{n-2}\sum_{k=q+1}^{n}Ric(e_k)-\frac{q(n-q-1)}{n-2}R\nonumber\\
&& +\frac{n^2(n-q-1)(q-1)}{n-2}(c+H^2)\nonumber\\
&\leq& \frac{n-2q}{n-2}Ric^{[n-q]}_{\max}-\frac{q(n-q-1)}{n-2}R
\nonumber\\&& +\frac{n^2(n-q-1)(q-1)}{n-2}(c+H^2)\nonumber\\
&<&0.\label{3.8}\end{eqnarray}
 Then the assertion follows from (\ref{3.6}), (\ref{3.8}) and Theorem 2.1.\qed \\\\
\textbf{ Proof of Theorem 1.1.} If $R>n(n-2)(c+H^2)$, and $Ric_M\geq \frac{n}{2}(c+H^2)$, then $$R+2Ric_{\min}>n(n-1)(c+H^2).$$
 It follows from Theorem 2.2 that $H_{1}(M;\mathbb{Z})=H_{n-1}(M;\mathbb{Z})=0,$ and $\pi_1(M)=0$.

For $q\in[2,\frac{n}{2}]$,  we consider the following three
    cases:

Case I. $q=\frac{n}{2}$. If $R>n(n-2)(c+H^2)$,  then
 $(n-q)(q-1)R+(n-2q)Ric^{[q]}_{\min}
> n^2(n-q-1)(q-1)(c+H^2)$. Hence we get from Theorem 3.1 that $H_{\frac{n}{2}}(M;Z)=0$.

Case II. $q<\frac{n}{2}$. If $R>n(n-2)(c+H^2)$ and $Ric^{[q]}_{\min}\geq n(q-1)(c+H^2)$, then
 $(n-q)(q-1)R+(n-2q)Ric^{[q]}_{\min}
> n^2(n-q-1)(q-1)(c+H^2),$ for some $q\in[2,\frac{n}{2})$. Therefore, we get from Theorem 3.1 that $H_{q}(M;Z)=H_{n-q}(M;Z)=0$ for some $q\in[2,\frac{n}{2})$.

Since   $Ric^{[l]}_{\min}\geq lRic_{\min}$, we have if
$n=even$, and $$Ric_M\geq (n-2-\frac{4}{n-2})(c+H^2),$$ then $Ric^{[q]}_{\min}\geq n(q-1)(c+H^2)$ for all $q\in[2,\frac{n}{2})$.
If $n=odd$, and $$Ric_M\geq (n-2-\frac{2}{n-1})(c+H^2),$$  then $Ric^{[q]}_{\min}\geq n(q-1)(c+H^2)$ for all $q\in[2,\frac{n}{2})$.

Moreover, we have $$\frac{n}{2}(c+H^2)\leq (n-2)(c+H^2)$$ for $n\geq4$,
 $$\frac{n}{2}(c+H^2)\leq (n-2-\frac{2}{n-1})(c+H^2)$$ for $n\geq 5$,
  and $$\frac{n}{2}(c+H^2)\leq (n-2-\frac{4}{n-2})(c+H^2)$$ for $n\geq 6$. Hence it follows from the assumption that $H_{q}(M;Z)=H_{n-q}(M;Z)=0$ for all $q\in[1,\frac{n}{2}]$ and $\pi_1(M)=0$. It follows from $H_{i}(M;Z)=0$ for $i=1,2,\cdots,n-1$ that $M$ is a homology sphere.
 Since $M$ is simply connected, $M$ is a homotopy sphere. This together with the generalized Poincar\'{e}
 conjecture implies that $M$ is a topological sphere. This completes the proof of Theorem 1.1. \qed\\

