text stringlengths 14 5.77M | meta dict | __index_level_0__ int64 0 9.97k ⌀ |
|---|---|---|
Q: Install python package Can we install python packages in IDLE (Python Shell)?
I know how to do it in windows cammand prompt. But I want to install packages in python shell.
If yes how?
A: from subprocess import Popen
package_name = input("Enter package name to install")
Popen(["pip","install",package_name])
Using Popen will be effectively the same as running the command in a terminal or command prompt, but without another window opening. It will all take place in your shell.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 7,496 |
★ ★ ★ ★★ ...here, thanks to Harry Potter actor Miriam Margolyes's artistry and Simon Callaghan's excellent pianism, is Poulenc's delightful musical response. And as I listened to this recording, I found the original drawings reappearing in my mind with all their detail intact – extraordinary. It lasts just 30 minutes, but my god does it resonate.
★ ★ ★ ★★
Anyone who grew up, as I did, on Jean de Brunhoff's Babar books will find this CD entrancing. For those unfortunates who missed this experience in their early childhood, Babar was a warm-hearted, public-spirited elephant whose elevation to the throne of a realm suspiciously similar to France, and whose gently socialist leadership of that country in war and peace, was made to seem the most natural thing in the world. Brunhoff was a French children's book author and illustrator whose wife Cécile concocted the story of Babar as a bedtime treat for her two sons; the first Babar book was published in 1931, and it was followed by six further titles in the series.
With their charmingly witty drawings, the books became a cult, and the composer Francis Poulenc was one of the army of fans. What we have here, thanks to Harry Potter actor Miriam Margolyes's artistry and Simon Callaghan's excellent pianism, is Poulenc's delightful musical response. And as I listened to this recording, I found the original drawings reappearing in my mind with all their detail intact – extraordinary. It lasts just 30 minutes, but my god does it resonate.
Michael Church
Independent - 25 November 2020
With Miriam Margolyes (narrator)
A delightful set of circumstances combined to produce the beloved masterpiece, Babar. The journey began in 1930 when Laurent and Mathieu, sons of French author and illustrator Jean de Brunhoff were told an enchanting bedtime story by their mother, Cécile. So moved were the young boys by the curious tale of the young elephant's adventures, that they asked their father to create illustrations. The resulting book initiated a series that was to be the crowning achievement of Jean de Brunhoff's short professional life, and that of his son Laurent, who added further volumes following his father's death in 1937. The children have acknowledged that the story originated with Cécile de Brunhoff, who, feeling that her contribution was too small to be credited, requested that her name be removed from the publications.
In a heart-warmingly similar situation ten years later, Poulenc was spending time with the granddaughter of one of his cousins. Noting that she became bored with the music he was playing, Poulenc put Brunhoff's Babar on the piano and began to improvise, to the great delight of the young girl. The musical ideas born that day were to simmer away at the back of Poulenc's mind until he completed the work in 1945. It was premiered on French radio the following year.
Scala Radio Article (4 December 2020) | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 9,374 |
namespace WebApp
{
public interface IWidget
{
}
public class GreenWidget : IWidget { }
public class RedWidget : IWidget { }
} | {
"redpajama_set_name": "RedPajamaGithub"
} | 4,715 |
{"url":"https:\/\/answers.yahoo.com\/question\/index?qid=20120428173643AA5Db99","text":"Anonymous\nAnonymous asked in Education & ReferenceHomework Help \u00b7 9 years ago\n\n# How would i simplify this?? Trying to understand this concept..?\n\nx^2 - 10x +16\n\ndivided by\n\nx^2 - 4.\n\nI got 10\/4x - 4 but I feel like it might be wrong\n\nRelevance\n\u2022 9 years ago\n\nI got ( X-8 ) \/ ( X+2 )\n\nthe numerator simplifies to (X-2)(X-8)\n\nthe denominator simplifies to (X-2)(X+2)\n\nthe common multiplier is (X-2) and so it cancels out, leaving us with the above answer.\n\n\u2022 9 years ago\n\nI can't say for sure but often with problems like that, they mean factor and simplify:\n\n(x - 8)(x - 2)\n\n------------------\n\n(x + 2)(x - 2); the (x - 2)'s cancel so\n\n(x - 8)\n\n--------\n\n(x + 2)","date":"2020-10-21 23:15:49","metadata":"{\"extraction_info\": {\"found_math\": false, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8409038186073303, \"perplexity\": 3031.065913227719}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2020-45\/segments\/1603107878633.8\/warc\/CC-MAIN-20201021205955-20201021235955-00336.warc.gz\"}"} | null | null |
let _ = require('lodash');
let async = require('async');
let restify = require('restify');
let assert = require('chai').assert;
import { ConfigParams, MultiString } from 'pip-services3-commons-node';
import { Descriptor } from 'pip-services3-commons-node';
import { References } from 'pip-services3-commons-node';
import { QuoteV1 } from '../../../src/data/version1/QuoteV1';
import { QuoteStatusV1 } from '../../../src/data/version1/QuoteStatusV1';
import { QuotesMemoryPersistence } from '../../../src/persistence/QuotesMemoryPersistence';
import { QuotesController } from '../../../src/logic/QuotesController';
import { QuotesHttpServiceV1 } from '../../../src/services/version1/QuotesHttpServiceV1';
let httpConfig = ConfigParams.fromTuples(
"connection.protocol", "http",
"connection.host", "localhost",
"connection.port", 3000
);
let QUOTE1: QuoteV1 = {
id: '1',
text: new MultiString({ en: 'Text 1' }),
author: new MultiString({ en: 'Author 1' }),
status: QuoteStatusV1.Completed,
tags: null,
all_tags: null
};
let QUOTE2: QuoteV1 = {
id: '2',
text: new MultiString({ en: 'Text 2' }),
author: new MultiString({ en: 'Author 2' }),
status: QuoteStatusV1.Completed,
tags: ['TAG 1'],
all_tags: null
};
suite('QuotesHttpServiceV1', ()=> {
let service: QuotesHttpServiceV1;
let rest: any;
suiteSetup((done) => {
let persistence = new QuotesMemoryPersistence();
let controller = new QuotesController();
service = new QuotesHttpServiceV1();
service.configure(httpConfig);
let references: References = References.fromTuples(
new Descriptor('pip-services-quotes', 'persistence', 'memory', 'default', '1.0'), persistence,
new Descriptor('pip-services-quotes', 'controller', 'default', 'default', '1.0'), controller,
new Descriptor('pip-services-quotes', 'service', 'http', 'default', '1.0'), service
);
controller.setReferences(references);
service.setReferences(references);
service.open(null, done);
});
suiteTeardown((done) => {
service.close(null, done);
});
setup(() => {
let url = 'http://localhost:3000';
rest = restify.createJsonClient({ url: url, version: '*' });
});
test('CRUD Operations', (done) => {
let quote1, quote2;
async.series([
// Create one quote
(callback) => {
rest.post('/v1/quotes/create_quote',
{
quote: QUOTE1
},
(err, req, res, quote) => {
assert.isNull(err);
assert.isObject(quote);
assert.equal(quote.author.en, QUOTE1.author.get('en'));
assert.equal(quote.text.en, QUOTE1.text.get('en'));
quote1 = quote;
callback();
}
);
},
// Create another quote
(callback) => {
rest.post('/v1/quotes/create_quote',
{
quote: QUOTE2
},
(err, req, res, quote) => {
assert.isNull(err);
assert.isObject(quote);
assert.equal(quote.author.en, QUOTE2.author.get('en'));
assert.equal(quote.text.en, QUOTE2.text.get('en'));
quote2 = quote;
callback();
}
);
},
// Get all quotes
(callback) => {
rest.post('/v1/quotes/get_quotes',
{},
(err, req, res, page) => {
assert.isNull(err);
assert.isObject(page);
assert.lengthOf(page.data, 2);
callback();
}
);
},
// Update the quote
(callback) => {
quote1.text = { en: 'Updated Content 1' };
rest.post('/v1/quotes/update_quote',
{
quote: quote1
},
(err, req, res, quote) => {
assert.isNull(err);
assert.isObject(quote);
assert.equal(quote.text.en, 'Updated Content 1');
assert.equal(quote.author.en, QUOTE1.author.get('en'));
quote1 = quote;
callback();
}
);
},
// Delete quote
(callback) => {
rest.post('/v1/quotes/delete_quote_by_id',
{
quote_id: quote1.id
},
(err, req, res, result) => {
assert.isNull(err);
//assert.isNull(result);
callback();
}
);
},
// Try to get delete quote
(callback) => {
rest.post('/v1/quotes/get_quote_by_id',
{
quote_id: quote1.id
},
(err, req, res, result) => {
assert.isNull(err);
//assert.isNull(result);
callback();
}
);
}
], done);
});
}); | {
"redpajama_set_name": "RedPajamaGithub"
} | 9,152 |
De hertshoornweegbree (Plantago coronopus) is een in België en Nederland van nature voorkomende, eenjarige of vaste, tweeslachtige, 5–20 cm hoge plant met een penwortel.
Kenmerken
De grondstandige bladeren staan in bladrozetten. Het blad is vlezig of zachtbehaard, enkelvoudig, lijn- tot lintvormig en/of veervormig ingesneden en toegespitst. De bladrand is gaaf of getand. De basis van de bladschijf is gevleugeld. De nerven zijn in een veernervig patroon gerangschikt. De bladsteel is zeer kort of afwezig.
De bloemstengel is rechtopstaand of opstijgend, behaard, rond en massief.
De bloeiwijze is een aar. De witte bloemen zijn buis- of stervormig en radiair symmetrisch. De bloemen bestaan uit vier kroonbladen en vier kelkbladen. De kroon is even lang of korter dan de kelk. De bloemen hebben vier meeldraden en één stijl met één stempel. Hertshoornweegbree bloeit van juni tot september. Na bevruchting worden er doosvruchten gevormd.
Verspreiding
De hertshoornweegbree komt voor in de kustregio's van Europa, Noord-Afrika en Voor-Azië.
De plant groeit op ziltige, zandige plekken in duinen, op zeedijken, hoge schorren en groene stranden. Deze plant kan goed tegen vertrapping en groeit op plaatsen waar andere planten niet goed gedijen. Bij het opkomen van andere soorten verdwijnt de hertshoornweegbree.
Gebruik
Hertshoornweegbree is al vierhonderd jaar als groente in cultuur. Hij kan als groente van maart tot augustus worden gezaaid en na zeven tot acht weken worden geoogst. De jonge bladeren zijn geschikt voor consumptie. Ze kunnen worden gekookt of rauw in salades worden verwerkt. De groente bevat vitamine C, β-caroteen, vitamine B1 en voedingsvezels. Italië is de voornaamste producent van de hertshoornweegbree als groente.
Externe link
hertshoornweegbree
Bladgroente
hertshoornweegbree | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 1,155 |
In defence of staff reforms
Your article 'Kinnock staff reforms "not working" claim unions' (European Voice 20-26 November) gives the floor to sensationalist claims.
By From Eric Mamer
November 26, 2003 5:00 pm CET
Accuracy implies the recognition that the introduction of the Commission's new system of staff appraisal and promotion demands a significant change of behaviour from the various actors.
Such change, in any organization, obviously takes more than one year. This is why its implementation explicitly includes annual reviews so that corrections and gradual adjustments can be made if necessary.
The first round of promotions is only just finishing and the results have not yet been published or analysed. It is thus more than premature to draw conclusions now. However, a number of statements in the article are plainly incorrect and should be put right.
Contrary to what is claimed, the machinery put in place does effectively "protect" the vast majority of deserving officials while enabling the best among them to advance more quickly. The unions quoted always claimed that such safeguards could not be obtained in the Commission's negotiations with Council and they are obviously disappointed to be proved wrong. The new arrangements facilitate promotion based on merit. The results of the promotion exercise will show that, as staff members will see for themselves.
Carbon markets are a license for continued pollution
By Nnimmo Bassey, Akinbode Oluwafemi and Ndivile Mokoena
Greek government is no friend of the far right
By Giannis Oikonomou
Biden's European tilt is nothing to be happy about
By Dalibor Rohac
Confusing the system set up in the Commission with the forced-distribution practices of some private companies seeking to lay-off a set a proportion of employees shows deep ignorance about both approaches.
Under the Commission's approach, which is based on a reasonable harmonization and reasonable comparability of the scores given in the individual assessments, it is not possible to leave a completely free rein. That is why the system is based on wide-ranging coordination between the directorates-general to standardize practice. The staff representatives are also involved in this exercise.
However, since the article stresses this point, it is worth emphasizing that no instruction has been given calling on directorates-general to limit their scores to 17 so that people working in the cabinets can have higher marks and to give 'good' people in the directorates-general no more than 14.
What is true is that a few marks have exceeded 17. This is a direct result of the high standards of quality and performance required of officials. When a mark of 13 or 14 already indicates a good to very good level, it is evident that 17 or above is exceptional and – by definition – rare. There will always be very few such marks.
Eric Mamer
Spokesman for administrative reform
More from ... From Eric Mamer
Letter to the editor Ethiopia is not Yugoslavia
Letter to the Editor Germany and Europe are better off without Nord Stream 2
Letter to the Editor Regulated encryption isn't possible — here's what is
Letter to the Editor We need more protection from government surveillance — not less
Letter to the editor Biden's 'summit of democracies' can rally allies against autocracies
The war in Ukraine is not 'a picnic'
By Petr Kratochvíl
Russia said the Snake Island soldiers were alive
March 8, 2022 4:44 pm CET
By Sergey Kovalevskiy
Though in need of reform, the watchdog is still watching
By Massimo Condinanzi, Jacopo Alberti and Camilla Burelli
Cyber warfare — call it what it is
By Fabrice Pothier | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 9,793 |
Jutro idziemy do kina – polski film telewizyjny z 2007 roku w reżyserii Michała Kwiecińskiego.
Zdjęcia rozpoczęto 21 maja 2007.
Obsada
Mateusz Damięcki – Andrzej Skowroński
Antoni Pawlicki – Piotr Dołowy
Jakub Wesołowski – Jerzy Bolesławski
Grażyna Szapołowska – matka Krysi
Olgierd Łukaszewicz – ojciec Krysi
Sylwia Oksiuta-Warmus – prostytutka
Anna Gzyra – Krysia Włosowska, ukochana Jerzego
Julia Pietrucha – Zosia Paluch, narzeczona Piotra
Maria Niklińska – Basia
Marta Ścisłowicz – Ania, córka starosty, ukochana Andrzeja
Krzysztof Banaszyk – porucznik Zaborowski
Krzysztof Stelmaszyk – starosta, ojciec Ani
Katarzyna Gniewkowska – matka Bolesławskiego
Jacek Romanowski – ojciec Bolesławskiego
Krzysztof Skonieczny – Bzowski
Adam Krawczyk – Zawada
Piotr Żurawski – Natan Cymertopf
Andrzej Szenajch – Generał
Daniel Olbrychski – właściciel kabrioletu
Grzegorz Łaguna – maturzysta
Sebastian Cybulski – maturzysta
Monika Jakowczuk – restauratorka Marina
Magdalena Lamparska – Małgosia
Karol Stępkowski – matematyk
Zofia Tomaszewska-Charewicz – Rózia
Jerzy Słonka – urzędnik
Michał Rolnicki – kolega Skowrońskiego
Bożena Adamek
i inni.
Opis fabuły
Scenariusz filmu powstał na kanwie debiutanckiej powieści Dawida Bieńkowskiego Jest. Film opowiada o losach trzech przyjaciół, maturzystów z 1938. Opisuje tragiczne losy pokolenia ludzi urodzonych już w świeżo odzyskanej ojczyźnie. Wojna tym ludziom odebrała u progu dorosłości bliskich, nadzieję, przyszłość, cały dobytek, szanse na normalne życie, a bardzo często pozbawiła ich życia.
Jest rok 1938, coraz poważniej mówi się o wojnie. Polska młodzież wychowana w duchu romantyzmu, a przede wszystkim będąca pierwszym pokoleniem urodzonym w wolnym kraju, za punkt honoru bierze sobie wojskową służbę. Troje przyjaciół: Andrzej Skowroński, Jerzy Bolesławski i Piotr Dołowy kończą gimnazjum. Zaraz po maturze stają przed pierwszym poważnym życiowym krokiem – co dalej?
Do wojska dostają się Andrzej – do szkoły podchorążych kawalerii w Grudziądzu i Jerzy – do szkoły podchorążych łączności w Zegrzu. Piotr, mający wadę serca, pozostaje w cywilu i rozpoczyna studia na medycynie. Każdy z trójki przyjaciół, tak naprawdę jeszcze chłopców, wciąż traktuje życie jako ekscytującą przygodę, więc zapowiedzi coraz bardziej nieuchronnej wojny podświadomie starają się przyjmować bez emocji. Ponieważ są młodzi, z gorącymi i otwartymi sercami, w kręgu zainteresowań są kobiety. Chłopcy są pełni wiary w to, że nic złego nie może ich spotkać, bo przecież jeszcze tyle chcą w życiu zrobić i doświadczyć.
Andrzej zakochał się w pięknej Ani (Marta Ścisłowicz), córce starosty szykującej się do studiów malarskich we Lwowie i umawia się z nią w końcu na randkę. Mają się spotkać szóstego września, gdy on wróci z ćwiczeń z Pomorza, a ona ze Lwowa. Spotkanie pierwszego września wyznaczają sobie z kolei Jurek i Krysia (Anna Gzyra), któremu udało się wreszcie przełamać opory dziewczyny. Mają iść do kina, a później do niej. Piotr i Zosia (Julia Pietrucha) są w tym samym czasie nad morzem, na pierwszych wspólnych wakacjach.
Nadchodzi wrzesień 1939 roku. Wieść o wybuchu wojny spada na nich wszystkich jak grom. Dla niektórych jej pierwszy dzień będzie zarazem ostatnim. Jerzy, Andrzej i Piotr nigdy więcej się nie spotkali. Żaden z nich nie przeżył wojny.
Zobacz też
Może kiedyś innym razem...
Przypisy
Linki zewnętrzne
Filmy kręcone w Górze Kalwarii
Filmy kręcone w Kazuniu Nowym
Filmy kręcone w Konstancinie-Jeziornie
Filmy kręcone w Milanówku
Filmy kręcone w Modlinie
Filmy kręcone w Warszawie
Filmy w reżyserii Michała Kwiecińskiego
Polskie dramaty filmowe
Polskie filmy obyczajowe
Polskie filmy telewizyjne
Polskie filmy z 2007 roku | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 2,388 |
<?php
declare(strict_types=1);
namespace Eelly\SDK\Log\Api;
use Eelly\SDK\EellyClient;
use Eelly\SDK\Log\Service\UserHandleInterface;
use Eelly\DTO\UserHandleDTO;
/**
*
* @author shadonTools<localhost.shell@gmail.com>
*/
class UserHandle implements UserHandleInterface
{
/**
* @author eellytools<localhost.shell@gmail.com>
*/
public function getUserHandle(int $UserHandleId): UserHandleDTO
{
return EellyClient::request('log/userHandle', __FUNCTION__, true, $UserHandleId);
}
/**
* @author eellytools<localhost.shell@gmail.com>
*/
public function getUserHandleAsync(int $UserHandleId)
{
return EellyClient::request('log/userHandle', __FUNCTION__, false, $UserHandleId);
}
/**
* @author eellytools<localhost.shell@gmail.com>
*/
public function addUserHandle(array $data): bool
{
return EellyClient::request('log/userHandle', __FUNCTION__, true, $data);
}
/**
* @author eellytools<localhost.shell@gmail.com>
*/
public function addUserHandleAsync(array $data)
{
return EellyClient::request('log/userHandle', __FUNCTION__, false, $data);
}
/**
* @author eellytools<localhost.shell@gmail.com>
*/
public function updateUserHandle(int $UserHandleId, array $data): bool
{
return EellyClient::request('log/userHandle', __FUNCTION__, true, $UserHandleId, $data);
}
/**
* @author eellytools<localhost.shell@gmail.com>
*/
public function updateUserHandleAsync(int $UserHandleId, array $data)
{
return EellyClient::request('log/userHandle', __FUNCTION__, false, $UserHandleId, $data);
}
/**
* @author eellytools<localhost.shell@gmail.com>
*/
public function deleteUserHandle(int $UserHandleId): bool
{
return EellyClient::request('log/userHandle', __FUNCTION__, true, $UserHandleId);
}
/**
* @author eellytools<localhost.shell@gmail.com>
*/
public function deleteUserHandleAsync(int $UserHandleId)
{
return EellyClient::request('log/userHandle', __FUNCTION__, false, $UserHandleId);
}
/**
* @author eellytools<localhost.shell@gmail.com>
*/
public function listUserHandlePage(array $condition = [], int $limit = 10, int $currentPage = 1): array
{
return EellyClient::request('log/userHandle', __FUNCTION__, true, $condition, $limit, $currentPage);
}
/**
* @author eellytools<localhost.shell@gmail.com>
*/
public function listUserHandlePageAsync(array $condition = [], int $limit = 10, int $currentPage = 1)
{
return EellyClient::request('log/userHandle', __FUNCTION__, false, $condition, $limit, $currentPage);
}
/**
* @return self
*/
public static function getInstance(): self
{
static $instance;
if (null === $instance) {
$instance = new self();
}
return $instance;
}
} | {
"redpajama_set_name": "RedPajamaGithub"
} | 9,020 |
Q: How to avoid bash command substitution to remove the newline character? To speed up some bash script execution, I would like to keep the result of a command in a variable using command substitution, but the command substitution replaces the 0x0A newline character by a space. For example:
a=`df -H`
or
a=$( df -H )
When I want to process further $a, the newline characters are replaced by a space and all the lines are now on one line, which is much harder to grep:
echo $a
What would be the easy tricks to avoid the newline character being removed by the command substitution?
A: Non-trailing newlines are not removed
The newlines you are looking for are there, you just don't see them, because you use echo without quoting the variable.
Validation:
$ a=$( df -H )
$ echo $a
Filesystem Size Used Avail Use% Mounted on /dev/sda3 276G 50G 213G 19% / udev 2.1G 4.1k 2.1G 1% /dev tmpfs 832M 820k 832M 1% /run none 5.3M 0 5.3M 0% /run/lock none 2.1G 320k 2.1G 1% /run/shm
$ echo "$a"
Filesystem Size Used Avail Use% Mounted on
/dev/sda3 276G 50G 213G 19% /
udev 2.1G 4.1k 2.1G 1% /dev
tmpfs 832M 820k 832M 1% /run
none 5.3M 0 5.3M 0% /run/lock
none 2.1G 320k 2.1G 1% /run/shm
$
Trailing newlines are removed
As @user4815162342 correctly pointed out, although newlines within the output are not removed, trailing newlines are removed with command substitution. See experiment below:
$ a=$'test\n\n'
$ echo "$a"
test
$ b=$(echo "$a")
$ echo "$b"
test
$
In most cases this does not matter, because echo will add the removed newline (unless it is invoked with the -n option), but there are some edge cases where there are more that one trailing newlines in the output of a program, and they are significant for some reason.
Workarounds
1. Add dummy character
In these case, as @Scrutinizer mentioned, you can use the following workaround:
$ a=$(printf 'test\n\n'; printf x); a=${a%x}
$ echo "$a"
test
$
Explanation: Character x is added to the output (using printf x), after the newlines. Since the newlines are not trailing any more, they are not removed by the command substitution. The next step is to remove the x we added, using the % operator in ${a%x}. Now we have the original output, with all newlines present!!!
2. Read using process substitution
Instead of using command substitution to assign the output of a program to variable, we can instead use process substitution to feed the output of the program to the read built-in command (credit to @ormaaj). Process substitution preserves all newlines. Reading the output to a variable is a bit tricky, but you can do it like this:
$ IFS= read -rd '' var < <( printf 'test\n\n' )
$ echo "$var"
test
$
Explanation:
*
*We set the internal field separator for the read command to null, with IFS=. Otherwise read would not assign the entire output to var, but only the first token.
*We invoke read with options -rd ''. The r is for preventing the backslash to act as a special character, and with d '' set the delimiter to nothing, so that read reads the entire output, instead of just the first line.
3. Read from a pipe
Instead of using command or process substitution to assign the output of a program to variable, we can instead pipe the output of the program to the read command (credit to @ormaaj). Piping also preserves all newlines. Note however, that this time we set the lastpipe shell optional behavior, using the shopt builtin. This is required, so that the read command is executed in the current shell environment. Otherwise, the variable will be assigned in a subshell, and it will not be accessible from the rest of the script.
$ cat test.sh
#!/bin/bash
shopt -s lastpipe
printf "test\n\n" | IFS= read -rd '' var
echo "$var"
$ ./test.sh
test
$
A: I was trying to wrap my head around this because I was using bash to stream in the result of running the interpreter on an F# script. After some trial and error, this turned out to solve the problem:
$ cat fsi.ch
#!/bin/bash
echo "$(fsharpi --quiet --exec --nologo $1)"
$ fsi.ch messages.fsx
Welcome to my program. Choose from the menu:
new | show | remove
Assuming, of course that you need to run a terminal program.
Hope this helps.
A: Another "neat trick" is to use the carriage return character, which prevents the newline from being stripped but doesn't add anything to the output:
$ my_func_1 () {
> echo "This newline is squashed"
> }
$ my_func_2 () {
> echo "This newline is not squashed"
> echo -n $'\r'
> }
$ echo -n "$(my_func_1)" && echo -n "$(my_func_2)" && echo done
This newline is squashedThis newline is not squashed
done
$
But buyer beware: as mentioned in the comments this can work nicely for output that is simply going to the terminal, but if you are passing this on to another process you might confuse it as it probably won't be expecting the weird terminating '\r'.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 3,459 |
Count Enrico Marone Cinzano (born April 5, 1963) is a financier, entrepreneur and artist from Turin, Italy. His work is based on the integration of Nature and technology, and his designs incorporate reclaimed, third party-certified sustainable as well as recycled materials and are built using eco-friendly construction techniques, with the support of advanced well-being focused technologies.
Early life
Marone Cinzano was born to Count Alberto Paolo Marone Cinzano who died in 1989 is a car crash on his way to visiting King Juan Carlos of Spain and Cristina Marone Cinzano, (born Countess Camerana). He attended boarding school at Aiglon College and Mount Kelly School, a military academy in the UK. Sub-sequentially, he graduated with a BS business administration at Babson College, Massachusetts. He then worked in advertising, banking, and became a real-estate developer, before co-founding the environmentally friendly fashion company Project Alabama in the year 2000, which he sold in 2010. Project Alabama's accolades include winner of the 2003 Ecco Domani award and runner up for the 2005 CFDA award.
Career
In 2007, Marone Cinzano established ToTheCube LLC, under which he presented the Enrico Marone Cinzano Collection www.enrico.art. The company focuses on ethical and sustainable homes and home products. He creates living spaces and sculptural art utilizing natural, recycled, third party certified sustainable materials,
as well as avant-garde technologies, with the intention of providing well being in home environments. His work has been exhibited in group and solo shows, is represented by Pearl Lam Galleries in Hong Kong and Shanghai, Spazio Rossana Orlandi in Milan, FriedmanBenda in New York City and also collaborated with Shanghai-based Stellar Works, and has shown at Art Basel HK, PAD London and Design Miami in Miami. In 2011, Marone Cinzano became a member of ADI, the industrial design organization based in Milan, Italy that brings professionals, researchers, teachers, critics, journalists, and the like together to examine the foremost topics of design. His museum shows have included a presence at the
Museo Bagatti Valsecchi in Milan and the Museo Nazionale della Scienza e della Tecnologia di Milano.
In 2012, Marone Cinzano applied the same ethos to architecture and interior design, developing property in New York and London, based on recovered non-toxic materials to create healthy living spaces: in 2017, he was invited as a speaker to a TED conference in Kings College, London, to discuss health and well-being in the home.
Family history
His family founded the epymous Cinzano wine company in 1757, his maternal great grandfather, Giovanni Agnelli, was the founder of FIAT Automobiles in 1899. Other notable relatives include Argentinian diplomat Amancio Alcorta and former President of Argentina Jose Figueroa Alcorta. His step-grandmother, Infanta Maria Cristina de Borbon y Battenberg, a member of the Spanish royal family, was the daughter of Spanish King Alfonso XIII and Spanish Queen Victoria Eugenie of Battenberg, who in turn was the grand-daughter of Queen Victoria. His grandfather, Enrico, was named the 1st Count Marone by King Victor Emmanuel III of Italy on May 13, 1940, to mark his marriage to the HRH Infanta Maria Cristina.
Personal life
On July 8, 1989, Marone Cinzano married Princess Mafalda of Hesse-Cassel, also a descendant of Queen Victoria, the couple have since separated. He lived in the Bacchus House in New York City, where he was known for his glamorous and extravagant lifestyle. He sold the property to Napster founder and Facebook executive Sean Parker. He currently lives in London and has a daughter.
Ancestry
References
Living people
1963 births
Babson College alumni
Counts of Italy
Italian nobility | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 3,078 |
\section{Introduction}
Words and their interactions (as sentences) are the basic units of natural language. Although words are readily modeled as discrete atomic units, this is unable to capture the relation between the words. Recent distributional real-valued representations of words (examples: word2vec, GloVe) have transformed the landscape of NLP applications -- for instance, text classification \citep{socher2013recursive,maas2011learning,kim2014convolutional}, machine translation \citep{sutskever2014sequence,bahdanau2014neural} and knowledge base completion \citep{bordes2013translating,socher2013reasoning}. The success comes from the geometry of the representations that efficiently captures linguistic regularities: the semantic similarity of words are well captured by the similarity of the corresponding vector representations; the latent relation between two words is well captured by the difference vector between the corresponding two vector representations.
A variety of approaches have been proposed in recent years to learn the word representations: \citet{collobert2011natural,turian2010word} learn the representations via semi-supervised learning by jointly training the language model and downstream applications; \citet{bengio2003neural,mikolov2010recurrent,huang2012improving} do so by fitting the data into a neural network language model; \citet{mikolov2013efficient,mnih2007three} by log-linear models; and \citet{dhillon2012two,pennington2014glove,levy2014neural,stratos2015model,arora2015rand} by producing a low-dimensional representation of the cooccurrence statistics. Despite the wide disparity of algorithms to induce word representations, the performance of several of the recent methods is roughly similar on a variety of intrinsic and extrinsic evaluation testbeds.
In this paper, we find that a {\em simple} processing renders the off-the-shelf existing representations {\em even stronger}. The proposed algorithm is motivated by the following observation.
\paragraph{Observation} {\em Every} representation we tested, in many languages, has the following properties:
\begin{itemize}
\item The word representations have {\em non-zero mean} -- indeed, word vectors share a large common vector (with norm up to a half of the average norm of word vector).
\item After removing the common mean vector, the representations are {\em far from} isotropic -- indeed, much of the energy of most word vectors is contained in a very low dimensional subspace (say, 8 dimensions out of of 300).
\end{itemize}
\paragraph{Implication} Since all words share the same common vector and have the same dominating directions, and such vector and directions strongly influence the word representations in the same way, we propose to eliminate them by: (a) removing the nonzero mean vector from all word vectors, effectively reducing the energy; (b) projecting the representations {\em away} from the dominating $D$ directions, effectively reducing the dimension. Experiments suggest that $D$ depends on the representations (for example, the dimension of the representation, the training methods and their specific hyperparameters, the training corpus) and also depends on the downstream applications. Nevertheless, a rule of thumb of choosing $D$ around $d/100$, where $d$ is the dimension of the word representations, works uniformly well across multiple languages and multiple representations and multiple test scenarios.
We emphasize that the proposed postprocessing is {\em counter intuitive} -- typically denoising by dimensionality reduction is done by eliminating the {\em weakest} directions (in a singular value decomposition of the stacked word vectors), and {\em not} the dominating ones. Yet, such postprocessing yields a ``purified'' word representation as seen in our elaborate experiments.
\paragraph{Experiments} By postprocessing the word representation by eliminating the common parts, we find the processed word representations to capture stronger linguistic regularities. We demonstrate this quantitatively, by comparing the performance of both the original word representations and the processed ones on three canonical lexical-level tasks:
\begin{itemize}
\item {\em word similarity} task tests the extent to which the representations capture the similarity between two words -- the processed representations are consistently better on seven different datasets, on average by 2.3\%;
\item {\em concept categorization} task tests the extent to which the clusters of word representations capture the word semantics -- the processed representations are consistently better on three different datasets, by 2.8\%, 4.5\% and 4.3\%;
\item {\em word analogy} task tests the extent to which the difference of two representations captures a latent linguistic relation -- again, the performance is consistently improved (by 0.5\% on semantic analogies, 0.2\% on syntactic analogies and 0.4\% in total). Since part of the dominant components are inherently canceled due to the subtraction operation while solving the analogy, we posit that the performance improvement is not as pronounced as earlier.
\end{itemize}
Extrinsic evaluations provide a way to test the goodness of representations in specific downstream tasks. We evaluate the effect of postprocessing on a standardized and important extrinsic evaluation task on sentence modeling:
{\em semantic textual similarity} task -- where we represent a sentence by its averaged word vectors and score the similarity between a pair of sentences by the cosine similarity between the corresponding sentence representation. postprocessing improves the performance consistently and significantly over 21 different datasets (average improvement of 4\%).
Word representations have been particularly successful in NLP applications involving supervised-learning, especially in conjunction with neural network architectures. Nodes in all neural network architectures have enough flexiblity to conduct the postprocessing step (i.e., subtract a bias term and null out a subspace) and thus one conjectures that perhaps postprocessing does not significantly affect the end-to-end performance. Indeed, we see this in an experiment on a standard {\em text classification} task on five datasets using a well established convoluntional neural network (CNN) classifier \cite{kim2014convolutional}. The performance improves only mildly with postprocessing over all five datasets.
Towards verifying our hypothesis that the neural network ``effectively learns to postprocess" within its nodes, we conduct the following experiement. We append an extra {\em linear} layer to the beginning of a recurrent neural network (RNN) where the classifier is built on the last hidden unit of the RNN. The bias term of the linear layer is jointly trained with the parameters of the RNN to best fit the training set (of the same five different datasets of the text classification task). We find that the bias term of the linear layer when inferred with the postprocessed word representations plus the mean vector of word representations is {\em very close} to the bias term of the linear layer when inferred with the original word representations (the cosine similarity is roughly 0.69, across the five datasets and three different RNN architectures -- the vanilla RNN, GRU-RNN \cite{chung2015gated} and LSTM-RNN \cite{greff2016lstm}). This allows us to conclude that the neural networks ``learn'' to postprocess the word representations.
We summarize our contributions below:
\begin{itemize}
\item We observe a novel geometric property of word representations on every representation learning algorithm we tested;
\item We propose a {\em very simple} postprocessing algorithm to purify the word representations so that they can capture stronger linguistic regularities;
\item We quantitatively demonstrate the superiority of the processed word representations using intrinsic and extrinsic evaluations, over a very wide range of standardized datasets, word representation methods with varying hyperparameters and in multiple languages.
\end{itemize}
\paragraph{Related Work} Our work is directly related to word representation algorithms, most of which have been elaborately cited in the introduction and also throughout the entire paper, at the appropriate locations. To the best of our knowledge, this is the first paper discussing a postprocessing algorithm on {\em pre-trained} word representations.
\section{Postprocessing}
We test our observations on various word representations: four publicly available word representations (WORD2VEC\footnote{\url{https://code.google.com/archive/p/word2vec/}} \cite{mikolov2013efficient} trained using Google News, GLOVE\footnote{\url{https://github.com/stanfordnlp/GloVe}} \cite{pennington2014glove} trained using Common Crawl, RAND-WALK \cite{arora2015rand} trained using Wikipedia and TSCCA\footnote{\url{http://www.pdhillon.com/code.html}} trained using English Gigaword) and two self-trained word representations using CBOW and Skip-gram \cite{mikolov2013efficient} on the 2010 Wikipedia corpus from \cite{al2013polyglot}. For completeness, we also consider the representations on other languages: we use the publicly available TSCCA representations \cite{dhillon2012two} on German, French, Spanish, Italian, Dutch and Chinese. The detailed statistics for all representations are listed in Table~\ref{tb:embedding}.
\begin{table*}
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|}
\hline
& \bf Language & \bf Corpus & \bf dim & \bf vocab size & \bf avg. $\|v(w)\|_2$ & $\bf \|\mu\|_2$\\
\hline
WORD2VEC & English & Google News & 300 & 3,000,000 & 2.04 & 0.69 \\
GLOVE & English & Common Crawl & 300 & 2,196,017 & 8.30 & 3.15\\
RAND-WALK & English & Wikipedia & 300 & 68, 430 & 2.27 & 0.70\\
CBOW & English & Wikipedia & 300 & 1,028,961 & 1.14 & 0.29 \\
Skip-Gram & English & Wikipedia & 300 & 1,028,961 & 2.32 & 1.25 \\
\hline
TSCCA-En & English & Gigawords & 200 & 300,000 & 4.38 & 0.78 \\
TSCCA-De & German & Newswire & 200 & 300,000 & 4.52 & 0.79 \\
TSCCA-Fr & French & Gigaword & 200 & 300,000 & 4.34 & 0.81 \\
TSCCA-Es & Spanish & Gigaword & 200 & 300,000 & 4.17 & 0.79 \\
TSCCA-It & Italian & Newswire+Wiki & 200 & 300,000 & 4.34 & 0.79 \\
TSCCA-Nl & Dutch & Newswire+Wiki & 200 & 300,000 & 4.46 & 0.72 \\
TSCCA-Zh & Chinese & Gigaword & 200 & 300,000 & 4.51 & 0.89 \\
\hline
\end{tabular}
\end{center}
\caption{A detailed description for the embeddings in this paper.}
\label{tb:embedding}
\end{table*}
Let $v(w) \in \mathbb{R}^d $ be a word representation for a given word $w$ in the vocabulary $V$. We observe the following two phenomena in each of the word representations listed above:
\begin{itemize}
\item $\{v(w): w\in V\}$ are not of zero-mean: i.e., all $v(w)$ share a non-zero common vector,
\begin{align*}
v(w) = \tilde{v}(w) + \mu,
\end{align*}
where $\mu$ is the average of all $v(w)$'s, i.e., $\mu = {1}/{|V|} \sum_{w\in V} v(w)$. As reported in Table~\ref{tb:embedding}, the norm of $\mu$ is approximately 1/6 to 1/2 of the average norm of all $v(w)$.
\item $\{\tilde{v}(w): w\in V\}$ are not isotropic: Let $u_1,..., u_d$ be the first to the last components recovered by the principal component analysis (PCA) of $\{\tilde{v}(w): w\in V\}$, and $\sigma_1,...,\sigma_d$ be the corresponding normalized variance ratio. Each $\tilde{v}(w)$ can be written as a linear combinations of $u$:
\begin{align*}
\tilde{v}(w) = \sum_{i=1}^d \alpha_i(w) u_i.
\end{align*}
As shown in Figure~\ref{fig:decay}, we observe that $\sigma_i$ decays near exponentially for small values of $i$ and remains roughly constant over the later ones. This suggests there exists $D$ such that $\alpha_i \gg \alpha_j$ for all $i \le D$ and $j \gg D$; from Figure~\ref{fig:decay} one observes that $D$ is roughly 10 with dimension $d=300$.
\end{itemize}
\begin{figure}[!h]
\centering
\subfigure[English]
{
\includegraphics[width = 0.4\textwidth]{figures/decay.pdf}
}
\subfigure[Multilingual]
{
\includegraphics[width = 0.4\textwidth]{figures/TSCCA-decay.pdf}
}
\caption{The decay of the normalized singular values of word representation.}
\label{fig:decay}
\end{figure}
Since all word representations share the same common vector $\mu$ and have the same dominating directions and such vector and directions strongly influence the word representations in the same way, we propose to eliminate them, as formally achieved in Algorithm~\ref{algo:representation}.
\begin{algorithm}[!h]
\SetKwInOut{Input}{Input}
\SetKwInOut{Output}{Output}
\Input{Word representations $\{v(w), w\in V\}$, a threshold parameter $D$, }
Compute the mean of $\{v(w), w\in V\}$,
$$ \mu \leftarrow \frac{1}{|V|} \sum_{w\in V} v(w); \quad \tilde{v}(w) \leftarrow v(w) - \mu$$, \\
Compute the PCA components:
$$u_1,...,u_d \leftarrow {\rm PCA}(\{\tilde{v}(w), w\in V\}).$$ \\
postprocess the representations:
$$v'(w) \leftarrow \tilde{v} - \sum_{i=1}^D \left(u_i^{\rm T} v(w)\right) u_i$$ \\
\Output{Processed representations $v'(w)$.}
\caption{All-But-The-Top: a postprocessing algorithm on word representations.}
\label{algo:representation}
\end{algorithm}
\paragraph{Discussion} In our proposed processing algorithm, the number of components to be nulled, $D$, is the only hyperparameter that needs to be tuned. We find that a good rule of thumb is to choose $D$ approximately to be $d/100$, where $d$ is the dimension of a word representation. This is empirically justified in the experiments of the following section where $d=300$ is standard for published word representations. We trained word representations for higher values of $d$ using the WORD2VEC and GLOVE algorithms and
repeat these experiments; we see corresponding consistent improvements due to postprocessing in Appendix \ref{app:extra}.
\section{Experiments}
Given the popularity and widespread use of WORD2VEC \cite{mikolov2013efficient} and GLOVE \cite{pennington2014glove}, we use their publicly available pre-trained reprepsentations in the following experiments. We choose $D=3$ for WORD2VEC and $D=2$ for GLOVE.
The key underlying principle behind word representations is that similar words should have similar representations. Following the tradition of evaluating word representations \citep{schnabel2015evaluation,baroni2014don}, we perform three canonical {\em lexical-level} tasks: (a) word similarity; (b) concept categorization; (c) word analogy; and one {\rm sentence-level} task: (d) semantic textual similarity. The processed representations consistently improve performance on all three of them, and especially strongly on the first two.
\subsection{Word Similarity}
\label{sec:similarity}
The word similarity task is as follows: given a pair of words, the algorithm assigns a ``similarity" score -- if the pair of words are highly related then the score should also be high and vice versa. The algorithm is evaluated in terms of Spearman's rank correlation compared to (a gold set of) human judgements.
For this experiment, we use seven standard datasets: the first published RG65 dataset \cite{rubenstein1965contextual}; the widely used WordSim-353 (WS) dataset \cite{finkelstein2001placing} which contains 353 pairs of commonly used verbs and nouns; the rare-words (RW) dataset \cite{luong2013better} composed of rarely used words; the MEN dataset \cite{bruni2014multimodal} where the 3000 pairs of words are rated by crowdsourced participants; the MTurk dataset \cite{radinsky2011word} where the 287 pairs of words are rated in terms of relatedness; the SimLex-999 (SimLex) dataset \cite{hill2016simlex} where the score measures ``genuine" similarity; and lastly the SimVerb-3500 (SimVerb) dataset \cite{gerz2016simverb}, a newly released large dataset focusing on
similarity of verbs.
In our experiment, the algorithm scores the similarity between two words by the cosine similarity between the two corresponding word vectors (${\rm CosSim}(v_1,v_2) = v_1^{\rm T}v_2 / \|v_1\|\|v_2\|$).
The detailed performance on the seven datasets is reported in Table~\ref{tb:similarity}, where we see a consistent and significant performance improvement due to postprocessing, across all seven datasets. These statistics (average improvement of 2.3\%) suggest that by removing the common parts, the remaining word representations are able to capture stronger semantic relatedness/similarity between words.
\begin{table}[!h]
\centering
\begin{tabular}{|r||c|c||c|c|}
\hline
\multirow{2}{*}{} & \multicolumn{2}{c||}{WORD2VEC} & \multicolumn{2}{c|}{GLOVE} \\ \cline{2-5}
& orig. & proc. & orig. & proc. \\ \hline
RG65 & 76.08 & \bf 78.34 & \bf 76.96 & 74.36 \\ \hline
WS & 68.29 & \bf 69.05 & 73.79 & \bf 76.79 \\ \hline
RW & 53.74 & \bf 54.33 & 46.41 & \bf 52.04 \\ \hline
MEN & 78.20 & \bf 79.08 & 80.49 & \bf 81.78 \\ \hline
MTurk & 68.23 & \bf 69.35 & 69.29 & \bf 70.85\\ \hline
SimLex & 44.20 & \bf 45.10 & 40
83 & \bf 44.97 \\ \hline
SimVerb & 36.35 & \bf 36.50 & 28.33 & \bf 32.23 \\ \hline
\end{tabular}
\caption{Before-After results (x100) on word similarity task on seven datasets.}
\label{tb:similarity}
\end{table}
\subsection{Concept Categorization}
This task is an indirect evaluation of the similarity principle: given a set of concepts, the algorithm needs to group them into different categories (for example, ``bear'' and ``cat'' are both animals and ``city'' and ``country'' are both related to districts). The clustering performance is then evaluated in terms of purity \cite{manning2008introduction} -- the fraction of the total number of the objects that were classified correctly.
We conduct this task on three different datasets: the Almuhareb-Poesio (ap) dataset \cite{almuhareb2006attributes} contains 402 concepts which fall into 21 categories; the ESSLLI 2008 Distributional Semantic Workshop shared-task dataset \cite{baroni2008bridging} that contains 44 concepts in 6 categories; and the Battig test set \cite{baroni2010distributional} that contains 83 words in 10 categories.
Here we follow the setting and the proposed algorithm in \cite{baroni2014don,schnabel2015evaluation} -- we cluster words (via their representations) using the classical $k$-Means algorithm (with fixed $k$).
Again, the processed vectors perform consistently better on all three datasets (with average improvement of 2.5\%); the full details are in Table~\ref{tb:categorization}.
\begin{table}[!h]
\centering
\begin{tabular}{|r||c|c||c|c|}
\hline
\multirow{2}{*}{} & \multicolumn{2}{c||}{WORD2VEC} & \multicolumn{2}{c|}{GLOVE} \\ \cline{2-5}
& orig. & proc. & orig. & proc. \\ \hline
ap & 54.43 & \bf 57.72 & 64
.18 & \bf 65.42 \\ \hline
esslli & 75.00 & \bf 84.09 & 81.82 & \bf 81.82 \\ \hline
battig & 71.97 & \bf 81.71 & 86.59 & \bf 86.59 \\ \hline
\end{tabular}
\caption{Before-After results (x100) on the categorization task.}
\label{tb:categorization}
\end{table}
\subsection{Word Analogy}
The analogy task tests to what extent the word representations can encode latent linguistic relations between a pair of words. Given three words $w_1$, $w_2$, and $w_3$, the analogy task requires the algorithm to find the word $w_4$ such that $w_4$ is to $w_3$ as $w_2$ is to $w_1$.
We use the analogy dataset introduced in \citep{mikolov2013efficient}. The dataset can be divided into two parts: (a) the {\em semantic} part containing around 9k questions, focusing on the latent semantic relation between pairs of words (for example, what is to {\rm Chicago} as {\rm Texas} is to {\rm Houston}); and (b) the {\em syntatic} one containing roughly 10.5k questions, focusing on the latent syntatic relation between pairs of words (for example, what is to ``amazing'' as ``apprently'' is to ``apparent'').
In our setting, we use the original algorithm introduced in \citep{mikolov2013efficient} to solve this problem, i.e., $w_4$ is the word that maximize the cosine similarity between $v(w_4)$ and $v(w_2) - v(w_1) + v(w_3)$.
The detailed performance on the analogy task is provided in Table~\ref{tb:analogy}. It can be noticed that while postprocessing continues to improve the performance, the improvement is not as pronounced as earlier. We hypothesize that this is because the mean and some dominant components get canceled during the subtraction of $v(w_2)$ from $v(w_1)$, and therefore the effect of postprocessing is less relevant.
\begin{table}[!h]
\centering
\begin{tabular}{|r||c|c||c|c|}
\hline
\multirow{2}{*}{} & \multicolumn{2}{c||}{WORD2VEC} & \multicolumn{2}{c|}{GLOVE} \\ \cline{2-5}
& orig. & proc. & orig. & proc. \\ \hline
syn. & 73.46 & \bf 73.50 & 74.95 & \bf 75.40 \\ \hline
sem. & 72.28 & \bf 73.36 & 79.22 & \bf 79.25 \\ \hline
all & 72.93 & \bf 73.44 & 76.89 & \bf 77.15 \\ \hline
\end{tabular}
\caption{Before-After results (x100) on the word analogy task.}
\label{tb:analogy}
\end{table}
\vspace{-0.05in}
\subsection{Semantic Textual Similarity}
Extrinsic evaluations measure the contribution of a word representation to specific downstream tasks; below, we study the effect of postprocessing on a standard sentence modeling task -- {\em semantic textual similarity} which aims at testing the degree to which the algorithm can capture the semantic equivalence between two sentences. For each pair of sentences, the algorithm needs to measure how similar the two sentences are. The degree to which this measure matches with human judgment (in terms of Pearson's correlation) is an index of the algorithm's performance.
We test the word representations on the 20 textual similarity datasets from the 2012-2015 SemEval semantic textual similarity (STS) shared tasks \cite{agirre2012semeval,agirre2013sem,agirre2014semeval,agirrea2015semeval}, and the 2012 SemEval Semantic Related task (SICK) \cite{marelli2014sick}.
Representing sentences by the average of their constituent word representations is surprisingly effective in encoding the semantic information of sentences
\cite{wieting2015paraphrase,adi2016fine} and close to the state-of-the-art in these datasets. We follow this rubric and represent a sentence $s$ based on its averaged word representation, i.e., $v(s) = 1/|s|\sum_{w\in s} v(w)$, and then compute the similarity between two sentences via the cosine similarity between the two
representations.
The detailed performance of the original and processed representations is itemized in Table~\ref{tb:sts} -- we see a consistent and significant improvement in performance because of postprocessing (on average 4\% improvement).
\begin{table}[!h]
\centering
\begin{tabular}{|r||c|c||c|c|}
\hline
\multirow{2}{*}{} & \multicolumn{2}{c||}{WORD2VEC} & \multicolumn{2}{c|}{GLOVE} \\ \cline{2-5}
& orig. & proc. & orig. & proc. \\ \hline
2012 & 57.22 & \bf 57.67 & 48.27 & \bf 54.06 \\ \hline
2013 & 56.81 & \bf 57.98 & 44.83 & \bf 57.71 \\ \hline
2014 & 62.89 & \bf 63.30 & 51.11 & \bf 59.23 \\ \hline
2015 & 62.74 & \bf 63.35& 47.23 & \bf 57.29 \\ \hline
SICK & 70.10 & \bf 70
20 & 65.14 & \bf 67.85 \\ \hline
all & 60.88 & \bf 61.45 & 49.19 & \bf 56.76 \\ \hline
\end{tabular}
\caption{Before-After results (x100) on the semantic textual similarity tasks.}
\label{tb:sts}
\end{table}
\section{Neural Networks Learn to Postprocess}
Supervised downstream NLP applications have greatly improved their performances in recent years by combining the discriminative learning powers of neural networks in conjunction with the word representations. All neural network architectures, ranging from feedforward to recurrent (either vanilla or GRU or LSTM), implement at least linear processing of hidden/input state vectors at each of their nodes; thus the postprocessing operation suggested in this paper can in principle be automatically ``learnt'' by the neural network, if such internal learning is in-line with the end-to-end training examples. Following this line of thought, the natural conjecture would be that both the original and postprocessed word representations should yield similar end-to-end results.
\subsection{Postprocessing does not Affect Neural Network Performance}
We evaluate the performance of a variety of neural network architectures on a standard and important NLP application: {\em text classification}, with
sentiment analysis being a particularly important and popular example. The task is defined as follows: given a sentence, the algorithm needs to decide which category it falls into. The categories can be either binary (e.g., positive/negative) or can be more fine-grained (e.g. very positive, positive, neutral, negative, and very negative).
We evaluate the word representations (with and without postprocessing) using four different neural network architectures (CNN, vanilla-RNN, GRU-RNN and LSTM-RNN) on five benchmarks: (a) the movie review (MR) dataset \cite{pang2005seeing};
(b) the subjectivity (SUBJ) dataset \cite{pang2004sentimental};
(c) the TREC question dataset \cite{li2002learning};
(d) the IMDb dataset \cite{maas2011learning};
(e) the stanford sentiment treebank (SST) dataset \cite{socher2013reasoning}.
A detailed description of these standard datasets, their training/test parameters and the cross validation methods adopted is in Appendix~\ref{app:sentiment}.
The performance of the four neural network architectures with the now-standard CNN-based text classification algorithm \cite{kim2014convolutional} (implemented using tensorflow\footnote{\url{https://github.com/dennybritz/cnn-text-classification-tf}}) is itemized in Table~\ref{tb:sentiment-analysis} -- the key observation is that the two performances, with and without postprocessing, are comparable (although postprocessing is mildly better in a majority of the instances). We hypothesize that this happens because the weight and bias in the neural network automatically encode the postprocessing; this is verified next.
\begin{table*}
\centering
\resizebox{1\textwidth}{!}{
\begin{tabular}{|r|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
\multirow{3}{*}{} & \multicolumn{4}{c|}{CNN} & \multicolumn{4}{c|}{vanilla-RNN} & \multicolumn{4}{c|}{GRU-RNN} & \multicolumn{4}{c|}{LSTM-RNN} \\ \cline{2-17}
& \multicolumn{2}{c|}{WORD2VEC} & \multicolumn{2}{c|}{GLOVE} & \multicolumn{2}{c|}{WORD2VEC} & \multicolumn{2}{c|}{GLOVE} & \multicolumn{2}{c|}{WORD2VEC} & \multicolumn{2}{c|}{GLOVE} & \multicolumn{2}{c|}{WORD2VEC} & \multicolumn{2}{c|}{GLOVE} \\ \cline{2-17}
& orig. & proc. & orig. & proc. & orig. & proc. & orig. & proc. & orig. & proc. & orig. & proc. & orig. & proc. & orig. & proc. \\ \hline
MR & 70.80 & \bf 71.27 & 71.01 & \bf 71.11 & 77.30 & \bf 77.37 & 77.69 & \bf 77.87 & \bf 78.72 & 78.65 & 78.38 & \bf 78.78 & 79.07 & \bf 79.20 & 79.02 & \bf 79.81 \\ \hline
SUBJ & 87.14 & \bf 87.33 & 86.98 & \bf 87.25 & 90.38 & \bf 90.70 & \bf 91.34 & 91.05 & 91.35 & \bf 91.48 & 91.79 & \bf 91.92 & 91.81 & \bf 92.16 & \bf 92.18 & 91.86 \\ \hline
TREC & 87.80 & \bf 89.00 & 87.60 & \bf 89.00 & \bf 89.60 & \bf 89.60 & \bf 89.60 & 88.20 & 90.80 & \bf 92.20 & 91.00 & \bf 91.40 & 90.60 & \bf 91.60 & 91.60 & \bf 92.40 \\ \hline
SST & \bf 38.46 & 38.33 & \bf 38.82 & 37.83 & 41.54 & \bf 41.90 & 42.58 & \bf 43.57 & \bf 44.57 & 44.16 & 43.76 & \bf 43.94 & 44.62 & \bf 45.43 & 44.16 & \bf 44.89 \\ \hline
IMDb & 86.68 & \bf 87.12 & \bf 87.27 & 87.10 & 78.84 & \bf 82.17 & 81.91 & \bf 82.72 & 83.83 & \bf 83.92 & 83.89 & \bf 84.08 & 84.72 & \bf 85.25 & 84.96 & \bf 85.47 \\ \hline
\end{tabular}}
\caption{Before-After results (x100) on the text classification task using CNN \cite{kim2014convolutional} and vanilla RNN, GRU-RNN and LSTM-RNN.}
\label{tb:sentiment-analysis}
\end{table*}
\subsection{An Appended-Layer Neural Network}
We construct a modified neural network by explicitly adding a ``postprocessing unit" as the first layer of the RNN architecture (as in Figure~\ref{fig:rnn}). Such an appended layer can be used to test the first step (i.e., remove the mean vector) of the postprocessing algorithm; setting up a corresponding test of the second step (nulling the top components) is unclear (see Appendix~\ref{app:neural_discussion} for a discussion).
In the modified neural network, the input word vectors are now $v(w) - b$ instead of $v(w)$. Here $b$ is a bias vector trained {\em jointly} with the rest of the neural network parameters. Note that this is only a relabeling of the parameters from the perspective of the RNN architecture: the nonlinear activation function of the node is now operated on $A(v(w)-b) + b' = Av(w) + (b'-Ab)$ instead of the previous $Av(w) + b'$. Let $b_{\rm proc.}$ and $b_{\rm orig.}$ be the inferred biases when using the processed and original word representations, respectively.
We itemize the cosine similarity between $b_{\rm proc.} + \mu$ and $b_{\rm orig.}$ in Table~\ref{tb:biases} for the 5 different datasets and 3 different neural network architectures. In each case, the cosine similarity is remarkably large (on average 0.69, in 300 dimensions) -- in other words, trained neural networks implicitly postprocess the word vectors nearly exactly as we proposed.
\begin{figure}
\centering
\includegraphics[width=0.4\textwidth]{./figures/rnn.pdf}
\caption{Time-expanded RNN architecture with an appended layer involving linear bias. \label{fig:rnn}}
\end{figure}
\begin{table}[]
\centering
\resizebox{0.49\textwidth}{!}{
\begin{tabular}{|r||c|c||c|c||c|c|}
\hline
& \multicolumn{2}{c||}{vanilla} & \multicolumn{2}{c||}{GRU} & \multicolumn{2}{c|}{LSTM} \\ \hline
& W. & G. & W. & G. & W. & G. \\ \hline
MR & 64.8 & 57.9 & 42.4 & 82.8 & 35.1 & 62.2 \\ \hline
SUBJ & 45.2 & 49.6 & 47.7 & 64.9 & 32.9 & 48.2 \\ \hline
TREC & 82.8 & 92.4 & 86.4 & 94.0 & 87.1 & 84.5 \\ \hline
SST & 94.8 & 88.9 & 75.3 & 93.1 & 73.1 & 73.2 \\ \hline
IMDb & 18.9 & 94.9 & 71.3 & 91.8 & 93.6 & 43.5 \\ \hline
\end{tabular}}
\caption{The cosine similarity (x100) between $b_{\rm proc.} + \mu$ and $b_{\rm orig.}$, where W. and G. stand for WORD2VEC and GLOVE respectively.}
\label{tb:biases}
\end{table}
\section{Discussion}
The postprocessing operation is extremely simple and yet, quite effective across a very wide range of both intrinsic and extrinsic evaluations of the word representations. A key question, unaddressed thus far, is whether the dominant singular directions signify anything, and if so what those aspects could be. Our preliminary results are discussed next; we begin by noting that the mean vector is strongly correlated with the first principle component direction (cosine similarity of more than 0.8, in 300 dimensions).
\subsection{Significance of Nulled Vectors}
Consider the representation of the words as viewed in terms of the top $D$ PCA coefficients $\alpha_{\ell}(w)$, for $1\leq \ell \leq D$.
We find that these few coefficients encode the {\em frequency} of the word to a significant degree; Figure~\ref{fig:frequency} illustrates the relation between the ($\alpha_{1}(w),\alpha_2(w))$ and the unigram probabilty $p(w)$, where the correlation is geometrically visible.
\begin{figure*}[]
\centering
\includegraphics[width=1.1\textwidth]{./figures/freq.png}
\vspace{-20pt}
\caption{The top two PCA directions (i.e, $\alpha_1(w)$ and $\alpha_2(w)$) encode frequency.}
\label{fig:frequency}
\end{figure*}
\subsection{Angular Asymmetry of Representations}
A modern understanding of word representations involves either PMI-based (including word2vec \cite{mikolov2010recurrent,levy2014neural} and GloVe \cite{pennington2014glove}) or CCA-based spectral factorization approaches. While CCA-based spectral factorization methods have long been understood from a probabilistic (i.e., generative model) view point \cite{browne1979maximum,hotelling1936relations} and recently in the NLP context \cite{stratos2015model}, a corresponding effort for the PMI-based methods has only recently been conducted in an inspired work \cite{arora2015rand}.
\citet{arora2015rand} propose a generative model (named RAND-WALK) of sentences, where every word is parameterized by a $d$-dimensional vector. With a key postulate that the word vectors are angularly uniform (``isotropic"), the family of PMI-based word representations can be explained under the RAND-WALK model in terms of the maximum likelihood rule.
We have observed that word vectors learnt through PMI-based approaches are not of zero-mean and are not isotropic (c.f.\ Section 2) contradicts with this postulate. The isotropy conditions are relaxed in Section~2.2 of \cite{arora2015rand}, but the match with the spectral properties observed in Figure~\ref{fig:decay} is not immediate.
In this section, we resolve this by explicitly relaxing the constraints on the word vectors to directly fit the observed spectral properties. The relaxed conditions are: the word vectors should be isotropic around a point (whose distance to the origin is a small fraction of the average norm of word vectors) lying on a low dimensional subspace. Our main result is to show that even with this enlarged parameter-space, the maximum likelihood rule continues to be close to the PMI-based spectral factorization methods. A formal proof that integrates with the technical results of \cite{arora2015rand}, and a discussion where we interpret the postprocessing operation as a ``rounding" or ``projection" towards isotropy (with better self-normalization properties \cite{andreas2015and}) is in Appendix~\ref{app:iso}.
\section{Conclusion}
We present a simple postprocessing operation that renders word representations even stronger. Due to their popularity, we have used the published representations of WORD2VEC and GLOVE in English in the main text of this paper; postprocessing continues to be successful for other representations and in multilingual settings -- the detailed empirical results are tabulated in Appendix~\ref{app:extra}.
\newpage
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 5,173 |
\section{Introduction}
The interstellar medium (ISM) is commonly envisioned as a
self-regulating system in statistical quasi-equilibrium. Multiple
components of gas with varying densities and temperatures coexist
\citep{1969ApJ...155L.149F,1974ApJ...189L.105C,1977ApJ...218..148M},
animated by turbulence that pervades the whole volume
\citep{2004ARA&A..42..211E}.
Different components of gas play different roles in the ISM ecosystem,
with the coldest and densest portions responsible for star formation.
Massive stars, when they are born, energize the ISM through the HII
regions and supernova blasts they create \citep{1978ppim.book.....S};
this energy input is important in replenishing continual losses
through turbulent dissipation. UV radiation from young massive stars
is also crucial in heating the gas. The rate of star formation is
determined by the available supply of dense gas, which in turn is
regulated by the interplay between dynamics and thermodynamics in the
ISM, and is affected by the galactic environment in which the ISM is
contained \citep{2004RvMP...76..125M,2007ARA&A..45..565M}. While this
overall framework is generally accepted and is supported by existing theory
and observations, much work remains on both fronts to quantify the
dependence of statistical properties on the global system parameters,
and to establish when and how self-regulated quasi-steady states are
achieved.
Given the importance of time-dependent processes and interdependencies
in the ISM, complex theoretical models are needed in order to address
even rather basic questions. For example, what sets the relative
proportions of the different gas components? In an idealized
classical picture such as that of \citet{1969ApJ...155L.149F} for the
atomic medium, given a pressure and a mean density $\bar n$, thermal
equilibrium defines a density for each of two stable phases, $n_{\rm
warm}$ and $n_{\rm cold}$, and the ratio of cold to warm gas is given
by a simple algebraic relation: $M_{\rm cold}/M_{\rm warm}= (n_{\rm
warm}^{-1} - \bar n^{-1})/ (\bar n^{-1} - n_{\rm cold}^{-1})$. In the
real ISM, however, which is a time-dependent system, thermal
equilibrium only holds to the extent that the radiative times are
short compared to dynamical times for compressions and rarefactions.
Furthermore, the value of the mean density $\bar n$ and pressure
(averaged over large scales) are not even known {\it a priori} for a
given ISM surface density, since
the vertical distribution of gas is sensitive to its dynamical state.
This dynamical state itself depends on the (unknown) dense gas
fraction, since more dense gas produces more feedback from star
formation, and hence more turbulence that inflates the disk vertically
to reduce $\bar n$
(and also produces local variations in density and pressure through
compressions and rarefactions). Multidimensional effects (the ISM is
not simply stratified perpendicular to the galactic plane, but is
composed of filamentary clouds) and self-gravity additionally
complicate the situation.
In recent years, a number of groups have begun developing models
of the turbulent, multiphase ISM using time-dependent computational
hydrodynamics simulations that include feedback from star formation
\citep[e.g.,][]{
1999ApJ...514L..99K,
2004A&A...425..899D,
2005A&A...436..585D,
2007ApJ...665L..35D,
2005ApJ...626..864M,
2006ApJ...653.1266J,
2006ApJ...638..797D},
self-gravity of the gas
\citep[e.g.,][]{
1999ApJ...516L..13W,
2002ApJ...577..197W},
and both of these effects
\citep[e.g.,][]{
2001ApJ...547..172W,
2007ApJ...660..276W,
2005MNRAS.356..737S,
2006ApJ...641..878T,
2008ApJ...673..810T,
2008ApJ...680.1083R}.
The treatment of feedback in these simulations is to inject thermal energy
in regions identified as sites of star formation; most models focus
on the energy input from supernovae.
In very large scale simulations that
have minimum resolution of only 50-100 pc, feedback implemented via
thermal energy deposition is not in
practice very effective, because the input energy is easily
radiated away. With finer numerical resolution, feedback regions
expand adiabatically at first to make hot diffuse bubbles,
driving shocks that sweep up surrounding gas and
ultimately generate turbulence throughout the computational domain.
A number of different issues have been addressed by these recent
simulations, including investigating departures from thermal equilibrium in
density and temperature PDFs, measuring the relative velocity
dispersions of various gas components, and testing whether
relationships between star formation and gas surface density emerge
that are similar to empirical Kennicutt-Schmidt laws.
Even before the advent of supernovae, massive stars photoionize their
surroundings, creating HII regions within molecular clouds that are
highly overpressured and expand. HII regions may in fact be the most
important dynamical agents affecting the properties of dense gas in
giant molecular clouds (GMCs), since the original turbulence inherited
from the diffuse ISM is believed to dissipate within a flow crossing
time over the cloud \citep{1998ApJ...508L..99S,1998PhRvL..80.2754M},
while GMCs are thought to live for at least a few crossing times
\citep{2007prpl.conf...81B}.
Analytic and semi-analytic treatments find that
GMCs with realistic sizes, masses,
and star formation rates can indeed be maintained by the energy input
from HII regions for a few crossing times, ultimately being destroyed
through a combination of
photoevaporation and kinetic energy inputs that unbind the remaining mass
\citep{1979MNRAS.186...59W,
1994ApJ...436..795F,
1997ApJ...476..166W,
2002ApJ...566..302M,2006ApJ...653..361K}.
Recent three-dimensional
numerical studies have begun to address this process in detail
\citep[e.g.][]{2006ApJ...647..397M,2007ApJ...668..980M,2007ApJ...671..518K},
focusing on
regions within
GMCs.
In the present work, we consider how the large-scale dynamical state of the
ISM is affected by star formation feedback in the form of expanding
HII regions. Our main interests are in exploring how the turbulence
driven by HII regions affects the properties of dense gas
(we measure statistics of density, temperature, and velocity), in testing
ideas of global self-regulation by feedback (we evaluate Toomre $Q$
parameters and virial ratios), and in exploring how galactic
environment systematically affects the character of the ISM, including
its ability to form stars.
Complete ISM models should of
course include feedback from supernovae as well as those from HII
regions, and it is our intention to do this in future work. However,
we consider it useful to adopt a sequential approach, independently
testing the effects of HII region feedback to provide a baseline for
more comprehensive simulations.
In addition to developing a physical
understanding of the ways in which feedback affects the ISM, another
goal of our work is to investigate the sensitivity of numerical
results to prescriptions that are a necessary -- but not always fully
tested -- aspect of galactic-scale studies of star formation. In
particular, we examine how the choice of density threshold in
commonly-adopted recipes for star formation affects the resulting
dependence of the star formation rate on ISM surface density.
Our approach to exploring the effects of galactic environment is to
conduct a large suite of local simulations that cover a range of
values for three basic parameters: the total surface density of gas in
the disk ($\Sigma$),
the local midplane stellar density ($\rho_*$), and the local rate
of galactic rotation ($\Omega$). The parameter range covers
a factor of six in gas surface density and galactic angular rotation
rate, and a factor of 30 in stellar density.
Our suite is divided into four
series, each of which has one independent parameter that is
systematically varied.
We also include comparisons with hydrostatic
models that are identical in terms of their input parameters to the
fully-dynamic models, but do not include feedback and hence are not
turbulent. For this first set of pilot studies, we have not
implemented full radiative transfer to evaluate the extent of HII
regions (we intend to do so in the future), but instead introduce a
simple prescription in which the boundaries of HII regions are determined by
the gravitational potential. Using this approach (rather than, for
example, adopting a single fixed outer radius) has the advantage
that the volume of the heated region expands as the density
surrounding the source drops. Since our treatment of HII regions does
not attempt to be exact, we do not consider our specific
results for e.g. velocity dispersions to be more than approximate
(although in fact we find similar values for velocity dispersions in
dense gas to those that are observed in GMCs).
Instead, we shall emphasize the general properties of a multiphase ISM
system in which turbulence is driven from within the dense phase.
This paper is organized as follows: In \S 2 we describe our numerical
methods, and in particular the recipe for star formation feedback.
The control parameters for our disk models, and the properties of each
model series, are presented in \S 3. Section 4 gives an overview of
evolution based on our fiducial model.
In \S 5 we present the statistical properties of the gas in each
model, and test environmental influences by
intercomparing the model series.
The implications of our results for star formation, both in real
galaxies and in numerical simulations, is analyzed in \S 6.
We conclude with a summary and discussion in \S \ref{Summary}.
\section{Numerical Methods}
\subsection{Basic Equations}
We study the evolution of rotating, self-gravitating, galactic gas
disks, including local heating and cooling terms. We solve the
hydrodynamic equations in a local Cartesian reference frame whose
center lies at a galactocentric radius $R_0$ and orbits the galaxy with
a fixed angular velocity $\Omega_0 =\Omega(R_0)$. In this local frame,
radial, azimuthal, and vertical coordinates are represented by
$x\equiv R-R_0$, $y\equiv R_0(\phi-\Omega_0 t)$, and $z$, respectively,
and terms associated with coordinate curvature are neglected
\citep{1965MNRAS.130..125G,1966ApJ...146..810J}.
The local-frame equilibrium background velocity relative to the center
of the box at $x=y=z=0$ is given by
${\bf v}_0=-q\Omega_0 x\hat{y}$, where
\begin{eqnarray}
q=-\left.
\frac{d\ln\Omega}{d\ln R}
\right|_{R_0}
\end{eqnarray}
is the local dimensionless shear rate.
In terms of $q$, the local epicyclic frequency $\kappa$ is given by
\begin{eqnarray}
\kappa^2\equiv \frac{1}{R^3}\frac{d}{dR}(R^4\Omega^2)
=(4-2q)\Omega^2.
\end{eqnarray}
We shall choose $q=1$ for all models, representing a flat background rotation
curve $V_c=\Omega R= const.$ for the unperturbed motion.
In addition to the tidal gravity and Coriolis terms from the
``shearing sheet'' local formulation, we also include
terms for the vertical gravity of the stellar disk, gas self-gravity,
radiative heating and cooling, and thermal conduction.
The resulting equations
\citep[see e.g.,][]{1995ApJ...440..742H,2004ApJ...601..905P}
are:
\begin{eqnarray}
\frac{\partial \rho}{\partial t}+\nabla\cdot(\rho {\bf v})&=&0,\\
\frac{\partial {\bf v}}{\partial t}+{\bf v}\cdot \nabla {\bf v}
&=&-\frac{1}{\rho}\nabla P+2q\Omega x\hat{\bf x}
-2{\bf \Omega}\times{\bf v}
-\nabla \Phi+{\bf g}_{\ast},\\
\frac{\partial e}{\partial t}+\nabla\cdot(e{\bf v})
&=&-P\nabla\cdot{\bf v}
+n\left[\Gamma(t,{\bf x})-n\Lambda(T)\right]
+ K\nabla^2T,
\end{eqnarray}
where
$P=f n k_{\rm B} T$ and $n =\rho/\mu$.
With $n$ the number of hydrogen nuclei per unit volume,
$f$ varies from 0.6 to 1.1 depending on whether the gas is
predominantly molecular or atomic; we simply adopt $f=1.1$.
We adopt $\mu=1.4 m_p$ to include the contribution of helium to
the mass density.
Here $e=P/(\gamma-1)$ is the internal energy per unit volume (we adopt
$\gamma=5/3$), $K$ is the thermal conductivity,
$\Phi$ is the self-gravitational potential due to gas,
and the vertical gravitational force due to stellar disk is
\begin{eqnarray}
{\bf g}_{\ast}=-4\pi G\rho_{\ast}z\hat{\bf z},
\end{eqnarray}
where $\rho_{\ast}$ is the stellar density and $z$ is the vertical
coordinate relative to the midplane.
In the above expressions and elsewhere in the remainder of the text,
we have dropped the ``0'' subscript on $\Omega$; $\kappa$ also refers
to the value evaluated at the center of the domain.
Computation of the gas self-gravity is discussed below.
In this paper, we present results of two-dimensional simulations of
the above set of equations. The two independent spatial coordinates
in our models are $x$ and $z$; thus, we follow evolution in
radial-vertical slices through a galactic disk. Although the
azimuthal ($y$) direction is not an independent spatial
variable for the current set of models, we do include azimuthal
velocities, and their variation with $x$ and $z$. Inclusion of $v_y$
is important because angular momentum can strongly affect the ability
of self-gravitating perturbations to grow. Radial motions that are
required for gas to become concentrated are coupled to azimuthal
motions through the Coriolis force; perturbations in the azimuthal
velocity with respect to the mean background shear correspondingly
lead to radial motions via the Coriolis force. Although our
two-dimensional models do capture some of the effects of galactic
rotation (i.e. epicyclic oscillations), they miss some of the effects
associated with shear. In three dimensions (or in the
height-integrated $R-\phi$ plane), azimuthal shear can make
it more difficult for self-gravitating concentrations to grow. Of
course, in three dimensions, self-gravity also increases more rapidly
as the density increases, which enhances the ability of dense
concentrations to grow. Although it will be important to revisit the
present models with fully three-dimensional simulations, we do not
anticipate large changes based on dimensionality. Previous
three-dimensional simulations of shearing, rotating disks have found
similar (within a factor 2)
nonlinear instability thresholds for self-gravitating cloud
formation to reduced-dimensional models (see
e.g.
\citealt{2002ApJ...581.1080K,2003ApJ...599.1157K,2007ApJ...660.1232K},
and references therein). Thermal
instability also develops similarly in the two-dimensional and
three-dimensional case to create a cold cloud/warm intercloud
structure (e.g.
\citealt{2004ApJ...601..905P,2005ApJ...629..849P}).
\subsection{Hydrodynamic Code and Boundary Conditions}
The numerical solutions to the two-dimensional
dynamical equations are obtained using
a temporally and spatially second order finite volume method
which includes TVD Runge-Kutta integration in time
\citep{Shu_Osher_Efficient_ENO_JCP1988},
with a
directionally unsplit flux update and
piecewise linear reconstruction with slope limiter
\citep[see, e.g.,][]{Hirsch_2nd}.
We use Roe's approximate Riemann solver with an entropy fix
\citep{1981JCoPh..43..357R}.
The heating and cooling terms in the
energy equation are separated out in an operator-split fashion and
updated using implicit time integration
(see \S \ref{sec:cooling}),
because the
cooling times are frequently much shorter than the other timescales.
The code is parallelized using MPI.
For the
hydrodynamic update, the time step is set to
$\Delta t={\rm min}(t_{\rm HD}, t_{\rm cond}, t_{\rm cool},
t_{\rm heat})$
where
\begin{eqnarray}
t_{\rm HD}&=&C_{\rm CFL}\,{\rm min}\left(
\frac{1}{\frac{c_s+|v_x|}{\Delta x}+\frac{c_s+|v_z|}{\Delta
z}}\right)_{\rm all\ zones},\\
t_{\rm cond}&=&C_{K}\,{\rm min}\left(
\frac{n k_{\rm B}(\Delta x)^2}{4K(\gamma-1)}\right)_{\rm all\ zones},\\
t_{\rm cool}&=&C_{T}\,{\rm min}\left(
\frac{k_{\rm B} T}{(\gamma-1)n\Lambda(T)}\right)_{\rm all\ zones},\\
t_{\rm heat}&=&C_{T}\,{\rm min}\left(
\frac{k_{\rm B} T}{(\gamma-1)\Gamma({\bf x})}\right)_{\rm all\ zones},
\end{eqnarray}
and we adopt $C_{\rm CFL}=0.8$, $C_{K}=0.5$ and $C_{T}=50$.
Here, $c_s=(\gamma kT/\mu)^{1/2}$ is the sound speed in any zone.
With a large value of $C_{T}$, the explicit hydrodynamic timestep is
not strongly limited by the cooling time in dense gas.
The adopted $C_{T}$ is chosen such that the solution
agrees with tests of individual expanding ``HII regions'' (for our
feedback model) that have $C_T\sim 1$ (equivalent to explicit
cooling); if a much larger value of $C_T$ is allowed, this expansion is
not accurately reproduced.
At the $x$ (radial) boundaries, we implement shearing-periodic
boundary conditions \citep{1995ApJ...440..742H}, in which the azimuthal
(angular) velocity term
$v_y\equiv R_0 (\dot \phi - \Omega_0)=(R_0/R) v_\phi -V_c$
is incremented or decremented by $\pm L_x \Omega_0$ in
mapping from the right$\rightarrow$ left or left$\rightarrow$ right
boundary, consistent with the equilibrium velocity field.
In the $z$-direction, we adopt periodic boundary
conditions for the hydrodynamic variables, such that the total mass in
the domain is conserved.\footnote{We have found that except
for mass loss, the overall evolution is similar when we apply outflow
boundary conditions in the vertical direction. Adopting periodic boundary
conditions for hydrodynamic variables makes it possible to maintain the
gas surface density $\Sigma$ at a constant value without devoting significant
computational resources to following the evolution of a tenuous corona
at large $|z|$.} The gravitational potential solver applies open
(i.e. vacuum) boundary conditions in $z$, as we next discuss.
\subsection{Poisson Solver}
We have developed a new method for obtaining the gravitational
potential of a disk in Cartesian geometry using Fast Fourier
Transforms (FFTs). Since the discrete Fourier Transformation allows only
periodic functions, a special approach is needed to solve for a disk
potential with vacuum boundary conditions outside the simulation domain.
Let us consider a simple case, consisting of an a uniform, isolated
gas sheet in the $z=0$ plane which has density
$\rho(z)=\sigma\delta(z)$. The corresponding gravitational potential
$\Phi(z)=2\pi G\sigma|z|$ is obtained by solving the Poisson equation,
\begin{equation}
\nabla^2 \Phi = 4 \pi G \rho,
\end{equation}
with vacuum boundary conditions. If we
have a finite domain of size $L_z$ and suppose that the gas sheet lies
somewhere within the domain, then we only would require values of the potential
at locations within $\pm L_z$ of the sheet. Thus, we may find the
potential within $z=(-L_z,L_z)$ in terms of discrete
Fourier components as
\begin{eqnarray}
\hat{\Phi}_{\ell}&=&2\pi G\sigma C_{\ell},
\label{eq:Phi_ell1}
\end{eqnarray}
\begin{eqnarray}
C_{\ell}&=&
\int_{-L_z}^{L_z}|z|e^{2\pi i \ell \left(\frac{z}{2L_z}\right)}dz=
-\frac{2(1-e^{\pm i\pi \ell})}{\left(\frac{\pi \ell}{L_z}\right)^2}.
\label{eq:C_ell}
\end{eqnarray}
In Fourier space, the Poisson equation for one independent variable is
\begin{equation}
-k^2 \hat \Phi_k = 4 \pi G \hat\rho_k.
\end{equation}
Thus, in terms of discrete Fourier components with
$k_\ell=\ell (2\pi/2L_z)$, we have
\begin{eqnarray}
\hat \Phi_\ell = - \frac{4\pi G \hat{\rho}_{\ell}}
{\left(\frac{\pi \ell}{L_z}\right)^2}.
\label{eq:Phi_ell2}
\end{eqnarray}
Equating the right-hand sides of equations (\ref{eq:Phi_ell1}) and
(\ref{eq:Phi_ell2}) and inserting the expression from equation
(\ref{eq:C_ell}), this implies $\hat \rho_\ell = \sigma (1 - e^{\pm i \pi
\ell})$ for the isolated sheet, so that the density
in real space is obtained by taking an inverse transform:
\begin{eqnarray}
\rho(z)
&=&\frac{1}{2N_z}\sum_{\ell=0}^{2N_z-1}e^{-2\pi i \ell \frac{z}{2L_z}}
\hat{\rho}_{\ell},\\
&=&\frac{1}{2N_z}\sum_{\ell=0}^{2N_z-1}e^{-2\pi i \ell \frac{z}{2L_z}}
\sigma(1-e^{\pm i\pi \ell}),\\
&=&\frac{\sigma}{2N_z}\sum_{\ell=0}^{2N_z-1}e^{-2\pi i \ell \frac{z}{L_z}}
-\frac{\sigma}{2N_z}\sum_{\ell=0}^{2N_z-1}e^{-2\pi i \ell \frac{z\mp
L_z}{2L_z}},\\
&=&\sigma\delta(z)-\sigma\delta(z\mp L_z).
\end{eqnarray}
We see that the first term corresponds to the original density.
However, a second term has appeared as an image density with the
opposite sign from the real (physical) density, located a domain
length away. This means that to obtain the correct solution for
$\Phi$ on the original domain $z=(-L_z/2,L_z/2)$, we need to
prepare twice as large a box in the vertical direction, and implement
the required image density within the augmented domain, at a distance
$\pm L_z$ from the physical slab. Thus, a density slab at
$0\le z\le L_z/2$
would require an image slab at $z_{\rm image}=z-L_z$ in
$(-L_z,-L_z/2)$, and a
density slab at $-L_z/2\le z \le 0$
would require an image slab at $z_{\rm image}=z+L_z$
in $(L_z/2, L_z)$.
Using a similar procedure, we have extended this idea to
the three dimensional case with an arbitrary density distribution
$\rho(x,y,z)$. The details are described in the Appendix. We note
that the numerical solution agrees with the solution obtained via
Green functions \citep{1987PThPh..78.1273M}, and is much faster to
compute because only FFTs (no direct sums) are needed.
\subsection{Cooling Function}\label{sec:cooling}
To allow for multiphase interstellar gas components, we must solve a
thermal energy equation that allows a wide range of conditions. We
use a cooling function for the diffuse ISM derived by
\cite{2002ApJ...564L..97K},
which includes atomic gas cooling for the warm and
cold neutral medium (WNM, CNM), as well as cold molecular-phase
cooling (H$_2$, CO, and dust cooling). We include a constant
volumetric heating rate to represent photoelectric heating by diffuse FUV.
This yields a standard
(cf. \citealt{1969ApJ...155L.149F,1995ApJ...443..152W})
thermal equilibrium curve in which there is a
maximum density and pressure for the warm phase given by
$1.0 ~ {\rm cm}^{-3}$ and $5.5\times 10^3 ~k_{\rm B}~ {\rm cm}^{-3}{\rm K}$,
and a minimum density and pressure for the cold phase given by
$8.7 ~ {\rm cm}^{-3}$ and $1.75 \times 10^{3} ~k_{\rm B}~ {\rm cm}^{-3}{\rm K}$
(see Fig. \ref{fig:phase-1}).
HII regions in the real ISM
include photoionization of atoms and dissociation of molecules, and
radiative cooling of photoionized gas and warm molecular gas. These
effects depend on chemical fractions, as well as dust evaporation.
For this work, we are interested primarily in dynamics of the neutral
media, rather than the details of photoionized gas -- including the
complexities of ionization front propagation at sub-parsec scales.
The main requirement for capturing the large-scale
dynamical effects of feedback
is thus to incorporate photoheating of gas in star-forming regions.
The simple but expedient approach we have chosen is to expose gas in
targeted regions to enhanced heating, while simply applying the same
cooling function we use for neutral gas. The enhanced heating we apply
yields thermal equilibrium for the ``photoheated'' gas at $\approx
10,000$K (see below), which is consistent with the temperatures that
would be attained if we had implemented realistic (but much more
computationally complex and expensive) photoprocesses.
Cooling and heating timescales often become much shorter than the hydrodynamic
time step (i.e. the flow or sound crossing time of a grid zone),
especially in HII regions, which have a high heating rate. For
efficiency, we adopt implicit time integration for the heating and
cooling operators. In a given zone, the integral from the (j) to the
(j+1) time step is formally expressed as
\begin{eqnarray}
\int_{T_{j}}^{T_{j+1}}\frac{C_vdT}{\Gamma_{j}-n\Lambda(T)}
=\Delta t_{j},
\end{eqnarray}
where $C_v=k_{\rm B}/(\gamma-1)$ is the heat capacity per particle.
This is a nonlinear equation with respect to $T_{j+1}$, with $T_j$ and
$\Delta t_j$ treated as parameters. For this integral, we adopt
Simpson's rule and solve using the Newton-Raphson method.
\subsection{Thermal Conduction}
Thermal conduction determines the thickness of interfaces between
phases in the ISM, and proper incorporation of conduction is essential
in numerical simulations of a multiphase medium which is subject to
thermal instability
\citep{2004ApJ...601..905P, 2004ApJ...602L..25K,2008ApJ...681.1148K}.
The characteristic length scale set by conduction is
the Field length,
\begin{equation}
\lambda_{\rm F}=2\pi\sqrt{\frac{KT}{n^2\Lambda(T)}},
\end{equation}
\citep{1965ApJ...142..531F,1990ApJ...358..375B},
which corresponds to the critical wavelength of thermal instability.
For realistic values of the conductivity at $T\aplt 10^4 {\rm K}$,
$K=2.5\times 10^3\sqrt{T}$ erg cm$^{-1}$ s$^{-1}$ K$^{-1}$
\citep{1953ApJ...117..431P}, the
Field length of 0.19 pc (at density $1 ~ {\rm cm}^{-3}$ and temperature 1,000 K)
is much smaller than the size of interstellar clouds
($\sim$ 1 -- 10 pc)
and we would
need extremely high spatial resolution to resolve it -- and a
correspondingly high computational cost.
Instead, we adopt an approach somewhat analogous to the use of artificial
viscosity (far exceeding the true physical viscosity in magnitude)
in resolving shocks on a numerical grid. Namely, we
adopt a sufficiently large numerical conduction coefficient
to resolve the Field length on our chosen grid. We choose $K$ such that
for any simulation with resolution $\Delta x$,
the Field number $N_{\rm F}\equiv\lambda_{\rm F}/\Delta x$ is equal to
1.7
at density and temperature typical of thermally unstable gas
(we use $n=1 \ {\rm cm}^{-3}$, $T=1,000$ K).
For example, the fiducial model Q11 has $\Delta x=1.5$ pc and
the artificial conductivity gives $\lambda_{\rm F}=2.57$ pc for thermally
unstable gas.
For low density gas, given our typical values of $K$
the thermal conduction term can become greater than
cooling/heating terms.
In order to limit the conduction in these regions, we adopt
\begin{eqnarray}
\tilde{K}&=&\frac{K}{1+n_{\rm crit}/n},
\end{eqnarray}
with $n_{\rm crit}=0.05 ~ {\rm cm}^{-3}$.
\subsection{Stellar Feedback Activity}
The primary focus of this work is to explore the dynamical effects of
strong, localized heating by OB stars in dense regions of the ISM.
Since this heating produces $T\sim 10^4$ K gas that is initially
overpressured by a factor $\sim 100$ or more compared to its
surroundings, HII regions expand rapidly. This process is believed to
be an important source of turbulence both within self-gravitating GMCs
and in the surrounding diffuse ISM. To study this process, ideally
one would implement (a) formation of OB stars from dense gas,
distributed throughout the space-time domain of the simulation; (b)
radiative transfer of ionizing photons from the OB stars through the
surrounding gas, with potentially multiple ionization sources
throughout the domain; (c) detailed ionization and heating of the gas
within HII regions.
In this first exploration, rather than attempting to model all of
these processes in an exact fashion, we instead adopt an idealized
approach, with the goal of gaining physical understanding. First, we
apply certain criteria to determine when and where
``star zones'' on the grid
will be turned on. Then, we apply strong heating to the gas in the
vicinity of each
``star zone'' for the duration of its lifetime.
All
``star zones''
have the same lifetime, $t_{\rm ms}$, which is set
to $3.7 \times 10^6$ yr, the typical lifetime of OB associations
in clouds whose mass is $10^5{\rm M}_\odot\ $ \citep{1997ApJ...476..144M}.
Within HII regions, we assume a constant gas heating rate, set via a
control parameter $G_{\rm HII}$. Each
``star zone'' is therefore essentially a control flag
for whether or not strong
photoheating is locally applied near that zone (which does not move).
Rather than solving a radiative transfer problem, we use a
simple criterion based on the gravitational potential to determine
whether gas is subject to strong heating.
Because our goal is to
identify gas localized around star-forming regions, it is necessary to
subtract out the background gravitational potential and retain just
the potential component due to an individual self-gravitating cloud.
The background potential is the potential averaged over horizontal
planes. In terms of Fourier components, the relative gravitational
potential $\Phi^{(1)}=\Phi - \Phi_{\rm background}$ is defined as
\begin{eqnarray}
\Phi^{(1)}({\bf x})=\sum_{k_z\ne 0}
\hat{\Phi}_{k}\hat{f}_{\bf k}\exp(-i{\bf k\cdot
x}). \label{eq:relative potential}
\end{eqnarray}
Here, $\hat{f}_{\bf k}$ is the Fourier component of a smoothing window
function,
\begin{eqnarray}
f({\bf x})=\frac{3}{4\pi r_0^3}
\frac{1}{1+\exp\left[\frac{10}{\Delta x}(\sqrt{x^2+z^2}-r_0)\right]},
\end{eqnarray}
where $r_0$ is a smoothing length. This window function
smooths the HII region within a radius $\approx r_0$.
Convolution of the relative potential with a smoothing function (or,
equivalently, multiplication in Fourier space as above) is desirable
so that any heating that is applied is resolved on the grid. We have
adopted $r_0=3\Delta x$ as providing adequate resolution.
HII photoheated regions are identified as regions where the relative
potential, $\Phi^{(1)}$, falls below some specified level:
$\Phi^{(1)}<\Phi_{\rm HII}$. We also employ the relative potential
$\Phi^{(1)}$ for setting one of the criteria for turning on feedback:
$\Phi^{(1)}<\Phi_{\rm SF}$ and $\rho>\rho_{\rm SF}$ must both be met
in a given zone for a ``star zone'' to be created at that
location. Thus, our recipe ensures that feedback will only occur in
dense and self-gravitating regions, consistent with the fact that OB
stars are observed to form only under these conditions.
For the feedback prescription we have adopted, there are
five control parameters: $\rho_{\rm SF}$, $\Phi_{\rm SF}$, $t_{\rm ms}$,
$\Phi_{\rm HII}$, $G_{\rm HII}$.
The detailed estimation of those parameters is described in the
remainder of this section.
\subsubsection{OB Star Formation Criterion}
We choose a density threshold for star formation as
\begin{eqnarray}
\rho_{\rm SF}&=& 10^3 \mu\, {\rm cm}^{-3}.
\end{eqnarray}
This density is comparable to that of clumps of gas within GMCs.
The free fall time at this density,
1.4 Myr, is
sufficiently small compared to the orbital time that structures
satisfying this threshold evolve independently of the global
environment. Note that the local Jeans length
\begin{eqnarray}
\lambda_{\rm J}=\left(\frac{\pi c_s^2}{G\rho_{\rm SF}}\right)^{1/2}
=4.1\,{\rm pc}\left(\frac{c_s}{0.9 ~{\rm km\ s^{-1}}}\right)
\end{eqnarray}
must be resolved by a few zones in order to prevent fragmentation
occurring as a consequence of numerical artifacts
\citep{1997ApJ...489L.179T}.
To obtain an estimate for the potential threshold for star formation,
we consider a cloud
with uniform number density $\bar{n}$ and radius $R_{\rm cl}$. For a
spherical cloud, the radius is related to the cloud mass using
\begin{eqnarray}
R_{\rm cl}&=&\left(\frac{3M_{\rm cl}}{4\pi\bar\rho}\right)^{1/3}
=19 ~ {\rm pc} \times
\bar{n}_{2}^{-1/3}M_{\rm cl,5}^{2/3},
\end{eqnarray}
where the fiducial value of
$\bar{n}_2\equiv \bar{n}/(10^2~{\rm cm}^{-3})$ is chosen using a
typical mean density within GMCs,
and $M_{\rm cl,5}\equiv M_{\rm cl}/(10^5 ~ {\rm M}_\odot\ )$.
Since our grid is
two-dimensional, the control parameter
$\Phi_{\rm SF}$ must be based on a
cylindrical regions of a given density. For a cylinder of radius
$R_{\rm cl}$, the potential difference between the center and a
distance $r$ ($>R_{\rm cl}$) is $\Phi(0)=\Phi(r) - 2 \pi G \bar\rho
R_{\rm cl}^2 [\ln(r/R_{\rm cl}) + 1/2]$.
The logarithmic term corresponds to the potential difference between
$r$ and $R_{\rm cl}$, and the $1/2$ is the contribution between the
cloud's surface and its center.
If the cloud is created out
of all of the mass originally within a disk of surface density
$\Sigma$ within a range $2r\equiv L_{\rm eqv}$ of the cloud center,
then $\pi R_{\rm cl}^2 \bar\rho = \Sigma L_{\rm eqv}$, and a fiducial
distance for defining the effective zero of the potential is
\begin{eqnarray}
L_{\rm eqv}&=&\frac{\pi\bar\rho R_{\rm cl}^2}{\Sigma},\\
&=&393~ {\rm pc} \times \bar{n}_{2}M_{\rm cl,5}^{2/3}
\left(\frac{\Sigma}{10{\rm M}_\odot\ {\rm pc}^{-2}}\right)^{-1}.
\end{eqnarray}
Here, the
radius is expressed
in terms of that of an equivalent spherical cloud with a given
mass.
If we set the potential at $r=L_{\rm eqv}/2$ to zero, then the
potential at the center of the cloud will be
\begin{equation}
\Phi(0)= -\Phi_{\rm SF}\equiv - 2 \pi G \bar\rho R_{\rm cl}^2
\left[\ln\left(\frac{L_{\rm eqv}}{2R_{\rm cl}}\right)
+ \frac{1}{2}\right].
\label{eq:Phi_0}
\end{equation}
For
$\bar n = 100~ {\rm cm}^{-3}$ and $R_{\rm cl}=19$ pc, the value of
$\Phi(0)$ is $3.4 \times 10^{11} {\rm cm}^2 ~{\rm s}^{-2}$
times an order-unity
factor that varies only logarithmically in the ratio of cloud surface
density to mean ISM surface density.
We choose to adopt a potential threshold for
star formation
$\Phi_{\rm SF} = 3.4 \times 10^{11} {\rm cm}^2 ~{\rm s}^{-2}
=(5.8 {\rm km} ~{\rm s}^{-1})^2$; we have
tested sensitivity to the value of $\Phi_{\rm SF}$, and found that
results are insensitive to the exact choice, except as described below.
\subsubsection{Definition of Photoheated Regions}
First, consider an HII region centered on the origin of a uniform cloud
with spherical crossection.
If we assume the radius of the HII region is $R_{\rm HII}$ ($<R_{\rm
cl}$), then if the center of the cloud has potential
$\Phi(0)$,
the potential at the ionization front for a cylindrical cloud with
uniform density
is $\Phi_{\rm HII}=\Phi(0) + \pi G\bar\rho R_{\rm HII}^2$. (For a
spherical cloud, the second term is smaller by a factor $2/3$.)
Taking the difference with $\Phi(L_{\rm eqv}/2)$ in order to represent
a relative potential, using equation (\ref{eq:Phi_0}), and
substituting $\Phi(0) \rightarrow
-\Phi_{\rm SF}$ (since the criterion for star formation must be
satisfied if feedback has turned on)
this implies that the relative potential at the location of the HII
region would be
\begin{equation}
\Phi_{\rm HII}=-\Phi_{\rm SF}\left[
1-\frac{\frac{1}{2}}
{\ln\left(\frac{L_{\rm eqv}}{2R_{\rm cl}}\right)+\frac{1}{2}}
\left(\frac{R_{\rm HII}}{R_{\rm cl}}\right)^2
\right].
\label{eq:R_HII_init}
\end{equation}
$R_{\rm HII}$ is given, for example, by the Str\"omgren radius in a
uniform medium:
\begin{eqnarray}
R_{\rm HII}&=&\left(\frac{3S}{4\pi \bar{n}^2\alpha^{(2)}}\right)^{1/3}
= 2.97~{\rm pc}~
\bar{n}_{2}^{-2/3}
S_{49}^{1/3},
\end{eqnarray}
where $S$ is the number ionizing photons per
unit time and $S_{49}\equiv S/(10^{49}{\rm s}^{-1})$, and
$\alpha^{(2)}=3.09\times 10^{-13} ~ {\rm cm}^3 {\rm s}^{-1}$
is the case B hydrogen recombination coefficient at
$T=8,000$ K \citep{1978ppim.book.....S}.
$S_{49}$ is equal to unity
for the ionizing luminosity in a
typical $10^5 {\rm M}_\odot\ $ GMC \citep{1997ApJ...476..144M}. From
equation (\ref{eq:R_HII_init}), when the density is comparable to the
mean density within a GMC, $R_{\rm HII}/R_{\rm cl}\ll 1$, and the HII
region is buried deep within the GMC at
$||\Phi_{\rm HII}|-\Phi_{\rm SF}|/\Phi_{\rm SF}\ll 1$.
HII regions are initially highly overpressured, however, and
will expand rapidly until breaking out of the surrounding GMC, creating
a blister HII region.
For the purposes of considering the momentum input to the system, the
$R_{\rm HII}\rightarrow R_{\rm cl}$ limit is most appropriate for
defining the photoheated region.
Thus, we suppose that the HII region has expanded, leaving a very low
density interior and a shell of radius $R_{\rm HII}<R_{\rm cl}$
in which most of the mass has piled up. The potential in the interior
of the shell is then given by
\begin{equation}
\Phi_{\rm HII}= -\Phi_{\rm SF}
\left[1- \frac{\frac{R_{\rm HII}^2}{R_{\rm cl}^2-R_{\rm HII}^2}
\ln(\frac{ R_{\rm cl}}{R_{\rm HII}})}
{\ln\left(\frac{L_{\rm eqv}}{2R_{\rm cl}}\right)+\frac{1}{2}}
\right].
\label{eq:R_HII_shell}
\end{equation}
When $R_{\rm HII}\rightarrow R_{\rm cl}$,
this expression is of course the same as if we had taken
$R_{\rm HII}/R_{\rm cl}\rightarrow 1$ in equation
(\ref{eq:R_HII_init}), that is, the potential near the surface
of the initially-uniform cloud.
For convenience, we
introduce a dimensionless parameter $\epsilon$:
\begin{equation}
\Phi_{\rm HII} \equiv -\Phi_{\rm SF}(1-\epsilon),
\label{eq:HII_eps}
\end{equation}
where
$\epsilon=(1/2)
[\ln\left(\frac{L_{\rm eqv}}{2R_{\rm cl}}\right)+\frac{1}{2}]^{-1}$
when $R_{\rm HII} \rightarrow R_{\rm cl}$.
For the fiducial parameter values discussed above, $\epsilon$ is equal
to 0.18.
We therefore
adopt $\epsilon =0.2$ as our ``standard'' value, although we have
tested how the results differ for much smaller values.
\subsubsection{Heating Rate in HII Regions}
During the period that ``star zones'' are alive, UV photons
enhance the heating rate within HII regions, defined as described
above. For the heating rate in any zone due to UV photons, we adopt
the on-the-spot approximation given by:
\begin{eqnarray}
\Gamma &=&
2.16\times 10^{-26}G_{\rm FUV} ~ {\rm erg} ~ {\rm cm}^{-3} {\rm s}^{-1},\\
G_{\rm FUV}&=&
\left\{
\begin{array}{ll}
G_{\rm HII} \quad &
\mbox{($\Phi^{(1)}<\Phi_{\rm HII}$)},\\
G_0 & \mbox{(otherwise)},\\
\end{array}
\right.
\end{eqnarray}
where $G_0=1.0$ is the background FUV field in the Galaxy.
We choose $G_{\rm HII}=10^3$ throughout this paper,
although we have also tested
results with lower and higher $G_{\rm HII}$.
In practice, the exact value of $G_{\rm HII}$ is not important, since the
purpose of this added heating is simply to increase the maximum
density at which a warm phase is present. From our
Fig. \ref{fig:phase-1} (see also e.g. Fig. 7 of
\citealt{1995ApJ...443..152W}), a value of $G_{\rm HII}=10^3$ boosts
this to $n \sim 10^3 ~ {\rm cm}^{-3}$.
\section{Model Parameters}
The models in this paper are characterized by three main parameters:
the total gas surface density $\Sigma$, the orbital angular velocity
$\Omega$ in the center of the domain, and the stellar density at the
midplane $\rho_{\ast}$. The first parameter defines the amount of
raw material
available for distribution between the dense and diffuse ISM phases,
while the second two parameters define the galactic environment in
which the gas evolves in response to the galactic tidal, rotational,
and shear stresses. The effectiveness of self-gravity in forming
massive clouds depends on all of these parameters, as well as on the
turbulent state that develops as a consequence of star formation
feedback.
The simulation domain we model is a radial-vertical slice through a
disk. For the vertical direction, we require a domain $L_z$ that
encloses most of the mass of the ISM, which is confined by a
combination of stellar gravity and gas self-gravity. The largest
scale height (excluding the hot ISM, absent in these models)
is that of the WNM, and an upper
limit is obtained by neglecting self-gravity, which yields a Gaussian
distribution with scale height:
\begin{eqnarray}
H_{w}&=&\frac{c_{w}}{\sqrt{4\pi G\rho_{\ast}}}
=95 ~{\rm pc}
\left(\frac{c_{w}}{7 ~ {\rm km\ s^{-1}}}\right)
\left(\frac{\rho_{\ast}}{0.1 ~ {\rm M}_\odot\ {\rm pc}^{-3} }\right)^{-1/2}.
\end{eqnarray}
Here, $c_{w}=(k T_{w}/\mu)^{1/2}$ is the thermal speed of the warm
medium, which typically has $T_w=8,000-10,000$ K.
We require $L_z > 2H_w$ for our numerical models.
If the ISM consisted only of WNM, then with $n_{w,0}=\rho_{w,0}/\mu$
the central volume
density, the total surface density would be
\begin{eqnarray}
\Sigma_w=\sqrt{2\pi} \rho_{w,0} H_w
=1.6 ~ {\rm M}_\odot\ {\rm pc}^{-2}
\left(\frac{n_w}{0.2~ {\rm cm}^{-3}}\right)
\left(\frac{c_w}{7 ~ {\rm km\ s^{-1}}}\right)
\left(\frac{\rho_{\ast}}{0.1 ~{\rm M}_\odot\ {\rm pc}^{-3} }\right)^{-1/2}.
\end{eqnarray}
The maximum density for which the warm phase is possible (when
$G_0=1$) is
$n_{w, {\rm max}}=P_{w, {\rm max}}/(\mu c_w^2)\approx 1 ~{\rm cm}^{-3}$;
this implies a maximum
possible total surface density of warm gas
$\Sigma_{w, {\rm max}}=(2G\rho_{\ast})^{-1/2} P_{w, {\rm max}}/c_w$.
In practice, the midplane
density of the warm medium is closer to the value $0.23 ~ {\rm cm}^{-3}$,
which represents
the warm medium density that is in pressure equilibrium with cold gas
at $P_{c, {\rm min}}/k_{\rm B} =1.75 \times 10^3 {\rm cm}^{-3}{\rm K}$.
We are interested in multiphase disks
similar to those observed in the Solar neighborhood and at smaller
radii in normal spirals; hence we choose surface density of at least
several ${\rm M}_\odot\ {\rm pc}^{-2}$ such that the pressure is high enough
to permit a cold phase to form, i.e.
$\Sigma > (2G\rho_{\ast})^{-1/2} P_{c, {\rm min}}/c_w$.
The radial domain should be sufficient to capture the
largest-scale gravitational instabilities, which in a disk galaxy are
limited by angular momentum.
The Toomre length \citep{1964ApJ...139.1217T} is the maximum scale for
axisymmetric modes in a thin, cold disk;
\begin{eqnarray}
\lambda_{T}&=&
\frac{4\pi^2 G\Sigma}{\kappa^2}=\frac{4\pi^2G\Sigma}{(4-2q)\Omega^2},\\
&=&1.36 ~ {\rm kpc}
\left(\frac{\Omega}{25 ~{{\rm km\ s^{-1}} ~ {\rm kpc}^{-1}} }\right)^{-2}
\left(\frac{\Sigma}{10 ~ {\rm M}_\odot\ /{\rm pc}^{2} }\right).
\end{eqnarray}
We require $L_x>\lambda_{T}$ for our numerical models,
typically by a factor 1.3.
The parameters of our models are summarized in Table \ref{tbl:models}.
In order to cover the three-dimensional parameter space and explore
environmental dependences systematically, we consider four
series of models: Q, K, S, and R.
For each Series, we hold two quantities fixed and vary a third
quantity, as follows:
\begin{itemize}
\item Series Q: $\kappa/\Sigma$ and
$\sqrt{\rho_{\ast}}/\Sigma$ are constant while $\Sigma$ varies;
\item Series K: $\kappa$ and $\sqrt{\rho_{\ast}}/\Sigma$ are constant
while $\Sigma$ varies;
\item Series R: $\kappa/\Sigma$ and $\rho_{\ast}$ are constant while
$\Sigma$ varies;
\item Series S: $\Sigma$ and $\rho_{\ast}$ are constant while $\kappa$
(and $\Omega$) varies.
\end{itemize}
The value of the Toomre $Q$ parameter for
the gas component, for a radial velocity dispersion
$\sigma_R$, is defined as
\begin{eqnarray}
Q&\equiv &\frac{\kappa\sigma_R}{\pi G\Sigma}\\
&=&1.82
\left(\frac{\Omega}{25 {\rm km~s^{-1}~kpc^{-1}}}\right)
\left(\frac{\sigma_R}{7\,{\rm km~s^{-1}}}\right)
\left(\frac{\Sigma}{10 {\rm M}_\odot\ {\rm pc}^{-2}}\right)^{-1}.
\label{eq:ToomreQdef}
\end{eqnarray}
Since Toomre's parameter is proportional to $\kappa/\Sigma$, Series Q
and R would have constant thermal $Q=\kappa c_s/(\pi G \Sigma)$ for
the gas if the
sound speed $c_s$ were constant (which is true for warm gas). The Q
and R series correspond to values of $Q=2.1 (\sigma_R/7 ~ {\rm km\ s^{-1}})$.
Assuming a
constant stellar velocity dispersion, $\Sigma_{\ast} \propto
\sqrt{\rho_{\ast}}$, so the stellar Toomre parameter
($Q_{\ast}\propto \kappa/\Sigma_{\ast}$)
would also have the same value for all members of Series Q.
As a fiducial model, we choose $\Sigma=10.6 ~ {\rm M}_\odot\ /{\rm pc}^2$,
$\Omega=31.2 ~ {\rm km\ s^{-1}} ~ {\rm kpc}^{-1}$,
and $\rho_{\ast}=0.07 ~ {\rm M}_\odot\ /{\rm pc}^3$, similar to conditions
in the Solar neighborhood; this is denoted as the
Q11 model in Table \ref{tbl:models}.
Relative to conditions similar to those in the Solar neighborhood, we
may think of the members of Series Q as representing conditions
ranging from slightly larger radii down to radii of a few kpc, for a
disk that has constant $Q$ and $Q_{\ast}$ -- i.e. larger gas and stellar
densities at small radii. We may think of Series R as models spanning
a similar range of radii, except for a disk that has stellar density
(and the corresponding vertical gravity) independent of radius, while
the gas surface density increases inward (such that the gas
self-gravity can become dominant). We may think of Series S as
relocation of the Solar neighborhood conditions of gas and stellar
density to either further in or further out in the galaxy's potential
well, where rotation and shear are stronger or weaker, respectively.
We may think of Series K as choosing a fixed location in the galactic
potential well (dominated by dark matter), and then varying the gas
and stellar surface densities in tandem.
To initialize our models, we set the density to a uniform value (given
by $\bar n$ in Table \ref{tbl:models})
throughout the domain, and set the pressure to
$P/k_{\rm B}=4,860 ~ {\rm cm}^{-3}{\rm K}$ which is in the thermally unstable regime.
The value of the initial pressure in fact does not matter, since the
gas rapidly separates into WNM and CNM due to thermal instability.
We also impose on the initial conditions isobaric density
perturbations
(with a flat spectrum
at wavenumbers smaller than $kL_z/2\pi = 32$, and 10 \% total amplitude).
The results are also not sensitive to the amplitude or power spectrum
of the initial perturbation spectrum; this is simply a convenience to
seed the initial evolution into thermal instability and then Jeans
fragmentation of cold gas.
In order to reach a quasi-steady state with repeated feedback cycles,
we run our models for two orbital periods in Series Q, R and K
and up to $t_{\rm final}=5.57\times 10^8$ year for Series S.
\subsection{Hydrostatic Models}
Because an important focus of this work is to assess the effects of
turbulence, it is important to ascertain how our dynamical models
differ from the situation in which there are no motions other than
background sheared rotation. For these comparison models, we
calculate the one-dimensional hydrostatic equilibria in the vertical
direction. These models include heating and cooling, but no feedback
from star formation. We consider two series, HSP and HSC which have
stellar volume density $\rho_*$
either {\it proportional} (P) to the square of
the gas
surface density $\Sigma$ or {\it constant} (C), respectively. Note that
Series HSP corresponds to the dynamical Series Q and K, while Series
HSC corresponds to the dynamical Series R. Details of these model
parameters are listed in Table \ref{tbl:HS}. Uniform 1024 grids are
used for all hydrostatic models. Note that the hydrostatic models
only allow one-dimensional vertical structure, and require higher
spatial resolution than the dynamic models because the scale height in
the cold layer near the midplane becomes very small when there is no feedback.
\section{Evolution of the Fiducial Model}
This section describes the overall evolution of the fiducial model.
After the initial thermal instability, the cold gas collapses into the
midplane, due to the vertical gravitational field. The cold,
high-density midplane gas then fragments rapidly, due to self-gravity.
The time scale for this
fragmentation is characterized by the Jeans time for a thin disk,
$t_{\rm J}=c_s/G\Sigma\approx 6 ~{\rm Myr}(c_s/0.3~{\rm km~s^{-1}})$.
The dense fragments collapse, with some of them coagulating, until the
feedback criteria are met, and heating is turned on to create
local HII regions. These HII regions expand, causing the dense gas to
be swept outward, forming shells that then break up into filaments.
New overdense regions continue to form, collapse, and
be dispersed by feedback.
Figure \ref{fig:snap-1} shows a snapshot from the fiducial model
(Model Q11) at a point after the system has reached a quasi-steady
state, in terms of the statistical distributions of density,
temperature, and velocity. The two panels show the temperature and
density throughout the domain. The contours in the lower panel denote
relative potential $\Phi^{(1)}$: solid and dashed lines show positive
and negative values, respectively. At the time of this snapshot,
there are three large clouds consisting of collections of dense filaments
that create minima in the gravitational potential
(dashed lines in the lower panel). Most of the dense filaments and
clumps in the lower panel correspond to cold gas in the upper panel.
In the upper panel, the orange circle associated with the cloud
near $x=250$ pc shows an active HII region, in which the gas is both
warm and dense and hence is expanding rapidly. Expansion of shells
slows at later times (after the pressure drops to ambient values and
the enhanced heating turns off). Since most of the time for any given
shell is spent near the maximum expansion, the widely-expanded
structure of the middle cloud is typical, in terms of the
time-averaged state.
Figure \ref{fig:snap-2} shows a snapshot of the density and
temperature in the same model Q11 at a time 38 Myr later. Overall,
the structure is qualitatively similar, although details change
because the state is highly dynamic. There are still three main
collections of filaments; the middle cloud has a large shell while the
left- and right-side clouds have contracted onto the midplane and have
nearly reached the point at which new HII regions will be born.
Three large ``clouds'' within the 1.16 kpc horizontal length of the
domain corresponds to mean separations of 390 pc.
One might expect the number of cloud entities to be related to the
properties of star formation feedback, and for our adopted
prescription to the parameter $\epsilon$, which effectively determines
the maximum volume of an HII region: large $\epsilon$ corresponds to
large HII regions, whereas small $\epsilon$ corresponds to HII regions
only in the immediate vicinity of a potential minimum defined by a
high-density clump of gas. In a situation with multiple local minima
in the gravitational potential, if $\epsilon$ is large then a single
HII region could engulf what would be multiple HII regions in the case
of small $\epsilon$. Expansion and shell collision of many small HII
regions would produce more (but smaller) clouds than expansion and
collision of a few large HII regions. In fact, when we run the same
model but set $\epsilon=0.02$, we find that there are typically 4-5
clouds in the same domain. We conclude that the number of large
clouds is not {\it very} sensitive to $\epsilon$, but since this
control parameter can only approximately model the effects of real HII
regions, the current study cannot provide an exact prediction for the
size or mass of GMCs. We note that the typical separation is,
however, in the same range as the two-dimensional Jeans length at the
typical velocity dispersion, $\lambda_{\rm J,2D}\equiv \sigma^2/(G\Sigma)=
340 {\rm pc} (\sigma/4 ~{\rm km~ s^{-1}})^2$ for this model.
Figure \ref{fig:phase-1} shows the phase diagram ($\rho-P$ plane) for
the same snapshot in Figure \ref{fig:snap-1} and \ref{fig:snap-2}. The gray scale is
proportional to the fraction of the total mass in the domain that is
found at a given point in the phase plane. We overlay the thermal
equilibrium curves for both the cases of ``normal'' heating (solid
curve) and the enhanced heating in HII regions (dashed curve).
Clearly, most of the gas resides near thermal equilibrium, due to the
short cooling times compared to the longer hydrodynamical times.
The range of properties of the gas can also be seen in the Figure
\ref{fig:PDF-1}, which shows the probability distribution functions (PDFs)
of gas density. Solid and dashed lines show mass and volume weighted
probabilities, respectively. The volume PDF shows that
the volume is mainly occupied by warm and diffuse gas (WNM) at
densities of a few $\times 0.1~ {\rm cm}^{-3}$.
In terms of the mass PDF, there are two peaks: one corresponds to the WNM,
and the other to a cold component at density above $10~ {\rm cm}^{-3}$.
Figure \ref{fig:time} shows the time evolution of thermal, kinetic and
potential energies averaged over the domain, for Model Q11. For the
potential energy, the background disk potential is subtracted out;
i.e. we use $\Phi^{(1)}$ (see eq. \ref{eq:relative potential}). There
are many sharp spikes in both thermal and kinetic energies, which
correspond to times when new HII regions are born and then rapidly
expand. The number of spikes corresponds to the number of stellar
generations in the model; note that this number must be proportional
to the domain size. In the second rotational period (i.e., $1 \le t
\le 2$ rotation), there are 18 generations per 1.16 kpc, or 6
generations per massive cloud per rotation period
(i.e. an interval of $3.3 \times 10^7$ yr)
if the mean number of clouds is 3 in this fiducial model.
Other models show similar overall behavior in terms of the evolution
of physical structure (consisting of clumps and filaments that
disperse and re-collect), as well as the distribution of mass in the
phase plane. As environmental parameters vary along a sequence,
however, there are some characteristic changes in structure. The most
notable difference is that for high gaseous surface density cases,
clouds are often more physically concentrated
(i.e., more compact and dense) because of the higher
stellar and gaseous gravity.
For example, Figure \ref{fig:snap-3} shows a snapshot
from Model Q42, which has $\Sigma$ and
$\rho_{\ast}$ four and 16 times larger, respectively, than the values
in the
fiducial model Q11. The three clouds that are seen in the figure are
more compact than in the lower-$\rho_{\ast}$ case. For a given $\Phi_{\rm
SF}$ and $\epsilon$, increasing the mass of a cloud implies that
HII region cannot break out as easily.
\section{Parameter Dependence of Statistical Properties}
All of our models show a turbulent, multiphase ISM with several
generations of feedback from photoheating. In this section, we
analyze how the statistical properties of those models depend on the
environmental parameters, both along a given series and from one
series to another. The statistical properties that we study are based
on averages of the fluid variables over space and time. First we
describe how these averages are defined in general, and then we turn
to the particular statistics.
\subsection{Space and Time Averaging Procedure}
For a variable of $A$ that is averaged over both space and time,
we use mass-weighted averages defined by:
\begin{eqnarray}
\langle A(a)\rangle &\equiv&
\frac{\int_{a} A\,dm}
{M(a)},\\
M(a)&\equiv& \int_{a} \,dm,
\end{eqnarray}
where the argument `$a$' denotes a given phase or component of the gas
(such as WNM).
The time averaging is then defined via:
\begin{eqnarray}
\overline{\langle A(a)\rangle}&\equiv&
\int_{t_1}^{t_2} \langle A(a)\rangle \,dt \Big/ S,
\label{eq:space-time-av}
\\
S&\equiv&
\int_{t_1}^{t_2} \theta ( M(a) )dt,
\end{eqnarray}
where the step function $\theta$ gives $1$
if $M(a)>0$
and $\theta=0$ otherwise,
so that only intervals in which the component is present are included
in the averaging.
Note that $S=t_2-t_1$ if the component `$a$' is present somewhere in
the domain at every interval during the simulation.
Thus, we denote spatial and temporal averages using angle brackets and
overlines, respectively.
To avoid the initial transients at the beginning of the simulation,
we adopt the interval between $t_1=t_{\rm final}/2$ and $t_2=t_{\rm final}$
for our time average.
For the purpose of averaging, our sampling rate is
$\Delta t=0.1 t_{\rm ms}$, where $t_{\rm ms}=3.7 \times 10^6$ yr
is the adopted lifetime
of ``star zone'' flags that control photoheating feedback.
\subsection{Mass Fractions}
The models of this paper focus on dynamics rather than chemistry, so
rather than dividing the gas into distinct phases, we simply bin it
according to density. The neutral gas (i.e. the gas that is not
within the limits defining HII regions) is separated into four bins.
The first bin ($n<1 ~ {\rm cm}^{-3}$)
corresponds approximately to the WNM, with densities
below the maximum for which a warm phase is possible in thermal
equilibrium. The second bin ($1~ {\rm cm}^{-3} <n \le 100 ~ {\rm cm}^{-3}$)
extends up to the maximum density that is in pressure equilibrium with
WNM gas, and corresponds approximately to the CNM (phase diagrams
show that the thermally-unstable regime is not highly populated for
our models; see e.g. Figs \ref{fig:phase-1} and \ref{fig:PDF-1}).
The dense medium (hereafter DM) is all the gas at
$n \ge 100 ~ {\rm cm}^{-3}$,
which in thermal equilibrium is above the maximum pressure
for the warm phase and therefore only exists in regions that are
internally stratified due to gravity. The DM gas corresponds
approximately to the molecular component of the ISM; we divide it into two
bins, DM2 ($100<n \le 10^3$) and DM3 ($10^3<n $). Gas that is within
the limits defined for enhanced heating is labeled as ionized gas
(hereafter HII). So that the mass fractions $f(a)$
of all components add to unity, $\sum_a f(a)=1$,
we use a slightly different definition from that of equation
(\ref{eq:space-time-av}). The
mass fraction of each component $a$ in (WNM, CNM, DM2, DM3 or HII) is
defined as
\begin{eqnarray}
f(a)=\frac{\displaystyle\int_{t_1}^{t_2}\frac{M(a)}{M_{\rm tot}}\,dt}{t_2-t_1}.
\label{eq:fraction}
\end{eqnarray}
Figure \ref{fig:mass} shows the mass fraction of the various
components either as a function of surface density (Series Q, R and K)
or angular velocity (Series S). For Series Q and R (which most closely
correspond to the radial variations found within normal spiral
galaxies), at low $\Sigma$ the diffuse (WNM+CNM) components dominate,
while at high $\Sigma$ the dense (gravitationally-confined) components
(DM2+DM3) dominate. The behavior is somewhat different in Series K,
which is highly gravitationally unstable and thus extremely active
when $\Sigma$ is large (since $\kappa$ is constant), leading to larger
HII and CNM mass fractions at high $\Sigma$. At low $\Sigma$, the
behavior in Series K is similar to that in Series Q and R. For all
series, the HII mass fraction increases at higher $\Sigma$ or lower
$\Omega$, corresponding to lower Toomre $Q$ (see Figure
\ref{fig:Qvalue}) and hence higher rates of stellar feedback
activity. The mass fraction of the WNM component secularly declines
with increasing $\Sigma$ in model series Q, K, and R.
Even though the models of Series S are most gravitationally unstable at
low $\Omega$, they remain dominated by diffuse gas (CNM) rather than
dense gas, because the total surface density is relatively modest for
this series ($\Sigma=15 ~ {\rm M}_\odot\ {\rm pc}^{-2}$).
\subsection{Surface Density}
The simulation domain for our two-dimensional models represents a
radial-vertical ($x-z$) slice through a galactic disk, such that if we
viewed the corresponding galaxy face-on, the surface density as
a function of radius would be given by $\Sigma(x)=\int dz \rho(x,z) $. The
area-weighted
mean surface density in any model is equal to
$\langle\Sigma\rangle_A \equiv \int
dx dz\, \rho/L_x$; this is a conserved quantity for any simulation, and
is one of the basic model parameters (see column 2 of Table
\ref{tbl:models}; in general we omit the angle brackets and ``A'' subscript).
The value of the surface density weighted by mass rather than by area
better represents the ``typical'' surface density of clouds found in the disk.
This is defined as
\begin{eqnarray}
\Sigma_{\rm cloud} =
\frac{\int \Sigma(x)^2\,dx}{\int \Sigma(x)\,dx}.
\label{eq:S}
\end{eqnarray}
Figure \ref{fig:S} shows $\Sigma_{\rm cloud}$ as a function of
$\Sigma$ for all model series. Interestingly, we find that
$\Sigma_{\rm cloud}$ does not strongly depend on parameters (either
$\Sigma$ or $\Omega$) throughout the four model series.
The largest value of $\Sigma_{\rm cloud}$ is $300 {\rm M}_\odot\ {\rm pc}^{-2}$ and
the smallest is $50 {\rm M}_\odot\ {\rm pc}^{-2}$, although for most cases the range
is even smaller: $\Sigma_{\rm cloud}=70-150{\rm M}_\odot\ {\rm pc}^{-2}$. This
factor-of-two range for $\Sigma_{\rm cloud}$ is significantly smaller
than the factor-of-six range of mean surface densities, $\Sigma = 7.5
- 42{\rm M}_\odot\ {\rm pc}^{-2}$. The range of $\Sigma_{\rm cloud}$ is also
similar to the typical observed surface densities of giant molecular clouds
(see discussion in \S \ref{Summary}).
This weak variation of $\Sigma_{\rm cloud}$ among the various series
suggests that it is star formation feedback, rather than the
feedback-independent parameters, that determines the typical surface
density of clouds.
In particular, other tests we have performed suggest that it is
the gravitational potential threshold for star formation
$\Phi_{\rm SF}$
that most
influences $\Sigma_{\rm cloud}$.
A model in which $\Phi_{\rm SF} \to 0$ (with any $\epsilon$) would
have photoheating events independent of the local ISM properties;
the consequent expansion of HII regions would thoroughly mix gas so
that $\Sigma_{\rm cloud} \to \Sigma$. This is indeed what we find
when we run models with $|\Phi_{\rm SF}|$ a factor 10 below our adopted
value. On the other hand, larger values of $|\Phi_{\rm SF}|$ require
more massive and compact clouds in order to have star formation, which
would raise $\Sigma_{\rm cloud}$. Tests with $|\Phi_{\rm SF}|$ a
factor 10 above our adopted value indeed result in larger $\Sigma_{\rm
cloud}$ (although only by a factor $\sim 2$).
The dependence on $\epsilon$ is much weaker than the dependence on
$\Phi_{\rm SF}$; reducing $\epsilon$ by a factor 10 changes
$\Sigma_{\rm cloud}$ by only tens of percent, at our standard $\Phi_{\rm SF}$.
The comparison between Series Q and Series R is also interesting, in
this regard. The difference between these two series is that the
stellar density $\rho_{\ast}$ increases with $\Sigma$ in Series Q, while
$\rho_{\ast}$ is constant throughout Series R.
Based on the larger
$\Sigma_{\rm cloud}$ value for the largest $\Sigma$ in Series Q
compared to Series R, when the stellar density is increased, the
surface density required in order to form clouds also increases.
\subsection{Temperatures}
Figure \ref{fig:temp} shows the space-time averages of temperature
for the components we have defined via density bins,
$\overline{\langle T(a)\rangle}$,
where the argument `$a$' denotes WNM, CNM, DM2, DM3 and HII.
Throughout the model series, the temperatures for the most dense and
most diffuse components are fairly constant; we find
$T=6,000-7,000$ K for WNM, $T=20-40$ K for DM2, and $T=10-20$ K
for DM3. For the CNM component (which in fact includes
thermally-unstable gas when it exists), the range is somewhat larger,
$T=100-400$ K, reflecting the larger range of conditions for this gas.
The gas which is subject to enhanced heating
has mean temperatures for most models of 4,000 -- 8,000 K.
The link between density and temperature in our models implies that
the components we have defined via density bins also approximately
correspond to natural ISM phases. This is because much of the gas
mass is close to thermal equilibrium (see Figure \ref{fig:phase-1}),
and we have chosen the bin edges so as to match up to points in the
phase plane with physical significance. Temperature PDFs that we have
constructed show a bimodal distribution, as is expected based on the
cooling function.
\subsection{Turbulent Velocities}
Expansion of HII regions feeds kinetic energy into the ISM.
This kinetic energy is not imparted solely to expanding HII bubbles
and shells surrounding them, but is shared throughout the ISM as turbulence.
Our models provide a first look at the results of this form of
turbulent driving. It is interesting to examine how the
turbulent amplitudes vary from one component to another in a given
model, and how the overall levels vary between models
with different feedback rates as a consequence of different system parameters.
Figure \ref{fig:velocity} shows the turbulent velocity dispersions for
all series, defined for each component as:
\begin{eqnarray}
v(a)&\equiv&\sqrt{\overline{
\langle v_x(a)^2+\tilde{v}_y(a)^2+v_z(a)^2\rangle}},
\end{eqnarray}
where the argument `$a$' denotes WNM, CNM, DM2, DM3 and HII, and
$\tilde{v}_{y}=v_y + \Omega x$ in order to subtract out the
velocity of unperturbed (sheared) rotation about the
galactic center.
The azimuthal velocities are excited by Coriolis forces
so that
the relation for epicyclic motions
\begin{equation}
\frac{{\tilde v}_{y}}{{v}_x}\simeq
\frac{\kappa}{2\Omega}=\frac{1}{\sqrt{2}}
\end{equation}
should apply \citep{1987gady.book.....B},
and we have checked that this is in fact satisfied.
We note that the velocity dispersion for each component is computed
by summing over the whole domain. Thus, the measured velocity
dispersions are larger than they would be within smaller-scale clouds
in the system. However, we have found that there
are not large contributions to the velocity dispersion from
velocity differences of widely-separated regions; this is because
turbulence is driven by the expanding HII regions, such that the
maximum correlation scale is comparable to the effective thickness of
the disk. For example, if we
divide the domain in Fig. \ref{fig:snap-1}
horizontally into eight equal parts, the mean
velocity dispersion of all gas
within these sub-domains is $\sim 98\%$ of that of
the domain as a whole. Considering just the dense gas, the velocity
dispersion for subdomains is $\sim 75\%$ of that in dense gas
for the whole domain. For the three large clouds seen in
Fig. \ref{fig:snap-3}, the mean internal velocity dispersions are an order
of magnitude larger than the dispersion in mean velocities.
In general, the densest component (DM3) has the lowest velocity
dispersion, with the next-densest (DM2) the next-lowest. The value of
the velocity dispersions for the dense components are highly
supersonic, and are similar to (or slightly below) those that are
observed within real GMCs (see \S \ref{Summary}).
The CNM component in our models typically
has higher turbulence levels than the WNM component, because the
former is in closer (space-time) contact with energy sources.
Because turbulent motions in our models
are driven by the pressure of photoheated gas,
$P=\rho c_s^2 \sim \rho v^2$,
the turbulent velocities have an upper limit of the
sound speed in gas heated to 8,000 K, $c_{\rm HII}\sim 7 ~ {\rm km\ s^{-1}}$.
Since the driving is intermittent, this upper limit is not
usually reached; mean values for the diffuse gas are closer to $\sim 5~{\rm km\ s^{-1}}$.
The diffuse-gas velocity dispersions in our models are lower by about 50\%
compared to observed levels, indicating (consistent with expectations)
that other turbulence sources are important in the real diffuse ISM.
The model series Q and R, which have $\Sigma \propto \Omega$ (and thus
effectively constant gaseous Toomre $Q$ if the velocity dispersion is
constant) show velocity dispersions that are insensitive to the value
of $\Sigma$. Series K, on the other hand, has much higher turbulence
levels for large $\Sigma$. This is because, with constant $\Omega$
($=\kappa/\sqrt{2}$) the high-$\Sigma$ models are quite susceptible to
gravitational instability (in terms of $Q$); this leads to very active
feedback which then raises the velocity dispersion. A similar
physical effect is seen in Series S: the velocity dispersion is
highest at low $\Omega$, since these are the most
gravitationally-susceptible models among the series. We discuss
measurements of the effective Toomre $Q$ values that account for
turbulence, in the next subsection.
\subsection{Effective Toomre $Q$ Parameters
\label{sec:Toomre}}
For a rotating disk that contains only thermal pressure,
susceptibility to growth of self-gravitating perturbations depends on
the Toomre $Q$ parameter, defined by setting $\sigma_R$ equal to the
thermal sound speed in equation (\ref{eq:ToomreQdef}). An
infinitesimally-thin gas disk is unstable to axisymmetric
perturbations if the value of this thermal $Q$-parameter is $<1$
\citep{1964ApJ...139.1217T}. Nonaxisymmetry, magnetic fields, and the
presence of active stars enhance gravitational instability
\citep{1965MNRAS.130..125G,1984ApJ...276..114J,2001MNRAS.323..445R,
2001ApJ...559...70K,2002ApJ...581.1080K,2003ApJ...599.1157K,
2007ApJ...660.1232K,2005ApJ...620L..19L}
allowing growth at higher $Q$, while nonzero disk thickness suppresses
gravitational instability
\citep{1965MNRAS.130...97G,2002ApJ...581.1080K,2007ApJ...660.1232K},
lowering the the critical $Q$ value. Allowing for all of these
effects, threshold levels measured from
simulations are $Q \approx 1.5$.
Turbulence at scales below the wavelength of gravitational instability
can also help to suppress the growth of large-scale density
perturbations, by contributing to the effective pressure. Since the
original Toomre parameter is arrived at based on effects of radial
pressure gradients, only the radial component of the velocity
dispersion should be added to the thermal velocity dispersion in
defining an effective $Q$ (see eq. \ref{eq:ToomreQdef}). It is
natural to expect galactic disks to self-regulate the values of the
effective $Q$: growth of self-gravitating instabilities subsequently leads to
star formation and energetic stellar feedback, which drives
turbulence, raises $Q$, and tends to suppress further GMC formation.
Indeed, the suggestion that galactic star formation is self-regulated through
turbulent feedback dates back to the earliest work on large-scale
instabilities in galactic gas disks \citep{1965MNRAS.130..125G},
with \citet{1972ApJ...176L...9Q} making the related suggestion that
galaxies deplete their gas until they reach marginal stability.
The self-regulation processes are complex, but they have begun to be
studied in recent numerical simulations
(e.g.\citealt{2002ApJ...577..197W,2008ApJ...673..810T}).
We use the results of our models to measure the values of the
effective Toomre parameter in the saturated state. We compare four
different measurements of $Q$ in each model. The first is closest to
Toomre's original definition for a gaseous medium
in that it is based on thermal velocity
dispersion; since our medium has components at differing
temperatures, we use a mass-weighted thermal velocity dispersion:
\begin{equation}
Q_T({\rm total}) = \frac{\kappa
}{\pi G\Sigma}\sqrt{\frac{\gamma k_{\rm B}
}{\mu}
\overline{\langle T\rangle}}.
\end{equation}
The second measurement incorporates turbulence, again including all gas
and weighting by mass:
\begin{equation}
Q({\rm total}) = \frac{\kappa}{\pi G\Sigma}
\sqrt{\frac{\gamma k_{\rm B}
}{\mu} \overline{\langle T \rangle}
+\overline{\langle v_R^2\rangle}}.
\end{equation}
For the third and fourth measurements,
we consider only the dense gas for both the
numerator and denominator:
\begin{equation}
Q_T(n>100) = \frac{\kappa}{\pi G\Sigma} \frac{\sqrt{ \frac{\gamma k_{\rm B}
}{\mu} \overline{\langle T(n>100)\rangle}}} {f(n>100)}.
\end{equation}
and
\begin{equation}
Q(n>100) = \frac{\kappa}{\pi G\Sigma} \frac{\sqrt{
\frac{\gamma k_{\rm B}}{\mu} \overline{\langle T(n>100)\rangle} +\overline{\langle
v_R^2(n>100) \rangle}}} {f(n>100)}.
\end{equation}
Here, the mass fraction of dense gas $f(n>100)$ is given
in eqn (\ref{eq:fraction}). The turbulent velocities dominate
the dense gas when they are included, since the
thermal sound speed is $<0.5~ {\rm km\ s^{-1}}$ whereas turbulent velocities
are several times larger; see Fig. \ref{fig:velocity}.
Note that $\kappa$ and $\Sigma$ are constant in time for any simulation.
Figure \ref{fig:Qvalue} shows the measured value of these four
quantities $Q_T({\rm total})$, $Q({\rm total})$, $Q_T(n>100)$, and $Q(n>100)$,
for all of our models. In general, we find that the
saturated-state values when turbulence is included
are near unity. The only significantly larger
values are for the low-$\Sigma$ models in Series K, which have large
$\kappa$ and hence the thermal value $Q_T({\rm total})$ is large; when
turbulence is included this is raised even more. Since Series Q and R
have $\kappa/\Sigma$ constant, the value of $Q_T({\rm total})$ is simply
proportional to the mass-weighted thermal velocity dispersion. The
increased fraction of cold gas at high $\Sigma$ leads to a
corresponding decrease in $Q_T({\rm total})$. Since the thermal and
turbulent velocity dispersions of the dense gas are small compared to
those of the diffuse components, the dense
components contribute to $Q_T({\rm total})$ and $Q({\rm total})$
mostly by lowering the mass fraction of the diffuse components, in the
numerator. Because turbulent contributions are positive-definite,
$Q({\rm total})\ge Q_T({\rm total})$.
The strongest evidence of self-regulation by feedback-driven
turbulence is seen
in the saturated-state results for $Q(n>100)$. With the low values of
the temperature in the dense component, the thermal-only values for
the dense gas are mostly $Q_T(n>100)<0.5$.
When turbulence is included, however, the saturated-state value of
$Q(n>100)$ is between 1 and 2 for almost all models. This is
consistent with expectations for marginal instability. We note in
particular that velocity dispersions in Series K
(see Fig. \ref{fig:velocity})
vary strongly with
$\Sigma$ (by a factor $\sim 5$), while $Q(n>100)$ varies weakly with
$\Sigma$ (by a factor $<2$); feedback evidently
self-adjusts in these models so as to maintain a state of marginal
gravitational instability.
\subsection{Virial Ratios}
In a self-gravitating system that approaches a statistical steady
state, the Virial Theorem predicts that the specific kinetic and gravitational
energies $E$ and $W$ will be related by $2E+W=0$; this is modified
when magnetic terms are present
\citep{1953ApJ...118..116C,1956MNRAS.116..503M,1992ApJ...399..551M}.
Classically, the Virial Theorem has often been assumed to hold within
individual GMCs in order to obtain estimates of their masses, and
indeed this yields values that are consistent (within a factor $\sim
2$) with other measures of the mass
(e.g. \citealt{1987ApJ...319..730S}). If individual GMCs are
short-lived, however, they may not satisfy the Virial Theorem because
the moment of inertia tensor is changing rapidly enough, and/or
surface terms are large enough, to be comparable to the kinetic and
gravitational energy integrated over the cloud volume
\citep{1999ApJ...515..286B, 2007ARA&A..45..565M,2007ApJ...661..262D}.
When averaged over an ensemble of clouds, there will be (partial)
cancellation of surface and time-dependent terms, as they appear with
opposite signs for forming and dispersing clouds. An added
complication is that self-gravitating GMCs form out of diffuse gas, and when
they are destroyed (whether after a short or long time) they return to diffuse
gas; thus the different terms in the Virial Theorem may be observed in
different tracers depending on whether diffuse gas is primarily atomic
or molecular.
Here, we consider virial ratios
\begin{eqnarray}
{\cal R}\equiv \frac{2E}{|W|}
\label{eq:vir}
\end{eqnarray}
separately for each component of the gas in our models.
The term
$E$ includes both thermal and bulk kinetic energy, computed via a
space-time average as:
\begin{equation}
E=\frac{1}{2}\overline{\langle v_x^2+\tilde{v}_y^2+v_z^2\rangle}+
\frac{1}{\gamma-1}\overline{\langle P/\rho \rangle},
\end{equation}
and for $W$ only the perturbed gravitational potential is used in
computing the space-time averaged value of the energy:
\begin{equation}
W=\frac{1}{2}\overline{\langle\Phi^{(1)}\rangle}.
\label{eq:Wdef}
\end{equation}
As for the other statistical properties we have considered, we measure
$\cal R$ separately for each component (separated into density bins)
of the system.
Figure \ref{fig:virial1} shows the virial ratio of each component, for
all models in all series. Note that ${\cal R}<2$ and ${\cal R}>2$
imply gravitationally bound and unbound states, respectively for any
component. As we do not separate the contributions to the potential
from the different density ranges, a given component may be bound
within a potential well that is created by more than one component.
Strictly speaking, the factor 1/2 in equation (\ref{eq:Wdef}) applies
only for self-potentials.
As expected, the lowest-density WNM component ($n<1~{\rm cm}^{-3}$) has $\cal R$
very large (above 100), and the intermediate-density CNM component
($1~{\rm cm}^{-3}<n<100 ~{\rm cm}^{-3}$) also is non-self-gravitating, with ${\cal R}$ in the range
$\sim 5-10$. The HII (photoheated) component generally has values of
${\cal R}\sim 10$, similar to that of the CNM component.
Although it has very large thermal energy (much greater than the CNM),
the photoheated gas by definition resides within deep parts of the
gravitational potential well.
The two dense
components, DM2 ($10^2~{\rm cm}^{-3}<n<10^3~{\rm cm}^{-3}$) and
DM3 ($10^3~{\rm cm}^{-3}<n$), on the other hand,
are consistent with being marginally or strongly gravitationally
bound, with ${\cal R} \aplt 2$ and ${\cal R} \aplt 1$, respectively.
For the majority of models, the value of ${\cal R}$ for the densest
component, DM3, is quite near unity, indicating consistency with
virial equilibrium for the component as a whole. For a few models,
${\cal R}$ is as low as 0.3 for the DM3 component; this indicates that
the dense gas is transient, with dense regions being rapidly
dispersed into lower-density gas by the feedback process. Overall, we
find no significant differences in the trends for ${\cal R}$ between
different series or different models within any series. There is a
weak correlation between $Q$ and ${\cal R}$, with lower-$Q$ (more
unstable) models having slightly higher virial ratios.
\subsection{Vertically-Averaged Density and Free-Fall Time
\label{sec:rhoave}}
Although the ISM consists of many phases at different densities, all
of this gas resides within a common potential well which tends to
confine material near the galactic midplane. The scale height of each phase
depends on the support provided by thermal and kinetic pressure (plus
support by magnetic stresses and cosmic rays, although these may be
less significant). In \citet{PaperII} we consider in detail
the vertical distribution of gas within our models, and show that
vertical equilibrium is a good approximation for the system as a
whole, provided that appropriate accounting is made for the differing
velocity dispersions of different components. We also discuss dependence
of the mean scale height on model parameters.
For the purpose of assessing gravitational timescales of the
overall ISM system, it is useful to measure the density when averaged
over large scales (i.e. a volume at least comparable to the scale height).
To evaluate this volume average in our models, we first compute the
vertical scale height, defined using the following averaging:
\begin{equation}
H_{\rm ave}=\sqrt{
\frac{\sum_{\rm all\ zones}\rho z^2 }{\sum_{\rm all\ zones}\rho}
}
\end{equation}
where $z$ is the vertical coordinate relative to the midplane.
We further average the values of $H_{\rm ave}$ over time.
For a Gaussian density profile,
$\rho(z)=\rho_0 \exp(-z^2/2H^2)$, the midplane density is
related to the surface density and scale height by
$\rho_0=\Sigma/(\sqrt{2\pi}H)$, and
the mass-weighted mean value of the average density is given by
$\rho_0/\sqrt{2}$. We therefore define an average density in our
models as:
\begin{equation}
\rho_{\rm ave}\equiv
\frac{\Sigma}{2\sqrt{\pi}H_{\rm ave}}
\end{equation}
(see also Appendix in \cite{PaperII}).
Figure \ref{fig:rave} shows the vertically-averaged density for all
models in all series. In general, we find that the average density
increases with the total surface density of gas in the disk.
A slightly shallower increase of $\rho_{\rm ave}$ with $\Sigma$
is obtained in Series K compared to Series Q,
which can be attributed to the large velocity
dispersions in strongly unstable (small $Q$) disk models.
Series R also has a shallower slope than in Series Q, because the
stellar gravity does not increase at large $\Sigma$ in the former.
For reference, we also plot in Figure \ref{fig:rave} the values of the
vertically-averaged density from our comparison hydrostatic model
series. The slope of Series HSC (lower-left panel) is shallower than that in
Series HSP (top panels), again because the stellar density does not
increasingly compress the gas at large $\Sigma$ in Series HSC.
The volume-averaged densities of the dynamic models are
lower than those of the hydrostatic models by up to an order of
magnitude; the difference increases at large surface density where
turbulence plays an increasingly important role (see also
\cite{PaperII}).
Using the mean density and the definition of the free-fall time,
\begin{eqnarray}
t_{\rm ff}(\rho)&=& \left(\frac{3\pi}{32G \rho}\right)^{1/2},
\label{eq:tff}
\end{eqnarray}
we can calculate the free-fall time for the
system as a whole, $t_{\rm ff}(\rho_{\rm ave})$.
Since $\rho_{\rm ave}$ increases with $\Sigma$ in our models,
$t_{\rm ff}(\rho_{\rm ave})$ will decrease with increasing $\Sigma$.
Because star formation requires gas to become self-gravitating, a
widespread notion is that the star formation timescale, when averaged
over large scales in a galaxy, will be proportional to the large-scale
average of $t_{\rm ff}$, i.e. $t_{\rm ff}(\rho_{\rm ave})$.
Since star formation takes place within GMCs
that have much higher density than the mean value in the ISM, the
conditions that control star formation where it actually takes place
are not those of the large-scale ISM. Thus, implicit in the notion
that star formation times should be related to the large-scale mean
$t_{\rm ff}(\rho_{\rm ave})$
is the idea that the formation of GMCs (on timescales
closer to $t_{\rm ff}(\rho_{\rm ave})$)
is the principal means of regulating star
formation. If the star formation efficiency per GMC is constant, then the GMC
formation rate would control the star formation rate. Alternatively,
the star formation rate might be related to the large-scale $t_{\rm
ff}(\rho_{\rm ave})$
if the densities within GMCs are proportional to the
large-scale mean densities of the ISM,
$\langle \rho_{\rm GMC}\rangle \propto \rho_{\rm ave}$.
Another important dynamical timescale in disk galaxies is the orbital
time, $t_{\rm orb}=2 \pi/\Omega$. Growth of large-scale
self-gravitating perturbations in disks in fact occurs at timescales
longer than $t_{\rm ff}(\rho_{\rm ave})$ (provided pressure limits small-scale
collapse), and more comparable to $t_{\rm orb}=2 \pi/\Omega$.
Observations \citep{1998ApJ...498..541K}
show that empirically-measured star formation
timescales in disk galaxies tend to be correlated with the orbital
time, with $\sim 10\%$ of gas being converted to stars per galactic
orbit. It is useful to compare $t_{\rm ff}(\rho_{\rm ave})$
with $t_{\rm orb}$ in our
models. Figure \ref{fig:tfftorb} shows the ratio of $t_{\rm
ff}(\rho_{\rm ave})/t_{\rm orb}$
for all hydrodynamic and hydrostatic series. For the
hydrodynamic series, the typical ratio is
$0.06-0.2$; for the
hydrostatic models, the densities are much higher at large $\Sigma$
so that
$t_{\rm ff}(\rho_{\rm ave})/t_{\rm orb}\sim 0.02-0.2$.
For series Q and R, the ratio $t_{\rm ff}(\rho_{\rm ave})/t_{\rm orb}$
varies relatively weakly with $\Sigma$, and lies in the range $0.1-0.2$.
The comparison hydrostatic models for these series also show
$t_{\rm ff}(\rho_{\rm ave})/t_{\rm orb}$ varying only modestly with $\Sigma$.
For these series, $t_{\rm orb}\propto 1/\Sigma$. Since turbulent velocity
dispersions do not depend strongly on $\Sigma$ for Series Q and R,
$\rho_{\rm ave}$ does not strongly depart from a scaling $\propto \Sigma^2$,
yielding behavior similar to $t_{\rm ff} \propto \Sigma \propto t_{\rm
orb}$.
Interestingly, the K series, which has constant $t_{\rm
orb}$, shows a smaller range of $\rho_{\rm ave}$ than
the Q series. This is indicative of self-regulation: high
feedback activity in the highest-$\Sigma$ models of series K yield
high turbulent amplitudes, which lead to lower values of $\rho_{\rm ave}$.
As a consequence, the ratios of $t_{\rm ff}/t_{\rm orb}$ are more
modulated in the hydrodynamic models for Series K than in the corresponding
hydrostatic series.
\section{Implications for Star Formation}
In the present work, we do not explicitly follow star formation.
Nevertheless, it is interesting to explore the
consequences of our statistical results, within the context of recipes
that are commonly adopted for star formation in numerical models.
We compare estimates of the implied star formation timescale both to
observations and to various fiducial dynamical times.
\subsection{Star Formation Rates and Timescales
\label{sec:SFR}}
A common practice in numerical simulations of galactic evolution is to
assume that the star formation rate per unit volume (in a
computational region) is proportional to the gas density per unit
volume divided by the free-fall time at that density. When a minimum
density threshold for star formation is imposed, the total star
formation rate (SFR, in mass of new stars per unit time) takes the
form
\begin{eqnarray}
\dot{M}_{\ast}\equiv
\frac{\epsilon_{\rm ff}(\rho_{\rm th})
M(\rho>\rho_{\rm th})}{t_{\rm ff}(\rho_{\rm th})}
\end{eqnarray}
provided that the density PDF decreases above the threshold, so that
most of the star forming activity is in gas near $\rho_{\rm th}$.
Here, the star formation efficiency per free-fall time,
$\epsilon_{\rm ff}(\rho_{\rm th})$,
is an arbitrary constant parameter that is adopted,
generally by comparing to observations.
In practice, the parameter $\epsilon_{\rm ff}(\rho_{\rm th})$ in this
sort of
recipe enfolds many different effects that limit star formation
compared to the fastest possible rate.
Within GMCs, turbulence and magnetic fields limit the rate of core formation
and collapse, and feedback from star formation limits GMC lifetimes;
at larger scales, dynamical processes in the diffuse ISM limit GMC formation.
Depending on the value of the threshold density, either more (low
$\rho_{\rm th}$) or fewer (high $\rho_{\rm th}$) processes are implicitly
packaged in the single efficiency parameter $\epsilon_{\rm ff}(\rho_{\rm th})$.
For a given star formation rate, the star formation timescale is
defined by dividing the total gas mass by the total SFR,
$\dot{M}_{\ast}$:
\begin{eqnarray}
t_{\rm SF}&\equiv& \frac{M_{\rm tot}}{\dot{M}_{\ast}}
=\frac{M_{\rm tot}}{M(\rho>\rho_{\rm th})}
\frac{t_{\rm ff}(\rho_{\rm th})}{\epsilon_{\rm ff}(\rho_{\rm th})},\\
&=&\frac{1}{f(\rho>\rho_{\rm th})}
\frac{t_{\rm ff}(\rho_{\rm th})}{\epsilon_{\rm ff}(\rho_{\rm th})}
\equiv \frac{\tau_{\rm SF}}{\epsilon_{\rm ff}(\rho_{\rm th})}.
\end{eqnarray}
The latter expression uses the mass fraction $f$ as defined in
equation (\ref{eq:fraction}); $M_{\rm tot}$ is the total gas mass.
Because $f$ and $\epsilon_{\rm ff}$ are, by definition,
less than 1, the star formation time always exceeds the free-fall time at
the threshold density.
Since the efficiency per free-fall time is arbitrary
(from the point of view of simulations), it is
convenient to introduce
$\tau_{\rm SF}\equiv \epsilon_{\rm ff}(\rho_{\rm th}) t_{\rm SF}$ such that
$\tau_{\rm SF}=t_{\rm ff}(\rho_{\rm th})/f(\rho>\rho_{\rm th})$.
This scaled
star formation time then depends only on the choice of density threshold
and the fraction of the total gas mass above this threshold.
In numerical simulations, the density threshold for star formation is
an arbitrary parameter; what difference does the choice of this value
make to the resulting SFR? To address this question, we first compare
values of $\tau_{\rm SF}$ using two different thresholds,
$n=10^2 ~{\rm cm}^{3}$ and $n=10^3 ~{\rm cm}^{3}$. Both threshold values are large
enough that the gas at these densities is in gravitationally-bound
structures, based on the results shown in Fig. \ref{fig:virial1}.
Figure \ref{fig:tSF} shows the values of the scaled star formation
time, $\tau_{\rm SF}$, for all models.
For our chosen density thresholds, $\tau_{\rm SF}$ is in the range
$3\times 10^6- 10^7$ yr for all models.
The true star formation time, $t_{\rm SF}$, exceeds $\tau_{\rm SF}$ by a
factor $\epsilon_{\rm ff}^{-1}$; the value of
$\epsilon_{\rm ff}(\rho_{\rm th})$ must
then be quite small ($<0.01$) for $t_{\rm SF}$ to be $>10^9$ yr.
Also, since $\tau_{\rm SF}$ is larger for the threshold
choice $n=10^2 ~{\rm cm}^{3}$ than $n=10^3 ~{\rm cm}^{3}$, the value of
$\epsilon_{\rm ff}(\rho_{\rm th})$
would have to be smaller for the higher density
threshold, in order to yield the same value of $t_{\rm SF}$ at a given
$\Sigma$. Note,
however, that while the thresholds differ by a factor 10, the values
of $\tau_{\rm SF}$ (and hence required
$\epsilon_{\rm ff}(\rho_{\rm th})$) differ by
less than a factor 2. This reflects the fact that $f$ decreases with
increasing $n$; between $n=10^2 ~{\rm cm}^{3}$ and $n=10^3 ~{\rm cm}^{3}$, our
results imply a dependence $f(>n)\propto n^{-s}$ with the range of
$s=0.2-0.5$.
Alternatively, we can think of our results
requiring a choice for
$\epsilon_{\rm ff}(\rho_{\rm th})\propto \rho_{\rm th}^{-r}$
with the range of $r=0-0.3$
in order for the SFR to be independent of the choice
of threshold at high densities.
Other aspects of the results shown in Figure \ref{fig:tSF} are also
interesting. First, it is evident that $\tau_{\rm SF}$ depends only
weakly on both surface density (Series Q, K, R) and angular velocity
(Series S). For Series Q and S, $\tau_{\rm SF}$ decreases with
increasing $\Sigma$. Interestingly, the hydrostatic models show a
similar range of $\tau_{\rm SF}$ to the dynamic, turbulent models.
The fact that $\tau_{\rm SF}$ is not strongly sensitive to environmental
conditions (total available gas content, local shear rate, level of
turbulence, etc.) may help to explain why empirical SFRs show such a
regular character in observed galaxies, in spite of widely-varying
local conditions. Conversely, the insensitivity of $\tau_{\rm SF}$ to
conditions within a model has implications for evaluating theoretical
results: successfully reproducing levels of star formation similar to
observations is
not a critical discriminant of how well a simulated galaxy resembles a
real system. Our hydrostatic models bear minimal resemblance to real
galaxies, yet for a choice of $\epsilon_{\rm ff}(\rho_{\rm th})\sim 0.01$
consistent with
observed efficiencies in CO-emitting gas in GMCs (which have
densities in the range $n=10^2-10^3 ~{\rm cm}^{3}$), the resulting star
formation times are $\sim 4\times 10^8-10^9$ yr, similar to the
observed range of $t_{\rm SF}$
for $\Sigma$ comparable to the range in our models.
To connect more directly to the way observed SFRs are normally presented,
in Figure \ref{fig:SFR} we show results for scaled surface density of
star formation as a function of
surface density of gas (Series Q, K, R) and angular velocity (Series S).
The scaled SFR per unit area is defined as
$\Sigma_{\rm SFR}/\epsilon_{\rm ff}(\rho_{\rm th})
=\Sigma_{\rm tot}/\tau_{\rm SF}$, where
the SFR is taken to follow
\begin{eqnarray}
\Sigma_{\rm SFR}\equiv \frac{\Sigma_{\rm tot}}{t_{\rm SF}}
=\frac{\epsilon_{\rm ff}(\rho_{\rm th})
\Sigma_{\rm tot}f(\rho>\rho_{\rm th})}{t_{\rm ff}(\rho_{\rm th})}.
\end{eqnarray}
As before, we compare results based on two different threshold density
choices, and also show the results from hydrostatic models.
Observations are typically fitted to power laws of the form
$\Sigma_{\rm SFR} \propto \Sigma^{1+p}$. For reference, we show
slopes with $1+p=1$ and $1.5$.
For each model series and each value of $\rho_{\rm th}$,
we fit a power-law index. We find $1+p$ equal to
1.32, 1.43 (Series Q for $n=10^2,\, 10^3 ~{\rm cm}^{3}$),
0.94 (Series K for $n=10^2~{\rm cm}^{3}$), and
1.24, 1.19 (Series R for $n=10^2,\, 10^3 ~{\rm cm}^{3}$).
For the hydrostatic cases, the indices are 1.38 (Series HSP) and 1.39
(Series HSC) at $n=10^2~{\rm cm}^{3}$.
As we shall discuss further in \S \ref{Summary},
these results are similar to the
observed ranges of power-law indices that have been reported.
We note
that Series Q and R show more regular behavior than Series K. This
reflects the different environmental parameters that are inputs to the
models: in Series K, the epicyclic frequency is held constant, while
in series Q and R we adopt a scaling $\Omega \propto \Sigma$.
For $\epsilon_{\rm ff}\sim 0.001-0.01$, both the magnitude and scaling
of the $\Sigma_{\rm SFR}$ vs. $\Sigma$ results in Series Q and R
are similar to observations.
\subsection{Comparison of Timescales}
In \S \ref{sec:SFR}, we investigated the relationship between the mean
large-scale surface density $\Sigma$ and the star formation time based
on the amount of high-density gas (at $n>10^2 ~{\rm cm}^{-3}$ within a
zone). This gas may be considered immediately eligible for star
formation, since it is cold and found in self-gravitating systems. As
noted in \S \ref{sec:rhoave}, if formation of massive, cold,
gravitationally bound systems is the principal throttle for star
formation, then star formation times would also be expected to vary
with the timescales for GMC formation.
GMC formation is a complex process, and to date no simple formula has
been obtained for the formation rate. Instead, several different
``large-scale'' dynamical times are commonly invoked to obtain
estimates of the GMC formation time. These include the free-fall time
at the large-scale mean density, $t_{\rm ff}(\rho_{\rm ave})$, the
Jeans time based on the surface density and the gas velocity
dispersion, $t_{\rm J}$, and the orbital time $t_{\rm orb}$, which is
generally related to the epicyclic and shear times. It is interesting
to explore how our measurement of $\tau_{\rm SF}$ compares to each of
these times, as a function of the independent parameter in each series.
We begin with the orbital time, $t_{\rm orb}=2\pi/\Omega$. Figure
\ref{fig:tSFtorb} shows the ratio between $\tau_{\rm SF}=
\epsilon_{\rm ff}(\rho_{\rm th})t_{\rm SF}$ (for the two
different density thresholds $\rho_{\rm th}$)
and the orbital time. In Series K,
$\Sigma$ is the independent parameter, but $t_{\rm orb}$ is
independent of $\Sigma$; thus the ratio is simply a rescaled version
of $\tau_{\rm SF}$ shown in Fig. \ref{fig:tSF}. In Series Q and R,
the independent parameter is $\Sigma$, and $t_{\rm orb} \propto
\Sigma^{-1}$, so $\tau_{\rm SF}/t_{\rm orb} \propto \tau_{\rm SF}
\Sigma$. In Series S, $t_{\rm orb}$ is the independent parameter.
Although the variation with the independent parameter is
moderate in all the series, the ratio is not constant, and for some
series shows secular trends. Namely, for
Series Q and R, which showed a trend of decreasing $\tau_{\rm
SF}$ at increasing $\Sigma$, $\tau_{\rm SF}/t_{\rm orb}$ increases
at larger $\Sigma$. Thus, assuming that the star formation time is
$\propto t_{\rm orb}$ would increasingly overestimate the true star
formation rate (presumed to depend on the amount of dense gas) as
$\Sigma$ increases.
We next consider the Jeans time for a disk,
$t_{\rm J} \equiv \sigma/(G \Sigma)$,
where $\sigma$ is either the thermal or the total
(radial) turbulent velocity dispersion, $c_s$ or $\sqrt{c_s^2 + v_R^2}$.
We note that the ratio $t_{\rm J}/t_{\rm orb}$ is given by
$Q/(2\sqrt{2})$, where the Toomre parameter $Q$ is either based on
mean sound speed or the total velocity dispersion (see \S
\ref{sec:Toomre}). Figure \ref{fig:tSFtJ} shows the ratio between
$\tau_{\rm SF}$ and the Jeans time, using
the total velocity dispersion.
Again, strong secular trends with $\Sigma$ are evident; $t_J$ is not a
good predictor of the star formation time.
Finally, in Figure \ref{fig:tSFtff} we show the ratio between
$\tau_{\rm SF}$ and and the free fall time at the vertically-averaged
large-scale mean density (\S \ref{sec:rhoave}; see Fig.
\ref{fig:tfftorb}). Although the values of this ratio are closer to
unity than $\tau_{\rm SF}/t_{\rm orb}$ and $\tau_{\rm SF}/t_{\rm J}$, we still
see that $\tau_{\rm SF}/t_{\rm ff}(\rho_{\rm ave})$ is not constant as
a function of $\Sigma$.
When we compare $\Sigma/t_{\rm ff}(\rho_{\rm
ave})$ to the scaled star formation rates based on high density gas
shown in Fig. (\ref{fig:SFR}), we find
a steeper rise with $\Sigma$, close to $\propto \Sigma^2$ in Series Q
and R and slightly shallower in Series K.
If star formation is regulated by the collection of diffuse gas into
self-gravitating regions, then strictly speaking one would expect specific star
formation rates to vary proportional to the fraction of diffuse gas
divided by the GMC formation time (estimated just including the
diffuse gas). The above comparisons between $\tau_{\rm SF}$ and
timescale estimates based on mean large-scale properties can be corrected
to account for this, yielding the ratios
$\tau_{\rm SF} (1-f_{\rm dense})/t_{\rm orb}$,
$\tau_{\rm SF} (1-f_{\rm dense})^{3/2}/t_{\rm ff}(\rho_{\rm ave})$,
and
$\tau_{\rm SF} (1-f_{\rm dense})^2/t_{\rm J}$. We find, however, that
these ratios also are non-constant in any series,
although the correction factor does tend to flatten out the secular rise
with increasing $\Sigma$ in the comparison to the free-fall time.
Taking all of our results together, we conclude that several commonly-used
estimates for galactic star formation timescales based on large-scale
mean galactic properties may have only limited
utility for making detailed predictions of star formation rates. That
is, the orbital time, the Jeans time, and the free-fall time based on
the vertically-averaged density are not
proportional to the star formation time based on the amount of dense,
gravitationally bound gas that is present. Simulations with
insufficient resolution or limited physics may therefore not be able
to provide accurate predictions of star formation rates, if they do
not capture processes at small enough scales to represent dense,
gravitationally-bound structures.
\section{Summary and Discussion
\label{Summary}
}
We have developed a numerical hydrodynamic code to study the
life-cycle of multiphase, turbulent interstellar gas in disk galaxies;
our model includes gas self-gravity, the vertical gravity associated
with a fixed stellar disk, radiative cooling and heating in the
temperature range of $10 ~\mbox{K} \le T \le 10,000 ~\mbox{K}$,
sheared rotation in the background galactic potential (we adopt a flat
galactic rotation curve), and a prescription for feedback in the form
of HII regions that originate within massive, dense clouds. Our
simulation domains represent slices in the radial-vertical plane
of a galactic disk. We focus on scales large enough to include
vertical stratification ($L_z$ up to 410 pc) and significant shear of the
disk angular velocity ($L_x$ up to 1.6 kpc), but small enough
to resolve sub-structure within dense, self-gravitating clouds that
form (typical zone resolution is $\sim 1$pc). Our models are
2.5-dimensional, in the sense that all three components of the
velocity are time-dependent functions. For feedback to occur, we
impose thresholds on both the local volume density and on the
gravitational potential, so that HII regions only occur within massive
clouds (consistent with observations). The expansion of HII regions
drives turbulence in all the components of the gas. We have performed
a large suite of simulations, covering a factor of six in gas surface
density $\Sigma$. In order to explore the dependence of ISM properties on
galactic environment (in particular, the stellar vertical gravity and
the angular momentum content of the gas), we have considered four
different model series. In Series Q, we vary $\Sigma$, stellar
volume density $\rho_*$, and the disk rotation rate $\Omega$ in tandem.
In Series K, we vary $\Sigma$ and $\rho_*$ together while holding
$\Omega$ fixed. In Series R, we vary $\Sigma$ and $\Omega$ together
while holding $\rho_*$ fixed. Finally, in Series S the values of
$\Sigma$ and $\rho_*$ are held constant while $\Omega$ is varied.
Our main conclusions, and their relation to other recent work, are as follows:
1. {\it Density, temperature, and pressure distributions}
We find that in spite of time-dependent effects, the density and
temperature distributions of the gas retain bimodal profiles
reminiscent of the classical \citet{1969ApJ...155L.149F} two-phase
model of the ISM.
Although large-amplitude turbulence
heats and cools via $P dV$ work and entropy production in shocks,
most of the gas (by mass) remains near the
curve in the pressure-density phase plane that is defined
by radiative equilibrium: $n^2 \Lambda - n \Gamma=0$. This is
possible because the cooling time is generally shorter than the
turbulent dynamical times, for our models.
If turbulent compressions or expansions
were more extreme in magnitude and also rapid in time compared to radiative
times, then these adiabatic changes would lead to density-temperature
pairs that more strongly departed from thermal equilibrium
(see e.g. \cite{2002ApJ...577..768S,2005A&A...433....1A,2008ApJ...683..786H}
for a discussion of the dependence on various physical timescales
involved).
\citet{2005A&A...433....1A},
\citet{2005ApJ...629..849P}, and \citet{2007ApJ...663..183P} similarly
found that even with large-amplitude turbulence --
and no direct heating of the gas -- mass-weighted
density and temperature PDFs have two (broadened) peaks, although
for very high amplitude turbulence the trough tends to fill in
(e.g. \citealt{2007A&A...465..431H}).
\citet{2008ApJ...681.1148K} have emphasized the importance of
maintaining sufficient
numerical resolution, as numerical diffusion associated with flow
across the grid broadens warm/cold interfaces, populating the
thermally-unstable regime in the phase diagram.
We note that bimodal character is most easily seen in mass-weighted
rather than volume-weighted PDFs, although many of the results in the
literature show only volume-weighted PDFs.
Our results on the bimodal thermal distribution of gas
are consistent with observations of atomic HI
in the midplane of the Milky Way \citep{2003ApJ...586.1067H}, which at
the same time show interesting evidence of out-of-equilibrium gas,
particularly at high latitudes.
2. {\it Turbulence}
We find that appreciable turbulence can be excited in all components
of the gas. The values of the velocity dispersion in dense gas
($n>10^2 {\rm cm}^{-3}$) of $\sim 2-4\ {\rm km\ s^{-1}}$ are similar to, though slightly
lower than, those observed in Milky Way and other local-group
GMCs (e.g.
\citealt{1987ApJ...319..730S,
2008ApJ...675..330S,
2008arXiv0807.0009B,
2008arXiv0809.1397H}). For lower-density gas in
our models, velocity dispersions are slightly higher, but still lower
by a factor two compared to observed velocity dispersions of
$\sim 7-10~{\rm km\ s^{-1}}$ of Solar-neighborhood
warm and cold HI seen in 21 cm emission and absorption
\citep[e.g.,][]{2003ApJ...586.1067H,2004JApA...25..185M}.
It is not surprising that the turbulence levels in our simulations are
only moderate, given that we have included only one of the many sources of
turbulence that is present in the ISM. Turbulent driving in the ISM
has been reviewed by e.g.
\citet{2004RvMP...76..125M,2004ARA&A..42..211E}.
Supernova are widely-believed to be the most important
source of turbulence for diffuse gas, and this has been demonstrated
by numerical simulations
\citep[e.g.,][]{
1999ApJ...514L..99K,
2005A&A...436..585D}.
In the outer galaxy
where star formation rates are low, driving by
the magnetorotational instability may, however, dominate
\citep{1999ApJ...511..660S,2005ApJ...629..849P,2007ApJ...663..183P}.
Spiral shocks are
also effective in driving turbulence in the warm ISM,
especially at high latitudes \citep{2008ApJ...681.1148K}.
The interaction between large-scale self-gravity, rotation, and shear
can drive near-sonic turbulence at large scales
\citep{2002ApJ...577..197W,2007ApJ...660.1232K},
although the amplitude of this at
scales less than the disk scale height may be modest.
It is not known how effective these other mechanisms are for driving
turbulence within GMCs, however, which are very dense and therefore present a
small effective crossection to the diffuse ISM.
3. {\it Feedback and the Toomre $Q$ parameter}
We have measured the Toomre parameter for each of our models, considering
both the entire medium and just the dense gas,
and comparing ``thermal-only'' with ``turbulent+thermal'' values.
For all models, we find that the ``turbulent+thermal'' $Q$-value for
dense gas is in the range $1-2$, and is
much greater than the thermal-only value.
A further interesting point is that the turbulence level evidently adjusts
with surface density in order to reach a marginally-stable state.
In Series Q and R, which have $\Omega/\Sigma=const.$,
the velocity dispersions are relatively independent
of $\Sigma$, yielding marginally-stable $Q$ in the cold, dense gas.
In Series K, which has constant $\Omega$ and
therefore is highly unstable at large $\Sigma$ in the absence of
turbulence, the velocity
dispersions strongly increase with $\Sigma$, as a consequence of much
higher levels of feedback activity. These higher velocity dispersions
lift $Q$ to near unity. Thus, our simulations give
direct evidence of feedback leading to a self-regulated quasi-steady
state. We note that $Q$ values vary for different
components; \citet{2002ApJ...577..197W} similarly found a large range
of $Q$ when measured in local patches within their disk simulations.
Depending on what exactly is included in a model, the
threshold for gravitational instability in previous non-turbulent
simulations is measured
to be at $Q\sim 1.5$ (see \citealt{2007ARA&A..45..565M}), which is
similar to the values we find here when turbulence is included in $Q$.
In the actively-star-forming regions of galaxies, measured values of the
Toomre parameter are not constant, but show a limited range
\citep{2001ApJ...555..301M}.
Evidence for star formation thresholds tied to $Q$ are more mixed
\citep{2001ApJ...555..301M,
2003MNRAS.346.1215B,
2007ApJS..173..524B}, possibly
because star formation in outer disks primarily takes place in spiral
arms
\citep{1998ApJ...506L..19F,2007ApJS..173..538T,2008ApJ...683L..13B}
which strongly compress the gas above ambient densities.
4. {\it The virial ratio}
We measure the virial ratio $\cal R$ (eq. \ref{eq:vir}) for all our models,
separating into different density regimes. We find that dense gas
($n>100\ {\rm cm}^{-3}$) generally has $\cal R$ between $1 - 2$, whereas
lower-density gas has large values of $\cal R$.
$\cal R$ does not vary strongly with $\Sigma$ in any of the series.
In particular, we note that in
spite of the large range of velocity dispersions in the dense gas in
Series K at different $\Sigma$, $\cal R$ varies only weakly with
$\Sigma$. This indicates that feedback can
effectively regulate the dynamics within massive, dense clouds,
independent of the larger-scale galactic environment.
This is consistent with both older studies based on $^{12}$CO
observations \citep{1987ApJ...319..730S}, and recent studies
based on $^{13}$CO observations \citep{2008arXiv0809.1397H}, both of
which find $\cal R$ near $1-2$ for Milky Way GMCs. Although masses based
on CO are less certain in external galaxies, virial ratios also likely
near unity for GMCs observed in the Local Group \citep{2008arXiv0807.0009B}
5. {\it Cloud surface densities}
We estimate the surface density of typical clouds by measuring the
mass-weighted vertically-integrated column of gas; we define this
as $\Sigma_{\rm cloud}$. For the values of the feedback threshold
that we adopt, we find that $\Sigma_{\rm cloud}$ is in the range
$70-150 ~{\rm M}_\odot\ {\rm pc}^{-2}$ for most models. This is comparable to the
typical GMC surface densities that are observed in the Milky Way and
in Local Group galaxies
\citep{1987ApJ...319..730S,2008ApJ...675..330S,
2008arXiv0807.0009B,
2008arXiv0809.1397H}. Whether observational selection effects or
physical processes impose a limited range of column densities for GMCs
is an open question. For example, magnetic fields may impose a minimum surface
density for formation of gravitationally-bound structures in the ISM,
at a value $\Sigma = B/(2 \pi\sqrt{G})= 30 ~ {\rm M}_\odot\ {\rm pc}^2 (B/10 \mu {\rm
G})$ \citep{2007ARA&A..45..565M}. Here, we find that
altering the volume heated in our HII region
prescription does not appreciably change $\Sigma_{\rm cloud}$, but
changing the gravitational potential threshold for star formation
feedback does: when potential thresholds are low, $\Sigma_{\rm
cloud}$ is also low. Comparison of cloud properties with
observations may turn out to be a much
more critical test of whether an ISM model is realistic than some
other measures, such as the star formation rate.
6. {\it Dependence of $\Sigma_{\rm SFR}$ on $\Sigma$}
We obtain estimates of the dependence of surface star formation rate
$\Sigma_{\rm SFR}$ on gas surface density $\Sigma$
in our models by assuming that the timescale for star
formation is proportional to the free-fall time in gas above some
density threshold $n_{\rm th}=\rho_{\rm th}/\mu$. We
compare results for two different threshold densities, $n_{\rm th} = 100\
{\rm cm}^{-3}$ and $n_{\rm th} = 10^3\ {\rm cm}^{-3}$. For Series Q and R
(which most resemble real galaxies), we find (for
either choice of threshold)
relations that are well-described by power laws:
$\Sigma_{\rm SFR}\propto \Sigma^{1+p}$ with $1+p=1.2-1.4$. In Series
K, a power law is a less-good fit; the slope is also shallower (closer
to unity).
These results are consistent with empirical Kennicutt-Schmidt laws
\citep{1959ApJ...129..243S,1963ApJ...137..758S,1998ARA&A..36..189K},
which show similar values of $1+p$ when all the gas is included in $\Sigma$
(e.g.
\citealt{1989ApJ...344..685K,1998ApJ...498..541K,2002ApJ...569..157W,
2007A&A...461..143S,2007ApJ...671..333K}).
Recent work has suggested that $1+p$
is close to unity if just CO-emitting molecular gas is included
\citep{Bigiel2008}
; this implies all CO-emitting gas (most of which
is at $n=10^2-10^3 {\rm cm}^{-3}$)
has the same star formation rate independent of galactic environment.
Our prescription that the star formation rate
per unit mass of dense gas is constant is equivalent to empirically
finding $1+p=1$ if only molecular gas is included.
It is encouraging that the results we find for Kennicutt-Schmidt
relations in our disk models with feedback are compatible with observations.
We also find, however, that star formation rates predicted from
hydrostatic models are in fact similar to those predicted from the
hydrodynamic models,
with similar slopes at large $\Sigma$. This is true even though the
hydrostatic models are not at all like real galaxies. Thus, one must
be cautious in considering a numerical model successful if it yields
reasonable star formation rates, since this can simply be a
consequence of choosing reasonable initial conditions
in a simulation. Indeed, a number of recent
numerical studies have found results similar to observed
Kennicutt-Schmidt laws, regardless of the detailed physics that they
included in the models (e.g.
\citealt{2006ApJ...639..879L,2006ApJ...641..878T,
2008ApJ...673..810T,2008ApJ...680.1083R}).
\citet{2008MNRAS.383.1210S} have also recently emphasized that
reproducing empirical star formation scaling laws is not by itself a
critical test of an ISM model.
7. {\it Density-dependence of star formation efficiency}
The star formation efficiency per free-fall time can be defined
locally as a
function of threshold density by $\epsilon_{\rm ff}(\rho_{\rm th}) =
t_{\rm ff}(\rho_{\rm th}) \Sigma_{\rm SFR}/\Sigma(\rho> \rho_{\rm
th})$; corresponding global measures can also be obtained. For a
given true star formation rate,
$\epsilon_{\rm ff}(\rho_{\rm th}) \propto \tau_{\rm SF}$
where the scaled star formation time
$\tau_{\rm SF}$ is shown for two different threshold densities
in Fig. \ref{fig:tSF}. We find that $\tau_{\rm SF}$
(or $\epsilon_{\rm ff}(\rho_{\rm th})$)
decreases with increasing $\rho_{\rm th}$. This is not a strong
effect, however: it is less than a factor 2 for an order of magnitude
difference in $\rho_{\rm th}$. In Series Q and R, the
ratio of efficiencies at different
threshold densities are also independent of $\Sigma$ (although this is
not true for Series K).
\citet{2007ApJ...654..304K} recently compiled a range of observations
of $\epsilon_{\rm ff}$ (which they refer to as SFR$_{\rm ff}$). They
point out that for threshold densities above $\sim 100 {\rm cm}^{-3}$, the
value of $\epsilon_{\rm ff}$ does not vary strongly with density. They also
find a smaller value for gas traced by HCN than for gas traced by
CO, which since HCN has a higher critical density than CO is
consistent with our finding that $\epsilon_{\rm ff}$ decreases with
increasing $\rho_{\rm th}$.
Even though scaled star formation times
at high density vary together independent of $\Sigma$, the free-fall time at
the vertically-averaged density is {\it not} proportional to
$\tau_{\rm SF}$. Instead, we find that $\tau_{\rm SF}/t_{\rm
ff}(\rho_{\rm ave})$ increases at increasing $\Sigma$. This implies
that a prescription for star formation based on the mean density
within one ``average''
scale height in a disk (or from a simulation that does not
resolve high-density gas)
would increasingly overestimate the star formation rate at high $\Sigma$.
The same is true for the orbital time and the Jeans time: an
assumption that the star formation rate varies $\propto \Sigma/t_{\rm
orb}$ or $\propto \Sigma/t_{\rm J}$ would increasingly overestimate
$\Sigma_{\rm SFR}$ at high $\Sigma$ compared to the value obtained
from measuring the mass of gas in dense, gravitationally-bound
regions. Thus, ISM models must resolve self-gravitating structures at
scales less than the disk thickness in
order to make accurate predictions of the star formation rate.
\begin{acknowledgements}
\vspace{12pt}
We are grateful to the referee for a thorough reading and thoughtful
set of comments that have helped us to clarify our presentation.
This research was supported by grant NNG-05GG43G from NASA. Numerical
computations were carried out on the OIT High Performance Computing
Cluster (HPCC) and CTC cluster in the Department of Astronomy, at
the University of Maryland.
\end{acknowledgements}
\bibliographystyle{apj}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 2,963 |
"I came in with no belly dancing experience at all and have felt increasingly confident each class. Our class is so fun with lots of laughter. I hope to take more belly dancing when this is done!"
We're not currently offering belly dance classes; but we do offer lots of other really super fun dance classes!
Belly dancing is one of the oldest forms of dance. The original name of belly dance is "Raqs Sharqi" which means "Dance of the Orient." It gained the term "belly dance" when it made its way from East to West. In France, the dance got dubbed "danse du ventre" which literally means "belly dance" due to the flutter of the dancer's stomach. It is believed that belly dancing was created to help prepare women for birth, since during labor, a woman's stomach undulates, much like the undulation used during belly dance. Belly dancing was also used a form of ancient Mother Goddess worship.
Every part of the body is involved in the dance; the most featured body part usually is the hips. Belly dance takes many different forms depending on country and region, both in costume and dance style, and new styles have evolved in the West as its popularity has spread globally. Although contemporary forms of the dance have generally been performed by women, some of the dances, particularly the cane dance, have origins in male forms of performance.
Wikipedia reports: Belly dance is a non-impact, weight-bearing exercise and is thus suitable for all ages, and is a good exercise for the prevention of osteoporosis in older people. Many of the moves involve isolations, which improves flexibility of the torso. Dancing with the veil can help build strength in the upper-body, arm and shoulders. Playing the zills trains fingers to work independently and builds strength. The legs and long muscles of the back are strengthened by hip movements. Paffrath also researched the effect of belly dance on women with menstruation problems. The subjects reported a more positive approach toward their menstruation, sexuality, and bodies.
Belly dancing definitely improves your freestyle club dancing. Some performers who have added belly dance moves to their include Christina Aguilera, Jessica Simpson, Beyonce, and Ciara.
Check out more of our current reviews.
Our two OUT to Dance studio locations, West Roxbury, MA and Roslindale, MA, are within twenty minutes of downtown Boston, Dorchester, Jamaica Plain, Hyde Park, Brighton, Allston, Brookline, Newton, Chestnut Hill, Dedham, Norwood, Needham, Westwood, Milton and Quincy; and within 25 to 35 minutes of Cambridge, Somerville, Arlington, Wellesley, Natick, Waltham, Braintree, Brockton, Stoughton, Canton, Foxboro, Weymouth and surrounding towns. We are also less than an hour from Providence, Rhode Island and New Hampshire.
Want to try our dance classes at West Roxbury School of Dance? See our current class schedule under Classes for Everyone. And we have LGBT dance classes on Sundays.
Discover the ancient art of belly dance! From hip circles to shimmies you will learn the basic moves universal in all styles of this beautiful art. People of all shapes, sizes and age welcome!
What to wear: for belly dance class: you may wear some form of dance or fitness attire. This can include a leotard, unitard, sports bra or fitted top, comfortable fitted yoga pants or gym pants. Bare feet or dance shoes or clean sneakers are fine. Not sure about attire? Just come as you are and ask questions at the first belly dance class! | {
"redpajama_set_name": "RedPajamaC4"
} | 9,607 |
Q: Gatsby build deployed on digital ocean give 404 on navigation I have deployed the build version of my project using "gatsby build" on Digital Ocean. When I try to navigate using menu, it gives 404 page. And even some of the Bootstrapped CSS and Other CSS & JS file get missing after the build deployed.
The project works fine on local machine by "gatsby serve". But not on Digital ocean
repo : https://github.com/mahayash/SE_Frontend.git
url : http://www.squareeducation.in/
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 3,963 |
package com.sequenceiq.cloudbreak.structuredevent.event.cdp;
import java.io.Serializable;
import com.fasterxml.jackson.annotation.JsonIgnoreProperties;
import com.fasterxml.jackson.databind.annotation.JsonDeserialize;
import com.fasterxml.jackson.databind.annotation.JsonSerialize;
import com.sequenceiq.cloudbreak.event.ResourceEvent;
import com.sequenceiq.cloudbreak.structuredevent.json.AnonymizingBase64Serializer;
import com.sequenceiq.cloudbreak.structuredevent.json.Base64Deserializer;
@JsonIgnoreProperties
public class CDPStructuredNotificationDetails implements Serializable {
private ResourceEvent resourceEvent;
private String resourceCrn;
private String resourceType;
@JsonSerialize(using = AnonymizingBase64Serializer.class)
@JsonDeserialize(using = Base64Deserializer.class)
private String payload;
public CDPStructuredNotificationDetails(ResourceEvent resourceEvent, String resourceCrn, String resourceType, String payload) {
this.resourceEvent = resourceEvent;
this.resourceCrn = resourceCrn;
this.resourceType = resourceType;
this.payload = payload;
}
public String getResourceCrn() {
return resourceCrn;
}
public void setResourceCrn(String resourceCrn) {
this.resourceCrn = resourceCrn;
}
public String getResourceType() {
return resourceType;
}
public void setResourceType(String resourceType) {
this.resourceType = resourceType;
}
public String getPayload() {
return payload;
}
public void setPayload(String payload) {
this.payload = payload;
}
public ResourceEvent getResourceEvent() {
return resourceEvent;
}
public void setResourceEvent(ResourceEvent resourceEvent) {
this.resourceEvent = resourceEvent;
}
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 5,669 |
Insurance Companies Headquartered in Indianapolis
Indianapolis is a great city to live in. It has a lot of history, and it's also home to many successful companies that have been around for decades. In this blog post, we will talk about insurance companies headquartered in Indianapolis.
What insurance companies are headquartered in Indianapolis and what do they offer?
Two of the most well-known insurance companies headquartered in Indianapolis are Anthem and Simon Property Group.
Simon Property Group is a real estate investment trust that invests in properties such as malls, shopping centers, warehouses, manufacturers' outlets, and distribution facilities. It offers commercial property & casualty insurance to its customers for their buildings and contents insured at these properties.
Anthem is one of the largest insurance companies in the US, and it offers products such as health services, dental care, vision benefits, and other medical coverage to individuals who are insured by them through their employer or union group plan. They also offer life insurance protection for families that have a breadwinner with Anthem-provided life insurance protection.
How did these companies come to be headquartered in Indianapolis, and what is their history in the city?
Anthem was founded by Mr. John C. Schilling during the Great Depression in Indiana, and initially only offered life insurance protection to individuals who were insured through their employer or union group plan. It wasn't until the 1990s that Anthem became a health benefits company as well after they acquired several hospitals along with other healthcare providers.
Simon Property Group was founded by Mr. Melvin Simon in 1957 and is headquartered in Indianapolis, Indiana. The company owns or has an interest in more than 325 retail real estate properties comprising over 190 million square feet of gross leasable area in the U.S., Europe, Asia, and South America.
Both companies have a long history in Indianapolis and continue to be successful businesses in the city.
How to Start a Retail Shop in Kenya Ultimate Guide
Best Way to Finance A Car in Australia: 9 Tips
What are some of the challenges faced by insurance companies headquartered in Indianapolis, and how are they overcoming them?
The insurance industry is constantly changing and evolving, so it can be a challenge for companies to keep up with the latest trends and changes.
Anthem has been successful in overcoming this challenge by investing in new technology and data analytics that allow them to better understand their customers' needs. They have also acquired other healthcare providers over the years to help expand their healthcare offerings.
Simon Property Group has been successful in overcoming this challenge by continuing to build retail properties and expanding outside the US, which allows them to generate a lot of revenue from other parts of the world as well. The company is also planning on building more properties within the city limits over the next few years that will help expand its portfolio.
What is the future of the insurance industry in Indianapolis, and how will these companies continue to thrive there?
The insurance industry is expected to grow significantly in the next decade, and these companies will continue to thrive in Indianapolis.
Anthem has been expanding its healthcare offerings into new markets, and they are also planning on investing in more technology and data analytics to help them better understand their customers' needs.
Simon Property Group is expected to continue expanding its business globally and building more properties within the city limits to help increase its portfolio.
car insuranceinsurance
The 5 Best Hospitals in Palm Beach County
Top 10 Cheapest Places to Buy a House in Texas
How to Start a Small Clothing Business from Home
How to Get Rid of Fruit Flies With White Vinegar | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 35 |
Q: Unity: Get resource path I work in Unity3D.
I use Resources.LoadAll(path); to load all items in the folder and subfolders. After I do this I would like to get the subfolder's name of the objects, or the complete path. Is this possible?
And don't suggest AssetDatabase, because it is an editor class, and I need it in my build.
Thanks in advance.
A: What I do is write out a "Resources/manifest.txt" file in the editor that I load in the game if I need to find files/subfolders in the Resources folder at runtime. Creating this manifest.txt can be automated in the editor so that it's always up to date.
Then at run time instead of Resources.LoadAll, I load the manifest.txt, look for the folder/asset in there and Resources.Load it.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 6,213 |
Круц (Круць) Іван Семенович ( 1906, Миколаїв — 1968, с. Жовтневе, Миколаївська область) — український цирковий артист, борець, важкоатлет.
Народився І. С. Круць у 1906 р. в місті Миколаєві. Походив з бідної української родини. Родина Круців рятуючись від голоду часів громадянської війни виїхала у село Роздол Голованіського повіту Єлизаветградської губернії (тепер Кіровоградська область), де і минуло дитинство майбутнього чемпіона.
У 1920 році родина Круців повернулася до Миколаєва, Іван навчався у робітничій школі, а у 1922 році влаштувався працювати у порт. Відтоді своє життя Іван Семенович пов'язав з Богоявленськом (В 1938 році Президія Верховної Ради УРСР перейменувала Богоявленськ на Жовтневе.), де прожив майже усе своє життя.
В порту, де майбутній атлет працював вантажником, хлопця примітили і запропонували виступати в цирку.
Наприкінці двадцятих років Іван Круць був уже відомим і популярним артистом циркової арени, одним з найсильніших спортсменів класичної боротьби. За порадою «продюсерів» атлет прибрав зі свого прізвища м'який знак і став тепер Круц.
Іван Круц боровся з Іваном Піддубним, виступав на гастролях спільно з богатирем Василем Дзюбою (1899—1945рр).
Іван Семенович легко крутив пальцями із цвяхів «спіраль», в'язав із сталевих прутів «краватки». У «силовий атракціон» входили такі трюки: атлет підкидав над собою 45-кілограмову кулю і ловив її на шию. Скріплював пальці рук у «замок», а з обох боків під лікті йому начіплювали міцні посторонки, в які запрягали по коняці. Хоч як не старалися коні, проте розірвати рук богатиря не могли. До речі цей номер вимагав неабиякої фізичної сили і його могли демонструвати лише кілька атлетів у Європі, серед яких був і Іван Круц.
Старожили згадують, що Іван Семенович мав почуття гумору та самоіронії, мав добру і щиру душу.
Початок Другої світової війни застав Івана Круца на гастролях селами Одещини, червона армія стрімко відступала під ударами вермахту та Румунської армії, тож безліч людей опинилися в окупації, не маючи змоги евакуюватися. Як наслідок Круц залишився в Одесі і після вступу німецьких військ до Миколаєва Іван Семенович повернувся додому. Заробляв на життя атлетичними номерами, керував цирковою трупою. Доводилося часто виступати на манежі перед румунськими та німецькими солдатами.
Після війни Іван Семенович мешкав у власноруч збудованому у Жовтневому (селище Жовтневе — передмістя Миколаєва) ще до війни домі.
І. Круца добре знали в тогочасних атлетичних колах як України, так і Радянського Союзу. Іван Семенович налагоджував зв'язки з атлетами, котрі лишилися живі після війни і з ними далі гастролював Україною.
Згадує В. Свіренко, український гирьовик: «У 1945 році в Жовтневму районний комітет НКВД очолив якийсь майор — уралець. Він не мав житла, і хтось порекомендував йому винайняти квартиру у Івана Семеновича . Іван Семенович до війни їздив з гастролями по всьому Радянському Союзу, заробляв добру копійку і мав можливість побудуватися як належить. На той час це було добротнее і велике помешкання.
Енкеведист уподобав будинок одразу. Іван Семенович, мешкав на той час сам, погодився взяти його на квартиру, надав енкаведисту одну кімнату… За якийсь час майор просить Івана Семеновича уступити йому ще одну кімнату, бо, мовляв, хоче викликати дружину з Уралу.
Круц дав згоду- у нього залишилося ще дві кімнати. Іван Семенович і не здогадувався, що добрий з виду майор вже готував йому кайдани.
Майор сфабрикував папір про те, що ніби-то Круц під час війни співпрацював з окупантами, приклав фотографію на якій богатир в одній руці тримав румунського офіцера а у іншій важкоатлетичний снаряд…»
Круца заарештували 29 січня 1945 року і звинуватили у державній зраді. Згідно з вироком трибуналу військ НКВД у Миколаївській області від 4 червня 1945 р його засудили до 6 років таборів, обмеження у правах на три роки і конфіскацією майна.
Майор став власником будинку Івана Круца.
Ось протоколи допитів на котрих Іван Семенович свідчив: "Народився у Миколаєві у 1906 р. Батько був моряк торгового флоту. У 1920 р, почав відвідувати спортивну секцію у трудовій школі. У 1922 р. закінчив трудову школу і почав виступати на сцені і в цирку як борець, одночасно працював вантажником у порту. З 1926 року і до окупації працював артистом-борцем — їздив з виступами містами та селами. На килимі змагався з відомими атлетами — Загоруйком, Криловим, Чемяком, Несуйком та ін. У перші дні окупації був заарештований фельдкомендатурою за наклепом — начебто при відході червоних я набрав 60 мішків борошна. Та німці швидко розібралися і на другий день мене звільнили. За наказом пішов і зареєструвався у драмтеатрі як артист.
Нами керував у міській Управі Клименко (до війни — керівник заводської хорової капели). Він добився для нас дозволу їздити по селах з цирковими номерами. У вересні 1941 р. в Управі отримав дозвіл на оренду приміщення для атракціону, витратив 5 тисяч рублів на обладнання, але німці забрали його під казино.
Працювали у приватному балагані, навесні 1942 р. я відкрив свій на Військовому ринку, (Район вулиці Степової) який невдовзі закрив, бо було замало публіки. Директор драмтеатру запропонував організувати в нього групу борців і ми до нього перейшли.
З вересня 1942 р. я їздив із своєю групою. 20 серпня 1942 року на честь 20-річчя моєї спортивної і артистичної діяльності був мій виступ у саду Петровського. Я ліг і через мене проїхала вантажна машина з людьми. Опісля народ кинувся до мене, а я заскочив на машину і подякував всім глядачам. Я пообіцяв показати номер по розтягуванню рук кіньми.
Загалом в мене було 6-7 артистів, іноді ще й оркестр з 5-6 людей. Я виступав з силовими номерами: танго з дамою в зубах, забивання кулаком цвяхів у дошку і витягування їх зубами, ламання руками кінських підков, боротьба з биком та ін. На виступах були також німці і румуни. Моя дочка Тамара грала на акордеоні попурі з українських пісень.
У жовтні 1942 р. були гастролі у Братському і я познайомився з Харченком — головою сільуправи (бургомістром). Він сказав, що є таємним керівником партизанського загону і пов'язаний з одеським підпіллям. Від нього в подальшому я отримував листівки і розповсюджував скрізь по селах і в дорозі. Я прийняв до себе втікача з полону Юсуфа (з ним виступав ще до війни), Пижа Фана — з єврейської родини, Простякову Валентину прийняв у сім'ю, спас від Німеччини. У 1965 році, коли справа переглядалася, Харченко знову підтвердив свідчення про співпрацю Круца з радянським підпіллям, надав список з 37 прізвищ, адреси конспіративних квартир, керівників одеського підпілля. Але відлига скінчилася і довіри не було, тобто не було реабілітації. У Круца так і залишилася стаття 58-10, тобто антирадянська агітація.
Йому пригадували кілька випадків на виступах, коли він підкидав угору двопудову гирю, потім приймав її на груди і скидав на поміст з вигуком «Капут!», а хтось з залу питав: «Кому?» — і він відповідав: «Більшовикам!». А — треба було казати: «Слава КПРС!», коли у залі повно німців? І наче немає такого поняття, як конспірація. Між тим, при слідстві було серйозно порушена процедура — не було рішення про відкриття карної справи — а слідство проводилося і був вирок.
Згідно запису № 59654 Національного банку репресованих, Круц І. С. 1906 року народження, народжений у м. Миколаєві, українець, із службовців, до політичних партій не належав, з початковою освітою, безробітний, про склад сім'ї і подальшу долю зазначено — «нема даних». Реабілітований у 1991 р.
Після повернення із заслання родина Круців повернулася у Голованівськ. Іван Семенович одружився у таборах (його майбутня дружина Оксана теж відбувала термін), тож повернувся він з дружиною Оксаною та сином Юрієм, котрий народився у засланні. На Кіровоградщині Круц знову почав виступати на арені, був тренером і одночасно капітаном команди на обласних, республіканських та всесоюзних змаганнях сільських спортсменів.
Однак Іван Степанович завжди хотів повернутися у Жовтневе і нарешті змінив Голованівськ на Жовтневе. Повернувшись у Жовтневе І. Круц збудував новий будинок на вул. Петровського (тепер Литовченка) у якому прожив із сім'єю до самої смерті.
Варто згадати цікавий епізод, котрий стався під час побудови будинку.
Згадує Дмитро Петрович Клименко: «Я будував хату і привіз на пилораму колоди для розпилювання. Тут приїхав Круц. З'ясувалося, що він теж будується і йому потрібно було розпустити колоди на балки. Робітники сказали Круцу: "Розпиляємо без черги, якщо сам покладеш колоду на візок пилорами". Круц підійшов до краю колоди, обхопив її обома руками знизу, різко видихнув і закинув колоду на візок. Потім так само завів і другий кінець. Це було неймовірним видовищем бо для такої роботи потрібно було кілька міцних чоловіків».
Іван Семенович не припиняв гастролі Жовтневим районом та Миколаївщиною, купив старенького «ЗІМа», найняв шофера і їздив селами з показовими виступами. Авто теж брало участь у виступах. Атлет клав собі на груди дві товсті дошки, на які в'їзджала машина. Раз на рік у клубі Жовтневого району Іван Семенович влаштовував показові виступи. Варто зауважити, що українець Іван Круц першим у світі підняв над собою 200 кг.
Як згадують старожили, на сцену викликали шістьох хлопців, їх зв'язували рушниками, і атлет піднімав усіх собі на плечі і крутив «вітряка». Цвях 150 мм забивав у підлогу рукою замотаною у хусточку по саму головку, після чого витягував його зубами.
Людина безмежно закохана у спорт, Іван Семенович з великою любов'ю ставився до людей, жив для них. Разом зі своїми вихованцями сприяв популяризації важкоатлетичних занять серед молоді . Як згадують старожили борцівський клуб, де І. С. Круц займався з молоддю, містився в приміщинні Жовтневого районного будинку культури на розі вулиць Бузької (тепер вулиця Пилипа Орлика) та Млинної (тепер вулиця Клечова балка). Завідувачем будинку культури довгий час був Іван Круц . Хоч Іван Семенович у 1962 році і подавав заяву про реабілітацію, проте він її так і не дочекався.
Помер видатний український спортсмен у 1968 році у селищі Жовтневе (зараз Корабельний район м. Миколаєва) і похований на місцевому цвинтарі. На пам'ятнику атлетові напис: «Іван Семенович Круц (1906—1968) український богатир, атлет, екс — чемпіон СРСР по класичній боротьбі».
Силами Вітовського Українського Товариства та Миколаївського обласного товариства «Меморіал» в архіві СБУ віднайдена карна справа І. Круца. За фактами справи Миколаївське обласне телебачення зняло документальний фільм про долю видатного спортсмена.
Джерела
Зайцев Ю. (Голова обл. товариства «Меморіал» . Лауреат премії ім. М. Аркаса) " Сповідь з-за ґрат ", сайт «Корабелів.інфо» 7.01.2016 р.
Миколаївська обласна телерадіокомпанія «ТМ», справа № …http://v4.korabelov.info/index.php?option=com_k2&view=item&id=11651:povernennya-iz-zabuttya-legendarnikh-imen-podij-minulikh-dniv-film-pro-nashogo-vidatnogo-zemlyaka-video&Itemid=126
Мних В. « Богатир Іван Круц — незаслужено забутий атлет з Корабельного району» . Сайт «Корабелів .інфо» 7.10.2015р .
Свідчення старожилів , — Россошинського П. Г. та Кожева В. С.
Особлива подяка Ясько О.І, завідувачці філії № 18 обласної бібліотеки ім. Кропивницького, за наданий архів світлин .
Борці XX століття
Українські важкоатлети
Українські борці
Посмертно реабілітовані
Померли в Миколаєві
Поховані на Балабанівському кладовищі
Репресовані в СРСР | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 3,485 |
Q: Función AJAX segura en Django tengo la siguiente funcion AJAX que me retorna los datos del profesional con el id que le paso por GET
$.get('/get_professional_info/', {idProfessional: id}, function(data){
var professional = JSON.parse(data)[0].fields;
# Trabajo con el profesional
});
Después declaro en my urls.py:
urlpatterns = [
# ...
url(r'^get_professional_info/$', get_professional_info),
# ...
]
Y por último mi función en views.py:
def get_professional_info(request):
idPro = request.GET['idProfessional']
profesional = Profesional.objects.filter(id_profesional = int(idPro))
# Lo formateo en JSON
return HttpResponse(json.dumps(data), content_type = "application/json")
Mi duda es como hago para hacer esta función segura, ya sea ocultandola o bloqueandola solo para que sea accesible desde mi función de javascript. Ya que si voy a la dirección desde el navegador enviando la información(ejemplo: www.genarito.com/idProfessional=1) me arroja todo en pantalla, dejando el escenario propicio para SQL injections de todo tipo.
Así se ve desde la URL:
A: Integrarlo en alguna carpeta static de tu proyecto
Hacer referencia al script en tu plantilla html de la siguiente manera:
Listo! Ahora puedes hacer tus peticiones sin tener que preocuparte en el token :)
/**
* Este script de javascript permite trabajar transparentemente solicitudes que requieren
* protección del token CSRF por medio de AJAX JQUERY.
* Esto te permitirá hacer solcitudes a web Services de Django por medio de AJAX Jquery.
* Para utilizarlo basta con integrar el archivo DjangoAjax.js en tu directorio de JS y hacer referencia a él en tus templates
* que requieren del uso de AJAX por POST o algún otro que requiera el token CSRF.
* Este script está basado en la documentación oficial de Django https://docs.djangoproject.com
*/
$(function(){
//Obtenemos la información de csfrtoken que se almacena por cookies en el cliente
var csrftoken = getCookie('csrftoken');
//Agregamos en la configuración de la funcion $.ajax de Jquery lo siguiente:
$.ajaxSetup({
beforeSend: function(xhr, settings) {
if (!csrfSafeMethod(settings.type) && sameOrigin(settings.url)) {
// Send the token to same-origin, relative URLs only.
// Send the token only if the method warrants CSRF protection
// Using the CSRFToken value acquired earlier
xhr.setRequestHeader("X-CSRFToken", csrftoken);
}
}
});
function sameOrigin(url) {
// test that a given url is a same-origin URL
// url could be relative or scheme relative or absolute
var host = document.location.host; // host + port
var protocol = document.location.protocol;
var sr_origin = '//' + host;
var origin = protocol + sr_origin;
// Allow absolute or scheme relative URLs to same origin
return (url == origin || url.slice(0, origin.length + 1) == origin + '/') ||
(url == sr_origin || url.slice(0, sr_origin.length + 1) == sr_origin + '/') ||
// or any other URL that isn't scheme relative or absolute i.e relative.
!(/^(\/\/|http:|https:).*/.test(url));
}
// usando jQuery
function getCookie(name) {
var cookieValue = null;
if (document.cookie && document.cookie != '') {
var cookies = document.cookie.split(';');
for (var i = 0; i < cookies.length; i++) {
var cookie = jQuery.trim(cookies[i]);
// Does this cookie string begin with the name we want?
if (cookie.substring(0, name.length + 1) == (name + '=')) {
cookieValue = decodeURIComponent(cookie.substring(name.length + 1));
break;
}
}
}
return cookieValue;
}
function csrfSafeMethod(method) {
// estos métodos no requieren CSRF
return (/^(GET|HEAD|OPTIONS|TRACE)$/.test(method));
}
});
El ejemplo de consulta seria
$.ajax({
url: 'mi_url',
type: "POST",
data: {'variable': value},
success: function (response) {
//lo que haces si es exitoso
}
});
Y tu archivo para que sea solo por consultas ajax para una api-rest
def ajax_test(request):
if request.is_ajax():
message = "This is ajax"
else:
message = "Not ajax"
return HttpResponse(message)
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 9,142 |
<?php
namespace SP\DataModel;
defined('APP_ROOT') || die();
/**
* Class UserBasicData
*
* @package SP\DataModel
*/
class UserData extends UserPassData implements DataModelInterface
{
/**
* @var string
*/
public $user_login = '';
/**
* @var string
*/
public $user_name = '';
/**
* @var string
*/
public $user_email = '';
/**
* @var string
*/
public $user_notes = '';
/**
* @var int
*/
public $user_groupId = 0;
/**
* @var int
*/
public $user_profileId = 0;
/**
* @var bool
*/
public $user_isAdminApp = 0;
/**
* @var bool
*/
public $user_isAdminAcc = 0;
/**
* @var bool
*/
public $user_isDisabled = 0;
/**
* @var bool
*/
public $user_isChangePass = 0;
/**
* @var bool
*/
public $user_isChangedPass = 0;
/**
* @var bool
*/
public $user_isLdap = 0;
/**
* @var int
*/
public $user_count = 0;
/**
* @var string
*/
public $user_lastLogin = '';
/**
* @var string
*/
public $user_lastUpdate = '';
/**
* @var bool
*/
public $user_isMigrate = 0;
/**
* @var
*/
public $user_preferences;
/**
* @var string
*/
public $usergroup_name = '';
/**
* @return int
*/
public function getUserCount()
{
return (int)$this->user_count;
}
/**
* @param int $user_count
*/
public function setUserCount($user_count)
{
$this->user_count = (int)$user_count;
}
/**
* @return string
*/
public function getUserLastLogin()
{
return $this->user_lastLogin;
}
/**
* @param string $user_lastLogin
*/
public function setUserLastLogin($user_lastLogin)
{
$this->user_lastLogin = $user_lastLogin;
}
/**
* @return string
*/
public function getUserLastUpdate()
{
return $this->user_lastUpdate;
}
/**
* @param string $user_lastUpdate
*/
public function setUserLastUpdate($user_lastUpdate)
{
$this->user_lastUpdate = $user_lastUpdate;
}
/**
* @return boolean
*/
public function isUserIsMigrate()
{
return (int)$this->user_isMigrate;
}
/**
* @param boolean $user_isMigrate
*/
public function setUserIsMigrate($user_isMigrate)
{
$this->user_isMigrate = (int)$user_isMigrate;
}
/**
* @return mixed
*/
public function getUserPreferences()
{
return $this->user_preferences;
}
/**
* @param mixed $user_preferences
*/
public function setUserPreferences($user_preferences)
{
$this->user_preferences = $user_preferences;
}
/**
* @return string
*/
public function getUserEmail()
{
return $this->user_email;
}
/**
* @param string $user_email
*/
public function setUserEmail($user_email)
{
$this->user_email = $user_email;
}
/**
* @return string
*/
public function getUserNotes()
{
return $this->user_notes;
}
/**
* @param string $user_notes
*/
public function setUserNotes($user_notes)
{
$this->user_notes = $user_notes;
}
/**
* @return int
*/
public function getUserGroupId()
{
return (int)$this->user_groupId;
}
/**
* @param int $user_groupId
*/
public function setUserGroupId($user_groupId)
{
$this->user_groupId = (int)$user_groupId;
}
/**
* @return int
*/
public function getUserProfileId()
{
return (int)$this->user_profileId;
}
/**
* @param int $user_profileId
*/
public function setUserProfileId($user_profileId)
{
$this->user_profileId = (int)$user_profileId;
}
/**
* @return boolean
*/
public function isUserIsAdminApp()
{
return (int)$this->user_isAdminApp;
}
/**
* @param boolean $user_isAdminApp
*/
public function setUserIsAdminApp($user_isAdminApp)
{
$this->user_isAdminApp = (int)$user_isAdminApp;
}
/**
* @return boolean
*/
public function isUserIsAdminAcc()
{
return (int)$this->user_isAdminAcc;
}
/**
* @param boolean $user_isAdminAcc
*/
public function setUserIsAdminAcc($user_isAdminAcc)
{
$this->user_isAdminAcc = (int)$user_isAdminAcc;
}
/**
* @return boolean
*/
public function isUserIsDisabled()
{
return (int)$this->user_isDisabled;
}
/**
* @param boolean $user_isDisabled
*/
public function setUserIsDisabled($user_isDisabled)
{
$this->user_isDisabled = (int)$user_isDisabled;
}
/**
* @return boolean
*/
public function isUserIsChangePass()
{
return (int)$this->user_isChangePass;
}
/**
* @param boolean $user_isChangePass
*/
public function setUserIsChangePass($user_isChangePass)
{
$this->user_isChangePass = (int)$user_isChangePass;
}
/**
* @return boolean
*/
public function isUserIsLdap()
{
return (int)$this->user_isLdap;
}
/**
* @param boolean $user_isLdap
*/
public function setUserIsLdap($user_isLdap)
{
$this->user_isLdap = (int)$user_isLdap;
}
/**
* @return string
*/
public function getUserLogin()
{
return $this->user_login;
}
/**
* @param string $user_login
*/
public function setUserLogin($user_login)
{
$this->user_login = $user_login;
}
/**
* @return string
*/
public function getUserName()
{
return $this->user_name;
}
/**
* @param string $user_name
*/
public function setUserName($user_name)
{
$this->user_name = $user_name;
}
/**
* @return string
*/
public function getUsergroupName()
{
return $this->usergroup_name;
}
/**
* @param string $usergroup_name
*/
public function setUsergroupName($usergroup_name)
{
$this->usergroup_name = $usergroup_name;
}
/**
* @return int
*/
public function getId()
{
return (int)$this->user_id;
}
/**
* @return string
*/
public function getName()
{
return $this->user_name;
}
/**
* @return bool
*/
public function isUserIsChangedPass()
{
return (int)$this->user_isChangedPass;
}
/**
* @param bool $user_isChangedPass
*/
public function setUserIsChangedPass($user_isChangedPass)
{
$this->user_isChangedPass = (int)$user_isChangedPass;
}
} | {
"redpajama_set_name": "RedPajamaGithub"
} | 3,912 |
{"url":"https:\/\/meta.stackexchange.com\/questions\/289107\/how-do-i-write-a-good-edit-summary\/289108","text":"# How do I write a good edit summary?\n\nAs a user, I don't often ask questions (in fact, I believe I have only asked two questions in total across the entire network, excluding meta). However I do a lot of answering, flagging, commenting and of course, editing.\n\nBecause I like things to be consistent, intuitive, efficient and definite, I've developed a system of structure and keywords for my edit summaries.\n\nOther discussions on Big Meta and per-site metas make it clear that the quality of edit summaries can be the difference between acceptance and rejection, or lead to grief when somebody hasn't explained their intent or actions quite so effectively.\n\nFinally, if one writes a summary, they can confirm to themselves what they have done, and whether it's (all of) what they were aiming for.\n\nHow do you write the summary when you make an edit?\n\nI will write an answer that describes the system and keywords I use. Other answers that extend this system, or better yet describe alternatives, will help people to understand what others are trying to say when they write an edit summary, as well as how they might write a better summary themselves.\n\nYour edit summary is aimed at three slightly different audiences:\n\n\u2022 reviewers, if you can only suggest edits. You want to convince them to approve your edit. If you are adding material that was in the comments, say so - normally just adding material doesn't get approved. If your change is minor but critical, this is where you explain that criticality so the suggestion isn't rejected as too minor.\n\u2022 the OP, either in their role as reviewer of your suggested edit, or as a notification-clicker who wants to know what happened to their post. A comment that teaches the site norms or explains a seemingly arbitrary change isn't strictly required, but if you're taking the time to fix someone's post, you can probably also take the time to explain to them why this is a fix, not vandalism\n\u2022 posterity, when people look over the revision history. By far the smallest case even though it's a long tail. It probably doesn't matter what you say here, but I would encourage focusing on why over what - we can all see what is changed, the comment is a chance to add something more. I typically only care about posterity when I'm editing my own posts, and then I explain my thinking.\n\nUse caution when your comment involves enforcing a norm newcomers are unaware of. Say you remove \"Hey everyone, this is my first post, I hope it's ok\" at the start of a post and \"Thanks in advance, hope you can help, this is really urgent for me\" at the end. Your edit comments could be\n\nThe first is likely to spark an argument or at least hurt feelings from the OP. The second doesn't help since people who include such content in their posts don't generally recognize the phrase \"meta content\". The third is backed up by site policy, though that may not reduce hurt feelings and instead encourage \"site policy is stupid and rude\" first meta postings. The fourth tells the OP what's in it for them and why the edit made their post better. (It's also easier to do than the third since you don't need to go find a link to prove you're right.)\n\nI put the most care into edit summaries on sites where my edits are reviewed. I try to explain a why and a benefit to the OP on my other edits, but I don't always do so. And if I have nothing better to say than \"spelling and formatting\" on a site where the edit summary is optional, I leave it out. Such summaries add no value, so I spare myself the trouble of typing them.\n\nFrom my personal experience, be precise. There is no definitive guide for putting the comments, you just need to communicate to the reviewer (including OP) in case, the edit is not so easy to understand.\n\nTry to put the things you edited in two or three words and use a comma separated list, like\n\nFixed broken links, corrected grammar & spelling, re-tagged.\n\n\nAvoid writing overly-long edit messages. If the edit is good, it generally speaks for itself.\n\nI think writing a good edit summary that is concise and self-explanatory is important.\n\nPersonally, I would usually describe what I changed in short sentences, for example:\n\n\u2022 Formatted code\n\n\u2022 Formatted error messages\n\n\u2022 Corrected spelling\/ grammar\n\n\u2022 Removed noise\n\n\u2022 Improved general formatting\n\nI do combine them sometimes:\n\n\u2022 Formatted code & corrected spelling\n\nBasically, the summary should summarise what you have edited, so that the OP would understand why the post was edited and such keep in mind the mistakes they made.\n\nI don't think edit summaries should be very long or very precise and definitely not in complete sentences, for example:\n\n\u2022 I have edited paragraph 2 to display image inline and also edited paragraph 3 to remove blank lines.\n\nBy just editing without leaving a edit summary, it would sometimes create a misunderstanding on why the post was edited.\n\nSo, I've described how I usually craft my edit summaries and what I think is useful to the general community.\n\nKeywords and keyphrases are listed in the order they are most commonly used. This is also the typical order in which they are included in an actual summary. The less common shorthand summaries may also be extended for precision.\n\n\u2022 Sp\/Gr or Sp\/Gr\/Synt - spelling and grammar; spelling, grammar and syntax. Used when making corrections to the language used in a question or answer. This includes misspellings, punctuation and general minor typography. Almost all my edits include this.\n\n\u2022 Readability - for when the structure of the question or answer creates difficulty for the reader in following its progression of thought, or when superfluous commentary is included which is not necessary to the question and makes identifying the actual question difficult. Often used to counter the possibility of flags and close-votes for \"unclear what you're asking\".\n\n\u2022 Tags - for when inappropriate tags are removed or reasonable tags are added.\n\n\u2022 Title - for when the title has been adjusted, either to make it more precise or to make it consistent with the question body, or when correcting issues that would be considered Sp\/Gr\/Synt or Readability changes.\n\n\u2022 Clarity or Clarification - for when a question is initially unclear but comments or discussion have found out what is meant to be asked about, and this explanation is being incorporated into the question.\n\n\u2022 Formatting - for when the markdown, including links and images, in a question or answer are being removed or added or fixed.\n\n\u2022 MathJax - for when MathJax or mhchem formatting is being altered, in order to differentiate from general markdown changes.\n\n\u2022 Terminology - for when the correct phrases or words are missing or the wrong phrases or words have been used. This occurs especially in a technical context or where the commonly understood meanings lead to a question or answer making no sense.\n\n\u2022 Removed [brief name of content] - for when text or other content is removed due to policy or best practise.\n\n\u2022 [specific comment in context] - for when a specific change has been made, which nay not fit well under a previous description, that would still be a significant change to the content of the question or answer.\n\n\u2022 Explicitly saying \"spelling\/grammar\" is much better than using abbreviations reviewers are unlikely to be familiar with \u2013\u00a0Cai Jan 1 '17 at 13:47","date":"2020-04-01 04:41:51","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.5009317398071289, \"perplexity\": 1411.5274074347342}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 20, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2020-16\/segments\/1585370505366.8\/warc\/CC-MAIN-20200401034127-20200401064127-00215.warc.gz\"}"} | null | null |
Q: is there any way to make the platform independent files just like Dlls? i am looking for some files like DLLs that can be used on windows as well as Mac OS is there any way of doing this will c++ or any other language that must be higher level than c++
need help regards.
A: Java seems like an obvious choice. Java's variant of DLL files are JAR files, which can be used on any system that has a Java Runtime Environment. This includes Windows, Mac OS, Linux and various other operating systems.
And, Java is percieved as being "higher level" than C++, although language wars might arise once in a while about topics such as this one.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 9,574 |
Ignez Francisco da Silva (Araras, — São Paulo, ), mais conhecida pelo nome artístico Dona Inah, foi uma cantora brasileira.
Biografia
Nascida em Araras, interior de São Paulo. Seu pai, Ireno, era trompetista, bem como tinha outros tios músicos. Estudou bandolim por três anos e cantou ao lado de seu pai, de quem aprendeu grande parte de seu conhecimento musical, e chegou a integrar a Orquestra de Araras quando jovem. Durante a infância, foi boia-fria, trabalhando em plantações de café, algodão e laranja, e conciliava o trabalho com os estudos.
Mudou-se com a família para a cidade de Santo André, na região metropolitana de São Paulo, em 1954, mesmo ano no qual participou de um programa de calouros comandado por Hélio Araújo na Rádio Cultura. Nessa época, venceu o concurso "Peneira Rodini", da mesma emissora, o que lhe abriu as portas para se apresentar com algumas orquestras – como as dos maestros Cyro Pereira e André Beer –, bem como para participar do programa "Só para Mulheres" da Rádio Record.
Também passou a cantar em bailes de gafieira de diversos clubes e salões de São Paulo e até gravou um disco de 78 rotações em 1958, pelo selo Trovador, mas todo material foi perdido. Sem o incentivo da família, Inah abandonou o sonho de viver de música.
Com filhos – Em vida, ela teve cinco – e dificuldades financeiras, passou um longo período afastada da vida artística e trabalhou como faxineira, lavadeira, cozinheira, babá, entre outros, e morou em Araras, Santos, Taboão da Serra e São Paulo. Naquela fase, trabalhava de segunda a sexta como empregada doméstica e aos finais de semana no guarda-volumes de um clube em Taboão da Serra. Nas noites de quinta a domingo, cantava em bares e casas noturnas na capital paulista. Embora tenha sido cantora da casa de show Rádio Clube, em Santo André entre 1972 a 1980, foram raras as oportunidades nos palcos. Em 1982, voltou para Santo André, onde cantou em um bar chamado Pedacinho do Céu e começou a se aproximar do pessoal do choro e do samba. Inah conciliou por três anos a música com o segundo emprego como faxineira da prefeitura de São Caetano do Sul, tendo se aposentado aos 65 anos de idade.
Cantando em pequenos espaços, como o Bar do Cidão, em Pinheiros, pensou em desistir, mas em 2002 foi convidada para participar do musical "Rainha Quelé", em homenagem a Clementina de Jesus, ao lado de Marília Medalha e Fabiana Cozza. A partir de então, sua carreira artística ganhou um novo impulso com a gravação de seu primeiro álbum de estúdio – "Divino Samba Meu", produzido por Hermínio Bello de Carvalho e lançado em 2004 por um selo independente – e a realizar apresentações regulares no circuito alternativo e em pequenas casas de espetáculo, como Ó do Borogodó, também em Pinheiros, onde se tornou uma atração semanal que frequentemente lotava o bar. Com a boa repercussão do disco de estreia, Inah chegou até mesmo a realizar apresentações na Espanha, na França e no Marrocos, país onde onde cantou para mais de 14 mil pessoas no "Festival Mawazine", em Rabat. Também ganhou o Prêmio da Música Brasileira de 2005 na categoria "Revelação".
Gravou ainda outros dois álbuns de estúdio. Em 2008, foi lançado "Olha quem chega", trabalho onde interpretou composições exclusivamente de Eduardo Gudin e contou com as participações especiais de músicos dos grupos paulistas Samba Novo e Quinteto em Branco e Preto, além de intérpretes como Oswaldinho da Cuíca e Juliana Amaral. Como este trabalho, Inah recebeu uma indicação no Prêmio da Música Brasileira de 2009 categoria "Cantora de Samba". Quatro anos depois, foi lançado "Fonte de Emoção", seu derradeiro disco, que contou com arranjos e direção musical de Zé Barbeiro e as participações de Monarco e Delcio Carvalho.
Diagnosticada com a doença de Alzheimer em meados da década de 2010, Inah afastou-se progressivamente dos palcos até se despedir em definitivo em setembro de 2016 com uma apresentação no Sesc Pompeia. Viveu seus últimos anos em um asilo para idosos na cidade de São Paulo e, em 8 de agosto de 2022, morreu de causas naturais aos 87 anos. Seu corpo foi enterrado no Cemitério Sagrado Coração de Jesus, em Santo André.
Discografia
2004 – Divino Samba Meu
2008 – Olha quem chega
2013 – Fonte de Emoção
Ligações externas
Sítio oficial
Entrevista com Dona Inah ao Almanaque Brasil, acesso 25 de janeiro de 2014
I
I
Cantores de samba
Pessoas com doença de Alzheimer | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 7,818 |
Q: Create 16 byte key for AES Encryption from Salt and Password in Node JS Crypto I'm trying to match an AES 256 CBC encryption implemented in C# by using node JS crypto module.
This is my C# code
using System;
using System.Security.Cryptography;
using System.Text;
public class Program
{
public static void Main()
{
Console.WriteLine(EncryptExt("Hello World"));
Console.WriteLine(DecryptExt(EncryptExt("Hello World")));
}
public static string EncryptExt(string raw)
{
using (var csp = new AesCryptoServiceProvider())
{
ICryptoTransform e = GetCryptoTransformExt(csp, true);
byte[] inputBuffer = Encoding.UTF8.GetBytes(raw);
byte[] output = e.TransformFinalBlock(inputBuffer, 0, inputBuffer.Length);
string encrypted = Convert.ToBase64String(output);
return encrypted;
}
}
public static string DecryptExt(string encrypted)
{
using (var csp = new AesCryptoServiceProvider())
{
var d = GetCryptoTransformExt(csp, false);
byte[] output = Convert.FromBase64String(encrypted);
byte[] decryptedOutput = d.TransformFinalBlock(output, 0, output.Length);
string decypted = Encoding.UTF8.GetString(decryptedOutput);
return decypted;
}
}
private static ICryptoTransform GetCryptoTransformExt(AesCryptoServiceProvider csp, bool encrypting)
{
csp.Mode = CipherMode.CBC;
csp.Padding = PaddingMode.PKCS7;
var passWord = Convert.ToString("AvbSkj3BVbf4o6mdlAofDp0/SD0susEWo0pKdmqas");
var salt = Convert.ToString("ABj4PQgf3j5gblQ0iDp0/Gb07ukQWo0a");
String iv = Convert.ToString("aAB1jhPQ89o=f619");
var spec = new Rfc2898DeriveBytes(Encoding.UTF8.GetBytes(passWord), Encoding.UTF8.GetBytes(salt), 65536);
byte[] key = spec.GetBytes(16);
csp.IV = Encoding.UTF8.GetBytes(iv);
csp.Key = key;
if (encrypting)
{
return csp.CreateEncryptor();
}
return csp.CreateDecryptor();
}
}
And this is my Node JS implementation
const crypto = require('crypto'),
algorithm = 'aes-128-cbc',
password = 'AvbSkj3BVbf4o6mdlAofDp0/SD0susEWo0pKdmqas',
salt = 'ABj4PQgf3j5gblQ0iDp0/Gb07ukQWo0a',
iv = 'aAB1jhPQ89o=f619',
inputEncoding = 'utf8',
outputEncoding = 'base64';
function encrypt(text) {
let cipher = crypto.createCipheriv(algorithm,createHashPassword(), iv);
let encrypted = cipher.update(text, inputEncoding, outputEncoding)
encrypted += cipher.final(outputEncoding);
return encrypted;
}
function createHashPassword(){
let nodeCrypto = crypto.pbkdf2Sync(Buffer.from(password), Buffer.from(salt), 65536, 16, 'sha1');
return nodeCrypto || nodeCrypto.toString('hex');
};
function decrypt(encrypted) {
let decipher = crypto.createDecipheriv(algorithm, Buffer.from(createHashPassword(),"hex"), iv)
let dec = decipher.update(encrypted, outputEncoding, inputEncoding)
dec += decipher.final(inputEncoding);
return dec;
}
console.log(encrypt('Hello World'));
console.log(decrypt(encrypt('Hello World')));
The encrypted data from both this options are coming different hence, not able to work this out.
So far what I have seen is,
*
*node crypto createCipheriv method takes only 32 byte buffer and if I
pass it a 16 byte buffer it says, invalid length.
*If I convert the 16 byte key to hex encoded string, the encrypted value changes and does not match with the C# implementation.
*I can't change the C# implementation as its already in production and been used by multiple applications.
*So there seems to be an issue with generating key from salt and password in node js, matching what is done in C# and I'm not able to figure that out.
Code can be tested in the below link:
C# Implementation: https://dotnetfiddle.net/bClrpW
Node JS Implementation: https://runkit.com/a-vi-nash/5c062544509d8200156f6111
A: It seems that you are creating a AES-128 instance in your C# code, because you are using 16 bytes keylen.
AES-256 keylen is 32 bytes, not 16 bytes.
Bugs in code:
*
*Since you set 16 bytes for key in C#, it uses AES-128, not AES-256. So you need to change node.js to AES-128 or change generated key to 32 bytes in both sides.
*Since you are using text string salt and password(not base64 encoded), your node.js side uses incorrect pbkdf2Sync parameters.
*IV len for AES algorithm in 16 bytes and you cannot use shorter ones.
