text:
Peace is not an option for the occupiers of the West Bank
Following up on the story posted below about the Global Peace index and how low Israel ranks comes this story that underscores that ranking and demonstrates how a country is sometimes its own worse enemy.
Two Israeli officers have testified that troops in the West Bank beat, bound and blindfolded Palestinian civilians as young as 14.
Both the soldiers, from the Harub battalion, highlighted the tight tying of the plastic hand restraints placed on detainees. "There are people who think you need to tighten the restraints all the way, until no drop of blood will pass from here to there," one soldier said. "It doesn't take much time until the hands turn blue. There were a lot of people that you know weren't feeling anything."
"The soldiers who took [detainees] to the toilet just exploded [over] them with beatings; cursed them with no reason. When they took one Arab to the toilet so that he could urinate, one of them gave him a slap that brought him to the ground. He had been handcuffed from behind with a nylon restraint and blindfolded. He wasn't insolent, he didn't do anything to get on anyone's nerves … [it was] just because he's an Arab. He was something like 15 years old." The soldier said he saw a lot of soldiers "just knee [Palestinians] because it's boring, because you stand there 10 hours, you're not doing anything, so they beat people up.
There were a lot of reservists that participated, and they totally had a celebration on the Palestinians: curses, humiliation, pulling hair and ears, kicks, slaps. These things were the norm."
This is systematic violence against people for the most part are under the authority of the Israeli government and it is this type of government sanctioned violence that keeps Israel in much the same company as Sudan, Somali, and Iraq, and it is all done with the knowledge and blessings of the US government.
Filed under Uncategorized Tagged with assault, IDF, Israel, military occupation, occupiers, Palestinians, West Bank
Should we take this kinda' stuff seriously or not?
It's pretty amusing reading in a way, but I also wonder if it was written to be that way in order to obscure what is really going on; the New York Daily News is as mainstream a media outlet as they come, and we all know what that means, don't we? Nevertheless, this article on one Daniel James Murray, aka the Capeman, threatening President Obama after insisting in a bank if he didn't receive the money he wanted to withdraw (read that steal) without producing the necessary identification would result in someone's death reads almost like a comic book. Capeman claims he was on a mission to kill the President, and thankfully he was caught gambling his heist away in a Nevada casino before accomplishing that mission.
Two things about this; first the kook nature of it. I believe all murderers are kooks, mentally unhinged, unstable, but the tendency to consistently portray those who plot to kill a president or actually do so in that way is conspiratorial in my view, because murderers are also calculating, and methodical, two characteristics that embody saneness. That said, why make references to this guy muttering while walking down the streets of his upstate New York neighborhood with a cape on in order to underscore that point, and why is he withdrawing money from a bank in Utah an area which houses unsavory movements some of which vehemently oppose a President Obama and a continent removed from his home state of New York? Had he moved there to become a part of the community of like minded people in Utah? That relationship has not been explored or revealed by the corporate, read mainstream, media. Finally the reference in almost all news stories on this to how Murray owns guns is telling, all too predictable and typical. It is media's attempt to get people to recoil at the idea of private gun ownership; anyone who owns weapons must by nature be suspicious and plotting towards a violent goal. My retort to the anti-gun bias in Murray's story is if he is as mentally unbalanced as some assert in news reports, how was he able to legally possess so many firearms? In many municipalities sound and emotional well being are key in order to legally purchase firearms. Was the guy mentally unbalanced or not and does the area where he purchased his firearms prohibit people who are unbalanced from owning guns? If yes, someone else is also responsible for his illegal procurement of weapons. Perhaps this isn't the lone gunman plot we are so accustomed to reading about when it comes to attempts against politicians as it may seem?
Filed under Uncategorized Tagged with assassination attempt, assassination squad, Daniel James Murray, Gun control, gun ownership, President Obama, the Capeman, Utah


text:
Kim Kardashian Calls Kanye West "A Genius"
The socialite also claims she would be "honoured" to work with Yeezy...
Kim Kardashianhas spoken out about her attempt at a pop career, revealing that she would be honoured to work withKanye West.
The Jennifer Lopez inspired single Turn It Up on New Year's Eve, also claimed that Kanye, who allegedly features on the single, is "a genius."
SEE SEXY PICS OF KIM KARDASHIAN HERE!
When asked about Yeezy's involvement on the track, the socialite told our US Team: "[Kanye's] a musical genius, so I would be honored to work with Kanye,"
Speaking of her possible collaboration with the Power rapper, Kardashian claimed: "It's all in fun. We're just trying to have a good time."
Kim, whose new music video will apparently be directed by legend Hype Williams, also teased that those who want further details about her musical project will have to tune into her reality TV show, Kourtney & Kim Take New York.
The former Playboy model claimed: "You're going to have to wait and watch the show. It's kind of in [Ciara's] world and not mine. ... I'm having fun — that's all I'll say, but you'll definitely see it all unfold on the show."
CHECK OUT MORE SNAPS OF KIM KARDASHIAN HERE!
Meanwhile, Kim looks set to become an aunt again following reports that her younger sister, Khloe, is pregnant.
A source told the New York Post that the 26-year old: "has been hiding a tell-tale bump behind flowing dresses and oversized handbags."
Khloe and hubby, Lamar Odoom, are expected to document the birth of their first child on a new programme, which will reportedly follow her pregnancy.
More From Kim Kardashian
Kim Kardashian Reveals Her 'Soul Was Broken' By People Trolling Her Pregnancy Body
Kim Kardashian Wins $2.7 Million From Missguided After They 'Ripped Off' Her Outfits
Kim Kardashian Is Renaming Her Shapewear Line After Cultural Appropriation Claims
Kylie Jenner Called Kim Kardashian And Begged Her To Not "Bully" Jordyn Woods
Kim Kardashian Releases Statement About Her Kimono Trademark Controversy
Kim Kardashian's Body Make-Up Line Is Receiving Backlash For This Reason
An Inside Look At North West and Penelope Disick's Lavish AF Birthday Party
Why Kim Kardashian Is Getting Backlash For Her Latest Instagram Post
Kim Kardashian's Ancestral Line Reveals She Is Actually Scottish Royalty
Why Does Kim Kardashian Have Beef With Fast Food Chain 'Jack In The Box'?