Moreover, when $M$ is oriented, we get the following sphere theorem.\\\\
\textbf{Theorem 3.2.} \emph{Let $M$ be an $n(\geq4)$-dimensional oriented compact submanifold in
$F^{n+p}(c)$ with  $c\geq0$. If  $R>n(n-2)(c+H^2)$, and $$Ric^{[2]}_{\min}\geq 2(n-2-\frac{2\sigma_n}{2n-\sigma_n})(c+H^2),$$
then $M$ is homeomorphic to a sphere. Here $\sigma_n=sgn(n-4)((-1)^n+3)$, and $sgn(\cdot)$ is the standard sign function.}\\\\
\textbf{Proof.}  We first prove that
$Ric_M>0$. We consider the following three cases:

$(i)$ If $n=4$, and if there exists a point $x_0\in M$ such that $Ric_{\min}(x_0)\leq 0$, then without loss of generality,
we assume that $Ric(e_1)\leq 0$ at point $x_0$. Hence it follows from the assumption that $Ric(e_i)\geq 2(n-2)(c+H^2)$ for $i=2,\cdots, n$.
Then we get \begin{eqnarray}R(x_0)&\geq& 2(n-2)(c+H^2)+(n-2)2(n-2)(c+H^2)\nonumber\\
&=&2(n-1)(n-2)(c+H^2)\nonumber\\
&=&n(n-1)(c+H^2).\end{eqnarray}
On the other hand, we know from the Gauss equation that $R\leq n(n-1)(c+H^2)$. Therefore $R(x_0)=n(n-1)(c+H^2)$, and $S(x_0)=nH^2$, i.e., $x_0$ is a totally umbilical point. Then we get from the Gauss equation that $Ric_M\equiv(n-1)(c+H^2)$ at that point, which contradicts with the assumption. Therefore, $Ric_M>0$.

$(ii)$ If $n$ is even, $n\geq 6$, and if there exists a point $x_1\in M$ such that $Ric_{\min}(x_1)\leq 0$, then without loss of generality,
we assume that $Ric(e_1)\leq 0$ at point $x_1$. Hence it follows from the assumption that $Ric(e_i)\geq 2(n-2-\frac{4}{n-2})(c+H^2)$ for $i=2,\cdots, n$.
Then we get \begin{eqnarray}R(x_1)&\geq& 2(n-2-\frac{4}{n-2})(c+H^2)+2(n-2)(n-2-\frac{4}{n-2})(c+H^2)\nonumber\\
&=&2(n-1)(n-2-\frac{4}{n-2})(c+H^2)\nonumber\\
&\geq&n(n-1)(c+H^2),\end{eqnarray}
and the equality holds only if $n=6$. This together with the Gauss equation implies that $R(x_1)=n(n-1)(c+H^2)$, and $n=6$.  But a similar argument as in Case (i) shows that $Ric_M\equiv(n-1)(c+H^2)$ at point $x_1$, and it contradicts with the assumption. Therefore, $Ric_M>0$.

(iii) If $n$ is odd, and if there exists a point $x_2\in M$ such that $Ric_{\min}(x_2)\leq 0$, then without loss of generality,
we assume that $Ric(e_1)\leq 0$ at point $x_2$. Hence it follows from the assumption that $Ric(e_i)\geq 2(n-2-\frac{2}{n-1})(c+H^2)$ for $i=2,\cdots, n$.
Then we get \begin{eqnarray}R(x_2)&\geq& 2(n-2-\frac{2}{n-1})(c+H^2)+2(n-2)(n-2-\frac{2}{n-1})(c+H^2)\nonumber\\
&=&2(n-1)(n-2-\frac{2}{n-1})(c+H^2)\nonumber\\
&\geq&n(n-1)(c+H^2),\end{eqnarray}
and the equality holds only if $n=5$. This together with the Gauss equation implies that $R(x_2)=n(n-1)(c+H^2)$, and $n=5$.  But a similar argument as in Case (i) shows that $Ric_M\equiv(n-1)(c+H^2)$ at point $x_2$, and it contradicts with the assumption. Therefore, $Ric_M>0$.