Since you wanted AES-256 here are your changed to both sides:
C# side:
String iv = Convert.ToString("SOME_IV_SOME_IV_"); // 16 bytes IV
....
byte[] key = spec.GetBytes(32); // 32 bytes key
node.js side:
iv = 'SOME_IV_SOME_IV_' // 16 bytes IV similar to C#
...
// Bugs in this function
function createHashPassword(){
// Change parameters to `base64` only if salt and password are base64. it may be true for salt, but it is can rarely be correct for password.
let nodeCrypto = crypto.pbkdf2Sync(Buffer.from(password), Buffer.from(salt), 65536, 32, 'sha1');
return nodeCrypto;
};
IMPORTANT NOTES:
*
*Remember that IV must be selected as a random buffer(neither fixed not text) and since it seems that you are sending it over network, you need to send IV with it too.
*SALT must be a random buffer(not text) and fixed on both sides.
*i suggest to use over 100000 iteration for PBKDF2 at least.
A: If you operate with passwords of 41 characters length, why don't you instead use an actual key? A base64 encoded 256 bit key would be 44 characters long.
The purpose of the salt and the iteration for the deviation is to counter the common problem of too short passwords. But why go through all the bother of implementering this in both ends with no added benefits, but more than one disadvantages - like more code & slower solution.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 4,407 |
I'm a big believer in staying unbranded; I'm about as likely to to wear a logo as a disco suit. Even my smartphone and laptop have logo-covering cases.
Well, according to new research, I could be missing out. A recent study conducted by researchers at Notre Dame, the University of Kentucky, and Penn State found that using brand- name gear can provide a noticeable placebo effect that could boost performance. In other words: If you've ever felt like you give better presentation when wearing an expensive designer suit, it might not be your imagination.
In the study, researchers found that simply being told you were using a Nike golf putter over a no-name club improved participants' performance by about 20 percent. And this effect isn't limited to the physical: The researchers also found that subjects who wore earplugs while taking a math quiz did better when they were told they were using high-performance 3M earplugs. The performance bump for this part of the study was also about 20 percent.
This isn't the first study to suggest workers and their bosses can lean on marketing buzz to boost their performance. In a 2008 study published in the Journal of Marketing Research, researchers at Duke University and the University of Waterloo found that exposure to logos can actually cause people to take on traits they might associate with a brand's corporate identity. For example, test subjects who were asked to complete a task did so with more creativity when exposed to Apple branding—what with its years of "Think Different" marketing—over an IBM logo. Likewise, the study found that exposure to a Disney Channel logo caused participants to behave with more honesty than those who saw one from E! Entertainment.
Similar results were found in a 2011 Boston College study published in the Journal of Consumer Psychology. The study found that exposure to a Red Bull logo primed test subjects to act more, well, extreme. Participants took control of a driving simulator that featured a logo-splashed vehicle. Those who drove a virtual Red Bull vehicle did so with more speed, power, and risk-taking behaviors than those piloting a Coca-Cola, Tropicana, or Guinness car.
Bottom line: The placebo effect is real. And just as a sugar pill may magically make your headache go away, surrounding yourself with brands that have positive connotations may actually boost your performance and creativity. | {
"redpajama_set_name": "RedPajamaC4"
} | 637 |
\section{Introduction}
In past many infectious diseases hampered humankind, like the Whooping cough outbreak in U.S. during the summer of 2012, Periodic outbreaks of H7N9 in China during 2013-2014 and emerging MERS outbreak in Saudi Arabia, Severe Acute Respiratory Syndrome (SARS), West Nile Virus and drug resistant tuberculosis \cite{zhao2012national}. Diseases are spread exponentially within a population. Peoples and products travel across the border in high volume so that an infection can rapidly attain a global scale.\\
A pandemic HiNi ("Swine Flu") appeared in 2009 \cite{szomszor2011twitter}. To increase this outbreak, a globally open online web-based framework gave warning to everyone concerned. The disease was catered for in response to this warning, and public understanding of emerging infection has increased. Online web-based early warning system's traffic rate was increased about 100 \% in 2-3 days. It gives birth to online early warning globally infection awareness. Hence, globally feels that there must be a system that alerts the early warning infectious disease timely vaccinate and cure the victims. Presently, nearly all major outbreaks, investigated by the World Health Organization (WHO) are first identified through these informal online sources.
The emergence of ubiquitous internet service and build up different internet-based resources like blogs, online newspapers (Google News), social media (Twitter), and discussion forums shape the world dynamic fast and very efficiently. Such services are highly useful in helping to diagnose, assess, incorporate and forecast outbreaks of infectious diseases or diseases spreading to new regions. Intelligent modeling of such sources using text-mining methods such as topic models, deep learning and dependency parsing can lead to automated generation of the mentioned surveillance tools..\\
The above resources produced big data at very high speed, and the quantity of that data is increasing day by day. Despite big data problem, these resources have the unstructured type of data, which is difficult to process, and extract useful information. Reading and assimilating a large number of reports daily basis is a hectic and burdensome task. .\\
. With BIO PAK FLASHER, a web-based solution focusing primarily on URDU \cite{jabbar2016analysis}, Pakistani Urdu news outlets, and covering Pakistani rulers and urban areas, we plan to solve these challenges with the automated framework for extraction, analysis, query, screening, incorporation, and visualization of unstructured and bulky disease outbreaks. Our system collects outbreak data from many sources such as news through the Google News aggregator, Pakistani local news channels published in Urdu, expert-curetted accounts such as ProMED \cite{morse1996promed} Mail, and validated official alerts such as World Health Organization announcements. By using the text processing algorithms, the system classifies outbreak in location and disease, after that overlay such data on an interactive geographic map. We measure the accuracy of the classification algorithms based on the level of human curation necessary to correct misclassification and examine geographic coverage. With state-of-the-art NLP technology, there is a need to compare all Urdu terms with a disease database called ontology to figure out the disease when the sentence is tokenized and all POS deleted. There is no disease ontology \cite{schriml2012disease} exist before to do such work. Here we also build Urdu-based Ontology with different patterns for further reference. .\\
Google Maps API consists of a set of classes within a JavaScript container that opens within an XHTML page. Each time the Google Maps web page opened, these classes are loaded from Google, and it has embedded JavaScript object on which all the functionality of the Google Maps is based. To visualize outbreaks, we use Google Maps.BioPak Flasher has a user- friendly interface and providing important disease outbreak information with text processing algorithms and also referred user to view full article with web link integration.
The rest of the paper is composed of Section 2 Literature review presents the background of infectious disease outbreaks and basic terminologies used in this research. Section 3 is about the Methodology used in this research, Section 4 Describes the implementation and Analysis of the proposed methodology. Section 5 presents the conclusion and future work for this research.
\section{Literature Review}
There are few traditional disease surveillance systems like Centers for Disease Control (CDC) \cite{berrios2017centers}, ProMED \cite{morse1996promed}, which released data of infectious disease outbreaks with delay due to their implemented setup (take news mail, get expert opinion then release outbreak announcement). In the modern period, the epidemic must be published in real-time. To cater to evolving diseases, take control steps quickly, and avoid them from spreading further, an early warning/detection of the outbreak system is important. These surveillance systems have some limitations. These programs do not protect all languages, so the epidemic news reported in state and local news would be missed. The system must be capable to process these languages so that any news published in theses languages will also be covered to give authenticated real-time system. Urdu is the national language of Pakistan and speaks in many other countries as well in non-official manners. Urdu is also not covered by these systems. Many local publishers like VOA-Urdu, Geo, Express, Jang, etc. published their news in Urdu. Moreover, these systems have some geographical limitations. These systems are not covering the ruler/ urban areas of some countries, especially Pakistan.\\
Infectious disease is increasing day by day due to unhygienic environment, people are in close contact with livestock or wildlife and chemicals, etc. Infectious diseases lead to mortality so awareness is a very essential part of the current era. Delayed detection of the outbreak may cause some mortality and may be created hinders in controlling and eradication process. Such mortality rate can be minimized if timely recognized. If we know in real- time, then the control and eradication measures also taken time to reduce mortality rate. \\
Moreover, when we want to visit somewhere or required to do trade with some area, we must consider emerging pathogens that causes the epidemic. Shifting of emerging product or human can cause the spreading of that epidemic in the hosting area the traditional outbreak system has a manual alarm that often requires a delay, but our, alert system, swift, and covering Pakistani news is our disease outbreak system written in Urdu. This system also helps in general awareness of infectious outbreaks and its remedies measures timely.
\subsection{Epidemic Breakdowns In Pakistan}
A government of Pakistan Official Body releases the statistics in 44th dated Mar 2019 to June 2019. According to these, Figure 1 alerts has been generated in the country, Figure 2 shows the statistical results of diseases outbreaks in Pakistan.
\subsubsection{Epidemic (A Communicate-able Disease) }
An epidemic \cite{newman2002spread} occurs when an infectious disease spreads rapidly in contact with too many people. An epidemic is an outbreak of a disease among members of a specific population that exceeds the extent of occurrence of the disease normally found in that population. Epidemics affect those members of the population who do not have an acquired or inherent immunity to the disease. Although infectious organisms cause most epidemics, the term can be applied to an outbreak of any chronic disease, such as lung cancer or heart disease. For example, in 2003, the severe acute respiratory syndrome (SARS) epidemic took the lives of nearly 800 people worldwide. Statistics and big outbreaks in history with casualty.
\subsubsection{RSS Feed: }
RSS stands for "Really Simple Syndication". It is a way to easily distribute a list of headlines, update notices, and sometimes content to a wide number of people \cite{johnson2009developing}. Computer programs that organize those headlines and notices for easy reading use it. Most people are interested in many websites whose content changes on an unpredictable schedule. News sites, community, and religious group information pages, product information pages, medical websites, and weblogs are examples of such websites. Repeatedly checking each website to see if there is any new content can be very tedious.
\begin{figure}
\centering
\includegraphics[width=0.75\textwidth]{Fig1.png}
\caption{Recent tutbreaks in Pakistan.}
\label{fig:fig1}
\end{figure}
\begin{figure}
\centering
\includegraphics[width=0.75\textwidth]{fig2.png}
\caption{Statistics of different disease outbreak in Pakistan.}
\label{fig:fig2}
\end{figure}
\subsubsection{How It Works:}
RSS works by having the website author maintain a list of notifications on their web-site in a standard way. This list of notifications is called an "RSS Feed". People who are interested in finding out the latest headlines or changes can check this list. Special computer programs called "RSS aggregators" have been developed that automatical-ly access the RSS feeds of websites you care about on your behalf and organize the results for you. (RSS feeds and aggregators are sometimes called "RSS Channels" and "RSS Readers".)
\subsubsection{RSS Feed Structure: }
An XML File provides very basic information to do its notification. It is made up of a list of items presented in order from newest to oldest. Each item usually consists of a simple title describing the item along with a more complete description and a link to a web page with the actual information being described. Sometimes this description is the full information you want to read (such as the content of a weblog post) and sometimes it is just a summary.
\section{Methodology}
The different aspects of this research including data collection were the compulsory requirement for further utilization of the data. The essentials tasks are explained figure 3.\\
In this figure, we presented the layer view of our model. We preferred to work in layer because it is easy to wok easily if we have to do any changing. We divide our model into four layers; each layer is dependent on its precedence layer. Each layer work is defined clearly. The first layer is the Data acquisition layer, which retrieves data from the RSS, feed and puts it into the database for further processing. The second layer is the Data filtering and preprocessing layer in which we eliminate and filter out unnecessary diacritics and punctuations then performing some NLP (Natu-ral Language Processing) techniques like tokenization, stemming, and stop words removal of Urdu language. Then our data has epidemic, city, date, and some garbage or unnecessary data. The third layer of Event Recognition is designed to find out the epidemic disease and concerned area with latitude and longitude. Our last layer is the Visualization layer, which deals with visualization of epidemic outbreak in the con-cerned area will show on Google map with marker view and store events in database for displaying it next 90 days. User can also view source evidence by simply clicking on a pinpoint map as well as associated headlines for the event.
\begin{figure}
\centering
\includegraphics[width=0.75\textwidth]{fig3.png}
\caption{Abstract overview of the BioPak flasher.}
\label{fig:fig3}
\end{figure}
\subsection{Data Acquisition and Storage Layer: }
In this layer. We acquire the data from Pakistani news channels and newspapers that are offering online resources (RSS feed) in Urdu and store these data into a database for further processing. Many online actors can provide news like RSS, tweeter, LinkedIn, Facebook, etc. However, our choice is RSS feed. RSS stand for Really Simple Syndication or sometimes refer as Rich Site Summary. Actually, this technol-ogy was introduced to avoid bulky and uninterested news, which is being pushed by a provider. This is also used to keep updated user with upcoming news, blogs, financial information, health, sports, classified sites, and government alerts. RSS feed is written in XML code which is easy to process by computer. It is a structured data type.
Figure 4 shows a sample of newsreader who acquired the news from Voice of America in Urdu. We can select any topic in which we are interested, in our case health topic is most prominent and maintained by almost every RSS feed provider. So we subscribe to the health RSS feed.
When we click on any news feed, it will refer us to its description web page shows in Figure 5, where we can see all detail of the news. In this figure, we refer to the voice of America page where we can find the detail of said news.
\begin{figure}
\centering
\includegraphics[width=0.75\textwidth]{fig4.png}
\caption{Sample overview of RSS feeds.}
\label{fig:fig5}
\end{figure}
\begin{figure}
\centering
\includegraphics[width=0.75\textwidth]{fig5.png}
\caption{Sample screenshot of the News.}
\label{fig:fig5}
\end{figure}
Our focus of research covered Pakistan and its national language Urdu Pakistan has several outlets that include printed media, news channels, several online podcasts, and internet-based TV channels used for consumer information. These all channels are published in different languages according to a location like English, Urdu, Sindhi, etc. Therefore, Urdu RSS feed newspaper or channels are our main actor for getting information. However, very few channels are maintained their RSS feed in Urdu. Because our research is focused on the RSS mechanisms so that we only focus that news, channels, which have their RSS, feed in Urdu and store it.
\subsubsection{Data filtering and Pre-Processing: }
There are many types of data which is published in the RSS feed. Input news items are not considered to be cleaned so making them ready for further processing is im-portant. Raw data, if obtained from the web normally includes HTML and or XML tags so they should be removed for further processing. In addition, text format (utf-8) for special characters is important so that no error occurs while processing.
\subsubsection{Removal of Diacritics: }
The Urdu language is using diacritics to clarify the pronunciation of Urdu words. Commonly used diacritics are Zabar, Zer, Pesh, Jazam, and shad, etc. Our raw data received from the database has many unnecessary elements that are required to be filtered shows in figure 6. In this stage of filtering, it is necessary that data must be filtered from diacritics. If it is not done then in processing, different result will be seen and our final output is very different from the actual. \\
\centering
\includegraphics[width=0.55\textwidth]{a.png}
\subsubsection{Removal of Punctuation}
Urdu language sentences have different punctuations, which used in the construction of sentences. These punctuations are undesired in our processing so that we removed these. Some punctuations which mostly used in the Urdu language are " " " '' – [](),.;: "
\subsubsection{Data Preparation (Tokenization with NLTK Library)}
Urdu tokenization is not easy as English due to its structure. However, Unicode recommends using Zero Width Non-Joiner character in this context \cite[4]. There would be an alignment of tags with the terms present at the simple stage, so tokenization is also an essential aspect of sorting. The text will be tokenized into sentences and phrases into words in the process.
Tokenization is a very common feature of NLP. Tokenization is a process to break a text string into single entities that may include words, symbols, phrases, etc. In our case, we apply this technique to divide the string into words, e.g.
\begin{figure}
\centering
\includegraphics[width=0.55\textwidth]{fig7.png}
\label{fig:fig7}
\end{figure}
Further processing is taken place on the resulted array after tokenization.
\subsubsection{Building up of Urdu Stop Words (USW) Dictionary }
After tokenization, now many unwanted words have no meaning when we tokenize but these words have more worth when used in combination and clear the concept of sentences. Unfortunately, there is not much work done in Urdu, so we used stop word dictionary Stop Words Removal (Based on USW Dictionary)
These dictionaries are used as a reference for the removal of stop words. This is another technique of NLP, which are tags, or separates various parts of speech in the text. For filtering, stop words are also attached with this technique to filter indifferent data. Epidemic disease, city, and date of occurrences are needed to be extracted from the text string. Everything else in the string is irrelevant and should not be extracted or in other words, should be deleted from the string.
Tokenization has been done in the previous step. Keeping in view the resulted string, now the irrelevant data should be filtered out from the string. At this stage, we have the tokenized string as:\\
\centering
\includegraphics[width=0.55\textwidth]{a1.png}\\
After deletion of Stop word from string, we will be left with the following array:\\
\centering
\includegraphics[width=0.55\textwidth]{a2.png}\\
Now the leftover data included only disease, city, and date or in some cases some garbage or irrelevant data.
\subsubsection{Data Correction (Data Stemming on Tokenize word to their basic form)}
It is also known as lemmatization in which the word is changed to its very basic form by removing suffixes and will be converted to its root form. This makes the handling of words very easily. The significance of stem is very well explained e.g. \includegraphics[width=0.09\textwidth]{a3.png}may be written as \includegraphics[width=0.09\textwidth]{a4.png} Failed to do so sometimes create problems in processing which ultimately affects our result.
\subsubsection{Data Correction (Different Format of Single Word)}
Urdu is a versatile language and words can be written in different styles. Like \includegraphics[width=0.06\textwidth]{a5.png} to \includegraphics[width=0.07\textwidth]{a6.png}, \includegraphics[width=0.04\textwidth]{a7.png} to \includegraphics[width=0.04\textwidth]{a8.png}
in this case our dictionary is unable to find these types of word and failed to produce results. Then we refer these cases to our format pattern dictionary where we describe the different patterns of the same word. If a word is not recognized by "Urdu based epidemic and city" dictionaries then we will move to this dictionary for finding the diversion of said word. This dictionary is flexible and regularly updated on the feed-back and custom detection by users. If we checked every word in this dictionary and found that maximum words exist in their standard shape, then it is useless and resource-consuming that we checked every word for its reformatting. This dictionary is optional.
\subsection{Event Recognition Layer}
After necessary filtering and preprocessing, we have now reduced the array of string. In this layer, we recognize the epidemic and also its occurrence city with the date stamp. This layer is also eliminating duplicate news from different channels and newspapers.
\subsubsection{Epidemic Directory / Ontology in Urdu (Creation and updating of said Directory / Ontology on regular basis)}
To find out disease from the preprocessed array, we must have a dictionary or data-base from where we can match and find the result. These types of dictionaries are available in English but in the Urdu language, it does not exist. We prepared URDU Directory / Ontology Dictionary which contains initially approx. 50 diseas-es/Pathogens/ Infections with its English version.
\subsubsection{Event Recognition (Epidemic) }
In this stage, we extracted the disease from the news feed in connection with epidem-ic disease ontology. There may be some confusion regarding disease nomenclature like \includegraphics[width=0.03\textwidth]{a9.png} or \includegraphics[width=0.03\textwidth]{a10.png} or fever, we removed this confusion with us reformat dictionary. We made them in simple and basic level, so that the exact disease is to be identified.
\subsubsection{City Database in Urdu (Enlisted the all cities of Pakistan in URDU) }
Pakistan has 374 cities. There is no database exist which describe the Pakistan cities names in Urdu, so for our purpose of city recognition, we will make a database. This database/ dictionary will use to detect the city of an epidemic outbreak. Urdu pub-lishers are not always following standard patterns and published their news in very random ways, sometimes the name of the city is written as \includegraphics[width=0.06\textwidth]{a11.png} and sometimes \includegraphics[width=0.05\textwidth]{a12.png} or sometimes "Burewala". This versatility in city names creates confusion for the program to find out the exact city. We will make our dictionary/ database in such a way that caters all these problems. We put different patterns of a city in Urdu and English also. So, when querying this dictionary, it will return us the basic and standard format of the city. We will also make this database relaxant so that names of cities or their different formats may include later on. After determining the disease, it is neces-sary to find out the epidemic area so that we can visualize it on a map.
\subsubsection{Entity Recognition (City) (Find City of concerned Epidemic)}
In the light of the city named dictionary, we extract city names from the RSS title. If the city is not identified in the RSS feed title, then we moved on to the description to find out the city. There may be variants of city names published in RSS feed that caters as described above. When we query the city database it will return our city name in that format which is recognized by our latitude and longitude stage.
\subsubsection{City Lat-Long Database in Urdu (Enlisted Latitude and Longitude of Pakistan cities in Urdu) }
The world is divided into latitude and longitude. Latitude and Longitude are the units that represent the coordinates at the geographic coordinate system. To pinpoint any location or refer to this on the map, we must know the latitude and longitude of said area. The map only understands the latitude and longitude. So that to visualize the city on the map we must know the latitude and longitude of said city. Before this, there is no database that describes the latitude and longitude of Pakistan cities in Urdu. We prepared a database that will return the latitude and longitude of each city of Pakistan when inquired. These are the fixed parameters and not changed with the passage of time so that we make it robust.
\subsubsection{Recognition of City Lat-Long (Find Latitude and Longitude of Concerned City)}
After identification of city, we will extract its latitude and longitude from the Lat-Long database and will store this information in a database for further processing of JSON to visualize the epidemic on the map.
\subsection{Visualization Layer (JSON)}
In this layer, we visualize the epidemic outbreak \cite{newman2002spread} with the concerned area and also used the cluster marker to simplify the map vision which removed the so many pins clutter and combined the same area with a marker. Epidemic disease, city, city lat-long, date, and link are stored in a dedicated separate database.
\subsubsection{Database Storage (All IDs, Epidemic, City, Lat-Long stored in Database with spe-cific ID)}
After extracting all necessary information regarding the disease, city, lat-long, date, link and system generated ID will be stored in a dedicated database. This infor-mation is accessed by JSON and will be used to view on map, we will set this data-base frequency up to 90 days which will be reduced on user-customized query.
\subsubsection{Web Site Backend (Fetch data from DB with respect to Location/ Disease)}
Website required certain types of data to be used on a map. These data included the epidemic disease and area with lat long. The Database is store in the list and is used in the visualization.
\subsubsection{Web site Backend Fetching Types:}
\begin{itemize}
\item Fetch data from DB with respect to Location/ Disease
\item Fetch customized data as per query: database can be customized as per user requirements.
\end{itemize}
\subsubsection{Visualization (In term of pathogen, syndrome, or text type)}
For visualization purpose, we have used the JSON. First, we module our data into JSON pattern and then this data are used by map to visualize. Wherever the disease outbreak occurred, a marker is placed there and a short note will appear there.
\subsubsection{Marker Clustering Google Map Marker Clustering (For more clarity in visuali-zation)}
If any place has many markers \cite{netek2019performance} then these markers are combined and displayed as a single marker, which simplified the view.
\subsubsection{Source Evidence (URL will open detail for that epidemic)}
Source evidence is a feature to facilitate the user with full detail of the news. If some-one wants to know further details of the concerned epidemic, then it will just click and web refers him to the concerned page for more detail. Different colors are used to make it more comprehensive for visualization.
\section{Experiments \& Results}
Our developed model has four layers and each layer has different stages. Here we describe the output of each stage.
\subsection{RSS Feed Output:}
We used almost all available resources that are being offered the Urdu RSS Feed. Here we only see the results of SUCHTV news. We automate it to run after the calendar day has been changed. SUCHTV news almost issued 10-20 RSS feed daily when the calendar date changed. We also tune our system so that it will not miss any feed. If we want to run manually or mistakenly run the developed system then it captured the RSS feed but did not put it in the database because there is already that feed available with access No control. Figure 7 shows the RSS feed which is being captured by our system.
\begin{figure}
\centering
\includegraphics[width=0.75\textwidth]{fig6.png}
\caption{SACH TV News RSS Feed.}
\label{fig:fig6}
\end{figure}
\subsection{RSS Feed in Database}
Here we captured the title, date of news which is being provided by RSS provider (SUCHTV news), description, access No, and flag. The flag is placed here to lessen the burden of the system. When our system placed the new RSS feed in the database then the flag is set to "1". But when the system processed that news and found the results then it assigned it "0" so that next time when the system approaches the data-base then only a new RSS feed will be processed, hence we less the burden of the system. Figure 8 shows the RSS feeds are placed in the database.
\begin{figure}
\centering
\includegraphics[width=0.75\textwidth]{fig7.png}
\caption{RSS Feed stored in Database.}
\label{fig:fig7}
\end{figure}
\subsection{Epidemic Outbreak Detection }
The accuracy of our system is depending on the detection of epidemic diseases from every RSS feed. For this purpose, we maintained and directory of epidemic disease as reference because Urdu based ontology is not exist before. Figures 9 shows the detected disease from RSS feed at system run time, figure 10 show the summary of detected disease with their database ID for easy tracking.
\begin{figure}
\centering
\includegraphics[width=0.75\textwidth]{fig8.png}
\caption{Run time Epidemic Detection.}
\label{fig:fig8}
\end{figure}
\begin{figure}
\centering
\includegraphics[width=0.75\textwidth]{Fig9.png}
\caption{Summary of detected disease with database ID.}
\label{fig:fig9}
\end{figure}
\subsection{City Detection and city latitude City Latitude and Longitude Detection:}
Figure 11 shows the latitude and longitude of the detected city.
\begin{figure}
\centering
\includegraphics[width=0.75\textwidth]{fig10.png}
\caption{Detection of Latitude and Longitude of Detected City.}
\label{fig:fig10}
\end{figure}
\subsection{JSON Data for Visualization}
To give geolocation parameters to Google Map, We convert our data in JSON so that map can take latitude and longitude from there. Figure 12 shows the JSON file with all detected disease, their location name, latitude, longitude, and Data-base ID for easy tracking and source verification.
\begin{figure}
\centering
\includegraphics[width=0.75\textwidth]{fig11.png}
\caption{JSON File.}
\label{fig:fig11}
\end{figure}
\subsection{Google Map Visualization}
Finally, all the epidemic outbreaks and locations are shown in the Google Map with marker feature which is used to give more clarity on a map, sometime also refers as clustering, when many diseases is spread in nearby areas then clustering/ marker feature marked that area with a single marker and shows numeric to give more clarity shows in figure 13.
\begin{figure}
\centering
\includegraphics[width=0.75\textwidth]{fig12.png}
\caption{Google Map with Clustering Marker.}
\label{fig:fig12}
\end{figure}
\subsection{Removal of Diacritics:}
Not all the RSS feed has diacritics but very few, So identify it with Expert Identi-fication and figure it out that 55 RSS feeds which have diacritics. Now we analy-sis our system performance with Expert Identification, and In Table 1, we can see how many accuracies, precision, recall and F-Score will obtain
\begin{table}[h]
\centering
\caption{{Performance Evaluation of Diacritics Removal}}
\label{tab:tab1}
\begin{tabular}{cccccccccccccc}
\toprule
\multicolumn{2}{c}{Removal of Diacritics (Confusion Matrix)} \\
\cmidrule(r){1-2}
\multicolumn{0}{c}{\textbf{Expert Identification}} &
\multicolumn{2}{c}{\textbf{BIOPAK Flasher}} &
\multicolumn{0}{c}{\textbf{Total}} &
\\
\boldmath{}&\boldmath{Removed} & \boldmath{Not Removed} \\
\midrule
$-$Removed & $-$55 (TP) & $-$0(FN) & $-$55\\
Not Removed & $-$0(FP) & $-$0(TN) & $-$0 \\
$-$Total & $-$55 & 2 & {$-$55} \\
\bottomrule
\end{tabular}
\end{table}
\subsection{Removal of Punctuations: }
Every sentence ends up with (Full stop). Others are used in sentences. It is too much difficult and vague to quantify this performance test on the whole existing database, for this purpose we choose 50 RSS feeds that have (90) punctuations. Now we analysis our system performance with Expert Identification, and we can see in Table 2 that how many accuracy, precision, recall, and F-Score will obtain.
\begin{table}[h]
\centering
\caption{Performance Evaluation of Punctuation Removal}
\label{tab:tab2}
\begin{tabular}{cccccccccccccc}
\toprule
\multicolumn{2}{c}{Removal of Punctuation (Confusion Matrix)} \\
\cmidrule(r){1-2}
\multicolumn{0}{c}{\textbf{Expert Identification}} &
\multicolumn{2}{c}{\textbf{BIOPAK Flasher}} &
\multicolumn{0}{c}{\textbf{Total}} &
\\
\boldmath{}&\boldmath{Removed} & \boldmath{Not Removed} \\
\midrule
$-$Removed & $-$88 (TP) & $-$2(FN) & $-$90\\
Not Removed & $-$0(FP) & $-$0(TN) & $-$0 \\
$-$Total & $-$88 & 2 & {$-$90} \\
\bottomrule
\end{tabular}
\end{table}
\subsection{Tokenization Correctness: }
A very important step in NLP \cite{mullen2018fast}, we have 301 RSS Feed for evaluation purposes. Now we analysis our system performance with Expert Identification, and see how much accuracies, precision, recall, and F-Score will obtain shows in Table 3.
\begin{table}[h]
\centering
\caption{Tokenization Correctness (Confusion Matrix)}
\label{tab:tab3}
\begin{tabular}{cccccccccccccc}
\toprule
\multicolumn{2}{c}{Removal of Diacritics (Confusion Matrix)} \\
\cmidrule(r){1-2}
\multicolumn{0}{c}{\textbf{Expert Identification}} &
\multicolumn{2}{c}{\textbf{BIOPAK Flasher}} &
\multicolumn{0}{c}{\textbf{Total}} &
\\
\boldmath{}&\boldmath{Tokenize} & \boldmath{Not Tokenize} \\
\midrule
$-$Tokenize & $-$273 (TP) & $-$16(FN) & $-$289\\
Not Tokenize & $-$11(FP) & $-$1(TN) & $-$12 \\
$-$Total & $-$284 & 17 & {$-$301} \\
\bottomrule
\end{tabular}
\end{table}
\subsection{Overall System Accuracy}
We conclude all the results and find the overall performance of our system. Starting from first layer shows in Table 4:
\begin{table}
\caption{In terms of overall performance of BIOPAK Flasher, it performed above 80\% for almost all stages.}
\centering
\begin{tabular}{lllllll}
\toprule
\cmidrule(r){1-4}
BIOPAK Flasher Stages & Accuracy & Precision & Recall & F- Score\\
\midrule
Removal of Diacritics & 100 \% & 100 \% & 100 \% & 100 \% \\
Removal of Punctuation & 97.8\% & 100\% & 97.8\% & 98.9 \% \\
Tokenization Correctness & 91\% & 96.1\% & 94.5\% & 95.3\% \\
Stop Word Removal & 89.6\% & 92.6\% &89.8\% & 91.2\% \\
Epidemic outbreak detection & 93.9\% &97.4\% & 95.7\% & 96.5\% \\
City detection & 96.4\% &97.8\% &92.3\% &98.1\%\\
City Lat Long Detection & 99.4\% & 100\% & 99.4\% &99.7\% \\
Overall BIOPAK Flasher & 95.4 \% &97.7\% &95.6\% &97.1\% \\
\bottomrule
\end{tabular}
\label{tab:table}
\end{table}
Pathogen \cite{lazcka2007pathogen} is the main cause of the epidemic outbreaks. These pathogens are required objects to survive and can spread from human to human. These pathogens can also spread through trade. It is necessary to stop these types of the pathogen to spread more and confined them. A carrier of such pathogen may cause sources of spread for others if travel other places. To escape and confined this outbreak, it is necessary to stop other people to visit that place or stop trade with that place, howev-er, if necessary, to continue relations then proper vaccination will be carried out.
\subsection{Conclusion}
Our research points out the areas where the epidemic outbreak has occurred using Urdu language. To know such type of information, many actors are playing a vital role e.g., RSS Feed, Twitter, Social media, Printed Media, e-paper and News channels, etc. These resources are available in printed form (hard copy) or the e-electronic form (soft copy) in English and Urdu. To get news of the epidemic outbreaks, we choose the RSS feed in Urdu as a resource.\\
Urdu language is ignored and not used for epidemic detection. Many web sites are working, but most of the sites are in English. Whereas, Urdu is widely spoken language in Asia and also the national language of Pakistan. Majority people can better understand if the information is spread in Urdu. Very few media groups are maintained their RSS feeds in Urdu, it may be increased with the passage of time. \\
Our proposed model is using state of the art technologies of NLP for the ex-traction of data from these feeds. To implement this, we used the different NLP li-braries. Some libraries have good support for Urdu but some have less support, which creates a problem for us. There is no full matured library is available for the NLP tasks on Urdu language. Every library has basic level support for Urdu, restriction of these libraries lies on the minimum level of Urdu data corpus for training different models. However, we extracted the epidemic diseases and cities from these RSS feed with the help of these libraries and some custom developed algorithm. \\
We divided our model into four layers, so that maintenance, troubleshoot-ing, and updating is so easy, we can add, modify or delete any layer. The first layer is used to acquire RSS feed from RSS provider and saved it into the database. RSS feed duplication from same or different sources removed. In the second layer, we performed NLP techniques to filter the data, we take a single RSS feed's title, removed diacritics, and punctuations in the very first stage. After these basic filtering, we to-kenized the data and removed stop words. At the end of the second layer, we have filtered unnecessary data, now it is easy to perform the next steps with great perfor-mance. Our third layer is used to extract epidemic diseases and their area of occur-rences. For this purpose, we developed three customized dictionaries with the en-hanced features of correction, and updating. These dictionaries have epidemic dis-ease ontology, Pakistani cities in Urdu, and their geo locations. These dictionaries are used as a reference, if any epidemic pathogen is detected after deployment of our system, then it is very easy to include this pathogen in our database. On detection of epidemic disease, our system then detect city from feed, if city is not found here then move our pointer to description portion for finding occurrence area. After finding the city, it is necessary to get its geo location which is required by Google Map to display occurrence areas. In the fourth layer, we visualized the disease on Google Map. The Map will be too messy if we display all the information there, to clutter down and give better visualization, we used the marker technique. In this layer, we also stored detected disease, city, and geo location in a database for displaying it up to 90 days. \v
We checked the performance of each layer with accuracy, precision, recall, and F-score. The overall system accuracy is 95.4\%, Precision 97.7\%, Recall 95.6\% and F-score is 97.1\%. In terms of the performance of BIO PAK Flasher, it performed above 80\% for almost all stages. We also give a comparison of different tools and libraries, which are available to perform said task. BIOPAK Flasher also gives re-source evidence, if any viewer is required to see the full news, our system can direct him to the concerned web page for a detail study. Our research gives a breakthrough in the field of Urdu language when working with online actors. However, more work and efforts are needed to mature Urdu NLP.
\bibliographystyle{unsrtnat}
\section{Introduction}
In past many infectious diseases hampered humankind, like the Whooping cough outbreak in U.S. during the summer of 2012, Periodic outbreaks of H7N9 in China during 2013-2014 and emerging MERS outbreak in Saudi Arabia, Severe Acute Respiratory Syndrome (SARS), West Nile Virus and drug resistant tuberculosis \cite{zhao2012national}. Diseases are spread exponentially within a population. Peoples and products travel across the border in high volume so that an infection can rapidly attain a global scale.\\
A pandemic HiNi ("Swine Flu") appeared in 2009 \cite{szomszor2011twitter}. To increase this outbreak, a globally open online web-based framework gave warning to everyone concerned. The disease was catered for in response to this warning, and public understanding of emerging infection has increased. Online web-based early warning system's traffic rate was increased about 100 \% in 2-3 days. It gives birth to online early warning globally infection awareness. Hence, globally feels that there must be a system that alerts the early warning infectious disease timely vaccinate and cure the victims. Presently, nearly all major outbreaks, investigated by the World Health Organization (WHO) are first identified through these informal online sources.
The emergence of ubiquitous internet service and build up different internet-based resources like blogs, online newspapers (Google News), social media (Twitter), and discussion forums shape the world dynamic fast and very efficiently. Such services are highly useful in helping to diagnose, assess, incorporate and forecast outbreaks of infectious diseases or diseases spreading to new regions. Intelligent modeling of such sources using text-mining methods such as topic models, deep learning and dependency parsing can lead to automated generation of the mentioned surveillance tools..\\
The above resources produced big data at very high speed, and the quantity of that data is increasing day by day. Despite big data problem, these resources have the unstructured type of data, which is difficult to process, and extract useful information. Reading and assimilating a large number of reports daily basis is a hectic and burdensome task. .\\
. With BIO PAK FLASHER, a web-based solution focusing primarily on URDU \cite{jabbar2016analysis}, Pakistani Urdu news outlets, and covering Pakistani rulers and urban areas, we plan to solve these challenges with the automated framework for extraction, analysis, query, screening, incorporation, and visualization of unstructured and bulky disease outbreaks. Our system collects outbreak data from many sources such as news through the Google News aggregator, Pakistani local news channels published in Urdu, expert-curetted accounts such as ProMED \cite{morse1996promed} Mail, and validated official alerts such as World Health Organization announcements. By using the text processing algorithms, the system classifies outbreak in location and disease, after that overlay such data on an interactive geographic map. We measure the accuracy of the classification algorithms based on the level of human curation necessary to correct misclassification and examine geographic coverage. With state-of-the-art NLP technology, there is a need to compare all Urdu terms with a disease database called ontology to figure out the disease when the sentence is tokenized and all POS deleted. There is no disease ontology \cite{schriml2012disease} exist before to do such work. Here we also build Urdu-based Ontology with different patterns for further reference. .\\
Google Maps API consists of a set of classes within a JavaScript container that opens within an XHTML page. Each time the Google Maps web page opened, these classes are loaded from Google, and it has embedded JavaScript object on which all the functionality of the Google Maps is based. To visualize outbreaks, we use Google Maps.BioPak Flasher has a user- friendly interface and providing important disease outbreak information with text processing algorithms and also referred user to view full article with web link integration.
The rest of the paper is composed of Section 2 Literature review presents the background of infectious disease outbreaks and basic terminologies used in this research. Section 3 is about the Methodology used in this research, Section 4 Describes the implementation and Analysis of the proposed methodology. Section 5 presents the conclusion and future work for this research.
\section{Literature Review}
There are few traditional disease surveillance systems like Centers for Disease Control (CDC) \cite{berrios2017centers}, ProMED \cite{morse1996promed}, which released data of infectious disease outbreaks with delay due to their implemented setup (take news mail, get expert opinion then release outbreak announcement). In the modern period, the epidemic must be published in real-time. To cater to evolving diseases, take control steps quickly, and avoid them from spreading further, an early warning/detection of the outbreak system is important. These surveillance systems have some limitations. These programs do not protect all languages, so the epidemic news reported in state and local news would be missed. The system must be capable to process these languages so that any news published in theses languages will also be covered to give authenticated real-time system. Urdu is the national language of Pakistan and speaks in many other countries as well in non-official manners. Urdu is also not covered by these systems. Many local publishers like VOA-Urdu, Geo, Express, Jang, etc. published their news in Urdu. Moreover, these systems have some geographical limitations. These systems are not covering the ruler/ urban areas of some countries, especially Pakistan.\\
Infectious disease is increasing day by day due to unhygienic environment, people are in close contact with livestock or wildlife and chemicals, etc. Infectious diseases lead to mortality so awareness is a very essential part of the current era. Delayed detection of the outbreak may cause some mortality and may be created hinders in controlling and eradication process. Such mortality rate can be minimized if timely recognized. If we know in real- time, then the control and eradication measures also taken time to reduce mortality rate. \\
Moreover, when we want to visit somewhere or required to do trade with some area, we must consider emerging pathogens that causes the epidemic. Shifting of emerging product or human can cause the spreading of that epidemic in the hosting area the traditional outbreak system has a manual alarm that often requires a delay, but our, alert system, swift, and covering Pakistani news is our disease outbreak system written in Urdu. This system also helps in general awareness of infectious outbreaks and its remedies measures timely.
\subsection{Epidemic Breakdowns In Pakistan}
A government of Pakistan Official Body releases the statistics in 44th dated Mar 2019 to June 2019. According to these, Figure 1 alerts has been generated in the country, Figure 2 shows the statistical results of diseases outbreaks in Pakistan.
\subsubsection{Epidemic (A Communicate-able Disease) }
An epidemic \cite{newman2002spread} occurs when an infectious disease spreads rapidly in contact with too many people. An epidemic is an outbreak of a disease among members of a specific population that exceeds the extent of occurrence of the disease normally found in that population. Epidemics affect those members of the population who do not have an acquired or inherent immunity to the disease. Although infectious organisms cause most epidemics, the term can be applied to an outbreak of any chronic disease, such as lung cancer or heart disease. For example, in 2003, the severe acute respiratory syndrome (SARS) epidemic took the lives of nearly 800 people worldwide. Statistics and big outbreaks in history with casualty.
\subsubsection{RSS Feed: }
RSS stands for "Really Simple Syndication". It is a way to easily distribute a list of headlines, update notices, and sometimes content to a wide number of people \cite{johnson2009developing}. Computer programs that organize those headlines and notices for easy reading use it. Most people are interested in many websites whose content changes on an unpredictable schedule. News sites, community, and religious group information pages, product information pages, medical websites, and weblogs are examples of such websites. Repeatedly checking each website to see if there is any new content can be very tedious.
\begin{figure}
\centering
\includegraphics[width=0.75\textwidth]{Fig1.png}
\caption{Recent tutbreaks in Pakistan.}
\label{fig:fig1}
\end{figure}
\begin{figure}
\centering
\includegraphics[width=0.75\textwidth]{fig2.png}
\caption{Statistics of different disease outbreak in Pakistan.}
\label{fig:fig2}
\end{figure}
\subsubsection{How It Works:}
RSS works by having the website author maintain a list of notifications on their web-site in a standard way. This list of notifications is called an "RSS Feed". People who are interested in finding out the latest headlines or changes can check this list. Special computer programs called "RSS aggregators" have been developed that automatical-ly access the RSS feeds of websites you care about on your behalf and organize the results for you. (RSS feeds and aggregators are sometimes called "RSS Channels" and "RSS Readers".)
\subsubsection{RSS Feed Structure: }
An XML File provides very basic information to do its notification. It is made up of a list of items presented in order from newest to oldest. Each item usually consists of a simple title describing the item along with a more complete description and a link to a web page with the actual information being described. Sometimes this description is the full information you want to read (such as the content of a weblog post) and sometimes it is just a summary.
\section{Methodology}
The different aspects of this research including data collection were the compulsory requirement for further utilization of the data. The essentials tasks are explained figure 3.\\
In this figure, we presented the layer view of our model. We preferred to work in layer because it is easy to wok easily if we have to do any changing. We divide our model into four layers; each layer is dependent on its precedence layer. Each layer work is defined clearly. The first layer is the Data acquisition layer, which retrieves data from the RSS, feed and puts it into the database for further processing. The second layer is the Data filtering and preprocessing layer in which we eliminate and filter out unnecessary diacritics and punctuations then performing some NLP (Natu-ral Language Processing) techniques like tokenization, stemming, and stop words removal of Urdu language. Then our data has epidemic, city, date, and some garbage or unnecessary data. The third layer of Event Recognition is designed to find out the epidemic disease and concerned area with latitude and longitude. Our last layer is the Visualization layer, which deals with visualization of epidemic outbreak in the con-cerned area will show on Google map with marker view and store events in database for displaying it next 90 days. User can also view source evidence by simply clicking on a pinpoint map as well as associated headlines for the event.
\begin{figure}
\centering
\includegraphics[width=0.75\textwidth]{fig3.png}
\caption{Abstract overview of the BioPak flasher.}
\label{fig:fig3}
\end{figure}
\subsection{Data Acquisition and Storage Layer: }
In this layer. We acquire the data from Pakistani news channels and newspapers that are offering online resources (RSS feed) in Urdu and store these data into a database for further processing. Many online actors can provide news like RSS, tweeter, LinkedIn, Facebook, etc. However, our choice is RSS feed. RSS stand for Really Simple Syndication or sometimes refer as Rich Site Summary. Actually, this technol-ogy was introduced to avoid bulky and uninterested news, which is being pushed by a provider. This is also used to keep updated user with upcoming news, blogs, financial information, health, sports, classified sites, and government alerts. RSS feed is written in XML code which is easy to process by computer. It is a structured data type.
Figure 4 shows a sample of newsreader who acquired the news from Voice of America in Urdu. We can select any topic in which we are interested, in our case health topic is most prominent and maintained by almost every RSS feed provider. So we subscribe to the health RSS feed.
When we click on any news feed, it will refer us to its description web page shows in Figure 5, where we can see all detail of the news. In this figure, we refer to the voice of America page where we can find the detail of said news.
\begin{figure}
\centering
\includegraphics[width=0.75\textwidth]{fig4.png}
\caption{Sample overview of RSS feeds.}
\label{fig:fig5}
\end{figure}
\begin{figure}
\centering
\includegraphics[width=0.75\textwidth]{fig5.png}
\caption{Sample screenshot of the News.}
\label{fig:fig5}
\end{figure}
Our focus of research covered Pakistan and its national language Urdu Pakistan has several outlets that include printed media, news channels, several online podcasts, and internet-based TV channels used for consumer information. These all channels are published in different languages according to a location like English, Urdu, Sindhi, etc. Therefore, Urdu RSS feed newspaper or channels are our main actor for getting information. However, very few channels are maintained their RSS feed in Urdu. Because our research is focused on the RSS mechanisms so that we only focus that news, channels, which have their RSS, feed in Urdu and store it.
\subsubsection{Data filtering and Pre-Processing: }
There are many types of data which is published in the RSS feed. Input news items are not considered to be cleaned so making them ready for further processing is im-portant. Raw data, if obtained from the web normally includes HTML and or XML tags so they should be removed for further processing. In addition, text format (utf-8) for special characters is important so that no error occurs while processing.
\subsubsection{Removal of Diacritics: }
The Urdu language is using diacritics to clarify the pronunciation of Urdu words. Commonly used diacritics are Zabar, Zer, Pesh, Jazam, and shad, etc. Our raw data received from the database has many unnecessary elements that are required to be filtered shows in figure 6. In this stage of filtering, it is necessary that data must be filtered from diacritics. If it is not done then in processing, different result will be seen and our final output is very different from the actual. \\
\centering
\includegraphics[width=0.55\textwidth]{a.png}
\subsubsection{Removal of Punctuation}
Urdu language sentences have different punctuations, which used in the construction of sentences. These punctuations are undesired in our processing so that we removed these. Some punctuations which mostly used in the Urdu language are " " " '' – [](),.;: "
\subsubsection{Data Preparation (Tokenization with NLTK Library)}
Urdu tokenization is not easy as English due to its structure. However, Unicode recommends using Zero Width Non-Joiner character in this context \cite[4]. There would be an alignment of tags with the terms present at the simple stage, so tokenization is also an essential aspect of sorting. The text will be tokenized into sentences and phrases into words in the process.
Tokenization is a very common feature of NLP. Tokenization is a process to break a text string into single entities that may include words, symbols, phrases, etc. In our case, we apply this technique to divide the string into words, e.g.
\begin{figure}
\centering
\includegraphics[width=0.55\textwidth]{fig7.png}
\label{fig:fig7}
\end{figure}
Further processing is taken place on the resulted array after tokenization.
\subsubsection{Building up of Urdu Stop Words (USW) Dictionary }
After tokenization, now many unwanted words have no meaning when we tokenize but these words have more worth when used in combination and clear the concept of sentences. Unfortunately, there is not much work done in Urdu, so we used stop word dictionary Stop Words Removal (Based on USW Dictionary)
These dictionaries are used as a reference for the removal of stop words. This is another technique of NLP, which are tags, or separates various parts of speech in the text. For filtering, stop words are also attached with this technique to filter indifferent data. Epidemic disease, city, and date of occurrences are needed to be extracted from the text string. Everything else in the string is irrelevant and should not be extracted or in other words, should be deleted from the string.
Tokenization has been done in the previous step. Keeping in view the resulted string, now the irrelevant data should be filtered out from the string. At this stage, we have the tokenized string as:\\
\centering
\includegraphics[width=0.55\textwidth]{a1.png}\\
After deletion of Stop word from string, we will be left with the following array:\\
\centering
\includegraphics[width=0.55\textwidth]{a2.png}\\
Now the leftover data included only disease, city, and date or in some cases some garbage or irrelevant data.
\subsubsection{Data Correction (Data Stemming on Tokenize word to their basic form)}
It is also known as lemmatization in which the word is changed to its very basic form by removing suffixes and will be converted to its root form. This makes the handling of words very easily. The significance of stem is very well explained e.g. \includegraphics[width=0.09\textwidth]{a3.png}may be written as \includegraphics[width=0.09\textwidth]{a4.png} Failed to do so sometimes create problems in processing which ultimately affects our result.
\subsubsection{Data Correction (Different Format of Single Word)}
Urdu is a versatile language and words can be written in different styles. Like \includegraphics[width=0.06\textwidth]{a5.png} to \includegraphics[width=0.07\textwidth]{a6.png}, \includegraphics[width=0.04\textwidth]{a7.png} to \includegraphics[width=0.04\textwidth]{a8.png}
in this case our dictionary is unable to find these types of word and failed to produce results. Then we refer these cases to our format pattern dictionary where we describe the different patterns of the same word. If a word is not recognized by "Urdu based epidemic and city" dictionaries then we will move to this dictionary for finding the diversion of said word. This dictionary is flexible and regularly updated on the feed-back and custom detection by users. If we checked every word in this dictionary and found that maximum words exist in their standard shape, then it is useless and resource-consuming that we checked every word for its reformatting. This dictionary is optional.
\subsection{Event Recognition Layer}
After necessary filtering and preprocessing, we have now reduced the array of string. In this layer, we recognize the epidemic and also its occurrence city with the date stamp. This layer is also eliminating duplicate news from different channels and newspapers.
\subsubsection{Epidemic Directory / Ontology in Urdu (Creation and updating of said Directory / Ontology on regular basis)}
To find out disease from the preprocessed array, we must have a dictionary or data-base from where we can match and find the result. These types of dictionaries are available in English but in the Urdu language, it does not exist. We prepared URDU Directory / Ontology Dictionary which contains initially approx. 50 diseas-es/Pathogens/ Infections with its English version.
\subsubsection{Event Recognition (Epidemic) }
In this stage, we extracted the disease from the news feed in connection with epidem-ic disease ontology. There may be some confusion regarding disease nomenclature like \includegraphics[width=0.03\textwidth]{a9.png} or \includegraphics[width=0.03\textwidth]{a10.png} or fever, we removed this confusion with us reformat dictionary. We made them in simple and basic level, so that the exact disease is to be identified.
\subsubsection{City Database in Urdu (Enlisted the all cities of Pakistan in URDU) }
Pakistan has 374 cities. There is no database exist which describe the Pakistan cities names in Urdu, so for our purpose of city recognition, we will make a database. This database/ dictionary will use to detect the city of an epidemic outbreak. Urdu pub-lishers are not always following standard patterns and published their news in very random ways, sometimes the name of the city is written as \includegraphics[width=0.06\textwidth]{a11.png} and sometimes \includegraphics[width=0.05\textwidth]{a12.png} or sometimes "Burewala". This versatility in city names creates confusion for the program to find out the exact city. We will make our dictionary/ database in such a way that caters all these problems. We put different patterns of a city in Urdu and English also. So, when querying this dictionary, it will return us the basic and standard format of the city. We will also make this database relaxant so that names of cities or their different formats may include later on. After determining the disease, it is neces-sary to find out the epidemic area so that we can visualize it on a map.
\subsubsection{Entity Recognition (City) (Find City of concerned Epidemic)}
In the light of the city named dictionary, we extract city names from the RSS title. If the city is not identified in the RSS feed title, then we moved on to the description to find out the city. There may be variants of city names published in RSS feed that caters as described above. When we query the city database it will return our city name in that format which is recognized by our latitude and longitude stage.
\subsubsection{City Lat-Long Database in Urdu (Enlisted Latitude and Longitude of Pakistan cities in Urdu) }
The world is divided into latitude and longitude. Latitude and Longitude are the units that represent the coordinates at the geographic coordinate system. To pinpoint any location or refer to this on the map, we must know the latitude and longitude of said area. The map only understands the latitude and longitude. So that to visualize the city on the map we must know the latitude and longitude of said city. Before this, there is no database that describes the latitude and longitude of Pakistan cities in Urdu. We prepared a database that will return the latitude and longitude of each city of Pakistan when inquired. These are the fixed parameters and not changed with the passage of time so that we make it robust.
\subsubsection{Recognition of City Lat-Long (Find Latitude and Longitude of Concerned City)}
After identification of city, we will extract its latitude and longitude from the Lat-Long database and will store this information in a database for further processing of JSON to visualize the epidemic on the map.
\subsection{Visualization Layer (JSON)}
In this layer, we visualize the epidemic outbreak \cite{newman2002spread} with the concerned area and also used the cluster marker to simplify the map vision which removed the so many pins clutter and combined the same area with a marker. Epidemic disease, city, city lat-long, date, and link are stored in a dedicated separate database.
\subsubsection{Database Storage (All IDs, Epidemic, City, Lat-Long stored in Database with spe-cific ID)}
After extracting all necessary information regarding the disease, city, lat-long, date, link and system generated ID will be stored in a dedicated database. This infor-mation is accessed by JSON and will be used to view on map, we will set this data-base frequency up to 90 days which will be reduced on user-customized query.
\subsubsection{Web Site Backend (Fetch data from DB with respect to Location/ Disease)}
Website required certain types of data to be used on a map. These data included the epidemic disease and area with lat long. The Database is store in the list and is used in the visualization.
\subsubsection{Web site Backend Fetching Types:}
\begin{itemize}
\item Fetch data from DB with respect to Location/ Disease
\item Fetch customized data as per query: database can be customized as per user requirements.
\end{itemize}
\subsubsection{Visualization (In term of pathogen, syndrome, or text type)}
For visualization purpose, we have used the JSON. First, we module our data into JSON pattern and then this data are used by map to visualize. Wherever the disease outbreak occurred, a marker is placed there and a short note will appear there.
\subsubsection{Marker Clustering Google Map Marker Clustering (For more clarity in visuali-zation)}
If any place has many markers \cite{netek2019performance} then these markers are combined and displayed as a single marker, which simplified the view.
\subsubsection{Source Evidence (URL will open detail for that epidemic)}
Source evidence is a feature to facilitate the user with full detail of the news. If some-one wants to know further details of the concerned epidemic, then it will just click and web refers him to the concerned page for more detail. Different colors are used to make it more comprehensive for visualization.
\section{Experiments \& Results}
Our developed model has four layers and each layer has different stages. Here we describe the output of each stage.
\subsection{RSS Feed Output:}
We used almost all available resources that are being offered the Urdu RSS Feed. Here we only see the results of SUCHTV news. We automate it to run after the calendar day has been changed. SUCHTV news almost issued 10-20 RSS feed daily when the calendar date changed. We also tune our system so that it will not miss any feed. If we want to run manually or mistakenly run the developed system then it captured the RSS feed but did not put it in the database because there is already that feed available with access No control. Figure 7 shows the RSS feed which is being captured by our system.
\begin{figure}
\centering
\includegraphics[width=0.75\textwidth]{fig6.png}
\caption{SACH TV News RSS Feed.}
\label{fig:fig6}
\end{figure}
\subsection{RSS Feed in Database}
Here we captured the title, date of news which is being provided by RSS provider (SUCHTV news), description, access No, and flag. The flag is placed here to lessen the burden of the system. When our system placed the new RSS feed in the database then the flag is set to "1". But when the system processed that news and found the results then it assigned it "0" so that next time when the system approaches the data-base then only a new RSS feed will be processed, hence we less the burden of the system. Figure 8 shows the RSS feeds are placed in the database.
\begin{figure}
\centering
\includegraphics[width=0.75\textwidth]{fig7.png}
\caption{RSS Feed stored in Database.}
\label{fig:fig7}
\end{figure}
\subsection{Epidemic Outbreak Detection }
The accuracy of our system is depending on the detection of epidemic diseases from every RSS feed. For this purpose, we maintained and directory of epidemic disease as reference because Urdu based ontology is not exist before. Figures 9 shows the detected disease from RSS feed at system run time, figure 10 show the summary of detected disease with their database ID for easy tracking.
\begin{figure}
\centering
\includegraphics[width=0.75\textwidth]{fig8.png}
\caption{Run time Epidemic Detection.}
\label{fig:fig8}
\end{figure}
\begin{figure}
\centering
\includegraphics[width=0.75\textwidth]{Fig9.png}
\caption{Summary of detected disease with database ID.}
\label{fig:fig9}
\end{figure}
\subsection{City Detection and city latitude City Latitude and Longitude Detection:}
Figure 11 shows the latitude and longitude of the detected city.
\begin{figure}
\centering
\includegraphics[width=0.75\textwidth]{fig10.png}
\caption{Detection of Latitude and Longitude of Detected City.}
\label{fig:fig10}
\end{figure}
\subsection{JSON Data for Visualization}
To give geolocation parameters to Google Map, We convert our data in JSON so that map can take latitude and longitude from there. Figure 12 shows the JSON file with all detected disease, their location name, latitude, longitude, and Data-base ID for easy tracking and source verification.
\begin{figure}
\centering
\includegraphics[width=0.75\textwidth]{fig11.png}
\caption{JSON File.}
\label{fig:fig11}
\end{figure}
\subsection{Google Map Visualization}
Finally, all the epidemic outbreaks and locations are shown in the Google Map with marker feature which is used to give more clarity on a map, sometime also refers as clustering, when many diseases is spread in nearby areas then clustering/ marker feature marked that area with a single marker and shows numeric to give more clarity shows in figure 13.
\begin{figure}
\centering
\includegraphics[width=0.75\textwidth]{fig12.png}
\caption{Google Map with Clustering Marker.}
\label{fig:fig12}
\end{figure}
\subsection{Removal of Diacritics:}
Not all the RSS feed has diacritics but very few, So identify it with Expert Identi-fication and figure it out that 55 RSS feeds which have diacritics. Now we analy-sis our system performance with Expert Identification, and In Table 1, we can see how many accuracies, precision, recall and F-Score will obtain
\begin{table}[h]
\centering
\caption{{Performance Evaluation of Diacritics Removal}}
\label{tab:tab1}
\begin{tabular}{cccccccccccccc}
\toprule
\multicolumn{2}{c}{Removal of Diacritics (Confusion Matrix)} \\
\cmidrule(r){1-2}
\multicolumn{0}{c}{\textbf{Expert Identification}} &
\multicolumn{2}{c}{\textbf{BIOPAK Flasher}} &
\multicolumn{0}{c}{\textbf{Total}} &
\\
\boldmath{}&\boldmath{Removed} & \boldmath{Not Removed} \\
\midrule
$-$Removed & $-$55 (TP) & $-$0(FN) & $-$55\\
Not Removed & $-$0(FP) & $-$0(TN) & $-$0 \\
$-$Total & $-$55 & 2 & {$-$55} \\
\bottomrule
\end{tabular}
\end{table}
\subsection{Removal of Punctuations: }
Every sentence ends up with (Full stop). Others are used in sentences. It is too much difficult and vague to quantify this performance test on the whole existing database, for this purpose we choose 50 RSS feeds that have (90) punctuations. Now we analysis our system performance with Expert Identification, and we can see in Table 2 that how many accuracy, precision, recall, and F-Score will obtain.
\begin{table}[h]
\centering
\caption{Performance Evaluation of Punctuation Removal}
\label{tab:tab2}
\begin{tabular}{cccccccccccccc}
\toprule
\multicolumn{2}{c}{Removal of Punctuation (Confusion Matrix)} \\
\cmidrule(r){1-2}
\multicolumn{0}{c}{\textbf{Expert Identification}} &
\multicolumn{2}{c}{\textbf{BIOPAK Flasher}} &
\multicolumn{0}{c}{\textbf{Total}} &
\\
\boldmath{}&\boldmath{Removed} & \boldmath{Not Removed} \\
\midrule
$-$Removed & $-$88 (TP) & $-$2(FN) & $-$90\\
Not Removed & $-$0(FP) & $-$0(TN) & $-$0 \\
$-$Total & $-$88 & 2 & {$-$90} \\
\bottomrule
\end{tabular}
\end{table}
\subsection{Tokenization Correctness: }
A very important step in NLP \cite{mullen2018fast}, we have 301 RSS Feed for evaluation purposes. Now we analysis our system performance with Expert Identification, and see how much accuracies, precision, recall, and F-Score will obtain shows in Table 3.
\begin{table}[h]
\centering
\caption{Tokenization Correctness (Confusion Matrix)}
\label{tab:tab3}
\begin{tabular}{cccccccccccccc}
\toprule
\multicolumn{2}{c}{Removal of Diacritics (Confusion Matrix)} \\
\cmidrule(r){1-2}
\multicolumn{0}{c}{\textbf{Expert Identification}} &
\multicolumn{2}{c}{\textbf{BIOPAK Flasher}} &
\multicolumn{0}{c}{\textbf{Total}} &
\\
\boldmath{}&\boldmath{Tokenize} & \boldmath{Not Tokenize} \\
\midrule
$-$Tokenize & $-$273 (TP) & $-$16(FN) & $-$289\\
Not Tokenize & $-$11(FP) & $-$1(TN) & $-$12 \\
$-$Total & $-$284 & 17 & {$-$301} \\
\bottomrule
\end{tabular}
\end{table}
\subsection{Overall System Accuracy}
We conclude all the results and find the overall performance of our system. Starting from first layer shows in Table 4:
\begin{table}
\caption{In terms of overall performance of BIOPAK Flasher, it performed above 80\% for almost all stages.}
\centering
\begin{tabular}{lllllll}
\toprule
\cmidrule(r){1-4}
BIOPAK Flasher Stages & Accuracy & Precision & Recall & F- Score\\
\midrule
Removal of Diacritics & 100 \% & 100 \% & 100 \% & 100 \% \\
Removal of Punctuation & 97.8\% & 100\% & 97.8\% & 98.9 \% \\
Tokenization Correctness & 91\% & 96.1\% & 94.5\% & 95.3\% \\
Stop Word Removal & 89.6\% & 92.6\% &89.8\% & 91.2\% \\
Epidemic outbreak detection & 93.9\% &97.4\% & 95.7\% & 96.5\% \\
City detection & 96.4\% &97.8\% &92.3\% &98.1\%\\
City Lat Long Detection & 99.4\% & 100\% & 99.4\% &99.7\% \\
Overall BIOPAK Flasher & 95.4 \% &97.7\% &95.6\% &97.1\% \\
\bottomrule
\end{tabular}
\label{tab:table}
\end{table}
Pathogen \cite{lazcka2007pathogen} is the main cause of the epidemic outbreaks. These pathogens are required objects to survive and can spread from human to human. These pathogens can also spread through trade. It is necessary to stop these types of the pathogen to spread more and confined them. A carrier of such pathogen may cause sources of spread for others if travel other places. To escape and confined this outbreak, it is necessary to stop other people to visit that place or stop trade with that place, howev-er, if necessary, to continue relations then proper vaccination will be carried out.
\subsection{Conclusion}
Our research points out the areas where the epidemic outbreak has occurred using Urdu language. To know such type of information, many actors are playing a vital role e.g., RSS Feed, Twitter, Social media, Printed Media, e-paper and News channels, etc. These resources are available in printed form (hard copy) or the e-electronic form (soft copy) in English and Urdu. To get news of the epidemic outbreaks, we choose the RSS feed in Urdu as a resource.\\
Urdu language is ignored and not used for epidemic detection. Many web sites are working, but most of the sites are in English. Whereas, Urdu is widely spoken language in Asia and also the national language of Pakistan. Majority people can better understand if the information is spread in Urdu. Very few media groups are maintained their RSS feeds in Urdu, it may be increased with the passage of time. \\
Our proposed model is using state of the art technologies of NLP for the ex-traction of data from these feeds. To implement this, we used the different NLP li-braries. Some libraries have good support for Urdu but some have less support, which creates a problem for us. There is no full matured library is available for the NLP tasks on Urdu language. Every library has basic level support for Urdu, restriction of these libraries lies on the minimum level of Urdu data corpus for training different models. However, we extracted the epidemic diseases and cities from these RSS feed with the help of these libraries and some custom developed algorithm. \\
We divided our model into four layers, so that maintenance, troubleshoot-ing, and updating is so easy, we can add, modify or delete any layer. The first layer is used to acquire RSS feed from RSS provider and saved it into the database. RSS feed duplication from same or different sources removed. In the second layer, we performed NLP techniques to filter the data, we take a single RSS feed's title, removed diacritics, and punctuations in the very first stage. After these basic filtering, we to-kenized the data and removed stop words. At the end of the second layer, we have filtered unnecessary data, now it is easy to perform the next steps with great perfor-mance. Our third layer is used to extract epidemic diseases and their area of occur-rences. For this purpose, we developed three customized dictionaries with the en-hanced features of correction, and updating. These dictionaries have epidemic dis-ease ontology, Pakistani cities in Urdu, and their geo locations. These dictionaries are used as a reference, if any epidemic pathogen is detected after deployment of our system, then it is very easy to include this pathogen in our database. On detection of epidemic disease, our system then detect city from feed, if city is not found here then move our pointer to description portion for finding occurrence area. After finding the city, it is necessary to get its geo location which is required by Google Map to display occurrence areas. In the fourth layer, we visualized the disease on Google Map. The Map will be too messy if we display all the information there, to clutter down and give better visualization, we used the marker technique. In this layer, we also stored detected disease, city, and geo location in a database for displaying it up to 90 days. \v
We checked the performance of each layer with accuracy, precision, recall, and F-score. The overall system accuracy is 95.4\%, Precision 97.7\%, Recall 95.6\% and F-score is 97.1\%. In terms of the performance of BIO PAK Flasher, it performed above 80\% for almost all stages. We also give a comparison of different tools and libraries, which are available to perform said task. BIOPAK Flasher also gives re-source evidence, if any viewer is required to see the full news, our system can direct him to the concerned web page for a detail study. Our research gives a breakthrough in the field of Urdu language when working with online actors. However, more work and efforts are needed to mature Urdu NLP.
\bibliographystyle{unsrtnat}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 367 |
{"url":"http:\/\/hal.in2p3.fr\/in2p3-01518704","text":"# $^{96}_{36}$Kr$_{60}$\u2013Low-Z Boundary of the Island of Deformation at N =60\n\nAbstract : Prompt \u03b3-ray spectroscopy of the neutron-rich $^{96}$Kr, produced in transfer- and fusion-induced fission reactions, has been performed using the combination of the Advanced Gamma Tracking Array and the VAMOS++ spectrometer. A second excited state, assigned to $J^\u03c0$ = $4^+$, is observed for the first time, and a previously reported level energy of the first 2+ excited state is confirmed. The measured energy ratio R4\/2 = E($4^+$)\/E($2^+$) = 2.12(1) indicates that this nucleus does not show a well-developed collectivity contrary to that seen in heavier N = 60 isotones. This new measurement highlights an abrupt transition of the degree of collectivity as a function of the proton number at Z = 36, of similar amplitude to that observed at N = 60 at higher Z values. A possible reason for this abrupt transition could be related to the insufficient proton excitations in the g9\/2, d5\/2, and s1\/2 orbitals to generate strong quadrupole correlations or to the coexistence of competing different shapes. An unexpected continuous decrease of R4\/2 as a function of the neutron number up to N = 60 is also evidenced. This measurement establishes the Kr isotopic chain as the low-Z boundary of the island of deformation for N = 60 isotones. A comparison with available theoretical predictions using different beyond mean-field approaches shows that these models fail to reproduce the abrupt transitions at N = 60 and Z = 36.\nDocument type :\nJournal articles\n\nCited literature [27 references]\n\nhttp:\/\/hal.in2p3.fr\/in2p3-01518704\nContributor : Michel Lion <>\nSubmitted on : Friday, May 5, 2017 - 11:24:21 AM\nLast modification on : Friday, September 10, 2021 - 1:50:16 PM\nLong-term archiving on: : Sunday, August 6, 2017 - 12:28:37 PM\n\n### File\n\nPhysRevLett.118.162501.pdf\nFiles produced by the author(s)\n\n### Citation\n\nJ. Dudouet, A. Lemasson, G. Duch\u00eane, M. Rejmund, E. Cl\u00e9ment, et al.. $^{96}_{36}$Kr$_{60}$\u2013Low-Z Boundary of the Island of Deformation at N =60. Physical Review Letters, American Physical Society, 2017, 118, pp.165201. \u27e810.1103\/PhysRevLett.118.162501\u27e9. \u27e8in2p3-01518704\u27e9\n\nRecord views","date":"2021-09-24 04:15:14","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.35879045724868774, \"perplexity\": 3303.698280042217}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-39\/segments\/1631780057496.18\/warc\/CC-MAIN-20210924020020-20210924050020-00086.warc.gz\"}"} | null | null |
package protocol
import (
"errors"
"github.com/SmartPool/smartpool-client"
"github.com/ethereum/go-ethereum/common"
"math/big"
"time"
)
type testContract struct {
Registered bool
Registerable bool
SubmitFailed bool
VerifyFailed bool
SubmitTime *time.Time
IndexRequestedTime *time.Time
claim *testClaim
DelayedVerification bool
}
func newTestContract() *testContract {
return &testContract{false, false, false, false, nil, nil, nil, false}
}
func (c *testContract) Version() string {
return "1.0.0"
}
func (c *testContract) IsRegistered() bool {
return c.Registered
}
func (c *testContract) CanRegister() bool {
return c.Registerable
}
func (c *testContract) Register(paymentAddress common.Address) error {
c.Registered = true
return nil
}
func (c *testContract) SubmitClaim(claim smartpool.Claim, lastClaim bool) error {
c.claim = claim.(*testClaim)
if c.SubmitFailed {
return errors.New("fail")
}
t := time.Now()
c.SubmitTime = &t
return nil
}
func (c *testContract) GetShareIndex(claim smartpool.Claim) (*big.Int, *big.Int, error) {
t := time.Now()
c.IndexRequestedTime = &t
return big.NewInt(0), big.NewInt(100), nil
}
func (c *testContract) VerifyClaim(claimIndex *big.Int, shareIndex *big.Int, claim smartpool.Claim) error {
if c.VerifyFailed {
return errors.New("fail")
}
if c.DelayedVerification {
time.Sleep(50 * time.Millisecond)
}
return nil
}
func (c *testContract) GetLastSubmittedClaim() *testClaim {
return c.claim
}
func (c *testContract) NumOpenClaims() (*big.Int, error) {
return big.NewInt(0), nil
}
func (c *testContract) ResetOpenClaims() error {
return nil
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 970 |
Parker Technology Announces 27 New Customer Service Agreements in Q2
Indianapolis, IN—Parker Technology, the leading customer service platform for parking operators, signed 27 new deals in Q2. The new agreements include parking management companies across 13 states, in the following verticals: real estate, higher education, healthcare, and private operations.
"It's telling that in a time when the entire parking industry is under unprecedented financial pressure, we've had so many customers vote with their scarce dollars to sign with us in the last three months," said Brian Wolff, President & CEO of Parker Technology. "We are proud that our customers view our service as being essential to their operations in these challenging times."
Parker will implement its industry-leading customer service solution at these 27 parking facilities, to provide virtual 24/7 call center support. When parking patrons have difficulty completing parking transactions, they can instantly reach one of Parker's trained customer service professionals to resolve their issues quickly and efficiently. While operators' costs and revenue remain uncertain, Parker's solution helps them complete more successful transactions, and staff efficiently.
"We are excited to be partnering with Parker Technology in three new facilities, to provide the best possible customer service to our guests," said Chuck Stills, Vice President of Operations for Platinum Parking. "In our increasingly automated parking industry, it's more important than ever to deliver a human interaction when our guests need help. We've worked with Parker for several years now, and their service and technology platform are the best in the industry."
About Parker Technology
Parker Technology is a software and services company that helps parking facilities provide an excellent customer experience, by resolving issues for parking guests when they fail in the face of automated payment kiosks. Parker's patient, well-informed specialists virtually answer and resolve intercom calls 24/7 and is the only company in the parking industry that can deliver this service with face-to-face, two-way video communication. Putting this personal, human touch back into an automated situation enhances the customer experience, provides metrics to improve operational efficiencies, increases successful payments, and ensures customer service calls are answered. Learn more at www.helpmeparker.com. | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 3,286 |
\section{Introduction}
\label{sec.Intro}
In order to reduce as much as possible the numerical errors due to nonlinear convective terms,
it is possible to exploit the power of Lagrangian methods:
with this kind of algorithms, the new position and configuration of each element of the mesh is
recomputed at each timestep according to the local fluid velocity, so that we can closely
follow the fluid flow in a Lagrangian fashion.
In this way the nonlinear convective terms disappear and Lagrangian schemes
exhibit \textit{negligible numerical dissipation} at \textit{contact} waves and
material \textit{interfaces};
moreover they results to be \textit{Galilean and rotational invariant},
and they provide, without any additional effort, an automatic \textit{mesh refinement} feature
even when the cell count is maintained constant, simply by transporting the mesh elements wherever needed.
The use of Lagrangian methods dates back to the works of
\cite{Neumann1950, Wilkins1964} and then many further improvements have been introduced in literature;
we cite here only some few relevant historical examples and
review papers~\cite{munz94,CaramanaShashkov1998,LoubereShashkov2004,Despres2009,Maire2010,chengshu2,scovazzi2,Dobrev3,despres2017numerical,morgan2021origins}.
However, ensuring the high quality of a \textit{moving mesh}
over long simulation times is difficult, therefore a certain degree of flexibility should be
allowed in order to avoid mesh distortion, for example a slightly relaxed choice of the actual mesh
velocity w.r.t the real fluid velocity, as well as the freedom of not only \textit{moving} the
control volumes, but really \textit{evolving} their shapes and allowing topology and neighborhood changes.
This led to the introduction of Arbitrary-Lagrangian-Eulerian~(ALE) schemes
of \textit{direct}~\cite{ALELTS1D,ALELTS2D,boscheri2013arbitrary}
and \textit{indirect}~\cite{ReALE2010,ReALE2011,ReALE2015} type.
In particular, as stated and shown in~\cite{springel2010pur}, connectivity changes between
different time level constitute a valid
alternative to remeshing~\cite{ReALE2011, ReALE2015, ReALE2010} for preserving or restoring mesh quality
in a Lagrangian setting.
With this in mind, we present here a novel family of very \textit{high-order} direct
Arbitrary-Lagrangian-Eulerian (ALE) Discontinuous Galerkin (DG) schemes for
the solution of general nonlinear hyperbolic PDE systems on moving Voronoi meshes that are
\textit{regenerated} at each timestep and which explicitly allow \textit{topology changes} in time,
in order to benefit simultaneously from high-order methods, high quality grids and substantially
reduced numerical dissipation; this method has been introduced for the first time by the two authors in~\cite{gaburro2020high}.
The \textit{key ingredient} of our approach is the integration of a \textit{space-time conservation
formulation} of the governing PDE system over closed, non-overlapping \textit{space-time} control
volumes~\cite{boscheri2013arbitrary} that are constructed from the moving, \textit{regenerated},
Voronoi-type polygonal meshes which are centroid-based dual grids of the Delaunay triangulation of a
set of \textit{generator} points:
this leads to also consider what we refer to in this work as \textit{crazy degenerate} control
volumes, or \textit{space-time sliver} elements, that only arise when adopting a space-time framework,
and would not exist from a purely spatial point of view! Such degenerate elements provide a
clear formal way of handling mesh connectivity changes while preserving the high-order of accuracy of the numerical
method.
\subsection{Goals}
The goal of this book chapter is to briefly review this novel and promising approach based on high
order direct ALE schemes with topology changes, and to provide numerical evidence regarding the utility of
Lagrangian methods in general, and in particular of the new technique object of this work, by means of
simple and illustrative benchmarks.
\subsection{Structure}
The rest of this chapter is organized as follows.
We first introduce the class of equations of interest in Section~\ref{sec.pde}.
Next, we briefly summarize the main characteristics of the employed direct Arbitrary-Lagrangian-Eulerian scheme,
focusing in particular on the space-time approach and its extension to \textit{crazy} sliver elements,
whose formation, caused by topology
changes, will be also addressed in Section~\ref{sec.crazy}.
Then, the core of this work consists in providing numerical evidence
for \textit{(i)}~the key role of topology changes and sliver elements in a high-order moving mesh code,
and \textit{(ii)}~the clear advantages of Lagrangian schemes on widely adopted benchmark problems.
Finally, we give some conclusive remarks and an outlook towards future work in Section~\ref{sec.conclusion}.
\section{Hyperbolic partial differential equations}
\label{sec.pde}
In order to model a wide class of physical phenomena, we consider
a very general formulation of the governing equations, namely all those which can be described by
\begin{equation}
\label{eq.generalform}
\partial_t{\mathbf{Q}} + \nabla \cdot \mathbf{F}(\mathbf{Q}) + \mathbf{B}(\mathbf{Q}) \cdot \nabla \mathbf{Q} = \mathbf{S}(\mathbf{Q}),
\end{equation}
where $\mathbf{Q}$ is the vector of the conserved variables, $\mathbf{F}$ the non linear flux,
$\mathbf{B}\cdot \nabla \mathbf{Q}$ the nonconservative products, and $\mathbf{S}$ a nonlinear algebraic source term.
Many physical models can be cast in this form, from the simple shallow water
system, some multiphase flow models, the magnetohydrodynamics equations,
up to the Einstein field equations of general relativity (with appropriate reformulation) or the GPR
unified model of continuum mechanics, see for
example~\cite{fambri2017space,gaburro2018wellb,kemm2020simple,gaburro2021well,chiocchetti2021high,tavelli2020space,chiocchetti2020solver,chiocchetti2021high,gabriel2021unified,peshkov2021simulation,boscheri2022cell};
in this work, we will present illustrative results concerning the Euler equations of gasdynamics.
\section{Numerical method}
\label{sec.method}
In this Section we presents a concise description of our direct Arbitrary-Lagrangian-Eulerian (ALE)
Discontinuous Galerkin (DG) scheme on moving Voronoi-type meshes with topology changes; for any
additional details we refer to our recent paper~\cite{gaburro2020high}.
\medskip
At the beginning of the simulation, we discretize our moving domain by a centroid-based Voronoi-type
tessellation built from a set of generators (the orange points in Figure~\ref{fig.controlvolumes}),
and we represent our data, the conserved variables $\mathbf{Q}$,
via discontinuous high-order polynomials in each mesh polygon (we
indicate the degree of the polynomial representation by $P_N$).
Then, we let the generators move
with a velocity chosen to be as close as possible to the local fluid velocity, computed mainly from
a high-order approximation of their pure Lagrangian trajectories, with small corrections obtained
from a flow-adaptive mesh optimization technique. The positions of the generators are being
continuously updated, and thus their Delaunay triangulation may change at any timestep
and the same will hold for the dual polygonal tessellation.
Then, a space-time connection between two polygonal tessellations corresponding to two successive time levels
has to be established in order to
evolve the solution in time locally and integrate the governing PDE.
\subsection{Direct Arbitrary-Lagrangian-Eulerian schemes}
\begin{figure}[!bp]
{\includegraphics[width=0.33\linewidth]{illustration_3db1_grids.pdf}}%
{\includegraphics[width=0.33\linewidth]{illustration_3db2_spacetimecv.pdf}}%
{\includegraphics[width=0.33\linewidth]{illustration_3db3_spacetimesubtri.pdf}}%
\caption{Space time connectivity \textit{without} topology changes, main space-time control volume (middle) and a standard sub-space-time control volume (right).}
\label{fig.controlvolumes}
\end{figure}
The key idea of \textit{direct} ALE methods (in contrast to
\textit{indirect} ones) consists in connecting two tessellations by means of so-called
\textit{space-time control volumes} $C_i^n$, and recover the unknown solution at the new
timestep $\u_h^{n+1}$ \textit{directly} inside the new polygon $P_i^{n+1}$,
from the data available at the previous timestep $\u_h^{n}$ in $P_i^n$.
This is achieved through the \textit{integration}, over such control volumes, of the fluxes, the nonconservative
products and the source terms, by means of a high-order fully discrete predictor-corrector ADER
method~\cite{dumbser2008unified,gaburro2021posteriori}.
In this way, the need for any further remapping/remeshing steps is
totally eliminated. By adopting the tilde symbol for referring to space-time quantities,
our direct ALE scheme~\cite{gaburro2020high,gaburro2021unified} reads
\begin{equation}
\label{eqn.ALE-ADER}
\begin{aligned}
\int_{P_i^{n+1}}
\tilde{\phi}_k \u_h^{n+1} &=
\int_{P_{i}^n}
\tilde{\phi}_k \u_h^n
- \sum_{j} \int_{\partial C_{ij}^n}
\tilde{\phi}_k \mathcal{F}(\mathbf{q}_h^{n,-},\mathbf{q}_h^{n,+}) \cdot \mathbf{\tilde n}
+ \int_{C_i^n}
\tilde \nabla \tilde{\phi}_k \cdot \tilde{\mathbf{F}} (\mathbf{q}_h^n) \\
&+ \int_{C_i^n} \tilde{\phi}_k \left ( \mathbf{S}(\mathbf{q}_h^n) - \tilde{\mathbf{B}}(\mathbf{q}_h^n)\cdot \nabla \mathbf{q}_h^n \right),
\end{aligned}
\end{equation}
where $\tilde{\phi}_k$ is a set of moving space-time basis functions, while $\mathbf{q}_h^{n,+}$ and $\mathbf{q}_h^{n,-}$ are
high-order space-time extrapolated data computed through the ADER predictor.
Finally, $\mathcal{F}(\mathbf{q}_h^{n,-},\mathbf{q}_h^{n,+})$ is an ALE numerical flux function which takes into
account fluxes across space-time cell boundaries $\partial C_{ij}^n$ as well as jump terms
related to nonconservative products.
In particular, we adopt here a two-point path-conservative numerical flux function of Rusanov-type~\cite{Rusanov:1961a,pares2006numerical} ALE approximate flux function,
\begin{equation}
\begin{aligned}
\mathcal{F}(\mathbf{q}_h^{n,-},\mathbf{q}_h^{n,+}) \cdot \mathbf{\tilde n} = &
\frac{1}{2} \left( \tilde{\mathbf{F}}(\mathbf{q}_h^{n,+}) + \tilde{\mathbf{F}}(\mathbf{q}_h^{n,-}) \right) \cdot \mathbf{\tilde n}_{ij} -
\frac{1}{2} s_{\max} \left( \mathbf{q}_h^{n,+} - \mathbf{q}_h^{n,-} \right) \\
&+ \frac{1}{2} \left(\int \limits_{0}^{1} \tilde{\mbf{B}} \left(\mbf{\Psi}(\mathbf{q}_h^{n,-},\mathbf{q}_h^{n,+},s) \right)\cdot\mbf{n} \, d\mbf{x} \right)
\cdot\left(\mathbf{q}_h^{n,+} - \mathbf{q}_h^{n,-}\right),
\label{eq.rusanov}
\end{aligned}
\end{equation}
where $s_{\max}$ is the maximum eigenvalue of the \textit{ALE Jacobian matrices} evaluated
on the left and right of the space-time interface and the path $\mbf{\Psi}=\mbf{\Psi}(\mathbf{q}_h^-,\mathbf{q}_h^+, s)$ is a straight-line segment path
connecting $\mathbf{q}_h^{n,-}$ and $\mathbf{q}_h^{n,+}$.
We emphasize that the ALE Jacobian matrix is obtained by subtracting the local normal mesh velocity
from the diagonal entries
of the system matrix of the quasilinear form of the governing equations~\cite{toro2013riemann}
(the Jacobian of the interface-normal flux for conservative systems), thus, when the mesh
velocity is sufficiently close to the local fluid velocity,
the wavespeed estimates obtained from the eigenvalues are significantly reduced,
leading to a lower associated numerical dissipation than what would be mandated in the Eulerian context.
This, especially but not exclusively, in conjunction with complete approximate Riemann solvers~\cite{hllem},
explains the capability of tracking material interfaces and capturing contact
discontinuities proper of Lagrangian-type schemes.
\begin{figure}[!bp]
\includegraphics[width=0.33\linewidth]{illustration_3da1_edgeflip_nodes_big.pdf}%
\includegraphics[width=0.33\linewidth]{illustration_3da2_crazy1_big.pdf}%
\includegraphics[width=0.33\linewidth]{illustration_3da4_sliver_big.pdf}%
\caption{Space time connectivity \textit{with} topology changes, degenerate sub-space-time control volumes (middle) and \textit{crazy} sliver element (right).}
\label{fig.crazy}
\end{figure}
Next, in order to compute the integrals with high order of accuracy, complete knowledge of the
\textit{space-time connectivity} between two consecutive timesteps is required,
as opposed to only the \textit{spatial} information at the two time levels,
which would be enough for a low order scheme~\cite{springel2010pur}
or for indirect schemes~\cite{ReALE2010,ReALE2011,ReALE2015}.
When no topology changes occur, the space-time
geometrical information is easily constructed by connecting via straight line segments the
corresponding vertexes of each polygon, obtaining an oblique prism than can be further subdivided
into a set of triangular oblique sub-prisms on which quadrature points are readily available (see
Figures~\ref{fig.controlvolumes} and~\ref{fig.quadraturepoints}).
\subsection{Topology changes and \textit{crazy} sliver elements}
\label{sec.crazy}
\begin{figure}[!bp]
{\includegraphics[width=0.33\linewidth]{illustration_3dc1_quadraturevolume.pdf}}%
{\includegraphics[width=0.33\linewidth]{illustration_3dc2_quadraturesurface.pdf}}%
{\includegraphics[width=0.33\linewidth]{illustration_3dc3_quadraturesliver.pdf}}%
\caption{Space-time quadrature points for third order methods on standard elements (left), lateral faces (middle) and \textit{crazy} sliver elements (right).}
\label{fig.quadraturepoints}
\end{figure}
On the contrary, when a topology change occurs, as in Figure~\ref{fig.crazy}, i.e. the number of
edges, the shape, and the neighbors of a polygon evolve within two consecutive timesteps, the
space-time connection between the mesh elements gives raise to degenerate elements of two types:
\textit{(i)}~\textit{degenerate} \textit{sub}-space-time control volumes, where either the top or bottom
faces are degenerate triangles that are collapsed to a segment; \textit{(ii)}~and also
\textit{crazy sliver} space-time elements $S_i^n$.
The first type of degenerate elements does not
pose any problems, and was already treated in~\cite{gaburro2017direct}.
Instead, space-time sliver elements are a completely new type of control volume.
In particular, they do not exist neither at time $t^n$, nor
at time $t^{n+1}$, since they coincide with an edge of the tessellation at the old \textit{and} at the new
time levels, and for this reason have zero area in space at the two bounding time levels.
However, they have a \textit{non-negligible volume} in space-time.
The difficulties related to this kind of elements are due to the fact that
for them an initial condition is not clearly defined at time $t^n$, and that contributions across
these elements should not be lost at time $t^{n+1}$, in order to ensure conservation. All the
details on how to successfully extend our direct ALE scheme also to \textit{crazy} elements can be found in
our recent paper~\cite{gaburro2020high}.
We would like to emphasize that topology changes are fundamental for long time simulations in the
ALE framework, in order to avoid explicit data remap steps, and our \textit{crazy} sliver
elements represent a \textit{novel and formally grounded}
way to allow for a relatively simple space-time connection around a change of connectivity.
The numerical results shown in Section~\ref{sec.shu} provide a clear proof of the necessity of
topology changes already on the simple situation of the long time evolution of a stationary isentropic smooth vortex.
\subsection{ADER-ALE algorithm: the predictor step}
The \textit{predictor} step represents an essential ingredient
for obtaining high-order in time in a fully-discrete one-step procedure: it yields a \textit{local}
solution of the governing equations~\eqref{eq.generalform} \textit{in the small} $\mathbf{q}_h^n$, inside each
space-time element, including the \textit{crazy} elements. The solution is \textit{local} in the sense that
it is obtained by only considering the initial data in each polygon, the governing equations and the
geometry of $C_i^n$, without taking into account interactions between $C_i^n$ and its neighbors.
Such local solution is computed for each standard space-time control volume $C_i^n$
and for each \textit{crazy} control volume $S_i^n$, in the form of a high
order polynomial \textit{in space and in time}, which serves as a predictor solution,
to be used for evaluating all the integrals in the \textit{corrector} step~\eqref{eqn.ALE-ADER}, i.e. the
final update of the solution from $t^n$ to $t^{n+1}$.
\subsection{\textit{A posteriori} sub-cell FV limiter}
High-order schemes that can be seen as linear in the sense of Godunov~\cite{godunov}, may develop
spurious oscillations in presence of discontinuities. In order to prevent this phenomenon, in the
case of a DG discretization we adopt an \textit{a~posteriori} limiting procedure based on the MOOD paradigm
\cite{CDL1,ADERMOOD,gaburro2021posteriori}: we first apply our unlimited ALE-DG scheme everywhere, and then (\textit{a
posteriori}), at the end of each timestep, we check the reliability of the obtained solution in
each cell against physical and numerical admissibility criteria, such
as floating point exceptions, violation of positivity or violation of a relaxed discrete maximum
principle (and see~\cite{guermond2018second,KENAMOND2021110254} for further criteria).
Next, we mark as \textit{troubled} those cells where the DG solution cannot be accepted.
For the troubled cells we now repeat the time evolution by employing, instead of the DG scheme,
a more robust finite volume (FV) method. Moreover, in
order to maintain the accurate resolution of our original high-order DG scheme, which would be lost
when switching to a FV scheme, the FV scheme is applied on a \textit{finer sub-cell grid}
that accounts for recovering the optimal accuracy of the numerical method performing a reconstruction step.
\section{Numerical examples}
\label{sec.examples}
In order to provide simple and clear numerical evidence of the effectiveness of the proposed ALE scheme with
topology changes we consider here the well known Euler equations, that can be cast in the form~\eqref{eq.generalform} by choosing
\begin{equation}
\label{eulerTerms}
\mathbf{Q} = \left( \begin{array}{c} \rho \\ \rho u \\ \rho v \\ \rho E \end{array} \right), \quad
\mathbf{F} = \left( \begin{array}{ccc} \rho u & \rho v \\
\rho u^2 + p & \rho u v \\
\rho u v & \rho v^2 + p \\
u(\rho E + p) & v(\rho E + p)
\end{array} \right), \qquad \mathbf{B} = \mathbf{0}, \qquad \S = \mathbf{0}.
\end{equation}
The vector of conserved variables $\mathbf{Q}$ is composed of the fluid density $\rho$,
the momentum density vector $\rho \v=(\rho u, \rho v)$ and the total energy density $\rho E$;
next, the fluid pressure $p$ is computed using the equation of state for an ideal gas
\begin{equation}
\label{eqn.eos}
p = (\gamma-1)\left(\rho E - \frac{1}{2} \rho \mathbf{v}^2 \right),
\end{equation}
where $\gamma$ (in this work taken to be $\gamma = 7/5$) is the ratio of specific heats. For this choice
of equation of state, the adiabatic
speed of sound takes the form $c=\sqrt{{\gamma p}/{\rho}}$.
\medskip
In what follows we will present numerical results regarding the following notable features of Lagrangian schemes
and of our direct Arbitrary-Lagrangian-Eulerian method with variable topology:
\begin{description}
\item[{i.}] Flows characterized by strong differential rotations, for example vortices,
can be studied over very long periods only by conceding to the element motion the
additional freedom of introducing topology changes, see Section~\ref{sec.shu};
\item[{ii.}] The use of sliver elements allows to clearly define the space-time evolution of
the solutions in-between discrete time levels and achieves high-order of accuracy also
in presence of many topology changes, see Section~\ref{sec.order};
\item[{iii.}] Lagrangian schemes sharply capture \textit{shock}
waves thanks to the automatic refinement obtained at the shock locations without
needing to increment the number of mesh elements but simply because
the element density increases wherever needed, see Section~\ref{sec.sedov};
\item[{iv.}] Lagrangian schemes minimize dissipation of \textit{contact} discontinuities, by applying
reduced numerical dissipation when using approximate Riemann solvers. In a pure Lagrangian
context, schemes capable of capturing stationary discontinuities exactly will do the same
also for moving interfaces (since the mesh motion is specified to follow such features).
Moreover, even when such hard constraints are relaxed in
Arbitrary-Lagrangian-Eulerian methods and even using
simpler solvers like the Rusanov flux,
the bounding wavespeed estimates and the associated numerical dissipation can be much lower
than what would be mandated in the Eulerian context,
see Section~\ref{sec.sod};
\item[{v.}] Lagrangian schemes discretely preserve the Galilean and rotational invariance properties
of the governing equations,
so that they can better capture any events (such as the explosion-type problems reported in this work)
that may occur in superposition to a high-speed background flow, see Section~\ref{sec.sod}.
\end{description}
\subsection{Long time evolution of a Shu-type vortical equilibrium}
\label{sec.shu}
\begin{figure}[!bp]
\centering
\includegraphics[width=0.33\linewidth]{Shu_notopc_000}%
\includegraphics[width=0.33\linewidth]{Shu_notopc_005}%
\includegraphics[width=0.33\linewidth]{Shu_notopc_010}\\
\includegraphics[width=0.33\linewidth]{Shu_notopc_020}%
\includegraphics[width=0.33\linewidth]{Shu_notopc_030}%
\includegraphics[width=0.33\linewidth]{Shu_notopc_040}\\
\caption{Mesh evolution corresponding to the solution of the stationary rotating vortex of Section~\ref{sec.shu}
solved on a moving grid with \textit{fixed} topology. The mesh quality rapidly deteriorates: elements
are stretched, the timestep size is reduced, and even mesh-tangling occurs, which means that the simulation
may stop entirely.}
\label{fig.shu_rho_ie_notopc}
\end{figure}
\begin{figure}[!bp]
\centering
\includegraphics[width=0.33\linewidth]{Shu_rho_0}%
\includegraphics[width=0.33\linewidth]{Shu_iE_0}%
\includegraphics[width=0.33\linewidth]{Shu_iE_2}\\
\includegraphics[width=0.33\linewidth]{Shu_iE_5}%
\includegraphics[width=0.33\linewidth]{Shu_iE_10}%
\includegraphics[width=0.33\linewidth]{Shu_iE_100}\\
\includegraphics[width=0.33\linewidth]{Shu_iE_300}%
\includegraphics[width=0.33\linewidth]{Shu_iE_500}%
\includegraphics[width=0.33\linewidth]{Shu_rho_500}%
\caption{Stationary rotating vortex solved with our fourth order ALE-DG scheme.
Density contours at $t=0$ and $t=500$ and position of a bunch of highlighted
elements at different times. Note that the solution is well preserved for more than
\textit{eighty} complete rotation periods of the yellow elements and generator trajectories are perfectly circular.}
\label{fig.shu_rho_ie}
\end{figure}
\begin{figure}[!bp]
\centering
\includegraphics[width=0.495\linewidth]{Shu_traj}
\includegraphics[width=0.495\linewidth]{Shu_traj_r_y}
\caption{Stationary rotating vortex solved with our fourth order $P_3$ ALE-DG scheme on a moving Voronoi-type mesh of $957$
elements with dynamical change of connectivity and with the generators trajectories computed with fourth order of accuracy.
Left: We depict the trajectories (in Cartesian coordinates) of the generators of $3$ mesh elements
(those highlighted respectively in blue, violet and red) from time $t=0$ up to time $t=250$. During this
time interval the red mesh element completes $30$ revolutions about the origin. Right: we depict the $y$
coordinates of the $3$ generators (top) and their radial coordinates (bottom). We would like to emphasize that the
trajectories are circular (their radius is almost constant) for a very long evolution time.}
\label{fig.trajectory}
\end{figure}
As a first test we consider a smooth isentropic vortex flow defined as similarly to~\cite{HuShuVortex1999}.
The initial computational domain is the square $\Omega=[0;10]\times[0;10]$ and boundary conditions are of wall (slip) type everywhere.
The initial condition is given by some perturbations $\delta$ that are superimposed
onto a homogeneous background state $\mathbf{Q}_0=(\rho,u,v,p)=(1,0,0,1)$, assuming that the entropy perturbation is zero, i.e. $\delta S= 0$.
The perturbations for density and pressure are
\begin{equation}
\label{rhopressDelta}
\delta \rho = (1+\delta T)^{\frac{1}{\gamma-1}}-1, \quad \delta p = (1+\delta T)^{\frac{\gamma}{\gamma-1}}-1,
\end{equation}
with the temperature fluctuation $\delta T = -\frac{(\gamma-1)\epsilon^2}{8\gamma\pi^2}e^{1-r^2}$ and the vortex strength $\epsilon=5$.
The velocity field is specified by
\begin{equation}
\label{ShuVortDelta}
\left(\begin{array}{c} \delta u \\ \delta v \end{array}\right) = \frac{\epsilon}{2\pi}e^{\frac{1-r^2}{2}} \left(\begin{array}{c} -(y-5) \\ \phantom{-}(x-5) \end{array}\right).
\end{equation}
This is a stationary equilibrium of the system so the exact solution coincides with the initial condition at any time.
Preserving this kind of vortical solution over long simulation times with
minimal dissipation is a nontrivial task in a moving-mesh context.
To achieve this result, we propose the use of a very high-order scheme (here an ADER-DG method of order~$4$) in a
Lagrangian framework. We remark that the combination cannot be used with fixed topology,
or advanced remapping techniques,
because the quality of the moving mesh subject to this constraint quickly deteriorates,
as is clearly apparent in Figure~\ref{fig.shu_rho_ie_notopc},
where the simulation has to be stopped after about half a vortex rotation period.
This highlights the well-known fact that, for long time evolution, the mesh connectivity must be somehow updated.
In this work this is naturally achieved by means of space-time topology changes.
Further, Figure~\ref{fig.shu_rho_ie} demonstrates that the treatment of topology changes via high-order
integration over \textit{crazy} sliver elements is actually quite effective. Indeed one can note that
the solution is visually the same at the beginning of the simulation and $500$ seconds after, even on a
rather coarse mesh of only $957$ polygonal elements.
Moreover, we take advantage of this test case to also emphasize the high precision of the mesh
movement. The Voronoi-type polygonal cells, as well as the generator points, are
in fact can be observed to orbit along perfectly circular trajectories, as evidenced
in Figures~\ref{fig.shu_rho_ie} and~\ref{fig.trajectory}.
\subsubsection{Order of convergence}
\label{sec.order}
\begin{table}[!tp]
\caption{Stationary vortex test case with final time $t_f=10$.
We report here the order of convergence, on the variable $\rho$ in the $L_2$ norm, for our ALE-DG scheme
up to order 4 and the total number of \textit{crazy} sliver elements $S$ that have been originated during the simulations.
We would like to underline that the high-order of convergence is maintained also when many sliver elements appear in the mesh.}
\label{tab.Euler_order}
\begin{center}
\begin{tabular}{cccc | cccc | cccc }
\hline\noalign{\smallskip}
\multicolumn{4}{c}{$P_1 \rightarrow \mathcal{O}2$} & \multicolumn{4}{c}{$P_2\rightarrow \mathcal{O}3$} & \multicolumn{4}{c}{$P_3\rightarrow \mathcal{O}4$} \\
\noalign{\smallskip}\svhline\noalign{\smallskip}
Tot $S$ & $h$ & $\epsilon(\rho)$ & $\mathcal{O}(L_2)$\ \ & \ \ Tot $S$ & $h$ & $\epsilon(\rho)$ & $\mathcal{O}(L_2)$\ \ & \ \ Tot $S$ & $h$ & $\epsilon(\rho)$
& $\mathcal{O}(L_2)$ \\
\noalign{\smallskip}\hline\noalign{\smallskip}
7433\ \ & 2.8e-2\ \ & 1.6e-3 & - & 566\ \ & 1.0e-1\ \ & 8.4e-2 & - & 138 & 3.7e-1\ \ & 8.2e-3 & - \\
12748\ \ &2.1e-2\ \ & 8.5e-4 & 2.3 & 1316\ \ & 6.9e-2\ \ & 2.5e-2 & 2.9 & 192 & 3.2e-1\ \ & 4.3e-3 & 4.4 \\
20121\ \ &1.7e-2\ \ & 5.4e-4 & 2.1 & 2149\ \ & 5.2e-2\ \ & 1.1e-2 & 3.1 & 260 & 2.8e-1 & 2.7e-3\ \ & 3.3 \\
28802\ \ &1.4e-2\ \ & 3.7e-4\ & 2.1 & 3258\ \ & 4.2e-2\ \ & 5.6e-3\ & 2.9 & 328 \ & 2.5e-1\ \ & 1.7e-3 \ & 4.2 \\
\noalign{\smallskip}\hline\noalign{\smallskip}
\end{tabular}
\end{center}
\end{table}
Finally, this stationary test case allows to show numerically the order of convergence of the proposed ALE-DG scheme, reported here up to $4$.
We would like to emphasize that in Table~\ref{tab.Euler_order} we show the numerical errors obtained at time $t_f=10$,
when computations have been carried out for thousands of timesteps and thousands of \textit{crazy} sliver elements have appeared.
\subsection{Sedov explosion problem}
\label{sec.sedov}
\begin{figure}[!bp]
\centering
\includegraphics[width=0.495\linewidth]{Sedov_AlevsEul_M2
\includegraphics[width=0.495\linewidth]{Sedov_AlevsEul_M1
\caption{Sedov explosion problem. Comparison between the exact solution (black), the solution
obtained with a fourth order Eulerian $P_3$ DG method on the fine mesh $M2$
(red) and with our $P_3$ ALE-DG scheme both on $M2$ (blue) and $M1$ (green).
Our ALE scheme is more accurate than the Eulerian one even using coarser meshes.}
\label{fig.sedov_alevseul}
\end{figure}
This test problem is a classic benchmark in the literature~\cite{LoubereSedov3D} and describes the evolution of a
strong blast wave that is generated at the origin $\mathbf{O}=(x,y)=(0,0)$ of the computational
domain $\Omega(0)=[0;1.2]\times[0;1.2]$.
The difficulty of this benchmark is mainly due to the near zero pressure outer state
that may induce positivity-preservation problems. An exact solution based on self-similarity
arguments is available from~\cite{Sedov}.
The initial condition consists in a uniform density $\rho_0=1$ and a near zero pressure $p_0$ imposed
everywhere except in the cell $V_{or}$ containing the origin $\mathbf{O}$ where it is given by
\begin{equation}
p_{or} = (\gamma-1)\rho_0 \frac{E_{tot}}{|V_{or}|}, \quad \textnormal{ with } \
E_{tot} = 0.979264,
\label{eqn.p0.sedov}
\end{equation}
with $E_{tot}$ being the total energy concentrated in the cell containing the coordinate $\mbf{x}=\mathbf{0}$.
We set $p_0=10^{-9}$ and solve this numerical test with a fourth order $P_3$ DG scheme;
we employ a coarse mesh $M1$ made of $1345$ polygonal cells and a finer mesh $M2$ of $6017$ polygonal elements.
\begin{figure}[!bp]
\centering
\includegraphics[width=0.33\linewidth]{Sedov_P3_33_010}%
\includegraphics[width=0.33\linewidth]{Sedov_P3_33_050}%
\includegraphics[width=0.33\linewidth]{Sedov_P3_33_100}\\
\includegraphics[width=0.33\linewidth]{Sedov_P3_71_010}%
\includegraphics[width=0.33\linewidth]{Sedov_P3_71_050}%
\includegraphics[width=0.33\linewidth]{Sedov_P3_71_100}\\
\caption{Sedov explosion problem. In this figure we show the density evolution and the corresponding
mesh movement at different output times computed with our $P_3$ ALE-DG scheme on the mesh $M1$ (top) and $M2$ (bottom).}
\label{fig.sedov_evolution}
\end{figure}
\begin{figure}[!bp]
\centering
\includegraphics[width=0.495\linewidth]{Sedov_P3_71_rho
\includegraphics[width=0.495\linewidth]{Sedov_P3_71_3D
\caption{Sedov explosion problem. $3D$ density profile computed with our $P_3$ ALE-DG scheme
on $M2$. In particular, on the right, we highlight in red the so-called \textit{troubled} cells
marked by our detector on which the \textit{a posteriori} FV limiter has been employed.
We make use of this image to further emphasize the robustness of our ALE schemes with topology
changes also in the presence of strong shock waves and near-zero pressure outer states.}
\label{fig.sedov_3d}
\end{figure}
The density profiles are shown in Figure~\ref{fig.sedov_alevseul} for various output times $t~=~0.1, 0.5, 1.0$.
The obtained results are in perfect agreement with the reference solution and the symmetry is very good despite using an unstructured grid,
as opposed to a regular one built in polar coordinates.
Also, one can note that the regularization procedure applied to the mesh elements does not compromise the natural
expansion of the central cells expected in such an explosion problem.
Moreover, one can refer to Figure~\ref{fig.sedov_3d} for a comparison between our numerical solution (scatter plot) and the exact solution:
the position of the shock wave and the density peak are perfectly captured.
\begin{figure}[!bp]
\centering
\includegraphics[width=0.495\linewidth]{Sod_P2_EUL}%
\includegraphics[width=0.495\linewidth]{Sod_P2_ALE}\\
\includegraphics[width=0.495\linewidth]{Sod_P3_EUL}%
\includegraphics[width=0.495\linewidth]{Sod_P3_ALE}
\caption{Traveling Sod explosion problem. $3D$ density profile (z-axis) and limiter
activation (red cells), over a domain located in $[8.9;11.1]\times[-1.1;1.1]$ at the final time $t_f=0.25$,
obtained with $P_2$ and $P_3$ ADER-DG schemes run on a fixed Eulerian mesh (left) and our
direct ALE framework with topology changes (right). The difference on the numerical dissipation
between the Eulerian and the Lagrangian schemes is quite evident. We clarify that these
results are obtained solving the classical Sod explosion problem over a
high speed moving background.}
\label{fig.sod_alevseul3D}
\end{figure}
In particular, we have chosen this test case in order to emphasize that Lagrangian schemes show a superior resolution
w.r.t. Eulerian ones even when both are compared at very high-order of accuracy,
and furthermore that our direct ALE scheme results more accurate then the Eulerian method,
even on a mesh ($M1$) that is coarser by a factor of two
with respect to the finer mesh $M2$.
Finally, we refer to Figure~\ref{fig.sedov_3d} for the behavior of our \textit{a posteriori} sub-cell finite volume limiter,
which activates only where the shock wave is located and is able to avoid any spurious oscillations or positivity problems,
as can be noticed from the precise $3D$ density profile shown in Figure~\ref{fig.sedov_3d}.
\subsection{Traveling Sod-type explosion problem}
\label{sec.sod}
\begin{figure}[!bp]
\includegraphics[width=0.495\linewidth]{Sod_p3FIXED_scatter
\includegraphics[width=0.495\linewidth]{Sod_p3_scatter}\\
\caption{Sod explosion problem: fixed background (left) and high speed traveling background (right).
Comparison between the $P_3$ DG schemes on fixed Eulerian meshes (red) and in the moving-mesh ALE framework (blue).
This numerical results clearly explain that the Lagrangian schemes allow to obtain minimally
dissipative results not only in a vanishing background flow, but even in a high speed one, and therefore
that Lagrangian
methods discretely preserve the Galilean invariance of the equations.
On the contrary the influence of strong background flows on the solution obtained with Eulerian schemes is immediately apparent.}
\label{fig.sod_alevseul}
\end{figure}
The explosion problem can be seen as a multidimensional extension of the classical Sod test case.
Here, we consider as computational domain a square of dimension $[-1.1;1.1]\times[-1.1;1.1]$ covered
with a mesh made of $4105$ Voronoi-type elements, and the initial condition is composed of two
different states, separated by a discontinuity at radius $r_d=0.5$
\begin{equation}
\begin{cases}
\rho_L = 1, \quad \mathbf{u}_L, \quad p_L = 1, \quad &\left\| \mbf{x} \right\| \leq r_d \\
\rho_R = 0.125, \quad \mathbf{u}_R, \quad p_R = 0.1, \quad & \left\| \mbf{x} \right\| > r_d .
\end{cases}
\end{equation}
In addition, we aim at capturing the evolution of this explosion over a very high speed moving
background (much higher than the speed of sounds): we impose $\mathbf{u}_L=\mathbf{u}_R=40$, so that
at the final simulation time $t_f=0.25$ the square $[-1;1]\times[-1;1]$ will have been displaced by $5$ times its initial size.
We would like to underline that this test problem involves three different waves, therefore it allows
each ingredient of our Lagrangian scheme to be properly checked. Indeed, we have \textit{(i)}~one cylindrical
shock wave that is running towards the external boundary: our high-order scheme does not exhibit spurious
oscillations thanks to the \textit{a posteriori} sub-cell finite volume limiter;
\textit{(i)}~a rarefaction fan traveling in the opposite direction, which is well captured
thanks to the high-order of accuracy of the DG scheme; and
\textit{(iii)}~an outward-moving contact wave, which is well resolved thanks to the Lagrangian nature
of our scheme, in which, since the mesh moves together with the fluid flow, we can introduce a
minimal dissipation when computing approximate Riemann fluxes.
In addition, the high speed moving background allows to show the \textit{translational invariance}
property of the Lagrangian schemes that indeed perfectly captures the three waves even when the
explosion solution is moving at high speed, while the Eulerian scheme is heavily affected by the increased numerical dissipation.
Numerical evidence of the above statements can be found in Figure~\ref{fig.sod_alevseul3D};
moreover, in Figure~\ref{fig.sod_alevseul} we show that for this mild explosion
it is really the background motion that requires the use of Lagrangian schemes,
which, while still useful, would be instead not fundamental on a fixed background.
Finally, we want to remark that, despite the very high dissipation associated with the high base convective speed,
the overall symmetry of the solution, even in the Eulerian case, is not entirely compromised,
thanks to the use of polygonal elements (see~\cite{boscheriAFE2022} for further discussion on the benefits of adopting polygonal meshes).
\section{Conclusion and outlook}
\label{sec.conclusion}
The accuracy of our results clearly show that the new combination of very high-order schemes with regenerated meshes,
that allow topology changes, may open new perspectives in the fundamental research field of Lagrangian methods.
We would like to remark that the chosen simple test cases can be seen as prototypes of classical
difficulties in astrophysical applications. Indeed, we have proposed here a method able to deal
with long time simulation of vortical phenomena, as those necessary for the study of gas clouds evolving around black holes and neutron stars,
and events, like explosions or interactions with near zero pressure states,
occurring in superposition with high speed background flows, as for supersonic or relativistic jets originating from
proto-planetary nebulae, binary stars or nuclei of active galaxies.
Future developments of this work will mainly concern the improvements of its robustness and effectiveness
through mesh optimization and smoothing techniques~\cite{re2017interpolation,anderson2018high,morgan2018reducing,dobrev2020simulation,wang2015high} and structure preserving algorithms~\cite{klingenberg2019arbitrary,Kapelli2014,castro2008well,dumbser2019glm,dumbser2020numerical,abgrall2022reinterpretation,gaburro2022high} so that future applications will effectively target in particular supersonic flows
in aerodynamics~\cite{antoniadis2012high,tsoutsanis2015comparison} and
astrophysics~\cite{gaburro2021well,Torsion2019,olivares2022new}, as well as
fluid-structure interaction problems~\cite{basting2017extended,kikinzon2018establishing,durrwachter2021efficient}.
\begin{acknowledgement}
E.~Gaburro is member of the CARDAMOM team at the Inria center of the University of Bordeaux in France and S.~Chiocchetti is member of the INdAM GNCS group in Italy.
E.~Gaburro gratefully acknowledges the support received from the European Union's Horizon 2020 Research and Innovation Programme under the Marie Skłodowska-Curie Individual Fellowship \textit{SuPerMan}, grant agreement No. 101025563. S.~Chiocchetti acknowledges the support obtained by the Deutsche Forschungsgemeinschaft (DFG) via the project DROPIT, grant no. GRK 2160/2.
\end{acknowledgement}
\bibliographystyle{plain}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 1,084 |
Q: Adding social authentication to site Django==1.10.5. i have isntalled - pip install python-social-auth==0.2.12. Then addes social.apps.djang_app.default to the INSTALLED_APP settings. After wanted to sync python-social-auth model with Database
python projectname\manage.py migrate
But i got an error: AppRegistryNotReady:Apps aren't loaded yet
A: 0.2.12 is outdated and the library itself is deprecated.
Consider trying newer version or switching to django-allauth instead.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 6,713 |
Veep producer Stephanie Laing parlayed a job as a bank teller into an Emmy Award–winning career in TV. We asked what her secrets are to finding—and seizing—career-changing breaks.
MARIE CLAIRE: You were a bank teller in Ohio when you got your big break in TV. How did that happen?
STEPHANIE LAING: One day one of my regulars, who worked in film and TV, said he had a truck commercial coming up and asked if I would like to work on it. So I drove the truck from the dealer to the set, washed it, and drove it back. That was it. But I was hooked. When I quit my job at the bank, they tried to keep me with an 11-cents-anhour raise. I said, "No, thanks!"
MC: How did you get from there to Hollywood?
SL: I worked on three or four commercials, and then a Joe Pesci movie. After that, I pretty much followed the circus to Los Angeles. Relocating was a huge gamble. I had $1,200 to my name and a six-week job paying $200 a week. I remember buying gas with dimes! But I stuck it out because I really loved what I was doing. And I wasn't afraid to go home to Cincinnati for a few months when things got tight. That never felt like failure.
MC: What advice would you give to anyone weighing a career change— but scared to make a move?
SL: It's OK to be afraid—that's when good stuff happens. Keep your eye on the future, but remember to stay in the present. That energy will attract the person who'll bring you your next career opportunity. It's about the journey as much as the destination.
MC: But what if you feel stuck?
SL: Think about why you feel stuck. Is it the job or the people around you? I've had jobs running lowly errands but had a blast because of the atmosphere. I've also had high-up producer gigs that felt toxic because the people there were unpleasant. Don't try and find a thing you'd rather do but people you'd rather be around. That way, no matter what you do, you'll do it well, and that will lead to more opportunities.
Stephanie Cutter: "I Know How to Throw a Punch" | {
"redpajama_set_name": "RedPajamaC4"
} | 1,897 |
{{Taxobox début | animal | Adelpha serpa | Adelpha serpa.jpg| Chenille dAdelpha serpa}}
Adelpha serpa''' est une espèce de papillons de la famille des Nymphalidae, sous famille des Limenitidinae et du genre des Adelpha.
Dénomination Adelpha serpa a été décrit par Jean-Baptiste Alphonse Dechauffour de Boisduval en 1836 sous le nom initial d' Heterochroa serpa.
Sous-espèces Adelpha serpa serpa; présent au Paraguay, en Argentine et dans le sud-est du Brésil.Adelpha serpa celerio (Bates, 1864); présent au Mexique, eu Guatemala et au VenezuelaAdelpha serpa diadochus Fruhstorfer, 1915; présent au Pérou, au Brésil, en Guyana et en Guyane.Adelpha serpa duiliae Fruhstorfer, 1913; présent en Équateur.
Noms vernaculaires Adelpha serpa celerio se nomme Celerio Sister en anglais.
Description Adelpha serpa est un papillon d'une envergure de à au Costa Rica et de à en Équateur. Le dessus, marron, est orné d'une tache orange postdiscale aux ailes antérieures proche du bord costal et d'une bande blanche allant aux ailes postérieures de l'angle anal au bord costal et se continuant aux ailes antérieures sur plus de la moitié de l'aire discale.
Le revers est marron clair avec la même bande blanche que sur le dessus, entourée d'une étroite bande ocre aux ailes postérieures. Le bord costal des ailes antérieures est orné dans l'aire discale d'un groupe de trois marques bleutées bordés de noir.
Chenille
La chenille est verte tachée de noir et de cuivré et développe de grandes épines.
Biologie
Écologie et distribution Adelpha serpa'' est présent au Mexique, eu Guatemala, au Venezuela, présent en Équateur, au Paraguay, en Argentine, au Pérou, au Brésil, en Guyana et en Guyane.
Biotope
Protection
Pas de statut de protection particulier.
Notes et références
Annexes
Articles connexes
Lépidoptère
Limenitidinae
Lépidoptère (nom scientifique)
Limenitidinae
Lépidoptère d'Amérique | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 2,344 |
import json
from flask import (Flask, render_template, redirect,
url_for, make_response, request,
flash)
from options import DEFAULTS
app = Flask(__name__)
app.secret_key = 'fsdaf24j32kjf@@$)$#*#lfdf'
def get_saved_data():
try:
data = json.loads(request.cookies.get('character'))
except TypeError:
data = {}
return data
@app.route('/')
def index():
return render_template('index.html', saves=get_saved_data())
@app.route('/builder')
def builder():
return render_template(
'builder.html',
saves=get_saved_data(),
options=DEFAULTS
)
@app.route('/save', methods=['POST'])
def save():
flash("Great! Looks awesome!")
response = make_response(redirect(url_for('builder')))
data = get_saved_data()
data.update(dict(request.form.items()))
response.set_cookie('character', json.dumps(data))
return response
app.run(debug=True, host='0.0.0.0')
| {
"redpajama_set_name": "RedPajamaGithub"
} | 1,180 |
module Jekyll
class RemoveTag < Liquid::Tag
def initialize(tag_name, text, tokens)
super
end
def render(context)
"<i class='icon-remove-sign' style='color:#D04437;'></i>"
end
end
end
Liquid::Template.register_tag('remove', Jekyll::RemoveTag)
| {
"redpajama_set_name": "RedPajamaGithub"
} | 7,499 |
{"url":"https:\/\/www.isr-publications.com\/jnsa\/articles-6678-a-natural-selection-of-a-graphic-contraction-transformation-in-fuzzy-metric-spaces","text":"# A natural selection of a graphic contraction transformation in fuzzy metric spaces\n\nVolume 11, Issue 2, pp 218--227\nPublication Date: January 23, 2018 Submission Date: July 02, 2017 Revision Date: November 14, 2017 Accteptance Date: December 08, 2017\n\u2022 1799 Views\n\n### Authors\n\nHanan Alolaiyan - Department of Mathematics, King Saud University, Saudi Arabia. Naeem Saleem - Department of Mathematics, University of Management and Technology, Lahore, Pakistan. Mujahid Abbas - Department of Mathematics, Government College University, Lahore, Pakistan. - Department of Mathematics, King Abdulaziz University, P. O. Box 80203, Jeddah 21589, Saudi Arabia.\n\n### Abstract\n\nIn this paper, we study sufficient conditions to find a vertex $v$ of a graph such that $Tv$ is a terminal vertex of a path which starts from $v,$ where $T$ is a self graphic contraction transformation defined on the set of vertices. Some examples are presented to support the results proved herein. Our results widen the scope of various results in the existing literature.\n\n### Share and Cite\n\n##### ISRP Style\n\nHanan Alolaiyan, Naeem Saleem, Mujahid Abbas, A natural selection of a graphic contraction transformation in fuzzy metric spaces, Journal of Nonlinear Sciences and Applications, 11 (2018), no. 2, 218--227\n\n##### AMA Style\n\nAlolaiyan Hanan, Saleem Naeem, Abbas Mujahid, A natural selection of a graphic contraction transformation in fuzzy metric spaces. J. Nonlinear Sci. Appl. (2018); 11(2):218--227\n\n##### Chicago\/Turabian Style\n\nAlolaiyan, Hanan, Saleem, Naeem, Abbas, Mujahid. \"A natural selection of a graphic contraction transformation in fuzzy metric spaces.\" Journal of Nonlinear Sciences and Applications, 11, no. 2 (2018): 218--227\n\n### Keywords\n\n\u2022 Graphic contraction\n\u2022 fuzzy metric space\n\u2022 natural selection\n\n\u2022 \u00a047H10\n\u2022 \u00a047H04\n\u2022 \u00a047H07\n\u2022 \u00a054H25\n\u2022 \u00a054C60\n\n### References\n\n\u2022 [1] M. Abbas, T. Nazir , Common fixed point of a power graphic contraction pair in partial metric spaces endowed with a graph, Fixed Point Theory Appl., 2013 (2013 ), 8 pages.\n\n\u2022 [2] M. Abbas, T. Nazir, S. Radenovi\u0107 , Common fixed points of four maps in partially ordered metric spaces, Appl. Math. Lett., 24 (2011), 1520\u20131526.\n\n\u2022 [3] M. Abbas, B. E. Rhoades , Fixed point theorems for two new classes of multivalued mappings , Appl. Math. Lett., 22 (2009), 1364\u20131368.\n\n\u2022 [4] A. Azam , Coincidence points of mappings and relations with applications , Fixed Point Theory Appl., 2012 (2012 ), 9 pages.\n\n\u2022 [5] A. George, P. Veeramani , On some results in fuzzy metric spaces, Fuzzy Sets and Systems, 64 (1994), 395\u2013399.\n\n\u2022 [6] A. George, P. Veeramani , On some results of analysis for fuzzy metric spaces , Fuzzy Sets and Systems, 90 (1997), 365\u2013368.\n\n\u2022 [7] M. Grabiec , Fixed points in fuzzy metric spaces, Fuzzy Sets and Systems, 27 (1988), 385\u2013389.\n\n\u2022 [8] V. Gregori, S. Romaguera , Some properties of fuzzy metric spaces, Fuzzy Sets and Systems, 115 (2000), 485\u2013489\n\n\u2022 [9] V. Gregori, A. Sapena , On fixed-point theorems in fuzzy metric spaces , Fuzzy Sets and Systems, 125 (2002), 245\u2013252.\n\n\u2022 [10] G. Gwozdz-Lukawska, J. Jachymski , IFS on a metric space with a graph structure and extensions of the Kelisky-Rivlin theorem, J. Math. Anal. Appl., 356 (2009), 453\u2013463.\n\n\u2022 [11] H. B. Hosseini, R. Saadati, M. Amini , Alexandroff theorem in fuzzy metric spaces, Math. Sci. Res. J., 8 (2004), 357\u2013361.\n\n\u2022 [12] J. Jachymski , The contraction principle for mappings on a metric space with a graph , Proc., Amer. Math. Soc., 136 (2008), 1359\u20131373.\n\n\u2022 [13] J. Jachymski, I. J J\u00f3\u017awik , Nonlinear contractive conditions: a comparison and related problems, Banach Center Publ., 77 (2007), 123\u2013146.\n\n\u2022 [14] I. Kramosil, J. Mich\u00e1lek, Fuzzy metric and statistical metric spaces, Kybernetika, 11 (1975), 336\u2013344.\n\n\u2022 [15] J. J. Nieto, R. L. Pouso, R. Rodr\u00edguez-L\u00f3pez , Fixed point theorems in ordered abstract spaces , Proc. Amer. Math. Soc., 135 (2007), 2505\u20132517.\n\n\u2022 [16] A. C. M. Ran, M. C. B. Reurings, A fixed point theorem in partially ordered sets and some applications to matrix equations, Proc. Amer. Math. Soc., 132 (2004), 1435\u20131443.\n\n\u2022 [17] J. Rodr\u00edguez-L\u00f3pez, S. Romaguera , The Hausdorff fuzzy metric on compact sets , Fuzzy Sets and Systems, 147 (2004), 273\u2013283.\n\n\u2022 [18] N. Saleem, B. Ali, M. Abbas, Z. Raza , Fixed points of Suzuki type generalized multivalued mappings in fuzzy metric spaces with applications , Fixed Point Theory Appl., 2015 (2015), 18 pages.\n\n\u2022 [19] B. Schweizer, A. Sklar , Statistical metric spaces , Pacific J. Math., 10 (1960), 314\u2013334.\n\n\u2022 [20] L. A. Zadeh , Fuzzy Sets, Informations and control, 8 (1965), 338\u2013353.","date":"2020-09-23 10:33:56","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.7572946548461914, \"perplexity\": 4096.670875024641}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2020-40\/segments\/1600400210616.36\/warc\/CC-MAIN-20200923081833-20200923111833-00518.warc.gz\"}"} | null | null |
\section{Insertion Channel}
%
\subsection{Proof of Lemma \ref{lem:ins_H_I_Y}} \label{proof:ins_H_I_Y}
We begin by noting that $\frac{M_n}{n} \to (1+i)$ almost surely, due to the strong law of large numbers.
We have
\begin{equation}
\label{eq:split_pIY_ub}
\begin{split}
&\frac{1}{n} H_P(I^{M_n}|Y^{M_n}) = \expec\left[ -\frac{1}{n} \log P(I^{M_n}|Y^{M_n})
\cdot \left(\mathbf{1}_{\left\{\frac{M_n}{n} \in (1+i-\e, 1+i+\e)\right\}} + \mathbf{1}_{\left\{\frac{M_n}{n} \notin (1+i-\e, 1+i+\e)\right\}}\right) \right]\\
& = \expec\left[ -\frac{1}{n} \log \frac{P(I^{M_n}, Y^{M_n})}{P(Y^{M_n})} \cdot \mathbf{1}_{\left\{\frac{M_n}{n} \in (1+i-\e, 1+i+\e)\right\}}\right] +\expec\left[-\frac{1}{n} \log P(I^{M_n}|Y^{M_n}) \cdot \mathbf{1}_{\left\{\frac{M_n}{n} \notin (1+i-\e, 1+i+\e)\right\}} \right] \\
&\leq \expec\left[ -\frac{1}{n} \log \frac{P(I^{n(1+i+\e)}, Y^{n(1+i+\e)})}{P(Y^{n(1+i-\e)})} \cdot \mathbf{1}_{\left\{\frac{M_n}{n} \in (1+i-\e, 1+i+\e) \right\}}\right] +\expec\left[-\frac{1}{n} \log P(I^{M_n}|Y^{M_n}) \cdot \mathbf{1}_{\left\{ \frac{M_n}{n} \notin (1+i-\e, 1+i+\e) \right\}} \right].\\
&= \expec\left[ -\frac{1}{n} \log \frac{P(I^{n(1+i+\e)}, Y^{n(1+i+\e)})}{P(Y^{n(1+i-\e)})} \right] - \expec\left[ -\frac{1}{n} \log \frac{P(I^{n(1+i+\e)}, Y^{n(1+i+\e)})}{P(Y^{n(1+i-\e)})} \cdot \mathbf{1}_{\left\{\frac{M_n}{n} \notin (1+i-\e, 1+i+\e) \right\}} \right]\\
& \qquad \qquad \qquad \qquad \qquad + \expec\left[-\frac{1}{n} \log P(I^{M_n}|Y^{M_n}) \cdot \mathbf{1}_{\left\{ \frac{M_n}{n} \notin (1+i-\e, 1+i+\e) \right\}} \right].
\end{split}
\end{equation}
We first examine the third term in \eqref{eq:split_pIY_ub}. The size of the support of $-\frac{1}{n} \log P(I^{M_n}|Y^{M_n})$ is at most $2^{2n}$, since $I^{M_n}$
is a binary sequence of length at most $2n$. Hence, from Lemma \ref{lem:support}, $\left\{ -\frac{1}{n} \log P(I^{M_n}|Y^{M_n}) \right\}_{n \geq 1}$
is uniformly integrable. From Lemma \ref{lem:unif_equiv}, for any $\epsilon >0$, there exists some $\delta>0$
\be \label{eq:pIY_small}
\expec\left[-\frac{1}{n} \log P(I^{M_n}|Y^{M_n}) \cdot \mathbf{1}_{\left\{ \frac{M_n}{n} \notin (1+i-\e, 1+i+\e) \right\}} \right] < \epsilon
\ee
whenever $\text{Pr}\left( \left\{ \frac{M_n}{n} \notin (1+i-\e, 1+i+\e) \right\} \right) < \delta$. Since $\frac{M_n}{n} \to (1+i)$ almost surely,
$\text{Pr}\left( \left\{ \frac{M_n}{n} \notin (1+i-\e, 1+i+\e) \right\} \right)$ is less than $\delta$ for all sufficiently large $n$.
Thus \eqref{eq:pIY_small} is true for all sufficiently large $n$. Similarly, the third term can be shown to be smaller than $\e$ for all sufficiently large $n$. Therefore, for all sufficiently large $n$, \eqref{eq:split_pIY_ub} becomes
\be
\begin{split}
\frac{1}{n} H_P(I^{M_n}|Y^{M_n}) &\leq \frac{H_P(I^{n(1+i+\e)}, Y^{n(1+i+\e)})- H(Y^{n(1+i-\e)})}{n} + \e \\
&= (1+i- \e) \frac{H_P(I^{n(1+i-\e)}| Y^{n(1+i-\e)} )}{n(1+i-\e)} +
\frac{1}{n}H_P(I_{n(1+i-\e)+1}^{n(1+i+\e)}, Y_{n(1+i-\e)+1}^{n(1+i+\e)}| I^{n(1+i-\e)}, Y^{n(1+i-\e)} ) + \e\\
& \stackrel{(a)}{\leq} (1+i- \e) \frac{H_P(I^{n(1+i-\e)}| Y^{n(1+i-\e)} )}{n(1+i-\e)} + 4\epsilon + \epsilon.
\end{split}
\ee
where $(a)$ holds because $I_{n(1+i-\e)+1}^{n(1+i+\e)}$ and $Y_{n(1+i-\e)+1}^{n(1+i+\e)}$ can each take on at most $2^{2n\e}$ different values.
Hence
\[
\limsup_{n \to \infty} \frac{1}{n} H_P(I^{M_n}|Y^{M_n}) \leq 5\epsilon+ (1+i+\e)\limsup_{m \to \infty} \frac{1}{m} H_P(I^m|Y^m).
\]
Since $\epsilon>0$ is arbitrary, we let $\epsilon \to 0$ to obtain
\be \label{eq:limsup_ub}
\limsup_{n \to \infty} \frac{1}{n} H_P(I^{M_n}|Y^{M_n}) \leq (1+i) \limsup_{m \to \infty} \frac{1}{m} H_P(I^m|Y^m).
\ee
Using steps similar to \eqref{eq:split_pIY_ub}, we have
\begin{equation}
\label{eq:split_pIY_lb}
\begin{split}
&\frac{1}{n} H_P(I^{M_n}|Y^{M_n}) = \expec\left[-\frac{1}{n} \log \frac{P(I^{M_n}, Y^{M_n})}{P(Y^{M_n})} \cdot \mathbf{1}_{\left\{\frac{M_n}{n} \in (1+i-\e, 1+i+\e)\right\}}\right] +\expec \left[-\frac{1}{n} \log P(I^{M_n}|Y^{M_n}) \cdot \mathbf{1}_{\left\{\frac{M_n}{n} \notin (1+i-\e, 1+i+\e)\right\}} \right] \\
&\geq \expec\left[ -\frac{1}{n} \log \frac{P(I^{n(1+i-\e)}, Y^{n(1+i-\e)})}{P(Y^{n(1+i+\e)})} \cdot \mathbf{1}_{\left\{\frac{M_n}{n} \in (1+i-\e, 1+i+\e) \right\}}\right] +\expec\left[-\frac{1}{n} \log P(I^{M_n}|Y^{M_n}) \cdot \mathbf{1}_{\left\{ \frac{M_n}{n} \notin (1+i-\e, 1+i+\e) \right\}} \right].\\
&= \expec\left[ -\frac{1}{n} \log \frac{P(I^{n(1+i-\e)}, Y^{n(1+i-\e)})}{P(Y^{n(1+i+\e)})} \right] - \expec\left[ -\frac{1}{n} \log \frac{P(I^{n(1+i-\e)}, Y^{n(1+i-\e)})}{P(Y^{n(1+i+\e)})} \cdot \mathbf{1}_{\left\{\frac{M_n}{n} \notin (1+i-\e, 1+i+\e) \right\}} \right] \\
& \qquad \qquad \qquad \qquad \qquad + \expec\left[-\frac{1}{n} \log P(I^{M_n}|Y^{M_n}) \cdot \mathbf{1}_{\left\{ \frac{M_n}{n} \notin (1+i-\e, 1+i+\e) \right\}} \right] .
\end{split}
\end{equation}
Using arguments identical to the ones following \eqref{eq:split_pIY_ub}, one can show that the last two terms of \eqref{eq:split_pIY_lb} are smaller than $\epsilon$ in absolute value for all sufficiently large $n$, leading to
\be
\label{eq:limsup_lb}
\limsup_{n \to \infty} \frac{1}{n} H_P(I^{M_n}|Y^{M_n}) \geq (1+i) \limsup_{m \to \infty} \frac{1}{m} H_P(I^m|Y^m).
\ee
Combining \eqref{eq:limsup_ub} and \eqref{eq:limsup_lb} completes the proof of the lemma.
\qed
\subsection{Proof of Lemma \ref{lem:lim_HI_Y}} \label{proof:lim_HI_Y}
We have
\be
\frac{1}{m} H_P(I^{m}|Y^{m}) = \frac{1}{m} \sum_{j=1}^m H_P(I_j|I^{j-1}, Y^m) \leq \frac{1}{m} \sum_{j=1}^m H_P(I_j|I_{j-1}, Y_{j}, Y_{j-1}, Y_{j-2})
\ee
where the inequality holds because conditioning cannot increase the entropy.
Therefore
\be
\limsup_{m \to \infty} \frac{1}{m} H_P(I^{m}|Y^{m}) \leq \limsup_{m \to \infty} \frac{1}{m} \sum_{j=1}^m H_P(I_j|I_{j-1}, Y_{j}, Y_{j-1}, Y_{j-2})=
\lim_{j \to \infty} H_P(I_j|I_{j-1}, Y_{j}, Y_{j-1}, Y_{j-2}),
\ee
provided the limit exists. We now show that $\lim_{j \to \infty} H_P(I_j|I_{j-1}, Y_{j}, Y_{j-1}, Y_{j-2})$ exists and is given by \eqref{eq:hI_limit}.
From Proposition \ref{prop:ins_sy}, the process $\{\mathbf{I}, \mathbf{Y}\}$ is characterized by a Markov chain with
state at time $j$ given by $(I_{j}, Y_{j}, Y_{j-1})$. For any $\e >0$, the distribution $P(I_{j}, Y_{j}, Y_{j-1})$
is at most $\e$ (in total variation norm) from the stationary joint distribution $\pi$ given by \eqref{eq:stat_pi} for all sufficiently large $j$.
The conditional distribution $P(I_j, Y_j|I_{j-1}, Y_{j-2}, Y_{j-1})$ is given by \eqref{eq:IY_dist}. Due to the continuity of the entropy function,
this implies
\[ \lim_{j \to \infty} H_P(I_j|I_{j-1}, Y_{j}, Y_{j-1}, Y_{j-2}) = \lim_{j \to \infty} H_\pi(I_j|I_{j-1}, Y_{j}, Y_{j-1}, Y_{j-2}).\]
where $\pi$ refers to the stationary joint distribution on $(Y_{j-2}, Y_{j-1}, I_{j-1}, I_{j}, Y_j)$, given by \eqref{eq:stat_pi} and \eqref{eq:IY_dist}.
$H_\pi(I_j|I_{j-1}, Y_{j}, Y_{j-1}, Y_{j-2})$ can be computed as follows. First, we note that
$I_j=0$ whenever $I_{j-1}=1$. Therefore
\be \label{eq:ins_HIj_cond}
\begin{split}
H_\pi(I_j|I_{j-1}, Y_{j}, Y_{j-1}, Y_{j-2}) = & \sum_{y=0}^1 \pi(I_{j-1}=0, Y_{j}=y, Y_{j-1}=y, Y_{j-2}=y) H(I_j|I_{j-1}=0, Y_{j}=y, Y_{j-1}=y, Y_{j-2}=y)\\
&+ \pi(I_{j-1}=0, Y_{j}=y, Y_{j-1}=y, Y_{j-2}=\bar{y}) H(I_j|I_{j-1}=0, Y_{j}=y, Y_{j-1}=y, Y_{j-2}=\bar{y})\\
&+ \pi(I_{j-1}=0, Y_{j}=y, Y_{j-1}=\bar{y}, Y_{j-2}=\bar{y}) H(I_j|I_{j-1}=0, Y_{j}=y, Y_{j-1}=\bar{y}, Y_{j-2}=\bar{y})\\
&+ \pi(I_{j-1}=0, Y_{j}=y, Y_{j-1}=\bar{y}, Y_{j-2}={y}) H(I_j|I_{j-1}=0, Y_{j}=y, Y_{j-1}=\bar{y}, Y_{j-2}={y})
\end{split}
\ee
From \eqref{eq:stat_pi} and \eqref{eq:IY_dist}, we have
\be
\begin{split}
\pi(I_{j-1}=0, Y_{j}=y, Y_{j-1}=y, Y_{j-2}=y) =& \pi(I_{j-1}=0, Y_{j-1}=y, Y_{j-2}=y) \left[ P(Y_{j}=y, I_j=1|I_{j-1}=0, Y_{j-1}=y, Y_{j-2}=y)\right.\\
&\hspace{1in} \left.+ P(Y_{j}=y, I_j=0|I_{j-1}=0, Y_{j-1}=y, Y_{j-2}=y)\right]\\
=&\frac{\bar{i}\gamma+ i \alpha \gamma + i \bar{\alpha} \bar{\gamma}}{2(1+i)} \cdot (i\alpha + \bar{i} \gamma),
\end{split}
\ee
and
\be H(I_j|I_{j-1}=0, Y_{j}=y, Y_{j-1}=y, Y_{j-2}=y) = h\left( \frac{i\alpha}{i\alpha + \bar{i}\gamma}\right).\ee
The remaining terms in \eqref{eq:ins_HIj_cond} can be similarly calculated to obtain \eqref{eq:hI_limit}.
\qed
\subsection{Proof of Lemma \ref{lem:lb1_liminf}} \label{proof:lb1_liminf}
We have
\be \label{eq:HIYX_expand}
\begin{split}
& H_P(I^{n(1+i)}|Y^{n(1+i)}, X^n) = \sum_{j=2}^{n(1+i)} H_P(I_{j+1}|I^{j}, Y^{n(1+i)}, X^n) \\
& \stackrel{(a)}{\geq} \sum_{j=1}^{n(1+i)} \sum_{y \in \{0,1\}} P(I^{j-1}, I_{j}=0)P(Y^{j-1}, (Y_j, Y_{j+1}, Y_{j+2}, Y_{j+3})=(y, \bar{y}, \bar{y}, y), \ (X_{k_j}, X_{k_j+1}, X_{k_j+2})=(y, \bar{y}, y)|I_j=0, I^{j-1})\\
& \hspace{0.8in} \cdot H_P(I_{j+1}|Y^{j-1}, I^{j-1}, I_{j}=0, (Y_j, Y_{j+1}, Y_{j+2}, Y_{j+3})=(y, \bar{y}, \bar{y}, y), \ (X_{k_j}, X_{k_j+1}, X_{k_j+2})=(y, \bar{y}, y))
\end{split}
\ee
where $k_j$ is the index of the input bit that corresponds to $Y_j$. ($k_j$ is uniquely determined given $I_{1},\ldots, I_j$.)
$(a)$ is obtained as follows. Since $I_{j+1}=0$ whenever $I_j=1$, the entropy terms in the sum are non-zero only when $I_j=0$. The inequality appears because we sum only over those indices $j$ that satisfy
\[I_j=0, \; (Y_j, Y_{j+1}, Y_{j+2}, Y_{j+3})=(y, \bar{y}, \bar{y}, y), \; (X_{k_j}, X_{k_j+1}, X_{k_j+2})=(y, \bar{y}, y). \]
For such indices, the input bit $X_{k_j}$ corresponds to $Y_j$, and $X_{k_{j}+2}$ to $Y_{j+3}$, the uncertainty in $I_{j+1}$ being whether $X_{k_j+1}$ corresponds to $Y_{j+1}$
or $Y_{j+2}$. We have
\be \label{eq:Ip1}
\begin{split}
&P((Y_j, Y_{j+1}, Y_{j+2}, Y_{j+3})=(y, \bar{y}, \bar{y}, y), \ (X_{k_j}, X_{k_j+1}, X_{k_j+2})=(y, \bar{y}, y)|I_j=0)\\
& = P((X_{k_j}, X_{k_j+1}, X_{k_j+2})=(y, \bar{y}, y)) \cdot
P((Y_j, Y_{j+1}, Y_{j+2}, Y_{j+3})=(y, \bar{y}, \bar{y}, y)|(X_{k_j}, X_{k_j+1}, X_{k_j+2})=(y, \bar{y}, y), I_j=0)\\
&= \frac{1}{2} \bar{\gamma}^2 \cdot(i\bar{\alpha} + (1-i)i\alpha)
\end{split}
\ee
where term $i\bar{\alpha}$ corresponds to the case where $Y_{j+1}$ is a complementary insertion ($I_{j+1}=1, I_{j+2}=0$), and the term $(1-i)i\alpha$
to the case where $Y_{j+2}$ is a duplication ($I_{j+1}=0, I_{j+2}=1$). Therefore,
\be \label{eq:Ip2}
H_P(I_{j+1}|I^{j-1}, I_{j}=0, (Y_j, Y_{j+1}, Y_{j+2}, Y_{j+3})=(y, \bar{y}, \bar{y}, y), \ (X_{k_j}, X_{k_j+1}, X_{k_j+2})=(y, \bar{y}, y))
=h\left(\frac{i\bar{\alpha}}{i\bar{\alpha} + (1-i)i\alpha} \right).
\ee
As explained in Section \ref{sec:insertion}, $\{I_j\}_{j\geq 1}$, is a Markov chain with $P(I_{j}=0)$ converging to $\frac{1}{1+i}$ as $j \to \infty$.
Substituting this along with \eqref{eq:Ip1} and \eqref{eq:Ip2} in \eqref{eq:HIYX_expand}, we obtain
\[
\liminf_{n \to \infty} \frac{1}{n} H_P(I^{n(1+i)}|Y^{n(1+i)}, X^n) \geq n(1+i) \cdot \frac{1}{1+i} \bar{\gamma}^2 \cdot(i\bar{\alpha} + (1-i)i\alpha)
h\left(\frac{\bar{\alpha}}{\bar{\alpha} + (1-i)\alpha} \right).
\]
\qed
\subsection{Proof of Lemma \ref{lem:ins_lim_HT_Y}} \label{proof:ins_lim_HT_Y}
We have
\be
\frac{1}{m} H_P(T^{m}|Y^{m}) = \frac{1}{m} \sum_{j=1}^m H_P(T_j|T^{j-1}, Y^m) \leq \frac{1}{m} \sum_{j=1}^m H_P(T_j|T_{j-1}, Y_{j}, Y_{j-1}).
\ee
Therefore
\be
\limsup_{m \to \infty} \frac{1}{m} H_P(T^{m}|Y^{m}) \leq \limsup_{m \to \infty} \frac{1}{m} \sum_{j=1}^m H_P(T_j|T_{j-1}, Y_{j}, Y_{j-1})=
\lim_{j \to \infty} H_P(T_j|T_{j-1}, Y_{j}, Y_{j-1}),
\ee
provided the limit exists. We now show that $\lim_{j \to \infty} H_P(T_j|T_{j-1}, Y_{j}, Y_{j-1})$ exists and is given by \eqref{eq:hT_limit}.
Note that $T_j=0$ whenever $T_{j-1}=1$ since we cannot have two consecutive insertions. Also, $T_j=0$ whenever $Y_j=Y_{j-1}$ since $T_j=1$
only when $Y_j$ is a complementary insertion. Thus we have for all $j \geq 2$:
\be \label{eq:HTjTj1YjYj1}
\begin{split}
H(T_j|T_{j-1}, Y_{j}, Y_{j-1})= & P(T_{j-1}=0, Y_{j}=1, Y_{j-1}=0)H(T_j|T_{j-1}=0, Y_{j}=1, Y_{j-1}=0)\\
&+ P(T_{j-1}=0, Y_{j}=0, Y_{j-1}=1)H(T_j|T_{j-1}=0, Y_{j}=0, Y_{j-1}=1).
\end{split}
\ee
Note that for all $j\geq 1$, $P(T_j=1) = P(I_j=1)\bar{\alpha}$, where $I_j=1$ if $Y_j$ is an inserted bit, and $I_j=0$ otherwise.
Therefore,
\be \label{eq:IjTj} P(T_{j}=0) =1- P(I_j=1)\bar{\alpha} \ ,j\geq 1.\ee
Note that the binary-valued process $\{I_j\}_{j \geq 1}$ is a Markov chain with transition probabilities
\be
\text{Pr}(I_j=1|I_j=0)=1-\text{Pr}(I_j=0|I_j=0)=i, \quad \text{Pr}(I_j=1|I_j=1)=1-\text{Pr}(I_j=0|I_j=1)=0.
\ee
For $i \in (0,1)$, this is an irreducible, aperiodic Markov chain. Hence a unique stationary distribution $\pi$ exists, which is given by
\be \pi(I_j=1)=1-\pi(I_j=0)=\frac{i}{1+i}. \ee
Hence for any $\e >0$,
\be \label{eq:i_stat}
\left|P(I_j=1) - \frac{i}{1+i}\right|<\epsilon \text{ and } \left|\text{Pr}(I_j=0) - \frac{1}{1+i}\right|<\epsilon
\ee
for all sufficiently large $j$. Using this in \eqref{eq:IjTj}, for all sufficiently large $j$, the distribution $P(T_{j})$
is within total variation norm $\e$ of the following stationary distribution.
\be \label{eq:Tj_asymp}
\pi(T_{j}=0) = 1-\frac{i\bar{\alpha}}{1+i} = \frac{1+i\alpha}{1+i}, \quad
\pi(T_{j}=1) = \frac{i\bar{\alpha}}{1+i}.
\ee
Further, we have $P(Y_{j}=1|T_{j}=0)= P(Y_{j}=\bar{y}|T_{j}=0)=0.5$, for $y \in \{0,1\}$ since both the input distribution and the insertion process are symmetric in $0$ and $1$. Hence the stationary distribution for $(T_{j-1}, Y_{j-1})$ is
\be \label{eq:Tj1Yj1_asymp}
\pi(T_{j-1}=0, Y_{j-1}=y) = \frac{1+i\alpha}{2(1+i)}, \ \
\pi(T_{j-1}=1, Y_{j-1}=y) = \frac{i\bar{\alpha}}{2(1+i)}.
\ee
Next, we determine the conditional distribution $P(Y_j, T_j|Y_{j-1}=y, T_{j-1}=0)$ for $y \in \{0, 1\}$.
We have
\be \label{eq:tjyj_cond1}
\begin{split}
&P(T_j=0, Y_j=y|Y_{j-1}=y, T_{j-1}=0)\\
&= P(T_j=0, Y_j=y, I_{j-1}=1|Y_{j-1}=y, T_{j-1}=0) + P(T_j=0, Y_j=y, I_{j-1}=0|Y_{j-1}=y, T_{j-1}=0) \\
&= P(I_{j-1}=1| T_{j-1}=0) \cdot P(T_{j}=0, Y_j=y|I_{j-1}=1, T_{j-1}=0, Y_{j-1}=y) \\
& \qquad + P(I_{j-1}=0| T_{j-1}=0) \cdot P(T_{j}=0, Y_j=y|I_{j-1}=0, T_{j-1}=0, Y_{j-1}=y)\\
&\stackrel{(a)}{=} \frac{P(I_{j-1}=1) P(T_{j-1}=0|I_{j-1}=1)} {P(T_{j-1}=0)} \gamma + \frac{P(I_{j-1}=0) P(T_{j-1}=0|I_{j-1}=0)} {P(T_{j-1}=0)} ((1-i)\gamma + i \alpha)\\
&\stackrel{(b)}{=} \frac{P(I_{j-1}=1) \alpha}{1- \bar{\alpha}P(I_{j-1}=1)} \gamma
+ \frac{P(I_{j-1}=0)}{1- \bar{\alpha}P(I_{j-1}=1)} ((1-i)\gamma + i \alpha).
\end{split}
\ee
In the above, $(b)$ is obtained using \eqref{eq:IjTj}. $(a)$ is obtained as follows. The event $(I_{j-1}=1, T_{j-1}=0, Y_{j-1}=y)$ implies
$Y_{j-1}$ is a duplication, and hence $Y_{j-2}=y$ corresponds to an input bit (say $X_a$),
and $Y_{j}$ is the next input bit $X_{a+1}$. The probability that $X_{a+1}=X_a$ is $\gamma$.
Hence $P(T_{j}=0, Y_j=y|I_{j-1}=1, T_{j-1}=0, Y_{j-1}=y)=\gamma$.
When $(I_{j-1}=0, T_{j-1}=0, Y_{j-1}=y)$, $Y_{j-1}$ corresponds to an input bit, say $X_b$. Conditioned on this, the event $(T_{j}=0, Y_j=y)$ can occur in two ways:
\begin{itemize}
\item $Y_j$ is the next input bit $X_{b+1}$ and is equal to $y$. This event has probability $(1-i)\gamma$.
\item $Y_j$ is a duplication of $Y_{j-1}$. This event has probability $i \alpha$.
\end{itemize}
Hence $P(T_{j}=0, Y_j=y|I_{j-1}=0, T_{j-1}=0, Y_{j-1}=y)= ((1-i)\gamma + i \alpha)$. Similarly, we calculate
\be \label{eq:tjyj_cond2}
\begin{split}
&P(T_j=0, Y_j=\bar{y}|Y_{j-1}=y, T_{j-1}=0)= P(I_{j-1}=1| T_{j-1}=0) P(T_{j}=0, Y_j=\bar{y}|I_{j-1}=1, T_{j-1}=0, Y_{j-1}=y) \\
&\hspace{2.5in}+ P(I_{j-1}=0| T_{j-1}=0) P(T_{j}=0, Y_j=\bar{y}|I_{j-1}=0, T_{j-1}=0, Y_{j-1}=y)\\
&\; {=} \frac{P(I_{j-1}=1) P(T_{j-1}=0|I_{j-1}=1)} {P(T_{j-1}=0)} \bar{\gamma} + \frac{P(I_{j-1}=0) P(T_{j-1}=0|I_{j-1}=0)} {P(T_{j-1}=0)} (1-i)\bar{\gamma} \\
&\; {=} \frac{P(I_{j-1}=1)}{1- \bar{\alpha}P(I_{j-1}=1)} \alpha \bar{\gamma}
+ \frac{P(I_{j-1}=0)}{1- \bar{\alpha}P(I_{j-1}=1)} (1-i)\bar{\gamma},
\end{split}
\ee
\be \label{eq:tjyj_cond3}
\begin{split}
&P(T_j=1, Y_j=\bar{y}|Y_{j-1}=y, T_{j-1}=0)= P(I_{j-1}=1| T_{j-1}=0) P(T_{j}=1, Y_j=\bar{y}|I_{j-1}=1, T_{j-1}=0, Y_{j-1}=y) \\
&\hspace{2.5in} + P(I_{j-1}=0| T_{j-1}=0) P(T_{j}=1, Y_j=\bar{y}|I_{j-1}=0, T_{j-1}=0, Y_{j-1}=y)\\
& \; {=} \frac{P(I_{j-1}=1) P(T_{j-1}=0|I_{j-1}=1)} {P(T_{j-1}=0)} 0 + \frac{P(I_{j-1}=0) P(T_{j-1}=0|I_{j-1}=0)} {P(T_{j-1}=0)} i\bar{\alpha}\\
& \; {=} \frac{P(I_{j-1}=0)}{1- \bar{\alpha}P(I_{j-1}=1)} i\bar{\alpha},
\end{split}
\ee
and
\be \label{eq:tjyj_cond4}
P(T_j=1, Y_j=y|Y_{j-1}=y, T_{j-1}=0)=0.
\ee
Using \eqref{eq:i_stat} in equations \eqref{eq:tjyj_cond1}-\eqref{eq:tjyj_cond4}, we see that for all sufficiently large $j$,
the distribution $P(T_j, Y_j|Y_{j-1}=y, T_{j-1}=0)$ is within a total variation norm $\e$ from the following stationary distribution
\be \label{eq:cond_tjyj_asymp}
\begin{split}
\pi(T_j=0, Y_j=y|Y_{j-1}=y, T_{j-1}=0) &=\frac{i\alpha(1+ \gamma) + (1-i)\gamma}{1+i\alpha} ,\\
\pi(T_j=0, Y_j=\bar{y}|Y_{j-1}=y, T_{j-1}=0) &= \frac{\bar{\gamma} (1-i \bar{\alpha})}{1+i\alpha},\\
\pi(T_j=1, Y_j=\bar{y}|Y_{j-1}=y, T_{j-1}=0) &= \frac{i\bar{\alpha}}{1+i\alpha},\\
\pi(T_j=1, Y_j=y|Y_{j-1}=y, T_{j-1}=0) &=0.
\end{split}
\ee
Due to the continuity of the entropy function in the joint distribution, we therefore have
\[ \lim_{j \to \infty} H_P(T_j|T_{j-1}, Y_{j-1}, Y_j) = H_\pi(T_j|T_{j-1}, Y_{j-1}, Y_j),\]
where the joint distribution $\pi(T_{j-1}, Y_{j-1}, Y_j, T_j)$ is given by \eqref{eq:Tj1Yj1_asymp} and \eqref{eq:cond_tjyj_asymp}.
Using this in \eqref{eq:HTjTj1YjYj1}, one can compute $H_\pi(T_j|T_{j-1}, Y_{j-1}, Y_j)$ to obtain the result in the lemma.
\qed
\subsection{Proof of Lemma \ref{lem:ins_x_ytil}}\label{proof:ins_x_ytil}
Due to \eqref{eq:ins_bits_to_runs}, it is enough to show that $\frac{1}{n} H_P(L^X_1, \ldots, L^X_{R_n}|L^{\tilde{Y}}_1, \ldots, L^{\tilde{Y}}_{R_n})$
converges to $(1-\gamma) H_P(L^X_1|L^{\tilde{Y}}_1)$.
Since $\{(L^X_1, L^{\tilde{Y}}_1), (L^X_2, L^{\tilde{Y}}_2), \ldots \}$ is an i.i.d process, from the strong law of large numbers, we have
\be
\lim_{m \to \infty} -\frac{1}{m} \log \text{Pr}(L^X_1, \ldots, L^X_{m} | L^{\tilde{Y}}_1, \ldots, L^{\tilde{Y}}_{m}) = H_P(L^X_1|L^{\tilde{Y}}_1) \quad a.s.
\ee
Further, we have the normalized number of input runs $\frac{R_n}{n} \to (1-\gamma)$ almost surely. Using the above in Lemma \ref{lem:entropyrate},
we obtain
\be \label{eq:ins_run_conv}
\lim_{n \to \infty} -\frac{1}{n} \log \text{Pr}(L^X_1, \ldots, L^X_{R_n} | L^{\tilde{Y}}_1, \ldots, L^{\tilde{Y}}_{R_n}) = H_P(L^X_1|L^{\tilde{Y}}_1) \quad a.s.
\ee
We now argue that $-\frac{1}{n} \log \text{Pr}(L^X_1, \ldots, L^X_{R_n} | L^{\tilde{Y}}_1, \ldots, L^{\tilde{Y}}_{R_n})$ is uniformly integrable.
Supp$(L^X_1, \ldots, L^X_{R_n} | L^{\tilde{Y}}_1, \ldots, L^{\tilde{Y}}_{R_n})$ can be upper bounded by $2^{n}$ since the random sequence
$(L^X_1, \ldots, L^X_{R_n})$ is equivalent to $X^n$, which can take on at most $2^n$ values. Hence, from Lemma \ref{lem:support}, $-\frac{1}{n} \log \text{Pr}(L^X_1, \ldots, L^X_{R_n} | L^{\tilde{Y}}_1, \ldots, L^{\tilde{Y}}_{R_n})$ is uniformly integrable.
Using this together with \eqref{eq:ins_run_conv} in Lemma \ref{lem:exchange_lim}, we conclude that
\be
\begin{split}
\lim_{n \to \infty}\frac{1}{n} H_P(L^X_1, \ldots, L^X_{R_n} | L^{\tilde{Y}}_1, \ldots, L^{\tilde{Y}}_{R_n})
&= \lim_{n \to \infty} \expec\left[ -\frac{1}{n} \log \text{Pr}(L^X_1, \ldots, L^X_{R_n} | L^{\tilde{Y}}_1, \ldots, L^{\tilde{Y}}_{R_n}) \right]\\
&=\expec\left[ \lim_{n \to \infty} -\frac{1}{n} \log \text{Pr}(L^X_1, \ldots, L^X_{R_n} | L^{Y'}_1, \ldots, L^{\tilde{Y}}_{R_n}) \right]\\
&= H_P(L^X_1|L^{\tilde{Y}}_1).
\end{split}
\ee
\qed
\subsection{Proof of Lemma \ref{lem:lb2_liminf}} \label{proof:lb2_liminf}
We have
\be \label{eq:HTYX_expand}
\begin{split}
& H_P(T^{n(1+i)}|Y^{n(1+i)}, X^n) = \sum_{j=2}^{n(1+i)} H_P(T_{j+1}|T^{j}, Y^{n(1+i)}, X^n) \\
& {\geq} \sum_{j=2}^{n(1+i)} H_P(T_{j+1}|I_j, T^{j}, Y^{n(1+i)}, X^n) \\
& \stackrel{(a)}{\geq} \sum_{j=1}^{n(1+i)} \sum_{y \in \{0,1\}} P(T^{j-1}, I_{j}=0)P(Y^{j-1},(Y_j, Y_{j+1}, Y_{j+2}, Y_{j+3})=(y, \bar{y}, \bar{y}, y), \ (X_{k_j}, X_{k_j+1}, X_{k_j+2})=(y, \bar{y}, y)|I_j=0, T^{j-1})\\
& \hspace{1in} \cdot H_P(T_{j+1}|Y^{j-1}, T^{j-1}, I_{j}=0, (Y_j, Y_{j+1}, Y_{j+2}, Y_{j+3})=(y, \bar{y}, \bar{y}, y), \ (X_{k_j}, X_{k_j+1}, X_{k_j+2})=(y, \bar{y}, y)).
\end{split}
\ee
$(a)$ is obtained as follows. Since $T_{j+1}=0$ whenever $I_j=1$, the entropy terms in the second line are non-zero only when $I_j=0$. The inequality appears because we only sum over indices $j$ such that
\[I_j=0, \; \ (Y_j, Y_{j+1}, Y_{j+2}, Y_{j+3})=(y, \bar{y}, \bar{y}, y), \; (X_{k_j}, X_{k_j+1}, X_{k_j+2})=(y, \bar{y}, y), \]
where $k_j$ is the index of the input bit that corresponds to $Y_j$. This is uniquely determined because given $T^{j}$, there is a one-to-one correspondence between the input and output runs until bit $Y^{j}$. Then $Y_{j}=y$ and $Y_{j+1}=\bar{y}$ implies $k_j$ is the last input bit in the run containing $Y_{j}$.
Therefore, input bit $X_{k_j}$ corresponds to $Y_j$, and $X_{k_{j}+2}$ corresponds to $Y_{j+3}$, the uncertainty being whether $X_{k_j+1}$ corresponds to $Y_{j+1}$
or $Y_{j+2}$. We have
\be \label{eq:Tp1}
\begin{split}
&P((Y_j, Y_{j+1}, Y_{j+2}, Y_{j+3})=(y, \bar{y}, \bar{y}, y), \ (X_{k_j}, X_{k_j+1}, X_{k_j+2})=(y, \bar{y}, y)|I_j=0)\\
& = P((X_{k_j}, X_{k_j+1}, X_{k_j+2})=(y, \bar{y}, y)|I_j=0) \cdot P((Y_j, Y_{j+1}, Y_{j+2}, Y_{j+3})=(y, \bar{y}, \bar{y}, y)|(X_{k_j}, X_{k_j+1}, X_{k_j+2})=(y, \bar{y}, y), I_j=0)\\
&= \frac{1}{2} \bar{\gamma}^2 \cdot(i\bar{\alpha} + (1-i)i\alpha)
\end{split}
\ee
where term $i\bar{\alpha}$ corresponds to the case where $Y_{j+1}$ is a complementary insertion ($T_{j+1}=1$), and the term $(1-i)i\alpha$
to the case where $Y_{j+2}$ is a duplication ($T_{j+1}=0$). Consequently,
\be \label{eq:Tp2}
H_P(T_{j+1}|T^{j-1}, I_{j}=0, (Y_j, Y_{j+1}, Y_{j+2}, Y_{j+3})=(y, \bar{y}, \bar{y}, y), \ (X_{k_j}, X_{k_j+1}, X_{k_j+2})=(y, \bar{y}, y))
=h\left(\frac{i\bar{\alpha}}{i\bar{\alpha} + (1-i)i\alpha} \right).
\ee
Substituting \eqref{eq:Tp1} and \eqref{eq:Tp2} in \eqref{eq:HTYX_expand} and using the fact that $P(I_{j}=0) \to \frac{1}{1+i}$, we obtain
\[
\liminf_{n \to \infty} \frac{1}{n} H_P(T^{n(1+i)}|Y^{n(1+i)}, X^n) \geq n(1+i) \cdot \frac{1}{1+i} \bar{\gamma}^2 \cdot(i\bar{\alpha} + (1-i)i\alpha)
h\left(\frac{\bar{\alpha}}{\bar{\alpha} + (1-i)\alpha} \right).
\]
\qed
\section{Deletion Channel}
\subsection{Proof of Lemma \ref{lem:del_s_y}} \label{proof:del_s_y}
We first show that almost surely
\[
\lim_{n \to \infty} -\frac{1}{n} \log P(S^{M_n}|Y^{M_n}) = (1-d) H_P(S_2|Y_1 Y_2).
\]
From Propositions \ref{prop:y} and \ref{prop:sy}, $\{Y_m\}_{m \geq 1}$ and $\{(S_m, Y_m)\}_{m \geq 1}$ are
both ergodic Markov chains with stationary transition probabilities. Therefore, from the Shannon-McMillan-Breiman theorem \cite{AlgoetCover88},
we have
\begin{align}
\lim_{m \to \infty} -\frac{1}{m} \log P(Y^m) & = H_P(Y_2|Y_1) \: a.s., \label{eq:hy2y1} \\
\lim_{m \to \infty} -\frac{1}{m} \log P(S^m, Y^m) & = H_P(S_2, Y_2|Y_1)\: a.s.\label{eq:s2y2y1}
\end{align}
Subtracting \eqref{eq:s2y2y1} from \eqref{eq:hy2y1}, we get
\be \label{eq:hs2_y2y1}
\lim_{m \to \infty} -\frac{1}{m} \log P(S^m|Y^m) = H_P(S_2|Y_2 Y_1) \: a.s.
\ee
Further, we have $\lim_{n \to \infty}\frac{M_n}{n} = 1-d$ almost surely. Using this with \eqref{eq:hs2_y2y1} in Lemma \ref{lem:entropyrate},
we conclude that
\be \label{eq:lim_s1y1y2}
\lim_{n \to \infty} -\frac{1}{n} \log P(S^{M_n}|Y^{M_n}) = (1-d) H_P(S_2|Y_1 Y_2) \ a.s.
\ee
We now argue that $-\frac{1}{n} \log P(S^{M_n}|Y^{M_n})$ is uniformly integrable.
The Supp$(S^{M_n}|Y^{M_n})$ can be upper bounded by representing
$S^{M_n}$ as
\[ \underbrace{xx\ldots x}_{S_1} \ Y \ \underbrace{xx\ldots x}_{S_2} \ Y \ \underbrace{x\ldots x}_{S_3} \ \ \ldots Y \ \underbrace{xx\ldots x}_{S_{M_n}} \ Y
\underbrace{xx\ldots x}_{S_{M_n+1}} \]
where the $Y$'s represent the bits of the sequence $Y^{M_n}$, and each $x$ represents a missing run. Since the maximum length of the above binary sequence is $n$,
we have Supp$(S^{M_n}|Y^{M_n}) \leq 2^n$. Hence, from Lemma \ref{lem:support}, $-\frac{1}{n} \log \text{Pr}(S^{M_n}|Y^{M_n})$ is uniformly integrable.
Using this together with \eqref{eq:lim_s1y1y2} in Lemma \ref{lem:exchange_lim}, we conclude that
\[
\lim_{n \to \infty}\frac{1}{n} H_P({S}^{M_n}|{Y}^{M_n})
= \lim_{n \to \infty} \expec\left[ -\frac{1}{n} \log P({S}^{M_n}|{Y}^{M_n}) \right]
=\expec\left[ \lim_{n \to \infty} -\frac{1}{n} \log P({S}^{M_n}|{Y}^{M_n}) \right]
=(1-d) H_P(S_2|Y_1 Y_2).
\]
Thus we have
\[
\lim_{n \to \infty}\frac{1}{n} H_P({S}^{M_n+1}|{Y}^{M_n}) = \lim_{n \to \infty}\frac{1}{n} H_P({S}^{M_n}|{Y}^{M_n})
+ \lim_{n\to \infty} \frac{1}{n} H_P({S}_{M_n+1}|{Y}^{M_n}, S^{M_n})=
(1-d) H_P(S_2|Y_1 Y_2)+0.
\]
\qed
\subsection{Proof of Lemma \ref{lem:del_x_sy}} \label{proof:del_x_sy}
Due to \eqref{eq:bits_to_runs}, it is enough to show that $\frac{1}{n} H_P(L^X_1, \ldots, L^X_{R_n}|L^{Y'}_1, \ldots, L^{Y'}_{R_n})$
converges to $(1-\gamma) H_P(L^X_1|L^{Y'}_1)$.
Since $\{(L^X_1, L^{Y'}_1), (L^X_2, L^{Y'}_2), \ldots \}$ is an i.i.d process, from the strong law of large numbers, we have
\be
\lim_{m \to \infty} -\frac{1}{m} \text{Pr}(L^X_1, \ldots, L^X_{m} | L^{Y'}_1, \ldots, L^{Y'}_{m}) = H_P(L^X_1|L^{Y'}_1) \quad a.s.
\ee
Further, we have the normalized number of input runs $\frac{R_n}{n} \to (1-\gamma)$ almost surely. Using the above in Lemma \ref{lem:entropyrate},
we obtain
\be \label{eq:run_conv}
\lim_{n \to \infty} -\frac{1}{n} \text{Pr}(L^X_1, \ldots, L^X_{R_n} | L^{Y'}_1, \ldots, L^{Y'}_{R_n}) = H_P(L^X_1|L^{Y'}_1) \quad a.s.
\ee
We now argue that $-\frac{1}{n} \log \text{Pr}(L^X_1, \ldots, L^X_{R_n} | L^{Y'}_1, \ldots, L^{Y'}_{R_n})$ is uniformly integrable.
Supp$(L^X_1, \ldots, L^X_{R_n} | L^{Y'}_1, \ldots, L^{Y'}_{R_n})$ can be upper bounded by $2^n$ since since the random sequence
$(L^X_1, \ldots, L^X_{R_n})$ is equivalent to $X^n$, which can take on at most $2^n$ values. Hence, from Lemma \ref{lem:support}, $-\frac{1}{n} \log \text{Pr}(L^X_1, \ldots, L^X_{R_n} | L^{Y'}_1, \ldots, L^{Y'}_{R_n})$ is uniformly integrable.
Using this together with \eqref{eq:run_conv} in Lemma \ref{lem:exchange_lim}, we conclude that
\be
\begin{split}
\lim_{n \to \infty}\frac{1}{n} H_P(L^X_1, \ldots, L^X_{R_n} | L^{Y'}_1, \ldots, L^{Y'}_{R_n})
&= \lim_{n \to \infty} \expec\left[ -\frac{1}{n} \log \text{Pr}(L^X_1, \ldots, L^X_{R_n} | L^{Y'}_1, \ldots, L^{Y'}_{R_n}) \right]\\
&=\expec\left[ \lim_{n \to \infty} -\frac{1}{n} \log \text{Pr}(L^X_1, \ldots, L^X_{R_n} | L^{Y'}_1, \ldots, L^{Y'}_{R_n}) \right]\\
&= H_P(L^X_1|L^{Y'}_1).
\end{split}
\ee
\qed
\section{Deletion+Insertion Channel}
\subsection{Proof of Lemma \ref{lem:delins_4}} \label{proof:delins_lem4}
We have
\[
\frac{1}{m} H(S^{m}| T^{m}, Y^{m}) = \sum_{j=1}^{m} \frac{1}{m} H(S_j|S^{j-1}, T^m, Y^m) \leq \sum_{j=1}^{m} \frac{1}{m} H(S_j|Y_{j-1}, Y_j, T_j)
\]
We will show that $\lim_{j \to \infty} H(S_j|Y_{j-1}, Y_j, T_j)$ exists and obtain an analytical expression for it.
For all $j$,
\be \label{eq:delins_sjyyt_split}
\begin{split}
& H(S_j|Y_{j-1}, Y_j, T_j) = P(Y_{j-1}, Y_j, T_j=0) H(S_j|Y_{j-1}, Y_j, T_j=0) = \\
& \sum_{y \in \{0,1\}} P(Y_{j-1}=Y_j=y, T_j=0) H(S_j|Y_{j-1}=Y_j=y, T_j=0) + P(Y_{j-1}=\bar{y}, Y_j=y, T_j=0) H(S_j|Y_{j-1}=\bar{y}, Y_j=y, T_j=0)
\end{split}
\ee
The first equality above holds since $T_j=1$ implies $Y_j$ is an inserted bit, and so no deleted runs occur between $Y_{j-1}$ and $Y_j$.
$P(Y_{j-1}, Y_j, T_j=0)$ can be computed as follows.
\be \label{eq:delins_yyt}
\begin{split}
& P( Y_{j-1}={y}, Y_j=y, T_j=0) = P(I_{j-1}=0, Y_{j-1}={y}, T_j=0, Y_j=y) + P(I_{j-1}=1, T_{j-1}=0, Y_{j-1}={y}, T_j=0, Y_j=y) \\
& \hspace{3in} + P(I_{j-1}=1, T_{j-1}=1 , Y_{j-1}={y}, T_j=0, Y_j=y)\\
&\stackrel{(a)}{=} \frac{1}{2} P(I_{j-1}=0) (P(I_j=0|I_{j-1}=0) q + P(I_j=1|I_{j-1}=0) \alpha) + \frac{1}{2} P(I_{j-1}=1) \alpha q +
\frac{1}{2} P(I_{j-1}=1) \bar{\alpha} (1-q)\\
&\stackrel{j \to \infty}{\to} \frac{1}{2}\left[\frac{1}{1+i'}((1-i')q + {i'}\alpha) +
\frac{i'}{1+i'} \alpha q + \frac{i'}{1+i'}\bar{\alpha} (1-q)\right].
\end{split}
\ee
The last two terms in (a) are obtained by noting that $I_{j-1}=1$ implies $Y_{j-1}$ is an insertion and hence $T_j=0$. In this case, $Y_{j-2}=y$ corresponds to the last non-inserted bit before $Y_j$. The last line is due to the fact that $\{I_j\}_{j \geq 1}$ converges is a Markov chain that converges to the
stationary distribution $P(I_j=1)=\frac{i'}{1+i'}$, $P(I_j=0)=\frac{1}{1+i'}$. Thus for sufficiently large $j$, $P( Y_{j-1}={y}, Y_j=y, T_j=0)$ is at most $\e$
in total variation norm from the stationary distribution
\[ \pi( Y_{j-1}={y}, Y_j=y, T_j=0) = \frac{1}{2}\left[\frac{1}{1+i'}((1-i')q + {i'}\alpha) +
\frac{i'}{1+i'} \alpha q + \frac{i'}{1+i'}\bar{\alpha} (1-q)\right], \quad y\in \{0,1\}. \]
Similarly, $P( Y_{j-1}={y}, Y_j=\bar{y}, T_j=0)$ converges to
\be \label{eq:delins_ybaryt}
\begin{split}
& \pi( Y_{j-1}={y}, Y_j=\bar{y}, T_j=0) = \pi(I_{j-1}=0, Y_{j-1}={y}, T_j=0, Y_j=\bar{y}) + P(I_{j-1}=1, T_{j-1}=0, Y_{j-1}={y}, T_j=0, Y_j=\bar{y}) \\
& \hspace{3in} + \pi(I_{j-1}=1, T_{j-1}=1 , Y_{j-1}={y}, T_j=0, Y_j=\bar{y})\\
&\stackrel{(a)}{=} \frac{1}{2} \pi(I_{j-1}=0) P(I_j=0|I_{j-1}=0) (1-q) + \frac{1}{2} P(I_{j-1}=1) \alpha (1-q) +
\frac{1}{2} \pi(I_{j-1}=1) \bar{\alpha} q\\
&= \frac{1}{2}\left[\frac{1}{1+i'}(1-i')(1-q) + \frac{i'}{1+i'} \alpha(1-q) + \frac{i'}{1+i'} \bar{\alpha}q \right].
\end{split}
\ee
We next determine the joint distributions $\pi(S_j, Y_{j-1}=\bar{y}, Y_j=y, T_j=0)$ and $\pi(S_j, Y_{j-1}=\bar{y}, Y_j=y, T_j=0)$
to compute $H_\pi(S_j|Y_{j-1}=\bar{y}, Y_j=y, T_j=0)$ and $H_\pi(S_j|Y_{j-1}={y}, Y_j=y, T_j=0)$ in \eqref{eq:delins_sjyyt_split}. For $k=0,1,\ldots$, we have
\be
\begin{split}
&\pi(S_j=k, Y_{j-1}={y}, Y_j=y, T_j=0) = \\
& \pi(I_{j-1}=0, Y_{j-1}={y}, T_j =0, Y_j=y, S_j=k) + \pi(I_{j-1}=1, T_{j-1}=0, Y_{j-1}={y}, T_{j=0}, Y_j=y, S_j=k) \\
& \quad + \pi(I_{j-1}=1, T_{j-1}=1, Y_{j-1}={y}, T_{j=0}, Y_j=y, S_j=k).
\end{split}
\ee
The first term corresponds to $Y_{j-1}$ being an original input bit, the second term to $Y_{j-1}$ being a duplication, and the third to
$Y_{j-1}$ being a complementary insertion, respectively. Each of these terms can be calculated in a manner very similar to
equations \eqref{eq:del_pys_y} and \eqref{eq:del_pbarys_y} of Proposition \ref{prop:sy}. Hence we obtain,
\be \label{eq:delins_sjyyt}
\pi(S_j=k, Y_{j-1}={y}, Y_j=y, T_j=0)=
\left\{
\begin{array}{ll}
\frac{1}{2(1+i')} \left[ i'\alpha + (1-i'\bar{\alpha}) \frac{\gamma (1-d)}{1-\gamma d} + i' \bar{\alpha} \frac{(1-\gamma)(1-d)}{(1-\gamma d)^2} \right], & k=0\\
\frac{1-i'\bar{\alpha}}{2(1+i')} \frac{(1-\gamma)(1-d)}{(1-\gamma d)^2} \left(\frac{d(1-\gamma)}{1-\gamma d}\right)^k, & k=1,3, \ldots\\
\frac{i'\bar{\alpha}}{2(1+i')} \frac{(1-\gamma)(1-d)}{(1-\gamma d)^2} \left(\frac{d(1-\gamma)}{1-\gamma d}\right)^k, & k=2,4,\ldots
\end{array}
\right.
\ee
Similarly, we also determine
\be \label{eq:delins_sjbaryyt}
\pi(S_j=k, Y_{j-1}=\bar{y}, Y_j=y, T_j=0)=
\left\{
\begin{array}{ll}
\frac{1}{2(1+i')} \left[ (1-i'\bar{\alpha})\frac{(1-\gamma)(1-d)}{(1-\gamma d)^2} + i' \bar{\alpha} \frac{\gamma (1-d)}{1-\gamma d} \right], & k=0\\
\frac{i'\bar{\alpha}}{2(1+i')} \frac{(1-\gamma)(1-d)}{(1-\gamma d)^2} \left(\frac{d(1-\gamma)}{1-\gamma d}\right)^k, & k=1,3, \ldots\\
\frac{1- i'\bar{\alpha}}{2(1+i')} \frac{(1-\gamma)(1-d)}{(1-\gamma d)^2} \left(\frac{d(1-\gamma)}{1-\gamma d}\right)^k, & k=2,4,\ldots
\end{array}
\right.
\ee
From \eqref{eq:delins_yyt} and \eqref{eq:delins_sjyyt}, we can compute
\be \label{eq:delins_final1}
\begin{split}
&\pi(Y_{j-1}={y}, Y_j=y, T_j=0) H_\pi(S_j|Y_{j-1}={y}, Y_j=y, T_j=0) \\
= & \frac{1}{2(1+i')}
\left[ (i'\alpha + (1-i'\bar{\alpha}) \frac{\gamma (1-d)}{1-\gamma d} + i' \bar{\alpha} \beta)
\log_2 \left(\frac{i'\alpha + (1-i'\bar{\alpha})q + i'\bar{\alpha}\bar{q}}{i'\alpha + (1-i'\bar{\alpha}) \frac{\gamma (1-d)}{1-\gamma d} + i' \bar{\alpha} \beta}\right) + i'\bar{\alpha}\beta \frac{2\theta^2}{(1-\theta^2)^2} \log_2\left(\frac{1}{\theta}\right) \right. \\
&\left. +(1- i'\bar{\alpha})\beta \frac{\theta (1+\theta^2)}{(1-\theta^2)^2} \log_2\left(\frac{1}{\theta}\right)
+\frac{\theta \beta(1-i' \bar{\alpha})}{1-\theta^2} \log_2\left(\frac{i'\alpha + (1-i'\bar{\alpha})q + i'\bar{\alpha}\bar{q}}{\beta (1-i'\bar{\alpha})}\right)
+\frac{\theta^2 \beta i' \bar{\alpha}}{1-\theta^2} \log_2\left(\frac{i'\alpha + (1-i'\bar{\alpha})q + i'\bar{\alpha}\bar{q}}{\beta i'\bar{\alpha}}\right)
\right]
\end{split}
\ee
where $\theta$ and $\beta$ are defined in the statement of the lemma. Similarly, from \eqref{eq:delins_ybaryt} and \eqref{eq:delins_sjbaryyt}, one can compute
\be \label{eq:delins_final2}
\begin{split}
&\pi(Y_{j-1}={y}, Y_j=y, T_j=0) H_\pi(S_j|Y_{j-1}={y}, Y_j=y, T_j=0) \\
= & \frac{1}{2(1+i')}
\left[ ( i' \bar{\alpha} \frac{\gamma (1-d)}{1-\gamma d} + (1-i'\bar{\alpha})\beta)
\log_2 \left(\frac{ (1-i'\bar{\alpha})\bar{q} + i'\bar{\alpha}{q}}{i' \bar{\alpha} \frac{\gamma (1-d)}{1-\gamma d} + (1-i'\bar{\alpha})\beta}\right) + i'\bar{\alpha}\beta \frac{\theta (1+\theta^2)}{(1-\theta^2)^2} \log_2\left(\frac{1}{\theta}\right) \right. \\
&\left. +(1- i'\bar{\alpha})\beta \frac{2\theta^2}{(1-\theta^2)^2} \log_2\left(\frac{1}{\theta}\right)
+\frac{\theta^2 \beta(1-i' \bar{\alpha})}{1-\theta^2} \log_2\left(\frac{(1-i'\bar{\alpha})\bar{q} + i'\bar{\alpha}{q}}{\beta (1-i'\bar{\alpha})}\right)
+\frac{\theta \beta i' \bar{\alpha}}{1-\theta^2} \log_2\left(\frac{(1-i'\bar{\alpha})\bar{q} + i'\bar{\alpha}{q}}{\beta i'\bar{\alpha}}\right)
\right].
\end{split}
\ee
Substituting \eqref{eq:delins_final1} and \eqref{eq:delins_final2} in \eqref{eq:delins_sjyyt_split} completes the proof of the lemma. \qed
\section{Introduction}
\label{sec:intro}
Consider a binary input channel where for each bit (denoted $x$), the output is generated in one of the following ways:
\begin{itemize}
\item The bit is deleted with probability $d$,
\item An extra bit is inserted after $x$ with probability $i$. The extra bit is equal to $x$ (a \emph{duplication}) with probability $\alpha$, and equal to
$1-x$ (a \emph{complementary insertion}) with probability $1-\alpha$,
\item No deletions or insertions occur, and the output is $x$ with probability $1-d-i$.
\end{itemize}
The channel acts independently on each bit. We refer to this channel as the deletion+insertion channel with parameters $(d,i,\alpha)$.
If the input to the channel is a sequence of $n$ bits, the length of the output sequence
will be close to $n(1+i-d)$ for large $n$ due to the law of large numbers.
Channels with synchronization errors can be used to model timing mismatch in communication systems.
Channels with deletions and insertions also occur in magnetic recording \cite{ISiegelWolf}.
The problem of synchronization also appears in file backup and file sharing \cite{MaKTse11,ITASynch11},
where distributed nodes may have different versions of the same file which differ by a small number of edits.
The edits may include deletions, insertions, and substitutions. The minimum communication rate required to synchronize the remote sources is
closely related to the capacity of an associated synchronization channel. This connection is discussed at the end of this paper.
The above model with $i=0$ corresponds to the deletion channel, which has been studied in several recent papers, e.g., \cite{DiggaviG06,DrineaM07,DrineaK10,DMP07,Mitzenmacher09,FertonaniD10,KalaiMM10,KanoriaM10}.
When $d=0$, we obtain the insertion channel with parameters $(i,\alpha)$. The insertion channel with $\alpha=1$ is the sticky channel \cite{Mitz_sticky}, where all insertions are duplications.
In this paper, we obtain lower bounds on the capacity of the deletion+insertion channel. Our starting point is the result of Dobrushin \cite{Dobrushin67} for general synchronization channels which states that the capacity is given by the maximum of the mutual information per bit between the input and output sequences. There are two challenges to computing the capacity through this characterization. The first is evaluating the mutual information, which is a difficult task because of the memory inherent in the joint distribution of the input and output sequences. The second challenge is to optimize the mutual information over all input distributions.
In this work, we choose the input distribution to be the class of first-order Markov processes and focus on the problem of evaluating the mutual
information. It is known that first-order Markov input distributions yield good capacity lower bounds for the deletion channel \cite{DiggaviG06,DrineaM07} and the sticky channel \cite{Mitz_sticky}, both special cases of the deletion+insertion channel. This suggests they are likely to perform well on the deletion+insertion channel as well. But our approach is quite general, and can be extended to obtain capacity lower bounds with other input distributions such as higher order Markov distributions, and distributions with independent run-lengths.
For a synchronization channel, it is useful to think of the input and output sequences in terms of runs of symbols rather than individual symbols.
(The runs of a binary sequence are its alternating blocks of contiguous zeros and ones.) If there were a one-to-one correspondence between the runs of the input sequence $\un{X}$ and those of the output sequence $\un{Y}$, we could write the conditional distribution $P(\un{Y}|\un{X})$ as a product distribution of run-length transformations; computing the mutual information would then be straightforward. Unfortunately, such a correspondence is not possible since deletions can lead to some runs being lost, and insertions to new runs being inserted.
The main idea of the paper is to use auxiliary sequences which indicate the positions (in the output sequence) where \emph{runs} were deleted and inserted. Consider a decoder that first decodes the auxiliary sequences, and then the transmitted codeword. Conditioned on the knowledge of the auxiliary sequences, the mutual information between the input and output sequences is a single-letter quantity which can be calculated easily.
However, decoding the auxiliary sequences in addition to the codeword is sub-optimal and incurs a rate penalty, which we upper bound. The challenge is to define auxiliary sequences that lead to analytical lower bounds on the capacity with minimal rate penalty.
To gain insight, we first consider the special cases of the insertion channel and the deletion channel separately.
The insertion channel with parameters $(i,\alpha)$ introduces approximately $n i$ insertions in a sufficiently long input sequence of length $n$. A fraction nearly $\alpha$ of these insertions are duplications, and the rest are complementary insertions. Note that new runs can only be introduced by complementary insertions. We consider a decoder that first decodes the positions of the complementary insertions. The decoder can then flip the bits at these positions to obtain a one-to-one correspondence between input and output runs.
For the deletion channel, we use a decoder that first decodes an auxiliary sequence whose symbols indicate the number of runs deleted between each pair of adjacent bits in the output sequence. Augmenting the output sequence with the positions of deleted runs results in a one-to-one correspondence between input and output runs. For the deletion+insertion channel, we consider a decoder that decodes both auxiliary sequences described above.
The main contributions of the paper are the following:
\begin{enumerate}
\item Theorems \ref{thm:ins_lb1} and \ref{thm:ins_lb2} together provide the first characterization of achievable rates for the general insertion channel.
For the special case of the `sticky' channel ($\alpha=1$, i.e., only duplications), the rates of Theorem \ref{thm:ins_lb2} are very close to the near-optimal lower bound given in \cite{Mitz_sticky}.
\item Theorem \ref{thm:delins_thm} provides the first characterization of achievable rates for the deletion+insertion channel. For the special case of the deletion channel ($i=0$), these rates are close to the best-known lower bounds given in \cite{DrineaM07}.
\item Our approach provides a general framework to compute the capacity of channels with synchronization errors, and suggests several directions
to obtain sharper capacity bounds. For example, results on the structure of optimal input distributions for these channels (in the spirit of \cite{KanoriaM10,KalaiMM10}) could be combined with our approach to improve the lower bounds.
One could also obtain upper bounds on the capacity by assuming that the auxiliary sequences are available `for free' at the decoder, as done in \cite{DiggaviG06} for the deletion channel.
\end{enumerate}
For clarity, we only consider the binary deletion+insertion channel in this paper. The results presented here can be extended to channels with any finite alphabet.
\subsection{Related Work}
Dobrushin's capacity characterization was used in \cite{DrineaK10} to obtain bounds on the deletion capacity. The `jigsaw' decoding approach of \cite{DrineaM07} is interpreted in \cite{DrineaK10} in terms of an auxiliary sequence that associates each run of the output
sequence with one or more runs of the input codeword. Using this, the mutual information is decomposed such that one part of it represents the rate achieved
by the jigsaw decoder, and the other part represents the rate loss due to the jigsaw decoder. This rate loss is hard to compute, and is estimated via simulation for a few values of the deletion probability in \cite{DrineaK10}.
Our approach is motivated by the observation that synchronizing the output with the transmitted sequence is difficult mainly due to runs being completely deleted and new runs being inserted. Accordingly, the auxiliary sequences we use indicate the positions of these runs in the output sequence, leading to a mutual information decomposition that is quite different from \cite{DrineaK10}. The approach of \cite{DrineaM07,DrineaK10} provides the best-known lower bounds on the deletion capacity, but is specific to channels with only deletions and duplications. Our techniques are general and apply to channels with both deletions and insertions, while giving bounds very close to the best-known ones for deletion and duplication channels.
Dobrushin's capacity characterization was also used in \cite{KanoriaM10} to estimate the deletion capacity and the structure of the optimal input distribution for small values of deletion probability. In \cite{DMP07}, a genie-aided decoder with access to the locations of deleted runs was used to upper bound the deletion capacity using an equivalent discrete memoryless channel (DMC). In \cite{FertonaniD10}, bounds on the deletion capacity were obtained by considering a decoder equipped with side-information specifying the number of output bits corresponding to successive blocks of $L$ input bits, for any positive integer $L$. This new channel is equivalent to a DMC with an input alphabet of size $2^L$, whose capacity can be numerically computed using the Blahut-Arimoto algorithm (for as large a value of $L$ as computationally feasible). The upper bound in \cite{FertonaniD10} is the best known for a wide range of deletion probabilities, but the lower bound is weaker than that of\cite{DrineaM07} and the one proposed here.
Finally, we note that a different channel model with bit flips and synchronization errors was studied in \cite{Gallager61, FDE11}. In this model, an insertion is defined as an input bit being replaced by two random bits.
We have only mentioned the papers that are closely related to the results of this work. The reader is referred to \cite{Mitzenmacher09}
for an exhaustive list of references on synchronization channels.
After laying down the formal definitions and technical machinery in Section \ref{sec:prelim}, we describe two coding schemes in Section \ref{sec:coding_scheme} which give intuition about our bounding techniques.
In Section \ref{sec:insertion}, we consider the insertion channel ($d=0$) and derive two lower bounds on its capacity.
For this channel, previous bounds exist only for the special case of sticky channels ($\alpha=1$) \cite{Mitz_sticky}.
We derive a lower bound on the capacity of the deletion channel ($i=0$) in Section \ref{sec:deletion}
and compare it with the best known lower bound.
In Section \ref{sec:delins_channel}, we combine the ideas of Sections \ref{sec:insertion} and \ref{sec:deletion} to obtain a lower bound for the deletion+insertion channel. Section \ref{sec:conc} concludes the paper with a discussion of open questions.
\section{Preliminaries}
\label{sec:prelim}
\emph{Notation}: $\mathbb{N}_0$ denotes the set of non-negative integers, and $\mathbb{N}$ the set of natural numbers. $h(.)$ is the binary entropy function, and $\mathbf{1}_{\mc{A}}$ is the indicator function of the set $\mc{A}$. For any $0<\alpha\leq 1$, $\bar{\alpha} \triangleq 1-\alpha$. Logarithms are with base $2$, and entropy is measured in bits. We use uppercase letters to denote random variables, bold-face letters for random processes, and superscript notation to denote random vectors. Thus the channel input sequence of length $n$ is denoted $X^n \triangleq (X_1, \ldots, X_n)$. The corresponding output sequence at the decoder has length $M_n$ (a random variable determined by the channel realization), and is denoted $Y^{M_n}$.
For brevity, we sometimes use underlined notation for random vectors when we do not need to be explicit about their length. Thus $\un{X}\triangleq X^n=(X_1,X_2,\ldots,X_n)$, and $\un{Y} \triangleq Y^{M_n}=(Y_1,\ldots,Y_{M_n})$.
The communication over the channel is characterized by three random processes defined over the same probability space: the input process $\mathbf{X}=\{X_n\}_{n \geq 1}$, the output process $\mathbf{Y}=\{Y_n\}_{n \geq 1}$, and $\mathbf{M}=\{M_n\}_{n \geq 1}$,
where $M_n$ is the number of output symbols corresponding to the first $n$ input symbols.
If the underlying probability space is $(\Omega, \mc{F}, P)$, each realization $\omega \in \Omega$ determines the sample paths $\mathbf{X}(\omega)=\{X_n(\omega)\}_{n \geq 1}$, $\mathbf{Y}(\omega)=\{Y_n(\omega)\}_{n \geq 1}$, and $\mathbf{M}(\omega)=\{M_n(\omega)\}_{n \geq 1}$.
\begin{defi}
An $(n,2^{nR})$ code with block
length $n$ and rate $R$ consists of
\begin{enumerate}
\item An encoder mapping $e: \{1, \ldots, 2^{nR} \} \to \{0,1\}^n$, and
\item A decoder mapping $g: \Sigma \to \{1, \ldots, 2^{nR} \}$
where $\Sigma$ is $\cup_{k=0}^n \{0,1\}^k$ for the deletion channel,
$\cup_{k=n}^{2n} \{0,1\}^k$ for the insertion channel, and $\cup_{k=0}^{2n} \{0,1\}^k$
for the deletion+insertion channel.
\end{enumerate}
\end{defi}
Assuming the message $W$ is drawn uniformly on the set $\{1, \ldots, 2^{nR}\}$, the probability of error of a $(n,2^{nR})$ code is
\[
\begin{split}
P_{e,n}= \frac{1}{2^{nR}}\sum_{l=1}^{2^{nR}}\text{Pr}(g({Y}^{M_n}) \neq l | W=l)
\end{split}
\]
A rate $R$ is achievable if there exists a sequence of $(n,2^{nR})$ codes
such that $P_{e,n} \to 0$ as $n \to \infty$. The supremum of all achievable rates is the capacity $C$.
The following characterization of capacity follows from a result proved for a general class of synchronization channels by Dobrushin \cite{Dobrushin67}.
\begin{fact}
Let $C_n = \max_{P_{X^n}} \frac{1}{n} I(X^n; Y^{M_n}).$
Then $C \triangleq \lim_{n \to \infty} C_n$ exists, and is equal to the capacity of the deletion+insertion channel.
\label{fact:dob}
\end{fact}
\begin{proof}
Dobrushin proved the following general result in \cite{Dobrushin67}.
Consider a channel with $\mc{X}$ and $\mc{Y}$ denoting the alphabets of possible symbols at the input and output, respectively. For each input symbol in $\mc{X}$, the output belongs to $\mc{\bar{Y}}$, the set of all finite sequences of elements of $\mc{Y}$, including the empty sequence. The channel is memoryless and is specified by the stochastic matrix $\{P(\bar{y}|x), \ \bar{y} \in \bar{\mc{Y}}, {x} \in \mc{X} \}$.
Also assume that for each input symbol $x$, the length of the (possibly empty) output sequence has non-zero expected value. Then $\lim_{n \to \infty} C_n$ exists, and is equal the capacity of the channel.
The deletion+insertion channel is a special case of the above model with $\mc{X} = \mc{Y} = \{0,1 \}$, and the length of the output corresponding to any input symbol has a maximum value of two and expected value equal to $(1-d+i)$, which is non-zero for all $d<1$. Hence the claim is a direct consequence of Dobrushin's result.
\end{proof}
In this paper, we fix the input process to be the class of binary symmetric first-order Markov processes and focus on evaluating
the mutual information. This will give us a lower bound on the capacity.
The input process $\mathbf{X}=\{X_n\}_{n \geq 1}$ is characterized by the following distribution for all $n$:
\[ P(X_1, \ldots, X_n) = P(X_1) \prod_{j=2}^n P(X_j|X_{j-1}),\]
with
\be \label{eq:inp_def}
\begin{split}
P(X_1=0)=P(X_1=1)=0.5,\quad P(X_j=1|X_{j-1}=1)=P(X_j=0|X_{j-1}=0)=\gamma, \ j\geq 1.
\end{split}
\ee
A binary sequence may be represented by a sequence of positive integers representing the lengths of its runs, and
the value of the first bit (to indicate whether the first run has zeros or ones).
For example, the sequence $0001100000$ can be represented as $(3,2,5)$ if we know that the first bit is $0$.
The value of the first bit of $\mathbf{X}$ can be communicated to the decoder with vanishing rate, and we will assume this has been done at the outset.
Hence, denoting the length of the $j$th run of $\mathbf{X}$ by $L^{X}_j$ we have the following equivalence:
$\mathbf{X} \leftrightarrow (L^{X}_1, L^{X}_2, \ldots).$
For a first-order Markov binary source of \eqref{eq:inp_def}, the run-lengths are independent and geometrically distributed, i.e.,
\be \label{eq:run_dist}
\text{Pr}(L^{X}_j =r)= \gamma^{r-1} (1-\gamma), \qquad r=1, 2, \ldots
\ee
The average length of a run in $\mathbf{X}$ is $\frac{1}{1-\gamma}$, so the number of runs in a sequence of length $n$ is close to $n(1-\gamma)$ for large $n$. Our bounding techniques aim to establish a one-to-one correspondence between input runs and output runs. The independence of run-lengths of $\mathbf{X}$ enables us to obtain analytical bounds on the capacity.
We denote by $I_P({X}^n; Y^{M_n}), H_P(X^n), H_P(X^n|Y^{M_n})$ the mutual information and entropies computed with the channel input sequence
${X^n}$ distributed as in \eqref{eq:inp_def}.
For all $n$, we have
\be
\begin{split}
C_n= \max_{P_{X^n}} \frac{1}{n} I(X^n;Y^{M_n}) & > \frac{1}{n} I_P(X^n;Y^{M_n}).
\end{split}
\ee
Therefore
\be \label{eq:mut_decomp}
\begin{split}
C > \liminf_{n \to \infty} \frac{1}{n} I_P(X^n;Y^{M_n})
&= h(\gamma) - \limsup_{n \to \infty} \frac{1}{n}H_P(X^n|Y^{M_n}).
\end{split}
\ee
We will derive upper bounds on $ \limsup_{n \to \infty} \frac{1}{n}H_P(X^n|Y^{M_n})$ and use it in \eqref{eq:mut_decomp}
to obtain a lower bound on the capacity.
\subsection{Technical Lemmas}
To formally prove our results, we will use a framework similar to \cite{DrineaK10}. The notion of uniform integrability will play an important role, and
we list the relevant definitions and technical lemmas below.
\begin{defi} \cite{Billingsley}
A family of random variables $\{Z_n\}_{n\geq 1}$ is uniformly integrable if
\[ \lim_{a \to \infty} \sup_{n} \expec[|Z_n| \mathbf{1}_{\{|Z_n| \geq a\}}] =0.\]
\end{defi}
\begin{lem} \cite{Billingsley}
\label{lem:unif_equiv}
A family of random variables $\{Z_n\}_{n\geq 1}$ is uniformly integrable if and only if:
\begin{enumerate}
\item $\sup_n \expec[|Z_n|] < \infty$, and
\item For any $\e >0$, there exists some $\delta >0$ such that for all $n$ and any event $\mc{A}$ with $\text{Pr}(\mc{A}) <\delta$,
we have $\expec[|Z_n| \ \mathbf{1}_{\mc{A}}] < \e$.
\end{enumerate}
\end{lem}
Let Supp$(W|Z)$ denote the random variable whose value is the size of the support of the conditional distribution of $W$ given $Z$.
\begin{lem} \cite[Lemma $4$]{DrineaK10} \label{lem:support}
Let $\{W_n, Z_n\}_{n \geq 1}$ be a sequence of pairs of discrete random variables with Supp$(W_n|Z_n) \leq c^n$ for some constant $c \geq 1$.
Then $\sup_n \expec \left[ \left(\frac{1}{n} \log \text{Pr}(W_n|Z_n)\right)^2 \right] < \infty$. In particular, the sequence
$\left\{ -\frac{1}{n} \log \text{Pr}(W_n|Z_n)\right\}_{n\geq 1}$ is uniformly integrable.
\end{lem}
\begin{lem}\cite{Billingsley}
\label{lem:exchange_lim}
Suppose that $\{Z_n: n\geq 1\}$ is a sequence of random variables that converges to $Z$ in probability. Then the following are equivalent.
\begin{enumerate}
\item $\{Z_n: n\geq 1\}$ is uniformly integrable.
\item $\expec[|Z_n|] < \infty$ for all $n$, and $\expec[|Z_n|] \to \expec[|Z|]$ as $n \to \infty$.
\end{enumerate}
\end{lem}
\begin{lem} \label{lem:entropyrate}
Let $\mathbf{Z}=\{Z_n\}_{n \geq 1}$ be a process for the asymptotic equipartition property (AEP) holds, i.e.,
\[ \lim_{n \to \infty} -\frac{1}{n} \log \text{Pr}(Z_1, \ldots, Z_{n}) \to H(Z) \quad a.s.\]
where $H(Z)$ is the (finite) entropy rate of the process $\mathbf{Z}$. Let $\{M_n\}_{n\geq 1}$ be a sequence of
positive integer valued random variables defined on the same probability space as the $Z_n$'s, and suppose that
$\lim_{n \to \infty} \frac{M_n}{n}=x$ almost surely for some constant $x$. Then
\[ \lim_{n \to \infty} -\frac{1}{n} \log \text{Pr}(Z_1, \ldots, Z_{M_n}) = H(Z)x \quad a.s. \]
\end{lem}
\begin{proof}
Fix any $\epsilon> 0$. Since $\lim_{n \to \infty} \frac{M_n}{n} =x$, there exists an $L(\epsilon)$ such that
for all $n > L(\epsilon)$, we have almost surely
\[ a(n, \epsilon) \triangleq \lceil n(x-\epsilon)\rceil \leq M_n \leq \lceil n(x+\epsilon)\rceil \triangleq b(n, \epsilon). \]
It follows that for all $n>L(\epsilon)$,
\be
-\frac{1}{n} \log \text{Pr}(Z_1, \ldots, Z_{M_n}) \geq -\frac{1}{n} \log \text{Pr}(Z_1, \ldots, Z_{a(n,\epsilon)})
= - \frac{a(n,\epsilon)}{n} \cdot \frac{ \log \text{Pr}(Z_1, \ldots, Z_{a(n,\epsilon)})}{a(n,\epsilon)}.
\ee
Hence,
\be \label{eq:inf_hz}
\begin{split}
\liminf_{n \to \infty} -\frac{1}{n} \log \text{Pr}(Z_1, \ldots, Z_{M_n}) &\geq \liminf_{n \to \infty} -\frac{a(n,\epsilon)}{n} \cdot \frac{ \log \text{Pr}(Z_1, \ldots, Z_{a(n,\epsilon)})}{a(n,\epsilon)}. \\
&{=} \lim_{n \to \infty} \frac{a(n,\epsilon)}{n} \cdot \lim_{n \to \infty}\frac{-\log \text{Pr}(Z_1, \ldots, Z_{a(n,\epsilon)})}{a(n,\epsilon)}\\
&= (x-\epsilon)H(Z).
\end{split}
\ee
Similarly, one can show that
\be \label{eq:sup_hz}
\limsup_{n \to \infty} -\frac{1}{n} \log \text{Pr}(Z_1, \ldots, Z_{M_n}) \leq (x+\epsilon)H(Z).
\ee
Since $\epsilon >0$ is arbitrary, combining \eqref{eq:inf_hz} and \eqref{eq:sup_hz}, we get the result of the lemma.
\end{proof}
\input{sec_coding_scheme}
\input{sec_insertion_channel}
\input{sec_deletion_channel}
\input{sec_delins_channel}
\section{Conclusion} \label{sec:conc}
The framework used in this paper suggests several directions for further progress on computing the capacity of channels with synchronization errors:
\begin{itemize}
\item There are a few different ways to sharpen the bounds for the insertion channel and the deletion+insertion channel. One target is the inequality in \eqref{eq:ins_lb2_decomp} - creating the sequence $\tilde{Y}^{M_n}$ ensures one-to-one correspondence between input and output runs, but is not an optimal way to use the positions of complementary insertions. Is there a better way to use the knowledge of $(T^{M_n}, Y^{M_n})$?
When the input distribution is Markov, the deletion channel ensures that the output distribution is also Markov. But the presence of insertions results in an output process that is not Markov, which is the reason an exact expression for the limiting behavior of $\frac{1}{m} H(T^{m}|Y^{m})$ could not be obtained in Lemma \ref{lem:ins_lim_HT_Y}. A better bound for this term would improve the capacity lower bound.
\item In the decomposition of $H(X^n|Y^{M_n})$ in \eqref{eq:del_decomp} and \eqref{eq:xyts}, the penalty terms - $H(S^{M_n}|X^n, Y^{M_n})$ for the deletion channel, and $H(T^{M_n}, S^{M_n}|X^n, Y^{M_n})$ for the deletion+insertion channel - were dropped to obtain the capacity bounds.
These terms are hard to compute precisely, but any non-trivial lower bound for these terms would improve the capacity bounds.
\item Another direction is to investigate the performance of more general input distributions with i.i.d runs. For example, a distribution that is constant for small values and then decays geometrically may be a good run-length distribution for deletion channels since it decreases the probability of a run being completely deleted. A result on the structure of the optimal input distribution for small values of $i$ and $d$ (in the spirit of \cite{KanoriaM10,KalaiMM10}) would be very useful. Such a result could be combined with the approach used here to obtain good estimates of the capacity for small insertion and deletion probabilities.
\item For the insertion channel, if we fix the insertion probability $i$ and vary $\alpha$, intuition suggests that the channel with
$\alpha=1$ has the largest capacity since there is always one-to-one correspondence between input and output runs, and no auxiliary sequences are needed.
The capacity lower bound plotted in Figure \ref{fig:ins_lb_combined} seems to verify this intuition. Formally proving this conjecture would
yield an upper bound on the insertion capacity $C(i, \alpha)$ for $\alpha <1$ since very tight bounds are known for the case of $\alpha=1$ \cite{Mitz_sticky}.
\item One could also obtain upper bounds on the insertion capacity by considering a genie-aided decoder with access to the auxiliary sequence
$T^{M_n}$, as done in \cite{DiggaviG06} for the deletion channel. The task then is obtain an upper bound on the capacity per unit cost of the equivalent DMC.
\item The framework used here can be extended to derive bounds for channels with substitution errors in addition to deletions and insertions. For this, we would need an additional auxiliary sequence, e.g., a sequence that indicates the positions of the bit flips.
\item The problem of synchronization also appears in file backup and file sharing \cite{MaKTse11,ITASynch11}, where distributed nodes with different versions of the same file want to synchronize their versions. For example, consider two nodes, with the first node having source $\un{X}$ and the second having source $\un{Y}$, which is an edited version of $\un{X}$. The edits may include deletions, insertions, and substitutions. A basic question is: To update $\un{Y}$ to $\un{X}$, what is the minimum communication rate needed from the first node to the second? This is a distributed source coding problem, and it can be shown that the optimal rate is given by the limiting behavior of $H(\un{X}|\un{Y})$. The results derived in this paper provide bounds on this optimal rate for the case where $\un{X}$ is Markov, and the edit model $P(\un{Y}|\un{X})$ is one with i.i.d deletions and insertions. Extension of these results to edit models with substitution errors would yield rates to benchmark
the performance of practical file synchronization tools such as rsync \cite{rsync}.
\end{itemize}
\bibliographystyle{ieeetr}
\section{Coding Schemes} \label{sec:coding_scheme}
In this section, we describe coding schemes to give intuition about the auxiliary sequences used to obtain the bounds. The discussion is informal; the capacity bounds are rigorously proved in the following sections using a different technique: the auxiliary sequences are used to directly bound the limiting behavior of $\frac{1}{n} H(X^n|Y^{M_n})$ using information-theoretic inequalities and elementary tools from analysis.
\subsection{Insertion Channel} \label{subsec:insertion_scheme}
Consider the insertion channel with parameters $(i,\alpha)$. For $0<\alpha<1$, the inserted bits may create new runs, so we cannot associate each run of $\un{Y}$ with a run in $\un{X}$. For example, let \be \label{eq:ins_example} \un{X}={000111000} \; \ \text{ and } \; \ \un{Y}=00{\Large\emph{1}}0111 {\Large\emph{0}}000 {\Large\emph{0}},\ee where the inserted bits are indicated in large italics. There is one duplication (in the third run),
and two complementary insertions (in the first and second runs). While a duplication never introduces a new run, a complementary
insertion introduces a new run, except when it occurs at the end of a run of $\un{X}$ (e.g., the $0$ inserted at the end of the second run in \eqref{eq:ins_example}).
For any input-pair $(X^n, Y^{M_n})$, define an auxiliary sequence $T^{M_n}=(T_1, \ldots, T_{M_n})$ where $T_j=1$ if $Y_j$ is a
\emph{complementary} insertion, and $T_j=0$ otherwise. The sequence $T^{M_n}$ indicates the positions of the complementary insertions in $Y^{M_n}$.
In the example of \eqref{eq:ins_example}, $T^{M_n}=(0,0,1,0,0,0,0,1,0,0,0,0)$.
Consider the following coding scheme. Construct a codebook of $2^{nR}$ codewords of length $n$, each chosen independently
according to the first-order Markov distribution\eqref{eq:inp_def}. Let $X^n$ denote the transmitted codeword, and $Y^{M_n}$ the channel output. From $Y^{M_n}$, the decoder decodes (using joint typicality) the positions of the complementary insertions, in addition to the input sequence.
The joint distribution of these sequences is determined by the input distribution \eqref{eq:inp_def} and the channel parameters $(i,\alpha)$.
Such a decoder is sub-optimal since the complementary insertion pattern $T^{M_n}$ is not unique given an input-output pair $(X^n, Y^{M_n})$. This is discussed in Section \ref{sec:insertion}.
The maximum rate achievable by this decoder is obtained by analyzing the probability of error. Assuming all sequences satisfy the asymptotic equipartition property \cite{CoverThomas}, we have for sufficiently large $n$
\be\text{Pr(error) } \leq 2^{n(R + H(T^{M_n}|X^n))} \cdot 2^{-nI(X^n T^{M_n};Y^{M_n})}. \ee
The second term above is the probability that $(X^n,T^{M_n}, Y^{M_n})$ are jointly typical when $Y^{M_n}$ is picked independently from $(X^n,T^{M_n})$.
The first term is obtained by taking a union bound over all the codewords and all the typical complementary insertion patterns for each codeword.
Hence the probability of error goes to zero if
\be R < \frac{1}{n} \left( I(X^n T^{M_n};Y^{M_n}) - H(T^{M_n}|X^n) \right) = \frac{1}{n} \left( H(X^n) - H(T^{M_n}|Y^{M_n}) - H(X^n|T^{M_n},Y^{M_n}) \right). \ee
We obtain a lower bound on the capacity in Section \ref{subsec:lb2} by obtaining good single-letter bounds on the limiting behavior of both
$\frac{1}{n} H(T^{M_n}|Y^{M_n})$ and $\frac{1}{n} H(X^n|T^{M_n},Y^{M_n})$.
\subsection{Deletion Channel} \label{subsec:deletion_scheme}
For the deletion channel with deletion probability $d$, consider the following pair of input and output sequences
$\un{X} = 000 111 000 , \ \un{Y}= 00 1 0$.
For this pair, we can associate each {run} of $\un{Y}$ uniquely with a run in $\un{X}$.
Therefore, we can write
\ben
\begin{split}
&P(\un{Y}=0010|\un{X}=000111000)
= P(L^Y_1=2|L^X_1=3) P(L^Y_2=1|L^X_2=3) P(L^Y_3=1|L^X_3=3)
\end{split}
\een
where $L^X_j, L^Y_j$ denote the lengths of the $j$th runs of $X$ and $Y$, respectively.
We observe that if {no} runs in $\un{X}$ are completely deleted, then the conditional distribution
of $\un{Y}$ given $\un{X}$ may be written as a product distribution of run-length transformations:
\be
\begin{split}
P(\un{Y}|\un{X})= P(L^Y_1|L^X_1) P(L^Y_2|L^X_2) P(L^Y_3|L^X_3) \ldots
\end{split}
\ee
where for all runs $j$, $P(L^Y_j=s|L^X_j=r) = {r \choose s}d^{r-s} (1-d)^s$ for $1\leq s\leq r$.
In general, we do have runs of $\un{X}$ that are completely deleted. For example,
if $\un{X}=000111000$ and $\un{Y}= 000$, we cannot associate the single run in
$\un{Y}$ uniquely with a run in $\un{X}$.
For any input-output pair $(X^n, Y^{M_n})$, define an auxiliary sequence $S^{M_n+1}=(S_1, S_2,\ldots, S_{M_n+1})$,
where $S_j \in \mathbb{N}_0$ is the number of {runs} \emph{completely} deleted in ${X^n}$ between the bits corresponding to
$Y_{j-1}$ and $Y_{j}$. ($S_1$ is the number of runs deleted before the first output symbol $Y_1$, and $S_{M_n+1}$ is the number of runs deleted after
the last output symbol $Y_{M_n}$.)
For example, if $ \un{X} = 00\underbrace{\emph{011100}}0 $ and the bits shown in italics were deleted to give $\un{Y}= 000$,
then $\un{S}=(0,0,1,0)$. On the other hand, if the last six bits were all deleted, i.e.,
$ \un{X} = 000\underbrace{\emph{111000}}$, then $\un{S}=(0, 0,0,2)$. Thus $\un{S}$ is not uniquely determined given
$(\un{X}, \un{Y})$. The auxiliary sequence $\un{S}$ enables us to augment $\un{Y}$ with the positions of missing runs.
As will be explained in Section \ref{sec:deletion}, the runs of this augmented output sequence are in one-to-one correspondence with the runs
of the input sequence.
Consider the following coding scheme. Construct a codebook of $2^{nR}$ codewords of length $n$, each chosen independently
according to \eqref{eq:inp_def}. The decoder receives $Y^{M_n}$, and decodes (using joint typicality) both the auxiliary sequence and the input sequence. Such a decoder is sub-optimal since the auxiliary sequence $S^{M_n+1}$ is not unique given a codeword $X^n$ and the output $Y^{M_n}$. Assuming all sequences satisfy the asymptotic equipartition property, we have for sufficiently large $n$
\be\text{Pr(error) } \leq 2^{n(R + H(S^{M_n+1}|X^n))} \cdot 2^{-nI(X^n S^{M_n+1};Y^{M_n})}. \ee
The second term above is the probability that $(X^n,S^{M_n+1}, Y^{M_n})$ are jointly typical when $Y^{M_n}$ is picked independently from $(X^n,S^{M_n+1})$.
The first term is obtained by taking a union bound over all the codewords and all the typical auxiliary sequences for each codeword.
Hence the probability of error goes to zero if
\be R < \frac{1}{n} \left( I(X^n S^{M_n+1};Y^{M_n}) - H(S^{M_n+1}|X^n) \right) = \frac{1}{n} \left(H(X^n) - H(S^{M_n+1}|Y^{M_n}) - H(X^n|S^{M_n+1},Y^{M_n}) \right). \ee
In Section \ref{sec:deletion}, we show that the above expression converges as $n \to \infty$, and obtain an analytical expression for the limit.
For the deletion+insertion channel, we use both the auxiliary sequences, $T^{M_n}$ and $S^{M_n+1}$. The decoder decodes both these sequences in addition to the
codeword $X^n$, and the maximum achievable rate is given by
\be \label{eq:delins_coding_rate}
R < \frac{1}{n} \left(H(X^n) - H(T^{M_n}, S^{M_n+1}|Y^{M_n}) - H(X^n|S^{M_n+1},T^{M_n}, Y^{M_n}) \right).
\ee
We obtain a lower bound on the capacity of the deletion+insertion channel in Section \ref{sec:delins_channel} by analyzing the limiting behavior of \eqref{eq:delins_coding_rate}.
\section{Deletion Channel}\label{sec:deletion}
In this channel, each input bit is deleted with probability $d$, or retained with probability $1-d$.
For any input-output pair $(X^n, Y^{M_n})$, define the auxiliary sequence $S^{M_n+1}$,
where $S_j \in \mathbb{N}_0$ is the number of {runs} \emph{completely} deleted in ${X^n}$ between the bits corresponding to
$Y_{j-1}$ and $Y_{j}$. ($S_1$ is the number of runs deleted before the first output symbol $Y_1$, and $S_{M_n+1}$ is the number of runs deleted after
the last output symbol $Y_{M_n}$.) Examples of $S^{M_n}$ for the input-output pair $(\un{X} = 000111000, \ \un{Y}=000)$ were given in Section \ref{subsec:deletion_scheme}.
The auxiliary sequence $\un{S}$ enables us to augment $\un{Y}$ with the positions of missing runs.
Consider $\un{X} = 00{{011100}}0$. If the decoder was given $\un{Y}=000$ and $\un{S}=(0,0,0,2)$, it can form the
augmented sequence $ \un{Y}'=000- -$, where a $-$ denotes a missing run, or equivalently a `run of length $0$' in $\un{Y}$.
With the ``$-$'' markers indicating deleted runs,
we can associate each run of the augmented sequence $\un{Y}'$ uniquely with a run in $\un{X}$. Denote by
$L^{Y'}_1, L^{Y'}_2, \ldots $ the run-lengths of the augmented sequence $\un{Y}'$, where $L^{Y'}_j = 0$ if the run is a $-$.
Then we have
\be
P(\un{X}, \un{Y}') = P(L^X_1)P(L^{Y'}_1|L^X_1) \cdot P(L^X_2) P(L^{Y'}_2|L^X_2) \ldots
\ee
where $\forall j$:
\be \label{eq:aug_joint_dist}
\begin{split}
P(L^X_j=r)& = \gamma^{r-1}(1-\gamma), \quad r=1,2,\ldots \\
P(L^{Y'}_j=s|L^X_j=r) & = {r \choose s}d^{r-s} (1-d)^s, \quad 0\leq s\leq r.
\end{split}
\ee
Using the auxiliary sequence $S^{M_n +1}$, we can decompose $H_P(X^n | Y^{M_n})$ as
\be \label{eq:del_decomp}
H_P({X}^n | {Y}^{M_n})= H_P({X}^n , {S}^{M_n+1}| {Y}^{M_n}) - H_P({S}^{M_n+1}|{X}^n, {Y}^{M_n}).
\ee
We therefore have
\be \label{eq:del_lb1}
\limsup_{n \to \infty} \frac{1}{n} H_P({X}^n | {Y}^{M_n}) \leq \limsup_{n \to \infty} \frac{1} {n} H_P({X}^n, {S}^{M_n+1}| {Y}^{M_n})
\ee
We will show that $\lim_{n \to \infty} \frac{1}{n} H_P({X}^n, {S}^{M_n+1}| {Y}^{M_n})$ exists, and obtain an analytical expression for this limit.
Using this in \eqref{eq:mut_decomp}, we obtain a lower bound on the deletion capacity. We remark that it has been shown in \cite{DrineaK10} that for any input distribution with independent runs, $\lim_{n \to \infty} \frac{1}{n}H(X^n|Y^{M_n})$ exists for the deletion channel. Hence the $\limsup$ on the left hand side of \eqref{eq:del_lb1} is actually a limit.
\begin{prop} \label{prop:y}
The process $\mathbf{Y} =\{Y_1,Y_2, \ldots\}$ is a first-order Markov process characterized by the following joint distribution for
all $m \in \mathbb{N}$.
\[P(Y^m) = P(Y_1) \prod_{j=2}^m P(Y_j|Y_{j-1})\]
where for $y \in \{0,1\}$
\be \label{eq:del_ydist} P(Y_j=y)=0.5, \qquad P(Y_j=y|Y_{j-1}=y)=1-\ P(Y_j=\bar{y}|Y_{j-1}=y)= \frac{\gamma+d-2\gamma d}{1+d-2\gamma d}. \ee
\end{prop}
\proof The proof of this lemma can be found in \cite{Mitzenmacher09}.
\begin{prop} \label{prop:sy}
The process $\{\mathbf{S}, \mathbf{Y}\} \triangleq \{(S_1,Y_1), (S_2,Y_2), \ldots\}$ is a first-order Markov process characterized by the following joint distribution for all $m \in \mathbb{N}$:
\[ P(S^{m}, Y^m) = P(Y_1, S_1) \prod_{j=2}^{m} P(Y_j, S_j|Y_{j-1}),
\]
where for $y \in \{0,1\}$ and $j \geq 2$:
\begin{align}
P(Y_j={y}, S_j=k|Y_{j-1}=y)&= \left\{
\begin{array}{ll}
\frac{\gamma(1-d)}{(1-\gamma d)}, & k=0\\
\frac{(1-d)(1-\gamma)}{(1-\gamma d)^2} \left(\frac{d(1-\gamma)}{1-\gamma d}\right)^k, & k=1,3, \ldots\\
0, & \text{ otherwise}
\end{array}
\right. \label{eq:del_pys_y}\\
P(Y_j=\bar{y}, S_j=k|Y_{j-1}=y)&= \left\{
\begin{array}{ll}
\frac{(1-d)(1-\gamma)}{ (1-\gamma d)^2}\left(\frac{d(1-\gamma)}{1-\gamma d}\right)^k, & k=0,2, \ldots\\
0, & \text{ otherwise}
\end{array}
\right. \label{eq:del_pbarys_y}
\end{align}
\end{prop}
\begin{proof}
We need to show that
\[ P(Y_j=y, S_j=k |Y_{j-1}=y_{j-1}, S_{j-1}=s_{j-1}, Y_{j-2}=y_{j-2}, S_{j-2}=s_{j-2}, \ldots) = P(Y_j=y, S_j=k |Y_{j-1}=y_{j-1}), \]
for all $y,y_{j-1},y_{j-2}, \ldots \in \{0,1\}$ { and } $k,s_{j-1},\ldots \in \mathbb{N}_0$.
Let the output symbols $Y_j, Y_{j-1}, Y_{j-2}, \ldots$ correspond to input symbols $X_{a_j}, X_{a_{j-1}}, X_{a_{j-2}}, \ldots$ for some positive integers
$a_j> a_{j-1} > a_{j-2}> \ldots$. $S_{j-1}$ is the number of runs between the input symbols $X_{a_{j-2}}$ and $X_{a_{j-1}}$, not counting the runs containing
$X_{a_{j-2}}$ and $X_{a_{j-1}}$. Similarly, $S_{j-2}$ is the number of runs between the input symbols $X_{a_{j-3}}$ and $X_{a_{j-2}}$, not counting the runs containing
$X_{a_{j-3}}$ and $X_{a_{j-2}}$ etc.
First consider the case where $Y_{j}=Y_{j-1}=y$. When $Y_j=X_{a_j}=y$ and $Y_{j-1}=X_{a_{j-1}}=y$, note that $S_j$, the number of completely deleted runs between $X_{a_{j-1}}$ and $X_{a_j}$, is either zero or an odd number. We have
\be
\begin{split}
P(Y_j=y, S_j=0 |Y_{j-1}=y, S_{j-1}=s_{j-1}, Y_{j-2}=y_{j-2}, S_{j-2}=s_{j-2}, \ldots)
\stackrel{(a)}{=} \sum_{m=1}^{\infty} \gamma^{m} (1-\gamma) (1-d^m) = \frac{\gamma (1-d)}{(1-\gamma d)}
\end{split}
\ee
where $(a)$ is obtained as follows. $\gamma^{m} (1-\gamma)$ is the probability that the input run containing $X_{a_{j-1}}$ contains $m$ bits after $a_{j-1}$, and $(1-d^m)$ is the probability that at least one of them is not deleted. This needs to hold for some $m\geq 1$ in order to have $S_j=0$ and $Y_j=Y_j-1$.
By reasoning similar to the above, we have for $k=1,3,5,\ldots$:
\be
\begin{split}
& P(Y_j=y, S_j=k |Y_{j-1}=y, S_{j-1}=s_{j-1}, Y_{j-2}=y_{j-2}, S_{j-2}=s_{j-2}, \ldots) \\
&\stackrel{(b)}{=} \left(\sum_{m=0}^{\infty} \gamma^{m} (1-\gamma) d^m\right) \ \left(\sum_{m=1}^{\infty} \gamma^{m-1} (1-\gamma) d^m \right)^k
\left(\sum_{m=1}^{\infty} \gamma^{m-1} (1-\gamma) (1-d^m)\right)\\
&=\frac{(1-\gamma)(1-d)} {(1-\gamma d)^2} \left[\frac{d (1-\gamma)}{(1-\gamma d)}\right]^k
\end{split}
\ee
where the first term in $(b)$ is the probability that the remainder of the run containing $X_{a_{j-1}}$ is completely deleted, the second term is the probability that the next $k$ runs are deleted, and the last term is the probability that the subsequent run is \emph{not} completely deleted.
When $Y_{j}=y$ and $Y_{j-1}=\bar{y}$, the number of deleted runs $S_j$ is either zero or an even number.
For $k=0,2,4, \ldots$ we have
\be
\begin{split}
&P(Y_j=y, S_j=k |Y_{j-1}=\bar{y}, S_{j-1}=s_{j-1}, Y_{j-2}=y_{j-2}, S_{j-2}=s_{j-2}, \ldots)\\
&\stackrel{(c)}{=} \left(\sum_{m=1}^{\infty} \gamma^{m} (1-\gamma) d^m \right) \left(\sum_{m=1}^{\infty} \gamma^{m-1} (1-\gamma) d^m\right)^k
\left(\sum_{m=0}^{\infty} \gamma^{m-1} (1-\gamma) (1-d^m) \right)\\
& =\frac{(1-\gamma)(1-d)} {(1-\gamma d)^2} \left[\frac{d (1-\gamma)}{(1-\gamma d)}\right]^k.