redpajama_set_name: RedPajamaCommonCrawl


text:
Clara and Roget, who left a mark on the world
November 30, 2014 | hgrange9
CLARA AND DAVIE
The True Story of Young Clara Barton Founder of the American Red Cross
by Patricia Polacco
Scholastic Press, 2014
Picture Book Nonfiction
Grades 2 and up
All opinions expressed are solely my own.
From bestselling author Patricia Polacco's family tree — the true story of young Clara Barton.
Animals and flowers were Clara's best friends. She had a special way with critters and found joy in the beauty that sprang from the soil. But whenever Clara talked, her words didn't come out right. As hard as she tried, she could not get over her lisp.
Clara's older brother Davie understood that his sister was gifted. When folks made fun of Clara's stilted words, Davie was always at her side reminding her that she had a talent for healing creatures.
Davie told his sister, "Some day you are going to be a very great lady." And that's exactly what happened. Clara Barton became one of the most famous medical practitioners of all time, and founded the American Red Cross.
Another winner from a prolific and talented author/illustrator, Clara and Davie introduces the reader to Clara Barton as a young girl. I especially enjoyed reading about her relationship with her brother, Davie. Not many brothers ten years older would take the time to help and befriend his younger sister the way Davie does. It's clear that Clara had a gift for healing from the time she was very young. But like everyone else she faced her own challenges including a prominent lisp. This lisp led many of those around her to make fun of her causing her to withdraw, but thanks to the efforts of her brother and other family members she was educated at home. When Davie suffers a devastating injury, Clara is just the one to help him face his own heart-wrenching challenges. Like all her other family stories, Polacco shares the experiences that helped people grow and become the people they were. Keep in mind that biographical picture books like this one often have made up dialogue in them, after all nobody was around to record everything that someone may have said at some point in his/her life.
Roget and His Thesaurus
by Jen Bryant, illustrated by Melissa Sweet
Eerdman's Books for Young Readers, 2014
For shy young Peter Mark Roget, books were the best companions — and it wasn't long before Peter began writing his own book. But he didn't write stories; he wrote lists. Peter took his love for words and turned it to organizing ideas and finding exactly the right word to express just what he thought. His lists grew and grew, eventually turning into one of the most important reference books of all time.
Readers of all ages will marvel at Roget's life, depicted through lyrical text and brilliantly detailed illustrations. This elegant book celebrates the joy of learning and the power of words.
One of my favorite books of 2014, The Right Word, takes a look at the creation of Roget's Thesaurus and the man who created it. Not only are the illustrations remarkable (which I would expect from Melissa Sweet), the text is beautifully integrated with them to present a striking book that is both informative and appealing. This book would be a great tool for helping children learn how to use a thesaurus and to develop an understanding of the power of having just the right word to use. Roget strikes me as a rather interesting man to study with his word fascination as well as interest in many aspects of science so I found the author's and illustrator's notes at the end thoroughly intriguing. The list of references and resources is also helpful and indicates the large amount of work that went into making this book. It was also really interesting to see a copy of one of the pages from Roget's notebook. A great book and a great resource and very possibly a soon to be award winner.
Nonfiction Monday Round-up	historypicture book biographies
Gus & Me
November 24, 2014 | Shelf-employed
Richards, Keith. 2014. Gus & Me: The Story of my Granddad and my First Guitar. Hachette Audio.
Keith Richards, the rough-edged, raspy-voiced, Rolling Stones guitarist, is hardly the man that comes to mind for a picture book writer and narrator, but then again, who better to tell the story of his first guitar?
Richards wins the listener over immediately with his folksy, working class Estuary English accent (think dropped h's and "intrusive" r's) and unmistakable fondness for his topics – his first guitar and his beloved Granddad, Gus. It was the musically talented Gus who introduced a young Keith Richards to the guitar, teaching him how to 'old it, and suggesting the classical Malagueña(r) as the pinnacle of guitar mastery.
I have yet to see the print version of this story, but I don't believe it could surpass the audio book. A story with music at its heart needs music to be understood. Richards plays bits from Malagueña in appropriate spots throughout the story, and during a visit to a music shop in London, we hear Steve Jordan on drums. Once, the listener even hears a little chuckle – not musical, but surprisingly sincere. Richards collaborated with other authors, but this is obviously his story, and he delights in telling it.
(Run time: about 7 minutes)
My review of Gus & Me for AudioFile Magazine appears here with a small excerpt. Take a listen!
See all of my reviews at Shelf-employed. Or follow me on Twitter @shelfemployed
Copyright © 2014 L Taylor All Rights Reserved.
Nonfiction Monday Round-up	memoirmusic
Microbes, butterflies, and prairies
An Unusual Butterfly Journey
by Loree Griffin Burns, photographs by Ellen Harasimowicz
Millbrook Press, 2014
Some farms grow vegetables or grains, and some raise cows, sheep, chickens, or pigs. But have you ever heard of a butterfly farm? How do you raise a butterfly?
There are many different kinds of farms out there, but this is the first time I've heard of butterfly farms. It makes a great deal of sense that there would be such a thing, I've just never thought about it before now. This book beautifully describes what happens on a Costa Rican butterfly farm, how the butterflies are raised and transported to places around the world. The photographs do a wonderful job of illustrating what the author is describing. The extra information at the end is very useful, explaining the insect life cycle in terms of butterflies, beetles, and flies. The reference lists and glossary also make this a great resource as well as the link to further resources from the publisher. A wonderful book that would be very useful in teaching children about the life cycle of butterflies.
The World of Microbes
by Nicola Davies, illustrated by Emily Sutton
Find out how the smallest things on the planet do some of the biggest jobs in this intriguing introduction to the world of microbes.
All around the world — in the sea, in the soil, in the air, and in your body — there are living things so tiny that millions could fit on an ant's antenna. They're busy doing all sorts of things, from giving you a cold and making yogurt to eroding mountains and helping to make the air we breathe. If you could see them with your eye, you'd find that they all look different, and that they're really good at changing things into something else and at making many more microbes like themselves! From Nicola Davies comes a first exploration for young readers of the world's tiniest living organisms.
Microbes as organisms too small to be seen by the human eye are a topic that would be hard for children to understand since it's a rather abstract concept. Nicola Davies has done an excellent job in explaining the concept of these small creatures that out number people by many millions to one. The amazing world of microbes comes to life in Davies' words and Sutton's illustrations. There is just enough detail to be interesting without becoming overly complicated for the intended audience. The illustrations show the differences in sizes between different microbes and how different they look as well as how quickly they can multiply. It's a subject that can be rather overwhelming yet made palatable by the excellent work of Davies and Sutton.
PLANT A POCKET OF PRAIRIE
by Phyllis Root, illustrations by Betsy Bowen
University of Minnesota Press, 2014
Author Phyllis Root and illustrator Betsy Bowen last explored the vast, boggy peatlands of northern Minnesota in their book Big Belching Bog. Now, in Plant a Pocket of Prairie, Root and Bowen take young readers on a trip to another of Minnesota's important ecosystems: the prairie.
Once covering almost 40 percent of the United States, native prairie is today one of the most endangered ecosystems in the world. Plant a Pocket of Prairie teaches children how changes in one part of the system affect every other part: when prairie plants are destroyed, the animals who eat those plants and live on or around them are harmed as well. Root shows what happens when we work to restore the prairies, encouraging readers to "plant a pocket of prairie" in their own backyards.
By growing native prairie plants, children can help re-create food and habitat for the many birds, butterflies, and other animals that depend on them. "Plant cup plants," Root suggests. "A thirsty chickadee might come to drink from a tiny leaf pool. Plant goldenrod. A Great Plains toad might flick its tongue at goldenrod soldier beetles." An easy explanation of the history of the prairie, its endangered status, and how to go about growing prairie plants follows, as well as brief descriptions of all the plants and animals mentioned in the story.
With Betsy Bowen's beautiful, airy illustrations capturing the feel of an open prairie and all its inhabitants, readers of all ages will be inspired to start planting seeds and watching for the many fascinating animals their plants attract. What a marvelous transformation could take place if we all planted a pocket of prairie!
While the focus of this book is on Minnesota prairie and wildlife, the principles in regard to building a small prairie in your backyard certainly apply to a lot of other places. I loved how the author and illustrator go through the building of a prairie from one plant and animal to more and more until a full-blown prairie has taken root. I enjoyed hearing about the different animals and plants and how interconnected their lives were. It was sad though to think how little prairie is left in its natural state. This book would make a great teaching tool for teaching about the prairie habitat as well as teaching about environmental repair. A great resource and enjoyable look at the beauty of the natural world that we would do well to take care of before it's gone altogether.
Nonfiction Monday Round-up	Handle with Care: An Unusual Butterfly Journeyscience
Hands-on Science at Its Best: The Kid's Guide to Exploring Nature
November 17, 2014 | Roberta
Today at Wrapped in Foil blog we have a discussion of a 2015 finalist for the AAAS/SubaruSB&F Prize for Excellence in Science Books in the Hands-on Science Book category, The Kid's Guide to Exploring Nature (BBG Guides for a Greener Planet) by a team of educators from the Brooklyn Botanic Garden and illustrated by László Veres.
What a lovely title! First of all, the book is beautifully illustrated. In fact the illustrations are so detailed and complex, it could easily be used as a seek-and-find book for the youngest set.
It is also packed with information. Roughly organized by season, it has hands-on activities, explanations of careers (such as nature educator and field biologist), explanations of habitats, and identification guides.
The Kid's Guide to Exploring Nature is an exciting new book for young nature lovers and scientists. Every nature educator is going to want a copy, as well.
We have a summary of all the AAAS/SubaruSB&F Prize for Excellence in Science Books finalists at Growing With Science blog, many with links to reviews.
Nonfiction Monday Round-up	Brooklyn Botanic GardenHands-on science for kidsnatureScience BooksThe Kid's Guide to Exploring Nature
Strike!: The Farm Workers' Fight for Their Rights
November 17, 2014 | Anastasia Suen
by Larry Dane Brimner (Author)
Booktalk: In 1965, as the grapes in California's Coachella Valley were ready to harvest, migrant Filipino American workers—who picked and readied the crop for shipping—negotiated a wage of $1.40 per hour, the same wage growers had agreed to pay guest workers from Mexico. But when the Filipino grape pickers moved north to Delano, in the Central Valley, and again asked for $1.40 an hour, the growers refused. The ensuing conflict set off one of the longest and most successful strikes in American history.
Snippet: The Delano grape workers wanted better wages. Growers only paid them 90 cents an hour, plus 10 cents a log, or box, of grapes picked. At the end of the day, the average picker earned about $1.20 per hour, while some other farm workers were earning more.
See more booktalks at the Booktalking #kidlit blog.
Tuesday Tucks Me In
November 10, 2014 | bargerj
written by Luis Carlos Montalvan and Bret Witter
photographs by Dan Dion
2014 (Roaring Brook Press)
Source: Mebane Public Library
Look at that cover. How in the world was I going to pass this book up? Tuesday is a service dog for Luis Montalvan, a veteran of the Iraq War. The book is told from Tuesday's point of view. He helps Luis navigate through the day. Luis has been diagnosed with post-traumatic stress disorder (PTSD) and traumatic brain injury (TBI). Luis suffers from nightmares both day and night. Tuesday walks beside Luis to help him walk along the streets of Brooklyn. He helps calm Luis's nerves when there are a lot of people around. Walking down stairs is challenging for Luis as he has difficulty with his balance. Grabbing Tuesday's handle keeps him upright. It's a wonderful relationship between man and beast.
This is a compelling story, but the star of the book is the photographs. Children will love seeing Tuesday with a toothbrush in his mouth or attempting to slurp an ice cream cone. I like the attention this book will bring to service dogs. Before Tuesday Tucks Me In, my experience with service dogs in children's literature was through dry informational texts. This book will also allow young readers to begin to understand that war affects more than the body. Tuesday Tucks Me In celebrates the bond between a service dog and a veteran.
Janet Halfmann's Animal Teachers
Our featured picture book today, Animal Teachers by Janet Halfmann and illustrated by Katy Hudson has been been off to a great start since it was released in September. It has been nominated in the 2014 Cybils elementary/middle grade nonfiction category and has already won the Oppenheim Toy Portfolio 2014 Gold Best award.
Have you ever wondered how baby animals learn? Are they born knowing everything they need to survive or do they learn from their parents and peers like we humans do? Animal Teachers will answer those questions with some surprising examples. The reader finds out that chicks need to be taught what proper food is and young cheetahs need to be taught how to run. Who would have guessed?
For the text Janet has used what could be described as a "reverse Q & A." Each two-page spread first tells what a baby animal learns from its parents and then asks a few questions about similar things humans might do, drawing the reader in. It is a wonderful way to inspire conversations and deeper understanding.
You will definitely want to share this fascinating and sweet book with the animal lovers. It would also be a great resource for the classroom, allowing students to discover new things about animals and learning.
If you would like to learn more, try our full review at Wrapped in Foil, as well as some related information and activities at Growing With Science.
Nonfiction Monday Round-up	Animal TeachersJanet Halfmann


text:
Q: Implement back navigation with DrawerLayout and AppCompatActivity without DrawerToggle I've been searching and thinking how to implement the desire behaviour that I want when switching back and forth between fragments when using a Navigation Drawer. Actually the documentation says:
When using fragments in your app, individual FragmentTransaction objects may represent context changes that should be added to the back stack. For example, if you are implementing a master/detail flow on a handset by swapping out fragments, you should ensure that pressing the Back button on a detail screen returns the user to the master screen
So in my app I have a MainActivity that controls everything and a navigation layout within you can change between predefined options. This is the view when you launch the app

When you click on an item in the navigation drawer it opens a new fragment that replaces the main_content as follows:

At this point the behaviour is the correct one, so if you want to change the options you need to open the navigation drawer again to toggle between the menu options.
This is the Main Activity (notice that there is no Drawer Toggle)
MainActivity.java
public class MainActivity extends AppCompatActivity {

ActionBarDrawerToggle mDrawerToggle;
private DrawerLayout drawerLayout;

protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);

        mSnackBarView = findViewById(R.id.myCoordinatorLayout);

        setToolbar(); // Set Toolbar como action bar

        if (savedInstanceState != null) {
            mCurrentSelectedPosition = savedInstanceState.getInt(STATE_SELECTED_POSITION);
            mFromSavedInstanceState = true;
        }
        drawerLayout = (DrawerLayout) findViewById(R.id.drawer_layout);
        NavigationView navigationView = (NavigationView) findViewById(R.id.nav_view);
        if (navigationView != null) {
            setupDrawerContent(navigationView);
        }

        drawerTitle = getResources().getString(R.string.app_name);
        if (savedInstanceState == null) {
            selectItem(drawerTitle, mCurrentSelectedPosition);
        }

    }

    private void setToolbar() {
        Toolbar toolbar = (Toolbar) findViewById(R.id.toolbar);
        setSupportActionBar(toolbar);
        final ActionBar ab = getSupportActionBar();
        if (ab != null) {
            // Poner ícono del drawer toggle
            ab.setHomeAsUpIndicator(R.drawable.ic_menu);
            ab.setDisplayHomeAsUpEnabled(true);

        }
    }

    private void setupDrawerContent(final NavigationView navigationView) {
        navigationView.setNavigationItemSelectedListener(
                new NavigationView.OnNavigationItemSelectedListener() {
                    @Override
                    public boolean onNavigationItemSelected(MenuItem menuItem) {
                        // Marcar item presionado
                        menuItem.setChecked(true);
                        // Crear nuevo fragmento
                        String title = menuItem.getTitle().toString();
                        int id = menuItem.getItemId();
                        selectItem(title, id);
                        return true;
                    }
                }
        );
    }

    @Override
    public boolean onCreateOptionsMenu(Menu menu) {
        if (!drawerLayout.isDrawerOpen(GravityCompat.START)) {
            getMenuInflater().inflate(R.menu.menu_main, menu);
            return true;
        }
        return super.onCreateOptionsMenu(menu);
    }

    @Override
    public boolean onOptionsItemSelected(MenuItem item) {
        switch (item.getItemId()) {
            case android.R.id.home:
                drawerLayout.openDrawer(GravityCompat.START);
                return true;
        }
        return super.onOptionsItemSelected(item);
    }

    private void selectItem(String title, int id) {
        Bundle args = new Bundle();
        args.putString(PlaceholderFragment.ARG_SECTION_TITLE, title);

