text:
Ed Miliband: The surprise sex symbol
Ed Miliband during a Q&A sessio, exhibiting the sex symbol qualities loved by women across Britain.n
By Ian Dunt Tuesday, 13 November 2012 9:40 AM Follow iandunt
Ed Miliband built an unlikely new career as a sex symbol yesterday, after surprised journalists watched him be mobbed by admiring female voters on a campaign trip.
The Labour leader was being interviewed by the Mirror during a tour of Wolverhampton when he was approached by an infatuated supporter.
"You can come back to mine for coffee and dumplings – can I snog you?" asked 29-year-old teaching assistant Wendy Cooper.
The Labour leader replied: "Definitely not, I've got a wife."
Humanists launch Northern Ireland's first billboards advertising legal same-sex marriages
Humanists welcome 'extraordinary' day as NI gets legal abortion, same-sex marriage
At this point reporters observed that Miliband had his bum pinched.
"I can't help it, I've been stalking you, watching you on TV," Wendy said.
"He is fit actually. He's got the nice model legs. If he didn't have a wife you would be taking a photo with my tongue down his throat," she told a nearby journalist.
Back in his car, Miliband later asked his aides who paid the woman and added: "There's no accounting for taste."
The Labour leader is currently trying to boost support for the police commissioners election this week, but turnout is expected to be low.
Young people's votes are changing the political landscape - this election proves it
The tactical voting dilemma: Canterbury goes down to the wire
Labour Day: Miliband shocks Westminster with 10p tax announcement
Ed Miliband gets gay vote
Ed Miliband: Hague treatment 'depressing'


text:
This ideal work/live Bartlett Building loft has been tastefully remodeled Kitchen features all new cabinets, sink and faucet, plus stainless-steel appliances and coffee bar. Lighting has been updated to luxurious fixtures adding dramatic appeal to the 10ft ceilings. The eighth floor corner placement, polished concrete floors throughout and towering original double-hung windows make for a light-filled urban space. In unit washer/dryer and tons of storage make this the perfect starter loft investment. Originally built in 1911 and reimagined into 139 lofts in 2004, the A.G. Bartlett building was designed in the Beaux-Arts style by legendary architects John Parkinson and Edwin Bergstrom. Formerly the Union Oil Building, when completed, it was the tallest building in the city. The tasteful re-envisioning of the building retains many original details like the intricate tile floors, restored elevator doors and iron staircases. The building includes a 24-hour doorman, communal rooftop space with 360-degree views of the LA city skyline and most importantly Mills Act Status (ensuring a savings of 50-70% off on owner property taxes). Situated in the heart of the DTLA Historic Core, The Bartlett is walking distance to farmer's markets, Whole Foods, Restaurant Row, Fashion District, Grand Central Market and LA Live / Crypto Arena. Easy to show.
SF : 742SF
Lot SF : 0SF
Lot PPSF : $inf
DOM : 0 Days

redpajama_set_name: RedPajamaCommonCrawl


text:
21. January 2022 | Minister Tajik: Serious companies will adapt to stricter rules for the construction industry in Norway
Princess Charlotte becomes two years old
TOPICS:BritainNorwayPrincess CharlotteTodayTwo years
In this photo released Monday May 1, 2017, by the Duke and Duchess of Cambridge, showing their daughter Princess Charlotte taken in April 2017 by her mother Kate the Duchess of Cambridge at Anmer Hall in Norfolk, England. The Duke and Duchess of Cambridge have said they are "delighted" to share this new photograph of Princess Charlotte enjoying the outdoors to mark their only daughter's second birthday on Tuesday May 2. (The Duchess of Cambridge via AP)
Posted By: Ali Pourramedani 1. May 2017
Royal celebration
Tuesday, Britain's Princess Charlotte becomes two years old and the British court has issued a new picture of her for the occasion.
The photographer is the princess' mother, Duchess Kate and was taken in April this year, it is informed in a statement from Kensington Palace.
Princess Charlotte is Duchess Kate and Prince Williams youngest child. They also have a son named George, who will be four years in July.
News, Sport
Skistad with harsh criticism of NSF
Eriksen Søreide becomes Foreign Minister, Bakke-Jensen takes over as Defense Minister
Yara is building the world's first electric and self-propelled Container Ship
Strong winds and cold weather this weekend
Minister Tajik: Serious companies will adapt to stricter rules for the construction industry in Norway
Minister of Finance Trygve Slagsvold Vedum (SP) told the newspaper Dagbladet that he does not want to rush when it comes to appointing a new…


To start, Parrish and Norlander open with Duke barely escaping — again — thanks to a bunny missed at the buzzer. Virginia Tech almost had it. But you know who's to thank for Duke moving on? Tre Jones. From there, the guys preview Duke-Michigan State (8:45), then talk the West/Gonzaga vs. Texas Tech (13:20), the South/Virginia-Purdue (21:00) and the Midwest/Kentucky vs. Auburn (33:40).


text:
Ariana Grande Didn't Go to Grammys But Wore Custom Dress Anyway
Entertainment 7:44 PM PST, February 10, 2019 - Inside Edition Staff
7:44 PM PST, February 10, 2019 - Inside Edition Staff
Ariana Grande may have decided not to go to the Grammy Awards Sunday, but she put her fancy dress to use anyway.
It was revealed Thursday that Grande would not perform at or attend the ceremony over a disagreement with the Recording Academy.
mhmmm here it is ! "too late for her to pull something together......" pic.twitter.com/YwuBOUkZjr
— Ariana Grande (@ArianaGrande) February 7, 2019
Then on Sunday before the event began, the academy announced several award winners -- including Grande's win for Best Pop Vocal Album for "Sweetener."
It was the 25-year-old pop stars's first Grammy win. And she didn't let the big night go to waste.
when @zacposen makes u a custom gown it doesn't matter if you're singing or not 🌫 .... thank u 🖤
A post shared by Ariana Grande (@arianagrande) on Feb 10, 2019 at 5:49pm PST
Grande posted a series of photos on Instagram posing in a shiny strapless ballgown. She wrote, "when @zacposen makes u a custom gown it doesn't matter if you're singing or not .... thank u."
The singer, who was also nominated in the Best Pop Solo Performance category for "God Is a Woman," made sure to thank her fans for her win.
Grande wrote on Twitter, "i know i'm not there tonight (trust, i tried and still truly wished it had worked out tbh) and i know i said i try not to put too much weight into these things .... but f*** ....... this is wild and beautiful. thank you so much."
She added, "but mostly thank y'all for being my main source of joy and inspiration always."
i know i'm not there tonight (trust, i tried and still truly wished it had worked out tbh) and i know i said i try not to put too much weight into these things .... but fuck ....... 🥺 this is wild and beautiful. thank you so much.
— Ariana Grande (@ArianaGrande) February 10, 2019
but mostly thank y'all for being my main source of joy and inspiration always
Grammy Awards 2019: The Complete List of Nominees and Winners
A Guide to Getting Tattoos Following Ariana Grande's 'Tiny BBQ Grill' Faux-Pas
Ariana Grande Tears Up While Performing on 'Ellen'
Ariana Grande Chokes Up During Performance on 'Ellen'
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Coaching and mentoring are old professions. Soldiers, artists and athletes have traditionally used teachers to learn skills, coaches to apply those skills and mentors to keep life progressing in balance with history and experience and "holistic" perspectives. Modern teachers, coaches and mentors synthesize the best skills and practices for use in business, leadership training, academics, life styles, exercise, health and nutrition and finance to assist students, individuals, executives, entrepreneurs and professionals.


Bring your own bottle of wine and enjoy an unforgettable evening filled with friends, fun and fabulous art! You supply your cocktails, and we'll supply the rest. At Wine & Design, all you have to do is sign up, show up and sip up! Your canvas, paints, paintbrushes and corkscrew will be waiting for you as a local artist provides you with step-by-step instructions on creating that night's featured painting.


See the opening hours for the Civic Offices in Bridgend.
View our concerns and formal complaints policy.
'Our promise to you' sets out what you can expect from us and what we expect from you in return.
Information on how we manage our social media accounts and advice for followers who wish to talk to us through social media.
Find the contact details for our chief officers.


Put the sentences in the passive voice.
1. Human beings are destroying the rainforests = The rainforests by human beings.
2. The woman is pouring water on the pavement = Water on the pavement by the woman.
3. The boys are throwing papers on the road = Papers on the road by the boys.
4. The fishermen are putting a liquid in the lake = A liquid in the lake by the fishermen.
5. The sellers are using plastic papers to cover foods = Plastic papers to cover foods by the sellers.
6. The child is breaking the table = The table by the child.
7. The students are learning English = English by the students.
8. The scientists are catching the butterflies with a net = The butterflies with a net by the scientists.
9. The girl is washing two babies = Two babies by the girl.
10. The father is hugging the child = The child by the father.


Gien was founded in 1821 by Thomas Hall, and Englishman who wanted to introduce fine English earthenware manufacturing techniques in France.
Production first began with functional tableware, and then went on to make fancier dinnerware, decorative pieces and tableware emblazoned with notable families' coats of arms.
The factory works non-stop on improving quality and developing artistic creativity. By following this philosophy, Gien has produced exceptional pieces, which it has presented at the various Universal Exhibitions from 1855 to 1900.
In the 20th Century, in spite of wars and other crises, modernization and creation continue.
Today, Gien is both the number one producer of the highest quality earthenware in France, and a top-selling company both at the national and international levels. Its product lines are highly unique and distinguish themselves from other faience-earthenware with rich, refined and varied shapes and styles. Louis Grandchamp has headed the Gien Factory since January 2002.
Gien Earthenware is without any doubt the most prestigious French manufacturer of earthenware. Its fame comes from flawless know-how, and from striving to blend craftsmanship and industrial processes, bringing tradition and creativity together. With this very philosophy, Mr. Louis Grandchamp and Mr. Bertrand Dambrine acquired this world-renowned factory in January of 2002.


text:
Q: Using JQuery to create a forward button and a backward button to go through data by ID in a mysql database I wish for one button to go forward through data by ID from a Mysql database and another button to go backward through the same data using JQuery. I do not want to display the entire database at once.  I would like to display, item ID 1, and then click a forward button and then item ID 2 is displayed. If I want to go forward again, I can click the same button and Item number 3 is displayed. If I want to go backward, I can click the backward button and now item number 2 is displayed again.
Currently, It displays all the items at once and with each item duplicating 6 times, which is the number of columns in the database. This is because the variable station_dataP +=.  If I remove the += and put the = sign, it only displays the last ID of 7, all by itself which is what I would like to see but it does not cycle back and forth through them.
ID: 1
Station: Carmichael Rd.
Address: 54 Myers Rd.
Sales: undefined
Operator: Michael Sears
Top SKU: undefined
ID: 1
Station: Carmichael Rd.
Address: 54 Myers Rd.
Sales: undefined
Operator: Michael Sears
Top SKU: undefined
ID: 1
Station: Carmichael Rd.
Address: 54 Myers Rd.
Sales: undefined
Operator: Michael Sears
Top SKU: undefined
ID: 1
Station: Carmichael Rd.
Address: 54 Myers Rd.
Sales: undefined
Operator: Michael Sears
Top SKU: undefined
ID: 1
Station: Carmichael Rd.
Address: 54 Myers Rd.
Sales: undefined
Operator: Michael Sears
Top SKU: undefined
ID: 1
Station: Carmichael Rd.
Address: 54 Myers Rd.
Sales: undefined
Operator: Michael Sears
Top SKU: undefined
ID: 2
Station: Baillou Hill
Address: 564 Jackson Ave.
Sales: undefined
Operator: Marla Pikes
Top SKU: undefined
ID: 2
Station: Baillou Hill
Address: 564 Jackson Ave.
Sales: undefined
Operator: Marla Pikes
Top SKU: undefined
etc..
I am able to do this in vanilla Javascript but I am not able to do so in JQuery. I will display the Javascript code below:

     <button type = "button" id="previous"><<<<</button>
     <button type = "button" id="next">>>>></button>

   <JQUERY> - DOES NOT WORK PROPERLY

  $(document).ready(() => {
  $('#previous').click(()=> {
  $.ajax({
   url:'http://localhost:5000/alldata',
   type:'GET',
   datatype:'json',
   success:(data) => {

   var station_dataP ='';

   $.each(data, function (key,value){
      for(var i in value){
        if(i > 0){
        i--;
        }

   station_dataP +='
ID: '+value.ID+'
' +
                      '
Station: '+value.Station+'
' +
                      '
Address: '+value.Address+'
' +
                      '
Sales: '+value.Monthly_CStore_Sales+'
' +
                      '
Operator: '+value.Operator+'
' +
                      '
Top SKU: '+value.Top_SKU+'
' +
                      '</
'+'
'; */