\end{split}
\ee
In the above, the first term in $(c)$ is the probability that the remainder of the run containing $X_{a_{j-1}}$ is completely deleted,
the second term is the probability that the next
$k$ runs are deleted ($k$ may be equal to zero), and the third term is the probability that the subsequent run is not completely deleted.
This completes the proof of the lemma.
\end{proof}
We now show that $\lim_{n \to \infty} \frac{1}{n} H_P({S}^{M_n +1}|{Y}^{M_n})$ and
$\lim_{n \to \infty} \frac{1}{n} H_P(X^n|Y^{M_n}, S^{M_n+1})$ each exist, thereby proving the existence of
$\lim_{n \to \infty} \frac{1}{n} H_P(X^n, {S}^{M_n +1}|{Y}^{M_n})$.
\begin{lem} \label{lem:del_s_y}
$\lim_{n \to \infty} \frac{1}{n} H_P(S^{M_n+1}|Y^{M_n}) = (1-d) H_P(S_2|Y_1 Y_2)$ where the joint distribution of $(Y_1,Y_2, S_2)$ is given by \eqref{eq:del_ydist}, \eqref{eq:del_pys_y}, and \eqref{eq:del_pbarys_y}.
\end{lem}
\begin{proof}
See Appendix \ref{proof:del_s_y}.
\end{proof}
To determine the limiting behavior of $\frac{1}{n}H({X}^n| {S}^{M_n +1}, Y^{M_n})$, we recall that $X^n$ can be equivalently represented in terms of its
run-lengths as $(L^X_1, \ldots, L^X_{R_n})$, where $R_n$, the number of runs in $X^n$, is a random variable. Also recall from the discussion at the beginning
of this section that the pair of sequences $({S}^{M_n +1}, Y^{M_n})$ is equivalent to an augmented sequence $\un{Y}'$ formed by adding the positions of the deleted runs to $\un{Y}=Y^{M_n}$. $\un{Y}'$ can be equivalently represented in terms of its run-lengths as $(L^{Y'}_1, \ldots, L^{Y'}_{R_n})$, where we emphasize that $L^{Y'}_1,L^{Y'}_2, \ldots$ can take value $0$ as well. To summarize, we have
\be
\begin{split}
X^n & \leftrightarrow (L^X_1, \ldots, L^X_{R_n}), \\
({S}^{M_n +1}, Y^{M_n}) & \leftrightarrow (L^{Y'}_1, \ldots, L^{Y'}_{R_n}).
\end{split}
\ee
Thus, for all $n$
\begin{equation}
\label{eq:bits_to_runs}
H_P({X}^n| {S}^{M_n +1}, Y^{M_n}) = H_P(L^X_1, \ldots, L^X_{R_n}|L^{Y'}_1, \ldots, L^{Y'}_{R_n}).
\end{equation}
\begin{prop} \label{prop:lxy'}
The process $\{\mathbf{L^X, L^{Y'}}\} \triangleq \{(L^X_1, L^{Y'}_1), (L^X_2, L^{Y'}_2), \ldots \}$ is an i.i.d process characterized by the following joint distribution for all $j \geq 1$:
\be P(L^X_j=r, L^{Y'}_j=s) = \gamma^{r-1} (1-\gamma) \cdot {r \choose s} d^{r-s} (1-d)^s, \quad r=1,2,\ldots, 0\leq s \leq r.
\label{eq:del_xy'_joint}
\ee
\end{prop}
\begin{proof} Since $\mathbf{X}$ is a Markov process, $\{L^X_j\}_{j\geq 1}$ are independent with
\[ P(L^X_j=r)= \gamma^{r-1} (1-\gamma), \:r=1,2,\ldots\]
Since the deletion process is i.i.d, each $L^{Y'}_j$ can be thought of being obtained by passing a run of length $L^X_j$ through a discrete memoryless channel with transition probability
\[ P(L^{Y'}_j=s|L^X_j=r) = {r \choose s} d^{r-s} (1-d)^s, \ 0\leq s \leq r. \]
\end{proof}
\begin{lem} \label{lem:del_x_sy} $\lim_{n \to \infty} \frac{1}{n} H_P({X}^n| {S}^{M_n +1}, Y^{M_n}) = (1-\gamma) H_P(L^X|L^{Y'})$
where the joint distribution of $(L^X, L^{Y'})$ is given by \eqref{eq:del_xy'_joint}.
\end{lem}
\begin{proof}
See Appendix \ref{proof:del_x_sy}.
\end{proof}
Using Lemmas \ref{lem:del_s_y} and \ref{lem:del_x_sy}, we obtain the following lower bound on the capacity of the deletion channel.
\begin{thm} \label{thm:del_prop}
The deletion channel capacity $C(d)$ can be lower bounded as
\[C(d) \geq \max_{0<\gamma<1} h(\gamma) - (1-d) H(S_2|Y_1 Y_2) - (1-\gamma) H(L^{X}|L^{Y'}) \]
where
\be \label{eq:del_HS2_y1y2}
\begin{split}
H(S_2|Y_1 Y_2) = {\gamma \bar{\theta}} \log_2 \frac{q}{\bar{\theta}}
+ \frac{\beta \theta }{{\bar{\theta}}^2} \log_2 \frac{1}{\theta} + \frac{\beta \theta }{1-\theta^2} \log_2 \frac{q}{\beta}
+\frac{\beta }{1-\theta^2}\log_2 \frac{\bar{q}}{\beta},
\end{split}
\ee
\[ q = \frac{\gamma + d - 2\gamma d}{1 + d - 2\gamma d}, \quad \theta = \frac{(1-\gamma) d}{1-\gamma d} , \quad
\beta = \frac{(1-\gamma)(1-d)}{(1-\gamma d)^2}\]
and
\be \label{eq:del_HLXLY}
\begin{split}
H(L^{X}|L^{Y'})=&\left(\frac{d}{\bar{\gamma}} -\frac{d \bar{\gamma}}{(1-\gamma d)^2}\right)\log_2 \frac{1}{\gamma d} + \frac{d \bar{\gamma} h(d\gamma)}{(1-d\gamma)^2 }
-\frac{\bar{d}(2-\gamma - \gamma d)\log_2 ({1-\gamma d})}{\bar{\gamma}(1-\gamma d)}\\
& - \frac{\bar{\gamma}}{\gamma} \sum_{k=1}^\infty \sum_{j=1}^\infty (\bar{d}\gamma)^k \, (d\gamma)^j\, {j+k \choose k} \, \log_2{j+k \choose k}.
\end{split}
\ee
\end{thm}
\begin{proof}
Combining \eqref{eq:mut_decomp} and \eqref{eq:del_lb1}, we obtain
\be
C(d)> h(\gamma) - \limsup_{n \to \infty} \frac{H_P(X^n, S^{M_n+1}|Y^{M_n})}{n}.
\ee
From Lemmas \ref{lem:del_s_y} and \ref{lem:del_x_sy}, we have
\be
\begin{split}
\lim_{n \to \infty} \frac{1}{n} H_P(X^n, S^{M_n+1}|Y^{M_n}) & = \lim_{n \to \infty} \frac{1}{n} H_P(S^{M_n+1}|Y^{M_n})
+ \lim_{n \to \infty} \frac{1}{n} H_P(X^n|S^{M_n+1},Y^{M_n}) \\
&= (1-d) H(S_2|Y_1 Y_2) + (1-\gamma) H(L^{X}|L^{Y'}).
\end{split}
\ee
$ H(S_2|Y_1 Y_2)$ can then be computed using the joint distribution given by \eqref{eq:del_ydist}, \eqref{eq:del_pys_y}, and \eqref{eq:del_pbarys_y}.
$H(L^{X}|L^{Y'})$ can be computed using the joint distribution given in Proposition \ref{prop:lxy'}. Finally, we optimize the lower bound
by maximizing over the Markov parameter $\gamma \in (0,1)$.
\end{proof}
\begin{figure}
\centering
\includegraphics[width=5.1in]{del_channel}
\caption{Lower bound of Theorem \ref{thm:del_prop} on the deletion capacity. The lower bound from \cite{DrineaM07}
is shown in dashed lines.}
\label{fig:del_fig}
\vspace{-1pt}
\end{figure}
Figure \ref{fig:del_fig} shows the lower bound of Theorem \ref{thm:del_prop} as well as the lower bound of \cite{DrineaM07}
for various values of $d$. We observe that our bound is close to, but slightly smaller than that of \cite{DrineaM07}, which is the
best known lower bound on the deletion capacity.
In the decomposition of $H_P(X^n|Y^{M_n})$ in \eqref{eq:del_decomp}, we dropped the term $H_P({S}^{M_n+1}|{X}^n, {Y}^{M_n})$ to obtain the bound in \eqref{eq:del_lb1}. $H_P({S}^{M_n+1}|{X}^n, {Y}^{M_n})$ is the uncertainty in the positions of the deleted runs given both the input and output sequences. For example, if $X^n=001100$ and $Y^{M_n}=000$, then $S^{M_n+1}$ equals
\begin{itemize}
\item $(0,1,0,0)$ if the deletion pattern is either $0 \ \emph{0} \ \emph{1} \ \emph{1} \ 0 \ 0$ or $\emph{0} \ 0 \ \emph{1} \ \emph{1} \ 0 \ 0$
(bits in italics are deleted).
\item $(0,0,1,0)$ if the deletion pattern is either $0 \ 0 \ \emph{1} \ \emph{1} \ \emph{0} \ 0$ or $0 \ 0 \ \emph{1} \ \emph{1} \ 0 \ \emph{0}$.
\end{itemize}
Computing $\lim_{n \to \infty} H_P({S}^{M_n+1}|{X}^n, {Y}^{M_n})$ precisely is hard, but obtaining a non-trivial lower bound for this quantity would improve the lower bound of Theorem \ref{thm:del_prop}.
\section{The Deletion+Insertion Channel}
\label{sec:delins_channel} %
Recall that this channel is defined by three parameters $(d,i,\alpha)$.
Each input bit undergoes a deletion with probability $d$, a duplication with probability.
$i\alpha$, a complementary insertion with probability $i\bar{\alpha}$.
\begin{figure}
\centering
\includegraphics[width=5.5in]{two_ch_decomp}
\caption{Cascade channel equivalent to the deletion+insertion channel.}
\label{fig:two_ch_decomp}
\vspace{-1pt}
\end{figure}
Note that each input bit is deleted with probability $d$; given that a particular bit is \emph{not} deleted, the probability that it undergoes an insertion is $\frac{i}{1-d}$. Therefore, one can think of the channel as a cascade of two channels, as shown in Figure \ref{fig:two_ch_decomp}. The first channel is a deletion channel that deletes each bit independently with probability $d$. The second channel is an insertion channel with parameters
$(i', \alpha)$, where $i' \triangleq \frac{i}{1-d}$. We prove the equivalence of this cascade decomposition below.
\emph{Claim}: The deletion+insertion channel is thus equivalent to the cascade channel in the sense that both have the same transition probability $P(\un{Y}|\un{X})$.
\begin{proof}
For an $n$-bit input sequence, define the deletion-insertion pattern $\Lambda^n=(\Lambda_1, \Lambda_2,\ldots, \Lambda_n)$ of the channel as the sequence where $\Lambda_i$ indicates whether the channel introduces a deletion/duplication/complementary insertion/no modification in bit $i$ of the input.
Note that if the underlying probability space is $(\Omega, \mc{F}, P)$, the realization $\omega \in \Omega$ determines the deletion-insertion pattern $\Lambda^n(\omega)$. We calculate the probability of any specified pattern occurring in a)the deletion+insertion channel, and b)the cascade channel.
Consider a deletion-insertion pattern $\lambda^n$ with $k$ deletions at positions
$a_1, a_2, \ldots, a_k$, $l$ duplications at positions $b_1, \ldots, b_l$, and $m$ complementary insertions at positions $c_1,\ldots,c_m$.
The probability of this pattern occurring in the deletion+insertion channel is
\[P_{delins}(\Lambda^n(\omega)=\lambda^n)= d^k (i\alpha)^l (i\bar{\alpha})^m (1-d-i)^{n-k-l-m}. \]
The probability of this pattern occurring in the cascade channel of Figure \ref{fig:two_ch_decomp} is
\be
\begin{split}
P_{casc}(\Lambda^n(\omega)=\lambda^n)&\stackrel{(a)}{=} \left[d^k (1-d)^{n-k} \right] \left[(i'\alpha)^l (i'\bar{\alpha})^m (1-i')^{(n-k)-l-m}\right]\\
&= \left[ d^k (1-d)^{n-k} \right] \left[\left(\frac{i\alpha}{1-d}\right)^l \left(\frac{i\bar{\alpha}}{1-d}\right)^m \left(\frac{1-d-i}{1-d}\right)^{n-k-l-m}\right] \\ &=d^k (i\alpha)^l (i\bar{\alpha})^m (1-d-i)^{n-k-l-m}.
\end{split}
\ee
where the first term in $(a)$ is the probability of deletions occurring in the specified positions in the first channel, and the second term is the probability of the insertions occurring in the specified positions in the second channel. Hence every deletion-insertion pattern has the same probability in both the deletion+insertion channel and the cascade channel. This implies that the two channels have the same transition probability.
\end{proof}
To obtain a lower bound on the capacity, we use two auxiliary sequences $T^{M_n}=(T_1, \ldots, T_{M_n})$, and
$S^{M_n +1} = (S_1, \ldots, S_{M_n+1})$. As in Section \ref{subsec:lb1}, $T^{M_n}$ indicates the complementary insertions in $Y^{M_n}$: $T_j=1$ if $Y_j$
is a complementary insertion, and $T_j=0$ otherwise. As in Section \ref{sec:deletion}, $S^{M_n+1}$ indicates the positions of the missing runs: $S_j=k$, if $k$ runs
were completely deleted between $Y_{j-1}$ and $Y_j$. Using these auxiliary sequences, we can decompose $H_P(X^n|Y^{M_n})$ as
\be \label{eq:xyts}
\begin{split}
H_P(X^n|Y^{M_n}) & = H_P(X^n, T^{M_n}, S^{M_n +1}|Y^{M_n}) - H_P(T^{M_n}, S^{M_n +1}| X^n, Y^{M_n})\\
&= H_P(T^{M_n}|Y^{M_n}) + H_P(S^{M_n +1}| T^{M_n}, Y^{M_n}) + H_P(X^n|S^{M_n +1}, T^{M_n}, Y^{M_n})- H_P(T^{M_n}, S^{M_n +1}| X^n, Y^{M_n})\\
&\leq H_P(T^{M_n}|Y^{M_n}) + H_P(S^{M_n +1}| T^{M_n}, Y^{M_n}) + H_P(X^n|S^{M_n +1}, \tilde{Y}^{M_n})- H_P(T^{M_n}, S^{M_n +1}| X^n, Y^{M_n})
\end{split}
\ee
where $\tilde{Y}^{M_n}$ is the sequence formed by flipping the complementary insertions in $Y^{M_n}$. The inequality in the last line of
\eqref{eq:xyts} holds because $\tilde{Y}^{M_n}$ is a function of $(T^{M_n}, Y^{M_n})$.
We therefore have
\be \label{eq:delins_ineq}
\begin{split}
\limsup_{n \to \infty} \frac{1}{n} H_P(X^n|Y^{M_n}) & \leq \limsup_{n \to \infty} \frac{1}{n} (H_P(T^{M_n}|Y^{M_n}) + H_P(S^{M_n +1}| T^{M_n}, Y^{M_n}) + H_P(X^n|S^{M_n +1}, \tilde{Y}^{M_n}) )\\
& \leq \limsup_{n \to \infty} \frac{H_P(T^{M_n}|Y^{M_n})}{n} + \limsup_{n \to \infty} \frac{H_P(S^{M_n +1}| T^{M_n}, Y^{M_n})}{n}
+ \limsup_{n \to \infty} \frac{H_P(X^n|S^{M_n +1}, \tilde{Y}^{M_n})}{n}.
\end{split}
\ee
Using this upper bound for $\limsup_{n \to \infty} \frac{H_P(X^n|Y^{M_n})}{n}$ in \eqref{eq:mut_decomp}, we obtain a lower bound on the capacity
of the deletion+insertion channel.
\begin{lem} \label{lem:delins_1}
$\limsup_{n \to \infty} \frac{1}{n} H_P(T^{M_n}|Y^{M_n}) = (1-d+i) \limsup_{m \to \infty} \frac{1}{m} H_P(T^{m}|Y^{m})$.
\end{lem}
\begin{proof}
The proof follows the same steps as that of Lemma \ref{lem:ins_H_I_Y}, with two changes: $T^{M_n}$ replaces $I^{M_n}$, and we note that $\frac{M_n}{n}$ converges almost surely to $(1-d+i)$ for the deletion+insertion channel.
\end{proof}
\begin{lem} \label{lem:delins_2}
$\limsup_{m \to \infty} \frac{1}{m}H_P(T^m|Y^m) \leq \lim_{j \to \infty} H_P(T_j|T_{j-1}, Y_{j}, Y_{j-1})$, where
\[ \lim_{j \to \infty} H_P(T_j|T_{j-1}, Y_{j}, Y_{j-1}) = \frac{(\bar{q}\bar{d} + q i \bar{\alpha})}{(1-d+i)} h\left(\frac{i \bar{\alpha}}{\bar{q}\bar{d} + q i \bar{\alpha}}\right), \; \text{ and } q= \frac{\gamma+d-2\gamma d}{1+d-2\gamma d}.\]
\end{lem}
\begin{proof}
We have
\be
\frac{H_P(T^{m}|Y^{m})}{m} = \frac{\sum_{j=1}^m H_P(T_j|T^{j-1}, Y^m)} {m} \leq \frac{\sum_{j=1}^m H_P(T_j|T_{j-1}, Y_{j}, Y_{j-1})} {m}.
\ee
Therefore
\be
\limsup_{m \to \infty} \frac{H_P(T^{m}|Y^{m})}{m} \leq \limsup_{m \to \infty} \frac{\sum_{j=1}^m H_P(T_j|T_{j-1}, Y_{j}, Y_{j-1})} {m}=
\lim_{j \to \infty} H_P(T_j|T_{j-1}, Y_{j}, Y_{j-1}),
\ee
provided the limit exists.
From the cascade representation in Figure \ref{fig:two_ch_decomp}, we see that the insertions are introduced by the second channel in the cascade,
an insertion channel with parameters $(i', \alpha)$. The input to this insertion channel is
a process $\mathbf{Z}=\{Z_m\}_{m \geq 1}$, which is the output of the first channel in the cascade. From Propositon \ref{prop:y}, $\mathbf{Z}$
is a first-order Markov process with parameter $q=\frac{\gamma+d-2\gamma d}{1+d-2\gamma d}$.
Therefore, we need to calculate $\lim_{j \to \infty} H(T_j|T_{j-1}, Y_{j}, Y_{j-1})$ where $\mathbf{Y}$ is the output when a first-order Markov process with parameter $q$ is transmitted through an insertion channel with parameters $(i',\alpha)$. But we have already computed
$\lim_{j \to \infty} H(T_j|T_{j-1}, Y_{j}, Y_{j-1})$ in Lemma \ref{lem:ins_lim_HT_Y} for an insertion channel with parameters $(i,\alpha)$ with a
first-order Markov input with parameter $\gamma$. Hence, in Lemma \ref{lem:ins_lim_HT_Y}, we can replace $\gamma$ by $q$, and $i$ by $i'$ to obtain
\[ \lim_{j \to \infty} H_P(T_j|T_{j-1}, Y_{j}, Y_{j-1})
= \frac{(1-q+ q i'\bar{\alpha})}{(1+i')} h\left(\frac{i' \bar{\alpha}}{1-q+ q i' \bar{\alpha}}\right).
\]
Substituting $i'=\frac{i}{1-d}$ and simplifying gives the statement of the lemma.
\end{proof}
\begin{lem} \label{lem:delins_3}
$\limsup_{n \to \infty} \frac{1}{n} H_P(S^{M_n +1}| T^{M_n}, Y^{M_n}) = (1-d+i) \limsup_{m \to \infty} \frac{1}{m} H_P(S^{m}| T^{m}, Y^{m})$.
\end{lem}
\begin{proof}
The proof is along the same lines as that of Lemma \ref{lem:ins_H_I_Y}; here we use the uniform integrability of the sequence
$\left\{ -\frac{1}{n} \log P(S^{M_n +1}| T^{M_n}, Y^{M_n}) \right\}$ along with the fact that $\frac{M_n}{n} \to (1-d+i)$ almost surely. The uniform integrability follows from Lemma \ref{lem:support} since Supp$(S^{M_n+1}|T^{M_n}, Y^{M_n})$ is upper bounded by $2^n$ for the reasons explained in Section \ref{proof:del_s_y}.
\end{proof}
\begin{lem} \label{lem:delins_4}
$\limsup_{m \to \infty} \frac{1}{m} H_P(S^{m}| T^{m}, Y^{m}) \leq \lim_{j \to \infty} H_P(S_j|Y_{j-1}, Y_j, T_j)
= \frac{1}{1+i'}(A_1 + A_2 - \frac{\theta \beta}{(1-\theta)^2} \log_2\theta)$,
where
\[ i' = \frac{i}{1-d}, \quad q= \frac{\gamma+d-2\gamma d}{1+d-2\gamma d}, \quad \theta = \frac{(1-\gamma) d}{1-\gamma d} , \quad
\beta = \frac{(1-\gamma)(1-d)}{(1-\gamma d)^2},\]
\ben
\begin{split}
A_1 &= \frac{\theta \beta(1-i' \bar{\alpha})}{1-\theta^2} \log_2\left(\frac{i'\alpha + (1-i'\bar{\alpha})q + i'\bar{\alpha}\bar{q}}{\beta (1-i'\bar{\alpha})}\right)
+\frac{\theta^2 \beta i' \bar{\alpha}}{1-\theta^2} \log_2\left(\frac{i'\alpha + (1-i'\bar{\alpha})q + i'\bar{\alpha}\bar{q}}{\beta i'\bar{\alpha}}\right)
\\
& \quad
+((1-i'\bar{\alpha}) \gamma \bar{\theta} + i' \bar{\alpha} \beta+ i'\alpha)
\log_2 \left(\frac{i'\alpha + (1-i'\bar{\alpha})q + i'\bar{\alpha}\bar{q}}{i'\alpha + (1-i'\bar{\alpha}) \gamma \bar{\theta} + i' \bar{\alpha} \beta}\right), \\
A_2&= \frac{\theta^2 \beta(1-i' \bar{\alpha})}{1-\theta^2} \log_2\left(\frac{(1-i'\bar{\alpha})\bar{q} + i'\bar{\alpha}{q}}{\beta (1-i'\bar{\alpha})}\right)
+\frac{\theta \beta i' \bar{\alpha}}{1-\theta^2} \log_2\left(\frac{(1-i'\bar{\alpha})\bar{q} + i'\bar{\alpha}{q}}{\beta i'\bar{\alpha}}\right)
\\
& \quad
+ (i' \bar{\alpha} \gamma \bar{\theta} + (1-i'\bar{\alpha})\beta)
\log_2 \left(\frac{ (1-i'\bar{\alpha})\bar{q} + i'\bar{\alpha}{q}}{i' \bar{\alpha} \gamma \bar{\theta} + (1-i'\bar{\alpha})\beta}\right).
\end{split}
\een
\end{lem}
\begin{proof}
See Appendix \ref{proof:delins_lem4}
\end{proof}
We now determine the limiting behavior of $\frac{1}{n} H(X^n|S^{M_n +1}, \tilde{Y}^{M_n})$. By flipping the complementary insertions in $Y^{M_n}$ to
obtain $\tilde{Y}^{M_n}$, we have removed the {extra} runs introduced by the channel. Using $S^{M_n+1}$, we can augment $\tilde{Y}^n$ by adding the positions of the {deleted} runs to obtain a sequence $Y'^{M_n}$ which contains the same number of runs as $X^n$. $Y'^{M_n}$ can be represented in terms of its run-lengths as $(L^{Y'}_1, \ldots, L^{Y'}_{R_n})$, where we emphasize that $L^{Y'}_1,L^{Y'}_2, \ldots$ can take value $0$ as well.
To summarize, we have
\be
\begin{split}
X^n & \leftrightarrow (L^X_1, \ldots, L^X_{R_n}), \\
({S}^{M_n +1}, \tilde{Y}^{M_n}) & \leftrightarrow (L^{Y'}_1, \ldots, L^{Y'}_{R_n}).
\end{split}
\ee
Thus, for all $n$
\begin{equation}
\label{eq:insdel_bits_runs}
H_P({X}^n| {S}^{M_n +1}, \tilde{Y}^{M_n}) = H_P(L^X_1, \ldots, L^X_{R_n}|L^{Y'}_1, \ldots, L^{Y'}_{R_n}).
\end{equation}
\begin{prop} \label{prop:delins_lxy'}
The process $\{\mathbf{L^X, L^{Y'}}\} \triangleq \{(L^X_1, L^{Y'}_1), (L^X_2, L^{Y'}_2), \ldots \}$ is an i.i.d process characterized by the following joint distribution for all $j \geq 1$:
\be \label{eq:LY'_LX}
\begin{split}
P(L^X_j=r) &= \gamma^{r-1} (1-\gamma), \quad r=1,2,\ldots \\
P(L^{Y'}_j=s|L^X_j=r)&= \sum_{n_i\in \mathcal I}{r \choose {n_i ; r+n_i-s}} i^{n_i} d^{r+n_i-s} (1-d-i)^{s-2n_i}, \quad 0\leq s \leq 2r
\end{split}
\ee
where $\mathcal{I}$, the set of possible values for the number of insertions $n_i$, is given by
\ben
\mathcal{I} =
\{0,1,\ldots, \lfloor{\frac{s}{2}}\rfloor \} \: \text{ for } s \leq r, \text{ and }
\{s-r,\ldots, \lfloor{\frac{s}{2}}\rfloor \} \: \text{ for } s>r.
\een
\end{prop}
\begin{proof} Since $\mathbf{X}$ is a Markov process, $\{L^X_j\}_{j\geq 1}$ are independent with
\[ P(L^X_j=r)= \gamma^{r-1} (1-\gamma), \:r=1,2,\ldots\]
Since there is a one-to-one correspondence between the runs of $\mathbf{X}$ and the runs of $\mathbf{Y'}$, we can think of each $L^{Y'}_j$ being obtained by passing a run of length $L^X_j$ through a discrete memoryless channel. For a pair $(L^{X}_j=r, L^{Y'}_j=s)$, if the number of insertions is $n_i$, the number of deletions is easily seen to be $r+n_i-s$. Since there can be at most one insertion after each input bit, no more than half the bits in an output run can be insertions; hence the maximum value of $n_i$ is $\lfloor{\frac{s}{2}}\rfloor$. The minimum value of $n_i$ is zero for $s\leq r$, and $s-r$ for $s>r$. Using these together with the fact that each bit can independently undergo an insertion with probability $i$, a deletion with probability $d$, or no change with probability $1-d-i$, the transition probability of the memoryless run-length channel is given by the second line of \eqref{eq:LY'_LX}.
\end{proof}
\begin{lem} \label{lem:delins_5}
$\limsup_{n \to \infty} \frac{1}{n} H_P({X}^n| {S}^{M_n +1}, \tilde{Y}^{M_n}) = (1-\gamma) H_P(L^X|L^{Y'})$
where the joint distribution of $(L^X,L^{Y'})$ is given by \eqref{eq:LY'_LX}.
\end{lem}
\begin{proof}
The proof is identical to that of Lemma \ref{lem:del_x_sy}.
\end{proof}
\begin{thm} \label{thm:delins_thm}
The capacity of the deletion+insertion channel can be lower bounded as
\[ C(d, i, \alpha) \geq \max_{0 < \gamma <1} h(\gamma) - (\bar{q}(1-d) + q i \bar{\alpha}) h\left(\frac{i \bar{\alpha}}{\bar{q}(1-d) + q i \bar{\alpha}}\right) - (1-d)(A_1 + A_2 - \frac{\theta \beta}{(1-\theta)^2} \log_2\theta) - (1-\gamma) H_P(L^X_{1}| L^{Y'}_1)\]
where $q, \beta, \theta, A_1, A_2$ are defined in Lemma \ref{lem:delins_4}, and $H_P(L^X_{1}| L^{Y'}_1)$
is computed using the joint distribution in \eqref{eq:LY'_LX}.
\end{thm}
\begin{proof}
The result is obtained by using Lemmas \ref{lem:delins_1}-\ref{lem:delins_5} in \eqref{eq:delins_ineq}, and substituting the resulting bound in \eqref{eq:mut_decomp}.
\end{proof}
\begin{figure}
\centering
\includegraphics[width=5.1in]{del_ins_channel}
\caption{Lower bound on the deletion+insertion capacity $C(d,i,\alpha)$ for $d=i$.}
\label{fig:delins_fig}
\vspace{-2pt}
\end{figure}
The lower bound is plotted in Figure \ref{fig:delins_fig} for various values of $d=i$,
for $\alpha=0.8$ and for $\alpha=1$. For Theorem \ref{thm:delins_thm}, we used the sequence $T^{M_n}$ to indicate the positions of complementary insertions together with $S^{M_n}$ to indicate deleted runs. We can obtain another lower bound on the deletion+insertion capacity by using the sequence $I^{M_n}$ instead of $T^{M_n}$, in Section \ref{subsec:lb1}. This bound can be derived in a straightforward manner by combining the techniques of Sections \ref{subsec:lb1} and \ref{sec:deletion}, and is omitted.
\section{Insertion Channel}\label{sec:insertion}
In this channel, an extra bit may be inserted after each bit of $\un{X}$ with probability $i \in (0,1)$. When a bit is inserted after $X_j$, the inserted bit is equal to ${X}_j$ (a duplication) with probability $\alpha$, and equal to $\bar{X}_j$ (a complementary insertion) with probability $1-\alpha$.
When $\alpha=1$, we have only duplications - this is the sticky channel studied in \cite{Mitz_sticky}. In this case, we can associate each run of $\un{Y}$ with a unique run in $\un{X}$, which leads to a computable single-letter characterization of the best achievable rates with a first-order Markov distribution.
We derive two lower bounds on the capacity of the insertion channel, each using a different auxiliary sequence.
\subsection{Lower Bound $1$} \label{subsec:lb1}
For any input-pair $(X^n, Y^{M_n})$, define an auxiliary sequence $I^{M_n}=(I_1, \ldots, I_{M_n})$ where $I_j=1$ if $Y_j$ is an inserted bit, and $I_j=0$ otherwise. The sequence $I^{M_n}$ indicates the positions of all the inserted bits in $Y^{M_n}$, and is not unique for a given $(X^n, Y^{M_n})$. Using $I^{M_n}$, we can decompose $H_P(X^n|Y^{M_n})$ as
\[ {H_P(X^n|Y^{M_n})}= H_P(X^n, I^{M_n}|Y^{M_n}) - H_P(I^{M_n}| X^n, Y^{M_n}) = H_P(I^{M_n}|Y^{M_n}) - H_P(I^{M_n}| X^n, Y^{M_n}) \]
since $H(X^n|Y^{M_n}, I^{M_n})=0$. Therefore,
\be \label{eq:HXY_HIY} \begin{split}
\limsup_{n \to \infty} \frac{1}{n}H_P(X^n|Y^{M_n}) &= \limsup_{n \to \infty} \frac{1}{n} \left(H_P(I^{M_n}|Y^{M_n}) - H_P(I^{M_n}| X^n, Y^{M_n}) \right)\\
&\leq \limsup_{n \to \infty} \frac{1}{n} H_P(I^{M_n}|Y^{M_n}) - \liminf_{n \to \infty} \frac{1}{n}H_P(I^{M_n}| X^n, Y^{M_n}).
\end{split}
\ee
We will derive an upper bound on $\limsup \frac{1}{n}H_P(I^{M_n}|Y^{M_n})$, and a lower bound on $\liminf \frac{1}{n}H_P(I^{M_n}| X^n, Y^{M_n})$. Using this in \eqref{eq:HXY_HIY}, we get an upper bound for $\limsup \frac{1}{n}H_P(X^n|Y^{M_n})$, which can then be used in \eqref{eq:mut_decomp}.
\begin{prop} \label{prop:ins_sy}
The process $\{\mathbf{I}, \mathbf{Y}\} \triangleq \{(I_1,Y_1), (I_2,Y_2), \ldots\}$ is a second-order Markov process characterized by the following joint distribution for all $m \in \mathbb{N}$:
\[ P(I^{m}, Y^m) = P(I_1, Y_1) P(I_2,Y_2|I_1, Y_1) \prod_{j=3}^{m} P(I_j, Y_j|I_{j-1}, Y_{j-1}, Y_{j-2})
\]
where for $x,y \in \{0,1\}$ and $j \geq 3$:
\be \label{eq:IY_dist}
\begin{split}
P(I_j=1, Y_j=y| I_{j-1}=0, Y_{j-1}=y, Y_{j-2}=x)= i\alpha, & \ \ P(I_j=1, Y_j=\bar{y}| I_{j-1}=0, Y_{j-1}=y, Y_{j-2}=x)= i\bar{\alpha} \\
P(I_j=0, Y_j=y| I_{j-1}=0, Y_{j-1}=y, Y_{j-2}=x)= \bar{i} \gamma, & \ \ P(I_j=0, Y_j=\bar{y}| I_{j-1}=0, Y_{j-1}=y, Y_{j-2}=x)= \bar{i} \bar{\gamma}\\
P(I_j=0, Y_j=x| I_{j-1}=1, Y_{j-1}=y, Y_{j-2}=x)= \gamma, &\ \ P(I_j=0, Y_j=\bar{x}| I_{j-1}=1, Y_{j-1}=y, Y_{j-2}=x)= \bar{\gamma}
\end{split}
\ee
\end{prop}
\begin{proof}
We need to show that for all $j\geq 3$, the following Markov relation holds: $(I_j, Y_j) - (I_{j-1}, Y_{j-1}, Y_{j-2}) - (I^{j-2}, Y^{j-3})$.
First consider $P(I_j, Y_j|I_{j-1}=0, Y_{j-1}=y, I^{j-2}, Y^{j-2})$. Since $I_{j-1}=0$, $Y_{j-1}$ is the most recent input bit (say $X_a$) before $Y_j$.
$P(I_j=0, Y_j=y|I_{j-1}=0, Y_{j-1}=y, I^{j-2}, Y^{j-2})$ is the probability that the following independent events both occur: $1$) the input bit $X_{a+1}$ equals $X_a$, \emph{and} $2$) there was no insertion after input bit $X_a$. Since the insertion process is i.i.d and is independent of the first-order Markov input process $\mathbf{X}$, we have
\[P(I_j=0, Y_j=y|I_{j-1}=0, Y_{j-1}=y, I^{j-2}, Y^{j-2}) = (1-i) \gamma. \]
Similarly, we obtain
\ben
\begin{split}
P(I_j=0, Y_j=\bar{y}|I_{j-1}=0, Y_{j-1}=y, I^{j-2}, Y^{j-2}) &= (1-i) (1-\gamma), \\
P(I_j=1, Y_j=y|I_{j-1}=0, Y_{j-1}=y, I^{j-2}, Y^{j-2}) &= i\alpha, \\
P(I_j=1, Y_j=\bar{y}|I_{j-1}=0, Y_{j-1}=y, I^{j-2}, Y^{j-2}) &= i(1-\alpha). \\
\end{split}
\een
Next consider $P(I_j, Y_j|I_{j-1}=1, Y_{j-1}=y, Y_{j-2}=x, I^{j-2}, Y^{j-3})$. Since $I_{j-1}=1$, $Y_{j-2}$ is the most recent input bit (say, $X_a$) before
$Y_j$. Also note that $Y_j$ is the input bit $X_{a+1}$ since $Y_{j-1}$ is an insertion. (At most one insertion can occur after each input bit.)
Hence $P(I_j=0, Y_j=x|I_{j-1}=1, Y_{j-1}=y, Y_{j-2}=x, I^{j-2}, Y^{j-2})$ is just the probability that $X_{a+1}=X_a$, which is equal to $\gamma$.
Similarly,
\[ P(I_j=0, Y_j=\bar{x}|I_{j-1}=1, Y_{j-1}=y, Y_{j-2}=x, I^{j-2}, Y^{j-2}) = 1-\gamma.\]
\end{proof}
\emph{\textbf{Remark}}: Proposition \ref{prop:ins_sy} implies that the process $\{\mathbf{I}, \mathbf{Y}\}$ can be characterized as a Markov chain with
state at time $j$ given by $(I_{j}, Y_{j}, Y_{j-1})$. This is an aperiodic, irreducible Markov chain. Hence a stationary distribution $\pi$
exists, which can be verified to be
\be \label{eq:stat_pi}
\begin{split}
\pi(I_{j}=1, Y_{j}=y, Y_{j-1}=y)= \frac{i\alpha}{2(1+i)}, & \quad \pi(I_{j}=1, Y_{j}=\bar{y}, Y_{j-1}=y)= \frac{i\bar{\alpha}}{2(1+i)}, \\
\pi(I_{j}=0, Y_{j}=y, Y_{j-1}=y)= \frac{\bar{i}\gamma+ i \alpha \gamma + i \bar{\alpha} \bar{\gamma}}{2(1+i)},
& \quad \pi(I_{j}=0, Y_{j}=\bar{y}, Y_{j-1}=y)= \frac{\bar{i} \bar{\gamma} + i \alpha \bar{\gamma} + i \bar{\alpha} \gamma }{2(1+i)}\\
\end{split}
\ee
for $y\in\{0,1\}$.
\begin{lem}
\label{lem:ins_H_I_Y}
$\limsup_{n \to \infty} \frac{1}{n} H_P(I^{M_n}|Y^{M_n}) = (1+i) \limsup_{m \to \infty} \frac{1}{m} H_P(I^{m}|Y^{m})$.
\end{lem}
\begin{proof}
See Appendix \ref{proof:ins_H_I_Y}.
\end{proof}
\begin{lem}
\label{lem:lim_HI_Y}
$\limsup_{m \to \infty} \frac{1}{m} H_P(I^{m}|Y^{m}) \leq \lim_{j \to \infty} H_P(I_j|I_{j-1}, Y_{j}, Y_{j-1}, Y_{j-2})$,
and
\be \label{eq:hI_limit}
\lim_{j \to \infty} H_P(I_j|I_{j-1}, Y_{j}, Y_{j-1}, Y_{j-2})= \frac{(i\alpha + \bar{i}\gamma)}{1+i} h\left(\frac{i\alpha}{i\alpha + \bar{i}\gamma}\right) + \frac{(i\bar{\alpha} + \bar{i}\bar{\gamma})}{1+i} h\left(\frac{i\bar{\alpha}}{i\bar{\alpha} + \bar{i}\bar{\gamma}}\right). \ee
\end{lem}
\begin{proof}
See Appendix \ref{proof:lim_HI_Y}.
\end{proof}
Next, we focus on $H(I^{M_n}|Y^{M_n}, X^n)$, which is the uncertainty in the positions of the insertions given both the channel input and output sequences.
For example, given input $\un{X} = 0 \ 0 \ 0 \ 1 \ 0$, and output $\un{Y}= 0 \ 0 \ 0 \ 1 \ 1 \ 0$,
we know that there is either a complementary insertion after the third bit of $\un{X}$
or a duplication after the fourth bit; so there is uncertainty in the values of $I_4$ and $I_5$ (one of them is zero, and the other is one.). But there is no uncertainty in $I_1, I_2,I_3, I_6$, which are all zero.
We use this intuition to obtain a lower bound on the limiting behavior of $\frac{1}{n}H(I^{M_n}|Y^{M_n}, X^n)$.
\begin{lem} \label{lem:insI_YX}
$\liminf_{n \to \infty} \frac{1}{n}H_P(I^{M_n}|Y^{M_n}, X^n) = \liminf_{n \to \infty}\frac{1}{n}H_P(I^{n(1+i)}|Y^{n(1+i)}, X^n)$.
\end{lem}
\begin{proof}
The proof of this lemma is similar to that of Lemma \ref{lem:ins_H_I_Y}, and is omitted.
\end{proof}
\begin{lem} \label{lem:lb1_liminf}
$ \liminf_{n \to \infty}\frac{1}{n} H_P(I^{n(1+i)}|Y^{n(1+i)}, X^n) \geq \bar{\gamma}^2 i (\bar{\alpha} + \bar{i} \alpha) h\left(\frac{\bar{\alpha}}{\bar{\alpha} + \bar{i} \alpha} \right).$
\end{lem}
\begin{proof}
See Appendix \ref{proof:lb1_liminf}.
\end{proof}
\begin{thm}
\label{thm:ins_lb1}
(LB $1$) The capacity of the insertion channel with parameters $(i,\alpha)$ can be lower bounded as
\[ C(i,\alpha) \geq \max_{0 < \gamma <1} \ h(\gamma)
- (i\alpha + \bar{i}\gamma) h\left(\frac{i\alpha}{i\alpha + \bar{i}\gamma}\right) -
(i\bar{\alpha} + \bar{i} \bar{\gamma}) h\left(\frac{i\bar{\alpha}}{i\bar{\alpha} + \bar{i}\bar{\gamma}}\right)
+ \bar{\gamma}^2 i (\bar{\alpha} + \bar{i} \alpha) h\left(\frac{\bar{\alpha}}{\bar{\alpha} + \bar{i} \alpha} \right).\]
\end{thm}
\begin{proof}
The result is obtained by using Lemmas \ref{lem:ins_H_I_Y}, \ref{lem:lim_HI_Y}, \ref{lem:insI_YX}, and \ref{lem:lb1_liminf}
in \eqref{eq:HXY_HIY}, and substituting the resulting bound in \eqref{eq:mut_decomp}.
We optimize the lower bound by maximizing over the Markov parameter $\gamma \in (0,1)$.
\end{proof}
\subsection{Lower Bound $2$} \label{subsec:lb2}
For any input-pair $(X^n, Y^{M_n})$, define an auxiliary sequence $T^{M_n}=(T_1, \ldots, T_{M_n})$ where $T_j=1$ if $Y_j$ is a
\emph{complementary} insertion, and $T_j=0$ otherwise. The sequence $T^{M_n}$ indicates the positions of the complementary insertions in $Y^{M_n}$.
Note that $T^{M_n}$ is different from the sequence $I^{M_n}$, which indicates the positions of \emph{all} the insertions.
Using $T^{M_n}$, we can decompose $H_P(X^n|Y^{M_n})$ as
\be \label{eq:ins_lb2_decomp}
\begin{split}
{H_P(X^n|Y^{M_n})}& = H_P(X^n, T^{M_n}|Y^{M_n}) - H_P(T^{M_n}| X^n, Y^{M_n}) \\
&= H_P(T^{M_n}|Y^{M_n}) + H_P(X^n|T^{M_n}, Y^{M_n}) - H_P(T^{M_n}| X^n, Y^{M_n})\\
& \stackrel{(a)}{\leq} H_P(T^{M_n}|Y^{M_n}) + H_P(X^n| \tilde{Y}^{M_n}) - H_P(T^{M_n}| X^n, Y^{M_n}),
\end{split}
\ee
where $\tilde{Y}^{M_n}$ is the sequence obtained from $(T^{M_n}, Y^{M_n})$ by flipping $Y_j$ whenever $T_j=1$, for $1\leq j \leq M_n$.
$(a)$ holds in \eqref{eq:ins_lb2_decomp} because $\tilde{Y}^{M_n}$ is a function of $(T^{M_n}, Y^{M_n})$, and hence
$H_P(X^{n}| T^{M_n}, Y^{M_n}) \leq H(X^n| \tilde{Y}^{M_n})$.
Therefore, we have
\be \label{eq:ins_lb2a}
\begin{split}
\limsup_{n \to \infty} \frac{1}{n} H_P(X^n|Y^{M_n}) \leq \limsup_{n \to \infty} \frac{1}{n}\left( H_P(T^{M_n}|Y^{M_n}) + H_P(X^n| \tilde{Y}^{M_n}) - H_P(T^{M_n}| X^n, Y^{M_n})\right) .
\end{split}
\ee
We will show that $\lim_{n \to \infty} \frac{1}{n}H(X^n| \tilde{Y}^{M_n})$ exists, and therefore \eqref{eq:ins_lb2a} becomes
\be \label{eq:ins_lb2}
\begin{split}
\limsup_{n \to \infty} \frac{1}{n} H_P(X^n|Y^{M_n})
&\leq \limsup_{n \to \infty} \frac{1}{n}\left(H_P(T^{M_n}|Y^{M_n})- H_P(T^{M_n}| X^n, Y^{M_n})\right)+ \lim_{n \to \infty} \frac{1}{n}H_P(X^n|\tilde{Y}^{M_n})\\
& \leq \limsup_{n \to \infty} \frac{1}{n} H_P(T^{M_n}|Y^{M_n})- \liminf_{n \to \infty} \frac{1}{n} H_P(T^{M_n}| X^n, Y^{M_n}) + \lim_{n \to \infty} \frac{1}{n} H_P(X^n|\tilde{Y}^{M_n})
\end{split}
\ee
We will use \eqref{eq:ins_lb2} in \eqref{eq:mut_decomp} to obtain a lower bound on the insertion capacity.
\begin{lem}
\label{lem:ins_H_T_Y}
$\limsup_{n \to \infty} \frac{1}{n}H_P(T^{M_n}|Y^{M_n}) = (1+i) \limsup_{m \to \infty} \frac{1}{m} H_P(T^{m}|Y^{m})$.
\end{lem}
\begin{proof}
The proof of this lemma is identical to that of Lemma \ref{lem:ins_H_I_Y}, and can be obtained $T^{M_n}$ replacing $I^{M_n}$.
\end{proof}
\begin{lem}
\label{lem:ins_lim_HT_Y}
$\limsup_{m \to \infty} \frac{1}{m} H_P(T^{m}|Y^{m}) \leq \lim_{j \to \infty} H_P(T_j|T_{j-1}, Y_{j}, Y_{j-1})$,
and
\be \label{eq:hT_limit}
\lim_{j \to \infty} H_P(T_j|T_{j-1}, Y_{j}, Y_{j-1})
= \frac{(1-\gamma+ \gamma i \bar{\alpha})}{(1+i)} h\left(\frac{i \bar{\alpha}}{1-\gamma+ \gamma i \bar{\alpha}}\right ). \ee
\end{lem}
\begin{proof}
See Appendix \ref{proof:ins_lim_HT_Y}.
\end{proof}
We now determine the limiting behavior of $\frac{1}{n}H({X}^n| \tilde{Y}^{M_n})$. Recall that $\tilde{Y}^{M_n}$ is obtained
by flipping the complementary insertions in $Y^{M_n}$. In other words, $\tilde{Y}^{M_n}$ has insertions in the same locations as $Y^{M_n}$, but the insertions are all duplications. Hence $\tilde{Y}^{M_n}$ has the same number of runs as $X^n$. Recall from Section \ref{sec:prelim} that we can represent both binary sequences in terms of their run-lengths as
\[
X^n \leftrightarrow (L^X_1, \ldots, L^X_{R_n}), \qquad
\tilde{Y}^{M_n} \leftrightarrow (L^{\tilde{Y}}_1, \ldots, L^{\tilde{Y}}_{R_n}),
\]
where $R_n$, the number of runs in $X^n$ (and $\tilde{Y}^n$) is a random variable. Therefore, for all $n$, we have
\begin{equation}
\label{eq:ins_bits_to_runs}
H_P({X}^n| \tilde{Y}^{M_n}) = H_P(L^X_1, \ldots, L^X_{R_n}|L^{\tilde{Y}}_1, \ldots, L^{\tilde{Y}}_{R_n}).
\end{equation}
\begin{prop} \label{prop:lxytil}
The process $\{\mathbf{L^X, L^{\tilde{Y}}}\} \triangleq \{(L^X_1, L^{\tilde{Y}}_1), (L^X_2, L^{\tilde{Y}}_2), \ldots \}$ is an i.i.d process characterized by the following joint distribution for all $j \geq 1$:
\[ P(L^X_j=r, L^{\tilde{Y}}_j=s) = \gamma^{r-1} (1-\gamma) \cdot {r \choose {s-r}} i^{s-r} (1-i)^{2r-s}, \quad r=1,2,\ldots, r\leq s \leq 2r. \]
\end{prop}
\begin{proof} Since $\mathbf{X}$ is a Markov process, $\{L^X_j\}_{j\geq 1}$ are independent with
\[ P(L^X_j=r)= \gamma^{r-1} (1-\gamma), \:r=1,2,\ldots\]
$\tilde{Y}^{M_n}$ is generated from $X^n$ by independently duplicating each bit with probability $i$. Hence $L^{\tilde{Y}}_j$ can be thought of being obtained by passing a run of length $L^X_j$ through a discrete memoryless channel with transition probability
\[ P(L^{\tilde{Y}}_j=s|L^X_j=r) = {r \choose {s-r}} i^{s-r} (1-i)^{2r-s}, \quad r\leq s \leq 2r. \]
\end{proof}
\begin{lem} \label{lem:ins_x_ytil}
$\lim_{n \to \infty} \frac{1}{n} H_P({X}^n| \tilde{Y}^{M_n}) = (1-\gamma) H_P(L^X_1|L^{\tilde{Y}}_1)$, where the joint distribution of $(L^X_1, L^{\tilde{Y}}_1)$ is given by Proposition \ref{prop:lxytil}.
\end{lem}
\begin{proof}
See Appendix \ref{proof:ins_x_ytil}.
\end{proof}
Finally, we need to analyze $H(T^{M_n}|Y^{M_n}, X^n)$, the uncertainty in the positions of the complementary insertions given both the channel input and output sequences.
For example, given input $\un{X} = 0 \ 0 \ 0 \ 1 \ 0$, and output $\un{Y}= 0 \ 0 \ 0 \ 1 \ 1 \ 0$,
we know that there is either a complementary insertion after the third bit of $\un{X}$
or a duplication after the fourth bit; so there is uncertainty in the value of $T_4$. There is no uncertainty in $T_1, T_2, T_3, T_5, T_6$, which are all zero.
We use this intuition to obtain a lower bound on the limiting behavior of $\frac{1}{n}H(T^{M_n}|Y^{M_n}, X^n)$.
\begin{lem} \label{lem:insT_YX}
$\liminf_{n \to \infty} \frac{1}{n}H_P(T^{M_n}|Y^{M_n}, X^n) = \liminf_{n \to \infty}\frac{1}{n}H_P(T^{n(1+i)}|Y^{n(1+i)}, X^n)$.
\end{lem}
\begin{proof}
The proof of this lemma is similar to that of Lemma \ref{lem:ins_H_I_Y}, and is omitted.
\end{proof}
\begin{lem} \label{lem:lb2_liminf}
$ \liminf_{n \to \infty}\frac{1}{n} H_P(T^{n(1+i)}|Y^{n(1+i)}, X^n) \geq \bar{\gamma}^2 i (\bar{\alpha} + \bar{i} \alpha) h\left(\frac{\bar{\alpha}}{\bar{\alpha} + \bar{i} \alpha} \right).$
\end{lem}
\begin{proof}
See Appendix \ref{proof:lb2_liminf}.
\end{proof}
\begin{figure}
\centering
\includegraphics[width=5.1in]{two_ins_lbs}
\caption{Comparison of the two lower bounds for different values of $i$ with $\alpha=0.8$}
\label{fig:ins_lbs_compare}
\vspace{-2pt}
\end{figure}
\begin{figure}
\centering
\includegraphics[width=5.1in]{ins_channel_improv}
\caption{Lower bound $\max\{LB1, LB2\}$ on the insertion capacity $C(i,\alpha)$. For $\alpha=1$, the lower bound of \cite{Mitz_sticky} is shown using `*'.}
\label{fig:ins_lb_combined}
\vspace{-1pt}
\end{figure}
\begin{thm} \label{thm:ins_lb2} (LB $2$)
The capacity of the insertion channel with parameters $(i,\alpha)$ can be lower bounded as
\[ C(i,\alpha) \geq \max_{0 < \gamma <1} \ h(\gamma) - \left(\bar{\gamma}+ \gamma i \bar{\alpha}\right) h\left(\frac{i \bar{\alpha}}{\bar{\gamma} + \gamma i \bar{\alpha}}\right) - \bar{\gamma} H({L_{X}}_1|{L_{\tilde{Y}}}_1) + \bar{\gamma}^2 i (\bar{\alpha} + \bar{i} \alpha) h\left(\frac{\bar{\alpha}}{\bar{\alpha} + \bar{i} \alpha} \right) \]
where $H({L_{X}}_1|{L_{\tilde{Y}}}_1)$ is computed using the joint distribution given in Proposition \ref{prop:lxytil}.
\end{thm}
\begin{proof}
The result is a direct consequence of using Lemmas \ref{lem:ins_H_T_Y} - \ref{lem:lb2_liminf} in \eqref{eq:ins_lb2}, and substituting the resulting upper bound for $\limsup_{n \to \infty} \frac{1}{n} H(X^n|Y^{M_n})$ in \eqref{eq:mut_decomp}.
\end{proof}
Figure \ref{fig:ins_lbs_compare} compares Lower bound $1$ with Lower bound $2$ for different values of $i$ with $\alpha$ fixed at $0.8$.
We observe that $LB \ 2$ is generally a better bound that $LB \ 1$, except when $i$ is large. For large $i$, it is more efficient to decode the positions of all the insertions (since $i$ is large) rather than just the complementary insertions. Specifically, comparing Lemmas \ref{lem:lim_HI_Y} and \ref{lem:ins_lim_HT_Y},
\[ \lim_{j \to \infty} H(I_j|I_{j-1}, Y_j, Y_{j-1}, Y_{j-2}) \leq \lim_{j \to \infty} H(T_j|T_{j-1}, Y_j, Y_{j-1})\] for large values of $i$.
Combining the bounds of Theorems \ref{thm:ins_lb1} and \ref{thm:ins_lb2}, we observe that $\max\{LB\ 1, LB\ 2\}$ is a lower bound to the insertion capacity. This is plotted in Figure \ref{fig:ins_lb_combined} for various values of $i$, for $\alpha=0.5,0.8,1$. For $\alpha=1$, the bound is very close to
the near-optimal lower bound in \cite{Mitz_sticky}. The gap occurs because we have used a Markov input distribution instead of the numerically optimized input distribution in \cite{Mitz_sticky}.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 3,720 |
\section{Introduction}
Three classical random matrix ensembles on the real line, Gaussian beta ensembles, Wishart beta ensembles and MANOVA beta ensembles, are now realized as eigenvalues of certain random Jacobi matrices. For instance, the following random Jacobi matrices whose components are independent and distributed as
\[
H_{n,\beta} = \begin{pmatrix}
a_1 &b_1 \\
b_1 &a_2 &b_2\\
&\ddots &\ddots &\ddots\\
& &b_{n - 1} &a_n
\end{pmatrix}
\sim\frac{1}{\sqrt{n\beta}}\begin{pmatrix}
{\mathcal N}(0,2) &\chi_{(n-1)\beta} \\
\chi_{(n - 1)\beta} &{\mathcal N}(0,2) &\chi_{(n - 2)\beta}\\
&\ddots &\ddots &\ddots\\
& &\chi_{\beta} &{\mathcal N}(0,2)
\end{pmatrix}
\]
are matrix models of (scaled) Gaussian beta ensembles for any $\beta > 0$. Here ${\mathcal N}(\mu, \sigma^2)$ denotes the normal (or Gaussian) distribution with mean $\mu$ and variance $\sigma^2$, and $\chi_k$ denotes the chi distribution with $k$ degrees of freedom. Namely, the eigenvalues of $H_{n,\beta}$ are distributed as Gaussian beta ensembles,
\[
(\lambda_1, \dots, \lambda_n) \propto |\Delta(\lambda)|^{\beta} \exp\left(-\frac{n\beta}{4} \sum_{i = 1}^n \lambda_i^2 \right),
\]
where $\Delta(\lambda) = \prod_{i<j}(\lambda_j - \lambda_i)$ denotes the Vandermonde determinant.
Three special values of beta, $\beta = 1,2$ and $4$, correspond to Gaussian orthogonal, unitary and symplectic ensembles (GOE, GUE and GSE) in which the above formula describes the joint distribution of eigenvalues of random matrices with real, complex and quaternion entries, respectively. As a generalization, Gaussian beta ensembles were originally defined as ensembles of points on the real line whose joint density function is given as above. They can be also viewed as the equilibrium measure of a one dimensional Coulomb log-gas at the inverse temperature $\beta$. Using the idea of tridiagonalizing a GOE matrix, Dumitriu and Edelman \cite{Dumitriu-Edelman-2002} introduced the model $H_{n, \beta}$ for Gaussian beta ensembles. In the same paper, they also gave a matrix model for Wishart beta ensembles. A model for MANOVA beta ensembles was discovered later by Killip and Nenciu \cite{Killip-Nenciu-2004}.
One of main objects in random matrix theory is to study the limiting behavior of the empirical distribution of eigenvalues
\[
L_n = \frac{1}{n} \sum_{i = 1}^n \delta_{\lambda_i},
\]
where $\delta$ denotes the Dirac measure. For Gaussian beta ensembles, as $n$ tends to infinity, the empirical distributions converge weakly, almost surely, to the semicircle distribution, which is well known as Wigner's semicircle law. The convergence means that for any bounded continuous function $f$ on ${\mathbb R}$,
\[
\langle L_n, f\rangle = \frac{1}{n} \sum_{i = 1}^n f(\lambda_i) \to \langle sc, f\rangle \text{ almost surely as } n\to \infty,
\]
with $sc$ denoting the semicircle distribution, a probability measure supported on $[-2,2]$ with density $sc(x) = (2\pi)^{-1} \sqrt{4 - x^2}$. A fluctuation around the limit was also investigated. To be more precise, it was shown that for a `nice' test function $f$,
\[
n(\langle L_n, f\rangle - \langle sc, f\rangle ) = \sum_{i = 1}^n (f(\lambda_i) - \langle sc, f\rangle) \overset{d}{\to } {\mathcal N}(0, a_f^2),
\]
where $a_f^2$ can be written as a quadratic functional of $f$. There are several ways to prove those results. See Johansson~\cite{Johansson-1998} for an approach based on joint density function, Dumitriu and Edelman \cite{Dumitriu-Edelman-2006} and Dumitriu and Paquette \cite{Dumitriu-Paquette-2012} for a combinatorial approach based on the random Jacobi matrix models.
Since GOE and GUE have their original matrix models, we can see more approaches in books \cite{Anderson-book,Pastur-book}. Note that the idea in the last section of this paper is also applicable to study such Gaussian fluctuation. Using the idea, we can show that the class of `nice' test functions for which the above central limit theorem holds contains at least differentiable functions whose derivative is continuous of polynomial growth.
The spectral measures of random Jacobi matrices associated with those beta ensembles have been investigated recently. The weak convergence to a limit distribution, a central limit theorem for moments and large deviations have been established \cite{Dette-Nagel-2012,Gamboa-Rouault-2011,Nagel-2013}. The spectral measure of a finite Jacobi matrix, a symmetric tridiagonal matrix of the form,
\[
J = \begin{pmatrix}
a_1 &b_1 \\
b_1 &a_2 &b_2\\
&\ddots &\ddots &\ddots\\
& &b_{n - 1} &a_n
\end{pmatrix},
(a_i \in {\mathbb R}, b_i > 0),
\]
is defined to be a unique probability measure $\mu$ on ${\mathbb R}$ satisfying
\[
\langle \mu, x^k\rangle = \langle J^k e_1, e_1\rangle = J^k(1,1), k = 0,1,\dots,
\]
where $e_1 = (1,0,\dots,0)^t \in {\mathbb R}^n$. Let $\{\lambda_1, \dots, \lambda_n\}$ be the eigenvalues of $J$ and $\{v_1, \dots, v_n\}$ be the corresponding eigenvectors which are chosen to be an orthonormal basis of ${\mathbb R}^n$. Then the spectral measure $\mu$ can be written as
\[
\mu = \sum_{i = 1}^n q_i^2 \delta_{\lambda_i},\quad q_i = |v_i(1)|.
\]
Note that the eigenvalues $\{\lambda_i\}$ are distinct and the weights $\{q_i^2\}$ are all positive. Moreover, a finite Jacobi matrix of size $n$ is one-to-one correspondence with a probability measure supported on $n$ real points.
Let $\mu_n$ be the spectral measure of $H_{n, \beta}$,
\[
\mu_n = \sum_{i = 1}^n q_i^2 \delta_{\lambda_i}.
\]
In this case, and in all three beta ensembles in the paper, the weights $\{q_i^2\}$ are independent of eigenvalues and have Dirichlet distribution with parameters $(\beta/2, \dots, \beta/2)$ (or symmetric Dirichlet distribution with parameter $\beta/2$). The distribution of $\{q_i^2\}$ is the same as that of the vector
\[
\left(\frac{\chi_{\beta,1}^2}{\sum_{i = 1}^n \chi_{\beta,i}^2} , \dots, \frac{\chi_{\beta,n}^2}{\sum_{i = 1}^n \chi_{\beta,i}^2} \right),
\]
where $\{\chi_{\beta,i}^2\}_{i = 1}^n$ is an i.i.d.~sequence of random variables having chi-squared distributions with $\beta$ degrees of freedom. In connection with empirical distributions, Nagel~\cite{Nagel-082013} showed that as $n$ tends to infinity, the Kolmogorov distance between $L_n$ and $\mu_n$ converges almost surely to zero. Thus the spectral measures and the empirical distributions converge to the same limit. Moreover, the limiting behavior of spectral measures of Jacobi matrices can be read off from the convergence of their entries. For Gaussian beta ensembles, it is clear that
\[
H_{n, \beta}
\to
\begin{pmatrix}
0 & 1\\
1 & 0 & 1\\
&\ddots &\ddots &\ddots
\end{pmatrix} =: J_{free}, \text{ almost surely as } n \to \infty.
\]
Here the convergence means the piecewise convergence of entries. The non-random Jacobi matrix $J_{free}$ is called the free Jacobi matrix whose spectral measure is nothing but the semicircle distribution \cite[Section~1.10]{Simon-book}. Consequently the spectral measures $\mu_n$, and hence, the empirical distributions $L_n$, converge weakly, almost surely, to the semicircle distribution. This result may be regarded as a strong law of large numbers for spectral measures.
A natural problem now is to study the fluctuation of spectral measures around the limit, or a central limit theorem for $\langle \mu_n, f \rangle$ with a `nice' function $f$. In a work which is not so related to random matrix theory, Dette and Nagel \cite{Dette-Nagel-2012} derived a central limit theorem for moments $\{\langle \mu_n, x^k \rangle\}$ of spectral measures. The result covers Gaussian, Wishart beta ensembles and MANOVA beta ensembles with fixed parameters. The aim of this paper is to reconsider the central limit theorem. We propose a universal approach which can be easily applied to all three beta ensembles. The idea is that for a polynomial test function, when $n$ is large enough, $
\langle \mu_n, p\rangle = p(H_{n,\beta}) (1,1)
$
is a polynomial of finite variables. Then the central limit theorem follows from the limiting behavior of entries of Jacobi matrices. Furthermore, by a relation between spectral measures and empirical measures, we obtain an explicit formula for the limit variance and can extend the central limit theorem to a large class of test functions. Our main result for Gaussian beta ensembles can be stated as follows.
\begin{theorem}
\begin{itemize}
\item[\rm (i)] The spectral measures $\mu_n$ converge weakly, almost surely, to the semicircle distribution as $n \to \infty$, that is, for any bounded continuous function $f$,
\[
\langle \mu_n, f\rangle = \sum_{i = 1}^n q_i^2 f(\lambda_i) \to \langle sc, f\rangle \text{ almost surely as }n \to \infty.
\]
\item[\rm (ii)] For a function $f$ with continuous derivative of polynomial growth,
\[
\frac{\sqrt{n\beta}}{\sqrt{2}} (\langle \mu_n, f \rangle - {\mathbb E}[\langle \mu_n, f \rangle]) \overset{d}{\to } {\mathcal N}(0, \sigma^2(f)) \text{ as } n \to \infty,
\]
where $\sigma^2(f) = \langle sc, f^2 \rangle - \langle sc, f \rangle^2 = \Var_{sc}[f]$. Here `$\overset{d}{\to }$' denotes convergence in distribution or weak convergence of random variables.
\end{itemize}
\end{theorem}
\noindent The results for Wishart and MANOVA beta ensembles are analogous where the semicircle distribution is replaced by Marchenko-Pastur distributions and Kesten-Mckey distributions, respectively.
The Kolmogorov distance, the metric which implies the weak convergence, between two measures $\mu$ and $\nu$ on the real line with distribution functions $F_\mu$ and $F_\nu$, respectively, is defined by
\[
d_K(\mu, \nu) = \sup_{x \in {\mathbb R}} |F_\mu(x) - F_\nu(x)|.
\]
Here $F_\mu(x) = \mu((-\infty, x])$. As mentioned above, for fixed $\beta > 0$, the Kolmogorov distance between the empirical distributions $L_n$ and the spectral measures $\mu_n$ converges to zero almost surely as $n$ tends to infinity. For the proof, we only need properties that both measures are support on the set of eigenvalues and that the weights $\{q_i^2\}$ have symmetric Dirichlet distribution with parameter $\beta/2$ \cite[Theorem~4.2]{Nagel-082013}. Consequently, the following strong law of large numbers for empirical distributions holds.
\begin{corollary}
For all three beta ensembles in this paper, as $n \to \infty$, the empirical distributions $L_n$ converge weakly, almost surely, to the same limit as the spectral measures.
\end{corollary}
The paper is organized as follows. In the next section, we consider general random Jacobi matrices and derive the weak convergence of spectral measures as well as the central limit theorem for polynomial test functions. Applications to Gaussian, Wishart and MANOVA beta ensembles are then investigated in turn. The last section is devoted to extend the central limit theorem to a larger class of test functions.
\section{Limiting behavior of spectral measures of random Jacobi matrices}
Let us begin by introducing some spectral properties of (non random) Jacobi matrices. A semi-infinite Jacobi matrix is a symmetric tridiagonal matrix of the form
\[
J = \begin{pmatrix}
a_1 &b_1 \\
b_1 &a_2 &b_2\\
&\ddots &\ddots &\ddots
\end{pmatrix},
\text{ where }a_i \in {\mathbb R}, b_i > 0.
\]
To a Jacobi matrix $J$, there exists a probability measure $\mu$ such that
\[
\langle \mu, x^k\rangle = \int_{\mathbb R} x^k d\mu = \langle J^k e_1, e_1\rangle, k = 0,1,\dots,
\]
where $e_1 = (1,0,\dots,)^t\in \ell^2$. Then $\mu$ is unique, or $\mu$ is determined by its moments, if and only if, $J$ is an essentially self-adjoint operator on $\ell^2$. If the parameters $\{a_i\}$ and $\{b_i\}$ are bounded, or more generally, if $\sum b_i^{-1} = \infty$, then $J$ is essentially self-adjoint \cite[Corollary 3.8.9]{Simon-book}. In case of uniqueness, we call $\mu$ the spectral measure of $J$, or of $(J, e_1)$. See \cite[Chapter~2]{Deift-book} or \cite[Section~3.8]{Simon-book} for more details on Jacobi matrices.
When $J$ is a finite Jacobi matrix of order $n$, then the spectral measure $\mu$ is supported on the eigenvalues $\{\lambda_i\}$ of $J$ with weights $\{q_i^2\} = \{v_i(1)^2\}$,
\[
\mu = \sum_{i = 1}^n q_i^2 \delta_{\lambda_i}.
\]
Here $\{v_1, \dots, v_n\}$ are the corresponding eigenvectors which are chosen to be an orthogonal basis of ${\mathbb R}^n$. The eigenvalues $\{\lambda_i\}_{i = 1}^n$ are distinct and the weights $\{q_i^2\}_{i = 1}^n$ are all positive \cite[Proposition~2.40]{Deift-book}.
We are now in a position to study the convergence of spectral measures of Jacobi matrices. Recall that spectral measures are defined by their moments. Does the convergence of moments imply the weak convergence of probability measures? The following lemma gives us the answer. It is a classical result which can be found in some textbooks in probability theory.
\begin{lemma}\label{lem:deterministic}
Assume that $\{\mu_n\}_{n=1}^\infty$ and $\mu$ are probability measures on ${\mathbb R}$ such that for all $k=0,1,\dots,$
\[
\langle \mu_n, x^k\rangle \to \langle \mu, x^k\rangle \text{ as } n\to \infty.
\]
Assume further that the measure $\mu$ is determined by its moments. Then $\mu_n$ converges weakly to $\mu$ as $n \to \infty$. Moreover, if $f$ is a continuous function of polynomial growth, that is, there is a polynomial $p$ such that $|f(x)|\le p(x)$ for all $x\in {\mathbb R}$, then we also have
\[
\langle \mu_n, f\rangle \to \langle \mu, f\rangle \text{ as } n\to \infty.
\]
\end{lemma}
\begin{proof}
The first part of this lemma is a well-known moment problem \cite[Theorem~30.2]{Billingsley-PnM}. The second part follows by a truncated argument. For the sake of completeness, we give proof here. Assume that the sequence $\{\mu_n\}$ converges weakly to $\mu$ and that $\langle \mu_n, p \rangle$ converges to $\langle \mu, p\rangle$ for all polynomials $p$. Let $f$ be a continuous function which is dominated by a polynomial $p$, $|f(x)| \le p(x)$ for all $x \in {\mathbb R}$. For $M > 0$, write $f_M$ for the truncated function
\[
f_M(x) = \begin{cases}
-M, &\text{if } f(x) \le -M,\\
f(x), &\text{if } |f(x)| \le M,\\
M, &\text{if } f(x) \ge M.\\
\end{cases}
\]
Then it is clear that $|f - f_M| \le p - p_M$, where $p_M$ is the truncated function of $p$. Thus by the triangle inequality,
\begin{align*}
\left| \langle \mu_n, f \rangle - \langle \mu, f \rangle\right| &\le \left| \langle \mu_n, f - f_M \rangle \right| + \left| \langle \mu_n, f_M \rangle - \langle \mu, f_M \rangle\right| + \left| \langle \mu, f - f_M \rangle\right| \\
&\le \langle \mu_n, p - p_M \rangle + \left| \langle \mu_n, f_M \rangle - \langle \mu, f_M \rangle\right| + \langle \mu, p - p_M \rangle \\
&= \langle \mu_n, p \rangle - \langle \mu_n, p_M \rangle + \left| \langle \mu_n, f_M \rangle - \langle \mu, f_M \rangle\right| + \langle \mu, p - p_M \rangle.
\end{align*}
As $n \to \infty$, the first term converges to $\langle \mu, p\rangle$ by the assumption, the second term converges to $\langle \mu, p_M\rangle$ and the third term converges to $0$ because $p_M$ and $f_M$ are bounded continuous functions. Therefore
\[
\limsup_{n \to \infty} \left| \langle \mu_n, f \rangle - \langle \mu, f \rangle\right| \le 2 \langle \mu, p - p_M \rangle.
\]
Finally, by letting $M \to \infty$, $ \langle \mu, p - p_M \rangle \to 0$ by the monotone convergence theorem. The lemma is proved.
\end{proof}
To random probability measures, we deal with two types of convergence, almost sure convergence and convergence in probability. A result for almost sure convergence is a direct consequence of the above deterministic result. However, it is not the case for convergence in probability. When the limit measure has compact support, the following result on convergence in probability of random measures can be derived by a method of polynomials approximation, see subsection~2.1.2 in \cite{Anderson-book}, for instance.
\begin{lemma}\label{lem:convergence-of-moments}
Let $\{\mu_n\}_{n = 1}^\infty$ be a sequence of random probability measures and $\mu$ be a non-random probability measure which is determined by its moments. Assume that any moment of $\mu_n$ converges almost surely to that of $\mu$, that is,
for any $k = 0,1,\dots,$
\[
\langle \mu_n, x^k\rangle \to \langle \mu, x^k\rangle \text{ a.s.~as } n\to \infty.
\]
Then as $n\to \infty$, the sequence of measures $\{\mu_n\}$ converges weakly, almost surely, to $\mu$, namely, for any bounded continuous function $f$,
\[
\langle \mu_n, f\rangle \to \langle \mu, f\rangle \text{ a.s.~as } n\to \infty.
\]
The convergence still holds for a continuous function $f$ of polynomial growth. An analogous result holds for convergence in probability.
\end{lemma}
\begin{proof}
The case of almost sure convergence is a direct consequence of Lemma~\ref{lem:deterministic}. Indeed, for $k \ge 1$, let
\[
A_k = \{\omega : \langle \mu_n(\omega) , x^k\rangle \to \langle \mu, x^k\rangle \text{ as } n \to \infty\}.
\]
Then ${\mathbb P}(A_k) = 1$ by the assumption. Therefore ${\mathbb P}(A:= \bigcap_{k=1}^\infty A_k) = 1$. Applying Lemma~\ref{lem:deterministic} to the sequence of probability measures $\{\mu_n(\omega)\}$, for $\omega \in A$, yields the desired result.
Next we consider the case of convergence in probability. The idea here is to use the following criterion for convergence in probability \cite[Theorem~20.5]{Billingsley-PnM}: a sequence $\{X_n\}$ converges to $X$ in probability if and only if for every subsequence $\{X_{n(m)}\}$, there is a further subsequence $\{X_{n(m_k)}\}$ that converges almost surely to $X$. Let $f$ be a continuous function which is dominated by some polynomial. Given a subsequence $\{n(m)\}$, the aim now is to find a subsequence $n(m_k)$ such that $\{\langle \mu_{n(m_k)}, f\rangle \}$ converges almost surely to $\langle \mu, f\rangle$. Let $\{n(0,m) = n(m)\}$. For $k \ge 1$, using the necessary condition in the criterion, we can find a subsequence $\{n(k,m)\}$ of $\{n(k-1,m)\}$ such that
\[
\langle \mu_{n(k, m)}, x^k\rangle \to \langle \mu, x^k\rangle \text{ a.s.~as } n\to \infty.
\]
By selecting the diagonal, we get a subsequence $\{n(m_k) = n(k, k)\}$ for which all moments of $\{\mu_{n(m_k)}\}$ converge almost surely to the corresponding moments of $\mu$. Consequently, the sequence $\{\mu_{n(m_k)}\}$ converges weakly, almost surely, to $\mu$ by the first part of this lemma, which implies that $\{\langle \mu_{n(m_k)}, f\rangle \} \to \langle \mu, f\rangle$ almost surely. The proof is complete.
\end{proof}
Let us now explain the main idea of this paper. Consider the sequence of random Jacobi matrices
\[
J_n = \begin{pmatrix}
a_1^{(n)} &b_1^{(n)} \\
b_1^{(n)} &a_2^{(n)} &b_2^{(n)} \\
&\ddots &\ddots &\ddots \\
&& b_{n - 1}^{(n)} &a_n^{(n)}
\end{pmatrix},
\]
and let $\mu_n$ be the spectral measure of $(J_n, e_1)$. Assume that each entry of $J_n$ converges almost surely to a non random limit as $n \to \infty$, that is, for any fixed $i$, as $n \to \infty$,
\begin{equation}\label{almost-sure-assumption}
a_i^{(n)} \to \bar a_i; \quad
b_i^{(n)} \to \bar b_i \text{ a.s.}
\end{equation}
Here we require that $\bar a_i$ and $\bar b_i$ are non random and $\bar b_i >0$. Assume further that the spectral measure of $(J_\infty, e_1)$, denoted by $\mu_\infty$, is unique, where $J_\infty$ is the infinite Jacobi matrix consisting of $\{\bar a_i\}$ and $\{\bar b_i\}$,
\[
J_\infty = \begin{pmatrix}
\bar a_1 &\bar b_1 \\
\bar b_1 &\bar a_2 &\bar b_2 \\
&\ddots &\ddots &\ddots
\end{pmatrix}.
\]
Then the measure $\mu_\infty$ is determined by its moments, and hence we get the following result.
\begin{theorem}
The spectral measures $\mu_n$ converge weakly, almost surely, to the limit measure $\mu_\infty$ as $n \to \infty$. If in the assumption \eqref{almost-sure-assumption}, convergence in probability is assumed instead of almost sure convergence, then the spectral measures $\mu_n$ converge weakly, in probability, to $\mu_\infty$ as $n \to \infty$.
\end{theorem}
\begin{remark}
These results may be referred to as weak and strong laws of large numbers for spectral measures of random Jacobi matrices. They are natural results which may be found somewhere. For instance, the strong law was mentioned in \cite{Nagel-082013}.
\end{remark}
\begin{proof}
Let $p$ be a polynomial of degree $m$. When $n$ is large enough,
$
\langle \mu_n, p\rangle = p(J_n) (1,1)
$
is a polynomial of $\{a_i^{(n)}, b_i^{(n)}\}_{i = 1, \dots, \lceil \frac m2 \rceil}$. Therefore, as $n \to \infty$,
\[
\langle \mu_n, p\rangle \to \langle \mu_\infty, p\rangle \text{ almost surely (resp.~in probability)},
\]
which implies the weak convergence of $\mu_n$ by Lemma~\ref{lem:convergence-of-moments}.
\end{proof}
Next, we consider the second order of the convergence of spectral measures, or a type of central limit theorem. It turns out that the central limit theorem for polynomial test functions is a direct consequence of a joint central limit theorem for entries of Jacobi matrices. Indeed, assume that there are random variables $\{\eta_i\}$ and $\{\zeta_i\}$ defined on the same probability space such that for some fixed $r > 0$, for any $i$, as $n \to \infty$,
\begin{equation}\label{weak-convergence-assumption}
\begin{aligned}
\tilde a_i^{(n)} = n^{r}(a_i^{(n)} - \bar a_i) \overset{d}{\to } \eta_i, \\
\tilde b_i^{(n)} =n^{r}(b_i^{(n)} - \bar b_i) \overset{d}{\to } \zeta_i.
\end{aligned}
\end{equation}
Moreover, we assume that the joint weak convergence holds. This means that any finite linear combination of $\tilde a_i^{(n)}$ and $\tilde b_i^{(n)}$ converges weakly to the corresponding linear combination of $\eta_i$ and $\zeta_i$ as $n \to \infty$, namely, for any real numbers $c_i$ and $d_i$,
\[
\sum_{finite} (c_i \tilde a_i^{(n)} + d_i \tilde b_i^{(n)}) \overset{d}{\to } \sum_{finite} (c_i \eta_i + d_i \zeta_i).
\]
From now on, both conditions~\eqref{almost-sure-assumption} and \eqref{weak-convergence-assumption} will be written in a compact form
\[
J_n \approx \begin{pmatrix}
\bar a_1 &\bar b_1 \\
\bar b_1 &\bar a_2 &\bar b_2 \\
&\ddots &\ddots &\ddots
\end{pmatrix}
+\frac{1}{n^r} \begin{pmatrix}
\eta_1 &\zeta_1 \\
\zeta_1 &\eta_2 &\zeta_2 \\
&\ddots &\ddots &\ddots
\end{pmatrix},
\]
or in term of entries
\begin{align*}
a_i^{(n)} \approx \bar a_i + \frac{1}{n^r} \eta_i,\\
b_i^{(n)} \approx \bar b_i + \frac{1}{n^r} \zeta_i.\\
\end{align*}
Let $f$ be a polynomial of $2k$ variables $(a_1,\dots, a_k, b_1, \dots, b_k)$. For simplicity, we write $f(a_i, b_i)$ instead of $f(a_1, \dots, a_k, b_1, \dots, b_k)$.
\begin{lemma}\label{lem:polynomial}
\begin{itemize}
\item [\rm (i)] As $n \to \infty$,
\begin{align*}
n^r \left( f(a_i^{(n)}, b_i^{(n)}) - f(\bar a_i, \bar b_i)\right) - \sum_{i = 1}^k \left(\frac{\partial f}{\partial a_i}(\bar a_i, \bar b_i) \tilde a_i^{(n)} + \frac{\partial f}{\partial b_i}(\bar a_i, \bar b_i) \tilde b_i^{(n)} \right) &\to 0 \\
&\text{in probability}.
\end{align*}
\item[\rm (ii)] As $n \to \infty$,
\[
n^r \left( f(a_i^{(n)}, b_i^{(n)}) - f(\bar a_i, \bar b_i)\right) \overset{d}{\to } \sum_{i = 1}^k \left(\frac{\partial f}{\partial a_i}(\bar a_i, \bar b_i) \eta_i + \frac{\partial f}{\partial b_i}(\bar a_i, \bar b_i) \zeta_i \right).
\]
\end{itemize}
\end{lemma}
\begin{proof}
Write
\[
a_i^{(n)} = \bar a_i + \frac{1}{n^r} \tilde a_i^{(n)}; b_i^{(n)} = \bar b_i + \frac{1}{n^r} \tilde b_i^{(n)}.
\]
Then use the Taylor expansion of $f(a_i^{(n)}, b_i^{(n)})$ at $(\bar a_i, \bar b_i)$ with noting that the Taylor expansion of a polynomial consists of finitely many terms,
\begin{align*}
f(a_i^{(n)}, b_i^{(n)}) = f(\bar a_i, \bar b_i) + \frac{1}{n^r} \sum_{i = 1}^k \left(\frac{\partial f}{\partial a_i}(\bar a_i, \bar b_i) \tilde a_i^{(n)} + \frac{\partial f}{\partial b_i}(\bar a_i, \bar b_i) \tilde b_i^{(n)} \right) + \sum{}^*.
\end{align*}
Each term in the finite sum $\sum{}^*$ has the following form,
\[
c(\alpha, \beta) \prod_{i = 1}^k (a_i^{(n)} - \bar a_i)^{\alpha_i}(b_i^{(n)} - \bar b_i)^{\beta_i},
\]
where $\{\alpha_i\}$ and $\{\beta_i\}$ are non negative integers and $\sum_{i = 1}^k(\alpha_i + \beta_i) \ge 2$. Therefore, when that term is multiplied by $n^r$, it converges to $0$ in distribution, and hence, in probability by Slutsky's theorem.
By using Slutsky's theorem again, we see that (ii) is a consequence of (i). The proof is complete.
\end{proof}
Let $p$ be a polynomial of degree $m>0$. Then there is a polynomial of $2\lceil \frac m2 \rceil$ variables such that for $n > m/2$,
\[
\langle \mu_n, p\rangle = p(J_n) (1,1) = f(a_1^{(n)},\dots, a_{\lceil \frac m2 \rceil}^{(n)}, b_1^{(n)}, \dots, b_{\lceil \frac m2 \rceil}^{(n)}).
\]
Therefore, by Lemma~\ref{lem:polynomial}, we obtain the central limit theorem for polynomial test functions.
\begin{theorem}\label{thm:weak-convergence}
For any polynomial $p$,
$
n^{r} \left( \langle \mu_n, p \rangle- \langle \mu_\infty, p\rangle\right)
$
converges weakly to a limit $\xi_\infty(p)$ as $n \to \infty$.
\end{theorem}
Since we do not assume that all moments of $\{a_i^{(n)}\}$ and $\{b_i^{(n)}\}$ are finite, even the expectation of $\langle \mu_n, p \rangle$, ${\mathbb E}[\langle \mu_n, p \rangle]$ may not exist. Thus we need further assumptions to ensure the convergence of mean and variance in the central limit theorem above. Our assumptions are based on the following basic result in probability theory (the corollary following Theorem~25.12 in \cite{Billingsley-PnM}).
\begin{lemma}\label{lem:convergence-of-variance}
Assume that the sequence $\{X_n\}$ converges weakly to a random variable $X$. If for some $\delta > 0$,
\[
\sup_{n} {\mathbb E}[|X_n|^{2 + \delta}] < \infty,
\]
then ${\mathbb E}[X_n] \to {\mathbb E}[X]$ and $\Var[X_n] \to \Var[X]$ as $n \to \infty$. In general, if $X_n$ converges to $X$ in probability and for $q \ge 1$ and $\delta > 0$,
\[
\sup_{n} {\mathbb E}[|X_n|^{q + \delta}] < \infty,
\]
then $X_n$ converges to $X$ in $L_q$.
\end{lemma}
We make the following assumptions
\begin{itemize}
\item[(i)] all moments of $\{a_i^{(n)}\}$ and $\{b_i^{(n)}\}$ are finite and in addition, the convergences in \eqref{almost-sure-assumption} also hold in $L_q$ for all $q <\infty$, which is equivalent to the following conditions
\begin{equation}\label{Lq-bounded}
\sup_n {\mathbb E}[|a_i^{(n)}|^k] < \infty, \quad \sup_n {\mathbb E}[|b_i^{(n)}|^k] < \infty, \text{ for all $k=1,2,\dots$};
\end{equation}
\item[(ii)] ${\mathbb E}[\eta_i] = 0, {\mathbb E}[\zeta_i] = 0$, and for some $\delta > 0$,
\begin{equation}\label{variance-bounded}
\sup_{n} {\mathbb E}[|\tilde a_i^{(n)}|^{2 + \delta}] < \infty, \quad \sup_{n} {\mathbb E}[|\tilde b_i^{(n)}|^{2 + \delta}] < \infty.
\end{equation}
\end{itemize}
\begin{lemma}\label{lem:mean-convergence}
As $n \to \infty$,
\begin{align}
n^{2r}{\mathbb E}\left [\left (f(a_i^{(n)}, b_i^{(n)}) - f(\bar a_i, \bar b_i) \right)^2 \right] &\to \Var \left[ \sum_{i = 1}^k \left(\frac{\partial f}{\partial a_i}(\bar a_i, \bar b_i) \eta_i + \frac{\partial f}{\partial b_i}(\bar a_i, \bar b_i) \zeta_i \right)\right], \\
n^r \left({\mathbb E}[f(a_i^{(n)}, b_i^{(n)})] - f(\bar a_i, \bar b_i) \right) &\to 0,\\
n^{2r}\Var\left [f(a_i^{(n)}, b_i^{(n)}) \right] &\to \Var \left[ \sum_{i = 1}^k \left(\frac{\partial f}{\partial a_i}(\bar a_i, \bar b_i) \eta_i + \frac{\partial f}{\partial b_i}(\bar a_i, \bar b_i) \zeta_i \right)\right].
\end{align}
\end{lemma}
\begin{proof}
It is just a direct consequence of Lemma~\ref{lem:convergence-of-variance}.
\end{proof}
We state now a slightly different form of the central limit theorem for polynomial test functions.
\begin{theorem}\label{thm:general}
Under assumptions~\eqref{almost-sure-assumption}--\eqref{variance-bounded},
for any polynomial $p$, as $n \to \infty$,
\[
\langle \mu_n, p \rangle \to \langle \mu_\infty, p \rangle \text{ almost surely and in $L^q$ for all $q <\infty$};
\]
\[
n^{r} ( \langle \mu_n, p \rangle- {\mathbb E}[\langle \mu_n, p\rangle]) \overset{d}{\to } \xi_\infty(p).
\]
Moreover, ${\mathbb E}[\xi_\infty(p)] = 0$ and
\[
n^{2r}\Var[\langle \mu_n, p \rangle] \to \Var[\xi_\infty(p)] \text{ as $n \to \infty$}.
\]
\end{theorem}
\section{Gaussian beta ensembles or $\beta$-Hermite ensembles}
Let $H_{n,\beta}$ be a random Jacobi matrix whose elements are independent (up to the symmetric constraint) and are distributed as
\[
H_{n,\beta} = \frac{1}{\sqrt{n\beta}}\begin{pmatrix}
{\mathcal N}(0,2) &\chi_{(n-1)\beta} \\
\chi_{(n-1)\beta} &{\mathcal N}(0,2) &\chi_{(n-2)\beta} \\
&\ddots &\ddots &\ddots \\
&& \chi_{\beta} &{\mathcal N}(0,2)
\end{pmatrix}.
\]
Then the eigenvalues $\{\lambda_i\}$ of $H_{n,\beta}$ have Gaussian beta ensembles \cite{Dumitriu-Edelman-2002}, that is,
\[
(\lambda_1,\lambda_2, \dots, \lambda_n) \propto |\Delta(\lambda)|^\beta \exp \left(-\frac{n\beta}{4}\sum_{i = 1}^n \lambda_j^2 \right).
\]
The weights $\{w_i\} = \{q_i^2\}$ are independent of $\{\lambda_i\}$ and have Dirichlet distribution with parameters $(\beta/2, \dots, \beta/2)$, that is,
\[
(w_1, \dots, w_{n - 1}) \propto \prod_{i = 1}^n w_i^{\frac\beta 2 - 1} \mathbf 1_{\{w_1+\cdots + w_{n - 1} < 1, w_i > 0\}}, \quad w_n = 1 - (w_1 + \cdots + w_{n - 1}).
\]
Recall that the distribution of $\{w_i\}$ is the same as that of the vector
\[
\left(\frac{\chi_{\beta,1}^2}{\sum_{i = 1}^n \chi_{\beta,i}^2} , \dots, \frac{\chi_{\beta,n}^2}{\sum_{i = 1}^n \chi_{\beta,i}^2} \right),
\]
with $\{\chi_{\beta,i}^2\}_{i = 1}^n$ being i.i.d.~sequence of random variables having chi-squared distributions with $\beta$ degrees of freedom.
\begin{lemma}\label{lem:chi-distribution}
\begin{itemize}
\item[\rm(i)] As $k \to \infty$,
\[
\frac{\chi_k}{\sqrt{k}} \to 1 \text{ in probability and in $L_q$ for all $q<\infty$.}
\]
A sequence $\{\chi_{k_n}/\sqrt{k_n}\}$ converges almost sure to $1$, if $\sum_{n = 1}^\infty e^{-\varepsilon k_n} < \infty$ for any $\varepsilon > 0$.
\item[\rm(ii)] As $k \to \infty$,
\[
\sqrt{k}\left(\frac{\chi_k}{\sqrt{k}} - 1 \right) = (\chi_k - \sqrt{k})\overset{d}{\to } {\mathcal N}(0, \frac 12).
\]
\end{itemize}
\end{lemma}
\begin{proof}
The convergence in probability and a central limit theorem are standard results. Let us prove the almost sure convergence part. For almost sure convergence, usually there is some relation between random variables in the sequence. If there is no relation, the following criterion, a direct consequence of the Borel-Cantelli lemma, becomes useful. The sequence $X_n$ converges almost surely to $x$ as $n \to \infty$, if for any $\varepsilon > 0$,
\[
\sum_{n = 1}^{\infty} {\mathbb P}(|X_n - x| > \varepsilon) < \infty.
\]
For the proof here and later for Beta distributions, we use the following bounds (see Lemma~4.1 in \cite{Nagel-082013})
\[
{\mathbb P}\left(\left| \frac{\chi_k^2}{k} - 1 \right| > \varepsilon \right) \le 2 e^{-\frac{k \varepsilon^2}{8}},
\]
\[
{\mathbb P}\left(\left| {\rm Beta}(x, y) - {\mathbb E}[{\rm Beta}(x, y)] \right| > \varepsilon \right) \le 4 e^{-\frac{\varepsilon^2}{128}\frac{x^3 + y^3}{xy}}.
\]
Here ${\rm Beta} (x,y)$ denotes the beta distribution with parameters $x$ and $y$. Then the proof follows immediately from the criterion above.
\end{proof}
Since the Jacobi parameters $\{a_i, b_i\}$ of $H_{n,\beta}$ are independent, it follows that joint convergence in distribution holds. Thus we can write
\[
H_{n,\beta} \approx
\begin{pmatrix}
0 &1 \\
1 &0 &1 \\
&\ddots &\ddots &\ddots
\end{pmatrix}
+ \frac{1}{\sqrt{\beta n}}
\begin{pmatrix}
{\mathcal N}(0,2) &{\mathcal N}(0,\frac12) \\
{\mathcal N}(0,\frac12) &{\mathcal N}(0,2) &{\mathcal N}(0,\frac12)\\
&\ddots &\ddots &\ddots
\end{pmatrix}.
\]
The nonrandom Jacobi matrix in the above expression is called the free Jacobi matrix, denoted by $J_{free}$, whose spectral measure is the semicircle distribution
\[
sc(x) = \frac{1}{2\pi} \sqrt{4 - x^2}, (-2 \le x \le 2).
\]
There are several ways to derive that fact. For example, in the theory of orthogonal polynomials on the real line, it is derived from the relation of Chebyshev polynomials of the second kind \cite[Section~1.10]{Simon-book}. One can also calculate moments and show that they match moments of the semicircle distribution. However, we introduce here a method to find the spectral measure by calculating its Stieltjes transform, which is applicable to the next two cases.
Let $J$ be a Jacobi matrix with bounded coefficients,
\[
J = \begin{pmatrix}
a_1 &b_1 \\
b_1 &a_2 &b_2 \\
&\ddots &\ddots &\ddots
\end{pmatrix}.
\]
Then the spectral measure $\mu$ is unique and has compact support. Let $S_\mu$ be the Stieltjes transform of $\mu$,
\[
S_\mu(z) = \int \frac{d\mu(x)}{x - z}.
\]
In the theory of Jacobi matrices, the Stieljes transform $S_\mu$ is called an $m$-function,
\[
S_\mu(z) = m(z) = \langle (J - z)^{-1}e_1, e_1 \rangle = (J-z)^{-1}(1,1).
\]
For bounded Jacobi matrix $J$, the $m$-function is an analytic function on ${\mathbb C} \setminus I$, for some bounded interval $I \subset {\mathbb R}$, $\Image m(z) > 0,$ if $\Image z > 0$, and $m(\bar z) = \overline{m(z)}$.
Let $J_1$ be a Jacobi matrix obtained by removing the top row and left-most column of $J$ and let $m_1(z)$ be the $m$-function of $J_1$. Then the following relation holds (see \cite[Theorem~3.2.4]{Simon-book})
\begin{equation}\label{m-function-relation}
-\frac{1}{m(z)} = z - a_1 + b_1^2 m_1(z).
\end{equation}
When $a_i \equiv a$ and $b_i \equiv b$, then $m_1(z) = m(z)$, and hence $m(z)$ satisfies the following equation
\[
-\frac{1}{m(z)} = z - a + b^2 m (z),
\]
which is equivalent to a quadratic form
\[
b^2 m(z) ^2 + (z - a) m(z) + 1 = 0.
\]
In particular, by solving the above equation with $a = 0$ and $b = 1$, we obtain the explicit formula for the $m$-function of the free Jacobi matrix,
\[
m_{free}(z) = - \frac{1}{2} \left( z - \sqrt{z^2 - 4}\right),
\]
where the branch of the square-root is chosen to ensure that $\Image m(z) > 0$, if $\Image z > 0$.
Then the spectral measure $\mu$ of $J_\infty$ can be easily calculated by an inverse formula (\cite[Theorem~2.4.3]{Anderson-book})
\[
\mu(I) = \lim_{\varepsilon \searrow 0} \frac{1}{\pi} \int_I \Image m(x + i \varepsilon) dx,
\]
provided that $I$ is an open interval with neither endpoint on an atom of $\mu$. Moreover, when the limit $\lim_{\varepsilon \searrow 0} \Image m(x + i \varepsilon)$ exists and is finite for all $x \in {\mathbb R}$, the measure $\mu$ has density given by
\[
\mu(x) = \lim_{\varepsilon \searrow 0} \frac{1}{\pi} \Image m(x + i \varepsilon).
\]
The spectral measure of the free Jacobi matrix is now easily derived from the above formula.
Back to the case $a_i \equiv a$ and $b_i \equiv b$, and denote by $m_{a,b}(z)$ its $m$-function. Then
\begin{equation}\label{m-ab}
m_{a,b} (z) = (J_{a,b} - z)^{-1} (1,1) = (b J_{free} + a - z )^{-1} (1,1) =\frac{1}{b} m_{free} (\frac{z - a}{b}).
\end{equation}
Here
\[
J_{a,b} = \begin{pmatrix}
a &b \\
b &a &b \\
&\ddots &\ddots &\ddots
\end{pmatrix}, \quad J_{free} = J_{0,1}.
\]
Here is the main result in this section.
\begin{theorem}
The spectral measure $\mu_n$ of $H_{n,\beta}$ converges weakly, almost surely, to the semicircle distribution as $n \to \infty$. Moreover, for any polynomial $p$, as $n \to \infty$,
\[
\frac{\sqrt{n\beta}}{\sqrt2} (\langle \mu_n, p\rangle - \langle sc, p\rangle ) \overset{d}{\to } {\mathcal N}(0, \sigma_p^2),
\]
where $\sigma_p^2$ is a constant.
\end{theorem}
\begin{remark}
Here we implicitly assume that the parameter $\beta$ is fixed. However, the result still holds if $\beta$ varies and $n\beta \to \infty$. In this case, by our method, the spectral measures $\mu_n$ converge weakly, in probability, to the semicircle distribution because we do not require the rate of convergence of $n\beta$ to infinity.
\end{remark}
\section{Wishart beta ensembles or $\beta$-Laguerre ensembles}
Let $G$ be an $m \times n$ matrix whose entries are i.i.d.~standard Gaussian ${\mathcal N}(0,1)$ random variables. Then for $m \ge n$, $(m^{-1} G^t G)$ is a Wishart real matrix whose eigenvalues are distributed as
\[
(\lambda_1, \dots, \lambda_n) \propto |\Delta(\lambda)| \prod_{i = 1}^n \lambda_i^{\frac12(m - n + 1) - 1} \exp\left( - \frac {m}2\sum_{i = 1}^n \lambda_i \right), \quad \lambda_i > 0.
\]
Note that Wishart complex matrices are defined similarly by considering i.i.d.~complex Gaussian matrices $G$. The eigenvalues also a nice density formula similar as above. Then Wishart beta ensembles are defined to be ensembles of $n$ points on $(0, \infty)$ with the following joint probability density function
\[
(\lambda_1, \dots, \lambda_n) \propto |\Delta(\lambda)|^\beta \prod_{i = 1}^n \lambda_i^{\frac{\beta}{2}(m - n + 1) - 1} \exp\left( - \frac {m \beta}2\sum_{i = 1}^n \lambda_i \right).
\]
Here we only require that $n-1 < m \in {\mathbb R}$.
Dumitriu and Edelman \cite{Dumitriu-Edelman-2002} gave the following matrix model for Wishart beta ensembles. For $n \in {\mathbb N}$ and $m > n-1$, let $B_{\beta}$ be a bidiagonal matrix whose elements are independent and are distributed as
\[
B_\beta = \frac{1}{\sqrt{m\beta}}\begin{pmatrix}
\chi_{\beta m} \\
\chi_{\beta(n - 1)} &\chi_{\beta m - \beta} \\
&\ddots &\ddots \\
& &\chi_\beta &\chi_{\beta m - \beta(n - 1)}
\end{pmatrix}.
\]
Let $L_{m,n,\beta} = B_\beta B_\beta^t$. Then the eigenvalues of $L_{m,n,\beta}$ are distributed as Wishart beta ensembles. Moreover, the weights $\{w_i\}$ in the spectral measures of $L_{m, n, \beta}$ have Dirichlet distribution with parameters $(\beta/2, \dots, \beta/2)$ and are independent of $\{\lambda_i\}$. For empirical distributions, the weak convergence to Marchenko-Pastur distributions and a central limit theorem for polynomial test functions were established, see \cite{Dumitriu-Edelman-2006} and the references therein.
Denote by $\{c_i\}_{i = 1}^n$ and $\{d_j\}_{j = 1}^{n - 1}$ the diagonal and the sub-diagonal of the matrix $\sqrt{m\beta}B_\beta$. Then
\[
L_{m,n,\beta} = \frac{1}{m\beta} \begin{pmatrix}
c_1^2 & c_1 d_1 \\
c_1 d_1 & c_2^2 + d_1^2 & c_2 d_2 \\
& \ddots & \ddots & \ddots \\
& & c_{n - 1}d_{n - 1} & c_{n}^2 + d_{n - 1}^2
\end{pmatrix}.
\]
\begin{lemma}
For fixed $i$, as $n \to \infty$ with $n/m \to \gamma \in (0,1)$,
\begin{align*}
\frac{c_i}{\sqrt{m\beta}} &= \frac{\chi_{\beta(m - i + 1)}}{\sqrt{m\beta}} \approx 1 + \frac{\sqrt{\gamma}}{\sqrt{n\beta}} \eta_i, \quad \eta_i \sim {\mathcal N}(0,\frac12),\\
\frac{d_i}{\sqrt{m\beta}} &= \frac{\chi_{\beta(n - i)}}{\sqrt{m\beta}} \approx \sqrt{\gamma} + \frac{\sqrt{\gamma}}{\sqrt{n\beta}} \zeta_i, \quad \zeta_i\sim {\mathcal N}(0,\frac12),\\
\frac{c_i^2}{{m\beta}} &\approx 1 + \frac{\sqrt{\gamma}}{\sqrt{n\beta}} 2\eta_i,\quad
\frac{d_i^2}{{m\beta}} \approx {\gamma} + \frac{\sqrt{\gamma}}{\sqrt{n\beta}}2 \zeta_i, \\
\frac{c_i d_i}{m\beta} &\approx \sqrt{\gamma} + \frac{\sqrt{\gamma}}{\sqrt{n\beta}}(\sqrt{\gamma} \eta_i + \zeta_i).
\end{align*}
\end{lemma}
Consequently, as $n \to \infty$ and $n/m \to \gamma \in (0,1)$, we can write
\[
L_{m,n,\beta} \approx \begin{pmatrix}
1 &\sqrt{\gamma} \\
\sqrt{\gamma} &1+\gamma &\sqrt{\gamma}\\
&\ddots &\ddots &\ddots
\end{pmatrix}
+
\frac{\sqrt{\gamma}}{\sqrt{n \beta }}
\begin{pmatrix}
2 \eta_1 & \sqrt{\gamma} \eta_1 + \zeta_1 \\
\sqrt{\gamma} \eta_1 + \zeta_1 & 2(\eta_2 + \zeta_1) & \sqrt{\gamma} \eta_2 + \zeta_2 \\
&\ddots &\ddots &\ddots
\end{pmatrix}.
\]
Here $\{\eta_i\}$ and $\{\zeta_i\}$ are two i.i.d.~sequences of ${\mathcal N}(0, \frac 12)$ random variables.
For $\gamma \in (0,1)$, let
\[
MP_\gamma =\begin{pmatrix}
1 &\sqrt{\gamma} \\
\sqrt{\gamma} &1+\gamma &\sqrt{\gamma}\\
&\ddots &\ddots &\ddots
\end{pmatrix} ,
\]
and denote its $m$-function by $m_\gamma(z)$. When the first row and the left-most column are removed, we obtain the Jacobi matrix $J_{1+\gamma, \sqrt{\gamma}}$. Then it follows from the relations~\eqref{m-function-relation} and \eqref{m-ab} that
\begin{align*}
m_\gamma(z) &= -\frac{1}{z - 1 + \sqrt{\gamma}m_{free} (\frac{z - (1 + \gamma)}{ \sqrt{\gamma}})} \\
&= -\frac{1}{z - 1 - \frac{\sqrt{\gamma}}{2} \left(\frac{z - (1 +\gamma)}{\sqrt{\gamma}} - \sqrt{(\frac{z - (1+ \gamma)}{\sqrt\gamma})^2 - 4} \right)} \\
&= \frac{1 - \gamma - z + \sqrt{(z - \lambda_+)(z - \lambda_-)}}{2\gamma z},
\end{align*}
where $\lambda_{\pm} = (1 \pm \sqrt{\gamma})^2$. By taking the limit $\lim_{\varepsilon \searrow 0} \pi^{-1} \Image m_\gamma(x + i \varepsilon)$, we get the density of the spectral measure of $MP_\gamma$, which coincides with the density of the Marchenko-Pastur distribution with parameter $\gamma \in (0,1)$,
\[
mp_\gamma(x) =
\frac{1}{2 \pi \gamma x} \sqrt{ (\lambda_+ - x)(x - \lambda_-)}, (\lambda_-<x< \lambda_+).
\]
\begin{theorem}
The spectral measure $\mu_n$ of $L_{m,n,\beta}$ conveges weakly, almost surely, to the Marchenko-Pastur distribution with parameter $\gamma$ as $n \to \infty$, $n/m \to \gamma \in (0,1)$. Moreover for any polynomial $p$,
\[
\frac{\sqrt{n\beta}}{\sqrt2} (\langle \mu_n, p\rangle - \langle mp_\gamma, p\rangle ) \overset{d}{\to } {\mathcal N}(0, \sigma_p^2),
\]
where $\sigma_p^2$ is a constant.
\end{theorem}
\section{MANOVA beta ensembles or $\beta$-Jacobi ensembles}
Let $W_1 = G_1^t G_1$ and $W_2 = G_2^t G_2$ be two independent real Wishart matrices with $G_1$ (resp.~$G_2$) being an $m_1\times n$ (resp.~$m_2\times n)$ matrix whose entries are i.i.d.~standard Gaussian ${\mathcal N}(0,1)$ random variables. Assume that $m_1, m_2 \ge n$. Then the eigenvalues $(\lambda_1, \dots, \lambda_n)$ of the matrix $W_1(W_1 + W_2)^{-1}$ (or of the Hermitian matrix $(W_1 + W_2)^{-1/2} W_1 (W_1 + W_2)^{-1/2}$) are distributed according to the following joint probability density function with $\beta = 1$ (\cite[Theorem~3.3.4]{Muirhead-book})
\begin{align}
(\lambda_1, \dots, \lambda_n) &\propto |\Delta(\lambda)|^\beta \prod_{i = 1}^n \lambda_i^{\frac{\beta}{2}(m_1 - n + 1) - 1} (1 - \lambda_i)^{\frac{\beta}{2}(m_2 - n + 1) - 1}, \nonumber\\
&= |\Delta(\lambda)|^\beta \prod_{i = 1}^n \lambda_i^{a} (1 - \lambda_i)^{b}, \quad \lambda_i \in [0,1]. \label{Jacobi-ensembles}
\end{align}
For general $\beta>0$, the above joint density is referred to as MANOVA beta ensembles or $\beta$-Jacobi ensembles with parameters $(a,b)$, where $a, b>-1$.
A matrix model for $\beta$-Jacobi ensembles is given as follows. Let $a, b > -1$ be fixed. Let $p_1, \dots, p_{2n - 1}$ be independent random variables distributed as
\[
p_k \sim \begin{cases}
{\rm Beta}\left(\frac{2n - k}{4}\beta, \frac{2n - k -2}{4}\beta + a + b + 2 \right), &\text{$k$ even},\\
{\rm Beta}\left(\frac{2n - k - 1}{4}\beta + a + 1, \frac{2n - k -1}{4}\beta + b + 1 \right), &\text{$k$ odd}.
\end{cases}
\]
Here recall that ${\rm Beta}(x, y)$ denotes the beta distribution with parameters $x, y$. Define
\begin{align*}
a_k &= p_{2k - 2} (1 - p_{2k - 3}) + p_{2k - 1}(1 - p_{2k - 2}), \\
b_k &=\sqrt{p_{2k - 1}(1 - p_{2k - 2}) p_{2k} (1 - p_{2k - 1})},
\end{align*}
where $p_{-1} = p_0 = 0$, and form a random Jacobi matrix $J_{n,\beta}(a,b)$ as
\[
J_{n,\beta}(a,b) = \begin{pmatrix}
a_1 & b_1 \\
b_1 & a_2 &b_2\\
&\ddots &\ddots &\ddots\\
&& b_{n - 1} &a_n
\end{pmatrix}.
\]
Then the eigenvalues $(\lambda_1, \dots, \lambda_n)$ of $J_{n,\beta}(a, b)$ are distributed as the $\beta$-Jacobi ensembles~\eqref{Jacobi-ensembles} (cf.~\cite{Killip-Nenciu-2004}).
The weights $\{w_i\} = \{q_i^2\}$ are independent of $\{\lambda_i\}$ and have Dirichlet distribution with parameters $(\beta/2, \dots, \beta/2)$. We are interested in studying $\beta$-Jacobi ensembles in the regime that $a(n)/(\frac{n \beta}{2}) \to \kappa_a \in [0, \infty)$ and $b(n)/(\frac{n \beta}{2}) \to \kappa_b \in [0, \infty)$. The limiting behavior of empirical distributions in that regime was studied in \cite{Dumitriu-Paquette-2012}.
We need the following properties of beta distributions.
\begin{lemma}\label{lem:Beta-distribution}
Let $\{x_k\}$ and $\{y_k\}$ be two sequences of positive real numbers such that as $k \to \infty$
\[
\frac{x_k}{k} \to x > 0, \frac{y_k}{k} \to y > 0.
\]
Then the following asymptotic behaviors of the beta distribution ${\rm Beta}(x_k, y_k)$ hold.
\begin{itemize}
\item[\rm(i)] As $k \to \infty$,
\[
{\rm Beta}(x_k,y_k) \to \frac{x}{x + y} \text{ in probability and in $L_q$ for all $q<\infty$.}
\]
The almost sure convergence also holds.
\item[\rm(ii)] As $k \to \infty$,
\[
\sqrt{k}\left({\rm Beta}(x_k, y_k) - \frac{x_k}{x_k +y_k}\right) \overset{d}{\to } {\mathcal N}(0,x y(x +y)^{-3}).
\]
Moreover, if $(x_k y - y_k x)/\sqrt{k} \to 0$ as $k \to \infty$, then
\[
\sqrt{k}\left({\rm Beta}(x_k, y_k) - \frac{x}{x + y}\right) \overset{d}{\to } {\mathcal N}(0,xy(x + y)^{-3}).
\]
\end{itemize}
\end{lemma}
\begin{proof}
Let $X_k$ and $Y_k$ be two independent random variables distributed as $\chi^2_{2x_k}$ and $\chi^2_{2y_k}$, respectively. Then
\[
{\rm Beta}(x_k, y_k) \overset{d}{=} \frac{X_k}{X_k + Y_k}.
\]
For chi-squared distribution, we have
\[
\frac{\chi^2_k}{k} \to 1 \text{ in probability as }k \to \infty.
\]
Therefore
\[
\frac{X_k}{X_k + Y_k} = \frac{\frac{X_k}{2x_k}\frac{2x_k}{2k}}{\frac{X_k}{2x_k} \frac{2x_k}{2k} + \frac{Y_k}{2y_k} \frac{2y_k}{2k}} \to \frac{x}{x + y} \text{ in probability as } k \to \infty.
\]
The convergence in $L_q$ is clear because beta distributions are bounded by $1$. For the almost sure convergence, see the proof of Lemma~\ref{lem:chi-distribution}.
Next we consider the central limit theorem for beta distributions. It also follows from the following central limit theorem for chi-squared distribution
\[
\sqrt{\frac k2} \left( \frac{\chi_k^2}{k} - 1 \right) \overset{d}{\to } {\mathcal N}(0, 1) \text{ as }k \to \infty.
\]
Indeed, if we write
\begin{align*}
&\sqrt{k}\left(\frac{X_k}{X_k + Y_k} - \frac{x_k}{x_k + y_k}\right) \\
&{=}
\frac{\left(\frac{y_k}{k} \frac{\sqrt{x_k}}{\sqrt{k}} \right) \sqrt{x_k} \left( \frac{X_k}{2x_k} - 1 \right) - \left(\frac{x_k}{k} \frac{\sqrt{y_k}}{\sqrt{k}} \right) \sqrt{y_k} \left( \frac{Y_k}{2y_k} - 1 \right)} {\left(\frac{x_k} k + \frac{y_k}k \right)\left(\frac{X_k}{2k} + \frac{Y_k}{2k} \right)},
\end{align*}
then as $k \to \infty$, the numerator converges in distribution to ${\mathcal N}(0,xy(x + y))$ because $X_k$ and $Y_k$ are independent while the denominator converges in probability to $(x + y)^2$. Thus we obtain (ii).
\end{proof}
\begin{lemma}
As $n \to \infty$ with $a(n) = \frac{n\beta}{2} \kappa_a + o((n\beta)^{1/2})$ and $b(n) = \frac{n\beta}{2} \kappa_b + o((n\beta)^{1/2})$,
\begin{align*}
p_{2k} & \approx {\rm Beta} (\frac{n \beta}{2}, \frac{n\beta}{2}(1 + \kappa_a + \kappa_b)) \approx \frac{1}{2 + \kappa_a + \kappa_b} + \frac{1}{\sqrt{n \beta}} {\mathcal N}(0, \sigma_{even}^2),\\
p_{2k - 1} &\approx {\rm Beta} (\frac{n \beta}{2}(1 + \kappa_a), \frac{n\beta}{2}(1 + \kappa_b)) \approx \frac{1+\kappa_a}{2 + \kappa_a + \kappa_b} + \frac{1}{\sqrt{n \beta}} {\mathcal N}(0, \sigma_{odd}^2).
\end{align*}
As a consequence,
\begin{align*}
a_1^{(n)} &= p_1 \approx \frac{1+\kappa_a}{2 + \kappa_a + \kappa_b} + \frac{1}{\sqrt{n\beta}} {\mathcal N} =: A_1 + \frac{1}{\sqrt{n\beta}} {\mathcal N},\\
b_1^{(n)} &= \sqrt{p_1 p_2 (1 - p_1)} \approx \frac{\sqrt{(1 + \kappa_a) (1 + \kappa_b)}}{(2 + \kappa_a + \kappa_b)^{3/2}}+ \frac{1}{\sqrt{n\beta}} {\mathcal N} =: B_1 + \frac{1}{\sqrt{n\beta}} {\mathcal N} ,\\
a_k^{(n)} &= p_{2k - 2}(1 - p_{2k-3}) + p_{2k - 1}(1 - p_{2k - 2}) \\
&\approx \frac{1 + \kappa_b + (1 + \kappa_a)(1 + \kappa_a + \kappa_b)}{(2 + \kappa_a + \kappa_b)^2}
+ \frac{1}{\sqrt{n\beta}} {\mathcal N}
=:A + \frac{1}{\sqrt{n\beta}} {\mathcal N}, \quad k\ge 2,\\
b_k^{(n)} &= \sqrt{p_{2k - 1}(1 - p_{2k - 2}) p_{2k}(1 - p_{2k - 1})} \\
&\approx \frac{\sqrt{(1 + \kappa_a) (1 + \kappa_b)(1 + \kappa_a + \kappa_b)}}{(2 + \kappa_a + \kappa_b)^{2}} + \frac{1}{\sqrt{n\beta}}{\mathcal N}
=: B + \frac{1}{\sqrt{n\beta}} {\mathcal N}, \quad k\ge 2.
\end{align*}
Here ${\mathcal N}$ denotes a normal distribution with mean $0$ and positive variance.
The joint asymptotic also holds.
\end{lemma}
\begin{proof}
For fixed $k$, as $n \to \infty$, the asymptotic for $p_k$ follows from Lemma~\ref{lem:Beta-distribution}.
The asymptotic for $a_k^{(n)}$ follows from Lemma~\ref{lem:polynomial} because it is a polynomial of $\{p_{2k-3},p_{2k-2}, p_{2k-1}\}$.
The asymptotic for $b_k^{(n)}$ is a consequence of the following fact. If $X_n \to c \ne 0$ in probability and $\sqrt{n}(X_n - c) \overset{d}{\to } {\mathcal N}(0, \sigma^2)$ as $n \to \infty$, then
\[
\sqrt{n}(\sqrt{X_n} - \sqrt{c}) = \frac{\sqrt{n}(X_n - c)}{\sqrt{X_n} + \sqrt{c}} \overset{d}{\to } \frac{{\mathcal N} (0, \sigma^2)}{2\sqrt{c}} = {\mathcal N} (0, \frac{\sigma^2}{4c}).
\]
The joint asymptotic is clear.
\end{proof}
Therefore, in the regime that $a(n) = \frac{n\beta}{2} \kappa_a + o((n\beta)^{1/2})$ and $b(n) = \frac{n\beta}{2} \kappa_b + o((n\beta)^{1/2})$, the limit Jacobi matrix is of the form
\[
J_\infty = \begin{pmatrix}
A_1 &B_1 \\
B_1 &A &B \\
&B &A &B \\
&&\ddots &\ddots &\ddots
\end{pmatrix},
\]
where $A_1, B_1, A$ and $B$ are defined in the above lemma.
We use relations~\eqref{m-function-relation} and \eqref{m-ab} again to derive an explicit formula for the $m$-function of $J_\infty$,
\[
m(z) = \frac{\kappa_a}{2z} + \frac{\kappa_b}{2(z - 1)} -\frac{(2+\kappa_a + \kappa_b) \sqrt{(z- u_-)(z - u_+)}}{2 z ( z -1)},
\]
where
\begin{equation}
u_{\pm} = A \pm 2B = \left(\frac{\sqrt{(1 + \kappa_a)(1 + \kappa_a + \kappa_b)} \pm \sqrt{1 + \kappa_b}}{2 + \kappa_a + \kappa_b} \right)^2.
\end{equation}
Note that $0 \le u_- < u_+ \le 1$. The density of the spectral measure of $J_\infty$ can be easily calculated by the inverse formula,
\[
km_{u_, u_+}(x) = \frac{2 + \kappa_a + \kappa_b}{2 \pi} \frac{\sqrt{(x - u_-)(u_+ - x)}}{ {x(1-x)}}, (u_- < x < u_+).
\]
It coincides with the density of the Kesten-McKey distribution with parameters $(u_-, u_+)$ (see \cite[Section~7.6]{Gamboa-Rouault-2011}) because
\[
\frac{1}{2}(1 - \sqrt{u_- u_+} - \sqrt{(1 - u_-)(1- u_+)}) = \frac{1}{2 + \kappa_a + \kappa_b}.
\]
When $\kappa_a = \kappa_b = 0$, then $u_-=0$ and $u_+=1$, we get the arcsine distribution.
Therefore, we obtain the limiting behavior of the spectral measures of $\beta$-Jacobi ensembles. \begin{theorem}
As $n \to \infty$ with $a(n) = \frac{n\beta}{2} \kappa_a + o((n\beta)^{1/2})$ and $b(n) = \frac{n\beta}{2} \kappa_b + o((n\beta)^{1/2})$, the spectral measure $\mu_n$ of $J_{n,\beta}(a(n), b(n))$ converges weakly, almost surely, to the Kesten-McKey distribution with parameters $(u_-, u_+)$. For any polynomial $p$,
\[
\frac{\sqrt{n\beta}}{\sqrt2} ( \langle \mu_n, p\rangle - \langle km_{u_-, u_+}, p\rangle) \to {\mathcal N}(0, \sigma_p^2),
\]
where $\sigma_p^2$ is a constant.
\end{theorem}
\begin{remark}
In different regimes, the weak convergence of both empirical distributions and spectral measures was considered in \cite{Nagel-082013}, in which the limit distribution is the Marchenko-Pastur distribution or the semicircle distribution.
\end{remark}
\section{Extend the central limit theorem to a large class of test functions}
For all three beta ensembles in this paper, the spectral measure $\mu_n$ can be written as
\[
\mu_n = \sum_{i = 1}^n w_i \delta_{\lambda_i},
\]
where the weights $\{w_i\} $ are independent of the eigenvalues $\{\lambda_i\}$ and have Dirichlet distribution with parameters $(\beta/2, \dots, \beta/2)$. Recall that as a consequence of Theorem~\ref{thm:general}, for any polynomial $p$, as $n \to \infty$,
\[
\langle \mu_n, p\rangle \to \langle \mu_\infty, p\rangle \text{ almost surely and in $L^q$ for $q\ge 1$},
\]
\[
\frac{\sqrt{n\beta}}{{\sqrt 2}} (\langle \mu_n, p\rangle - {\mathbb E}[\langle \mu_n, p\rangle]) \overset{d}{\to } {\mathcal N} (0, \sigma^2_p),
\]
where
\[
\sigma^2_p = \lim_{n \to \infty} \frac{n \beta}{2} \Var[\langle \mu_n, p\rangle ],
\]
and $\mu_\infty$ denotes the limit distribution.
One can easily show that
\begin{align*}
{\mathbb E}[w_i] &= \frac 1n,
{\mathbb E}[w_i^2] = \frac{\beta+2}{n(n\beta + 2)},
{\mathbb E}[w_i w_j] = \frac{\beta}{n(n \beta +2)}, (1\le i \neq j \le n).
\end{align*}
Therefore for any test function $f$, the following relations hold
\begin{align}
{\mathbb E}[\langle \mu_n, f\rangle] &= {\mathbb E}[\langle L_n, f\rangle], \label{mean-measure}\\
\Var[\langle \mu_n, f\rangle]&=\frac{\beta n}{\beta n + 2} \Var[\langle L_n, f \rangle] + \frac{2}{n\beta + 2} \left ({\mathbb E}[\langle \mu_n, f^2 \rangle] - {\mathbb E}[\langle \mu_n, f \rangle]^2 \right).\label{variance-relation}
\end{align}
The mean of a random measure $\mu$, denoted by $\bar \mu$, is defined to be a probability measure satisfying
\[
\langle \bar \mu, f \rangle = {\mathbb E}[\langle \mu, f \rangle], \]
for all bounded continuous function $f$. Moreover, the above relation still holds for any continuous function $f$ with ${\mathbb E}[\langle \mu, |f|\rangle] < \infty$. It follows from \eqref{mean-measure} that $\bar L_n = \bar \mu_n$, and hence, $\bar L_n = \bar\mu_n$ converges weakly to $\mu_\infty$ as $n \to \infty$. Thus for any continuous function of polynomial growth,
\[
\langle \bar \mu_n, f\rangle \to \langle \mu_\infty, f\rangle \text{ as } n \to \infty.
\]
Denote by $C({\mathbb R})$ the set of continuous functions on ${\mathbb R}$ and let
\[
\mathcal D = \{f \in C({\mathbb R}): n\Var[\langle L_n, f \rangle] \to 0, \langle \bar \mu_n, f \rangle \to \langle \mu_\infty, f\rangle, \langle \bar\mu_n, f^2\rangle \to \langle \mu_\infty, f^2\rangle\}.
\]
Then $\mathcal D$ is a linear space containing all polynomials. It follows from the variance relation \eqref{variance-relation} that for $f \in \mathcal D$,
\[
\lim_{n \to \infty} \frac{n \beta}{2} \Var[\langle \mu_n, f\rangle] = \langle \mu_\infty, f^2\rangle - \langle \mu_\infty, f\rangle^2 =: \sigma^2(f).
\]
Next, we use the following result to extend the central limit theorem to any test function in $\mathcal D$.
\begin{lemma}[{\cite[Theorem 25.5]{Billingsley-PnM}}]\label{lem:triangle}
Let $\{Y_n\}_n$ and $\{X_{n,k}\}_{k,n}$ be real-valued random variables. Assume that
\begin{itemize}
\item[\rm(i)]
$
X_{n,k} \overset{d}{\to } X_k \text{ as }n \to \infty;
$
\item[\rm(ii)]
$
X_k \overset{d}{\to } X \text{ as } k \to \infty;
$
\item[\rm(iii)] for any $\varepsilon > 0$,
$
\lim_{k \to \infty} \limsup_{n \to \infty} {\mathbb P}(|X_{n,k} - Y_n| \ge \varepsilon) =0.
$
\end{itemize}
Then $Y_n \overset{d}{\to } X$ as $n \to \infty$.
\end{lemma}
\begin{theorem}
For $f \in \mathcal D$,
\[
\frac{\sqrt{n\beta}}{\sqrt{2}} \big(\langle \mu_n, f\rangle - {\mathbb E}[\langle \mu_n, f\rangle] \big) \overset{d}{\to } {\mathcal N}(0, \sigma(f)^2),
\]
where $\sigma^2(f) = \langle \mu_\infty, f^2\rangle - \langle \mu_\infty, f\rangle^2 = \Var_{\mu_\infty}[f]$.
\end{theorem}
\begin{proof}
Let $f \in \mathcal D$. Since $\mu_\infty$ has compact support, we can find a sequence of polynomials $\{p_k\}$ converging to $f$ uniformly in the support of $\mu_\infty$. Thus
\begin{equation*}\label{variance-approximation}
\sigma^2(p_k) \to \sigma^2(f) \text{ as }k \to \infty.
\end{equation*}
Let
\begin{align*}
Y_n &= \frac{\sqrt{n\beta}}{\sqrt{2}} (\langle \mu_n, f \rangle - {\mathbb E}[\langle \mu_n, f \rangle] ),\\
X_{n,k} &= \frac{\sqrt{n\beta}}{\sqrt{2}} (\langle \mu_n, p_k \rangle - {\mathbb E}[\langle \mu_n, p_k \rangle] ).
\end{align*}
We only need to check three conditions in Lemma~\ref{lem:triangle}. Conditions (i) and (ii) are clear. For the
condition (iii), note that $(f - p_k) \in \mathcal D$, and thus
\[
\lim_{n \to \infty}\Var[X_{n,k} - Y_n] = \langle \mu_\infty, (f-p_k)^2\rangle - \langle \mu_\infty, (f-p_k)\rangle^2,
\]
which tends to zero as $k \to \infty$.
Therefore, for any $\varepsilon > 0$,
\begin{align*}
\lim_{k \to \infty} \limsup_{n \to \infty} {\mathbb P}(|X_{n,k} - Y_n| \ge \varepsilon) \le \lim_{k \to \infty} \limsup_{n \to \infty} \frac{1}{\varepsilon^2} \Var[X_{n,k} - Y_n] = 0.
\end{align*}
The theorem is proved.
\end{proof}
\begin{lemma}\label{lem:variance-control}
For the Gaussian and Wishart beta ensembles, the class $\mathcal D$ contains at least all functions whose derivative is continuous of polynomial growth. For the MANOVA beta ensembles, the class $\mathcal D$ contains at least all differentiable functions with continuous derivative on $[0,1]$, provided that the parameters $a(n)$ and $b(n)$ are positive and $a(n) + b(n) \to \infty$ as $n \to \infty$.
\end{lemma}
The idea of proof is taken from \cite{Dumitriu-Paquette-2012}. The key tool is the following result.
\begin{lemma}[{\cite[Proposition~2.1]{Bobkov-Ledoux-2000}}]
Let $d\nu = e^{-V} dx$ be a probability measure supported on an open convex set $\Omega \subset {\mathbb R}^n$. Assume that $V$ is twice continuously differentiable and strictly convex on $\Omega$. Then for any locally Lipschitz function $F$ on $\Omega$,
\begin{align*}
\Var_\nu[F] = \int F^2 d\nu - \left( \int F d\nu \right)^2 &\le \int \langle \Hess(V)^{-1} \nabla F, \nabla F \rangle d\nu \\
&\le \int \frac{1}{\lambda_{\min} (\Hess(V))} | \nabla F|^2 d\nu,
\end{align*}
where $\Hess(V)$ denotes the Hessian of $V$ and $\lambda_{\min} (A)$ denote the smallest eigenvalue of $A$.
\end{lemma}
\begin{proof}[Proof of Lemma~{\rm\ref{lem:variance-control}}]
Let us consider the Gaussian beta ensembles case first. Here we consider the ordered eigenvalues $\lambda_1 < \cdots < \lambda_n$. Let $\Omega = \{(\lambda_1, \dots, \lambda_n) : \lambda_1 < \lambda_2 < \cdots < \lambda_n \} \subset {\mathbb R}^n$. Then the joint probability density function can be written in the form $e^{-V}$ on $\Omega$ with
\[
V = const + \frac{n \beta}{4} \sum_{i = 1}^n \lambda_i^2 - \frac{\beta}{2}\sum_{i \ne j}{\log| \lambda_j - \lambda_i|}.
\]
It follows that
\[
\frac{\partial V}{\partial \lambda_i} = \frac{n\beta}{2} \lambda_i - \beta \sum_{j \ne i} \frac{1}{\lambda_i - \lambda_j},
\]
and hence
\begin{align*}
\frac{\partial^2 V}{\partial\lambda_i^2} &= \frac{n \beta}{2} + \beta \sum_{j \ne i} \frac{1}{(\lambda_i - \lambda_j)^2},\\
\frac{\partial^2 V}{\partial\lambda_i \partial \lambda_j} &= - \beta \frac{1}{(\lambda_i - \lambda_j)^2}.
\end{align*}
By the Gershgorin circle theorem, the smallest eigenvalue of $\Hess(V)$ is at least $n\beta /2$,
\[
\lambda_{\min} (\Hess(V)) \ge \frac{n\beta}{2}.
\]
Therefore for any locally Lipschitz function $F$,
\begin{equation}\label{general-F}
\Var_\nu[F] \le \frac{2}{n\beta} \int |\nabla F|^2 d\nu.
\end{equation}
Now let $f$ be a continuous function on ${\mathbb R}$ with continuous derivative and let
\[
F(\lambda_1, \dots, \lambda_n) = \frac{1}{n} \sum_{i = 1}^n f(\lambda_i) (= \langle L_n, f\rangle).
\]
Then it follows from \eqref{general-F} that
\begin{equation}\label{GE}
\Var[\langle L_n, f\rangle] \le \frac{2}{n^2 \beta} \int \langle L_n, (f')^2\rangle d\nu = \frac{2}{n^2 \beta} \langle \bar L_n, (f')^2\rangle = \frac{2}{n^2 \beta}\langle \bar \mu_n, (f')^2\rangle.
\end{equation}
Therefore when $f$ has continuous derivative of polynomial growth, as $n \to \infty$,
\[
\langle \mu_n, (f')^2\rangle \to \langle sc, (f')^2\rangle.
\]
Consequently, $n\Var[\langle L_n, f\rangle] \to 0$, and thus the class $\mathcal D$ contains all functions $f$ which have continuous derivative of polynomial growth.
For MANOVA beta ensembles, by similar argument we arrive at the following inequality
\[
\Var[\langle L_n, f\rangle] \le \frac{1}{n(a(n) + b(n)) } \langle \bar \mu_n, (f')^2\rangle,
\]
provided that the parameters $a(n)$ and $b(n)$ are positive. Thus if $a(n) + b(n)$ tends to infinity, we also have
\[
n \Var[\langle L_n, f\rangle] \to 0 \text{ as } n \to \infty,
\]
for all functions $f$ which have continuous derivative on $[0,1]$.
The Wishart beta ensembles are little different because we do not have a uniform estimate for $\lambda_{\min}(\Hess(V))$. In this case we are working on $\Omega = \{(\lambda_1, \dots, \lambda_n) : 0< \lambda_1 < \lambda_2 < \cdots < \lambda_n \} \in {\mathbb R}^n$ with
\[
V = const + \frac{m\beta}{2} \sum_{i = 1}^n \lambda_i - a \sum_{i = 1}^n\log \lambda_i - \frac{\beta}{2}\sum_{i \ne j}{\log| \lambda_j - \lambda_i|}.
\]
Here $a = \frac{\beta}{2} (m - n + 1) - 1$.
Therefore
\begin{align*}
\frac{\partial^2 V}{\partial\lambda_i^2} &= \frac{a}{\lambda_i^2} + \beta \sum_{j \ne i} \frac{1}{(\lambda_i - \lambda_j)^2},\\
\frac{\partial^2 V}{\partial\lambda_i \partial \lambda_j} &= - \beta \frac{1}{(\lambda_i - \lambda_j)^2}.
\end{align*}
By the Gershgorin circle theorem again, we get the following bound
\[
\lambda_{\min} (\Hess(V)) \ge \frac{a}{ \lambda_n^2}.
\]
Consequently,
\[
\Var_\nu[F] \le \frac{1}{a} \int \lambda_n^2 |\nabla F|^2 d\nu,
\]
and hence,
\[
\Var[\langle L_n, f \rangle] \le \frac{1}{a} {\mathbb E} \left[\frac{\lambda^2_{\max}}{n} \langle L_n, (f')^2\rangle \right] .
\]
Here $\lambda_{\max}$ denotes the largest eigenvalue.
Assume that $f'$ is dominated by some polynomial. Then $\langle L_n, (f')^2 \rangle $ is bounded in $L^2$ because of \eqref{variance-relation}. In addition,
\[
{\mathbb E} \left[\frac{\lambda_{\max}^4}{n^2}\right] \le \frac{1}{n} {\mathbb E}[\langle L_n, x^4\rangle] = \frac{1}{n} \langle \bar \mu_n, x^4 \rangle \to 0 \text{ as } n \to \infty.
\]
Finally, recall that $a = (\beta/2) (m - n + 1) -1$, which behaves like $\frac{\beta}{2}(\frac{1}{\gamma} - 1) n$ in the regime that $n/m \to \gamma \in (0,1)$. Thus
\[
n \Var[\langle L_n, f \rangle] \to 0 \text{ as } n \to \infty.
\]
The proof is complete.
\end{proof}
\bigskip
{\bf Acknowledgement. } The author would like to thank the referee for valuable comments.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 2,039 |
Ad hoc teachers stay for now: DUTA after MHRD meeting
Thousands of ad-hoc teachers gherao VC office (credit: Areesh Ahmad) Team Careers360 | Dec 5, 2019 - 8:22 p.m. IST
NEW DELHI: Following two days of protest by hundreds of Delhi University teachers, the DU Teachers' Association, DUTA, said it has been "promised" that a controversial circular instructing colleges to hire guest teachers against "new vacancies" instead of ad hoc teachers will be "amended".
The announcement came after a meeting with the Ministry of Human Resource Development on Thursday evening. The meeting was attended also by officials from the higher education regulator, the University Grants Commission.
A brief statement was issued by the DUTA president, Rajib Ray who also said "no ad hoc teacher will lose job on the basis of August 28 letter".
Ad hoc to guest
The statement refers to the circular issued by the DU administration on August 28, 2019. While the circular was relevant to "new vacancies" arising in the 2019-20 academic sessions, the Delhi University Principals' Association had cited in that to put all ad hoc appointments on hold last week. As a result, around 4,500 ad hoc teachers were set to be removed from their posts or converted into guest teachers who are paid by the lecture and hold no other academic responsibility in a college.
This was the immediate provocation for the massive protests at the DU Vice Chancellor's office since Wednesday morning. Teachers had also boycotted exam invigilation and evaluation duty.
On Thursday morning, after a night-long meeting between the teachers and DU authorities failed, the MHRD hastened to intervene. Now the DUTA has announced that the amended August 28 letter will "allow provision of ad hoc appointments against substantive posts" but adds that it "waits to see the hard copy of this new notice".
While this staves off the immediate threat of dismissal, it is only a partial victory for teachers. As protesting teacher and elected member of the DU executive council, Rajesh K Jha points out, it does not resolve the larger problem of absorption of ad hoc teachers. For that, DUTA has been demanding a "one-time regulation" from the UGC. Alok Ranjan Pandey, DUTA joint secretary, issued a separate statement saying that "nothing less than absorption is acceptable"
At the time of publishing, it was still not clear whether the protests will continue or even if there was consensus among the various factions that make up DUTA on the direction of their protests. This copy will be updated when there is more clarity. | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 1,180 |
PLYMOUTH.GOV.UK
Search this site ×
Search PLYMOUTH.GOV.UK
Problems paying your Council Tax
Council Tax attachment of earnings orders
If you do not pay your Council Tax, we can apply to a Magistrates' Court for a liability order. If a court grants a liability order, one of the options for recovering the outstanding amount is a Council Tax Attachment of Earnings Order (CTAEO).
A CTAEO is sent to your employer to take money directly from your earnings. A copy of the order will be sent to you and your employer confirming the liability order details and the total debt to be repaid. Payment will be set up on a daily, weekly or monthly basis depending on how you're paid, until the debt is settled.
Submit a General Enquiry (opens new tab)
Guidance for employers
Any CTAEOs you receive will confirm:
the name and payroll number (if known) of the employee that owes Council Tax
the amount that has to be paid
a local authority reference
Deductions should begin as soon as possible after the order has been received. We must receive the deducted amounts by the 18th day of the month following the month in which the deduction was made.
As well as the deducted amount you can deduct £1 per transaction from your employee's salary towards administrative costs. A statement of the total amount deducted (including the £1 administrative costs) should be given to your employee.
Deductions should be made each pay day until the total amount has been paid.
If the employee has moved on or has never been in your employment, you need to let us know within 14 days and then you no longer need to do anything.
Earnings include:
wages or salary (including any fees, bonus, commission, overtime pay or other additions to wages or salary payable under a contract of service)
statutory sick pay
Earnings do not include:
public departments of the Government of Northern Ireland or of a territory outside the United Kingdom sums
armed forces pay and allowances
social security benefits or allowances (including maternity pay)
disability benefits or allowances
wages payable to a person as a seaman, other than as a seaman of a fishing boat
youth training allowances
The amount to be deducted depends on the total net earnings received by the employee. See the tables below for the percentage to be deducted according to the amount of net earnings and the frequency of the pay period.
Table A: Deductions from weekly earnings
Net earnings
Deduction rate (per cent)
Not exceeding £75 0
Exceeding £75 but not exceeding £135 3
Exceeding £135 but not exceeding £185 5
Exceeding £225 but not exceeding £355 12
Exceeding £505 17 in respect of the first £505 and 50 in respect of the remainder
Table B: Deductions from monthly earnings
Not exceeding £300 0
Exceeding £900 but not exceeding £1420 12
Exceeding £1420 but not exceeding £2020 17
Exceeding £2020 17 in respect of the first £2020 and 50 in respect of the remainder
Table C: Deductions based on daily earnings
Exceeding £11 but not exceeding £20 3
Exceeding £33 but not exceeding £52 12
Exceeding £72 17 in respect of the first £72 and 50 in respect of the remainder
Column 1 details pay bands which correspond to net earnings. Column 2 details the percentage of earnings to be deducted. Locate the earnings band in column 1 and then read across to column 2 to find the percentage and then calculate the amount to be deducted.
If the person is paid:
Table A should be use to work out deductions.
Table B should be use to work out deductions.
In weekly intervals
Net earnings should be divided by the number of months in the pay period.
Table B should then be used to work out the appropriate monthly deduction and the resulting amount multiplied by the number of months in the period.
In regular intervals (not whole weeks or months)
Net earnings should be divided by the number of days.
Table C should then be used to work out the appropriate daily rate, which should then be multiplied by the number of days in the period.
In two or more series of payments at regular intervals
Select the series with the shortest interval between payments and use the tables as described above.
In addition, deduct 20 per cent of the net earnings payable in every other series.
If the person is paid in two or more series and all the intervals are the same length, then select one of these, make deductions as described above, and in addition deduct 20 per cent of the net earnings payable in every other series.
An employee's net pay is £150 weekly and £600 monthly.
A deduction of £7 is made for the weekly pay and £120 for the monthly pay (20 per cent of £600).
In irregular intervals
Net earnings should be divided by the number of days since the last payment and Table C should be used to work out the appropriate daily deductions, which in turn should be multiplied by the number of days in the period.
An employee's net pay:
(a) £90 (from 1 April to 9 April - nine days)
(b) £120 (from 10 April to 19 April - 10 days)
(c) £176 (from 20 April to 30 April - 11 days)
The deductions to be made would be:
(a) 90/9 = £10
Daily deduction = £10 x 3 per cent (30p)
Deduction to be made for period = 9 x 0.30 = 2.70
(b) 120/10 = £12
Deduction to be made for period = 10 x 0.36 = £3.60
(c) 176/11 = £16
In regular and irregular and intervals
If on the same pay day the person is to be paid regular period earnings and irregular period earnings these amounts should be added together and treated as earnings payable at the regular interval, the appropriate table being used.
An employee receives £250 as normal net weekly pay. In addition £350 is received every 15 days for a different task.
The deductions to be made would be for weekly earnings (Table A) of £250 = £350 = £600.
The deduction rate for £600 is 17 per cent of the first £370 plus 50 per cent of the remainder (£62.90 = £115 = £177.90).
Multiple orders
If an order is already in place, the new CTAEO is still applied (in date sequence) with the later order being applied to the remaining earnings. If there are two or more orders in place, then no further CTAEOs can be added.
Full details on how to deal with multiple orders:
Priorities between attachment of earnings orders
Regulations that apply:
The Council Tax (Administration and Enforcement) Regulations 1992
Schedule 3 including regulations 32 and 38 to 42 to the Council Tax (Administration and Enforcement) Regulations 1992
View your Council Tax account
Council Tax (energy) rebate
Register or tell us about a change
Your Council Tax bill
Council Tax appeals and complaints
Contact Council Tax
Council Tax collection framework
Council Tax information for landlords
Social media acceptable use policy | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 4,884 |
Wadersloh ist eine Gemeinde im südöstlichen Teil des Kreises Warendorf in Nordrhein-Westfalen. Die Gemeinde Wadersloh besteht aus den drei Dörfern Wadersloh, Diestedde und Liesborn.
Geografie
Im Südosten grenzt die Gemeinde Wadersloh an die Stadt Lippstadt, im Südwesten an die Gemeinde Lippetal. Größere Städte in der Umgebung der Gemeinde Wadersloh sind neben Lippstadt das westlich gelegene Beckum und die Großstadt Hamm, im Nordwesten die Großstadt Münster, im Nordosten Rheda-Wiedenbrück und die Großstadt Bielefeld, welche gleichzeitig die größte Stadt im näheren Umfeld der Gemeinde darstellt, und im Südwesten Soest.
Gewässer
Die südöstliche Gemeindegrenze bildet die Lippe und die östliche Gemeindegrenze ihr Nebenfluss Glenne. Das Hauptgewässer der Gemeinde ist die Liese, die im Ortsteil Diestedde Liesebach und Mühlenbach, im Ortsteil Wadersloh Rottbach und im Ortsteil Liesborn Liese genannt wird und in die Glenne mündet. Zuflüsse zum Liesebach (bzw. Mühlenbach, Rottbach, Liese) sind Maybach, Oenkhausgraben, Boxelbach, Halsterbirke, Biesterbach mit Biestergraben und Krähenbach sowie Bergwiesenbach mit Merschbach und Risselbach. Weitere Bäche sind Baagebach, Kleybach und Landgraben. Die genannten Flüsse und Bäche gehören alle zum Einzugsgebiet des Rheins. Die nördliche Gemeindegrenze wird vom Fortbach gebildet, der zum Einzugsgebiet der Ems gehört.
Gemeindegliederung
Zur Gemeinde Wadersloh werden folgende Ortsteile gezählt: Liesborn mit den Bauerschaften Göttingen, Hentrup, Osthusen, Suderlage und Winkelhorst, Diestedde mit Altendiestedde, Düllo und Entrup und die Kernstadt Wadersloh mit Ackfeld, Basel, Bornefeld, Geist und Vahlhaus.
Urgeschichte
Bei einem Baggersee in Wadersloh wurden vor allem zwischen 1997 und 2001 etwa 900 steinzeitliche Artefakte zutagegefördert. Der Standort wurde wahrscheinlich von Neandertalern benutzt, entweder wiederholt oder über einen längeren Zeitraum, um eiszeitliche Säugetiere zu zerlegen, Rohstoffe zu sammeln und Steinwerkzeuge herzustellen. Die strategische Lage des Standortes zwischen einem Hügel (Wadersloher Platte) und der Ur-Lippe, einem Flusssystem der Weichsel-Kaltzeit, zeigt, dass die Umgebung ein wichtiges Jagdrevier war.
Geschichte
Im Jahr 1187 wurde Wadersloh erstmals urkundlich erwähnt. Wadersloh als Kirchspiel mit seinen Bauerschaften ist allerdings wesentlich älter. Man kann wohl davon ausgehen, dass die Wadersloher Bauerschaften und auch der Standort der Kirche in die sächsischfränkische Zeit (9. Jahrhundert) zurückreichen.
Das Münsterland war bis weit ins Mittelalter hinein ein großes Waldland mit vielen kleinen Siedlungsinseln. In der Regel waren es einige Höfe, die in Sichtweite zueinander lagen. Der Haupthof gab einer solchen Ansiedlung in der Regel den Namen. Zwischen diesen Hofgruppen und ihren Ackerflächen lag der Wald – dieses Waldland steckt auch hinter der Namensendung -lo oder -loe, die sich auch bei vielen Namen dieses Raumes findet, so auch im Namen Wadersloh.
Die Namen Bozo und Bardo gelten als die Gründer des adeligen Damenstifts und späteren Klosters Liesborn. Es spricht einiges dafür, dass sie gegenüber dem Stromberg einen Haupthof besaßen, der einer Bauerschaft den Namen Bardesloe gab. Dieser Name (durch Lautverschiebung) Wardesloe muss auch bei der Kirchgründung eine Rolle gespielt haben, denn der Kirchort hieß von Anfang (der Schriftlichkeit) an Warsloe (1193), Wadersloe (1217), Wardesloe (1498). Wadersloh dürfte von Anfang an ein Kirchdorf gewesen sein, mit der Kirche als Zentrum, von der aus die Straßen und Kirchwege radialstrahlig in die Bauerschaften liefen.
Die Menschen waren Handwerker und Händler, die in der Regel auch selbst eine kleine Landwirtschaft betrieben. Als es im 11. und 12. Jahrhundert langsam wärmer wurde (Klimaoptimum), wurden die Ernten besser und die Bevölkerung nahm erheblich zu; auch die Erfindungen in der Landwirtschaft trugen dazu bei. Städte (wie Lippstadt) wurden gegründet und die Kirchdörfer wuchsen.
So kam es Ende des 12. Jahrhunderts zu einer Neuordnung in der Verwaltung des Bistums Münster. Bischof Hermann II. von Katzenelnbogen ordnete die Kirchspiele neu und unterstellte Wadersloh dem Archidiakonat des Propstes von St.Martini in Münster. In diese Zeit fiel auch die Errichtung einer Steinkirche.
Dem Klimaoptimum folgte im 14. Jahrhundert eine "kleine Eiszeit", die Ernten wurden schlechter und die Pest dezimierte die Einwohnerschaft und suchte Dörfer und Städte in den folgenden Jahrhunderten immer wieder heim. Viele Höfe verfielen und Armut und Hunger machten sich breit. Fehden und Kriege taten ein Übriges. Die Grenzlage Waderslohs war ein großer Nachteil, durchziehende Truppen nahmen, was sie bekommen konnten, plünderten und brandschatzten. Zu Beginn des 19. Jahrhunderts waren große Teile des Münsterlandes ein Armenhaus.
Diese Situation besserte sich mit der Industrialisierung und der Entstehung der städtischen Ballungsräume. Durch den Bahnanschluss 1898 wurden auch die Arbeitsplätze in Beckum und Lippstadt erreichbar. Bis 1945 blieb Wadersloh aber ein von der Landwirtschaft geprägtes Dorf.
Am 1. April 1898 wurde die Gemeinde Benteler durch Ausgliederung aus Wadersloh neu gebildet.
Die heutige Gemeinde Wadersloh wurde am 1. Januar 1975 aus den ehemals eigenständigen Gemeinden Wadersloh, Liesborn (ohne den Gemeindeteil Bad Waldliesborn) und Diestedde gebildet. Zuvor hatten diese zum 1975 aufgelösten Amt Liesborn-Wadersloh gehört.
Archiv
1972 ging das Archiv des damaligen Amtes Wadersloh in das Kreiszentralarchiv Warendorf über. Dort liegt es bis heute.
Politik
Gemeinderat
Der Gemeinderat setzt sich seit der Kommunalwahl 2020 wie folgt zusammen:
Bürgermeister
Im August 2009 wurde Christian Thegelkamp mit knappem Vorsprung zum neuen Bürgermeister gewählt, er setzte sich dabei gegen Amtsinhaber Theo Westhagemann durch. Im Mai 2014 und im September 2020 wurde Thegelkamp wiedergewählt.
Gemeindepartnerschaften
Marcillat-en-Combraille, Frankreich
Néris-les-Bains, Frankreich
Faulungen, Thüringen
Wappen
Banner
Architektur
Im Ort gibt es die denkmalgeschützte Pfarrkirche St. Margareta in Wadersloh sowie die Abtei Liesborn in Liesborn und das Wasserschloss Crassenstein in Diestedde.
Wirtschaft
Ein weltweit bekanntes Produkt aus Wadersloh sind die roten Gloria-Feuerlöscher. Die Gloria GmbH mit Sitz im Hauptort Wadersloh ist der größte Hersteller von Feuerlöschgeräten in Europa. Firma Gödde-Beton-Liesborn (GBL) und die Firma Laukötter Druckguss sowie die Firma Westag & Getalit AG, die mit Hauptsitz in Rheda-Wiedenbrück ihr größtes Zweitwerk in Wadersloh betreibt.
Schulen
In Wadersloh gibt es zwei weiterführende Schulen. Das Gymnasium Johanneum Wadersloh sowie die Sekundarschule Wadersloh. In jedem der drei Ortsteile befindet sich eine Grundschule. Im Schloss Crassenstein befindet sich ein Internationales Internat.
Persönlichkeiten
Franz Bornefeld-Ettmann (1881–1961), Politiker (Zentrum), Reichstagsabgeordneter
Hermann Stammschröer (1890–1957), katholischer Geistlicher und im KZ Dachau inhaftiert
Anton Bornefeld (1898–1980), katholischer Priester und ein entschiedener Gegner des Nationalsozialismus
Rita Süssmuth geb. Kickuth (* 1937), deutsche Politikerin (CDU), wuchs in Wadersloh auf
Heinz Knüwe (* 1956), ehemaliger deutscher Fußballprofi und Fußballtrainer
Josef Klenner (* 1949), Präsident des Deutschen Alpenvereins (DAV)
Bernhard Gerwert (* 1953), Vorstand EADS, CEO von Airbus Defence and Space und Präsident des Bundesverband der Deutschen Luft- und Raumfahrtindustrie
Herbert Gövert, ehemaliger Bürgermeister und Träger des Verdienstordens des Landes Nordrhein-Westfalen
Weblinks
Gemeindeverwaltung Wadersloh
Wadersloh für Familienforscher im GenWiki
Kinderstadtplan der Gemeinde Wadersloh
Einzelnachweise
Ort im Kreis Warendorf
Ort im Münsterland
Ort an der Lippe
Ersterwähnung 1187 | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 4,094 |
There was pandemonium in parliament on Tuesday when angry MDC Youths clad in party regalia came and beat up some petitioners whom they accused of belonging to one of the factions in the MDC-Alliance.
Members of the Civil society groups and churches had stormed Parliament to petition the Speaker, Advocate Jacob Mudenda to punish rowdy Members of Parliament who are taking parliament business for granted.
The citizens under the banner of Civic Society and Churches Joint Forum were led by self styled Bulawayo based pastor Anglistone T. Sibanda.
We the citizens of Zimbabwe who turned out in large numbers across the country to exercise our constitutional right and mandate to elect representatives into the August house on the 31st of July 2018.
Concerned by the wanton transformation of our esteemed decision making house into a caricature and house of theatric performances.
Worried by the abuse of our hard earned money that we pay as taxes to support the welfare of our elected officials that is being wasted because they are not serving the purpose of instead citizens feel shortchanged if our representatives prove to be priotizing their party interest in parliament ahead of national interests.
Further disturbed that our communities are being disenfranchised, neglected and marginalized by the behavior of the people we elected to represent us who are wasting our votes, our trust and indeed our efforts towards building a democratic society by not representing us.
Perturbed by the denigration, insults, abuse of each other by the members of the August house while we grappling with Gender Based Violence and other forms of violence that make us a primitive society.
The abuse of each other right on camera by our legislators reflects badly on our societal values of tolerance, respect and dignity and projects us as a rowdy and lawless society.
We note with concern the legitimacy illegitimacy question that has taken the centre stage in Parly and wish to remind our Honourables that parliamentary seats and infact parliament has no such battles on record ,its unfair to hold the Parly over a borrowed battle. What made us send you to Parly is policy business ,reforms and to focus on delivering you election promises not otherwise.
Mr Speaker sir. May we remind each other that the issue of legitimacy was closed by the highest court in the land when a ruling was made after the disputed Presidential election, the parliamentarians in unity moved on to take oaths of office in terms of the very constitution which endorsed the President.
We as citizens take it that the decision by the Constitutional court is binding as directed by the constitution, hence anyone who is rejecting the decision is by default in contempt of the Constitutional court.
In the spirit of promoting the rule of law, Mr Speaker sir, we note that some members of the august house are technically in violation of the law while purporting to be fighting for the restoration of the rule of law, according to us, they are breaking the law in the very place where laws are made thus contradicting themselves.
We are concerned that the people we elected to represent us are sacrificing our developmental challenges and issues on the alter in preference to lamenting on behalf of their Presidential election losing candidate and as such feel that our rights are being violated and we are taken for granted.
Mr Speaker sir. We are aware that party regalia is not allowed in parliament, but we note with concern that some members of the August house come into parliament putting on party regalia, not in the physical sense but in the attitude and behavioral sense.
When we, the citizens watch the debate we easily tell who is from which party and that is against the standing rules and orders Mr Speaker Sir.
With respect to the whipping system, we believe that the concept is working against us and fueling parliamentary polarization and tensions, promoting party position over our issues as the electorate and therefore violating our right to be represented.
1. The Committee on standing orders speedily review the code of conduct for parliamentarians and come up with strict punitive measures against rowdy behavior in Parliament.
2. The Committee on Standing Rules and orders protect the public interest by whipping all MPs into National Agenda as expected by the constitution .
3. Redefine Party regalia that not allowed in parliament to include partisan behavior that is disruptive to the proceedings in the house.
4. To revisit the whipping system that takes away the independence of our elected officials and capture them into party surrogates and infringe on our right to be represented as citizens.
We have strong belief that Zimbabwe can become a leading nation with a growing economy when we depoliticize and depolarize our institutions, transform our mindsets, change our politics into development economics and move into a dispensation of a developmental trajectory for the benefit us the ordinary citizens.
We demand that our lawmakers immediately put off their political party jackets and re commit to the national cause and desist from deriving non Parliament Agendas/battles into the August house.
We anticipate a radical and robust intervention and response on our concerns stated above. | {
"redpajama_set_name": "RedPajamaC4"
} | 3,125 |
STAMFORD, CT. – HARMAN, the global premium audio and infotainment group (NYSE:HAR) is pleased to announce that it was awarded Ferrari's prestigious Innovation Award on July 20 in Maranello, Italy.
In presenting the Innovation Award, Ferrari Chairman Luca di Montezemolo and CEO Amedeo Felisa applauded HARMAN for its "high-end solutions regarding audio, navigation and telematics" and customizing them for Ferrari. HARMAN introduced the QuantumLogic™ surround-sound technology with Ferrari, making the new FF model the first production vehicle to include this unique technology.
QuantumLogic surround technology uses audio algorithms to extract signal streams and impulse responses from the original recording. Individual voices and instruments, as well as embedded reverberant spatial information, are identified and then re-authored into a precise multi-channel soundstage. Unique to QuantumLogic is its "aesthetic engine," which combines the individual signal streams and patented spatial filter bank technology with psycho-acoustic modeling for transparent digital processing and perfect acoustic reconstruction. The result is a stunningly immersive playback system with refined clarity and detail. | {
"redpajama_set_name": "RedPajamaC4"
} | 8,791 |
\section{Introduction}
\label{sec:Intro}
During the past two decades, Asymptotic Safety \cite{W80} matured from a hypothetical scenario to a theory with a
realistic chance to describe the structure of spacetime and the gravitational interaction consistently and predictively
even on the shortest length scales possible. In particular, there is mounting evidence supporting the existence of the
decisive nontrivial renormalization group (RG) fixed point. Yet, a number of immediate questions are still open. The
most obvious one is about the precise nature of the action functional which describes this fixed point. In which way
exactly does it depend on the metric, the background metric, and the Faddeev--Popov ghosts? Is it local? What are the
structural properties of the fixed point theory, i.e.\ the one defined directly at the fixed point rather than by a
trajectory running away from it. Is this theory a conformal field theory?
In $2$ dimensions we are indeed used to the picture that the conformal field theories correspond to points in theory
space that are fixed points of the RG flow \cite{Na15}. In $4$ dimensions, however, Quantum Einstein Gravity (QEG) has
a scale invariant fixed point theory but it is unclear whether it is conformal.
While conformality is not known to be indispensable, there exist several other properties an asymptotically safe theory
\emph{must} possess in addition to its mere nonperturbative renormalizability, that is, the existence of a suitable
non-Gaussian fixed point. The two most important ones are clearly \emph{Background Independence} and \emph{unitarity}.
While there are by now first promising results which indicate that the requirements of Background Independence and
Asymptotic Safety can be met simultaneously in sufficiently general truncations of the RG flow \cite{BR14}, little
is known about the status of unitarity.
In this connection the somewhat colloquial term ``unitarity'' is equivalent to ``Hil\-bert space positivity'' and is
meant to express that the state space of the system under consideration contains no vectors having a negative scalar
product with itself (``negative norm states''). If it does so, it is not a Hilbert space in the mathematical sense of
the word and cannot describe a \emph{quantum} system as the probability interpretation of
quantum mechanics would break down then.
At least on (nondynamical) flat spacetimes the criterion of Hilbert space positivity, alongside with the spectral
condition can be translated from the Lorentzian to the Euclidean setting where it reappears as the requirement of
reflection-, or Osterwalder--Schrader, positivity \cite{OS73,S93,GJ87}.
Unitarity is indeed a property that is not automatic and needs to be checked in order to demonstrate the viability of
the Asymptotic Safety program based upon the Effective Average Action (EAA). The EAA for gravity \cite{R98} is a scale
dependent effective action for a BRST gauge-fixed theory. Its operator formulation amounts to an indefinite metric
(Krein space) quantization, and so the negative norm states it contains should be ``factored out'' ultimately in order
to obtain a positive (``physical'') state space, a true Hilbert space. While this procedure is standard and familiar
from perturbative quantum gravity and Yang--Mills theory, for instance, the situation is much more involved in
Asymptotic Safety. The reason is that, implicitly, this indefinite metric quantization is applied to a bare action
which is essentially given by the fixed point functional \cite{MR09,VZ11,M94,MS15}. As such it is already in itself the
result of a technically hard nonperturbative computation which in practice can be done only approximately, for the time
being.
In the present paper we explore the question of Hilbert space positivity together with a number of related issues such
as locality by analyzing the situation in $2$ dimensions where a number of technical simplifications occur. We start
out from the Einstein--Hilbert truncation of the EAA in $d=2+\ve>2$ dimensions, investigate the nature of its
$\ve\rightarrow 0$ limit, and finally construct a \emph{manifestly $2$-dimensional} action which describes 2D
Asymptotic Safety without reverting to ``higher'' dimensions in any way. In this manner we shall see that the
non-Gaussian fixed point (NGFP) underlying Asymptotic Safety is governed by a conformal field theory (CFT) which is
interesting in its own right, and whose properties we shall discuss. Interestingly enough, it turns out to possess a
positive central charge, thus giving rise to a unitary representation of the Virasoro algebra and a ``positive''
Hilbert space in the above sense.
\medskip
\noindent
\textbf{(1)}
A framework which allows to systematically search for asymptotically safe theories is the Effective Average Action, and
in particular its functional RG equation (FRGE): The RG flow of the EAA, $\Gamma_k$, is governed by the exact
nonperturbative evolution equation
\cite{RW93,W93,RW94,R98}
\begin{equation}
k\p_k\Gamma_k = \frac{1}{2}\STr\left[\big(\Gamma_k^{(2)}+\Rk\big)^{-1}\,k\p_k\Rk\right] .
\label{eq:FRGE}
\end{equation}
It describes the dependence of $\Gamma_k$ on the RG scale $k$ which plays the role of a
regularization scale for infrared (IR) fluctuations. The suppression of IR modes is realized by the cutoff operator
$\Rk$, satisfying $\Rk\rightarrow k^2$ for IR and $\Rk\rightarrow 0$ for UV modes, respectively. Besides,
$\Gamma_k^{(2)}$ denotes the Hessian of the EAA with respect to the dynamical field. In
particular, in the case of gravity when the background field method is employed, it is the second functional derivative
of $\Gamma_k[g,\gb]$ w.r.t.\ $g$ at fixed background $\gb$. By construction, $\Gamma_k$ approaches the full quantum
effective action $\Gamma$ in the limit $k\rightarrow 0$. Although its $k\rightarrow\infty$ limit is \emph{formally}
related to the bare action \cite{W93,RW94}, in the Asymptotic Safety program no bare action is posited a priori; it is
rather ``reconstructed'' from the condition that this limit actually exists nonperturbatively
\cite{MR09,VZ11,M94,MS15}.
In order to find solutions to the FRGE \eqref{eq:FRGE} one usually resorts to truncations, implying a reduction of the
infinite dimensional theory space that consists of all possible action monomials compatible with the underlying
symmetry. By now a large variety of truncations have been studied to support the existence of a NGFP of metric gravity,
including for instance terms of higher orders in the curvature or actions that couple gravity to matter
\cite{NR06,P09,CPR09,RS12}.
\medskip
\noindent
\textbf{(2)}
In this paper we will focus on a gravity+matter system where the purely gravitational sector consists of the Euclidean
Einstein--Hilbert truncation,
\begin{equation}
\Gamma_k^\text{grav}[g] = \frac{1}{16\pi G_k} \int \dd x \sg \,\big( -R + 2\mku\Lambda_k \big),
\label{eq:EHfunctional}
\end{equation}
and the matter contribution is given by a multiplet of $\ns$ scalar fields $A=(A^i)$, with $i=1,\cdots,\ns$, minimally
coupled to the full, dynamical metric:
\begin{equation}
\GM[g,A]\equiv \Gamma^\text{M}[g,A]=\frac{1}{2}\sum\limits_{i=1}^{\ns}\int\dd x\sg\, g^\mn\,\p_\mu A^i\,\p_\nu A^i\,.
\label{eq:matter}
\end{equation}
Note that the matter action contains no running parameters in the present truncation(s).
Supplemented by appropriate gauge fixing and ghost terms, the sum of $\Gamma_k^\text{grav}$ and $\GM$ can be inserted
into the flow equation \eqref{eq:FRGE} in order to determine the running of Newton's constant $G_k$ and the
cosmological constant $\Lambda_k$.
The RG studies of this system for $d=2+\ve$ dimensions ($\ve\searrow 0$) are of particular importance for our
analysis. It is possible then to argue on general grounds that the $\beta$-function of the dimensionless Newton
constant $g_k\equiv G_k k^{d-2}=G_k k^\ve$ must be of the form
\begin{equation}
\beta_g = \ve g - b g^2 + \mO(g^3)\,.
\end{equation}
This implies the existence of a nontrivial fixed point which is located at
\begin{equation}
g_* = \ve/b\,,
\label{eq:NGFPWithB}
\end{equation}
up to higher $\ve$-orders. Therefore, we have $g_k\propto\ve$ and, in turn, $G_k\propto\ve$ in the vicinity of the
NGFP, having far-reaching consequences for the dimensional continuation of the Einstein--Hilbert action to two
dimensions.
\medskip
\noindent
\textbf{(3)}
In \emph{exactly} $2$ dimensions the integral $\int\td^2 x\sg\,R$ is a purely topological term according to the
Gauss--Bonnet theorem. In particular, it is independent of the metric and does not imply any local dynamics. Thus, one
might expect that \eqref{eq:EHfunctional} becomes trivial up to the cosmological constant term when $d$ approaches $2$.
However, the fact that the prefactor $1/G_k$ entails a $1/\ve$ pole gives so much weight to $\int\td^{2+\ve} x\sg\,R$
that the limit $\ve\rightarrow 0$ remains indeed nontrivial.
We will present a new argument in this work showing that
the (local) Einstein--Hilbert action turns into a \emph{non-local} action in the limit $d\rightarrow 2$. The essential
part of this limit action will be seen to be given by Polyakov's induced gravity action.
Our proof will confirm recurring speculation \cite{CD15} that the induced gravity action is the natural
$2$-dimensional analogue of the Einstein--Hilbert action in $d>2$ as both actions determine field equations for the
metric in their respective spacetime dimension.
\medskip
\noindent
\textbf{(4)}
Here we go one step further: We do not require that one action has to be replaced
by the other one when switching between $d=2$ and $d>2$. The idea is rather to say that there is \emph{only one} common
origin, the Einstein--Hilbert action in a general dimension $d$, and that \emph{the induced gravity action emerges
automatically when $d$ approaches $2$}.
It is this latter 2D action, analyzed at the NGFP, that establishes the contact between the Asymptotic Safety studies
within the Einstein--Hilbert truncation and $2$-dimensional conformal field theory. It will form the basis of our
investigations concerning central charges and unitarity.
\medskip
\noindent
\textbf{(5)}
It turns out that the crucial fixed point value $g_*$ depends on the way the metric is parametrized.
The background field technique which underlies the Asymptotic Safety studies expresses the dynamical metric $g_\mn$ in
terms of a fixed background metric $\gb_\mn$ and the fluctuations $h_\mn$. There are, however, different possible
choices of how these fields are related, in particular the standard linear parametrization $g_\mn=\gb_\mn+h_\mn$ and
the exponential parametrization $g_\mn=\gb_{\mu\rho}(\e^h)^\rho{}_\nu$. These two parametrizations
give rise to different $\beta$-functions and different central charges.
As we will argue, this disagreement does not necessarily mean that one choice is correct while the other one is wrong,
and we advocate the possibility that the two parametrizations might refer to different universality classes.
In the 2D limit, however, we provide strong evidence that the exponential parametrization is more appropriate for a
consistent description of the gravitational interactions. It is this choice that leads to a NGFP theory with the value
$c_\text{grav}=25$ for the central charge of the pure gravity sector.
\medskip
\noindent
\textbf{(6)}
This paper is organized as follows. We review Background Independence and the special role of self-consistent
backgrounds in \emph{section \ref{sec:BackgroundIndependence}}. In particular, we re-interpret the effective Einstein
equation as a tadpole condition and the trace of the stress energy tensor due to metric fluctuations as a kind of
classical ``trace anomaly''. Here, all calculations are performed in $2+\ve$ dimensions, and the 2D limit is taken
at the very end only.
This leads us to the question if the same trace anomaly can be obtained when starting out from a strictly 2D action.
The answer to this question will be given in \emph{section \ref{sec:IndGravityFromEH}} where we compute the 2D limit of
the Einstein--Hilbert action at the NGFP and argue that it results indeed in an action with the sought-for properties.
We demonstrate in \emph{section \ref{sec:NGFPCFT}} that this 2D gravitational NGFP action gives rise to a unitary
conformal field theory. Particular attention is paid to the relation between the crucial sign of its central charge,
the occurrence of a conformal factor instability, and unitarity. \emph{Section \ref{sec:Emergence}} is dedicated to
the impact of different metric parametrizations and establishes the special status of the exponential parametrization
in the 2D limit. Finally, \emph{section \ref{sec:recFI}} is devoted to the properties of the fixed point CFT, and
a comparison of Asymptotic Safety to other approaches to 2D gravity. For that purpose we construct a functional
integral which reproduces the effective average action related to the CFT behind the NGFP. This ``reconstructed''
functional integral is used to investigate the gravitational dressing of matter field operators when conformal matter
is coupled to asymptotically safe gravity. We demonstrate that, contrary to what one would have expected, there is
no KPZ scaling in our setting as a consequence of Asymptotic Safety. We also discuss similarities and differences
compared with non-critical string theory and Monte--Carlo simulations in the CDT approach.
\emph{Section \ref{sec:Conc}} contains our conclusions and an outlook.
In the appendix we catalog various useful identities for Weyl transformations, and we include a detailed discussion of
the induced gravity action in the presence of zero modes.
\section{Background Independence via background fields}
\label{sec:BackgroundIndependence}
In this preparatory section we collect a number of results concerning the implementation of Background Independence in
the EAA framework which actually does employ (unspecified!) background fields. In particular, we introduce the energy
momentum tensor of metric fluctuations in a background, as well as an associated ``trace anomaly''. The latter will be
used later on in order to identify the conformal field theory at the heart of Asymptotic Safety in $2$ dimensions.
\subsection{The effective Einstein equation re-interpreted}
\label{sec:EEE}
Let us consider a generic effective average action $\Gamma_k[\Phi,\bP]\equiv \Gamma_k[\vp;\bP]$ involving a multiplet of
dynamical fields $\big\langle\hP^i\big\rangle\equiv \Phi^i$, associated background fields $\bP^i$, and fluctuations $\vp^i
\equiv \langle \hvp^i\rangle = \Phi^i - \bP^i$. The effective average action implies a source $\leftrightarrow$ field
relationship which contains an explicit cutoff term linear in the fluctuation fields:
\begin{equation}
\frac{1}{\sgb}\frac{\delta\Gamma_k[\vp;\bP]}{\delta\vp^i(x)}+\Rk[\bP]_{ij}\,\vp^j(x)=J_i(x) \,.
\label{eq:so-fi}
\end{equation}
By definition, self-consistent backgrounds are field configurations $\bP(x)\equiv \bP_k^\text{sc}(x)$ which allow
$\vp^i=0$ to be a solution of \eqref{eq:so-fi} with $J_i=0$. A self-consistent background is particularly ``liked'' by
the fluctuations, in the sense that they leave it unaltered on average: $\langle\hP\rangle = \bP+\langle\hvp\rangle =
\bP^\text{sc}$. These special backgrounds are determined by the tadpole condition $\langle\hvp^i\rangle=0$, which reads
explicitly
\begin{equation}
\frac{\delta}{\delta\vp^i(x)}\Gamma_k[\vp;\bP]\Big|_{\vp=0,\,\bP=\bP_k^\text{sc}} = 0 \,.
\label{eq:tapo}
\end{equation}
Equivalently, in terms of the full dynamical field,
\begin{equation}
\frac{\delta}{\delta\Phi^i(x)}\Gamma_k[\Phi,\bP]\Big|_{\Phi=\bP=\bP_k^\text{sc}} = 0 \,.
\label{eq:tapo-full}
\end{equation}
In this paper we consider actions of the special type
\begin{equation}
\Gamma_k[g,\xi,\xb,A,\gb] = \Gg[g,\gb] +\GM[g,A,\gb] +\Gamma_k^\text{gf}[g,\gb] +\Gamma_k^\text{gh}[g,\xi,\xb,\gb] .
\end{equation}
These functionals include a purely gravitational piece, $\Gg$, furthermore a (for the time being) generic matter action
$\GM$, as well as gauge fixing and ghost terms, $\Gamma_k^\text{gf}$ and $\Gamma_k^\text{gh}$, respectively. Concerning the
latter, only the following two properties are needed at this point:
(i) The $h_\mn$-derivative of the gauge fixing functional $\Gamma_k^\text{gf}[h;\gb] \equiv
\Gamma_k^\text{gf}[\gb+h,\gb]$ vanishes at $h_\mn=0$. This is the case, for example, for classical gauge fixing terms
$S_\text{gf} \propto \int(\mF h)^2$ which are quadratic in $h_\mn$.
(ii) The functional $\Gamma_k^\text{gh}$ is ghost number conserving, i.e.\ all terms contributing to it have an equal
number of ghosts $\xi^\mu$ and antighosts $\xb_\mu$. Again, classical ghost kinetic terms
$\propto \int\xb\mathcal{M}\xi$ are of this sort.
Thanks to these properties, $\Gamma_k^\text{gf}$ drops out of the tadpole equation \eqref{eq:tapo-full}, and it
follows that $\xi=0=\xb$ is always a consistent background for the Faddeev--Popov ghosts. Adopting this background
for the ghosts, \eqref{eq:tapo-full} boils down to the following conditions for self-consistent metric and matter
field configurations $\gsc$ and $\Asc$, respectively:
\begin{align}
0 &= \frac{\delta}{\delta g_\mn(x)}\Big\{\Gg[g,\gb]+\GM[g,\Asc,\gb]\Big\}\Big|_{g=\gb=\gsc} \;,
\label{eq:selfcon} \\
0 &= \frac{\delta}{\delta A(x)}\GM[g,A,\gb]\Big|_{g=\gb=\gsc,\, A=\Asc} \;.
\label{eq:selfconmatter}
\end{align}
Introducing the energy-momentum tensor of the matter field,
\begin{equation}
T^\text{M}[\gb,A]^\mn(x) \equiv \frac{2}{\sgbx} \frac{\delta}{\delta g_\mn(x)} \GM[g,A,\gb]\Big|_{g=\gb} \;,
\end{equation}
the first condition, equation \eqref{eq:selfcon}, becomes
\begin{equation}
0 = \frac{2}{\sgbx} \frac{\delta}{\delta g_\mn(x)} \Gg[g,\gb]\Big|_{g=\gb=\gsc} + T^\text{M}[\gsc,\Asc]^\mn(x) .
\label{eq:EFE}
\end{equation}
This relation plays the role of an effective gravitational field equation which, together with the matter
equation \eqref{eq:selfconmatter}, determines $\gsc$ and $\Asc$. Structurally, eq.\ \eqref{eq:EFE} is a generalization
of the classical Einstein equation to which it reduces if $\Gg[g,\gb]\equiv \Gg[g]$ happens to have no ``extra
$\gb$-dependence'' \cite{MR10} and to coincide with the Einstein--Hilbert action; then the
$\delta/\delta g_\mn$-term in \eqref{eq:EFE} is essentially the Einstein tensor $G_\mn$.
In this very special background-free case we recover the familiar setting of classical General Relativity where there is
a clear logical distinction between matter fields and the metric, meaning the full one, $g_\mn$, while none other appears
in the fundamental equations then. It is customary to express this distinction by putting $G_\mn$ on the LHS of Einstein's
equation, the side of gravity, and $T_\mn^\text{M}$ on the RHS, the side of matter.
In the effective average action approach where, for both deep conceptual and technical reasons \cite{MR10,BR14}, the
introduction of a background is unavoidable during the intermediate calculational steps, this categorical distinction of
matter and gravity, more precisely, matter fields and metric fluctuations, appears unmotivated. It is much more natural to
think of $h_\mn$ as a \emph{matter} field which propagates on a background spacetime furnished with the metric $\gb_\mn$.
Adopting this point of view, we interpret the $\delta/\delta g_\mn$-term in \eqref{eq:EFE} as the energy-momentum tensor of
the $h_\mn$-field, and we define
\begin{equation}[b]
T^\text{grav}[\gb]^\mn(x) \equiv \frac{2}{\sgbx} \frac{\delta}{\delta g_\mn(x)} \Gg[g,\gb]\Big|_{g=\gb}
= \frac{2}{\sgbx} \frac{\delta}{\delta h_\mn(x)} \Gg[h;\gb]\Big|_{h=0} \;.
\end{equation}
The tadpole equation \eqref{eq:EFE} turns into an Einstein equation with zero LHS then:
\begin{equation}[b]
0 = T_\mn^\text{grav}[\gsc] + T_\mn^\text{M}[\gsc,\Asc] .
\label{eq:Tadpole}
\end{equation}
It says that for a background to be self-consistent, the total energy-momentum tensor of matter and metric
fluctuations, in this background, must vanish. (In the general case there could be a contribution from the
ghosts also.)
\subsection[The stress tensor of the \texorpdfstring{$h_\mn$}{h}-fluctuations]
{The stress tensor of the \texorpdfstring{$\bm{h_\mn}$}{h}-fluctuations}
\label{sec:Stresshmunu}
Note that in general, $T_\mn^\text{grav}$ is not conserved, $\bar{D}_\mu T^\text{grav}[\gb]^\mn\neq 0$, since due to the
presence of two fields in $\Gg$ the standard argument does noes not apply. Of course, it is conserved in the special
case $\Gg[g,\gb]\equiv\Gg[g]$ when there is no extra $\gb$-dependence.
For example, choosing $\Gg[\gb]$ to be the single-metric Einstein--Hilbert functional \eqref{eq:EHfunctional},
the corresponding energy-momentum tensor of the $h_\mn$-fluctuations is given by the divergence-free expression
\begin{equation}
T_\mn^\text{grav}[\gb] = \frac{1}{8\pi G_k} \Big(\bar{G}_\mn + \Lambda_k\, \gb_\mn \Big) ,
\label{eq:EHEMTensor}
\end{equation}
with $\bar{G}_\mn$ the Einstein tensor built from $\gb_\mn$. The trace of the energy-momentum tensor
\eqref{eq:EHEMTensor} reads
\begin{equation}
\Theta_k[\gb] \equiv \gb^\mn\, T_\mn^\text{grav}[\gb] = \frac{1}{16\pi G_k}\Big[ -(d-2)\bar{R}+2d\,\Lambda_k \Big],
\end{equation}
where $\bar{R}\equiv R(\gb)$. A remarkable feature of this trace is that it possesses a completely well defined,
unambiguous limit $d\rightarrow 2$ if $G_k$ and $\Lambda_k$ are of first order in $\ve=d-2$. In terms of the finite
quantities $\bGk \equiv G_k/\ve$ and $\bLk \equiv \Lambda_k/\ve$ which are of order $\ve^0$, we have
\begin{equation}
\begin{split}
\Theta_k[\gb] &= \frac{1}{16\pi\bGk} \Big[-\bar{R}+4\bLk\Big] +\mO(\ve) \\
&= \frac{1}{16\pi\bgk} \Big[-\bar{R}+4k^2\blk\Big] +\mO(\ve).
\end{split}
\label{eq:Theta2}
\end{equation}
In the second line of \eqref{eq:Theta2} we exploited that in exactly two dimensions the dimensionful and
dimensionless Newton constant are equal, so $g_k = G_k$ and $\bgk = \bGk$, while, as always,
$\lambda_k \equiv \Lambda_k/k^2$, hence $\blk=\bLk/k^2$.
When the underlying RG trajectory is in the NGFP scaling regime the dimensionless couplings are scale
independent, and
\begin{equation}
\Theta_k^\text{NGFP}[\gb] = \frac{1}{16\pi\breve{g}_*} \Big[-\bar{R}+4\breve{\lambda}_* k^2 \Big].
\end{equation}
Using the parametrization $g_*\equiv \ve/b$ as in references \cite{NR13,N15,BR14} we obtain
\begin{equation}[b]
\Theta_k^\text{NGFP}[\gb] = \Big({\textstyle\frac{3}{2}}b\Big) \, \frac{1}{24\pi}
\Big[-\bar{R}+4\breve{\lambda}_* k^2 \Big].
\label{eq:Theta3}
\end{equation}
Here and in the following we consider $\Theta_k$ and $\Theta_k^\text{NGFP}$ as referring to \emph{exactly $2$
dimensions}, in the sense that the limit has already been taken, and we omit the ``$\mO(\ve)$'' symbol.
\subsection{The intrinsic description in exactly 2 dimensions}
\label{sec:descTwoD}
In this paper we would like to describe the limit $d\rightarrow 2$ of QEG in an intrinsically $2$-dimensional
fashion, that is, in terms of a new functional $\Gamma_k^\text{grav,2D}$ whose arguments are fields in strictly
$2$ dimensions, and which no longer makes reference to its ``higher'' dimensional origin. Since the Einstein--Hilbert
term is purely topological in exactly $d=2$, it is clear that the sought-for action must have a different structure.
\medskip
\noindent
\textbf{(1)}
One of the conditions which we impose on $\Gamma_k^\text{grav,2D}$ is that it must reproduce the trace $\Theta_k$
computed in $d>2$, since we saw that this quantity has a smooth limit with an immediate interpretation in $d=2$ exactly:
\begin{equation}
2 g_\mn \frac{\delta}{\delta g_\mn} \Gamma_k^\text{grav,2D}[g,\gb]\Big|_{g=\gb}=\sgb\, \Theta_k[\gb].
\label{eq:RelEMTensors}
\end{equation}
Furthermore, if $\Gg$ is a single-metric action, we assume that $\Gamma_k^\text{grav,2D} \equiv \Gamma_k^\text{grav,2D}[g]$
has no extra $\gb$-dependence either. The condition \eqref{eq:RelEMTensors} fixes its response to an infinitesimal Weyl
transformation then:
\begin{equation}
2 g_\mn(x) \frac{\delta}{\delta g_\mn(x)} \Gamma_k^\text{grav,2D}[g]
\equiv \frac{\delta}{\delta \sigma(x)} \Gamma_k^\text{grav,2D}[\e^{2\sigma} g]\Big|_{\sigma=0}
= \sqrt{g(x)}\,\Theta_k[g](x).
\label{eq:InfWeyl}
\end{equation}
For the example of the Einstein--Hilbert truncation, $\Theta_k$ is of the form
\begin{equation}
\Theta_k[g]=a_1(-R+a_2),
\label{eq:ThetaEH}
\end{equation}
with constants $a_1,a_2$ which can be read off from \eqref{eq:Theta2} -- \eqref{eq:Theta3} for the various cases.
\medskip
\noindent
\textbf{(2)}
It is well known how to integrate equation \eqref{eq:InfWeyl} in the conformal gauge \cite{P81}. By setting
\begin{equation}
g_\mn(x) = \hg_\mn(x)\,\e^{2\phi(x)},
\end{equation}
with a fixed reference metric $\hg_\mn$ (conceptually unrelated to $\gb_\mn$), one for each topological sector, and
taking advantage of the identities in appendix \ref{app:Weyl}, eq.\ \eqref{eq:InfWeyl} with \eqref{eq:ThetaEH} is
turned into
\begin{equation}
\frac{\delta}{\delta \phi(x)} \Ggd[\e^{2\phi} \hg] = a_1\shgx \Big[2\hD_\mu\hD^\mu\phi(x)-\hR(x)
+a_2 \,\e^{2\phi(x)}\Big].
\end{equation}
The general solution to this equation is easy to find:
\begin{equation}[b]
\Ggd[\hg\e^{2\phi}] = \GL[\phi;\hg]+U_k[\hg].
\label{eq:GravToLiou}
\end{equation}
Here $U_k$ is a completely arbitrary functional of $\hg$, independent of $\phi$, and $\GL$ denotes the Liouville action
\cite{RW97}:
\begin{equation}
\begin{split}
\GL[\phi;\hg] &= (-2a_1)\int\td^2x\shg \left({\frac{1}{2}}\hD_\mu\phi\hD^\mu\phi
+ {\frac{1}{2}}\hR\phi - {\frac{a_2}{4}}\e^{2\phi} \right) \\
&= (-2a_1)\,\Delta I[\phi;\hg] + \frac{1}{2}a_1 a_2\int\td^2x\shg\,\e^{2\phi} \,.
\end{split}
\label{eq:LiouvilleAction}
\end{equation}
In the last line we employed the normalized functional
\begin{equation}
\Delta I[\phi;g] \equiv \frac{1}{2}\int\td^2x\sg\,\big(D_\mu\phi D^\mu\phi+R\phi\big).
\label{eq:DeltaIDef}
\end{equation}
While this method of integrating the trace ``anomaly'' applies in all topological sectors, it is \emph{unable to find
the functional} $U_k[\hg]$. Usually in conformal field theory or string theory this is not much of a disadvantage,
but in quantum gravity where Background Independence is a pivotal issue it is desirable to have a more complete
understanding of $\Ggd$. For this reason we discuss in the next section the possibility of taking the limit
$\ve\rightarrow 0$ directly at the level of the action.
\section{How the Induced Gravity Action emerges from the Einstein--Hilbert action}
\label{sec:IndGravityFromEH}
In this section we reveal a mechanism which allows us to regard Polyakov's induced gravity action in $2$ dimensions as
the $\ve\rightarrow 0$ limit of the Einstein--Hilbert action in $2+\ve$ dimensions. Here and in the following
we always consider the case $\ve>0$, i.e.\ the limit $\ve\searrow 0$. This will confirm the point of view that the
induced gravity action is fundamental in describing $2$-dimensional gravity, while it is less essential for
$d>2$ where gravity is governed mainly by an (effective average) action of Einstein--Hilbert type. The
dimensional limit exhibits a discontinuity at $d=2$, producing a non-local action out of a local one.
\medskip
\noindent
\textbf{(1)}
The crucial ingredient for a nontrivial limit $\ve\rightarrow 0$ is a prefactor of the Einstein--Hilbert action
proportional to $1/\ve$. This occurs whenever the Newton constant is proportional to $\ve$. As mentioned in the
introduction, such a behavior was found in the Asymptotic Safety related RG studies. The renormalization group flow has
a non-Gaussian fixed point with a Newton constant of order $\ve$; a result which is independent of the underlying
regularization scheme and which is found in both perturbative and nonperturbative investigations.
Employing a reconstruction formula \cite{MR09,MS15,NR16} we shall see that this property holds not only for the
effective, but also for the \emph{bare} action: Using an appropriate regularization prescription the bare Newton
constant is of first order in $\ve$, too.
This is our motivation for considering a generic Einstein--Hilbert action with a Newton constant proportional to $\ve$.
For the discussion in this section it is not necessary to specify the physical role of the action under
consideration -- the arguments apply to both bare and effective (average) actions. In both cases our aim is eventually
to make sense of, and to calculate
\begin{equation}
\frac{1}{\ve}\int\td^{2+\ve}x\sg\,R
\label{eq:LimitInt}
\end{equation}
in the limit $\ve\rightarrow 0$.
\medskip
\noindent
\textbf{(2)}
It turns out helpful to consider the transformation behavior of the Einstein--Hilbert action under Weyl
rescalings. In this way an expansion in powers of $\ve$ is more straightforward. Loosely speaking, the reason why Weyl
variations are related to the 2D limit resides in the fact that the conformal factor is the only dynamical part of the
metric that ``survives'' when the limit $d\rightarrow 2$ is taken, i.e.\ the conformal sector captures the essential
information in $d=2+\ve$. This idea is detailed in subsection \ref{sec:ConfGauge}.
Weyl transformations are defined by the pointwise rescaling
\begin{equation}
g_\mn(x) = \e^{2\sigma(x)} \hg_\mn(x) \,,
\label{eq:WeylTrans}
\end{equation}
with $\sigma$ a scalar function on the spacetime manifold. In appendix \ref{app:Weyl} we list the transformation
behavior of all quantities relevant in this section.
From \eqref{eq:WeylTrans} it follows that $g_\mn$ is invariant under the split-symmetry transformations
\begin{equation}
\hg_\mn\rightarrow \e^{2\chi}\hg_\mn \, ,\qquad \sigma\rightarrow \sigma - \chi \, .
\label{eq:SplitSymmetry}
\end{equation}
Thus, any functional of the full metric $g_\mn$ rewritten in terms of $\hg_\mn$ and $\sigma$ is invariant under
\eqref{eq:SplitSymmetry}. On the other hand, a functional of $\hg_\mn$ and $\sigma$ which is not split symmetry
invariant cannot be expressed as a functional involving only $g_\mn$, but it contains an ``extra $\hg_\mn$-dependence''
\cite{MR10}.
Before actually calculating the 2D limit of \eqref{eq:LimitInt} in section \ref{sec:EpsilonLimit} and
\ref{sec:LimitFullEH} in a gauge invariant manner, we illustrate the situation in section \ref{sec:ConfGauge} by
employing the conformal gauge, and we give some general arguments in section \ref{sec:genRem} why and in what sense
the limit is well defined.
\subsection{Lessons from the conformal gauge}
\label{sec:ConfGauge}
In exactly $2$ spacetime dimensions any metric $g$ can be parametrized by a diffeomorphism $f$ and a Weyl scaling
$\sigma$:
\begin{equation}
f^* g = \e^{2\sigma}\,\hg_{\{\tau\}} \,,
\label{eq:MetricDiffWeyl}
\end{equation}
where $f^*g$ denotes the pullback of $g$ by $f$, and $\hg_{\{\tau\}}$ is a fixed reference metric that depends only on
the Teichm\"uller parameters $\{\tau\}$ or ``moduli'' \cite{IT92}. Stated differently, a combined Diff$\times$Weyl
transformation can bring any metric to a reference form. Thus, the moduli space is the remaining space of inequivalent
metrics, $\mathcal{M}_h= \mathcal{G}_h/ (\text{Diff}\times\text{Weyl})_h$, where $\mathcal{G}_h$ is the space of all
metrics on a genus-$h$ manifold.\footnote{For the topology of a sphere $\mathcal{M}_h=\mathcal{M}_0$ is trivial, while
for a torus there is one complex parameter, $\tau$, assuming values in the fundamental region, $F_0$. Apart from such
simple examples it is notoriously involved to find moduli spaces \cite{IT92}.} Its precise form is irrelevant for the
present discussion. Accordingly, if not needed we do not write down the dependence on $\{\tau\}$ explicitly in the
following. Here we consider $\hg$ a reference metric for a fixed topological sector.
In order to cope with the redundancies stemming from diffeomorphism invariance we can fix a gauge by picking one
representative among the possible choices for $f$ in eq.\ \eqref{eq:MetricDiffWeyl}, the most natural choice being
the conformal gauge:
\begin{equation}
g_\mn = \e^{2\sigma}\,\hg_\mn \,.
\label{eq:ConfGauge}
\end{equation}
Equation \eqref{eq:ConfGauge} displays very clearly the special role of $2$ dimensions: The metric depends only on the
conformal factor and possibly on some topological moduli parameters. Since the latter are global parameters, we see
that \emph{locally} the metric is determined only by the conformal factor.
\medskip
\noindent
\textbf{(1) Conformal flatness}.
At this point a comment is in order. By choosing an appropriate coordinate system it is always possible to bring a
2D metric to the form
\begin{equation}
g_\mn = \e^{2\sigma} \delta_\mn\,,
\label{eq:NotConfGauge}
\end{equation}
in the neighborhood of an arbitrary spacetime point, where $\delta_\mn$ is the flat Euclidean metric (see ref.\
\cite{DFN92} for instance). However, this is a local property only. \emph{For a general metric on a general 2D manifold
there exists no scalar function $\sigma$ satisfying \eqref{eq:NotConfGauge} globally}.\footnote{This can be understood
by means of the following counterexample. Consider the standard sphere $S^2\subset \mathbb{R}^3$ with the induced
metric. Upon stereographic projection the sphere is parametrized by isothermal coordinates, say $(u,v)$, where the
metric assumes the form $g=\frac{4}{(1+u^2+v^2)^2} (\td u^2+\td v^2)$. Setting $\sigma\equiv\ln\left(\frac{2}{1+
u^2+v^2}\right)$ we have $g=\e^{2\sigma}\hg$ with $\hg=\delta$. If we assumed that $g=\e^{2\sigma}\hg$ holds globally
for a valid scalar function $\sigma$, we could make use of identity \eqref{eq:WeylgR} to arrive at a contradiction for
the Euler characteristic $\chi=2\mku$, namely: $8\pi=4\pi\chi\equiv\int\!\sg\,R = \int\!\shg\,(\hR-2\,
\hB\mku\sigma)=-2\int\!\shg\;\hB\mku\sigma=0$, since $\hR=0$ for the flat metric, and since the sphere has vanishing
boundary. A resolution to this contradiction is to take into account that we need (at least) two coordinate patches all
of which have a boundary contributing to $\int\!\sg R$. Decomposing $S^2$ into two half spheres, $H_+$ and $H_-$, for
instance, and using $\hB\mku\sigma=-4/(1+u^2+v^2)^2$, we obtain $\int\!\sg\,R
=-2\int_{H_+}\shg\;\hB\mku\sigma-2\int_{H_-}\shg\;\hB\mku\sigma =8\pi=4\pi\chi$, as it should be.}
Rather must the reference metric in eq.\ \eqref{eq:ConfGauge} be compatible with all topological constraints, e.g.\ the
value of the integral $\int\shg\,\hR$ which is fixed by the Euler characteristic, a topological invariant that measures
the number of handles of the manifold. As a consequence, we cannot restrict our discussion to a globally conformally
flat metric in general.
\medskip
\noindent
\textbf{(2) \bm{$\text{Diff}\times\text{Weyl}$} invariant functionals}.
This has a direct impact on diffeomorphism and Weyl invariant functionals $F:g\mapsto F[g]$. The naive argument
claiming that diffeomorphism invariance can be exploited to make $g_\mn$ conformally flat, and then Weyl invariance to
bring it to the form $\delta_\mn$ such that $F[g]=F[\delta]$ would be independent of the metric, i.e.\ constant, is
wrong actually. The global properties of the manifold thwart this argument.
When choosing appropriate local coordinates
to render $g$ flat up to Weyl rescaling, there is some information of the metric implicitly encoded in the coordinate
system, e.g.\ in the boundary of each patch, giving rise to a remaining metric dependence in $F$. A combined
Diff$\times$Weyl transformation can bring the metric to unit form, but it changes boundary conditions (like periodicity
constraints for a torus) as well (see e.g.\ \cite{P98}). Therefore, $F$ is in fact constant with respect to local
properties of the metric, while it can still depend on global parameters. According to eq.\ \eqref{eq:MetricDiffWeyl}
these are precisely the moduli parameters. Hence, \emph{the metric dependence of any 2D functional which is both
diffeomorphism and Weyl invariant is reduced to a dependence on} $\{\tau\}$, and we can write $F[g]=f\big(\{\tau\}
\big)$ where $f$ is a function (not a functional).
\medskip
\noindent
\textbf{(3) Calculating 2D limits}.
Let us come back to the aim of this section, simplifying calculations by employing the conformal gauge
\eqref{eq:ConfGauge}. Following the previous discussion we should not rely on the choice \eqref{eq:NotConfGauge}.
Nevertheless, as an example we may assume for a moment that the manifold's topology is consistent with a metric $\hg$
that corresponds to a flat space, where -- for the above reasons -- conformal flatness is not expressed in local
coordinates as in \eqref{eq:NotConfGauge} but by the coordinate free condition $\hR=0$, which is possible iff the Euler
characteristic vanishes. The general case with arbitrary topologies will be covered in section \ref{sec:EpsilonLimit}.
We now aim at finding a scalar function $\sigma$ which is compatible with eq.\ \eqref{eq:ConfGauge} with $g_\mn$ given.
Exploiting the identities \eqref{eq:WeylLaplace} and \eqref{eq:WeylR} given in the appendix with $\hR=0$ we obtain
\begin{equation}
R=-2\,\Box\mku\sigma \,.
\label{eq:CondSigma}
\end{equation}
Once we have found a solution $\sigma$ to eq.\ \eqref{eq:CondSigma}, it is clear that $\sigma'=\sigma+
\text{(\emph{zero modes of} }\Box\text{)}$ defines a solution, too. In particular, we can subtract from $\sigma$ its
projection onto the zero modes. This way, we can always obtain a solution to \eqref{eq:CondSigma} which is free of zero
modes. Thus, we can assume that $\sigma$ does not contain any zero modes before actually having computed it. In doing
so, relation \eqref{eq:CondSigma} can safely be inverted (cf.\ appendix \ref{app:Weyl} for a more detailed discussion
of zero modes):
\begin{equation}
\sigma = -\frac{1}{2}\,\Box^{-1}R
\label{eq:SigmaSolved}
\end{equation}
Note that the possibility of performing such a direct inversion is due to the simple structure of eq.\
\eqref{eq:CondSigma} which, in turn, is a consequence of $\hR=0$.
Now we leave the strictly $2$-dimensional case and try to ``lift'' the discussion to $d=2+\ve$. For this purpose we
make the assumption that we can still parametrize the metric by \eqref{eq:ConfGauge} with a reference metric $\hg$
whose associated scalar curvature vanishes, $\hR=0$. (Once again, the general case will be discussed in section
\ref{sec:EpsilonLimit}.) In this case, by employing equations \eqref{eq:EHexpanded} and \eqref{eq:WeylLaplace} we
obtain the following relation for the integral \eqref{eq:LimitInt}:
\begin{equation}
\frac{1}{\ve}\int\td^{2+\ve}x\sg\,R=\frac{1}{\ve}\int\td^2 x\shg\,\Big[\ve\,\sigma\big(-\hB\big)\sigma\Big]+\mO(\ve).
\end{equation}
This expression can be rewritten by means of the $2+\ve$-dimensional analogues of eqs.\ \eqref{eq:CondSigma} and
\eqref{eq:SigmaSolved} which read $R=-2\,\Box\mku\sigma+\mO(\ve)$ and $\sigma = -\frac{1}{2}\,\Box^{-1}R+\mO(\ve)$,
respectively, and we arrive at the result
\begin{equation}
\phantom{(\hR=0)}\quad
\fbox{$\displaystyle
\frac{1}{\ve}\int\td^{2+\ve}x\sg\,R = -\frac{1}{4}\int\td^2 x\sg\, R\,\Box^{-1} R +\mO(\ve).
$}\quad (\hR=0)
\end{equation}
Clearly, the assumption $\hR=0$ is quite restrictive. But already in this simple setting we make a crucial
observation: the emergence of a non-local action from a purely local one in the limit $d\rightarrow 2$. More
precisely, \emph{in the 2D limit the Einstein--Hilbert type action} $\frac{1}{\ve}\int\td^{2+\ve}x\sg\,R$ \emph{becomes
proportional to the induced gravity action}. As we will see below, a similar result is obtained for general topologies
without any assumption on $\hR$.
\subsection{General properties of the limit}
\label{sec:genRem}
\textbf{(1) Existence of the limit}. In the following we argue that $\lim_{\ve\rightarrow 0}\left(\frac{1}{\ve}
\int\td^{2+\ve}x\,\sg\,R\right)$ is indeed a meaningful quantity without restricting ourselves to a particular
topology or gauge. For convenience let us set
\begin{equation}
\mS_\ve[g] \equiv \int\td^{2+\ve}x\,\sg\, R .
\label{eq:SEpsilon}
\end{equation}
We would like to establish that $\mS_\ve[g]$ has a Taylor series in $\ve$ whose first nonzero term which is sensitive
to the \emph{local} properties of $g_\mn$ is of order $\ve$.
For the proof we make use of the relation $R_\mn=\frac{1}{2}g_\mn R$, valid in $d=2$ for any metric, so that the
Einstein tensor vanishes identically in $d=2$,
\begin{equation}
G_\mn\big|_{d=2}=0 \, .
\end{equation}
Going slightly away from $2$ dimensions, $d=2+\ve$, we assume continuity and thus
conclude that $G_\mn\big|_{d=2+\ve}=\mO(\ve)$. Furthermore, the order $\ve^1$ is really the first
nonvanishing term of the Taylor series with respect to $\ve$ in general, i.e.\ $G_\mn\big|_{d=2+\ve}$ is not of
order $\mO(\ve^2)$ or higher. This can be seen by taking the trace of $G_\mn$,
\begin{equation}
g^\mn G_\mn=g^\mn\left(R_\mn-\frac{1}{2}g_\mn R\right)=R-\frac{d}{2}R = -\frac{1}{2}R\,\ve.
\end{equation}
Therefore, we have $G_\mn \propto \ve$ (of course we assume $R\neq 0$ since $\mS_\ve$ would vanish identically
otherwise), and, similarly $G^\mn\propto\ve$. But even the non-trace (tensor) parts
of $G_\mn$ can be expected to be of order $\ve$ in general, as the following argument suggests. Let us consider a Weyl
transformation of the metric, $g_\mn=\e^{2\sigma}\hg_\mn$. The corresponding transformation of the Einstein tensor is given
by equation \eqref{eq:WeylEinstein} in the appendix. Now, let us assume that $\hg_\mn$ belongs to an Einstein manifold,
i.e.\ to a maximally symmetric spacetime.\footnote{In $d>2$ it is always possible to find $\sigma$ for a given metric $g_\mn$
such that $\hg_\mn=\e^{-2\sigma}g_\mn$ leads to a space with constant curvature provided that the manifold is compact. This
is known as the Yamabe problem \cite{Y60} (while the case $d=2$ is covered by Poincar\'{e}'s uniformization theorem).
However, this statement does not imply that the manifold is Einstein. In fact, there are known examples of metrics which are
not conformal to any Einstein metric \cite{NP01}. On the other hand, in $d=2$ any Riemannian manifold is of Einstein type.} In
this case the Ricci tensor is proportional to the scalar curvature, $\hR_\mn = \frac{1}{d}\hg_\mn\hR$. Then the Einstein
tensor reads
\begin{equation}
G_\mn = (d-2)\left[-\frac{1}{2d}\hg_\mn\hR -\hD_\mu \hD_\nu \sigma + \hg_\mn\hB\sigma + \hD_\mu\sigma \hD_\nu\sigma
+ \frac{d-3}{2}\hg_\mn \hD_\alpha\sigma \hD^\alpha\sigma \right],
\end{equation}
so we find $G_\mn\propto\ve$ again.
This $\ve$-proportionality is exploited now to make a statement about the Taylor series of $\mS_\ve$. For that purpose we
consider the variation of $\mS_\ve$ with respect to $g_\mn$ (assuming vanishing surface terms):
\begin{equation}
\begin{split}
\frac{\delta \mS_\ve[g]}{\delta g_\mn(x)}= \int\td^{2+\ve}y\,\sg\left[\frac{1}{2}g^\mn R -R^\mn\right]\delta(x-y)
= -\sg \, G^\mn = \mO(\ve).
\end{split}
\label{eqn:SVariation}
\end{equation}
As a result we obtain $\mS_\ve[g]=C+\mO(\ve)$, where the constant $C$ is independent of $g_\mn$. Clearly, $C$
is obtained by computing $\mS_\ve$ in $d=2$, which is known to lead to the Euler characteristic $\chi\,$:
\begin{equation}
C=\mS_\ve\big|_{\ve=0}=4\pi\chi.
\end{equation}
That is, we have $\mS_\ve=4\pi\chi+\mO(\ve)$. (This result differs from ref.\ \cite{CK80}, but it is in agreement with
refs.\ \cite{MR93,J06,GJ10}). As a consequence, the integral \eqref{eq:LimitInt} amounts to
\begin{equation}[b]
\frac{1}{\ve}\int\td^{2+\ve}x\sg\, R = \frac{4\pi\chi}{\ve} + \text{finite}
= \text{top.} + \text{finite},
\label{eq:expansion}
\end{equation}
where 'top.' is a field independent (up to topological information) and thus irrelevant contribution to the action.
The terms in \eqref{eq:expansion} that contain the interesting information about the dynamics of the field are of order
$\mO(\ve^0)$, so the ``relevant'' part of $\frac{1}{\ve}\int\td^{2+\ve}x\,\sg\, R$ has indeed a
meaningful limit $\ve\rightarrow 0$.
\medskip
\noindent
\textbf{(2) The role of the volume form}. Next we argue that the important part of the $\ve$-dependence of
$\mS_\ve$ originates from the scalar density $\sg\,R\,$ in the integrand of \eqref{eq:SEpsilon} alone, i.e.\ loosely
speaking it is sufficient to employ the a priori undefined fractional integration element $\td^{2+\ve}x$ at $\ve=0$.
Stated differently, all consistent definitions of ``$\td^{2+\ve}x$'' away from $\ve=0$ that one might come up with are
equivalent. The reason for that is the following.
Any integration over a scalar function on a manifold involves a volume form, i.e.\ a nowhere vanishing $d$-form (or a
density in the non-orientable case), in order to define a measure. This volume form is given by $\dd x\sg$, where $\sg$
is the square root of the corresponding Gramian determinant. If an integral is to be evaluated, the unit vectors of the
underlying coordinate system are inserted into the volume form. Since, for any $d$, these unit vectors produce a factor
of $1$ when inserted into $\dd x$, we see that it is the remaining part of the volume element that contains its
complete $d$-dependence, namely $\sg$. In particular, $\sg$ carries the canonical dimension of the volume
element.\footnote{Our conventions for the canonical mass dimensions are such that all coordinates are dimensionless,
$[x^\mu]=0$, while the metric components have $[g_\mn]=-2$, giving $\td s^2=g_\mn \td x^\mu\td x^\nu$ the canonical
dimension of an area, $[\td s^2]=-2$, whatever is the value of $d$. Hence $[\td x^\mu]=0$ and $[\sg]=-d$.
As a consequence,
the symbolic integration over the remaining ``fraction of a dimension'', $\td^\ve x$, is irrelevant even for the
dimension of $\mS_\ve[g]$.}
To summarize, for the evaluation of $\lim_{\ve\rightarrow 0} \frac{1}{\ve}\mS_\ve$ it is sufficient to consider the
$\ve$-depen\-dence of $\sg R$, while the integration can be seen as an integration over $\td^2x$. This
prescription can be considered our \emph{definition} for taking the $\ve$-limit in a well behaved way. Clearly,
the details of the domain of integration contribute some $\ve$-dependence, too. However, as we have seen in
\textbf{(1)} in equation \eqref{eqn:SVariation}, the first relevant non-constant, i.e.\ metric dependent, part of the
action comes from $\sg R$ alone, and any further $\ve$-dependent contributions would be of order $\ve^2$. This
makes clear that our argument is valid in the special case of an integral over $\sg R$, but not for arbitrary
integrands.
\medskip
\noindent
\textbf{(3) Comment and comparison with related work}. As an aside we note that in ref.\ \cite{MR93} it is argued that
the irrelevant divergent term in \eqref{eq:expansion} can be made vanish by subtracting the term $\frac{1}{\ve}
\int\dd x\sqrt{\tilde{g}}\,\tilde{R}$ from $\frac{1}{\ve}\int\dd x\sg \,R$ where the metric $\tilde{g}_\mn$ is assumed
to be $g_\mn$-dependent but chosen in such a way that the resulting field equations for $g_\mn$ do not change when $d$
approaches $2$. That means, the $g_\mn$-variation of the subtracted term (and, in turn its variation w.r.t.\
$\tilde{g}$) must vanish for $d\rightarrow 2$, leading to the requirement $\lim_{\ve\rightarrow 0} \big(\frac{1}{\ve}
\tilde{G}_\mn\big) = 0$ for the corresponding Einstein tensor.
This subtraction term would cancel the $\ve$-pole in \eqref{eq:expansion}. It is assumed in \cite{MR93} that such a
term exists for some metric $\tilde{g}_\mn$ which is conformally related to $g_\mn$. However, it remains unclear if
this is possible at all. According to the above argument in \textbf{(1)} we would rather expect $\frac{1}{\ve}
\tilde{G}_\mn$ to remain finite in the limit $\ve\rightarrow 0$.
Unlike ref.\ \cite{MR93}, we do not need to subtract further $g_\mn$-dependent terms from the action here, and our
discussion is valid for all metrics.
\subsection{Establishing the 2D limit}
\label{sec:EpsilonLimit}
Next we determine the first relevant order of the Taylor series, providing the basis for our main statements. Let us
define the $\ve$-dependent action functional
\begin{equation}
\Ye[g] \equiv \frac{1}{\ve}\int\dex\sg\, R\, - \frac{4\pi\chi}{\ve}\,.
\label{eq:Ye}
\end{equation}
Here $\chi$ denotes again the metric independent Euler characteristic defined in strictly $2$ dimensions. Corresponding
to the arguments of section \ref{sec:genRem}, $\Ye$ is well defined in the limit $\ve\rightarrow 0$ since it is
of order $\ve^0$. Therefore, $Y[g]$ defined by
\begin{equation}
Y[g] \equiv \lim_{\ve\rightarrow 0} \Ye[g]
\label{eq:YDef}
\end{equation}
is a finite functional.
To expand the integral in \eqref{eq:Ye} in powers of $\ve$ we make use of the general transformation law of
$\int\dd x\sg R$ under Weyl rescalings $g_\mn=\e^{2\sigma}\hg_\mn$, given by equation \eqref{eq:EHexpanded} in the appendix.
This yields
\begin{equation}
\begin{split}
\Ye[g] &= \frac{1}{\ve}\int\dex\shg\,\e^{\ve\sigma}\left[\hR+(1+\ve)\ve
\big(\hD_\mu\sigma\big) \big(\hD^\mu\sigma\big)\right] - \frac{4\pi\chi}{\ve}\\
&= \frac{1}{\ve}\int\dex\shg\, \hR - \frac{4\pi\chi}{\ve}
+ \int\td^2 x\shg\big(\hR\sigma+\hD_\mu\sigma\hD^\mu\sigma \big) + \mO(\ve).
\end{split}
\label{eq:YeExp1}
\end{equation}
We observe that the first two terms of the second line of \eqref{eq:YeExp1} can be combined into $\Ye[\hg]$.
Furthermore, the terms involving the parameter of the Weyl transformation, $\sigma$, are seen to agree with the
definition in \eqref{eq:DeltaIDef} and can be written as $\int\td^2x\shg\big[\hD_\mu\sigma\hD^\mu\sigma+\hR\sigma\big]
\equiv 2\, \Delta I[\sigma;\hg]$.
This, in turn, can be expressed by means of the (normalized) induced gravity functional \cite{P81}, defined
by\footnote{If the scalar Laplacian $\Box$ has zero modes, then $\Box^{-1}$ is defined as the inverse of $\Box$ on the
orthogonal complement to its kernel, that is, before $\Box^{-1}$ acts on a function it implicitly projects onto nonzero
modes. For the arguments presented in this section we may assume that $\Box$ does not have any zero modes, although a
careful analysis shows that the inclusion of zero modes does not change our main results (see detailed discussion in
appendix \ref{app:Weyl}, in particular section \ref{app:Zero}).}
\begin{equation}
I[g] \equiv \int \td^2 x \sg\, R\, \Box^{-1}R \,.
\label{eq:IndGrav}
\end{equation}
As shown in appendix \ref{app:Weyl}, the change of $I$ under a finite Weyl transformation of the metric in its argument
equals precisely $-8\,\Delta I$ which therefore has the interpretation of a Wess--Zumino term, a $1$-cocycle related to
the Abelian group of Weyl transformations \cite{MM01}:\footnote{As a consequence of identity \eqref{eq:decomp1}, the
Liouville action \eqref{eq:LiouvilleAction} can be rewritten as $\GL[\phi;\hg]=\frac{a_1}{4}I[\e^{2\phi}\hg]+
\frac{1}{2}a_1 a_2\int\td^2x\sqrt{\det(\e^{2\phi}\hg)} - \frac{a_1}{4} I[\hg]$. Note that the first two terms on the
RHS of this equation depend on $\phi$ and $\hg_\mn$ only in the combination $\e^{2\phi}\hg_\mn = g_\mn$.}
\begin{equation}
I[\e^{2\sigma}\hg] - I[\hg] = -8\,\Delta I[\sigma;\hg] \,.
\label{eq:decomp1}
\end{equation}
Inserting \eqref{eq:decomp1} into \eqref{eq:YeExp1} leads to
\begin{equation}
\begin{split}
\Ye[g] = \Ye[\hg]+2\,\Delta I[\sigma;\hg] +\mO(\ve)
= \Ye[\hg] +\frac{1}{4}I[\hg]-\frac{1}{4}I[g] +\mO(\ve).
\end{split}
\end{equation}
Rearranging terms and taking the limit $\ve\rightarrow 0$ results in the important identity
\begin{equation}
Y[g]+\frac{1}{4}I[g]=Y[\hg]+\frac{1}{4}I[\hg].
\label{eq:YplusI}
\end{equation}
Note that the LHS of eq.\ \eqref{eq:YplusI} depends on the full metric $g=\e^{2\sigma}\hg$ while the RHS depends on
$\hg$ only.
For the further analysis it is convenient to introduce the functional
\begin{equation}
F[g] \equiv Y[g] + \frac{1}{4} I[g].
\label{eq:FDef}
\end{equation}
By construction $F$ has the following properties:
\renewcommand{\theenumi}{(\roman{enumi})
\setlist{nolistsep}
\begin{enumerate}
\item It is diffeomorphism invariant since it has been constructed from diffeomorphism invariant objects only.
\item It is a functional in $d=2$ precisely since the $\ve$-limit has already been taken.
\item It is insensitive to the conformal factor of its argument since from eq.\ \eqref{eq:YplusI} follows
Weyl invariance:
\begin{equation}
F[\e^{2\sigma}\hg] = F[\hg].
\end{equation}
\end{enumerate
\noindent
Thanks to our preparations in section \ref{sec:ConfGauge} we can conclude immediately that $F$ is constant apart from
a remaining dependence on some moduli $\{\tau\}$ possibly. Here it is crucial that the moduli are \emph{global}
parameters of purely \emph{topological} origin. They are insensitive to the local properties of the metric, in
particular they do not depend on a spacetime point. These arguments show that the \emph{functional} $F[g]$ becomes a
\emph{function} of the moduli, say $C\big(\{\tau\}\big)$. The precise dependence of $F$ on these moduli is irrelevant
for the present discussion since they encode only topological information. We thus have
\begin{equation}
F[g] = C\big(\{\tau\}\big) ,
\end{equation}
i.e.\ $F$ is a metric independent constant functional, up to topological terms.
\bigskip
For the functional $Y[g]$ defined in eq.\ \eqref{eq:YDef} we obtain, using eq.\ \eqref{eq:FDef},
\begin{equation}
Y[g]=-\frac{1}{4}I[g]+C\big(\{\tau\}\big) \, ,
\end{equation}
which leads to our final result:
\begin{equation}[b]
\frac{1}{\ve}\int\td^{2+\ve}x\sg\, R= -\frac{1}{4}\int\td^2x\sg\,R\,\Box^{-1}R +\frac{4\pi\chi}{\ve}
+C\big(\{\tau\}\big)+\mO(\ve).
\label{eq:LimitResult}
\end{equation}
The terms $4\pi\chi/\ve$ and $C\big(\{\tau\}\big)$ are topology dependent but independent of the local properties of
the metric, and thus they may be considered irrelevant for most purposes.
So we have established that the limit $d\rightarrow 2$ of the Einstein--Hilbert action equals precisely the induced gravity
action up to topological terms. Clearly the most remarkable aspect of this limiting procedure is that it leads from a local
to a non-local action.
A similar mechanism has been discussed earlier in the framework of dimensional regularization \cite{MM01}. The result
\eqref{eq:LimitResult} is in agreement with the one of reference \cite{J06} where it has been obtained by means of a different
reasoning based on the introduction of a Weyl gauge potential.
\subsection[The full Einstein--Hilbert action for \texorpdfstring{$\ve\rightarrow 0$}{epsilon to zero}]
{The full Einstein--Hilbert action for \texorpdfstring{\bm{$\ve\rightarrow 0$}}{epsilon to zero}}
\label{sec:LimitFullEH}
Including also the cosmological constant term, the Einstein--Hilbert truncation of the (gravitational part of
the) effective average action reads
\begin{equation}
\Gamma_k^\text{grav}[g] = \frac{1}{16\pi G_k} \int \dd x \sg \,\big( -R + 2\Lambda_k \big),
\label{eq:EHTruncation}
\end{equation}
with the dimensionful Newton and cosmological constant, $G_k$ and $\Lambda_k$, respectively.
\medskip
\noindent
\textbf{(1)}
As we have mentioned already, RG studies in $d=2+\ve$ show that the $\beta$-functions of the dimensionless versions
of these couplings, $g_k\equiv k^{d-2}G_k$ and $\lambda_k\equiv k^{-2}\Lambda_k$, possess a nontrivial fixed point
which is proportional to $\ve$ \cite{R98} (see also \cite{W80,T77,B77,GKT78,CD78,KN90,JJ91,KKM93,NR13,N15,CD15,F15}),
\begin{equation}
g_*\propto\ve\quad\text{and}\quad \lambda_*\propto\ve.
\end{equation}
Thus, at least in the vicinity of this non-Gaussian fixed point the dimensionful couplings are of the form
\begin{equation}
G_k \equiv \ve\,\breve{G}_k \, ,\quad \Lambda_k \equiv \ve\,\breve{\Lambda}_k \,,
\end{equation}
where $\breve{G}_k$ and $\breve{\Lambda}_k$ are of order $\mO(\ve^0)$. Making use of eq.\ \eqref{eq:LimitResult} in the
limit $\ve\rightarrow 0$ we arrive at the $2$-dimensional effective average action
\begin{equation}[b]
\Gamma_k^\text{grav,2D}[g] = \frac{1}{64\pi \breve{G}_k} \int\td^2x\sg\,R\,\Box^{-1}R
+ \frac{\breve{\Lambda}_k}{8\pi\breve{G}_k}\int\td^2x\sg
+ \text{top}.
\label{eq:Gamma}
\end{equation}
Here 'top' refers again to topology dependent terms which are insensitive to the local properties of the metric. The
result \eqref{eq:Gamma} is quite general; it holds for any RG trajectory provided that the couplings $G_k$ and
$\Lambda_k$ in $d=2+\ve$ are of first order in $\ve$.
As an aside we note that the topological terms in \eqref{eq:Gamma} include a contribution proportional to
$\int\!\sg\,R=4\pi\mku\chi$. Thus, eq.\ \eqref{eq:Gamma} contains the induced gravity action, a cosmological constant
term, and the $\chi$-term. These are precisely the terms that were included in the truncation ansatz in ref.\
\cite{CD15}. By contrast, in our approach they are not put in by hand through an ansatz, but they rather emerge as a
result from the Einstein--Hilbert action in the 2D limit.
\medskip
\noindent
\textbf{(2)}
If we want to consider $\Gamma_k$ exactly at the NGFP, we can insert the known fixed point values, where the one of
Newton's constant is given by $g_*=\ve/b$ according to eq.\ \eqref{eq:NGFPWithB}. The coefficient $b$ is independent of
the cutoff scheme underlying the computation. It depends, however, on the parametrization of the metric. In the linear
parametrization, $g_\mn=\gb_\mn+h_\mn$, it is given by \cite{R98,W80,T77,B77,KN90,JJ91,NR13,N15,CD15}\footnote{When the
running of the Gibbons--Hawking surface term instead of the pure Einstein--Hilbert action is computed, the result reads
$b=\frac{2}{3}(1-\ns)$ \cite{GKT78,CD78}. See ref.\ \cite{NR13} for a discussion.}
\begin{equation}
b=\frac{2}{3}\big(19-\ns\big),
\end{equation}
while the exponential parametrization \cite{DN15}, $g_\mn=\gb_{\mu\rho}(\e^h)^\rho{}_\nu$, leads to
\cite{KKM93,N15,CD15,F15}
\begin{equation}
b=\frac{2}{3}\big(25-\ns\big).
\label{eq:bExp}
\end{equation}
While we will argue in section \ref{sec:Emergence} that the two parametrizations might possibly describe different
universality classes, to make contact to the results known from $2$-dimensional conformal field theory the exponential
parametrization turns out to be more appropriate. Therefore, we will mostly state the results based on eq.\
\eqref{eq:bExp} in the following, although the analogues for the linear parametrization can be obtained simply by
replacing $25\rightarrow 19$. Using the definition \eqref{eq:IndGrav} and combining \eqref{eq:Gamma} with
\eqref{eq:bExp} we obtain the NGFP action
\begin{equation}[b]
\Gamma_k^\text{grav,2D,NGFP}[g] = \frac{(25-\ns)}{96\pi} \,I[g]
+ \frac{(25-\ns)}{12\pi}\,k^2\breve{\lambda}_*\int\td^2x\sg
+ \text{top}\,,
\label{eq:GammaNGFP}
\end{equation}
where $\breve{\lambda}_*\equiv\lambda_*/\ve$ is cutoff dependent and thus left unspecified here. The actions
\eqref{eq:Gamma} and \eqref{eq:GammaNGFP} will be the subject of our discussion in section \ref{sec:NGFPCFT}.
\medskip
\noindent
\textbf{(3)}
Finally, let us briefly establish the connection with Liouville theory. For this purpose we separate the conformal
factor from the rest of the metric. Inserting
\begin{equation}
g_\mn = \e^{2\phi}\hg_\mn
\label{eq:DefConfFac}
\end{equation}
into eq.\ \eqref{eq:Gamma} for $\Gamma_k^\text{grav,2D}[g]$ and using \eqref{eq:ItoDeltaI} and \eqref{eq:DeltaI} from
the appendix yields
\begin{equation}[b]
\begin{aligned}
\Gamma_k^\text{grav,2D}[\phi;\hg] =\; &\frac{1}{64\pi\breve{G}_k}\int\td^2x\shg\,\hR\,\hB^{-1}\hR \\
& -\frac{1}{16\pi\breve{G}_k}\int\td^2x\shg\,\Big[\hD_\mu\phi\,\hD^\mu\phi+\hR\phi-2\breve{\Lambda}_k \e^{2\phi}\Big]
+ \text{top} \,,
\end{aligned}
\end{equation}
where $\hg_\mn$ is a fixed reference metric for the topological sector (i.e.\ a point in moduli space) under
consideration. Hence, the effective average action for the conformal factor in precisely $2$ dimensions is nothing but
the Liouville action.
Of course, this is well known to happen if one starts from the induced gravity action, an object that lives already in
2D. It is quite remarkable and nontrivial, however, that \emph{Liouville theory can be regarded the limit of the higher
dimensional Einstein--Hilbert theory}. Note that this result is consistent with the discussions in
refs.\ \cite{MR93,GJ10} (cf.\ also \cite{LS94}).
\subsection{Aside: Is there a generalization to 4D?}
\label{sec:GenTo4D}
For the sake of completeness we would like to comment on a generalization of our results to $4$ dimensions. At first
sight there seems to be a remarkable similarity. Dimensional analysis suggests that the role of the $R$-term in the
Einstein--Hilbert action near $2$ dimensions is now played by curvature-square terms in $d=4+\ve$. The gravitational
part of the action assumes the form
$\Gamma_k[g]=\Gamma_k^\text{EH}+\int\td^{4+\ve}x\sg\left\{\frac{1}{a_k}E+\frac{1}{b_k}F+\frac{1}{c_k}R^2\right\}$
where $F\equiv C_{\mn\rs}C^{\mn\rs}$ is the square of the Weyl tensor, and
$E\equiv R_{\mn\rs}R^{\mn\rs}-4R_\mn R^\mn+R^2+\frac{d-4}{18}R^2$ gives rise to the Gauss--Bonnet--Euler topological
invariant when integrated over in $d=4$. Considerations of nontrivial cocycles of the Weyl group show that the
corresponding Wess--Zumino action in $d=4$ is generated by the $E$- and the $F$-term \cite{MM01}, analogous to the
generation of $\Delta I$ in sec.\ \ref{sec:EpsilonLimit} due to the $R$-term. It may thus be expected that there would
be a mechanism to take the 4D limit, similar to the one of sec.\ \ref{sec:EpsilonLimit} but now for $E$ and $F$ instead
of $R$, if the couplings $a_k$ and $b_k$ were of first order in $\ve$.
At one-loop level the $\beta$-functions in $d=4+\ve$ have indeed a fixed point with $a_*=\mO(\ve)$, $b_*=\mO(\ve)$ and
$c_*$ finite \cite{OP14}. There are, however, two crucial differences in comparison with the $2$-dimensional case:
(i) The term $\int\td^4 x\sg \,F$ is not a topological invariant, i.e.\ there is no appropriate subtraction analogous
to definition \eqref{eq:Ye}, and the limit $\ve\rightarrow 0$ remains problematic. (ii) Even if we managed to define
some 4D-functional similar to \eqref{eq:FDef} which is both diffeomorphism and Weyl invariant, this would not be
sufficient to conclude that the functional is constant since in $d=4$ the space of metrics modulo
$\text{Diff}\times\text{Weyl}$-transformation is too large and cannot be classified in terms of topological
parameters. Roughly speaking, if we found a way to circumvent problem (i), the 4D limit of the above action computed
with our methods might lead to the same non-local action as found in \cite{MM01}, but this would not represent the
general 4D limit since the latter must certainly contain further terms that do not originate from a variation of the
conformal factor alone.
In summary, in spite of many similarities to the 2D case there seems to be no direct generalization of our
approach of computing a non-local limit action to $4$ spacetime dimensions. Nevertheless, we expect that the 4D fixed
point action contains non-local terms, too.