On the other hand, it follows from the proof of Theorem 1.1 and the assumption that $H_{q}(M;Z)=H_{n-q}(M;Z)=0$ for all $q\in[2,\frac{n}{2}]$.
This together with the universal
coefficient theorem implies that
 $H^{n-1}(M;\mathbb{Z})$ has no torsion, and hence neither does $H_{1}(M;\mathbb{Z})$
 by the Poincar$\acute{{\rm e}}$ duality. We also get from $Ric_M>0$ and
the Bonnet-Myers theorem that the fundamental group $\pi_{1}(M)$ is
finite. Then we have $H_{1}(M;\mathbb{Z})=0$. Hence,
 $H_{n-1}(M;\mathbb{Z})=0$. Denote by $\widetilde{M}$ the universal Riemannian covering of
$M$. We may consider $\widetilde{M}$ be a Riemannian submanifold of
$F^{n+p}(c)$, and hence $\widetilde{M}$ is a homology sphere.
Since $\widetilde{M}$ is simply connected, it is a topological
sphere, which together with a result of Sjerve \cite{Sjerve} implies
that $M$ is simply connected. Then $M$ is a homotopy sphere. This together with the generalized Poincar\'{e}
 conjecture implies that $M$ is a topological sphere. \qed \\

It's easy to get that if $Ric^{[2]}_{\min}>2(n-2)(c+H^2)$, then $R>n(n-2)(c+H^2)$. Hence we have the following corollary.\\\\
\textbf{Corollary 3.3.} \emph{Let $M$ be an $n(\geq4)$-dimensional oriented compact submanifold in
$F^{n+p}(c)$ with  $c\geq0$. If $Ric^{[2]}_{\min}>2(n-2)(c+H^2)$,
then $M$ is homeomorphic to a sphere.}\\