    });
   }

   $('#storesNav').append(station_dataP);

     }
   });
 });

 //(4)
 $('#next').click(()=> {
  $.ajax({
   url:'http://localhost:5000/alldata',
   type:'GET',
   datatype:'json',
   success:(data) => {

      var station_dataP ='';

      $.each(data, function (key,value){
      for(var i in value){
        len = value.length;
        if(i < len - 1){
        i++;
        }
        station_dataP +='
ID: '+value.ID+'
' +
                      '
Station: '+value.Station+'
' +
                      '
Address: '+value.Address+'
' +
                      '
Sales: '+value.Monthly_CStore_Sales+'
' +
                      '
Operator: '+value.Operator+'
' +
                      '
Top SKU: '+value.Top_SKU+'
' +
                      '</
'+'
'; */

    });
   }

   $('#storesNav').append(station_dataP);

     }
   });
 });

       <JAVASCRIPT> IT WORKS HERE BUT NOT THE JQUERY CODE.

       function previous(xhttp){
        //var xhttp;
        //xhttp = new XMLHttpRequest();
        var users = JSON.parse(xhttp.responseText);
        len = users.length
        if(i > 0){
        i--;
        }

        var usersText2 = '';

        usersText2 +=
       '
' +
       'ID: ' + users[i].id + '
' +
       'Name: ' + users[i].name + '
' +
       'Email: '+ users[i].email;

       document.getElementById("usersNav").innerHTML = usersText2;

       }
       function next(xhttp){

      //var xhttp;
      //xhttp = new XMLHttpRequest();
      var users = JSON.parse(xhttp.responseText);
      len = users.length;
      if(i < len - 1){
      i++;
      }

      var usersText = '';
      usersText +=
       '
' +
       'ID: ' + users[i].id + '
' +
       'Name: ' + users[i].name + '
' +
        'Email: '+ users[i].email;

       document.getElementById("usersNav").innerHTML = usersText;

        }

A: The brace input is invalid
Also, there should be no jquery source in ajax other than the parameter
$(document).ready(() => {
    $('#previous').click(() => {
        $.ajax({
            url: 'http://localhost:5000/alldata',
            type: 'GET',
            datatype: 'json',
            success: (data) => {

                var station_dataP = '';

                $.each(data, function (key, value) {
                    for (var i in value) {
                        if (i > 0) {
                            i--;
                        }

                        station_dataP += '
ID: ' + value.ID + '
' +
                            '
Station: ' + value.Station + '
' +
                            '
Address: ' + value.Address + '
' +
                            '
Sales: ' + value.Monthly_CStore_Sales + '
' +
                            '
Operator: ' + value.Operator + '
' +
                            '
Top SKU: ' + value.Top_SKU + '
' +
                            '</
' + '
';

                    }
                });
                $('#storesNav').append(station_dataP);
            }
        })
    });

    //(4)
    $('#next').click(() => {
        $.ajax({
            url: 'http://localhost:5000/alldata',
            type: 'GET',
            datatype: 'json',
            success: (data) => {

                var station_dataP = '';

                $.each(data, function (key, value) {
                    for (var i in value) {
                        var len = value.length;
                        if (i < len - 1) {
                            i++;
                        }
                        station_dataP += '
ID: ' + value.ID + '
' +
                            '
Station: ' + value.Station + '
' +
                            '
Address: ' + value.Address + '
' +
                            '
Sales: ' + value.Monthly_CStore_Sales + '
' +
                            '
Operator: ' + value.Operator + '
' +
                            '
Top SKU: ' + value.Top_SKU + '
' +
                            '</
' + '
';
                    }
                });
                $('#storesNav').append(station_dataP);
            }
        });
    });
});

A:

$(document).ready(() => {
    $('#previous').click(() => {

      $.ajax({
        url: 'http://localhost:5000/alldata',
        type: 'GET',
        datatype: 'json',
        success: (data) => {
          $.each(data, function (i) {

          });

          data[i]
          console.log(i);
          if (i > 0) {
            i--;
            var station_loop = '';
            console.log(JSON.stringify(data[i].Station));

            station_loop += '
ID: ' + JSON.stringify(data[i].ID) + '
' +
              '
Station: ' + JSON.stringify(data[i].Station) + '
' +
              '
Address: ' + JSON.stringify(data[i].Address) + '
' +
              '
Sales: ' + JSON.stringify(data[i].Monthly_CStore_Sales) + '
' +
              '
Operator: ' + JSON.stringify(data[i].Operator) + '
' +
              '
Top SKU: ' + JSON.stringify(data[i].Top_SKU) + '
' +
              '</
' + '
';

            $('#storesNav').html(station_loop);
          }
        }
      })
    });
  });

  $(document).ready(() => {
    $('#next').click(() => {
      //var url = 'localhost:5000/alldata';
      //var url = 'https://api.github.com/users/topjavatutorial/repos';
      $.ajax({
        url: 'http://localhost:5000/alldata',
        type: 'GET',
        datatype: 'json',
        success: (data) => {

          $.each(data, function (i) {

          });

          data[i]
          console.log(i);
          len = data.length;
          if (i < (len - 1)) {
            i++;
            var station_loop = '';
            console.log(JSON.stringify(data[i].Station));

            station_loop += '
ID: ' + JSON.stringify(data[i].ID) + '
' +
              '
Station: ' + JSON.stringify(data[i].Station) + '
' +
              '
Address: ' + JSON.stringify(data[i].Address) + '
' +
              '
Sales: ' + JSON.stringify(data[i].Monthly_CStore_Sales) + '
' +
              '
Operator: ' + JSON.stringify(data[i].Operator) + '
' +
              '
Top SKU: ' + JSON.stringify(data[i].Top_SKU) + '
' +
              '</
' + '
';

            $('#storesNav').html(station_loop);
          }
        }
      })
    });
  });

<button type="button" id="previous">
      <<<<</button> <button type="button" id="next">>>>>
    </button>

redpajama_set_name: RedPajamaStackExchange


Here it is…my first attempt at a YouTube video.
Hope you enjoyed our story 🙂 Check us out on Facebook at Caregiver's Storyboard and remember: sometimes comic relief is what's needed in times of adversity.
Some people might call that silent voice their conscience, but is it really?
In Don Miguel Ruiz's best selling book, The Four Agreements, he proposes the theory that the voices we hear in our head, are really the agreements (conscious or subconscious) we made as children with our parents and other people in authority.
What does that mean you ask? Let me explain.
Developmentally speaking, when an infant is less than four months old, he can't conceptualize the fact that there are other people beside himself. In other words he thinks that whoever is holding him, is part of himself. In addition, he believes that food appears in his mouth and diapers are changed spontaneously.
Around four months of age this changes and the child becomes aware he is not alone. He also senses if he wants his needs to be adequately met, it is best if the person taking care of him is happy; and that's where this conscience/critic thing all begins.
As the child grows older he understands he must comply by making agreements with his all-powerful parents if he wants to be happy.
I have to brush my teeth and take a bath or I will get in trouble.
If I don't clean the bathroom spotless, my mom will be mad and I will have to do it again.
If I get bad grades, I will be grounded.
Ruiz's message is that we carry these agreements into adulthood.
The disturbing thing is, they become so ingrained into our unconscious mind that we often unknowingly accept them as fact.
If as adults we are not careful, the critical voices in our heads (agreements we made with authority figures in the past) can continue to rule our actions through fear of abandonment. Even worse, they can define our identity.
Before we know it, we are making choices based out of fear rather than love.
One day I was walking in my upstairs hallway when a little miracle happened. I had the unexpected, uninvited gift of looking into my mind as an objective bystander. I was able to differentiate my own voice from the subtle yet powerful critical chatter that was calling the shots.
Startled by my observation, I asked myself, "Where are those voices coming from?" That's when a fuzzy picture of my parents, former teachers, husband, children, magazine covers, and a lot of unknown blurry faces, began to come into focus.
Wait a minute! I didn't give permission for those people (love them though I do) to take up residence in my head, let alone whisper orders from the recesses of my subconscious.
…and that's when I decided to thank them for visiting, but tell them I would no longer be needing their services. From here on out I will be calling the shots; thank you very much.
So what is the difference between your conscience and your critic, or is there a difference?
In psychological terms conscience is described as leading to feelings of remorse when one does things that go against his/her moral values, and to feelings of goodness or integrity when actions conform to his/her moral values.
Notice I highlighted his/her. I believe this is the key factor in determining whether the voice is your conscience speaking, or your critic.
Being an authentic individual requires us to choose our own values and morals.
As adults, if we abdicate our right to choose, we become subject to the critic. It's perfectly okay to choose to keep some of the agreements we made as children, as long as it's our choice.
As for agreements made through the influence of peers, and the media, do a mental inventory and decide if they are congruent with your moral values. If they aren't, and they are inviting negativity into your life; KICK THEM OUT!
P.S. This post is not an endorsement for The Four Agreements, by Don Miguel Ruiz. I love the concept, but dislike the book. The four agreements are good principles to live by but the book takes it so far, in my opinion the message loses its' power.
Yesterday, Malala Yousafzai received Harvard University's 2013 Peter J Gomes Humanitarian Award. She is truly a Pioneer of Peace.
Are you more a football or a book?
Is "Simple" the New Black?
I have always been intrigued by the concept of "Simple," so when I came across the minimalist blog, Be More With Less I was delighted. Courtney Carver, former advertising executive, now Simplicity Guru is the mastermind behind Project 333 which challenges us to reduce our wardrobes to 33 items for 3 months.
There is just something about a good challenge that makes me want to say: I CAN DO IT! So over the past month I've been mulling this over in my mind. Could I really pare my wardrobe down to only 33 pieces? After all, I probably have close to 33 pairs of shoes. Shameful I know, but I am not the only one, and I know there are those with a lot more; so I don't feel too bad.
Austerity… Self Control… Diet… Self Mastery… Boring… 33 really great pieces that I love… Head space to think about other things than the way I look… Simplicity… Improved priorities… Space… Lower carbon footprint… Suffering?
Well, I did it yesterday. In preparation I mind mapped what my wardrobe should include. Categories were: pants, shirts, sweaters, skirts/dresses, outerwear, accessories and footwear. I didn't include pajamas, underwear, or workout clothes.
Next I thought of a fall color scheme which included: grey, black, ivory/white, denim blue, camel and mustard and then started to fill in the blanks.
I have a lot of clothes! My bedroom looked like a department store after I separated what I was going to keep, and what I was going to pack away.
The boundary of 33 (which is a metaphor for 40 in my case,) puts a halt to impulse buying. I have no room in my life for anything frivolous or excessive.
I have a very small wardrobe so I can afford quality. This is inline with my value of quality over quantity. It is a personal truth. I know it because I feel my heart-strings singing like a beautiful violin.
I feel less greedy. I have enough for my needs. I feel like I'm living with more respect for those who are less fortunate.
I can take good care of 33 pieces that I love. My life is filled with "to do's," and I am often overwhelmed with taking care of "things." I love the feeling of being released from duty to inanimate objects. I am free to use my precious energy on more important things.
From the movie Gone With the Wind… Are you more Scarlett O'Hara or Melanie Hamilton?
This morning I was checking out Facebook and saw this video shared by an old friend of mine. I am so impressed and touched, that I must share this Pioneer of Peace's story. I hope it brings happiness to your day as it did mine.
Are you more a fireplace or an air conditioner?
South Jordan, Ut. At 7:02 am., the birds on my block convened a town hall meeting in the pine tree at the bend in the road. Their rhetoric was fierce as their chirping was heard as far as two blocks away. What were they discussing? Plans for how they will manage the scheduled water shut off between 9am and 5pm today, or perhaps the conflict in Syria and what President Obama will tell us in his address to the nation tonight.


text:
Q: Text flowing outside div in IE with angular material I am having these issues using angular material on IE (i'm using latest version edge and same result for other versions (9,10,11)

*

*Text is flowing outside div tag

The below is snippet code for the current div containing 'Task' item, This is the html and css only (the screenshot will demostrate the issue).
(note that multilingual-message is a custom directive for multi-language texts support.)

.comp-insp-th .th-cell {
  background: #1e9a9f;
  padding: 10px 5px;
  margin: 5px 0px;
}

  <multilingual-message message-id="Home.Task" class="ng-isolate-scope">
    Task
  </multilingual-message>

*Divs (tags) not taking paddings and margins, the below is snippet code with styles, using angular material directive  'layout-align' and assigned as "space-between stretch" wich will add space between items in column, still the tags are overlapping and paddings/margins not working even if added manually.

