text stringlengths 14 5.77M | meta dict | __index_level_0__ int64 0 9.97k ⌀ |
|---|---|---|
var moment = require('moment')
var loremIpsum = require('lorem-ipsum')
var _ = require('lodash')
module.exports = function (env) {
/**
* Instantiate object used to store the methods registered as a
* 'filter' (of the same name) within nunjucks. You can override
* gov.uk core filters by creating filter methods of the same name.
* @type {Object}
*/
var filters = {}
filters.pickDescriptor = function(nuggets, descriptor)
{
return _.filter(nuggets, function(o)
{
return _.includes(o.descriptors,descriptor);
})
}
/*
ascdes is to say whether it's
*/
filters.sortBy = function(collection,type,des) {
var newArray = _.sortBy(collection,type);
if (!des) return newArray
else return _.reverse(newArray)
}
filters.formatDate = function(str,format) {
var d = moment(str).format(format);
if (d !== 'Invalid date') return d;
else return '';
}
filters.fmeOverdue = function(date){
return moment(date).diff(moment(), "days") < -18;
}
filters.formatDateAdd2Days = function(str,format) {
var d = moment(str).add(2, 'days').format(format);
if (d !== 'Invalid date') return d;
else return '';
}
filters.monthsAgo = function(format,num) {
var d = moment().subtract(num,"months").format(format)
if (d !== 'Invalid date') return d
else return ''
}
filters.daysAgo = function(format,num) {
var d = moment().subtract(num,"days").format(format)
if (d !== 'Invalid date') return d
else return ''
}
filters.randMonthsAgo = function(format,num) {
var num = Math.ceil(Math.random()*num);
var d = moment().subtract(num,"months").format(format)
if (d !== 'Invalid date') return d
else return ''
}
filters.randDaysAgo = function(format,num) {
var num = Math.ceil(Math.random()*num);
var d = moment().subtract(num,"days").format(format)
if (d !== 'Invalid date') return d
else return ''
}
filters.baseDate = function(format,num) {
var num = Math.ceil(num);
var d = moment().subtract(num,"days").format(format)
if (d !== 'Invalid date') return d
else return ''
}
filters.baseDateAdd = function(format,num) {
var num = Math.ceil(num);
var d = moment().add(num,"days").format(format)
if (d !== 'Invalid date') return d
else return ''
}
filters.formatInPast = function(date){
var today = moment(new Date()).hours(0).minutes(0).seconds(0).milliseconds(0);
var testDate = moment(date).hours(0).minutes(0).seconds(0).milliseconds(0);
if(date.getDay() === 0 || date.getDay() === 6){
return 'calendar__day--closed';
} else if (today > testDate){
return 'calendar__day--past';
} else if(today.toString() == testDate.toString() ){
}
}
filters.formatInPastPicker = function(date){
var today = moment(new Date()).hours(0).minutes(0).seconds(0).milliseconds(0);
var testDate = moment(date).hours(0).minutes(0).seconds(0).milliseconds(0);
if(date.getDay() === 0 || date.getDay() === 6){
return 'calendar-picker__day--closed';
} else if (today > testDate){
return 'calendar-picker__day--closed';
} else if(today.toString() == testDate.toString() ){
}
}
filters.formatAvailable = function(date1, date2){
if(moment(date1).diff(moment(date2), "days") > 0){
return 'calendar-picker__day--closed'
}
}
filters.formatCalendarFull = function(date){
console.log(moment(date).diff(moment(), "days"));
if(moment(date).day() !== 6 &&
moment(date).day() !== 0 && (
moment(date).diff(moment(), "days") == 0 ||
moment(date).diff(moment(), "days") == 3 ||
moment(date).diff(moment(), "days") == 5 ||
moment(date).diff(moment(), "days") == 6 ||
moment(date).diff(moment(), "days") == 8 ||
moment(date).diff(moment(), "days") == 9 ||
moment(date).diff(moment(), "days") == 10 ||
moment(date).diff(moment(), "days") == 11 ||
moment(date).diff(moment(), "days") == 12 )){
return 'calendar-picker__day--full';
}
}
filters.formatSelected = function(date1, date2){
if(!date1 || !date2){
return '';
}
if(moment(date1).format('YYYYMD') === moment(date2).format('YYYYMD')){
return 'calendar-picker__day--selected';
} else {
return '';
}
}
filters.formatRole = function(date){
if(date.getDay() === 1){
return 'calendar__day--scrutiny'
} else if(date.getDay() !== 0 && date.getDay() !== 6){
return 'calendar__day--appointments'
}
}
filters.formatSessionClass = function(date){
if(date.getDay() === 1){
return 'calendar__session--unavailable'
} else if(date.getDay() !== 0 && date.getDay() !== 6){
return 'calendar__session--available'
}
}
filters.formatSessionValue = function(date){
if(date.getDay() === 1){
return 'unavailable'
} else if(date.getDay() !== 0 && date.getDay() !== 6){
return 'available'
}
}
filters.formatSelectable = function(date){
var today = moment(new Date(2018,7,9)).hours(0).minutes(0).seconds(0).milliseconds(0);
var testDate = moment(date).hours(0).minutes(0).seconds(0).milliseconds(0);
if(date.getDay() === 0 || date.getDay() === 6){
return 'calendar-picker__day--closed';
} else if(today.toString() === testDate.toString()) {
return 'calendar-picker__day--selectable'
} else if(testDate.isBefore(today)){
return 'calendar-picker__day--closed';
} else if(testDate.isAfter(today)) {
return 'calendar-picker__day--selectable'
}
}
filters.formatClosed = function(date){
return '';
}
filters.lorum = function(pars) {
var str = loremIpsum({
count: pars,
units: "paragraphs",
format: 'plain',
suffix: "<br />"
})
return lowerFirstLetter(str);
}
filters.lorumsent = function(pars) {
var str = loremIpsum({
count: pars,
units: "sentences",
format: 'plain',
suffix: "<br />"
})
return lowerFirstLetter(str);
}
filters.lorumwords = function(pars) {
var str = loremIpsum({
count: pars,
units: "words",
format: 'plain',
suffix: "<br />"
})
return lowerFirstLetter(str);
}
filters.randArray = function(o) {
for(var j, x, i = o.length; i; j = parseInt(Math.random() * i), x = o[--i], o[i] = o[j], o[j] = x);
return o;
}
filters.randNumber = function(num) {
return Math.ceil(Math.random()*num);
}
filters.slug = function(str) {
return slugify(str);
}
function slugify(text)
{
return text.toString().toLowerCase()
.replace(/\s+/g, '') // Replace spaces with -
.replace(/[^\w\-]+/g, '') // Remove all non-word chars
.replace(/\-\-+/g, '') // Replace multiple - with single -
.replace(/^-+/, '') // Trim - from start of text
.replace(/-+$/, ''); // Trim - from end of text
}
filters.limit = function(arr, limit) {
return arr.slice(0, limit);
}
filters.nl2br = function(str) {
return str.replace(/\r|\n|\r\n/g, '<br>')
}
filters.checkedIfAvailable = function(date){
if(moment(date).day() === 1){
return ''
} else {
return 'checked'
}
}
/* ------------------------------------------------------------------
add your methods to the filters obj below this comment block:
@example:
filters.sayHi = function(name) {
return 'Hi ' + name + '!'
}
Which in your templates would be used as:
{{ 'Paul' | sayHi }} => 'Hi Paul'
Notice the first argument of your filters method is whatever
gets 'piped' via '|' to the filter.
Filters can take additional arguments, for example:
filters.sayHi = function(name,tone) {
return (tone == 'formal' ? 'Greetings' : 'Hi') + ' ' + name + '!'
}
Which would be used like this:
{{ 'Joel' | sayHi('formal') }} => 'Greetings Joel!'
{{ 'Gemma' | sayHi }} => 'Hi Gemma!'
For more on filters and how to write them see the Nunjucks
documentation.
------------------------------------------------------------------ */
/* ------------------------------------------------------------------
keep the following line to return your filters to the app
------------------------------------------------------------------ */
return filters
}
function lowerFirstLetter(string) {
return string.charAt(0).toLowerCase() + string.slice(1);
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 4,694 |
Donington Park – tor wyścigowy w hrabstwie Leicestershire w centralnej Anglii. Obiekt otwarto w 1931 roku. Dotychczas tylko raz zorganizowano tutaj Grand Prix Formuły 1 – wyścig o Grand Prix Europy w sezonie 1993.
Tor posiada dwie konfiguracje:
Grand Prix – 4,003 km
National – 3,185 km
4 lipca 2008 roku podjęto decyzję, że od sezonu 2010 na tym obiekcie ponownie rozgrywane będą zawody Grand Prix Formuły 1, jednak ostatecznie, z powodów finansowych tor Donington Park stracił umowę na organizację GP Wielkiej Brytanii.
Zwycięzcy GP Europy na torze Donington Park
Przypisy
Linki zewnętrzne
Tory w Formule 1 w Wielkiej Brytanii
Leicestershire
Tory wyścigowe w Wielkiej Brytanii | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 7,148 |
Q: Is it possible to page through a view in phpmyadmin? Running phpmyadmin version 3.4.8.
I just noticed that there are no "paging" buttons when displaying a view that allows you to jump to the next page or the last page like when browsing a table.
I know you can set $cfg['MaxRows'] and I tried putting a "ShowAll" button on the page with $cfg['ShowAll'] = TRUE. I also know I can fill out the input boxes to the right of the "Show" button.
Phpmyadmin offers so many helpful features I am a little surprised I can't do this, but I haven't used it that long so maybe I am missing something.
A: I think $cfg['MaxExactCountViews'] is the variable you are looking for: http://docs.phpmyadmin.net/en/latest/config.html#cfg_MaxExactCountViews
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 624 |
Q: Is there a way to unpack a dictionary into an f-string **template**? With .format we can unpack a dictionary into a string template. That's the only thing I haven't found how to do with f-strings.
Is there a way to do it?
E.g. how to do this with f-strings?
template = "my values: {number} {item}"
values = dict(number=1, item='box')
print(template.format(**values))
Edit:
I thought you can have f-string templates for undefined values with something like this, but it doesn't work either:
template2 = "my values: {(number := None)} {(item := None)}" # values are immutable after defining
values = dict(number=1, item='box')
# how to unpack `values` into `template2`, is it possible?
I'm aware of the longer ways to do it, what I'm looking for is only whether it's possible to use unpacking. Cheers!!
A: Not that I am aware of with f-strings, but there are string templates:
from string import Template
template = Template("my values: $number $item")
values = dict(number=1, item='box')
print(template.substitute(**values)) # my values: 1 box
A: What's wrong with these:
values = dict(number=1, item='box')
print(f"{values=}")
print(f"values= {values['number']} {values['item']}{values.get('missing')}")
which gets you:
values={'number': 1, 'item': 'box'}
values= 1 box None
Or do you want to declare your f-string before you've set up the dict and then plug in the dict? Sorry, no can do, f-strings get evaluated right away, they are not really pre-defined templates.
Pre-defined templates hack
I've worked a lot with templates, usually the %(foo)s kind or else Jinja2 templates and have missed the capability to pass templates around, when using f-strings.
This below is a hack, but it does allow doing that, after a fashion. Thought of doing that a while back, never tried it though.
def template_1(**values):
number = values["number"]
item = values["item"]
missing = values.get("missing")
return f"values1= {number} {item} {missing}"
def template_2(**values):
return f"values2= {values['number']} {values['item']} {values.get('missing')}"
print(template_1(**values))
def do_it(t, **kwargs):
return t(**kwargs)
print(do_it(template_2,missing=3, **values))
Output
values1= 1 box None
values2= 1 box 3
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 2,128 |
Joseph Tapine, né le à Wellington (Nouvelle-Zélande), est un joueur de rugby à XIII néo-zélandais évoluant au poste de deuxième ligne. Il fait ses débuts en National Rugby League (« NRL ») avec les Knights de Newcastle lors de la saison 2014 puis rejoint en 2016 les Raiders de Canberra. Il est appelé sous le maillot de la Nouvelle-Zélande pour participer au Tournoi des Quatre Nations 2016.
Biographie
Palmarès
Collectif :
Finaliste de la Coupe du monde de rugby à neuf : 2019 (Nouvelle-Zélande).
Finaliste du Tournoi des Quatre Nations : 2016 (Nouvelle-Zélande).
Finaliste de la National Rugby League : 2019 (Canberra).
Individuel :
Elu meilleur pilier de la National Rugby League : 2022 (Canberra).
En club
Statistiques
Références
Lien externe
Joueur néo-zélandais de rugby à XIII
Naissance en mai 1994
Naissance à Wellington | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 7,008 |
**Hidden Fires**
**Sandra Brown**
**Begin Reading**
Table of Contents
A Preview of _Friction_
Newsletters
Copyright Page
In accordance with the U.S. Copyright Act of 1976, the scanning, uploading, and electronic sharing of any part of this book without the permission of the publisher constitute unlawful piracy and theft of the author's intellectual property. If you would like to use material from the book (other than for review purposes), prior written permission must be obtained by contacting the publisher at permissions@hbgusa.com. Thank you for your support of the author's rights.
# Chapter 1
The heat from the September sun was like a physical assault to the young woman who stepped down from the train at the Austin depot. Her ivory cheeks were slightly flushed, and a few vagrant tendrils of raven-black hair escaped the chignon under her hat. She fanned a lacy handkerchief in front of her face as she eagerly scanned the crowd for a familiar brown Stetson, and the tall, white-haired man who would be wearing it.
A sizable throng had gathered at the depot for the arrival of the noon train from Fort Worth. Families embraced returning prodigals, while others waved goodbye to passengers boarding the train. Commissions to write soon and be careful were issued in a cacophonous blend of English and Spanish, with the train's hissing white steam and sharp whistle providing the percussion for this discordant orchestra. With amazing alacrity, porters wheeled long, flatbed carts loaded with luggage, managing to skirt old ladies, businessmen, and young children.
Mexican women dressed in bright, full skirts strolled the platform hawking homemade candy, flowers, and Texas souvenirs. _Vaqueros_ leaned lazily against the depot wall, toying with lariats, rolling cigarettes, or squinting at the train they were reluctant to board, for they preferred open spaces and the cerulean ceiling of the Texas sky to the narrow confines of a railroad car.
Many of these cowboys noticed the young woman who watched each approaching carriage expectantly. Her gray eyes, which had been so full of excitement only minutes ago, became clouded with anxiety as the crowd began to diminish. The folds of her skirt swished behind her enticingly as she walked the length of the platform and back again. Dainty, high-button shoes tapped on the smooth boards with each step.
One by one, the _vaqueros_ sauntered toward the train bound back to Fort Worth. Most cast one last, longing look at the girl who, despite the heat and her obvious agitation, maintained a cool appearance.
With a screech of steel on steel, a geyser of steam, and a long blast of the whistle, the train slowly inched away from the depot, gained momentum, and finally chugged out of sight.
The platform emptied of people. The Mexican vendors covered the wares in their baskets, and the porters parked their carts in the shade of the building.
The girl in the navy-blue serge suit, white shirtwaist, and tan felt hat stood beside her meager luggage looking forlorn and lonely.
Ed Travers bustled out the depot door, sighted the girl, and, tugging his vest over his rotund stomach, hurried toward her.
"Miss Holbrook?" he inquired politely. "Miss Lauren Holbrook?"
The dismayed eyes brightened at the sound of her name and she smiled, parting perfectly formed lips to reveal small white teeth. "Yes," she answered breathlessly. "Yes, I'm Lauren Holbrook. Did Ben... uh... Mr. Lockett send you for me?"
Ed Travers covered his bafflement with a reassuring smile. "No, Miss Holbrook, not exactly. I'm Ed Travers, the depot manager. I'm sorry I kept you waiting, but the telegraph machine—" He broke off, impatient with himself for bungling what was already a delicate situation. "Forgive me for rambling and forcing you to stand in this heat. Come with me and I'll explain everything." He signaled to a lounging porter, who reluctantly came forward to carry Lauren's luggage.
Mr. Travers indicated the end of the platform by tipping his bowler hat. Still Lauren hesitated. "But Mr. Lockett told me—"
"Mr. Lockett did come for you, Miss Holbrook, but he fell ill and asked—"
"Ben is ill?" she asked quickly, paling and clutching the station manager's arm in alarm.
Her reaction stunned Ed Travers. Why did she keep referring to Ben Lockett? What was this girl to that old buzzard? She was beautiful. No question about that. And Ben had always had an eye for the ladies. Everyone in Texas knew what kind of marriage Ben had with Olivia, but even so, this girl was perplexing. Where did she come from? Why had she come to Texas to see Ben Lockett? She could be no more than twenty, and Ben was in his sixties. Maybe she was a relative. She certainly didn't look like a doxy. And why would Ben be setting up a mistress? He had—
"Mr. Travers, please." Lauren was anxiously waiting for an explanation, and the pleasant, kindly man was studying her with an unsettling intensity. Having arrived after an arduous trip from her home in North Carolina only to find that Ben was not here to meet her was disconcerting enough. Of course, he had warned her that if he couldn't leave Coronado, he would send someone else to greet her. "Is Mr. Lockett ill?"
"Ben?" Travers asked distantly. Then, clearing his throat, he said, "No, not Ben. I guess he sent Jared after you, and he's the one who's sick."
He was leading her down the platform with an encouraging hand under her elbow.
"Jared?" she asked.
My God! She didn't even know Jared! But then, it would be distressing to think that this lovely young woman had anything to do with him. It all came back to Ben. What was his game this time? He had a reputation for practical jokes and surprises, usually embarrassing for the recipient. But would Ben's legendary humor extend to victimizing an innocent like Miss Holbrook? In the few moments he had spent with her, Ed Travers had inferred that Lauren Holbrook was trusting and naive to a fault, uncommon as that was in this third year of the twentieth century.
"Jared is Ben's son, Miss Holbrook," he answered patiently. "Didn't Ben ever mention him to you?"
Lauren laughed easily. "Oh, yes. He told me he had a son. I don't recall if he told me his name, though." Her smile faded into an expression of genuine concern. "He's ill?"
"In a manner of speaking," Travers said gruffly, taking her arm more firmly as they descended the steps to the ground below. Lauren saw a long, flatbed wagon parked several yards ahead of them. The green paint on its sideboards was faded and peeling, the wheels mud-splattered. Its two horses were grazing at a tuft of grass under the enormous pecan tree.
Another horse, a palomino of magnificent proportions, was tied to the end of the wagon. Proudly he tossed his blond mane as if protesting the indignity of being hitched to such a lowly vehicle.
"Apparently, Miss Holbrook, Ben sent young Jared for you, and he came from Coronado last night. This morning, when he became incapacitated, he asked me to escort you to his home. I'm afraid the trip won't be very comfortable. I apologize, but this was the best conveyance I could find on short notice."
"I'm sure I'll be fine." She smiled. Ed Travers became dizzy under the radiance of her face and gentle voice. Then he cursed himself for being an old fool and hastened toward the wagon.
The depot manager assisted Lauren onto the rickety seat. As the porter dropped her bags unceremoniously onto the rough floor of the wagon bed, she heard a muffled moan.
She gasped in surprise when she saw the long figure sprawled on his back in the wagon. "Mr. Travers!" she exclaimed. "Is he seriously injured?"
"No," he answered. "Only a little indisposed. He'll live, though he may soon wish he were dead." He mumbled the last few words, and his meaning escaped Lauren.
She settled herself as best she could on the uncomfortable seat. The brown leather was cracked. At intervals where it had ripped open, the stuffing poked through in hard lumps. The rusted springs groaned under her slight weight. She kept her gaze focused on the road ahead.
"I must run back inside for a moment, Miss Holbrook, and speak to my assistant. If you'll indulge me, we'll be on our way without further delays." Ed Travers doffed his hat again and turned back toward the depot. The porter shuffled after him.
Lauren sighed. Well, it's not the greeting I expected, but it's novel, she thought. Then she smiled with the sheer joy of being in Texas and almost at the end of her journey. Had it been only three weeks since she last saw Ben? It seemed like eons. So much had happened since he had visited her guardians and issued the impulsive invitation for her to come to Texas.
They had all been in the parlor of the parsonage. Lauren was pouring tea, which was one of her chores when Reverend Abel Prather and his wife, Sybil, entertained. Guests visited often with the middle-aged couple, who had opened their home to Lauren when her clergyman father died eight years ago. She loved the Prathers, though she realized they were unenlightened about anything outside their sphere. Most of their callers were either other ministers or parishioners.
Their guest on that particular day had been unique. Ben Lockett had served in the Confederate Army with the young Chaplain Prather during the last three years of the war. Their philosophies differed greatly, but the two men enjoyed each other's company and found pleasure in taking opposing sides of any debate, whether over the strength of the Union Army or predestination.
After the war, Ben Lockett had left his native Virginia for unknown parts of Texas. He was of a breed of ambitious, angry young men who defiantly carved empires out of the vast plains of Texas. In the forty years since the War Between the States, Ben Lockett had become an influential cattle baron.
Lauren was intrigued by the imposing Texan. He stood tall and lean, with only the slightest paunch to indicate his advancing years. His hair was thick and snowy white, brushed back from his wide, deep forehead like a crest. Blue eyes twinkled merrily from under shaggy white eyebrows, as if he were perpetually amused by the world. But Lauren observed that Ben was capable of a piercing, glacial stare if his emotions dictated it.
His voice was deep and mellow when he said to her, "Tell me, Miss Holbrook, what you think of Texas. Like most Texans, I feel that everyone should be as enthralled with my country as I am." He stared at her from under the shaggy brows, but it was a friendly look.
"I... I don't know that much about it, Mr. Lockett," she replied honestly. "I've read about the Alamo, and I know that the state was once a republic. The rest of my knowledge is confined to the penny-novel book covers that I see on display at the general store. They depict train robberies, cattle rustling, and saloons. I don't know if that is a true characterization or not."
Ben threw back his head with its shock of white hair and roared with laughter. The booming sound rattled the china figurines that cluttered every conceivable space in Sybil Prather's overdecorated parlor.
"Well, we have our share of train robberies, and I've frequented a few saloons myself, begging your pardon, Abel. I've even chased a few rustlers all the way to Mexico." He paused. "Maybe the pictures you've seen are accurate at that, Miss Holbrook." He studied her for a moment longer, then challenged, "Why don't you come back to Texas with me and see it for yourself?"
There were several startled exclamations.
"Ben, you're joking, of course! I'd forgotten what a tease you are." Abel laughed.
"Let my Lauren go to Texas where Indians live!" Sybil cried. The ruffles covering her ample bosom quivered with distress.
"What an utterly preposterous suggestion!" came from William.
William. Yes, William Keller had been there, too.
Lauren shuddered, even in the stifling heat. She pushed the thought of William out of her mind. She wasn't going to let the memory of him ruin her reunion with Ben Lockett.
Another groan, accompanied this time by a mumbled curse, diverted her from her reverie. Hesitantly she swiveled her head to look at the ailing man. Her eyes lighted first on an ornately tooled saddle, with filigreed silver decorations glittering against the black leather. Her bags were at the back of the wagon, near the man's feet.
He must be very tall, Lauren thought as she quickly scanned the length of the prone body. Her initial impression was that he was lean and well proportioned. After that first hasty appraisal, she began at his boots and studied the figure with increasing fascination.
The black boots were of smooth leather and came to just under his knees. Tight black chinos were tucked into the tops of them. Lauren blushed at the perfect fit of the pants, which contoured the long, muscled things like a second skin.
Lauren's breath caught in her throat, and she stared as one hypnotized at the bulge between his thighs. The tight pants emphasized and detailed his anatomy. To Lauren, who was raised in deliberate ignorance of the opposite sex, it was a bold display. How could anyone be so flagrantly nonchalant about his... person? she wondered.
Her palms grew moist within her gloves.
She forced her eyes to move from his crotch. The buff-colored shirt was shoved sloppily into his belted waistband. Only the last two buttons of the shirt were closed, and the soft fabric fell away from a broad chest that rose and fell with his even breathing. The wide chest tapered to a flat belly and was covered with light brown hair that glinted with golden highlights as the sun filtered through the branches of the pecan tree and shone on him.
Lauren had never seen a man shirtless before. Once a member of Reverend Prather's congregation had caught a deadly fever and she had glimpsed his upper torso as one of the married women in attendance had bathed him. The sufferer was fat; his skin was pink; and his chest was smooth and hairless. No, he had looked nothing like this.
Lauren swallowed hard and pressed her hand against the fluttering in her stomach.
Jared Lockett groaned again, and she held her breath, afraid that he would awaken and find her looking at him with this shameful temerity. But he only sighed, making a deep hollow of his stomach under his rib cage. His hand moved onto his chest, where it stirred restlessly before remaining still. The hand was tanned and large, with strong, slim fingers. The same sun-bleached hair that covered his chest sprinkled the back of his hand.
A strong column of throat extended from the powerful shoulders. Lauren raised her eyes to his face and was crushingly disappointed. His features were covered by a black, flat-crowned, wide-brimmed hat. Her curiosity was piqued by this son of Ben's, and she wanted to view the face that belonged to this long, hard body.
Lauren almost jumped when Ed Travers said briskly, "There. I think we can leave now."
So engrossed was she with Jared Lockett's form that she hadn't noticed the man returning from his errand.
"You are extremely kind to do this, Mr. Travers." Lauren's level voice surprised her. The tickling sensation in her stomach had spread into her chest and throat. These symptoms of "the vapors" were uncharacteristic of the usually serene Lauren Holbrook.
"No problem at all," Travers hastened to assure her.
He clucked to the bedraggled horses and began maneuvering them through the traffic on the streets of the state's capital. They dodged trolleys, buggies, and horseback riders as they made their way through the city. There were no motorcars, which Lauren had seen on recent trips to Raleigh.
She enjoyed looking at the capitol building from the different angles their route afforded her. "I think you're justifiably proud of your capitol building. I've read about it. It's very impressive."
Travers smiled. "The red granite came from a quarry near the Lockett ranch."
"Keypoint," Lauren said. She remembered Ben's proud voice as he told her about the ranch. Her comment on its clever name, which used a play on words with _Lockett_ and _Key_ point, caused him to beam at her astuteness. "You'd be surprised at how few people catch that," he said. As he grinned broadly, the furrows on either side of his mouth deepened into facsimiles of dimples.
Lauren smiled at the memory, and Travers glanced at her out of the corner of his eye. So she knew about Keypoint. Did she also know who lived there? Conversationally he asked, "Have you ever been to Texas before, Miss Holbrook?"
"No, I haven't. That's why I was delighted to accept Ben's invitation to come and stay with his family for a while."
The wagon lurched when Travers suddenly jerked on the reins. She was going to stay with them? In the house in Coronado? Or at Keypoint? Either place was inconceivable. This girl was as innocent as the day was long. Had Ben Lockett gone mad?
They were outside the city now and heading west on the well-traveled road. When Lauren pulled the long pins out of her hat, Travers warned her, "I wouldn't take that off if I were you, Miss Holbrook. Our sun is hot. You might get a burn on that pretty nose."
Lauren agreed and readjusted her hat, but slipped out of her jacket. The slight breeze stirred by the movement of the wagon cooled her damp skin somewhat.
When she was settled again, Travers returned to his thoughts. That wild buck in the back of the wagon was enough reason not to keep any decent woman under the same roof with him.
Jared Lockett was notorious throughout the state for whoring and drinking. When he was younger, his activities had been deemed "sowing wild oats," but since he had passed his thirtieth birthday, they had become a matter of public scorn. When was Jared going to start acting responsibly? No time soon, Travers mused glumly.
Just last month, Jared had caused a big disturbance at the Rosenburg depot. He and some of his feckless cronies had gone into the Harvey House there and had spent the afternoon drinking and gambling. They had made their presence known in the restaurant by behaving like a pack of wild dogs. Jared made an unseemly proposition to one of the more winsome Harvey girls. The girls who worked as waitresses in the restaurant chain that served the Santa Fe Railroad were known for their scrupulous morals. If a man proposed anything to one of those young ladies, it had better be nothing less than marriage and a vine-covered cottage.
When the girl summarily rejected his suggestion, Jared had become more aggressive. The management had ejected him from the place, but not before Jared, fighting like a demon out of hell, had wreaked havoc on furniture, dishes, and a few of the patrons. It had taken six men to subdue him.
Well, sighed Travers mentally, it was probably just as well that this young woman didn't know about Jared Lockett's antics. They would no doubt scare her to death.
"Is it always this hot in September?" Lauren asked, trying to draw the station manager into conversation. She had had years of practice making small talk in the Prathers' parlor. Mr. Travers had been kind to her, but she was made uneasy by the wrinkled brow and the puzzled expression that would cross his face whenever he looked at her. Was she that different from the women in Texas?
"Yeah," he answered, reassuring her with his easy, open smile. "We usually get our first norther about the end of October. Most years, September is hotter than June or even July. Is it this warm in...?" He let the question trail off suggestively, and she didn't disappoint him.
"North Carolina. I lived—live—in Clayton. It's a small town not too far from Raleigh. And no, it's not this hot there in September."
"Is that where you met Ben?" he asked curiously. At her affirmative nod, he prodded, "And what was Ben doing in Clayton, North Carolina?"
Lauren explained the friendship between her guardian and the rancher. "For years, they corresponded, but the letters had lagged for the past decade or so. Still, on his way home from a business trip to New York, Ben decided to pay his old friend a visit."
"How long have you lived with this guardian?" Was he being too nosy? He didn't want to offend her, and no man in his right mind would cross Ben Lockett. However, she answered him readily enough and without self-consciousness.
"My father was a clergyman, too. Abel Prather was his bishop. I was twelve when my father died. The Prathers gave me a home with them."
"Your mother?" Travers asked quietly.
"I was three when she died giving birth. The baby—a boy—was stillborn." Her voice was suddenly soft and pensive. Travers noted that she touched the brooch watch pinned to her shirtwaist just above the gentle swell of her breast.
The small brooch was all she had of her mother's possessions. That and a picture taken of her parents on their wedding day. She vainly tried to remember moments she had shared with the pretty, petite woman in the picture, but no memories would come. Lauren had no inkling of the personality that had lived behind the shy eyes captured in the photograph. In stressful times, or when she longed for the parent she couldn't remember, she touched the watch with her fingertips as if the action brought her in contact with her mother. But this was a habit Lauren wasn't conscious of.
After his young wife's death, Gerald Holbrook had totally dedicated himself to his work. He delved into religious dogma and contemplated theological doctrines in the hours when he wasn't actively serving his congregation or preparing his inspired sermons. If the care of his young daughter fell to his current housekeeper, that was the price one had to pay for absolute commitment to Christ. Lauren knew that, in his way, her father loved her and wasn't bitter over his neglect—though she felt it. She would have welcomed a more demonstrative relationship, but knew her father lived on a higher plane—like God.
She was a well-behaved child, quiet and unobtrusive as she sat near her father when he studied in his library. She learned to read at an early age, and books and the characters in them became her playmates and confidantes. Her classmates weren't particularly inclined to include the "preacher's kid" in their pranks. Out of loneliness, Lauren acquired a talent for creating her own diversions.
When Gerald Holbrook died, Lauren barely missed him. She moved into the Prathers' house and assumed their routine without question. They were kind and, because of their childlessness, welcomed the adolescent girl into their home. Their generosity extended to giving Lauren piano lessons. She was musically gifted, and the piano became a passion along with literature.
No one ever left the Prathers' gaudy, crowded house without knowing their pride in Lauren. She had never betrayed their trust or disappointed them.
Except with William. How unfair was their changed attitude toward her! She was blameless!
"Miss Holbrook?" Ed Travers asked for the third time, and finally succeeded in gaining her attention.
"I'm sorry, Mr. Travers. What did you say?" Lauren flushed under her hat at being caught so deep in her own thoughts.
"I asked if you would like a drink of water," he said, reaching under the seat for a canteen, which he had filled before leaving the depot.
"Oh, yes, thank you." Lauren reached for the canteen. Never having drunk from one, she felt like a pioneer as she tipped it back and took a tentative, ladylike sip.
Just then, the wagon hit a deep rut in the road, and some of the water sloshed onto her shirtwaist. She wiped her dripping chin and laughed delightedly. Her merriment was checked when the figure in the back of the wagon groaned and cursed vehemently.
"Sonofabitch!"
# Chapter 2
Lauren whirled her head around so quickly that the motion hurt her neck. Jared's hand came up and clamped the hat more firmly over his face. He adjusted his long body to another position, contracting and relaxing muscles that Lauren didn't know existed. But then, she had never seen a masculine physique like this before. His languid movements were repelling and thrilling at the same time. It was like watching some pagan god who was beautiful even in his decadence.
She looked at Ed Travers, who was blushing furiously. "I'm sorry about that, Miss Holbrook. Don't pay any attention to his language. He—"
She interrupted with a question. "What's the matter with him?" She was afraid that Ben's son was seriously ill.
"He... uh... must've tied one on last night." When Travers realized her total lack of comprehension, he reluctantly explained. She might as well learn about Jared now. "He drank too much, don't you see," he said anxiously, "and got—"
"Drunk?" she asked incredulously. "He's got a hangover?" She stared with fixed horror at the prone figure. Never in her twenty years had she witnessed intoxication. A cordial glass of sherry and wine with Thanksgiving and Christmas dinners were the extent of alcohol consumption in the parsonage.
Jared had apparently slipped back into unconsciousness. Gentle snores were coming from under the black hat.
"Yes. Please don't fret about it, Miss Holbrook. It happens all the time. We're just lucky the sheriff didn't pick him up and take him to jail to sleep it off. Fortunately he made it to my office early this morning and asked me to meet your train and drive the two of you to Coronado. He passed out about an hour before you arrived."
"Ben told me that if he couldn't come to Austin himself, he'd send someone else. I imagine that Jared wasn't too happy over being appointed the emissary," Lauren commented.
"Whether he liked it or not, he knew he'd better do what his daddy told him to. Despite their differences, Jared respects his father."
Lauren sniffed as she cast one last reproachful glance over her shoulder. "I can't see that Jared Lockett has much respect for anyone or anything."
Ed Travers chuckled as he diverted the wagon around another collection of deep ruts. "You're probably right, Miss Holbrook."
He turned his attention to private musings, and conversation between them waned. Lauren gazed at the landscape around her.
Ben had told her he lived in the hill country, and her eyes could testify to that. Gently rolling hills covered with grass turning brown in the last days of summer surrounded them. They were driving west out of Austin, and on the right a cypress tree–lined river cheerfully wended its way through the rocky ground. Cattle grazed among small cedar trees.
As the sun slipped lower on the horizon, it became hotter. Lauren could feel rivulets of perspiration coursing down her scalp. She longed to whisk off her hat, release her heavy hair from its restricting pins, and allow what little breeze there was to blow through it.
Her hair had been the scourge of every housekeeper who had worked for Gerald Holbrook. Its washing and combing had been a constant source of muttered grumblings. Mrs. Dorothea Harris, an embittered widow who had been housekeeper from the time Lauren was seven until her father died, had declared that the girl had enough hair for six children. Each morning, she roughly pulled it into braids that were so tight they brought tears to Lauren's eyes. Lauren's father had said in a rare compliment that her thick black hair was like her mother's. In this Lauren took secret pride.
Of course, it was out of the question to take her hair down now. It wouldn't do at all to arrive at the Locketts' house without a hat, let alone with unbound hair.
Dismally she looked at the fine layer of dust on her navy skirt and agonized over the disheveled appearance she would present when she arrived at her destination. What would Ben think? Would he be ashamed of her and regret his invitation? Lauren wanted so badly to impress his family.
She flicked away what she could of the settling dust. It was instantly replaced, and she sighed resignedly.
Ed Travers said, "It does get a mite dry and dusty. Ben must have done quite a sales job to get you to leave the green hills of North Carolina and come all the way out here." His curiosity over Lauren Holbrook's future status in the Lockett household hadn't yet been satisfied.
Lauren laughed. "He did sell me on Texas, and I haven't been disappointed. It's wonderful."
"How long will you be here?" He couldn't help asking.
She averted her head quickly and clenched her hands into fists. "I... I'm not sure." She managed to control her initial agitation and go on. "It will depend on Mrs. Lockett. You see, I'm to be her secretary."
Ed Travers almost fell off his seat. Olivia Lockett with a secretary? What was old Ben trying to pull?
He swallowed hard before he asked squeakily, "What are you going to do for her?"
"I've spent years helping my guardians entertain. Ben thought that I might relieve Mrs. Lockett of some of those responsibilities. I can help her with her correspondence, for instance. The length of my visit will depend on how well we get along and whether she likes me or not," Lauren answered. As she explained her future to him, she tried to assimilate it in her own mind.
Poor lass, thought Travers. If it were a case of Olivia Lockett liking a young, pretty girl living under her roof, then the innocent Lauren Holbrook would be on the next train out of Austin heading anywhere. Olivia could freeze the balls off any man with one icy blast from those hard, green Creole eyes of hers. What would she do to this poor child?
Intuitively Lauren sensed Travers's bewilderment. She had felt that same incredulity at Ben's offer. It had come so suddenly, and she was completely unprepared for it.
* * *
They had dined on overdone lamb and bland vegetables, the typical fare that came out of the Prathers' kitchen. Lauren was always painfully aware of the unpalatable meals that Sybil Prather served her guests. She was grateful to Ben Lockett for doing justice to his plate, though he graciously declined a refill.
After dinner, Lauren played the piano for the guests at the persistent urgings of her guardians. The recital was well received but, as usual, the Prathers' gushing praise embarrassed their ward.
Sybil, her plump figure swathed in pink ruffles, sat beside her husband on a garishly upholstered settee. Unfortunately Sybil's taste in clothes extended to her house as well. Her motto was: "More is better." The house was dark and heavy with brocades and velvets. Chandeliers and vases of dark-colored glass added to the gloom. Wallpaper in overgrown prints and a maroon carpet splashed with large orange and yellow flowers vied for supremacy.
The pastor's wife simpered as Abel boasted of her prizewinning roses. Much to their surprise, Lauren's relief, and William Keller's aggravation, Ben asked Lauren to show him this noteworthy garden.
The evening had been warm and still, and cicadas serenaded them as Lauren led Ben to the small rose garden and sat down on a low bench.
"Do you grow roses in Texas, Mr. Lockett?"
"Indeed we do. I have a Mexican gardener who tends to the grounds around the house in Coronado, and he grows them much sweeter and much larger than these prizewinning flowers of Sybil's. I think his secret is horse manure."
There was a momentary pause. Lauren wasn't sure what her reaction should be. Then they both laughed spontaneously. She chided herself for condoning his indelicacy but, somehow, it didn't seem to matter.
"Thank you for inviting me out here with you," she said. "Abel and Sybil usually contrive for me to be alone with William."
"And you don't want to be alone with William?"
She shuddered and said, "No. I don't."
William Keller was a serious, thirty-five-year-old preacher who had accepted the pastorate of a small church on the outskirts of Clayton. Lauren sensed that, beneath the guise of piety, he was ambitious and shrewd. He was continually trying to impress the bishop with the strength of his moral fiber and his undying love for humanity.
Much to Lauren's dismay, the Prathers considered William a superb candidate to relieve her of the state of spinsterhood. They extolled William's virtues to her at least three times a day, and she was forced to take these doses of him much as one is forced to take bad-tasting medicine at regular intervals.
Lauren had only a vague conception of what the intimacies of marriage implied, but the idea of even sharing the same room with the preacher convinced her that spinsterhood would be preferable to a lifetime spent with William Keller
Normally Lauren's impressions of people were charitable, but she found William physically unattractive, intellectually boring, and socially bigoted. His entire person repulsed her. He had an annoying habit of talking to one's chest rather than one's eyes. Tall and stoop-shouldered, he had thin, lank blond hair which was perpetually falling into colorless eyes fringed by equally colorless lashes. His nose was the most prominent and unfortunate feature of his face. Lauren thought he greatly resembled illustrations of Ichabod Crane, Washington Irving's main character in _The Legend of Sleepy Hollow._
Ben Lockett had brought her attention back to him with a brusque clearing of his throat. He didn't pursue the subject of William. Instead, he asked her, "What do you do, Miss Holbrook?" Lauren looked at him, puzzled by the question. He clarified, "What keeps you busy around here? Are you happy?"
She answered him frankly. "The Prathers are dear people, and it was kind of them to take me into their home when my father died. I had no relatives. Father had a small annuity, which they have refused to accept for my expenses. I had hoped to teach music or perhaps tutor students in literature or grammar and earn my own money, but the Prathers adamantly reject the idea of my working outside the house."
"So you entertain their guests. That's all?" He smiled at her kindly, and she didn't take offense.
"It's not a very ambitious undertaking, is it?" she asked ruefully. "Oh, I do charity work in the church, sit with shut-ins and sick people, help new mothers with the rest of their families while they're in confinement. I play the organ for the Sunday services and teach a children's Sunday school class." Even to her own ears, these accomplishments sounded dreary.
"Did you ever think of having a family of your own? Marrying?" He fixed her with a blue gaze that was penetrating and compelling.
"I... well, not really," she said shyly, and shifted her eyes away from him.
"When I'm trying to make some decision and sort things out, I ride the line for a few days by myself. I like being alone with no company except my horse and Mother Nature."
"'Ride the line'?" she asked with quickening interest.
"Yes. We ride along the fences to make sure none have been knocked down or cut down. Sometimes rustlers try to steal Lockett cows, or maybe a sheep farmer wants to water his flock and not pay for it, so the sheep just make themselves at home on Lockett land."
Lauren drew a deep breath and held it a long time before releasing it. "It sounds... oh... beautiful, primeval, exciting. I don't know the word to use."
"It's all those things." He studied his knuckles for a moment, then asked, "Why don't you come to Texas with me?" His tone was no longer bantering, as it had been when he made the same offer, publicly, earlier that afternoon.
"You're teasing me about this, Mr. Lockett." The statement contained only a hint of query.
"No, I'm not, Miss Holbrook. I'm just an old cowboy who believes in saving time and getting right to the point."
"But what in the world would I do in Texas?" She had thought at the time that it was an impossibility for her to have such an adventure but hadn't wanted to give up the idea just yet.
"My wife is very active in social and civic affairs in Coronado. That's the town we live in. It's about half a day's ride from Austin. I'm at Keypoint or away on business so often that she can't always count on me. I think she could use someone with your abilities to help her. You've had a lot of experience in arranging social functions. You are an accomplished musician and well read, both of which would be helpful. You could handle her correspondence and such. What do you think?"
When she didn't respond, he pressed his point.
"Of course, we would pay you a salary and give you a room in the house. My son has never married, so we have a lot of space that I was hoping would one day be taken up with grandchildren." He paused for a long moment and, when Lauren looked at him, he was staring unseeingly at the rose-bushes. Then he seemed to shake himself loose from the thought and continued, "I want you to feel like one of the family. You would in no way be considered a servant." He grinned engagingly.
"But why are you asking _me_? I'm sure if Mrs. Lockett were looking for a secretary, she could find one."
He shrugged negligently, dismissing her logic. "I'm sure she could, too, but it would probably never occur to her to look for one. My reasons are my own, but I promise you that they are above reproach." He smiled down at her, and his eyes twinkled like blue lights under the bushy white brows.
"Mr. Lockett, I appreciate your liking me enough to invite me," Lauren said earnestly, "but my place is here. This is where my father wanted me to be."
"Your father is dead. You're alive, but you'll be as good as dead if you don't get out now."
Lauren had been startled when he stood up abruptly, almost angrily, and took several impatient steps away from her. When he turned back, he had looked at her with tenderness and spoken more gently.
"Lauren," she noticed his switch to her first name, "I know you have been taught to obey without questioning. You have a keen sense of duty that is admirable. But I think I see in you a restlessness, an eagerness for life, that needs to be unleashed. You could come to stay awhile, and if things didn't work out, or if you hated Texas and the Locketts, I would see that you were sent home right away. No hurt feelings."
What a fool she had been not to accept his invitation right then! Instead, with her head bowed, she had responded softly, "Mr. Lockett, your invitation is overwhelming, and I would love to accept. But I can't go anywhere." She shook her head sadly. "I _have_ been taught duty and responsibility, you see. I will probably live with the Prathers for the rest of their lives. They depend on me. It would destroy them for me to leave."
"And what happens to you when they die? If you haven't been pawned off on William or someone like him, what will you do then?"
"I'm sure that some provision will have been made for me."
He sighed heavily and, seeing him so deflated, Lauren was almost prompted to change her mind. He seemed to lose some of his vibrancy. His age was suddenly more apparent on the chiseled features, and there was a mute appeal in the deep blue eyes.
"If there is ever the slightest possibility that you might change your mind, wire me immediately. I mean it. You have a standing invitation."
"Thank you, Mr. Lockett," she replied graciously. She wanted his understanding and she said, "I don't want to be like them." She had been horrified at her admission. "No, no, I don't mean—"
"I know what you mean, Miss Holbrook. I'm sure that you have very few unkind thoughts, but you would like to have a broader horizon than the Prathers do, am I correct?"
"Yes! That's what I was trying to say."
"Remember, if you ever change your mind..." he reiterated quietly as they walked back toward the front door.
* * *
The sun beat down on the wagon relentlessly. Lauren was becoming weary. The muscles of her back and shoulders ached with fatigue from maintaining her erect posture on the uncomfortable leather seat. Though she had taken numerous sips from the canteen, her throat was parched, and she was covered with dust from the road. Just when she was despairing of ever reaching their destination, Ed Travers nodded his head forward and said, "Coronado."
The wagon topped a hill, allowing Lauren a panoramic view of the small town where Ben lived when he wasn't at his ranch. As the horses picked up their pace on the downward side of the hill, she asked eagerly, "How many people live here?"
"Ummm, about three thousand," Travers replied.
"And how far are we from the ranch? From Keypoint?"
"About a three-hour ride west."
Lauren's disappointment was covered by her interest in Coronado as they drove down the main thoroughfare. She realized that people on the street recognized the large palomino tied to the back of the wagon. Several whispered conjectures were exchanged behind screening hands. Lauren resolutely ignored the man behind her and the persons speculating on his condition.
Her only purposes now were to see Ben again and to meet Mrs. Lockett. Examining her feelings for the man who had come to mean so much to her in such a short space of time, Lauren concluded that Ben Lockett represented the father she had never had. He was merry while her own father had been austere; he was big and robust while the pastor had been slight and less than hearty; he was warm while Gerald Holbrook had been reserved, even toward his own daughter. Ben's deep voice and sharp sense of humor had attracted her to him, and she was breathless with excitement to see him in his own element.
Travers turned the wagon onto a wide, tree-shaded avenue that led south from the center of town. Through the trees, Lauren glimpsed the large house long before Mr. Travers directed the horses up the shell lane.
The house was a credit to whomever had designed it. It was Victorian in design, but not overly ornate, with only a minimum amount of trim. Graceful but sturdy railings outlined the porch that surrounded the house on three sides. On each of the front corners of the second floor were circular rooms domed by onion-shaped cupolas. The tall windows, three on each side of the front porch, were framed in brick-red shutters which contrasted beautiful with the cream-colored frame house. The front door was the same brownish-red and flanked by urns which sported a profusion of red geraniums. The lateness of the summer season didn't hinder the zinnias, petunias, and roses from blooming in the lush beds that lined the front porch. The grass inside the iron picket-fenced yard was still green and clipped to perfection.
"Oh, how lovely," Lauren whispered as she gazed at the house in awe. She sat for several moments relishing the fact that she was finally here at Ben's house.
Travers eased his aching body out of the wagon and went toward the rear of it. He lifted out Lauren's bags and set them at the end of the sidewalk that led up to the steps in front of the house. He then returned to the back of the wagon and, none too gently, nudged Jared Lockett with his fist. "Come on, Jared, wake up. You're home."
Lauren barely noticed the disgruntled groan that issued from under the black hat. She was distracted from her joy over the house only by Ed Travers coming to her side of the wagon and offering his assistance as she alighted. She straightened her hat as best she could without a mirror, shook some of the dust from her navy skirt, and was about to pull on her jacket when the body in the back of the wagon finally climbed down.
She stopped to stare at the rumpled figure. It leaned against the sideboards of the wagon and held its head as if in an effort to keep the head on its shoulders.
Impatient fingers were raked haphazardly through sun bleached brown hair that disobediently fell back into wavy disarray. The man bent from the waist and supported his upper body by placing his hands on his knees as he drew in several long, shuddering breaths. Lauren was fearful of seeing him plunged into the throes of nausea, but he slowly straightened up to his full height. Only then did he turn and see the young woman who was staring at him in fascination.
The deepening afternoon shadows prevented Lauren from having a clear look at his face. She thought his eyes must be dark, but his constant blinking to focus them made their color impossible to discern.
A sardonic smirk lifted one corner of his sensual mouth before he straightened his shoulders a trifle and took three stumbling steps. He stood within an arm's length of her. She was entranced by this man and his barbaric behavior, and couldn't find it within herself to move away from him.
Jared placed a hand over the left side of his chest, which lay bare under the loose, unbuttoned shirt, and said with a slur, "Your servant, Miss Ho... Hol... Holberk."
He bent from the waist again, this time in a travesty of a courtly bow. Executing the gesture was beyond him in his present state. To Lauren's horror, he continued on his way downward until he grasped her around the waist with two strong hands and leaned upon her bosom to break his fall. She gasped in mortification as he found what he considered to be a haven of repose. His head nestled between her breasts and he sighed contentedly, not knowing or caring what a comical picture he made. Instinctively his hands slid around her narrow waist to her back, and he pressed her closer.
His breath was warm on her skin through the thin linen of her shirt. For an instant, when his nose nuzzled the inside curve of her breast, Lauren felt certain she would faint. Even more staggering to her was a fleeting, overpowering urge to clasp his head into the soft depth of her cleavage.
Suddenly Mr. Travers circled the wagon, angrily grabbed Jared by the shoulders, and hauled him off her.
"Lockett! My God, man, you're an animal."
The animal seemed oblivious to the insult as he slumped once more against the wagon, a stupid grin on his face.
A Mexican man came running from the back of the house to lend his assistance as the front door opened and a woman stepped onto the porch.
Lauren's head was spinning. Things were happening too quickly, and she couldn't take them all in. She wanted to see Ben and rely on his sturdy presence to restore some measure of sanity to this situation. Hurriedly she shrugged into her jacket before facing the woman who stood on the edge of the porch looking down at her.
Lauren smiled shyly and walked through the iron gate and up the sidewalk. She halted in front of the bottom step and looked up at the woman. Instinct warned her she should go no further. The figure at the top of the steps had the aspect of a sentinel protecting an inviolable domain.
"Mrs. Lockett, I've brought—"
"Yes, thank you, Mr. Travers," Olivia Lockett sharply interrupted the man. "Can you find accommodations for the night? We will compensate you, of course, for your time and trouble."
Ed Travers was being dismissed, and he knew it. He nodded his silent acquiescence but didn't leave Lauren's side.
"You are Miss Holbrook," Olivia said.
The statement was clipped, and Lauren answered with like brevity, "Yes, Lauren Holbrook."
The woman appeared to be slightly taller than Lauren. Her hair was dark but, around her face, it was streaked most attractively with silver. She was slender, but held herself straight and rigid. This militant stance made her seem larger than she was. Her face was unlined; her complexion was olive. It was impossible to see the color of her eyes in the poor light, but Lauren was uncomfortably aware of their hawklike penetration.
She wore a green dress of some stiff fabric, and it would have been incongruous with her character to see one wrinkle, one piece of lint, any flaw that would mar her impeccability. Because of her disheveled appearance, Lauren felt at a distinct disadvantage. The woman's face betrayed neither approval nor disapproval of her guest.
"I'm Olivia Lockett. I trust your trip was uneventful." She didn't wait for a reply but continued in the same crisp tone, "I think you may have made the trip in vain, Miss Holbrook. I cannot conceive what my husband had in mind when he invited you here."
Lauren was stunned by Olivia Lockett's harsh words. Where was Ben? Obviously Mrs. Lockett had expected her. So why this instant hostility? She stammered, "I... I'm sure that if we could all sit down and talk about it, Ben would explain—"
"Is Jared all right, Pepe?" Olivia interrupted Lauren imperiously.
" _Sí,_ Señora Lockett," the man who had come to Jared's aid answered her quickly, still supporting the younger man, who slumped against him unconsciously.
"He really got drunk last night, I guess," Olivia said. Lauren thought she saw the hint of a smile in the corners of Olivia's mouth, but then it was gone, and she was sure she had imagined it. What mother would be pleased to see her son in such a condition?
Continuing to ignore Lauren, Olivia addressed Pepe again. "Take Jared to the stable and sober him up." Her tone was caustic. "Miss Holbrook, I'll send someone out for your bags."
Lauren took that to be the only invitation to enter the house she was likely to get. Where was Ben—at Keypoint? Why had he deserted her this way?
She lifted her skirt and climbed the steps until she was level with Olivia. The woman looked at her coldly, and a premonition of disaster coiled in Lauren's stomach. She found the courage to say, "If you would summon Mr. Lockett, I'm sure—"
"That's impossible, Miss Holbrook. My husband died early this morning."
# Chapter 3
Lauren was struck dumb. Was Olivia Lockett mad? The aristocratic face looking back at her with implacable eyes conveyed no emotion.
"That's impossible," Lauren breathed. The words were barely audible.
"I'm afraid it's true, Miss Holbrook. He hasn't been well for some time. He told me upon his return from New York that he had gone there to consult with a heart specialist." She paused for a moment and looked toward her son who, still supported by Pepe, was disappearing around the corner of the house. "Last night, Ben and Jared argued. After Jared left, Ben had a seizure. He died this morning," she repeated.
Lauren's eyes filled with tears. "I'm sorry," she said. What was she going to do now? "I didn't know anything about his illness. You must believe that, Mrs. Lockett."
Olivia looked at Lauren closely and then said in the same brisk tones she had used to address Pepe, "We can discuss this another time. For the next few days, make yourself comfortable in your room. Elena will be assigned to you. I must ask you to stay to yourself as much as possible. It would be awkward to explain your being here at this difficult time. Do you understand?"
Lauren simply nodded.
Turning, she looked down at Ed Travers, who remained standing at the iron gate, holding his hat at his chest. His slack mouth and wide eyes revealed his astonishment at the news of Ben Lockett's death.
"Thank you for your help, Mr. Travers. You have been most kind," Lauren called down to him.
The depot manager said humbly, "I'm always at your service, Miss Holbrook. If there's ever anything I can do for you, you have only to ask." Travers raised his hands expressively as he spoke.
"Thank you," Lauren mumbled.
"Mrs. Lockett, with your permission, I'll help notify the public of Be—Mr. Lockett's demise."
"The funeral will be the day after tomorrow at two o'clock. Your help is always appreciated, Mr. Travers. As is your discretion."
Her last words sounded almost like a threat. Ed Travers nodded and, replacing his bowler hat on his head, returned to the wagon.
Lauren followed Olivia into the house.
The impressions she drew from what brief glances of Ben's home she was allowed were delightful ones. A wide foyer with rooms leading off either side of it ran the length of the house. The staircase was directly opposite the front door and, at the top, hallways extended in three directions.
Lauren and Olivia ascended the stairs, turned right, and passed down a long, well-lit hall with doors, probably opening into bedrooms, on either side. At the end of the hall, Olivia opened a door. Like all of the woodwork in the house, it was painted a pristine white. Lauren followed Olivia into the room and looked around the chamber in which she would be sequestered for the next few days.
Well, if I have to be imprisoned, this is a pleasant cell, she thought. The small room was one of the round ones she had seen from the front of the house. It was beautifully furnished. The floors were stained oak, relieved now and then by small throw rugs. An intricate ecru lace spread covered the full-sized, four-poster bed. The walls were papered with a pale yellow flower print that was subtle and tasteful. There were a dresser and washstand, a bookcase, a rocking chair, a round table next to the chair, and a smaller table at the bedside. Fresh flowers filled several vases scattered around the room and, though the windows were curtained now, Lauren knew the morning sun would flood the room with even more cheerfulness. _Someone_ had planned on receiving her graciously.
"It's lovely, Mrs. Lockett. Thank you."
"Then you won't mind staying in here for a few days until the funeral is over."
Lauren wanted to attend the funeral. But something in the woman's manner clearly indicated that she would strenuously object to an appearance by Lauren.
"The bathroom is through there." She indicated a door. "There's a door on the other side of it, but it remains locked. You needn't worry about anyone disturbing you."
Or me disturbing anyone, Lauren thought.
"Elena will be here soon with your supper. If you need anything, ask her. She is solely responsible to you." She was about to leave the room when Lauren halted her.
"Mrs. Lockett, I'm sorry about your husband. He was—"
"Yes," Olivia broke in. "Goodnight, Miss Holbrook."
* * *
Lauren sat down in the rocking chair and tried to absorb the events that had taken place since her arrival.
Ben Lockett dead? It wasn't possible. For weeks, she had envisioned his kind face and heard his voice compelling her to come here. Now, he was dead, and her future was, at best, uncertain.
Having sat down, she realized how tired she was. The endless days and cramped nights on the trains, the rough, dusty drive from Austin, that horrible man sprawled in a drunken stupor even as his own father lay dead, and then the confrontation with Olivia. It was all too much. Lauren rested her head on the small pillow attached to the back of the rocking chair and fell into a deep sleep.
She was awakened by a persistent voice and someone shaking her arm. Go away, she thought. I don't want to wake up, because something terrible has happened. I don't want to remember.
The pest wouldn't go away. Lauren awakened to meet the blackest, most liquid eyes she had ever seen. She took in the rest of the face. It was dark, smooth, unblemished, and beautiful. The smile was gentle and warm. The voice was soothing and sympathetic.
"Poor _señorita._ You are so tired that you fall asleep in the chair. With your hat on! No supper? No bath? Elena will help you, _sí_?"
"Elena? I'm Lauren. How do you do?" Lauren grasped the girl's friendliness like a lifeline.
"You're so beautiful, _señorita._ I think you be prettier and feel better after a bath. I run the water for you. You get undressed, _sí_?"
Elena stood back from the chair, and Lauren saw her protruding stomach, announcing the late stages of pregnancy. Was Olivia hiding Elena by "assigning" the maid to her? Her condition would no doubt be an embarrassment to the family and their expected callers.
Elena couldn't be more than sixteen or seventeen, and seemed unaffected by being seen when her confinement was so near. Her breasts, the dark nipples readily apparent, were almost as large as her stomach, and hung unrestrained under an embroidered white blouse.
She waddled into the bathroom, keeping up lively chatter in a mixture of English and Spanish. The topics of conversation she chose switched as quickly as her languages. When she returned to the bedroom and saw that Lauren had not moved, she scolded her.
" _Señorita,_ your water will get cold, not to mention your supper. Come, let Elena help you."
Lauren was shocked when Elena turned her around and began undoing her buttons with deft fingers. She wanted to object but was too tired to force the words through her lips. Swiftly Elena divested her of her clothes.
When all that remained were Lauren's pantalets, corset, and camisole, Elena shook her head from side to side and made a _tsk_ ing sound.
"A corset! And you are so slim. You can't even breathe." She loosened the laces, and soon the offending garment was lying in the heap of soiled clothes at Lauren's feet.
Lauren caught Elena's hands when the maid tried to remove her other underwear. Hastily she stepped into the bathroom, which was decorated as tastefully as the bedroom. She looked gratefully at the bathtub of scented water, stepped into it, and eased her tired, sore body into the steamy water. She finished bathing, and was luxuriating in the first relaxation she had known for days, when Elena bustled in. Lauren gasped in surprise for, since early adolescence, no one had seen her naked.
" _La señorita_ is ready for me to wash her hair, _sí_?"
"No!" Lauren protested, desperately trying to cover herself. When she saw the hurt expression on Elena's face, she added hurriedly, "I can do it myself."
"But why should you? I'm here," Elena said with happy logic. "Señor Lockett say, 'Take care of the young lady,' and so I do." She made the sign of the cross across her enormous breasts at her mention of Ben.
Elena had already begun to take the pins from Lauren's hair, which needed no encouragement to cascade down her back to her waist. The Mexican girl continued to chatter as she poured pitcher after pitcher of warm water over Lauren's head. She lathered the thick tresses in a massage that was hypnotic. Lauren felt her nerves dissipating under Elena's capable hands.
"Señor Lockett look so forward to you coming. He tell all of us about the pretty lady who come to live with us. He order the room be made ready. He check it himself to make sure everything okay." Before Lauren could protest, Elena pulled her out of the tub.
Lauren's efforts to cover herself were ineffectual, but Elena didn't seem to notice her embarrassment. The bright pink blush that suffused Lauren's body was due only in part to her warm bath.
It was necessary for her to change the subject away from herself and Ben. She couldn't think of him now. Her grief would be saved for a more private time. She asked companionably, "When is your baby coming?"
_"Quién sabe?"_ Elena shrugged. "When he get ready to come, he come." She smiled.
"What does your husband do?"
"Oh, he one fine _vaquero_ on the Lockett ranch. His name is Carlos. He one fine man." She rolled her expressive eyes at Lauren, who blushed instinctively. She didn't want Elena to elaborate.
"Isn't it late for you to still be working? Feel free to go anytime."
Elena's laughter bubbled forth again. " _Señorita,_ I live here. Carlos stay at the ranch, and I stay here. We get together when we can at his mamma's house in Pueblo."
Lauren was aghast. "But surely you would rather have your own home and live together!"
" _Sí,_ but we would also like to eat. With no money, we could do neither." She giggled.
"I see," Lauren murmured, although she didn't see at all. Thus far, she understood nothing of this alien land and its people.
They were back in the bedroom. Elena took a nightgown out of one of Lauren's bags and slipped it over her head. Lauren stood in the middle of the room feeling lost and helpless as Elena arranged dishes on a large tray. Apparently she had carried it up with her and deposited it on the table before she had awakened Lauren. Delicious aromas filled the room as Elena lifted the lids of the dishes, and Lauren's mouth began to water. She hadn't eaten since... when?
Elena set the tray on Lauren's lap. On it were a beautifully grilled steak, potatoes, a salad, and two kinds of bread. One was a yeast roll, and the other was a flat, round bread that was totally foreign to Lauren. There was also a bowl of beans with a tomato sauce ladled over them.
"What is this?" she asked, pointing to the bread.
" _Tortilla._ Bread made of corn," Elena explained.
Lauren took a bite and found that it had very little flavor. Then Elena scooped some butter on it, salted it lightly, and rolled it like a cigar. It was delicious. _"Tortilla?"_ Lauren repeated the word, and Elena nodded, clapping her hands.
Lauren then pointed to the bowl of beans.
_"Frijoles,"_ Elena said. "With _picante._ "
Lauren had lost her timidity now, and took a generous mouthful. She knew instantly that she had made a grave error. Her mouth was on fire! She quickly swallowed what she couldn't manage to spit out, appalled that she could do such an unladylike thing. Elena was laughing so hard that her breasts and stomach were bouncing.
"Water," Lauren croaked. She gulped the glass that Elena gave her and asked for more. Finally the fire was out, but she tentatively tasted the other foods before taking any more big bites. The rest of the meal was delicious, and she finished everything—except the _frijoles._
Despite Lauren's objections, Elena braided her hair into one long plait. The maid then hurried to pull down the bedcovers.
"Go to bed now, _señorita,_ and rest. It has been a hard day, _sí_?"
"Yes, it has." She climbed into the bed as Elena quietly loaded the tray and went around the room turning off the gas lamps.
_"Buenas noches, señorita,"_ she whispered as she left the room.
"Goodnight, Elena."
Lauren burrowed between the sheets. The house was quiet, though she could pick up muted and indistinguishable voices that wafted up the stairs.
"Ben Lockett, how could you do this to me?" she asked into her pillow, and was immediately ashamed of her thoughts.
After the dreadful scenes she'd been subjected to before leaving North Carolina, Ben's strength, affection, and warmth had been her salvation. She had hoped to start a new life with Ben's family. Now, all those hopes were dashed. Ben was dead. This splendid house seemed to swallow her. And what of the cold, formidable woman who dominated it?
It occurred to Lauren then that Ben's widow hadn't shown any signs of emotion. Maybe Olivia was one of those people who expressed their grief privately. Maybe. The thought was disturbing.
What would Jared Lockett think when he learned of his father's death? Why would a man who had money and position get blind drunk and make a public spectacle of himself? Ed Travers had intimated that it wasn't at all unusual to see Jared in such a condition.
Well, it's none of my concern, Lauren thought as she resolutely closed her eyes. She wouldn't have any dealings with _him._
He was very tall, wasn't he? She wished she could forget the tremors that had coursed through her when his hands had closed around her waist and caressed her back. The heavy pressure of his head against her breasts hadn't been an altogether unpleasant sensation. His hair was light brown. Did sunlight bring out streaks of gold as she knew it did in the down on his chest?
* * *
Lauren awoke languidly, after ten hours of sleep. The room was awash with sunlight, which filtered through the airy, yellow drapes in the east windows.
She flung off the covers and padded into the bathroom. Desolation over Ben's death and uncertainty over her future still weighed heavily on her mind. She couldn't stay here now. And she definitely couldn't return to North Carolina, either.
Elena came in just as Lauren finished dressing.
_"Buenos días, señorita,"_ she greeted her cheerfully.
"Good morning, Elena," said Lauren, continuing to brush her thick, black hair.
"Did you sleep well?" Elena asked conversationally as she spread the covers smoothly over the bed. She busied herself with straightening the spotless room, watering the plants and flowers, and arranging the breakfast dishes on the same tray that had held Lauren's dinner the previous night.
"Yes, very well." Lauren swallowed uncomfortably when she recalled some of her dreams. They had been unsettling. Tall men stalked her. One man had white hair and Ben's smiling face. The other's face was shadowed by a large black hat, but she recognized the physique. It was imprinted on her mind with indelible clarity.
After eating the large dinner last night, she didn't think she could be hungry. But the fresh melon slices were delicious and juicy. She drank the hot coffee, though she would have preferred tea. Timidly she asked Elena if she could have tea from now on.
"Oh, _sí, sí._ My mamma, she is the cook." She laughed at the startled expression on Lauren's face. "She work for the Locketts since before I was born. You call her Rosa."
"I hate to think of you carrying that heavy tray upstairs for my meals, Elena, but Mrs. Lockett made it clear that I am to stay as close to this room as possible during the funeral preparations and while they're receiving callers." Her gaze drifted to the open windows as another wave of sadness ebbed over her. "Is the funeral still scheduled for tomorrow?"
_"Sí,"_ Elena answered softly. "Lots of people coming from places far away."
"Well, I guess I shall busy myself somehow," Lauren sighed.
She managed to while away the long hours reading and embroidering a sampler she had brought with her from Clayton. She was denied Elena's company; the girl explained that she was needed to help her mother in the kitchen.
The day passed slowly. To Lauren, who was accustomed to activity and would seek out chores if none were apparent, it seemed interminable.
Late in the afternoon, she paused in her reading when she heard a heavy tread in the hallway. Whoever it was entered a room before reaching the end of the hall where her room was located. Slipping off her eyeglasses to rest her eyes, she listened closely to the sounds of drawers opening and shutting, of wardrobe doors swinging back, of heavy shoes or boots dropping with a thud onto the floor. Stockinged footsteps shuffled back and forth.
Lauren heard the clink of glass on glass, water splashing, a few low mumbled words, the scraping of furniture against wood floors, the rustle of clothing.
Some minutes later, the person was finished with his toilette and left the room. A door closed quietly and footsteps receded down the hall. Someone occupied the room on the other side of the bathroom. Lauren hadn't been aware of anyone being in there since she had moved in.
That evening, Lauren was embroidering as Elena gathered up the dinner tray and said goodnight.
"Elena," Lauren asked, "who occupies that room through the bathroom?"
"Ah! That is Señor Jared's room." Elena's eyes widened expressively. "My Carlos threaten me never to go near it." She giggled as she adjusted the tray around her expanded belly. "He say Señor Jared can please any woman." She winked broadly as she closed the door.
Lauren's gray eyes stared unseeingly at the bright bead of blood on her pricked finger.
# Chapter 4
The sun refused to shine on the day of Ben Lockett's funeral. It, too, seemed to be mourning the man who had spent long hours under its hot rays, worshiping the land and its elements.
For two days, Lauren had watched from her window as all types of people came to pay tribute to Ben. There were wealthy visitors, their affluence evidenced by their clothes and their conveyances. Others looked to be farmers or ranchers wearing clean but worn clothes. Their wives tagged after them staring at the beautiful house in awe. _Vaqueros_ in dusty leather chaps rode up to the house on trail-weary horses. The mourners came singly, in pairs, or in groups, but there was a continuous parade of them. Lauren couldn't imagine the woman who had greeted her with such hostility graciously welcoming the humblest of these visitors.
The hearse came down the shell lane, glistening blackly. With its tassel-trimmed, fringed drapes, plumed horses, and driver who wore a cutaway coat and top hat, it looked like some sort of circus vehicle. Ben would hardly have chosen such an ostentatious, frivolous conveyance to carry him to his grave, Lauren thought, feeling another pang of grief for the rugged, virile man.
Lauren watched from her window as Olivia was escorted down the front walk by a man no taller than she. From Lauren's perspective, his bald head seemed on a level with Olivia's veiled black hat. His black coat fit snugly across a portly torso. Shyly he touched Olivia's elbow as he assisted her. It was hard to tell if his hesitancy in touching her was in deference to her grief or in fear that she would turn on him. His attitude toward her seemed to be almost subservient.
Lauren inhaled sharply when she saw the figure behind the other two. His height and the breadth of his shoulders gave away his identity, though she still couldn't see his face under the wide-brimmed black hat. His black suit bore no distinguishing details. He appeared to be withdrawn, oblivious to the sympathetic friends who watched him with pity as he followed his mother and the other man to the covered carriage which waited behind the hearse.
The coffin was ceremoniously lifted into the hearse. Lauren thought Ben would have scoffed at all this pomp and circumstance. She was sure he was somewhere watching all of them, his blue eyes twinkling in amusement. She offered a prayer for his soul as the hearse led the procession away from the house.
As the family's carriage rolled by, she noticed a strong, lean, tanned hand dangling negligently against the door.
* * *
The summons came so suddenly that Lauren was unprepared for it. Elena had flung open the door to the room and, with colorful skirts swirling around her bare legs, and breasts bobbing like lanterns suspended on a wire, she sputtered the message.
" _La señora_ wants to see you _pronto, señorita._ Quickly she say. Quickly. She is with Señor Wells in Señor Lockett's office."
She was a flurry of motion as she helped Lauren button her shirtwaist, which had been discarded when she had prepared for a nap. Her hair was hastily pulled into its usual bun. Elena knelt down to button her shoes. Lauren would have thought it impossible for Elena to fold herself into that position, but didn't have time to wonder about it now. She was breathing rapidly, her heart was pounding, and her palms were sweating. In all her life, she had never been this nervous.
They left the room after Lauren grabbed a lace handkerchief. Whether it was to dry her hands or to have something to hold on to, she didn't know. Elena also seemed jittery as she led Lauren down the hall to the wide staircase. They descended quickly and walked toward a large sliding door. Elena gave Lauren a quick nod of encouragement and pulled aside one panel of the door. Lauren drew a deep breath.
She stepped into the room and was again surprised at the simple beauty of the house. There were floor-to-ceiling book-shelves on one side of the room. Other shelves flanked a large fireplace. The mantel was intricately and masterfully carved. Wide, full-length windows composed the fourth wall.
An Aubusson rug covered most of the hardwood floor. Leather chairs and small tables, strategically placed, lent themselves to private conversations. There was a long sideboard loaded with decanters and glasses of cut crystal. The window drapes had been completely opened, allowing the afternoon sun to stream in and reflect on the glass surfaces.
In front of the windows was a massive desk littered with ledgers and papers of various shapes, sizes, and colors. Olivia sat in the high-backed leather chair behind the desk. The short, stout man Lauren had seen with her as she left for the funeral was seated in a chair in front of the desk. He stood and walked toward her.
"Miss Holbrook, this is indeed a pleasure. I'm sorry that circumstances have prevented me from meeting you before now. I hope you haven't been too uncomfortable since your arrival." He seemed to expect no answer as he continued, "I am Carson Wells, an old friend of Ben's and Olivia's, and also their lawyer. How do you do?"
"How do you do, Mr. Wells." Lauren's nerves were calmed by his graciousness. She replied steadily, "I have been most comfortable. I'm only sorry that I was an intruder at an unhappy time."
"No one blames you." He spoke to her gently, and she was glad for his presence in the room. He was bald except for a skimpy fringe of nondescript brown hair which adorned the back of his head. As if to compensate for his bald-headedness, bushy sideburns grew, in an outdated fashion, to within inches of his fleshy nose. His eyes were kind and smiling, and he seemed aware of her awkward situation.
Olivia had not uttered a sound. Now, she said in level tones, "Mr. Wells and I wish to speak to you, Miss Holbrook. Will you sit down? Would you care for some sherry?"
Lauren accepted the chair Mr. Wells held for her and declined the sherry. Olivia's position in front of the glare of the windows outlined her frame, but kept her features dark and inscrutable. Lauren wondered if Ben had thought how advantageous this placement of his desk would be to the person sitting behind it. She almost had to squint to see Olivia clearly.
"I will get directly to the point of this discussion, Miss Holbrook. I'm ignorant of my husband's reasons for inviting you here. I had construed some, though upon meeting you, I realize that they were wrong." She didn't explain and her meaning eluded Lauren. Olivia continued, "In any case, he was determined that you stay for at least two months. The night he had his seizure, as ill as he was, he asked me to allow you to stay for that length of time. Your being here was obviously important to him."
Lauren moistened her lips nervously with her tongue. She wasn't sure she could speak. "Your husband told me that you might find me helpful in handling your correspondence, entertaining, things like that. I envisioned myself as a sort of secretary." Lauren's heart was pounding so loudly that she could barely hear her own words.
Olivia came the closet to smiling that Lauren had ever seen. Carson Wells reached over and patted her hand as he said quietly, "Miss Holbrook, Ben liked to surprise folks and joke with them. Olivia is an astute businesswoman and has a number of clerks at the bank at her disposal. Ben may have told you his wife needed a secretary, but he had an ulterior motive, I assure you."
Bank? She didn't know anything about a bank. She was grasping at straws and she knew it, but she stammered, "I... I play the piano quite well. Maybe he thought I could give small concerts for your guests or something."
Olivia lifted a derisive eyebrow. "That would be lovely, I'm sure, but we don't even own a piano."
Lauren was stunned, and had nothing else to say. She looked first at one and then the other. Humiliated beyond endurance, she bowed her head and stared at the soggy, twisted handkerchief clenched in her lap by white, trembling fingers. "I'm sorry. I didn't know any of this. You must think... I was so sure... He didn't tell me..." The tears that had been clouding her vision finally flooded her eyes and spilled onto her cheeks.
"Now, now, no need for that," Carson said quickly. "I'm afraid old Ben was just playing one of his notorious tricks on someone at your expense, and didn't live to see it through. You can stay for a while. Olivia and I will try to make this an enjoyable visit for you. Come now, stop crying." Carson sounded genuinely distressed and was patting her hand so vigorously that it stung.
"Will you join us in the dining room at seven-thirty for dinner, Miss Holbrook?" Olivia sounded annoyed at this display of emotion.
Lauren took her cue of dismissal and stood as she said, "Yes, thank you, Mrs. Lockett."
She summoned all of her poise as she nodded to them in turn and then glided to the heavy panels of the door. Olivia called her name sharply. "Miss Holbrook."
"Yes?" Lauren said tremulously as she turned back to face them.
"There's something I must know."
"Olivia, please—" Carson interjected. He was ignored.
"Were you my husband's mistress?"
Mistress! The word screamed at her, echoing in her head and ricocheting off the walls of the room. Had Olivia thrown stones at her, Lauren couldn't have felt more abused. Her cheeks flamed with color, but her whole body had turned cold.
"No!" she gasped. "Whatever...? No, no." She was too astonished by the question to deny the allegation more eloquently.
"I didn't think so," was Olivia's only reply. "We'll see you at dinner."
Retracing her way to her room, Lauren barely held on to a tenuous thread of composure. When she was in her room, she collapsed on the bed and cried. She felt mortification for her naïveté and Olivia's surmise. All too recently similar words had been flung at her, and had been equally unjust. Why was she suspect?
She grieved for a man whom she had trusted and who had deceived her. Trepidation for an ominous future consumed her.
Two months! What had Ben expected to happen in that time? And at the end of those sixty days, what was she to do?
* * *
She dressed carefully for dinner, wearing one of the two nice dresses in her wardrobe. It was made of soft lilac voile. Slender pleats and tiny pearl buttons adorned the bodice, and a high, lace-lined collar reached to just under her jaw. The skirt fell in soft folds to the instep of her white leather slippers.
Elena assisted her with her toilette. It seemed almost natural now for the Mexican girl to help her bathe and dress. Lauren had always been more or less alone, but these last few days seemed the loneliest of her life, and she was grateful for her new companion.
The dining room was decorated with the same elegant understated taste evident in the rest of the house. If either Carson or Olivia noticed Lauren's reddened eyes, they didn't mention them as they took their seats.
The meal was served by an obese Mexican woman who Lauren supposed was Rosa, Elena's mother. Each time she carried in a platter of food, she looked at Lauren and smiled with open friendliness. Lauren smiled back thankfully.
The food was sumptuous and she ate everything except the beans and _picante_ sauce, which seemed to be a staple at every meal except breakfast.
The conversation was limited to trivial, everyday topics, and since Lauren had supervised many dinners similar to this in the Prathers' parsonage, she was at ease. She wondered at the absence of Jared Lockett, and started violently the first time Carson made reference to him. His not being there seemed of no concern to them. Olivia mentioned to Carson in passing that Jared would be at Keypoint for several days.
Olivia was relieved to see that Lauren Holbrook exhibited good manners at least. If someone came to call, she wouldn't have to explain a gauche, stupid strumpet, which was how she had pictured the girl when Ben had first told her about Lauren. She was obviously well read, and maintained her composure this afternoon even when she resorted to that weak, feminine trait of crying. How sweet, how captivating, thought Olivia sarcastically. Carson had naturally succumbed to the tears, as all men did. They couldn't resist a vulnerable woman.
Carson Wells had indeed felt compassion for Lauren's pitiful state this afternoon. She wasn't the scheming, devious woman he had feared she would be. He had expected a floozy who would drop a baby on the Locketts' doorstep, declaring Ben's paternity and demanding a sizable purse.
Lauren Holbrook was an innocent, a victim of circumstances. Olivia could have spared the girl the question about her relationship to Ben. He hadn't wanted her for a lover, Carson was certain of that. Ben had liked them lusty, naughty, and buxom. This fragile, doe-eyed young woman could melt a man's heart with her delicate beauty, but she would never have stirred the loins of Ben Lockett.
As for Carson, there was only one woman for him. Always had been. But no man was ever going to possess her. No man. Still, Carson loved Olivia Lockett. After all this time, after all the pain she had put him through, after having to bear the guilt of betraying his best friend, Carson loved her.
The talk turned to business, and Lauren listened distractedly, not really grasping or caring about the subject. Instead, she speculated on which chair Jared sat in when he ate in this dining room.
A few facts did manage to arrest her attention. The Locketts owned the Coronado Bank. They were trying to get a railroad trunk to Coronado, though there were some obstacles involved. Keypoint was managed by someone named Mendez, though the Locketts were apprised of the profits and liabilities to the penny.
Carson was speaking in an emphatic voice. "We've got to get Vandiver out here and wine and dine him, Olivia. He's the power behind it. Without him, we don't get the railroad. Now that Ben... well, now is the time to approach him again."
"We'll have to come to terms with the water rights, you know," Olivia said coolly.
"That's something we'll face when the time comes. The important thing now is for us to let them know we're interested. What about Jared, Olivia? Do you think he'll object?"
"Jared will do whatever we tell him to," she snapped. "There may be some resistance but he knew that Ben wanted the railroad. I think that's the point to stress to him."
They were quiet for a moment, and Lauren looked timidly at their faces. Both were wearing expressions of deep concentration.
* * *
Jared was tired, dirty, and in dire need of a drink as he trudged up the stairs. The ride from the ranch had been hot even on this early-October day. The trails were dusty, choking off a man's breath. It hadn't rained since the day of the funeral. He stopped abruptly in mid-stride and forcibly thrust that thought from his mind.
As he reached the top of the stairs, he noticed that the door at the end of the hall was slightly ajar. For some reason he couldn't name, he stepped lightly as he approached his room. So stealthy were his footsteps that his spurs didn't even jingle. Thorn would be proud of him, he thought with a smile.
He stopped outside his room and put his hand on the doorknob, but an overwhelming curiosity compelled him to continue down the hall until he stood in front of the door to the guest bedroom.
Why not open it? She probably wasn't in there. And if she were, what the hell? This was _his_ house, wasn't it?
He pushed open the door. Thanks to its well-oiled hinges, it opened silently. Lauren sat at a small desk. She was writing acknowledgments to letters of condolence. She had insisted on a project, and Olivia had grudgingly, and with a secret respect for the girl, assigned her this tedious job.
Well, well. The old boy hadn't done so badly for himself, Jared thought. All he could see was her back, but Lauren turned her head slightly as she bent over the letter she was composing.
It was completely silent in the room except for her pen scratching across the paper and a small clock ticking on the bedside table. Dust motes danced in the warm afternoon sunlight that projected into the room at slanted angles.
Lauren straightened her shoulders slightly and sighed deeply as she dipped her pen in the inkwell. Jared held his breath lest he be discovered, but she bent over the paper once again. From his position at the door, Jared had only a partial view of her face—a smooth ivory complexion with a hint of blush at the cheekbone, and a pair of small eyeglasses perched on the straight, slender nose.
Her clothes were too proper to be true, he thought. The white, high-necked shirtwaist and the maroon skirt could have belonged to a schoolmarm. It was fetching the way the buttons on her blouse followed her spine as if inviting a man to trace his fingers along that graceful back up to a slim column of neck above which was piled a luxuriant mass of hair. God, what hair! It was coal-black with dark blue highlights that added to its richness. A few tendrils had escaped the heavy bun on the top of her head and lay coyly on her neck. Jared wondered what those curls would feel like between his fingers.
She was slender. Maybe too slender. Skinny.
Moving slowly, hoping not to attract her attention until just the right moment, he reached into his breast pocket and took out a cheroot and a match. He clamped the cheroot between his teeth and, putting on the face he showed the world, struck the match against the doorjamb.
The sound was like a cannon shot in the small, quiet room, and Lauren bolted out of her chair. She drew herself up sharply against the desk when she saw Jared, and clutched a dainty, tight fist to her breast.
Jared's eyes flitted to her chest, and he amended his first speculation. No. She wasn't skinny.
She looked at him in terror from over the top of her spectacles, and Jared was momentarily taken aback by his first full look at Lauren's face. What in hell color eyes were those? Blue? No, gray. Goddam. He had to hand it to his old man. She wasn't bad at all.
Lauren felt like a cornered animal as she leaned against the desk for support. Jared lit the cigar, his eyes never leaving her. The smoke wreathed his face as he lazily pushed the flat-crowned hat back off his head with his thumb. It caught on a thin leather cord tied around his neck and hung against his shoulders.
He squinted at her through narrowed eyes in an insolent and lascivious fashion, raking her body up and down until her cheeks were on fire with embarrassment.
Lauren did not move or speak as she returned Jared's scrutiny. His hair was brown, with sun-bleached strands giving it the golden highlights she had expected. His complexion was dark, a combination of heredity and long hours in the sun. His eyes, though brown, were light, amber-colored. Like two perfect topazes.
There was a lot of Ben in him, particularly in his physique, but his face showed none of Ben's merriment. The stance, the face, the expression communicated arrogance, conceit, and contempt.
He leaned negligently against the doorframe, ankles crossed, dressed in much the same manner as the first time she had seen him in the back of the wagon, except this time there was a colorful bandana around his throat. The silver spurs on his boots fascinated her, and she stared at them for a moment before her eyes traveled up the long body to catch his amber eyes, which were still fixed on her in an unsettling appraisal.
"Miss Holbrook, I came to offer my humblest apologies. I understand that on our first meeting, I was somewhat indisposed and behaved abominably." His voice was Ben's. It held the same soft timbre and low pitch, but was full of sarcasm. Lauren wondered what she had done to earn this disdain. _He_ was the one who deserved ridicule. "What can I do to redeem myself?"
"You might start by apologizing for entering my room without invitation," she commented.
He was surprised at her aplomb and cocked a skeptical eyebrow. He recovered quickly, however, and said in soft, conspiratorial tones, " _Would_ you invite me into your room, _Miss_ Holbrook?"
She flushed at his emphasis on the "miss." Realizing that she still held her fist against her chest, she lowered it quickly, touching her watch fleetingly. At the same time, she took off her eyeglasses with the other hand. Smiling wickedly at her discomfiture, Jared watched her hands, particularly the one that fondled the watch.
"It's still there," he said quietly. "I'm not a thief."
She was furious at having drawn attention to her body.
He pushed himself away from the doorjamb with a shove of his shoulder and crossed the room with the slow, predatory gait of a stalking cat. His spurs jingled on the hardwood floor.
Lauren's throat closed completely when he stood only inches in front of her. He towered over her, and she had to tilt her head back to look into his face. It required a tremendous amount of courage to do so, but she instinctively knew that it would be to her disadvantage if he thought she were afraid of him.
Her false bravado evaporated as he raised his hand and extended it toward her. The long fingers reached out and, by an act of will, she didn't recoil.
"What is this, anyway?" he asked softly. His fingers closed around the watch pinned to her shirtwaist. His breath stirred the fine hairs that framed her face, and she caught the pungent fragrance of tobacco.
He held the watch in the palm of his hand and stared at it in a silent pensiveness that contrasted with the fierce emotional explosions that erupted from deep within Lauren's body.
She was on fire. Every cell burned with an unnamed compulsion to move even closer to this man who tormented her with his nearness.
The brooch was laid back in its original position, but not without the firm pressure of Jared's hand on her breast to assure its security.
For long moments, time ceased to exist. Amber eyes locked with gray, and the cynicism in the amber ones was replaced by wonderment. Jared's head descended toward Lauren's with imperceptible motion. For one heartbeat, she thought he was about to kiss her. Her moist lips parted of their own volition.
She didn't know that it was that involuntary gesture of welcome which jerked him back into the shell of scorn he used for protection. Mockery cooled the eyes that had been clouded with warmth, and Lauren was sensitive to the change. The pressure on her breast increased, but without the former tenderness.
She swatted his hand in a lightning reaction.
He chuckled deep in his throat. "What's the matter, Miss Holbrook? I was only checking the time of day," he sneered.
She ignored his sardonic words and tried desperately to restore balance to the spinning world. Still, she was gasping when she said, "Please, Mr. Lockett, I have a lot of work to do." Why was her heart thumping this way? Her whole chest was hurting and congested. She could no longer look directly into his handsome face or those brown-gold eyes. Why didn't he leave? Why didn't she want him to?
He stepped away from her and took a long draw on the cheroot which he had been holding at his side in his inactive hand. "I'll see you at dinner, I guess," he drawled.
He never looked back as he sauntered down the hall to his room. Lauren walked, entranced, to her door, and closed it.
# Chapter 5
They were all in the formal parlor adjacent to the dining room. Lauren could hear their muffled voices as she descended the wide staircase. Not only would she have to endure her first dinner with Jared Lockett at the table, but Elena had informed her there were three guests tonight. One was Mr. Wells, whom she felt moderately comfortable with. She had shared most dinners with him at Olivia's table. The other two guests were important men from Austin.
Lauren's dress rustled against her legs as she walked across the wide foyer. She dreaded entering the dining room. Meeting these powerful men from the state capital would be intimidating enough, but the real cause of her consternation was having to face Jared Lockett. Her private introduction to him in her room this afternoon had left her flustered.
She wore her best dress, a peacock-blue crepe. The high collar and straight, tight sleeves were trimmed with cream-colored lace. The cummerbund was of the same cream color and adorned with one pink silk rose that was pinned to the left side of her waist.
Nervously she stood framed in the doorway, watching the others.
Olivia and Carson were bent in concentration over some charts that the two men from the capital had spread before them on a low table. All four were poring over the diagrams with rapt attention.
Jared was slouched in a chair, his long legs stretched out in front of him with booted ankles crossed. He was examining the contents of a crystal tumbler with the exactitude of a chemist. The amber liquid matched the color of his eyes.
The room was another expression of Olivia's perfect taste. It was softly lit by glass lamps shaded with frosted globes. The sofas and chairs, arranged in harmonious order, were upholstered in pastel damasks and blended beautifully with the drapes that were drawn across the large windows. A Persian rug woven with the same muted greens, golds, and beiges evidenced in the furniture covered a large portion of the floor.
Each vase, ashtray, and picture had been chosen and positioned with utmost care. It was a peaceful room. But, like the woman who had decorated it, it lacked warmth and cheer.
Carson Wells was the first to notice Lauren, and he immediately jumped up from his chair and came toward her, both hands extended.
"Miss Holbrook, you are indeed looking beautiful tonight." He always met her with chivalry that was as overdone and outdated as his muttonchop sideburns, but Lauren appreciated his welcoming smile and returned it tremulously.
"Good evening, Mr. Wells."
"Come and let me introduce you to our guests." He took her arm and escorted her toward the two men standing near the table where they had previously been involved in their discussion.
"Miss Lauren Holbrook, may I present Mr. Parker Vandiver and his son, Kurt. Gentlemen, Miss Holbrook."
Lauren nodded to each of them as they acknowledged the introduction. Kurt Vandiver took her hand in his, raised it, and kissed the air inches above it.
"It is indeed an honor to meet so lovely a woman here in Coronado. An honor and a surprise," said Kurt, his blue eyes glinting in the soft light.
A rude, derisive sound came from across the room. Everyone chose to ignore it, but the embarrassed tint in Lauren's cheeks deepened.
"We're proud of our beautiful Texas women, Miss Holbrook, but it seems they've all been transplanted from places like North Carolina. I believe Carson said that was your home?" The younger Vandiver hadn't released her hand and, as he spoke the flowery compliment, Lauren gently pulled it away from his firm grasp.
"Yes. Clayton, North Carolina. I thank you, Mr. Vandiver."
"Good evening, Lauren. Would you care for some sherry?" Olivia spoke to her for the first time. She was dressed in black, the first evidence of mourning Lauren had seen her display since the funeral. Jet earrings dropped from her ears and matching beads glittered darkly from the bodice of her dress. She was beautiful, but in a dangerous sort of way. Lauren looked at her much as one would look at a deadly beast—with admiration, but caution.
"Yes, thank you. I'll get it myself," Lauren replied to Olivia's question.
"No, allow me, please, Miss Holbrook. Sit down and I'll bring it to you." Kurt lightly took her elbow, guided her to one of the small sofas, and went to the sideboard that was well stocked with liquor and glasses.
Lauren's eyes moved involuntarily to the silent man lounging in the chair and she was disconcerted when they locked into his golden gaze. He had not even stood when she came into the room. How very rude! His stare was almost menacing.
She held her breath, afraid that he would make reference to that afternoon, but he only held his glass to the light and studied it carefully as he said in a bored fashion, "Good evening, Miss Holbrook." He made her name sound like an insult.
Kurt handed her the sherry, and she took a quick sip, determined to divert her attention from Jared. Kurt sat next to her on the sofa and began asking questions about her visit to Texas. She kept her responses as general and vague as she could. Kurt's interest made her uneasy. His heavy body took up much of the space on the sofa, and she had to make an effort to avoid touching him. Deftly she steered the conversation away from herself.
"What line of business are you in, Mr. Vandiver?" Hadn't Olivia and Carson referred to the Vandivers one evening at dinner? She couldn't remember what had been said.
"Investments." She looked puzzled, and he laughed slightly. "All kinds of investments—railroads, lumber, cattle. Currently we're interested in getting electricity into these smaller towns."
"I see," murmured Lauren, although she didn't see at all.
Carson picked up the conversation by launching into what he considered a humorous tale. Out of the corner of her eye, Lauren watched Jared get up from the chair, saunter to the sideboard, and pour himself another neat whiskey. Abruptly Olivia suggested they all go into dinner. To Lauren, the suggestion sounded like a command. Or a threat.
Lauren looked toward Olivia quickly and saw her gimlet eyes boring into her son. As if to be deliberately provoking, Jared gulped his drink and poured another, which he carried into the dining room along with the crystal decanter. If anyone but Lauren had noted this hostile exchange between mother and son, they didn't show it. The Vandivers laughed heartily at Carson's story as they crossed to the dining room.
Olivia and Carson sat at opposite ends of the table, with Lauren and Kurt on one side, and Jared and Parker on the other. Jared was directly across from Lauren. He gave her a slow appraisal as he took his seat, but his face was expressionless.
As dinner was served by Rosa, who had forsaken her bright skirts and loose blouse for a starched white uniform, Lauren studied the Vandivers.
Parker had a pugnacious face, almost brutal in its strength of feature. His piercing blue eyes darted around the room in quick movements, as if looking for hidden secrets. His voice and manner were polite and conversational, though Lauren suspected that he absorbed only the facts he considered pertinent and sloughed off the rest of what was said as inconsequential. His body was thick and solid. His fat hands, with fingers like tight, pink sausages, rested folded on his stomach whenever he was not using them. This relaxed posture was contradicted by his busy eyes.
Kurt was taller, though built in the same solid way. His eyes were as aggressive as his father's, but he had deep dimples which appeared and disappeared at will, relieving the belligerence of his face. The cropped blond hair, also like his father's, was crisp and wiry, fitting his head like a snug cap. His ruddy complexion made his eyebrows look white against his beefy forehead.
Though the Teutonic-featured Vandivers had exhibited perfect manners toward her, Lauren was instinctively wary of them. Their etiquette was too polished, their conversation too eloquent, their attitude too humble. Their entire demeanor seemed too rehearsed to be sincere. When she caught Kurt's eyes on her, she shivered involuntarily. His rapacious expression was reminiscent of William's.
Jared spoke very little, ate almost nothing, drank quite a lot. He responded in low, disinterested mumbles when anyone directed a question to him, and initiated no conversation himself. Lauren was uneasy at his careful, persistent scrutiny of her. It seemed that his implacable eyes never left her face for the hour they were at the table.
If Lauren found the meal tedious, to Jared it was interminable. He despised the Vandivers and the grasping ambition that they unsuccessfully tried to camouflage with sleek conversation and courtly manners. Jared hated all forms of deception and pretense. And this China doll opposite me is an expert at it, he thought cynically.
He tried to ignore Lauren, but finally gave up the effort and studied her, wanting to figure her out. He admitted to himself that she wasn't what he had expected. Not at all. And Jared didn't like surprises. That's why he had been furious with Ben the night before he died.
No, I refuse to think of that, he told himself.
Lauren handled her cutlery with graceful ease, and Jared was intrigued by her hands. They looked soft and smooth. The fingers were long and slim and tapered to pink oval nails that were well kept. What had he expected to see? Red nail lacquer?
What had Ben said? Oh, yes. She was a pianist. Jared chuckled to himself and thought ribaldly of a more pleasant activity for those hands. Then he wondered fleetingly if this woman even knew about things like that, and quickly decided that she didn't. He had seen the fearful caution in her eyes this afternoon when he had touched her. It was genuine. That was the rub.
She raised the velvety lashes that hid her eyes and glanced in his direction. For a moment, she returned his steady gaze, then rapidly shifted her eyes away from him. He could see why Ben had been hoodwinked by the little tart. A man could drown in those dove-colored eyes before he even knew he was sinking. She knew how to use them, too. You were never graced with a full look, only an elusive glance. It was enough to make you go crazy if she didn't look at you directly.
She had too much hair, he decided. It was too heavy for her finely boned face and figure. Yet the knot on top of her head was softened by the bouffant fullness around her face. She didn't need any "rats" like other women used to achieve this style. The wispy tendrils that framed her temples seemed more translucent than the china from which they ate.
His eyes moved to the gold watch pinned to her breast. He shifted in his chair uncomfortably and unconsciously moistened his lips with his tongue when he looked at the gentle rise and fall of her bosom. Whatever else was fake about her—her quiet voice, circumspect manners, her reasons for trailing Ben halfway across the country—one thing was real. He could still feel the firm cushion of her breast against his palm. His hand trembled slightly as he poured another glass of whiskey, and his manhood refused to relax.
He should leave now. He should mount Charger and go into Pueblo for a nice, uncomplicated toss in the hay with a whore. But he didn't really want to. He knew his constant staring was making Lauren ill at ease. If it weren't such a damned entertaining exercise in itself, her discomfort was incentive enough for him to stay.
* * *
As they returned to the parlor after the meal, Carson commented, "It's a pity that you don't have a piano, Olivia. Lauren could play for us."
"That is a pity. I would love to hear you play," Kurt said at Lauren's elbow.
"What brings you to Texas, Miss Holbrook?" Parker Vandiver asked bluntly as they took seats.
Lauren was momentarily at a loss for words. The cold, incisive blue eyes seemed to challenge her.
"Lauren is the sister of one of Jared's friends at Harvard. Jared had visited the Holbrooks on some of his holidays from school, and we wanted to repay their hospitality. Since her brother has married, we issued the invitation to Lauren. Ben brought her back with him on his way home from New York."
Lauren stared incredulously as Olivia lied so glibly. The older woman smiled radiantly at the stunned girl as she continued, "She has been such a comfort to me since Ben passed away. I don't know how I could have managed without her."
Lauren's astonishment turned to anger. How dare Olivia lie about her that way! She had nothing to be ashamed of or to apologize for!
"I don't recall seeing her at the funeral," Parker observed shrewdly.
"She was overcome with grief. She and Ben had become quite attached on their trip. I wouldn't allow her to go through such a public ordeal," Olivia said simply.
"That's perfectly understandable. I felt that way when my mother died," Kurt whispered as he reached out to pat Lauren's hand. She pulled it back quickly.
Carson redirected the conversation. "Tell us about this power plant you envision, Parker."
"It's no vision, Carson. We intend to build it, one way or another. Of course, we would like to have the help of the Locketts."
"What will happen to the electric company that's already here?" inquired Olivia, all business now. She referred to a small power plant owned and operated by Orville Kendrick. Lauren had heard Olivia and Carson discussing it. The plant provided electric power to the citizens of Coronado each evening from six until ten. Oddly enough, the Lockett house was not yet wired for electric power.
"It will no doubt go out of business," Parker answered Olivia's question brusquely. "We will provide twenty-four-hour service. Kendrick won't be able to survive the competition. One day soon, everything will be powered by electricity."
Kurt leaned toward Lauren and asked, "What do you think of electric lighting, Miss Holbrook?"
All eyes turned to her, and she hesitated before timidly stating, "I think it's ugly." At their shocked expressions, she hastened to explain. "I see the need for it, and I agree with Mr. Vandiver that we'll become more reliant on it, but I do think it's ugly. I prefer the softness of gas lighting."
"Spoken like a true romantic," Kurt said, and nodded his approval.
"Hear, hear."
The two caustic words fell like stones into the room. Everyone looked at Jared, who had resumed his slouching position in the chair. This was his first utterance since dinner.
"I'm interested in your opinions on all of this, Jared," Parker broke the silence.
"My opinion," Jared snarled, "is that we cut all the bull and get to the heart of the matter, Mr. Vandiver." His voice was hard and quiet.
"And what do you perceive to be the 'heart of the matter'?" Parker fired back.
Jared unfolded himself and stood up slowly, strolled to the sideboard, poured a full glass of whiskey, and only then turned to face Parker. Lauren noted the impeccable fit of the black wool suit on his trim frame. His white shirt collar was a startling contrast to his dark face, which had hardened into a sinister scowl.
"The heart of the matter is that you want to build a power plant that will destroy another man's business. In order to generate that power plant, you must have water. The most accessible source of water is on Lockett land. Am I right so far?"
"Your assessment of the facts is somewhat distorted, but it conveys the gist of the plan." Parker spoke calmly, though his face was flushed once again, and his fat fingers were furiously working with the gold watch fob stretched across his stomach.
"What happens to all of the people who depend on that water for their livelihood when you come along and dam it all up? The Locketts have prided themselves on being generous with their water. It's a source of revenue, yes, but sometimes my father took a lamb or two as payment, sometimes as little as a basket of corn from a farmer who'd had a bad crop. Even before the land acts were effected several years ago, Ben allowed cattle or sheep ranchers with smaller herds to water their stock on our land. What will happen to those people when the water is no longer available?"
Lauren listened avidly. Jared's speech was eloquent. The thick brows that reminded her so much of Ben were drawn together in determination. There was no insolence in his manner now.
"Mr. Lockett... Jared," Parker said with condescending patience, "perhaps I understand the business world better than you do. After all, I've got about thirty years more experience. In every business venture, there are those who gain and those who lose. It's a basic fact of economics."
"I'm not an idiot, Vandiver, so you needn't talk to me as if I were," Jared cut in. "I have a business degree from Harvard of which my mother is very proud. Please continue. I think I can keep up with you." Jared mocked the other man with a salute of his glass.
"Very well, I'll be blunt," Parker said. "You want something, and we want something. We make a trade. Your rail-road for our power plant. We both stand to profit tremendously."
"Parker, I don't think it's necess—"
"Carson, don't interrupt him," Jared commanded sharply. Lauren was surprised when Carson obeyed without question. "We're getting to the good part. Mr. Vandiver and _son,_ " he said the last word with a sneer, "have come all the way from Austin bringing a detailed TransPlains Railroad diagram showing a straight track into Coronado." Jared paused and took a long drink of the whiskey. "I'd hate to think they've wasted their time."
Lauren viewed this whole scene with fascination. The man became more of an enigma every time he opened his mouth. Cowboy, Harvard graduate, businessman?
"Vandiver," Jared continued, "the railroad has been in Kerrville for several years. Comfort and Fredericksburg are in the process of negotiating one. With Lockett cattle, the granite quarries, and the cedar posts business, why do you imagine we need your help in obtaining a railroad?"
"Shut up, Jared. You're drunk and you're offending our guests," Olivia barked. Her face was a mask of fury.
"Quite all right, Olivia. He asked a simple question, to which I will give a simple answer." Parker bowed to her slightly before facing Jared again. "Ben Lockett was a respected man in this state. A very powerful man. I don't have to tell you that. Yet for years he was unable to secure a railroad into Coronado."
"That's because he wasn't going to exploit the people who respected him!" Jared shouted.
"Whatever the reasons for his failure, you still have no railroad. I happen to have many friends on the Railroad Commission. If I tell them that Coronado is a bad risk..." He shrugged expressively. "On the other hand, if I say that it's a potential money maker, they'll jump at the chance to build train tracks here. If a man as influential as your father failed to achieve this goal without my help, how do _you_ propose to do it?"
Some of the arrogance and fire went out of Jared then. He placed his glass on a small table and faced Parker, staring at him for several long, silent moments. Parker stared back levelly, clearly evaluating a foe.
Calmly, hardly above a whisper, Jared pronounced, "You _are_ a sonofabitch."
"Yes, I am," Parker agreed grimly.
Jared turned his alcohol-brightened eyes on Kurt, who had remained silent throughout the exchange. "I wonder what that makes you," Jared said contemptuously. Then his eyes swept across Lauren to his mother and Carson. His face registered disgust and resignation before he turned and strode from the room, the heels of his polished black boots emphasizing his anger with each long, hurried step on the parqueted floor. A few seconds later, the front door slammed.
Always the diplomat, Carson conciliated, "Parker, Kurt, you must be patient with the boy. He just lost his father. Ben's death hit him hard."