 For $n=4$, we also get the following differentiable theorem.\\\\
 \textbf{Theorem 3.4.} \emph{Let $M$ be a $4$-dimensional  compact oriented submanifold in
$F^{n+p}(c)$ with $c\geq0$. If  $R>8(c+H^2)$, and $Ric_M>0$, then $M$ is diffeomorphic to a sphere.} \\\\
 \textbf{Proof.} Since $R>8(c+H^2)$ is equal to $S<4c+8H^2$, it follows from a theorem due to Xu-Zhao \cite{XZ} that $M$ is diffeomorphic to a spherical space form.
 Moreover, we know from Lemma 3.1 that $H_{2}(M;\mathbb{Z})=0$. Using the same argument as in the proof of Theorem 3.2 implies that
  $H_{1}(M;Z)=H_{n-1}(M;Z)=0$, and $M$ is simply connected. Therefore, $M$ is diffeomorphic to a sphere.\qed \\\\
\textbf{Theorem 3.5.} \emph{Let $M$ be an $n(\geq5)$-dimensional
oriented compact submanifold in $F^{n+p}(c)$ with  $c\geq0$. If $n$ is odd,
and
$$Ric^{[2]}_{\min}>2(n-2-\epsilon_n)(c+H^2),$$  then M
is homeomorphic to a sphere. Here
$\epsilon_n =\frac{4}{n^3-2n^2-n-2}.$} \\\\
\textbf{Proof.} Since $R\geq\frac{n}{l}Ric^{[l]}_{\min}$, we get from Lemma 3.1 that
 if $$Ric^{[q]}_{\min}>\Big[q(n-1)-\frac{q^2(n-q)(n-2)}{q(n-2q)+n(q-1)(n-q)}\Big](c+H^2),$$
where $q$ is some integer in $[2, \frac{n}{2}]$, then $H_{q}(M;Z)=H_{n-q}(M;Z)=0$.
On the other hand,  we know that $\frac{Ric^{[l]}_{\min}}{l}\geq\frac{ Ric^{[2]}_{\min}}{2}$ for $l\geq 2$, and the function $$f(s)=\frac{s(n-s)}{s(n-2s)+n(s-1)(n-s)}$$ is strictly monotone decreasing for $2\leq s\leq  \frac{n}{2}$. Hence, if $$Ric^{[2]}_{\min}>\Big[2(n-1)-\frac{2s(n-s)(n-2)}{s(n-2s)+n(s-1)(n-s)}\Big](c+H^2),$$ for
some integer $s\in[2, \frac{n}{2}]$, then $H_{q}(M;Z)=H_{n-q}(M;Z)=0$ for all
 $q\in[2, s].$ Let $s=\frac{n-1}{2}$. Then we get from the assumption that $H_{q}(M;Z)=H_{n-q}(M;Z)=0$ for $q\in[2,\frac{n}{2}]$.
 Moreover, it follows from the proof of Theorem 3.2 that $Ric_M>0$. Using a same argument as in the proof of Theorem 3.2, we get that
  $H_{1}(M;Z)=H_{n-1}(M;Z)=0$, and $M$ is simply connected. Then $M$ is a homotopy sphere. This together with the generalized Poincar\'{e}
 conjecture implies that $M$ is a topological sphere. This proves Theorem 3.5.\qed\\\\
\textbf{Theorem 3.6.} \emph{Let $M$ be an $n(\geq5)$-dimensional compact submanifold in
$F^{n+p}(c)$ with  $c\geq0$.  Assume that the scalar curvature of M satisfies  $R>n(n-2)(c+H^2)$. Then we have\\
$(i)$ if $n=5, 6, 8$, and  $\frac{Ric^{[s]}_{\max}}{s}\leq[n-2+\frac{n-4}{2(n-1)}](c+H^2),$
then $M$ is homeomorphic to a sphere.\\
$(ii)$ if $n=7$,  $n\geq 9$,  and $\frac{Ric^{[s]}_{\max}}{s}\leq (n-2+\frac{2\tau_n}{2n+\tau_n})(c+H^2)$,
then $M$ is homeomorphic to a sphere. \\
Here $s$ is some integer in $[1, \frac{n+2}{2}]$, and $\tau_n=(-1)^n+3$.}\\\\
\textbf{Remark.} It follows from (ii) of  Example 1.3 that $\frac{Ric^{[s]}_{\max}}{s}=(n-2+\frac{4}{n+2})(c+H^2)$, for the submanifold $S^{m-1}\big(\sqrt{\frac{m-1}{n(c+H^2)}}\big)\times
S^{m+1}\big(\sqrt{\frac{m+1}{n(c+H^2)}}\big)$ with $n=2m$. It follows from (iii) of  Example 1.3 that  $\frac{Ric^{[s]}_{\max}}{s}=(n-2+\frac{2}{n+1})(c+H^2),$ for the submanifold $S^{m}\big(\sqrt{\frac{m}{n(c+H^2)}}\big)\times
S^{m+1}\big(\sqrt{\frac{m+1}{n(c+H^2)}}\big)$ with $n=2m+1$. \\\\
\textbf{ Proof.}  If $R>n(n-2)(c+H^2)$, and $Ric^{[n-1]}_{\max}\leq (n^2-2.5n)(c+H^2)$, then $$3R-2Ric^{[n-1]}_{\max}>n(n-1)(c+H^2).$$
 It follows from Theorem 2.2 that $H_{1}(M;\mathbb{Z})=H_{n-1}(M;\mathbb{Z})=0,$ and $\pi_1(M)=0$.

For $q\in[2,\frac{n}{2}]$,  we consider the following three
    cases:

Case I. $q=\frac{n}{2}$. If $R>n(n-2)(c+H^2)$,  then
  $q(n-q-1)R-(n-2q)Ric^{[n-q]}_{\max}
> n^2(n-q-1)(q-1)(c+H^2).$ Hence we get from Lemma 3.1 that $H_{\frac{n}{2}}(M;Z)=0$.