A: I tried angular material Troubleshooting tips but it did not help me that much,
The two issues were the effect of flexbox layout which is not totally supported on IE-11, then after a deep inspection -and with the help of a friend- I figure it out, the solution is adding -ms-flex: 1 0 30px to the .flex class of the element where 30px indicates the prefferedHeight for the flex item, then change the value based on each element height you preffer.
Note to mention that you should always add a selector before .flex class for your current element, in my example:
.comp-insp-th .th-cell .flex{
  -ms-flex:1 0 30px !important;
}


Army of the North, Ardennes and Moselle.
Capture of Charleroy after a surrender, and 50 cannon; 3,000 garrison men prisoners.
Capture of Wilstett; 150 cuirassiers taken prisoners; capture of 1 cannon and 60 horses.


We do everything from bug fixing to coding high performance, multinational shopping carts. No job is too small or too large and we are passionate about building high quality, successful solutions.
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In the mid-1960's I had the pleasure of attending a talk by Lotfi Zadeh at which he presented some of his basic (and at the time, recent) work on fuzzy sets. Lotfi's algebra of fuzzy subsets of a set struck me as very nice; in fact, as a graduate student in the mid-1950's, I had suggested similar ideas about continuous-truth-valued propositional calculus (inffor "and", sup for "or") to my advisor, but he didn't go for it (and in fact, confused it with the foundations of probability theory), so I ended up writing a thesis in a more conventional area of mathematics (differential algebra). I especially enjoyed Lotfi's discussion of fuzzy convexity; I remember talking to him about possible ways of extending this work, but I didn't pursue this at the time. I have elsewhere told the story of how, when I saw C. L. Chang's 1968 paper on fuzzy topological spaces, I was impelled to try my hand at fuzzi- fying algebra. This led to my 1971 paper "Fuzzy groups", which became the starting point of an entire literature on fuzzy algebraic structures. In 1974 King-Sun Fu invited me to speak at a U. S. -Japan seminar on Fuzzy Sets and their Applications, which was to be held that summer in Berkeley.


text:
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Daily Devotional for Wednesday April 2, 2003
The ultimate reality TV program
( Genesis 6:5, Genesis 6:12, Psalm 52:3, Proverbs 2:14 )
The ultimate reality TV program. Over the past several years, reality TV programs like "Survivor" have become the hottest and top-rated programs on television. The appeal was in watching real people living in real life situations. They are cheap to produce and garner high ratings which equate to a financial bonanza for the networks that air them. With the initial success of these types of programs, came a flood of new ones. They all started to look the same, so to make themselves stand out, they had to come up with more bizarre concepts that began to push the limits of good taste and decency.
Pretty soon it was a race to see who could be the raunchiest, who could degrade people the most, and just how low they could get before the public said enough. That brought us wonderful television programs that made a mockery of God's institution of marriage, let us see people eat living insects and raw animal parts for a few dollars, and allowed us to cheer for a race between an elephant and dozens of midgets who were harnessed to a 747. Sadly, the public watched this garbage in great numbers. Never saying it was enough, only wanting more. The only thing left to do was actually have someone die, right there on LIVE television. Even the sleaze merchants of reality TV couldn't justify crossing that line.
But wait, along comes the war. I have long contended that the news has quit being about news, but is actually a form of entertainment. Why do you think Dan Rather, Tom Brokaw, and Peter Jennings are all paid Hollywood-type salaries? They are ENTERTAINERS! TV could now give the public the ultimate reality TV program, complete with death and destruction, under the guise of news. Let me assure you, as I have clearly laid out in the past I am in full support of this war for the purpose of our national defense, with a byproduct being the liberation of the Iraqi people from the tyrannical quarter century reign of Saddam Hussein. I have done several Devotionals on the Biblical support for war, so if you are NOT in favor of the war, please, I am not interested in having those debates again. My point today is NOT about whether or not we should be engaged in the war with Iraq, but about the role that television is playing in this conflict and how we are viewing it.
Sadly, we are living in a time when we want everything NOW! Cell phones, pagers, faxes, wireless Internet connections, FedEx, drive-thru food. WE WANT IT NOW!!! Many thought that we could come home at night, pop a bowl of popcorn, sit down and "watch the war" like we would watch a sporting event. We had FOX News, MSNBC, CNN all bringing us 24/7 coverage. We had their expert analysts breaking it all down for us, we had their reporters embedded in with the troops to bring us LIVE updates from the front, and we even had cameras strategically placed in Baghdad to record our bombing raids on that city. This was going to be great. With all of the live coverage available, we were going to get to see in real- time, bombs blowing up targets and our troops fighting the enemy, complete with all of the death and destruction that war entails. However, the war is lasting much longer than the general public expected, we really don't see much in those blurry satellite-phone video feeds, and we could see more "shock and awe" at most neighborhood 4th of July celebrations. The war has not proved to be the sensational real-life video game many expected it to be, and after just two weeks the ratings are already dropping dramatically.
I love you and care about you so much. One of the things that I attempt to do each day is take the real life issues that we all face, take the real life events happening today, and put them in the context of our faith. The larger issue of reality television is a societal one. The ongoing moral decay in our culture has brought us to where we are at today. The problem is NOT with television, but with US! I don't blame the TV networks. They are simply providing what the public demands to make a dollar. GE and Microsoft that own MSNBC are billion dollar publicly traded companies whose only goal is the bottom line. CNN is owned by AOL/Time Warner, another media giant driven strictly by profits. FOX News is owned by Rupert Murdoch who has become a billionaire by selling sleaze worldwide. While I realize that FOX News is the favorite of most conservatives, never forget that it is owned by Murdoch who is responsible for much of the unGodly reality programming that I referred to earlier. Like the others, his ONLY goal is profits. The television networks are simply giving the public what it demands
I urge you today to pray for this sin-filled world that we live in. A world that is so lost in sin that it looks to escape in the most perverse ways. As richly blessed as we are in the United States, we have allowed our culture to decay to the point that we are really no better than the ancient Romans who spent their recreation time in the Coliseum watching Christians being fed to the lions. THAT was their entertainment!!! THAT is where we find OUR culture today!!!I am saying that we, the people of God, the blood-bought sons and daughters of the King, better start to get serious before WE are the ones being fed to the lions. CAN'T HAPPEN? 40 years ago if you told someone that you could watch the most perverted sex acts, watch the most vile humor, watch the most graphic acts of violence imaginable right in the comfort of your living room on TV, they would have told you, CAN'T HAPPEN!
In His love and service, Your friend and brother in Christ, Bill Keller
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Description: Lila Gerais granite is a fine to medium-grained, pink-violet to red gneiss (orthogneiss) of the Precambrian period. This stone internationally may be nominated as a granite but in the area of application of the European Standard this stone must be nominated as a gneiss. The veining and colour varies and a small sample of Lila Gerais granite may not be representative of the whole slab therefore swatch samples must be approved for large projects to ensure that the blocks are all extracted from the same quarry face for matching purposes. During the polishing process a clear epoxy resin filler may be used to fill any micro fissures or tiny pitting however this does not affect the integrity of the stone, it merely provides an even smoother surface finish. Lila Gerais Granite is suitable for both interior and exterior use futhermore is frost resistant and with a constant polish.
Surface Finish: Lila Gerais Granite looks best with a gloss polished surface but is also available as honed, leather/satin, river-washed/antique, thermal/flamed, brushed and sandblast finish.
Sealing: Test with a small sample. It is possible that a sealant can be used on Lila Gerais granite however excessive attempts at sealing it could actually create a residue film build-up.
Quarry Location: Lila Gerais granite is quarried at quarry RN 000210, Fazenda Cachoeirinha, Pedra do Indaia, Minas Gerais, Brazil.
Gneiss: A high grade metamorphic rock. This means that gneiss has been subjected to more heat and pressure than schist. Gneiss is coarser than schist and has distinct banding. This banding has alternating layers that are composed of different minerals. The minerals that compose gneiss are the same as granite. Feldspar is the most important mineral that makes up gneiss along with mica and quartz. Gneiss can be formed from a sedimentary rock such as sandstone or shale, or it can be formed from the metamorphism of the igneouse rock grantite. Gneiss can be used by man as paving and building stone.


On-demand membership airline Surf Air has hired Text100 as its agency of record in the US, as it prepares to launch the Californian brand in Europe in October, starting this weekend with a paid social campaign.
The agency, part of Next Fifteen, was engaged as an agency of record for the company in London at the start of the year. The agency was tasked with strategy, PR, influencer engagement, events, social media and other roles and has undertaken research and background work ahead of Surf Air's European launch in October, following its US debut in 2013.
Following a competitive pitch in June, the agency has started working with the airline in the US, again as an agency of record.
Part of the agency's work has been pinpointing the key messages with which to take Surf Air to market – UK managing director Tara O'Donnell told PRWeek UK that while time-saving had emerged as the offer most likely to resonate with European consumers, luxury would also be a key message.
Time-saving is therefore at the heart of a paid campaign (images above and below) running on Facebook and Instagram from Friday (19 August) to launch the brand in Europe. The campaign's second phase will include a competition element.
Text100's French and German offices are also working on the launch.
Surf Air has about 200 staff globally, and had previously had freelance PR support in the US, making Text100 its first agency of record.
Text100 has more than 600 staff across 21 countries, and is the world's 31st largest agency, according to PRWeek's Global Agency Busines Report. Other major clients include Adobe, British Airways, Four Seasons, MTV, Skype and Vodafone.


text:
Nearly half of births in England and Wales are to mothers over 30
49% of women are over 30 when their baby arrives, as average age of mothers continues 40-year climb, ONS figures show
James Meikle
Thu 24 Jan 2013 08.33 EST First published on Thu 24 Jan 2013 08.33 EST
Births among young mothers have plunged since the 1970s. Photograph: Vivid/Getty
Nearly half of all births in England and Wales are to mothers aged 30 or older, according to the Office for National Statistics.
The average age of mothers has continued to increase for almost four decades. It is currently 29.7, but 49% of women are over 30 when their baby arrives. However, the overall percentages for the three main age groups for mothers are similar to the level they were just before and after the second world war. This is because births among young mothers have plunged from their late-60s to early-70s peak, a time when there were significantly fewer older mums.
The ONS bulletin detailing the latest figures for 2011 says that between the mid-1940s and mid-1970s the average age of mothers decreased by just under three years, from 29.3 in 1944 to 26.4 in 1973.
It said on Thursday: "The overall rise since 1973 reflects the increasing numbers of women who have been delaying childbearing to later years. Possible influences include increased participation in higher education, increased female participation in the labour force, the increasing importance of a career, the rising opportunity costs of childbearing, labour market uncertainty, housing facts and instability of partnerships."
The average age for a mother's first birth over the past 70 years mirrors the all-birth trends since 1940 – it is estimated at 27.9 compared with 27.7 in 2010 and 26.6 in 2001.
Actual numbers for women giving birth aged 45 and older are comparable to the 1940s, despite the rise in infertility treatment. In 2011, 1,832 babies were born to women of that age, out of 723,913 births. Before, during and after the second world war, it was common for that figure to top 2,000 in years when the total number of births varied from 579,091 in 1941 to 881,426 in 1947. Numbers sometimes fell to well below 500 between the 70s and mid-90s.
Marriage, or civil partnership in recent years, remains the most common family setting for births, despite the steady fall in births registered to married couples since the 1960s, down to 53% compared with 60% in 2001 and 94% in 1961. Mothers between 30 and 39 are most likely to be married or in a civil partnership, with only 31% of babies born outside these arrangements.
Office for National Statistics
Parents and parenting


text:
class Mod < ActiveRecord::Base
end

redpajama_set_name: RedPajamaGithub


i am nearing the end of my 21st novel, the fourth in a young adult series about shapeshifters entitled bloodline. my goal is to finish in the next few days so i have a break before i begin the next book on november 1st for nanowrimo. i wrote the initial book in the series last november, so i am pleased that i am circling around to the fifth book this year. this may be the last in the series, although i am not certain. it was supposed to be a trilogy, but that didn't seem to work out. this is my fourth year doing nanowrimo, and i only wish i'd discovered it sooner.
i have been researching stones and gems and their meanings for this book, and came across some interesting facts about the soladite stone. it is rumored to [and please don't roll your eyes too hard here, CJT] boost the immune system, clarify the mind, and bring good luck to writers and artists. well, at this point i can use all the good luck [real or imagined] that i can get my hands on. needless to say, fedex is delivering mine today.