"If he were a boy, I could tolerate his behavior," Parker said. "As it is, he's a thirty-year-old man acting like a boy. Olivia, you'd better get things straight with him. If he can't be counted on for his support in this venture, then the deal is off. Your son is Ben's heir, and everyone will be watching to see how he handles himself. If word gets around that he can't control his temper and his personal habits," he paused and looked significantly toward the liquor cabinet, "then I couldn't endorse a joint venture for fear of losing my own credibility."
"I understand, Parker." Olivia's green eyes were as cold and hard as emeralds. "Jared will come around to our way of thinking. He always does."
Lauren was unaccountably irritated by Olivia's self-assured guarantee. Jared was a grown man who had made some astute observations of his own. She felt a compulsion to defend him but, of course, she could not.
"He has a terrible reputation for activities that are unmentionable in Miss Holbrook's presence," Kurt contributed sanctimoniously.
"I don't need you to tell me my son's virtues or vices, Mr. Vandiver," Olivia snapped.
"No, Mrs. Lockett. I only meant—"
"I think it's time for us to take our leave," Parker interrupted his son. "We've had a thorough discussion. We should give the proposal some further study and weigh all the elements involved." He stood, walked over to Olivia, took her hand, and held it in both of his. "Thank you for a lovely evening, Olivia. The dinner was excellent. That Lockett beef can't be beat."
"I'm sure we can look forward to working together in the future, Parker. Your plans will proceed without interference, I assure you."
"I hope so."
Kurt murmured a personal goodnight to Lauren. This time when he raised her hand, his fleshy lips brushed across the back of it. It took all her composure not to jerk her hand away. She was grateful when the wide oak door with the etched and beveled glass closed behind the Vandivers.
Carson, Olivia, and Lauren stood in the foyer. The rest of the house was silent. Lauren turned to Olivia and faced her squarely. "Mrs. Lockett, why did you lie to them about me? You made me an unwilling accomplice in that lie." She was astounded at her own temerity, but honesty was an integral part of her nature.
"Unwilling?" Olivia asked. Her brows arched over her eyes like two black wings. "You could have denied it right then and told them the truth. But you didn't. I think you saw, as I did, that my story was more plausible and less... compromising."
Lauren looked at Carson, who was staring at his shoes and offering no help. She clenched her hands tightly at her waist and gnawed on her bottom lip. Her initial instinct was to deny again the insinuation that she and Ben Lockett had meant anything more than friends to each other, but she refrained.
Two months. She must stay at least two months. Then...
Quickly she excused herself and went upstairs.
* * *
Carson lay in the tester bed and watched Olivia as she stepped from behind the decorative screen and walked naked across her bedroom.
She never ceased to amaze him. He knew her to be in her midfifties, but her excellent body belied her age. As she took down her hair, he could see her high, firm, full breasts reflected in the cheval glass. Her stomach was flat, and her thighs were without the heaviness that cursed most middle-aged women. Her hips were slender and taut. The skin on her buttocks was smooth and unwrinkled.
Each time he saw her thus, he was made painfully aware of his own unattractive physique. Out of the confines of his tight vest, his chest and stomach sagged, and his short legs had thickened with age. Carson had always envied his friend Ben his lithe, tall body. That powerful build and thick white hair had turned the heads of many women even as his years advanced.
Unperturbed by his careful observance of her, Olivia walked to the bedside and turned down the gas lamp. She sighed tiredly as she lay down and rested her head against the scented pillowcase.
"You were marvelous with them tonight, darling. I'm sure the evening was exhausting for you," Carson said as he reached over to stroke Olivia's luxuriant hair with his stubby fingers.
"Those bastards," she hissed. "They know they have us over a barrel, and they're making full use of our position to kick us while we're down. If I didn't want that railroad so desperately, I'd never give that goddam German sonofabitch the time of day." Carson was used to her explicit language. He was gratified that she spoke this candidly only with him. He saw it as an indication of trust.
"I know, my dear. We'll just have to play their game for a while. We've had to make sacrifices like this before, but they've always worked in our favor."
"Yes. But this time it's particularly galling."
"Forget them for now and try to relax." Carson moved closer and settled his stocky body along hers. He stroked her cheek before raising himself to kiss her briefly on the mouth; he knew she didn't enjoy ardent kisses.
Laying his head as close to Olivia's as space would allow, Carson trailed his hand down her throat and chest to cup her breast. Her one pregnancy hadn't darkened her nipples, and they were almost as pink as a young girl's. He continued to enjoy the feel of her warm flesh and the tender peaks of her breasts until she shifted impatiently. Her restlessness was his signal to go about his business.
He mounted her and met no resistance when he entered her quickly. His passion rose and climaxed in a matter of minutes. He never tried to sustain the pleasure. Olivia had been taught by the nuns at the Ursuline Academy in her native New Orleans that ladies didn't enjoy the sexual act, but tolerated it out of love. Carson understood. If he ever wished he could coax a warmer response from Olivia, it was a fleeting fancy. His own cries of ecstasy were muzzled by the thick pillow in which he buried his face.
He was treated to a brief caress on his shoulder and a brush of her lips across his before she extricated herself from his embrace. Since that day over twenty years ago when she had unemotionally invited him to be her lover, he had never been allowed to linger inside her or enjoy her nakedness afterward.
Tonight, as usual, she left the bed and went directly into her bathroom. He heard the sounds of her washing. When she came back into the room, she was clad in a nightgown and robe.
"Carson, I have an idea." She paced the expensive rug at the foot of the bed. He was never allowed to spend the entire night with her, and he begrudged the time she spent out of bed.
"Yes, dear?" he asked resignedly. He could tell by her agitated posture and the intent expression on her lovely face that she was enmeshed in thought. Tonight he would have to content himself with what lovemaking she had already permitted.
He listened with unconcealed astonishment as she related her plan. It was audacious and dangerous, clever and manipulative, impossible yet feasible. He objected to her motives, protested her means, but, as he always did, he agreed to her scheme.
# Chapter 6
While Olivia outlined the plan that would have a dramatic impact on Lauren's life, the girl was in her own bed trying vainly to sleep. She tossed restlessly, myriad thoughts darting through her mind and upsetting the modicum of serenity she had managed to preserve since her arrival at the Lockett household.
Olivia and Carson were a puzzle she couldn't decipher. One moment she felt they accepted her for what she was, and the next moment she felt they posed a threat to her. Carson treated her kindly, but he was Olivia's chattel. And Olivia's attitude toward Lauren was reserved and cool to say the least.
The Vandivers had frightened Lauren. She was unaccustomed to hearing business deals discussed, and Parker's callousness had appalled her. And Kurt was no doubt as greedy and ambitious as his father.
Thoughts of that young man made Lauren shiver even under the warmth of her bedcovers. He was handsome in a brutal sort of way, but his thick, powerful body repelled her, and his unctuous voice and conciliatory manner made her uneasy. She felt threatened by him, but it was a different kind of alarm than she felt when she looked at Jared Lockett.
Jared. Ben's son was a rake and scoundrel, a drunkard and a womanizer, so why did she continue to dwell on him? Why did Jared's long, lean body intrigue her so? Why did she feel a compulsion to touch him?
Since she had been initiated into the rites of womanhood at age eleven and given a very rudimentary explanation of her body's workings by the embittered Dorothea Harris, Lauren's education on the subject of sexuality had been sadly deficient.
She was fifteen before she realized something mysterious, some strange chemistry, attracted the bodies of men and women to each other. She was at a picnic held in the city park in honor of the veterans returning home from the Spanish-American War in Cuba. As she sat under a shade tree, her attention was diverted from her book to a young soldier and his pretty young wife. Lauren knew them both. They had been married only a few weeks before he joined the army.
They were sitting close together under another tree. They weren't talking, but were nonetheless communicating. They gazed steadfastly into each other's eyes. The young woman rested her hand on her husband's thigh and lightly caressed it with her fingertips. Lauren watched covertly from behind her eyeglasses as he raised her hand to his lips and kissed the palm ardently. He then returned her hand to his thigh, pressing it gently.
For some inexplicable reason, Lauren's heart started pounding, and she felt hot and flushed all over. She noticed a strange sensation in the lower part of her body. Her breasts were tingling, and the nipples became taut and pointed under her camisole. She was uncomfortable and ashamed to have such distinct physical reactions in these private parts of her body.
The man leaned down and whispered into his wife's ear. She smiled, nodded. He stood, extending his hand to help pull her to her feet, then kissed her fervently on the mouth. Lauren was finding it difficult to draw a full breath.
They smiled at each other and, glancing around, clandestinely left the picnic. They said goodbye to no one, and apparently Lauren was the only one to witness their leaving.
Those disturbing but exquisite sensations she had experienced years ago when she watched the intimacy between the young couple had almost been forgotten. They had come back to her with stunning clarity when she saw Jared Lockett leaning negligently against her doorjamb that afternoon. Why?
She read his scorn for her in the amber lights of his eyes, and was deeply hurt. What had she done to evoke such disdain? Not even the hateful, ugly words William had flung at her had pierced her like that knowing, twisted smirk on Jared's sensuous mouth.
He had stared at her all during dinner. Kurt had watched her, too. But his stare was cold and calculating, while Jared's eyes had burned into her like tongues of golden flame.
Her whole body trembled under her nightgown. She closed her eyes, but Jared's image was imprinted on the back of her lids.
Once again, she relived the moment when his hand had pressed against her breast. She felt his breath on her face, and tried to imagine how his lips would feel against hers. A long, shuddering sigh escaped her and she moaned into the pillow. She wanted to know what it was like.
And she knew that finding out would bring her perilously close to the brink of hell... or heaven.
* * *
The pendulum wall clock in the office chimed the hour of eight. It was the morning following the dinner with the Vandivers. To Lauren's ears, the tolling chimes sounded ominous as she sat in silence with Carson and Olivia, waiting for Jared to join them.
Olivia looked full of resolve and purpose as she sat upright and grim in her chair behind the desk. Carson was nervous and uneasy, periodically wiping his forehead with a linen handkerchief. Lauren daintily sipped a cup of tea.
She had awakened early after her restless night. Her heavy maroon skirt and ecru shirtwaist were donned hurriedly, and without Elena's help. Lauren had pinned her hair into a haphazard chignon at the nape of her neck and left her room. Her haste in making an appearance at breakfast was probably unnecessary, but she didn't want her hostess to think her lazy.
Before she entered the dining room, Carson intercepted her and asked if she would please join him and Olivia in the office where they had interviewed her previously. She would always think of it as Ben's office, for an aura of the man still clung to the atmosphere.
Lauren went in behind Carson and quickly poured a cup of tea, sweetening it liberally, intuitively guessing she would need some sustenance during this mysterious meeting. Carson's full cheeks reddened considerably, and he wouldn't look her fully in the eyes when he extended the invitation, which Lauren knew was not an invitation at all, but a command. Carson's jitters were communicated to her.
Something was in the wind. And it obviously affected her. But how? She couldn't imagine. Maybe in light of the events of last evening, they were going to ask her politely to leave. They would have their hands full for a while with the business of the railroad, and she couldn't blame them for not wanting an outsider cluttering up their lives.
Why was it necessary for Jared to hear her fate? She'd prefer that he didn't witness the interview. He, however, would undoubtedly relish any misfortune that befell her, she thought dismally.
Lauren started as she heard his boots echoing on the parquet floor in the hall. He stomped into the room and looked darkly at his mother.
"This better be damned important for you to root me out of bed this early when I got in late and feel as godawful as I do. I'm going to get some coffee." He strode out of the room and no one spoke until he returned a few moments later carrying a steaming mug of coffee. He sipped it and cursed under his breath when it burned his tongue.
In contrast to Olivia's military neatness, Jared was disheveled—his brown hair mussed, the wrinkled shirt hastily and sloppily stuffed into rumpled pants, the boots which had shone last night now scuffed and dull. He slouched in a chair in a posture Lauren was coming to know. He ignored her and Carson completely.
"If it weren't important, Jared, I would not have disturbed you." Olivia spoke as if no time had elapsed since he first came in and rudely addressed her. "Carson and I had a long talk last night after the Vandivers left. We have arrived at some conclusions and want to tell you our course of action that will resolve our problems with Parker."
Carson once again mopped his perspiring brow, and licked his lips nervously as he watched Jared from across the room. If they were going to rehash the same argument, Lauren would prefer to return to her room and the correspondence she was doing for Olivia. Or start packing her bags.
"Why do I suspect that I'm not going to like our 'course of action'?" Jared asked stonily. "You know how I feel about those jackasses. I don't want them on one square inch of Lockett land."
"I don't particularly like them, either, and I trust them less; but I want that railroad, Jared. Carson does, too, and so did your father."
"Ben didn't want it bad enough to let a thief like Vandiver make the deals for him."
"Jared, what your mother is trying to say is that we've got to give a little in order to gain a lot." Carson looked at the younger man almost pleadingly. "I know why you hesitate about damming up even a small tributary of Rio Caballo. Some of the smaller farmers and ranchers will suffer setbacks, but we'll work with them all we can. We're not going to leave them high and dry." He laughed skittishly at his own play on words, but Jared didn't even smile.
Lauren had watched Jared since the discussion began and saw the same expression of conviction that he had shown last night. His jaw worked convulsively as he clenched his teeth.
Then he hung his head and swirled the coffee around in his mug. He studied it hard, a deep groove forming between the thick brows. When he looked up again, his face had completely changed. He looked at his mother with accusation, at Carson with disgust, and then assumed an attitude of complete indifference.
He shrugged insolently. "Do whatever you like. I don't give a damn. The two of you will make great partners with Vandiver."
He set the mug on the table in front of him and stood to leave, but Olivia checked him. "Jared, wait. I'm afraid it's not as simple as that. Sit down." Impatience was written on his surly face, but he slumped into the chair again and restlessly propped one ankle on the other knee.
"It seems, Jared, that some of the investors in this railroad of ours are concerned about your attitude. After word of your behavior last night gets around, I'm sure it will confirm their low opinion of you. You must be, or at least appear to be, behind this project one hundred percent. The power plant, as well as the railroad, needs your public endorsement now that you are taking over your father's businesses."
"Taking over? That's a laugh," he muttered caustically.
Olivia ignored the interruption. "Of course, Carson and I will be running things for you until you feel ready to assume responsibility. But to the world, you must present a credible air of authority and maturity." She let all of that sink in, pausing dramatically before she said, "That's why we feel you should marry as soon as possible. Marry Miss Holbrook."
The words hung in the air, suspended on the palpable currents of differing emotions evoked in those who had heard them.
Rivers of blood rushed to Lauren's head, causing a great roaring, and a fire consumed her eyes and ears even as perspiration covered her body in a chilling film.
Olivia remained unperturbed. She sat calmly and regally, waiting for her subjects to do her bidding so she could get on with the affairs of state.
Carson's eyes darted from Lauren to Jared to Olivia, and then back to Lauren. He had no idea what the girl was thinking. She stared in front of her as if she had lost all her senses.
Jared's reaction surprised them all. He burst out laughing. He stood and stumbled around the room holding his sides until, completely spent, he collapsed against the windowsill. Drawing in several gulps of air, he said with total incredulity, "You can't be serious! Marry Miss Holbrook! That's the best laugh I've had in weeks." He wiped tears of mirth from his eyes and Lauren was poignantly reminded of Ben. Only that memory penetrated her shocked brain.
Olivia stated quietly, "I'm not joking, Jared. I'm serious, and I think you'll see why if you let me explain it to you."
"I don't need any explanations, Mother. It's absurd," Jared cried. He pointed a finger at Lauren. "She's still wet behind the ears."
"Miss Holbrook is twenty. Ben told me."
"I wasn't referring to her age, dammit, I was refer—"
"I know what you meant," Olivia said. "It's a far cry from your earlier opinion of her. I remember your accusing Ben of bringing his 'southern belle whore' into the house when he told us about her."
Lauren swelled with anger at Jared's unjust and unwarranted comment. He hadn't even met her! Before she could fill her constricted lungs with enough air to protest, Olivia continued, "We made a plausible story of her being a classmate's sister. We'll embellish it and say that you fell in love on one of your visits with her family. Your father surprised you by bringing her here, and you can't bear for her to go back. You'll be married right away." She held up a palm to stave off the objection she saw coming. "We expect you to be more discreet in your... other interests, Jared, but you need not change your lifestyle. Lauren is educated and refined and will add a convincing touch to your new, responsible image."
Jared leaned against the wall, crossed his ankles, and folded his arms across his chest. He sounded amused when he drawled, "We weren't exactly a loving couple last night. How the hell do you expect those shrewd Vandivers to fall for your little farce?"
"We'll say Lauren thought it improper to marry so soon after Ben's death and that you had quarreled about it. You were having a lover's spat." Then, dismissing any further arguments, she said, "You will marry immediately. Have I made myself clear?"
Jared looked at his mother levelly for a few moments. When he spoke, all traces of humor were gone. His voice was low and menacing. "Every goddam decision in my life, you have made for me. I'll go along with your scheme for the railroad, but there's no way you'll saddle me with an unwanted wife. No way in hell."
Olivia smiled smugly. "I think there is, Jared. I happened to overhear your argument with Ben the night before he died. Do you wish to tell Miss Holbrook why Ben arranged for her to come here?" Jared's face paled alarmingly, even considering his hungover pallor. His eyes looked pained. He stood up straight and balled his hands into fists at his side, but he didn't speak. "No?" Olivia turned to Lauren. "Well, Lauren, it seems that Ben picked you out for Jared. He was bringing you here in the hopes that before he died, he would see his son married and settled down with you."
She faced her son again. "It's ironic that for once my objectives are the same as your father's, isn't it, Jared? And it was your vehement objection to his matchmaking and your promise to make Lauren's visit as abominable as possible that caused your father's heart attack. I think you owe him his last wish, don't you?" She leaned back in her chair and smiled dangerously. "Yes, I think you owe him this."
Jared's jaw had turned to iron. His teeth clenched and unclenched; his hands worked at his sides. He turned and faced the window, looking out with eyes blinded by impotent rage.
Had Lauren not been so stunned by the events of the past few minutes, she would have resented being discussed as if she were deaf and mute, an object rather than a person. Ben had invited her here for that wastrel to marry! How could he have been so callous? And Olivia and Jared had known all along, probably Carson had, too. They had conspired to deceive her, had even humiliated her with spurious doubts of her virtue.
Olivia turned a hard, green stare on her and said, "Lauren, we've not heard your opinion." Clearly her statement was a mere formality. Olivia would brook no opposition.
At Lauren's obvious inability to speak, Carson hesitantly cautioned, "Olivia, maybe we're being unrealistic. Let's give them time—"
"No," Olivia said. "The sooner the better, Carson. I'm sure when Lauren hears the rest of our proposal, she'll agree fast enough."
Lauren met Olivia's eyes again and felt a spurt of courage. This woman could never force her into marriage. Never.
Before Lauren could tell Ben's widow exactly what she thought of the entire family, Olivia said, "Lauren, we don't expect you to do this without compensation. We're only asking that you stay married until the railroad is complete and running to our satisfaction. Then, when all the fireworks are over, you may go wherever you wish. For your time and trouble, we'll give you twenty thousand dollars in cash when you leave."
"I don't want any money from you, Olivia!" Lauren declared. She wasn't even aware of having used Olivia's first name. This final insult was too humiliating. Olivia had actually offered her money to marry Jared!
She ignored the derisive snort that came from the direction of the window. Her eyes were riveted on Olivia as she whispered hoarsely, "This is impossible. Surely you are joking?" Olivia only stared back at her stoically. Lauren looked toward Carson, who was wiping his moist hands with his handkerchief. The man at the window remained motionless, brooding. He was leaving her to fight this battle alone, when they should have been allies. Anger made her brave.
"I won't marry anyone. I don't want to marry anyone," she declared with a defiant lift of her small, pointed chin.
Olivia laughed indulgently. "My dear Lauren, I'm not suggesting that you marry in the biblical sense of the word. The marriage need never be consummated." Again Lauren heard a scoffing sound from near the window. "I would think our offer would be attractive. You don't want to return to that dismal parsonage right away, do you? When you leave here, you'll be a woman of independent means."
"I would also be a married woman," Lauren protested.
With a hint of impatience in her voice, Olivia replied, "An unconsummated marriage can be annulled expeditiously. Don't worry about that now. As Jared's wife, things will be very pleasant for you here."
The green eyes narrowed on Lauren and Olivia asked pointedly, "Was there someone you were involved with in North Carolina? Perhaps a former love is the reason for your resistance."
"No," Lauren rasped. She shuddered when William's gloating face invaded her thoughts. "No," she said with finality. Feeling bolder now, she challenged, "If you've known all along why Ben brought me here, why didn't you say so?"
"That's a lesson for you to learn, Lauren. Gather what information you can, and then save it until just the right moment. Had this occasion for us to use you not arisen, you would have been sent away after two months none the wiser." She smiled tightly. "No doubt Ben's romantic soul hoped you and Jared would develop an affection for each other during your visit. Ben was always a fool about such things."
Lauren was shocked by the hatred in Olivia's voice, and could only stare at the woman in lieu of a response.
Olivia stood up and briskly crossed the room. "Then if there is no further discussion, I need to start making preparations." She looked at Jared, then at Lauren. When no one made an objection, she motioned for Carson to follow. He patted Lauren quickly on the shoulder before he left the room in the wake of Olivia's stiff, rustling skirts.
Lauren's mind was in turmoil. Why was she still sitting here? She should be upstairs furiously packing. She should run away from this house at once. She stared vacantly as her thoughts ran rampant.
What were her options? She couldn't go back to North Carolina and face William and her former guardians. She had closed that chapter of her life. Or rather, it had been closed for her.
She would live with a loveless marriage to a stranger. But not forever. She would have the means to start over when she left the Locketts and she would somehow survive the stigma of divorce. Maybe she could even pass herself off as a widow if she moved far enough away from Texas. In the meantime, she would live comfortably.
What else could she do? Briefly she thought of Ed Travers and his kind offer to help her. Maybe he could secure her a position as a Harvey House girl. Swiftly Lauren rejected the idea. Her experience at entertaining parlor guests would not have prepared her for the dining room of a restaurant where she would have to serve food from heavily laden trays. And she would despise living in a dormitory where moments of privacy were scarce, if they existed at all.
The pros and cons paraded back and forth in her head, but the basis of her decision was the same one that Olivia had so cleverly used on Jared. Ben had wanted it.
It always came back to that. Ben had chosen her for his son. _Why?_ It was an agonizing thought that she would never know his reasons.
Lauren had been so engrossed in her own musings that she didn't remember until now that Jared was still in the room. He stood rigidly at the window in the same position as before, with his back to her. Why hadn't he said anything?
She looked at the man and tried to analyze what he must be feeling. Would marriage to him be so terrible? Olivia made it clear that his bride would be only a figurehead. She would remain a bride, and never become a wife. For the marriage was not to be consummated. Lauren's heart skipped a beat and fluttered in her breast. The mechanics of such intimacy still eluded her, but she knew that the word implied a physical avowal of the marriage.
They must speak to each other. Before she consented to this preposterous action, she must know his feelings. Timorously she vacated her chair and approached him. She cleared her throat. "Mr. Lockett?"
At the sound of her voice, his body tensed automatically. When the tremor passed, he straightened his shoulders and pivoted slowly on the heels of his boots until he faced her. He didn't say anything, only looked at her with an aloof, cold expression in his eyes. The full, sensuous lips were set in a grim, hard line.
"I... I want to know—" she stammered before he interrupted her.
"You took their bait, hook, line, and sinker, didn't you? You couldn't wait for this chance, could you? All that money! And a husband! My, my, the old maid preacher's kid has come a long way up the ladder today."
His words were cruel and vicious and delivered with intention to hurt. Is that what he thought of her? Did he think she had known about Ben's plans for her? Anger and shame forced unwanted tears to gather in the corners of her eyes, and she looked at him imploringly. "Mr. Lockett, you must—"
He lunged, grabbing her shoulders with his strong hands. Her head snapped back painfully and the bun at her nape began to uncoil.
Through clenched teeth, but with a voice oozing feigned charm, he growled, "I think that under the circumstances, we can dispense with the Mr. and Miss. My name is Jared. Say it!" he commanded.
His hands were unrelentingly tight on her arms, and her teeth chattered in fright, but she managed to gasp, "Jared," before one huge tear escaped her lower lid and rolled like a sparkling gem down her cheek.
That one tear infuriated Jared further and he hissed, "I don't know how you so completely hoodwinked and bewitched a man as smart as my father, but don't think you can resort to tears and vapors to get to _me._ Those fathomless gray eyes won't work on me, understand?" He gave her a little shake. "We're in this together. Just stay out of my way, and possibly we'll be able to stand it. Old Ben was famous for his clever tricks, and he seems to have played his last big joke on me. He sure found a willing partner in you."
"No!" she cried. "I didn't know what he intended. He only mentioned you once in passing. I didn't—"
"Then you're more of a fool than I thought. Did you think Ben wanted you for himself? Well, then, the joke's on you, too, isn't it? Were you after a nice, rich, elderly husband who would more than likely make you a rich widow very soon. _Were you?_ " He shouted the last two words, his face inches from hers. He held her tight against him, each muscle straining and pressing into her.
His amber eyes flickered for an instant when he looked down at her appealing face. His surprised expression mirrored hers as they simultaneously realized that their bodies were touching chest to toes. It was softness against hardness, weakness against strength, femininity against masculinity. The contrast was too compelling to ignore.
Jared hadn't planned it, had never even thought about it, but he couldn't control taking complete possession of her mouth with a bruising kiss. He wanted to insult her, to further humiliate her, to shatter her damned poise. But her body was so female, her lips so soft, warm, virginal, that what had been hurtful and brutal became tender, seeking, questioning.
His arms went around her slowly, drawing her into an even closer embrace. He disregarded the heels of her hands on his shoulders making a weak, futile attempt to push him away. One hand went to the back of her head and held it immobile until her lips parted under the increased pressure of his. His fingers unconsciously entwined in the thick strands of hair that tumbled further down her neck.
Sweetly his tongue plundered the hollows of her mouth, unique in taste and feel from any other he had kissed. His arms tightened around her until her breasts were flattened against his chest. The feel of her taut nipples through the soft linen of her blouse clouded his judgment, and his mouth became even more avaricious.
Lauren's mind fed on the new sensations—the scratching stubble of his beard, the coffee taste in his mouth, the faint aroma of tobacco, the feel of his lips, teeth, and tongue as they explored, the sound of a moan that came from... where?
Jared pushed her away from him so suddenly that she almost fell backward. She regained her balance and tried to assimilate what had happened. Both her hands came up and covered her mouth. Over her trembling fingertips, she looked at Jared, who appeared none too composed himself.
He stared at her, breathing rapidly, then swallowed convulsively. When he was moderately restored, he dropped the mask of indifference over his features and sneered contemptuously, "Very nice, Lauren, but I told you it wouldn't work on me." His lips curled cruelly and his thick-fringed eyelids lowered speculatively over the cynical eyes. "You were no doubt comparing my sexual prowess to Ben's." He barked a short laugh. "Well, I'll be damned before I'll have any of my father's leftovers!"
It seemed that every blood vessel in her head had burst, for her whole world was washed in red fury. She advanced on him like a warrior bent on vengeance and slapped him smartly on the cheek.
# Chapter 7
Jared was too astonished to react. She glared up at him, eyes wide and dark, chest heaving with agitation. At any other time, he would have found her anger ludicrous and would have been greatly amused. Now he stood statuelike as she said scathingly, "You are insufferable, Jared Lockett. Never, _never,_ " she stamped a small foot for emphasis, "insult me as you just have." She spun around and marched from the room, leaving Jared nonplussed.
As soon as she closed the door of the office behind her, Lauren ran toward the staircase and, quite unladylike, raised her skirt above her ankles and sped up the stairs. By the time she reached her room, her face was bathed with tears. "I hate him!" she professed in a grating whisper. "He's abominable. They all are," she cried. The pillows caught her when she fell across the bed. They absorbed her tears and muffled sounds of frustration and anger.
Why!? Why had he thought such a thing about her? What had she done in innocence that made them all suspect her of something evil? Olivia and Carson had been half-convinced that she and Ben had been lovers. Jared must have thought the same thing, and even attributed mercenary motives to her.
Even Abel and Sybil, who had loved her like a daughter, were ready to accept William's vicious lies about her. Their first expressions of shocked disbelief when William had told his tale had turned to the same contemptuous frown that Jared had worn moments before. She had done nothing to warrant their severe judgment.
William. Even now, when his memory was beginning to fade from her mind, the pain he had inflicted came back with terrific force.
* * *
The day Ben Lockett left early after breakfast, the Prathers had decided on the spur of the moment to travel to Raleigh for the day. Abel had invited Lauren to accompany them, but the prospect of spending the day in close company with the parson and his wife, much as she loved them, was unappealing. Lauren gave the excuse of a headache and begged them not to change their plans on her account. Indeed, Sybil was already enumerating the stores she wished to visit and naming possible places for a late luncheon.
Lauren waved to the Prathers cheerfully when they finally departed. She was immensely grateful for the day she could spend alone. She needed to nurse her feelings about having given up the opportunity to seek a new life for herself.
The day passed too quickly. She became involved in a new piece of music and spent hours practicing until she played it to her satisfaction. She spent a quiet hour lying on her bed with a novel, but found her thoughts kept drifting back to Ben Lockett. She would never see him again, but he would forever be in her mind. He had been so kind. How she wished her father had been like that.
Resolutely, she shoved Ben to the further reaches of her mind as she cooked an omelet for her dinner. Since Abel and Sybil were to stay the night in Raleigh, she viewed the privacy as an unexpected but welcome gift.
After her light and, for once, well-seasoned meal, she was on her way upstairs when the bell at the front door rang. Accustomed to people calling at all hours seeking Abel for one emergency or another, she opened the door without hesitation.
William Keller stood on the porch.
She was tempted to slam the door in his pale face, but her innate good manners made such an action impossible.
Hoping to rid herself of his company, she said, "Hello, William. Abel isn't at home." She barred the door with her body and purposely didn't ask him in.
The Prathers' absence was no news to William. Abel had called him about a hospital patient who needed visiting before they left for Raleigh, casually mentioning that Lauren was staying home.
He looked at Lauren smugly and took a step through the door. She was forced to move aside or have her body come into contact with his, and that she would avoid at all costs.
"Good," William oozed. "I'm glad that I'll have a chance to speak to you frankly and privately."
He deposited his hat and coat on the hall tree and proceeded into the parlor, unaffected by Lauren's cool greeting.
Lauren hadn't been in this room all day. Apparently neither had anyone else, for the drapes remained unopened and only narrow slits of violet dusk outlined them. The parlor was dim, stuffy, and close, and William's presence made the atmosphere seem even more stifling.
"What do you want to speak to me about, William? I'm very busy," she said in a shaky voice. She knew then that the one emotion William had always stimulated in her was fear.
That was ridiculous! What did she have to fear from him? She tilted her chin higher, determined that he would not see her nervousness at being alone with him in an empty house.
William stood in the middle of the room with his hands at his back as he faced her. "Abel has given me permission to ask for your hand in marriage," he stated pedantically "Before you find yourself unable to resist the advances of men like this recent visiting cowboy. I have decided that we should marry as soon as possible."
She was aghast at his words, and her nervousness gave way to anger. "Ben Lockett did not make _advances_ to me. But whether he did or not is of no concern to you, William, for I have no intention of marrying you." She paused to draw in a ragged breath and clasped her hands together at her waist. "This discussion is over. I'll see you out."
She turned and walked toward the portiere. Before she reached it, William's cold hand gripped her arm above the elbow and whirled her around to face him. She was so surprised at his accosting her in this manner that she didn't try to extricate herself, but only stared at him incredulously.
"Not so fast, Miss Priss," he snarled. "I'm not finished with you yet." His voice was a feral growl, and she leaned away from him in revulsion. "You may have everyone else fooled, but not me. What kind of proposition did Lockett make you out there in the rose garden?"
She tried to jerk her arm away, but his grip became painfully more restricting and she winced. "I don't know what you're talking about. He asked me to go to Texas for a visit, that's all."
"Oh, I just bet he did." William smirked. "The only visits you'd be paying him would be to his bedroom."
Lauren gasped. "I... I have _no_ idea what you're talking about." This was true. She had only a vague notion of what his words implied, but she knew instinctively that the implication was ugly.
William squinted at her carefully. "Maybe you don't. Well, then, I'm going to be the one to teach you. Not some old, overgrown oaf."
Terror gave Lauren the strength to free her arm and, turning, she tried to run from him. He was right behind her and grabbed her again before she had taken two steps. His arms went around her in a viselike grip and pulled her around to face him. His fleshy, wet lips mashed against hers.
Lauren couldn't believe this was happening. Her mind screamed silently while his mouth over hers made actual screaming impossible. He held her even closer, his legs straddling hers as he bent over her. Then, placing his wet mouth against her ear, he muttered, "Don't fight it, Lauren. I've watched you move in that maddening way of yours. I'm not fooled by the perfect lady act." All this time, she could feel his cold hands working with the buttons on the back of her shirtwaist. She did scream when she felt his bloodless fingers against her flesh.
He stopped the scream with his mouth again and, as Lauren's mouth was open, he thrust his tongue into it deeply. She fought even harder, scratching his face, pulling his hair, kicking his bony shins. An instinct of self-preservation drove her to do things she would never have thought herself capable of.
She was horrified to think about what all this awful pawing and slobbering culminated in, and she knew she could not allow it. Repulsion and fear gave her one last surge of strength, and she shoved against his chest with all her might. He staggered and fell backward over a petit-point cushioned footstool in front of Sybil's easy chair. While he made an ungainly attempt to stand, Lauren lunged toward the fireplace and seized the iron poker, brandishing it in front of her.
"Get out of here," she managed to croak between gulping breaths. "If you try to touch me again, I'll kill you."
Standing there with bleeding scratches on his face, his thin hair sticking out at varying angles, his clothes in disarray, William bore little resemblance to the stiff, circumspect minister who delivered hair-raising sermons to his congregation on the consequences of pursuing the lusts of the flesh.
"What would you tell everyone, Lauren darling, when they came in and found my skull crushed if, in fact, you succeeded? Your reputation would be irreparably damaged. People would believe that you'd invited me here while your watchdogs were out of town." He took a tentative step toward her.
He stopped when she raised the poker higher. "I would never have thought myself capable of murder, but so help me, I will do it, William," she threatened. "You're a hypocrite and a parody of a man. Now get out of my sight. At once!"
He snickered nastily. "I'm not giving up. You probably delivered the goods to Lockett, anyway." Having flung this final insult at her, he walked past her warily and paused in the hall only long enough to straighten his clothing and slip on his coat and hat. Lauren heard the front door open and close softly. She stood with the poker raised and only when the weight of it began to make her arms ache did she lower it.
She moved with stunned, dreamlike slowness. After climbing the stairs as if she wore lead shoes, she opened the door to her room and then locked it behind her. She crossed to her dresser and looked with dismay at her reflection in the mirror.
Her chin was still shiny with William's saliva. Her hair hung in tangled knots down her back.
She removed her clothing as she drew water into the deep tub. She rinsed her mouth out several times with antiseptic, and then stepped into the hot water to soak away the degradation. The bruises on her upper arms were painful.
Lauren had awakened the next morning still shattered from the experience of the night before. Pacing the floor, she tried to arrange her thoughts. How was she going to tell the guileless Prathers what had happened? Their disillusionment with the young pastor would be shocking. Certainly Abel would have to relieve him of his position and forbid him to ever enter this house again. Lauren wished she could spare them this hurt, but she couldn't remain silent. William Keller was a menace.
It never occurred to her that her guardians wouldn't believe her.
When they came home, she welcomed them happily. Sybil's chattering helped dispel the gloom in the house, which had seemed to become even more oppressive since last evening. Lauren was presented with a box of lace handkerchiefs. She thanked the Prathers profusely while consciously ushering them into the parlor.
They had just sat down when the doorbell rang. Lauren was astounded to hear William's voice when Abel answered it. They spoke quietly for a few moments before a perplexed Abel stood under the portiere and said, "If you ladies will excuse us, William has an urgent matter to discuss with me. We'll no doubt join you shortly."
He disappeared in the direction of his study and a nameless premonition pricked Lauren's mind. She was nervous and apprehensive as she listened to Sybil's detailed account of the trip. Lauren's mouth went dry and her agitation grew as the interview in the study stretched into a half-hour.
Her heart lurched when she heard the study door opening and the two men came into the parlor. Abel's face was an alarming red. He shook his head as he looked disbelievingly at Lauren. William stood several humble steps behind him. He seemed contrite, but Lauren caught a victorious gleam in his reptilian eyes when they lighted on her.
"Abel, what—" Sybil's voice quivered when she saw her husband's obvious distress.
"My dear, I wish I could spare you this, but I'm afraid you'll have to know of our shame sooner or later."
Abel crossed the room with a heavy tread and sat down beside his wife, taking her hand. William stood just inside the doorway and studied the ugly carpet under his serviceable shoes.
Was it possible William thought she had told the Prathers about his shameful behavior last night and had come to apologize? Her surmise was rejected when she saw the expression on Abel's face as he turned toward her. It was sad. It was censoring. It was sanctimonious.
He sighed before he said, "Fornication is a grievous sin, Lauren." Her lips parted in astonishment. Sybil gasped and crammed a handkerchief against her lips.
"What—" Lauren started to speak, but Abel continued.
"It is an abomination unto the Lord. William has come to me like a man and confessed that the two of you have for several months yielded to your lusts."
Sybil collapsed against the back of the sofa, and tiny sobs escaped from her trembling lips. Lauren opened her mouth to protest, but again Abel anticipated her.
"A man's drives are stronger than a woman's. Even a man of God like William isn't free from the cravings of his flesh. However," here his voice became more stern, "it is up to the woman to keep a tight rein on those cravings. William told me that you enticed him to the point where he succumbed."
Sybil cried out loud now, the tears flowing copiously down her fat cheeks. All the blood drained from Lauren's face. The wild pounding of her heart seemed to stop. I must be dreaming, she told herself.
"Somehow we, who loved you as our own, have failed. You were entrusted into our care by your sainted father. We have betrayed that trust just as surely as you have betrayed our love."
Lauren's heart ached at seeing her guardians suffer so, but she made no effort to speak then. She glanced at William, who continued to stare at the floor in abject repentance.
Abel closed his eyes for a moment, then said, "Lauren, William told me that he couldn't live with himself another day without arranging for an immediate wedding. He offers you marriage, not out of guilt, but out of deep and abiding love. I, for one, am grateful—" He broke off and buried his face in his hands.
Witnessing his misery spurred Lauren into action. She flew out of her chair and crouched beside Abel. She placed both hands on his and waited until he raised his head to look at her before she spoke.
"He's lying, Abel," she stated simply. "The only time I have been alone with William was last night. He came here while you were gone and tried to kiss and... touch me." Tears rolled down her face as she recounted the indignities she had experienced.
When she finished her story, she felt William's hands on her shoulders as he pulled her conciliatorily to her feet. "Lauren, dear, we don't have to hide anymore. Don't you see? We will be married and live together. We have sinned. But I've confessed my sin to God and Man. If you confess your transgression, you, too, will feel the peace that now suffuses my spirit."
She jumped away from him, her eyes flashing. "Are you mad? The only thing I'll confess to is my loathing of you."
William smiled sweetly at Abel. "I'm afraid she's overwrought. She wanted to prevent you from finding out about our illicit relations. She wanted to spare you that."
"Of course," Abel agreed, staring at Lauren as if he'd never seen her before. "I think the wedding should take place as quickly as possible. Lauren, I appreciate your charity in trying to spare Sybil and me. I will bless this marriage. You will be restored to my family and to the family of God."
William had won. With satanic subtlety, he had conceived this counterattack, and the Prathers believed him. Abel had spoken of betrayal, but he couldn't know how deeply she felt betrayed.
She could show them the purple bruises on her arms, but why should she? If William's seeds of deception had found such fertile soil in the minds of those who should know her and love her better than to suspect her of such depravity, then she wasn't going to plead her case. Abel was God's servant, but he wasn't God. And her conscience was clear.
"I'm not going to marry William. He is lying. And even if he weren't, I would never marry William Keller."
Sybil shrieked, and fell back against the cushions once again.
"Lauren, do you wish to heap burning coals of shame upon our heads? Child, please consider us if you don't consider your immortal soul," Abel pleaded.
"I have done nothing. I will not be condemned to a life of unhappiness with a man I despise," Lauren said firmly. "But you needn't dread the shame my continued presence in your house will bring you. I intend to leave."
And she had. She left within the week, taking out of the bank what small funds belonged to her.
Her telegram to Ben Lockett sent the morning after William's conference with Abel caught the cattle baron just as he was arriving in Austin. He hastily replied, and his evident joy at her imminent visit restored her high hopes. She left the Prathers' parsonage several days later, unforgiven, an anathema. But she was on her way to Texas, and a new life with the family of Ben Lockett.
* * *
The family of Ben Lockett. She was about to become a member of that family. Was she doing the right thing?
Yes, she told herself. She was doing the only thing she could do, short of leaving penniless and without direction. Yes. Her best course was to marry Jared.
His kiss had stunned her, offended her. But where was the disgust she should be feeling? Nausea had kept her awake the night William had kissed her. Why hadn't she been sickened by Jared's mouth? All she could recall of his kiss was the sensuous persuasion of his lips and tongue and the protective warmth of his firm embrace.
She rolled onto her back on the bed and covered her face with her hands against the shame she felt. She had enjoyed that kiss.
And she would never forget it. Could she marry the man, live with him, and constantly be reminded of that one fleeting moment of passion? Added to all his other insults, could she tolerate that final humiliation, knowing that he would remember that kiss, too?
On the other hand, she had not yet thought of a palatable alternative. At least by marrying Jared, she would have a goal to accomplish, an obligation to fulfill.
The money seemed insignificant now, but in two years, it could be extremely important. The twenty thousand dollars could mean the difference between living comfortably and destitution. Looking at it strictly from a financial point of view, did she dare pass up an opportunity like this?
And then there was the man, Jared Lockett. No! She wouldn't think about him, for thoughts of him clouded her ability to reason logically. Somehow they would learn to coexist peaceably. That would be a major accomplishment, but she would take it a step at a time.
She made her decision.
For the remainder of the day, Lauren stayed in her room, trying to still the turbulence of her mind. At dinnertime, she changed her shirtwaist and recombed her hair, securing it better than she had that morning. A cool, damp cloth pressed to her eyes had relieved them of the redness and puffiness of weeping. She felt restored, but her heart pounded painfully as she descended the stairs.
At the dinner table, Olivia stated, almost as an afterthought, "Lauren, the wedding will take place here a week from tomorrow. There will be a select number of invited guests."
"Very well," Lauren answered.
Jared wasn't at dinner. His absence was never mentioned or explained to his fiancée.
# Chapter 8
For the next few days, Lauren was thrust into such a beehive of activity that she had little time for introspection. The sheer importance of the transpiring events prevented her from examining them too closely. It was far easier to be swept along on the tide. So busy were the days that at night she fell into bed exhausted, hoping for rapid and complete oblivion. But her brain refused to slumber, and she was forced to dwell on what she was about to do until her mind finally relinquished its control over her body, and her burning, gritty eyes closed in restless sleep.
Elena and Rosa were at first surprised that Señor Jared was marrying the lovely Miss Holbrook. But soon they were riding the crest of excitement and got caught up in the flurry of activity. They nearly drove Lauren to distraction with their petting and fawning over her.
Jared's reaction to their upcoming marriage was one of cool acceptance. He neither feigned affection, nor treated her with the resentment she knew he must feel. Each time they were together, he regarded her with the same aloof indifference that he did everyone else. He was polite, but not effusively so. He conversed with her when necessary, but didn't initiate any private dialogues between them. He could have loved her passionately or despised her with equal fervor. His remote expression gave away nothing.
Olivia organized the wedding in the brief span of time allotted. Lauren was consulted on little in regard to the arrangements. She was told it would be a private civil ceremony held in the large parlor. There would be a small reception afterward for the few invited guests. The following day, Lauren and Jared would go to Keypoint for a "honeymoon."
Lauren was fitted several times a day for the trousseau which Olivia insisted on despite Lauren's protests to the contrary. Olivia's gesture wasn't motivated by generosity or any blossoming maternal affection for Lauren, but by concern that everything appear proper and above suspicion.
She had spread the word of Lauren's background through several famed gossips. The story was that Lauren's parents couldn't attend the wedding because her father was suffering from a heart condition that made travel impossible. Lauren's dear mother had far too many responsibilities to oversee the wedding, so Olivia had graciously offered to handle it. If anyone was suspicious, they feared the indomitable Olivia too much to say so.
Mrs. Gibbons, the seamstress who had been commissioned to provide Lauren with a complete wardrobe in an impossible amount of time, reflected the general shock and thrill that Texas's most eligible bachelor, Jared Lockett, had finally been snared.
Lauren was amazed at the quantity and quality of clothes being made for her. There were skirts and shirtwaists of the finest fabrics. All of the blouses were trimmed in delicate, weblike lace. Dresses for daytime and evening, cloaks, coats, hats, gloves were strewn around her room in varying stages of completion. Mrs. Gibbons worked around Lauren like a sculptor around a masterpiece, measuring, twisting, pulling, turning, lifting, pinching, all the time murmuring to herself in appreciation of her subject. She made undergarments of the finest linen trimmed with blue satin ribbons and fine lace.
Lauren stood with flushed cheeks as Mrs. Gibbons deftly pinned a new chemise onto her. "They are very... sheer, aren't they?" Lauren asked shyly, glancing down at her breasts so clearly revealed through the fabric.
Mrs. Gibbons chuckled softly. "Mr. Lockett will love these underthings, but he'll be impatient to see you out of them, too. You have a beautiful body, Lauren. After he sees you without your clothes, he may never let you dress again!"
Lauren was appalled at such a thought. She was still alarmed over her reaction to Jared's kiss. Though she had fought William with all her strength, she hadn't resisted Jared, at least not strenuously. Of course, one couldn't compare the two men. William was repulsive, while Jared was handsome and virile, and his eyes...
No! She wasn't going to think about him. He obviously didn't think of her. Where before he had stared at her relentlessly, now he studiously ignored her. Of the two, Lauren couldn't have said which disturbed her most.
* * *
To Olivia's credit, everything was completed in time. Lauren's new wardrobe hung in her closet except for the clothes Elena had packed for her to take to Keypoint.
Rosa and several extra helpers had been cooking and baking for days in preparation for the "small reception." The house was bedecked with flowers and potted ferns. How Olivia had managed to have them transported from Austin without their wilting would remain a mystery to the bride.
She watched the dawn of her wedding day from the upstairs window where only weeks before she had watched Ben Lockett's funeral cortège commence.
"Are you pleased, Ben?" Lauren asked in a whisper just as the sun broke over the horizon. She tried to convince herself that the timely sunrise was a good omen.
Elena arrived with a breakfast tray. Anticipation glowed in her liquid eyes and she chattered cheerfully. When Lauren had eaten all her nervous stomach would tolerate, she bathed leisurely and Elena began helping her dress.
Lauren's wedding gown spurned tradition and was beautiful in its uniqueness. Ecru lace was lined with a silk slip of the same color. The leg-of-mutton sleeves and the bodice, to the top curve of her breasts, were left unlined. The collar stood high and flared slightly at her jaw. One deep flounce accented the bottom of the slim skirt and barely brushed the vamp of her bone kid slippers. She pinned a nosegay of deep purple violets at her waist. Her brooch watch was invisibly secured to her petticoat.
Her hair was pulled up in its usual pompadour, but Elena insisted on curling a few loose tendrils to hang around her face and on her neck. Lauren looked back at the girl in the mirror and wondered if that vision in the costly gown, with the pale face and cautious eyes, could really be she.
Elena stared at her idol with reverence. "Señorita Lauren, you are beautiful," she whispered. "Like the Blessed Virgin Mary." Shyly she kissed Lauren on the cheek, her huge eyes filled with tears.
"Thank you, Elena. I wish you could be there during the ceremony. You're my best friend."
"I would like that, but..." Elena gave a characteristic shrug. Then she giggled and said mischievously, "I would rather be a witness to the wedding night and see if all the stories about Señor Jared are true. They say he is as big as a stallion. You are very lucky, no?" Still giggling, she pushed Lauren toward the stairs. The bride's face had whitened significantly.
It was prearranged that Carson would escort her down the aisle. As she met him at the foot of the grand staircase, Lauren was terrified that her knees wouldn't be able to support her much longer. Carson spoke to her softly, smiled, patted her arm reassuringly, and led her toward the formal parlor.
They stood under the portiere while the fifty or more guests who had been whispering animatedly suddenly ceased all conversation. The organ, borrowed for the afternoon from the Methodist church, filled the room with soft, slightly wheezing strains.
This was the first time anyone except Parker and Kurt Vandiver had seen the mysterious Miss Holbrook, and all were instantly captivated by her breathtaking beauty. It was easy to see how she had lassoed the mustang heart of Jared Lockett.
Had the groom not had years of experience in keeping the rigid muscles of his face from showing any emotion, he might very well have gasped in pleasure at the sight of his bride. He was adamantly against this farce of a marriage, but how could any red-blooded man remain indifferent to the woman who would soon bear his name?
Dammit, under different circumstances he might even... That kind of thinking won't do, ol' boy, he cautioned himself. However, it didn't hurt to look, did it? Isn't that what bridegrooms were supposed to do? The woman walking toward him on Carson's arm was exquisite. She was like, like... what?... whom? No one he'd ever seen.
He studied her as she moved closer, her gray eyes chastely downcast. Quite different from a few days ago, when she had slapped him. Then her eyes had been a dark blue-gray, like the most ominous of storm clouds that gathered over the plains. And they had been just as threatening, just as exciting, just as electric.
Lauren's slim figure was accentuated by the perfectly fitted gown. Her full bust was noticeably defined above the narrow waist encircled by the high cummerbund. Jared imagined he saw faint shadows where the crowns of those high, proud breasts would be, but told himself it was a foolish fantasy and clenched his fingers, which longed to touch those shadows. Swallowing hard, he dragged his gaze away from her chest to her face.
She raised her eyes, and the fear in them was evident. Had he been the devil incarnate, she could not have looked more apprehensive. Why did he feel a sudden urge to reassure her?
Carson delivered Lauren to Jared, and she graced the older man with a timid smile before he left her and took a seat next to Olivia on the front row of chairs temporarily set up for the ceremony.
Jared extended his arm and Lauren slipped her hand through the crook of his elbow as they faced the judge. Jared hadn't planned to, and Lauren certainly hadn't expected it, but his opposite hand closed over hers. The warm strength of his flesh contrasted with the cold fragility of hers. Her fingers were firmly pressed against the fabric of his sleeve. She risked looking at him through her lashes, keeping her eyes lowered. Her heart turned over when she met glowing amber eyes filled with an emotion she had never seen in them before. Was it understanding? Approval? Even admiration? She could almost imagine that his lips softened into the semblance of a smile. She wanted to continue looking at this transformed Jared, but the judge was speaking. Reluctantly she freed her eyes from her bridegroom's possession.
The ceremony was brief and to the point, not like any of the weddings Lauren had attended. She felt a twinge of disappointment. In the ceremonies he performed, Abel had talked about the sanctity of marriage and God's blessing on the institution, about Jesus performing his first public miracle during the celebration of a wedding feast. Of course, this was the service for a civil ceremony. Yet when the judge said, "I now pronounce you man and wife," the words rang hollow and held no meaning for her.
Judge Andrews beamed his congratulations on the attractive couple in front of him.
"You may kiss your bride, Jared." He smiled.
Jared arranged his features into a detached expression. He prepared to face Lauren perfunctorily, to place his hands on her shoulders, and to kiss her coolly on the cheek. His plan went awry. His composure slipped when his eyes riveted on her pink, trembling mouth.
He remembered that other kiss. The kiss that had shaken him to the core. The kiss that had been earthshattering in its violence and enlightenment. The kiss whose impact on him had not yet been diminished.
The recollection that came unbidden to his mind brought with it both intolerable pain and immense pleasure. A now-familiar heat radiated from the center of his body to its every extremity. He longed to taste those lips again and either quench the fire in his veins or increase it to a pitch where he could not be held accountable for seeking its source and extinguishing it. He called upon every particle of self-restraint when he drew her to him and woodenly placed his lips against those that promised such sweetness.
He was rescued from further torment when Olivia and Carson and their friends, who were eager to meet Lauren, surrounded him with hearty congratulations. He surprised himself by possessively tucking Lauren's hand under his arm.
For the rest of the afternoon, the Locketts entertained their guests with an abundance of food and drink. Only a month ago, many of these guests had been here to pay their respects to Ben Lockett, who had lain in state in this very room. All gloom was now dispelled. A festive mood prevailed.
Carson made a flowery wedding toast, and glasses of champagne were lifted in salute. As Lauren raised her glass to her lips, her eyes met Jared's over the crystal rim. The lights in his eyes shone with golden effervescence and were much more intoxicating than the wine. Her sip was more frugal and daintier than his, but when she would have lowered her glass, he reached out and tipped it again to her mouth. The champagne trickled down her throat and heightened the fluttering in her stomach. _They say he is as big as a stallion!_ The exact meaning of Elena's words escaped Lauren. All she knew was that as she looked at Jared, thoughts of his physique shocked and thrilled her.
Just as he was lowering his glass, someone jostled his elbow and champagne dribbled down his chin and onto his lapel.
"Oh!" Lauren exclaimed, then couldn't help the laugh that bubbled out of her throat.
"You think it's funny, do you? Having a soggy bridegroom?" He was smiling, too, as he set down his glass and shook off his wet hand.
Acting instinctively, Lauren whisked away the droplets on his lapel. Obeying a subconscious command, her fingers settled under his chin and her thumb gently removed the drops of wine from beneath his lips. "There, that's better," she said, smiling, and raised her eyes to his. She recoiled from the bitterness she read there.
Jared's body had betrayed him. At the moment her fingers made contact with his face, he felt himself swell and harden with arousal. He had begun to think that her innocence was real. Hell. No woman could touch a man, look at a man the way she did and not know exactly what she was doing. Ben should have been smarter than to fall for her act. Jared was determined not to fall for it himself.
"Jared...?" she asked waveringly.
"We'd better mingle with our guests, Mrs. Lockett," he said curtly as he gripped her elbow and guided her toward a group of well-wishers.
Lauren's heart sank. Just when she thought she was making headway, Jared's true feelings for her surfaced. How could they possibly survive the next two years? Inwardly she sighed. It was too late now to dwell on that. She would have to cope with the future later.
Today she could concentrate on nothing except the man beside her. His charisma made him almost impossible to ignore. Indeed, Jared looked exceptionally handsome in his black suit and white shirt. His eyes sparkled with amber lights and brilliant teeth flashed a dazzling smile on his tanned face. He won the confidence of everyone present. He charmed the women shamelessly as their approving husbands looked on. He drank only the two traditional glasses of champagne, and used affectionate terms and gestures when introducing Lauren to people whom he had known all his life, accepting with aplomb their jibes on his cunning in finding such a delightful bride.
There was a noticeable pause in his effusiveness when the Vandivers came over to express their congratulations. Lauren saw a tightening around Jared's mouth and felt his body, ever near hers, grow tense. He was coolly polite, but his voice lacked conviction when he thanked them for attending the wedding.
"You sure pulled a fast one, Jared." Parker patted him on the shoulder and Lauren detected the tension under the black broadcloth. "We had no idea that you and Lauren were planning to marry."
Jared met Parker's piercing stare without flinching. "I'm afraid you caught us in the midst of a tiff. She thought it was improper to have the wedding so soon after my father's death. Fortunately, I was able to talk her out of her scruples."
He put a possessive arm around Lauren's shoulders and looked deeply into her eyes as he pulled her closer to him. He was playing his part well, she thought. This proximity to his body and that probing gaze made breathing difficult for her.
"Mrs. Lockett, may I wish you much happiness." Kurt looked at Jared with a challenge on his smug face. "I _do_ have your permission to kiss the bride, don't I?"
Without waiting for an answer, Kurt leaned toward her and pressed thick, hard lips against hers. She recoiled even as Jared's fingers gripped her shoulder so hard that she almost cried out from the pressure.
Kurt smiled a slow, provoking smile and moved away.
Jared watched with a deadly glint in his eyes as Kurt sat down beside one of the local girls who was heartbroken over this sudden marriage. The ever-watchful eyes of Parker Vandiver saw the animosity on Jared's face. He chuckled to himself and thought how stimulating the next few months would be. He relished dissension.
Jared was unaware that he still held Lauren in a deathlike grip until she shifted her weight uncomfortably. He withdrew his arm quickly and muttered under his breath, "No one can resist your charms, can they, Mrs. Lockett?" In the same undertone he warned her, "Smile, Lauren. You're the radiant bride and don't you forget it."
One of the other guests came over then and Jared resumed his character of loving bridegroom with the ease of an experienced actor. Lauren's head spun with the effects of the champagne and the emotional clamor of the past week. She longed to excuse herself and seek the cool serenity of her room. But, of course, that was out of the question.
Ben's widow reigned over the reception with her customary regal bearing. She looked beautiful today, Lauren noted. The silver streaks in her dark hair caught the afternoon sunlight, and her smooth face was flushed with excitement. Lauren suspected that this was as close as Olivia could ever come to being happy. She wore a dress of sea-green georgette, and it floated around her tall, imposing figure like a mist.
Carson paid court to her in his pitiful, humble way. Lauren felt great tenderness and sorrow for this man. He had never shown her anything but kindness, obviously a rare sentiment in this house. He was pathetic in his devotion to Olivia, who either ignored, patronized, or berated him with equal aptitude.
Lauren's glance around the crowded room chanced to fall on Kurt Vandiver. His blue eyes were penetrating under the white-blond brows, and she felt unnerved by his stare. Instinctively, and totally unaware that she did so, she moved closer to Jared, welcoming the protection his large physique intimated.
He looked down at her quickly, distracted by her touch from his conversation with one of the bank directors' wives. He followed Lauren's anguished gaze across the room to Vandiver, who was lounging against the wall. Impulsively Jared slid his arm around Lauren's slender waist and rested his hand just under her breast.
The movement wasn't lost on Kurt. Like a voyeur wandering through a bordello, he licked his lips lasciviously. Lauren shuddered, but didn't know if she were reacting to the lewd expression on Kurt's face or the warm pressure of Jared's hand seemingly burning through the fabric of her dress. Kurt winked at Jared, pushed away from the wall, and strolled through a door leading to the wide porch outside.
Jared and Lauren didn't move. The banker's wife gushed and simpered, totally unaware of the drama being played out beside her.
Lauren could smell the starch that kept Jared's shirtfront crisp, which blended intoxicatingly with tobacco and champagne. When he spoke in confidential tones to the silly woman, Lauren could feel the vibration of his voice in his chest. The bank director's wife moved away, and still Jared retained his possessive hold on her. His hand trembled slightly as his thumb moved upward and lightly stroked the side of her breast. Or did she only imagine it? Lauren thought she would die from the constriction in her chest that pounded up into her throat and sought release in a small moan.
Another guest walked toward them. Slowly, reluctantly, the strong fingers were withdrawn, leaving behind an imprint on Lauren's skin as scorching as a brand.
* * *
The afternoon had waned into an indigo twilight. The guests were gone. Carson and Olivia sat together on one of the sofas. They were weary and relieved that, at last, the ordeal was over.
"Well, we pulled it off beautifully. Everyone was convinced that it was a marriage made in heaven. And word will get around," Olivia gloated.
"We owe it all to Lauren," Carson said generously. "You are a beautiful bride, my dear."
"Yes. She's the perfect little wife, all right." Jared smirked at Lauren, who was helping Rosa load dishes onto a large tray. He tossed down a tumbler of whiskey.
Olivia coughed to cover a malicious smile that rose unbidden to her lips. Obviously her son disliked his wife. That could be most advantageous in the future. "You performed your part well, too, Jared," she said. "Treat Lauren courteously in public, and no one will be the wiser."
Lauren kept her eyes lowered. She was still no part of them, and they cared nothing for her feelings.
"You can stay at Keypoint for as long as you like. We'll pass that off as your honeymoon. Not very glamorous, but if anyone inquires, we'll say that Lauren was eager to see the ranch," Olivia remarked, stifling a yawn.
"Good," Jared said. "I can't wait to get back. When I left—"
"We know your unwarranted enthusiasm for the ranch, Jared. Please don't bore us with it now," Olivia said curtly. Her laziness of the moment before was gone.
Jared's lips pressed together in a hard line before he said brusquely, "Lauren, we'll leave first thing in the morning. And I mean first thing. You can ride, can't you?" The question sounded almost like a dare.
"Yes, I can ride, Jared," she said. He scowled at her. Had he wished she couldn't? Would he have liked to find another flaw to ridicule?
"Then we'll go on horseback," he said impatiently. "Pepe'll bring our bags in a flatbed." Then, looking at her, he said sharply, "We're traveling light, so don't pack everything you own."
"I hadn't intended to," she shot back, her own ire rising under his imperiousness. She continued in spite of his dark look, "Since we are to leave early, I think I'll go upstairs. Goodnight, Carson. Goodnight, Olivia." Her head was held high and her back was straight as she crossed the room. At the doorway, she paused and looked at her mother-in-law. She found it difficult to say what she knew she should. "I know you went to tremendous effort and expense, Olivia. Though it wasn't for my sake, I appreciate your doing it just the same. The flowers, the food, the clothes, everything was lovely. Thank you."
The other three didn't say anything for several moments after Lauren left them.
Then Carson coughed uneasily and said quietly, "Jared, treat her kindly. We have our ulterior motives, but she is an innocent party in all this. Be gentle with her."
Jared resented being instructed on how to treat a woman. Intending to make a retort to that effect, he turned away from the liquor cabinet where he was getting a large bottle of whiskey. Carson's face was guileless. He hadn't issued a directive; he had made a plea. Jared stifled the rejoinder already on his lips, mumbled his goodnights, and plodded up the stairs.
* * *
For an hour, he had been in his room drinking steadily and listening to the light taps that small slippers made on the floor next door.
"To my wedding night," he scornfully saluted himself in the mirror over his dresser. He was shocked by the reflection. He didn't remember discarding his vest and coat, but a swift glance over his shoulder revealed them to be carelessly draped over the back of a chair. When had he taken out his cuff links and unbuttoned his shirt? In a characteristic gesture, he raked his fingers through his hair.
His bride was undoubtedly just as immaculate and cool as she had been when she met him at the altar. Or maybe she was already dressed for bed. What did she sleep in? Nothing provocative, he scoffed silently. Something chaste and...
Why not? Why not see for himself? Why should he be the only one to suffer through this hellish night? He was her husband after all and, by God, he had some rights!
He didn't consciously decide to disturb her but, propelled by some mystical and obsessive force, he found himself before the door that connected their rooms through the bathroom.
He knocked sharply. No answer, but the movements in the adjoining room halted abruptly. He knocked again, this time saying her name. It came out as a tremulous sigh. He cleared his throat, shook his head, and repeated it with more force. Silence.
"Yes?" Apprehensive. Tentative.
"Open the door." It was a command. He hoped.
Long pause.
Finally, calmly, "What do you want, Jared?"
He laughed mirthlessly, muttered a few unintelligible obscenities, and raised his voice another decibel. "Open this door!"
He heard her footsteps and the rustle of her clothing coming closer until he knew she was in the bathroom just beyond the door.
"We can talk from here, Jared."
"If you don't open this goddam door," he growled, "I'll kick it down. Do you want a ruckus? It won't embarrass me, because I don't give a damn."
There was a momentary hesitation, then the key was turning in the lock made rusty by disuse. The knob rattled as she pulled the door open.
Her hair was down, framing her face in a black cloud and cascading down her back in heavy waves to her waist. She wore a rose-colored dressing gown cut in a deep V in the front and buttoning from her breast to her knees. Lace spilled over her delicate hands at the end of the wide sleeves. Her trembling lips were parted to allow agitated, quick breaths to escape. She tried unsuccessfully to mask her fear.
The sight of her stupefied him, and her flowery scent filled his head with a greater vertigo than had the vast amount of alcohol he had consumed. He longed to taste the smooth skin at the base of her throat, which fluttered with the frantic beating of her pulse. He wanted to investigate what treasures lay beyond that first button of her dressing gown.
By an act of will, he regained his self-control and said thickly, "You needn't look so frightened, Mrs. Lockett. I have no intention of forcing you to give me my conjugal rights."
Her only response was to moisten her lips with a dainty pink tongue. Jared swallowed hard, stifling an animal groan, and said, "I demand only one thing. There will be no locked doors. Separate bedrooms are not that uncommon, but a locked door invites speculation. Maids gossip, you know. One locked door between us and this whole farce is shot to hell. So no locked doors. Is that understood?"
"Yes, Jared," she answered levelly.
Dammit! Why didn't she scream or swoon or something? She was so damned composed, while he stood here like an adolescent idiot with his sweating palms and pounding heart and aching loins.
Not trusting himself any longer, he reached for the door and closed it quickly. He didn't hear the key turn in the lock before her footsteps receded into the bedroom beyond, but he lacked the nerve to turn the knob and test her obedience.
"I guess I showed her who's boss," he boasted as he flung himself upon the bed, wondering why he felt no satisfaction in his victory. All he felt was a deep longing which he tried to obliterate with sleep.
# Chapter 9
Elena gently shook Lauren's shoulder and whispered, "Señora Lockett, wake up. It's time to get ready for your trip."
Lauren opened her eyes. She was greeted by a dark room and mumbled a protest into her pillow. She didn't want to give up the sleep that had been so long in coming. Elena's persistent needling finally penetrated her slumberousness, and with sudden clarity she remembered where she was going today. She threw back the covers and rolled out of bed. Soon she was wide awake and, in spite of her misgivings toward Jared, was excited about seeing Keypoint for the first time.
As she performed her toilette, Elena chattered about how lovely the wedding had been, how beautiful Lauren looked in her bride's dress, how handsome Jared was, and how lucky Lauren was to have such a husband.
The maid had been surprised only minutes earlier when she had knocked on Jared's door to awaken him.
"Señor Jared, are you awake? Time to get ready for your trip. Señora Lauren, do you hear me?"
Only silence greeted her until Jared mumbled a sleepy, "I'm awake."
"Señora Lauren, do you want me to help you?" Elena twittered, thinking that the new husband would probably take over some of her former duties.
There was a rustle of bedcovers, a muttered curse, and then Jared said, "She's in her room. Go wake her up."
Elena had stood outside the door staring at it in a puzzled fashion. "But, _señor_ —"
"She's in her room," he growled.
Now, as she packed last-minute additions to Lauren's bags, Elena shrugged. Why wasn't Lauren sleeping with her new husband? The ways of the _gringos_ had always been a mystery to her.
The one possession that Lauren had that kept her from coming to Jared as a pauper bride was her riding habit. It had been a gift to her from the Prathers, who had insisted that she attend a riding academy. The blue velvet habit with its long, trailing skirt and tightly tailored jacket fit her figure to perfection. It had taken up an exorbitant amount of space in one of her valises, but she couldn't bear to part with the finest garment she owned when she left North Carolina.
Elena eyed the riding habit dubiously and asked Lauren tentatively if she wouldn't prefer wearing one of the split skirts that Mrs. Gibbons had made for her.
"No," Lauren adamantly refused as she thrust long pins into the smart matching hat with its decorative veil. "I want Jared to see me in something of my own. Something his mother didn't buy for me."
The two young women were snapping shut the fastenings of her bags when Pepe tapped lightly on the door. Elena opened it and he bowed swiftly and said, "Señor Jared, he is waiting." Pepe picked up Lauren's bags and preceded the women down the stairs. He looked at her attire skeptically and muttered to himself in Spanish, shaking his head in bewilderment.
He carried her valises through the large front door. Stubbornly Lauren detoured into Rosa's kitchen for a quick cup of tea. The morning was chilly, and the tea warmed her body, but nothing could warm the chill in her heart. She had not yet seen Jared, and after their confrontation at the bathroom door, she dreaded facing him again. What would his mood be today?
Rosa was bustling around the kitchen even at this early hour. As Lauren was eating fresh, hot _tortillas_ dripping with butter, Rose saw the sadness on the young _gringa_ 's face. Rosa knew everything that went on in the house, and she wasn't fooled into thinking this sudden marriage was based on love. Maternally she reached out and patted Lauren's arm. "Señora Lockett, everything will be all right. Señor Jared, he... he hurt inside. Here." She placed a plump hand over her enormous breasts. "But he is a good man. He like you very much." Lauren moved to protest, but Rosa went on quickly, "Rosa knows the boy since he is born. I can tell." She smiled radiantly, reassuringly, and squeezed Lauren's hand. _"Vaya con Dios,"_ she whispered.
Pepe poked his head through the kitchen door, cleared his throat, and said apologetically, "Señor Jared, he..." and indicated with his head that Lauren should follow him without delay.
Before she left the kitchen, Lauren turned to Rosa and hugged her, her arms barely encompassing the woman's girth. Elena was standing by the front door, tears glistening in her eyes. Lauren hugged her as close as her protuberant stomach would allow.
"I'll be upset if the baby comes while I'm away. Can you send word to me? I hope you'll be all right."
"I will let you know, but don't concern yourself. The _niño,_ he will be born fine." Elena laughed.
"Goodbye, Elena." The two women clung to each other for a few seconds, then Lauren stepped through the door
Jared was sitting on the large palomino that had been tied to the back of the wagon the day Lauren arrived. He was a _vaquero_ again. He wore tall black boots, the customary tight black pants tucked into the tops of them. A leather jacket protected him against the coolness of the October morning. A blue shirt was under that, and a red bandana had been tied negligently around his throat. The black hat was pulled down low over his brows, and he was casually smoking one of his thin cigars.
Jared looked her up and down, only his eyes moving, and in the early-morning darkness his expression was inscrutable. Pepe held the reins of a saddled mare. Both the mare and Jared's stallion pranced skittishly when Jared's deep laugh roared through the still morning.
"Where the hell do you think you're going in that getup?"
Lauren was stunned. She thought she looked quite fetching in the riding habit with its rich fabric and matching bonnet. "Th-this is a riding habit," she stuttered lamely.
"I know what it is," Jared said witheringly. "It's just going to be entertaining as hell to see how you get up on that horse in it, that's all." He chuckled.
Lauren looked at the beautiful sorrel mare. She seemed to be placid enough. Then she saw the saddle and swallowed convulsively.
"I would prefer a sidesaddle, Jared," she said with all the poise she could muster.
"You would?" he drawled, securing the cheroot in the corner of his lips. "Well, that's too bad, because all we have are western saddles. Can't you ride astride?"
Again the gauntlet was thrown down. "Of course I can," she retorted.
"Then go change into some of those new clothes you have and get your... rear... out here quick. We're wasting time." As she turned back to the front door, he added, "And do something with your hair. You can't wear a proper hat over that..." He made a descriptive motion with his hands around his own head. "And if you don't wear a hat, you'll scorch that buttermilk complexion of yours," he said scathingly.
Lauren lifted her heavy skirt and stumbled back into the house. Elena, who had been standing inside the door and had heard everything, sympathetically took Lauren's arm and led her back upstairs.
Silently the Mexican girl divested Lauren of her habit and redressed her in a brown split skirt which Lauren thought disgracefully tight across her hips and much too short. A white cotton shirt that buttoned down the front much like a man's went on next. Elena ignored the tears Lauren sniffed back and the quivering shoulders over which she slipped a soft leather jacket. Brown kid boots molded to Lauren's calves, and she was modestly grateful that they covered her legs to just below the knee where they met the bottom of the culottes.
The tears began to roll down her cheeks as Lauren thought of the humiliation Jared had subjected her to. He had stripped her of all dignity in front of the servants, her friends, and he had enjoyed it.
In silent sympathy, Elena removed Lauren's hat, took the pins out of the heavy black hair, and brushed it hurriedly. She braided the ebony tresses into one long plait that hung to Lauren's waist like a silken rope. Then she handed Lauren a brown, flat-crowned hat that looked like the one Jared wore, and Lauren placed it on her head, securing the thin leather cord under her chin. Finally Elena handed her a pair of brown kid gloves.
They had been in the room no more than ten minutes, but the transformation was astounding.
As they went down the broad stairs, Elena whispered, "Lauren, can you ride astride?" In the crisis, all formality was dropped.
Lauren swallowed hard. "I don't know. I've never tried." Elena looked at her compassionately, but saw only determination on Lauren's face. The tears had vanished.
Lauren strode out the front door without so much as a glance in Jared's direction. Expectantly she stood beside the sorrel mare. She placed one small, booted foot in the stirrup and grasped the pommel. Pepe cupped his hands and boosted her up by her other foot. She landed in the saddle with a plop, almost crying out in shock as her tender thighs slapped against the leather. She immediately composed her face and took the reins Pepe offered up to her.
Jared watched her with interest and smiled a sardonic, knowing smile. This was going to be some ride!
They didn't speak as they rode out the lane to the house and followed a road leading west out of town. Lauren scanned the countryside.
The sun was only now rising behind them, and its rays gradually illuminated the breathtaking scenery. Tall cypress trees lined the bends of the Rio Caballo, which paralleled the road on the right. On the left were gently rolling hills glistening with frost, which sparkled like a mantle of diamonds as the sun reflected off of it. Oak and elm trees were tinged with the russet tones of fall, and the cedars provided a dark evergreen contrast. White limestone formations jutting out of the hillside caught the morning sun and dazzled the eye.
They rode side by side, Lauren guiding her horse away from Jared's any time it came within a few feet of the larger animal. Her hat fell back against her shoulders, and Jared looked at the top of her head as the sun crowned it with highlights.
In spite of her declaration to the contrary, he knew she had never ridden astride before. She's got some spunk, he conceded silently. She was riding well, but God, she was going to hurt later on.
He broke the silence. "You don't look quite so comical now. Isn't that outfit more comfortable than that contraption you had on before?" he goaded.
"I'm fine, thank you, Jared."
_Damn!_ Always so cool. I'll just bet she's comfortable, he thought snidely. That cute little butt he'd noticed as she mounted the horse was probably screaming in pain. Why didn't she complain?
He deliberately spurred his horse and increased their pace.
Lauren did likewise in order to keep up with him, and the throbbing in her thighs and bottom was almost unbearable. But she would rather die than reveal her discomfort to that superior, arrogant, hateful man!
In spite of her mounting anger, she couldn't keep her eyes away from Jared. She wanted to hate him, but that was hard in light of his handsomeness. No picture she had ever seen of the dashing western men depicted anyone as exciting as Jared Lockett.
Try as she might, she couldn't forget how the sight of him last night with his chest bare and his hair mussed had caused her heart to pound. She had been terrified of opening that door, but it wasn't only his threat to break it down and rouse the house that had constrained her to obey. She had to admit that she was curious about what would happen when she did. Tremors had coursed through her body, setting up strange sensations as the topaz eyes traveled over her. Lauren almost imagined that Jared had been unnerved himself, but that would be out of character for him.
Objectively she studied horse and rider now. They moved together as one being. The stallion's honey-gold coat was almost the same color as the sun-gilded hair that covered Jared's chest.
The sun rose higher behind them, and they continued at a canter. Finally, when Lauren thought she couldn't keep from crying in pain much longer, Jared slowed down and led her off the road toward the swiftly running river. They reined in under the enormous cypress trees.
"I'm ready for a break," he said as he lithely dismounted. He led his horse to the stream and the palomino lowered his head to drink.
Lauren still sat on her horse. She had done exceptionally well in riding, she thought, but she was unsure about mounting and dismounting without a block to stand on.
Jared looked back, then walked over and offered his hands up to her. Painfully she pulled her leg from the far side of the saddle and timidly placed a hand on each of his broad shoulders. His hands encircled her waist, and he lifted her gently to the ground. She didn't look at him, but kept her head lowered until he released her. She felt his breath against her cheek. It was warm.
The mare needed no encouragement to join Jared's mount at the river. Taking a canteen from his saddlebag, Jared uncapped it and handed it to Lauren. She took a few swallows and gave it back to him. "Are you hungry?" he asked.
"Yes, a little." She was trying not to wince as she slowly and stiffly lowered herself onto a large, flat rock.
He pulled a few wrapped sandwiches out of the saddlebag and offered her one. "I made these this morning, and I can't vouch for their quality."
"It's fine," Lauren said, biting into the thick, dry ham sandwich.
"I know how to cook over a campfire, but a kitchen makes me nervous." His mouth quirked in the semblance of a smile.
She had never made small talk with him before. Since he had initiated the conversation, she was eager to continue it. Strained though it was, it was a beginning. "I like to cook. My guardians in Clayton had a cook/housekeeper who ruled her kitchen like a despot, but sometimes she would let me experiment with a recipe."
"Maybe Gloria will let you putter around in her kitchen. If you can stand all the kids underfoot."
"Who's Gloria?" she asked with interest.
"She's Rudy's wife." He saw her eyebrows raise in another question. "Rudy's my... uh... he's the foreman at the ranch. He and Gloria live there with... his mother. They have a baby every year or so." He smiled, and Lauren noticed that his eyes crinkled at the corners. She had glimpsed only a few unguarded genuine smiles at their wedding reception. They made him appear younger.
"Ben..." she hesitated over that name. Why should she? She continued doggedly, "Ben told me that one of your _vaqueros_ is an Indian."
Jared laughed. "You'd better believe it. Right down to his boots—which he usually forsakes for moccasins. Thorn is a Comanche. My father found him when he and a few Rangers raided a village to rescue some white captives. Thorn was half-dead with a gunshot wound and starvation, either of which should have killed him."
He swallowed the last bite of his sandwich and dusted his hands free of bread crumbs. "Anyway, Ben brought him back to Keypoint. Thorn was about eleven or twelve, I guess. It was long before I was born. He's been there ever since and is one of our best hands. He and Ben were very attached to each other. Thorn used to take us—Rudy and me—out on the plains. He taught us to stalk deer, read the stars, the weather, things like that."
Lauren was amazed, not only at the story, but at Jared's telling it. He had never been so loquacious. "Then you and Rudy grew up together?" she asked.
He was made uneasy by the question and answered laconically, "Yeah."
She tried again. "I think your land is beautiful, Jared. I really do." She said it impulsively, but emphatically.
He looked at her strangely, then away. He squinted his eyes against the late-morning sun's glare as they swept the panorama before them. "Yes, it is beautiful." He seemed entranced with the view for long, silent minutes, then stood up abruptly as if embarrassed by having spoken so freely. "If you need some privacy, go behind those rocks over there."
It took her a few seconds to comprehend his meaning. Then she lowered her head in confusion and stammered, "No, I... I'm fine."
"Then if you'll excuse me," he said with exaggerated gallantry. He loomed over her with characteristic impudence, and she blushed to the roots of her hair. He laughed out loud as he sauntered off, his spurs jingling against the small rocks lining the riverbank.
Lauren rewrapped the remainder of her sandwich and put it back into Jared's saddlebag, standing warily beside the huge palomino. A rifle was sheathed in a scabbard strapped to the back of his saddle. She had noted earlier that Jared was wearing a holstered pistol. Never in her life had she been around firearms, and they terrified her. Yet her husband seemed not to think anything at all about having them so close at hand.
When he had come back and gathered the reins of the horses, she commented, "I noticed some uprooted trees over there. Is someone clearing that land?"
"No, that's what the Rio Caballo can do if it gets angry enough. That happened two years ago and was one of the worst flash floods in recent history. Trees, cattle, houses, even bridges, were washed away."
"But the river seems so tranquil."
"Most of the time, it is. But if it rains hard enough and fills up the streams in the hills, the river can become an entity to reckon with. It takes back some of what it's given to the land." It was a poetic philosophy for this usually taciturn man.
Mechanically he gave her a boost up to her horse. She couldn't contain a gasp as she resumed that torturous position. To cover it, she asked, "What is my horse's name?"
"Name? I think the _vaqueros_ call her Flame because of her color."
"And yours?"
"This is Charger," he said proudly, patting the beautiful stallion's neck. "He's not his usual self this morning. I think... Flame has got him excited." The golden-brown eyes slanted toward Lauren and were rewarded with a high blush that stained her cheeks. Involuntarily her eyes were drawn to the part of the stallion that manifested his maleness.
_They say he is as big as a stallion!_ Lauren nearly choked when Elena's words once again came back to her.
Jared roared with laughter. "Don't worry, Lauren. He's too much of a gentleman to mount her _in public._ But I'll keep a tight rein on him, in case her attraction proves to be too much for him." When his laughter subsided he said, "He's quite an animal, isn't he? Ben gave him to me when I came home from Cuba." In an instant, Lauren's embarrassment vanished and was replaced by astonishment.
"You fought in the war?" she exclaimed in surprise.
He nodded curtly in affirmation. He obviously didn't appreciate reminders of the war. His eyes turned as hard and cold as agates. Distressed that she had spoiled his civil mood she turned her head away from him and became intensely interested in the horizon.
* * *
They rode for another hour before cresting a hill and reining in. On the other side, the land spread out like a large, shallow bowl, creating an incredible vista. A sizable herd of cattle, mostly Hereford, was grazing in the pasture and standing in the shallows of the river, which wound through the meadow. A few cattle, whose curly red coats contrasted with the colors of the verdant pasture, lazed in the shadows of the cedar trees dotting the valley.
Lauren was so taken by the sight that she wasn't aware of the thundering hooves approaching her and Jared until the horses were almost upon them. She screamed when she saw about ten riders, bandanas pulled up over their noses, hats pulled down low, brandishing pistols and rifles and shouting at the top of their voices.
They rode toward her and Jared pell-mell, leaning forward over their saddle horns. She whirled toward Jared in fright, and was astounded to see him adjusting his red scarf over his nose in the same manner as the bandits. Faster than she could follow his movements, he whipped the rifle from the scabbard and cocked it. Then, spurring Charger, he galloped toward the attackers giving a bloodcurdling Comanche yell.
The bandits fired their guns into the air and effortlessly surrounded Jared. He reined in on Charger so hard that the horse reared and pawed the air. Miraculously Jared maintained his seat.
Lauren's heart pounded in her ears in pure terror. Why had Jared left her? Surely he didn't think he could fight off all of these desperadoes single-handedly? After they killed him, what would they do to her? Flame pranced excitedly beneath her. She couldn't concentrate on holding the horse steady, so fixed was her attention on the scene being played out before her.
Jared slid off Charger, and the leader of the gang dismounted his own horse with a similar graceful motion. The others remained in their saddles and formed a tight circle around their leader and Jared, who faced each other squarely. Everything was deathly quiet.
Lauren was terrified as she realized that Jared and the bandit were going to draw on each other. The men stood a few feet apart, their legs wide, arms loose at their sides, every muscle tense. They were much alike in build and height. They stared at each other over their bandanas. Lauren held her breath.
They moved with lightning speed, and the pistols exploded simultaneously in the silence, the blast echoing off the surrounding hills.
# Chapter 10
Either the sudden crack of the pistols or Lauren's kneejerk reaction to it startled Flame. She bolted and raced uncontrolled across the pasture. Lauren was too frightened to scream. Only instinct forced her to hold on to the saddle as the ground rushed under her in a blur. The thudding of Flame's hooves was joined by another's, but Lauren didn't risk turning around for fear of unseating herself and greater fear of what she would see in pursuit.
From the corner of her eye, she saw a flash of blond mane seconds before she felt an iron arm grab her around the waist and drag her from Flame's back. The mare streaked out from under her. Lauren's legs danced in midair like a puppet's as she was held against the heaving sides of the horse. She closed her eyes tightly. Her arms closed around the waist of the man in the saddle as he hauled her up in front of him.
After what seemed like an eternity, they began to slow down, and then were completely still. The heart in the hard chest beneath her head beat loudly against her ear. Lauren raised her head and looked into amber eyes over a red bandana.
"Lauren?"
She recognized the voice and slumped against Jared in relief that they were both still alive. He tightened his arms around her. She was content to leave her head against his chest with her eyes closed. Sounds of approaching horses invaded her serenity. She had forgotten their attackers!
"Is she all right, Jared?"
"I think so. Just a little shaken up." Jared's reassuring voice reverberated in her head. "Lauren," he repeated quietly. Reluctantly she opened her eyes and raised her head.
Jared lowered his bandana and looked down at her. Was it concern she saw in his face? Charger impatiently tossed his head and brought her out of the hypnotic state that this unplanned closeness to Jared had induced.
She looked at the other rider. His features were much like Jared's, except he was darker of complexion and hair. He, too, had pulled the bandana from his face and was smiling at her congenially. Shyly her gaze traveled around the semicircle of _vaqueros_ nearly surrounding them. They appeared not in the least malevolent, but only curious and a trifle chagrined.
"Lauren, this is Rudy Mendez and these are Keypoint's _vaqueros_ —some of them. Did we frighten you?" Jared's tenderness amazed her.
She nodded dumbly; then, remembering the draw contest, said, "But I saw you shoot each other." Her lips trembled.
Rudy laughed and white teeth flashed in his swarthy face. "Sometimes I think Jared needs shooting, but we try to miss when we're only playing." He winked at her. "Rudy Mendez at your service, Mrs. Lockett. Welcome to Keypoint. Jared should have warned you of our rough games. Our welcoming committee got overzealous today. Can you forgive us?"
His smile was so engaging that Lauren smiled tremulously and murmured, "Yes. I'm sorry I caused such a fuss." Suddenly she was conscious that she must look a fright. Her hat had been torn from her head and was resting on her shoulders. Her hair had loosened from the tight braid, and wisps of it blew around her face. Worst of all, she was being held in a most unladylike way across Jared's lap, his strong arms supporting her. When she realized she still had her arms wrapped around him, she withdrew them instantly.
But she hadn't counted on Charger choosing that time to stamp the ground impatiently. She came close to being unbalanced most ignominiously. Grasping desperately for whatever handhold was available, her fingers groped around the tight bulge between Jared's thighs. His expletive echoed in her ear just before he hissed unevenly, "For godsake, don't do that. Put your arms around my waist and keep them there." She obeyed. Luckily Charger had continued his prancing, so no one had witnessed her folly. With a flick of his wrist, Jared brought the horse around. He was unaware of the silly grin on his face that was so obvious to everyone else.
The _vaqueros_ and Rudy stared unabashedly at Jared's bride. She was so beautiful! When word had gotten back to the ranch that their Jared was hurriedly marrying a girl from back East, there had been some unflattering speculations on her charms. Now they saw why their boss had been so eager to get this girl into his bed.
Lauren was becoming increasingly uncomfortable under their stares, and Rudy noticed it. "Doesn't anyone have work to do?" he asked the _vaqueros._ They caught his drift and one by one tipped their hats to Lauren and pulled their horses around, heading back toward the herd. "I'll fetch your horse, Mrs. Lockett."
"No, Rudy," Jared said. "I'll take her the rest of the way with me. That mare is probably into her second bag of oats by now."
"No trouble, Jared," Rudy taunted. Jared's preference for the current arrangement wasn't lost on him.
The _vaqueros_ whistled and hooted back at them as Jared nudged Charger with his knee and they started across the pasture. "Dammit," Jared muttered close to her ear. Oh, God, he groaned silently as she shifted her hips, looking for a more comfortable position.
Rudy rode up beside them and watched as Lauren eased herself away from Jared's chest. His arms seemed reluctant to relieve their firm hold on her, but Rudy saw the muscles slowly relax. There appeared to be a tangible tension between these two.
Jared had cursed this Lauren Holbrook when he came to Keypoint after Ben's funeral. He insulted her heritage, intelligence, and morality. Rudy had teased him about her then but, to his surprise, Jared informed him he hadn't even seen the girl.
"Then how do you know she's so disreputable? Anyway, I thought you said you had gone to Austin to pick her up."
"I did!" Jared exclaimed angrily. "But I... oh, hell. Just drop it, will you?"
Rudy knew better than to press further. Scarcely a week later, Pepe came out to the ranch with supplies and the news that Jared was marrying the very girl he had claimed never to have seen and yet had called a "conniving little tart." Well, Rudy thought, he's seen her now. But good.
Gloria and Mamma were thrilled about the marriage, so Rudy had kept his reservations about its future to himself. If Jared had a brain in his head, which Rudy sometimes doubted, he would keep this woman under lock and key. She was gorgeous, and a lady, and, by the tortured look on Jared's face, he wasn't oblivious to her charms. He chuckled to himself. Good for you, Lauren.
He spoke out loud, "Gloria, my wife, is looking forward to meeting you, Mrs. Lockett. She's been more excited than the kids."
"Please call me Lauren. How many children do you have, Mr. Mendez?"
"I'm Rudy." He grinned. "Well, let's see," he said, silently ticking off names on his fingers. "Jared, help me. Is it six or seven?"
Jared snorted. "I think when I left it was six, with number seven well on the way. Of course, if Gloria has delivered early, I'm sure you've already started number eight."
"Don't be crude in front of your bride," Rudy scolded, but his eyes were twinkling. He looked several years older than Jared, and there was something familiar about him.
Lauren was trying to decide what it was when she saw the ranch house. It was large and sprawling, a one-story building of limestone blocks with a cedar shingle roof. Four cedar posts supported the roof over a wide front porch running the breadth of the house. There were several outbuildings, all built of the same materials. Well-constructed corrals dotted the compound. Laughing, squealing children played about the yard, the only area where grass was growing. The two horses were led into the yard and halted before a hitching post. Rudy dismounted.
Jared and Lauren remained as they were for a moment before Jared looked down at her and said softly, "You were frightened. I'm sorry."
His face was very close to hers. There was no measurable expression on it, but the voice was so unlike any he had formerly used that she was held spellbound. Emotion swelled her throat, but she said, "It's all right. I've recovered." Again she witnessed that quick lifting of the corners of his mouth that might be considered a smile.
He took a firm hold around her waist with one arm and lowered her gently to the ground. On her descent, his hand slid from her waist to her armpit. His fingers lightly grazed her breast.
They were both shaken by the contact. Lauren tried to recover herself by straightening her clothing. Jared swung down from his horse as if irritated. Rudy, who had seen the whole thing and noted the reactions, caught Jared's eye and winked slyly. Jared glowered at him.
The door of the house was flung open and Gloria flew out of it. The children, who had spotted their father and Jared, rushed over to them, grabbing them around the legs, pushing and shoving and vying for their attention.
"Lauren, Lauren, welcome. I'm Gloria."
The uninhibited Gloria hugged Lauren in a warm, encompassing embrace. Lauren laughed at the Mexican woman's exuberance. She was older than Lauren and, though she lacked the classic beauty of Elena, was very pretty. Her olive complexion glowed with health and happiness, which were also reflected in her dark eyes. Like Elena, she was expecting a child, though her time wasn't as imminent. She wore a dark skirt, with a full but tailored shirtwaist over it in deference to her condition. Her glossy black hair was pulled back from a center part into a chignon at the back of her head.
"Thank you, Gloria. I'm happy to be here. I hope our visit hasn't put you to any trouble."
"No, no. Besides, this is Jared's home. I'm just glad the scoundrel has finally brought a wife to it. I've been after him for years to get married. Now I'm glad he waited. I bet he is, too." She laughed and pinched Jared on both cheeks, having to stand on tiptoe to reach him. He leaned down and kissed her soundly on the mouth, his arms encircling her waist.
"If Rudy would let you out of bed long enough, we'd run off together, wouldn't we, Gloria?" He hugged her tight and patted her bottom. She pushed him away with feigned annoyance.
"You! Here with your bride, hugging a fat, pregnant lady! Shame on you, Jared Lockett." She grinned at him affectionately.
Such familiarity appalled Lauren, but then, what about this culture hadn't shocked her? Rudy shepherded the children toward her.
"Lauren, may I present James, John, Maria, Anna, and Lucy. Who's missing?" he asked.
One of the boys piped, "Consuelo. She's little. She's in the house taking a nap," he explained to Lauren.
"Children, this is Jared's wife, Lauren."
Lauren looked at the faces turned up to her and met five steady, curious stares. One of the girls, five-year-old Anna, whispered reverently, "You're beautiful."
Lauren smiled. Leaning down to the child's level, she said, "I was going to say the same thing about you."
"Lunch is ready. Come on in." Gloria took Lauren's arm and hurried her toward the house. Lauren winced with each painful step as the abused muscles in her thighs and bottom throbbed.
They stepped into a large room that was as wide as the house. At one end was a stone wall with a fireplace large enough for a man to stand in. At the other end was a dining alcove containing a table, sideboard, and china cabinet. In between was a comfortable living area, which exuded an aura of warmth and welcome. There were several long sofas, easy chairs, and ottomans, colorful rugs scattered over the quarry tile floors, paintings depicting Western culture, bookcases filled to capacity, and a roll-top desk. Lauren loved it.
Two hallways led off the big room, and she assumed they led to bedrooms. Another door off the dining area, she knew, must lead to the kitchen. Gloria beamed as Lauren complimented her on the decor.
"I'll show you to Jared's room. I've tried to tidy it up for you, and add a few feminine touches. You know how barren a bachelor's room can look."
Lauren had no idea how a man's bedroom looked. She blanched at the words "Jared's room."
They walked down the hall and Gloria opened the door to a large bedroom with wide windows opening onto a view of rolling hills and the river in the distance. It was magnificent.
The furnishings consisted of a chest of drawers with a shaving mirror on it, an overstuffed chair with an ottoman, and a wardrobe which Gloria opened to show that she had pushed all Jared's clothes to one side.
Jared's clothes!
A double bed dominated the room. A small table stood beside it.
"I moved this in here for you, Lauren," Gloria said. "It was all I could find in the storage shack." She indicated a small vanity table with a washbowl and pitcher on it. The mirror was wavy and cloudy. "Pepe will pick up something else for you and bring it out on his next trip." She took Lauren's dismay for dislike. "Is it all right?" she asked shyly.
"No!... I mean yes. Everything is beautiful, Gloria. You have gone to a lot of trouble, and I appreciate it more than you know, but..." Her voice trailed off when she didn't know what else to say. She looked helplessly toward her husband. He and Rudy stood in the doorway. Jared seemed as disconcerted as she.
Rudy leaned against the jamb with his arms and ankles crossed. He was enjoying himself immensely. A mischievous light glinted in his eyes as he looked at Jared and said, "Gloria and I will leave you two alone so you can wash up. Then we'll eat. I know you must be starved, but take all the time you want."
"I... uh... think I'll go out to the bunkhouse and say hello to some of the hands. I'll wash up out there." Before anyone could object, Jared rushed out of the room, leaving in his wake a very surprised Gloria, a mildly surprised Rudy, and a relieved Lauren.
Damn! Jared cursed as he crossed the yard. He hadn't even thought about the bedrooms at the ranch. Rudy and his brood used every goddam one of them except his. He liked his room. He didn't want to give it up, but there was no way he could ask the romantic Gloria to move his wife—his bride—out of his room.
Even more nerve-racking to consider was spending the night in the same room with Lauren. He had made a vow to himself not to touch her. But it would take a blind, feeble ninety-year-old monk to stay the night in such intimate confines with Lauren and not—
He'd have to think of something. Fast.
When, after a short toilette, Lauren went into the large room for lunch, she saw a woman sitting alone at the dining table. She was beautiful and of Mexican heritage, and Lauren was struck by her bearing. She held herself straight and proud, yet serene.
She was small, her figure dainty. Her black hair was streaked with gray and pulled back into a glossy knot on the nape of her neck. She wore a high-necked black dress. There was no jewelry or other adornment to relieve its severity. Her face had cameo perfection. Smooth, well-shaped brows framed dark, sad eyes. Her nose was straight and narrow. Above a delicate chin was a sweet, well-shaped mouth.
Her smile was genuine as Lauren shyly walked toward her. "Lauren Holbrook Lockett. I am Maria Mendez. Welcome to Keypoint."
"Señora Mendez, it is a pleasure to meet you. You are Rudy's mother?"
"Yes," she answered abstractedly as she studied Lauren's face. "You are just as beautiful as he said. He was very eager to have you here. I think he did the right thing."
"You mean Ben?" Lauren was puzzled as to how this woman knew so much about Ben's plans for her.
She nodded. "He told me all about you the last time I saw him."
She seemed about to say more when Jared came through the front door carrying a Mendez offspring under each arm and one on his shoulders. The one on top was holding on for dear life with tight fistfuls of Jared's hair. All were laughing and shouting as the other children ran in behind them.
"Do it to me, Jared!"
"No, to me."
"Please, Jared, me next."
Gloria came through the door that led to the kitchen clapping her hands. "Leave Jared alone. He's just arrived. There will be plenty of time to play later. Come to the table, and I'd better not see any dirty hands."
The children reluctantly released their hold on Jared and took their places at the table. Jared raked his hair with his fingers and tucked in his shirttail, which had come out during the scuffle.
Lauren would never have imagined that Jared could enjoy child's play, but he was laughing as he smoothed the cotton shirt over the muscles of his wide chest. When he sucked in his stomach, in order to shove his shirttail into the front of his pants, Lauren's eyes flickered lower. She tried to swallow the increasingly familiar congestion in her throat.
Jared's head came up with a snap and his eyes locked with hers as if she had called out to him. Her hair had been rebraided. Jacket, gloves, and hat had been discarded. The shirt molded against her breasts, making them look softer, more touchable than ever. The tight fit of the split skirt announced her femininity like a banner. Jared wiped damp palms down his thighs and looked away from that intriguing spot where the legs of her culottes came together. As he moved farther into the room, he cursed his physical susceptibility.
His eyes lit up with affection as he sighted Maria Mendez at the table, and softened as Lauren had never seen them do before. "Maria, how are you?" He came around the table and took the lady in his arms, giving her a big hug. He had never addressed Olivia in such a warm manner.
"Jared, I have just met Lauren, and I think your bride is lovely."
Jared darted a glance at Lauren and muttered under his breath, "Yes... well, thank you."
Rudy came in and Gloria carried the last platter from the kitchen and placed it on the table. The meal began. It was a loud, confusing time, but Lauren enjoyed it immensely after all of the lonely meals at her father's house, the stuffy, dull dinners at the Prathers', and the silent, tense meals at Olivia's table.
Everyone chattered, relating things of great importance or no importance. Rudy told them about the incident of the "gunfight." As the children laughed at Lauren's mistake, Gloria and Maria chided the men on their recklessness.
The table was crowded despite its length, and Gloria had assigned Lauren and Jared side-by-side places. Their elbows were constantly touching as one or the other lifted their cutlery or glasses. Several times, their knees or thighs would press together under the table. If anyone noticed two forks suspended motionless en route to two surprised mouths, they didn't mention it.
Rudy noticed, and chuckled each time. They were fighting it, all right. It was going to be interesting to watch this—the downfall of Jared the Great—at the hands of a slip of a girl who barely reached his shoulder.
When the meal was over, the men left to attend to the never-ending work of running a ranch. Lauren watched as they strolled companionably toward one of the corrals. They were dressed almost identically and walked with the same graceful swagger. Each sported broad shoulders, a narrowness of hip, and long, lean legs. Each had buckled on flopping chamois chaps and pulled on leather gloves. "They look alike from the back," Lauren mused aloud.
Gloria was standing at her shoulder. "Yes, I think anyone could tell they are brothers."
Lauren whirled her head around to Gloria in astonishment. "Brothers?" she gasped.
Gloria was nonplussed. "Why, yes. I thought you knew that Rudy is your brother-in-law." She was amazed that Jared hadn't told his wife of the kinship.
"Mrs. Mendez and..." whispered Lauren, still trying to fit the puzzle together.
"Maria and Ben," Gloria finished for her. At Lauren's shattered look, Gloria's features closed coldly and she said, "Don't judge them too harshly. They loved each other very much. Maria has lived here with Ben for almost forty years."
"But... but Olivia was his wife," Lauren protested weakly. Ben had lived in adultery? _Her_ Ben?
"You've been with Olivia," Gloria was saying. "You know what kind of woman she is. Ever since Ben brought her here from New Orleans, she made his life miserable. She insisted on living in Coronado and refused to have anything to do with Keypoint. She assumed the responsibility of the bank. Maria was the daughter of one of the first _vaqueros_ Ben hired to help him run the ranch. He fell in love with her on sight, and she with him."
Yes. Lauren could see how that could be. The gentle, serene Maria and the robust, virile Ben. They would have complemented each other. He had given her loving protection and she had given him comfort and a son. Happiness. Where one can find it. Was that wrong? A few weeks ago, Lauren would have been scandalized to hear of such a sordid arrangement, but somehow, now...
Was this land to rob her of her convictions, too?
"The two sons of Ben Lockett have a deep affection for each other," Gloria said, glancing out the window as the brothers rode away together. "Rudy accepts the fact that he cannot bear the family name, and that he is not the legal heir. Ben loved him and he knew that. Ben's will made it clear that as long as there is a Keypoint, the sons of Rudolfo Mendez will share an equal partnership with the sons of Jared Lockett.
"It's sad, isn't it?" she went on when Lauren remained silent, watching the two men on horseback disappear over the horizon. "Maria could not even go to Ben's funeral. Nor could his first son. But Ben came to the ranch just a few days before he died. He spoke of you, Lauren, and promised to bring you to us as soon as possible. That was the last night he spent with Maria. I hope they made love all night long. He had been gone for over a month, so if I know old Ben, they did."
"I... I'm sure they did," Lauren said, blushing.
Gloria took her bowed head as a sign of fatigue. "You should have stopped my ramblings. You're probably tired. Why don't you take a nap?"
"I am a little tired," Lauren confessed. "But thank you for telling me the circumstances. I understand things better now." Gloria's face softened and she leaned forward to kiss Lauren lightly on the cheek.
Lauren went into the bedroom that belonged to Jared. It _did_ belong to him. His ownership was stamped on everything. As she moved about the room, Lauren could feel his presence like a palpable force. Disrobing slowly, she almost felt as if she were stripping before the man himself.
She stretched out on the bed, having the strangest sensation that Jared was on the bed, too. She moved her hand over the heavy spread and wondered how many nights he had lain in this exact spot. It was an unsettling thought. Her eyes closed and she dropped off into a deep sleep, imagining that amber eyes were watching her.
# Chapter 11
Lauren took a bath just before dinner. Gloria told her Ben had piped in water from a cistern several years ago, so the kitchen and two bathrooms had running water. There was, however, no water heater. A small brazier was stationed in a corner of each bathroom, with a large copper kettle kept constantly simmering.
"The only rule," Gloria said, smiling, "is that when you empty the kettle, you refill it for the next person. Oh, and Jared asked me to give you this," she added as she handed Lauren a brown bottle.
"What is it?" Lauren asked.
"Liniment," was Gloria's amused reply.
Pepe had arrived earlier and deposited her luggage in the bedroom. Jared's bags stood beside hers. One of them, she noticed, had been opened. What would she do if he followed her into the room tonight? Apparently the Mendezes expected him to. She calmed her nerves, and went into the dining area.
Strangely, and in contrast to the noon meal, everything was quiet. The table was candlelit and set with china and crystal, replacing the pottery which had been used at the earlier meal.
"Where are the children?" Lauren inquired as she brought in a tray laden with serving dishes.
"Bathed and in bed," Gloria sighed. "A quiet dinner is the one luxury Rudy and I allow ourselves."
Jared and Rudy came in from the front porch where they had been enjoying glasses of whiskey. Jared crushed out his cheroot in the nearest ashtray. He was washed and brushed and wearing a clean shirt. He must have taken it out of his suitcase while she was in the bathroom.
Rudy went to Gloria and took her in his arms, kissing her in a thorough manner that stunned Lauren. They murmured to each other privately, oblivious to the other two people in the room.
Jared stood looking out the window, his thumbs hooked in the waistband of his tight pants. Lauren stood in the center of the room feeling self-conscious and nervously fingering the watch on her breast. Seeing Rudy and Gloria be so affectionate made her long for such closeness with Jared. Something inside her wanted to cry out to him, to move toward him.