Case II. $q<\frac{n}{2}$. If $R>n(n-2)(c+H^2)$ and $Ric^{[n-q]}_{\max}\leq n(n-q-1)(c+H^2)$, then
$q(n-q-1)R-(n-2q)Ric^{[n-q]}_{\max}
> n^2(n-q-1)(q-1)(c+H^2),$ for some $q\in[2,\frac{n}{2})$. Therefore, we get from Lemma 3.1 that $H_{q}(M;Z)=H_{n-q}(M;Z)=0$ for some $q\in[2,\frac{n}{2})$.

Since   $\frac{Ric^{[l_1]}_{\max}}{l_1}\leq \frac{Ric^{[l_2]}_{\max}}{l_2}$ for $l_1\geq l_2$, we have if
$n=even$, and $$ Ric^{[\frac{n+2}{2}]}_{\max}\leq \frac{n^2}{2}(c+H^2),$$ then $Ric^{[n-q]}_{\max}\leq n(n-q-1)(c+H^2)$ for all $q\in[2,\frac{n}{2})$.
If $n=odd$, and $$Ric^{[\frac{n+1}{2}]}_{\max}\leq \frac{n(n-1)}{2}(c+H^2),$$  then $Ric^{[n-q]}_{\max}\leq n(n-q-1)(c+H^2)$ for all $q\in[2,\frac{n}{2})$.

Moreover, we have $$\frac{n(2n-5)}{2(n-1)}<\frac{n(n-1)}{(n+1)}$$ for $n=5$, and $$\frac{n(2n-5)}{2(n-1)}\geq\frac{n(n-1)}{(n+1)}$$ for $n\geq7$;
$$\frac{n(2n-5)}{2(n-1)}<\frac{n^2}{n+2}$$ for $n=6,8$, and $$\frac{n(2n-5)}{2(n-1)}\geq\frac{n^2}{n+2}$$ for $n\geq10$.
Hence it follows from the assumption that $H_{q}(M;Z)=H_{n-q}(M;Z)=0$ for all $q\in[1,\frac{n}{2}]$ and $\pi_1(M)=0$. It follows from $H_{i}(M;Z)=0$ for $i=1,2,\cdots,n-1$ that $M$ is a homology sphere.
 Since $M$ is simply connected, $M$ is a homotopy sphere. This together with the generalized Poincar\'{e}
 conjecture implies that $M$ is a topological sphere. This completes the proof of Theorem 3.6.\qed\\\\
 \textbf{Theorem 3.7.} \emph{Let $M$ be an $n$-dimensional compact oriented  submanifold in
$F^{n+p}(c)$ with  $c\geq0$. Assume that the scalar curvature of M satisfies  $R>n(n-2)(c+H^2)$. We have that \\
$(i)$ if $n=5$, and  $\frac{Ric^{[s]}_{\max}}{s}\leq(n-2+\frac{2}{n+1})(c+H^2),$
then $M$ is homeomorphic to a sphere.\\
$(ii)$ if $n=4,6,8$,  and $\frac{Ric^{[s]}_{\max}}{s}\leq(n-2+\frac{4}{n+2})(c+H^2)$, then $M$ is homeomorphic to a sphere.\\
Here $s$ is some integer in $[1, \frac{n+2}{2}]$.}\\\\
\textbf{Proof.} It follows from the assumption and the proof of Theorem 3.6 that  $H_{2}(M;Z)=H_{3}(M;Z)=0$ for $n=5$, and $$H_{q}(M;Z)=H_{n-q}(M;Z)=0$$ for $n=4$, $6$, $8$, where $q\in[2,\frac{n}{2}]$. Moreover, we have
\begin{eqnarray}
Ric_{\min}&\geq& R-(n-1)\frac{Ric^{[l]}_{\max}}{l}\nonumber\\
&>& n(n-2)(c+H^2)-(n-1)(n-2+\frac{4}{n+2})(c+H^2)\nonumber\\
&=&\frac{n(n-4)}{n+2}(c+H^2)\nonumber\\
&>&0.
\end{eqnarray}
Then a similar argument as in the proof of Theorem 3.2 shows that $$H_{1}(M;Z)=H_{n-1}(M;Z)=0,$$ and $M$ is simply connected. Therefore, $M$ is homeomorphic to a sphere. This proves Theorem 3.7.\qed\\\\
\textbf{Proof of Theorem 1.2.}  Choosing $s=1$, we get the conclusion from Theorems 3.6 and 3.7.\qed\\