Nah it was just a snafu with our hosting/billing. However, somehow a DNS change got involved while they marked our site unavailable. Even though we were down for a very short time, it took a long time for people's DNS cache to clear.
I was surprised to see Chrome has its own DNS cache, I had to clear it and restart chrome.


text:
WORLD CHAMPION SPEAKER
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Be a Sponge: "I Lost? What? But, Everyone Said…"
Sponge Articles
Be a Sponge: "I Lost?…
Ouch. It's frustrating and painful when you compete at anything and don't win. It is more painful when you take the competition very seriously, work hard and fall short of your expectations. This pain is magnified when well-meaning, biased friends in an attempt to console you by saying, "You should have won!" That seemingly comforting advice can fuel your pain and delay your healing. This wisdom I learned from World Champion, Jim Key. It came to me when recording our recent podcast. Jim came in second in the World Championship twice and came back and won the contest the following year. That is an inspiring story that he lived. Think about it.
When was your most painful loss or bomb? What was the situation? What was your mindset back then? Let me take you back to my journey in 1998. At the time I was a telemarketer at Bose Corporation, an opening act standup comic, and a Toastmaster. I entered the Humorous speech contest with high hopes, after all, I was an emerging comedian. I had lots of comedy club stage time, so I felt like I had a huge advantage. I've got this! I worked painstakingly hard on a routine. It was solid. I won the first two levels of the contest and went on to the Division level. I nailed it! And then they announced, "And the second-place winner is Darren LaCroix!" What? But? The contestant that won had taken one-liners from famous comedians and put them together as an original speech. It was funny, and it was creative; but it wasn't original. In the comedy world if you use anyone else's material, you are looked down on. This wasn't the comedy world. I put on my game face and fake smile and accepted second place. Ouch.
Here is some truth behind that moment. I was mad, hurt and very upset. Now, I could have protested, but honestly my self-esteem was so low at that point in my life; and I wanted so badly for people to like me, so I did NOT protest. I see now, that was on me. My bad. I was hurt, I decided that my protest would take place the following year in the form of a boycott of Toastmasters speech contests. That will show them! Well, if you aren't familiar with the history of the speech contest, the following year Toastmasters had the contest without me. Can you actually call it a protest though when only one person doesn't compete? I can laugh about it now, and I do; but at the time it hurt, and I didn't realize I needed to grow up a bit and learn a few more lessons.
Steven Tyler, of Aerosmith fame and an American Idol judge had a brilliant thought when he was being interviewed by Oprah. He said that when he sent people home from the show, he was not saying that they will never be the American Idol. All that he was saying was that they were not the American Idol that day. He went on to say that all it meant was they had to go back into the clubs and learn a few more lessons. I love that! It's deep. How does that apply to you?
You may have heard me say in my speech, "I called my mentor, Rick, and said, 'I bombed, I died; they hated me!" Rick said, "So?" How do you argue with so? That was a bomb in my professional career for a paid speech. I thought my paid speaking career was over. It never dawned on me that my mentors, my heroes, the greatest speakers who ever took the stage have also bombed. We just didn't see it. We see their success from the lessons they learned after they bombed! What if I didn't call my mentor? What would I be doing now? I would not be a speaker today; that I know.
The reason I'm writing this article now is, it's that time of year again. The time when the other World Champions and me get emails from around the world saying, "Everyone told me I should have won." Everyone? Really? Well, I don't think you spoke to a judge then. Or we hear, "The contest is fixed," "It's political," or "They have it out for me," or "The judging was horrible." Guess what, any of the above could be correct, other than "everyone one said." So? I heard of the new concept that with virtual contests people are taking screenshots of chats to try and back up their point that they should have won. Augh.
I remember the advice that I got from Otis Williams, Jr, "Be so good that the only question is who came in second." And maybe you were that good. David Brooks said, "To win you have to be talented and lucky." Guess what. It is a contest. Judging is subjective in a speech contest. When you enter, so do 35,000 other people. Which also means 34,999 other competitors don't win either.
Here is my advice to people when I get those emails. I tell them that right now they are in the pain portion of the moment. They are too close to step back and look at what really happened. I say, "Wait two weeks and then watch the recording of your speech with a coaching checklist and then ask yourself if there were things you could have done better. Then come back to me." In all my years of answering that email, I've only ever had one person reach back out to me two weeks later. I was so impressed when that one young man did and said, "I watched the recording of my contest speech. You were right." He knew he had more to learn. That is the response of a future Champion.
One of our Stage Time Members, Darryl Pace, who competed this year, just sent me this email.
Your and your team's advice helped me turn a major FAILURE last year into SUCCESS this year.
Last year I BOMBED at my Area International Speech Contest. It was a spectacular failure. After that, I remembered that someone, or maybe multiple people from the Stage Time team advised that our best speech material comes from our "Firsts, Flaws, Failures, and Frustrations".
The way I bombed was a First for me, it exposed some Flaws in my speech preparation, it was definitely a Failure, and it was quite Frustrating. So, I turned it into a speech that told of what happened and the lesson I learned. Sharing that embarrassing experience has taken me through the Club, Area, Division, and District level competitions so far.
In addition, I've received a couple of invitations to speak as a result of the speech — one at a Nissan Toastmasters annual celebration and another at a Women's Prison Ministry.
Anyway, just wanted to share that with you.
Hope you're doing well. Best regards,
Darryl Pace
Love that! Please think way beyond the contest. When you get invited to speak elsewhere that's gold! It's usually more likely because of your message more as a result of your victory. I love Darryl's mindset, in hindsight of his spectacular failure. Few people can see that instantly. I was competing to improve stories from my professional keynote. Mark Brown ended up taking his seven-minute Toastmaster winning speech and created a keynote about bullying that he delivered to over a million youth around the country.
Great speeches aren't written; they are rewritten. After I grew up and looked back at my second place speech Humorous Speech, I rewrote it, tightened it up and delivered it to my NSA Chapter, it went so well I got invited to deliver it at a National Convention during a fundraiser. I also took the same routine and customized it for the organization I was speaking to. It became my signature opening of my paid keynote speech. After winning the speech contest and getting invites around the world, I offered to do a humor night at Toastmasters conventions before I would keynote the next morning. I recorded one of those events and posted the video of that which now has over 308K views on YouTube Video. Remember, that was the speech that didn't win. It became a foundation that I reworked many times and helped grow my career. Maybe I did win, I just wasn't the winner that night as Steven Tyler said. There were more lessons I needed to learn.
I have an incredible admiration for Jim Key and his transparency. If you get a chance to listen to Episode 38 on our podcast you can hear him for yourself. I think every speech contestant that did not win, should listen. Please pass it on. He says that in 2002 he competed with the wrong mindset and a sense of entitlement. In 2003 he realized his kids were watching his actions and how he responded to his two second place finishes. Children hear what you say, but your actions are what teach them. In 2003, with lessons learned and the right attitude to serve he became the World Champion. What example do you set? What could your children and grandchildren learn from your actions? Respond to not winning today, like a champion of tomorrow. What will everyone say about how you responded to not winning?
When you compete to win there is an amazing chance you won't. Las Vegas would take the house odds on that one any day. When you compete to grow, you, the audience and all of your future audiences all win. Ask yourself, "Why am I really competing?"
What do you take from this?
Funniest Speech:
https://youtu.be/qovsP1aaEQY
Please add your comments/read other comments, on this blog post.
Category: Sponge ArticlesBy Darren LaCroix
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Be a Sponge: "I Wanna Win the Speech Contest!"
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text:
\section{Introduction}

Introduced by Jacobi~\cite{Jacobi} in 1848, harmonic morphisms are maps which pull back local harmonic functions onto
harmonic functions and, more recently, they were characterized by Fuglede~\cite{Fuglede} and Ishihara~\cite{Ishihara} as horizontally weakly
conformal harmonic maps. Their dual nature of analytical and geometrical objects has led to a rich theory (cf.~\cite{B-W}) which has encouraged the study of various other morphisms, that is maps preserving germs of certain differential operators.
The central role of the Dirac operator in differential geometry and mathematical physics called for this approach to be applied to
harmonic spinors. Unlike previous cases, the first hurdle is to make sense of a notion of pull-back of spinors by a map. This requires the identification of the spinor bundles involved, necessarily restricting our investigation to horizontally conformal maps between Riemannian manifolds and even-dimensional targets (cf. Section~\ref{section2}). Combining a chain rule for the Dirac operator and a local existence lemma, we show that a horizontally conformal submersion between spin manifolds is a Dirac morphism if and only if its horizontal distribution is integrable and the mean curvature of the fibres is related to the dilation factor, in a manner reminiscent of the fundamental equation for harmonic morphisms.
We conclude with some simple examples between Euclidean spaces and explicit our results in the set-up of~\cite{moro}, which inspired
initially our construction.

\section{Pull-back of a spinor}~\label{section2}

Let $(M^{m}, g)$ be a spin Riemannian manifold, the two-sheeted covering $\mathrm{Spin}(m) \stackrel{\rho}{\longrightarrow}
\mathrm{SO(m)}$ induces a double cover $\chi: P_{\mathrm{Spin}(m)}M \longrightarrow P_{\mathrm{SO(m)}}M$ of the bundle of positively
oriented orthonormal frames by the principal $\mathrm{Spin}(m)$-bundle over $M$, such that $\chi(s \cdot g)=\chi(s)\cdot \rho(g)$,
$\forall  s \in P_{\mathrm{Spin}(m)}M, g \in \mathrm{Spin}(m)$. The associated bundle $\mathrm{Cl}(M) = P_{\mathrm{SO(m)}}M \times _{cl_{m}}
\mathrm{Cl}_{m}$ is the {\em Clifford bundle}, where $\mathrm{Cl}_{m}$ is the Clifford algebra and $\mathrm{cl}_{m}$ the representation of $\mathrm{SO(m)}$ into
$\mathrm{Aut}(\mathrm{Cl}(\mathbb{R}^m))$, and the {\em spinor bundle} is $\mathcal{S} M = P_{\mathrm{Spin}(m)}M \times _{\gamma} \mathcal{S}_{m}$, with $\gamma$ the
spinorial representation of $\mathrm{Spin}(m)$ on the Clifford module $\mathcal{S}_{m}=\mathbb{C}^{2^{[m/2]}}$ (cf.~\cite {law}).

A  {\em spinor field} is a (smooth) section of $\mathcal{S} M$, $\Psi: U \subset M \longrightarrow \mathcal{S} M$, $\Psi(x)=[s_{x}, \psi(x)]$,
where $s_{x} \in P_{\mathrm{Spin}(m)}M$ is a spinorial frame at $x \in M$ and
$\psi: U \longrightarrow \mathcal{S}_{m}$, the equivalence class being defined by
$$[s, \psi]=[s \cdot g^{-1}, \gamma(g) \psi],$$
for all $g \in \mathrm{Spin}(m)$. The covariant derivative is
$$\nabla_{e_{j}} \Psi = \left[s, \ \ \mathrm{d} \psi (e_{j}) + \tfrac{1}{2}\sum_{k<l=1}^{m} \Gamma_{jk}^{l}\ e_{k} \cdot e_{l} \cdot \psi \right],$$
where $\chi(s)=\{e_{1},...,e_{m}\}$ is an orthonormal frame of $TM$, ``$\cdot$'' is the Clifford multiplication
and the symbols $\{\Gamma_{ij}^{k}\}_{i,j,k=1,\dots,m}$ are defined by $\nabla_{e_{i}}e_{j}=\Gamma_{ij}^{k}e_{k}$.