\section{The NGFP as a conformal field theory}
\label{sec:NGFPCFT}
We can summarize the previous sections by saying that every trajectory $k \mapsto (g_k,\lambda_k)\equiv (\bgk,\blk)\ve$,
i.e.\ every solution to the RG equations of the Einstein--Hilbert truncation in $2+\ve$ dimensions, induces the
following intrinsically $2$-dimensional running action:
\begin{equation}
\Ggd[g] = \frac{1}{96\pi}\left(\frac{3}{2}\,\frac{1}{\bgk}\right)\left[ I[g]+8\blk\,k^2\int\td^2 x\sg \,\right].
\label{eq:GammaGravSummarized}
\end{equation}
In this section we discuss the main properties of this RG trajectory, in particular its fixed point.
\medskip
\noindent
\textbf{(1) The fixed point functional}. Strictly speaking, the theory space under consideration comprises functionals
which depend on the \emph{dimensionless} metric $\tg_\mn\equiv k^2 g_\mn$. For any average action $\Gamma_k[g]$ we
define its analog in the dimensionless setting by $\mA_k[\tg]\equiv\Gamma_k[\tg\mku k^2]$. Thus, equation
\eqref{eq:GammaGravSummarized} translates into
\begin{equation}
\mA_k[\tg] = \frac{1}{96\pi}\left(\frac{3}{2}\,\frac{1}{\bgk}\right)\left[ I[\tg]+8\blk \int\td^2 x\stg \,\right].
\end{equation}
It is this functional that becomes strictly constant at the NGFP: $\mA_k\rightarrow \mA_*\mku$, with
\begin{equation}
\mA_*[\tg] = \frac{1}{96\pi}\left(\frac{3}{2}\,\frac{1}{\bgs}\right)\left[ I[\tg]+8\bls \int\td^2 x\stg \,\right].
\label{eq:AStar}
\end{equation}
For the exponential field parametrization we find the fixed point functional
\begin{equation}[b]
\mA_*[\tg] = \frac{(25-\ns)}{96\pi} \int\td^2 x\stg\,\Big(\tR\,\tB^{-1}\tR+8\bls\Big) .
\end{equation}
Here and in the following we usually present the results for the exponential parametrization. The corresponding
formulae for the linear parametrization can be obtained by replacing $(25-\ns)\rightarrow(19-\ns)$. See also section
\ref{sec:Emergence} for a discussion of the different metric parametrizations.
While the NGFP is really a point in the space of $\mA$-functionals, it is an entire \emph{line}, parametrized by $k$,
in the more familiar dimensionful language of the $\Gamma_k$'s. Let us refer to the constant map $k\mapsto(g_*,
\lambda_*)$ $\forall\, k\in[0,\infty)$ as the ``\emph{FP trajectory}''. Moving on this trajectory, the system is never
driven away from the fixed point. According to eq.\ \eqref{eq:GammaNGFP}, it is described by the following EAA:
\begin{equation}[b]
\Gamma_k^\text{grav,2D,NGFP}[g] = \frac{(25-\ns)}{96\pi} \left[ I[g]+8\mku\bls\, k^2\int\td^2 x\sg \,\right].
\label{eq:GgAtNGFP}
\end{equation}
As always in the EAA framework, the EAA at $k=0$ equals the standard effective action, $\Gamma =
\lim_{k\rightarrow 0}\Gamma_k$. So, letting $k=0$ in \eqref{eq:GgAtNGFP}, we conclude that the ordinary effective action
related to the FP trajectory has vanishing ``renormalized'' cosmological constant and reads
\begin{equation}
\Gamma^\text{grav,2D,NGFP}[g] = \frac{(25-\ns)}{96\pi} \int\td^2 x\sg\,R\,\Box^{-1} R\,.
\label{eq:GgZeroAtNGFP}
\end{equation}
\medskip
\noindent
\textbf{(2) The 2D stress tensor}. Differentiating $\Ggd$ of equation \eqref{eq:GammaGravSummarized} with respect to the
metric leads to the following energy-momentum tensor in the gravitational sector \cite{CR89}:
\begin{equation}
\begin{split}
T_\mn^\text{grav}[g] = \frac{1}{96\pi}\left(\frac{3}{2}\,\frac{1}{\bgk}\right)\bigg[ g_\mn\, D_\rho\big(\Box^{-1}R\big)
D^\rho\big(\Box^{-1}R\big) - 2\, D_\mu\big(\Box^{-1}R\big) D_\nu\big(\Box^{-1}R\big) & \\
+ 4\, D_\mu D_\nu\big(\Box^{-1}R\big) -4\, g_\mn R + 8\, \blk\,k^2 g_\mn & \bigg].
\end{split}
\label{eq:Tgrav}
\end{equation}
It is easy to see that taking the trace of this tensor yields $\Theta_k[g]=\left(\frac{3}{2}\,\frac{1}{\bgk}\right)
\frac{1}{24\pi}\big[-R+4\blk\,k^2\big]$ which, as it should be, agrees with the result from the Einstein--Hilbert action in
$d>2$, see equations \eqref{eq:Theta2} and \eqref{eq:Theta3}.\footnote{Note that in string theory or conformal field theory
one would usually redefine the stress tensor and employ $T_\mn'\equiv T_\mn-\frac{1}{2}g_\mn\Theta$ which is traceless at
the expense of not being conserved. It is the modes of $T_\mn'$ that satisfy a Virasoro algebra whose central extension
keeps track of the anomaly coefficient then.} As for the non-trace parts of $T_\mn^\text{grav}$, the comparatively
complicated non-local structures in \eqref{eq:Tgrav} can be seen as the 2D replacement of the Einstein tensor in
\eqref{eq:EHEMTensor}.
In absence of matter ($\GM=0$) the tadpole equation \eqref{eq:Tadpole} boils down to $T_\mn^\text{grav}[\gsc]=0$ with
the above stress tensor. Hence, self-consistent backgrounds have a constant (but $k$-dependent) Ricci scalar:
\begin{equation}
\Theta_k[\gsc]=0 \quad\Leftrightarrow\quad R\big(\gsc\big) = 4 \blk\,k^2 \,.
\label{eq:Rsc}
\end{equation}
In terms of the dimensionless metric, $R\big(\tilde{\bar{g}}_k^\text{sc}\big)=4\blk$, in this case.
\medskip
\noindent
\textbf{(3) Intermezzo on induced gravity}. As a preparation for the subsequent discussion consider an arbitrary conformal
field theory on flat Euclidean space, having central charge $c_\mS$, and couple this theory to a gravitational background
field $g_\mn$, comprised in an action functional $\mS[g]$. Then the resulting (symmetric, conserved) stress tensor,
\begin{equation}
T^{(\mS)}[g]^\mn \equiv \frac{2}{\sg}\frac{\delta\mS[g]}{\delta g_\mn} \,,
\label{eq:StressTensorGeneral}
\end{equation}
will acquire a nonzero trace in curved spacetimes, of the form
\begin{equation}
g_\mn\, T^{(\mS)}[g]^\mn = -c_\mS\, \frac{1}{24\pi}R+\text{const} \,,
\label{eq:TCentralCharge}
\end{equation}
where ``const'' is due to a cosmological constant possibly.
\medskip
\noindent
\textbf{(3a)} Above $\mS[g]$ can stand for either a classical or an effective action.
In the first case, $\mS[g]$ might result from a CFT of fields $\chi^I$ governed by an action $S[\chi,g]$ upon solving the
equations of motion for $\chi$, and substituting the solution $\chi_\text{sol}(g)$ back into the action:
$\mS[g] = S[\chi_\text{sol}(g),g]$. If $c_\mS \neq 0$ then the system displays a ``classical anomaly'', and Liouville
theory is the prime example \cite{D91,GM92,N04,AADZ94}.
In the ``effective'' case, $\mS[g]$ could be the induced gravity action $S_\text{ind}[g]$ which we obtain from $S[\chi,g]$
by integrating out the fields $\chi^I$ quantum mechanically:
\begin{equation}
\e^{-S_\text{ind}[g]} = \int\mD\chi^I\,\e^{-S[\chi,g]}\,.
\end{equation}
Then $S_\text{ind}$ is proportional to the central charge $c_\mS$,
\begin{equation}
S_\text{ind}[g] = +\frac{c_\mS}{96\pi} I[g] + \cdots \,,
\label{eq:Sind}
\end{equation}
and by \eqref{eq:StressTensorGeneral} the action $S_\text{ind}$ gives rise to a stress tensor whose trace is precisely
of the form \eqref{eq:TCentralCharge}. (The dots represent a cosmological constant term.)
\medskip
\noindent
\textbf{(3b)} It is important to observe that the functional $I[g]$ is \emph{negative}, i.e.\ for any metric $g$ we
have $\int\td^2 x\sg\, R\,\Box^{-1}R <0$ . (Recall that $\Box^{-1}$ acts only on nonzero modes while it ``projects
away'' the zero modes. Since $-\Box$ is non-negative, we conclude that $-\Box^{-1}$ has a strictly positive spectrum.)
Leaving the cosmological constant term in \eqref{eq:Sind} aside, this entails that for a positive central charge
$c_\mS>0$ the (non-cosmological part of the) induced gravity action is negative, $S_\text{ind}[g]<0$.
The implications are particularly obvious in the conformal parametrization $g=\e^{2\phi} \hg$, yielding
\begin{equation}
S_\text{ind}[\phi;\hg] = -\frac{c_\mS}{24\pi} \int\td^2 x\shg\Big(\hD_\mu\phi \hD^\mu\phi + \hR\phi\Big)
+ \frac{c_\mS}{96\pi} I[\hg] + \cdots \,.
\label{eq:instab}
\end{equation}
When $c_\mS$ is positive, the field $\phi$ is unstable, it has a ``wrong sign'' kinetic term. Stated differently,
\emph{integrating out unitary conformal matter induces an unstable conformal factor of the emergent spacetime metric.}
The 4D Einstein--Hilbert action is well known to suffer from the same conformal
factor instability, that is, a negative kinetic term for $\phi$ if the overall prefactor of $\int\!\sg R$ is adjusted
in such a way the concomitant kinetic term for the transverse-traceless (TT) metric fluctuations comes out positive,
as this befits propagating physical modes. Irrespective of all questions about the conventions in which the equations
are written down, the crucial signs are always such that
\begin{equation}
c_\mS>0 \quad \stackrel{\mathclap{d=2}}{\Longleftrightarrow} \quad \phi \text{ unstable} \quad
\stackrel{\mathclap{d>3}}{\Longleftrightarrow} \quad h_\mn^\text{TT} \text{ stable}.
\end{equation}
We shall come back to this point in a moment.
\medskip
\noindent
\textbf{(4) Central charge of the NGFP}. The fixed point action $\mA_*^\text{grav,2D}$ given by \eqref{eq:AStar}
describes a conformal field theory with central charge
\begin{equation}
\cgr = \begin{cases} \;25 - \ns , \qquad\text{exponential parametrization}\\
\;19-\ns , \qquad\text{linear parametrization.} \end{cases}
\label{eq:CentChParams}
\end{equation}
This follows by observing that for the two field parametrizations, directly at the NGFP, the trace of the stress tensor
is given by
\begin{equation}[b]
\Theta_k[g] = \frac{1}{24\pi}\Big(-R+4\bls k^2\Big)\times\begin{cases}\;25-\ns\qquad\text{(exp.)}\\
\;19-\ns\qquad\text{(lin.)\,.}\end{cases}
\label{eq:ThetaParams}
\end{equation}
Applying the rule \eqref{eq:TCentralCharge} to eq.\ \eqref{eq:ThetaParams}, we see indeed that, first, the fixed point
theory is a CFT, and second, its central charge is given by \eqref{eq:CentChParams}.\footnote{Reading off the central
charge according to \eqref{eq:TCentralCharge} and \eqref{eq:Sind} is consistent with refs.\ \cite{CDP14,CD15} where the
relation between the central charge and the $\beta$-function of Newton's constant is discussed in the FRG framework,
implying a relation between $\cgr$ and $g_*$.}
According to eq.\ \eqref{eq:GgAtNGFP}, the EAA related to the FP trajectory, $\Gamma_k^\text{grav,2D,NGFP}$, happens
to have exactly the structure of the induced gravity action \eqref{eq:Sind} with the corresponding central charge, for
all values of the scale parameter.
At the $k=0$ endpoint of this trajectory, the dimensionful cosmological constant $\bLk=\bls k^2$ runs to zero without
any further ado, and $\Gamma_{k\rightarrow 0}^\text{grav,2D,NGFP}$ becomes the standard effective action
\eqref{eq:GgZeroAtNGFP}. At this endpoint, by eq.\ \eqref{eq:Rsc}, self-consistent backgrounds have vanishing curvature
in the absence of matter: $R(\gb_{k=0}^\text{sc})=0$. Therefore, we have indeed inferred a central charge pertaining to
flat space by comparing \eqref{eq:ThetaParams} to \eqref{eq:TCentralCharge}.
\medskip
\noindent
\textbf{(5) Auxiliary ``matter'' CFTs}. Since the 2D gravitational fixed point action is of the induced gravity type,
we can, if we wish to, introduce a conformal matter field theory which induces it when the fluctuations of those
auxiliary matter degrees of freedom are integrated out (although such auxiliary fields are not required by our
formalism). Denoting the corresponding fields by $\chi^I$ again, and their ($k$ independent) action by
$S_\text{aux}[\chi;g]$, we have then
\begin{equation}[b]
\e^{-\Gamma_k^\text{grav,2D,NGFP}[g]} \equiv \int\mD\chi\;\e^{-S_\text{aux}[\chi;g]}\cdot\e^{-N[g]} \; .
\label{eq:GammaSAux}
\end{equation}
Here $N[g]\propto\int\td^2 x\sg$ is a inessential correction term to make sure that also the nonuniversal cosmological
constant terms agree on both sides of \eqref{eq:GammaSAux}; it depends on the precise definition of the functional
integral.
Clearly the auxiliary matter CFT can be chosen in many different ways, the only constraint is that it must have the
correct central charge, $c_\text{aux}=\cgr$, that is, $c_\text{aux}=25-\ns$ or
$c_\text{aux}=19-\ns$, respectively. Here are two examples of auxiliary CFTs:
\medskip
\noindent
\textbf{(5a) Minimally coupled scalars}.
For $c_\text{aux}>0$ the simplest choice is a multiplet of minimally coupled scalars $\chi^I(x)$,
$I=1,\cdots,c_\text{aux}$. These auxiliary fields may not be confused with the physical matter fields $A^i(x)$,
$i=1,\cdots,\ns$. The $\chi$'s and $A$'s have nothing to do with each other except that their respective numbers must
add up to $25$ (or to $19$).
\medskip
\noindent
\textbf{(5b) Feigin--Fuks theory}. The induced gravity action $I[g]$ being a non-local functional, it is natural to
introduce one, or several fields in addition to the metric that render the action local. The minimal way to achieve
this is by means of a single scalar field, $B(x)$, as in Feigin--Fuks theory \cite{CT74}, which has a nonminimal
coupling to the metric. Consider the following local action, invariant under general coordinate transformations applied
to $g_\mn$ and $B$:
\begin{equation}
I_\text{loc}[g,B] \equiv \int\td^2 x\sg\,\big(D_\mu B\,D^\mu B + 2\mku R\mku B\big)\,.
\label{eq:FFAction}
\end{equation}
The equation of motion $\delta I_\text{loc}/\delta B = -2\sg\,(\Box B-R) = 0$ is solved by $B=B(g)\equiv\Box^{-1}R$
which when substituted into $I_\text{loc}$ reproduces precisely the non-local form of the induced gravity action:
$I_\text{loc}[g,B(g)] = \int\sg\,R\,\Box^{-1}R\equiv I[g]$.
As $I_\text{loc}$ is quadratic in $B$, the same trick works
also quantum mechanically when we perform the Gaussian integration over $B$ rather than solve its field equation. Hence,
the exponentiated $\Gamma_k^\text{grav,2D,NGFP}$ has the representation
\begin{equation}
\e^{-\frac{(25-\ns)}{96\pi}\, I[g] + \cdots} = \int\mD B\; \e^{-\frac{(24-\ns)}{96\pi}\int\td^2 x\sg\,(
D_\mu B\,D^\mu B + 2\mku R\mku B + \cdots ) }
\end{equation}
Here again the dots stand for a cosmological constant which depends on the precise definition of the functional measure
$\mD B$. It is well known that thanks to the $R\mku B$-term the CFT of the $B$-field (in the limit $g_\mn\rightarrow
\delta_\mn$) has a shifted central charge \cite{FMS97,M10}; in the present case this reproduces the values
\eqref{eq:CentChParams}.
So the conclusion is that while the fixed point action is a non-local functional $\propto\int\sg\,R\,\Box^{-1}R$ in
terms of the metric alone, one may introduce additional fields such that the same physics is described by a local
(concretely, second-derivative) action. In particular, $\Gamma_k^\text{grav,2D,NGFP}$ and the local functional
\begin{equation}
\Gamma_k^\text{loc}[g,B]\equiv \frac{(24-\ns)}{96\pi}\int\td^2 x\sg\,\big(D_\mu B\,D^\mu B + 2\mku R\mku B+\cdots\big)
\end{equation}
are fully equivalent, even quantum mechanically.
\medskip
\noindent
\textbf{(6) Positivity in the gravitational sector}. Pure quantum gravity ($\ns=0$) and quantum gravity coupled to
less than $25$ (or $19$) scalars are governed by a \emph{fixed point CFT with a positive central charge}.
Clearly, this is good news concerning the pressing issue of unitarity (Hilbert space positivity) in asymptotically safe
gravity. The theories with with $\cgr>0$, continued to Lorentzian signature, do indeed admit a quantum
mechanical interpretation and have a state space which is a Hilbert space in the mathematical sense (no negative norm
states), supporting a unitary representation of the Virasoro algebra \cite{S08}.
\medskip
\noindent
\textbf{(6a) Schwinger term}. Leaving the analytic continuation to the Lorentzian world aside, it is interesting to
note that already in Euclidean space the simple-looking induced gravity action ``knows'' about the fact that
$c_\text{grav}^\text{NGFP}<0$ would create a problem for the probability interpretation. By taking two functional
derivatives of the standard effective action \eqref{eq:GgZeroAtNGFP} we can compute the $2$-point function
$\langle\, T_\mn^\text{grav}(x) T_{\rho\sigma}^\text{grav}(y)\,\rangle$ and, in particular, its contracted form
$\langle\,\Theta_0(x)\Theta_0(y)\, \rangle$. Setting thereafter $g_\mn=\delta_\mn$, which as we saw is a
self-consistent background, and choosing a suitable coordinate chart, we obtain the following Schwinger term:
\begin{equation}[b]
\langle\,\Theta_0(x)\Theta_0(y)\,\rangle = -\frac{\cgr}{12\pi}\, \p^\mu\p_\mu \delta(x-y)\,.
\label{eq:SchwingerTerm}
\end{equation}
Let us smear $\Theta_0$ with a real valued test function $f$ that vanishes at the boundary and outside of the chart
region, or, in the case where the chart is the entire Euclidean plane, falls off rapidly at infinity:\footnote{Note
that in the latter case the function $f$ has support on the entire Euclidean plane, hence we are clearly not testing
Osterwalder--Schrader \cite{OS73} reflection positivity here \cite{S93,GJ87}.}
$\Theta_0[f]\equiv\int\td^2 x\, f(x)\Theta_0(x)$. Then, applying $\int\td^2 x\,\td^2 y\,f(x)f(y)\cdots$ to both sides
of \eqref{eq:SchwingerTerm}, we find after an integration by parts:
\begin{equation}
\langle\,\Theta_0[f]^2\,\rangle =
+ \, \cgr\,\frac{1}{12\pi}\int\td^2 x\,(\p_\mu f)\delta^\mn(\p_\nu f)\,.
\label{eq:ThetaSquared}
\end{equation}
Since the integral on the RHS of \eqref{eq:ThetaSquared} is manifestly positive, we conclude that if
$\cgr<0$ the expectation value of the square $\Theta_0[f]^2$ is negative. Clearly this would be
problematic already in the context of statistical mechanics (at least with real field variables).
\medskip
\noindent
\textbf{(6b) Induced gravity approach in 4D: a comparison}. Note that one can extract the central charge from the
Schwinger term by performing an integral $\int\td^2 x\, x^2(\cdots)$ over both sides of eq.\ \eqref{eq:SchwingerTerm}.
Since Newton's constant is dimensionless in 2D, and $\breve{G}^{-1}=\bgs^{-1}=b=\frac{2}{3}\,\cgr$, this leads to the
following integral representation for the Newton constant belonging to the 2D world governed by the FP trajectory
\cite{A82}:
\begin{equation}[b]
\breve{G}^{-1} = -2\pi\int\td^2 x\; x^2\langle\,\Theta_0(x)\Theta_0(0)\,\rangle \,.
\end{equation}
It is interesting to note that this representation is of precisely the same form as the relations that had been derived
long ago within the induced gravity approach in 4D, the hope being that ultimately one should be able to compute its RHS
from a matter field theory, assumed known, the Standard Model, say, and would then predict the value of Newton's
constant in terms of matter-related constants of Nature.
For a review and a discussion of the inherent difficulties we refer to \cite{A82}. We see that, in a sense, Asymptotic
Safety was successful in making this scenario work, producing a positive Newton constant in particular, but with one
key difference: The underlying matter field theory, here the `aux' system, is no longer an arbitrary external input,
but is chosen so as to reproduce the NGFP action, an object computed from first principles.
\medskip
\noindent
\textbf{(7) Complete vs.\ gauge invariant fixed point functional}. So far we mainly focused on the gravitational part
of the NGFP functional. The complete EAA, namely $\Gamma_k = \Gg +\GM +\Gamma_k^\text{gf} +\Gamma_k^\text{gh}$ contains
matter, gauge fixing and ghost terms in addition. But since the present truncation neglects the running of the latter
three parts, they may be considered always at their respective fixed point. Also, they have an obvious interpretation
in 2D exactly. Furthermore, our truncation assumes that neither $\Gg$ nor $\GM$ as given in \eqref{eq:matter} has an
``extra'' $\gb$-dependence.
As a result, the sum of gravity and matter (`GM') contributions,
\begin{equation}
\Gamma_k^\text{GM,2D}[g,A] \equiv
\Ggd[g]+\frac{1}{2}\sum\limits_{i=1}^{\ns}\int\td^2 x\sg\;g^\mn\p_\mu A^i\p_\nu A^i \,,
\end{equation}
enjoys both Background Independence, here meaning literally independence of the background metric, and Gauge Invariance,
i.e.\ it does not change under diffeomorphisms applied to $g_\mn$ and $A^i$.
Thanks to the second property\footnote{Which might not be realized in more complicated truncations!},
we may adopt the point of view that it is actually the gauge invariant functional $\Gamma_k^\text{GM,2D}$ alone which
contains all information of interest and was thus ``handed over'' alone from the higher dimensional Einstein--Hilbert
world to the intrinsically $2$-dimensional induced gravity setting.
Therefore, if in 2D the necessity of gauge fixing arises, we can in principle pick a new gauge, different from
the one employed in $d>2$ for the computation of the $\beta$-functions.\footnote{This could not be
done if one wants to combine loop or RG calculations from $d>2$ with others done in $d=2$ exactly. However, in the
present paper all dynamical calculations are done in $d>2$, i.e.\ before the 2D limit is taken}
\medskip
\noindent
\textbf{(8) Unitarity vs.\ stability: the conformal factor ``problem''}.
Next we take advantage of the particularly convenient conformal gauge available in strictly $2$ dimensions (cf.\
section \ref{sec:ConfGauge}), and evaluate $\GgdN[g]$ as given explicitly by eq.\ \eqref{eq:GgAtNGFP} for metrics of
the special form $g_\mn = \hg_\mn\,\e^{2\phi}$. The result is a Liouville action as before in eqs.\
\eqref{eq:LiouvilleAction}, \eqref{eq:GravToLiou}, this time without any undetermined piece such as $U_k[\hg]$,
however:
\begin{align}
\GgdN[\hg\mku\e^{2\phi}] &\equiv \frac{\cgr}{96\pi}\,I[\hg] + \GL[\phi;\hg]\,,
\label{eq:GammaGravILiou}\\
\GL[\phi;\hg] &= \frac{\cgr}{12\pi}\int\td^2 x\shg\,\left\{-\frac{1}{2}\,\hD_\mu\phi\,\hD^\mu\phi-\frac{1}{2}\hR\phi
+\bls\,k^2\mku\e^{2\phi} \right\} \,.
\label{eq:LiouvilleNGFP}
\end{align}
Since $\cgr=25-\ns$ (or $\cgr=19-\ns$ with the linear parametrization), we observe that for pure gravity, and gravity
interacting with not too many matter fields, the conformal factor has a ``wrong sign'' kinetic term that might seem
to indicate an instability at first sight. If we think of the fixed point action as induced by some auxiliary CFT
with central charge $c_\text{aux}=\cgr=25-\ns>0$, we see that this is exactly the correlation mentioned in paragraph
\textbf{(3b)} above: bona fide unitary CFTs generate ``wrong sign'' kinetic terms for the conformal factor.
We emphasize that the unstable $\phi$-action is neither unexpected, nor ``wrong'' from the physics point of view,
nor in contradiction with the positive central charge of the fixed point CFT. Let us discuss these issues in turn
now.
\medskip
\noindent
\textbf{(8a) The importance of Gauss' law}. Recall the standard count of gravitational degrees of freedom in
Einstein--Hilbert gravity: In $d$ dimensions, the symmetric tensor $g_\mn$ contains $\frac{1}{2}d(d+1)$ unknown
functions which we try to determine from the $\frac{1}{2}d(d+1)$ field equations $G_\mn=\cdots\,$. Those are not
independent, but subject to $d$ Bianchi identities. Moreover, we need to impose $d$ coordinate conditions due to
diffeomorphism invariance. This leaves us with $\NEH(d)\equiv\frac{1}{2}d(d+1)-d-d=\frac{1}{2}d(d-3)$ gravitational
degrees of freedom, meaning that by solving the Cauchy problem for $g_\mn$ we can predict the time evolution of
$\NEH(d)$ functions that, (i), are related to ``physical '' (i.e.\ gauge invariant) properties of space, (ii), are
algebraically independent among themselves, and (iii), are \emph{independent of the functions describing the evolution
of matter}.
With $\NEH(4)=2$ we thus recover the gravitational waves of 4D General Relativity, having precisely $2$
polarization states. Similarly, $\NEH(3)=0$ tells us that there can be no gravitational waves in $3$
dimensions since all independent, gauge invariant properties described by the metric can be inferred already from
the matter evolution. No extra initial conditions can, or must, be imposed.
Finally $\NEH(2)=-1$ seems to suggest that ``gravity has $-1$ degree of freedom in $2$ dimensions''. Strange as it
might sound, the meaning of this result is quite clear: the quantum metric with its ghosts removes one degree of
freedom from the matter system.
If, in absence of gravity, the Cauchy problem of the matter system has a unique solution after specifying $N_\text{M}$
initial conditions, then this number gets reduced to $N_\text{M}-1$ by coupling the system to gravity.
Quantum mechanically, on a state space with an indefinite metric, the removing of degrees of freedom happens upon
imposing ``Gauss law constraints'', or ``physical state conditions'' on the states. As a result, the potentially
dangerous negative-norm states due to the wrong sign of the kinetic term of $\phi$ are not part of the actual
(physical) Hilbert space. The latter can be built using matter operators alone, and it is smaller in fact than without
gravity.\footnote{See Polchinski \cite{P89} for a related discussion.}
The situation is analogous to Quantum Electrodynamics (QED) in Coulomb gauge, for example. The overall sign of the
Maxwell action $\propto F_\mn F^\mn$ is chosen such that the spatial components of $A_\mu$ have a positive kinetic term,
and so it is unavoidable that the time component $A_0$ has a negative one, like the conformal factor in
\eqref{eq:LiouvilleNGFP}. However, it is well known \cite{BD65} that the states with negative $(\text{norm})^2$
generated by $A_0$ do not survive imposing Gauss' law $\bm{\nabla}\cdot\bm{E}=e\,\psi^\dagger\psi$ on the states. This
step indeed removes one degree of freedom since $A_0$ and $\rho_\text{em}\equiv e\,\psi^\dagger\psi$ get coupled
by an \emph{instantaneous} equation, $\bm{\nabla}^2 A_0(t,\bm{x})=-\rho_\text{em}(t,\bm{x})$.
\medskip
\noindent
\textbf{(8b) Instability and attractivity of classical gravity}. To avoid any misunderstanding we recall that in
constructing realistic 4D theories of gravity it would be quite absurd, at least in the Newtonian limit, to ``solve''
the problem of the conformal factor by manufacturing a positive kinetic term for it in some way. In taking the
classical limit of General Relativity this kinetic term descends essentially to the $\bm{\nabla}\vp_\text{N}\cdot
\bm{\nabla}\vp_\text{N}$-part of the classical Lagrangian governing Newton's potential $\vp_\text{N}$ and therefore
fixes the positive sign on the RHS of Poisson's equation, $\bm{\nabla}^2\vp_\text{N}=+4\pi\mku G\rho$. However, this
latter plus sign expresses nothing less than the universal attractivity of classical gravity, something we certainly
want to keep.
This simple example shows that the conformal factor instability is by no means an unmistakable sign for a
physical deficiency of the theory under consideration. The theory can be perfectly unitary if there are appropriate
Gauss' law-type constraints to cut out the negative norm states of the indefinite metric state space.
\medskip
\noindent
\textbf{(8c) Central charge in Liouville theory}. Finally, we must discuss a potential source of confusion concerning
the correct identification of the fixed point's central charge. Let us pretend that the Liouville action
$\GL[\phi;\hg]$ describes a matter field $\phi$ in a ``background'' metric\footnote{Recall, however, that the reference
metric $\hg_\mn$ that enters only the conformal parametrization of 2D metrics is to be distinguished carefully from the
true background metric $\gb_\mn$ which is at the heart of the entire gravitational EAA setting. In this conformal
parametrization, a generic bi-metric action $F[g,\gb]$ translates into a functional of \emph{two} conformal factors,
$F\big[\phi,\bar{\phi};\hg\big] \equiv F\big[\mku\hg\,\e^{2\phi},\mku\hg\,\e^{2\bar{\phi}}\mku\big]$.} $\hg_\mn$.
It would then be natural to ascribe to this field the stress tensor
\begin{equation}
T_k^\text{L}[\phi;\hg]^\mn \equiv \frac{2}{\shg}\frac{\delta\GL[\phi;\hg]}{\delta\hg_\mn}\,.
\label{eq:TLiouvilleDef}
\end{equation}
Without using the equation of motion (i.e.\ ``off shell'') its trace is given by
\begin{equation}
\Theta_k^\text{L}[\phi;\hg]\equiv \hg_\mn\,T_k^\text{L}[\phi;\hg]^\mn = \frac{\cgr}{12\pi}
\left(\hB\phi + 2\mku\bls\,k^2\mku\e^{2\phi} \right).
\label{eq:TLiouville}
\end{equation}
Concerning \eqref{eq:TLiouville}, several points are to be noted.
\begin{enumerate}
\item Varying $\GL$ with respect to $\phi$ yields Liouville's equation $\hB\phi+2\mku\bls\, k^2\mku\e^{2\phi}
= \frac{1}{2}\hR$. With $\phi_\text{sol}$ denoting any solution to it we obtain ``on shell'' the following
$k$-independent trace:
\begin{equation}
\Theta^\text{L}[\phi_\text{sol};\hg] = +\cgr \frac{1}{24\pi}\hR \,.
\label{eq:ThetaLiouville}
\end{equation}
If we now compare \eqref{eq:ThetaLiouville} to the general rule \eqref{eq:TCentralCharge} we conclude that the
Liouville field represents a CFT with the central charge
\begin{equation}
c^\text{L} = -\cgr\,,
\end{equation}
which is \emph{negative} for pure asymptotically safe gravity, namely $c^\text{L} = -25$, or $-19$, respectively.
\end{enumerate}
Does this result indicate that the fixed point CFT is non-unitary, after all? The answer is a clear `no', and the
reason is as follows.
\begin{enumerate}[resume]
\item The Liouville theory governed by $\GL$ of \eqref{eq:LiouvilleNGFP} is not a faithful description of the
NGFP. According to eq.\ \eqref{eq:GammaGravILiou}, the full action contains the ``pure gravity'' term
$\frac{\cgr}{96\pi}I[\hg]$ in addition. In order to correctly identify the central charge of the NGFP it is essential
to add the $\hg_\mn$-derivative of this term to the Liouville stress tensor. Hence, the trace \eqref{eq:TLiouville}
gets augmented to
\begin{align}
\frac{2\mku\hg_\mn}{\shg}\frac{\delta}{\delta\hg_\mn}\left(\frac{\cgr}{96\pi}I[\hg]\right)+\Theta_k^\text{L}[\phi;\hg]
&= -\frac{\cgr}{24\pi}R(\hg)+\Theta_k^\text{L}[\phi;\hg]
\label{eq:ThetaFull}\\
&= \frac{\cgr}{24\pi}\left[-R(\hg)+2\mku\hB\phi+4\mku\bls\, k^2\mku\e^{2\phi}\right]
\nonumber\\
&= \frac{\cgr}{24\pi}\left[-\e^{-2\phi}\big(R(\hg)-2\mku\hB\phi\big)+4\mku\bls\, k^2\right] \e^{2\phi}
\nonumber\\
&= \frac{\cgr}{24\pi}\left[-R\big(\hg\mku\e^{2\phi}\big)+4\mku\bls\, k^2\right] \e^{2\phi}
\nonumber\\
&= \e^{2\phi}\,\Theta_k\big[\mku\hg\mku\e^{2\phi}\big].
\nonumber
\end{align}
In the 2${}^\text{nd}$ line of \eqref{eq:ThetaFull} we inserted \eqref{eq:TLiouville}, in going from the 3${}^\text{rd}$
to the 4${}^\text{th}$ line we exploited the identity \eqref{eq:WeylR} from the appendix, and in the last line we used
\eqref{eq:ThetaParams}. So with this little calculation we have checked that the Liouville stress tensor makes physical
sense only when combined with the pure gravity piece.\footnote{In isolation, $\Theta^\text{L}[\phi;\hg]$ is not
invariant under the split-symmetry transformations \eqref{eq:SplitSymmetry}, i.e.\ not a function of the combination
$\hg\mku\e^{2\phi}$ only.} If this is done, the total gravitational trace from which the correct central charge is
inferred, eq.\ \eqref{eq:ThetaParams}, is indeed recovered, as it should be. It satisfies the relation\footnote{The
explicit factor $\e^{-2\phi}$ in \eqref{eq:ThetaAdded} is simply due to the different volume elements $\shg$ and
$\sg=\shg\,\e^{2\phi}$ appearing in the definitions of the stress tensors \eqref{eq:TLiouvilleDef} and
\eqref{eq:StressTensorGeneral}, respectively.}
\begin{equation}
\Theta_k[\mku g] \equiv \Theta_k\big[\mku\hg\mku\e^{2\phi}\big]
= \e^{-2\phi}\bigg(-\frac{\cgr}{24\pi}\,\hR+\Theta_k^\text{L}[\phi;\hg]\bigg),
\label{eq:ThetaAdded}
\end{equation}
which holds true even off shell.
\item If we take $\phi$ on shell, eq.\ \eqref{eq:ThetaLiouville} applies and so the two terms in the brackets
of eq.\ \eqref{eq:ThetaAdded} cancel precisely. This, too, is as it should be since from eq.\ \eqref{eq:Rsc} we know
already that $\Theta_k[g]$ vanishes identically when $g\equiv\gb$ is a self-consistent background, and this is exactly
what we insert into \eqref{eq:ThetaAdded} when $\phi$ is a solution of Liouville's equation.
\end{enumerate}
Thus, taking the above points together we now understand that nothing is wrong with $c^\text{L} = -\cgr$. In fact,
$c^\text{L}<0$ for pure gravity is again a reflection of the Liouville field's ``wrong-sign'' kinetic
term\footnote{Hence, at the technical level, the wrong-sign kinetic term requires special attention (regularization,
analytic continuation, or similar) at intermediate steps of the calculation at most.} and its perfectly correct
property of reducing the total number of degrees of freedom.
\section{Status of different field parametrizations}
\label{sec:Emergence}
To fully establish the properties of the NGFP in exactly $2$ dimensions, it remains to understand the status and
reliability of the exponential and linear field parametrizations, respectively. Why is it the former that reproduces the
results of standard conformal field theory, including the bosonic critical dimension, while the latter fails to do so
\cite{N15,CD15,F15}?
\subsection{Different universality classes?}
\label{sec:UnivClasses}
It might be appropriate to begin with a word of warning: For the time being, it is not clear whether the exponential
and the linear parametrization, respectively, describe the same physics at the exact level. We cannot fully exclude the
possibility that both of them are equally correct, but probe instead two different universality classes.
It is a notoriously difficult question in virtually all functional integral based approaches\footnote{It is well known
that standard 1D configuration space functional integrals are dominated by non-differentiable paths since the set of
differentiable ones has measure $0$. The basic laws of quantum mechanics, noncommutativity of positions and momenta,
force us to include these classically forbidden non-differentiable trajectories in the path integral \cite{GCP15}.
Similarly, a consistent gravitational path integral might require integrating over ``metrics'' which have further
nonclassical features to a degree that is to be found out.} to quantum gravity
whether, or to what extent, \emph{degenerate, wrong-signature} or even \emph{vanishing} tensor fields should be
included \cite{W88}. Since the set of nondegenerate metrics with fixed signature forms a nonempty open subset in the
space of all covariant symmetric $2$-tensor fields \cite{DN15}, there is no a priori reason to expect that it has
vanishing measure, and so this question has no obvious answer.
It is known, however, that ``sufficiently different'' choices will lead to inequivalent theories \cite{AL98}.
In this context it might well be that a NGFP in a theory which is based only on nondegenerate fixed-signature metrics
falls into another universality class than a fixed point in a theory which includes all covariant symmetric tensors
without restrictions concerning degeneracy and signature. If so, the properties of the respective NGFP, in particular
the values of the central charge, could be different then.
Interestingly enough, the two parametrizations considered here give rise to these two different options, namely genuine
metrics vs.\ unconstrained symmetric tensors:
\noindent
\textbf{(a)}
In ref.\ \cite{DN15} it was argued that the exponential parametrization is adapted to the nonlinear structure of the
field space $\mF$ of all pure metrics, i.e.\ objects in $\mF$ are nondegenerate and have a prescribed signature. The
nondegeneracy and signature requirements prevent $\mF$ from exhibiting vector space character. As a consequence,
it is not possible to simply \emph{add} an arbitrary fluctuation to a given metric in order to obtain another metric
since it may happen that one leaves the set $\mF$ owing to the addition. To reach another metric one rather has to start
from a given metric and \emph{move along geodesics}. Only then it is ensured that all metrics parametrized this way
satisfy the signature and nondegeneracy constraint. It can be shown that there is a canonical connection on $\mF$
adjusted to this constraint, and that the resulting geodesics are parametrized precisely by the exponential relation
$g_\mn=\gb_{\mu\rho} (\e^h)^\rho{}_\nu$ with a symmetric tensor field $h_\mn$ \cite{DN15}. This explains the special
status of the exponential parametrization.
\noindent
\textbf{(b)}
The linear parametrization, $g_\mn=\gb_\mn+h_\mn$, on the other hand, allows integrating over a much larger space of
field configurations including degenerate, wrong-signature and vanishing tensor fields:\footnote{This would be in the
spirit of ref.\ \cite{W88}, and one might expect to find a phase of unbroken diffeomorphism invariance, among others.}
Since $h_\mn$ can be any symmetric tensor field, it is clear that the set of pure metrics can be left by adding it to
$\gb_\mn$.
Thus, the two parametrizations fall indeed into qualitatively different categories, and there is at least the
possibility that future RG studies might show that the respective NGFPs refer to different universality classes. We
conjecture that these classes would then be represented by $\cgr=19$ for the linear parametrization and by $\cgr=25$ if
the metric is parametrized by an exponential. In any case, the issue of parametrization dependence \cite{GKL15} should
always be reconsidered when a better truncation becomes technically manageable.\footnote{A first indication pointing
towards the possibility of different classes might be contained in recent results from the $f(R)$-truncation in 4D
where an apparently parametrization dependent number of relevant directions was observed \cite{OPV15}.}
We believe that although at this stage no parametrization of the metric should be preferred over the other one, the
compatibility of EAA-based Asymptotic Safety results with other approaches to quantum gravity can in principle
discriminate between different parametrizations. The critical value `25' for the central charge indicates indeed that
our calculations based on the exponential parametrization are closer to those of conformal field theory.
In the next subsection we aim at presenting another argument suggesting that the exponential parametrization is
particularly appropriate in the 2D limit.
\subsection{The birth of exponentials in 2D}
\label{sec:GoodVsBad}
The argument in this subsection considers only such dynamical metrics $g_\mn$ that are conformally related to a fixed
reference metric $\hg_\mn$, and only their relative conformal factor is quantized. The resulting ``conformally
reduced'' setting \cite{RW09,RW09b} amounts to the exact theory in 2D, but it is an approximation in higher dimensions.
(Accordingly, ``exponential parametrization'' refers to the form of the conformal factor in the following.)
Among all possible ways of parametrizing the conformal factor there exists a distinguished parametrization in each
dimension $d$.
\medskip
\noindent
\textbf{(1) Distinguished parametrizations.} Let us consider the conformal reduction of the Einstein--Hilbert action
$\SEH[g]\equiv -\frac{1}{16\pi G}\int\dd x\sg\,(R-2\Lambda)$ in any number of dimensions $d>2$. That is, we evaluate
$\SEH$ only on metrics which are conformal to a given $\hg$ consistent with the desired topology. But how should we
write the factor relating $g$ and $\hg$ now? Assume, for instance, the reduced $\SEH$ plays the role of a bare action
under a functional integral over a certain field $\Omega$ representing the conformal factor, how then should the latter
be written in terms of $\Omega\,$? Clearly, infinitely many parametrizations are possible here, and
depending on our choice the reduced $\SEH$ will look differently.
There exists a distinguished parametrization, however, which is specific to the dimensionality $d$, having the property
that $\int\!\sg\,R$ \emph{becomes quadratic in} $\Omega$. Starting out from a power ansatz, $g_\mn=\Omega^{2\nu}\,
\hg_\mn$, the integral $\int\!\sg\,R$ will in general produce a potential term $\propto \hR$ times a particular power
of $\Omega$, and a kinetic term $\propto \big( \hD\Omega \big)^2$ times another power of $\Omega$. The exponent of the
latter turns out to be zero, yielding a kinetic term quadratic in $\Omega$, precisely if \cite{JNP05}
\begin{equation}
\nu = 2/(d-2)\,,\qquad g_\mn = \Omega^{4/(d-2)}\,\hg_\mn \,.
\label{eq:ExponentChoice}
\end{equation}
In this case the potential term is found to be quadratic as well, and one obtains \cite{JNP05,RW09}
\begin{equation}[b]
\SEH[g = \Omega^{4/(d-2)}\,\hg] = -\frac{1}{8\pi G}\int\dd x\shg\left[ \frac{1}{2\,\xi(d)}\,\hD_\mu\Omega\,
\hD^\mu \Omega + \frac{1}{2}\hR\,\Omega^2-\Lambda\,\Omega^{2d/(d-2)}\right].
\label{eq:SCREH}
\end{equation}
Here we introduced the constant
\begin{equation}
\xi(d) \equiv \frac{(d-2)}{4(d-1)}\,.
\end{equation}
Usually one employs $\Omega(x)-1\equiv\omega(x)$ rather than $\Omega$ itself as the dynamical field that is
quantized, i.e.\ integrated over if $\SEH$ appears in a functional integral. Then there will be no positivity issues
as long as $\omega(x)$ stays small. We emphasize, however, that the derivation of neither \eqref{eq:SCREH} nor the
related action for $\omega$,
\begin{equation}
\SEH[\omega;\hg] = -\frac{1}{8\pi G}\int\dd x\shg\left[ \frac{1}{2\,\xi(d)}\,\hg^\mn \p_\mu\omega\,\p_\nu \omega
+ \frac{1}{2}\hR\,(1+\omega)^2-\Lambda\,(1+\omega)^{2d/(d-2)}\right],
\label{eq:omegaAction}
\end{equation}
involves any (small field, or other) expansion. (It involves an integration by parts though, hence there could be
additional surface contributions if spacetime has a boundary.)
\medskip
\noindent
\textbf{(2) Metric operators.} The exponent appearing in the conformal factor $\Omega^{2\nu}$ is
non-integer in general, exceptions being $d=3,4$, and $6$, see Table \ref{tab:Dimensions}.
{\renewcommand{\arraystretch}{1.2}
\begin{table}[tp]
\centering
\begin{tabular}{cccc}
\hline
$d$ & $\quad 3\quad$ & $\quad 4\quad$ & $\quad 6\quad$\\
\hline
$\;$ Conformal factor $\;$ & $\Omega^4$ & $\Omega^2$ & $\Omega$\\
Volume operator & $\Omega^6$ & $\Omega^4$ & $\Omega^3$\\
\hline
\end{tabular}
\caption{Conformal factor and volume operator for the distinguished parametrization.}
\label{tab:Dimensions}
\end{table
The virtue of a quadratic action needs no mentioning, of course. As long as the cosmological constant plays no
role --- $\Lambda$ will always give rise to an interaction term --- the computation of the RG flow will be easiest and
\emph{most reliable} if we employ the distinguished parametrization.\footnote{The RG flow of
the conformally reduced Einstein--Hilbert truncation (``CREH'') with the distinguished parametrizations has been
computed in \cite{RW09}, an LPA-type extension was considered in \cite{RW09b}, see also \cite{MP09}.}
One should be aware that there is a conservation of difficulties also here. Generically the conformal factor depends on
the quantum field \emph{nonlinearly}. Hence, canonically speaking, even if the action is trivial (Gaussian), the
construction of a \emph{metric operator} amounts to defining $\Omega^{2\nu}$ or $(1+\omega)^{2\nu}$ as a composite
operator. And in fact, the experience with models such as Liouville theory \cite{OW86,DO94,KN93} shows how extremely
difficult this can be.
At present we are just interested in comparing the relative degree of reliability of two truncated RG flows, based upon
different field parametrizations. For this purpose it is sufficient to learn from the above argument that the ``most
correct'' results should be those from the distinguished parametrization \eqref{eq:ExponentChoice} since then the
theory is free (for $\Lambda=0$). But what is the distinguished parametrization in $2$ dimensions?
\medskip
\noindent
\textbf{(3) The limit \bm{$d\rightarrow 2$}}. As we lower $d$ towards two dimensions, the distinguished form of the
conformal factor, $(1+\omega)^{4/(d-2)}$, develops into a function which increases with $\omega$ faster than any
power. At the same time the constant $\xi(d)$ goes to zero, and \eqref{eq:omegaAction} becomes
\begin{equation}
\begin{split}
\SEH[\omega;\hg] = -\frac{1}{16\pi \bG}\int\td^{2+\ve}x\shg\,\bigg[\frac{4}{\ve^2}\,\hg^\mn \p_\mu\omega\,
\p_\nu\omega\,\big\{1+\mO(\ve)\big\} & \\
+ \frac{1}{\ve}\,\hR\,(1+\omega)^2 - 2\bL\,(1+\omega)^{2(2+\ve)/\ve}& \bigg].
\end{split}
\label{eq:omegaAction2}
\end{equation}
Here we wrote $G\equiv \bG\,\ve$ and $\Lambda\equiv \bL\,\ve$ again, assuming that $\bG,\bL=\mO(\ve^0)$. We see that in
order to obtain a meaningful kinetic term we must rescale $\omega$ by a factor of $\ve$ prior to taking the limit
$\ve\searrow 0$.
Introducing the new field $\phi(x) \equiv 2\omega(x)/\ve$, its kinetic term $\hg^\mn \p_\mu\phi\,
\p_\nu\phi\,\big\{1+\mO(\ve)\big\}$ will have a finite and nontrivial limit. The concomitant conformal factor
$\Omega^{2\nu}$ has the limit
\begin{equation}
\lim\limits_{\ve\rightarrow 0} \, (1+\omega)^{4/\ve}
= \lim\limits_{\ve\rightarrow 0} \Big(1+{\textstyle\frac{1}{2}}\ve\phi\Big)^{4/\ve}
= \lim\limits_{n\rightarrow\infty} \Big(1+{\textstyle\frac{2\phi}{n}}\Big)^n = \e^{2\phi} \,.
\end{equation}
This demonstrates that \emph{the exponential parametrization} $g_\mn = \e^{2\phi}\hg_\mn$ \emph{is precisely the 2D
limit of the distinguished (power-like) parametrizations in} $d>2$.
The cosmological term in \eqref{eq:omegaAction2} involves the same exponential for $d\rightarrow 2$, and the originally quadratic potential
$\hR(1+\omega)^2$ turns into a linear one for $\phi$. Taking everything together the Laurent series of $\SEH$ in $\ve$
looks as follows:
\begin{equation}
\SEH[\phi;\hg] = -\frac{1}{16\pi\bG}\bigg\{ \frac{1}{\ve}\int\!\td^{2+\ve} x\shg\,\hR + \int\!\td^2 x\shg\,
\Big(\hg^\mn\,\p_\mu\phi\,\p_\nu\phi + \hR\,\phi - 2\bL\,\e^{2\phi}\Big)\bigg\} + \mO(\ve).
\label{eq:SEHphiAction}
\end{equation}
The first term on the RHS is $\phi$-independent and involves a purely topological contribution proportional to the
Euler characteristic, which was already encountered in section \ref{sec:IndGravityFromEH}. Obviously from
\eqref{eq:SEHphiAction} we reobtain Liouville theory as the intrinsically 2D part of the Einstein--Hilbert action, but
this is perhaps not too much of a surprise.
What is important, though, is that in this derivation, contrary to the discussion in the previous sections, the
exponential field dependence of the conformal factor was not put in by hand, we rather \emph{derived} it.
Here our input
were the following two requirements: First, the scaling limit of $\SEH$ should be both non-singular and nontrivial, and
second, it should go through a sequence of actions which, apart from the cosmological term, are at most quadratic in the
dynamical field. Being quadratic implies that when $\SEH[\omega;\hg]$ is used as the (conformal reduction of the)
Einstein--Hilbert truncation, this truncation is ``perfect'' at any $\ve$.
Therefore, we believe that using the exponential parametrization already in ``higher'' dimensions $d>2$ yields
more reliable results for the $\beta$-functions and their 2D limits than using the linear parametrization in $d>2$ and
taking the 2D limit of the corresponding $\beta$-functions.
(There is still a minor source of uncertainty due to the ghost sector. In either parametrization there are
ghost-antighost-graviton interactions which are not treated exactly by the truncations we use.)
The basic difference between the two parametrizations can also be seen quite directly. If we insert $g=\e^{2\phi}\hg$
into $\SEH$, the resulting derivative term reads exactly, i.e.\ without any expansion in $\ve$ and/or $\phi$ and
rescaling of $\phi$:
\begin{equation}
-\frac{(d-1)}{16\pi\bG}\int\dd x\shg\;\e^{(d-2)\phi}\big(\hD\phi\big)^2 \,.
\end{equation}
For $d\rightarrow 2$ this term has a smooth limit (we did use $G=\bG\,\ve$ after all) and this limit is quadratic in
$\phi$.
On the other hand, inserting the linear parametrization $g=(1+\omega)\hg$ into $\SEH$ we obtain again exactly, i.e.\
without expanding in $\ve$ and/or $\omega$ and rescaling $\omega$:
\begin{equation}
-\frac{(d-1)}{64\pi\bG}\int\dd x\shg\;(1+\omega)^{(d-2)/2}\,\frac{\big(\hD\omega\big)^2}{(1+\omega)^2} \;.
\label{eq:KinTermOmega}
\end{equation}
The term \eqref{eq:KinTermOmega}, too, has a smooth limit $d\rightarrow 2$, but it is not quadratic in the dynamical
field. This renders the $\omega$-theory interacting and makes it a nontrivial challenge for the truncation.
\medskip
\noindent
\textbf{(4) The dimension \bm{$d=6$}}. As an aside we mention that according to Table \ref{tab:Dimensions} the case
$d=6$ seems to be easiest to deal with since in the preferred field parametrization the conformal factor is linear in
the quantum field, and so there is no need to construct a composite operator. The kinetic term \eqref{eq:KinTermOmega}
becomes quadratic exactly at $d=6$.
It is intriguing to speculate that this observation is related to the following rather surprising property enjoyed by
the $\beta$-functions derived from the bi-metric Einstein Hilbert truncation (see Appendix A.1 of ref.\ \cite{BR14}):
If $d=6$, and if in addition the dimensionful dynamical cosmological constant $\Lambda^\text{Dyn}$ is zero, then
\emph{the gravity contributions to the $\beta$-functions of both $\Lambda^\text{Dyn}$ and the dimensionful dynamical
Newton constant $G^\text{Dyn}$ vanish exactly}. (There are nonzero ghost contributions, though.)
\medskip
\noindent
\textbf{(5) Summary}. On the basis of the above arguments we conclude that \emph{most probably the exponential
pa\-ra\-me\-tri\-za\-tion is more reliable in 2D than the linear one}. We believe in particular that $\cgr=25$ is more
likely to be a correct value of the central charge at the pure gravity fixed point than its competitor `19'. Depending
on the reliability of the linear parametrization, the `19' could be a poor approximation to `25', or a hint at another
universality class.
\section{The reconstructed functional integral}
\label{sec:recFI}
For the following it is important to keep in mind that the ``derivation'' of the FRGE from a functional integral is
only formal as it ignores all difficulties specific to the UV limit of quantum field theories. In fact, rather than the
integral, the starting point of the EAA based route to a fundamental theory is the mathematically perfectly well
defined, UV cutoff-free flow equation \eqref{eq:FRGE}. In this setting, the problem of the UV limit is shifted from the
properties of the equation itself to those of its \emph{solutions}, converting renormalizability into a condition on
the existence of fully extended RG trajectories on theory space. The Asymptotic Safety paradigm is a way of achieving
full extendability in the UV and, barring other types of (infrared, etc.) difficulties, it leads to a well-behaved
action functional $\Gamma_k$ at each $k\in[0,\infty)$.
\subsection{The reconstruction process}
While every complete RG trajectory defines a quantum field theory (with the cutoff(s) removed), it does \emph{not} give
rise to a functional integral description of this theory a priori. Nevertheless, one may ask for a functional integral
reproducing a given complete $\Gamma_k$-trajectory. This ``reconstruction'' step has been motivated and discussed in
detail in ref.\ \cite{MR09}.
The association of a functional integral, i.e.\ a bare theory, to a $\Gamma_k$-trajectory is highly non-unique. The
first decision to be taken concerns the variables of integration: They may or may not be fields of the same sort as
those serving as arguments of $\Gamma_k$. From the practical point of view the most important situation is when the
integration variables are no (discretized) fields at all, but rather belong to a certain statistical mechanics model
whose partition function at criticality is supposed to reproduce the predictions of the EAA trajectory. Besides the
nature of the integration variables, a UV regularization scheme, a correspondingly regularized functional integration
measure, and an associated bare action $\SB$ are to be chosen. Then the information encapsulated in
$\Gamma_{k\rightarrow\infty}$ can be used in order to find out how the bare parameters contained in $\SB$ must depend
on the UV cutoff $\UV$ in order to give rise to a well-defined path integral reproducing the EAA-trajectory in the
limit $\UV\rightarrow\infty$.
In \cite{MR09} this reconstruction has been carried out explicitly for metric gravity in the Einstein--Hilbert
truncation for the following choices:
\textbf{(i)}
The integration variable is taken to be the metric fluctuation $h_\mn$, i.e.\ (accidentally) the same sort of
variable as in the argument of $\Gamma_k$.
\textbf{(ii)}
The UV regularization is implemented by means of a sharp mode cutoff.\footnote{When the cosmological constant is
an issue a higher-derivative regularization in the UV similar to the one in the IR is known to be problematic
\cite{MR09}. For a purely scalar theory without gauge and gravity fields the implications of a sharp cutoff were
discussed in ref.\ \cite{MS15} recently.}
\textbf{(iii)}
The truncated bare action $\SB$ is of Einstein--Hilbert type with bare couplings $\cg_\UV$ and $\cl_\UV$,
respectively. Under these conditions an explicit map relating $(g_k,\lambda_k)$ for $k\rightarrow\infty$ to
$(\cg_\UV,\cl_\UV)$ when $\UV\rightarrow\infty$ was derived.
This map depends on a parameter $M$ which labels a certain 1-parameter family of measures. This $M$-dependence reflects
the fact that it is not the bare action alone which is uniquely determined but rather the combination of measure and
bare action: Certain redefinitions of the measure can be absorbed by redefinitions of the bare action and vice versa,
signaling the ``unphysicalness'' of the bare action.
The $M$-dependence can be exploited to conveniently adjust the bare coupling constants. In $d=2+\ve$ dimensions there
is one particular value of $M$ that leads to an exactly vanishing bare cosmological constant,
$\cl_*\equiv \cl_{\UV\rightarrow\infty}=0$, and a bare Newton constant $\cg_\UV$ which equals precisely the effective
one at the NGFP \cite{NR16}:
\begin{equation}
\cg_*=g_*\,.
\label{eq:BareEqualsEffective}
\end{equation}
After having reconstructed the gravitational functional integral in $d=2+\ve$, we take its 2D limit employing the
method of section \ref{sec:IndGravityFromEH}. In combination with eq.\ \eqref{eq:BareEqualsEffective} we obtain
\begin{equation}[b]
\SB^\text{grav}[g] = \frac{(25-\ns)}{96\pi} \,I[g] +\cdots
\label{eq:SBareGrav}
\end{equation}
The dots indicate that there might appear additional terms originating from the zero modes, according to eq.\
\eqref{eq:LimitResultGen} in the appendix.
Further details about the reconstruction and the derivation of \eqref{eq:BareEqualsEffective} will be
presented elsewhere \cite{NR16}. For our present purposes it suffices to know that the bare action has the structure
\eqref{eq:SBareGrav}, displaying a cutoff independent term $\propto I[g]$ and possibly ``zero mode terms''.
The bare action of the matter system too can be reconstructed according to the results of ref.\ \cite{MS15}: For
cutoffs satisfying certain constraints the bare action equals precisely the EAA when the respective cutoff scales are
identified. Thus, the bare matter action is given by the RHS of eq.\ \eqref{eq:matter}.
We would like to point out that, as a consequence, the number $\ns$ enters \emph{both the matter and the gravitational
part of the bare action}. (Note that the fixed point value $g_*$ depends on $\ns$, cf.\ sec.\ \ref{sec:LimitFullEH}.)
\subsection{A functional integral for 2D asymptotically safe gravity}
\textbf{(1) The partition function}. Based on the above considerations we obtain the full reconstructed partition
function
\begin{equation}
Z = \int[\td\tau] \int\mD_{\gfull}\phi\;\, Z_\text{gh}\big[\gfull\big]\,\Zm\big[\gfull\big]\, \Yg\big[\gfull\big]\,.
\label{eq:PartFunc}
\end{equation}
The integrand of \eqref{eq:PartFunc} comprises the following factors:
the exponential of the gravitational part of the fixed point action,
\begin{equation}
\Yg[g] \equiv \exp\!\left(-\frac{(25-\ns)}{96\pi}\,I[g]+\cdots\right),
\label{eq:Ygrav}
\end{equation}
the partition function of the matter system,
\begin{equation}
\begin{split}
\Zm[g] &\equiv \int\mD A\;\exp\!\Bigg(-\frac{1}{2}\sum\limits_{i=1}^{\ns} \int \td^2 x\sg\,
g^\mn\mku\p_\mu A^i\,\p_\nu A^i\Bigg)\\
&= {\det}^{-\ns/2}\big(-\Box_g\big)
= \exp\left(-\frac{\ns}{96\pi}\,I[g]+\cdots\right),
\end{split}
\label{eq:Zmatter}
\end{equation}
the partition function of the $b$-$c$ ghost system, $Z_\text{gh}$, the split symmetry invariant measure for the
integration over the Liouville field, $\mD_{\gfull}\phi$, and finally the measure $[\td\tau]$ for the integration over
the moduli that are implicit in the reference metric pertaining to a given topological type of the spacetime manifold
(cf.\ sec.\ \ref{sec:ConfGauge}). In eqs.\ \eqref{eq:Ygrav} and \eqref{eq:Zmatter} we suppressed possible contributions
to the bare cosmological constant. Here and in the following we indicate them by the dots.
The behavior under Weyl transformations of the various factors is well known. Using in particular
eq.\ \eqref{eq:decomp1} with the (non-cosmological constant part of the) renormalized Liouville action, $\Delta I$, as
defined in \eqref{eq:DeltaIDef}, we have
{\allowdisplaybreaks
\begin{subequations}
\begin{align}
\Yg\big[\gfull\big] &= \Yg[\mku\hg]\;\exp\!\bigg(\!+\frac{(25-\ns)}{12\pi}\,\Delta I[\phi;\hg]\bigg), \\
\Zm\big[\gfull\big] &= \Zm[\mku\hg]\;\exp\!\bigg(\!+\frac{\ns}{12\pi}\,\Delta I[\phi;\hg]\bigg), \\
Z_\text{gh}\big[\gfull\big] &= Z_\text{gh}[\mku\hg]\;\exp\!\bigg(\!+\frac{(-26)}{12\pi}\,\Delta I[\phi;\hg]\bigg),
\label{eq:WeylGhosts}\\
\mD_{\gfull}\phi &= \mD_{\hg}\phi\;\exp\!\bigg(\!+\frac{1}{12\pi}\,\Delta I[\phi;\hg]\bigg).
\label{eq:WeylDPhiMeasure}
\end{align}
\label{eq:WeylPartition}
\end{subequations}
As before, possible (measure dependent) terms involving the bare cosmological constant are suppressed in eqs.\
\eqref{eq:WeylPartition}. On the RHS of \eqref{eq:WeylDPhiMeasure}, $\mD_{\hg}\phi$ is the translational invariant
measure now.
Up to this point, the discussion is almost the same as in non-critical string theory \cite{P81}. The profound
difference lies in the purely gravitational part of the bare action, $\Yg$. Contrary to what happens in any
conventional field theory, whose bare action is a \emph{postulate} rather than the result of a \emph{calculation},
asymptotically safe gravity in $2$ dimensions is based upon a gravitational action \emph{which depends explicitly on
properties of the matter system}. In the example at hand, this dependence is via the number $\ns$ of $A^i$-fields that
makes its appearance in the fixed point action and hence in the ``Boltzmann factor'' \eqref{eq:Ygrav}.
\medskip
\noindent
\textbf{(1a) Matter refuses to matter: a compensation mechanism}. Remarkably enough, the integrand of
\eqref{eq:PartFunc} depends on $\ns$ only via the product $\Zm\cdot\Yg$ in which the $\ns$-dependence cancels between
the two factors. Multiplying \eqref{eq:Ygrav} and \eqref{eq:Zmatter} we obtain a result which, for any $\ns$, equals
that of pure gravity. It is always the same no matter how many scalar fields there are:
\begin{equation}
\Zm[g]\,\Yg[g]= \exp\!\left(-\frac{25}{96\pi}\,I[g]+\cdots\right),
\end{equation}
which transforms as $\Zm\big[\gfull\big]\mku\Yg\big[\gfull\big]=\Zm[\hg]\mku\Yg[\hg]\,
\exp\left(+\frac{25}{12\pi}\Delta I[\phi;\hg]\right)$ under a Weyl rescaling. As a consequence, the reconstructed
functional integral coincides always with that of \emph{pure gravity} (as long as we do not evaluate the expectation
value of observables depending on the $A$'s and as long as cosmological constant terms do not play a role):
\begin{equation}[b]
\!Z = \int[\td\tau]\;\Zm[\hg]\,\Yg[\hg] \int\!\mD_{\gfull}\phi\;\, Z_\text{gh}\big[\gfull\big]\,
\exp\left(\!+\frac{25}{12\pi}\,\Delta I[\phi;\hg]+\cdots\mkern-1mu\right).\!\!
\end{equation}
\medskip
\noindent
\textbf{(1b) Zero total central charge}. Over and above the specific form of its matter dependence, the fixed point
action displays a second miracle: Its central charge equals precisely the critical value $25$. Up to a cosmological
constant term possibly, this leads to a complete cancellation of the entire $\phi$-dependence of the integrand once the
ghost contribution \eqref{eq:WeylGhosts} and the ``Jacobian'' factor in \eqref{eq:WeylDPhiMeasure} are taken into
account:
\begin{equation}[b]
Z = \int[\td\tau]\;Z_\text{gh}[\hg]\,\Zm[\hg]\,\Yg[\hg] \int\mD_{\hg}\phi \; \exp(0+\cdots)\,.
\label{eq:ZNonCritical}
\end{equation}
Hence, for every choice of the matter sector, the total system described by the reconstructed functional integral of
asymptotically safe 2D gravity is a conformal field theory with central charge zero. The various sectors of this system
contribute to the total central charge as follows:
\begin{equation}
c_\text{tot} = \underbrace{(25-\ns)}_{\text{NGFP, grav.\ part}}+\underbrace{\phantom{|}\ns\phantom{|}}_{\text{matter}}
+ \underbrace{\phantom{|}1\phantom{|}}_{\text{Jacobian}} + \underbrace{(-26)}_{\text{ghosts}} = 0 \,.
\label{eq:ctotZero}
\end{equation}
Actually the result \eqref{eq:ctotZero} is even more general than we indicated so far. In addition to the scalar matter
fields underlying our considerations up to this point we can also bring massless free Dirac fermions into play and
couple them (minimally) to the dynamical metric by adding a corresponding term to the matter action \eqref{eq:matter}.
The contribution of each of such fermions to the $\beta$-function of Newton's constant in $d=2+\ve$ is the same as for
a scalar field \cite{DP13}, that is, fermions and scalars enter the central charge in the same way. Hence, in all above
equations for $\beta$-functions and central charges we may identify $\ns$ with
\begin{equation}
\ns\equiv N_\text{S}+N_\text{F}\,,
\end{equation}
where $N_\text{S}$ and $N_\text{F}$ denote the number of real scalars and Dirac fermions, respectively.
In particular, we recover the same cancellation in the total central charge as in eq.\ \eqref{eq:ctotZero}: The central
charge of the matter system, $+\ns$, removes exactly a corresponding piece in the pure gravity contribution enforced by
the fixed point, $25-\ns$.
\medskip
\noindent
\textbf{(2) Observables}. By inserting appropriate functions $\bar{\ob}[\phi,A;\hg]$ into the path integral
\eqref{eq:PartFunc} we can in principle evaluate the expectation values of arbitrary observables $\ob[\mku g,A]=
\ob[\gfull,A]$. In the case when the observables do not involve the matter fields, their expectation
values read
\begin{equation}
\langle\ob\rangle=\frac{1}{Z}\int[\td\tau]\;\Zm[\hg]\,\Yg[\hg] \int\mD_{\gfull}\phi\;\, Z_\text{gh}\big[\gfull\big]\,
\bob[\phi;\hg]\,\exp\left(\frac{25}{12\pi}\,\Delta I[\phi;\hg]\right).
\label{eq:expObs}
\end{equation}
Without actually evaluating the $\phi$-integral we see that when the cosmological constant term is negligible \emph{the
expectation value of purely gravitational observables does not depend on the presence or absence of matter and its
properties}, provided the background factor $\Zm[\hg]$ in \eqref{eq:expObs} cancels against the corresponding piece in
the denominator of \eqref{eq:expObs}. At the very least, this happens if one considers expectation values at a fixed
point of moduli space.
\medskip
\noindent
\textbf{(3) Gravitational dressing}. As it is well known \cite{DDK88,Wa93}, it is not completely straightforward to
find the functional $\bob[\phi;\hg]$ which one must use under a conformally gauge-fixed path integral in order to
represent a given diffeomorphism (and, trivially, Weyl) invariant observable $\ob[g]=\ob[\gfull]$. The association
$\ob\rightarrow\bob$ should respect the following conditions \cite{Wa93}: $\bob[\phi;\hg]$ must be invariant under
diffeomorphisms, it must approach $\ob[\gfull]$ in the classical limit and $\ob[\hg]$ in the limit $\phi\rightarrow 0$,
and most importantly it must be such that the expectation value computed with its help is independent of the reference
metric chosen, $\hg_\mn$.
Let us briefly recall the David--Distler--Kawai (DDK) solution to this problem \cite{DDK88}. For this purpose we
consider 2D gravity coupled to an arbitrary matter system described by a CFT with central charge $c$ and partition
function $\ZMc[g]$. First we want to evaluate the partition function for a fixed volume (area) of spacetime, $V$:
\begin{equation}
Z_V=\int\frac{\mD g}{\text{vol(Diff)}}\;\ZMc[g]\;\delta\!\left(V-\int\td^2 x\sg\right).
\end{equation}
This integral involves the observable $\ob[g]\equiv\int\td^2 x\sg\equiv\int\td^2 x\shg\,\exp(2\phi)$. The associated
$\bob$ satisfying the above conditions turns out to require only a ``deformation'' of the prefactor of $\phi$ in the
exponential:
\begin{equation}
\bob[\phi;\hg] = \int\td^2 x\shg\,\exp(2\mku\alpha_1 \phi)\,.
\label{eq:AreaOp}
\end{equation}
The modified prefactor $\alpha_1$ depends on the central charge of the matter CFT according to
\begin{equation}
\alpha_1 = \frac{2\mku\sqrt{25-c}}{\sqrt{25-c}+\sqrt{1-c}} = \frac{1}{12}\left[25-c-\sqrt{(25-c)(1-c)}\right]\,.
\label{eq:AlphaOne}
\end{equation}
Thus, in the conformal gauge, $Z_V$ reads as follows:
\begin{equation}
Z_V = \int[\td\tau]\;Z_\text{gh}[\hg]\,\ZMc[\hg] \int\mD_{\hg}\phi\;\delta\!\left(
V-\int\td^2 x\shg\,\e^{2\mku\alpha_1\phi}\right)\;\exp\!\left(\!-\frac{(25-c)}{12\pi}\,\Delta I[\phi;\hg]\right).
\label{eq:PartFuncZV}
\end{equation}
Similarly the expectation value of an arbitrary observable $\ob[g]$ at fixed volume is given by $\langle\ob[g]\rangle
= Z_V^{-1}\langle\bob[\phi;\hg]\rangle'$. Here $\langle\cdots\rangle'$ is defined by analogy with \eqref{eq:PartFuncZV}
but with the additional factor $\bob[\phi;\hg]$ under the $\phi$-integral.
The DDK approach to the gravitational dressing of operators from the matter sector was developed as a conformal
gauge-analogue to the work of Knizhnik, Polyakov and Zamolodchikov (KPZ) \cite{KPZ88} based upon the light cone gauge.
To study gravitational dressing, let us consider an arbitrary spinless primary field $\ob_n[g] \equiv \int\td^2 x\sg\;
\mP_{n+1}(g)$, where $\mP_n(g)$ is a generic scalar involving the matter fields with conformal weight $(n,n)$, that is,
it responds to a rescaling of the metric according to $\mP_n(\e^{-2\sigma}g) = \e^{2 n \sigma}\,\mP_n(g)$. Under the
functional integral, the observables $\ob_n$ are then represented by
\begin{equation}
\bob_n[\phi;\hg] = \int\td^2 x\shg\;\exp(2\mku\alpha_{-n}\mku\phi)\,\mP_{n+1}(\hg)\,,
\label{eq:OpN}
\end{equation}
where the $c$-dependent constants in the dressing factors generalize eq.\ \eqref{eq:AlphaOne}:
\begin{equation}
\alpha_n = \frac{2\mku n\mku\sqrt{25-c}}{\sqrt{25-c}+\sqrt{25-c-24n}}
\label{eq:AlphaN}
\end{equation}
Using \eqref{eq:AlphaN} it is straightforward now to write down the modified conformal dimensions corrected by the
quantum gravity effects.
The results of the DDK approach reproduce those of KPZ (valid for spherical topology) and generalize them for
spacetimes of arbitrary topology. Within the framework of the EAA and its functional RG equations, the KPZ relations
were derived from Liouville theory in ref.\ \cite{RW97}; for a review see \cite{CD15}.
\medskip
\noindent
\textbf{(4) Quenching of the KPZ scaling}. Let us apply the general DDK--KPZ formulae to the NGFP theory of
asymptotically safe gravity. We must replace then
\begin{equation}
c\;\longrightarrow\; \cgr + N \equiv (25-N)+N = 25\,,
\end{equation}
since the relevant bare action arises now from both the integrated-out matter fluctuations and the pure-gravity NGFP
contribution, $\Yg$. Setting $c=25$ in eqs.\ \eqref{eq:AlphaOne} and \eqref{eq:AlphaN} we obtain
\begin{equation}
\alpha_1=0\qquad \text{and}\qquad \alpha_n=0 \mku ,
\label{eq:AlphaZero}
\end{equation}
respectively. This implies that \emph{the Liouville field completely decouples from the area operator \eqref{eq:AreaOp}
and any of the observables \eqref{eq:OpN}}.
As a consequence, the dynamics of the matter system is unaffected by its coupling to quantum gravity. In particular,
its critical behavior is described by the properties (critical exponents, etc.) of the matter CFT defined on a
non-dynamical, rigid background spacetime. Thus, the specific properties of the NGFP lead to a perfect ``quenching''
of the a priori expected KPZ scaling.
\medskip
\noindent
\textbf{(5) Relation to non-critical string theory}. The functional integral \eqref{eq:ZNonCritical} is identical to
the partition function of non-critical string theory in 25 Euclidean dimensions. This theory is equivalent to the usual
critical bosonic string living in a (25+1)-dimensional Minkowski space whereby the Liouville mode plays the role of
the time coordinate in the target space \cite{DG87,DNW89}. Whether we consider pure asymptotically safe gravity in two
dimensions, or couple any number of scalar and fermionic matter fields to it, the resulting partition function equals
always the one induced by the fluctuations of precisely 25 string positions $X^m(x^\mu)$.
There is, however, a certain difference between asymptotically safe gravity and non-critical string theory in the way
the special case of vanishing total central charge, i.e.\ of precisely 25 target space dimensions, is approached.
To see this note that in the present paper we related the Liouville field to the metric by eq.\ \eqref{eq:DefConfFac},
and at no point we redefined $\phi$ by absorbing any constant factors in it. In this connection the Liouville action
for a general central charge $c$ has the structure $\GL = -\frac{c}{24\pi} \int \big( \hD_\mu \phi \mku \hD^\mu \phi
+ \hR\phi \big) + \cdots$.
\noindent
\textbf{(i)} In order to combine $\GL$ with the action of the string positions, $+\frac{1}{8\pi}\int \hD_\mu X^m
\mku \hD^\mu X^m$, it is natural to introduce the redefined field
\begin{equation}
\phi' \equiv Q\mku\phi\qquad \text{with}\qquad Q\equiv \sqrt{\frac{c}{3}} \,,
\end{equation}
in terms of which $\GL = -\frac{1}{8\pi}\int \big( \hD_\mu \phi' \mku \hD^\mu \phi' + Q\hR\phi' \big) + \cdots$.
It is this new field $\phi'$ that plays the role of time in target space and combines with the $X^m$'s in the
conventionally normalized action $\frac{1}{8\pi}\int \big(-\hD_\mu \phi' \mku \hD^\mu \phi' + \hD_\mu X^m
\mku \hD^\mu X^m - Q\hR\phi' \big) + \cdots$ which enhances the original $\mathrm{O}(25)$ symmetry to the full Lorentz
group in target space, $\mathrm{O}(1,25)$ \cite{DNW89}.
In string theory conformal invariance requires the total central charge to vanish, $c_\text{tot}=0$. Hence, arguing
that the combined $(X^0\equiv\phi',X^m)$-quantum system is equivalent to the usual bosonic string theory in the
critical dimension involves taking the limit $c\equiv c_\text{tot} \rightarrow 0$ in the above formulae. Obviously this
requires some care in calculating correlation functions as the relationship $\phi'\equiv \sqrt{c/3}\,\phi$ breaks down
in this limit. Considering vertex operators for the emission of a tachyon of 26-dimensional momentum $(P_0,P_m)$, say,
this involves combining the rescaling $\phi\rightarrow\sqrt{c/3}\,\phi$ with a corresponding rescaling of $P_0$ with
the inverse factor, $P_0\rightarrow\sqrt{3/c}\,P_0$, rendering their product $P_0 X^0\equiv P_0\phi'$ independent of
$c$. The vertex operator $\exp\mkern-1mu\big\{i(-P_0 X^0+P_m X^m)\big\}$, too, displays the full $\mathrm{O}(1,25)$
invariance. (See \cite{DG87} for a detailed discussion.)
\noindent
\textbf{(ii)} In 2D asymptotically safe quantum gravity, too, the total central charge was found to vanish, albeit for
entirely different reasons than in string theory. However, here there is no obvious reason or motivation for any
rescaling before letting $c\rightarrow 0$. In all of the above equations, including \eqref{eq:AreaOp} and
\eqref{eq:OpN}, $\phi$ still denotes the Liouville field introduced originally. In quantum gravity we let
$c\rightarrow 0$ in the most straightforward way, setting in particular $c=0$ directly in \eqref{eq:AlphaOne} and
\eqref{eq:AlphaN}. This is what led us to \eqref{eq:AlphaZero}, that is, the disappearance of $\phi$ from the
exponentials $\exp(2\mku\alpha_{-n}\phi)$ multiplying the matter operators and the ``quenching'' of the KPZ-scaling.
\subsection{Comparison with Monte Carlo results}
In earlier work \cite{DR09,LR05,RS11} indications were found that suggest that Quantum
Einstein Gravity in the continuum formulation based upon the EAA might be related to the discrete approach employing
Causal Dynamical Triangulation \cite{AL98,AGJL12}. In particular, the respective predictions for the fractal dimensions
of spacetime were compared in detail and turned out similar \cite{LR05,RS11}. It is therefore natural to
ask whether the quenching of the KPZ-scaling due to the above compensation mechanism can be seen in 2D CDT simulations.
And in fact, the Monte-Carlo studies indeed seem to suggest a picture that looks quite similar at first sight:
Coupling several copies of the Ising model \cite{AAL00} or the Potts model \cite{AALP09} to $2$-dimensional
Lorentzian quantum gravity in the CDT framework, there is strong numerical evidence that the critical behavior of the
combined system, in the matter sector, is described by \emph{the same} critical exponents as on a fixed, regular
lattice. Under the influence of the quantum fluctuations in the geometry the critical exponents do not get shifted
to their KPZ values.
While this seems a striking confirmation of our Asymptotic Safety-based prediction, one should be careful in
interpreting these results. In particular, it is unclear whether the underlying physics is the same in both cases.
In CDT, the presence (absence) of quantum gravity corrections of the matter exponents is attributed to the presence
(absence) of baby universes in Euclidean (causal Lorentzian) dynamical triangulations. In our approach instead, the
quantum gravity corrections that could in principle lead to the KPZ exponents are exactly compensated by the
\emph{explicit matter dependence} of the \emph{pure gravity}-part in the bare action. This matter dependence is an
immediate consequence of the very Asymptotic Safety requirement.
As yet we considered conformal matter only which was exemplified by massless, minimally coupled scalar fields. In the
non-conformal case when those fields are given a mass for instance, the compensation between the matter contributions
to the bare NGFP action and those resulting from integrating them out will in general no longer be complete.
On the EAA side, this situation is described by a trajectory $k\mapsto\Gamma_k$ that runs away from the fixed point
as $k$ decreases, and typically the resulting ordinary effective action of the gravity+matter system, $\Gamma_{k=0}$,
will indeed be affected by the presence of matter.
This expected behavior seems to be matched by the results of very recent 2D Monte-Carlo simulations of CDT coupled to
more than one massive scalar field \cite{AGJZ14}. It was found that, above a certain value of their mass, the dynamics
of the CDT+matter system is significantly different from the massless case. In particular, a characteristic
``blob + stalk'' behavior was observed, well known from 4D pure gravity CDT simulations, but absent in 2D with
conformal matter.
\section{Conclusions}
\label{sec:Conc}
In this paper we started from the Einstein--Hilbert truncation for the effective average action of metric quantum
gravity in $d>2$ dimensions and constructed its intrinsically $2$-dimensional limit. Contrary to earlier work on the
$\ve$-expansion of $\beta$-functions this limit was taken directly at the level of the action functional. We saw that
it turns the (local, second-derivative) Einstein--Hilbert term into the non-local Polyakov action.
Using this result we were able to conclude that in 2D the non-Gaussian fixed point underlying Asymptotic Safety gives
rise to a \emph{unitary} conformal field theory whose gravitational sector possesses the central charge $+25$. We
analyzed the properties of the fixed point CFT using both a gauge invariant description and a calculation based on the
conformal gauge where it is represented by a Liouville theory.
Finding that the complete fixed point action amounts to a CFT with vanishing total central charge, we compared and
contrasted asymptotically safe quantum gravity in $2$ dimensions with non-critical string theory.
Furthermore, exploring the gravitational dressing in 2D asymptotically safe gravity coupled to conformal matter we
discovered a rather surprising compensation mechanism that leads to a complete quenching of the KPZ-scaling,
that is, of the behavior we would have expected to occur a priori. Remarkably enough, this quenching is precisely what
is observed in Monte-Carlo simulations of analogous systems in the framework of causal dynamical triangulation. We
argued that these observations can possibly be interpreted as a reflection of the compensating effect displayed by
the integrated-out matter fluctuations and the explicit dependence of the bare gravitational action on properties of
the matter sector.
We close with a number of further comments.
\medskip
\noindent
\textbf{(1)}
An important step in proving the viability of the Asymptotic Safety program consists in demonstrating that Hilbert
space positivity can be achieved together with Background Independence and the nonperturbative renormalizability. While
we consider our present result on the unitarity of the pertinent CFT as an encouraging first insight, it is clear,
however, that the 2D case is not yet a crucial test since the gravitational field has no independent degrees of
freedom, and so there is no pure-gravity subspace of physical states whose positivity would be at stake. To tackle
the higher dimensional case additional techniques will have to be developed. Nevertheless, it is interesting that at
least at the purely geometric level the remarkable link between the Einstein--Hilbert and the Polyakov action which we
exploited has an analogue in all even dimensions $d=2n$. Each nontrivial cocycle of the Weyl cohomology yields, in an
appropriate limit $d\rightarrow 2n$, a well-defined non-local action that is known to be part of the standard effective
action in $2n$ dimensions \cite{MM01}.
\medskip
\noindent
\textbf{(2)}
A number of general lessons we learned here will be relevant in higher dimensions, too. We mention in particular that
the issue of unitarity cannot be settled by superficially checking for the stability of some bare action and ruling out
``wrong sign'' kinetic terms as this is implied sometimes. We saw that the CFT which is at the heart of the NGFP is
unitary \emph{even though} in conformal gauge it entails a negative kinetic energy of the Liouville field.
As we explained in section \ref{sec:NGFPCFT}, the background field, indispensable in our approach to quantum gravity,
plays an important role in reconciling these properties.
\medskip
\noindent
\textbf{(3)}
We determined the crucial central charge $\cgr$ from the leading term in the $\beta$-function of Newton's constant, and
we saw that depending on which parametrization of the metric is chosen the pure gravity result is either $25$ or $19$
for the exponential and the linear parametrization, respectively. In section \ref{sec:Emergence} we found convincing
evidence for accepting the result of the former, $+25$, as the correct one in the present context. Nevertheless, the
issue of parametrization dependence is not fully settled yet, and one should still be open towards the possibility that
the two sets of results, obtained from the same truncation ansatz but different choices of the fluctuating field,
actually might refer to \emph{different universality classes}.
\medskip
\noindent
\textbf{(4)}
Regarding different universality classes, it is perhaps not a pure coincidence that
the ``$19$'' is also among the ``critical dimensions for non-critical strings'' which were found by Gervais
\cite{G90}:
\begin{equation}
D_\text{crit}=7,13,19.
\end{equation}
They correspond to gravitational central charges $c_\text{grav}=19,13,7$,
respectively. For these special values the Virasoro algebra admits a unitary truncation, that is, there exists a
subspace of the usual state space on which a corresponding chiral algebra closes, and which is positive (in the sense
that it contains no vectors $|\psi\rangle$ with $\langle\psi|\psi\rangle<0$). The associated string theories were
advocated as consistent extensions of standard Liouville theory, which is valid only for $c<1$ and $c>25$ when gravity
is weakly coupled, into the strongly coupled regime, $1<c<25$, in which the KPZ formulae would lead to meaningless
complex answers.
Thus, for the time being, we cannot exclude the possibility that a better understanding of the RG flow computed with
the linear parametrization (but with more general truncations than those analyzed in the present paper) will lead to
the picture that there exists a second pure gravity fixed point compatible with Hilbert space positivity, namely at
$c_\text{grav}=19$, and that this fixed point represents another, inequivalent universality class.
We know already that this picture displays the following correlation between pa\-ra\-me\-tri\-za\-tion and universality
class, which we would then indeed consider the natural one: The exponential parametrization, i.e.\ the ``conservative''
one in the sense that it covers only nonzero, nondegenerate, hence ``more classical'' metrics having a fixed signature,
leads to $c_\text{grav}=25$ which is located just at the boundary of the strong coupling interval. In the way it is
employed, the linear parametrization, instead, gives rise to an integration also over degenerate, even vanishing tensor
field configurations not corresponding to any classical metric; typically enough, it
is this parametrization that would be linked to the hypothetical, certainly quite non-classical theory with
$c_\text{grav}=19$ deep in the strong coupling domain.
Whatever the final answer will be it seems premature, also in more than $2$ dimensions, to regard the exponential
parametrization merely as a tool to do calculations in a more precise or more convenient way than this would be
possible with the linear one. It might rather be that in this manner we are actually computing \emph{something else}.
\vspace{4em}
\noindent
\textbf{Acknowledgments}:
We are grateful to J.~Ambj\o{}rn, J.~Gizbert-Studnicki, R.~Loll, C.~Pagani and R.~Percacci for helpful discussions.
\clearpage
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 4,467 |
{"url":"https:\/\/socratic.org\/questions\/how-do-you-combine-the-like-terms-to-make-a-simpler-expression-4p-2-2p-3","text":"# How do you combine the like terms to make a simpler expression: -4p+(-2)+2p+3?\n\nApr 5, 2018\n\n$- 2 p + 1$\n\n#### Explanation:\n\n$- 4 p + \\left(- 2\\right) + 2 p + 3$\n\nAdding a negative number is the same thing as the negative number itself, so:\n$- 4 p - 2 + 2 p + 3$\n\nNow you combine like terms:\n$- 4 p + 2 p = - 2 p$\n$- 2 + 3 = 1$\n\nAnd when you put them together you get:\n$- 2 p + 1$","date":"2019-09-15 10:13:02","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 6, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.71262526512146, \"perplexity\": 650.8540829926361}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 20, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2019-39\/segments\/1568514571027.62\/warc\/CC-MAIN-20190915093509-20190915115509-00195.warc.gz\"}"} | null | null |
Ела́нь-Чишма́ () — село в Ермекеевском районе Башкортостана, входит в состав Суккуловского сельсовета.
История
Административный центр упразднённого в 2008 году Елань-Чишминского сельсовета.
Деревня основана чувашами по договору 1788 года о припуске на вотчинных землях башкир Кыр-Еланской волости Белебеевского уезда.
Жители занимались земледелием, скотоводством, пчеловодством. В 1906 году зафиксированы церковно-приходская школа, водяная мельница, 2 бакалейные лавки, хлебозапасный магазин.
Население
В 1865 году в 79 дворах проживало 454 человека, в 1906 году в деревне насчитывалось 904 человека, в 1911 году — 543 мужчины, 536 женщин, в 1920 году — 1081 человек, в 1939 году — 984, в 1959 году — 839 человек, в 1989 году — 400 человек.
Национальный состав
Согласно переписи 2002 года, преобладающая национальность — чуваши (93 %).
Название
Происходит от башкирского «ялан» — луг, «шишмэ» — река.
Географическое положение
Расстояние до:
районного центра (Ермекеево): 16 км,
ближайшей ж/д станции (Приютово): 47 км.
Находится в низменности в долине реки Стивензя; рядом расположен комплексный объект природы «Белые камни» (чув. «Шур чул») — уступ, сложенный известняками, с вершины которого, образуя водопады высотой 1,5—2 м, стекают родники.
Инфраструктура
Есть основная школа (основана в 1895 году), детский сад, фельдшерско-акушерский пункт, клуб, библиотека.
Улицы: Кирова, Кооперативная, Крестьянская, Луговая, Молодежная, Пролетарская, Речная, Садовая, Советская, Угловая, Чапаева, Школьная.
Памятники и памятные места
Обелиск павшим в годы Великой Отечественной войны.
Известные уроженцы
Корина Прасковья Тихоновна (девичья фамилия Петрова; 1900—1992) — художник-реставратор станковой масляной живописи, жена П. Д. Корина.
Примечания
Ссылки
Совет муниципальных образований Республики Башкортостан.
Елань-Чишма на портале «Генеалогия и Архивы»
Населённые пункты Ермекеевского района | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 23 |
{"url":"https:\/\/www.edaboard.com\/threads\/avr-code-vision-or-iar.2564\/#post-10099","text":"AVR Code Vision or IAR ????\n\nStatus\nNot open for further replies.\n\nrampo\n\nJunior Member level 1\nWhat's the better C compiler for AVR Code Vision or IAR???\nThanks\n\nkarabas\n\nJunior Member level 2\nLook through the forum at h**p:\/\/www.avrfreaks.net\/phorum\/list.php?f=3. It was descussed many times.\n\nmarie65\n\nSuper Member\nCodevision is easy to use and produce good code. The best part it's the code generator for setting various parts of MCU (timers, initerrupts, ADC etc). Also have good support from the producer. Same low cost range is also ImageCraft with almost similar features. There is also GNU C compiler with the advantage to be free.\nIAR is 10 times more expensive then Codevison. It's a little more difficult to acomodate with all the stuffy settings but it worth. Produce shorter code then other compilers (don't forget that AVR core have been designed together with the IAR guys).\n\ncancel\n\nFull Member level 5\nre:iar\n\nhi ...\n\nfor advanced project codevision is not a choice. I would choose from iar or gcc. Do not forget iar is c++. You will write small code faster and easier in codevision but as project gets more complex than it gets complicated. In IAR you have to write everything from the scratch or use other sources but the final code forks very fine. You can see it on some complex projects like AVR EIT(embeded internet toolkit) where atmel used IAR for their code. Other complex project like Ethernut or yampp are using GCC.\n\nregards\n\ncancel\n\nvenz\n\nFull Member level 2\nmarie65 said:\nIAR is 10 times more expensive then Codevison. It's a little more difficult to acomodate with all the stuffy settings but it worth. Produce shorter code then other compilers (don't forget that AVR core have been designed together with the IAR guys).\n\nHi marie65,\n\nIAR is actually 20 times more expensive then Codevision depending on what version you are looking for.\n- EW for AVR Light Edition EURO 3000,- ex. VAT.\n- EW for AVR Base Line(8k limit) EURO 1000,- ex. VAT.\n- EW & MakeApp for AVR full suite EURO 3625,- ex. VAT.\n- Codevision Light $60 Node-Locked - Codevision Standard$150 Node-Locked\n- Codevision Standard \\$250 Floating\n\ngreetz, venz.\n\nStatus\nNot open for further replies.","date":"2021-10-25 09:28:02","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.20412585139274597, \"perplexity\": 10243.61681635627}, \"config\": {\"markdown_headings\": false, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-43\/segments\/1634323587659.72\/warc\/CC-MAIN-20211025092203-20211025122203-00555.warc.gz\"}"} | null | null |
\section{Introduction}
In recent years there has been much work in improving lattice actions.
The idea behind all these attempts is to modify the lattice
action with the addition of irrelevant operators in order to reduce lattice
artifacts: in this way one hopes to have scaling (and finite-size
scaling) at smaller correlation lengths.
There have been many different approaches to the problem of
improving lattice actions. In this letter we will address the problem in the
context of two-dimensional $N$-vector models, trying to point out differences
and similarities of the various approaches.
The first systematic study of improvement of lattice actions is due to
Symanzik \cite{Symanzik}.
The idea is very simple. Consider on a square lattice the standard action
\begin{equation}
S^{std} = N \sum_{x\mu} \mbox{\protect\boldmath $\sigma$}_x\cdot\mbox{\protect\boldmath $\sigma$}_{x+\mu}
\end{equation}
where the fields $\mbox{\protect\boldmath $\sigma$}_x$ are $N$-component spins;
the partition function is
\begin{equation}
Z = \int \prod_x d\mbox{\protect\boldmath $\sigma$}_x \delta(\mbox{\protect\boldmath $\sigma$}_x^2 - 1)\, e^{\beta S} \;\; .
\end{equation}
Then consider the
one-particle irreducible Green's functions
$\Gamma^{(n)}(p_1,\ldots,p_n)$. It is easy to see that at tree level
\begin{equation}
\Gamma^{(n)}(p_1,\ldots,p_n) =
\Gamma^{(n)}_{cont}(p_1,\ldots,p_n) + O(a^2)
\label{eq1}
\end{equation}
where $\Gamma^{(n)}_{cont}(p_1,\ldots,p_n)$ is the
{\em continuum} $n$-point function.
The strategy proposed by Symanzik consists in modifying the action so as
to cancel the terms of $O(a^2)$ in \reff{eq1}. The simplest action
which satisfies this condition at tree-level is the improved action
\begin{equation}
S^{Sym} =
N \sum_{x\mu} \left( {4\over3} \mbox{\protect\boldmath $\sigma$}_x \cdot \mbox{\protect\boldmath $\sigma$}_{x+\mu} \, -\,
{1\over12} \mbox{\protect\boldmath $\sigma$}_x \cdot \mbox{\protect\boldmath $\sigma$}_{x+2\mu} \right) \;\; .
\label{SSymanzik}
\end{equation}
In the Symanzik approach one can proceed further in two different directions:
first of all one can improve the action to order $O(a^4)$, $O(a^6)$ and
so on. This does not seem to be particularly interesting: indeed,
even if the action is tree-level improved to order $O(a^{2k})$, $k>1$,
then corrections of order $O(a^2)$ will again appear at one- and higher-loop
order.
The second important characteristic of the Symanzik approach is that it can
be systematically extended to higher loops: in other words
one can remove terms of order $O(g^2 a^2)$, $O(g^4 a^2)$ and so
on within perturbation theory.
A second approach to the improvement of lattice actions (the so-called
on-shell improvement) has been put forward by L\"uscher and Weisz
\cite{LW}.
The idea here is to improve only spectral quantities like the masses
of stable particles. In the $O(N)$ $\sigma$-model,
in a strip $L\times\infty$,
one can consider the mass gap $\mu(\beta,L)$ defined by
\begin{equation}
\mu(\beta,L) = - \lim_{y\to +\infty} {1\over y}
\log \left[ \sum_{x=1}^L\< \mbox{\protect\boldmath $\sigma$}_0\cdot \mbox{\protect\boldmath $\sigma$}_{(x,y)}\>\right]
\;\; .
\end{equation}
Then one considers the asymptotic expansion of $\mu(\beta,L)$ for
$\beta\to\infty$ at $L$ fixed which will have generically the form
\begin{equation}
\mu(\beta,L) L\, =\, \sum_{n=1}^\infty {\mu_n(L)\over \beta^n} \;\; .
\label{mupert}
\end{equation}
Each coefficient $\mu_n(L)$ has an expansion in powers of $1/L^2$
(with additional logarithms of $L$). The $O(a^2)$ improved action is then
chosen by requiring, order by order in perturbation theory,
that $\mu_n(L)$ does not have $1/L^2$ corrections.
It should be noticed that $O(a^2)$ improvement of the one-loop term
$\mu_2(L)$ gives the $O(a^2)$ tree-level improved action \cite{LW}.
Let us now discuss the two different approaches in the soluble case of
$N=\infty$.
Let us consider the generalized action
\begin{equation}
S =\, N \sum_{x,y} J(x-y) \mbox{\protect\boldmath $\sigma$}_x \cdot \mbox{\protect\boldmath $\sigma$}_y \;\; .
\label{eq2}
\end{equation}
We will assume the interaction to be local and parity invariant:
if $\widehat{J}(p)$ is
the Fourier transform of $J(x)$, we will require that
$\widehat{J}(p)$ is a continuous function of $p$, even under $p\to -p$.
We will also
require invariance under rotations of $\pi/2$, that is we will
assume that $\widehat{J}(p)$ is symmetric under exchange of ${p}_1$ and
${p}_2$. Redefining $\beta$ we can normalize the
couplings so that
\begin{equation}
\widehat{J}(q) = \widehat{J}(0) - {q^2 \over 2} + O(q^4)
\end{equation}
for $q\to 0$. We also introduce the function
\begin{equation}
w(q) = - 2 (\widehat{J}(q) - \widehat{J}(0))
\end{equation}
which behaves as $q^2$ for $q\to 0$.
Finally we will require the theory to have
the usual (formal) continuum limit: we will assume that
the equation $w(q) = 0$ has only one solution for
$-\pi \le q_i \le \pi$, namely $q=0$.
We will need the small-$q$ behaviour of $w(q)$: we will assume in this limit
the form
\begin{equation}
w(q) =\, \hat{q}^2 + \alpha_1\, \sum_\mu \hat{q}^4_\mu +
\alpha_2 \ (\hat{q}^2)^2 + O(q^6)
\end{equation}
where $\alpha_1$ and $\alpha_2$ are arbitrary constants. Here
$\hat{q}^2 = \hat{q}^2_1 + \hat{q}^2_2$, $\hat{q}_\mu = 2 \sin (q_\mu/2)$.
Notice that for \reff{SSymanzik} we have $w(q)=q^2 + O(q^6)$ i.e.
$\alpha_1 = {1\over12}$ and $\alpha_2 = 0$.
Let us now study the improvement program \`a la Symanzik. Let us consider
the two-point function in infinite volume. A trivial calculation gives
\begin{equation}
\<\mbox{\protect\boldmath $\sigma$}_0\cdot\mbox{\protect\boldmath $\sigma$}_x\> \ \equiv G_V(x) =\ 1 + {1\over \beta}
\int {d^2p\over (2\pi)^2} {e^{ipx} - 1\over w(p)} +
O(e^{-4 \pi \beta}) \;\; .
\end{equation}
Notice that only the tree-level term appears, all successive ones being
zero. The Fourier transform $\widehat{G}_V(q;\infty)$ for $q\not=0$ is then
given by
\begin{equation}
\widehat{G}_V^{-1}(q;\infty) = \beta w(q) + O(e^{-4 \pi \beta})
\end{equation}
Improvement \`a la Symanzik requires then $w(q) = q^2 + O(q^6)$ ,
i.e. $\alpha_1 = {1\over 12}$ and $\alpha_2 = 0$. When these two
conditions are satisfied the action is improved to all orders in
perturbation theory.
Let us now consider the mass-gap $\mu(\beta,L)$ in a strip and its perturbative
expansion \reff{mupert}.
At one loop it is easy to compute \cite{CP_LAT96,CPprep}
\begin{eqnarray}
\mu(\beta,L) L &=& {1\over 2\beta} +
{1\over \beta^2}\left[ {1\over 4\pi} \log L +
{1\over2} (\overline{F}_{00} + \Lambda_0) \right.
\nonumber \\
&& \qquad \left . + {\pi\over 144 L^2} (1 - 12 \alpha_1) + O(L^{-4})\right]
+ O(\beta^{-3})
\end{eqnarray}
where
\begin{eqnarray}
\overline{F}_{00} &=& {1\over 2\pi} \left(\gamma_E - \log \pi +
{1\over2} \log 2 \right) \\
\Lambda_0 &=& \int {d^2p\over (2\pi)^2} \left( {1\over w(p)} -
{1\over \hat{p}^2} \right)
\end{eqnarray}
and $\gamma_E \approx 0.577215665$ is the Euler constant.
Perturbative improvement requires the cancellation of the $1/L^2$ term
and thus gives the condition $1-12\alpha_1 = 0$, i.e.
$\alpha_1 = \smfrac{1}{12}$.
Now consider the next order. A simple calculation gives
\begin{eqnarray}
\hskip -20pt
\mu_3(L) &=& {1\over 8\pi^2} \log^2 L + {1\over 2\pi}
(\overline{F}_{00} + \Lambda_0) \log L +
{1\over2} (\overline{F}_{00} + \Lambda_0)^2
\nonumber \\
&& \hskip -20 pt
+ {1\over L^2}\left[ {1\over144}(1 -12 \alpha_1) \log L +
{\pi\over72} (1 - 12\alpha_1) (\overline{F}_{00} + \Lambda_0)
+ {1\over48} (12 \alpha_1 + 12\alpha_2 - 1)\right]
\nonumber \\ [-2mm]
&& {}
\end{eqnarray}
Requiring the cancellation of the $1/L^2$
term we get $\alpha_1 = \smfrac{1}{12}$
and $\alpha_2 = 0$. Thus only Symanzik tree-level improved
actions are improved at
this level. We can go further and compute $\mu_4(L)$: for
$\alpha_1 = \smfrac{1}{12}$
and $\alpha_2 = 0$ the coefficient of $1/L^2$ is
\begin{equation}
{1\over 8} \left( {1\over 96\pi} - \Lambda_1\right)
\end{equation}
where
\begin{equation}
\Lambda_1 =\, \int {d^2p\over (2\pi)^2}
\left[ {1\over w(p)^2} - {1\over (\hat{p}^2)^2} +
{2\over (\hat{p}^2)^3} \left( \alpha_1 \sum_\mu \hat{p}^4_\mu
+ \alpha_2 (\hat{p}^2)^2 \right) \right] \;\; .
\label{Lambda1testo}
\end{equation}
Thus improving the three-loop contribution to $\mu(\beta,L)$ gives
\begin{equation}
\Lambda_1 = {1\over 96 \pi} \;\; .
\label{conditionLambda1}
\end{equation}
Notice that this condition is global, that is it does not only fix the
small $q^2$ behaviour of $w(q)$ but it depends on the behaviour of $w(q)$
over all the Brillouin zone.
A particular action satisfying \reff{conditionLambda1} is given by
\begin{eqnarray}
\hskip -7pt
S^{Sym2} = N\sum_{x\mu}
\left[ \left( {4\over3} + 15 a\right) \mbox{\protect\boldmath $\sigma$}_x\cdot\mbox{\protect\boldmath $\sigma$}_{x+\mu} + \,
\left(- {1\over12} - 6 a\right) \mbox{\protect\boldmath $\sigma$}_x\cdot\mbox{\protect\boldmath $\sigma$}_{x+2\mu} + \,
a \mbox{\protect\boldmath $\sigma$}_x\cdot\mbox{\protect\boldmath $\sigma$}_{x+3\mu} \right] \;\; .
\label{Symanzik2}
\end{eqnarray}
where $a = 0.00836533968(1)$.
A peculiarity of the large-$N$ limit
is the fact that, once \reff{conditionLambda1} is satisfied,
all subsequent coefficients $\mu_n(L)$ do not have $1/L^2$ corrections:
improvement at three-loops is equivalent to improvement to
all orders in perturbation theory: Symanzik actions
satisfying \reff{conditionLambda1} are improved to all perturbative
orders. Of course this feature will not be true
for finite values of $N$.
In conclusion, within Symanzik approach only the two conditions
$\alpha_1 = {1\over12}$ and $\alpha_2=0$ are required to improve
the action to order $O(a^2)$ and to all orders in perturbation theory.
In the on-shell program instead an additional condition
(equation \reff{conditionLambda1}) is obtained. Notice that naively
one would have expected the opposite result since Symanzik approach
is intended to improve all Green's functions while the on-shell
program considers only spectral quantities. This result is due to the
large $N$ limit~\footnote{We thank Giorgio Parisi who told us that this point was
already known to K.~Symanzik.}.
Indeed, one can see that condition
\reff{conditionLambda1} appears in the Symanzik approach at order $1/N$. At this
order one should consider, beside the two-point function also the four-point
function. If $\Gamma_{\alpha \beta
\gamma \delta}(p,q,r,s)$ is the one-particle irreducible four-point function, we have
\begin{equation}
\Gamma_{\alpha \beta \gamma \delta}(p,q,r,s) =
-{1\over N} \left[ \delta_{\alpha \beta} \delta_{\gamma \delta} \Delta(p+q) +
\delta_{\alpha \gamma} \delta_{\beta \delta} \Delta(p+r) +
\delta_{\alpha \delta} \delta_{\gamma \beta} \Delta(p+s) \right] + O(1/N^2)
\end{equation}
where
\begin{equation}
\Delta^{-1}(p) = {1\over 2} \int {d^2k\over (2 \pi)^2} {1\over \left[ w(k) + m_0^2
\right] \left[w(p+k) + m_0^2 \right]}
\end{equation}
In the perturbative limit
\begin{equation}
\Delta^{-1}(p) = {1\over \beta w(p)} \left[ 1 + {1\over 2 \beta}
\int {d^2 k\over (2 \pi)^2} {w(p) - w(k) - w(p+k) \over w(p+k) w(k)} \right] +
O\left(e^{-4 \pi \beta} \right)
\end{equation}
If $\alpha_1 = {1\over12}$ and $\alpha_2=0$ we have then~\cite{CPprep}
\begin{equation}
\Delta^{-1}(p) = {1\over \beta p^2} \left\{ 1 + {1\over \beta} \left[
{1\over 4 \pi} \log {p^2\over 32} - \Lambda_0 + {p^2\over 2} \left( \Lambda_1 -
{1\over 96 \pi}\right)\right] + O(p^4 \log p^2) \right\}
\end{equation}
Thus $O(a^4)$-improvement required the condition \reff{conditionLambda1}.
In the Symanzik approach \reff{conditionLambda1} is thus a one-loop improvement
condition for the four-point function. Notice again the peculiarity of the large $N$
limit: improvement of the one-loop result is enough to improve all orders of
perturbation theory.
We have checked that $\alpha_1 = {1\over12}$, $\alpha_2=0$ and
\reff{conditionLambda1} are sufficient conditions to improve both the scaling and the
finite-size-scaling behaviour of the theory. Indeed, consider for instance the ratio
(in infinite volume)
\begin{equation}
R(\beta) =\, \left({\xi^{(2)}_V(\beta) \over \xi^{(2)}_T(\beta)}\right)^2
\end{equation}
where
\begin{equation}
\xi^{(2)}_\#(\beta) = {1\over4}
{\sum_x |x|^2 G_\# (x) \over \sum_x G_\# (x)}
\end{equation}
and
\begin{equation}
G_T(x) = \< \sigma_0^a \sigma_0^b; \sigma_x^a \sigma_x^b \> \;\; .
\end{equation}
Then an easy calculation gives
\begin{eqnarray}
R(\beta) &=& 6 - 24 e^{-4\pi(\beta-\Lambda_0)}
\left[4\pi (12\alpha_1 + 16\alpha_2 - 1)(\beta - \Lambda_0)
\right. \nonumber \\
&& \qquad \left. - (8 \alpha_1 + 8 \alpha_2 - 1 + 32 \pi\Lambda_1)\right]
+ O(e^{-8\pi\beta}\beta) \;\; .
\end{eqnarray}
Improvement requires then
\begin{eqnarray}
12 \alpha_1 + 16 \alpha_2 - 1 &=& 0 \;\; ,\\
8\alpha_1 + 8\alpha_2 - 1 + 32 \pi \Lambda_1 &=& 0 \;\; .
\end{eqnarray}
The first term clearly vanishes for the Symanzik action for which
$\alpha_1 = {1\over12}$ and $\alpha_2 =0$. The second one however
requires the additional condition \reff{conditionLambda1}.
Let us now consider finite-size scaling. On a strip of width $L$, we have
computed the corrections to the finite-size-scaling function\footnote{The
leading term was already computed in \cite{Lu}.}
for $\mu(\beta,L)$.
In the limit $L\to\infty$, $\beta\to\infty$ with
$\mu(\beta,L) L \equiv x$ fixed we find \cite{CP_LAT96,CPprep}
\begin{equation}
{\mu(\beta,\infty)\over \mu(\beta,L)} \, =\,
f_\mu(x) \left(1 + {\Delta_{\mu,1}(x)\over L^2} \log L +
{\Delta_{\mu,2}(x)\over L^2} \right)
\label{FSSformula}
\end{equation}
with corrections of order $\log^2 L/L^4$.
Cancellation of the terms of
order $\log L/L^2$ requires $12 \alpha_1 + 16\alpha_2 - 1 = 0$, while
cancellation of the terms $1/L^2$ requires $\alpha_1 = {1\over12}$,
$\alpha_2 = 0$ and the condition \reff{conditionLambda1}.
Again \reff{conditionLambda1} is necessary to completely eliminate the
corrections of order $1/L^2$.
We want to make an additional remark on the on-shell improvement program.
For $N=\infty$ tree-level improvement requires the one-loop contribution
to $\mu(\beta,L)$ to have corrections of order $O(L^{-4})$ and thus
it requires only $\alpha_1 = {1\over12}$; the second coefficient
$\alpha_2$ is arbitrary to this order. This is a peculiarity
of the large-$N$ limit. For finite values of $N$, using the results
of \cite{RV,CPprep} , we have
\begin{equation}
L \mu(\beta,L) = {N-1\over 2N\beta} \left[ 1 +
{1\over N \beta} (r_1(L) + (N-2) r_2(L) ) + O(\beta^{-2}) \right]
\label{LmuLpert}
\end{equation}
with
\begin{eqnarray}
\hskip -10pt
r_1(L) &=& - \int {d^2 k\over (2\pi)^2} {1\over w(k)}
\left[ {1\over4} \sum_\mu {\partial^2 w(k)\over \partial k_\mu^2} -
1\right]
- {\pi\over 12 L^2} (12 \alpha_1 + 8 \alpha_2 -1) + O(L^{-4})
\;\; , \nonumber \\ [-2mm]
{} \label{r1L} \\
\hskip -10pt
r_2(L) &=& {1\over 2\pi} \log L
+ \overline{F}_{00} + \Lambda_0
+ {\pi\over 72 L^2} (1 - 12\alpha_1) + O(L^{-4})
\;\; .
\label{r2L}
\end{eqnarray}
Clearly the $1/L^2$ corrections cancel only if $\alpha_1 = {1\over12}$
and $\alpha_2 = 0$ except for $N=\infty$. Let is notice again \cite{LW}
that the one-loop computation fixes the improved action
at tree-level\footnote{On-shell improvement to order $a^{2k}$, $k>1$,
was discussed in \cite{RV}.
It was shown numerically that if one chooses
the generalization of \reff{SSymanzik} improved to order
$O(a^{2k})$, then $\mu_2(L)$ has corrections of order $O(L^{-2k-2})$.
Assuming this result it is trivial to prove rigorously that for
{\em any} action such that $w(q)=q^2 + O(q^{2k+4})$
(i.e. for any $O(a^{2k})$ tree-level Symanzik-improved action)
$\mu_2(L)$ has corrections of order $O(L^{-2k-2})$
(i.e. the action is also $O(a^{2k})$ tree-level on-shell improved in
the sense of L\"uscher and Weisz \cite{LW}).
Notice that the opposite result is not true: it is {\em not necessary}
that $w(q)=q^2 + O(q^{2k+4})$ to have $\mu_2(L) =
\hbox{\rm leading} + O(L^{-2k-2})$. For instance consider
$w(q)$ such that for $q\to 0$
\begin{equation}
w(q) = q^2 + \beta_1 \sum_\mu q^8_\mu +
\beta_2 (q^2)^2 \sum_\mu q^4_\mu + \beta_3 (q^2)^4
+ O(q^{10}) \;\; .
\end{equation}
Requiring that $\mu_2(L)$ does not have terms of order $1/L^4$ we get two
equations (corresponding to $r_1(L)$ and $r_2(L)$) which will not fix
completely the three parameters $\beta_i$. Thus at order $O(a^6)$ one can have an
on-shell action which is a Symanzik-improved one. }.
We want now to discuss a third type of actions which we will call
classically spectrum-improved actions. The idea
which has been put forward by the Bern group
\cite{Niedermayer_LAT96,perfect} consists in changing the action
so that to improve the dispertion relations.
For instance an action is $O(a^2)$-spectrum improved if the
equation $w(iE,p)=0$ gives the continuum dispersion relation
$E^2 - p^2 = 0$ modulo terms of order $O(a^4)$. In practice
$w(q)$ must have an expansion with $\alpha_1 = {1\over12}$ and $\alpha_2$
arbitrary, i.e.
\begin{equation}
w(q) = q^2 + \alpha_2 (q^2)^2 + O(q^6) \;\; .
\end{equation}
An example is the action proposed in \cite{Niedermayer_LAT96}:
\begin{equation}
S^{diag} =
N \sum_{x} \left( {2\over3} \sum_\mu \mbox{\protect\boldmath $\sigma$}_x \cdot \mbox{\protect\boldmath $\sigma$}_{x+\mu}
\, +\, {1\over6} \sum_{\hat{d}}\mbox{\protect\boldmath $\sigma$}_x \cdot \mbox{\protect\boldmath $\sigma$}_{x+\hat{d}} \right)
\label{Symanzikdiag}
\end{equation}
where $\hat{d}$ are the two diagonal vectors $(1,\pm1)$. For this action
$\alpha_1 = {1\over 12}$ and $\alpha_2 = - {1\over 12}$.
We will now show that these actions do not show any improved behaviour.
First of all \cite{RV} these actions are not
on-shell tree-level improved in the sense of L\"uscher and Weisz \cite{LW}.
Indeed tree-level improvement requires $\alpha_2=0$ (see equations \reff{r1L}
and \reff{r2L}).
For $N=\infty$
it is also easy to see that these actions do not show any improved
behaviour compared to the standard action \cite{CP_LAT96}.
Consider for instance the finite-size-scaling behaviour of the ratio
$\mu(\beta,\infty)/\mu(\beta,L)$.
In table \reff{table_FSS} we report the value of
the deviations from finite-size-scaling
(see equation \reff{FSSformula})
\begin{equation}
R(L,x) = \left( {\mu(\beta,\infty)\over \mu(\beta,L)}\right)
{1\over f_\mu(x)} -\, 1
\end{equation}
for the various actions for $L\mu(\beta,L) \equiv x=2$
(this is the value of $x$ for which
the deviations are larger). Clearly the action
\reff{Symanzikdiag} has larger corrections that the standard action.
On the other hand,
as expected, the Symanzik actions \reff{SSymanzik} and \reff{Symanzik2}
show much smaller deviations from finite-size scaling, especially
\reff{Symanzik2} which is Symanzik improved to all orders in perturbation
theory and for which the corrections to finite-size scaling behave
as $\log L/L^4$.
Notice that the improvement is working even for $L=4$--6 in spite
of the large spatial extent of the Symanzik actions.
\begin{table}
\begin{center}
\begin{tabular}{|l|rrrr|}
\hline
$L$ & $S^{std}$ & $S^{diag}$ & $S^{Sym}$ & $S^{Sym2}$ \\
\hline
4 & 0.022022 & 0.039225 & 0.0002840 & $-$0.0008378 \\
6 & 0.012721 & 0.021359 & 0.0007414 & $-$0.0001953 \\
10 & 0.005758 & 0.009302 & 0.0003948 & $-$0.0000246 \\
20 & 0.001806 & 0.002825 & 0.0001141 & $-$0.0000011 \\
\hline
\end{tabular}
\end{center}
\caption{Values of
$R(L;x)$ for $x=2$ for the various actions we have introduced in the
text. Here $N=\infty$.
}
\label{table_FSS}
\end{table}
We want now to show that if $\alpha_2\not=0$ an on-shell improved
action can be obtained by adding a new four-spin coupling.
Indeed consider the continuum action
\begin{equation}
S = - N \int d^2 x \left[ {1\over2} (\mbox{\protect\boldmath $\sigma$}\cdot \Box\mbox{\protect\boldmath $\sigma$}) +
{\alpha_2 a^2\over2} (\mbox{\protect\boldmath $\sigma$}\cdot \Box^2 \mbox{\protect\boldmath $\sigma$} -
(\mbox{\protect\boldmath $\sigma$}\cdot \Box \mbox{\protect\boldmath $\sigma$})^2 ) + O(a^4) \right]
\label{Honshellimp}
\end{equation}
where $\Box = \sum_\mu \partial_\mu^2$. We will now show that, by means
of an isospectral transformation, one can get rid of the $O(a^2)$
terms\footnote{Equivalently one can show that the term proportional
to $a^2$ in \reff{Honshellimp} vanishes because of the equations
of motion \reff{eqmotion}.}.
Setting $\mbox{\protect\boldmath $\sigma$} = (\mbox{\protect\boldmath $\pi$}, \sqrt{1 - \mbox{\protect\boldmath $\pi$}^2})$, the equation of motion
can be written as
\begin{equation}
\Box \mbox{\protect\boldmath $\pi$} - \mbox{\protect\boldmath $\pi$} (\mbox{\protect\boldmath $\sigma$}\cdot \Box\mbox{\protect\boldmath $\sigma$}) + O(a^2) = 0 \;\; .
\label{eqmotion}
\end{equation}
Now consider the isospectral change of variable
\begin{equation}
\mbox{\protect\boldmath $\pi$}' = \mbox{\protect\boldmath $\pi$} +
{A a^2\over2} (\Box \mbox{\protect\boldmath $\pi$} - \mbox{\protect\boldmath $\pi$} (\mbox{\protect\boldmath $\sigma$}\cdot \Box\mbox{\protect\boldmath $\sigma$}))\;\; .
\end{equation}
where $A$ is a free parameter. It is immediate to verify that, setting
$A= -\alpha_2$ we can rewrite
\begin{equation}
S = -{N\over2} \int d^2 x\, (\mbox{\protect\boldmath $\sigma$}'\cdot \Box\mbox{\protect\boldmath $\sigma$}') +\,
O(a^4)
\end{equation}
where $\mbox{\protect\boldmath $\sigma$}' = (\mbox{\protect\boldmath $\pi$}',\sqrt{1 - (\mbox{\protect\boldmath $\pi$}')^2})$. As a final check
we compute at one loop $\mu(\beta,L)$ for an action of the form
\reff{Honshellimp}. On the lattice we consider
\begin{equation}
S = N \sum_{xy} J(x-y) \mbox{\protect\boldmath $\sigma$}_x \cdot \mbox{\protect\boldmath $\sigma$}_y +
{\alpha_3 N\over2} \sum_{xyz}K(x-y) K(x-z)
(\mbox{\protect\boldmath $\sigma$}_x \cdot \mbox{\protect\boldmath $\sigma$}_y) (\mbox{\protect\boldmath $\sigma$}_x\cdot \mbox{\protect\boldmath $\sigma$}_z)
\label{Slattice-onshell}
\end{equation}
where
$\hat{K}(q) = q^2 + O(q^4)$ for $q\to 0$ and $\alpha_3$ is a free
parameter. For $a\to0$ the action \reff{Slattice-onshell}
has the continuum limit \reff{Honshellimp} if
$\alpha_1 = {1\over 12}$ and $\alpha_2 = \alpha_3$.
Using the results of
\cite{RV,CPprep} we get the expansion \reff{LmuLpert} where
$r_1(L)$ and $r_2(L)$ are given by
\begin{eqnarray}
r_1(L) &=& - \int {d^2 k\over (2\pi)^2} {1\over w(k)}
\left[ {1\over4} \sum_\mu {\partial^2 w(k)\over \partial k_\mu^2} -
{\alpha_3\over2}
\sum_\mu \left({\partial \widehat{K} (k) \over \partial k_\mu}
\right)^2 - 2 \alpha_3 \widehat{K}(k) -
1\right] \nonumber \\
&& \qquad - {\pi\over 12 L^2}
(12 \alpha_1 + 8 \alpha_2 - 8 \alpha_3 -1) + O(L^{-4}) \;\; ,
\\
r_2(L) &=& {1\over 2\pi} \log L + \overline{F}_{00} +
\Lambda_0 + 2 \alpha_3 \int {d^2k\over (2 \pi)^2}
{\widehat{K}(k) \over w(k)} \nonumber \\
&& \qquad + {\pi\over 72 L^2} (1 - 12 \alpha_1) + O(L^{-4}) \;\; .
\end{eqnarray}
As expected the $1/L^2$ corrections vanish for $\alpha_1={1\over12}$ and
$\alpha_2 = \alpha_3$.
An explicit example with all couplings lying in a plaquette is given by
\begin{equation}
S^{onshell} = S^{diag} - {N\over24} \sum_x \sum_{i=1}^4
\left(\sum_{{\mu_i}}
(\mbox{\protect\boldmath $\sigma$}_x\cdot\mbox{\protect\boldmath $\sigma$}_{x+{\mu_i}} -1) \right)^2
\end{equation}
where $\mu_1$ runs over the vectors $(1,0)$ and $(0,1)$,
$\mu_2$ over $(-1,0)$ and $(0,1)$,
$\mu_3$ over $(-1,0)$ and $(0,-1)$ and
$\mu_4$ over $(1,0)$ and $(0,-1)$ (the sum over $i$ symmetrizes
over the four plaquettes stemming from the point $x$).
This action should be equivalent to $S^{Sym}$ for on-shell
quantities. However $S^{onshell}$ enjoys an additional property:
it is reflection positive for reflections with respect to the line
$x_1 = 0$; it is then possible, with a standard construction, to
define a transfer matrix
\cite{reflectionpositivity1,reflectionpositivity2}.
It is also trivial to prove that this action has the correct continuum
limit, in the sense that the ordered configuration is the unique
absolute maximum of $S^{onshell}$.
The appearance of a four-spin coupling in $S^{onshell}$ suggests a connection
between the on-shell improvement program and the perfect actions
\cite{perfect}. The perfect
action is indeed an action which is {\em tree-level} on-shell
improved\footnote{
It is somewhat misleading the statement \cite{Niedermayer_LAT96,Farchioni}
that perfect actions are {\em one-loop quantum perfect} because
$\mu_2(L)$ has exponentially small corrections.
Indeed cancellation of the $1/L^{2k}$ corrections in $\mu_2(L)$ fixes
the on-shell improved action only at tree-level \cite{LW,RV}.}
to all orders in $a^2$ \cite{Farchioni}. Work in this direction is in progress.
\bigskip
We thank Ferenc Niedermayer, Giorgio Parisi, Paolo Rossi and Ettore Vicari for
useful discussions.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 7,104 |
\section{Introduction}
\label{sec:introduction}
The most common procedure to regularize ultraviolet (UV) and infrared
(IR) singularities of scattering amplitudes is to apply conventional
dimensional regularization ({\scshape cdr}), whereby all relevant quantities are
treated as $D=4-2\epsilon$ dimensional. In {\scshape cdr}, IR singularities of
next-to-next-to leading order (NNLO) scattering amplitudes in massless
QCD have a remarkably simple structure~\cite{Becher:2009cu, Gardi:2009qi,
Becher:2009qa, Gardi:2009zv}. Key ingredients are the
cusp anomalous dimension $\gamma_{\text{cusp}}$ and the anomalous
dimensions of quarks and gluons, $\gamma_{q}$ and $\gamma_{g}$,
respectively.
For practical computations it is sometimes advantageous to apply
certain variants of {\scshape cdr}, such as the 't~Hooft-Veltman scheme
({\scshape hv})~\cite{'tHooft:1972fi}, dimensional reduction
({\scshape dred})~\cite{Siegel:1979wq} or the four-dimensional helicity scheme
({\scshape fdh})~\cite{BernZviKosower:1992}. This leads to the question how
virtual amplitudes computed in these schemes are related to the
corresponding amplitudes computed in {\scshape cdr}. In the massless case at
NNLO, this question has been answered in Ref~\cite{Broggio:2015dga},
where, drawing on earlier results~\cite{ Kunszt:1994,
Catani:1996pk, Catani:1998bh, Stockinger:2005gx, Signer:2005, Signer:2008va,
Kilgore:2011ta, Kilgore:2012tb, Gnendiger:2014nxa, Broggio:2015ata},
it has been shown that the IR structure of {\scshape cdr}\ is only modified through
changes in the anomalous dimensions. We indicate this regularization-scheme
({\scshape rs}) dependence by the shifts
$\gamma_{\text{cusp}}\to\gamma_{\rm cusp}^{{\rm \scriptscriptstyle RS}}$,
$\gamma_{q}\to\gamma_{q}^{{\rm \scriptscriptstyle RS}}$ and $\gamma_{g}\to\gamma_{g}^{{\rm \scriptscriptstyle RS}}$.
The explicit expressions of the anomalous dimensions as well as the
$\beta$~functions of the various couplings in the different schemes
have been determined at least up to NNLO.
In the presence of massive quarks there are additional structures in
the IR singularities of QCD amplitudes~\cite{Becher:2009kw}. Hence,
the scheme dependence will also have to be generalized.
At NLO the scheme-dependence has been discussed in
Ref.~\cite{Catani:2000ef}.
The generalization of the scheme dependence at NNLO to QCD amplitudes
including massive quarks is the main result of this paper. As we will
show, once the scheme-dependent UV renormalization has been carried
out, this scheme dependence is contained entirely in two additional
anomalous dimensions, the velocity-dependent cusp anomalous dimension
$\gamma_{\rm cusp}^{{\rm \scriptscriptstyle RS}}(\beta)$ and the anomalous dimension of a
heavy quark $\gamma_Q^{{\rm \scriptscriptstyle RS}}$.
In fact, the scheme dependence of
$\gamma_{\rm cusp}^{{\rm \scriptscriptstyle RS}}(\beta)$ itself is induced solely through the
scheme dependence of the cusp anomalous dimension
$\gamma_{\rm cusp}^{{\rm \scriptscriptstyle RS}}$ from the massless case.
With the results presented here it is possible to convert any NNLO QCD
amplitude between the four schemes {\scshape cdr}, {\scshape hv}, {\scshape fdh}, and {\scshape dred}. This
allows for using whatever scheme is most convenient in the computation
of the virtual amplitude and then combine this with the real
corrections, typically computed in {\scshape cdr}. In fact, for the
generalization to the massive case it is sufficient to consider the
difference between the {\scshape fdh}\ and the {\scshape hv}\ (or {\scshape cdr}) scheme. If there
are no external gluons, {\scshape fdh}\ is equivalent to {\scshape dred}.
Hence, the IR anomalous dimensions are the same, e.\,g.\
$\gamma^{\text{{\scshape fdh}}}_{\text{cusp}}(\beta)
=\gamma^{\text{{\scshape dred}}}_{\text{cusp}}(\beta)$ and
$\gamma^{\text{{\scshape fdh}}}_{Q}=\gamma^{\text{{\scshape dred}}}_{Q}$.
Furthermore, {\scshape cdr}\ and {\scshape hv}\ also have the same anomalous dimensions,
$\gamma^{\text{{\scshape cdr}}}_{\text{cusp}}(\beta)
=\gamma^{\text{{\scshape hv}}}_{\text{cusp}}(\beta)$ and
$\gamma^{\text{{\scshape cdr}}}_{Q}=\gamma^{\text{{\scshape hv}}}_{Q}$.
These schemes differ simply in the dimension of the polarization sum of
external gluons.
Apart from the four schemes treated in this paper, other possibilities
to regularize virtual amplitudes have been considered. The
{\scshape fdh}\ scheme has been adapted to the so-called {\scshape fdf} scheme
(four-dimensional formulation) for using unitary-based methods to
compute NLO amplitudes~\cite{Fazio:2014xea,Bobadilla:2015wma}. There
are also proposals to abandon dimensional regularization altogether
and perform computations completely in four dimensions in the context
of implicit regularization~\cite{Pontes:2006br, Dias:2008iz,
Fargnoli:2010ec, Cherchiglia:2010yd}, {\scshape fdr}
(four-dimensional regularization/renormalization)~\cite{Pittau:2012zd,
Donati:2013voa, Zirke:2015spg}, and using loop-tree duality to deal
with IR singularities at the integrand
level~\cite{Hernandez-Pinto:2015ysa, Sborlini:2015uia,
Sborlini:2016fcj}.
While this list is by no means exhaustive it shows that despite the
impressive technical advances in computing higher-order corrections in
{\scshape cdr}\ there is considerable interest in exploring alternative methods.
The results presented here complete the description at NNLO of a first
step away from a fully $D$ dimensional treatment of the problem.
Apart from allowing to perform computations in {\scshape fdh}\ and {\scshape dred}, we
hope it also helps to understand better the relation between {\scshape cdr}\ and
the different four-dimensional approaches mentioned above. The
ultimate goal is, of course, to develop efficient methods to
explicitly perform ever more accurate computations.
The paper is organized as follows: in Section~\ref{sec:schemes} we
briefly review the various schemes, the IR structure of amplitudes and
its extension to the massive case. We also discuss the UV
renormalization, emphasizing the special features of {\scshape fdh}\ in the
presence of massive quarks. Section~\ref{sec:ad} is devoted to the
computation of $\gamma_Q^{{\rm \scriptscriptstyle RS}}$ and $\gamma_{\rm cusp}^{{\rm \scriptscriptstyle RS}}(\beta)$
at NNLO in the {\scshape fdh}\ scheme. These results are obtained by direct
computations using soft-collinear effective theory. In order to
obtain an independent test of the scheme dependence of NNLO
amplitudes, in Section~\ref{sec:examples} we compare the heavy-quark
and heavy-to-light form factors in the {\scshape fdh}\ and {\scshape cdr}\ schemes and
verify that the results are in agreement with the expected scheme
dependence obtained from the anomalous dimensions. We also provide a
guide on how to actually perform computations in the {\scshape fdh}\ scheme and
show that the modifications compared to {\scshape cdr}\ are minimal. Finally we
present our conclusion in Section~\ref{sec:conc}.
\section{UV and IR structure of massive QCD}
\label{sec:schemes}
\subsection{DRED and FDH}
As has been shown in a series of papers \cite{Jack:1994bn,
Stockinger:2005gx, Gnendiger:2014nxa,
Broggio:2015ata,Broggio:2015dga}, a consistent formulation of the
dimensional reduction ({\scshape dred}) and the four-dimensional helicity ({\scshape fdh})
scheme in the framework of massless QCD requires the introduction of
three vector spaces. In this work we investigate how this can be
extended to the case of massive partons. In doing so we do not
consider processes including external vector fields. The names
{\scshape fdh}\ and {\scshape dred}\ are in the following therefore used synonymously,
meaning that whenever a statement about the {\scshape fdh}\ schemes is made, the
same argument also applies in {\scshape dred}. For a detailed discussion and a
precise definition of the schemes, of the related vector spaces and
their algebraic relations we refer to Ref.~\cite{Signer:2008va}. Here
we only provide the most important characteristics.
In {\scshape fdh}, the underlying quasi $4$-dimensional space $Q4S$ with metric
$g^{\mu\nu}$ is split into a direct sum of the quasi $D$-dimensional
space of {\scshape cdr}\ with metric ${\hat{g}}^{\mu\nu}$ and a disjoint space
$Q2\epsilon S$ with metric ${\tilde{g}}^{\mu\nu}$:
\begin{align}
g^{\mu\nu}={\hat{g}}^{\mu\nu}+{\tilde{g}}^{\mu\nu} \, .
\end{align}
In order to have full control over the contributions originating from
$Q2\epsilon S$, we \textit{define} complete contractions of the corresponding
metric tensors as
\begin{align}
{\tilde{g}}^{\mu\nu}{\tilde{g}}_{\mu\nu}\defeqN_\epsilon \, .
\end{align}
As a consequence, arbitrary {\scshape fdh}\ quantities in general depend on $N_\epsilon$.
They are in the following denoted by a bar.
At the level of the Lagrangian, the structure of the different vector
spaces is reflected in a split of the quasi $4$-dimensional gluon
field into a $D$-dimensional gluon field and an $\epsilon$-scalar
field: $A^\mu={\hat{A}}^\mu+{\tilde{A}}^\mu$. The 'particles' associated with
these fields are in the following denoted by $g$ and ${\tilde{g}}$,
respectively. In
Refs.~\cite{Jack:1993ws,Harlander:2006rj,Kilgore:2011ta}, it has been
shown that because of this split in principle five different couplings
need to be distinguished in the bare theory: the gauge coupling
$\alpha_s=g_s^2/(4\pi)$, the ${\tilde{g}} q \bar{q}$ coupling
$\alpha_e=g_e^2/(4\pi)$, and three different quartic
${\tilde{g}}$-couplings. However, for the calculations presented in this
work it is sufficient to consider only $\alpha_s$ and $\alpha_e$.
For later purposes it turns out to be useful to include repeatedly
occurring universal factors in the definition of the bare couplings
\begin{equation}
a_i(m^2)\mathrel{\mathop:}=
e^{-\epsilon \gamma_{E}}(4\pi)^{\epsilon}\Big{(}\frac{1}{m^{2}}\Big{)}^{\epsilon}
\Big(\frac{\alpha_{i}^{0}}{4\pi}\Big)
=\Big{(}\frac{\mu^{2}}{m^{2}}\Big{)}^{\epsilon}
\bar{Z}_{\alpha_{i}}\Big(\frac{\alpha_{i}}{4\pi}\Big)\, ,
\label{eq:asdef}
\end{equation}
where $\gamma_E$ is the Euler-Mascheroni constant, $m$ is the mass of
a heavy fermion, and $a_i\in\{a_s,a_e\}$. As renormalization
prescription for the couplings we use the $\overline{\text{MS}}$
scheme throughout this work. The constants $\bar{Z}_{\alpha_{i}}$ in
{\scshape fdh}\ are given in e.\,g.\ Ref.~\cite{Gnendiger:2014nxa}. The
perturbative expansion of {\scshape fdh}/{\scshape dred}\ quantities in terms of the UV
renormalized couplings is in the following written as
\begin{align} \label{eq:generalexApp}
X^{\text{{\scshape fdh}/{\scshape dred}}}(\{\alpha\},N_\epsilon) =
\bar{X}(\{\alpha\},N_\epsilon) \equiv \sum^{\infty}_{m,n}
\left(\frac{\alpha_s}{4 \pi}\right)^{m}
\left(\frac{\alpha_e}{4 \pi}\right)^{n}\,
\bar{X}_{mn}(N_\epsilon) \, .
\end{align}
\subsection{IR factorization at NNLO in the FDH scheme}
In {\scshape cdr}, the IR divergence structure of scattering amplitudes
including massive external partons has been investigated up to the
two-loop level in Ref.~\cite{Becher:2009kw}. Using a combination of
soft-collinear effective theory (SCET) (for an introduction see
e.\,g.\ Ref.~\cite{Becher:2014oda}) and heavy-quark effective theory
(HQET) (for an introduction see e.\,g.\ Ref.~\cite{Neubert:1993mb}) it has
been shown that amplitudes with an arbitrary number of massive and
massless legs factorize into a hard and a soft function, where the
latter depends on both massive and massless Wilson lines. For
amplitudes including massive partons, the corresponding IR anomalous
dimension has less constraints compared to the massless case and
additional color structures arise.
Starting from the {\scshape cdr}\ expression for the IR anomalous dimension, we
write the two-parton correlation terms of the respective quantity in
{\scshape fdh}\ as \begin{eqnarray} \label{eq:G2parmass} \bar{\mathbf{\Gamma}}\left( \{
\underline{p} \} , \{ \underline{m} \} ,\mu \right)
\Big|_{\text{2-parton}} &=& \sum_{(i,j)} \frac{\mathbf{T}_i
\cdot\mathbf{T}_j}{2} \bar{\gamma}_{\mbox{{\tiny cusp}}} (\{\alpha\})
\ln{\frac{\mu^2}{-s_{ij}}} + \sum_i \bar{\gamma}_i (\{\alpha\}) \nonumber
\\* && - \sum_{(IJ)} \frac{\mathbf{T}_I
\cdot\mathbf{T}_J}{2}\bar{\gamma}_{\mbox{{\tiny cusp}}} (\beta_{IJ},\{\alpha\}) +
\sum_I \bar{\gamma}_I (\{\alpha\}) \nonumber \\* &&+\sum_{(Ij)}
\frac{\mathbf{T}_I \cdot\mathbf{T}_j}{2} \bar{\gamma}_{\mbox{{\tiny cusp}}}
(\{\alpha\})\ln{\frac{m_I\, \mu}{-s_{Ij}}} \, , \end{eqnarray}
where the capital indices $I,J$ correspond to the massive partons and
the angle $\beta_{IJ}$ is defined as
\begin{equation}
\beta_{IJ} \mathrel{\mathop:}= \text{arcosh}\left(\frac{-s_{IJ}}{2\,m_{I}m_{J}}\right) \, .
\label{eq:betaIJ}
\end{equation}
For the definition of the color generators $\mathbf{T}_{i}$, of the
kinematic variable $s_{ij}$, and of the sets $\{ \underline{p} \}$,
$\{ \underline{m} \}$ we refer to \cite{Becher:2009kw}.
In Eq.~\eqref{eq:G2parmass}, the first line corresponds to
contributions from massless partons, already discussed in
Refs.~\cite{Gnendiger:2014nxa, Broggio:2015ata,Broggio:2015dga}; the
remainder is given by additional terms arising in the massive theory.
Suppressing the dependence on the couplings, the complete set of IR
anomalous dimensions in {\scshape fdh}/{\scshape dred}\ is given by
\begin{subequations}
\label{eq:IRanomDim}
\begin{align}
&\bar{\gamma}_{\mbox{{\tiny cusp}}},\qquad\phantom{(\beta)}
\bar{\gamma}_i\in\{\bar{\gamma}_{q},\bar{\gamma}_{g},\bar{\gamma}_{{\tilde{g}}}\},
\phantom{\Big|}
\label{eq:IRanomDimMassless}
\\
&\bar{\gamma}_{\mbox{{\tiny cusp}}}(\beta_{IJ}),\qquad
\bar{\gamma}_I\in\{\bar{\gamma}_{Q}\} \, ,
\label{eq:IRanomDimMassive}
\end{align}
\end{subequations}
where $\bar{\gamma}_{{\tilde{g}}}$ only appears in {\scshape dred}.
The quantities in the first line have been computed up to the two-loop
level in Refs.~\cite{Gnendiger:2014nxa,
Broggio:2015ata,Broggio:2015dga}; the values of
$\bar{\gamma}_{\mbox{{\tiny cusp}}}(\beta_{IJ})$ and $\bar{\gamma}_{Q}$
are so far unknown and will be given in Section~\ref{sec:ad}.
Since there is no difference between the IR anomalous dimensions appearing
both in {\scshape fdh}\ and {\scshape dred}, relation~\eqref{eq:G2parmass} also holds in {\scshape dred}.
In the massive theory, the IR anomalous dimension also
contains three-parton correlation terms which we write in {\scshape fdh}\ as
\begin{eqnarray}
\bar{\mathbf{\Gamma}}
\left( \{\underline{p}\}, \{\underline{m}\}, \mu \right)\Big|_{\text{3-partons}}
&=& i f^{abc} \sum_{(I,J,K)} \mathbf{T}_I^a\mathbf{T}_J^b\mathbf{T}_K^c\,
F_1\left(\beta_{IJ}, \beta_{JK}, \beta_{KI}\right)
\nonumber \\*
& &
+ i f^{abc} \sum_{(I,J)}\sum_{k} \mathbf{T}_I^a\mathbf{T}_J^b\mathbf{T}_k^c\,
f_2\left(\beta_{IJ},
\ln\frac{-\sigma_{Ik} \, v_I \cdot p_k}{-\sigma_{Jk} \, v_J \cdot p_k}
\right) \, ,
\label{eq:3part}
\end{eqnarray}
including the four-velocities of the massive partons
\begin{align}
v^\mu_I \mathrel{\mathop:}= \frac{p_I^\mu}{m_I} \, , \qquad\quad v_I^2 \equiv 1 \, .
\end{align}
In Refs.~\cite{Ferroglia:2009ep, Ferroglia:2009ii}, the functions
$F_1$ and $f_2$ are given for the case of {\scshape cdr}. Since in {\scshape fdh}\ these
functions do not receive evanescent contributions from the
$\epsilon$-scalar up to NNLO, Eq.~\eqref{eq:3part} is a
scheme-independent quantity at this order. Its value in {\scshape fdh}\ is
therefore the same as in {\scshape cdr}.
In analogy to the massless case \cite{Gnendiger:2014nxa,
Broggio:2015dga}, we subtract all IR divergences of QCD loop
amplitudes by means of a factor $\bar{\mathbf{Z}}$ which is given by a
path-ordered integral over $\bar{\mathbf{\Gamma}}$ (compare with
Eqs.~(2.8) and (2.12) of Ref.~\cite{Broggio:2015dga}). This
renormalization factor is given in the effective theory where the
heavy quarks have been integrated out. Hence, it is written in terms
of $\alpha_i$, the couplings defined in the massless theory. In the
massive case, however, we also need to take into account contributions
from heavy-quark loops. To reproduce the correct IR behavior of the
effective low-energy theory we therefore have to perform a matching of
the couplings between the full and the effective theory. For an
amplitude describing a process with $n$ external partons then the
following relation holds:
\begin{align}
\lim_{\epsilon\to 0}\
\bar{\mathbf{Z}}^{-1}(\{\alpha\})\,\Bigg[
\big|{\cal M}_n(\{\alpha\}^{f})\rangle
\Bigg]_{\alpha_{i}^{f}\to\,\zeta_{\alpha_i}\alpha_i} =\
\mbox{finite } .
\label{eq:ZmassiveFDH}
\end{align}
As mentioned above, $\alpha_i$ is a coupling in the effective theory,
meaning that the heavy quark flavors have been integrated out. It is
related to the corresponding coupling of the full theory via the
decoupling relation $\alpha_{i}^{f}=\zeta_{\alpha_i}\alpha_i$. Explicit
results for the decoupling constants in the {\scshape fdh}\ scheme will be given
in Section~\ref{sec:dec}.
\subsection{Mass renormalization of the $\epsilon$-scalar}
\label{sec:epsilonmass}
\begin{figure}
\begin{center}
\scalebox{.9}{
\begin{picture}(135,50)(0,25)
\DashLine(0,45)(35,45){4}
\DashLine(85,45)(120,45){4}
\DoubleArc[arrow](60,45)(25,0,180){2}
\DoubleArc[arrow](60,45)(25,180,360){2}
\Vertex(35,45){2}
\Vertex(85,45){2}
\end{picture}
}
\end{center}
\caption{\label{epsilontwopoint}
One-loop diagram that effectively generates an $\epsilon$-scalar mass
at the one-loop level. Massive quarks are depicted by double lines.}
\end{figure}
In the case of massive fermions there is no symmetry that protects the
propagator of the $\epsilon$-scalar from acquiring a mass term
$\propto m^{2}{\tilde{g}}^{\mu\nu}$ where $m$ is a fermion mass. As a
consequence, the $\epsilon$-scalar mass is effectively shifted away
from zero, even if the $\epsilon$-scalar is massless at the
tree-level. Therefore we have to introduce a mass counterterm $\delta
m^2_{\epsilon}$ in the Lagrangian to restore the initial 'on-shell'
condition of a vanishing $\epsilon$-scalar mass~\cite{Jack:1994rk}.
At the one-loop level there is only one diagram that effectively
generates a mass term in the $\epsilon$-scalar propagator, see
Fig.~\ref{epsilontwopoint}. To obtain the mass counterterm we need to
compute the full one-particle irreducible (1PI) two-point function of
the $\epsilon$-scalar, whose tensor structure is given by
\begin{align}
-i\tilde{\Pi}^{\mu\nu}\,
=-i\tilde{\Pi}\,p^2\,\tilde{g}^{\mu\nu}
=-i\Big(A+\frac{m^2}{p^2}B\Big)\,p^{2}\,\tilde{g}^{\mu\nu} \, ,
\end{align}
including the dimensionless quantities $A$ and $B$. The mass
counterterm can be extracted by writing the propagator of the
$\epsilon$-scalar as
\begin{align}
\frac{-i\tilde{g}_{\mu\nu}}{p^{2}\,\big(1+\tilde{\Pi}\big)
+\delta m^2_{\epsilon}}
=\frac{-i\tilde{g}_{\mu\nu}}{p^{2}\,\big(1+A\big)+m^2\,B+\delta m^2_{\epsilon}}\,.
\end{align}
In order to maintain the $\epsilon$-scalar massless we then require
\begin{equation}
\delta m^2_{\epsilon}
\mathrel{\mathop:}= -\,m^{2}\,B
=-\,a_e(m^{2})\,m^{2}\,N_{H}\,\Bigg[
\frac{2}{\epsilon}
+2
+\epsilon\Big(2+\frac{\pi^{2}}{6}\Big)
+\mathcal{O}(\epsilon^2)
\Bigg]
+\mathcal{O}(a^2)\, ,
\label{eq:epsMassCT}
\end{equation}
where $N_{H}$ denotes the number of heavy quark flavors and the
coupling is defined in Eq.~\eqref{eq:asdef}. As a consequence, any
time we encounter a massive loop diagram insertion as in
Fig.~\ref{epsilontwopoint}, we add the mass counterterm
\eqref{eq:epsMassCT} in order to impose the on-shell condition of a
massless $\epsilon$-scalar.
\subsection{Decoupling transformations}
\label{sec:dec}
The decoupling transformation needed in Eq.~\eqref{eq:ZmassiveFDH} is
well known for the gauge coupling. In order to extend it to $\alpha_e$
we apply the procedure described in Ref.~\cite{Chetyrkin:1997un} and
build an effective Lagrangian in which the heavy quark flavors have
been integrated out. As a consequence, the parameters and fields of
the effective theory are in general different from the ones of the
full theory. To relate the two theories we introduce decoupling
constants in the following way:
\begin{align}
g^{0,f}=\zeta^{0}_{g}\,g^{0}, \qquad
X^{0,f}=\sqrt{\zeta^{0}_{X}}\,X^{0} \, ,
\end{align}
where $g$ and $X$ stand for parameters and fields of the theory,
respectively. In this way we are able to relate the full and the
effective bare QCD Lagrangian in terms of the re-scaled parameters and
fields
\begin{equation}
\mathcal{L}^{f}\,(
g_{s}^{0,f},\,g_{e}^{0,f},\,{\hat{A}}^{0,f},\,{\tilde{A}}^{0,f},\psi^{0,f},
\dots)
=\mathcal{L}\,(
g_{s}^0,\,g_{e}^0,\,{\hat{A}}^0,\,{\tilde{A}}^0,\,\psi^{0},
\dots,\{\zeta^{0}_{g}\},\{\zeta_{X}^{0}\})\, .
\end{equation}
The decoupling constants can be obtained from a matching calculation.
For $\zeta_{{\hat{A}}}^{0}$ which is related to the gluon field decoupling,
for example, we get
\begin{subequations}
\label{eq:decgluFull}
\begin{align}
\frac{
-{\hat{g}}_{\mu\nu
}{p^2\left(1+\hat{\Pi}^{0,f}\right)}
&=
i\int {\rm d}^4x\, e^{i\, p x}\, \langle\,
T\, {\hat{A}}^{0,f}_{\mu}(x)\, {\hat{A}}^{0,f}_{\nu}(0)
\rangle
\\*
&=
i\, \zeta_{{\hat{A}}}^0 \int {\rm d}^4x\, e^{i\, p x}\,\langle\,
T\, {\hat{A}}^{0}_\mu(x)\, {\hat{A}}^{0}_\nu(0)
\rangle
= \zeta_{{\hat{A}}}^0\,\frac{
-{\hat{g}}_{\mu\nu
}{p^2\left(1+\hat{\Pi}^0\right)}\, ,\phantom{\Bigg|}
\label{eq:decglu}
\end{align}
\end{subequations}
where $\hat{\Pi}^{0}$ only contains light degrees of freedom and
$\hat{\Pi}^{0,f}$ receives virtual contributions from the heavy quarks.
From Eqs.~\eqref{eq:decgluFull} we then get
\begin{align}
\zeta_{{\hat{A}}}^0 = \frac{1+\hat{\Pi}^0}{1+\hat{\Pi}^{0,f}} \, .
\label{eq:decglu2}
\end{align}
Since the l.\,h.\,s\ does not depend on the kinematics of the process
it is possible to consider the special case $p=0$. The
renormalization of the decoupling constant is done in the usual way by
means of the gluon field renormalization constants in the effective
and the full theory:
$\zeta_{{\hat{A}}}=\bar{Z}^{\phantom{f}}_{{\hat{A}}}/\bar{Z}^{f}_{{\hat{A}}}\,\zeta_{{\hat{A}}}^0$.
The same method also applies to the decoupling of the
$\epsilon$-scalar field where, however, according to the discussion in
Sec.~\ref{sec:epsilonmass} a mass counterterm has to be added in order
to maintain the $\epsilon$-scalar massless. In fact, this counterterm
is even required to ensure that
\begin{align}
\zeta_{{\tilde{A}}}^{0}
=\frac{1+\tilde{\Pi}^0}{1+\tilde{\Pi}^{0,f}
+\delta m^2_{\epsilon}}\Bigg|_{p\to 0}
\label{eq:decglu3}
\end{align}
is properly defined.
For the calculations in this work we need the decoupling transformations
for $\alpha_{s}$ and $\alpha_{e}$ at the one-loop level which
can be obtained from a matching of the $g q\bar{q}$ and
${\tilde{g}} q\bar{q}$ vertices, in analogy to Eqs.~\eqref{eq:decgluFull}
\begin{equation}
\zeta^0_{g_{s}}
=\frac{1}{\zeta^0_{\psi}\sqrt{\zeta^0_{{\hat{A}}}}}\,
\frac{1+\Gamma_{{\hat{g}} q\bar{q}}^{0,f}}{1+\Gamma_{{\hat{g}} q\bar{q}}^{0}\phantom{\Big|}}\, ,\qquad\qquad
\zeta^0_{g_{e}}
=\frac{1}{\zeta^0_{\psi}\sqrt{\zeta_{{\tilde{A}}}^{0}}}\,
\frac{1+\Gamma_{{\tilde{g}} q\bar{q}}^{0,f}}{1+\Gamma_{{\tilde{g}} q\bar{q}}^{0}\phantom{\Big|}}\, .
\end{equation}
Since $\zeta_{\psi}^0$, $(\Gamma_{{\hat{g}} q\bar{q}}^{0,f}-\Gamma_{{\hat{g}} q\bar{q}}^{0})$, and
$(\Gamma_{{\tilde{g}} q\bar{q}}^{0,f}-\Gamma_{{\tilde{g}} q\bar{q}}^{0})$ are of
$\mathcal{O}(\alpha^2)$, the (bare) one-loop decoupling constants for
$g_{s}$ and $g_{e}$ are entirely given by $\zeta_{{\hat{A}}}^0$ and
$\zeta_{{\tilde{A}}}^0$, respectively. Using
$(\zeta^0_{g_{s}})^2=\zeta^0_{\alpha_{s}}$ and $\zeta_{\alpha_{s}}=
\bar{Z}^{\phantom{f}}_{\alpha_s}/\bar{Z}^{f}_{\alpha_s}\,\zeta_{\alpha_{s}}^0$
and similar for the evanescent coupling we finally obtain
\begin{subequations}
\begin{align}
\zeta_{\alpha_s}
&=1+\Big(\frac{\alpha_{s}}{4\pi}\Big)\,N_{H}\,
\frac{2}{3}\,\text{ln}\left(\frac{\mu^{2}}{m^{2}}\right)
+\mathcal{O}(\alpha^2) \, ,\\
\zeta_{\alpha_e}
&=1+\Big(\frac{\alpha_{e}}{4\pi}\Big)\,N_{H}\,
\text{ln}\left(\frac{\mu^{2}}{m^{2}}\right)
+\mathcal{O}(\alpha^2)
\end{align}
\end{subequations}
for the renormalized decoupling constants of $\alpha_s$ and $\alpha_e$.
\subsection{Field and mass renormalization of the heavy quarks}
\label{sec:onshell}
\begin{figure}[t]
\begin{center}
\scalebox{.9}{
\begin{picture}(135,50)(0,0)
\DoubleLine[arrow](-10,0)(20,0){2}
\DoubleLine[arrow](20,0)(100,0){2}
\DoubleLine[arrow](100,0)(130,0){2}
\DashCArc(60,0)(40,0,62){4}
\DashCArc(60,0)(40,118,180){4}
\DoubleArc[arrow](60,36)(18,0,180){2}
\DoubleArc[arrow](60,36)(18,180,360){2}
\Vertex(20,0){2}
\Vertex(100,0){2}
\Vertex(42,36){2}
\Vertex(78,36){2}
\end{picture}
\qquad\qquad
\begin{picture}(135,50)(0,0)
\DoubleLine[arrow](-10,0)(20,0){2}
\DoubleLine[arrow](20,0)(100,0){2}
\DoubleLine[arrow](100,0)(130,0){2}
\DashCArc(60,0)(40,0,180){4}
\Vertex(20,0){2}
\Vertex(100,0){2}
\Text(60,32.5)[b]{\scalebox{2}{\ding{53}}}
\end{picture}
}
\end{center}
\caption{\label{epsiloninsertion}
Sample two-loop contributions to the field renormalization of the heavy quark.
The diagram on the r.\,h.\,s.\ shows the insertion of the mass
counterterm $\delta m_{\epsilon}^2$.}
\end{figure}
To obtain UV-finite Green functions in the {\scshape fdh}\ scheme we need to
perform a renormalization of the heavy quark field and mass, where the
corresponding renormalization constants are defined by
\begin{align}
\psi^0=\sqrt{\bar{Z}_{2,h}}\,\psi,\qquad\quad
m^0=\bar{Z}_m\, m \, .
\end{align}
Extending the standard {\scshape cdr}\ procedure for obtaining renormalization constants
in the on-shell (OS) scheme, we write the 1PI self-energy of the heavy quark
in {\scshape fdh}\ as
\begin{align}
\bar{\Sigma}(p,m,N_\epsilon)
= m\,\bar{\Sigma}_{1}(p^2,m,N_\epsilon)
+ (\slashed{p}-m)\,\bar{\Sigma}_{2}(p^2,m,N_\epsilon)\, .
\end{align}
The renormalization constants are then given by
\begin{subequations}
\begin{align}
\big(\bar{Z}_{2,h}\big)^{-1}
&=1+2m^2 \frac{\partial}{\partial p^2}\bar{\Sigma}_{1}\big|_{p^2=m^2}
+\bar{\Sigma}_{2}\big|_{p^2=m^2}\, ,
\label{eq:Z2OS}
\\
\bar{Z}_m
&=1+\bar{\Sigma}_{1}\big|_{p^2=m^2}\, .
\label{eq:ZmOS}
\end{align}
\end{subequations}
To obtain their values we calculated the quantities
$\bar{\Sigma}_{1}$ and $\bar{\Sigma}_{2}$ up to the two-loop level,
with sample diagrams shown in Fig.~\ref{epsiloninsertion}. One point
of major importance is that apart from genuine two-loop diagrams we
have to include contributions originating from UV
(sub)renormalization. This in particular comprises the mass
counterterm for the $\epsilon$-scalar given in
Eq.~\eqref{eq:epsMassCT}, see the r.\,h.\,s.\ of
Fig.~\ref{epsiloninsertion}. In terms of the bare couplings we then
get
\begin{align}\label{z2os}
\bar{Z}_{2,h}&=
1+
a_{s}(m^2)\,C_{F}\,\Bigg[
-\frac{3}{\epsilon}
-4
-\epsilon\Big(8+\frac{\pi^2}{4}\Big)
\Bigg]
+a_e(m^2 )\,C_{F}\,N_\epsilon\,\Bigg[
-\frac{1}{2\epsilon}
-\frac{1}{2}
-\epsilon\Big(\frac{1}{2}+\frac{\pi^2}{24}\Big)
\Bigg]
\nonumber\\&\quad
+a_s^2(m^2)\,\Bigg\{
C_{F}^{2}\,\Bigg[
\frac{9}{2\epsilon^{2}}
+\frac{51}{4\epsilon}
+\frac{433}{8}
-\frac{49}{4}\pi^{2}
+16\pi^{2}\ln(2)
-24\zeta(3)
\Bigg]
\nonumber\\&\qquad\qquad\qquad
+C_{A}C_{F}\,\Bigg[
-\frac{11}{2\epsilon^2}
-\frac{101}{4\epsilon}
-\frac{803}{8}
+\frac{49}{12}\pi^2
-8\pi^2\ln(2)
+12\zeta(3)
\nonumber\\&\qquad\qquad\qquad\qquad\qquad\
+N_\epsilon\,\Big(
\frac{1}{4\epsilon ^2}
+\frac{11}{8\epsilon}
+\frac{5}{24}\pi^2
+\frac{81}{16}
\Big)
\Bigg]
\nonumber\\&\qquad\qquad\qquad
+C_{F}N_{F}\,\Bigg[
\frac{1}{\epsilon^2}
+\frac{9}{2\epsilon}
+\frac{59}{4}
+\frac{5}{6}\pi^2
\Bigg]
+C_{F}N_{H}\,\Bigg[
\frac{2}{\epsilon^2}
+\frac{19}{6\epsilon}
+\frac{1139}{36}
-\frac{7}{3}\pi^2
\Bigg]
\Bigg\}
\nonumber\\&\quad
+a_e^2(m^2)\,N_\epsilon\,\Bigg\{
C_{F}^{2}\,\Bigg[
\frac{1}{\epsilon^2}
+\frac{2}{\epsilon}
+\frac{\pi^2}{2}
-3
+N_\epsilon\Big(
-\frac{1}{8\epsilon^2}
-\frac{3}{16\epsilon}
-\frac{13}{48}\pi^2
+\frac{91}{32}
\Big)
\Bigg]
\nonumber\\&\qquad\qquad\qquad\quad
+C_{A}C_{F}\,\Bigg[
\Big(
-\frac{1}{2\epsilon^2}
-\frac{1}{\epsilon}
-\frac{\pi^2}{4}
+\frac{3}{2}
\Big)\Big(
1-\frac{N_\epsilon}{2}
\Big)\Bigg]
\nonumber\\&\qquad\qquad\qquad\quad
+C_{F}N_{F}\,\Bigg[
\frac{1}{4\epsilon^2}
+\frac{7}{8\epsilon}
+\frac{21}{16}
+\frac{5}{24}\pi^2
\Bigg]
+C_{F}N_{H}\,\Bigg[
\frac{1}{4\epsilon^2}
+\frac{7}{8\epsilon}
-\frac{3}{16}
+\frac{\pi^2}{24}
\Bigg]
\Bigg\}
\nonumber\\&\quad
+a_{s}(m^2)\,a_{e}(m^2)\,N_\epsilon\,\Bigg\{
C_{F}^{2}\,\Bigg[
\frac{3}{2\epsilon}
+\frac{47}{4}
-\pi^2
\Bigg]
+C_{A}C_{F}\,\Bigg[
-\frac{9}{4\epsilon}
-\frac{77}{8}
+\frac{\pi^2}{6}
\Bigg]
\Bigg\}
+\mathcal{O}(a^3) \, .
\end{align}
For later purposes it is convenient to introduce a mass counterterm
$\delta m=m-m^{0}=m\,(1-\bar{Z}_{m})$ for the heavy quarks.
Using Eq.~\eqref{eq:ZmOS}, a direct calculation of $\bar{\Sigma}_{1}$
yields
\begin{align}\label{eq:epsMassRen}
\frac{\delta m}{m}&=
a_{s}(m^2)\,C_{F}\,\Bigg[
\frac{3}{\epsilon}
+4
+\epsilon\Big(8+\frac{\pi^2}{4}\Big)
\Bigg]
+a_e(m^2 )\,C_{F}\,N_\epsilon\,\Bigg[
\frac{1}{2\epsilon}
+\frac{1}{2}
+\epsilon\Big(\frac{1}{2}+\frac{\pi^2}{24}\Big)
\Bigg]
\nonumber\\&\quad
+a_s^2(m^2)\,\Bigg\{
C_{F}^{2}\,\Bigg[
- \frac{9}{2\epsilon^{2}}
-\frac{45}{4\epsilon}
-\frac{199}{8}
+\frac{17}{4}\pi^{2}
-8\pi^{2}\ln(2)
+12\zeta(3)
\Bigg]
\nonumber\\&\qquad\qquad\qquad
+C_{A}C_{F}\,\Bigg[
\frac{11}{2\epsilon^2}
+\frac{91}{4\epsilon}
+\frac{605}{8}
-\frac{5}{12}\pi^2
+4\pi^2\ln(2)
-6\zeta(3)
\nonumber\\&\qquad\qquad\qquad\qquad\qquad\
+N_\epsilon\,\Big(
-\frac{1}{4\epsilon ^2}
-\frac{9}{8\epsilon}
-\frac{5}{24}\pi^2
-\frac{63}{16}
\Big)
\Bigg]
\nonumber\\&\qquad\qquad\qquad
+C_{F}N_{F}\,\Bigg[
-\frac{1}{\epsilon^2}
-\frac{7}{2\epsilon}
-\frac{45}{4}
-\frac{5}{6}\pi^2
\Bigg]
+C_{F}N_{H}\,\Bigg[
- \frac{1}{\epsilon^2}
-\frac{7}{2\epsilon}
-\frac{69}{4}
+\frac{7}{6}\pi^2
\Bigg]
\Bigg\}
\nonumber\\&\quad
+a_e^2(m^2)\,N_\epsilon\,\Bigg\{
C_{F}^{2}\,\Bigg[
-\frac{1}{\epsilon^2}
-\frac{3}{\epsilon}
+\frac{\pi^2}{6}
-6
+N_\epsilon\Big(
\frac{1}{8\epsilon^2}
+\frac{13}{16\epsilon}
-\frac{11}{48}\pi^2
+\frac{75}{32}
\Big)
\Bigg]
\nonumber\\&\qquad\qquad\qquad\quad
+C_{A}C_{F}\,\Bigg[
\Big(
\frac{1}{2\epsilon^2}
+\frac{3}{2\epsilon}
-\frac{\pi^2}{12}
+3
\Big)\Big(
1-\frac{N_\epsilon}{2}
\Big)\Bigg]
\nonumber\\&\qquad\qquad\qquad\quad
+C_{F}N_{F}\,\Bigg[
-\frac{1}{4\epsilon^2}
-\frac{5}{8\epsilon}
-\frac{11}{16}
-\frac{5}{24}\pi^2
\Bigg]
\nonumber\\&\qquad\qquad\qquad\quad
+C_{F}N_{H}\,\Bigg[
-\frac{1}{4\epsilon^2}
-\frac{5}{8\epsilon}
-\frac{3}{16}
-\frac{\pi^2}{24}
\Bigg]
\Bigg\}
\nonumber\\&\quad
+a_{s}(m^2)\,a_{e}(m^2)\,N_\epsilon\,\Bigg\{
C_{F}^{2}\,\Bigg[
\frac{3}{2\epsilon}
+\frac{23}{4}
-\pi^2
\Bigg]
+C_{A}C_{F}\,\Bigg[
\frac{3}{4\epsilon}
+\frac{11}{8}
+\frac{\pi^2}{2}
\Bigg]
\Bigg\}
+\mathcal{O}(a^3) \, .
\end{align}
up to the two-loop level. The pure $\alpha_s$ terms for $N_\epsilon=0$
correspond to the {\scshape cdr}\ result.
\subsection{Field renormalization of the light quarks}
In analogy to the previous section we determine the field renormalization of
the light quark fields where the corresponding renormalization constant is
in the following denoted by $\bar{Z}_{2,l}$.
As in the case of heavy quarks, $\bar{Z}_{2,l}$ receives contributions from
heavy quark loops, see Fig.~\ref{lightZ2}.
\begin{figure}
\begin{center}
\scalebox{.9}{
\begin{picture}(135,50)(0,0)
\Line[arrow](-10,0)(20,0)
\Line[arrow](20,0)(100,0)
\Line[arrow](100,0)(130,0)
\GlueArc(60,0)(40,0,62){4}{5}
\GlueArc(60,0)(40,118,180){4}{5}
\DoubleArc[arrow](60,40)(18,0,180){2}
\DoubleArc[arrow](60,40)(18,180,360){2}
\Vertex(20,0){2}
\Vertex(100,0){2}
\Vertex(42,37){2}
\Vertex(78,37){2}
\end{picture}
\qquad\qquad
\begin{picture}(135,50)(0,0)
\Line[arrow](-10,0)(20,0)
\Line[arrow](20,0)(100,0)
\Line[arrow](100,0)(130,0)
\DashCArc(60,0)(40,0,62){4}
\DashCArc(60,0)(40,118,180){4}
\DoubleArc[arrow](60,36)(18,0,180){2}
\DoubleArc[arrow](60,36)(18,180,360){2}
\Vertex(20,0){2}
\Vertex(100,0){2}
\Vertex(42,36){2}
\Vertex(78,36){2}
\end{picture}
}
\end{center}
\caption{\label{lightZ2}
Two-loop contributions to the field renormalization of the light quark.}
\end{figure}
However, there is no one-loop contribution since in dimensional regularization
all corresponding loop integrals are scaleless. This also implies that up to the
two-loop level there is no contribution from the $\epsilon$-scalar mass
counterterm. The explicit calculation then yields for the field renormalization
of the light quark in the {\scshape fdh}\ scheme
\begin{equation}
\bar{Z}_{2,l}=1+
C_{F}N_{H}\Bigg[
a_s^2(m^2)\Big(
\frac{1}{2\epsilon }
-\frac{5}{12}
\Big)
+a^2_e(m^2)\,N_\epsilon\,\Big(
-\frac{1}{4\epsilon^2}
+\frac{3}{8\epsilon}
-\frac{13}{16}
-\frac{\pi^2}{24}
\Big)
\Bigg]
+\mathcal{O}(a^3) \, .
\end{equation}
As for the mass counterterm, the pure $\alpha_s$ terms are of course
not new.
\section{IR anomalous dimensions in the massive case}
\label{sec:ad}
The aim of this section is to provide all so far unknown IR anomalous
dimensions present in the general IR factorization
formula~\eqref{eq:G2parmass}, i.\,e.\ $\bar{\gamma}_Q$ and
$\bar{\gamma}_{\text{cusp}}(\beta)$. As in the massless
case~\cite{Broggio:2015dga}, for this we use the SCET framework.
\subsection{Scheme dependence of the heavy-to-light soft function and
$\gamma_{Q}$ }
\label{sec:gQ}
In Ref.~\cite{Gao:2012ja}, it has been shown that the top quark decay
factorizes into regions where only soft radiation and (or) radiation
collinear to the massless partons are present. More precisely, the
factorization consists of a hard function whose renormalization
group equation (RGE) depends on the heavy-quark anomalous dimension, a
quark jet function, and a soft function. In {\scshape cdr}, the jet and soft
functions have been calculated up to the two-loop level in
Refs.~\cite{Becher:2006qw} and \cite{Becher:2005pd}, respectively. In
{\scshape fdh}, so far only the jet function is known \cite{Broggio:2015dga}.
The general relation between the corresponding IR anomalous dimensions
is given by
\begin{align}\label{gamQ}
\gamma_{Q}^{{\rm \scriptscriptstyle RS}}=\gamma_{S}^{{\rm \scriptscriptstyle RS}}+\gamma_{J}^{{\rm \scriptscriptstyle RS}}-\gamma_{q}^{{\rm \scriptscriptstyle RS}}\, ,
\end{align}
where $\gamma_{S}^{{\rm \scriptscriptstyle RS}}$ and $\gamma_{J}^{{\rm \scriptscriptstyle RS}}$ are the ({\scshape rs}-dependent)
anomalous dimensions of the soft and
jet function. Eq.~(\ref{gamQ}) is a direct consequence of the fact
that the RGE of the factorization formula does not depend on the
factorization scale. The values of
$\bar{\gamma}_{J}=\gamma_{J}^{\text{{\scshape fdh}/{\scshape dred}}}$ and $\bar{\gamma}_{q} =
\gamma_{q}^{\text{{\scshape fdh}/{\scshape dred}}}$ have been calculated in
Ref.~\cite{Broggio:2015dga} up to the two-loop level. In order to
obtain $\bar{\gamma}_Q = \gamma_{Q}^{\text{{\scshape fdh}/{\scshape dred}}}$ we therefore have
to compute $\bar{\gamma}_S = \gamma_{S}^{\text{{\scshape fdh}/{\scshape dred}}}$.
Extending the approach of Ref.~\cite{Becher:2005pd}, we define the
scheme-dependent (bare) soft function as
\begin{align}
S_{\text{bare}}^{{\rm \scriptscriptstyle RS}}\Big( \ln\frac{\Omega}{\mu},\mu \Big)
\mathrel{\mathop:}= \int_0^{\Omega} d\omega\,
\langle b_v|\,\bar h_v\,\delta(\omega+in\cdot D)\,h_v\,|b_v\rangle \,,
\end{align}
where $h_v$ are effective quark fields in
HQET~(see e.\,g.\ Ref.~\cite{Neubert:1993mb}),
$b_v$ are on-shell $b$-quark states with velocity $v$,
and $n$ is a light-like 4-vector with $n\cdot v=1$ and $n^2=0$.
The normalization is fixed by $\langle b_v|\,\bar h_v\,h_v\,|b_v\rangle=1$.
For explicit calculations it is useful to express the soft function as a contour
integral
\begin{align}\label{eq:softdisp}
S_{\text{bare}}^{{\rm \scriptscriptstyle RS}}\Big( \ln\frac{\Omega}{\mu},\mu \Big)
= \frac{1}{2\pi i} \oint\limits_{|\omega|=\Omega} d\omega\,
\langle b_v|\,\bar h_v\,\frac{1}{\omega+in\cdot D+i0}\,h_v\,|b_v\rangle
=\frac{1}{2\pi i} \oint\limits_{|\omega|
=\Omega} d\omega\,\mathcal{S}_{\text{bare}}^{{\rm \scriptscriptstyle RS}}\Big(\omega\Big)
\end{align}
and to work in Laplace space
\begin{align}\label{soft:Lap}
s_\text{bare}^{{\rm \scriptscriptstyle RS}}(\Omega)
\mathrel{\mathop:}= \int_0^\infty d \omega\,
\exp\left(-\frac{\omega}{\Omega\,e^{\gamma_E}}\right)
\frac{1}{\pi}\text{Im}\Big{[}\mathcal{S}_{\text{bare}}^{{\rm \scriptscriptstyle RS}}(\omega)\Big{]}\, .
\end{align}
Since $h_v$ and $b_v$ are Heisenberg fields, the usual perturbative expansion
results in loop diagrams contributing to the heavy quark propagator.
As in the massless case, the scheme dependence is related to the
UV singularities of such diagrams.
\begin{figure}[t]
\begin{center}
\scalebox{.9}{
\begin{picture}(135,50)(-7,0)
\DoubleLine[arrow](-10,0)(20,0){2}
\DoubleLine[arrow](20,0)(100,0){2}
\DoubleLine[arrow](100,0)(130,0){2}
\GlueArc(60,0)(40,0,62){4}{5}
\GlueArc(60,0)(40,118,180){4}{5}
\DashCArc(60,40)(18,0,360){4}
\Vertex(20,0){2}
\Vertex(100,0){2}
\Vertex(42,37){2}
\Vertex(78,37){2}
\Text(20,-7.5)[b]{\scalebox{2}{\ding{53}}}
\end{picture}
\qquad\
\begin{picture}(135,50)(-7,0)
\DoubleLine[arrow](-10,0)(20,0){2}
\DoubleLine[arrow](20,0)(100,0){2}
\DoubleLine[arrow](100,0)(130,0){2}
\GlueArc(60,0)(40,0,62){4}{5}
\GlueArc(60,0)(40,118,180){4}{5}
\DashCArc(60,40)(18,0,360){4}
\Vertex(20,0){2}
\Vertex(100,0){2}
\Vertex(42,37){2}
\Vertex(78,37){2}
\Text(100,-7.5)[b]{\scalebox{2}{\ding{53}}}
\end{picture}
\qquad\
\begin{picture}(135,50)(-7,0)
\DoubleLine[arrow](-10,0)(20,0){2}
\DoubleLine[arrow](20,0)(60,0){2}
\DoubleLine[arrow](60,0)(100,0){2}
\DoubleLine[arrow](100,0)(130,0){2}
\GlueArc(60,0)(40,0,62){4}{5}
\GlueArc(60,0)(40,118,180){4}{5}
\DashCArc(60,40)(18,0,360){4}
\Vertex(20,0){2}
\Vertex(100,0){2}
\Vertex(42,37){2}
\Vertex(78,37){2}
\Text(60,-7.5)[b]{\scalebox{2}{\ding{53}}}
\end{picture}
}
\end{center}
\caption{\label{fig:massivesoft} Evanescent two-loop contributions to
the heavy-to-light soft anomalous dimension in the {\scshape fdh}\ scheme. The
crosses denote the insertion of the operator $(\omega+in\cdot
D+i0)^{-1}$.}
\end{figure}
At the one-loop level there are no evanescent contributions since the
$\epsilon$-scalar does not couple to heavy quark lines, see also
Ref.~\cite{Broggio:2015dga}. There are exactly three diagrams that
induce a scheme dependence of the soft function at the two-loop
level. They are shown in Fig.~\ref{fig:massivesoft}. For the
explicit computation we generated the diagrams with
QGRAF~\cite{Nogueira:1991ex} and applied a tensor reduction of the
integrals with Reduze~2~\cite{vonManteuffel:2012np}, where the master
integrals needed in {\scshape fdh}\ are identical to the ones of
{\scshape cdr}\ given in Ref.~\cite{Becher:2005pd}.
In {\scshape fdh}\ we then get up to the two-loop level
\begin{align}\label{sbare}
\bar{s}_\text{bare}(\Omega)
=
1 + a_{s}(\Omega^2)\, C_F &\Bigg[
\!-\! \frac{2}{\epsilon^2}
\!+\! \frac{2}{\epsilon}
\!-\! \frac{5}{6}\pi^2
+ \epsilon\Big(
\frac{5}{6}\pi^2
- \frac{14}{3}\,\zeta_3
\Big)
\!-\!\epsilon^2\Big(
\frac{193}{720}\pi^4
-\frac{14}{3}\,\zeta_3
\Big)
+{\cal O}(\epsilon^3)
\Bigg]
\nonumber\\*
+\, a^2_{s}(\Omega^2)
\, C_F&
\Bigg[ C_F \bar{K}_F(\epsilon) + C_A \bar{K}_A(\epsilon)
+ \frac{1}{2} N_F \bar{K}_f(\epsilon)
\Bigg]+{\cal O}(a^3)\,,
\end{align}
with
\begin{subequations}
\label{eq:sbareCoeff}
\begin{align}
\bar{K}_{F}(\epsilon)&=
\frac{2}{\epsilon ^4}
-\frac{4}{\epsilon ^3}
+\frac{2+\frac{5 \pi ^2}{3}}{\epsilon ^2}
+\frac{-\frac{10}{3}\pi^2+\frac{28}{3}\zeta(3)}{\epsilon }
+\frac{5}{3}\pi^2-\frac{56}{3}\zeta(3)+\frac{53}{60}\pi^4\, , \\
\bar{K}_{A}(\epsilon)&=
-\frac{11}{6 \epsilon ^3}
+\frac{-\frac{1}{18}+\frac{\pi^2}{6}}{\epsilon ^2}
+\frac{-\frac{55}{27}-\frac{37}{12}\pi^2+9\zeta(3)}{\epsilon }
-\frac{326}{81}-\frac{41}{12}\pi^2-\frac{437}{9}\zeta(3)+\frac{107}{180}\pi^4
\nonumber\\*
&\quad\,+N_{\epsilon}\Big{(} \frac{1}{12 \epsilon ^3}+
\frac{1}{18 \epsilon ^2}
+\frac{\frac{1}{27}+\frac{\pi ^2}{8}}{\epsilon }
+\frac{2}{81}+\frac{\pi^2}{12}+\frac{25}{18}\zeta(3)\Big)\, ,\\
%
\bar{K}_{f}(\epsilon)&=
\frac{2}{3 \epsilon ^3}
-\frac{2}{9 \epsilon ^2}
+\frac{-\frac{4}{27}+\pi^2}{\epsilon }
-\frac{8}{81}-\frac{\pi^2}{3}+\frac{100}{9}\zeta(3) \,.
\end{align}
\end{subequations}
Taking the limit $N_\epsilon\to 0$ in Eq.~\eqref{sbare} we obtain
the {\scshape cdr}\ result which is in agreement with the one given in
Ref.~\cite{Becher:2005pd}.
As for the quark and gluon jet functions~\cite{Broggio:2015dga},
all divergences of the soft function can be removed multiplicatively
by means of a $Z$ factor
\begin{equation}\label{eq:rensoft}
s_{\text{sub}}^{{\rm \scriptscriptstyle RS}}(\Omega,\mu)=
Z_{S}^{{\rm \scriptscriptstyle RS}}(\Omega,\mu)\,
s_{\text{bare}}^{{\rm \scriptscriptstyle RS}}(\Omega)\, .
\end{equation}
To relate $Z_{S}^{{\rm \scriptscriptstyle RS}}(\Omega,\mu)$ with $\gamma_{S}^{{\rm \scriptscriptstyle RS}}$ we compare the RGE
of the soft function,
\begin{align}
\label{rge:softZ}
\frac{d}{d\ln\mu}\, s_{\text{sub}}^{{\rm \scriptscriptstyle RS}}(\Omega,\mu)=
\Bigg[
\Big(\frac{d}{d\ln\mu}\,Z^{{\rm \scriptscriptstyle RS}}_{S}(\Omega,\mu)\Big)\,
\Big(Z^{{\rm \scriptscriptstyle RS}}_{S}(\Omega,\mu)\Big)^{-1}
\Bigg]\,
s_{\text{sub}}^{{\rm \scriptscriptstyle RS}}(\Omega,\mu)\, \, ,
\end{align}
with the RGE written in terms of $\gamma_{S}^{{\rm \scriptscriptstyle RS}}$,
\begin{align}
\label{rge:softG}
\frac{d}{d \ln \mu}s_{\text{sub}}^{{\rm \scriptscriptstyle RS}}(\Omega,\mu)&=
\Bigg[
C_F\,\gamma_{\text{cusp}}^{{\rm \scriptscriptstyle RS}}\, L_\Omega
- 2\gamma_{S}^{{\rm \scriptscriptstyle RS}}
\Bigg]\,
s_{\text{sub}}^{{\rm \scriptscriptstyle RS}}(\Omega,\mu) \, ,
\end{align}
where $L_\Omega= \ln(\Omega^{2}/\mu^2)$ and the cusp anomalous dimension is
known from the massless
case~\cite{Gnendiger:2014nxa, Broggio:2015ata,Broggio:2015dga}.
In {\scshape fdh}, the factor $\bar{Z}_{S}$ is given by
\begin{align}
\ln\bar{Z}_{S}
&=\Big(\frac{\alpha_{s}}{4\pi}\Big)\Bigg[
\frac{C_F\,\bar{\gamma}^{\text{cusp}}_{10}}{2\epsilon^{2}}
-\frac{1}{\epsilon}\Bigg(\frac{C_F\,\bar{\gamma}^{\text{cusp}}_{10}}{2}\,
L_\Omega-\bar{\gamma}_{10}^{S}\Bigg)\Bigg]
\nonumber\\
&\quad+\Big{(}\frac{\alpha_{s}}{4\pi}\Big{)}^{2}
\Bigg[\,
-\frac{3\,C_F\,\bar{\gamma}^{\text{cusp}}_{10}\,\bar\beta^s_{20}}{8\epsilon^{3}}
+\frac{\bar\beta^s_{20}}{2\,\epsilon^{2}}
\Bigg(\frac{C_F\,\bar{\gamma}^{\text{cusp}}_{10}}{2}\,L_\Omega
-\bar{\gamma}_{10}^{S}\Bigg)
+\frac{C_F\,\bar{\gamma}^{\text{cusp}}_{20}}{8\,\epsilon^{2}}
\nonumber\\
&\qquad\qquad\quad
-\frac{1}{2\,\epsilon}
\Bigg(\frac{C_F\,\bar{\gamma}^{\text{cusp}}_{20}}{2}\,
L_\Omega-\bar{\gamma}_{20}^{S}\Bigg)\Bigg]
+{\cal O}(\alpha^3)
\label{eq:logZS}
\end{align}
and the coefficients of the $\beta$ function
can be found e.\,g.\ in Ref.~\cite{Broggio:2015dga}.
Imposing minimal subtraction with $N_\epsilon$ as an independent quantity
we can read off the soft anomalous dimension
\begin{align}
\bar{\gamma}_{S
=\ &
\Big{(}\frac{\alpha_{s}}{4\pi}\Big{)}
\big( -2 C_{F}\big)
\nonumber\\*
&+\Big{(}\frac{\alpha_{s}}{4\pi}\Big{)}^{2}\Bigg\{
C_{A}C_{F}\Bigg[
\frac{110}{27}
+\frac{\pi^2}{18}
-18 \zeta(3)
-N_\epsilon\Big(\frac{2}{27}-\frac{\pi^2}{36}\Big)\Bigg]
+C_{F}N_{F}\Bigg[
\frac{4}{27}
+\frac{\pi^2}{9}\Bigg] \Bigg\}
\nonumber\\*
&+\mathcal{O}(\alpha^3) \, ,
\end{align}
which is scheme independent at the one-loop level.
Apart from $\bar{\gamma}_{S}$ it is also possible to extract the already known
values of the cusp anomalous dimension as well as the $\beta$ functions in the
{\scshape fdh}\ scheme,
which provides a strong consistency check on the applied procedure.
Using the obtained results together with Eq.~\eqref{gamQ} we then find
\begin{align}\label{anomres}
\bar{\gamma}_{Q
=\ &
\Big(\frac{\alpha_{s}}{4\pi}\Big)
\big(-2C_{F}\big)\nonumber\\*
&+\Big(\frac{\alpha_{s}}{4\pi}\Big{)}^{2}\Bigg\{
C_{A} C_{F}\Bigg[
-\frac{98}{9}
+\frac{2}{3}\pi ^2
-4\zeta(3)
+\frac{8}{9}N_\epsilon\Bigg]
+C_{F} N_{F}\,\frac{20}{9}\Bigg\}
+\mathcal{O}(\alpha^3)
\end{align}
for the IR anomalous dimension of the heavy quarks in the {\scshape fdh}\ scheme.
Like $\bar{\gamma}_S$, at NLO it does not depend on $N_\epsilon$ and is therefore
scheme independent, as already found in Ref.~\cite{Catani:2000ef}.
However, at NNLO it receives {\scshape rs}-dependent contributions.
Eq.~\eqref{anomres} is the main result of this section. However, for the sake of
completeness we give the result of the finite and scheme-independent soft function
by setting $N_\epsilon=2\epsilon$ and taking the subsequent limit $\epsilon\to 0$
\begin{align}
s_{\text{fin}}(\Omega,\mu) &=
\lim_{N_\epsilon,\epsilon\to\,0} s_{\text{sub}}^{{\rm \scriptscriptstyle RS}}(\Omega,\mu)
= 1+
\Big(\frac{\alpha_{s}}{4\pi}\Big) \Bigg[
-C_F\gamma_{10}^{\text{cusp}}\frac{L_\Omega^{2}}{4}
+\gamma_{10}^{S}L_\Omega+c_{1}^{S} \Bigg]
\nonumber\\
&\quad +\Big{(}\frac{\alpha_{s}}{4\pi}\Big{)}^{2} \Bigg[
C_F^{2} (\gamma_{10}^{\text{cusp}})^{2}\frac{L_\Omega^{4}}{32}
+\Big(
2\gamma_{10}^{S}\big(\gamma_{10}^{S}-\beta_{20}^{s}\big)
-C_F(\gamma_{20}^{\text{cusp}}+\gamma_{10}^{\text{cusp}}c_{1}^{S})
\Big) \frac{L_\Omega^{2}}{4}\phantom{\Bigg|}
\nonumber\\*& \qquad\qquad\quad\ \
+\Big(\beta^s_{20}-3\gamma_{10}^{S}\Big)\,
C_F\gamma_{10}^{\text{cusp}}\frac{L_\Omega^{3}}{12}
+\Big(
c_{1}^{S}\big(\gamma_{10}^{S}-\beta_{20}^{s}\big)
+\gamma_{20}^{S}
\Big)L_\Omega+c_{2}^{S}\Bigg]\, ,
\end{align}
with
\begin{subequations}
\begin{align}
c_{1}^{S}&=C_{F}\Big{(} -\frac{5\pi^{2}}{6}\Big{)}\, ,\\
c_{2}^{S}&=C^2_{F}\Big{(} \frac{25 \pi ^4}{72} \Big{)}
+C_{F}C_{A}\Big(
-\frac{326}{81}
-\frac{233 \pi ^2}{36}
-\frac{283 \zeta (3)}{9}
+\frac{107 \pi ^4}{180}\Big)\nonumber\\
&\quad+C_{F}N_F\Big(
-\frac{4}{81}+\frac{7}{18}\pi^2+\frac{22}{9}\zeta(3)\Big) \, .
\end{align}
\end{subequations}
This result is in agreement with the one given in Ref.~\cite{Becher:2005pd}.
\subsection{Determination of $\bar{ \gamma}_{\rm cusp}(\beta)$
}
\label{sec:gC}
\begin{figure}[t]
\begin{center}
\scalebox{.9}{
\begin{picture}(135,65)(0,10)
\DoubleLine[arrow](0,65)(50,65){2}
\DoubleLine[arrow](50,65)(100,65){2}
\DoubleLine(0,75)(0,55){3}
\Gluon( 50, 0)( 50, 65){5}{6}
\Vertex( 50, 65){2}
\Text(90,73)[l]{\scalebox{1.11}{$p_I$}}
\LongArrow( 60, 45)( 60, 25)
\Text( 65, 35)[l]{\scalebox{1.11}{$k\rightarrow 0$}}
\end{picture}
\quad
\begin{picture}(135,65)(0,10)
\DoubleLine[arrow](0,65)(50,65){2}
\DoubleLine[arrow](50,65)(100,65){2}
\DoubleLine(0,75)(0,55){3}
\DashLine( 50, 65)( 50, 0){4}
\Vertex( 50, 65){2}
\LongArrow( 60, 45)( 60, 25)
\Text( 65, 35)[l]{\scalebox{1.11}{$k\rightarrow 0$}}
\end{picture}
}
\end{center}
\caption{\label{fig:eikonalFeynmanRules} Coupling of a gluon (left)
and an $\epsilon$-scalar (right) to a heavy quark propagator. In the
eikonal approximation the latter vanishes. }
\end{figure}
The velocity-dependent cusp anomalous dimensions can be extracted from the
heavy-to-heavy soft anomalous dimension $\Gamma_{hh}$ for the pair production of
massive quarks. Using {\scshape cdr}, $\Gamma_{hh}$ has been calculated in
Ref.~\cite{Kidonakis:2009ev} in the framework of the eikonal approximation.
This method can also be used to derive the respective quantity in {\scshape fdh}.
In general, the eikonal approximation is suited for describing the emission
of soft gluons from partons in a hard scattering process,
see the l.\,h.\,s.\ of Fig.~\ref{fig:eikonalFeynmanRules}.
For a vanishing gluon momentum, the Feynman rule for the coupling of a gluon to
a massive quark propagator can be reduced to
\begin{subequations}
\begin{align}
\bar{u}(p_I)(-ig_{s} T^a)\,\hat{\gamma}^\mu
\left[i \frac{\slashed{p}_I+\slashed{k}+m_{I}^{\phantom{|}}}
{(p_I+k)^2-m_{I}^{2}}\right]
%
\ \rightarrow & \
\bar{u}(p_I)\,g_{s} T^a\,\hat{\gamma}^\mu
\left[ \frac{\slashed{p}_I+m_{I}^{\phantom{|}}}{2\, p_I\cdot k}\right]\\
%
\ = & \
\bar{u}(p_I)\,g_{s} T^a\,
\left[ (p_I)_{\nu}\frac{\{\hat{\gamma}^\mu,\hat{\gamma}^\nu\}}{2\, p_I\cdot k}\right]\\
%
\ = & \
\bar{u}(p_I)\,g_{s} T^a\,
\left[\frac{v_I^\mu}{v_I\cdot k}\right]\, ,
\label{eq:CouplingEikonal}
\end{align}
\end{subequations}
where in the second line the Dirac equation
$\bar{u}(p_I)(\slashed{p}_I-m_I^{\phantom{|}})=0$ has been
used.
Since the Feynman rule~\eqref{eq:CouplingEikonal} does not contain a
Dirac matrix anymore, the evaluation of loop contributions is much simpler compared
to ordinary QCD.
Extending this to the case of an $\epsilon$-scalar we get
\begin{align}
\bar{u}(p_I)(-ig_{e} T^a)\,\tilde{\gamma}^\mu
\left[i \frac{\slashed{p}_I+\slashed{k}+m_{I}^{\phantom{|}}}
{(p_I+k)^2-m_{I}^{2}}\right]
%
&\ \rightarrow \
\bar{u}(p_I)\,g_{e} T^a\,
\left[ (p_I)_{\nu}\frac{\{\tilde{\gamma}^\mu,\hat{\gamma}^\nu\}}{2\, p_I\cdot k}\right]
=0\, .
\label{eq:evCouplingEikonal}
\end{align}
Due to the vanishing anticommutator, a direct coupling of $\epsilon$-scalars to
massive quark propagators does not exist in the eikonal approximation.
\begin{figure}[t]
\begin{center}
\scalebox{.8}{
\begin{picture}(135,90)(20,0)
\DoubleLine(30,45)(100,80){2}
\DoubleLine(100,80)(130,95){2}
\DoubleLine(30,45)(100,10){2}
\DoubleLine(100,10)(130,-5){2}
\Gluon(100,10)(100,80){5}{7}
\Text(140,100)[t]{\scalebox{1.25}{$p_I$}}
\Text(140, -3)[t]{\scalebox{1.25}{$p_J$}}
\Vertex(30,45){4}
\Vertex(100,80){2}
\Vertex(100,10){2}
\end{picture}
\quad
\begin{picture}(135,90)(20,0)
\DoubleLine(30,45)(100,80){2}
\DoubleLine(100,80)(130,95){2}
\DoubleLine(30,45)(100,10){2}
\DoubleLine(100,10)(130,-5){2}
\Gluon(100,60)(100,80){4}{2}
\Gluon(100,10)(100,30){5}{2}
\DashCArc(100,45)(15,-90, 90){2}
\DashCArc(100,45)(15, 90,270){2}
\Vertex(30,45){4}
\Vertex(100,60){2}
\Vertex(100,30){2}
\Vertex(100,80){2}
\Vertex(100,10){2}
\end{picture}
\quad
\begin{picture}(135,90)(20,0)
\DoubleLine(30,45)(55,57){2}
\DoubleLine(55,57)(100,80){2}
\DoubleLine(100,80)(130,95){2}
\DoubleLine(30,45)(130,-5){2}
\DashCArc(67,88.5)(13.5, 15,195){2}
\DashCArc(67,88.5)(13.5,195,375){2}
\Gluon(55,57)(55,82.5){4}{3}
\Gluon(79,94)(100,80){4}{3}
\Vertex(30,45){4}
\Vertex(55,57){2}
\Vertex(100,80){2}
\Vertex(55,82.5){2}
\Vertex(79,94){2}
\end{picture}
}
\end{center}
\caption{\label{fig:eikonalDiagrams} One- and two-loop contributions
to the heavy-to-heavy soft anomalous dimension in the eikonal
approximation. Since there is no direct coupling of
$\epsilon$-scalars to massive quark propagators there is no
evanescent contribution at the one-loop level. }
\end{figure}
Following the approach of Ref.~\cite{Kidonakis:2009ev}, the soft anomalous
dimension for heavy-quark pair production can be obtained from the UV poles of
corresponding eikonal diagrams with one- and two-loop examples shown in
Fig.~\ref{fig:eikonalDiagrams}.
Since there is no direct coupling of $\epsilon$-scalars to massive
quarks, the soft anomalous dimension is scheme independent at the one-loop level.
At the two-loop level, however, closed $\epsilon$-scalar loops yield evanescent
contributions $\propto\alpha_s N_\epsilon$.
In the following, the scalar product of the two outgoing velocity vectors is fixed
by $v_I\cdot v_J \mathrel{\mathop:}= -\,\text{cosh}\,\beta_{IJ}$ with $\beta_{IJ}$ given
in Eq.~\eqref{eq:betaIJ}, and the indices of $\beta$ are suppressed:
$\beta_{IJ}=\mathrel{\mathop:}\beta$.
Generalizing Eq.~(14) of Ref.~\cite{Becher:2009kw} to the case of {\scshape fdh},
the result of the soft anomalous dimension can then be written as
\begin{align}
\bar\Gamma_{hh}(v_I,v_J)
= C_F\,\bar{ \gamma}_{\rm cusp}(\beta)+2\,\bar{\gamma}_Q \, .
\end{align}
Using Eq.~\eqref{anomres}, it is now possible to extract the velocity-dependent
cusp anomalous dimension in {\scshape fdh}\ which in terms of the renormalized
couplings reads
\begin{align}
\bar{ \gamma}_{\rm cusp}(\beta)
=\ & \bar{\gamma}_{\rm cusp}\,\beta\coth\beta
+8\,C_A \left( \frac{\alpha_s}{4\pi} \right)^2\Bigg\{
\beta^2 + \frac{\pi^2}{6} + \zeta_3
\nonumber\\*
&+\coth\beta^{\phantom{2}} \Bigg{[}
\mbox{Li}_2(e^{-2\beta}) - 2\,\beta\,\ln(1-e^{-2\beta})
-\frac{\pi^2}{6}\,(1+\beta) - \beta^2 - \frac{\beta^3}{3}\Bigg{]}
\nonumber\\
&+\coth^2\beta \Bigg{[} \mbox{Li}_3(e^{-2\beta})
+ \beta\,\mbox{Li}_2(e^{-2\beta})
-\zeta_3 + \frac{\pi^2}{6}\,\beta + \frac{\beta^3}{3} \Bigg{]}
\Bigg\}+\mathcal{O}(\alpha^3) \, .
\label{eq:gCuspBeta}
\end{align}
Since the terms in the curly brackets do not depend on $N_\epsilon$,
the scheme dependence of $\bar{ \gamma}_{\rm cusp}(\beta)$
is entirely governed by the scheme dependence of the cusp anomalous dimension
in the massless case, i.\,e.\ $\bar{ \gamma}_{\rm cusp}$.
\section{Guideline for FDH calculations and checks of the results
}
\label{sec:examples}
In order to check the obtained results for the scheme dependence of IR
divergences in massive QCD we compute the heavy and the heavy-to-light quark
form factor in {\scshape fdh}\ up to the two-loop level. Apart from a pure check this
section is also intended to provide a guideline how practical calculations in
the {\scshape fdh}\ scheme can actually be done. For the two-loop calculations we
therefore use the following approach:
\begin{itemize}
\item At the one-loop level we distinguish the $\epsilon$-scalar from the
$D$-dimensional gluon since the related couplings $\alpha_s$ and $\alpha_e$
renormalize differently.
\item At the two-loop level we use a (quasi) $4$-dimensional Lorentz algebra
for the evaluation of genuine two-loop diagrams and do not distinguish the
$\epsilon$-scalar from the $D$-dimensional gluon.
\item After having applied the UV renormalization we set equal the couplings
$\alpha_s$ and $\alpha_e$ in contributions from one-loop counterterm diagrams.
\item Throughout the calculations we identify $N_\epsilon=2\epsilon$.
\end{itemize}
Using this setup it turns out that practical calculations in the {\scshape fdh}\ scheme
are not significantly more complicated than the respective ones in {\scshape cdr}.
\subsection{Heavy quark form factor}
\begin{figure}[t]
\begin{center}
\scalebox{.9}{
\begin{picture}(135,90)(0,10)
\Photon(0,45)(30,45){3}{3}
\DoubleLine[arrow](30,45)(100,80){2}
\DoubleLine[arrow](100,80)(120,90){2}
\DoubleLine[arrow](120,0)(100,10){2}
\DoubleLine[arrow](100,10)(30,45){2}
\Gluon(100,10)(100,80){4}{9}
\Vertex(30,45){2}
\Vertex(100,80){2}
\Vertex(100,10){2}
\LongArrow(105,90)(115,95)
\Text(110,102)[c]{\scalebox{1.1}{$p_1$}}
\LongArrow(110,12)(120,7)
\Text(115,20)[c]{\scalebox{1.1}{$p_2$}}
\end{picture}
\qquad
\begin{picture}(135,90)(0,10)
\Photon(0,45)(30,45){3}{3}
\DoubleLine[arrow](30,45)(100,80){2}
\DoubleLine[arrow](100,80)(120,90){2}
\DoubleLine[arrow](120,0)(100,10){2}
\DoubleLine[arrow](100,10)(30,45){2}
\DashLine(100,10)(100,80){4}
\Vertex(30,45){2}
\Vertex(100,80){2}
\Vertex(100,10){2}
\end{picture}
}
\end{center}
\caption{\label{fig1}
One-loop diagrams contributing to the heavy-quark form factor in {\scshape fdh}.}
\end{figure}
\begin{figure}[t]
\scalebox{.9}{
\begin{picture}(135,90)(0,0)
\Photon(0,45)(30,45){3}{3}
\DoubleLine[arrow](30,45)(62.5,61.5){2}
\DoubleLine[arrow](62.5,61.5)(100,80){2}
\DoubleLine[arrow](100,80)(120,90){2}
\DoubleLine[arrow](120,0)(100,10){2}
\DoubleLine[arrow](100,10)(62.5,28.5){2}
\DoubleLine[arrow](62.5,28.5)(30,45){2}
\Gluon(62.5,28.5)(62.5,61.5){4}{4}
\Gluon(100,10)(100,80){4}{9}
\Vertex(30,45){2}
\Vertex(62.5,61.5){2}
\Vertex(62.5,28.5){2}
\Vertex(100,80){2}
\Vertex(100,10){2}
\end{picture}
\quad
\begin{picture}(135,90)(0,0)
\Photon(0,45)(30,45){3}{3}
\DoubleLine[arrow](30,45)(50,55){2}
\DoubleLine[arrow](50,55)(80,70){2}
\DoubleLine[arrow](80,70)(100,80){2}
\DoubleLine[arrow](100,80)(120,90){2}
\DoubleLine[arrow](120,0)(100,10){2}
\DoubleLine[arrow](100,10)(30,45){2}
\Gluon(100,10)(100,80){4}{9}
\GlueArc(65,62.5)(17, 20, 200){4}{6}
\Vertex(30,45){2}
\Vertex(50,55){2}
\Vertex(80,70){2}
\Vertex(100,80){2}
\Vertex(100,10){2}
\end{picture}
\quad
\begin{picture}(135,90)(0,0)
\Photon(0,45)(30,45){3}{3}
\DoubleLine[arrow](30,45)(100,80){2}
\DoubleLine[arrow](100,80)(120,90){2}
\DoubleLine[arrow](120,0)(100,10){2}
\DoubleLine[arrow](100,10)(30,45){2}
\Gluon(100,60)(100,80){4}{2}
\Gluon(100,10)(100,30){4}{2}
\DoubleArc[arrow](100,45)(15,90,270){2}
\DoubleArc[arrow](100,45)(15,270,90){2}
\Vertex(30,45){2}
\Vertex(100,80){2}
\Vertex(100,60){2}
\Vertex(100,30){2}
\Vertex(100,10){2}
\end{picture}
}
\caption{\label{fig2}
Sample two-loop diagrams contributing to the heavy-quark form factor in {\scshape fdh}.
All gluons belong to the quasi 4-dimensional space $Q4S$.}
\end{figure}
In {\scshape cdr}, the heavy quark form factor has been calculated up to NNLO in
Ref.~\cite{Bernreuther:2004ih}.
In {\scshape fdh}, the Green function for the interaction of a
virtual photon and two massive quarks can be written as
\begin{subequations}
\begin{align}
\hspace{-5mm}
\bar{V}^{\mu}_{c_{1}c_{2}}(p_1,p_2) & =
\bar{u}_{c_{1}}(p_1)\,
\bar{\Gamma}^{\mu}_{c_{1}c_{2}}(p_1,p_2)\,
v_{c_{2}}(p_2) \, ,
\end{align}
with
\begin{align}
\bar{\Gamma}^{\mu}_{c_{1}c_{2}}(p_1,p_2)
& = - i \,v_Q \, \delta_{c_{1}c_{2}} \Biggl[
\bar{F}_{1}(x) \, \hat{\gamma}^{\mu}
+ \frac{1}{2m} \bar{F}_{2}(x) \, i \, \hat{\sigma}^{\mu \nu} q_{\nu}
\Biggr] \,.
\label{b0002}
\end{align}
\end{subequations}
Here and in the following, $p_1$ and $p_2$ denote the (outgoing) momenta of the two
external quarks with $p_{1}^{2}=p_{2}^{2}=m^2$ and $s = (p_{1}+p_{2})^2/m^2$.
In general, the $\gamma$ matrices appearing in Eq.~\eqref{b0002} are
scheme-dependent. However, since we are only interested in the structure of
$\bar{F}_1$ their dimensionality can be chosen arbitrarily.
Here and in the following we therefore use $D$-dimensional $\gamma$ matrices
in the Lorentz decomposition.
The IR anomalous dimensions can be obtained from the heavy-quark form factor,
$\bar{F}_1$, which can be extracted from Eq.~\eqref{b0002} by using an appropriate
projection operator. For the proper definition of the projection and other
details we refer to Ref.~\cite{Bernreuther:2004ih}.
In the {\scshape fdh}\ scheme, only two diagrams contribute to the form factor at the
one-loop level, see Fig.~\ref{fig1}. Using 1-dimensional harmonic polylogarithms
\cite{Remiddi:1999ew,Gehrmann:2001pz} of the variable
\begin{equation}
x = \frac{\sqrt{-s+4} - \sqrt{-s} }{\sqrt{-s+4} + \sqrt{-s} }
\qquad ( 0 \leq x \leq 1 )
\label{xvar}
\end{equation}
and notation \eqref{eq:asdef} for the couplings, we represent the
bare one-loop coefficients of the form factor as%
\footnote{Note that $\bar{F_1}$ denotes the (all-order) form factor in {\scshape fdh}\ whereas
its perturbative coefficients are written using a calligraphic form,
$\big[\bar{F_1}\big]_{mn}=\bar{\mathcal{F}}_{mn}$.}
\begin{eqnarray}
\hspace{-5mm}
\bar{F}_{1}(x) & = &
1 + a_{s}(m^{2})\,\bar{\mathcal{F}}_{10}(x)
+ a_{e}(m^{2})\,\bar{\mathcal{F}}_{01}(x
+ {\mathcal O}(a^{2})\, ,
\label{b00020}
\end{eqnarray}
with
\begin{subequations}
\begin{align}
\bar{\mathcal{F}}_{10}(x) & =
2\,C_{F} \, \Biggl\{\frac{1}{\epsilon} \,\Biggl[
\frac{1}{2}
+H(0;x)\,\frac{x^2+1}{x^2-1}
\Biggr]
+\frac{1}{2} H(0;x)\,\frac{x+1}{x-1}
\nonumber\\&\qquad\qquad
-\Bigg(
\frac{\pi^2}{6}
-H(0;x)
-H(0,0;x)
+2 H(\!-1,0;x)
\Bigg)\,\frac{x^2+1}{x^2-1}\phantom{\Bigg|}
\nonumber\\&\qquad\qquad
+ \epsilon \,\Bigg[\,
\frac{\pi^2}{24}
-\Bigg(
\frac{\pi^2}{12}
- \frac{H(0,\!0;x)}{2}
+ H(\!-1,\!0;x)
\Bigg)\frac{x+1}{x-1}
-\Bigg(
\frac{\pi^2}{6}
-\big(4\! - \! \frac{\pi^2}{12}\big) H(0;x)
\nonumber\\&\qquad\qquad\qquad
+ 2\,\zeta(3)
- \frac{\pi^2}{3}\,H(\!-1;x) \!
- H(0,\!0;x)
+ 2 H(\!-1,\!0;x)
- H(0,\!0,\!0;x)\phantom{\Bigg|}
\nonumber\\*&\qquad\qquad\qquad
+ 2 H(\!-1,\!0,\!0;x)
+ 2 H(0,\!-1,\!0;x) \!
- 4 H(\!-1,\!-1,\!0;x)
\Bigg)\frac{x^2+1}{x^2-1}\!\Bigg]
+\mathcal{O}(\epsilon^2)\Bigg\},
\label{1loopF1}
\\
\bar{\mathcal{F}}_{01}(x) & =
\, C_{F} \,\Bigg\{
1+\epsilon\,\Big[
1+\frac{1-x}{1+x}H(0;x)\Big]
+\,\mathcal{O}(\epsilon^2)\Bigg\} \, .
\label{1loopF2}
\end{align}
\end{subequations}
To obtain the result at the two-loop level we evaluate the Feynman
diagrams (see Fig.~\ref{fig2}) using a quasi $4$-dimensional Lorentz
algebra. This in particular means that the absolute number of
diagrams and master integrals~\cite{Bonciani:2003te, Bonciani:2003hc} we have
to evaluate is exactly the same as in {\scshape cdr}. In line with that we do
not have to introduce evanescent couplings like $\alpha_e$ in the
computation of the genuine two-loop diagrams.
In the following we give the difference between the UV renormalized form factors
in {\scshape fdh}\ and {\scshape cdr}\ at the two-loop level.
For the renormalization of the couplings, the quark mass, and the fields we use
Eqs.~\eqref{eq:asdef}, \eqref{eq:epsMassRen}, and \eqref{z2os}, respectively,
and set $\alpha_s=\alpha_e$ after renormalization.
Because of the appearing $\epsilon$-scalar propagator in the right diagram of
Fig.~\ref{fig1} we also have to add the mass counterterm~\eqref{eq:epsMassCT}
of the $\epsilon$-scalar. Combining all results we finally get
\begin{align}
\bar{F}_1(x)-F_{1}(x)&=
\Big(\frac{\alpha_s}{4\pi}\Big)^{2}\Bigg\{
C_A C_F \Bigg[
\frac{1}{3\epsilon}-\frac{8}{9}\Bigg]
\Big(-1+\frac{x^2+1}{x^2-1}H(0;x)\Big{)}
\nonumber\\
&\qquad\qquad\quad
+C_F^{2}\Bigg[
\frac{2 \left(x^2-1\right) H(0;x)
-4 \left(x^2+1\right) H(0,0;x)}{(x+1)^2}
\Bigg]
+\mathcal{O}(\epsilon^1)
\Bigg\}
\nonumber\\
&\qquad
+\mathcal{O}(\alpha_s^3) \, .
\end{align}
This difference can be expressed in terms of the IR anomalous dimensions and
$\beta$ functions through Eqs.~\eqref{eq:G2parmass} and \eqref{eq:ZmassiveFDH},
in a similar way as shown in Ref.~\cite{Broggio:2015dga} for the case of massless
partons:
\begin{align}\label{massdiff}
\bar{F_1}(x)-F_1(x)&=
\Big{(}\frac{\alpha_s}{4\pi}\Big{)}^{2}\Bigg{\{ }
-\frac{1}{\epsilon^2}\,C_F\,
\Big(\bar{\beta}_{20}^s-\beta_{20}^s\Big)
\Big(-1+\frac{x^2+1}{x^2-1} H(0;x)\Big)
\nonumber\\ &\qquad\qquad\quad
+\frac{1}{4\epsilon} \Bigg[
C_F \Big(
\bar{\gamma}^{\text{cusp}}_{20}(\beta)
-\gamma^{\text{cusp}}_{20}(\beta)
-8\, \mathcal{F}^{\text{diff}}_1
\Big)
+2\,\Big(\bar{\gamma}^{Q}_{20}-\gamma^{Q}_{20}\Big)
\nonumber\\ &\qquad\qquad\qquad\qquad
+8\,C_F\,\mathcal{F}^{\text{diff}}_{1}\,\frac{x^2+1}{x^2-1}\,H(0;x)
\Bigg]+\mathcal{O}(\epsilon^1)
\Bigg{\} }
+\mathcal{O}(\alpha_s^3)\, ,
\end{align}
where $\mathcal{F}^{\text{diff}}_{1}=
\bar{\mathcal{F}}_{10}^{\text{ren}}
+\bar{\mathcal{F}}_{01}^{\text{ren}} -\mathcal{F}_{1}^{\text{ren}}$ is
the difference of the UV renormalized one-loop coefficients,
i.\,e.\ including a field renormalization of the heavy quarks. The
fact that the scheme dependence of the IR divergences related to the
heavy-quark form factor can be predicted with the results from
Secs.~\ref{sec:schemes} and \ref{sec:ad} constitutes a strong
consistency check of the results obtained so far.
\subsection{Heavy-to-light form factor}
\begin{figure}[t]
\scalebox{.9}{
\begin{picture}(135,90)(0,0)
\Photon(0,45)(30,45){3}{3}
\Line[arrow](30,45)(62.5,61.5)
\Line[arrow](62.5,61.5)(100,80)
\Line[arrow](100,80)(120,90)
\DoubleLine[arrow](120,0)(100,10){2}
\DoubleLine[arrow](100,10)(62.5,28.5){2}
\DoubleLine[arrow](62.5,28.5)(30,45){2}
\Gluon(62.5,28.5)(62.5,61.5){4}{4}
\Gluon(100,10)(100,80){4}{9}
\Vertex(30,45){2}
\Vertex(62.5,61.5){2}
\Vertex(62.5,28.5){2}
\Vertex(100,80){2}
\Vertex(100,10){2}
\end{picture}
\quad
\begin{picture}(135,90)(0,0)
\Photon(0,45)(30,45){3}{3}
\Line[arrow](30,45)(50,55)
\Line[arrow](50,55)(80,70)
\Line[arrow](80,70)(100,80)
\Line[arrow](100,80)(120,90)
\DoubleLine[arrow](120,0)(100,10){2}
\DoubleLine[arrow](100,10)(30,45){2}
\Gluon(100,10)(100,80){4}{9}
\GlueArc(65,62.5)(17, 20, 200){4}{6}
\Vertex(30,45){2}
\Vertex(50,55){2}
\Vertex(80,70){2}
\Vertex(100,80){2}
\Vertex(100,10){2}
\end{picture}
\quad
\begin{picture}(135,90)(0,0)
\Photon(0,45)(30,45){3}{3}
\Line(30,45)(120,90)
\Line[arrow](30,45)(100,80)
\Line[arrow](100,80)(120,90)
\DoubleLine[arrow](120,0)(100,10){2}
\DoubleLine[arrow](100,10)(30,45){2}
\Gluon(100,60)(100,80){4}{2}
\Gluon(100,10)(100,30){4}{2}
\DoubleArc[arrow](100,45)(15,90,270){2}
\DoubleArc[arrow](100,45)(15,270,90){2}
\Vertex(30,45){2}
\Vertex(100,80){2}
\Vertex(100,60){2}
\Vertex(100,30){2}
\Vertex(100,10){2}
\end{picture}
}
\caption{\label{lifig1}
Sample two-loop diagrams contributing to the heavy-to-light form factor in {\scshape fdh}.}
\end{figure}
The {\scshape cdr}\ result for the decay process $b\to u\,W^{*}\to u\,l\,\bar{\nu}_{l}$
has been computed at NNLO in Refs.~\cite{Bonciani:2008wf,
Asatrian:2008uk, Beneke:2008ei, Bell:2008ws}.
Applying the procedure of the previous section we here extend the calculation
to the case of {\scshape fdh}, with sample two-loop diagrams shown in Fig.~\ref{lifig1}.
In {\scshape fdh}, the tensor structure of the heavy-to-light form factor can be
written as
\begin{eqnarray}
\bar{\Gamma}^{\mu}(p_1,p_2) &=&
\bar{F}_1(q^2)\,\hat{\gamma}^{\mu}
+ \frac{1}{2 m} \bar{F}_2(q^2)\,\hat{\sigma}^{\mu \nu}\,q_{\nu}
+ \frac{i}{2 m} \bar{F}_3(q^2)\,q^{\mu}
+ \bar{G}_1(q^2)\,\hat{\gamma}^{\mu}\,\gamma_{5}
\nonumber\\* & &
+ \frac{i}{2 m} \bar{G}_2(q^2)\,\gamma_{5}\,q^{\mu}
+ \frac{i}{2 m} \bar{G}_3(q^2)\,\gamma_{5}\,(p_1^\mu-p_2^\mu) \, ,
\label{Vmu}
\end{eqnarray}
with $q=p_1+p_2$.
Again, we are interested in the form factor $\bar{F}_{1}$ which can be extracted
by means of a projection operator. Accordingly, the matrix $\hat{\gamma}^{\mu}$
is treated in $D$ dimensions.
We compute the bare diagrams up to NNLO and perform the UV renormalization exactly
in the same way as described in the previous section, taking into account that here
only one leg is massive. Again we have to add a counterterm to subtract
the $\epsilon$-scalar mass shift.
Using Eq.~\eqref{b00020} for the perturbative expansion of the form factor
and expressing the result in terms of the dimensionless quantity
\begin{align}
y\mathrel{\mathop:}=\frac{q^2}{m^2} \, ,
\end{align}
we get for the bare one-loop coefficients
\begin{subequations}
\begin{align}
\hspace*{-5mm} \bar{\mathcal{F}}_{10}(y) & =
-\,C_{F} \Bigg[
\frac{1}{\epsilon ^2}
+\frac{1+2\,H(1;y)}{\epsilon }
+4
+\frac{\pi^2}{12}
+3\,H(1;y)
+2\,H(0,1;y)
+4\,H(1,1;y)
\nonumber\\*&\qquad\qquad
+\epsilon\, \Bigg(
8
+\frac{\pi^2}{12}
-\frac{\zeta(3)}{3}
+\Big(8+\frac{\pi^2}{6}\Big)\,H(1;y)
+3\,H(0,1;y)
+6\,H(1,1;y)
\nonumber\\&\qquad\qquad\qquad\
+8\,H(1,1,1;y)
+4\,H(-1,0,-1;-y)
+4\,H(0,-1,-1;-y)
\phantom{\Bigg|}
\nonumber\\*&\qquad\qquad\qquad\
+2\,H(0,0,1;y)
\Bigg)\Bigg]
+ \, {\mathcal O} \left( \epsilon^2 \right) \, ,
\label{he1loopF1} \\
\hspace*{-5mm} \bar{\mathcal{F}}_{01}(y) & =
C_{F} \Bigg[
1
+\epsilon\Big(1+H(1;y)\Big)
\Bigg]
+ \, {\mathcal O} \left( \epsilon^2 \right) \, .
\label{he1loopF2}
\end{align}
\end{subequations}
As in the previous section we give the difference between the
UV renormalized form factors in {\scshape fdh}\ and {\scshape cdr}\ up to the two-loop level:
\begin{align}
\bar{F}_{1}(y)&-F_{1}(y)=
\Big{(}\frac{\alpha_s}{4\pi}\Big{)}\frac{C_F}{2}\nonumber\\
&+\Big{(}\frac{\alpha_s}{4\pi}\Big{)}^2\Bigg\{
C_A C_F\, \Bigg[
-\frac{1}{4 \epsilon^2}
+\frac{-\frac{25}{36}-\frac{1}{3} H(1;y)-\frac{L}{6}}{\epsilon }
+\frac{965}{216}+\frac{\pi^2}{24}+\frac{8}{9} H(1;y)+\frac{4}{9}L\Bigg]
\nonumber\\&\qquad\qquad\quad
-C_F^2\,\Bigg[
\frac{1}{2 \epsilon ^2}
+\frac{\frac{9}{4}+2 H(1;y)+L}{\epsilon }
+\frac{49}{8}
+\frac{\pi^2}{4}
+\big(6+4 L\big) H(1;y)
\nonumber\\&\qquad\qquad\qquad\qquad\
+8 H(1,1;y)
+2 H(0,1;y)
+\frac{7}{2}L
+L^2\Bigg]
\nonumber\\&\qquad\qquad\quad
+C_F N_F\,\Bigg[\frac{1}{4 \epsilon }-\frac{3}{8}\Bigg]
-C_F N_H\,\frac{ L}{2}
+\mathcal{O}(\alpha_s^3) \, ,
\label{eq:HtoL}
\end{align}
where $L$ is defined as $L=\ln\Big{(}\frac{\mu^{2}}{m^{2}}\Big{)}$.
In terms of the IR anomalous dimensions, the $\beta$ functions, and the
factor $\bar{\mathbf{Z}}$ defined in Eq.~\eqref{eq:ZmassiveFDH} this difference
is given by
\begin{align}\label{eq:tranhtl}
\bar{F_1}(y) & -F_1(y)=
\Big{(}\frac{\alpha_{s}}{4\pi}\Big{)}
\frac{\bar{\gamma}^{q}_{01}}{2\epsilon}
\nonumber\\&
+\Big{(}\frac{\alpha_{s}}{4\pi}\Big{)}^{2}\Bigg{\{ }
\frac{3}{16\epsilon^{3}}\,C_F\gamma^{\text{cusp}}_{10}
\Big(\bar{\beta}^{s}_{20}-\beta^{s}_{20}\Big)
-\frac{1}{16\epsilon^{2}} \Bigg{[}
\bar{\gamma}^{q}_{01}\Big(
4(\bar{\beta}^{e}_{11}+\bar{\beta}^{e}_{02})
+2\bar{\gamma}^{q}_{01}
-4\gamma^{Q}_{10}
\Big)
\nonumber\\&\qquad\qquad\quad\ \
+\Big(\bar{\beta}^{s}_{20}-\beta^{s}_{20}\Big)\Big(
4\,\big(\gamma^{Q}_{10}+\gamma^{q}_{10}\big)
-2\,C_F\gamma^{\text{cusp}}_{10}\big(2\,H(1;y)+L\big)
\Big)
\nonumber\\&\qquad\qquad\quad\ \
+C_F \Big(
\bar{\gamma}^{\text{cusp}}_{20}
\!-\gamma^{\text{cusp}}_{20}
\!-8 \bar{\gamma}^{q}_{01}
\Big)
+4\,C_F \gamma^{\text{cusp}}_{10} \mathcal{F}^{\text{diff}}_{1}
\Bigg]
\nonumber\\&\qquad\qquad\ \
+\frac{1}{4\epsilon} \Bigg{[}
-\frac{1}{2} C_F \big(2\,H(1;y)+L\big)
\Big(
\bar{\gamma}^{\text{cusp}}_{20}
-\gamma^{\text{cusp}}_{20}
+2\,\gamma^{\text{cusp}}_{10}\,\mathcal{F}^{\text{diff}}_{1}
\Big)
\nonumber\\&\qquad\qquad\quad\ \
+(\bar{\gamma}^{Q}_{20}-\gamma^{Q}_{20})
+(\bar{\gamma}^{q}_{20}-\gamma^{q}_{20})
+\bar{\gamma}^{q}_{11}
+\bar{\gamma}^{q}_{02}
-2\,N_H\,\bar{\gamma}^{q}_{01}\,L\phantom{\Big|}
\nonumber\\&\qquad\qquad\quad\ \
+2\,\mathcal{F}^{\text{diff}}_{1}\,\Big(
\gamma^{Q}_{10}
+\gamma^{q}_{10}
+\bar{\gamma}^{q}_{01}
\Big)
+2\,\bar{\gamma}^{q}_{01}\,\mathcal{F}^{\text{fin}}_{1}
\Bigg{]}
+\mathcal{O}(\epsilon^1)\Bigg{\} }
+\mathcal{O}(\alpha_s^3)\, ,
\end{align}
with
\begin{subequations}
\begin{align}
\mathcal{F}^{\text{diff}}_{1}&=
\bar{\mathcal{F}}_{10}^{\text{ren}}
+\bar{\mathcal{F}}_{01}^{\text{ren}}
-\mathcal{F}_{1}^{\text{ren}} \, ,
\\*
\mathcal{F}^{\text{fin}}_{1} &=
\lim_{\epsilon\to 0}\bigg{[}
\bar{\mathcal{F}}_{10}^{\text{ren}}
+\delta\bar{\textbf{Z}}_{10}
\bigg{]}=
\lim_{\epsilon\to 0}\bigg{[}
\mathcal{F}_{10}^{\text{ren}}
+\delta\textbf{Z}_{1}
\bigg{]}\, .
\end{align}
\end{subequations}
The fact that Eq.~\eqref{eq:HtoL} matches with Eq.~\eqref{eq:tranhtl}
constitutes an additional and independent check of our results for the
IR anomalous dimensions.
\section{Conclusions}
\label{sec:conc}
The scheme dependence of massless QCD amplitudes at NNLO had been
discussed in Ref.~\cite{Broggio:2015dga}. In this paper we complete
this study by extending it to amplitudes containing massive quarks.
This requires modifications in the UV and IR sector. For the UV part,
the presence of heavy quarks modifies the renormalization. In
particular, the $\epsilon$-scalar field requires a mass
counterterm. Also, the decoupling of $\alpha_e$ (the coupling of the
$\epsilon$-scalars to the quarks) has to be determined. Furthermore,
we have computed the additional contributions required in {\scshape fdh}\ in
the quark mass and the quark wave-function renormalization.
Regarding the IR part, the important result is that the IR structure
of massive QCD amplitudes in {\scshape fdh}\ (and {\scshape dred}) is the same as in
{\scshape cdr}\ (and {\scshape hv}). The only change is in the explicit scheme-dependent
expressions of the various anomalous dimensions. In the massive case,
there are two additional anomalous dimensions, the
velocity-dependent cusp anomalous dimension and the heavy-quark anomalous
dimension. We have computed them in the {\scshape fdh}\ scheme, using a SCET
approach.
We have checked our results by computing the heavy-quark and
heavy-to-light form factor in {\scshape fdh}\ at NNLO. These results differ from
the corresponding expressions in {\scshape cdr}. After UV renormalization, the
difference can be reproduced by the scheme dependence of the IR
factorization formula. This provides us with a strong consistency
check and establishes {\scshape fdh}\ as a consistent regularization scheme also
in the massive case, at least to NNLO.
\section*{Acknowledgments}
It is a pleasure to thank Alessandro Broggio and Dominik
St$\breve{\text o}$ckinger for useful discussions and comments on the
manuscript. We are grateful to Pierpaolo Mastrolia, Thomas Gehrmann
and Andrea Ferroglia for providing assistance with the master
integrals needed for the computation of the form factors and to Thomas
Becher for clarifications about the computation of the soft
function. A.~Visconti is supported by the Swiss National Science
Foundation (SNF) under contracts 200021-144252 and 200021-163466.