She was about to take one tentative step in his direction when Maria, who had just entered, said, "What is this? The newlyweds acting like strangers, and the old married people loving like doves!"
"I was about to remark on that myself." Rudy crossed to Lauren and took both her hands in his. "If that sorry husband of yours isn't going to kiss you, I will. Welcome, Lauren, into our family." He kissed her lightly on each cheek.
There was a tense, expectant silence among them. Jared, however, seemed unaffected and went to the table. With a curt nod of his head, he offered Lauren the chair he had pulled out for her. With flaming cheeks, she walked toward it. The others, awkwardly quiet, followed her.
The conversation was easy; the Mendezes were able to talk freely about anything. Some of their topics embarrassed Lauren. Castration, branding, breeding. Sybil would have had vapors for a month.
Lauren and Gloria did the dishes. Gloria adamantly refused Maria's offer to help, but secretly confided to Lauren that she wished she had someone to assist her with the large household and the children.
When they joined the men and Maria in the large living room, Lauren was glad to note that Jared wasn't drinking. His face had lost the hostile arrogance that it usually bore like a shield. He seemed relaxed and... yes, happy. He was a different person from the one who lived in that beautiful but cold house in Coronado.
He was stretched lazily in a chair, his long legs extended in front of him. She traveled the length of them with her eyes, much as she had done the first time she had seen him, in the back of the wagon. The bulge at his crotch still intrigued her. She looked away quickly.
Gloria sat down beside Maria on one of the sofas and launched into a tale about one of the children's escapades. The men were talking ranch business. Left to her own devices, Lauren went to one of the bookcases. She took her eyeglasses out of her skirt pocket and put them on, leaning down to inspect the loaded shelves. She found several volumes that were interesting to her and withdrew them, taking a seat in a chair under a lamp.
She soon became engrossed in a book which described the lifestyle of the Comanche as witnessed firsthand by a Texas Ranger. She was vaguely aware of Rudy getting up and going toward the kitchen asking if anyone else wanted more coffee.
She read a few more pages before something compelled her to raise her eyes above the rim of her spectacles. With a jolt, she realized Jared was staring at her through the blue smoke of his cigar. He wasn't wearing his usual cold, implacable expression. He was looking at her almost tenderly, with a slight smile on his lips.
His gaze traveled down to her hand, which was absently fondling her watch. Why did those slim fingers always fuss with that watch? What significance did it hold for her? He really knew very little about her, he realized.
When he had held her against him today on his horse, it had taken every ounce of his willpower to keep from pressing a kiss on the nape of her neck where tendrils of ebony hair lay. He could have ridden for a hundred miles with her on his lap. What sweet agony it had been every time her body had been jostled against his.
The memory of that brief touch to her breast was vivid in his mind. He convinced himself that he hadn't meant to do it. It _was_ an accident. But his curiosity had been gratified to feel a firm fullness under his palm. If only he could see...
Jared stiffened when it occurred to him where his thoughts were leading. Just remember how she got here, he cautioned himself. She's not the innocent, frail flower she looks with those ridiculous eyeglasses on. She's a scheming little bitch. What I need is a rousing roll in the hay with some obliging whore. A pair of eager, parted thighs and it wouldn't take long to get this preacher's kid out of my mind.
Lauren saw that the soft look had fled his face. Dispiritedly she put the book away and took off her eyeglasses.
"Goodnight, everyone," Maria stood as Rudy came back into the room carrying a mug of coffee. "I'm going to bed and leave you young people alone. Lauren," she came and stood in front of the girl, cupping her face between her soft hands, "I'm so glad you are here with us. Ben would have loved to have been here. As a matter of fact, I think he was." She kissed Lauren on the cheek and walked slowly down the hall.
"She'll never get over losing him," Rudy said quietly, as the slender figure moved out of sight.
"Yes," Gloria sighed. "I'm going to bed, too," she said, standing and going to Rudy's chair. She leaned on the arms of it, her gaping blouse giving him an unobstructed view of the generous breasts barely contained within her chemise. They kissed long and hard, his hands resting lightly on her thick waist.
"Get my place warm for me. I'll be a few more minutes," he said when she finally pulled away.
"Don't keep Jared up late. Remember, this is his honeymoon," Gloria teased, rolling her eyes at Jared.
He shifted uneasily in his chair. "That's all right, Gloria. I'm joining a hot poker game over at the bunkhouse. Some of the boys invited me, and you know these things can go all night sometimes." He tried to sound jocular, but failed.
"A poker game!" Gloria exploded. "What in the world—" A sharp look from Rudy halted her. She looked at Lauren, who was standing stiffly in the shadows next to the door.
"Lauren, I'll see that you are settled for the night," Gloria said with commiseration.
"Goodnight, ladies," Rudy said gently.
"Goodnight, Rudy. Goodnight, Jared," Lauren whispered hoarsely.
"Goodnight, Lauren," Jared said nonchalantly as he studied a fingernail.
Gloria shot her brother-in-law a murderous look before she went with Lauren down the hall.
* * *
Lauren had been tossing on the wide bed for several hours. She had heard Rudy going into the room across the hall and Gloria's welcoming murmur as he closed the door behind him. That had been a long while ago, and the house was silent. The moon was bright, illuminating the bedroom in an ethereal glow.
She started when she heard footsteps stealthily approaching the door to the room. She turned to the far side of the bed and feigned sleep as the door opened.
Her whole body tensed as Jared walked into the room. He picked up one of the bags left standing in the middle of the floor, for she heard the buckle on it jangle. There was a long pause, and then the bag was lowered again to the floor.
He moved on Indian-trained feet to the side of the bed. She could smell the faint aroma of tobacco and the musky scent of leather as he leaned over her. He stood there for torturous minutes, immobile and silent. Lauren was aware of each breath as he inhaled and exhaled rhythmically. She felt a butterfly touch made by strong, lean fingers against her cheekbone. Her throat constricted and her heart pounded as if it would burst. Finally he turned and went back to the bags, picked them up, and then carried them from the room, closing the door softly.
The scents remained to tease her senses.
* * *
"Gone?"
"Yes, Lauren. He left before dawn this morning." Gloria looked piteously at her new sister-in-law. Her heart went out to the girl whose husband treated her so abominably. When Rudy had joined her in their room the night before, they had speculated on the strange relationship between Jared and his bride. It appeared to be a marriage of convenience, but Rudy couldn't guess the reasons behind it. He only knew that they must be damned essential to Jared's well-being to have forced him into any marriage.
"Wh-where did he go?" Lauren's heart had plummeted when Gloria told her that Jared had ridden out with enough supplies in his saddlebags to last him several days.
"He went to check out some problems on the far western side of the ranch. There have been reports of marauding bobcats killing our cattle. Don't worry about him, Lauren. He'll be back soon, I'm sure." She didn't tell her that any one of the _vaqueros_ who worked at Keypoint could have handled this job.
"Yes, I'm sure he will," Lauren mumbled. A few days ago, she had dreaded the sight of Jared. Now the prospect of not seeing him every few hours seemed dismal. What was the matter with her? She felt rejected, abandoned. This was supposed to be her honeymoon!
The liniment hadn't worked out all the soreness in her legs and hips, but when Rudy asked her if she would like to ride with him that afternoon, she agreed. The ranch life was exhilarating. She wanted to absorb every aspect of it, savor its energy, its vigor. This was Ben's land. Keypoint was his conception. And like him, it was vital and alive. At Keypoint Lauren felt even closer to the man who had brought her here wanting her to become a part of it.
As Jared's absence lengthened, it became her habit to ride with Rudy or some of the _vaqueros_ in the afternoons. In the mornings, she played with the Mendez children, who were teaching her Spanish. They would burst into peals of laughter when she got a word wrong or had trouble pronouncing it. Sometimes she read to them before bedtime. Lauren also loved being with Gloria, and the two women soon shared a deep friendship. Lauren had had so few women friends in her life that she treasured this new relationship.
She relished her visits with Maria Mendez, too. The older woman spent a great deal of time secluded in her room, as had been her inclination ever since Ben's death. Even when she was with the rest of them, she seemed withdrawn, dreamlike, separated from reality. Lauren thought she looked at peace during these times, and rather imagined that she was communicating with Ben on a plane where no one else could intrude.
The first week passed quickly. Lauren's heart raced each time she heard pounding hooves, but she continued to be disappointed. She kept searching the horizon for signs of the large palomino and his mount in the wide-brimmed black hat, but to no avail.
She acclimated to the ranch life so enthusiastically that Gloria and Rudy were amazed. The _vaqueros_ would tip their hats to her with a humble, "Good morning, Mrs. Lockett," to which she replied, calling them by name. They all respected and liked her.
One day she was timidly approached by a Mexican cowboy with dark, dancing eyes. "Señora Lockett, I am Carlos Rivas, Elena's husband." He smiled shyly and twisted his sombrero in his hands.
"It's so nice to meet you, Carlos!" Lauren cried. "How is Elena?"
"She thinks the baby come soon."
"Please let me know when it does. Tell her hello for me."
_"Sí, señora."_
She had also seen the enigmatic Comanche, Thorn. He had never spoken to her, but he tipped his hat whenever they met. His expression never changed, but Lauren felt that his eyes missed nothing. She hoped his assessment of her was favorable. His dark, austere face and long braids intimidated her, though she felt instinctively that he was a friend.
Ten days after his departure, Jared came home. He rode in one evening just before dinnertime, looking tired and dirty as he clumped through the large front door. From his boots to his hat, he was covered with a fine layer of trail dust.
"Well, look what just dragged in." Rudy stood with his hands on his hips eyeing Jared as if he were something distasteful.
"Am I in time for dinner? I've ridden like hell for the past few hours. I'm sick of camp food," he said sheepishly.
"We'll wait for you, Jared. Just go wash up first and turn Charger over to someone in the stable." Gloria spoke to him coolly, and he looked in turn at the faces staring at him reproachfully. Lauren didn't look at him at all. Her dark head was bowed as she stared down into her plate. _He_ felt like the outsider.
"I'll be right back," he muttered as he stepped out the door.
Lauren's heart was in her throat. He had barely glanced in her direction, yet she felt his presence in the room as strongly as if he had touched her. Conversation went on around her as everyone waited patiently for Jared to come back.
He returned, having discarded his chaps, leather vest, bandana, and spurs. He wore a clean shirt and his hair was still damp from the recent dunking in the bunkhouse washbowl.
He crossed the room to Maria and kissed her proffered cheek. Her greeting was the only warm one he received. His kiss intended for Gloria's cheek landed somewhere in the air as she turned away quickly, and Rudy shook his hand with none of the usual banter between them. He took his seat beside Lauren and only then did he turn to her. "Hello, Lauren."
"Hello, Jared. Was your... trip... successful?"
"I shot two bobcats and visited some of the nesters we allow to use our water. It was basically uneventful."
There seemed to be nothing more to say, and everyone commenced eating. The food stuck in Lauren's throat. She was jittery and breathless, and when one of the children called from the bedrooms, she jumped up to go to him, anxious to get away from the dominating individual beside her.
"I don't know what we'll do when Lauren has to go back to Coronado with you, Jared. We have all come to love her so much." Gloria wanted to slap her brother-in-law as he shrugged indifferently. She continued undaunted, "The children adore her and she has been such a help around here, hasn't she, Maria?"
"She's a wonderful girl, Jared. You're lucky that Ben brought her here." Jared growled deep in his throat.
Rudy chuckled. "I know at least a dozen of the hands who would love for you to have a fatal accident, Jared. They'd whisk her away in no time flat."
Jared scowled at him. "When the hell has she been around any _vaqueros_?"
"Every day when she goes out riding. You'd be amazed how eager they've been to answer her questions. She's a fast learner."
"I'll bet," Jared grumbled around the food in his mouth.
Lauren came back and sat down. Rudy stood up for her; Jared stared sullenly at the bowl of chrysanthemums in the center of the table.
"One place Lauren hasn't seen is Pecan Creek. You really should take her up there before the weather turns too cold."
Gloria took her cue from Rudy. "Why don't you go tomorrow? You could take a picnic and enjoy being alone."
"I don't think—" Lauren began.
"Not tomorrow," Jared interrupted. "I've got too much work to do around here."
"Nonsense," Maria interjected. She wasn't going to sit idly by and see one of Ben's best plans thwarted by Jared's obstinacy. "You've been gone for almost two weeks. You deserve a day off. You'll leave first thing in the morning, and I'll supervise packing your lunch. I know just what you like.
"Be sure to take a gunnysack and gather some pecans for me. We'll be needing them for Thanksgiving and Christmas baking. Remember when I used to send you and Rudy there each fall? Ben would give you a penny for each nut you brought back. Those were such happy days," she said wistfully. She cleared her throat. "Yes, you'll go tomorrow."
The matter seemed settled. Jared glared at Rudy with a withering look. Rudy smiled back, the epitome of innocence. Maria and Gloria smiled at each other conspiratorially. Lauren fumbled with her watch, which trembled on her breast.
* * *
"Goddammit, what do you expect me to do, Rudy? I've tried every argument. They want that power plant and we want a railroad. They've got us by the balls."
"I don't know, but you've got to do something! Ben would have gone to war before letting those sonsofbitches on his land."
The ladies in the main room looked at each other as the voices, raised in frustration and anger, reached them from the front porch where Jared and Rudy had retired after dinner to smoke a cigar. Maria put aside her mending, Gloria paused in stringing some beads she had promised Lucy, and Lauren dropped the book she was reading into her lap. She alone knew what subject had brought such an outburst from the brothers.
The voices outside returned to normal, and the women resumed their activities. Every once in a while, a word or phrase would carry back to them, punctuated in emphasis or urgency.
When Rudy and Jared came back inside, their mouths were set and grim. Each looked immediately toward Lauren, Jared with hostility, Rudy with something akin to pity.
So, Lauren thought, Jared has told his brother the circumstances of our marriage. For a moment, she feared that Rudy's opinion of her would alter, that he would condemn her for the decision she had made. But one look at his face—open, friendly, compassionate—assured her that this wouldn't be the case.
"If we have to go to Pecan Creek tomorrow, be ready early." Jared stalked out the door after addressing Lauren with all the enthusiasm of a prisoner awaiting execution at dawn. No one challenged his sleeping arrangements this time.
Gloria and Rudy went to their room. Lauren placed her book back in the shelf and, taking off her eyeglasses, walked to the window. She saw the tall, lean figure striding toward the bunkhouse with his broad shoulders hunched defensively.
She didn't know that Maria was behind her until the woman placed a reassuring arm around her slender waist. "Ben fought his love for me like a man possessed, Lauren. They are both tough, strong men. Tenderness doesn't come easy to Jared. Or even kindness. Be patient with him."
Lauren couldn't speak for fear of weeping. She turned to Maria and hugged her quickly before seeking the privacy of her room. Jared's room.
* * *
Gloria helped Lauren braid her hair in the style now familiar to her. She wore the same suit she had worn on the morning she and Jared left Coronado for Keypoint. The ensemble that had seemed scandalously indecent at the time now felt quite comfortable. She had become accustomed to many changes in her life.
Maria was in the kitchen making good her promise to prepare their lunch. Jared strolled in and without a word handed Lauren a dark blue bandana. She looked at it and then at him with puzzlement.
"It's clean," he said testily. "I borrowed it from one of your many admirers and washed it myself. You may need it today."
She took the scarf and folded it into a triangle. Placing it around her neck, she tried to tie it as the _vaqueros_ wore theirs, but her fingers were unaccountably clumsy.
"Here," Jared said in exasperation, batting her hands away. He stepped closer to her and wound the ends of the bandana into a perfect knot. Deft as he was at this, it seemed to take an inordinate amount of time to get it right. He moved closer still, and his fingers found it necessary to brush against the warm, smooth skin of her throat as he adjusted the scarf.
"Thank you," she said when he finally stepped away. He only shrugged in response.
After a hurried breakfast, she and Jared departed. Lauren waved goodbye to Gloria and Maria, who stood framed in the doorway. Flame had now become known as "Mrs. Lockett's mount," and a rapport had developed between Lauren and the mare. Jared was mildly surprised when he spurred Charger into a gallop and Lauren followed suit, keeping pace with him effortlessly. Well, she's learned _something,_ he thought grudgingly. She had lost her eastern pallor, too, and her complexion had taken on a healthy, rosy glow.
He wouldn't admit to anyone, even to himself, that he had missed her while he had been away. He wouldn't define the sense of longing that had plagued him from the time he left. Scattered over Lockett land were sheep ranchers and nesters whom he visited, and their daughters always welcomed a pat on the bottom or a stolen kiss. They had all been disappointed this time. Jared spent his time in serious conversation with the menfolk. He hadn't consciously avoided the women. He just wasn't interested, and therefore didn't give them a thought.
At night, rolled up in his blankets, he tossed and turned in an effort to rid himself of disturbing mental images. Lauren in her dressing gown, her hair spilling over her shoulders. Lauren sleeping in his bed at the ranch, moonlight caressing her cheeks. Lauren in deep concentration over a book, her eyeglasses resting on her nose. Lauren. Lauren. Lauren.
He cursed himself for being a fool as his imagination drifted and he pictured himself lifting a stray lock of hair from her shoulder and kissing it. He was caressing her cheek resting against his pillow. He was sliding the spectacles off her nose in order to kiss her soft mouth.
Sleep eluded him night after night. He sat before his dying campfire smoking cheroots and cursing his intense physical discomfort and the conniving wench who had manipulated his father and was now trying to do the same thing to him. Well, he'd be damned before he'd let her get to him!
But as he'd approached Keypoint last night, his heartbeat had accelerated as he spurred Charger into a mad gallop. Jared swore that his eagerness to get home had nothing to do with the woman he had left there. Now, as he watched her from under the protection of his hat brim, he wasn't so sure.
They rode in silence for half an hour. Jared slowed Charger to a trot and led the way to the riverbank where cypress roots snaked along the ground, knotted and ropelike. On the other side of the river, a rock formation formed a wall, a backdrop, looming up fifty feet. About midway up, jutting out of the rock wall, was the strangest structure Lauren had ever seen.
It was barely more than a wooden shingle façade a few feet deep. A black metal flue extended a few inches out of the roof, emitting a thin wisp of smoke. The only door, in the center of the structure, was made of rough planks. A square window was on either side of it. Over these had been nailed cowhides, which stirred slightly in the breeze. Various antlers of deer and cattle adorned the exterior walls. The small shelf of rock on which the house was perched was barely wide enough for a man to stand on, but it was littered with all types of utensils; pails and washtubs, bridles and harnesses, rope, plows in sad disrepair, a stack of nondescript pelts, metal objects that Lauren couldn't identify from this distance.
"What is that?" she asked Jared in awe as he reined in and began to dismount.
"Just stay where you are. We'll only be here a minute. Crazy Jack doesn't like company."
"Wh—"
"Just sit still, Lauren," he said crisply.
She watched him as he untied a bundle from behind his saddle and casually walked to the riverbank. He knelt down and scooped several handfuls of the clear water into his mouth. Then he placed the package on a flat-surfaced rock and returned to Charger, mounting with studied nonchalance.
Lauren stifled her curiosity as they rode away from the strange scene in silence. She glanced nervously back over her shoulder to steal a final look at the bizarre sight.
They had covered about a mile before Jared once again led the horses near the Rio Caballo, this time nudging them down to the bank to drink. He handed Lauren a canteen and crossed his leg across his saddle, lighting a cigar.
"What was that house, Jared? Does someone live there?" She couldn't contain her curiosity any longer.
"Yes, someone lives there." His manner was irritatingly casual. "His name is Jack Turner, though everyone has nicknamed him Crazy Jack. He built a façade over a dry cave for his house. He's a hermit and not at all crazy."
"A hermit!?" she exclaimed. "How long has he lived there? Where did he come from? Is he dangerous?"
With annoying slowness, Jared retrieved his canteen, recapped it, and took a long pull on his cheroot before he replied. "Jack and his brother Bill came to Texas in the late fifties from God knows where and settled in a small deserted cabin. They either didn't have the initiative or the capital to ranch or farm, but they grew staple crops. They did odd jobs when they needed money, otherwise they were pretty reclusive. The German settlers around here were so industrious that they shunned anyone who didn't share their proclivity for work." He shifted in his saddle and drew again on the cigar.
"In 1872, the Comanche went on a rampage and raided the smaller farms. Jack and Bill were both captured, their cabin burned. They were held captive for six months or so, but then Jack was rescued. If they had been the only hostages, no one would have bothered, but some women and children had been taken at the same time. So a rescue party had been formed. Brother Bill had been killed by the Indians. Tortured and killed. Jack was... injured... and when he came back to civilization he was scorned by all his 'Christian' neighbors."
Jared's lip curled in a derisive sneer. "Jack built his house there in the cave and, though it was on our land, Ben looked the other way. Jack cuts out a couple of beeves each year, but they're never the best ones, and he doesn't waste them. He uses every bit of the carcass. We bring him staples every few months. All he asks is to be left alone. His house is somewhat like a fortress. God knows how he gets in and out of it. You can bet we were sighted in his rifle as soon as we got into range."
Lauren was quiet for a moment as she absorbed the story. "Why was he shunned by everyone? He couldn't help being taken by the Indians." She was immediately sympathetic to the eccentric hermit.
Jared watched her closely as he said slowly, "He'd had his nose and ears cut off. That's what the Indians did to him while they mutilated and killed his brother. He's not too pretty to look at, and people don't like their sensibilities insulted by the sight of him."
# Chapter 12
Lauren's hands flew to her mouth. The horrors inflicted on Jack Turner by the Comanche were incomprehensible. But the torture inflicted on him by his own people was even worse. Ben and Jared had treated him kindly. She lowered her eyes as she said softly, "You're very charitable, Jared, to do these things for him."
"It's not charity. We'll find a jar of his home-brewed corn liquor on the front porch in a day or two. It's always left for us after we bring him something. Of course, I wouldn't drink it for the world. It's pure rotgut. But I'd never ignore it, either." Ever since he could remember, Ben had ridden out to take supplies to Crazy Jack. The man must be in his seventies by now. "I wonder if he knows Ben is dead," Jared mused aloud. "He probably does. I think he knows everything that goes on around here." He tossed down his cigar butt and placed his boot back in the stirrup. "Ready?"
Lauren nodded and they took off again. Jared raised his bandana over his nose and indicated that she should do the same. Moving away from the river, the grass became sparse and dry, and their horses kicked up clouds of dust. Lauren was grateful for Jared's thoughtfulness in bringing her the bandana.
A short while later, Jared slowed their horses to a leisurely walk as he entered a pecan grove. The old, massive trees, gradually losing their foliage in the change of seasons, umbrellaed the gently rising hill.
At this point, the river was wide. The bank to which Jared now led her was grassy before becoming littered with pebbles. Those tiny rocks grew into giant limestone boulders that rose like smooth tables out of the river. The swift water rushed over them, crystal-clear and gurgling.
"How lovely!" she cried. In her excitement, she swung her leg off the saddle and dropped to the ground, rushing to the riverbank.
On the opposite side of the river, there was a wall of rock much like the one Crazy Jack had built his house into. With the natural screening of the rock wall and the protective covering of the pecan trees, the setting was intimate and private despite its primitive nature.
She didn't realize Jared had dismounted and come to stand behind her until he spoke. "The water here is fed by underground springs. That's why it's so clear. Come on."
She was surprised when he took her hand and pulled her out onto the rock formations in the river. The leather gloves they both wore did nothing to dilute the warmth of the hand tightly holding hers. They walked together over the white boulders, which had been polished smooth by water washing over them year after year. When they reached the point where the water rushed over the rock, Jared knelt down. Lauren followed suit and took off her glove to place her hand in the water. "Oh! It's so cold," she exclaimed, laughing.
"Until you get used to it," he said with a smile. "When Rudy and I were kids, we came up here to swim. Ben brought us until we were old enough to look out for ourselves. You see, when it rains, this tributary of the Caballo becomes a torrent. Where we're standing now would be covered with water coming down from the hills." They had pulled the bandanas away from their mouths, and she watched the way his chin caressed the soft cloth beneath it as he spoke. "In the spring, this looks completely different. The redbuds bloom and the bluebonnets cover the hills like a carpet."
She listened intently and watched his hands as he gestured. He had said Ben's name without the haunted expression that usually crossed his face whenever his father was mentioned.
She leaned over the water and cupped a handful, bringing it to her mouth. The brackish taste was terrible. She didn't know she had made a face until she heard Jared's chuckle near her ear.
"Tastes bad, doesn't it? The water is pure, but it has to be filtered through charcoal before it tastes good enough to drink," he explained. "See where the water is bubbling up from under that rock?" He pointed and she nodded. "That's one of the springs."
They walked back across the rocks until they regained the riverbank where their mounts were standing docilely, nibbling the grass. As Jared went about the business of unpacking the saddlebags that held their lunch, Lauren walked up the hill to the crest. Her breath caught in her throat: The entire valley opened up beneath her. It was a breathtaking sight.
"Luncheon is served, Madam," Jared called to her and made a sweeping bow over the blanket that served as their table.
Feeling free and uninhibited, she ran down the hill to join him. The fallen pecans and autumn leaves on the ground crunched under her boots.
Maria had packed enough food for an army, but Lauren was relieved to see that for once there were no beans. Thin slices of cold roast beef, potato salad packed in a jar, spiced peaches, fresh bread, _tortillas,_ and sugar cookies composed their menu. They ate off tin plates the men used when on the trail. Incongruously, Maria had also packed snowy white linen napkins.
"It's beautiful here, Jared," Lauren said after a long, awkward pause which they filled by concentrating on their food.
"Yeah." He munched on a piece of bread before he said offhandedly, "This is where I want to build a house someday. Right up there on top of the hill." He indicated the place with an inclination of his chin. "I'd have the house facing the valley, and this," he swept his hand in a broad gesture, "would be my backyard. Even if the river overflowed the banks, the house would be high enough to be protected."
"That would be perfect," she enthused. "I'd love living in a setting like this."
The moment the words left her mouth she would have given heaven and earth to bring them back. His head whipped around and his eyes bore into hers, hard and uncompromising. She hadn't meant to imply they would be living together. She had only been speaking rhetorically. Mortified, she lowered her head.
Each was painfully aware of the other and their isolated surroundings. The silence was palpable. Using her best conversational voice, acquired from years of practice entertaining guests in the Prathers' parlor, Lauren asked, "Why didn't you tell me Rudy was your brother, Jared?"
The question took him completely off-guard, and he stopped chewing his mouthful of food. Finally he swallowed, took a long gulp of beer from one of the bottles Maria had packed for him, and asked, "Would it have mattered?"
"His illegitimacy?" He looked at her sharply, but saw only understanding in her eyes. "No. That doesn't matter to me, Jared."
"Well, it does to a lot of people. That and his being half Mexican," he said bitterly. "No one understands about Ben and Maria."
"I do."
Again she had surprised him, and his eyes studied her briefly before he looked away. He reclined, stretching his long legs in front of him and supporting himself on one elbow. Lauren was reminded of the first time she had seen him and wished he would sit erect. She found it hard to keep her eyes diverted from the body so unabashedly displayed.
To cover her flustered state, she commented, "You always refer to your father by his first name. Why?"
He seemed momentarily irritated by her myriad questions, but then he laughed softly and said, "That's what everyone else called him." Jared shrugged. "He didn't like titles. Didn't need them. I felt the same way when I came back from Cuba and suddenly I was Lieutenant Lockett." His muscles bunched in agitation.
"It must have been terrible there," she offered quietly. "I read that our army fought the climate as much as they did the Spanish."
"That's an understatement," he said. "I never drew a deep breath the whole time I was there. It was godawful. No matter how hard you tried to suck that heavy air into your lungs, you could never get enough. Most of us got a good case of malaria, and we went into battle with the fever and sweating weakening us until it was an effort to crawl. It got to where I didn't care if we took the goddam hill or not."
"There was a girl at home who was married to a soldier, a marine. We prayed for him and were so thankful when he returned with only a slight leg wound." She shifted her gaze away from his belt buckle and plucked at a napkin spread over her thighs.
His eyes, narrowed to slits, traveled from the part in her hair to the toe of her soft boots. "What about you, Lauren? Didn't you pine away for some sweetheart to come home to you?"
She flushed as much from his scrutiny as from his words. "No," she said into her lap. "I had no admirers or... anything. Besides, I was too young then."
"Oh. But what about later? Didn't any of the deacon's sons try to steal a kiss behind the church door? No hanky-panky in the choir loft under those voluminous robes?" As he spoke, his hand moved to her chest. His dexterous fingers worked the buttons of her jacket until it fell open. She was dizzy with emotion when she felt him fingering the pearl buttons on her shirt, though he didn't try to undo them.
"Surely someone has made a pass at you." His tone was teasing. He couldn't know that his mockery conjured up abhorrent memories of William Keller. She squeezed her eyes tight and shook her head violently, trying to dispel the hateful recollection.
Jared was alarmed. He had meant to shake her cool reserve, but her reaction was far stronger than he'd expected. His hand stilled, though he didn't withdraw it. She composed herself slowly and finally raised her eyes to meet his. "No," she whispered, "I never had any sweethearts."
Of its own volition, his hand moved up so his fingers could settle lightly on her cheek. It just wasn't possible that anyone could be as innocent as she appeared to be. No one that naive would leave the security of a parsonage for an adventure in Texas with a man, a stranger, as virile as Ben Lockett.
Why _had_ she come with Ben? He was on the verge of putting the question to her, but stopped himself. Maybe he didn't want to know the answer. The realization that the truth might hurt caused him to turn his frustration on himself. He looked away from the gray eyes that were now watching him closely. He wasn't going to be a fool over his old man's doxy. He jerked his hand away as though he had reached for something desirable and realized too late that it was decayed and hideous.
Lauren felt his withdrawal immediately. The course their conversation had taken was disturbing, but it _was_ conversation, and she hated to give it up. Nevertheless, she was glad he was no longer touching her. His touch, no matter how slight, did strange things to her, set off reactions both alarming and embarrassing.
"We'd better start gathering those damn pecans," he said tersely, and strode toward Charger to get the gunnysack he had brought along for that purpose.
Lauren put their eating implements away after rinsing them in the river. Then she repacked the remaining food. Jared had the sack half-filled with pecans when she bent down to help him.
"I can do it," he said gruffly. "No sense in your getting dirty."
She looked up the length of his body and met the amber eyes glaring down at her. What had she done to make him angry? "I want to help," she said simply.
"Suit yourself," he answered indifferently and turned away, looking for an area that hadn't been harvested.
By the time he came stamping back to her, Lauren had gathered a pile of the nuts. He held the mouth of the sack open while she scooped her hoard into it.
"All done," she said cheerfully, dusting off her hands. Licking her lips quickly, she asked, "Do you think we have enough pecans?"
He didn't answer. He was too intrigued by the tongue which had raked across incredibly sexy lips and disappeared behind them to hide from him. Then he spun away from her, saying over his shoulder, "Let's go. If I read my weather signs right, we're in for a norther before long."
They mounted their horses. He spoke only once. "We'll go down the other side. It's not as scenic and we have to go by the charcoal burners' camp, but it's closer. I'm afraid of getting caught out here without warmer clothes."
They followed the spring-fed tributary that tripped over limestone until, at the bottom of the hill, it flowed into the Rio Caballo.
Lauren sniffed the air and caught the smell of wood smoke. As they rode around a bend, a derelict encampment, like an ugly sore marring the scenery's beauty, came into sight. Tents and dilapidated shacks were scattered around pits from which the dark smoke rose. Ragged children ran among the fires with heart-stopping recklessness. Mangy dogs came running out from under various covers, barking ferociously. Several dirty, bewhiskered men ambled out of the lean-tos to see who the intruders were.
The women, dirty and as ragged as their children, scowled at Lauren as they squatted around campfires stirring pots of foul-smelling stew. One of the dirtiest men separated himself from the rest and shuffled toward them. Lauren suspected his nonchalant swagger was deceptive. His beady, deep-set eyes didn't miss anything, and were bright under shaggy brows.
Jared looked at her out of the corner of his eye, never averting his head from the man. "Whatever happens, don't get off your horse." He had barely opened his lips to say the words, rasping them from behind his teeth.
Jared reined in their horses and waited for him to amble toward them. The man was short and stocky with powerful-looking arms that were too long for his body, giving him an apelike physique. He wore dirty, patched overalls, with only his red, faded long johns under them. Lauren shrank in disgust at the stained and moist armholes of the garment. He had several days' stubble on his face, and his oily black hair was matted to his head when he scooped off a battered hat in feigned humility.
"Well, lookey here. If it ain't Mr. Jared come to pay us a call with his new lady." His teeth were yellow and broken, covered with thick dark scum. Lauren had never seen anyone so repulsive. Or menacing.
"Duncan," Jared said curtly.
"We sure was sorry to hear about yore pa, Mr. Jared. That's a real shame now, ain't it?"
Jared ignored the comment. "How's your business?"
"Well," he whined, "it could always be better. If'n you'd let us clear some of the land where them damned nesters is, we could both be better off."
"You know that land is off-limits to you, and it always will be. You stay on this side of the river, or you're off for good, understand?"
"Now, Mr. Jared, you wouldn't run us off. What with our famblies and all." He paused and split his lips in a sickening parody of a smile. "You couldn't see June no more, either."
Jared swung down from his saddle and stood facing the man, his body as tense as a coiled snake ready to strike. Only common sense and the grim consequences of such a stupid action kept him from grinding his fist into Duncan's insolent face.
The charcoal burner read the hesitation and continued with a leer, "You hadn't forgotten Juney now, hadja, Mr. Jared?" He inclined his head and Lauren followed his indication to the cabin where a young woman leaned against the doorjamb. Her expression was as insolent as the man's. She pushed away from the door and sauntered closer to them, her hips swinging suggestively. She was barefoot and her feet were caked with dirt. Her dress barely covered her knees and the bodice was stretched across pendulous breasts. Lauren realized that she was naked under the thin cotton dress and was stunned at the girl's immodesty. Her hair was almost white and her eyes were piercing blue. She might have been pretty, even beautiful, if it weren't for the sullen mouth that drooped at the corners and her lack of personal hygiene.
The slinking walk brought her to within a few inches of Jared. She swayed slightly as she said huskily, "Hello, Jared."
Jared turned on his heels and walked over to Flame and her rider. He raised his voice. "This is my wife." He placed a gloved hand on Lauren's thigh, and if she had not been frightened by this strange camp and the gypsylike people who lived here, she would have wondered why she trembled and felt like melting at his touch. "If anyone from this camp comes near her, I'll kill him. You have been warned." It could have been her imagination that he applied more pressure to her leg just before he released it. He walked around Charger and mounted in one fluid movement.
"Uh... Mr. Jared, we was wonderin' what's gonna happen to all of the goddam nesters and sheepherders when you dam up the river." Duncan stood with his squat legs spread, arms akimbo, his chin thrust out belligerently. Gone was the groveling attitude he had assumed at first.
Jared riveted his amber eyes on the man. "Where in hell did you ever hear that?"
"I don't rightly recall." He scratched his head in mock-puzzlement, and Lauren was nauseated to see startled lice crawling in his hair. "Word just got around, that's all."
"Well, it's only gossip. Understand? I don't want to hear any more about it."
"If'n they was to move on like, could we work that land then?"
"I'm going to say it one more time." Jared's voice was hard and even, as sharp as a rapier. "You work only where I or Rudy tell you you can. Nowhere else. And anything else that happens on Lockett land is none of your business." He rested his hand lightly on his pistol holster.
He nudged Charger with his knees and Lauren did the same to Flame. They rode out of the camp slowly, though she would have liked to gallop, so malevolent were the looks that June had given her. When she had passed close to the girl, Lauren heard her hiss, "Bitch!"
When the camp was well behind them, Jared pulled up and listened for a moment before he spoke. "I think it's all right now."
"What in the world is that place? I was frightened."
"I was, too." He laughed. "That riffraff back there are charcoal burners. Wat Duncan is more or less their leader. Ben made a deal with him years ago that they could cut down the cedar and burn it into charcoal. There's a market for it in San Antonio. They use it to purify the water and make it taste better." Lauren remembered the bitter-tasting water she had drunk from the spring and Jared's explanation that it had to be filtered. "We let them keep all their profits and, in turn, they keep the ranges cleared of excess cedar. The only problem is that they are mean, dishonest, and completely amoral."
Lauren looked away from him as she murmured, "The girl was pretty in a way."
A grin twitched his lips as he studied her. He said, "One day in my reckless youth, Ben caught June and me giving each other a biology lesson. He beat me to within an inch of my life. I never went near her again, especially after he impressed upon me what can befall a young man who fools around with sluts like her. She and Wat must have been insulted, because he never fails to make reference to her when I'm around."
"Are they related?"
"Yes. She's his sister." He paused significantly. "And his mate."
Lauren felt ill as she spurred Flame into a gallop behind Jared's lead.
* * *
They were only a few miles from the house when the wind suddenly shifted to northerly and the gusts of cold air stung Lauren's cheeks. Her eyes began to water.
Jared shouted for her to pull her bandana over her nose as he did, and it offered a little protection from the biting wind.
They rode a few more minutes and then he signaled for her to follow him. He led her to a group of boulders and rode into a pocket formed by the enormous rocks.
Lauren was shivering from the cold, but at least now they were out of the fierce wind. Jared came around to Flame's side and offered up his arms to help her down. She placed her hands on his shoulders as he lowered her gently to the ground.
She welcomed the warmth of his arms as they slowly enclosed her in a hesitant embrace. Her hat slipped from her head as she rested her cheek against the hard chest. She looked up, laughing when she realized that she still had the bandana pulled up over the lower half of her face.
Her laughter was choked off as she met Jared's eyes over the top of his scarf. They impaled her with the intensity of their gaze. The brows were dark and the lashes were tipped with gold. The laugh lines at the corners of his eyes were white where his squinting had kept them from tanning like the rest of his face. The brown irises were flecked with gold. Sherry or amber or topaz.
Slowly he reached out and took the bottom of her bandana in his fingers and lowered it. It was a caress. He stroked her lips with his thumb. Only then did he lower his gaze from her eyes to study her lips as his fingers traced their shape, savored their texture, marveled at their softness. They trembled beneath his touch. Leisurely he lowered his own bandana.
She leaned into him as his arms drew her closer. Without conscious thought, her hands went to his waist, then around to meet at his back.
His lips came to hers gently, softly. Barely touching them, he whispered, "Lauren." Then his mouth closed over hers. His lips moved over hers expertly, persuasively. His tongue teased them until they were shyly parted and she met the tip of his tongue with her own. A low moan escaped Jared's throat and his hand went to the small of her back to press her against him. Her fingers splayed over the muscles of his back. Losing all timidity, she opened her mouth to the hungry plea of his.
"Now that _is_ a touching sight," Rudy said, laughing.
# Chapter 13
Lauren and Jared jumped so violently at the sound of the amused voice that their movement startled their horses.
"Having a nice time?" Rudy queried innocently. He had seen them riding toward the boulders and had followed them into the shelter.
Jared snarled at him. "Lauren was cold and I remembered that I had a poncho in one of my saddlebags. I was getting it out for her." He was infuriated with himself for explaining their kiss like a guilty schoolboy.
"Yeah, you were warming her up all right. Looks like you were getting pretty hot yourself." Rudy would have loved to continue taunting Jared, but Lauren was the greater sufferer. When he saw her distress, he softened. "We'd better get home. Gloria was worried when the Norther blew in and sent me out to check on you."
Jared had extracted an old, worn woolen poncho from his saddlebag and unceremoniously pulled it over Lauren's head. Roughly plopping her hat back onto her head, he remounted Charger. She could tell by the way he sat rigidly on the stallion's back as he rode away that he was angry. Probably with her.
She and Rudy followed him at a distance. Her brother-in-law gave her a reassuring smile.
Jared was seething. It didn't matter to him that Rudy had seen him in a tender embrace. He wasn't shy. He and some of his cronies had even shared whores, cheering each other on. What bothered him was that Rudy had witnessed his susceptibility to Lauren.
She had lured him into kissing her. Those damn captivating eyes, full of tears produced by the cold wind, were hard for any man to resist. The raven tendrils whipping cheeks rosy from the exertion of their ride cried out to be caressed. That body which had tormented him all day with its nearness had been impossible to release once he held it against him.
Above all, he resented her composure. She retained that superior coolness no matter what happened to them. She hadn't panicked at the charcoal burners' camp. She hadn't been violently ill when he told her about the atrocities inflicted on Crazy Jack. Everyone at the ranch adored her. She fit right in.
Damn her!
* * *
Dinnertime was subdued. Everyone ate almost silently. Gloria, Maria, and Rudy recognized one of Jared's black moods and spoke to him with the care of one walking on thin ice. He growled his responses.
Lauren was completely withdrawn. She spoke to no one except for an occasional please or thank you. Her head bowed, she stared into her lap as they sat in the living area after dinner. Her fingers would wander to her watch subconsciously and draw comfort from it as she always did in times of stress.
Outside the wind howled, and the fireplace, cheerful though it was with the crackling fire in it, could not lighten the atmosphere of gloom.
"We had visitors today, Jared," Rudy said cautiously.
"Who was that?" Jared appeared completely bored.
"The Vandivers. Father and son."
"Goddammit," he cursed. "What did they want?" Rudy had his undivided attention now.
"They said they were over at the site of their new power plant and just stopped in to say hello."
"Like hell. The power plant will be fifteen miles from here." Jared stood up and crossed to the fireplace. He stared into it for several seconds, then glanced smugly at Lauren. "It's too bad we missed them. Lauren has developed quite an attraction to Kurt." The words were deliberately provoking.
She jerked up her head and met his challenging stare. She was the first to avert her gaze, humiliated and angry with herself for being such a coward and not rebuking him.
There was an embarrassed silence in the room for long minutes. The pendulum clock ticked loudly and the logs in the fireplace shifted, showering Jared's boots with sparks as he maintained his stance on the hearth.
He tossed the remainder of his cigar into the flames and, taking down his shearling coat from the hall tree, muttered a sullen goodnight.
"Jared, it's cold out there. Why don't you stay here in the house tonight?" Gloria was vexed that her scheme to throw these two stubborn people together for a whole day had not broken down any barriers between them. Instead, it seemed as though more had been raised.
"If I could have my own room back, I would gladly stay. As it is, I don't care to sleep there. If Lauren gets cold, I'm sure one of the _vaqueros_ would be delighted to warm her bed."
Lauren bolted out of her chair and flew across the room so quickly that it surprised even her when she stood trembling directly in front of Jared.
She raised her hand as if to strike him, but the arrogant tilt of his chin stopped her. He was daring her to show a bit of temper, and she would not give him the satisfaction. She clenched her fist, but lowered it to her side.
_"Why?"_ she asked insistently. "Why do you persist in tormenting me so? I don't like this 'arrangement' any better than you do. But I don't forget my manners."
She turned and marched into the hall leading to the bedrooms. Jared, in spite of his anger, admired the undaunted way she held her head.
Lauren's limbs felt heavy as she climbed onto the wide bed. She was tired from the long ride to Pecan Creek, and mentally fatigued as well. She was weary of trying to adjust herself to Jared's moods, of warding off his verbal attacks. His inconsistency had her totally baffled. He was vindictive and abusive one minute, and tender the next.
She wished she had resisted when he had kissed her. What had she been thinking? Nothing. That was the problem. When his arms went around her, warming her, she had ceased to think. She had allowed her senses to take over, relinquished all control to them. He had been so gentle... almost loving.
She buried her face in the pillow—his pillow—and groaned as she recalled the feel of his tongue against hers, the strong hands that had just begun caressing her back when Rudy interrupted them. What would have happened if they had not been seen?
It was no use speculating, for she was still not positive where kisses led. Having been around Gloria and listening to her talk about her relationship with Rudy, Lauren had inferred that whatever it was, it was pleasant. She had fought that mysterious culmination with William. She couldn't quite imagine it being enjoyable.
But if it had been Jared's hands on her back pulling away her clothing, how would she have reacted? She blushed in the darkness. She was restless and physically unsatisfied and her body was transmitting strange, unrelenting impulses to her brain.
* * *
Rudy and Jared would leave each morning after breakfast, and return just in time to wash before dinner. The mood in the house cheered somewhat, though Jared and Lauren still treated each other with polite indifference. He asked her permission when he went into his room to retrieve one possession or another. Lauren felt guilty about using the room, but when she suggested that Gloria make a bed for her in one of the children's rooms, the other woman adamantly refused. No argument could convince her, so Lauren dropped the subject.
She continued to play with and read to the children, constantly delighted by them. With Maria, Lauren shared many quiet moments, listening to stories about Ben and about Jared and Rudy as boys. Maria carefully stayed away from the subject of Lauren's marriage. The condition of that union grieved her.
"Jared is so reticent about his early life," Lauren admitted to the older woman one day. "He rarely speaks of his childhood, or schooling, or anything. He won't tell me anything personal," she sighed. "I mentioned Cuba to him once. He was reluctant to talk about it."
Marie shook her small, sleek head sadly. "I'm glad you weren't here when he came home. Ben, I think, was secretly proud of him for joining the army. Olivia was furious and tried to keep Jared from going. She had friends of her family trying to pull strings, but when Ben found out about it, he stopped her. After the war, when the town made Jared a hero, she acted as if the whole idea had been hers." Maria sipped the tea they were enjoying while sitting on the front porch. "She's a very sad, lonely woman, you know," she added quietly.
"Did he catch malaria while he was there?"
"Yes, but his main injury was psychological. He had a friend whose father owns a ranch west of Kerrville. Alex Craven and Jared had been friends since boyhood. They joined at the same time and were in the same battalion. Alex was killed. Jared felt his death was the result of an error in judgment by the company commander, that his friend was sacrificed for no reason. He still has nightmares about that day in battle. Alex's death hurt him deeply, but Jared keeps everything inside. He reveals his true self to no one."
"I'm beginning to think he has no 'true self.' Whenever I think I've figured him out, I discover some new enigmatic facet of his character," Lauren said. Who was this man she had married?
"A very good man is there, Lauren. One day you will know him. I'm certain of that." Maria patted Lauren's hand as she stood up and went into the room she had shared with Ben, closing out the world by shutting her door.
* * *
"Oh, no! Look what I've done!" Lauren exclaimed. She was helping Gloria prepare the evening meal. Opening a can of tomatoes, she was about to empty it into a pot of stew when some of the contents splashed onto her sleeve, spreading a dark stain on the fine fabric of her shirtwaist.
"Better go change quickly and let me put that in cold water to soak," Gloria said unperturbedly.
"I'll be right back," Lauren promised as she hurriedly left the kitchen and rushed into her room. No sooner had she unbuttoned all the buttons down her back and slipped off the blouse than she heard a commotion outside in the hallway. Before she could reach for something to cover herself, the door to the room was flung open and Jared was being pushed inside by Gloria and Rudy.
Lauren forgot her state of deshabille when she saw her husband's torn and blood-smeared shirt. She suppressed a gasp of horror. "What happened?" she asked on a high, anxious note.
Jared was staring at her, too stunned by her appearance to speak. Or was he numbed by pain? It was left to Rudy to answer her. "Jared and I were restringing some fences that had come down. An ornery length of barbed wire backlashed and caught Jared across the chest. It needs to be seen to."
"I told you it's all right," Jared growled as he was shoved from behind into the room.
"Nonsense," said Gloria in the same tone she used to her children. "Get out of that shirt and I'll get the medicine. Lauren, you'd better help him."
A conspiratorial smile was passed from husband to wife as Gloria rushed out to retrieve a bottle of antiseptic and cotton. Lauren shyly walked toward Jared's back and settled her hands on his shoulders. She eased the shirt off his back as he unbuttoned it and painfully pulled it away from the skin on his chest. The drying, congealed blood made the fabric stick, and the wounds reopened and bled profusely.
Gloria virtually threw the medicine at Jared when she returned. Pulling a gawking Rudy out of the room, she said, "Take your time. Maria and I will get dinner. Jared, you should lie down and rest. You've lost a lot of blood."
The door was drawn shut and the two were left alone.
Lauren dropped the blood-soaked shirt on top of her own soiled blouse. Silently, together, they watched fabric settle against fabric.
Stirred into action, Lauren crossed to the wardrobe, intending to get another blouse and don it quickly when she turned her head slightly over her shoulder and asked, "Does it hurt too much?" She drew her breath in sharply when she saw the oozing puncture marks on Jared's chest and the rivulets of blood that ran through the thick mat of hair. "Oh, Jared," she cried, rushing toward him, suddenly not caring that she was clothed from the waist up in only her sheer, lace-edged chemise. She had taken Elena's advice and stopped wearing a corset every day.
"Here, sit down," she directed, taking his hand and leading him to the small vanity stool Gloria had provided for her. "Let me wash you off so we can see how bad it is."
"It's really nothing," he said again, and she wondered at the low, uneven sound of his voice. Was he in that much pain?
She poured fresh water into her washing bowl and dipped a clean towel into it. Her hand paused over his naked chest. She drew a long, shuddering breath and squeezed her eyes tightly shut. Then she touched him, dabbing at the hair-matted skin with the absorbent towel. "I don't want to hurt you," she murmured.
Jared gritted his teeth, not in pain from his chest wounds, but in agony from having her this close to him. The skin of her creamy white throat was scented with lavender and the fragrance intoxicated him, making him dizzy. Or was that light-headedness caused by loss of blood? It didn't matter, the result was the same. He could feel her breath, like a cooling balm, fanning his face as she exhaled.
Of their own volition, his eyes lowered to her breasts. His jaw clenched reflexively when he saw them swaying slightly under the soft fabric. It took every ounce of willpower he could garner not to reach out and touch them, peel away the diaphanous cloth and learn the true color of her nipples, which were only vague shadows, elusive and bewitching.
His physical desire was becoming painfully manifested in the tight pants. He diverted his eyes to her hands, which dipped the towel into the washbasin and wrung out the excess water, staining the bowl with his blood. Think about the blood, he commanded himself. Think about the pain you felt when the wire slapped into you. Think about anything but—
"There, I think that will do for now," she was saying. Her voice was soft and low, caressing his ears. "Those punctures are deep. Didn't you have on that cowhide vest you usually wear?"
"No," he answered, glad they were talking. Anything helped. "I took it off because it got too hot. If I had left it on, I probably wouldn't have been cut."
"This is going to hurt, too," she apologized softly as she soaked a piece of cotton with the foul-smelling medicine out of the corked blue bottle.
"I'm tough," he said, and looked up at her with a mischievous smile.
Both were momentarily mesmerized by the other's nearness. Their eyes locked in a silent communication and the message transmitted struck each of them in the heart and was startling in its impact. Lauren tore her eyes away first.
"I'll try not to hurt you anymore," she said as she delicately touched one of the wounds. He sharply sucked in air between his teeth, making a whistling sound. Beads of perspiration popped out of his forehead.
"I'm sorry," she said as she quickly dabbed the other punctures. Then, to his agonized delight, she began to blow on the stinging wounds. From the angle at which he looked down at her as she leaned over him, her eyes appeared to be closed. Violet shadows tinted her lids, and her black lashes contrasted with the delicate cheek on which they seemed to rest. He looked along the slender length of her nose to her mouth. The lips were moist, pink, slightly parted, and bow-shaped as her breath passed through them and teased his fevered skin. It stirred the hair on his chest and cooled the burning sensation from the punctures, but ignited a fire in another part of his body.
"Oh, God," he groaned deep in his throat. He stood up abruptly, upsetting the cushioned stool, and clasped her in his arms. He pulled her toward him with such force that the breath whooshed out of her body from the impact. The mouth that took hers was avaricious, parting her surprised lips with a thrusting tongue. Yet when the treasure had been discovered, that plundering invader became gentle and savored what it had found.
His hands slid along the bare skin of her arms, as smooth and cool as satin, and lifted them around his own neck. For a moment, they lay there, unpracticed and still. Jared's breath was expelled in a relieved sigh when he felt her hands lock behind his neck and her fingers plow through his thick, unruly hair. He brought her closer by applying pressure with the hand on the small of her back. His legs straddled hers, molding them into an ageless position as she curved up against him.
Lauren became aware of a foreign hardness spearing into her belly and was both alarmed and intrigued. Responsively she moved against the intruder and felt a melting warmth in the pit of her stomach that rendered the rest of her weak.
Jared's hand moved between their bodies and fear and desire combated in her brain when she realized the direction it was taking. She was wanting something she couldn't name. Did it have anything to do with his hand that hovered near her breast?
He wouldn't touch her there. Would he? No. She didn't want him to. Did she? Yes. Yes, please, she cried silently, not examining where such a wicked thought had originated, but dimly aware that it had something to do with the hard strength of his body pressed against the harmonizing softness of hers.
Jared's mind was reeling. No one felt like this. No mouth had ever tasted this good, he thought as he savored her lips. His hand slid over the soft mound of her breast and pressed gently. It was as firm and full in his palm as he remembered from their brief contact the day he had surprised her in her room. Under the steady, coaxing movements of his fingers, her nipple responded, becoming a firm bud of passion, eager to bloom.
In unguarded and introspective moments, Jared had hoped that somewhere there would be a woman like this. Uniquely his. Different from all others. Hadn't Ben told him—
Ben!
The name screamed through his mind, ricocheting off the walls of his brain with a cacophonous echo. Ben! Had his father held her this way? Had she responded in kind, murmuring that low purr deep in her throat?
He pushed her away from him with such force that she fell across the bed, looking up at him with rapidly blinking and uncomprehending eyes. Her hair tumbled across her shoulders and fell onto the creamy, rose-tipped breasts left partially bare from his caress.
He pointed an accusing finger at her. "I told you to stay away from me!" he shouted. His breathing was a harsh rasp. "You're beautiful. I'll give you that. And you're softer and taste sweeter..." His voice dwindled to an anguished whisper. "God!" He slammed his fist into his palm. The pounding demand in his loins was unbearable. With her sprawled across the bed, looking up at him with such absolute innocence, his organ answered with an excruciating throbbing.
He ought to take her now, conquer her once and for all. He longed to flip up her skirts and see if her thighs were as smooth as he had imagined them to be. Then he would ram himself between them, thrusting until he found release from the desire that had stalked and haunted him for so long.
Had Lauren known his thoughts, she would have been terrified. Instead, as she lay on the bed, watching the misery radiate from every pore of his body, she knew only compassion for her husband. She sat up and timidly extended her hand toward him, an offer to soothe away the pain inflicted by whatever devil tormented him.
He recoiled. "I don't want any part of you," he declared unsteadily. "Do you understand?" He whirled away from her, flung open the wardrobe door, tore a shirt from the hanger, and stamped across the room to the door. It slammed with a resounding crash when he went out.
Lauren fell back and rolled over onto her stomach, burying her face into the mattress of the bed. She sobbed brokenly, her tears absorbed by the bedspread.
Was she crying because he had kissed her so insultingly or because she had responded so wantonly? Because he had stopped kissing her? Or because of his abusive words? Was her biggest fear that he would soon grow tired of the trap he was in, pay her the twenty thousand dollars, and send her packing?
The questions tumbled in her mind. And for none of them did she have an answer.
* * *
The next morning, Lauren and Maria were returning from a ride when they heard shouting from one of the corrals. The _vaqueros_ were gathered around the fence.
Lauren spotted the tall, lean figure of her husband. She hadn't seen him since the day before when she had tended his wounds. After he had slammed out of the room, she had lain there for a while before restoring herself enough to go in to dinner. The Mendezes were at the table, patiently awaiting her.
Jared didn't appear. After everyone had started eating, Rudy quietly and inconsequentially stated that Jared was needed at the bunkhouse. No one commented, and Lauren had pretended indifference to his absence.
As they dismounted and tied their horses to the hitching rail in front of the house, Lauren said to Maria, "I think I'll stay out for a while." Her curiosity was piqued by the commotion at the corral.
"Very well," Maria said, smiling. "I enjoyed our ride. Ben and I used to ride early in the mornings. I've missed the exercise."
"We'll do it whenever you want." Lauren patted the older woman's arm before Maria climbed the steps to the front door.
Strolling in the direction of the corral, Lauren told herself she wasn't going there to see Jared. As she approached, twenty or so _vaqueros_ were driving a bull into a chute.
"What's going on, Rudy?" she asked her brother-in-law as she reached the fence.
He jerked his head around to face her. "It's... uh... we're going to... uh... castrate this bull."
"Oh," Lauren replied, red-faced. She turned to go, but was blocked by the sudden appearance of her husband. He put out a restraining arm.
"Why don't you stay and watch? Your being so interested in Keypoint and the ranching business and all, I'm sure you'll enjoy it."
"Jared—" Rudy began.
"No, Rudy. Lauren is dying to learn everything she can about the cattle and the _vaqueros._ "
His words were biting and harsh. Lauren wished desperately that she had followed Maria into the house. Suddenly Jared gripped her shoulders and turned her toward the corral, holding her against him with deceptive gentleness. His hands felt like iron bands around her upper arms.
"Why... why do you do that to a particular bull?"
She hoped the question would have a calming effect on him. She really didn't care to know anything about the procedure and certainly didn't want to watch it.
Jared drawled around his cigar, "Well, there could be any number of reasons. Better beef. Or maybe he can't please the cows anymore. It may be only because he's a mean sonofabitch."
"Then maybe that's what we should do to you," Rudy said in a deadly voice. He wasn't about to let Lauren witness the bloody procedure.
Jared spun around and glared at his brother. He tossed his cigar away with a negligent flick of his wrist. "Is that a fact? Well, just who in hell is going to try?"
Without preamble, Rudy lowered his head and charged into Jared's stomach, knocking the younger man down.
Lauren gasped and cowered against the fence as they rolled in the dust, arms and legs thrashing, blood spurting from busted lips and smashed noses. They stood and circled each other warily, then Jared counterattacked and they crashed to the ground again. The _vaqueros_ had stopped their work and stood in a wide circle around the fighting brothers. The only sounds were the thuds of landed blows and the grunts of pain and effort.
Gloria came running from the house, skirts flying. A few of the children stood in awe of the spectacle. They had seen their father and Uncle Jared fight before, but it was always playful wrestling. Their young, precocious minds perceived that this was different.
Gloria grabbed the revolver out of one of the cowboy's holsters before he had time to react. Since she knew that the first chamber was always empty, she cocked it twice and then fired into the air.
The two bodies on the ground fell apart and gasped for needed breath. When they had recovered somewhat, they sheepishly wiped away blood from their faces and bodies with dusty, tattered sleeves. The abrasions on Jared's chest had reopened and were staining his shirt with bright red blood. Embarrassed, they looked at each other.
Jared grinned at Rudy, grimacing with the pain of moving his swollen lips. "Just as I thought. You're getting soft and out of shape, old man."
"Like hell. A few more seconds and you would have been begging for mercy."
Jared rose painfully to his feet, swayed until he was not so dizzy, and then extended his hand to Rudy, who took it gratefully and pulled himself up. They supported each other for a few moments and then burst out laughing.
Everyone joined in their laughter, relieved that the fracas had been in fun after all. Only Lauren knew differently.
Without thinking of the consequences, only wanting to put distance between herself and these barbarians, she dashed toward Flame, who was still tied to the rail. Placing her booted foot in the stirrup, she vaulted into the saddle.
She kneed the mare into a gallop and raced past the others, who were staring at her in temporary stupefaction. Her hat sailed off her head and landed to within inches of Jared's feet.
"What the—" he started.
"You'd better go after her, Jared," Rudy suggested tentatively. "She was upset."
"Why?" Gloria demanded.
"She... uh... Jared wanted her to watch a castration," Rudy said.
"My God! How could you even have considered such a thing?" Gloria asked angrily. "Go after her. Both of you."
The two men saw the wisdom of her words as they sought the horizon and barely made out the tiny figure of horse and rider as they went over the hill.
"Come on," Jared ordered tersely as he started in the direction of Charger at a lope. Rudy mounted his own horse and soon they were thundering across the plain in the direction Lauren had taken. Gloria muttered imprecations under her breath about the immature behavior of men as she gathered up her children and shooed them into the house.
Lauren's eyes were stinging from the cold wind, but the wind wasn't responsible for the tears that clouded her eyes and ran unchecked down her face. Why had she ever consented to marry Jared Lockett? He was a brute, the most callous, abusive man she had ever met.
Heedlessly, she raced over the rocky ground. Usually even at a gallop, she handled the obliging mare with gentleness. Today she was too caught up in her own problems to see the prairie dog hole before they were upon it. Flame's hoof caught in the indentation and Lauren heard the fatal snap of the bone a fraction of a second before she went sailing through the air.
# Chapter 14
She landed on her back. Lying still a moment, she tried to determine if she were injured. Deciding she wasn't, she sat up gingerly. Nothing appeared to be broken, though she was sure to have bruises the next day.
At Flame's piteous whinnying, Lauren stood up and scrambled toward the mare. Flame's eyes were gaping wide in fright and pain. Lauren saw the awkward angle at which her front leg lay.
"Oh, no," Lauren murmured as she fell to her knees and stroked the mare's neck. "I'm sorry, girl," she sobbed. "I didn't mean to punish you. I'll see that you're fixed. You'll get well. You must." Tears ran down her face and she wiped them away, creating a dirty smear across her cheek.
Dimly she heard approaching horses, but she didn't take her eyes from the mare, who still screamed in pain while Lauren spoke to her in low, soothing tones.
Jared and Rudy reined in and assessed the situation in an instant. Jared didn't want to acknowledge the relief he felt when he saw that Lauren was apparently intact. He and Rudy glanced at each other and nodded in unison. They dismounted together, as though choreographed.
Lauren looked up when she saw their boots close to Flame's head. Bounding to her feet, she ran to her husband, gripping his arms. Her eyes were full of pleading tears. "Jared, it was my fault. She stumbled in a gopher hole. She's... It will be all right... Help her... She'll get well."
With deadly calm, Jared ignored her eyes as his hand sought his pistol in its holster and withdrew it. "No," she rasped. "No!"
"Rudy," was all he said.
Lauren felt herself hauled out of the way as Jared aimed his pistol and fired. The mare's scream ceased immediately, only to be replaced by the echoing of the pistol shot. Then another scream bounced off the surrounding hills, but Lauren didn't recognize it as hers as she flew into Jared.
"You monster! You killed her. Beast! Animal! Killer! Killer, killer." Her small fists pummeled his chest and her feet kicked at his shins. She wanted only to hurt him, to avenge her own pain. He stood passively and took the punishment, not raising a hand to protect himself. "I hate you!" she screamed. "You're vile and savage. Cruel." Her voice began to lose some of its impetus, as did her pounding fists. "I hate you." The words were barely a whisper now and they came out as a sob. She dropped to the ground in a heap, like a wind-up toy which had suddenly wound down. Great racking sobs shook her shoulders.
Rudy squatted down on his haunches and laid a solicitous hand on her shoulder. "He did what he had to do, Lauren. It was hopeless. Jared only shot her to save her pain." His voice became gentler. "I think you know that."
The crying stopped, but her head remained bowed. Rudy stood up. "I'll get her home," he said quietly. "You see to the horse." He had never felt so impotent in his life. He wasn't sure what had transpired between his brother and his new wife in the past few days. All he knew was that both were suffering, and he was powerless to do anything for either of them.
Jared raised his eyes, which had been riveted on the weeping figure at his feet, to his brother, and spoke resolutely. "No. She's my wife, and no man except me is going to see to her. If she goes home, she goes with me."
Rudy bit off an argument even as he watched Jared lean down and grasp Lauren under her arms, pulling her to her feet. She yanked her arms free and stared up at him defiantly. Then without a word, she walked toward Charger and pushed herself up into the saddle. She sat stiffly as Jared mounted behind her. Rudy watched them until they rode out of sight. He shook his head in puzzlement and despair for these two people whom he loved. Then he set about gathering brush for Flame's funeral pyre.
Lauren held herself rigid on the saddle in front of Jared, who seemed as determined not to touch her as she him. What he was feeling after her attack remained a mystery. When they finally cantered into the compound and up to the front of the ranch house, Jared spoke the first words that had passed between them.
"The honeymoon is over," he said sarcastically. "Congratulations, Lauren. No one has ever been able to come between my brother and me. Before you cause more friction, I want to leave here. In the morning. Be ready early."
"Very well," she replied as she slid to the ground and strode into the house without once glancing back at him.
* * *
After the misty-eyed goodbyes and warm farewell embraces from the Mendezes, it was hard for Lauren to return to Olivia's house in Coronado where she knew only disquietude. Had it not been for the arrival of Elena's baby girl, which a beaming Carlos had announced to her a few days ago, leaving Keypoint would have been unbearable.
Upon their unexpected arrival, Olivia pointedly didn't ask about their visit. Lauren now knew why she held such a great deal of contempt for the family who lived at the ranch. Carson Wells politely inquired into Lauren's well-being on their first night back. He was, it now seemed, a regular at dinner.
Jared was defensively sullen and drank steadily throughout the meal. Olivia heaped acclaim on the Vandivers for being successful in securing a trunk of the railroad to Coronado. The date for the groundbreaking was to be announced soon.
"With any kind of luck with the weather, labor, and so on, we should have a railroad by this time next year."
"Swell," Jared mumbled into his glass.
"I should think that would make you happy," Olivia snapped at him.
He pushed his chair back and rose unsteadily to his feet. "I'll tell you what makes me _unhappy._ Those damn Vandivers were snooping around Keypoint the other day. I wasn't there or they wouldn't have gotten anywhere near it. I want them restricted to the area designated for the power plant. Is that clear?" His face was flushed and his amber eyes glowed like animal eyes in the dark.
"Yes, Jared, I'll mention that to them. I'm sure they'll comply," Olivia mollified him.
He only snorted as he refilled his glass, sloshing whiskey over his unsteady hand.
The only joy in the house for Lauren was Elena's baby. Her introduction to Isabela had been made as the baby sucked greedily at her mother's milk-laden breast. Lauren was shocked at Elena's immodesty as she bared her breast, but the new mother wasn't at all embarrassed. Rosa looked on with grandmotherly pride. Isabela's hair was coal-black and her black eyes were lightly fringed with dark lashes.
The baby stayed in the small room off the kitchen that Rosa shared with Elena. When Elena was busy somewhere in the house, Rosa was nearby to answer the demanding cries of her granddaughter. As long as Olivia wasn't disturbed, everything would be fine.
Lauren hated for Carlos to be separated from his family. She intended to talk to Jared about them living together at Keypoint. Surely with another Mendez baby coming, Gloria could use Elena's help.
Lauren was amazed that Elena recovered from her child-birthing so quickly. The new mothers whom she had attended while living in the parsonage had taken weeks to get out of bed, but Elena resumed her duties in the house right away, pausing periodically and only long enough to feed Isabela.
So it was a matter of deep concern to Lauren when she found Elena leaning against the bannister one afternoon, unable to continue upstairs.
"Elena, what's wrong?" she cried as she rushed toward the girl and lent her support.
"I'm just tired, I think." Elena's voice contained none of its usual animation.
"Why don't you go lie down for a while? I'll explain to Olivia."
She took the girl's elbow and steered her toward her room. Her alarm increased even more when Elena didn't argue with her as she was wont to do. Without protest, she lay on her bed and Lauren covered her with a light blanket. The baby was sleeping quietly across the room in her crib. Lauren left them, hoping that Rosa would soon return from her marketing.
After dinner, Lauren surreptitiously slipped into the kitchen while Olivia and Jared discussed some banking business. Rosa was sitting at the work table, her fingers sifting through the beads of her rosary. When the door closed softly behind Lauren, Rosa opened her eyes. "Rosa? What's wrong?" she asked quickly. "Is it Elena?"
The woman clasped large hands over her broad cheeks and bobbed her head up and down in affirmation. Tears pooled in her chocolate eyes.
" _Está enferma._ She has the fever bad."
Lauren tiptoed into the darkened room and knelt down to feel Elena's forehead. It was burning. Rosa had undressed her and the young woman lay under the blanket clad only in her thin chemise. Lauren turned up the gas light nearest the bed and immediately saw the rash. Red eruptions covered Elena's throat and chest. Lauren unbuttoned her chemise, but knew before she looked that the rash extended down Elena's torso. With an aching heart, she returned to the kitchen.
"Rosa," Lauren said calmly, swallowing the bile that rose in her throat, "did Elena vomit last night or this morning? Did she complain of being chilled?"
_"Sí, señora,"_ Rosa answered dismally. The woman's ravaged face confirmed Lauren's suspicions. Rosa knew the gravity of her daughter's illness.
"Her throat is sore?" Rosa only nodded.
Lauren closed her eyes briefly and prayed for strength. The next several days would be a trial for them all. The task facing her was unpleasant, but she would do it. These people were her friends and they needed her. If she didn't help them, no one would.
Her voice showed no trace of the panic she felt as she began issuing instructions. "Brew some tea and keep the kettle on at all times. Move the baby out of that room at once and don't let anyone else near it. Scald all of the kitchen utensils and don't go into the sickroom again. Where has Elena been today?"
"Nowhere, _señora._ She feels too bad to do much. She was in Pueblo a few days ago showing off Isabela." Rosa's voice wavered as she asked, "She has the scarlet fever, _señora_?"
"Yes, she does." Lauren remained calm despite the turbulence inside her as she went again into the dim room.
Scarlet fever. Isabela. She hated to see. Please, God, no. The baby had been sleeping peacefully all day. Unusual. Lauren forced herself to go to the crib. She pulled up the tiny sacque and cried out in anguish as she saw the rapidly rising and falling chest covered with the telltale rash.
_"Madre de Dios,"_ Rosa murmured behind her.
"Has the fever been going around in Pueblo?" Lauren asked.
" _Si, señora._ Many have been sick. Elena didn't think she would catch it. No one in the family had it when she visited."
"Go do what I told you to, Rosa. I'll stay here with her and the baby."
When the woman had retreated to the kitchen to carry out her instructions, Lauren sat down on the edge of the bed and took Elena's hand. The girl's eyes fluttered open and she offered a weak smile. When she tried to speak, she could only croak.
"Don't try to talk, Elena, I'm here to make you feel better." Lauren pushed back a few strands of lank hair resting on fevered temples.
"Baby?" Elena asked.
"The baby is... sleeping. Everything will be all right. You go back to sleep. I'll give you some tea when it's made." Elena closed her eyes apathetically and her breathing was soon even if somewhat shallow.
Lauren left the room, went through the kitchen, and walked in slow, measured steps into the parlor where Olivia and Carson were playing cards. Jared was slumped in a chair, a whiskey decanter near his hand.
Not quite believing her temerity, she asked for their attention, and when the three had turned startled eyes toward her, she told them about Elena and the baby.
"You can't be serious!" Olivia exploded when Lauren made clear her intention to nurse them.
"I'm quite serious, Olivia," she said levelly. "They need constant care and, since I have no other responsibilities, I'm the one to do it. I only came to tell you that you might want to have food catered in, as the kitchen is so close to the sickroom. And keep everyone out of the house. Go nowhere that isn't absolutely necessary. We should quarantine ourselves for the sake of others."
She spoke with such authority that the other three were momentarily stilled. But the respite was brief. Olivia unleashed her fury in full force.
"If you think I'm going to let a Mexican girl and her brat lie sick and possibly die in my house, contaminating the rest of us, you are very much mistaken. Get Pepe to remove them at once, Jared. Let them take care of their own."
Lauren turned to Jared, who had sobered considerably and was watching her closely through clear eyes. "Jared, if they go, I do, too. Would you have it said that Jared Lockett banished his wife to Pueblo?" she challenged.
He glanced toward his mother and said uneasily, "Lauren, those people are accustomed to epidemics. They die by the hundreds in San Antonio every few years from yellow fever. Pueblo doesn't have proper sanitation to protect them from these diseases, and once one gets started, it runs rampant."
"Then someone who has a lot of money and power should improve their sanitation system, shouldn't he?" Her voice was an accusation. She wondered why these Locketts had ever intimidated her. Right now, she felt very strong.
Jared tried another tack. "It's highly contagious, Lauren. Did you think of that? What's to prevent you from catching it?"
She looked at him steadily. "I had it. When I was ten years old. I was ignored by a father terrified of disease and by a housekeeper angry with me for causing her so much extra work. It's a wonder that I lived. I have not forgotten the misery and fear. I won't let Elena suffer that way. Now am I to nurse her here or somewhere else?"
Olivia opened her mouth to speak, but Jared ordered, "Shut up, Mother." His eyes never left Lauren's face. They stared at each other long and hard. Her hand found its way to his arm and rested there as she gazed up at him suppliantly.
"All right," he said finally. "Is there anything I can do?"
"No. Stay away from the rooms in the back of the house. I'll have Rosa scrub everything with disinfectant as soon as possible. Thank you." It was only when she tried to pull away that either of them became aware of his strong fingers trapping hers against his arm. Slowly, regretfully, they were released.
She didn't look at Olivia or Carson as she moved out of the parlor. At the portiere, she turned and looked back at her husband. "I don't think the baby will live." He saw tears shining in the luminous eyes.
* * *
The days and nights blended together in a montage of pain, suffering, exhaustion, and despair. Isabela died the afternoon of the second day. Lauren tried valiantly to spoon sweetened tea through the tiny lips, but the swollen, red tongue and obstructed throat strangled on it, and the baby couldn't get the fluids essential for her life.
Lauren watched the tiny chest as it shuddered one final time and, without so much as a cry, Isabela ended her short sojourn on earth. Lauren wanted to grieve the loss, but she needed to focus her attention on saving Elena.
Lauren spooned gallons of tea into her patient despite Elena's unwillingness to accept it. Her tongue was covered with painful red blisters that made it look like a strawberry. Her fever rose drastically each night. Rosa and Lauren would strip her and bathe her body with cool water. They didn't tell her about Isabela, and she was too delirious to ask.
Pepe made a tiny coffin, and the infant's grandmother laid her out for burial. Carlos was summoned, but he remained in the stables in compliance with Lauren's orders. It was not only for his protection, but also for those she loved at Keypoint. Pepe ran messages back and forth to the anxious young man who mourned the death of his daughter and feared for the life of his wife.
Lauren never left the sickroom. She sent Rosa to her room for fresh clothing, but barely had time to change into it during her vigil over the sufferer. At night, after they managed to keep Elena's fever from rising further, she would sleep fitfully in a chair near the bed. She prayed constantly for the life of her friend and for continued strength. She prayed, too, that Jared would not contract the disease. The words had formed on her lips, coming straight from her soul before she gave them conscious thought.
The fever literally burned the skin off of Elena's palms and fingers and the soles of her feet. While the girl slept, Lauren gently peeled it away so Elena would not be frightened if she should see the dead tissue hanging like cobwebs from her hands.
Five days after Lauren had gone into the stifling room, she woke from a cramped position in the chair to hear regular breathing instead of the labored, shallow respiration she had listened to for long days and nights. She hurried to Elena's bed and put her hand on a cool forehead. Forcing apart the relaxed lips, she saw that the tongue was less swollen and the blisters had all but disappeared. The rash was fading. She could have laughed aloud. Instead, she sank back into the chair and offered a prayer of thanksgiving.
The next morning, when she told Rosa the news, the old woman wept openly. For the rest of that day, they allowed Elena to sleep a healing sleep. They changed her linens and, at noon, spoonfed her some beef broth until she slipped once again into slumber. Lauren stayed with her to make sure the fever wasn't going to return.
She was exhausted but happy and relieved when she stumbled into the kitchen late that evening. She was surprised to find Jared standing at the back door, staring out over the yard through the window. Rosa had informed him of Elena's recovery earlier.
He turned when he heard her enter. "Lauren, this has gone on long enough," he said without preamble. "I will not let you quarantine yourself in that room one more minute without some rest."
"I'm fine, really I am," Lauren sighed. "I don't think Elena needs me anymore, though. Only plenty of liquids and sleep. I'll let Carlos see her in the morning."
_"Si, señora."_ Rosa came to Lauren and took both of her hands in hers, kissing them in turn. "Señor Jared, she is an angel."
"Yeah, she's an angel all right, but she looks like hell right now," he said grimly.
Through her fatigue-muddled mind Lauren noted absently that he didn't look all that wonderful himself. Stubble covered his chin and upper lip. His cheeks were gaunt and sunken under red-rimmed eyes.
Rosa could have told her that for days he had paced, cursed, threatened, and pleaded. He was like a wild man in his worry. His only source of nourishment was a shot of whiskey taken at regular intervals.
Lauren tried to focus her eyes, but images began to blur, recede infinitesimally, then loom hugely. The kitchen was spinning crazily. "Jared—" she cried hoarsely before she collapsed into the strong arms reaching out for her.
"She's unconscious," he said. "And hungry, from the feel of her. I'll bet she's lost ten pounds. First thing in the morning, Rosa, fix her a big breakfast and serve it to her in her room. Stay with her until she eats every bite. I think she needs rest first."
He swept the inert figure into his arms and carried her upstairs to her room, kicking the door shut behind him. He stood for a moment, allowing his eyes to grow accustomed to the darkness, then moved toward the bed. There was just enough light coming in through the windows for him to see without lighting the lamp.
Lauren murmured unintelligibly as he put her feet back on the floor, supporting her with his body. She leaned heavily against him and he muttered imprecations at her foolishness for totally exhausting herself like this. He tried to keep his mind off the body pressed close to his. How could she stay in a sickroom for a week and come out smelling like lavender? He didn't know that Lauren had asked Rosa to fetch a bottle of cologne from her room which she added to the water she washed with each day.
Well, I can't just dump her on the bed, Jared reasoned. With trembling fingers, he began unfastening the buttons on the back of her shirtwaist. Her head lolled against his chest. It took a long time for him to get to the last button because he used only one hand, supporting her with the other. His trembling fingers lacked their usual dexterity.
He pulled the blouse out of the waistband of her skirt and then began undoing the fastener. He untied the ribbons of several petticoats, cursing as they knotted in his fingers. Why do women wear so damned many clothes anyway? he thought. Finally he was able to push the skirt and petticoats down over her hips and they fell to the floor in a ruffled froth at her ankles.
He paused, drawing deep breaths in an effort to supply oxygen to his brain, which was whirling like a maelstrom. If she woke up now, he thought ruefully, she would probably scream the house down.
With meticulous care, he supported her against one of his arms and, leaning her back, slowly pulled the shirtwaist from her shoulders and slid the sleeves down her arms.
It was off. She still slept. He was perspiring and trembling. He pressed her against him, postponing the moment when he would look at her, savoring the anticipation.
He reached up and began hunting for the pins that held her hair, removing them gently as he found them among the thick tresses. Her hair tumbled down her back and over her shoulders, spilling into his hands. Then, as he had wanted to do ever since he had first seen her, he ran his hands through the black silk, caressing each strand, rubbing the smooth curls between his fingers, delighting in the feel of them the way a miser loves the feel of gold. He buried his face in her hair and whispered accolades to its glory.
Jared lowered her gently onto the bed, slipping the skirt and petticoats from around her ankles. She lay on the pillow and sighed contentedly, her hair fanning out behind her on the snowy linen.
Jared sat on the side of the bed, easing himself down in order not to wake her. God! She was exquisite. Even the lines of fatigue around her mouth and the hollows in her cheeks added to her beauty. Long black lashes rested on alabaster cheeks. He followed the column of her neck to the base of her throat, where he saw the flutter of her pulse. Her shoulders were white and sloped into a flawless bosom.
He hesitated, but his fingers moved of their own volition and reached out to the top of her camisole. He untied the blue satin ribbon that was threaded through the eyelet lace and slowly, leisurely unbuttoned the first few buttons. Again he wanted to prolong the anticipation.
His eyes traveled down to her waist, which was wasp-thin, then to the slight flair of her hips. He usually preferred more voluptuous curves, but she was perfectly proportioned, he noticed as he took in the long, slender thighs and shapely calves.
Damn! He had forgotten the shoes. The tedious buttons were almost impossible to handle in the dark, but he finally succeeded in undoing them all.
When he had slipped her shoes off silk-encased feet, he returned his gaze to her face as his fingers slowly parted the camisole. She didn't stir. His eyes wandered aimlessly, until he rewarded his patience and looked down at her.
He had imagined how she would look, but the fantasies were inadequate and he wasn't prepared for the vision that greeted his eyes. Two perfectly shaped breasts, round, high, and firm, were displayed. Her skin was as creamy and white as a magnolia blossom. The nipples that crowned each soft mound were virginally pink. Botticelli would have adored her. Her ethereal beauty was definitely _quattrocento._ Rosa wasn't far from wrong. In her naked loveliness, she appeared to be an angel.
But Jared was mortal, and he wanted her as he had never wanted a woman before. He carefully lowered his head and kissed the pulse in her throat. Then his lips traveled with a blissful laziness over her breasts, nibbling and licking lightly so she wouldn't ever know that he had worshiped at this temple of her body. She was forbidden to him. It was a self-imposed denial, but that made it even more binding.
But now, now...
He raised his head and touched one rosy nipple. Gently rolling it between his fingers, he watched in fascination as it responded to his touch and became pointed and inviting. Unable to resist, his mouth opened around it. It melted against his tongue like a piece of sugar candy, and tasted even sweeter.
His eyes lifted once again to the face that lay in peaceful repose in the cape of black hair. "Lauren, forgive me," he whispered as he lowered his mouth once again.
* * *
When Lauren awakened not long before noon, she couldn't remember the events of the past few days or why she was sleeping so late and wearing her underwear and stockings. She stretched cramped muscles as she pieced together the fragments flashing through her memory.
Elena! She recalled the girl's illness and jumped out of bed, throwing back the single blanket that covered her. Why was she sleeping on top of the bedspread? Disoriented, she stood in the middle of her room, her hands on either side of her head, trying to banish the yellow spots that danced against the black curtain which seemed to have fallen over her eyes. She reeled dizzily. She had stood up too fast for the blood to flow into her head. Lack of proper food for the past six days had rendered her weak.
Rosa bustled in as Lauren was feeling her way back to the edge of the bed.
"Señora Lauren, you are awake! You have been sleeping like a baby."
"How is Elena?" she asked quickly. The radiant look on Rosa's face dispelled any fears.
"She is weak and sleepy, but she ate some toast this morning and talked to Carlos." The happy features drooped a little. "We had to tell her about the _niña,_ and she is very sad. But seeing Carlos made her feel better. She is grateful to you, _señora._ I thank you, too." Her lip began to tremble.
"I'm glad I could help Elena. I only wish we could have saved the baby, though after the high fever, Isabela might never have been completely healthy."
" _Sí,_ her little soul is in heaven and she is well now. If I know Elena and Carlos, they will make another _niño_ soon." She grinned broadly. "I brought your breakfast." She stepped out to the hall and retrieved a tray she had left on a hall table. The delicious aromas emanating from the dishes made Lauren's mouth water. When had she last eaten a complete meal?
"Señor Jared told me to see that you stayed in bed and ate everything on the tray."
"Is he... I mean, are the rest of the family healthy? I haven't seen them for almost a week."
"You saw Señor Jared last night, _señora._ Don't you remember? You fainted in the kitchen. He brought you up to bed. Carried you in his arms."
The room was spinning again and in her ears was a great roaring. She sipped the scalding tea and tried to keep her hand from shaking as she placed the cup back on the saucer.
"No... no, I... uh... don't remember that. I know I was very tired."
"He was angry I think for you to be so tired. You would like a bath, _señora, sí_?" Without waiting for an answer, Rosa waddled into the bathroom, gathering up the discarded clothes lying on the floor beside the bed.
Her clothes! Jared had put her to bed. Jared had undressed her!
She had dreamed of him during her deep sleep. She recalled the dreams vividly now. Jared was leaning over her and looking at her tenderly. His head was resting against her flesh. If only her arms hadn't been so heavy, she could have reached up and run her fingers through the sun-bleached hair that tickled her throat. In one dream, Jared was whispering Spanish words against her ear. In another, he had been doing something with his mouth that spread a delicious warmth through her.
She finished eating slowly, distracted by the memory of her disturbing dreams. When she began undressing, she noticed the ribbon from her camisole was missing. She would have to tell Rosa to be more careful when laundering her underwear.
She stepped into the bathtub and welcomed the soothing warmth of the water. It had been days since she had enjoyed the luxury of a bath. Cupping her hands and ladling a handful of suds over herself, she was shocked when the soapy water caused a stinging sensation on her flesh. Involuntarily she gasped. Examining herself self-consciously, she noticed her nipples were slightly sore, though when she touched them, they tingled and a thrilling shiver coursed through her body. Her breasts were chafed, as if abraded by something scratchy. Whatever could have—
Her eyes dilated with the horror of the thought. Trembling fingers crammed against tight, compressed lips gone white with shock.
No! No! That could not be. Her dreams and this actuality had nothing to do with each other. It was unthinkable. Still, she quaked at the possibility. Jared, unshaven, seeing her, touching her, kissing...
Hastily she got out of the tub, dried quickly, and wrapped a robe around her before returning to the bedroom. "Where is Jared?" she asked Rosa timidly as she prepared to go downstairs and visit Elena. She wanted to avoid him. Meeting him face to face with her speculations vividly intact would be too humiliating.
"He went to Austin this morning. He say he may be gone several weeks. Something to do with business. He left a package for you." She bustled out the door and returned immediately with a small box. "He say you need this next time you go to Keypoint."
Lauren untied the ribbon around the box, lifted the lid, and saw a blue silk scarf nestled in the tissue paper. A bandana. He had remembered that she needed one. Tears misted her eyes, but aware of Rosa's keen perception, she shoved the scarf into one of her lingerie drawers with feigned indifference.
"I must remember to thank him when he comes home." She walked out her door leaving Rosa shaking her head, baffled and disgusted.
# Chapter 15
"Have you heard from Jared, Olivia?" Carson asked one night at dinner. "Seems he's been in Austin an uncommon length of time."
"Come, Carson," Olivia said with a laugh, "you know why Jared is 'detained' in Austin. I'm amazed that he abstained as long as he did. Having a wife of convenience drives a man to find some release from his physical needs, doesn't it, Lauren?"
Lauren's fork clattered to her plate. How dare the woman be so blatantly crude! "Since it's Jared's 'physical needs' you're referring to, you should save that question for him," she retorted angrily. But underlying her anger was heart-sickness. Her mother-in-law was most probably right. Why else would Jared extend his stay in Austin? She hadn't guessed she could miss him so much. The longer he was gone, the more the tension mounted in the house.
The daily routine in Coronado was different from that at Keypoint, but was routine nonetheless. Elena gained strength each day. If the weather permitted, she would walk with Lauren in the gardens, which boasted only a few chrysanthemum blooms. The wind, when it whistled in unpredictably from the north, was biting and bitter, and there were a few days in early November when it rained incessantly and Lauren thought she would go mad at the tedium of her life in the large house.
She longed for the hectic days at Keypoint with Rudy and Gloria, their loving banter, and the children with their pranks. She missed her quiet talks with Maria, too. A few months ago, Lauren would have been scandalized by a woman living with someone else's husband, but she now condoned Maria's life with Ben. Their love had been pure. Lauren's uncharacteristic broadmindedness was due in part to her knowledge of the kind of life Ben must have had with Olivia.
Olivia went to the bank most days, and when she stayed at home, Carson telephoned often from his office. He was always at the house for dinner, and Lauren was grateful for his kindly presence. She could not have borne being alone with Olivia. The woman looked at her with such hostility that it startled Lauren each time she caught Olivia's emerald glare on her.
Carson continued to be obsequious to Olivia. They talked constantly about the railroad and the plans for the power plant. Apparently any hesitation Olivia had once had about forming an alliance with the Vandivers no longer existed. She commended their decisiveness and ability to manipulate others to achieve their goals. Olivia mentioned Kurt often, and always with a sly look in Lauren's direction, which made her feel distinctly uncomfortable.
One evening at dinner, Olivia observed her daughter-in-law balefully through a forest of lowered lashes. She would never forgive her for convincing Jared to allow that sick Mexican and her baby to remain under her roof. The fact that Lauren gave shelter to an ailing Mexican was bad enough, but her real transgression was that she had triumphed over Olivia in an argument with Jared. More threatening than that, she had forced Jared to make a decision. That could be dangerous to Olivia's plans. If Lauren gained any control over him, it could jeopardize everything.
Olivia also resented that, for appearance's sake, she had to act the doting mother-in-law. God! Friends pressed her for news of approaching grandchildren. They simpered over their teacups about how handsome Jared and Lauren looked together, and ladies coming into the bank stopped to remark how wonderful it was that Jared, after having waited so long, had finally found the perfect mate. Olivia bore it all with a stiff smile and the proper responses. She dutifully attended church with Lauren every Sunday. The young woman insisted on going, and Olivia knew it would never do for Jared's wife to be seen there alone.
It was evident that Lauren had formed quite an attachment to the family at Keypoint. Whenever Carson drew her into conversation, she ended up relating some anecdote that had happened at the ranch. She never mentioned names, though Olivia knew each character in these narratives intimately. When you hate someone for as long as she had hated Maria Mendez, her son, and his brood, you come to know them quite well. She had made it an obsession to learn all she could about them.
Unable to punish the Mendezes just yet for the humiliation they had caused her, she found a perfect scapegoat in Lauren. She had been raised to honor her elders, to respect family, and to bear persecution with forbearance and forgiveness.
Olivia was quite sure that the marriage between Jared and Lauren had not been consummated, though she was well aware of her son's strong sexual appetite. Her lips tightened into a bitter line when she equated it with her late husband's.
She was certain Jared's fierce pride would have kept him from loving any woman chosen for him by his father. Still, the girl had a look, a way about her that Olivia knew was irresistible to men. That gentleness, that vulnerability shook them to their very core. She herself had never had that quality, and she loathed it in other women.
In every glance, gesture, and spoken word, Olivia tormented Lauren. She must keep Jared and his wife at odds with each other. To think that they might reverse her whole scheme and from a lasting affection was an abhorrent possibility.
It wasn't jealousy, Olivia assured herself. Jealousy was such a petty emotion, and far beneath her dignity. All she demanded in any relationship was absolute loyalty. Ben had betrayed her love with disloyalty. Olivia was determined that Jared would not.
"How is Elena, Lauren? All well by now, I hope." Dinner was over and Carson had followed the ladies into the parlor for coffee. Lauren knew that Olivia's question didn't stem from concern. It was merely to remind her again that she had gone against Olivia's commands.
"Yes, she's recovered. The death of her baby was a blow she'll not soon get over, but that is to be expected."
"Most unfortunate that the child couldn't live, Lauren. You did a miraculous job of pulling the young woman out of it."
"Thank you, Carson, but I had self-interested motives. Scarlet fever nearly cost me my life. This was my only opportunity to fight back."
"It's a shame those Mexicans in Pueblo can't keep their community clean and sanitary." Olivia's face was ugly.
"I'm sure they do their best. I don't think they choose for their children to die," Lauren said quietly but with conviction.
"Nonsense. You haven't been here long enough to know how dirty those people are. They're vile."
"How can you say that Rosa is vile and then eat the meals she prepares? She's the cleanest person I've ever met," Lauren said heatedly.
Olivia tossed her head in anger. Her green eyes were brilliant and her slender fingers gripped the arms of the chair like talons. "It won't be long until we will be able to relieve some of their worries in Pueblo. I can't wait—"
"Olivia!" Carson interrupted abruptly. "I don't think we should discuss this any further. I can see that it upsets you." He gave her a telling look.
Olivia drew herself up sharply, then let her breath out slowly. She had almost made a blunder and was grateful that she had been stopped. "You are quite right, Carson, it does upset me. More coffee?"
Lauren excused herself and went upstairs. Unable to stop herself, she paused outside the door of Jared's room, staring at it intently. What did she expect to see? Would the door dissolve and reveal him sitting in a chair smoking one of his inevitable cheroots? Would he materialize before her as he did in her dreams night after night?
No. He wasn't here. He was in Austin on "business." Surmising what kind of business he was involved in brought a heavy pain to her heart. Was he with another woman? Women? Why should she care? Yet she did.
Her hand went instinctively to the watch pinned on her breast. Without conscious thought, her fingers lowered and lightly brushed her nipple through the soft cotton of her blouse. Her whole body flushed hotly and she caught her breath at the sensations that assailed her.
Did you really touch me here, Jared? she asked the darkness. She didn't want to know, didn't want to care, but couldn't refrain from adding, What did you think of me?
Anguish and loneliness accompanied her into her room.
* * *
Jared came home the day before Thanksgiving. Lauren was in her room reading. Her heart leaped in her chest when she heard the familiar clump of boots and the ring of spurs in the hallway. She gripped the book tightly and held it against her. The door to his room opened and closed quietly, and she heard the thump of valises on the floor. He moved around the room awhile, then the bed creaked as his weight was lowered onto it. All became quiet.
Try as she might, Lauren couldn't regain her interest in the book.
She took a great deal of care in dressing for dinner, donning a flattering dress and soft kid slippers. Her hair had been washed and brushed. She had a wild impulse to leave it hanging loose but, of course, that would be unseemly. Instead, she let Elena arrange it in a glossy pompadour.
She heard Jared leave his room and go downstairs, and followed a few minutes later. She was incredibly nervous. How could she face him after he had taken her to her room, undressed her, and put her to bed? And if he had...
She walked into the parlor and her heart turned over when she saw his tall, broad back leaning over the sideboard as he poured a drink.
"Lauren, you make the evening glow. How are you, my dear?" Carson came to her and kissed her cheek.
She watched Jared over Carson's shoulder. His back stiffened and he took a long gulp of whiskey before he turned around. He is so handsome, she thought mournfully. She could bear his indifference and cruelty, his desertion, if he were ugly or repellent to her. But from the first she had felt a strange chemical attraction to this man.
He met her eyes over Carson's pudgy form and lifted his glass in a mocking salute.
"Jared, would you pour Lauren a sherry, please, and I'll have another." Lauren noticed that Olivia's eyes were especially brilliant. The news that Jared had brought from Austin must have pleased her.
"I'm so glad the Vandivers are coming for Thanksgiving tomorrow. Thank you for delivering my invitation, Jared." Olivia watched closely as Jared handed Lauren her glass of sherry.
"You're welcome," he replied. He looked down at Lauren from his towering height as she took the drink. Their fingers touched briefly. The contact was electrifying and she thought she heard his breath rush past his teeth in a soft gasp. But as her eyes traveled up the long body to meet his glowing eyes, she saw them frost over immediately.
"I knew Lauren would want to see Kurt again. It's been such a long time." The slurring words were said low and for her hearing only. Why was he doing this again? He was as arrogant and hateful as the first night she had seen him in this room.
All her excitement at seeing him drained away as dinner dragged on interminably. Jared drank more than he ate. What had happened to the healthy appetite that couldn't be appeased at Keypoint? Maria and Gloria were always teasing him about his hollow leg.
He was sullen and erratic. One minute he was strenuously criticizing the Railroad Commission and their ineptitude, and the next minute, he was muttering into his glass.
Lauren went into the library to read by the fire after dinner. She had stood all the tension she could in one evening. The others retired to the office to discuss the results of Jared's trip to the capital.
Hours later, Lauren was still in the library, seated in an overstuffed chair with her shoes lying on the hearth and her feet curled up under her legs.
She didn't know how long he had been there before she noticed Jared standing in the doorway. When she looked up at him, he stumbled wearily into the room and collapsed into the twin chair beside hers. His head fell back against the plump cushions and he closed his eyes. Lauren sat still. As the minutes ticked by, she thought that he must have fallen asleep, but eventually he opened his eyes, though he didn't move his head or body.
"It's awfully late, Lauren. Why are you still up?" He sounded exhausted.
"I got absorbed in my book. Sometimes I can read well into the night if a book is particularly interesting." Or if I can't get you off my mind, she added to herself.
She looked like a vision in the firelight. Its shimmer caught on the black hair and seemed to ignite each strand. Her dress was a soft rose wool with pearl buttons at the neck. With the firelight, it enhanced the enticing blush on her cheeks, which he knew were velvet-soft. Behind her eyeglasses, her eyes were wide and deep and clear.
Jared cleared his throat, rousing himself from his perusal. "What do you read? You are quite a bookworm, you know," he teased softly.
"Everything," she said lightly, indulging his need for small talk. "Books were my best friends when I was growing up. Other children didn't want to play with the minister's daughter. It made them uncomfortable, you see. My mother died when I was three." He noticed her reach up and touch her watch. "So my friends were the characters I found in print. I read the classics, history, and philosophy. But for fun, I like Dickens, the Brontës, and Jane Austen."
His eyes were closed again, and it surprised her when he picked up the conversation. "When I was a boy, I was enthralled with Poe. Ben used to rile me about reading such 'rubbish,' as he called it. 'If you want to hear ghost stories, have Thorn tell you some.' Thorn's were pretty good, too." He laughed. "He used to raise the hair on my neck telling me about Indian legends, taboos, and secret rites." He stared reflectively into the fire, the flames dancing in his amber eyes. Putting his thumb and middle finger against his lids, he rubbed them in fatigue. "I don't have time to read anymore."
She hesitated only a moment before asking gently, "Was your trip a taxing one?"
He sighed heavily. "Yes. Some of the men in public office are frighteningly stupid. I'm sick to death of having to pander to them. I want..."
When he didn't continue, she urged him softly. "Yes. You want...?"
It was the only prodding he needed to voice his innermost thoughts. "I want the railroad because it's the only way we can operate a successful, profitable ranch in the twentieth century. Ben wanted it so bad he could taste it, but I hate having to go through so much red tape and catering to idiots in order to get it." He leaned forward and clasped his hands between his knees.
Lauren remained quiet. This was a time for listening.
"I want to live on Keypoint and ride the fences like any other _vaquero_ and let someone else do all of this politicking."
Lauren swallowed her caution, stood, and went to him timorously. She placed her hands lightly on his shoulders and massaged the knotted, tense muscles. "Maybe when the railroad is finished, you can do that, Jared. I hope so, for your sake."
He leaned back in the chair again, noticeably relaxing under the magic touch of her fingers. After a few moments, she said, "Thank you for the scarf. It's lovely."
He looked up at her standing behind him. His eyes were tired and bloodshot, but he read the encouragement and compassion in hers as she smiled down at him. He covered one small hand on his shoulder with his, then grasped it and squeezed tightly. He brought it up to his face and pressed her palm against his hard cheek.
"Your hands are beautiful, Lauren. I noticed—" He broke off, feeling that he was probably making a fool of himself, but then went on, "I noticed them that first night we were at dinner together." His fingers smoothed down her slender counterparts.
"If they are, it's from hours at the piano, I suppose. My father once told me my hands were like my mother's. She played, too."
"Do you miss it?"
"Yes," she admitted readily. "I suppose my music is for me what riding the line is for you."
He studied the hand in his with the appreciation of an art connoisseur for a masterpiece. Bringing it back to his mouth his tongue brushed each fingertip. Lauren's eyes closed.
His lips moved to her wrist, and when her ruffled cuff obstructed him, his thumb slipped between the buttons and buttonholes and laid bare her translucent skin.
"You're so soft," he murmured as his lips caressed her wrist. "You make everything seem so uncomplicated, so..." His words trailed off as he buried his lips in the soft cushion of her palm. Her heart fluttered erratically when she felt the moist warmth of his tongue sliding sensuously over her flesh.
He raised his hand and trailed a finger down her cheek as his eyes traveled her face beseechingly. "Lauren, I—"
"Jared." Whatever he was going to say was arrested by Olivia's imperious interruption. "Carson is waiting to go over that last group of figures before we call it a night. Lauren, dear, there's no need for you to wait up."
Jared's mouth tensed into a thin, hard line and the muscles of his face became rigid as he pushed himself out of the chair and strode from the room.
Lauren retrieved her shoes and her book and, after turning off the lamp, faced the door. Olivia's tall figure was still silhouetted in the doorway. As Lauren came closer, she saw that one of Olivia's black eyebrows was raised speculatively.
"Goodnight, Lauren," she said coldly.
"Goodnight." As she mounted the stairs, Lauren could feel her mother-in-law's piercing eyes boring into her back.
* * *
Lauren went down to the kitchen the following morning to ask if she could help Rosa with her preparations for the Thanksgiving meal. She was assured that things were well underway. Elena was sorting laundry at a work table. When Lauren was about to leave the kitchen, she offered to take her things upstairs and save the girl the steps.
"I'll take Jared's things, too," she offered as Elena piled her arms with fresh-smelling clothes. The stiff, starched white dress shirts and the soft-colored ones Jared wore on the trail were added to Lauren's load.
She went up the stairs quickly and tapped on Jared's door. There was no response. She was about to call out to him, when she heard splashing sounds coming from one of the bathrooms down the hallway. To her knowledge, he had never used the bathroom that connected their rooms.
Indecisively she stood outside his slightly opened door. What could it hurt? she asked herself as she eased the door ajar far enough for her to slip inside.
The room was simply furnished. A bed with a tall oak headboard occupied one wall. A massive wardrobe filled another. A bureau, complete with shaving mirror, washbowl, and pitcher, stood in the corner. The only other piece of furniture was a tall wing chair.
The brown striped curtains at the window had been pushed aside to let in the sunlight. The bed had been neatly made with a spread that matched the curtains. There was no sign of disorder around her as she crossed the decidedly masculine room toward the bureau. She carefully lowered the stack of shirts onto its glossy surface and was turning to leave when something caught her eye.
Apparently Jared had emptied his pockets onto the top of the bureau the night before. She studied the items curiously. A tortoiseshell pocket comb. Did he ever use it? she asked herself with a fond smile. Coins of every denomination. A roll of bills. She recognized his pocket watch with its gold chain. Three pieces of paper, doubtless receipts of some kind, each folded neatly in two. A key ring with six brass keys. A tiny box of matches. And—
Her heart stopped for a moment. When it started again, it beat so rapidly that she covered it with her hand, enfolding her watch in trembling fingers.
There, winding its way through the other items, glaringly out of place, was a slender, blue satin ribbon. A ribbon much like the ones which laced through the chemises that Mrs. Gibbons had made for her before her wedding. Much like the one she had lost only a few weeks ago, the night Jared had undressed her and put her to bed.
She didn't know she had spoken his name until the soft, wondering whisper vibrated through the still room and jarred her out of her stupor. Just as well it did, for she heard the bathroom door down the hall opening. Spinning around, she stared at the door through which he would walk any moment. She mustn't let him see her!
She dashed toward the door opening into the bathroom connecting to her room and pulled it open. She had just closed it after her when she heard him enter his room. He was humming softly, quite unaware that he had caused an emotional avalanche in the breast of his young wife.
# Chapter 16
Olivia, your bountiful table never ceases to amaze me," beamed Parker Vandiver. He had eaten several helpings of traditional Thanksgiving food, and was enjoying one last glass of wine.
They were sitting at the table in the same arrangement as on the first occasion the Vandivers had taken a meal with the Locketts. Olivia and Carson sat at either end, Parker and Jared on one side, Lauren and Kurt on the other.
"Yes," Kurt chimed in. "Everything was delicious, made even more enjoyable by the beautiful company."
The hand wrapped around Jared's wineglass clenched, the knuckles turned white, and Lauren was amazed that the glass didn't shatter under the pressure.
She moved away from Kurt, his presence at her side becoming more unbearable each minute. Since they had taken their chairs at the table, his thick, heavy thigh had pressed against hers, and he used every opportunity to lean toward her, touching her in some way.
She hadn't lost her aversion for the Vandivers. Both father and son repelled her with the brutal strength apparent in each move of their husky bodies. Their polite conversation and demeanor, she was sure, stemmed from some secret ulterior motive.
She was grateful to Carson when he suggested they retire to the parlor for coffee, thereby relieving her of sitting next to Kurt.
"We were sorry we missed you when we visited Keypoint, Mrs. Lockett," Kurt said as everyone was seated in the formal room. Lauren had taken a chair next to the fireplace, forsaking the couches for fear that Kurt would sit beside her.