The {\em Dirac operator} is the first-order differential operator defined by
\begin{align*}
D^M:\Gamma (\mathcal{S} M) &\to \Gamma (\mathcal{S} M) \\
 \Psi &\mapsto D^M \Psi = \sum_{j=1}^{m} e_{j} \cdot \nabla_{e_{j}} \Psi .
\end{align*}

The above constructions extend to any oriented Riemannian bundle and, as for
orientation, given three vector bundles $\mathcal{E}^{\prime}$, $\mathcal{E}^{\prime \prime}$ and
$\mathcal{E} = \mathcal{E}^{\prime} \oplus \mathcal{E}^{\prime \prime}$ over $M$, a choice of spin-structure
on any two of them uniquely determines a spin structure on the third (\cite{law}).

\begin{de}
A smooth map $\pi : (M^m , g) \to (N^n ,h)$ between Riemannian manifolds is a {\em horizontally conformal map} if,
at any point $x\in M$, $d\pi_{x}$ maps the {\em horizontal space} $\mathcal{H}_{x} = (\ker d\pi_{x})^{\perp}$ conformally onto
$T_{\pi(x)}N$, i.e. $d\pi_{x}$ is surjective and there exists a number $\lambda (x) \neq 0$ such that
$$(\pi^{*}h)_{x}\Big|_{\mathcal{H}_{x} \times \mathcal{H}_{x}} = \lambda^{2}(x) g_{x}\Big|_{\mathcal{H}_{x} \times \mathcal{H}_{x}}.$$
The function $\lambda$ is the {\em dilation} of $\pi$ and the orthogonal complement of $\mathcal{H}_{x}$ is the {\em vertical distribution}
$\mathcal{V}_{x} = \ker d\pi_{x}$.\\
The mean curvatures of the distributions $\mathcal{H}$ and $\mathcal{V}$ are denoted $\mu^{\mathcal{H}}$ and $\mu^{\mathcal{V}}$ and $I^{\mathcal{H}}$
is the integrability tensor of $\mathcal{H}$.\\
A frame $\{V_{a},X_{i}^{*}\}^{a=1,\dots,m-n}_{i=1,\dots,n}$ of $TM$ will be called {\em adapted} if $V_{a} \in \mathcal{V}, a =
1,\dots,m-n$
and $\{X_{i}^{*}\}_{i=1,\dots,n}$ is the horizontal lift by $\pi$ of an orthonormal frame $\{X_{i}\}_{i=1,\dots,n}$ on $N$. \\
Note that $\lambda \equiv 1$ corresponds to Riemannian submersions.\\
We call the map $\pi : (M^m , g_{1}) \to (N^n ,h)$, where $g_{1}=\pi^{*}h+g^{\mathcal{V}}$, the {\em associated} Riemannian submersion of $\pi : (M^m , g) \to (N^n ,h)$.
\end{de}

For the remainder of the article, we will make the blanket assumption that the dimension $n$ of the manifold $N$ is even.
Since a general submersion $\pi : (M^m , g) \to (N^n ,h)$, between spin Riemannian manifolds, splits the tangent bundle
$TM$ into $\mathcal{H} \oplus \mathcal{V}$, if $\mathcal{H}$ admits a spin structure, so does $\mathcal{V}$ and
\begin{equation} \label{splitc}
\mathrm{Cl}(M) \ = \ \mathrm{Cl}(\mathcal{H}) \ \hat{\otimes} \ \mathrm{Cl}(\mathcal{V}) .
\end{equation}
The spin structures $P_{\mathrm{Spin}(n)}\mathcal{H}$ and $P_{\mathrm{Spin}(m-n)}\mathcal{V}$ induce
a spin structure $P_{\mathrm{Spin}(m)}M$ by prolongation of the principal bundle
$P_{\mathrm{Spin}(n) \times \mathrm{Spin}(m-n)}M$ (cf.~\cite{hus}). General
properties of associated bundles of reduced principal bundles (\cite[Theorem
3.1]{hus}) together with a dimension count yield the following isomorphisms of
(associated) vector bundles
\begin{align} \label{splits2}
\mathcal{S} M \ &= \ \mathcal{S} \mathcal{H} \otimes \mathcal{S} \mathcal{V} .
\end{align}

For any map $\pi : M \to N$ into a spin manifold, consider the pull-back spinor bundle
$$\pi^{-1}\mathcal{S} N = \{ (x, [s, \psi]) \in M \times \mathcal{S} N \ | \ \varpi_{\mathcal{S}}([s, \psi])=\pi(x) \},$$
where $\varpi_{\mathcal{S}}$ is the projection map of $\mathcal{S} N$.\\
If $\pi$ is a Riemannian submersion, then the isomorphism $\pi^{-1}\mathcal{S} N  =  \mathcal{S}
\mathcal{H}$, due to the identification of orthonormal frames, simplifies \eqref{splits2} into (cf. \cite{bis})
\begin{equation}
 \mathcal{S} M =\pi^{-1}\mathcal{S} N \otimes \mathcal{S} \mathcal{V}.
\end{equation}

\begin{re}
When $\pi$ is a Riemannian submersion with totally geodesic fibres, $\mathcal{H}$
complete and $N$ connected then the fibres are isometric to a Riemannian manifold $F$. If
$N$ and $F$ are spin manifolds, consider on $M$ the induced spin structure and, via
the isomorphism $\pi^{-1}\mathcal{S} N  =  \mathcal{S} \mathcal{H}$, \eqref{splits2} reads (see \cite{moro})
$$ \mathcal{S} M =\pi^{-1}\mathcal{S} N \otimes \mathcal{S} F.$$
\end{re}

\begin{re}
Since $n$ is even, the Clifford algebra $\mathrm{Cl}_n$ possesses an irreducible complex module
$\mathcal{S}_{n}$ of complex dimension $2^{n/2}$, the complex spinor module. When
restricted to $\mathrm{Cl}^{0}_{n}$ the spinor module decomposes into $\mathcal{S}_{n} =
\mathcal{S}^{+}_{n}\oplus \mathcal{S}^{-}_{n}$, the submodules of spinors of positive and  negative chirality,
characterized by the action of the volume element, once an
orientation is given.
In particular, the spin group $\mathrm{Spin}(n) \subset \mathrm{Cl}^{0}_{n}$ acts on
$\mathcal{S}^{+}_{n}$ and on $\mathcal{S}^{-}_{n}$ (the spinor representations).\\
Moreover, $\mathcal{S}_{n+1}$ pulls back to $\mathcal{S}_{n}$ under the
algebra isomorphism $\mathrm{Cl}_{n} = \mathrm{Cl}^{0}_{n+1}$. In other words, we can regard
$\mathcal{S}_{n+1}$ as the spinor representation of $\mathrm{Cl}_{n}$, provided we define the
action of $\mathrm{Cl}_{n}$ on $\mathcal{S}_{n+1}$ by
$v \otimes \sigma \rightarrow e_0 \cdot v \cdot \sigma$.
\end{re}

When $\pi : (M^{n+1},g) \to (N^{n},h)$ is a Riemannian submersion with
one-dimensional fibres into a spin manifold $N^{n}$, the manifold $M^{n+1}$ inherits
a natural spin structure, cf~\cite{amoro}. Moreover, $\mathcal{S} M = \pi^{-1}\mathcal{S} N$.

These identifications justify the following definition.
\begin{de}[Riemannian submersions]
Let $\pi : (M^m , g) \to (N^n ,h)$ be a Riemannian submersion between spin manifolds
and endow the vertical bundle with the induced spin structure (if $m=n+1$, we
consider on $M$ the natural spin structure inherited from $N$). Let $\Psi=[s, \psi]$
be a (local) spinor field on $N$. Since a local spin frame
$s=\{X_{i}\}_{i=1,\dots,n}$ on $N$ lifts to an adapted spin frame on $M$
$$\tilde{s}=\{X_{i}^{*}, V_{a}\}_{i=1,\dots,n}^{a=1,\dots,m-n}\in
P_{\mathrm{Spin}(n) \times _{\mathbb{Z}_{2}} \mathrm{Spin}(m-n)}M \vert _{\pi^{-1}U},$$
where $X_{i}^{*}$ is the horizontal lift of $X_{i}$ and $\{V_{a}\}_{a=1,\dots,m-n}$ is an orthonormal frame of $\mathcal{V}$,
we define the {\em pull-back} $\widetilde \Psi$ of $\Psi$ to be
\begin{itemize}
\item If $m-1=n$, the section $\widetilde \Psi=[\Tilde{s}, \psi \circ \pi]$ of the bundle
$\pi^{-1}\mathcal{S} N$, identified with $\mathcal{S} M$.\\
\item If $m-n \geq 2$, the section
$\widetilde{\Psi}=[\widetilde{s}, \widetilde \psi=(\psi \circ \pi) \otimes \alpha]$
in $\pi^{-1}\mathcal{S} N \otimes \mathcal{S} \mathcal{V}$, identified with $\mathcal{S} M$, where $\alpha$ is a fixed (non-zero) section of $\mathcal{S} \mathcal{V}$.
\end{itemize}
\end{de}

\begin{re}
When $m-n \geq 2$, this notion of pull-back depends on the choice of the section $\alpha$. In general, there exists no such non-vanishing global section.
\end{re}

\begin{re}
Note that Clifford multiplication with this kind of spinor fields is given by
$$
X^* \cdot \widetilde{\psi}  = \widetilde{X \cdot \psi}; \quad V \cdot
\widetilde{\psi} = \mathrm{i} \widetilde{\overline{\psi}},
$$
when $m=n+1$, and by
$$
X^* \cdot ((\psi\circ\pi) \otimes \alpha) = (( X \cdot \psi)\circ\pi) \otimes \alpha;
\quad V \cdot ((\psi\circ\pi) \otimes \alpha) = (\overline \psi\circ\pi) \otimes V \cdot
\alpha,
$$
when $m-n \geq 2$, where $X^*$ is the horizontal lift of the vector field $X \in
\Gamma(TN)$, $V$ is a unit vertical vector field and $\overline \psi$ is the conjugate of $\psi$ with respect to the $\mathbb{Z}
_2$-graduation (see~\cite{moro}).
\end{re}

\begin{re}
For a horizontally conformal submersion $\pi : (M^m , g) \to (N^n ,h)$, we deform the metric on $M$ into
$g_{1}=\pi^{*}h+g^{\mathcal{V}}$ and denote by (cf.~\cite{law})
\begin{align*}
& \xi_{\gamma} : \mathcal{S} M_{1} \to \mathcal{S} M \\
& \xi_{\gamma}([s, \psi]) = [\xi(s), \psi],
\end{align*}
where  $\xi(s)=\{ E_i=\lambda E_i ^1, V_a \}$ if $s=\{ E_i ^1, V_a \}$, the bundle isometry induced by the Spin-equivariant map
$$\xi: P_{\mathrm{Spin}(n) \times _{\mathbb{Z}_{2}} \mathrm{Spin}(m-n)}M_{1}
\to P_{\mathrm{Spin}(n) \times _{\mathbb{Z}_{2}} \mathrm{Spin}(m-n)}M$$
given by the natural correspondence between adapted orthogonal frames with respect to the two metrics:
$E_{i}^{1}=\lambda^{-1} E_{i}, V^{1}_{a}=V_{a}$. \\
The Clifford multiplication will be given by
$$
E_i \cdot \Psi = \xi_{\gamma}\left(E_i ^1 \cdot \Psi_1\right), \qquad
V_a \cdot \Psi = \xi_{\gamma}\left(V_a \cdot \Psi_1\right),
$$
where $\Psi=\xi_{\gamma} \circ \Psi_{1}$.
\end{re}

\begin{de}[Horizontally conformal submersions]
Let $\pi : (M^m , g) \to (N^n ,h)$ be a horizontally conformal submersion between spin manifolds and
endow the vertical bundle with the induced spin structure. Let $\Psi=[s, \psi]$ be a (local) spinor field on $N$.
The {\em pull-back} of $\Psi$ is
$\widetilde{\Psi}=\xi_{\gamma} \circ \widetilde{\Psi}_{1}$, where
$\widetilde{\Psi}_{1}$ is the pull-back of $\Psi$ by the associated Riemannian submersion $\pi : (M^m , g_{1}) \to (N^n ,h)$
and $\xi_{\gamma}$ the bundle isometry between $\mathcal{S} M_{1}$ and $\mathcal{S} M$.
\end{de}

\section{Dirac morphisms with high dimensional fibres}

Throughout this section $\pi$ has fibres of dimension at least two.

\begin{de}\label{def4}
A horizontally conformal submersion $\pi : (M^m ,g) \to (N^n ,h)$ between spin manifolds is called a {\em Dirac morphism} if
there exists a section $\alpha \in \Gamma(\mathcal{S} \mathcal{V})$, $\nabla^{\mathcal{V}}$-parallel in horizontal directions with $D^{\mathcal{V}}\alpha - \frac{n}{2} \mu^{\mathcal{H}} \cdot \alpha= 0$, and if for any local harmonic spinor $\Psi$ defined on $U\subseteq N$, such that $\pi^{-1}(U)\neq \emptyset$,
the pull-back of $\Psi$ (with respect to $\alpha$) $\widetilde{\Psi}=\xi_{\gamma} \circ \widetilde{\Psi}_{1}$ is a harmonic spinor on $\pi^{-1}(U) \subseteq M$, where $\widetilde{\Psi}_{1}$ is the pull-back (with respect to $\alpha$) of $\Psi$ by the associated Riemannian submersion.
\end{de}

\begin{re}
We assume, in Definition~\ref{def4}, the existence of the section $\alpha$ in order to construct pull-backs of spinors. Though these pull-backs will depend on the choice of $\alpha$ (if any), the two conditions on $\alpha$ will make the notion of Dirac morphism independent of the choice of such a section $\alpha$.
\end{re}

We first need some lemmas.