        Fragment fragment = PlaceholderFragment.newInstance(title);
        fragment.setArguments(args);
        FragmentManager fragmentManager = getSupportFragmentManager();
        fragmentManager
                .beginTransaction()
                .replace(R.id.main_content, fragment)
                .commit();

        switch (id) {
            case R.id.nav_localizacion:
                //Snackbar.make(mSnackBarView, R.string.menu_localization, Snackbar.LENGTH_SHORT).show();
                mCurrentSelectedPosition = 0;
                LocalizacionFragment fragment_localizacion = new LocalizacionFragment();
//                fragmentManager = getSupportFragmentManager();
                Snackbar.make(mSnackBarView, R.string.menu_localization, Snackbar.LENGTH_SHORT).show();
                fragmentManager
                        .beginTransaction()
                        .replace(R.id.main_content, fragment_localizacion)
                        .commit();
                break;
            case R.id.nav_productos:
                Snackbar.make(mSnackBarView, R.string.menu_productos, Snackbar.LENGTH_SHORT).show();
                mCurrentSelectedPosition = 1;
                fragmentManager
                        .beginTransaction()
                        .replace(R.id.main_content, fragment)
                        .commit();
                break;
            case R.id.nav_consejos:
                Snackbar.make(mSnackBarView, R.string.menu_consejos, Snackbar.LENGTH_SHORT).show();
                mCurrentSelectedPosition = 3;
                ConsejosFragment fragment_consejo = new ConsejosFragment();
                fragmentManager
                        .beginTransaction()
                        .replace(R.id.main_content, fragment_consejo)
                        .commit();
                break;
            default:
                break;
        }

        drawerLayout.closeDrawers(); // Cerrar drawer

        setTitle(title); // título actual

    }

}

I don't use the hamburger icon because it hides under the Navigation Layout, but here is the thing. When you click on "Recetas" in the recycle view it open a new fragment (replace) but now I want to show the up caret icon and give proper back navigation to my app.
Here is the code of the "Consejos" fragment class
Consejos.java
public class ConsejosFragment extends Fragment {

    RecyclerView mRecycler;
    ConsejosAdapter mAdapter;
    FragmentActivity mActivity;

    @Override
    public void onAttach(Activity activity) {
        super.onAttach(activity);
        this.mActivity = (FragmentActivity) activity;
        setRetainInstance(true);
    }

    @Override
    public void onActivityCreated(Bundle savedInstanceState) {
        super.onActivityCreated(savedInstanceState);
        setHasOptionsMenu(true);

    }

    @Override
    public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState) {

        List items = new ArrayList();
        items.add(new ConsejosInfo("Recetas", R.drawable.icon_recetas));

        /* Inflamos el layout */
        View v = inflater.inflate(R.layout.consejos_layout_recycler, container, false);
        /* Obtenemos el Recycle */
        mRecycler = (RecyclerView) v.findViewById(R.id.recycler_consejos);

        /* Creamos el adaptador */
        mAdapter = new ConsejosAdapter(mActivity, items);

        /* Set click en adapter */

        return v;
    }

    @Override
    public void onViewCreated(View view, Bundle savedInstanceState) {
        super.onViewCreated(view, savedInstanceState);
        mRecycler.setHasFixedSize(true);

        mRecycler.setLayoutManager(new LinearLayoutManager(getActivity()));

        mRecycler.setAdapter(mAdapter);
    }
}

And this is the adapter that the Recyclerview use and handles the click on the item inside:
Adapter.java
public class ConsejosAdapter extends RecyclerView.Adapter<ConsejosAdapter.ConsejosViewHolder> {

    private List<ConsejosInfo> _items = new ArrayList<ConsejosInfo>();
    private final FragmentActivity mActivity;
    private Context context;

    public ConsejosAdapter(FragmentActivity mActivity, List<ConsejosInfo> items) {
        this._items = items;
        this.mActivity = mActivity;
    }

    @Override
    public int getItemCount() {
        return _items.size();
    }

    @Override
    public ConsejosViewHolder onCreateViewHolder(ViewGroup viewGroup, int viewType) {
        View v = LayoutInflater.from(viewGroup.getContext()).inflate(R.layout.cardview_consejos, viewGroup, false);
        return new ConsejosViewHolder(v);
    }

    @Override
    public void onBindViewHolder(ConsejosViewHolder viewHolder, int position) {
        viewHolder.imagen.setImageResource(_items.get(position).get_imagen());
        viewHolder.nombre.setText(_items.get(position).get_nombre());
    }

    public class ConsejosViewHolder extends RecyclerView.ViewHolder implements View.OnClickListener {

        public ImageView imagen;
        public TextView nombre;

        public ConsejosViewHolder(final View itemView) {
            super(itemView);
            imagen = (ImageView) itemView.findViewById(R.id.consejos_imagen);
            nombre = (TextView) itemView.findViewById(R.id.consejos_nombre);
            itemView.setOnClickListener(this);
        }

        @Override
        public void onClick(View v) {
            RecetasFragment recetasFragment = new RecetasFragment();
            FragmentManager fragmentManager = mActivity.getSupportFragmentManager();
            fragmentManager
                    .beginTransaction()
                    .replace(R.id.main_content, recetasFragment)
                    .addToBackStack(null)
                    .commit();

        }
    }
}

In the Consejos Adapter I added to backstack the fragment before calling the new fragment (inside fragment) and that changes the behaviour of the back button, so before this if you press the back button is closes the app but now it takes you to the Consejos fragment, but now I want to add the up caret and make the exactly same operation as when you click the back button but I don't know how to achieve it.
Please feel free to ask for more code
Thank you very much

A: I will give you the full answer:
We will use the interfaces to communicate with the MainActivity from RecetasFragment so we can enable and disable navigation back button.
RecetasFragment class:
import android.app.Activity;
import android.os.Bundle;
import android.support.v4.app.Fragment;
import android.view.LayoutInflater;
import android.view.View;
import android.view.ViewGroup;

/**
 * Created by hema on 8/19/2015.
 */
public class RecetasFragment extends Fragment {

    private CommunicateWithActivity mWithActivity;

    public RecetasFragment() {
    }

    @Override
    public void onAttach(Activity activity) {
        super.onAttach(activity);

        try {

            mWithActivity = (CommunicateWithActivity) activity; // MainActivity must implement CommunicateWithActivity interface.

            mWithActivity.enableNavigationBack(true); // We enable navigation back here.

        } catch (ClassCastException e) {

            throw new ClassCastException(activity.toString()
                    + " must implement CommunicateWithActivity");

        }

    }

    @Override
    public View onCreateView(LayoutInflater inflater, ViewGroup container,
                             Bundle savedInstanceState) {
        return inflater.inflate(R.layout.fragment_recetas, container, false);
    }

    @Override
    public void onDestroy() {
        super.onDestroy();

        mWithActivity.enableNavigationBack(false); // We disable navigation back here, restore menu icon.

    }

    public interface CommunicateWithActivity {

        void enableNavigationBack(boolean enable);// we will implement this method in MainActivity class.

    }

}

And in MainActivity class we will implement CommunicateWithActivity interface, Modify your MainActivity class (only required modifications):
import android.support.v7.app.AppCompatActivity;
import android.os.Bundle;
import android.support.v7.widget.Toolbar;
import android.view.Menu;
import android.view.MenuItem;
import android.view.View;

public class MainActivity extends AppCompatActivity implements RecetasFragment.CommunicateWithActivity {

    private Toolbar mToolbar;
    private DrawerLayout drawerLayout;

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);

        mToolbar = (Toolbar) findViewById(R.id.toolbar);
        setSupportActionBar(mToolbar);

        mToolbar.setNavigationIcon(R.drawable.ic_menu);

        mSnackBarView = findViewById(R.id.myCoordinatorLayout);

        if (savedInstanceState != null) {
            mCurrentSelectedPosition = savedInstanceState.getInt(STATE_SELECTED_POSITION);
            mFromSavedInstanceState = true;
        }
        drawerLayout = (DrawerLayout) findViewById(R.id.drawer_layout);
        NavigationView navigationView = (NavigationView) findViewById(R.id.nav_view);
        if (navigationView != null) {
            setupDrawerContent(navigationView);
        }

        drawerTitle = getResources().getString(R.string.app_name);
        if (savedInstanceState == null) {
            selectItem(drawerTitle, mCurrentSelectedPosition);
        }

    }

    private void setupDrawerContent(final NavigationView navigationView) {
        navigationView.setNavigationItemSelectedListener(
                new NavigationView.OnNavigationItemSelectedListener() {
                    @Override
                    public boolean onNavigationItemSelected(MenuItem menuItem) {
                        // Marcar item presionado
                        menuItem.setChecked(true);
                        // Crear nuevo fragmento
                        String title = menuItem.getTitle().toString();
                        int id = menuItem.getItemId();
                        selectItem(title, id);
                        return true;
                    }
                }
        );
    }

    @Override
    public boolean onCreateOptionsMenu(Menu menu) {
        if (!drawerLayout.isDrawerOpen(GravityCompat.START)) {
            getMenuInflater().inflate(R.menu.menu_main, menu);
            return true;
        }
        return super.onCreateOptionsMenu(menu);
    }

    @Override
    public boolean onOptionsItemSelected(MenuItem item) {
        switch (item.getItemId()) {
            case android.R.id.home:
                drawerLayout.openDrawer(GravityCompat.START);
                return true;
        }
        return super.onOptionsItemSelected(item);
    }

    private void selectItem(String title, int id) {
        Bundle args = new Bundle();
        args.putString(PlaceholderFragment.ARG_SECTION_TITLE, title);

        Fragment fragment = PlaceholderFragment.newInstance(title);
        fragment.setArguments(args);
        FragmentManager fragmentManager = getSupportFragmentManager();
        fragmentManager
                .beginTransaction()
                .replace(R.id.main_content, fragment)
                .commit();

        switch (id) {
            case R.id.nav_localizacion:
                //Snackbar.make(mSnackBarView, R.string.menu_localization, Snackbar.LENGTH_SHORT).show();
                mCurrentSelectedPosition = 0;
                LocalizacionFragment fragment_localizacion = new LocalizacionFragment();
//                fragmentManager = getSupportFragmentManager();
                Snackbar.make(mSnackBarView, R.string.menu_localization, Snackbar.LENGTH_SHORT).show();
                fragmentManager
                        .beginTransaction()
                        .replace(R.id.main_content, fragment_localizacion)
                        .commit();
                break;
            case R.id.nav_productos:
                Snackbar.make(mSnackBarView, R.string.menu_productos, Snackbar.LENGTH_SHORT).show();
                mCurrentSelectedPosition = 1;
                fragmentManager
                        .beginTransaction()
                        .replace(R.id.main_content, fragment)
                        .commit();
                break;
            case R.id.nav_consejos:
                Snackbar.make(mSnackBarView, R.string.menu_consejos, Snackbar.LENGTH_SHORT).show();
                mCurrentSelectedPosition = 3;
                ConsejosFragment fragment_consejo = new ConsejosFragment();
                fragmentManager
                        .beginTransaction()
                        .replace(R.id.main_content, fragment_consejo)
                        .commit();
                break;
            default:
                break;
        }

        drawerLayout.closeDrawers(); // Cerrar drawer

        setTitle(title); // título actual

    }

    @Override
    public void enableNavigationBack(boolean enable) {

        if(enable) {

            // We enable the navigation back button here

            mToolbar.setNavigationIcon(R.drawable.abc_ic_ab_back_mtrl_am_alpha);

            mToolbar.setNavigationOnClickListener(new View.OnClickListener() {
                @Override
                public void onClick(View v) {

                    // On navigation back button clicked.

                    if(getSupportFragmentManager().getBackStackEntryCount() > 0) { // Check if there is fragments in BackStack.

                        getSupportFragmentManager().popBackStack(); // PopBackStack.

                    } else {

                        // You can implement this part as you want.