Lauren didn't want to lie and say she was sorry she had missed seeing them, too. Not knowing what else to say, she replied truthfully, "Jared and I went on an outing that day."
No one noticed Olivia's shocked face, for Jared was saying caustically, "It was our honeymoon, remember?"
Kurt maintained his poise. "Yes, so it was. Did you like life on the ranch, Mrs. Lockett?"
"It was a different experience for me, Mr. Vandiver. I enjoyed the uniqueness of it." While she responded to Kurt's question, she was remembering the night before with a pang of wistfulness. She had felt so close to her husband for the first time. What had happened since then to make him cold and unapproachable? Because it was a holiday, she had worn a new dress in a violet-gray shade of georgette, knowing that the color highlighted her eyes. If Jared had taken note of her appearance, he hadn't deigned to comment.
"You obviously like ranching, Jared." Parker studied him shrewdly.
"Yes," Jared retorted shortly and went to the sideboard. He poured a large glass of whiskey and walked to the wide windows, staring outside, his attitude one of total indifference to the rest of them. Yet all through dinner, his mood had been one of constrained violence, and Lauren was apprehensive.
"I've never had the time to learn the ins and outs of ranching. Maybe you could teach me all about it sometime, Jared."
Jared took a swallow of his drink. "I've never had time for the ins and outs of school teaching, Mr. Vandiver. If you want to learn about ranching, I suggest you learn the way everyone else does—by trial and error. That's the way Ben did, and it seemed to work for him." He turned on his heels and fixed Parker with a menacing stare. "And learn to ranch some place other than Keypoint," he warned.
"Jared, how rude!" Olivia chided him. "We're business partners with the Vandivers now." She smiled reassuringly at Parker.
"Not in ranching we're not." Jared tossed down the last of the whiskey and turned to the window, putting his back to them again.
An awkward silence descended over the room. Olivia fidgeted with her coffee cup. Carson covered a pretend yawn. Lauren looked bleakly at Jared. She knew he felt that Keypoint, his true love, was threatened by these unscrupulous men.
"I, for one, am looking forward to the groundbreaking ceremonies," Carson interjected heartily. "We've arranged festivities for the entire population to enjoy. It should be quite a celebration."
"That's the spirit, Carson," Parker agreed. "We want the townspeople to know that this railroad is something they can all be proud of."
"Will you be going to the ceremonies, Mrs. Lockett?" Kurt asked.
Lauren floundered. "Well, I suppose so." Then she added, "With my husband, of course."
No one had a rejoinder to that, so silence descended again.
"I would love a game of bridge. Would anyone else be interested?" Olivia's green eyes sparkled, and she looked younger and gayer than Lauren had ever seen her. She must have stopped the heart of many a New Orleans blade when she turned on the charm.
"That would be wonderful. Lauren, would you be my partner?" Kurt walked over to her and extended his beefy hand. Loath to touch it, Lauren shrank from him. And when had he started calling her by her first name?
With a broad yawn, Carson said, "Go ahead, Lauren. Olivia will play with Parker. I dislike bridge myself." He walked toward the door, saying over his shoulder, "I'm going into the library for a nap. Wake me when the game is over."
Lauren looked helplessly toward Jared as he stormed out of the room without speaking to anyone.
Olivia smiled to herself. She hated to admit to a flash of concern when Lauren had mentioned their "outing." Surely Jared was smart enough not to get involved with the chit any more than necessary. But if he wasn't, Olivia planned to make such an involvement difficult, if not entirely impossible.
They gathered around the game table in the corner of the room and the rubber was soon started. Lauren was able to go through the motions without concentrating too hard. Her mind was on Jared and his whereabouts. She didn't think he had left the house. What was he doing?
The rubber went on for hours. Carson eventually joined them and, diplomatically, cheered first one team and then the other.
When Rosa announced a cold buffet had been laid out in the dining room, everyone protested but strolled into the room and began filling their plates with leftover turkey, salad, and relishes. Olivia asked Rosa to fetch Jared.
They moved back into the parlor toting their plates. Lauren sat on one of the small couches in the shadows of the room. Realizing her mistake too late to move, she saw Kurt striding over to her.
"You didn't get much to eat, Lauren," he said as he sat close beside her.
She murmured, "I'm not very hungry." His looming closeness would have taken away any appetite she might have had.
"You didn't get any of the pickled peppers. Here, try one." He lifted one of the peppers from his plate and offered it to her.
She remembered vividly the first night she spent in Coronado and the innocent-looking tomatoes garnishing the beans. "No, thank you. I don't like them." She shook her head.
"This isn't one of the hot variety. It's very mild. Look." He bit into the pepper and chewed it slowly and deliberately to show her that he suffered no ill effects. "Come on, Lauren."
Hoping he would go away if she complied, she leaned forward and took a tiny bite between her teeth. She sat up and pulled away from him as he smiled down at her.
When she looked away from that victorious smile, she saw Jared standing under the portiere watching them.
His fists were clenched at his sides, the muscles of his jaw working furiously. His eyes held a demonic gleam as he took in the scene. She realized how intimate she and Kurt must look to him, sitting together on a couch in the shadows.
"Jared." Her voice was a strangled whisper.
Kurt whipped his head around, following her gaze. He spotted Jared and noted his fury. Turning back to Lauren, he chucked her under the chin playfully. "See, it didn't hurt a bit."
She practically pushed him from her as she stood and hurriedly followed Jared, who had spun on his heels and stalked out of the room.
She caught up with him in the hall and reached out a hand to restrain him. "Jared." She cleared her throat. "Jared, I—"
"Shut up!" he barked at her and jerked his arm away. His face was contorted with rage. She wanted to shout at him, plead with him, and strike him all at the same time. Anything to erase the angry, accusing look on his face.
Rosa was cowering against the wall wishing she could escape unseen. Lauren saw her and, knowing that conciliation was hopeless, went to her. "Rosa, I'm going up to my room. I have a headache. Would you please extend my apologies to our guests?"
_"Sí, señora,"_ Rosa whispered, hoping that Jared wouldn't say anything.
He didn't, but only glared at his wife as she retreated.
* * *
Jared lit another cigar and fanned out the match as he stood at the window of his upstairs room. He watched Lauren as she walked slowly along the fence that encircled the front yard. Wasn't she cold out there? Yes, she was, he concluded when he saw her shiver slightly. What in hell was she doing walking around in the front yard in the middle of the night?
The Vandivers had finally departed. It had taken every ounce of self-control Jared possessed not to kill Kurt Vandiver with his bare hands. He didn't remember ever being so consumed with anger and hatred just at the mere sight of a person.
As he left, Kurt had turned to Jared and said, "I'll look forward to seeing you at the groundbreaking. I hope Lauren is feeling better by then. I would hate for her not to be there." His smile was mocking, and Jared's fists clenched at his sides to keep from smashing them into that smirking face.
After that, he had spent several hours getting progressively drunker. The sound of Lauren's slippers tapping on the floor of the hall had penetrated his alcohol-befuddled mind, and he listened, following their progress.
It surprised him when he heard the front door opening and closing. From his front window, he saw her step from the porch and walk toward the iron gate. She was wearing the dressing gown he had seen before, her hair trailing down her back. She folded her arms tightly to ward off the chill November air.
Bitch! he thought vindictively. She was playing it so cozy with that Vandiver buffoon. He had been right about her all along. She was a schemer, an opportunist, a whore who teased and tormented but never came across.
Well, I don't give a damn, he swore. But he did. That was what rankled. He did care, and it ate at his gut every time he saw that Vandiver bastard go anywhere near her.
Jared watched her now with the moonlight shining silver on her hair. She leaned wearily on the gate and bowed her head, some of her hair falling forward over her cheeks. Her slender back was outlined by the trim, snug fit of the dressing gown, and with a growing ache in his groin, Jared remembered how her body looked partially clothed. He gulped the whiskey in his glass.
If he had to have a wife of convenience, why couldn't she be ugly? Why did she have to be Lauren?
His sexual exploits were well known in the capital city, and in others as well. Whores vied for the chance to offer him their services. His ardent lovemaking was followed by a nonchalant, take-it-or-leave-it attitude that challenged every woman's innate feminine instincts. Perversely they loved him for it.
But when he had been in Austin, he had had no desire to frequent any of his usual haunts. Lauren's image was constantly in his mind, leaving no room for others. Her body was the one he saw in his fantasies, the one he craved.
Furious with the monklike existence he was leading, he had finally forced himself to go to one of the most exclusive "clubs" in the city. He was greeted enthusiastically. Everyone had missed him. Had marriage spoiled Jared Lockett? they asked.
He had drunk whiskey. He had gambled. But when it came time to choose a woman and retire with her upstairs, he was tired of pretending to enjoy himself.
Striving for objectivity, he surveyed the women displayed provocatively before him. This one was too heavy. That one's hair was too brassy. Another one was too coarse. And so it went.
Finally, disgusted with the place and more so with himself, he mumbled some lame excuse and returned to his hotel room. Lying alone on the bed, the hard throbbing between his thighs painfully demanded assuagement. He resorted to a means that had been unnecessary since early adolescence.
Afterward, as he was drifting off to sleep, he convinced himself that it was purely an accident that it had been Lauren's name he had cried into his pillow when the tumult came.
I ought to leave right now. If I did, I could tumble several good whores before morning. But he didn't want anyone else. No, he decided. What I ought to do is act like a man and tumble my own wife.
Lauren was walking slowly back toward the house. Why not? he thought. She _is_ my wife, isn't she? She flirts with everyone from the lowliest _vaquero_ at Keypoint to Kurt Vandiver, and God knows how many others when I'm not around.
Why not?
He took one last long swallow of the liquor before he staggered out into the hall.
* * *
Lauren had been miserable when she retired to her room pleading a headache. She threw herself across her bed and cried as she had not done in a long time. The tears were bitter, angry ones, and had no healing properties. She had drenched her handkerchief and the pillowcase before the tears ran out and then she cried in dry sobs.
Elena had knocked and inquired about her, but Lauren sent her away with reassurances that she was feeling better, and only needed a good night's sleep.
She heard Jared when he came upstairs and went into his room. The house had been quiet for some time when she undressed and climbed into bed. Sleep eluded her. Every time she closed her eyes, Jared's angry face rose up before her. The lips that had curled in contempt were so different from those that had kissed her. The eyes that had looked at her with such enmity were not the same ones that had looked at her tenderly over the top of his red bandana.
Seeking respite from the agonizing images, she donned her robe and crept downstairs and outside to get some fresh air. The night was quiet and beautiful. Everything was bathed in silver moonlight. The stars were brilliant and close.
Lauren was never sure when the thought took form and solidified in her mind, but all of a sudden it was there. I am in love with Jared Lockett. I love Jared.
Never before had she known the meaning of the word in all its scope. Never had she experienced this all-consuming passion.
Every thought related to Jared. Each word she spoke was weighed against what he would think of it. With everything she did, no matter how trivial, she secretly sought his approval. He dominated her mind. She wanted to share his torments as well as his joys. Was this love? Did it always bring so much pain?
She loved Jared. Smiling to herself, she basked in her secret knowledge as she went back into the house and climbed the stairs.
Jared was standing behind the door in her room, so she didn't see him until she shut it. She stifled a startled scream. "Jared, you scared me out of my wits," she gasped, holding a hand against her thumping heart. "What do you want?"
He was balanced on the balls of his feet as if about to pounce. She noticed for the first time that he reeked of whiskey and his golden eyes shone maniacally. His shirt was opened and the shirttail hung around his hips.
"Jared?" she said tremulously, and took a step backward.
"What do you think I want, Mrs. Lockett?" His voice was raspy and harsh, the inflection on her name ugly. He lunged at her, cornering her against the wall.
His lips swooped down on hers in a brutal kiss that tore her lips apart. A rapacious tongue plunged into her mouth as he pressed his hard body against hers. The buckle on his belt gouged her stomach. His knee plowed between her thighs, thrusting them apart.
Lauren had been so stunned at first that she hadn't reacted. Now she panicked. She fought him ineffectually with her fists, beating him about his head and shoulders.
She twisted her face free, sobbing, "No, Jared. Please, no!"
"No? Why not?" he growled through clenched teeth. "You are my wife, Mrs. Lockett. So do your duty by me. You've given it away to everyone else, and I won't be denied any longer."
She recognized the illogical reasoning induced by alcohol, but part of what he said was the truth. She was his wife. His words reverberated in her head like an echo. She was his wife.
She stopped struggling immediately, and he almost lost his balance. She made no effort to stop him as he jerked open her robe and ripped the sheer nightgown from neck to waist. He feasted his fevered eyes on her breasts, covering them roughly with his hands and squeezing with the intention of hurting her. Expecting a reaction and getting none, it finally registered with him that she wasn't fighting. He looked into her eyes.
She returned his gaze steadily, levelly, without fear. It was the look of a small animal that offers up the jugular to its predator when it admits that struggling is futile.
If she had poured cold water on him, she couldn't have extinguished the fires of passion more thoroughly. He stood motionless before her, breathing heavily. After long moments rife with suppressed emotion, he ran his hand through his hair, making a valiant effort to regain a modicum of dignity.
He leaned against her, but not with the lust of a moment before. His head rested against the wall above hers and he rolled his forehead from one side to another as if in agony. She felt his hands at her breasts, but realized he was pulling together the front of her dressing gown to hide her nakedness.
When his breathing had returned to normal, he backed away, holding a lock of her hair between his fingers. As he moved backward, he kept the strand in his hand until it was extended its full length, and then he let it fall from his fingers a little at a time. He watched every hair as it filtered through his fingers, falling to lie against her shoulder. Then he turned and left the room, closing the door softly behind him.
Lauren slumped to the floor and muffled her sobs with the fabric of her gown.
Jared was gone the next morning. He had returned to Austin. A week later, Lauren accepted the delivery of a brand-new, baby grand piano. It had been purchased for her by one Mr. Jared Lockett.
# Chapter 17
The date of the groundbreaking ceremony for the new railroad was set for December fifteenth. Everyone hoped that the Texas weather would cooperate. The collective spirit of the townspeople was soaring with excitement and anticipation. The municipal band gathered for joint rehearsals with the high school band. Speechmakers wrote and rewrote their speeches. Games were organized and yards of bunting were taken out of storage to be used in decorating the platform that was being erected in the center of the park.
Lauren had spent considerable time alone while Jared was away. She hadn't yet recovered from the shock of his attack Thanksgiving night. Again he had shown her that side of his character that was violent and frightening. She knew that most of his anger that night had been brought on by the Vandivers' presence in his house. Compound that with his seeing her with Kurt, and then by the alcohol he had consumed, and the results weren't too surprising.
Lauren still trembled as she recalled his face as he lowered it to hers in a parody of romantic passion. He had intended to punish her physically, but had wounded her spirit instead. If she could hate him, things would be easier. But loving him, his insults hurt even more.
The tender way he had touched her hair before he left that night had almost been her undoing. The anguish and suffering she had read in his eyes were more than she could bear. Had he wanted to apologize? Did masculine pride prevent him from expressing his regret?
When the piano was being unloaded from the wagon sent to deliver it, she had cried. Was this supposed to be a substitute for his respect and affection? It was a generous, beautiful gift, but she would have preferred one kind, gentle, caring word from Jared's lips.
After she had thought about it for a long while, she realized that Jared didn't know how to ask her forgiveness. She, more than anyone, had come to know how fierce his pride was. He would never verbalize an apology, so he had sent her the expensive piano as his peace offering.
Lauren played the instrument every day. Having not played for months, her fingers had become stiff and her touch lacked the fluidity that she had been capable of before. She practiced for a week before she felt she had regained some semblance of her former talent.
She was playing on the afternoon Jared rode into the yard on Charger. He slowed his horse when he first heard the music. He slipped from the saddle and nodded absently as Pepe came running from the stable.
He relinquished care of the animal to Pepe and walked on light footsteps up the steps to the front door. He didn't want to disturb her playing, and he was anxious as to what her attitude toward him would be. He had never had to resort to rape in his life. And then to try to rape his own wife! Godalmighty! He was brimming with self-loathing and disgust. What would she do when she saw him? Probably clutch her clothes to her body and flee in terror. He couldn't blame her.
Slipping quietly into the hall and shutting the front door, he crept on tiptoe toward the parlor, raising his heels high off the floor to keep his spurs from jingling.
Lauren saw him the instant he stood under the portiere. He looked much as he had when he had invaded her bedroom that first time. The clothes, hat, holster, boots, everything was the same except for his demeanor. Then he had been sardonic and arrogant. Today he looked like a shy little boy. He melted her heart, banished her fearful reserve.
"Jared!" she cried. Her face was wreathed in smiles as she stood quickly and went to him. "Thank you for the piano. I can't tell you how much I love it. Thank you." She stood on her toes and kissed his cheeks in turn, lightly.
He was so taken aback by her reception that he stood mute, staring into the sparkling gray eyes. They held no accusation, no anger, no revulsion. He was completely baffled.
Her hands still rested lightly on his shoulders. The fragrance he had come to associate with her wafted up to him. Her complexion looked warm and rosy. Her lips were softly parted, inviting, expectant. It was all too much.
When he drew her to him, it was with extreme caution, as though she might rebuff his embrace. He moaned with gratitude when, all too willingly, she fit her body along his. His arms went around her carefully as he buried his face in the rich glossiness of her hair. When his lips met hers, they were suppliant. At her immediate, sweet, acquiescent response, he grew bolder and traced her lower lip with the tip of his tongue.
"Jared," she breathed, before his invading tongue prohibited speech. He kissed her frantically, like a man who had been doomed to die and then had been granted a reprieve. Thirstily he drank of her mouth. Finally they drew apart, regretfully.
"Why don't you take off your hat and stay awhile?" she asked shakily. She reached up and pulled the hat from his head, clutching it quickly and tightly to her bosom before handing it back to him. "Would you like something drink?"
"N-no, thank you, Lauren. I'm not... not thirsty."
"Would you like to hear me play something? Maybe you don't think my talents warrant such a magnificent piano."
"I heard you from outside. It was... You play very well."
"Sit down and relax," she invited gently as she returned to the piano.
He sat on the edge of the sofa, aware of his dusty clothes. He felt gauche and awkward. What the hell was the matter with him? She played several selections, and he stared in fascination at her hands as they flitted over the ivory keys. Her back was straight, her head tilted, tendrils of black curls had escaped the confines of her chignon during their kiss and were resting on her cheeks and neck.
For just a moment, Jared felt a hard lump forming in his throat. He was going to make a damn fool of himself if he didn't get out of here. He stood abruptly and said, "It's beautiful, Lauren. You play better than anyone I've ever heard. I've got to go upstairs now. Clean up and all." He fled the room.
Lauren's fingers were poised over the keys where they had halted when he took his hasty exit. Thoughtfully she began to play again. The music followed Jared upstairs and into his room.
* * *
One adage Lauren had heard often since she had come to Texas was that if there was anything predictable about Texas weather, it was that it was unpredictable. No one would have bet that a December day in Coronado would have dawned clear and crisp, perfect for the activities planned for that auspicious day. For Carson Wells's sake, Lauren was glad this was so. He had worried, worked, and planned until she thought he would explode before the day of the groundbreaking ceremonies finally arrived.
The Locketts and Carson gathered in the dining room to share an enormous breakfast before they left for the site of the new depot. Jared looked dashing in black britches tucked into his high, black boots, black coat over a white shirt, and butter-colored leather vest. His black hat had been brushed and hung on the hall tree. Olivia murmured that she wished he had worn dress clothes. He ignored her.
Lauren wore a wool suit of deep burgundy with a cream lace blouse underneath. The skirt had a high waistband that came to a point in the center just under her breasts. Jared found it hard to keep from staring at the spot and wished she could leave off the jacket to the suit.
Carson was nervous and kept clattering his china and cutlery until Olivia berated him and said that he was driving her to distraction. He humbly apologized and settled down somewhat.
Lauren pinned on her veiled hat. There was a flurry of gathering up gloves, extra coats in case a Norther should blow in, blankets, flasks, and, at Rosa's insistence, a basket of sandwiches. At last, they departed.
Jared handed Lauren and his mother up into the buggy while Carson climbed in with them. Jared mounted Charger, whose golden coat gleamed from the brisk brushing Pepe had given him that morning at Jared's instructions.
"Why don't you ride with us, Jared?"
"And let Charger miss all the fun today? No, thank you, Mother."
Lauren looked at him and he met her gaze. She understood his purpose. He was showing everyone that despite the new business undertaking, Jared Lockett, like his father, was first and foremost a rancher.
Pepe guided their buggy through the traffic. There was already a large gathering of people at the groundbreaking site but they moved aside and deferentially made room for the prominent family who had made the dream of a railroad a reality.
Lauren was aware of curious stares. She felt self-conscious as Jared came around to her side of the buggy and lifted her down, taking her hand lightly as he led her to where Carson and Olivia had joined the other dignitaries. The mayor of Coronado was there, a county judge, the state representative from the district, several clergymen, and the Vandivers.
Jared held Lauren's hand throughout the mayor's speech, and released it only when it was his turn to spade a shovelful of dirt.
Lauren was proud to be standing next to him. He was playing his part well, she thought ruefully. If she was aware that the townsfolk watched every move of their favorite son and his new wife, she was certain that he was cognizant of it, too. He treated her with respectful politeness, but there was new warmth in the amber eyes that glanced at her with embarrassing frequency. She stifled a soft gasp when his elbow pressed against her breast as he held her hand under his arm. He looked down at her quickly, but he made no effort to remove his arm.
Moving en masse to the park, they filed onto the platform that had been erected in the center of it for the official proceedings. The state representative introduced Parker Vandiver, who stepped forward to deliver a flowery speech which extolled the advantages of a community committed to growth and expansion. Coronado, he proclaimed, was such a community.
"This railroad will pave the way for more commerce, more opportunity, more profits for all the citizens of this great city. We are proud to have a small part in bringing this about. We will watch with eager interest to see the progress you make. We want each one of you to realize the potential this offers you, your business, your family. We will be even more pleased when we can return to announce the completion of the track." There was a smattering of applause and a chuckle or two.
"There is a man not with us today who I'm sure would have shared in this celebration. Mr. Ben Lockett made strides for years to bring this railroad to his community."
Lauren risked a quick glance at Jared, whose face had turned as hard as stone. She prayed he could restrain himself.
"Ben Lockett would have been proud to have stood here today and seen the fruits of all his labors. We grieve his passing and are grateful to his widow and son, who endorse this venture wholeheartedly."
There was a burst of applause and cheers from the crowd for Ben Lockett and his family.
After a few more minutes of overblown oratory, Parker Vandiver took his seat. The mayor graciously offered the podium to Jared. He politely refused. The audience was disappointed, but some remembered when he had been honored after the war. He had refused to speak then out of sorrow for those not fortunate enough to come home. There were murmurs of approval and nods of understanding. They had been worried over some of his wild escapades, but anyone with eyes could see that he had married a real lady and was devoted to her. Marriage had settled him down. No doubt about it.
It was a good thing that the attention of the crowd was riveted on Jared and his wife, for had they been watching Olivia, they would have seen the perturbed expression on her face. She was angry with her son for not publicly commending the Vandivers and the railroad venture.
The formalities were out of the way, and everyone was ready to have a good time. Barrels of iced-down beer were heaved onto wagon beds, relay races were organized, the band assembled in the bandstand and began playing their limited repertoire.
As Jared was helping Lauren off the platform, one of the local youths came rushing up to him. The boy's freckled face was flushed, and his carrot-red hair radiated from his head like a burnished halo.
"Jared, Jared, there's gonna be a shootin' contest and they sent me to fetch you. Come on. They're waitin'."
Jared smiled at the boy's exuberance. "Lauren, may I present Billy Holt. Billy, Mrs. Lockett." The boy gave her a perfunctory nod. "Shooting contest, huh?" Jared continued. "Why did they send you after me, Billy?"
"Ah, hellfire and damnation! You know—" Realizing what he'd said, Billy turned scarlet cheeks toward Lauren. "Oh, pardon me, Miz Lockett," he gulped. Not able to meet her eyes, he turned back to Jared. "Hell, Jared, you know you're the best goddam shot anywhere around. It won't be no kinda contest atall if you ain't in it." He was so excited that the other expletives escaped unnoticed.
"Lauren, what do you think? Would you like to see a shooting contest?"
She smiled up at Jared. "It sounds as though everyone will be disappointed if you don't enter it."
"I'll do it only if you'll go with me and watch."
Billy was hopping first on one foot and then the other, barely able to contain his enthusiasm. "Please, Miz Lockett?"
"Yes! Lead the way." She laughed.
Billy leapt into the air and whooped, then raced off to let the others know that the star attraction was on his way.
Jared yelled for Pepe to bring his holster and pistol from the back of the buggy.
Lauren looked up at him in feigned exasperation. "You knew all along there would be a contest, didn't you? And you planned to enter it."
"Well, it's nice to be begged every once in a while." He grinned engagingly. "I just hope I don't humiliate myself. Obviously I have a reputation to maintain."
They strolled over to the men who were gathering for the contest. Some were checking out their revolvers and others were placing bets on the outcome of the match. The odds were strongly in Jared's favor.
Billy and some of his cohorts had assembled bottles and cans for targets, and were lining them up on a fence rail about thirty yards away from the large oak tree that served as the base.
Jared took off his coat and handed it to Lauren as Pepe brought him his holster. He checked the Colt pistol, twirling the chambers and nodding in satisfaction. Lauren remembered being told that _vaqueros_ loaded only five of the chambers in the Colts which most of them toted. The first chamber was always kept empty, preventing excited cowboys from shooting their own knees, toes, or friends in a stressful situation.
The ground rules were laid down by Carson whom, it appeared, was the accepted referee. "Each man takes three shots. One miss and he's disqualified," he intoned.
The ten men entering the contest lined up. Jared was last in line to shoot. He caught a glimpse of Kurt Vandiver leaning negligently against a tree. He wasn't watching the contest. He was watching Lauren. Jared looked toward her and she gave him an encouraging smile. He turned his attention back to the contest.
It became boring as all of the gunmen proved to be expert marksmen. But slowly, one by one, they began to miss shots. There were three entries left, Jared among them, when someone suggested that they "fan" their hammers and try to hit three of the targets.
The first man stepped to the line that had been drawn on the ground and, when given the signal, fired rapidly while fanning the hammer of his pistol. He hit three of the targets. The second man only hit two of the bottles.
Jared stepped to the line. He picked a cheroot out of his breast pocket, nonchalantly struck a match, and lit the cigar, drawing on it for several seconds while everyone stood stock-still in anticipation and awe at his insouciance. His pistol was in his holster, though Lauren had seen him loading it while the others were shooting.
"Call it when you're ready, Carson," he said over his shoulder.
"You're going to draw?" Carson asked in amazement.
"Yes. Call it." He spoke calmly, though Lauren could sense his excitement.
Carson shrugged and gave the call. "Draw!"
With the speed of lightning, Jared whipped his pistol from the holster and loud retorts spewed from the barrel so fast that they sounded like one continuous blast. When the smoke cleared, the witnesses saw only one bottle left on the fence. He had hit five out of six!
A great roar went up from the onlookers and other contestants. Billy was turning somersaults in the winter grass. Lauren clapped her hands in delight. Jared took all of the slaps on his back with casual aplomb.
"Goddam shootin' machine, that's what."
"Quicker'n a mad rattler."
"Heard that Rudy Mendez is just as fast. You can tell that sonofabitchin' Ben taught them boys how to shoot."
Lauren was so wrapped up in the commotion that she didn't notice Kurt standing near her.
"It's a pity he can't control his drinking as well as he can that Colt."
She whirled on him, furious. But Jared's approach prevented her from snapping a rejoinder to Kurt's petty observation. Instead of wasting thought on him, she congratulated Jared as she helped him put on his coat. "Jared, you were brilliant."
Undaunted, Kurt interrupted any response Jared might have made. "Lauren, I have a surprise for you. Come over here. I want to show you something."
"Jared?" Lauren looked up at him. She didn't know how to rebuff Kurt without making a scene for everyone in town to see.
"It's all right, Lauren. I'd like to see this surprise, too." Jared took her arm possessively and followed Kurt's stocky figure as he plowed his way through the throng. There was a small cluster of people gathered around something, and only when Kurt rudely shoved them aside did Lauren see it was a motorcar.
She had seen them in North Carolina, particularly on her trips to Raleigh, but she had never seen one in Coronado. Apparently no one else had either, because the automobile was causing quite a stir.
"It's an Oldsmobile. One of the gasoline-powered models," Kurt boasted. "I want to take you for a ride in it, Lauren."
The automobile was beautiful. The sides and motor casing were glossy black, trimmed in red. The tires had white sides, and the hubs of the wheels were red. The upholstery was black leather. There were two brass lanterns mounted on the front of the car, and brass accents shone on the steering stick and a mounted horn.
"I... un—"
"I'm sure your _husband,_ " he stressed the word, "won't mind. Would you, Jared?"
Aware of the people standing by listening to every word, Jared smiled expansively and said, "I think Lauren would enjoy that very much. Here, dear, let me help you." She saw the tight lines around his mouth as he lifted her carefully into the vehicle. She wanted to protest, but knew she couldn't.
Kurt went to the front of the car and made a big show of cranking the motor. It churned to life with a terrific racket. Running around to the driver's side, he vaulted into the seat beside Lauren. Unconsciously she moved her skirt away from his heavy thigh. The gears were engaged and the automobile moved forward. Lauren cast an anxious look back at Jared, but he was staring at Kurt's back with a threatening scowl.
Kurt took the road that led down to the river and followed near its banks. They crossed a narrow bridge and continued on the road to the opposite side. They were virtually out of sight, hidden by the trees that grew along the riverbank on both sides. Under other circumstances, she would have enjoyed the ride, but the ominous presence of Kurt made Lauren terribly uncomfortable.
"Are you having a good time?" he asked close to her ear.
"It's very pleasant."
She shifted farther away from him on the narrow seat. He continued to maneuver the car along the rough road, and Lauren was relieved that he didn't try to engage her in conversation.
"I'd like to go back now," she said. "Jared will be worrying about me."
He laughed humorlessly. "I'm not fooled by this so-called marriage, Lauren. Separate bedrooms, isn't it?"
"My married life is none of your business, Mr. Vandiver." Her voice was hard and cold, but her cheeks flushed hotly. His swift, derisive look told her he was not convinced. He did, however, turn the automobile around, recross the bridge, and retrace their way back to where they had started. The crowd was still there. Jared was leaning against a tree, casually smoking a cigar. Only his flashing eyes indicated his anger.
In his haste to get out and help Lauren down, Kurt didn't take time to cut the motor. Jared strolled to the car with studied leisure, then jumped onto the seat vacated by Kurt.
"You call that a ride, Vandiver?" He rammed the car into gear. Lauren clutched the seat as the car lurched, almost running over Kurt, who scrambled out of the way just in time. It had all happened so quickly that he was stunned as his Oldsmobile sped away at a speed he would never have dared to push it. The crowd went wild with hysterical laughter at his expense.
Jared handled the steering stick like a magic wand, turning corners at breakneck speed, managing to hit every chuckhole in the road. They drove around the square in the middle of town several times, until Lauren's head was spinning. She clung to Jared's arm for fear of being thrown off the seat as the car bounced over the streets. Finally they left the center of town, taking one of the lanes that led away from the city.
The wind lashed her face, and her bonnet blew back, painfully pulling against the pins that held it on. She would have reached up to take it off, but dared not release her grip on Jared.
He was laughing uproariously, like a naughty child who had pulled off a tremendously funny joke on his school-teacher.
"Did you see his face, Lauren? Did you? That sonofabitch! I'll show him how to drive a car." His hat had blown off, and his hair tumbled around his head. His face was flushed and his eyes glittered in delight over his own devilry.
Before she realized it, Lauren was joining his laughter. They were like two children let loose from restrictions for the first time. Jared risked looking at her when he heard her laugh. He realized his error when his eyes returned to the road. It had curved and they were headed for the ditch. "Hold on!" he shouted as he applied the brakes. They slowed then, but the automobile was going too fast. The wheels locked and the car spun crazily until it slid off the road and the front wheel on the driver's side sank up to the hub in the soft ground. Jared cut the chugging, choking motor.
They were gasping for breath, momentarily shaken over their narrow escape from death or severe injury. The car sat at a perilous angle and Lauren had been thrown against Jared. He looked down at her and asked, "Are you all right?"
She took stock of herself. "Yes, I'm fine," she answered breathlessly. "I'm not so sure about Kurt's Oldsmobile."
Then they burst into spontaneous laughter. They laughed so hard that tears gathered in their eyes. It was the first time they had ever laughed together. There had been times when one of the Mendez children had reduced the family to laughter with an antic, but this was different. This was a personal moment that they were sharing.
Lauren dabbed at her eyes with her gloved hands and drew the long, lethal-looking hatpins from her hair. The hat came off and her heavy bun drooped almost to her shoulders; strands of hair escaped it completely.
Jared's merriment subsided and, of its own volition, his hand touched the knot of hair resting against her neck. She raised her eyes to his and they held only briefly before she was in his arms. It happened quietly, naturally, instinctively. He held her to him and whispered her name repeatedly against her ear, his breath sending shivers of pleasure down her spine.
He drew back and searched her eyes for some sign of rejection, but saw only invitation. His mouth took hers in a telling kiss. Their lips sought and found and celebrated each other. The kiss was tender, but held promise of restrained passion. It was committing, but left room for reservation, for caution. It was a kiss for the moment. For now. Only now.
Jared pulled away from her and studied her face, startled by the intensity of his feelings. They stared at each other in a silent communication. Then Jared rasped, "Lauren, kiss me again. Kiss me—"
"Lauren, are you injured?" Kurt was galloping toward them pell-mell. He reined up before his car and dismounted, running to Lauren's side and offering his hands up to her. Before she could assure him she didn't want to be rescued from Jared's embrace, his beefy hands gripped her around the waist and swung her down to the ground. "Lauren, are you all right? That idiot could have killed—"
The breath gushed out of Kurt's body when Jared's fist landed squarely in his stomach after first spinning him around. Kurt landed on his back in the ditch and, before he could regain his breath, Jared had straddled his chest and pushed the barrel of his pistol against Kurt's nose.
"If you ever, _ever,_ touch my wife again, I'll kill you, Vandiver." His face was an inch from Kurt's and the words were strained through his teeth. "If you even look like you're thinking of touching her, I'll kill you. Do you understand me?"
"Get that goddam gun out of my face," Kurt said with misplaced bravery. "It's not loaded. You emptied it on that last round." He struggled but couldn't budge Jared.
The clicking sound of metal on metal as Jared cocked the pistol arrested Kurt's futile movements. "Are you so sure it's empty, Vandiver?" Jared taunted.
Kurt laughed nervously. "There's no way in hell you could have kept from firing that sixth shot as fast as you were fanning."
"You forgot that there was one target left," Jared said smoothly.
"Even so," Kurt persisted, though his voice was beginning to waver and he was perspiring profusely, "everyone knows cowboys leave the first chamber empty."
Jared shrugged negligently. "Some do. Those not confident in their abilities. In all humility, I'm not one of those."
"The damn gun isn't loaded!" Kurt screamed as Jared shoved the barrel further into his fleshy nose.
"Wanna bet?" Jared drawled.
With the merest movement, he turned his hand and fired the pistol only inches from Kurt's head. The bullet embedded itself in one of the rubber tires of the car.
The blood drained from Kurt's face and he began making a whimpering sound.
"I could have hit that last target, Vandiver, but no fool carries around an unloaded pistol." With disdain, Jared stood up and stepped over the groveling, inert form.
"Señor Jared!" Pepe came riding up on Charger with several other riders trailing him. "Are you all right? Is the _señora_ all right?"
He reined in the palomino stallion and slid off his back, running up to them.
"Yes, yes, we're fine," Jared assured him. "Had a helluva good time, too. Can't say much for the car." Lauren knew that he was speaking for the benefit of the curiosity seekers who had followed them. "I don't think it'll ever be as thrilling as galloping on Charger. And he sure as hell has better sense than to go off in a ditch."
All of them laughed, relieved that no one was hurt. Mr. Vandiver looked a little peaked, but he was just worried about his car, they thought.
"Lauren." Jared extended his hand and she took it. He led her to Charger and lifted her to the saddle. He mounted behind her, putting his arms firmly around her.
"Pepe, can you ride with someone back to the park?"
" _Sí,_ Señor Jared."
"Fine. Gentlemen." He tipped his hat, which Pepe had retrieved for him, turned Charger around, and headed for town at a slow trot.
# Chapter 18
Their escapade had created quite a commotion. They were asked hundreds of times about their well-being, despite their continued assurances to everyone that they were fine.
Sitting at a picnic table with Olivia and Carson, they ate the sandwiches that Rosa had packed for them. Jared drank locally brewed beer with the men gathered around the barrels. Lauren watched him from her place near Olivia, and was thrilled when he glanced her way and smiled. She tried hard to keep her mind on the conversations going on around her and to give the proper answers to the myriad questions being asked by the ladies of Coronado about how she liked her new life in Texas. But her mind was on Jared's mouth, how warm it was against hers. His hands, strong, demanding, yet gentle.
While the band played a Christmas concert to mark the conclusion of the festivities, Jared sat close beside her on the blanket he had spread on the grass. His breath was on her cheek. She could smell the aroma of his cheroots, the leather of his vest. If only we could stay like this forever, she thought.
Olivia hadn't missed the looks and the "accidental" touches between them after the madcap drive in the car. After the Vandivers had left, pleading that they had commitments in Austin, she focused all her attention on Jared and his wife. She didn't like what she saw. Something was simmering, and it must be cooled before it came to a boil. It was dangerous to all of her plans. It must not happen!
They left late in the afternoon. Rosa had chili and cornbread waiting for them when they got home. They ate tiredly, but two of the people at the table were too exhilarated to have much of an appetite.
Carson left directly afterward, and Olivia pleaded fatigue and suggested they retire early. Lauren and Olivia walked up the stairs together, leaving Jared in the library with a nightcap.
* * *
Sometime during the night, Lauren awoke with a start. It took a moment for her to get her bearings. She listened. The house was still. She lay back down and, just as she did, she heard a groan. When a sharp cry followed it, she jumped out of bed, alarmed.
The sound came from Jared's room. Without even pausing to put on a robe, she crept through the bathroom and tentatively knocked on the connecting door. He didn't answer, but again she heard the rasping cry and louder moans. What if he were ill? Should she go in? She paused only an instant before she opened the door a crack and peered into the room.
Jared was thrashing on his bed, twisting and tossing in the agony of a nightmare. Lauren quickly crossed to the bed and saw his bare chest heaving, his face beaded with perspiration. The words coming out of his mouth were unintelligible, but conveyed a terrible torment. He murmured the name Alex over and over. "Jared, wake up." Reaching out a cautious hand, she touched his shoulder and shook him slightly. "Jared, please wake up. You're having a nightmare."
He only became more violent. He thrashed his arms and tossed his head on the pillow, his teeth bared as he gnashed them. Lauren dodged the flailing arms, but managed to capture both wrists and, leaning over, pinioned them on either side of his head.
He struggled for release, but somehow she was able to hold him. "Jared, wake up."
He opened his eyes and, as if hinged at the waist, bolted upright. The sheet fell around his middle. Oh, my Lord, he's naked! Lauren realized. He gulped in great amounts of air and shook his head in an effort to clear it of the tormenting dream. Shaky fingers raked through damp, tousled hair before covering his face.
Lauren slipped through the darkness to his shaving mirror on the bureau and poured fresh water from a pitcher into the basin. She moistened a towel and brought it back to the bed.
"Jared, you were having a nightmare," she said softly, comfortingly. "Are you all right now?"
He nodded dumbly as she sponged his forehead with the cool cloth. When he lowered his hands, she bathed the rest of his face and neck.
"Thank you, Lauren. I'm fine now." He moved away from her hands.
"Were you dreaming about Cuba? Alex?"
He looked at her sharply, then away. "Yes," he said hoarsely. Was he trembling?
"Would you like to talk about it?" Her voice was a faint whisper. Until her fingers touched the sun-bleached curls on his head, she hadn't realized she had reached for them.
"No," he answered gruffly. Then, desperately, "Yes. Lauren?" His arms went around her waist in a viselike clasp, and he drew her to him, burying his face between her breasts. Her knees bumped against the bed frame.
She hesitated only a moment before cradling his head in her arms. His ragged breath was warm and moist on her skin through the thin batiste nightgown. She was grateful to the darkness for lending her a modicum of modesty. Murmuring words of comfort, she stroked his head, weaving the coarse curls through her fingers. His hands moved over her back, tracing her spine with sensitive fingers.
Long minutes passed, and still he didn't release her. Almost imperceptibly, he moved his head between her breasts, and began nuzzling her with his nose and mouth, pressing small, hot kisses on her flesh.
A longing deeper and more potent than any emotion she had felt in her life pierced her to the core. A warm flush washed over her skin as her heartbeat accelerated.
Her limbs seemed to have turned to water, but with surprising strength, she drew his head closer yet. A soft, ecstatic cry escaped her open lips when he took her nipple into his mouth. He tugged on it gently with a sweet warmth before his tongue swept across it, wetting the fabric of her gown.
Lauren's body was swimming with sensations and she almost melted to the floor when the strong support of his arms was suddenly withdrawn. Jared sat with his knees raised, his head hanging between them, his face in his hands.
"Leave me, Lauren," he grated. His voice was so low that she could barely hear the words.
"Jared, I—"
"Leave me, please," he repeated with a groan.
"Why? Why must I leave you... now?" She was on the verge of tears. Her emotions were running so high, her voice cracked under the pressure.
"Because, dammit," he swore vehemently, "I can't stand having you this close, this willing, this... naked... and not... Just go back to your room, please."
"No, Jared," she breathed.
He looked up then. "No?"
She swallowed hard. "I... I want to be a real wife to you, Jared." She couldn't help the tears that spilled down her cheeks. "I want to stay with you."
"Lauren," he said, shaking his head. His voice was sympathetic, like an adult speaking to a child. "You don't know what you're saying."
"I realize that! I don't know anything about... this. But I want to know. I want to be a wife like Gloria is to Rudy. Like Maria was to Ben." Neither of them noticed the incongruity of that statement. "Please let me stay with you tonight."
He looked at her. Tears were rolling down her flawless cheeks, her hair was tumbling over her shoulders onto the breasts he longed to caress again, her slim figure was outlined under the sheer nightgown, and the pounding of his heart thundered through his head. His body was on fire with desire for her, and his manhood stood proudly, painfully.
He didn't dare move when she raised one knee onto the bed and sat down close to him. She leaned forward and, resting timid hands on his shoulders, placed her lips against his.
He moaned in helplessness as he clasped her to him and fell back against the pillows on the wide bed. Drawing her beneath him, he kissed her with a searing fervor, thrusting his tongue into her mouth and exploring it with unleashed passion.
Frantically he fumbled with the buttons of her nightgown. Lauren didn't have time to be embarrassed by her nakedness as he flung the offending garment over her head, and quickly covered her body with his.
Difficult as it was, he forced himself to practice self-control. This wasn't a whore. This was Lauren. His wife. He must go slowly, mustn't frighten her.
Lauren felt the long, hard body against hers relax somewhat as Jared kissed her again, leisurely this time, as if memorizing her lips and tongue and teeth. He traced kisses over her cheeks, and nibbled at her earlobe until she felt her own body turning languorous and pliant beneath him.
One large, tanned hand moved over the tops of her breasts. He cupped one gently. Her nipple became hard as the palm moved in slow, easy circles over it. He took the swollen bud between his fingers and caressed it tenderly until Lauren wanted to cry out from the pleasure he brought her, pleasure she had never known existed.
Then he enfolded her nipple in his wet, warm mouth, and drew on her with exquisite tenderness. As he moved to the other breast, his beard stubble scraped against her smoothness, heightening her awareness of their physical differences. He gave the same attention to her other breast as his hands moved to her abdomen, stroking, caressing, arousing.
It was a sensation like warm liquid being poured over her as he lowered his head and kissed her navel with an ardent, hungry mouth. His palm rested lightly on the tight nest of raven curls at the top of her legs and she wondered why she wasn't repulsed or afraid. The feelings were so... what? She had nothing to compare this to.
His hand moved between her thighs, and Lauren was alarmed when he encountered a moistness. Apparently he found that strange wetness gratifying. He sighed and whispered, "God, Lauren, you're ready for me. Oh how sweet you are." The movement of Jared's hand then took away all conscious thought. The delicious explorations of his fingers released a primitive instinct and she rotated her hips against his powerful body. When I think of this later, I'll be so ashamed, she thought, but now, I can't help it.
He was kissing her mouth again, tenderly, deeply, murmuring against her lips a mixture of English and Spanish. He moved on top of her and she welcomed the crushing weight by wrapping her arms around the breadth of his back. His knees gently urged her thighs apart as he settled himself between them. She felt his masculine strength as he probed the opening of her body.
A flash of panic seized her, and he was instantly aware of it. He raised his head and looked into her wide, fearful eyes, searching for the answer he hoped was there. "It's true then? You've never been with a man?" His inflection was one of awe and carried with it a need to know. And she knew why her reply would be so important to him. Ben. She didn't speak, but her lips formed the word _no_ as she shook her head. "Lauren." Every trace of emotion that made up the spirit of Jared Lockett went into that speaking of her name. He kissed her quickly, hotly, passionately. Then he beseeched her, "Forgive me. I'm going to hurt you, Lauren. I'm sorry." He thrust himself inside her and she would have screamed if he hadn't held her head protectively against the hollow of his shoulder.
The burning pain was ripping her apart, searing her insides. "I'm sorry, my darling. Relax as much as you can."
The words were husky against her ear, and she forced muscles she didn't know she had to relax into acceptance. The pain abated, but how long did this last?
Jared hadn't moved. She heard his breath rushing in her ear like a strong wind. She shifted under him, seeking a more comfortable position, and heard his sharp intake of breath. "Oh, God, you feel so good," he ground out, pressing his face into the pillow. "So tight. Perfect, perfect."
Slowly he began to move inside her. The pain came back in rhythmic waves to match his thrusts, but somewhere amidst the pain was a promise of pleasure. Jared's murmurings in her ear were indiscernible, but their meaning was clear.
Jared was inside her! It was a thrilling thought—his body and hers fused together in the most intimate and unifying way. Without her knowing it, her body had taken control and she was responding to his thrusts with answering movements. Suddenly his whole body tensed and she felt a shower of life flowing into her. Instinctively she squeezed her knees against his hips.
He lay spent on top of her until his breathing returned to normal. She ran her hands over his broad back, marveling at the contour of muscles, bone, and skin. Eventually he raised himself onto his elbows and looked into her eyes.
"My God, Lauren," he whispered in wonder. "What did you do to me?"
He rolled off her and drew her to him, nestling her against his chest. They lay for a long while without speaking. His hands traveled lazily over her back, hips, and legs. The place between her thighs was still on fire, but she was content. She sighed. Placing a finger under her chin, he lifted her face to him.
"My little virgin. Did it hurt too much?"
"No," she lied.
He tried to keep from smiling, couldn't, and even laughed quietly. "Like hell. You are a lady to the bitter end, aren't you?" He lowered his head and kissed her sweetly, almost chastely.
He swung his long legs over the side of the bed and walked unabashedly across the room to the dresser. Lauren studied his physique. No artist could have captured the sensual grace of Jared's walk or the texture of his skin covered lightly with soft, tawny hair. He was a beautiful male animal and, in spite of her new awareness, she blushed.
He came back to the bed carrying a damp cloth. Kneeling beside her, he moved to open her legs. She shrank back instinctively.
"I'm not going to hurt you," he said gently. "This may make you feel more comfortable."
His voice was so soothing and his hands so tender that she allowed him to separate her legs and press the cloth to her. Avoiding his eyes, she stared up at the ceiling, astonished that these things were transpiring. Never in her uneducated fantasies had she imagined that such familiarity could exist between a man and a woman. Or maybe such things only happened with Jared. She risked a glance at her husband. He was gazing at her as though he could read her mind and knew each personal thought.
"I promise, Lauren, that it will never hurt as much again." He smiled slightly. "You may even learn to enjoy it."
The cooling cloth had reduced the stinging considerably, and she whispered, "Thank you," as he took it back to the dresser.
She sat up, clutching the twisted sheet to her, and reached for her nightgown lying at the foot of the bed. "What are you doing?" he asked as he crawled back beside her.
"I thought—"
"You thought what?" he interrupted, taking the garment away from her and tossing it out of her reach. He began nibbling at her shoulder as he drew her back onto the pillows. "What did you think, Lauren?" he asked thickly. His lips were at her breasts now, and she couldn't think at all.
"I thought... uh..." Oh, Lord.
He chuckled softly. "Go to sleep." Laying his head close to hers on the same pillow, he closed his eyes. His arm rested heavily on her stomach. His hand cupped her breast lightly.
Sleep? Not tonight. She had too much to think about. She and Jared, who had fought, argued, ignored each other, and hurt each other, were lying here side by side completely naked in his bed after experiencing the most splendid coupling, and he wanted her to sleep. Impossible.
She would never be able to sleep.
But she did.
* * *
"Good morning."
"Hmm?"
"I said 'good morning,' Mrs. Lockett."
Lauren sleepily opened one eye and saw her husband's smiling face close to hers. The room was still wrapped in dark gray shadows. "Jared," she mumbled in complaint, "it's not morning. It's still the middle of the night." She buried her face against his hairy chest and yawned broadly.
"I love getting up first thing in the morning." He laughed at his own double-entendre, but Lauren looked up at him with naive eyes. He realized again just how innocent she was. Still. "Lauren." He stroked her cheek and leaned down to kiss her softly on the lips.
She snuggled against his warmth as he pulled a blanket higher over them. He caressed the hair that spilled across his chest. As his eyes roamed the soft form cuddled against him, he laughed ruefully to himself. Who would have thought this of Jared Lockett? That he had spent the entire night with a woman was unusual in itself. As soon as he had satisfied his initial lust, he was always anxious to leave any woman he had been with before. Last night he had wanted to stay. With this woman.
Never had he lost himself so completely with a female as he had with Lauren. He always enjoyed the act, certainly. But his thoughts were often elsewhere: on a card game, business, another woman.
Last night, however, he had been aware only of the woman beneath him, responding with only a trace of virginal shyness. The feel of her, the scent, the texture, the taste had all combined to totally capture his senses. His absorption had been so tremendous that he hadn't wanted it to end.
In a million years, he would never admit that last night had been a first for him, too. It was her initiation into the rites of loving. She had been a virgin. And it was the first time Jared Lockett had ever been with a virgin. It was a gift he had never expected to receive and one he didn't feel he deserved, yet she had given herself to him.
Why? After the abuse he had heaped on her, why had she come to him, offering herself? He bent his head and brushed his lips across her forehead as the question continued to haunt him.
Her fingertips smoothed the crease between his thick, brown brows. They'll be shaggy like Ben's when he gets older, she thought. Twenty-four hours ago, she hadn't even known that it was possible to be as intimate with another human body as they had been. Now they were lying with arms and legs entwined, and she felt no shyness or modesty or guilt. What had happened to all her rigid scruples? No matter. She didn't want them back. Even now, she wanted to discover what further delights his body could offer hers.
"Jared, when?... I mean, how long?... Does it?...
He smiled. In answer to the question she couldn't quite bring herself to ask, his mouth sought hers. That first real kiss since the night before generated the same stirrings, the same exciting anticipation of things to follow. She allowed him unlimited access to her body. He found her secret places and caressed or kissed, eliciting moans of pleasure from them both.
Shyly at first, then encouraged by his soft groans of ecstasy, her fingers traversed the forested chest. Lightly she fanned the crinkly hair. His skin was warm and stretched tautly over contoured muscles which she massaged with fingers suddenly grown talented in the art. Air hissed through his teeth when she encountered the turgid brown nipples. She withdrew quickly, but his hand went to the back of her head and brought her back. Nuzzling her face in his neck, she traced it with her tongue. As she leaned over him, his strong fingers splayed over her hips and pressed her against him to meet his virile force.
"I want you again, Lauren. But I don't want to hurt you," he said, anguished.
"You won't, you won't."
Strands of her hair were still wound around his neck as he almost fiercely rolled her to her back. Poised above her, he sought her eyes and demanded, "Say my name, Lauren. Let me see my name on your lips. Say my name, please."
Reaching up, her fingertips played upon his cheekbones as she whispered, "Jared. Jared. Jared." The last syllable was a plea.
When he entered her, it was slowly and deliberately. There was a remnant of soreness, but none of the burning pain she had experienced before. Jared moved against her like a well-trained machine whose only purpose was to bring sensuous pleasure.
His hands gently kneaded the fullness of her hips as he lifted her up and forward to meet his sublime invasion. He plunged more deeply than ever before, stroking the walls of her femininity, the gate of her womb, withdrawing to tease and tantalize the portal, only to bury himself within her again.
Lauren felt her spirit rising, floating above the surface of the bed, climbing higher and higher toward some pinnacle still nebulous and mysterious but desirable. It frightened her, this abyss she hovered over. She squeezed her eyes shut.
"No. Lauren, come with me," he urged breathlessly. "Come... with... me..." He held her tight as his passion, beyond the point of restraint, emptied inside her, laving her with living fire.
He rested only a moment. Then, with a hand on the small of her back, he eased them to their sides. Still entrapped within that honeyed cave, Jared stirred himself arousingly.
"Can you feel me, Lauren?"
"Yes," she whispered. "Yes."
"Do I hurt you?" He moved inside her again.
Oh, Lord. "No, it's... it feels nice."
"Then why did you stop?" he asked in a solicitous whisper. "You were on the brink of incredible joy. Are you afraid of it?"
Unable to meet his probing eyes, she nodded into his shoulder.
"I see," he said quietly. Confident of his own abilities, he knew he could bring her to completion even now, but decided she wasn't ready to accept it. He eased himself away from her, aware that, in spite of her denial, she must feel bruised and battered.
He brushed the hair away from her temples and kissed her forehead and eyelids. "Let's take another nap." Turning her away from him, he drew her back against his chest.
He fell asleep breathing in the perfume of her hair.
# Chapter 19
Olivia knew immediately upon seeing Jared and Lauren together at lunch that her worst fear had come about. The household staff was buzzing about something, but she had never taken it upon herself to learn Spanish. She didn't understand their excited whispers.
At the meal, however, the secret smiles, the oblivion to everything around them, and the dark smudges under their eyes testified to her how Jared and Lauren had spent the night.
Damn! How could her son succumb to that hothouse flower? He always preferred boisterous, blowsy girls, girls with scandalous reputations. What did he see in this fragile, ladylike paragon of virtue? Well, it didn't matter if he slept with the chit as long as he didn't form a lasting affection. Or—God forbid—if she should conceive a child. Hopefully, Jared was smart enough to prevent this from happening.
She watched them as they excused themselves and started upstairs for a "nap."
"Jared, could I have a word with you?" she asked quickly, folding her napkin beside her plate as she stood. "Yesterday Parker and I talked over some plans amidst all that insanity at the groundbreaking, and I think you should be kept apprised of the developments."
Jared looked wistfully toward Lauren, but begrudgingly complied. "All right, Mother. I'll see you later, Lauren."
After her daughter-in-law left the dining room, Olivia suggested that Jared follow her into the office. When he was relaxed and seated in a deep leather chair and smoking a cigar, she told him their new scheme.
* * *
"Really, Jared, I think you overreacted this afternoon."
They were at dinner, and Olivia was speaking in well-modulated tones that belied the underlying turbulence in the room.
"I'd rather not discuss business now, Mother." Jared's voice was clipped and terse. Lauren had been dismayed to find that all his lightheartedness of the morning had disappeared after the meeting with his mother. They had been sequestered in the office for over an hour and, when Jared had returned upstairs, he made no effort to see her. He had gone straight to his room and then later left the house. He had returned just before dinner. She hadn't seen him alone since before lunch.
"We certainly have no secrets from Lauren, do we?" Olivia asked sweetly, glancing at her daughter-in-law. "After all, she is your wife, Jared."
Lauren blushed and looked unseeingly at her plate. Did Olivia know about last night? How would she react to the consummation of their marriage?
"Do you think some of our plans will shock her?" Olivia asked her son coolly.
"Dammit, Mother!" The lines around Jared's mouth seemed to be carved of granite.
"You see, Lauren," Olivia continued, undisturbed by Jared's anger, "the Vandivers and I have reconsidered the optimum location for the power plant. The new site will require damming up the river above Pueblo instead of below it. Naturally this will greatly affect the community. We fear an outcry of public disapproval, since the community's source of water will be cut off. Therefore, we have taken measures to assure that Pueblo's destruction be seen as a blessing."
Lauren had placed her fork on her plate and was staring from mother to son with blank, uncomprehending eyes. Olivia returned her stare levelly. Jared wouldn't look at her.
"I don't think I understand." Lauren licked her lips. " _What_ measures are you taking?" She didn't want to know. But she had to know.
"Lauren, stay out of this. It's none of your concern," Jared barked.
"Of course it concerns her, Jared. My dear," Olivia said, addressing her again in that pleasant, conversational voice, "we are going to round up a gang of desperadoes, mercenaries, whatever name you choose to call them. I think you get the idea. They will go into Pueblo and, posing as citizens, cause a ruckus. They will be instructed to burn, loot, injure, anything they deem necessary to bring about a riot. The ghetto will destroy itself. Our mercenaries will help it along by igniting some well-set fires. I thought it up, of course. Parker thinks it's a brilliant idea. He's asked Jared to come to Austin to organize the men. Jared frequents places where such characters might be found."
Lauren's face had drained of color. She couldn't believe what she was hearing. When she spoke, her voice was barely more than a hoarse croak. "But people could get killed! And think of all the homes that would be destroyed."
Olivia shrugged. "I suppose they will, but it will be no great loss, will it?"
"But it's unnecessary! Why can't the power plant be built where originally planned?"
"It could. It's just that I don't wish it there."
"Then it _is_ unnecessary. You would have a whole town destroyed on a whim. Why?" She slumped back in her chair and stared at the woman across the table from her with perplexity. Olivia's face was hard and ugly, filled with hate. Suddenly Lauren understood. "Revenge," she wheezed. "You're taking out your hate of one woman on a whole people."
Olivia glared at her. "I don't know what you're babbling about, Lauren," she said.
"Of course you do. Maria Mendez. You can't bear it that Ben loved—"
"Shut up!" Olivia screamed and thumped her fists on the table. Glassware and china clattered.
"No, I will not shut up. You would have let Elena and Isabela die in the streets before caring for them. At the time, I thought it was the epitome of heartlessness and cruelty, but now you surpass even that. Perhaps I can see why you would feel the way you do about Maria, but to unnecessarily—"
"It's necessary if I say it is," Olivia broke in. Her chest was heaving with emotion. "I don't have to justify myself to you, or to anyone else on earth."
Lauren was struck then by the passion in the woman. She was driven to hate, to destroy. It was useless to try to reason with someone so obsessed. "No, you don't have to justify yourself to anyone on _earth,_ " Lauren said, stressing the last word. She knew then that she would never fear Olivia again. Distrust her, dislike her, but never fear her. Olivia was a lost cause, but there was still hope for Jared.
During the heated exchange between his wife and his mother, he had continued to stare into the candle flame lighting the dining table.
Lauren turned toward him. "Jared? Jared, you can't endorse such a horrendous scheme?" The words came out as an incredulous question.
"I told you to stay out of it, Lauren," he growled. "You don't understand these things."
"I understand everything!" she shouted. "I understand that what you propose is heinous and criminal and—"
Jared jumped from his chair, knocking it over and upsetting a glass of red wine onto the table. "Goddammit, get off my back." He strode toward the door leading into the hall, but Lauren was right behind him. She stepped in front of him, placing both hands on his chest, searching his eyes.
"Tell me that you will have no part in this. Please tell me that." When he didn't move, she went on, "Jared, _think._ There are families there who will be hurt. Elena, Rosa, and Gloria have friends and relatives who live there. Surely you would not condone anything that would harm them?"
On top of what his mother had ordered him to do, he didn't need this. He didn't need Lauren looking up at him with an expression that was both pleading and accusing.
Last night he had wanted her with a passion that surpassed anything he had ever experienced before, and the culmination had been earthmoving. Memories of the hours they had spent in his bed still stirred him.
He didn't need that, either. He didn't want her to be special. To feel the way he was coming to feel toward her would be dangerous with any woman, but with her, it was insanity. She had come to Texas for Ben Lockett, not his son. And she had only married him for the twenty thousand dollars; no doubt she'd leave as soon as she got the money. His face became an ugly sneer and he threw off her hands, sending her reeling backward.
"Who the hell are you to tell me anything about it? Have you lived here all your life? Have you ever been down to see the way those people live? It's a cesspool, Lauren. The dregs of society. Whores and gamblers and thieves. You don't know what you're talking about when you plead with me to save it."
"I'm sure that element of society is there. But there are innocent people who will suffer needlessly." To emphasize her words, she gripped his upper arms again.
"Don't presume to interfere in my life," he hissed and impatiently flung off her hands. His gesture carried more impetus with it than he had intended and he watched in bitter remorse when his hand flew up and caught her in the lip, cutting the tender flesh on her teeth and drawing blood.
They were held suspended in surprise, stunned by what had happened.
Jared was the first to rouse himself. He reached into his breast pocket and withdrew a handkerchief, extending it toward her. "I'm sorry, Lauren. Here—" He reached to blot up the thread of blood.
"Don't touch me!" She jerked away from him and slapped his hand away. The handkerchief fluttered to the floor. "I don't want anything from you. You're just like the rest of them. Leave me alone."
Animosity lay between them like a great gulf. Her gray eyes, stormy and hostile, met his hard and unyielding gaze. "So be it," he said after a long silence. "You won't be bothered with my touching you ever again."
She fled the room and rushed for the stairs. She was halfway up when Jared halted her. "I'm going to Austin tomorrow. I don't know when I'll be back."
She looked down at him. In spite of her disillusionment, she loved him. During their argument, his hair had become mussed and hung over his forehead, shadowing his eyes, making their expression indiscernible. His booted foot was resting on the bottom step, his arm draped over the bannister. His dishevelment only added to his handsomeness. Her heart cried, "Jared, I love you!"
But she said nothing. Not even goodbye.
* * *
"I'm leaving for Keypoint tomorrow and I'm taking Elena with me." Lauren faced Olivia across the wide expanse of Ben's desk. She had made a decision and was now daring Olivia to protest her action. "I think Gloria will need all the help she can get with her new baby coming. Elena and Carlos need to live together, like the family they are. I shall ask Rudy to give them one of the cabins on the ranch."
"The _foreman_ doesn't make decisions for Keypoint, Lauren." Olivia was seething. How dare she mention that bastard and his brood to her? Lauren was deliberately provoking her as she had done for the past several days.
The morning Jared left, Lauren had launched her campaign. Before Olivia was privy to her intentions, Lauren had organized a church committee strictly for the purpose of aiding and abetting the community of Pueblo.
Lauren's head wasn't in the clouds. She realized that if the citizenry of Coronado had really been concerned about their Mexican counterparts, they would have done something long before now. She used her new last name to its full extent. Sweetly she asked for help as Jared Lockett's wife on a project they had conceived. Her victims were powerless to refuse.
She had coaxed a reluctant Pepe to drive her through the streets of the town and she had been appalled at what she saw. In the bright sunlight, all of Pueblo's ugliness and deprivation were displayed like open wounds. Lauren had been shocked at the lack of sanitation, the poverty, the sickness, the squalor.
Under Lauren's direction, the church ladies commissioned other social groups to start charitable projects of their own. Old clothing was collected and distributed. Lumber—albeit used—was donated for construction projects. And Lauren wrote to the university in Austin, asking them if medical students would consider establishing clinics for the treatment of the sick and to teach basic lessons on hygiene.
When Olivia heard about Lauren's efforts from a guileless bank customer, she barely made it home before her temper erupted with the impetus of a volcano.
She began the interview in the office with, "You will desist from your ridiculous project immediately."
Lauren didn't pretend to misunderstand. She had come into the office well armed for combat. "I will not sit here and discuss anything with you until the drapes behind the desk are drawn." Whether Olivia was shocked by Lauren's bravery or too stunned by anger to object, she sat silently while Lauren calmly went to the windows and drew the drapes. When she returned to the chair opposite the desk she said, "Now, Olivia, I believe you have an opinion on my activities in Pueblo."
"Your interference in that community is sheer lunacy!" Olivia fairly screamed. "Whatever 'committees' you have organized will die a sudden death, starting now. Am I understood?"
"You are, but no, none of the projects I've started will be stopped."
"I'll see to it that they are," she threatened with a tone of voice that would have intimidated many a brawny man.
Lauren didn't even blink. "I don't think you will," she stated calmly. "What would everyone think if my 'noble' efforts were halted by my own mother-in-law?"
"No one would know," Olivia said with a trace of amusement. Was the girl simple?
"Yes they would. I'd tell them."
"Oh, I see. It's your intention to disgrace me?"
"Since when is it a disgrace to aid one's fellow man?" Lauren considered the interview over then, and left Olivia alone with her hatred.
If Olivia planned to stymie Lauren's actions, she found herself ill-equipped and ill-trained to do so. This was the type of work Lauren had done all her life. She was a good administrator. It was a rare talent to be able to talk someone into doing an unpleasant task and leave them thinking that it had been their idea in the first place.
Everyone was enamored with the courageous Mrs. Lauren Lockett. Olivia's subtle, tentative suggestions that her daughter-in-law might be a bit ambitious met only high praise and enthusiasm for the projects.
When all her committees were in full swing and Pueblo already showed signs of improvement, Lauren decided to return to Keypoint. For the present, she had done all she could do. The cold hostility in the house in Coronado had become untenable. Now, as she faced Olivia purposefully, unafraid of the woman's imposing mien, she said, "I spoke to Jared about Elena and Carlos, and he agreed." That was the first time that Lauren Holbrook Lockett had told a lie, but she felt that she would be forgiven. "Besides, with both Jared and me away, I'm sure you can spare her. Pepe can escort us to Keypoint and bring the wagon back. He should be able to return the day after tomorrow."
"You have it all planned, I see. What do you hope to accomplish with this little escapade? Do you intend to tell certain persons our plans in hopes that they will be aborted?" Olivia's green eyes were daggers as she glared at Lauren.
"If I told Rudy," she paused after emphasizing his name, "it could endanger him and his family. I wouldn't do anything to jeopardize their safety."
"How thoughtful."
Lauren ignored the sarcasm and continued, "I don't think your barbarous plans will ever see fruition, Olivia. I don't think Jared will take part. I've come to know him, and it would be against his nature to do such a despicable thing. It's your soul I fear for, Olivia, not Jared's."
Olivia laughed with genuine pleasure. "What a fool you are. You think you have redeemed my son. I warrant that he would be a challenge for any missionary to try to save." Then all amusement left her cold, beautiful face. "Don't count on Jared. He belongs to me and will always do as I tell him to."
Lauren rose gracefully from her chair and, totally composed, left the room.
* * *
The next morning dawned cold, rainy, and dreary, in perfect harmony with Lauren's mood. Rosa bid a tearful goodbye to Elena, but the younger girl was jubilant. When Lauren had told her about taking her to Keypoint so she could live with Carlos, Elena had been overcome with gratitude. She couldn't believe that it was really going to come about. Despite the dreadful weather, she was chattering merrily when they left.
The trip took longer than usual because of the rain and the muddy roads. Lauren sat on the seat of the wagon with Pepe, while Elena sat huddled under a tarp in the wagon bed.
They were cold and tired and hungry when they finally reached the ranch house late in the afternoon. Gloria ran out to meet them and hugged Lauren to her in a sisterly embrace.
"We've missed you so much. It's good to see you. Where is Jared? Is he coming later?"
The mention of his name brought a lump to Lauren's throat, but she answered calmly enough, "I don't think he's coming out this time. He has business in Austin."
Gloria would have said more, but Lauren's obvious reluctance to talk about her husband stifled any questions she would have asked. Would these two never get their differences settled?
"Gloria, this is Elena, Carlos Rivas's wife and my friend. I've brought her with me to help you here in the house. I hope Rudy can arrange for them to live in one of the cabins. Do you think that's possible?"
"We'll make it possible. Welcome, Elena," Gloria said, taking both of the girl's hands in her own. "We think a lot of Carlos and I'm happy to finally meet you. We were sorry to hear about your baby. Will you help me take care of my children? They are more than I can handle."
Gloria was reacting just as Lauren knew she would. Elena would be happy here. Lauren sighed and went into the house to greet the children and Maria. As each child was hugged, he or she had a special story to relate to Aunt Lauren. She listened to them avidly, jealous of their carefree innocence. They were full of questions about their hero, Uncle Jared, and she told them about his winning the shooting contest. They listened with wide, round eyes to the entire episode with the automobile.
Gloria shooed them into their rooms and Maria stepped forward to embrace Lauren, who welcomed the succor she found in those slender arms. Maria pulled away from her and looked deeply into the sad gray eyes. "I think you are unhappy, Lauren, no?" Lauren hung her head and nodded miserably. "We'll talk later." Maria patted her arm and turned to welcome Elena.
When he came in that evening, Rudy gave her a hearty kiss. At a signal from Gloria, he refrained from asking too many questions about Jared or the situation in Coronado.
Pleading exhaustion, Lauren retired early to her room. It was painfully reminiscent of Jared—his things, his clothes, his scent. Everything was a grim reminder of their parting.
She knew he hadn't meant to strike her. That had been an accident. What she considered to be his betrayal was the harsh, angry words he had flung at her. Could those words have come from the same mouth that had kissed her with such tenderness? Kisses that even now as she remembered them made her body tremble?
It was futile to deny the sensations that rocketed through her whenever she recalled his lovemaking. Her body became as malleable as melting butter when she recalled his hands and lips and how they had aroused and pleased her. The intricacies of her body and his had been revealed to her under Jared's practiced touch, yet she longed to know more. She wanted to feel again that sublime ecstasy that had encompassed her the moment his masculinity filled her and made her complete.
Love words he had chanted in her ear came back to haunt and mock her as she tossed restlessly on her pillow, which was already damp with tears.
* * *
Rudy gave the use of a cabin about a mile away from the large house to Carlos and Elena. He even permitted Carlos to take a day off to set up housekeeping. Very little got accomplished, but Elena was radiant the next morning when Carlos dropped her off at the ranch house before he reported to work. She fell right in with the routine in the kitchen and herded the children around with the patience of an experienced schoolmarm. The children adored her and minded her far better than they did either their mother or their indulgent Aunt Lauren or their grandmother, whom they manipulated unmercifully.
Lauren wasn't in the Christmas spirit, but the holiday came nonetheless. On Christmas morning, James and John were delighted with a new pair of suspenders and play guns with holsters. The girls squealed when they found new petticoats and hair ribbons in their boxes from Aunt Lauren.
There was so much confusion amidst the packages and presents that Santa Claus had left behind, that no one noticed when Lauren slipped back into her bedroom and shut the door.
Some compulsion she couldn't name moved her toward the closet and she opened it slowly. She looked at Jared's clothes hanging there, an old pair of boots flung negligently onto the floor, a leather vest hanging on a nail inside the door.
"Jared," she groaned, and pressed her face into one of the shirts that, even though it had been laundered, still retained his scent, the aroma of his tobacco and the faint smell of leather. She sobbed for several minutes into the cloth until she felt gentle hands on her shoulders.
"Lauren, are you ready to talk about it now? I'll listen if you are."
"Oh, Maria, I don't want to burden anyone with my problems, especially not you, when Ben's death is so heavy on your mind."
Maria led her to the bed and they both sat down. The children's exuberant exclamations came from the next room. Maria patted Lauren's hand, giving her time to assemble her thoughts.
"I... this... my marriage is a farce," she blurted out. Unevenly, brokenly, she told the whole story from the time she had met Ben to William Keller's attempted rape and the Prathers' censorious acceptance of his lies. She was ashamed to tell Maria about the terms of her deal with Olivia, but she related them dispassionately, omitting any mention of the Vandivers. "You probably think I'm terrible to sell myself that way."
"I'm in no position to pass judgment on anyone, Lauren. But under the same circumstances, I would have done the same thing. Sometimes our choices in life are between the lesser of two evils. You did what you hoped would be the right thing. No one can condemn you for that. Besides, I think your motivations included something besides Ben's wishes and the money, didn't they?"
This was the question that Lauren had asked herself time and again. Had she been in love with Jared even then? Was that why she had gone along with this outlandish scheme? Nothing was clear to her anymore.
"You fell in love with Jared, didn't you?" Maria asked softly. Lauren nodded and Maria continued, "And this marriage of convenience has become something else, hasn't it? You've... been with Jared?"
Lauren covered her face with her hands and sobbed, "Yes, oh, yes. Not until just a few days ago and... oh, I don't know, Maria. All my life I've been conditioned to believe that men did 'bad' things to women and that ladies protected themselves from that debasement. Even though we're married, I know that Jared doesn't love me. Is it sinful for me to feel what I do? To find pleasure in..."
"Did you think it was 'bad'?"
"No," Lauren answered vehemently, and Maria suppressed a smile. "When I first came to Ben, I was young and unsure, too. I tried to deny the joy he could bring me. But when I saw that I was giving him great happiness, I could share it without guilt or shame. I, too, was brought up to think that the woman should expect nothing but pain and degradation from lovemaking. God, not Man, created sex, Lauren. And even though Man perverts it and misuses it, it is still a gift to two people who love each other."
"But Jared doesn't love me. I'm terrified of the day he'll grow tired of this farcical marriage and send me away."
"I think Jared is fighting his own battle of feelings, Lauren. I don't believe he could have been coerced into marrying anyone, no matter how high the stakes, if he hadn't been attracted to her. He's far too headstrong. And I think sending you away is the last thing he wants to do." She looked at the tearful young woman and her heart went out to her. Ben had wanted them to be happy. He'd had such high hopes that their relationship would bloom into love.
"Lauren," she said gently, "don't be afraid of loving Jared. If I hadn't loved Ben, think of the useless life I would have had. Think of the misery he would have had to bear alone. I don't think you will regret loving your Lockett any more than I do having loved mine. The only regret I ever had was not being able to give Ben more children."
Lauren sniffed and dabbed at her eyes with a lace-edged handkerchief Maria had extended her. "Thank you, Maria. You go on and join your... our... family. I'll be out later." She smiled tremulously.
Maria stroked the tear-stained cheek and then left Lauren alone to wonder where Jared was spending Christmas.
# Chapter 20
Two weeks after Christmas, the residents of Keypoint enjoyed a few days of unseasonably mild weather. The old-timers predicted that the worst of the winter wasn't over if the weather was this warm in January, but everyone was glad to see a respite from the howling winds that sometimes brought icy rains and sleet.
One morning, Lauren rode out by herself. Maria had a cold and she hadn't wanted to bother Rudy or one of the _vaqueros._ It was amazing how much she missed Flame. But any recollection of the mare reminded her, too, of Jared. A Jared vicious and violent. She pushed those thoughts from her mind.
Never losing sight of the compound and judging her direction by the Rio Caballo, she cantered up and down several hills, grateful for the exercise. She had been gone about an hour and was on her way back to the house when she noticed her mount's ears pull back sharply. About the same time, Lauren heard a mumbled curse followed by a moan. She reined in the gelding and listened. The moan was repeated, coming from the direction of a clump of cedars. She nudged the horse nearer. When she was a few feet from the grove, she could barely make out a form lying on the ground.
She dismounted cautiously and took a tentative step forward.
"Stay right where you're at." She heard the unmistakable click of a rifle being cocked and she froze, her heart jumping to her throat.
"Don'tcha come no closer." The voice was sinister, but clipped as if the effort to speak was painful.
"Are you hurt? I heard you moaning." Lauren was quaking on the inside, but she felt this person needed help.
"You Jared Lockett's woman?"
"Yes, I'm Mrs. Lockett. Who are you?" She took one more step forward.
"I said don't come no closer." The last word raised an octave as the sentence dissolved into a long, heart-constricting wail.
Throwing caution aside, Lauren ran into the thicket. She pulled up sharply and covered her mouth to stifle the scream that rose from her throat.
The man was ragged and dirty, his ankle caught grotesquely in a trap of some kind, blood oozing out around the steel teeth that were biting into his flesh.
His face was a hideous nightmare. This was Crazy Jack, the hermit. It was a death mask this poor creature wore instead of a face. Red ugly scars adorned the sides of his head where his ears should have been. Two open holes that gaped eerily served as his nose.
Lauren swallowed the bile that flooded into her mouth. "Mr. Turner, let me help you." She crept closer to him.
What was left of his face was contorted in pain. His lips were pulled back in agony over a scarcity of teeth, and his eyes were squeezed shut. Lauren noted that the fingers which had held the rifle were now clenched around the injured leg. The firearm had been abandoned on the ground.
"Don't want no help," he hissed.
"You may not want it, but I think you should have some." The firmness in her voice surprised him. He opened his eyes and looked at her suspiciously, searching for some threat. He saw none.
"Can you get this goddam contraption off my leg?"
"I... I don't know." She looked at the ominous thing and shuddered. "I can try."
"Well, quit jawin' then and do it afore I bleed to death," he grumbled. "Take aholt on either side and pull as hard as you can."
"Won't it hurt when I lift up your foot?" she asked timorously.
"Yes, goddammit, but it's hurtin' like hell now, and I got to git it off, ain't I?"
"Very well," said Lauren decisively, removing her gloves. Obviously the man was determined to be rude.
Her heart was thudding as she knelt down beside the disfigured hermit and gently closed her fingers around his shin above where the trap had sprung on his ankle. He gasped even at this slight pressure and she looked at him with pity. "I'm sorry, I know it's excruciating."
"Go on and git it over with," he rasped.
She placed her fingers on either side of the trap, finding as good a hold as she could on the blood-slick metal. Tentatively she tried to pull the trap apart. It didn't budge and Crazy Jack's breath sucked into the vacuum of his mouth as the pain increased. "Harder, lady."
Lauren tried again, exerting tremendous pressure. Just as she was about to give up, she felt the metal beneath her fingers give way a fraction. The muscles of her arms ached with the effort she was demanding of them. Finally the sides of the trap sprang apart, tearing into the poor victim's flesh before coming free of it.
Jack screamed. The trap's teeth had left deep puncture wounds around his ankle. They were bleeding profusely. Lauren went to her saddlebags and retrieved a canteen of water. She knelt down again beside him and poured the liquid onto the wounds. Jack actually laughed at her.
"Water won't do no good, Missy. Get that canteen off my horse. He's around here somewheres." She looked around until she saw a mangy animal nibbling on the short grass under the trees. She approached him timidly, afraid that he might be as shy of people as his owner, but he stood docilely as she lifted the canteen from where it hung around the saddle horn. She uncapped it and the unmistakable odor of whiskey assailed her nostrils. This must be the rotgut that Jared had told her the old recluse distilled.
She paused only an instant before generously bathing the punctures with the liquor. Jack winced and his eyes began to water, but he didn't scream again. He gestured for her to take the scarf from his neck and wrap it around his leg. It was grimy and dark with grease.
"Why... why don't I use mine? It's..." she suppressed the word cleaner and substituted, "larger."
"Ain't takin' no charity—"
"No, no, nothing like that." She didn't give him time to protest further as she whipped her bandana from around her neck. She formed a silent, selfish prayer of thanksgiving that she wasn't wearing the blue silk one Jared had given her, but one of cotton print she had bought for herself in Coronado. Not allowing herself to think of the pain she must be causing the poor man, she hastily tied the scarf around his oozing wounds.
"There. That should hold you until we can get you back to Keypoint and summon the doctor. Can you ride?"
"Hold on just a goddam minute, Missy. I ain't agoin' nowhere but to my house, and no stinkin' sawbones is goin' to touch any part of Jack Turner."
"But, Mr. Turner, those wounds are serious. Your ankle may be broken." She couldn't let him return to that cave he lived in without medical attention. "Please, if you don't want to go to Keypoint, let me get Rudy, you know Rudy Men—"
"Hell, yes, I know who Rudy Mendez is, and he or no one else is goin' to take care of this ankle 'ceptin' me. I've had more broken bones than you've had years."
"But you may need to be sutured."
He raised his scornful eyes to her then and cursed imaginatively. "Who you think sewed up my face when them Injuns did this to it, huh?" He didn't expect an answer and Lauren was too mortified to make one. "Now get outa' my way."
Jack struggled to his feet and shrugged off her attempts to help him. He leaned down and picked up his trap, condemning it for being empty. He damned as well his own clumsiness at having stepped on it. He limped to his horse and took a long pull on the canteen before he hoisted himself into the saddle.
"Would you like for me to follow you and see you home? You may need _some_ help."
"No, ma'am. You seen more of Jack Turner than any other human has in twenty years or so. I'd be obliged if you was to forget what you seen." He looked away shyly and said, "You seem like a real decent sort of woman, Miz Lockett."
Lauren knew that he would resent the pity she felt inclined to show him, so she said, "Thank you. It has been a pleasure to meet you, Mr. Turner."
He tipped his hat to her and rode away. She didn't follow him so as not to invade the privacy he coveted. She waited until he was out of sight before she mounted her horse and rode back to the ranch house, after stopping once at the river to wash the blood from her hands.
She went into the kitchen and gathered up a basket of food, then returned swiftly to the riverbank across from the house growing out of the rock wall. She left the basket on the flat rock as she had seen Jared do. She didn't tarry, but rode away without looking back.
* * *
As Gloria's delivery date drew near, the women sewed and knitted, talking over names, deciding first on one and then discarding it for another. Elena happily announced one morning that she, too, was pregnant again. She radiated good health and energy, while Carlos wore a perpetually sappy grin. Lauren was touched by their obvious happiness with each other and their new home.
The Mendez baby chose the twenty-third day of January to make its appearance. Gloria had been listless since getting up that morning. She, Lauren, and Maria were sitting in the large living room near the fire enjoying a cup of midmorning tea when she clutched the arms of her chair.
"I think I'm going into labor. That's about the third pain, and it's the strongest."
Lauren nearly dropped her cup, but Maria went to her daughter-in-law and supported her as they walked into the bedroom she shared with Rudy.
"Lauren, will you come help Gloria into bed, please?"
Lauren jumped in alarm, but she followed the other two into the room. At Maria's instructions, Lauren turned down the spread and covers on the bed, and Maria eased Gloria down onto the sheet.
"I'll undress her, Lauren, if you'll go tell Elena to take care of the children for the rest of the day," Maria said placidly. "Send word to Rudy by one of the _vaqueros._ They'll know where he is. Then come back. I'm sure we can use another pair of hands."
Lauren sped out of the room, grateful for any task that took her away from the birthing bed. She didn't relish witnessing that secret rite which had killed her own mother.
Elena was elated, and assured Lauren that she would take care of the house and the children and would stand by if needed for anything else. Rudy came bounding into the house a few minutes later, looking gray under his dark features.
"Rudy," Lauren cried. "I was counting on you to calm my nerves. I thought you'd be used to this by now. You look like a _new_ father."
He grinned abashedly. "I guess every time is like the first time. Is she all right? Can I go see her?"
"Well, I suppose so." Lauren wasn't sure what etiquette dictated in this instance. "Let me check."
She crept into the dim room to find Gloria propped up primly in the middle of the bed talking amiably to Maria. Lauren hadn't expected that. She thought she would find her writhing in pain. "Is it all right if Rudy comes in to see you?"
Gloria laughed. "It's his fault I'm in this predicament, so I guess that entitles him."
Rudy entered after Lauren bade him to and crossed to the bed in three long strides. He sat down close to Gloria and put both of his large hands on her abdomen.
"So by nightfall we'll have another mouth to feed, hey?"
"I'll be the only one feeding it for a while, remember?"
"I'm sure you can handle it." He grinned and cupped her swollen breasts in his hands.
She swatted them away playfully. "Rudy Mendez, even at a time like this you're a lecher. And in front of your mother!"
"She knows I take after my father." He laughed. Then he leaned down and kissed his wife tenderly on the forehead. "I'll be outside if you need me. I love you."
Gloria kissed both his hands before he got up and left the room. Lauren's eyes filled with tears and a lump in her throat prevented her from replying as Maria asked her to stoke the fire in the grate.
The afternoon dragged by. Gloria's pains became more insistent, and Lauren watched in horror as the bed was flooded with water tinged pink with blood. She thought something was dreadfully wrong until Maria assured her that this was normal, and that the baby would be coming soon.
About an hour later, Gloria's face twisted in pain, but with Maria's gentle urging to push harder, she was delivered of a baby boy. Lauren watched as Maria drew him from his mother's body and cut the cord that had bound them together. Her mother must have suffered in the same way to deliver her and the little brother whose birth had killed her. She felt a great loneliness for the woman she had never known. She wished she could remember telling her mother that she loved her. Surely she had.
Maria was wrapping the squirming, squealing baby in a warm blanket when Gloria cried out, "I don't think that's all!"
Maria and Lauren rushed back to her and saw another dark head emerging from the opening between her legs.
"Lauren, help her," Maria commanded as she held the infant boy closer, trying to muffle his loud cries.
Lauren turned pale and started to object, but Gloria moaned again. She looked back to see the baby's shoulders trying to push their way into the world. Trembling, Lauren took the child's head in her hands as she had seen Maria do and gently pulled. The baby didn't move, but it set up a howl even with mucus still clogging its small throat. Lauren was perspiring and shaking as she pulled more firmly on the slippery head. The baby almost popped out into her waiting hands. It was a girl.
She was laughing and crying all at once as she announced, "It's a girl," to the anxious mother whose face then relaxed into peaceful repose.
"Here, Lauren, you take this one and I'll cut the cord." Lauren laid the baby girl on the sheet as Maria shoved the other one at her before cutting the cord for his sister.
"Gloria, you have twins. One of each." Maria was giggling like a young girl.
"Tell Rudy," Gloria whispered weakly from the pillows. Lauren turned and left the room, still carrying the new boy who was making his presence known.
"Rudy, it's twins! A boy and a girl!"
Rudy came to her quickly and looked down at his new son. "Twins?" he asked stupidly.
"Yes, come and see." She led him back into the room where Maria held up the baby girl.
"Twins!" He laughed, then whooped so loudly that the babies started screaming even louder.
"Now see what you've done," scolded Maria as she laid the baby girl on Gloria's shoulder. Rudy took his son from Lauren and sat down on the bed next to his wife. They oohed and aahed over the babies even while Maria was tending to Gloria between her raised knees. Lauren felt like an interloper, and left the room.
She didn't realize how tired she was. It was late in the evening and she had spent most of the day in the room with Gloria and Maria. She hadn't eaten since breakfast, but the emptiness she felt inside wasn't hunger.
Crossing to the wardrobe, she opened the door. She took one of Jared's shirts from its hanger and held it close to her. It had been over a mouth since she had seen her husband. She longed to share the experience of the birth with him, the wonder of it.
She lay down on the bed and pulled the shirt over her. Closing her eyes, she could see Jared's face just as it was when he had gazed down at her in awe after he had made love to her. There had been no cynicism or bitterness on his face then. Only tenderness. Where are you now, Jared? What are you doing? Do you ever think of me?
She pulled off her shirtwaist and skirt and slipped Jared's shirt over her head. Pulling a pillow against her, she fell asleep.
* * *
For the next few days, the arrival of the twins upset the normal routine of the household. The other children were continually underfoot trying to get glimpses of their new brother and sister. Lauren was amazed at the patience that Gloria exhibited. She listened to their chatter and managed to have a private time with each of them every day. Lauren knew she was fatigued after the birth and from the necessity of feeding two healthy infants, yet she didn't neglect her other children.
Benjamin was the name given to the baby boy, in honor of his paternal grandfather, and Lauren was deeply moved when they named the girl after her.
"After all, you brought her into the world. And I hope she will grow up to be a lady like you." Gloria had hugged her sister-in-law and ignored the tears rolling down her cheeks. She was quick to cry these days.
Young Benjamin and Lauren were one week old when something greatly disturbing happened. Rudy was in the house for the noon meal when one of the _vaqueros_ came to the door and told him someone outside wanted to see him. The ranch hand shifted his eyes uneasily and looked ready to stand behind Rudy if support was needed.
Lauren followed Rudy out to the porch and saw the charcoal burner Wat Duncan and his sister June sitting on a derelict horse which was almost as dirty as his riders.
Duncan dismounted and strode toward them. "Howdee do, Seeñor Mendez. Heerd you had a new set of twins." Duncan smiled his insolent smile. He had not changed clothes since Lauren had first seen him, only added layers in deference to the cold.
"What are you doing here, Duncan? You know only certain areas of Keypoint are open to you, and this is definitely not one of them." Rudy's voice was firm and cold.
"Well, I jes' came to give you my congrajulashuns. It 'pears your babies is the only younguns that'll be born in the Lockett fambly. Seein' as how Jared deserted his purty bride an' all." He grinned sardonically at Lauren and she shivered under his lascivious scrutiny. She averted her gaze to June, whose shapely legs, bare and seemingly impervious to the cold wind, straddled her mount.
June was staring at Lauren spitefully and licked her dry, chapped lips when she met the other's eye. "Reckon Jared had to go to Austin to find a warm bed for the winter," she drawled.
"My brother's business is none of your concern," Rudy snapped, after a furtive glance at Lauren. It worried him that they knew so much about the happenings at Keypoint and Jared's whereabouts. "He is engaged in important business with the railroad, but keeps in constant contact with us. Now, you'll state your business and leave."
Wat Duncan assumed an obsequious attitude. "Now don'tcha go gettin' all riled up, Rudy. I come in good faith. Vandiver and some of his heavies been snoopin' around and axin' a lot of questions. I don't give a good goddam what happens to them Mexies—no offense intended. I jes' don't want nothin' to happen to my deal with Lockett. Y'all understand. It's my bisness I'm worried about."
The muscles of Rudy's jaw had hardened to granite and Lauren saw him stroke the holster that held his six-shooter. "Get out of here, Duncan. Don't ever come near this house again, or I'll personally kill you. As for your _business,_ it's intact as long as Jared and I say so. Vandiver has nothing to do with it. Now get off this property."
"Awright. I'm agoin'. Jes' tryin' to be neighborly like." He sauntered to the mangy horse and mounted in front of his sister. She hooked her thumbs into his belt. Her fingers brushed the front of his trousers as she cooed, "Who's Rudy been sleepin' with, Miz Lockett, now that his wife is laid up havin' babies?"
Rudy reached for his gun, but Lauren put out a restraining arm. "No, Rudy," she whispered, for she had seen Duncan's hand moving to the far side of the horse where a shotgun was strapped. She was grateful to see that the word had spread among the _vaqueros_ that the disreputable pair were there and that many of them were moving into place, virtually surrounding the miscreants.
The girl tossed her long white hair and laughed, confident that Rudy wouldn't draw on her brother. "Tell Jared I come axin' about him. I'd like to see him when he gets back." She looked at Lauren and snorted. Duncan pulled the reins of the horse around and they clopped out of the yard, obviously in no hurry to leave.
The _vaqueros_ and Rudy watched until they were out of sight, then Rudy commissioned two hands to follow them and make sure that they returned to their camp.
When he came back into the large room, Lauren was sitting staring into the fire. He crossed to her and took both of her cold hands into his as he squatted down in front of her. "Lauren, Jared has never had anything to do with that dirty slut."
Lauren smiled at his kind, sympathetic face. "I know that. He has no affection for me," she confessed wryly, "but I know his taste in women would be more discriminating than that." Maria and Gloria had moved over to them and were listening anxiously to the conversation. "What worries me," Lauren continued, "is what Duncan said about Vandiver and his heavies being in Pueblo."
"Yes, that bothers me, too, but not as much as his knowing about everything that is going on in this house. He is trouble and no doubt about it. Ben would have had him shot on sight if he had come here before, and I may regret that I didn't. How did he know you?"
Lauren told them about the day she and Jared had gone to Pecan Creek and passed through the charcoal burners' camp on their way home.
"Rudy, I'm frightened," Gloria said.
He stood and put his arm around his wife, who had been up out of bed only one day. "I'm sure he's just testing our authority now that Ben's gone. There's no need to worry. Jared and I'll talk it over and decide what's to be done about them. I don't like having that scum anywhere on Keypoint."
The lines around his mouth were grim and he stayed near the house for the rest of the day, though he tried not to appear nervous. Lauren noticed that for the next few days, three or four _vaqueros_ were posted around the house. Despite his reassurances, Rudy was still worried about Wat Duncan.
* * *
Holding true to the dire predictions, the unseasonably mild weather of January gave way to blindingly cold storms in February. There was little that could be done on the ranch in such weather, and Lauren felt sorry for the _vaqueros_ whose turn it was to ride out and check on the vast acreage of Keypoint. They always took plenty of provisions, planning to spend days at a time in one of the line cabins built for just that purpose.
Those in the house confined themselves to entertaining the children, caring for the demanding twins, and sewing and baking for the family and the idle cowboys in the bunkhouse.
It was on one such late-evening excursion to the bunkhouse to deliver a batch of cookies that Thorn approached Lauren. The basket had been gratefully received by the _vaqueros_ and she was scurrying back across the compound toward the house when the Comanche loomed out of the deep shadows to stand directly in front of her.
She managed to stifle a startled scream by covering her lips with her hand.
Without preamble or apology, he said, "Mrs. Lockett, I found this on the gate this morning."
She hadn't known what to expect his voice to sound like, but it was low and deep, almost cultured. She found herself staring up into the implacable mask of his face. Then she looked down at the crude package he had extended to her.
The bundle was wrapped in brown paper and tied with a string. On it had been childishly scrawled _Miz Lokit._ "What...?" she said, looking into Thorn's face again.
"I believe it's something that belongs to you."
She slipped the string away and opened the paper. The kerchief she had wrapped around Crazy Jack's ankle was caught by the cold wind and nearly ripped from her hand. When she had grabbed it back, she saw that it had been washed and folded. All traces of blood were gone. Her lips tilted into a secret smile. Who would have expected the old hermit to meticulously launder her scarf? Had he chanced being seen to return it to her? He must have.
Lauren was suddenly aware that Thorn was staring at her closely. "I... I must have lost this somewhere," she stuttered. "I guess someone found it and... returned it. Thank you."
The Indian's eyes didn't waver, and she sensed that he knew more than that stoic face revealed. He didn't speak again, but acknowledged her thanks by a quick jerk of his chin. Lauren didn't realize he had moved away from her until his form was swallowed up by the descending darkness.
* * *
The third week of February, they saw the first snowfall. It had threatened for about a week with fierce north winds bringing rain, freezing drizzle, and enough sleet to coat the ground.
As the storm increased its fury after nightfall, Rudy, Gloria, Maria, and Lauren sat around the fireplace enjoying the peace and quiet that descended as soon as the children were put to bed. Gloria had nursed little Lauren and handed her to her namesake. Lauren held the baby on her chest, stroking the small, dark head under her chin. Gloria was appeasing Benjamin's hearty appetite while Rudy and Maria looked on lovingly.
They all jumped in startled reaction when they heard heavy boots thudding across the porch. Still edgy about the charcoal burners, Rudy reached for his holster, which was hanging over the mantel, pulled out the Colt, and had almost reached the door when it was flung open, accompanied by a gust of snow-laden wind.
# Chapter 21
The looming figure entered, closing the door quickly behind him. He turned slowly and Lauren gasped when she recognized her husband under his heavy clothing. He followed the sound of her reaction and almost repeated it as he saw her sitting in the glow of the fire, hair tumbled around her shoulders and down her back, holding the baby at her breast. He stood as if struck dumb.
"Good Godalmighty, Jared, aren't you full of surprises!" Rudy clapped his brother on the back. "You almost got shot, you stupid sonofabitch. Why did you plan your homecoming on the night we have the season's only blizzard?"
Jared shook his head as if to clear it. "I... I didn't know it was going to get so bad until I got halfway here."
"We're glad you made it safely, Jared." Maria looked at him fondly and he returned her affectionate smile.
"I think it was a crazy thing to do, but I'm glad to see you anyway." Gloria went to him with her arms extended. She was still piqued at his abandoning Lauren, but her natural fondness for him temporarily overcame it.
"Well, look at you, Sister. You've got your figure back. I'd better hug you before Rudy pumps you up again." He took her in his arms in a bear hug despite her protestations.
"Come see what we've done," she said, extricating herself from his embrace.
She had laid Benjamin in one of the two cradles set near the fire where the babies slept during the day when their brothers and sisters permitted it. Jared bent over the cradle and hesitantly stroked the baby's cheek. "Who is this?" he whispered.
"That is Benjamin," Maria said proudly.
"And this is Lauren," Gloria said, turning Jared toward the twin still held by his wife.
Lauren hadn't been able to move or speak, his presence in the room stunning her into silence. She couldn't take all of him in at once and was glad to have had time while the others greeted him to look at him closely. He untied the bandana that held on his hat and ran his hands through the long, unruly hair still damp with snow before shrugging out of the shearling coat as he crossed to the cradle to view Benjamin. He looked gaunt and tired. The stubble on his face was twenty-four hours old at least.
But he was Jared. And he was here.
He dropped down to his haunches in front of her chair. He met her swimming eyes over the top of the baby's head. A silent communication more puissant than words passed between them.
"Lauren delivered her, so we named her after her aunt," Maria said.
"You delivered the baby?" Jared asked softly, incredulously.
Lauren nodded as she turned the small bundle toward him. He took the tiny fist in his and smiled as the baby made a sucking motion with her mouth.
He looked closely at Lauren once more before he straightened to his full height. He eyed his brother derisively, spread his arms wide, looked heavenward, and pleaded, "Is there no end? Twins!" Then he broke into a broad grin and slapped Rudy on the back as he congratulated him. "Does this entitle me to a drink?"
"You bet. I haven't even celebrated properly. I've been waiting for you to get here."
"Are you hungry, Jared?" Maria asked him.
"Yeah, but let me warm up a little first. It's cold enough to freeze your... it's cold out there," he finished lamely, and everyone laughed.
He and Rudy shared a couple of glasses of whiskey while they caught up on general ranching business.
Gloria and Lauren took the babies into the bedroom that they shared with their parents for the time being. Maria kissed Rudy and Jared in turn and excused herself for the night.
A short while later, Gloria said, "Jared, please forgive me, but you have no idea how exhausting twins can be. I'll see you in the morning and you can tell me everything that's going on in Austin." She leaned over him and kissed his cheek. He smacked her bottom soundly with the palm of his hand. "Jared Lockett, my husband is sitting right there," she said indignantly.
"Yeah! Let's do something that will make him really jealous."
"You! You're incorrigible."
"Yes, but you love me." He grinned winningly.
"A little bit," she conceded, suppressing a laugh. "Coming, Rudy?"
"In a minute." He ignored her exasperation as she stalked out of the room.
"I'm hungry, Lauren. Could you get me something to eat?" Jared's voice was curt, and Lauren felt that she had been summarily dismissed. Rather than make a scene in front of Rudy, she nodded just as curtly and went into the kitchen.
She warmed the soup that was still on the stove, cut thick slices of bread baked that afternoon, poured a steaming cup of coffee, and, almost as an afterthought, added a large slice of apple pie that she had baked to the tray.
The men were talking low with their heads close together when she came back, and stopped abruptly when they saw her. A knowing glance passed between them. She read it to mean that they would continue their conversation later.
"Rudy." The plaintive cry came from the direction of his bedroom. "Please come to bed. I'm cold."
Rudy stood and tossed his cigar into the fireplace. He stretched his long frame and gave an exaggerated yawn. "The babies are just barely three weeks old and already that woman can't keep her hands off me." He shrugged in feigned helplessness and sighed, "What's a guy to do?" He winked at Jared and swaggered into the hall toward his waiting wife.
Jared chuckled as he turned his attention to the tray. Lauren had practically dropped it onto the low table in front of the easy chair by the fire. If he had noticed the loud, angry clatter of dishes, he didn't show it. He took several bites of the scalding soup, ignoring her completely. Angered by his calculated indifference, she whirled around and headed toward the hallway.
"Lauren."
It was hard for her to face him, but she forced herself to stifle her anger and pivoted toward him. "Yes?"
He studied her a moment as she stood framed against the darkness of the hall. She was poised for an attack, but her militant stance was belied by the vulnerability she conveyed in her white woolen robe and slippers. No warrior Jared had ever seen had hair that cascaded in a riot of thick waves and curls.
"How have you been?"
She folded her arms across her chest and laughed mirthlessly. "I don't believe for one moment that you care about my well-being, but as was taught me, I'll answer politely. I've been well, and you?"
He raised one eyebrow in quizzical surprise at her tone. "I've been fine. But please refrain from doing my thinking for me. I _do_ care about... about you."
"Then I can only surmise that all your messages and letters were diverted." She loathed the sarcasm in her voice, but she was angry, had a right to be, and he deserved this. "I assume your business in Austin went well."
He glanced back down at the tray quickly. "Some of it, yes," he replied in clipped tones. His own anger wasn't far from the surface.
"No doubt you're pleased. I think I'll go to bed now. We took the children out in the snow today, and I'm tired."
"Yes, go on. I'll clear this up when I'm finished."
"I'm sure Gloria will appreciate that. Goodnight."
He didn't look at her as he mumbled a response. He seemed dejected, the hollows of his cheeks and the lines around his mouth and eyes emphasized by the shadows the firelight cast on his face. Lauren steeled herself against the temptation to go to him. Instead, she walked down the dark hall to the bedroom.
She had just warmed a spot under the covers where her body huddled when she heard the bedroom door open. Jared came in, closing the door behind him.
She sat up quickly, pulling the blankets up to her chin. "What do you think you're doing?" she demanded.
He didn't even look toward her as he sat down on the ottoman and began tugging off his boots. "If I recall correctly, this is _my_ bedroom in _my_ house. It is a very cold night out, and I have no inclination to seek another place to sleep. If it offends your sensibilities to sleep with me—and I stress _sleep_ —then I suggest you find yourself another bed. This one belongs to me."
He had pulled off his socks, shirt, and the top of his underwear, and was working on his belt buckle. The firelight picked up golden tints on the hair that furred his chest.
Lauren flounced back against the pillows and scooted to the far side of the bed, putting her back to him. She heard his pants drop to the floor followed by the soft swish of his underwear. No! He couldn't sleep like that on such a cold night! He padded across the floor and tossed a few more logs onto the fire in the fireplace, then went to the trunk at the foot of the bed. He raised the lid, which squeaked slightly, and took something out. She dared not look. He flung whatever it was over the bed.
"Thorn made this for me. It'll keep us warm as toast."
She opened her eyes to slits and saw that it was some sort of fur blanket. She closed her eyes quickly when the cold air rushed in under the raised covers as the bed sagged with his weight.
"Goodnight, Lauren," he said. She lay perfectly still and didn't answer. He laughed and turned away from her, settling himself in the warm cocoon of the bed. It wasn't too many minutes before she heard the even breathing of his sleep.
She didn't sleep for a long while.
* * *
At some point during the night, they turned to each other. Whether it was for warmth or something Lauren didn't want to name, she awoke to find herself lying against Jared's chest, his heavy arm imprisoning her, their legs wrapped together.
She lay still, savoring the nearness of the body next to hers. The hairs tickled her nose as his chest rose and fell gently beneath her head. She could feel his breath on the top of her head. The dull thudding of his heart echoed in her ear.
Afraid to move for fear she would wake him, her eyes wandered as far as they could and delighted in their perusal. The fire in the grate had all but burned itself out, but one small log caught and flickered in the dark room, illuminating it briefly. Lauren saw Jared's broad chest under her head, the hair fanning out at his throat and tapering to a slender, silken column on his stomach.