\begin{lm}[Chain rule]
Let $\pi : (M^m ,g) \to (N^n ,h)$ be a horizontally conformal submersion of dilation $\lambda$ ($m-n \geq 2$) and $\psi$ a (local) spinor field on $N$.
If $ \widetilde{\Psi}$
is the pull-back of $\Psi$ by $\pi$, with respect to some section $\alpha \in \Gamma(\mathcal{S} \mathcal{V})$, then

\begin{equation} \label{chainhwc}
\begin{split}
D^{M} \widetilde{\psi}  =
&\lambda \widetilde{D^{N}\psi}
-\tfrac{1}{2} \left[(m-n) \mu^{\mathcal{V}}+(n-1)\mathrm{grad}^{\mathcal{H}}(\mathrm{ln}\lambda) \right] \cdot \tilde \psi  \\
&+ \sum _{i=1}^{n} E_{i} \cdot (\psi \circ \pi) \otimes \nabla^{\mathcal{V}}_{E_{i}}\alpha
+\tfrac{1}{4} \ I^{\mathcal{H}} \cdot \tilde \psi  \\
&+(\overline \psi \circ \pi) \otimes \left[D^{\mathcal{V}}\alpha-\tfrac{n}{2} \mu^{\mathcal{H}} \cdot \alpha \right],
\end{split}
\end{equation}
where  $\{E_{i}\}_{i=1\dots,n}$ is a local orthonormal horizontal frame on $M$ and $I^{\mathcal{H}} \cdot \widetilde \psi$ denotes
$\sum _{i < j = 1}^{n} E_{i} \cdot E_{j} \cdot I^{\mathcal{H}}(E_{i} , E_{j}) \cdot \widetilde \psi$
(the standard action of vector-valued 2-forms on spinor fields).
\end{lm}

\begin{proof}
Let $\pi$ be a horizontally conformal submersion of dilation $\lambda$. Let $\{X_i \}_{i=1,\dots,n}$ be an orthonormal frame on $(N,h)$ and
$\{V_a, X^{*}_i \}_{i=1,\dots,n}^{a=1,\dots,m-n}$ an orthonormal adapted frame on $(M,g_1)$, where $g_1=\pi^* h+g^{\mathcal{V}}$. With respect to the metric $g$, $\{V_a, \lambda  X^{*}_i \}_{i=1,\dots,n}^{a=1,\dots,m-n}$ is an orthonormal adapted frame. Denote by $\nabla$ and $\nabla ^1$ the (spinorial) connections corresponding to $g$ and $g_1$, and
note $E_i ^1 = X_i ^*$, $E_i = \lambda X_i ^*$.

As $D^{M} \widetilde{\Psi} =[\tilde s, D^{M} \tilde \psi]$, for the pull-back spinor field $\tilde \psi$

\begin{align*}
D^{M} \widetilde{\psi} & =  \sum _{i=1}^{n} E_{i} \cdot \nabla_{E_{i}} \widetilde{\psi}+
 \sum _{a=1}^{m-n} V_{a} \cdot \nabla_{V_{a}} \widetilde{\psi}  \\
&= \sum _{i=1}^{n} E_{i} \cdot E_{i}((\psi \circ \pi) \otimes \alpha) \qquad (H_{0}) \\
&+ \tfrac{1}{4}\sum _{i, j, k=1}^{n} E_{i} \cdot g(\nabla_{E_{i}}E_{j}, E_{k}) \ E_{j} \cdot E_{k} \cdot \widetilde{\psi} \qquad (H_{1}) \\
&+ \tfrac{1}{2}\sum _{i, j, a=1}^{n,m-n} E_{i} \cdot g(\nabla_{E_{i}}E_{j}, V_{a}) \ E_{j} \cdot V_{a} \cdot \widetilde{\psi} \qquad (H_{2}) \\
&+ \tfrac{1}{4}\sum _{i, a, b=1}^{n,m-n} E_{i} \cdot g(\nabla_{E_{i}}V_{a}, V_{b}) \ V_{a} \cdot V_{b} \cdot \widetilde{\psi} \qquad (H_{3})
\end{align*}

\begin{align*}
\quad & +\sum _{a=1}^{m-n} V_{a} \cdot  V_{a}((\psi \circ \pi) \otimes \alpha) \qquad (V_{0}) \\
&+ \tfrac{1}{4}\sum _{i, j , a=1}^{n,m-n} V_{a} \cdot g(\nabla_{V_{a}}E_{i}, E_{j}) \ E_{i} \cdot E_{j} \cdot \widetilde{\psi} \qquad (V_{1}) \\
& + \tfrac{1}{2}\sum _{i, a, b=1}^{n,m-n} V_{a} \cdot g(\nabla_{V_{a}}E_{i}, V_{b}) \ E_{i} \cdot V_{b} \cdot \widetilde{\psi} \qquad (V_{2}) \\
&+ \tfrac{1}{4}\sum _{a, b, c=1}^{m-n} V_{a} \cdot g(\nabla_{V_{a}}V_{b}, V_{c}) \ V_{b} \cdot V_{c} \cdot \widetilde{\psi} \qquad (V_{3}).
\end{align*}
Note that
\begin{align*}
\widetilde{D^{N}\psi}&=\xi_{\gamma}(\widetilde{D^{N}\psi}^1)\\
&= \xi_{\gamma}\left[(X_{i}^{*}\cdot X_{i}^{*}(\psi \circ \pi) +
g_1(\nabla^{1}_{X_{i}^{*}}X_{j}^{*}, X_{k}^{*}) X_{i}^{*} \cdot X_{j}^{*}
\cdot X_{k}^{*} \cdot (\psi \circ \pi)) \otimes \alpha\right],
\end{align*}
where $g_1(\nabla^{1}_{X_{i}^{*}}X_{j}^{*}, X_{k}^{*})=
h(\nabla^N _{X_{i}}X_{j}, X_{k})$.\\
The computation breaks down into five steps:

\textbf{Step 1}:
\begin{equation}\label{eq6}
\begin{split}
(H_{0})+(H_{1})+(H_{3})=&\lambda \widetilde{D^{N}\psi}
- \tfrac{n-1}{2} \mathrm{grad}^{\mathcal{H}}(\mathrm{ln}\lambda)\cdot \widetilde \psi
+ \sum _{i=1}^{n} E_{i} \cdot (\psi \circ \pi) \otimes \nabla^{\mathcal{V}}_{E_{i}}\alpha.
\end{split}
\end{equation}
As
\begin{align*}
(H_{0})+(H_{1})=&\sum _{i=1}^{n} E_{i} \cdot \left[E_{i}(\psi \circ \pi) \otimes \alpha+ (\psi \circ \pi) \otimes E_{i}(\alpha)\right] \\
&+ \tfrac{1}{4}\sum _{i, j, k=1}^{n} E_{i} \cdot g(\nabla_{E_{i}}E_{j}, E_{k}) \ E_{j} \cdot E_{k} \cdot \widetilde{\psi},
\end{align*}
in order to recognize the lift of $D^N \psi$, first observe that
\begin{align*}
g(\nabla_{E_{i}}E_{j}, E_{k}) &=
g(\nabla_{\lambda X_{i}^{*}} \lambda X_{j}^{*}, \lambda X_{k}^{*}) \\
&= \lambda g_1(\nabla^{1}_{X_{i}^{*}}X_{j}^{*}, X_{k}^{*}) +
\left[X_{k}^{*}(\lambda)\delta_{i}^{j} - X_{j}^{*}(\lambda)\delta_{i}^{k} \right].
\end{align*}
Whilst
\begin{align*}
E_{i} \cdot \ E_{j} \cdot E_{k} \cdot \widetilde{\psi} &=
\xi_{\gamma}(E_{i}^1 \cdot \ E_{j}^1 \cdot E_{k}^1 \cdot \widetilde{\psi} ^1)\\
&= \xi_{\gamma}(X_{i}^{*} \cdot X_{j}^{*} \cdot X_{k}^{*} \cdot \widetilde{\psi} ^1),
\end{align*}
and
\begin{align*}
E_{i} \cdot E_{i}(\psi \circ \pi) \otimes \alpha &=
\lambda E_{i} \cdot E_{i}^1(\psi \circ \pi) \otimes \alpha \\
&= \lambda \xi_{\gamma}(X_{i}^{*}\cdot X_{i}^{*}(\psi \circ \pi) \otimes \alpha).
\end{align*}
Hence
\begin{align*}
(H_{0})+(H_{1}) =& \lambda \xi_{\gamma}(\widetilde{D^{N}\psi}^1)
+ \sum _{i=1}^{n} E_{i} \cdot (\psi \circ \pi) \otimes E_{i}(\alpha)  \\
&+ \xi_{\gamma}\left(\tfrac{1}{4}\sum _{i,j,k=1}^{n} \left[X_{k}^{*}(\lambda)\delta_{i}^{j} - X_{j}^{*}(\lambda)\delta_{i}^{k} \right] X_{i}^{*} \cdot X_{j}^{*} \cdot X_{k}^{*} \cdot \widetilde \psi ^1 \right).
\end{align*}
The last term can be rewritten
\begin{align*}
\tfrac{1}{4}\sum _{i,j,k=1}^{n} \left[X_{k}^{*}(\lambda)\delta_{i}^{j} - X_{j}^{*}(\lambda)\delta_{i}^{k} \right]
X_{i}^{*} \cdot X_{j}^{*} \cdot X_{k}^{*} \cdot \widetilde \psi ^1 =
-\tfrac{n-1}{2}\mathrm{grad}^{\mathcal{H} _1}(\lambda)\cdot \widetilde \psi ^1 ,
\end{align*}
and, since $\mathrm{grad}^{\mathcal{H}}(\lambda) =\lambda ^2 \mathrm{grad}^{\mathcal{H} _1}(\lambda)$, it becomes
\begin{align*}
-\tfrac{n-1}{2}\xi_{\gamma}\left(\mathrm{grad}^{\mathcal{H} _1}(\lambda)\cdot \widetilde \psi ^1\right) &=
-\tfrac{n-1}{2} \lambda \mathrm{grad}^{\mathcal{H} _1}(\lambda)\cdot \widetilde \psi \\
&= -\tfrac{n-1}{2} \mathrm{grad}^{\mathcal{H}}(\mathrm{ln}\lambda)\cdot \widetilde \psi.
\end{align*}
Summing up with ($H_{3}$), we obtain~\eqref{eq6}.

\textbf{Step 2}:
\begin{equation}
(V_{2})= -\tfrac{m-n}{2} \ \mu^{\mathcal{V}} \cdot \widetilde{\psi} .
\end{equation}

As $g(\nabla_{V_{a}}E_{j}, V_{b})= - g(E_{j}, \nabla_{V_{a}} V_{b})$ and $\mathcal{V}$ is integrable
\begin{align*}
(V_{2})&=-\tfrac{1}{2}\sum _{a, b=1}^{m-n} V_{a} \cdot (\nabla_{V_{a}} V_{b})^{\mathcal{H}} \cdot V_{b} \cdot \widetilde{\psi}\\
&=-\tfrac{1}{2}\left( \sum _{a < b=1}^{m-n} V_{a} \cdot (\nabla_{V_{a}} V_{b})^{\mathcal{H}} \cdot V_{b}
+ \sum _{a > b=1}^{m-n} V_{a} \cdot (\nabla_{V_{a}} V_{b})^{\mathcal{H}} \cdot V_{b} \right) \cdot \widetilde{\psi}
-\tfrac{1}{2} \sum _{a=1}^{m-n} (\nabla_{V_{a}} V_{a})^{\mathcal{H}} \cdot \widetilde{\psi}\\
&=-\tfrac{1}{2}\left( \sum _{a < b=1}^{m-n} V_{a} \cdot [V_{a}, V_{b}]^{\mathcal{H}} \cdot V_{b} \right) \cdot \widetilde{\psi}
-\tfrac{m-n}{2}  \mu^{\mathcal{V}} \cdot \widetilde{\psi} \\
&= -\tfrac{m-n}{2} \ \mu^{\mathcal{V}} \cdot \widetilde{\psi}.
\end{align*}

\textbf{Step 3}:
\begin{equation}
(V_{0})+(V_{3}) = (\overline \psi \circ \pi) \otimes D^{\mathcal{V}} \alpha,
\end{equation}
since
\begin{align*}
(V_{0})+(V_{3})=&\sum _{a=1}^{m-n}  (\overline \psi \circ \pi) \otimes V_{a} \cdot V_{a}(\alpha)
+ \tfrac{1}{4}\sum _{a, b, c=1}^{m-n} V_{a} \cdot g(\nabla_{V_{a}}V_{b}, V_{c}) \ V_{b} \cdot V_{c} \cdot \widetilde{\psi} \\
=&(\overline \psi \circ \pi) \otimes V_{a} \cdot \nabla^{\mathcal{V}}_{V_{a}}\alpha.
\end{align*}

\textbf{Step 4}:
\begin{equation}
(H_{2})= \tfrac{1}{2}I^{\mathcal{H}} \cdot \widetilde \psi - \tfrac{n}{2}\mu^{\mathcal{H}}\cdot \widetilde \psi.
\end{equation}
As for Step 2, we have
\begin{align*}
(H_{2})=\tfrac{1}{2}\sum _{i, j, a=1}^{n,m-n} E_{i} \cdot g(\nabla_{E_{i}}E_{j}, V_{a}) \ E_{j} \cdot V_{a} \cdot \widetilde{\psi}= \tfrac{1}{2}\sum _{i < j=1}^{n} E_{i} \cdot E_{j} \cdot I^{\mathcal{H}}(E_{i}, E_{j}) \cdot \widetilde{\psi} - \tfrac{n}{2}\mu^{\mathcal{H}}\cdot \widetilde \psi,
\end{align*}
where the terms $i=j$ give $\mu^{\mathcal{H}}$.