\bigskip
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 9,521 |
{"url":"https:\/\/tex.stackexchange.com\/questions\/264598\/centering-doesnt-work-for-figure-with-tabularx","text":"# Centering doesn't work for figure with tabularx\n\nI have a figure with 3 images in it, searching TeX I found out that the best way to accomplish that was with a tabularx inside a figure, like this:\n\n\\begin{figure}[t]\n\\centering\n\\def\\tabularxcolumn#1{m{#1}}\n\\begin{tabularx}{\\linewidth}{c}\n%\n\\begin{tabular}{ccc}\n\\end{tabular}\n\\end{tabularx}\n\n\\caption{Pantalla de Buz\u00f3n con opciones de mensaje. Elaboraci\u00f3n propia.}\\label{fig:pantallaMulti2}\n\\end{figure}\n\n\nIt shows the multiple figures, but the whole figure is positioned to the left, not the center, like is ignoring the \\centering instruction.\n\nAny ideas? First time I use LaTex.\n\n\u2022 Where did you read to use a tabular for that? And what is tabular ... ? Where did you read that\/ \u2013\u00a0Johannes_B Sep 3 '15 at 17:11\n\u2022 Your best bet seems to be to ckick out everything related to tabular material; or use \\centering inside the tabularx. \u2013\u00a0Johannes_B Sep 3 '15 at 17:11\n\u2022 Welcome to TeX.SX! I'm afraid you're overcomplicating things. :) \u2013\u00a0egreg Sep 3 '15 at 17:28\n\nAnother option using subfig.\n\n\\documentclass{article}\n\\usepackage[T1]{fontenc}\n\\usepackage[utf8]{inputenc}\n\\usepackage[spanish]{babel}\n\\usepackage{subfig,graphicx}\n\\begin{document}\n\n\\begin{figure}\n\n\\subfloat[Opciones de mensaje]{\\includegraphics[width=3cm]{example-image-a}}\n\\hfill\n\\subfloat[Confirmaci\u00f3n de eliminar (Android)]{\\includegraphics[width=3cm]{example-image-b}}\n\\hfill\n\\subfloat[Confirmaci\u00f3n de eliminar (iOS)]{\\includegraphics[width=3cm]{example-image-c}}\n\n\\caption{Pantalla de Buz\u00f3n con opciones de mensaje. Elaboraci\u00f3n propia.}\\label{fig:pantallaMulti2}\n\n\\end{figure}\n\n\\end{document}\n\n\nAs stated by @egreg, you're overcomplicating things. You can just put the figures one beside another and \\hfill will distribute them horizontally.\n\n\u2022 You should add the instructions \\usepackage[T1]{fontenc} and \\usepackage[utf8]{inputenc} to the preamble, in order to properly process accented characters (such as \u00f3). Loading the babel package with the option spanish would also be a good idea. \u2013\u00a0Mico Sep 3 '15 at 17:51\n\u2022 I used your answer but with \\quad from the other answer instead of \\hfill, it looks better. Thanks. \u2013\u00a0David Prieto Sep 3 '15 at 18:58\n\u2022 @AboAmmar, if you use \\hfill to separate sub figures, than \\centering is superfluous :-). With use of centering\u02dbthe use of \\hfil (one l!) has sense. With use of both (\\centering and \\hfil) the sub figures are equally horizontally spaced. \u2013\u00a0Zarko Sep 3 '15 at 19:17\n\nYou're overcomplicating things: you don't need tabularx at all, because as you're using it it's just for getting a normal paragraph as wide as the normal text width.\n\nAlso the subcaptions should be aligned, which is much easier using subcaption instead of subfig (that's recommended only in some special cases) and its \\subcaptionbox command.\n\nThere's no need to specify everytime type=png,ext=.png,read=.png,angle=0, because LaTeX will pick up a PNG file if it finds it by the specified name.\n\n\\documentclass{article}\n\\usepackage[T1]{fontenc}\n\\usepackage[utf8]{inputenc}\n\\usepackage[spanish]{babel}\n\n\\usepackage{graphicx}\n\\usepackage{subcaption}\n\n\\usepackage{lipsum} % for mock text\n\n\\begin{document}\n\n\\lipsum[2]\n\n\\begin{figure}[htp]\n\\centering\n\n\\subcaptionbox{Opciones de mensaje}{%\n\\includegraphics[width=3cm]{example-image}%\n\\subcaptionbox{Confirmaci\u00f3n de eliminar (Android)}{%\n\\includegraphics[width=3cm]{example-image-9x16}%\n`","date":"2020-03-29 16:12:19","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8867824077606201, \"perplexity\": 5225.599531232534}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2020-16\/segments\/1585370494349.3\/warc\/CC-MAIN-20200329140021-20200329170021-00012.warc.gz\"}"} | null | null |
{"url":"https:\/\/blog.csdn.net\/buutterfly\/article\/details\/5630192","text":"# Apache 2 and PHP 5 (mod_php) on Linux\n\n## Apache 2 and PHP Installation\n\nThe following notes are how I got Apache 2 and PHP 5 (or PHP 4) working together on Linux. These instructions also apply, mostly, for any UNIX-like system, especially other Linux distributions. If you have a recent Linux distribution (say since 2002), you already have Apache 2 and PHP, so you don't need to do this unless you want the latest Apache 2 or PHP release or need to customize the Apache or PHP software. Don't forget to remove (or at least disable) the Apache rpm package if you install your own custom Apache.\n\nApache 2 Version Tip:\u00a0Beginning with Apache 2.0.42 the API will be kept stable (yeah!). That means you will\u00a0NOT\u00a0have to recompile modules (and possibly upgrade\/fix source) every time you install a new Apache release. This assumes you stay in the same Apache release series. For example, upgrading from 2.2.0 to 2.2.2 should work. This will not apply to upgrading to the next series (e.g., \"development\" 2.3.x or \"stable\" 2.4.x).\n\nI was only able to get PHP working with Apache 2 as a *.so DSO (\"dynamic shared object\"), as opposed to compiled into Apache (with a static *.a library). I think DSO is the only way PHP is supported now on Apache 2.\n\nI first used httpd-2.0.43 and php-4.3.0 with RedHat 7.3. I am now using httpd-2.2.3 and php-5.2.0 with SUSE 10.1.\n\nNote: If you have problems with PHP and think it's a recent bug, you may want to consider using the latest\u00a0http:\/\/snaps.php.net\/\u00a0snapshot. Beware that snapshots frequently have regression and are not for production use. Usually problems are because of mis-configuration, not bugs, so snapshots will probably\u00a0hurt more than help.\n\n2. In the Apache 2 source directory, create a\u00a0Makefile\u00a0by typing:\n .\/configure --prefix=\/usr\/local\/apache \/ --enable-so \/ --enable-cgi \/ --enable-info \/ --enable-rewrite \/ --enable-speling \/ --enable-usertrack \/ --enable-deflate \/ --enable-ssl \/ --enable-mime-magic \nYou only need the\u00a0enable-so\u00a0line above. For information on other options, type\u00a0.\/configure --help\u00a0and see \"Compiling and Installing\" in the\u00a0Apache 2 Documentation, http:\/\/httpd.apache.org\/\n3. Make Apache from the just-created\u00a0Makefile:\n make \n4. If make is successful, install Apache as root:\n make install \nPick the latest from the 4.x series or 5.x series.\n6. In the PHP source directory, create a\u00a0Makefile\u00a0by typing:\n .\/configure \/ --with-apxs2=\/usr\/local\/apache\/bin\/apxs \/ --with-mysql \/ --prefix=\/usr\/local\/apache\/php \/ --with-config-file-path=\/usr\/local\/apache\/php \/ --enable-force-cgi-redirect \/ --disable-cgi \/ --with-zlib \/ --with-gettext \/ --with-gdbm \nYou only need the\u00a0--with-apxs2, and\u00a0prefix\u00a0lines.\u00a0--with-mysql\u00a0adds MySql (you need to specify the directory if it's in a unusual location (e.g., --with-mysql=\/usr\/local ),\u00a0--with-config-file\u00a0moves the php.ini file location,\u00a0disable-cgi\u00a0disables the CGI version, which is not needed if you use Apache modules. It also enables and installs the command line interface (CLI) version.\u00a0--with-zlib\u00a0allows use of gzip-type compression,\u00a0--with-gettext\u00a0is for internationalization, and\u00a0--with-gdbm\u00a0allows access to GDBM databases. For more information, type\u00a0.\/configure --help\u00a0and see the \"Installation\" chapter in the\u00a0PHP Manual, http:\/\/ww.php.net\/docs.php\n7. Make PHP from the just-created\u00a0Makefile:\n make \n8. If make is successful, type this as root to install PHP:\n make install \nIf you are not root (I do not perform makes while root, for security and safety reasons), become root and type the following:\n make install-su \n9. If file\u00a0\/usr\/local\/apache\/modules\/libphp5.so\u00a0does not exist or is an older version, type this (change this to\u00a0libphp4.so\u00a0for PHP 4):\n cp -p .libs\/libphp5.so \/usr\/local\/apache\/modules \n10. Install the php.ini file:\n cp -p php.ini-recommended \/usr\/local\/apache\/php\/php.ini \n11. Add these directives are in\u00a0\/usr\/local\/apache\/conf\/httpd.conf\u00a0(if already there, verify they are correct):\n # Make sure there's only **1** line for each of these 2 directives: # Use for PHP 4.x: #LoadModule php4_module modules\/libphp4.so #AddHandler php-script php # Use for PHP 5.x: LoadModule php5_module modules\/libphp5.so AddHandler php5-script php # Add index.php to your DirectoryIndex line: DirectoryIndex index.html index.php AddType text\/html php # PHP Syntax Coloring # (optional but useful for reading PHP source for debugging): AddType application\/x-httpd-php-source phps \n\nNote 1: The php documentation recommends\u00a0AddType application\/x-httpd-php php\u00a0instead of the above. However, it causes problems with the \"MultiViews\" feature of HTTP. That is, if the .php extension is left off a URL, and with certain browser\u00a0Accept\u00a0headers, Apache will not know .php (application\/x-httpd-php) is HTML and will return a\u00a0406 Not Acceptable\u00a0error. Using the\u00a0AddType\u00a0and\u00a0AddHandler\u00a0as shown above fixes this problem. For details see Mark Tranchant's webpage, \"Using Apache's MultiViews with PHP whilst avoid 406 errors,\" and\u00a0PHP bug 28023.\n\nNote 2: PHP Syntax coloring isn't required, but it's very nice for looking at your php source while debugging.\u00a0Here's an example.\n\nNote 3: just for completeness I'll mention that you should be able to use\u00a0SetOutputFilter\u00a0\/\u00a0SetInputFilter\u00a0instead of\u00a0AddType, but you can't use both. However,\u00a0SetOutputFilter\u00a0\/\u00a0SetInputFilter\u00a0no longer works for me. It used to work with an earlier PHP 4.x or Apache 2 version, but not with Apache 2.0.47\/PHP 4.3.3. I understand this (PHP as an Apache 2 filter) is experimental, so I don't use it anymore:\n\n SetOutputFilter PHP SetInputFilter PHP \n12. You're now ready to try it out. Start Apache (httpd) as root:\n \/usr\/local\/apache\/bin\/apachectl start \n13. Perform these sanity checks to verify your install went OK:\n $\/usr\/local\/apache\/bin\/httpd -t Syntax OK$ \/usr\/local\/apache\/bin\/httpd -v Server version: Apache\/2.2.2 Server built: May 29 2006 12:40:55 $\/usr\/local\/apache\/bin\/httpd -V Server version: Apache\/2.2.2 Server built: May 29 2006 12:40:55 Server's Module Magic Number: 20051115:2 Server loaded: APR 1.2.7, APR-Util 1.2.7 Compiled using: APR 1.2.7, APR-Util 1.2.7 Architecture: 32-bit Server MPM: Prefork threaded: no forked: yes (variable process count) Server compiled with.... -D APACHE_MPM_DIR=\"server\/mpm\/prefork\" -D APR_HAS_SENDFILE -D APR_HAS_MMAP -D APR_HAVE_IPV6 (IPv4-mapped addresses enabled) -D APR_USE_SYSVSEM_SERIALIZE -D APR_USE_PTHREAD_SERIALIZE -D SINGLE_LISTEN_UNSERIALIZED_ACCEPT -D APR_HAS_OTHER_CHILD -D AP_HAVE_RELIABLE_PIPED_LOGS -D DYNAMIC_MODULE_LIMIT=128 -D HTTPD_ROOT=\"\/usr\/local\/apache\" -D SUEXEC_BIN=\"\/usr\/local\/apache\/bin\/suexec\" -D DEFAULT_PIDLOG=\"logs\/httpd.pid\" -D DEFAULT_SCOREBOARD=\"logs\/apache_runtime_status\" -D DEFAULT_LOCKFILE=\"logs\/accept.lock\" -D DEFAULT_ERRORLOG=\"logs\/error_log\" -D AP_TYPES_CONFIG_FILE=\"conf\/mime.types\" -D SERVER_CONFIG_FILE=\"conf\/httpd.conf\"$ \/usr\/local\/apache\/bin\/httpd -S VirtualHost configuration: . . . $\/usr\/local\/apache\/bin\/httpd -l Compiled in modules: core.c . . . mod_so.c$ \/usr\/local\/apache\/bin\/httpd -M Loaded Modules: . . . php5_module (shared) Syntax OK (the above works for Apache 2.2.x and higher only) $ps -ef |grep httpd root 24069 1 0 09:17 ? 00:00:08 \/usr\/local\/apache\/bin\/httpd -k s apache 29917 24069 0 15:30 ? 00:00:00 \/usr\/local\/apache\/bin\/httpd -k s . . . Note: on BSD-based UNIX systems, you need to use \"ps -aux\" instead of \"ps -ef\". 14. Access your webserver with telnet. Type HEAD \/ HTTP\/1.0 followed by a blank line: $ telnet localhost 80 Trying 127.0.0.1... Connected to localhost (127.0.0.1). Escape character is '^]'. HEAD \/ HTTP\/1.0 HTTP\/1.1 200 OK Date: Mon, 29 May 2006 23:28:18 GMT Server: Apache\/2.2.2 (Unix) mod_ssl\/2.2.2 OpenSSL\/0.9.7a PHP\/5.1.4 X-Powered-By: PHP\/5.1.4 Last-Modified: Wed, 15 Mar 2006 06:53:17 GMT Vary: Accept-Encoding Connection: close Content-Type: text\/html; charset=ISO-8859-1 Content-Language: en Connection closed by foreign host. \n15. Access your webserver with your favorite browser. The following is a good test page to use for PHP. You only need the one line in\u00a0bold\u00a0is needed to display PHP configuration information. Name the file anything you want, but it\u00a0must\u00a0end with\u00a0.php, such as\u00a0phpinfo.php, and move the file to your web server content directory (for me\u00a0\/usr\/local\/apache\/htdocs), with read permission set:\n PHP Test\n\nPHP Test\n\nAn Example of PHP in Action\n\"; echo date(\"g:i A l, F j Y.\");?>\n\nPHP Information\n\n\n\n## Tips and Notes\n\n\u2022 Disabling Existing Apache Software\n\nIf Apache is already installed with your Linux distribution you have to disable it before installing your own custom version. To do this, type the following under RedHat Linux:\n\n \/sbin\/chkconfig --del httpd \/etc\/init.d\/httpd stop \nYou may also remove the\u00a0httpd\u00a0and\u00a0php\u00a0rpm packages, and dependent packages, if you wish.\n\n\u2022 Automatic Startup of Apache\n\nTo start Apache automatically, follow these steps:\n\n\u2022 Copy Apache startup file to the startup directory:\n cp \/usr\/local\/apache\/bin\/apachectl \/etc\/init.d\/ \n\n\u2022 Edit\u00a0\/etc\/init.d\/apachectl\u00a0by inserting these 2 lines in\u00a0bold:\n #!\/bin\/sh # # chkconfig: - 85 15 # description: Apache is a Web server used to serve HTML files and CGI. # # Copyright 2000-2005 The Apache Software Foundation or its licensors, as # applicable. . . . # Apache control script designed to allow an easy command line interface # to controlling Apache. Written by Marc Slemko, 1997\/08\/23 . . . \n\u2022 Enable httpd to startup automatically:\n \/sbin\/chkconfig --add apachectl \/sbin\/chkconfig --level 2 apachectl on \nThis works for RedHat Linux, Mandrake, SuSE, and Sun Java Desktop System. The equivalent of\u00a0chkconfig\u00a0for Debian is:\nupdate-rc.d apachectl defaults\nFor other distributions, make appropriate softlinks to directories\u00a0\/etc\/rc*.d\/\n\u2022 Reboot Linux to verify Apache starts\n\n[Thanks to Beriah Dutcher and others]\n\n\u2022\u00a0Make\u00a0doesn't make dynamic library\u00a0libphp5.so, just static library\u00a0libphp5.a\n\nFirst make sure libphp5.so isn't there. It's usually in the libs or .libs subdirectory. Run find from the PHP source directory root:\u00a0find . -name libphp5.soIf that doesn't work, make sure you use the\u00a0--with-apxs2\u00a0option with the correct directory in\u00a0.\/configure. For example:\n.\/configure --with-mysql --with-apxs2=\/path\/to\/apache\/apxs\nWhen re-running\u00a0.\/configure\u00a0always start with a fresh, clean source directory (no object).\n\n(For PHP 4, substitute\u00a0libphp4.so\u00a0and\u00a0libphp4.a\u00a0for\u00a0libphp5.so\u00a0and\u00a0libphp5.a, respectively.)\n\n\u2022\u00a0PHP 5: My PHP scripts that worked with PHP 4 now ignores input variables\n\nIf you have old PHP 4 scripts you don't want to convert to PHP 5, change this line in your php.ini file:\n\n register_long_arrays = On \n\nFor the long term, you may want to upgrade your PHP software or convert the PHP source code to use new-style variables. E.g., use\u00a0$_REQUEST[i] instead of$i\n\n\u2022\u00a0PHP 5.2.0 Upgrade from 5.1.x\u00a0(November 2006)\n\nI had one problem with upgrading from PHP 5.1.x to 5.2.0. The Phorum 3 software I use has a function\u00a0date_format\u00a0that now conflicts with a new PHP 5.2 internal function, also called\u00a0date_format. This new function is used for PHP 5.2.x's new \"datetime support\". I fixed this by renaming Phorum's\u00a0date_formatfunction to not conflict with PHP's function. See also\u00a0PHP 5.2 Update Info\u00a0for more PHP 5.2 upgrade information.\n\n\u2022\u00a0PHP 5.x: you may need to upgrade libxml2 to use XML\n\nTo use PHP 5 and XML (--with-xml), you need the libxml2 library version 2.6.0 or greater. You may need to upgrade if you have an older Linux distribution. Check with\u00a0rpm -q libxml2\nIf it's not installed, check your installation disks for the libxml2 package. If it's too old, you can get the library from\u00a0XMLSoft.org. You will probably need to compile and install it yourself.\n\n\u2022\u00a0PHP 4.3.5 \/ 4.3.6 \/ 5.0RC1 PCRE Regression\u00a0(March 2004)\n\nIf you use PHP 4.3.5, 4.3.6, or PHP 5.0 RC1 with PCRE (Perl Regular Expressions, the default), you get a core dump when you try and restart (SIGHUP) Apache (at least with Apache 2.0.49). The cause is from adding a new version of PCRE, 4.5, in PHP. This is fixed in PHP 4.3.7 and 5.0.\n\nThe symptom are these messages in error_log (which is also a symptom for lots of other memory-type bugs in Apache):\n\n [Wed Mar 31 17:14:43 2004] [notice] SIGHUP received. Attempting to restart [Wed Mar 31 17:14:43 2004] [notice] seg fault or similar nasty error detected in the parent process [Wed Mar 31 17:14:48 2004] [warn] pid file \/var\/run\/httpd.pid overwritten -- Unclean shutdown of previous Apache run? \n\nTo fix, upgrade to PHP 4.3.7 or higher, which fixes it (PHP 5.0 or higher also fixes the bug). A workaround is to stop\/start Apache, instead of restarting Apache. For details, see\u00a0PHP bug 27810.\n\n\u2022\u00a0ServerRoot\u00a0Installation Directory\n\nBy default, Apache is installed in ServerRoot, as specified in httpd.conf or the httpd -d option:\nServerRoot \"\/usr\/local\/apache\"\nSome packagers and distributions move the location to\u00a0\/usr\/local\/httpd\u00a0or\u00a0\/usr\/local\/apache2\u00a0so check that Apache is installed where you say it is, and check that you don't have two separate Apache installations by mistake.\n\n\u2022\u00a0PHP-required Apache files must be present\n\nPHP expects certain files to be present. They may be missing for some Linux distributions, which tend to separate files among the\u00a0\/var\/etc, and other directories. The\u00a0apxs\u00a0program expects to find httpd under the\u00a0bin\u00a0directory where apache is installed (by default\u00a0\/usr\/local\/apache\/bin). PHP expects to find the\u00a0os.h\u00a0and\u00a0unixd.h\u00a0header files under the\u00a0include\u00a0directory where apache is installed (by default\u00a0\/usr\/local\/apache\/include).\n\n\u2022\u00a0MySQL\n\nUse the\u00a0--with-mysql\u00a0configure option to build PHP with PHP's version of MySql. If you want the system's MySQL, or are also using MySql with another apache module (e.g., mod_perl), use this:\u00a0with-mysql=\/usr\n\nMake sure the MySQL developent headers are installed. This is usually package\u00a0mysql-devel\u00a0(thanks to Dev for the tip)\n\n\u2022\u00a0Custom OpenSSL and SSL\n\nWhen using \"--enable-ssl\" (i.e, build with OpenSSL), the configure script will try to guess the location of OpenSSL. If you're using your own custom SSL, you must specify the location. For example, \"--with-ssl=\/usr\/local\/ssl\" (of course, you can use the SSL and PHP that comes with Redhat 9 and other distributions).\n\nPlease note that mod_ssl is only for Apache 1.x. Apache 2.x does not use mod_ssl\u2014SSL is built into Apache 2.x.\n\nJason reports that with RedHat 9 on his system, the make failed at mod_ssl. It needed a missing krb5.h header file. The fix is to Add an environment variable, \\$CFLAGS, before making PHP. That is:\n\"-I\/usr\/kerberos\/include\/ -L\/usr\/kerberos\/lib\"\n\n\u2022\u00a0gdbm\nIf you get this error when using \"--with-gdbm\":\nconfigure: error: DBA: Could not find necessary header file(s).\nthen you are probably missing the gdbm-devel rpm package.\n\n\u2022\u00a0Apache 1.x and Apache 2 Configuration Directives\nThe \"AddModule,\" \"AgentLog,\" \"ReferrerLog,\" \"RefererIgnore,\" \"BindAddress,\" \"ClearModuleList,\" \"Port,\" \"FancyIndexing,\" \"ServerType,\" \"AccessConfig,\" and \"ResourceConfig\" directives in Apache 1.x\u00a0no longer exist\u00a0in Apache 2. Also, \"HAVE_*\" httpd.conf definitions no longer exist. Comment them out of your httpd.conf for Apache 2. Most of these directives have been replaced by other directive(s).\n\n\u2022\u00a0Error after glibc library upgrade\n\nApache should still work after upgrading glibc with security and bug fixes for the same release. If you use another architecture (say from or to i386\/i686), you need to recompile. One symptom is an error like this in your Apache error_log:\n[error] (38)Function not implemented:\n\n\u2022\u00a0make clean\n\nK\u00e5re Olaussen adds this note: If you're compiling php4 for both Apache 1.3 and Apache 2, use a separate php source directory for each version of Apache. \"The module wouldn't load into Apache 2, because I had forgotten to run make clean in between.\" [Ed. note: I recommend untaring the source from scratch in another directory, that way you are sure no old state information is left behind from previous makes. Do this for any configuration or version changes.]\n\n\u2022\u00a0Enabling\u00a0register_globals\u00a0for Apache 2\n\nFrom Duke: For those of you that are desperately trying to enable\u00a0register_globals\u00a0on a per-directory basis using Apache 2 with PHP 4 [or PHP 5] using.htaccess\u00a0files, the syntax is different from Apache 1.x. In the directory you want to enable\u00a0register_globals\u00a0(or pretty much any other on\/off setting), add an\u00a0.htaccess\u00a0file containing:\n\n php_value register_globals 1 \nNote that\u00a0php_flag\u00a0does\u00a0not\u00a0work with Apache 2. Neither does\u00a0on\u00a0\/\u00a0off\u00a0\u2014 you must use\u00a00\u00a0(off) or\u00a01\u00a0(on).\n\n[Editor's note: To enable globally, add\u00a0register_globals=off\u00a0to your\u00a0php.ini\u00a0file.]\n\nTo enable\u00a0register_globals\u00a0on a per-directory basis, you can add to a\u00a0<Directory>\u00a0stanza or create a\u00a0.htaccess\u00a0configuration file in the directory. To use.htaccess, you must first enable it by adding\u00a0AllowOverride All\u00a0(or some combination of\u00a0AllowOverride\u00a0options within the correct\u00a0<Directory>\u00a0stanza. Also newer PHP software, such as SquirrelMail, no longer require\u00a0register_globals\u00a0to be enabled. Keeping\u00a0register_globals\u00a0disalbed makes the system much more secure (as it's harder to set internal php variables through submitting html forms). ]\n\n\u2022\u00a0Multi-Processing Module (MPM): don't use multi-threading MPMs\n\nMPMs is how Apache 2 handles multiple web server requests. That is, with multiple processes or multiple threads or some combination. For now, on Linux (and even UNIX) you should only use the (default) prefork module with PHP. This is specified at compile time. Other MPM modules (any involving threads) break PHP. This is partly because PHP uses a great number of external libraries, and many or most of them are not thread-safe or thread-aware. In any case, Linux 2.4 doesn't handle threads efficiently yet\u2014multiple processes are better (this changes with Linux 2.6, or RedHat 9 with 2.6 threads backported to Linux 2.4).\n\n\u2022\u00a0API module structure php5_module' in file . . . is garbled\n\nIf you get a message like this:\n httpd: Syntax error on line 195 of \/usr\/local\/apache\/conf\/httpd.conf: API module structure php5_module' in file \/usr\/local\/apache\/modules\/libphp5.so is garbled - perhaps this is not an Apache module DSO \nit's because your version of PHP wasn't built for the version of Apache you are using (For PHP 4, substitute\u00a0php4_module\u00a0. . .\u00a0libphp4.so). Recompile PHP in a \"fresh\" source directory (where you haven't built PHP before) and make sure\u00a0--prefix--with-config-file\u00a0and\u00a0--with-apxs2\u00a0points to the correct Apache directories you are using and the\u00a0libphp4.so\u00a0file has been copied to your\u00a0\/usr\/local\/apache\/modules\u00a0directory.\n\n\u2022\u00a0PATH_INFO\n\nDan Fitzpatrick notes that he uses PATH_INFO for many PHP scripts like\u00a0\/index.php\/go\/do\/something\u00a0(where parameters are passed to PHP as \"fake\" subdirectories). He received a\u00a0404 Not Found\u00a0errors with Apache 2 but not Apache 1.3.27. He had to add \"AcceptPathInfo On\" to file\u00a0httpd.conf. For details, seehttp:\/\/httpd.apache.org\/docs-2.2\/mod\/core.html#acceptpathinfo\n\n\u2022\u00a0Using\u00a0<? \u00a0 ?>\u00a0style tags in addition to\u00a0<?php \u00a0 ?>\u00a0style tags\n\nIf you want to also use the old-style (or MS ASP-like) tags\u00a0<?\u00a0and\u00a0?>\u00a0then add this line to your\u00a0php.ini\u00a0file. This is the default starting with PHP 5:\n\n short_open_tag = On \n\n\u2022\u00a0GD library now built-in PHP\n\nPHP 4.3.0 has gd built-in, just pass\u00a0--with-gd\u00a0to PHP configure.\n\n\u2022\u00a0--with-apache\n\nIf you don't have Apache source at the default directory at\u00a0\/usr\/local\/apache, then use\u00a0--with-apache=DIR\u00a0with PHP configure, where DIR is the root source directory, This tells PHP where to look for the Apache source.\n\n\u2022\u00a0Redhat 9\nApache 2 and PHP is built-in RedHat 9 (packages httpd 2.0.40, and php 4.2.2). Apache is mostly under\u00a0\/usr\/lib\/httpd\u00a0and\u00a0\/usr\/sbin\/httpd. PHP is at\/usr\/lib\/httpd\/modules\/libphp4.so\u00a0For a complete file listing, type:\u00a0rpm -ql httpd php\n\nFor RedHat 9, PHP 4.3.1 gives a compiler errors with RedHat 9 similar to:\u00a0my_malloc.c:24: undefined reference to `errno'\nThe problem is with mysql and will have to wait for a fix from them. Until then, add this line:\n#include <errno.h>\nto the beginning of file\u00a0.\/ext\/mysql\/libmysql\/mysql.h\u00a0in your PHP source and remake PHP (from scratch).\n\n\u2022\u00a0Notes about Gentoo Linux and Apache2 with PHP\n\nbecomes:\n\nTo start automatically under Gentoo:\n(change\u00a0php5\u00a0to\u00a0php4\u00a0for PHP 4)\n\n[Thanks to Phillip Temple]\n\n\u2022 \u5e7f\u544a\n\u2022 \u6284\u88ad\n\u2022 \u7248\u6743\n\u2022 \u653f\u6cbb\n\u2022 \u8272\u60c5\n\u2022 \u65e0\u610f\u4e49\n\u2022 \u5176\u4ed6\n\n120","date":"2019-01-16 06:17:45","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.30373236536979675, \"perplexity\": 7746.92128716567}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 20, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2019-04\/segments\/1547583656897.10\/warc\/CC-MAIN-20190116052151-20190116074151-00037.warc.gz\"}"} | null | null |
\section{Introduction}
Let $X$ be a smooth complex projective curve of genus $g\ge2$ and let $G$ be $\mathrm{GL}(r,\mathbb{C})$ or $\mathrm{SL}(r,\mathbb{C})$. Let $\mathbf{M}_{\mathrm{Dol}}^{d}(G)$ be the moduli space of $G$-Higgs bundles $(E,\phi)$ of rank $r$ and degree $d$ on $X$ (with fixed $\det E$ and traceless $\phi$ in the case $G=\mathrm{SL}(r,\mathbb{C})$) and let $\mathbf{M}_{\mathrm{B}}^{d}(G)$ be the character variety for $G$ defined by
$$\mathbf{M}_{\mathrm{B}}^{d}(G)=\{(A_{1},B_{1},\cdots,A_{g},B_{g})\in G\,|\,[A_{1},B_{1}]\cdots[A_{g},B_{g}]=e^{2\pi\sqrt{-1}d/r}I_{r}\}/\!/G.$$
By the theory of harmonic bundles (\cite{Co88}, \cite{Simp92}), we have a homeomorphism $\mathbf{M}_{\mathrm{Dol}}^{d}(G)\cong\mathbf{M}_{\mathrm{B}}^{d}(G)$ as a part of the nonabelian Hodge theory. If $r$ and $d$ are relatively prime, these moduli spaces are smooth and their underlying differentiable manifold is hyperk\"{a}hler. But the complex structures do not coincide under this homeomorphism.
Under the assumption that $r$ and $d$ are relatively prime, motivated by this fact, there have been several works calculating invariants of these moduli spaces on both sides over the last 30 years. The Poincar\'{e} polynomial of the ordinary cohomology is calculated, for $\mathbf{M}_{\mathrm{Dol}}^{d}(\mathrm{SL}(2,\mathbb{C}))$ by N. Hitchin in \cite{Hit87}, and for $\mathbf{M}_{\mathrm{Dol}}^{d}(\mathrm{SL}(3,\mathbb{C}))$ by P. Gothen in \cite{Go94}. The compactly supported Hodge polynomial and the compactly supported Poincar\'{e} polynomial for $\mathbf{M}_{\mathrm{Dol}}^{d}(\mathrm{GL}(4,\mathbb{C}))$ can be obtained by the motivic calculation in \cite{GHS14}. By counting the number of points of these moduli spaces over finite fields with large characteristics, the compactly supported Poincar\'{e} polynomials for $\mathbf{M}_{\mathrm{Dol}}^{d}(\mathrm{GL}(r,\mathbb{C}))$ and $\mathbf{M}_{\mathrm{B}}^{d}(\mathrm{GL}(r,\mathbb{C}))$ are obtained in \cite{Sch16}. By using arithmetic methods, T. Hausel and F. Rodriguez-Villegas expresses the E-polynomial of $\mathbf{M}_{\mathrm{B}}^{d}(\mathrm{GL}(r,\mathbb{C}))$ in terms of a simple generating function in \cite{HR08}. By the same way, M. Mereb calculates the E-polynomial of $\mathbf{M}_{\mathrm{B}}^{d}(\mathrm{SL}(2,\mathbb{C}))$ and expresses the E-polynomial of $\mathbf{M}_{\mathrm{B}}^{d}(\mathrm{SL}(r,\mathbb{C}))$ in terms of a generating function in \cite{Me15}.
Without the assumption that $r$ and $d$ are relatively prime, there have been also some works calculating invariants of $\mathbf{M}_{\mathrm{B}}^{d}(\mathrm{SL}(r,\mathbb{C}))$. For $g=1,2$ and any $d$, explicit formulas for the E-polynomials of $\mathbf{M}_{\mathrm{B}}^{d}(\mathrm{SL}(2,\mathbb{C}))$ are obtained by a geometric technique in \cite{LMN13}. The E-polynomial of $\mathbf{M}_{\mathrm{B}}^{d}(\mathrm{SL}(2,\mathbb{C}))$ is calculated, for $g=3$ and any $d$ by a further geometric technique in \cite{MM16-1}, and for any $g$ and $d$ in \cite{MM16-2}.
When we deal with a singular variety $\mathbf{M}_{\mathrm{Dol}}^{d}(G)$ under the condition that $r$ and $d$ are not relatively prime, the intersection cohomology is a natural invariant. Our interest is focused on the intersection cohomology of $\mathbf{M}:=\mathbf{M}_{\mathrm{Dol}}^0(\mathrm{SL}(2,\mathbb{C}))$.
For a quasi-projective variety $V$, $IH^{i}(V)$ and $\mathbf{IC}^{\bullet}(V)$ denote the $i$-th intersection cohomology of $V$ of the middle perversity and the complex of sheaves on $V$ whose hypercohomology is $IH^{*}(V)$ respectively. $IP_{t}(V)$ denotes the Poincar\'{e} polynomial of $IH^{*}(V)$ defined by
$$IP_{t}(V)=\sum_{i}\dim IH^{i}(V).$$
Recently, $IP_{t}(\mathbf{M})$ was calculated in \cite{Ma21} by using ways different from ours. First of all, the $E$-polynomial of the compactly supported intersection cohomology of $\mathbf{M}$ was calculated in \cite{Fel18} over a smooth curve of genus $2$ and in \cite{Ma21} over a smooth curve of genus $g\ge 2$. Then the author of \cite{Ma21} proved the purity of $IH^{*}(\mathbf{M})$ from the observation of the semiprojectivity of $\mathbf{M}$. He used the purity of $IH^{*}(\mathbf{M})$ and the Poincar\'{e} duality for the intersection cohomology (Theorem \ref{GPD}) to calculate $IP_{t}(\mathbf{M})$.
From now on, $\mathrm{GL}(n,\mathbb{C})$, $\mathrm{SL}(n,\mathbb{C})$, $\mathrm{PGL}(n,\mathbb{C})$, $\mathrm{O}(n,\mathbb{C})$ and $\mathrm{SO}(n,\mathbb{C})$ will be denoted by $\mathrm{GL}(n)$, $\mathrm{SL}(n)$, $\mathrm{PGL}(n)$, $\mathrm{O}(n)$ and $\mathrm{SO}(n)$ respectively for the simplicities of notations.
\subsection{Main result}
In this paper, we prove that a blowing-up formula for $IH^{*}(\mathbf{M})$ holds.
It is known that $\mathbf{M}$ is a good quotient $\mathbf{R}/\!/\mathrm{SL}(2)$ for some quasi-projective variety $\mathbf{R}$ (Theorem \ref{M is a good quotient}, Theorem \ref{R and M are quasi-projective}). $\mathbf{M}$ is decomposed into
$$\mathbf{M}^{s}\bigsqcup(T^{*}J/\mathbb{Z}_{2}-\mathbb{Z}_{2}^{2g})\bigsqcup\mathbb{Z}_{2}^{2g},$$
where $\mathbf{M}^{s}$ denotes the stable locus of $\mathbf{M}$ and $J:=\mathrm{Pic}^0(X)$. The singularity along the locus $\mathbb{Z}_{2}^{2g}$ is the quotient $\Upsilon^{-1}(0)/\!/\mathrm{SL}(2)$, where $\Upsilon^{-1}(0)$ is the affine cone over a reduced irreducible complete intersection of three quadrics in $\mathbb{P}(\mathbb{C}^{2g}\otimes sl(2))$ and $\mathrm{SL}(2)$ acts on $\mathbb{C}^{2g}\otimes sl(2)$ as the adjoint representation. The singularity along the locus $T^{*}J/\mathbb{Z}_{2}-\mathbb{Z}_{2}^{2g}$ is $\Psi^{-1}(0)/\!/\mathbb{C}^{*}$, where $\Psi^{-1}(0)$ is the affine cone over a smooth quadric in $\mathbb{P}((\mathbb{C}^{g-1})^{4})$ and $\mathbb{C}^{*}$ acts on $(\mathbb{C}^{g-1})^{4}$ with weights $-2,2,2$ and $-2$. Let us consider the Kirwan's algorithm consisting of three blowing-ups $\mathbf{K}:=\mathbf{R}_{3}^{s}/\mathrm{SL}(2)\to\mathbf{R}_{2}^{s}/\mathrm{SL}(2)\to\mathbf{R}_{1}^{ss}/\!/\mathrm{SL}(2)\to\mathbf{R}/\!/\mathrm{SL}(2)=\mathbf{M}$ induced from the composition of blowing-ups $\pi_{\mathbf{R}_{1}}:\mathbf{R}_{1}\to\mathbf{R}$ along the locus $\mathbb{Z}_{2}^{2g}$, $\pi_{\mathbf{R}_{2}}:\mathbf{R}_{2}\to\mathbf{R}_{1}^{ss}$ along the strict transform $\Sigma$ of the locus over $T^{*}J/\mathbb{Z}_{2}-\mathbb{Z}_{2}^{2g}$ and $\mathbf{R}_{3}\to\mathbf{R}_{2}^{s}$ along the locus of points with stabilizers larger than the center $\mathbb{Z}_{2}$ in $\mathrm{SL}(2)$ (Section \ref{Kirwan desing}).
We also have local pictures of the Kirwan's algorithm mentioned above. For any $x\in\mathbb{Z}_{2}^{2g}$, we have $\pi_{\mathbf{R}_{1}}^{-1}(x)=\mathbb{P}\Upsilon^{-1}(0)$ which is a subset of $\mathbb{P}\mathrm{Hom}(sl(2),\mathbb{C}^{2g})$ and $\pi_{\mathbf{R}_{1}}^{-1}(x)\cap\Sigma$ is the locus of rank $1$ matrices (Section \ref{local picture}). Thus the strict transform of $\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{PGL}(2)$ in the second blowing-up of the Kirwan's algorithm is just the blowing-up $$Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss}/\!/\mathrm{PGL}(2)\to\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{PGL}(2)$$
along the image of the locus of rank $1$ matrices in $\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{PGL}(2)$.
In this setup, we have the following main result.
\begin{theorem}[Theorem \ref{computable intersection blowing-up formula}]\label{main theorem}\label{thm1}
Let $I_{2g-3}$ be the incidence variety given by
$$I_{2g-3}=\{(p,H)\in\mathbb{P}^{2g-3}\times\breve{\mathbb{P}}^{2g-3}|p\in H\}.$$
Then we have
(1) $\dim IH^{i}(\mathbf{R}_{1}^{ss}/\!/\mathrm{SL}(2))=\dim IH^{i}(\mathbf{M})$
$$+2^{2g}\dim IH^{i}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{PGL}(2))-2^{2g}\dim IH^{i}(\Upsilon^{-1}(0)/\!/\mathrm{PGL}(2))$$
for all $i\ge0$.
(2) $\dim H^{i}(\mathbf{R}_{2}^{s}/\mathrm{SL}(2))=\dim IH^{i}(\mathbf{R}_{1}^{ss}/\!/\mathrm{SL}(2))$
$$+\sum_{p+q=i}\dim[H^{p}(\widetilde{T^{*}J})\otimes H^{t(q)}(I_{2g-3})]^{\mathbb{Z}_{2}}$$
for all $i\ge0$, where $t(q)=q-2$ for $q\le\dim I_{2g-3}=4g-7$ and $t(q)=q$ otherwise, where $\alpha:\widetilde{T^{*}J}\to T^{*}J$ is the blowing-up along $\mathbb{Z}_{2}^{2g}$.
(3) $\dim IH^{i}(Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss}/\!/\mathrm{SL}(2))=\dim IH^{i}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))$
$$+\sum_{p+q=i}\dim[H^{p}(\mathbb{P}^{2g-1})\otimes H^{t(q)}(I_{2g-3})]^{\mathbb{Z}_{2}}$$
for all $i\ge0$, where $t(q)=q-2$ for $q\le\dim I_{2g-3}=4g-7$ and $t(q)=q$ otherwise.
\end{theorem}
It is an essential process to apply this blowing-up formula to calculate $IP_{t}(\mathbf{M})$.
\subsection{Method of proof of Theorem \ref{main theorem}}
We follow the same steps as in the proof of \cite[Proposition 2.1]{K86}, but we give a proof in each step because our setup is different from that of \cite{K86}. We start with the following formulas coming from the decomposition theorem (Proposition \ref{3 consequences}-(1)) and the same argument as in the proof of \cite[Lemma 2.8]{K86} :
$$\dim IH^{i}(\mathbf{R}_{1}^{ss}/\!/\mathrm{SL}(2))=\dim IH^{i}(\mathbf{M})+\dim IH^{i}(\tilde{U}_{1})-\dim IH^{i}(U_{1}),$$
$$\dim IH^{i}(\mathbf{R}_{2}^{s}/\mathrm{SL}(2))=\dim IH^{i}(\mathbf{R}_{1}^{ss}/\!/\mathrm{SL}(2))+\dim IH^{i}(\tilde{U}_{2})-\dim IH^{i}(U_{2})$$
and
$$\dim IH^{i}(Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss}/\!/\mathrm{SL}(2))=\dim IH^{i}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))+\dim IH^{i}(\tilde{U})-\dim IH^{i}(U),$$
where $U_{1}$ is a disjoint union of sufficiently small open neighborhoods of each point of $\mathbb{Z}_{2}^{2g}$ in $\mathbf{M}$, $\tilde{U}_{1}$ is the inverse image of the first blowing-up, $U_{2}$ is an open neighborhood of the strict transform of $T^{*}J/\mathbb{Z}_{2}$ in $\mathbf{R}_{1}/\!/\mathrm{SL}(2)$, $\tilde{U}_{2}$ is the inverse image of the second blowing-up, $U$ is an open neighborhood of the locus of rank $1$ matrices in $\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{PGL}(2)$ and $\tilde{U}$ is the inverse image of the blowing-up map $Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss}/\!/\mathrm{PGL}(2)\to\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{PGL}(2)$. By Proposition \ref{normal slice is the quadratic cone}, we can see that $U_{1},\tilde{U}_{1},U_{2},\tilde{U}_{2},U$ and $\tilde{U}$ are analytically isomorphic to relevant normal cones respectively. By Section \ref{Kirwan desing} and Lemma \ref{geometric descriptions of second cones}, these relevant normal cones are described as free $\mathbb{Z}_{2}$-quotients of nice fibrations with concrete expressions of bases and fibers. By the calculations of the intersection cohomologies of fibers (Lemma \ref{int coh of affine cone of git quotient} and Lemma \ref{suffices-to-show-on-projectivized-git}) and applying the perverse Leray spectral sequences (Proposition \ref{3 consequences}-(2)) of intersection cohomologies associated to these fibrations, we complete the proof.
\subsection{Towards a formula for the Poincar\'{e} polynomial of $IH^{*}(\mathbf{M})$}
For a topological space $W$ on which a reductive group $G$ acts, $H_{G}^{i}(W)$ and $P_{t}^{G}(W)$ denote the $i$-th equivariant cohomology of $W$ and the Poincar\'{e} series of $H_{G}^{*}(W)$ defined by
$$P_{t}^{G}(W)=\sum_{i}\dim H_{G}^{i}(W).$$
We start with the formula of $P_{t}^{\mathrm{SL}(2)}(\mathbf{R})$ that comes from that of \cite{DWW11}. Then we use a standard argument to obtain the follows.
\begin{lemma}[Lemma \ref{prototype of equivariant cohomology blowing-up formula}]
\begin{enumerate}
\item $P_{t}^{\mathrm{SL}(2)}(\mathbf{R}_{1})=P_{t}^{\mathrm{SL}(2)}(\mathbf{R})+2^{2g}(P_{t}^{\mathrm{SL}(2)}(\mathbb{P}\Upsilon^{-1}(0))-P_{t}(\mathrm{BSL}(2)))$.
\item $P_{t}^{\mathrm{SL}(2)}(\mathbf{R}_{2})=P_{t}^{\mathrm{SL}(2)}(\mathbf{R}_{1})+P_{t}^{\mathrm{SL}(2)}(E_2)-P_{t}^{\mathrm{SL}(2)}(\Sigma)$.
\end{enumerate}
\end{lemma}
Then we set up the following conjecture.
\begin{conjecture}[Conjecture \ref{equivariant blowing up formula conjecture}]
\begin{enumerate}
\item $P_{t}^{\mathrm{SL}(2)}(\mathbf{R}_{1}^{ss})=P_{t}^{\mathrm{SL}(2)}(\mathbf{R})+2^{2g}(P_{t}^{\mathrm{SL}(2)}(\mathbb{P}\Upsilon^{-1}(0)^{ss})-P_{t}(\mathrm{BSL}(2)))$.
\item $P_{t}^{\mathrm{SL}(2)}(\mathbf{R}_{2}^{s})=P_{t}^{\mathrm{SL}(2)}(\mathbf{R}_{1}^{ss})+P_{t}^{\mathrm{SL}(2)}(E_2^{ss})-P_{t}^{\mathrm{SL}(2)}(\Sigma)$.
\end{enumerate}
\end{conjecture}
We use this conjectural blowing-up formula for the equivariant cohomology to get $P_{t}^{\mathrm{SL}(2)}(\mathbf{R}_{2}^{s})$ from $P_{t}^{\mathrm{SL}(2)}(\mathbf{R})$. Since $\mathbf{R}_{2}^{s}/\mathrm{SL}(2)$ has at worst orbifold singularities (Section \ref{Kirwan desing}), $P_{t}^{\mathrm{SL}(2)}(\mathbf{R}_{2}^{s})=P_{t}(\mathbf{R}_{2}^{s}/\mathrm{SL}(2))$ (Section \ref{strategy}). Now we use the blowing-up formula for the intersection cohomology (Theorem \ref{thm1}) to get $IP_{t}(\mathbf{M})$ from $P_{t}(\mathbf{R}_{2}^{s}/\mathrm{SL}(2))$.
\begin{proposition}[Proposition \ref{A conjectural formula for IP(M)}]
Assume that Conjecture \ref{equivariant blowing up formula conjecture} holds. Then
$$IP_{t}(\mathbf{M})=\frac{(1+t^3)^{2g}-(1+t)^{2g}t^{2g+2}}{(1-t^2)(1-t^4)}$$
$$-t^{4g-4}+\frac{t^{2g+2}(1+t)^{2g}}{(1-t^2)(1-t^4)}+\frac{(1-t)^{2g}t^{4g-4}}{4(1+t^2)}$$
$$+\frac{(1+t)^{2g}t^{4g-4}}{2(1-t^2)}(\frac{2g}{t+1}+\frac{1}{t^2-1}-\frac{1}{2}+(3-2g))$$
$$+\frac{1}{2}(2^{2g}-1)t^{4g-4}((1+t)^{2g-2}+(1-t)^{2g-2}-2)$$
$$+2^{2g}\big[\big(\frac{1-t^{12}}{1-t^2}-\frac{1-t^6}{1-t^2}+(\frac{1-t^6}{1-t^2})^{2}\big)\frac{(1-t^{4g-8})(1-t^{4g-4})(1-t^{4g})}{(1-t^2)(1-t^4)(1-t^6)}$$
$$-\frac{1-t^6}{1-t^2}\frac{(1-t^{4g-4})(1-t^{4g})}{(1-t^2)(1-t^4)}\frac{t^2(1-t^{2(2g-5)})}{1-t^2}$$
$$-\frac{(1-t^{4g-4})^{2}}{(1-t^2)(1-t^4)}\cdot\frac{1-t^{4g}}{1-t^2}+\frac{1}{1-t^4}\frac{1-t^{4g}}{1-t^2}\big]-\frac{2^{2g}}{1-t^4}$$
$$+(\frac{1}{2}((1+t)^{2g}+(1-t)^{2g})+2^{2g}(\frac{1-t^{4g}}{1-t^2}-1))\frac{(1-t^{4g-4})^{2}}{(1-t^2)(1-t^4)}$$
$$+\frac{1}{2}((1+t)^{2g}-(1-t)^{2g})\frac{t^2(1-t^{4g-4})(1-t^{4g-8})}{(1-t^2)(1-t^4)}$$
$$-\frac{1}{(1-t^4)}(\frac{1}{2}((1+t)^{2g}+(1-t)^{2g})+2^{2g}(\frac{1-t^{4g}}{1-t^2}-1))$$
$$-\frac{t^2}{(1-t^4)}\frac{1}{2}((1+t)^{2g}-(1-t)^{2g})$$
$$-\frac1 2(
(1+t)^{2g}+(1-t)^{2g})+2^{2g}(\frac{1-t^{4g}}{1-t^2}-1))\frac{t^2(1-t^{4g-4})(1-t^{4g-6})}{(1-t^2)(1-t^4)}$$
$$-\frac1 2(
(1+t)^{2g}-(1-t)^{2g})
(\frac{t^4(1-t^{4g-4})(1-t^{4g-10})}{(1-t^2)(1-t^4)}+t^{4g-6})$$
$$-2^{2g}\big[\frac{(1-t^{8g-8})(1-t^{4g})}{(1-t^{2})(1-t^{4})}-\frac{1-t^{4g}}{1-t^{4}}\big]$$
which is a polynomial with degree $6g-6$.
\end{proposition}
This conjectural formula for $IP_{t}(\mathbf{M})$ coincides with that of \cite{Ma21}.
\subsection*{Notations}
Throughout this paper, $X$ denotes a smooth complex projective curve of genus $g\ge2$ and $K_X$ the canonical bundle of $X$.
\subsection*{Acknowledgements}
I would like to thank Young-Hoon Kiem for suggesting problem, and for his help and encouragement. This work is based and developed on the second topic in my doctoral dissertation \cite{Y09}.
\section{Higgs bundles}
In this section, we introduce two kinds of constructions of the moduli space of Higgs bundles on $X$. For details, see \cite{Hit87}, \cite{Simp94I} and \cite{Simp94II}.
\subsection{Simpson's construction}
An \textbf{$\mathrm{SL}(2)$-Higgs bundle} on $X$ is a pair of a rank $2$ vector bundle $E$ with trivial determinant on $X$ and a section $\phi\in H^0(X,\mathrm{End}_{0}(E)\otimes K_{X})$, where $\mathrm{End}(E)$ denotes the bundle of endomorphisms of $E$ and $\mathrm{End}_0(E)$ the subbundle of traceless endomorphisms of $\mathrm{End}(E)$. We must impose a notion of stability to construct a separated moduli space.
\begin{definition}[\cite{Hit87}, \cite{Simp94I}]
An $\mathrm{SL}(2)$-Higgs bundle $(E,\phi)$ on $X$ is \textbf{stable} (respectively, \textbf{semistable}) if for any $\phi$-invariant line subbundle $F$ of $E$, we have
$$\deg(F)<0\text{ (respectively, }\le\text{)}.$$
\end{definition}
Let $N$ be a sufficiently large integer and $p=2N+2(1-g)$. We list C.T. Simpson's results to construct a moduli space of $\mathrm{SL}(2)$-Higgs bundles.
\begin{theorem}[Theorem 3.8 of \cite{Simp94I}]
There is a quasi-projective scheme $Q$ representing the moduli functor which parametrizes the isomorphism classes of triples $(E,\phi,\alpha)$ where $(E,\phi)$ is a semistable $\mathrm{SL}(2)$-Higgs bundle and $\alpha$ is an isomorphism $\alpha:\mathbb{C}^{p}\to H^0(X,E\otimes\mathcal{O}_{X}(N))$.
\end{theorem}
\begin{theorem}[Theorem 4.10 of \cite{Simp94I}]
Fix $x\in X$. Let $\tilde{Q}$ be the frame bundle at $x$ of the universal object restricted to $x$. Then the action of $\mathrm{GL}(p)$ lifts to $\tilde{Q}$ and $\mathrm{SL}(2)$ acts on the fibers of $\tilde{Q}\to Q$ in an obvious fashion. Every point of $\tilde{Q}$ is stable with respect to the free action of $\mathrm{GL}(p)$ and
$$\mathbf{R}=\tilde{Q}/\mathrm{GL}(p)$$
represents the moduli functor which parametrizes triples $(E,\phi,\beta)$ where $(E,\phi)$ is a semistable $\mathrm{SL}(2)$-Higgs bundle and $\beta$ is an isomorphism $\beta:E|_{x}\to\mathbb{C}^{2}$.
\end{theorem}
\begin{theorem}[Theorem 4.10 of \cite{Simp94I}]
Every point in $\mathbf{R}$ is semistable with respect to the action of $\mathrm{SL}(2)$. The closed orbits in $\mathbf{R}$ correspond to polystable $\mathrm{SL}(2)$-Higgs bundles, i.e. $(E,\phi)$ is stable or $(E,\phi)=(L,\psi)\oplus(L^{-1},-\psi)$ for $L\in\mathrm{Pic}^0(X)$ and $\psi\in H^0(K_{X})$. The set $\mathbf{R}^{s}$ of stable points with respect to the action of $\mathrm{SL}(2)$ is exactly the locus of stable $\mathrm{SL}(2)$-Higgs bundles.
\end{theorem}
\begin{theorem}[Theorem 4.10 of \cite{Simp94I}]\label{M is a good quotient}
The good quotient $\mathbf{R}/\!/\mathrm{SL}(2)$ is $\mathbf{M}$.
\end{theorem}
\begin{theorem}[Theorem 11.1 of \cite{Simp94II}]\label{R and M are quasi-projective}
$\mathbf{R}$ and $\mathbf{M}$ are both irreducible normal quasi-projective varieties.
\end{theorem}
\subsection{Hitchin's construction}
Let $E$ be a complex Hermitian vector bundle of rank $2$ and degree $0$ on $X$. Let $\mathcal{A}$ be the space of traceless connections on $E$ compatible with the Hermitian metric. $\mathcal{A}$ can be identified with the space of holomorphic structures on $E$ with trivial determinant. Let
$$\mathcal{B}=\{(A,\phi)\in\mathcal{A}\times\Omega^0(\mathrm{End}_0(E)\otimes K_{X}):d''_{A}\phi=0\}.$$
Let $\mathcal{G}$ (respectively, $\mathcal{G}^{\mathbb{C}}$) be the gauge group of $E$ with structure group $SU(2)$ (respectively, $SL(2)$). These groups act on $\mathcal{B}$ by
$$g\cdot(A,\phi)=(g^{-1}A'' g+g^{*}A'(g^{*})^{-1}+g^{-1}d'' g-(d' g^{*})(g^{*})^{-1},g^{-1}\phi g),$$
where $A''$ and $A'$ denote the $(0,1)$ and $(1,0)$ parts of $A$ respectively.
The cotangent bundle $T^{*}\mathcal{A}\cong\mathcal{A}\times\Omega^0(\mathrm{End}_0(E)\otimes K_{X})$ admits a hyperk\"{a}hler structure preserved by the action of $\mathcal{G}$ with the moment maps for this action
$$\begin{matrix}\mu_{1}=F_{A}+[\phi,\phi^{*}]\\
\mu_{2}=-i(d''_{A}\phi+d'_{A}\phi^{*})\\
\mu_{3}=-d''_{A}\phi+d'_{A}\phi^{*}.\end{matrix}$$
$\mu_{\mathbb{C}}=\mu_{2}+i\mu_{3}=-2id''_{A}\phi$ is the complex moment map. Then
$$\mathcal{B}=\mu_{2}^{-1}(0)\cap\mu_{3}^{-1}(0)=\mu_{\mathbb{C}}^{-1}(0).$$
Consider the hyperk\"{a}hler quotient
$$\mathcal{M}:=T^{*}\mathcal{A}/\!/\!/\mathcal{G}=\mu_{1}^{-1}(0)\cap\mu_{2}^{-1}(0)\cap\mu_{3}^{-1}(0)/\mathcal{G}=\mu_{1}^{-1}(0)\cap\mathcal{B}/\mathcal{G}.$$
Let $\mathcal{B}^{ss}=\{(A,\phi)\in\mathcal{B}:((E,d''_{A}),\phi)\text{ is semistable}\}$.
\begin{theorem}[Theorem 2.1 and Theorem 4.3 of \cite{Hit87}, Theorem 1 and Proposition 3.3 of \cite{Simp88}]
$$\mathcal{M}\cong\mathcal{B}^{ss}/\!/\mathcal{G}^{\mathbb{C}}.$$
\end{theorem}
\section{Intersection cohomology theory}
In this section, we introduce some basics on the intersection cohomology (\cite{GM80}, \cite{GM83}) and the equivariant intersection cohomology (\cite{BL94}, \cite{GKM98}) of a quasi-projective complex variety. Let $V$ be a quasi-projective complex variety of pure dimension $n$ throughout this section.
\subsection{Intersection cohomology}
It is well-known that $V$ has a Whitney stratification
$$V=V_{n}\supseteq V_{n-1}\supseteq\cdots\supseteq V_{0}$$
which is embedded into a topological pseudomanifold of dimension $2n$ with filtration
$$W_{2n}\supseteq W_{2n-1}\supseteq\cdots\supseteq W_{0},$$
where $V_{j}$ are closed subvarieties such that $V_{j}-V_{j-1}$ is either empty or a nonsingular quasi-projective variety of pure dimension $j$ and $W_{2k}=W_{2k+1}=V_{k}$.
Let $\bar{p}=(p_{2},p_{3},\cdots,p_{2n})$ be a perversity. For a triangulation $T$ of $V$, $(C_{\bullet}^{T}(V),\partial)$ denotes the chain complex of chains with respect to $T$ with coefficients in $\mathbb{Q}$. We define $I^{\bar{p}}C_{i}^{T}(V)$ to be the subspace of $C_{i}^{T}(V)$ consisting of those chains $\xi$ such that
$$\dim_{\mathbb{R}}(|\xi|\cap V_{n-c})\le i-2c+p_{2c}$$
and
$$\dim_{\mathbb{R}}(|\partial\xi|\cap V_{n-c})\le i-1-2c+p_{2c}.$$
Let $IC_{i}^{\bar{p}}(V)=\displaystyle\varinjlim_{T}I^{\bar{p}}C_{i}^{T}(V)$. Then $(IC_{\bullet}^{\bar{p}}(V),\partial)$ is a chain complex. The $i$-th \textbf{intersection homology} of $V$ of perversity $\bar{p}$, denoted by $IH_{i}^{\bar{p}}(V)$, is the $i$-th homology group of the chain complex $(IC_{\bullet}^{\bar{p}}(V),\partial)$. The $i$-th \textbf{intersection cohomology} of $V$ of perversity $\bar{p}$, denoted by $IH_{\bar{p}}^{i}(V)$, is the $i$-th homology group of the chain complex $(IC_{\bullet}^{\bar{p}}(V)^{\vee},\partial^{\vee})$.
When we consider a chain complex $(IC_{\bullet}^{cl,\bar{p}}(V),\partial)$ of chains with closed support instead of usual chains, we can define the $i$-th \textbf{intersection homology with closed support} (respectively, \textbf{intersection cohomology with closed support}) of $V$ of perversity $\bar{p}$, denoted by $IH_{i}^{cl,\bar{p}}(V)$ (respectively, $IH_{cl,\bar{p}}^{i}(V)$)
There is an alternative way to define the intersection homology and cohomology with closed support. Let $\mathbf{IC}_{\bar{p}}^{-i}(V)$ be the sheaf given by $U\mapsto IC_{i}^{cl,\bar{p}}(U)$ for each open subset $U$ of $V$. Then $\mathbf{IC}_{\bar{p}}^{\bullet}(V)$ is a complex of sheaves as an object in the bounded derived category $D^{b}(V)$. Then we have $IH_{i}^{cl,\bar{p}}(V)=\mathcal{H}^{-i}(\mathbf{IC}_{\bar{p}}^{\bullet}(V))$ and $IH_{cl,\bar{p}}^{i}(V)=\mathcal{H}^{i-2\dim(V)}(\mathbf{IC}_{\bar{p}}^{\bullet}(V))$, where $\mathcal{H}^{i}(\mathbf{A}^{\bullet})$ is the $i$-th hypercohomology of a complex of sheaves $\mathbf{A}^{\bullet}$.
When $\bar{p}$ is the middle perversity $\bar{m}$, $IH_{i}^{\bar{m}}(V)$, $IH_{\bar{m}}^{i}(V)$, $IH_{i}^{cl,\bar{m}}(V)$, $IH_{cl,\bar{m}}^{i}(V)$ and $\mathbf{IC}_{\bar{m}}^{\bullet}(V)$ are denoted by $IH_{i}(V)$, $IH^{i}(V)$, $IH_{i}^{cl}(V)$, $IH_{cl}^{i}(V)$ and $\mathbf{IC}^{\bullet}(V)$ respectively.
\subsection{Equivariant intersection cohomology}
Assume that a compact connected algebraic group $G$ acts on $V$ algebraically. For the universal principal bundle $\mathrm{E} G\to \mathrm{B} G$, we have the quotient $V\times_{G}\mathrm{E} G$ of $V\times \mathrm{E} G$ by the diagonal action of $G$. Let us consider the following diagram
$$\xymatrix{V&V\times \mathrm{E} G\ar[l]_{p}\ar[r]^{q}&V\times_{G}\mathrm{E} G}.$$
\begin{definition}[2.1.3 and 2.7.2 in \cite{BL94}]
The \textbf{equivariant derived category} of $V$, denoted by $D_{G}^{b}(V)$, is defined as follows:
\begin{enumerate}
\item An object is a triple $(F_{V},\bar{F},\beta)$, where $F_{V}\in D^{b}(V)$, $\bar{F}\in D^{b}(V\times_{G}\mathrm{E} G)$ and $\beta:p^{*}(F_{V})\to q^{*}(\bar{F})$ is an isomorphism in $D^{b}(V\times \mathrm{E} G)$.
\item A morphism $\alpha:(F_{V},\bar{F},\beta)\to(G_{V},\bar{G},\gamma)$ is a pair $\alpha=(\alpha_{V},\bar{\alpha})$, where $\alpha_{V}:F_{V}\to G_{V}$ and $\bar{\alpha}:\bar{F}\to\bar{G}$ such that $\beta\circ p^{*}(\alpha_{V})=q^{*}(\bar{\alpha})\circ\beta$.
\end{enumerate}
\end{definition}
$\mathbf{IC}_{G,\bar{p}}^{\bullet}(V)$ (respectively, $\mathbb{Q}_{V}^{G}$) denotes $(\mathbf{IC}_{\bar{p}}^{\bullet}(V),\mathbf{IC}_{\bar{p}}^{\bullet}(V\times_{G}\mathrm{E} G),\beta)$ (respectively, $(\mathbb{Q}_{V},\mathbb{Q}_{V\times_{G}\mathrm{E} G},\mathrm{id} )$) as an object of $D_{G}^{b}(V)$. The $i$-th equivariant cohomology of $V$ can be obtained by $H_{G}^{i}(V)=\mathcal{H}^{-i}(\mathbb{Q}_{V\times_{G}\mathrm{E} G})$. The \textbf{equivariant intersection cohomology} of $V$ of perversity $\bar{p}$, denoted by $IH_{G,\bar{p}}^{*}(V)$, is defined by $IH_{G,\bar{p}}^{i}(V):=\mathcal{H}^{-i}(\mathbf{IC}_{\bar{p}}^{\bullet}(V\times_{G}\mathrm{E} G))$.
When $\bar{p}$ is the middle perversity $\bar{m}$, $IH_{G,\bar{m}}^{i}(V)$ and $\mathbf{IC}_{G,\bar{m}}^{\bullet}(V)$ are denoted by $IH_{G}^{i}(V)$ and $\mathbf{IC}_{G}^{\bullet}(V)$ respectively.
The equivariant cohomology and the equivariant intersection cohomology can be described as a limit of a projective limit system coming from a sequence of finite dimensional submanifolds of $\mathrm{E} G$. Let us consider a sequence of finite dimensional submanifolds $\mathrm{E} G_{0}\subset \mathrm{E} G_{1}\subset\cdots\subset \mathrm{E} G_{n}\subset\cdots$ of $\mathrm{E} G$, where $G$ acts on all of $\mathrm{E} G_{n}$ freely, $\mathrm{E} G_{n}$ are $n$-acyclic, $\mathrm{E} G_{n}\subset \mathrm{E} G_{n+1}$ is an embedding of a submanifold, $\dim \mathrm{E} G_{n}<\dim \mathrm{E} G_{n+1}$ and $\displaystyle \mathrm{E} G=\bigcup_{n}\mathrm{E} G_{n}$. Since $G$ is connected, such a sequence exists by \cite[Lemma 12.4.2]{BL94}. Then we have a sequence of finite dimensional subvarieties $V\times_{G}\mathrm{E} G_{0}\subset V\times_{G}\mathrm{E} G_{1}\subset\cdots\subset V\times_{G}\mathrm{E} G_{n}\subset\cdots$ of $V\times_{G}\mathrm{E} G$. Hence we have $\displaystyle H_{G}^{*}(V)=\varprojlim_{n}H^{*}(V\times_{G}\mathrm{E} G_{n})$ and $\displaystyle IH_{G,\bar{p}}^{*}(V)=\varprojlim_{n}IH_{\bar{p}}^{*}(V\times_{G}\mathrm{E} G_{n})$.
\subsection{The generalized Poincar\'{e} duality and the decomposition theorem}
In this subsection, we state two important theorems. One is the generalized Poincar\'{e} duality and the other is the decomposition theorem.
\begin{theorem}[The generalized Poincar\'{e} duality]\label{GPD}
If $\bar{p}+\bar{q}=\bar{t}$, then there is a non-degenerate bilinear form
$$IH_{i}^{\bar{p}}(V)\times IH_{2\dim(V)-i}^{cl,\bar{q}}(V)\to\mathbb{Q}.$$
\end{theorem}
\begin{theorem}[The decomposition theorem]\label{DT}
(1) Suppose that $\varphi:W\to V$ is a projective morphism of quasi-projective varieties. Then there is an isomorphism
$$R\varphi_{*}\mathbf{IC}^{\bullet}(W)\cong\bigoplus_{i}\leftidx{^p}\mathcal{H}^{i}(R\varphi_{*}\mathbf{IC}^{\bullet}(W))[-i]$$
in the derived category $D^{b}(V)$ and closed subvarieties $V_{i,\alpha}$ of $V$, local systems $L_{i,\alpha}$ on the non-singular parts $(V_{i,\alpha})_{nonsing}$ of $V_{i,\alpha}$ for each $i$ such that there is a canonical isomorphism
$$\leftidx{^p}\mathcal{H}^{i}(R\varphi_{*}\mathbf{IC}^{\bullet}(W))\cong\bigoplus_{\alpha}\mathbf{IC}^{\bullet}(V_{i,\alpha},L_{i,\alpha})$$
in $Perv(V)$, where $\leftidx{^p}\mathcal{H}$ is the perverse cohomology functor and $\mathbf{IC}^{\bullet}(V_{i,\alpha},L_{i,\alpha})$ is the complex of sheaves of intersection chains with coefficients in $L_{i,\alpha}$.
(2) Suppose that $\varphi:W\to V$ is a projective $G$-equivariant morphism of quasi-projective varieties. Then there exist closed subvarieties $V_{\alpha}$ of $V$, $G$-equivariant local systems $L_{\alpha}$ on the non-singular parts $(V_{\alpha})_{nonsing}$ of $V_{\alpha}$ and integers $l_{\alpha}$ such that there is an isomorphism
$$R\varphi_{*}\mathbf{IC}_{G}^{\bullet}(W)\cong\bigoplus_{\alpha}\mathbf{IC}_{G}^{\bullet}(V_{\alpha},L_{\alpha})[l_{\alpha}]$$
in the derived category $D_{G}^{b}(V)$, where $\mathbf{IC}_{G}^{\bullet}(V_{\alpha},L_{\alpha})$ is the complex of equivariant intersection cohomology sheaves with coefficients in $L_{\alpha}$.
\end{theorem}
There are three special important consequences of the decomposition theorem.
\begin{proposition}\label{3 consequences}
(1) Suppose that $\varphi:W\to V$ is a resolution of singularities. Then $\mathbf{IC}^{\bullet}(V)$ (respectively, $IH^{*}(V)$) is a direct summand of $R\varphi_{*}\mathbf{IC}^{\bullet}(W)$ (respectively, $IH^{*}(W)$).
(2) Suppose that $\varphi:W\to V$ is a projective surjective morphism. Then there is a perverse Leray spectral sequence $E_{r}^{ij}$ converging to $IH^{i+j}(W)$ with $E_2$ term $E_{2}^{ij}=IH^{i}(V,\leftidx{^p}\mathcal{H}^{j}R\varphi_{*}\mathbf{IC}^{\bullet}(W))$. The decomposition theorem for $\varphi$ is equivalent to the degeneration of $E_{r}^{ij}$ at the $E_{2}$ term.
(3) Suppose that $\varphi:W\to V$ is a $G$-equivariant resolution of singularities. Then $\mathbf{IC}_{G}^{\bullet}(V)$ (respectively, $IH_{G}^{*}(V)$) is a direct summand of $R\varphi_{*}\mathbf{IC}_{G}^{\bullet}(W)$ (respectively, $IH_{G}^{*}(W)$).
\end{proposition}
\begin{proof}
\begin{enumerate}
\item Applying the decomposition theorem to $\varphi$ and to the shifted constant sheaf $\mathbb{Q}_{W}[\dim W]$, we get the result. The details of the proof can be found in \cite[Corollary 5.4.11]{Dim04}.
\item The degeneration follows from the decomposition $$R\varphi_{*}\mathbf{IC}^{\bullet}(W)\cong\bigoplus_{j}\leftidx{^p}\mathcal{H}^{j}(R\varphi_{*}\mathbf{IC}^{\bullet}(W))[-j]$$
that comes from Theorem \ref{DT}-(1).
\item We know that $\mathbf{IC}_{G}^{\bullet}(V)=(\mathbf{IC}^{\bullet}(V),\mathbf{IC}^{\bullet}(V\times_{G}\mathrm{E} G),\alpha)$ and $\mathbf{IC}_{G}^{\bullet}(W)=(\mathbf{IC}^{\bullet}(W),\mathbf{IC}^{\bullet}(W\times_{G}\mathrm{E} G),\beta)$. It follows from item (1) that $\mathbf{IC}^{\bullet}(V)$ is a direct summand of $R\varphi_{*}\mathbf{IC}^{\bullet}(W)$ and that $\mathbf{IC}^{\bullet}(V\times_{G}\mathrm{E} G_{n})$ is a direct summand of $R\varphi_{*}\mathbf{IC}^{\bullet}(W\times_{G}\mathrm{E} G_{n})$ for all $n$. Since $\mathbf{IC}^{\bullet}(V\times_{G}\mathrm{E} G)=\displaystyle\varprojlim_{n}\mathbf{IC}^{\bullet}(V\times_{G}\mathrm{E} G_{n})$ and $\mathbf{IC}^{\bullet}(W\times_{G}\mathrm{E} G)=\displaystyle\varprojlim_{n}\mathbf{IC}^{\bullet}(W\times_{G}\mathrm{E} G_{n})$, $\mathbf{IC}^{\bullet}(V\times_{G}\mathrm{E} G)$ is a direct summand of $R\varphi_{*}\mathbf{IC}^{\bullet}(W\times_{G}\mathrm{E} G)$.
Let $i:\mathbf{IC}^{\bullet}(V)\hookrightarrow R\varphi_{*}\mathbf{IC}^{\bullet}(W)$ and $\bar{i}:\mathbf{IC}^{\bullet}(V\times_{G}\mathrm{E} G)\hookrightarrow R\varphi_{*}\mathbf{IC}^{\bullet}(W\times_{G}\mathrm{E} G)$ be the inclusions from the decomposition theorem. It is easy to see that the following diagram
$$\xymatrix{&p_{V}^{*}\mathbf{IC}^{\bullet}(V)\ar[r]^{\alpha}\ar[d]_{p^{*}(i)}&q_{V}^{*}\mathbf{IC}^{\bullet}(V\times_{G}\mathrm{E} G)\ar[d]^{q^{*}(\bar{i})}&\\
R\varphi_{*}p_{W}^{*}\mathbf{IC}^{\bullet}(W)\ar@^{=}[r]&p_{V}^{*}R\varphi_{*}\mathbf{IC}^{\bullet}(W)\ar[r]_{R\varphi_{*}(\beta)\quad\quad}&q_{V}^{*}R\varphi_{*}\mathbf{IC}^{\bullet}(W\times_{G}\mathrm{E} G)\ar@^{=}[r]&R\varphi_{*}q_{W}^{*}\mathbf{IC}^{\bullet}(W\times_{G}\mathrm{E} G)}$$
commutes, where $p_{V}:V\times \mathrm{E} G\to V$ (respectively, $p_{W}:W\times \mathrm{E} G\to W$) is the projection onto $V$ (respectively, $W$) and $q_{V}:V\times \mathrm{E} G\to V\times_{G}\mathrm{E} G$ (respectively, $q_{W}:W\times \mathrm{E} G\to W\times_{G}\mathrm{E} G$) is the quotient.
\end{enumerate}
\end{proof}
\begin{remark}\label{Leray ss for intersection cohomology with smooth target}
Assume that $V$ is smooth in Proposition \ref{3 consequences}-(2). Then
$$E_{2}^{ij}=IH^{i}(V,\leftidx{^p}\mathcal{H}^{j}R\varphi_{*}\mathbf{IC}^{\bullet}(W))=IH^{i}(V,R^{j}\varphi_{*}\mathbf{IC}^{\bullet}(W)).$$
\end{remark}
\section{Kirwan's desingularization of $\mathbf{M}$}\label{Kirwan desing}
In this section, we briefly explain how $\mathbf{M}$ can be desingularized by three blowing-ups by the Kirwan's algorithm introduced in \cite{K85-2}. For details, see \cite{KY08} and \cite{O99}.
We first consider the loci of type (i) of $(L,0)\oplus(L,0)$ with $L\cong L^{-1}$ in $\mathbf{M}\setminus\mathbf{M}^{s}$ and in $\mathbf{R}\setminus\mathbf{R}^{s}$, where $\mathbf{R}^{s}$ is the stable locus of $\mathbf{R}$. The loci of type (i) in $\mathbf{M}$ and in $\mathbf{R}$ are both isomorphic to the set of $\mathbb{Z}_2$-fixed points $\mathbb{Z}_{2}^{2g}$ in $J:=\mathrm{Pic}^0(X)$ by the involution $L\mapsto L^{-1}$. The singularity of the locus $\mathbb{Z}_{2}^{2g}$ of type (i) in $\mathbf{M}$ is the quotient
$$\Upsilon^{-1}(0)/\!/\mathrm{SL}(2)$$
where $\Upsilon:[H^0(K_{X})\oplus H^1(\mathcal{O}_{X})]\otimes sl(2)\to H^1(K_{X})\otimes sl(2)$ is the quadratic map given by the Lie bracket of $sl(2)$ coupled with the perfect pairing $H^0(K_{X})\oplus H^1(\mathcal{O}_{X})\to H^1(K_{X})$ and the $\mathrm{SL}(2)$-action on $\Upsilon^{-1}(0)$ is induced from the adjoint representation $\mathrm{SL}(2)\to\mathrm{Aut} (sl(2))$.
Next we consider the loci of type (iii) of $(L,\psi)\oplus (L^{-1},-\psi)$ with $(L,\psi)\ncong (L^{-1},-\psi)$ in $\mathbf{M}\setminus\mathbf{M}^{s}$ and in $\mathbf{R}\setminus\mathbf{R}^s$. It is clear that the
locus of type (iii) in $\mathbf{M}$ is isomorphic to
$$J\times_{\mathbb{Z}_2}H^0(K_X)-\mathbb{Z}_2^{2g}\cong T^*J/\mathbb{Z}_2-\mathbb{Z}_2^{2g}$$
where $\mathbb{Z}_2$ acts on $J$ by $L\mapsto L^{-1}$ and on $H^0(K_X)$
by $\psi\mapsto -\psi$. The locus of type (iii) in $\mathbf{R}$ is a $\mathbb{P}
\mathrm{SL}(2)/\mathbb{C}^*$-bundle over $T^*J/\mathbb{Z}_2-\mathbb{Z}_2^{2g}$ and in particular
it is smooth. The singularity along the
locus of type (iii) in $\mathbf{M}$ is the quotient
\[ \Psi^{-1}(0)/\!/\mathbb{C}^*,\]
where $\Psi:[H^0(L^{-2}K_X)\oplus H^1(L^2)]\oplus[H^0(L^{2}K_X)\oplus H^1(L^{-2})]\to H^1(K_X)$ is the quadratic map given by the sum of perfect pairings $H^0(L^{-2}K_X)\oplus H^1(L^2)\to H^1(K_X)$ and $H^0(L^{2}K_X)\oplus H^1(L^{-2})\to H^1(K_X)$ over $(L,\psi)\oplus (L^{-1},-\psi)\in T^*J/\mathbb{Z}_2-\mathbb{Z}_2^{2g}$ and the $\mathbb{C}^{*}$-action on $\Psi^{-1}(0)$ is induced from the $\mathbb{C}^{*}$-action on $[H^0(L^{-2}K_X)\oplus H^1(L^2)]\oplus[H^0(L^{2}K_X)\oplus H^1(L^{-2})]$ given by
$$\lambda\cdot(a,b,c,d)=((\lambda^{-2}a,\lambda^{2}b),(\lambda^{2}c,\lambda^{-2}d)).$$
Since we have identical singularities as in \cite{O99}, we can follow K.G. O'Grady's arguments to construct the Kirwan's desingularization $\mathbf{K}$ of $\mathbf{M}$. Let $\mathbf{R}_{1}$ be the blowing-up of $\mathbf{R}$ along the locus $\mathbb{Z}_{2}^{2g}$ of type (i). Let $\mathbf{R}_{2}$ be the blowing-up of $\mathbf{R}_{1}^{ss}$ along the strict transform $\Sigma$ of the locus of type (iii), where $\mathbf{R}_{1}^{ss}$ is the locus of semistable points in $\mathbf{R}_{1}$. Let $\mathbf{R}_{2}^{ss}$ (respectively, $\mathbf{R}_{2}^{s}$) be the locus of semistable (respectively, stable) points in $\mathbf{R}_{2}$. Then it follows from the same argument as in \cite[Claim 1.8.10]{O99} that
\begin{enumerate}
\item[(a)] $\mathbf{R}_{2}^{ss}=\mathbf{R}_{2}^{s}$,
\item[(b)] $\mathbf{R}_{2}^{s}$ is smooth.
\end{enumerate}
In particular, $\mathbf{R}_{2}^{s}/\mathrm{SL}(2)$ has at worst orbifold singularities. When $g=2$, this is smooth. When $g\ge3$, we blow up $\mathbf{R}_{2}^{s}$ along the locus of points with stabilizers larger than the center $\mathbb{Z}_{2}$ of $\mathrm{SL}(2)$ to obtain a variety $\mathbf{R}_{3}$ such that the orbit space $\mathbf{K}:=\mathbf{R}_{3}^{s}/\mathrm{SL}(2)$ is a smooth variety obtained by blowing up $\mathbf{M}$ along $\mathbb{Z}_{2}^{2g}$, along the strict transform of $T^*J/\mathbb{Z}_2-\mathbb{Z}_2^{2g}$ and along a nonsingular subvariety contained in the strict transform of the exceptional divisor of the first blowing-up. $\mathbf{K}$ is called the \textbf{Kirwan's desingularization} of $\mathbf{M}$.
Throughout this paper, $\pi_{\mathbf{R}_{1}}:\mathbf{R}_{1}\to\mathbf{R}$ (respectively, $\pi_{\mathbf{R}_{2}}:\mathbf{R}_{2}\to\mathbf{R}_{1}^{ss}$) denotes the first blowing-up map (respectively, the second blowing-up map). $\overline{\pi}_{\mathbf{R}_{1}}:\mathbf{R}_{1}^{ss}/\!/\mathrm{SL}(2)\to\mathbf{R}/\!/\mathrm{SL}(2)$ and $\overline{\pi}_{\mathbf{R}_{2}}:\mathbf{R}_{2}^{ss}/\!/\mathrm{SL}(2)\to\mathbf{R}_{1}^{ss}/\!/\mathrm{SL}(2)$ denote maps induced from $\pi_{\mathbf{R}_{1}}$ and $\pi_{\mathbf{R}_{2}}$ respectively.
\section{Local pictures in Kirwan's algorithm on $\mathbf{R}$}\label{local picture}
In this section, we list local pictures that appear in Kirwan's algorithm on $\mathbf{R}$ for later use. For details, see \cite[1.6 and 1.7]{O99}.
We first observe that $\pi_{\mathbf{R}_{1}}^{-1}(x)=\mathbb{P}\Upsilon^{-1}(0)$ for any $x\in\mathbb{Z}_{2}^{2g}$. We identify $\mathbb{H}^{g}$ with $T_{x}(T^{*}J)=H^{1}(\mathcal{O}_X)\oplus H^{0}(K_X)$ for any $x\in T^{*}J$, where $\mathbb{H}$ is the division algebra of quaternions. Since the adjoint representation gives an identification $\mathrm{PGL}(2)\cong \mathrm{SO}(sl(2))$, $\mathrm{PGL}(2)$ acts on both $\Upsilon^{-1}(0)$ and $\mathbb{P}\Upsilon^{-1}(0)$. Since $\mathrm{PGL}(2)=\mathrm{SL}(2)/\{\pm\mathrm{id} \}$ and the action of $\{\pm\mathrm{id} \}$ on $\Upsilon^{-1}(0)$ and $\mathbb{P}\Upsilon^{-1}(0)$ are trivial,
$$\Upsilon^{-1}(0)/\!/\mathrm{SL}(2)=\Upsilon^{-1}(0)/\!/\mathrm{PGL}(2)\text{ and }\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2)=\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{PGL}(2).$$
We have an explicit description of semistable points of $\mathbb{P}\Upsilon^{-1}(0)$ with respect to the $\mathrm{PGL}(2)$-action as following.
\begin{proposition}[Proposition 1.6.2 of \cite{O99}]\label{ss-local-first-blowup}
A point $[\varphi]\in\mathbb{P}\Upsilon^{-1}(0)$ is $\mathrm{PGL}(2)$-semistable if and only
if:
$$\mathrm{rk}\,\varphi\begin{cases}\geq 2,&\text{or}\\
=1&\text{and }[\varphi]\in\mathrm{PGL}(2)\cdot\mathbb{P}\{\left(\begin{array}{cc}\lambda&0\\0&-\lambda\end{array}\right)|\lambda\in\mathbb{H}^{g}\setminus\{O\}\}.
\end{cases}$$
\end{proposition}
Let $\mathrm{Hom}^{\omega}(sl(2),\mathbb{H}^{g}):=\{\varphi:sl(2)\rightarrow\mathbb{H}^{g}|\varphi^{*}\omega=0\}$, where $\omega$ is the Serre duality pairing on $\mathbb{H}^{g}$. Let $(m,n)=4\mathrm{Tr}(mn)$ be the killing form on $sl(2)$. The killing form gives isomorphisms
$$\mathbb{H}^{g}\otimes sl(2)\cong\mathrm{Hom}(sl(2),\mathbb{H}^{g})\text{ and }sl(2)\cong\wedge^{2}sl(2)^{\vee}.$$
By the above identification, $\Upsilon:\mathrm{Hom}(sl(2),\mathbb{H}^{g})\to\wedge^{2}sl(2)^{\vee}$ is given by $\varphi\mapsto\varphi^{*}\omega$. Then we have
$$\Upsilon^{-1}(0)=\mathrm{Hom}^{\omega}(sl(2),\mathbb{H}^{g}).$$
Let
$$\mathrm{Hom}_{k}(sl(2),\mathbb{H}^{g}):=\{\varphi\in\mathrm{Hom}(sl(2),\mathbb{H}^{g})|\mathrm{rk}\,\varphi\leq k\}$$
and
$$\mathrm{Hom}_{k}^{\omega}(sl(2),\mathbb{H}^{g}):=\mathrm{Hom}_{k}(sl(2),\mathbb{H}^{g})\cap\mathrm{Hom}^{\omega}(sl(2),\mathbb{H}^{g}).$$
We have a description of
points of $E_{1}\cap\Sigma$ as following.
\begin{proposition}[Lemma 1.7.5 of \cite{O99}]\label{intersection-of-1st-exc-and-2nd-center}
Let $x\in\mathbb{Z}_2^{2g}$. Then
$$\pi_{\mathbf{R}_{1}}^{-1}(x)\cap\Sigma=\mathbb{P}\mathrm{Hom}_{1}(sl(2),\mathbb{H}^{g})^{ss},$$
where $\mathbb{P}\mathrm{Hom}_{1}(sl(2),\mathbb{H}^{g})^{ss}$ denotes the set of semistable points of $\mathbb{P}\mathrm{Hom}_{1}(sl(2),\mathbb{H}^{g})$ with respect to the $\mathrm{PGL}(2)$-action.
\end{proposition}
Assume that $\varphi\in\mathrm{Hom}_{1}(sl(2),\mathbb{H}^{g})$. Since the Serre duality pairing is skew-symmetric, we can choose bases $\{e_{1},\cdots,e_{2g}\}$ of $\mathbb{H}^{g}$ and $\{v_{1},v_{2},v_{3}\}$ of $sl(2)$ such that $\varphi=e_{1}\otimes v_{1}$ and so that
$$<e_{i},e_{j}>=\begin{cases}1 &\text{if }i=2q-1,j=2q,q=1,\cdots,g,\\
-1&\text{if }i=2q,j=2q-1,q=1,\cdots,g,\\
0&\text{otherwise}.\end{cases}$$
Every element in $\mathrm{Hom}(sl(2),\mathbb{H}^{g})$ can be written as
$\sum_{i,j}Z_{ij}e_{i}\otimes v_{j}$. Then we have a description of the normal cone $C_{\mathbb{P}\mathrm{Hom}_1(sl(2),\mathbb{H}^{g})}\mathbb{P}\Upsilon^{-1}(0)$ to $\mathbb{P}\mathrm{Hom}_{1}(sl(2),\mathbb{H}^{g})$ in $\mathbb{P}\Upsilon^{-1}(0)$.
\begin{proposition}\label{normalcone-localmodel}
Let $[\varphi]\in\mathbb{P}\mathrm{Hom}_1(sl(2),\mathbb{H}^{g})$ and let
$\omega^{\varphi}$ be the bilinear form induced by $\omega$ on
$\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi$. There is a
$\mathrm{Stab}([\varphi])$-equivariant isomorphism
$$(C_{\mathbb{P}\mathrm{Hom}_1(sl(2),\mathbb{H}^{g})}\mathbb{P}\Upsilon^{-1}(0))|_{[\varphi]}\cong\mathrm{Hom}^{\omega^{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)$$
where
$$\mathrm{Hom}^{\omega^{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)=\{\chi\in\mathrm{Hom}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)|\chi^*\omega^{\varphi}=0\}$$
\end{proposition}
\begin{proof}
Following the idea of proof of \cite[Lemma 1.7.13]{O99}, both sides are defined by the equation
$$\sum_{2\leq q\leq 2g}(Z_{2q-1,2}Z_{2q,3}-Z_{2q,2}Z_{2q-1,3})=0.$$
under the choice of basis as above.
\end{proof}
We now explain how $\mathrm{Stab}([\varphi])$ acts on $(C_{\mathbb{P}\mathrm{Hom}_{1}(sl(2),\mathbb{H}^{g})}\mathbb{P}\Upsilon^{-1}(0))_{[\varphi]}$. If we add the condition that
$$\left.\begin{array}{ccc}(v_{1},v_{i})=-\delta_{1i}\\(v_{j},v_{j})=0,&j=2,3 \\(v_{2},v_{3})=1, \end{array}\right.$$
and $v_{1}\wedge v_{2}\wedge v_{3}$ is the volume form, where
$\wedge$ corresponds to the Lie bracket in $sl(2)$, then $\mathrm{Stab}([\varphi])=\mathrm{O}(\ker\varphi)=\mathrm{O}(2)$ is generated by
$$\{\theta_{\lambda}:=\left(\begin{array}{ccc}1&0&0\\0&\lambda&0\\0&0&\lambda^{-1}\end{array}\right)|\lambda\in\mathbb{C}^{*}\}\text{ and }\tau:=\left(\begin{array}{ccc}-1&0&0\\0&0&1\\0&1&0\end{array}\right)$$
as a subgroup of $\mathrm{SO}(sl(2))$. $\mathrm{O}(2)$ can be also realized as the
subgroup of $\mathrm{PGL}(2)$ generated by
$$\mathrm{SO}(2)=\big\{\theta_{\lambda}:=\left(\begin{array}{cc}\lambda&0\\0&\lambda^{-1}\end{array}\right)|\lambda\in\mathbb{C}^{*}\big\}/\{\pm\mathrm{id} \},\quad \tau=\left(\begin{array}{cc}0&1\\1&0\end{array}\right).$$
The action of $\mathrm{Stab}([\varphi])$ on
$(C_{\mathbb{P}\mathrm{Hom}_{1}(sl(2),\mathbb{H}^{g})}\mathbb{P}\Upsilon^{-1}(0))_{[\varphi]}$
is given by
$$\theta_{\lambda}(\sum_{i=3}^{2g}(Z_{i,2}e_{i}\otimes v_{2}+Z_{i,3}e_{i}\otimes v_{3}))=\sum_{i=3}^{2g}(\lambda Z_{i,2}e_{i}\otimes v_{2}+\lambda^{-1}Z_{i,3}e_{i}\otimes v_{3}),$$
$$\tau(\sum_{i=3}^{2g}(Z_{i,2}e_{i}\otimes v_{2}+Z_{i,3}e_{i}\otimes v_{3}))=\sum_{i=3}^{2g}(-Z_{i,3}e_{i}\otimes v_{2}-Z_{i,2}e_{i}\otimes v_{3}).$$
Let us consider the blowing-up
$$\pi:Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss}\to\mathbb{P}\Upsilon^{-1}(0)^{ss}$$
of $\mathbb{P}\Upsilon^{-1}(0)^{ss}$ along $\mathbb{P}\mathrm{Hom}_{1}(sl(2),\mathbb{H}^{g})^{ss}$ with the exceptional divisor $E$, where $\mathbb{P}\Upsilon^{-1}(0)^{ss}$ is the locus of semistable points of $\mathbb{P}\Upsilon^{-1}(0)$ with respect to the $\mathrm{PGL}(2)$-action. It is obvious that $(\pi_{\mathbf{R}_{1}}\circ\pi_{\mathbf{R}_{2}})^{-1}(x)=Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss}$ for any $x\in\mathbb{Z}_{2}^{2g}$.
\begin{proposition}[Lemma 1.8.5 of \cite{O99}]\label{smoothness of 2nd local blowing-up}
$Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss}$ is smooth.
\end{proposition}
\section{Blowing-up formula for intersection cohomology}\label{blowing-up formula for intersection cohomology}
In this section, we prove that a blowing-up formula for the intersection cohomology holds in Kirwan's algorithm introduced in Section \ref{Kirwan desing}.
Let $E_{1}$ (respectively, $E_{2}$) be the exceptional divisor of $\pi_{\mathbf{R}_{1}}$ (respectively, $\pi_{\mathbf{R}_{2}}$). Let $\mathcal{C}_{1}$ be the normal cone to $\mathbb{Z}_{2}^{2g}$ in $\mathbf{R}$, $\tilde{\mathcal{C}}_{1}$ the normal cone to $E_{1}^{ss}:=E_{1}\cap\mathbf{R}_{1}^{ss}$ in $\mathbf{R}_{1}^{ss}$, $\mathcal{C}_{2}$ the normal cone to $\Sigma$ in $\mathbf{R}_{1}$, $\tilde{\mathcal{C}}_{2}$ the normal cone to $E_{2}^{ss}:=E_{2}\cap\mathbf{R}_{2}^{ss}$ in $\mathbf{R}_{2}^{ss}$, $\mathcal{C}$ the normal cone to $\mathbb{P}\mathrm{Hom}_{1}(sl(2),\mathbb{H}^{g})^{ss}$ in $\mathbb{P}\Upsilon^{-1}(0)^{ss}$ and $\tilde{\mathcal{C}}$ the normal cone to $E^{ss}:=E\cap(Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss})^{ss}$ in $(Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss})^{ss}$, where $(Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss})^{ss}$ is the locus of semistable points of $Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss}$ with respect to the lifted $\mathrm{PGL}(2)$-action. Note that
$$(Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss})^{ss}=(Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss})^{s},$$
where $(Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss})^{s}$ is the locus of stable points of $Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss}$ with respect to the $\mathrm{PGL}(2)$-action (\cite[Lemma 1.8.6]{O99}). Then we have the following formulas.
\begin{lemma}\label{intersection blowing-up formula}
(1) $\dim IH^{i}(\mathbf{R}_{1}^{ss}/\!/\mathrm{SL}(2))=\dim IH^{i}(\mathbf{R}/\!/\mathrm{SL}(2))$
$$+\dim IH^{i}(\tilde{\mathcal{C}}_{1}/\!/\mathrm{SL}(2))-\dim IH^{i}(\mathcal{C}_1/\!/\mathrm{SL}(2))$$
$$=\dim IH^{i}(\mathbf{R}/\!/\mathrm{SL}(2))+2^{2g}\dim IH^{i}(Bl_{0}\Upsilon^{-1}(0)/\!/\mathrm{PGL}(2))-2^{2g}\dim IH^{i}(\Upsilon^{-1}(0)/\!/\mathrm{PGL}(2))$$
for all $i\ge0$, where $Bl_{0}\Upsilon^{-1}(0)$ is the blowing-up of $\Upsilon^{-1}(0)$ at the vertex.
(2) $\dim IH^{i}(\mathbf{R}_{2}^{s}/\mathrm{SL}(2))=\dim IH^{i}(\mathbf{R}_{1}^{ss}/\!/\mathrm{SL}(2))$
$$+\dim IH^{i}(\tilde{\mathcal{C}}_{2}/\!/\mathrm{SL}(2))-\dim IH^{i}(\mathcal{C}_2/\!/\mathrm{SL}(2))$$
for all $i\ge0$.
(3)\label{local second intersection blowing-up formula} $\dim IH^{i}((Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss})^{s}/\!/\mathrm{SL}(2))=\dim IH^{i}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))$
$$+\dim IH^{i}(\tilde{\mathcal{C}}/\!/\mathrm{SL}(2))-\dim IH^{i}(\mathcal{C}/\!/\mathrm{SL}(2))$$
for all $i\ge0$.
\end{lemma}
For the proof, we need to review a useful result by C.T. Simpson. Let $A^i$ (respectively, $A^{i,j}$) be the sheaf of smooth $i$-forms
(respectively, $(i,j)$-forms) on $X$. For a polystable Higgs bundle
$(E,\phi)$, consider the complex
\begin{equation}\label{e4.1}
0\to \mathrm{End}_{0}(E)\otimes A^0\to \mathrm{End}_{0}(E)\otimes A^1\to \mathrm{End}_{0}(E)\otimes
A^2\to 0\end{equation} whose differential is given by
$D''=\overline{\partial}+\phi$. Because $A^1=A^{1,0}\oplus
A^{0,1}$ and $\phi$ is of type $(1,0)$, we have an exact sequence
of complexes with \eqref{e4.1} in the middle
$$\xymatrix{
& 0\ar[d] & 0\ar[d] &0\ar[d] & \\
0\ar[r] & 0\ar[r]\ar[d] &\mathrm{End}_{0}(E)\otimes
A^{1,0}\ar[r]^{\overline{\partial}}\ar[d] &\mathrm{End}_{0}(E)\otimes
A^{1,1}\ar[r]\ar[d]^= &0\\
0\ar[r]& \mathrm{End}_{0}(E)\otimes A^0\ar[r]^{D''}\ar[d]_= & \mathrm{End}_{0}(E)\otimes
A^1\ar[r]^{D''}\ar[d] & \mathrm{End}_{0}(E)\otimes A^2\ar[r]\ar[d] & 0\\
0\ar[r] & \mathrm{End}_{0}(E)\otimes A^{0,0}\ar[r]^{\overline{\partial}}\ar[d]
&
\mathrm{End}_{0}(E)\otimes A^{0,1}\ar[r]\ar[d] & 0\ar[r]\ar[d] & 0 \\
& 0 & 0 &0 }$$
This gives us a long exact sequence
\[\xymatrix{ 0\ar[r] & T^0\ar[r] &H^0(\mathrm{End}_{0}(E))\ar[r]^(.42){[\phi,-]}
& H^0(\mathrm{End}_{0}(E)\otimes K_X)\ar[r]& }\]
\[\xymatrix{ \ar[r]& T^1\ar[r] &H^1(\mathrm{End}_{0}(E))\ar[r]^(.42){[\phi,-]}
& H^1(\mathrm{End}_{0}(E)\otimes K_X)\ar[r] &T^2\ar[r] &0 }
\]
where $T^i$ is the $i$-th cohomology of \eqref{e4.1}. The Zariski
tangent space of $\mathbf{M}$ at polystable $(E,\phi)$ is isomorphic to
$T^1$.
\begin{proposition}[Theorem 10.4 and 10.5 of \cite{Simp94II}]\label{normal slice is the quadratic cone}
Let $C$ be the quadratic cone in $T^1$
defined by the map $T^1\to T^2$ which sends an $\mathrm{End}_{0}(E)$-valued 1-form $\eta$ to
$[\eta,\eta]$. Let $y=(E,\phi,\beta:E|_{x}\to\mathbb{C}^{2})\in \mathbf{R}$ be a point with
closed orbit and $\bar{y}\in\mathbf{M}$ the image of $y$. Then the formal completion ${(\mathbf{R},y)}^\wedge$ is
isomorphic to the formal completion ${(C\times
\mathfrak{h}^\perp,0)}^\wedge$ where $\mathfrak{h}^\perp$ is the
perpendicular space to the image of $T^0\to H^0(\mathrm{End}_{0}(E))\to
sl(2)$. Furthermore, if we let $Y$ be the \'etale slice at $y$ of
the $\mathrm{SL}(2)$-orbit in $\mathbf{R}$, then
$(Y,y)^\wedge \cong (C,0)^\wedge$ and $(\mathbf{M},\bar{y})^\wedge=(Y/\!/\mathrm{Stab}(y),\bar{y})^\wedge\cong(C/\!/\mathrm{Stab}(y),v)^\wedge$ where $\mathrm{Stab}(y)$ is the stabilizer of $y$ and $v$ is the cone point of $C$.
\end{proposition}
\begin{proof}[Proof of Lemma \ref{intersection blowing-up formula}]
\begin{enumerate}
\item Let $U_{x}$ be a sufficiently small open neighborhood of $x\in\mathbb{Z}_{2}^{2g}$ in $\mathbf{R}/\!/\mathrm{SL}(2)$, let $\displaystyle U_{1}=\sqcup_{x\in\mathbb{Z}_{2}^{2g}}U_{x}$ and $\tilde{U}_{1}=\overline{\pi}_{\mathbf{R}_{1}}^{-1}(U_{1})$. By the same argument as in the proof of \cite[Lemma 2.8]{K86}, we have
$$\dim IH^{i}(\mathbf{R}_{1}/\!/\mathrm{SL}(2))=\dim IH^{i}(\mathbf{R}/\!/\mathrm{SL}(2))+\dim IH^{i}(\tilde{U}_{1})-\dim IH^{i}(U_{1})$$
for all $i\ge0$. By \cite[Theorem 3.1]{GM88} and Proposition \ref{normal slice is the quadratic cone}, there is an analytic isomorphism $U_{1}\cong\mathcal{C}_1/\!/\mathrm{SL}(2)$. Since $\tilde{\mathcal{C}}_1/\!/\mathrm{SL}(2)$ is naturally isomorphic to the blowing-up of $\mathcal{C}_1/\!/\mathrm{SL}(2)$ along $\mathbb{Z}_{2}^{2g}$, we also have an analytic isomorphism $\tilde{U}_{1}\cong\tilde{\mathcal{C}}_1/\!/\mathrm{SL}(2)$. Since $\mathcal{C}_1/\!/\mathrm{SL}(2)$ (respectively, $\tilde{\mathcal{C}}_1/\!/\mathrm{SL}(2)$) is the $2^{2g}$ copy of
\begin{center}$\Upsilon^{-1}(0)/\!/\mathrm{PGL}(2)$ (respectively, of $Bl_{0}\Upsilon^{-1}(0)/\!/\mathrm{PGL}(2)$),\end{center}
we get the formula.
\item Let $U_{2}$ be a sufficiently small open neighborhood of the strict transform of $T^*J/\mathbb{Z}_2$ in $\mathbf{R}_{1}/\!/\mathrm{SL}(2)$ and let $\tilde{U}_{2}=\overline{\pi}_{\mathbf{R}_{2}}^{-1}(U_{2})$. By the same argument as in the proof of \cite[Lemma 2.8]{K86}, we have
$$\dim IH^{i}(\mathbf{R}_{2}/\mathrm{SL}(2))=\dim IH^{i}(\mathbf{R}_{1}/\!/\mathrm{SL}(2))+\dim IH^{i}(\tilde{U}_{2})-\dim IH^{i}(U_{2})$$
for all $i\ge0$. By \cite[Theorem 3.1]{GM88} and Proposition \ref{normal slice is the quadratic cone} and Hartogs's Extension Theorem, there is an analytic isomorphism $U_{2}\cong\mathcal{C}_2/\!/\mathrm{SL}(2)$. Since $\tilde{\mathcal{C}}_2/\!/\mathrm{SL}(2)$ is naturally isomorphic to the blowing-up of $\mathcal{C}_2/\!/\mathrm{SL}(2)$ along the strict transform of $T^*J/\mathbb{Z}_2$ in $\mathbf{R}_{1}/\!/\mathrm{SL}(2)$, we also have an analytic isomorphism $\tilde{U}_{2}\cong\tilde{\mathcal{C}}_2/\!/\mathrm{SL}(2)$. Hence we get the formula.
\item Let $\overline{\pi}:Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss}/\!/\mathrm{PGL}(2)\to\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{PGL}(2)$ be the map induced from $\pi$. Since $\mathcal{C}=\mathcal{C}_{2}|_{\Sigma\cap\mathbb{P}\Upsilon^{-1}(0)^{ss}}$ and $\tilde{\mathcal{C}}=\tilde{\mathcal{C}}_{2}|_{E_{2}\cap Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss}}$, it follow from the argument of the proof of item (2) that $\mathcal{C}/\!/\mathrm{SL}(2)$ (respectively, $\tilde{\mathcal{C}}/\!/\mathrm{SL}(2)$) can be identified with an open neighborhood $U$ of $\mathbb{P}\mathrm{Hom}_{1}(sl(2),\mathbb{H}^{g})/\!/\mathrm{PGL}(2)$ (respectively, with $\overline{\pi}^{-1}(U)$). Again by the same argument as in the proof of \cite[Lemma 2.8]{K86}, we get the formula.
\end{enumerate}
\end{proof}
We give a computable fomula from Lemma \ref{intersection blowing-up formula} by more analysis on $Bl_{0}\Upsilon^{-1}(0)/\!/\mathrm{PGL}(2)$, $\mathcal{C}_{2}/\!/\mathrm{SL}(2)$, $\tilde{\mathcal{C}}_{2}/\!/\mathrm{SL}(2)$, $\mathcal{C}/\!/\mathrm{SL}(2)$ and $\tilde{\mathcal{C}}/\!/\mathrm{SL}(2)$.
We first give explicit geometric descriptions for $\mathcal{C}_{2}/\!/\mathrm{SL}(2)$, $\tilde{\mathcal{C}}_{2}/\!/\mathrm{SL}(2)$, $\mathcal{C}/\!/\mathrm{SL}(2)$ and $\tilde{\mathcal{C}}/\!/\mathrm{SL}(2)$. Let $\alpha:\widetilde{T^{*}J}\to T^{*}J$ be the blowing-up along $\mathbb{Z}_{2}^{2g}$. Let $(\mathcal{L},\psi_{\mathcal{L}})$ be the pull-back to $\widetilde{T^{*}J}\times X$ of the universal pair on $T^{*}J\times X$ by $\alpha\times 1$ and let $p:\widetilde{T^{*}J}\times X\to\widetilde{T^{*}J}$ the projection onto the first factor.
\begin{lemma}\label{geometric descriptions of second cones}
(1) $\mathcal{C}_{2}|_{\Sigma\setminus E_{1}}/\!/\mathrm{SL}(2)$ is a free $\mathbb{Z}_{2}$-quotient of $\Psi^{-1}(0)/\!/\mathbb{C}^{*}$-bundle over $\widetilde{T^{*}J}\setminus\alpha^{-1}(\mathbb{Z}_{2}^{2g})$.
(2) $\tilde{\mathcal{C}}_{2}|_{\Sigma\setminus E_{1}}/\!/\mathrm{SL}(2)$ is a free $\mathbb{Z}_{2}$-quotient of $Bl_{0}\Psi^{-1}(0)/\!/\mathbb{C}^{*}$-bundle over $\widetilde{T^{*}J}\setminus\alpha^{-1}(\mathbb{Z}_{2}^{2g})$, where $Bl_{0}\Psi^{-1}(0)$ is the blowing-up of $\Psi^{-1}(0)$ at the vertex.
(3) $\mathcal{C}_{2}|_{\Sigma\cap E_{1}}/\!/\mathrm{SL}(2)$ is a free $\mathbb{Z}_{2}$-quotient of $\mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)/\!/\mathbb{C}^{*}$-bundle over $\alpha^{-1}(\mathbb{Z}_{2}^{2g})$, where $[\varphi]\in\Sigma\cap E_{1}$.
(4) $\tilde{\mathcal{C}}_{2}|_{\Sigma\cap E_{1}}/\!/\mathrm{SL}(2)$ is a free $\mathbb{Z}_{2}$-quotient of $Bl_{0}\mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)/\!/\mathbb{C}^{*}$-bundle over $\alpha^{-1}(\mathbb{Z}_{2}^{2g})$, where $[\varphi]\in\Sigma\cap E_{1}$ and $Bl_{0}\mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)$ is the blowing-up of $\mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)$ at the vertex.
(5) $\mathcal{C}/\!/\mathrm{SL}(2)$ is a free $\mathbb{Z}_{2}$-quotient of $\mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)/\!/\mathbb{C}^{*}$-bundle over $\mathbb{P}^{2g-1}$, where $[\varphi]\in\mathbb{P}\mathrm{Hom}_{1}(sl(2),\mathbb{H}^{g})^{ss}$.
(6) $\tilde{\mathcal{C}}/\!/\mathrm{SL}(2)$ is a free $\mathbb{Z}_{2}$-quotient of $Bl_{0}\mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)/\!/\mathbb{C}^{*}$-bundle over $\mathbb{P}^{2g-1}$, where $[\varphi]\in\mathbb{P}\mathrm{Hom}_{1}(sl(2),\mathbb{H}^{g})^{ss}$.
\end{lemma}
\begin{proof}
Let $x$ be a point of $X$.
\begin{enumerate}
\item Consider the principal $\mathrm{PGL}(2)$-bundle
$$q:\mathbb{P}\mathrm{Isom}(\mathcal{O}_{\widetilde{T^{*}J}}^{2},\mathcal{L}|_{x}\oplus\mathcal{L}^{-1}|_{x})\to\widetilde{T^{*}J}.$$
$\mathrm{PGL}(2)$ acts on $\mathcal{O}_{\widetilde{T^{*}J}}^{2}$ and $\mathrm{O}(2)$ acts on $\mathcal{L}|_{x}\oplus\mathcal{L}^{-1}|_{x}$. By the same argument as in the proof of \cite[Proposition 1.7.10]{O99},
$$\Sigma\cong\mathbb{P}\mathrm{Isom}(\mathcal{O}_{\widetilde{T^{*}J}}^{2},\mathcal{L}|_{x}\oplus\mathcal{L}^{-1}|_{x})/\!/\mathrm{O}(2).$$
$\mathcal{C}_{2}|_{\Sigma\setminus E_{1}}/\!/\mathrm{SL}(2)$ is the quotient of $q^{*}\Psi_{\mathcal{L}_{III}}^{-1}(0)/\!/\mathrm{O}(2)$ by the $\mathrm{PGL}(2)$-action, where $\mathcal{L}_{III}=\mathcal{L}|_{\widetilde{T^{*}J}\setminus\alpha^{-1}(\mathbb{Z}_{2}^{2g})}$ and
$$\Psi_{\mathcal{L}_{III}}:[p_{*}(\mathcal{L}_{III}^{-2}K_X)\oplus R^{1}p_{*}(\mathcal{L}_{III}^2)]\oplus[p_{*}(\mathcal{L}_{III}^{2}K_X)\oplus R^{1}p_{*}(\mathcal{L}_{III}^{-2})]\to R^{1}p_{*}(K_X)$$
is the sum of perfect pairings $p_{*}(\mathcal{L}_{III}^{-2}K_X)\oplus R^{1}p_{*}(\mathcal{L}_{III}^2)\to R^{1}p_{*}(K_X)$ and $p_{*}(\mathcal{L}_{III}^{2}K_X)\oplus R^{1}p_{*}(\mathcal{L}_{III}^{-2})\to R^{1}p_{*}(K_X)$. Since the actions of $\mathrm{PGL}(2)$ and $\mathrm{O}(2)$ commute and $q$ is the principal $\mathrm{PGL}(2)$-bundle, $$\mathcal{C}_{2}|_{\Sigma\setminus E_{1}}/\!/\mathrm{SL}(2)=\Psi_{\mathcal{L}_{III}}^{-1}(0)/\!/\mathrm{O}(2)=\frac{\Psi_{\mathcal{L}_{III}}^{-1}(0)/\!/\mathrm{SO}(2)}{\mathrm{O}(2)/\mathrm{SO}(2)}=\frac{\Psi_{\mathcal{L}_{III}}^{-1}(0)/\!/\mathbb{C}^{*}}{\mathbb{Z}_{2}}.$$
Hence we get the description.
\item Since $\tilde{\mathcal{C}}_{2}|_{\Sigma\setminus E_{1}}/\!/\mathrm{SL}(2)$ is isomorphic to the blowing-up of $\mathcal{C}_{2}|_{\Sigma\setminus E_{1}}/\!/\mathrm{SL}(2)$ along $T^{*}J/\mathbb{Z}_{2}\setminus\mathbb{Z}_{2}^{2g}$, it is isomorphic to $\displaystyle\frac{\widetilde{\Psi_{\mathcal{L}_{III}}^{-1}(0)}/\!/\mathbb{C}^{*}}{\mathbb{Z}_{2}}$, where $\widetilde{\Psi_{\mathcal{L}_{III}}^{-1}(0)}$ is the blowing-up of $\Psi_{\mathcal{L}_{III}}^{-1}(0)$ along $\widetilde{T^{*}J}\setminus\alpha^{-1}(\mathbb{Z}_{2}^{2g})\cong T^{*}J\setminus\mathbb{Z}_{2}^{2g}$.
\item Note that $E_{1}$ is a $2^{2g}$ disjoint union of $\mathbb{P}\Upsilon^{-1}(0)$. It follows from Proposition \ref{intersection-of-1st-exc-and-2nd-center} that $\Sigma\cap E_{1}$ is a $2^{2g}$ disjoint union of $\mathbb{P}\mathrm{Hom}_{1}(sl(2),\mathbb{H}^{g})^{ss}$. By Proposition \ref{ss-local-first-blowup}, we have
$$\mathbb{P}\mathrm{Hom}_{1}(sl(2),\mathbb{H}^{g})^{ss}=\mathrm{PGL}(2)Z^{ss}\cong\mathrm{PGL}(2)\times_{\mathrm{O}(2)}Z^{ss}$$
and
$$\mathbb{P}\mathrm{Hom}_{1}(sl(2),\mathbb{H}^{g})/\!/\mathrm{PGL}(2)\cong Z/\!/\mathrm{O}(2)=Z_{1}=\mathbb{P}^{2g-1},$$
where $Z=Z_{1}\cup Z_{2}\cup Z_{3}$, $Z^{ss}$ is the set of
semistable points of $Z$ for the action of $\mathrm{O}(2)$,
$Z_{1}=\mathbb{P}\{v_{1}\otimes\mathbb{H}^{g}\}=Z^{ss}$,
$Z_{2}=\mathbb{P}\{v_{2}\otimes\mathbb{H}^{g}\}$ and
$Z_{3}=\mathbb{P}\{v_{3}\otimes\mathbb{H}^{g}\}$ for the basis $\{v_{1},v_{2},v_{3}\}$ of $sl(2)$ chosen in Section \ref{local picture}. Then we have
$$\mathcal{C}_{2}|_{\Sigma\cap\mathbb{P}\Upsilon^{-1}(0)}=\mathrm{PGL}(2)\times_{\mathrm{O}(2)}\mathcal{C}_{2}|_{Z^{ss}}$$
and
$$\mathcal{C}_{2}|_{\Sigma\cap\mathbb{P}\Upsilon^{-1}(0)}/\!/\mathrm{SL}(2)=\mathcal{C}_{2}|_{\Sigma\cap\mathbb{P}\Upsilon^{-1}(0)}/\!/\mathrm{PGL}(2)=\mathcal{C}_{2}|_{Z^{ss}}/\!/\mathrm{O}(2)=\frac{\mathcal{C}_{2}|_{Z^{ss}}/\!/\mathrm{SO}(2)}{\mathbb{Z}_{2}}.$$
Since $\mathcal{C}_{2}|_{Z^{ss}}$ is a $\mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)$-bundle over $Z^{ss}$ by Proposition \ref{normalcone-localmodel},
$$\mathcal{C}_{2}|_{Z^{ss}}/\!/\mathrm{SO}(2)$$
is a $\mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)/\!/\mathbb{C}^{*}$-bundle over $Z/\!/\mathrm{SO}(2)=Z_{1}=\mathbb{P}^{2g-1}$. Since $\alpha^{-1}(\mathbb{Z}_{2}^{2g})$ is a $2^{2g}$ disjoint union of $\mathbb{P}^{2g-1}$, we get the description.
\item Since $\tilde{\mathcal{C}}_{2}|_{\Sigma\cap E_{1}}/\!/\mathrm{SL}(2)$ is isomorphic to the blowing-up of $\mathcal{C}_{2}|_{\Sigma\cap E_{1}}/\!/\mathrm{SL}(2)$ along $2^{2g}$ disjoint union of $\mathbb{P}\mathrm{Hom}_{1}(sl(2),\mathbb{H}^{g})/\!/\mathrm{PGL}(2)\cong\mathbb{P}^{2g-1}$, it is isomorphic to $2^{2g}$ disjoint union of $\displaystyle\frac{\widetilde{\mathcal{C}_{2}|_{Z^{ss}}}/\!/\mathrm{SO}(2)}{\mathbb{Z}_{2}}$, where $\widetilde{\mathcal{C}_{2}|_{Z^{ss}}}$ is the blowing-up of $\mathcal{C}_{2}|_{Z^{ss}}$ along $Z^{ss}$.
\item Since $\mathcal{C}=\mathcal{C}_{2}|_{\Sigma\cap\mathbb{P}\Upsilon^{-1}(0)^{ss}}$, we get the description from (3).
\item Since $\tilde{\mathcal{C}}=\tilde{\mathcal{C}}_{2}|_{E_{2}\cap Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss}}$, we get the description from (4).
\end{enumerate}
\end{proof}
We next explain how to compute the terms
$$\dim IH^{i}(Bl_{0}\Upsilon^{-1}(0)/\!/\mathrm{PGL}(2))-\dim IH^{i}(\Upsilon^{-1}(0)/\!/\mathrm{PGL}(2)),$$
$$\dim IH^{i}(\tilde{\mathcal{C}}_{2}/\!/\mathrm{SL}(2))-\dim IH^{i}(\mathcal{C}_2/\!/\mathrm{SL}(2))$$
and
$$\dim IH^{i}(\tilde{\mathcal{C}}/\!/\mathrm{SL}(2))-\dim IH^{i}(\mathcal{C}/\!/\mathrm{SL}(2))$$
that appear in Lemma \ref{intersection blowing-up formula}. We start with the following technical lemma.
\begin{lemma}[Lemma 2.12 in \cite{K86}]\label{int coh of quotient by finite group}
Let $V$ be a complex variety on which a finite group $F$ acts. Then
$$IH^{*}(V/F)\cong IH^{*}(V)^{F}$$ where $IH^{*}(V)^{F}$ denotes the
invariant part of $IH^{*}(V)$ under the action of $F$.
\end{lemma}
Now recall that $\Upsilon^{-1}(0)/\!/\mathrm{SL}(2)=\Upsilon^{-1}(0)/\!/\mathrm{PGL}(2)$ and $\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2)=\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{PGL}(2)$ from section \ref{local picture}.
We can compute $IH^{*}(\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))$ (respectively, $IH^{*}(\Psi^{-1}(0)/\!/\mathbb{C}^*)$ and $$IH^{*}(\mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)/\!/\mathbb{C}^*)\text{)}$$
in terms of $IH^{*}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))$ (respectively, $IH^{*}(\mathbb{P}\Psi^{-1}(0)/\!/\mathbb{C}^*)$ and $$IH^{*}(\mathbb{P}\mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)/\!/\mathbb{C}^*)\text{)}.$$
In order to explain this, we need the following lemmas. The first lemma shows the surjectivities of the Kirwan maps on the fibers of normal cones and exceptional divisors.
\begin{lemma}\label{Kir-Map}
(1) The Kirwan map
$$IH_{\mathrm{SL}(2)}^{*}(\mathbb{P}\Upsilon^{-1}(0)^{ss})\rightarrow IH^{*}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))$$
is surjective.
(2) The Kirwan map
$$IH_{\mathrm{SL}(2)}^{*}(\Upsilon^{-1}(0))\rightarrow IH^{*}(\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))$$
is surjective.
(3) The Kirwan map
$$H_{\mathbb{C}^*}^{*}(\mathbb{P}\Psi^{-1}(0)^{ss})\rightarrow IH^{*}(\mathbb{P}\Psi^{-1}(0)/\!/\mathbb{C}^*)$$
is surjective.
(4) The Kirwan map
$$IH_{\mathbb{C}^*}^{*}(\Psi^{-1}(0))\rightarrow IH^{*}(\Psi^{-1}(0)/\!/\mathbb{C}^*)$$
is surjective.
(5) The Kirwan map
$$H_{\mathbb{C}^*}^{*}(\mathbb{P}\mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)^{ss})\rightarrow IH^{*}(\mathbb{P}\mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)/\!/\mathbb{C}^*)$$
is surjective.
(6) The Kirwan map
$$IH_{\mathbb{C}^*}^{*}(\mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi))\rightarrow IH^{*}(\mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)/\!/\mathbb{C}^*)$$
is surjective.
\end{lemma}
\begin{proof}
\begin{enumerate}
\item Consider the quotient map
$$f:\mathbb{P}\Upsilon^{-1}(0)^{ss}\to\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2).$$
In \cite[section 6]{BL94}, Bernstein and Lunts define a functor
$$Qf_{*}:\mathbf{D}_{\mathrm{SL}(2)}(\mathbb{P}\Upsilon^{-1}(0)^{ss})\to\mathbf{D}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))$$
that extends the pushforward of sheaves $f_{*}$.
By the same arguments as those of \cite[$\S2$ and
$\S3$]{Woo03}, we can obtain morphisms
$$\lambda_{\mathbb{P}\Upsilon^{-1}(0)}:\mathbf{IC}^{\bullet}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))[3]\to Qf_{*}\mathbf{IC}^{\bullet}_{\mathrm{SL}(2)}(\mathbb{P}\Upsilon^{-1}(0)^{ss})$$
and
$$\kappa_{\mathbb{P}\Upsilon^{-1}(0)}:Qf_{*}\mathbf{IC}^{\bullet}_{\mathrm{SL}(2)}(\mathbb{P}\Upsilon^{-1}(0)^{ss})\to\mathbf{IC}^{\bullet}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))[3]$$
such that $\kappa_{\mathbb{P}\Upsilon^{-1}(0)}\circ\lambda_{\mathbb{P}\Upsilon^{-1}(0)}=\mathrm{id} $. $\lambda_{\mathbb{P}\Upsilon^{-1}(0)}$ induces a map
$$IH^{*}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))\to IH_{\mathrm{SL}(2)}^{*}(\mathbb{P}\Upsilon^{-1}(0)^{ss})$$
which is an inclusion.
Hence $\kappa_{\mathbb{P}\Upsilon^{-1}(0)}$ induces a map
$$\tilde{\kappa}_{\mathbb{P}\Upsilon^{-1}(0)}:IH_{\mathrm{SL}(2)}^{*}(\mathbb{P}\Upsilon^{-1}(0)^{ss})\to IH^{*}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))$$
which is split by the inclusion $IH^{*}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))\to IH_{\mathrm{SL}(2)}^{*}(\mathbb{P}\Upsilon^{-1}(0)^{ss})$.
\item Let $R:=\mathbb{C}[T_0,T_1,\cdots,T_{6g-1}]$. For an
$\mathrm{SL}(2)$-invariant ideal $I\subset R$ generated by three quadratic
homogeneous polynomials in $R$ defining $\Upsilon^{-1}(0)$, we can
write
$$\Upsilon^{-1}(0)=\mathrm{Spec}(R/I).$$
Let $\overline{\Upsilon^{-1}(0)}$ be the Zariski closure of
$\Upsilon^{-1}(0)$ in $\mathbb{P}^{6g}$. Since the homogenization of
$I$ equals to $I$, we can write
$$\overline{\Upsilon^{-1}(0)}=\mathrm{Proj}(R[T]/I\cdot R[T])$$
where $\mathrm{SL}(2)$ acts trivially on the variable $T$. Thus
$$\Upsilon^{-1}(0)/\!/\mathrm{SL}(2)=\mathrm{Spec}(R^{\mathrm{SL}(2)}/I\cap R^{\mathrm{SL}(2)})$$
and
$$\overline{\Upsilon^{-1}(0)}/\!/\mathrm{SL}(2)=\mathrm{Proj}(R[T]/I\cdot R[T])^{\mathrm{SL}(2)}.$$
Since $\mathrm{SL}(2)$ acts trivially on the variable $T$,
$$\overline{\Upsilon^{-1}(0)}/\!/\mathrm{SL}(2)=\mathrm{Proj}(R^{\mathrm{SL}(2)}[T]/(I\cap R^{\mathrm{SL}(2)})\cdot R^{\mathrm{SL}(2)}[T]).$$
Hence we have an open immersion
$\Upsilon^{-1}(0)/\!/\mathrm{SL}(2)\hookrightarrow\overline{\Upsilon^{-1}(0)}/\!/\mathrm{SL}(2)$
given by $\mathfrak{p}\mapsto\mathfrak{p}^{hom}$ where
$\mathfrak{p}^{hom}$ is the homogenization of $\mathfrak{p}$.
Note that
$$\overline{\Upsilon^{-1}(0)}\setminus\Upsilon^{-1}(0)=\mathrm{Proj}(R[T]/I\cdot
R[T]+(T))\cong\mathrm{Proj}(R/I)=\mathbb{P}\Upsilon^{-1}(0)$$ and
$$[\overline{\Upsilon^{-1}(0)}/\!/\mathrm{SL}(2)]\setminus[\Upsilon^{-1}(0)/\!/\mathrm{SL}(2)]$$
$$=\mathrm{Proj}(R^{\mathrm{SL}(2)}[T]/(I\cap R^{\mathrm{SL}(2)})\cdot R^{\mathrm{SL}(2)}[T]+((T)\cap R^{\mathrm{SL}(2)}[T]))$$
$$\cong\mathrm{Proj}(R^{\mathrm{SL}(2)}/I\cap R^{\mathrm{SL}(2)})=\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2)$$
where $(T)$ is the ideal of $R[T]$ generated by $T$.
Consider the quotient map
$$f:\overline{\Upsilon^{-1}(0)}^{ss}\to\overline{\Upsilon^{-1}(0)}/\!/\mathrm{SL}(2).$$
In \cite[section 6]{BL94}, Bernstein and Lunts define a functor
$$Qf_{*}:\mathbf{D}_{\mathrm{SL}(2)}(\overline{\Upsilon^{-1}(0)}^{ss})\to\mathbf{D}(\overline{\Upsilon^{-1}(0)}/\!/\mathrm{SL}(2))$$
that extends the pushforward of sheaves $f_{*}$.
By the same arguments as those of \cite[$\S2$ and
$\S3$]{Woo03}, we can obtain morphisms
$$\lambda_{\overline{\Upsilon^{-1}(0)}}:\mathbf{IC}^{\bullet}(\overline{\Upsilon^{-1}(0)}/\!/\mathrm{SL}(2))[3]\to Qf_{*}\mathbf{IC}^{\bullet}_{\mathrm{SL}(2)}(\overline{\Upsilon^{-1}(0)}^{ss})$$
and
$$\kappa_{\overline{\Upsilon^{-1}(0)}}:Qf_{*}\mathbf{IC}^{\bullet}_{\mathrm{SL}(2)}(\overline{\Upsilon^{-1}(0)}^{ss})\to\lambda_{\overline{\Upsilon^{-1}(0)}}:\mathbf{IC}^{\bullet}(\overline{\Upsilon^{-1}(0)}/\!/\mathrm{SL}(2))[3]$$
such that $\kappa_{\overline{\Upsilon^{-1}(0)}}\circ\lambda_{\overline{\Upsilon^{-1}(0)}}=\mathrm{id} $. $\lambda_{\overline{\Upsilon^{-1}(0)}}$ induces a map
$$IH^{*}(\overline{\Upsilon^{-1}(0)}/\!/\mathrm{SL}(2))\to IH_{\mathrm{SL}(2)}^{*}(\overline{\Upsilon^{-1}(0)}^{ss})$$
which is an inclusion.
Hence $\kappa_{\overline{\Upsilon^{-1}(0)}}$ induces a map
$$\tilde{\kappa}_{\overline{\Upsilon^{-1}(0)}}:IH_{\mathrm{SL}(2)}^{*}(\overline{\Upsilon^{-1}(0)}^{ss})\to IH^{*}(\overline{\Upsilon^{-1}(0)}/\!/\mathrm{SL}(2))$$
which is split by the inclusion $IH^{*}(\overline{\Upsilon^{-1}(0)}/\!/\mathrm{SL}(2))\to IH_{\mathrm{SL}(2)}^{*}(\overline{\Upsilon^{-1}(0)}^{ss})$.
Consider the following commutative diagram:
\begin{equation}\label{Seq-Cpx}\xymatrix{\vdots\ar[d]&\vdots\ar[d]\\IH_{\mathrm{SL}(2)}^{i-2}(\mathbb{P}\Upsilon^{-1}(0)^{ss})\ar[r]^{\tilde{\kappa}_{\mathbb{P}\Upsilon^{-1}(0)}}\ar[d]&IH^{i-2}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))\ar[d]\\
IH_{\mathrm{SL}(2)}^i(\overline{\Upsilon^{-1}(0)}^{ss})\ar[r]^{\tilde{\kappa}_{\overline{\Upsilon^{-1}(0)}}}\ar[d]&IH^{i}(\overline{\Upsilon^{-1}(0)}/\!/\mathrm{SL}(2))\ar[d]\\
IH_{\mathrm{SL}(2)}^i(\Upsilon^{-1}(0))\ar[r]^{\tilde{\kappa}_{\Upsilon^{-1}(0)}}\ar[d]&IH^{i}(\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))\ar[d]\\
\vdots&\vdots}\end{equation}\\
Vertical sequences are Gysin sequences and $\tilde{\kappa}_{\Upsilon^{-1}(0)}$ is induced from $\tilde{\kappa}_{\mathbb{P}\Upsilon^{-1}(0)}$ and $\tilde{\kappa}_{\overline{\Upsilon^{-1}(0)}}$. Since
$\tilde{\kappa}_{\mathbb{P}\Upsilon^{-1}(0)}$ and
$\tilde{\kappa}_{\overline{\Upsilon^{-1}(0)}}$ are surjective, $\tilde{\kappa}_{\Upsilon^{-1}(0)}$ is surjective.
\item Following the idea of the proof of item (1), we get the result.
\item Following the idea of the proof of item (2), we get the result.
\item Following the idea of the proof of item (1), we get the result.
\item Following the idea of the proof of item (2), we get the result.
\end{enumerate}
\end{proof}
The second lemma shows how to compute the intersection cohomologies of the fibers of the normal cones of the singularities of $\mathbf{M}$ via those of the projectivizations of the fibers.
It is well known that there is a very ample line
bundle $\mathcal{L}$ (respectively, $\mathcal{M}_{1}$ and $\mathcal{M}_{2}$) on
\begin{center}$\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2)$ (respectively, $\mathbb{P}\Psi^{-1}(0)/\!/\mathbb{C}^*$ and $\mathbb{P}\mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)/\!/\mathbb{C}^*$),\end{center}
whose
pullback to $\mathbb{P}\Upsilon^{-1}(0)^{ss}$ (respectively, $\mathbb{P}\Psi^{-1}(0)^{ss}$ and $\mathbb{P}\mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)^{ss}$) is the $M$th (respectively, $N_{1}$th and $N_{2}$th) tensor
power of the hyperplane line bundle on $\mathbb{P}\Upsilon^{-1}(0)$ (respectively, $\mathbb{P}\Psi^{-1}(0)$ and $\mathbb{P}\mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)$) for some $M$ (respectively, $N_{1}$ and $N_{2}$).
Let $C_{\mathcal{L}}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))$ (respectively, $C_{\mathcal{M}_{1}}(\mathbb{P}\Psi^{-1}(0)/\!/\mathbb{C}^*)$ and
$$C_{\mathcal{M}_{2}}(\mathbb{P}\mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)/\!/\mathbb{C}^*)\text{)}$$
be the affine cone on $\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2)$ (respectively, $\mathbb{P}\Psi^{-1}(0)/\!/\mathbb{C}^*$ and
$$\mathbb{P}\mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)/\!/\mathbb{C}^*\text{)}$$
with respect to the projective embedding induced by the sections of $\mathcal{L}$ (respectively, $\mathcal{M}_{1}$ and $\mathcal{M}_{2}$).
\begin{lemma}\label{int coh of affine cone of git quotient}
(1) $IH^{*}(\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))=IH^{*}(C_{\mathcal{L}}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2)))$ and $$IH^{*}(Bl_{0}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))=IH^{*}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2)),$$
(2) $IH^{*}(\Psi^{-1}(0)/\!/\mathbb{C}^{*})=IH^{*}(C_{\mathcal{M}_{1}}(\mathbb{P}\Psi^{-1}(0)/\!/\mathbb{C}^{*}))$ and $$IH^{*}(Bl_{0}\Psi^{-1}(0)/\!/\mathbb{C}^{*})=IH^{*}(\mathbb{P}\Psi^{-1}(0)/\!/\mathbb{C}^{*}).$$
(3) $IH^{*}(\mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)/\!/\mathbb{C}^{*})=IH^{*}(C_{\mathcal{M}_{2}}(\mathbb{P}\mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)/\!/\mathbb{C}^{*}))$ and $$IH^{*}(Bl_{0}\mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)/\!/\mathbb{C}^{*})=IH^{*}(\mathbb{P}\mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)/\!/\mathbb{C}^{*}).$$
\end{lemma}
\begin{proof}
\begin{enumerate}
\item We first follow the idea of the proof of \cite[Lemma 2.15]{K86} to see that
$$C_{\mathcal{L}}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))\cong\Upsilon^{-1}(0)/\!/\mathrm{SL}(2)\times
F\cong(\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))/F,$$
where $F$ is the finite subgroup of $\mathrm{GL}(6g)$ consisting of all diagonal matrices
$diag(\eta,\cdots,\eta)$ such that $\eta$ is an $M$th root of unity.
The coordinate ring of $C_{\mathcal{L}}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))$ is the subring $(\mathbb{C}[Y_{0},\cdots,Y_{6g-1}]/I)_{M}^{\mathrm{SL}(2)}$ of the coordinate ring $\mathbb{C}[Y_{0},\cdots,Y_{6g-1}]/I$ of $\Upsilon^{-1}(0)$ which is generated by homogeneous polynomials fixed by the natural action of $\mathrm{SL}(2)$ and of degree $M$. Since
$$(\mathbb{C}[Y_{0},\cdots,Y_{6g-1}]/I)_{M}=\mathbb{C}[Y_{0},\cdots,Y_{6g-1}]_{M}/I\cap\mathbb{C}[Y_{0},\cdots,Y_{6g-1}]_{M},$$
we have
$$(\mathbb{C}[Y_{0},\cdots,Y_{6g-1}]/I)_{M}^{\mathrm{SL}(2)}=\mathbb{C}[Y_{0},\cdots,Y_{6g-1}]_{M}^{\mathrm{SL}(2)}/I\cap\mathbb{C}[Y_{0},\cdots,Y_{6g-1}]_{M}^{\mathrm{SL}(2)}$$
$$=\mathbb{C}[Y_{0},\cdots,Y_{6g-1}]^{\mathrm{SL}(2)\times F}/I\cap\mathbb{C}[Y_{0},\cdots,Y_{6g-1}]^{\mathrm{SL}(2)\times F}=(\mathbb{C}[Y_{0},\cdots,Y_{6g-1}]/I)^{\mathrm{SL}(2)\times F}.$$
Thus we get
$$C_{\mathcal{L}}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))\cong\Upsilon^{-1}(0)/\!/\mathrm{SL}(2)\times
F\cong(\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))/F$$
and then
$$IH^{*}(\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))^{F}=IH^{*}(C_{\mathcal{L}}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2)))$$
by Lemma \ref{int coh of quotient by finite group}.
It remains to show that the action of $F$ on
$IH^{*}(\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))$ is trivial. Since the Kirwan map
$$IH_{\mathrm{SL}(2)}^{*}(\Upsilon^{-1}(0))\rightarrow IH^{*}(\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))$$
is surjective by Lemma \ref{Kir-Map}-(2), it suffices to show that
the action of $F$ on $IH_{\mathrm{SL}(2)}^{*}(\Upsilon^{-1}(0))$ is trivial.
Let
$$\pi_1:Bl_0\Upsilon^{-1}(0)\to \Upsilon^{-1}(0)$$
be the blowing-up of $\Upsilon^{-1}(0)$ at the vertex and let
$$\pi_2:Bl_{\mathrm{Hom}_1}Bl_0\Upsilon^{-1}(0)\to Bl_0\Upsilon^{-1}(0)$$
be the blowing-up of $Bl_0\Upsilon^{-1}(0)$ along
$\widetilde{\mathrm{Hom}_1(sl(2),\mathbb{H}^g)}$, where $\widetilde{\mathrm{Hom}_1(sl(2),\mathbb{H}^g)}$ is the strict transform of $\mathrm{Hom}_1(sl(2),\mathbb{H}^g)$. By the universal property of blowing-up, the action of $F$ on $\Upsilon^{-1}(0)$ lifts to an action of $F$ on $Bl_{\mathrm{Hom}_1}Bl_0\Upsilon^{-1}(0)$. Since $\pi_1\circ\pi_2$ is proper and $Bl_{\mathrm{Hom}_1}Bl_0\Upsilon^{-1}(0)$ is smooth (See the proof of Lemma 1.8.5 in \cite{O99}), by Proposition \ref{3 consequences}-(3), $IH_{\mathrm{SL}(2)}^{*}(\Upsilon^{-1}(0))$ is a direct summand
of
$$IH_{\mathrm{SL}(2)}^{*}(Bl_{\mathrm{Hom}_1}Bl_0\Upsilon^{-1}(0))=H_{\mathrm{SL}(2)}^{*}(Bl_{\mathrm{Hom}_1}Bl_0\Upsilon^{-1}(0)).$$
Since $Bl_{\mathrm{Hom}_1}Bl_0\Upsilon^{-1}(0)$ is homotopically equivalent
to $Bl_{\mathbb{P}\mathrm{Hom}_1}\mathbb{P}\Upsilon^{-1}(0)$, it suffices to
show that the action of $F$ on
$H_{\mathrm{SL}(2)}^{*}(Bl_{\mathbb{P}\mathrm{Hom}_1}\mathbb{P}\Upsilon^{-1}(0))$ is
trivial. But this is true because the action of $F$ on
$\mathbb{P}\Upsilon^{-1}(0)$ is trivial and it lifts to the trivial action of $F$ on $Bl_{\mathbb{P}\mathrm{Hom}_1}\mathbb{P}\Upsilon^{-1}(0)$. Hence $F$ acts trivially on $IH^{*}(\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))$.
Similarly, we next see that $Bl_{v}(C_{\mathcal{L}}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2)))$ is naturally isomorphic to
$$(Bl_{0}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))/F,$$
where $v$ is the vertex of $C_{\mathcal{L}}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))$.
Let $J$ be the ideal of $\mathbb{C}[Y_{0},\cdots,Y_{6g-1}]/I$ corresponding to the vertex $O$ of $\Upsilon^{-1}(0)$. Then we have $Bl_{0}\Upsilon^{-1}(0)=\mathbf{Proj}(\oplus_{m\ge 0}J^{m})$. Then
$$(Bl_{0}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))/F=Bl_{0}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2)\times F=\mathbf{Proj}(\oplus_{m\ge 0}(J^{m})^{\mathrm{SL}(2)\times F}).$$
Since $(J^{m})^{\mathrm{SL}(2)\times F}=J^{m}\cap(\mathbb{C}[Y_{0},\cdots,Y_{6g-1}]/I)^{\mathrm{SL}(2)\times F}=(J\cap(\mathbb{C}[Y_{0},\cdots,Y_{6g-1}]/I)^{\mathrm{SL}(2)\times F})^{m}$
$$=(J^{\mathrm{SL}(2)\times F})^{m}$$
and $J^{\mathrm{SL}(2)\times F}$ is the ideal corresponding to $v=O/\!/\mathrm{SL}(2)\times F$,
we have
$$\mathbf{Proj}(\oplus_{m\ge 0}(J^{m})^{\mathrm{SL}(2)\times F})=\mathbf{Proj}(\oplus_{m\ge 0}(J^{\mathrm{SL}(2)\times F})^{m})=Bl_{v}(\Upsilon^{-1}(0)/\!/\mathrm{SL}(2)\times
F)$$
$$=Bl_{v}(C_{\mathcal{L}}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))).$$
Thus
$$IH^{*}(Bl_{0}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))^{F}=IH^{*}(Bl_{v}(C_{\mathcal{L}}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2)))).$$
By the same idea of the proof of the first statement, $F$ acts trivially on $IH^{*}(Bl_{0}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))$ and then
$$IH^{*}(Bl_{0}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))^{F}=IH^{*}(Bl_{0}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2)).$$
Since $Bl_{v}(C_{\mathcal{L}}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2)))$ is homeomorphic to the line bundle $\mathcal{L}^{\vee}$ over $\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2)$, there is a Leray spectral sequence $E_{r}^{pq}$ converging to
$$IH^{*}(Bl_{v}(C_{\mathcal{L}}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))))$$
with
$$E_{2}^{pq}=IH^{p}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2),IH^{q}(\mathbb{C}))=\begin{cases}IH^{p}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))&\text{if }q=0\\
0&\text{otherwise}.\end{cases}$$
Hence we get
$$IH^{*}(Bl_{0}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))=IH^{*}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2)).$$
\item Following the idea of the proof of item (1), we get the result.
\item Following the idea of the proof of item (1), we get the result.
\end{enumerate}
\end{proof}
By the standard argument of \cite[Proposition 4.7.2]{KW06}, we get the third lemma as follows. It gives a way to compute the intersection cohomology of affine cones of projective GIT quotients.
\begin{lemma}\label{int coh of affine cone of git quotient II}
(1) Let $n=\dim_{\mathbb{C}}C_{\mathcal{L}}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))$. Then
$$IH^{i}(C_{\mathcal{L}}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2)))\cong\begin{cases}0&\text{for }i\ge n\\
IH^{i}(C_{\mathcal{L}}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))-\{0\})&\text{for }i<n.\end{cases}$$
(2) Let $n=\dim_{\mathbb{C}}C_{\mathcal{M}_{1}}(\mathbb{P}\Psi^{-1}(0)/\!/\mathbb{C}^{*})$. Then
$$IH^{i}(C_{\mathcal{M}_{1}}(\mathbb{P}\Psi^{-1}(0)/\!/\mathbb{C}^{*}))\cong\begin{cases}0&\text{for }i\ge n\\
IH^{i}(C_{\mathcal{M}_{1}}(\mathbb{P}\Psi^{-1}(0)/\!/\mathbb{C}^{*})-\{0\})&\text{for }i<n.\end{cases}$$
(3) Let $n=\dim_{\mathbb{C}}C_{\mathcal{M}_{2}}(\mathbb{P}\mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)/\!/\mathbb{C}^{*})$. Then
$$IH^{i}(C_{\mathcal{M}_{2}}(\mathbb{P}\mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)/\!/\mathbb{C}^{*}))$$
$$\cong\begin{cases}0&\text{for }i\ge n\\
IH^{i}(C_{\mathcal{M}_{2}}(\mathbb{P}\mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)/\!/\mathbb{C}^{*})-\{0\})&\text{for }i<n.\end{cases}$$
\end{lemma}
The following lemma explains how $IH^{*}(\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))$ (respectively, $IH^{*}(\Psi^{-1}(0)/\!/\mathbb{C}^*)$ and $IH^{*}(\mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)/\!/\mathbb{C}^*)$) can be computed in terms of $IH^{*}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))$ (respectively, $IH^{*}(\mathbb{P}\Psi^{-1}(0)/\!/\mathbb{C}^*)$ and $IH^{*}(\mathbb{P}\mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)/\!/\mathbb{C}^*)$) as desired.
\begin{lemma}\label{suffices-to-show-on-projectivized-git}
(1) $\begin{cases}
IH^{i}(\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))=0&\text{for }i\geq\dim \Upsilon^{-1}(0)/\!/\mathrm{SL}(2)\\
IH^{i}(\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))\cong\mathrm{coker} \lambda&\text{for }i<\dim \Upsilon^{-1}(0)/\!/\mathrm{SL}(2),
\end{cases}$
where $\lambda:IH^{i-2}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))\rightarrow IH^{i}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))$ is an injection.
(2) $\begin{cases}
IH^{i}(\Psi^{-1}(0)/\!/\mathbb{C}^{*})=0&\text{for }i\geq\dim \Psi^{-1}(0)/\!/\mathbb{C}^{*}\\
IH^{i}(\Psi^{-1}(0)/\!/\mathbb{C}^{*})\cong\mathrm{coker} \lambda&\text{for }i<\dim \Psi^{-1}(0)/\!/\mathbb{C}^{*},
\end{cases}$
where $\lambda:IH^{i-2}(\mathbb{P}\Psi^{-1}(0)/\!/\mathbb{C}^{*})\rightarrow IH^{i}(\mathbb{P}\Psi^{-1}(0)/\!/\mathbb{C}^{*})$ is an injection.
(3) $\begin{cases}
IH^{i}(\mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)/\!/\mathbb{C}^{*})=0&\text{for }i\geq\dim \mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)/\!/\mathbb{C}^{*}\\
IH^{i}(\mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)/\!/\mathbb{C}^{*})\cong\mathrm{coker} \lambda&\text{for }i<\dim \mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)/\!/\mathbb{C}^{*},
\end{cases}$
where $\lambda:IH^{i-2}(\mathbb{P}\mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)/\!/\mathbb{C}^{*})\rightarrow IH^{i}(\mathbb{P}\mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)/\!/\mathbb{C}^{*})$ is an injection.
\end{lemma}
\begin{proof}
We follow the idea of the proof of \cite[Corollary 2.17]{K86}. We only prove item (1) because the proofs of item (2) and item (3) are similar to that of item (1).
By Lemma \ref{int coh of affine cone of git quotient}-(1),
$$IH^{*}(\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))=IH^{*}(C_{\mathcal{L}}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))).$$
Let $n=\dim_{\mathbb{C}}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2)$. By Lemma \ref{int coh of affine cone of git quotient II}-(1), $$IH^{i}(C_{\mathcal{L}}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2)))\cong\begin{cases}0&\text{if }i\ge n\\
IH^{i}(C_{\mathcal{L}}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))-\{0\})&\text{if }i<n.\end{cases}$$
Since $C_{\mathcal{L}}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))-\{0\}$ fibers over
$\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2)$ with fiber $\mathbb{C}^{*}$, there is a Leray spectral sequence
$E_{r}^{pq}$ converging to
$$IH^{*}(C_{\mathcal{L}}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))-\{0\})$$
with
$$E_{2}^{pq}=IH^{p}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2),IH^{q}(\mathbb{C}^{*}))=\begin{cases}IH^{p}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))&\text{if }q=0,1\\
0&\text{otherwise}.\end{cases}$$
It follows from \cite[5.1]{CGM82} and \cite[p.462--p.468]{GH78} that the differential
$$\lambda:IH^{i-2}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))\rightarrow IH^{i}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))$$
is given by the multiplication by $c_{1}(\mathcal{L})$. By the Hard
Lefschetz theorem for intersection cohomology, $\lambda$ is injective for $i<n$. Hence we get the result.
\end{proof}
The quotients $\mathbb{P}\Psi^{-1}(0)/\!/\mathbb{C}^{*}$ and $\mathbb{P}\mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)/\!/\mathbb{C}^{*}$ can be identified with some incidence variety.
\begin{lemma}\label{isomorphic to incidence variety}
Let $I_{2g-3}$ be the incidence variety given by
$$I_{2g-3}=\{(p,H)\in\mathbb{P}^{2g-3}\times\breve{\mathbb{P}}^{2g-3}|p\in H\}.$$
\begin{enumerate}
\item\label{description of projectivized middle singularity} $\mathbb{P}\Psi^{-1}(0)/\!/\mathbb{C}^{*}\cong I_{2g-3}$,
\item $\mathbb{P}\mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)/\!/\mathbb{C}^{*}\cong I_{2g-3}$.
\end{enumerate}
\end{lemma}
\begin{proof}
\begin{enumerate}
\item Consider the map $f:\mathbb{P}\Psi^{-1}(0)\to I_{2g-3}$ given by
$$(a,b,c,d)\mapsto((b,c),(-a,d)).$$
Since $f$ is $\mathbb{C}^{*}$-invariant, we have the induced map
$$\bar{f}:\mathbb{P}\Psi^{-1}(0)/\!/\mathbb{C}^{*}\to I_{2g-3}.$$
We claim that $\bar{f}$ is injective. Assume that $\bar{f}([a_{1},b_{1},c_{1},d_{1}])=\bar{f}([a_{2},b_{2},c_{2},d_{2}])$ where $[a,b,c,d]$ denotes the closed orbit of $(a,b,c,d)$. Then there are nonzero complex numbers $\lambda$ and $\mu$ such that $(b_{1},c_{1})=\lambda(b_{2},c_{2})$ and $(-a_{1},d_{1})=\mu(-a_{2},d_{2})$. Then
$$[a_{1},b_{1},c_{1},d_{1}]=[\mu a_{2},\lambda b_{2},\lambda c_{2},\mu d_{2}]=[(\lambda\mu)^{1/2}a_{2},(\lambda\mu)^{1/2}b_{2},(\lambda\mu)^{1/2}c_{2},(\lambda\mu)^{1/2}d_{2}]=[a_{2},b_{2},c_{2},d_{2}].$$
Thus $\bar{f}$ is injective.
Since the domain and the range of $\bar{f}$ are normal varieties with the same dimension and the range $I_{2g-3}$ is irreducible, $\bar{f}$ is an isomorphism.
\item Consider the map $g:\mathbb{P}\mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)\to I_{2g-3}$ given by $$(Z_{12},\cdots,Z_{2g,2},Z_{13},\cdots,Z_{2g,3})\mapsto((Z_{12},Z_{22},\cdots,Z_{2g-1,2},Z_{2g,2}),(Z_{23},-Z_{13}\cdots,Z_{2g,3},-Z_{2g-1,3})).$$
Since $g$ is $\mathbb{C}^{*}$-invariant, we have the induced map
$$\bar{g}:\mathbb{P}\mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)/\!/\mathbb{C}^{*}\to I_{2g-3}.$$
We can see that $\bar{g}$ is injective by the similar way as in the proof of (\ref{description of projectivized middle singularity}). Since the domain and the range of $\bar{g}$ are normal varieties with the same dimension and the range $I_{2g-3}$ is irreducible, $\bar{g}$ is an isomorphism.
\end{enumerate}
\end{proof}
By the proof of Lemma \ref{geometric descriptions of second cones}, $\mathcal{C}_{2}/\!/\mathrm{SL}(2)=(Y/\!/\mathbb{C}^{*})/\mathbb{Z}_{2}$ and $\tilde{\mathcal{C}}_{2}/\!/\mathrm{SL}(2)=(Bl_{\widetilde{T^{*}J}}Y/\!/\mathbb{C}^{*})/\mathbb{Z}_{2}$, where $Y$ is either a $\Psi^{-1}(0)$-bundle or a $\mathrm{Hom}^{\omega_{\varphi}}(\ker\varphi,\mathrm{im} \varphi^{\perp}/\mathrm{im} \varphi)$-bundle over $\widetilde{T^{*}J}$.
To give computable fomulas from Lemma \ref{intersection blowing-up formula}, we need the following technical statements for $Y/\!/\mathbb{C}^{*}$ and $Bl_{\widetilde{T^{*}J}}Y/\!/\mathbb{C}^{*}$.
\begin{lemma}\label{second cone gives a constant sheaf}
Let $g:Y/\!/\mathbb{C}^{*}\to\widetilde{T^{*}J}$ be the map induced by the projection $Y\to\widetilde{T^{*}J}$ and let $h:Bl_{\widetilde{T^{*}J}}Y/\!/\mathbb{C}^{*}\to\widetilde{T^{*}J}$ be the map induced by the composition of maps $Bl_{\widetilde{T^{*}J}}Y\to Y\to\widetilde{T^{*}J}$. Then $R^{i}g_{*}\mathbf{IC}^{\bullet}(Y/\!/\mathbb{C}^{*})$ and $R^{i}h_{*}\mathbf{IC}^{\bullet}(Bl_{\widetilde{T^{*}J}}Y/\!/\mathbb{C}^{*})$ are constant sheaves for each $i\ge0$.
\end{lemma}
\begin{proof}
Following the idea of proof of \cite[Proposition 2.13]{K86}, we can see that $R^{i}g_{*}\mathbf{IC}^{\bullet}(Y/\!/\mathbb{C}^{*})$ and $R^{i}h_{*}\mathbf{IC}^{\bullet}(Bl_{\widetilde{T^{*}J}}Y/\!/\mathbb{C}^{*})$ are locally constant sheaves for each $i\ge0$. We get the conclusion from Lemma \ref{int coh of affine cone of git quotient}-(2), (3), Lemma \ref{suffices-to-show-on-projectivized-git}-(2), (3) and Lemma \ref{isomorphic to incidence variety}.
\end{proof}
Then we have the following computable blowing-up formula.
\begin{theorem}\label{computable intersection blowing-up formula}
(1) $\dim IH^{i}(\mathbf{R}_{1}^{ss}/\!/\mathrm{SL}(2))=\dim IH^{i}(\mathbf{R}/\!/\mathrm{SL}(2))$
$$+2^{2g}\dim IH^{i}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{PGL}(2))-2^{2g}\dim IH^{i}(\Upsilon^{-1}(0)/\!/\mathrm{PGL}(2))$$
for all $i\ge0$.
(2) $\dim IH^{i}(\mathbf{R}_{2}^{s}/\mathrm{SL}(2))=\dim IH^{i}(\mathbf{R}_{1}^{ss}/\!/\mathrm{SL}(2))$
$$+\sum_{p+q=i}\dim[H^{p}(\widetilde{T^{*}J})\otimes H^{t(q)}(I_{2g-3})]^{\mathbb{Z}_{2}}$$
for all $i\ge0$, where $t(q)=q-2$ for $q\le\dim I_{2g-3}=4g-7$ and $t(q)=q$ otherwise.
(3) $\dim IH^{i}((Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss})^{s}/\!/\mathrm{SL}(2))=\dim IH^{i}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{SL}(2))$
$$+\sum_{p+q=i}\dim[H^{p}(\mathbb{P}^{2g-1})\otimes H^{t(q)}(I_{2g-3})]^{\mathbb{Z}_{2}}$$
for all $i\ge0$, where $t(q)=q-2$ for $q\le\dim I_{2g-3}=4g-7$ and $t(q)=q$ otherwise.
\end{theorem}
\begin{proof}
\begin{enumerate}
\item Since it follows from Lemma \ref{int coh of affine cone of git quotient}-(1) that $$IH^{i}(Bl_{0}\Upsilon^{-1}(0)/\!/\mathrm{PGL}(2))=IH^{i}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{PGL}(2)),$$
we get the formula.
\item Let $g:Y/\!/\mathbb{C}^{*}\to\widetilde{T^{*}J}$ and $h:Bl_{\widetilde{T^{*}J}}Y/\!/\mathbb{C}^{*}\to\widetilde{T^{*}J}$ be the maps induced by the projections $Y\to\widetilde{T^{*}J}$ and $Bl_{\widetilde{T^{*}J}}Y\to\widetilde{T^{*}J}$. By Proposition \ref{3 consequences}-(2) and Remark \ref{Leray ss for intersection cohomology with smooth target}, the perverse Leray spectral sequences of intersection cohomology associated to $g$ and $h$ have $E_2$ terms given by
$$E_{2}^{pq}=IH^{p}(\widetilde{T^{*}J},R^{q}g_{*}\mathbf{IC}^{\bullet}(Y/\!/\mathbb{C}^{*}))$$
and
$$E_{2}^{pq}=IH^{p}(\widetilde{T^{*}J},R^{q}h_{*}\mathbf{IC}^{\bullet}(Bl_{\widetilde{T^{*}J}}Y/\!/\mathbb{C}^{*})).$$
Then we have
$$E_{2}^{pq}=IH^{p}(\widetilde{T^{*}J})\otimes IH^{q}(\widehat{I_{2g-3}})$$
and
$$E_{2}^{pq}=IH^{p}(\widetilde{T^{*}J})\otimes IH^{q}(I_{2g-3})$$
by Lemma \ref{int coh of affine cone of git quotient}, Lemma \ref{isomorphic to incidence variety} and Lemma \ref{second cone gives a constant sheaf}, where $\widehat{I_{2g-3}}$ is the affine cone of $I_{2g-3}$. It follows from Proposition \ref{3 consequences}-(2) that the decomposition theorem for $h$ implies that the perverse Leray spectral sequence of intersection cohomology associated to $h$ degenerates at the $E_2$ term. Since $IH^{q}(\widehat{I_{2g-3}})$ embeds in $IH^{q}(I_{2g-3})$ by Lemma \ref{suffices-to-show-on-projectivized-git}-(2), (3) and Lemma \ref{isomorphic to incidence variety}, the perverse Leray spectral sequence of intersection cohomology associated to $g$ also degenerates at the $E_2$ term. Since $\mathcal{C}_{2}/\!/\mathrm{SL}(2)=(Y/\!/\mathbb{C}^{*})/\mathbb{Z}_{2}$ and $\tilde{\mathcal{C}}_{2}/\!/\mathrm{SL}(2)=(Bl_{\widetilde{T^{*}J}}Y/\!/\mathbb{C}^{*})/\mathbb{Z}_{2}$, we have
$$IH^{i}(\mathcal{C}_{2}/\!/\mathrm{SL}(2))=\bigoplus_{p+q=i}[H^{p}(\widetilde{T^{*}J})\otimes H^{q}(\widehat{I_{2g-3}})]^{\mathbb{Z}_{2}}$$
and
$$IH^{i}(\tilde{\mathcal{C}}_{2}/\!/\mathrm{SL}(2))=\bigoplus_{p+q=i}[H^{p}(\widetilde{T^{*}J})\otimes H^{q}(I_{2g-3})]^{\mathbb{Z}_{2}}$$
by Lemma \ref{int coh of quotient by finite group}. Applying Lemma \ref{suffices-to-show-on-projectivized-git}-(2), (3) again, we get the formula.
\item Note that $\mathcal{C}/\!/\mathrm{SL}(2)=(Y|_{\mathbb{P}^{2g-1}}/\!/\mathbb{C}^{*})/\mathbb{Z}_{2}$ and $\tilde{\mathcal{C}}_{2}/\!/\mathrm{SL}(2)=(Bl_{\mathbb{P}^{2g-1}}Y|_{\mathbb{P}^{2g-1}}/\!/\mathbb{C}^{*})/\mathbb{Z}_{2}$. Let $g':Y|_{\mathbb{P}^{2g-1}}/\!/\mathbb{C}^{*}\to\mathbb{P}^{2g-1}$ and $h':Bl_{\mathbb{P}^{2g-1}}Y|_{\mathbb{P}^{2g-1}}/\!/\mathbb{C}^{*}\to\mathbb{P}^{2g-1}$ be the maps induced by the projections $Y|_{\mathbb{P}^{2g-1}}\to\mathbb{P}^{2g-1}$ and $Bl_{\mathbb{P}^{2g-1}}Y|_{\mathbb{P}^{2g-1}}\to\mathbb{P}^{2g-1}$. Since $\mathbb{P}^{2g-1}$ is simply connected, $R^{i}g'_{*}\mathbf{IC}^{\bullet}(Y|_{\mathbb{P}^{2g-1}}/\!/\mathbb{C}^{*})$ and $R^{i}h'_{*}\mathbf{IC}^{\bullet}(Bl_{\mathbb{P}^{2g-1}}Y|_{\mathbb{P}^{2g-1}}/\!/\mathbb{C}^{*})$ are constant sheaves for each $i\ge0$ and then the perverse Leray spectral sequences of intersection cohomology associated to $g'$ and $h'$ have $E_2$ terms given by
$$E_{2}^{pq}=IH^{p}(\mathbb{P}^{2g-1})\otimes IH^{q}(\widehat{I_{2g-3}})$$
and
$$E_{2}^{pq}=IH^{p}(\mathbb{P}^{2g-1})\otimes IH^{q}(I_{2g-3})$$
by Lemma \ref{int coh of affine cone of git quotient} and Lemma \ref{isomorphic to incidence variety}. By the same argument as in the remaining part of the proof of item (2), we get the formula.
\end{enumerate}
\end{proof}
\section{A strategy to get a formula for the Poincar\'{e} polynomial of $IH^{*}(\mathbf{M})$}\label{strategy}
Since $\mathbf{R}_{2}^{s}/\mathrm{SL}(2)$ has an orbifold singularity, we have $H^{i}(\mathbf{R}_{2}^{s}/\mathrm{SL}(2))\cong H_{\mathrm{SL}(2)}^{i}(\mathbf{R}_{2}^{s})$ for each $i\ge0$. If we have a blowing-up formula for the equivariant cohomology that can be applied to get $\dim H_{\mathrm{SL}(2)}^{i}(\mathbf{R}_{2}^{s})$ from $\dim H_{\mathrm{SL}(2)}^{i}(\mathbf{R})$ for each $i\ge0$, Theorem \ref{computable intersection blowing-up formula} can be used to calculate $\dim IH^{i}(\mathbf{M})$ from $\dim H^{i}(\mathbf{R}_{2}^{s}/\mathrm{SL}(2))$ for each $i$.
\subsection{Towards blowing-up formula for the equivariant cohomology}
In this subsection, we give a strategy to get a blowing-up formula for the equivariant cohomology in Kirwan's algorithm, and prove that the blowing-up formula for the equivariant cohomology on the blowing-up $\pi:Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss}\to\mathbb{P}\Upsilon^{-1}(0)^{ss}$ holds. Assume that $G$ is a compact connected algebraic group throughout this subsection.
\begin{proposition}\label{equiv-coh to equiv-int-coh}
For a normal quasi-projective complex variety $Y$ on which $G$ acts algebraically, we have an injection
$$i_{G}:H_{G}^{i}(Y)\hookrightarrow IH_{G}^{i}(Y).$$
\end{proposition}
\begin{proof}
For each $j$, we have a morphism
$$i_{j}:H^{i}(Y\times_{G}\mathrm{E} G_{j})=IH_{2\dim(Y\times_{G}\mathrm{E} G_{j})-i}^{\bar{0}}(Y\times_{G}\mathrm{E} G_{j})\to IH_{2\dim(Y\times_{G}\mathrm{E} G_{j})-i}^{\bar{m}}(Y\times_{G}\mathrm{E} G_{j})=IH^{i}(Y\times_{G}\mathrm{E} G_{j})$$
induced from the inclusion $IC_{2\dim(Y\times_{G}\mathrm{E} G_{j})-i}^{\bar{0}}(Y\times_{G}\mathrm{E} G_{j})\hookrightarrow IC_{2\dim(Y\times_{G}\mathrm{E} G_{j})-i}^{\bar{m}}(Y\times_{G}\mathrm{E} G_{j})$ given by $\xi\to\xi$.
We claim that $i_{j}$ is injective for each $j$. Assume that $\xi\in IC_{2\dim(Y\times_{G}\mathrm{E} G_{j})-i}^{\bar{0}}(Y\times_{G}\mathrm{E} G_{j})$ and $\xi=\partial\eta$ for some $\eta\in IC_{2\dim(Y\times_{G}\mathrm{E} G_{j})-i+1}^{\bar{m}}(Y\times_{G}\mathrm{E} G_{j})$. Then
$$\dim_{\mathbb{R}}(|\xi|\cap Y_{n-c})\le(2\dim(Y\times_{G}\mathrm{E} G_{j})-i)-2c.$$
Since $\dim_{\mathbb{R}}(|\eta|\cap Y_{n-c})-\dim_{\mathbb{R}}(|\partial\eta|\cap Y_{n-c})\le1$,
$$\dim_{\mathbb{R}}(|\eta|\cap Y_{n-c})\le\dim_{\mathbb{R}}(|\partial\eta|\cap Y_{n-c})+1\le (2\dim(Y\times_{G}\mathrm{E} G_{j})-i+1)-2c,$$
that is,
$$\eta\in IC_{2\dim(Y\times_{G}\mathrm{E} G_{j})-i+1}^{\bar{0}}(Y\times_{G}\mathrm{E} G_{j}).$$
Then $\xi=\partial\eta=0$ in $IH_{2\dim(Y\times_{G}\mathrm{E} G_{j})-i}^{\bar{0}}(Y\times_{G}\mathrm{E} G_{j})$. Thus $i_{j}$ is injective for each $j$.
Since $i_{G}=\displaystyle\varprojlim_{j}i_{j}$, we get the result.
\end{proof}
\begin{proposition}[\cite{Web99}]\label{commutativity}
Let $Y_{1}$ and $Y_{2}$ be normal irreducible quasi-projective complex varieties on which $G$ acts algebraically. For any $G$-equivariant morphism $f:Y_{1}\to Y_{2}$, there exists $\lambda_{f}:IH_{G}^{i}(Y_{2})\to IH_{G}^{i}(Y_{1})$ such that the following diagram
$$\xymatrix{H_{G}^{i}(Y_{2})\ar@^{(->}[r]^{i_{G}}\ar[d]_{f^{*}}&IH_{G}^{i}(Y_{2})\ar[d]^{\lambda_{f}}\\
H_{G}^{i}(Y_{1})\ar@^{(->}[r]^{i_{G}}&IH_{G}^{i}(Y_{1}).}$$
commutes.
\end{proposition}
\begin{proof}
We follow the idea of \cite{Web99}. Note that $i_{G}:H_{G}^{i}(Y)\hookrightarrow IH_{G}^{i}(Y)$ is induced from the canonical morphism of $G$-equivariant sheaves $\omega_{Y}:\mathbb{Q}_{Y}^{G}\to\mathbf{IC}_{G}^{\bullet}(Y)$. Let $\pi_{Y_{2}}:\tilde{Y}_{2}\to Y_{2}$ be a resolution of $Y_{2}$. $\tilde{Y}_{1}$ denotes the fiber product $Y_{1}\times_{Y_{2}}\tilde{Y}_{2}$. Then the following diagram
$$\xymatrix{\tilde{Y}_{1}\ar[r]^{p_{2}}\ar[d]_{p_{1}}&\tilde{Y}_{2}\ar[d]^{\pi_{Y_{2}}}\\Y_{1}\ar[r]^{f}&Y_{2}}$$
commutes. Here $p_{1}$ and $\pi_{Y_{2}}$ are proper.
We have only to show the following diagram of sheaves over $Y_{2}$
$$\xymatrix{R\pi_{Y_{2}*}\mathbf{IC}_{G}^{\bullet}(\tilde{Y}_{2})\ar@^{=}[r]&R\pi_{Y_{2}*}\mathbb{Q}_{\tilde{Y}_{2}}^{G}\ar[r]^{R\pi_{Y_{2}*}(p_{2}^{*})}&Rf_{*}Rp_{1*}\mathbb{Q}_{\tilde{Y}_{1}}^{G}\ar[r]^{Rf_{*}Rp_{1*}(\omega_{\tilde{Y}_{1}})}&Rf_{*}Rp_{1*}\mathbf{IC}_{G}^{\bullet}(\tilde{Y}_{1})\ar[d]^{Rf_{*}(p)}\\
\mathbf{IC}_{G}^{\bullet}(Y_{2})\ar[u]^{i}&\mathbb{Q}_{Y_{2}}^{G}\ar[l]_{\omega_{Y_{2}}}\ar[r]^{f^{*}}\ar[u]_{\pi_{Y_{2}}^{*}}&Rf_{*}\mathbb{Q}_{Y_{1}}^{G}\ar[r]^{Rf_{*}(\omega_{Y_{1}})\quad}\ar[u]_{Rf_{*}(p_{1}^{*})}&Rf_{*}\mathbf{IC}_{G}^{\bullet}(Y_{1})}$$
commutes, where the inclusion $i:\mathbf{IC}_{G}^{\bullet}(Y_{2})\hookrightarrow R\pi_{Y_{2}*}\mathbf{IC}_{G}^{\bullet}(\tilde{Y}_{2})$ and the projection $p:Rp_{1*}\mathbf{IC}_{G}^{\bullet}(\tilde{Y}_{1})\to\mathbf{IC}_{G}^{\bullet}(Y_{1})$ are induced from Proposition \ref{3 consequences}-(3). It suffices to show that
$$\xymatrix{\mathbb{Q}_{Y_{2}}^{G}\ar[r]^{\omega_{Y_{2}}\quad}&\mathbf{IC}_{G}^{\bullet}(Y_{2})\ar[r]^{i\quad\quad\quad\quad}&R\pi_{Y_{2}*}\mathbf{IC}_{G}^{\bullet}(\tilde{Y}_{2})=R\pi_{Y_{2}*}\mathbb{Q}_{\tilde{Y}_{2}}^{G}}$$
and
$$\xymatrix{\mathbb{Q}_{Y_{2}}^{G}\ar[r]^{\pi_{Y_{2}}^{*}\quad}&R\pi_{Y_{2}*}\mathbb{Q}_{\tilde{Y}_{2}}^{G}}$$
$$\text{(respectively, }\xymatrix{\mathbb{Q}_{Y_{1}}^{G}\ar[r]^{p_{1}^{*}\quad}&Rp_{1*}\mathbb{Q}_{\tilde{Y}_{1}}^{G}\ar[r]^{Rp_{1*}(\omega_{\tilde{Y}_{1}})\quad}&Rp_{1*}\mathbf{IC}_{G}^{\bullet}(\tilde{Y}_{1})\ar[r]^{\quad p}&\mathbf{IC}_{G}^{\bullet}(Y_{1})}$$
and
$$\xymatrix{\mathbb{Q}_{Y_{1}}^{G}\ar[r]^{\omega_{Y_{1}}\quad}&\mathbf{IC}_{G}^{\bullet}(Y_{1})}\text{)}$$
coincide over $Y_{2}$ (respectively, $Y_{1}$).
Let $U$ (respectively, $V$) be the regular part of $Y_{2}$ (respectively, $Y_{1}$). Note that these morphisms are equal on $U$ (respectively, $V$) after multiplication by a constant if necessary. Thus we have only to show that the restriction morphisms
$$\xymatrix{\mathrm{Hom}(\mathbb{Q}_{Y_{2}}^{G},R\pi_{Y_{2}*}\mathbb{Q}_{\tilde{Y}_{2}}^{G})\ar[r]^{\rho_{U}\quad}&\mathrm{Hom}(\mathbb{Q}_{Y_{2}}^{G}|_{U},R\pi_{Y_{2}*}\mathbb{Q}_{\tilde{Y}_{2}}^{G}|_{U})}$$
and
$$\xymatrix{\mathrm{Hom}(\mathbb{Q}_{Y_{1}}^{G},\mathbf{IC}_{G}^{\bullet}(Y_{1}))\ar[r]^{\rho_{V}\quad}&\mathrm{Hom}(\mathbb{Q}_{Y_{1}}^{G}|_{V},\mathbf{IC}_{G}^{\bullet}(Y_{1})|_{V})}$$
are injective.
Since the restriction morphisms
$$\xymatrix{\mathrm{Hom}(\mathbb{Q}_{Y_{2}\times_{G}\mathrm{E} G_{j}},R(\pi_{Y_{2}}\times 1_{\mathrm{E} G_{j}})_{*}\mathbb{Q}_{\tilde{Y}_{2}\times_{G}\mathrm{E} G_{j}})\ar[r]^{\rho_{U,j}\quad}&\mathrm{Hom}(\mathbb{Q}_{Y_{2}\times_{G}\mathrm{E} G_{j}}|_{U},R(\pi_{Y_{2}}\times 1_{\mathrm{E} G_{j}})_{*}\mathbb{Q}_{\tilde{Y}_{2}\times_{G}\mathrm{E} G_{j}}|_{U})}$$
and
$$\xymatrix{\mathrm{Hom}(\mathbb{Q}_{Y_{1}\times_{G}\mathrm{E} G_{j}},\mathbf{IC}^{\bullet}(Y_{1}\times_{G}\mathrm{E} G_{j}))\ar[r]^{\rho_{V,j}\quad}&\mathrm{Hom}(\mathbb{Q}_{Y_{1}\times_{G}\mathrm{E} G_{j}}|_{V},\mathbf{IC}^{\bullet}(Y_{1}\times_{G}\mathrm{E} G_{j})|_{V})}$$
are injective for all $j$ by \cite{Web99}, $\rho_{U}$ and $\rho_{V}$ are also injective.
Hence $\lambda_{f}$ is defined as the morphism on hypercohomologies induced from the composition $Rf_{*}(p)\circ Rf_{*}Rp_{1*}(\omega_{\tilde{Y}_{1}})\circ R\pi_{Y_{2}*}(p_{2}^{*})\circ i:\mathbf{IC}_{G}^{\bullet}(Y_{2})\to Rf_{*}\mathbf{IC}_{G}^{\bullet}(Y_{1})$.
\end{proof}
\begin{corollary}\label{commutativity for res sing}
Let $Y$ be an irreducible normal quasi-projective complex variety on which $G$ acts algebraically and let $f:Y'\to Y$ be a $G$-equivariant blow up of $Y$. Then we have the following commutative diagram
$$\xymatrix{H_{G}^{i}(Y)\ar@^{(->}[r]^{i_{G}}\ar[d]_{f^{*}}&IH_{G}^{i}(Y)\ar[d]^{\lambda_{f}}\\
H_{G}^{i}(Y')\ar@^{=}[r]^{i_{G}}&IH_{G}^{i}(Y')}$$
and $\lambda_{f}:IH_{G}^{i}(Y)\to IH_{G}^{i}(Y')$ is an inclusion induced from Proposition \ref{3 consequences}-(3).
\end{corollary}
By using Corollary \ref{commutativity for res sing}, we can use a standard argument to get the following formula.
\begin{lemma}\label{prototype of equivariant cohomology blowing-up formula}
(1) $P_{t}^{\mathrm{SL}(2)}(\mathbf{R}_{1})=P_{t}^{\mathrm{SL}(2)}(\mathbf{R})+2^{2g}(P_{t}^{\mathrm{SL}(2)}(\mathbb{P}\Upsilon^{-1}(0))-P_{t}(\mathrm{BSL}(2)))$.
(2) $P_{t}^{\mathrm{SL}(2)}(\mathbf{R}_{2})=P_{t}^{\mathrm{SL}(2)}(\mathbf{R}_{1})+P_{t}^{\mathrm{SL}(2)}(E_2)-P_{t}^{\mathrm{SL}(2)}(\Sigma)$.
(3)
$P_{t}^{\mathrm{SL}(2)}(Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss})=P_{t}^{\mathrm{SL}(2)}(\mathbb{P}\Upsilon^{-1}(0)^{ss})+P_{t}^{\mathrm{SL}(2)}(E)-P_{t}^{\mathrm{SL}(2)}(\mathbb{P}\mathrm{Hom}_{1}(sl(2),\mathbb{H}^{g})^{ss})$.
\end{lemma}
\begin{proof}
Since the action of the center $\{\pm\mathrm{id} \}\subset\mathrm{SL}(2)$ on $\mathbf{R}$ is trivial by \cite[Remark below Lemma 9.4]{Simp94I}, the action of $\mathrm{SL}(2)$ and that of $\mathrm{PGL}(2)$ on $\mathbf{R}$ coincide. Further, since the lifted actions of $\{\pm\mathrm{id} \}\subset\mathrm{SL}(2)$ on $\mathbf{R}_{1}$ and $\mathbf{R}_{2}$ are trivial by \cite[Proof of Corollary 3.7]{K85-2}, the actions of $\mathrm{SL}(2)$ and those of $\mathrm{PGL}(2)$ on $\mathbf{R}_{1}$ and $\mathbf{R}_{2}$ coincide. Thus $\mathrm{SL}(2)$-equivariant cohomologies of $\mathbf{R}$, $\mathbf{R}_{1}$, $\mathbf{R}_{2}$ and their $\mathrm{SL}(2)$-invariant subvarieties are equal to $\mathrm{PGL}(2)$-equivariant cohomologies of $\mathbf{R}$, $\mathbf{R}_{1}$, $\mathbf{R}_{2}$ and their $\mathrm{PGL}(2)$-invariant subvarieties respectively. Let $G=\mathrm{PGL}(2)$.
\begin{enumerate}
\item Let $U_{x}$ be a sufficiently small open neighborhood of $x\in\mathbb{Z}_{2}^{2g}$ in $\mathbf{R}$, let $U_{1}=\sqcup_{x\in\mathbb{Z}_{2}^{2g}}U_{x}$ and let $\tilde{U}_{1}=\pi_{\mathbf{R}_{1}}^{-1}(U_{1})$. Let $V_{1}=\mathbf{R}\setminus\mathbb{Z}^{2g}$. We can identify $V_{1}$ with $\mathbf{R}_{1}\setminus E_{1}$ under $\pi_{\mathbf{R}_{1}}$. Then we have the following commutative diagram
$$\xymatrix{\cdots\ar[r]&H_{G}^{i-1}(U_{1}\cap V_{1})\ar[r]^{\alpha}\ar@^{=}[d]&H_{G}^{i}(\mathbf{R})\ar[r]\ar[d]&H_{G}^{i}(U_{1})\ar@<-4ex>[d]\oplus H_{G}^{i}(V_{1})\ar[r]^{\beta}\ar@<4ex>@^{=}[d]&H_{G}^{i}(U_{1}\cup V_{1})\ar[r]\ar@{=}[d]&\cdots\\
\cdots\ar[r]&H_{G}^{i-1}(U_{1}\cap V_{1})\ar[r]^{\tilde{\alpha}}&H_{G}^{i}(\mathbf{R}_{1})\ar[r]&H_{G}^{i}(\tilde{U}_{1})\oplus H_{G}^{i}(V_{1})\ar[r]^{\tilde{\beta}}&H_{G}^{i}(U_{1}\cup V_{1})\ar[r]&\cdots,}$$
where the horizontal sequences are Mayer-Vietoris sequences and the vertical maps are $\pi_{\mathbf{R}_{1}}^{*}$. It follows from Corollary \ref{commutativity for res sing} that the vertical maps are injective. So $\ker\alpha=\ker\tilde{\alpha}$ and then $\mathrm{im} \beta=\mathrm{im} \tilde{\beta}$. Thus we have
$$P_{t}^{G}(\mathbf{R}_{1})=P_{t}^{G}(\mathbf{R})+P_{t}^{G}(\tilde{U}_{1})-P_{t}^{G}(U_{1}).$$
By \cite[Theorem 3.1]{GM88} and Proposition \ref{normal slice is the quadratic cone}, $U_{1}$ is analytically isomorphic to $2^{2g}$ copies of $\Upsilon^{-1}(0)$ and then $\tilde{U}_{1}$ is analytically isomorphic to $2^{2g}$ copies of $Bl_{0}\Upsilon^{-1}(0)$. Since $\mathbb{P}\Upsilon^{-1}(0)$ is a deformation retract of $Bl_{0}\Upsilon^{-1}(0)$ and $P_{t}(\mathrm{B}G)=P_{t}(\mathrm{BSO}(3))=P_{t}(\mathrm{BSL}(2))$, we get
$$P_{t}^{G}(\mathbf{R}_{1})=P_{t}^{G}(\mathbf{R})+2^{2g}(P_{t}^{G}(\mathbb{P}\Upsilon^{-1}(0))-P_{t}(\mathrm{BSL}(2))).$$
\item Let $U_{2}$ be a sufficiently small open neighborhood of $\Sigma$ and let $\tilde{U}_{2}=\pi_{\mathbf{R}_{2}}^{-1}(U_{2})$. Let $V_{2}=\mathbf{R}_{1}\setminus\Sigma$. We can identify $V_{2}$ with $\mathbf{R}_{2}\setminus E_{2}$ under $\pi_{\mathbf{R}_{2}}$. By Corollary \ref{commutativity for res sing} and the same way as in the proof of item (1), we have
$$P_{t}^{G}(\mathbf{R}_{2})=P_{t}^{G}(\mathbf{R}_{1})+P_{t}^{G}(\tilde{U}_{2})-P_{t}^{G}(U_{2}).$$
By \cite[Theorem 3.1]{GM88} and Proposition \ref{normal slice is the quadratic cone}, $U_{2}$ is analytically isomorphic to $C_{\Sigma}\mathbf{R}_{1}$ and then $\tilde{U}_{2}$ is analytically isomorphic to $Bl_{\Sigma}(C_{\Sigma}\mathbf{R}_{1})$. Since $\Sigma$ is a deformation retract of $C_{\Sigma}\mathbf{R}_{1}$ and $E_{2}$ is a deformation retract of $Bl_{\Sigma}(C_{\Sigma}\mathbf{R}_{1})$, we get
$$P_{t}^{G}(\mathbf{R}_{2})=P_{t}^{G}(\mathbf{R}_{1})+P_{t}^{G}(E_{2})-P_{t}^{G}(\Sigma).$$
\item Recall that $\pi:Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss}\to\mathbb{P}\Upsilon^{-1}(0)^{ss}$ is the blowing-up of $\mathbb{P}\Upsilon^{-1}(0)^{ss}$ along $\mathbb{P}\mathrm{Hom}_{1}(sl(2),\mathbb{H}^{g})^{ss}$. By Corollary \ref{commutativity for res sing} and the same way as in the proof of item (1), we have
$$P_{t}^{G}(Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss})=P_{t}^{G}(\mathbb{P}\Upsilon^{-1}(0)^{ss})+P_{t}^{G}(\tilde{U})-P_{t}^{G}(U)$$
for some sufficiently open neighborhood $U$ of $\mathbb{P}\mathrm{Hom}_{1}(sl(2),\mathbb{H}^{g})^{ss}$ and $\tilde{U}=\pi^{-1}(U)$.
By \cite[Theorem 3.1]{GM88} and Proposition \ref{normal slice is the quadratic cone}, $U$ is analytically isomorphic to
$$C_{\mathbb{P}\mathrm{Hom}_{1}(sl(2),\mathbb{H}^{g})^{ss}}\mathbb{P}\Upsilon^{-1}(0)^{ss}$$
and then $\tilde{U}$ is analytically isomorphic to
$$Bl_{\mathbb{P}\mathrm{Hom}_{1}(sl(2),\mathbb{H}^{g})^{ss}}(C_{\mathbb{P}\mathrm{Hom}_{1}(sl(2),\mathbb{H}^{g})^{ss}}\mathbb{P}\Upsilon^{-1}(0)^{ss}).$$
Since $\mathbb{P}\mathrm{Hom}_{1}(sl(2),\mathbb{H}^{g})^{ss}$ is a deformation retract of $C_{\mathbb{P}\mathrm{Hom}_{1}(sl(2),\mathbb{H}^{g})^{ss}}\mathbb{P}\Upsilon^{-1}(0)^{ss}$ and $E$ is a deformation retract of $Bl_{\mathbb{P}\mathrm{Hom}_{1}(sl(2),\mathbb{H}^{g})^{ss}}(C_{\mathbb{P}\mathrm{Hom}_{1}(sl(2),\mathbb{H}^{g})^{ss}}\mathbb{P}\Upsilon^{-1}(0)^{ss})$, we get
$$P_{t}^{G}(Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss})=P_{t}^{G}(\mathbb{P}\Upsilon^{-1}(0)^{ss})+P_{t}^{G}(E)-P_{t}^{G}(\mathbb{P}\mathrm{Hom}_{1}(sl(2),\mathbb{H}^{g})^{ss}).$$
\end{enumerate}
\end{proof}
The blowing-up formula for the equivariant cohomology on the blowing-up
$$\pi:Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss}\to\mathbb{P}\Upsilon^{-1}(0)^{ss}$$
follows from the same argument as in \cite{K85-2}
\begin{proposition}\label{equivariant cohomology local blowing-up formula}
$P_{t}^{\mathrm{SL}(2)}((Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss})^{ss})$
$$=P_{t}^{\mathrm{SL}(2)}(\mathbb{P}\Upsilon^{-1}(0)^{ss})+P_{t}^{\mathrm{SL}(2)}(E^{ss})-P_{t}^{\mathrm{SL}(2)}(\mathbb{P}\mathrm{Hom}_{1}(sl(2),\mathbb{H}^{g})^{ss}).$$
\end{proposition}
\begin{proof}
By Proposition \ref{smoothness of 2nd local blowing-up}, $Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss}$ is a smooth projective variety. Thus the same argument as in \cite{K85-2} can be applied. We sketch the proof briefly.
There is a Morse stratification $\{S_{\beta}|\beta\in\mathbf{B}\}$ of $Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss}$ associated to the lifted action of $\mathrm{SL}(2)$. Then $\{S_{\beta}\cap E|\beta\in\mathbf{B}\}$ is the Morse stratification of $E$. By \cite[Section 5 and 8]{K85-1}, $\{S_{\beta}|\beta\in\mathbf{B}\}$ and $\{S_{\beta}\cap E|\beta\in\mathbf{B}\}$ are equivariantly perfect. Thus we have
$$P_{t}^{\mathrm{SL}(2)}(Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss})=P_{t}^{\mathrm{SL}(2)}((Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss})^{ss})+\sum_{\beta\ne0}t^{2d'(\beta)}P_{t}^{\mathrm{SL}(2)}(S_{\beta})$$
and
$$P_{t}^{\mathrm{SL}(2)}(E)=P_{t}^{\mathrm{SL}(2)}(E^{ss})+\sum_{\beta\ne0}t^{2d(\beta)}P_{t}^{\mathrm{SL}(2)}(S_{\beta}\cap E),$$
where $d'(\beta)$ (respectively, $d(\beta)$) is the codimension of $S_{\beta}$ (respectively, $S_{\beta}\cap E$) in $Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss}$ (respectively, $E$). We also have
$$P_{t}^{\mathrm{SL}(2)}(S_{\beta})=P_{t}^{\mathrm{SL}(2)}(\mathrm{SL}(2)Z_{\beta}^{ss})$$
and
$$P_{t}^{\mathrm{SL}(2)}(S_{\beta}\cap E)=P_{t}^{\mathrm{SL}(2)}(\mathrm{SL}(2)Z_{\beta}^{ss}\cap E),$$
where $Z_{\beta}^{ss}$ denotes the set of points of $S_{\beta}$ fixed by the one-parameter subgroup generated by $\beta$. Since $Z_{\beta}^{ss}\subseteq E$ by \cite[Lemma 7.6]{K85-2} and $S_{\beta}\not\subseteq E$ for any $\beta\in\mathbf{B}$ by \cite[Lemma 7.11]{K85-2}, we have
$$P_{t}^{\mathrm{SL}(2)}((Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss})^{ss})=P_{t}^{\mathrm{SL}(2)}(\mathbb{P}\Upsilon^{-1}(0)^{ss})+P_{t}^{\mathrm{SL}(2)}(E^{ss})-P_{t}^{\mathrm{SL}(2)}(\mathbb{P}\mathrm{Hom}_{1}(sl(2),\mathbb{H}^{g})^{ss})$$
$$+\sum_{\beta\ne0}(t^{2d'(\beta)}-t^{2d(\beta)})P_{t}^{\mathrm{SL}(2)}(\mathrm{SL}(2)Z_{\beta}^{ss})$$
$$=P_{t}^{\mathrm{SL}(2)}(\mathbb{P}\Upsilon^{-1}(0)^{ss})+P_{t}^{\mathrm{SL}(2)}(E^{ss})-P_{t}^{\mathrm{SL}(2)}(\mathbb{P}\mathrm{Hom}_{1}(sl(2),\mathbb{H}^{g})^{ss}).$$
\end{proof}
The following blowing-up formulas for the blowing-ups $\pi_{\mathbf{R}_{1}}:\mathbf{R}_{1}\to\mathbf{R}$ and $\pi_{\mathbf{R}_{2}}:\mathbf{R}_{2}\to\mathbf{R}_{1}^{ss}$ are what we desire.
\begin{conjecture}\label{equivariant blowing up formula conjecture}
(1) $P_{t}^{\mathrm{SL}(2)}(\mathbf{R}_{1}^{ss})=P_{t}^{\mathrm{SL}(2)}(\mathbf{R})+2^{2g}(P_{t}^{\mathrm{SL}(2)}(\mathbb{P}\Upsilon^{-1}(0)^{ss})-P_{t}(\mathrm{BSL}(2)))$.
(2) $P_{t}^{\mathrm{SL}(2)}(\mathbf{R}_{2}^{s})=P_{t}^{\mathrm{SL}(2)}(\mathbf{R}_{1}^{ss})+P_{t}^{\mathrm{SL}(2)}(E_2^{ss})-P_{t}^{\mathrm{SL}(2)}(\Sigma)$.
\end{conjecture}
Since $\mathbf{R}_{1}$ and $\mathbf{R}_{2}$ are neither smooth nor projective, we cannot directly apply the Morse theory of \cite{K85-1} developed by F. Kirwan for a proof of Conjecture \ref{equivariant blowing up formula conjecture}.
\subsection{Intersection Poincar\'{e} polynomial of the deepest singularity of $\mathbf{M}$}\label{intersection poincare polynomial of the deepest singularity}
In order to use Theorem \ref{computable intersection blowing-up formula}-(1), we must calculate $IP_{t}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{PGL}(2))$ and $IP_{t}(\Upsilon^{-1}(0)/\!/\mathrm{PGL}(2))$.
Recall that it follows from Proposition \ref{ss-local-first-blowup} that
$$\mathbb{P}\mathrm{Hom}_{1}(sl(2),\mathbb{H}^{g})^{ss}=\mathrm{PGL}(2)Z^{ss}\cong\mathrm{PGL}(2)\times_{\mathrm{O}(2)}Z^{ss}$$
and
$$\mathbb{P}\mathrm{Hom}_{1}(sl(2),\mathbb{H}^{g})/\!/\mathrm{PGL}(2)\cong Z/\!/\mathrm{O}(2)=Z/\!/\mathrm{SO}(2)=Z_{1}=\mathbb{P}^{2g-1},$$
where $Z=Z_{1}\cup Z_{2}\cup Z_{3}$, $Z^{ss}$ is the set of
semistable points of $Z$ for the action of $\mathrm{O}(2)$,
$Z_{1}=\mathbb{P}\{v_{1}\otimes\mathbb{H}^{g}\}=Z^{ss}$,
$Z_{2}=\mathbb{P}\{v_{2}\otimes\mathbb{H}^{g}\}$ and
$Z_{3}=\mathbb{P}\{v_{3}\otimes\mathbb{H}^{g}\}$. Here $\{v_{1},v_{2},v_{3}\}$ is the basis of $sl(2)$ chosen in Section \ref{local picture}.
We see that
$$P_{t}^{+}(Z/\!/\mathrm{SO}(2))=P_{t}(\mathbb{P}^{2g-1})=1+t^{2}+\cdots+t^{2(2g-1)}=\frac{1-t^{4g}}{1-t^{2}}$$
and
$$P_{t}^{-}(Z/\!/\mathrm{SO}(2))=0,$$
where $P_{t}^{+}(Z/\!/\mathrm{SO}(2))$ and $P_{t}^{-}(Z/\!/\mathrm{SO}(2))$ are Poincar\'{e} polynomials of the invariant part and variant part of $H^{*}(Z/\!/\mathrm{SO}(2))$ with respect to the action of $\mathbb{Z}_{2}=\mathrm{O}(2)/\mathrm{SO}(2)$ on $Z/\!/\mathrm{SO}(2)$.
By \cite[Proposition 3.10]{Ma21} and Theorem \ref{computable intersection blowing-up formula}-(3),
$$IP_{t}(Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss}/\!/\mathrm{PGL}(2))=IP_{t}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{PGL}(2))$$
$$+\sum_{p+q=i}\dim[H^{p}(\mathbb{P}^{2g-1})\otimes H^{t(q)}(I_{2g-3})]^{\mathbb{Z}_{2}}t^{i}$$
$$=IP_{t}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{PGL}(2))+\sum_{p+q=i}\dim[H^{p}(\mathbb{P}^{2g-1})^{\mathbb{Z}_{2}}\otimes H^{t(q)}(I_{2g-3})^{\mathbb{Z}_{2}}]t^{i}$$
\begin{equation}\label{2nd local blowing-up}=IP_{t}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{PGL}(2))+\frac{1-t^{4g}}{1-t^{2}}\cdot\frac{t^2(1-t^{4g-6})(1-t^{4g-4})}{(1-t^2)(1-t^4)}.\end{equation}
Let
$\widetilde{\mathbb{P}\mathrm{Hom}_{2}^{\omega}(sl(2),\mathbb{H}^{g})}=Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\mathrm{Hom}_{2}^{\omega}(sl(2),\mathbb{H}^{g})$
be the blowing-up of
$\mathbb{P}\mathrm{Hom}_{2}^{\omega}(sl(2),\mathbb{H}^{g})$ along
$\mathbb{P}\mathrm{Hom}_{1}(sl(2),\mathbb{H}^{g})^{ss}$ and let
$$Bl_{\widetilde{\mathbb{P}\mathrm{Hom}_{2}^{\omega}}}Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss}$$
be the blowing-up of
$Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss}$ along
$\widetilde{\mathbb{P}\mathrm{Hom}_{2}^{\omega}(sl(2),\mathbb{H}^{g})}^{ss}$.
Assume that $g\ge3$. Denote
$D_{1}=Bl_{\widetilde{\mathbb{P}\mathrm{Hom}_{2}^{\omega}}}Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss}/\!/\mathrm{PGL}(2)$.
By \cite[Proposition 4.2]{KY08}, $D_{1}$ is a
$\widehat{\mathbb{P}}^5$-bundle over $Gr^{\omega}(3,2g)$ where
$\widehat{\mathbb{P}}^5$ is the blowing-up of $\mathbb{P}^5$
(projectivization of the space of $3\times 3$ symmetric matrices)
along $\mathbb{P}^2$ (the locus of rank $1$ matrices). Since $D_1$
is a nonsingular projective variety over $\mathbb{C}$,
$$\dim_{\mathbb{C}}H^{k}(D_{1};\mathbb{C})=\sum_{p+q=k}h^{p,q}(H^{k}(D_{1};\mathbb{C})).$$
Thus it follows from \cite[Proposition 5.1]{KY08} that
$$IP_{t}(D_{1})=P_{t}(D_{1})=E(D_{1};-t,-t)=(\frac{1-t^{12}}{1-t^{2}}-\frac{1-t^{6}}{1-t^{2}}+(\frac{1-t^{6}}{1-t^{2}})^{2})\cdot\prod_{1\leq i\leq 3}\frac{1-t^{4g-12+4i}}{1-t^{2i}}.$$
Moreover by the proof of \cite[Proposition 3.5.1]{O97}
$$\widetilde{\mathbb{P}\mathrm{Hom}_{2}^{\omega}(sl(2),\mathbb{H}^{g})}/\!/\mathrm{PGL}(2)\cong
\mathbb{P}(S^{2}\mathcal{A})$$ where $\mathcal{A}$ is the
tautological rank $2$ bundle over $Gr^{\omega}(2,2g)$. Following the proof of \cite[Lemma 3.5.4]{O97}, we can see that the exceptional divisor of $D_1$ is a $\mathbb{P}^{2g-5}$-bundle over $\mathbb{P}(S^{2}\mathcal{A})$.
By the usual blowing-up formula mentioned in \cite[p.605]{GH78}, we have
$$\dim H^{i}(D_1)=\dim H^{i}(Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss}/\!/\mathrm{PGL}(2))$$
\begin{equation}\label{3rd local blowing-up}+\big(\sum_{p+q=i}\dim[H^{p}(\mathbb{P}(S^{2}\mathcal{A}))\otimes H^{q}(\mathbb{P}^{2g-5})]-\dim H^{i}(\mathbb{P}(S^{2}\mathcal{A}))\big).\end{equation}
Since
$\mathbb{P}(S^{2}\mathcal{A})$ is the $\mathbb{P}^2$-bundle over
$Gr^{\omega}(2,2g)$,
$$P_{t}(\mathbb{P}(S^{2}\mathcal{A}))=P_{t}(\mathbb{P}^2)P_{t}(Gr^{\omega}(2,2g))=\frac{1-t^{6}}{1-t^{2}}\cdot\prod_{1\leq i\leq 2}\frac{1-t^{4g-8+4i}}{1-t^{2i}}$$
by Deligne's criterion (see \cite{D68}).
Therefore it follows from (\ref{3rd local blowing-up}) that
$$IP_{t}(Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss}/\!/\mathrm{PGL}(2))$$
$$=(\frac{1-t^{12}}{1-t^{2}}-\frac{1-t^{6}}{1-t^{2}}+(\frac{1-t^{6}}{1-t^{2}})^{2})\cdot\prod_{1\leq
i\leq
3}\frac{1-t^{4g-12+4i}}{1-t^{2i}}-\frac{1-t^{6}}{1-t^{2}}\cdot\prod_{1\leq
i\leq 2}\frac{1-t^{4g-8+4i}}{1-t^{2i}}$$
$$\times\frac{t^{2}(1-t^{2(2g-5)})}{1-t^{2}}.$$
Assume that $g=2$. In this case, we know from \cite[Proposition 2.0.1]{O97} that
$$Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss}/\!/\mathrm{PGL}(2)$$
is already nonsingular and that it is a $\mathbb{P}^2$-bundle over
$Gr^{\omega}(2,4)$. Then by Deligne's criterion (See \cite{D68}),
$$IP_{t}(Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss}/\!/\mathrm{PGL}(2))=P_{t}(Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss}/\!/\mathrm{PGL}(2))$$
$$=P_{t}(\mathbb{P}^2)P_{t}(Gr^{\omega}(2,4))=\frac{1-t^{6}}{1-t^{2}}\cdot\prod_{1\leq i\leq 2}\frac{1-t^{4i}}{1-t^{2i}}=\frac{(1-t^{6})(1-t^{8})}{(1-t^{2})^2}.$$
Combining these with (\ref{2nd local blowing-up}), we obtain
\begin{proposition}
$$IP_{t}(\mathbb{P}\Upsilon^{-1}(0)/\!/\mathrm{PGL}(2))=\frac{(1-t^{8g-8})(1-t^{4g})}{(1-t^{2})(1-t^{4})}.$$
\end{proposition}
By Lemma \ref{suffices-to-show-on-projectivized-git}-(1), we also obtain
\begin{proposition}\label{intersection-Betti-of-normal-cone}
$$IP_{t}(\Upsilon^{-1}(0)/\!/\mathrm{PGL}(2))=\frac{1-t^{4g}}{1-t^{4}}$$
\end{proposition}
\subsection{Intersection Poincar\'{e} polynomial of $\mathbf{M}$}
In this subsection, we compute a conjectural formula for $IP_{t}(\mathbf{M})$.
\subsubsection{Computation for $P_{t}^{\mathrm{SL}(2)}(\mathbf{R})$}\label{eqiv coh of R}
We start with the following result.
\begin{theorem}[Corollary 1.2 in \cite{DWW11}]\label{equiv coh of moduli space of Higgs bundles}
$$P_{t}^{\mathcal{G}_{\mathbb{C}}}(\mathcal{B}^{ss})=\frac{(1+t^3)^{2g}-(1+t)^{2g}t^{2g+2}}{(1-t^2)(1-t^4)}$$
$$-t^{4g-4}+\frac{t^{2g+2}(1+t)^{2g}}{(1-t^2)(1-t^4)}+\frac{(1-t)^{2g}t^{4g-4}}{4(1+t^2)}$$
$$\frac{(1+t)^{2g}t^{4g-4}}{2(1-t^2)}(\frac{2g}{t+1}+\frac{1}{t^2-1}-\frac{1}{2}+(3-2g))$$
$$\frac{1}{2}(2^{2g}-1)t^{4g-4}((1+t)^{2g-2}+(1-t)^{2g-2}-2).$$
\end{theorem}
In this subsection, we show that $P_{t}^{\mathrm{SL}(2)}(\mathbf{R})=P_{t}^{\mathcal{G}_{\mathbb{C}}}(\mathcal{B}^{ss})$. To prove this, we need some technical lemmas.
Choose a base point $x\in X$. Let $E$ be a complex Hermitian vector bundle of rank $2$ and degree $0$ on $X$. Let $p:E\to X$ be the canonical
projection. Let $(\mathcal{G}_{\mathbb{C}})_0$ be the normal subgroup of
$\mathcal{G}_{\mathbb{C}}$ which fixes the fiber $E|_{x}$.
We first claim that $(\mathcal{G}_{\mathbb{C}})_0$ acts freely on $\mathcal{B}^{ss}$. In
fact, assume that $g\cdot(A,\phi):=(g^{-1}Ag,\phi)=(A,\phi)$ for $g\in(\mathcal{G}_{\mathbb{C}})_0$. For an arbitrary point $y\in X$ and for any smooth path
$\gamma:[0,1]\to X$ starting at $\gamma(0)=x$ and ending at
$\gamma(1)=y\in X$, there is a parallel transport mapping
$P_{\gamma}:E|_{x}\to E|_{y}$ defined as follows. If $v\in E|_{x}$,
there exists a unique path $\gamma_v:[0,1]\to E$ such that
$p\circ\gamma_v=\gamma$, $\gamma_v(0)=v$ given by $A$. Define
$P_{\gamma}(v)=\gamma_v(1)$. By the assumption, $P_{\gamma}\circ
g|_{x}=g|_{y}\circ P_{\gamma}$. Since $g|_{x}$ is the identity on
$E|_{x}$, $g|_{y}$ is also the identity on $E|_{y}$. Therefore $g$
is the identity on $E$.
Since the surjective map $\mathcal{G}_{\mathbb{C}}\to\mathrm{SL}(2)$ given by $g\mapsto g|_{x}$ has the kernel $(\mathcal{G}_{\mathbb{C}})_{0}$, we have
\begin{equation}\label{G mod G0 isom SL2}
\mathcal{G}_{\mathbb{C}}/(\mathcal{G}_{\mathbb{C}})_{0}\cong\mathrm{SL}(2).
\end{equation}
Let $\mathcal{G}_{\mathbb{C}}/(\mathcal{G}_{\mathbb{C}})_0\times_{\mathcal{G}_{\mathbb{C}}}\mathcal{B}^{ss}$ be the quotient space of
$\mathcal{G}_{\mathbb{C}}/(\mathcal{G}_{\mathbb{C}})_0\times\mathcal{B}^{ss}$ by the action of
$\mathcal{G}_{\mathbb{C}}$ given by
$$h\cdot(\overline{g},(A,\phi))=(\overline{g}\overline{h}^{-1},h\cdot(A,\phi))$$
where $\overline{f}$ is the image of $f\in\mathcal{G}_{\mathbb{C}}$ under the
quotient map $\mathcal{G}_{\mathbb{C}}\to\mathcal{G}_{\mathbb{C}}/(\mathcal{G}_{\mathbb{C}})_0$. Since
$(\mathcal{G}_{\mathbb{C}})_0$ acts freely on $\mathcal{B}^{ss}$, $\mathcal{G}_{\mathbb{C}}$ acts freely
on $\mathcal{G}_{\mathbb{C}}/(\mathcal{G}_{\mathbb{C}})_0\times\mathcal{B}^{ss}$.
\begin{lemma}\label{framing}
There exists a homeomorphism between
$\mathrm{SL}(2)\times_{\mathcal{G}_{\mathbb{C}}}\mathcal{B}^{ss}$ and $\mathcal{B}^{ss}/(\mathcal{G}_{\mathbb{C}})_0$.
\end{lemma}
\begin{proof}
By (\ref{G mod G0 isom SL2}), it suffices to show that there exists a homeomorphism between
$\mathcal{G}_{\mathbb{C}}/(\mathcal{G}_{\mathbb{C}})_0\times_{\mathcal{G}_{\mathbb{C}}}\mathcal{B}^{ss}$ and $\mathcal{B}^{ss}/(\mathcal{G}_{\mathbb{C}})_0$.
Consider the continuous surjective map
$$q:\mathcal{G}_{\mathbb{C}}\times\mathcal{B}^{ss}\to\mathcal{G}_{\mathbb{C}}/(\mathcal{G}_{\mathbb{C}})_0\times\mathcal{B}^{ss}$$
given by $(g,(A,\phi))\mapsto(\overline{g},(A,\phi))$. If $\mathcal{G}_{\mathbb{C}}$ acts on $\mathcal{G}_{\mathbb{C}}\times\mathcal{B}^{ss}$ by
$h\cdot(g,(A,\phi))=(gh^{-1},h\cdot(A,\phi))$, $q$ is
$\mathcal{G}_{\mathbb{C}}$-equivariant.
Taking quotients of both spaces
by $\mathcal{G}_{\mathbb{C}}$, $q$ induces the continuous surjective map
$$\overline{q}:\mathcal{G}_{\mathbb{C}}\times_{\mathcal{G}_{\mathbb{C}}}\mathcal{B}^{ss}\to\mathcal{G}_{\mathbb{C}}/(\mathcal{G}_{\mathbb{C}})_0\times_{\mathcal{G}_{\mathbb{C}}}\mathcal{B}^{ss}$$
given by $[g,(A,\phi)]\mapsto[\overline{g},(A,\phi)]$.
If $(\mathcal{G}_{\mathbb{C}})_0$ acts on $\mathcal{G}_{\mathbb{C}}\times_{\mathcal{G}_{\mathbb{C}}}\mathcal{B}^{ss}$ by
$h\cdot[g,(A,\phi)]=[g,h\cdot(A,\phi)]$, $\overline{q}$ is
$(\mathcal{G}_{\mathbb{C}})_0$-invariant. Precisely for $g_0\in(\mathcal{G}_{\mathbb{C}})_0$,
$\overline{q}([g,g_0\cdot(A,\phi)])=[\overline{g},g_0\cdot(A,\phi)]=[\overline{g}\overline{g_0},(A,\phi)]=[\overline{g},(A,\phi)]=\overline{q}([g,(A,\phi)])$.
Thus $\overline{q}$ induces the continuous surjective map
$$\tilde{q}:\frac{\mathcal{G}_{\mathbb{C}}\times_{\mathcal{G}_{\mathbb{C}}}\mathcal{B}^{ss}}{(\mathcal{G}_{\mathbb{C}})_0}\to\mathcal{G}_{\mathbb{C}}/(\mathcal{G}_{\mathbb{C}})_0\times_{\mathcal{G}_{\mathbb{C}}}\mathcal{B}^{ss}$$
given by $\overline{[g,(A,\phi)]}\mapsto[\overline{g},(A,\phi)]$.
Furthermore $\tilde{q}$ is injective. In fact, assume that
$$\tilde{q}(\overline{[g_1,(A_1,\phi_1)]})=\tilde{q}(\overline{[g_2,(A_2,\phi_2)]}),$$
that is,
$$[\overline{g_1},(A_1,\phi_1)]=[\overline{g_2},(A_2,\phi_2)].$$
Then there is $k\in\mathcal{G}_{\mathbb{C}}$ such that
$(\overline{g_1},(A_1,\phi_1))=(\overline{g_2}\overline{k}^{-1},k\cdot(A_2,\phi_2))$.
Then $g_1=g_2 k^{-1}l$ for some $l\in(\mathcal{G}_{\mathbb{C}})_0$. Thus
$\overline{[g_1,(A_1,\phi_1)]}=\overline{[g_2
k^{-1}l,k\cdot(A_2,\phi_2)]}=\overline{[g_2,
k^{-1}lk\cdot(A_2,\phi_2)]}=\overline{[g_2,(A_2,\phi_2)]}$
because $(\mathcal{G}_{\mathbb{C}})_0$ is the normal subgroup of $\mathcal{G}_{\mathbb{C}}$.
On the other hand, since both $q$ and the quotient map $\mathcal{G}_{\mathbb{C}}/(\mathcal{G}_{\mathbb{C}})_0\times\mathcal{B}^{ss}\to\mathcal{G}_{\mathbb{C}}/(\mathcal{G}_{\mathbb{C}})_0\times_{\mathcal{G}_{\mathbb{C}}}\mathcal{B}^{ss}$ are open, $\overline{q}$ is open. Moreover since the quotient map $\displaystyle\mathcal{G}_{\mathbb{C}}\times_{\mathcal{G}_{\mathbb{C}}}\mathcal{B}^{ss}\to\frac{\mathcal{G}_{\mathbb{C}}\times_{\mathcal{G}_{\mathbb{C}}}\mathcal{B}^{ss}}{(\mathcal{G}_{\mathbb{C}})_0}$ is open,
$\tilde{q}$ is also open.
Hence $\tilde{q}$ is a homeomorphism. Since there is a homeomorphism $\xymatrix{\mathcal{G}_{\mathbb{C}}\times_{\mathcal{G}_{\mathbb{C}}}\mathcal{B}^{ss}\ar[r]^{\quad\quad\cong}&\mathcal{B}^{ss}}$ given by $[g,(A,\phi)]\mapsto g\cdot(A,\phi)$, we get the conclusion.
\end{proof}
\begin{lemma}\label{from-conn-to-bdl}
There is an isomorphism of complex analytic spaces
$$\mathrm{SL}(2)\times_{\mathcal{G}_{\mathbb{C}}}\mathcal{B}^{ss}\cong\mathbf{R}.$$
\end{lemma}
\begin{proof}
There is a bijection between
$\mathrm{SL}(2)\times_{\mathcal{G}_{\mathbb{C}}}\mathcal{B}^{ss}$ and
$\mathbf{R}$. In fact, consider a map
$$f:\mathrm{SL}(2)\times\mathcal{B}^{ss}\to\mathbf{R}$$
given by
$(\beta,(A,\phi))\mapsto(((E,\overline{A}^{0,1}),\overline{\phi}),\overline{\beta})$,
where $(\overline{\beta},(\overline{A},\overline{\phi}))$ is the
image of $(\beta,(A,\phi))$ of the quotient map
$\mathrm{SL}(2)\times\mathcal{B}^{ss}\to
\mathrm{SL}(2)\times_{\mathcal{G}_{\mathbb{C}}}\mathcal{B}^{ss}$. Since $f$
is surjective and $\mathcal{G}_{\mathbb{C}}$-invariant, $f$ induces
a bijection between
$\mathrm{SL}(2)\times_{\mathcal{G}_{\mathbb{C}}}\mathcal{B}^{ss}$ and
$\mathbf{R}$.
Further, the family $E\times(\mathrm{SL}(2)\times_{\mathcal{G}_{\mathbb{C}}}\mathcal{B}^{ss})$ over $X\times(\mathrm{SL}(2)\times_{\mathcal{G}_{\mathbb{C}}}\mathcal{B}^{ss})$ gives a complex
analytic map $g:\mathrm{SL}(2)\times_{\mathcal{G}_{\mathbb{C}}}\mathcal{B}^{ss}\to\mathbf{R}$ by \cite[Lemma 5.7]{Simp94I}, and $f((\overline{\beta},(\overline{A},\overline{\phi})))=g((\overline{\beta},(\overline{A},\overline{\phi})))$ for all $(\overline{\beta},(\overline{A},\overline{\phi}))\in\mathrm{SL}(2)\times_{\mathcal{G}_{\mathbb{C}}}\mathcal{B}^{ss}$.
Hence $f$ is an isomorphism of complex analytic spaces $\mathrm{SL}(2)\times_{\mathcal{G}_{\mathbb{C}}}\mathcal{B}^{ss}\cong\mathbf{R}$.
\end{proof}
There is a technical lemma for equivariant cohomologies.
\begin{lemma}\label{quot-in-stage}
Let $H$ be a closed normal subgroup of $G$ and $M$ be a $G$-space on
which $H$ acts freely. Then $G/H$ acts on $M/H$ and
$$H_{G}^*(M)=H_{G/H}^*(M/H).$$
\end{lemma}
\begin{proof}
Use the fibration $\mathrm{E} G\times_{G}M\cong(\mathrm{E} G\times \mathrm{E}(G/H))\times_{G}M\to
\mathrm{E}(G/H)\times_{G}M\cong \mathrm{E}(G/H)\times_{G/H}(M/H)$ whose fibers $\mathrm{E} G$ is
contractible.
\end{proof}
The following equality is an immediate consequence from Lemma
\ref{framing}, Lemma \ref{from-conn-to-bdl} and Lemma \ref{quot-in-stage}.
\begin{proposition}
$$P_{t}^{\mathrm{SL}(2)}(\mathbf{R})=P_{t}^{\mathcal{G}_{\mathbb{C}}}(\mathcal{B}^{ss})$$
\end{proposition}
Thus we get the same formula for $P_{t}^{\mathrm{SL}(2)}(\mathbf{R})$ as Theorem
\ref{equiv coh of moduli space of Higgs bundles}.
\subsubsection{Computation for $P_{t}^{\mathrm{SL}(2)}(\Sigma)$}\label{eqiv coh of Sigma}
In the proof of Lemma \ref{geometric descriptions of second cones}-(1), we observed that
$$\Sigma\cong\mathbb{P}\mathrm{Isom}(\mathcal{O}_{\widetilde{T^{*}J}}^{2},\mathcal{L}|_{x}\oplus\mathcal{L}^{-1}|_{x})/\!/\mathrm{O}(2).$$
Since $\{\pm\mathrm{id} \}\subset\mathrm{SL}(2)$ acts trivially on $\mathbb{P}\mathrm{Isom}(\mathcal{O}_{\widetilde{T^{*}J}}^{2},\mathcal{L}|_{x}\oplus\mathcal{L}^{-1}|_{x})$, $\{\pm\mathrm{id} \}\subset\mathrm{SL}(2)$ also acts trivially on $\Sigma$. Then
$$\mathrm{ESL}(2)\times_{\mathrm{SL}(2)}\Sigma\cong\mathrm{EPGL}(2)\times_{\mathrm{PGL}(2)}\Sigma.$$
Since $\mathrm{O}(2)$ acts on $\mathbb{P}\mathrm{Isom}(\mathcal{O}_{\widetilde{T^{*}J}}^{2},\mathcal{L}|_{x}\oplus\mathcal{L}^{-1}|_{x})$ freely and both actions of $\mathrm{PGL}(2)$ and $\mathrm{O}(2)$ commute,
$$\mathrm{EPGL}(2)\times_{\mathrm{PGL}(2)}\Sigma\sim\mathrm{EPGL}(2)\times_{\mathrm{PGL}(2)}(\mathrm{EO}(2)\times_{\mathrm{O}(2)}\mathbb{P}\mathrm{Isom}(\mathcal{O}_{\widetilde{T^{*}J}}^{2},\mathcal{L}|_{x}\oplus\mathcal{L}^{-1}|_{x}))$$
$$\cong\mathrm{EO}(2)\times_{\mathrm{O}(2)}(\mathrm{EPGL}(2)\times_{\mathrm{PGL}(2)}\mathbb{P}\mathrm{Isom}(\mathcal{O}_{\widetilde{T^{*}J}}^{2},\mathcal{L}|_{x}\oplus\mathcal{L}^{-1}|_{x}))\sim\mathrm{EO}(2)\times_{\mathrm{O}(2)}\widetilde{T^{*}J}$$
$$\cong(\mathrm{ESO}(2)\times_{\mathrm{SO}(2)}\widetilde{T^{*}J})/(O(2)/SO(2))\cong(\mathrm{BSO}(2)\times\widetilde{T^{*}J})/\mathbb{Z}_{2},$$
where $\sim$ denotes the homotopic equivalence. Thus
$$P_{t}^{\mathrm{SL}(2)}(\Sigma)=P_{t}^{+}(\mathrm{BSO}(2))P_{t}^{+}(\widetilde{T^{*}J})+P_{t}^{-}(\mathrm{BSO}(2))P_{t}^{-}(\widetilde{T^{*}J}),$$
where $P_{t}^{+}(W)$ (respectively, $P_{t}^{-}(W)$) denotes the Poincar\'{e} polynomial of the invariant (respectively, variant) part of $H^{*}(W)$ with respect to the action of $\mathbb{Z}_{2}$ on $W$ for a $\mathbb{Z}_{2}$-space $W$.
\begin{lemma}
$$P_{t}^{\mathrm{SL}(2)}(\Sigma)=\frac{1}{(1-t^4)}(\frac{1}{2}((1+t)^{2g}+(1-t)^{2g})+2^{2g}(\frac{1-t^{4g}}{1-t^2}-1))+\frac{t^2}{(1-t^4)}\frac{1}{2}((1+t)^{2g}-(1-t)^{2g}).$$
\end{lemma}
\begin{proof}
Note that $\mathrm{BSO}(2)\cong\mathbb{P}^{\infty}$. Since the action of $\mathbb{Z}_{2}\setminus\{\mathrm{id} \}$ on $H^*(\mathrm{BSO}(2))$ represents reversing of orientation and $\mathbb{P}^{n}$ possess an orientation-reversing self-homeomorphism only when $n$ is odd, we have $\displaystyle P_{t}^{+}(\mathrm{BSO}(2))=\frac{1}{1-t^{4}}$ and $\displaystyle P_{t}^{-}(\mathrm{BSO}(2))=\frac{t^{2}}{1-t^{4}}$.
Further, by the computation mentioned in \cite[Lemma 4.3]{CK06} and \cite[Section 5]{CK07}, we have $$P_{t}^{+}(\widetilde{T^{*}J})=\frac{1}{2}((1+t)^{2g}+(1-t)^{2g})+2^{2g}(\frac{1-t^{4g}}{1-t^2}-1)$$
and
$$P_{t}^{-}(\widetilde{T^{*}J})=\frac{1}{2}((1+t)^{2g}-(1-t)^{2g}).$$
\end{proof}
\subsubsection{Computation for $P_{t}^{\mathrm{SL}(2)}(\mathbb{P}\Upsilon^{-1}(0)^{ss})$}\label{eqiv coh of deepest sing}
Since $E/\!/\mathrm{SL}(2)$ has an orbifold singularity and $E/\!/\mathrm{SL}(2)\cong\mathbb{P}\mathcal{C}/\!/\mathrm{SL}(2)$ is a free $\mathbb{Z}_{2}$-quotient of $I_{2g-3}$-bundle over $\mathbb{P}^{2g-1}$ by Lemma \ref{geometric descriptions of second cones}-(5) and Lemma \ref{isomorphic to incidence variety}, we use \cite[Proposition 3.10]{Ma21} to have
$$P_{t}^{\mathrm{SL}(2)}(E^{ss})=P_{t}(E/\!/\mathrm{SL}(2))=P_{t}^{+}(I_{2g-3})P_{t}(\mathbb{P}^{2g-1})$$
$$=\frac{(1-t^{4g-4})^{2}}{(1-t^2)(1-t^4)}\cdot\frac{1-t^{4g}}{1-t^2}.$$
By Proposition \ref{equivariant cohomology local blowing-up formula},
$$P_{t}^{\mathrm{SL}(2)}((Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss})^{s})$$
$$=P_{t}^{\mathrm{SL}(2)}(\mathbb{P}\Upsilon^{-1}(0)^{ss})+P_{t}^{\mathrm{SL}(2)}(E^{ss})-P_{t}^{\mathrm{SL}(2)}(\mathbb{P}\mathrm{Hom}_{1}(sl(2),\mathbb{H}^{g})^{ss})$$
$$=P_{t}^{\mathrm{SL}(2)}(\mathbb{P}\Upsilon^{-1}(0)^{ss})+\frac{(1-t^{4g-4})^{2}}{(1-t^2)(1-t^4)}\cdot\frac{1-t^{4g}}{1-t^2}-\frac{1}{1-t^4}\frac{1-t^{4g}}{1-t^2}$$
On the other hand
$$P_{t}^{\mathrm{SL}(2)}((Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss})^{s})=P_{t}((Bl_{\mathbb{P}\mathrm{Hom}_{1}}\mathbb{P}\Upsilon^{-1}(0)^{ss})^{s}/\!/\mathrm{SL}(2))$$
$$=\big(\frac{1-t^{12}}{1-t^2}-\frac{1-t^6}{1-t^2}+(\frac{1-t^6}{1-t^2})^{2}\big)\frac{(1-t^{4g-8})(1-t^{4g-4})(1-t^{4g})}{(1-t^2)(1-t^4)(1-t^6)}$$
$$-\frac{1-t^6}{1-t^2}\frac{(1-t^{4g-4})(1-t^{4g})}{(1-t^2)(1-t^4)}\frac{t^2(1-t^{2(2g-5)})}{1-t^2}$$
from the subsection \ref{intersection poincare polynomial of the deepest singularity} for any $g\ge2$.
Hence
$$P_{t}^{\mathrm{SL}(2)}(\mathbb{P}\Upsilon^{-1}(0)^{ss})$$
$$=\big(\frac{1-t^{12}}{1-t^2}-\frac{1-t^6}{1-t^2}+(\frac{1-t^6}{1-t^2})^{2}\big)\frac{(1-t^{4g-8})(1-t^{4g-4})(1-t^{4g})}{(1-t^2)(1-t^4)(1-t^6)}$$
$$-\frac{1-t^6}{1-t^2}\frac{(1-t^{4g-4})(1-t^{4g})}{(1-t^2)(1-t^4)}\frac{t^2(1-t^{2(2g-5)})}{1-t^2}$$
$$-\frac{(1-t^{4g-4})^{2}}{(1-t^2)(1-t^4)}\cdot\frac{1-t^{4g}}{1-t^2}+\frac{1}{1-t^4}\frac{1-t^{4g}}{1-t^2}.$$
\subsubsection{Computation for $P_{t}^{\mathrm{SL}(2)}(E_{2}^{ss})$}\label{eqiv coh of E2}
Since $E_2/\!/\mathrm{SL}(2)$ has an orbifold singularity and $E_2/\!/\mathrm{SL}(2)\cong\mathbb{P}\mathcal{C}_{2}/\!/\mathrm{SL}(2)$ is a free $\mathbb{Z}_{2}$-quotient of a $I_{2g-3}$-bundle over $\widetilde{T^{*}J}$ by Lemma \ref{geometric descriptions of second cones}-(1) and Lemma \ref{geometric descriptions of second cones}-(3), we use \cite[Proposition 3.10]{Ma21} to have
$$P_{t}(E_2/\!/\mathrm{SL}(2))=P_{t}^{+}(\widetilde{T^{*}J})P_{t}^{+}(I_{2g-3})+P_{t}^{-}(\widetilde{T^{*}J})P_{t}^{-}(I_{2g-3})$$
$$=(\frac{1}{2}((1+t)^{2g}+(1-t)^{2g})+2^{2g}(\frac{1-t^{4g}}{1-t^2}-1))\frac{(1-t^{4g-4})^{2}}{(1-t^2)(1-t^4)}$$
$$+\frac{1}{2}((1+t)^{2g}-(1-t)^{2g})\frac{t^2(1-t^{4g-4})(1-t^{4g-8})}{(1-t^2)(1-t^4)}.$$
\subsubsection{A conjectural formula for $IP_{t}(\mathbf{M})$}
Combining Theorem \ref{computable intersection blowing-up formula}, Conjecture \ref{equivariant blowing up formula conjecture}, Proposition \ref{intersection-Betti-of-normal-cone}, section \ref{eqiv coh of R}, section \ref{eqiv coh of Sigma}, section \ref{eqiv coh of deepest sing} and section \ref{eqiv coh of E2}, we get a conjectural formula for $IP_{t}(\mathbf{M})$ as following. The residue calculations show that the coefficients of the terms of $t^{i}$ are zero for $i>6g-6$ and the coefficient of the term of $t^{6g-6}$ is nonzero.
\begin{proposition}\label{A conjectural formula for IP(M)}
Assume that Conjecture \ref{equivariant blowing up formula conjecture} holds. Then
$$IP_{t}(\mathbf{M})=\frac{(1+t^3)^{2g}-(1+t)^{2g}t^{2g+2}}{(1-t^2)(1-t^4)}$$
$$-t^{4g-4}+\frac{t^{2g+2}(1+t)^{2g}}{(1-t^2)(1-t^4)}+\frac{(1-t)^{2g}t^{4g-4}}{4(1+t^2)}$$
$$+\frac{(1+t)^{2g}t^{4g-4}}{2(1-t^2)}(\frac{2g}{t+1}+\frac{1}{t^2-1}-\frac{1}{2}+(3-2g))$$
$$+\frac{1}{2}(2^{2g}-1)t^{4g-4}((1+t)^{2g-2}+(1-t)^{2g-2}-2)$$
$$+2^{2g}\big[\big(\frac{1-t^{12}}{1-t^2}-\frac{1-t^6}{1-t^2}+(\frac{1-t^6}{1-t^2})^{2}\big)\frac{(1-t^{4g-8})(1-t^{4g-4})(1-t^{4g})}{(1-t^2)(1-t^4)(1-t^6)}$$
$$-\frac{1-t^6}{1-t^2}\frac{(1-t^{4g-4})(1-t^{4g})}{(1-t^2)(1-t^4)}\frac{t^2(1-t^{2(2g-5)})}{1-t^2}$$
$$-\frac{(1-t^{4g-4})^{2}}{(1-t^2)(1-t^4)}\cdot\frac{1-t^{4g}}{1-t^2}+\frac{1}{1-t^4}\frac{1-t^{4g}}{1-t^2}\big]-\frac{2^{2g}}{1-t^4}$$
$$+(\frac{1}{2}((1+t)^{2g}+(1-t)^{2g})+2^{2g}(\frac{1-t^{4g}}{1-t^2}-1))\frac{(1-t^{4g-4})^{2}}{(1-t^2)(1-t^4)}$$
$$+\frac{1}{2}((1+t)^{2g}-(1-t)^{2g})\frac{t^2(1-t^{4g-4})(1-t^{4g-8})}{(1-t^2)(1-t^4)}$$
$$-\frac{1}{(1-t^4)}(\frac{1}{2}((1+t)^{2g}+(1-t)^{2g})+2^{2g}(\frac{1-t^{4g}}{1-t^2}-1))$$
$$-\frac{t^2}{(1-t^4)}\frac{1}{2}((1+t)^{2g}-(1-t)^{2g})$$
$$-\frac1 2(
(1+t)^{2g}+(1-t)^{2g})+2^{2g}(\frac{1-t^{4g}}{1-t^2}-1))\frac{t^2(1-t^{4g-4})(1-t^{4g-6})}{(1-t^2)(1-t^4)}$$
$$-\frac1 2(
(1+t)^{2g}-(1-t)^{2g})
(\frac{t^4(1-t^{4g-4})(1-t^{4g-10})}{(1-t^2)(1-t^4)}+t^{4g-6})$$
$$-2^{2g}\big[\frac{(1-t^{8g-8})(1-t^{4g})}{(1-t^{2})(1-t^{4})}-\frac{1-t^{4g}}{1-t^{4}}\big]$$
which is a polynomial with degree $6g-6$.