                        return;

                    }

                }
            });

        } else {

            mToolbar.setNavigationIcon(R.drawable.ic_menu);

            mToolbar.setNavigationOnClickListener(new View.OnClickListener() {
                 @Override
                 public void onClick(View v) {
                     if(drawerLayout != null) {
                         drawerLayout.openDrawer(Gravity.LEFT);
                     }
                 }
            });

        }

    }
}

Please let me know if there is something you don't understand.

redpajama_set_name: RedPajamaStackExchange


Millions of Moments of Pride!
Mark your calendar now…you won't want to miss a moment of San Luis Obispo County's 23rd annual pride celebration.
The Gay and Lesbian Alliance of the Central Coast is in constant need of volunteer support and financial contribution.


text:
\section{Introduction}

Symmetric orbifold CFTs are widely used in AdS$_3$/CFT$_2$ as boundary theories to understand the bulk physics \cite{Maldacena:1997re,Strominger:1996sh,Maldacena:1999bp,Seiberg:1999xz,Dijkgraaf:1998gf,Larsen:1999uk,Jevicki:1998bm,deBoer:1998kjm}. Consider a seed CFT with target space $M$. The symmetric orbifold CFT has the target space
\be
M^N/S_N
\ee
where $N$ is the number of copies of the seed CFT and $S_N$ is the permutation group of $N$ elements.

Due to the $S_N$ orbifold, there exist twist operators. If we circle the insertion of a twist operator, different copies of the seed CFT permute into each other. The traditional way to compute correlation functions of twist operators is to map the 2-d base space to the covering space \cite{Lunin:2000yv,Lunin:2001pw,Pakman:2009zz,Pakman:2009ab}. On this covering space, the ramification caused by the twist operator in the base space is resolved, such that the target space becomes a single copy of $M$.
The correlation functions on the base space can be computed from a combination of the correlation functions in the covering space and a Liouville factor which takes into account the covering map.

The effect of a twist operator can be studied by using the covering map \cite{Avery:2010er,Avery:2010hs,Avery:2010qw,Carson:2014yxa,Carson:2014ena}.
To be specific, take two copies of the seed CFT of a free boson, such that $M=R$. Suppose around $z=0$ that the state is in the untwisted sector as shown in fig \ref{fig}. Around this point, there are two separate copies of the seed CFT. Let us put a twist operator $\sigma_2$ at $z_0$, which is the unique twist operator for two copies. This twist operator generates a branch cut from $z_0$ to infinity. The two separate copies join into a single doubly wound copy for $|z|>|z_0|$. If an initial state is given at $z=0$ for the two separate copies, what is the state after the twist operator, e.g. the state at infinity? The result of this question tells us all the three-point functions involving a twist operator $\sigma_2$.

\begin{figure}
\centering
        \includegraphics[width=9.5cm]{fig_twist.pdf}
\caption{The effect of a twist operator. The red and green circles represent the states living on two separate copies located before the twist operator. The twist operator produces a branch cut from $z_0$ to infinity. The dashed circle represents the state living on the joined copy with two sheets.}
\label{fig}
\end{figure}

It has been observed that the effect of a twist operator is in the form of a Bogoliubov transformation between the modes before and after the twist operator. From the covering space method point of view, the map from the base space to the covering space leads to this linear transformation of modes. The effect of a twist operator is encoded in this linear transformation. This can be derived from the covering map or by matching the modes just before and after the twist \cite{Carson:2014xwa,Carson:2017byr}. In the covering map method, it seems that the details of the map are necessary. In the latter method, the covering map is not needed but in practice not all effects can be obtained since it requires one to invert an infinite-dimensional matrix.

In this paper, we will develop a method that does not involve the covering map and can obtain the effects completely. To do that, we will use the `weak' Bogoliubov ansatz\footnote{ `Weak' means that we don't require some relations among the coefficients in the original Bogoliubov ansatz. For more details, see section \ref{sec B}. These relations come out naturally. As a result, this method does not involve the inversion of infinite-dimensional matrices.}
and conformal symmetry
\be
\text{Weak Bogoliubov ansatz} ~+~ \text{Conformal symmetry}~ \Rightarrow ~\text{Effect of a twist operator}
\ee
For earlier work in this direction see \cite{Burrington:2014yia} which incorporates the use of the covering map and conformal symmetry.
There have also been works to compute correlation functions of twist operators using conformal symmetry, see e.g. \cite{Dixon:1986qv,Arutyunov:1997gt,Arutyunov:1997gi,Jevicki:1998bm}. For recent work, see \cite{Dei:2019iym}.

The plan of the paper is as follows: In section \ref{sec mode} we outline the orbifold CFT of one boson. In section \ref{sec ansatz} we describe the effect of the twist operator. In section \ref{sec B} we discuss the Bogoliubov transformation. In section \ref{bootstrap} we bootstrap the effect of the twist operator using the weak Bogoliubov ansatz and conformal symmetry. In section \ref{discussion} we discuss our results and future work.

\section{Orbifold CFT of one boson}\label{sec mode}

Symmetric orbifold CFTs are obtained by orbifolding $N$ copies of a seed CFT by the permutation group $S_{N}$, which results in the target space
\be
M^{N}/S_N
\ee
where $M$ is the target space of the seed CFT.
In this paper, we will consider the simplest case where the seed CFT is a free boson with target space $M=R$.
The base space is the complex $z$ plane.

The $N$ copies of the free boson are labeled by $X^{(i)}$ with $i=1,\dots,N$. In the untwisted sector, the fields have the boundary condition
\begin{eqnarray}\displaystyle\label{untwist b}
X^{(i)}\to X^{(i)},~~~~z\to z e^{2\pi i}
\end{eqnarray}
There also exist twist sectors where the $N$ copies can join into many linked copies in all possible ways. For example, a $k$-wound linked copy has the boundary condition
\begin{eqnarray}\displaystyle
X^{(1)}\to X^{(2)}\to\ldots\to X^{(k)}\to X^{(1)},~~~~z\to z e^{2\pi i}
\end{eqnarray}
It is convenient to define a single field $X$ living on the $k$-wound copy. On the $i$-th segment of the $k$-wound copy, the field $X$ equals to $X^{(i)}$, such that the field $X$ has the boundary condition
\be
X \to X,~~~~z\to z e^{2\pi k i}
\ee
Notice that the field $X$ is multi-valued in the base space and should be thought of as a single-valued field living on a Riemann surface with $k$ sheets.

In radial quantization, the modes of the holomorphic part in the untwisted sector are defined as
\be
\a^{(i)}_{n} = {1\over2\pi}\oint_{C_0}dz \, z^{n} \partial X^{(i)} (z)
\ee
where $C_0$ is a contour centered around $z=0$.
The $n$ is an integer as required by the boundary condition (\ref{untwist b}).
The commutation relation is
\be\label{com untwist}
[\a^{(i)}_{m},\a^{(j)}_{n}]=m\delta^{ij}\d_{m+n,0}
\ee
We also define
\be
\a^{(i)\dagger}_n = \a^{(i)}_{-n}
\ee
The vacuum $|0\rangle^{(i)}$ of copy $i$ is defined by the condition
\be\label{v untwist}
\a^{(i)}_{n}|0\rangle^{(i)}=0, ~~~~n\geq 0
\ee
The Virasoro generators can be expanded in terms of a sum over bilinears of the modes
\be\label{L untwist}
L_{m} = {1\over2}\sum_{i}\sum_n\a^{(i)}_{n}\a^{(i)}_{m-n}
\ee
with implicit normal-ordering.
Using the commutation relation (\ref{com untwist}), we have
\be\label{4-9}
[L_m,\a^{(i)}_n]=-n \a^{(i)}_{m+n}
\ee

For the field $X$ living on the $k$-wound copy, the modes are defined as
\be
\a_{n} = {1\over2\pi}\oint_{C_0^{(2\pi k)}}dz \, z^{n} \partial X (z)
\ee
where the contour of the integral $C_0^{(2\pi k)}$ is again centered around $z=0$ but now from angle $0$ to $2\pi k$. The boundary condition for the field $X$ requires that $n=m/k$ where $m$ is an integer. The commutation relation is given by
\be\label{comm twist}
[\a_{m},\a_{n}]=k m\d_{m+n,0}
\ee
We also define
\be
\a^{\dagger}_n = \a_{-n}
\ee
The vacuum $|0^k\rangle$ of the $k$-wound copy is defined by the condition
\be
\a_{n}|0^k\rangle = 0, ~~~~~~~n\geq 0
\ee
The Virasoro generators can be expanded in terms of a sum over bilinears of modes
\be\label{4-7}
L_{m} = {1\over2k}\sum_n\a_{n}\a_{m-n}
\ee
again with implicit normal-ordering.
Using the commutation relation (\ref{comm twist}), we have
\be\label{L com}
[L_m,\a_n]=-n \a_{m+n}
\ee

The $k$-wound copy can be produced by applying the twist operator $\sigma_k$ to the untwisted sector. The twist operator $\sigma_k$ has dimension \cite{Lunin:2000yv}
\be\label{h}
h(\sigma_k)=\frac{c}{24}(k-\frac{1}{k})
\ee
For a single free boson, $c=1$.

\section{The effect of a twist operator}\label{sec ansatz}

In this section, we will briefly review the effect of a twist operator. In this paper, we restrict ourselves to the simplest case where there are only two copies of the seed CFT, such that $N=2$.
Suppose at $z=0$ an initial state in the untwisted sector is given by
\be\label{i}
\a^{(i_1)}_{-n_1}\a^{(i_2)}_{-n_2}\dots \a^{(i_m)}_{-n_m}|0\rangle^{(1)}|0\rangle^{(2)}
\ee
where $n_k>0$ and $i_k=1,2$ is the copy label. Let us apply the twist operator $\sigma_2$ at $z_0$. The question is to find out the state $\phi$ at $|z|>|z_0|$ which is after the twist operator. The state $\phi$ is defined as
\be
|\phi\rangle=\sigma_2(z_0)\a^{(i_1)}_{-n_1}\a^{(i_2)}_{-n_2}\dots \a^{(i_k)}_{-n_k}|0\rangle^{(1)}|0\rangle^{(2)}
\ee
which lives on a doubly wound copy since the twist operator has joined the two singly wound copies in the initial state.
This question has been addressed completely by the covering map method, which will be reproduced for a single boson in appendix \ref{App cover}. The effect of a twist operator can be summarized by the following three basic rules:

\bigskip

(i) Contraction: Two modes $\a^{(i)}_{-m}$ and $\a^{(j)}_{-n}$ in the initial state (\ref{i}) can `Wick contract', giving a number
\be
C^{ij}[m,n] \equiv C[\a^{(i)}_{-m}\a^{(j)}_{-n}]
\ee
For the process of Wick contraction, we consider all possible pairs of modes. For each such pair, we get a term where the pair contracts to the above number, and a term where the pair does not contract but will pass through the twist as shown in step (ii) below.