Hesitating only a moment, she lifted her hand and, placing it against him, began slowly tracing the pattern of hair on his muscular chest, down the corded, flat stomach, until she felt it grow thick and coarse on his abdomen. She rested her hand on the wiry mat, unable to bring herself to explore further. Only then did she notice that the breathing above her head was no longer steady and the heartbeat beneath her ear was more rapid. She raised her head quickly and met the amber eyes glowing in the fading firelight.
"Ah, Lauren, Lauren." Her name was half-sigh, half-groan before his mouth melted into hers. He kissed her hungrily, wildly, while his hands sought the hem of her nightgown and raised it past her waist, her quivering breasts, over her head, and flung it away.
Raising his head, he looked deeply into her bright eyes as he lifted her hand. He kissed the palm ardently, teasing it with his tongue. Without taking his eyes away from hers, he drew her hand beneath the covers and placed it over his awakened manhood. He studied her reaction, fearing that she would be repelled. Jealously he watched the tip of her tongue disappear between her lips after nervously wetting them.
Don't be afraid of loving this man, Maria had told her. Don't be afraid. Her slender fingers closed around the warm shaft with its velvet skin stretched smooth. Gently her fingers played over him, curious, wondering fingers, fingers made exultant by their discoveries.
Reflexively Jared arched his back. His head went back in a gesture of exquisite feeling. Then his chin lowered and he was searching her face again. His golden eyes shone bright with emotion. "Touch me, Lauren. Touch me until I die from the pleasure of it. Know all of me." His voice was breathy and uneven.
Emboldened by his impassioned plea, she stroked and caressed until she found the smooth spearhead lubricated with the precious nectar of his desire. "Oh, God," he groaned as he lowered himself over her and took her mouth under his. His hands found her breasts and massaged them in rhythm to her own caresses. He squeezed the soft mounds gently while his thumbs appreciated the aroused centers.
For Lauren, every vestige of reluctance, doubt, and mistrust disappeared as she thrilled to the mysteries of her husband's body. Instinct instructed her in the best ways to show her admiration, and she was rewarded with his urgent, whispered words of praise and encouragement. Her hands glided over the firm muscles of his buttocks, down the hard thighs, up the sinewy back. She touched him unafraid. Imitating him, she kissed him passionately, using her mouth and tongue to explore his thoroughly.
His mouth and fingers were gentle stimulants that tormented her mercilessly. Relentlessly they trailed her neck, chest, breasts, and stomach, until she was making small whimpering sounds that surprised him as much as they shocked her.
"Put your hands around my neck," he instructed as he rose above her. His fingers found her feminine threshold moist and pliant and trembling. She tightened around his fingers like warm, closing petals as they entered that haven. He withdrew a fraction and stroked her lightly, but the mere touch struck her like a lightning bolt.
Her eyes opened wide in astonishment as she began to writhe uncontrollably. "Jared—" she gasped.
He replaced the seeking fingers with his tumescent shaft. Guided by his own hand, it rubbed against her, that magic spot, until she didn't think she could bear the pleasure any longer. She felt herself swelling, reaching out to him, opening, squeezing, dying brief little deaths to know his magnificence fully.
Jared, who before now had boasted of his sexual prowess, learned from the woman moving with him in such perfect tempo that he had known nothing of lovemaking. Not until he saw her face radiant with joy as she reached the peak of fulfillment under his manipulation did he realize the immense satisfaction of giving. Then he filled her completely, giving her all of himself, leaving no room for the frustration and fear that had come between them.
She clung to him tenaciously, matching his ardor, his fervent kisses that deepened even as he delved into her. In one shattering instant, they met on a plane where joy replaced sorrow, trust reduced uncertainty to insignificance, unity conquered loneliness, and indecision became commitment.
After the tumult, they held each other tightly, still unable to comprehend the upheaval of emotions that continued to race through them. Jared looked into her face and smoothed the ebony tendrils from her temples. Satiated, he slid down her body to cushion his head on her breasts. He kissed them lightly in turn, flicking his tongue over the rosy nipples, swollen and agitated with their recent lovemaking. "Beautiful, beautiful woman," he sighed.
He laid his head on that welcoming pillow. He was almost asleep, drugged by the fragrance of her dewy skin, when he heard her voice coming from far away and whispering, "Jared, I love you."
* * *
It continued to snow until noon the next day. The accumulation was in excess of six inches, which was unusual for that part of Texas. The world, from the view of those in the ranch house at Keypoint, appeared to be covered by a vast blanket, white, clean, pristine, and soft.
The bedroom occupied by Lauren and Jared was off-limits to the other occupants of the house. When the two failed to come to breakfast, and Gloria noted that Jared's coat still hung on the bracket by the door—evidence that he had not gone out to the bunkhouse the night before—she was thrilled. She forbade Rudy or any of the children to go anywhere near the bedroom. Rudy was amused by her protectiveness, but at the same time glad that his brother was finally sleeping with his beautiful, neglected wife. He would tease him later, away from Gloria's hearing.
The two people in the closed bedroom were totally unconcerned about any of the others in the house. In fact, they had not given them a conscious thought, so absorbed were they in each other. After sleeping for a while, they talked long hours about themselves. Lauren told him of her lonely childhood with a remote, undemonstrative father. Jared, in turn, reminisced about Ben, and about his slain friend Alex.
In the months they had known each other, they had never discussed personal things, except for that one brief conversation in the library in Coronado. Now they talked of trivialities—food preferences, favorite things, aversions and fears, birthdays—revealing the bits and pieces of themselves that made them what they were.
Early in the afternoon, there was a light tap on the door. The entwined figures on the wide bed moved but slightly, resentful of anything that separated them by more than inches. Jared muttered to himself as he crept out of the covers and crossed to the door. He was quite unashamed of his nakedness, and Lauren gloried in his physique with equally unashamed interest.
"What is it?" he asked through the heavy door.
There was no answer. He opened the door a crack and peeped around it. No one was there. Then he began to chuckle.
Lauren sat up, puzzled by his amusement. He knelt down and picked up a tray laden with food and drink. He closed the door with his foot before bearing the repast to the bed.
Lauren saw a platter of light, fluffy eggs, thick slices of ham, biscuits and _tortillas_ dripping with butter, a pot of coffee, and even a decanter of whiskey, along with plates, napkins, cutlery, and glasses.
"Remind me to thank Gloria later," Jared said as he bit into a _tortilla._ They ate until they were filled, and he removed the tray from the bed. They had opened the drapes earlier to enjoy the sight of the snow-covered hills. Now he went to the windows and closed the drapes, dimming the room.
He stretched like a lazy mountain cat and yawned broadly.
"Am I boring you?" Lauren asked mischievously as she twirled a curl around her finger then dropped it precariously close to a pouting nipple.
His footsteps, hurried because of the cold, slowed as he came closer to the bed. He placed one knee on the mattress, eyed her provocatively, and drawled, "Could be. What are you going to do about it?"
Lauren's flushed face became impish as she retorted, "Nothing!" and flopped to the opposite side of the bed, providing him an unrestricted view of her smooth back, tapering to the gentle swell of her hips.
He laughed before falling down beside her and grabbing a handful of hair. He wound it around his fist, pulling on it inexorably until she was forced to roll against him. Not quite sure how it came about, she discovered herself atop his chest, her legs straddling him.
"Jared!" she exclaimed, trying to disengage herself from the arms that clamped across her back. Her struggles only widened his grin. He risked taking one arm away from her back to cup her head in his hand and force her face down to receive his burning kiss.
At last, she pulled away from him and sat up. He was dazzled by the breasts that were enticingly suspended in front of him. With an index finger, he skimmed across the tops of her breasts, circled them leisurely, then teased the tips. He watched her immediate response in awe, intrigued and bewitched.
"Jared, I... oh... what do I... please..."
"Do whatever feels right," he said before he lifted his head and raked his tongue over her distended nipples.
"I—"
"Whatever feels right, Lauren," he breathed as his hands slid down her ribs to settle on her hips. His thumbs met at the dark nest in the center of her abdomen and impressed hypnotic circles onto it. Impossible to ignore, the hard shaft pressed up into her. His hands stroked her thighs as she rose to her knees, poised over him, then impaled herself on his strength.
"My God," he hissed through gritted teeth. His head tossed on the pillow. She rocked upon him, moving up and down, reveling in the feel of him inside her. She combed her fingers through the mat of hair on his chest, teasing the hard brown nipples. When she grew tired, she lay upon his chest. Her lips sought out the sensitive areas of his face and neck, nibbling, caressing them with light kisses.
His hands clasped the backs of her thighs even as his fingers stroked the moist folds between them. Neither could stand any more, and the eruption came. He filled her with the lava of his loins, which seemed to flow into her veins and scorch each nerve ending until it exploded under the onslaught.
Afterward, she knelt beside him, bathing the residue of their lovemaking from his body with a warm, damp towel.
He yawned again as she dropped the towel to the floor. Laughing as she brushed her lips across his, she asked, "Bored again?"
He smiled. "No, I'm sleepy," he confessed. "Come here." He pulled her down deep into the covers and snuggled against her. The hair on his chest tickled her back and his arm weighed on her waist. She moved closer still. He kissed her shoulder before they both slipped into dreamless sleep. Dreams had ceased to be necessary.
* * *
Jared knelt in front of the fireplace coaxing the dying embers into a red glow, adding small logs until they flared and ignited. He had wrapped the fur blanket around him to ward off the chill. Lauren could see no light seeping in around the edges of the drapes. It was after nightfall.
She took up her discarded dressing gown and slipped into it as she stepped from the bed. Her bare feet touched the cold floor and she scurried to the hearth and squatted down beside Jared.
"Hey, you're going to freeze. Why didn't you stay in bed?" He ran his hands briskly up and down her arms to warm her.
"Because you weren't there," she replied honestly, her eyes reflecting the glowing fire. Impulsively she kissed him on the lips.
He wrapped his arms around her and drew her down on the rug in front of the hearth. Staring into the flames, neither spoke for a moment. He stroked the length of her hair absently while his chin rested on the crown of her head.
"Jared?"
"Hm?"
"You can't imagine how terrified I was of you the first time I saw you."
He pulled his head back. "Terrified of _me_?" he asked with mock-dismay. There was a glint of mischief in his eyes.
"You were sprawled out, unconscious, in the back of that wagon. I had never been near a man who was so... virile... threatening... I don't know. You intrigued me, though. And that day you sneaked into my room, I was afraid I was going to faint."
He laughed quietly. "I was intrigued by you, too, though up until then I hadn't even seen you. Pepe told me later how I fell against you most ignominiously." He chuckled and hugged her tight. "I wanted to put you on the defensive. You were just as threatening to me."
"I? Threatening? How?" She stared at him incredulously.
Jared reached up to the small table where he had emptied his pockets the night before and picked up a cigar. He struck a match against the stones of the fireplace and lit the cheroot, drawing on it slowly and blowing the smoke over their heads.
Now was the time. She had to know.
"I resented you like hell, Lauren. Not you specifically, but any woman suddenly thrust upon me the way you were. The night Ben died, we had a fierce argument. I don't think he intended to tell me his plan for our marriage at first. He started the lecture by reminding me of my responsibilities and warning me that my indiscretions would eventually catch up with me. 'You're a man now, Jared. It's time you settled down and started behaving like an adult and not some wild bronco bent on destroying yourself and everything I'm leaving you as a legacy.'"
Jared drew once more on the cigar and flicked the gray ash at its tip into the fireplace. Lauren didn't speak. She wanted to know what had transpired that night, why Jared had hated her for so long.
"Mostly out of belligerence, I countered every point he made, until he lost his temper and hit me with both barrels. Our houseguest, whom I had been commissioned to fetch off the train in Austin, was the woman he had selected for me to marry. Since I had dragged my heels, he had taken matters into his own hands, and provided me with a wife. He stressed the fact that you knew nothing of this when I referred to you as an opportunist, along with some very ungentlemanly epithets."
He tore his eyes away from the dancing flames and looked at her closely. His hand cradled the back of her neck as his thumb stroked along her jaw. "You see, Lauren, my whole life, my parents had used me as a pawn to hurt each other. If I did something to please one, it infuriated the other. If I came to Keypoint, which I loved, my mother would be on a rampage for weeks after I returned to Coronado. My childhood and adolescence were one big battle to see who had the most influence over me. The older I got, I just didn't give a damn anymore. I sought to please myself and to hell with everyone else. I didn't relish having a wife chosen for me, especially when I wasn't sure what kind of relationship she had had with my father."
Lauren was suffused with love and pity for this complex man who was her husband. No wonder he had reacted to her intrusion with hate and resentment. "I think I see how confused you must have been by my coming here." Then, with an insight that surprised him, she said quietly, "You want to know why I followed Ben here. Is that it?" He didn't answer, but she sensed by his silence that this was the crux of the matter.
She sighed and stared at her hands as they pleated the skirt of her robe. "Jared, there was never anything illicit between your father and me. You couldn't be more wrong if you think there was. I was attracted to him because he was dashing and exciting and interesting. To someone raised in a parsonage, living with a sweet, but naive couple interested only in their small world, he was like a character out of one of my books. Of course, accepting his invitation to come to Texas to visit was unthinkable, and I never would have done it except for something that happened within days of Ben's departure."
Her lips trembled slightly as she remembered William's attack. Quickly, nervously, she said, "Believe me, it... those reasons for my leaving were justified."
He cupped her chin in his fist and asked, "What happened to make you want to leave your home?"
She tried to turn away, but he was adamant. He didn't release her chin, but forced her to meet his probing eyes. "It... I... Does it matter so much?" she asked piteously.
"Yes."
Again she tried to lower her eyes, and again he held her fast. "Please," she begged in a whisper. Slowly he eased the pressure of his fingers and his hand fell away. She turned her back on him to stare into the fire.
"There was a man," she said. "His name was William Keller. He... the Prathers thought that we should marry. I told them countless times that I couldn't abide him, but..." Her voice trailed away to nothingness and she breathed in deeply. Dare she tell Jared the rest of it? Would he turn away from her in disgust, blame her as her guardians had done? "Go on," he said from behind her.
Haltingly, she related the full tale of William's attack, his deception of the Prathers, her inability to persuade her guardians that the minister was lying. "After that, I had to leave," she finished hoarsely.
For at least a full minute, a heavy silence hung over them. She had raised her knees and rested her forehead against them. She didn't want to know what Jared thought. Yet she had had to tell him the truth.
His movement was so sudden that she started when he bounded to his feet. Whipping her head around, she saw him reach for his hat and clamp it on his head. Then, as she watched in stupefaction, he grabbed his gun belt with the deadly pistol safely ensconced in it and strapped it around his hip.
"Jared? What... what are you doing?" she sputtered.
He had reached the door by now and his hand gripped the knob. Glancing over his shoulder, she could see the rigid planes of his face set with determination. His amber eyes shone with resolve. "I'm going after that sonofabitch to kill him."
Despite the severity of his pronouncement, a smile twitched Lauren's lips and then burst forth in delighted laughter. "Like that?" she asked, her eyes shining with new love. He cared! He wasn't angry at her, but at William.
Stunned by her laughter enough to shake him out of his fury, Jared suddenly realized the picture he must present. He glanced down at himself. The gun belt was his only garment. He smiled at her sheepishly from the shadows under his hat. "You reckon he's worth going out buck naked in a snowstorm for?"
Her eyes still twinkled, but she answered seriously, "He's not worth going after at all."
He rid himself of the ludicrous attire and, before she realized, was beside her, gathering her into his arms. "I'll kill anyone who touches you again. I swear it." He spoke in a low rumble in her ear as he pressed her head against his chest.
His mouth descended and captured hers in a blazing kiss that branded her as his own. Their mouths fused together hotly. Tongues sought, found, took, gave.
Lauren's fingers tangled in the curls lying in wild disarray around his head and pulled her mouth free. "Jared, you must believe that my attraction to Ben was not sexual. To me, he represented the affectionate father I never had. Since that day I saw you framed in the doorway of my room, looking like the meanest desperado ever to roam the plains of Texas, you have dominated my thoughts. Until you kissed me in the office that morning Olivia said we were to marry, I had no idea how a woman was supposed to feel toward a man."
He took her head between his strong, lean hands, sliding his fingers under her hair and drawing her mouth back to his. The kiss was tender, soft, and her lips yielded to the slight pressure of his. But, as all their kisses had done in the past intimate hours, the tenderness turned to passion, and when he lowered her to the fur blanket covering the floor, she was more than ready to comply. Her hair spread out like a dark fan behind her head.
He slid his hands into the opening of her robe and separated the folds, revealing her breasts. His hands closed over them, gently, but with no question of his possession. His mouth gave them the high praise he thought them worthy of. First his lips closed around each crest, then his tongue paid them homage until they glistened wetly in the firelight.
Lauren moaned ecstatically under his adoration while her own hands smoothed over the bunched muscles of his shoulders and back.
His lips left a fiery path across her breasts, down her ribs, back onto her stomach, and then lower to her abdomen. He kissed her navel and then lower still until—
"Jared—" She covered the dark thatch his lips had just reached with the palm of her hand.
He raised passion-clouded eyes to hers, opened wide in alarm. "Lauren," he said huskily, "you must know that I would never do anything to hurt you. Trust me." When she remained silent, staring at him with fearful, pleading eyes, he repeated, "Trust me."
Slowly she nodded her head, and didn't resist when he leaned down and kissed the back of her hand where it lay. His lips were warm. Pleasure-giving. In spite of her modest wariness, she felt her muscles surrendering to the diplomacy of his mouth.
Without haste, he lifted her hand and pressed a kiss into the palm. Then he kissed her where it had lain on the dark delta. An intense heat washed over her and a cry of joy bubbled out of her throat. His mouth brought her a pleasure so exquisite that she was helpless to think of anything except his lips and tongue as they moved lower to explore, probe, and please.
The storm mounted and ebbed as he alternately teased and demanded. The pressure within her increased, and when it crashed around her, she called to him plaintively and was gratified when his body covered hers to protect her from the assault of emotions and senses. Still, she wasn't satisfied until he was inside her, deep, hard, fulfilling, touching her womb with the essence of himself.
Her fingers dug into the flesh of his hips as his own passion peaked. His face was buried in her neck and her skin felt his rapid, moist breath as he chanted her name.
He didn't leave her. He couldn't forsake the paradise just yet. Nestled within her body, he raised himself on his elbows and looked down at her. Tenderly he kissed each feature of her face.
"Is this possible?" she breathed, referring to the enormity of her rapture.
"Yes, yes," he murmured against her lips.
He raised his head and his eyes searched her face once again. His expression was difficult to define, but it closely resembled love.
# Chapter 22
The next morning, everyone looked up in surprise as Jared and Lauren strolled arm in arm into the dining alcove for breakfast. Lauren flushed in embarrassment under the curious stares, but Jared assumed his casual, aloof attitude and questioned why there were no place-settings in front of their chairs. Gloria immediately ran toward the kitchen.
Jared held the chair for his wife as Rudy asked him teasingly, "Hey, Brother, why did you risk a snowstorm to get here when all you've done is spend the entire time closeted in your room?"
He took the plate from Gloria, filled it to his satisfaction, and only then answered, "Well, I'd been away a long time. When I got to Coronado and saw that Laur... I... uh... just figured that it was time for me to get back out here to check on things. I was worried about Keypoint."
"Your concern is touching," Rudy teased. His black eyes flashed wickedly as he said, "It couldn't be because you were, uh, restless, and wanted to see Lauren, could it?"
"Rudy! You're going to embarrass Lauren," Gloria scolded.
Jared laid his fork in his plate. "You know something? I think you might be right." He grabbed Lauren and leaned her backward over his supporting arm so quickly that she didn't have time to protest. He kissed her full on the mouth with exaggerated passion. The children began squealing with laughter and were soon joined by their initially startled parents and a delighted Maria, who smiled on them fondly.
Breathlessly Jared released Lauren and they joined the laughter. Eventually Gloria resettled the children and everyone resumed eating.
A few minutes later, Jared caught Lauren's eye, winked at her, and, finding her knee under the table, squeezed it.
* * *
The snow melted quickly under a crystal-blue Texas sky. The brothers rode out each day to check on various sections of the ranch. The snow had been heavy and wet, and some of their fences had suffered damage. _Vaqueros,_ restless for something to do, were sent out to repair them and to keep an eye out for animals that had not survived the storm.
Lauren rode out with them one afternoon. They were only a mile or so from the main house when they came upon the carcass of one of the finest Lockett beeves. It had been mutilated. A few chunks of meat had been carved out. The rest had been left to rot in the warming temperatures.
"Goddammit!" Jared cursed. "Who in hell did this?"
"It couldn't have been Crazy Jack. He never wastes any part of a carcass." Rudy said.
"No. Mr. Turner wouldn't do this," Lauren said quietly.
The men looked at her in surprise and she told about meeting the hermit. Humbly she recounted how she had taken the trap off his foot when he had been injured. "Fortunately I took him some food, too, before the weather got so inclement. But he was in no shape to do a thing like this even if he had been so inclined."
"You mean you actually saw him? Had a conversation with him?" Rudy asked in astonishment. He had never seen the hermit, only traces of him.
"Yes."
"I've learned that my wife has a great capacity to get involved with people, usually the desperate and helpless." Jared spoke teasingly, but there was respect in his eyes. During his brief stop in Coronado, Olivia had filled his ears with a tirade against Lauren's charity work in Pueblo. Little did she know that she was making her son all the more lonesome for his wife. He had been shocked and disappointed to learn that Lauren had gone to Keypoint. His eyes warmed as he looked at her now and said, "If anyone could worm their way into Jack Turner's life, it would be Lauren."
"That still doesn't answer who killed our cow," reflected Rudy.
"It was probably one of Duncan's gang," Jared said bitterly. "It looks like something they'd do."
"Did Lauren tell you about the visit they paid us a while back?" Rudy asked hesitantly.
"What?!" Jared exploded.
Rudy related the conversation he'd had with Duncan. Lauren was relieved that he wisely omitted the slur that June and her brother had made to her. She feared Jared would seek Duncan out immediately. Hadn't he been ready to charge naked out of their bedroom in search of William Keller? She smiled at the memory, but his loud, angry voice brought her back to the present.
"Why didn't you tell me about this sooner?" Jared demanded.
"Because I knew you'd react exactly as you're reacting," Rudy replied calmly. "I think we should exercise caution and keep an eye on them, but I don't want to provoke them into any more meanness."
"Okay," Jared said grudgingly. "But by the end of next summer, I want them off of our property. When the railroad is finished, the _vaqueros_ won't have to be driving the cattle to Austin. They can assume more duties toward maintaining the ranch. One of those duties can be clearing the cedar. We won't need that scum anymore." He gave one more disgusted glance to the carcass, then spun Charger around and spurred him to a gallop.
* * *
When Wat Duncan struck again, it was swift and sure and deadly. To say the least, it got the attention of the Locketts and Mendezes and proved to them what a powerful enemy the man and his gang were.
Lauren and Maria had arranged the night before to meet in the stables the following morning for a sunrise ride. They had grown accustomed to riding together at that time of day.
Lauren crossed the yard and strolled toward the stable. She was wearing a black suede riding skirt and coat. Her boots were soft black leather, as were her gloves. The ensemble had been a belated Christmas present from Jared. She had wrapped her head in a long, woolen _serape_ loaned to her by Gloria and, of course, she was wearing the blue silk bandana. The vapor of her breath hung in the cold morning air. The door to the stable was closed.
Strange, she thought to herself. Maybe it was so cold that Maria had chosen to leave the door shut against the wind. But it wasn't windy, Lauren argued with herself.
The door was heavy, and she had to tug hard several times before it came open. The interior of the stable was dark. It was quiet except for the restlessness of the horses.
"Maria?" A sinister chill, having nothing to do with the weather, crept up Lauren's spine, and she was suddenly afraid to enter the building. Glancing over her shoulder, she saw that no one was stirring in the house. She had left Jared asleep under the covers of their bed. Elena hadn't yet arrived with Carlos to begin her day's duties.
"Maria?" Lauren called again, praying for the sound of Maria's soft voice. Swallowing a lump of fear, she stepped into the stable. She didn't need to go far.
Maria's body was sprawled out in front of her. Even in the darkness, Lauren could see the pool of bright blood forming beneath her.
Her scream ripped through the morning air. Fists clenched at her sides, then rose to cover her mouth, but didn't stifle the screams of terror that kept coming. She was vaguely aware of cursing and nonsensical mutterings as the doors to the bunkhouse were thrown open and the _vaqueros_ stumbled out in various stages of undress, their eyes bleary from sleep. Running footsteps came pounding across the yard.
Her screams had dwindled to faint whimpers as she heard someone say, _"Madre de Dios!"_
Rudy pushed her aside and moved cautiously toward his mother, disbelieving his eyes. Strong arms gripped her shoulders. "Don't look, Lauren," Jared said in her ear as Rudy knelt down to turn Maria's body over. Jared's warning came too late. She saw the gaping windpipe with its gurgling fountain of blood where Maria's throat had been neatly sliced. She screamed again, but the sound was muffled against Jared's bare chest as he held her head against him, supporting her wilting body with his.
He led her out of the stable so she wouldn't have to witness Rudy's grief. They could hear his deep, soul-piercing animal wail. Lauren sobbed dryly as they walked past the _vaqueros_ standing awkwardly, their eyes averted, instinctively knowing what had happened.
* * *
Gloria and the sleepy-eyed children were huddled together on the front porch. Gloria's lips were white, her eyes questioning.
"Maria," Jared said tersely. Gloria squeezed her eyes shut, intuitively understanding as she heard her husband's racking sobs.
"Come into the house, children." To Jared, she said, "I'll get some coffee."
He only nodded as he ushered Lauren into the house. She stood mutely just inside the door as he went into the bedroom to pull on pants and a shirt. When he came back, he knelt down on the hearth and began rekindling the fire.
It came as a great surprise to Lauren when Rudy's shadow blocked the dawn light coming in through the door that had been left open.
There were no tears. Instead, his eyes were hard and cold, devoid of any emotion except bitter hatred. He tossed something onto the floor and Lauren jumped back from it, staring in horrified comprehension. Jared looked at the object, too. There was no mistaking the battered, greasy hat that was usually worn on Wat Duncan's head.
"You going with me?" Rudy asked his brother.
"I'm going," Jared said quietly.
Without another word between them, they went through the hall toward their bedrooms. Gloria came out of the kitchen carrying a coffee pot and three tin cups. When her eyes lit on the hat on the floor, she set the coffee on the dining table and went to the gun rack.
While Lauren watched in stupefied astonishment, Gloria methodically took down the rifles, checked them, loaded them, and set them aside.
When they were dressed, Jared and Rudy joined her and, like well-trained soldiers preparing for battle, they moved without wasting words or motions.
When all was ready, Rudy drew Gloria to him and held her tightly. "One of the _vaqueros_ will bring her in after we've left. See to her." He kissed his wife quickly on the lips and strode from the room.
Lauren was spun around from behind. Jared kissed her fiercely, almost angrily, before he released her and followed his brother out the door. She rushed after him.
The _vaqueros_ had formed a semicircle in the yard. Thorn, his hawklike features fearsome, stood holding the reins of the waiting mounts. Charger pawed the hard-packed earth beneath his hooves. Rudy and Jared mounted in unison. Disdaining the stirrups of his own saddle, the Comanche vaulted astride his horse. Rudy nodded briefly to his small army and then jerked hard on the reins of his horse, turned him, and galloped out the gate, Thorn and Jared on either side, his men behind him.
Lauren whirled around and faced Gloria, who had come out onto the porch. "Gloria, you're not going to let—"
"It's something they have to do, Lauren," she said with quiet assurance. "Come. We have things to do, too."
The hours ticked by with a slow monotony. To Lauren, the horror of seeing Maria lying in her own fresh blood had been dimmed by the realization that Jared might never come back. The brothers had left seeking revenge and she knew the battle would be bloody. No, God, no, she prayed as she mechanically went through the tasks Gloria assigned her.
Maria's body had been carried in by one of the older hands. She was laid on the bed she had shared with Ben. Gloria prepared her for burial. Lauren didn't think she could have looked at the body again, but felt that Gloria would be offended if she didn't accompany the children into the room to pay their last respects to their grandmother.
Lauren was shocked. She didn't know how Gloria had managed it, but Maria's wound didn't show under the high collar of her dress. Her hair was smoothed back in its usual bun. Her face was unlined and her lips were relaxed into a semblance of a smile. Her hands, beautiful, gentle hands, rested on her breast with a rosary entwined in the slender fingers.
The same cowboy who had carried Maria inside constructed the coffin. Lauren took the children out of the room when he came in to lift Maria into the wooden box.
For the rest of the afternoon, Lauren and Elena, who had come to the house upon learning of the tragedy, tried to keep the children quiet while Gloria tended to the twins. Even after the children were all abed, and the twins were sleeping peacefully in their cribs, the women kept vigil, waiting tensely for their men to return.
Finally, long after sundown, they heard the thunder of horses and raced to the porch. The figures were too small to distinguish at first in the fading light, but each woman sighed relief when she saw her man among those returning.
Rudy and Jared rode up into the yard and tiredly got off their horses, turning the reins over to the _vaquero_ who would care for their exhausted mounts.
Gloria didn't say a word, only went down the steps to greet her husband by wrapping her arms around him. He held her close, as if to absorb her strength. When she raised her head and looked into his weary face, he said, "Not a trace. Nothing."
They all went into the house and the men collapsed at the dining table. Gloria and Lauren scurried to set the food, which had been simmering on the stove, on the table. Elena had seen to it that the tired _vaqueros_ had a pot of the savory stew delivered to the bunkhouse before she left with Carlos.
When his plate was empty, Rudy wiped his mouth with a napkin and scooted his chair back. Jared pulled Lauren down onto his lap and rested his head against her breasts as Rudy began to speak.
"We went to their camp first. Deserted. Not a sign except for the rubbish they left behind. We combed the hills all day, looking in every nook and cranny, and didn't see a trace of any of them." He paused to take a drink of the whiskey Gloria had poured for him. "We found one old nester, about half-crazy. He said he'd seen Duncan and a few others at the river just above the Fredericksburg Road. Day before yesterday, he thought. Duncan was talking to a 'fancy man.'"
"Vandiver?" Gloria asked. Lauren gasped.
"Probably," Jared answered.
They all became quiet then, each lost in his own thoughts. Rudy broke the silence. "I'll find him," he said. "Murdering sonofabitch. I'll find him." The level tone of his voice was frightening to hear. He raised his eyes to Gloria. "Where is she?"
"In her room."
He nodded and stared at the flame of the gas lamp on the table in front of him. "One of the hands offered to ride into Pueblo and bring a priest back in the morning. We'll bury her then." He paused, then said, "I was thinking today that she would never have gotten over losing Ben. Ever since he died, she's been unhappy. Maybe... maybe this was... She'll be happy..." His voice broke and Gloria rushed to his side. He came to his feet under her support and they left the room.
* * *
"Jared, that's impossible! Even if you could pull off such a thing, do you realize the lives that could be lost? The property that would be destroyed? How could you suggest such a... harebrained scheme?"
"What choices do I have? Try to understand this from my point of view."
Lauren heard the voices raised in heated argument coming from the front porch. Dinner was over and the brothers had gone outside. Gloria was caring for the twins. Lauren had been reading in front of the fire in the large living room when she heard Rudy's harsh words.
It had been a week since Maria's burial in a cottonwood grove overlooking the Rio Caballo. Each morning, Rudy and Jared rode out with their men to search for Wat Duncan. Each evening, they returned disappointed in not having seen a trace of their quarry. Maria's death had affected them all, Jared included. But even before that, since the night of his arrival during the blizzard, he hadn't been the sarcastic, angry man he was in Coronado. The man she slept with every night made tender, passionate love to her on the wide bed that had been his since childhood. He told her of his plans to build a house of his own at Pecan Creek. He related to her the circumstances surrounding the death of his friend Alex, in Cuba, and clung to her, tormented by visions of the atrocities of war even as he related them. She had come to love Jared in a new dimension. She loved him fiercely, passionately, and protectively. Strong as he was, virile, stubborn and proud, she had discerned a shred of vulnerability. She loved that most of all.
Lauren longed to share with her husband the rewarding news coming out of Pueblo. Pepe, whenever the weather permitted, brought her news of the projects she had initiated. All were going well. A clinic had been set up two days a week. Remedial construction on public buildings was underway and more was planned for the spring. Warm clothes were being distributed to those who needed them the most. Pepe left with a detailed list of instructions for the committee chairwomen and a personal note of gratitude and praise from Lauren.
She wished to tell Jared about all of this, but she remained silent. She wanted to do or say nothing that would remind him of the events taking place in Austin and Coronado. For that reason, she had not broached the subject of the railroad and the Vandivers. It seemed that Rudy had.
"I know you feel that you have to go through with this, but there has to be some alternative, Jared," he argued.
"I don't see any other way. I've gone over every single aspect of it, and unless I carry out this plan, everything will go up in smoke."
"Everything will go up in smoke if you do. Literally," Rudy countered.
They were silent for a while. Lauren didn't move. Jared was still planning to bring on a riot in Pueblo. He was selling out to the Vandivers and his mother for the railroad.
"At least promise me this." Rudy spoke quietly. "Don't do anything until you've given me some warning. Let me see what can be done from this angle."
"All right, Rudy. I promise. But I don't know how long I can hold them off. They're ready to go. Just be forewarned. When they do it, I'll have to be there. You understand that."
Rudy hesitated for just a moment. "Yes," was the curt reply.
Lauren was crushed. How could he? Maybe he wouldn't. Maybe he could convince them not to destroy the community. She heard their boots shuffling toward the front door and forced her face into a smile before they could see the distress on it.
Jared came to stand in front of her and said quietly, "Lauren, I must leave for Coronado in the morning. You're to stay here until I can come back and get you."
"No. I want to be with you." Her voice trembled, but she didn't give in to the tears she felt gathering in her eyes. He was running away from her again!
"I... I'm going to be very busy with the railroad and you'd be bored in town. Here you have Gloria and the kids to keep you busy."
Her eyes beseeched Rudy for support, but he was concentrating on lighting a cigar and wouldn't meet her eyes. She turned back toward Jared. "I'm going back with you, Jared. I don't care how busy you are. If you don't take me with you, I'll just follow on my own." She raised her chin a fraction, and he saw the resolve in her steady, blue-gray gaze.
"Dammit!" he cursed, and slammed his fist into his palm. He turned toward his brother as if seeking an ally. Rudy had become even more fascinated by his cheroot. Jared muttered sourly, "All right. Get packed tonight."
* * *
Besides Rosa, Pepe, and her piano, Lauren was not happy about seeing anything in the house in Coronado. It was truly one of the most beautiful houses she had ever been in, but her mother-in-law cast such a cold foreboding on the atmosphere that it could never be considered a comfortable home.
Olivia's greeting had been polite, if not exactly warm. Carson complimented Lauren on the healthy bloom in her cheeks. Lauren met Olivia's green eyes over his shoulder as he hugged her, and wondered if Olivia knew the reason for the glow she had taken on. She reasoned that the older woman did. It was after Jared's first night with her that Olivia had provoked the argument between them that culminated in their separation.
Lauren recognized the subtlety Olivia had used to drive the wedge between her and her husband. She was capable of anything to ensure that her own greedy plans come to fruition. She would even jeopardize the happiness of her son. Perhaps she did love him. But it was a jealous, self-gratifying love. Olivia Lockett had to be in control. Wasn't that the reason her marriage to Ben had been so disastrous? Ben was not a man who could be controlled. Olivia was learning that Jared wasn't so easily manipulated, either. Not like Carson Wells.
"Carson, thank you. I can always rely on you to make me feel beautiful even when I'm covered with trail dust." Lauren laughed and hugged the plump man again. He evoked her pity, and she couldn't quite decide why.
The next few weeks went by smoothly and uneventfully. Olivia went to the bank every day. Jared went out on pursuits of his own, sometimes riding out to check on the progress of the railroad. The track moved closer to Coronado daily. If the spring weather held and there wasn't too much rainfall, it would probably be completed sometime in the early fall.
Lauren spent long hours at the piano.
She missed Elena more than she could have imagined. There were no laughing children to break the staid atmosphere of this house. There were no disruptive calamities that reduced everyone to a state of mirth. There wasn't the serene presence of Maria... Maria. Her friend. Ben's love. Perhaps Rudy was right. Maybe her death had mercifully brought them together again.
There was no more discussion of the Pueblo riot. Lauren could almost imagine that she had dreamed the whole ugly episode. Was it even possible that Olivia had changed her mind?
* * *
One afternoon in early March, Lauren sat at a small table in Ben's office composing a letter of commendation to the Ladies of Texas Freedom who had so generously donated fifty pounds of cornmeal to be distributed to the needier citizens of Pueblo.
She heard Jared's spurs jingling on the parquet floor in the hall just before he stood framed in the doorway. The sight of him never failed to accelerate her heartbeat. Never had anyone loved as much as she did. Of that she was positive.
He was wearing his cowboy garb and looked much as he had the first time she saw him. She put her fountain pen down and started to get up and go to him.
"No, stay there." Puzzled by his words, she sat back down and watched him close and lock the door.
"Jared?" She laughed a bit nervously. His expression was so intent it was almost frightening. "What are you doing?"
"Do you know how many times I've fantasized about you looking just the way you do now? You nearly drive me crazy with that prim and proper countenance, the eyeglasses perched on your nose, back straight as you bent over some damn thing or another. It's become a driving ambition of mine to ruffle those calm feathers."
He advanced into the room, pausing only to fling his black hat into a chair. He strode purposefully to the large picture windows and pulled the drawcord on the heavy drapes, plunging the room into deep shadows.
As though stalking prey, he came toward her chair with measured steps and drew her out of it. He sat down where she had been and pulled her onto his lap facing away from him.
"Can your feathers be ruffled, Mrs. Lockett? Since that day I sneaked into your room and spied on you, I've wanted to do this." He placed his lips against the nape of her neck and traced a path of warm kisses from there to her earlobe, tantalizing it with a capricious tongue.
"And this." Her watch pin was covered by a hand that came around her and pressed against her breast. His hands slid over her breasts and met at her waist where he began to pull her blouse from the high waistband of her skirt.
"Then I was going to do this." His hands moved slowly to her back and began unbuttoning the bottom buttons on her shirtwaist. He only released about half of them before his hands slipped under the blouse and moved to her front. Brushing past eager, quivering breasts, he untied the decorative bow at the top of her chemise. Feeling his way, he unbuttoned the tiny buttons, pulling the diaphanous garment down with agonizing slowness until her breasts spilled into his hands.
Lauren had not spoken, but leaned back against his hard chest and purred pleasurably as his fingers caressed her, bringing her nipples to hard peaks by rotating his thumbs over them.
"Tell me when you lose all composure," he whispered challengingly. His breath became uneven. The lips that nibbled at her neck became more impassioned, the tongue more adventuresome.
Hands, too, ceased to be teasing and became imploring. "You feel so good, Lauren," he rasped as he stroked her. "Silk... no, satin. Cool. Warm. God, I don't know," he groaned, as he gently rolled the dusky pink crests of her breasts between his fingers.
A fumbling, clumsy hand finally waded through the material of her skirt and petticoats, over a silk stocking and a lacy garter, to find her linen-covered thigh. The skin beneath the sheer covering trembled as the searching fingers smoothed up the length of her thigh until, even through the light fabric, he discovered her prepared for his love. "Oh, God," he groaned.
One hand left her to unfasten the remaining buttons on her back and slip the blouse from her shoulders. Then he pulled the pins from her heavy hair and buried his face in its cascading waves, drinking in the lavender-water fragrance of it.
Turning her slowly toward him, he rested her shoulder against his chest and looked down at her disheveled state. "Just as I imagined. You're ravishing," he whispered huskily.
She suddenly realized she was seeing him through the lenses of her eyeglasses and raised a hand to take them off. He trapped her hand in his and said, "Uh-uh. They're part of the fantasy."
His fingers followed her collarbone and moved down her chest, adoring the tips of her breasts, tormenting her by not touching what she craved to be touched. She arched her back at the same time she tangled her fingers in his thick hair and drew his head to her.
He cupped one of her breasts, brought it up to his descending face, and nuzzled it with his nose and beard-roughened chin before closing his lips around the center bud and raking it lightly with his tongue. When she moaned into his hair, he raised his head and smiled in devilish satisfaction before melting her lips with an ardent kiss.
He pulled back in shock as he felt her slender fingers working with the buttons of his shirt. Playful lips and a darting tongue tormented his nipples until they were turgid. Then her mouth followed the path down his chest and stomach that her fingers charted. She snuggled down his body until she dropped to her knees between his thighs. Staring up at him boldly, she peeled away the chemise and slipped her arms free, completely baring her breasts for his avid inspection. Her raven hair cloaked his thighs as she rested her cheek against his lap.
"I've had some fantasies of my own, Mr. Lockett," she whispered as her fingers deftly unfastened his pants.
She said something else as her hand closed around the swollen shaft, but he couldn't hear her over the pounding of his heart. And when the love-moistened tip of his sex felt the sweet brush of her tongue, his ragged breathing drowned out every other sound.
* * *
Much later, they lay on the rug before the fireplace where Jared had struck a match to the logs already stacked there. He lay on his back, hands folded under his head, a cheroot clenched in his teeth, brazenly unconcerned by his nakedness.
Lauren was curled up on her side, staring into the fire, her cheek resting on folded hands. He had covered her with his shirt, long ago discarded along with the rest of their clothes.
"You're very quiet, Lauren. Is something wrong?"
She was glad that he was sensitive to her mood, but reluctant to disclose the worry niggling at the back of her mind. She felt him turn on his side toward her, felt his eyes on her, though she didn't look at him. "What is it? Tell me."
He could barely hear her, she spoke so softly. "I enjoy... the things we do, Jared. I... it's wonderful, but..." She stopped speaking, closed her eyes in embarrassment, and continued, "I don't think ladies are supposed to... to participate. I'm afraid you'll think me wanton if I do... if I..."
His laughter boomed in her ear as he drew her around to face him. Between guffaws, he covered her face with light kisses. When his amusement subsided, he said tenderly, "Lauren, you'll always be a lady. You couldn't be anything but a lady. And no matter how _often_ we make love, or _how_ we make love, or how _much_ you enjoy it, you'll still retain that aura of innocence that first attracted me. It set you apart from any other woman I'd ever met."
He traced her high cheekbone with a gentle finger. "Wanton? I'm surprised you even know the meaning of the word." He chuckled again before his mouth claimed hers.
The kiss was deep and telling, and when it ended, his lips remained on hers as he said, "However..." The shirt was moved aside. His index finger began at the base of her throat and traveled down the length of her torso, between her lush breasts, over the smooth skin of her stomach, past her navel and mons to disappear between her thighs. "As long as we're on the subject of wantonness..." He touched her knowingly and was rewarded by her ready response.
She sighed in mock-despair. "I'm no better than a common prostitute."
He smiled even as he kissed her. "Yes, you are. Much better."
She wanted to admonish him for his impudence, but his swift and certain possession robbed her of the initiative.
# Chapter 23
They came out of the office arm in arm into the wide hall. There they met Olivia. She looked at their wrinkled clothing and mussed hair and assessed the situation correctly.
"I heard you had come home early today, Jared."
"Yes, Mother. You might say I took the afternoon off."
She chose to ignore his bantering tone as she did his holding Lauren close to him with an arm firmly around her waist. She could feel her control over him slipping and it both angered and terrified her. "You have some mail, Lauren," she said tightly.
Lauren looked at her with questioning eyes and took the white envelope that was extended to her. A tiny gasp escaped her lips when she read the return address. "It's from the Prathers," she said. "My guardians," she clarified to Jared, who was looking curiously over her shoulder at the letter. She glanced up at him significantly. He knew the story behind her leaving North Carolina, knew why she would be surprised to receive the letter.
"Open it," he said gently.
She inserted her finger under the flap and withdrew the two sheets of white paper. Lauren had read enough of Abel's sermons to recognize his neat, careful handwriting. Eagerly, somewhat apprehensively, her eyes scanned the page.
"They don't know of Ben's death. They send him their regards." Her eyes roved lower on the page. "Oh!" she exclaimed. Her hand flew to her throat. "William Keller is dead!"
"Good. How did he die?" Jared asked harshly.
Stammeringly she explained as she read, "A big scandal. He was murdered... a woman's husband shot him... she confessed to their being lovers." She paused in her recitation to read more. The words blurred on the page, seen through a lake of tears. "He... they are sorry now for not believing me." She folded the paper and looked up at Jared. "They beg my forgiveness and say that I have a home with them if I ever want to come back."
Jared was looking at her, but he was thinking of the fate of William Keller and not the sentiments of the couple who had condemned Lauren. "That bastard! I wish I had killed him."
"Someone from Lauren's past?" cooed Olivia, who had watched the whole scene with growing interest. She was ignored.
Lauren grabbed Jared's sleeve and shook him slightly. "Don't say that, please."
At her touch and remonstration, he snapped out of his temper. His gaze was warm and compelling as he looked down at her. "In a way, I guess I have Mr. Keller to thank, don't I?"
She smiled, understanding his meaning. Shyly she murmured, "I guess maybe I do, too."
* * *
Lauren would remember that afternoon and evening in the weeks that followed. She treasured those hours with Jared in the seclusion of the office. For after that day, everything changed.
The next night was the first time the men came to the house. Jared had left early that morning and didn't come in until after Olivia and Lauren had shared a silent and tense meal.
Jared's mood was surly and rude. He ate little of the food Rosa had kept warm for him, but drank incessantly. When the men began to arrive, Olivia suggested that Lauren might be more comfortable in her room upstairs. Lauren took the hint. She looked toward Jared, expecting him to intercede, but his back was to her as he poured himself another drink at the sideboard.
She watched from the front windows of her bedroom as more and more men arrived. They came in groups of twos or threes, but they all had the same characteristics. They looked mean, disreputable, and vexatious. These were the mercenaries who had been hired to excite trouble in Pueblo, trouble that would be blamed on the inhabitants of that community.
Loud, ribald talking and laughter came from the rooms below. Lauren saw Parker and Kurt Vandiver when they arrived, and a cheer rose to greet them as they entered the parlor.
That was the only occasion when the mercenaries all came at once. As the weeks went by, a few of them would come to the door almost nightly asking for Jared. He would leave with them and Lauren would hear him return to his room in the early-morning hours. Sometimes he would ride out alone late at night and be gone for hours before she heard Charger galloping up to the stables at the back of the house. Was he holding meetings to plan the attack on Pueblo?
To help relieve her anxiety and alleviate her boredom, Lauren took a more visible role in the projects abetting Pueblo. Her frequent trips to that community filled her committee workers with renewed zeal. The townspeople soon grew accustomed to seeing Pepe drive her down their dusty streets in a buggy. Some of the less shy even presented her with handmade gifts. She accepted each one graciously and with a gratitude disproportionate to its value. If Olivia or Jared knew of or cared about her work in the Mexican settlement, neither said so.
Just as dawn was breaking one morning, Lauren awakened to the heavy thumping of Jared's boots in the hallway. Excitement welled in her breast when the footsteps neared her door. There they stopped. Expectantly she sat up in bed. Long moments passed. Once she even thought the doorknob rattled slightly. But she was crushed with disappointment as Jared's tread retreated toward his room.
Flinging the covers aside, she flew out of bed, grabbed her wrapper, and dashed to her door. Opening it, she called faintly, "Jared."
The large silhouette halted abruptly. Dejection and weariness were etched along every angle on his body. He turned toward her slowly. "I'm sorry I awakened you, Lauren. Go back to sleep."
She clung to her door frame, her knuckles white with anxiety. "D... did you want, need, anything?" She hated the pleading sound of her voice, but she longed for that closeness they had shared for even a brief time.
"No," he said harshly. "Go back to bed." He took a step away from her.
"Jared," she said with more force. "Tell me what you're doing, where you're going. Tell me you're not having anything to do with—"
"Lauren," he barked, cutting her off. His voice echoed through the still house. In agitation, he whipped off his hat and slapped it against his thigh as he stared at the floor. Finally he raised his head. His tone was softer, almost apologetic. "You're my wife, but don't expect me to account to you for everything I do. Some things you'll either have to overlook or... or accept on trust. Do you understand?"
Trust? Could she trust him? She wanted to. Never had she thought Jared could carry through with Olivia's plot. She wanted to believe that still. "Yes," she answered softly. "I understand." Silently she begged _come to me._
"Then we won't speak of this again," Jared dismissed her, entering his room alone.
From then on, Jared avoided Lauren completely. If they should chance to meet each other, he inquired politely about her well-being. That was all. He never came to her room. She never went to his. It was as if the intimacies that had been established between them existed only in her vivid imaginings. His indifference was as hard for her to accept as the reason for it.
Olivia sparkled radiantly during these weeks. Her smooth cheeks were flushed and her eyes glittered with excitement. She looked far younger than her years. The tight lines on her face relaxed. She was in her element.
Carson was at the house constantly. He was extremely nervous.
Lauren found the presence of the Vandivers the hardest cross to bear. Three or more times a week, they had dinner with the Locketts and Carson. For Lauren, the meals were an ordeal. Jared sat picking at his food, drinking too much, snarling if Kurt so much as spoke to her, answering anyone brave enough to speak to him in monosyllables.
Kurt, as if sensing Jared's animosity, provoked it at every opportunity. He was unctuously courteous to Lauren. Each time his hand closed around her elbow to lead her into another room or seat her in a chair, it took all her control to keep from snatching her arm away. She worried, too, that Jared might make good his threat to kill the man one day. Nor had Kurt forgotten Jared's pistol being pointed into his face.
For all his bravado, Lauren knew that Kurt was afraid of Jared. Her husband's malevolent looks were too blackly threatening to be taken lightly.
One evening, Jared was called away for one of his "secret meetings," as Lauren termed them to herself. Olivia had just suggested that she and her guests take their coffee in the parlor. Lauren watched forlornly as Jared went out the front door without so much as a nod in her direction. Instead of following the others, she excused herself and went into the library. She loved that room of the house and often sought refuge there, for no one else used it much.
She had been reading in one of the overstuffed easy chairs for about twenty minutes when she heard the door open and close quietly. She turned to see Kurt standing just inside the room. His thick, bulky body was repugnant to her, as was the insinuating expression on his ruddy face.
"Lauren, I missed your piano playing this evening. You deprive us of your company. Why? Are my father and I so offensive?"
She knew he was deliberately baiting her and she refused to rise to it.
"Of course not, Mr. Vandiver. I was overly tired tonight and knew that I would not be very good company."
"I'm sorry you are unwell." He approached her and took the chair closest to hers, his knees inches from her own. She pulled back quickly and the action wasn't wasted on Kurt. He was not at all perturbed. Rather, he seemed to enjoy her uneasiness. Again she felt that there were undercurrents of cruelty in this man.
"Your husband shouldn't neglect you this way. You're far too tempting to be left alone for long."
"Jared will be home shortly," she said hurriedly, furious with herself for showing him her nervousness.
He laughed. "I happen to know that he'll be gone most of the night, Lauren." He fixed her with a sinister stare that caused her to jump from her chair.
"If you'll excuse me, Mr. Vandiver, I'll—"
She all but ran past him, but he reached out and grabbed her arm, spinning her around and pulling her against him.
"You're not being very friendly, Lauren, with your husband's business partner. Haven't you learned anything from your mother-in-law? She has always been _nice_ to Carson, Ben's partner."
He laughed and it was an ugly sound. By now, Lauren had figured out for herself the relationship between Carson and Olivia. It would have taken a fool not to see it. Kurt's snide comment maligned Ben and she wanted to slap his face in Ben's defense. But he held her arms painfully just above her elbows.
"Of course, Ben had Maria Mendez. It's a shame what happened to her, isn't it?" he asked in a lilting voice that suggested he didn't think it was a shame at all.
Lauren's struggles ceased abruptly as she stared open-mouthed into his cold, blue eyes. "Wh-what do you know about Maria? How did you know—"
"I make it my business to know everything about the Locketts. Did you learn anything from Maria's unfortunate demise? Hmm?" he went on smoothly. "See what happens to whores who sell themselves cheaply to the first bidder?"
"Let me go," she grated, and renewed her efforts to escape his grasp.
"Be nice to me, Mrs. Lockett, and I'll take care of you. You won't end up like that Mexican slut."
His thick lips were inches from hers and she was near screaming when Parker's voice thundered down the hallway. "Kurt! Let's go. I'm tired and need to get to bed early."
Kurt cursed under his breath and the hands on Lauren's arms increased their pain-inducing pressure. "I promise you, Lauren, there will be a time when I won't be interrupted by your cowboy husband or anyone else. I won't be disappointed again."
"Kurt!"
"Coming," Kurt called back. Then, lowering his voice again, he whispered near her face, "You won't be disappointed, either, because you've never seen one as big as mine. It's going to rip into you like a battering ram. And when it's finished, you'll be begging for more." To emphasize his vulgar promise, he ground his hips against her middle.
"Kurt!" the voice down the hallway boomed out.
Cursing expansively, Kurt released her and was gone. Lauren leaned weakly against the back of one of the chairs, her head spinning, her knees trembling and threatening to buckle beneath her.
When she heard the front door closing after Olivia and Carson had bade the Vandivers goodbye, she crept upstairs to the sanctuary of her room. She was violently ill in the bathroom.
* * *
Lauren hung her head over the basin in the bathroom and retched dryly, her stomach having been emptied the night before. The muscles in her throat constricted and painfully urged something to come up and relieve the racking nausea. When the spasms finally subsided, she fell weakly back into her bed.
She thought the illness which had overcome her the night before had been the direct result of her encounter with Kurt. That her nausea had carried over to the morning must mean that she had a minor ailment of the stomach. In fact, she hadn't felt well for several days, she realized now.
Each morning, an uncharacteristic lethargy accompanied her out of bed. It pressed upon her head and grew heavier as it shrouded the length of her entire body. The simple chores of arising and dressing seemed insurmountable burdens. The weight of her hairbrush as she drew it through her hair caused her arms to fall weakly to her sides. When she brushed her teeth each morning, bile rose up in the back of her throat, and the smell of breakfast wasn't at all appetizing. Though she ate sparingly, her stomach felt full and bloated and continued to feel like that even when she was hungry.
Lauren's spirit was ill because of the rift that had come between her and Jared. The idea that soon the heinous plan dreamed up by Olivia and seconded by Parker Vandiver would be put into action was sickening, and she decided that her physical ailments must be manifestations of her mental upheaval.
Rosa's smooth brown face gazed down at Lauren with concern. "The _señora_ is not well this morning?" She brushed a few stray tendrils of hair away from Lauren's pale cheek as she lay back on her pillows.
"No, I don't feel very well. I don't know what's wrong with me. I have no energy, food makes me sick, even the thought of it makes me sick. I feel so puffy and..." Her voice trailed off, lacking the energy to continue.
Rosa scrutinized her mistress pensively. "When did you last bleed?" she asked softly.
Lauren blushed furiously, but she tried to remember. The process of thinking seemed not worth the effort it required. "I... I don't recall. I was still at Keypoint. It was sometime in late January. I remember because Gloria had just had the babies and she and I—"
" _Señora,_ don't you see?" Rosa interrupted her excitedly, "It's been two months. You are going to have a baby."
The words fell like stones on her ears, rolled to her aching stomach and almost caused it to revolt again. A baby! That was impossible. She tried to sit up, as if negating her weakness, ignoring the symptoms, could eliminate the malady.
"No, Rosa. I couldn't be with child. It's something else, I'm certain."
As she looked at her friend for confirmation, she saw only the beaming smile, the gladness at their discovery. Yet Lauren felt an inexplicable wave of sadness, for she knew one day she would have to leave Jared. To Rosa's consternation, she burst into tears and buried her face in the cook's plump bosom, weeping uncontrollably.
It was a long time before the tears stopped and, when they did, Lauren was embarrassed by her sudden show of emotion.
"I'm sorry, Rosa. I couldn't help it."
"It is another symptom of women in your condition, _Señora._ The tears will make you feel better. Would you like some tea?"
"Yes, that sounds nice," Lauren mumbled absently as she rose from the bed and walked to the window. Rosa was shuffling out of the room to fetch the tea when Lauren called her back. She didn't turn around and her voice was soft as she requested, "Rosa, don't tell anyone about the... the baby... just yet. Please."
"I understand, Señora Lauren." Rosa closed the door behind her.
* * *
Lauren kept her secret, though she longed to talk to Gloria. She took long walks around the gardens outside even when she didn't feel like it, but her face remained pale and wan. Dark shadows circled her eyes. If Jared noticed her listlessness and lack of appetite when he saw her in brief snatches, he didn't mention it. Olivia acted as if she weren't there.
When Lauren saw Jared for the first time after she had learned of her pregnancy, her heart warmed with love for a few moments before the chill of gloom settled over it again. She had given little thought to their "arrangement" since the marriage had been consummated. Now, she was forced to think about it.
Jared had said that the railroad would be completed by the end of the summer or early fall. The baby would come, if she calculated correctly, around the first of November. She couldn't hide her pregnancy until the railroad was finished and then leave according to the bargain. What would happen to her baby? She would take him with her, of course, but finding work to support herself would be harder to do with a baby. She could live for several years on the twenty thousand dollars Olivia had promised her if she were frugal, but what then?
The Prathers had urged her in their letter to return to them. She had written back telling them of Ben's death, her marriage to Jared—omitting the details—of her family and friends at Keypoint. The missive had been warm and loving, but she knew she could never return to their staid, dull life. Where could she live with her baby?
The one thought that plagued her was that she would never be able to take her baby with her. She might well be providing the next Lockett heir. Try as she might, she couldn't decide how Jared and Olivia would feel about her child. Of one thing she was sure—nothing and no one would separate her from her baby. She already loved it, was protective of it, and it was probably the only part of Jared she would have after her usefulness to the Locketts' enterprise had ended.
A small glimmer of hope refused to dim on the horizon of her mind. While Jared had never spoken of love, she had read tenderness in his eyes, seen an affection there as he watched her. Surely he felt some fondness for her. It wasn't much to go on, but it was all she had.
But as she met his closed, remote face at each of their fleeting encounters, that hope began to diminish.
* * *
The tension in the house mounted. The chasm between Jared and his wife grew wider. His midnight rides became a nightly ritual. The muscles of his face, the nervous darting eyes, and the continual clenching of his fists indicated a restrained violence that had to erupt soon or destroy the man from within. His abrupt, curt attitude toward Lauren forbade her to approach him. He'd asked for her trust. Now, in the dawn light, that appeal seemed unreal. His behavior certainly didn't inspire trust.
She ventured into the stables one morning and was alarmed to see boxes of guns and ammunition stacked against one wall. It was an arsenal of immense proportions. Her heart quaked. Before, she had been concerned about other people getting hurt. Now she realized there was a very real possibility that her own husband could be injured or killed. She prayed fervently that something would happen to prevent this entire fiasco.
Nothing did.
Unable to tolerate her own passivity, Lauren decided that if she couldn't keep the tragedy from happening, she could at least erect obstacles for the perpetrators. Late one evening after Jared had left the house, she let herself quietly out the front door and ran to the stable. Pepe was working by lantern light mending a bridle.
"Señora Lockett," he said in surprise as Lauren swiftly shut the door behind her.
"Pepe, do you know what is in those boxes? What they're for?"
He licked his lips nervously and looked away quickly. " _Sí,_ but Señor Jared told me not to tell."
"Well, we're not going to let it happen, Pepe. You and I are going to do something to slow them down."
"But, Señora Lockett, he—"
"How can you keep a gun from firing, Pepe?" she asked, ignoring his discomposure.
"Señora, the guns—"
"You're right. We'd never be able to handle all of the guns without them noticing. What can we do?" she wailed, twisting her hands. Then her eyes lighted on the boxes of shells. "The bullets! That's it. Without a bullet, a gun is useless, isn't it?" She was talking rapidly, musing aloud. "We'll hide them. Of course, they might bring their own on the night of the raid, but at least they won't have these."
"You want to hide the bullets?" Pepe's voice had risen an octave as he stared at her incredulously. His dark eyes were wide and his mouth hung agape.
She placed a comforting hand on his arm. "Don't worry, Pepe. I'll take full responsibility if Jared or anyone else should ever find out." Then her gentling tone changed and became brisk and businesslike. "Now, where are the shovels? We'll bury the boxes out there," she said, pointing to the back of the building. "Please hurry. I may be missed any minute."
"Señora—"
"Please, Pepe," she said impatiently. "Don't be afraid of reprisals. Don't you want to help your own people?"
He turned away from her, muttering to himself in Spanish and shaking his head, but he did as he was instructed. By the time they hauled the heavy boxes to the rear of the stable, dug the holes, and buried them to her satisfaction, Lauren was dirty and tired and her back ached abominably.
"If I can, I'll try to get word to you when the raid is going to take place. Can you warn the people of Pueblo?"
_"Si,"_ he said with the weary attitude of one who is ready to agree to anything.
"If I can't get word to you, take it upon yourself. Ride into Pueblo and alert as many as you can. Tell them to take cover. Anything to protect themselves."
"I will, Señora Lockett."
"Thank you, Pepe. You've been a tremendous help. You'll be a hero to your people." She smiled at him before leaving the stable. Undetected, she made her way upstairs, where she washed away the damp soil clinging to her hands and clothes. When she fell into bed, her limbs were as heavy and sore as her spirit.
* * *
It was fortunate that she and Pepe had done their work the night before, because the next day it began to rain and it rained for several days. Large, heavy pellets of water fell relentlessly from the sky. The air was close and stifling, adding to Lauren's pregnancy-related discomfort.
Confined to the house, Lauren paced back and forth in her room between the bed and the windows, unable to read or sew, or to concentrate on anything except her unforeseeable future and that of her child.
Then the rain stopped. The clouds hung low in the sky and the air was still thick with humidity, but the rain ceased.
And at dusk one evening, the mercenaries converged on the house.
Lauren heard the first clump of boots as she sat at the dinner table with Olivia, Jared, Carson, and the Vandivers. She toyed with her food, every now and then putting a bite into her mouth and forcing herself to swallow it, praying that it would stay down.
She was deathly afraid of Kurt Vandiver now. Since he had accosted her and made his lewd threats, the sight of him sickened her and made her tremble in fear.
When the sound of the heavy treads was heard on the porch, the others at the table started, glancing at each other quickly and tensely. Kurt rose from the table and rushed out of the room to the front door even before the knock echoed through the still rooms.
Lauren heard the low mumblings of several voices before Kurt closed the door and returned to his place at the table.
"I told them to wait in the stable. They'll gather in there and... ready themselves."
Parker nodded, satisfied. His cold blue eyes flashed in anticipation. His look was predatory. Lauren shivered despite the warm, muggy air.
She looked at her husband, unable to continue the pretense of eating. He was dressed like a _vaquero._ The red scarf he had been wearing the day he kissed her in the shelter of the boulders was tied negligently around his throat. He usually wore formal attire for dinner in Coronado, but she had been too absorbed in her own thoughts to notice this incongruity before now. She knew he felt her eyes on him, but he refused to meet them.
Carson's normally hearty appetite had waned and he sat sipping his wine. Olivia ate calmly, as though the interruption had never happened. Kurt leaned back in his chair indolently, studying Lauren. He seemed unaffected by the tension around the table.
"How... how long before we know something, Jared? I mean, how long will it take?" Carson asked nervously.
Jared shrugged and took a long swallow of whiskey.
"You have a big night ahead of you, Jared. I'd lay off the whiskey if I were you," Parker said.
Jared fixed a golden stare on him, then tipped the glass to his lips once again. Parker's face flooded with anger, which purpled the veins in his nose and cheeks.
"Send us word when the plan has gone into effect. Carson and I will be eager to know what is happening." Olivia's face was shining. She might well have been talking about a Mardi Gras ball. To Lauren, her eagerness to destroy was obscene.
"Well, never put off until tomorrow what you can do today, or something to that effect," Kurt said lightly, rising from the table and going to the window. "Looks like they're all here. It's a good thing you don't live in the center of town, Lockett. People might wonder what the hell was going on."
Parker, Olivia, and Carson stood, pushing their chairs back and moving toward the door. Except for Lauren, Jared was the last one to rise. He did so slowly, deliberately. His Colt, in its holster, lay against his hip. A leather cord tied around his thigh held it secure. He wasn't entering a harmless shooting match. His gun was loaded and primed for a deadly purpose.
Lauren was out of the chair in a flash, her former sluggishness forgotten. She placed herself between Jared and the door, grabbing his forearms with her slender fingers.
Breathlessly but incisively, she said, "Jared, I beg you, don't be a part of this thing. Please." His face was cold, implacable, his eyes impenetrable. When he didn't speak, she went on, "Think, Jared! Think of Rudy and his family."
"Rudy is a fool," he lashed out. "He wants to solve things peacefully. He thinks everything Ben ever said is chiseled in stone, but I've learned that Ben could be wrong. He was certainly wrong about you and me and this 'grand love' between us."
The barb hit home and she floundered, aware of the others listening to their dialogue. She tightened her grip on him. "I believe in you, Jared. Once, just a few weeks ago, you asked me to trust you—"
"You heard only what you wanted to hear. I also strongly hinted that you'd do well to accept things as they are and to keep your moralistic opinions to yourself. Apparently you don't take hints too well. I don't give a damn whether you _trust_ or approve of me. I do as I please."
She wanted to scream in frustration. "Do you realize how this might jeopardize your future... our future?"
His eyes darted to the interested witnesses behind her, then came back to her face. His lips curled mockingly. " _Our_ future? We have no future, Lauren. You know what your future with us is. When you have fulfilled your part of the bargain, you're gone. A lot richer, but gone. Do you now regret selling yourself so cheaply? What do you want? More money?"
Lauren swallowed the congestion in her throat. She stuttered as she tried to speak. "At... at Keypoint, you... we... it was... different."
He laughed at her mirthlessly, his face ugly in its contempt. "Do you think because I've slept with you that things have changed between us? That I've developed an 'attachment' to you?" He snorted derisively. "You were a pretty good lover once you learned how. And you were handy. What did you expect me to do? You were the only warm body available in a snowstorm. If you place any more importance on my bedding you than that, you're even more stupid than I thought."
Her disillusionment turned to pain, then to anger. "Not too stupid to see you for what you really are. I placed far too much faith in you, Jared Lockett. I thought you were growing up, becoming the kind of son your father deserved, the kind of husband I wanted." She paused and gulped for air. "Now I see that you're as avaricious, as cruel as they are. Jared, I—" She wanted to tell him that she loved him, to beg him not to do this thing. Instead, she said, "I curse you to perdition if you do this thing."
"Then to hell I shall go." He laughed again as he shoved her away from him and shouted, "Let's go," to Kurt and Parker.
His words struck her like physical blows. The breath had literally been knocked out of her. She heard the shouts and thudding of horses' hooves as a score of men mounted and rode out of the stableyard. She stood rooted to the spot, oblivious to everything except the constricting pain around her heart.
Carson came up behind her and touched her shoulder solicitously. "Jared's a little keyed up, Lauren. He spouts off things he doesn't really mean."
His kindness penetrated the protective shell she had drawn around herself. It was strange, but she wasn't humiliated or embarrassed that they had heard what Jared said to her. Her wounds went too deep to be bothered by superficial lacerations.
She felt only the desolate loss. They had shared something beautiful and the product of that sharing lay sleeping in her womb. She touched her stomach as if to reassure herself that his words hadn't ripped the seed out of her body. Scalding tears rolled down her cheeks as she was ravaged by despair.
Passing Olivia on her way out of the room, she met the smug, triumphant face of her mother-in-law fully. Lauren wondered at the hate that so consumed her that she sought to destroy those she supposedly loved. Maria had once said that Olivia was a sad, lonely woman. Lauren thought she was most likely right. In spite of her aversion, she felt a moment of pity for Olivia. The woman was incapable of real love. Her destructive selfishness wouldn't permit it.
Olivia must have perceived her thought, for her emerald eyes narrowed with loathing. Lauren knew then that she had scored a small victory over this domineering woman. That knowledge emboldened her as she swept past Olivia and went up the stairs.
* * *
The hours passed with interminable slowness. Once Rosa had finished clearing the dinner things away, a tangible silence settled over the house. Carson and Olivia sat in the office, she in her usual place behind the large desk, he in a nearby chair. They talked little.
Had Jared noticed the missing ammunition? Surely. It hadn't stopped them. Had Pepe gotten away with his warning in time? The questions came at Lauren out of a dark void which provided no answers.
She lay on her bed wondering how she would survive this nightmare. Visions of Jared wounded and bleeding crowded those of him with the mocking sneer on his face, his eyes cold and hard, resembling those of his mother. Dimly she saw him tender and loving. But only dimly.
The pounding hooves were heard from far off, so quiet was the house. Lauren held her breath as she heard the single horse being reined in just outside the fence-enclosed yard. When she heard the rapid knocking on the front door, she rolled off her bed and fled to the top of the stairs where she could see Carson hurrying to open the door. Kurt Vandiver lunged into the hall.
"Jared's been shot!"
# Chapter 24
"It looks bad," Kurt told the startled Carson. "Where is Lauren?" Just then he spotted her at the top of the stairs, gripping the bannister so hard her knuckles were as white as her face.
He stepped quickly to the bottom of the staircase and looked up at her. "Can you get into some riding clothes? He's asking for you."
She didn't even answer him, but whirled away and ran toward her room.
Olivia was standing next to Carson when Kurt turned back to them. "The bastards were waiting for us. Somehow they were tipped off. There was a lot of shooting. I don't know how many of our men were killed or wounded. None of the fires we planned really got started because of the damp weather. It's raining now."
"Are Jared's wounds serious, Kurt?" Carson asked anxiously.
Kurt's eyes darted to Olivia, then he answered slowly, "It's hard to tell just yet, Carson. I thought his wife ought to be with him."
Olivia's features remained calm. She asked no unnecessary questions. Carson had always respected her stoicism in the face of trouble. He patted her arm reassuringly.
Lauren came racing down the stairs. She had donned one of her split riding skirts and boots, but left on the lace-trimmed shirtwaist she had been wearing. She had raided the bathroom for first-aid implements and stuffed them into a small cloth bag.
"I'm ready." She hurried out the front door, never even glancing at the others.
Kurt glanced hastily at Olivia before he followed Lauren swiftly out the door. "I told your man in the stable to saddle up a horse. I'll get it," he called to her.
His stocky legs carried him quickly across the yard and he turned down the side of the house toward the stable. Lauren was unmindful of the rain that fell gently around her. She saw lightning flash and light up the sky to the west in the direction of Keypoint, but it was far away. She clasped her hands in front of her and prayed silently for the life of her husband. Don't let him die. Please, God, don't let him die.
She had changed her clothes with trembling fingers, shaken to the depths of her being at Kurt's news. Judd brought the gelding around and offered to give her a boost up. She hesitated only an instant. Would it hurt the baby for her to ride? But she had to get to Jared. She mounted quickly.
"Where is he? Jared. Where did you take him?"
Kurt shouted over the crunching sound of the horses' hooves as they sped down the lane, "He was shot up pretty bad. We took him to a cave one of the men knew about. I think they were going to try to fetch a doctor. They were afraid to move him anymore. He was bleeding pretty bad."
Lauren clung to the reins and shut her eyes momentarily. Shot up pretty bad. What did that mean? Bleeding. Oh, God!
* * *
Carson never ceased to be amazed at Olivia's composure. She sat with her eyes closed, her head leaning against the high back of the leather chair behind the desk. It had been an hour since Kurt had ridden in with the news of Jared's injury and the terrible turn of events in Pueblo. Soon after that, Parker had arrived and stormed into the office in a rage.
"The whole goddam thing blew up in our faces. I tell you, they were like animals hiding in the buildings where we couldn't even see to shoot them. They picked off our men one by one. They knew we were coming. They knew!"
"Parker, please calm down. There's nothing we can do now." Olivia's voice was dispassionate. "In a way, we can turn it to our advantage. We can say that the degenerate citizens of Pueblo went on a shooting spree and shot a few cowboys who drifted into town after dark. We'll think of something."
"Well, it better be damned good. I'm getting tired of all of this muck." The Teutonic features were congested and contorted.
"We'll work it out, Parker," Carson said with more assurance than he felt.
"You'd goddam better. I could call off this whole thing just like that." He snapped his pink, sausage fingers loudly and then turned his bulky body and strode out the door.
Even after that altercation, Olivia had remained tranquil. Carson paced the rug in front of the desk as if hypnotized by the pattern woven into it.
Suddenly they heard hurried footsteps and jingling spurs coming across the porch and through the front door.
"Maybe that's word of Jared," Carson said hopefully and rushed into the hall. He was struck dumb.
Jared was striding purposefully toward the office. He wasn't wounded at all. In fact, Carson had rarely seen him so exhilarated. Behind him was Rudy Mendez, his white teeth flashing against the dark complexion of his face, his black eyes dancing. Carson was shocked as always by their resemblance. Standing in the doorway but not coming into the house was Thorn, the taciturn Comanche who had been the boys' childhood friend and their father's before them.
"What—" Carson started before Jared interrupted him.
"Carson, you look as if you've seen a ghost," Jared said heartily. "I believe you know my brother and Thorn." Jared pushed past the flabbergasted man and tramped to the desk, confronting his mother.
"It didn't work, Mother. Your friends from Austin ran off with their tails between their legs. It's all over."
Olivia had fastened her eyes on Rudy and her face had paled significantly. "What is he doing in my house?" she gasped. Her voice rose a note or two on the last words. Her agitation was apparent. "Get that bastard out of my sight."
"Rudy is my brother," Jared said levelly. "He stays in my house if I say so." His eyes never wavered from her face, daring her to challenge him.
She looked at him closely then, the truth dawning on her. "It was _you,_ wasn't it? _You_ were the traitor. You ruined everything for us."
"No, Mother, I didn't ruin anything. Hopefully, I helped to save a lot of property and lives."
"You sound just like your father," she spat. "Always so full of goodwill and nobility. You were against me all along. Don't you see, Jared, that you've probably set us back for years by this insane high-mindedness?"
Jared shook his head. "No. No, I didn't. We'll get our railroad, but not through exploitation and violence. And without the help of Vandiver and his lot." Jared turned to the stunned man standing behind him. "Carson, you see why I had to go against you, don't you? We were being manipulated. I had to take matters in my own hands. I think I did what Ben would have done."
Carson looked deflated and tired and old. He smiled kindly at the young men before him. They were both good men. Honorable. Strong. Ben's sons. He was proud of them for their absent father, his best friend. "Yes, Jared, I think you did the right thing," he said, clamping him on the shoulder. The two men stared long at each other. Finally the younger man turned away in embarrassed humility. It was an emotion completely new to him and he covered it quickly.
"Rudy, wait here just a minute. I'll go tell Lauren the news and then we'll bed you and Thorn down for the night. I know you're both exhausted." Jared was already headed for the door when Olivia halted him in mid-stride with her level, quiet words. "Lauren's not here, Jared."
For some reason, her tone and the calm manner in which she spoke set off an alarm buzzing in his head. A great foreboding squeezed his chest.
He turned back slowly and faced his mother. "Not here? Where is she?" His voice was low. Lethal.
"She's with Kurt Vandiver. They left on horseback about an hour ago."
"She wouldn't go anywhere with Vandiver. What the hell are you talking about?" His anger was building.
Olivia smiled sweetly at her son. "Jared, have you forgotten your first impressions of Miss Lauren Holbrook? You thought she was a trollop, an adventuress. I think your hunch has proved to be correct."
Carson butted in, sputtering, "Olivia, tell the boy why she left. Tell him!" He stared incredulously at the woman whom he had worshiped for over a score of years, as if he were now seeing her for the first time.
"This is none of your concern, Carson," she declared sharply.
Jared faced Carson, his whole body tense. "Where is she?" he asked hoarsely.
"I don't know, Jared," Carson answered honestly, baffled by this twist of circumstances. "Kurt came in here hell-bent for leather about an hour ago. He said that you were hurt critically, and that Lauren should go with him. He hinted that if she didn't, she might never again see you alive." He left Jared and walked over to where Olivia sat with her hands folded on the desk. "You knew that Jared wasn't hurt." It wasn't a question. It was a statement.
Jared pushed Carson out of the way and spread his hands wide on the desk as he leaned over it, his face a few inches from his mother's. "Why did you let her go off with that sonofabitch? Why, goddam you?" He slapped the desk with his palms and the sound exploded like a shot in the air already fraught with tension.
When she answered, it was with the same even tones she had used before. "Lauren has served her purpose. It was expedient for her to be a victim of tonight. That way, no one could blame us if anything went wrong. As it did." She looked malevolently at Rudy. "Who could blame us if one of our own had been kidnapped and... hurt as a result of the fracas?"
Jared was trying hard to keep the coldness in his stomach from spreading to his entire body and freezing him on the spot. His teeth were clenched as he rasped, "Where did he take her?"
"I don't know."
"Where?!" he screamed.
"I don't know!" she screamed back.
Rudy hadn't spoken once since entering the house. Now he grabbed Jared by the sleeve. "Come on. We're wasting precious time here. Thorn can track them, but they've got an hour's head start in the rain. We'd better hurry."
Jared still stared at his mother, thinking all sorts of crude insults to fling at her. But she wasn't worth it. She was defeated and she knew it. Bitter disappointment that this woman who had given him life could never give him love filled his lungs, drowning him. The innumerable times in his life he had been hurt and rejected by her flashed through his mind in kaleidoscopic fashion. All his efforts at attempting to please her had been scorned and for naught. He never quite met her expectations. That rejection had been the crux of his bitterness, his contempt, his anger at the world. If he wasn't loved, then, by God, he wasn't going to love anyone!
But it hadn't worked. He had loved Ben. Yes. He admitted it now. He loved his father and had been devastated by his death. And in spite of Ben's ill health, Jared was still haunted by the argument that had precipitated the final seizure. He had loved Ben. He loved Rudy and Gloria. And Maria. And he loved Lauren.
Lauren! It had only taken a few seconds for these soul-rending admissions to pass through his mind. Rudy's hand was still on his arm. He gave his mother one last regretful look and then turned on his heels and fled the room.
"Thorn, we need to track two horses that left here an hour ago. Vandiver's got my wife." He shouted all of this as the three men dashed out to the yard and mounted their horses.
"We'll have to hurry. The trail could be washed away soon," Thorn said matter-of-factly as they picked up the faint trace of horseshoes impressed in the soft mud.
* * *
Lauren clung to the pommel of her saddle with stiff, cold fingers, trying to navigate her horse up the slippery, muddy incline. The rain that had started as a fine drizzle had now increased to a steady downpour. She had not taken the time to put on her hat after Kurt had informed them of Jared's injury, nor had she put on a jacket or any kind of protective covering. The hard raindrops fell like lead balls on the top of her head. Her hair, even heavier with the weight of the rain in it, had pulled loose from its pins and hung down her back, making her neck ache. She was soaked through to the skin and shivered with cold.
Vivid flashes of lightning had spooked their horses several times and the thunder rolled over the plains and off the shallow hills like giant bowling balls striking a stone wall.
Cold, wet, and miserable as she was, one thought kept reverberating through her head like the thunder: Please, God, let Jared be alive.
It seemed to her that it was taking an inordinate amount of time to reach their destination. They had ridden off in the opposite direction from Pueblo and had been riding for what she calculated to be a couple of hours. But they may not have been riding anywhere near that long. Time had stood still for her when she heard that her husband's life was in danger.
She questioned Kurt about Jared's condition when the trail had widened enough to allow them to ride abreast. "Where did they take him, Kurt? We've come miles from Pueblo and you said they were afraid to let him travel too far."
He avoided her eyes. "Well, one of the men knew about this cave. They wanted to get Jared away from the scene of the trouble for his own safety, so this cave seemed ideal. Not too many people know about it."
His words didn't make much sense to her, but she didn't argue. She only wanted to get to Jared as quickly as possible.
She had been hearing a loud roaring for the past several minutes, and when the sky was illuminated by a lightning flash, she saw that the river was about fifty feet ahead of them. The Rio Caballo, which was usually so placid even as it formed small rapids over the limestone that lined its bed, was raging and boiling out of its banks.
Kurt rode toward it and cursed loudly as he saw their predicament. "We have to cross it, Lauren," he shouted over the roaring of the river and the crashes of thunder. In daylight, the prospect would have been grim, but in this darkness, with stinging rain in their faces, it seemed suicidal.
"Isn't there another way?" Her throat hurt in the effort to make herself heard over the din. "The horses can't swim this. Even if they could, look at all the debris."
There were large trees, barrels, wagon wheels, lumber, and sundry other objects being carried by the swift, churning water. Lauren didn't think they could make it to the other side without seriously injuring either themselves or their mounts.
"Do you want to get to Jared or not?" Kurt demanded, frustrated by her caution.
"They couldn't have brought him this way and gotten him across the river with it flooded like this," she argued.
"It's been hours since they must have crossed it. I'm sure it didn't flood until it started raining hard again."
He was right and she knew it. With all the rainfall of the past few days, the possibility of a flash flood was great. When the new rain had begun to fall, the saturated ground at the top of the hills had refused to hold any more, causing it to funnel downstream into the river. It wouldn't have taken long for the river to rise to these proportions.
She nodded at Kurt as she gripped the pommel more firmly.
"You go first, and I'll be right behind you," he shouted. "If the horse won't swim, just try to ride the current until you get a chance to make it to the other side. Okay?"
She nodded again, dumbly, her heart pounding, and nudged the horse's flanks. The gelding shied away from the water, tossing his head, and for a moment, Lauren thought he would refuse to go into it. Obedience won out, however, and he stepped into the boiling river. He had just gotten all four feet into the water, when he was almost immediately swept into the middle of the river by the swift current.
Lauren held on for dear life, thinking suddenly that if she died, her and Jared's baby would die, too. Why didn't I tell him I carried his child? She berated herself. She tightened her knees, gripping the sides of the horse. She risked looking back to see if she could spot Kurt and saw him urging his horse away from the bank into the water.
Darkness and rain surrounded them, but when the lightning flashed again, she saw the tree being swept toward her with alarming impetus. Oh, God, no! her mind screamed, and she braced herself for the impact.
The main trunk of the tree floated in front of them, but apparently a branch that was under the surface struck her horse in the forelegs, for he suddenly buckled and Lauren was all but thrown from his back. She gripped the pommel tighter, her fingers slipping on the muddy water that had doused the saddle. The horse was screaming in pain, and Lauren knew his legs must have been broken. The tree had not flowed past them, but was circling wildly as if seeking direction. In his agony, the gelding thrashed violently and, this time, Lauren couldn't maintain her hold.
She was hurled into the dark murky waters of the raging river. The current lifted her up long enough for her to gasp a breath before once again she was sucked under. Beneath the surface, she struggled to propel herself upward. She could feel herself growing lighter, knowing that the surface and the much-needed oxygen were near, when her head struck something above her, and a blinding pain shot through her body. She floundered helplessly, felt the drawstring of the cloth bag with the medical supplies slip from her wrist, and began to sink into the deep oblivion of unconsciousness.
Then her head was stung with a million needles as she was lifted by her hair out of the river. She swallowed the brackish water in her mouth and began sucking in great mouthfuls of precious air. Kurt held her under the chin as she clutched his saddle trying desperately not to lose her hold on either it or consciousness.
Somehow, miraculously, Kurt's horse was able to make it to the other side of the bank. He dropped Lauren gently onto the ground, where she lay choking and spitting. Kurt dismounted and lifted her limp form against him. He shook her slightly as he asked, "Are you all right?"
It took her a few moments to regain enough strength to answer, "Y-yes. I think so."
"Come on, it's not far now. Just at the top of the hill there." He lifted her onto the horse, and she would have fallen forward if he hadn't mounted quickly behind her and supported her with his sturdy arms.
She must have slipped out of consciousness then, because when she came to, they had stopped and Kurt was dragging her off the horse. He steered her toward what appeared to be a solid wall of rock. As they got nearer, she saw that there was a small opening practically hidden by a clump of bushes. Kurt drew them aside and pushed her into the cave. She had to bend from the waist in order to get through the low opening. She walked in the crouched position with dread of the unseen terrors lurking in the walls of the cavern.
Finally she saw light ahead and, knowing that she would soon see Jared, she quickened her pace. The pale light came from a small lantern hanging on the wall within the cave. She stepped into it and stood upright. She was looking directly into the hooded, reptilian eyes of Wat Duncan!
Behind him stood June, her pale hair forming a halo around her head as the light from the lantern reflected on it. Lauren's eyes darted swiftly to all corners of the room, and nowhere did she see Jared.
Myriad horrors flashed through her mind, so quickly that her brain couldn't catalogue them. She felt Kurt behind her as he stepped into the room.
"Well, well, Miz Lockett. I see you made it to our little party. Not a very nice night for a swim, though." Duncan grinned indolently and his eyes roved her body.
Lauren whirled on Kurt. "Where is my husband? You said you were bringing me to my husband!" She clenched her fists and pounded his bulky chest ineffectually. Grabbing her wrists, he held her away from him.
"I had to tell you something to get you here. And it worked."
Stunned, she stammered, "B-but why? You planned all of this? Why?" She was too shocked at the situation she found herself in to be frightened. That would come later. Now she was angry and puzzled.
"I don't think I have to tell you _why,_ Lauren. Surely you can guess my motivation." His eyes moved from her face to her chest and all the way down her body, taking in every detail. Lauren shivered, but not from the cold. She looked down at the dripping garments that were molded to her and revealed every secret of her body.
"We done just what you axed us to, Mr. Vandiver. I got the lantern and brung some food. It's all ready for you." Duncan's tone was sycophantic, and Lauren knew he must have been promised money for his generosity.
"Everything is fine, but what's she doing here?" questioned Kurt, indicating June, who as yet had not spoken.
"June and me, we got a lot in common, and one of them common interests, you might say, is bringing down Mr. Jared Lockett." At the mention of his name, Lauren started and June looked at her smugly, haughtily. "So she begged me to bring her along. 'Sides, she kinda took a shine to you, Mr. Vandiver. She wanted to see you again."
Kurt smiled seductively and looked through half-closed eyes at the girl who stood with her feet planted far apart, hands on her hips, displaying her body shamelessly in the thin, low-cut dress. Lauren looked away in disgust as June's tongue slid slowly across her bottom lip, making promises to Kurt with her eyes.
"You're not bad, Miss June." Kurt walked farther into the cave and stood directly in front of her. "Yes, not bad at all. I may have you for seconds." Then he turned to Lauren. "After I've satisfied Mrs. Lockett."
June shot Lauren a look full of jealous hatred and resentment, but Lauren didn't see it. June posed no threat at all compared to Kurt Vandiver.
* * *
"Why did you do it, Olivia? It wasn't necessary. Why did you endanger that girl?" Carson had not quite recovered himself after having faced the fact that Olivia had betrayed her own son. He still couldn't believe that she was capable of such treachery.
Jared and his brother, along with Thorn, had ridden out of the yard only minutes ago. Carson hoped they would find Lauren before any harm could come to her at the hands of Vandiver. Carson stood over the woman he had loved for years and stared down at her black hair. The silver strands fanned through it, away from her aristocratic brow. He had always thought her hair was beautiful.
Olivia appeared not at all unnerved either by the events that had happened earlier or by Carson's close scrutiny.
"Answer me, please, Olivia. Why did you do this?" His voice was raised and held more censure than she had heard since she had known him.
"Carson, really, I find this sudden streak of conscience in you surprising and tiresome. You, of all people, should know that sometimes in business we must do things that are not always appealing."
"I have done things, yes, to manipulate people. As far as I know, however, I have never sacrificed an innocent young woman, who has done nothing to hurt me, to someone as ruthless as Vandiver."
"Innocent? Ha! You men, you're all alike," she accused. "And I wouldn't exactly call the way she turned Jared against us doing _nothing._ That was no small feat."
"Lauren isn't responsible for Jared's convictions, and you know it. He has mistrusted the Vandivers from the beginning. He felt that his whole legacy from Ben was threatened by them. He protected it as he saw fit. Lauren has been good for him, I think. She has made him take a real hard look at himself."
Olivia slammed her hands down on the desk much the way Jared had done earlier. "I don't want to hear any more about Lauren and her virtues. I'm sick to death of everyone showering her with accolades."
Carson considered her closely. Her green eyes were flashing fire and her nostrils flared with the heavy breaths she took. Then he understood. "You're jealous of the girl, aren't you, Olivia? She came between you and Jared. I showed her some paternal affection. And Ben—"
She pounced on the name. "Yes, Ben! He had humiliated me for years with that whore of his. People laughed behind my back because my husband preferred that Mexican bitch to me. He loved her son as much as he did mine. His liaison with her was a continual insult to me. _Me!_ " She punctuated the pronoun by pointing to her chest. "The belle of New Orleans society. Me, from one of the most respected, affluent, influential families in the history of that city. He brought me to this barbaric, godforsaken country and expected me to live like one of his Mexican peons." Tears were streaming down her cheeks. In all the years he had known her, Carson had never seen Olivia cry.
"And he didn't love me," she added miserably, pitifully. "He didn't love me." She was still for a moment and then she tilted her chin in that autocratic way she had. "But I wasn't going to stand by and let him deposit his latest doxy here in my own house. I'd be damned first!"
Carson spoke softly, "Olivia, you know that Ben brought Lauren here in hopes that she and Jared—"
Once again, she interrupted. "That's what he _said._ But you heard the way he talked about her. Lauren was so beautiful. She was so kind, so innocent, so sweet, so polished." She buried her face in her hands as her elbows supported them on the desk. "Why didn't he love me? _Why?_ "
The words tore at Carson's heart. He didn't want to ask, but he did. "You've loved Ben all this time, haven't you?"
"No!" she cried. "I hated him. I hated him!" She pounded on the desktop with her fists.
"No," he said quietly. "You loved him."
Olivia looked up at him suddenly as if remembering that he was still there. Her eyes were bright with tears, the lashes spiky and black. "You!" she snarled with contempt. "Don't you see that you were used as a means to hurt Ben? I carried on a shabby affair all these years with his best friend and business partner as revenge for all the humiliation he had heaped on me." She laughed a bitter laugh. "But he just turned his back and forced me to tolerate _you_ all these years. I look so forward to the day when the railroad is completed, Vandiver is satisfied, and I can throw you out of my house, my life, and never have to look at you again. Didn't you ever ask yourself why any woman would want a short, fat, balding man who made love like a sloppy adolescent when she was married to a hard, virile stallion like Ben Lockett? You are a fool."
The words really didn't have any meaning for Carson. He had already been shattered to learn that he had given the prime of his life to a selfish, shallow woman. He had given up having a wife, and children, and the respect of his best friend for a chimera, an illusion. Worst of all, he had sacrificed his self-esteem. He looked at himself now and found himself wanting in every facet of his being. He was a shell of a man. Olivia's words had not hurt him. He was too empty to hurt anymore.
Olivia sat staring straight ahead, her eyes glazed. She didn't see Carson remove the derringer from his inside breast pocket. He moved the gun close to her head, and when at last she saw it in her peripheral vision, she looked up at him and laughed.
Her mouth was wide, her eyes streaming tears, and her head was thrown back in laughter as he pulled the trigger. He watched sadly as her head fell forward and thudded onto the desk. She was so beautiful. So beautiful.
He was still looking at her as he raised the pistol to his own temple.
# Chapter 25
The heavy rainfall wasn't making the tracking any easier. Had it not been for Thorn's innate ability, Jared would have been even more desperate than he already was. Thorn had taught him and Rudy in their youth to track with precision but, though they were keen pupils, they still wouldn't have been able to follow this mud-obscured trail. The darkness and the blinding rain made it nearly impossible to find.
They had been riding for an hour and a half when the tall Indian pulled the reins of his horse. "They're headed toward the river."
Jared heard him over the loud rumble of thunder and said, "Well, let's get on with it." He was puzzled by Thorn's hesitation.
Rudy intervened. "Jared, do you know what the Rio Caballo will look like? It will be raging. I don't think they could have crossed it, and even if they did, we couldn't pick up the trail until morning. Shouldn't we go on to Keypoint and wait out this storm? We can start again at dawn."
"Hell, no!" Jared's roar challenged the thunder. "If you aren't coming with me, I'll go by myself. No telling what that bastard has in mind for her. And she's afraid of him. I know it."
He wore the determined look he had inherited from two very forceful personalities. Rudy and Thorn exchanged looks. Rudy's was exasperated. Thorn's was totally noncommittal, as if it were of supreme indifference to him that he was riding around the countryside in the middle of the night in a fierce spring thunderstorm. Without speaking, he turned his horse toward the river and focused on the rapidly dissolving hoof prints.
"I don't know how we'll cross that damned river when we get there," Rudy muttered. He was surprised when Jared answered, thinking he couldn't have heard him over the sounds of the storm.
"I don't know either, but we have to go on. I've got to find her. We'll proceed with Thorn's plan—it's our only hope."
The trio rode on in silence.
* * *
The dull throbbing in Lauren's head had increased to excruciating proportions. She suspected that when she was seeking the surface of the water and banged her head on the tree branch, she had been hurt more seriously than she had first thought. Waves of nausea threatened to make her vomit, and she struggled not to give in to them. She wanted to keep the focus of attention off herself. The other three in the confines of the cave with her were entertaining themselves with a bottle of whiskey. She had time to think.
Lauren's main concern was for the baby she carried. She was still somewhat ignorant about pregnancy, but she was sure that an exhausting horseback ride, falling into a raging river, and receiving a severe blow to her head were not good for the embryo in her womb. She prayed fervently that it had not been injured.
What was going to happen to her at the hands of Kurt and Wat Duncan? Had she not been in such pain, she would have been more afraid. As it was, she more or less accepted her fate with a passive resignation. She couldn't fight them. She lacked the strength and the ability. She couldn't escape them. Where would she go in the middle of the night in a fearsome storm without anyone to guide her over the rough terrain? Just don't hurt my baby, was her only lucid thought. That and the slim possibility that Jared might rescue her.
Kurt hadn't mentioned him, and she was afraid to ask. He hadn't been shot. At least, he hadn't been brought here as Kurt had told her. He could have been wounded or... killed. No! She wouldn't even think that. He might—just might—be alive and searching for her. He might not be motivated by any grand passion for her. He had made his feelings clear earlier this evening. But he might be motivated by pride. He wouldn't let Kurt Vandiver take away anything belonging to him. Lauren grabbed on to that thought and clung to it.
Too much thinking had caused her head to ache even more, so she held that one thought and kept repeating it to herself as though to will it into a reality.
Wat Duncan and Kurt Vandiver were having an argument and, as their voices rose, Lauren raised her head and looked across the cave floor from where she sat on an old blanket toward the two men. It was hard to focus on their figures no matter how she squinted.
"I tell you, we got sumpum' comin' outa this too, Mr. Vandiver." Duncan was standing a few inches from Kurt and looking defiantly up into the taller and more powerfully built man's face. "Juney and me, we been waitin' a long time to bring Jared Lockett down. And you owe us for heppin' you these past weeks. Ain't we done everythin' you axed us to? Didn't we do right by you with that Mendez woman?" He thrust his face even closer to Kurt's.
Kurt spoke softly. "I don't care what you do to her after I've had mine, but right now you're not touching her. Now the two of you get lost. When I need you again, I'll let you know." His superiority and carelessness was a mistake. He underestimated the Duncans' feral instincts.
At a small, quick signal from Wat, June threw her arms around Kurt's neck and toppled him to the ground. He landed facedown in the soft earth of the cave, June lying on his back, pinioning him on the ground with surprising strength.
Duncan laughed bitterly. "That'll teach you to cross a Duncan, Mr. Vandiver." He all but spat the name. "You hold him, Juney, while I service Miz Lockett here. Vandiver, you watch. You may pick up a few pointers."
June laughed into Kurt's ear, a soft, seductive laugh, and ground her hips against his buttocks. Kurt lay perfectly still, chagrined that he had let this lowlife get the best of him.
"Now, Missy, I'm gonna show you what a real man is like."
Lauren watched with terror-glazed eyes as Wat Duncan approached her. His lips were pulled back in a lubricious grin, revealing putrid teeth. His black eyes raked her body and she shrank back against the damp stone wall. He licked his lips as he knelt down in front of her and reached out to the top of her shirtwaist and began unbuttoning it.
Lauren's hands tried to push his away, but they seemed capable of only useless flutterings. His efforts grew more frantic as he ripped her chemise and laid bare her breasts, made fuller by pregnancy.
"Lookey here, Juney," he whistled. "Did you ever see any tits purtier than these? Soon's I'm done, you can play with 'em ifn' you've a mind to."
Lauren hadn't intended to do it, but she opened her mouth and a scream that originated in her deepest self propelled its way out of her throat and pierced the dim enclosure.
June jumped involuntarily at the startling cry and Kurt used that split second to roll from under her and unsheathe his gun from its holster. He couldn't imagine why Duncan and his sister had overlooked it. He pointed the Colt directly at the center of Duncan's back and pulled the trigger. Lauren felt the thudding impact as the bullet entered Duncan's body. If a bone hadn't stopped it, it could have exited his body and entered hers.
He fell heavily against her, a puzzled expression on his ugly face. Blood bubbled out of his mouth. Lauren screamed again, fighting frantically to push his weight from her. She managed to lift him enough for her to slide from beneath him, and almost fainted when she saw his blood soaking into her clothes.
June stared transfixed at the body of her brother, emptying its blood onto the earth. She growled a savage sound as she attacked Kurt, who was still lying on the ground. They rolled together, punching wildly, thrashing arms and legs. June's thighs flashed whitely in the lantern-lit cave as she struggled for possession of Kurt's gun. Then another shot blasted through the cave. Lauren watched breathlessly as the two forms locked in combat lay still.
Finally Kurt moved, extricating himself from June's death grip on the front of his shirt. He slung her limp arms from his impatiently, pulling his legs from their entanglement with the dead girl's. He practically crawled to the heap of supplies that Duncan had gathered for him. Finding another bottle of whiskey, he uncorked it and tilted it to his lips, watching Lauren all the while.
He drank deeply, then lowered the bottle and wiped his mouth with the back of his hand. Several bloody scratches ribboned down his cheeks. June had managed to inflict her own brand of pain before she had died at the hands of this ruthless man.
Lauren was encapsulated in a mild stupor, viewing the atrocities happening before her apathetically. Her head was pounding with a cadence that was deafening to her ears and agonizing in its intensity. The walls of the cave were slowly tipping first one way and then another, and her eyes refused to focus on any one thing.
"You goddam well better be worth it," Kurt snarled at her as he stood above her. As she tried to focus her eyes, his form swayed in a sickening rhythm. He reached down to her and, with hands magnified to the size of hams by Lauren's distorted vision, grabbed her shoulders and hauled her suddenly to her feet. She closed her eyes as pain shot through her head. Dizzily she tried to stand upright.
"I've been waiting a long time for this," grunted Kurt as he feasted his eyes on her. He seemed unaware that her eyes were glazed and, if not for his support, she would have collapsed. One of the large hands closed over a tender breast and squeezed her nipple roughly, while with the other hand, he worked feverishly with the fastenings of his pants.
Lauren recoiled, and for the first time began to struggle against him. "That's right, Lauren, fight me a little. I don't want you completely docile."
She saw the swinging movement out of the corner of her eye just an instant before the rifle butt cracked into Kurt's skull. His eyes rolled back into his head before he fell heavily to the ground. He dropped Lauren in his fall and she collapsed beside him.
"Miz Lockett!? Is that you?" Crazy Jack Turner knelt down beside her cautiously. She focused on his hideous face, but never had anyone looked so beautiful to her. "What in hell is goin' on here? Where's your husband?" Lauren had a fleeting impulse to laugh. Of all things to ask at a time like this!
"Mr. Turner," she croaked. It was difficult to get her thick tongue to form any words. "Help me. Where are we?"
"You're in the back of my house, that's where. I thought no one else on earth knew about the rear door to my cave, 'ceptin' me. I heard the screamin' and shootin' and came through the tunnel to see what in hell the racket was all about."
He surveyed the bodies lying sprawled on the cave's floor. "Whoever shot those two should be decorated," he said without emotion.
Lauren realized that this must be an alternate entrance to Crazy Jack's house that jutted out of the rock wall above the river. The mutilated face gazed down at her with kindness in the eyes. He looked away, embarrassed, as she vainly tried to cover herself with the bloody remnants of her chemise. She tried to talk, but the words would not come. Her brain seemed incapable of forming a coherent thought. He sensed this and leaned over her. "Let's get out of here before that—"
He was interrupted when his breath was expelled in a great _whoosh_ as Kurt's booted foot caught him in the stomach. Crazy Jack fell backward and rolled to his side, reaching for the rifle he had set down on the ground as he knelt over Lauren. Kurt was quick to note the action, raised his pistol, and fired. Lauren's scream was stifled by shock as Jack's body jerked when the bullet struck it. She looked at the prone figure. He was bleeding from a hole in his chest. "Oh God," she groaned. When does this nightmare end? she wondered.
Kurt kicked the sole of Jack's boot. "Godalmighty! Have you ever seen anything so ugly? I'd heard of this crazy old hermit, but thought he was only a legend that mothers used to scare naughty children. Surely, my dear, you don't prefer his company to mine?" he questioned as he noticed her whimpering and edging away from him. "Come—"
"If you touch her, you're dead." The words came out of the shadows on the other side of the cave near the entrance.
Lauren knew Jared's voice immediately. He had spoken, but she was as confused as Kurt, who whirled toward the voice and then stood stock-still, frightened and bewildered.
Two figures stood side-by-side. Both had on the clothes of _vaqueros,_ their hats pulled down low and dripping raindrops from the brims. Both had pulled their pistols from their holsters and the barrels glinted in the light of the lantern. Both guns were aimed with deadly accuracy on Kurt's chest.
But what was so startling was that the figures were mirror images of each other. Bandanas were pulled up over their noses. They were of the same size and build and, in the dim light of the cave, their hair appeared to be the same dark color. In the shadows cast on their faces by the faint light, their eye-color was indiscernible.
Kurt's heart pounded and rose to his throat as he stared at what seemed to be twin apparitions, a figment of drunkenness, when everything seen is doubled. Slowly sanity returned, and he knew he was facing Jared and his half-breed brother. But they looked so much alike he couldn't tell which was which.
"Move away from her slowly, or by God, I'll kill you, Vandiver." Even as Jared spoke, it was impossible to tell which figure the voice belonged to. It was so controlled that the bandana over his mouth didn't move with the expulsion of his breath. "You've only got three shots left, unless you've used one we don't know about, which makes your position even more precarious. We have twelve shots between us. No matter how you add it up, you die. Move away from Lauren." There was steel in the level voice.
Lauren was finding it hard enough to focus, and now she was seeing four Jareds instead of two. His voice sounded faraway and indistinct, but somewhere in the back of her mind, it registered that he was here. Regardless of what he had said in the past, or what he had done, he was here to save her from Kurt Vandiver. When Jared spoke her name, she reacted by jumping slightly, and this drew Kurt's attention. With uncanny speed, he turned his gun onto the figure crouching at his feet.
"You aren't going to shoot anybody, Lockett, unless you want your bride to die. Even if you got me, I would kill her, too, before I died. No way I could miss. I suggest that you and that bastard brother of yours put down your guns and stop playing masquerade games." He laughed as he saw them glance at each other out of the corner of their eyes. "Now!" he commanded.
Reluctantly the men let their pistols drop from their hands. Kurt moved with cautious steps closer to the brothers, but even at a distance of a few feet, he couldn't make out which was which. He kicked the Colts dropped in front of their boots out of their reach, and stepped back hurriedly. He wanted to yank the bandanas from their faces, but he wasn't quite that brave. The dangerous stance of their bodies and the fierce hatred glowing out of their eyes were identical.