\end{proposition}
In low genus, we have $IP_{t}(\mathbf{M})$ as follows :
\begin{itemize}
\item $g=2$ : $IP_{t}(\mathbf{M})=1+t^{2}+17t^{4}+17t^{6}$
\item $g=3$ : $IP_{t}(\mathbf{M})=1+t^{2} +6t^{3} +2t^{4} +6t^{5} +17t^{6} +6t^{7} +81t^{8} +12t^{9} +396t^{10} +6t^{11} +66t^{12}$
\item $g=4$ : $IP_{t}(\mathbf{M})=1+t^{2} +8t^{3} +2t^{4} +8t^{5} +30t^{6} +16t^{7} +31t^{8} +72t^{9} +59t^{10} +72t^{11} +385t^{12}+ 80t^{13} + 3955t^{14} + 80t^{15} + 3885t^{16} + 16t^{17} + 259t^{18}$
\item $g=5$ : $IP_{t}(\mathbf{M})=1+t^{2} +10t^{3} +2t^{4} +10t^{5} +47t^{6} +20t^{7} +48t^{8} +140t^{9} +93t^{10} +150t^{11}+ 304t^{12} + 270t^{13} + 349t^{14} + 522t^{15} + 1583t^{16} + 532t^{17} + 29414t^{18} + 532t^{19}+ 72170t^{20} + 280t^{21}+ 28784t^{22} + 30t^{23} + 1028t^{24}$.
\end{itemize}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 5,081 |
namespace base {
namespace internal {
namespace {
constexpr int kWarmupRuns = 1;
constexpr TimeDelta kTimeLimit = Seconds(1);
constexpr int kTimeCheckInterval = 100000;
constexpr char kMetricPrefixLock[] = "PartitionLock.";
constexpr char kMetricLockUnlockThroughput[] = "lock_unlock_throughput";
constexpr char kMetricLockUnlockLatency[] = "lock_unlock_latency_ns";
constexpr char kStoryBaseline[] = "baseline_story";
constexpr char kStoryWithCompetingThread[] = "with_competing_thread";
perf_test::PerfResultReporter SetUpReporter(const std::string& story_name) {
perf_test::PerfResultReporter reporter(kMetricPrefixLock, story_name);
reporter.RegisterImportantMetric(kMetricLockUnlockThroughput, "runs/s");
reporter.RegisterImportantMetric(kMetricLockUnlockLatency, "ns");
return reporter;
}
class Spin : public PlatformThread::Delegate {
public:
Spin(MaybeLock<true>* lock, uint32_t* data)
: lock_(lock), data_(data), should_stop_(false) {}
~Spin() override = default;
void ThreadMain() override {
started_count_++;
// Local variable to avoid "cache line ping-pong" from influencing the
// results.
uint32_t count = 0;
while (!should_stop_.load(std::memory_order_relaxed)) {
lock_->Lock();
count++;
lock_->Unlock();
}
lock_->Lock();
(*data_) += count;
lock_->Unlock();
}
// Called from another thread to stop the loop.
void Stop() { should_stop_ = true; }
int started_count() const { return started_count_; }
private:
MaybeLock<true>* lock_;
uint32_t* data_ GUARDED_BY(lock_);
std::atomic<bool> should_stop_;
std::atomic<int> started_count_{0};
};
} // namespace
TEST(PartitionLockPerfTest, Simple) {
LapTimer timer(kWarmupRuns, kTimeLimit, kTimeCheckInterval);
uint32_t data = 0;
Lock lock;
do {
lock.Acquire();
data += 1;
lock.Release();
timer.NextLap();
} while (!timer.HasTimeLimitExpired());
ALLOW_UNUSED_LOCAL(data);
auto reporter = SetUpReporter(kStoryBaseline);
reporter.AddResult(kMetricLockUnlockThroughput, timer.LapsPerSecond());
reporter.AddResult(kMetricLockUnlockLatency, 1e9 / timer.LapsPerSecond());
}
TEST(PartitionLockPerfTest, WithCompetingThreads) {
uint32_t data = 0;
MaybeLock<true> lock;
// Starts a competing thread executing the same loop as this thread.
Spin thread_main(&lock, &data);
std::vector<PlatformThreadHandle> thread_handles;
constexpr int kThreads = 4;
for (int i = 0; i < kThreads; i++) {
PlatformThreadHandle thread_handle;
ASSERT_TRUE(PlatformThread::Create(0, &thread_main, &thread_handle));
thread_handles.push_back(thread_handle);
}
// Wait for all the threads to start.
while (thread_main.started_count() != kThreads) {
}
LapTimer timer(kWarmupRuns, kTimeLimit, kTimeCheckInterval);
do {
lock.Lock();
data += 1;
lock.Unlock();
timer.NextLap();
} while (!timer.HasTimeLimitExpired());
thread_main.Stop();
for (int i = 0; i < kThreads; i++) {
PlatformThread::Join(thread_handles[i]);
}
auto reporter = SetUpReporter(kStoryWithCompetingThread);
reporter.AddResult(kMetricLockUnlockThroughput, timer.LapsPerSecond());
reporter.AddResult(kMetricLockUnlockLatency, 1e9 / timer.LapsPerSecond());
}
} // namespace internal
} // namespace base
| {
"redpajama_set_name": "RedPajamaGithub"
} | 1,732 |
"""
Server Density plugin
MySQL
https://www.serverdensity.com/plugins/mysql/
https://github.com/serverdensity/sd-agent-plugins/
version: 0.1
"""
import traceback
import re
try:
import MySQLdb
except ImportError:
pass
# com commands.
COMMANDS = [
'Com_select',
'Com_delete',
'Com_update',
'Com_commit',
'Questions',
'Com_rollback',
'Handler_commit',
'Handler_delete',
'Handler_update',
'Handler_write',
'Handler_rollback',
'Handler_read_first',
'Handler_read_rnd',
]
class MySQL(object):
def __init__(self, agent_config, checks_logger, raw_config):
self.agent_config = agent_config
self.checks_logger = checks_logger
self.raw_config = raw_config
self.connection = None
self.datastore = {}
def version_is_above_5(self, status):
if (int(status['version'][0]) >= 5 and
int(status['version'][2]) >= 2):
return True
else:
return False
def get_db_results(self, db, query):
cursor = db.cursor()
try:
cursor.execute(query)
results = float(cursor.fetchone()[1])
except ValueError:
cursor.execute(query)
results = cursor.fetchone()[1]
return results
def run_query(self, db, query):
"""Run a query and returns a dictionary with results"""
try:
cursor = db.cursor()
cursor.execute(query)
metric = {}
for entry in cursor:
try:
metric[entry[0]] = float(entry[1])
except ValueError as e:
metric[entry[0]] = entry[1]
return metric
except MySQLdb.OperationalError as message:
self.checks_logger.debug(
'mysql: MySQL query error when getting metrics = '.format(
message)
)
def calculate_per_s(self, command, result):
if (not self.datastore.get(command) and
self.datastore.get(command) != 0):
self.checks_logger.debug(
'mysql: Datastore unset for '
'{0}, storing for first time'.format(command))
self.datastore[command] = result
com_per_s = 0
else:
com_per_s = (result - self.datastore[command]) / 60
if com_per_s < 0:
com_per_s = 0
self.datastore[command] = result
return com_per_s
def preliminaries(self):
if ('MySQLServer' not in self.raw_config and
'mysql_server' not in self.raw_config['MySQLServer'] or
self.raw_config['MySQLServer']['mysql_server'] == '' or
self.raw_config['MySQLServer']['mysql_user'] == '' or
self.raw_config['MySQLServer']['mysql_pass'] == ''):
self.checks_logger.debug('mysql: config not set')
return False
if not self.raw_config['MySQLServer'].get('mysql_port'):
self.raw_config['MySQLServer']['mysql_port'] = "3306"
self.checks_logger.debug('mysql: config set')
try:
import MySQLdb
except ImportError:
self.checks_logger.error('mysql: unable to import MySQLdb')
return False
# Note, code here doesn't really make sense. See what I copied.
if self.raw_config['MySQLServer'].get('mysql_server'):
# Connect
try:
MySQLdb.connect(
host=self.raw_config['MySQLServer']['mysql_server'],
user=self.raw_config['MySQLServer']['mysql_user'],
passwd=self.raw_config['MySQLServer']['mysql_pass'],
port=int(self.raw_config['MySQLServer']['mysql_port'])
)
except MySQLdb.OperationalError as message:
self.checks_logger.error(
"mysql: MySQL connection error: {0}".format(message))
return False
elif (self.raw_config['MySQLServer'].get('mysql_ssl_cert') and
self.raw_config['MySQLServer'].get('mysql_ssl_key')):
ssl = {
'cert': self.raw_config['MySQLServer']['mysql_ssl_cert'],
'key': self.raw_config['MySQLServer']['mysql_ssl_key']
}
MySQLdb.connect(
host=self.raw_config['MySQLServer']['mysql_server'],
user=self.raw_config['MySQLServer']['mysql_user'],
passwd=self.raw_config['MySQLServer']['mysql_pass'],
port=int(self.raw_config['MySQLServer']['mysql_port']),
ssl=ssl
)
else:
# Connect
try:
MySQLdb.connect(
host='localhost',
user=self.raw_config['MySQLServer']['mysql_user'],
passwd=self.raw_config['MySQLServer']['mysql_pass'],
port=int(self.raw_config['MySQLServer']['mysql_port']))
except MySQLdb.OperationalError as message:
self.checks_logger.error(
'mysql: MySQL connection error: {0}'.format(message)
)
return False
return True
def get_connection(self):
try:
# connection
if (self.raw_config['MySQLServer'].get('mysql_ssl_cert') and
self.raw_config['MySQLServer'].get('mysql_ssl_key')):
self.checks_logger.debug('mysql: Trying to connect via SSL')
ssl = {
'cert': self.raw_config['MySQLServer']['mysql_ssl_cert'],
'key': self.raw_config['MySQLServer']['mysql_ssl_key']
}
db = MySQLdb.connect(
host=self.raw_config['MySQLServer']['mysql_server'],
user=self.raw_config['MySQLServer']['mysql_user'],
passwd=self.raw_config['MySQLServer']['mysql_pass'],
port=int(self.raw_config['MySQLServer']['mysql_port']),
ssl=ssl
)
self.connection = db
self.checks_logger.error('mysql: Connected to DB via SSL')
else:
self.checks_logger.debug(
'mysql: Trying to connect via password')
db = MySQLdb.connect(
host=self.raw_config['MySQLServer']['mysql_server'],
user=self.raw_config['MySQLServer']['mysql_user'],
passwd=self.raw_config['MySQLServer']['mysql_pass'],
port=int(self.raw_config['MySQLServer']['mysql_port'])
)
self.connection = db
self.checks_logger.debug(
'mysql: Connected to DB with password')
# note, how do I take into account the socket?
except Exception:
self.checks_logger.error(
'Unable to connect to MySQL server {0}'
' - Exception: {1}'.format(
self.raw_config['MySQLServer']['mysql_server'],
traceback.format_exc())
)
return False
return True
def run(self):
self.checks_logger.debug('mysql: started gathering data')
if not self.preliminaries():
return False
if not self.get_connection():
return False
try:
db = self.connection
# setup
status = {}
# Get MySQL version
try:
self.checks_logger.debug('mysql: getting mysqlversion')
cursor = db.cursor()
cursor.execute('SELECT VERSION()')
result = cursor.fetchone()
version = result[0].split('-')
# Case 31237. Might include a description e.g. 4.1.26-log.
# See http://dev.mysql.com/doc/refman/4.1/en/
# information-functions.html#function_version
version = version[0].split('.')
status['version'] = []
for version_item in version:
number = re.match('([0-9]+)', version_item)
number = number.group(0)
status['version'].append(number)
except MySQLdb.OperationalError as message:
self.checks_logger.error(
(
'mysql: MySQL query error when getting version: '
'{0}'
).format(
message)
)
return False
# get show status metrics
status_metrics = self.run_query(db, 'SHOW GLOBAL STATUS')
status_variables = self.run_query(db, 'SHOW VARIABLES')
# get Uptime
status['Uptime'] = status_metrics['Uptime']
self.checks_logger.debug('mysql: getting Uptime - done')
# Slow queries
# Determine query depending on version. For 5.02 and above we
# need the GLOBAL keyword (case 31015)
# note, update with slow queries store. making it per second?
# ask jordi about that.
status['Slow queries'] = status_metrics['Slow_queries']
self.checks_logger.debug('mysql: getting Slow_queries - done')
# Note, check for which version of mysql?
# try:
# if self.version_is_above_5(status):
# query = 'SHOW GLOBAL STATUS LIKE "Slow_queries"'
# else:
# query = 'SHOW STATUS LIKE "Slow_queries'
# QPS - Queries per second.
status['Queries per second'] = self.calculate_per_s(
'qps', status_metrics['Queries']
)
# Note check for which version of mysql
self.checks_logger.debug('mysql: getting QPS - done')
# Connection pool
status['threads connected'] = status_metrics['Threads_connected']
status['threads running'] = status_metrics['Threads_running']
status['max connections'] = status_variables['max_connections']
status['max used connections'] = status_metrics[
'Max_used_connections']
status['Connection usage %'] = (
(status['threads running'] /
status['max connections'])*100
)
self.checks_logger.debug('mysql: getting connections - done')
# Buffer pool
status['buffer pool pages total'] = status_metrics[
'Innodb_buffer_pool_pages_total']
status['buffer pool pages free'] = status_metrics[
'Innodb_buffer_pool_pages_free']
status['buffer pool pages dirty'] = status_metrics[
'Innodb_buffer_pool_pages_dirty']
status['buffer pool pages data'] = status_metrics[
'Innodb_buffer_pool_pages_data']
self.checks_logger.debug('mysql: getting buffer pool - done')
# Query cache items
status['qcache hits'] = status_metrics['Qcache_hits']
status['qcache hits/s'] = self.calculate_per_s(
'qcache_ps', status['qcache hits'])
status['qcache free memory'] = status_metrics['Qcache_free_memory']
status['qcache not cached'] = status_metrics['Qcache_not_cached']
status['qcache in cache'] = status_metrics[
'Qcache_queries_in_cache']
self.checks_logger.debug('mysql: getting Qcache data - done')
# writes, reads, transactions
writes = (status_metrics['Com_insert'] +
status_metrics['Com_replace'] +
status_metrics['Com_update'] +
status_metrics['Com_delete'])
status['Writes/s'] = self.calculate_per_s('writes', writes)
# reads
reads = status_metrics['Com_select'] + status['qcache hits']
status['Reads/s'] = self.calculate_per_s('reads', reads)
try:
status['RW ratio'] = reads/writes
except ZeroDivisionError:
status['RW ratio'] = 0
# transactions
transactions = (status_metrics['Com_commit'] +
status_metrics['Com_rollback'])
status['Transactions/s'] = self.calculate_per_s(
'transactions', transactions)
self.checks_logger.debug(
'mysql: getting transactions, reads and writes - done')
# Aborted connections and clients
status['aborted clients'] = status_metrics['Aborted_clients']
status['aborted connects'] = status_metrics['Aborted_connects']
self.checks_logger.debug(
'mysql: getting aborted connections - done')
# Replication - Seconds Behind Master
secondsBehindMaster = None
try:
cursor = db.cursor(MySQLdb.cursors.DictCursor)
cursor.execute('SHOW SLAVE STATUS')
result = cursor.fetchone()
except MySQLdb.OperationalError as message:
self.checks_logger.error(
'getMySQLStatus: MySQL query error when '
'getting SHOW SLAVE STATUS = %s', message)
result = None
if result is not None:
try:
# Handle the case when Seconds_Behind_Master is NULL
if result['Seconds_Behind_Master'] is None:
secondsBehindMaster = -1
else:
secondsBehindMaster = result['Seconds_Behind_Master']
self.checks_logger.debug(
'getMySQLStatus: '
'secondsBehindMaster = %s', secondsBehindMaster
)
except IndexError as e:
self.checks_logger.debug(
'getMySQLStatus: secondsBehindMaster empty. %s', e
)
else:
self.checks_logger.debug(
'getMySQLStatus: secondsBehindMaster empty. Result = None.'
)
# Created temporary tables in memory and on disk
status['created tmp tables'] = status_metrics['Created_tmp_tables']
status['created tmp tables on disk'] = status_metrics[
'Created_tmp_disk_tables']
# Note check mysql version?
self.checks_logger.debug(
'mysql: getting temporary tables data - done')
# select_full_join
status['select full join'] = status_metrics['Select_full_join']
# note check for mysql version?
self.checks_logger.debug('mysql: getting select_full_join - done')
# slave_running
result = status_metrics['Slave_running']
if result == 'OFF':
result = 0
else:
result = 1
status['slave running'] = result
self.checks_logger.debug(
'mysql: getting slave_running - done')
# open files
status['open files'] = status_metrics['Open_files']
status['open files limit'] = status_variables['open_files_limit']
self.checks_logger.debug('mysql: getting open_files - done')
# table_locks_waited
status['table locks waited'] = status_metrics['Table_locks_waited']
self.checks_logger.debug(
'mysql: getting table_locks_waited - done')
# checkpoint age
# note this needs to be changed.
try:
cursor = db.cursor()
cursor.execute('SHOW ENGINE INNODB STATUS')
results = cursor.fetchone()[2]
log_loci = results.find('Log sequence number')
checkpoint_loci = results.find('Last checkpoint at')
log_nr = int(re.search(r'\d+', results[log_loci:]).group(0))
cp_nr = int(re.search(
r'\d+', results[checkpoint_loci:]).group(0))
cp_age = cp_nr - log_nr
status['Checkpoint age'] = cp_age
except MySQLdb.OperationalError as message:
self.checks_logger.error(
'mysql: MySQL query error when '
'getting checkpoint age = {0}'.format(
message)
)
return False
self.checks_logger.debug(
'mysql: getting checkpoint age - done')
# note remove this.
try:
# Key cache hit ratio
# http://www.percona.com/blog/2010/02/28/why-you-should-ignore-mysqls-key-cache-hit-ratio/
key_read = self.get_db_results(
db, 'SHOW STATUS LIKE "Key_reads"')
key_requests = self.get_db_results(
db, 'SHOW STATUS LIKE "Key_read_requests"')
# status['Key cache hit ratio'] = (
# 100 - ((key_read * 100) / key_requests))
status['Key reads/s'] = self.calculate_per_s(
"Key_reads", key_read)
except MySQLdb.OperationalError as message:
self.checks_logger.error(
'mysql: MySQL query error when '
'getting key cache = {0}'.format(
message)
)
return False
self.checks_logger.debug(
'mysql: getting key cache hit ratio - done')
# com commands per second
com = self.raw_config['MySQLServer'].get('mysql_include_per_s')
if com:
user_com_ps = com
user_com_ps = user_com_ps.split(',')
user_com_ps = [command.strip() for command in user_com_ps]
user_com_ps = user_com_ps + COMMANDS
else:
user_com_ps = COMMANDS
for command in user_com_ps:
try:
com_per_s = self.calculate_per_s(
command, status_metrics[command])
status[command.replace('_', ' ')+'/s'] = com_per_s
except KeyError, e:
self.checks_logger.exception(e)
if self.raw_config['MySQLServer'].get('mysql_include'):
user_com = self.raw_config['MySQLServer']['mysql_include']
user_com = user_com.split(',')
user_com = [command.strip() for command in user_com]
user_com = user_com + COMMANDS
else:
user_com = COMMANDS
for command in user_com:
status[command.replace('_', ' ')] = status_metrics[
command]
self.checks_logger.debug(
'mysql: getting com_commands - done')
except Exception:
self.checks_logger.error(
'mysql: unable to get data from MySQL - '
'Exception: {0}'.format(traceback.format_exc())
)
self.checks_logger.debug('mysql: completed, returning')
return status
if __name__ == "__main__":
"""Standalone test"""
import logging
import sys
import json
import time
host = 'localhost'
port = '3306'
raw_agent_config = {
'MySQLServer': {
'mysql_server': host,
'mysql_port': port,
'mysql_user': 'jonathan',
'mysql_pass': 'password',
'mysql_include_per_s': 'Com_check, Com_checksum, Com_begin',
# 'mysql_ssl_cert': '/etc/mysql-ssl/client-cert.pem',
# 'mysql_ssl_key': '/etc/mysql-ssl/client-key.pem'
}
}
main_checks_logger = logging.getLogger('MySQLplugin')
main_checks_logger.setLevel(logging.DEBUG)
main_checks_logger.addHandler(logging.StreamHandler(sys.stdout))
mysql_check = MySQL({}, main_checks_logger, raw_agent_config)
while True:
try:
result = mysql_check.run()
print(json.dumps(result, indent=4, sort_keys=True))
except:
main_checks_logger.exception("Unhandled Exception")
finally:
time.sleep(60)
| {
"redpajama_set_name": "RedPajamaGithub"
} | 1,807 |
\section{Instructions}
This document has been adapted
by Steven Bethard, Ryan Cotterrell and Rui Yan
from the instructions for earlier ACL and NAACL proceedings, including those for
ACL 2019 by Douwe Kiela and Ivan Vuli\'{c},
NAACL 2019 by Stephanie Lukin and Alla Roskovskaya,
ACL 2018 by Shay Cohen, Kevin Gimpel, and Wei Lu,
NAACL 2018 by Margaret Michell and Stephanie Lukin,
2017/2018 (NA)ACL bibtex suggestions from Jason Eisner,
ACL 2017 by Dan Gildea and Min-Yen Kan,
NAACL 2017 by Margaret Mitchell,
ACL 2012 by Maggie Li and Michael White,
ACL 2010 by Jing-Shing Chang and Philipp Koehn,
ACL 2008 by Johanna D. Moore, Simone Teufel, James Allan, and Sadaoki Furui,
ACL 2005 by Hwee Tou Ng and Kemal Oflazer,
ACL 2002 by Eugene Charniak and Dekang Lin,
and earlier ACL and EACL formats written by several people, including
John Chen, Henry S. Thompson and Donald Walker.
Additional elements were taken from the formatting instructions of the \emph{International Joint Conference on Artificial Intelligence} and the \emph{Conference on Computer Vision and Pattern Recognition}.
\section{Introduction}
The following instructions are directed to authors of papers submitted to ACL 2020 or accepted for publication in its proceedings.
All authors are required to adhere to these specifications.
Authors are required to provide a Portable Document Format (PDF) version of their papers.
\textbf{The proceedings are designed for printing on A4 paper.}
\section{Electronically-available resources}
ACL provides this description and accompanying style files at
\begin{quote}
\url{http://acl2020.org/downloads/acl2020-templates.zip}
\end{quote}
We strongly recommend the use of these style files, which have been appropriately tailored for the ACL 2020 proceedings.
\paragraph{\LaTeX-specific details:}
The templates include the \LaTeX2e{} source (\texttt{\small acl2020.tex}),
the \LaTeX2e{} style file used to format it (\texttt{\small acl2020.sty}),
an ACL bibliography style (\texttt{\small acl\_natbib.bst}),
an example bibliography (\texttt{\small acl2020.bib}),
and the bibliography for the ACL Anthology (\texttt{\small anthology.bib}).
\section{Length of Submission}
\label{sec:length}
The conference accepts submissions of long papers and short papers.
Long papers may consist of up to eight (8) pages of content plus unlimited pages for references.
Upon acceptance, final versions of long papers will be given one additional page -- up to nine (9) pages of content plus unlimited pages for references -- so that reviewers' comments can be taken into account.
Short papers may consist of up to four (4) pages of content, plus unlimited pages for references.
Upon acceptance, short papers will be given five (5) pages in the proceedings and unlimited pages for references.
For both long and short papers, all illustrations and tables that are part of the main text must be accommodated within these page limits, observing the formatting instructions given in the present document.
Papers that do not conform to the specified length and formatting requirements are subject to be rejected without review.
The conference encourages the submission of additional material that is relevant to the reviewers but not an integral part of the paper.
There are two such types of material: appendices, which can be read, and non-readable supplementary materials, often data or code.
Additional material must be submitted as separate files, and must adhere to the same anonymity guidelines as the main paper.
The paper must be self-contained: it is optional for reviewers to look at the supplementary material.
Papers should not refer, for further detail, to documents, code or data resources that are not available to the reviewers.
Refer to Appendices~\ref{sec:appendix} and \ref{sec:supplemental} for further information.
Workshop chairs may have different rules for allowed length and whether supplemental material is welcome.
As always, the respective call for papers is the authoritative source.
\section{Anonymity}
As reviewing will be double-blind, papers submitted for review should not include any author information (such as names or affiliations). Furthermore, self-references that reveal the author's identity, \emph{e.g.},
\begin{quote}
We previously showed \citep{Gusfield:97} \ldots
\end{quote}
should be avoided. Instead, use citations such as
\begin{quote}
\citet{Gusfield:97} previously showed\ldots
\end{quote}
Please do not use anonymous citations and do not include acknowledgements.
\textbf{Papers that do not conform to these requirements may be rejected without review.}
Any preliminary non-archival versions of submitted papers should be listed in the submission form but not in the review version of the paper.
Reviewers are generally aware that authors may present preliminary versions of their work in other venues, but will not be provided the list of previous presentations from the submission form.
Once a paper has been accepted to the conference, the camera-ready version of the paper should include the author's names and affiliations, and is allowed to use self-references.
\paragraph{\LaTeX-specific details:}
For an anonymized submission, ensure that {\small\verb|\aclfinalcopy|} at the top of this document is commented out, and that you have filled in the paper ID number (assigned during the submission process on softconf) where {\small\verb|***|} appears in the {\small\verb|\def\aclpaperid{***}|} definition at the top of this document.
For a camera-ready submission, ensure that {\small\verb|\aclfinalcopy|} at the top of this document is not commented out.
\section{Multiple Submission Policy}
Papers that have been or will be submitted to other meetings or publications must indicate this at submission time in the START submission form, and must be withdrawn from the other venues if accepted by ACL 2020. Authors of papers accepted for presentation at ACL 2020 must notify the program chairs by the camera-ready deadline as to whether the paper will be presented. We will not accept for publication or presentation the papers that overlap significantly in content or results with papers that will be (or have been) published elsewhere.
Authors submitting more than one paper to ACL 2020 must ensure that submissions do not overlap significantly (>25\%) with each other in content or results.
\section{Formatting Instructions}
Manuscripts must be in two-column format.
Exceptions to the two-column format include the title, authors' names and complete addresses, which must be centered at the top of the first page, and any full-width figures or tables (see the guidelines in Section~\ref{ssec:title-authors}).
\textbf{Type single-spaced.}
Start all pages directly under the top margin.
The manuscript should be printed single-sided and its length should not exceed the maximum page limit described in Section~\ref{sec:length}.
Pages should be numbered in the version submitted for review, but \textbf{pages should not be numbered in the camera-ready version}.
\paragraph{\LaTeX-specific details:}
The style files will generate page numbers when {\small\verb|\aclfinalcopy|} is commented out, and remove them otherwise.
\subsection{File Format}
\label{sect:pdf}
For the production of the electronic manuscript you must use Adobe's Portable Document Format (PDF).
Please make sure that your PDF file includes all the necessary fonts (especially tree diagrams, symbols, and fonts with Asian characters).
When you print or create the PDF file, there is usually an option in your printer setup to include none, all or just non-standard fonts.
Please make sure that you select the option of including ALL the fonts.
\textbf{Before sending it, test your PDF by printing it from a computer different from the one where it was created.}
Moreover, some word processors may generate very large PDF files, where each page is rendered as an image.
Such images may reproduce poorly.
In this case, try alternative ways to obtain the PDF.
One way on some systems is to install a driver for a postscript printer, send your document to the printer specifying ``Output to a file'', then convert the file to PDF.
It is of utmost importance to specify the \textbf{A4 format} (21 cm x 29.7 cm) when formatting the paper.
Print-outs of the PDF file on A4 paper should be identical to the hardcopy version.
If you cannot meet the above requirements about the production of your electronic submission, please contact the publication chairs as soon as possible.
\paragraph{\LaTeX-specific details:}
PDF files are usually produced from \LaTeX{} using the \texttt{\small pdflatex} command.
If your version of \LaTeX{} produces Postscript files, \texttt{\small ps2pdf} or \texttt{\small dvipdf} can convert these to PDF.
To ensure A4 format in \LaTeX, use the command {\small\verb|\special{papersize=210mm,297mm}|}
in the \LaTeX{} preamble (below the {\small\verb|\usepackage|} commands) and use \texttt{\small dvipdf} and/or \texttt{\small pdflatex}; or specify \texttt{\small -t a4} when working with \texttt{\small dvips}.
\subsection{Layout}
\label{ssec:layout}
Format manuscripts two columns to a page, in the manner these
instructions are formatted.
The exact dimensions for a page on A4 paper are:
\begin{itemize}
\item Left and right margins: 2.5 cm
\item Top margin: 2.5 cm
\item Bottom margin: 2.5 cm
\item Column width: 7.7 cm
\item Column height: 24.7 cm
\item Gap between columns: 0.6 cm
\end{itemize}
\noindent Papers should not be submitted on any other paper size.
If you cannot meet the above requirements about the production of your electronic submission, please contact the publication chairs above as soon as possible.
\subsection{Fonts}
For reasons of uniformity, Adobe's \textbf{Times Roman} font should be used.
If Times Roman is unavailable, you may use Times New Roman or \textbf{Computer Modern Roman}.
Table~\ref{font-table} specifies what font sizes and styles must be used for each type of text in the manuscript.
\begin{table}
\centering
\begin{tabular}{lrl}
\hline \textbf{Type of Text} & \textbf{Font Size} & \textbf{Style} \\ \hline
paper title & 15 pt & bold \\
author names & 12 pt & bold \\
author affiliation & 12 pt & \\
the word ``Abstract'' & 12 pt & bold \\
section titles & 12 pt & bold \\
subsection titles & 11 pt & bold \\
document text & 11 pt &\\
captions & 10 pt & \\
abstract text & 10 pt & \\
bibliography & 10 pt & \\
footnotes & 9 pt & \\
\hline
\end{tabular}
\caption{\label{font-table} Font guide. }
\end{table}
\paragraph{\LaTeX-specific details:}
To use Times Roman in \LaTeX2e{}, put the following in the preamble:
\begin{quote}
\small
\begin{verbatim}
\usepackage{times}
\usepackage{latexsym}
\end{verbatim}
\end{quote}
\subsection{Ruler}
A printed ruler (line numbers in the left and right margins of the article) should be presented in the version submitted for review, so that reviewers may comment on particular lines in the paper without circumlocution.
The presence or absence of the ruler should not change the appearance of any other content on the page.
The camera ready copy should not contain a ruler.
\paragraph{Reviewers:}
note that the ruler measurements may not align well with lines in the paper -- this turns out to be very difficult to do well when the paper contains many figures and equations, and, when done, looks ugly.
In most cases one would expect that the approximate location will be adequate, although you can also use fractional references (\emph{e.g.}, this line ends at mark $295.5$).
\paragraph{\LaTeX-specific details:}
The style files will generate the ruler when {\small\verb|\aclfinalcopy|} is commented out, and remove it otherwise.
\subsection{Title and Authors}
\label{ssec:title-authors}
Center the title, author's name(s) and affiliation(s) across both columns.
Do not use footnotes for affiliations.
Place the title centered at the top of the first page, in a 15-point bold font.
Long titles should be typed on two lines without a blank line intervening.
Put the title 2.5 cm from the top of the page, followed by a blank line, then the author's names(s), and the affiliation on the following line.
Do not use only initials for given names (middle initials are allowed).
Do not format surnames in all capitals (\emph{e.g.}, use ``Mitchell'' not ``MITCHELL'').
Do not format title and section headings in all capitals except for proper names (such as ``BLEU'') that are
conventionally in all capitals.
The affiliation should contain the author's complete address, and if possible, an electronic mail address.
The title, author names and addresses should be completely identical to those entered to the electronical paper submission website in order to maintain the consistency of author information among all publications of the conference.
If they are different, the publication chairs may resolve the difference without consulting with you; so it is in your own interest to double-check that the information is consistent.
Start the body of the first page 7.5 cm from the top of the page.
\textbf{Even in the anonymous version of the paper, you should maintain space for names and addresses so that they will fit in the final (accepted) version.}
\subsection{Abstract}
Use two-column format when you begin the abstract.
Type the abstract at the beginning of the first column.
The width of the abstract text should be smaller than the
width of the columns for the text in the body of the paper by 0.6 cm on each side.
Center the word \textbf{Abstract} in a 12 point bold font above the body of the abstract.
The abstract should be a concise summary of the general thesis and conclusions of the paper.
It should be no longer than 200 words.
The abstract text should be in 10 point font.
\subsection{Text}
Begin typing the main body of the text immediately after the abstract, observing the two-column format as shown in the present document.
Indent 0.4 cm when starting a new paragraph.
\subsection{Sections}
Format section and subsection headings in the style shown on the present document.
Use numbered sections (Arabic numerals) to facilitate cross references.
Number subsections with the section number and the subsection number separated by a dot, in Arabic numerals.
\subsection{Footnotes}
Put footnotes at the bottom of the page and use 9 point font.
They may be numbered or referred to by asterisks or other symbols.\footnote{This is how a footnote should appear.}
Footnotes should be separated from the text by a line.\footnote{Note the line separating the footnotes from the text.}
\subsection{Graphics}
Place figures, tables, and photographs in the paper near where they are first discussed, rather than at the end, if possible.
Wide illustrations may run across both columns.
Color is allowed, but adhere to Section~\ref{ssec:accessibility}'s guidelines on accessibility.
\paragraph{Captions:}
Provide a caption for every illustration; number each one sequentially in the form:
``Figure 1. Caption of the Figure.''
``Table 1. Caption of the Table.''
Type the captions of the figures and tables below the body, using 10 point text.
Captions should be placed below illustrations.
Captions that are one line are centered (see Table~\ref{font-table}).
Captions longer than one line are left-aligned (see Table~\ref{tab:accents}).
\begin{table}
\centering
\begin{tabular}{lc}
\hline
\textbf{Command} & \textbf{Output}\\
\hline
\verb|{\"a}| & {\"a} \\
\verb|{\^e}| & {\^e} \\
\verb|{\`i}| & {\`i} \\
\verb|{\.I}| & {\.I} \\
\verb|{\o}| & {\o} \\
\verb|{\'u}| & {\'u} \\
\verb|{\aa}| & {\aa} \\\hline
\end{tabular}
\begin{tabular}{lc}
\hline
\textbf{Command} & \textbf{Output}\\
\hline
\verb|{\c c}| & {\c c} \\
\verb|{\u g}| & {\u g} \\
\verb|{\l}| & {\l} \\
\verb|{\~n}| & {\~n} \\
\verb|{\H o}| & {\H o} \\
\verb|{\v r}| & {\v r} \\
\verb|{\ss}| & {\ss} \\
\hline
\end{tabular}
\caption{Example commands for accented characters, to be used in, \emph{e.g.}, \BibTeX\ names.}\label{tab:accents}
\end{table}
\paragraph{\LaTeX-specific details:}
The style files are compatible with the caption and subcaption packages; do not add optional arguments.
\textbf{Do not override the default caption sizes.}
\subsection{Hyperlinks}
Within-document and external hyperlinks are indicated with Dark Blue text, Color Hex \#000099.
\subsection{Citations}
Citations within the text appear in parentheses as~\citep{Gusfield:97} or, if the author's name appears in the text itself, as \citet{Gusfield:97}.
Append lowercase letters to the year in cases of ambiguities.
Treat double authors as in~\citep{Aho:72}, but write as in~\citep{Chandra:81} when more than two authors are involved. Collapse multiple citations as in~\citep{Gusfield:97,Aho:72}.
Refrain from using full citations as sentence constituents.
Instead of
\begin{quote}
``\citep{Gusfield:97} showed that ...''
\end{quote}
write
\begin{quote}
``\citet{Gusfield:97} showed that ...''
\end{quote}
\begin{table*}
\centering
\begin{tabular}{lll}
\hline
\textbf{Output} & \textbf{natbib command} & \textbf{Old ACL-style command}\\
\hline
\citep{Gusfield:97} & \small\verb|\citep| & \small\verb|\cite| \\
\citealp{Gusfield:97} & \small\verb|\citealp| & no equivalent \\
\citet{Gusfield:97} & \small\verb|\citet| & \small\verb|\newcite| \\
\citeyearpar{Gusfield:97} & \small\verb|\citeyearpar| & \small\verb|\shortcite| \\
\hline
\end{tabular}
\caption{\label{citation-guide}
Citation commands supported by the style file.
The style is based on the natbib package and supports all natbib citation commands.
It also supports commands defined in previous ACL style files for compatibility.
}
\end{table*}
\paragraph{\LaTeX-specific details:}
Table~\ref{citation-guide} shows the syntax supported by the style files.
We encourage you to use the natbib styles.
You can use the command {\small\verb|\citet|} (cite in text) to get ``author (year)'' citations as in \citet{Gusfield:97}.
You can use the command {\small\verb|\citep|} (cite in parentheses) to get ``(author, year)'' citations as in \citep{Gusfield:97}.
You can use the command {\small\verb|\citealp|} (alternative cite without parentheses) to get ``author year'' citations (which is useful for using citations within parentheses, as in \citealp{Gusfield:97}).
\subsection{References}
Gather the full set of references together under the heading \textbf{References}; place the section before any Appendices.
Arrange the references alphabetically by first author, rather than by order of occurrence in the text.
Provide as complete a citation as possible, using a consistent format, such as the one for \emph{Computational Linguistics\/} or the one in the \emph{Publication Manual of the American
Psychological Association\/}~\citep{APA:83}.
Use full names for authors, not just initials.
Submissions should accurately reference prior and related work, including code and data.
If a piece of prior work appeared in multiple venues, the version that appeared in a refereed, archival venue should be referenced.
If multiple versions of a piece of prior work exist, the one used by the authors should be referenced.
Authors should not rely on automated citation indices to provide accurate references for prior and related work.
The following text cites various types of articles so that the references section of the present document will include them.
\begin{itemize}
\item Example article in journal: \citep{Ando2005}.
\item Example article in proceedings, with location: \citep{borschinger-johnson-2011-particle}.
\item Example article in proceedings, without location: \citep{andrew2007scalable}.
\item Example arxiv paper: \citep{rasooli-tetrault-2015}.
\end{itemize}
\paragraph{\LaTeX-specific details:}
The \LaTeX{} and Bib\TeX{} style files provided roughly follow the American Psychological Association format.
If your own bib file is named \texttt{\small acl2020.bib}, then placing the following before any appendices in your \LaTeX{} file will generate the references section for you:
\begin{quote}\small
\verb|\bibliographystyle{acl_natbib}|\\
\verb|
\section{Introduction}
We present a web-based system for diacritization of Hebrew text, which caters to both casual and expert users. The diacritization engine driving the system combines manually curated
linguistic resources with modern machine learning models.
\paragraph{Diacritization}
In Hebrew writing, the letters are almost entirely consonantal; the vowels are indicated by diacritic marks, generally positioned underneath the letters. However, in most cases, printed Hebrew omits the diacritic marks and includes only the letters, resulting in a highly ambiguous text, in which any given non-diacritized word can represent a host of different Hebrew words, each with a different meaning and pronunciation. For example, the form \<b.sl> can be diacritized as \<b*A.sAl> (noun, ``onion''), \<b*:.sel> (prefix+noun, ``in a shadow''), \<b*a.s*el> (prefix+definitive+noun, ``in the shadow'') and others. The task of diacritization is thus a task of disambiguation: choosing from among the valid word possibilities for each non-diacritized word, and then adding in the diacritic marks accordingly. The multiple possibilities for diacritizing any given word often represent different morphological possibilities. Thus, to an extent, choosing the correct diacritization entails morphological disambiguation; conversely, prior morphological disambiguation greatly reduces the total possible diacritization possibilities.We provide further details in \S\ref{sec:diacritization}.
\paragraph{Hybrid Neural and Rule-based Approach}
Our approach, described in \S\ref{sec:approach}, uses several bi-LSTM-based
deep-learning modules for disambiguating the correct diacritization in context.
However, it is also supplemented by comprehensive inflection tables and
lexicons, when appropriate.
\begin{figure*}[t]
\includegraphics[width=1\textwidth]{nakdan-main-chopped3.png}
\caption{The main web interface of our diacritization tool, showing the automatic diacritized text (A) and allowing the user to proofread and potentially correct the text. The user can navigate the words using the mouse or the left/right keys, and can select an alternate diacritization option from the listbox on the left (B) using either the mouse or the up/down keys. Changes for a given word can be marked for application over the entire text (C), and are marked in color (not shown in this example). The user can also choose to see the morphological analysis of each form (D). The resulting diacritized text can be exported to various formats (E).}
\label{fig:main}
\end{figure*}
\paragraph{Web Interface}
We provide a web interface for the user to input a text for diacritization and refine the resulting diacritized text (Figure \ref{fig:main}).
Our system parses the text and automatically adds diacritics throughout. Afterward, the user can proofread the text in the interface. For each word, all alternate diacritization possibilities are provided for immediate selection, ordered according to their predicted probability. Keyboard shortcuts allow efficient navigation of the text and fast selection of alternate options. Users can choose to see morphological analyses for each of the diacritization options, to assist in distinguishing between options.
\paragraph{Diacritics in Scientific Editions}
We aim to provide a tool that is useful to casual users and language
enthusiasts, but also to experts and professionals who may use it to set
scientific editions of historical Hebrew texts. This latter requirement poses several challenges: handling of editorial sigla interspersed within the words; flexible handling of matres lectionis (letters which function as semi-vowels); and dealing with the orthography of medieval Hebrew, which often diverges widely from that of Modern Hebrew. Our tool meets scholarly requirements on all these fronts, as detailed in \S\ref{sec:advanced}.
\section{The Hebrew Diacritics System}
\label{sec:diacritization}
The diacritics system of modern Hebrew marks vowels and gemination, and includes 12 primary diacritic symbols:
\begin{center}
\includegraphics[width=0.4\textwidth]{nakdan-vowelchart2.jpg}
\end{center}
Additionally, a dot in the middle of a letter indicates gemination. For the case of the 'shin' letter, an upper dot distinguishes between pronunciation as 's' or as 'sh'.
Diacritized Hebrew aims to position a diacritic on every single letter of the word, with the exception of final letters and matres lectionis.
\paragraph{Ambiguity}
In our tests, knowing the correct diacritics reduces the
full-morphological-analysis ambiguity from 9.1 to 2.4 average analyses per word form,
while knowing the full-morphological-analysis reduces the diacritization ambiguity
from 6.2 to 1.4 average options per word form. Note that these numbers reflect fine-grained morphological tagging. If we utilize coarse-grained tagging, sufficing with the part of speech for each word, then knowing the correct diacritization reduces the average morphological ambiguity from 3.2 options to 1.97, while knowing the correct POS tag reduces the average diacritization ambiguity from 6.2 options to 2.75. Thus, the need for an automated diacritization utility is particularly crucial in order to properly disambiguate a Hebrew text.
\section{Approach}
\label{sec:approach}
Recent trends in NLP suggest moving towards machine-learned models that
automatically learn to extract the regularities in the data.
Such approaches have also been applied to diacritization of Arabic \cite{Belinkov2015ArabicDW, rashwan-arabic-diacritization, abandah-arabic-diacritization, mubarak-etal-2019-highly}.
However, while these generally provide
very strong results, they also often make mistakes that contradict our prior
knowledge of the linguistic system. While the machine-learned models generalize very well and
can learn to perform tasks in which humans cannot articulate the underlying
regularities, there are also many cases that language-experts \emph{can} articulate
precisely, and these tend to correlate with the cases that the learned models fail on.
We therefore take a hybrid approach. Similar to traditional diacritization
systems \cite{choueka1995nakdan}, we use our explicit knowledge about the language and the diacritization system whenever we can.
However, we also supplement our
knowledge with learned model predictions for the challenging cases for which we
cannot articulate the rules and regularities: selecting the appropriate
diacritization in context, and providing diacritization for out-of-vocabulary words.
This methodology departs from recent diacritization works that rely on HMM and
neural-network methods \cite{gal2002hmm,Belinkov2015ArabicDW}, while
ignoring forms of explicit linguistic knowledge.
We use such a combination of machine-learned and human-specified knowledge in
all the components of the system, either by supplementing the predictor with
manually constructed options, or by filtering its output space.
Of course, a prerequisite for an effective machine-learned system is
high-quality training data. Our system is trained on a collection of 1,5M
diacritized tokens which we annotated in-house.
\section{High-quality Data Sources}
We make use of the following language resources and corpora, which we collected.
\paragraph{Language Resources}
\label{wordlists}
Our main resource is a high-coverage and accurate lexicon of Hebrew word forms,
their diacritization and their corresponding morphological analyses. Employing a staff of language experts, we began by assembling a list of all nouns, adjectives and verbal roots in the Hebrew language. This list includes 50K lexemes altogether (10K roots, 30,5K nouns, and 9,5K adjectives). We then built comprehensive inflection tables to generate all possible inflected forms from each of these lexemes, including all valid combinations of possessive and accusative suffixes, with full diacritization. Altogether, this process generated some 5,5 million inflected forms (3,8M verbal forms; 1,3M nominal forms; and 460K adjectival forms).
We also added 1,7K adverbs, and another 4,5K function words (conjunctions, prepositions, existentials, quantifiers, etc., including all possible suffix combinations).
Finally, we collected a set of 17,5K frequent proper nouns (countries and major cities; heads of state and other notable people; and frequently-mentioned companies and organizations), and our language experts diacritized these as well.
These tables suffice for modern Hebrew; however, in historical Hebrew texts, we often find Aramaic terms interspersed within the Hebrew. Therefore, we also built a similarly comprehensive and diacritized wordlist for Babylonian Aramaic. Our Aramaic wordlist contains 750K verbal inflected forms; 200K nominals; 1,5K adjectives; and another 2K adverbs and function words.
We additionally assembled an exhaustive list of non-diacritized Hebrew names of persons
and locations (including collections of both street names and city names).
\paragraph{Annotated Corpora}
For morphological tagging, we make use of a corpus of 200K tokens of modern Hebrew, composed of
Hebrew fiction, news, wikipedia, and blogs. These tokens were manually annotated with fine-grained morphological information according to the scheme of \cite{elhadad2005hebrew}. Additionally, as noted, we annotated a 1,5M word diacritized modern Hebrew corpus, consisting of Hebrew prose (both fiction and non-fiction), newspapers (both news and op-ed), wikipedia, blogs (including many female-dominant blogs, to ensure coverage of feminine word forms), law protocols, Parliament proceedings, TV transcripts, academic texts, and biographical sketches.
We have similarly collected and annotated corpora of historical Hebrew, consisting of Jewish legal writings and commentaries from the 3rd-12th centuries: 110K words with fine-grained morphological tagging, and 2M words with diacritization. Finally, regarding poetic Hebrew, we collected and annotated a corpus of 1,3M words, containing Hebrew poetry from both medieval and modern periods.
The undiacritized base texts were collected largely through partnerships with cooperating organizations in Israel; the morphological tagging and diacritization was done primarily in-house by our Hebrew language experts.
\section{System Architecture}
On a high level, our system works in the following stages, which we will
elaborate on below. Each stage combines engineered linguistic information and a
trained neural model.
\begin{enumerate}
\item POS-tagging and morphological disambiguation.
\item Filtering the possible diacritization analyses based on high coverage
accurate tables and the output of stage (1).
\item Ranking the possible diacritizations for each word, in context.
\end{enumerate}
\paragraph{Part-of-speech tagging and morphological disambiguation}
As diacritic marks closely interact with the morphological analysis and
part-of-speech (POS) of the token, we first perform POS-tagging and
morphological disambiguation, using a two-stage process. In the first stage,
each word is assigned its core part-of-speech, and in the second stage it is
enriched with additional morphological properties, where the set of
considered morphological properties is determined based on the coarse-grained
POS (e.g., nouns take gender, number and definiteness, while verbs do not take
definiteness but do take tense and person).\footnote{We consider the
following POS-tags: \textit{Adj, Adv, Conj, At\_Prep, Neg, Noun, Num, Prep, Pron, ProperNoun, Verb, Interrogative, Interj, Quantifier, Existential, Modal, Prefix, Participle, Copula, Titular, Shel\_Prep},
\normalfont
and the following morphological properties: \textit{Gender, Number, Person, Construct/Absolute, Suffix (possessive / accusative / pronominal)}.}
Training is performed on our annotated corpus of 200K tokens. The resulting tagger has an accuracy
of 92\% for the coarse-grained part-of-speech, and 79\% for full morphological
disambiguation.\footnote{While these numbers may seem low, we note that they are (a)
on-par with other Hebrew systems \cite{DBLP:conf/acl/AdlerE06,moretsarfatycoling2016} and (b) are only intended to support the diacritization process, where we find they do well.}
Both taggers are 2-layer bi-LSTM transducers \cite{goldberg-book}, where the
first stage coarse-grained tagger maps each token $w_i$ to a coarse POS-tag $t_i$, while the
second stage morphological tagger adds additional morphological properties
$m_i^1,...,m_i^k$. Each bi-LSTM takes as inputs vectors $x_1,...,x_n$
corresponding to tokens $w_1,...,w_n$ and produces vectors $h(x_1),...,h(x_n)$. These
vectors are then fed into multi-layer perceptrons (MLP) for predicting the POS-tags and
morphological properties, where each property is predicted by a different MLP:
\[
t_i = \arg\max_j \text{softmax}(MLP_{\text{pos}}(h(x_i))_{[j]}
\]
\[
m_i^k = \arg\max_j \text{softmax}(MLP_{m_k}(h(x_i))_{[j]}
\]
The set of MLPs $m_i^k$ for a word is determined based on its predicted coarse-grained
POS-tag.
In the coarse-grained tagger, each token $w_i$ is mapped to an input vector $x_i$ which encodes character
level information, distributional word-level information, possible morphological
analyses of $w_i$,\footnote{We find that providing the coarse-grained tagger with
information about possible fine-grained analyses of neighbouring words helps to disambiguate cases where a given word can be resolved as more than one POS. For instance, a given word may be resolvable as a noun or adjective; however, if the adjective possibility involves a feminine conjugation, and the preceding noun is a masculine noun, then the probability of the adjectival POS is severely reduced.} and lexicon-based features of $w_i$.
Specifically, $x_i$ is a concatenation of: (a) for a word $w_i$ made of
characters $c_1^{w_i},...,c_m^{w_i}$ the sum of
bi-LSTM states $\sum_j h(c_j^{w_i})$ from a char-level bi-LSTM that runs over the
entire sentence; (b) bi-LSTM state at $w_i$, for a word-level bi-LSTM that runs on pre-trained word2vec vectors for all of the words in the sentence; (c) a vector representing the possible fine-grained morphological analyses for the
word,\footnote{We assign trainable embeddings of 3-5 dimensions to each morphological category (gender, number, person, etc.), and we concatenate these together to form the input vector.} according to our wide-coverage lexicon; (d) bits indicating whether $w_i$
is in our comprehensive list of proper-nouns (names of streets, cities and people),
and whether it is in our wide-coverage lexicon at all (the latter is used to
mark rare and unknown words). In the fine-grained tagger, $x_i$ is a
concatenation of vector (b) above and: for a word $w_i$ where the predicted POS tag is $t_i$, and the possible fine-grained morphological analyses for $w_i$ limited by $t_i$ is represented by $m_i$, the bi-LSTM state for a bi-LSTM that runs on the concatenation of ($t_i$;$m_i$). Significantly, note that in the fine-grained tagger, $x_i$ does not include the information of the word form on the character-level. We find
this to be more accurate, because it removes bias in cases where a specific character form happens to appear in the training corpus in only one configuration. This is particularly relevant regarding verbs which can be resolved as either a masculine or a feminine verb, each with a distinct diacritization. In many cases, the training corpus contains the verb only in one stereotypical gender configuration. By hiding the character-level information, we force the system to make a more logical morphological determination, because it is not able to mechanically set the feature equal to what was seen in the training corpus.
\paragraph{Constraints} The tagger predictions are constrained by a wide-coverage lexicon that maps word forms to their possible morphological analyses.
When a word is not in the lexicon, we allow all POS-tags for the word. We also
apply additional filters to rule out POS-tags for words that participate in a
hand-crafted list of about 10K word collocations, and in all of their possible inflected forms (e.g., in the context of the
tokens (\<byt mrq.ht>) \textit{byt mrk\^ht}, the word \<byt> \textit{byt} should not be tagged as the absolute form \<b*ayit> \textit{bayit}, but rather as the construct form \<b*eyt> \textit{beyt}. And thus too for the plural inflection of the same collocation - \<bty mrq.ht> \textit{bty mrk\^ht}, the word \<bty> \textit{bty} should not be tagged as \<b*it*iy> \textit{byty} (feminine noun with possessive suffix), but rather as the plural-construct form \<b*At*ey> \text{batey}).
\paragraph{Filtering}
For each word $w_i$ in the text, we retrieve from our wordlists (see
\S\ref{wordlists}) a set of possible diacritizations $D_i = d^i_1,...,d^i_\ell$ and their corresponding
morphological analyses. This set is then further refined by intersecting it with
the predicted morphological analysis for the word. Words that are not in our
list get an empty set, indicating that their diacritization is not
constrained. This stage leaves us with an average of 1.2 diacritic sequences
for each known word. If we were to perform random selection from this list, we
would achieve 87.1\% exact-match word-level diacritization accuracy on our Modern Hebrew test corpus.
\paragraph{Diacritization Ranking}
Finally, we run an LSTM-based diacritization module to rank the possible
diacritization sequences from the previous stage, and to assign
diacritics to unknown words.
The LSTM-based module assigns a diacritic mark for each character in the
sequence.\footnote{Combinations of gemination with an additional diacritic mark are considered distinct diacritic symbols for prediction. An independent MLP predicts the position of the upper dot for the 'shin' character.}
The diacritics for each word $w_i$ are predicted separately, using beam-search over
the predictions of the diacritic for each letter with the word, to ensure word-level consistency. For known words, the
beam-search is constrained to valid diacritic predictions from the set $D_i$,
while for unknown words it is unconstrained.
Note that when predicting the diacritics for a letter $c^{w_i}_j$ in token $w_i$ the model is
aware of the other diacritic assignments in that word, but not of diacritic
assignments for the other words of the sentence. However, the model \emph{is}
context-aware, as it considers the character-level and word-level information
from the entire sentence via a sentence-level bi-LSTM layer.
To be more precise, each letter $c^{w_i}_j$ is mapped to a vector $h'(c^{w_i}_j)$ which is a concatenation of the followings two items: (a) bi-LSTM state at $c^{w_i}_j$ for a char-level bi-LSTM that runs over the entire sentence; (b) bi-LSTM state at $w_i$ for a word-level bi-LSTM that runs on the pre-trained word2vec vectors for all words in the sentence.
Then, for a given word $w_i$ we have a list of vectors representing each letter $h'(c^{w_i}_1)...h'(c^{w_i}_m)$. We then predict the diacritization sequence as follows. If this is a known word, then we have a list of $k$ possible diacritization sequences, and we choose the one with the highest score:
\[
s=\arg\max_k score(c^{w_i}_{1:m},t^{k}_{1:m}) \\[-2mm]
\]
where $t^k_{1:m}$ is the $k$th diacritic sequence, and $score(c^{w_i}_{1:m},t^{k}_{1:m})$ is calculated as:
\[
\sum_{j=1}^m MLP(h'(c_j^{w_i});LSTM(t^k_{0:j-1}))_{[t_j]} \\[-2mm]
\]
For unknown words, we run beam-search with $k=8$ to predict the $k$ most likely diacritization sequences, and we choose the top beam-ray.
\section{Evaluation}
We evaluate the system quantitatively against two commercial Hebrew diacritization
systems, Morfix\footnote{\url{https://nakdan.morfix.co.il/}} and Snopi\footnote{\url{http://www.nakdan.com/}}, considered state-of-the-art.
We also provide qualitative evaluation, demonstrating the ability to diacritize unknown words, and to produce context-sensitive
diacritization.
\paragraph{Quantitative Evaluation}
We use two quantitative measures to evaluate our model. (1) Word-level accuracy: for a given word\footnote{For this calculation, punctuation and non-Hebrew words or symbols are ignored.}, we consider the prediction correct if and only if all the diacritic marks on the word are correct, including gemination and the 'shin' dot, with all matres lectionis removed. (2) Character-level accuracy: For each Hebrew letter in the input text we check if the model predicted the correct set of diacritic marks for the letter (and, for matres lectionis, we check that the model predicted their removal).
We evaluated the system on a 6,000-word unseen gold-test corpus, manually diacritized by a professional linguist (Table \ref{tab:eval-results}). The corpus consists of a random selection of Hebrew wiki articles. We have made the test corpus publicly available.\footnote{The test corpus can be downloaded at this link: \url{http://tiny.cc/hebrew-test-git}}
\begin{table}[]
\centering
\scalebox{0.8}{
\begin{tabular}{|c|c|c|}
\hline
\multicolumn{1}{|l|}{} & \textbf{Letter Accuracy} & \textbf{Word Accuracy} \\ \hline
\textbf{Dicta} & \textbf{95.12\%} & \textbf{88.23\%} \\ \hline
\textbf{Morfix} & 90.32\% & 80.9\% \\ \hline
\textbf{Snopi} & 78.96\% & 66.41\% \\ \hline
\end{tabular}}
\caption{Accuracy on Modern Hebrew Test Corpus}
\label{tab:eval-results}
\end{table}
\paragraph{Qualitative Evaluation}
For the qualitative evaluation, we demonstrate that the system knows how to handle diacritization for unknown words, and this, in a context-sensitive manner.
For this example we choose an invalid word which conforms to Hebrew letter patterns but which does not actually exist in modern Hebrew:
\<srdynwt>. No such word exists in Hebrew dictionaries, nor in our wordlist.
We put the word into a sentence in two contexts - in the first, it fills the role of an adverb, and in the second, it fills the role of a noun. Hebrew diacritization norms would dictate two different diacritizations for these two usages: for the adverb, the final vowel should be \textit{'u'}, while for the noun, it should be \textit{'o'}. Our system handles both correctly (Figure \ref{fig:sardinot}).
\begin{figure}[h]
\includegraphics{nakdan-sardinot-chopped.jpg}
\caption{Diacritization of the fictional word \<srdynwt> in two different contexts, with two different prefixes; the word is diacritized as expected in both contexts.}
\label{fig:sardinot}
\end{figure}
\section{Additional Text Genres}
In addition to modern Hebrew, we also support Rabbinic Hebrew and poetic Hebrew. These genres require specialized handling. Firstly, we cannot use our modern Hebrew morphology model, because the morphological and syntactic norms of these genres differ from those of modern Hebrew. Secondly, we cannot use our modern Hebrew wordlist filters. There is no standardized orthography for Rabbinic Hebrew, nor for medieval poetic Hebrew. Additionally, poets often specifically choose less common words in order to meet prosodic constraints; thus, our rare-word filters are not relevant. Finally, many words which would be considered invalid in modern Hebrew are found within these other genres. Rabbinic Hebrew includes many Aramaic words, as well as Hebrew words with Aramaic prefixes. Poetic Hebrew includes oddities such as past-tense verbs with temporal prefixes.
For Rabbinic Hebrew, we train a specialized morphology model based on our tagged historical Hebrew corpus. For poetry, where morphological sequences are less constrained and less predictable, we skip the morphology layer and diacritize the text directly based on the diacritization LSTM.
In order to test our performance, we created test corpora for each of the genres. The poetry test corpus includes a set of liturgical poems of the 'yotzer' genre, transcribed from Cairo Genizah manuscripts.\footnote{Full data on these texts is available here: \url{http://weekdayyotzrot.com}} The Rabbinic Hebrew test corpus is taken from the 'Bet Yosef', a 16th century commentary on Jewish law.\footnote{Both test corpora are available for download here: \url{http://tiny.cc/hebrew-test-git}} In Tables \ref{tab:rab-results} and \ref{tab:poetry-results} we display our quantitative results on these two corpora.
\begin{table}[]
\centering
\scalebox{0.8}{
\begin{tabular}{|c|c|c|}
\hline
\multicolumn{1}{|l|}{} & \textbf{Letter Accuracy} & \textbf{Word Accuracy} \\ \hline
\textbf{Dicta} & \textbf{94.94\%} & \textbf{87.94\%} \\ \hline
\textbf{Morfix} & 80.25\% & 68.1\% \\ \hline
\textbf{Snopi} & 72.53\% & 58.39\% \\ \hline
\end{tabular}}
\caption{Accuracy on Rabbinic Hebrew Test Corpus}
\label{tab:rab-results}
\end{table}
\begin{table}[]
\centering
\scalebox{0.8}{
\begin{tabular}{|c|c|c|}
\hline
\multicolumn{1}{|l|}{} & \textbf{Letter Accuracy} & \textbf{Word Accuracy} \\ \hline
\textbf{Dicta} & \textbf{85.76\%} & \textbf{70.23\%} \\ \hline
\textbf{Morfix} & 80.9\% & 65.3\% \\ \hline
\textbf{Snopi} & 69.24\% & 52\% \\ \hline
\end{tabular}}
\caption{Accuracy on Poetic Hebrew Test Corpus}
\label{tab:poetry-results}
\end{table}
\section{Advanced Features}
\label{sec:advanced}
{\bf 1.} Scientific Editions: In scientific editions, editorial sigla are interspersed throughout the text. For instance, letters which are rubbed out in the textual witnesses will be supplied within brackets (\</s>[\<r>]\<md>). Existing diacritization tools fail here because they parse such sigla as word separators. Secondly, normative Hebrew diacritization entails the omission of matres lectionis, and indeed existing tools omit these letters when returning the diacritized text. However, in scientific editions, matres lectionis must be maintained in order to represent the manuscript evidence. Finally, the orthography of medieval Hebrew manuscripts can diverge wildly from modern norms; for example, we often find a yod inserted after the initial letter of a hitpael construction (e.g. \<hytlb/s>), a phenomenon which would never occur in a modern Hebrew text. Our tool meets all of these needs, and allows the user to either remove or maintain matres lectionis.
\noindent{\bf 2.} The web interface automatically highlights Biblical quotes within the Hebrew text. Biblical phrases are often incorporated into Hebrew texts, whether as explicit prooftexts or as rhetorical flourishes. We automatically identify such quotes, diacritize them according to the canonized diacritization of the Hebrew Bible, and display them in the distinctive Koren font (a font well-known for its use in modern Hebrew Bibles). See figure \ref{fig:advanced-features-2} for an example.
\begin{figure}[h]
\begin{center}
\includegraphics[width=0.5\textwidth]{nakdan-makor-chopped.png}
\end{center}
\caption{Integrated Biblical quote marked with font.}
\label{fig:advanced-features-2}
\end{figure}
\section{Conclusion}
We are pleased to release our Hebrew diacritization system for free unrestricted use. It is powered by a combination of advanced machine learning and manually curated linguistic resources, and thus succeeds in setting a new state of the art for Hebrew diacritization. We have released also our diacritized test corpora for benchmarking.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 9,198 |
{"url":"https:\/\/brilliant.org\/practice\/exponents-misconceptions\/","text":"Algebra\n\n# Exponents Misconceptions\n\n$\\big(x^2\\big)^3 =\\, ?$\n\nIf $x^5 = 3$, then $(2x)^5 =$\n\nIf $x = 5$, then $\\sqrt{x^2} =$\n\nWhich of the following is equal to $2^{-3}\\text{?}$\n\nFor which choices of $a$ and $b$ is the equation $(ab)^\\frac{1}{2} = a^\\frac{1}{2}\\cdot b^\\frac{1}{2}$\n\nnot true?\n\n\u00d7","date":"2020-12-04 02:44:10","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 23, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.9880099892616272, \"perplexity\": 410.21833626123447}, \"config\": {\"markdown_headings\": true, \"markdown_code\": false, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2020-50\/segments\/1606141733120.84\/warc\/CC-MAIN-20201204010410-20201204040410-00136.warc.gz\"}"} | null | null |
define({
"chartTasks": "Diagrammi toimingud",
"addNew": "Lisa uus",
"addNewTip": "Lisa uus diagrammi toiming",
"name": "Nimi",
"set": "Määra",
"setDataSource": "Määra andmeallikas",
"generalSettingsTip": "Saate määrata diagrammide andmeallikad, pealkirjad ja kirjeldused.",
"chartModeSettingTip": "Saate valida andmete analüüsimise ja diagrammina kuvamise viisi.",
"detailsSettingTip": "Saate muuta diagrammi tüüpi, värvi ja valida kuvatavad väärtused:",
"symbolSettingTip": "Saate määrata diagrammile kantud andmete kaardil esitamise viisi.",
"addFilterTip": "Saate lisada nendele andmetele filtri.",
"dataSource": "Andmeallikas",
"noResults": "Objekte ei leitud.",
"chartTitle": "Diagrammi pealkiri",
"description": "Kirjeldus",
"chartingMode": "Diagrammi kuva",
"featureOption": "Kuva väärtused objektide kaupa",
"categoryOption": "Kuva väärtused kategooriate kaupa",
"countOption": "Kuva objektide kogused kategooriate kaupa",
"fieldOption": "Kuva atribuutide väärtused diagrammidena",
"category": "Kategooria",
"axisLabel": "Kategooria märgis",
"categoryField": "Kategooria väli",
"operation": "Toiming",
"originalOption": "Kasuta algväärtusi",
"sumOption": "Summa",
"averageOption": "Keskmine",
"minOption": "Miinimum",
"maxOption": "Maksimaalne",
"valueFields": "Väärtuste väljad",
"moveUp": "Liiguta ülespoole",
"moveDown": "Liiguta allapoole",
"sortOrder": "Sordi diagrammid kategooria märgiste põhjal",
"ascending": "Kasvav",
"descending": "Kahanev",
"chartType": "Diagrammi tüüp",
"column": "Tulpdiagramm",
"bar": "Lintdiagramm",
"line": "Joondiagramm",
"pie": "Sektordiagramm",
"chartColor": "Diagrammi värv",
"showDataLabels": "Kuva andmete märgised",
"setSymbol": "Valiku sümbol",
"setHighLightColor": "Esiletõstmise värv",
"settings": "Seaded",
"mapInterationOptions": "Kaardi interaktsiooni valikud",
"preview": "Eelvaade",
"ok": "OK",
"cancel": "Tühista",
"close": "Sulge",
"setFilterTip": "Määrake filter õigesti. ",
"setChartTitleTip": "Määrake diagrammi pealkiri.",
"setChartTypeTip": "Valige vähemalt üks diagrammitüüp.",
"setDataFieldTip": "Valige vähemalt üks väärtuse väli.",
"setOutputFieldTip": "Määrake väljundi väljad.",
"count": "Kogus",
"label": "Märgis",
"horizontalAxis": "Horisontaaltelg",
"verticalAxis": "Vertikaaltelg",
"dataLabels": "Andmete märgised",
"color": "Värv",
"colorful": "Värviline",
"monochromatic": "Ühevärviline",
"noTaskTip": "Uue diagrammitoimingu lisamiseks klõpsake käsku Lisa uus",
"info": "Teave",
"charts": "Graafikud",
"options": "Seaded",
"taskName": "Ülesande nimi",
"summary": "Kokkuvõte",
"resultDescription": "Tulemuse kirjeldus",
"spatialFilterTip": "Valige vidinas tulemuste piiramiseks ruumiline filter (filtrid).",
"defaultOption": "Vaikimisi",
"useCurrentExtentTip": "Osaliselt või täielikult praeguses kaardi kuvaulatuses",
"useDrawGraphicTip": "Osaliselt või täielikult kaardile joonistatud kujundis",
"useFeaturesTip": "Vastavalt nende asukohale teise kihi objektide suhtes",
"noSpatialLimitTip": "Ära piira tulemusi ruumiliselt",
"drawingTools": "Joonistamise töövahendid",
"bufferSettings": "Puhvri seaded",
"defaultDistance": "Vaike-vahemaa",
"bufferDistance": "Puhvri ulatus",
"defaultUnit": "Vaikeühik",
"miles": "Miilid",
"kilometers": "Kilomeetrid",
"feet": "Jalad",
"meters": "Meetrid",
"yards": "Jardid",
"nauticalMiles": "Meremiilid",
"availableSpatialRelationships": "Saadaolevad ruumilised seosed",
"setSpatialRelationships": "Määra ruumilised seosed",
"availability": "Saadavus",
"relationship": "Seos",
"actions": "Toimingud",
"intersect": "lõikuvad",
"contain": "sisaldavad",
"areContainedIn": "sisalduvad",
"cross": "rist",
"envelopeIntersect": "lõikuvad ümbrisega",
"overlap": "kattuvad",
"areOverlapped": "kaetakse",
"touche": "puutuvad kokku",
"within": "vahemikus",
"areWithin": "on vahemikus",
"indexIntersect": "lõikuvad indeksiga"
}); | {
"redpajama_set_name": "RedPajamaGithub"
} | 8,574 |
Carriera
Club
Giocò sempre nel campionato uruguaiano.
Nazionale
Con la maglia della Nazionale ha collezionato nove presenze.
Note
Collegamenti esterni
Calciatori della Nazionale uruguaiana | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 1,261 |
La Souche és un municipi francès situat al departament de l'Ardecha i a la regió d'Alvèrnia-Roine-Alps. L'any 2007 tenia 332 habitants.
Demografia
Població
El 2007 la població de fet de La Souche era de 332 persones. Hi havia 159 famílies de les quals 56 eren unipersonals (32 homes vivint sols i 24 dones vivint soles), 63 parelles sense fills i 40 parelles amb fills.
La població ha evolucionat segons el següent gràfic:
Habitants censats
Habitatges
El 2007 hi havia 437 habitatges, 160 eren l'habitatge principal de la família, 225 eren segones residències i 53 estaven desocupats. 422 eren cases i 13 eren apartaments. Dels 160 habitatges principals, 127 estaven ocupats pels seus propietaris, 26 estaven llogats i ocupats pels llogaters i 7 estaven cedits a títol gratuït; 2 tenien una cambra, 14 en tenien dues, 36 en tenien tres, 37 en tenien quatre i 71 en tenien cinc o més. 117 habitatges disposaven pel capbaix d'una plaça de pàrquing. A 79 habitatges hi havia un automòbil i a 68 n'hi havia dos o més.
Piràmide de població
La piràmide de població per edats i sexe el 2009 era:
Economia
El 2007 la població en edat de treballar era de 185 persones, 114 eren actives i 71 eren inactives. De les 114 persones actives 102 estaven ocupades (53 homes i 49 dones) i 14 estaven aturades (3 homes i 11 dones). De les 71 persones inactives 36 estaven jubilades, 13 estaven estudiant i 22 estaven classificades com a «altres inactius».
Ingressos
El 2009 a La Souche hi havia 178 unitats fiscals que integraven 378 persones, la mediana anual d'ingressos fiscals per persona era de 15.198 €.
Activitats econòmiques
Dels 13 establiments que hi havia el 2007, 1 era d'una empresa alimentària, 1 d'una empresa de fabricació d'altres productes industrials, 2 d'empreses de construcció, 2 d'empreses de comerç i reparació d'automòbils, 1 d'una empresa de transport, 2 d'empreses d'hostatgeria i restauració, 1 d'una empresa d'informació i comunicació, 1 d'una empresa immobiliària, 1 d'una empresa de serveis i 1 d'una empresa classificada com a «altres activitats de serveis».
Dels 5 establiments de servei als particulars que hi havia el 2009, 1 era una oficina de correu, 1 funerària, 1 paleta, 1 electricista i 1 restaurant.
L'únic establiment comercial que hi havia el 2009 era una fleca.
L'any 2000 a La Souche hi havia 19 explotacions agrícoles.
Equipaments sanitaris i escolars
El 2009 hi havia una escola elemental.
Poblacions més properes
El següent diagrama mostra les poblacions més properes.
Referències
Résumé statistique Fitxa resum de dades estadístiques de La Souche a l'INSEE.
Évolution et structure de la population Fitxa amb el detall de dades de La Souche a l'INSEE
France par commune Dades detallades de tots els municipis de França accessibles a partir del mapa.
Municipis de l'Ardecha | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 7,354 |
St Stephen's Church is in Balcarres Avenue, Whelley, Wigan, Greater Manchester, England. It is an active Anglican parish church in the deanery of Wigan, the archdeaconry of Wigan and West Lancashire, and the diocese of Liverpool. Its benefice is united with that of St John, New Springs. The church is recorded in the National Heritage List for England as a designated Grade II listed building.
History
The first phase of the church was built between 1928 and 1930, the foundation stone being laid in November 1928, and the church being consecrated on 9 April 1930. It was designed by the Lancaster architect Henry Paley of Austin and Paley, and the church was built on land given by Lord Crawford. The first phase consisted of the east end of the church, and the first two bays of the nave and the aisles: this cost £9,863 (equivalent to £ in ). The church was completed in 1937–38, and a choir vestry was added, the cost of these additions being £5,253.
Architecture
St Stephen's is constructed in red and brown sandstone with green slate roofs, and is in Free Perpendicular style. Its plan consists of a nave with a clerestory and a south porch, north and south aisles, a chancel with a canted east end, a south vestry, and a single gabled bellcote standing at right angles to the south side of the chancel. At the west end are broad buttresses, with a canted baptistry between them. The west window has four lights, the east window has five lights, and the clerestory windows have three lights. Pollard and Pevsner in the Buildings of England series comment that it is "an odd time" for the architect to be continuing to use the style of the practice during the 1880s. The two-manual pipe organ, the third to be installed in the church, was made in 1964 by J. W. Walker & Sons Ltd. In 2011 the pipe organ was replaced with an electronic imitation produced by Johannus of Holland.
See also
List of churches in Greater Manchester
Listed buildings in Wigan
List of ecclesiastical works by Austin and Paley (1916–44)
Bibliography
Notes
References
Church of England church buildings in Greater Manchester
Grade II listed churches in the Metropolitan Borough of Wigan
Churches completed in 1938
20th-century Church of England church buildings
Gothic Revival church buildings in Greater Manchester
Austin and Paley buildings
Anglican Diocese of Liverpool | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 7,150 |
Q: Why a segmentation fault for changing a non-const char*? With this code, I get a segmentation fault:
char* inputStr = "abcde";
*(inputStr+1)='f';
If the code was:
const char* inputStr = "abcde";
*(inputStr+1)='f';
I will get compile error for "assigning read-only location".
However, for the first case, there is no compile error; just the segmentation fault when the assign operation actually happened.
Can anyone explain this?
A: Here is what the standard says about string literals in section [2.13.4/2]:
A string literal that does not begin with u, U, or L is an ordinary string literal, also referred to as a narrow string literal. An ordinary string literal has type "array of n const char", where n is the size of the string as defined below; it has static storage duration (3.7) and is initialized with the given characters.
So, strictly speaking, "abcde" has type
const char[6]
Now what happens in your code is an implicit cast to
char*
so that the assignment is allowed. The reason why it is so is, likely, compatibility with C. Have a look also at the discussion here: http://learningcppisfun.blogspot.com/2009/07/string-literals-in-c.html
Once the cast is done, you are syntactically free to modify the literal, but it fails because the compiler stores the literal in a non writable segment of memory, as the standard itself allow.
A: This gets created in the code segment:
char *a = "abcde";
Essentially it's const.
If you wish to edit it, try:
char a[] = "abcde";
A: The standard states that you are not allowed to modify string literals directly, regardless of whether you mark them const or not:
Whether all string literals are distinct (that is, are stored in nonoverlapping
objects) is implementation-defined. The effect of attempting to modify a string literal is undefined.
In fact, in C (unlike C++), string literals are not const but you're still not allowed to write to them.
This restriction on writing allows certain optimisations to take place, such as sharing of literals along the lines of:
char *ermsg = "invalid option";
char *okmsg = "valid option";
where okmsg can actually point to the 'v' character in ermsg, rather than being a distinct string.
A: String literals are typically stored in read-only memory. Trying to change this memory will kill your program.
Here's a good explanation: Is a string literal in c++ created in static memory?
A: It is mostly ancient history; once upon a long time ago, string literals were not constant.
However, most modern compilers place string literals into read-only memory (typically, the text segment of your program, where your code also lives), and any attempt to change a string literal will yield a core dump or equivalent.
With G++, you can most certainly get the compilation warning (-Wall if it is not enabled by default). For example, G++ 4.6.0 compiled on MacOS X 10.6.7 (but running on 10.7) yields:
$ cat xx.cpp
int main()
{
char* inputStr = "abcde";
*(inputStr+1)='f';
}
$ g++ -c xx.cpp
xx.cpp: In function 'int main()':
xx.cpp:3:22: warning: deprecated conversion from string constant to 'char*' [-Wwrite-strings]
$
So the warning is enabled by default.
A: What happened is that the compiler put the constant "abcde" in some read-only memory segment. You pointed your (non-const) char* inputStr at that constant, and kaboom, segfault.
Lesson to be learned: Don't invoke undefined behavior.
Edit (elaboration)
However, for the first case, there is no compile error, just segmentation fault when the assign operation actually happened.
You need to enabled your compiler warnings. Always set your compiler warnings as high as possible.
A: Even though "abcde" is a string literal, which should not be modified, you've told the compiler that you don't care about that by having a non-const char* point to it.
The compiler will happily assume that you know what you're doing, and not throw an error. However, there's a good chance that the code will fail at runtime when you do indeed try to modify the string literal.
A: String literals, while officially non-const, are almost always stored in read-only memory. In your setup, this is apparently only the case if it is declared as const char array.
Note that the standard forbids you to modify any string literal.
A: a little bit of history of string literals in Ritchie's words.
mostly about the orgin and the evolution of string literals from K&R 1.
Hope this might clarify a thing or two about const and string literals.
"From: Dennis Ritchie
Subject: Re: History question: String literals.
Date: 02 Jun 1998
Newsgroups: comp.std.c
At the time that the C89 committee was working, writable
string literals weren't "legacy code" (Margolin) and what standard
there existed (K&R 1) was quite explicit (A.2.5) that
strings were just a way of initializing a static array.
And as Barry pointed out there were some (mktemp) routines
that used this fact.
I wasn't around for the committee's deliberations on the
point, but I suspect that the BSD utility for fiddling
the assembler code to move the initialization of strings
to text instead of data, and the realization that most
literal strings were not in fact overwritten, was more
important than some very early version of gcc.
Where I think the committee might have missed something
is in failure to find a formulation that explained
the behavior of string literals in terms of const.
That is, if "abc" is an anonymous literal of type
const char [4]
then just about all of its properties (including the
ability to make read-only, and even to share its storage
with other occurrences of the same literal) are nearly
explained.
The problem with this was not only the relatively few
places that string literals were actually written on, but much
more important, working out feasible rules for assignments
to pointers-to-const, in particular for function's actual
arguments. Realistically the committee knew that whatever
rules they formulated could not require a mandatory
diagnostic for every func("string") in the existing world.
So they decided to leave "..." of ordinary char array
type, but say one was required not to write over it.
This note, BTW, isn't intended to be read as a snipe
at the formulation in C89. It is very hard to get things
both right (coherent and correct) and usable (consistent
enough, attractive enough).
Dennis
"
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 7,161 |
\section{Introduction}
The eddy parameterization problem (how to account for the effect of unresolved small scales onto the resolved large scales)
is one of the most challenging problems in the ocean modelling, counting decades of active research.
Despite a number of parameterizations (computationally affordable and physically justified mathematical models of
unresolved processes) have been proposed to solve the problem
(e.g.,~\citet{GentMcwilliams1990,DuanNadiga2007,Frederiksen_et_al2012,
PortaMana_Zanna2014,CooperZanna2015,
Grooms_et_al2015,Berloff_2015,Berloff_2016,Berloff_2018,Ryzhov_etal_2019,
CCHWS2019_1,Ryzhov_etal_2020,
CCHWS2019_3,CCHWS2020_4,CCHPS2020_J2}), it remains largely unresolved.
The main general point is that most of the parameterization approaches are physics-based rather than data-driven, and the former has obvious advantage of being valid in situation when the underlying physics changes.
However, in the situation when physics-based parameterizations remain in their infancy, there is a great practicality in considering data-driven parameterizations,
which can reproduce nominally-resolved flow structures within their obvious limitations.
For example, for many research questions a computationally-cheap data-driven solution can replace a computationally-expensive dynamical ocean simulation in climate-type models and predictions.
Advancing the hyper-parameterization approach, within broader context of data-driven parameterizations, provides the main motivation for our study, in which we continue the series of precursor works~\citep{SB2021_J1,SB2022_J1,SB2022_J2}.
The main novelty is to adapt the methodology for fully comprehensive and realistic ocean models, and demonstrate its success with full confidence.
In turn this paves the way for broad use of hyper-parameterizations across the ocean modelling community.
\section{The method}
In this work, we consider the method called ``Advection of the image point'';
it is called so, because the image point is advected by the flow in phase space.
The method falls into the category of data-driven methods, and has been successfully tested on
a multi-layer quasi-geostrophic model and showed promising results~\citep{SB2021_J1}.
The method is based on the fact that the first-order ordinary differential equation
\begin{equation}
\mathbf{x}'(t)=\mathbf{F}(\mathbf{x}),\quad \mathbf{x}\in\mathbb{R}^n
\label{eq:ode1}
\end{equation}
can be interpreted as a vector field $\mathbf{F}(\mathbf{x})$ in the phase space of equation~\eqref{eq:ode1}.
If $\mathbf{F}(\mathbf{x})$ is known, it can be used to advect the image point $\mathbf{x}$ in the phase space. Evolution of
an image point can be described by the equation:
\begin{equation}
\mathbf{y}'(t)=\frac{1}{N}\sum\limits_{i\in\mathcal{U}(\mathbf{y}(t))}\mathbf{F}(\mathbf{x}(t_i))+
\eta\left(\frac{1}{M}\sum\limits_{i\in\mathcal{U}(\mathbf{y}(t))}\mathbf{x}(t_i)-\mathbf{y}(t)\right),\quad
\mathbf{y}(t_0)=\mathbf{x}(t_0)\, ,
\label{eq:evolution_eq_no_nudging}
\end{equation}
where $\mathcal{U}(\mathbf{y}(t))$ is a neighbourhood of solution $\mathbf{y}(t)$,
and $i$ is the timestep of the reference solution $\mathbf{x}(t_i)$. The neighbourhood is computed as the average over
$N$ (and $M$ for the nudging term) nearest, in $l_2$ norm, to the solution $\mathbf{y}(t)$ points, and $\eta$ is the nudging strength;
we will return to the choice of $N$, $M$, and $\eta$ in the next section; we refer to these parameters as hyper-parameters.
The hyper-parameters can be set based on the chosen metric
and available data. Getting a bit ahead, we report that in our case the neighbourhood consists of $N=15$ and $M=5$ points, and the nudging strength is $\eta=0.001$.
Originally, the method was supposed to have the same size of the neighborhood for both the vector field $\mathbf{F}(\mathbf{x})$ and the nudging term (i.e., $N=M$). However,
our experiments showed that using neighborhoods of different sizes can prevent the so-called build-up effect which we discuss later.
We refer the reader to~\citep{SB2021_J1} for a more detailed discussion of the method.
\section{Model configuration and numerical results}
For the purpose of this study, we consider the Massachusetts Institute of Technology general circulation model (MITgcm)~\citep{Marshall_etal_1997} in the North Atlantic configuration~\citep{SB2022_J1}.
It is a 46-layer oceanic model coupled with an atmospheric boundary model~\citep{Marshall_etal_1997,Deremble_et_al2013}.
The coupled model is initially spun up for 5 years and then integrated for another 2 years. Although,
for this work it is not essential that the initial state of
the ocean circulation is in the statistically equilibrated regime.
The model is integrated at two different horizontal resolutions ($1/12^{\circ}$ and $1/3^{\circ}$);
the oceanic and atmospheric models are implemented with the same horizontal resolution.
We refer to the $1/12^{\circ}$-solution projected onto the $1/3^{\circ}$-grid (Figure~\ref{fig:rv_12_3}a)
as the reference solution, and to the solution computed on the $1/3^{\circ}$-grid as the modelled solution (Figure~\ref{fig:rv_12_3}c).
The hyper-parameterized solution computed with equation~\eqref{eq:evolution_eq_no_nudging} is presented in Figure~\ref{fig:rv_12_3}b.
\begin{figure}[H]
\hspace*{-3.25cm}
\begin{tabular}{cccc}
& \hspace*{0.5cm}\begin{minipage}{0.33\textwidth} \hfill{$t=1$ year} \end{minipage} &
\hspace*{-0.125cm}\begin{minipage}{0.33\textwidth} \hfill{$t=2$ years} \end{minipage} &
\hspace*{-0.125cm}\begin{minipage}{0.33\textwidth} \hfill{2-year average} \end{minipage}
\end{tabular}
\hspace*{2cm}
\begin{tabular}{cccc}
\hspace*{-5.75cm}\begin{minipage}{0.02\textwidth}\rotatebox{90}{\bf (a)}\end{minipage} &
\hspace*{-3cm}\begin{minipage}{0.24\textwidth}\includegraphics[scale=0.45]{RVref_11.jpg}\end{minipage} &
\hspace*{1.25cm}\begin{minipage}{0.24\textwidth}\includegraphics[scale=0.45]{RVref_21.jpg}\end{minipage} &
\hspace*{1.3cm}\begin{minipage}{0.24\textwidth}\includegraphics[scale=0.45]{RVref_31.jpg}\end{minipage}\\
& & & \\[-0.35cm]
\hspace*{-5.75cm}\begin{minipage}{0.02\textwidth}\rotatebox{90}{\bf (b)}\end{minipage} &
\hspace*{-3cm}\begin{minipage}{0.24\textwidth}\includegraphics[scale=0.45]{RVbar_13.jpg}\end{minipage} &
\hspace*{1.25cm}\begin{minipage}{0.24\textwidth}\includegraphics[scale=0.45]{RVbar_23.jpg}\end{minipage} &
\hspace*{1.3cm}\begin{minipage}{0.24\textwidth}\includegraphics[scale=0.45]{RVbar_33.jpg}\end{minipage}\\
& & & \\[-0.35cm]
\hspace*{-5.75cm}\begin{minipage}{0.02\textwidth}\rotatebox{90}{\bf (c)}\end{minipage} &
\hspace*{-3cm}\begin{minipage}{0.24\textwidth}\includegraphics[scale=0.45]{RVmodelled_12.jpg}\end{minipage} &
\hspace*{1.25cm}\begin{minipage}{0.24\textwidth}\includegraphics[scale=0.45]{RVmodelled_22.jpg}\end{minipage} &
\hspace*{1.3cm}\begin{minipage}{0.24\textwidth}\includegraphics[scale=0.45]{RVmodelled_32.jpg}\end{minipage}\\
& & & \\[-0.35cm]
\multicolumn{4}{c}{\hspace*{-0.95cm}\includegraphics[width=6cm,height=0.75cm]{colorbar_bwr.jpg}}\\
\end{tabular}
\caption{Shown are snapshots of
the surface relative vorticity $\zeta=v_x-u_y$ {\rm[1/s]} for {\bf (a)} the reference solution (computed
at horizontal resolution $1/12^{\circ}$ and then projected on the $1/3^{\circ}$-grid),
{\bf (b)} hyper-parameterized solution computed at horizontal resolution $1/3^{\circ}$ for $N=15$, $M=5$, $\eta=0.001$,
{\bf (c)} modelled solution computed at horizontal resolution $1/3^{\circ}$,
and the 2-year time-average (last column).
Snapshots are taken after 1 year (left column) and 2 years (middle column) of simulations.
Note the modelled solution {\bf (c)} fails to reproduce important large-scale
(the Gulf Stream eastward jet extension) and small-scale (vortices) structures of the flow dynamics in both
instantaneous and time-averaged fields.}
\label{fig:rv_12_3}
\end{figure}
As it follows from the results in Figure~\ref{fig:rv_12_3}, the hyper-parameterized solution computed on the coarse grid
(Figure~\ref{fig:rv_12_3}b) reproduces both the large scales (the Gulf Stream flow) and small scales (vortices) of the reference solution
(Figure~\ref{fig:rv_12_3}a) in
instantaneous and time-averaged fields, while the solution computed on the coarse grid without the hyper-parameterization
(Figure~\ref{fig:rv_12_3}c) leads to no Gulf Stream or small-scale flow features.
We remark that the hyper-parameterized solution is a part of a 2-year simulation from which we use only
the first year of the reference solution; over the second year model~\eqref{eq:evolution_eq_no_nudging} runs on its own.
The hyper-parameterized solution with a bad choice of hyper-parameters $N$, $M$, $\eta$ (Figure~\ref{fig:rv_12_3_buildup})
still reproduces both large- and small-scales flow features but only over the period in which the reference solution is available (it is one year in our case).
After that the solution almost stops evolving and eventually settles to a constant in time field like the one
in the middle plot of Figure~\ref{fig:rv_12_3_buildup}. It can be seen from
the time-mean over the second year (the right subplot in Figure~\ref{fig:rv_12_3_buildup}), which is very similar to the snapshot
taken at year 2 (middle subplot in Figure~\ref{fig:rv_12_3_buildup}), thus showing that the solution evolution is stalled over the second year.
\begin{figure}[H]
\hspace*{-2.5cm}
\begin{tabular}{ccc}
\multicolumn{3}{c}{\hspace*{2.5cm} \bf The build-up effect for the hyper-parameterized solution}\\
& & \\[-0.35cm]
\hspace*{0.5cm}\begin{minipage}{0.33\textwidth} \hfill{$t=1$ year} \end{minipage} &
\hspace*{-0.125cm}\begin{minipage}{0.33\textwidth} \hfill{$t=2$ years} \end{minipage} &
\hspace*{-0.125cm}\begin{minipage}{0.43\textwidth} \hspace*{1.5cm} Time-average over the second year \end{minipage}
\end{tabular}
\hspace*{2.75cm}
\begin{tabular}{ccc}
\hspace*{-3.05cm}\begin{minipage}{0.24\textwidth}\includegraphics[scale=0.45]{RVbar_buildup1_1460.jpg}\end{minipage} &
\hspace*{1.25cm}\begin{minipage}{0.24\textwidth}\includegraphics[scale=0.45]{RVbar_buildup1_2915.jpg}\end{minipage} &
\hspace*{1.25cm}\begin{minipage}{0.24\textwidth}\includegraphics[scale=0.45]{RVbar_buildup1_time_av_second_half.jpg}\end{minipage}\\
& & \\[-0.35cm]
\multicolumn{3}{c}{\hspace*{-1.35cm}\includegraphics[width=6cm,height=0.75cm]{colorbar_bwr.jpg}}\\
\end{tabular}
\caption{Shown are snapshots of the surface relative vorticity $\zeta=v_x-u_y$ {\rm[1/s]} demonstrating the build-up effect
for the hyper-parameterized solution computed at horizontal resolution $1/3^{\circ}$
for $N=M=5$, $\eta=0.001$.
Snapshots are taken after 1 year (left column) and 2 years (middle column) of simulations.}
\label{fig:rv_12_3_buildup}
\end{figure}
{\bf The build-up effect and solution degradation}.
The lack of reference data for the hyper-parameterized model and/or bad choice
of hyper-parameters can steer the trajectory to leave the reference phase space (the phase space occupied by the reference solution).
This escape may result in a significant degradation of the hyper-parameterized solution and even lead to a numerical blow-up. We refer to this as the build-up effect, meaning
that after a period of time, say $T$, the neighbourhood of the nearest points stalls (Figure~\ref{fig:rv_12_3_buildup}), i.e. the points in the neighbourhood become
the closest ones to the image point $\mathbf{y}(t)$ for $\forall t>T$ (in the present case $T=1$ year); therefore, the same points are used again and again
during the integration
of equation~\eqref{eq:evolution_eq_no_nudging} thus driving the image point away from the reference phase space.
In principle, building up numerical errors may terminate this runaway, and the solution can return back to the reference phase space region, but this is case dependent and should be kept in mind.
However, if the return time is relatively long (longer than the characteristic time of the reference solution)
then the flow dynamics can be seriously distorted over the period of the trajectory injection.
In order to prevent the build-up effect, one should properly set up the hyper-parameters $N$, $M$, and $\eta$.
After some experiments we have found that $N=15$, $M=5$, and $\eta=0.001$ lead to no build-up effect (it is not necessarily the only choice, and other sets of hyper-parameters providing no build-up may exist).
As an alternative, one can also change the algorithm on how to pickup points from the neighbourhood.
It is worth noting that these experiments are very fast and computationally cheap, even relative to a single run of the coarse-grid model.
We leave for the future work any further optimization of the image point advection algorithm.
{\bf The measure of goodness and evolution in phase space}.
The measure of goodness (i.e., the proximity of the modelled or parameterized solution to the reference one) in a given metric depends on the specific purpose.
Our measure of goodness is how close the hyper-parameterized solution
is to the reference phase space.
We use this measure to allow the hyper-parameterized solution to evolve in the neighborhood of the reference phase space, since
the failure of the coarse-grid model (blue dots in Figure~\ref{fig:phase_space}) to reproduce large- and small-scale features of the flow dynamics is because it steers away from where it should be
(black dots in Figure~\ref{fig:phase_space}, i.e. the reference phase space).
Measuring the proximity of individual trajectories in phase space is of no value,
because small initial perturbations will grow exponentially due to the inherently chaotic dynamics.
\begin{figure}[h]
\hspace*{-0.7cm}
\begin{tabular}{cc}
\hspace*{-0.5cm}\begin{minipage}{0.1\textwidth} {\bf (a)} \end{minipage} & \hspace*{-0.5cm}\begin{minipage}{0.1\textwidth} {\bf (b)} \end{minipage}\\
\begin{minipage}{0.5\textwidth}\includegraphics[scale=0.175]{phase_space.jpg}\end{minipage} &
\hspace*{-1cm}\begin{minipage}{0.5\textwidth}\includegraphics[scale=0.165]{phase_space_buildup_zoomed.jpg}\end{minipage}\\
\end{tabular}
\caption{Shown is {\bf (a)} 3D projection of the reference phase space (black dots), modelled phase space (blue dots), hyper-parameterized
solution (red trajectory); in other words, we plot 3 random coordinates from the multi-dimensional phase space of surface relative vorticity,
{\bf (b)} build-up effect for the second year of the hyper-parameterized solution (short green line near the red circle)
for $N=M=5$, and $\eta=0.001$; the green circles are those stalled five points in the neighborhood.
The time-means for every solution are denoted by bold circles of corresponding colours.}
\label{fig:phase_space}
\end{figure}
The build-up effect is demonstrated in Figure~\ref{fig:phase_space}b;
the short green line near the time-mean over the second year (red circle) is the 1-year evolution of the hyper-parameterized solution
affected by the bad choice of hyper-parameters, namely $N=M=5$, and $\eta=0.001$.
Evolution of the Euclidean distance from the reference time mean (the time-mean of the reference solution)
to the hyper-parameterized solution (Figure~\ref{fig:dist_to_time_mean}) shows that the hyper-parameterized solution
affected by the build-up effect (green line) stops to evolve after one year (the period over which the reference solution is available).
In this case, we observe no numerical blow-up as the solution quickly settles to a constant in time field.
When the build-up effect is prevented, the hyper-parameterized solution (red line) continues to evolve with the reference phase space.
\begin{figure}[h]
\centering
\includegraphics[scale=0.175]{dist_to_time_mean2.jpg}
\caption{Shown is evolution of the Euclidean distance (vertical axis) from the reference two-year time-mean
to the hyper-parameterized solution (red line)
and to the hyper-parameterized solution affected by the build-up effect (green line).
The horizontal axis shows the simulation time in time steps (6 hours in our case).}
\label{fig:dist_to_time_mean}
\end{figure}
\clearpage
{\bf Coupled fields}. Comprehensive ocean models have several prognostic fields (velocity, temperature, etc.), and a good choice of
hyper-parameters ($N$, $M$, and $\eta$) for one field is not necessarily applicable to another.
For example, $N=15$, $M=5$, and $\eta=0.001$ choice works well only for the relative vorticity. However, it leads to a build up for the coupled fields (relative vorticity and temperature, not shown).
Thus, a set of hyper-parameters for a coupled case has to be found separately.
Namely, for the coupled fields (relative vorticity and temperature) we have found that $N=18$ and $M=4$ remove
the build up effect, hence, the hyper-parameterized solution is of high quality (Figure~\ref{fig:rv_T_3}).
The nudging strength in this case remains unchnged from the single relative vorticity case (Figure~\ref{fig:rv_12_3}), $\eta=0.001$.
\begin{figure}[H]
\hspace*{-3.25cm}
\begin{tabular}{cccc}
\multicolumn{4}{c}{\hspace*{4.25cm} \bf Surface relative vorticity}\\
& \hspace*{0.5cm}\begin{minipage}{0.33\textwidth} \hfill{$t=1$ year} \end{minipage} &
\hspace*{-0.125cm}\begin{minipage}{0.33\textwidth} \hfill{$t=2$ years} \end{minipage} &
\hspace*{-0.125cm}\begin{minipage}{0.33\textwidth} \hfill{2-year average} \end{minipage}
\end{tabular}
\hspace*{2cm}
\begin{tabular}{cccc}
\hspace*{-5.75cm}\begin{minipage}{0.02\textwidth}\rotatebox{90}{\bf (a)}\end{minipage} &
\hspace*{-3cm}\begin{minipage}{0.24\textwidth}\includegraphics[scale=0.45]{RVref_11.jpg}\end{minipage} &
\hspace*{1.25cm}\begin{minipage}{0.24\textwidth}\includegraphics[scale=0.45]{RVref_21.jpg}\end{minipage} &
\hspace*{1.3cm}\begin{minipage}{0.24\textwidth}\includegraphics[scale=0.45]{RVref_31.jpg}\end{minipage}\\
& & & \\[-0.35cm]
\hspace*{-5.75cm}\begin{minipage}{0.02\textwidth}\rotatebox{90}{\bf (b)}\end{minipage} &
\hspace*{-3cm}\begin{minipage}{0.24\textwidth}\includegraphics[scale=0.45]{RVbar_withTbar_13.jpg}\end{minipage} &
\hspace*{1.25cm}\begin{minipage}{0.24\textwidth}\includegraphics[scale=0.45]{RVbar_withTbar_23.jpg}\end{minipage} &
\hspace*{1.3cm}\begin{minipage}{0.24\textwidth}\includegraphics[scale=0.45]{RVbar_withTbar_33.jpg}\end{minipage}\\
& & & \\[-0.35cm]
\multicolumn{4}{c}{\hspace*{-0.95cm}\includegraphics[width=6cm,height=0.75cm]{colorbar_bwr.jpg}}\\
\end{tabular}
\vspace*{0.25cm}
\hspace*{-3.25cm}
\begin{tabular}{cccc}
\multicolumn{4}{c}{\hspace*{4.25cm} \bf Sea surface temperature}\\
& \hspace*{0.5cm}\begin{minipage}{0.33\textwidth} \hfill{$t=1$ year} \end{minipage} &
\hspace*{-0.125cm}\begin{minipage}{0.33\textwidth} \hfill{$t=2$ years} \end{minipage} &
\hspace*{-0.125cm}\begin{minipage}{0.33\textwidth} \hfill{2-year average} \end{minipage}
\end{tabular}
\hspace*{2cm}
\begin{tabular}{cccc}
\hspace*{-5.75cm}\begin{minipage}{0.02\textwidth}\rotatebox{90}{\bf (a)}\end{minipage} &
\hspace*{-3cm}\begin{minipage}{0.24\textwidth}\includegraphics[scale=0.45]{Tref_11.jpg}\end{minipage} &
\hspace*{1.25cm}\begin{minipage}{0.24\textwidth}\includegraphics[scale=0.45]{Tref_21.jpg}\end{minipage} &
\hspace*{1.3cm}\begin{minipage}{0.24\textwidth}\includegraphics[scale=0.45]{Tref_31.jpg}\end{minipage}\\
& & & \\[-0.35cm]
\hspace*{-5.75cm}\begin{minipage}{0.02\textwidth}\rotatebox{90}{\bf (b)}\end{minipage} &
\hspace*{-3cm}\begin{minipage}{0.24\textwidth}\includegraphics[scale=0.45]{Tbar_withRVbar_13.jpg}\end{minipage} &
\hspace*{1.25cm}\begin{minipage}{0.24\textwidth}\includegraphics[scale=0.45]{Tbar_withRVbar_23.jpg}\end{minipage} &
\hspace*{1.3cm}\begin{minipage}{0.24\textwidth}\includegraphics[scale=0.45]{Tbar_withRVbar_33.jpg}\end{minipage}\\
& & & \\[-0.35cm]
\multicolumn{4}{c}{\hspace*{-0.75cm}\includegraphics[width=6cm,height=0.75cm]{colorbar_temperature.jpg}}\\
\end{tabular}
\caption{Shown are snapshots of the coupled fields
(sea surface relative vorticity $\zeta=v_x-u_y$ {\rm[1/s]} and surface temperature [$^\circ{\rm C}$])
for {\bf (a)} the reference solution (computed at horizontal resolution $1/12^{\circ}$ and then projected on the $1/3^{\circ}$-grid),
{\bf (b)} hyper-parameterized solution computed at horizontal resolution $1/3^{\circ}$ for $N=18$, $M=4$, $\eta=0.001$,
and the 2-year time-average (last column).
Snapshots are taken after 1 year (left column) and 2 years (middle column) of simulations.
Note that the hyper-parameterized solution {\bf (b)}
reproduces both large- (the Gulf Stream) and small-scale (vortices) features of the flow dynamics in both fields.}
\label{fig:rv_T_3}
\end{figure}
This example of the coupled fields demonstrates that one should exercise caution when it comes to setting up the hyper-parameters.
On the other hand, it also shows that the hyper-parameterization method works for coupled fields (even with huge differences between the fields, which is 7 orders of magnitude for the relative
vorticity and temperature, see colorbars in Figure~\ref{fig:rv_T_3}).
\section{Conclusions and discussion\label{sec:conclusions}}
In this work we have applied the hyper-parameterization method ``Advection of the image point'' to
the Massachusetts Institute of Technology general circulation model in the North Atlantic configuration and, thus, tested the method in a significantly
more complicated setting, as compared to the earlier idealized-model tests. Our results show that the hyper-parameterization
method significantly improves a non-eddy-resolving solution (on a $1/3^{\circ}$-grid) towards the reference eddy-resolving solution
(on a $1/12^{\circ}$-grid and then projected onto the $1/3^{\circ}$-grid) for both single and coupled fields (even with large difference between
the fields, it is 7 orders of magnitude in our case) by reproducing both the large-scale (the Gulf Stream eastward jet extension) and small-scale (vortices) flow features.
It is important to note that the hyper-parameterization method reproduces both large- and small-scale flow features not only over the period for which the reference solution is available
(1 year in our case), but also over the second year for which there is no reference data.
We have also explained the build-up effect and showed that a bad choice of hyper-parameters leads to the build-up effect,
and eventually to a significant degradation of the hyper-parameterized solution.
In the worst-case scenario, the build-up can lead
to a numerical blow up (which we did not observe in our experiments though).
The build-up can be avoided by a proper setup of hyper-parameters which we have found through a series of numerical experiments.
In addition, the hyper-parameters have to be found separately
for hyper-parameterizing single and coupled fields, as well as for different single fields
(for instance, the hyper-parameters that work well for relative vorticity may not be optimal for temperature, and vice versa).
It is important to keep in mind that the proposed method is data-driven and, therefore, can suffer from lack of data as any data-driven method.
The hyper-parameterization approach has other methods~\citep{SB2022_J1,SB2022_J2} that
can be used to reproduce effects of mesoscale oceanic eddies on the large-scale ocean circulation, but demonstration of their skills on the level of comprehensive models is left for the future.
In other words, staying complimentary to the mainstream physics-based parameterization approach, we propose to work in phase space of the corresponding dynamical system and to interpret the lack of eddy effects as persistent tendency of phase space trajectories
(representing the modelled solution) to escape the correct reference phase-space region.
The proposed method is much faster than even a single run of the coarse-grid ocean model,
requires no modification of the model, and is easy to implement. The method can take as input data not only the reference solution
but also real measurements from different sources (drifters, weather stations, etc.), or combination of both.
All this offers a great flexibility not only to ocean modellers working with mathematical models but also to those working with measurements.
\section{Acknowledgments}
The authors thank The Leverhulme Trust for the support of this work through the grant RPG-2019-024.
\bibliographystyle{apalike}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 6,584 |
La prueba de eslalon K-1 masculino de piragüismo en los Juegos Olímpicos de Londres 2012, se llevó a cabo entre el 29 al 1 de agosto, en el Lee Valley White Water Centre en Hertfordshire.
Horario
Todas las horas están en horario de verano (UTC+1)
Resultados
Series
Calificado para semifinales
Semifinales
calificado para la final
Final
Referencias
Enlaces externos
Sitio oficial de Londres 2012 (en inglés)
Eslalon K-1 masculino | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 4,418 |
Would you like to send flowers to Morocco, using online florists? Then you've come to the right flower shop! Monday to Saturday we can take orders for flowers, chocolates, wine, balloons and teddies through our online flower shop for next day flower delivery to Morocco.
Our online flower shop sends flowers to Morocco on every occasion.
If you order flowers for next day delivery, your flowers will be delivered by our distribution partner. Our Morocco florists are highly skilled and talented florists and our flower shop is expert at sending flowers for special occasions. So if you want a florist to send Valentine's Day, Mother's Day flowers then our florists team would love to help.
We can deliver your flowers in Morocco and surroundings! Our skilled florists in Morocco will prepare your flower arrangement using the finest and freshest flowers available in Morocco, so your floral gift is as beautiful as the thought behind it.
Through our flower delivery network of online florists and flower shops you can order online and have fresh flowers delivered to Morocco. We are ready to assist you in flower delivery to Morocco!
Our online florists make every effort to guarantee delivery of your gift of flowers into the lucky recipient's hands. Every international flower delivery is special and we take great care to guarantee the flower shops service to send flowers that are on time, fresh, and delivered to the right person at the right place at the right time! In the event the recipient is not home the florist will leave a 'called today' card, inviting them to call the florist and arrange a more convenient delivery time.
Our florists can also deliver beautiful Christmas flowers to Morocco. We hope you enjoy our Morocco flower shop and our beautiful fresh flowers as much as our Morocco florists enjoy creating them.
We provide delivery to the following cities: Agadir, Ait, Melloul, Asilah Azemmour, Ben-Slimane, Bouznika, Casablanca, Essaouira, Fes, Fnideq, Inezgane, Kenitra, Larache, Marrakech, Martil, M'Diq, Meknes, Mohammedia, Nador, Oujda, Rabat, Saleh, Sidi-Kacem, Sidi-Slimane, Skhirate, Tanger, Temara and Tetouan.
Please send us an email with your enquiry if the city or town you wish to deliver in is not listed above. | {
"redpajama_set_name": "RedPajamaC4"
} | 2,407 |
Rowan Williams is Not a "Liberal"
One often finds the talking heads on the BBC and op-eds in various papers referring to the "sharia row" as another indication of Rowan Williams' "liberal" tendencies (surely one of the slipperiest and equivocal epithets we have in religious circles). But if one actually attends to his argument–and his corpus–I think one finds that Williams' is, in fact, a critic of liberalism. Indeed, the kernel of his argument at the Royal Courts of Justice was calling into question the liberal monopoly of identity that characterizes the (supposedly) "secular" state. One of the hallmarks of liberalism (fostered here in England, as well as the States, by John Locke) is a secularization of the "public" sphere of politics, economics, and the common good, along with a corresponding privatization of religious identity as an affair of the heart–a private and interior matter of one's "personal relationship" to God. In other words, religion is fine for the weekends, "if you're into that." But don't bring it to work. Don't let it affect how you function "in public." In short, you're welcome to let religion be one of your private pursuits, a kind of hobby. It's fine to let religion be "part" of who you are, but that religious faith can't shape or influence you in such a way that it would make a difference in how you pursue life in public.
But for any integral confession (whether Jewish, Christian, or Muslim), such a liberal directive would amount to idolatry. God's election of a people in Abraham and the liberation of his people in the Exodus were not undertaken with the goal of creating a late modern hobby. It was divine action meant to constitute a people who pursue the kingdom of God as their highest and most fundamental vision of human flourishing. The Incarnation, Cross and Resurrection were not just the means for securing a weekend hobby for good liberal citizens; they called into being a cruciform people whose very identity is constituted by their calling to be image bearers of this humiliated God.
I hear in Williams' argument a refusal of these two aspects of liberalism: a "secular" democratic monopoly on identity along with its corresponding privatization (and therefore triviliazation) of religious faith. In short, the Archbishop is no "liberal."
Of course rejecting such liberalism does not thereby make him a "conservative" either (recall John Ruskin's rejection of such binary alternatives). In fact, many "conservatives" are all too happy to accommodate their faith to the shape of the secular state, retreating to some kind of a-political "Jesus-in-my-heart" privatism (which the state is all too happy to permit) and letting their identity as "British" (or "American," or whatever the case may be) trump their calling as Christians. In this respect, liberal and conservative Anglicans often exhibit the same patterns. [And for the record, I think Tariq Ramadan might just be a good "liberal" Muslim–but I'm suspending judgment on that until I can read more of his work, and have an opportunity to hear him here in York.]
Ruskin, Work, and 'The Nature of the Gothic'
Rowan Williams, Sharia Law, and the End of the Liberal State: Take 2 | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 8,689 |
él è un'etichetta discografica indipendente fondata a Londra da Mike Alway, oggi sussidiaria della Cherry Red Records. I suoi musicisti erano caratterizzati da una forte sensibilità inglese, nonché dall'influenza francese derivante dallo scrittore / produttore interno Louis Philippe. Durante il suo primo periodo, él ricevette molto interesse da parte della stampa, ma registrò poche vendite, tranne che in Giappone, dove l'etichetta ebbe un'enorme influenza sul J-pop, pubblicando Cornelius e Pizzicato Five. Chiuse quindi nel 1989, per poi appunto riaprire nel 2005.
Lista parziale di artisti della él
Artisti originali
Mayfair Charm School (Featuring Victor Armada)
Shock Headed Peters
The Monochrome Set (Bid)
Vic Godard
Felt
Momus
Would-be-goods
James Dean Driving Experience
Marden Hill
The King Of Luxembourg
Bad Dream Fancy Dress
Louis Philippe
The Cavaliers
Cagliostra
Anthony Adverse
Artisti pubblicati
Andrés Segovia
Antônio Carlos Jobim
Baden Powell/Vinícius de Moraes
Brigitte Bardot
David Axelrod
Edda Dell'Orso
Edgard Varèse
Elis Regina
Ennio Morricone
Erik Satie
Four Freshmen
Gábor Szabó
Gary McFarland
Gilberto Gil
Henry Mancini
Jacques Brel
Juan García Esquivel
Nino Rota
Orson Welles
Quarteto em Cy
Ravi Shankar
Roy Budd
Sabicas
Ustad Ali Akbar Khan
Note
Collegamenti esterni
Etichette discografiche britanniche | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 4,900 |
{"url":"https:\/\/math.stackexchange.com\/questions\/3185301\/is-the-sheaf-of-rings-mathscro-the-sheafification-of-a-presheaf","text":"# Is the sheaf of rings $\\mathscr{O}$ the sheafification of a presheaf?\n\nThis a paragraph from Hartshorne's Algebraic Geometry:\n\nNext we will define a sheaf of rings $$\\mathscr{O}$$ on $$\\text{Spec }A$$. For each prime ideal $$\\mathfrak{p}\\subseteq A$$, let $$A_{\\mathfrak{p}}$$ be the localization of $$A$$ at $$\\mathfrak{p}$$. For an open set $$U\\subseteq \\text{Spec }A$$, we define $$\\mathscr{O}(U)$$ to be the set of functions $$s:U\\to \\coprod_{\\mathfrak{p}\\in U}A_{\\mathfrak{p}}$$, such that $$s(\\mathfrak{p})\\in A_{\\mathfrak{p}}$$ for each $$\\mathfrak{p}$$, and such that $$s$$ is locally a quotient of elements of $$A$$: to be precise, we require that for each $$\\mathfrak{p}\\in U$$, there is a neighborhood $$V$$ of $$\\mathfrak{p}$$, contained in $$U$$, and elements $$a, f\\in A$$, such that for each $$\\mathfrak{q}\\in V$$, $$f\\notin \\mathfrak{q}$$, and $$s(\\mathfrak{q})=a\/f$$ in $$A_{\\mathfrak{q}}$$. (Note the the similarity with the definition of the regular functions on a variety. The difference is that we consider functions into the various local rings, instead of to a field.)\n\nMy Question: $$\\mathscr{O}$$ seems to be the sheafification of a presheaf. Is it? What is it?\n\nI have compared the construction of the sheafification of a presheaf:\n\nWe construct the sheaf $$\\mathscr{F}^+$$ as follows. For any open set $$U$$, let $$\\mathscr{F}^+(U)$$ be the set of functions $$s$$ from $$U$$ to the union $$\\bigcup_{P\\in U}\\mathscr{F}_P$$ of the stalks of $$\\mathscr{F}$$ over points of $$U$$, such that\n\n(1) for each $$P\\in U$$, $$s(P)\\in \\mathscr{F}_P$$, and\n\n(2) for each $$P\\in U$$, there is a neighborhood $$V$$ of $$P$$, contained in $$U$$, and an element $$t\\in \\mathscr{F}(V)$$, such that for all $$Q\\in V$$, the germ $$t_Q$$ of $$t$$ at $$Q$$ is equal to $$s(Q)$$.\n\nI think it should have $$\\varinjlim_{\\mathfrak{p}\\in U}\\mathscr{F}(U)=\\mathscr{F}_{\\mathfrak{p}}=A_{\\mathfrak{p}}=\\varinjlim_{f\\in A\\backslash \\mathfrak{p}}A_f.$$ But I don't know how to defin $$\\mathscr{F}(U)$$...\n\nEdit. I know that every sheaf is a presheaf itself. However, we have not proven that $$\\mathscr{O}$$ is a sheaf. It is the task of the following paragraph in the book.\n\nNow it is clear that sums and products of such funtions are again such, and that the element $$1$$ which gives $$1$$ in each $$A_{\\mathfrak{p}}$$ is an identity. Thus $$\\mathscr{O}(U)$$ is a commutative ring with identity. If $$V\\subseteq U$$ are two open sets, the natural restriction map $$\\mathscr{O}(U)\\to \\mathscr{O}(V)$$ is a homomorphism of rings. It is then clear that $$\\mathscr{O}$$ is a presheaf. Finally, it is clear from the local nature of the definition that $$\\mathscr{O}$$ is a sheaf.\n\nThat is why I am looking for a presheaf which induces $$\\mathscr{O}$$.\n\n\u2022 Every sheaf is a presheaf, and is the sheafification of itself. \u2013\u00a0Lord Shark the Unknown Apr 12 at 16:45\n\u2022 \u2013\u00a0Andr\u00e9 3000 Apr 12 at 17:25","date":"2019-05-21 06:46:12","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 55, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.9658887982368469, \"perplexity\": 54.959085329078896}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2019-22\/segments\/1558232256281.35\/warc\/CC-MAIN-20190521062300-20190521084300-00300.warc.gz\"}"} | null | null |
Each and every individual has three bodies; gross, subtle and causal i.e. sthula, sukshma and karana sareeras. Sthula sareera is visible to oneself as well as to others. Sukshma sareera is not visible to onself but one is conscious of it and it is known to him but it is neither visible nor known to others. Karana sarrera is not known or visible either to oneself or to others. Now we shall see each one of them in detail.
It is made out of gross elements of Pancha bhuthas i.e. space, air, fire, water and earth. This can be seen from a few instances. The body is sustained by food from the earth and also it needs minerals like sodium and pottassium in the body for survival. The water in the body gives it the shape. The fire in the body keeps the body temperature at 98.4 F in all environs. The air in the body helps blood circulation and the space in the body allows intake of food and water. This is the general common cause for all bodies. But there is a special cause, visesha karanam, which is unique for each individual body and varies from individual to individual, which is Purva janma karma. The physical body in human form is itself acquired as a result of Punya karmas in the past.
Asti (potential existence) - In the womb it exists as a foetus.
Jayate (birth) - After the gestation period it emerges to experience the outer world as child.
Viparinamate (matures) - After adolescence, it stops growing structurally and only undergoes minor structural modifications, functioning in its peak vitality.
Apakshiyate (decays) - After middle age the joints weaken, limbs ache, hair whitens/falls, skin wrinkles and there is decline in all functions.
Vinasyati (dies) - the decline culminates in death, when the body disintegrates into the constituent elements.
The general common cause is the subtle elements of Pancha bhutas. The special cause is Purva janma karma. While the sthula sareera houses the external and internal gross physiological systems, the sukshma sareera, is the counterpart that performs all physiological functions and operates the jnanendriyas and karmendriyas. For instance for the ear the golakam is in the sthula sareera and the hearing power operates from sukshma sareera so much so the golakam will stand as an ornamental piece only if the hearing power does not operate. The sthula sareera will die when the sukshma sareera leaves it. So while sthula sareera is called भोगायतनं ( bhogayatanam) i.e. abode of experiences, the sukshma sareera is called भोगसधनम् (bhoga sadhanam) i.e. instruments for the experience. Sukshma sareera is also subject to change but it enjoys a longer life than sthula sareera as sukshma sareera is dissolved only in pralaya or at the time of Videha Mukthi.
From the above it can be seen that jnanendriyas are input organs and karmendriyas are output organs. Let me remind once again that what is referred to here is only the functioning capacity behind the organs and not these organs as seen from outside. Those organs die with sthula sareera while these interior powers survive with sukshma sareera but their new powers will depend upon the new body acquired through karma phala.
It is beginningless and it survives pralayam also but it comes to an end in videha mukthi. As it has no parts it is nirvikalpaka rupam. Ignorance of a thing one can state whether it exists or not. But pure ignorance itself one cannot. So it is neither sat nor asat, and is inexplicable. In short causal body is the one in the form of ignorance, which is inexplicable, which is beginning-less, which veils the Real nature of Self, which is the cause of the other two bodies and which has no parts.
Ahter CHATHUSHTAYAM, it is GREAT THRAYAM-nice explanation-- next is it something DHWAYAM? | {
"redpajama_set_name": "RedPajamaC4"
} | 1,741 |
\section{Introduction}
\label{sec:intro}
Among different theories of origin of life, one recurrent conundrum is the abiotic polymerization of amino acids since it requires ribosomes, macromolecular machines containing ribonucleic acid (RNA) and proteins. How could the first proteins form if they were needed to synthesize others? Following the ideas of
\citet[][]{foden2020prebiotic}, a possible solution involves a thiol-based scenario in which SH-bearing molecules, together with the family of thioacids (R-C(O)SH) and thioesters (R-S-R'), have important properties as energy carriers and catalysts \citep[][]{chandru2016,leman2017}. Although these types of compounds could be created in-situ by an H$_2$S-mediated chemistry under prebiotically plausible conditions on early Earth \citep[][]{shalayel2020}, they also could have been delivered exogenously. Hence, observations of thiol-based molecules in space could shed some light on the availability of such compounds on a primitive Earth, and on their role in the prebiotic synthesis of proteins.
More than 220 molecules have been detected in the interstellar medium (ISM) and circumstellar shells\footnote{ \url{https://cdms.astro.uni-koeln.de/classic/molecules}} to date. However, sulfur-containing species only account for 20 of them. Furthermore, while molecules detected carrying C, H or N range from 2 up to 13 atoms, the vast majority of S-bearing molecules have, at most, 4 atoms \citep[such as H$_2$CS;][]{sinclair1973}. This could be due to the relatively low cosmic abundance of atomic sulfur \citep[$\approx\,$10$^{-5}$ with respect to H$_2$, that is, more than 10 times lower than C or O;][]{asplund2009} together with its ability to have many different oxidation states and allotropes when compared to the more abundant elements$\,$\citep[][]{jm2011,shingledecker2020sulphur} and its capacity of depleting fast in dense molecular clouds$\,$\citep[][]{laas2019}.
As a consequence, a very few S-bearing molecules containing more than 4 atoms have firmly been detected in the ISM so far. One example is methyl mercaptan (hereafter CH$_3$SH), which has been detected in several environments, such as pre-stellar cores$\,$\citep[][]{gibb2000chemistry}, massive star-forming regions like Sagittarius B2$\,$\citep[][]{linke1979,muller2016}, and Solar-like protostars$\,$\citep[][]{majumdar2016}. The other one is ethyl mercaptan (hereafter C$_2$H$_5$SH) which was tentatively detected toward Orion KL$\,$\citep{kolesnikova2014}. Other searches for complex S-bearing molecules were unsuccessful, such as CH$_3$CHS \citep[thioacetaldehyde;][]{margules2020}, NH$_2$CHS \citep[thioformamide;][]{motiyenko2020} and CH$_3$SC(O)H \citep[S-methyl thioformate;][]{jabri2020}.
In this Letter, we report the first detection in the ISM of the trans isomer of monothioformic acid (hereafter HC(O)SH), the simplest thiol acid. We also report a solid confirmation of the gauche isomer of C$_2$H$_5$SH, together with the detection of CH$_3$SH. These molecules are found toward the quiescent Giant Molecular Cloud G+0.693-0.027 located in the Galactic Center (hereafter G+0.693).
This source shows a very rich chemistry with up to 40 different complex organic molecules (COMs\footnote{These are usually referred as carbon-based molecules that have 6 or more atoms \citep[][]{herbst2009}.}) detected \citep[see, for example:][]{requena2008galactic,zeng2018complex,rivilla2018,rivilla2019_cyanomehtanimine,rivilla2020prebiotic,ijimenez2020}. Studies suggest that this cloud could be undergoing a cloud-cloud collision \citep[][]{zeng2020}, which induces large-scale shocks that sputter dust grains and that enhance the gas-phase abundance of molecules by several orders of magnitude \citep[][]{requenatorres2006}.
\section{Observations} \label{sec:observations}
We have used a spectral line survey toward G+0.693 covering several windows between 32$\,$GHz to 172$\,$GHz with an average resolution of 1.5$\,$km$\,$s$^{-1}$, although the final spectral resolution employed in the figures has been smoothed up to 3$\,$km$\,$s$^{-1}$. We stress that this is for a proper line visualization and does not alter in any form the analysis done.
For the observations, we used both the IRAM 30m telescope located at Pico Veleta (Granada, Spain) and the Yebes 40m telescope\footnote{Yebes Observatory is operated by the Spanish Geographic Institute (IGN, Ministerio de Transportes, Movilidad y Agencia Urbana)}
(Guadalajara, Spain). The equatorial coordinates of the molecular cloud G+0.693 are $\alpha$(J2000.0)$\,$=$\,$17$^h$47$^m$22$^s$, and $\delta$(J2000.0)$\,$=$\,$−28$^{\circ}$21$'$27$''$. The position switching mode was used in all the observations with the off position located at (-885$''$, 290$''$) from the source. The line intensity of the spectra was measured in units of $\mathrm{T_A^{\ast}}$ as the molecular emission toward G+0.693 is extended over the beam \citep[][]{requenatorres2006,martin2008,rivilla2018}. In all the observations, each frequency setup was repeated shifting the central frequency by 20-100 MHz in order to identify spurious lines or contamination from the image band.
The IRAM 30m observations were performed during three different sessions in 2019: 10-16th of April, 13-19th of August and 11-15th of December. The dual polarization receiver EMIR was used connected to the fast Fourier transform spectrometers (FFTS), which provided a channel width of 200 kHz in the 3 and 2$\,$mm radio windows. The observations with the Yebes 40m radiotelescope were carried out in February 2020: from the 3rd to 9th and from the 15th to 22th. In this case, the Nanocosmos Q-band (7$\,$mm) HEMT receiver was used which enables ultra broad-band observations in two linear polarizations$\,$\citep[][]{tercero2020}. The receiver was connected to 16 FFTS providing a channel width of 38 kHz and an instantaneous bandwidth of 18.5 GHz per polarization, covering the frequency range between 31.3 GHz and 50.6 GHz.
\section{Analysis and results}\label{sec:analysis}
We used the software \textsc{madcuba}\footnote{MAdrid Data CUBe Analysis is a software developed at the Center of Astrobiology in Madrid: \url{https://cab.inta-csic.es/madcuba/Portada.html}}$\,$\citep{madcuba} to perform the data analysis and line identification
The Spectral Line Identification and Modelling (SLIM) tool of \textsc{madcuba} uses the spectroscopic data entries from different molecular catalogs, and generates a synthetic spectra based under the assumption of Local Thermodynamic Equilibrium (LTE) conditions, and considering line opacity effects.
The fitted parameters used to reproduce the molecular emission are: column density ($N$), excitation temperature ({\it $T_{\mathrm{ex}}$}), local standard of rest velocity ({\it v$_\mathrm{LSR}$}) and full width at half maximum ({\it FWHM}).
Both HC(O)SH and C$_2$H$_5$SH have two rotamers, associated with the rotation of their respective C-S bond.
HC(O)SH has one cis (c-HC(O)SH) and one trans (t-HC(O)SH) rotamer, being the former $\sim\,$330$\,$K higher in energy \citep[][]{hocking1976rotational}. C$_2$H$_5$SH has two degenerated $\pm\,$gauche isomers (g-C$_2$H$_5$SH) and one antiperiplanar.
For our analysis we have used the following spectroscopic entries from the the CDMS\footnote{Cologne Database for Molecular Spectroscopy$\,$\citep[][]{CDMS}. \url{https://cdms.astro.uni-koeln.de/classic/}} catalog:
entries 062515/062516 for trans/cis isomers of HC(O)SH \citep[][]{hocking1976rotational}, 062523/062524 for gauche/anti isomers of C$_2$H$_5$SH \citep[][]{kolesnikova2014,muller2016}, and 048510 for CH$_3$SH \citep[][]{xu2012,zakharenko2019}.
\subsection{Detection of t-HC(O)SH}
\begin{figure*}
\centering
\includegraphics[width=1\textwidth]{thioformic_final.pdf}
\caption{Cleanest and brightest lines of t-HC(O)SH detected toward G+0.693 labeled with their corresponding quantum numbers in red. The red line shows the best LTE fit to the observed spectra (represented by the black lines). The data has been smoothed up to 3 km s$^{-1}$ for an optimal line visualization. The blue lines show the spectra including the emission of all the molecules searched toward G+0.693. Note that these lines are tagged with their corresponding molecular compound in pink.}
\label{fig:monothioformic}
\end{figure*}
\begin{deluxetable*}{ccccccccl}[htp!]
\tablecaption{Lines of t-HC(O)SH detected toward G+0.693 with their corresponding quantum numbers (QNs), logarithm of the Einstein coefficients ($\mathrm{\log A_{ul}}$) , degeneracy ($\mathrm{g_u}$) and energy ($\mathrm{E_u}$) of the upper state. The integrated signal ($\int T_A^* d\nu$) and root mean square (rms) noise level are also provided and used to calculate the signal to noise ratio (SNR) of the detections.\label{tab:thioformicacid}}
\tablecolumns{9}
\tablewidth{0pt}
\tablehead{
\colhead{Rest frequency} & \colhead{QNs\tablenotemark{a}} & \colhead{g$_u$} & \colhead{E$_u$} & \colhead{$\mathrm{\log A_{ul}}$} & \colhead{rms}& \colhead{$\int T_A^* d\nu$\tablenotemark{b}} & \colhead{SNR\tablenotemark{c}} & \colhead{Comments} \\
\colhead{(MHz)} & \colhead{} & \colhead{} & \colhead{(K)} & \colhead{(s$^{-1}$)} & \colhead{(mK)} & \colhead{($\mathrm{mK\,km\,s^{-1}}$)} & \colhead{} & \colhead{}}
\startdata
$34248.82$ & $3_{1,3}\rightarrow2_{1,2}$ & 7 & 4.3 & -6.4784 & 1.4 & 56 & 7.3 & clean transition \\
$35915.48$ & $3_{1,2}\rightarrow2_{1,1}$ & 7 & 4.4 & -6.4164 & 1.4 & 58 & 7.6 & blended with HOCH$_2$CH$_2$OH \\
$45659.99$ & $4_{1,4}\rightarrow3_{1,3}$ & 7 & 6.0 & -6.0647 & 2.4 & 88 & 13.4 & clean transition \\
$46737.73$ & $4_{0,4}\rightarrow3_{0,3}$ & 7 & 3.4 & -6.0064 & 2.6 & 122 & 8.6 & clean transition \\
$71800.18$ & $6_{1,5}\rightarrow5_{1,4}$ & 13 & 11.3 & -5.4428 & 3.5 & 121 & 6.3 & clean transition \\
$81630.08$ & $7_{0,7}\rightarrow6_{0,6}$ & 15 & 11.8 & -5.2587 & 3.4 & 153 & 8.2 & slightly blended with CH$_3$CH$_2$CHO \\
$83749.29$ & $7_{1,6}\rightarrow6_{1,5}$ & 15 & 14.8 & -5.2342 & 3.4 & 116 & 11.8 & slightly blended with CH$_3$COOH \\
$91251.84$ & $8_{1,8}\rightarrow7_{1,7}$ & 17 & 18.1 & -5.1166 & 1.7 & 104 & 11.2 & slightly blended with (CH$_2$OH)$_2$ and unidentified species \\
$93505.09$ & $8_{2,7}\rightarrow7_{2,6}$ & 17 & 26.5 & -5.1061 & 1.4 & 46 & 6.0 & blended with CH$_3$\isotope[18]{O}H
\enddata
\tablenotetext{a}{$\mathrm{J^{''}_{K^{''}_{a}K^{''}_{c}}\rightarrow J^{'}_{K^{'}_{a}K^{'}_{c}}}$ (where the double prime indicates the upper state).}
\tablenotetext{c}{Signal to noise ratio is calculated from the integrated signal and noise level $\mathrm{\sigma=rms\,\sqrt{\delta v\,FWHM}}$, where $\mathrm{\delta v}$ is the velocity resolution of the spectra.}
\end{deluxetable*}
The fitting procedure was performed considering the total emission of any other identified molecule in the spectral survey. This evaluation was carried out introducing all the compounds already detected in the ISM and in G+0.693 \citep[see][]{requenatorres2006,requena2008galactic,zeng2018complex,rivilla2019_cyanomehtanimine,ijimenez2020}. For the analysis, it was also assumed a cosmic microwave background temperature (T$_\mathrm{bg}$) of 2.73$\,$K and no background continuum source$\,$\citep{zeng2020}.
The global fit of all rotational lines to the observed data is shown in blue lines in Figure$\;$\ref{fig:monothioformic}, while in red we show the fit of the individual lines of t-HC(O)SH, and the observational data in black. As shown in both Figure$\;$\ref{fig:monothioformic} and Table$\;$\ref{tab:thioformicacid}, we have detected a total of nine a-type transitions in the 7$\,$mm and 3$\,$mm bands, each one of them detected above the 5$\sigma$ level in integrated intensity. Note that the SLIM synthetic spectrum shows small deviations with respect to the observations. These deviations are larger that the ones obtained for CH$_3$SH and g-C$_2$H$_5$SH (Appendix \ref{ap:A}). This is due to the fact that the t-HC(O)SH lines are significantly weaker than those of CH$_3$SH and g-C$_2$H$_5$SH (3-7$\,$mK versus 5-12$\,$mK and $>$40$\,$mK, respectively). However, the fit of all clean t-HC(O)SH lines is consistent with the noise. We also note that for the three slightly blended lines of t-HC(O)SH, the contribution of the other molecules is less than 10\% of the total intensity. Despite that the remaining two appear blended, the predicted LTE intensities are consistent with the observed lines. Note that the rest of the t-HC(O)SH lines covered within the observed frequency range are not shown due to strong blending issues.
The physical parameters obtained from the fit are listed in Table$\;$\ref{tab:physical_parameters_all}. The values obtained for $T_\mathrm{ex}$ and $FWHM$ are consistent with those obtained previously for other molecular species toward this cloud \citep[$T_{\mathrm{ex}}\sim\,$5 - 20$\;$K and linewidths around 20 km s$^{-1}$;][]{zeng2018complex,rivilla2020prebiotic}. The low $T_{\mathrm{ex}}$ indicates that the emission of the molecules in G+0.693 is sub-thermally excited as a result of the low H$_2$ densities of this source \citep[$T_{\mathrm{kin}}\sim$150$\,$K;][]{requenatorres2006,zeng2018complex}.
For t-HC(O)SH, the fitted column density gives (1.6$\,\pm\,$0.1)$\times$10$^{13}$\,cm$^{-2}$. In addition, we derived a 3$\,\sigma$ upper limit of $\leq$3$\times$10$^{12}\,$cm$^{-2}$ for c-HC(O)SH (Table$\;$\ref{tab:physical_parameters_all}), as no clear transition has been detected within our dataset. For the calculation, we have assumed the same $T_{\mathrm{ex}}$, $v_{\mathrm{LSR}}$ and $FWHM$ as for t-HC(O)SH. A ratio c-HC(O)SH$\,$/$\,$t-HC(O)SH$\,\leq\,$0.2 was obtained.
\begin{deluxetable*}{cccccc}
\tablecaption{Physical parameters of the species derived by LTE analysis in \textsc{madcuba}. \label{tab:physical_parameters_all}}
\tablecolumns{6}
\tablewidth{0pt}
\tablehead{\colhead{Molecular formula} & \colhead{N} & \colhead{$\mathrm{T_{ex}}$} & \colhead{$\mathrm{v_{LSR}}$} & \colhead{FWHM} & \colhead{Abundance} \\
\colhead{} & \colhead{$(\times$ 10$^{13}$ cm$^{-2})$} & \colhead{(K)} & \colhead{$\mathrm{(km\,s^{-1})}$} & \colhead{$\mathrm{(km\,s^{-1})}$} & \colhead{($\tfrac{N(X)}{N(H_2)}\times10^{-10}$)}}
\startdata
$\mathbf{trans-HC(O)SH}$ & $1.6 \pm 0.1$ & $10 \pm 1$ & $69.0^{\,\mathrm{a}}$ & $21.0^{\,\mathrm{a}}$ & $1.2\pm 0.2$\\
$\mathbf{cis-HC(O)SH}$ & $\leq 0.3$ & $10.0$ & $69.0^{\,\mathrm{a}}$ & $21.0$\tablenotemark{a} & $\leq 0.2$\\
$\mathbf{g-C_2H_5SH}$ & $4\pm2$ & $10\pm 5$ & $69.0^{\,\mathrm{a}}$ & $20.0$\tablenotemark{a} & $3\pm1$ \\
$\mathbf{a-C_2H_5SH}$ & $\leq 2.3$ & $9.91 $ & $69.0^{\,\mathrm{a}}$ & $20.0^{\,\mathrm{a}}$ & $\leq 1.7$\\
$\mathrm{CH_3SH,\, K_a=0}$ & $46.8\pm0.5$ & $8.5\pm0.1$ & $68.0\pm0.1$ & $21.2\pm0.3$ & -\\
$\mathrm{CH_3SH,\, K_a=1}$ & $18.6\pm 0.7$ & $14.9\pm0.5$ & $68.8\pm0.3$ & $22.0\pm0.7$ & -\\
$\mathbf{CH_3SH}$ & $65\pm2$ & - & - & - & $48\pm 5$\\
$\mathbf{t-HCOOH}$ & $20\pm4$ & $10\pm2$ & $68\pm2$ & $22\pm5$ & $15\pm4$ \\
\enddata
\tablenotetext{a}{Value fixed in the fit.}
\end{deluxetable*}
\subsection{Detection of g-C$_2$H$_5$SH and CH$_3$SH}
This molecule was tentatively detected in Orion KL \citep{kolesnikova2014}, based on a few isolated transitions assigned to this isomer. We report here an unambiguous detection of this isomer that confirms it presence in the ISM. We have listed in Table$\;$\ref{tab:gC2H5SH} (Appendix A) the transitions measured toward G+0.693. Note that eight of the targeted lines are totally clean (Figure$\;$\ref{fig:c2h5sh}, Appendix A) and above the 5$\sigma$ level in integrated intensity. We derived a total column density of (4$\,\pm\,$2)$\times$10$^{13}\,$cm$^{-2}$ (Table$\;$\ref{tab:physical_parameters_all}).
In the case of CH$_3$SH, the the fit was carried out separating the targeted lines into its $K_a=0$ and $K_a=1$ levels\footnote{The moments of inertia of CH$_3$SH frame it in the limiting prolate case, $\kappa=(2B-A-C)/(A-C)\approx-0.988$.}.
The brightest transitions fall into the 3$\,$mm and 2$\,$mm bands as presented in Figure$\;$\ref{fig:ch3sh} and Table$\;$\ref{tab:ch3sh} (Appendix B). Note that the agreement between the predicted and observed spectra is excellent for all clean transitions. The column density using the $K_a=0$ and $K_a=1$ levels is (6.5$\,\pm\,$0.2)$\times$10$^{14}\,$cm$^{-2}$ (Table$\;$\ref{tab:physical_parameters_all}). This gives a ratio of CH$_3$SH$\,$/$\,$g-C$_2$H$_5$SH$\;$=$\;$16$\,\pm\,$7. Note that the column density of $K_a=2$ and remaining transitions in CH$_3$SH contribute less than a 10\% in the total determined.
Finally, we note that both \isotope[13]{C} and \isotope[34]{S} isotopologues of CH$_3$SH have also been detected with a few clean lines and will be presented in a forthcoming paper (Colzi et al. in prep.).
\subsection{Molecular Abundances and comparison with their O-bearing analogues} \label{sec:abundances}
To derive the fractional abundances of these species relative to H$_2$, we have assumed an H$_2$ column density of $N_{H_2}=1.35\times10^{23}\,$cm$^{-2}$ \citep[][]{martin2008}. This gives $\sim\,$3$\,\times\,$10$^{-10}$, $\sim\,$1$\,\times\,$10$^{-10}$ and $\sim\,$5$\,\times\,$10$^{-9}$ for g-C$_2$H$_5$SH, t-HC(O)SH and CH$_3$SH, respectively (Table$\,$\ref{tab:physical_parameters_all}).
In Figure$\;$\ref{fig:fractional_abundances} we have plotted these values and compared them with the abundances obtained for their OH molecular analogues, namely: C$_2$H$_5$OH (ethanol), HC(O)OH (formic acid) and CH$_3$OH (methanol). We obtained a ratio CH$_3$OH$\,$/$\,$CH$_3$SH$\,$=$\,$23, C$_2$H$_5$OH$\,$/$\,$C$_2$H$_5$SH$\,$=$\,$15 and HC(O)OH$\,$/$\,$HC(O)SH$\,$=$\,$13. Although these OH-analogues are more abundant by a factor$\,\geq$10, the resulting trend is strikingly similar. Note, however, that this trend is lost when we compare molecules such as carbon monosulfide (CS) and thioformaldehyde (H$_2$CS) with their O-bearing analogues, resulting in H$_2$CO$\,$/$\,$H$_2$CS$\,$=$\,$3.5 and CO$\,$/$\,$CS$\,$=$\,$3.5$\times$10$^3$ (Figure \ref{fig:fractional_abundances}).
\begin{figure}[htp!]
\centering
\includegraphics[width=0.45\textwidth]{abundances_GRAPH.PNG}
\caption{Molecular abundances wrt H$_2$ for the reported detections and their OH-analogues (CH$_3$OH, C$_2$H$_5$OH and HC(O)OH) including CS, CO, H$_2$CO and H$_2$CS. The lines that connect each dot do not have any physical meaning but are just a visual aid.}
\label{fig:fractional_abundances}
\end{figure}
Since CH$_3$OH, C$_2$H$_5$OH, H$_2$CO, CO and CS are optically thick toward G+0.693, the column density of these molecules were inferred from CH$_3$\isotope[18]{O}H, \isotope[13]{C}H$_3$CH$_2$OH, H$_2$C\isotope[18]{O}, C\isotope[18]{O} and \isotope[13]{C}\isotope[34]{S}. It was assumed \isotope[16]{O}$\,$/$\,$\isotope[18]{O}$\,$=$\,$250$\;$ and \isotope[12]{C}$\,$/$\,$\isotope[13]{C}$\,$=$\,$21 \citep[][]{armijos2014} and \isotope[32]{S}$\,$/$\,$\isotope[34]{S}$\,$=$\,$22 \citep[][]{wilson1999}. For HC(O)OH, the main isotopologue of HC(O)OH was employed as the ratio H\isotope[12]{C}(O)OH$\,$/ H\isotope[13]{C}(O)OH is consistent with the \isotope[12]{C}$\,$/$\,$\isotope[13]{C} ratio.
For the sulfur-analogues, except CS, we do not expect opacity issues due to their much lower abundances. The information about the fit employed for the molecules presented in Figure \ref{fig:fractional_abundances} or their isotopologues is included in Appendix \ref{ap:B}.
\section{Discussion}
\label{sec:discussion}
\subsection{Comparison with previous observations}
We have compared different column density ratios obtained between the S-bearing compounds measured toward G+0.693 and two other sources, namely: Orion KL \citep[][]{kolesnikova2014} and Sgr B2(N2) \citep[][see Table \ref{tab:abundances}]{muller2016}.
From this table, we find that all abundance ratios measured in G+0.693 are strikingly similar. For instance, when we compare SH-bearing molecules with their OH-analogues (CH$_3$OH / CH$_3$SH and C$_2$H$_5$OH / C$_2$H$_5$SH) we recover the trend already found in Figure \ref{fig:fractional_abundances}, which might indicate a similar chemistry between OH- and SH-bearing species.
The CH$_3$OH$\,$/$\,$CH$_3$SH ratio measured in G+0.693 is a factor of 5 lower than those found in Sgr B2(N2) and Orion KL, which indicates that G+0.693 is richer in sulfur-bearing species than the two massive hot cores. This may be related to the fact that the chemistry of this cloud is affected by large-scale shocks \citep[][]{requenatorres2006,zeng2018complex}. Since sulfur is heavily depleted on grains \citep[possibly in the form of S$_8$ and other sulfur allotropes;][]{shingledecker2020sulphur}, the sputtering of dust grains induced by shocks could liberate a significant fraction of the locked sulfur.
The C$_2$H$_5$OH$\,$/$\,$C$_2$H$_5$SH ratio is comparable to the value determined toward Orion KL while there is a difference by a factor of 10 between G+0.693 and the upper limit toward Sgr B2(N2). All these ratios suggest that sulfur is less incorporated into molecules than oxygen, which again might be due to the fact that a significant fraction of sulfur is locked up in grains.
\begin{deluxetable*}{cccccc}[htp!]
\tablecaption{Relative abundances of molecules and comparison with other sources.\label{tab:abundances}}
\tablecolumns{5}
\tablewidth{0pt}
\tablehead{
\colhead{Source} &
\colhead{CH$_3$SH$\,$/$\,$C$_2$H$_5$SH} &
\colhead{CH$_3$OH$\,$/$\,$C$_2$H$_5$OH} &
\colhead{CH$_3$OH$\,$/$\,$CH$_3$SH} &
\colhead{C$_2$H$_5$OH$\,$/$\,$C$_2$H$_5$SH} &
}
\startdata
G+0.693 & 16$\pm$7 & 24$\pm$4\tablenotemark{c,d} & 23$\pm$3\tablenotemark{c} & 15$\pm$7\tablenotemark{d} \\
Sgr B2(N2)\tablenotemark{a} & $\geq$21 & 20 & 118 & $\geq$125 \\
Orion KL\tablenotemark{b} & 5 & 31 & 120 & 20 \\
\enddata
\tablenotetext{a}{Data taken from \citet[][]{muller2016}.}
\tablenotetext{b}{Data taken from \citet[][]{kolesnikova2014}.}
\tablenotetext{c}{Data inferred from CH$_3$\isotope[18]{O}H assuming \isotope[16]{O} / \isotope[18]{O} = 250 \citep[][]{armijos2014}.}
\tablenotetext{d}{Data inferred from \isotope[13]{C}H$_3$CH$_2$OH assuming \isotope[12]{C} / \isotope[13]{C} = 21 \citep{armijos2014}.}
\end{deluxetable*}
\subsection{Interstellar chemistry of thiols}
\label{sec:chemicalpathways}
In order to understand the chemistry of S-bearing compounds, different models have been proposed based on the simple molecules detected in the gas phase \citep[][]{muller2016,gorai2017,lamberts2018,laas2019}.
CH$_3$SH is thought to be formed on grain surfaces by sequential hydrogenations starting from CS:
$$
\mathrm{CS + H \rightarrow HCS }
$$
$$
\mathrm{HCS + H \rightarrow H_2CS}
$$
$$
\mathrm{H_2CS + H \rightarrow CH_3S}
$$
$$
\mathrm{CH_3S + H \rightarrow CH_3SH.}
$$
The last step could also yield other products but theoretical calculations show a branching ratio (br) of $\sim$75\% for CH$_3$SH \citep[][]{lamberts2018}. This is a viable mechanism since approximately half of the CS present in the ices is available to undergo hydrogenation, while the other half is converted to OCS \citep[][]{palumbo1997}. Once formed, CH$_3$SH could remain stored in the ices until released by grain sputtering in the large-scale shocks present in G+0.693 \citep[][]{requenatorres2006,zeng2020}.
Likewise, ethyl mercaptan is proposed to be formed by radical-radical reactions \citep[][]{gorai2017,muller2016} such as:
$$
\mathrm{C_2H_5 + SH \rightarrow C_2H_5SH}
$$
$$
\mathrm{CH_3 + CH_2SH \rightarrow C_2H_5SH.}
$$
Observations of the reactants could give us a hint of the dominant reaction based on the measured column densities. However, chemical modeling is needed to understand the efficiency of these formation routes. Note that CH$_2$SH is the main product of the hydrogenation of H$_2$CS \citep[][]{lamberts2018}.
To our knowledge, there is no information available in the literature about the chemistry of HC(O)SH. However, we can make a guess and assume similar astrochemical pathways for the SH-based species as the ones for their OH-analogues. A possible pathway to HC(O)SH could mimic the formation of HC(O)OH. \citet[][]{ioppolo2011} showed that the formation of HC(O)OH starts from CO and the OH radical in the ice. The thiol-equivalent reaction would be:
$$
\mathrm{CO + SH \rightarrow HSCO \xrightarrow[]{\text{H}} HC(O)SH.}
$$
The first step has been found to be efficient by \citet[][]{adriens2010}, but further experimental and/or theoretical work is needed to investigate whether HSCO could be hydrogenated further.
Another possibility could be:
$$
\mathrm{HCO + SH \rightarrow HCOSH}
$$
or
$$
\mathrm{OCS \xrightarrow[]{2H} HC(O)SH.}
$$
The first reaction was initially proposed for HC(O)OH by \citet[][]{garrod2006} and could be a viable mechanism since SH can be formed on surfaces from S + H or H + H$_2$S via tunneling \citep[][]{vidal2017}. The second involves the sequential hydrogenation of OCS, which is a molecule detected in ices \citep[][]{palumbo1997} and it is moderately abundant in the gas phase in G+0.693 \citep[N$>$10$^{15}$cm$^{-2}$;][]{armijos2014}. Theoretical studies about these chemical networks are currently under work and will be presented in a forthcoming paper (Molpeceres et al. in prep).
\subsection{Implications for theories on the origin of life}
As it was previously stated, thioacids and thioesters have been proposed as key agents in the polymerization of amino acids into peptides and proteins \citep{foden2020prebiotic,muchowska2020}. In some of these works, it is stressed the importance of cystein, HSCH$_2$CH(NH$_2$)COOH, as the primary organic source of sulfide in biology and a key catalyst in the abiotic polymerization of peptides. Smaller thiols such as CH$_3$SH and C$_2$H$_5$SH are also believed to play a key role in prebiotic chemistry and in theories about the origin of life, since they are precursors for the synthesis of the aminoacids methionine and ethionine \citep[][]{parker2011}. In turn, HC(O)SH could be an important ingredient in the phosphorylation of nucleosides as demonstrated by \citet{lohrmann1968}.
The idea of an extraterrestrial delivery of SH-compounds onto Earth, is supported by the detection of CH$_3$SH in Murchinson carbonaceous chondrite \citep[][]{tingle1991} and in the coma of the 67P/Churyumov-Gerasimenko comet \citep[][]{calmonte2016}. In the latter, C$_2$H$_6$S was also detected (either in the form of C$_2$H$_5$SH or CH$_3$SCH$_3$ -dimethyl sulfide-) with an abundance ratio CH$_3$SH$\,$/$\,$C$_2$H$_6$S$\,\sim\,$10 which is consistent with the value obtained in G+0.693 (Table$\;$\ref{tab:abundances}). This resemblance could be possibly indicating a pre-solar origin of these compounds. Still, further studies are required within other regions in the ISM and planetary bodies to make a proper connection.
In summary, not only have our observations confirmed the presence of sulfur-bearing complex organics such as CH$_3$SH and C$_2$H$_5$SH, but they also have revealed the existence of the simplest thioacid known, HC(O)SH, in the ISM.
\acknowledgments
\section{Acknowledgments}
L.F.R.-A. acknowledges support from a JAE-intro ICU studentship funded by the Spanish National Research Council (CSIC). L.F.R.-A., V.M.R. and L.C. also acknowledge support from the Comunidad de Madrid through the Atracci\'on de Talento Investigador Modalidad 1 (Doctores con experiencia) Grant (COOL: Cosmic Origins Of Life; 2019-T1/TIC-15379; PI: V.M.Rivilla). I.J.-S. and J.M.-P. have received partial support from the State Research Agency (AEI) through project numbers PID2019-105552RB-C41 and MDM-2017-0737 Unidad de Excelencia "María de Maeztu" - Centro de Astrobiología (CSIC-INTA). PdV and BT thank the support from the European Research Council through Synergy Grant ERC-2013-SyG, G.A. 610256 (NANOCOSMOS) and from the Spanish Ministerio de Ciencia e Innovación (MICIU) through project PID2019-107115GB-C21. BT also thanks the Spanish MICIU for funding support from grants AYA2016-75066-C2-1-P and PID2019-106235GB-I00.
\vspace{5mm}
\facilities{IRAM 30m, Yebes 40m}
\software{\textsc{madcuba}}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 2,769 |
Visiting Card Design For Stone Crusher Plant Sand Stone Crushing Plant At Brekke Quarries Plant 3 Stock Photo Congratulations and thank you for your business.
Nov 16, 2016 Stone Crusher Plant Layout Design The crushing project geographical loion, Stone visiting card design for stone crusher plant Crusher.
visiting card design for stone crusher plant. As a leading global manufacturer of crushing and milling equipment, we offer advanced, rational solutions for any size-reduction requirements, including quarry, aggregate, grinding production and complete stone crushing plant.
Crusher Plant Visiting Card Design rrwaterproofing . design of stone crusher plant visiting card design for stone crusher plant Jaw crusher,Jaw crushers,Jaw rockers,Stone It is the first choice for Stone crusher plant .
Stone Crushing Plant At Brekke Quarries Plant 3 Stock Photo Congratulations and thank you for your business. Stone crushing plant which crushes the stone to shingle, gravel and fine sand and their work is in the extraction of gravel and sand-pits and mining of clays and kaolin.
free stone crusher business card maker Dream Kitchen Gallery Free coloring pages, visiting card design for stone crusher plant 94 (Total: Read more.
Design Stone Crusher Plant Stone Pulverizing Machinery. Crushing Plant Design and Layout Considerations Ken Boyd, Manager, Ore characteristics are a critical element in visiting card design for stone crusher plant.
visiting card design for stone crusher plant [crusher and mill] custom logo design, crusher plant visiting card design [crusher and mill] jaw crusher, mobile.
visiting card design for stone crusher plant Logo Stock Photos, Logo Stock Photography,, Ajay Stone Crusher in Karveer, Kolhapur 416122 Sulekha Kolhapur. | {
"redpajama_set_name": "RedPajamaC4"
} | 9,752 |
{"url":"http:\/\/fkpd.anjadaube.de\/separable-differential-equations-problems-and-solutions.html","text":"8: Separable Equations But I usually like to have the solution to a differential equation just y equal something. Stiffness is an efficiency issue. Solutions by Substitutions. Exact Equations and Integrating Factors. In earlier parts, we described symbolic solutions of particular differential equations. Given a linear first\u2014order differential equation in standard form then where ma:) is the integrating factor For an exact first\u2014order differential equation an implicit solution is given by where Ax) f (c) dx, o, o, = M y) and Given a homogeneous first\u2014order differential equation make the substitution y =. (b) Find the particular solution yfx= ( ) to the differential equation with the initial condition f (\u221211)= and state its domain. This is a linear equation. And if I can, I would like to conclude the series by reaching partial differential equations. 18 Problems: Heat Equation 255 5. This equation arises from Newton's law of cooling where the ambient temperature oscillates with time. Find the general solution of the differential equation. Separable differential equations are equations that can be separated so that one variable is on one side, and the other variable is on the other side. Linear Differential Equations. Differential Equation of first Order and first Degree OF A differential equation of the first order and. When reading a sentence that relates a function to one of its derivatives, it's important to extract the correct meaning to give rise to a differential equation. 2 Solutions of differential equations. \"For he who knows not mathematics cannot know any other sciences; what is more, he cannot discover his own ignorance or find its proper remedies. Advanced Math Solutions - Ordinary Differential Equations. and so y = 25+15e\u22122t is a solution to the initial value problem. Numerical Approximations; Use the Euler or tangent line method to find an approximate solution to a linear differential equation. Numerical methods. First-Order Differential Equations 1 1. We will now look at some examples of solving separable differential equations. In fact, because the differential equation is separable, we can define the solution curve implicitly by a function in the form. 4 Direction Fields 5 1. Then we attempt to solve for y as an explicit function of x, if possible. Chapter 7 - Linear Algebra and Linear Systems of Equations: Matrices. The distinction between the singular and the general solution is just an algebraic distinction. 3 The heat equation; separation of variables 483 5. Separable Differential Equations. (x\u00a1y)dx+xdy = 0:Solution. Here is an example: Example problem 1: Example problem 2:. 2 Solutions of differential equations. From the series: Differential Equations and Linear Algebra Gilbert Strang, Massachusetts Institute of Technology (MIT) A second order equation can change its initial conditions on y(0) and dy\/dt(0) to boundary conditions on y(0) and y(1). 06) Particular solution (5. These conditions used to develop a calculational procedure for determining whether any given equation of this type can be transformed into a separable equation and also to develop a procedure for determining the various changes of variable which will lead to separable equations. Classify differential equations according to their type and order. Separable Differential Equations Problem. Also talked about directions fields , and how to find solutions to first-order separable and homogeneous differential equations. In earlier parts, we described symbolic solutions of particular differential equations. Watch It links provide step-by-step instruction with short, engaging videos that are ideal for visual learners. Advanced Math Solutions \u2013 Ordinary Differential Equations. Definitions and Terminology. Initial Value Problem An thinitial value problem (IVP) is a requirement to find a solution of n order ODE F(x, y, y\u2032,,())\u2208 \u2282\\ () \u2208: = =. The Wolfram Language's differential equation solving functions can be applied to many different classes of differential equations, automatically selecting the appropriate algorithms without needing preprocessing by the user. Solution methods for ordinary and partial differential equations, usually seen in university mathematics courses. 4 Separable Equations and Applications 32 1. Inhomogeneous Problems. In this section, we will try to apply differential equations to real life situations. Separable equations have the form dy\/dx = f(x) g(y), and are called separable because the variables x and y can be brought to opposite sides of the equation. So the previous method will not work because we will be unable. Chapter 1 in Review. The initial value problem in Example 1. A separable differential equation is a common kind of differential calculus equation that is especially straightforward to solve. Exact Equations and Integrating Factors. We obtained a particular solution by substituting known values for x and y. De nition 1. Be able to find the general and particular solutions of separable first order ODEs. Find the solution of with initial conditions y (0) = 1 and y' (0) = 0. Sufficient Condition of Existence and Uniqueness: If and its partial derivative with respect to are continuous in the neighborhood region , the solution of this initial value problem in the region exists and is unique. Of these, the separable variables case is usually the simplest, and solution by an inte- grating factor is usually a last resort. Separable differential equations Calculator Get detailed solutions to your math problems with our Separable differential equations step-by-step calculator. If is some constant and the initial value of the function, is six, determine the equation. A separable differential equation is of the form y0 =f(x)g(y). Multiply the equation by integrating factor: ygxf 12 1 2. Differential equations are separable if you can separate the variables and integrate each side. A series of free Calculus 2 Video Lessons. A clever method for solving differential equations (DEs) is in the form of a linear first-order equation. If we can symbolically compute these integrals, then we can solve for. Separable Equations The Simplest Differential Equations Separable differential equations Mixing and Dilution Models of Growth Exponential Growth and Decay The Zombie Apocalypse (Logistic Growth) Linear Equations Linear ODEs: Working an Example The Solution in General Saving for Retirement Parametrized Curves Three kinds of functions, three. Solutions by Substitutions. Variable Separable DE Solved Problems - Duration:. Most first order linear ordinary differential equations are, however, not separable. These applications use Clickable Calculus methods to solve problems interactively. Over the last hundred years, many techniques have been developed for the solution of ordinary differential equations and partial differential equations. We obtained a particular solution by substituting known values for x and y. Solution Curves Without a Solution. This solution is part of the general solution family, obtained by plugging in the initial value condition. In this article, we show how to apply this to ordinary differential equations. 8: Separable Equations But I usually like to have the solution to a differential equation just y equal something. Solving Separable First Order Differential Equations - Ex 1 Solving Separable First Order Differential Equations - Ex 1. Chapter 1 in Review. In earlier parts, we described symbolic solutions of particular differential equations. Differential Equations \u2013 Sample Exam I (g) )Suppose (0=4. Differential Equations and Linear Algebra, 1. 5 Even and odd functions 493 5. \" Make sure you remember what proportionality and inverse proportionality are, because these words come up a lot around differential equations. 5 Linear First-Order Equations 45 1. Find differential Equations course notes, answered questions, and differential Equations tutors 24\/7. The Wolfram Language's differential equation solving functions can be applied to many different classes of differential equations, automatically selecting the appropriate algorithms without needing preprocessing by the user. Initial-Value Problems. dydx=6x+86y2+14y+4. Numerical Approximations; Use the Euler or tangent line method to find an approximate solution to a linear differential equation. Download Citation on ResearchGate | Solution of Differential Equations with Applications to Engineering Problems | Over the last hundred years, many techniques have been developed for the solution. Would my solution be acceptable? Would my solution be acceptable? I also have to solve the ivp for this problem which is y(0)=3. Integrate each side. Solving separable first order ODE\u2019s 1. I know that this differential equation is not separable, but is there a way to solve it? dy\/dx=y+x I've tried a substitution of y=vx Nonseparable Differential Equation | Physics Forums. 4 Separable Equations and Applications 32 1. In this article, we show how to apply this to ordinary differential equations. Solutions by Substitutions. Substituting these values in the general solution gives A = 1. Let us try to \ufb01gure out this adaptation using the differential equation from the \ufb01rst example. ap calculus ab: q302: differential equations and slope fields A slope field is a lattice of line segments on the Cartesian plane that indicate the slope of a function or other curve at the designated points if the curve were to go through the point. Find the solution of y0 +2xy= x,withy(0) = \u22122. 6 Substitution Methods and Exact Equations 60 CHAPTER 2 Mathematical Models and Numerical Methods 79. In the present section, separable differential equations and their solutions are discussed in greater detail. Understand the concept of mass balance, and half-life. Separable differential equation example #1 9. (b) Replace y with and x with in the ODE to get: tx dx dy ty 2 You can divide both sides by t to recover the original ODE and so this is a homogeneous ODE. Euler\u2019s method gives approximate solutions to differential equations, and the smaller the distance between the chosen points, the more accurate the result. Setting 1 \u2212 u 50 = 0 1 \u2212 u 50 = 0 gives u = 50 u = 50 as a constant solution. Solving Exact Differential Equations. We will \ufb01rst give a quick review of the solution of separable and linear \ufb01rst order equations. 6 Substitution Methods and Exact Equations. So I'll just write some partial differential equations here, so you know what they mean. 3y 2y yc 0 3. Here is an example: Example problem 1: Example problem 2:. An equilibrium solution is a constant solution, i. Suppose that the system of ODEs is written in the form y' f t, y, where y represents the vector of dependent variables and f represents the vector of right-hand-. General First-Order Differential Equations and Solutions A first-order differential equation is an equation (1) in which \u0192(x, y) is a function of two variables defined on a region in the xy-plane. We now begin an analytical study of these differential equations by devel-oping some solution techniques that enable us to determine the exact solution to certain types of differential equations. Solving differential equation(separable) Thread starter aerograce; Start date Feb 20, 2014; Feb 20, 2014. The reason we care about separable differential equations is that: Separable differential equations help model many real-world contexts. A(x) dx + B(y) dy = 0, where A(x) is a function of x only and B(y) is a function of y only. Introduction to Differential Equations Date_____ Period____ Find the general solution of each differential equation. What are Separable Differential Equations? 1. Note that y is never 25, so this makes sense for all values of t. This technique is called separation of variables. 4 Fourier series 487 5. An ode is an equation for a function of. 2 Integrals as General and Particular Solutions. The course provides an introduction to ordinary differential equations. Chapters 2, 3, 6 - First-Order Equations and Applications: Solution techniques for linear, separable and exact equations. General and Particular Solution of Differential Equation (in Hindi) Mixed Problems on Differential. Separable Differential Equations Introduction. Separable Equations The Simplest Differential Equations Separable differential equations Mixing and Dilution Models of Growth Exponential Growth and Decay The Zombie Apocalypse (Logistic Growth) Linear Equations Linear ODEs: Working an Example The Solution in General Saving for Retirement Parametrized Curves Three kinds of functions, three. 4 Differences Between Linear and Nonlinear Equations. Initial conditions are also supported. with g(y) being the constant 1. (1) Note that in order for a differential equation to be separable all the y's in the differential equation must be multiplied by the derivative and all the x's in the differential equation must be on the other side of the equal sign. We'd have to resort to numeric techniques to estimate the solutions. We'll also start looking at finding the interval of validity from the solution to a differential equation. Hence the solution to the initial value problem is. The first type of nonlinear first order differential equations that we will look at is separable differential equations. A series of free Calculus 2 Video Lessons. Occasionally, you are given a differential equation in which, if you are allowed to commit a horrendous notational atrocity, nice things can happen. differential equations have exactly one solution. 4 Direction Fields 5 1. It is now time. Double check if the solution works. An equilibrium solution is a constant solution, i. Step-by-step solutions to separable differential equations and initial value problems. 2 Separable Equations. Separable differential equations: This is right out of the first day of an ODE course. We now begin an analytical study of these differential equations by devel-oping some solution techniques that enable us to determine the exact solution to certain types of differential equations. General Solution of a Differential Equation. Separable Equations Recall the general differential equation for natural growth of a quantity y(t) We have seen that every function of the form y(t) = Cekt where C is any constant, is a solution to this differential. So by convention, the solutions of differential equations are defined on one single interval. The general solution of (1) can also be written as ?. All of the topics are covered in detail in our Online Differential Equations Course. Separable Equations - Identifying and solving separable first order differential equations. Be able to find the general and particular solutions of separable first order ODEs. De nition 1. Solutions: There are solutions to selected problems from HW2 WeBWorK: LinSep and solutions to the written part turned in. Practice your math skills and learn step by step with our math solver. Solution techniques for differential equations (des) depend in part upon how many independent variables and dependent variables the system has. And if I can, I would like to conclude the series by reaching partial differential equations. Also talked about directions fields , and how to find solutions to first-order separable and homogeneous differential equations. a separable equation are given. If an initial condition is provided, you can solve the implicit solution for an explicit solution, and determine the interval of validity, the range of x where the solution is valid. In the present section, separable differential equations and their solutions are discussed in greater detail. Description. Linear Equations. 3 Separable differential equations 1. Note that y is never 25, so this makes sense for all values of t. For briefer traditional courses in elementary differential equations that science, engineering, and mathematics students take following calculus. This facilitates solving a homogenous differential equation, which can be difficult to solve without separation. First order linear differential equation example #2 7. We will now look at some examples of solving separable differential equations. Determine the interval(s) (with respect to the independent variable) on which a solution to a separable di erential equation is de ned. Course Hero has thousands of differential Equations study resources to help you. A tank has pure water \ufb02owing into it at 10 l\/min. The first three worksheets practise methods for solving first order differential equations which are taught in MATH108. The study on the methods of solution to second order linear differential equation with variable coefficients will be of immense benefit to the mathematics department in the sense that the study will determine the solution around the origin for homogenous and non-homogenous second order differential equation with variable coefficients, the. Mathematical Models and Numerical. Lecture videos and image slides are available as a textbook resource. 4 Historical Remarks 26. Hi guys, I'm currently taking Differential Equations this semester and I have a theory question that my professor wasn't able to answer. Differential Equation of first Order and first Degree OF A differential equation of the first order and. 5 Linear First-Order Equations 48 1. 3 Slope Fields and Solution Curves. differential equations have exactly one solution. The ultimate test is this: does it satisfy the equation?. Online Solving Separable First Order Differential Equations Practice and Preparation Tests cover Differential Equations - 2, Differential Equations - 3, Differential For full functionality of this site it is necessary to enable JavaScript. 2 Integrals as General and Particular Solutions. Student Solutions Manual for Zill's Differential Equations with Boundary-Value Problems, 9th, 9th Edition Student Solutions Manual for Zill's A First Course in Differential Equations with Modeling Applications, 11th, 11th Edition. Find the solution of with initial conditions y (0) = 1 and y' (0) = 0. To solve a separable differential equation, follow these three steps: Separate the variables. All Differential Equations Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks. They're word problems that require us to create a separable differential equation based on the concentration of a substance in a tank. DIFFERENTIAL EQUATIONS PRACTICE PROBLEMS 1. The criterion for the equation M(x,y) + N(x,y)dy dx = 0 to be exact is for. These known conditions are called boundary conditions (or initial conditions). If M(x, y) and N(x, y) are both homogeneous and of the same degree, the function or N\/M is of degree 0. 1 Problem 13E. Sufficient Condition of Existence and Uniqueness: If and its partial derivative with respect to are continuous in the neighborhood region , the solution of this initial value problem in the region exists and is unique. The following examples show how to solve differential equations in a few simple cases when an exact solution exists. written as. You often get. y ' = f(x) \/ g(y) Examples with detailed solutions are presented and a set of exercises is presented after the tutorials. Mixture Problems Leading to Separable Differential Equations Solutions 1. What are Separable Differential Equations? 1. The differential equation has no explicit dependence on the independent variable x except through the function y. edu This book has been judgedto meet theevaluationcriteria set. A di\ufb00erential equation has in\ufb01nitely many solutions. 2012 Q5 - second derivative of a DE and separable equation. 5 Linear First-Order Equations. Understand how to solve differential equations in the context of chemical kinetics. 1102 CHAPTER 15 Differential Equations EXAMPLE2 Solving a First-Order Linear Differential Equation Find the general solution of Solution The equation is already in the standard form Thus, and which implies that the integrating factor is Integrating factor A quick check shows that is also an integrating factor. To solve a separable differential equation, follow these three steps: Separate the variables. Free PDF download of NCERT Solutions for Class 12 Maths Chapter 9 - Differential Equations solved by Expert Teachers as per NCERT (CBSE) Book guidelines. Separable Differential Equations Date_____ Period____ For each problem, find the particular solution of the differential equation that satisfies the initial. For most applications, the two kinds of solutions su\ufb03ce to determine all possible solutions. Order of a differential equation Exact vs numerical solutions to a differential equation Existence and uniqueness of first order ODEs Direction fields Euler's method First Order Differential Equations Separable DEs Linear DEs. DIFFERENTIAL EQUATIONS PRACTICE PROBLEMS: ANSWERS 1. Most of the solutions that we will get from separable differential equations will not be valid for all values of x. (i) (Now suppose the initial condition is 25253)=4, and let 1( ) be the solution to this new IVP. Solve differential equations by finding a general solution. If M(x, y) and N(x, y) are both homogeneous and of the same degree, the function or N\/M is of degree 0. (This should not be confused with the case of a separable ODE, which refers to a somewhat different class of problems that can be broken into a pair of. In this answer, we do not restrict ourselves to elementary functions. (1) Note that in order for a differential equation to be separable all the y's in the differential equation must be multiplied by the derivative and all the x's in the differential equation must be on the other side of the equal sign. This solution is part of the general solution family, obtained by plugging in the initial value condition. Separable differential equations are nice because it is possible to separate the two variables on either side of the equation: In the above example, dividing both sides by separates the variables Separating the variables to individual sides of the equation then allows one to compute a solution to the DE by finding antiderivatives. An initial value problem in Elementary Di\ufb00erential Equations by Boyce and DiPrima is the following: solve y0 + 2 x y = 4x y(1) = 2 and determine the interval in which the solution is valid. The Wolfram Language's differential equation solving functions can be applied to many different classes of differential equations, automatically selecting the appropriate algorithms without needing preprocessing by the user. Example: Free fall. 2: Solutions of Some Differential Equations \u2022 Recall the free fall and owl\/mice differential equations: \u2022 These equations have the general form y' = ay - b \u2022 We can use methods of calculus to solve differential equations of this form. Over the last hundred years, many techniques have been developed for the solution of ordinary differential equations and partial differential equations. SAMPLE APPLICATION OF DIFFERENTIAL EQUATIONS 3 Sometimes in attempting to solve a de, we might perform an irreversible step. when y or x variables are missing from 2nd order equations. Separable Variables. Separable Equations. Types of Problems There are six types of problems in this exercise: Which of the following is the. These problems require the additional step of translating a statement into a differential equation. FIRST-ORDER DIFFERENTIAL EQUATIONS. 3 Slope Fields and Solution Curves; 1. This might introduce extra solutions. Initial Value Problems Example equation of the solution curve. An example of a separable equation is yy0 +4xyy0 \u2212y2 \u22121=0:. ASMAR\u00b4 University of Missouri. Find such a solution and then give the related functions requested. 2 Solutions of Some Differential Equations. What is the solution of the differential. The integrating factor is e R 2xdx= ex2. It's A Problem In Differential Equations. Course Outcome(s):. Solution:-Step1Given thatWe have to classify the given equation as separable, linear, exact, or none of these. For instance, consider the equation. 519 # 1-21 odd In Section 7. 5 Linear First-Order Equations 45 1. The general solution of (1) can also be written as ?. FIRST-ORDER DIFFERENTIAL EQUATIONS. The following video provides an outline of all the topics you would expect to see in a typical Differential Equations class (i. In general every linear combinations of a separable solution is still a solution (superposition principle), so you can take the simplest separable solution, put it in a linear combination with arbitrary coefficient (complex numbers, phases) and you obtained a solution of Sc. It explains how to integrate the function to find the general solution and how. Linear Equations \u2013 Identifying and solving linear first order differential equations. Over the last hundred years, many techniques have been developed for the solution of ordinary differential equations and partial differential equations. Course Hero has thousands of differential Equations study resources to help you. Solve separable di erential equations and initial value problems. and how you can get Solutions Manual for Differential Equations Computing and Modeling and Differential Equations and Boundary Value Problems Computing and Modeling, sixth Edition Edwards, Penney & Calvis in most effective way? download solution manual for Differential Equations Computing and Modeling sixth editor. That is, a differential equation is separable if the terms that are not equal to y0 can be factored into a factor that only depends on x and another factor that only depends on y. Elementary Differential Equations with Boundary Value Problems is written for students in science, en-gineering,and mathematics whohave completed calculus throughpartialdifferentiation. order separable ODE and you can use the separation of variables method to solve it, see study guide: Separable Differential Equations. We shall continue our study of differential equations in Chapter 12 after we have learned more calculus. Separable equation is a first-order differential. Afterwards, we will find the general solution and use the initial condition to find the particular solution. That is, dy dx = g(x) f(y) The challenge now is to solve this di erential equation so that we get yas function of x. 2 Integrals as General and Particular Solutions. Integrating the equation (1), the general solution is int f(x)dx - int g(y)dy = c where c is an arbitrary constant. Di\ufb00erential equations play a central role in modelling a huge number of di\ufb00erent phenomena. > with( DEtools ) :. is a 3rd order, non-linear equation. 3y 2y yc 0 3. STUDENT SOLUTIONS MANUAL FOR ELEMENTARY DIFFERENTIAL EQUATIONS AND ELEMENTARY DIFFERENTIAL EQUATIONS WITH BOUNDARY VALUE PROBLEMS William F. 1 separable equations; We are an online community that gives free mathematics help any time of the day about any problem, no. In other words, it can be written in the form dy dx = g(x)f(y): Orthogonal trajectory. Example 1: Solve the following separable differential equations. Since the initial amount of salt in the tank is 4 4 kilograms, this solution does not apply. Separation of variables is one of the most important techniques in solving differential equations. This proven and accessible book speaks to beginning engineering and math students through a wealth of pedagogical aids, including an abundance of examples, explanations, \"Remarks\" boxes, definitions, and group projects. In this video lesson we will discuss Separable Differential Equations. General and Particular Solution of Differential Equation (in Hindi) Mixed Problems on Differential. Thus, $\\ds y=25+Ae^{-2t}$ describes all solutions to the differential equation $\\ds\\dot y = 2(25-y)$, and all solutions to the associated initial value problems. Linear Equations \u2013 Identifying and solving linear first order differential equations. Most first order linear ordinary differential equations are, however, not separable. The values of y(x) at a single point( , )x y0 0 are called initial conditions. This is a linear equation. Separable differential equations Introduction (9. Find all solutions to the di erential equations (1), (2) and (3) from Example 1. Exact Equations. 5 Even and odd functions 493 5. Understand the concept of mass balance, and half-life. Solutions: There are solutions to selected problems from HW3 WeBWorK: Exact-EU and solutions to the written part turned in. dydx=6x+86y2+14y+4. We will look more into this later. Euler\u2019s method is a way of approximating solutions to differential equations by assuming that the slope at a point is the same as the slope between that point and the next point. Multiply the equation by integrating factor: ygxf 12 1 2. Solving Exact Differential Equations. Definitions and Terminology. You can find the general solution to any separable first order differential equation by integration, (or as it is sometimes referred to, by \"quadrature\"). An equilibrium solution is a constant solution, i. Given the frequency with which differential equations arise in the world around us, we would like to have some techniques for finding explicit algebraic solutions of certain initial value problems. Finally, we will learn about systems of linear differential equations, including the very important normal modes problem, and how to solve a partial differential equation using separation of variables. Solutions by Substitutions. In this post, we will talk about separable. How this Differential Equations course is set up to make complicated math easy: This approximately 40-lesson course includes video and text explanations of everything from Differential Equations, and it includes more than 55 quiz questions (with solutions!) to help you test your understanding along the way. Equations of this kind are called separable equations (or autonomous equations), and they fit into the following form. For example, in the equation + + =, the largest derivative is the second, so the order is 2. This equation arises from Newton's law of cooling where the ambient temperature oscillates with time. Separable equations have the form dy\/dx = f(x) g(y), and are called separable because the variables x and y can be brought to opposite sides of the equation. Integrating the equation (1), the general solution is int f(x)dx - int g(y)dy = c where c is an arbitrary constant. Find the solution of with initial conditions y (0) = 1 and y' (0) = 0. Solve first order differential equations that are separable, linear, homogeneous, exact, as well as other types that can be solved through different substitutions. It's A Problem In Differential Equations. But it is separable. 8 A first order differential equation is separable if it can be written in the form \u02d9y=f(t)g(y). On that note, a solution curve is the graph of a general solution (or many general solutions) for a first-order differential equation. A solution is then a function y(x) that passes through the slopes. In fact, many of the solutions we present are only defined on a specific interval. Why could we solve this problem?. The sixth line gives the final solution to this separable differential equation (this is also an initial value problem). (f) You cannot separate the variables here. Course Outcome(s):. 4 Separable Equations and Applications 32 1. Free practice questions for AP Calculus AB - Solving separable differential equations and using them in modeling. These worked examples begin with two basic separable differential equations.","date":"2020-01-25 00:09:42","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.7712337374687195, \"perplexity\": 344.4409541787211}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2020-05\/segments\/1579250626449.79\/warc\/CC-MAIN-20200124221147-20200125010147-00221.warc.gz\"}"} | null | null |
Home Blogs Don Holding – The Most Inspirational Person In Wrestling
Don Holding – The Most Inspirational Person In Wrestling
Christopher Annino
There are many people in athletics and in life who have astonished many by overcoming
the odds. People who have been given an obstacle to overcome often times are stronger than an average person. Strength doesn't always mean physical but it does mean in this case how much one can endure and improve. When I was younger I watched a one-armed man major league baseball player who went by the name of Jim Abbott pitch a no-hitter against Cleveland. This is the story of Don Holding.
"A strongly desired goal or purpose" is one of the definitions of what a dream is. Don
Holding is a walking inspiration and miracle to everyone he has met because right now he is living his dream. When Don was younger growing up in Texas he idolized the Von Erich's.
People like Bam Bam Terry Gordy and my favorite of all time David Von Erich were my
inspirations growing up.
Don was living in the Dallas Fort Worth area in 1987. He went to Gentleman Chris
Adams School then to pursue his dream. However, the reality of the situation was the school was not within his budget, and being the ethical person that he is Don decided to take care of his family instead. "That was a long drive home that day but I love my kids," Holding said.
In 2005 Don went to a pro wrestling event with his kids. "I was disappointed with the
wrestling. There were only 25 people there and was no financial investment into the company" said Holding.
Don began to do research on what it would take to have a pro wrestling promotion.
Within weeks he became a promoter and opened up his own promotion Southern Championship Wrestling. Pro Wrestling icon Dusty Wolfe became a mentor to Don "I became a sponge I put my focus on running the business, and we were successful for a number of years until my health got bad and we decided to close it down. My weight kept climbing until it got to 350 lbs."
Holding was having a difficult time walking and decided to get healthy for himself and
for everyone else. He went to the gym religiously and carved off 100 lbs off him. This helped build his confidence to pursue his dream of being a pro wrestler. He began training at the now legendary pro wrestling training facility Dogg Pound Dojo ran by Rodney Mack and Jazz. "It was difficult and for a while, I struggled but I am so thankful for both Rodney and Jazz for not giving up on me" Don expressed.
After a long year of disciplined, vigorous training Don transformed himself into a hard-hitting outlaw cowboy. Don has the barroom brawling skills of black Jack mulligan, the
psychology of Terry Gordy, and the passion of his hero David Von Erich. Each of the men he idolized when he was younger. Each wrestler has always been in his mind and his heart which helped fuel his desire to pursue his dream.
Don Holding recently won his first title in a gauntlet-style match. Half of the match he was blinded when Stewie the Clown sprayed green mist in his eyes. This was a tremendous milestone for Don because not only did he win the Dogg Pound Championship Title but he was able to live a dream. Part of his passion and desire came from a speech by Art Williams called "Just Do It" Art Williams was a football coach and a multi-millionaire businessman.
Don is incredibly happy despite recently being diagnosed with prostate cancer however he has the following to say to his new fans "Never lose focus of your dream and don't be discouraged if you can't do it at the time and make a plan. Secure what it is, work hard, and make it right. Never let age be a deterrent. Many said I was too old and because of that, I used that as fuel to uplift me clear the path and be positive" in regards to the prostate cancer "don't be afraid to get these things checked out because there are people who depend and love you when you invest in yourself your health is the best investment. "
Pro Wrestling needs more people like Don Holding and it is his dream to one day either work or tag with the NWA's Trevor Merdoch.
Rodney Mack and Jazz are part of the reason why this man is living a dream them along
with the amazing people from Dogg Pound Dojo have built and trained numerous talents such as Soda Hendrix. What makes them great trainers is the fact that they are great people that care about their students.
On January 29th, 2022 Don Holding will be wrestling at the first-ever Dogg Pound Championship Wrestling's Teddy Long Tag Team Classic at 1103 Cincinnati Ave, San Antonio, TX. The event is named after WWE Hall of Famer Teddy Long as he has helped inspire numerous people in the wrestling community.
WCW SuperBrawl VI | WCW Sunday Night RetroView
5 Biggest AEW Stories of the Week for 01/08/23 | Revolution Main Event Set and New TNT Champion
AJPW Giant Series 2015 Night 1 | AJPW Saturday Night RetroView | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 4,906 |
Уо́лтер Клемент А́льварес (; 1884, Сан-Франциско, Калифорния, США — 18 июня 1978, США) — американский врач, профессор Калифорнийского университета, получивший известность среди широкой североамериканской публики благодаря своим многочисленным выступлениям на медицинские темы в газетах, по радио и телевидению.
Семья
Отец Уолтера, выходец из Испании, Луис Ф. Альварес (; 1853—1937), был известным американским врачом, занимающимся, кроме лечебной, также и исследовательской деятельностью. В 1878 году состоялся его брак с Клементиной Шульц, будущей матерью Уолтера. Сестра Уолтера — известная художница Мэйбл Альварес (; 1891—1985). Детство Уолтера прошло на Гавайях, куда переехала семья и где его отец работал правительственным медиком в местном лепрозории.
У Уолтера и его жены, Харриет Скидмор, в девичестве Смит (), было четверо детей: Глэдис, Луис (1911—1988, физик, нобелевский лауреат), Роберт и Береника (, р. 1913, американская писательница, фотограф и поэтесса). Внук Уолтера Клемента Альвареса, сын Луиса, также Уолтер Альварес (р. 1940), известный геолог, профессор Калифорнийского университета.
Учёба и научная деятельность
Окончив в 1910 году Стенфордский университет, Альварес начал работать практическим врачом. С 1913 и до конца 1925 года имел практику в Сан-Франциско и занимался медицинским исследованиями в Калифорнийском университете (Беркли). В 1934 году стал профессором медицины в Университете Миннесоты. Длительное время был консультантом Клиники Майо (), специализируясь на физиологии и неврогенных расстройствах пищеварительной системы.
Альварес — автор нескольких десятков книг по медицинской тематике. Награждён медалью Фрайденвальда Американской гастроэнтерологической ассоциации (1951).
Электрогастрография
Альварес заслуженно считается основателем электрогастрографии. Он первым (в 1921—1922 годах) провёл электрогастрографические исследования и дал имя новому методу («electrogastrogram»). Современная эпигастральная электрогастрография и в настоящее время делается по методике Альвареса: запись гастрографического сигнала выполняют электродами, устанавливаемыми накожно на передней брюшной стенке пациента; измерения производят в диапазоне от примерно одного до нескольких колебаний в минуту и по отклонениям записанного сигнала от частоты 3 колебания в минуту судят о наличии и виде моторных расстройств желудка.
Именем Альвареса названы
В честь Уолтера Клемента Альвареса названы:
«Синдром Альвареса» — синдром истеричного или невротического раздувания живота без какой-либо клинической причины и при отсутствии избытка газов в пищеварительном тракте.
«Волны Альвареса» — безболезненные утробные сокращения, происходящие во время всей беременности.
«Мемориальная премия Уолтера К. Альвареса», присуждаемая за лучшие подходы и методы донесения проблем здравоохранения до широкой аудитории.
«Награда Альвареса по электрогастрографии» (), присуждаемая за лучшую работу в области электрогастрографии на конференциях «Международного электрогастрографического общества» (IEGGS).
«Американский домашний доктор»
C 1950 года Альварес начал писать медицинскую колонку, которая скоро стала распространяться по всей Северной Америке в сотнях ежедневных и еженедельных газет. Благодаря этому и многочисленным выступлениям по радио и телевидению он стал одним из самых известных врачей в США и его стали называть «Американский домашний доктор».
Примечания
Внешние ссылки
Ранняя фотография Уолтера Клемента Альвареса
Первая электрогастрограмма человека, полученная Альваресом в 1922 году
Учёные США
Медики США
Родившиеся в Сан-Франциско | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 7,600 |
Here are some old websites for crafting that went offline, here are the new links, copy the info while you can!
Here is a better spreadsheet that was created by Shena, Bobthebuilder, and a few others including me. It is posted under the FAQ section. | {
"redpajama_set_name": "RedPajamaC4"
} | 2,358 |
IR INSIDER | IR SOCIETY
IR Society and IR INSIDER Alumni
Eastern-Central Europe
Southeast Asia - Oceania
Economics - Business
Music - Culture
UN | State Dept. | White House
Foreign News Monitoring
Op-Eds and Cartoons
Powered by IR Society at NYU
IR Insider is a production of NYU's International Relations Society. Our goal is to explain and discuss issues in IR in an engaging and thought-provoking fashion. We are written by students, for students, about issues students care about.
Sri Lankan President Adopts Hardline Stance on Drug Trafficking
February 14, 2019 / Sarika Bhattacharjee
An advertisement attempting to recruit two hangmen in Colombo, Sri Lanka. Photo: Reuters, Dinuka Liyanawatte
Sri Lanka's president, Maithripala Sirisena, recently announced his intention to continue utilizing capital punishment in drug trafficking cases in the next two months. Sirisena's new hardline approach is inspired by the Philippines' war on drugs, led by President Rodrigo Duterte.
This week, Sri Lanka's state-run newspaper, Daily News, ran an advertisement recruiting two hangmen, or executioners. The applicants must be male and between 18 to 45 years old with "mental strength" and "excellent moral character" in order to be eligible for the two advertised vacancies. The indicated monthly salary for the designated position is of 36, 310 rupees (or $203.99), which surpasses Sri Lanka's standard pay for governmental positions.
Sri Lanka has not used the death penalty in any drug trafficking case since 1976, when a moratorium on capital punishment was announced. Since then, criminals originally sentenced to death row have served life sentences in jail instead.
However, after four decades of not executing prisoners convicted of drug-related crimes, Sri Lanka is being taken in a new direction by its president. President Sirisena's concerns about drug trafficking come from a view that Sri Lanka is at risk of becoming a "transit hub" for the Asian narcotics trade.
President Sirisena is following in the footsteps of controversial and outspoken President Rodrigo Duterte of the Philippines. During President Sirisena's state visit to the Philippines in January, the Sri Lankan leader told President Duterte: "the war against crime on drugs carried out by you is an example to the whole world, and personally to me. Drug menace is rampant in my country and I feel that we should follow your footsteps to control this hazard."
President Duterte responded by acknowledging the potential for a partnership between the Philippines and Sri Lanka in fighting against international drug trafficking.
However, other international figures are much more wary about President Duterte's war on drugs, which has already taken the lives of approximately 12,000 people, according to the Human Rights Watch. HRW stated that Duterte's campaign has inspired violence that could amount to "crimes against humanity."
In the next few months, Sri Lanka's President Sirisena may become President Duterte's partner in his efforts to combat drug crimes.
Sri Lankan President Maithripala Sirisena standing with President Rodrigo Duterte of the Philippines. Photo: AFP, Express Tribune
Former Maldives President Under Investigation for Money Laundering
Board for Sabarimala Temple Withdraws From Challenging the Ban on Women
HEADLINES EVERY MORNING | MONTHLY EVENT BRIEFS | UPDATES ON THE GO | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 6,373 |
The dashboard is the hub of your feedback; you get an aggregated view of line graphs, pie-charts, and maps. Instantly see changing feedback trends & discover more about your visitors.
Central to this is the emotional trendline where you can see how visitor sentiment has changed over time. Discover the impact of site changes, good or bad to make informed decisions for future modifications.
Further graphing options can be found in your surveys. Each survey element (radio buttons, dropdowns, NPS, etc.) has its own graph. Immediately see an aggregation of your results, making it easier to see where you need to take action. | {
"redpajama_set_name": "RedPajamaC4"
} | 3,207 |
Q: Issue with passing global "external" variables to Angular 2 application I've been setting up a global external object which I can then pass to my Angular application. I'm doing this as the application will be provided to different websites and so each website can then pass different configurations.
This is my setup:
index.html
<script type="text/javascript">
var appConfig = {
welcome_msg: 'Welcome to Test App'
};
</script>
Exporting the appConfig object in the application (test-export.ts)
import { OpaqueToken } from '@angular/core';
export let APP_CONFIG = new OpaqueToken('appConfig');
Injecting appConfig in a component
...
import { APP_CONFIG } from '../../app/test-export';
@Component({
selector: `<home></home>`,
providers: [
{ provide: APP_CONFIG, useValue: appConfig }
],
styleUrls: [ './home.style.css' ],
templateUrl: './home.template.html'
})
export class HomeComponent implements OnInit {
constructor(@Inject(APP_CONFIG) private appConfig: any) { }
}
Using value in the template
<h1>{{ appConfig.welcome_msg }}</h1>
All of the above works fine and I get no errors when running the application, however after I build the code I'm getting the following error:
Cannot find name 'appConfig'.
In fact even in Visual Studio Code it's being marked as "incorrect":
I'm using WebPack and when I'm running in "dev mode" I simply get the error when I build but the application runs successfully. However the issue is that when I try to build in "prod mode" the application does not build because of this error.
As references I was following this question on StackOverflow and this article. Am I missing an export, or could it somehow be a configuration issue?
A: You need to import the file that contains appConfig. If the script is in index.html this is not possible.
What you can do is to make it global by assigning it to window
<script type="text/javascript">
window.appConfig = {
welcome_msg: 'Welcome to Test App'
};
</script>
providers: [
{ provide: APP_CONFIG, useValue: (<any>window).appConfig }
// or
{ provide: APP_CONFIG, useValue: (window as any).appConfig }
// or
{ provide: APP_CONFIG, useValue: window['appConfig'] }
],
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 483 |
Perynea viridicincta är en fjärilsart som beskrevs av George Francis Hampson 1897. Perynea viridicincta ingår i släktet Perynea och familjen nattflyn. Inga underarter finns listade i Catalogue of Life.
Källor
Nattflyn
viridicincta | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 3,118 |
Category Archives: Media Center
Category "Media Center"
2020-12-15 Media CenterBy osang
A South Korean medical diagnostics maker has ramped up production and is aiming to sell millions of virus test kits to the U.S., where cash-strapped states are scrambling for federal funds to buy them.
Osang Healthcare, a South Korean manufacturer and distributor of medical devices, announced that after receiving approval from the FDA on March 18, the cumulative order quantities of COVID-19 test kits had surpassed 10 million.
Osang Healthcare said it has won the U.S. Food and Drug Administration's approval for emergency use of the test kit for the new coronavirus, which was the first for a Korean firm.
Osang Healthcare's coronavirus test kits have won an emergency use authorization from regulators in the United States – a first for a Korean company.
In response to an appeal from President Trump, South Korea is set to deliver 600,000 coronavirus testing kits to the U.S. on Tuesday, according to a new report.
Government has delayed order of South Korean kits as they have not been tested, it is claimed
Bulgaria to buy PCR tests for COVID-19 from S. Korea's Osang Healthcare
SOFIA (Bulgaria), April 10 (SeeNews) – Bulgaria's government said on Friday that it has empowered health minister Kiril Ananiev to sign an agreement for the purchase of PCR tests for novel coronavirus disease (COVID-19) testing from South Korea's Osang Healthcare.
Morocco to receive 100,000 testing kits from South Korea's Osang Healthcare
The Health Ministry has announced having allocated two billion dirhams to upgrade its medical devices, including the acquisition of 100,000 testing kits for the coronavirus.
Italian diagnostics firm EliTech Group said on Tuesday that South Korea's OsangHealthcare has received CE marking for its.. | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 2,291 |
@extends('front.templatelogin')
@section('main')
<div class="cd-user-modal"> <!-- this is the entire modal form, including the background -->
<div class="cd-user-modal-container"> <!-- this is the container wrapper -->
<ul class="cd-switcher">
<li>{!! link_to('auth/login', trans('front/login.connection') ) !!}</li>
<li class="selected"> <a href="#">{!! trans('front/login.register') !!}</a></li>
</ul>
<div id="cd-signup"> <!-- sign up form -->
{!! Form::open(['url' => 'auth/register', 'method' => 'post', 'role' => 'form','class' => 'cd-form','id'=>'form']) !!}
<h2 class="intro-text text-center">{{ trans('front/register.title') }}</h2>
@if(session()->has('error'))
@include('partials/error', ['type' => 'danger', 'message' => session('error')])
@endif
<p class="fieldset">
{!! Form::control('text', 6, 'username', $errors, trans('front/register.pseudo'), null, [trans('front/register.warning'), trans('front/register.warning-name')],'Username','full-width has-padding has-border','signup-username') !!}
</p>
<p class="fieldset">
{!! Form::control('email', 6, 'email', $errors, trans('front/register.email'),'','','E-mail','full-width has-padding has-border','signup-email') !!}
</p>
<p class="fieldset">
{!! Form::control('password', 6, 'password', $errors, trans('front/register.password'), null, [trans('front/register.warning'), trans('front/register.warning-password')],'Password','full-width has-padding has-border','signup-password') !!}
</p>
<p class="fieldset">
{!! Form::control('password', 6, 'password_confirmation', $errors, trans('front/register.confirm-password'),'','','Password Confirm','full-width has-padding has-border','signup-password2') !!}
</p>
{!! Form::text('address', '', ['class' => 'hpet']) !!}
<p class="fieldset">
{!! Form::submit(trans('front/form.send')) !!}
</p>
<div class="row">
<div class="col-lg-12">
{!! link_to('password/email', trans('front/login.forget')) !!}
</div>
</div>
<p class="fieldset">
<a href="{!! url('/') !!}">{{ trans('front/missing.button') }}</a>
</p>
{!! Form::close() !!}
</div>
</div>
</div>
@stop
@section('script')
{!! HTML::script('js/mainregister.js') !!}
<script src="http://ajax.aspnetcdn.com/ajax/jquery.validate/1.11.1/jquery.validate.js" charset="utf-8"></script>
<script src="http://ajax.aspnetcdn.com/ajax/jquery.validate/1.11.1/additional-methods.js" charset="utf-8"></script>
<script src="http://netdna.bootstrapcdn.com/twitter-bootstrap/2.3.2/css/bootstrap-combined.min.css" charset="utf-8"></script>
<script type="text/javascript">
$('#form').validate({
rules: {
username: {
minlength: 5,
maxlength: 30,
required: true
},
email: {
required: true,
email: true
},
password: {
minlength: 8,
required: true
},
password_confirmation: {
equalTo: "#signup-password"
}
},
highlight: function (element) {
$(element).closest('input').removeClass('success').addClass('error');
},
success: function (element) {
element.addClass('valid')
.closest('input').removeClass('error').addClass('success');
}
});
</script>
@stop
| {
"redpajama_set_name": "RedPajamaGithub"
} | 2,807 |
{"url":"https:\/\/socratic.org\/questions\/the-world-future-society-predicts-that-by-the-year-2020-airplanes-will-be-able-t","text":"# The World Future Society predicts that by the year 2020, airplanes will be able to carry 1400 passengers. Today\u2019s biggest jets can carry 600 passengers. What will be the percent of increase of airplane passengers?\n\nJan 21, 2017\n\nThe will be a 133.3% increase rounded to the nearest tenth.\n\n#### Explanation:\n\nThe formula for calculate the rate or percentage of change across a period of time is:\n\n$p = \\frac{N - O}{O} \\cdot 100$\n\nWhere:\n\n$p$ is the percent change\n$N$ is the New Value, 1400 for this problem\n$O$ is the Old Value, 600 for this problem\n\nSubstituting and calculating for $p$ gives:\n\n$p = \\frac{1400 - 600}{600} \\cdot 100$\n\n$p = \\frac{800}{600} \\cdot 100$\n\n$p = \\frac{80000}{600}$\n\n$p = 133.3$","date":"2020-02-18 23:06:28","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 9, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.6086969375610352, \"perplexity\": 1833.831795758183}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2020-10\/segments\/1581875143815.23\/warc\/CC-MAIN-20200218210853-20200219000853-00183.warc.gz\"}"} | null | null |
{"url":"https:\/\/crypto.stackexchange.com\/questions\/24240\/koblitz-encoding-a-message-to-a-point-what-is-the-associated-auxiliary-base-pa","text":"# Koblitz encoding a message to a point, what is the \u201cassociated auxiliary base parameter\u201d?\n\nI am looking at the Koblitz method for encoding a message as an elliptic curve point. The first step given in the paper I'm reading is:\n\n\"Choose an elliptic curve and its associated auxiliary base parameter, k.\"\n\nI have not heard the term \"associated auxiliary base parameter\" before. Does it mean the order of the curve, or something else?\n\nBased on a quick Google search, I assume you're reading the paper \"Modified Koblitz Encoding Method for ECC\" by Kodali and Sarma (Int.J.Rec.Tr.Eng.Tech., vol. 8, no. 1, Jan 2013).\n\nThe fact that the paper doesn't actually cite any source for the encoding scheme, or even cite anything by Koblitz, makes me a bit skeptical of the paper's quality. For that matter, so does the fact that they seem to be using tiny curve sizes and encoding individual ASCII characters as curve points. The security levels they seem to be claiming in their last figure also look absurdly low (9 bits, really??), which at least seems honest, if not particularly useful.\n\nAll in all, while I admit to only skimming the paper, and that I may have missed some crucial detail, at least at a glance it certainly trips my bullshit detector. I suggest you, at the very least, read it with a highly critical eye.\n\nIn any case, I assume that, by \"Koblitz's encoding method\", they're referring to one of the three encoding schemes described in section 3 of Koblitz's original 1987 paper, \"Elliptic Curve Cryptosystems\" (Koblitz, N.; Mathematics of Computation 48(177), January 1987, pp. 203\u2013209). It's not 100% clear to me which one they're describing, but I suspect it's most likely the second one, where the plaintext $m$ is mapped to a curve point by multiplying it by some constant $k$ (1,000 in the example) and testing all $mk \\le x < (m+1)k$ by brute force to (hopefully) find an $x$ that corresponds to a curve point.\n\nPs. You may also find these earlier related questions interesting:\n\nPps. I took a closer look at the Kodali & Sarma paper to see which EC cryptosystem they're actually using, and I couldn't make any sense of it \u2014 it looks as if they're effectively just running a symmetric Caesar cipher over an elliptic curve (after first doing ECDH key agreement). If so, that still makes absolutely no sense to me; it's not semantically secure, and anyway they'd be much better off just feeding the ECDH secret (computed over a secure curve, not the tiny one they seem to be using) to a KDF and using it to key a standard symmetric cipher, like normal people do.\n\n\u2022 Oh of course ECIES is the way to go. But this is academic I am just trying to learn how it works. and how to implement it. \u2013\u00a0colobusgem Mar 4 '15 at 13:08\n\u2022 Also the method decodePoint here docjar.org\/docs\/api\/org\/bouncycastle\/math\/ec\/ECCurve\\$Fp.html, what is this method doing? case case 0x03 seems to look a bit like method 2 in Koblitz's paper. \u2013\u00a0colobusgem Mar 4 '15 at 15:15\n\u2022 @colobusgem There are serious academic papers and pseudo (crappy) academic papers. The one you are looking at seems to fall in the latter category ;) \u2013\u00a0DrLecter Mar 4 '15 at 19:12\n\u2022 That makes sense. Either way I have managed to implement what I think is algorithm 2 from koblitz's paper in java and have found that (obviously) some integers can easily be encoded on to a point, and others not so much. Is there a way to figure out the probability an integer will be successfully encoded before starting out and brute forcing all possible x's? Also on a probability note: I assume it is better to randomly select from the pool of possible X's as opposed to starting at the lowest and working up to the highest. \u2013\u00a0colobusgem Mar 5 '15 at 16:34","date":"2019-12-08 00:33:45","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.5506158471107483, \"perplexity\": 684.1743867186677}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2019-51\/segments\/1575540503656.42\/warc\/CC-MAIN-20191207233943-20191208021943-00019.warc.gz\"}"} | null | null |
Северный комбайн Анвельта и Григорьева. Модель № 5А (СКАГ-5А) — советский прицепной безмоторный зерноуборочный комбайн, предназначенный для работы в северных районах СССР, где выпадает в 2-3 раза больше осадков, ниже температура и выше влажность зерна во время уборки, убираемые участки характеризуются чаще всего небольшими размерами. Впервые изготовлен в количестве 15 штук в 1935 году на ленинградском заводе «Красная вагранка» (сейчас Ленпродмаш), крупносерийно производился на Люберецком заводе сельскохозяйственных машин им. Ухтомского с 1936 по 1941 года.
Предшественники
Испытания и производство
Впервые СКАГ-5А изготовлен в количестве 15 штук в 1935 году на ленинградском заводе «Красная вагранка» посредством модернизации комбайна СКАГ-5, выпущенного в количестве 5 штук на заводе Института механизации сельского хозяйства в 1934 году. От своего предшественника новый комбайн отличался увеличенной шириной платформы жнеи, регулировкой мотовила на ходу, цепной передачей к мотовилу (СКАГ-5 имел карданную передачу), регулировкой положения подбарабанника эксцентриком, конструкцией решёток и барабанов, сменным решетом, трёхшарнирной карданной передачей от трактора, большей шириной обода главного колеса и т. д. Технологическая схема работы обоих комбайнов — поперечно-прямоточная.
Испытания СКАГ-5А проходили в Ленинградской, Московской, Омской областях, в Западносибирском и Дальневосточных краях. В неблагоприятном 1935 году потери зерна у СКАГ-5А оказались ниже, чем у СКАГ-5, СКАГ-7А и южных комбайнов Коммунар и Сталинец-4, сезонная выработка также оказалась лучше. А вот на испытаниях в Дальневосточном крае оказалось, что для уборки риса СКАГ-5А не приспособлен из-за малой проходимости на поливных полях. Однако у комбайна имелись более значительные недостатки, как то: примеси в зерне составляли 5 — 10 %, сложность перевода машины из рабочего положения в транспортное. Для решения первой проблемы было сконструировано 4 образца (СКАГ-5АУ-№) с различными очистками, но эти образцы так и не пошли в серию, хотя испытания на чистоту зерна прошли успешно. Конструированием очисток для СКАГ-5А занимались и отдельные механизаторы, известна, например, очистка С. А. Долинкина из Ярославской области, которая была сделана наподобие второй очистки комбайна Коммунар. Очистка Долинкина давала меньше примесей, но при этом 2-4 % зерна терялось в отходах.
Но для северных условий СКАГ-5А был незаменим, что и показали испытания в северо-западных районах, поэтому с некоторыми конструктивными улучшениями он в 1936 году начал крупносерийно выпускаться на Люберецком заводе сельскохозяйственных машин им. Ухтомского. Помимо зерновых культур, комбайн работал на уборке семенного клевера (созданием оборудования для уборки клевера занималась опытная станция механизации с/х Ленобласти в 1936—1940 годах), гороха, вики, корпандра с установленным дополнительно оборудованием. В 1941 году в связи с началом Великой Отечественной войны выпуск СКАГ-5А прекратился, всего было выпущено 7000 машин. Высокое количество примесей в зерне и сложность перевода комбайна в транспортное положение стали главными причинами невозобновления его производства после войны. СКАГ-5А работали в северо-западных районах СССР, Сибири, Средней Азии, поставлялись за границу и были награждены дипломом Парижской выставки в 1937 году.
Принцип работы
Комбайн СКАГ-5А состоит из крепко связанных между собой хедера и молотилки. При рабочем положении комбайн опирается на два колеса, на прицеп трактора. У комбайна отсутствует мотор, а движение передаётся от трактора СТЗ-ХТЗ. В конструкции хедера использовались механизмы сноповязалки Люберецкого завода, например, режущий аппарат аналогично аппарату сноповязалки состоит из пальцевого бруса, пальцев со вкладышами, прижимных пластинок, пластинок трения и спинки ножа с приклёпанными сегментами. Высота резания прямо во время работы изменяется наклонением комбайна при помощи вращения рукоятки. Также высоту резания можно регулировать механизмом подъёма, который меняет положение полевого и ходового колёс относительно рамы комбайна. Положение мотовила относительно режущего аппарата регулируется посредством двух рычагов. Работа режущего аппарата, мотовила и полотен аналогична работе соответствующих механизмов обычного комбайна Регулировка натяжения обоих полотен осуществляется ремнями. Срезанные и уложенные на большое полотно стебли транспортируются в приёмную камеру 3. В приёмной камере имеется подаватель 4. За барабаном 5 расположены три соломочёса 6, 7 и 8. Значительная часть зерна и мелких частей вороха (до 75 %) проваливается через решётку деки барабана на скатную доску, и оттуда попадает на решето грохота 9. Солома с оставшимся в ней невыделенным зерном последовательно проходит через все три соломочёса. Здесь проходит выделение остальной части зерна из соломы. Зерно и мелкие примеси, проваливаясь через отверстия решёток соломочёсов, также попадают на решето грохота. Последний соломочёс выбрасывает солому в копнитель из молотилки. Скорость лопостей 10 регулируется степенью открытия входных окон с помощью заслонок. Струя воздуха, идущая от вентилятора, препятствует забиванию решета грохота и одновременно выносит из молотилки часть наиболее лёгких примесей. Крупные примеси вороха, не провалившиеся через отверстия решета, выбрасываются из молотилки. Зерно, сорняки и наиболее тяжёлые мелкие примеси, а также часть мякины проваливаются через решето и по скатной доске попадают в шнек 11. Из шнека вся эта масса подаётся элеватором 12 в один из двух мешков, подвешенных к хоботку элеватора. Направление перемещающегося элеватором в ящик продукта регулируется вращающейся заслонкой.
Источники
В. Г. Антипин, С. М. Григорьев, Я. Д. Абрамов, С. М. Коган. Работа северных комбайнов и их усовершенствование, М.-Л.: Государственное издательство сельскохозяйственной литературы, 1951
Библиотека по агрономии. Комбайн
Зерноуборочные комбайны
Появились в 1935 году в СССР | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 6,475 |
With 23 yrs experience we are standing ready to convey our skills and knowledge. Affordable.
We travel as well.We also sell Puppet-show boxes as well as hand-made Puppets. Or you can order a Manuel and a 90 minute DVD.
my name is Meike Wiechert and Iam working as a german volunteer at the Inkululeko Community Centre in Salvokop, Pretoria. The Inkululeko Community Center has a Day Care Center for Childre from 4 to 6 years old, a homeworkcenter for the preschool children and a holiday program called vana va hina. The vana va hina program is creating a holiday program for the children during their holidays. In our next holiday program we would like to show our 6-13 year old children a puppet show. The holiday program will be at theend of march till the beginning of april. We would like to do the puppet show on the 5th of april. Could you please send me a email, if it could be possible to offer our children one of your nice puppet shows for a small amount or for free?
Please send me a email with your answear. | {
"redpajama_set_name": "RedPajamaC4"
} | 4,087 |
ALSO BY CHARLES KAISER
_1968 in America: Music, Politics, Chaos, Counterculture, and the Shaping of a Generation_
_The Gay Metropolis: The Landmark History of Gay Life in America_
Copyright © 2015 by Charles Kaiser
Production editor: Yvonne E. Cárdenas
Text designer: Julie Fry
All rights reserved. No part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or by any information storage and retrieval system, without written permission from Other Press LLC, except in the case of brief quotations in reviews for inclusion in a magazine, newspaper, or broadcast. For information write to Other Press LLC, 2 Park Avenue, 24th Floor, New York, NY 10016.
Or visit our Web site: www.otherpress.com
The Library of Congress Cataloging-in-Publication Data
was unavailable at the time of this book's printing.
Hardcover ISBN 978-1-59051-614-0
E-book ISBN 978-1-59051-615-7
v3.1
For Joe
_&_
For Christiane
_naturellement_
_If mankind lasts because of the masses of people for whom enduring has a higher value than acting, its fate is determined by those who choose, act, and decide._
— Stanley Hoffmann, paraphrasing Paul Valéry
_This is such a story of coincidences — or luck, or destiny._
— Eric Katlama
# CONTENTS
_Cover_
_Other Books by This Author_
_Title Page_
_Copyright_
_Dedication_
_Epigraph_
_Prologue_
Part I
Part II
_Afterword_
ACKNOWLEDGMENTS
PRINCIPAL ACTORS
NOTES
SELECT BIBLIOGRAPHY
PHOTO CREDITS
# _Prologue_
THE FIRST TIME I SAW PARIS, I FELL IN LOVE.
It was 1962, and I was eleven years old.
Her name was Christiane Boulloche-Audibert.
She was a beguiling thirty-eight-year-old brunette, the mother of four children, a campaigner for women's rights, and a hero of the French Resistance. She was strong and warm, percolating with life and love, and ten years younger than my own mother.
Her husband was Jean Audibert, a brilliant, high-spirited engineer with a belly-shaking laugh, a passion for fast cars, and a gigantic joie de vivre.
Jean had fought in the Free French Navy, and he witnessed the Normandy invasion from the sea—a good war, but not nearly as fraught as Christiane's. Their sons and daughters, each just a little bit older or younger than I was, were smart, political, precocious, and bilingual, which made them very much like everyone in my family.
I was the third of three sons. My father had just become a diplomat, and we were living in Dakar, in West Africa, where John F. Kennedy had made him his ambassador.
As a child, I was obsessed with the black-and-white images of World War II. To a young American who hadn't lived it, the war glowed with the romance of victorious history. When I met Christiane, the war was seventeen years behind her—an expanse that felt like a lifetime and a half to an eleven-year-old.
My uncle, Henry Kaiser, who moved in with Christiane and Jaqueline Boulloche at the end of 1944, when he was an American lieutenant stationed in Paris. He quickly learned everything that had happened to them while they were in the Resistance.(photo credit 1.1)
It did not occur to me that for Christiane, it felt more like the day before yesterday. And, as she admonished me twenty years later, to her it had never been romantic _at all._
In the fall of 1944, just after the Liberation of Paris and the triumphant return of Charles de Gaulle, Christiane and her sister, Jacqueline, saw an ad in the newspaper seeking housing for U.S. Army officers. That ad created a bond that continues after seven decades.
My uncle, Henry Kaiser, was the lieutenant they took in, rent free, "in gratitude to the Allies," as Christiane always explained it. They installed him in an empty bedroom in their parents' sprawling apartment in the 16th arrondissement.
The most dramatic movie about the war was one I learned by heart but had seen only in my head. All of its images came from my uncle Henry, a charismatic storyteller with animated eyebrows and magnetic good looks. In an implausibly deep, tobacco-tinged baritone, he re-created the movie's layers of suspense for me, over and over again.
The film starred Christiane, Jacqueline, and their brother André, and its plot was exhilarating.
During the twelve months my uncle Henry lived in their apartment, he learned about almost everything the two sisters had endured during the four years the Nazis occupied Paris. There had been many narrow escapes. I reveled in their brave adventures and their incredible grace under pressure.
Years later, as the red wine flowed freely around the dining room table at their apartment in square Alboni in Paris, or the picnic table at their country house in Fontainebleau, where Christiane and Jean were joined almost every weekend by Christiane's extended family, Kaisers and Audiberts and Boulloches dissected everything from de Gaulle and Kennedy to _Jules and Jim._ But I don't remember anyone telling the stories I had heard from my uncle about the war. I may have assumed that it was not discussed because the grown-ups had already talked about it so often. But there was also the number tattooed on André's forearm—the first one I had ever seen—and André's grim demeanor, hinting at unvanquished demons.
Nevertheless, for a very long time, I did not realize that World War II was a taboo subject within Christiane's family. The ones who had been so magnificent in the Resistance never discussed their bravery with their own children.
They actually avoided anything that might remind them of those piercing years. Christiane had had many close friends in the Resistance, but after the war ended, she never saw any of them again. She blotted out that part of her life, as much as she could, after she and her sister decided that "it was necessary to turn the page."
I grew up inspired by the story of my remarkable French cousins, whom I thought of as a branch of my own extended family. Christiane was like a beacon. Her life proved that you could do the right thing, the most difficult thing, if you were determined to do it. Perhaps one wouldn't, but one always could—even under the most chilling circumstances.
But their own children were never really nourished by their parents' bravery. They admired it, they appreciated it, they were intimidated by it — but it never felt nurturing.
What they experienced most of the time was an amorphous black cloud, never fully visible, hovering somewhere above their parents' past.
It would take me five decades, including two and a half years living in France, to unravel the reasons for the heroes' silence.
The answer is _The Cost of Courage._
# _Part I_
# _One_
PARIS — JANUARY 12, 1944
IT IS a few minutes before four on a gray Paris afternoon when the black Citroën Traction Avant pulls up in front of a drab apartment building in the rue de la Santé on the Left Bank. The low-slung, front-wheel-drive Citroën is famous as the getaway car for French gangsters, but now it has acquired a more menacing pedigree: It is the official automobile of the German secret police.
Two Gestapo agents in black leather raincoats jump out onto the sidewalk. They pull a single prisoner, a short twenty-year-old Frenchman named Jacques, out of the car after them. The youth's nearly limp body broadcasts defeat, but he shows no obvious marks of a beating.
Two and a half miles away, a swastika sways in the wind atop the Eiffel Tower. It is the one thousand three hundred and eighth day of the Nazi Occupation of Paris. Dozens of other swastikas defile the French capital. Below them, street signs written in German punctuate the avenues with unfamiliar accents, humiliating Parisians at every carrefour.
The city's best restaurants, like Maxim's and La Tour d'Argent, are still flourishing, but now their customers are mostly German officers and their young French companions.
German street signs punctuate Paris avenues with unfamiliar accents, humiliating the French at every carrefour.(photo credit 1.2)
Starvation rations for the French have transformed apartment terraces into rabbit farms, as the urban dawn is oddly heralded by roosters. More fortunate Parisians rely upon the generosity of country cousins, who have much more access to food.
Daytime Paris echoes to the sound of shoes with wooden soles clip-clopping down its narrow side streets and grand avenues. "If an old pair of shoes needs a new sole, you can't do anything about it, because there is no leather," said Pierre Mendès-France. "It's really very difficult to describe what life is like in a country where everyone spends all their time looking for things."
The nightly blackout means the only authorized light outdoors is the eternal flame under the Arc de Triomphe. Electricity and gas are both erratic. Heated apartments are a dimly remembered luxury from 1940. Only seven thousand cars circulate on the streets of the City of Light — many of them converted to run on wood. They are called _gazogènes._ Two million bicycles are the best way to get around aboveground. But a good bicycle can cost 10,000 francs — almost as much as a car did before the war. The only taxis are pedicabs pedaled by bicycle riders, or "taxis hippomobiles," pulled by a single horse. The fastest pedicab is propelled by veterans of the Tour de France.
Bicycle power also keeps the movie theaters open: four men pedaling a generator at thirteen miles an hour for six hours can produce enough reliable electricity for two full shows.
Jews and Communists are the first victims of the Occupation, but the dangers of resisting the Nazis are escalating for everyone at the beginning of 1944. Huge yellow posters plastered on the walls of the Métro proclaim that members of the Resistance are no longer the only ones facing German firing squads. A new edict ordains that their fathers, cousins, and in-laws will be executed as well.
And yet, in January, there is a new uncertainty bubbling underneath the humid winter air. All across Nazi Europe, the occupied are buoyed, and the occupiers menaced, by the event that everyone knows is coming, but no one knows exactly when or where: the Allied invasion of the Nazi-ruled continent.
Across the English Channel, massive numbers of British, American, Canadian, and French troops are gathering on the southern coast of England, where General Dwight Eisenhower is making the plans for a spectacular invasion of France. Thanks to a huge disinformation campaign, its location remains a secret six months before the assault begins. After four years of war, a cautiously optimistic Churchill believes that the biggest danger now facing the Allies is stalemate rather than defeat.
YESTERDAY, the young prisoner accompanying his German captors was a proud member of the French Resistance. Today, he is leading the German agents to the secret address he had sworn to conceal, so that they can arrest his boss, André Boulloche — a man he worships. If André is really there, the Germans have promised Jacques that he will be rewarded with his freedom.
But can he believe them?
Jacques is young looking, even for his age; especially today, he looks practically like a little boy. Deeply religious, he has joined the Resistance just three months earlier, after being recruited by his Sorbonne classmate, André's sister Christiane Boulloche. Christiane has no trouble persuading him to join their cause. When she asks him if he wants to work for her brother, the boy signs up immediately, without hesitation or reflection.
Jacques is the same age as Christiane, who turned twenty at the end of 1943. Christiane is smart, strong, and attractive. She also has a prominent nose that she thinks is unattractive. She worries, perhaps, that it makes her less glamorous looking than her older sister, Jacqueline, who has joined the Resistance with her.
Christiane's clandestine duties require her to ride her bicycle all over Paris, sometimes as much as sixty miles in a day. She picks up telegrams from secret drop-off points and decodes them, transports forbidden radio equipment, and sometimes smuggles guns through the capital, usually in a basket underneath eggs or vegetables.
All Boulloches share an innate sense of duty. When Christiane returns from the countryside after the armistice to find German soldiers goose-stepping through Paris, she is consumed by a single thought: "This is wrong." Before the war started, she had been certain: "We wouldn't just resist them, we would beat them. That's why the Occupation was a thunderclap."
Coupled with youthful fearlessness, and hero worship of her brother, that simple notion — "This is wrong" — propels her into the underground fight against the Germans. She is hypnotized and horrified by the Occupation. It swallows all of her attention.
The Boulloche sisters' very first act of resistance occurs when they are stopped by two German soldiers on avenue du Président Wilson. When the young Germans ask for directions to Place de la Concorde, the girls cheerfully dispatch them in the opposite direction.
There is no heat at her lycée, and Christiane wears gloves to turn the pages of the classroom dictionary. She is upset when one of her Jewish teachers loses her job, but she does not consider the plight of the Jews to be the most important thing. More than anything else, it is instinctive patriotism that pushes her into battle.
When the Germans are finally driven out of France, everyone's nightmare will be over.
Or so she believes.
As 1944 begins, her brother André François Roger Jacques Boulloche has been back in France for only four months. He is an engineer, a lawyer, and something of an adventurer. He and his sisters come from many generations of Catholic judges and prominent civil servants.
Iconoclasm is a leitmotiv in their family: Two Boulloche ancestors were members of the Cour de Cassation, the highest court in France, at the turn of the century. Both of them, remarkably, had been pro-Dreyfus: a belief that had made them strangers to their class — because they were partisans of the truth.
FOUR MONTHS EARLIER, in the second week of September 1943, André has taken off from England under a nearly full moon, with seven other passengers in a single-engine Westland Lysander. Many underground fighters are being parachuted into occupied France, but the plane carrying this group touches down on a secret airstrip in the Loire Valley, near Tours.
These landings are dangerous, because there is always the possibility that the Germans have been tipped off. This one has been organized by Jean-François Clouet des Pesruches, who has arrived the night before from London, and it goes off without a hitch. Resistance members outline the tiny runway with flashlights pointed straight up at the sky. Foil extends over the tops of their torches, to make them invisible to everyone except the airplane circling above them.*
André is a handsome twenty-eight-year-old with brown hair and thick eyebrows that hover over a permanent glint in his eyes. Nearly six feet tall, he walks with a tempered, youthful swagger. Before the war, he was considered something of a dandy.
André has been ordered back to occupied France by Charles de Gaulle, to be the general's personal military delegate in Paris. Pseudonym, Armand, code name, Hypotenuse, André's charge from the renegade general† is to bring some order to the burgeoning Resistance movements now operating in eleven different departments in northern France.
During André's absence from France in 1943, there has been a dramatic increase in the membership of the Resistance. In the fall of 1942, the collaborationist Vichy government has taken one of its most unpopular steps, shipping off two hundred thousand Frenchmen to work in Germany. But with so many German soldiers fighting on so many different fronts, that isn't nearly enough slave labor to satisfy the voracious appetite of the Nazi war machine.
In February 1943, Vichy makes an even bigger blunder: It inaugurates the Service du Travail Obligatoire. The STO requires all Frenchmen between the ages of eighteen and twenty to work in Germany for two years. Faced with the prospect of forced deportation, thousands of these young men simply disappear into the mountains, where, by June, they have vastly increased the number of _Résistants._ They and their place of refuge both become known by a Corsican word for mountainous scrubland: Maquis.
As one historian put it, "The concept did not exist in January 1943; it was everywhere by June."
In September 1943, after a nine-month absence, André has returned to France on that Lysander. He is carrying 500,000 French francs in cash. Like everyone in the Resistance arriving from England, he also carries a cyanide pill in his pants pocket. It will stay there, always, unless he is arrested. When he touches it with his index finger, it feels like his insurance against torture. Or, perhaps, like his destiny. Either way, he knows he will swallow it if he is captured by the Germans. In this period, the average life of an agent is just four months before he is arrested.
A certain fatalism fuels André's fearlessness. "I never felt the slightest hesitation on his part," said his sister Christiane. But there is one irony that probably escapes him: The only thing that could impinge on his heroism might be his own survival.
NOW, IN JANUARY 1944, André is inside his secret headquarters on the top floor of an apartment building on the Left Bank. With him are his "right hand," Charles Gimpel, and his assistant, Geneviève.
Gimpel has arrived one month earlier, carrying another 100,000 francs from England. During his brief time in Paris, Gimpel has already set up an excellent liaison organization with the southern zone, and he has arranged for a large number of radios to be brought up from the south to transmit messages from Paris to London.
The ranks of the Maquis jump sharply after the forced deportation of French workers to Germany. The poster says "French and German workers unite!"(photo credit 1.3)
Because their junior aide, Jacques, has been absent for only a day, neither Boulloche nor Gimpel suspects that he has been arrested.
They certainly haven't imagined that Jacques has violated the cardinal principle of the Resistance. No matter how badly you are tortured, you must try with all your might not to divulge anything important for forty-eight hours after your arrest.