\bigskip

(ii) Propagation: Any modes left after the contraction will pass through the twist and become modes after the twist operator, which are modes on the doubly wound copy.
\be\label{p rule}
\alpha^{(i)}_{-n}~~\longrightarrow~~  \sum_{p>0}f_i[-n,-p] \alpha_{-p},~~~~~i=1,2
\ee
where $\alpha_{-p}$ is a mode on the doubly wound copy and $p=m/2$ where $m$ is a positive integer. In section \ref{sec global}, we will show that the nontrivial $f_i$'s are
\begin{eqnarray}\displaystyle\label{f mode}
f_i[-n,-n]&=&1/2 \nn
f_i[-n,-p]&\neq&  0,~~~~~\text{when}~~~ p\neq n ~~~\text{and}~~~ p ~\text{is a positive half integer}
\end{eqnarray}

\bigskip

(iii) Pair creation: After the previous two steps, the modes in the initial state have either been contracted or passed through the twist. We are left with the twist operator acting on the untwisted vacuum
\begin{eqnarray}\displaystyle\label{chi}
|\chi\rangle \equiv \s_2(z_0) |0\rangle^{(1)}|0\rangle^{(2)}=\text{exp}\big(\sum_{m,n>0}\gamma_{mn}\a_{-m}\a_{-n}\big)|0^2\rangle
\end{eqnarray}
where the dimension of the twist operator $\s_2$ is (\ref{h})
\be
h=h(\sigma_2)=1/16
\ee
which takes into account the difference of dimensions between the vacuum of doubly wound copy $|0^2\rangle$ and the vacuum of two singly wound copies $|0\rangle^{(1)}|0\rangle^{(2)}$.
In section \ref{sec global}, we will show that
\be\label{gamma mode}
\gamma_{mn}\neq 0 ~~~~~ \text{only if $m,n$ are positive half integers}
\ee

\bigskip

\begin{figure}
\centering
        \includegraphics[width=10cm]{fig_twist1.pdf}
\caption{The three basic rules to compute the effect of a twist operator.}
\label{fig_1}
\end{figure}

The above rules are shown in fig.\ref{fig_1}.
To better understand these rules, let us consider an example with one initial mode
\begin{eqnarray}\displaystyle\label{one}
\sigma_2(z_0)\a^{(i)}_{-n}|0\rangle^{(1)}|0\rangle^{(2)}
= \sum_{p>0}f_i[-n,-p]\a_{-p}
\, \text{exp}\big(\sum_{m,n>0}\gamma_{mn}\a_{-m}\a_{-n}\big)|0^2\rangle
\end{eqnarray}
We first use the propagation rule (\ref{p rule}) to pass the initial mode through the twist. Then the twist operator acts on the untwisted vacuum to produce pairs using the rule (\ref{chi}).
Let us now consider an example of two initial modes
\begin{eqnarray}\displaystyle\label{two}
\sigma_2(z_0)\a^{(i)}_{-n_1}\a^{(j)}_{-n_2}|0\rangle^{(1)}|0\rangle^{(2)}
\!\!&=&\!\! \Big(\sum_{p_1>0}f_i[-n_1,-p_1]\a_{-p_1}\sum_{p_2>0}f_j[-n_2,-p_2]\a_{-p_2}+C^{ij}[n_1,n_2]\Big)\nn &&~\times\text{exp}\big(\sum_{m,n>0}\gamma_{mn}\a_{-m}\a_{-n}\big)|0^2\rangle
\end{eqnarray}
The first term in the parentheses comes from the propagation of the two initial modes while the second term is from the contraction. The exponent in the last line comes from the pair creation.

If the effect of an operator satisfies the above three rules but with independent and undetermined coefficients
$f_i$, $C^{ij}$, and $\gamma$, we call it the weak Bogoliubov form. As will be explained in the next section, in the `normal' Bogoliubov transformation these coefficients are not independent of each other.

\section{Bogoliubov transformation}\label{sec B}

The Bogoliubov transformation is a linear transformation that mixes creation and annihilation operators
\begin{eqnarray}\displaystyle
\hat a&=&\a\,\hat b + \beta\,\hat b^\dagger \nn
\hat a^\dagger&=&\a^*\,\hat b^\dagger + \beta^*\,\hat b
\end{eqnarray}
where $|\a|^2-|\beta|^2 = 1$ to make both sets of operators canonical
\be
[\hat a,\hat a^\dagger]= 1,~~~~~~[\hat b,\hat b^\dagger]= 1
\ee
The `$a$' vacuum and `$b$' vacuum are defined by the conditions
\be
\hat a|0\rangle_a =0,~~~~~\hat b|0\rangle_b =0
\ee
Since the Bogoliubov transformation mixes the creation and annihilation operators, the `$a$' vacuum is no longer the `$b$' vacuum
\be
\hat a |0\rangle_a = (\a\,\hat b + \beta\,\hat b^\dagger) |0\rangle_a = 0
~~\longrightarrow~~ |0\rangle_a =  e^{\frac{1}{2}\gamma\, \hat b^\dagger \hat b^\dagger} |0\rangle_b
\ee
where
\be\label{b r}
\gamma = -\frac{\beta}{\alpha}
\ee
Consider an example with two initial modes on the `$a$' vacuum
\begin{eqnarray}\displaystyle
\hat a^\dagger \hat a^\dagger |0\rangle_a
&=& (\a^*\,\hat b^\dagger + \beta^*\,\hat b)(\a^*\,\hat b^\dagger + \beta^*\,\hat b)  e^{\frac{1}{2}\gamma \,\hat b^\dagger \hat b^\dagger} |0\rangle_b\nn
&=& \left[(\a^* + \beta^*\,\gamma)\hat b^\dagger (\a^* + \beta^*\,\gamma)\hat b^\dagger + \beta^*\alpha^* \right]  e^{\frac{1}{2}\gamma\, \hat b^\dagger \hat b^\dagger} |0\rangle_b
\end{eqnarray}
It is similar to the effect of a twist operator with two initial modes (\ref{two}).
The first term in the bracket comes from the propagation of the two initial modes while the second term is from their contraction. The exponent in the last line comes from the pair creation. Similarly, we can compute the case with multiple initial modes
\be
\hat a^\dagger\dots \hat a^\dagger|0\rangle_a
\ee
using the following three rules that are similar to the rules in section \ref{sec ansatz}.

\bigskip

(i) Contraction:
\be
C[\hat a^\dagger,\hat a^\dagger] = \beta^*\alpha^*
\ee

\bigskip

(ii) Propagation:

\be
\hat a^\dagger~ \rightarrow~ (\a^* + \beta^*\,\gamma)\hat b^\dagger
\ee

\bigskip

(iii) Pair creation:

\be
\hat a |0\rangle_a = e^{\frac{1}{2}\gamma\, \hat b^\dagger \hat b^\dagger} |0\rangle_b
\ee

\bigskip

The coefficients in these expressions are not independent of each other. They are related through (\ref{b r}).
We will call the above three rules with the constraint (\ref{b r}) the `normal' Bogoliubov ansatz and the same rules without the constraint the `weak' Bogoliubov ansatz.

In \cite{Carson:2014xwa}, it has been shown that the effect of a twist operator explained in section \ref{sec ansatz} satisfies the `normal' Bogoliubov ansatz with the generalization: the creation and annihilation operators are generalized to infinite-dimensional vectors, i.e. $\hat a$ becomes $\hat a_i$; the coefficients $\a$, $\beta$, $\gamma$ are generalized to infinite-dimensional matrices, i.e. $\a$ becomes $\a_{ij}$. In \cite{Carson:2014xwa}, by matching the modes before and after the twist, they obtained the matrices $\a$ and $\beta$. However, there is no systematic way to invert the infinite-dimensional matrices $\a$ to obtain $\gamma$ through the constraint (\ref{b r}).
In this paper, we will take the weak Bogoliubov ansatz which is without the constraint.
We will show that it is enough to determine all the coefficients without inverting any infinite-dimensional matrices. The constraint (\ref{b r}) is satisfied automatically.

\section{Bootstrapping the effect of a twist operator}\label{bootstrap}

In this section, we will take the weak Bogoliubov ansatz and apply the Virasoro generators to obtain recursion relations for all the coefficients in the ansatz. The solutions to these recursion relations match the results from the covering map method.

\subsection{Global modes}\label{sec global}

In this section we will show (\ref{f mode}) and (\ref{gamma mode}). Let us start with
\begin{eqnarray}\displaystyle\label{global 1}
\alpha_{n}\sigma_2(z_0) - \sigma_2(z_0) (\alpha_{n}^{(1)}+\alpha_{n}^{(2)})
&=& \frac{1}{2\pi}\oint_{C_{z_0}^{(4\pi)}} dz~z^{n}\partial X(z)\sigma_2(z_0)\nn
&=& z_0^{n} \frac{1}{2\pi}\oint_{C_{z_0}^{(4\pi)}} dz~\big(1+(z-z_0)/z_0\big)^{n}\partial X(z)\sigma_2(z_0)\nn
&=& z_0^{n} \big(\alpha^{(z_0)}_0+n\alpha^{(z_0)}_{1}/z_0+\dots\big)\sigma_2(z_0)
\end{eqnarray}
where $n$ is an integer.
To obtain the first equality, we join the two contours, one before and one after the twist operator, into a single contour centered around the twist operator.
We note that modes with superscript $(z_0)$, i.e. $\alpha^{(z_0)}_0$, are the modes centered around $z_0$, while modes without it, i.e. $\alpha_{n}$ and $\alpha_{n}^{(i)}$, are the modes centered around $z=0$. Since the twist operator $\sigma_2$ is defined as the lowest dimension operator that produces the twisted sector, we have
\be
\alpha^{(z_0)}_{n}\sigma_2(z_0)=0,~~~n\geq 0
\ee
Thus (\ref{global 1}) becomes
\be
\alpha_{n}\sigma_2(z_0) - \sigma_2(z_0) (\alpha_{n}^{(1)}+\alpha_{n}^{(2)}) = 0
\ee

Consider the following state where $m$ and $n$ are positive integers
\begin{eqnarray}\displaystyle
\langle 0^2|\alpha_{m}\sigma_2(z_0) \alpha_{-n}^{(i)}|0\rangle^{(1)}|0\rangle^{(2)}
&=&\langle 0^2|\sigma_2(z_0)(\alpha^{(1)}_{m}+\alpha^{(2)}_{m}) \alpha_{-n}^{(i)}|0\rangle^{(1)}|0\rangle^{(2)}\nn
&=&\delta_{mn}n
\end{eqnarray}
which is nonzero only if $m$ and $n$ are equal.
From $\langle 0^2|\alpha_{n} \alpha_{-n}|0^2\rangle = 2n $, we find
\be
\sigma_2(z_0) \alpha_{-n}^{(i)}|0\rangle^{(1)}|0\rangle^{(2)} = \left(\frac{1}{2} \alpha_{-n}+ \text{half integer modes}\right)|0^2\rangle
\ee
which is (\ref{f mode}). To show (\ref{gamma mode}), consider the following state where $m$ is a positive integer
\be
\alpha_{m} |\chi\rangle=
\alpha_{m} \sigma_2(z_0) |0\rangle^{(1)}|0\rangle^{(2)}
=\sigma_2(z_0)(\alpha^{(1)}_{m}+\alpha^{(2)}_{m}) |0\rangle^{(1)}|0\rangle^{(2)} =0
\ee
Thus the state from the pair creation $\chi$ can not have any excitations of integer modes, which is stated in (\ref{gamma mode}).

\subsection{Pair creation}

In this subsection, we will derive the pair creation coefficients $\gamma_{m n}$.

\subsubsection{Relations from $L_{-1}$}

Starting with
\be
L_{-1}|0\rangle^{(1)}|0\rangle^{(2)} = 0
\ee
where we have used (\ref{v untwist}) and (\ref{L untwist}), we have
\begin{eqnarray}\displaystyle\label{4-5}
0&=&\s_2(z_0)L_{-1}|0\rangle^{(1)}|0\rangle^{(2)}\cr
&=&\big(L_{-1}\s_2(z_0)-[L_{-1},\s_2(z_0)]\big)|0\rangle^{(1)}|0\rangle^{(2)}
\end{eqnarray}
Let's compute the commutator. We have
\be\label{L-1 s}
[L_{-1},\s_2(z_0)]=\oint_{C^{(4\pi)}_{z_0}} {dz \over2\pi i}\,T(z)\s_2(z_0)=L^{(z_0)}_{-1} \s_2(z_0)
= \partial \s_2(z_0)
\ee
The mode $L^{(z_0)}_{-1}$ with superscript $(z_0)$ is the mode centered around $z_0$, while the mode without it, i.e. $L_{-1}$, is the mode centered around $z=0$.
Therefore (\ref{4-5}) becomes
\begin{eqnarray}\displaystyle\label{pair L}
0&=&(L_{-1} - \partial)\s_2(z_0)|0\rangle^{(1)}|0\rangle^{(2)}\cr
&=&(L_{-1} - \partial)\text{exp}\big(\sum_{m,n\geq0}\gamma_{m+1/2,n+1/2}\,\a_{-(m+1/2)}\a_{-(n+1/2)}\big)|0^2\rangle
\end{eqnarray}
Here we have relabeled $m\rightarrow m+1/2$ and $n\rightarrow n+1/2$ in (\ref{chi}) where $m$ and $n$ are now integers.
Our aim is to find relations which will help us solve for $\gamma_{m+1/2,n+1/2}$.
There are four cases we will look at it.