\textbf{Step 5}:
\begin{equation}
(V_{1})= - \tfrac{1}{4}I^{\mathcal{H}} \cdot \widetilde \psi,
\end{equation}
since
\begin{align*}
(V_{1})=& \tfrac{1}{4}\sum _{i, j, a=1}^{n,m-n} g(\nabla_{V_{a}}E_{i}, E_{j}) V_{a} \cdot \ E_{i} \cdot E_{j} \cdot \widetilde{\psi}   \\
&= \tfrac{1}{4}\sum _{i, j, a=1}^{n,m-n} \left[g([V_{a}, E_{i}], E_{j}) -
g(\nabla_{E_{i}}E_{j}, V_{a}) \right] V_{a} \cdot \ E_{i} \cdot E_{j} \cdot \widetilde{\psi}  \\
&= \tfrac{1}{4}\sum _{i, j, a=1}^{n,m-n} g([V_{a}, E_{i}], E_{j}) V_{a} \cdot \ E_{i} \cdot E_{j} \cdot \widetilde{\psi} - \tfrac{1}{2}(H_2).
\end{align*}
But $g([V_{a}, E_{i}], E_{j})=g([V_{a}, \lambda X_{i}^{*}], \lambda X_{j}^{*})
= \tfrac{V_{a}(\lambda)}{\lambda} \delta_i ^j$, since $[V_{a}, X_{i}^{*}]\in \mathcal{V}$. \\
Therefore
\begin{align*}
(V_{1})=&-\tfrac{n}{4}\sum _{a=1}^{m-n} V_{a}(\mathrm{ln}\lambda) V_{a} \cdot \widetilde{\psi} - \tfrac{1}{2}(H_2)\\
&=- \tfrac{n}{4}\mathrm{grad}^{\mathcal{V}}(\mathrm{ln}\lambda) \cdot \widetilde{\psi} - \tfrac{1}{2}(H_2)\\
&=- \tfrac{n}{4}\mu^{\mathcal{H}} \cdot \widetilde{\psi} - \tfrac{1}{2}(H_2) \\
&=- \tfrac{1}{4}I^{\mathcal{H}} \cdot \widetilde \psi.
\end{align*}
Summing up these five steps yields the chain rule.
\end{proof}

A generalization of \cite[Proposition 2.4]{A-G} to vector-valued functions yields local existence of harmonic spinors with prescribed value.

\begin{lm}[Local existence]\label{localexistence}
For any point $p\in M$ and $\psi_0 \in \mathcal{S}_p M$, there exists an open neighbourhood $U$ of $p$ and a harmonic spinor
$\psi : U \to \mathcal{S} M$ such that $\psi(p)=\psi_0$.
\end{lm}

\begin{te}[Characterization for horizontally conformal submersions]\label{th3.2}
Let $\pi: (M,g) \longrightarrow (N,h)$ be a horizontally conformal submersion between spin manifolds and
assume there exists a section $\alpha$ satisfying the conditions of Definition~\ref{def4}.\\
Then $\pi$ is a Dirac morphism if and only if its horizontal distribution is integrable and
\begin{equation}\label{F-Eq}
(m-n) \mu^{\mathcal{V}}+(n-1)\mathrm{grad}^{\mathcal{H}}(\mathrm{ln}\lambda) = 0,
\end{equation}
where $\mu^{\mathcal{V}}$ is the mean curvature of the fibres.
\end{te}

\begin{proof}
Let $\pi$ be a horizontally conformal submersion with integrable horizontal distribution and such that \eqref{F-Eq} is satisfied.
In this case, the Chain rule \eqref{chainhwc} simplifies to
\begin{equation*}
D^{M} \widetilde{\psi}  =
\lambda \widetilde{D^{N}\psi}
+ \sum _{i=1}^{n} E_{i} \cdot (\psi \circ \pi) \otimes \nabla^{\mathcal{V}}_{E_{i}}\alpha
+(\overline \psi \circ \pi) \otimes \left[D^{\mathcal{V}}\alpha-\tfrac{n}{2} \mu^{\mathcal{H}} \cdot \alpha \right].
\end{equation*}
Let $\psi$ be a local harmonic spinor and $\alpha$ a section of $\Gamma(\mathcal{S} \mathcal{V})$, $\nabla^{\mathcal{V}}$-parallel in horizontal directions and with
$D^{\mathcal{V}}\alpha - \frac{n}{2} \mu^{\mathcal{H}} \cdot \alpha= 0$. From the above formula, it follows that $D^{M} \widetilde{\psi}=0$ and therefore $\pi$ is a Dirac morphism.

Conversely, suppose that $\pi$ is a Dirac morphism. Then, by Definition~\ref{def4}, there exists a horizontally parallel section $\alpha \in \Gamma(\mathcal{S} \mathcal{V})$
satisfying $D^{\mathcal{V}}\alpha - \frac{n}{2} \mu^{\mathcal{H}} \cdot \alpha= 0$, and, for a harmonic spinor field $\psi$ on $N$, we have, according to \eqref{chainhwc}
\begin{align} \label{Chain-Split}
0 = & - \tfrac{1}{2} \left[(m-n) \mu^{\mathcal{V}}+(n-1)\mathrm{grad}^{\mathcal{H}}(\mathrm{ln}\lambda) \right] \cdot \widetilde{\psi} \\
& + \tfrac{1}{4}\sum_{i < j=1}^{n} X_{i}^{*} \cdot X_{j}^{*} \cdot (\psi \circ \pi) \otimes I^{\mathcal{H}}(X_{i}^{*} , X_{j}^{*}) \cdot \alpha \notag.
\end{align}
Putting $X=(m-n) \mu^{\mathcal{V}}+(n-1)\mathrm{grad}^{\mathcal{H}}(\mathrm{ln}\lambda)$ and
$V^{ij}=I^{\mathcal{H}}(X_{i}^{*} , X_{j}^{*})$, \eqref{Chain-Split}  becomes
$$
0= - \tfrac{1}{2} X \cdot \widetilde{\psi}
+ \tfrac{1}{4}\sum _{i < j=1}^{n} X_{i}^{*} \cdot X_{j}^{*} \cdot V^{ij} \cdot \widetilde{\psi}.
$$
As the value of $\widetilde{\psi}$ at $p \in M$ can be prescribed, the above equation implies
$$
0= - \tfrac{1}{2} X + \tfrac{1}{4}\sum _{i < j=1}^{n} X_{i}^{*} \cdot X_{j}^{*} \cdot V^{ij}.
$$
But since $V^{ij}$ is vertical, $X$ and $X_{i}^{*} \cdot X_{j}^{*} \cdot V^{ij}$ have different degrees, necessarily
$X=0$ and $V^{ij}=0$. Therefore, if a horizontally conformal submersion is Dirac morphism, it must satisfy Equation~\eqref{F-Eq} and have integrable horizontal distribution.
\end{proof}

\begin{re}
\begin{enumerate}
\item Note the analogy between Equation~\eqref{F-Eq} and the fundamental equation for harmonic morphisms in~\cite{B-W}.
\item Compare Formula~\eqref{chainhwc} with \cite[(4.26)]{bis} and \cite[1.1.1]{moro}.\\
\item If the fibres are totally geodesic, the integrability of the horizontal distribution
makes the section $\alpha$ ``basic transversally harmonic'', as introduced
in~\cite{glaz}.
\end{enumerate}
\end{re}

\begin{co}
A Riemannian submersion $\pi : (M^m ,g) \to (N^n ,h)$ between spin manifolds is a Dirac morphism if and only if
its fibres are minimal and its horizontal distribution is integrable.
\end{co}

Recall that if $\pi$ is a Riemannian submersion then $\mu^{\mathcal{H}}=0$.

\begin{re}
If the dilation function $\lambda$ is a projectable function (i.e. $V(\lambda)=0$), the conformal invariance of the Dirac operator
(\cite{law}) allows a correspondence between harmonic spinors of the spaces involved in the commutative diagram below.
\begin{center}
\setlength{\unitlength}{1mm}
\begin{picture}(50,40)
\put(41,27){\vector(0,-3){27}}
\put(3,27){\vector(4,-3){35}}
\put(15,30){\vector(4,0){12}}
\put(8,-4){\vector(4,0){29}}
\put(0,27){\vector(0,-3){27}}
\put(42,33){\makebox(0,0)[t]{$(M, \pi^{*}h+g^{\mathcal{V}})$}}
\put(0,33){\makebox(0,0)[t]{$(M,  \lambda^{2} \pi^{*}h+g^{\mathcal{V}})$}}
\put(44,13){\makebox(0,0)[t]{$\pi$}}
\put(0,-2){\makebox(0,0)[t]{$(N, \tilde \lambda^{2}h)$}}
\put(20,0){\makebox(0,0)[t]{\bf 1$_N$}}
\put(20,34){\makebox(0,0)[t]{\bf 1$_M$}}
\put(2,13){\makebox(0,0)[t]{$\pi$}}
\put(43,-2){\makebox(0,0)[t]{$(N, h)$}}
\end{picture}
\end{center}
\end{re}

\bigskip

\section{Dirac morphisms with one-dimensional fibres}

In this section $m=n+1$.

\begin{de}
A horizontally conformal submersion $\pi : (M^{n+1}, g) \rightarrow (N^n, h)$ between spin manifolds is called a Dirac morphism if for any local harmonic spinor $\Psi$ defined on $U \subset N$, such that $\pi^{-1}(U) \neq \emptyset$, the pullback $\widetilde \Psi = \xi_{\gamma} \circ \widetilde \Psi _1$ is a harmonic spinor on $\pi^{-1}(U) \subset M$, where  $\widetilde \Psi _1$ is the pullback of $\Psi$ by the associated Riemannian submersion.
\end{de}

\begin{lm} \ \emph{(Chain rule)}. Let $\pi : (M^{n+1}, g) \rightarrow (N^n, h)$ be a horizontally conformal submersion of dilation $\lambda$ and $\psi$ a (local) spinor field on $N$, then
\begin{equation}\label{Chain-Split1}
\begin{split}
D^{M} \widetilde{\psi} =
&\lambda \widetilde{D^{N}\psi} -
\frac{1}{2} \left[\mu^{\mathcal{V}}+(n-1)\mathrm{grad}^{\mathcal{H}}(\mathrm{ln}\lambda) +n \mu^{\mathcal{H}}\right] \cdot \widetilde \psi +
\frac{1}{4} I^{\mathcal{H}}\cdot \widetilde \psi.
\end{split}
\end{equation}
\end{lm}

\begin{proof} Take $\{X_i \}_{i=1,...,n}$
an orthonormal frame on $(N^n, h)$ and
$\{ V, E_i=\lambda X_i ^*\}_{i=1,...,n}$ an adapted frame on $(M^m, g)$. Let $\Psi$ be a (local) spinor field on $N$
and $\widetilde \Psi$ its pullback by $\pi$.
The proof is similar to the proof of Lemma 1, except that $(H_0)= \sum E_i \cdot E_i(\psi \circ \pi)$, $(V_0)=0$ and the terms $(H_3), (V_3)$ do not appear.
\end{proof}

Note that $\mu^{\mathcal{H}} \cdot \widetilde \psi =  i \norm{\mu^{\mathcal{H}}}^2 \widetilde{\overline{\psi}}$ as $V \cdot \widetilde \psi = i \widetilde{\overline \psi}$.

\begin{te}
A horizontally conformal submersion
$\pi : (M^{n+1} ,g) \to (N^n ,h)$ between spin manifolds is a Dirac morphism if and only if its
horizontal distribution is integrable and minimal, and
$$\mu^{\mathcal{V}}+(n-1)\mathrm{grad}^{\mathcal{H}}(\mathrm{ln}\lambda)=0 ,$$
where $\mu^{\mathcal{V}}$ is the mean curvature of the fibres.
\end{te}

\begin{proof}
The argument is similar to the one of Theorem 1, except that
$$
X=\mu ^{\mathcal{V}} + (n - 1)\mathrm{grad}^{\mathcal{H}}(\ln \lambda)  + n \mu^{\mathcal{H}}.
$$
Observe that $X=0$ if and only if $\mu ^{\mathcal{V}} + (n - 1)\mathrm{grad}^{\mathcal{H}}(\ln \lambda)=0$ and $\mu^{\mathcal{H}}=0$, as they belong to orthogonal distributions.
\end{proof}

\begin{co}
A Riemannian submersion $\pi : (M^{n+1}, g) \rightarrow (N^n, h)$ between spin manifolds is a Dirac morphism if and only if its horizontal distribution is integrable and the fibres are minimal.
\end{co}

\begin{re} Suppose that $\pi$ is a Riemannian submersion.