After you have been missing for two days, your comrades are supposed to assume that you have been arrested, and relocate immediately to a new clandestine location. Only then, after the captured agent has endured two days of unimaginable affliction, is he authorized to tell everything he knows.
If the system works the way it's supposed to, by then his information should be largely worthless anyway.
Jacques adores his boss, André. But almost immediately after he is grabbed by the Germans, the young agent realizes that he will never be able to remain silent after his captors begin to beat him, or semidrown him, since a form of waterboarding is one of the Gestapo's favorite methods of torture.
He sees only one way out. In a small cell with another Frenchman, he whispers, "Strangle me so I won't talk! If you don't, I will tell them everything."
But the boy's cell mate is incapable of providing such grisly mercy.
Soon after that, the young Sorbonne student begins to give up all of his secrets, including the location of André's secret apartment.
Barely an hour later, Jacques is squeezing into the narrow elevator in the apartment house on rue de la Santé on the Left Bank with two Gestapo agents. When it reaches the fifth floor, the three men exit silently onto the landing.
Jacques has been brought here to perform the secret knock. The young Frenchman points at the door of the doomed apartment, then walks toward it to carry out the sordid duty: one sharp rap of his fist, a beat, then two softer knocks on the door.
Inside the apartment, André recognizes the cloak-and-dagger sequence and stands up from his desk to acknowledge it. When he opens the door, he sees two Germans in black leather raincoats pointing identical Walther PPK pistols at his heart.
"HANDS UP!" they shout.
"WHAT'S GOING ON?" André screams back.
Instinct propels the Frenchman toward the staircase as the Germans open fire. Two bullets strike the Resistance fighter just below the chest. One agent rushes toward him to check for a weapon as the other one storms the apartment to capture his confederates.
If he hadn't been wounded, André thinks, this part would have been easy: He would have swallowed the fatal pill right away. But now he is writhing on the floor, with blood spurting out of his stomach — and the cyanide never leaves his pocket.
For a very long time, he will wonder whether this has been the right decision.
BACK ON THE SIDEWALK in front of the apartment building, the German agents shove André into the backseat of the black Citroën, next to Jacques. Then they speed away through the darkening streets of the Paris dusk.
Jacques is in a state of total collapse; he is weeping and howling and begging his boss for forgiveness. For the rest of his life, André will carry a hideous memory of these moments.
This part of André's trauma ends a few minutes later, when he is dumped off at La Salpêtrière, which the Germans have taken over as a prison hospital, where his wounds will be operated on.
For the moment, the Gestapo men prefer to keep André alive because they won't be able to torture him unless they first make an effort to nurse him back to some semblance of health.
Despite his promised reward, Jacques is soon shipped off to Germany, where he will die in a concentration camp.
FOR JUST ONE HOUR, the Gestapo leaves the secret apartment unoccupied and unwatched.
Forty-five minutes have passed since the first German team has departed with André and his betrayer. Now Jacqueline, André's sister and confederate, is at the apartment's front door. She is there to make dinner for everyone, because — naturally — the women are expected to cook. (In prewar France women don't even have the right to vote — a privilege de Gaulle will finally bestow upon them after the Liberation.)
Jacqueline performs the secret knock. Hearing no answer, she lets herself in with her own key. At first, the empty apartment doesn't make her anxious. Gradually, she senses a certain disarray, but if there is blood on the floor of the hallway outside, she hasn't noticed it. She walks into the kitchen and begins to peel some potatoes. Then she strikes a match to light a fire beneath them.
Suddenly, she realizes she is missing a vital part of the meal. One immutable fact of Parisian life has not been altered by the Nazi Occupation: Frenchmen still require wine with their dinner. So Jacqueline leaves the potatoes simmering on the stove and walks out the front door.
She skips the elevator and bounces down five flights of stairs to the street. Then she walks around the corner to the neighborhood _épicerie._ There she buys a few more groceries and the vital bottle of _vin rouge._
She will always remember this as the best-timed shopping trip of her life. When she returns to the apartment house, she notices something out front that had not been there four minutes earlier: another black Citroën Traction Avant.
Jacqueline tiptoes into the apartment house. The passengers from the car out front have just closed the door of the elevator behind them. Now it is only one floor above her. Her body stiffens as she watches the wooden cabin rise slowly through the narrow open shaft. Second, third, fourth... finally, it stops at the fifth floor.
Where her potatoes are cooking.
Later, she will wonder what the Gestapo men think when they discover her potatoes on top of the kitchen stove.
But right now, she is already sprinting toward the Métro.
_I have to save my sister._ That is her only thought.
Most of the time, Jacqueline and Christiane don't tell each other where they will be during the day. But today Jacqueline happens to know that her little sister has taken the afternoon off to listen to some Bach at the Palais de Chaillot.
But the concert hall is on the Right Bank, at Trocadéro — fourteen Métro stops away.
Jacqueline sprints eight hundred yards to the Métro Glacière. She is pretty enough to turn heads anytime, but the sight of her racing through the Métro, her arms full of groceries, makes her uncomfortably conspicuous.
She is out of breath when she reaches the platform — just as the train DIRECTION ÉTOILE pulls into the station.
The train's doors close behind her; then the trip seems to take forever: Denfert-Rochereau, Raspail, Montparnasse, Pasteur — still seven more stops to go.
Jacqueline is hardly religious, but she is praying anyway — praying that she will somehow be able to intercept her sister before she returns to the secret apartment, where the second wave of the Gestapo is now waiting to capture both of them.
But can she possibly get there before the concert ends?
And even if she arrives before the final notes, how will she ever be able to pick her sister out of the crowd? She has no idea where Christiane is sitting, and the Palais de Chaillot is one of the largest concert halls in Paris.
Now the train is rumbling out of its underground tunnel to travel over the Seine, to Passy. Just one more stop to go. Then it dives back underground to pull into Trocadéro.
Jacqueline knows every inch of this station: It is the one she grew up with, the one closest to her parents' apartment.
She runs toward the exit for Palais de Chaillot. Now she is dashing past the statues of Apollo and Hercules, still carrying her groceries. As she enters the lobby, she can just make out the sounds of Bach still seeping out of the hall.
_I haven't missed the end of the performance!_
But how will she ever be able to snatch Christiane out of the crowd?
Happy accident? Or intuition?
To the end of her life, Christiane will never be able to answer that question. Before today, she has never left a concert early. But this afternoon, something suddenly makes her stand up to leave the hall — _ten minutes before the concert has ended!_
When she reaches the empty lobby, she walks straight into the arms of her frantic sister.
"We can't leave by the front door!"
That is Jacqueline's only greeting: She thinks that the Gestapo may have followed her here.
Deciding that they have nothing left to lose, they approach the box office.
Jacqueline tells the ticket seller that she needs to speak to the manager. He may betray them, but they see no other choice.
When the manager appears, Jacqueline exclaims, "We can't leave by the front door." Nothing more.
The manager looks at her silently, his face revealing nothing.
Now he will save them... or turn them over to the Germans.
He turns around and walks out of the box office.
Then he leads the terrified (but still very attractive) young women to the stage door.
Jacqueline repays his kindness by thrusting the groceries and the bottle of wine she has carried halfway across Paris into his hands. They will only slow them down now anyway. Outside, they scour the street for Gestapo men, but no one looks particularly menacing.
BACK ON THE MÉTRO PLATFORM at Trocadéro, Christiane tells her sister she has another problem: She is carrying dozens of coded telegrams in her purse, having spent most of the day retrieving them from secret drop-off points all across Paris. Just three days earlier, André had finally managed to open a direct communications link to his handlers in London.
Christiane never considers throwing away the incriminating cables; she must find somewhere to hide them. She settles on the apartment building of a sympathetic cousin, who lives on avenue Marceau. When they get there, Christiane runs inside and shoves the telegrams under the carpet in the elevator cabin. Without telling her cousin, of course. Later, she will retrieve them, after the immediate danger has passed.
By now it is after eight o'clock, and the sisters have no idea whether André, or anyone else, has been arrested. They decide to return to the Glacière Métro stop. They think they might still catch their brother before he returns to the secret apartment, so they plant themselves at the station's two exits.
Then they stay there until the last Métro.
Exhausted and depressed, they finally return to their parents' apartment across the Seine, which is where they are still living, right next to the Palais de Chaillot. Fortunately, their parents are asleep, so they don't have to decide whether they should tell them anything right away about what has happened.
* Of the seven other men on the plane, one would be arrested eight days later, another was shot, and a third would die in a German concentration camp.
† A renegade is certainly what de Gaulle was considered when he left France in 1940, "when Vichy dominated the climate of opinion" and Germany's collaborators argued that to "remain in France was itself proof of honor," as Ian Ousby put it. "To go to Britain and accept British favors made the desertion far worse." ( _Occupation,_ p. 234) In 1942, the Institut d'étude des Questions Juives published posters depicting the general as a puppet of the Jews — "The Real Face of Free France." (Peschanski, _Collaboration and Resistance,_ p. 173)
# _Two_
_Dignity is incompatible with submission... I am a man who feels the necessity of engagement._
— André Boulloche
THE NEXT DAY, the Boulloche sisters are in shock. They still don't know the fate of their brother and the rest of their confederates. They don't even know if any of them was in the secret apartment when the Gestapo arrived, but they do know that André, his right-hand man, Charles Gimpel, his assistant, Geneviève, and Jacques may now be under arrest.
And if the Germans have taken them, today they are almost certainly being tortured.
In any case, the sisters must assume the worst and immediately evacuate all the clandestine locations known to their brother and his assistants. After two days, their co-conspirators are supposed to have found new hiding places, and the knowledge of the prisoners should be useless to their captors.
In the afternoon, the sisters finally learn the dreadful news from Marie-Hélène Lefaucheux, a friend who is on good terms with the chaplain at La Salpêtrière. Marie-Hélène suspects that the chaplain is playing a double game with the Germans, but his information is usually reliable.
The chaplain tells Marie-Hélène that André, Gimpel, and Geneviève have all been arrested, and that André has been shot. Marie-Hélène immediately relays the terrible news to André's sisters.
In a family full of secrets since the war began, the biggest secret of all is the fact that André has been back in Paris since September. For four months, the two girls have been spending their days with André, and sleeping at their parents' home at night, without ever telling their mother and father that their youngest son has returned — because that is what André has ordered them to do.
But their brother has also said that if he is arrested, they must tell their parents immediately, because his capture could jeopardize their safety.
So now they finally reveal that André has been in Paris for four months, working for the Resistance. And today he is in a prison hospital, with a gunshot wound in his stomach.
Their parents are horrified.
But their mother also says this about her children's decision to fight the Germans: "That's what I would have done."
Perhaps she means that is how she would have acted if she had been sixteen or twenty when the war began, instead of fifty-one.
The last time the whole family had been together was a dinner at the parents' apartment in the spring of 1942. That evening, twenty months earlier, Christiane had felt terrible anxiety when she said goodbye to her two brothers.
_When will I see them again?_ she wondered.
And even earlier, during the German invasion of France in 1940, her father had thought that his whole family would never be reunited in the same room again.
For the next three weeks, Christiane and Jacqueline decamp to the apartment of a sympathetic woman friend who lives on boulevard Saint Michel. Her husband is a mining engineer who is away, working in the north most of the time. When he returns home one weekend, he makes it clear that he completely disapproves of what the two sisters are doing, because he's not in the Resistance himself. But he also does not betray them.
After three weeks at their friend's house, they briefly return home to their parents' sprawling apartment, because they think the danger has disappeared. "In general," Christiane recalled, "things happened very quickly." Then they decamp permanently to a series of secret apartments on rue de Lille and avenue Mozart.
Her parents don't feel that they are in danger, partly because they assume the Gestapo has been unable to learn their son's real identity.
# _Three_
_I think I was marked by the debacle of 1940 for my whole life. I saw all the egotism and all the cowardice. I had an unreasoning patriotism. For me war is fought to the end — to the death._
— André Boulloche
WHEN WORLD WAR II BEGINS with Hitler's invasion of Poland, on September 1, 1939, André and his older brother, Robert, have already been mobilized to serve in the French Army. Christiane is just fifteen years old. She has spent the summer across the channel in Kent, living with an English family so that she can learn the language. Before she leaves France, her father warns her that she may have to come home early. A couple of days before the war begins, he telephones her in England to summon her back to Paris.
Remembering how the capital was bombarded during World War I, Jacques Boulloche decides to move his wife and daughters to their country house in Fontainebleau, forty-five miles south of Paris. The fifty-one-year-old father is still in his government job in the capital, which puts him in charge of all French highways. Since 1937 he has been _inspecteur général des ponts et chaussées_ and _directeur des routes du ministère des travaux publics._
IN 1939, the catastrophe of the First World War is still a scorching memory for every French person over twenty-five. More than 1,357,000 French soldiers died in the "Great War" — and a staggering 76 percent of the 8,410,000 Frenchmen who fought it were killed, wounded, taken prisoner, or missing in action.
That was more than one-fifth of the population of France.
Those casualties transform postwar France into a nation of old people and cripples and women without men in the 1920s. Every surviving municipality has a monument commemorating its losses in the Great War, but scores of villages are wiped out altogether, after a whole generation of their young men has been annihilated.
The worst legacy of World War I is the obsequious approach Britain and France adopt toward Hitler's Germany. Between 1935 and 1939, British and French leaders bungle every opportunity to rein in the Nazi leader before another all-out war becomes inevitable.
In 1935, a year after promoting himself to the new office of Führer, Hitler abrogates the military provisions of the Treaty of Versailles and begins Germany's massive rearmament program. In March 1936, he violates the treaty again, by reoccupying the Rhineland on France's border.
At that moment, Britain and France could easily have rolled over Hitler's troops, but both countries are paralyzed — still reeling from the Great Depression, and their losses in World War I.
The depth of antiwar feeling in France and Britain is suggested by the reaction of a clergyman in Liverpool after Germany reoccupies the Rhineland. When the British government declares that the Rhineland should be occupied by an international force until the dispute with Hitler can be resolved, the canon gives orders that no more prayers are to be offered up for the British government. "To continue an enforcement of the spirit of inequality upon Germany is a proposal unworthy of our creed and of our country," the canon thunders. "To renew an occupation of their homeland is a proposal monstrous and unjustifiable; it is an unnecessary degradation of the soul of a great people. To add to it that we shall do it again at the dictation of France is to lend aid to malice and to surrender right to vengeance. We cannot pray a blessing upon such proposals."*
The views of a large portion of the British upper classes are reflected in a repellant ode to Hitler that appears in _Homes & Gardens_ in November 1938. The article oozes admiration of the Führer's modesty: His gardeners, his chauffeur, and his pilot are "not so much servants as loyal friends." Air-Marshall Göring is a delightful "bon viveur" [ _sic_ ] and the Führer himself is a "droll raconteur" who prefers the "society of brilliant foreigners, especially painters, singers and musicians."
The dread of anything that might provoke a new world war is equally powerful in France. In 1938, the famous anti-Semitic novelist Louis-Ferdinand Céline predicts that a new conflict with Germany will bring twenty-five million casualties, and the "end of the breed... We'll disappear body and soul from this place like the Gauls."
These overwhelming feelings of fearfulness culminate in the tragedy of Munich. The British and French prime ministers, Neville Chamberlain and Édouard Daladier, meet there in September 1938. They agree to the dismemberment of Czechoslovakia and give Hitler the right to occupy the Sudetenland and all of its fortifications.
When Chamberlain returns to London, he is hailed as a conquering hero when he calls the accord "peace for our time."
The _Manchester Guardian_ is one of the very few British newspapers to identify the agreement's real significance: "Politically Czecho-Slovakia is rendered helpless, with all that that means to the balance of forces in Eastern Europe, and Hitler will be able to advance again, when he chooses, with greatly increased power."
Four days after the agreement is announced, Winston Churchill rises in the House of Commons to give one of the most brutal and prescient speeches of his entire career. After "immense exertions," said Churchill, the most that Chamberlain had "been able to gain for Czechoslovakia in the matters which were in dispute has been that the German dictator, instead of snatching the victuals from the table, has been content to have them served to him course by course." Churchill continued:
_All is over. Silent, mournful, abandoned, broken, Czechoslovakia recedes into the darkness... Not only are they politically mutilated, but, economically and financially, they are in complete confusion. Their banking, their railway arrangements, are severed and broken, their industries are curtailed, and the movement of their population is most cruel..._
_When I think of the fair hopes of a long peace which still lay before Europe at the beginning of 1933 when Herr Hitler first obtained power, and of all the opportunities of arresting the growth of the Nazi power which have been thrown away, when I think of the immense combinations and resources which have been neglected or squandered, I cannot believe that a parallel exists in the whole course of history..._
_We are in the presence of a disaster of the first magnitude which has befallen Great Britain and France. Do not let us blind ourselves to that. It must now be accepted that all the countries of Central and Eastern Europe will make the best terms they can with the triumphant Nazi power. The system of alliances in Central Europe upon which France has relied for her safety has been swept away, and I can see no means by which it can be reconstituted. The road down the Danube Valley to the Black Sea, the road which leads as far as Turkey, has been opened..._
_Many people, no doubt, honestly believe that they are only giving away the interests of Czechoslovakia, whereas I fear we shall find thatwe have deeply compromised, and perhaps fatally endangered, the safety and even the independence of Great Britain and France..._
_Do not suppose that this is the end. This is only the beginning of the reckoning. This is only the first sip, the first foretaste of a bitter cup which will be proffered to us year by year unless by a supreme recovery of moral health and martial vigour, we arise again and take our stand for freedom as in the olden time._
THE CONVICTIONS EXPRESSED in the home where the Boulloche children grow up are closer to Churchill's than to Chamberlain's. The most important intellectual influence on the Boulloche brothers and sisters is their father, Jacques, who was born in 1888.
Jacques's reputation for resilience begins when he is still a child. One August night, eleven-year-old Jacques boards a train in Paris with his family to travel to the fashionable beach at Dinard, 250 miles to the west on the Côte d'Émeraude in Brittany.
At one o'clock in the morning, he pushes on an unlocked door and tumbles off the train into the summer darkness. Miraculously, he lands in a hedge, which cushions him from serious injury.
Jacques brushes himself off and walks along the tracks for fifteen minutes, until he stops to knock on a stranger's door to request assistance. The next day, his terrified mother is immensely relieved when she learns that her son is alive and unharmed. Partly because of the prominence of the boy's father, who sat on the Cour de Cassation, Paris newspapers duly report the event for posterity.
As the family patriarch, Jacques sees the Second World War coming much sooner than most of his contemporaries. He was brought up by a German governess, he speaks German fluently, and his highway work takes him to Germany frequently during the 1930s. As early as 1937, he is convinced that another war with Germany is inevitable — and he is pessimistic about the outcome.
One of his favorite sayings is well known to his children: "One German is okay. Five or six? That's a catastrophe!" (Churchill had his own version of this sentiment: "A Hun alive is a war in prospect," he told visitors to Chequers in October 1940.)
Jacques won the Legion of Honor for his service as an officer, fighting the Germans for four years during World War I. Often he was so tired that he fell asleep in the saddle of his horse.
At three o'clock in the afternoon of November 11, 1918, the day the armistice was declared, he celebrated with a letter to his mother: "The great day has finally arrived — the day of glory and happiness. Everyone is more or less crazy with joy!" Four days later, he joked in a letter to his father that he is about to get Occupation duty in "Bochie" — the land of the Boches, an epithet for the hated Germans.†
Jacques returned home for leave from time to time, and two of his children, André and Jacqueline, were born while he was at the front. Jacques almost never speaks about his wartime experiences, except to say that he loathes all career military officers. He regards them as "imbeciles," and he is proud that he knows of none among his ancestors.
The Boulloches are upper-middle-class Catholics, fierce republicans whose forebears have been prominent lawyers, judges, and civil servants in Paris since the revolution. The family traces its ancestors back to Normandy in the sixteenth century. They migrated there from Scotland, where they had been "Bullocks" or "Bullochs."
Whenever the Boulloches stand up for what they believe in, they often stand apart. The Boulloche children are all mindful of their family's dual pedigrees of patriotism and iconoclasm.
Their grandfather was the chief magistrate of France, the president of France's highest court, the Cour de Cassation. His brother, their great-uncle, was his colleague on the same supreme court. After the Jewish army officer named Alfred Dreyfus was convicted of treason in 1894, these two Boulloche judges maintained the tradition of the Boulloche "exception," by insisting that Dreyfus was innocent.
While most of the haute bourgeoisie joined in the anti-Semitic frenzy that produced the Dreyfus prosecution, these Boulloches made themselves traitors to their class by defending the Jewish officer. Characteristically, there was no self-congratulation when Dreyfus was finally exonerated. "We triumph," declared the chief judge, "but in such sad company."
THE CHILDREN OF JACQUES AND HÉLÉNE BOULLOCHE enjoy an unusual amount of freedom. A young friend often at their dinner table remembered Jacques as a "calm and liberal man," while his wife was "a lover of fine music."
Unlike their cousins, who go to Catholic schools, Robert, André, Jacqueline, and Christiane all attend public schools. Their parents encourage intellectual independence, and outsiders are startled by the outspokenness of the children, whose exuberance is sometimes more than their charming mother can control.
When she is still very young, Christiane wonders out loud, "How will we feel if we find ourselves in front of Buddha instead of God when we get to heaven?" It is the sort of question that delights her parents.
André excels at mathematics at the Lycée Janson de Sailly in Paris, and he is one of the only students who can hold his own discussing philosophy with the teacher after class.
Precocious but unpretentious, his contemporaries give him the nickname "Boull." He wins most of the academic prizes, while keeping his classmates amused with risqué magazines like _La Vie Parisienne_ , illustrated with photos of scantily clad young women. Boull even provides forbidden alcohol, the perfect accompaniment for the magazines. Trains, planes, and automobiles are some of his other passions.
As a teenager, André is struck by the power of money to corrupt, and the young idealist even dreams of returning France to a barter system. At eighteen, he follows his father's footsteps into the French elite, when he is admitted to the country's most celebrated _grande école,_ the Ecole polytechnique, or l'X, and, again, like his father, becomes an engineer.
After that he spends two years at Ecole Nationale des Ponts et Chaussées (National School of Bridges and Highways), while simultaneously studying for a law degree — because every Boulloche male is still expected to continue centuries of family tradition by studying the law, whether or not he plans to practice it.
THE BOULLOCHES are "social Catholics," but they are not particularly observant. They blend a soft anticlericalism with a sharp republican spirit. No priest ever joins them at the dinner table, and by her seventeenth birthday, Christiane is certain there is no God.
The children grow up surrounded by books and music. Their parents' favorite writers are Proust and Rimbaud, literary tastes that make them slightly avant-garde. Jacques plays the piano, Hélène is an accomplished violinist, and each of the children also plays an instrument.
Jacques is kind but serious — almost austere — while Hélène is more playful. Christiane thinks her mother has only one real passion in life: her husband.
When Christiane is just sixteen months old, her parents leave for a long trip to Indochina, depositing their children with their very stern grandparents. It is a dramatic and difficult change for the older children, who are suddenly forbidden to laugh at the dinner table.
College chums: André Boulloche is on top smoking a pipe. On his left, André Rondenay steadies himself by holding on to a lamp. Everyone is wearing the uniform of L'Ecole polytechnique. Several years later, Rondenay succeeded Boulloche as de Gaulle's military delegate in Paris, after Boulloche was arrested.(photo credit 1.4)
In a letter home during their stay in colonial Saigon, Hélène jokes to her ten-year-old son André that she is disappointed when a Chinese man hosting a party is not smoking opium when she meets him, although "he did have his pipe next to him!"
When the parents finally return, Hélène is furious when she hears Christiane calling her nurse "Mother."
Their father is famous for his genuineness, and his children inherit that trait, together with a profound sense of duty. "It was not taught," Christiane remembered. "It was implicit. There was no discussion at all. But it was in the air — that's for sure."
THE FINAL FRANCO-BRITISH BLUNDER in the run-up to World War II is the failure of the two countries to forge an alliance with the Soviet Union against Hitler. After Britain and France bungle the Russians' feelers, Hitler seizes the opportunity to secure his eastern flank. On August 23, 1939, archenemies Hitler and Stalin astound the world by announcing a nonaggression pact.
Ten-year-old Stanley Hoffmann, who would grow up to become a brilliant political scientist and an expert in French history, was vacationing in the Alps with his mother when news of the pact arrived in the tiny village where they were staying. "It was as if the plague had suddenly struck," Hoffmann remembered. Even at his young age, he "understood at once that this meant war — that Hitler had free hands for his next aggression and that Britain and France could no longer weasel out."
The West considers Stalin's decision to make temporary peace with Hitler typical "Communist duplicity." But as the historian James Stokesbury points out, for Stalin it was a "simple either- or proposition": "Either he could ally with Britain and France, in which case there would be a war, a war that Russia was expected to fight while Britain and France sat and waited it out, after which, when Germany and Russia had destroyed each other, France and Britain would move in and pick up the pieces; or, he could make a deal with Hitler, they could divide Poland between them, Hitler would (probably) turn west, and Germany, France, and Britain would fight it out, after which Stalin would move in and pick up the pieces."
The secret protocol of the Hitler-Stalin pact — widely rumored but publicly denied — divides Finland, Estonia, Latvia, Lithuania, Poland, and Romania into separate spheres of influence for Germany and the Soviet Union. Freed of fears of a battle with the Communist monolith, Hitler invades Poland nine days later, on September 1. Blackout conditions begin in London the same day. On September 3, Britain and France both declare war on Germany. The _New York Times_ proclaims, "Germans Rush Gayly to Arms, Believing Poland Will Be Crushed in 10 Days."
The prediction is nearly correct. By the end of the month, Germany has conquered Poland, and Hitler divides up the country with Stalin. Organized Polish resistance is over by the first week in October.
* The British diplomat and politician Duff Cooper, who quotes this passage in his autobiography, points out that by this time "anybody who read the newspapers" was already aware "not only of the hideous persecution of the Jews which [Hitler] had initiated, but also of the blood-bath of June 1934 [Night of the Long Knives] in which he had slaughtered without trial so many of his own closest associates." ( _Old Men Forget,_ p. 197)
† At the end of World War I, Canadian general Andrew McNaughton said of the Germans, "We have them on the run. That means we will have to do it over again in another 25 years." Or, as satirical songwriter Tom Lehrer put it in 1965, "We taught them a lesson in 1918 — and they've hardly bothered us since then."
# _Four_
_Throughout the '30's, France and Britain had been slow, often indecisive and hopelessly outmaneuvered by Hitler's devilish strokes. The ponderous bourgeois gentlemen in heavy suits and hats who led the Western democracies didn't have a chance against the cunning gangster._
— Stanley Hoffmann
THE CONCLUSION of the German campaign in Poland in the fall of 1939 marks the beginning of the Phony War. This period of relative calm will last seven months on most of the European continent. Churchill calls it the Twilight War. In France it is _drôle de guerre_ (funny or strange war); in Germany, _Sitzkrieg_ (sitting war — a pun on _Blitzkrieg._ )
While Russia invades Finland at the end of November, and Germany and Britain engage in occasional skirmishes at sea, French and German troops remain mostly silent, as they gape "at each other from behind their rising fortifications" on France's eastern front, throughout the winter of 1939.
After her father moves his wife and daughters out of Paris in the fall of 1939, Christiane Boulloche embraces her new life at the lycée in Fontainebleau. Her classes are smaller than they were in the capital, and she excels with very little effort. Her parents agree to board a French officer who teaches at the artillery school in Fontainebleau, and his twenty-something students are frequent guests at the Boulloche dinner table — a beguiling circumstance for a fifteen-year-old girl.
At the end of 1939, most French officers still believe in France's military superiority, and that is the opinion Christiane hears at her dinner table. Like most French people (except de Gaulle), the Boulloches still hope that they will be protected by the Maginot Line, the massive fortifications France has built above- and belowground along the German border between 1929 and 1940. They also think that when the war finally comes to France, it will be a short one.
Christiane enjoys the period before Germany invades France. It feels like a "new and exciting adventure." But she is jealous of her brothers, because they are actually able to _do something,_ by being in the army. Predictably, her brother André is dissatisfied with his calm life as a lieutenant with the 6th _régiment du génie_ (engineers) on the eastern front. Starving for action, he snares a transfer to the aviation officers' school in Dinard.
As the Phony War continues through the winter, more and more children are evacuated from the northeast corner of France in anticipation of the expected invasion. While their parents stay behind to work, the children are relocated to places like the golf club in Fontainebleau. The children's presence, and a nightly blackout, are the only things Christiane experiences that make the possibility of an actual war seem real. Jacqueline, who has recently graduated from Sciences Po (the institute of Political Science in Paris), spends her days working with the refugee children from the north.
The eight-month-long lull between the time Britain and France declare war on Germany and the moment when full-scale hostilities begin does not work to the Allies' advantage. After Hitler and Stalin announce their nonaggression pact in August 1939, the French Communists take their cue from Moscow and end their active opposition to the Nazis. They denounce the war as "an imperialist and capitalist crime against democracy."
Meanwhile, morale plunges among French troops, who are undermined by shortages of socks and blankets, as well as simple inaction. Even the weather conspires against the French: The winter of 1939–1940 is the coldest France has shivered through since 1893.
In November 1939, Jean-Paul Sartre writes in his diary that the men who were mobilized with him were "raring to go at the outset," but three months later they are "dying of boredom."
WHEN THE GERMANS finally storm into Holland and Belgium on May 10, 1940, the impact is felt almost instantly in Fontainebleau. Christiane experiences the effects of the crushing advance of the enemy, as thousands flee in front of the German blitzkrieg, traveling in horse-drawn carriages and cars overflowing with exhausted women and children.
No one has expected the Germans to roll over the French Army so quickly, not even Germany's own generals. In England, Neville Chamberlain resigns as prime minister, as his dream of "peace for our time" evaporates, and the king summons Winston Churchill to Buckingham Palace to form a new government to fight the war.
Churchill is appalled by the speed of the German onslaught. The new prime minister learns that German tanks are advancing at least thirty miles a day through the French countryside, passing through "scores of towns and hundreds of villages without the slightest opposition, their officers looking out of the open cupolas and waving jauntily to the inhabitants. Eyewitnesses [speak] of crowds of French prisoners marching along with them, many still carrying their rifles, which were from time to time collected and broken under the tanks... The whole German movement was proceeding along the main roads, which at no point seemed to be blocked."
With hundreds of thousands of refugees now surging southward, Jacques Boulloche once again tries to move his wife and daughters out of danger. This time he sends them to an aunt's house in Perros-Guirec, in Brittany, in the northwest corner of France, far from the Germans' invasion route. Christiane is miserable because she has a new puppy that she isn't allowed to bring with her, and she never sees that dog again.
But when the three Boulloche women flee Fontainebleau at the end of May, they are much more fortunate than the refugees who had been housed at the golf club. The cars carrying those children are bombarded from the air, and scores of them are killed.
As France's army is being pulverized at the end of May, a gigantic crowd gathers in front of Sacré-Coeur on the hill above Paris to pray for victory. At the end of the emergency service, fifty thousand voices belt out "The Marseillaise." More than six million French citizens have already abandoned their homes.
Across the channel in London, a thirty-year-old foreign correspondent named James Reston writes in the _New York Times_ that a German invasion of the British Isles is now considered nearly certain.* Looking down from a plane, the writer-pilot Antoine de Saint-Exupéry observes that the mobs of refugees below look like a massive anthill, kicked by a giant.
ON JUNE 3, more than one hundred German bombers attack Paris. The French authorities claim they have shot down twenty-five of the planes, but the bombardment kills more than 250 Parisians and wounds 600 more.
"This is what we dreaded for so long, and what we hoped might never happen," the _New York Times_ declares on its editorial page the next day.
It seems worse, somehow, than any of the other crimes perpetrated by Germany in this war... Great cities... are more than aggregations of men, women and children. They are the treasure-houses of the Western spirit. Whoever strikes at them strikes at all that man has built through ages of sacrifice and suffering...
Free men will not endure these things without a new resolve to destroy the forces of evil that have sent the German bombers on their errand. The great columns of smoke and flame that rose above Paris yesterday... were also the first fires of a wrath such as our world has never known. If this kind of fiendishness continues, if Paris and London are to become shambles of ruined buildings and murdered civilians, the fires of hate will not be quenched in our time. The anger of civilized peoples will burn so fiercely that it will consume the hateful German system which has loosed these horrors upon the world.
The _Times_ is almost alone in its prescience. Certainly no one in France is optimistic about the eventual defeat of the Nazis the day after the bombardment of Paris.
SIX DAYS LATER, the French government prepares to evacuate Paris. Jacques Boulloche, the director of the national bureau of highways, is ordered to leave the capital for Royan, three hundred miles to the south on the western coast of France. That evening he heads for his country house in Fontainebleau, but his car breaks down on the way, and he doesn't get there until four o'clock in the morning.
The next day he opens his fountain pen to write a letter to his wife:
The Boulloche country house in Fontainebleau, where Jacques Boulloche retreated as the Germans advanced on Paris in 1940.(photo credit 1.5)
_I can't describe my feelings when I got here, considering all that we've had to abandon. And yet, we are among the lucky ones... The peace of the garden and the fragrant smells make the unfolding tragedy seem like nothing more than a bad dream..._
The same day as Jacques writes to his wife, Norway surrenders to the Nazis, and Italy declares war on Britain and France.
The capital Jacques has left behind is filled with smoke from burning archives and incendiary bombs. On June 10, the French government declares Paris an open city, meaning it will no longer be defended — after the Germans are already inside the city's gates. There are twenty thousand people jammed outside the doors of Gare d'Austerlitz, trying to force themselves onto trains leaving the capital.
The _New York Times_ reports on June 13, "The German guns are battering at the hearts and minds of all of us who think of Paris when we try to define what we mean by civilization... Of all cities it expresses best the aspiration of the human spirit."
Paris falls the next day, on June 14. By then, its population of three million has shrunk to eight hundred thousand.
Jacques Boulloche is distraught. His boys are at the front, and his wife and daughters are hundreds of miles away in Brittany. He is terrified that he may never see any of them again. He and his wife have an exceptionally strong bond. They are so close, their children sometimes complain that it's hard to find any room for themselves in between them.
On the afternoon of June 15, Jacques sits down at the desk in his room at the Grand Hôtel de Paris in Royan to write another anguished letter to his wife:
_My Beloved,_
_Everything is finished. Will we ever see each other again?_
_In case you are reunited someday with the children, I have included a message for them in this letter._
_I don't know what to say, except that I adore you, and I can't stop thinking about you. Stay calm, courageous and proud._
_I love you, and I still hope that someday we will be reunited._
_Jacques_
And this is what he writes to his children:
_If I never see you again, know that my last thoughts will have been about the four of you._
_I hope you will once again see a free and joyous France._
_I love you with all my heart._
_Your Father_
_Jacques_
In a letter to his family, Jacques Boulloche pours out his fears about what the German invasion will bring.(photo credit 1.6)
* In the underground bedroom Churchill slept in during much of the war, his bed faced a map that highlighted all the possible invasion points of the British Isles.
# _Five_
_Had all of us in France meekly, lawfully carried out the orders of the German master, no Frenchman could have ever looked another man in the face. Such submission would have saved the lives of many — some very dear to me — but France would have lost its soul._
— Commandant le Baron de Vomécourt
ALTHOUGH the German campaign in France lasts less than six weeks before the armistice is declared, the French still suffer enormous casualties. There are ninety thousand French soldiers killed and two hundred thousand wounded. Another two million soldiers are taken prisoner, and a million and a half of them are sent to Germany. These French prisoners will remain on the far side of the Rhine until the war ends, almost exactly five years later.
The same day Jacques Boulloche sends his farewell letters to his family, President Roosevelt rejects the Allies' plea to America to enter the war against Germany at once. Meanwhile, French prime minister Paul Reynaud, who had been an early opponent of appeasement, informs the British that he intends to split his government and lead half of it abroad.
On June 16, Churchill makes a dramatic gesture to try to convince the French to continue the fight against the Germans. The new British prime minister proposes the merger of France and Britain into an "indissoluble" Franco-British union. A single War Cabinet will direct the affairs of the new nation, to "concentrate its whole energy against the power of the enemy, no matter where the battle may be."
But defeatist members of the French cabinet recoil at the idea of a union with their historic rival. They predict that within three weeks, England will "have her neck wrung like a chicken."* A French minister of state even declares, "Better to be a Nazi province. At least we know what that means."
Marshal Henri-Philippe Pétain, the eighty-four-year-old hero of World War I whom Reynaud had summoned out of retirement a few weeks earlier, leads the fight against a merger with Britain. When Reynaud is unable to convince his colleagues to embrace Churchill's bold idea, the French prime minster resigns, and Pétain succeeds him as the head of the French government.
Charles de Gaulle had been wounded three times during World War I, and he spent almost three years in German prison camps. Since the 1930s, he has been a voice in the wilderness, warning of France's unreadiness to confront the Germans. He has published a book advocating the mechanization of the army and the offensive deployment of tanks, but his fellow officers ignore his advice. A prescient Reynaud has been the only politician to support him.
De Gaulle has fought as long as he can to keep France in the war. But after Pétain becomes France's new leader, de Gaulle finally slips away from his office in Bordeaux, on June 17, to be driven to a nearby airfield.
At nine A.M., he takes off for England in a plane provided by the British. Churchill observed that the solitary French general "carried with him in the small airplane the honor of France."†
That evening, Marshal Pétain goes on the radio to tell the French Army to surrender. Christiane listens to Pétain's broadcast with her aunt and five of her aunt's six children. The young teenager feels like the sky is falling on her head. She is especially angry when her elderly uncle declares that England will never be able to continue the fight alone. She has no idea that he is echoing the majority view inside the French government.
Christiane considers her uncle appallingly defeatist: "In spite of everything, I never stopped believing in a miracle — that somehow our army would rise again." But she is also aware of the state of the French Army: "Not defeated. Crushed."
The following evening, Churchill allows de Gaulle to use the BBC to broadcast his first appeal to the French people. This time Christiane is mesmerized. The general's words have a profound effect on everyone who still believes that the Germans may someday be defeated:‡
_Has the last word been said? Must we abandon all hope? Is our defeat final and irremediable? To those questions, I answer No!... I ask you to believe me when I say that the cause of France is not lost._
And then the prickly iconoclast minted the phrase that made it possible for all unvanquished French citizens to continue to fight for the honor of their fallen nation:
_Whatever happens, the flame of French Resistance must not, and shall not, die!_
Christiane has never heard of de Gaulle before tonight. Although he had become undersecretary of state for war on June 6, he had only been a colonel before the war.
When he is introduced on the BBC, Christiane asks herself, "General of Gaul — what rank of officer is that?" As soon as she hears him, she immediately agrees with her male cousins: We must all decamp to England! But none of them is old enough to cross the channel without an adult.
AT THIS POINT, her older brother André really isn't any more political than Christiane is. But he shares all of his sister's instinctive patriotism.
On the night of June 17 — the same day de Gaulle decamps for London — André distinguishes himself by helping Lieutenant Jean-Pierre Berger blow up the bridge at Marcigny-sur-Loire, between Digoin and Roanne, which slows the German advance a bit.
As the French government is suing for peace, André retains an unquenchable appetite for action. He decides that he must leave France to avoid surrendering to the enemy. Together with ten of his comrades-in-arms, he sneaks onto a boat leaving Port-Vendres on the Mediterranean coast the day before the armistice is announced.
From the moment he heard Pétain on the radio saying, "We must stop fighting," he categorically refused to accept defeat and had only one desire: "to continue and then to resume combat." André doesn't base his actions "on a critical analysis of the situation, or a particular political belief... but simply on an elementary conviction: _that dignity is incompatible with submission._ " Like so many of those who are praying that the Germans' victory is only temporary, he blends optimism with fatalism. He thinks that "we [will] win in the end, and that it [is] the duty of all Frenchmen to fight for this victory" — but he also thinks it's unlikely that he will live long enough to witness the German defeat himself.
André and his compatriots decide that North Africa is the best place to continue the battle against the Germans. But when their cargo freighter reaches Oran in French Algeria, the welcome they receive from the French colonial authorities is not at all what they expect. Instead of being greeted as heroes, they are treated practically like traitors.
That is because, between the time they leave France and the time they reach North Africa, the new government headed by Pétain has discarded Reynaud's idea of sending some officials abroad to wage the war in exile, as Belgium, Holland, Poland, and Norway already have — to continue the fight against the Germans from London. Mostly because the Pétain government wants to rid itself of some unwelcome dissenters, on June 21, twenty-four deputies and one senator are allowed to sail away from France to North Africa on an armed auxiliary cruiser, the _Massilia._
When news of the armistice is picked up on the _Massilia's_ radio two days later, the anti-Nazi legislators on board plead with the captain to change his course for England. But the captain is taking his orders from the new French government, and he continues on to Morocco. When the _Massilia_ reaches Casablanca on June 24, the whole party of government officials is put under ship arrest for nearly three weeks, while Pétain's cronies debate what to do with them. Finally, they are sent back to France.
"I embarked on the _Massilia_ never dreaming the _Massilia_ would become a trap," remembered Pierre Mendès-France, a French prime minister in postwar France. "But quite soon the politicians who had remained in Bordeaux realized they could exploit this and present the departure of the _Massilia,_ with a number of politicians on board, as a sign of panic — an escape, a surrender... And paradoxically a certain number of them — Viénot, Jean Zay and myself — were charged with desertion, when our idea had been to go on fighting."
Churchill noted with disgust that these patriots were "disposed of as the Vichy Government thought convenient to themselves, and agreeable to their German masters."
By now the new French government has accepted the terms that Hitler's generals have dictated to them in a railway car at Rethondes, near Compiègne. With revenge at the heart of his brutal campaign, Hitler chooses the exact spot where French General Ferdinand Foch had accepted the surrender of the Germans at the end of World War I, just a quarter century earlier.
To encourage their rapid acceptance by the French in 1940, the Germans do not make the terms especially harsh. The southern portion of the country will be left unoccupied while the Germans are granted an occupied zone in the north that includes Paris. (The unoccupied zone will also free up more German troops for the expected invasion of Britain.) The French Army will be demobilized, except for a force of one hundred thousand to ensure internal order. The fleet will be disarmed and the ships will be docked in their home ports, and the Germans promise they will not touch them. One and a half million French prisoners will remain in captivity until a peace treaty replaces the armistice (which never happens). The cost of keeping German troops in France will be paid for by the French government.
THREE DAYS after the armistice is declared, Hitler sneaks into Paris for an early-morning visit. He flies into Le Bourget airport with a small entourage that includes Albert Speer, the Third Reich's official architect. The civilians accompanying him all wear borrowed uniforms, because Hitler has demanded that they dress as soldiers for the trip.
Thirteen years earlier, a hundred thousand Parisians had mobbed Charles Lindbergh at the same airfield to celebrate his triumphant arrival from New York.§ But today, there are no crowds to greet the Führer.
Three large Mercedes sedans whisk his party to the ornate Paris Opera House. This is the German leader's favorite building in the French capital. As a student he even studied its architectural plans, and his French guide is impressed when the dictator asks him what has happened to a particular room. The guide explains that it no longer exists, because of a renovation.
Fascist sympathizers show their enthusiasm inside the French Chamber of Deputies after the rebroadcast of a speech by Hitler in July 1940.(photo credit 1.7)
At the end of the opera tour, one of Hitler's aides offers the guide a 50-mark tip. Albert Speer watches as the Frenchman quietly refuses the gift — twice.
Choosing the obvious tourist stops, Hitler continues on to the Tour Eiffel, the Arc de Triomphe, and Les Invalides. Inside the monument to the French emperor, he stands for a long time gazing silently at Napoleon's tomb. Perhaps he is already contemplating a Russian invasion — and vowing to make his more successful than Napoleon's.
Hitler poses in front of the Eiffel Tower, but he never gets to the top, because the French have cut the cables to the elevators before his arrival. The elevators will remain out of commission until the end of the war.
Three hours after he lands, Hitler climbs back onto his plane to return to Germany. He has spent less than half a day inspecting his most glorious conquest.
He will never enter the City of Light again.
ANDRÉ AND HIS COMRADES reach Algeria the same day Hitler visits Paris. They are disgusted when they are greeted like deserters instead of patriots. Two of them make contact with the British consul. But when they ask the diplomat for help to get to England, he urges them to wait out the war in Morocco instead.
André is completely baffled by his predicament. He is desperate to continue the fight, but he has no idea how to do that. Should he stay in Morocco until the war is over? Or should he return to France to confront the enemy?
On July 6, 1940, the twenty-four-year-old lieutenant sits down at his desk to describe his anguish in a letter from Rabat. The letter smolders with his devotion to duty.
_My Dear Father,_
_I can finally write to France with a small chance that my letter will actually reach you. I won't try to tell you how I've experienced the past month — I can't find the words to describe it. Once again, it seems as though we have reached the depths of degradation and debasement, but day after day, our descent continues. What a struggle it will be to pull our country out of this abyss!_
_All I can tell you is that I came to North Africa so that I would be able to battle the enemy, by joining an organized resistance, but you know as well as I do how vain that hope was... Ignorance is a formidable thing at a time like this. Whatever one decides to do, it's a descent into the unknown. I desperately want your advice right now. I've been reduced to imagining what it might be, but I can't be sure you would approve of the course that I've chosen._
_Now I am posted at a base at Rabat, without any real responsibilities. I'm suffering terribly from inaction... There is one question in particular for which I need your advice... Should I return to France?... Everyone here agrees our country is in terrible disarray, and people like me may not be experienced enough to help put it back together. Some people think I should stay here to gain some seasoning, so that I can be more effective when I return home._
_Personally, I prefer the opposite course. The ghastly state my country was in when I left makes me want to return as soon as possible. It can only be saved by a complete moral resurrection, something that will require the work of all men of good will... I think I can contribute a great deal. And if more troubles lie ahead, isn't it my duty to be present?_
_This is the question that has really gotten under my skin. I never thought it would be so difficult to determine one's duty, once one had put aside all personal considerations. And yet, for the last two weeks, I have been at war with myself._
_I am impatient for news about all of you... If the only thing that Frenchmen still have is the affection of our families, at least ours won't be the most badly divided._
_André_
André's last sentiment did not turn out to be prescient.
His father's response has not survived. But ten weeks after writing his letter, André has returned to France to resume the struggle against the Germans.
* This led, of course, to Churchill's famous riposte eighteen months later: "Some chicken. Some neck!"
† Later, de Gaulle said of France's collapse in 1940, "We staggered, it is true. But was this not, first of all, a result of all the blood we had shed twenty years before in others' defense as much as our own?" ( _Complete Wartime Memoirs_ , p. 461).
‡ De Gaulle spoke on June 18, but Christiane thinks she didn't hear the broadcast until it was repeated the following evening.
§ At the height of his fame, Lindbergh became one of Hitler's most naïve supporters.
# _Six_
_The Resistance was irresistible._
— André Postel-Vinay
THE FEARS Jacques Boulloche described in his letters of a permanent separation from his family were premature. At the end of the summer of 1940, his wife and daughters return to the French capital from Brittany, and they move back into the family's spacious apartment.
Jacques's youngest daughter, Christiane, has a visceral reaction to what she sees in the newly occupied capital. The French _tricolore_ has been banned. In its place there are huge swastikas swaying in the wind, freshly painted street signs in German — "black on yellow," she remembers clearly — and German drummer boys in front of Wehrmacht soldiers goose-stepping down the Champs-Élysées. "There would be parades in the morning and they would sing. And they sang well — that was especially annoying!"
Christiane is stunned by all of this. She sees it as "the visible proof of our defeat. Seeing the Germans in Paris is ghastly. You feel like you are no longer at home. We were touching the reality of the Occupation with our fingers. It was a succession of shocks."
A German soldier complained to an attractive Parisian girl that her city seemed sad. "You should have been here before you got here," she replied.
This is when Christiane begins to ride a bicycle, when there is no longer any heat in her parents' apartment, and when she begins to feel hungry all the time. Jacqueline goes to work for an organization that sends packages to French soldiers who are imprisoned in Germany.
Jacques and Hélène Boulloche share their children's revulsion at the German Occupation. As early as July 1940, Jacques writes to his wife that the new anti-Semitic campaign is going to make life difficult for one of his colleagues. Three months later, the government publishes its first "Jewish law," which excludes Jews from the higher levels of public service, as well as professions like teaching and the press, where they might influence public opinion.
Back at her Paris lycée in the fall of 1940, Christiane and her friends take up a collection for Mademoiselle Klotz, a much-loved history teacher, who is fired after the publication of the new law. Christiane is shocked by the persecution of the Jews, especially when one of her classmates, Janine Grumbach, is forced to start wearing a yellow star.
Beginning in October 1940, all foreign Jews can be interned at the discretion of prefects. On November 1, even in the unoccupied zone, every Jewish-owned business must display a yellow poster in the window: ENTREPRISE JUIVE.
By the start of 1941, some forty thousand Jews are held in seven main camps, in atrocious conditions. Some three thousand Jews perish in the French camps even before the Final Solution has begun.
In June 1942, every Jew over the age of six in the occupied zone is ordered to wear a star over the heart, and they are forced to buy three of them from their local gendarmerie. Adding insult to humiliation, those purchases are even deducted from their clothing coupons. In July, they are banned from all public places: cinemas, main roads, libraries, parks, cafés, restaurants, swimming pools, and phone booths, and they can only ride in the last car of the Métro.
By 1941 the swastika was everywhere in Paris. Here the former (British) WHSmith had become a German bookstore on the rue de Rivoli.(photo credit 1.8)
Christiane considers all of this outrageous, and she thinks that her parents are helping some of their Jewish friends to escape. Her family is also directly affected by the new law, because one of her cousins, a surgeon named Funck-Brentano, has a Jewish wife, and she is forced to go into hiding.*
IN SPITE OF their shared hatred of the Boches, the two halves of the Boulloche family choose very different paths after the fall of France. Unlike their three youngest children, neither Jacques nor Hélène will ever join the Resistance.
Jacques and Hélène both turn fifty-two in 1940, and they share the caution of middle age. Jacques helps some Jewish friends to go into hiding, but he is always discreet. He is also careful not to do anything that might jeopardize his family's safety, or weaken the nation, to which he and his ancestors have devoted decades of service.
As the British historian Julian Jackson observed, "Conduct which might be described as collaboration could incorporate a myriad of motives including self-protection, the protection of others, even patriotism."
Christiane doesn't consider her father "pro-German at all." She thinks he merely wants to make France work for the French, "not for the Germans." The Boulloches defiantly listen to the BBC at home. They realize that if the British don't stay in the fight against Hitler, their only hope for the future will be extinguished.
Jacques Boulloche never confronts the German occupiers directly. He keeps his government job, and he commutes to Vichy, the seat of the collaborationist government. His family notices he is always particularly depressed when he returns from Vichy. He never tells his children whether he has signed an oath of loyalty to the Vichy government, but he is almost certainly required to do so.
Jacques and Hélène's oldest son, Robert, serves in the army during the German invasion, but he avoids capture after the armistice. Now he has returned to his job as an inspector for the Finance Ministry, in Toulouse.
Robert shares his parents' prudence and their commitment to the French government. Like his father, Robert probably sees some value in keeping France functioning for the French, despite the invasion of the Germans.
Jacques and Hélène Boulloche. Jacques was the family patriarch, who kept his job as the director of the national bureau of highways after the Occupation began. Like her husband, Hélène never joined the Resistance. But when her daughters told her that they had, she said, "That's what I would have done."(photo credit 1.9)
A twenty-seven-year-old bachelor, Robert is not as good-looking as his younger brother, André. But he has a terrific sense of humor, a passion for art, and an impressive collection of original paintings.
At the end of 1940, Robert becomes the first member of his family to be asked to join the Resistance. The invitation comes from André Postel-Vinay, who had met Robert two years earlier, when the two of them took the exam to become finance inspectors together.
Postel-Vinay is a strikingly handsome twenty-nine-year-old, with delicate features set off by a broad forehead and an aquiline nose. As a lieutenant in the 70th Régiment d'Artillerie in 1940, he is celebrated for his bravery and his exceptionally accurate shooting. On June 17, he is captured by the enemy, but he manages to escape a week later. After three days on foot, he makes it back to his parents' home in Paris, utterly exhausted.
Robert Boulloche was the oldest son in the family, and shared some of his parents' caution. He declined André Postel-Vinay's invitation to join the Resistance at the end of 1940 — but he predicted that his younger brother would be eager to join the fight against the Germans.(photo credit 1.10)
The apartment is empty, because his parents are in Brittany. The next morning, he is awakened by a German military ceremony taking place in the street below. When he goes to the window, he has the same reaction as Christiane and thousands of others. He is appalled by the savage sight of German soldiers holding gilded crosses in the air as they parade down the street.
LIKE JACQUES BOULLOCHE, Postel-Vinay had seen the war coming many years earlier. After the Night of the Long Knives in 1934, when Hitler ordered the execution of Ernst Röhm, the gay commander of the Nazi SA, two army generals, and at least a hundred others, Postel-Vinay decided that "war is the only way to communicate with the Nazis — because even among themselves, they behave like butchers."
(In a speech to the Reichstag after the bloody massacre, Hitler freely admitted that what he had done was completely illegal. The legislature quickly passed a law that retroactively legalized his butchery.)
After France's collapse in 1940, Postel-Vinay thinks that any victory over the Nazis will take a very long time, if it ever comes at all. But he is propelled by the conviction he shares with the small number of very early _Résistants_ : He believes that his life will have no meaning until he finds a way to fight for the Germans' defeat.
By October 1940, he has linked up with one of the earliest British Resistance groups in the Paris region, which is helping downed British airmen escape back to England through Spain. Then he connects with a group of French officers from the army's intelligence service, the Deuxième Bureau, who are collecting information on German troop movements and relaying it to London.
Postel-Vinay also makes contact with a group of anti-Fascist intellectuals at the Musée de l'Homme in Paris, who have begun to organize in the summer of 1940. Most of them are leftists, and some of them aren't French, including the group's most active member, Boris Vildé, a linguist with anarchist leanings and Estonian origins.
The group at the Musée de l'Homme is one of the very first Resistance groups. In December 1940 it starts to publish the newspaper called _Résistance._ During the next three years, dozens of other clandestine newspapers will appear all over France. Delivering some of these publications is one of Christiane Boulloche's earliest acts of resistance.
When Postel-Vinay asks Robert, his former classmate, if he will join the Resistance to collect intelligence about the Germans for the British, Robert shows no interest. He doesn't seem to think the work will be very useful, and as the oldest son in his family, he probably shares some of his parents' caution.
André Postel-Vinay in his army uniform. He convinced André Boulloche to join the Resistance at the end of 1940.(photo credit 1.11)
Robert tells Postel-Vinay he won't become a clandestine enemy of the Germans. Then he makes a fateful prediction: "I know someone who will jump right in."
"Who's that?" asks his friend.
"My little brother, André," Robert replies.
ANDRÉ BOULLOCHE has managed to return to France from Morocco at the beginning of September 1940. He is demobilized in Marseille, then quickly makes his way back to Paris, where a family reunion takes place, as joyful as it is unexpected.
He rejoins the Department of Bridges and Highways, where he worked as an engineer before he was mobilized, and he is posted to Soissons, where he is named _adjoint ingenieur-en-chef_ (deputy chief engineer). This puts him sixty miles northeast of Paris.
When Robert takes him to meet Postel-Vinay, André responds just as Robert had predicted: He immediately joins the underground. For the first time since the armistice, he has finally found a way to fight the Germans.
On the surface, André doesn't seem very emotional. But Postel-Vinay quickly discovers that beneath a placid exterior, André is full of zeal. The two of them agree about everything that matters at this ominous moment. Neither can bear France's defeat, they share a profound horror of Nazism, and they both feel a compulsion to do something about it. "It was absolutely unbreathable," said Postel-Vinay. "André was very passionate, and he couldn't sit still."
Although Postel-Vinay is four years older than his new recruit, the two men share another quality that pushes them into this treacherous adventure: the impetuousness of youth.
"For the two of us, the Resistance was a kind of lifesaver," Postel-Vinay explained, "because without it, life no longer had any meaning." Their decision to join the Resistance is so instinctive, and so immediate, they barely consider the possible consequences — for themselves, or for anyone else.
THE DECISION of these two young technocrats to join the secret war against the Germans at the end of 1940 is very unusual for Frenchmen of their class. There is hardly any other milieu more unprepared for clandestine activity than bourgeois civil servants, and at this point there is only a small number of Frenchmen actively challenging the German Occupation.
Unlike so many early members of the Resistance, these fiercely committed young men are not outsiders at all: they are neither Jewish nor foreign-born nor Communist. But they share a larger idea about human progress, which makes them passionate about the horror and the absurdity of Nazism — and the perils it poses for everyone.
By now the Third Reich has conquered Austria, Poland, Czechoslovakia, Holland, Belgium, Luxembourg, Denmark, and Norway, as well as France, and most people consider it invincible at this stage of the war.
However, there remain a couple of skeptics. When the Russian foreign minister, Vyascheslav Molotov, visits Hitler in Berlin in November 1940, the Führer tells him the British are finished. "Then whose bombers are those overhead?" Molotov asks. "And why are we in this bomb shelter?"
Over the next eighteen months, André Boulloche convinces his boss, Pierre Pène, and a fellow engineer, Jean Bertin, to join him in collecting information about German troop movements in the region. He reports on the work of the Germans who are constructing a secret headquarters in a tunnel at Margival, outside Soissons. This is supposed to become Hitler's headquarters when he invades Britain, but he will not visit it for the first time until 1944, after the Normandy invasion.
Through his own contacts and those of his colleagues with local builders, André obtains the plans of more than 150 structures being built by the Germans, and he believes that they are reaching London.
He also marks off parachute fields and sets up arms depots. To make himself a better secret agent, he memorizes a book that interprets every insignia of the German Army. The book is easy to get when the Germans first arrive in France. It disappears when the Germans realize how useful it can be to their enemies.
Employing primitive spy craft, he sends Postel-Vinay letters written with lemon juice, which only becomes legible when the pages are heated over a candle. Twice a month he goes to Paris to give his information to one of Postel-Vinay's contacts, who is supposed to transmit it to London. These trips also make it possible for him to visit his parents. But he never discusses his clandestine activities with his family, partly because he doesn't want to influence his sisters.
MEANWHILE, in London, Charles de Gaulle is cementing his position as the leader of the Free French. Once France has signed an armistice, the British no longer worry about offending the new French government, and Churchill is ready to grant de Gaulle formal recognition. At this point de Gaulle is still a little-known, recently promoted general, and Churchill hopes that he will attract other, more famous French personalities to his cause in London. But that never happens.
At the end of May 1940, eight hundred small boats had loaded 338,000 men into larger ships during the legendary evacuation of Allied troops from Dunkirk, including 500 French officers and 18,000 French sailors, to prevent them from being captured or killed by the Germans. But all but 50 of these officers and 200 of these sailors will return home to occupied France, rather than stay in Britain to fight the Germans.
"Their idea was to get out of the war no matter what, as quickly as possible," recalled Sir Edward Spears, the wartime liaison between Churchill and de Gaulle. "We had 15,000 French sailors at Liverpool. I went to speak to them. I tried to persuade them to continue the fighting. Impossible... As for what might happen to England, they couldn't care less. That was the way it was — we were defeated, and if the French army was defeated, it was impossible to imagine that the English would survive."
Only one deputy, one admiral, and one leading academic remain with the Free French in London, and de Gaulle notices that all of his earliest supporters are either Jews or Socialists.
A man of mythic pride, de Gaulle is infuriated by his total dependence on the British. His relationship with Churchill vacillates between prickliness and open hostility. But both men share "a love of drama and a deep sense of history" — and they recognize that they need one another. Like André Boulloche and Postel-Vinay, de Gaulle experienced the defeat of 1940 as a searing humiliation. Some thought de Gaulle felt like a man who had been skinned alive.
There was one other thing de Gaulle had in common with the Boulloches' ancestors — the general's father had also believed that Dreyfus was innocent.
One thing that provokes the suspicion of antifascist Frenchmen in London is de Gaulle's initial reluctance to publicly embrace republicanism. This hesitancy makes some people doubt his commitment to democracy in a postwar France. Early Free French broadcasts from London are introduced with the motto _"Honneur et Patrie_ " (Honor and Country), rather than the traditional republican " _Liberté, Egalité, Fraternité."_
But the general sees his position as tactical: Especially at the beginning, he tries to avoid all political labels so that he can attract the widest possible support. Not until November 1941 does he finally embrace "Liberty, Equality, and Fraternity" — to remain "faithful to the democratic principles... which are at stake in this war of life and death."
The BBC broadcasts quickly become a vital part of the Allied propaganda effort aimed at France. For millions of French people, listening to the outlawed BBC is the main act of rebellion they engage in.
The British give the Free French five minutes on the BBC every night. At the same time, French-language broadcasts of the BBC expand gradually from two and half hours daily in 1940 to five hours in 1942. As the size of the organized Resistance increases in France, these broadcasts also include a growing number of coded messages, which communicate everything from the location of new arms drops by parachute to the launching of the Allied invasion in Normandy.
AT THE END OF 1940, just as André Boulloche starts collecting information for the Resistance, de Gaulle creates a department in London responsible for "action in the occupied territories." The agents who are eventually recruited are hardly professionals. Almost anyone who volunteers is accepted, once he has satisfied British intelligence that he isn't a double agent.
Most of de Gaulle's earliest recruits are from French units that were evacuated from Norway or Dunkirk. At the end of June, his ranks are swelled by the residents of Sein, a rocky island off the western tip of Brittany near Audierne. The Germans don't reach the island until July, and by then two small fleets of fishing boats have put to sea with 133 men aged fourteen to fifty-one — virtually all the able-bodied men from the island.
Each of the emigrants carries a little food, a liter of wine, whatever money their family has — and the family shotgun, if there is one. After they dock at Falmouth, de Gaulle welcomes them in London. They will become some of the earliest recruits of the Free French. "The island of Sein stands watch duty for France," the general proclaims.
A week later, a French Army captain named André Dewavrin, who had fought in Norway in the spring, presents himself to de Gaulle at his temporary headquarters at St. Stephen's House on the Victoria Embankment in London. Nearly all of what Dewavrin knows about spying he has learned by reading thrillers.
De Gaulle is impressed anyway — and at this point he doesn't have a lot of alternatives. He promotes Dewavrin to major and puts him in charge of what will eventually become the Bureau Central de Renseignements et d'Action (BCRA) or Central Bureau of Intelligence and Action. The BCRA is the product of a merger of two organizations de Gaulle has started after his retreat to London: the Deuxième Bureau (intelligence) and the Troisième Bureau (operations). The original task of the Deuxième Bureau is to gather as much information as possible about German preparations for what is considered an almost inevitable invasion of England.
Dewavrin gives himself the code name of "Col. Passy," and he eventually dispatches more than 350 agents to occupied France. Because he depends on the British for transportation and radio equipment, he has to work with the newly created Special Operations Executive, which has an RF (République Française) section to work with the Free French, as well as its own French section (section F), which carries out independent operations in France. There is constant tension among all three bureaucracies, but by the end of 1941, Col. Passy has already managed to send twenty-nine of his own agents to France.
* The doctor joined the Resistance in 1943. But when he was interviewed by the government after the war about his clandestine activities, he never mentioned the fact that he had a Jewish wife. (French National Archives, box 72AJ80)
# _Seven_
_Never in the field of human conflict was so much owed by so many to so few._
— Winston Churchill, addressing Parliament, August 20, 1940
_And while I am talking to you mothers and fathers, I give you one more assurance. I have said this before, but I shall say it again and again and again: Your boys are not going to be sent into any foreign wars._
— President Franklin D. Roosevelt, campaign address, October 30, 1940
Throughout 1941, André continues his work as a highway engineer and a secret agent. The two jobs go well together, because his official responsibilities make it easy for him to travel without arousing suspicion. In his clandestine life, he alternates between diverting supplies destined for the Germans and accumulating a private stock of gasoline for his Resistance unit. He also continues to collect intelligence about German troop movements, to forward to London.
The war news since the fall of France has been relentlessly bleak. The Battle of Britain begins immediately after the Nazis conquer France. Between July and September 1940, Hitler's Luftwaffe targets Royal Air Force airfields and radar stations, to soften up Britain for what many people still think of as a certain German invasion.
On July 3, the great British iconoclast George Orwell writes in his journal: "Everywhere a feeling of something near despair among thinking people because of the failure of the government to act and the continuance of dead minds and pro-Fascists in positions of command. Growing recognition that the only thing that would certainly right the situation is an unsuccessful invasion; and coupled with this a growing fear that Hitler won't after all attempt the invasion but will go for Africa and the Near East."
Enraged by an RAF bombing attack on Berlin, Hitler switches his targets in the fall to London, Coventry, Birmingham, Liverpool, Manchester, and other British cities and ports. Beginning on September 7, London is bombed every day (or night) for fifty-seven days in a row. On September 12, the British government issues an invasion alert, but the scare fades quickly.
On November 14, St. Michael's Cathedral in Coventry, mostly built in the fourteenth and fifteenth centuries, is almost completely demolished by a German bombardment. By this time the British have already broken the German codes. Some historians believe that no extraordinary measures were taken to protect Coventry, to prevent the Germans from realizing that the British were reading their most secret messages.
Between July and December, German bombs kill 23,002 and wound 32,138 in Britain. Nearly 3,000 Britons are wiped out in a single day at the end of December. But despite the loss of 1,173 RAF planes and 500 pilots, Britain survives the German onslaught. In the spring of 1941, the bombing campaign finally tapers off. Britain's spirit, stoked by Churchill's extraordinary oratory, is still intact. As Orwell and others had predicted, Hitler has turned his sights elsewhere, abandoning his cherished plan to conquer the British Isles.
THE BATTLE OF BRITAIN is broadcast directly into American living rooms by Edward R. Murrow. The newly minted radio correspondent for CBS News becomes the most celebrated broadcaster of his generation practically overnight. Just thirty-two in 1940, Murrow is a master of vivid images, all of them rendered in the rich baritone of a Broadway actor. Fearless and theatrical, Murrow transfixes his listeners with live reports delivered from London rooftops, as German bombs fall all around him:
_This is Trafalgar Square. The noise that you hear at the moment is the sound of the air raid sirens... The searchlight just burst into action off in the distance. One single beam sweeping the sky above me... There's another searchlight... You see them reach straight up into the sky, and occasionally they catch a cloud and seem to splash on the bottom of it... One of the strangest sounds one can hear in London these days — or rather these dark nights — just the sound of footsteps walking along the street, like ghosts shod with steel shoes._
Blessed with a story that doesn't require objectivity, Murrow becomes a good friend of Winston Churchill and a lover of his daughter-in-law, Pamela Digby Churchill.
When he returns to America for a visit in 1941, he is greeted at the dock by a crowd of fans and reporters. CBS celebrates him with a banquet for eleven hundred at the Waldorf-Astoria. Three years later, Murrow will dine with FDR at the White House on the night of the Normandy invasion.
In the fall of 1940, Murrow's reports inspire considerable sympathy in America for the beleaguered British. But when Franklin Roosevelt decides to seek an unprecedented third term that year, he feels compelled to promise a reluctant country that he will stay out of the growing European war. "Your boys are not going to be sent into any foreign wars," he pledges to a Boston audience a few days before he is reelected in November.
At the same time, Roosevelt continues to withhold American recognition of the Free French and de Gaulle. Churchill's relationship with de Gaulle is always prickly; Roosevelt actually loathes the Frenchman. As Eisenhower puts it delicately in his memoirs, Roosevelt "could not agree to forcing De Gaulle upon anyone else."
Or as the historian Ian Ousby slyly summarized their relationships, "The familiar slur of enemy propaganda that [de Gaulle] was merely the tool of Britain or the Allies certainly found no answering echo in the hearts of Churchill or Roosevelt."
At the beginning of 1941, Roosevelt still sees some value in maintaining relations with Vichy France, and he names Admiral William D. Leahy to be his ambassador there. Leahy even arrives with a 1941 Cadillac limousine to present to the Vichy president, Marshal Pétain. The historian Robert Paxton called the fancy American automobile "a very, very explicit act of support."*
American public opinion begins to rally to de Gaulle long before the president does. In a rift with Churchill, Roosevelt hopes to keep France neutral by cozying up to the Vichy regime. His actions are befuddling to the budding Resistance movement.
THE MORALE of the French Resistance gets a huge boost from the rupture of the Hitler-Stalin nonaggression pact on June 22, 1941. That day the Führer announces that he is invading the Soviet Union, on a line stretching from Norway to Romania. Hitler's announcement includes what the _New York Times_ calls one "vitally interested statement," which is also a tiny source of hope — a public suggestion that German military forces will not be strong enough in the west to conquer the British Isles, as long as so many Soviet troops are stationed on Germany's eastern flank.
The invasion transforms the attitude of Communists in France, most of whom have refrained from joining the Resistance up until now, because of the nonaggression pact. Now they will become some of the Nazis' fiercest enemies.
The new military campaign also buoys every French citizen who remembers the importance of June 22, 1812, from history class at the lycée. That was the date Napoleon launched _his_ invasion of Russia. Now, millions are praying that Hitler will replicate Napoleon's disastrous experience on his journey toward the Urals.
* When de Gaulle liberates Paris in August 1944, he makes sure that he enters the city in a Hotchkiss, a large French limousine. (Collins and Lapierre, _Is Paris Burning?,_ p. 182)
# _Eight_
_This was our obsessive fear: that we would be tortured into giving names if we were captured by the Germans. Compared to that nightmare, death hardly seemed like a menace at all._
— Christiane Boulloche
ANDRÉ POSTEL-VINAY, the man who has recruited André Boulloche into the Resistance, is lucky because his two bosses at the Finance Ministry know about his work as a secret agent, and never object to it or betray him. Thanks to their complicity, he is able to work practically full time against the Germans.
By the middle of 1941, Postel-Vinay has begun to wonder whether the information he is collecting is actually reaching London. To find out, he includes a request in one of his coded radio messages, asking the BBC to confirm the arrival of his dispatches by broadcasting "CBA-321." One night, just back from a trip, he flips on the radio. Through the garbled sounds of the jammed transmission, he manages to make out the magic combination: "CBA-321." At that moment, those syllables feel like a miracle "from the great beyond."
Most of the newly organized Resistance units are dangerously porous organizations, easily infiltrated by double agents. Even the Resistance leaders who are trained in Britain before their repatriation to France receive only the most rudimentary instruction in the dark arts of espionage.
Pierre d'Harcourt introduced Postel-Vinay to the first two Resistance units he works with. In July 1941, d'Harcourt is captured by the Gestapo in a Paris Métro station. When he is cornered by the Germans, d'Harcourt tries to run away and tumbles down a stairway. As he falls, he tries desperately to destroy the secret documents he is carrying with him. At the bottom of the stairs, the Germans fire on him, shooting him through the foot, the leg, and the lung.
Postel-Vinay learns the identity of the "charming accomplice" of the man who has denounced d'Harcourt. Postel-Vinay thinks that this accomplice is also aware of his own work in the Resistance, so now he feels like he is perilously balanced on a tightrope. And yet he still doesn't want to "interrupt" himself.
His behavior suggests a kind of fatalism that is familiar to his co-conspirators.
Two months after d'Harcourt is shot, the Gestapo raids the apartment of Captain d'Autrevaux, the number-two man in the French military intelligence unit Postel-Vinay has been working with. D'Autrevaux happens to be away when the Germans arrive, and he manages to escape to the unoccupied zone in the south. After that, three of his associates ask Postel-Vinay to start relaying their information to London. Postel-Vinay considers this a wonderful development: He finally feels like he is making a difference.
But then another danger sign appears. In November, an agent named Wiltz, who has been d'Autrevaux's closest collaborator, misses an appointment with Postel-Vinay. Almost immediately, Postel-Vinay receives word that Wiltz has been arrested. Now he begins to feel like he is back in the infantry, surrounded by artillery, the sounds of their explosions steadily approaching.
LATE IN THE MORNING OF DECEMBER 13, Postel-Vinay arrives home at his parents' apartment on avenue de Villars, adjacent to Les Invalides, which houses Napoleon's tomb. As he walks through the front door of the building, he spots two young men coming down the stairs in front of him.
He immediately recognizes the first one: a tall, thin, blond Englishman he knows only as Paul. Paul has been an aide to Patrick O'Leary, the head of one of two Resistance groups Postel-Vinay has been working with all year.
O'Leary pretends to one and all that he is a French-Canadian officer. In fact, he is a former surgeon in the Belgian Army, whose real name is Albert-Marie Edmond Guérisse.
After serving with the Belgian Army during the eighteen-day campaign of 1940, he escaped to England, where he secured a British Navy commission as a lieutenant commander. On April 25, O'Leary was on the HMS _Fidelity_ when it overturned in a squall off the French coast near the eastern end of the Pyrenees, but he managed to swim to shore.
Identifying himself to the gendarme who arrested him as Albert O'Leary, an evading Canadian airman, he was sent to St. Hippolyte du Fort near Nîmes, to be with British officers. There he met Ian Garrow, a tall, dark-haired captain in the Highlanders in his early twenties, who quickly helped O'Leary escape.
O'Leary then joined one of the most effective networks of the war devoted to the repatriation of downed Allied pilots. After Garrow is arrested in October 1941 O'Leary takes charge of the operation, which becomes known as the PAO line (for his initials) and, more famously, as the Pat or O'Leary line. It eventually helps an astonishing six hundred pilots to escape from occupied France.
Postel-Vinay has met Paul once before, in Marseille. Although Paul is introduced to him as a fellow _Résistant_ who is working for one of the best Resistance organizations, Postel-Vinay is immediately suspicious of him. He listens as the Englishman delivers a speech filled with beautiful principles — but everything he says rings false in the Frenchman's skeptical ears.
Now, in Paris, Paul is accompanied by someone Postel-Vinay has never seen before: a stocky, gray, sinister-looking fellow, who keeps his hands shoved deep into the pockets of his shabby raincoat.
"That's him," says Paul, motioning toward Postel-Vinay.
"My chauffeur," Paul explains, indicating his unpleasant companion.
Postel-Vinay has other reasons to suspect the Englishman. A few weeks earlier he had heard about a fistfight between Paul and his boss, O'Leary. The word on the street is that the fight was about money. Postel-Vinay knows that O'Leary has an impeccable reputation, so he assumes that Paul must have been in the wrong.
He also knows that Paul usually works in the unoccupied zone. So it feels odd to see him here in the north, where his pidgin French and pronounced English accent can hardly provide him with much cover if he is detained by the Germans.
Pondering all of this, Postel-Vinay realizes that these two have just been knocking on the door of his parents' apartment.
"Let's talk outside," Postel-Vinay suggests, hoping to get them away as quickly as possible. He leads the way, as Paul and the other man follow silently.
On the sidewalk, Paul explains that Patrick O'Leary's organization has been split into two parts: one to look after downed British airmen, and the other to collect intelligence for the British. Paul says he is working with the intelligence unit, and the stranger accompanying him is actually his new boss.
"Bring all the information you've collected to tomorrow's meeting," says Paul. "Even the stuff you've already sent to Marseille. We have a new radio post here in the occupied sector, so you won't need to go to Marseille anymore.
"I'll be back to pick you up at nine tomorrow morning," Paul continues, "with a beautiful fake identity card on the windshield of my car." Since Germans are almost the only people allowed to drive in Paris now, this boast hardly bolsters Postel-Vinay's confidence. Yet he agrees to meet Paul anyway.
In the afternoon, Postel-Vinay gets a visit from Bernard Vernier-Palliez, a young man who has studied for a job in the Foreign Ministry and has been recruited into the Resistance by André Boulloche, after André is signed up by Postel-Vinay. Vernier-Palliez is transporting a case filled with weapons, including a German Mauser — a job often consigned to Resistance women, because women are less likely to be suspected as arms smugglers.
Vernier-Palliez asks Postel-Vinay if he can keep the weapons for him for a while. Once again, Postel-Vinay agrees, even though he strongly suspects that this is not the ideal moment to perform this favor.
Then he spends the rest of the day gathering all the information he can for tomorrow's meeting.
Of course Postel-Vinay knows that Paul may be planning to betray him. But his judgment is clouded by exhaustion, and he continues to behave as if he isn't in serious danger. Part of him thinks it's a terrible idea to meet Paul tomorrow. But is that because he has real reasons to fear a trap? Or is it paranoia, the product of constant danger? During the last twelve months, he has taken many risks, and he has often thought that he was in danger. But he has always come through okay. Perhaps this is making him believe too much in his own luck.
Postel-Vinay thinks it is difficult to believe in the perfect treason until you have experienced one yourself. He imagines the ideal traitor would do you in with grinning flair. But he finds it hard to imagine such a person in real life. Paul's story about the organization being split in two does sound plausible. And the new radio transmitter he described in the occupied zone is exactly what Postel-Vinay has been looking for.
There is one other large question weighing on him, the same one that troubles many of his fellow young _Résistants_ : If he runs away, will his parents be arrested in his place? Those big posters in the Métro are constant reminders that every relative of a _Résistant_ is now subject to arrest.
Postel-Vinay sleeps badly. The next morning, his sister, Marie-Hélène, is the first person to arrive at their parents' apartment, at eight thirty. Like the Boulloche sisters who work with their brother André, Marie-Hélène participates in her brother's clandestine activities. Until today she has always managed to bury her fears about her brother's fate. But after he tells her about Paul's visit, she begs him not to meet with him again.
Postel-Vinay does his best to calm her down. Then his sister leaves the room to speak with their parents. At that moment, Postel-Vinay turns around to remove a loaded six-shot Enfield pistol from the closet. He slides it into the inside pocket of his overcoat. In his other pocket, he places a bulky envelope with all the intelligence he has gathered for Paul.
When his sister returns, he tries to mock her fears. But then he blurts out, "If Paul betrays me, I'll kill him! And then I'll kill myself." He thinks this is the first really good idea he has had all morning.
As soon as he walks out into the street, a black car pulls up beside him. Paul's "chauffeur" is in the driver's seat, with Paul next to him. When he tells Postel-Vinay to climb into the empty backseat, that somehow feels reassuring.
Now they are driving through the place de la Concorde, then past Madeleine and Saint-Lazare. They pause for a moment in front of a Métro entrance. The perfect moment to jump out, Postel-Vinay thinks to himself. But his hand never touches the door handle.
A few minutes later they are passing the cemetery in Montmartre. Finally, they arrive at the Terrass Hotel, a venerable institution with panoramic views of Paris. Paul and Postel-Vinay climb out of the car, leaving the chauffeur behind.
As Paul guides him up to the second floor, Postel-Vinay thinks he sees the hotel clerk giving them an odd look. But then they walk into an empty, ordinary-looking hotel room, and Postel-Vinay decides it's a good sign that the chauffeur hasn't accompanied them. When the door is closed behind them, he hands Paul the fat envelope filled with intelligence reports.
The documents include information about German troop movements, detailed blueprints of airports used by the Germans, a plan of the port of Brest, and descriptions of the results of British bombardments of military targets.
He asks Paul about the location of the new radio transmitter, but Paul deflects his question.
Suddenly the door bursts open, and the chauffeur runs in with three accomplices. All four of them are pointing their pistols at Postel-Vinay. Just like André Boulloche, who is carrying a cyanide pill when he is arrested and has always planned to kill himself if the Germans capture him, Postel-Vinay is carrying a revolver for the same purpose. He reaches into his pocket so that he can shoot himself. But first he thinks, _I must shoot Paul!_
Before he can fire a single shot, four gun barrels are pressed against his chest. One of the men grabs his Enfield revolver, which is still in his pocket, as another one slams on the handcuffs.
Postel-Vinay feels himself entering a new universe, separated by a vast distance from the world where he lived before — somewhere dangerously close to hell.
He is impressed when one of the men throws Paul on the bed and starts slapping him. These men obviously take pride in their work, because they are trying to make it look as though Paul isn't the one who has betrayed him. The man attacking Paul seems to be enjoying the violent pantomime, but Postel-Vinay doesn't think any less of him for that.
PAUL'S REAL NAME is Harold Cole. Trained as an engineer, he was known in England before the war as a con man and a burglar. As a sergeant with the British Expeditionary Force in France, he had absconded with the sergeant's mess funds. When he turned up in Lille after the armistice, he identified himself as "Captain Harold Cole" of the British Secret Service.
During the fall of 1941, Cole had actually helped thirty-five British airmen escape. But on December 6 — one week before he met Postel-Vinay in Paris — Cole had been arrested by the Germans in Lille. Probably to avoid a threatened execution, it was at this moment that he switched sides to the Nazis.*
ALTHOUGH HE IS UNABLE to shoot himself at the moment of his arrest, suicide remains Postel-Vinay's urgent priority. Three days after entering the Prison de la Santé, he breaks away from the guards and flings himself over a railing, down two stories into the center of the prison courtyard. The violent plunge breaks most of the bones in his body, but he survives his own suicide attempt.
Now he decides to feign madness, and the Germans transfer him to the Quentin Pavilion in l'hôpital de la Pitié.
André Boulloche is tipped off to Postel-Vinay's new location by André's cousin, Funck-Brentano, his only relative with a Jewish wife. His cousin also happens to be the chief surgeon at l'hôpital de la Salpêtrière next door to the one where Postel-Vinay is a prisoner. With the help of a young doctor, the surgeon locates an underground passage that connects the two hospitals.
After months of work, which includes help from his brother Robert, Jacques Postel-Vinay (a close cousin of the prisoner), André's fellow _Résistants_ Bernard Vernier-Palliez and Hubert Rousselier, as well as the fabrication of a number of keys, they finally manage to reach the cellar beneath Postel-Vinay's cell. There they remove a floorboard above them and try to communicate with Morse code. But Postel-Vinay only remembers enough of the code to respond with a simple SOS.
Postel-Vinay considers this attempt to free him crazier than his own decision to feign madness. Even if his rescuers manage to open a hole for him to escape through, he will never be able to perform the gymnastics required for an escape, since he can barely walk.
Before they can proceed to the next step, he is moved again, and the escape effort is abandoned. Postel-Vinay thinks it's a miracle that his would-be rescuers haven't themselves been captured during their attempts to free him.
* After the war, Cole was shot and killed after exchanging gunfire with a French police inspector in January 1946. "A Soldier in Four Armies: He Betrayed Them One After Another," reported France-Soir. (Murphy, _Turncoat,_ p. 258)
# _Nine_
POSTEL-VINAY has undergone two surprisingly civil interrogations by his German captors after his attempt to kill himself. He is completely baffled by the gentleness of his interrogators. But he concludes that logic is just as irrelevant as justice in his present circumstances.
Put into multiple casts by the Germans after his misadventure, he endures excruciating pain. But after his second interrogation, Postel-Vinay feels gigantic relief. He deduces from his interrogators' questions that his parents probably haven't been arrested, and he concludes that the Germans have failed to find his blue notebook — the one that includes the names of some of his confederates — because he is never asked any questions about it.
His ultimate nightmare — that he might have endangered his family or his comrades — has "miraculously disappeared." As a result, he regains some of his will to live, even as he remains certain that the Germans will eventually execute him. "The fear of talking stopped haunting me. Now I was sure I would be able to die in peace."
A month or two after the arrest, his mother makes contact with a German officer who is a chaplain to find out if her son is alive. The next day, the chaplain comes to Postel-Vinay's cell and extends his hand, covered in a field-gray glove, and declares, "Take this hand. Your mother touched it yesterday. She loves you very much. She is really very courageous."
Even more remarkably, a month later, one of Postel-Vinay's interrogators authorizes one package from his parents, every fifteen days, containing food, books, pencils, and paper. He begins to write poems, deep into the night, and goes to sleep with "the joys of an author's vanity, and duty accomplished."
Then around April 15, 1942, he receives a visit from a German officer and a white-haired man whom Postel-Vinay at first mistakes for an artist. In fact, his visitor is Clovis Vincent, the prewar head of neurology at the hospital who has retained his post during the German Occupation. Gradually, Postel-Vinay realizes the famous doctor has been sent there by his family to urge him to feign insanity, to avoid a firing squad.
Part of him remains eager to die. But is it fair to spurn his parents' attempt to save him? He knows their hope isn't completely far-fetched: Even one of his interrogators has mentioned the possibility that he will be tried by a tribunal rather than face summary execution.
Finally he decides it is too selfish to reject his parents' plan. He has already made them suffer too much. So he embarks upon "a ship of fools" to try to deceive his captors.
For the first stage of his fake madness, he pretends to suffer from terrible migraines. Then, toward the end of April, the casts are finally removed from his legs. They are skeletal, with no trace of calves, and his ankles are frozen in place. Postel-Vinay is certain he will never again be able to walk more than a few yards.
Realizing that his fake migraines and fraudulent tics will never get the attention of guards already numbed by the genuine traumas of his fellow prisoners, Postel-Vinay decides his only option is to attempt suicide — again.
At the beginning of May, he begins to search for the right instrument of destruction. The only one he can find is a short nail at the end of his bed. Blunt, rusty, and slightly twisted, it is hardly ideal, but it's all he has.
His first thought is to plunge the nail into his right eye (which doesn't see very well anyway), but he quickly realizes that he lacks the resolve to mutilate himself that way.
Once a week he is given a Gillette razor for ten minutes so that he can shave himself. On June 20, he shoves the handle of the razor into a bar of soap, then plunges the naked edge of the blade into his left forearm. He means only to cut some veins, but instead he hits an artery and severs some tendons.
When his captors return to his room a few minutes later, they instantly take him away to the operating room. There the surgeon spends an hour and a half repairing the artery and the surrounding tendons — without offering any anesthetic.
On August 1, he is ordered to get dressed. Then he is removed from his cell and taken outside. Waiting for him in the street is the familiar Citroën Traction Avant used by the Gestapo. Postel-Vinay is pushed into the backseat. Then he watches the streets go by — boulevard de l'Hôpital, boulevard Saint-Marcel, boulevard Arago, rue de la Santé. Now he knows he is being sent back to the place where his misfortune began — the Prison de la Santé.
After a month back in his old prison, on August 31, a guard opens his cell and tells him to gather all of his belongings, which at this moment consist of a single toothbrush. Is he being sent to a concentration camp? Or to the "next world"? He has no idea where he is going, but — to his own surprise — Postel-Vinay feels no fear.
This time, a windowless gray van awaits to take him to his next destination. Ten minutes later, he knows there will be no deportation or execution today. He has been returned to the Quentin Pavilion in the l'hôpital de la Pitié — to a cell just down the hall from the one he left four weeks earlier.
The following afternoon, on September 1, he is moved again. This time there is an ambulance downstairs, attended by a male nurse and a German soldier. Postel-Vinay lies down on the ambulance bed and the soldier closes the door. Ten minutes later, he has arrived at a new, unfamiliar building. He is led to something that looks like a shower room, and the nurse locks the door behind him.
Through the window he peers into a garden, where he sees other inmates walking around, who are obviously crazy. He doesn't know it yet, but he is actually in a psychiatric institution: l'hôpital Sainte-Anne, in the 14th arrondissement of Paris.
Nothing happens for two more days, and Postel-Vinay becomes increasingly nervous about the performance he has planned. He must convince his captors that he is truly crazy. Meanwhile, he is praying with all his might: "Here I am at the end of my strength. Help me God!"
Finally, on the afternoon of his third day at Sainte-Anne, a nurse leads him into the office of a German psychiatrist.
"So," the doctor asks, "what's wrong?"
"What's wrong?" he replies. "I know where I am. I'm in an insane asylum! My parents always thought I belonged here, but they are the ones who are crazy!"
Postel-Vinay continues his charade for several minutes, but to no avail. Finally, his German interrogator speaks again. "Monsieur Postel-Vinay, now I'm going to tell you what I think. You have played your role very well. But you are not crazy."
"Of course I'm not crazy," Postel-Vinay replies. "It's only my parents..." But suddenly the doctor's face darkens, and he stands up. Postel-Vinay stands up with him. At last the prisoner lets down his guard: "Whatever I did, I did for my country."
"Ah, yes," the German replies. "That is exactly the way I understand things."
The doctor walks him back out into the hallway, noticing that he is still suffering tremendously because of his barely healed ankles. "You really walk as badly as that?" he asks. "I will order an ambulance to take you back to Quentin."
The doctor deposits him on a bench and returns to his office. Postel-Vinay looks around him. He is in a narrow hallway, surrounded by other patients, all wearing the blue uniforms provided by the hospital. One door off the hall leads to a room where he can hear German being spoken — that must be a guard post. At the far end, another door leads to the garden — and possible freedom? But a soldier is guarding that exit.
After all he has been through, the odds against an escape seem overwhelming. But a primal instinct propels Postel-Vinay off the bench, toward the door to the garden. He pauses at the exit for a couple of seconds, like a man looking to see if his car has arrived.
Then the miracles begin: The guard makes no attempt to bar his way.
Why not? The soldier is only supposed to stop the patients in blue uniforms, and no one else? Postel-Vinay is very disheveled, but he _is_ dressed like a civilian. _Oh, saintly German discipline!_ he thinks silently to himself.
As he continues down the steps into the garden, he thinks he hears the soldier being called back into the guard room. But he does not turn around. He spots an archway in front of him and a workman — a Frenchman — coming toward him from that direction.
"Where's the exit?" Postel-Vinay asks, as casually as possible.
"Take your first left and follow the long alley to the end."
Surely there will be another guard at the end of the alley, but Postel-Vinay isn't about to stop now. And when he reaches the street, there is no soldier.
He listens for someone running behind him and watches for the ambulance that has been dispatched to return him to his cell. But there is no one and nothing, except for two small children walking toward him on rue Cabanis, to his right.
Now seconds matter.
"Listen, kids, I've just escaped. The Germans will shoot me if they catch me. Give me enough money for the Métro. I'm wounded, without a cent, and I can barely walk."
The children are wide-eyed, astonished — and speechless. But they immediately start searching their pockets for change. Between them they have 23 centimes, just enough for one Métro ticket. They give him all of their money and point him to the closest Métro station — Glacière .*
Postel-Vinay is self-conscious about the way he walks and looks; surely his appearance will draw unwelcome attention. He is unshaven, his suit is unpressed, there are no suspenders to hold up his pants, and no laces in his shoes. He looks like a young hobo, and the unhealed bones in his ankles are incredibly painful.
If his life did not depend on it, he could not possibly walk more than a couple of blocks by himself. But his life _does_ depend on it, and somehow, he summons a supernatural force from within that propels him down into the Métro Glacière.
He thinks, _If I were them, I would certainly alert all the subway station chiefs to look for me, and I would stop the trains on the nearest line_. But the trains keep running.
Underground there are throngs of Parisians — and many German soldiers. Postel-Vinay gets on the train DIRECTION ÉTOILE. He travels five stops, then gets off at Montparnasse. Randomly, he decides to go north, DIRECTION PORTE DE LA CHAPELLE — anything to get off the line he started on.
The passages between the Métro lines suddenly feel incredibly long. When he reaches the platform, he thinks he's Jean Valjean, the forever-on-the-run protagonist of _Les Misérables_ — and the Nazis are all Javerts. When the train arrives, it is full of more German soldiers.
He changes lines two more times and hobbles out at Trocadéro. He decides to telephone an old friend and fellow _Résistant_ , Pierre Heeley. But he has no coins in his pocket. So he settles his gaze on a "perfectly suitable older woman."
"Excuse me, madame. I am an escaped prisoner of war. My feet are wounded from walking and I want to telephone a friend. Could you give me just enough for one call?"
Like the children on the street a half hour earlier, she says nothing; just hands over the necessary coins and disappears. Postel-Vinay makes his call, but there is no answer. Then, wretched luck: His coins are not returned to him.
Now he sees a man in a Métro uniform approaching him, and this time he is certain he is done for.
"Excuse me, sir," says the Métro man. "A woman I know told me who you are. I can help you. First of all, take twenty francs — in coins, because it will be more useful to you that way. If you can't reach one of your friends, come back here around eleven thirty tonight. I'll be here until midnight. I won't take you to my house, because of the curfew; I'll put you in a broom closet. You'll have a bad night, but when the first Métro comes in the morning, I'll come and get you."
Postel-Vinay thanks the stranger with all his heart, and just as quickly, his benefactor disappears. In a city filled with noncombatants — Frenchmen who are neither resisting the Germans nor collaborating with them — there are a remarkable number of secret heroes. And when a downed British or American pilot knocks on a stranger's door in the countryside, 99 percent of the time, he will be hidden rather than betrayed.
Postel-Vinay takes the Métro to rue de la Pompe to see if there is anyone in the apartment of his friend Pierre Heeley. But the shutters are closed; he must be away on vacation, or on a voyage.
Now he is so exhausted and so thirsty that he commits what he knows is a "folly": He steps into a bistro in his hobo garb to order _un demi._ It tastes like "the most delicious beer ever drunk."
He must call someone who will recognize his voice so that he won't have to say his name over the telephone, especially if he's speaking to someone in the Resistance. His friends Henry and Suzanne Rollet were part of the underground fight long before Postel-Vinay was arrested, so he settles on them. He returns to the phone booth at Trocadéro.
"Hello! Henry! Yes, my friend, I am completely cured, and I'm spending a few days in Paris. Could I come see you tonight?"
"What a great surprise," Henry replies. "But of course — with joy. We'll wait dinner."
"Thank you so much. I'll be there, but it will probably take me an hour."
The Rollets live at 68, rue Nollet, in the 17th arrondissement. Postel-Vinay will have to change at Étoile, then take the train to Rome. In between Boissière and Kléber — back on the line he first jumped on after his escape — the train rumbles to a halt. Again, he's sure the Germans have discovered him, but again, the interruption means nothing.
When he steps out of the Rome Métro, he proceeds in the opposite direction from his destination. He confuses rue des Dames, rue de Saussure, and rue Lebouteux, and suddenly he is in a new nightmare. Then he finds himself back in front of the Rome Métro. Parisians sitting outside a café stare at him. If they notice him this time, it's because they noticed him a few minutes earlier. "Who is this young tramp who can barely stand up and doesn't even have a cane?" he imagines them wondering.
Without his glasses, he can barely read the street names on the sides of the darkening buildings. He takes a left on rue Boursault, and a right on rue de La Condamine. Finally he is going in the right direction, and he spies his destination: 68, rue Nollet.
Utterly exhausted, he must still climb five flights of stairs. With one more supreme act of will, he makes it up to his friends' apartment.
"What a joy!" Henry declares.
"No! Look at the shape I'm in." Then he adds the caution every escapee seems to offer whenever he reaches safety: "I'm not even sure the Gestapo hasn't followed me here."
"Ridiculous," Henry says. "I've opened my last bottle of champagne, it's cold and we will drink it with dinner."
That night, Postel-Vinay has the soundest sleep of his life. The next morning, he wakes up with an incredible feeling: "Beloved liberty! Liberty miraculously reconquered, but still so fragile..."
TWELVE DAYS after Postel-Vinay's escape, his friend Henry Rollet makes contact with Patrick O'Leary, the head of one of the two Resistance organizations Postel-Vinay worked with before his arrest. O'Leary has always promised that he would get him out of the country if it became absolutely necessary, and he is a man of his word. His message to Postel-Vinay is that he must reach Marseille in four days, on September 18, where he will contact Georges Zarifi, at 12, allée Léon Gambetta.
Henry organizes Postel-Vinay's escape from Paris meticulously. The first leg is a bicycle ride to Pont Saint-Michel station — a departure point chosen to avoid the larger stations, which are more dangerous because they are more heavily patrolled by the police.
At the end of a bridge over the Seine, he is met by Henry and another co-conspirator, Jean Vialla, who takes Postel-Vinay's bicycle and rides away on it. Then Henry guides him down the stairs into the station.
They change trains at Juvisy and Brétigny. At six the next morning they arrive at Coutras. Now, the hard part: He must cross the line between occupied and unoccupied France. He has no idea when or how he will do it; he only knows that he must meet a guide in the café in the train station at ten o'clock. Then he is supposed to ride a bicycle for twelve miles. But will he have the strength to do that?
The café isn't open yet, so he and Henry sit in the waiting room, a dirty, poorly lit room, with a stone-cold stove in the corner on a black and chilly morning. Suddenly two policemen walk in. Is _this_ the end of his voyage? Postel-Vinay remembers the name on his false identity card in his pocket: Fernand François Claude André Duval, an engineer, born July 4, 1912, in Algiers. But the policemen never approach him. Instead, they sit down and talk quietly.
Fifteen minutes later, they stand up to leave.
Exactly on time, the young guide arrives at the café with two bicycles. Postel-Vinay and Henry say their goodbyes, and Postel-Vinay begins to pedal painfully behind his guide. Often he has to get off and walk to climb a hill. The weather is clear and fresh and dry. After about two hours, his guide finally dismounts and tells Postel-Vinay to do the same.
They walk through a silent forest until they reach a clearing; a farmhouse lies beyond it. The guide tells Postel-Vinay to wait outside while he goes in to meet the residents. A moment later he signals him to follow him into the house.
The farmer and his wife greet him like a son. On the dining room table, among other marvels there is a huge smoked ham. They explain to him that he will be crossing the border between the two zones that evening — in an oxcart with a secret compartment!
After a magnificent lunch, Postel-Vinay retires for a nap in a large bedroom upstairs. Three hours later, he is awakened by the farmer's wife: Time to go.
The farmer removes a plank from the oxcart to reveal his hiding place. It's so narrow he's grateful that his months in prison have made him so thin. After a short ride, the oxcart comes to a halt, and the plank beneath him disappears so that he can climb out.
The farmer kisses him on both cheeks and sends him on his way. After a couple of hundred yards on foot, he reaches the zone libre. And his next guide — "small, round, and dark" — appears out of the darkness to lead him to the automobile he has borrowed from a local doctor.
Another two hundred yards and they reach the car. "I will take you to Ribérac," the new guide explains. "You will spend the night in a hotel there. At seven o'clock in the morning you'll catch a bus for Brive, and from there, the train to Marseille."
The next day, he travels for sixteen hours, reaching Marseille Saint-Charles station at eleven fifteen, just forty-five minutes before the curfew. Outside the station, there is a taxi driving by — and it stops for him!
Two more miracles.
"I've sprained my ankle and I need a hotel for the night," Postel-Vinay tells the driver.
"Well," he replies, "all the hotels in Marseille are full. But I'll help you out. I know a woman who will put you up in her house. And it will be cheaper than a hotel! Will that suit you?"
"Perfect!"
The driver delivers him to a mansion that shows no sign of being a commercial establishment. The proprietress welcomes him and gives him a room with a big bed and a lovely bathroom. Not even a registration form to fill out!
He leaves the house the next morning and walks out into a beautiful Mediterranean light "that chases away unhappiness." It is Friday, September 18, 1942. He asks another "suitable-looking" passerby for directions to allée Léon Gambetta, the home of his fellow _Résistant_ Georges Zarifi.
Georges is an old friend, who has been in the Resistance as long as Postel-Vinay. He is about twenty-five, thin and athletic, a member of the French national tennis team. He comes from a wealthy family from Greece.
When Postel-Vinay reaches Georges's house, he is welcomed by his father. As with so many other young _Résistants,_ it is never entirely clear how much his parents know about what Georges is doing. So Postel-Vinay assumes his father's ignorance and repeats his usual story about having just sprained his ankle. The father greets him warmly and tells him Georges will be back home for lunch.
When Georges arrives a few hours later, he tells Postel-Vinay that he has arranged for him to leave for Gibraltar — with forty other escapees!
The next morning when he arrives at nine o'clock at Saint-Charles station, Postel-Vinay is reunited with Patrick O'Leary, a man he particularly admires for his judgment and his organizational talent — qualities that he did not find often enough among many other Resistance leaders. O'Leary himself is in charge of the operation. The escapees break off in groups of three or four to take their places on the train. The RAF pilots with him have fake papers identifying them as deaf-mutes, so they won't betray themselves with their accents. The others include French and Polish agents, and even one German, Paula, who has been Patrick's secretary.
The train takes them to Perpignan in the south of France, not far from the border with Spain. Postel-Vinay's ankles begin to give out on their walk to the beach, and two big Canadian airmen carry him between them.
Hiding behind a dune, he is approached by an RAF pilot from New Zealand — whom Postel-Vinay has actually rescued one year earlier from the farmhouse where he had been hiding after his plane was shot down. And this pilot also had stayed with Suzanne and Henry Rollet in Paris — the same couple who harbored Postel-Vinay for a few days right after his escape.
At two A.M., the boat that will take them to Gibraltar finally appears offshore, and a smaller boat reaches the beach to pick them up. A Polish sailor loads Postel-Vinay into the first vessel, then helps him on the larger ship a couple hundred yards later. The trip takes three days, and Postel-Vinay suffers from terrible seasickness until the weather finally calms.
A fellow passenger tells him he must make contact with de Gaulle when he reaches London. Some Frenchmen are working directly for the English, but Postel-Vinay must work for the Free French! Postel-Vinay says he had already decided to do that.
On the morning of the third day, they are met by the British vessel that will take them on their last leg to Gibraltar. Calm seas ensure an easy transfer. One of the British officers on board immediately offers Postel-Vinay his cabin; the Frenchman is moved by this "instant act of kindness."
The following afternoon, he finally reaches Gibraltar — "English territory." He quickly makes contact with Captain Jacques Vaudreuil (real name: François Thierry-Mieg), who has been de Gaulle's personal representative in the British colony since October. One of Vaudreuil's aides warns Postel-Vinay that a vetting awaits him at Patriotic School in England — a prospect that fills him with disgust. After all he has suffered at the hands of the Germans, he must now endure a new challenge from the British.
He understands the necessity of trying to ferret out double and triple agents, but he still resents the idea that he will have to prove himself again.
THREE WEEKS LATER, on October 15, 1942. Postel-Vinay takes off from Gibraltar in a twin-engine DC-3, which delivers him to an airfield in the English countryside. From there, a bus takes him to a police station in a London suburb, where a charming British officer listens to Postel-Vinay's unbelievable tale of escape.
The Englishman listens politely — and doesn't seem to believe a single word Postel-Vinay tells him. Then he dispatches him to Patriotic School for further interrogation.
This is the first stop for every self-described _Résistant_ arriving in England during the war. Patriotic School has been created to distinguish between genuine Allied sympathizers ("sheep") and double agents who are really working for the Germans ("goats").
The London Reception Centre had been established at the Royal Victoria Patriotic School in Wandsworth by the British intelligence agency MI5 at the beginning of 1941. During the course of the war, thirty-three thousand refugees will pass through the center, where they are "questioned about their methods of escape, the routes they had followed, safe houses, couriers, helpers and documentation. Their statements [are] meticulously indexed and cross-checked against those of their companions and earlier arrivals. Intelligence [is] extracted and circulated to Whitehall departments."
Those identified as "goats" are shipped off to Camp 020, which oversees 440 prisoners during the course of the war. After an early instance of a violent, unauthorized interrogation, a strict rule against torture is enforced — because the British believe non-coercive interrogations are the ones most likely to produce accurate information. This is also the conclusion of more sophisticated Allied interrogators almost everywhere during World War II.
Patriotic School is an austere place, lightened a bit by a library that reminds Postel-Vinay of a London club. It is filled with refugees from every country invaded by the Nazis, many with escape stories just as implausible as his own. After eight days of waiting, his interrogation by the British finally begins.
The examination lasts five days: two or three hours in the morning and another two or three hours in the afternoon. He has to recount all of his experiences in the Resistance twice — first chronologically, then divided up among the various branches he has served in.
When he describes his escape from the mental hospital in Paris, he mentions the 23 centimes he received from the boys in the street. "And then I took my second left —"
"That's impossible!" his interrogator interrupts. "The rue Cabanis, where the hospital exit is, ends at the rue de la Glacière. So there is no 'second left'!"†
They agree to examine a map of the neighborhood together.
"Look here," says Postel-Vinay. "There is another street — the street that starts right in front of the exit of the hospital. When I turned right into the street, I left this other street on my left. So for me, that was my first left. Rue La Glacière was my _second left._ "
"Good answer," says the British officer. "Quite plausible. We will meet once more tomorrow morning, but it should be brief."
The Frenchman has climbed over his final hurdle before freedom.
At nine o'clock the following morning, Postel-Vinay walks into his inquisitor's office. "I have nothing else to ask you," says the Englishman. "But I have three things to tell you. First, you will leave us this morning. The BCRA is sending a car for you. Second, the head of Patriotic School, Major Y, is waiting to meet you in his office. And third... I am proud to know you!"
Major Y offers a parting glass of champagne — "a good champagne." Then Postel-Vinay climbs into the waiting car, which delivers him to BCRA headquarters in London at 10 Duke Street. There he meets Pierre Brossolette, and then, Charles de Gaulle himself.
The general asks just one question: "What would the effect be on the French if they learned there was a grave difference between me and the English?"
Postel-Vinay replies that this would be very dangerous, because it would reinforce German propaganda that England is France's real enemy.
Then de Gaulle tells him that he is making him associate director general of the newly created Central Bank for Free French colonies. Postel-Vinay is not enthusiastic about this, but de Gaulle has made his decision. Although he will no longer be fighting the Germans directly, the young Frenchman comforts himself with the thought that at least he will be working for the greater good of France. And de Gaulle has guaranteed that he will spend the rest of the war out of danger.
* The same station Jacqueline ran to when she had to rescue Christiane from the Palais de Chaillot.
† Postel-Vinay's daughter, Claire Andrieu, points out that this is either a mistake by the interrogator, or something misremembered by her father, because "the rue Cabanis does not end at the rue de la Glacière, it ends at the rue de la Santé. But my father's argument remains correct — he did indeed take his second left." (e-mail to the author from Claire Andrieu, September 30, 2014)
# _Ten_
LESS THAN A MONTH after Postel-Vinay arrives in London, General Dwight Eisenhower oversees an invasion of North Africa, with 110,000 troops, who land near Casablanca, Oran, and Algiers on November 8, 1942.
Franklin Roosevelt favors the invasion because he thinks it's a political imperative to engage German troops as soon as possible, and North Africa is the only feasible place to do that at the end of 1942. The big question is whether the French troops in the African colonies will fight the Allied invasion. As Operation Torch begins, Roosevelt issues a statement saying that he hopes it will "prove the first historic step to the liberation and restoration of France."
As André considers where to go at the end of 1942, he is buoyed by the news that the French troops led by Vichy generals in North Africa have offered only a day or two of token resistance to the American assault. On November 11 — which happens to be Christiane's nineteenth birthday — Admiral François Darlan, the senior French officer in North Africa, signs a cease-fire ending French fighting in the area.
Hitler reacts to the cease-fire by instantly ordering the German Occupation of the free zone in the south of France — a clear violation of the armistice he had signed in 1940. After that the free zone becomes known as the southern zone. De Gaulle observed that by not firing a single shot to resist the Occupation of the south, the Vichy regime "dissipated the lying pretense of independence which [it] had claimed in order to justify its capitulation" in 1940.
The arrival of the Germans means that the French fleet at Toulon, commanded by Admiral Jean de Laborde, is now in danger of falling into the hands of the Nazis. It would be a huge prize for the Germans: three battleships, eight cruisers, seventeen destroyers, sixteen torpedo boats, sixteen submarines, and seven dispatch vessels, as well as some sixty transport ships, tankers, mine sweepers, and tugs.
For the next two weeks, Admiral Laborde is lobbied fiercely by the Germans, by de Gaulle, and by Admiral Darlan, who wants Laborde to sail his fleet to North Africa, where Darlan has just reached an armistice with the Allies. But Laborde remains paralyzed, and on November 26, the Germans storm Toulon to seize all of his ships.
To prevent the ships from falling into the hands of the Germans, Laborde orders what de Gaulle describes as "the most pitiful and sterile suicide imaginable": Laborde commands his sailors to scuttle the entire fleet. Just one destroyer, one torpedo boat, and five tankers are still intact by the time the Germans gain control of the port.
Now the Allies must decide which Frenchman will be allowed to lead the French overseas territories. De Gaulle is the only general who has sided with the British from the start, but he has been kept in the dark about the North African invasion, and he certainly has no friends among former Vichy officials.
The Americans flirt with General Henri Giraud but then settle on Admiral Darlan. Eisenhower agrees to recognize him as head of the French state in return for committing French troops to the war against the Germans. But de Gaulle's representative in Algiers reports back that despite the announced collaboration, Eisenhower's general staff has "stressed their desire to enter into direct relations with General de Gaulle."
There is an immediate uproar in Britain and the United States over the decision to embrace Darlan, a man who has openly connived with the Germans. "There was a tremendous outcry in the American press," said Robert Paxton. "Why are we working with these collaborators? That was one of the things that Roosevelt did that was most openly and bitterly criticized" in America.
The conflict is resolved when Darlan is murdered on Christmas eve. His assassin is a passionate young man named Fernand Bonnier de la Chapelle, acting on behalf of either the Count of Paris or de Gaulle's allies, depending on which version of events you choose to believe. The killer is almost immediately executed by French authorities, thus guaranteeing a permanent mystery.
In his memoirs, de Gaulle gives such a convoluted description of the assassin's possible motives that he (perhaps unconsciously) encourages the idea that his allies were responsible for the assassination. That is particularly so when he writes, "If the tragic character of Darlan's disappearance from the scene could not fail to be condemned by many, the very fact that he was forced from the stage seemed in accord with the harsh logic of events. For history, in its great moments, tolerates in positions of authority only those men capable of directing their own course."
Suspicion falls on de Gaulle because Darlan's assassination makes it possible for de Gaulle to immediately outmaneuver Giraud — and seize the reins of French power in North Africa.
# _Eleven_
AT THE BEGINNING OF 1944, just over two years after Postel-Vinay's capture, André Boulloche has landed in the same prison hospital his boss was held in — after he has fumbled his own suicide at the time of his arrest. André has lost a great deal of blood when he is shot, but he survives the drive to l'hôpital de la Salpêtrière. The Germans operate on him immediately. They want him alive — but only so that they can make him talk. After the operation, they put him alone in a prison hospital room, without any postoperative care.
André thinks that the gunshot wound has probably saved his life. At the moment he was fired on, he had the cyanide pill in his pocket. He had always promised himself that he would swallow the pill when he was arrested, a denouement he had expected.
But in the instant after the Germans shoot him, he decides he will not kill himself.
_Everything is ruined anyway_ , he says to himself. _Let's see what the next twenty-four hours bring._ Afterward, he remains certain that he would have swallowed the cyanide if he had been arrested unhurt.
Lying in bed, alone in his prison hospital room, André begins to ponder all the peregrinations that have led him to this wretched outcome.
First there was the abortive trip to North Africa, immediately after the armistice, where he found it was impossible to fight against the Germans. Then his return to France and his civilian job with the government, his recruitment into the Resistance by Postel-Vinay at the end of 1940, and Postel-Vinay's arrest at the end of 1941.
He had taken over for Postel-Vinay in the northern region after his boss was captured by the Germans. He made contacts with other leaders of the secret army and recruited three more of his government colleagues to fight the enemy. Throughout 1942, he continued to smuggle arms and to collect information about German troop movements to send to London.
Toward the end of 1942, André learns that the Gestapo is about to arrest him. When the news of the Allied invasion of North Africa reaches France in November through the always-jammed transmissions of the BBC, he tells his family that he has decided to return to Morocco.
AFTER SAYING GOODBYE to his family in Paris on December 1, 1942, André boards a train for Nevers in Burgundy, 150 miles south of the capital. His Resistance colleague Bernard Vernier-Palliez puts him in touch with a sympathetic garage owner who helps him cross the demarcation line into the newly christened southern zone on foot.
André arrives in Toulouse on December third. From there he travels to Pau, in the foothills of the Pyrenees, thirty miles from the Spanish border. Then he makes his way to Tardets-Sorholus, where an inn owner is another Resistance sympathizer.
At seven o'clock in the evening on the day after Christmas, André and twelve others meet up with a guide who leads them through the snow, on an unmarked path, over a six-thousand-foot mountain in the Pyrenees. When they reach the Spanish border seventeen hours later, the guide leaves them.
The Frenchmen continue on their own, finally reaching a hotel in the Irati forest at four in the afternoon. They are arrested almost immediately by the Spanish authorities. André pretends that he is a Canadian named Nicolas Boulloche. His fellow _Résistant_ Vernier-Palliez identifies himself as an American.*
On January 1, they are transferred to a prison in Pamplona. There André meets other escapees, who convince him that he should try to make it to England, where de Gaulle is, instead of North Africa.
Toward the end of January, two pieces of encouraging war news reach the French refugees: British forces in North Africa have begun an assault on the German troops under the command of General Erwin Rommel ("the Desert Fox"). At almost the same moment, Soviet Marshal Georgi Zhukov is leading an overwhelming force against the German defenses south of Lake Ladoga. Within hours, the 872-day siege of Leningrad is broken.
On March 13, German officers make their first attempt to assassinate Hitler. A bomb disguised as two bottles of brandy is put on board the Führer's personal Focke-Wulf 200 Condor plane. The detonator activates, but cold temperatures prevent the plastic explosives from blowing up.
On March 20, André is moved to Jaraba, where he reconnects with André Rondenay, who had been his fellow student at the Lycée Janson de Sailly, as well as at the elite Ecole polytechnique, a decade earlier.†
The twenty-nine-year-old Rondenay is the son of a French general who served as the commandant of the military school at Saint-Maixent. The son is a charismatic fighter and a legendary forger. Fifteen months earlier he had created fake documents that enabled him to walk out the front gate of a German prison camp where he was incarcerated. Now he presents André with the impeccable identity card of a German officer, to assist in _his_ escape.
From Jaraba, André is once again transferred to Pamplona. On the evening of April 5, the Spanish guards there are distracted long enough to allow André and a fellow prisoner named Jean Martin to escape. It takes them several days to walk more than a hundred miles to Ariza. There they board a train to Madrid. In the Spanish capital, they locate another contact given to them by Rondenay, who introduces them to the British consul.
Ten days later, they are driven to an address that has been provided by their British contact. At the beginning of May, they cross the Spanish frontier into Portugal, reaching Lisbon on May 3. Two days later, they board a plane for Bristol, England.
There is exciting news from North Africa as soon as André reaches England. On May 7, American and British forces capture Tunis and Bizerte, and 160,000 German and Italian soldiers surrender. Six days later, German Afrika Korps commander General Dietloff Jürgen von Arnim surrenders another 275,000 troops.
Like Postel-Vinay before him, André is dispatched to Patriotic School as soon as he lands in England. Once he has convinced the British that he is indeed a "sheep" — a genuine enemy of the Germans — he seeks out Charles de Gaulle at his headquarters in London.
André is excited at the prospect of the massive invasion to be led by the Americans, although his British handlers have noted that his "deep satisfaction" with the growing strength of the Allies "is mingled with a fear of eventual blunders by American authorities, due to a lack of understanding of aspirations in the occupied countries."
André tells de Gaulle that he wants to return to France with the forthcoming Allied invasion. But the French general is still desperately short of men to send back to occupied France. The fifty-two-year-old general tells the twenty-seven-year-old engineer that he requires him in occupied Paris as soon as possible, so that he can be the general's military delegate, coordinating all Resistance activity in northern France.
Flattered by the general's invitation, André immediately accepts his commission. André is assigned to the action division of the intelligence section of BCRA, which coordinates with the British foreign intelligence agency, MI6, while the action section focuses on sabotage operations in France. To prepare him for his return to Paris, the French send him to the British espionage school, where he is given the agent name Roger Doneau.
The first officer to work with him is Lieutenant Colonel Hutchison. He describes André as a "capable type, enthusiastic, straightforward, intelligent. Has a pleasing personality and is helpful, in a quiet way, towards his less brilliant comrades. Is modest. Not very communicative about himself. Should turn out to be A.1 from security point of view."
His next instructor echoes Hutchison's assessment: "An outstanding man. He has considerable previous experience but has shown himself very keen to learn. He is very pleasant to work with." The base commandant also finds him "very quick to learn. Is intelligent and possesses powers of Leadership. Should do well."
André goes on to win superlatives in almost every clandestine competition: "physical training, very fit and enthusiastic; field-craft, excellent; weapon training, a good shot; explosives & demolitions, very competent, previous experience stands him in good stead; signaling & communications, very good; he has an incisive brain and completely understands the requirements of any given situation."
He earns a General Agent Grading of "A" for "Outstanding," and his intelligence rating is 9, which also puts him in the top category: "Superior Intelligence."
BUT NONE OF THESE QUALITIES is enough to prevent André's capture by the Germans less than four months after his return to Paris.
Lying in his prison hospital bed in January 1944, recovering from his gunshot wounds, André reflects on his many narrow escapes. Still delirious in his postoperative state, he decides that he still has the strength to liberate himself.
Unlike Postel-Vinay, who tried to plunge to his death, André believes that he can ascend to freedom. The wounded man will use his bare hands to scale the curtains in his room, which lead to the skylight above him — and a faint possibility of escape, if he can shatter the glass to reach the roof.
André starts slowly, gripping the curtain with one fist right above the other. Superhuman determination carries him up a couple of yards; then he loses his grip and crashes down on the floor.
As soon as he hits the ground, he realizes that he has popped most of the stitches in his stomach. Despite excruciating pain, he manages to pull himself back onto the bed.
The Germans don't discover his ruptured wound until their next examination, eight days later. Then they do a very poor job of suturing him a second time. For a long time afterward, his organs are covered only by a frighteningly thin layer of skin.
A week after his arrest, his sisters decide to organize André's escape. Their cousin, the surgeon Funck-Brentano, works in a hospital right next to the one where André is a prisoner. He is the surgeon who had assisted André in his unsuccessful attempt to liberate Postel-Vinay from the same prison hospital two years before.
Funck-Brentano has worked with André ever since his return to Paris, sometimes performing forbidden operations on Resistance members who have suffered gunshot wounds. Gradually he becomes so involved in the secret war that he acquires his own code name: Paulin.
With the surgeon's help, a group of André's friends locate the underground passage that connects the two hospitals, and an escape attempt is organized at the end of January. Christiane and Jacqueline rent an apartment near the hospital so that they can receive their wounded brother after his escape.
The two sisters spend a horrible night together in the apartment. With their stomachs in knots, they are so scared the neighbors will hear them that they don't dare even to turn on a water tap.
Their surgeon-cousin takes the first Métro at dawn so that he will be at the secret apartment when André gets there. But by the time he arrives, the sisters already know that the escape has failed.
The plotters have been unable to penetrate the floor of his prison hospital room from the basement below. Hearing the awful news, Christiane and Jacqueline fall into each other's arms and dissolve into tears.
Their cousin, fearing a trap, immediately flees the apartment.
ANDRÉ'S FIRST INTERROGATION by the Gestapo occurs at the hospital at the end of January 1944.
This is the baffling part of his story.
André and his chief aide, Charles Gimpel, have been arrested together on January 12, 1944.
Gimpel is brutally tortured by the Germans, but he survives the torture — and never talks.
Unlike Gimpel, André is shot at the time of his arrest, and that may explain why he never suffered the same way his deputy did — if André has told the truth about the way he was treated.
It is possible that by the time he had recovered enough from his gunshot wounds to be interrogated, whatever information he had was no longer worth much, because by then his unarrested comrades had all moved on to new secret locations.
Or else the Germans felt they had already learned whatever they needed to know from Jacques, the Sorbonne student who had betrayed him and brought the Gestapo to his door.
Later, André tells his sister that his interrogators read him excerpts of their interview with Jacques, including this passage: "I was recruited by Christiane Boulloche. She lives with her parents and her sister Jacqueline at 28, avenue d'Eylau, third floor on the right. Her brother is my boss. He is living in a clandestine apartment on rue de la Santé in the Thirteenth."
When his German interrogators demanded confirmation of Jacques's words, André said that he used a formula that failed for scores of others but somehow worked for him: He identified himself as a French officer on a military mission. Therefore, under the rules of war, he cannot speak.
"If you found yourself in the same circumstances," he tells the German officers standing over him, "you would certainly do the same, and remain silent."
After that, André always said, the Germans stopped trying to interrogate him that day.‡
The strongest counterevidence to this claim appears in the secret British file about André's wartime activities, which was declassified almost six decades later, at my request. There was only one item in the file that contradicted anything that he had told his family.
It is this entry from 1946: "in spite of [his wounds] _and of cruel tortures_ [emphasis supplied], he gave away no compromising information." (Later, for his valor he was awarded the King's Medal for Courage.)
In February, André is moved to Fresnes, one of the largest prisons in France, in the town of Fresnes, Val-de-Marne, just outside Paris. Most of the other prisoners are fellow Resistance members, and many of them have already been tortured. On February 15, he undergoes his second interrogation. By then he has been a prisoner for just over a month. The Germans read him a list of names and ask him to identify the ones he knows. Once again he refuses to cooperate.
By now his information is so old, most of it is probably worthless anyway. That could be why he again avoids being tortured.
"The Gestapo wasn't always logical." That is his sister Christiane's only explanation for the way her brother says he was treated. Or, as Postel-Vinay put it, pondering his own survival without ever being tortured, "Perhaps I was wrong to look for logic in an organization in which ability, incoherence, chance, barbarism and political rivalries were constantly warring for the upper hand."
TIPPED OFF that her brother has been moved to Fresnes, Christiane tries to deliver packages to him there. Occasionally they are accepted, but usually they are rejected.
One night in March, André reconnects with fellow prisoner Gilbert Farges, a former rugby player whom he had met a few times before. They stay up all night talking about their families, their arrests, and the "loving attention" they have both received from the Gestapo. Despite their circumstances, they also talk about the future, and their determination to escape. About this André is "inflexible, inexhaustible — and full of imagination."
By dawn, Farges and Boulloche have become good friends. For André, this will quickly become a lifesaving friendship.
On April 7, André is moved to the transit camp of Royallieu, in Compiègne, fifty miles north of Paris. Compiègne has been the hunting grounds of the kings of France since the eighth century. In 1944, it becomes the final stop for thousands of French patriots hunted down by the enemy, before they are deported to Germany.
Despite his badly tended wound, André is already becoming a leader of his fellow prisoners, through his dignity, his serenity, and his intellectual strength. In Fresnes, he had been kept in an underground cell. Now, for the first time in months, he can occasionally see the sun.
To Gilbert Farges, Compiègne feels like "semiliberty" compared to Fresnes: "The simple fact of seeing the sky and the sun, of being able to walk around and talk, feel like precious gifts. But we were barely nourished and many of us were already in terrible health."
André reconnects with another important person here: Charles Gimpel, his deputy in Paris before their arrest, now a fellow prisoner. As Gimpel's British handlers noted before his return to France, Gimpel is "an excellent man," "physically well above average," with "an excellent brain," an "outstanding" personality and a "keenness on the job" that is "infectious." (After the war, the British also award Gimpel the King's Medal for Courage.)
During their walks outside in the prison courtyard, the men are buoyed by glimpses of squadrons of American B-17 Flying Fortresses and B-24 Liberators filling the sky on their way to Germany.
Another fellow prisoner, Michel Bommelaer, finds it incredibly moving "to watch these heads from a mortuary... following the planes as they flew east, where a promised land of unimaginable cruelty awaits us."
In the third week of April, Christiane gets a tip that André is about to be shipped off to Germany. Hoping to catch a final glimpse of him, Christiane and Jacqueline travel with their parents to Compiègne, where they spend the night in a squalid hotel. The next morning, they spot André in the courtyard, hunched over because of his badly tended wound. When André sees them, Christiane tries to throw him a loaf of bread with a file hidden in the middle of it, but the Gestapo shout at her and push her away.
At this moment it occurs to André that this may be the last time that he will ever see his family. But despite his family's horror over his departure, there is also an element of hope: "When we learned one of our comrades would be deported, we were relieved," Christiane remembered. "Because it meant that they hadn't been tortured to death."
André boards the deportation train on April 27, 1944 — forty days before the Allied invasion at Normandy. Gilbert Farges pushes him into a corner of the car to make it easier to protect him. Then Farges asks Rémy, a beefy man from the Basque region, to stand next to both of them. From then on, André leans on one of these two men, concentrating all the time to conserve his energy.
Seventeen hundred prisoners are put in seventeen cattle cars: one hundred crammed into each car. Their destination is Auschwitz. Among them are thirty-nine employees of the French national railroad, two poets, twenty priests, and members of sixty-four different Resistance organizations.
Most of them are younger than thirty-five. The youngest prisoner is fifteen, the oldest, seventy-one. They range from laborers and journalists to businessmen, politicians, and policemen.
Among the more famous are Pierre Johnson, inspector general of Compagnie Transatlantique shipping company, who is a great-great-nephew of the nineteenth-century American president Andrew Johnson, and Count Paul Chandon-Möet, a champagne magnate, who will die a few days before his prison camp is liberated by the Allies.
Also on board is Charles Porte. Porte was the police commissioner of Chartres when Jean Moulin, the head of the Resistance, was prefect of Eure-et-Loir. Porte had been responsible for security for the inaugural meeting of the Conseil national de la Résistance, which Moulin presided over on May 27, 1943. Porte will survive deportation and return to France after the war.
During four horrific days and three terrible nights, the prisoners struggle to survive without food or water. Dozens of them succumb before the train reaches the death camp. The cars are so crammed with humanity, no one ever has enough room to sit down. André is guarded the whole time by Gilbert Farges and Rémy, both of whom are big enough to protect him; fortunately, Farges is as wide as he is tall. André's organs are kept inside by a very thin layer of skin, and a blow to his stomach would surely be fatal.
When thirst and hunger take hold, many of the prisoners become delirious, touching off terrifying scenes of madness. "As time passed, the temperature mounted, asphyxiation was winning, and a devouring thirst installed itself with dementia and violence," Farges remembered.
As one prisoner after another dies, their corpses are piled on top of one another in a corner to make more room for the survivors. "The crazy people become dangerous to the point where they try to open their own veins or those of others to drink their blood."
Rémy tells Farges, "We will not let your friend André die because of these maniacs." Somehow, through it all, André retains a strong spirit. "That was essential," said Farges. Occasionally, André is able to snatch a breath of fresh air from the cracks in the car. Just once, the Germans offer a bucket of water, but the prisoners are so frantic that it gets turned over before anyone can take a drink.
As their journey approaches the end, the survivors are nearly naked after stripping themselves because of the scorching heat. As the train reaches Silesia, the temperature drops, and the fetid air inside becomes slightly fresher. By the time they reach Auschwitz, it is freezing, and the ground is covered with snow.
When the doors of the cars are finally flung open, the prisoners are greeted by Gestapo men armed with revolvers and machine guns, others with vicious dogs pulling at their leashes, and the notorious capos, prisoners who are violent criminals used by the Germans to assist them with their punishments.
As he climbs out of the train, one of the Frenchmen jumps on a guard, causing his gun to misfire; the man is immediately shot dead by another German.
The prisoners are put in two separate barracks in Birkenau, the main extermination camp at Auschwitz. There each one is tattooed with a number on his forearm. Finally, the day after their arrival, they are given their first drink of water in five days.
After that, under the watchful eyes of the capos — "precarious survivors, brutal, hostile thieves who also know fear in this atmosphere of death" — all of the hair on their bodies is shaved.
Auschwitz is an unusual destination for these political prisoners: The convoy of April 27 is one of only three trains of non-Jews to be sent there from France during the entire war. André believed that these seventeen hundred prisoners were brought to Auschwitz to be exterminated in retaliation for the death of Pierre Pucheu. A former minister of the interior in the collaborationist Vichy government, he had been executed by the Free French in Algiers one month earlier.
The personal intervention of Marshal Pétain may be the only reason that everyone on the train isn't executed immediately.
DETERMINATION, DIGNITY, AND LUCK are the three requirements for survival. Remarkably, those are the attributes André still has in abundance. "After three weeks of murderous drills, the survivors still hadn't given up," Gilbert Farges recalled. The fearful scene reminded him of a saying of Montaigne: "They bent their knees, but held their souls high."
The prisoners watch the smoke spewing from the chimneys of the crematoria of Auschwitz all day and all night. After a few weeks, the survivors of the first convoy are put onto a new train of cattle cars, this one destined for Buchenwald. The transfer means a tiny improvement in André's chances for survival. Ninety-five percent of the deportees to Auschwitz died at that camp, compared to 85 percent of the inmates at Buchenwald.
The capos at Auschwitz tell them their new home will be practically a "sanitarium" compared to their present one.
* An ironic choice for Vernier-Palliez: Four decades later he will become France's ambassador to the United States.
† Neither man can imagine that one year from now, Rondenay will succeed André as de Gaulle's military delegate in Paris, after André is arrested.
‡ André's brother-in-law, Alex Katlama, told his son, Michel, that André hadn't been tortured because he had been wounded: "My father told me that they didn't torture him because they were worried he would die, or he wouldn't be able to take it." (author's interview with Michel Katlama, March 14, 1999)
# _Twelve_
_Of course the two years through which Free France had endured had also been filled with reversals and disappointments, but then we had to stake everything to win everything; we had felt ourselves surrounded by a heroic atmosphere, sustained by the necessity of gaining our ends at any price._
— Charles de Gaulle
BACK IN PARIS, Christiane and Jacqueline have finally moved out of their parents' home into a new secret apartment. Apparently, it doesn't occur to anyone that Jacques, or André, might have been tortured into giving up their parents' address. In any case, their father never considers resigning his job as the director of the bureau of highways or going into hiding with his wife. His main concern, Christiane believes, is to keep his country working, "for the French."
As James L. Stokesbury observed, men like Christiane's father "were servants of the state, sworn to defend it; the accidents of war had made the legitimate French government a tool of the conqueror, yet it was still the official French government. It took a special kind of person to throw aside the norms of a lifetime, to make the lonely decision that there was a higher loyalty, a higher duty, than that he had always acknowledged, and in the name of some abstract moral principle to work against his own government."
Robert Paxton, the preeminent American historian of Vichy France, points out that "as the state came under challenge by Resistance vigilantism, a commitment to the ongoing functioning of the state reinforced the weight of routine... Resistance was not merely personally perilous. It was also a step toward social revolution."
After André Boulloche's arrest in January 1944, de Gaulle replaces him as military delegate in Paris with André Rondenay, André's college classmate, whom he had run into in Spain a year earlier, when Boulloche was on his way to London.
Rondenay is a man of parts. His thick brown hair and horn-rim glasses give him the look of an intellectual, and he favors three-piece suits with double-breasted vests.
Three and a half years into the war, Rondenay is a famous escape artist. First taken prisoner in the Vosges region of France near the German border in June 1940, he spends time as a prisoner in Saarburg and Westphalia before being transferred to Mainz in October.
After escaping and being recaptured several times, he is taken to Colditz Castle, a prisoner-of-war camp for officers, in January 1942. In May, he is moved again, this time to Lübeck.* There he manages to manufacture the perfect papers of a German officer. On December 19, 1942, he and a comrade walk out unchallenged through the front gate of the camp.
His fake identity continues to protect him during an incredible journey through Hamburg, Frankfurt, Mayence, Ludwigshafen, and Strasbourg. After a brief stay in France, he crosses the Spanish frontier on January 23, 1943.
André Rondenay was a master forger and a famous escape artist. When he became Christiane's boss in the Resistance in 1944, she was mesmerized by him.(photo credit 1.12)
Arrested by the Spanish police, he is interned at Pamplona, where he runs into his old classmate André Boulloche, who had arrived in Spain four weeks before him.
Managing to fabricate the fake papers of a German officer for himself and a comrade named Noël Palaud, Rondenay escapes yet again with Palaud and a third conspirator. Within a few weeks, they reach Portugal, and Rondenay finally makes it to England on April 4, 1943.
IN THE FACE of mounting danger in the spring of 1944, Rondenay remains surprisingly cheerful. His fearlessness is lightened by an infectious sense of humor and a beguiling ability to make fun of himself. To boost Christiane's spirits, he writes a song that makes fun of her youth and the fact that she has never told her parents exactly what she is doing. And he continues to concoct everything from a train ticket to a German officer's identity card at a moment's notice. His wife, Solange, is one of his deputies.
Christiane knows Rondenay as "Jarry" — the latest of his many wartime aliases. The others are Lemniscate, Sapeur, Jean-Louis Lebel, and Francis Courtois. He gives Christiane "very large sums" of money, which arrive from London, and she distributes it to their confederates. "I was twenty," she said. "It was completely crazy. He had total confidence in me. I think he had more confidence in me than he did in himself."
Now Monsieur and Madame Rondenay are living with the Boulloche girls, plus Georges, another force of nature, who has escaped Germany three times. This motley crew is rounded out by their radio operator, Riquet, and his wife, Gaby. All of them live together in the same apartment, in total secrecy.
CHRISTIANE had become a student at Sciences Po in the fall of 1941, and André told her to continue to show up there occasionally, after she went to work for him in the fall of 1943. She also continued to live with her parents. Only after André's arrest does she stop going to school and move out of her parents' apartment, to begin a completely clandestine life.
One part of Christiane's job is to take care of Riquet's radio equipment. His transmissions are in constant danger of being discovered by the Gestapo, which deploys "gonios" — trucks equipped with homing devices to locate the secret transmitters.
To reduce the chances of being discovered, they never transmit from the same location for very long. At one point, Christiane's aunt, Françoise Farcot, allows her niece to use her modest country house to make a few broadcasts. One day, when Christiane isn't there, the Gestapo arrives at her aunt's house.
Seeing them at the front door, one of Christiane's cousins snatches the radio and manages to hide it in the attic. If he hadn't been so quick, everyone in the house would have been arrested.
Even as a teenager, Christiane is already a person of tremendous self-control. One night in Paris, she is in a secret apartment with four confederates, waiting for an important phone call. One of her comrades, a boy who has just been parachuted in from England, soothes his nerves by downing one cognac after another. Christiane is quietly disdainful. She thinks to herself, _At least I am courageous_ — _even though I am a girl!_
When the phone finally rings, they are all ordered outside. Christiane is carrying their radio equipment. Suddenly her suitcase bursts open, and all of the electronic contraband spills out into the street. Christiane scrambles to put everything back, then disappears into the night as quickly as possible.
Christiane is also responsible for coding and decoding the telegrams that go to and from London. It is exacting work, and her least favorite task. Sometimes she delivers weapons to assist in sabotage missions. Whenever possible, London prefers to get things blown up in France on the ground, rather than bombed from the air, because it reduces the chances of civilian casualties.†
She continues to follow all the rudimentary rules of spy craft that the British have taught her brother. She particularly loves the "Métro trick": In the first car there is a conductor who controls the doors. "You put yourself in the first car, close to the conductor. The Métro stops, and you don't move. And then when you see he is about to push the button to close the doors, you jump out. Then you can't be followed."
She always chooses a secret apartment with two exits, and she never waits more than ten minutes for someone to show up for a rendezvous. She trains herself to be suspicious of everyone, all the time: That is the hardest part of her job.
She has to memorize dozens of addresses and telephone numbers, because she is not allowed to write anything down. Once, when she is carrying important telegrams in her purse, she is stopped at a Wehrmacht checkpoint. The German soldier who searches her never recognizes what she is carrying, even though the telegrams don't look like normal documents, with their coded, disjointed letters. "He looked at them. He saw them. And he let me go." She rode her bicycle two hundred yards from the checkpoint — until her legs didn't work anymore. Her terror had paralyzed her. She knew she had "been incredibly lucky," and she "couldn't count on that kind of luck all the time."
AT THE BEGINNING OF 1944, the Germans are starting to feel threatened on all fronts. In January, the German siege of Leningrad is finally lifted by the Soviets, almost nine hundred days after it began. That victory comes exactly one year after the Germans lost the crucial battle of Stalingrad, where the Soviets encircled nearly all of the German 6th Army under General Friedrich Paulus. The Germans and Soviets suffered gigantic losses there: nearly two million military and civilian casualties. The historian Ian Ousby argues that the Reich's failure on the eastern front was a main impetus of the French Resistance. The other was the Reich's brutality in France.
In July 1943, more than three thousand ships had arrived off the beaches of Sicily, and almost half a million Allied soldiers invaded — a larger force than the one used in the initial stages of the assault at Normandy eleven months later. By July 22, American troops had entered Palermo, the major city of western Sicily. On July 25, the Italian king summoned Mussolini and dismissed him as prime minister. His successor, Pietro Badoglio, announced publicly that Italy would stay in the war on the side of the Germans. Privately, he immediately opened negotiations in Portugal with the Allies to switch sides.
At the end of the third week of January 1944, the Allies launch Operation Shingle. Fifty thousand American and British troops land at Anzio, south of Rome. The Allies will finally enter Rome on June 4.
Meanwhile, the Germans occupying France are nervously awaiting the main Allied invasion, still not knowing where it will occur, as the American and British air forces continue a brutal bombardment of Germany. The Americans carry out their first raid on Germany on January 27, 1944. Three and a half weeks later, 823 British bombers attack Leipzig, and 73 British planes are lost in the raid. The next day, 200 American bombers continue that attack, as the British make a massive assault on Stuttgart.
Four days after that, bombers from the U.S. 8th Air Force attack the German cities of Augsburg, Stuttgart, Regensburg, and Braunschweig. On March 4, the Americans make their first attack on Berlin. Two days later, British bombers attack the French national railway complex at Trappes, southwest of Paris, and inflict enormous damage without losing any planes. These are the opening salvos of the Allies' Transportation Plan, designed to disrupt German reinforcement routes prior to the Normandy invasion. By the time of the final attack of this campaign on June 2 — once again at Trappes — the Allies have launched 9,000 sorties in 69 attacks, with a loss of 198 planes.
At the end of the third week of March, Hitler makes one of his most accurate predictions. He tells his commanders in the west that if they are able to prevent a successful Allied landing, that will decide the outcome of the war. At almost the same moment as Hitler's prediction, the Allies cancel Operation Anvil, which would have meant an invasion in southern France at the same moment as the onslaught at Normandy in the north.
On April 17, the British government suspends all diplomatic pouches leaving Britain, except those going to the United States and the Soviet Union, to try to prevent a leak of the location of the forthcoming invasion at Normandy.
CHRISTIANE AND JACQUELINE are aware of the general course of the war, through the broadcasts of the BBC and the stream of new confederates who continue to arrive from England. The life of one of the sisters is about to take a dramatic turn, with the arrival of a dashing confederate, another fresh graduate of Patriotic School.
His name is Alex Katlama. He is thirty-three years old when he is parachuted into a secret reception area, northwest of Dijon, on the night of March 15, 1944. A strapping young man with elegant, delicate features, he reaches Paris two days later. His mission is to teach others how to carry out sabotage.
Born in Moscow in 1910, Alex is the third and last child of Russian-Estonian parents. His father had been a well-known owner of racehorses, an "elitist" avocation that made him conspicuously unwelcome in Russia after the success of the Bolshevik Revolution.
In 1918, the Katlamas escaped Russia through Odessa, the southern Ukrainian port on the Black Sea. Then they spent eighteen months in Constantinople, waiting for the revolution to fail. When the Bolsheviks prevail, the Katlamas move on to France in 1920. Alex is homeschooled until the age of thirteen, before attending the Lycée Russe in Paris. He is brought up by a French governess, who teaches him that Alsace is part of France, and that it was a scandal when the Germans "stole" it in 1871.‡ He studies higher mathematics at the Lycée Louis le Grand for two years. There he also becomes a star of the basketball team. Just shy of six feet, Alex is much taller than most of his contemporaries; before the war, the mean height for Frenchmen is closer to five foot six.
In 1933, he graduates from the Ecole Supérieure Aéronautique, with an aircraft engineer's diploma. Two years later, Alex becomes a French citizen, which makes him immediately eligible for military service. He is drafted as a corporal and sent to the air base at Rheims. After the Germans reoccupy the Rhineland in March 1936, Alex is thrilled when bombs are suddenly put on the wings of French airplanes. But the bombers never take off. He assumes the inertia of the French is due to a fear of repeating the catastrophe of World War I.
Because of his knowledge of Russian and of aeronautics, he is promoted to sergeant and transferred to the Air Force intelligence bureau in Paris. There he translates the Russian technical press on aircraft topics, until his army service ends in 1937. Upon his return to civilian life, he gets a job as a sound engineer at Metro-Goldwyn-Mayer in Paris. He remains there until war is declared in 1939, when he is mobilized again.
He is assigned to Air Force intelligence, where he alternates between studying aviation developments in the Soviet Union and choosing targets there for future bombardment, since the Soviets are still in a nonaggression pact with the Nazis when the war begins.
When he hears Pétain's surrender speech in June 1940, Alex has exactly the same reaction as André Boulloche: He is determined to find a way to continue the fight against the Germans. As a Russian immigrant who chose French citizenship, he is especially patriotic.
First he tries to commandeer a plane to fly to Britain, but his superior officers prevent him from taking off. Then he and a friend convince a captain to fly them to Oran in North Africa. But — again like André — he discovers there is no way to fight the Germans in Europe from the French colony, which has remained loyal to the collaborators in Vichy.
Alex Katlama was a star of the French national basketball team. He is second from the right in the back row.(photo credit 1.13)
Demobilized in August, he rejoins his family in France. At the beginning of 1941, he redoubles his efforts to reach England. He learns that the easiest way to escape to England is from North Africa, but he needs an excuse to return there. Opportunity knocks when he is invited to join the French national basketball team, which is about to play in Algeria, Tunisia, and Morocco.
But during the basketball tour he fails to find a passage to England, and so once again he returns to France. In November 1942, he finally decides to go over the Pyrenees to Spain.
Katlama meets three Jews in Osséja, on the eastern border with Spain, and persuades them to accompany him to Spain. In a café, he finds two guides, who are Spanish Republican refugees. Alex and his Jewish companions agree to pay the Spaniards 7,500 francs to lead them across the border.
They set off at nine P.M. on November 11, 1942.§ The guides take them over the Pyrenees, and they arrive in Berga twenty-four hours later. From there, Alex takes the train to Barcelona, where he is immediately arrested.
Alex pretends that he is an American named Glen Boylen, who was born in Bedford, England. His English is excellent, thanks to his regular visits to the American church in Paris before the war — first to play basketball, and then to listen to jazz. He also happens to love being with Americans. And before the war, he had spent six weeks in England with a family he found through an ad in the _Times_ of London.
Somehow, he manages to get his identity papers mailed to George Favre, a Spanish reporter who had accompanied the Catalonian basketball team to France, and to whom Katlama had confided his plan to escape.
Favre has friends at the British consulate in Barcelona, who forward Alex's papers to the British military attaché in Madrid. In January, he is imprisoned at the camp at Miranda, where foreign prisoners are mixed in with Spanish Republicans from the losing side of the Spanish Civil War. Alex is shocked that Republicans are still being executed in the camp, three years after Franco's Fascists defeated them.
The foreigners at Miranda are assigned their nationality on the basis of their birthplace, and Alex continues to insist that he was born in England. After five months, British diplomats finally convince his captors that Alex really is a British citizen. At the end of May, he is put on a train from Barcelona to Gibraltar, the ancient British colony on the Mediterranean coast. On May 30, he sets sail for England, where he arrives one week later.
After a month of vetting at Patriotic School, he is given the alias of Lieutenant Alex Daniel. In August, he begins training to become a secret agent in France. Because he is still recovering from the harsh treatment he received in the Spanish prison camp, the initial assessment of his British handlers is mixed:
A quiet intelligent intellectual type with ideas of his own, but these are not always very practical... Conscientious and serious of purpose. His mind works slowly and he has shown a lack of self-assurance under stress. Is at present in a poor state of mind and body, which has probably spoilt his performance. He is... abnormally slow, lacking in physical assurance, nervous and worried about his shortcomings. Under these conditions the board has failed him but strongly recommends that he should receive treatment and be reconsidered at a later date.
Four months later, he has regained his strength and impressed his trainers. They have noticed that Katlama is "motivated by gratitude to the country of his adoption." He is the kind of man who works hard to succeed at every goal he sets for himself, and by now, the British understand that. His affection for the British also helps with his rapid adjustment.
His new evaluation is much more enthusiastic:
Intelligent, both practical and academic. He is quick, clearheaded, logical and has plenty of common sense and imagination. He is keen and worked hard, took great pains, and was neat and accurate. He is determined and conscientious. He has plenty of self confidence but is not conceited nor over sure of himself. His personality is pleasant but not very forceful. He is a good mixer and has a good sense of humor. He was generally popular. He should make a competent and loyal assistant.
Alex receives his final training certificate on March 3, 1944. Twelve days later, he is parachuted into France. Two weeks after that, he is in Paris. He carries with him the vital _Plan Vert,_ which he will deliver to Rondenay. The plan details how the Resistance is supposed to disrupt the French railroad system when the invasion at Normandy begins.
As soon as he reaches the French capital, Alex is spared by another accident of fate — the kind that rescues so many of the Resistance members who are lucky enough to survive the war. For Alex, it is an Allied bombardment near Paris that halts the Métro train he is traveling on, for half an hour.
Sometime during those thirty minutes, the contact he is supposed to meet is arrested. Thus, when Alex reaches the location of his rendezvous, there is no longer anyone there to greet him.
If Alex had been on time, he too would certainly have been seized by the Germans.
The next day, Alex's other contact in the Resistance in Paris is arrested. After that, he loses all contact with the underground for a month. In the third week of April, he goes to an emergency address he has been given to use in case of such a catastrophe: a clothing store on rue de Flandre.
Rondenay is always worried about sending Christiane into a trap. Concealing his habitual anxiety as well as he can, he dispatches her to the clothing store, with a password to identify herself to the clerk. When Christiane says the secret word, she has a moment of panic: At first the saleswoman doesn't seem to recognize the code. After Christiane repeats the password, the clerk finally leads her into a back room, where Alex Katlama is waiting for her.
CHRISTIANE likes the handsome Alex "very much — but no more." She feels too consumed by her duties to pursue a romance with him, or anyone else. But when Alex meets Jacqueline, sparks begin to fly quickly.
It is Jacqueline who becomes Alex's "precious collaborator, the only person in my service on whom I could count completely" — and, soon, his lover. Even if Christiane is a bit jealous, Jacqueline's romance does nothing to diminish the sisters' extraordinary closeness.
About a month after making contact with Christiane and Jacqueline, Alex begins a secret course on sabotage for a cell of _Résistants_ on rue des Entrepreneurs.
On May 19, Alex leads three confederates on a sabotage mission. Their target is a ball-bearing factory, and they are so successful, production there is never resumed. For some reason, the twenty Germans responsible for guarding the factory never engage them during the attack.
By now everyone knows the Allied invasion is imminent, but the Germans still have no clear idea of when or where it will occur. On May 8, Dwight Eisenhower secretly chooses June 5 as the tentative date for the attack at Normandy. The next day, the British bomber command makes its first attack on German coastal batteries near Pas de Calais.
CHRISTIANE AND JACQUELINE know that their brother has been shipped off to Germany. But they don't know which camp he is in, or what it really means to be in a concentration camp. They bury themselves in their clandestine duties to distract them from a debilitating terror about what could happen to him in Germany.
Meanwhile, their boss, Rondenay, is leading sabotage missions against many factories outside Paris. To prepare for the Allied invasion, he meets with Resistance members working for the national telephone company, to plan how to blow up underground long-distance telephone lines (code name: _Plan Violet_ ) and with railroad workers, who are supposed to blow up strategic railroad lines ( _Plan Vert_ ) _._
* Two months before Rondenay got there, on March 28, Lübeck was subjected to a horrendous attack by the RAF, when 234 British bombers dropped 400 tons of bombs, including 25,000 incendiary devices, which created a firestorm that burned one-third of the city. The intent was to demoralize the civilian population. Nazi propaganda minister Joseph Goebbels wrote in his diary: "The damage is really enormous, I have been shown a newsreel of the destruction. It is horrible. One can well imagine how such a bombardment affects the population."
† On April 20, 1944, an Allied bombing raid on Paris leaves 651 dead and 461 wounded, provoking a minirebound in the popularity of General Pétain, the collaborationist chief of state. (Jackson, _France,_ p. 535)
‡ Alsace became French again at the end of World War I, reverted to Germany after the conquest of France in 1940, and then returned to being part of France in 1945.
§ Which happens to be Christiane's nineteenth birthday.
# _Thirteen_
_Rather than just being liberated by foreigners... France herself would rise up to take an honorable part in her own liberation. That, really, was what resistance was all about._
— Ian Ousby
_I opened the window_ [ _in London on the morning of June_ 6] _and the noise became deafening... It was possible to see the aircraft flying in massed formation above the sleeping capital. They flew over in a never-ending stream. Holding my breath and looking steadily in the direction of Nazi Germany, I could see, beyond the barbed wire sealing the frontiers, beyond the prisons, the dawn that was bringing to our enslaved friends the first glimmer of their victory._
— Marie-Madeleine Fourcade
ACROSS THE CHANNEL, on the southeastern coast of England, the Allies have assembled a gigantic collection of airplanes, tanks, trucks, jeeps, and more than two million men. "All southern England was one vast military camp," Eisenhower recalled.
By the eve of the invasion, fifty-two million square feet are filled with supplies, including almost half a million tons of ammunition. Soldiers joke that if the invasion doesn't happen soon, the southern edge of England will sink into the sea beneath the weight of their preparations.
"The southernmost camps where assault troops were assembled were all surrounded by barbed-wire entanglements to prevent any soldier leaving the camp," Eisenhower remembered. "The mighty host was tense as a coiled spring, and indeed that is exactly what it was — a great human spring, coiled for the moment when its energy... would vault the English Channel in the greatest amphibious assault ever attempted."
The Germans have fifty-five divisions assigned to the defense of France, while the Allies have only thirty-five divisions assembled for the invasion. But not all the German soldiers can be used to defend the northern coast of France. Some of them must remain in the south to defend the Mediterranean coast, and the Russians have promised to begin a new offensive on the eastern front when the invasion begins, to discourage the Germans from sending any reinforcements.
In the weeks before the invasion, the Allies do their best to cripple communications in the north of France, to isolate the Normandy beaches from the rest of the country. Between May 24 and June 6, British and American bombers destroy all nine railroad crossings and a dozen highway bridges in the region. The Allies keep them closed by repeatedly bombing the boats and the temporary bridges the Germans try to put in their place. This will severely limit the Germans' ability to send in reinforcements after the invasion begins.
Meanwhile, the Allies are continually leaking false invasion plans to prevent Hitler from figuring out that Normandy is their real target. They make a special effort to convince the Germans that they will land near Boulogne, Calais, and Dunkirk, and many of Hitler's generals are taken in by the ruse.
The British even create a fictitious 4th Army headquarters at Edinburgh Castle, which pours out phony radio messages to convince the Nazis that the Allies are also about to invade Norway, where the Germans have stationed twenty-seven divisions. Again, the deception works: Those German divisions never leave Norway to defend France.
Realizing that it will be crucial for the French to participate in the liberation of their own country, Churchill makes a special appeal to Eisenhower to find enough transport to allow the 2nd French Armored Division to land in Normandy during the summer of 1944. Eisenhower agrees to his request. This will eventually make it possible for French soldiers commanded by General Jacques-Philippe Leclerc to spearhead the Liberation of Paris.
On the last day of May, 245 minesweepers begin clearing the English coast. Then they start to create paths to the landing sites on the French coast. Two days later, two British submarines leave Scotland to head for the Normandy coast. Their task is to mark the approaches for the landing craft that will arrive there four days later.
On June 3, Dwight Eisenhower gives de Gaulle his first briefing on Operation Overlord, the code name for the Normandy invasion. This is the first official information the Frenchman has received about the gigantic operation that will liberate his country — a good indication of the "warmth" between the two generals.
IN THE WEEKS before the Normandy invasion, Resistance leaders in Paris are notified how they will be alerted to the great event. The Allies apparently believe this is necessary because so much depends on the extensive sabotage of railroad tracks and other communication channels that the Resistance has been ordered to carry out just as the invasion begins.
At the moment when the French should begin to ready themselves for the invasion, the BBC will broadcast the first lines of "Chanson d'automne," one of France's best-loved poems, by Paul Verlaine, one of its most celebrated nineteenth-century poets. On June 1, the chosen French listeners who know the code are riveted when they hear these seven words traveling across the channel from the BBC:
_Les sanglots longs_
_Des violons_
_De l'automne_
(The long sobs of autumn violins)
Someone with a literary sensibility, a decent knowledge of French culture, and a fine sense of history has clearly made this selection. The melancholy poem comes from the "paysages tristes" (sad countrysides) section of Verlaine's first volume of poetry. When the next stanza is broadcast, the Normandy countryside will be drenched with the blood of hundreds of thousands of Allied and German soldiers, and at least fifteen thousand French civilians, most of whom will be victims of Allied bombardments.
After the first lines are broadcast, Christiane, Jacqueline, and scores of their confederates in the Resistance know that the moment they have been waiting for since the bleakest days of 1940 has finally arrived. Within days, Allied troops will return to French soil, to try to expel the Germans.
From Belgium and the Netherlands to Poland, Czechoslovakia, and Greece, millions of freedom lovers are praying for the success of the invasion, without knowing yet exactly when or where it will begin.
Churchill, Roosevelt, de Gaulle, Eisenhower, and Hitler* all agree about the importance of this moment. Each believes that this will be _the_ crucial event of the war. The fate of the "Thousand-Year Reich" is about to pivot on the skill, the bravery, and the good fortune of 156,000 mostly British, American, and Canadian soldiers, sailors, and airmen. Additional troops will hail from Belgium, Poland, Norway, the Netherlands, Czechoslovakia, Australia, New Zealand, and, of course, France.
They will use nearly 7,000 ships, 2,395 aircraft, and 867 gliders to land in and around French beaches, which have been given the code names Utah, Omaha, Gold, Juno, and Sword. It will be the largest naval armada in the history of the world.
After eleven years of Nazi atrocities metastasizing across Europe, this is the moment of truth in a global contest between democracy and tyranny. The main thing that muddles the idea that this is simply a conflict between good and evil is the role of Stalin. Like Hitler, Stalin is responsible for genocides that have killed millions of innocent people. But now, destiny has made him _the_ essential ally of the Western democracies.
No one but Stalin is willing to sacrifice so much to defeat the Nazis: a staggering twenty million men and women. World War II will kill ten million Soviet soldiers and ten million Soviet civilians.† That is nearly five times the losses sustained by the Germans — and _fifty times_ the losses of the United States.
Ten million civilians is also ten million more than were killed at home in America during the entire war. The United States and Canada are the only major combatants that suffer no losses at home, apart from those killed at Pearl Harbor. In France, on the other hand, 350,000 civilians will perish between 1940 and 1945.
The worst per capita civilian losses of all are in Poland, where 5.7 million will die, including 2.8 million Jews, out of a prewar population of just under 35 million — more than one-seventh of all Polish citizens.
ON JUNE 4, Eisenhower's meteorological committee predicts low clouds, high winds, and formidable waves, and the American general decides to reject a June 5 invasion date, against the advice of British field marshal Montgomery. "The tension continued to mount as prospects for decent weather became worse and worse," Eisenhower remembered.
Driving to the next meeting of the meteorological committee, at three thirty on the morning of June 5, "our little camp was shaking and shuddering under a wind of almost hurricane proportions and the accompanying rain seemed to be traveling in horizontal streaks." Under those circumstances, Ike thought it wasn't even worth discussing the situation. But when the meeting began a half hour later, the weathermen had a welcome surprise.
They were now confident of thirty-six hours of relatively good weather, beginning on the morning of June 6. At four fifteen A.M. on June 5, Eisenhower tells his colleagues he has decided to proceed with the massive invasion the following day.
With the fate of the entire war weighing on his shoulders, Eisenhower pauses to write a resignation letter, to be released in the event that Operation Overlord is a failure: "If any blame or fault attaches to the attempt it is mine alone." Then he orders the gigantic operation to go forward. "I hope to God I know what I'm doing," he tells his staff.
At eleven fifteen on the evening of June 5, Christiane is sitting in her secret apartment with her confederates, listening to the "personal messages" on the BBC, when the magic words crackle out of the radio:
_Blessent mon coeur_
_D'une langueur_
_Monotone._
(Wound my heart with a monotonous languor)
Everyone in the room jumps up to embrace one another. Then Christiane's boss, Rondenay, dispatches her into the blacked-out streets of Paris, to deliver the joyful tidings to their colleagues. She rides her bicycle without any lights. It is thrilling to be the bearer of such happy tidings.
Some days earlier, a leader of the Maquis, as the Resistance is known in the countryside, has been arrested — after he has learned that the invasion will be heralded by the Verlaine poem. Unbeknownst to the Allies, after being tortured by the Germans, the Maquis member has revealed the existence of the code.
As a result, when the poem's lines are read on the night of June 5, the commander of Germany's 15th Army in the Pas de Calais immediately puts his troops on alert.
But the Allies' vast disinformation campaign has a completely unpredictable — and quite miraculous — effect on the German officers at Army Group B's headquarters at La Roche–Guyon. There have been so many false reports of so many imminent invasions before, for them the truth is transformed into a mirage. And thus they decide to pay no attention at all to the secret gleaned from the tortured Maquis leader.
_Why would the BBC possibly broadcast the time of the invasion in advance?_ these Germans ask themselves.
_Obviously this is just another feint on the part of the Allies!_
So the commanders of the 7th Army — the one defending Normandy — are never warned on the night of June 5 of the imminent attack. Two other factors contribute to the lackadaisical attitude of many of the German commanders. During May there have been eighteen days when the weather, the sea, and the tides have been perfect for a landing, and the Germans obviously noticed that Eisenhower has not taken advantage of them.
And on June 4, the German Air Force meteorologist in Paris has predicted that inclement weather means that no Allied action could be expected for at least a fortnight.
Field Marshal Erwin Rommel had been shifted from Italy to northern France by Hitler in January 1944, where he became commander of Army Group B, which has the main responsibility for repelling an Allied invasion. On the basis of everything he has been told, the Desert Fox writes a situation report on June 5 stating that the invasion is not imminent. Then he sets off to Germany, first to celebrate his wife's birthday and then to meet with the Führer. The second meeting will not occur.
Around one A.M. on June 6, one British and two American airborne divisions begin descending by parachute and glider right into the middle of the German 7th Army. But even this isn't enough to convince Field Marshal Gerd von Rundstedt, the German commander in chief in the west. At two forty he sends word to the chief of staff of the 7th Army that he does "not consider this a major operation."
Hitler himself has been up until three A.M. with his deputy, Heinrich Himmler, "reminiscing, and taking pleasure in the many fine days... we have had together," Himmler recorded. When three of his generals call his headquarters to beg for permission to rush two tank divisions to the front, word comes back that Hitler wants to wait and see what develops. Then he goes to bed and sleeps undisturbed until three o'clock the following afternoon.
Even after the gigantic Allied armada starts arriving later that morning, the chief German general in the west remains convinced that the main invasion is going to happen elsewhere.
At five fifty on the morning of June 6, the Allies open a massive naval bombardment on beach fortifications and nearby Normandy villages. The main American landings take place forty minutes later at Utah and Omaha beaches.
There is heavy fighting everywhere, but the Allies are particularly successful at Utah, where 23,000 men get ashore with only 210 dead and wounded on the first day. After the 101st Airborne manages to block four exits from the beach, the only regiment facing them from the German 709th division surrenders in large numbers.
At Omaha, the situation is catastrophically different. This is the site of by far the largest landing of D-day by Americans. Here, 34,250 troops face the Germans dug in on bluffs 150 feet above the beach, and the inward curvature of the coast also allows German fields of fire to overlap.
The disasters for the Allies begin at six A.M., when waves of American B-24 bombers drop thirteen hundred tons of bombs intended for German defenses at Omaha and completely miss their targets, bombing too far inland.
The official history of the 116th Infantry, 29th division, was written by S. L. A. Marshall, a World War I veteran who rejoined the army in 1942 as a combat historian. In his notebook he recorded the horrific conditions faced by the men who hit the beach at Omaha at six thirty that morning:
ABLE Company riding the tide in seven Higgins boats is still five thousand yards from the beach when first taken under artillery fire. The shells fall short. At one thousand yards, Boat No. 5 is hit dead on and foundered. Six men drown before help arrives...
At exactly 6:36 a.m.... the men jump off in water anywhere from waist deep to higher than a man's head. This is the signal awaited by the Germans atop the bluff. Already pounded by mortars, the floundering line is instantly swept by crossing machine-gun fires from both ends of the beach...
The first men out... are ripped apart before they can make five yards. Even the lightly wounded die by drowning, doomed by the waterlogging of their overloaded packs. From Boat No. 1, all hands jump off in water over their heads. Most of them are carried down... All order has vanished from Able Company before it has fired a shot.
Already the sea runs red. Even among some of the lightly wounded who jumped into shallow water the hits prove fatal. Knocked down by a bullet in the arm or weakened by fear and shock, they are unable to rise again and are drowned by the onrushing tide. Other wounded men drag themselves ashore and, on finding the sands, lie quiet from total exhaustion, only to be overtaken and killed by the water. A few move safely through the bullet swarm to the beach, then find that they cannot hold there. They return to the water to use it for body cover. Faces turned upward, so that their nostrils are out of water, they creep toward the land at the same rate as the tide. That is how most of the survivors make it. The less rugged or less clever seek the cover of enemy obstacles moored along the upper half of the beach and are knocked off by machine-gun fire.
... From the cliff above, the German gunners are shooting into the survivors as from a roof top.
In spite of the huge initial casualties, the terrible handicaps of Omaha's topography, and the almost total lack of cover, seven hours after the first troops hit the beach, General Leonard Gerow signals General Omar Bradley that "troops formerly pinned down on the beaches" are finally "advancing up heights behind the beaches."
At a cost of two thousand Americans killed at Omaha, by the end of the first day more than thirty thousand men have made it ashore. Two Ranger battalions scale the hundred-foot-high Pointe du Hoc with rope ladders, only to discover that the Germans have already dismantled their big cannon.
It isn't until four o'clock that afternoon (one hour after he was finally awake) that Hitler agrees to send two more Panzer divisions into the battle to bolster the 12th SS and 21st Panzer Divisions. "But the reinforcements dribbled into the invasion front were never enough," writes the historian Gerhard Weinberg, "and the Allied air forces as well as the sabotage efforts of the French resistance and Allied special teams slowed down whatever was sent."
Indeed, the actions of the Resistance were probably just as important to the success of the invasion as the incredible bravery of the men storming the beaches. "We were depending on considerable assistance from the insurrectionists in France," Eisenhower reported.
During the first twenty-fours of the assault, nearly one thousand acts of sabotage paralyze the French railways. Locomotives are destroyed, trains are derailed, and more bridges are blown up, reducing rail traffic by 50 percent. For a week after the invasion, every train leaving Marseille for Lyon is derailed at least once, and in the department of Indre, which includes the line from Toulouse to Paris, there are eight hundred acts of railroad sabotage in June alone.
This is vital to the success of the Allies, because 90 percent of the German Army is still transported by train or horse. The disruptions achieved by the Resistance give the troops on the beaches crucial additional hours, and then days, to prevail, before significant German reinforcements can arrive.
Across the French coast, by the end of the first day, there are 9,000 Allied casualties, of whom one-half were killed: 2,500 Americans, 1,641 Britons, 359 Canadians, 37 Norwegians, 19 Free French, 13 Australians, 2 New Zealanders, and 1 Belgian. British air chief marshal Arthur Tedder had predicted a casualty rate of 80 percent for the airborne troops, but the actual number was 15 percent.
Rommel finally makes it back to the front from his wife's birthday party at the end of the first day of the invasion, after canceling his meeting with Hitler. By the time he returns, one of his earlier predictions is well on its way to coming true. Unlike General von Rundstedt, who thought it was impossible to prevent an Allied landing and hoped to fling the invaders back into the sea with a counterattack, Rommel had been certain they had to be prevented from coming ashore at all. "The first twenty-four hours will be decisive," he said.
On the night after the invasion, Roosevelt goes on the radio to ask one hundred million Americans to pray with him:
_Almighty God: Our sons, pride of our Nation, this day have set upon a mighty endeavor, a struggle to preserve our Republic, our religion, and our civilization, and to set free a suffering humanity..._
_They fight not for the lust of conquest. They fight to end conquest. They fight to liberate. They fight to let justice arise, and tolerance and good will among all Thy people..._
_With Thy blessing, we shall prevail over the unholy forces of our enemy. Help us to conquer the apostles of greed and racial arrogancies. Lead us to the saving of our country, and with our sister Nations into a world unity that will spell a sure peace, a peace invulnerable to the schemings of unworthy men. And a peace that will let all of men live in freedom, reaping the just rewards of their honest toil._
_Thy will be done, Almighty God._
By the end of June 11 (D-day plus five), an astounding 326,547 troops, 54,186 vehicles, and 104,428 tons of supplies have been landed on the beaches. While there would be several more serious setbacks on the way to victory — and hundreds of thousands of additional casualties — by now the Allies have clearly turned the tide of war.
The sabotage missions carried out by the Resistance in the immediate aftermath of the invasion make a huge contribution to the success of the Allies. They also come at a tremendous cost to the French civilian population.
When the 2nd SS Das Reich Panzer Division sets out on June 8 on a 450-mile journey from the south of France, they expect to reach Normandy a few days later. Instead, the trip takes three weeks, because of the heroism of the Maquis, who attack the Germans and destroy numerous bridges and railway tracks in their path.
On the second day of the trip, in retaliation for the deaths of forty German soldiers, the Panzers seize a hundred men at random in the town of Tulle in the Corrèze and massacre all of them. "I came home from shopping on June 9 to find my husband and son hanging from the balcony of our house," recalled a woman from the town.
On the third day, Major Adolf Diekmann's unit is responsible for a much greater atrocity in the village of Oradour-sur-Glane, where 642 citizens, including 205 children, are killed. The men are shot; the women and children are burned to death in a church.
Though still shocking in France in the fourth year of the Nazi Occupation, the German war crimes committed in these villages paled in comparison to what the Nazis had been doing on a vast scale in Eastern Europe ever since 1940. Referring to the latest massacre in France, an eastern front veteran who had become one of Diekmann's officers told a colleague, "In our circles, Herr Muller, it was _nothing._ "
* On November 3, 1943, Hitler had written in Führer Directive No. 51, "The threat from the East remains, but an even greater danger looms in the West: the Anglo-American landing!... It is there that the enemy has to attack, there — if we are not deceived — that the decisive landing battles will be fought." (Roberts, _The Storm of War,_ p. 462, and www.britannica.com/dday/article-9400228)
† That was the number given by the Soviets immediately after the war, but when Mikhail Gorbachev was president of the USSR, he said the total number of Soviet deaths could have been 29 million. (O'Neill, _The Oxford Essential Guide to World War II,_ p. vii)
# _Fourteen_
_Throughout France the Free French had been of inestimable value in the campaign. They were particularly active in Brittany, but on every portion of the front we secured help from them in a multitude of ways. Without their great assistance the liberation of France and the defeat of the enemy in western Europe would have consumed a much longer time and meant greater losses to ourselves._
— General Dwight Eisenhower
ON THE DAY BEFORE the Normandy invasion, Christiane, Jacqueline, and André and Solange Rondenay are ordered by London to leave Paris, to join up with the Maquis in the Morvan, 150 miles south of Paris, near the Château de Vermot in Dun-les-Places.
The two sisters go to their parents' apartment to say goodbye. Jacques and Hélène are very unhappy that they are leaving Paris. They have had no word about their youngest son since he was shipped off to a concentration camp in Germany at the end of April, and now they won't know where their daughters are either. Only their son Robert still seems relatively safe in his government job at the Finance Ministry.
Once again, Rondenay manufactures impeccable identification cards for everyone, and they reach the Morvan without incident. They are incredibly relieved to be out of Paris. In the weeks before the invasion, the Gestapo has intensified all its activities, and the pain of seeing more and more of their friends and relatives getting arrested has become overwhelming. Now, for the first time in years, they can speak out loud without worrying about being "overheard, suspected or unmasked." They finally feel safe, although that is not exactly what they are.
In the Morvan they join up with Jean Longhi, another legendary figure of the Resistance, whose nom de guerre is Grandjean. Christiane is captivated by him. A Communist and a veteran of the Spanish Civil War, Longhi flees Paris after learning that the Germans are after him. Longhi and his friend Paul Bernard founded the Maquis de Camille in the fall of 1941.
With his sister, Longhi started a hospital at the Château de Vermot to care for the wounded of the Maquis. At the end of 1943, the Service National Maquis, which coordinated the actions of the various individual groups, made him the head of all of the Maquis in the department of Nièvre in the center of France. Even though Grandjean is a Communist, Christiane does not sense any political tension with him: "We had a common enemy, and that was enough."
Christiane and her comrades sleep in leaky tents; the weather is terrible and nothing ever gets dry. On the afternoon of June 26, four hundred Nazi soldiers arrive at Dun-les-Places in cars and trucks to search for "terrorists." Everyone's identity papers are checked, and everyone is let go. But when the Germans leave the village, they are attacked by the Maquis.
The Germans respond by attacking the château housing the hospital. The counterattack is led by Longhi, Bernard, and Rondenay. Bernard is gravely (although not fatally) wounded. Not knowing how to fire a gun, Christiane spends the battle passing ammunition to a confederate with a machine gun, and then tending to the wounded in the infirmary.
The violent attack on the château lasts for twelve hours, until the Maquis are finally forced to abandon it. When the Germans capture it, they burn it down. The Maquis suffer two dead and five wounded, but the rest are able to escape.
The Maquis regroup easily after the attack, but once again, the nearby villages suffer terrible reprisals. The hamlet of Vermot is burned to the ground, and twelve houses in Dun-les-Places are destroyed. Then all the men of Dun-les-Places are arrested, and all twenty-seven of them are killed. Among the dead are the mayor, the village priest, and the headmaster of the school.
Four days later, Rondenay gets a telegram from London ordering him back to Paris.
"I'm coming back with you," Christiane tells him. "He didn't say no, because he knew he needed me." Jacqueline decides to stay behind with the Maquis.
As the Allies slowly fight their way south toward Paris, there is fear, near famine, and growing chaos in the still blacked-out City of Light. For Christiane, her worst nightmare is yet to come.
# _Fifteen_
ELEVEN DAYS AFTER the Normandy invasion, Hitler travels to Margival, northwest of Paris. Here the Germans have constructed the headquarters that André Boulloche had first described to London at the end of 1940. The bunker was supposed to have served as the Führer's headquarters when he invaded Britain, but of course that invasion never occurred.
Now Hitler is visiting for the first time to meet with Erwin Rommel and Field Marshal Gerd von Rundstedt, who had become the German commander in the west in the summer of 1942. The generals' purpose is to tell Hitler that the war has become hopeless.
Four months earlier, Rommel has joined the conspiracy to remove Hitler as Führer, although he is opposed to killing him, because he thinks that would make him a martyr.
To Rommel's chief of staff, General Hans Speidel, the Führer looks "pale and sleepless" and "his hypnotic powers seemed to have waned."* After a "curt and frosty greeting," Hitler speaks "bitterly of his displeasure at the success of the Allied landings, for which he tried to hold the field commanders responsible."
Emboldened "by the prospect of another stunning defeat," Rommel is remarkably frank — and accurate. He tells Hitler that "the German front in Normandy would collapse and that a breakthrough into Germany by the allies could not be checked... He doubted whether the Russian front could be held. He pointed to Germany's complete political isolation," and "concluded with an urgent request that the war be brought to an end."
Finally, Hitler cuts Rommel off. "Don't you worry about the future course of the war," the Führer tells his general, "but rather about your own invasion front."
One month after this meeting with Hitler — and three days before the next attempt on Hitler's life — Rommel's car is strafed by a low-flying Allied Spitfire. The general barely survives the attack and then recovers slowly.† Speidel believed that the anti-Hitler conspirators "felt themselves painfully deprived of their pillar of strength."
WHEN CHRISTIANE AND RONDENAY return to Paris at the end of June, the streets are still patrolled by German soldiers, but now many of them look like they are no more than fourteen. There are also German street signs that hadn't been there at the beginning of June: ZUR NORMANDIE FRONT.
After the initial successes of the invasion, the battle in Normandy is going much slower than the planners had anticipated. But the wide incursions of the Allies in the first days of the invasion have convinced a very large number of senior German Army officers that the war is essentially over: It is now only a matter of time before the Allies will cross the Rhine and obliterate what is left of Germany.
It now seems clear that right from the start most German generals knew that Hitler was psychotic. But as long as he was winning the war, almost all of them were happy to overlook that detail — as well as the massive war crimes they were committing at his behest.
By the middle of 1944, very few of them remain under the spell of the Führer's much diminished "magic powers." As a result, a remarkably large number of generals and colonels are recruited for the latest plot to assassinate their supreme leader. Many more are aware of the conspiracy without participating in it — or betraying it.‡
Seven decades after the war, German opposition to Hitler is barely remembered. But there were actually some two dozen unsuccessful attempts on his life.
In the summer of 1944, Major General Hening von Tresckow is the chief of staff of the 2nd Army on the rapidly deteriorating Russian front. He is also a longtime opponent of Hitler. As the latest conspiracy against the Führer takes shape, the general is asked for his advice by some of the nervous young plotters. He provides them with the most compelling and prescient reason for carrying out the latest plot to kill Hitler — the attempt that will come closest to success, six weeks after the Normandy invasion: "The assassination must be attempted at any cost. Even should it fail, the attempt to seize power in the capital must be undertaken. We must prove to the world and to future generations that the men of the German Resistance Movement dared to take the decisive step and to hazard their lives upon it. Compared with this object, nothing else matters."§
The historian William Shirer points to another reason for a growing sense of urgency among the conspirators: "The threatened collapse of the fronts in Russia, France and Italy impelled the plotters to act at once."
ON THE OTHER SIDE of the battle, the Allies are increasingly concerned by how much the Germans have delayed the Allies' progress — so much so that Eisenhower remembers late June as "a difficult period for all of us. More than one of our high-ranking visitors began to express the fear that we were stalemated and that those who had prophesied a gloomy fate for Overlord were being proved correct."
Seven weeks pass after the Normandy invasion before the Allies are finally able to launch a new offensive from the area they had hoped to reach on D-day plus five: a line of cities and towns stretching from Caen through Caumont to St. Lô.
THE LATEST PLOT to kill Hitler is led by Claus Schenk Graf von Stauffenberg, an aristocrat with many famous German generals among his ancestors. In the spring of 1943, Stauffenberg was attached to the 10th Panzer Division in Tunisia when his car drove into a minefield. It may also have been strafed by an Allied plane. Stauffenberg is gravely injured and recovers slowly — after losing one arm, one eye, and two fingers.
At end of June 1944, the anti-Hitler plotters get a boost when Stauffenberg becomes a full colonel and is appointed chief of staff to General Friedrich Fromm, commander in chief of the Home Army. This gives Stauffenberg the power to issue orders to the Home Army in his boss's name, which will be necessary to carry out the coup d'état that is supposed to take place after the assassination. It also gives him regular access to Hitler.
By July 1944, the conspiracy, code-named Valkyrie — named for the maidens in Norse-German mythology who hovered over battlefields, choosing who would die and who would survive — includes Stauffenberg's former boss, General Friedrich Olbricht; General Hemuth Stieff; General Eduard Wagner, first quartermaster general of the army; General Erich Fellgiebel, the chief of signals of the Wehrmacht (Supreme Command of the Armed Forces); and General Fritz Lindemann, head of the Ordnance Office. General Paul von Hase, chief of the Berlin Kommandantur (High Command), is important because he can provide the troops needed to take over Berlin. Colonel Freiherr von Roenne, head of the Foreign Armies Section, and his chief of staff, Captain Count von Matuschka, as well as Count von Helldorf, the head of the Berlin police, are also part of the plot.
On July 20, Stauffenberg drives to the airport outside Berlin to board a plane provided by General Wagner. In his suitcase he carries a British-made bomb that is set off by breaking a glass capsule. The capsule contains acid that eats away a small wire, which releases a firing pin against a percussion cap. British fuses are favored by the conspirators because they do not make a telltale hissing noise. The thickness of this particular wire should make Stauffenberg's bomb explode ten minutes after he shatters the capsule.
The plane delivers Stauffenberg to Rastenburg soon after ten A.M. From there he is driven to Hitler's secret Wolfsschanze (Wolf's Lair) headquarters, named after Hitler's longtime Nazi code name, Wolf. It is located in a gloomy forest in East Prussia. General Alfred Jodl describes the atmosphere there as "somewhere between a monastery and a concentration camp."
With a staff of two thousand, the complex sits in the middle of three security zones protected by mine fields, pillboxes, and an electrified barbed-wire fence, all patrolled by fanatical SS troops. The compound includes two airfields, a railway stop, a power station, saunas, cinemas, and tearooms. Hitler's own headquarters, the Führerbunker, has six-foot-thick concrete walls, electric heating, and air-conditioning.
At twelve thirty-two in the afternoon, Stauffenberg shatters the glass capsule in the bomb he has carried in a briefcase and marches into the conference barracks where Hitler is already being briefed about the eastern front. The Führer is sitting at the center of one side of a heavy oak table, eighteen feet long and five feet wide, which stands on two massive supports instead of legs. Its unusual construction will determine his fate.
Stauffenberg sits down at the table and slides his briefcase on the side of one of the two heavy supports — the side closer to Hitler, about six feet from the dictator's legs. Then Stauffenberg tells a colleague, Colonel Heinz Brandt, that he has to leave the room to make an urgent telephone call.
After Stauffenberg leaves, Brandt stands up to examine the map that is sitting on the table before him. Finding Stauffenberg's briefcase in his way, Brandt reaches down and moves it to the far side of the barrier supporting the table — farther away from Hitler. That random act will sharply change the history of the next twelve months of the war: When the bomb explodes ten minutes later, Colonel Brandt is killed — but Hitler survives.
Stauffenberg is about two hundred yards away when the bomb goes off, and he watches as bodies and debris fly out the shattered windows. He is certain everyone inside must be dead. After bluffing his way through three checkpoints, he boards the plane waiting to take him back to Berlin.
When the head of the conspiracy finally reaches the German capital late in the afternoon, he learns that his fellow conspirators know the explosion has occurred. But because they aren't certain that Hitler has been killed, they have done nothing to put the planned coup into effect.
Hitler's hair is singed, his legs pierced by a hundred splinters, his right arm paralyzed temporarily, his eardrums punctured, and his back is hit by a falling beam, but he manages to stumble out of the demolished building. Four of his colleagues are killed. Initially, the Führer thinks his headquarters are the victim of an Allied bombardment, but gradually the clues accumulate implicating Stauffenberg.
From Berlin, Stauffenberg telephones Paris, where more senior German army officers are involved in the conspiracy than anywhere else. He speaks to his cousin, Lieutenant Colonel Caesar von Hofacker, at General Carl-Heinrich von Stuelpnagel's headquarters in Paris, and tells him that the army is proceeding with a coup.
Before darkness has fallen in Paris on July 20, General Stuelpnagel has arrested all twelve hundred SS and SD officers and men in Paris, including their commander, SS major general Karl Oberg.
But back in Berlin, the conspiracy is in total disarray. It begins to fall apart altogether when Major Otto Remer is ordered to arrest Joseph Goebbels, the propaganda minister and the highest-ranking Nazi official in Berlin at this moment. When Remer confronts Goebbels in his office, the minister manages to get Hitler on the telephone and puts the major on the line with him. Realizing the Führer is still alive, the major reverses his loyalties on the spot.
At six thirty in the evening, Goebbels manages to get the Deutschlandsender, a radio station powerful enough to be heard all over Europe, to broadcast a brief announcement that an attempt to kill Hitler has failed. (The conspirators' failure to secure the radio station earlier in the afternoon is one of their many elementary blunders.) Goebbels initially blames the Allies for the bombing. Hearing the bulletin that Hitler is still alive, German Army generals in Prague and Vienna who have already started to arrest SS and Nazi party leaders begin to backtrack.
Stauffenberg's boss, General Friedrich Fromm, who has tolerated the conspiracy against Hitler for months without actively participating it, now turns decisively against the plotters, in the hope of saving his own skin. He pretends to carry out an instant courtmartial of Stauffenberg, as well as two generals and a lieutenant who were among his collaborators. Then he orders all of them shot immediately by a firing squad in the courtyard — to make sure they can't implicate him in the conspiracy.
By one A.M., Hitler is on the airwaves himself, describing a plot by "a very small clique of ambitious, irresponsible, and at the same time, senseless and stupid officers... It is a gang of criminal elements which will be destroyed without mercy."
By now the twelve hundred SS and SD officers and men who had been arrested in Paris have been released.
"Seized by a titanic fury and an unquenchable thirst for revenge," William Shirer wrote, Hitler "whipped Himmler and [Ernst] Kaltenbrunner to ever greater efforts to lay their hands on every last person who had dared to plot against him." During the next nine months, at least two thousand and perhaps as many as five thousand Germans are executed for their alleged roles in the plot, some of them in concentration camps, just days before the war is over.
One of the worst effects of the assassination's failure is the rekindling of Hitler's belief that he is protected by divine providence. Churchill believed that these were Hitler's first words after the attack: "Who says I am not under the divine protection of god?"
The other terrible effect was to convince Hitler that since so many of his enemies in the army had now been unmasked — men he decided must have been undermining him all along — Germany's success in the war was now assured. Minutes after the bombing, this is what he tells his private secretary: "Believe me, this is the turning point for Germany. From now on things will look up again. I'm glad the _Schweinehunde_ [bastards] have unmasked themselves."
* A nimble fellow who was also part of the conspiracy to assassinate Hitler, Speidel became commander of NATO land forces in Central Europe from 1957 to 1963.
† Three months after the failure of the plot to kill Hitler, the Führer sends two generals to Rommel's home to offer him the choice of committing suicide or facing the People's Court for his role in the conspiracy. Told that his family will be taken care of if he chooses the first option, Rommel goes upstairs to say goodbye to his wife. Then he gets in a car with the two generals and swallows a cyanide capsule. He is dead fifteen seconds later.
‡ In February 1944, the Swiss minister to Vichy met with Prime Minister Pierre Laval, who predicted the annihilation of Russia and said there was no question of a breach of the Atlantic Wall. When the Swiss diplomat repeated this assessment to one of Laval's close collaborators, he was told, "What do you expect! Laval has gambled and he knows he has lost. But he wants neither to believe nor admit it." (Jackson, _France,_ p. 527)
§ Von Tresckow had been involved in several previous plots to kill Hitler. When the one on July 20 failed, he committed suicide the next day.
# _Sixteen_
_Although the smell of retreat wafted on the summer air, the business of Nazi horror continued as usual._
— Matthew Cobb
_A great tide of popular enthusiasm and emotion seized me on my entry into Cherbourg, and bore me onward as far as Rennes, passing through Coutances, Avranches and Fougères. In the ruins of demolished cities and burned-out villages, the population gathered along the roads and burst out in jubilant demonstrations... The contrast was remarkable between the ardor of their spirit and the ravages endured by their persons and property. Certainly France would live, for she was equal to her suffering._
— Charles de Gaulle, describing his triumphant journey through the countryside toward Paris, days before its liberation
IN PARIS, the Nazi establishment reacts with horror to the slow but steady progress of the Allied invasion. The surge of hope Parisians feel after Normandy is miserably balanced by a huge upswing of anxiety. After the failure to assassinate Hitler, the Germans in the occupied French capital become completely unhinged: Their vindictive sadism knows no bounds.
"We had our hands at their throats," Christiane remembered, "and they were scared that our specialty — sabotage — might hinder their retreat. It wasn't funny before, but now it was horrible. Everything that they had done before was multiplied many times over." Sensing their imminent exile from the French capital, the Nazis lash out at the local population more viciously than ever, arresting, shooting, and deporting scores of Parisians who had never felt threatened before.
At the same time, the broader population of Paris is displaying growing public disdain for their Nazi oppressors. On July 14, a huge illegal demonstration of one hundred thousand people marches through the city to celebrate Bastille Day. German troops fire shots in the air to try to contain the crowds, but French policemen stand by and do nothing. The Occupation apparatus is plainly breaking down.
In the second half of July, Christiane lives through another catastrophe; at the same moment, she dodges another extremely close call. Then, unexpectedly, her parents get their first hopeful news about the fate of their younger son, André.
Jacqueline has stayed behind with Alex Katlama and the Maquis in the countryside, while Christiane is leading a clandestine life in Paris. She lives with the Rondenays and three other confederates in a secret apartment on avenue Mozart.
Nevertheless, Christiane continues to visit her parents, without ever giving any details about her secret life. The summer of 1944 is extremely difficult for Jacques and Hélène Boulloche: "They had a deported son," Christiane remembered, "and two daughters working in the Resistance at a time when the Gestapo seemed especially dangerous. It was hard for them to bear, even though they were in complete agreement with what we were doing. We never talked about it... but in a sense, our involvement was a direct result of everything they had taught us."
On July 27, Rondenay tells Christiane she will accompany him to a meeting with Alain de Beaufort, who has just arrived in Paris from the Morvan. But as the temperature inches toward eighty-four degrees on another sweltering Paris day, something makes Rondenay change his mind. At the last minute, he decides to send Christiane to a different rendezvous.
During June and July alone, de Beaufort has distributed more than 150 tons of arms and supplies for the Maquis in the departments of Aube, Yonne, and Nièvre. He and Rondenay have escaped numerous assassination attempts, including one just ten days earlier, organized by Henri Dupré, a double agent for the Abwehr, the German military intelligence service.
Rondenay has escaped from half a dozen German and Spanish prisons since the beginning of the war. He has carried out dozens of successful sabotage missions, crafted countless fake identity cards, and fought bravely with the Maquis in the French countryside. But on this steamy Thursday, just four weeks before the Liberation of Paris, the luck of these two remarkable _Résistants_ finally runs out.
Disaster strikes as soon as Rondenay greets de Beaufort at the Passy Métro station in the 16th arrondissement — and Gestapo agents with guns drawn swarm around them. De Beaufort tries to escape, and he is shot in the foot. For the next two weeks, both men are brutally tortured, but neither of them ever talks.
The news about her hero reaches Christiane very quickly. Immediately she goes into her standard routine after every arrest: "There was no question of giving in to panic; we had to act as quickly as possible. We spent the night burning all the incriminating documents in our possession. It was that night when I realized how _slowly_ things could burn!"
She still has a clandestine apartment on rue de Lille, but she is afraid it won't be safe there. So she goes to her parents' apartment for a couple of days instead.
"Despite everything," Christiane recalled, "there was no time to be discouraged. I didn't have time to be miserable because of Jarry's arrest. There was so much I had to do, as there always was after every arrest: change the contact places, change the mailboxes. At times like this, you feel incredibly useful, because you believe in what you're doing.
"At the same time I felt a trap was being set around us. After you've survived so many disasters, you say to yourself, 'I can't always be so lucky!' It was exhausting, especially because the atmosphere in Paris was so electric. The Allies were advancing, the Germans were standing their ground. We were truly in a waiting game. On top of everything else, it was incredibly hot!"
Now Christiane always thinks she is being followed. One day, she is waiting outdoors at a bistro in Saint-Germain-des-Prés to meet a contact, when a strange-looking man begins to stare at her. Gripped with terror, she gets on her bicycle — and the man follows her. Finally he runs straight into her. But it turns out he isn't trying to arrest her; he only wants to flirt!
"At times like this, we felt really odd," Christiane remembered. "We were living on another planet, but we had to act as naturally as we could. It was a completely different world of extraordinary intensity." Christiane won't really understand how she feels right now until after the Liberation. Then she will experience "a terrible void: one that was almost impossible to fill."
WHILE SHE IS BACK in Paris, her parents receive the first hopeful news they've had all year: a two-page letter from André. It is sent from Flossenbürg, a German concentration camp in Bavaria. This is the third camp he has been sent to in three months, after Auschwitz and Buchenwald.
Official German policy allows prisoners of the camps to write one letter a month,* as long as it is written in German, but for some reason, André has been forced to wait three months before he is permitted to send his first message:
_Dear Father: Finally I am allowed to write you a letter. My health is good, my wound is better and my morale is excellent. I hope to get letters from you soon. I will write you once a month._
André explains that he is allowed to receive packages, as long as they weigh "less than 5 or 6 kilos." Among other necessities, he asks for a pullover sweater.
His sister Jacqueline writes back immediately, describing the family's joy at receiving his letter and promising that Christiane will send him the requested package. "The conditions of our daily lives are more and more difficult, but they are bearable," Jacqueline writes. "Of course, there is no possibility of leaving Paris this summer... We embrace you from the bottom of our hearts."
A few days later, on August 2, 1944, Hélène writes to her imprisoned son: "We are happy to know you are healthy. We hope to see you again soon. The whole family kisses you very tenderly."
TWO DAYS AFER mailing her letter to André, Hélène gets another pleasant surprise: a forty-eight-hour visit from her older son, Robert. Now thirty-one, Robert has never joined the Resistance, although he has provided important information to the Free French from time to time. Like his father, he has kept his government job throughout the war, after being demobilized following the armistice in 1940. Now he is an inspector with the Finance Ministry, posted to Toulouse. A bachelor and a dutiful son, he has traveled to Paris this weekend to visit his beleaguered parents.
It is now Saturday, August 5, 1944. In the south, the Allies have begun a massive aerial bombardment to prepare for Operation Dragoon, the second Allied invasion of France, which will start in just ten days on the Mediterranean coast between Cannes and Toulon.†
And in exactly twenty days, the tanks of the 2nd French Armored Division, commanded by General Jacques-Philippe Leclerc, will finally roll into the exhausted capital, to the most rapturous welcome of any reconquering army of modern times.
But today, the Germans are still in control, food supplies are dwindling, and Parisians are increasingly nervous about the sluggish progress of the Allies toward their capital, two long and stifling months after their spectacular assault on Normandy.
On Saturday night, Christiane's aunt and uncle host a small dinner for all the Boulloches who are in Paris this weekend — Christiane and Robert, and fifty-six-year-old Jacques and Hélène. For Christiane, the dinner feels like a parenthesis in the middle of a life of permanent stress. Nothing important is discussed, and among the guests, only Christiane's parents know about her secret life as a _Résistante._ Christiane excuses herself around ten thirty and bicycles back to her clandestine apartment on avenue Mozart.
Her parents and her brother return to the family's large apartment on avenue d'Eylau. The Boulloches' servants, Marie and Simone, are asleep in the maids' rooms nearby when the family retires for the night. The apartment is a short walk from the Palais de Chaillot, the concert hall where Jacqueline had miraculously snatched Christiane from the Nazis' clutches eight months earlier.
At three A.M., the deep Sunday morning silence of the pitchblack 16th arrondissement explodes into terror, when the Gestapo storm the third-floor apartment. The Nazis rouse the horrified sleepers from their beds, then bellow the reason for their nocturnal invasion: "We are here to arrest Christiane Boulloche."
But Christiane is not here. She is sound asleep in her secret apartment on avenue Mozart.
So the Nazis seize her father, her mother, and her brother in her place.
JUST AS THEY HAD after André Boulloche's arrest, the Gestapo leave the apartment unguarded for an hour or two, with the servants all alone. Marie seizes the moment to telephone Christiane's aunt Ginette to tell her the terrible news — and to implore her to prevent Christiane's next visit to her parents, which she is planning to make that very morning.
But Ginette doesn't know where Christiane is or how to reach her on the telephone. There is only one thing she can do: At the crack of dawn she dispatches her maid, Odette, to the rond-point de Longchamps, and Christiane's cousin, Louis, to place du Trocadéro, to try to intercept her niece before she reaches 28 avenue d'Eylau. There the second wave of the Gestapo has camped out in the concierge's apartment and tacked Christiane's photograph to the wall.
Now there is one more miracle. Christiane sets out on her bicycle for her parents' house around eight thirty on Sunday morning. After a few minutes, to her great annoyance, she gets a flat tire. When she can't pedal anymore, she changes her route. Normally she would cycle up from Trocadéro to her parents' apartment. But today, because she is forced to walk her bicycle, she takes the route through the rond-point de Longchamps. Because Christiane is pushing her bicycle down the street, Odette easily intercepts her — seventy yards from her parents' front door. Christiane knows what Odette has done: "She saved my life."
FOR THE NEXT WEEK, the Germans, and then the Milice, the Vichy paramilitary force, terrorize her parents' apartment building. An elderly woman who tries to visit Hélène Boulloche is detained for three hours; a young girl who comes to visit someone else in the building is stripped and searched.
As soon as Odette tells Christiane about the arrests, her only concern is to alert the rest of her unit, and then to disappear as quickly as she can. Alone in Paris, without Jacqueline, or any other member of her immediate family, this twenty-one-year-old secret agent must summon all of her inner strength to escape confusion and despair.
Later that morning, Christiane runs into a close family friend, Dr. René Cler. When she tells him she is on the run from the Gestapo, the thirty-four-year-old doctor immediately invites her back to his apartment on avenue Sully-Prud'homme. What strikes him most about this brave young woman is her "sense of immediate spontaneity."‡
Eight months earlier, Cler had already collaborated with the movement, after Jacqueline Boulloche had arrived "very agitated" at his house for lunch.
"Our unit has just been arrested," she declared. "Can we hide our radio transmitter with you?"
"I had no choice," the doctor remembered. "So then I passed several unpleasant nights." He recognized "a remarkable simplicity" in the Boulloche sisters: "For us, it is our duty."
A year earlier, the doctor had had his own close call, when he answered a summons from the Gestapo, after they became suspicious of the activities at his apartment. "It was not pleasant. An attractive French girl interrogated me for two or three hours. And there was a sexual current between us. When she went next door to make her report in German, she said, 'This man came voluntarily to see us, and we should let him go.' " And they did.
Cler decides that it's too dangerous to lodge Christiane in his fourth-floor apartment. But there is an empty apartment on the sixth floor that belongs to an absent Swedish diplomat, and the concierge allows him to put Christiane in there. Now, not even the members of her own unit know where she is living.
The apartment building next door is filled with Germans. During the final days before the Liberation, there are German snipers on the roof — and a Senegalese sniper fighting with the Free French in an apartment across the way. Christiane spends most of her days in the doctor's apartment and her nights in the apartment of the Swedish diplomat. One night when she goes back upstairs, she discovers bullet holes from stray shots lodged in the wall exactly where she had been sleeping.
The next eighteen days will determine the fate of the City of Light. But Christiane will never leave her secret hiding place — until de Gaulle finally makes his triumphant return to the capital.
* The notation on a 1942 letter from Auschwitz stated, "Each prisoner in protective custody may receive from and send to his relatives two letters or two cards per month... Packages may not be sent, because the prisoners in the camp can purchase everything." (www.historyinink.com/935308_WWII_Auschwitz_letter.htm)
† Eight hundred eighty-five ships will disgorge 151,00 Allied troops in another huge operation, barely remembered because it was overshadowed by Normandy. Seven of the eleven divisions under American general Alexander Patch are actually Free French soldiers commanded by General Jean de Lattre de Tassigny. (Ousby, _Occupation,_ p. 279)
‡ "It was obvious," Christiane said many years later. "I never thought I was doing anything extraordinary. Never. _Never!_ " (author's interview with Christiane Boulloche-Audibert, March 11, 1999)
# _Seventeen_
_Even when motionless, which he often was, others could feel the volcano inside._
— Gregor Dallas describing Charles de Gaulle
ONE DAY AFTER Jacques, Hélène, and Robert Boulloche are arrested, Adolf Hitler summons General Dietrich von Choltitz to meet with him at Wolfsschanze. At the first of the three security rings, all of the general's luggage is removed from his car — a new precaution inaugurated after Stauffenberg's failed assassination attempt.
A fourth-generation Prussian soldier, Choltitz had led the Nazi assault on Rotterdam in 1939. Three years later, he captured Sebastopol in the Russian Ukraine. When the siege there began, Choltitz was leading a regiment of 4,800 men. When it ended, only 347 of his soldiers were still alive. But the Germans had won the battle.
Shortly, Choltitz will tell a Swedish diplomat, "Since Sebastopol, it has been my fate to cover the retreat of our armies and to destroy the cities behind them."
His flair for destruction and his fierce loyalty to the Führer are the reasons he is meeting with Hitler on August 7. Hitler has been told Choltitz is a man who never wavers in the execution of an order. Coming from France, where his corps has failed to halt the breakout of American forces into Brittany, the general hopes to be rejuvenated by his leader after his recent setback on the battlefield.
But when the general reaches Hitler's lair, he finds the same hollow man Erwin Rommel and Field Marshal Gerd von Rundstedt had encountered in France a few weeks earlier. "I went into the room and there he stood, a fat, broken-down old man with festering hands," Choltitz remembered two months later. "I was really almost sorry for him because he looked horrible."
Then Hitler "began reeling off a gramophone record like a man stung by a tarantula and spoke for three-quarters of an hour!" The Führer told him that "dozens of generals" had already "bounced at the end of a rope" since the assassination attempt, because they had tried to prevent him from fulfilling his destiny of leading the German people.*
Today Hitler is making Choltitz his new commander in Paris. The Führer has chosen the Prussian to make sure that he has someone who will leave Paris in ashes if the Germans are forced to abandon their greatest prize. He is making Choltitz a Befehlshaber, a title that gives him the widest possible powers for a commander of a German garrison. Hitler orders him to "stamp out without pity" any act of terrorism against the German armed forces in Paris.
But Choltitz leaves the meeting more pessimistic than ever about the future of the Third Reich. When the Allies reach the outskirts of Paris, he knows what the next order will be from the mad dictator: _Blow it all up._
THE URGENCY to liberate Paris felt by the Resistance fighters in Paris and the Free French Forces under de Gaulle is not shared at all by Eisenhower and the rest of the generals leading the Allied invasion. While they obviously understand the power of the capital's symbolism, they don't think it has much importance as a military objective.
De Gaulle is worried that Communists will try to seize control of the capital if there is a premature insurrection there. But the Allies believe that the liberation of the French capital will require the diversion of tremendous resources from the effort to defeat the Germans. Right now they are determined to save every ounce of fuel and food and ammunition for combat operations that will "carry our lines forward the maximum distance" to wipe out the Nazi armies.
A planning document that lands on Eisenhower's desk after the Normandy invasion warns that "Paris food and medical requirements alone are 75,000 tons for the first two months, and an additional 1,500 tons of coal daily are likely to be needed for the public utilities."
For all of these reasons, Eisenhower hopes to put off the "actual capture of the city" as long as possible — unless he receives "evidence of starvation or distress among its citizens."†
ALTHOUGH CHRISTIANE is confined to Dr. Cler's apartment building, she still manages to learn the fate of her newly arrested family members, because the doctor is in touch with some of her comrades. Her parents and Robert are going to be shipped off to Germany on a train that is scheduled to leave the Gare de l'Est in Paris on August 12.
But with the Allies pushing steadily forward, and the Liberation achingly near, the Resistance redoubles its efforts to prevent any more deportations. On August 10, national railroad workers in the Paris region declare a general strike. Their leaflet exhorts, "To make the Hun retreat, strike! For the complete and definitive liberation of our country, strike!" Within two days, half of the eighty thousand railroad workers have walked off the job, and the train that is supposed to deport Christiane's parents and brother is stranded at the station.
On the night of August 12, the Resistance launches another sabotage mission, which cripples the Gare de l'Est. But nothing will halt the demonic momentum of the Nazis.
Instead of bringing them to the Gare de l'Est, on August 15, General Choltitz orders more than two thousand prisoners assembled on the _"quai au bestiaux"_ (the animal platform) of the Pantin station.
Among those boarding the train are Jacques, Hélène, and Robert Boulloche, André Rondenay, and Alain de Beaufort — while Christiane remains cut off from all of them in her secret hiding place.
This will be the final train of French prisoners to depart Paris for Germany, the infamous _dernier convoi._ ‡
Unusually, the train carries 168 Allied airmen — Americans, Britons, and Canadians — who have been rescued by the Resistance only to be captured by the Nazis afterward.
The Red Cross arrives at the station before the train leaves and manages to distribute some food rations to the prisoners. Somehow, the Red Cross agents also convince the Germans to release thirty-six prisoners who are sick or pregnant. Then the train rumbles out of the station.
Half an hour after it leaves Paris, the long line of wooden cattle cars suddenly shudders to a halt. At the last minute, the Nazis have figured out the identity of two of their most important prisoners. A Gestapo officer gives the order to unlock the doors on one of the cars, and Rondenay and de Beaufort tumble off the train, along with a handful of others. Minutes later, they are driven to the Domont forest, where they are executed by a German firing squad. Christiane imagined the horror of their final moments: "They must have thought they were being taken off the train so that they could return to Paris."
The three Boulloches remain in their fetid cars to continue their wretched journey.
A survivor of the convoy remembered hearing railroad workers yelling at the train, "You won't go any further, the war is over! The Allies have landed at Saint-Tropez" — as indeed they had. But despite more attempts by the Resistance to halt the convoy — several prisoners try to escape and are immediately shot, and another insurrection forces the Germans to transfer all the prisoners to a new train at Nanteuil-Saâcy — the convoy continues its relentless progress.
On August 19, the train arrives at Weimar, Germany. The following day, the men are dispatched to Buchenwald and the women to Ravensbrück.
Much later, Christiane learns that her mother has been waterboarded by the Gestapo after her arrest and before her final train ride. But Hélène Boulloche never tells the Germans anything.
BACK IN PARIS, German demolition teams are planting the charges necessary to blow up every bridge, every factory, and every telephone exchange, as well as every famous Paris landmark, from the Palais du Luxembourg to Napoleon's tomb and the Quai d'Orsay, home of the French Foreign Ministry. The chief German engineer promises General Choltitz that the Allies won't find a single working factory when they reach the capital — the industry of Paris will be paralyzed for at least six months.
Since Allied bombers are continuing their obliteration of German cities every day, German generals see nothing unusual about their plan to level Paris. They have just finished demolishing Warsaw, after the uprising there.
After receiving an anonymous phone call, warning of the Germans' plans to blow up every bridge crossing the Seine, the capital's Vichy mayor, Pierre-Charles Taittinger, decides to pay a call on the German commander. Choltitz recites his plan to blow the whole city up "as indifferently as if it were a crossroads village in the Ukraine."
Taittinger decides there is nothing he can do except to try to convey his love for the city. "Often it is given to a general to destroy, rarely to preserve," Taittinger begins. "Imagine that one day it may be given to you to stand on this balcony as a tourist, to look once more on these monuments to our joys, our sufferings, and to be able to say, 'One day I could have destroyed all this, and I preserved it as a gift for humanity.' General, is not that worth all a conqueror's glory?"
"You are a good advocate for Paris," Choltitz replied. "You have done your duty well. And likewise I, as a German general, must do mine."
The general's response is disheartening. But the Frenchman has planted a powerful seed.
WHEN DE GAULLE returns to France from his headquarters in Algiers, he reaches General Eisenhower's headquarters on August 20. Once again, Eisenhower declares his intention to bypass Paris when his troops cross the Seine. De Gaulle says this strategy might be acceptable, if the Resistance had not already begun an uprising there.§ Eisenhower replies that the uprising has begun too soon, and against the Allies' wishes.
"Why too soon?" the French general replies. "Since at this very moment your forces are on the Seine?" Ultimately, Eisenhower commits himself to liberating the capital with General Leclerc's French troops, but he still refuses to specify a date. De Gaulle believes it is "intolerable that the enemy should occupy Paris even a day longer than it was necessary, from the moment we had the means to drive him out of it."
De Gaulle also suggests to Ike that if the Allied command delays too long, he will ignore the Allied chain of command and give the order to General Leclerc's armored division to take Paris himself.
The next day, Roger Gallois, an emissary of the Resistance in Paris, sneaks through German lines to make another appeal to the Americans to get to Paris as soon as possible. When he reaches the tent of George S. Patton, the famously gruff general tells the Frenchman that the Americans are in the business of "destroying Germans, not capturing capitals." The insurrection in the city has begun without permission from the Allies, and now the Resistance will have to accept the consequences; the Allies cannot accept "the moral responsibility of feeding the city."
But Eisenhower's gratitude to the Resistance for everything it has done to make the Normandy invasion successful will soon make it impossible for him to ignore the demands of the Free French.
BY NOW General Choltitz has received the order for the "neutralizations and destructions envisaged for Paris," but he does nothing to carry it out except to blow up a single telephone exchange. Eventually his troops will also set fire to the Grand Palais.
In the final week of the Occupation, the troops under Choltitz's command are increasingly skittish; now you can put your life in the Nazis' hands just by walking down the sidewalk.
One morning in the third week of August, Simone de Beauvoir leans out the window of her Left Bank apartment. This is what she remembers:
The swastika was still flying over the Sénat... Two cyclists rode past shouting, "The Préfectureǁ has fallen." At the same moment a German detachment emerged from the Sénat, and marched off toward the Boulevard St-Germain. Before turning the corner of the street the soldiers let loose a volley of machine gun-fire. Passersby... scattered, taking cover as best they could in doorways. But every door was shut; one man crumpled and fell in the very act of knocking... while others collapsed along the sidewalk.
By now the walls of Paris are plastered with posters reading À CHACUN SON BOCHE — meaning, roughly, that each Parisian should choose his or her own German to shoot.
THE NIGHT AFTER Roger Gallois's pleas are rejected by General George Patton, the Frenchman gets one more chance to make his case that the Allies must advance on Paris immediately. On August 22, he reaches the headquarters of General Omar Bradley, commander of the 12th Army Group, where Gallois is given an audience with Brigadier General Edwin Silbert, Bradley's intelligence chief. As soon as the meeting ends, Silbert and Bradley are scheduled to leave to meet with Eisenhower.
Albert Lebel, a French colonel who is a liaison officer to the U.S. Army, has already made his own written plea to General Bradley: "If the American Army, seeing Paris in a state of insurrection, does not come to its aid, it will be an omission the people of France will never be able to forget."
Now Silbert is accompanying the haggard envoy of the Paris Resistance, as Gallois begins to pour his heart out: "You must come to our help, or there is going to be terrible slaughter. Hundreds of thousands of Frenchmen are going to be killed."
By the time Silbert climbs into a Piper Cub to fly to Eisenhower's headquarters, he has already begun to reconsider his opposition to an immediate move on Paris.
When Silbert and Bradley give their report to Eisenhower, the supreme commander realizes that his hand has finally been "forced by the action of the Free French forces inside Paris." Because they had begun their uprising, "it was necessary to move rapidly to their support." Eisenhower also recognizes that de Gaulle "was always determined to get where he wanted to go, and he wasn't about to let anybody stop him."
When Silbert and Bradley fly back to their headquarters a few hours later, General Jacques-Philippe Leclerc, commander of the 2nd French Armored Division, is waiting for them on the tarmac. He rushes up to their airplane before the propeller has stopped turning. "You win," General Silbert tells him. "They've decided to send you straight to Paris."
ON THE SAME DAY that Eisenhower finally decides to move on the capital, General Choltitz has summoned Swedish consul general Raoul Nordling to a remarkable meeting. Nordling has already convinced the general to release 3,893 prisoners, including 1,482 Jews held at Drancy, although he has failed to halt the train that has carried the three Boulloches to concentration camps in Germany.
By now the German general has decided that he has nothing to gain by following Hitler's orders to blow up the City of Light. So Choltitz makes an extraordinary request of the Swedish diplomat: He asks him to cross German lines, so that he can tell the Allies they must advance on Paris immediately. If they don't, they will enter a city that is already in ruins.
Choltitz then hands the Swede a laissez-passer: "The Commanding General of Gross Paris authorizes the Consul General of Sweden R. Nordling to leave Paris and its line of defense." But before Nordling can leave Paris that night, he is stricken with a heart attack. In his place, he sends his brother, Rolf, accompanied by two Allied intelligence agents and two Gaullists the consul has selected, to improve the chances that the improbable story of the defecting German general will be believed by the Allies.
When the motley crew finally reaches General Bradley's headquarters the next day with their bizarre message, the American commander reacts immediately: "Have the French division hurry the hell in there," he declares. He also orders the American 4th Division to get ready to "get in there too. We can't take any chances on that general [Choltitz] changing his mind."
De Gaulle has tried hard to convince the Allies that there is a danger of a Communist takeover of the capital if his forces don't get there quickly enough. French Communists are indeed "the driving force" behind the on-again, off-again insurrection in Paris, but the historian Julian Jackson and others argue that this did not mean they were trying to seize power. If there was a strategy for that, it came from the Gaullists: They were the ones who decided to occupy the Préfecture on August 19 and the Hôtel de Ville (City Hall) the next day.
After fierce fighting with the German troops dug in on the outskirts of Paris, most of General Leclerc's 2nd Armored Division has reached the "immediate proximity" of the capital on the evening of August 24. When the spectral outline of the Eiffel Tower finally appears on the horizon, the troops are "galvanized by an electric current" that propels them forward.
SHORTLY AFTER NINE P.M., a tiny French detachment of three tanks and four half-tracks form a steel ring around the Hôtel de Ville. At last, Free French troops are back in Paris — four years and seventy-one days after the first German troops passed through the Porte de la Vilette to begin their odious Occupation.
A Paris radio station has been seized by the Resistance five days earlier. Pierre Schaeffer grabs the microphone and begins to shout:
_Parisians rejoice! We have come on the air to give you the news of our deliverance. The Leclerc Division has entered Paris. We are crazy with happiness!_
To mark the glorious moment, the station broadcasts the mystic chords of "The Marseillaise," the rousing anthem that had been banished in Paris ever since the Germans arrived here. Spontaneously, thousands of residents turn their radios up full blast and fling open their windows to make the still-darkened streets explode with the joyful sound of freedom.
Then Schaeffer returns to the microphone to command every Paris parish to start ringing its bells. Within minutes, every block is reverberating with the clanging noise, from the south tower of Notre-Dame to Sacré-Coeur high up on the hill.
IT WAS PARTICULARLY APPROPRIATE that this was all happening on the radio, because this was the medium that had done the most to stoke the fires of Resistance since the Occupation began. It was also the radio that transformed Charles de Gaulle from an unknown officer into the larger-than-life figure who was now being embraced as the nation's savior — practically a Joan of Arc for his time.
His celebrity was almost entirely the product of the regular broadcasts the British had allowed him to make on the BBC. Between 1940 and 1944, he delivered sixty-eight speeches. Gradually, his broadcasts became as beloved among his countrymen as Churchill's in Britain and Edward R. Murrow's and Roosevelt's in the United States. The Vichy government estimated that three hundred thousand French people were listening to de Gaulle at the beginning of 1941 — and _three million_ just one year later.
As Ian Ousby observes, de Gaulle used the radio to accomplish "precisely what Pétain had hoped but miserably failed to do as leader of Vichy: he had become France. Actually, the claim to be France in some indefinable but potent way had been implicit in de Gaulle's wartime utterances right from the moment of his arrival in Britain."
The six-foot-five general is mobbed by grateful Frenchmen in every town and city he travels through after the Normandy invasion. Yet no one is more aware of the deep canyons of division in France, where some have fought the Germans, some have collaborated — and the vast majority have simply kept their heads down and tried desperately to get enough to eat.
De Gaulle knows exactly how he will smooth these divisions. He understands that France has been infected by a terrible disease, and denial is a necessary part of the cure, an indispensable part of the healing. On the eve of his return to Paris, this is how he envisions his task: He will "mold all minds into a single national impulse, but also cause the figure and the authority of the state to appear at once."
DINING WITH HIS FELLOW OFFICERS at his headquarters at the opulent Hôtel Meurice hotel, facing the Tuileries, General Choltitz hears the clanging bells, and he knows exactly what they mean. "Gentlemen," he tells his guests, "I can tell you something that's escaped you here in your nice life in Paris. Germany's lost this war, and we have lost it with her." Later that evening, Choltitz's aide, Count Dankwart von Arnim, writes in his diary, "I have just heard the bells of my own funeral."
Shortly after midnight, Captain Werner Ebernach pays a visit to Choltitz. This is the German officer who has placed several hundred oxygen bottles, at a pressure of 180 atmospheres, to magnify the effect of the dynamite planted in the cellars of Les Invalides, two tons of explosives behind the pillars of the Chamber of Deputies, five tons of explosives under the Ministry of the Marine on the Place de la Concorde, mines on the southeast leg of the Eiffel Tower, and dynamite under more than forty Paris bridges spanning the Seine.
Ebernach has also heard the bells, and he understands what they mean just as well as the general does. Declaring his mission accomplished, Ebernach asks Choltitz if he has any further orders. The general does not. Then the captain asks for permission to withdraw, to avoid being captured by the advancing Allies. He explains that he will leave behind enough men to detonate all the bridges and monuments he has readied for destruction.
But Choltitz has another idea: "Take _all your men_ and leave us," he tells the colonel. Three hours later, all of the demolition experts of the 813th Pionierkompanie (combat engineers) have left the city, taking with them the principal menace to its most magnificent structures. From now on, the increasingly urgent query from Hitler and his minions — "Is Paris burning?" — will be greeted by nothing but silence.
Choltitz explained later that he had no fear of death, but he had begun to have nightmares in which he saw his own corpse suspended over the ruins of the City of Light.
FRENCH AND AMERICAN TROOPS roar into the city under a perfect summer sky on Friday, August 25, the Feast of St. Louis, which honors an unusually benevolent French king of the Middle Ages who had made his officials swear to give justice to all.
Choltitz is eager to surrender, but his sense of military honor compels him to put up a token defense. There is a brief but brutal battle outside his headquarters at the Hôtel Meurice and some heavy fighting near the Ecole Militaire and Les Invalides. German snipers increase Allied casualties, especially near the Tuileries.
French officers conduct the German general to the Préfecture, where General Leclerc has just started a celebratory lunch. Henri Rol-Tanguy, the Communist commander of the Forces Françaises de l'Intérieur, barges in to demand that his name appear next to Leclerc's on the document of surrender. Then the generals proceed to the Gare Montparnasse for the formal signing.a
German prisoners being marched down the rue de Rivoli on August 25, 1944.(photo credit 1.14)
At seven o'clock that evening, de Gaulle finally enters the capital. From the Hôtel de Ville, he delivers a radio address to the nation. In a speech that lasts less than five minutes, he sets the tone for all of his future efforts to bind up the wounds of his tortured nation.
_Why should we hide the emotion which seizes us all, men and women, who are here, at home, in Paris that stood up to liberate itself and that succeeded in doing this with its own hands?_
_No! We will not hide this deep and sacred emotion. These are minutes which go beyond each of our poor lives. Paris! Paris outraged! Paris broken! Paris martyred! But Paris liberated! Liberated by itself, liberated by its people with the help of the French armies,with the support and the help of all France, of the France that fights, of the only France, of the real France, of the eternal France!_
With those words, de Gaulle immortalizes all those who had fought the Germans as the only "real" Frenchmen of "eternal France."
_Well! Since the enemy which held Paris has capitulated into our hands, France returned to Paris. She has returned bleeding but resolute. She has returned, enlightened by this immense lesson, but more certain than ever of her responsibilities and her rights..._
_It would not even be enough, after what has happened, if with the help of our dear and admirable allies we chased him out of our country. We want to go to his country as we should, as conquerors._
_This is why the French advance guard has entered Paris with guns blazing. This is why the great French army from Italy has landed in the south and is advancing rapidly up the Rhône valley. This is why our brave and dear Forces of the Interior are going to arm themselves with modern weapons. It is for this revenge, this vengeance and justice, that we will keep fighting until the last day, until the day of total and complete victory._
_This duty of war, all the men who are here and all those who hear us in France know that it demands national unity. We, who have lived the greatest hours of our History, we have nothing else to wish than to show ourselves, up to the end, worthy of France._
_Long live France!_
Immediately after his speech, he reinforces his message of the one "real France" when he is asked to "proclaim the Republic before the people who have gathered here." De Gaulle refuses to do so: "The Republic has never ceased. Free France, Fighting France, the French Committee of National Liberation have successively incorporated it. Vichy always was and still remains null and void. I myself am the President of the government of the Republic. Why should I proclaim it now?" (Two days earlier, he had said, "France is a country which continues, not a country which begins.")
Shots were fired as de Gaulle approached Notre Dame, and continued as he strode down the aisle inside, but the general never flinched at the sounds of gunfire.(photo credit 1.15)
The next day, de Gaulle defies American commanders who want to use General Leclerc's troops to guard the northeast approaches to the city. The French general concedes that militarily, the Americans are correct, but he insists, "We must have this parade," and he needs Leclerc's troops to provide security for the huge festivity. "Today we were to revive, by the spectacle of its joy and the evidence of its liberty, the self-awareness of a people who yesterday were crushed by defeat and scattered in servitude."
At three o'clock in the afternoon, de Gaulle lays a huge wreath on the tomb of the unknown solider under the Arc de Triomphe and relights the eternal flame, the first Frenchman to do so without German minders watching since the Occupation had begun four years earlier. De Gaulle thinks Parisians are watching him "as though I were the materialization of a dream."
Shots ring out as de Gaulle leaves the Champs-Élysées to turn into the place de la Concorde, and thousands dive to the sidewalks, although de Gaulle thinks most of the fire is pointed up into the air. The same thing happens again when he arrives in front of Notre-Dame, and it even continues inside the cathedral. But no one has ever been more certain that he is a man of destiny than he is. He wrote about himself on this day, "Since each of all of those here had chosen Charles de Gaulle in his heart as the refuge against his agony and the symbol of his hopes, we must permit the man to be seen... so that the national unity should shine forth at this sight."
And so de Gaulle walks straight down the nave of the cathedral, never flinching at the sounds of gunfire.b
CHRISTIANE is among the millions of Parisians who descend upon the Champs-Élysées to celebrate their liberator. But she never really experiences "the euphoria of the liberation."
"Everyone seemed happy and relieved," she remembered. "There were many who greeted de Gaulle with acclaim, after they had supported Pétain. I viewed all of this from a certain distance. It was true that the war was almost over, but my parents and my brothers had been deported, my sister had stayed with the Maquis in the countryside, and I was completely alone in Paris."
At midnight on August 26, hours after de Gaulle's triumphant march through the city, the Nazis return one more time to terrorize the French capital. Ignoring the surrender, which had been signed by General Choltitz, German planes bomb Paris, destroying five hundred houses, setting fire to the wine market, and killing or wounding a thousand citizens.
* The Führer is not exaggerating. The day after this meeting with Choltitz, at least nine people are executed for their role in the assassination plot, including one field marshal, four generals, two captains, and Berthold von Stauffenberg, the brother of the ringleader. Four months later, Choltitz told his fellow generals, "This 'Putsch' of 20 July will be regarded as an event of historic significance. Those 1,500 men, hanged by these criminals, will all get a memorial dedicated to them, for they were the only patriotic, resolute and 'ready to act' men that we had." (Neitzel, _Tapping Hitler's Generals_ ) In fact, the widows of most of the anti-Hitler conspirators were initially denied pensions by the West German government, on the grounds that their husbands had been traitors. (www.dw.de/germany-remembers-operation-valkyrie-the-plot-to-kill-hitler/a-1271174)
† In fact, Patton's 3rd Army does run out of gasoline, one hundred miles from the Rhine, on August 30 — five days after Paris is liberated. (Collins and Lapierre, _Is Paris Burning?_ p. 219)
‡ Although this is the last such prisoner train from Paris to reach Germany, there are several later ones from other parts of France. (http://dora-ellrich.fr/les-hommes-du-convoi-du-15-aout-1944/)
§ De Gaulle doesn't know the exact numbers, but French casualties are mounting in Paris: 125 killed and 479 wounded on August 19, and another 106 killed and 357 wounded the following day.
ǁ Paris Police Headquarters.
a Choltitz was taken prisoner by the British, who sent him, along with many other captured German officers, to the Combined Services Detailed Interrogation Centre at Trent Park, near Enfield in Middlesex, where all of their conversations were secretly bugged. In October 1944, Choltitz declared, "We are also to blame. We have cooperated and have almost taken the Nazis seriously... We've let those stupid cattle talk and chatter to us... I feel thoroughly ashamed. Maybe we are far more to blame than those uneducated cattle who in any case never hear anything else at all. It wouldn't be so bad if we Generals, or the generation before us, for that matter, hadn't taken part. The trouble is that we participated without a murmur." The general also said, "The worst job I ever carried out — which however I carried out with great consistency — was the liquidation of the Jews. I carried out this order down to the very last detail." Choltitz was probably referring to actions he took in Crimea. (Neitzel, _Tapping Hitler's Generals_ ) From Britain, Choltitz was sent to Camp Clinton, Mississippi. He was released by the Allies in 1947. In 1950, he published a memoir, _Brennt Paris?_ (Is Paris Burning?). He wrote that he had refused to blow up Paris because he thought Hitler was crazy and the destruction of the French capital would make any future friendship between France and Germany impossible. His memoir was a principal source for _Is Paris Burning?_ by Larry Collins and Dominique Lapierre. He received a German general's pension of $675 a month. He died in Baden-Baden in 1966 at the age of seventy-one.
b None of the snipers was ever caught alive, or identified. De Gaulle believed they were agents provocateurs, who had fired "a few bullets into the air" to "create the impression that certain threats were still lurking in the shadows" and "that the resistance organizations must remain armed and vigilant." (de Gaulle, _Complete War Memoirs,_ p. 658)
# _Eighteen_
_We must not forget that we owe a great debt to the blunders — the extraordinary blunders — of the Germans._
— Winston Churchill, addressing Parliament, September 1944
TWO DAYS AFTER the Liberation, Christiane goes to a police station to request a gendarme to accompany her to her parents' apartment. When she walks through the front door, she discovers a hovel: the Germans have looted everything, from her mother's jewelry to her father's rare books.*
When she is alone in the ravaged apartment, a neighbor appears, the father of a young girl who is Christiane's age. He says he is there to comfort her, but when she lets him in, he tries to rape her. "I was shocked and quite undone. After everything else I had been through, this was really too much."
To escape her brutal neighbor, her loneliness, and her parents' decimated home, Christiane leaves Paris at the end of August by car to rejoin the Maquis and Jacqueline in the Morvan: "She was all I had left." At the end of September, the region is liberated. On October 21, Jean Longhi (Grandjean) presents Christiane with the Croix de Guerre. The self-effacing _Résistante_ is annoyed by this recognition of her bravery: "I considered all of my clandestine activity to be a matter of course, and now a decoration! After so many dramas and so many deaths, it seemed like a ridiculous gesture."
Christiane and Jacqueline return to Paris and their parents' apartment. Their lives are brightened by their new tenant, Lieutenant Henry Kaiser, the charismatic Brooklynite who is a labor lawyer and the labor adviser to the occupying American Army.
Then Christiane suffers one more serious scare. Jacqueline becomes ill, and the doctor telephones Christiane and asks to see her alone. He tells her Jacqueline has only a few months to live. After all the other catastrophes she has already endured, Christiane feels completely overwhelmed — until she takes the X-rays to her friend Dr. Cler for a second opinion.
After examining them, Cler explains that they actually show only a few benign traces of an old case of pleurisy. He promises Christiane her sister will recover before long — and she does.
THE SISTERS' GREATEST PREOCCUPATION is the fate of their parents and their brothers, about whom they have heard nothing since André's letter from Flossenbürg the previous summer. At first they don't even know that their father has been sent to Buchenwald, their mother to Ravensbrück, a women's concentration camp about ninety miles north of Berlin, and Robert to Ellrich, a subcamp of Dora-Mittelbau, about twelve miles away.