\subsubsection*{\underline{$m,n>0$}}

The first case is $m,n>0$. We will look for ways to obtain terms propotional to
\be
\a_{-(m+1/2)}\a_{-(n+1/2)}|0^2\rangle
\ee
There are three ways to get this. From (\ref{pair L}) we obtain
\begin{eqnarray}\displaystyle
0\!\!&=&\!\!\gamma_{m-1/2,n+1/2}\big[L_{-1},\a_{-(m-1/2)}\big]\a_{-(n+1/2)} + \gamma_{m+1/2,n-1/2} \a_{-(m+1/2)}\big[L_{-1},\a_{-(n+1/2)}\big]\cr
&&-\partial\, \gamma_{m+1/2,n+1/2}\a_{-(m+1/2)}\a_{-(n+1/2)}
\end{eqnarray}
Using (\ref{4-9}) we find the relation
\be\label{gamma e}
\gamma_{m-1/2,n+1/2}(m-1/2) + \gamma_{m+1/2,n-1/2}(n-1/2) = \partial\, \gamma_{m+1/2,n+1/2}
\ee

\subsubsection*{\underline{$m=0,n>0$}}

Next we consider the case where $m=0,n>0$. We need terms proportional to
\be
\a_{-1/2}\a_{-(n+1/2)}|0^2\rangle
\ee
Looking at (\ref{pair L}) we get
\be
0=\gamma_{1/2,n-1/2}\a_{-1/2}[L_{-1},\a_{-(n-1/2)}] -\partial\, \gamma_{1/2,n+1/2}\a_{-1/2}\a_{-(n+1/2)}
\ee
Again using (\ref{4-9}) we obtain the relation
\be\label{4-18}
\gamma_{1/2,n-1/2}(n-1/2)=\partial\,\gamma_{1/2,n+1/2}
\ee

\subsubsection*{\underline{$m>0,n=0$}}

Similar to the previous case, for $m>0,n=0$ we obtain the relation
\be \label{4-19}
\gamma_{m-1/2,1/2}(m-1/2)=\partial\,\gamma_{m+1/2,1/2}
\ee

\subsubsection*{\underline{$m=n=0$}}

For $m=n=0$ we need terms proportional to
\be
\a_{-1/2}\a_{-1/2}|0^2\rangle
\ee
For this we use (\ref{4-7}) with $k=2$ where
\be
L_{-1} |0^2\rangle={1\over 4}\a_{-1/2}\a_{-1/2}|0^2\rangle
\ee
Using this in (\ref{pair L}) we find the relation
\be \label{gamma 1/2}
{1\over 4}=\partial\,\gamma_{1/2,1/2}
\ee