"Which one of you _hombres_ is Jared?" Neither so much as blinked. "Which one is Jared?" Kurt's suddenly soprano voice betrayed his frayed nerves. The two figures could have been statues. "Then I guess I'll have to kill both of you. I can't very well rape this woman with one of you breathing down my neck, can I? And I'm in a great hurry to do just that." He paused, hoping that the threat would goad Lauren's husband into revealing himself, but both remained motionless, knowing well his intention. They knew, too, that Thorn was waiting...
Kurt Vandiver took advantage of having the mighty sons of Ben Lockett held at gunpoint. Since he was going to kill them, he might as well have some fun first. "I wonder if she's as hot as that Mendez woman. Took quite a bitch to keep a stallion like Ben satisfied all those years. I think Duncan was planning on getting a piece of that himself, but Lauren came out of the house too soon. Was that slice across her throat as clean as he bragged it was?"
A vile-tasting fluid filled Lauren's mouth and she almost gagged. The two men appeared unaffected by Kurt's taunts.
"All right, then," he said. "I gave you both a chance to stand up like men." He took careful aim at one of them. Lauren held her breath. She thought that Kurt was only bluffing. He would never kill a Lockett for fear of the reprisals. He was basically a coward.
Refuting her supposition, the pistol shot exploded in the room of the cave. She watched both men, her hands covering her mouth, trapping her scream and terror inside. For long moments, neither of them moved: then, as a dark stain began to spread on one's shirtfront, he fell backward against the rock wall and slumped to the ground.
Her husband or her brother-in-law had just been murdered. It was too much to grasp. Her head was pounding and the rock room was spinning, dancing crazily in front of her. It couldn't be Jared over there bleeding, motionless and unconscious... dead.
Something tapped against her knee and she flicked it away. She didn't want to look, didn't want anything to interrupt her grief, her disbelief, at what she had just witnessed. Finally she forced her eyes away from the inert body across the cave and looked down at the persistent nudging. It was the barrel of a rifle.
Her befuddled mind couldn't imagine how the weapon was moving of its own volition. Her eyes painfully focused and traveled down the length until she saw the gnarled hands of Jack Turner holding it by the butt, moving it just enough to poke against her knee and get her attention.
Crazy Jack wasn't dead! He looked at her through eyes glazed with pain, and tried to communicate a silent message to her. She glanced at Kurt's bulky silhouette, his back to her, as he threw disparaging remarks at either Rudy or Jared, whichever it was who remained alive.
Lauren knew what Jack wanted her to do. But she knew she couldn't do it. Everything she had ever been taught, every principle she held, forbade the action Jack was urging her to take. Even if she had the physical strength and the mental capacity to do it, she knew she could not. It wasn't right. Nothing justified it. Nothing.
Jared? Her baby? At what point did wrong become right?
Why didn't this dark, moist, dreary room cut out of rock stop moving? Her head throbbed. She couldn't swallow. Her stomach wasn't going to hold what was in it much longer. Splattered blood was drying on her skin.
Has my baby survived this? Yes, please, God, she prayed. And my husband? Is that him lying over there with his blood flowing onto the ground? No! Jared! she screamed, but made no sound. She was stunned, dazed, weak.
Metal clicked on metal as Kurt cocked his pistol again.
God, please don't ask this of me, she begged. What if I miss Kurt and hit someone I love instead? I don't know how to fire a gun. God, please, let there be another way.
It was too late. Kurt was taking careful aim. Of their own volition, her hands reached out for the rifle. The dizziness and blurred vision vanished. With heightened clarity, she pointed the barrel at the broad back and pulled the trigger even as she heard the blast from Kurt's pistol.
The rifle butt slammed into her chest with unbelievable force. The echo of the rifle shot joined that of the pistol and bounced off the stone walls, filling the small, dank chamber with a deafening racket.
Then there was another explosion. This one was in Lauren's head. It was louder and more terrifying than the ones preceding it, and reverberated in her mind, blocking out conscious thought. Bright yellow flashes burst in her brain with the rhythm of heartbeats. Then all went black as she surrendered to blissful oblivion.
# Chapter 26
Peace. Serenity. Silence. All welcome.
Dreams.
The Prathers' parlor. Lauren was sitting at the piano playing, though no notes sounded. Her dress was lacy and white, startling in its brightness.
Maria stood beside a smiling Ben. He was bigger than life and twice as robust as she had remembered. He glanced down at Maria and patted her arm fondly. Carson Wells was there, smiling at Lauren in his kind, sad way. What were these people doing in the Prathers' parlor?
She searched for another face, missing the notes she was trying to play. She tried vainly to spot a dark, lean face among the others. Whom was she looking for? She didn't remember. She only knew that she wanted to see that face more than any other. It was important to her, but why? Why? The vision vanished into a nebulous whirl around her and she was alone again.
Later, she was flying down a long corridor. There were pillars lining the sides of the hallway. Her hair was long, so long that if she flew between the pillars, it wound around them like a bolt of fine cloth. It trailed for miles behind her as she continued down the corridor, weaving in and out of the columns. At the end of the hall, she saw a figure dressed in a wedding gown. The face framed in the lace veil was like hers, but not hers. It was her mother.
Mother! she cried silently. Mother, I'm coming. She drew closer and closer to the apparition and gazed at the beautiful face that was almost transparent and yet real. Eyes exactly like Lauren's gazed back at her with apparent love, and she was suffused with joy. The lovely lips smiled and opened to speak. Though no sound came from them, Lauren understood what her mother was saying.
"Lauren, I'm so proud of the woman you have become. I loved you when you were a baby. I would come into your room at night while you were sleeping and watch you, kiss you, and touch you, marveling at your delicacy and praying for a brilliant future for you. You were such a sweet child. You never caused me any trouble. I'm sorry I had to leave you. I wanted to stay and see you grow up, but I was in such pain, Lauren. Please understand why I had to go when I did."
"Mother, Mother, I loved you. Did I ever tell you that I did? Did I?"
"Of course you did, darling. Over and over sometimes. I knew that you loved me."
"I wear your watch over my heart, Mother. Every day. And I look at the picture of you and Father on your wedding day." She spoke rapidly, trying to cover years in moments. Even as she spoke, her mother began to move away from her, and she wanted to reach out and take hold of her and never let her go away again.
"Mother, don't leave me, _please._ I need you."
"No, Lauren, you have someone else to take care of you now. We'll have time much later to talk. I'll be waiting for you, but now I must go." The image of the woman moved farther and farther away. She couldn't let her go. She had waited so long to see her, talk to her.
Her arms reached out for her mother, but something held them down. She struggled, but she couldn't shake off the force that held her back. "Mother!" she screamed as the beautiful lady vanished.
Once she tried to open her eyes, but the pain prevented her. She could hear muffled voices, but she didn't know what they were saying. It was so hard to concentrate.
Someone held her hand in both of his. Cool lips brushed her forehead lightly. A thumb caressed her palm. None of this was unpleasant, but she didn't want to cope with it yet. She slipped back into the region of dreams.
But the dreams were no longer pleasant. She recognized and knew Jared immediately, though he was strange and transformed. He stood in a blinding light, his whole body looking golden, shiny. He was naked except for sandals whose leather thongs wound around his muscular calves up to his knees. He carried a sword and a small, round shield. He looked like a Spartan warrior. His face was chiseled, hard. His sex stood erect and was the only part of him that was pulsing with life. He stood motionless. She approached him timidly, almost frightened by his stern countenance.
He was so handsome. She reached out to touch him, then recoiled in horror. He wasn't real, not human. He was made of stone, a beautiful carving reflecting the bright light around him.
Behind her came a ghastly laugh that made the hair on her neck rise and crawl. She turned and saw Olivia, her hair radiating from her head like Medusa's snakes. Her face was ugly and cruel. Lauren screamed, but Olivia only laughed harder, opening her mouth wide.
Lauren screamed again and again, thrashing her arms in an effort to escape those that reached for her. She turned her head wildly from side to side, trying not to look at the beast that stood before her.
Then, again, black oblivion.
* * *
The climb up was hard, but once she started, she couldn't stop. There, on the other side of the door, was the life she had left on the floor of the cave. She remembered it all now. She had only to open the door, and she would have to face all that she wished to forget. It wasn't possible. She wasn't strong enough yet. But she had to come back sometime. Now. She opened the door.
Her eyes took in the ceiling of Jared's room at Keypoint. The familiar window was there, but the drapes were drawn against bright sunlight she saw around the edges of them. The large wardrobe stood against the opposite wall. Slowly she turned her head. Reclining in the chair at her bedside was Jared.
He wasn't dead! He was alive. She wanted to shout with joy. She felt a momentary pang of guilt at her happiness, knowing that Rudy must have been the one shot in the cave. Gloria! The children!
But she couldn't help staring in wonder and love at the man to whom she was married. He was asleep, his hands dangling off the arms of the chair. Lauren remembered seeing one of those hands hanging outside the buggy that carried him to his father's funeral. It had looked just like that, casual, negligent, yet bespeaking strength. That was so long ago. She looked up at his face and was instantly alarmed. He looked haggard and worn. His cheeks were sunken. Deep lines furrowed across his forehead and down the sides of his sensuous mouth. His lashes rested on dark shadows under his eyes. Several days' growth of beard stood out from his chin and his clothes were rumpled and stained. What had happened to him? She wanted to smooth the blond-streaked curls away from his brow where they lay limply.
The door to the bedroom opened slowly, and Gloria came through it carrying a tray with a teapot on it. She closed the door behind her and stopped with a small cry when she saw that Lauren's eyes were open. " _Gracias a Dios!_ Lauren, you are awake."
Jared bolted out of the chair, trying to assimilate his surroundings after being startled out of an exhausted sleep. He stumbled to the bed and fell to his knees beside it, searching Lauren's face for signs of pain or delirium.
"Jared?" she sighed.
"Darling." He held both her hands tightly, as if he would never let them go. "How do you feel?"
Darling? Had he called her "darling"? Maybe she was still dreaming.
"How do you think she feels?" questioned Gloria with amusement, trying to ease the tension she felt building in the room. "She's hungry and weak, and probably still has a headache."
Lauren looked from one of them to another. "Rudy?" she asked with a trembling lip. A tear escaped her lid and trailed down her pale cheek.
"He's not near as nice an invalid as you are," Gloria said cheerfully. "He doesn't sleep all the time like you. Instead, he whines and complains and tries to sneak out of bed."
"He's alive?" Lauren was confused. Had she dreamed the incidents in the cave? Her head ached abominably.
Jared answered her quietly, smoothing back tendrils of hair from her cheek. "Yes, he's alive. It would take more than a wound in the shoulder to kill him."
"But I saw him die." She began crying in earnest as memory of all that had happened came flooding back.
Jared gathered her into his arms and shushed her. "It's all right, now, Lauren. It's all over. Don't cry. We're all safe. Shhhhh."
She rested against his strong chest, letting his strength flow into her. Then a terrifying thought occurred to her. The baby! She pushed herself away from him and looked into his eyes. "My baby? Is it all right?"
Jared looked puzzled and then smiled a gentle, patient smile. "Darling, you've been delirious for days and had some very confused dreams. That must have been one of them."
"No it wasn't." Gloria watched this tender scene with tear-blurred eyes. "Dr. Graham told me she was pregnant after he examined her. I... I thought you had enough to worry about, Jared, so I didn't tell you. Lauren, your baby is fine."
Jared stared at his sister-in-law stupidly before turning once again to Lauren. "You're going to have a baby? A baby?" he asked incredulously. "Why didn't you tell me?"
Lauren's lip began to quiver with emotion again and tears rolled down her cheeks as she remembered the last few weeks before that fateful night. Even now she could hear Jared's harsh voice and see his cold face as he denounced her in front of Olivia and the Vandivers.
When he saw the hurt so blatantly apparent on her face, he had a hard time keeping tears out of his own eyes. He cupped her face between his hands and whispered, "Why, indeed?" He wiped away her tears with his thumbs and looked deeply into the gray pools so full of disillusionment. "Can you ever forgive me for the hell I've put you through? Can you, Lauren?" His voice cracked with distress. His plea was desperate. Gloria didn't wait to hear any more. She crept out of the room and shut the door on the two who had so much to discuss.
* * *
Jared came out of the room about an hour later. He glanced over his shoulder at the bed one more time, assuring himself that Lauren was still there though he had just left her side. She was no longer in danger, but he wasn't taking any chances.
He found Gloria in the bedroom with her husband, trying to keep him in bed one more day. Rudy's shoulder was swathed in bandages, but it seemed to be mending properly. They looked up expectantly when Jared entered.
"She's sleeping again, but peacefully," he answered the question he read in their eyes. "God, but I'm tired." He collapsed into the nearest chair and buried his face in his hands, rubbing his eyes with the heels of his palms.
"She's been through a lot, Jared. We still don't know what happened to her before we got there. She's the only one who can tie up the loose ends, but we can't press her on it. It will take time for her mental injuries to heal." Rudy knew his brother had suffered since they had brought Lauren out of the cave.
* * *
Rudy had regained consciousness just in time to see Kurt Vandiver pitch forward after the rifle blast. Pandemonium broke loose. Jared leapt across the cave to Lauren's side, bending over her and shouting her name with a strangled cry. Thorn came in as stoic as ever. As had been arranged, he was waiting just beyond the room in the tunnel leading to the outside. It was his idea that the brothers go in looking like identical reproductions of each other. His Indian blood put a lot of stock in the psychological breakdown of one's enemy. Anything that would be unnerving to one's opponent was considered a weapon. He was to wait outside and make sure Kurt died if anything happened to Rudy or Jared. He understood and didn't question Jared's need to seek his own vengeance.
Thorn had come into the cave and assessed the situation quickly. He bent over Rudy and gave him a perfunctory examination. Fortunately it was a superficial wound. He hid the fact that an inch lower and the bullet could have punctured his friend's lung. Then he stood and walked to the bodies of Wat and June Duncan, giving them no more than a glance. As he knelt beside Crazy Jack and ascertained that the hermit was still alive, Thorn immediately began giving him first aid, because his wound was more serious than Rudy's.
* * *
Jared sighed and closed his eyes as he began to speak. "I filled her in on what happened. About how we left the cave with her and me on Charger and Cr... Jack riding with Thorn. She didn't think it was near as funny as I did that you fainted and nearly fell off your horse when we crossed the river."
A grin split his tired features and Rudy scowled at him. "Yeah, well, just let me get well and we'll see how funny you think it was." Rudy knew his brother was only teasing him, and they looked at each other with mutual love.
"Lauren's worried because Thorn didn't take out the bullet in Jack's chest," Jared continued. "But I told her it was too deep, too close to vital organs."
"I think he'll be all right," Gloria said. "Especially since Thorn's taking care of the wound."
"That's what I told Lauren," Jared said. "She's glad we left him with Elena." Elena had taken the old man under her wing as would a loving mother. With black eyes flashing and braids dancing as she tossed her head, she said, "He'll never go back to live in that old cave. I promise that on my daughter's grave!" Lauren had laughed when Jared told her about Elena's pledge.
He didn't tell her about the misery he had suffered. When they had brought her battered body back to Keypoint, he had been beside himself.
Her torn clothing and the blood splattered on her torso and face could only hint at the horrors she had been through. She had shown no outward signs of injury, but kept vomiting even in her unconscious state. Dr. Graham was dispatched, but it wasn't until noon that he finally arrived. He had diagnosed a concussion and showed them the lump on her scalp underneath her dark hair. He had also discovered that the lady was pregnant, but confided this information only to Gloria after taking one look at Jared's haunted face. Besides, the news about his mother would also be devastating to this young man. He feared for Jared's sanity.
But Jared surprised them all and received the news of his mother's death with grim resignation. The shock of seeing Lauren near death so absorbed him that it cushioned all other blows.
Dr. Graham followed Jared back to Keypoint after the expedient and private funeral for Olivia and Carson in Coronado. He was disturbed to find that Lauren hadn't yet regained consciousness, but said that he had done all he could.
"She'll have to come out of this by herself and in her own good time. I just hope she's... well, she received quite a blow on the head."
He left them with those grim words. That had been five days ago. Jared had barely left the room since then, pacing back and forth like a caged animal and snapping at anyone who even suggested that he rest.
Now, as he sat with his head hanging low over his knees, Gloria and Rudy felt compassion for him. He had come so close to losing the woman he had only recently realized that he loved.
He spoke in a low voice. "She remembered shooting Kurt, and it was horrible for her. She feels so damned guilty." He couldn't meet their eyes. "She said that he didn't... uh... do anything to her except... touch her. And Duncan... God! Will she ever be normal again?" He threw his head back and clenched his bared teeth, tormented by thoughts of the way Lauren had been mistreated. By everyone. By him.
"Did you tell her about Olivia and Carson?" Gloria asked.
Jared sighed wearily. "Yes. I wanted to spare her the details, but she wouldn't let me." He was still amazed that even in her own pain, she had consoled him for his mother's death. She had reached out pale, thin fingers and placed them against his lips as he told her about the grisly sight that Rosa had found in the office.
"Do you know what she said?" he asked them rhetorically. "She said that Mother was to be pitied. How can she be so generous after everything Mother said and did to her?" He shook his head in wonder. His voice was choked when he added, "She said that Mother had loved me in the only way she could, that it wasn't in her nature to love sacrificially."
Gloria was touched by his desire to believe that. "I think Lauren is probably right, Jared," she said.
"Yeah, well." He cleared his throat and sat up straighter, trying to control the emotions that were so near the surface. "She was shocked to hear about Parker leaving Austin when he found out about Kurt's death. Then I explained about the shady deals he was pulling off with some of his railroad buddies, getting kickbacks and so forth. She didn't know that all these months I've been skirting Vandiver and his yes-men and going to the top executives of the TransPlains Railroad and working out my own deals with them. Vandiver barely crossed the state line into Oklahoma Territory before he could be caught. I haven't finished with him yet."
"What about the railroad?" Gloria asked.
"Oh, we'll get it. On schedule, too." He smiled. "I told Lauren the bank would be subsidizing Kendrick and helping him expand his power plant's operation without jeopardizing anyone's water supply. I even promised to initiate some more community improvements in Pueblo." He laughed. "I feel like the Salvation Army."
"Jared, you jackass. Why did you let her think you were in cahoots with Vandiver?" Rudy was thoroughly disgusted at his brother's lack of openness with his wife.
"Because if we became too close, Mother and the Vandivers would have become suspicious. After I fell..." He blushed uncharacteristically. "After I fell in love with Lauren, I had to turn my back on her. Otherwise, they would have known I wasn't on their side and I would have been powerless to fight them. She had heard our conversation on the porch that night, Rudy, and thought I was talking about their plan to raid Pueblo."
"She didn't know you were planning to take a virtual army of _vaqueros_ and ranchers and storm the Vandiver office complex in Austin. Thank God I was able to talk you out of that!"
"She thought all these midnight rides of mine were for meetings with the toughs that Vandiver had hired and not the men we had gathered up to ambush that gang when they rode into Pueblo." His lips curled into a half-smile. "She even had poor Pepe bury boxes of ammunition, thinking it was for the mercenaries. Pepe, of course, was sworn to secrecy about my organizing Pueblo's defense. As soon as she went back to the house, he had to dig the bullets up again." He laughed softly, then stared at the floor and shook his head. "I just hope we'll be able to reconcile our differences after all that's happened."
"Lauren is strong, Jared. She has shown a lot of fortitude by putting up with you for so long." Gloria went to her brother-in-law and hugged him. "Go on now, and _please_ take a bath. And rest. I'll wake you later."
He looked down at her and then at his brother propped up on pillows in the wide bed. He smiled boyishly as he asked, "Does Rudy know about the baby?"
His brother winked at him broadly. "Yes, I told him," Gloria said and patted Jared on the back as he slumped out of the room.
* * *
Gloria was true to her word and awakened Jared when she had seen to Lauren's needs. But by the time Jared had dressed and gone into the bedroom, she was asleep again. He didn't have the heart to awaken her, but sat quietly in the chair beside the bed and watched her as she slept.
The next morning, Lauren awoke feeling much improved. Her brain wasn't fuzzy, even though she now realized the full weight of the past events. She felt stronger and had a much more active appetite.
"Well, that's no surprise," said Gloria when Lauren mentioned it. "You have lived on sweetened tea for almost a week now."
Rudy came in to visit her. She was so happy to see him alive, she felt tears gathering in her eyes. Embarrassed by her emotion, he made light of his injury, not wanting to remind her of the horrors of Crazy Jack's cave.
In the course of the day, everyone came to see her but Jared. Pride kept her from asking about him. She knew she shouldn't have expected him. He always deserted her. She should be accustomed to it by now.
Gloria was furious with her brother-in-law for riding out that morning before anyone else was up and about. She tried to ignore the hurt she read in Lauren's eyes.
Something quite unexpected did cheer her, however. The clear notes of her piano were instantly recognizable, no matter how discordantly the keys were being pounded upon by the children.
"My piano!" she exclaimed.
"Yes," Gloria said. "Jared had it brought out here. He's closed up the house in Coronado for a while. He knew you'd want it. Rosa is with Elena, helping her care for Jack. The stables are now under Pepe's supervision. Jared also has another surprise for you, but I'll let him tell you about it."
Lauren's eyes closed and Gloria asked softly, "How do you really feel?"
"I'm all right. Truly. But I'm worried about my baby."
"I've told you a thousand times, the doctor assured me that it was fine. He wants to see you in a couple of weeks to make sure that the baby is growing, but you didn't lose it. Please believe me. I wouldn't lie to you."
Lauren took Gloria's hand. "I know you wouldn't. I was just so afraid." She plucked at the bedspread with her other hand. "The baby may be all I have left," she mumbled.
* * *
Jared returned late. He took Charger to the stables and commissioned Pepe to take care of him. Then he went into the bunkhouse and washed and changed clothes. He had been riding all day and felt grimy.
When he was clean and brushed, he went into the house. With a fleeting hello to Rudy and Gloria, who sat in the living room surrounded by their children, he went directly to his bedroom.
The room was dim; only one lamp on the vanity across from the bed had been lit. In the shadows, he saw Lauren leaning against the fresh linens that Gloria had put on the bed.
Her hair had been brushed until it shone like the wings of a raven. She wore a snowy nightgown, the scooped neckline molding to the gentle rise of her breast. Her skin, dusted by talc after a sponge bath, shone with the iridescence of a pearl.
Jared closed the door quietly behind him and walked toward the bed, thinking she might be asleep. But as he drew nearer, he saw that her eyes were open and that she was watching him.
"Hello, Jared." An emotional whisper was all she could manage.
"Lauren." He asked her permission to sit down by raising his eyebrows, and she accommodated him by sliding over toward the middle of the bed. He lowered himself next to her and studied her face, taking in every lovely detail. Lauren had never seen such a tender expression on his face. Even during moments of passion, she had never captured so much... love... in his eyes.
"Today I rode out to Pecan Creek and looked over the place where I want to build the house. The way Rudy's family keeps multiplying. I think they will squeeze us out before too long." He took her hand. "I'm not sure if we can have the house completed by the time the baby comes, but we'll be able to provide our son—or daughter—with some kind of roof. I paced off a few of the rooms, so we can start construction right away." She looked so surprised that he said hurriedly, "If you like the site and the house plans, that is."
She stuttered, "I... I know I will, but are you sure you want me to stay?"
"Want you to stay?" he asked, genuinely puzzled. "What the hell are you talking about?"
"Well, you never... The marriage agreement and all..."
"Lauren, darling, when I thought I might lose you, I went crazy. I had never told you how much I love you. If you had died without knowing what you mean to my life, I..." He shrugged helplessly. Tracing her cheekbone lightly with his finger, he said, grinning, "Of course, you're twenty thousand dollars poorer than you could be. Am I worth it?"
"I'll reserve judgment. But it may take me forty or fifty years to decide."
He pinched her earlobe lightly. "I think I started loving you when you marched so proudly out of the house wearing that ridiculous riding getup." She too, laughed, at her own foolishness. "But you were haunting me even before then. I've no doubt that if Mother hadn't speeded up the process, I would have married you just to get you into my bed. Exactly as Ben had expected me to. He knew me pretty well." He smiled.
"Then you loved me just for my body."
"Well, it was a start," he said mischievously, his eyes glowing topaz. He kissed her then, sweetly and gently, on her mouth.
"Thank you for bringing out my piano," she said softly as she caressed his eyebrows with her finger.
"You're quite welcome."
"What's my other present?" she asked slyly.
He quirked one of the eyebrows so recently smoothed. "I can see that Gloria is great at keeping secrets." Lauren laughed. "How would you like a palomino mare for your very own? She's the color of honey and has a white mane and tail. Big brown eyes. Charger's hotter'n hell." He chuckled. "Come to think of it, I'm not sure I should entrust you with another horse."
Lauren was stricken. "Oh, Jared—"
Her distress was obvious and he was immediately sorry for his tactlessness. "I was only teasing. Truly. The accidents weren't your fault." He outlined the veins on the back of her hand with his thumb. "About Flame, Lauren, I—"
"No need." She covered his mouth with her fingers. "I understand you now."
"I almost died that night I slapped you. It was an accident. I—"
"I know all of that, Jared."
Silently they stared at each other, each dangerously close to tears.
"You're beautiful," he murmured. He hooked his thumbs under the straps of her nightgown and pulled it down over her breasts and lower until it bunched around her hips. He noticed the changes in her breasts brought on by motherhood and marveled at them. Her waist was still trim. He encircled it with his hands, his thumbs caressing her navel. Then he lowered his head and laid it on her stomach. His hair tickled, but she didn't move. She ran her fingers through the unruly waves.
"My baby," he whispered, and kissed her abdomen with an emotion deeper than passion.
"I love you, Jared," she vowed.
He raised his eyes to hers and smiled ruefully. "You said that to me months ago, and I more or less ignored it." His voice became gruff. "I'm listening now."
"Then come to me."
He seemed surprised. "You're sure? Duncan and Vandiver in the cave..."
"Was something else entirely. I want you."
He rose and undressed quickly. When he was settled under the blankets, he pulled her close and kissed her deeply. His hands traveled over her body, reacquainting themselves with the curves and textures he had missed for so long.
While still capable of rational thought, Lauren placed a cautious hand on the thick mat of hair covering his chest. "Jared, will it hurt the baby? Should we wait—"
"There's no way. No way," he said as he pressed her down deeper into the pillows, trying to capture her evasive lips.
"Jared," she said with more emphasis.
Patiently he raised his head and sighed. "Didn't I tell you once before that I would never do anything to hurt you? Would I lie about something as important as my own baby?"
She smiled teasingly. "Forgive me for thinking that at this moment your judgment might be somewhat clouded."
Her hands locked behind his neck as his mouth melded with hers. Her fingers moved over his back and ended up stroking his shoulders as he lowered himself to fondle and kiss her breasts.
"You taste so good," he murmured as his tongue flicked over her aroused nipples. "Our baby is going to be the fattest one around." He stayed at her breasts to tell her of his loneliness of the past few weeks. "I wanted you so badly, Lauren. But I had to stay away from you for your own protection." He raised his head and she saw the sincerity in his eyes. "Tell me you know that."
In answer, she took him in her hand and guided him to the gate of her womanhood. Bathing the pulsating tip with the moistness of her own loins, she led him further into the welcoming folds of her body.
He stretched himself along her length, covering her completely. "Hold me tight, Jared," she breathed.
"Entrap me, Lauren. Surround me."
The sweeping tide of loving was carrying them away when she heard the soft, husky cry which was most precious to her. "Lauren, thank you for loving me."
#
**From #1 _New York Times_ bestselling author Sandra Brown comes a gripping story of family ties and forbidden attraction.**
**Please see the next page for an exciting preview of**
**_Friction_**
# Prologue
The two stalwart highway patrolmen guarding the barricade stared at her without registering any emotion, but because of the media blitz of the past few days, she knew they recognized her and that, in spite of their implacable demeanor, they were curious to know why Judge Holly Spencer was angling to get closer to the scene of a bloodbath.
"... bullet hole to the chest..."
"... ligature marks on his wrists and ankles..."
"... half in, half out of the water..."
"... carnage..."
Those were the phrases that Sergeant Lester had used to describe the scene beyond the barricade, although he'd told her he was sparing her the "gruesome details." He'd also ordered her to clear out, go home, that she shouldn't be here, that there was nothing she could do. Then he'd ducked beneath the barricade, got into his sedan, and backed it into a three-point turn that pointed him to the crime scene.
If she didn't leave voluntarily, the pair of patrolmen would escort her away, and that would create even more of a scene. She started walking back to her car.
In the few minutes that she'd been away from it, more law enforcement and emergency personnel had converged on the area. There was a lengthening line of cars, pickups, and minivans forming along both shoulders of the narrow road on either side of the turnoff. This junction was deep in the backwoods and appeared on few maps. It was nearly impossible to find unless one knew to look for the taxidermy sign with an armadillo on it.
Tonight it had become a hot spot.
The vibe of the collected crowd was almost festive. The flashing lights of the official vehicles reminded Holly of a carnival midway. An ever-growing number of onlookers, drawn to the emergency like sharks to blood, stood in groups swapping rumors about the body count, speculating on who had died and how.
Overhearing one group placing odds on who had survived, she wanted to scream, _This isn't entertainment._
By the time she reached her car, she was out of breath, her mouth dry with anxiety. She got in and clutched the steering wheel, pressing her forehead against it so hard, it hurt.
"Drive, judge."
Nearly jumping out of her skin, she whipped her head around, gasping his name when she saw the amount of blood soaking his clothes.
The massive red stain was fresh enough to show up shiny in the kaleidoscope of flashing red, white, and blue lights around them. His eyes glinted at her from shadowed sockets. His forehead was beaded with sweat, strands of hair plastered to it.
He remained perfectly still, sprawled in the corner of the backseat, left leg stretched out along it, the toe of his blood-spattered cowboy boot pointing toward the ceiling of the car. His right leg was bent at the knee. His right hand was resting on it, holding a wicked-looking pistol.
He said, "It's not my blood."
"I heard."
Looking down over his long torso, he gave a gravelly, bitter laugh. "He was dead before he hit the ground, but I wanted to make sure. Dumb move. Ruined this shirt, and it was one of my favorites."
She wasn't fooled by either his seeming indifference or his relaxed posture. He was a sudden movement waiting to happen, his reflexes quicksilver.
Up ahead, officers had begun moving along the line of spectator vehicles, motioning the motorists to clear the area. She had to either do as he asked or be caught with him inside her car.
"Sergeant Lester told me that you'd—"
"Shot the son of a bitch? That's true. He's dead. Now drive."
# Chapter 1
_Five days earlier_
Crawford Hunt woke up knowing that this was the day he'd been anticipating for a long time. Even before opening his eyes, he felt a happy bubble of excitement inside his chest, which was instantly burst by a pang of anxiety.
It might not go his way.
He showered with customary efficiency but took a little more time than usual on personal grooming: flossing, shaving extra-close, using a blow dryer rather than letting his hair dry naturally. But he was no good at wielding the dryer, and his hair came out looking the same as it always did—unmanageable. Why hadn't he thought to get a trim?
He noticed a few gray strands in his sideburns. They, plus the faint lines at the corners of his eyes and on either side of his mouth, lent him an air of maturity.
But the judge would probably regard them as signs of hard living.
"Screw it." Impatient with his self-scrutiny, he turned away from the bathroom mirror and went into his bedroom to dress.
He had considered wearing a suit, but figured that would be going overboard, like he was trying too hard to impress the judge. Besides, the navy wool blend made him feel like an undertaker. He settled for a sport jacket and tie.
Although the small of his back missed the pressure of his holster, he decided not to carry.
In the kitchen, he brewed coffee and poured himself a bowl of cereal, but neither settled well in his nervous stomach, so he dumped them into the disposal. As the Cheerios vaporized, he got a call from his lawyer.
"You all right?" The qualities that made William Moore a good lawyer worked against him as a likable human being. He possessed little grace and zero charm, so, although he'd called to ask about Crawford's state of mind, the question sounded like a challenge to which he expected a positive answer.
"Doing okay."
"Court will convene promptly at two o'clock."
"Right. Wish it was earlier."
"Are you going into your office first?"
"Thought about it. Maybe. I don't know."
"You should. Work will keep your mind off the hearing."
Crawford hedged. "I'll see how the morning goes."
"Nervous?"
"No."
The attorney snorted with skepticism. Crawford admitted to experiencing a few butterflies.
"We've gone over it," the lawyer said. "Look everyone in the eye, especially the judge. Be sincere. You'll do fine."
Although it sounded easy enough, Crawford released a long breath. "At this point, I've done everything I can. It's now up to the judge, whose mind is probably already made up."
"Maybe. Maybe not. The decision could hinge on how you comport yourself on the stand."
Crawford frowned into the phone. "But no pressure."
"I have a good feeling."
"Better than the other kind, I guess. But what happens if I don't win today? What do I do next? Short of taking out a contract on Judge Spencer."
"Don't even think in terms of losing." When Crawford didn't respond, Moore began to lecture. "The last thing we need is for you to slink into court looking pessimistic."
"Right."
"I mean it. If you look unsure, you're sunk."
"Right."
"Go in there with confidence, certainty, like you've _already_ kicked butt."
"I've got it, okay?"
Responding to his client's testiness, Moore backed down. "I'll meet you outside the courtroom a little before two." He hung up without saying good-bye.
With hours to kill before he had to be in court, Crawford wandered through his house, checking things. Fridge, freezer, and pantry were well stocked. He'd had a maid service come in yesterday, and the three industrious women had left the whole house spotless. He tidied his bathroom and made his bed. He didn't see anything else he could improve upon.
Last, he went into the second bedroom, the one he'd spent weeks preparing for Georgia's homecoming, not allowing himself to think that from tonight forward his little girl wouldn't be spending every night under his roof.
He'd left the decorating up to the saleswoman at the furniture store. "Georgia's five years old. About to start kindergarten."
She asked, "Favorite color?"
"Pink. Second favorite, pink."
"Do you have a budget?"
"Knock yourself out."
She'd taken him at his word. Everything in the room was pink except for the creamy white headboard, chest of drawers, and vanity table with an oval mirror that swiveled between upright spindles.
He had added touches he thought Georgia would like: picture books with pastel covers featuring rainbows and unicorns and such, a menagerie of stuffed animals, a ballet tutu with glittery slippers to match, and a doll wearing a pink princess gown and gold crown. The saleswoman had assured him it was a five-year-old girl's fantasy room.
The only thing missing was the girl.
He gave the bedroom one final inspection, then left the house and, without consciously intending to, found himself driving toward the cemetery. He hadn't come since Mother's Day, when he and his in-laws had brought Georgia to visit the grave of the mother she didn't remember.
Solemnly, Georgia had laid a bouquet of roses on the grave as instructed, then had looked up at him and asked, "Can we go get ice cream now, Daddy?"
Leaving his parents-in-law to pay homage to their late daughter, he'd scooped Georgia into his arms and carried her back to the car. She'd squealed whenever he pretended to stumble and stagger under her weight. He figured Beth wouldn't take exception. Wouldn't she rather have Georgia laughing over an ice cream cone than crying over her grave?
Somehow, it seemed appropriate to visit today, although he came empty-handed. He didn't see what difference a bouquet of flowers would make to the person underground. As he stood beside the grave, he didn't address anything to the spirit of his dead wife. He'd run out of things to say to her years ago, and those verbal purges never made him feel any better. They sure as hell didn't benefit Beth.
So he merely stared at the date etched into the granite headstone and cursed it, cursed his culpability, then made a promise to whatever cosmic puppeteer might be listening that, if given custody of Georgia, he would do everything within his power to make amends.
* * *
Holly checked her wristwatch as she waited on the ground floor of the courthouse for the elevator. When it arrived and the door slid open, she stifled a groan at the sight of Greg Sanders among those onboard.
She stood aside and allowed everyone to get off. Sanders came only as far as the threshold, but there he stopped, blocking her from getting on.
"Well, Judge Spencer," he drawled. "Fancy bumping into you. You can be the first to congratulate me."
She forced a smile. "Are congratulations in order?"
He placed his hand on the door to prevent it from closing. "I just came from court. The verdict in the Mallory case? Not guilty."
Holly frowned. "I don't see that as cause for celebration. Your client was accused of brutally beating a convenience store clerk during the commission of an armed robbery. The clerk lost an eye."
"But my client didn't rob the store."
"Because he panicked and ran when he thought he'd beaten the clerk to death." She was familiar with the case, but since the defending attorney, Sanders, was her opponent in the upcoming election for district court judge, the trial had been assigned to another court.
Greg Sanders, flashed his self-satisfied smirk. "The ADA failed to prove his case. My client—"
Holly interrupted. "You've already argued the case at trial. I wouldn't dream of asking you to retry it for me here and now. If you'll excuse me?"
She sidestepped him into the elevator. He got out, but kept his hand against the door. "I'm chalking up wins. Come November..." He winked. "The big win."
"I'm afraid you're setting yourself up for a huge disappointment." She punched the elevator button for the fifth floor.
"This time 'round, you won't have Judge Waters shoehorning you in."
They were monopolizing one of three elevators. People were becoming impatient, shooting them dirty looks. Besides the fact they were inconveniencing others, she wouldn't be goaded into defending either herself or her mentor to Greg Sanders. "I'm due in court in fifteen minutes. Please let go of the door."
By now, Sanders was fighting the automation to keep it open. Speaking for her ears alone, he said, "Now what would a pretty young lawyer like you have been doing for ol' Judge Waters to get him to go to bat for you with the governor?"
The "pretty" was belittling, not complimentary.
She smiled, but with exasperation. "Really, Mr. Sanders? If you're resorting to innuendos suggesting sexual impropriety between the revered Judge Waters and me, you must be feeling terribly insecure about a successful outcome in November." Without a "please" this time, she enunciated, "Let go of the door."
He raised his hands in surrender and backed away. "You'll mess up. Matter of time." The door closed on his grinning face.
Holly entered her chambers to find her assistant, Mrs. Debra Briggs, eating a carton of yogurt at her desk. "Want one?"
"No thanks. I just had a face-to-face exchange with my opponent."
"If that won't spoil your appetite, nothing will. He reminds me of an old mule that my grandpa had when I was a kid."
"I can see the resemblance. Long face, big ears, toothy smile."
"I was referring to the other end of the mule."
Holly laughed. "Messages?"
"Marilyn Vidal has called twice."
"Get back to her and tell her I'm due in court. I'll call her after this hearing."
"She won't like being put off."
Marilyn, the powerhouse orchestrating her campaign, could be irritatingly persistent. "No, she won't, but she'll get over it."
Holly went into her private office and closed the door. She needed a few minutes alone to collect herself before the upcoming custody hearing. The encounter with Sanders—and she hated herself for this—had left her with an atypical uneasiness. She was confident that she could defeat him at the polls and retain the judgeship to which she'd been temporarily appointed.
But as she zipped herself into her robe, his parting shot echoed through her mind like a dire prediction.
* * *
Crawford?"
Having arrived early, he'd been trying to empty his mind of negative thoughts while staring through the wavy glass of a fourth-floor window of the venerable Prentiss County Courthouse.
His name brought him around. Grace and Joe Gilroy were walking toward him, their expressions somber, as befitted the reason for their being there.
"Hi, Grace."
His mother-in-law was petite and pretty, with eyes through which her sweet disposition shone. The outside corners tilted up slightly, a physical trait that Beth had inherited. He and Grace hugged briefly.
As she pulled back, she gave him an approving once-over. "You look nice."
"Thanks. Hello, Joe."
He released Grace and shook hands with Beth's dad. Joe's hobby was carpentry, which had given him a row of calluses at the base of his fingers. Indeed, everything about Joe Gilroy was tough for a man just past seventy.
"How are you doing?" he asked.
Crawford forced himself to smile. "Great."
Joe appeared not to believe the exaggeration, but he didn't comment on it. Nor did he return Crawford's smile.
Grace said, "I guess we're all a little nervous." She hesitated, then asked Crawford if he was feeling one way or the other about the hearing.
"You mean whether I'll win or lose?"
She looked pained. "Please don't think of the outcome in terms of winning or losing."
"Don't you?"
"We only want what's best for Georgia," Joe said. Interpreted, that meant it would be best for her to remain with them. "I'm sure that's what Judge Spencer wants, too."
Crawford held his tongue and decided to save his debate for the courtroom. Talking it over with them now was pointless and could only lead to antagonism. The simple fact was that today he and his in-laws were on opposing sides of a legal issue, the outcome of which would profoundly affect all of them. Somebody was going to leave the courthouse defeated and unhappy. Crawford wouldn't be able to congratulate them if the judge ruled in their favor, and he wasn't about to wish them luck. He figured they felt much the same way toward him.
Since both parties had agreed to leave Georgia out of the proceedings entirely, Crawford asked Grace what arrangements she'd made for her while they were in court. "She's on a play date with our neighbor's granddaughter. She was so excited when I dropped her off. They're going to bake cookies."
Crawford winced. "Her last batch were a little gooey in the center."
"She always takes them out of the oven too soon," Joe said.
Crawford smiled. "She can't wait to sample them."
"She needs to learn the virtue of patience."
In order to maintain his smile, Crawford had to clench his teeth. His father-in-law was good at getting in barbs like that, aimed at Crawford's character flaws. That one had been a zinger. Also well timed. Before Crawford could respond, the Gilroys' attorney stepped off the elevator. They excused themselves to confer with him.
Within minutes Crawford's attorney arrived. Bill Moore's walk was as brisk as his manner. But today his determined stride was impeded by dozens of potential jurors who had crowded into the corridor looking for their assigned courtroom.
The attorney plowed his way through them, connected with Crawford, and together they went into Judge Spencer's court.
The bailiff, Chet Barker, was a courthouse institution. He was a large man with a gregarious nature to match his size. He greeted Crawford by name. "Big day, huh?"
"Yeah it is, Chet."
The bailiff slapped him on the shoulder. "Good luck."
"Thanks."
Crawford's butt barely had time to connect with the seat of his chair before Chet was asking everyone to rise. The judge entered the courtroom, stepped onto the podium, and sat down in the high-backed chair that Crawford uneasily likened to a throne. In a way, it was. Here, the honorable Judge Holly Spencer had absolute rule.
Chet called court into session and asked everyone to be seated.
"Good afternoon," the judge said. She asked the attorneys if all parties were present, and when the formalities were out of the way, she clasped her hands on top of the lectern.
"Although I took over this case from Judge Waters, I've familiarized myself with it. As I understand the situation, in May of 2010, Grace and Joe Gilroy filed for temporary custody of their granddaughter, Georgia Hunt." She looked at Crawford. "Mr. Hunt, you did not contest that petition."
"No, Your Honor, I did not."
William Moore stood up. "If I may, Your Honor?"
She nodded.
In his rat-a-tat fashion, the lawyer stated the major components of Crawford's petition to regain custody and summarized why it was timely and proper that Georgia be returned to him. He ended by saying, "Mr. Hunt is her father. He loves her, and his affection is returned, as two child psychologists attest. I believe you have copies of their evaluations of Georgia?"
"Yes, and I've reviewed them." The judge gazed thoughtfully at Crawford, then said, "Mr. Hunt will have a chance to address the court, but first I'd like to hear from the Gilroys."
Their lawyer sprang to his feet, eager to get their objections to Crawford's petition on the record. "Mr. Hunt's stability was brought into question four years ago, Your Honor. He gave up his daughter without argument, which indicates that he knew his child would be better off with her grandparents."
The judge held up her hand. "Mr. Hunt has conceded that it was in Georgia's best interest to be placed with them at that time."
"We hope to persuade the court that she should remain with them." He called Grace to testify. She was sworn in. Judge Spencer gave her a reassuring smile as she took her seat in the witness box.
"Mrs. Gilroy, why are you and Mr. Gilroy contesting your son-in-law's petition to regain custody?"
Grace wet her lips. "Well, ours is the only home Georgia has known. We've dedicated ourselves to making it a loving and nurturing environment for her." She expanded on the healthy home life they had created.
Judge Spencer finally interrupted. "Mrs. Gilroy, no one in this courtroom, not even Mr. Hunt, disputes that you've made an excellent home for Georgia. My decision won't be determined by whether or not you've provided well for the child, but whether or not Mr. Hunt is willing and able to provide an equally good home for her."
"I know he loves her," Grace said, sending an uneasy glance his way. "But love alone isn't enough. In order to feel secure, children need constancy, routine. Since Georgia doesn't have a mother, she needs the next best thing."
"Her daddy." Crawford's mutter drew disapproving glances from everyone, including the judge.
Bill Moore nudged his arm and whispered, "You'll have your turn."
The judge asked Grace a few more questions, but the upshot of what his mother-in-law believed was that to remove Georgia from their home now would create a detrimental upheaval in her young life. She finished with, "My husband and I feel that a severance from us would have a damaging impact on Georgia's emotional and psychological development."
To Crawford the statement sounded scripted and rehearsed, something their lawyer had coached Grace to say, not something that she had come up with on her own.
Judge Spencer asked Crawford's attorney if he had any questions for Mrs. Gilroy. "Yes, Your Honor, I do." He strode toward the witness box and didn't waste time on pleasantries. "Georgia often spends weekends with Mr. Hunt, isn't that right?"
"Well, yes. Once we felt she was old enough to spend a night away from us, and that Crawford was... was _trustworthy_ enough, we began allowing him to keep her overnight. Sometimes two nights."
"When she's returned to you after these sleepovers with her father, what is Georgia like?"
"Like?"
"What's her state of mind, her general being? Does she run to you crying, arms outstretched, grateful to be back? Does she act intimidated, fearful, or traumatized? Is she ever in a state of emotional distress? Is she withdrawn and uncommunicative?"
"No. She's... fine."
"Crying _only_ when her father returns her to you. Isn't that right?"
Grace hesitated. "She sometimes cries when he drops her off. But only on occasion. Not every time."
"More often crying after a lengthier visit with him," the attorney said. "In other words, the longer she's with him, the greater her separation anxiety when she's returned to you." He saw that the Gilroys' lawyer was about to object and waved him back into his seat. "Conclusion on my part."
He apologized to the judge, but Crawford knew he wasn't sorry for having gotten his point across and on the record.
He addressed another question to Grace. "When was the last time you saw Mr. Hunt intoxicated?"
"It was a while ago. I don't remember exactly."
"A week ago? A month? A year?"
"Longer than that."
"Longer than that," Moore repeated. "Four years ago? During the worst of his bereavement over the loss of his wife?"
"Yes. But—"
"To your knowledge, has Mr. Hunt ever been drunk while with Georgia?"
"No."
"Lost his temper and struck her?"
"No."
"Yelled at her, used abusive or vulgar language in front of her?"
"No."
"Failed to feed her when she was hungry?"
"No."
"Failed to secure her in her car seat? Not shown up when she was expecting him? Has he _ever_ neglected to see to his daughter's physical or emotional needs?"
Grace dipped her head and spoke softly. "No."
Moore turned to the judge and spread his arms at his sides. "Your Honor, this proceeding is an imposition on the court's time. Mr. Hunt made some mistakes, which he readily acknowledges. Over time, he's reconstructed his life. He relocated to Prentiss from Houston in order to see his daughter regularly.
"He's undergone the counseling that your predecessor mandated twelve months ago. A year hasn't diminished his determination to regain custody of his child, and I submit that, except for their own selfish interests, there are no grounds whatsoever for Mr. and Mrs. Gilroy to be contesting my client's petition."
The Gilroys' lawyer surged to his feet. "Your Honor, my clients' grounds for contesting this petition are in the file. Mr. Hunt has proved himself to be unfit—"
"I have the file, thank you," Judge Spencer said. "Mrs. Gilroy, please step down. I'd like to hear from Mr. Hunt now."
Grace left the witness stand looking distraught, as though she had miserably failed their cause.
Crawford stood up, smoothed down his necktie, and walked to the witness box. Chet swore him in. Crawford sat down and looked at the judge—in the eye, as Moore had coached him to do.
"Mr. Hunt, four years ago some of your behavior brought your ability to be a good parent into question."
"Which is why I didn't contest Joe and Grace being awarded temporary custody of Georgia. She was only thirteen months old when Beth died. She needed constant care, which circumstances prevented me from providing. My obligations at work, other issues."
" _Serious_ other issues."
That wasn't a question. He kept his mouth shut.
The judge flipped through several official looking papers and ran her finger down one sheet. "You were arrested and pled guilty to DUI."
"Once. But I—"
"You were arrested for public indecency and—"
"I was urinating."
"—assault."
"It was a bar fight. Everyone who threw a punch was detained. I was released without—"
"I have the file."
He sat there seething, realizing that his past would devastate his future. Judge Holly Spencer was cutting him no slack. After giving him a long, thoughtful appraisal, she again shuffled through the pages of what she had referred to as his "file." He wondered how bad it looked with his transgressions spelled out in black and white. If her frown was any indication, not good.
Finally, she said, "You went to all the counseling sessions."
"Judge Waters made clear that each one was mandatory. All twenty-five of them. I made certain not to miss any."
"The therapist's report is comprehensive. According to her, you made remarkable progress."
"I think so. I _know_ so."
"I commend your diligence Mr. Hunt, and I admire your commitment to regaining custody of the daughter you obviously love."
_Here it comes_ , he thought.
"However—"
The door at the back of the courtroom burst open and a figure straight out of a horror movie ran up the center aisle, handgun extended. The first bullet struck the wall behind the witness box, splitting the distance between Crawford and Judge Spencer.
The second one got the bailiff Chet Barker square in the chest.
# A Letter from the Author
Dear Reader:
Early in my career I wrote two books under the pseudonym Laura Jordan. The name had no significance other than that I liked the sound of it! However, the books themselves were significant, each in its own way.
Prior to _The Silken Web_ , I had written only romances for various series, where word count was specified to fit a particular format. I had no such restrictions with _The Silken Web_ , so it became my first "long" book.
A year or so after it was published, the editor Star Helmer told me she had heard through the grapevine that I had written a western romance, set in Texas around the turn of the twentieth century. I had, but since I was focusing on contemporary romances, I had never submitted the manuscript.
Ms. Helmer asked to see it. I took it off the shelf, spent a month or so rewriting and revising, and sent it to her. She bought it. I've published only four books with a historical setting. _Hidden Fires_ was the first of them.
The commonality of _The Silken Web_ and _Hidden Fires_ ends with their having been originally published under the Laura Jordan pen name and both are love stories. I hope you enjoy them!
—Sandra Brown
# About the Author
Sandra Brown is the author of sixty-three _New York Times_ bestsellers. There are over 80 million copies of her books in print worldwide, and her work has been translated into thirty-four languages. She lives in Texas. For more information you can visit www.SandraBrown.net.
# Look for These Thrilling Sandra Brown Novels!
_Friction_
_Mean Streak_
_Low Pressure_
_Lethal_
_Mirror Image_
_Where There's Smoke_
_Charade_
_Exclusive_
_Envy_
_The Switch_
_The Crush_
_Fat Tuesday_
_Unspeakable_
_The Witness_
_The Alibi_
_Standoff_
_Best Kept Secrets_
_Breath of Scandal_
_French Silk_
# **LOOK FOR SANDRA BROWN'S THRILLERS**
### Thank you for buying this ebook, published by Hachette Digital.
To receive special offers, bonus content, and news about our latest ebooks and apps, sign up for our newsletters.
Sign Up
Or visit us at hachettebookgroup.com/newsletters
# Contents
1. Cover
2. Title Page
3. Welcome
4. Chapter 1
5. Chapter 2
6. Chapter 3
7. Chapter 4
8. Chapter 5
9. Chapter 6
10. Chapter 7
11. Chapter 8
12. Chapter 9
13. Chapter 10
14. Chapter 11
15. Chapter 12
16. Chapter 13
17. Chapter 14
18. Chapter 15
19. Chapter 16
20. Chapter 17
21. Chapter 18
22. Chapter 19
23. Chapter 20
24. Chapter 21
25. Chapter 22
26. Chapter 23
27. Chapter 24
28. Chapter 25
29. Chapter 26
30. A Preview of _Friction_
31. A Note from the Author
32. About the Author
33. Also by Sandra Brown
34. You Might Also Like...
35. Newsletters
36. Copyright
# Navigation
1. Begin Reading
2. Table of Contents
# Copyright
This book is a work of fiction. Names, characters, places, and incidents are the product of the author's imagination or are used fictitiously. Any resemblance to actual events, locales, or persons, living or dead, is coincidental.
Copyright © 1982 by Sandra Brown
Excerpt from _Friction_ copyright © 2015 by Sandra Brown Management, Ltd.
Cover design by Faceout Studio
Cover copyright © 2015 by Hachette Book Group, Inc.
All rights reserved. In accordance with the U.S. Copyright Act of 1976, the scanning, uploading, and electronic sharing of any part of this book without the permission of the publisher constitute unlawful piracy and theft of the author's intellectual property. If you would like to use material from the book (other than for review purposes), prior written permission must be obtained by contacting the publisher at permissions@hbgusa.com. Thank you for your support of the author's rights.
Grand Central Publishing
Hachette Book Group
1290 Avenue of the Americas
New York, NY 10104
hachettebookgroup.com
twitter.com/grandcentralpub
First published in mass market paperback by Warner Books
First ebook edition: December 2015
Grand Central Publishing is a division of Hachette Book Group, Inc.
The Grand Central Publishing name and logo is a trademark of Hachette Book Group, Inc.
The publisher is not responsible for websites (or their content) that are not owned by the publisher.
The Hachette Speakers Bureau provides a wide range of authors for speaking events. To find out more, go to www.hachettespeakersbureau.com or call (866) 376-6591.
ISBN 978-1-4555-4632-9
E3
| {
"redpajama_set_name": "RedPajamaBook"
} | 4,333 |
Аустис () — коммуна в Италии, располагается в регионе Сардиния, в провинции Нуоро.
Население составляет 787 человек (30-06-2019), плотность населения составляет 15,49 чел./км². Занимает площадь 51 км². Почтовый индекс — 8030. Телефонный код — 0784.
В коммуне 15 августа особо празднуется Успение Пресвятой Богородицы.
Демография
Динамика населения:
Администрация коммуны
Официальный сайт: http://www.comune.austis.nu.it/
Ссылки
Официальный сайт населённого пункта
Национальный институт статистики
Национальный институт статистики
Примечания | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 461 |
Q: Trying to send and receive data using telnet in C++ but received data is truncated. Why? I'm using the follow function to send and receive data using telnet. But the received data is truncated past a few hundred characters. Why is this happening and how can I fix this? I suspect, the data is not being sent "fast enough" from the other end and my code thinks it's the end of the telnet stream. If this is the case, how can I make the receive wait long enough?
std::string sendTelnet(std::string str)
{
const int default_buflen = 512;
const char *loopback_address = "127.0.0.1";
std::string retval;
WSADATA wsaData;
SOCKET ConnectSocket = INVALID_SOCKET;
struct addrinfo *result = NULL,
*ptr = NULL,
hints;
char recvbuf[default_buflen];
int iResult;
int recvbuflen = default_buflen;
for(int i=0; i<default_buflen; i++)
{
recvbuf[i] = '\0';
}
// Initialize Winsock
iResult = WSAStartup(MAKEWORD(2,2), &wsaData);
if (iResult != 0) {
std::cerr<<"WSAStartup failed with error: " + iResult;
}
ZeroMemory( &hints, sizeof(hints) );
hints.ai_family = AF_UNSPEC;
hints.ai_socktype = SOCK_STREAM;
hints.ai_protocol = IPPROTO_TCP;
// Resolve the server address and port
iResult = getaddrinfo(loopback_address, std::to_string((long double)jtagTerminalPort).c_str(), &hints, &result);
if ( iResult != 0 ) {
std::cerr<<"getaddrinfo failed with error: " + iResult;
WSACleanup();
}
// Attempt to connect to an address until one succeeds
for(ptr=result; ptr != NULL ;ptr=ptr->ai_next) {
// Create a SOCKET for connecting to server
ConnectSocket = socket(ptr->ai_family, ptr->ai_socktype,
ptr->ai_protocol);
if (ConnectSocket == INVALID_SOCKET) {
std::cerr<<"socket failed with error: " + WSAGetLastError();
WSACleanup();
}
// Connect to server.
iResult = connect( ConnectSocket, ptr->ai_addr, (int)ptr->ai_addrlen);
if (iResult == SOCKET_ERROR) {
closesocket(ConnectSocket);
ConnectSocket = INVALID_SOCKET;
continue;
}
break;
}
freeaddrinfo(result);
if (ConnectSocket == INVALID_SOCKET) {
std::cerr<<"Unable to connect to server!\n";
WSACleanup();
}
// Send an initial buffer
iResult = send( ConnectSocket, str.c_str(), (int)strlen(str.c_str()), 0 );
if (iResult == SOCKET_ERROR) {
std::cerr<<"send failed with error: " + WSAGetLastError();
closesocket(ConnectSocket);
WSACleanup();
}
//telnet_bytes_sent = iResult;
// Receive until the peer closes the connection
do {
iResult = recv(ConnectSocket, recvbuf, recvbuflen, 0);
if ( iResult > 0 )
{
//telnet_bytes_received = iResult;
}
} while( iResult == 0 );
// shutdown the connection since no more data will be sent
iResult = shutdown(ConnectSocket, SD_SEND);
if (iResult == SOCKET_ERROR) {
std::cerr<<"shutdown failed with error: " + WSAGetLastError();
closesocket(ConnectSocket);
WSACleanup();
}
// cleanup
closesocket(ConnectSocket);
WSACleanup();
retval = recvbuf;
Sleep(100);
return retval;
}
PS. What I mean by the data being truncated is that retval returns a truncated version of what is actually being sent over telnet. For example if the data sent over from the other end is "1,2,3,...,1000", by the end of the function, retval will only be holding "1,2,3,...,300".
EDIT: In application, I'm sending some commands over telnet and waiting for the response. Then, I take the received response and run some functions accordingly. Then I send the next set of commands over telnet, and wait for the response. And so on...
A: Your do/while loop condition is wrong. It should be while (iResult > 0). But you should rewrite it as a while loop, to avoid the double test.
A: // Receive until the peer closes the connection
do {
iResult = recv(ConnectSocket, recvbuf, recvbuflen, 0);
if ( iResult > 0 )
{
//telnet_bytes_received = iResult;
}
} while( iResult == 0 );
This is a really convoluted way to read 512 bytes from your socket (or block forever if it's been closed).
This should look something like this:
// Receive until the peer closes the connection
while ((iResult = recv(ConnectSocket, recvbuf, recvbuflen, 0)) > 0)
{
retval.append(recvbuf, iResult);
}
This will read from the socket until it's closed or an error occurs, and append the received data to your return string each time through.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 6,489 |
\section{Introduction}
\label{sec:Introduction}
Ethereum has recently undergone a major change in its protocol, successfully passing from proof-of-work to proof-of-stake. The change underpins an entirely new consensus protocol, which brings Byzantine fault-tolerance to Ethereum. The main design goal is to keep using a Nakamoto-style consensus, i.e., a protocol that constantly creates blocks in a tree-like form and selects a branch as the current chain using a fork-chain rule. However, a new mechanism (called finality gadget) incrementally finalizes blocks in the chain as opposed to pure Nakamoto-style consensus. A \textit{finalized block} is a block that is voted by at least two-thirds of validators\footnote{To become a validator, one needs to 'stake' an amount of 32 Eth (the native cryptocurrency of the blockchain).}. In a system with less than one-third of Byzantine validators, a finalized block is never revoked.
Interestingly, this design aims at guaranteeing, at the same time, the availability of the chain and consistency (uniqueness) of a finalized prefix. Note that classical BFT consensus protocols re-adapted to blockchains such as Tendermint \cite{buchman_latest_2018} for the Cosmos blockchain \cite{kwon_cosmos_2016} or Tenderbake \cite{astefanoaei_tenderbake_2021} for Tezos blockchains \cite{goodman_tezos_2014}, finalize one block at the time: for each height of the blockchain only one block is ever added. On the other hand, Ethereum PoS builds a common prefix, but the suffix can change: for a given height of the blockchain, different blocks can be seen at that height over time. The advantage of this approach is to always make progress, regardless of Byzantine behavior and network partitions, while classical BFT consensus protocols stop producing blocks during asynchronous periods and attacks.
Ethereum Proof-of-Stake tries then to provide consistency without renouncing availability. This seems to be in striking contrast with the CAP theorem \cite{gilbert_brewer_2002}, which states that it is impossible to guarantee progress and consistency in the case of network partitions.
The caveat here is that Ethereum maintains a data structure where the prefix is consistent and finalized only when possible, while the suffix can grow without being consistent. The resulting protocol, however, is quite involved. For this reason, we aim to provide a formal ground for analysis of the Ethereum Proof-of-Stake (PoS) protocol.
We first provide a novel formalization of the Ethereum Proof-of-Stake blockchain properties, which are a combination of properties of Nakamoto-style blockchains and BFT-style ones. We then provide a formalization of the protocol itself through pseudo-code. Formalisation allows to analyse the protocol in terms of its properties.
In this paper, we focus on the liveness of the finalization process, where liveness is the capacity to always finalize new blocks. Our analysis reveals a new possible attack on the protocol's liveness.
Indeed previous work has already pointed out the risks of a so-called \textit{bouncing attack} on liveness \cite{nakamura_analysis_2019}. A bouncing attack is an attack that prevents the chain from being finalized because the main chain selected through the fork chain rule continually bounces between two alternative branches. After the attack identification, the Ethereum community responded by implementing a patch to the protocol to prevent this attack. The patch aims at mitigating bouncing by forcing validators to stick to a chain after a while. Interestingly, we managed to find a bouncing attack on the patched version of the protocol. We found that the bouncing can be repeated over time but with decreasing probability of success. This shows that the liveness of the patched protocol is probabilistic. Note that the attack is plausible in a Byzantine environment since it only relies on the Byzantine validator's capacity to withhold votes and to release them at the right time to make honest validators change their mind on the chosen chain.
The paper is organised as follows: Section \ref{sec:systemModel} elaborates on the system model while Section \ref{sec:blockchainProperties} defines the properties essential to our formalization. In Section \ref{sec:ethereumConsensusProtocol}, we explain and formalize the Ethereum PoS protocol providing all the materials necessary for its understanding, including pseudo-code. Section \ref{sec:livenessAttack} presents a new liveness attack still possible in the current version of the Ethereum PoS protocol. We present the related works in Section \ref{sec:relatedWorks}, and we conclude in Section \ref{sec:conclusion}.
\section{System model}
\label{sec:systemModel}
We consider a system composed of a finite set $\Pi$ of processes called \emph{validators}\footnote{At the implementation level, validators are the processes with ETH staked that allow them to vote as part of the consensus protocol.}. There are a total of $n$ validators.
Each validator has an associated public/private key pair for signing and can be identified by its public key.
We assume that digital signatures cannot be forged.
Validators have synchronized clocks\footnote{Clocks can be offset by at most $\tau$,
this way, the offset can be captured as part of the network delay.}.
Time is measured by periods of 12 seconds called \emph{slot}s, a period of 32 slots is called an \emph{epoch}.
\paragraph*{Network} Processes communicate by message passing.
We assume the existence of an underlying broadcast primitive, which is a best effort broadcast. This means that when a correct process broadcasts a value, all the correct processes eventually deliver it. Messages are created with a digital signature
We assume a \emph{partially synchrony model} \cite{dwork_consensus_1988}, where after some unknown Global Stabilization Time (\texttt{GST}), the system becomes synchronous, and there is a finite known bound $\Delta$ on the message transfer delay.
Note that even if we have synchronized clocks, having an asynchronous network before \texttt{GST} still makes the system partially synchronous.
\paragraph*{Fault Model}
Validators can be \emph{correct} or \emph{Byzantine}. Correct validators follow the protocol,
while Byzantines ones may arbitrarily deviate from the protocol\footnote{Since in this paper we are only interested in the Consensus protocol, we only characterize validator's behavior. For clients submitting transactions, as in any blockchain, we assume they can be Byzantine.}.
We denote by $f$ the number of Byzantine validators, with $f<n/3$.
\section{Blockchain properties}
\label{sec:blockchainProperties}
In our analysis, we will continuously use the term Ethereum Proof-of-Stake as \cite{schwarz_three_2021} to name the new protocol of Ethereum\footnote{Other appellation such as Ethereum 2.0 or Consensus Layer can be found, we have chosen to stick with Ethereum Proof-of-Stake.}. To begin our analysis of Ethereum Proof-of-Stake, we start by defining the terms and properties that we will investigate.
Similarly to \cite{anceaume_abstract_2018} we formalize the data structure implemented by blockchain as a \emph{BlockTree}.
Indeed the blockchain takes the form of a tree in which every node is a block pointing to its unique parent, and the tree's root is the \emph{genesis block}.
Among the different branches of the BlockTree, the protocol indicates a unique branch, or chain, to build upon with a so-called fork choice rule (e.g., the longest chain rule in Bitcoin).
The selected chain is called the \emph{candidate chain}.
\begin{definition}[\textbf{Candidate chain}]
We call \textbf{candidate chain} the chain designated as the one to build upon by the fork choice rule.
Considering the view of the chain of an honest validator $i$, $i$'s associated candidate chain is noted $C_i$.
\end{definition}
The blocks in the candidate chain can be finalized or not.
\begin{definition}[\textbf{Finalized block}]
A block is finalized for a validator $i$ if and only if the block cannot be revoked, i.e., it permanently belongs to the candidate chain $C_i$ .
\end{definition}
\emph{Note:} It stems from the definition that all the predecessors of a finalized block are finalized
\begin{definition}[\textbf{Finalized chain}]
The finalized chain is the chain constituted of all the finalized blocks.
\end{definition}
\emph{Note:} The finalized chain $C_{fi}$ is always a prefix of any candidate chain $C_i$.
To analyze the protocol, one needs to analyze the capability of the Ethereum Proof-of-Stake protocol to construct a consistent blockchain (safety), to allow validators to add blocks despite network partitions and failures (availability), and to make constant progress on the finalization of new blocks (liveness). Safety, availability, and liveness are expressed as follows:
\begin{definition}[\textbf{Safety}]
A blockchain is consistent or \textbf{safe}
if for any two correct validators $i$ and $j$, having a finalized chain $C_{fi}$ and $C_{fj}$, respectively, then $C_{fi}$'s is the prefix of $C_{fj}$ or viceversa
\end{definition}
\begin{definition}[\textbf{Availability}]
A blockchain is \textbf{available} if the following two conditions hold: (1) any correct validator is able to append a block to the candidate chain in bounded time, regardless of the failures of other validators and the network partitions; (2) the candidate chain is eventually growing, i.e., given a block $b_k$ added to its candidate chain at a distance $d$ from the genesis block $b_0$, where the distance is the number of blocks separating $b_k$ from $b_0$, then eventually a block $b_l$ will be added to the candidate chain at a distance $d'>d$.
\end{definition}
\begin{definition}[\textbf{Liveness}]
A blockchain is \textbf{live} if the finalized chain is ever growing.
\end{definition}
The fundamental difference between the finalized and the candidate chain lies in the fact that blocks of the finalized chain can never be revoked, while the candidate chain can change from one branch to another in the tree so that a suffix of blocks of the previously selected branch might be revoked.
Availability, on the other hand, guarantees that adding blocks to the candidate chain is a wait-free operation whose time to complete does not depend on network failures or Byzantine behaviors. Availability also implies that blocks are constantly added in such as way that the height of the candidate chain eventually grows. This property avoids the pathological scenario in which all the blocks are added to the genesis block to form a star.
As in any distributed system, blockchains are faced with the dilemma brought by the CAP-Theorem \cite{gilbert_brewer_2002}. This theorem states that distributed systems cannot satisfy these three properties at the same time: \emph{consistency} \emph{availability}, and \emph{partition tolerance}. Indeed, if network partitions occur, either the system remains available at the expense of consistency, or it stops making progress until the network partition is resolved to guarantee consistency.
This means that no blockchain can simultaneously be available and safe. However, by maintaining the candidate and the finalized chain at the same time, Ethereum Proof-of-stake aims to offer both safety and availability. The candidate chain aims to be available but without guaranteeing consistency all the time, while the finalized chain falls on the other side of the spectrum, guaranteeing consistency without availability. Therefore, the finalized chain will finalize blocks only when it is safe to do so where the candidate chain will still be available during network partitions (caused by network failures or attacks). The only caveat here is that finalized chain grows by finalizing blocks of the candidate chain, which means that the properties of the two chains are interdependent. In particular to assure liveness it is necessary that the candidate chain steadily grows. This interdependence is a source of vulnerability that must be thoroughly analyzed.
\section{Ethereum Proof-of-Stake Protocol}
\label{sec:ethereumConsensusProtocol}
\subsection{Overview}
\label{subsec:overview}
The Ethereum Proof-of-Stake (PoS) protocol design is quite involved. We identify, similarly to \cite{neu_ebb_2021}, the objectives underlying its design as follows: (i) finalizing blocks and
(ii) having an available candidate chain that does not rely on block finality to grow. To this end, the Ethereum Proof-of-Stake protocol combines two blockchain designs: a Nakamoto-style protocol to build the tree of blocks containing the transactions and a BFT finalization protocol to progressively finalize blocks in the tree. The objective is to keep the blockchain creation process always available while guaranteeing the finalization of blocks through Byzantine-tolerant voting mechanisms. The finalization mechanism is a \emph{Finality Gadget} called \emph{Casper FFG}, and the fork choice rule to select candidate chains is \emph{LMD GHOST}.
Before introducing how the fork choice rule and the finality gadget work together, we will introduce the following basic concepts: (i) slots, epochs, and checkpoints, which set the pace of the protocol allowing validators to synchronize together on the different steps, (ii) committees formation and assignment of roles to validators as proposers and voters for each slot, and (iii) the different types of votes the validators must send in order to grow and maintain the candidate chain as well as the finalized chain.
In this section, we focus on providing a formal version of the protocol following the specification given by the Ethereum Foundation \cite{github_specs}. Every implementation of the protocol must be compliant with the specification. Note that a description of an initial plan of the protocol is proposed by Buterin and al. in \cite{buterin_combining_2020}. In this paper, we describe and formalize, through pseudo-code, the current implementation of the protocol.
\subsubsection{Slots, Epochs \& Checkpoints}
\label{subsubsec:time}
In proof of work protocols, such as originally in \cite{nakamoto_peer_2008}, the average frequency of the block creation is predetermined in the protocol, and the mining difficulty changes to follow that pace.
On the contrary, in Ethereum Proof-of-Stake, each validator uses the hypothesis of synchronized clocks to propose blocks at regular intervals. More specifically, in the protocol, time is measured in \emph{slots} and \emph{epochs}. A slot lasts 12 seconds.
Slots are assigned with consecutive numbers; the first slot is slot $0$.
Slots are encapsulated in \emph{epochs}. An epoch is composed of 32 slots, thus lasting 6 minutes and 24 seconds. The first epoch (epoch 0) contains from slot 0 to slot 31; then epoch 1 contains slot 32 to 63, and so on. These slots and epochs allow associating the validators' roles to the corresponding time frame.
An essential feature of epochs is the \emph{checkpoint}.
A checkpoint is a pair block-epoch $(b,e)$ where $b$ is the block of the first slot\footnote{In the event of an epoch without a block for the first slot, the block used for the checkpoint is the last block in the candidate chain, belonging then to a previous epoch.
On the contrary, if the proposer of the first slot proposes multiple blocks, this will make multiple checkpoints for the other validators to choose from using the fork choice rule.
} of epoch $e$.
\subsubsection{Validators \& Committees}
\label{subsubsec:validator&committee}
Validators have two main roles: \emph{proposer} and \emph{attester}. The proposer's role consists in proposing a block during a specific slot\footnote{The current protocol specifications \cite{github_specs} indicate that correct validators should send their block proposition during the first third of their designated slot.}. This role is pseudo-randomly given to 32 validators by epoch (one for each slot).
The attester's role consists in producing an attestation sharing the validator's view of the chain. This role is given once by epoch to each validator.
During every epoch, each validator is assigned to one committee.
A committee $C_j$ is a set of validators.
One validator belongs to exactly one committee, i.e., $\forall j\neq k, C_j \bigcap C_k = \emptyset $. Each committee is associated with a slot. During this slot, each member of the committee will have to perform an \emph{attestation} to indicate its view of the chain.
In short, during an epoch, validators are all attesters once and have a small probability of being proposers ($32/n$).
\subsubsection{Vote \& Attestation}
\label{subsubsec:vote}
There are two types of votes in Ethereum Proof-of-Stake, the \emph{block vote}\footnote{Also called GHOST vote.} and the \emph{checkpoint vote}\footnote{Also called FFG vote.}. The message containing these two votes is called \emph{attestation}.
During an epoch, each validator must make one attestation. The attestation ought to be sent during a specific slot. This slot depends on the committee of which the validator is a member.
The two types of votes,
checkpoint vote and block vote, have very distinct purposes. The checkpoint vote is used to finalize blocks to grow the finalized chain. The block vote is used to determine the candidate chain
Although validators cast their two types of votes in one attestation, an important distinction must be made between the two. Indeed, the two types of votes do not require the same condition to be taken into account. The checkpoint vote of an attestation is only considered when the attestation is included in a block. In contrast, the block vote is considered one slot after its emission, whether it is included in a block or not.
The code associated with the production of attestations is described in \autoref{algo:prepareAttestation} at \autoref{subsec:code}. We then describe in \autoref{algo:syncAttestation} how the reception of attestations is handled.
\subsubsection{Finality Gadget}
\label{subsubsec:finalityGadget}
The finality gadget is the mechanism that aims at finalizing blocks. The finality gadget grows the finalized chain disregarding the block production. This decoupling of the finality mechanism and the block production permits block availability even when the finalizing process is slowed down. This differs from protocols like Tendermint \cite{buchman_latest_2018}, where a new block can be added to the chain only after being finalized.
The finality gadget works at the level of epochs. Instead of finalizing blocks one by one, the protocol uses checkpoint votes to finalize entire epochs.
We now present in more detail how the finality gadget of Ethereum PoS grows the finalized chain.
Recall that to be taken into account, a checkpoint vote needs to be included in a block. The vote will then influence the behavior of validators regarding this particular branch.
Thus, in \autoref{algo:Casper} of \autoref{subsec:code} the function \texttt{countMatchingCheckpointVote} only count the matching checkpoint vote of attestation included in a block. More specifically, this function takes a source and a target as an argument and only count checkpoint vote with matching source and target.
\paragraph*{Justification}
The justification process is a step to achieve finalization\footnote{The genesis checkpoint (i.e., the checkpoint of the first epoch) is the exception to this rule: it is justified and finalized by definition.}. It operates on checkpoints at the level of epochs.
Justification occurs thanks to checkpoint votes. The checkpoint vote contains a pair of checkpoints: the checkpoint \emph{source} and the checkpoint \emph{target}. We can count with \texttt{countMatchingCheckpointVote} the sum of balances of the validator's checkpoint votes with the same source and target. If validators controlling more than two-thirds of the stake make the same checkpoint vote, then we say there is a \emph{supermajority link} from the checkpoint source to the checkpoint target. The checkpoint target of a supermajority link is said to be \emph{justified}.
More formally, a checkpoint vote is in the form of a pair of checkpoints: $\big((a,e_a),(b,e_b)\big)$, also noted $(a,e_a) \xrightarrow{} (b,e_b)$. For the checkpoint vote $(a,e_a) \xrightarrow{} (b,e_b)$ we call $(a,e_a)$ the checkpoint source and $(b,e_b)$ the checkpoint target. The checkpoint source is necessarily of an earlier epoch than the checkpoint target, i.e., $e_a<e_b$.
In line with \cite{buterin_combining_2020} we say there is a \emph{supermajority link} from checkpoint $(a,e_a)$ to checkpoint $(b,e_b)$ if validators controlling more than two-thirds of the stake cast an attestation
with checkpoint vote $(a,e_a) \xrightarrow{} (b,e_b)$.
In this case, we write $(a,e_a) \xrightarrow{\texttt{J}} (b,e_b)$ and the checkpoint $(b,e_b)$ is justified.
\paragraph*{Finalization}
The finalization process aims at finalizing checkpoints, thus growing the finalized chain. Checkpoints need to be justified before being finalized. Let us illustrate the finalization process with the two scenarios that can lead to finalization. The first case presents the main scenario in the synchronous setting. It shows how a checkpoint can be finalized in two epochs, the least amount of epochs needed for finalization.
\subparagraph*{Case 1:} The scenario is depicted in Figure \ref{fig:finalizationCase1}.
\begin{enumerate}
\item Let $A=(a,e)$ and $B=(b,e+1)$ be checkpoints of two consecutive epochs such that $A=(a,e)$ is justified.
\item A supermajority link occurs between checkpoints $A$ and $B$ where $A$ is the source and $B$ the target. This justifies checkpoint $B$.
Hence, we can write: $(a, e) \xrightarrow{\texttt{J}} (b, e+1)$ or equivalently $A \xrightarrow{\texttt{J}} B$.
\item This leads to $A$ being finalized.
\end{enumerate}
\begin{figure}[th]
\centering
\resizebox{.4\linewidth}{!}{
\begin{tikzpicture}[
finalizednode/.style={regular polygon, regular polygon sides=6,, draw=black, fill=green, minimum size=7mm}, finalizednode2/.style={regular polygon, regular polygon sides=6,, draw=black, minimum size=5.5mm},
justifiednode/.style={regular polygon, regular polygon sides=6,, draw=black, minimum size=7mm},
justifiednode2/.style={regular polygon, regular polygon sides=6,, draw=black, minimum size=5.5mm},
checkpointnode/.style={regular polygon, regular polygon sides=6,, draw=black, minimum size=7mm},
squarednode/.style={regular polygon, regular polygon sides=6,, draw=gray!60, fill=gray!5, minimum size=5mm},
]
\node (Dots1) {$\cdots$};
\node[justifiednode] (A) [right of= Dots1] {$A$};
\node[justifiednode2] (A2) [right of= Dots1] {};
\node[checkpointnode] (B) [right of= A] {$B$};
\node (Dots2) [right of= B] {$\cdots$};
\draw[-] (A.east) -- (B.west);
\draw[-] (Dots1.east) -- (A.west);
\draw[-] (B.east) -- (Dots2.west);
\end{tikzpicture}
}
\hspace{0cm}
\resizebox{.4\linewidth}{!}{
\begin{tikzpicture}[
finalizednode/.style={regular polygon, regular polygon sides=6,, draw=black, fill=green, minimum size=7mm}, finalizednode2/.style={regular polygon, regular polygon sides=6,, draw=black, minimum size=5.5mm},
justifiednode/.style={regular polygon, regular polygon sides=6,, draw=black, minimum size=7mm}, justifiednode2/.style={regular polygon, regular polygon sides=6,, draw=black, minimum size=5.5mm},
checkpointnode/.style={regular polygon, regular polygon sides=6,, draw=black, minimum size=7mm},
squarednode/.style={regular polygon, regular polygon sides=6,, draw=gray!60, fill=gray!5, minimum size=5mm},
]
\node (Dots1) {$\cdots$};
\node[justifiednode] (A) [right of= Dots1] {$A$};
\node[justifiednode2] (A2) [right of= Dots1] {};
\node[justifiednode] (B) [right of= A] {$B$};
\node[justifiednode2] (B2) [right of= A] {};
\node (Dots2) [right of= B] {$\cdots$};
\draw[-] (A.east) -- (B.west);
\draw[-] (Dots1.east) -- (A.west);
\draw[-] (B.east) -- (Dots2.west);
\draw[->,double] (A.north) to[out=45] (B.north);
\end{tikzpicture}
}
\hspace{0cm}
\resizebox{.4\linewidth}{!}{
\begin{tikzpicture}[
finalizednode/.style={regular polygon, regular polygon sides=6,, draw=black, fill=green, minimum size=7mm},
finalizednode2/.style={regular polygon, regular polygon sides=6,, draw=black, minimum size=5.5mm},
justifiednode/.style={regular polygon, regular polygon sides=6,, draw=black, minimum size=7mm}, justifiednode2/.style={regular polygon, regular polygon sides=6,, draw=black, minimum size=5.5mm},
checkpointnode/.style={regular polygon, regular polygon sides=6,, draw=black, minimum size=7mm},
]
\node (Dots1) {$\cdots$};
\node[finalizednode] (A) [right of= Dots1] {$A$};
\node[finalizednode2] (A2) [right of= Dots1] {};
\node[justifiednode] (B) [right of= A] {$B$};
\node[justifiednode2] (B2) [right of= A] {};
\node (Dots2) [right of= B] {$\cdots$};
\draw[-] (A.east) -- (B.west);
\draw[-] (Dots1.east) -- (A.west);
\draw[-] (B.east) -- (Dots2.west);
\draw[->,double] (A.north) to[out=45] (B.north);
\end{tikzpicture}
}
\caption{\small The figure depicts the finalization scenario of \textbf{case 1} with the 3 steps from top to bottom. We represent a checkpoint with a hexagon,
a justified checkpoint with a double hexagon,
and a finalized checkpoint with double hexagon coloured.
The arrow
between two checkpoints indicates a supermajority link.}
\label{fig:finalizationCase1}
\end{figure}
The second case illustrates the scenario in which two consecutive checkpoints are justified but not finalized. This means that the current highest justified checkpoint (e.g., $B$ in Figure \ref{fig:finalizationCase2}) was not justified with a supermajority link having the previous checkpoint $A$ as its source. Then occurs a new justification with the source and target being at the maximum distance (2 epochs) for the source to become finalized. An important note is that there is no limit on the distance between two checkpoints for justification to be possible. This limit only exists for finalization.
\subparagraph*{Case 2: The scenario is depicted in \autoref{fig:finalizationCase2}.}
\begin{enumerate}
\item Let $A=(a,e)$, $B=(b,e+1)$ and $C=(c, e+2)$ be checkpoints of consecutive epochs such that $A$ and $B$ are justified. There is no supermajority link between $A$ and $B$, $A$ cannot be finalized as in Case 1 above.
\item Now, a supermajority link occurs between checkpoints $A$ and $C$ where $A$ is the source and $C$ the target. This justifies checkpoint $C$, i.e., $A \xrightarrow{\texttt{J}} C$.
\item This leads to $A$ being finalized.
\end{enumerate}
\begin{figure}[th]
\centering
\resizebox{.5\linewidth}{!}{
\begin{tikzpicture}[
finalizednode/.style={regular polygon, regular polygon sides=6,, draw=black, fill=green, minimum size=7mm}, finalizednode2/.style={regular polygon, regular polygon sides=6,, draw=black, minimum size=5.5mm},
justifiednode/.style={regular polygon, regular polygon sides=6,, draw=black, minimum size=7mm}, justifiednode2/.style={regular polygon, regular polygon sides=6,, draw=black, minimum size=5.5mm},
checkpointnode/.style={regular polygon, regular polygon sides=6,, draw=black, minimum size=7mm},
squarednode/.style={regular polygon, regular polygon sides=6,, draw=gray!60, fill=gray!5, minimum size=5mm},
]
\node (Dots1) {$\cdots$};
\node[justifiednode] (A) [right of= Dots1] {$A$};
\node[justifiednode2] (A2) [right of= Dots1] {};
\node[justifiednode] (B) [right of= A] {$B$};
\node[justifiednode2] (B2) [right of= A] {};
\node[checkpointnode] (C) [right of= B] {$C$};
\node (Dots2) [right of= C] {$\cdots$};
\draw[-] (A.east) -- (B.west);
\draw[-] (Dots1.east) -- (A.west);
\draw[-] (B.east) -- (C.west);
\draw[-] (C.east) -- (Dots2.west);
\end{tikzpicture}
}
\hspace{0cm}
\resizebox{.5\linewidth}{!}{
\begin{tikzpicture}[
finalizednode/.style={regular polygon, regular polygon sides=6,, draw=black, fill=green, minimum size=7mm}, finalizednode2/.style={regular polygon, regular polygon sides=6,, draw=black, minimum size=5.5mm},
justifiednode/.style={regular polygon, regular polygon sides=6,, draw=black, minimum size=7mm}, justifiednode2/.style={regular polygon, regular polygon sides=6,, draw=black, minimum size=5.5mm},
checkpointnode/.style={regular polygon, regular polygon sides=6,, draw=black, minimum size=7mm},
squarednode/.style={regular polygon, regular polygon sides=6,, draw=gray!60, fill=gray!5, minimum size=5mm},
]
\node (Dots1) {$\cdots$};
\node[justifiednode] (A) [right of= Dots1] {$A$};
\node[justifiednode] (B) [right of= A] {$B$};
\node[justifiednode] (C) [right of= B] {$C$};
\node[justifiednode2] (A2) [right of= Dots1] {};
\node[justifiednode2] (B2) [right of= A] {};
\node[justifiednode2] (C2) [right of= B] {};
\node (Dots2) [right of= C] {$\cdots$};
\draw[-] (A.east) -- (B.west);
\draw[-] (Dots1.east) -- (A.west);
\draw[-] (B.east) -- (C.west);
\draw[-] (C.east) -- (Dots2.west);
\draw[->,double] (A.north) to[out=45] (C.north);
\end{tikzpicture}
}
\hspace{0cm}
\resizebox{.5\linewidth}{!}{
\begin{tikzpicture}[
finalizednode/.style={regular polygon, regular polygon sides=6,, draw=black, fill=green, minimum size=7mm}, finalizednode2/.style={regular polygon, regular polygon sides=6,, draw=black, minimum size=5.5mm},
justifiednode/.style={regular polygon, regular polygon sides=6,, draw=black, minimum size=7mm}, justifiednode2/.style={regular polygon, regular polygon sides=6,, draw=black, minimum size=5.5mm},
checkpointnode/.style={regular polygon, regular polygon sides=6,, draw=black, minimum size=7mm},
squarednode/.style={regular polygon, regular polygon sides=6,, draw=gray!60, fill=gray!5, minimum size=5mm},
]
\node (Dots1) {$\cdots$};
\node[finalizednode] (A) [right of= Dots1] {$A$};
\node[finalizednode2] (A2) [right of= Dots1] {};
\node[justifiednode] (B) [right of= A] {$B$};
\node[justifiednode] (C) [right of= B] {$C$};
\node[justifiednode2] (B2) [right of= A] {};
\node[justifiednode2] (C2) [right of= B] {};
\node (Dots2) [right of= C] {$\cdots$};
\draw[-] (A.east) -- (B.west);
\draw[-] (Dots1.east) -- (A.west);
\draw[-] (B.east) -- (C.west);
\draw[-] (C.east) -- (Dots2.west);
\draw[->,double] (A.north) to[out=45] (C.north);
\end{tikzpicture}
}
\caption{\small The figure depicts the finalization scenario of \textbf{case 2} with the 3 steps from top to bottom. We represent a checkpoint with a hexagon,
a justified checkpoint with double hexagon,
and a finalized checkpoint with a coloured double hexagon.
The arrow
between two checkpoints indicates a supermajority link.}
\label{fig:finalizationCase2}
\end{figure}
These two cases illustrate the fact that for a checkpoint to become finalized, it needs to be the source of a supermajority link between justified checkpoints. Once a checkpoint is finalized, all the blocks leading to it (including the block in the pair constituting the checkpoint) become finalized.
We now describe the condition for a checkpoint to be finalized more formally. Let $(a,e_A)$ and $(b,e_B)$ be two checkpoints such that $e_a<e_b$ and $e_b-e_a\leq 2$. \footnote{This last condition necessitating the two checkpoints to be at most 2 epochs away from each other is also called \emph{2-finality} \cite{buterin_combining_2020}.}
The checkpoint $(a,e_A)$ is finalized if the following conditions are respected:
\begin{itemize}
\item Checkpoint $(a,e_A)$ is justified.
\item There exists a supermajority link $(a,e_a) \xrightarrow{\texttt{J}} (b,e_b)$.
\item If $e_b-e_a=2$ then the checkpoint in between at epoch $e_a+1$($=e_b-1$) must be justified.
\end{itemize}
The importance of the last condition is illustrated by the \autoref{fig:notFinalizationCase}.
In practice, as explained in \cite{buterin_combining_2020}, for the checkpoint $(A,e_A)$ to become finalized, it must not be more than 4 epochs old. At the implementation level, checkpoints more than 4 epochs old are not considered for finalization. This is illustrated by the last 4 conditions of \autoref{algo:Casper} in \autoref{subsec:code}.
\begin{figure}[th]
\centering
\resizebox{.5\linewidth}{!}{
\begin{tikzpicture}[
finalizednode/.style={regular polygon, regular polygon sides=6,, draw=black, fill=green, minimum size=7mm}, finalizednode2/.style={regular polygon, regular polygon sides=6,, draw=black, minimum size=5.5mm},
justifiednode/.style={regular polygon, regular polygon sides=6, draw=black, minimum size=7mm},
justifiednode2/.style={regular polygon, regular polygon sides=6,, draw=black, minimum size=5.5mm},
checkpointnode/.style={regular polygon, regular polygon sides=6, draw=black, minimum size=7mm},
squarednode/.style={regular polygon, regular polygon sides=6, draw=gray!60, fill=gray!5, minimum size=5mm},
]
\node (Dots1) {$\cdots$};
\node[justifiednode] (A) [right of= Dots1] {$A$};
\node[justifiednode2] (A2) [right of= Dots1] {};
\node[checkpointnode] (B) [right of= A] {$B$};
\node[checkpointnode] (C) [right of= B] {$C$};
\node (Dots2) [right of= C] {$\cdots$};
\draw[-] (A.east) -- (B.west);
\draw[-] (Dots1.east) -- (A.west);
\draw[-] (B.east) -- (C.west);
\draw[-] (C.east) -- (Dots2.west);
\end{tikzpicture}
}
\hspace{0cm}
\resizebox{.5\linewidth}{!}{
\begin{tikzpicture}[
finalizednode/.style={regular polygon, regular polygon sides=6,, draw=black, fill=green, minimum size=7mm}, finalizednode2/.style={regular polygon, regular polygon sides=6,, draw=black, minimum size=5.5mm},
justifiednode/.style={regular polygon, regular polygon sides=6,, draw=black, minimum size=7mm},
justifiednode2/.style={regular polygon, regular polygon sides=6,, draw=black, minimum size=5.5mm},
checkpointnode/.style={regular polygon, regular polygon sides=6,, draw=black, minimum size=7mm},
squarednode/.style={regular polygon, regular polygon sides=6,, draw=gray!60, fill=gray!5, minimum size=5mm},
]
\node (Dots1) {$\cdots$};
\node[justifiednode] (A) [right of= Dots1] {$A$};
\node[justifiednode2] (A2) [right of= Dots1] {};
\node[checkpointnode] (B) [right of= A] {$B$};
\node[justifiednode] (C) [right of= B] {$C$};
\node[justifiednode2] (C2) [right of= B] {};
\node (Dots2) [right of= C] {$\cdots$};
\draw[-] (A.east) -- (B.west);
\draw[-] (Dots1.east) -- (A.west);
\draw[-] (B.east) -- (C.west);
\draw[-] (C.east) -- (Dots2.west);
\draw[->,double] (A.north) to[out=45] (C.north);
\end{tikzpicture}
}
\caption{\small This illustrate the case of two checkpoints respecting all the conditions but the one that stipulates that . We represent a checkpoint with a hexagon
and a justified checkpoint with double hexagon
The arrow
between two checkpoints indicates a supermajority link.}
\label{fig:notFinalizationCase}
\end{figure}
\subsubsection{Fork choice rule \& Block proposition}
\label{subsubsec:forkChoiceRule&blockProposition}
The fork choice rule is the mechanism that allows each validator to determine the candidate chain depending on their view of the BlockTree and the state of checkpoints. Ethereum PoS fork choice rule is LMD GHOST. The LMD GHOST fork choice rule stems from the Greedy Heaviest-Observed Sub-Tree (GHOST) rule \cite{sompolinsky_secure_2015}
which considers only each participant's most recent vote (Latest Message Driven).
During an epoch, each validator must make one \emph{block vote} on the block considered as the head of the candidate chain according to its view.
To determine the head of the candidate chain, the fork choice rule does the following:
\begin{enumerate}
\item Go through the list of validators and check the last block vote of each.
\item For each block vote, add a weight to each block of the chain that has the block voted as a descendent. The weight added is proportional to the stake of the corresponding validator.
\item Start from the block of the justified checkpoint with the highest epoch and continue the chain by following the block with the highest weight at each connection. Return the block without any child block. This block is the head of the candidate chain.
\end{enumerate}
The actual implementation is written in \autoref{algo:GHOST} in \autoref{subsec:code}. This algorithm is similar to the one already presented in \cite{buterin_combining_2020}.