Motivated by Theorem 1.1, Example 1.3, the convergence results of Brendle and Schoen for
Ricci flow \cite{Brendle, Brendle0, Brendle2}, and the differentiable sphere theorems for submanifolds with positive Ricci curvature proved in \cite{XG2, XT}, we propose the following conjecture.\\\\
\textbf{Conjecture 3.8.} \emph{Let $M$ be an $n(\geq 5)$-dimensional compact
submanifold in the space form $F^{n+p}(c)$ with $c+H^2>0$.
 If $n$ is odd, and  $$Ric_{M}>(n-2-\frac{2}{n-1})(c+H^2),$$ then
$M$ is diffeomorphic to $S^n$.}\\

To verify Conjecture 3.8,  we hope to prove the
following conjecture on the normalized Ricci flow.\\\\
\textbf{Conjecture 3.9.} \emph{Let $(M,g_0)$ be an
$n(\geq5)$-dimensional compact submanifold in an $(n+p)$-dimensional
space form $F^{n+p}(c)$ with $c+H^2>0$. If $n$ is odd, and the Ricci curvature of
$M$ satisfies
$$Ric_{M} >
(n-2-\frac{2}{n-1})(c+H^2),$$
then the normalized Ricci flow with initial metric
$g_0$
$$\frac{\partial}{\partial t}g(t) = -2Ric_{g(t)} +\frac{2}{n}
r_{g(t)}g(t),$$ exists for all time and converges to a constant
curvature metric as $t\rightarrow\infty$. Moreover, $M$ is
diffeomorphic to a spherical space form.}\\