\subsubsection{The solution}

To find the initial condition for these differential equations, we consider
\begin{eqnarray}\displaystyle
|0^2\rangle &=& \sigma_2 (z_0=0) |0\rangle^{(1)}|0\rangle^{(2)} \nn
&=& \text{exp}\big(\sum_{m,n\geq0}\gamma_{m+1/2,n+1/2}(z_0=0)\,\a_{-(m+1/2)}\a_{-(n+1/2)}\big)|0^2\rangle
\end{eqnarray}
The first line comes from the definition of the twist operator $\sigma_2$ which is the lowest dimension operator that changes the untwisted sector to the twisted sector. Thus we have
\be
\gamma_{m+1/2,n+1/2}(z_0=0) = 0
\ee
Using these initial conditions, we can solve the differential equations. The solution to (\ref{gamma 1/2}) is
\be
\gamma_{1/2,1/2} = {1\over 4}z_0
\ee
We can find all other $\gamma_{m+1/2,n+1/2}$'s by using the relations (\ref{gamma e}), (\ref{4-18}), and (\ref{4-19}). You can think of the $\gamma_{m+1/2,n+1/2}$'s as forming an inverted triangle with integer lattice spacing with $\gamma_{1/2,1/2}$ as the bottom lattice point. The relation (\ref{4-18}) moves you along the right edge, (\ref{4-19}) moves you along the left edge and (\ref{gamma e}) moves you within the interior.
\be
\begin{tikzcd}[row sep=16pt, column sep=-5pt]
~~~~~~& &~~~~~~~~ & &~~~~~~~~ & & ~~~~~~ \\
 &\arrow[ul,dotted]\gamma_{1/2,5/2}\arrow[ur,dotted]& &\arrow[ul,dotted]\gamma_{3/2,3/2} \arrow[ur,dotted] & &\arrow[ul,dotted]\gamma_{5/2,1/2}\arrow[ur,dotted] & \\
 & &\arrow[ul]\gamma_{1/2,3/2}\arrow[ur] &  &\arrow[ul]\gamma_{3/2,1/2}\arrow[ur] & &\\
 & & & \arrow[ul]\gamma_{1/2,1/2} \arrow[ur]& & &
\end{tikzcd}
\ee
The solution is
\be\label{gamma}
\gamma_{m+1/2,n+1/2}=\frac{z_0^{m+n+1}\Gamma[\frac{3}{2}+m]\Gamma[\frac{3}{2}+n]}{(2m+1)(2n+1)(1+m+n)\pi \Gamma[m+1]\Gamma[n+1]}
\ee
where $m$ and $n$ are non-negative integers.

\subsection{Propagation}\label{sec f}

Here we derive the propagation coefficients $f_i[-n,-p]$ which correspond to a mode passing through the twist.

\subsubsection{Relations from $L_0$}

Here we use the generator $L_0$ to derive a relation for $f_1[-1,-p]$. We begin with
\begin{eqnarray}\displaystyle
\s_2(z_0) \a^{(1)}_{-1}|0\rangle^{(1)}|0\rangle^{(2)}&=&\s_2(z_0) L_0\a^{(1)}_{-1}|0\rangle^{(1)}|0\rangle^{(2)}\nn
&=& (L_0\s_2(z_0) - [L_0,\s_2(z_0)]) \a^{(1)}_{-1}|0\rangle^{(1)}|0\rangle^{(2)}
\end{eqnarray}
where the initial mode is on the first copy.
Let us compute the commutator. We have
\begin{eqnarray}\displaystyle\label{L0}
[L_0,\s_2(z_0)] &=& \oint_{C^{(4\pi)}_{z_0}} {dz\over2\pi i}\,z\,T(z)\s_2(z_0)\cr
&=& \oint_{C^{(4\pi)}_{z_0}} {dz\over2\pi i}\,(z_0+z-z_0)\,T(z)\s_2(z_0)\cr
&=& (z_0L^{(z_0)}_{-1} + L^{(z_0)}_{0})\s_2(z_0)\cr
&=&(z_0\partial+h) \s_2(z_0)
\end{eqnarray}
Therefore
\be
\s_2(z_0) \a^{(1)}_{-1}|0\rangle^{(1)}|0\rangle^{(2)}= (L_0 - (z_0\partial+h))\s_2(z_0) \a^{(1)}_{-1}|0\rangle^{(1)}|0\rangle^{(2)}
\ee
which gives
\be
\sum_{p> 0}f_1[-1,-p]\a_{-p}|\chi\rangle
=(L_0 - (z_0\partial+h))\sum_{p> 0}f_1[-1,-p]\a_{-p}|\chi\rangle
\ee
where $\chi$ is state (\ref{chi}) from pair creation. Taking the term proportional to $\a_{-p}|0^2\rangle$ with $p$ being a positive half integer, we have
\be
f_1[-1,-p]\a_{-p}|0^2\rangle
=(L_0 - (z_0\partial+h))f_1[-1,-p]\a_{-p}|0^2\rangle
\ee
which gives
\be
f_1[-1,-p]
=(p - z_0\partial)f_1[-1,-p]
\ee
The solution is
\be\label{f pro}
f_1[-1,-p]\propto z_0^{p-1}
\ee
where $p$ is a positive half integer.

\subsubsection{Relations from $L_1$}

Here we use $L_1$ to determine the proportional coefficients in (\ref{f pro}).
Since
\be
\a^{(1)}_{0}|0\rangle^{(1)} = 0
\ee
we begin with the following relation
\be
0=\s_2(z_0) L_1\a^{(1)}_{-1}|0\rangle^{(1)}|0\rangle^{(2)}
\ee
Bringing $L_1$ through the twist we obtain
\be \label{4-25}
0= (L_1\s_2(z_0) - [L_1,\s_2(z_0)]) \a^{(1)}_{-1}|0\rangle^{(1)}|0\rangle^{(2)}
\ee
Let us compute the commutator. We have
\begin{eqnarray}\displaystyle
[L_1,\s_2(z_0)] &=& \oint_{C^{(4\pi)}_{z_0}} {dz\over2\pi i}\,z^{2}\,T(z)\s_2(z_0)\cr
&=& \oint_{C^{(4\pi)}_{z_0}} {dz\over2\pi i}\,(z_0+z-z_0)^{2}T(z)\s_2(z_0)\cr
&=&(z_0^2 L^{(z_0)}_{-1}+2z_0 L^{(z_0)}_0 + L^{(z_0)}_1)\s_2(z_0)\cr
&=&z_0(z_0\partial + 2 h)\s_2(z_0)
\end{eqnarray}
Inserting this into (\ref{4-25}) yields
\begin{eqnarray}\displaystyle \label{4-28}
0&=& (L_1 -z_0(z_0\partial + 2 h))\s_2(z_0) \a^{(1)}_{-1}|0\rangle^{(1)}|0\rangle^{(2)}\cr
&=& (L_1 -z_0(z_0\partial + 2 h))\sum_{p> 0}f_1[-1,-p]\a_{-p}\text{exp}\big(\sum_{m,n\geq0}\gamma_{m+1/2,n+1/2}\a_{-(m+1/2)}\a_{-(n+1/2)}\big)|0^2\rangle\nn
\end{eqnarray}
We again match terms to obtain relations which one solves to find $f_1[-1,-p]$. We want to keep terms which are $\a_{-(m+1/2)}|0^2\rangle$ with $m>0$. The terms which contribute are
\begin{eqnarray}\displaystyle
0&=&2f_1[-1,-1/2]\gamma_{m+1/2,1/2}\a_{-(m+1/2)}[[L_1,\a_{-1/2}],\a_{-1/2}]\cr
&& + f_1[-1,-(m+1/2)]\gamma_{1/2,1/2}\a_{-(m+1/2)}[[L_1,\a_{-1/2}],\a_{-1/2}]\cr
&&+f_1[-1,-(m+3/2)][L_1,\a_{-(m+3/2)}]-z_0(z_0\partial + 2 h)f_1[-1,-(m+1/2)] \a_{-(m+1/2)}\nn
\end{eqnarray}
Using the commutation relation (\ref{L com}), this gives a recursion relation for $m>0$
\begin{eqnarray}\displaystyle\label{f re1}
&&(m+3/2)f_1[-1,-(m+3/2)]\nn
&=&[z_0(z_0\partial + 2 h) -{1\over2} \gamma_{1/2,1/2}]f_1[-1,-(m+1/2)]-f_1[-1,-1/2]\gamma_{m+1/2,1/2}
\end{eqnarray}
Now we need to find a relation for $f_1[-1,-3/2]$. In order to do so we will need terms which are proportional to $\a_{-1/2}|0^2\rangle$. From (\ref{4-28}) we get
\begin{eqnarray}\displaystyle
0&=&3f_1[-1,-1/2]\gamma_{1/2,1/2}[[L_1,\a_{-1/2}],\a_{-1/2}]\a_{-1/2}\nn
&&+f_1[-1,-3/2][L_1,\a_{-3/2}]-z_0(z_0\partial + 2 h)f_1[-1,-1/2] \a_{-1/2}
\end{eqnarray}
where we have collected terms coming from appropriate commutators in order to leave only one $\a_{-1/2}$. Again using commutation relations we obtain the relation
\begin{eqnarray}\displaystyle\label{f re2}
{3\over2}f_1[-1,-3/2]=z_0(z_0\partial + 2 h)f_1[-1,-1/2]-{3\over2}f_1[-1,-1/2]\gamma_{1/2,1/2}
\end{eqnarray}
Therefore all $f_1[-1,-p]$ can be determined from $f_1[-1,-1/2]$.

\subsubsection{The solution}\label{sec f s}

The solution satisfying the recursion relations (\ref{f re1}) and (\ref{f re2}) is
\be\label{f1pc1}
f_1[-1,-p]=C\frac{\Gamma[p-1]}{\Gamma[p+1/2]}z_0^{p-1}
\ee
where $p$ is a positive half integer.
The constant $C$ will be derived in (\ref{C})
\be\label{C 0}
C=\pm \frac{i}{4\sqrt{\pi}}
\ee
To obtain $f_2[-1,-p]$ for an excitation on copy 2, we change the location of the twist by $z_0\to z_0 e^{2\pi i}$. It interchanges copy 1 and copy 2. Since $p$ is a half integer, we have $z_0^{p-1} \to -z_0^{p-1}$. Therefore, we have
\be\label{f2pc1}
f_2[-1,-p]=-C\frac{\Gamma[p-1]}{\Gamma[p+1/2]}z_0^{p-1}
\ee
Thus the two possible signs of (\ref{C 0}) correspond to two possible conventions of labeling copy 1 and copy 2. By applying $L_{-1}$ repeatedly, we can compute $f_i[-n,-p]$ for $n>1$, which will be shown in appendix \ref{App f}. The final expressions are given by
\begin{eqnarray}\displaystyle\label{final f int}
f_1[-n,-p]&=&{1\over2} \delta_{n,p}\cr
f_1[-n,-p]
&=&{iz_0^{p-n}\Gamma({1\over2}+n)\Gamma(p)\over\pi(2p - 2n)\Gamma(n)\Gamma(p+\frac{1}{2})},\quad n\neq p
\cr
f_2[-n,-p]&=&{1\over2} \delta_{n,p}\cr
f_2[-n,-p]&=&-f_1[-n,-p],\quad n\neq p
\end{eqnarray}

\subsection{Contraction}

Here we derive the expression for the contraction of two modes in the initial state under the effect of the twist.

\subsubsection{Relations from $L_{1}$}

Let's start with the following state with $n>1$
\begin{eqnarray}\displaystyle
\s(z_0)\a^{(i)}_{-1}\a^{(j)}_{-(n-1)}|0\rangle^{(1)}|0\rangle^{(2)}&=&\frac{1}{n}
\s(z_0)L_1\a^{(i)}_{-1}\a^{(j)}_{-n}|0\rangle^{(1)}|0\rangle^{(2)}\cr
&=&{1\over n}\big[L_1-z_0(z_0\partial + 2 h)\big]\s_2(z_0)\a^{(i)}_{-1}\a^{(j)}_{-n}|0\rangle^{(1)}|0\rangle^{(2)}
\end{eqnarray}
where we remind the reader that $i,j=1,2$ are copy labels.
So we have the relation
\begin{eqnarray}\displaystyle\label{C r1}
[L_1-z_0(z_0\partial + 2 h)]\s_2(z_0)\a^{(i)}_{-1}\a^{(j)}_{-n}|0\rangle^{(1)}|0\rangle^{(2)} = n\s(z_0)\a^{(i)}_{-1}\a^{(j)}_{-(n-1)}|0\rangle^{(1)}|0\rangle^{(2)}
\end{eqnarray}
which becomes
\begin{eqnarray}\displaystyle
&&[L_1-z_0(z_0\partial + 2 h)]\Big(\sum_{p> 0}f_i[-1,-p]\a_{-p}\sum_{p'> 0}f_j[-n,-p']\a_{-p'} + C^{ij}[1,n]\Big)|\chi\rangle\cr
&=& n\Big(\sum_{p> 0}f_i[-1,-p]\a_{-p}\sum_{p'> 0}f_j[-(n-1),-p']\a_{-p'} +C^{ij}[1,n-1]\Big)|\chi\rangle
\end{eqnarray}
We collect terms which carry no bosonic modes in order to easily isolate $C^{ij}$. Terms of this kind are given by
\begin{eqnarray}\displaystyle
&&\big(f_i[-1,-1/2]f_j[-n,-1/2]+C^{ij}[1,n]\gamma_{1/2,1/2} \big)[[L_1,\a_{-1/2}],\a_{-1/2}]\cr
&&-z_0(z_0\partial + 2 h)C^{ij}[1,n]=nC^{ij}[1,n-1]
\end{eqnarray}
Using the commutation relations yield for $n>1$
\be
{1\over2}f_i[-1,-1/2]f_j[-n,-1/2]+\frac{z_0}{8}C^{ij}[1,n] -z_0(z_0\partial + 2 h)C^{ij}[1,n]=nC^{ij}[1,n-1]
\ee
Setting $h=1/16$ (the dimension of the twist operator), the relation becomes
\be \label{C r2}
{1\over2}f_i[-1,-1/2]f_j[-n,-1/2] -z_0^2 \partial C^{ij}[1,n]=nC^{ij}[1,n-1]
\ee
where $n>1$.
To find $C^{ij}[1,1]$, notice that in (\ref{C r1}) if $n=1$, the RHS vanishes. Thus the RHS of (\ref{C r2}) vanishes if $n=1$, which gives
\be
{1\over2}f_i[-1,-1/2]f_j[-1,-1/2]=z_0^2\partial C^{ij}[1,1]
\ee
The solution is
\be\label{C11}
C^{ij}[1,1] =
-(-1)^{i+j}\pi C^2 z_0^{-2}
\ee
where we have used (\ref{f1pc1}) and (\ref{f2pc1}).

\subsubsection{Relations from $L_{-2}$}

Here we will derive the constant $C$ which appears in (\ref{f1pc1}), (\ref{f2pc1}), and (\ref{C11}).
Start with
\be\label{C 1}
\sigma (z_0)L_{-2} |0\rangle^{(1)}|0\rangle^{(2)}
=\sigma (z_0)\frac{1}{2}\big(\alpha^{(1)}_{-1}\alpha^{(1)}_{-1}+\alpha^{(2)}_{-1}\alpha^{(2)}_{-1}\big)|0\rangle^{(1)}|0\rangle^{(2)}
\ee
Using the contraction (\ref{C11}), the term without any modes on the RHS is
\be\label{R C}
\frac{1}{2}\big(C^{11}[1,1]+C^{22}[1,1]\big)|0^2\rangle = - \pi C^2 z_0^{-2}|0^2\rangle
\ee
Let us compute the LHS of (\ref{C 1}) in another way. We have
\be
\s_2(z_0)L_{-2}=L_{-2}\s_2(z_0) - [L_{-2},\s_2(z_0)]
\ee
where
\begin{eqnarray}\displaystyle
[L_{-2},\s_2(z_0)] &=& \oint_{C^{(4\pi)}_{z_0}} {dz\over2\pi i}z^{-1}T(z)\s_2(z_0)\cr
&=& z_0^{-1}\oint_{C^{(4\pi)}_{z_0}} {dz\over2\pi i} \big(1+\frac{z-z_0}{z_0}\big)^{-1}T(z)\s_2(z_0)\cr
&=& z_0^{-1}\oint_{C^{(4\pi)}_{z_0}} {dz\over2\pi i} \big(1-\frac{z-z_0}{z_0}+\dots\big)T(z)\s_2(z_0)\cr
&=& z_0^{-1}\big(L^{(z_0)}_{-1}-z_0^{-1}L^{(z_0)}_{0}\big)\s_2(z_0)\cr
&=& z_0^{-1}\big(\partial-z_0^{-1}h\big)\s_2(z_0)
\end{eqnarray}
Thus we have
\be
\sigma (z_0)L_{-2} |0\rangle^{(1)}|0\rangle^{(2)}\nn
=\big[L_{-2}-z_0^{-1}(\partial-z_0^{-1}h)\big]|\chi\rangle
\ee
The term without any modes is
\be
 z_0^{-2} h |0^2\rangle = \frac{z_0^{-2}}{16} |0^2\rangle
\ee
where we have used $h=1/16$.
Comparing to (\ref{R C}), we obtain
\be
- \pi C^2 = \frac{1}{16}
\ee
which gives
\be\label{C}
C=\pm \frac{i}{4\sqrt{\pi}}
\ee
As explained in section \ref{sec f s}, the two signs correspond to the two different conventions of labeling the copies.

\subsubsection{The solution}

Using (\ref{C}), the contraction (\ref{C11}) becomes
\be\label{C11 f}
C^{ij}[1,1] =
-(-1)^{i+j}\pi C^2 z_0^{-2}=(-1)^{i+j}{1\over16}z_0^{-2}
\ee
Solving (\ref{C r2}) recursively we find the following solution
\be
C^{ij}[1,n] = (-1)^{i+j}{z_0^{-(1+n)}\Gamma({1\over2}+n)\over4(1+n)\sqrt\pi\Gamma(n)} \label{C111n}
\ee
The contractions $C^{ij}[n_1,n_2]$ with $n_1,n_2>0$ are computed in appendix \ref{Cij}. The final expressions are
\be\label{final Cij}
C^{ij}[n_1,n_2]
= (-1)^{i+j}{z_0^{-(n_1+n_2)}\over2(n_1+n_2)\pi}{\Gamma({1\over2}+n_1)\Gamma({1\over2}+n_2)\over\Gamma(n_1)\Gamma(n_2)}
\ee
where $i,j=1,2$ are copy labels.

\section{Discussion}\label{discussion}

The traditional way to compute the effect of a twist operator is by using the covering map method.
In this paper, we have developed a new method using the Bogoliubov ansatz and conformal symmetry.
The Bogoliubov ansatz includes three quantities which characterize the final state uniquely. These three quantities correspond to three effects produced by the twist operator. One effect is pair creation in which the twist, when acting on the vacuum in the initial state, produces pairs of modes in the final state. This effect is encoded in the coefficients $\gamma_{mn}$ which are computed in (\ref{gamma}). Another effect, which arises by applying the twist operator to a single mode in the initial state, is propagation. This effect is encoded in the functions $f_i[-n,-p]$ which are computed in (\ref{final f int}).
The third effect produced by the twist is the contraction of two modes in the initial state. This process is encoded in the functions $C^{ij}[n_1,n_2]$. These expressions were computed in (\ref{final Cij}). Each of these quantities agrees with the results from the covering map method in appendix \ref{App cover}. Using the Bogoliubov ansatz along with conformal symmetry we have derived a new method for computing effects of the twist operator in orbifold CFTs.

Our results also answer an important question about the nature of the effect of a twist operator.
From the general idea of the covering map, we know that the twist operator generates a Bogoliubov transformation.  This raises the following question: if we know that the effect of a twist is a Bogoliubov transformation what else do we need to completely determine the coefficients in this transformation?