Albeit each \emph{block vote} being for a specific block, the fork choice rule considers all the chains leading to that block. This reflects the fact that a vote for a block is a vote for the chain leading to that block. The \autoref{fig:forkChoiceRuleExample} offers an explanation with an example of how attestations influence the fork choice rule. At each chain intersection, the fork choice rule favors the chain with the most attestations.
\begin{figure}[ht]
\centering
\resizebox{\linewidth}{!}{
\begin{tikzpicture}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1]
\draw [draw opacity=0][fill={rgb, 255:red, 162; green, 162; blue, 162 } ,fill opacity=1 ][blur shadow={shadow xshift=-0.75pt,shadow yshift=-0.75pt, shadow blur radius=1.5pt, shadow blur steps=4 ,shadow opacity=73}] (317,47.91) .. controls (317,46.43) and (318.2,45.22) .. (319.69,45.22) -- (330.31,45.22) .. controls (331.8,45.22) and (333,46.43) .. (333,47.91) -- (333,55.98) .. controls (333,57.46) and (331.8,58.67) .. (330.31,58.67) -- (319.69,58.67) .. controls (318.2,58.67) and (317,57.46) .. (317,55.98) -- cycle ;
\draw (325,67.94) -- (325,61.61) ;
\draw [shift={(325,58.61)}, rotate = 90] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (3.57,-1.72) -- (0,0) -- (3.57,1.72) -- cycle ;
\draw (325,106.39) -- (325,82.61) ;
\draw [shift={(325,79.61)}, rotate = 90] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (3.57,-1.72) -- (0,0) -- (3.57,1.72) -- cycle ;
\draw [color={rgb, 255:red, 0; green, 0; blue, 0 } ,draw opacity=1 ][fill={rgb, 255:red, 70; green, 137; blue, 214 } ,fill opacity=1 ] (359.94,51.89) .. controls (359.94,48.88) and (362.38,46.44) .. (365.39,46.44) .. controls (368.4,46.44) and (370.84,48.88) .. (370.84,51.89) .. controls (370.84,54.9) and (368.4,57.34) .. (365.39,57.34) .. controls (362.38,57.34) and (359.94,54.9) .. (359.94,51.89) -- cycle ;
\draw [color={rgb, 255:red, 0; green, 0; blue, 0 } ,draw opacity=1 ][fill={rgb, 255:red, 70; green, 137; blue, 214 } ,fill opacity=1 ] (362.12,51.89) .. controls (362.12,48.88) and (364.56,46.44) .. (367.57,46.44) .. controls (370.58,46.44) and (373.02,48.88) .. (373.02,51.89) .. controls (373.02,54.9) and (370.58,57.34) .. (367.57,57.34) .. controls (364.56,57.34) and (362.12,54.9) .. (362.12,51.89) -- cycle ;
\draw [dash pattern={on 0.84pt off 2.51pt}] (359.34,52.14) -- (335.97,52.08) ;
\draw [shift={(333.97,52.08)}, rotate = 0.15] [color={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.75] (4.37,-1.96) .. controls (2.78,-0.92) and (1.32,-0.27) .. (0,0) .. controls (1.32,0.27) and (2.78,0.92) .. (4.37,1.96) ;
\draw [color={rgb, 255:red, 0; green, 0; blue, 0 } ,draw opacity=1 ][fill={rgb, 255:red, 70; green, 137; blue, 214 } ,fill opacity=1 ] (360.34,92.69) .. controls (360.34,89.68) and (362.78,87.24) .. (365.79,87.24) .. controls (368.8,87.24) and (371.24,89.68) .. (371.24,92.69) .. controls (371.24,95.7) and (368.8,98.14) .. (365.79,98.14) .. controls (362.78,98.14) and (360.34,95.7) .. (360.34,92.69) -- cycle ;
\draw [color={rgb, 255:red, 0; green, 0; blue, 0 } ,draw opacity=1 ][fill={rgb, 255:red, 70; green, 137; blue, 214 } ,fill opacity=1 ] (381.1,92.69) .. controls (381.1,89.68) and (383.54,87.24) .. (386.55,87.24) .. controls (389.56,87.24) and (392,89.68) .. (392,92.69) .. controls (392,95.7) and (389.56,98.14) .. (386.55,98.14) .. controls (383.54,98.14) and (381.1,95.7) .. (381.1,92.69) -- cycle ;
\draw [dash pattern={on 0.84pt off 2.51pt}] (359.74,92.94) -- (310.54,92.94) ;
\draw [shift={(308.54,92.94)}, rotate = 360] [color={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.75] (4.37,-1.96) .. controls (2.78,-0.92) and (1.32,-0.27) .. (0,0) .. controls (1.32,0.27) and (2.78,0.92) .. (4.37,1.96) ;
\draw [dash pattern={on 0.84pt off 2.51pt}] (380.94,90.44) .. controls (372.02,86.18) and (353.31,81.16) .. (333.94,77.94) ;
\draw [shift={(332.14,77.64)}, rotate = 9.09] [color={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.75] (4.37,-1.32) .. controls (2.78,-0.56) and (1.32,-0.12) .. (0,0) .. controls (1.32,0.12) and (2.78,0.56) .. (4.37,1.32) ;
\draw [color={rgb, 255:red, 0; green, 0; blue, 0 } ,draw opacity=1 ][fill={rgb, 255:red, 70; green, 137; blue, 214 } ,fill opacity=1 ] (360.34,112.69) .. controls (360.34,109.68) and (362.78,107.24) .. (365.79,107.24) .. controls (368.8,107.24) and (371.24,109.68) .. (371.24,112.69) .. controls (371.24,115.7) and (368.8,118.14) .. (365.79,118.14) .. controls (362.78,118.14) and (360.34,115.7) .. (360.34,112.69) -- cycle ;
\draw [color={rgb, 255:red, 0; green, 0; blue, 0 } ,draw opacity=1 ][fill={rgb, 255:red, 70; green, 137; blue, 214 } ,fill opacity=1 ] (362.52,112.69) .. controls (362.52,109.68) and (364.96,107.24) .. (367.97,107.24) .. controls (370.98,107.24) and (373.42,109.68) .. (373.42,112.69) .. controls (373.42,115.7) and (370.98,118.14) .. (367.97,118.14) .. controls (364.96,118.14) and (362.52,115.7) .. (362.52,112.69) -- cycle ;
\draw [dash pattern={on 0.84pt off 2.51pt}] (359.74,112.94) -- (335.14,112.94) ;
\draw [shift={(333.14,112.94)}, rotate = 360] [color={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.75] (4.37,-1.96) .. controls (2.78,-0.92) and (1.32,-0.27) .. (0,0) .. controls (1.32,0.27) and (2.78,0.92) .. (4.37,1.96) ;
\draw (300.39,126.44) -- (300.49,102.28) ;
\draw [shift={(300.5,99.28)}, rotate = 90.24] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (3.57,-1.72) -- (0,0) -- (3.57,1.72) -- cycle ;
\draw (300.79,94.88) -- (315.54,81.28) ;
\draw [shift={(317.74,79.24)}, rotate = 137.33] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (3.57,-1.72) -- (0,0) -- (3.57,1.72) -- cycle ;
\draw [color={rgb, 255:red, 0; green, 0; blue, 0 } ,draw opacity=1 ][fill={rgb, 255:red, 70; green, 137; blue, 214 } ,fill opacity=1 ] (359.94,72.69) .. controls (359.94,69.68) and (362.38,67.24) .. (365.39,67.24) .. controls (368.4,67.24) and (370.84,69.68) .. (370.84,72.69) .. controls (370.84,75.7) and (368.4,78.14) .. (365.39,78.14) .. controls (362.38,78.14) and (359.94,75.7) .. (359.94,72.69) -- cycle ;
\draw [color={rgb, 255:red, 0; green, 0; blue, 0 } ,draw opacity=1 ][fill={rgb, 255:red, 70; green, 137; blue, 214 } ,fill opacity=1 ] (362.12,72.69) .. controls (362.12,69.68) and (364.56,67.24) .. (367.57,67.24) .. controls (370.58,67.24) and (373.02,69.68) .. (373.02,72.69) .. controls (373.02,75.7) and (370.58,78.14) .. (367.57,78.14) .. controls (364.56,78.14) and (362.12,75.7) .. (362.12,72.69) -- cycle ;
\draw [dash pattern={on 0.84pt off 2.51pt}] (359.34,72.94) -- (334.74,72.94) ;
\draw [shift={(332.74,72.94)}, rotate = 360] [color={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.75] (4.37,-1.96) .. controls (2.78,-0.92) and (1.32,-0.27) .. (0,0) .. controls (1.32,0.27) and (2.78,0.92) .. (4.37,1.96) ;
\draw [color={rgb, 255:red, 0; green, 0; blue, 0 } ,draw opacity=1 ][fill={rgb, 255:red, 70; green, 137; blue, 214 } ,fill opacity=1 ] (360.3,132.69) .. controls (360.3,129.68) and (362.74,127.24) .. (365.75,127.24) .. controls (368.76,127.24) and (371.2,129.68) .. (371.2,132.69) .. controls (371.2,135.7) and (368.76,138.14) .. (365.75,138.14) .. controls (362.74,138.14) and (360.3,135.7) .. (360.3,132.69) -- cycle ;
\draw [color={rgb, 255:red, 0; green, 0; blue, 0 } ,draw opacity=1 ][fill={rgb, 255:red, 70; green, 137; blue, 214 } ,fill opacity=1 ] (362.48,132.69) .. controls (362.48,129.68) and (364.92,127.24) .. (367.93,127.24) .. controls (370.94,127.24) and (373.38,129.68) .. (373.38,132.69) .. controls (373.38,135.7) and (370.94,138.14) .. (367.93,138.14) .. controls (364.92,138.14) and (362.48,135.7) .. (362.48,132.69) -- cycle ;
\draw [dash pattern={on 0.84pt off 2.51pt}] (359.7,132.94) -- (309.83,132.94) ;
\draw [shift={(307.83,132.94)}, rotate = 360] [color={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.75] (4.37,-1.96) .. controls (2.78,-0.92) and (1.32,-0.27) .. (0,0) .. controls (1.32,0.27) and (2.78,0.92) .. (4.37,1.96) ;
\draw (300.36,149.37) -- (300.36,143.04) ;
\draw [shift={(300.36,140.04)}, rotate = 90] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (3.57,-1.72) -- (0,0) -- (3.57,1.72) -- cycle ;
\draw [color={rgb, 255:red, 0; green, 0; blue, 0 } ,draw opacity=1 ][fill={rgb, 255:red, 70; green, 137; blue, 214 } ,fill opacity=1 ] (360.3,154.12) .. controls (360.3,151.11) and (362.74,148.67) .. (365.75,148.67) .. controls (368.76,148.67) and (371.2,151.11) .. (371.2,154.12) .. controls (371.2,157.13) and (368.76,159.57) .. (365.75,159.57) .. controls (362.74,159.57) and (360.3,157.13) .. (360.3,154.12) -- cycle ;
\draw [color={rgb, 255:red, 0; green, 0; blue, 0 } ,draw opacity=1 ][fill={rgb, 255:red, 70; green, 137; blue, 214 } ,fill opacity=1 ] (362.48,154.12) .. controls (362.48,151.11) and (364.92,148.67) .. (367.93,148.67) .. controls (370.94,148.67) and (373.38,151.11) .. (373.38,154.12) .. controls (373.38,157.13) and (370.94,159.57) .. (367.93,159.57) .. controls (364.92,159.57) and (362.48,157.13) .. (362.48,154.12) -- cycle ;
\draw [dash pattern={on 0.84pt off 2.51pt}] (359.3,154.37) -- (310.1,154.37) ;
\draw [shift={(308.1,154.37)}, rotate = 360] [color={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.75] (4.37,-1.96) .. controls (2.78,-0.92) and (1.32,-0.27) .. (0,0) .. controls (1.32,0.27) and (2.78,0.92) .. (4.37,1.96) ;
\draw [draw opacity=0][fill={rgb, 255:red, 162; green, 162; blue, 162 } ,fill opacity=1 ][blur shadow={shadow xshift=-0.75pt,shadow yshift=-0.75pt, shadow blur radius=1.5pt, shadow blur steps=4 ,shadow opacity=73}] (317,70.51) .. controls (317,69.03) and (318.2,67.82) .. (319.69,67.82) -- (330.31,67.82) .. controls (331.8,67.82) and (333,69.03) .. (333,70.51) -- (333,78.58) .. controls (333,80.06) and (331.8,81.27) .. (330.31,81.27) -- (319.69,81.27) .. controls (318.2,81.27) and (317,80.06) .. (317,78.58) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 162; green, 162; blue, 162 } ,fill opacity=1 ][blur shadow={shadow xshift=-0.75pt,shadow yshift=-0.75pt, shadow blur radius=1.5pt, shadow blur steps=4 ,shadow opacity=73}] (317,108.78) .. controls (317,107.29) and (318.2,106.09) .. (319.69,106.09) -- (330.31,106.09) .. controls (331.8,106.09) and (333,107.29) .. (333,108.78) -- (333,116.84) .. controls (333,118.33) and (331.8,119.53) .. (330.31,119.53) -- (319.69,119.53) .. controls (318.2,119.53) and (317,118.33) .. (317,116.84) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 162; green, 162; blue, 162 } ,fill opacity=1 ][blur shadow={shadow xshift=-0.75pt,shadow yshift=-0.75pt, shadow blur radius=1.5pt, shadow blur steps=4 ,shadow opacity=73}] (292.39,89) .. controls (292.39,87.51) and (293.59,86.31) .. (295.07,86.31) -- (305.7,86.31) .. controls (307.19,86.31) and (308.39,87.51) .. (308.39,89) -- (308.39,97.07) .. controls (308.39,98.55) and (307.19,99.76) .. (305.7,99.76) -- (295.07,99.76) .. controls (293.59,99.76) and (292.39,98.55) .. (292.39,97.07) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 162; green, 162; blue, 162 } ,fill opacity=1 ][blur shadow={shadow xshift=-0.75pt,shadow yshift=-0.75pt, shadow blur radius=1.5pt, shadow blur steps=4 ,shadow opacity=73}] (292.39,129.13) .. controls (292.39,127.65) and (293.59,126.44) .. (295.07,126.44) -- (305.7,126.44) .. controls (307.19,126.44) and (308.39,127.65) .. (308.39,129.13) -- (308.39,137.2) .. controls (308.39,138.69) and (307.19,139.89) .. (305.7,139.89) -- (295.07,139.89) .. controls (293.59,139.89) and (292.39,138.69) .. (292.39,137.2) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 162; green, 162; blue, 162 } ,fill opacity=1 ][blur shadow={shadow xshift=-0.75pt,shadow yshift=-0.75pt, shadow blur radius=1.5pt, shadow blur steps=4 ,shadow opacity=73}] (292.39,149.89) .. controls (292.39,148.4) and (293.59,147.2) .. (295.07,147.2) -- (305.7,147.2) .. controls (307.19,147.2) and (308.39,148.4) .. (308.39,149.89) -- (308.39,157.96) .. controls (308.39,159.44) and (307.19,160.64) .. (305.7,160.64) -- (295.07,160.64) .. controls (293.59,160.64) and (292.39,159.44) .. (292.39,157.96) -- cycle ;
\draw (117,68.94) -- (117,62.61) ;
\draw [shift={(117,59.61)}, rotate = 90] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (3.57,-1.72) -- (0,0) -- (3.57,1.72) -- cycle ;
\draw (117,107.39) -- (117,83.61) ;
\draw [shift={(117,80.61)}, rotate = 90] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (3.57,-1.72) -- (0,0) -- (3.57,1.72) -- cycle ;
\draw [color={rgb, 255:red, 0; green, 0; blue, 0 } ,draw opacity=1 ][fill={rgb, 255:red, 70; green, 137; blue, 214 } ,fill opacity=1 ] (151.94,52.89) .. controls (151.94,49.88) and (154.38,47.44) .. (157.39,47.44) .. controls (160.4,47.44) and (162.84,49.88) .. (162.84,52.89) .. controls (162.84,55.9) and (160.4,58.34) .. (157.39,58.34) .. controls (154.38,58.34) and (151.94,55.9) .. (151.94,52.89) -- cycle ;
\draw [color={rgb, 255:red, 0; green, 0; blue, 0 } ,draw opacity=1 ][fill={rgb, 255:red, 70; green, 137; blue, 214 } ,fill opacity=1 ] (154.12,52.89) .. controls (154.12,49.88) and (156.56,47.44) .. (159.57,47.44) .. controls (162.58,47.44) and (165.02,49.88) .. (165.02,52.89) .. controls (165.02,55.9) and (162.58,58.34) .. (159.57,58.34) .. controls (156.56,58.34) and (154.12,55.9) .. (154.12,52.89) -- cycle ;
\draw [dash pattern={on 0.84pt off 2.51pt}] (151.34,53.14) -- (126.74,53.14) ;
\draw [shift={(124.74,53.14)}, rotate = 360] [color={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.75] (4.37,-1.96) .. controls (2.78,-0.92) and (1.32,-0.27) .. (0,0) .. controls (1.32,0.27) and (2.78,0.92) .. (4.37,1.96) ;
\draw [color={rgb, 255:red, 0; green, 0; blue, 0 } ,draw opacity=1 ][fill={rgb, 255:red, 70; green, 137; blue, 214 } ,fill opacity=1 ] (152.34,93.69) .. controls (152.34,90.68) and (154.78,88.24) .. (157.79,88.24) .. controls (160.8,88.24) and (163.24,90.68) .. (163.24,93.69) .. controls (163.24,96.7) and (160.8,99.14) .. (157.79,99.14) .. controls (154.78,99.14) and (152.34,96.7) .. (152.34,93.69) -- cycle ;
\draw [color={rgb, 255:red, 0; green, 0; blue, 0 } ,draw opacity=1 ][fill={rgb, 255:red, 70; green, 137; blue, 214 } ,fill opacity=1 ] (173.1,93.69) .. controls (173.1,90.68) and (175.54,88.24) .. (178.55,88.24) .. controls (181.56,88.24) and (184,90.68) .. (184,93.69) .. controls (184,96.7) and (181.56,99.14) .. (178.55,99.14) .. controls (175.54,99.14) and (173.1,96.7) .. (173.1,93.69) -- cycle ;
\draw [dash pattern={on 0.84pt off 2.51pt}] (151.74,93.94) -- (102.54,93.94) ;
\draw [shift={(100.54,93.94)}, rotate = 360] [color={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.75] (4.37,-1.96) .. controls (2.78,-0.92) and (1.32,-0.27) .. (0,0) .. controls (1.32,0.27) and (2.78,0.92) .. (4.37,1.96) ;
\draw [dash pattern={on 0.84pt off 2.51pt}] (172.94,91.44) .. controls (164.02,87.18) and (145.31,82.16) .. (125.94,78.94) ;
\draw [shift={(124.14,78.64)}, rotate = 9.09] [color={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.75] (4.37,-1.32) .. controls (2.78,-0.56) and (1.32,-0.12) .. (0,0) .. controls (1.32,0.12) and (2.78,0.56) .. (4.37,1.32) ;
\draw [color={rgb, 255:red, 0; green, 0; blue, 0 } ,draw opacity=1 ][fill={rgb, 255:red, 70; green, 137; blue, 214 } ,fill opacity=1 ] (152.34,113.69) .. controls (152.34,110.68) and (154.78,108.24) .. (157.79,108.24) .. controls (160.8,108.24) and (163.24,110.68) .. (163.24,113.69) .. controls (163.24,116.7) and (160.8,119.14) .. (157.79,119.14) .. controls (154.78,119.14) and (152.34,116.7) .. (152.34,113.69) -- cycle ;
\draw [color={rgb, 255:red, 0; green, 0; blue, 0 } ,draw opacity=1 ][fill={rgb, 255:red, 70; green, 137; blue, 214 } ,fill opacity=1 ] (154.52,113.69) .. controls (154.52,110.68) and (156.96,108.24) .. (159.97,108.24) .. controls (162.98,108.24) and (165.42,110.68) .. (165.42,113.69) .. controls (165.42,116.7) and (162.98,119.14) .. (159.97,119.14) .. controls (156.96,119.14) and (154.52,116.7) .. (154.52,113.69) -- cycle ;
\draw [dash pattern={on 0.84pt off 2.51pt}] (151.74,113.94) -- (127.14,113.94) ;
\draw [shift={(125.14,113.94)}, rotate = 360] [color={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.75] (4.37,-1.96) .. controls (2.78,-0.92) and (1.32,-0.27) .. (0,0) .. controls (1.32,0.27) and (2.78,0.92) .. (4.37,1.96) ;
\draw (92.79,128.06) -- (92.79,104.39) ;
\draw [shift={(92.79,101.39)}, rotate = 90] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (3.57,-1.72) -- (0,0) -- (3.57,1.72) -- cycle ;
\draw (92.79,95.88) -- (107.54,82.28) ;
\draw [shift={(109.74,80.24)}, rotate = 137.33] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (3.57,-1.72) -- (0,0) -- (3.57,1.72) -- cycle ;
\draw [color={rgb, 255:red, 0; green, 0; blue, 0 } ,draw opacity=1 ][fill={rgb, 255:red, 70; green, 137; blue, 214 } ,fill opacity=1 ] (151.94,73.69) .. controls (151.94,70.68) and (154.38,68.24) .. (157.39,68.24) .. controls (160.4,68.24) and (162.84,70.68) .. (162.84,73.69) .. controls (162.84,76.7) and (160.4,79.14) .. (157.39,79.14) .. controls (154.38,79.14) and (151.94,76.7) .. (151.94,73.69) -- cycle ;
\draw [color={rgb, 255:red, 0; green, 0; blue, 0 } ,draw opacity=1 ][fill={rgb, 255:red, 70; green, 137; blue, 214 } ,fill opacity=1 ] (154.12,73.69) .. controls (154.12,70.68) and (156.56,68.24) .. (159.57,68.24) .. controls (162.58,68.24) and (165.02,70.68) .. (165.02,73.69) .. controls (165.02,76.7) and (162.58,79.14) .. (159.57,79.14) .. controls (156.56,79.14) and (154.12,76.7) .. (154.12,73.69) -- cycle ;
\draw [dash pattern={on 0.84pt off 2.51pt}] (151.34,73.94) -- (126.74,73.94) ;
\draw [shift={(124.74,73.94)}, rotate = 360] [color={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.75] (4.37,-1.96) .. controls (2.78,-0.92) and (1.32,-0.27) .. (0,0) .. controls (1.32,0.27) and (2.78,0.92) .. (4.37,1.96) ;
\draw [color={rgb, 255:red, 0; green, 0; blue, 0 } ,draw opacity=1 ][fill={rgb, 255:red, 70; green, 137; blue, 214 } ,fill opacity=1 ] (152.3,133.69) .. controls (152.3,130.68) and (154.74,128.24) .. (157.75,128.24) .. controls (160.76,128.24) and (163.2,130.68) .. (163.2,133.69) .. controls (163.2,136.7) and (160.76,139.14) .. (157.75,139.14) .. controls (154.74,139.14) and (152.3,136.7) .. (152.3,133.69) -- cycle ;
\draw [color={rgb, 255:red, 0; green, 0; blue, 0 } ,draw opacity=1 ][fill={rgb, 255:red, 70; green, 137; blue, 214 } ,fill opacity=1 ] (154.48,133.69) .. controls (154.48,130.68) and (156.92,128.24) .. (159.93,128.24) .. controls (162.94,128.24) and (165.38,130.68) .. (165.38,133.69) .. controls (165.38,136.7) and (162.94,139.14) .. (159.93,139.14) .. controls (156.92,139.14) and (154.48,136.7) .. (154.48,133.69) -- cycle ;
\draw [dash pattern={on 0.84pt off 2.51pt}] (151.7,133.94) -- (101.83,133.94) ;
\draw [shift={(99.83,133.94)}, rotate = 360] [color={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.75] (4.37,-1.96) .. controls (2.78,-0.92) and (1.32,-0.27) .. (0,0) .. controls (1.32,0.27) and (2.78,0.92) .. (4.37,1.96) ;
\draw (203.63,93) -- (218.85,93) -- (229,100.25) -- (218.85,107.5) -- (203.63,107.5) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 162; green, 162; blue, 162 } ,fill opacity=1 ][blur shadow={shadow xshift=-0.75pt,shadow yshift=-0.75pt, shadow blur radius=1.5pt, shadow blur steps=4 ,shadow opacity=73}] (109,49.24) .. controls (109,47.76) and (110.2,46.56) .. (111.69,46.56) -- (122.31,46.56) .. controls (123.8,46.56) and (125,47.76) .. (125,49.24) -- (125,57.31) .. controls (125,58.8) and (123.8,60) .. (122.31,60) -- (111.69,60) .. controls (110.2,60) and (109,58.8) .. (109,57.31) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 162; green, 162; blue, 162 } ,fill opacity=1 ][blur shadow={shadow xshift=-0.75pt,shadow yshift=-0.75pt, shadow blur radius=1.5pt, shadow blur steps=4 ,shadow opacity=73}] (109,71.24) .. controls (109,69.76) and (110.2,68.56) .. (111.69,68.56) -- (122.31,68.56) .. controls (123.8,68.56) and (125,69.76) .. (125,71.24) -- (125,79.31) .. controls (125,80.8) and (123.8,82) .. (122.31,82) -- (111.69,82) .. controls (110.2,82) and (109,80.8) .. (109,79.31) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 162; green, 162; blue, 162 } ,fill opacity=1 ][blur shadow={shadow xshift=-0.75pt,shadow yshift=-0.75pt, shadow blur radius=1.5pt, shadow blur steps=4 ,shadow opacity=73}] (109,109.91) .. controls (109,108.43) and (110.2,107.22) .. (111.69,107.22) -- (122.31,107.22) .. controls (123.8,107.22) and (125,108.43) .. (125,109.91) -- (125,117.98) .. controls (125,119.46) and (123.8,120.67) .. (122.31,120.67) -- (111.69,120.67) .. controls (110.2,120.67) and (109,119.46) .. (109,117.98) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 162; green, 162; blue, 162 } ,fill opacity=1 ][blur shadow={shadow xshift=-0.75pt,shadow yshift=-0.75pt, shadow blur radius=1.5pt, shadow blur steps=4 ,shadow opacity=73}] (84.79,90.24) .. controls (84.79,88.76) and (85.99,87.56) .. (87.47,87.56) -- (98.1,87.56) .. controls (99.59,87.56) and (100.79,88.76) .. (100.79,90.24) -- (100.79,98.31) .. controls (100.79,99.8) and (99.59,101) .. (98.1,101) -- (87.47,101) .. controls (85.99,101) and (84.79,99.8) .. (84.79,98.31) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 162; green, 162; blue, 162 } ,fill opacity=1 ][blur shadow={shadow xshift=-0.75pt,shadow yshift=-0.75pt, shadow blur radius=1.5pt, shadow blur steps=4 ,shadow opacity=73}] (84.79,130.13) .. controls (84.79,128.65) and (85.99,127.44) .. (87.47,127.44) -- (98.1,127.44) .. controls (99.59,127.44) and (100.79,128.65) .. (100.79,130.13) -- (100.79,138.2) .. controls (100.79,139.69) and (99.59,140.89) .. (98.1,140.89) -- (87.47,140.89) .. controls (85.99,140.89) and (84.79,139.69) .. (84.79,138.2) -- cycle ;
\draw (259,74.44) node [font=\scriptsize,xscale=1.25,yscale=1.25] {${\displaystyle \ 1\ }$};
\draw (259,53) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$\ 0\ $};
\draw (259,94.44) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$\ 2\ $};
\draw (259,154.44) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$\ 5\ $};
\draw (246,25.4) node [anchor=north west][inner sep=0.75pt] [font=\small,xscale=1.25,yscale=1.25] {$\mathrm{Slot}$};
\draw (324.71,52) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$10$};
\draw (300.39,133.32) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$\ 4\ $};
\draw (259,114.44) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$\ 3\ $};
\draw (259,134.44) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$\ 4\ $};
\draw (325,74.51) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$\ 8\ $};
\draw (325,112.87) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$\ 2\ $};
\draw (300.39,93.25) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$\ 6\ $};
\draw (300.39,154.23) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$\ 2\ $};
\draw (51,74.44) node [font=\scriptsize,xscale=1.25,yscale=1.25] {${\displaystyle \ 1\ }$};
\draw (51,54) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$\ 0\ $};
\draw (51,94.44) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$\ 2\ $};
\draw (51,154.44) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$\ 5\ $};
\draw (38,26.4) node [anchor=north west][inner sep=0.75pt] [font=\small,xscale=1.25,yscale=1.25] {$\mathrm{Slot}$};
\draw (117,53.67) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$\ 8\ $};
\draw (92.79,134.32) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$\ 2\ $};
\draw (51,114.44) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$\ 3\ $};
\draw (51,134.44) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$\ 4\ $};
\draw (117,75.33) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$\ 6\ $};
\draw (117,114.33) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$\ 2\ $};
\draw (92.79,94.31) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$\ 4\ $};
\end{tikzpicture}
}
\caption{ \small Fork choice rule example observed from a validator $i$'s point of view. We represent block votes with blue circles.
Block votes point to specific blocks indicating the block considered as the \emph{headblock} of the candidate chain at the moment of the vote.
Each block has a number representing the value attributed by the fork choice rule algorithm (cf. \autoref{algo:GHOST}) to determine the candidate chain - we assume for this example that each validator has the same stake of 1.
At the end of slot 4, $i$'s fork choice rule gives the block of slot 4 as $C_i$'s head.
At the end of slot 5, the fork choice rule designates the block of slot 5 as the head of the candidate chain.}
\label{fig:forkChoiceRuleExample}
\end{figure}
\subsection{Pseudo Code}
\label{subsec:code}
In this section, we dive into a practical understanding of the mechanism behind the Ethereum PoS protocol. According to the specifications \cite{github_specs} and various implementations (such as Prysm \cite{prysm_code} and Teku \cite{teku_code}) we formalize the main functions of the protocol through pseudo-codes for a better understanding and for analysis purposes.
Each validator $p$ runs an instance of this particular pseudo code.
For instance, when a validator $p$ proposes a block, he broadcasts the following message: $\langle PROPOSE, ( slot,$ $\texttt{hash}(headBlock_p) ,$ $content) \rangle$, where $slot$ is the slot at which the proposer proposes the block, the hash of the $headBlock_p$ is the hash of the block considered to be the head of the candidate chain according to the fork choice rule (see \autoref{algo:GHOST}), and \textit{content} contains data used for pseudo randomness, among other things that we will not detail here. We instead focus on the consensus protocol.
We describe in the following paragraphs the variables and functions used in the pseudo-code and the goal of these functions.
\paragraph*{Variables.} During the computation, each variable takes a value that is subjective and may depend on the validator. We indicate with $p$ the fact that the value of variables on each process. The variable $tree_p$ is considered to be a graph of blocks with each block linked to its predecessor and thus represents the view of the blockchain (more precisely, $tree_p$ represents the view of all blocks received by the validator since the genesis of the system).
Each $tree_p$ starts with the genesis block.
$role_p$ corresponds to the different roles a validator can have, which is possibly none (i.e., for each slot, the validator can be proposer, attester, or have no role). $role_p$ is a list containing the role(s) of the validator for the current slot.
The $slot_p$ is a measure of time. In particular, a slot corresponds to 12 seconds. $slot_p \in \mathbb{N}$. Slot 0 begins at the time of the genesis block and is incremented every 12 seconds. $headblock_p$ is the head of the candidate chain according to $p$'s local view and the fork choice rule.
A checkpoint $C$ is a pair block-epoch that is used for finalization. $C$ has two attributes which are $justified$ and $finalaized$ which can be true or false (e.g., if $C$ is only justified $C.justified= $ \texttt{true} and $C.finalized= $ \texttt{false}). $lastJustifiedcheckpoint_p$ is the \textit{justified} checkpoint with the highest epoch. $currentCheckpoint_p$ is the checkpoint of the current epoch. The list $attestation_p$ is a list of size $n$ (i.e., the total number of validators). This list is updated only to contain the latest messages of validators (of at most one epoch old).
$CheckpointVote_p$ is a pair of checkpoints, so a pair of pairs, used to make a checkpoint vote. Let us stress out the fact that all those variables are local, and at any time, two different validators may have different valuations of those variables.
\paragraph*{Functions.}
We describe the main functions of the protocol succinctly before giving the pseudo code associated and a more detailed explanation:
\begin{itemize}
\item \texttt{validatorMain} is the primary function of the validator, which launches the execution of all subsidiary functions.
\item \texttt{sync} is a function that runs in parallel of the \texttt{validatorMain} function and ensures the synchronization of the validator. It updates the slot, the role(s) and processes justification and finalization at the end of the epoch and when a new validator joins the system.
\item \texttt{getHeadBlock} applies the fork choice rule. This is the function that indicates the head block of the candidate chain.
\item \texttt{justificationFinalization} is the function that handles the justification and finalization of checkpoints.
\end{itemize}
We depict in \autoref{algo:validatorMain} the main of the validator. This function initializes all the values necessary to run a validator. In this paper, we consider the selection of validators already made to focus on the description of the consensus algorithm itself. The main starts a routine called \texttt{sync} to run in parallel. Then there is an infinite loop that handles the call to an appropriate function when a validator needs to take action for its role(s).
\begin{algorithm}[th]
\caption{Main code for a validator $p$}
\label{algo:validatorMain}
\begin{algorithmic}[1]
\setlength{\lineskip}{3pt}
\Procedure{validatorMain}{\ }
\State $tree_p := nil$ \Comment{The tree represents the linked received blocks}
\State $role_p := [\ ]$ \Comment{$role_p $ can be ROLE\_PROPOSER and/or ROLE\_ATTESTER when it is not empty}
\State $slot_p := 0$ \Comment{$slot_p \in \mathbb{N}$}
\State $lastJustifiedCheckpoint_p := (0, genesisBlock)$ \Comment{A checkpoint is a tuple (epoch, block)}
\State $attestation_p := [\ ]$ \Comment{List of latest attestations received for each validator}
\State $validatorIndex_p := $ index of the validator \Comment{Each validator has a unique index}
\State $listValidator := [p_0,p_1,\dots,p_{N-1} ]$ \Comment{A list of the validators index}
\State $balances := [\ ]$ \Comment{A list of the balances of the validators, their stake}
\State
\State \textbf{start} \hyperref[algo:sync]{\texttt{sync}($tree_p,$ $slot_p,$ $attestation_p,$ $lastJustifiedCheckpoint_p,$ $role_p,$
$balances$)}
\State
\While{\texttt{true}}
\If{$role_p\neq \emptyset$
\If{ROLE\_PROPOSER $\in role_p$ }
\State \texttt{prepareBlock}()
\EndIf
\If{ROLE\_ATTESTER $\in role_p$}
\State \texttt{prepareAttestation}()
\EndIf
\State $role_p \gets [\ ]
\Else
\State no role assigned \Comment{No action required}
\EndIf
\EndWhile
\EndProcedure
\end{algorithmic}
\end{algorithm}
The roles performed by the validator when proposer or attester are defined in \autoref{algo:prepareBlock} and \autoref{algo:prepareAttestation}, respectively.
The proposer of a block does the three following tasks:
\begin{enumerate}
\item Get the head of its candidate chain to have a block to build upon;
\item Sign a predefined pair to participate in the process of pseudo-randomness;
\item Broadcast a new block built on top of the head of the candidate chain.
\end{enumerate}
The attestation is composed of three parts: the slot, the block vote, and the checkpoint vote. The validator uses the fork choice rule presented in \autoref{algo:GHOST} to obtain the block chosen for the block vote. \autoref{algo:GHOST} and the one stemming from it, \autoref{algo:weight}, have already been defined in \cite{buterin_combining_2020}. We restated them here for the sake of completeness. For the checkpoint vote, an honest validator should always vote for the current epoch as the target and take the justified checkpoint with the highest epoch (i.e., \textit{lastJustifiedCheckpoint}) as the source.
In order to broadcast this attestation, the attester must wait for one of two things, either a block has been proposed for this slot or 1/3 of the slot (i.e., 2 seconds) has elapsed.
\begin{algorithm}[th]
\caption{broadcast block}
\label{algo:prepareBlock}
\begin{algorithmic}[1]
\setlength{\lineskip}{3pt}
\Procedure{prepareBlock}{ }
\State $headBlock_p :=$
\hyperref[algo:GHOST]{\texttt{getHeadBlock(
)}}
\State $randaoReveal := $ \texttt{sign}($slot$, \texttt{epochOf($slot$)})
\State \textbf{broadcast} $\langle PROPOSE, ( slot, \texttt{hash}(headBlock_p), randaoReveal,$ $ content) \rangle$
\EndProcedure
\end{algorithmic}
\end{algorithm}
\begin{algorithm}[th]
\caption{Broadcast Attestation}
\label{algo:prepareAttestation}
\begin{algorithmic}[1]
\setlength{\lineskip}{3pt}
\Procedure{prepareAttestation}{ }
\State \texttt{waitForBlockOrOneThird()} \Comment{wait for a new block in this slot or $\frac{1}{3}$ of the slo
}
\State $headBlock_p :=$ \hyperref[algo:GHOST]{\texttt{getHeadBlock
)}}
\State $currentCheckpoint_p := $ (first block of the epoch, \texttt{epochOf}($slot$) )
\State $CheckpointVote_p := \big(lastJustifiedCheckpoint_p, currentCheckpoint_p \big) $
\State \textbf{broadcast} $\langle ATTEST, (slot_p, \underbrace{\texttt{hash}(headBlock_p)}_\textrm{block vote}, \underbrace{CheckpointVote_p}_\textrm{checkpoint vote}) \rangle$
\EndProcedure
\end{algorithmic}
\end{algorithm}
The synchronization of the validator $p$ is handled by the function \texttt{sync} described in \autoref{algo:sync}. This algorithm allows the validator to update its view of the blockchain, in particular, the current slot, the list of attestations, the last justified checkpoint, the validator's role, and the balances of all validators.
This function also starts two other routines which are \texttt{sync Block} and \texttt{sync Attestation} corresponding to \autoref{algo:syncBlock} and \autoref{algo:syncAttestation}, respectively.
These routines are used to handle the broadcast of proposers and attesters. In both functions, upon receiving a block or an attestation, the validator $p$ verifies that it is valid thanks to the \texttt{isValid} function. It is important to note that upon receiving a block, a validator can update the last justified checkpoint only if the current epoch has not started more than 8 slots ago.
This particular condition is what the patch has brought to prevent a liveness attack (see \autoref{sec:livenessAttack}).
\begin{algorithm}[th]
\caption{Sync}
\label{algo:sync}
\begin{algorithmic}[1]
\setlength{\lineskip}{3pt}
\Procedure{sync}{$tree, slot, attestation,$ $role,$ $lastJustifiedCheckpoint,$}
\State \textbf{start} \hyperref[algo:syncBlock]{\texttt{syncBlock($slot, tree$)}}
\State \textbf{start} \hyperref[algo:syncAttestation]{\texttt{syncAttestation($attestation$)}}
\Repeat
\State $previousSlot \gets slot$
\State $slot \gets \lfloor$ time in seconds since genesis block / 12 $\rfloor$
\If{$previousSlot \neq slot$} \Comment{If we start a new slot}
\State \textit{roleSlotDone} $ \gets $ false
\If{\hyperref[algo:computeProposerIndex]{\texttt{computeProposerIndex}}(\hyperref[algo:getSeed]{\texttt{getSeed}}(current epoch)) = $validatorIndex$}
\State $role_p \gets $ ROLE\_PROPOSER
\EndIf
\EndIf
\If{$slot \pmod{32} = 0$} \Comment{First slot of an epoch}
\State \hyperref[algo:Casper]{\texttt{jutificationFinalization}($tree,$ $lastJustifiedCheckpoint$)}
\State \hyperref[algo:getCommitteeAssignement]{\texttt{getCommitteeAssignement}(next epoch, $validatorIndex, listValidator$)}
\EndIf
\Until{{validator exit}}
\EndProcedure
\end{algorithmic}
\end{algorithm}
\begin{algorithm}[th]
\caption{Sync Block}
\label{algo:syncBlock}
\begin{algorithmic}[1]
\setlength{\lineskip}{3pt}
\Procedure{syncBlock}{$slot, tree$}
\Upon{$\langle PROPOSE, ( slot_i, \texttt{hash}(headBlock_i) , randaoReveal_i,$ $ content_i) \rangle$}{validator $i$}
\State $block := \langle PROPOSE, ( slot_i, \texttt{hash}(headBlock_i) , $ $ randaoReveal_i,$ $ content_i) \rangle$
\If{\texttt{isValid}($block$)}
\State add $block$ to $tree$
\If{$slot \pmod{32} \leq 8$ }
\State update justified checkpoint if necessary
\EndIf
\EndIf
\EndUpon
\EndProcedure
\end{algorithmic}
\end{algorithm}
\begin{algorithm}[th]
\caption{Sync Attestation}
\label{algo:syncAttestation}
\begin{algorithmic}[1]
\setlength{\lineskip}{3pt}
\Procedure{syncAttestation}{$attestation$}
\Upon{$\langle ATTEST, (slot_i, headBlock_i, checkpointEdge_i) \rangle$}{validator $i$}
\State $attestation_i := \langle ATTEST, (slot_i, headBlock_i, checkpointEdge_i) \rangle$
\If{\texttt{isValid}($attestation$)
\State $attestation[i] \gets attestation_i$
\EndIf
\EndUpon
\EndProcedure
\end{algorithmic}
\end{algorithm}
\begin{algorithm}[th]
\caption{Get Head Block}
\label{algo:GHOST}
\begin{algorithmic}[1]
\setlength{\lineskip}{3pt}
\Procedure{getHeadBlock}{ }
\State $block \gets$ block of the justified checkpoint with the highest epoch
\While{$block$ has at least one child}
\State $block \gets \underset{b'\text{ child of }block}{\argmax}$ \texttt{weight($tree, Attestation, b'$)}
\State (ties are broken by hash of the block header)
\EndWhile
\State \Return $block$
\EndProcedure
\end{algorithmic}
\end{algorithm}
\begin{algorithm}[th]
\caption{Weight}
\label{algo:weight}
\begin{algorithmic}[1]
\setlength{\lineskip}{3pt}
\Procedure{weight}{$tree, Attestation,block$}
\State $w \gets 0$
\For{every validator $v_i$}
\If{$\exists a \in \ Attestation$ an attestation of $v_i$ for $block$ or a descendant of $block$}
\State $w \gets w + $ stake of $v_i$
\EndIf
\EndFor
\State \Return $w$
\EndProcedure
\end{algorithmic}
\end{algorithm}
\autoref{algo:Casper} can be considered the most intricate one. This algorithm is responsible for the justification or finalization of the checkpoints at the end of an epoch. To do so, it counts the number of checkpoint votes with the same source and target. If this number corresponds to more than 2/3 of the stake of all validators then the target is considered justified for the validator running this algorithm. The last four conditions concern finalization. They verify among the last four checkpoints which one fulfills the conditions to become finalized. The conditions to become finalized are formally described in \autoref{subsec:overview} and can be summarized by: the checkpoint must be the source of a supermajority link, and all the checkpoints between the source and target included must be justified.
\begin{algorithm}[th]
\caption{Justification and Finalization}
\label{algo:Casper}
\begin{algorithmic}[1]
\setlength{\lineskip}{3pt}
\Procedure{jutificationFinalization}{$tree,$ $ lastJustifiedCheckpoint$}
\State $source \gets lastJustifiedCheckpoint$
\State $target \gets $ the current checkpoint
\State $nbCheckpointVote \gets $ \texttt{countMatchingCheckpointVote}($source,$ $target$)
\\
\hskip\algorithmicindent {\color{gray} $\triangleright$ \emph{justification process}:}
\If{ $nbCheckpointVote \geq \frac{2}{3} *$ total balance of validators }
\State $target.justified = $ \texttt{true}
\State $lastJustifiedCheckpoint \gets target$
\EndIf
\\
\hskip\algorithmicindent {\color{gray} $\triangleright$ \emph{finalization process}:}
\State $A,B,C,D \gets $ the last 4 checkpoints \Comment{With $D$ being the current checkpoint.}
\If{$A.justified$ $\land$ $B.justified$ $\land$ ($A \xrightarrow{\texttt{J}} C$)}
\State $D.finalized = $ \texttt{true} \Comment{Finalization of $A$}
\EndIf
\If{$B.justified$ $\land$ ($B \xrightarrow{\texttt{J}} C$)}
\State $B.finalized = $ \texttt{true} \Comment{Finalization of $B$}
\EndIf
\If{$B.justified$ $\land$ $C.justified$ $\land$ ($B \xrightarrow{\texttt{J}} D$)}
\State $B.finalized = $ \texttt{true} \Comment{Finalization of $B$}
\EndIf
\If{$C.justified$ $\land$ ($C \xrightarrow{\texttt{J}} D$)}
\State $C.finalized = $ \texttt{true} \Comment{Finalization of $C$}
\EndIf
\EndProcedure
\end{algorithmic}
\end{algorithm}
\section{Liveness attack}
\label{sec:livenessAttack}
In this section, we describe a liveness attack called \emph{bouncing attack} that delays finality in a partially synchronous network after \texttt{GST}.
Previous works also exhibit liveness attacks against the protocol using the intertwining of the fork choice rule and the finality gadget \cite{nakamura_analysis_2019,neu_ebb_2021}.
To prevent this type of attack, the protocol now contains a ``patch'' \cite{pullRequest_bouncing_2022} that was suggested on the Ethereum research forum \cite{nakamura_prevention_2019}. We show that the implemented patch is insufficient and this type of attack is still possible if certain conditions are verified.
This is a probabilistic liveness attack against the protocol of Ethereum Proof-of-Stake.
Our attack can happen with less than 1/3 of Byzantine validators, as discussed in \autoref{subsec:probabilisticBouncingAttack}.
\subsection{Bouncing Attack}
\label{subsec:bouncingAttack}
The \emph{Bouncing Attack} \cite{nakamura_analysis_2019} describes a liveness attack where the suffix of the chain changes repetitively between two candidate chains, thus preventing the chain from finalizing any checkpoint.
The Bouncing Attack exploits the fact that the candidate chains should start from the justified checkpoint with the highest epoch.
It is possible for Byzantine validators to divide honest validator's opinion by justifying a new checkpoint once some honest validators have already cast their vote (made an attestation) during the asynchronous period before \texttt{GST}.
The bouncing attack becomes possible once there is a \emph{justifiable} checkpoint in a different branch from the one designated by the fork choice rule with a higher epoch than the current highest justified checkpoint.
A \emph{justifiable checkpoint} is a checkpoint that can become justified only by adding the checkpoint votes of Byzantine validators.
If this setup occurs, the malicious validators could make honest validators start voting for a different checkpoint on a different chain, leaving a justifiable checkpoint again for them to repeat their attack and thus making validators \emph{bounce} between two different chains and not finalizing any checkpoint.
Hence the name Bouncing attack.
Let us illustrate the attack with a concrete case. We show in \autoref{fig:simplifiedBoncingAttack} an oversimplified case with only 10 validators, among which 3 are Byzantines.
To occur, the attack needs to have a justifiable checkpoint with a higher epoch than the last justified checkpoint. We reach this situation before \texttt{GST}, which is presented in the left part of the figure. After reaching \texttt{GST}, Byzantine validators wait for honest validators to make a new checkpoint justifiable. When a new checkpoint is justifiable, the Byzantine validators cast their vote to justify another checkpoint, as shown by the right part of the figure.
This will lead honest validators to start voting for the left branch, thus reaching a situation similar to the first step allowing the bouncing attack to continue. The repetition of this behavior is the bouncing attack. We emphasize this example in more detail in \autoref{fig:complexBouncingAttack} by detailing the sequence of votes allowing a ``bounce'' to occur and leaving a justifiable checkpoint on the other branch.
\begin{figure}
\centering
\resizebox{.7\linewidth}{!}{
\begin{tikzpicture}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1]
\draw (22.18,89.35) -- (22.18,72.23) ;
\draw [shift={(22.18,69.23)}, rotate = 90] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (6.25,-3) -- (0,0) -- (6.25,3) -- cycle ;
\draw (62.18,89.35) -- (62.18,72.29) ;
\draw [shift={(62.18,69.29)}, rotate = 90] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (6.25,-3) -- (0,0) -- (6.25,3) -- cycle ;
\draw (34.59,100.32) -- (28.51,110.86) -- (16.34,110.86) -- (10.25,100.32) -- (16.34,89.78) -- (28.51,89.78) -- cycle ;
\draw (74.19,100.37) -- (68.08,110.96) -- (55.86,110.96) -- (49.75,100.37) -- (55.86,89.79) -- (68.08,89.79) -- cycle ;
\draw (34.44,59.18) -- (28.33,69.76) -- (16.11,69.76) -- (10,59.18) -- (16.11,48.59) -- (28.33,48.59) -- cycle ;
\draw (74.94,59.18) -- (68.83,69.76) -- (56.61,69.76) -- (50.5,59.18) -- (56.61,48.59) -- (68.83,48.59) -- cycle ;
\draw [fill={rgb, 255:red, 0; green, 255; blue, 0 } ,fill opacity=1 ] (54.87,18.67) -- (48.76,29.25) -- (36.54,29.25) -- (30.42,18.67) -- (36.54,8.08) -- (48.76,8.08) -- cycle ;
\draw (21.38,49.02) -- (34.61,31.64) ;
\draw [shift={(36.42,29.25)}, rotate = 127.27] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (6.25,-3) -- (0,0) -- (6.25,3) -- cycle ;
\draw (61.88,49.02) -- (50.31,31.74) ;
\draw [shift={(48.65,29.25)}, rotate = 56.21] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (6.25,-3) -- (0,0) -- (6.25,3) -- cycle ;
\draw [fill={rgb, 255:red, 255; green, 255; blue, 255 } ,fill opacity=0 ] (52.45,18.67) -- (47.54,27.17) -- (37.72,27.17) -- (32.81,18.67) -- (37.72,10.16) -- (47.54,10.16) -- cycle ;
\draw [fill={rgb, 255:red, 255; green, 255; blue, 255 } ,fill opacity=1 ] (72.49,59.16) -- (67.58,67.66) -- (57.76,67.66) -- (52.85,59.16) -- (57.76,50.66) -- (67.58,50.66) -- cycle ;
\draw (142.18,92.63) -- (142.18,72.23) ;
\draw [shift={(142.18,69.23)}, rotate = 90] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (6.25,-3) -- (0,0) -- (6.25,3) -- cycle ;
\draw (182.18,92.63) -- (182.18,72.3) ;
\draw [shift={(182.18,69.3)}, rotate = 90] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (6.25,-3) -- (0,0) -- (6.25,3) -- cycle ;
\draw (154.84,103.42) -- (148.76,113.96) -- (136.59,113.96) -- (130.5,103.42) -- (136.59,92.88) -- (148.76,92.88) -- cycle ;
\draw (194.44,103.87) -- (188.33,114.46) -- (176.11,114.46) -- (170,103.87) -- (176.11,93.29) -- (188.33,93.29) -- cycle ;
\draw (154.44,59.18) -- (148.33,69.76) -- (136.11,69.76) -- (130,59.18) -- (136.11,48.59) -- (148.33,48.59) -- cycle ;
\draw (194.94,59.18) -- (188.83,69.76) -- (176.61,69.76) -- (170.5,59.18) -- (176.61,48.59) -- (188.83,48.59) -- cycle ;
\draw [fill={rgb, 255:red, 0; green, 255; blue, 0 } ,fill opacity=1 ] (174.87,18.67) -- (168.76,29.25) -- (156.54,29.25) -- (150.42,18.67) -- (156.54,8.08) -- (168.76,8.08) -- cycle ;
\draw (141.38,49.02) -- (154.61,31.64) ;
\draw [shift={(156.42,29.25)}, rotate = 127.27] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (6.25,-3) -- (0,0) -- (6.25,3) -- cycle ;
\draw (181.88,49.02) -- (170.31,31.74) ;
\draw [shift={(168.65,29.25)}, rotate = 56.21] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (6.25,-3) -- (0,0) -- (6.25,3) -- cycle ;
\draw [fill={rgb, 255:red, 0; green, 0; blue, 0 } ,fill opacity=0 ] (172.45,18.67) -- (167.54,27.17) -- (157.72,27.17) -- (152.81,18.67) -- (157.72,10.16) -- (167.54,10.16) -- cycle ;
\draw (192.49,59.16) -- (187.58,67.66) -- (177.76,67.66) -- (172.85,59.16) -- (177.76,50.66) -- (187.58,50.66) -- cycle ;
\draw (195.34,148.92) -- (189.23,159.51) -- (177.01,159.51) -- (170.9,148.92) -- (177.01,138.34) -- (189.23,138.34) -- cycle ;
\draw (182.43,137.88) -- (182.43,117.55) ;
\draw [shift={(182.43,114.55)}, rotate = 90] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (6.25,-3) -- (0,0) -- (6.25,3) -- cycle ;
\draw (262.18,92.63) -- (262.18,72.23) ;
\draw [shift={(262.18,69.23)}, rotate = 90] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (6.25,-3) -- (0,0) -- (6.25,3) -- cycle ;
\draw (302.18,92.63) -- (302.18,72.3) ;
\draw [shift={(302.18,69.3)}, rotate = 90] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (6.25,-3) -- (0,0) -- (6.25,3) -- cycle ;
\draw (274.84,103.42) -- (268.76,113.96) -- (256.59,113.96) -- (250.5,103.42) -- (256.59,92.88) -- (268.76,92.88) -- cycle ;
\draw (314.44,103.87) -- (308.33,114.46) -- (296.11,114.46) -- (290,103.87) -- (296.11,93.29) -- (308.33,93.29) -- cycle ;
\draw (274.44,59.18) -- (268.33,69.76) -- (256.11,69.76) -- (250,59.18) -- (256.11,48.59) -- (268.33,48.59) -- cycle ;
\draw (314.94,59.18) -- (308.83,69.76) -- (296.61,69.76) -- (290.5,59.18) -- (296.61,48.59) -- (308.83,48.59) -- cycle ;
\draw [fill={rgb, 255:red, 0; green, 255; blue, 0 } ,fill opacity=1 ] (294.87,18.67) -- (288.76,29.25) -- (276.54,29.25) -- (270.42,18.67) -- (276.54,8.08) -- (288.76,8.08) -- cycle ;
\draw (261.38,49.02) -- (274.61,31.64) ;
\draw [shift={(276.42,29.25)}, rotate = 127.27] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (6.25,-3) -- (0,0) -- (6.25,3) -- cycle ;
\draw (301.88,49.02) -- (290.31,31.74) ;
\draw [shift={(288.65,29.25)}, rotate = 56.21] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (6.25,-3) -- (0,0) -- (6.25,3) -- cycle ;
\draw [fill={rgb, 255:red, 0; green, 0; blue, 0 } ,fill opacity=0 ] (292.45,18.67) -- (287.54,27.17) -- (277.72,27.17) -- (272.81,18.67) -- (277.72,10.16) -- (287.54,10.16) -- cycle ;
\draw (312.49,59.16) -- (307.58,67.66) -- (297.76,67.66) -- (292.85,59.16) -- (297.76,50.66) -- (307.58,50.66) -- cycle ;
\draw (315.34,148.92) -- (309.23,159.51) -- (297.01,159.51) -- (290.9,148.92) -- (297.01,138.34) -- (309.23,138.34) -- cycle ;
\draw (272.51,103.47) -- (267.6,111.97) -- (257.78,111.97) -- (252.87,103.47) -- (257.78,94.97) -- (267.6,94.97) -- cycle ;
\draw (302.43,137.88) -- (302.43,117.55) ;
\draw [shift={(302.43,114.55)}, rotate = 90] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (6.25,-3) -- (0,0) -- (6.25,3) -- cycle ;
\draw (90,76) -- (106.4,76) -- (117.33,83.81) -- (106.4,91.62) -- (90,91.62) -- cycle ;
\draw (210,76) -- (226.4,76) -- (237.33,83.81) -- (226.4,91.62) -- (210,91.62) -- cycle ;
\draw (62.18,58.83) node [xscale=1.25,yscale=1.25] {$7$};
\draw (21.38,100.35) node [xscale=1.25,yscale=1.25] {$4$};
\draw (62.18,100.45) node [xscale=1.25,yscale=1.25] {$3$};
\draw (42.65,18.7) node [xscale=1.25,yscale=1.25] {$7$};
\draw (21.38,59.2) node [xscale=1.25,yscale=1.25] {$3$};
\draw (182.43,59) node [xscale=1.25,yscale=1.25] {$7$};
\draw (142.18,103.47) node [xscale=1.25,yscale=1.25] {$4$};
\draw (182.43,103.95) node [xscale=1.25,yscale=1.25] {$3$};
\draw (162.63,18.7) node [xscale=1.25,yscale=1.25] {$7$};
\draw (142.85,59.2) node [xscale=1.25,yscale=1.25] {$3$};
\draw (182.43,150) node [xscale=1.25,yscale=1.25] {$4$};
\draw (302.43,59) node [xscale=1.25,yscale=1.25] {$7$};
\draw (262.18,103.47) node [xscale=1.25,yscale=1.25] {$7$};
\draw (302.43,103.95) node [xscale=1.25,yscale=1.25] {$3$};
\draw (282.65,18.2) node [xscale=1.25,yscale=1.25] {$7$};
\draw (262.18,59.2) node [xscale=1.25,yscale=1.25] {$3$};
\draw (296.93,141.7) node [anchor=north west][inner sep=0.75pt] [xscale=1.25,yscale=1.25] {$4$};
\end{tikzpicture}
}
\caption{\small A bouncing attack presented in 3 steps. We have 10 validators, of which 3 are Byzantines. Checkpoints are represented by hexagonal shapes. A double hexagon signifies that the checkpoint is justified. The number inside each hexagon corresponds to the number of validators who made a checkpoint vote with this checkpoint as target.
\textbf{1st step:} We start in a situation where two checkpoints are justified ($7> \frac{2n}{3}$) and there are two concurrent chain. We are at the end of the third epoch in which honest validators have divided their vote on each side. \textbf{2nd step:} We have reached \texttt{GST} at the beginning of the fourth epoch and 4 honest validators have already voted (rightfully so). \textbf{3rd step:} Here is the moment Byzantine validators take action and release their checkpoint vote for the concurrent chain, thus justifying the previously forsaken checkpoint and thereby changing the highest justifying checkpoint. By repeating this process, the bouncing attack can continue indefinitely.}
\label{fig:simplifiedBoncingAttack}
\end{figure}
\begin{figure}
\centering
\resizebox{1\linewidth}{!}{
\begin{tikzpicture}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1]
\draw (300.79,192.5) -- (300.45,137.09) ;
\draw [shift={(300.43,134.09)}, rotate = 89.65] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (5.36,-2.57) -- (0,0) -- (5.36,2.57) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 80; green, 227; blue, 194 } ,fill opacity=1 ][blur shadow={shadow xshift=-0.75pt,shadow yshift=-0.75pt, shadow blur radius=1.5pt, shadow blur steps=4 ,shadow opacity=73}] (332.83,174.16) .. controls (332.83,172.68) and (334.03,171.48) .. (335.52,171.48) -- (346.14,171.48) .. controls (347.63,171.48) and (348.83,172.68) .. (348.83,174.16) -- (348.83,182.23) .. controls (348.83,183.72) and (347.63,184.92) .. (346.14,184.92) -- (335.52,184.92) .. controls (334.03,184.92) and (332.83,183.72) .. (332.83,182.23) -- cycle ;
\draw (340.79,171.44) -- (340.49,137.49) ;
\draw [shift={(340.47,134.49)}, rotate = 89.51] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (5.36,-2.57) -- (0,0) -- (5.36,2.57) -- cycle ;
\draw [dash pattern={on 0.84pt off 2.51pt}] (273.5,234.25) -- (273.5,244.25) ;
\draw [draw opacity=0][fill={rgb, 255:red, 251; green, 193; blue, 96 } ,fill opacity=1 ][blur shadow={shadow xshift=-0.75pt,shadow yshift=-0.75pt, shadow blur radius=1.5pt, shadow blur steps=4 ,shadow opacity=73}] (292.83,194.16) .. controls (292.83,192.68) and (294.03,191.48) .. (295.52,191.48) -- (306.14,191.48) .. controls (307.63,191.48) and (308.83,192.68) .. (308.83,194.16) -- (308.83,202.23) .. controls (308.83,203.72) and (307.63,204.92) .. (306.14,204.92) -- (295.52,204.92) .. controls (294.03,204.92) and (292.83,203.72) .. (292.83,202.23) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 80; green, 227; blue, 194 } ,fill opacity=1 ][blur shadow={shadow xshift=-0.75pt,shadow yshift=-0.75pt, shadow blur radius=1.5pt, shadow blur steps=4 ,shadow opacity=73}] (332.83,214.16) .. controls (332.83,212.68) and (334.03,211.48) .. (335.52,211.48) -- (346.14,211.48) .. controls (347.63,211.48) and (348.83,212.68) .. (348.83,214.16) -- (348.83,222.23) .. controls (348.83,223.72) and (347.63,224.92) .. (346.14,224.92) -- (335.52,224.92) .. controls (334.03,224.92) and (332.83,223.72) .. (332.83,222.23) -- cycle ;
\draw (340.5,211.48) -- (340.5,187.81) ;
\draw [shift={(340.5,184.81)}, rotate = 90] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (3.57,-1.72) -- (0,0) -- (3.57,1.72) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 80; green, 227; blue, 194 } ,fill opacity=1 ][blur shadow={shadow xshift=-0.75pt,shadow yshift=-0.75pt, shadow blur radius=1.5pt, shadow blur steps=4 ,shadow opacity=73}] (332.87,259.16) .. controls (332.87,257.68) and (334.07,256.48) .. (335.56,256.48) -- (346.18,256.48) .. controls (347.67,256.48) and (348.87,257.68) .. (348.87,259.16) -- (348.87,267.23) .. controls (348.87,268.72) and (347.67,269.92) .. (346.18,269.92) -- (335.56,269.92) .. controls (334.07,269.92) and (332.87,268.72) .. (332.87,267.23) -- cycle ;
\draw (340.5,257) -- (340.5,227.81) ;
\draw [shift={(340.5,224.81)}, rotate = 90] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (5.36,-2.57) -- (0,0) -- (5.36,2.57) -- cycle ;
\draw (301.17,273.5) -- (301.17,207.81) ;
\draw [shift={(301.17,204.81)}, rotate = 90] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (5.36,-2.57) -- (0,0) -- (5.36,2.57) -- cycle ;
\draw (313.05,178.2) -- (306.94,188.78) -- (294.72,188.78) -- (288.61,178.2) -- (294.72,167.61) -- (306.94,167.61) -- cycle ;
\draw (353.05,178.2) -- (346.94,188.78) -- (334.72,188.78) -- (328.61,178.2) -- (334.72,167.61) -- (346.94,167.61) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 117; green, 214; blue, 11 } ,fill opacity=1 ] (416.12,172.42) -- (416.12,181.53) -- (412.12,181.53) -- (412.12,172.42) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 117; green, 214; blue, 11 } ,fill opacity=1 ] (410.4,172.42) -- (410.4,181.53) -- (406.4,181.53) -- (406.4,172.42) -- cycle ;
\draw (313.05,126.7) -- (306.94,137.28) -- (294.72,137.28) -- (288.61,126.7) -- (294.72,116.11) -- (306.94,116.11) -- cycle ;
\draw (353.55,126.7) -- (347.44,137.28) -- (335.22,137.28) -- (329.11,126.7) -- (335.22,116.11) -- (347.44,116.11) -- cycle ;
\draw (300.79,115.94) -- (300.79,80.62) ;
\draw [shift={(300.79,77.62)}, rotate = 90] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (5.36,-2.57) -- (0,0) -- (5.36,2.57) -- cycle ;
\draw (340.79,115.94) -- (340.79,81.62) ;
\draw [shift={(340.79,78.62)}, rotate = 90] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (5.36,-2.57) -- (0,0) -- (5.36,2.57) -- cycle ;
\draw (313.05,71.2) -- (306.94,81.78) -- (294.72,81.78) -- (288.61,71.2) -- (294.72,60.61) -- (306.94,60.61) -- cycle ;
\draw (353.55,71.2) -- (347.44,81.78) -- (335.22,81.78) -- (329.11,71.2) -- (335.22,60.61) -- (347.44,60.61) -- cycle ;
\draw [fill={rgb, 255:red, 0; green, 255; blue, 0 } ,fill opacity=1 ] (334.28,30.58) -- (328.17,41.17) -- (315.94,41.17) -- (309.83,30.58) -- (315.94,20) -- (328.17,20) -- cycle ;
\draw (300.43,64.42) -- (315.67,39.58) ;
\draw [shift={(317.23,37.02)}, rotate = 121.51] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (6.25,-3) -- (0,0) -- (6.25,3) -- cycle ;
\draw (341.63,64.82) -- (327.7,39.84) ;
\draw [shift={(326.23,37.22)}, rotate = 60.84] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (6.25,-3) -- (0,0) -- (6.25,3) -- cycle ;
\draw (331.86,30.58) -- (326.95,39.09) -- (317.13,39.09) -- (312.22,30.58) -- (317.13,22.08) -- (326.95,22.08) -- cycle ;
\draw (351.1,71.35) -- (346.19,79.85) -- (336.37,79.85) -- (331.46,71.35) -- (336.37,62.84) -- (346.19,62.84) -- cycle ;
\draw [color={rgb, 255:red, 255; green, 255; blue, 255 } ,draw opacity=1 ][fill={rgb, 255:red, 255; green, 255; blue, 255 } ,fill opacity=1 ] [dash pattern={on 0.84pt off 2.51pt}] (300.75,108.69) -- (300.75,93.69) ;
\draw [color={rgb, 255:red, 255; green, 255; blue, 255 } ,draw opacity=1 ][fill={rgb, 255:red, 255; green, 255; blue, 255 } ,fill opacity=1 ] [dash pattern={on 0.84pt off 2.51pt}] (340.75,108.98) -- (340.75,93.98) ;
\draw [color={rgb, 255:red, 255; green, 255; blue, 255 } ,draw opacity=1 ][fill={rgb, 255:red, 255; green, 255; blue, 255 } ,fill opacity=1 ] [dash pattern={on 0.84pt off 2.51pt}] (300.75,162.52) -- (300.75,147.52) ;
\draw [color={rgb, 255:red, 255; green, 255; blue, 255 } ,draw opacity=1 ][fill={rgb, 255:red, 255; green, 255; blue, 255 } ,fill opacity=1 ] [dash pattern={on 0.84pt off 2.51pt}] (340.75,162.72) -- (340.75,147.72) ;
\draw [draw opacity=0][fill={rgb, 255:red, 80; green, 227; blue, 194 } ,fill opacity=1 ][blur shadow={shadow xshift=-0.75pt,shadow yshift=-0.75pt, shadow blur radius=1.5pt, shadow blur steps=4 ,shadow opacity=73}] (333.5,67.5) .. controls (333.5,66.01) and (334.7,64.81) .. (336.19,64.81) -- (346.81,64.81) .. controls (348.3,64.81) and (349.5,66.01) .. (349.5,67.5) -- (349.5,75.56) .. controls (349.5,77.05) and (348.3,78.25) .. (346.81,78.25) -- (336.19,78.25) .. controls (334.7,78.25) and (333.5,77.05) .. (333.5,75.56) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 80; green, 227; blue, 194 } ,fill opacity=1 ][blur shadow={shadow xshift=-0.75pt,shadow yshift=-0.75pt, shadow blur radius=1.5pt, shadow blur steps=4 ,shadow opacity=73}] (333.53,122.5) .. controls (333.53,121.01) and (334.74,119.81) .. (336.22,119.81) -- (346.85,119.81) .. controls (348.33,119.81) and (349.54,121.01) .. (349.54,122.5) -- (349.54,130.56) .. controls (349.54,132.05) and (348.33,133.25) .. (346.85,133.25) -- (336.22,133.25) .. controls (334.74,133.25) and (333.53,132.05) .. (333.53,130.56) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 251; green, 193; blue, 96 } ,fill opacity=1 ][blur shadow={shadow xshift=-0.75pt,shadow yshift=-0.75pt, shadow blur radius=1.5pt, shadow blur steps=4 ,shadow opacity=73}] (292.83,67.16) .. controls (292.83,65.68) and (294.03,64.48) .. (295.52,64.48) -- (306.14,64.48) .. controls (307.63,64.48) and (308.83,65.68) .. (308.83,67.16) -- (308.83,75.23) .. controls (308.83,76.72) and (307.63,77.92) .. (306.14,77.92) -- (295.52,77.92) .. controls (294.03,77.92) and (292.83,76.72) .. (292.83,75.23) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 162; green, 162; blue, 162 } ,fill opacity=1 ][blur shadow={shadow xshift=-0.75pt,shadow yshift=-0.75pt, shadow blur radius=1.5pt, shadow blur steps=4 ,shadow opacity=73}] (314.16,26.5) .. controls (314.16,25.01) and (315.37,23.81) .. (316.85,23.81) -- (327.48,23.81) .. controls (328.96,23.81) and (330.17,25.01) .. (330.17,26.5) -- (330.17,34.56) .. controls (330.17,36.05) and (328.96,37.25) .. (327.48,37.25) -- (316.85,37.25) .. controls (315.37,37.25) and (314.16,36.05) .. (314.16,34.56) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 117; green, 214; blue, 11 } ,fill opacity=1 ] (416.79,214.92) -- (416.79,224.03) -- (412.79,224.03) -- (412.79,214.92) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 117; green, 214; blue, 11 } ,fill opacity=1 ] (411.07,214.92) -- (411.07,224.03) -- (407.07,224.03) -- (407.07,214.92) -- cycle ;
\draw [color={rgb, 255:red, 255; green, 255; blue, 255 } ,draw opacity=1 ][fill={rgb, 255:red, 255; green, 255; blue, 255 } ,fill opacity=1 ] [dash pattern={on 0.84pt off 2.51pt}] (301.15,247.42) -- (301.15,232.42) ;
\draw [color={rgb, 255:red, 255; green, 255; blue, 255 } ,draw opacity=1 ][fill={rgb, 255:red, 255; green, 255; blue, 255 } ,fill opacity=1 ] [dash pattern={on 0.84pt off 2.51pt}] (340.47,246.59) -- (340.47,231.59) ;
\draw [color={rgb, 255:red, 154; green, 209; blue, 33 } ,draw opacity=1 ] [dash pattern={on 0.84pt off 2.51pt}] (404.89,176.59) .. controls (387.34,177.24) and (375.49,176.62) .. (353.9,176.59) ;
\draw [shift={(352.22,176.59)}, rotate = 360] [fill={rgb, 255:red, 154; green, 209; blue, 33 } ,fill opacity=1 ][line width=0.08] [draw opacity=0] (7.2,-1.8) -- (0,0) -- (7.2,1.8) -- cycle ;
\draw [color={rgb, 255:red, 208; green, 2; blue, 27 } ,draw opacity=1 ] [dash pattern={on 0.84pt off 2.51pt}] (391,194.03) .. controls (378.39,173.01) and (331.59,142.27) .. (310.54,134.36) ;
\draw [shift={(308.67,133.7)}, rotate = 18.43] [fill={rgb, 255:red, 208; green, 2; blue, 27 } ,fill opacity=1 ][line width=0.08] [draw opacity=0] (7.2,-1.8) -- (0,0) -- (7.2,1.8) -- cycle ;
\draw [color={rgb, 255:red, 154; green, 209; blue, 33 } ,draw opacity=1 ] [dash pattern={on 0.84pt off 2.51pt}] (407.67,217.36) .. controls (387.72,213.63) and (365.21,196.28) .. (353.42,184.87) ;
\draw [shift={(352,183.48)}, rotate = 45] [fill={rgb, 255:red, 154; green, 209; blue, 33 } ,fill opacity=1 ][line width=0.08] [draw opacity=0] (7.2,-1.8) -- (0,0) -- (7.2,1.8) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 117; green, 214; blue, 11 } ,fill opacity=1 ] (411.79,235.59) -- (411.79,244.7) -- (407.79,244.7) -- (407.79,235.59) -- cycle ;
\draw [color={rgb, 255:red, 154; green, 209; blue, 33 } ,draw opacity=1 ] [dash pattern={on 0.84pt off 2.51pt}] (408.67,234.14) .. controls (390.65,224.49) and (360.84,199.94) .. (348.6,188.05) ;
\draw [shift={(347.33,186.81)}, rotate = 45] [fill={rgb, 255:red, 154; green, 209; blue, 33 } ,fill opacity=1 ][line width=0.08] [draw opacity=0] (7.2,-1.8) -- (0,0) -- (7.2,1.8) -- cycle ;
\draw [color={rgb, 255:red, 154; green, 209; blue, 33 } ,draw opacity=1 ] [dash pattern={on 0.84pt off 2.51pt}] (372.83,295.84) .. controls (349.33,287.34) and (339.56,287.81) .. (329.83,271.84) .. controls (320.44,256.42) and (320.33,201.38) .. (311.8,185.42) ;
\draw [shift={(310.83,183.84)}, rotate = 54.87] [fill={rgb, 255:red, 154; green, 209; blue, 33 } ,fill opacity=1 ][line width=0.08] [draw opacity=0] (7.2,-1.8) -- (0,0) -- (7.2,1.8) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 117; green, 214; blue, 11 } ,fill opacity=1 ] (416.79,260.25) -- (416.79,269.36) -- (412.79,269.36) -- (412.79,260.25) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 117; green, 214; blue, 11 } ,fill opacity=1 ] (411.07,260.25) -- (411.07,269.36) -- (407.07,269.36) -- (407.07,260.25) -- cycle ;
\draw [color={rgb, 255:red, 154; green, 209; blue, 33 } ,draw opacity=1 ] [dash pattern={on 0.84pt off 2.51pt}] (404.22,265.03) .. controls (364.78,273.44) and (350.42,212.22) .. (345.33,192.11) ;
\draw [shift={(344.89,190.36)}, rotate = 75.47] [fill={rgb, 255:red, 154; green, 209; blue, 33 } ,fill opacity=1 ][line width=0.08] [draw opacity=0] (7.2,-1.8) -- (0,0) -- (7.2,1.8) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 251; green, 193; blue, 96 } ,fill opacity=1 ][blur shadow={shadow xshift=-0.75pt,shadow yshift=-0.75pt, shadow blur radius=1.5pt, shadow blur steps=4 ,shadow opacity=73}] (292.83,274.66) .. controls (292.83,273.18) and (294.03,271.98) .. (295.52,271.98) -- (306.14,271.98) .. controls (307.63,271.98) and (308.83,273.18) .. (308.83,274.66) -- (308.83,282.73) .. controls (308.83,284.22) and (307.63,285.42) .. (306.14,285.42) -- (295.52,285.42) .. controls (294.03,285.42) and (292.83,284.22) .. (292.83,282.73) -- cycle ;
\draw [color={rgb, 255:red, 208; green, 2; blue, 27 } ,draw opacity=1 ] [dash pattern={on 0.84pt off 2.51pt}] (394.17,281) .. controls (365.67,259) and (380.67,262.5) .. (327.21,236.57) .. controls (274.56,211.02) and (275.62,169.92) .. (291.91,136.68) ;
\draw [shift={(292.67,135.17)}, rotate = 116.91] [fill={rgb, 255:red, 208; green, 2; blue, 27 } ,fill opacity=1 ][line width=0.08] [draw opacity=0] (7.2,-1.8) -- (0,0) -- (7.2,1.8) -- cycle ;
\draw (310.6,126.66) -- (305.69,135.16) -- (295.87,135.16) -- (290.96,126.66) -- (295.87,118.15) -- (305.69,118.15) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 251; green, 193; blue, 96 } ,fill opacity=1 ][blur shadow={shadow xshift=-0.75pt,shadow yshift=-0.75pt, shadow blur radius=1.5pt, shadow blur steps=4 ,shadow opacity=73}] (292.66,122.5) .. controls (292.66,121.01) and (293.87,119.81) .. (295.35,119.81) -- (305.98,119.81) .. controls (307.46,119.81) and (308.67,121.01) .. (308.67,122.5) -- (308.67,130.56) .. controls (308.67,132.05) and (307.46,133.25) .. (305.98,133.25) -- (295.35,133.25) .. controls (293.87,133.25) and (292.66,132.05) .. (292.66,130.56) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 251; green, 193; blue, 96 } ,fill opacity=1 ][blur shadow={shadow xshift=-0.75pt,shadow yshift=-0.75pt, shadow blur radius=1.5pt, shadow blur steps=4 ,shadow opacity=73}] (292.83,173.94) .. controls (292.83,172.46) and (294.03,171.25) .. (295.52,171.25) -- (306.14,171.25) .. controls (307.63,171.25) and (308.83,172.46) .. (308.83,173.94) -- (308.83,182.01) .. controls (308.83,183.49) and (307.63,184.7) .. (306.14,184.7) -- (295.52,184.7) .. controls (294.03,184.7) and (292.83,183.49) .. (292.83,182.01) -- cycle ;
\draw [dash pattern={on 0.84pt off 2.51pt}] (275.5,290.54) -- (275.5,297.79) -- (275.5,300.54) ;
\draw (300.83,303.42) -- (300.83,288.82) ;
\draw [shift={(300.83,285.82)}, rotate = 90] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (5.36,-2.57) -- (0,0) -- (5.36,2.57) -- cycle ;
\draw [color={rgb, 255:red, 255; green, 255; blue, 255 } ,draw opacity=1 ][fill={rgb, 255:red, 255; green, 255; blue, 255 } ,fill opacity=1 ] [dash pattern={on 0.84pt off 2.51pt}] (300.83,306.25) -- (300.83,290.03) ;
\draw [draw opacity=0][fill={rgb, 255:red, 251; green, 193; blue, 96 } ,fill opacity=1 ][blur shadow={shadow xshift=-0.75pt,shadow yshift=-0.75pt, shadow blur radius=1.5pt, shadow blur steps=4 ,shadow opacity=73}] (293.33,306.16) .. controls (293.33,304.68) and (294.53,303.48) .. (296.02,303.48) -- (306.64,303.48) .. controls (308.13,303.48) and (309.33,304.68) .. (309.33,306.16) -- (309.33,314.23) .. controls (309.33,315.72) and (308.13,316.92) .. (306.64,316.92) -- (296.02,316.92) .. controls (294.53,316.92) and (293.33,315.72) .. (293.33,314.23) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 117; green, 214; blue, 11 } ,fill opacity=1 ] (377.33,291.59) -- (377.33,300.7) -- (373.33,300.7) -- (373.33,291.59) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 117; green, 214; blue, 11 } ,fill opacity=1 ] (393.12,306.92) -- (393.12,316.03) -- (389.12,316.03) -- (389.12,306.92) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 117; green, 214; blue, 11 } ,fill opacity=1 ] (387.4,306.92) -- (387.4,316.03) -- (383.4,316.03) -- (383.4,306.92) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 117; green, 214; blue, 11 } ,fill opacity=1 ] (398.62,306.85) -- (398.62,315.96) -- (394.62,315.96) -- (394.62,306.85) -- cycle ;
\draw [dash pattern={on 4.5pt off 4.5pt}] (403.4,171.85) -- (403.4,323.85) ;
\draw [color={rgb, 255:red, 154; green, 209; blue, 33 } ,draw opacity=1 ] [dash pattern={on 0.84pt off 2.51pt}] (382.83,314.84) .. controls (359.33,306.34) and (337.06,296.81) .. (327.33,280.84) .. controls (317.84,265.26) and (315.45,211.61) .. (310.24,189.48) ;
\draw [shift={(309.83,187.84)}, rotate = 75.32] [fill={rgb, 255:red, 154; green, 209; blue, 33 } ,fill opacity=1 ][line width=0.08] [draw opacity=0] (7.2,-1.8) -- (0,0) -- (7.2,1.8) -- cycle ;
\draw (260,58.57) .. controls (253.29,58.44) and (260.55,83.72) .. (254.83,83.85) ;
\draw (254.57,83.84) .. controls (260.86,83.71) and (254.57,109.26) .. (260,108.99) ;
\draw (259.86,115.96) .. controls (253.14,115.83) and (260.29,140.29) .. (254.57,140.42) ;
\draw (254.57,140.55) .. controls (260.86,140.42) and (254.57,165.14) .. (260,164.88) ;
\draw (259.86,170.11) .. controls (253.15,169.64) and (260.29,241.89) .. (254.57,242.26) ;
\draw (254.57,242.64) .. controls (260.86,242.26) and (254.57,315.17) .. (260,314.41) ;
\draw [color={rgb, 255:red, 208; green, 2; blue, 27 } ,draw opacity=1 ] [dash pattern={on 0.84pt off 2.51pt}] (370.83,235.59) .. controls (360,216.67) and (338.5,204.67) .. (327.5,190.17) .. controls (316.89,176.18) and (313.25,155.67) .. (307.57,139.07) ;
\draw [shift={(306.94,137.28)}, rotate = 70.27] [fill={rgb, 255:red, 208; green, 2; blue, 27 } ,fill opacity=1 ][line width=0.08] [draw opacity=0] (7.2,-1.8) -- (0,0) -- (7.2,1.8) -- cycle ;
\draw (160.25,162.72) -- (160.28,135.89) ;
\draw [shift={(160.29,132.89)}, rotate = 90.07] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (5.36,-2.57) -- (0,0) -- (5.36,2.57) -- cycle ;
\draw (132.55,126.7) -- (126.44,137.28) -- (114.22,137.28) -- (108.11,126.7) -- (114.22,116.11) -- (126.44,116.11) -- cycle ;
\draw (173.05,126.7) -- (166.94,137.28) -- (154.72,137.28) -- (148.61,126.7) -- (154.72,116.11) -- (166.94,116.11) -- cycle ;
\draw (120.29,115.94) -- (120.29,80.62) ;
\draw [shift={(120.29,77.62)}, rotate = 90] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (5.36,-2.57) -- (0,0) -- (5.36,2.57) -- cycle ;
\draw (160.29,115.94) -- (160.29,81.62) ;
\draw [shift={(160.29,78.62)}, rotate = 90] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (5.36,-2.57) -- (0,0) -- (5.36,2.57) -- cycle ;
\draw (132.55,71.2) -- (126.44,81.78) -- (114.22,81.78) -- (108.11,71.2) -- (114.22,60.61) -- (126.44,60.61) -- cycle ;
\draw (173.05,71.2) -- (166.94,81.78) -- (154.72,81.78) -- (148.61,71.2) -- (154.72,60.61) -- (166.94,60.61) -- cycle ;
\draw [fill={rgb, 255:red, 0; green, 255; blue, 0 } ,fill opacity=1 ] (153.78,30.58) -- (147.67,41.17) -- (135.44,41.17) -- (129.33,30.58) -- (135.44,20) -- (147.67,20) -- cycle ;
\draw (119.93,64.42) -- (135.17,39.58) ;
\draw [shift={(136.73,37.02)}, rotate = 121.51] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (6.25,-3) -- (0,0) -- (6.25,3) -- cycle ;
\draw (161.13,64.82) -- (147.2,39.84) ;
\draw [shift={(145.73,37.22)}, rotate = 60.84] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (6.25,-3) -- (0,0) -- (6.25,3) -- cycle ;
\draw (151.36,30.58) -- (146.45,39.09) -- (136.63,39.09) -- (131.72,30.58) -- (136.63,22.08) -- (146.45,22.08) -- cycle ;
\draw (170.6,71.35) -- (165.69,79.85) -- (155.87,79.85) -- (150.96,71.35) -- (155.87,62.84) -- (165.69,62.84) -- cycle ;
\draw [color={rgb, 255:red, 255; green, 255; blue, 255 } ,draw opacity=1 ][fill={rgb, 255:red, 255; green, 255; blue, 255 } ,fill opacity=1 ] [dash pattern={on 0.84pt off 2.51pt}] (120.25,108.69) -- (120.25,93.69) ;
\draw [color={rgb, 255:red, 255; green, 255; blue, 255 } ,draw opacity=1 ][fill={rgb, 255:red, 255; green, 255; blue, 255 } ,fill opacity=1 ] [dash pattern={on 0.84pt off 2.51pt}] (160.25,108.98) -- (160.25,93.98) ;
\draw [color={rgb, 255:red, 255; green, 255; blue, 255 } ,draw opacity=1 ][fill={rgb, 255:red, 255; green, 255; blue, 255 } ,fill opacity=1 ] [dash pattern={on 0.84pt off 2.51pt}] (120.25,162.52) -- (120.25,147.52) ;
\draw [color={rgb, 255:red, 255; green, 255; blue, 255 } ,draw opacity=1 ][fill={rgb, 255:red, 255; green, 255; blue, 255 } ,fill opacity=1 ] [dash pattern={on 0.84pt off 2.51pt}] (160.25,162.72) -- (160.25,147.72) ;
\draw [draw opacity=0][fill={rgb, 255:red, 80; green, 227; blue, 194 } ,fill opacity=1 ][blur shadow={shadow xshift=-0.75pt,shadow yshift=-0.75pt, shadow blur radius=1.5pt, shadow blur steps=4 ,shadow opacity=73}] (153,67.5) .. controls (153,66.01) and (154.2,64.81) .. (155.69,64.81) -- (166.31,64.81) .. controls (167.8,64.81) and (169,66.01) .. (169,67.5) -- (169,75.56) .. controls (169,77.05) and (167.8,78.25) .. (166.31,78.25) -- (155.69,78.25) .. controls (154.2,78.25) and (153,77.05) .. (153,75.56) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 80; green, 227; blue, 194 } ,fill opacity=1 ][blur shadow={shadow xshift=-0.75pt,shadow yshift=-0.75pt, shadow blur radius=1.5pt, shadow blur steps=4 ,shadow opacity=73}] (153.03,122.5) .. controls (153.03,121.01) and (154.24,119.81) .. (155.72,119.81) -- (166.35,119.81) .. controls (167.83,119.81) and (169.04,121.01) .. (169.04,122.5) -- (169.04,130.56) .. controls (169.04,132.05) and (167.83,133.25) .. (166.35,133.25) -- (155.72,133.25) .. controls (154.24,133.25) and (153.03,132.05) .. (153.03,130.56) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 251; green, 193; blue, 96 } ,fill opacity=1 ][blur shadow={shadow xshift=-0.75pt,shadow yshift=-0.75pt, shadow blur radius=1.5pt, shadow blur steps=4 ,shadow opacity=73}] (112.33,67.16) .. controls (112.33,65.68) and (113.53,64.48) .. (115.02,64.48) -- (125.64,64.48) .. controls (127.13,64.48) and (128.33,65.68) .. (128.33,67.16) -- (128.33,75.23) .. controls (128.33,76.72) and (127.13,77.92) .. (125.64,77.92) -- (115.02,77.92) .. controls (113.53,77.92) and (112.33,76.72) .. (112.33,75.23) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 162; green, 162; blue, 162 } ,fill opacity=1 ][blur shadow={shadow xshift=-0.75pt,shadow yshift=-0.75pt, shadow blur radius=1.5pt, shadow blur steps=4 ,shadow opacity=73}] (133.66,26.5) .. controls (133.66,25.01) and (134.87,23.81) .. (136.35,23.81) -- (146.98,23.81) .. controls (148.46,23.81) and (149.67,25.01) .. (149.67,26.5) -- (149.67,34.56) .. controls (149.67,36.05) and (148.46,37.25) .. (146.98,37.25) -- (136.35,37.25) .. controls (134.87,37.25) and (133.66,36.05) .. (133.66,34.56) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 251; green, 193; blue, 96 } ,fill opacity=1 ][blur shadow={shadow xshift=-0.75pt,shadow yshift=-0.75pt, shadow blur radius=1.5pt, shadow blur steps=4 ,shadow opacity=73}] (112.16,122.5) .. controls (112.16,121.01) and (113.37,119.81) .. (114.85,119.81) -- (125.48,119.81) .. controls (126.96,119.81) and (128.17,121.01) .. (128.17,122.5) -- (128.17,130.56) .. controls (128.17,132.05) and (126.96,133.25) .. (125.48,133.25) -- (114.85,133.25) .. controls (113.37,133.25) and (112.16,132.05) .. (112.16,130.56) -- cycle ;
\draw (79.5,58.57) .. controls (72.79,58.44) and (80.05,83.72) .. (74.33,83.85) ;
\draw (74.07,83.84) .. controls (80.36,83.71) and (74.07,109.26) .. (79.5,108.99) ;
\draw (79.36,115.96) .. controls (72.64,115.83) and (79.79,140.29) .. (74.07,140.42) ;
\draw (74.07,140.55) .. controls (80.36,140.42) and (74.07,165.14) .. (79.5,164.88) ;
\draw (120.43,162.74) -- (120.53,136.37) ;
\draw [shift={(120.54,133.37)}, rotate = 90.2] [fill={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.08] [draw opacity=0] (5.36,-2.57) -- (0,0) -- (5.36,2.57) -- cycle ;
\draw [color={rgb, 255:red, 255; green, 255; blue, 255 } ,draw opacity=1 ][fill={rgb, 255:red, 255; green, 255; blue, 255 } ,fill opacity=1 ] [dash pattern={on 0.84pt off 2.51pt}] (120.5,163.02) -- (120.5,148.02) ;
\draw (181.89,190.51) -- (194.78,190.51) -- (203.38,196.65) -- (194.78,202.79) -- (181.89,202.79) -- cycle ;
\draw (401.13,167.17) .. controls (401.22,161.75) and (385.06,167.52) .. (384.98,162.91) ;
\draw (384.89,162.91) .. controls (384.98,167.98) and (368.66,162.91) .. (368.83,167.29) ;
\draw (436.53,166.97) .. controls (436.62,161.55) and (420.46,167.32) .. (420.38,162.71) ;
\draw (420.29,162.71) .. controls (420.38,167.78) and (404.06,162.71) .. (404.23,167.09) ;
\draw [draw opacity=0][fill={rgb, 255:red, 208; green, 2; blue, 27 } ,fill opacity=1 ] (393.99,192.78) -- (393.99,201.89) -- (389.99,201.89) -- (389.99,192.78) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 117; green, 214; blue, 11 } ,fill opacity=1 ] (393.99,201.89) -- (393.05,201.89) -- (389.99,197.11) -- (389.99,194.78) -- (393.99,201.05) -- (393.99,201.89) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 117; green, 214; blue, 11 } ,fill opacity=1 ] (389.99,201.89) -- (389.99,199.57) -- (391.47,201.89) -- (389.99,201.89) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 117; green, 214; blue, 11 } ,fill opacity=1 ] (393.99,199.05) -- (389.99,192.78) -- (391.47,192.78) -- (393.99,196.72) -- (393.99,199.05) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 117; green, 214; blue, 11 } ,fill opacity=1 ] (393.99,192.78) -- (393.99,194.84) -- (392.68,192.78) -- (393.99,192.78) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 208; green, 2; blue, 27 } ,fill opacity=1 ] (399.08,192.87) -- (399.08,201.98) -- (395.08,201.98) -- (395.08,192.87) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 117; green, 214; blue, 11 } ,fill opacity=1 ] (399.08,201.98) -- (398.14,201.98) -- (395.08,197.2) -- (395.08,194.87) -- (399.08,201.14) -- (399.08,201.98) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 117; green, 214; blue, 11 } ,fill opacity=1 ] (395.08,201.98) -- (395.08,199.66) -- (396.56,201.98) -- (395.08,201.98) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 117; green, 214; blue, 11 } ,fill opacity=1 ] (399.08,199.14) -- (395.08,192.87) -- (396.56,192.87) -- (399.08,196.81) -- (399.08,199.14) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 117; green, 214; blue, 11 } ,fill opacity=1 ] (399.08,192.87) -- (399.08,194.93) -- (397.77,192.87) -- (399.08,192.87) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 208; green, 2; blue, 27 } ,fill opacity=1 ] (374.99,235.8) -- (374.99,244.91) -- (370.99,244.91) -- (370.99,235.8) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 117; green, 214; blue, 11 } ,fill opacity=1 ] (374.99,244.91) -- (374.05,244.91) -- (370.99,240.13) -- (370.99,237.8) -- (374.99,244.07) -- (374.99,244.91) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 117; green, 214; blue, 11 } ,fill opacity=1 ] (370.99,244.91) -- (370.99,242.59) -- (372.47,244.91) -- (370.99,244.91) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 117; green, 214; blue, 11 } ,fill opacity=1 ] (374.99,242.06) -- (370.99,235.8) -- (372.47,235.8) -- (374.99,239.74) -- (374.99,242.06) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 117; green, 214; blue, 11 } ,fill opacity=1 ] (374.99,235.8) -- (374.99,237.86) -- (373.68,235.8) -- (374.99,235.8) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 208; green, 2; blue, 27 } ,fill opacity=1 ] (397.39,275.5) -- (397.39,284.61) -- (393.39,284.61) -- (393.39,275.5) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 117; green, 214; blue, 11 } ,fill opacity=1 ] (397.39,284.61) -- (396.45,284.61) -- (393.39,279.83) -- (393.39,277.5) -- (397.39,283.77) -- (397.39,284.61) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 117; green, 214; blue, 11 } ,fill opacity=1 ] (393.39,284.61) -- (393.39,282.29) -- (394.87,284.61) -- (393.39,284.61) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 117; green, 214; blue, 11 } ,fill opacity=1 ] (397.39,281.76) -- (393.39,275.5) -- (394.87,275.5) -- (397.39,279.44) -- (397.39,281.76) -- cycle ;
\draw [draw opacity=0][fill={rgb, 255:red, 117; green, 214; blue, 11 } ,fill opacity=1 ] (397.39,275.5) -- (397.39,277.56) -- (396.08,275.5) -- (397.39,275.5) -- cycle ;
\draw (275.83,197.92) node [font=\scriptsize,xscale=1.25,yscale=1.25] {${\displaystyle 1}$};
\draw (275.5,177.59) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$0$};
\draw (275.83,217.92) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$2$};
\draw (275.83,261.21) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$18$};
\draw (219.73,268.42) node [anchor=north west][inner sep=0.75pt] [font=\small,rotate=-270,xscale=1.25,yscale=1.25] {$\mathrm{Epoch} \ e$};
\draw (274.15,34.12) node [font=\small,xscale=1.25,yscale=1.25] {$\mathrm{Slot}$};
\draw (216.4,167.42) node [anchor=north west][inner sep=0.75pt] [font=\footnotesize,rotate=-270,xscale=1.25,yscale=1.25] {$ \begin{array}{l}
\mathrm{Epoch} \ \\
\ \ e-1
\end{array}$};
\draw (214.73,111.75) node [anchor=north west][inner sep=0.75pt] [font=\footnotesize,rotate=-270,xscale=1.25,yscale=1.25] {$ \begin{array}{l}
\mathrm{Epoch\ }\\
\ \ e-2
\end{array}$};
\draw (321.63,30.65) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$77$};
\draw (341.43,71.65) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$67$};
\draw (300.75,71.08) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$33$};
\draw (342,126.11) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$33$};
\draw (300.7,126.57) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$67$};
\draw (340.93,177.37) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$44$};
\draw (373.67,235.21) node [anchor=north west][inner sep=0.75pt] [font=\scriptsize,xscale=1.25,yscale=1.25] {$\mathnormal{\times } 20$};
\draw (411.33,235.54) node [anchor=north west][inner sep=0.75pt] [font=\scriptsize,xscale=1.25,yscale=1.25] {$\mathnormal{\times } 38$};
\draw (276.12,278.92) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$19$};
\draw (275.83,311.42) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$32$};
\draw (376.17,291.21) node [anchor=north west][inner sep=0.75pt] [font=\scriptsize,xscale=1.25,yscale=1.25] {$\mathnormal{\times } 30$};
\draw (300.93,177.87) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$33$};
\draw (93.65,34.12) node [font=\small,xscale=1.25,yscale=1.25] {$\mathrm{Slot}$};
\draw (35.9,167.42) node [anchor=north west][inner sep=0.75pt] [font=\footnotesize,rotate=-270,xscale=1.25,yscale=1.25] {$ \begin{array}{l}
\mathrm{Epoch} \ \\
\ \ e-1
\end{array}$};
\draw (34.23,111.75) node [anchor=north west][inner sep=0.75pt] [font=\footnotesize,rotate=-270,xscale=1.25,yscale=1.25] {$ \begin{array}{l}
\mathrm{Epoch\ }\\
\ \ e-2
\end{array}$};
\draw (141.13,30.65) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$77$};
\draw (160.93,71.65) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$67$};
\draw (120.25,71.08) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$33$};
\draw (161.5,126.11) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$33$};
\draw (120.2,126.57) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$44$};
\draw (160.72,175.76) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$\ A\ $};
\draw (121.02,175.63) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$\ B\ $};
\draw (404.03,132.65) node [font=\scriptsize,xscale=1.25,yscale=1.25] {${\textstyle \mathrm{checkpoint\ vote}}$};
\draw (406.56,152.67) node [anchor=west] [inner sep=0.75pt] [font=\scriptsize,xscale=1.25,yscale=1.25] {$\mathrm{{\textstyle chain}} \ A$};
\draw (398.94,152.98) node [anchor=east] [inner sep=0.75pt] [font=\scriptsize,xscale=1.25,yscale=1.25] {$\mathrm{{\textstyle chain}} \ B$};
\draw (340.9,330.4) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$\ A\ $};
\draw (300.83,330.43) node [font=\scriptsize,xscale=1.25,yscale=1.25] {$\ B\ $};
\end{tikzpicture}
}
\caption{\small This figure presents a detailed version of the bouncing attack. In this example, we have a total of 100 validators, of which 23 are Byzantines.
Checkpoints are represented by hexagonal shapes. A double hexagon signifies that the checkpoint is justified. A block in a checkpoint corresponds to the block associated with that checkpoint.
The number inside each hexagon (hovering a block) corresponds to the number of validators who made a checkpoint vote with this checkpoint as target. We distinguish between two sorts of checkpoint votes, the Byzantine ones, which are bi-color rectangles,
and the honest ones, which are uni-color rectangles.
We compile the 3 steps of \autoref{fig:simplifiedBoncingAttack} in 2 with more information on how justification's turning point is accomplished because of the Byzantine agents. \textbf{First step:} We begin from a situation where epoch $e-1$ just ended and we now reach \texttt{GST}. Notice that the candidate chain is chain $A$ because the checkpoint with the highest epoch is on chain $A$ but not chain $B$. \textbf{Second step:} In this step, the checkpoint vote released during epoch $e$ can change the last justified checkpoint to change the candidate chain for chain $A$ to chain $B$. Byzantine validators released their checkpoint vote from the previous epoch during epoch $e$. They send their last checkpoint vote at slot 23 once the checkpoint of epoch $e$ on chain $A$ has reached 44, thus becoming justifiable (i.e., not yet justified but with enough votes so that Byzantine validators can justify it). This trigger the candidate chain to change from chain $A$ to chain $B$ starting the \emph{bounce}.}
\label{fig:complexBouncingAttack}
\end{figure}
\subsection{Implemented patch}
The explication of the patch is described on the Ethereum research forum \cite{nakamura_prevention_2019}.
The solution that was found to mitigate the bouncing attack is to engrave in the protocol the fact that validators cannot change their minds regarding justified checkpoints after a part of the epoch has passed.
The goal of the solution proposed is to prevent the possibility of justifiable checkpoints being left out by honest validators.
To prevent honest validators from leaving a justifiable checkpoint, the patch must stop validators from changing their view of checkpoints before more than 1/3 of validators have cast their checkpoint vote.
This condition stems from the fact that we reckon the proportion of Byzantine validators to be at most $1/3-\epsilon$.
To apply this condition, the patch designates a number of slots after which honest validators cannot change their view of checkpoints.
Indeed since validators are scattered equally among the different slots to cast their vote (in attestations) in a specific time frame, stopping validators from changing their view after a number of slots is equivalent to stopping changing their view after a certain proportion of validators have voted.
This does look like a solution to prevent Byzantine validators from influencing honest validators into forsaking a checkpoint now \emph{justifiable} for them.
To enforce this behavior, called the ``fixation of view'', the protocol has a constant $j$ called \texttt{SAFE\_}\texttt{SLOTS\_}\texttt{TO}\texttt{\_UPDATE} \texttt{\_JUSTIFIED} in the code (cf. \autoref{algo:syncBlock} in \autoref{subsec:code}).
This constant is the number of slots\footnote{At the time of writing this paper, $j=8$ \cite{github_specs}.} until when validators can change their view of the justified checkpoints.
In the patch introducing this constant $j$, they mention a possible attack called \emph{splitting attack}. As they point out, the splitting attack relies on a ``last minute delivery" family of strategies whereby releasing a message late enough, some validators will consider it too late while others will not. This could thus split the validators into two different chains, not being able to conciliate their view before the end of the epoch. They consider the assumption of attackers being able to send a message at the right time to split honest validators too strong, we discuss in \autoref{sec:relatedWorks} whether it is the case or not.
In \ref{subsec:probabilisticBouncingAttack}, we present a new attack inspired by the splitting attack with more realistic assumptions.
\subsection{Probabilistic Bouncing attack - why the patch is not enough}
\label{subsec:probabilisticBouncingAttack}
In this part, we present our novel attack against the protocol of Ethereum Proof-of-Stake.
\subsubsection{Attack Conditions}
\label{subsubsec:attackConditions}
Our attack takes place during the synchronous period
and uses the power of \textit{equivocation} of Byzantine processes
Equivocation is caused by a Byzantine process that sends a message only to a subset of validators at a given point in time and potentially another message or none to another subset of validators.
The effect is that only a part of the validators will receive the message on time. More in detail, the bounded network delay is used by a Byzantine validator to convey a message to be read on a specific slot by some validators and read on the next slot by the other validators. Note that if a protocol is not tolerant to equivocation, then it is not BFT (Byzantine-Fault Tolerant), since equivocation is the typical action possible for Byzantine validators.
\subsubsection{Attack Description and Analysis}
Let $\beta \le f/n$ be the fraction of Byzantine validators in the system. We start by explaining the attack on the Ethereum PoS protocol before generalizing our analysis.
The attack setup is the following. First, as in the traditional bouncing attack, we start in a situation where the network is still partially synchronous. A fork occurs and results in the highest justified checkpoint being on chain $A$ at epoch $e$, and a justifiable checkpoint at epoch $e+1$ on chain $B$. Assume now that \texttt{GST} is reached, the attack can proceed\footnote{Note that before \texttt{GST}, no algorithm can ensure liveness since communication delays may not be bounded.} as follows:
\begin{enumerate}
\item Since \texttt{GST} is reached, the network is fully synchronous. Chain $A$ is the candidate chain for all validators.
\item Just before validators must stop updating their view concerning justified checkpoint (i.e., before reaching the limit of 8 slots in the epoch corresponding to the condition line 6 in \autoref{algo:syncBlock}), a Byzantine proposer proposes a block (cf. \autoref{algo:prepareBlock}) on chain $B$.
This block contains attestation with enough checkpoint votes to justify the justifiable checkpoint that was left by honest validators.
The attestations included in the block are attestations of Byzantine validators that were not issued in the previous epoch when they were supposed to.
The block must be released just in time, that is, right before the end of slot $j$, so that $(2/3-\beta)$ of the validators change their view of the candidate chain to be active on chain $B$ while the rest of honest validator continue on chain $B$. This is possible due to the patch preventing validators from changing their mind after $j$ slots.
\item Repeat the process.
\end{enumerate}
An important aspect to consider in the attack is the probability for Byzantine validators to become proposers. This is an important part since without the role of proposer, validators are not legitimate to propose blocks and thus cannot add new attestations containing checkpoint votes on top of the concurrent chain.\footnote{Note that Byzantine validators can not use their role of proposer during the previous epoch to release a block with the right attestations because it might not be the last block of the epoch. Indeed because a part of honest validators is on the concurrent chain, they also add blocks to it. For the checkpoint votes contained in the Byzantine attestations to justify the justifiable checkpoint, it must be on the same chain as the attestation making the checkpoint justifiable in the first place.}.
The probability of being selected to be a proposer directly impacts how long the bouncing attack can continue. In the following theorem, we establish the probability of a bouncing attack lasting for a specific number of epochs.
\begin{theorem}
The probability of a bouncing attack occurring during $k$ epochs after \texttt{GST} and a favorable setting is:
\begin{equation}
P(\textrm{bouncing }k \textrm{ times})=(1-\alpha^j)^k,
\end{equation}
with $\alpha \in [0,1]$ the proportion of honest validators and $j$ the number of slots before locking a choice for the justification.
\end{theorem}
\begin{proof}
We denote by $\alpha$ the proportion of honest validators and $j$ the number of slots before locking
the choice for justification.
We want to know the probability of delaying the finality for $k$ epochs.
Once we assume a setup condition sufficient to start a bouncing attack, the attack continues until it becomes impossible for Byzantine validators to cast a vote to justify the justifiable checkpoint.
To cast their vote, Byzantine validators need one of the $j$ first slots of the concurrent
chain to have a Byzantine validator as proposer. Considering the probability of choosing between each validator, the chance for a Byzantine validator to be a proposer for one of the first $j$ slots is $(1-\alpha^j)$, with $\alpha$ being the proportion of honest validators. For $k$ epochs, we take this result to the power of $k$.
\end{proof}
As we can see, the probability of the bouncing attack to continue for $k$ epochs depends on two factors: $\alpha$ the proportions of honest validators which cannot be controlled and $j$ the number of slots before which validators are allowed to switch branches.
Reducing $j$ to 0 would prevent the bouncing attack from happening (probability falls to 0), but it would mean that validators are never allowed to change their view of the candidate chain.
This naive solution would allow irreconcilable choices between the set of validators and prevent any new checkpoint from being justified, which is a more severe threat to the liveness of Ethereum PoS.
Reducing the number of slots where validators can change their view of the blockchain implies that different views cannot reconcile quickly. Or at least, the window of opportunity for doing so gets smaller.
In theory, the proportion of Byzantine validators necessary to perform this attack is $1/n$. This is because we assume a favorable setup and that Byzantine validators can send messages so that only a wished portion of honest validators receives it on time.
Our analysis focuses on the course of action of the attackers during the attack rather than the course of action necessary for it to appear.
\section{Related Works}
\label{sec:relatedWorks}
Blockchain protocol analyses can be divided into two main categories: those that specify or formalize protocols and those that identify vulnerabilities of the protocols. Our work is at the junction of the two categories because we
both formalize the protocol to permit its analysis, and we present
a novel attack.
The category of specification and formalization includes the ``white papers'' (e.g., Bitcoin \cite{nakamoto_peer_2008}, Ethereum \cite{wood_ethereum_2014}). It also contains academic papers providing formal specifications and demonstrating the properties guaranteed by the protocols. For example, \cite{garay_bitcoin_2015} formally describes, analyzes the Bitcoin protocol, and proves its security guarantees, \cite{amoussou_dissecting_2019} does the same for the Tendermint's protocol (the consensus protocol of the Cosmos blockchain\cite{buchman_latest_2018}). Our work lies in this category of formalization by formalizing the Ethereum Proof-of-Stake protocol. For what concerns protocol formalization of Ethereum Proof-of-Stake, \cite{buterin_combining_2020} is the first to propose a draft specification of the Ethereum PoS protocol and related properties. However, that specification is outdated and not complete. Our paper provides a complete formalization of the consensus mechanisms with respect to the current implemented code, along with a novel specification of its properties.
A famous example of papers identifying vulnerabilities is \cite{eyal_majority_2018}, which presents the selfish mining attack on Bitcoin. \cite{eyal_majority_2018} shows that in Bitcoin (and proof-of-work in general), miners can benefit from deviating from the prescribed protocol by withholding blocks for a while at the expense of honest miners.
\cite{amoussou_correctness_2018} points out a liveness vulnerability of the Tendermint protocol. \cite{neuder_defending_2020} presents an attack where nodes can reorganize the Tezos' Emmy+ chain and then do a double-spend attack.
Our work follows the line of research focusing on flaws of Ethereum Proof-of-Stake \cite{neu_ebb_2021,neu_two_2022,schwarz_three_2021}. Neu et al. \cite{neu_ebb_2021} exhibit a balancing attack, highlighting the shortcomings of a consensus mechanism being separated into two layers (finality gadget, fork choice rule). Mitigation against this attack was proposed, but \cite{neu_two_2022} overcame this mitigation with a new balancing attack. \cite{schwarz_three_2021} presented reorg attacks revealing that validators with the role of proposer could gain from disturbing the protocol by releasing their block late.
Our paper presents another flaw regarding the liveness of the current Ethereum Proof-of-stake protocol, on which some attacks do not seem feasible anymore, and thus insists on the importance of finding new ways to conciliate availability and finality.
Nakamura \cite{nakamura_prevention_2019} presents an attack called \textit{splitting attack} in which the adversary sends messages to split the set of validators. However, to achieve this attack, Nakamura assumes that the adversary needs to control and play with network delays. This is a strong assumption and can be considered unrealistic. More recently, \cite{schwarz_three_2021} showed through experiments that attackers can predict the proportion of validators receiving a given message within a specific time frame with sufficient accuracy. This contradicts Nakamura's claim that the attack necessitates the adversary to control the network delay.
In this work, we present a form of the splitting attack based on the weaker assumption that the adversary knows the network delay (in line with \cite{schwarz_three_2021}) but does not control it. Moreover, our attack is repeated, hence the name bouncing, being a threat to the liveness.
Outside of these two categories lie works that provide formal ground for blockchains. \cite{anceaume_finality_2021} describes the types of finality a blockchain can achieve,
\cite{anceaume_abstract_2018} proposes a formalization of blockchains and their evolutions as BlockTrees. We rely on the definition of BlockTree and finality to express the Ethereum protocol properties.
\section{Conclusion}
\label{sec:conclusion}
We described a framework for formal analysis of the Ethereum Proof-of-Stake (PoS) protocol. Our contribution provides a formal description of Ethereum PoS, which can be used for future analysis and comparison with future versions of the protocol.
We proposed a novel distinction between the definition of liveness and availability. This distinction is crucial to pinpoint the difference between Nakamoto-style and BFT consensus. It makes possible a comparison between the two. We outlined an attack against the liveness of the protocol, showing probabilistic liveness to the protocol under this attack.
This work explores whether it is possible to combine Nakamoto-style and BFT to ensure safety and liveness.
Another aspect is that our analysis did not consider the protocol's rewards and incentives. We leave this part, as well as analyzing rational behavior in the protocol, as future work.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 1,482 |
At Surrey Business School, we encourage our students to be enterprising and progressive — key abilities for all sorts of graduate employment. More broadly, the purpose is to develop students potential to be taught and manage their own learning, and to encourage efficient time management and personal resources planning. Dimata manajer, efficiency management merupakan tambahan beban kerja, disamping menjalankan tugas yang selama ini sudah dikerjakan. This module covers a variety of points related to the modules coated in a Management Studies diploma.
Key titles embrace Management Decision, (Emerald's first journal, founded again in 1963), and Cross Cultural & Strategic Management, which supplies reducing-edge analysis on all points of worldwide management. If you are interested in danger administration software I would recommend looking out the web and searching for the one which seems least sophisticated to you.
As part of Lao PDR's precedence deliverables below its ASEAN Chairmanship in 2016, ASEAN has developed this Report and Work Programme on Starting a Business" in ASEAN. You can select for which area you need to start your plastic enterprise from a variety of choices – vehicles, aircrafts, fiberglass boats, electrical and electronic gadgets, medical equipment, and machinery. Plan to keep purchases of workplace equipment and overheads to a minimum when beginning up. You do not need amazing workplace premises, the most recent in office chairs and pricey art work on the walls.
A market is the place consumers and sellers come together and exchange their merchandise for money. Content Marketing : Creating participating content that can appeal to prospective patrons to your services or products after which strategically distributing that content material on the web the place prospective clients will see and interact with it and get led to your products or services. The Isenberg MBA with a spotlight in advertising presents a mix of normal enterprise and advertising and marketing elective courses to fulfill your specific academic objectives. Referral Marketing : Building your business's buyer base by asking customers to refer other possible prospects to you.
As a part of our dedication to the expansion of small businesses nationwide, U.S. Bank is a number one participant in the lending applications of the U.S. Small Business Administration (SBA). To qualify, what you are promoting have to be in one of many 13 industries the bank funds: agriculture/poultry, dental, household leisure, funeral service, resorts, insurance coverage, funding advisory, medical, ophthalmic, pharmacy, self-storage, veterinary and wine/craft beverage. Find counseling, coaching, and business growth specialists providing free and low-cost providers in your space. The SBA licenses and regulates Small Business Investment Companies (SBICs), which are privately owned and managed investment funds. You can discover more details about these loans from Rosemary Peavler's article about SBA Loan Programs.
Click the following image if you're in search of the most effective business insurance coverage round, or read under for more helpful details about insuring your enterprise! The costs related to a business insurance coverage policy are small compared with the massive dangers that you will defend towards. It is especially important for small enterprise house owners to rigorously think about and evaluate their enterprise insurance coverage needs as a result of they may have extra personal financial exposure within the event of loss.
Consulting agency's services have come to form half and parcel of the strategic insurance policies of business entities. On 30 November there's the Service Management, Software and Service Ecosystems workshop and on 2-3 December there's a Service Management and Engineering track during ACIS. While this stuff come naturally with the development service, the development strategy maker must guarantee one of the best utilization of firm fund, refine the operation of management & completely different departments and handle any authorized difficulty that will arise because the business continues to grow. The enterprise organizing specialist will searched for buyer suggestions from the shopper care executives.
Many web sites supply the service of. Whether you are looking ahead to outsource jobs or you are looking for work, these sites are of nice assistance to you in either scenario. The average person can use social media to promote his small scale business to a wider neighborhood as nicely. Most younger folks for instance want to use social media to communicate quite than nose to nose. Please observe that Company won't be accountable for disclosures of your knowledge resulting from errors or unauthorized acts of third parties. One should be able to use it within the safest and most useful potential ways to realize the utmost profit from it.
JAKARTA/BANGKOK — Banks in Indonesia are boosting lending to operators of small and midsize corporations, who've traditionally relied on family and friends for loans, as demand for credit grows. An worldwide agriculture mission has been instrumental in improving the earnings and high quality of life for a lot of Indonesian women involved in agriculture-primarily based business enterprises, said project members and beneficiaries. One of the important benefits of those finances is that it allows all types of low credit score holders to get the cash assist without dealing with any restriction. They may ask private details, enterprise details and the main points about your Master or Visa Card. Unsecured enterprise loans are provided by many banks, financial institution or lending organizations. | {
"redpajama_set_name": "RedPajamaC4"
} | 2,614 |
\section{Introduction}
A study of photon propagation in gravitational field has revealed a surprising
phenomenon that quantum corrections modify the characteristics of photon
equation of motion in such a way that in some cases they may lay outside
the light cone. This effect was first found in ref. \cite{dh} for several
different geometries (gravitational wave, Schwarzschild, and Robertson-Walker).