(1) \ The Chain Rule \eqref{Chain-Split1} gives us the formula of \cite{amoro} (where the fibres are minimal)

\begin{equation}\label{dirak1}
D^{M} \widetilde{\Psi} =
\widetilde{D^{N}\Psi} -\frac{1}{2} \ \mu^{\mathcal{V}} \cdot \widetilde{\Psi}
-\frac{1}{4} i \sum_{j=1}^{n} X_{j}^{*} \cdot A_{X_{j}^{*}}V \cdot \widetilde{\overline{\Psi}},
\end{equation}
where $A$ denotes the second O'Neill tensor of $\pi$.

(2) \ If $\pi$ is a Dirac morphism, any pullback spinor field is parallel along the fibres and its \textit{metric Lie derivative} (\cite{bour}) vanishes.
\end{re}

\begin{re}
When the fibres are circles, a similar
chain rule was obtained by Ammann in~\cite{Am}.
\end{re}

\section{Examples}

\begin{ex}[Dirac morphisms from 3 to 2-dimensional Euclidean spaces]

With respect to the irreducible representation $\gamma$ of $\mathrm{Cl}_{2}$ given by Pauli matrices,
a spinor field on $(\mathbb{R}^{2}, \langle, \rangle _{\mathrm{standard}})$ will be
$\psi:\mathbb{R}^{2} \longrightarrow \mathbb{C}^{2}$,
$\psi =\left(
\begin{array}{c}
\psi^{+} \\
\psi^{-}
\end{array}
\right)$, in a global spin frame ($e_{1}=\tfrac {\partial}{\partial y_{1}}, e_{2}=\tfrac {\partial}{\partial y_{2}}$) on $\mathbb{R}^{2}$,
and the Dirac operator on $\mathbb{R}^{2}$
$$
D^{\mathbb{R}^{2}}=\gamma(e_{1})\frac {\partial}{\partial y_{1}}+\gamma(e_{2})\frac {\partial}{\partial y_{2}}
=\left(
\begin{array}{cc}
0 & -\tfrac{\partial}{\partial z} \\
\tfrac{\partial}{\partial \overline{z}} & 0
\end{array}
\right).
$$
Consider a Riemannian submersion $\pi:\mathbb{R}^{3} \longrightarrow \mathbb{R}^{2}$, the pull-back of $\psi$,
$\widetilde{\psi}=\psi \circ \pi:\mathbb{R}^{3}\longrightarrow \mathbb{C}^{2}$ is a spinor field on $\mathbb{R}^{3}$
with respect to the frame $\{e_{1}^{*}, e_{2}^{*}, v \}$,
$v \in \mathop\mathrm{Ker\,} \mathrm{d} \pi$. Supposing $\mathcal{H}$ integrable, this spin adapted frame can be chosen to be
$\{\tfrac {\partial}{\partial x_{1}}, \tfrac {\partial}{\partial x_{2}}, \tfrac {\partial}{\partial x_{3}} \}$, and
with the Pauli representation, the Dirac operator on $\mathbb{R}^{3}$ is
$$
D^{\mathbb{R}^{3}}=i\sigma_{k} \frac {\partial}{\partial x_{k}}
=\left(
\begin{array}{cc}
i \tfrac {\partial}{\partial x_{3}} & \tfrac {\partial}{\partial x_{2}}+i\tfrac {\partial}{\partial x_{1}} \\
-\tfrac {\partial}{\partial x_{2}}+i\tfrac {\partial}{\partial x_{1}} & - i \tfrac {\partial}{\partial x_{3}}
\end{array}
\right).
$$
Let $\psi$ be an arbitrary harmonic spinor on $\mathbb{R}^{2}$ (so $\psi^{+}$ a holomorphic function), then
$\widetilde{\psi}^{+}$ is also harmonic if and only if
$$
\frac{\partial \pi_{1}}{\partial x_{1}} =
\frac{\partial \pi_{2}}{\partial x_{2}},
\quad
\frac{\partial \pi_{1}}{\partial x_{2}} =
-\frac{\partial \pi_{2}}{\partial x_{1}},
\quad
\frac{\partial \pi_{1}}{\partial x_{3}} =
\frac{\partial \pi_{2}}{\partial x_{3}} = 0.
$$
The analogous question for $\widetilde{\psi}^{-}$ involves a change of sign in the first two equalities
(i.e. $\pi$ must be anti-holomorphic with respect to $x_{1}+ ix_{2}$).

These conditions are exactly the harmonicity of $\pi$, which is equivalent to the minimality of the fibres.

\end{ex}

\begin{ex}[Dirac morphisms from 4 to 2-dimensional Euclidean spaces]

A spinor field on $(\mathbb{R}^{4}, \langle, \rangle _{\mathrm{standard}})$ can be seen as a $\mathbb{C}^{4}$-valued function on $\mathbb{R}^{4}$ once a global spin frame is chosen. Using Pauli matrices, the Dirac operator on $\mathbb{R}^{4}$ can be described as
$$
D^{\mathbb{R}^{4}}=\gamma_{k}  \frac {\partial}{\partial x_{k}}
=\left(
\begin{array}{cccc}
0 & 0 & \tfrac {\partial}{\partial x_{3}}+i \tfrac {\partial}{\partial x_{0}} &
\tfrac {\partial}{\partial x_{1}}-i\tfrac {\partial}{\partial x_{2}} \\
0 & 0 & \tfrac {\partial}{\partial x_{1}}+i\tfrac {\partial}{\partial x_{2}} &
-\tfrac {\partial}{\partial x_{3}}+i \tfrac {\partial}{\partial x_{0}} \\
-\tfrac {\partial}{\partial x_{3}}+i \tfrac {\partial}{\partial x_{0}} &
-\tfrac {\partial}{\partial x_{1}}+i\tfrac {\partial}{\partial x_{2}} & 0 & 0\\
-\tfrac {\partial}{\partial x_{1}}-i\tfrac {\partial}{\partial x_{2}} &
\tfrac {\partial}{\partial x_{3}}+i \tfrac {\partial}{\partial x_{0}}  & 0 & 0
\end{array}
\right).
$$
Let $\pi:\mathbb{R}^{4} \longrightarrow \mathbb{R}^{2}$ be a Riemannian submersion and $\psi: \mathbb{R}^{2} \longrightarrow \mathbb{C}^{2}$
 a spinor field on $(\mathbb{R}^{2}, \langle, \rangle _{\mathrm standard})$. Then the pull-back $\widetilde{\psi}$ of $\psi$ is
$(\psi \circ \pi) \otimes \alpha :\mathbb{R}^{4}\longrightarrow \mathbb{C}^{4}$. With respect to $\{e_{1}^{*},e_{2}^{*},v,w\}$,
$v, w \in \mathop\mathrm{Ker\,} \mathrm{d} \pi$, assuming $\mathcal{H}$ integrable and choosing a frame
$\{\tfrac {\partial}{\partial x_{2}},\tfrac {\partial}{\partial x_{3}}, \tfrac {\partial}{\partial x_{0}},
\tfrac {\partial}{\partial x_{1}} \}$,
$$
\widetilde{\psi}
=\left(
\begin{array}{c}
\widetilde{\psi}^{+} \\
\widetilde{\psi}^{-}
\end{array}
\right)
=\left(
\begin{array}{c}
\psi^{+} \alpha^{+} \\
\psi^{-} \alpha^{-} \\
\psi^{+} \alpha^{-} \\
\psi^{-} \alpha^{+}
\end{array}
\right).
$$
The conditions of harmonicity and parallelism make $\alpha: \mathbb{R}^{2} \rightarrow \mathbb{C}^{2}$
a harmonic spinor field with respect to the variables $x_{0}, x_{1}$ (i.e. $\alpha^{+}$ is holomorphic
and $\alpha^{-}$ anti-holomorphic).
Take a harmonic spinor $\psi$ on $\mathbb{R}^{2}$
(i.e. $\psi^{+}$ is a holomorphic function), one can directly check that $\widetilde{\psi}^{+}$
is harmonic, for any $\alpha$, if and only if $\pi$ satisfies
$$
\frac{\partial \pi_{1}}{\partial x_{0}} =\frac{\partial \pi_{1}}{\partial x_{1}} =
\frac{\partial \pi_{2}}{\partial x_{0}}=\frac{\partial \pi_{2}}{\partial x_{1}}= 0,
\quad \frac{\partial \pi_{1}}{\partial x_{2}} =
\frac{\partial \pi_{2}}{\partial x_{3}}, \quad
\frac{\partial \pi_{1}}{\partial x_{3}} =
-\frac{\partial \pi_{2}}{\partial x_{2}}.
$$

The same question for $\widetilde{\psi}^{-}$ merely introduces a different sign. Again, this forces $\pi$ to be harmonic, i.e. its fibres are minimal.

\end{ex}

\begin{ex}[Moroianu's projectable spinors]\label{exmoro}

In \cite{moro}, Moroianu considers a principal fibre bundle $\pi: (M^{m},g)
\longrightarrow N$ with compact structural group $G$, over a compact spin manifold
$(N^{n},h)$, such that $\pi$ is a Riemannian submersion with totally geodesic fibres
and the horizontal distribution $\mathcal{H}$ is a principal connection.

Since its tangent space is trivial, $G$ admits a canonical spin structure, and a
spinor $(\psi \circ \pi) \otimes \alpha$ is called projectable if $\alpha: G
\longrightarrow \mathcal{S}_{m-n}$ is a constant function with respect to the canonical frame
of left-invariant vector fields. To have a $D^M$-invariant notion of projectable
spinor, it is necessary and sufficient to suppose $G$ commutative (\cite{moro})

Let $X^{*}$ be the horizontal lift of a vector field $X$ on $N$ and, using $[V,
X^{*}]=0$ for $V\in \mathcal{V}$, since $\mathcal{H}$ is a principal connection, we have
\begin{equation*}
\begin{split}
\nabla^{\mathcal{V}}_{X^{*}}\alpha=& X(\alpha)+ \tfrac{1}{2}\sum _{b<c=1}^{m-n} g(\nabla_{X^{*}}V_{b}, V_{c})V_{b} \cdot V_{c} \cdot \alpha\\
=&X^{*}(\alpha)+ \tfrac{1}{2}\sum _{b<c=1}^{m-n} g(\nabla_{V_{b}}X^{*}, V_{c})V_{b} \cdot V_{c} \cdot \alpha\\
=&X^{*}(\alpha)- \tfrac{1}{2}\sum _{b<c=1}^{m-n} g(X^{*}, \nabla_{V_{b}}V_{c})V_{b} \cdot V_{c} \cdot \alpha\\
=&X^{*}(\alpha),
\end{split}
\end{equation*}
since the fibres are totally geodesic. Therefore the condition of parallelism of $\alpha$ translates into constancy in horizontal directions. Moreover

\begin{equation*}
\begin{split}
&D^{\mathcal{V}}\alpha = V_{a} \cdot \nabla^{\mathcal{V}}_{V_{a}}\alpha \\
&=\sum _{a=1}^{m-n}  V_{a} \cdot V_{a}(\alpha)
+ \tfrac{1}{4}\sum _{a, b, c=1}^{m-n} V_{a} \cdot g(\nabla_{V_{a}}V_{b}, V_{c}) \ V_{b} \cdot V_{c} \cdot \alpha \\
&= \sum _{a=1}^{m-n}  V_{a} \cdot V_{a}(\alpha)
+ \tfrac{3}{4}\sum _{a < b < c=1}^{m-n} g([V_{a},V_{b}], V_{c}) \ V_{a} \cdot \ V_{b} \cdot V_{c} \cdot \alpha, \\
\end{split}
\end{equation*}
where, on a Lie group, $g([V,W], Z)=g(V, [W,Z])$ if $g$ is an invariant metric.
Clearly, if $\alpha$ is a constant function and $G$ is commutative (i.e. the
structural constants vanish) then $D^{\mathcal{V}}\alpha =0$.

Therefore a projectable spinor field is the pull-back of some spinor field on the
base, with respect to $\alpha$ satisfying the two conditions required for Dirac
morphisms. Hence, Theorem~\ref{th3.2} says that a principal bundle over a spin
manifold with commutative structural group is a Dirac morphism if and only if it is
flat.
\end{ex}

redpajama_set_name: RedPajamaArXiv