redpajama_set_name: RedPajamaArXiv


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Use this form to contact the KFH Canada Water Sales team.
Use this form to contact the KFH Canada Water Lettings team.
For most people Canada Water it is all about location: the proximity to the City and Canary Wharf attracts City workers, but as the nearby attractions of Bermondsey spill into historic Rotherhithe, it increasingly brings in young media professionals too.
We've had a great experience letting our property via KFH. You have been very helpful and responsive. You found great tenants for our property in a very short time. It's been a pleasure working with you.
I would highly recommend you. You were knowledgeable, helpful and available to chat to me about the process, from viewing properties all the way up to completion. You were particularly helpful in moving the property sale forward when things were going slowly.
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Stone Age Migrants Transform DNA!
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Article - History
Mar 28, 2019 By Aaditi P, Writer
Did you know that our ancestors could have changed the very genetic material in their descendants by crossing a mountain range!
This is exactly what researchers recently discovered as they studied the migration patterns and fossilized remains of people from the Bronze Age (beginning in 2500 BC) who navigated over the Pyrenees Mountains to live on the Iberian Peninsula. The Iberian Peninsula is located on the modern-day continent of Europe and is home to the countries of Spain, Portugal, and Andorra.
Researchers extracted DNA from many Bronze Age Iberians and found that many of them traced their ancestry to the few Neolithic farmers (people from the Stone Age) living in various places in Europe including Spain, while the rest of their genetic makeup was similar to people who lived on the Russian Steppe (grassland plains between Europe and Asia.
The analysis of these historical footprints aimed to uncover the full history of the widespread migration and better understand the evolution of humanity since the Neolithic Age.
What Caused the Pyrenees Crossing?
One scientific theory to explain the Pyrenees Crossing is the emergence of an ancient plague in Sweden. Scientists support this theory with the fact that Y. Pestis, a bacteria that causes the pneumonic, septicemic, and bubonic plagues, spread during the end of the Neolithic Age.
By comparing the genomes of Iberians and other early humans to analyze the presence of diseases in them, scientists concluded that an epidemic possibly sprung up in large establishments near the steppes and over trade routes.
Researchers also analyzed the presence of the Y chromosome, a spool of DNA that is typically passed down unchanged from father to son (in a paternal line). After analyzing the DNA data, scientists discovered that the original Y chromosome lineage disappeared from the local Iberians' genome before the arrival of the new migrants and was replaced with a Y chromosome that matched that of the migrants!
What Does the Research Tell Us?
Male domination was a key aspect of the migrants' society, especially given the replacement of the original Iberians' Y chromosome. Iñigo Olalde, a co-author of the genetic study report, said that "It would be a mistake to jump to the conclusion that Iberian men were killed or forcibly displaced," because there was no evidence of violence in the archaeological record.
In fact, the theory is that early clans were structured in such a way that only the first-born son inherited the clan's property, while the rest of the sons would establish new clans elsewhere to spread their male lineage.
Scientists also concluded that the people of the steppes moved west and integrated elements of other cultures into their own.
DNA can tell us so many incredible things about our past, and this is just one of them. Even after so much exploration, much of the long history of humanity remains unknown. Researchers continue to make more and more discoveries and one day, all of these discoveries may finally unlock a key to our past.
Sources: National Geographic, BBC, New York Times, archaeology.org
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Neanderthals: Quite The Artists! (0)
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Oldest Fossils Of Humans Discovered (4)
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The average pay for a Construction Foreman is €58,913 a year and €28 an hour in Finland. The average salary range for a Construction Foreman is between €42,177 and €72,569. On average, a High School Degree is the highest level of education for a Construction Foreman. This compensation analysis is based on salary survey data collected directly from employers and anonymous employees in Finland.
Coordinates and oversees activities of workers engaged in construction, installation, and repair of structures and fixtures.
Studies blueprints to determine dimensions of structure.
Selects materials and structural units, ensuring conformance with provisions of building code and local ordinance.
Determines sequence of activities concerned with fabrication, assembly, and erection of structure.
Assigns workers to such tasks as cutting material to size, building concrete forms, erecting wooden framework, installing doors, building stairs, and laying flooring.


I recently returned to New York from a trip back to Australia, where I was running some workshops with studiOH!. studiOH! is a pop-up workshop business I have in partnership with my friend Eko and we run several rounds of workshops each year in a range of crafty activities. This time our Summer Workshops series was held at the lovely Kookery cafe and included workshops in Water-slide Decals, Cork Coasters & Trivets, Concrete Planters, Kokedama, Terrariums and Felted Creatures.
These are just a few photos from some of the workshops (more photos are on our Facebook page). I hope if you made it along you enjoyed yourself as much as we did! I particularly loved seeing how beautifully everyone's concrete planters turned out!
If you're in Perth, keep an eye on the studiOH! website for new workshops being announced soon, and if you're in New York keep an eye on my website as I teach regular workshops at Brooklyn Brainery and have kokedama and concrete planter ones planned for the coming months!
05 Mar	This entry was published on March 5, 2014 at 11:24 pm. It's filed under make and tagged concrete, DIY, grow, ideas, kokedama, make, new york, perth, studiOH!, terrariums, workshops.	Bookmark the permalink.	Follow any comments here with the RSS feed for this post.
I did attend the workshops and they were terrific.


Hillandale (GNIS FID: 120116) is a populated place LOCATED WITHIN the City of Huntsville. In turn, the City of Huntsville is located within Madison County.
Hillandale is located at latitude 34.732 and longitude -86.616. The elevation of Hillandale is 666 feet above mean sea level.
Since Hillandale is located within the City of Huntsville, all of the census and demographic data for residents of Hillandale are included as part of the information reported for City of Huntsville in the "Profile" tab above.