By November, they have learned their father's whereabouts. A family friend, Pierre Lefaucheux, miraculously returns from Buchenwald, and Christiane immediately goes to see him. When she arrives at his house, he is at the dinner table with friends. She hears laughter from the dining room while she waits for them to finish. When dinner ends, he finally tells Christiane that despite the dire conditions at Buchenwald, her father is fine. But the friend is obviously uncomfortable, and she wonders if he is telling the truth or just doesn't want to alarm her.
The daughters know nothing about their mother's fate. But her husband, Jacques, actually hears from Hélène shortly after he arrives at Buchenwald. She is permitted to write one letter a month, and he receives her first one toward the end of September. Jacques's college classmate Etienne Audibert, who was with him in the camp, recorded what happened next.
Although letter writing from one camp to another continued to be authorized, after that [Jacques] never received another letter [from his wife]. He could not fail to understand what that meant. This was the greatest blow that could have struck him, and it crushed his indomitable energy. Once his moral resolve had been broken, there was nothing left for him but death; that is the law of the camp. He succumbed on February 18, 1945.
Jacques's deduction had been correct: Hélène had died at Ravensbrück on October 25, 1944, just two months after she had been waterboarded by the Nazis.
Etienne Audibert's remembrance continued:
It is something entirely different to die in a concentration camp. Death from action or during combat, consensual death accepted with one's full strength, has nothing in common with gradual physical and moral erosion, a progressive degradation, when the victim sees his faculties slowly disappear, his personality silently breaking up. Those who knew Jacques Boulloche, his wife, and their children could not fail to appreciate the extreme refinement of their milieu. Everything breathed elegance; everything carried the luster of high culture and good taste. Could they recognize these tattered people when they were dressed in rags, deprived of everything, famished, shivering from the cold, treated worse than any beast has ever been, struck by blows for no reason, weakened daily by humiliations that could only be conceived by the deranged imaginations of monstrous perverts? Can others imagine their agony, alone, far from the sky of France and everyone they loved, these spectral beings with a feverish glow, their eyes like concave sockets, with protruding ears and parched skin ready to crack, living their final weeks in a ghastly atmosphere polluted by the foul smell of the smoke of the crematorium? And added to their own pain, the desperate anxiety each of them felt about the other three? When they are finally dead, no friendly hand comes to close their eyes. And then the cinders of their flesh are dispersed to the wind.
AT THE BEGINNING OF 1945, the sisters still don't know any of this. Jacqueline decides to go to Switzerland, to see if there might be a way to ransom the freedom of the rest of the family. Her trip is a failure: She learns nothing about her parents or her brothers. When she returns to Paris, she tells her sister there is only an infinitesimal chance that any of them will survive.
Christiane briefly gets a job as the secretary for the Canadian ambassador to Paris, but the position is a poor fit. Christiane doesn't even know how to type, and she doesn't last there very long. Jacqueline goes to work for the Bureau Central de Renseignements et d'Action, the same organization André had been assigned to when he was de Gaulle's military delegate in occupied Paris.
The sisters enjoy going out with the American and Canadian troops who are now crowding the city. They take them to dinner at their mess, which has much better food than what is available to most Parisian civilians. The Occupation is over, but most store shelves remain empty.
At the same time, the sisters are very much aware of the Battle of the Bulge, the last major German counteroffensive, and the greatest surprise attack on American forces after Pearl Harbor. It begins in the middle of December 1944 in the Ardennes region of Belgium, France, and Luxembourg.
On December 16, two hundred thousand soldiers from three German armies suddenly launch themselves against the Allies. There has been a serious Allied intelligence failure — the day before the attack, British field marshal Montgomery promised Eisenhower that the Germans cannot "stage major offensive operations." Just as the Germans had partly ignored the explicit BBC warning of the impending invasion at Normandy, Allied intelligence officers had discounted the information from four captured German POWs, who had warned of a pre-Christmas offensive.
The German thrust is forty miles wide and fifty-five miles deep into the Allied line, creating a shape on the map that gave the battle its enduring name. On December 19, the 5th Panzer Army surrounds the American 106th Division in the center of the offensive at St. Vith, and forces eight thousand American soldiers to surrender — the largest defeat of American troops since the Civil War.
As the weather slowly improves, the Allies' overwhelming air superiority gradually reasserts itself, the Germans begin to run out of gasoline, and the systematic destruction of railroad lines makes it impossible to bring a single German train across the Rhine. When the battle finally ends in the fourth week of January, the Germans have lost 120,000 men killed, wounded, captured, or missing, while the Americans have suffered 19,000 killed, 48,000 wounded, and 21,000 captured or missing. "The great difference," wrote Andrew Roberts, "was that in material the Allies could make up these large losses, whereas the Germans no longer could."
In the end, the Ardennes offensive weakens the remaining German troops so much that its biggest effect is to hasten the progress of the Russian advance from the east.
IT IS APRIL 1945 before Jacqueline and Christiane receive confirmation of the first of three catastrophes. They are at lunch at the home of their aunt Ginette — the same aunt who had sent her maid into the street to intercept Christiane and save her life, hours after her parents had been arrested. The telephone rings, and they learn officially that their mother is dead. A short time later, they are notified of their father's death. But they still have no news of either of their brothers.
Two weeks later, Adolf Hitler, the father of all of Europe's agony, summons his longtime mistress, Eva Braun, to his underground bunker in Berlin. He marries her on April 29. At three thirty the following afternoon, Hitler shoots himself in the mouth with a revolver, and Braun — his wife of forty hours — swallows poison. Hitler's propaganda minister, Joseph Goebbels, is the last of his top collaborators to remain in the bunker with the Führer as the Russians swarm over Berlin. The day after Hitler dies, Goebbels and his wife poison their six children inside the bunker. Then they both commit suicide.
Mussolini and his mistress are caught by Italian partisans on April 26 while trying to escape into Switzerland. They are executed two days later and then strung up by their feet from lampposts in Milan. On April 29, the Germans sign an unconditional surrender of Italy and southern Austria; it takes effect on May 2, removing one million German troops from the conflict.
In Nuremberg, the site of gigantic Nazi rallies in the 1930s, American troops replace "Adolf-Hitler-Str." signs with new ones reading "Roosevelt Blvd." Then they blow up the huge stone swastika atop the Nuremberg stadium.
On May 5, Admiral Hans-Georg von Friedeburg, the new commander in chief of the German Navy, arrives in Reims, in northeastern France, where Eisenhower has established his headquarters. Two days later, at three forty-one in the morning, Germany surrenders unconditionally. At midnight on May 8, the guns stop firing and the bombs stop falling all across Europe. The "Thousand-Year Reich" is finally extinguished, after twelve years, four months, and eight days of mayhem, perversity, destruction, and death.
A FEW DAYS BEFORE the surrender is signed ending the war in Europe, Christiane and Jacqueline get a terrible scare. The doorbell rings at their parents' apartment. It is Gilbert Farges, one of the men who had shielded André on the train when he was deported to Auschwitz. When he introduces himself — "I was deported with your brother André" — the sisters are sure he is there to announce André's death. But Farges immediately adds that their brother is still alive — the first genuinely good news they have had in 1945.
Unlike the other members of his family sent to the camps, André has a vital cadre of friends at Flossenbürg. Somehow, everyone who survives with him is equipped with "the intimate conviction that we would still be alive after the war," remembered his friend Michel Bommelaer. "Among us, there was always a faithfully burning pocket of joy and hope."
André's most extraordinary morale booster for Michel, his fellow piano player, is to teach him "the Schumann piano concerto, certain Beethoven sonatas, and the Brandenburg Concertos" — all without a piano, of course. "In this way he shared a small piece of his very tender heart, his determination to fight," and his belief — remarkable, under the circumstances — in the "intellectual quest of a humanity that, no matter what, had a chance to better itself."
Gilbert Farges said, "Survival was a constant act of will and dignity. André applied himself with discipline, determination, and a ferocious courage at all moments. The influence of his example surely saved the lives of a number of his companions. This period of his existence tempered his character — tempered it the way one used to temper a sword in an earlier era."
A postcard Jacqueline wrote to André in German at the beginning of 1945, when he was still a prisoner at Flossenbürg.(photo credit 1.16)
André Boulloche in the summer of 1945, immediately after his return to Paris from three German concentration camps.(photo credit 1.17)
IT WILL REQUIRE one more miracle for André to return to his sisters' arms alive. At dawn on April 16, 1945, at an altitude of twenty-five hundred feet, the camp at Flossenbürg is bathed in "a pure and wonderfully soft light," Georges d'Argenlieu remembered. The sounds of American guns can be heard in the distance. And suddenly all the SS guards leave the camp.
"We had been saved!" d'Argenlieu exulted. "There will be no evacuation! Fourteen thousand deportees with no strength lay extended, offering their bodies to the rays of this first sun of resurrection." The sounds of American guns grow closer: ten miles, then seven miles away. But when the sun goes down that night, the SS returns, heaving everyone "back into the void."
"On April 20th, the sinister evacuation — the one we had so rejoiced at escaping — begins," d'Argenlieu continued. "Fourteen thousand leave the camp in five hours. Their rescue by the Americans is three days and eighty miles away. The six thousand who succumb on the road will never know that minute."
Through a final stroke of good fortune, André Boulloche, Charles Gimpel, Henri Lerognon, another alumnus of the Ecole polytechnique, and Georges d'Argenlieu manage to slip into the typhoid ward of the camp at the moment of the evacuation. That is the only way they can avoid the death march. Three days later, on April 23, a unit of George Patton's army finally arrives to liberate the camp.
IN THE CHAOS following the liberation, it takes André nearly four weeks to make his way back to Paris, a fearfully gaunt figure with the bulging eyes of a camp survivor. On May 19, he finally enters the lobby of his parents' apartment building on avenue d'Eylau. The concierge recognizes the ravaged twenty-nine-year-old as he walks in the front door. As André gets on the elevator, the concierge sprints up the stairs to warn his sisters that their brother has come home.
"Of course he was extremely thin," Christiane remembered. "But what was worse was the horrible look in his eyes." The sisters have agreed that Christiane will deliver the terrible news about their parents when André gets there. "But, when he was finally standing there before us," Christiane remembered, "with that ghastly appearance that all the deportees had, I could not utter a word."
When she finally gathers her strength to tell him what has happened, this is his immediate reply: "If I'd known that, I would not have come back. I would have died in the camps." When they tell him which camp Robert has been sent to, André says there is very little chance that he has survived there.
A couple of weeks after André's return, they are officially notified that Robert died at Ellrich, on January 20, 1945.
* During the time General Choltitz was a prisoner of the Allies, he declared, "I'm saying that we steal! We collect the stuff up into stores, like proper robbers... that's the frightful part of it. This revolting business of engaging in organized robbery of private property... Throughout the whole of France... Whole train-loads of the most beautiful antique furniture from private houses! It's frightful; it's an indescribable disgrace!" (Neitzel, _Tapping Hitler's Generals_ )
# _Part II_
# _Nineteen_
_My deportation to the camps is very largely what made me what I am today. And it was the war that led me to socialism. I am a man who engages in life — who feels the necessity to engage._
— André Boulloche
_André Boulloche had the longest service in the Resistance, the most audacious, the most important and the most challenging._
— André Postel-Vinay
_"Did your father André ever talk about the war?"_
_"Once, perhaps. But he didn't have to talk about it. It was always there. It's as if you said we're going to talk about the fact that the walls are white. Obviously they're white! You're not going to talk about them, because they're there, all the time."_
— Agnès Boulloche
THE THREE SURVIVORS chose very different paths to salvation when the war was over. Christiane and Jacqueline "turned the page" by getting married* and having children — and by never discussing the war with each other, or almost anyone else, for fifty years. Jacqueline and Alex Katlama were the first to marry, in the summer of 1945, shortly after André's return from the camps.
Jacqueline and Alex Katlama. They became wife and husband shortly after André Boulloche returned to Paris in 1945.(photo credit 1.18)
André took a very different approach. Although he also married quite quickly and started a family, unlike his sisters he blamed himself explicitly for everything that had happened. Beyond his austere personality — a mournful contrast to the cheerful young man he had been before the war — two things made his attitude clear to everyone: For the rest of his life, he kept his hair shorn to a crew cut, and he wore a black tie every day, in memory of the dead.
Jacqueline and Christiane after the war. By the middle of 1947 both of them had gotten married.(photo credit 1.19)
In the months after the Liberation, there was an orgy of retribution against those who had collaborated with the Nazis. Women who had slept with the Germans had their heads shaved in the streets, while ten thousand Frenchmen were the victims of summary executions. Another one hundred thousand were tried in civilian and military courts, and about fifteen hundred of them were executed.
When he got back to Paris, André had an operation to repair the badly tended gunshot wound to his stomach. But he quickly discovered that nearly giving his life for the liberty of France was not enough to guarantee him a warm welcome when he returned. Several of his relatives made it clear that they blamed him and his sisters for the deaths of their parents and their brother. That was something I learned from two of his nephews. Christiane never mentioned it to me. This was another taboo subject, within the larger taboo of silence the three of them embraced.
The hostility they faced from their family was probably one of the reasons he and Christiane decided to go to America for a year. André returned to his roots as a highway engineer and proposed a study of American traffic lights. He arranged for Christiane to accompany him as part of his team. "Being former _Résistants_ opened a lot of doors" in 1946, Christiane explained.
Before they left on their trip, they organized a funeral mass for their parents and Robert at Saint-Honoré d'Eylau, a church eight hundred yards from their parents' apartment. It was rare for half of a non-Jewish French family to have died in the German camps. Christiane remembered the service as a horrible event, an overflowing church with an endless parade of hands to shake afterward.
Because of heavy winter seas in the North Atlantic, their trip to America on a cargo ship took twenty-two days and came with plenty of seasickness. When they arrived in New York in December, they were astonished by the vitality of the city, especially after the drab postwar Paris they had left behind. But New York was freezing and covered in snow, and Christiane did not have anything to wear to cope with the weather.
"I went to Macy's to buy a coat, and I couldn't buy anything. Because there was too much! I was used to the stores in France, where there was three times nothing. So going to Macy's was horrible! My head was spinning. It really had a strange effect on me. The next day was better. I had to buy something, because I was cold!"
Their American base was Washington, D.C. Their apartment was in a black neighborhood, which shocked their American friends. Christiane was oblivious of her neighbors, except for the annoying owner of the liquor store below them, who seemed determined to sleep with her. They reconnected with Henry Kaiser in Washington, and met his wife, Paula. They also met my parents there for the first time. Christiane thought New York was much more interesting than Washington, and whenever she had a little money, she would take the train up to Manhattan.
The invitation for the memorial service for the family members who died in Germany. Christiane remembered it as a horrible event. Soon afterward she and her brother left for a trip to America.(photo credit 1.20)
When she got to the United States, Christiane realized she had a new duty: to educate Americans about what the Resistance had done in France. She traveled across the country, from New York to San Francisco, giving lectures about the Resistance to American college students.
Once again Christiane felt she was performing her duty. When she spoke about her experiences at a club in Washington, she got her picture in the _Washington Post._ "Fear of the Gestapo was transmitted to the children," the _Post_ reported her saying. "Nevertheless, the young people are eager to shake off that haunting fear, and eager to rebuild their lives." One of the young people she was talking about, of course, was herself.
Christiane also learned that Americans generally knew absolutely nothing about how the Resistance had operated in France during the war — something that has barely changed seventy years later.
The brother-and-sister team stayed in America through the summer. André continued to suffer terrible guilt about the fate of half his family. On August 5, 1946, he revealed his feelings in his diary:
_Why did my life have to be spared, when I was offering it so willingly, even cheerfully? And why did those three who wanted to live, and who loved life so passionately — why were their lives taken from them in the vilest, most brutal way imaginable? Why did I have to be left behind, I who had pushed the barge so far from the shore? Left behind without faith, without hope, but chained to life by my passionate love for my two sisters...._
_When this unlikely and idiotic thing happened, my nature was awakened beneath a kind of false shell of wisdom. When the war came, I had, with great effort, offered the complete sacrifice of my life — mine and mine alone. When I left France [at the end of 1942], I was the only one in danger. With the near certainty of my imminent death, I felt the compensations that only a profound determination can provide._
_Why didn't I have the courage to remain absolutely alone when I returned to Paris? Why did I have to add the possibility of the sacrifice of my own family members to the near certainty of my own sacrifice?_
AFTER CHRISTIANE spoke at Smith College, the administrators there offered her a full scholarship if she would live in the Maison Française. But by the end of 1946, both she and André were eager to get back to France. When they returned, Christiane decided to take advantage of the new vogue for plastic surgery. Ignoring the objections of her friends, she got a nose job: "I've always done pretty much whatever I wanted to."
André had become a Socialist because of his experience as an inmate in three concentration camps, where, in addition to the millions who perished in the gas chambers, thousands had died from forced labor. "It was when I was deported that I realized what life was like for a worker who is completely stupefied by his job, and whose only perspective is to keep working until he dies," he explained. For the rest of his life, he continued to believe that society was unjust to the average worker, barely allowing him to subsist and making it "very, very difficult" to improve his station in life.†
"We bourgeois learned some things during the war," Christiane explained. "We discovered that poverty existed, and injustice existed. We were young, you know."
In January 1947, there were two big events in the Boulloche family. That month another former Resistance member, Paul Ramadier, became the first prime minister of France's Fourth Republic. André Boulloche knew Ramadier through mutual friends in the Resistance, and he was tapped for Ramadier's cabinet, to handle economic issues. By the end of the year, André had become the prime minister's chief of staff, an accomplishment that even impressed André — "I was very young for that job: I was 32."
It was also in January that Jacques Boulloche's college classmate from the Ecole polytechnique, Etienne Audibert, wrote an article about their time together at Buchenwald. Composed for the school's alumni magazine, the article described how Jacques Boulloche had given up after he deduced that his wife had died in another camp. The piece included a harsh attack on their alma mater, because it had expelled anti-Nazi students from the class of 1941. Its tone was fiercely anticollaborator. It made the editor of the alumni magazine nervous because of the criticism of the school, and he initially refused to publish it.
Etienne Audibert telephoned Christiane and asked if he could meet with her to show her the piece. When he came over and she opened the door to greet him, she saw that he was accompanied by his dashing twenty-five-year-old son, Jean.
Another former student from Polytechnique, Jean Audibert had been able to get to England during the war and joined the Free French Navy. His ship engaged in antisubmarine warfare, and he had been offshore during the Normandy invasion. His mother had died in France while his father was imprisoned at Buchenwald.
Jean was taken with Christiane's strength and wit, and she was charmed by his intelligence and his roaring laugh. By the end of lunch it was obvious they had hit it off.
Both of them agreed with Etienne that nothing in the twenty-five-hundred-word article should be altered. Eventually, the editor of the alumni magazine backed down and agreed to publish all of it.
Christiane and Jean had a whirlwind romance throughout the winter. When they met, they already shared the tragedy of the loss of their mothers during the war. And almost immediately they shared another one. Jean's brother Pierre was a former _Résistant_ who had gone into the army after the war and got shipped off to Indochina. At the beginning of 1947, he was killed there in an ambush. So now they also shared the loss of a brother.
Their common grief deepened their bond. Just six months after they met, Christiane was pregnant with their first child. Soon after that, they were married, on June 18, 1947. "Obviously, we chose the date for a reason," said Christiane. It was the seventh anniversary of de Gaulle's famous appeal from England after France's collapse in 1940: the speech in which he declared "the flame of French Resistance must not, and shall not, die!"
The first member of the new generation arrived "two weeks early," but Christiane never bothered to pretend that she wasn't already pregnant when she got married. Catherine Hélène Julienne Jacqueline Audibert was born on February 3, 1948. (Naturally, just twenty-three years later, Catherine would become a judge.) Christiane considered her daughter's birth a big event for the whole family: her first step in "turning the page." Jacqueline's first child, Eric Katlama, quickly followed seven months later.
On September 24, 1949, André married Anne Richard, whom he had met shortly after returning from the concentration camp. She had played a big role in getting him readjusted to society when he first returned to Paris. Eight months after their wedding, Anne gave birth to Robert Boulloche, named for André's dead brother.
Then there was a torrent of natal activity. Noëlle Audibert joined Jean and Christiane's hearth in 1949, and Pierre followed in 1951; and the following year Jacqueline had her second son, Michel Katlama.
Jacqueline and her firstborn, Eric Katlama, in 1949.(photo credit 1.21)
Just five years after Catherine's birth, there were nine members of the new Boulloche-Audibert-Katlama generation. Jean and Christiane had their fourth and last, François, four years later.
FROM THE VERY BEGINNING, all of the children were marked by their parents' silence about the war. Most of them remember seeing books lying around the house with terrifying pictures of the inmates of the German camps, but if something about World War II suddenly came on television, their parents usually left the room.
Like their own father, who had never talked to them about World War I, André, Jacqueline, and Christiane almost never spoke to their own children about World War II. And just as Christiane had complained that her parents were so close that there was no room in between them for anyone else, the children of Christiane, Jacqueline, and André saw the three siblings as an equally tight unit, almost impenetrable to everyone else, including their own offspring.
"You've never seen two sisters who were so close," said Eric Katlama, Jacqueline's older son. "I've never seen it. They were sisters of incredible proximity and complicity." They talked on the phone every single day. "I think André and Jacqueline built a wall around Christiane," Eric continued. "My mother would not allow anyone to question what Christiane had done" during the war — or its effects on the rest of her family. "Jacqueline always defended Christiane — even when there wasn't even an attack, but just a hint of anything. That was instantaneous. I can't put myself in Christiane's place, but if I were Christiane, I would have completely put it out of my mind, without ever looking back at it again."
Eric remembered his uncle André as a "very very big personality. He was very impressive. There weren't a huge number of people who came out of Auschwitz. André was like the statue of the commander in the opera _Don Giovanni._ The commander is the one who is killed at the beginning of the opera, and he comes back as the statue. So it's true we had something of that picture.
"Of course we were proud that they had all been in the Resistance," Eric continued. "But it was a pride that we kept to ourselves. We never externalized it. Never."
When I asked André's oldest son, Robert, if he agreed with his cousins that the war had been a taboo subject, he replied, "It's true. We were always scared to ask any questions. And we were always scared that we wouldn't ask them the right way, that was tragic — or else, not tragic enough." We both laughed. "It was difficult."
Beginning in 1958, the family did deal with the war directly, once a year, with a ceremony that struck the children like a grim exclamation point. The young ones "camouflaged their unhappiness with fake laughs that were a little nervous," while the two sisters stood next to their brother and cried uncontrollably. Often André cried as well. Eric Katlama thought "it was clear that André was suffering even more than Christiane and Jacqueline. My mother made us understand that André had suffered even more than she had."
The ceremony occurred every year on October 25, the date the Germans had inscribed in their meticulous records for the death of Hélène at Ravensbrück. It took place at the family plot at Père-Lachaise, Paris's largest and most celebrated cemetery, built in 1804 by Napoleon on more than one hundred acres on a sloping hillside at the north end of the 20th arrondissement. In its first years, the cemetery was a bit of a flop, because the Catholic Church hadn't blessed its land, and the fashionable people thought it was too far from the center of the city. In a brilliant marketing gimmick, the proprietors transferred the remains of Jean de La Fontaine and Molière to the new burial ground. After that it was a huge success. Balzac, Oscar Wilde, Sarah Bernhardt, Chopin, Bizet, Proust, Yves Montand, Simone Signoret, Maria Callas, and Jim Morrison eventually joined them there.
And the Boulloches. At their plot, the names of their dead parents and their dead brother had been etched on a huge polished granite sarcophagus, which was actually empty. Attendance at the annual ceremony was required for all sixteen family members — ten children and six parents.
"I remember there was a feeling of meditation that began in the car on the way to the cemetery," said Eric Katlama. "They prepared us psychologically. They made us understand that it was something terribly painful for those who had lived through it — perhaps more for André than for the others, because he had been deported, so he had seen what it was like."
The black granite sarcophagus on which the family engraved the names of their dead.(photo credit 1.22)
André always presided. And he spoke only two sentences: "We are reunited in memory of your grandparents and your uncle, who were killed by the Nazis." ("He didn't say 'the Germans,' " his nephew, Michel Katlama, recalled. "He said 'the Nazis.' ")
"Remember that they died for the liberty and the liberation of France." Then the patriarch called every grandchild forward — Catherine, the oldest, was first — handing each of them one red rose to be placed on top of the granite box with a cross carved in the middle.
"This was the Boulloches," said Christiane's older son, Pierre Audibert. "There was an aspect of this ceremony which excluded: it was the pure solidarity of the Boulloche family. A pact with their dead."
As one of Christiane's children put it, "Just as there is original sin, the Boulloches had the opposite: original virtue. They had chosen the right side. They had done everything the way you were supposed to. This feeling of belonging to some kind of martyr's elite is quite heavy. I realized later that I had made many Jewish friends in school who had lost a lot of people in their families during the war. I didn't realize they were Jewish at the time. It was unconscious. The first value transmitted by my mother was, you must not accept — you must act."
The family reunited one last time at Père Lachaise in 2004 at my request. Christiane is second from left in the front row. On her left is her niece Véronique Katlama; on her right, her daughter Noëlle. Behind Véronique is her brother-in-law Eric Katlama.(photo credit 1.23)
Christiane's youngest, François, echoed his sibling: "It showed that you can do good things in life. But it's also a little crushing: It sets the bar very high. Christiane was lucky to make the right choice. Lucky — with the time to pay for it."
WHEN THE WAR ENDED, de Gaulle and the Boulloches felt much the same way: The only chance they had to survive was to avoid dwelling on the past. De Gaulle's task was more complicated, because he had to paper over the mixed record of his countrymen. Stanley Hoffmann, the great historian of modern France, explained de Gaulle's postwar attitude this way: "If one wants people to win victories over their very worst flaws, one must appeal to what is noble in them. If one wants to bring out the best in them, it is the best that one must celebrate." The general adopted a kind of therapeutic optimism, which he considered essential to France's recovery. De Gaulle told the novelist André Malraux, who became his minister of culture, that "man was not made to be guilty, sin is not interesting, the only ethics are those which lead man toward the greater things he carries in himself."
BUT DE GAULLE never literally said that "all the French were _Résistants,_ " explained Claire Andrieu, an associate professor of history at the Sorbonne, who is the author of a shelf of books about the war and an expert on the Occupation and the Liberation. She also happens to be André Postel-Vinay's daughter.
"Contrary to what some people say, no one has ever written that all of the French were in the Resistance," the professor continued. "On August 25, [1944], in liberated Paris, de Gaulle spoke of the 'only France [a clever adjective], the France who is fighting, the eternal France.' De Gaulle never said that all the French were _Résistants._ In fact, he often said the opposite, but not in public. For example, when he received the National Council of the Resistance on September 6, 1944, he said, 'You are the Resistance, but the Resistance is not the nation.' He knew well. He wasn't crazy."
"But obviously," I interjected, "most people were neither collaborators nor resisters."
"I don't think it's that either," she said. "The problem is that we had a quasi-totalitarian dictatorship. Because you had Vichy plus the Nazis — that was a lot. And a radical system of economic exploitation. So objectively, whatever the wishes of the French were, they collaborated. That is to say, we let freight cars full of cows and metal leave for Germany. We let six hundred thousand young Frenchmen leave to work in Germany. This is a collaboration. We allowed more than fifty thousand members of the Resistance to be deported to Germany. So objectively, it was a collaboration — without even speaking of seventy-six thousand Jews who were deported.
"The country functioned in the midst of the Vichy-Nazi system. Everyone, including the baker, was forced to do so. But elsewhere — outside of working hours — they listened to the BBC. And if a British or American aviator knocked at the door of a house, 99 percent of them were hidden. That is the figure the American Military Intelligence Service gave in 1943. Therefore, there are certain counterweights to this objective collaboration. It's why I look at things from this perspective.
"It's complicated, because it can be the same people who organized the convoys of looted goods — people who worked for the SNCF [the French national railroad] — and who also hid an Allied pilot at home. It can be the same person. So that's why I am personally not very satisfied with the existing theses on the behavior of the population. I think people forget everything that could be done outside of institutions."
While de Gaulle felt he had to disguise the history of France in public, the Boulloches merely remained mute about their own. At the beginning, this had seemed odd to me, since all three of them had been decorated for their bravery, and André was part of an elite of just a few hundred _compagnons de la Libération,_ the most revered Resistance fighters of all. But that was the cost of their courage: Because half of their family had been killed by the Nazis, they needed their own silence as much as France needed its myths.
FRANCE'S SEARCH for its own truth was reignited by the strikes and riots led by students and workers against de Gaulle's government in May 1968. Among the many issues pushed by the students who barricaded the streets of Paris was a demand that their parents reconsider the official version of how most Frenchmen had behaved during the German Occupation.
André's only daughter, Agnès, was arrested after getting into a fight with a policeman during one of the demonstrations. André told her that if she "wanted to fight, there were a couple of things you should pay attention to. If you're getting hit on the head, never put your hands there, because your fingers are much more fragile than your skull, and you'll just get your fingers broken. If you carry a weapon, it is always to kill. Do not think it is to defend yourself. If you draw your weapon, never get closer than three meters to the person you want to kill, because otherwise he can take your weapon from you. All of us also learned how to strangle someone, even if you weren't strong, by taking him from behind. So we did talk about stuff, and sometimes we played with him, pretending that we were trying to strangle him."
THE UPRISINGS also had a direct effect on three journalists working for the Office de Radiodiffusion-Télévision Française, the state-owned television network. These three men would make the movie that did more to change France's self-image than any other event in the postwar period.
Marcel Ophuls, André Harris, and Alain de Sedouy all worked together for the French TV network, and when de Gaulle called for a media blackout of the barricades in the streets, the three of them joined the strike. Ophuls had written part of the strike manifesto for the TV journalists. By the late '60s de Gaulle had moved sharply to the right, and Ophuls told me that he thought de Gaulle's conception of what television should be was "very much like Franco's." When the strike failed, he and his two colleagues were dismissed, which meant they had to figure out how to go on making a living. So at the moment when dozens of national institutions were under attack, the three of them decided to make a film about France during the Occupation.
With $160,000 — raised, ironically, in Germany and Switzerland — Ophuls became the director of the documentary, and he started filming in Clermont-Ferrand, the capital of the Auvergne province in the center of France, 242 miles from Paris. He chose Clermont-Ferrand partly because of its proximity to Vichy, the capital of the unoccupied zone during the war. The region was also the birthplace of the Maquis, the guerrilla army of the Resistance that fought in the countryside.
Ophuls thought de Gaulle's myth of an almost-universally resistant France had contaminated everyone who had lived through the war. "There's something unhealthy about asking all these millions of individuals to lie to each other, to lie to themselves," he told me. "I don't think this can possibly be good politics — now or then."‡ So he set out to correct the record.
The result was one of the greatest documentaries of the twentieth century. After it was — predictably — banned from French television, _The Sorrow and the Pity_ became a huge hit with the younger generation at the movie theaters on the Left Bank in the spring of 1971. Its honesty fit the revolutionary spirit of the '60s. The film featured interviews with everyone from the local butcher of Clermont-Ferrand to a Paris aristocrat who had served with the French division of the Waffen SS, as well as two former prime ministers of France and Great Britain: Pierre Mendès-France and Sir Anthony Eden.
The two-part, four-hour film filled a twenty-six-year vacuum of information with a brutal portrait of the stark divisions inside wartime France. Recalling a split reminiscent of the one in the Boulloche family, teachers from the lycée in Clermont-Ferrand described how their students were more active in the Resistance than their instructors, because "young people are generally much more sincere, and... more alive." It also included a searing section on the decision of the French gendarmerie to arrest 4,051 Jewish children in Paris in the summer of 1942 — even though the Germans had never asked them to arrest anyone younger than sixteen. After four days of indecision, all the children were shipped off to Germany — and every one of them was gassed as soon as they arrived at the camps.
All of the Boulloche children of my generation saw _The Sorrow and the Pity,_ and we all agreed it was an important event.§ A year later, the debate about France's wartime behavior intensified with the publication of _Vichy France_ by Robert O. Paxton, a young American historian who used German archives to document the extent of French collaboration during the war. Most of the Boulloches also read Paxton's book. But neither the film nor the book lifted the veil shrouding the family's wartime experiences.
IN THE LATE 1950s AND EARLY '60s, André, Christiane, her husband, Jean Audibert, and Jacqueline all embraced France's most progressive cause, by joining the Club Jean Moulin to battle the OAS (Organisation de l'armée secrète), a paramilitary terrorist group that used bombings and assassinations to try to prevent France from allowing Algeria to become independent.
Christiane around 1960, just before I met her for the first time in Paris.(photo credit 1.24)
France's withdrawal from its North African colony was a long, bloody process for Algeria and France alike. Once de Gaulle became convinced that it was necessary to leave Algeria, he was the main enemy of the OAS. After an OAS uprising failed in Algiers in April 1961, the organization turned to terror on both sides of the Mediterranean, including several unsuccessful attempts to murder de Gaulle.
André shared the widespread view that the Algerian quagmire posed a serious threat to French democracy. He decided very quickly that de Gaulle was the only politician powerful enough to bring about France's exit and prevent the return of a right-wing dictatorship in France. Indeed, that was the main reason he and his sisters supported de Gaulle's return to power at the head of a new Fifth Republic in 1958. After that, André lobbied the general to leave Algeria whenever he saw him.
André said the war in Algeria was "eating the national tissue away like an acid. Civilians are scared of the military, and the military doesn't trust the government."
Meanwhile, Christiane remained as fearless as ever. She had become the treasurer of the Association for the Defense of Djamila Boupacha. The young Algerian woman was a famous and beautiful member of the FLN (Front de libération nationale algérien) who was tortured by the French into confessing terrorist activities. She was only twenty-three when she was condemned to death on June 28, 1961. Part of Christiane probably identified with the undaunted young activist.
"It was a period when there were terrorist attacks in Paris," Christiane told me. "The OAS was carrying out terrorist attacks. And when I was the treasurer of the association, my name and address was on our literature. Which was a risk — there's no question about that. When they were still very small, I tried to make [my] children understand that if they saw a suspicious package, they had to tell me right away. They thought this was very exciting! They didn't realize that it was also very dangerous."
Fortunately, Christiane's apartment was never blown up, and she celebrated when Boupacha was granted amnesty, after France signed the Évian Accords, which gave Algeria independence, in Évian-les-Bains, on March 18, 1962. Boupacha was freed one month later — the same month that 90 percent of France voted to approve the accords in a referendum.
Jacqueline, who had worked for her brother when he was a minister, later became secretary-general of the UNESCO Clubs, which gave her a forum to lobby for the interests of the developing nations of Africa, Asia, and South America.
In the 1960s, Christiane turned her attention to women's issues. Since 1920, French law had banned abortion and all information about contraception. Under the banner of _"Si je veux, quand je veux"_ (If I want, when I want), Christiane campaigned for the legalization of both. In 1967, the Neuwirth Act finally legalized the sale of contraceptive devices, but abortion wasn't legalized until 1975, and advertisements for contraceptives remained banned until 2001, four decades after the pill was introduced in America.
Christiane also became a sought-after family therapist, who continued to see patients well into her seventies. "As you can see," she said, "we never lost the spirit of the Resistance."
IN 1978, André Boulloche was approaching the pinnacle of his political career. Twenty years earlier, he had shown his independence by accepting an appointment as de Gaulle's minister of education, even though the Socialist Party had voted against participating in de Gaulle's new government. But less than a year later, André had quit over a dispute about government aid to parochial schools. André was against it, while de Gaulle and his prime minister, Michel Debré, were for it. After André voiced his objection, the president beseeched him to stay on, but his minister resigned anyway.
His tenure was marred by harsh publicity when his marriage blew up while he was minister. It led to an ugly divorce and a trial to determine the custody of his three children. This was especially difficult, because divorce was still very much a scandal in Catholic France in the late 1950s (as indeed it was for any American politician in this period).
Less than four months after his divorce, he married the beautiful Odile Pathé, the daughter of Charles Pathé, a founder of Pathé pictures, the legendary French film company. Odile had first met André immediately after the war, and she had her own book-publishing company. When all the left-wing publishers in Paris ignored George Orwell's _Animal Farm_ because it was a barely veiled attack on Stalinism, Odile rushed to London and secured the book for her publishing house, after two memorable lunches with the author. _Les Animaux Partout_ was published by Éditions Odile Pathé in 1947.
President Charles de Gaulle at a ceremony commemorating the Resistance. André Boulloche, his education minister, is at the far right.(photo credit 1.25)
Following his resignation as de Gaulle's minister and his new marriage, André gradually rebuilt his career with the hyperactivity that was his trademark. His children ended up living most of the time with their mother and saw their father only on alternate weekends. The children sometimes found their father terrifying. His experiences in three concentration camps were not without their consequences.
"He had a terrible violence in him which he contained," said Jacques Boulloche, his youngest son — named, of course, for André's dead father. "But sometimes it got out of hand. And when he was driving, he was a tyrant. I was terrified. It was quite something. He raced with everyone — he would not allow himself to be passed. It was crazy. And then, when he got to [his constituency in] Montbéliard, it was the opposite. As soon as someone was trying to cross the street, he stopped. It was unbelievable. Because, you know, he would say, 'That's a voter!' It was very funny. He was the sort of person who could not allow any car to go faster than he was going."
He could also be a violent disciplinarian with his children — and his dog. He was "very rough" with Jacques, who had been a poor student when his father was minister of education — a record his father considered a personal embarrassment. "I got terrible, heavy spankings," Jacques said.
"On the other hand, when I was bigger, I had a lot in common with him through science. I have very lovely memories of when I was at his house in Montigny-sur-Loing. I had a little chemistry lab in the basement, and I did chemistry experiments. And every weekend, he would ask me, 'What do you need?' Then during the week there was a store next to the National Assembly which sold stuff for chemistry labs, and he would go there and buy me whatever I needed. That was really nice. That's how he encouraged my scientific spirit." So despite the beatings he had suffered as a child, Jacques [ _fils_ ] admired his father enough to sustain another ancient family tradition — he named his son André.
"My father was above all scientific," Jacques told me. "I don't know why he went into politics, because he wasn't someone who knew how to communicate at all. He was a very hard worker. He was a technician, and he was very methodical. When he had a file, he analyzed it. He was, first and foremost, a _polytechnicien._ "
But with hard work, André Boulloche methodically overcame his lack of the common touch. In 1965, he was elected mayor of Montbéliard, a small city on the eastern border of France next to Switzerland and Germany. In 1967, he added the position of deputy in the National Assembly. With two full-time jobs, he was constantly in motion between Paris and Montbéliard; he told a national news magazine that the secret to his success was a "good airplane." Two years later, he was a member of the directors' committee of the Socialist Party and vice president of the Socialist group in the National Assembly.
By 1978, André was mayor of Montbéliard, a deputy in the National Assembly, and a rising star in the Socialist Party.(photo credit 1.26)
These were remarkable achievements, especially because there had never been any warmth between him and François Mitterrand, the leader of the Socialist Party. Boulloche had been much closer to Mitterrand's earthier predecessor, Gaston Defferre, the celebrated mayor of Marseille. Raymond Forni, a young friend and colleague in the National Assembly, remembered his mentor's attitude toward Mitterrand this way: "Boulloche was certain that he was working for France — and Mitterrand was working for himself." But despite the coolness between them, Mitterrand recognized that Boulloche was a man to be reckoned with. By 1978, André had become his party's chief spokesman on economic affairs, and most people expected Mitterrand to make him the first Socialist finance minister since the war, if Mitterrand was elected president in 1981.
But a fateful plane trip during the 1978 parliamentary campaign changed all that.
FRANÇOIS MITTERRAND was invited to participate in a debate at Saint-Dié-des-Vosges by Christian Pierret, a thirty-two-year-old Socialist candidate for the National Assembly there.ǁ Pierret's district adjoined André's. When the party leader was too busy to participate in the debate, André Boulloche stepped in for him at the midday gathering in the provincial city in the Vosges Mountains.
André's noontime appointment was less than a hundred miles from his own city of Montbéliard. But if he took a regularly scheduled airline, he wouldn't be able to get back in time for his own campaign meeting that evening. So he asked his secretary to arrange another way to get there. Christian Pierret offered him an air taxi. "I don't like to take those little planes," André told his secretary. "But if I have to, I will."
The weather was calm, with ten miles of visibility, when André took off at two thirty in the afternoon of March 16 from the airstrip outside Saint-Dié-des-Vosges. He was alone with the twenty-three-year-old pilot, Renaud Mary, who was the son of the president of the commuter line they were traveling on.a Now André was returning to his own adopted city of Montbéliard, which was only a half an hour away by plane. He expected to address a thousand constituents at his own political meeting that night.
As he had all his life, the sixty-two-year-old politician was pushing himself as hard as he could. The first round of the election had gone badly for the Socialists, and this was no time to let up. The second, decisive round of the election was only four days away.
Montbéliard was the city that Boulloche had "parachuted" into in 1962 to create a new political base for himself. He had chosen it partly because it was a region of liberal Catholics and devout Protestants. Boulloche was a very liberal Catholic himself (none of his children had even been baptized) and his new constituents quickly embraced him. Three years after his arrival in Montbéliard, he was elected mayor.
His political accomplishments were the result of pure determination. A technocrat who had been an engineer of bridges and highways before the war, Boulloche didn't start out with any aptitude for politics. He certainly didn't fit the profile of a typical politician: One reporter wrote that the man with the gray crew cut and a somber countenance looked like a "secular monk." He once described politics as "an infernal life." But this was also someone who brought an unbreakable will to bear on everything he did.
"Are you ever discouraged?" a radio reporter asked him in 1976.
"Yes, but not very often."
"What do you do when that happens?"
"I wait for it to pass."
When he first arrived in Montbéliard, Boulloche had bonded with the workers of Peugeot, whose factories employed thirty-seven thousand workers in the region. His constituents recognized him as a gifted administrator. After thirteen years at City Hall, his adopted city boasted an improved public transportation system, a bustling cultural life, a new sanitation system, and a rebuilt city center.
After his first decade as mayor, André declared, "I think I can say without exaggeration that my team has completely transformed the place. People who come back after being away for fifteen years don't even recognize it." Now, whenever there was an election, the popular Socialist had the luxury of being able to spend much of his time assisting the campaigns of his less-well-established friends and allies.
AS THE LITTLE AIRPLANE headed south, it followed a flight path parallel with the Rhine. Suddenly, Boulloche's preoccupation with the election was replaced by a more palpable danger. Half an hour into the flight, cruising at 150 miles an hour, the small aircraft started bouncing in high winds. Without warning, they were at the center of a violent winter storm. As they approached Montbéliard, snow and sleet blanketed the windscreen and hail rattled the cockpit, cutting their visibility to less than three hundred yards. The provincial airstrip at Montbéliard had no radar, and it waved them away. A policeman on the ground spotted them circling overhead; later he remembered that it had looked as if they were searching for the proper path.
From there they headed for Belfort-Fontaine, which was ten miles away. Raymond Forni, who represented a neighboring constituency, spotted the red-and-white plane when it was trying to land at the second airport. But the storm was fierce there too, so they decided to make for Basel-Mulhouse, an international airport equipped with radar for all-weather landings. It serves Basel, Switzerland, and Mulhouse, France. By now they had been flying for nearly two hours. Boulloche had been trained as a pilot in Morocco after the war, but he never got to fly because of a shortage of planes. Now he climbed into the copilot's seat to try to help guide the plane to safety.
It was four twenty-three in the afternoon. The weather had cleared up at Montbéliard, but the pilot and his passenger didn't know that, and the storm had followed them to Basel-Mulhouse.
"I can't come down — I'll manage by myself."
Those were the last words the tower at Basel-Mulhouse heard from the plane's young pilot. As he headed east toward Frankfurt, air traffic controllers watched the plane disappear from their radar screens. The Piper started losing altitude because of the weight of the ice on its wings. At the same moment, its radio stopped working, because the antenna had been torn off by the storm.
A MOMENT LATER, buffeted by severe winds, or disoriented by the storm, the pilot changed direction by forty-five degrees — and slammed into the side of the Hochblauen, twenty-three hundred feet up the side of the thirty-eight-hundred-foot mountain.
This unconquerable man — the one who had survived three German concentration camps — had crashed in the Black Forest of Germany.
The little Piper was demolished, its shattered shell suspended upside down from the branches of a tree. Boulloche and the pilot were ejected from the plane at the moment of impact. They tumbled out of the cockpit into a foot-deep cushion of snow. The pilot's face was smashed in, and André had a punctured lung. They were dazed and bleeding and suffering enormous pain. And yet, somehow, they were still alive.
When they realized they were able to walk, they began to limp down a path leading into the valley. They were assaulted by sheets of wet, white snow dropping out of the black sky. After struggling toward the valley a few hundred yards, they reached a shed built of logs, which had been erected to shelter summer tourists.
There they stopped to rest. Two miles away in the distance, they could hear a faint church bell, chiming five o'clock in a nearby village.
BACK IN PARIS, it was Boulloche's secretary, Andrée Vauban, who was the first to raise the alarm. André had told her that he would be back in his office in Montbéliard by four thirty in the afternoon, and he would call her when he arrived. When she hadn't heard from him by five o'clock, she called Montbéliard to check up on him.
"The plane must have left late," Boulloche's aide in Montbéliard told her. The aide seemed unperturbed, but Vauban was immediately suspicious. Her next call was to Christian Pierret, the candidate in Saint-Dié-des-Vosges for whom Boulloche had been campaigning earlier that afternoon. Pierret told her that Boulloche had taken off on time. Then his secretary made another call to Montbéliard to inquire about the weather there.
"Snow! Terrible wind! Hail!"
Andrée Vauban immediately sensed that her boss was in danger. _Something is happening,_ she said to herself. _He should have arrived by now._ Her next call was to the Ministry of the Interior. Then she contacted Odile. Mme. Boulloche rushed off to Le Bourget to catch the last plane of the night to Belfort-Fontaine, one of the airports where André's plane had been unable to land a few hours earlier. As soon as she arrived, this formidable spouse started phoning civil and military officials to get them to expand the search.
André's younger son, Jacques, was now a twenty-four-year-old medical student in Paris. He was driving across the city when he heard a radio report that his father's airplane had disappeared. His first thought was that his father might have been the victim of some kind of attack.
When André's sister Christiane returned home to her Paris apartment on square Alboni, in the 16th arrondissement, it was her husband, Jean, who greeted her with the disquieting news: "André's plane hasn't arrived back in Montbéliard."
That was all they knew.
That night, François Mitterrand, the leader of the Socialist Party, issued a statement:
This is a very remarkable man — remarkable among the remark-ables — greatly loved by all who know him. I feel great pain and great concern. When one says those two words, "André Boulloche," they are a salute to someone quite exceptional, on every level. He is an exemplary man.
German and French helicopters swarmed above the Black Forest in search of the plane on Thursday night, but darkness and the snowstorm prevented them from finding anything. It was the following afternoon when a search party found the bodies of André and Renaud Mary, a hundred yards apart, down the hill from their fallen plane. Apparently the young pilot had kept searching for help after André could go no farther.
Their bodies were found by French troops of the 12th Regiment stationed in Mulheim, Germany. Years later, Jacqueline's daughter, Claudine Lefer, had a clear memory of her mother's reaction to the news: "And on top of everything else, the plane crashed in Germany."
FOUR DAYS AFER HIS DEATH, André's body lay in state in the Montbéliard City Hall, surrounded by a silent honor guard of local council members and city officials. "Men and women of every class and every age passed his coffin, sometimes depositing a small bouquet of flowers," the local newspaper, _L'Est Républicain_ , reported. "One had the impression of the entire city parading by, like a river flowing slowly and majestically. There was an extraordinary impression of grave sadness, but above all, of dignity, which was symbolic of the particular influence of André Boulloche, who had attracted such affection and respect." Seven special airplanes were flying dignitaries in for the funeral on March 21, plus an eighth for François Mitterrand.
At 9 o'clock in the morning, André's wooden coffin was taken from City Hall by six white-gloved firemen, followed by an army officer holding a pillow displaying the dead man's four most important decorations — Commandeur de la Légion d'Honneur, Compagnon de la Libération, Croix de Guerre, and Médaille de la Résistance. Outside, the firemen wrapped the coffin in the French Tricolor, then placed it in the center of a fire truck and covered it with a glass enclosure. As rain began to drench the slow procession, umbrellas of every color bobbed up and down above the huge crowd that followed the peculiar red funeral coach. The mourners included four delegations of deportees who had survived the German camps.
The funeral itself took place at the Fairgrounds Hall of Montbéliard. Because it could accommodate only three thousand people, an outdoor sound system was installed for thousands more who wanted to listen outside.
There were half a dozen speakers, including Mitterrand, André Postel-Vinay and, remarkably, two Germans: Volker Hauff, a prominent Social Democrat who was minister of science and technology, and Otfried Ulshöfer, the mayor of Ludwigsburg, the German twin city of Montbéliard.
Every speaker described the main postwar preoccupation of this survivor of three German concentration camps: André's unstinting efforts to foster friendship and reconciliation between France and Germany. In a remarkable act of intellectual jiu-jitsu, André took all of the ghastly energy from his wartime incarceration and turned it around to make sure that nothing like what had happened to him would ever happen to another Frenchman or German again.
In his funeral oration, Mitterrand, the future president of France, declared that André had wanted to "sublimate his sufferings, to give them meaning beyond this moment in history. It was as if he had found the capacity within Europe to construct peace and harmony among all peoples... He considered reconciliation with Germany a necessity. When he turned toward the Germans, he was the first among us who knew how to say, 'My friends.' "
Otfried Ulshöfer quoted André's speech from three years earlier, when the two mayors had celebrated the twenty-fifth anniversary of the twinning of their two cities. "Today the two of us maintain the flame of friendship between us. Tomorrow, others will have this duty, and I am convinced that they shall not fail."
Then the German mayor spoke for himself: "Today we consider that sentence as the duty he has left us. Now we must endeavor to carry it out."
* My uncle gave Jacqueline a bottle of expensive perfume as a wedding present, probably from the Army PX. It was dropped, and shattered — a small tragedy he still remembered decades later.
† André's second wife, Odile, added another reason for his conversion to socialism: "It was the feeling that if there is no solidarity, life is absolutely not possible." (author's interview with Odile Boulloche, March 20, 1999)
‡ Jacqueline's older son, Eric Katlama, felt the same way as Ophuls: "I think what de Gaulle did is fairly unforgivable, having sustained this kind of myth about a France united against its invaders, without really wanting to make the necessary effort of memory."
§ It was brought to America by Woody Allen and shown at the New York Film Festival in September 1971. It opened at the Beekman Theatre in Manhattan the following March.
ǁ Pierret was elected to the National Assembly that year and remained there for fifteen years.
a Although he was only twenty-three, the pilot had been flying since he was sixteen.
# _Twenty_
_Courage is more exhilarating than fear and in the long run it is easier. We do not have to become heroes overnight. Just a step at a time._
— Eleanor Roosevelt
_Don't make it too sad._
— parting words from André's son Jacques, after I interviewed him about his father
AFTER THE SHOCK of his death, Christiane and Odile went to work right away to assure André's legacy. Odile drew on her experience as a book publisher to produce two beautiful volumes, one illustrated, one not, with extraordinary stories from André's closest friends about every phase of his remarkable life, from his time at the lycée, to his life as de Gaulle's secret agent in Paris, to his survival at three concentration camps. Then there were his three interlocking postwar careers: the brilliant mayor of a provincial town on France's eastern border, a force to be reckoned with for more than four decades in the capital's corridors of power, and an indefatigable advocate of reconciliation between France and Germany, and the unity of Europe. One book was privately published; the other was a special issue of the municipal review of Montbéliard.
This book would not have been possible without dozens of contributions from those two volumes. André Postel-Vinay's memoir, _Un fou s'évade_ (A Fool Escapes), published in 1996, was another crucial resource for me. But I still needed one more actor's full cooperation before I could write my own account.
Ever since I became a reporter for the _New York Times_ at the age of twenty-four, and probably even before that, I knew that the Boulloche saga could be the most extraordinary story I would ever tell. I had been mesmerized at eleven when I first heard it recounted by my uncle Henry, and I have never been less than mesmerized ever since. Meeting the main actors when I was eleven — and falling in love with one of them — only made me more eager to write about them.
During many visits to France over the next three decades, I often shared my ambition to write this book with Christiane's children, especially François and Noëlle. But we agreed it would be impossible for me to do so as long as Jacqueline and Christiane maintained their silence.
Two more tragedies were necessary before the floodgates could open, even a little bit. In 1989, Jean Audibert, an exceptionally vigorous sixty-eight-year-old, died suddenly of a massive stroke, making Christiane a widow. Four years later, Jacqueline received a fatal diagnosis, and this time it was not a false alarm. She had leukemia, she was too old for a bone-marrow transplant, and she died one year later at the age of seventy-six.
Now Christiane was the unmistakable head of her family, the indomitable matriarch. The death of her sister acted as a release mechanism for her. For fifty years, she had considered her secrets too fraught to share with her children, because of the horrors suffered by her parents and her brothers.
Suddenly, she felt just as compelled to tell the story as she had felt required to remain silent about it. Realizing that it would disappear if she failed to record it, she forced herself to write a forty-five-page memoir — "for my grandchildren." Once again, Christiane acted out of a sense of obligation.
"It was obvious," she told me, using the same words she had used to describe her decision to join the Resistance. She had never wanted to write this book, but, at the age of seventy-one, she had to. "It was extremely painful for me to relive these black years. But it was also my duty."
With the help of Mathilde Damoisel, a brilliant young history student at the Sorbonne whose specialty was women in the Resistance, Christiane produced an amazing narrative. Mathilde described it as an "homage to the spirit of her family."
When Christiane was writing her book, her older daughter, Catherine, visited her every Monday evening. "She would read me what she had written the previous week," Catherine remembered. "And she cried, and she cried and she cried."
At the end there was still a great deal missing: Christiane's emotions were almost completely absent from her pages. The essential facts were all that she could manage. But when Catherine told her it was "too dry," Christiane refused to change her approach. "No," Christiane told her daughter. "I don't want to. It's enough this way. And there are things that I won't say. That I don't want to say and that I will not say. So I'm putting in what I want to put in, and that's all."
When I visited Paris again at the end of the 1990s, Christiane's younger son, François Audibert, met me at the Gare du Nord. He greeted me with the startling news of Christiane's book. When I reached his house, I devoured it all in a single sitting. Then I embarked upon my own, relying heavily on Christiane's for guidance.
I SPENT TWO AND HALF YEARS living in France, interviewing all of the surviving Boulloches, as well as many others who had worked with them or hidden them during the Occupation. An oral history of the Resistance in the French National Archives included the accounts of many others who had known the Boulloches during the war. At my request, the Public Record Office in London declassified all the files MI5 had compiled on André Boulloche, Alex Katlama, and Charles Gimpel when they were in Britain during the war.
Remarkably, everything I learned in the British and French archives confirmed and elaborated upon everything Christiane had told me.
Neither Christiane nor I enjoyed my efforts to force her to reveal as much as possible. But she never refused any request. She also urged everyone in her family to cooperate with me, and everyone did. One thing in particular surprised me. I hadn't expected to share any of Christiane's ambivalence about unearthing her secrets. But very gradually, I realized that it was also painful for me to part with the black-and-white version of her family's heroism that I had grown up with.
ALTHOUGH THE COST of their courage was gigantic, there are more triumphs than tragedies in this story. It is true that the _Résistants_ in the family conveyed a certain malaise to many of their children by never talking about war. But I don't believe they could have made any better choices about how to deal with their history. And despite their private agony, they managed to transmit all of the finest values of the Boulloches, the Audiberts, and the Katlamas.
Christiane had four fabulous children, twelve grandchildren, and seven great-grandchildren. All of them became righteous, rigorously informed, and deeply committed citizens of France, as did their cousins, and their offspring.
"We did not talk about the Resistance in terms of what each of them had done," said Michel Katlama, Alex and Jacqueline's younger son. "But I think the whole generation that followed was enormously marked by that. The whole family was extremely conscious of its political choices. Never Communist. But it was a family that always voted for the left. Almost everyone. The Audiberts and the Katlamas all voted for the left. And it wasn't just an accident. I can't imagine any member of our generation being anti-Semitic. Or a Fascist. Or not a democrat — and very attached to democracy."
WHEN CHRISTIANE had finished her book, she summoned all her children and grandchildren and nephews and nieces to her grand apartment in Passy.
"Christiane said she had done what she had to do," her niece, Claudine Lefer, remembered. "Because she was the last person who could tell this story. She didn't give a speech, she said it in tête-à-têtes with small groups of us."
Christiane's granddaughter, Hélène Dujardin, believed that "she had done her duty. And in the end — I'm going to say something terrible — she could leave now, knowing that she had done what she had to do. That touched me enormously."
I repeated those words to Christiane.
"Yes," she said. "That's true. Absolutely true."
From time to time, I suggested that Christiane's survival had been her destiny, rather than just the product of good fortune. A woman with no taste for superstition, she mostly turned away the idea that fate had played any role in her longevity. But in our final interview before I moved back to New York, she hedged a bit. "I was born on November 11, 1923, in Paris in the seventh arrondissement, during the minute of silence — at 11 a.m. on November 11th. That's what they always told me."
Then she laughed at the idea, but warmly: "So perhaps I was a little predestined."*
Christiane receives the galley of this book from the author. Square Alboni, December 23, 2014.(photo credit 1.27)
As I write these words, she is still flourishing in the elegant apartment she moved into with Jean Audibert and their children in 1958, the one where I first met her in 1962. She is on her own, but a steady stream of children, grandchildren, and great-grandchildren come to bask in her warmth and her wisdom. She and I remain in constant touch, by e-mail and on the telephone.
She still remembers everything.
* World War I had ended at the 11th hour of the 11th day of the 11th month of 1918; the minute of silence is in memory of its dead.
# _Afterword_
MOST AMERICANS are smugly dismissive of the way the French behaved during the Nazi Occupation. "Was there one?" That was the question I was asked most often — even by intelligent people — whenever I mentioned that I was writing about the French Resistance.
That reflexive condescension is coupled with a popular amnesia about the German sympathies of famous American appeasers, from Charles Lindbergh to Joseph P. Kennedy. Equally forgotten are the anti-Semitic organizations that flourished in the United States in the 1930s. The pro-Nazi German-American Bund counted eight thousand storm troopers among its members, and filled New York City's Madison Square Garden at the beginning of 1939 with twenty thousand supporters shouting, _"Heil Hitler."_
The bund worked closely with the Reverend Charles Coughlin's Christian Front. In the 1930s, Coughlin was one of the nation's most influential broadcasters, and his supporters organized Buy Christian rallies across the country. After the Nazis looted Jewish stores and burned down synagogues all across Germany in November 1938, Coughlin even defended the horrors of Kristallnacht on his national radio show, describing them as appropriate retaliation for Jewish persecution of Christians.
THE TRUTH is, there were hundreds of thousands of French men and women like the Boulloches who risked everything to liberate their country from the Nazis, while Americans at home never had to risk anything the way the French did during World War II. American servicemen and women made gigantic sacrifices, from Normandy to Iwo Jima, to free the world from the tyranny of Germany and Japan. But American civilians, living thousands of miles from the battlefields, never faced anything remotely resembling the choices that confronted everyone who lived in Nazi-occupied Europe.
However, these facts are not the main reason I cannot judge France harshly for its behavior during World War II. To me what is most persuasive is the attitude of the two men who did more than anyone else in the 1960s and the 1970s to bring about a more balanced view of France's record: Robert Paxton and Marcel Ophuls.
Both men understood that if you have never actually faced life-and-death decisions, it is easy to assume that you would have done the right thing if the Nazis had occupied your country. It is also a great mistake to do so.
In his brilliant book _Vichy France,_ Paxton wrote that "an American reader who honestly recreates the way the world looked from France [in 1940] cannot assume that he or she would easily have found the path to a 1944 hero's role." And Ophuls told me that former British prime minister Anthony Eden was also speaking for the filmmaker when the statesman made this crucial observation at the end of _The Sorrow and the Pity_ : "If one hasn't been through — as our people mercifully did not go through — the horror of an occupation by a foreign power, you have no right to pronounce upon what a country does which has been through all that."
That is one of the most important and least understood lessons of World War II.
# ACKNOWLEDGMENTS
_THE COST OF COURAGE_ would never have been written without the help of dozens of enthusiastic collaborators.
Although she was extremely reluctant at the beginning, Christiane has been unfailingly helpful at every stage of my research, and my life. Her children, grandchildren, nephews, and nieces all followed her lead. François Audibert, Noëlle Audibert, Catherine Dujardin, Pierre Audibert, Eric Katlama, Michel Katlama, Claudine Lerer, Robert Boulloche, Jacques Boulloche, Hélène Katlama, Hélène Dujardin, Laurence Dujardin, and Stephane Dujardin all shared their knowledge and their memories. Catherine Dujardin also provided dozens of news stories about her uncle André's untimely death.
Agnès Boulloche and Odile Boulloche shared hundreds of photographs, news clippings, and diary entries, some of them more than a hundred years old. Like Christiane, Odile offered her friendship, her affection, and her intelligence, and she did everything she could to make sure I got the story right. But any errors that crept into the manuscript are mine alone.
When we lived in Paris, Joe and I formed cherished bonds with Vincent Demongeot, Mark Trilling, Meredith Artley and Naka Nathaniel, Robert and Barbara McCartney, Jeffrey and Casey O'Brien Blondes, Alan Riding, Pascale Belzacq, David Tanis and Randal Breski. Laure de Gramont is the best friend anyone has ever had in Paris: she opened her home and shared her friends, and all of her passions. Laure and her sisters, Claire Shea and Isabelle de Lastours, also loaned me their amazing country house so that I could begin to write in solitary splendor. Andrew Jacobs, Dan Levin, Tom Donaghy, and Shaffiq Essajee were equally generous with their Napanoch paradise.
Bob Paxton and Marcel Ophuls freely offered their time and their wisdom. Paxton's extraordinary books and Ophuls's brilliant movies remain the essential touchstones for anyone writing about this period. Claire Andrieu is one of the great French historians of her generation. She became my good friend and my essential guide through the French National Archives and dozens of other mysteries of wartime and postwar France. She also gave her blessing for me to borrow long sections from her father's extraordinary memoir, _Un fou s'évade._
Early readers of the manuscript who spurred me on and offered dozens of corrections included my great friend Rick Whitaker, Janet Suzman, Victor Gurewich, Eugene Gregan, Frank Rich, Rick Hertzberg, Michael and Laura Kaiser, Peter Duchin, Judy Barnett, Walter Isaacson, Nick Rostow, Lisa Chase, Sarah Burke, Russell T. Davies, and Rebecca Kaiser Gibson. Since I first started writing books thirty years ago, Sal Matera, Stephanie Lane, Mark Polizzotti, and Renata Adler have been my essential readers and magnificent supporters.
Mathilde Damoisel was Claire Andrieu's student when she became Christiane's assistant on her memoir, and her interviews with Christiane were a source of dozens of insights. Now a brilliant documentarian, she made more contributions to this effort than I could possibly enumerate.
Steve and Nancy Shapiro, Sal Matera and Ann Jensen, Rich Meislin and Hendrik Uyttendaele, Beverly and Eugene Gregan, Katie Hustead, Rick Whitaker and Javier Molina, Michael Finnegan, Judy Knipe, Alice McGillion, Michelle Clunie, Jean Graham, Paul Goldberger and Susan Solomon, Bryan Lowder and Cam McDonald know that their importance in my life defies definition. The empathy and insights of Gerald Dabbs have sustained me for almost two decades.
I am especially grateful for constant encouragement from Eric Gelman, Maralee Schwartz, Steve Weisman, Mary Stouter, Linda Amster, David Korzenik, Syd Schanberg, Steve Marcus, Henry Bloomstein, Jackie Green, Hope Kostmayer, Craig Zadan, Heyden White Rostow, Will Parker and Stephen Schwalen, Lynn Goldberg, Gordon Wheeler, Peter Wittig and Huberta von Voss-Wittig, Ben Wheeler and Kate Cortesi, Nick and Judgie Graham, David Whitaker, Sam and Andi Shapiro, Roger and Kay Greeley, Steve Kay, Cyd Savage, Martha Fay, David Dunlap, Mary Murphy, Thatcher Barton, Judy Hottensen, Virginia Cannon, Wolf Hertzberg, François Fortin, Michael Butler, Tomas van Houtryve, Tree Adams, Ben Golberger, Mel Rothberg, Alex Goldberger, Anna Wainwright, Kirk Semple, Frank Clines, Alison Mitchell, Shelley Wanger, Steve Rattner, Steve Adler, Bill Carey, Jimmy Hayes, Peter Goldman, Margo and Garth Johnston, Arabella Kurtz and Nick Everett, Lucy Howard, and Zarrina and Antony Kurtz.
Arthur Gelb was my first newspaper editor and my lifelong friend. It is a great sadness that he did not live to see this book published, but we continue to cherish our friendship with his wonderful wife, Barbara.
It has been my great good fortune to have been born into a family of brilliant writers and voracious readers, starting with my magnificent parents, Hannah and Phil Kaiser. My thanks and love go to Charlotte, Emily, Dan, Tom, Josh Thelin, Nick Peterson, Lauren Langlois, Linus Peterson, David and Bob, Hannah and Patti, Tema, Mark, Abe, Ezra and Isaac Silk, Moss and Adelaide Kaiser, Tamara Kaiser, and Sarah Hyams.
This book would never have come about if I hadn't fallen in love with the story through its telling by my uncle Henry, known to most as Husky. He and my uncle Jerry Kaiser were two of the greatest men I have ever known.
Everyone at Other Press has been a pleasure to work with, especially Keenan McCracken, Yvonne Cárdenas, Bill Foo, Jessica Greer, Terrie Akers, Charlotte Kelly, and Iisha Stevens. Kathleen DiGrado found a brilliant image and designed a beautiful cover; Julie Fry worked the same sort of magic on the interior of the book.
All my life I have been looking for the perfect professional collaborator. When Judith Gurewich sat down next to me at a lunch she was hosting three years ago, I finally found her. She transformed my life. In short order she became my editor, my publisher, and my very close friend. Her warmth, her brilliance, and her passion are the greatest gifts an author could aspire to.
After thirty-six years my husband, Joe Stouter, continues to astonish me every day with his love, his imagination, and his art. It is hard to imagine feeling even closer to each other after nearly four decades, but we do. I have never done anything important without him.
# PRINCIPAL ACTORS
THE BOULLOCHE FAMILY
_The parents_|
---|---
Jacques Boulloche (b. 1888)| Director, French Bureau of Highways
Hélène Boulloche (b. 1888)| Wife of Jacques Boulloche
_The children_|
Robert Boulloche (b. 1913)| Inspector, Minister of Finance
André Boulloche (b. 1915)| _Résistant_ ; Charles de Gaulle's personal representative in Paris, 1943–1944; arrested by the Gestapo January 12, 1944
Jacqueline Boulloche (b. 1918)| _Résistante_
Christiane Boulloche (b. 1923)| _Résistante_
_The spouses_|
Alex Katlama (m. 1946)| _Résistant_ , Jacqueline's husband
Jean Audibert (m. 1947)| Christiane's husband, member of the Free French Navy
Anne Richard (m. 1949)| André's first wife, mother of his three children
Odile Boulloche (m. 1959)| André's second wife, George Orwell's French publisher ( _Animal Farm,_ 1947)
_The postwar generations_|
Eric Katlama (b. 1948) Michel Katlama (b. 1950) Claudine Lefer (b. 1953)| Children of Jacqueline and Alex
Catherine Dujardin (b. 1948) Noëlle Audibert (b. 1949) Pierre Audibert (b. 1951) François Audibert (b. 1957)| Children of Christiane and Jean
Robert Boulloche (b. 1949) Agnès Boulloche (b. 1951) Jacques Boulloche (b. 1953)| Children of André and Anne
Hélène Dujardin (b. 1979)| Daughter of Catherine and Hubert Dujardin
OTHERS|
Etienne Audibert| Fellow prisoner of Jacques Boulloche at Buchenwald, father of Jean Audibert
Harold Cole (b. 1906), code name "Paul"| Resistance member who became a notorious double agent for the Gestapo after he was arrested on December 6, 1941. Betrayed André Postel-Vinay on December 14, 1941; killed by a French police inspector in January 1946
Albert-Marie Edmond Guérisse (b. 1911), code name "Patrick O'Leary"| _Résistant,_ colleague of Postel-Vinay; father of the "O'Leary line," which rescued 600 allied pilots
André Postel-Vinay| Inspector, Finance Ministry; recruited André Boulloche for the Resistance in 1940; arrested by the Gestapo on December 14, 1941; escaped September 3, 1942
André Rondenay (b. 1913)| _Résistant_ ; classmate of André Boulloche who succeeded him as de Gaulle's military delegate in Paris in January 1944; arrested by the Nazis on July 27, 1944; executed on August 15, 1944
Henry and Suzanne Rollet| _Résistants_ who assisted André Postel-Vinay in his escape from France.
Bernard Vernier-Palliez (b. 1918)| _Résistant_ ; later CEO of Renault; French Ambassador to the USA, 1982–1984
GOVERNMENT/MILITARY|
Winston Churchill| British Prime Minister, 1940–1945 and 1951–1955 First honorary citizen of the United States, 1963
Charles de Gaulle| Leader of the Free French President of France, 1959–1969
Dwight Eisenhower| Supreme Commander, Allied Forces in Europe President of the United States, 1953–1961
Adolf Hitler| _Führer_ of German Third Reich, 1934–1945
François Mitterrand| Elected first secretary of the Socialist Party, 1971 President of France, 1981–1995
Henri-Philippe Pétain| French hero of World War I Chief of state of Vichy France, 1940–1944
Franklin Roosevelt| President of the United States, 1933–1945
Dietrich von Choltitz| Final German commander ( _Befehlshaber_ ) of German forces in occupied Paris, who refuses to carry out Hitler's order to blow the city up
Claus Schenk Graf von Stauffenberg| Leader of the plot to kill Hitler in July 1944
# NOTES
"If mankind lasts": Ophuls, _The Sorrow and the Pity,_ introduction by Stanley Hoffmann, pp. viii–ix.
"This is such": Author's interview with Eric Katlama March 13, 1999.
"it was necessary to turn the page": Boulloche-Audibert, _Souvenirs._
"If an old pair of shoes": Ophuls, _The Sorrow and the Pity,_ p. 71.
They are called _gazogènes_ : Collins and Lapierre, _Is Paris Burning?,_ p. 15.
a good bicycle can cost: Eparvier, _À Paris sous la botte des Nazis._
"taxis hippomobiles": Ophuls, _The Sorrow and the Pity,_ p. 71.
The fastest pedicab: Collins and Lapierre, _Is Paris Burning?,_ p. 16.
four men pedaling: Ibid., p. 18.
Huge yellow posters: Permanent exhibit, Paris Museum of World War II.
After four years of war: Churchill, _Second World War,_ V:ix.
Deeply religious: Interview with Raymond Jovignot, conducted by Melle Patrimonio, February 1, 1946, French National Archives, box 72AJ42.
Christiane's clandestine duties: Boulloche-Audibert, _Souvenirs._
"We wouldn't just resist them": Author's interview with Christiane Boulloche-Audibert, Fontainebleau, March 19, 1999.
hypnotized and horrified: Mathilde Damoisel's interview with Christiane Boulloche-Audibert, February 3, 1997.
There is no heat: Ibid.
Four months earlier: Madame Grenlet's interview with André Boulloche, February 1, 1950, French National Archives, box 72AJ68.
In the fall of 1942: Cobb, _Resistance,_ p. 160.
The STO requires _and_ Faced with the prospect: Ibid., p. 162.
As one historian put it: H. R. Kedward, _In Search of the Maquis: Rural Resistance in Southern France, 1942–1944,_ quoted ibid., p. 161.
"I never felt": Christiane Boulloche-Audibert interviewed by a French television reporter, March 22, 1999, on the occasion of the unveiling of a bust of André Boulloche.
During his brief time: British Intelligence file on Eric Katlama.
"Strangle me": Author's interview with Dr. René Cler, Paris, March 17, 1999. Cler told me he had heard this story from Jacques's cell mate. He no longer remembered his name.
Instinct propels the Frenchman: His son Jacques said, "I was told that one of them had a submachine gun, and he threw himself on it." Author's interview with Jacques Boulloche at this home in Le Havre, February 1, 2004.
Jacqueline performs the secret knock _and the rest of this section_ : Boulloche-Audibert, _Souvenirs._
"Dignity is incompatible with submission": Letter from André Boulloche to C. Hettier de Boislambert, Grand Chancelier de L'Ordre de la Libération, June 13, 1969, French National Archives, box 72AJ2056.
_"When will I see them again?":_ Boulloche-Audibert, _Souvenirs._
"In general," Christiane recalled: Author's interview with Christiane Boulloche-Audibert, March 19, 1999.
Her parents don't consider themselves: Author's interview with Christiane Boulloche-Audibert, March 11, 1999.
Those casualties transform: Paxton, _Vichy France,_ p. 12.
"To continue an enforcement": Cooper, _Old Men Forget,_ p. 199.
"Politically Czecho-Slovakia": _Manchester Guardian,_ October 1, 1938.
After "immense exertions": www.winstonchurchill.org/learn/speeches/speeches-of-winston-churchill/101-the-munich-agreement.
At one o' clock: _Argus de la Presse,_ Paris, August 15, 1900.
As early as 1937: Author's interview with Christiane Boulloche-Audibert, March 19, 1999.
"A Hun alive": Martin Gilbert, _Churchill: A Life,_ p. 680.
"The great day has finally arrived": Letter from Jacques Boulloche, November 1, 1918, collection of Agnes Boulloche.
He regards them as "imbeciles": Mathilde Damoisel's interview with Christiane Boulloche-Audibert, Paris, February 10, 1997; author's interview with Christiane Boulloche-Audibert, March 25, 1999.
"We triumph," declares the chief judge: This quote was famous among his grandchildren. Some thought his words referred to the inferior social class of their fellow _Dreyfusards,_ but Christiane Boulloche believed his comments were just about politics. "He was a republican, and most of Dreyfus's supporters were leftists," she explained. She thought her ancestors had abandoned the conventional wisdom of their social milieu because they were magistrates. "When you're a magistrate, you do have a particular point of view — you're in favor of justice." And to these magistrates, it had always been obvious that Dreyfus was innocent.
A young friend: Author's interview with Dr. René Cler, March 17, 1999.
Their parents encourage: Letter from Dr. Robert Desmond, collection of Agnès Boulloche.
Trains, planes, and automobiles: Dr. Robert Desmond, Témoignage, p. 17.
As a teenager: _André Boulloche, 1915–1978,_ p. 12, Maurice Bourgès-Maunoury.
No priest ever joins them: Mathilde Damoisel's interview with Christiane Boulloche-Audibert, Paris, February 3, 1997.
Christiane is certain: Author's interview with Christiane Boulloche-Audibert, March 19, 1999.
Their parents' favorite writers: Author's telephone interview with Christiane Boulloche-Audibert, November 21, 2003.
Jacques plays the piano: Mathilde Damoisel's interview with Christiane Boulloche-Audibert, February 3, 1997.
who are suddenly forbidden: Author's interview with Christiane Boulloche-Audibert, March 19, 1999.
In a letter home: Letter from Hélène Boulloche to André Boulloche, March 12, 1926, collection of Agnès Boulloche.
When the parents finally return: Author's interview with Christiane Boulloche-Audibert, March 19, 1999.
"It was implicit": Ibid.
"It was as if the plague": _New York Times,_ September 1, 1989.
"Either he could ally": Stokesbury, _Short History of World War II,_ p. 65.
"Germans Rush Gayly": _New York Times,_ September 3, 1939, p. 11.
"Throughout the 30's": _New York Times,_ September 1, 1989.
gape "at each other": Churchill, _Second World War,_ I:434.
most French officers still believe in: Ibid., II:32.
"new and exciting adventure": Mathilde Damoisel's interview with Christiane Boulloche-Audibert, February 10, 1997.
"an imperialist and capitalist crime": Churchill, _Second World War,_ I:511–12.
undermined by shortages: Jackson, _France,_ p. 116.
"raring to go": Quoted ibid., p. 117.
"scores of towns": Churchill, _Second World War,_ II:53.
At the end of the emergency: Ousby, _Occupation,_ p. 43.
Across the channel: _New York Times,_ June 1 and 2, 1940.
mobs of refugees: Quoted in Jackson, _France,_ p. 120.
"This is what we dreaded": _New York Times,_ June 4, 1940.
"I can't describe": Letter from Jacques Boulloche, collection of Agnès Boulloche.
There are twenty thousand people: Ousby, _Occupation,_ p. 46.
"The German guns": _New York Times,_ June 12, 1940.
"My Beloved": Letter from Jacques Boulloche, collection of Agnès Boulloche.
"Had all of us in France": Commandant le Baron de Vomécourt, who served with both the British Army and the French Resistance during the war, replies with an emphatic yes to Captain Liddell Hart's article, "Was the Maquis Worthwhile?" Peter [aka Pierre] de Vomécourt, [London] _Daily Mail,_ February 4, 1947.
There are ninety thousand: Ousby, _Occupation_ , p. 33.
Meanwhile, French prime minister Paul Reynaud: Churchill, _Second World War,_ II:176.
A French minister of state: Ibid., pp. 180, 184, 187. The minister was Jean Ybarnégaray.
A prescient Reynaud: Jackson, _France,_ p. 389.
At nine A.M.: de Gaulle, _Complete War Memoirs,_ p. 80.
"carried with him": Churchill, _Second World War,_ II:192.
André distinguishes himself: Biography of Jean-Pierre Berger, www.ordredelaliberation.fr/fr_compagnon/82.html. André gave different dates in different places for this departure date. In 1943, he told British intelligence officers that the ship had left France on June 22 and reached Algeria on the day of the Armistice. Seven years later, he thought he had left on June 24 and arrived two days later.
He thinks "that we [will] win": Letter from André Boulloche to C. Hettier de Boislambert, Grand Chancelier de L'Ordre de la Liberation, June 13, 1969, French National Archives, box 72AJ2056.
On June 21: Churchill says 21 deputies; Ousby says 19. In a speech before the National Assembly, its president, Raymond Fourni, said there had been 26 deputies and 1 senator aboard.
"I embarked on the _Massilia_ ": Ophuls, _The Sorrow and the Pity,_ pp. 59–60.
Churchill noted with disgust: Churchill, _Second World War,_ II:193–94.
One and a half million French prisoners: Jackson, _France,_ p. 127.
At the end of the opera tour: Albert Speer, _Inside the Third Reich,_ pp. 171–72.
"My Dear Father": Letter from André Boulloche, collection of Agnès Boulloche.
"And they sang well": Author's interview with Christiane Boulloche-Audibert, March 19, 1999.
She sees it as: Author's interview with Christiane Boulloche-Audibert, March 19, 1999.
"It was a succession of shocks": Mathilde Damoisel's interview with Christiane Boulloche-Audibert, February 10, 1997.
"You should have been here": Eparvier, _À Paris sous la botte des Nazis._
This is when Christiane: Mathilde Damoisel's interview with Christiane Boulloche-Audibert, February 3, 1997.
Three months later: Ousby, _Occupation_ , p. 99.
By the start of 1941: Jackson, _France,_ p. 356.
Adding insult to the humiliation: Ousby, _Occupation_ , p. 182.
As the British historian: Jackson, _France,_ p. 243.
Like his father: Author's interview with Christiane Boulloche-Audibert, March 19, 1999.
But he has a terrific sense of humor: Author's interview with Agnès Boulloche, December 8, 2001.
At the end of 1940: "Hommage à André Boulloche," p. 13.
He is appalled by the savage sight: Author's interview with André Postel-Vinay, Paris, February 2, 2004.
"war is the only way": Author's interview with Claire Andrieu (Postel-Vinay's daughter), January 31, 2004.
In December 1940: Jackson, _France,_ p. 403; _Ordre de la liberation_ website, biography of André Postel-Vinay, www.ordredelaliberation.fr/fr_compagnon/801.html.
"I know someone": Author's interview with André Postel-Vinay, February 2, 2004.
"Andre was very passionate": Ibid.
"For the two of us": Ibid.
"Then whose bombers are those": Stokesbury, _Short History of World War II,_ p. 150.
visit it for the first time: Shirer, _Rise and Fall of the Third Reich,_ p. 1039.
he believes that they are reaching London: When Pierre Pène is arrested, he sees in the Gestapo's dossier the plan for the works at Margival, which was supposed to have arrived in London.
never discusses his clandestine: Author's interview with Dr. René Cler, March 17, 1999.
Churchill hopes that he will attract: Jackson, _France_ , p. 389.
"Their idea was to get out": Ophuls, _The Sorrow and the Pity,_ pp. 56, 58.
Only one deputy: Paxton, _Vichy France,_ p. 42.
de Gaulle notices: Jackson, _France,_ p. 398.
felt like a man who had been skinned alive: Ibid., pp. 392–93.
There was one other thing: Ibid., p. 396.
But the general sees: Ibid., p. 397.
Not until November: Ibid., pp. 397–98.
At the same time, French-language: Ibid., p. 398.
Almost anyone who volunteers: Ibid., p. 399.
"The island of Sein stands watch": Ousby, _Occupation,_ p. 44.
Nearly all of what Dewavrin knows: Jackson, _France,_ p. 399.
an almost inevitable invasion: _New York Times_ , June 1 and 6, 1940.
"Everywhere a feeling": Orwell, _Diaries,_ p. 299.
Three years later: Nicholas Lemann, "The Murrow Doctrine," _New Yorker_ , January 23 and 30, 2006.
"could not agree to forcing De Gaulle": Eisenhower, _Crusade in Europe,_ p. 248.
"The familiar slur": Ousby, _Occupation,_ p. 236.
"a very, very explicit act": Author's telephone interview with Robert Paxton, January 29, 2004. In 2014, Paxton told me he had recently seen an interview on the Internet with Pétain's chauffeur saying the Cadillac had been left on the dock in Bordeaux by someone fleeing in June 1940, and Pétain had purchased it then, so Paxton was no longer certain that the automobile was a gift of the American ambassador. However, most other sources agree that it came from Admiral Leahy.
American public opinion begins to rally: Ibid.
"vitally interested statement": _New York Times,_ June 22, 1941.
"This was our obsessive fear": Boulloche Audibert, _Souvenirs._
Thanks to their complicity: Postel-Vinay, _Un fou s'évade,_ p. 8, used by permission of the author's estate.
At that moment: Ibid., pp. 8–9.
And yet he still doesn't want: Ibid., p. 9. Most of the rest of this chapter is taken from _Un fou s'évade._
Probably to avoid: www.rafinfo.org.uk/rafescape/guerisse.htm.
Postel-Vinay considers: Remarks of Postel-Vinay honoring André Boulloche, January 26, 1986.
His ultimate nightmare: This and most of chapter 9 is from Postel-Vinay, _Un fou s'évade._
And when a downed British or American: Author's interview with Claire Andrieu, January 31, 2004.
Patriotic School has been created: Andrew, _Defend the Realm,_ p. 250.
Those identified as "goats": Ibid., p. 251.
The big question is: Stokesbury, _Short History of World War II,_ p. 224.
As Operation Torch begins: www.ibiblio.org/pha/policy/1942/421107b.html.
De Gaulle observed that by not firing: de Gaulle, _Complete War Memoirs,_ p. 358.
It would be a huge prize: Ibid., p. 359.
Just one destroyer: Ibid.
There is an immediate uproar: Stokesbury, _Short History of World War II,_ pp. 228–29.
"There was a tremendous outcry": Author's telephone interview with Robert Paxton, January 29, 2004.
"If the tragic character": De Gaulle, _Complete War Memoirs_ , p. 379. This is the other key paragraph about Darlan's assassination in de Gaulle's book: _The man who had killed him, Fernand Bonnier de la Chapelle, had made himself the instrument of the aggravated passions that had fired the souls around him to the boiling point but behind which, perhaps,moved a policy determined to liquidate a "temporary expedient" after having made use of him. This young man, this child overwhelmed by the spectacle of odious events, thought his action would be a service to his lacerated country, would remove from the road to French reconciliation an obstacle shameful in his eyes. He believed, moreover, as he repeatedly said until the moment of his execution, that an intervention would be made in his behalf by some outside source so high and powerful that the North African authorities could not refuse to obey it. Of course no individual has the right to kill save on the field of battle. Moreover, Darlan's behavior as a governor and as a leader was answerable to national justice, not, certainly, to that of a group or an individual. Yet how could we fail to recognize the nature of the intentions that inspired his juvenile fury? That is why the strange, brutal and summary way the investigation was conducted in Algiers, the hasty and abbreviated trial before a military tribunal convened at night and in private session, the immediate and secret execution of Fernand Bonnier de la Chapelle, the orders given to the censors that not even his name should be known — all these led to the suspicion that someone wanted to conceal at any price the origin of his decision and constituted a kind of defiance of those circumstances which, without justifying the drama, explained and, to a certain degree, excused it._
Darlan's disappearance from the scene: Stokesbury, _Short History of World War II,_ p. 229.
"Everything is ruined anyway": Perrault, _La Longue Traque,_ p. 120.
At seven o'clock in the evening: Témoignage de M. André Boulloche.
impeccable identity card: André never actually uses this fake identity card, and by the time he reaches England he can no longer remember the name in which it was issued.
"deep satisfaction": British Public Record Office HS 9/190/6 114106.
the intelligence section of BCRA: Rossiter, _Women in the Resistance,_ pp. 13–14.
a "capable type": Public Record Office HS 9/190/6 114106. SECRET. Y box 3558.
For a long time afterward: Témoignage de M. André Boulloche.
"I was recruited": Boulloche-Audibert, _Souvenirs._ Raymond Jovignot, a member of the Resistance who knew André Boulloche after his arrest in 1944, told an interviewer in 1946 that Jacques (code name: Crassus) was in fact tortured by the Germans. Jovignot described him as "a very good boy, deeply religious, who loved his boss," André.
"Perhaps I was wrong": Postel-Vinay, _Un fou s'évade,_ p. 172n.
By dawn, Farges: Remarks of Gilbert Farges honoring André Boulloche, January 26, 1986.
As Gimpel's British handlers have noted: British Intelligence file on Charles Gimpel.
"Because it meant": Author's interview with Christiane Boulloche-Audibert, March 19, 1999.
Seventeen hundred _through_ Porte will survive deportation: All details about the prisoners on the train are from the catalog for an exhibit at the Musée Jean Moulin mounted in 2002, www.french-art.com/musees/jean_moulin/auschwitz.htm.
"The crazy people": Remarks of Gilbert Farges honoring André Boulloche.
Just once, the Germans offer: Farges quoted in _André Boulloche,_ p. 27.
As he climbs out of the train: Remarks of Gilbert Farges honoring André Boulloche.
given their first drink: Montbéliard, p. 24.
"precarious survivors": Remarks of Gilbert Farges honoring André Boulloche.
The personal intervention of Marshal Pétain: Author's interview with Odile Boulloche, March 20, 1999.
Ninety-five percent of the deportees: Gilbert Farges, "Hommage à André Boulloche," p. 25.
As James L. Stokesbury: _Short History of World War II,_ pp. 225–26.
"as the state came under challenge": Paxton, _Vichy France,_ p. 286.
His fake identity continues: www.ordredelaliberation.fr/fr_compagnon/855.html.
To boost Christiane's spirits: Boulloche-Audibert, _Souvenirs._
The others are Lemniscate: www.ordredelaliberation.fr/fr_compagnon/855.html.
"I was twenty": Author's interview with Christiane Boulloche-Audibert, March 19, 1999.
Seeing them at the front door: Boulloche-Audibert, _Souvenirs._
_At least I am courageous:_ Ibid.
That is the hardest part: Mathilde Damoisel's interview with Christiane Boulloche-Audibert, February 10, 1997.
"been incredibly lucky": Author's interview with Christiane Boulloche-Audibert, March 25, 1999; _Souvenirs._
The historian Ian Ousby: _Occupation,_ p. 245.
In July 1943: Stokesbury, _Short History of World War II,_ p. 293.
His successor, Pietro Badoglio: Ibid., p. 296.
Then they spent eighteen months: Author's interview with Eric Katlama, Hotel Des Deux Continents, March 23, 1999 (for the fact they were waiting for the revolution to fail). Most of this account of Katlama's early years comes from Alex Katlama's interviews with the British during the war, which were declassified at my request: Public Record Office HS 9/823/1.
He is brought up: Author's interview with Michel Katlama, March 14, 1999.
the mean height for Frenchmen: Timothy J. Hatton and Bernice E. Bray, "Long Run Trends in the Heights of European Men, 19th–20th Centuries," private www.essex.ac.uk/~hatton/Tim_height_paper.pdf.
He assumes the inertia: Author's interview with Eric Katlama, March 23, 1999.
As a Russian immigrant: Author's interview with Michel Katlama, March 14, 1999.
He also happens to love: Ibid.
"A quiet intelligent" _through_ "a competent and loyal assistant": British Intelligence file on Eric Katlama.
In the third week of April: Christiane's memoirs place this event in January, but Alex remembers it clearly as April.
Christiane likes the handsome Alex: Author's interview with Christiane Boulloche-Audibert, March 19, 1999.
he meets with Resistance members: www.ordredelaliberation.fr/fr_compagnon/855.html.
"All southern England": Eisenhower, _Crusade in Europe,_ p. 248.
By the eve of the invasion: Roberts _, Storm of War_ , p. 466.
Soldiers joke that if the invasion: Stokesbury, _Short History of World War II_ , p. 311.
"The southernmost camps": Eisenhower, _Crusade in Europe,_ p. 249.
This will severely limit: O'Neill, _Oxford Essential Guide to World War II,_ pp. 86–87.
They make a special effort: Shirer, _Rise and Fall of the Third Reich,_ p. 1037. Rundstedt and Rommel were certain it would be in the Pas-de-Calais area, where the channel was at its narrowest.
Again, the deception works: Stokesbury, _Short History of World War II,_ p. 211; O'Neill, _Oxford Essential Guide to World War II,_ p. 87.
Knowing that it's crucial: Churchill, _Second World War,_ V:544–46.
"The tension continued to mount": Eisenhower, _Crusade in Europe,_ p. 249.
At four fifteen A.M.: Ibid., p. 250.
"I hope to God": Roberts, _Storm of War,_ pp. 469–70.
It is thrilling: Boulloche-Audibert, _Souvenirs._
So the commanders: Jane Penrose, ed., _The D-Day Companion,_ quoted in Roberts, _Storm of War,_ p. 471.
And on June 4: Shirer, _Rise and Fall of the Third Reich,_ p. 1036.
On the basis of everything: Ibid., pp. 1036–37.
Around one A.M.: Ibid., p. 1038.
Hitler himself has been up: Roberts, _Storm of War,_ p. 472.
Then he goes to bed: Shirer, _Rise and Fall of the Third Reich,_ p. 1038. 133 At five fifty on the morning: Roberts, _Storm of War,_ p. 473.
After the 101st Airborne: Ibid., p. 473.
"ABLE company riding through from a rooftop": _The Atlantic,_ November 1960. www.theatlantic.com/magazine/archive/1960/11/first-wave-at-omaha-beach/303365/.
At a cost of two thousand Americans: Roberts, _Storm of War,_ p. 476.
"We were depending": Eisenhower, _Crusade in Europe,_ p. 248.
For a week after the invasion: Cobb, _Resistance,_ p. 245; Jackson, _France,_ pp. 544–45 (for Marseille and Toulouse).
This is vital: Cobb, _Resistance,_ p. 245.
British air chief marshal Arthur Tedder: Roberts, _Storm of War,_ p. 477.
"The first twenty-four hours": Quoted ibid., p. 459.
By the end of June 11: www.ddaymuseum.co.uk/d-day/d-day-and-the-battle-of-normandy-your-questions-answered. By July 2, those numbers had swelled to about 1,000,000 men, 171,532 vehicles, and 566,648 tons of supplies.
"In our circles": Roberts, _Storm of War,_ pp. 479–80.
"Throughout France the Free French": Eisenhower, _Crusade in Europe._
Now, for the first time in years: Boulloche-Audibert, _Souvenirs._
Christiane is captivated: Ibid.
"We had a common enemy": Ibid.
Christiane spends the battle: Author's interview with Christiane Boulloche-Audibert, March 25, 1999; Boulloche-Audibert, _Souvenirs._
The Maquis suffer two dead: www.memoiresvivantes.org/histoire_resistance_dunlesplacesetvermot.php; _Dictionnaire biographique de Paul-Camille DUGENE_ ; and www.morvan-des-lacs.com/images/actualites/le%20massacre%20de%20dun2.pdf.
"I'm coming back with you": Boulloche-Audibert, _Souvenirs._
Rommel has joined the conspiracy: Shirer, _Rise and Fall of the Third Reich_ , p. 1031.
To Rommel's chief of staff: General Hans Speidel, _Invasion_ , quoted ibid., p. 1039.
"Don't you worry": Ibid., p. 1040.
Speidel believed that: Ibid., p. 1042.
no more than fourteen: Photo and caption in Eparvier, _À sous la botte des Nazis_.
ZUR NORMANDIE FRONT: _La Libération de Paris,_ documentary, 1944.
"The assassination must be attempted": Shirer, _Rise and Fall of the Third Reich_ , p. 1043.
"The threatened collapse": Ibid.
Eisenhower remembers late June: Eisenhower, _Crusade in Europe,_ p. 263.
Seven weeks pass: Ibid., p. 272.
the anti-Hitler plotters get a boost: Shirer, _Rise and Fall of the Third Reich,_ pp. 1033–45.
By July 1944, the conspiracy: Ibid., pp. 1030, 1034.
The thickness of this particular wire: Ibid., p. 1049; Roberts, _Storm of War_ , p. 481.
"somewhere between a monastery": Ibid.
The compound includes: Ibid.
Before darkness has fallen: Shirer, _Rise and Fall of the Third Reich_ , p. 1060.
The conspirators' failure: Ibid., p. 1064.
Goebbels initially blames the Allies: _New York Times_ , July 21, 1944.
"Seized by a titanic fury": Shirer, _Rise and Fall of the Third Reich,_ p. 1070.
"Who says I am not": Churchill, _Second World War,_ VI:25.
"Believe me, this is the turning point": David Irving, _Hitler's War,_ pp. 662–64.
"Although the smell of retreat": Cobb, _Resistance,_ p. 258.
"A great tide of popular enthusiasm": De Gaulle, _Complete War Memoirs,_ p. 638.
"We had our hands": Boulloche-Audibert, _Souvenirs_ ; author's interview with Christiane Bulloche-Audibert, March 11, 1999.
On July 14, a huge: Cobb, _Resistance,_ pp. 258–59.
"They had a deported son": Boulloche-Audibert, _Souvenirs._
But as the temperature inches: www.meteo-paris.com/bibliotheque/documents/3403.txt.
He and Rondenay have escaped: www.ordredelaliberation.fr/fr_compagnon/426.html.
Disaster strikes: Christiane remembers the arrest taking place at Passy; Rondenay's official biography says it took place at La Muette, www.ordredelaliberation.fr/fr_compagnon/855.html. For de Beafort's wound, www.ordredelaliberation.fr/fr_compagnon/426.html.
"There was no question": Boulloche-Audibert, _Souvenirs._
"At the same time I felt a trap": Ibid.
"At times like this": Ibid.
"The conditions of our daily lives" _and_ "We are happy to know": I found these letters in the family archive maintained by Agnès Boulloche, André's daughter. André must have saved them during his ten remaining months at the camp and brought them back to France after he was liberated.
Nothing important is discussed: Boulloche-Audibert, _Souvenirs._
For the next week, the Germans: Ibid.
Alone in Paris: Ibid.
"I had no choice": Author's interview with Dr. René Cler, March 17, 1999.
"It was not pleasant": Ibid.
"Even when motionless": Dallas, _1945,_ p. 187.
At the first of the three security rings: Collins and Lapierre, _Is Paris Burning?,_ p. 34.
When it ended, only 347: Ibid., p. 32.
Then Hitler "began reeling off": Neitzel, _Tapping Hitler's Generals._
"dozens of generals": Collins and Lapierre, _Is Paris Burning?,_ p. 35.
Hitler orders him to "stamp out": Ibid., p. 36.
Right now they are determined: Eisenhower, _Crusade in Europe,_ p. 296.
"Paris food and medical requirements": Quoted in Collins and Lapierre, _Is Paris Burning?,_ p. 20.
For all of these reasons: Eisenhower, _Crusade in Europe,_ p. 296.
Their leaflet exhorts: Cobb, _Resistance,_ p. 259.
On the night of August 12: http://dora-ellrich.fr/les-hommes-du-convoi-du-15-aout-1944/.
Choltitz orders more than two thousand: Cobb, _Resistance,_ p. 258.
Half an hour after it leaves: Boulloche-Audibert, _Souvenirs._
"You won't go any further": http://memoiredeguerre.pagespro-orange.fr/convoi44/derniers-convois.htm#Pantin.
The chief German engineer promises: Collins and Lapierre, Is Paris Burning?, pp. 68–69.
Since allied bombers are continuing: Ibid., p. 68.
"Often it is given to a general": Ibid., pp. 89–90.
"Why too soon?": De Gaulle, _Complete Wartime Memoirs,_ pp. 636–37. 165 De Gaulle believes it is "intolerable": Ibid., p. 640.
De Gaulle also suggests: Ibid., p. 637.
By now General Choltitz: Collins and Lapierre, _Is Paris Burning?,_ p. 210.
"The swastika was still flying": quoted in Cobb, _Resistance,_ pp. 260–61.
A CHACUN SON BOCHE: Dallas, _1945,_ p. 194.
"If the American Army": Collins and Lapierre, _Is Paris Burning?,_ pp. 178–79.
By the time Silbert climbs: Ibid., pp. 178–80.
Because they had begun: Eisenhower, _Crusade in Europe_ , p. 296.
Eisenhower also recognizes: Quoted in Collins and Lapierre, _Is Paris Burning?,_ p. 181.
Nordling has already convinced: Dallas, _1945,_ p. 177.
In his place, he sends: Collins and Lapierre, _Is Paris Burning?,_ pp. 187–91.
"Have the French division hurry": Ibid., p. 208.
If there was a strategy: Jackson, _France,_ p. 566.
When the spectral outline: Collins and Lapierre, _Is Paris Burning?,_ pp. 236–37.
At last, Free French troops are back: Ibid., p. 255.
Within minutes, every block is reverberating: Ibid., p. 257.
The Vichy government estimated: Ousby, _Occupation,_ p. 237.
As Ian Ousby observes: Ibid., p. 238.
On the eve of his return: de Gaulle, _Complete Wartime Memoirs,_ pp. 645–46.
"Gentlemen," he tells his guests: Collins and Lapierre, _Is Paris Burning?,_ pp. 258–59.
This is the German officer: Dallas, _1945,_ p. 176.
he had begun to have nightmares: From Choltitz's memoirs, quoted ibid., p. 176.
German snipers increase Allied casualties: Ousby, _Occupation,_ p. 293.
"Why should we hide" _through_ "Long live France!": www.emersonkent.com/speeches/paris_liberated.htm.
"The Republic has never ceased": De Gaulle, _Complete Wartime Memoirs,_ p. 650.
Two days earlier: Dallas, _1945,_ p. 188.
The next day, de Gaulle defies: Collins and Lapierre, _Is Paris Burning?,_ p. 331.
The French general concedes: Ibid., p 333.
"Today we were to revive": De Gaulle, _Complete Wartime Memoirs,_ p. 653.
De Gaulle thinks Parisians: Ibid.
"Since each of all of those": Ibid.
"Everyone seemed happy and relieved": Bulloche-Audibert, _Souvenirs._
At midnight on August 26: De Gaulle, _Complete Wartime Memoirs_ , p. 659.
"I was shocked": Boulloche-Audibert, _Souvenirs._
Christian leaves Paris _through_ doesn't want to alarm her: Ibid.
"Although letter writing": _Bulletin de l'Association des anciens élèves de l'école polytechnique,_ September 1947.
When she returns to Paris: Boulloche-Audibert, _Souvenirs._
Just as the Germans had partly ignored: Roberts, _Storm of War,_ pp. 505–7.
the Germans have lost 120,000 men killed: Shirer, _Rise and Fall of the Third Reich,_ p. 1095.
"The great difference": Roberts, _Storm of War,_ p. 509.
In Nuremberg, the sight of gigantic Nazi rallies: Video of the explosion at www.ushmm.org/wlc/en/media_fi.php?MediaId=2048.
after twelve years, four months: Shirer, _Rise and Fall of the Third Reich,_ p. 1139.
"I was deported": Boulloche-Audibert, _Souvenirs._
Somehow, everyone who survives with him: _Andre Boulloche,_ p. 30.
"In this way he shared": Ibid., p. 31.
"Survival was a constant act": Ibid., p. 33.
At dawn on April 16 _through_ arrives to liberate the camp: Ibid., p. 32.
"Of course he was extremely thin": Author's interview with Christiane Boulloche-Audibert, March 19, 1999.
"But, when he was finally standing there": Boulloche-Audibert, Souvenirs.
"If I'd known that": Author's interview with Eric Katlama, March 23, 1999. "I remember very well that my mother [Jacqueline] told me," Katlama said. "I have no doubt about that memory at all."
"My deportation to the camps": Andre Boulloche F.R. 3 radio broadcast, November 23, 1976.
"André Boulloche had the longest service": Postel-Vinay, _Un fou s'évade,_ p. 128.
"Did your father André ever talk": Author's interview with Agnès Boulloche, December 8, 2001.
ten thousand Frenchmen were the victims: Author's interview with Claire Andrieu, January 31, 2004.
"I went to Macy's": Author's interview with Christiane Boulloche-Audibert, March 25, 1999.
Christiane thought New York: Ibid.
"Fear of the Gestapo": _Washington Post,_ April 23, 1946, p. 13.
"Why did my life have to be spared": Collection of Agnès Boulloche.
"I've always done": Mathilde Damoisel's interview with Christiane Boulloche-Audibert, April 7, 1997.
"It was when I was deported": André Boulloche interview on Radio FR 3, November 20, 1976.
"We bourgeois learned some things": Author's interview with Christiane Boulloche-Audibert, March 25, 1999.
"I was very young": André Boulloche radio interview, November 20, 1976.
killed there in an ambush: www.annales.org/archives/x/etienneaudibert.html.
"You've never seen two sisters": Author's interview with Eric Katlama, March 23, 1999.
"It's true. We were always scared": Author's interview with Robert Boulloche, March 21, 1999.
"camouflaged their unhappiness": Author's interview with Pierre Audibert, March 24, 1999.
"it was clear that André was suffering": Author's interview with Eric Katlama, March 23, 1999.
"I remember there was a feeling of meditation": Ibid.
"This was the Bolloches": Author's interview with Pierre Audibert, March 24, 1999.
"It showed that you can do good": Author's interview with François Audibert, March 18, 1999.
"If one wants people to win": Ophuls, _The Sorrow and the Pity,_ pp. xiii–xiv.
But de Gaulle never literally said _through_ outside of institutions: Author's interview with Claire Andrieu, January 31, 2004.
André told her that if she "wanted to fight": Author's interview with Agnès Boulloche, December 8, 2001.
Ophuls told me that he thought de Gaulle: Author's interview with Marcel Ophuls, March 23, 2004.
He chose Clermont-Ferrand: Elliot Wilhelm, "The Sorrow and the Pity," _VideoHound's World Cinema_ (Detroit: Visible Ink Press, 1999).
"There's something unhealthy": Author's interview with Marcel Ophuls, March 23, 2004.
teachers from the lycée: Ophuls, _The Sorrow and the Pity,_ p. 86.
"eating the national tissue": "À la mémoire d'Andre Boulloche."
"It was a period when there were terrorist attacks": Author's interview with Christiane Boulloche-Audibert, March 25, 1999.
In 1967, the Neuwirth Act: www.ined.fr/fichier/t_publication/1336/publi_pdf2_pesa439.pdf.
"As you can see": Boulloche-Audibert, _Souvenirs._
the Socialist Party had voted against participating: _New York Times,_ January 12, 1959.
Odile rushed to London: Author's interview with Odile Boulloche, March 20, 1999.
"He had a terrible violence" _through_ "first and foremost, a _polytechnicien_ ": Author's interview with Jacques Boulloche, April 4, 2004.
the secret to his success: _Le Point,_ January 9, 1978.
"Boulloche was certain that he was working for France": Author's interview with Raymond Forni, June 3, 2003.
most people expected Mitterrand: Author's interview with Andrée Vauban, April 23, 2003.
"secular monk": Ibid.
"an infernal life": André Boulloche radio interview, November 20, 1976. 217 "Are you ever discouraged?": Ibid.
"I think I can say without exaggeration": Ibid.
Now he climbed into the copilot's seat: Author's interview with Andrée Vauban, April 23, 2003. Investigators apparently determined this by examining the wreckage of the plane.
At the same moment: Ibid.
A moment later: _Le Monde,_ March 19–20; _Le Figaro,_ March 18–19, 1978.
"This is a very remarkable man": _L'Est Républicain,_ March 17, 1978.
Their bodies were found: _Le Monde,_ March 19–20, 1978.
"And on top of everything else": Author's interview with Claudine Lefer, March 24, 1999.
"Courage is more exhilarating": www.gwu.edu/~erpapers/abouteleanor/er-quotes/.
The death of her sister: Author's interview with Mathilde Damoisel, July 4, 2003.
"It was obvious": Author's interview with Christiane Boulloche-Audibert, March 19, 1999.
"It was extremely painful": Boulloche-Audibert, _Souvenirs._
"She would read me what she had written": Author's interview with Catherine (Audibert) Dujardin, March 21, 1999.
news of Christiane's book: Six years after she published it privately, Christiane's book became part of the collection _Femmes dans la guerre, 1940–1945._
"We did not talk about the Resistance": Author's interview with Michel Katlama, March 14, 1999.
"Christiane said she had done": Author's interview with Claudine Lefer, March 24, 1999.
"she had done her duty": Author's interview with Hélène Dujardin, March 24, 1999.
German-American Bund: www.ushmm.org/wlc/en/article.php?ModuleId=10005684.
After the Nazis looted Jewish stores: www.ushmm.org/wlc/en/article.php?ModuleId=10005516.
"an American reader who honestly recreates": Paxton, _Vichy France,_ p. xiv.
"If one hasn't been through": _The Sorrow and the Pity._ Ophuls told me, "I think that's why I put it there. It seems sort of pretentious to use Anthony Eden as a spokesman for the author, but he is expressing my sentiments there. The other people, not necessarily. But he does. I do associate with that statement. I'm glad that it made an impression on you." Author's telephone interview with Marcel Ophuls, March 23, 2004.
# SELECT BIBLIOGRAPHY
_André Boulloche: 1915–1978._ Paris: C. Boulloche, 1979.
Andrew, Christopher. _Defend the Realm: The Authorized History of MI5._ New York: Vintage Books, 2010.
Audibert, Etienne. "Notices sur nos morts, Jacques BOULLOCHE (1906)." _Bulletin de l'association des anciens élèves de l'école polytechnique,_ no. 9, September 1947.
Boulloche-Audibert, Christiane. _Souvenirs 1939–1945._ Privately published, 1998. Reprinted in Christiane Audibert-Boulloche et al., _Femmes dans la guerre, 1940–1945_ (Paris: éditions du Félin, 2004).
British Intelligence file on André Boulloche, National Archives, formerly Public Record Office HS 9/190/6 114106. Declassified at the request of the author.
British Intelligence file on Charles Gimpel, National Archives, formerly Public Record Office HS 9/586/1 114106. Declassified at the request of the author.
British Intelligence File on Eric Katlama, National Archives, formerly Public Record Office HS 9/823/1. Declassified at the request of the author.
Churchill, Sir Winston. _Great War Speeches._ London: Corgi Books, 1965.
———. _The Second World War._ 6 vols. London: Cassell, 1948–54.
Cobb, Matthew. _The Resistance: The French Fight Against the Nazis._ New York: Simon & Schuster, 2009.
Collins, Larry, and Dominique Lapierre. _Is Paris Burning?_ New York: Simon & Schuster, 1965.
Cooper, Duff. _Old Men Forget: The Autobiography of Duff Cooper (Viscount Norwich)._ New York: Dutton, 1954.
Dallas, Gregor. _1945: The War That Never Ended._ New Haven: Yale University Press, 2005.
de Gaulle, Charles. _The Army of the Future._ Philadelphia: Lippincott, 1941.
———. _The Complete War Memoirs._ New York: Simon & Schuster, 1964.
Eisenhower, Dwight D. _Crusade in Europe._ Garden City, NY: Doubleday, 1948.
Eparvier, Jean. _À Paris sous la botte des Nazis._ Paris: éditions Raymond Schall, 1944.
Foot, M. R. D. _SOE in France: An Account of the Work of the British Special Operations Executive in France, 1940–1944._ London: Her Majesty's Stationery Office, 1966.
Foot, M. R. D., and J. M. Langley. _MI9: Escape and Evasion, 1939–1945._ Boston: Little, Brown, 1980.
Gilbert, Martin. _Churchill: A Life._ New York: Henry Holt, 1991.
"Hommage à André Boulloche." _Revue Municipale Numero Special._ Montbéliard, March 1979.
Irving, David. _Hitler's War._ New York: Viking, 1977.
Jackson, Julian. _The Fall of France: The Nazi Invasion of 1940._ New York: Oxford University Press, 2003.
———. _France: The Dark Years, 1940–1944._ New York: Oxford University Press, 2001.
Marks, Leo. _Between Silk and Cyanide: A Codemaker's War 1941–1945._ New York: Free Press, 1998.
Marrus, Michael R., and Robert O. Paxton. _Vichy France and the Jews._ Stanford: Stanford University Press, 1995.
Marshall, S. L. A. "First Wave at Omaha Beach." _Atlantic,_ November 1, 1960.
"À la mémoire d'André Boulloche (34), compagnon de la Libération, 1915–1978." _La Jaune et la Rouge,_ no. 583, March 2003.
Murphy, Brendan. _Turncoat: The Strange Case of British Sergeant Harold Cole, "The Worst Traitor of the War."_ San Diego: Harcourt, 1987.
Neitzel, Sönke, ed. _Tapping Hitler's Generals: Transcripts of Secret Conversations 1942–45._ Barnsley, Yorkshire, UK: Frontline Books, 2007.
O'Neill, William L _. The Oxford Essential Guide to World War II._ New York: Berkley Books, 2002 _._
Ophuls, Marcel. _The Sorrow and the Pity: Chronicle of a French City under German Occupation._ Trans. Mireille Johnston. Intro. Stanley Hoffmann. St. Albans, UK: Paladin, 1975.
Orwell, George. _Diaries._ Ed. Peter Davison. New York: Liveright, 2012.
Ousby, Ian. _Occupation: The Ordeal of France, 1940–1944._ New York: St. Martin's Press, 1998.
Paxton, Robert O. _Vichy France: Old Guard and New Order, 1940–1944._ New York: Knopf, 1972.
Perrault, Gilles. _La Longue Traque._ Paris: J. C. Lattès, 1975.
Perrault, Gilles, and Pierre Azema. _Paris Under the Occupation._ New York: Vendome Press, 1989.
Peschanski, Denis, et al. _Collaboration and Resistance: Images of Life in Vichy France 1940–44._ Trans. Lory Frankel. New York: Harry N. Abrams, 2000.
Postel-Vinay, André. _Un fou s'évade: Souvenirs de 1941–42._ Paris: éditions du Félin, 1997.
Roberts, Andrew. _The Storm of War: A New History of the Second World War._ New York: Allen Lane, 2009.
Rossiter, Margaret L. _Women in the Resistance._ New York: Praeger, 1986.
Shirer, William L _. The Rise and Fall of the Third Reich: A History of Nazi Germany._ New York: Simon & Schuster, 1960.
Speer, Albert. _Inside The Third Reich: Memoirs._ Trans. Richard and Clara Winston. New York: Macmillan, 1970.
Stokesbury, James L. _A Short History of World War II._ New York: Morrow, 1980.
Témoignage de M. André Boulloche, Ingenieur des Ponts et Chaussées, 18 Av. D'Eylau, XVIeme, pseudo Armand et a Londres: Marin-Segment. Recueilli par Madame Granlet le 1 février 1950. French National Archives, box 72AJ68.
# PHOTO CREDITS
I gratefully acknowledge the following for permission to reproduce the photographs and documents in this book:
Front endpaper: Photographer unknown, from _A Paris sous la botte des Nazis_ (Paris: Éditions Raymond Schall, 1944).
1.1: Courtesy Rebecca Kaiser Gibson
1.2, 1.3, and 1.8: Photographs by Roger Schall, courtesy The Image Works.
1.4, 1.6, 1.16, and 1.20: Courtesy Agnès Boulloche
1.5, 1.9, 1.10, 1.19, and 1.25: Courtesy Christiane Boulloche-Audibert
1.7: Photographer unknown, from _Paris Under the Occupation_ by Gilles Perrault and Pierre Azema (New York: The Vendome Press, 1989).
1.11: Courtesy Claire Andrieu
1.12: Courtesy Musée de l'Ordre de la Libération
1.13, 1.18, and 1.21: Courtesy Eric Katlama
1.14 and 1.15: Photographs by Albert Seeberger, from _Paris Under the Occupation_. Copyright holder unknown.
1.17 and 1.26: Courtesy Odile Boulloche
1.22 and 1.23: © Joe Stouter
1.24: Courtesy Pierre Audibert
1.27: © Tomas van Houtryve
Back endpaper: Photograph by Maurice Jarnoux, from _A Paris sous la botte des Nazis._ Copyright holder unknown.
In cases where the photographer or copyright holder is unknown, every effort has been made to identify such parties, and I ask that anyone with information about these photographs contact the publisher, Other Press.
| {
"redpajama_set_name": "RedPajamaBook"
} | 8,730 |
function jqGridInclude()
{
var pathtojsfiles = "js/"; // need to be ajusted
// set include to false if you do not want some modules to be included
var modules = [
{ include: true, incfile:'i18n/grid.locale-en.js'}, // jqGrid translation
{ include: true, incfile:'grid.base.js'}, // jqGrid base
{ include: true, incfile:'grid.common.js'}, // jqGrid common for editing
{ include: true, incfile:'grid.formedit.js'}, // jqGrid Form editing
{ include: true, incfile:'grid.inlinedit.js'}, // jqGrid inline editing
{ include: true, incfile:'grid.celledit.js'}, // jqGrid cell editing
{ include: true, incfile:'grid.subgrid.js'}, //jqGrid subgrid
{ include: true, incfile:'grid.treegrid.js'}, //jqGrid treegrid
{ include: true, incfile:'grid.custom.js'}, //jqGrid custom
{ include: true, incfile:'grid.postext.js'}, //jqGrid postext
{ include: true, incfile:'grid.tbltogrid.js'}, //jqGrid table to grid
{ include: true, incfile:'grid.setcolumns.js'}, //jqGrid setcolumns
{ include: true, incfile:'grid.import.js'}, //jqGrid import
{ include: true, incfile:'grid.grouping.js'}, //jqGrid grouping
{ include: true, incfile:'jquery.fmatter.js'}, //jqGrid formater
{ include: true, incfile:'JsonXml.js'}, //xmljson utils
{ include: true, incfile:'jquery.searchFilter.js'} // search Plugin
];
var filename;
for(var i=0;i<modules.length; i++)
{
if(modules[i].include === true) {
filename = pathtojsfiles+modules[i].incfile;
if(jQuery.browser.safari) {
jQuery.ajax({url:filename,dataType:'script', async:false, cache: true});
} else {
IncludeJavaScript(filename);
}
}
}
function IncludeJavaScript(jsFile)
{
var oHead = document.getElementsByTagName('head')[0];
var oScript = document.createElement('script');
oScript.type = 'text/javascript';
oScript.charset = 'utf-8';
oScript.src = jsFile;
oHead.appendChild(oScript);
};
};
jqGridInclude(); | {
"redpajama_set_name": "RedPajamaGithub"
} | 2,167 |
const { isNil, length } = require('ramda')
const { type } = require('..')
/**
* @description This rule fails if the length of the provided string is less
* than the min specified. It will pass if value is undefined.
*/
module.exports = len => ({
name: 'string/minLength',
evaluator: x => isNil(x) || length(x) >= len,
errorMsg: x => `Expected '${x}' to be at least ${len} characters in length`,
prereqs: [type(String)],
})
| {
"redpajama_set_name": "RedPajamaGithub"
} | 6,567 |
use strict;
use warnings;
package MongoDB::BulkWriteView;
# ABSTRACT: Bulk write operations against a query document
use version;
our $VERSION = 'v2.2.2';
use Moo;
use MongoDB::Error;
use MongoDB::_Types qw(
Boolish
Document
IxHash
);
use Types::Standard qw(
Maybe
ArrayRef
InstanceOf
);
use boolean;
use namespace::clean -except => 'meta';
# A hash reference containing a MongoDB query document
has _query => (
is => 'ro',
isa => IxHash,
coerce => IxHash->coercion,
required => 1
);
# Originating bulk write object for executing write operations.
has _bulk => (
is => 'ro',
isa => InstanceOf['MongoDB::BulkWrite'],
required => 1,
handles => [qw/_enqueue_write/]
);
has _collation => (
is => 'ro',
isa => Maybe [Document],
);
has _array_filters => (
is => 'ro',
isa => Maybe [ArrayRef[Document]],
);
has _upsert => (
is => 'ro',
isa => Boolish,
default => 0,
);
sub collation {
my ($self, $collation) = @_;
return $self->new( %$self, _collation => $collation );
}
sub arrayFilters {
my ( $self, $array_filters ) = @_;
return $self->new( %$self, _array_filters => $array_filters );
}
sub upsert {
my ($self) = @_;
unless ( @_ == 1 ) {
MongoDB::UsageError->throw("the upsert method takes no arguments");
}
return $self->new( %$self, _upsert => true );
}
sub update_many {
push @_, "update_many";
goto &_update;
}
sub update_one {
push @_, "update_one";
goto &_update;
}
sub replace_one {
push @_, "replace_one";
goto &_update;
}
sub _update {
my $method = pop @_;
my ( $self, $doc ) = @_;
my $type = ref $doc;
unless ( @_ == 2 && grep { $type eq $_ } qw/HASH ARRAY Tie::IxHash BSON::Array/ ) {
MongoDB::UsageError->throw("argument to $method must be a single hashref, arrayref, Tie::IxHash or BSON::Array");
}
if ( ref $doc eq 'ARRAY' ) {
MongoDB::UsageError->throw("array reference to $method must have key/value pairs")
if @$doc % 2;
$doc = Tie::IxHash->new(@$doc);
}
elsif ( ref $doc eq 'HASH' ) {
$doc = Tie::IxHash->new(%$doc);
}
$self->_bulk->_retryable( 0 ) if $method eq 'update_many';
my $update = {
q => $self->_query,
u => $doc,
multi => $method eq 'update_many' ? true : false,
upsert => boolean( $self->_upsert ),
is_replace => $method eq 'replace_one',
(defined $self->_collation ? (collation => $self->_collation) : ()),
(defined $self->_array_filters ? (arrayFilters => $self->_array_filters) : ()),
};
$self->_enqueue_write( [ update => $update ] );
return;
}
sub delete_many {
my ($self) = @_;
$self->_bulk->_retryable( 0 );
$self->_enqueue_write(
[
delete => {
q => $self->_query,
limit => 0,
( defined $self->_collation ? ( collation => $self->_collation ) : () ),
(defined $self->_array_filters ? (arrayFilters => $self->_array_filters) : ()),
}
]
);
return;
}
sub delete_one {
my ($self) = @_;
$self->_enqueue_write(
[
delete => {
q => $self->_query,
limit => 1,
( defined $self->_collation ? ( collation => $self->_collation ) : () ),
(defined $self->_array_filters ? (arrayFilters => $self->_array_filters) : ()),
}
]
);
return;
}
1;
__END__
=head1 SYNOPSIS
my $bulk = $collection->initialize_ordered_bulk_op;
# Update one document matching the selector
bulk->find( { a => 1 } )->update_one( { '$inc' => { x => 1 } } );
# Update all documents matching the selector
bulk->find( { a => 2 } )->update_many( { '$inc' => { x => 2 } } );
# Update all documents matching the selector, with respect to a collation
bulk->find( { a => { '$gte' => 'F' } )->collation($collation)
->update_many( { '$inc' => { x => 2 } } );
# Update all documents
bulk->find( {} )->update_many( { '$inc' => { x => 2 } } );
# Replace entire document (update with whole doc replace)
bulk->find( { a => 3 } )->replace_one( { x => 3 } );
# Update one document matching the selector or upsert
bulk->find( { a => 1 } )->upsert()->update_one( { '$inc' => { x => 1 } } );
# Update all documents matching the selector or upsert
bulk->find( { a => 2 } )->upsert()->update_many( { '$inc' => { x => 2 } } );
# Replaces a single document matching the selector or upsert
bulk->find( { a => 3 } )->upsert()->replace_one( { x => 3 } );
# Remove a single document matching the selector
bulk->find( { a => 4 } )->delete_one();
# Remove all documents matching the selector
bulk->find( { a => 5 } )->delete_many();
# Update any arrays with the matching filter
bulk->find( {} )->arrayFilters([ { 'i.b' => 1 } ])->update_many( { '$set' => { 'y.$[i].b' => 2 } } );
# Remove all documents matching the selector, with respect to a collation
bulk->find( { a => { '$gte' => 'F' } )->collation($collation)->delete_many();
# Remove all documents
bulk->find( {} )->delete_many();
=head1 DESCRIPTION
This class provides means to specify write operations constrained by a query
document.
To instantiate a C<MongoDB::BulkWriteView>, use the L<find|MongoDB::BulkWrite/find>
method from L<MongoDB::BulkWrite>.
Except for L</arrayFilters>, L</collation> and L</upsert>, all methods have an
empty return on success; an exception will be thrown on error.
=method arrayFilters
$bulk->arrayFilters( $array_filters )->update_many( $modification );
Returns a new C<MongoDB::BulkWriteView> object, where the specified arrayFilter
will be used to determine which array elements to modify for an update
operation on an array field.
=method collation
$bulk->collation( $collation )->delete_one;
Returns a new C<MongoDB::BulkWriteView> object, where the specified
collation will be used to determine which documents match the query
document. A collation can be specified for any deletion, replacement,
or update.
=method delete_many
$bulk->delete_many;
Removes all documents matching the query document.
=method delete_one
$bulk->delete_one;
Removes a single document matching the query document.
=method replace_one
$bulk->replace_one( $doc );
Replaces the document matching the query document. The document
to replace must not have any keys that begin with a dollar sign, C<$>.
=method update_many
$bulk->update_many( $modification );
Updates all documents matching the query document. The modification
document must have all its keys begin with a dollar sign, C<$>.
=method update_one
$bulk->update_one( $modification );
Updates a single document matching the query document. The modification
document must have all its keys begin with a dollar sign, C<$>.
=method upsert
$bulk->upsert->replace_one( $doc );
Returns a new C<MongoDB::BulkWriteView> object that will treat every
update, update_one or replace_one operation as an upsert operation.
=cut
| {
"redpajama_set_name": "RedPajamaGithub"
} | 5,459 |
{"url":"https:\/\/www.reddit.com\/user\/luoyun","text":"Press J to jump to the feed. Press question mark to learn the rest of the keyboard shortcuts\nSort\nluoyun commented on\nPosted by\n1 point \u00b7 10 days ago\n1 point \u00b7 10 days ago\n\nTeam: Bikes, Skinner, Hurd, Wong.\n\nIndividual: Carey Nominative AA: McCusker.\n\nAlso okay with switching McCusker and Smith.\n\n1 point \u00b7 10 days ago\n\nForgot alternates: Lee\/McCallum\/Thomas.\n\n19 points \u00b7 11 days ago\n\nI think she\u2019s only 20... her hair, clothes, and makeup just really age her.\n\nsee more\n19 points \u00b7 11 days ago\n\nThis is not a new photo. She was between 17 and 18 when this photo was taken.\n\nOriginal Poster1 point \u00b7 17 days ago\n\nI am an analyst for a county health department. I look at youth risk behavior outcomes, chronic disease trends, and sometimes do more advanced analyses for smaller projects when the need arises. I love my work. I don't care that I'm making half as much compared to colleagues in private industry. I will always choose purpose over profit.\n\nI am very interested in working for a public health department. How hard is it to find a job in that space? I've looked, but it seems like openings are fairly limited.\n\nsee more\n2 points \u00b7 17 days ago\n\nDepends on where you are. I interviewed for a salaried position at this department that I did not get, but was hired as a contract analyst. Once there is enough funding, I will probably be made a more permanent staff member because they like the work I've done. This is in a medium sized metro area in Appalachia. I would reach out and see if they need contractors! Never hurts to get your foot in the door. My background is in biostatistics, and I am a graduate student at the local school of public health, so that might have helped me get the job.\n\nOriginal Poster3 points \u00b7 17 days ago\n\nI have a background in statistics and public health as well!\n\nIf you don't mind me asking - did you just go to their office and drop off a resume for the contract work? Or how did you start working with them?\n\nsee more\n2 points \u00b7 17 days ago\n\nI responded to an online job posting for a salaried position and was interviewed. A week later I got a rejection letter for that position along with an invite from HR to speak with one of the people who hired me about a contract position. I discussed the specifics of the work one-on-one with this person and was offered the position at the end of that discussion.\n\n5\nCrossposted by18 days ago\nPosted by18 days ago\n\nI am preparing for an MS comprehensive exam in biostatistics and am stuck on a review problem.\n\nWe have a random sample X_1 to X_n from the density f(x;\u03b8)=\u03b8e-\u03b8x, x> 0, \u03b8>0.\n\nWe are being asked to test the hypothesis H0: theta>= theta_0 versus H_a: theta< theta_0 based on the MLE of theta = 1\/X-bar\n\nI have constructed a reasonable critical region and am being asked to derive a power function for this critical region.\n\nThe critical region I have constructed is {1\/x-bar < c} where c is calculated such that the maximum allowable type I error is some alpha.\n\nIn order to write the power function for this problem, I have to find the distribution of 1\/x-bar. That\u2019s where I\u2019m having some trouble. I can see that x_1 follows an exponential distribution with mean 1\/theta and that the sum of X_i from 1 to n follows a gamma distribution with mean 1\/theta and kappa= n. I think from here I need to make a variable transformation to a chi-squared distribution, but I\u2019m having trouble seeing it. Any help greatly appreciated.\n\n2 points\n4 points \u00b7 18 days ago\n\nSince you are stuck on the distribution of $1\/\\bar{x}$, I'll focus on that and assume that you can derive the power function once you're beyond this hurdle. First, you should recognize that given the PDF of $X$, $X\\sim$Exponential$(\\theta)$; that is, $X$ is exponential with rate $\\theta$.\n\nWith this, you said your rejection region is ${\\frac{1}{\\bar{x} < c}$, which is equivalent to $\\sum_{i=1}n X_i > nc = c'$. Now there are two avenues of approach. You can use the fact that, given iid $X_1, \\dots, X_n$ with $X_i\\sim$Exponential$(\\theta)$, the r.v. $Y = \\sum_{i=1}n X_i \\sim$Gamma$(n, \\theta)$. Another approach is that you could use the fact that $2\\theta \\sum_{i=1}n X_i \\sim \\chi_{2n}2$, and this latter approach will likely be more useful when you're deriving a power function. If you can't take these facts as given, you can readily prove them via moment generating functions. Hope this helps.\n\nsee more\nOriginal Poster3 points \u00b7 18 days ago\n\nThanks! That does make sense. I can see where my initial error was in identifying the distribution of X_i as ~Exp(1\/theta) instead of EXP(theta). I appreciate your help.\n\n1\nCrossposted by18 days ago\nPosted by18 days ago\n\nI am preparing for an MS comprehensive exam in biostatistics and am stuck on a review problem.\n\nWe have a random sample X_1 to X_n from the density f(x;\u03b8)=\u03b8e-\u03b8x, x> 0, \u03b8>0.\n\nWe are being asked to test the hypothesis H0: theta>= theta_0 versus H_a: theta< theta_0 based on the MLE of theta = 1\/X-bar\n\nI have constructed a reasonable critical region and am being asked to derive a power function for this critical region.\n\nThe critical region I have constructed is {1\/x-bar < c} where c is calculated such that the maximum allowable type I error is some alpha.\n\nIn order to write the power function for this problem, I have to find the distribution of 1\/x-bar. That\u2019s where I\u2019m having some trouble. I can see that x_1 follows an exponential distribution with mean 1\/theta and that the sum of X_i from 1 to n follows a gamma distribution with mean 1\/theta and kappa= n. I think from here I need to make a variable transformation to a chi-squared distribution, but I\u2019m having trouble seeing it. Any help greatly appreciated.\n\n2 points\n1\nCrossposted by18 days ago\nPosted by18 days ago\n\nI am preparing for an MS comprehensive exam in biostatistics and am stuck on a review problem.\n\nWe have a random sample X_1 to X_n from the density f(x;\u03b8)=\u03b8e-\u03b8x, x> 0, \u03b8>0.\n\nWe are being asked to test the hypothesis H0: theta>= theta_0 versus H_a: theta< theta_0 based on the MLE of theta = 1\/X-bar\n\nI have constructed a reasonable critical region and am being asked to derive a power function for this critical region.\n\nThe critical region I have constructed is {1\/x-bar < c} where c is calculated such that the maximum allowable type I error is some alpha.\n\nIn order to write the power function for this problem, I have to find the distribution of 1\/x-bar. That\u2019s where I\u2019m having some trouble. I can see that x_1 follows an exponential distribution with mean 1\/theta and that the sum of X_i from 1 to n follows a gamma distribution with mean 1\/theta and kappa= n. I think from here I need to make a variable transformation to a chi-squared distribution, but I\u2019m having trouble seeing it. Any help greatly appreciated.\n\n2 points\n2\nPosted by18 days ago\n\nI am preparing for an MS comprehensive exam in biostatistics and am stuck on a review problem.\n\nWe have a random sample X_1 to X_n from the density f(x;\u03b8)=\u03b8e-\u03b8x, x> 0, \u03b8>0.\n\nWe are being asked to test the hypothesis H0: theta>= theta_0 versus H_a: theta< theta_0 based on the MLE of theta = 1\/X-bar\n\nI have constructed a reasonable critical region and am being asked to derive a power function for this critical region.\n\nThe critical region I have constructed is {1\/x-bar < c} where c is calculated such that the maximum allowable type I error is some alpha.\n\nIn order to write the power function for this problem, I have to find the distribution of 1\/x-bar. That\u2019s where I\u2019m having some trouble. I can see that x_1 follows an exponential distribution with mean 1\/theta and that the sum of X_i from 1 to n follows a gamma distribution with mean 1\/theta and kappa= n. I think from here I need to make a variable transformation to a chi-squared distribution, but I\u2019m having trouble seeing it. Any help greatly appreciated.\n\nu\/luoyun\n2,094\nAugust 21, 2013\nTrophy Case (7)\n\n2016\n\n2016","date":"2019-05-20 21:16:55","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.7350802421569824, \"perplexity\": 976.6912986568783}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2019-22\/segments\/1558232256147.15\/warc\/CC-MAIN-20190520202108-20190520224108-00425.warc.gz\"}"} | null | null |
Беляево — Деревня в Угранском районе Смоленской области России. Входит в состав Знаменского сельского поселения.
Население —2 жителя (2010 год).
Расположена в юго-восточной части области в 30 км к северо-востоку от Угры, в 9 км севернее автодороги Вязьма — Калуга — Тула — Рязань, .
История
В годы Великой Отечественной войны деревня была оккупирована гитлеровскими войсками в октябре 1941 года, освобождена в марте 1943 года.
Примечания
Населённые пункты Угранского района | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 8,143 |
package ezbake.example.ezmongo;
import ezbake.base.thrift.Visibility;
import ezbake.data.common.classification.ClassificationUtils;
import ezbake.data.mongo.redact.RedactHelper;
import ezbake.data.mongo.thrift.EzMongoBaseException;
import ezbake.example.ezmongo.springData.Name;
import ezbake.example.ezmongo.springData.User;
import org.apache.accumulo.core.security.VisibilityParseException;
import org.springframework.context.ApplicationContext;
import org.springframework.context.support.GenericXmlApplicationContext;
import org.springframework.data.mongodb.core.MongoOperations;
import org.springframework.data.mongodb.core.query.Criteria;
import org.springframework.data.mongodb.core.query.Query;
import ezbake.classification.ClassificationConversionException;
/**
* NOTE: Be sure to set the environment variable:
* EZCONFIGURATION_DIR=src/main/resources
* before running this class.
* <p/>
* NOTE: For packing & deploying to Production, your project would NOT include
* this project's files under src/main/resources - specifically, the
* ezbake-config.properties and the client directory.
* They are here just for testing purposes.
* <p/>
* NOTE: The markings.config file is on the classpath (src/main/resources) of this project -
* it has the DerivedFromMarking check disabled.
* If a markings.config file is not provided on the classpath or by an EzConfiguration property
* "markings.path"
* <p/>
* This is an example of using Spring Data with the ezmongo-java-driver.
* You can use this as you would use the regular Spring Data.
* By default, Client mode is enabled so that in the ezmongo-java-driver,
* it will call out to the ezmongo thrift service when making database calls.
* <p/>
* NOTE: Your POJO model classes should inherit from ezbake.data.mongo.EzMongoBasePojo -
* refer to the User.java in this project for an example.
* <p/>
* You can also try running in non-Client mode, by uncommenting out the
* "mongoClientMode=false" in the src/main/resources/ezbake-config.properties.
* That will make it use the default Mongo API with the URI & database listed in the code below,
* instead of hitting the ezmongo thrift service.
* For non-Client mode, make sure to have "mongod" running on your local machine first.
*/
public class EzMongoSpringDataSampleClient {
public static void main(String[] args) throws VisibilityParseException, ClassificationConversionException, EzMongoBaseException {
// For XML
ApplicationContext ctx = new GenericXmlApplicationContext("springDataConfig.xml");
// For Annotation
//ApplicationContext ctx =
// new AnnotationConfigApplicationContext(SpringMongoConfig.class);
MongoOperations mongoOperation = (MongoOperations) ctx.getBean("mongoTemplate");
User user = new User("mkyong", "password123");
// Here, we would insert the security tagging fields into the DBObject by calling a utility class (RedactHelper.java).
Visibility vis = new Visibility();
// Convert CAPCO to Accumulo-style boolean expression string and set it in the Visibility object.
String booleanExpressionString = ClassificationUtils.getAccumuloVisibilityStringFromCAPCO("SECRET");
vis.setFormalVisibility(booleanExpressionString);
RedactHelper.setSecurityFieldsInDBObject(user, vis, "testAppId");
// Also set the Name field in the User
Name name = new Name("testFirstName", "testLastName");
Visibility nameVis = new Visibility();
nameVis.setFormalVisibility(booleanExpressionString);
RedactHelper.setSecurityFieldsInDBObject(name, nameVis, "testAppId");
user.setName(name);
// Call the Provenance service to get a unique ID for the document -
// the unique ID would be used for the Purge feature.
// save
mongoOperation.save(user);
// now user object got the created id.
System.out.println("1. user : " + user);
// query to search user
Query searchUserQuery = new Query(Criteria.where("username").is("mkyong"));
// find the saved user again.
User savedUser = mongoOperation.findOne(searchUserQuery, User.class);
System.out.println("2. find - savedUser : " + savedUser);
}
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 1,190 |
{"url":"http:\/\/umairsidd.com\/publication\/content\/publication\/siddique-2018-formal\/","text":"# Formal Analysis of Discrete-Time Systems using z-Transform\n\nType\nPublication\nJournal of Applied Logic","date":"2020-10-28 05:54:24","metadata":"{\"extraction_info\": {\"found_math\": false, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.9091852307319641, \"perplexity\": 5073.978489067279}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2020-45\/segments\/1603107896778.71\/warc\/CC-MAIN-20201028044037-20201028074037-00390.warc.gz\"}"} | null | null |
\section{Introduction}
\label{s-intro}
Crowdsourcing, where a problem or task is broadcast to a crowd of potential participants for solution, is used for an increasingly wide variety of tasks on the Web. One particularly common application of crowdsourcing is in the context of making evaluations, or judgements--- when the number of evaluations required is too large for a single expert, a solution is to replace the expert by an evaluation aggregated from a `crowd' recruited on an online crowdsourcing platform such as Amazon Mechanical Turk. Crowdsourced judgement elicitation is now used for a number of applications such as image classification and labeling, judging the quality of online content, identifying abusive or adult content, and most recently for peer grading in online education, where Massively Open Online Courses (MOOCs) with enrollment in the hundreds of thousands crowdsource the problem of evaluating assignments submitted by students back to the class itself. While one issue in the context of crowdsourcing evaluations is how best to aggregate the evaluations obtained from the crowd, there is also a key question of {\em eliciting the best possible evaluations} from the crowd in the first place.
The problem of designing incentive mechanisms for such crowdsourced judgement elicitation scenarios has two aspects. First, suppose each worker has already evaluated, or formed a judgement on, the tasks allocated to her. Since the `ground truth' for each task is unknown to the system, a natural solution is to reward workers based on other workers' reports for the same task (this being the only available source of information about this ground truth)\footnote{It is of course infeasible for a requester to monitor every worker's performance on her task, since this would be a problem of the same scale as simply performing all the tasks herself. We also note that a naive approach of randomly checking some subset of evaluations, either via inserting tasks with known responses, or via random checking by the requester, turns out to be very wasteful of effort at the scale neccessary to achieve the right incentives.}. The problem of designing rewards to incentivize agents to truthfully report their observation, rather than, for example, a report that is more likely to agree with other agents' reports, is an {\em information elicitation} problem with unobservable ground truth. Information elicitation has been recently been addressed in the literature in the context of eliciting opinions online (such as user opinions about products, or experiences with service providers); see \S \ref{s-relwork}. However, in those settings, agents (users) have {\em already} formed an opinion after receiving their signal (for example a user who buys a product forms an opinion about it after buying it)--- so agents only need to be incentivized to incur the cost to report this opinion and find it more profitable to report their opinions {\em truthfully} than to report a different opinion.
In the crowdsourcing settings we consider, however, the user does not have such a pre-formed, or experiential, opinion anyway, but rather forms a judgement {\em as part of her task}--- further, the accuracy of this judgement depends on whether or not the agent puts in effort into it (for instance, a worker evaluating whether images contain objectionable content could put in no effort and declare all images to be clean, or put in effort into identifying which images are actually appropriate; a similar choice applies in other contexts like peer-grading). A key issue in these crowdsourced judgement elicitation scenarios is therefore {\em incentivizing effort}\footnote{We thank David Evans (VP Education, Udacity) for pointing out this issue in the context of peer-grading applications--- while students might put in their best efforts on grading screening assignments to ensure they demonstrate the minimum proficiency required to be allowed to grade, how can we be sure that they will continue to work with the same proficiency when grading homeworks outside of this screening set?}--- that is, ensuring that agents {\em make} the best judgements that they possibly can (in addition, of course, to ensuring that they then truthfully report this observation). This leads to a new kind of information elicitation problem where an agent's proficiency now {\em depends} on her effort choice, and so is {\em endogenous} and unknown to the system--- even if an agent's maximum proficiency is known, the actual proficiency with which she performs a task is an endogenous, strategic choice and therefore cannot be assumed as fixed or given.
A mechanism for information elicitation in this setting should make it `most beneficial', if not the only beneficial strategy, for agents to not just {\em report} their observations truthfully, but to also {\em make} the best observations they can in the first place. Also, it is even more important now to ensure that the payoffs from all agents always blindly reporting the same observation (for instance, declaring all content to be good) are strictly smaller than the payoffs from truthfully reporting what was actually observed, since declaring all tasks to be of some predecided type requires no effort and therefore incurs no cost, whereas actually putting in effort into making observations will incur a nonzero cost. Finally, unlike mechanisms designed for settings where a large audience is being polled for its opinion about a single event, a mechanism here must retain its incentive properties even when there are only a few reports per task--- this is because it can be infeasible, due to either monetary or effort constraints, to solicit reports from a large number of agents for each task. (For example, the number of tasks in peer grading scales {\em linearly} with the number of agents, limiting the number of reports available for each task since each student can only grade a few assignments; similarly, the total cost to the requester in crowdsourcing platforms such as Amazon Mechanical Turk scales linearly with the number of workers reporting on each task). How can we elicit the best possible evaluations from agents whose proficiency of evaluation depends on their strategically chosen effort, when the ground truth as well as the effort levels of agents are unobservable to the mechanism? \\
\noindent {\bf Our Contributions.} We introduce a model for information elicitation with {\em endogenous proficiency}, where an agent's strategic choice of whether or not to put in effort into a task endogenously determines her proficiency (the probability of correctly evaluating the ground truth) for that task. We focus on the design of mechanisms for binary information elicitation, \ie, when the underlying ground truth is binary (corresponding to eliciting `good' or `bad' ratings). While generalizing to an arbitrary underlying type space is an immediate direction for further work, we note that a number of interesting judgement and evaluation tasks, for example identifying adult content or correctness evaluation, are indeed binary; also, even very recent literature providing improved mechanisms for information elicitation (e.g.~\cite{witkowski2012robust,witkowski_ec12}), as well as experimental work on the performance of elicitation mechanisms~\cite{prelec2007algorithm,john2012measuring}, focuses on models with binary ground truth.
Our main contribution is a simple, new, mechanism for binary information elicitation for multiple tasks when agents have endogenous proficiencies. Our mechanism has the following incentive properties.
\begin{enumerate}
\item[(i)] Exerting maximum effort followed by truthful reporting of observations is a Nash equilibrium.
\item[(ii)] This is the equilibrium with {\em maximum payoff} to all agents, {\em even} when agents have different maximum proficiencies, can use {\em mixed} strategies, and can choose a different strategy for each of their tasks.
Showing that full-effort truthtelling leads to the maximum reward amongst all equilibria (including those involving mixed strategies) requires arguing about the rewards to agents in all possible equilibria that may arise. To do this, we use a matrix representation of strategies where every strategy can be written as a convex combination of `basis' strategies, so that maximizing a function over the set of all possible strategies is equivalent to a maximization over the space of coefficients in this convex combination. This representation lets us show that the reward to an agent over all possible strategy choices (by herself and other agents), and therefore over all equilibria, is maximized when all agents use the strategy of full-effort truthful reporting.
\item[(iii)] Suppose there is some positive probability, however small, that there is some `trusted' agent for each task who will report on that task truthfully with proficiency greater than half. Then the equilibrium where all agents put in full effort and report truthfully on all their tasks is essentially the only equilibrium of our mechanism, {\em even if} the mechanism does not know the identity of the trusted agents.
\end{enumerate}
We note that our mechanism requires only minimal bounds on the priors and imposes no conditions on a diverging number of agent reports per task to achieve its incentive properties--- to the best of our knowledge, previous mechanisms for information elicitation do not provide all these guarantees simultaneously, even when proficiency is not an endogenously determined choice (see \S\ref{s-relwork} for a discussion).
The main idea behind our mechanism $\M$ is the following. With just one task, it is difficult to distinguish between agreement arising from high-effort observations of the same ground truth, and `blind' agreement achieved by the low-effort strategy of always making the same report. We use the presence of {\em multiple} tasks and ratings to distinguish between these two scenarios and appropriately reward or penalize agents to incentivize high effort---
$\M$ rewards an agent $i$ for her report on task $j$ for agreeing with another `reference' agent $\ipj$'s report on the same task, but also {\em penalizes for blind agreement} by subtracting out a statistic term corresponding to the part of $i$ and $\ipj $'s agreement on task $j$ that is to be `expected anyway' given their reporting statistics estimated from other tasks. This statistic term is chosen so that there is no benefit to making reports that are independent of the ground truth; the incentive properties of the mechanism follow from this property that agents obtain positive rewards {\em only} when they put effort into their evaluations.
\subsection{Related Work}
\label{s-relwork}
The problem of designing incentives for crowdsourced judgement elicitation is closely related to the growing literature on information elicitation mechanisms. The key difference between this literature (discussed in greater detail below) and our work is that agents in the settings motivating past work have opinions that are {\em experientially formed} anyway--- independent, and outside of, any mechanisms to elicit opinions--- so that agents only need be incentivized to participate and truthfully report these opinions. In contrast, agents in the crowdsourcing settings we study {\em do not have such experientially formed opinions} to report--- an agent makes a judgement only because it is part of her task, expending effort to {\em form} her judgement, and therefore must be incentivized to both expend this effort and then to truthfully report her evaluation. There are also other differences in terms of the models and guarantees in previous mechanisms for information elicitation; we discuss this literature below.
The peer-prediction method, introduced by Miller, Resnick and Zeckhauser~\cite{miller2005eliciting}, is a mechanism for the information elicitation problem for general outcome spaces where truthful reporting is a Nash equilibrium, uses proper scoring rules to reward agents for reports that are predictive of other agents' reports. The main difference between our mechanism and~\cite{miller2005eliciting}, as well as other mechanisms based on the peer prediction method~\cite{jurcafaltings06,jurcafaltings07,witkowski_sc11,jurcafaltings09,witkowski_ec12}, is in the model of agent proficiency. In peer-prediction models, while agents can decide whether to incur the cost to {\em participate} (\ie, submit their opinion), an agent's proficiency--- the distribution of opinions or evaluations conditional on ground truth--- is {\em exogenously} determined (and common to all agents, and in most models, known to the center).
That is, an agent might not participate at all, but if she does participate she is assumed to have some known proficiency. In contrast, in our setting, agents can choose not just whether or not to participate, but also {\em endogenously} determine their proficiency conditional on participating through their effort choice. Thus, while peer prediction mechanisms do need to incentivize agents to participate (by submitting a report), they then know the proficiency of agents who do submit a report, and therefore can, and do, dispense rewards that crucially use knowledge of this proficiency. In contrast, even agents who do submit reports in our setting cannot be assumed to be using their maximum proficiency to make their evaluations, and therefore cannot be rewarded based on any assumed level of proficiency.
Additionally, truthtelling, while an equilibrium, is {\em not necessarily the maximum-reward equilibrium} in these existing peer-prediction mechanisms--- \cite{jurca2005enforce} shows that for the mechanisms in~\cite{miller2005eliciting,jurcafaltings06}, the strategies of always reporting `good' or always reporting `bad' both constitute Nash equilibria, at least one of which generates {\em higher} payoff than truthtelling. Such blind strategy equilibria can be eliminated and honest reporting made the unique Nash equilibrium by designing the payments as in~\cite{jurcafaltings09}, but this needs agents to be restricted to pure reporting strategies, and requires full knowledge of the prior and conditional probability distributions to compute the rewards.
The Bayesian Truth Serum (BTS)~\cite{prelec2004bayesian} is another mechanism for information elicitation with unobservable ground truth. BTS does not use the knowledge of a common prior to compute rewards, but rather collects two reports from each agent--- an `information' report which is the agent's own observation, as well as a `prediction' report which is the agent's prediction about the distribution of information reports from the population--- and uses these to compute rewards such that truthful reporting is the highest-reward Nash equilibrium of the BTS mechanism. In addition to the key difference of exogenous versus endogenous proficiencies discussed above, an important limitation of BTS in the crowdsourcing setting is that it requires the number of agents $n$ reporting on a task to diverge to ensure its incentive properties. This $n \rightarrow \infty$ requirement is infeasible in our setting due to the scaling of cost with number of reports as discussed in the introduction. ~\cite{witkowski2012robust} provides a robust BTS mechanism (RBTS) that works even for small populations (again in the same non-endogenous proficiency model as BTS and peer prediction mechanisms), and also ensures payments are positive, making the mechanism ex-post individually rational in contrast to BTS. However, the RBTS mechanism does not retain the property of truthtelling being the highest reward Nash equilibrium--- indeed, the `blind agreement' equilibrium via constant reports achieves the maximum possible reward in RBTS, whereas truthtelling might in fact lead to lower rewards.
There is also work on information elicitation in conducting surveys and online polling~\cite{lambert2008truthful,jurcafb08}, both of which are not quite appropriate for our crowdsourcing setting. The mechanism in~\cite{lambert2008truthful} is weakly incentive compatible (agents are indifferent between lying and truthtelling), while \cite{jurcafb08} presents a online mechanism that is not incentive compatible in the sense that we use and potentially requires a large (constant) number of agents to converge to the true result. For other work on information elicitation, albeit in settings very different from ours, see~\cite{goelrp09,chen2012predicting,papakonstantinou2011mechanism}.
We note also that we model settings where there is indeed a notion of a ground truth, albeit unobservable, so that proficient agents who put in effort are more likely than not to correctly observe this ground truth. Peer-prediction methods, as well as the Bayesian truth serum, are designed for settings where there may be no underlying ground truth at all, and the mechanism only seeks to elicit agents' true observations (whatever they are) which means that some agents might well be in the minority even when they truthfully report their observation--- this makes the peer prediction setting `harder' along the dimension of inducing truthful reports, but easier along the dimension of not needing to incentivize agents to choose to exert effort to make high-proficiency observations.
We note that our problem can also be cast as a version of a principal-agent problem with a very large number of agents, although the principal cannot directly observe an agent's `output' as in standard models. While there is a vast literature in economics on the principal-agent problem too large to describe here (see, eg, ~\cite{bolton2005contract} and references therein), none of this literature, to the best of our knowledge, addresses our problem. Finally, there is also a large orthogonal body of work on the problem of {\em learning} unknown (but {\em exogenous}) agent proficiencies, as well as on the problem of optimally combining reports from agents with differing proficiencies to come up with the best aggregate evaluation in various models and settings. These problems of learning exogenous agent proficiencies and optimally aggregating agent reports are orthogonal to our problem of providing incentives to agents with endogenous, effort-dependent proficiencies to elicit the best possible evaluations from them.
\section{Model}
\label{s-model}
We now present a simple abstraction of the problem of designing mechanisms for crowdsourced judgement elicitation settings where agents' proficiencies are determined by strategic effort choice.
{\em Tasks.} There are $m$ tasks, or objects, $j = 1, \ldots, m$, where each task has some underlying `true quality', or type, $\bar X_j$. This true type $\bar X_j$ is unknown to the system. We assume that the types are binary-valued: $\bar X_j$ is either $H$ (or $1$, corresponding to high-quality) or $L$ (or $0$, for low quality) for all $j$; we use $1$ and $H$ (resp. $0$ and $L$) interchangeably throughout for convenience. The prior probabilities of $H$ and $L$ for all tasks are denoted by $\ph$ and $\pl$. We assume throughout that $\max(\ph,\pl) < 1$, \ie, that there is at least some uncertainty in the underlying qualities of the objects.
{\em Agents.} There are $n$ workers or agents $i = 1, \ldots, n$ who noisily evaluate, or form judgements on, the qualities of objects. We say agent $i$ performs task $j$ if $i$ evaluates object $j$. Agent $i$'s judgement on task $j$ is denoted by $\hat X_{ij} \in \{0,1\}$, where $\hat X_{ij}$ is $0$ if $i$ evaluates $j$ to be of type $L$ and $\hat X_{ij}$ is $1$ if $i$ evaluates it to be $H$. Having made an evaluation $\hat X_{ij}$, an agent can choose to report any value $X_{ij} \in \{0,1\}$ either based on, or independent of, her actual evaluation $\hat X_{ij}$.
We denote the set of tasks performed by an agent $i$ by $J(i)$, and let $I(j)$ denote the set of agents who perform task $j$. We will assume for notational simplicity that $|J(i)| = D$ and $|I(j)| = T$ for all agents $i$ and tasks $j$.
{\em Proficiency.} An agent's {\em proficiency} at a task is the probability with which she correctly evaluates its true type or quality. We assume that an agent's proficiency is an increasing function of the {\em effort} she puts into making her evaluation. Let $e_{ij}$ denote agent $i$'s effort level for task $j$: we assume for simplicity that effort is binary-valued, $e_{ij} \in \{0,1\}$. Putting in $0$ effort has cost $c_{ij}(0) = 0$, whereas putting in full effort has cost $c_{ij}(1) \geq 0$ (we note that our results also extend to a linear model with continuous effort where $e_{ij} \in [0,1]$ and the probability $p_i(e_{ij})$ of correctly observing $\bar X_j$ as well as the cost $c_i(e_{ij})$ increase linearly with $e_{ij}$).
An agent who puts in zero effort makes evaluations with proficiency $p_{ij}(0) = 1/2$ and does no better than random guessing, \ie, $\Pr(\hat X_{ij} = \bar X_j|e_{ij} = 0) =1/2$. An agent who puts in full effort $e_{ij} = 1$ attains her {\em maximum proficiency}, $\Pr(\hat X_{ij} = \bar X_j|e_{ij} = 1) = p_{ij}(1) = p_i$. Note that this maximum proficiency $p_i$ can be different for individual agents modeling their different abilities, and need {\em not} be known to the center. We assume that the maximum proficiency $p_i \ge \frac{1}{2}$ for all $i$--- this minimum requirement on agent ability can be ensured in online crowdsourcing settings by {\em prescreening} workers on a representative set of tasks (Amazon Mechanical Turk, for instance, offers the ability to prescreen workers~\cite{paolacci2010running,harris11csdm}, whereas in peer-grading applications such as on Coursera, students are given a set of pre-graded assignments to measure their grading abilities prior to grading their peers, the results of which can be used as a prescreen.)
We note that our results also extend easily to the case where the maximum proficiency of an agent {\em depends} on whether the object is of type $H$ or $L$, \ie, the probabilities of correctly observing the ground truth when putting in full effort are different for different ground truths, $\Pr(\hat X_{ij} = \bar X_j| \bar X_j = H) \neq \Pr(\hat X_{ij} = \bar X_j| \bar X_j = L)$ (of course, different agents can continue to have different maximum proficiences).
{\em Strategies.} Agents strategically choose {\em both} their effort levels and reports on each task to maximize their total utility, which is the difference between the reward received for their reports and the cost incurred in making evaluations. Formally, an agent $i$'s strategy is a vector of $D$ tuples $[(e_{ij}, f_{ij})]$, specifying her effort level $e_{ij}$ as well as the function $f_{ij}$ she uses to {\em map} her actual evaluation $\hat X_{ij}$ into her report $X_{ij}$ for each of her tasks. Note that since an agent's proficiency on a task $p_{ij}$ is a function of her strategically chosen effort $e_{ij}$, the proficiency of agent $i$ for task $j$ is {\em endogenous} in our model.
For a single task, we use the notation $(1, \truth)$ to denote the choice of full effort $e_{ij} = 1$ and truthfully reporting one's evaluation (\ie, $f_{ij}$ is the identity function $X_{ij} = \hat X_{ij}$), $(1, \invert)$ to denote full effort followed by inverting one's evaluation, and $(0, r)$ to denote the choice of exerting no effort ($e_{ij} = 0$) and simply reporting the outcome of a random coin toss with probability $r$ of returning $H$. We use $[(1,\truth)]$ to denote the strategy of using full effort and truthtelling on all of an agent's tasks, and similarly $[(1, X^c)]$ and $[(0,r)]$ for the other strategies.
{\em Mechanisms.} A mechanism in this setting takes as input the set of all received reports $X_{ij}$ and computes a reward for each agent based on her reports, as well as possibly the reports of other agents. Note that the mechanism has no access\footnote{Crowdsourcing is used typically precisely in scenarios where the number of tasks is too large for the principal (or a set of trusted agents chosen by the principal) to carry out herself, so it is at best feasible to verify the ground truth for a tiny fraction of all tasks, which fraction turns out to be inadequate (a formal statement is omitted here) to incentivize effort using knowledge of the $\bar X_j$.} to the underlying true qualities $\bar X_j$ for any task, and so cannot use the $\bar X_j$ to determine agents' rewards. A set of effort levels and reporting functions $[(e_{ij},f_{ij})]$ is a full-information Nash equilibrium of a mechanism if no agent $i$ can strictly improve her expected utility by choosing either a different effort level $\hat e_{ij}$, or a different function $\hat f_{ij}$ to map her evaluation $\hat X_{ij}$ into her report $X_{ij}$. Here, the expectation is over the randomness in agents' noisy evaluation of the underlying ground truth, as well as any randomness in the mechanism.
We will be interested in designing mechanisms for which it is (i) an equilibrium for all agents to put in full effort and report their evaluations truthfully on all tasks, \ie, use strategies $[(1, \truth)]$, and (ii) for which $[(1, \truth)]$ is the maximum utility (if not unique) equilibrium. We emphasize here that we do not address the problem of how to optimally aggregate the $T$ reports $X_{ij}$ for task $j$ into a final estimate of $\bar X_j$: this is an orthogonal problem requiring application-specific modeling; our only goal is to elicit the best possible judgements to aggregate, by ensuring that agents find it most profitable to put in maximum effort into their evaluations and then report these evaluations truthfully.
\section{Mechanism}
\label{s-mech}
The main idea behind our mechanism $\M$ is following. Recall that a mechanism does not have access to the true qualities $\bar X_j$, and therefore must compute rewards for agents that do not rely on directly observing $\bar X_{j}$. Since the only source of information about $\bar X_{j}$ comes from the reports $X_{ij}$, a natural solution is to reward based on some form of agreement between different agents reporting on $j$, similar to the peer-prediction setting~\cite{miller2005eliciting}. However, an easy way for agents to achieve perfect agreement with no effort is to always report $H$ (or $L$). With just one task, it is difficult for a mechanism to distinguish between the scenario where agents achieve agreement by making accurate, high-effort, evaluations of the same ground truth, and the low-effort scenario where agents achieve agreement by always reporting $H$, especially if $\ph$ is high. However, in our setting, we have the benefit of {\em multiple} tasks and ratings, which could potentially be used to distinguish between these two strategies and appropriately reward agents to incentivize high effort.
$\M$ uses the presence of multiple ratings to subtract out a {\em statistic} term $B_{ij}$ from the agreement score, chosen so that there is no benefit to making reports that are independent of $\bar X_j$---
roughly speaking, $\M$ rewards an agent $i$ for her report on task $j$ for agreeing with another `reference' agent $\ipj$'s report on the same task, but {\em only beyond} what would be expected if $i$ and $\ipj$ were randomly tossing coins with their respective empirical frequencies of heads.
Let $d$ denote the number of other reports made by $i$ and $\ipj$ that are used in the computation of this statistic term $B_{ij}$ based on the observed frequency of heads for each pair $(i,j)$. We use $\M_d$ to denote the version of $\M$ which uses $d$ other reports from each of $i$ and $\ipj$ to compute $B_{ij}$. To completely specify $\M_d$, we also need to specify a reference rater $\ipj $ as well as this set of $d$ (non-overlapping) tasks performed by $i$ and $\ipj$, for which we use the following notation. (We require these $d$ other tasks to be non-overlapping so that the reports for these tasks $X_{ik}$ and $X_{\ipj l}$ are independent\footnote{We assume that co-raters' identities are kept unknown to agents, so there is no collusion between $i$ and $\ipj$.}, which is necessary to achieve the incentive properties of $\M_d$.)
\begin{definition}[$S_{ij}, S_{\ipj j}$] \label{d-Sij} Consider agent $i$ and task $j \in J(i)$, and a reference rater $\ipj$. Given a value of $d$ ($1 \leq d \leq D-1$), let $S_{ij}$ and $S_{\ipj j}$ be sets of $d$ non-overlapping tasks other than task $j$ performed by $i$ and $\ipj$ respectively, \ie,
\begin{align*}
& S_{ij} \subseteq J(i)\setminus j, \quad S_{\ipj j} \subseteq J(\ipj)\setminus j, \quad S_{ij}\cap S_{\ipj j} = \emptyset, \quad |S_{ij}|= |S_{\ipj j}| = d.
\end{align*}
\end{definition}
A mechanism $\M_d$ is completely specified by reference raters $\ipj$ and the sets $S_{ij}$ and $S_{\ipj j}$, and rewards agents as defined below. Note that $\M_d$ only uses agents' {\em reports} $X_{ij}$ to compute rewards and not their maximum {\em proficiencies} $p_i$, which therefore need not be known to the system.
\begin{definition}[Mechanism $\M_d$] \label{d-mech} $\M_d$ computes an agent $i$'s reward for her report $X_{ij} \in \{0,1\}$ on task $j$, $R_{ij}$, by comparing against a `reference rater' $\ipj$'s report $X_{\ipj j}$ for $j$, as follows:
\begin{align}
R_{ij} &= A_{ij} - B_{ij}, \quad \mbox{where}\\
A_{ij}&= X_{ij}X_{\ipj j} + (1-X_{ij})(1-X_{\ipj j}), \quad \mbox{and}\nonumber \\
B_{ij} &= (\frac{\sum_{k\in S_{ij}} X_{ik}}{d})(\frac{\sum_{l\in S_{\ipj j}} X_{\ipj l}}{d}) + (1-\frac{\sum_{k\in S_{ij}} X_{ik}}{d})(1-\frac{\sum_{l\in S_{\ipj j}} X_{\ipj l}}{d} ),
\end{align}
where the sets $S_{ij}$ and $S_{\ipj j}$ in $B_{ij}$ are as in Definition \ref{d-Sij}. The final reward to an agent $i$ is $\beta R_i$, where $R_i = \sum_{j \in J(i)} R_{ij}$ and $\beta$ is simply a non-negative scaling parameter that is chosen based on agents' costs of effort.
\end{definition}
The first term, $A_{ij}$, in $R_{ij}$ is an `agreement' reward, and is $1$ when $i$ and $\ipj $ both agree on their report, \ie, when $X_{ij} = X_{\ipj j} = 1$ or when $X_{ij} = X_{\ipj j} = 0$. The second term $B_{ij}$ is the `statistic' term which, roughly speaking, deducts from the agreement reward whatever part of $i$ and $\ipj $'s agreement on task $j$ is to be `expected anyway' given their reporting statistics, \ie, the relative frequencies with which they report $H$ and $L$. This deduction is what gives $\M$ its nice incentive properties--- while $\M$ rewards agents for agreement via $A_{ij}$,
$\M$ also {\em penalizes for blind agreement} that agents achieve without effort, by subtracting out the $B_{ij}$ term corresponding to the expected frequency of agreement if $i$ and $\ipj $ were randomly choosing reports corresponding to their estimated means.
For example, suppose all agents were to always report $H$. Then $A_{ij}$ is always $1$, but $B_{ij}=1$ as well so that the net reward is $0$; similarly if agents chose their reports according to a random cointoss, even one with the `correct' bias $\ph$, the value of $A_{ij}$ is exactly equal to $B_{ij}$ since there is no correlation between the reports for a particular task, again leading to a reward of $0$. The reward function $R_{ij}$ is designed so that it {\em only} rewards agents when they put in effort into their evaluations, which leads to the desirable incentive properties of $\M_d$. (We note that there are other natural statistics which might incentivize agents away from low-effort reports--- \eg, rewarding reports which collectively have an empirical mean close to $\ph$, or for variance. However, it turns out that appropriately balancing the agreement term (which is necessary to ensure agents cannot simply report according to a cointoss with bias $\ph$) with a term penalizing blind agreement to simultaneously ensure that $[(1, \truth)]$ is an equilibrium {\em and} the most desirable equilibrium is hard to accomplish.)
There are two natural choices for the parameter $d$, \ie, how many reports of $i$ and $\ipj $ to include for estimating the statistic term that we subtract from the agreement score in $R_{ij}$\footnote{Understanding the effect of the parameter $d$ in our mechanisms, which appears irrelevant to the mechanism's behavior when agents are risk-neutral, is an interesting open question.}.
(i) In $\M_{D-1}$, we set $d = D-1$ and include all reports of agents $i$ and $\ipj$, except those on their common task $j$. Here, the non-overlap requirement for sets $S_{ij}$ and $S_{\ipj j}$ says that an agent $i$ and her reference rater $\ipj$ for task $j$ have {\em only} that task $j$ in common.
(ii) In $\M_1$, we set $d = 1$, \ie, subtract away the correlation between the report of $i$ and $\ipj$ on exactly one other non-overlapping task.
In $\M_1$, the non-overlap condition only requires that for each agent-task pair, there is a reference agent $\ipj$ available who has rated one other task that is different from the remaining tasks rated by $i$, a condition that is much easier to satisfy than that in $\M_{D-1}$.
In \S~\ref{s-eq}, we will see that $\M_1$ will require that the choices of $(j,j')$, where $\{j'\} = S_{ij}$ is the task used in the statistic term of $i$'s reward for task $j$, are such that each task $j'$ performed by $i$ is used exactly once to determine $R_{ij}$ for $j \neq j'$. Note that this is always feasible, for instance by using task $j+1$ in the statistic term for task $j$ for $j = 1, \ldots, D-1$ and task $1$ for task $D$.
\section{Analyzing $\M$}
\label{s-eq}
In this section, we analyze equilibrium behavior in $\M_d$. We begin with some notation and preliminaries.
\subsection{Preliminaries}
Recall that proficiency is the probability of correctly evaluating the true quality. We use $\sph$ (respectively $\spl$) to denote the probability that an agent observes $H$ (respectively $L$) when making evaluations with proficiency $p$, \ie,
the probability that $\hat X_{ij}=H$ is $\sph = p\ph + (1 - p)\pl$.
Similarly, $q[H], q[L]$ and $p_i[H], p_i[L]$ correspond to the probabilities of seeing $H$ and $L$ when making evaluations with proficiencies $q$ and $p_i$ respectively. \\
\noindent{\em Matrix representation of strategies.} We will frequently need to consider the space of all possible strategies an agent may use in the equilibrium analysis of $\M_d$.
While the choice of effort level $e_{ij}$ in an agent's strategy $[(e_{ij}, f_{ij})]$ is easily described--- there are only two possible effort levels $1$ and $0$--- the space of functions $f_{ij}$ through which an agent can map her evaluation $\hat X_{ij}$ into her report $X_{ij}$ is much larger. For instance, an agent could choose $f_{ij}$ corresponding to making an evaluation, performing a Bayesian update of her prior on $\bar X_j$, and choosing the report with the higher posterior probability.
We now discuss a way to represent strategies that will allow us to easily describe the set of all reporting functions $f_{ij}$.
An agent $i$'s evaluation $\hat X_{ij}$ can also be written as a two-dimensional vector $o^{ij} \in {\mathbb R}^2$, where $o^{ij} = \begin{bmatrix} 1 & 0 \end{bmatrix}^T$ if $i$ observes a $H$, and $o^{ij} = \begin{bmatrix} 0 & 1 \end{bmatrix}^T$ if $i$ observes a $L$, where ${\bf a}^T$ denotes the transpose of ${\bf a}$. For the purpose of analyzing $\M_d$, any choice of reporting function $f_{ij}$ can then be described via a $2 \times 2$ matrix
\[
M^{ij} = \begin{bmatrix}
x & 1-y\\
1-x & y
\end{bmatrix},
\]
where $x$ is the probability with which $i$ chooses to report $H$ after observing $H$, \ie, $x = \Pr(X_{ij} = H | \hat X_{ij} = H)$, and similarly $y = \Pr(X_{ij} = L | \hat X_{ij} = L)$. Observe that the choice of effort $e_{ij}$ affects {\em only} $o^{ij}$ and its `correctness', or correlation with the (vector representing the) actual quality $\bar X_j$, and the choice of reporting function $f_{ij}$ {\em only} affects $M^{ij}$.
Any reporting matrix $M^{ij}$ of the form above can be written as a convex combination of four matrices--- one for each of the $f_{ij}$ corresponding to (i) truthful reporting ($X_{ij} = \hat X_{ij}$) (ii) inverting ($X_{ij} = \hat X_{ij}^c$), and (iii, iv) always reporting $H$ or $L$ independent of one's evaluation ($X_{ij} = H$ and $X_{ij} = L$ respectively):
\[
M_{X} = \begin{bmatrix}
1 & 0\\
0 & 1
\end{bmatrix},
M_{X^c} = \begin{bmatrix}
0 & 1\\
1 & 0
\end{bmatrix},
M_{H} = \begin{bmatrix}
1 & 1\\
0 & 0
\end{bmatrix},
M_{L} = \begin{bmatrix}
0 & 0\\
1 & 1
\end{bmatrix}.
\]
That is, $M^{ij} = \al_1 M_{X} + \al_2 M_{X^c} + \al_3 M_{H} + \al_4 M_{L}$, where $\al_1 = x-\al_3$, $\al_2 = 1-y-\al_3$, and $\al_3 = x-y$ and $\al_4 = 0$ if $x \geq y$, and $\al_3 = 0$ and $\al_4 = y-x$ if $y > x$. It is easily verified that $\al_i \geq 0$, and $\sum \al_i = 1$, so that this is a convex combination.
Since all possible reporting strategies $f_{ij}$ can be described by appropriately choosing the values of $x \in [0,1]$ and $y \in [0,1]$ in $M^{ij}$, every reporting function $f_{ij}$ can be written as a convex combination of these four matrices.
The agent's final report $X_{ij}$ is then described by the vector $M^{ij}o^{ij} \in {\mathbb R}^2$, where the first entry is the probability that $i$ reports $H$, \ie, $X_{ij} = H$, and the second entry is the probability that she reports $X_{ij}= L$. The expected reward of agent $i$ for task $j$ can therefore be written using the matrix-vector representation (where $^T$ denotes transpose and $\ones$ is the vector of all ones) as
\begin{align*}
E[R_{ij}] &= E[(M^{\ipj j}o^{\ipj j})^T M^{ij}o^{ij} + (\ones-M^{\ipj j}o^{\ipj j})^T(\ones-M^{ij}o^{ij})] \\
&- [(M^{\ipj j}E[o^{\ipj j}])^TM^{ij}E[o^{ij}] + (\ones-M^{\ipj j}E[o^{\ipj j}])^T(1-M^{ij}E[o^{ij}])],
\end{align*}
which is linear in $M^{ij}$. So the payoff from an arbitrary reporting function $f_{ij}$ can be written as the corresponding linear combination of the payoffs from each of the `basis' functions (corresponding to $M_X, M_{X^c}, M_H$ and $M_L$) constituting $f_{ij} = M^{ij}$. We will use this to argue that it is adequate to consider deviations to each of the remaining basis reporting functions and show that they yield strictly lower reward to establish that $[(1,X)]$ is an equilibrium of $\M_d$. \\
\noindent{\em Equivalent strategies.} For the equilibrium analysis, we will use the following simple facts.
(i) The strategy $(0,X)$ (\ie, using zero effort but truthfully reporting one's evaluation) is equivalent to the strategy $(0,r)$ with $r = 1/2$, \ie, to the strategy of putting in no effort, and randomly reporting $H$ or $L$ independent of the evaluation $\hat X_{ij}$ with probability $1/2$ each.
(ii) The strategy $(1,r)$ is equivalent to the strategy $(0,r)$, since the report $X_{ij}$ in both cases is completely independent of the evaluation $\hat X_{ij}$ and therefore of $e_{ij}$.\\
\noindent{\bf Cost of effort.} While agents do incur a higher cost when using $e_{ij} = 1$ as compared to $e_{ij} = 0$, we will not need to explicitly deal with the cost in the equilibrium analysis--- if the reward from using a strategy where $e_{ij}=1$ is strictly greater than the reward from any strategy with $e_{ij}=0$, the rewards $R_{ij}$ can always be scaled appropriately using the factor $\beta$ (in Definition \ref{d-mech}) to ensure that the net utility (reward minus cost) is strictly greater as well.
We remark here that bounds on this scaling factor $\beta$ could be estimated empirically without requiring knowledge of the priors by estimating the cost of effort $c_{ij}$ from the maximum proficiencies obtained from a pre-screening (\S \ref{s-model}), by conducting a series of trials with increasing rewards and then using individual rationality to estimate the cost of effort from observed proficiencies in these trials. \\
\noindent{\bf Individual rationality and non-negativity of payments.} The expected payments made by our mechanism to each agent are always nonnegative in the full-effort truthful reporting equilibrium, \ie, when all agents use strategies $[(1,X)]$. To ensure that the payments are also non-negative for every instance (of the tasks and reports) and not only in expectation, note that it suffices to add $1$ to the payments currently specified, since the penalty term $B_{ij}$ in the reward $R_{ij}$ is bounded above by $1$.
We also note that individual rationality can be achieved by using a value of $\beta$ large enough to ensure that the net utility $\beta R_{ij} - c(1)$ remains non-negative for all values of $\ph$--- while the expected payment $R_{ij}$ does go to zero as $\ph$ tends to $1$ (\ie, in the limit of vanishing uncertainty as the underlying ground truth is more and more likely to always be $H$ (or always be $L$)), as long as there is some bound $\epsilon > 0$ such that $\max\{\ph, \pl\} \leq 1- \eps$, a simple calculation can be used to determine a value $\beta^*(\eps)$ such that the resulting mechanism with $\beta = \beta^*$ leads to nonnegative utilities for all agents in the full-effort truth-telling Nash equilibrium of $\M$.
\subsection{Equilibrium analysis}
We now analyze the equilibria of $\M_d$. Throughout, we index the tasks $J(i)$ corresponding to agent $i$ by $j \in \{1,\ldots,D\}$.
First, to illustrate the idea behind the mechanism, we prove the simpler result that $[(1, \truth)]$ is an equilibrium of $\M$ when agents all have equal proficiency $p_i = p$, and are restricted to choosing one common strategy for all their tasks.
\begin{proposition}
\label{p-easy}
Suppose all agents have the same maximum proficiency $p$, and are restricted to choosing the same strategy for each of their tasks. Then, all agents choosing $[(1, \truth)]$ is an equilibrium of $\M_d$ for all $d$, if $p \neq 1/2$.
\end{proposition}
\begin{proof}
Consider an agent $i$, and suppose all other agents use the strategy $(1, \truth)$ on all their tasks. As discussed in the preliminaries,
an agent's reward is linear in her reports fixing the strategies of other agents, so
it will be enough to show that there is no beneficial deviation to $(1,\invert)$, or $(0, r)$ for any $r \in [0,1]$ to establish an equilibrium.
(as noted earlier, the choice of effort level is irrelevant when reporting $X_{ij}$ according to a the outcome of a random coin toss independent of the observed value of $X_{ij}$). The reward to agent $i$ when she uses strategy $[(1,\truth)]$ is
\begin{align*}
E[R_i((1,\truth))] &= \sum_{j=1}^D E\left[X_{ij}X_{\ipj j} + (1-X_{ij})(1-X_{\ipj j})\right]\\
&- E\left[\frac{\sum_{k \in S_{ij}} X_{ik}}{d}\frac{\sum_{l \in S_{\ipj j}} X_{\ipj l}}{d} + (1-\frac{\sum_{k \in S_{ij}} X_{ik}}{d})(1-\frac{\sum_{l \in S_{\ipj j}} X_{\ipj l}}{d} )]\right]\\
&= D\left[p^2 + (1-p)^2 - (\sph^2 + (1-\sph)^2)\right]\\
&= D(p- p(H))(p- p(L))\\
&= D(2p-1)^2 \ph\pl,
\end{align*}
which is strictly positive if $p \neq 1/2$ and $\min(\ph,\pl) > 0$ (as assumed throughout), where we use $p - \sph = (2p-1)\pl$ and $p - \spl = (2p-1)\ph$.
The expected reward from deviating to $(1,\invert)$, when other agents are using $(1,\truth)$ is
\begin{align*}
E[R_i((1,\invert))] &= D\left(2p(1-p)- 2\sph (1-\sph)\right) = -D(p-p(H))(p- p(L)).
\end{align*}
Therefore, the expected reward from deviating to $(1,\invert)$ is negative and strictly smaller than the reward from $(1,\truth)$ if $p \neq 1/2$.
Finally, suppose agent $i$ deviates to playing $(0,r)$, \ie, reporting the outcome of a random coin toss with bias $r$ as her evaluation of $X_{ij}$. Her expected reward from using this strategy when other agents play according to $(1,\truth)$ is
\begin{align*}
E[R_i((0,r))] &= D\left(r\sph + (1-r)\sph - (r\sph + (1-r)(1-\sph))\right) = 0.
\end{align*}
(In fact, if either agent reports ratings on her tasks by tossing a random coin with any probability $r \in [0,1]$, independent of the underlying true realization of $X_{ij}$, the expected reward to agent $i$ is $0$.)
Therefore, if $p \neq 1/2$, deviating from $(1,\truth)$ leads to a strict decrease in reward to agent $i$. Hence, the rewards $R_{ij}$ can always be scaled appropriately to ensure that $[(1,\truth)]$ is an equilibrium of $\M$ for any values of the costs $c_i$.
\end{proof}
We will now move on to proving our main equilibrium result for $\M_d$, where agents can have different maximum proficiencies, as well as possibly use different strategies for different tasks. We begin with a technical lemma and a definition.
\begin{lemma}
\label{l-fpq}
Let $f_{\alpha}(p,q) = pq+(1-p)(1-q) - \alpha(\sph\sqh + (1-\sph)(1-\sqh))$.
If $\alpha \leq 1$,
(i) $f_{\alpha}(p,q)$ is strictly increasing in $p$ if $q > 1/2$, and strictly increasing in $q$ if $p > 1/2$.
(ii) $f_{\alpha}(p,q)$ is nonnegative if $p, q \geq 1/2$, and positive if $p,q > 1/2$.
(iii) Denote $f(p, q) \triangleq f_1(p, q)$. Then, $f(p, q) = f(q, p) = f(1-p, 1-q)$. Also $f(p, 1-q) = f(1-p, q) = -f(p, q)$.
\end{lemma}
\begin{proof}
Recall that $\sph = p\ph + (1-p)\pl$, and similarly for $\sqh$.
{\small
\begin{align*}
f_{\alpha}(p,q) &= p(2q - 1) + (1-q) -\alpha(p\ph+(1-p)\pl)(2\sqh - 1) - (1-\sqh) \\
& = p\left[(2q-1)- \alpha(\ph-\pl)(2\sqh - 1)\right] + K_{-p}\\
&= p(2q-1)(1-\alpha(\ph-\pl)^2) + K_{-p},
\end{align*} }
where $K_{-p}$ is a term that does not depend on $p$, and we use $2\sqh - 1 = (2q-1)(\ph-\pl)$ in the last step.
Note that $\ph - \pl < \ph < 1$ if $\max(\ph,\pl) < 1$, so that $1-\alpha(\ph-\pl)^2 > 0$ if $\alpha \leq 1$. Therefore, $f_{\alpha}(p,q)$ is linear in $p$ with strictly positive coefficient when $q > 1/2$ and $\alpha \leq 1$. An identical argument can be used for $q$ since $f_{\alpha}(p,q)$ can be written as a linear function of $q$ exactly as for $p$:
\begin{align*}
f_{\alpha}(p,q) = q(2p-1)(1-\alpha(\ph-\pl)^2) + K_{-q}.
\end{align*}
This proves the first claim.
For nonnegativity of $f_{\alpha}(p,q)$ on $p \in [1/2,1]$, we simply argue that $f_{1}(p,q)$ is increasing in $q$ when $p \in [1/2,1]$, and $0$ at $q = 1/2$. So for any $q > 1/2$, $f_1(p, q) \geq 0$ for any $p \in [1/2,1]$. But $f_{\alpha}(p,q)$ is decreasing in $\alpha$, so $f_{\alpha}(p,q)$ is nonnegative for any $\alpha \leq 1$ as well.
The final claims about $f(p, q)$ and $f(1-p, q)$ can be verified just by substituting the definitions of $\sph$ and $\sqh$ and from symmetry in $p$ and $q$.
\end{proof}
\begin{definition}[$T_{ij}$, $d_{ij}$]\label{d-Tij} Let $T_{ij}$ be the set of all tasks $j' \neq j$ such that $j \in S_{ij'}$, \ie, $T_{ij}$ is the set of tasks $j'$ for which $i$'s report on task $j$ is used to compute the statistic term of $i$'s reward $R_{ij'}$ for task $j'$. We use $d_{ij} = |T_{ij}|$ to denote the number of such tasks $j'$.
\end{definition}
Our main equilibrium result states that under a mild set of conditions on the choice of reference raters $\ipj$ and sets $T_{ij}$, exerting full effort and reporting truthfully on all tasks is an equilibrium of $\M_d$--- {\em even when} agents have different maximum proficiencies and can choose a different strategy for each task (for instance, an agent could choose to shirk effort on some tasks and put in effort on the others). The main idea behind this result can be understood from the proof of Proposition {p-easy} above, where all agents had the same maximum proficiency $p_i = p$ and were restricted to using the same strategy for each task. There, the expected payoff from using $[(1,\truth)]$ is exactly $f(p,p)$ where $f$ is as defined in Lemma \ref{l-fpq}, while the payoff from playing $[(0,r)]$ is $0$ (independent of other agents' strategies); the payoff from deviating to $[(1,\invert)]$ when other agents play $[(1,\truth)]$ is $-f(p,p)$. Since $f(p,p) > 0$ for $p > 1/2$ and increases with $p$, it is a best response for every agent to attain maximum proficiency and truthfully report her evaluation.
Extending the argument when agents can have both different maximum proficiencies $p_i$ and use different strategy choices for each task requires more care, and are what necessitate the conditions on the task assignment in Theorem \ref{t-eqbm} below. We note that these conditions on $\M_d$ arise because of the generalization to {\em both} differing abilities $p_i$ {\em and} being allowed to choose a different strategy for each task--- if either generalization is waived, \ie, if agents can choose different strategies per task but all have equal ability ($p_i = p$), {\em or} agents can have different abilities $p_i$ but are restricted to choosing the same strategy for all their tasks, $[(1,X)]$ can be shown to be an equilibrium of $\M_d$ even without imposing these conditions.
\begin{theorem}
\label{t-eqbm}
Suppose $p_i > 1/2$ for all $i$, and for each agent $i$, for each task $j \in J(i)$, (i) $d_{ij} = d$, and (ii) $E[p_{\ipj}] = E_{j_l \in T_{ij}}[p_{\ipjl}] \triangleq \pib$, where the expectation is over the randomness in the assignment of reference raters to tasks and the sets $T_{ij}$.
Then, $[(1, \truth)]$ is an equilibrium of $\M_d$.
\end{theorem}
The first condition in Theorem~\ref{t-eqbm}, $d_{ij}=d$, says that each task $j$ performed by an agent $i$ must contribute to computing the reward via the statistic term for exactly $d$ other tasks in $J(i)$, where $d$ is the number of reports used to compute the `empirical frequency' of $H$ reports by $i$ in the statistic term. The second condition $E[p_{\ipj}] = E_{j_l \in T_{ij}}[p_{r_{j_l}(i)}]$ says that an agent $i$ should expect the average proficiency of her reference rater $\ipj$ to be equal for all the tasks that she performs, \ie, agent $i$ should not be able to identify any particular task where her reference raters are, on average, worse than the reference raters for her other tasks (intuitively, this can lead to agent $i$ shirking effort on this task being a profitable deviation). The first condition holds for each of the two specific mechanisms $\M_1$ and $\M_{D-1}$, and the second condition can be satisfied, essentially, by a randomization of the agents before assignment, as described in \S\ref{s-graph}. We now prove the result.
\begin{proof}
Consider agent $i$, and suppose all other agents use strategy $[(1, \truth)]$, \ie, put in full effort with truthtelling on all their tasks. It will be enough to consider pure strategy deviations, and show that there is no beneficial deviation to $(1,\invert)$, or $(0, r)$ for any $r \in [0,1]$ on any single task or subset of tasks.
First, consider a particular assignment of reference raters $\ipj$ and the sets $S_{ij}$ (and therefore $T_{ij}$). The total expected reward to agent $i$ from all her $D$ tasks in this assignment, when other agents all play according to $[(1, \truth)]$ is
{\small
\begin{align*}
& E[R_i] = \sum_{j=1}^D E[X_{ij}X_{\ipj j} + (1-X_{ij})(1-X_{\ipj j})] - \left[\frac{\sum_{k \in S_{ij}} E[X_{ik}]}{d}\sphipj + (1-\frac{\sum_{k \in S_{ij}} E[X_{ik}]}{d})(1-\sphipj)\right]\\
&= \sum_{j=1}^D E[X_{ij}X_{\ipj j} + (1-X_{ij})(1-X_{\ipj j})] - \left[\frac{\sum_{k \in S_{ij}} E[X_{ik}]}{d}(2\sphipj-1) + (1-\sphipj)\right]\\
&= \sum_{j=1}^D \Big[ E\left[X_{ij}X_{\ipj j} + (1-X_{ij})(1-X_{\ipj j})\right] - \sum_{j_{l}\in T_{ij}} \left(\frac{E[X_{ij}] }{d}(2p_{r_ {j_l}(i)}[H]-1)\right) - (1-\sphipj) \Big],
\end{align*} }
where the expectation is over any randomness in the strategy of $i$ as well as randomness in $i$ and $\ipj$'s evaluations for each task $j$, and we rearrange to collect $X_{ij}$ terms in the last step.
Now, agent $i$ can receive different reference raters and task sets $S_{ij}$ in different assignments.
So to compute her expected reward, agent $i$ will also take an expectation over the randomness in the assignment of reference raters to tasks and the sets $S_{ij}$, which appear in the summation above via $T_{ij}$.
Recall the condition that $E[p_{\ipj}] = E_{j_l \in T_{ij}}[p_{\ipjl}] \triangleq \pib$. Using this condition and taking the expectation over the randomness in the assignments of $\ipj $ and $S_{ij}$, the expected reward of $i$ is
{\small
\begin{align*}
E[R_i] &= \sum_{j=1}^D \Big[ E\left[X_{ij}X_{\ipj j} + (1-X_{ij})(1-X_{\ipj j})\right] - \sum_{j_{l}\in T_{ij}} \left(\frac{E[X_{ij}] }{d}(2\sphib-1)\right) - (1-\sphib) \Big]\\
&= \sum_{j=1}^D \Big[ E\left[X_{ij}X_{\ipj j} + (1-X_{ij})(1-X_{\ipj j})\right] - \frac{d_{ij}}{d}E[X_{ij}] (2\sphib-1) - (1-\sphib) \Big],
\end{align*} }
where $\sphib = E[p_{\ipj}[H]]=E_{j_l \in T_{ij}}[p_{\ipjl}[H]]$.
The expected reward to agent $i$, when she makes evaluations with proficiency $q_j$ for task $j$ and truthfully reports these evaluations ($X_{ij} = \hat X_{ij}$), is then
{\small
\begin{align}
\label{erew-pt}
E[R_i] & = \sum_{j=1}^D \Big[q_j\pib + (1-q_j)(1-\pib) -\frac{d_{ij}}{d}\sqjh(2\sphib-1) - (1-\sphib) \Big]\nonumber\\
&= \sum_{j=1}^D \Big[q_j\pib + (1-q_j)(1-\pib) - \frac{d_{ij}}{d}\left(\sqjh\sphib + (1-\sqh)(1-\sphib)\right) - (1-\frac{d_{ij}}{d})(1-\sphib)\Big]\nonumber\\
&= \sum_{j=1}^D \left[f_{\frac{d_{ij}}{d}}(q_j,\pib) + (\frac{d_{ij}}{d} - 1)(1-\sphib)\right].
\end{align} }
where the expectation is taken over randomness in all agents' evaluations, as well as over randomness in the choices of $\ipj $ and $S_{ij}$.
We can now show that choosing full effort and truthtelling on all tasks is a best response when all other agents use $[(1, \truth)]$ if $d_{ij} = d$. First, by Lemma \ref{l-fpq}, $f_{\frac{d_{ij}}{d}}(q_j,\pib)$ is increasing in $q_j$ provided $\frac{d_{ij}}{d} \leq 1$, so agent $i$ should choose full effort to achieve her maximum proficiency $p_i$ on all tasks.
Next, note that in terms of the expected reward, using proficiency $q_j$ and reporting $\invert$ is equivalent to using proficiency $1 - q_j$ and reporting $X$.
So again by Lemma \ref{l-fpq}, deviating to $\invert$, \ie, $(1-q_j)$, on any task is strictly dominated by $\truth$ for $q_j > 1/2$ and $\pib > 1/2$.
Finally, if agent $i$ chooses $f_{ij}$ as the function which reports the outcome of a random cointoss with probability $r$ of $H$ for any task $j$, the component of $E[R_i]$ contributed by the term corresponding to $X_{ij}$ becomes
{\small
\begin{align*}
& E\left[X_{ij}X_{\ipj j} + (1-X_{ij})(1-X_{\ipj j})\right] - \frac{d_{ij}}{d}\left(E[X_{ij}]p_{\ipj}[H] + (1-E[X_{ij}])(1-p_{\ipj}[H])\right)\\
&= rp_{\ipj}[H] + (1-r)(1-p_{\ipj}[H]) - (rp_{\ipj}[H] + (1-r)(1-p_{\ipj}[H]))) \\
&=0,
\end{align*} }
which is strictly smaller than the reward from $f_{ij} = \truth$ in (\ref{erew-pt}) if $q_j > 1/2$ and $\frac{d_{ij}}{d} \ge 1$, since $f(q_j, \pib)$ is strictly positive when $q_j, \pib > 1/2$ by Lemma \ref{l-fpq}.
Since we need $\frac{d_{ij}}{d} \leq 1$ to ensure that $(1, \invert)$ is not a profitable deviation, and $\frac{d_{ij}}{d} \geq 1$ to ensure that $(0,r)$ is not a profitable deviation, requiring $d_{ij} = d$ simultaneously satisfies both conditions. Therefore, if $d_{ij} = d$, deviating from $(1,\truth)$ on any task $j$ leads to a strict decrease in reward to agent $i$. Since the total reward to agent $i$ can be decomposed into the sum of $D$ terms which each depend only on the report $X_{ij}$ and therefore the strategy for the single task $j$, any deviation from $[(1,X)]$ for any single task or subset of tasks strictly decreases $i$'s expected reward.
Therefore, the rewards $R_i = \sum_{j \in J(i)} R_{ij}$ can always be scaled appropriately to ensure that $[(1,\truth)]$ is an equilibrium of $\M_d$.
\end{proof}
\noindent{\bf Other equilibria.} While $[(1,X)]$ is an equilibrium, $\M_d$ can have other equilibria as well--- for instance, the strategy $[(0,r)]$, where all agents report the outcome of a random cointoss with bias $r$ on each task, is also an equilibrium of $\M_d$ for all $r \in [0,1]$, albeit with $0$ reward to each agent. In fact, as we show in the next theorem, no equilibrium, symmetric or asymmetric, in pure or mixed strategies, can yield higher reward\footnote{
We note that another equilibrium which achieves the same maximum expected reward is $[(1, X^c)]$, where all agents put in full effort to make their evaluations, but then all invert their evaluations for their reports. However, $[(1, X^c)]$ is a rather unnatural, and risky, strategy, and one that is unlikely to arise in practice. Also, as we will see later, $[(1, X^c)]$ can also lead to lower rewards when there are some agents who always report truthfully.} to agents
than $[(1, \truth)]$, as long as agents `treat tasks equally' (for example, while an agent may choose to shirk effort on one task and work on all others, each of her tasks is equally likely to be the one she shirks on). We will refer to this as tasks being `apriori equivalent', so that agents cannot distinguish between tasks prior to putting in effort on them (or equivalently, the assignment of reference raters is such that agents will not find it beneficial (in terms of expected reward) to use a different strategy for a specific task). Note that this assumption is particularly reasonable in the context of applications where agents are recruited for a collection of similar tasks as in crowdsourced abuse/adult content identification, or in peer grading where each task is an anonymous student's solution to the same problem.
\begin{theorem}
\label{t-maxeq}
Suppose $p_i > 1/2$, and tasks are apriori equivalent. Then, the equilibrium where all agents choose $[(1, \truth)]$ yields maximum reward to each agent.
\end{theorem}
\begin{proof}
Consider a particular agent $i$ and task $j$, and a single potential reference rater $\ipj$ for $(i,j)$. Recall from the preliminaries that agent $i$'s choice of $f_{ij}$ can be described via a matrix $M = \al_1 M_{X} + \al_2 M_{X^c} + \al_3 M_{H} + \al_4 M_{L}$, and that we denote $i$'s evaluation via a vector $o$, where $o = [1 \quad 0]^T$ if $i$ observes $H$ and $o = [0 \quad 1]^T$ if $i$ observes $L$. Similarly, let us describe $\ipj$'s choice of reporting function via the matrix $M'$ with corresponding coefficients $\al'_i$, and denote $\ipj$'s evaluation by $o'$.
Since tasks are apriori equivalent, each player $i$ (hence $\ipj$ too) uses strategies such that $E[X_{ij}] = E[X_{ik}]$ for all $j,k \in J(i)$. Then, we can rewrite the expected reward for agent $i$ on task $j$, when paired with reference rater $\ipj$, as
{\small
\begin{align*}
E[R_{ij}] = 2(E[X_{ij}X_{\ipj j}] - E[X_{ij}]E[X_{\ipj j}]).
\end{align*}}
Using the matrix-vector representation, substituting $M, M'$ with their representations in terms of the basis matrices and expanding, and evaluating the matrix-matrix products, we have
\[
X_{ij}X_{\ipj j} = o'^TM'^T M o = o'^T R_M o,
\]
where
\begin{align*}
R_M &= (\al_1\al'_1 + \al_2\al'_2)I + \al_2\al'_1M_{X^c} + \al_1\al'_2M^T_{X^c} + (\al_3\al'_3 + \al_4\al'_4)\ones + (\al_3\al'_1 + \al_4\al'_2)M_H + (\al_1\al'_3 + \al_2\al'_4)M^T_H \\
&+ (\al_4\al'_1 + \al_3\al'_2)M_L + (\al_1\al'_4 + \al_2\al'_3)M^T_L,
\end{align*}
and $I, \ones$ denote the identity matrix and the matrix of all ones in $R^{2 \times 2}$ respectively, and we use $M_{X}^TM_{X} = M_{X^c}^TM_{X^c} = I$, $M_H^T M_H = M_L^T M_L = \ones$, $M_{X^c}^T M_H = M_L$, $M_{X^c}^T M_L = M_H$, and $M_H^T M_L = \zeros$. Similarly,
\begin{align*}
E[X_{ij}]E[X_{\ipj j}] = E[o'^TM'^T]E[M o] = E[o'^T]R_M E[o],
\end{align*}
where $R_M$ is as defined above.
Now, note that $M_Ho = [o_1 + o_2 \quad 0]^T = [1 \quad 0]^T$ since $o_1 + o_2 = 1$ for any evaluation vector $o$ by definition, so that $E[o'^T M_Ho] = E[o'^T]E[M_Ho]$, since $M_Ho$ is a constant. The same is the case for each of the terms $E[o'^T\ones o], E[o'^T M^T_H o], E[o'^T M^T_L o], E[o'^T M_L o]$. Therefore, these terms cancel out when taking the difference\\ $E[X_{ij}X_{\ipj j}] - E[X_{ij}]E[X_{\ipj j}]$ (corresponding to the reward from either agent choosing to report $X_{ij}$ independent of her evaluation being $0$). Also note that $E[o^{ij}] = \begin{bmatrix} \sph & \spl \end{bmatrix}^T$ if agent $i$ makes evaluations with proficiency $p$. Suppose the agents use effort leading to proficiencies $p$ and $p'$ respectively. Then, we have
\begin{align*}
E[X_{ij}X_{\ipj j}] - E[X_{ij}]E[X_{\ipj j}] &= (\al_1\al'_1 + \al_2\al'_2)(E[o'^To]- E[o']^TE[o])+~\al_2\al'_1(E[o'^TM_{X^c}o]-E[o'^T]M_{X^c}E[o]) \\
& \quad+~\al_1\al'_2(E[o'^TM^T_{X^c}o]-E[o'^T]M^T_{X^c}E[o])\\
&= (\al_1\al'_1 + \al_2\al'_2)(E[o'_1o_1 + o_2o_2'] - E[o'_1]E[o_1] - E[o_2]E[o_2']) \\
&\quad + ~ (\al_2\al'_1 + \al_1\al'_2)(E[o'_1o_2 + o_1o_2'] - E[o'_1]E[o_2] - E[o_1]E[o_2'])\\
&=(\al_1\al'_1 + \al_2\al'_2)\Big(pp' + (1-p)(1-p') - \sph p'[H] - \\
& \quad (1-\sph)(1-p'[H])\Big) + (\al_2\al'_1 + \al_1\al'_2)\Big(p(1-p')~ + \\
&\quad (1-p)p' - \sph(1- p'[H]) - (1-\sph)p'[H]\Big).
\end{align*}
Now, note that multiplier of $(\al_1\al'_1 + \al_2\al'_2)$ is precisely $f(p,p')$, which by Lemma \ref{l-fpq} is nonnegative if $p,p'\geq 1/2$, and strictly positive if $p, p' > 1/2$. Also, for $p, p' \geq 1/2$, note that $\sph \leq p$ and $p'[H] \leq p'$. Now, the function $g(x,y) = x(1 - y) + y(1 - x)$ is decreasing in both $x$ and $y$ for $x, y \in [\frac{1}{2},1]$ (taking derivatives), so the multiplier of $(\al_2\al'_1 + \al_1\al'_2)$ is non-positive, and negative if $p,p' > 1/2$.
So the maximum value that $E[X_{ij}X_{\ipj j}] - E[X_{ij}]E[X_{\ipj j}]$ can take for nonnegative coefficients with $\sum \al_i = \sum \al'_i = 1$, is $f(p,p')$, which is obtained by setting $\al_3 = \al_4 = 0$, $\al'_3 = \al'_4 = 0$ (\ie, with no weight on random independent reporting), and $a_1 = \al'_1 = 1$, $\al_2 = \al'_2 = 0$ (or viceversa): this is because the maximum value of term $(\al_1\al'_1 + \al_2\al'_2)$ when $\al_2 = 1-\al_1$ and $\al'_2 = 1-\al'_1$ is $1$ and is achieved with these values, which also minimize the value of the term $(\al_2\al'_1 + \al_1\al'_2)$ with the non-positive multiplier, since $(\al_2\al'_1 + \al_1\al'_2) \geq 0$ and is equal to $0$ for these values of $\al_i, \al'_i$. Also, since $f(p,p')$ increases with increasing $p$ and $p'$, it is maximized when agents put in full effort and achieve their maximum proficiencies $p_i, p_{\ipj}$.
Therefore the expected reward for the single component of $E[R_{ij}]$ coming from a specific reference rater achieves its upper bound when both agents use $[(1,X)]$. The same argument applies for each reference rater, and therefore to the expected reward $E[R_{ij}]$, and establishes the claim.
\end{proof}
We next investigate what kinds of Nash equilibria might exist where agents use low effort with any positive probability. Apriori, it is reasonable to expect that there would be mixed-strategy equilibria where agents randomize between working and shirking, \ie, put in effort (choose $e_{ij} =1$) sometimes and not (choose $e_{ij} =0$) some other times. However, we next show that as long as tasks are apriori equivalent and agents only randomize between reporting truthfully and reporting the outcome of an independent random cointoss (\ie, they do not invert evaluations), the {\em only} equilibrium in which any agent uses any support on $(0,r)$ is the one in which {\em all} agents {\em always} use $(0,r)$ on all their tasks. To show this, we start with the following useful lemma saying that an agent who uses a low-effort strategy any fraction of the time will always have a beneficial deviation as long as some reference agent plays $(1, \truth)$ with some positive probability.
Roughly speaking, this is because as long as there is some probability that an agent's reference rater plays $(1, \truth)$ rather than $(0,r)$, the agent strictly benefits by always playing $(1, \truth)$ to maximize the probability of both agents playing $(1, \truth)$, which is the only time the agent obtains a positive reward.
\begin{lemma}
\label{l-deviate}
Suppose the probability of agent $i$ using strategy $(1,\truth)$ is $\delta$ and strategy $(0,r_i)$ is $1-\delta$ for each task $j \in J(i)$. Suppose $i$'s potential reference raters $\ipj$ use strategies $(1, \truth)$ and $(0, r_{\ipj})$ with probabilities $\eps_{\ipj}$ and $1-\eps_{\ipj}$ respectively, for each task $j \in J(i)$. If $\eps_{\ipj} > 0$ for any reference rater with proficiency $p_{\ipj} > 1/2$, then agent $i$ has a (strict) profitable deviation to $\delta' = 1$, \ie, to always using strategy $(1, \truth)$, for all values of $r_i \in [0,1]$.
\end{lemma}
\begin{proof}
Consider a particular task $j$, and let $k = 1, \ldots, K$ be the potential reference rater for $(i,j)$. Let $a_k$ denote the probability that $k$ is the reference rater for agent $i$ for task $j$. By linearity of expectation, $i$'s expected reward for $j$ can be written as
{\small \begin{align*}
E[R_{ij}]&= \sum_{k=1}^K a_{k}\Big[\delta\eps_{k}(p_i p_k + (1-p_i)(1- p_k) - (p_i[H] p_k[H] + (1-p_i[H])(1-p_k))) \\
&\quad+~ (1-\delta)\eps_k(r_ip_k[H] + (1-r_i)(1- p_k[H]) - (r_ip_k[H] + (1-r_i)(1-p_k))) \\
&\quad+~ \delta(1-\eps_k)(p_i[H] r_k + (1-p_i[H])(1- r_k) - (p_i[H]r_k + (1-p_i[H])(1-r_k))) \\
&\quad+~ (1-\delta)(1-\eps_k)(r_i r_k + (1-r_i)(1- r_k) - (r_i r_k + (1-r_i)(1-r_k)))\Big] \\
&= \delta \sum_{k} a_k\eps_k(p_i p_k + (1-p_i)(1- p_k) - (p_i[H] p_k[H] + (1-p_i[H])(1-p_k[H]))) \\
&= \delta \sum_{k} a_k\eps_k f_1(p_i, p_k).
\end{align*}}
Now, $E[R_{ij}]$ is linear in $\delta$, and by Lemma \ref{l-fpq}, the coefficient of $\delta$ is nonnegative for all $\eps_k$ and $p_k \geq 1/2$, and strictly greater than $0$ if $\eps_k > 0$ for some $k$ with $p_k > 1/2$. Therefore, $i$ can strictly increase her expected reward $E[R_{ij}]$ by increasing $\delta$ for any $\delta < 1$, as long as there is some reference agent $k$ with $\eps_k > 0$ and $p_k > 1/2$.
The same argument holds for each task $j \in J(i)$, and therefore to strictly improve $i$'s total reward $E[R_{i}]$, we only need one reference rater across all tasks to satisfy $\eps_k > 0$ and $p_k > 1/2$ to obtain a strictly beneficial deviation (recall that we assumed $p_i \geq 1/2$ for all $i$).
\end{proof}
This lemma immediately allows us to show that the only low-effort equilibria of $\M$ that we reasonably\footnote{(We say reasonably because of the technical possibility of equilibria where some agents mix over $(1, \invert)$ as well.)} need to be concerned about is the pure-strategy equilibrium in which $e_{ij}=0$ for all $i,j$. Note that different agents could use different $r_i$ (or even $r_{ij}$) in such equilibria, but all agents will receive reward $0$ in {\em all} such equilibria.
\begin{theorem}
Suppose every agent can be a reference rater with some non-zero probability for every other agent, and tasks are apriori equivalent. Then, the only equilibria (symmetric or asymmetric) in which agents mix between $(1, \truth)$ and any low-effort strategy $[(0, r_{ij})]$ with non-trivial support on $[(0, r_{ij})]$ are those where all agents always use low effort on all tasks.
\end{theorem}
\noindent{\bf Eliminating low-effort equilibria.} Our final result uses Lemma \ref{l-deviate} to obtain a result about eliminating low-effort equilibria. Suppose there are some trusted agents (for example, an instructor or TA in the peer-grading context or workers with long histories of accurate evaluations or good performance in crowdsourcing platforms) who always report truthfully with proficiency $t > 1/2$. Let $\eps_t$ denote the minimum probability, over all agents $i$, that the reference rater for agent $i$ is such a trusted agent (note that we can ensure $\eps_t >0$ by having the trusted agent randomly choose each task with positive probability). Lemma \ref{l-deviate} immediately gives us the following result for $\M_d$, arising from the fact that the reward from playing a random strategy $(0,r)$ is exactly $0$---the presence of trusted agents with a non-zero probability, however small, is enough to eliminate low-effort equilibria altogether.
\begin{theorem}
\label{t-norand}
Suppose $\eps_t > 0$. Then $[(0,r_{ij})]$ is not an equilibrium of $\M$ for any $r_{ij} \in [0,1]$.
\end{theorem}
\begin{proof}
Suppose all agents except the trusted agent use the strategy $(0,r_{ij})$, and $\eps_t$ is the probability that the trusted agent is the reference rater for any agent-task pair. Then, since agent $i$ reports $X_{ij}$ according to a random coin toss independent of the actual realization of $j$, the payoff from any reference rater, whether the trusted agent or another agent playing $(0,r)$ is $0$. For notational simplicity, let $r = r_{ij}$, $r'=r_{\ipj j}$.
{\small \begin{align*}
E[R_{ij}]& = \eps_t(rt[H] + (1-r)(1-t[H]) - (rt[H] + (1-r)(1-t[H])) \\
&\quad +~ (1-\eps_t)(r r' + (1-r)(1-r') - (rr' + (1-r)(1-r')))\\
&=0.
\end{align*} }
By deviating to $(1,\truth)$, agent $i$ can strictly improve her payoff as long as $\eps_t > 0$ and $t, p > 1/2$, since her expected reward from this deviation is
{\small \begin{align*}
E[R_{ij}] &= \eps_t(pt + (1-p)(1-t) - (\sph t[H] + (1-\sph)(1-t[H])) \\
&\quad + ~(1-\eps_t)(r r' + (1-r)(1-r') - (rr' + (1-r)(1-r'))\\
&> 0,
\end{align*} }
since the coefficient of $\eps_t$ is positive for $t, p > 1/2$ by Lemma \ref{l-fpq}.
Therefore, there is a strictly beneficial deviation to $(1, \truth)$, so there is a choice of multiplier for the reward such that the payoff to agent $i$, which is the difference between the reward and the cost of effort $c$, is strictly positive as well. So $(0,r_{ij})$ is not an equilibrium of $\M$ when $\eps_t > 0$.
\end{proof}
This result, while simple, is fairly strong: as long as some positive fraction of the population can be trusted to always report truthfully with proficiency greater than $1/2$, the only reasonable\footnote{Again, we say reasonable rather than unique because $(1, \invert)$ does remain an equilibrium of $\M$ for all $\eps_t$ less than a threshold value--- however, in addition to being an unnatural and risky strategy, this equilibrium yields strictly smaller payoffs than $[(1, \truth)]$ when $\eps_t > 0$. Note also that the introduction of such trusted agents does not introduce new equilibria, and that $[(1, \truth)]$ remains an equilibrium of $\M$.)}
equilibrium of $\M$ is the high-effort equilibrium $[(1, \truth)]$, no matter {\em how small} this fraction. In particular, note that $\M$ does not need to assign a higher reward for agreement with a trusted agent to achieve this result, and therefore {\em does not need to know the identity} of the trusted agents. In contrast, the mechanism which rewards agents for agreement with a reference rater without subtracting out our statistic term must use a higher reward $w(\eps_t)$ for agreement with the trusted agents which increases as $\frac{1}{\eps_t}$ to eliminate low-effort equilibria\footnote{The same is the case for a mechanism based on rewarding for the `right' variance, which does retain $[(1,X)]$ as a maximum reward equilibrium, but still requires identifying the trusted agents and rewarding extra for agreement with them.}--- this, in addition to being undesirably large, also requires identification of trusted agents.
\section{Creating the Task Assignment}
\label{s-graph}
While in some crowdsourcing settings, agents choose tasks at will, there are also applications where a principal can potentially choose an assignment of a collection of her tasks among some assembled pool of workers. In this section, we present a simple algorithm to design assignment of tasks to agents such we can satisfy the condition in Theorem~\ref{t-eqbm} for mechanism $\M_{D-1}$, \ie, when $d = D-1$. We note that with this assignment of tasks to agents, choosing reference raters appropriately is trivially feasible for $d = 1$, \ie, for $\M_1$, and ensuring $d_{ij} = d$ is also easy as described in \S \ref{s-mech}.
We start out by randomly permuting all agents using a permutation $\pi$. For simplicity of presentation we assume that $\frac{m}{D}$ ($=\frac{n}{T}$) is an integer. The $m$ tasks are divided into $\frac{m}{D}$ task-blocks, each containing $D$ tasks. Similarly, the $n$ agents are divided into $T$ agent blocks, each containing $\frac{n}{T}$ agents. We number the task-blocks by $b=1,\ldots,\frac{m}{D}$ and the agent blocks by $a=1,\ldots,T$. The agents in block $a$ are thus $(a-1)\frac{n}{T} + 1,\ldots,a\frac{n}{T}$ and the tasks in block $b$ are $(b-1)D+1,\ldots,bD$.
We first describe the algorithm and then show that it produces an assignment that satisfies the conditions required in the definition of $\M_{D-1}$, in particular that for each agent-task pair, it is possible to choose a reference rater who has only that task in common with this agent. The algorithm works as follows: we assign tasks for agents starting from the agent block $a=1$ onwards. For block $1$, each agent $i'$ in the block is assigned all the tasks corresponding to the task block $i'$ (recall that number of agents in a block equals $\frac{n}{T} = \frac{m}{D}$, the number of task-blocks). This completes fills up the capacity of the agents in block $1$. For blocks $a = 2,\ldots, T$, consecutively, the agent $(a - 1)\frac{n}{T} + i'$ is assigned $D$ tasks $\{i', i' + \frac{m}{D},\ldots, i' + \frac{m}{D}(D-1)\}$, for $i' =1,\ldots,\frac{n}{T}$.
The above assignment completely describes the sets $J(i)$ and $I(j)$ for every agent $i$ and task $j$. For each task $j$, let $i^*_j$ denote the unique agent in block $1$ who works on task $j$. We define the reference raters as follows: for each agent-task pair $(i, j)$, if $i$ lies in blocks $\{2,\ldots,T\}$, define the reference rater $\ipj = i^*_j$. If $i$ lies in block $1$, define the reference rater to be any other user who is working on this task. Note that for $d= D-1$, the sets $S_{ij}$ and $S_{\ipj j}$ are exactly $S_{ij} = J(i) \setminus \{j\}$ and $S_{\ipj j}=J(\ipj) \setminus \{j\}$.
The following lemma proves two things--- first, the assignment above is actually feasible under fairly mild conditions, and second, that the choice of reference raters satisfies the conditions in the definition of $\M$ and those required by Theorem \ref{t-eqbm}.
\begin{lemma}
If $m \ge D^2$, the above algorithm generates a feasible assignment, i.e. every agent is assigned exactly $D$ tasks and every task to $T$ agents. Also, for agent-task pair $(i,j)$, the reference rater $\ipj$ satisfies $J(\ipj ) \cap J(i) = \{j\}$. Furthermore, $E[p_{\ipj }] = E_{j_l \in T_{ij}}[p_{i'_{j_l}}]$.
\end{lemma}
\begin{proof}
Agents in block $1$ are clearly assigned to their full capacity. For blocks $a = 2, \ldots, T$, for every agent $i = (a - 1)\frac{n}{T} + i'$, the set $I(i) = \{i', i' + \frac{m}{D},\ldots, i' + \frac{m}{D}(D-1)\}$. Note that for each $i$, the above values are all distinct, and that $i' + \frac{m}{D}(D-1) \leq \frac{n}{T} + \frac{m}{D}(D-1) = m$. Thus every agent's assignment is feasible. Since the total capacity of agents equals the total capacity of the tasks, the tasks are also assigned completely, and to distinct agents.
In order to see that the choice of reference raters is feasible, note that if $D^2 \le m$, then $\frac{m}{D} \ge D$, and hence the tasks for each agent belong to distinct blocks. For agent-task pairs $(i, j)$ where the agents are in blocks $2,\ldots,T$, the reference rater $\ipj = i^*_j$, the unique agent in block $1$ who worked only on the task-block that $j$ belongs to. By the above argument, $i$ does not work on any other task from this block, and hence $J(i^*_j) \cap J(i) = \{j\}$. By the same argument, $i$ is also a feasible reference rater for $i^*_j$ on task $j$. Thus, the choice of reference raters satisfies the condition for $\M_{D-1}$.
Finally, the expectation condition follows simply from the random permutation applied to the set of agents at the beginning of the construction.
\end{proof}
\section{Discussion}
\label{s-disc}
In this paper, we introduced the problem of information elicitation when agents' proficiencies are endogenously determined as a function of their effort, and presented a simple mechanism which uses the presence of multiple tasks to identify and penalize low-effort agreement to incentivize effort when tasks have binary types. Our mechanism has the property that maximum effort followed by truthful reporting is the Nash equilibrium with maximum payoff to all agents, including mixed strategy equilibria. In addition to handling endogenous agent proficiencies, to the best of our knowledge this is the first mechanism for information elicitation with this 'best Nash equilibrium' property over all pure and over mixed strategy equilibria that
requires agents to only report their own evaluations (\ie, without requiring `prediction' reports of their beliefs about other agents' reports), and does not impose any requirement on a diverging number of agent reports per task to achieve its incentive properties. Our mechanism provides a starting point for designing information elicitation mechanisms for several crowdsourcing settings where proficiency is an endogenous, effort-dependent choice, such as image labeling, tagging, and peer grading in online education.
We use the simplest possible model that captures the complexities arising from strategically determined agent proficiencies, leading to a number of immediate directions for further work. First, our underlying outcome space is binary ($H$ or $L$)--- modeling and extending the mechanism to allow a richer space of outcomes and feedback is one of the most immediate and challenging directions for further work. Also, our model of effort is binary, where agents either exert full effort and achieve maximum proficiency, or exert no effort to achieve the baseline proficiency. While our results extend to a model where proficiency increases linearly with cost, a natural question is how they extend to more general models, for example, with convex costs. Finally, a very interesting direction is that of {\em heterogenous} tasks with task-specific priors and abilities. In our model, tasks are homogenous with the same prior $\ph$, and agents have the same cost and maximum proficiency for each task. If tasks differ in difficulty, and agents can observe the difficulty of a task prior to putting in effort, there are clear incentives to shirk on harder tasks while putting in effort for the easier ones. While tasks are indeed apriori homogenous (or can be partitioned to be so) in some crowdsourcing settings, there are other applications where some tasks are clearly harder than others; also, agents may have task-specific abilities. Designing mechanisms with strong incentive properties for this setting is a very promising and important direction for further work. \\
{\bf Acknowledgements.} We thank David Evans, Patrick Hummel and David Stavens for helpful discussions and pointers to related literature, and anonymous referees for comments and suggestions that helped improve the presentation of the paper.
\bibliographystyle{abbrv}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 5,955 |
A sense of calm and tranquility will wash over you with luxurious lavender and rose floral waters.
In the morning the aromatherapy of of this blend is priceless. Lavender floral water helps to reduce acne, soothe the skin, and clear the mind all at the same time. Roses are natures anti-aging remedy.
In the evening, relax and take off your eye makeup wile deep cleaning pores. Prepare your skin to drink in your moisturizer.
Healthy glowing skin is evident almost immediately, instant gratification.
Cleanse with Cocolene's soap or your favorite cleanser.
Add a generous amount of toner to a cotton round and apply to your face and neck.
Continue with Cocolene's Rosehip+ for maximum age-defying or blemish reduction results. | {
"redpajama_set_name": "RedPajamaC4"
} | 2,348 |
VFW Posts
US Federal Prisons
Weed Towns
State of Rhode Island
Tap the red
Towns | Counties | Home | Points of Interest
10 Largest Cities in Rhode Island
South Kingstown
Rhode Island State Facts
List of Counties
List of Towns
Capital: Providence
Largest city: Providence
Highest point: Jerimoth Hill, 812 ft (247 m)
Lowest point: Atlantic Ocean, sea level
Admission to Union: May 29, 1790 (13th)
Rhode Island State Information
RI, the smallest state, has the largest state name, State of Rhode Island and Providence Plantations
A drive though RI is only 45 minutes border to border
Rhode Island enacted the first law prohibiting slavery in North America on May 18, 1652
The first NFL night game was held on November 6, 1929, at Providence's Kinsley Park
Ride Rhode Island's 100-Mile Network of Bike Paths and Nature Trails
Rhode Island never ratified the 18th Amendment. With its 400 miles of open coastline, rum running was a thriving industry for its residents.
The White Horse Tavern was built in 1673 and is the oldest operating tavern in the United States.
The Flying Horse Carousel is the nation's oldest carousel is in the town of Watch Hill.
Rhode Island Points of Interest
Points of Interest of Rhode Island, Things To Do, Places To Go, People To Meet
The state of Rhode Island is located within the New England area of the United States and is the least extensive state in the country with a total land area of only 1,214 square miles, as well as the 8th least populated with a total population of 1.05 million, however it is the 2nd ranked most densely populated state. Rhode Island is bordered by Connecticut, Massachusetts, and shares a water boundary with Long Island. Additional water boundaries include the Narragansett Bay, Atlantic Ocean, and Rhode Island Sound. It was also the first of the 13 original colonies which declared its independence from the British ruling, and was the last to ratify the U.S. Constitution. The capital of Rhode Island is Providence which is also the states largest city.
Rhode Island is said to have gotten its name either from the Isle of Rhodes located in the Mediterranean Sea or for the red clay prevalent in the state. Rhode Island is nicknamed "The Ocean State" which is a reflection of the geography of the state which consists of numerous large bays and water inlets which make up roughly 14% of the states total area. The climate in the state is recognized as humid continental and experiences warm and rainy summers with cold or chilly winters. The highest recorded temperature in the state was 104 degrees Fahrenheit on 1975, and the lowest was -25 degrees Fahrenheit in 1996.
The major industries in the state include jewelry, rubber products, textiles, tourism, and machinery. Rhode Island has a few major bodies of water which include the Sakonnet River and Scituate Reservoir. The highest point of elevation in the state is Jerimoth Hill which peaks at 812 feet above sea level. The state motto is "Hope" and the state song is "Rhode Island, It's For me". The state bird of Rhode Island is the Rhone Island Red, the state flower is the Violet, and the state tree is the Red Maple Tree. While Rhode Island may be the smallest state in the country, it has a busy economy which is comparable to larger states, and thus it is an important part of the overall U.S. Economy.
Home Page | Disclaimer | Copyright | Privacy Policy
© 2022 TownsOfTheUSA.Com - All Rights Reserved | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 3,011 |
__author__ = 'greg'
import numpy as np
class Node:
def __init__(self,min_x,min_y,max_x,max_y):
self.min_x = min_x
self.min_y = min_y
self.max_x = max_x
self.max_y = max_y
self.markings = []
self.users = []
self.children = None
self.split_x = None
self.split_y = None
def __give_to_child__(self,m,u):
assert(len(self.children) == 4)
if m[0] <= self.split_x:
if m[1] <= self.split_y:
self.children[0].__add_point__((m,u))
else:
self.children[1].__add_point__((m,u))
else:
if m[1] <= self.split_y:
self.children[2].__add_point__((m,u))
else:
self.children[3].__add_point__((m,u))
def __add_point__(self,(m,u)):
#does this node have children, if so we should pass this pt onto the appropriate child
if self.children is not None:
self.__give_to_child__(m,u)
#else we are doing our own insertion
else:
self.markings.append(m)
self.users.append(u)
#have we now exceeded our capacity - hard code for now to be 10
if len(self.markings) == 10:
#find out split points
self.split_x = np.median(zip(*self.markings)[0])
self.split_y = np.median(zip(*self.markings)[1])
#create the children
self.children = []
self.children.append(Node(self.min_x,self.min_y,self.split_x,self.split_y))
self.children.append(Node(self.min_x,self.split_y,self.split_x,self.max_y))
self.children.append(Node(self.split_x,self.min_y,self.max_x,self.split_y))
self.children.append(Node(self.split_x,self.split_y,self.max_x,self.max_y))
for m2,u2 in zip(self.markings,self.users):
self.__give_to_child__(m2,u2)
#just keep the memory clean
self.markings = []
self.users = []
def __ward_traverse__(self):
if self.children is None:
print self.markings
print self.users
else:
self.children[0].__ward_traverse__()
self.children[1].__ward_traverse__()
| {
"redpajama_set_name": "RedPajamaGithub"
} | 2,492 |
Q: ICS keyboard back,home buttons not taking touch when opened on overlay I am creating an Overlay Screen. Everything works perfect on it except the android 4.0+ default keyboard`s back(keyboard down),home & recent apps button when the keyboard is open. These keys do not take the touch and instead the keys above it are being tapped. For example when tapped on back the comma is getting tapped, when tapped home the space key is being tapped etc.
The issue is seen with the android default keyboard from 4.0 onwards only since the samsung devices which have their own keyboard implementation do not show this issue also when in landscape mode the default keyboard works normally but then the keyboard is in full screen mode.
I am not sure if the fault is in the keyboard implementation or in mine. In anyway the fix should be found since majority users are going to use the default keyboard.
For the Overlay Screen i am using following flags :
int flags = WindowManager.LayoutParams.FLAG_NOT_TOUCH_MODAL
| WindowManager.LayoutParams.FLAG_LAYOUT_IN_SCREEN
| WindowManager.LayoutParams.FLAG_LAYOUT_INSET_DECOR;
params.type = WindowManager.LayoutParams.TYPE_SYSTEM_ERROR;
params.flags |= flags;
params.format = PixelFormat.TRANSPARENT;
Please let me know if any more code snippet is required in order to understand my query.I have been through many already asked questions but didn`t found any matching to my query. Hope to find some solution.
A: there is an error correction introduced in 4.0 IME for touch to key.
could you check the LatinIME source for same.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 2,973 |
After years of drugs and counseling, Polly and Melba have developed a unique relationship-and to Crockett, this relationship represents something her people can work with. in exchange for release from prison, Crockett asks Melba (and Polly) to serve as IMP hunters. For Melba, it's a chance to prove that she's innocent, convinced to murder by a monster...a monster she must now unleash. | {
"redpajama_set_name": "RedPajamaC4"
} | 9,554 |
Q: Passing forEach loop data to href attribute in Vue I have an array that I'm looping over in Vue, and rendering an anchor element with each iteration. The array that i'm looping over is one of my props, I've tried several different things and while I do believe I could get it working, the way would seem very hacky to me and I'm wanting to make sure it's not simpler than I'm thinking
<template v-for="(lead, index) in leads">
<a target="_blank" v-bind:href="lead.present_url"><b>foobar</b></a>
<div>{{lead.name}}</div>
<div>{{lead.id}}</div>
</template>
I've tried all manner of combinations, the following is not a comprehensive list but just some examples of what I've tried: :href="lead.present_url" href='lead.present_url' :href='${lead.present_url}' (with tildas), several variations of these, including with and without v-bind, v-bind:href='leads[index].present_url'
props: {
leads: Array
},
My question is: what is the best way to go about doing this?
A: You actually want to use v-for on the element that you want multiple of, and you need to bind a :key as well:
<template>
<div class="lead-container">
<div v-for="lead in leads" :key="lead.id">
<a target="_blank" :href="lead.present_url">
<strong>foobar</strong>
</a>
<div>{{lead.name}}</div>
<div>{{lead.id}}</div>
</div>
</div>
</template>
<script>
export default {
props: {
leads: {
type: Array,
default: () => [],
}
}
}
</script>
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 9,601 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.