From the covering map method, it seems that the answer is the covering map. The covering map is essential to determine the effect of a twist operator. However, our results show that with conformal symmetry we do not need other input to determine its effect. The nature of the effect of a twist operator is captured by the form of the Bogoliubov transformation and conformal symmetry.

In the study of the perturbative D1D5 CFT, twist operators and their effects carry a wide variety of applications since a class of marginal deformations of the theory contain twist operators \cite{Gava:2002xb,Gaberdiel:2015uca,Burrington:2018upk,Hampton:2018ygz,DeBeer:2019oxm,Guo:2019pzk,Keller:2019yrr,Hampton:2019hya,Hampton:2019csz,Guo:2019ady,Guo:2020gxm,Lima:2020boh,Lima:2020nnx,Lima:2020urq,Lima:2020kek,Lima:2021wrz,Benjamin:2021zkn,AlvesLima:2022elo,Apolo:2022fya}.
The D1D5 CFT is realized by four free bosons and four free fermions.
Our results, which are for one boson, can be simply generalized to four bosons where each boson has its own propagation, contraction and pair creation. These effects for each of the four bosons of the D1D5 CFT are the same as those of one free boson computed in this paper, thus yielding the same expressions for the three relevant quantities.
The generalization to free fermions is straight forward but will contain some additional features. We will present these results in a future work. Furthermore, to compute higher order effects in the D1D5 CFT, which is relevant for holography, it is necessary to compute higher order twist correlation functions \cite{Carson:2016cjj,Carson:2016uwf,Carson:2015ohj,Guo:2021ybz,Guo:2021gqd}. In the covering map method, these computations become more challenging due to the growing complexity of the covering maps themselves. The method developed in this paper provides tools to possibly compute these higher order effects.

\section*{Acknowledgements}
We would like to thank Soumangsu Chakraborty, Nicolas Kovensky, Samir Mathur and Hynek Paul for helpful discussions. The work of B.G. and S.H. is supported ERC Grant 787320 - QBH Structure.

redpajama_set_name: RedPajamaArXiv


text:
Q: Recursive dependency found in a makefile for creating a static library I have 4 .c files hello.c,here.c,bye.c and main.c.
One header file mylib.h
The contents are as follows
hello.c
#include<stdio.h>

void hello()
{
    printf("Hello!\n");
}

here.c
#include<stdio.h>

void here()
{
     printf("I am here \n");
}

bye.c
#include<stdio.h>

void bye()
{
    printf("Bye,Bye");
}

main.c
#include<stdio.h>
#include "mylib.h"

int main()
{

  hello();
  here();
  bye();
  return 1;
}

mylib.h
#ifndef _mylib_
#define _mylib_

void hello();
void here();
void bye();

#endif

The makefile for creating a static lib is :
Makefile
all:    myapp

#Macros

#Which Compiler
CC = g++

#Where to install
INSTDIR = /usr/local/bin

#Where are include files kept
INCLUDE = .

#Options for developement
CFLAGS = -g -Wall -ansi

#Options for release
#CFLAGS = -O -Wall -ansi

#Local Libraries
MYLIB = mylib.a

myapp:  main.o $(MYLIB)
    $(CC) -o myapp main.o $(MYLIB)

$(MYLIB):       $(MYLIB)(hello.o) $(MYLIB)(here.o) $(MYLIB)(bye.o)
main.o:         main.c mylib.h
hello.o:        hello.c
here.o:         here.c
bye.o:          bye.c

clean:
    -rm main.o hello.o here.o bye.o $(MYLIB)

install:        myapp
    @if [ -d $(INSTDIR) ]; \
    then \
            cp myapp $(INSTDIR);\
            chmod a+x $(INSTDIR)/myapp;\
            chmod og-w $(INSTDIR)/myapp;\
            echo "Installed in $(INSTDIR)";\
    else \
            echo "Sorry, $(INSTDIR) does not exist";\
    fi

Problem: When I execute the command
make -f Makefile all

I get the following dependecy error:
make: Circular mylib.a <- mylib.a dependency dropped.
ar rv (hello.o) hello.o
/bin/sh: -c: line 0: syntax error near unexpected token `('
/bin/sh: -c: line 0: `ar rv (hello.o) hello.o'
make: *** [(hello.o)] Error 2

Questions : How do I resolve this? Which command is causing the cyclic dependency?

A: #Local Libraries
MYLIB = mylib.a

myapp:  main.o $(MYLIB)
    $(CC) -o myapp main.o $(MYLIB)

$(MYLIB):       $(MYLIB)(hello.o) $(MYLIB)(here.o) $(MYLIB)(bye.o)

It looks like the last rule is
mylib.a: mylib.a (hello.o) mylib.a (here.o) mylib.a (bye.o)

Which is a circular dependency.
The line should be
mylib.a: hello.o here.o bye.o

Without the parentheses.

A: Not positive, but:
$(MYLIB):       $(MYLIB)(hello.o) $(MYLIB)(here.o) $(MYLIB)(bye.o)

If I remember my makefile syntax correctly, that line says $(MYLIB) depends on $(MYLIB)... Which of course evaluates to: mylib.a:     mylib.a...

A: Drop the extraneous $(MYLIB) in the dependency list.  That is:
$(MYLIB):       $(MYLIB)(hello.o) $(MYLIB)(here.o) $(MYLIB)(bye.o)

should be:
$(MYLIB):       hello.o here.o bye.o


When Exodus Kodi comes to entertainment one thing that people just cannot miss out is Kodi. This is one of the easiest ways to watch movies and television. One of the most widely used addons with Kodi is the Exodus. This helps the users access their favourite online entertainment with ease.
This software is maintained by ace developers Lambada. It is one of the first choices in case of addons if you want to stream videos and watch movies. So, do you also want to enter a whole new world of entertainment where you can access your favourite television shows and movies with ease then you must first install kodi and then install the addon exodus.
Here are the steps with the help of which you can download kodi on the your computer.
Step 2: You now need to install the software. For this you need to click on the file that you have downloaded.
Step 3: You will now get a window which has the Run option. Click on this option.
Step 4: You will now be prompted to go to the next page. Click on the Next button and let the installation of Kodi get completed.
Step 5 : Now you will get the Finish tab. Click on the tab and now you are through with the installation of Kodi on your computer.
Step 1: You will first have to click on Kodi software that you have installed on your computer.
Step6: You can name this as Kdil. There are chances that this may get done automatically. Now click on Ok.
Step 7: It should be noted that if you have a firewall installed, then you may have to disconnect the firewall. Only then you will be able to access repo.
Step 8 : Click on Add ons.
Step 10: Now you need to click on the option of Install from zip.
Step 11: Next step is to locate the Kdil FTP connection. Once you locate it click on it.
Step 12: Now you will find the option Kodil.zip. Click on this option to install the repository.
Step 17: Locate Exodus and click on it and install it on your kodi.
Now that you are through with the installation of kodi and exodus all that you have to do is start using it for the streaming the best movies and television shows.
One of the biggest advantages that this add on gives is that it gives a wide range of streaming options. You can check out a wide range of movies and television shows of American and International producers.
As far as the menu and the browsing functions are concerned, you will find that gives plenty of options and it is the best. Like for Television you can see the content based on date, genre, the rating of show, the popularity of the show etc.
It also gives you the option to set up various accounts.
The other drawback is that it is not live television. It will provide only on demand streaming option.
It does not have any documentary section. If it is documentaries that you are looking for then you will find that they offer limited choice and you will have to look out for the same elsewhere.
Exodus Kodi- Making news recently!
Exodus enjoys massive popularity and due to this this add on and its developer were at the receiving end and some adversaries attacked them. It seems the developer of Exodus whose real identity is not known was threatened by the adversaries that they will reveal his true identity if their demands were not met. It seems that the developer who is only known as Lambada inserted a code into Exodus and launched a DDoS attack on the website of the adversaries. There was lots of discussion about this on the Kodi forum and Lambada said that he had to do it for self defence and if he is threatened legally then he will henceforth only focus on the legal add ons.
This may not actually be a problem associated with the add on. It may be that the website from where the streams are being sourced are down. It may also be that you are using an outdated version of the add on or there are some content filters due to which you are facing this issue. So, make sure that you use only the latest version of Exodus.
If you get this type of error then you must uninstall Exodus and then reinstall it again. Sometimes you will have to uninstall Kodi itself and then reinstall it again.
Sometimes you will only see movies of a certain time and date. This means that there is a problem with the time and date settings you need to update it.
This problem was mainly seen in the older versions of Exodus. This problem does not occur in the new versions. So what you need to do is ensure that you make use of only the latest version of Exodus.
These are some of the common issues that one may face with Exodus. There are chances that you may encounter some other problems which you will have to troubleshoot and find a solution for the same.
One of the best add ons that you can use for Kodi is Exodus. Once you install Exodus Kodi you can easily watch movies and television shows. It has very good streaming quality and is one of the preferred add ons. It does have some minor problems and issues. But most of the times it has been observed that this problem is due to the websites from where the video link is being taken or due to the use of older version of the add on or due to some system compatibility issues. Also you cannot watch live television with Exodus and it lacks documentaries for which a person may have to turn elsewhere.
But overall one can say that one of the best ways to watch movies and television shows on demand is with Exodus. The installation is easy and does not take much time and it opens the doors of a whole new world of entertainment. The identity of the developer of this superb add on is not known which has led to some controversies and attacks on Exodus and a lot of discussion about the same erupted on the online forums. But if you want a good quality streaming add on which provides you with some of the best options in movies which even includes pirated movies then you have to install Exodus on Kodi and start using it right away. Also get more information about kodi apps features.


text:
Trump orders Pentagon to cut US troops in S Korea
South security adviser to meet Bolton
Published: May 04, 2018 22:22:58
US President Donald Trump has ordered the Pentagon to prepare options for reducing the number of US troops in South Korea, the New York Times reported on Thursday, citing several people briefed on the deliberations, reports Reuters.
Reduced US troop levels are not intended to be a bargaining chip in Trump's planned summit in late May or early June with North Korean leader Kim Jong Un about Pyongyang's nuclear weapons program, the Times said.
The officials said, however, that a peace treaty between the two Koreas could diminish the need for the 23,500 US soldiers currently stationed on the peninsula, the newspaper said.
A full withdrawal of US troops was unlikely, the officials said, according to the paper.
But a US National Security Council official told a visiting South Korean official in Washington via telephone the report was false, the South Korean presidential office said in a statement.
The White House and Pentagon did not immediately respond to requests for comment.
Trump has said the United States should consider reducing the number of troops in South Korea unless South Korea shoulders more of the cost.
Then-CIA Director Mike Pompeo, before taking office as US secretary of state, met Kim last month and reported the North Korean leader was not demanding the withdrawal of all US forces as a precondition for a summit with Trump.
South Korea said on Wednesday the issue of US troops stationed in the South was unrelated to any future peace treaty with North Korea and that American forces should stay even if such an agreement is signed.
Earlier, South Korea's national security adviser is in Washington to meet his US counterpart, John Bolton, ahead of an expected summit between US President Donald Trump and North Korean leader Kim Jong Un, officials in Washington and Seoul said.
The US National Security Council asked Chung Eui-yong to fly to the United States to discuss matters related to the summit, a South Korean presidential official told reporters on Friday.
The United States had asked that the visit be kept quiet due to the issues to be addressed at meetings there, said the official, who declined to be identified.
A senior administration official in Washington confirmed Chung's visit and his meeting with Bolton.
South Korea has been working closely with old ally the United States on efforts to rein in North Korea's nuclear and missile programs, with an easing of tensions in recent months after threats of war from North Korea and Trump.