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Updated Nov-19-2020, 15:05:00 GMT
Why we should be funding more Solyndras
MIT Technology Review Nov-19-2020, 15:05:00 GMT
President-elect Joe Biden won the US election in part by running on an ambitious climate platform promising to invest heavily to avert climate catastrophe while creating millions of well-paying jobs. But the question of how Biden's proposed nearly $2 trillion in green investment will get spent, and what other measures the government will take to put the green economy on the fast track, is still up for debate. Some politicians are now championing industrial policy as the way forward. Under industrial policy, the government makes investments that the private sector is unwilling or unable to make, and which will help the country achieve certain socially desirable goals. In short, industrial policy is a form of government planning to create or support strategic industries.
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Busting the Myths about Artificial Intelligence
#artificialintelligence Aug-22-2017, 02:10:05 GMT
Manufacturers face a difficult task juggling the current "innovation agenda." Today, the Industrial Internet of Things (IIoT), robotic automation and artificial intelligence (AI) are all poised to be the next big thing. But those on front line of manufacturing are cautious to embrace innovation--and rightly so. Too often expectations are unfulfilled, capital investments are made in vain, and experimentation doesn't translate positively into profits. Instead, many enterprises take a wait-and-see approach.
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#artificialintelligence Feb-5-2017, 00:30:07 GMT
That term has changed over the years. As we enter the Fourth Industrial Revolution, analysts and researchers are categorizing the future technologies that will change our lives. I've grouped these categories together, and they include: "The best way to predict the future is to invent it." For those of you unfamiliar with my work, I write frequently about Fourth Industrial Revolution (#4IR) themes such as Tech Innovation, Agtech, and Industrial Disruption. If I had to pick one area I find most exciting, most potentially disruptive and thus the best opportunity for return on investment, it would be A.I.
industrial revolution, technology innovation
Capitalism Behaving Badly
MIT Technology Review Oct-12-2016, 05:00:09 GMT
Despite healthy corporate earnings, an employment rate that has slowly rebounded since the financial crisis of 2008, and the outpouring of high-tech distractions from Silicon Valley, many people have an aching sense that there is something deeply wrong with the economy. Slow productivity growth is stunting their financial opportunities; high levels of income inequality in the United States and Europe are fueling public outrage and frustration in those left behind, leading to unprecedentedly angry politics; and yet despite the obvious symptoms, economists and other policy makers have been largely befuddled in explaining the causes and, even more important, the cures for these problems. A series of essays by authors including Joseph Stiglitz, an economist at Columbia University who won a Nobel Prize in 2001, and Mariana Mazzucato, a professor of the economics of innovation at the University of Sussex and a rising voice in British politics, the book attempts to provide, as explained in its introduction, "a much better understanding of how modern capitalism works--and why in key ways it now doesn't." Together, the essays provide a compelling argument that we need more coherent and deliberate strategic planning in tackling our economic problems, especially in finding more effective ways to reduce greenhouse-gas emissions. In particular, Mazzucato, who also co-edited the book and co-wrote an introduction with Michael Jacobs, wants to counter the view that free markets inevitably lead to desirable outcomes and that freer markets are always better: the faith that "the'invisible hand' of the market knows best."
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Theresa May promises 'active' industrial strategy
BBC News Jan-22-2017, 23:50:06 GMT
The prime minister is to unveil a new, more interventionist, industrial strategy on Monday, designed to boost the post-Brexit UK economy. The government will be "stepping up to a new, active role," Mrs May said. She will launch the new strategy at her first regional cabinet meeting, to be held in the north-west of England. Business leaders welcomed the plan which will focus on "sector deals" in areas such as life sciences and low emission vehicles. A green paper will set out ways the government can provide support to businesses by addressing regulatory barriers, agreeing trade deals and helping to establish institutions that encourage innovation and skills development.
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