In subsequent papers it has been shown that photons may propagate "faster
than light" also in the Reissner-Nordstr{\H o}m \cite{ds1}
and the Kerr \cite{ds2} backgrounds. Similar effect was found for
propagation of massless neutrinos in gravitational field \cite{yo}. The only
essential difference is that photons can propagate with $v>c$ even in vacuum
where Riemann tensor is non-vanishing, $R_{\alpha\beta\mu\nu} \neq 0$, while
neutrinos may acquire superluminal velocity only in space-time with
$ R_{\mu\nu} \neq 0$. Superluminal propagation has been also found for
flat space-time with boundaries, e.g. for photon propagation between
conducting plates \cite{ks,gb,lpt} (see also \cite{sbm}). In what follows we
will discuss both possibilities and will show that in these cases one could
find a coordinate frame where photons would return to their source
before they were created there.
The photon effective action in vacuum in one loop approximation and in the
lowest order in the gravitational field strength is described by
the well known vacuum
polarization diagram. Because of gauge invariance of electrodynamics and
general covariance of gravity the result for the action is unique and
can be written immediately:
\begin{equation}
S = \int d^4 x \sqrt {-g} \left( -{1\over 4} F_{\mu\nu}F^{\mu\nu}
+ {\alpha C \over m_e^2} R_{\alpha\beta\mu\nu}F_{\mu\nu}F^{\alpha\beta}
\right)
\label{s}
\end{equation}
Here $F_{\mu\nu}=\partial_\mu A_\nu - \partial_\nu A_\mu$ is the
electromagnetic field tensor,
$\alpha=1/137$ is the fine structure constant, $m_e$ is the mass of
electron and the coefficient $C$, as calculated in ref. \cite{dh}, is
$C= - 1/360\pi$. For our purpose is essential just that $C\neq 0$, even its
sign is not important. The action may also contain other terms proportional to
the Ricci tensor $R_{\mu\nu}$ or to the curvature scalar $R$ but they both
vanish in vacuum and will be neglected in what follows.
It is evident that this action leads to the equation of motion with modified
highest derivative terms and generally speaking the coefficient at
$\partial^2 /\partial_t^2$ would not be equal to the one at
$\partial^2 /\partial_j^2$ (with the evident factor $g_{ij}/g_{tt}$).
In other words the characteristics of the photon wave equation with the
corrections (\ref{s}) generically would not coincide with the normal
light cone. It means that the velocity of the propagation of the {\it front}
of the signal would be modified.
One should not worry if this happened to be inside the light cone
but as we have already mentioned this is not the case. In particular, as was
shown in ref. \cite {dh}, in the Schwarzschild background photons going
in a non-radial direction could propagate either faster than $c$ or slower
depending on their polarization. This may imply serious problems for the
theory.
The velocity of light front propagation is known \cite{lm}
to be determined by the asymptotics of the refraction index $n(\omega)$ for
$\omega \rightarrow \infty$. Since the action (\ref{s}) is believed to be
valid only for sufficiently low photon frequency, $\omega \ll m_e$, the
problem of superluminal propagation might be resolved by possible
unaccounted for terms which vanish in the low frequency limit. It was
argued in ref. \cite{dh} that if the refraction index satisfy
dispersion relation with a positive definite $Im \, n(\omega)$, then
$n(0) > n(\infty )$ and the problem persists. However it was shown \cite{dk}
that positiveness of $Im \, n $ in gravitational fields is not necessarily
true and one may hope that high frequency contributions
would give rise to $n(\infty ) > 1$ and correspondingly $v<c$.
Another twist to
the problem was given by observation \cite{ibk} that the expression for
refraction index calculated from the effective action
(\ref{s}) is valid in the first order of gravitational interaction for
{\it any} value of photon frequency. Indeed refraction index is determined by
the forward scattering amplitude and the latter, in turn, is exactly given by
the second term because forward scattering amplitude of a photon in the lowest
order in external field can be only a function of photon 4-momenta squared,
$k_1^2 =k_2^2 =0$ and of the transferred momentum, $q^2 = (k_1 -k_2)^2$, which
also vanishes for forward scattering. This simple argument invalidates the
possibility of resolving the problem by higher order terms in $\omega/m_e$
mentioned in references \cite{dk,gms}.
Thus in the lowest order in external classical
gravitational field, in the first order of electromagnetic coupling $\alpha$,
and neglecting quantum gravity corrections the result~(\ref{s}) determines
photon refraction index for all frequencies. In this approximation photons
in a certain polarization state would propagate outside the normal light cone.
In this connection the following two questions arise. First, if such
superluminal propagation would create any problem with causality and, second,
if higher order corrections in electromagnetic and gravitational interactions
could return photons "back to normality".
The first question has been addressed
in the literature (in practically all quoted here papers) and the almost
unanimous conclusion is that, though there exist reference frames in which
superluminal photons would reach detector, by the clock of an observer in this
frame, earlier than they were emitted but it is impossible to send photons
back to their source prior to their emission in the proper time of the source.
In other words, non-causal signals would not be possible and, in particular,
closed space-like trajectories for superluminal photons
do not exist. This conclusion was based
on the absence of Poincare invariance of the theory due to presence of a fixed
gravitating center. We will show here that this conclusion is incorrect
and explicitly construct an example of a non-causal
closed space-like trajectory for a
signal propagating outside the light cone. In view of that the
problem seems to be considerably more grave than it was apprehended earlier.
Before presenting our example let us discuss what mechanisms or what
changes in physics may in principle cancel the effect of superluminal
propagation or prevent from traveling backward in time if the effect
persists. An evident possibility is a contribution from higher order
corrections in electromagnetic and gravitational interactions. Still it is
not easy to achieve. Higher order electromagnetic corrections do not help.
They could only give an extra power of $\alpha$ and no extra terms in the
action proportional to derivatives of $F_{\mu\nu}$. It follows from
dimensional consideration and from renormalizability of quantum electrodynamics
in classical gravitational background. The structure of correction to the
effective action remains the same, $\delta S \sim (RFF)/m_e^2$. Power
counting shows that the diagrams with several external graviton legs (higher
order corrections in external field) also do not give rise to derivative of
electromagnetic field. The only dangerous diagrams are those with virtual
gravitons, and only with at least two virtual gravitons. Such diagrams could
give the terms of the following form:
\begin{equation}
\delta_2 S = C_2 (\partial F \partial F R)/ m^4_{Pl}
\label{ds2}
\end{equation}
Here $m^4_{Pl}$ is the Planck mass and in the expression in the brackets
a proper contraction of indices is made.
Since the coefficient $C_2$ cannot be large (in fact it should be much
smaller than unity), this contribution is not dangerous too for photon
frequency below $m_{Pl}$. The loops with three
virtual gravitons can give terms similar to (\ref{ds2}) but now with the
diverging coefficient $C_3 \sim \Lambda^2 /m^4_{Pl}$. Strictly speaking one
cannot say anything about the magnitude of such terms but in superstring based
theories of consistent renormalizable quantum gravity one would expect that
$\Lambda^2 \sim m^2_{Pl}$ and in these theory (theories?) the contribution
of higher order corrections from virtual gravitons would be negligible. Still
it is worthwhile to understand more rigorously if the effect of
superluminal propagation exists in a well defined quantum gravity.
Another possibility mentioned in the literature \cite{dh} is a modification
of the usual causal light cone by an effective cone which corresponds to
the fastest possible signal propagation. This would require quite serious
changes in the usual physics and moreover it is not clear how it could be
realized. The theory even with the modified effective action (\ref{s})
remains Lorenz invariant as well as general covariant. It permits to
express time running in a new reference frame through time and coordinates in
the original frame. Introduction of a new causal cone with a non-constant
maximum speed would probably result in a theory quite different from
General Relativity. So we have either to admit that general relativity is
broken at velocities very close to speed of light or to live in the world
with non-causal signals.
\section{Non-causal signals in Special Relativity}
Let us remind how one can obtain acausal signal propagation in Special
Relativity with "normal" tachyons moving with a constant speed $u$,
exceeding speed of light $c$ (which we take to be equal to 1). We assume that
a tachyon is emitted at the point $x_1$ at the moment $t_1$ and is
registered at the point $x_2$ at the moment $t_2$ of some inertial frame.
In this reference frame the evident relation holds:
\begin{equation}
t_2 - t_1 = (x_2 -x_1)/u >0
\label{t2t1}
\end{equation}
Now let us make Lorenz transformation to another inertial frame moving
with respect to the first one with velocity $V$:
\begin{equation}
x' = \gamma (x-Vt), \,\,\,\, t' =\gamma (t-Vx)
\label{x'}
\end{equation}
where $\gamma = 1/\sqrt{1-V^2}$.
In the new frame the time interval between emission and registration of the
tachyon is given by:
\begin{equation}
t'_2 - t'_1 = \gamma (t_2 - t_1 ) (1 - V u)
\label{t'2}
\end{equation}
For $V= 1/u$ the time interval $(t'_2 - t'_1)$ can be zero, which
corresponds to tachyon propagating in the second frame with infinite velocity
and for a larger $V$ the interval $(t'_2 - t'_1)$ could be even negative, i.e.
the tachyon in this frame propagates backward in time. The absolute value of
the tachyon velocity is always bigger than $c$, approaching $\pm \infty$ when
$V$ tends to $1/u$ from above or from below. Let us assume now that
at the moment when tachyon reaches the detector at the point $x_2$ (or $x'_2$
in the second frame) this detector emits another tachyon back to the source.
Let us assume for simplicity that the picture is symmetric so that
the second source/detector emits a tachyon with the same velocity $u$
with respect to itself and that both the
first and the second emitters of tachyons move in the primed reference frame
in opposite directions with equal velocities $V=1/u$. Thus both tachyons
would have infinite velocities in this frame and the signal would instantly
return to the place of origin. It is evident that in the case of $V>1/u$
both tachyons would travel into the past and the signal would return back
to the first emitters "in less than no time".
Bearing in mind the examples presented below we will consider a slightly
modified and more "realistic" construction of gedanken experiment with
tachyons, which permits to avoid collision of tachyon emitters. Let us assume
that there are two identical tachyon emitters A and B moving in the opposite
directions along parallel straight lines (along $x$) separated by some distance
$\Delta y$ (see fig. 1).
A tachyon is emitted by the source A at the time moment $t_0$ and in the
chosen reference frame it moves backward in time. At the moment $t_1$
($t_1 <t_0$) the tachyon reflects in perpendicular direction (now it moves
along $y$) and at the moment $t_2$ ($t_1<t_2<t_0$)
it collides with the emitter B. Space-time
picture of its motion (in terms of $t$ and $x$ with fixed $y$) is presented in
fig. 2. The collision of the tachyon with B triggers the emission by the
latter of another tachyon. This new tachyon moves again backward in time
along the line of motion of B but in the opposite direction (all along $x$).
At the moment $t_3$ it reflects perpendicular to $x$ and moves to A and
hits the latter at the moment $t_4$. It is evident that the system can be
chosen so that $t_4< t_0$. Space-time picture of the motion of the second
tachyon is presented in fig. 3.
The possibility of sending signal, moving faster than light,
back into the past to the source of its origin, which we have just demonstrated,
is well known in the standard Special Relativity. It is normally assumed
that tachyons move with a constant speed $u>1$ independently of space points.
In the case of propagation across two conduction plates it is not so,
the velocity of photon can be bigger then $c$ between the plates and equal to
$c$ outside \cite{ks,gb}. Still it is evident that the conclusion of acausal
propagation would also survive in this case.
Let us consider one dimensional motion along $x$ and assume that
the velocity of light is $v_l =1 $ for $|x| > d$ and $v_l =u >1$ for $|x|<d$.
For a realization of the gedanken experiment which we will discuss, a small
hole should be made in the plates so that the photon could penetrate into
the inner space. The size of the hole should be large in comparison with the
wave length and simultaneously small not to destroy the effect of superluminal
propagation. We will not go into these subtleties and consider this model as a
toy model for illustration of a possible travel into the past. The
arguments go essentially along the same lines as for "normal" tachyons.
Let us assume as above that photon is emitted at the point $x_1$ at the moment
$t_1$ and registered at $x_2$ at the moment $t_2$ in some inertial frame.
On the way the photon passes
the region between the plates where it moves faster than light with velocity
$u>1$. In this reference frame the following evident relation
holds:
\begin{equation}
t_2 - t_1 = x_2 - x_1 - 2d + 2d/u
\label{t2plate}
\end{equation}
We can again make the Lorenz transformation (\ref{x'}). The time of arrival
of the signal in the primed coordinate frame to the point $x_2'$ is
given by the expression
\begin{equation}
t'_2 -t_1' = \gamma (t_2 - t_1 ) \left( 1-V - {2d \over t_2 - t_1}\,
{u-1 \over uV}\right)
\label{t2'plate}
\end{equation}
Again the difference $(t'_2 -t_1')$ may be negative for $V$ sufficiently close
to unity. It is interesting that the sign of the ratio
$(t'_2 -t_1')/(t_2 - t_1)$ depends upon the distance between the points 1 and 2.
The gedanken experiment with the return of the signal back to the emitter
prior to the emission can be constructed in the same way as above with
the standard tachyons which moves with a constant superluminal velocity.
The space-time picture for such an "experiment" is presented in fig. 4.
The world line of the signal in the primed
system looks as following: the photon is "produced" out of nothing somewhere
between his real birth place and the detector (the moment $t''$ in the
figure) and propagates both ways to the place of birth and to the detector.
When it reaches the moment $t_0$ another photon (original) is emitted from the
source and "annihilates" with the first one at the moment $t'$.
Returning to arranging the time machine for photons in this conditions we
can do essentially the same things as in the previously considered case with
the only difference that now we have to supply both source/detectors A and B
with their own plates with holes so that one set of plates
moves together with A while the other moves with B .
\section{Acausal signals in gravitational fields}
In the case of superluminal photons traveling in a gravitational field
the excess of the velocity over $c$ is not given by a step function but
continuously drops as $1/r^4$ with the increasing distance from the
gravitating center. Evidently
this does not create any serious difficulties. What is more important is the
curvature of space-time and the dependence of distance and time intervals
upon the space-time metric and also the bending of the tachyon trajectory and
the delay of signal due to interaction with gravitational field.
We will consider motion in spherically
symmetric Schwarzschild background created by a localized matter not
necessarily forming a black hole. The metric can be e.g. written in the form:
\begin{equation}
ds^2 = a^2 (r) dt^2 - b^2 (r) (dx^2 + dy^2 + dz^2)
\label{ds21}
\end{equation}
where $r^2 = x^2 + y^2 + z^2$ and $a^2 = (1- r_g/4r)^2/(1 + r_g/4r)^2$ and
$b^2 = (1 + r_g/4r)^4$. However the explicit form of the metric functions
$a$ and $b$ is not essential. A coordinate transformation to another reference
frame moving at infinity with respect to the original one with velocity $V$
has been considered in ref. \cite{tpm}. The choice of coordinates made in this
book is not convenient for our purposes so we will take another coordinate
system which essentially coincides with that of ref. \cite{tpm} at
asymptotically large distances from the gravitating center~\footnote{The
inconvenience of the coordinate choice made in ref. \cite{tpm} is related to
the fact that for large $V$ the coordinate frame has a physical singularity
due to peculiarity of its motion near the source of gravity.}.
As one of the spatial coordinate lines we will choose the trajectory of
the superluminal photon in this metric and the coordinate running
along this trajectory we denote as $l$. The other two, denoted $x_\bot$,
are assumed to be orthogonal (in three dimensional sense).
We will consider only motion along this trajectory so that assume that
$x_\bot =0$. The metric along this trajectory can be written in the form:
\begin{equation}
ds^2 = A^2 (l) dt^2 - B^2 (l) dl^2
\label{ds22}
\end{equation}
Now let us go to a different coordinate frame which moves with respect to the
original one with velocity $V$ along $l$ at large distances from the center.
The corresponding coordinate transformation can be chosen as
\begin{equation}
t' = \gamma \left[ t -Vl - V f(l) \right]
\nonumber
\end{equation}
\begin{equation}
l' = \gamma \left[l + f(l) -Vt \right]
\label{t'l'}
\end{equation}
where the function $f(l)$ is chosen in such a way so that the crossed terms
$dt'dl'$ do not appear in the metric. One can easily check that this can be
achieved if
\begin{equation}
f(l) = \int^l dl \left( {B\over A} -1 \right)
\label{fl}
\end{equation}
In this moving frame the metric takes the form:
\begin{equation}
ds^2 = A^2 \left( dt'^2 - dl'^2 \right)
\label{ds24}
\end{equation}
where $A$ should be substituted as a function of $l'$ and $t'$.
The motion of the tachyonic photon in the original frame satisfies the
condition
\begin{equation}
u= {Bdl \over Adt} = 1 + \delta u
\label{u}
\end{equation}
The quantum correction to the speed of light $\delta u$ is assumed to
be positive. As was shown in ref. \cite{dh} it is indeed the case for certain
photon polarization. In the case that the polarization turns out to
correspond to subluminal propagation we can always put (in our gedanken
experiment) a depolarizor to change the polarization to the
superluminal one. Radially moving photons in Schwarzschild background has
normal velocity, $u=c$, so that we have to choose a trajectory with a nonzero
impact parameter. However in dilaton gravity, as shown in ref. \cite{cho},
a superluminal photon velocity is possible even for radial trajectories. In
this case the arguments proving an existence of time machine would be slightly
simpler.
The correction to the velocity of light is extremely small,
roughly it is
\begin{equation}
\delta u = { \alpha C_v r_g \rho\over r^4 m^2_e }
\label{deltau}
\end{equation}
where $r$ is the distance to the gravitating source and $\rho$ is the impact
parameter.
The coefficient $C_v$ is numerically small,
about $\alpha /30\pi \approx 10^{-4}$, but the
main suppression comes form the enormously small factor $1/(rm_e)^2$.
Correspondingly the time interval between emission and
absorption of the superluminal photon is equal to
\begin{equation}
t_2 - t_1 = \int_{l_1}^{l_2} dl \left( 1 - \delta u \right) (B/A)
\label{t2grav}
\end{equation}
As follows from expression (\ref{t'l'}) this time interval in the primed system
is
\begin{equation}
t'_2 -t'_1 = \gamma \left[ (1-V) \int_{l_1}^{l_2} dl ( B / A)
- \int_{l_1}^{l_2} dl \delta u (B/A) \right]
\label{t'2grav}
\end{equation}
Clearly for sufficiently small $(1-V)$ this difference can be negative and
the signal can propagate into the past. Its
behavior as a function of the distance $(l_2-l_1)$ is similar to that found
above for the case of propagation between conduction plates (see
eq.~(\ref{t2plate}).
To arrange traveling backward in time we will do now the same trick as was
described above for the case of flat space-time. We assume that in the primed
system the photon detector moves symmetrically with respect to the photon source
i.e. it moves together with the attached to it another gravitating body
along the parallel path
with the same velocity but in the opposite direction.
The arguments proving the possibility
of traveling backward in time in this frame exactly repeat the previous ones
for the propagation across the plates.
It is worth noting that the gravitational field of the body moving together
with the detector influences the motion of the tachyon and its source
and vice versa.
However for the motion with ultrarelativistic velocities this gravitational
field is concentrated in a very thin plane moving together with
the body and perpendicular to the direction of the motion. Thus the influence
of this field on the magnitude of the tachyon velocity would be extremely short
and hence negligible. The direction of the tachyon motion might be changed quite
significantly but this change could be immediately corrected by a mirror
or by an emission of a new tachyon with the same properties from the place
of the collision with the field.
\section{Conclusion}
Thus we have shown that quantum corrections to the photon
propagation in a curved
space-time or across conducting plates, which lead to superluminal velocity,
would indeed permit travel backward in time, as one would naively expect.
This statement is in contradiction with the previously published
papers~\cite{dh,gms} where it was claimed that though it would be possible
in a certain coordinate frame to arrive to detector earlier than the photon
had been emitted, still a return to the original emitter prior to the birth
of the photon was impossible. This statement was based on the assumed absence
of the Poincare invariance in the system under consideration. However, as one
can see, the Poincare invariance still exists but in a slightly more complicated
form, namely one should consider the photon source or detector together with
the attached to each of them gravitating center as a single entity. After
this observation it becomes practically evident that traveling
backward in time due to quantum corrections is indeed possible.
Thus we face the following dilemma, either time machine is possible in principle
or something is wrong in the conclusion of superluminal propagation due to
quantum corrections (vacuum polarization). It was argued \cite{idn1,idn2} that
one can still has a consistent physics even if time travel is possible.
The other option that some unaccounted for effect may kill superluminal
propagation and permit to return to normality, looks more conservative and
for many people more natural. However at the moment it is not clear how this
cure can be achieved. The analysis of higher order corrections seems to
support the present conclusion of superluminal propagation.
{\bf Acknowledgment.} We thank
I. Khriplovich, E. Kotok, M. Marinov, H. Rubinstein, and
A. Vainshtein for discussion and criticism.
This work was supported in part by Danmarks Grundforskningsfond through its
funding of the Theoretical Astrophysical Center (TAC).
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 8,960 |
HomeOpinionIndia needs Muslim neuroscientists and Dalit Fields medalists among its STEM researchers
India needs Muslim neuroscientists and Dalit Fields medalists among its STEM researchers
Devesh Kapur
9 August, 2018 10:20 am IST
Representational image | Alyssa Banta/Newsmakers
A country that does not leverage the genius of half a billion of its own people is unlikely to go very far.
Amidst the continuing travails of Indian higher education, exemplified most recently by the botched attempt to create Institutions of Eminence, how has India been faring in research generally?
Among large countries, India is an outlier — its aggregate expenditures on research and development (R&D) are modest, both in absolute numbers as well as the low share of R&D in GDP. Furthermore, within this expenditure, public spending looms much larger than in other countries and the role of higher education is de minimis. (Table 1).
Table 1 | Source: NSF Science and Engineering Indicators, 2018.
While the low share of higher education is unsurprising given its tribulations, the comparatively low share of business in R&D speaks of the low priority given by Indian businesses to innovation as a driver of growth. Indeed, these numbers would be even lower but for the growing presence of so-called Global Innovation Centers (GICs) established by MNCs in India over the last 10-15 years.
Also read: Number of women enrolling in higher education rises 1,350 per cent in 7 years
For instance, in 1997 US MNCs' expenditure on R&D carried out within India was barely $22 million and India was not even among the top 20 countries where US firms conducted R&D. By 2014, US firms were spending more than $2.9 billion – more than a 100-fold increase – and India was ranked sixth for overseas R&D by US firms.
Despite these limited expenditures, the production of research papers coming out of India has sharply increased over the last decade (Table 2).
This increase needs three qualifications. First, quantity is not quality. Much of this is mediocre and India's share in 'high impact' or 'most cited' publications considerably lags the overall increase in publications output.
Also read: Academic research on India in the US: for whom does the bell toll?
Second, the increase in output is concentrated in more 'mature' areas of science and less in so in the cutting edge emerging areas such as artificial intelligence or quantum computing. India does relatively well in research in fields such as engineering, chemistry and computer science, while lagging in areas such as medical sciences, mathematics and social sciences.
Third, Indian researchers are relatively parochial in that their international collaborations are low. For instance, while 23 per cent of all international collaborations in research publications by researchers based in the United States are with collaborators in China, just 3.5 per cent are with researchers based in India. The difference is almost double than what would be predicted by the relative difference in the overall research publications output of the two countries.
Nonetheless, the substantial growth in publications is welcome progress. A key factor in building on this momentum is the training of researchers both within and outside India and in the case of the former, encouraging post-doctoral work overseas.
The largest number of Indian researchers training overseas is in the United States, and while their number has increased considerably over the last decade, the annual number – about 2,300 – is relatively small compared to the country's needs. While in the past the majority would stay behind in the US, that has dropped to about half more recently.
However, many others move from the US to other locales whether Europe, Canada or elsewhere, rather than necessarily moving back to India.
Given the paucity of quality faculty, India cannot easily ramp up doctoral training which lies at the heart of building robust long-term research prowess. The long-term solution obviously lies in much stronger schooling systems and build on those foundations by substantially improving college education, thereby deepening the pool from which more and better trained researchers can be drawn from. But the long-term is just that – a wishful aspiration for now. What can be done in the more immediate future?
One possibility is to draw from the post-war Japanese experience when the country was faced with a severe shortage of research capacity and training in universities. At that time doctoral students were allowed to pursue their research in corporate labs, which rapidly increased the pool of researchers and boosted applied research and innovation that propelled Japan's post-war economic miracle. Given the profusion of GICs, this would boost research and training in precisely the type of leading edge technologies that India needs.
A second possibility is for India to focus particularly in research areas with low capital intensity, unambiguous standards of quality and low language entry barriers, and where there is clear demand in the foreseeable future. The most obvious is mathematics and statistics. This would require developing a structured system of math camps and competitions at the bloc, district, state and national levels with clear entry pathways to India's best higher education institutions that are distinct from the current system. A century ago, other than mathematics, Srinivasa Ramanujan failed all his other subjects and therefore could not join the University of Madras. Somehow, he found a way to be recognised. Imagine how many young Indians have exceptional skills in this one area but have fallen by the wayside because of the system's inflexibility.
A third option would be to facilitate international collaborations of Indian researchers. Just the exposure itself would be rewarding to younger researchers. In this, India needs to get out of a mindset that is fixated on elite American institutions and reach out in specific areas to collaborate with counterparts in Brazil, Mexico, Cuba in Latin America, the Nordic countries, France and Germany in Europe, and Taiwan, Japan or South Korea in Asia, just to take some examples.
A fourth pathway is to award overseas fellowships to leading global higher education institutions for students from marginalised communities, conditional, of course, on their being admitted to these institutions. India needs Dalit Fields medalists, Muslim neuroscientists and Bhil artificial intelligence researchers, not only because they will contribute to global knowledge and serve as role models that excite younger researchers in their communities, but also because a country that does not leverage the genius of half a billion of its own people is unlikely to go very far.
Devesh Kapur is the Starr Foundation South Asia Studies Professor and Asia Programs Director at the Paul H. Nitze School of Advanced International Studies at Johns Hopkins University.
Institutes of Eminence'
Rameshwar Singh 11 August, 2018 At 9:27 pm
Heading of the article is funny. It is not discussed in the article at all, which is otherwise detailed covering relevant aspects. The titlereflects the conspiracy to create and fan dissentions where none exist.
Sanjiv 11 August, 2018 At 10:06 am
Very well analysed and states the axiom of the Indian education system. But not quite sure, why did he chose to specify Muslims and Dalits? When, in the general category also, the number of research papers, does not justify their contribution as they participate in the population numbers.
Nirmala 10 August, 2018 At 10:27 am
The great writing on the real need of the industry and its future of the industry and the society
Ankur Jairath 10 August, 2018 At 10:27 am
No India doesn't need any Hindu or Muslim or Dalit Scientists…but what India need is good platform for R&D where Scientists can work for the progress of the society….Religion or caste do not matter…Only intellectual knowledge matters…..
Savy 9 August, 2018 At 8:43 pm
It has become fashionable to demand reservation everywhere. There is only 1 govt college which has a research background where students get in based on KVPY, NEET and JEE scores. This is a central govt institute called IISC. There are scores of other private colkeges with the best being Manipal with 40 times the fees of IISC. So question is whether author wrote some paras on mariganalised society with no context to research in India, not surprising given the spate of secular liberal pieces that we see so disconnected from reality. | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 2,167 |
2021 Toyota Highlander Model Review - Arlington, TX
New Toyota Highlander Inventory
Compare Toyota Highlander
What's New For the 2021 Toyota Highlander?
As one of the most popular SUVs on the road, the Toyota Highlander continues to evolve and become more accommodating to drivers. The 2021 Toyota Highlander comes with a host of active safety features and boasts advanced technology in the cabin. The vehicle also stands out for the exterior redesign, which looks more modern and sporty. The SUV also comes with a new hybrid engine to ensure drivers can save money and conserve fuel while commuting or taking extended road trips with the family-friendly mode. The XSE trim is an exciting new addition to this model lineup.
Available Features in the 2021 Highlander
There are ten trims between the Highlander and Highlander Hybrid models. Both share some of the same trim levels, while the 2021 Highlander adds a few additional trims to its lineup. The Highlander's first trim is the L and includes Wi-Fi charging, folding seats in the back two rows, a sound system with six speakers, a power-operated driver's seat, and three-zone climate control. Its technology includes an eight-inch touchscreen interface that supports Amazon Alexa, Android Auto, and Apple CarPlay.
The second trim, the LE, includes fog lights and a leather steering wheel. With the XLE comes a sunroof and roof rails. The seats are upgraded in material and design with simulated leather upholstery and heating-the seating configuration changes to seven with captain's chairs. XSE trim adds ambient interior lighting and navigation. The Limited upgrades its seats again to leather and ventilated and heated seats.
The Platinum trim that tops the line gets a 12.3-inch infotainment display and a panoramic sunroof. The Hybrid LE, which is the base trim, combines the first two Highlander trims' features. The XLE trim follows with a wireless phone charger and heated seats. The Limited gets navigation and leather upholstery, while the Platinum includes a panoramic sunroof and 12.3-inch touchscreen.
Highlander Performance
When you want something quick with solid acceleration, the new Highlander delivers. The SUV comes standard with a V6 engine that produces 295 horsepower and is paired with an eight-speed transmission. A hybrid option is also available for those who want a more green vehicle. The gasoline engine and two electric motors produce a combined 243 horsepower. Despite its size, the SUV can reach 60 mph in 6.8 to 8.2 seconds and proves to feel peppy. The car also manages to stay composed with minimal body roll when taking sharp turns at higher speeds.
2021 Toyota Highlander Safety Technologies
One of the new Highlander's main selling points is its excellent crash test rating, as the NHTSA earns it five stars. It's also named a 2021 Top Safety Pick by Insurance Institute for Highway Safety. The model has a host of safety features with lane-keeping assist, emergency braking with pedestrian detection, and automatic high beam headlamps for more visibility at night.
Additional safety features include a rear-seat reminder, a rearview camera for increased visibility, traffic sign recognition, and lane trace to reduce the risk of a collision.
Visit Us in Arlington Today
Feel free to reach out to our team today to schedule a new 2021 Toyota Highlander test drive at our dealership. We'll be happy to answer your questions and assist you in viewing our large inventory of vehicles on our lot when you're ready to buy a new car.
Highlander Questions
Review & Compare Toyota
Used Toyota Specials
Friday, 01 October, 2021
Wednesday, 28 July, 2021
Wednesday, 10 March, 2021
*Get Today's Price is available to all customers and can also be obtained by calling or coming into the dealership today. Purchase prices do not include tax, title, license and 150.00 doc fee. Please verify all information. We are not responsible for typographical, technical, or misprint errors. Inventory is subject to prior sale. Contact us via phone or email for more details. | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 6,270 |
\section{Implementation Details}
We use the same architecture as \cite{Gabbay2020Demystifying} for the generator and the encoders. $G$ is a deep neural network consisting of $3$ fully connected layers and six convolutional layers, with the final output size being an image of size $64 \times 64$ with three colour channels. We use the default values for the noise power $\sigma = 1$ (added to the content code) and regularization coefficient $\lambda = 0.001$ (regularizing the content codes) to train $G$. The generative model is trained in a non-adversarial fashion, following \cite{bojanowski2018optimizing}.
For the adversarial training of $F$, we follow PGD-based adversarial training with random patch locations. At each step, a batch of $128$ samples is selected and a number of $n_m$ samples from different classes (persons) are selected for each samples, where $n_m$ represent an integer parameter that controls the level of semi-hard example mining. We mine for hard examples before any adversarial training, but after all static augmentations (including cutout) and we pick the negative pair with the lowest feature distance out of the $n_m$ candidates.
For all samples, we apply random mirroring, random shifts of at most $5$ pixels and Gaussian cutout with probability $p=0.5$, in which a rectangle with random surface between $0.02$ and $0.2$ of the image and aspect ratio between $0.5$ and $2$ is selected, and replaced with random pixels. This is similar to the random noise initialization used in PGD-based adversarial training (and valid for us, since we assume the adversary is unconstrained in norm).
Then, for the samples where cutout is applied, we perform adversarial training with a number of $\{7, 10, 20, 40\}$ steps with step sizes $\{30/255, 16/255, 12/255, 6/255\}$, respectively. Thus, each batch consists of (on average) $128$ clean pairs and $128$ augmented pairs (where we consider adversarial training a form of augmentation).
For weak adversarial training, we attack all samples in the batch and select a random patch location with random surface between $0.02$ and $0.1$ of the image and aspect ratio between $0.5$ and $2$. We use a smaller upper bound for the surface of the patch, since we find that training with $0.2$ does not decrease the test loss significantly after $2000$ steps but still exhibits the same overfitting phenomenon, thus a good solution cannot be obtained even with early stopping.
For DOA, we attack all samples in the batch and search for a square patch of size $20 \times 20$ pixels (which matches the average patch size and aspect ratio for the proposed approach), with a stride of $5$ pixels, since searching across a range of aspect ratios is computationally intractable. For both our method and DOA, we find that the training is not critically sensitive on the number of steps, thus we train with $10$ steps and step size $16/255$ for both. We train all models for $20000$ steps (or, for weak adversarial training, until divergence occurs) using an Adam optimizer with learning rate $0.001$ and default values $\beta_1 = 0.9$, $\beta_2 = 0.999$.
\section{Additional Results}
\subsection{Performance without Adversarial Training}
Table \ref{table:clean_performance} shows the results for our approach without any adversarial training (but with both test time augmentations included), where the impact of weak adversarial training can be measured and it is revealed as a necessary component for a robust solution (but not for clean performance).
\begin{table}[h!]
\caption{Performance of our proposed approach with and without added adversarial training. Note that only with $n_m = 2$, a very weak form of semi-hard example mining, our method converges to a stable solution (in terms of clean accuracy) without adversarial training.}
\label{table:clean_performance}
\centering
\begin{tabular}{cccccccc}
\toprule
&
\multicolumn{1}{c}{Clean}
&
\multicolumn{2}{c}{Eyeglasses}
&
\multicolumn{2}{c}{Square Patch}
&
\multicolumn{2}{c}{Eye Patch}\\\hline
& AU-ROC & AU-ROC & AU-PR & AU-ROC & AU-PR & AU-ROC & AU-PR \\
Without AT & $0.977$ & $0.590$ & $0.669$ & $0.695$ & $0.757$ & $0.657$ & $0.608$ \\
With AT & $0.978$ & $0.868$ & $0.896$ & $0.850$ & $0.882$ & $0.893$ & $0.903$ \\
\bottomrule
\end{tabular}
\end{table}
\subsection{Performance under Random Noise}
We evaluate the proposed approach against two types of random perturbations. In the first we consider a random noise attack, where an intruder adversary does not run a gradient descent algorithm to find the value of the perturbation, but instead generates a number of $1000$ random noise patterns, fills the perturbation mask with them and then selects the one that minimizes feature space distance to the target. The detection AU-ROC under for a random eyeglasses attack suffers a minor drop from $0.978$ (clean) to $0.973$ when accumulated over $100$ targets and $900$ intruders for each target, thus we consider this another sanity check against gradient masking.
The second perturbation is additive random noise applied to the entire image. This investigates the robustness of the model under non-adversarial global perturbations. We add uniform pixel noise with values between $[-10/255, 10/255]$ and obtain an AU-ROC of $0.965$ compared to a clean value of $0.978$, indicating that the proposed scheme is robust to non-adversarial uniform (across the entire image) noise as well.
\subsection{Weak Adversarial Training}
Figure \ref{fig:val_loss_weak} shows the critical overfitting phenomenon highlighted in the main text happening for all tested combinations of adversarial steps and step sizes. We use early stopping on a validation set of $100$ persons to pick the best performing model and use it as a reference (the \textit{Weak AT} line in our main results).
\begin{figure}[h!]
\centering
\includegraphics[width=0.9\textwidth]{val_loss_weak.png}
\caption{Test loss of weak adversarial training, where random patch locations are selected at each step. Critical overfitting occurs for all tried step sizes, indicating that random patch location is not sufficient to ensure convergence to a robust solution.}
\label{fig:val_loss_weak}
\end{figure}
\section{Details on Adversarial Attacks}
To implement our attacks, we adopt the hyperbolic tangent formulation of \cite{carlini2017towards}. That is, the perturbation $\delta$ is the solution to the optimization problem
$$
\min_\delta \norm{F(\tanh{(x + \delta \cdot \mathcal{M})}) - F(x_t)}_2^2
$$
\noindent where $x$ is the input image converted with the hyperbolic arc-tangent function, and we also use appropriate scaling factors to ensure the resulting image belongs to $[0, 1]$. For evasion attacks, we use the same objective, but maximize it instead (minimize its negative). For the eyeglasses and square patch attacks, we use a number of $1000$ iterations with learning rate $0.01$. To pick the optimal patch location, we first run a number of $20$ iterations at $0.1$ learning rate for all possible patch locations (with a stride of $5$ pixels), pick the location and then restart optimization. For the universal eye patch, we use $5000$ iterations at $0.005$ learning rate. For the distal attack, we use $2000$ iterations at learning rate $0.01$.
Figure \ref{fig:mask_types} shows examples for the three perturbation masks used in our evaluation. For the eyeglasses perturbation mask, we use a public domain image available at \href{https://pngio.com/images/png-a1902002.html}{https://pngio.com/images/png-a1902002.html}, which is reshaped to an image of size $32 \times 16$ and placed in the same location for all images.
For the square patch attack, we pick a $10 \times 10$ pixel mask in the region of the image, searching with a stride of $5$ pixels in both dimensions. Figure \ref{fig:best_mask_location} plots the search region, as well as the distribution of near-optimal locations for the attacks, where a preference for the eyes can be noticed. For the universal eye patch attack, we place a rectangular patch of size $32 \times 12$ in the eye region. To find the eye region, we average the eye center annotations in the CelebA dataset.
\begin{figure}[h!]
\centering
\includegraphics[width=1.\textwidth]{mask_types.png}
\caption{The three perturbation masks an adversary is allowed to operate on. \textbf{Left}: Eyeglasses. \textbf{Middle}: Square patch (location varies). \textbf{Right}: Universal eye patch.}
\label{fig:mask_types}
\end{figure}
Figure \ref{fig:examples_eyepatch} shows several examples of trained universal eye patches, the target images, and a set of test adversarial samples (not used for training the patch). It can be noticed that the eye patch is semantically meaningful and captures features of the target person.
\begin{figure}[h!]
\centering
\includegraphics[width=0.7\textwidth]{best_mask_location.png}
\caption{\textbf{Left}: Region in which the adversary searches for the best patch location. \textbf{Right}: Spatial distribution of optimal patch locations found. Lighter color indicates higher frequency.}
\label{fig:best_mask_location}
\end{figure}
\begin{figure}[h!]
\centering
\includegraphics[width=1.\textwidth]{examples_eyepatch.png}
\caption{\textbf{Left}: Target images used as anchors in feature space. \textbf{Middle}: Adversarial eye patches (up-sampled) after training on $100$ intruders. \textbf{Right}: Adversarial eye patches applied to other intruders (not in the training set, randomly picked).}
\label{fig:examples_eyepatch}
\end{figure}
\newpage
\section{More Examples of $x, y, t, u$ Quadruplets}
Figure \ref{fig:more_quadruplets} shows sets of random (not cherry picked) quadruplets. In some cases, it can be noticed that the class identity of $y$ (and, implicitly, $u$) is slightly distorted under human inspection, highlighting its use as a virtual anchor for learning similarity.
\begin{figure}[h!]
\centering
\includegraphics[width=1.\textwidth]{more_quadruplets.png}
\caption{More examples of quadruplets $x, y, t, u$. In each sub-block of $2 \times 2$ images: upper left -- $x$, upper right -- $y$, lower left -- $t$, lower right -- $u$. $(x, y$) is the positive pair and $(t, u)$ is the negative pair.}
\label{fig:more_quadruplets}
\end{figure}
\newpage
\section{Introduction}
Deep neural networks have become the \textit{de facto} model of choice for complex computer vision tasks such as large-scale image classification \citep{russakovsky2015imagenet}, face recognition \citep{Parkhi15, liu2017sphereface}, high quality image synthesis \citep{karras2019analyzing}, and denoising \citep{zhang2017beyond}. In particular, face recognition, re-identification, and verification have seen marked improvements when paired with deep learning \citep{masi2018deep}. At the same time, adversarial examples \citep{goodfellow2014explaining} have demonstrated to be a failure mode of deep neural networks not only in the digital domain but in the physical world as well \citep{kurakin2016adversarial, sharif2016accessorize, pmlr-v80-athalye18b}.
Adversarial training \citep{goodfellow2014explaining, madry2018towards} is one of the more successful empirical defense methods that has inspired numerous extensions \citep{zhang2019towards, xie2019feature}. Our work is carried out in this direction and introduces an architecture for adversarially robust face verification against \textit{physical attacks}. Our proposed method trains an embedding model using a similarity loss, a generative model to augment pairs of samples and a \textit{weak} adversarial training procedure --- where the location (but not the pattern) of the perturbation is chosen randomly.
As a motivating example, consider the following scenario, where we train a robust model that recognizes whether two images are of the same person or not. During training, we sample a negative pair of images and add an adversarial eye patch to one of them. If the two samples contain persons facing different directions, then pose also acts as a discriminator and prevents the model from learning robust features. If we would have pairs of samples that are \textit{aligned} in terms of content (pose, background, smile, etc.), then the network would focus on learning features that are meaningful when obscured by the patch. In our proposed model we use a generative model with disentangled latent factors as a form of online data augmentation to generate such pairs during training by \textit{transferring class information between samples}.
Our main contribution is, to the best of our knowledge, the first architecture that uses a generative model for data augmentation to reach convergence when trained with a \textit{weak} inner adversary. Our approach randomly selects a perturbation mask at each step. We show that this leads to higher clean \textit{and} robust accuracies over state-of-the-art methods, while being more computationally efficient. State-of-the-art approaches implement a non-augmented model trained using a near-optimal inner adversary. Furthermore, we find that non-augmented adversarial training with random patch locations -- as approximators to physical attacks -- does not converge to a robust solution and instead exhibits a critical overfitting phenomenon \citep{rice2020overfitting}.
\section{Proposed Method}
Let $G(z_\mathrm{cl}, z_\mathrm{co})$ be a deep generative model with a partitioned latent space, such that $z_\mathrm{cl}$ is responsible for modulating the class identity of the generated image $x$, whereas $z_\mathrm{co}$ carries all information about the content. Let $E_{cl}(x)$ and $E_{co}(x)$ be two deep neural networks (encoders) that extract the class and content information from $x$, respectively. Together, the composed model $G(E_{cl}(x), E_{co}(x))$ can be viewed as an autoencoder with a disentangled latent space. A formal notion of disentanglement is still an open problem, but we use the recent definitions introduced in \cite{Shu2020Weakly}: we consider a generator disentangled if modifying one of the two components does not change the other when re-extracted from the new sample by an oracle function. While we do not have a closed-form expression for the oracle functions, we think of them as equivalent to human-level perception.
Let $x$ and $t$ be two samples with potentially different ground truth class identities $c_x$ and $c_t$. Our goal is to learn a binary decision function $D$ that takes as input the pair $(x, t)$ and returns one if $c_x \neq c_t$ and zero otherwise. A zero output signifies $D$ perceiving that $x$ and $t$ have the same class identity. We use the feature embedding from a deep neural network $F$ and rewrite the detector that uses $x$ as \textit{target} and $t$ as \textit{candidate} image as
\begin{equation}
D(t; x) =
\begin{cases}
1, & \norm{F(t) - F(x)}_2 \geq \delta \\
0, & \text{otherwise} \\
\end{cases}
\end{equation}
\noindent where $\delta$ is a detection threshold and $x$ has class identity $c_x$, assumed to be unaltered by any adversary. In practice, for a face verification system, $x$ can be a single image taken of the target person in a physically secure environment. To efficiently train the embedding function $F$, we propose a \textit{two-pair contrastive loss} term \citep{hadsell2006dimensionality}, in the form of
\begin{equation}
L(x, y, t, u) = \norm{F(x) - F(y)}_2^2 + \max\{ m - \norm{F(t)-F(u)}_2^2, 0 \}
\end{equation}
\noindent where $m$ is the margin between the positive and negative pairs. A block diagram is shown in Figure \ref{fig:block_diagram}. $x, y, t, u$ are four distinct samples such that $x, y, u$ share the same class identity $c_x$ and $t$ has a different class identity $c_t \ne c_x$.
The four samples are obtained accordingly:
\begin{itemize}
\item $x$ and $t$ are sampled independently from the underlying distributions governing $c_x$ and $c_t$, respectively.
\item $y = G \left( E_\mathrm{cl}(x), E_\mathrm{co}(x) \right)$. That is, $y$ is the pass-through reconstruction of $x$ through the autoencoder.
\item $u = G \left( E_\mathrm{cl}(x), E_\mathrm{co}(t) \right)$. That is, $u$ is obtained by generating a sample with the same class identity as $x$ but with the content information of $t$.
\end{itemize}
\begin{figure}
\centering
\includegraphics[width=0.9\textwidth]{block_v4}
\caption{\textbf{Left}: Architecture of the proposed approach. Blocks containing the same sub-models are identical (weights are shared at all times). \textbf{Right}: Example of $x, y, t, u$ quadruplets.}
\label{fig:block_diagram}
\end{figure}
In practice, we only have access to a finite number of labeled samples for each class identity, each with different content information, e.g., different pictures of the same person in different poses. To sample $x$ and $t$, we first randomly sample a class $c_x$, then uniformly sample in- and out-of-distribution images. In the non-robust version, the weights $\theta$ of the network are shared and receive gradient updates from both pairs simultaneously towards the objective function
\begin{equation}
\min_{\theta} \mathbb{E}_{x, t} \left[ L(x, y, t, u; \theta) \right]
\end{equation}
\noindent where $\theta$ are the weights of $F$ optimized with stochastic gradient descent.
\subsection{Why Autoencode $x$?}
Notice that, if the autoencoder given by $G(E_\mathrm{cl}, E_\mathrm{co})$ has zero generalization error, then $y = x$ and $F(y) = F(x)$. This would lead to a collapse of our loss function, since it would only learn to maximize the loss between different samples, encouraging the learning of a chaotic mapping that does not actually take the person's identity into account. There are two key factors that help regularize against this type of collapse, outlined as follows.
\paragraph{The autoencoder is not ideal.} The reconstruction error is distributed asymmetrically between the class and content information. While there are slight variations in the class identity, the background is often reconstructed poorly, as shown for example in \cite{bau2019seeing}. We find that this also holds true for the generative models that we use in our experiments. This helps the embedding model by signaling that background information is not relevant for good discrimination.
\paragraph{Data augmentation.} We perform patch augmentation \citep{devries2017improved, lopes2019improving}, followed by patch adversarial training on the real images $x$ but not on the synthetic images $y$ and $u$. This encourages the embedding $F$ to learn robust features that signal facial similarity.
\subsection{Why Generate $u$?}
The main purpose of $G$ in the proposed scheme is to efficiently generate negative pairs closer to the decision boundary by transferring all content information between classes, allowing $F$ to learn discriminative features regarding the class identity. In particular, for face images, the learned features should be independent of the person's pose, expression, etc. This role is played even in the absence of any augmentation in the negative pair and is further magnified by adversarial training.
Together, the generated images $y$ and $u$ serve as virtual anchors for learning the embedding $F$. Furthermore, since they both have the same class identity, our proposed loss function is similar to a triplet loss, except that the anchor is different in the two pairs, but preserves class information. It could be possible to replace $y$ with a real sample with different content information, however, this causes the learning to quickly overfit, since $u$ is the only sample produced by the generator and has spurious artifacts that act as discriminators.
\subsection{Robust Optimization}
We adopt the framework of \cite{madry2018towards} and define the robust optimization objective as
\begin{equation}
\min_{\theta} \mathbb{E}_{x, t, \mathcal{M}_x, \mathcal{M}_t} \left[ \max_{\delta_x, \delta_t} L \left( x+\mathcal{M}_x \cdot \delta_x, \mathrm{sg}(y), t+\mathcal{M}_t \cdot \delta_t, \mathrm{sg}(u) ; \theta \right) \right]
\end{equation}
The variables $\mathcal{M}_x$ and $\mathcal{M}_t$ represent binary masks that restrict the perturbation to a subset of input pixels. Critical to our approach is the fact that we \textit{randomly} sample $\mathcal{M}_x$ and $\mathcal{M}_t$, instead of including them in the inner optimization objective. This allows us to reduce computational complexity but also significantly weakens the online adversary. The function $\mathrm{sg}$ represents the stop gradient operator, meaning that $y$ and $u$ are treated as constants for the inner optimization problem. Expanding the loss function allows us to decompose the optimization over both inputs as
\begin{equation}
\min_{\theta} \mathbb{E}_{x, t, \mathcal{M}_x, \mathcal{M}_t} \left[ \max_{\delta_x} \norm{F ( x+\mathcal{M}_x \cdot \delta_x) - \mathrm{sg}(F(y))}_2^2 - \min_{\delta_t} \norm{F(t+\mathcal{M}_t \cdot \delta_t) - \mathrm{sg}(F(u))}_2^2 \right]
\end{equation}
The first term can be viewed as a form of untargeted optimization: we optimize $\delta_x$ such that $x$ loses its original class identity (when viewed through $F$). Meanwhile, the second term is a form of targeted optimization: $\delta_t$ is optimized such that $t$ is viewed as belonging to a particular class identity. Note that we neglect $m$ in the above for brevity. From this formulation we derive another form of robust optimization that operates with real (non-augmented) image pairs and we call \textit{weak adversarial training}. Letting $u=y$ and sampling $y$ from the distribution of real images, we effectively remove the generator and both encoders from the optimization, recovering the optimization problem
\begin{equation}
\min_\theta \mathbb{E}_{x, y, t, \mathcal{M}_x, \mathcal{M}_t} \left[ \max_{\delta_x} \norm{F(x+\mathcal{M}_x \cdot \delta_x) - F(y)}_2^2 - \min_{\delta_t} \norm{F(t+\mathcal{M}_t \cdot \delta_t) - F(y)}_2^2 \right]
\label{eq:weak_at}
\end{equation}
\section{Physical Adversarial Attacks and Defenses}
We use mask-based attacks as an approximation for physically realizable attacks, similar to \cite{Wu2020Defending}. That is, we assume that an adversary can modify a contiguous region of the image without any restrictions on the norm of the perturbations. The two degrees of freedom an adversary has are related to the relative size of the perturbed region $\mathcal{M}$ and its location in the image.
We assume that the adversary has complete (white-box) knowledge of the detector $D$, including the weights and architecture of $F$, as well as the image of the target person $x_t$ and their corresponding features $F(x_t)$. We test our approach against impersonation attacks (an adversary starts from an image $x$ with class identity different than $c_t$ and attempts to bypass the detector $D$ so as to label them as $c_t$) and evasion attacks (the adversary attempts to evade identification). Given a white-box impersonating adversary and a fixed position for the allowed perturbations as $\mathcal{M}$, the optimization objective is straightforward to formalize as an unconstrained feature adversary \citep{sabour2015adversarial}
\begin{equation}
\min_{\delta} \norm{F(x+\mathcal{M} \cdot \delta) - F(x_t)}_2^2.
\label{eq:feature_adv}
\end{equation}
We evaluate our performance against the following threat models:
\begin{enumerate}
\item Adversarial eyeglass frame attack with a fixed location for the mask $\mathcal{M}$ covering $2.6\%$ of the image.
\item Square patch attack \citep{Wu2020Defending} covering $2.5\%$ of the image, in which the attacker performs a search to find the best patch location.
\item Universal eye patch attack \citep{brown2017adversarial} covering $10\%$ of the image, where a single adversarial pattern with a fixed location covering the eyes is trained for multiple intruders.
\end{enumerate}
The adversarial eyeglasses and square patch are single-image attacks for which an instantaneous perturbation is found, thus are only an approximation for physical conditions. We do not consider the expectation-over-transformation framework \citep{pmlr-v80-athalye18b}, which imposes more restrictions on the attacker. Doing this makes our attacks stronger, on average, but occupying a smaller area of the image.
\subsection{Adversarial Training against Occlusion Attacks}
\begin{figure}
\centering
\includegraphics[width=1.\textwidth]{training_dynamics.png}
\caption{Train/test loss during training for the three proposed methods, with the same hyper-parameters. Weak adversarial training on real images shows overfitting, while our approach converges to a robust solution.}
\label{fig:train_dynamics}
\end{figure}
\cite{Wu2020Defending} propose to extend the robust optimization framework of \cite{madry2018towards} to defend against occlusion attacks. This poses another important question: how to choose the patch location $\mathcal{M}$ during training? This is an issue, since the ideal solution would be to perform the attack for all patch locations and then pick the best result. However, even simply computing the initial loss function at all possible patch locations is computationally prohibitive, since it requires a large amount of forward passes each batch. To solve this issue, \cite{Wu2020Defending} propose an approximate search algorithm that integrates the magnitude of the gradient across different patch locations, picks the top $C$ candidates, and finally selects the location with the highest value of the loss function after in-painting.
We ask the following question: \textit{is it possible to train a robust classifier by randomly selecting the occlusion mask at each optimization step}? As baselines, we consider weak patch adversarial training directly on the triplet loss in (\ref{eq:weak_at}) with random mask selection and the gradient-based selection in \cite{Wu2020Defending}.
\section{Experimental Results}
We use the disentangled generative model introduced by \cite{Gabbay2020Demystifying}, trained on the CelebA dataset \citep{liu2015faceattributes}. We define the class code as the person identity and the content code as all other information included in the image and split the images in 9177 training persons and 1000 test persons. The cost of training $G$ is approximately $14$ hours for $200$ epochs. For the embedding network, we use the architecture in VGGFace \citep{Parkhi15} with the following modifications: we normalize the weights of the final embedding layer to unit $L_2$-norm and we remove the ReLU activation to help regularize the model during training and enable a more expressive feature space. For the two-pair contrastive loss, we set $m = 10$.
We train $F$ using a subset of images corresponding from 2000 samples from the data used to train the generative model and validate the performance on a set of 100 test persons. The test persons are never seen by $G, E_\mathrm{cl}, E_\mathrm{co}$ or $F$. We use the Adam optimizer \cite{kingma2014adam} with learning rate $0.001$ and PGD-based adversarial training \citep{madry2018towards} with a constraint of $\epsilon = 1$. We perform a hyper-parameter search for the best combination of step size and number of steps, and find that $10$ steps with step size $16/255$ produces the best results. For our approach, we measure validation loss exclusively on real pairs of images. For the feature adversary, we run an Adam optimizer on (\ref{eq:feature_adv}) for $1000$ iterations with learning rate $0.01$ and $5$ restarts. For the square patch attack we search for the best location by running $20$ iterations with a larger learning rate across all possible locations.
Training takes approximately 11 and 25 hours for our method (excluding the cost of training $G$) and DOA, $C=10$, respectively, on a single Nvidia RTX 2080Ti GPU.
\subsection{Weak Adversarial Training}
\begin{figure}
\centering
\includegraphics[width=1.\textwidth]{measuring_similarity.png}
\caption{Feature distance between a pair of real-real and real-generated images with different class identities. \textbf{Left}: Measured with a pretrained VGGFace \citep{Parkhi15} network. \textbf{Right}: Measured with a robust network trained with DOA \citep{Wu2020Defending}.}
\label{fig:measuring_similarity}
\end{figure}
Our first result is shown in Figure \ref{fig:train_dynamics} and indicates that the proposed approach convergences with random patch locations, whereas training on real image pairs does not. This validates our approach as an online data augmentation method that enables fast converge to a robust solution \citep{rice2020overfitting}. Furthermore, even when considering the Defense against Occlusion Attacks (DOA) approach as a baseline, which performs an approximate search for the best patch location, our approach shows comparable test performance while being less expensive.
The divergence of weak adversarial training is reminiscent of mode collapse in generative adversarial networks, where the generator overpowers a discriminator that is weakly trained (either through the number of steps or the architectural choice) and the loss function collapses, implying that randomly selecting a patch location significantly weakens the inner adversary. To illustrate the role of $G$ as a helper to the inner adversary Figure \ref{fig:measuring_similarity} plots the pairwise distance between persons with different class identities when the second image in the pair is either real or generated. To measure this distance we use two networks, independent of our approach: a pretrained VGGFace network \citep{Parkhi15} and a robust network trained with DOA \citep{Wu2020Defending}, since prior work \citep{ilyas2019adversarial} has shown that robust networks serve as good feature extractors.
Furthermore, even though the test losses are comparable, our approach achieves better test areas under the receiver operating characteristic curve (AU-ROC) and precision-recall (AU-PR) curve, thus is superior for face verification purposes. We use a weak form of semi-hard example mining to further regularize the convergence of all methods. For each negative pair, we sample a number of two persons with identity different than the target and pick the one that produces a lower feature distance when matched against the target. In the case of our approach, we first reconstruct both candidate intruders and measure similarity between $t$ and $u$, instead of comparing them with the real image $x$.
\subsection{Test Time Augmentations}
We augment $F$ at test time with face mirroring, which is routinely used to boost the performance of face recognition systems \citep{schroff2015facenet}. That is, we extract representations from both the candidate image and its symmetrically flipped version and then compute the average feature distance with respect to the ones extracted (or stored) of the target image. Additionally, when we have multiple samples of the target available, we perform target sample selection: we choose as target the image which minimizes the average feature space distance to all others. In practice, this means choosing a clear, non-obscured image as the target, as opposed to outliers. We apply these augmentations to both our baselines.
We test our approach on a set of samples corresponding to $27$ randomly selected test persons from the CelebA dataset, each with at least ten samples. The results are shown in Table \ref{table:results}, from which we see that our approach is superior to both weak adversarial training (with early stopping at the best validation AU-ROC) and DOA, even though training is done with random patch locations. As an \textit{a posteriori} justification for choosing eyes as the location for the universal patch attack, we inspect the best location found in the square patch attack and note that the locations mostly focus around the eye region, which agrees with how humans perform face recognition \citep{keil2009look}.
\begin{table}
\caption{Performance of the proposed approach with different test time augmentations. We compare against weak patch adversarial training and DOA \citep{Wu2020Defending} trained with a number of $C = \{4, 10\}$ mask candidates returned by the approximate search. Higher is better.}
\label{table:results}
\centering
\begin{tabular}{cccccccc}
\toprule
&
\multicolumn{1}{c}{Clean}
&
\multicolumn{2}{c}{Eyeglasses}
&
\multicolumn{2}{c}{Square Patch}
&
\multicolumn{2}{c}{Eye Patch}\\\hline
& AU-ROC & AU-ROC & AU-PR & AU-ROC & AU-PR & AU-ROC & AU-PR \\
Ours, Basic & $0.957$ & $0.773$ & $0.809$ & $0.742$ & $0.785$ & $0.775$ & $0.789$ \\
Ours, Mirroring & $0.959$ & $0.812$ & $0.845$ & $0.789$ & $0.825$ & $0.791$ & $0.804$ \\
Ours, Selection & $0.976$ & $0.831$ & $0.868$ & $0.809$ & $0.848$ & $0.884$ & $0.895$ \\
\textbf{Ours, Both } & $\mathbf{0.978}$ & $\mathbf{0.868}$ & $\mathbf{0.896}$ & $\mathbf{0.850}$ & $\mathbf{0.882}$ & $\mathbf{0.893}$ & $\mathbf{0.903}$ \\
\hline
Weak AT & $0.886$ & $0.704$ & $0.726$ & $0.700$ & $0.722$ & $0.571$ & $0.556$ \\
DOA, $C=4$ & $0.972$ & $0.830$ & $0.873$ & $0.836$ & $0.873$ & $0.826$ & $0.844$ \\
DOA, $C=10$ & $0.967$ & $0.843$ & $0.878$ & $0.831$ & $0.864$ & $0.879$ & $0.886$ \\
\bottomrule
\end{tabular}
\end{table}
\subsection{Evasion Attack}
We run a feature adversary that starts from images $t$ with the same class identity as $x$ and maximize the objective in (\ref{eq:feature_adv}) in order to evade verification. We set the detection threshold $\delta$ such that the false positive rate is equal to $5\%$ and obtain that the attack detection rates are $25.2\%$ for our proposed approach and $19.8\%$ for DOA. While this marks evasion attacks as a weakness for both methods, the proposed approach is still able to obtain an improvement. The drop in performance can be intuitively explained by the fact that the adversary is no longer constrained by a targeted attack (i.e., impersonating a specific person).
\subsection{Indirect Anchor Attack}
Recall that our model is never exposed to pairs of real samples during training and that the two generated samples $y$ and $u$ share the same person identity and act as virtual anchors. We introduce a new type of attack in which the feature adversary \textit{intentionally} attempts to copy the features of the auto-encoded target $y$, instead of $x$. We find that this attack is almost as successful as attacking the real target $x$ and that our scheme maintains its robustness, with the AU-ROC dropping from $0.868$ to $0.867$ when we consider the most successful attack across both the regular and anchor attacks.
Even though $F$ is fully differentiable --- because of the virtual anchors $y, u$ that are used for training --- there is a reasonable suspicion of gradient masking \citep{pmlr-v80-athalye18b} on the loss surface between two real images. Since our indirect attack is almost as successful as attacking the real image of the person, we conclude that the loss surface is non-degenerate and no gradient masking occurs.
Finally, we also perform a completely unrestricted feature copy attack (also referred to as \textit{distal} adversarial examples \citep{schott2018towards}) when starting from a pure noise image. Prior work \citep{ilyas2019adversarial} has shown that robust networks will produce samples that resemble the image from which the target features are obtained. Figure \ref{fig:distal_results} plots the results, where we notice that some distinctive features of the targets, such as eyes, mouth and forehead are present in the images. All images successfully converge to a low value of the feature distance, a further sign that gradient masking does not occur.
\begin{figure}
\centering
\includegraphics[width=1.\textwidth]{distals.png}
\caption{\textbf{Top}: Target images from which target features are extracted. \textbf{Bottom}: Images found by running unrestricted feature copy attack when starting from random noise images.}
\label{fig:distal_results}
\end{figure}
\section{Discussion}
Patch adversarial training using an optimal adversary at every step leads to an increase of complexity for the inner solver and raises an important practical issue: which spatial location should be chosen? The proposed approach exhibits higher clean \textit{and} adversarial accuracy than prior work that approximately searches for the patch location. Furthermore, from the results we have obtained, it is at least apparent that randomly picking the patch location is not sufficient to ensure convergence to an accurate and robust solution by training on non-augmented data. This raises the important question of if there is a fundamental trade-off between robustness and accuracy \citep{zhang2019theoretically} when defending against patch attacks, which we consider a promising direction for future work.
Recent work \citep{Wong2020Fast, zhang2020attacks, cai2018curriculum} has investigated the adjustments that can be made to weak (single-step optimization) adversarial training in order to converge to a robust solution and has found that particular schedules of the learning rate and tuning the strength of the attacker over time improve convergence. We believe that this is a very promising topic of future research when it comes to defending against physically realizable attacks and our work contributes in this direction by enabling fast adversarial training against patches and removing one layer of complexity, i.e., searching for the optimal location. Finally, in this work we did not perform any fine tuning of the learning rates for the weight optimizer in the style of \cite{Wong2020Fast}, but we believe a larger-scale study to understand the convergence properties of patch adversarial training is an interesting research topic.
\section{Related Work}
\paragraph{Adversarial Attacks and Defenses} The closest related work to ours is \cite{Wu2020Defending}, which introduces adversarial training \citep{madry2018towards} for patch-based attacks. However, their setting is different: they apply their method to a ten-class face recognition (classification) problem, where each person identity is seen during training. In contrast, we train our model at a larger scale and test on a completely different subset of persons. \cite{Yin2020GAT:} propose an integrated training procedure for robust classification and adversarial sample detection. Our approach is similar, in that we apply adversarial training to a detection criteria, but we do it symmetrically. Specializing to patch attacks, several certified robustness methods exist for classification \citep{Chiang*2020Certified, alex2020derandomized}, but these methods still must pay a price in clean accuracy. In particular, \cite{Chiang*2020Certified} discuss the issue of providing a certificate by picking a subset of random patch locations, similar to what we do for training. Recent work \citep{Wong2020Fast, rice2020overfitting, zhang2020attacks} also investigates the dynamics of adversarial training and has shown that using a weak inner solver without extra precautions, such as manually tuning the learning rate, leads to overfitting. Our proposed approach converges without any additional tuning.
\paragraph{Face Recognition and Verification} The tasks of face recognition and verification have seen significant improvements in the recent years, starting with \cite{schroff2015facenet}. In particular, hyperspherical embeddings have seen increased popularity recently \citep{liu2017sphereface, deng2019arcface}, where they have been found to be favorable for different types of margin-based losses. \cite{zhang2016joint} tackle the problem of joint representation learning and face alignment by using the same embedding function. We decouple the two tasks and use the generative model $G$ for virtual alignment. The quadruplet loss \citep{chen2017beyond} is an extension of the triplet loss \citep{hoffer2015deep} that also uses four terms for training a similarity metrics but in a conceptually different way, since they form three pairs of samples. Regarding adversarial attacks on face recognition systems, there exists a plethora of work that shows the vulnerability of deep learning-based models \citep{sharif2016accessorize, dong2019efficient, carlini2020evading}.
\paragraph{Disentangled Latent Representations} \cite{Shu2020Weakly} recently introduced a framework for formal definitions of disentanglement, which we use to provide intuition for our approach. Learning disentangled representations has been proposed, among others, with variational autoencoders \citep{mathieu2018disentangling} and generative networks \citep{tran2017disentangled, Gabbay2020Demystifying}, in both unsupervised and supervised settings. We use the particular construction of \cite{Gabbay2020Demystifying}, but our approach works with any architecture that achieves disentanglement and can benefit from future improvements in the field.
\newpage
\medskip
\small
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 8,891 |
\section{Introduction}
Binarity is widely present in all kinds of stars \citep[25\% of low-mass stars and more than 80\% of high-mass stars have at least one companion;][]{Duchene2013} and binary research constitutes a main domain of stellar astrophysics.
Binary evolution gives birth to diverse phenomena such as thermonuclear novae, supernovae type Ia, sub-luminous supernovae, merger events generating detectable gravitational waves, and objects with lower initial mass such as sub-dwarf B-stars, barium stars, cataclysmic variables, and asymmetric planetary nebulae (PNe).
Understanding the diverse impact of binarity in stellar evolution is therefore crucial but is, also, still poorly understood.
In this paper we focus on observations of post-asymptotic giant branch (pAGB) binaries, objects in fast transition ($\sim$10$^{5}$ years) between the AGB and the PNe stages, that are surrounded by a circumbinary disk \citep[][]{VanWinckel2018}.
Binarity is playing a central role in the formation of the dusty disks, which were first postulated from the infrared excess in the spectral energy distribution (SED).
These excesses cannot be attributed to expanding detached shells \citep[e.g.,][]{deRuyter2006}.
Most of the disk sources were then discovered to be binaries through radial velocity measurements \citep[][]{VanWinckel2003,vanwinckel2009,Oomen2018}.
Those observations lead to the conclusion that pAGB disks originate from the evolved star's mass loss via a poorly understood binary interaction mechanism that happens at the end of the AGB phase for low- and intermediate-mass stars (0.8 - 8\,M$_\odot$).
Millimeter observations of CO lines with the Plateau de Bure interferometer and the Atacama Large Millimeter/submillimeter Array (ALMA) resolved the outer parts of these disks and showed them to be in Keplerian rotation, and thus stable \citep[][]{Bujarrabal2013,Bujarrabal2015,Bujarrabal2016,Bujarrabal2017,Bujarrabal2018}.
The CO observations also revealed a disk-wind component, suggesting angular momentum transport in the disk.
Dust grains are inferred to have large sizes \modif{(ranging from a few microns to millimeter sizes)} and a high crystallinity fraction, based on analyses of the mid-infrared (mid-IR) spectral features \modif{and sub-mm spectral slopes} \citep{deRuyter2006,Gielen2008,Gielen2011,Hillen2015}.
The dust masses found in these disks are of the order of 10$^{-4}$-10$^{-3}$\,M$_\odot$ \citep{Sahai2011,Hillen2014}, but are highly model dependent.
Despite very different forming processes, pAGB disks are in many ways (infrared excess, Keplerian rotation, winds, dust mass, dust mineralogy and grain sizes) similar to protoplanetary disks (PPDs) around young stellar objects (YSOs).
Radiative transfer models of PPDs are able to successfully reproduce both the SED and infrared interferometric measurements on the few pAGB targets studied so far \citep{Hillen2014,Hillen2015,Hillen2017,Kluska2018}.
As the PPDs are well studied both observationally and theoretically, the very close similarity with the disks around pAGB binaries points toward a potential universality of physical processes in dusty circumstellar disks occupying a different parameter space (i.e. different formation process, presumably shorter lifetime, high stellar luminosity).
Also, such a similarity between those two types of disks raises the question of the planet formation efficiency in pAGB disks \citep[e.g.,][]{Schleicher2014}, especially as several planets are candidates of being formed in such second-generation disks \citep[e.g., NN Ser;][]{Volschow2014,Marsh2014,Parsons2014,Hardy2016}.
The interaction between the binary and the disk gives rise to several complex physical processes.
First\modif{ly}, these systems show indirect signs for re-accretion from the circumbinary disk onto the central system: the primary's photospheric spectrum shows depletion in elements that have the highest condensation temperature \citep[][Kamath \& Van Winckel, 2019, in press]{Maas2003,Gezer2015,Oomen2018}.
The scenario explaining this depletion is that the condensed elements are subject to radiative pressure and remain in the disk while the gas is re-accreted onto the central star(s) \citep[][]{Waters1992}.
However, this scenario needs to be confirmed by direct observations of the re-accretion.
Second\modif{ly}, spectral time series observations allowed the detection of bipolar jets linked to the secondary star \citep[e.g.,][]{Thomas2013,Gorlova2015,Bollen2017}.
These jets have their origin around the secondaries which should also be surrounded by an accretion disk.
Interestingly, the Very Large Telescope Interferometer (VLTI) observations of one of the most studied pAGB-binaries, IRAS08544-4431, with the Precision Integrated-Optics Near-infrared Imaging ExpeRiment (PIONIER) in the near-infrared, detected a point-source emission at the position of the secondary.
It should not be detectable if the emission was coming from a photosphere alone and it was tentatively linked with the circum-secondary accretion disk \citep[][]{Hillen2016,Kluska2018}.
The existence of this unexpected continuum emission from the secondary needs to be investigated in other systems as well.
Third, the orbits of the binaries disagree with predictions of theoretical models.
At the end of the AGB phase we expect the period distribution of the binaries to be bimodal: the systems that went through common envelope evolution should result in a shrinkage of their orbital period and wider systems should have larger orbits because of the mass loss of the primary \citep{Nie2012}.
However, radial velocity monitoring of the binary orbits revealed that the detected orbital distribution falls between these two modes and show periods that are not predicted by population studies.
Moreover, tidal circularization of the orbits is expected when the primary evolves on the giant branches, whereas observations show orbits with nonzero eccentricity values ($\sim$0.2-0.3) pointing at an eccentricity pumping mechanism \citep[][]{Oomen2018}.
Interactions between the circumbinary disk and the binary could explain some of the observed eccentricities \citep[e.g.,][]{Dermine2013,Vos2015}.
However, this mechanism is still debated as strong assumptions were made about the disk radial and vertical structure, disk viscosity and lifetime \citep[][]{Rafikov2016}.
Spatially resolved observations of the disk inner rim from which we could infer the radius, height, eccentricity, perturbation will therefore help to constrain hydrodynamic models of disk-binary interactions.
The binary eccentricity can also disturb the circumbinary disk \citep[e.g.,][]{Thun2017}.
Another possibility is that the orbital eccentricity is pumped-up by an increased mass exchange between the two stars at the periastron passage \citep[grazing envelope evolution, e.g.,][]{Kashi2018}.
This mechanism could also delay the common envelope phase extending the final orbital period \citep{Shiber2017}.
Our previous high-spectral resolution time series and high-angular resolution interferometric data enabled us to build an archetype of a pAGB binary system.
\modif{In our current state of knowledge, i}ts building blocks are \modif{likely to be}:
\begin{itemize}
\item a pAGB primary
\item a main-sequence secondary surrounded by an accretion disk which launches a wide bipolar jet
\item a binary orbit that is eccentric and not predicted by population synthesis models
\item a circumbinary disk with a dust inner rim at a radius of several astronomical units, likely ruled by dust sublimation physics and that is azimuthally perturbed \citep[e.g.,][]{Kluska2018}
\item a near-infrared extended flux from a yet unknown origin.
\end{itemize}
Here, we present an interferometric snapshot survey in the near-infrared of 23 systems in order to test this archetype.
We focus on the general properties of the different components contributing to the $H$-band flux in those systems.
The paper is organized as follows.
We describe the photometric and interferometric observations in Sect.\,\ref{sec:obs} and the geometric models in Sect.\,\ref{sec:models}.
We show the results in Sect.\,\ref{sec:results}, discuss them in Sect.\,\ref{sec:discussion} before concluding in Sect.\,\ref{sec:conclusions}.
\section{Observations}
\label{sec:obs}
\subsection{Photometry}
\modiff{The best photosphere fit to the visible part of the photometry of the sources was used to extrapolate the stellar spectrum to the $H$-band (1.65$\mu$m).}
We used the targets photometry of \citet[][]{Hillen2017,Oomen2018}.
We have compiled archival photometry \modif{on all our targets except V494\,Vel (Figs.\,\ref{fig:SED1}, \ref{fig:SED2} and \ref{fig:SED3})}.
The photometric data is coming from Johnson-Cousins bands \citep[][]{1966CoLPL...4...99J,1975RMxAA...1..299J,1978A&AS...34..477M,1997A&AS..124..349M,2001KFNT...17..409K,2002yCat.2237....0D,2006yCat.2168....0M,2007PASP..119.1083R,2008AJ....136..735L,2008PASP..120.1128O,2012yCat.5137....0A,2016MNRAS.463.4210N,2016yCat.2336....0H}, Geneva photometry \citep[][]{1997A&AS..124..349M} and Str\"omgren photometry \citep[][]{1998A&AS..129..431H,2001A&A...373..625P}.
For some targets, we also used photometry from TYCHO \citep[][]{2000A&A...355L..27H,1997A&A...323L..57H,1998AJ....115.1212U} and GAIA \citep[][]{2016yCat.1337....0G}.
For near-infrared photometry we used 2MASS \citep[][]{2003yCat.2246....0C} while for mid and far-infrared we used AKARI \citep[][]{2007PASJ...59S.369M,2010A&A...514A...1I}, WISE \citep[][]{2012yCat.2311....0C}, IRAC \citep[][]{2009yCat.2293....0S}, IRAS \citep[][]{1988SSSC..C......0H} and MSX \citep[][]{2003yCat.5114....0E}.
\modif{To fit the SED we have first derived stellar parameters (such as effective temperature, $T_\mathrm{eff}$, surface gravity, $\log g$; see Table\,\ref{table:1}) from existing spectra of the stars \citep[][]{Waelkens1991,Giridhar1994,VanWinckel1997,VanWinckel1998,Giridhar2000,Dominik2003,Maas2002,Maas2003,Maas2005,deRuyter2006} using Kurucz models \citep[][]{Kurucz1993}.}
\modif{We then defined allowed ranges around those values that the fitting algorithm can explore ($T_\mathrm{eff}\pm250\,K$; $\log g \pm0.3$ or between 0 and 2.5 if not constrained by the spectrum).
We then minimized the $\chi^2$ via a Levenberg-Marquardt algorithm between the photometric data and the reddened model (the results are displayed in Table\,\ref{tab:SEDfit}).}
\subsection{Interferometry}
The interferometric observations were obtained with PIONIER located at the VLTI at Mount Paranal in Chile.
PIONIER recombines light from four telescopes in the near-infrared $H$-band (\modif{between 1.5 and 1.85}$\mu$m).
The interferometric observables are the squared visibility amplitude (V$^2$), that is a measure of the \modif{degree of} spatial resolution of the source \modif{by a given baseline at a given wavelength}, and closure phase (CP), that is a measure of the departure from point-symmetry of the target.
The target selection (see Table\,\ref{table:1}) was based on 1) the identification of the object as a post-AGB star, 2) the presence of an H-band excess \citep[][]{deRuyter2006,Gezer2015} and 3) observability with PIONIER on the VLTI.
\begin{table*}
\caption{Binary post-AGB stars in our sample.}
\label{table:1}
\centering
\begin{tabular}{c l l c c c c c c c c}
\hline\hline
\# & Target & IRAS & RA & DEC & dist & Teff & $\log g$ & P &Ref. \\
& & & & & [pc] & [K] & & [days] & \\% table heading
\hline
1 &\object{AC\,Her} &18281+2149 & 18 30 16.2 & +21 52 00 & $ 1231^{+ 46}_{ -43}$ & $ 5500^{+ 250}_{- 250}$ & $ 0.5^{+ 2.0}_{- 0.5}$ & 1189$\pm$1.2 & (4)\\[1pt]
2 &\object{AI\,Sco} &17530-3348 & 17 56 18.5 & -33 48 43 & $11886^{+ 4870}_{-3412}$ & $ 5000^{+ 250}_{- 250}$& $ 1.8^{+ 0.7}_{- 1.8}$ & 977&(2)\\[1pt]
3 &\object{EN\,TrA} &14524-6838 & 14 57 00.6 & -68 50 22 & $ 2751^{+ 386}_{ -303}$ & $ 6000^{+ 250}_{- 250}$ & $ 1.0^{+ 0.0}_{- 0.5}$ & 1493$\pm$7&(1)\\[1pt]
4 &\object{HD\,93662} &10456-5712 & 10 47 38.3 & -57 28 02 & $ 1045^{+ 94}_{ -80}$ & $ 4250$ & $ 0.5$ & 572$\pm$6&(3)\\[1pt]
5 &\object{HD\,95767} &11000-6153 & 11 02 04.3 & -62 09 42 & $ 3820^{+ 503}_{ -401}$ & $ 7600^{+ 250}_{- 250}$ & $ 2.0^{+ 0.5}_{- 1.0}$ & 1989$\pm$61&(4)\\[1pt]
6 &\object{HD\,108015} &12222-4652 & 12 24 53.5 & -47 09 07 & $ 3867^{+ 795}_{ -582}$ & $ 7000^{+ 250}_{- 250}$ & $ 1.5^{+ 1.0}_{- 0.5}$ & 906.3$\pm$5.9&(4)\\[1pt]
7 &\object{HD\,213985} &22327-1731 & 22 35 27.5 & -17 15 26 & $ 644^{+ 27}_{ -25}$ & $ 8250^{+ 250}_{- 250}$ & $ 1.5^{+ 0.5}_{- 0.5}$ & 259.6$\pm$0.7&(4)\\[1pt]
8 &\object{HR\,4049} &10158-2844 & 10 18 07.5 & -28 59 31 & $ 1574^{+ 487}_{ -315}$ & $ 7500^{+ 250}_{- 250}$ & $ 1.0^{+ 1.5}_{- 0.5}$ & 430.6$\pm$0.1&(4)\\[1pt]
9 & &05208-2035 & 05 22 59.4 & -20 32 53 & $ 1480^{+ 78}_{ -71}$ & $ 4000^{+ 100}_{- 170}$ & $ 0.5^{+ 2.0}_{- 0.5}$ & 234.38$\pm$0.04&(4)\\[1pt]
10 & &08544-4431 & 08 56 14.1 & -44 43 10 & $ 1470^{+ 193}_{ -154}$ & $ 7250^{+ 250}_{- 250}$ & $ 1.5^{+ 1.0}_{- 1.0}$ & 501.1$\pm$0.1&(4)\\[1pt]
11 & &10174-5704 & 10 19 16.8 & -57 19 25 & $ 2613^{+ 668}_{ -452}$ & $ 6000^{+ 250}_{- 250}$ & $ 1.0^{+ 1.5}_{- 0.5}$ & 323$\pm$50&(3)\\[1pt]
12 & &15469-5311 & 15 50 43.8 & -53 20 43 & $ 3179^{+ 613}_{ -449}$ & $ 7500^{+ 250}_{- 250}$ & $ 1.5^{+ 1.0}_{- 1.0}$ & 390.2$\pm$0.7&(4)\\[1pt]
13 & &17038-4815 & 17 07 36.6 & -48 19 08 & $ 4330^{+ 1254}_{ -823}$ & $ 4750^{+ 250}_{- 250}$ & $ 1.0^{+ 1.5}_{- 0.5}$ & 1394$\pm$12&(4)\\[1pt]
14 & &18123+0511 & 18 14 49.3 & +05 12 55 & $ 6196^{+ 2256}_{-1443}$ & $ 5000^{+ 250}_{- 250}$ & $ 1.0^{+ 1.5}_{- 0.5}$ & -&-\\[1pt]
15 & &19125+0343 & 19 15 01.1 & +03 48 42 & $ 4131^{+ 905}_{ -645}$ & $ 7750^{+ 250}_{- 250}$ & $ 1.5^{+ 1.0}_{- 0.5}$ & 519.7$\pm$0.7&(4)\\[1pt]
16 &\object{IW\,Car} &09256-6324 & 09 26 53.3 & -63 37 48 & $ 1811^{+ 107}_{ -96}$ & $ 6700^{+ 500}_{- 500}$ & $ 2.0^{+ 1.0}_{- 1.5}$ & 1449&(2)\\[1pt]
17 &\object{LR\,Sco} &17243-4348 & 17 27 53.6 & -43 50 46 & $ 7325^{+ 3025}_{-1946}$ & $ 6250^{+ 250}_{- 250}$ & $ 1.0^{+ 1.5}_{- 0.5}$ & $\sim475$&(3)\\[1pt]
18 &\object{PS\,Gem} &07008+1050 & 07 03 39.6 & +10 46 13 & $ 2835^{+ 385}_{ -306}$ & $ 6000^{+ 250}_{- 100}$ & $ 1.0^{+ 1.5}_{- 0.5}$ & -&-\\[1pt]
19 &\object{R\,Sct} &18448-0545 & 18 47 28.9 & -05 42 18 & $ 1273^{+ 1079}_{ -410}$ & $ 4500^{+ 500}_{- 500}$ & $ 0.9^{+ 1.6}_{- 0.9}$ & -&-\\[1pt]
20 &\object{RU\,Cen} &12067-4508 & 12 09 23.8 & -45 25 34 & $ 1822^{+ 190}_{ -158}$ & $ 6000^{+ 250}_{- 250}$ & $ 1.5^{+ 1.0}_{- 1.5}$ & 1489$\pm$10&(4)\\[1pt]
21 &\object{SX\,Cen} &12185-4856 & 12 21 12.5 & -49 12 41 & $ 3870^{+ 721}_{ -538}$ & $ 6000^{+ 250}_{- 250}$ & $ 1.0^{+ 0.5}_{- 1.0}$ & 564.3$\pm$7.6&(4)\\[1pt]
22 &\object{U\,Mon} &07284-0940 & 07 30 47.4 & -09 46 36 & $ 1067^{+ 120}_{ -98}$ & $ 5000^{+ 250}_{- 250}$ & $ 0.0^{+ 2.5}_{- 0.0}$ & 2550$\pm$143&(4)\\[1pt]
23 & \object{V 494 Vel} & 09400 -4733 & 09 41 51.9 & -47 46 57.8 & $ 2018^{+ 147}_{ -129}$ & - & - & - & -\\[1pt]
\hline
\end{tabular}
\tablefoot{
The distances are from \citet[][]{BJ2018}
}
\tablebib{
(1)~\citet{vanwinckel2009}; (2) \citet{KissBodi2017}; (3) \citet{Hillen2017}; (4) \citet{Oomen2018}.
}
\end{table*}
Most targets were observed as part of a dedicated observing programme (European Southern Observatory (ESO) program 093.D-0573, PI: Hillen), but some were also observed as a supplement to the imaging campaign on \object{IRAS\,08544-4431} (ESO program 094.D-0865; PI: Hillen).
This explains, to some extent, the diversity in uv-coverage among the sample (see Table\,\ref{tab:log}).
The observations of IRAS08544-4431 are described in \citet[][]{Hillen2016}.
Each observation of the science star was bracketed with two calibrator stars to well interpolate the transfer function and calibrate the squared visibilities (V$^2$) and closure phases (CP).
The observations were taken using the small (A1-B2-C1-D0), the intermediate (D0-G1-H0-I1) and the extended (A1-G1-J3-K0) configurations depending on the expected size and luminosity of the object.
Therefore, not all the targets have observations on the three configurations (see Fig.\,\ref{fig:uv1} and \ref{fig:uv2}) causing an in-homogeneous (u, v)-coverage throughout the whole dataset.
All the targets were observed with a grism, leading to spectrally dispersed data with a low resolution ($R\sim$30).
We are therefore sensitive to the continuum only.
The data was reduced using the \texttt{pndrs} software \citep{lebouquin2011}.
The entire dataset is available on the Optical interferometry DataBase (OiDB)\footnote{accessible at oidb.jmmc.fr}
All the targets have squared visibilities significantly below unity (see Figs.\,\ref{fig:data1}, \ref{fig:data2} and \ref{fig:data3}) meaning that at least a fraction of the near-infrared emission is spatially resolved in all of them.
For several targets the closure phases are showing a nonzero signal indicating departure from point symmetry.
\section{Model fitting}
\label{sec:models}
In order to analyze the dataset we performed a fitting with geometrical models with increasing complexity.
We stress that the dataset is diverse in both the sizes of our targets and the obtained (u, v)-coverage.
The challenge is that the observational constraints differ from object to object.
Because of the sparsity of the (u, v)-coverages, our models do not take into account any intrinsic variability of the inner source as there is orbital motion and/or large amplitude pulsations.
We define several classes of models in Sect.\,\ref{sec:s}, \ref{sec:sr}, \ref{sec:br} and \ref{sec:b}.
We also describe our fitting strategy in Sect.\,\ref{sec:strategy}.
We start by describing in Sect.\,\ref{sec:modspec} the way these models attribute spectral dependence between the different components.
\subsection{Model definition and spectral dependence}
\label{sec:modspec}
Thanks to the linearity property of the Fourier transform, the analytic models are defined in the Fourier plane as a linear combination of different geometrical components.
The weights of this combination are the relative fluxes of the components.
Those flux contributions are defined as:
\begin{eqnarray}
\sum_\mathrm{i} f^\mathrm{i}_0 &=& 1 \label{eqn:norm}\\
f^\mathrm{i}_0 &\geqslant& 0 \\
f^\mathrm{i}_0 &\leqslant& 1 ,
\end{eqnarray}
where $f^\mathrm{i}_0$ is the flux ratio of the $i$-th component at 1.65$\mu$m.
In the models there are four possible components: the primary star, the secondary star, the \modiff{circumbinary} ring and the background.
Not all components are present in all the models.
To extrapolate the flux ratios over the observed wavelength range the components are assigned with a spectral dependence law ($f^\mathrm{i}$).
The spectral dependence of the primary star is taken from the photospheric flux in the $H$-band from the best-fit from the SED\footnote{\modif{The used parameters to model for the photospheric flux are shown Table\,\ref{tab:SEDfit}.}} and is normalized to unity at 1.65$\mu$m.
\modiff{This is possible as the contribution from the secondary and the ring are negligible in the visible because of their lower temperature and high contrast with the primary.}
The fluxes of the secondary star and the background are defined as a power-law with wavelength and a spectral index ($d_\mathrm{i} = \frac{d \log F_\lambda}{d \log \lambda}$) such that:
\begin{equation}
f^\mathrm{i} = f^\mathrm{i}_0 \Bigg(\frac{\lambda}{1.65\mu\mathrm{m}}\Bigg)^\mathrm{d_\mathrm{i}},
\end{equation}
where $\lambda$ is the wavelength of an observation and $i$ is either $sec$ or $bg$ if it is the secondary star or the background respectively.
Finally, the ring spectral dependence is defined as a black body function at a given temperature $T_\mathrm{ring}$ that is normalized to unity at 1.65$\mu$m:
\begin{equation}
f^\mathrm{ring} = f^\mathrm{ring}_0 \Bigg(\frac{BB(\lambda,T_\mathrm{ring})}{BB(1.65\mu\mathrm{m},T_\mathrm{ring})}\Bigg),
\end{equation}
where $BB$ is the black body function.
In the following sections the model geometrical descriptions are presented as well as the full model equations.
\subsection{Single star and background flux: s\texttt{0-1}}
\label{sec:s}
This first set of models includes two components: the star and the background flux.
\begin{itemize}
\item \textbf{The star:}
The star is geometrically defined by the diameter of its uniform disk ($UD_\mathrm{prim}$).
The stellar visibility $V^*(u,v)$ is therefore:
\begin{equation}
V^*(u,v) = 2\frac{J_1(\pi UD_\mathrm{prim} \sqrt{u^2+v^2})}{\pi UD_\mathrm{prim} \sqrt{u^2+v^2}}
\label{eqn:Vdisk},
\end{equation}
where $u$ and $v$ are the coordinates in the Fourier domain and $J_1$ is the first order Bessel function.
\item \textbf{The background:}
The background is the over-resolved flux which means that this component is fully resolved even for the smallest baseline.
Its visibility ($V^\mathrm{bg}$) equals 0 for all baselines.
\item \textbf{The final model} is a linear combination of the visibilities of the star and the background with as factors their flux contribution that depend on the observed wavelength (see Sect.\,\ref{sec:modspec}).
The spectral dependence of the background is either assumed to be a black-body (model \texttt{s0}) or a power-law (model \texttt{s1}).
We made this choice as the \texttt{s0} model is the starting point to all the models and, in the absence of the ring component, it is already giving a good indication for the temperature of the environment whereas in all the other models the background is modeled as a power-law.
The final visibility is expressed as:
\begin{eqnarray}
V^\mathrm{tot,s}(u,v) &=& \frac{f^\mathrm{prim} V^*(u,v) }{f^\mathrm{prim} + f^\mathrm{bg}}.
\end{eqnarray}
Thanks to Eqn.\,\ref{eqn:norm} we have:
\begin{equation}
f^\mathrm{bg}_0 = 1 - f^\mathrm{prim}_0.
\end{equation}
\end{itemize}
\subsection{Single star and a ring: sr\texttt{0-6}}
\label{sec:sr}
In this set of models there are three components: the primary star, the ring and the background.
The star is modeled as a point source and the circumstellar matter is modeled by a Gaussian ring that can be inclined and modulated, depending on the complexity of the model.
\begin{itemize}
\item \textbf{The star:}
The visibility of a point source is $V^*=1$. The position of the star with respect to the cent\modiff{e}r of the ring is defined by $x_0$ and $y_0$. Because of this shift, the complex visibility of the star is:
\begin{equation}
V^*(u,v) = \exp-2 i \pi (x_0 u + y_0 v),
\label{eqn:Vstar}
\end{equation}
where $u$ and $v$ are the spatial frequencies in the West-East and South-North directions.
\item \textbf{The ring:}
The ring is first defined as an infinitesimal ring distribution. Its visibility ($V^\mathrm{ring0}(u,v)$) equals to:
\begin{equation}
V^\mathrm{ring0}(u,v) = J_\mathrm{0}(\pi\rho'\theta),
\end{equation}
where $J_\mathrm{0}$ is the Bessel function of the 0$^\mathrm{th}$ order, $\theta$ the diameter of the ring and $\rho'$ is the spatial frequency of a data point corrected for inclination ($inc$) and position angle ($PA$) of the ring such as:
\begin{eqnarray}
\rho' &=& \sqrt{u'^2 + v'^2}\\
u' &=& u \cos PA + v \sin PA \\
v' &=& (-u \sin PA + v \cos PA) \cos inc.
\end{eqnarray}
This ring is then azimuthally modulated using a set of sinusoidal functions having a period 2$\pi$ and $\pi$ called first order and second order modulations respectively.
The ring visibility with first order modulation ($V^\mathrm{ring1}(u,v)$) and with first and second order modulations ($V^\mathrm{ring2}(u,v)$) can be written as:
\begin{eqnarray}
V^\mathrm{ring1}(u,v) &=& V^\mathrm{ring0} -i (c_\mathrm{1} \cos\alpha + s_\mathrm{1} \sin\alpha) J_\mathrm{1}(\pi\rho'\theta) \\
V^\mathrm{ring2}(u,v) &=& V^\mathrm{ring1} - (c_\mathrm{2} \cos2\alpha + s_\mathrm{2} \sin2\alpha) J_\mathrm{2}(\pi\rho'\theta),
\end{eqnarray}
where $c_1$ and $s_1$ are the first order modulation coefficients, $c_2$ and $s_2$ are the second order modulation coefficients, $J_\mathrm{1}$ and $J_\mathrm{2}$ are the first and second order Bessel functions and $\alpha$ is the azimuthal angle of the ring, starting at the major-axis (see Fig.\,\ref{fig:modpedago}).
\begin{figure}
\centering
\includegraphics[width=8.5cm]{Fig/ModulationPedago.pdf}
\caption{Illustration of the four azimuthal modulation coefficients on the geometry of the ring. For each quadrant all modulation parameters are kept to zero apart from the one displayed.}
\label{fig:modpedago}
\end{figure}
Finally, to have a Gaussian ring, one needs to convolve the infinitesimal ring in the image space by a Gaussian, which is equivalent to a multiplication in the Fourier domain:
\begin{equation}
V^\mathrm{ring}(u,v) = V^\mathrm{ring2}(u,v) \exp \frac{\pi \frac{\theta}{2} \delta\theta \sqrt{u^2 + v^2} }{4 \ln 2}, \label{eqn:ring}
\end{equation}
where $\delta\theta$ is the ratio \modiff{of the ring full width at half maximum to its radius}.
\item \textbf{The background:}
The extended flux is modeled by an over-resolved emission that has a null visibility.
\item \textbf{The final model:}
As the flux ratios of the three components are normalized to 1 at 1.65$\mu$m (Eqn.\ref{eqn:norm}), $f^\mathrm{ring}_0$ is defined as:
\begin{equation}
f^\mathrm{ring}_0 = 1 - f^\mathrm{prim}_0 - f^\mathrm{bg}_0.
\end{equation}
The final visibility can therefore be written as a linear combination of the three components of the model such as:
\begin{equation}
V^\mathrm{tot}(u,v) = \frac{f^\mathrm{prim} V^*(u,v) + f^\mathrm{ring} V^\mathrm{ring}(u,v) }{ f^\mathrm{prim} + f^\mathrm{ring} + f^\mathrm{bg} }.
\end{equation}
\end{itemize}
\subsection{Binary: b\texttt{1-2}}
\label{sec:b}
This set of models is made of two stars and a background flux.
\begin{itemize}
\item \textbf{The binary:}
The two stars are defined as uniform disks (see Equation\,\ref{eqn:Vdisk}).
The position of the primary star is defined by the two position parameters ($x_0,y_0$) as in Eqn.\,\ref{eqn:Vstar}.
The visibility of the primary star is therefore defined as:
\begin{equation}
V^\mathrm{prim}(u,v) = 2\frac{J_1(\pi UD_\mathrm{prim} \sqrt{u^2+v^2})}{\pi UD_\mathrm{prim} \sqrt{u^2+v^2}} \exp-2 i \pi (x_0 u + y_0 v).
\end{equation}
The secondary star is centered at position (0,0) and its visibility equation is identical to Eqn.\,\ref{eqn:Vdisk}:
\begin{equation}
V^\mathrm{sec}(u,v) = 2\frac{J_1(\pi UD_\mathrm{sec} \sqrt{u^2+v^2})}{\pi UD_\mathrm{sec} \sqrt{u^2+v^2}}.
\end{equation}
\item \textbf{The final model:}
The normalization of the fluxes (Eqn.\,\ref{eqn:norm}) gives:
\begin{equation}
f^\mathrm{prim}_0 = 1 - f^\mathrm{sec}_0 - f^\mathrm{back}_0.
\end{equation}
In this set of models $f^\mathrm{prim}_0$ is therefore not fitted and is computed from the two other flux ratios.
The final visibility is therefore:
\begin{equation}
V^\mathrm{tot,b\#}(u,v) = \frac{f^\mathrm{prim} V^\mathrm{prim}(u,v) + f^\mathrm{sec} V^\mathrm{sec}(u,v) }{ f^\mathrm{prim} + f^\mathrm{sec} + f^\mathrm{back} }.
\end{equation}
\end{itemize}
\subsection{Binary and a ring: br\texttt{1-5}}
\label{sec:br}
The last set of models have four components: the primary, the secondary, a ring and a background.
\begin{itemize}
\item \textbf{The binary:}
The primary star is defined as a uniform disk which is shifted w.r.t. to the cent\modiff{e}r of the disk by ($x_0,y_0$) such as:
\begin{equation}
V^\mathrm{prim}(u,v) = 2\frac{J_1(\pi UD_\mathrm{prim} \sqrt{u^2+v^2})}{\pi UD_\mathrm{prim} \sqrt{u^2+v^2}} \exp-2 i \pi (x_0 u + y_0 v). \label{eqn:prim}
\end{equation}
The coordinates of the secondary ($x_\mathrm{sec},y_\mathrm{sec}$) is defined in reference to the coordinates of the primary such as the coordinates of the secondary are:
\begin{eqnarray}
x_\mathrm{sec} &=& -rM x_0 \\
y_\mathrm{sec} &=& -rM y_0,
\end{eqnarray}
where $rM$ is the mass ratio between the primary and the secondary.
The visibility of the secondary is:
\begin{equation}
V^\mathrm{sec}(u,v) = \exp-2 i \pi (x_\mathrm{sec} u + y_\mathrm{sec} v). \label{eqn:sec}
\end{equation}
The binary separation is limited to be lower or equal to one third of the ring diameter projected along the binary separation.
\item \textbf{The full model:}
The flux ratio normalization (Eqn.\,\ref{eqn:norm}) is as follows:
\begin{equation}
f^\mathrm{ring}_0 = 1 - f^\mathrm{prim}_0 - f^\mathrm{sec}_0 - f^\mathrm{back}_0.
\end{equation}
The ring-to-total flux ratio is therefore not fitted and is recovered from this equation.
The total visibility is therefore:
\begin{align}
V^\mathrm{tot,br\#}(u,v,\lambda) = &\frac{ f^\mathrm{prim}V^\mathrm{prim}(u,v) }
{ f^\mathrm{prim} + f^\mathrm{sec} + f^\mathrm{ring} + f^\mathrm{back} }\\[8pt]
+&\frac{ f^\mathrm{sec} V^\mathrm{sec}(u,v) + f^\mathrm{ring} V^\mathrm{ring}(u,v) }
{ f^\mathrm{prim} + f^\mathrm{sec}+ f^\mathrm{ring} + f^\mathrm{back} }\nonumber,
\end{align}
\end{itemize}
where $V^\mathrm{ring}(u,v)$, $V^\mathrm{prim}(u,v)$ and $V^\mathrm{sec}(u,v)$ are defined from Eqn.\,\ref{eqn:ring}, \ref{eqn:prim} and \ref{eqn:sec} respectively.
\subsection{Strategy}
\label{sec:strategy}
\begin{figure}
\centering
\includegraphics[width=8.5cm]{Fig/Sketch_models.pdf}
\caption{The tree of models. The single star and background models are in orange, the single star and a ring models are in blue, the binary models are in light green and the binary and ring models are in dark green.}
\label{fig:tree}
\end{figure}
\begin{table*}
\caption{Matrix of parameters used in the different models. s indicates that the background and the secondary have the same spectral behavior \modiff{as} the primary as fitted from the SED.}
\label{tab:modvar}
\centering
\begin{tabular}{c c c c c c c c c c c c c c c c c c c c c c}
\hline
\hline
Model & \modif{n$_{par}$} & $f_{prim,0}$ & $f_{sec,0}$ & $f_{bg0}$ & $d_{sec}$ & $d_{bg}$ & $T_\mathrm{ring}$ & $\theta$ & $\delta \theta$ & $inc$ & $PA$ & $c_1$ & $s_1$ & $c_2$ & $s_2$ & $x_0$ & $y_0$ & $rM$ & UD$_\mathrm{prim}$ & UD$_\mathrm{sec}$ \\
\hline
s0 & \modif{2} & \checkmark & - & - & - & - & \checkmark & - & - & - & - & - & - & - & - & - & - & - & 0 & - \\
s1 & \modif{3} & \checkmark & - & - & - & \checkmark & - & - & - & - & - & - & - & - & - & - & - & - & \checkmark & - \\
\hline
sr0 & \modif{5} & \checkmark & - & \checkmark & - & $s$ & \checkmark & \checkmark & \checkmark & 0 & - & 0 & 0 & 0 & 0 & 0 & 0 & - & 0 & - \\
sr1 & \modif{7} & \checkmark & - & \checkmark & - & $s$ & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & 0 & 0 & 0 & 0 & 0 & 0 & - & 0 & - \\
sr2 & \modif{8} & \checkmark & - & \checkmark & - & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & 0 & 0 & 0 & 0 & 0 & 0 & - & 0 & - \\
sr3 & \modif{9} & \checkmark & - & \checkmark & - & s & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & 0 & 0 & 0 & 0 & - & 0 & - \\
sr4 & \modif{11} & \checkmark & - & \checkmark & - & s & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & 0 & 0 & \checkmark & \checkmark & - & 0 & - \\
sr5 & \modif{11} & \checkmark & - & \checkmark & - & s & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & 0 & 0 & - & 0 & - \\
sr6 & \modif{13} & \checkmark & - & \checkmark & - & s & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & - & 0 & - \\
\hline
b1 & \modif{7} & - & \checkmark & \checkmark & \checkmark & \checkmark & - & - & - & - & - & - & - & - & - & \checkmark & \checkmark & - & \checkmark & 0 \\
b2 & \modif{8} & - & \checkmark & \checkmark & \checkmark & \checkmark & - & - & - & - & - & - & - & - & - & \checkmark & \checkmark & - & \checkmark & \checkmark \\
\hline
br1 & \modif{11} & \checkmark & \checkmark & \checkmark & $s$ & $s$ & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & 0 & 0 & 0 & 0 & \checkmark & \checkmark & \checkmark & 0 & 0 \\
br2 & \modif{13} & \checkmark & \checkmark & \checkmark & $s$ & $s$ & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & 0 & 0 & \checkmark & \checkmark & \checkmark & 0 & 0 \\
br3 & \modif{15} & \checkmark & \checkmark & \checkmark & $s$ & $s$ & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & 0 & 0 \\
br4 & \modif{16} & \checkmark & \checkmark & \checkmark & $s$ & $s$ & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & 0 \\
br5 & \modif{17} & \checkmark & \checkmark & \checkmark & \checkmark & $s$ & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & \checkmark & 0 \\
\hline
\end{tabular}
\end{table*}
\begin{table}
\caption{Table of allowed parameter ranges}
\label{tab:paramranges}
\centering
\begin{tabular}{c c c c }
\hline
\hline
\modif{Parameter} & \modif{unit} & \modif{minimal value} & \modif{maximal value} \\
$f_{prim,0}$ & - & 0 & 1 \\
$f_{sec,0}$ & - & 0 & 1\\
$f_{bg0}$ & - & 0 & 1 \\
$d_{sec}$ & - & 0 & 1 \\
$d_{bg}$ & - & 0 & 1 \\
$T_\mathrm{ring}$ & [K] & 500 & 10000 \\
$\theta$ & [mas] & 0.01 & 500 \\
$\delta \theta$ & - & 0 & 20 \\
$inc$ & [$^\circ$] & 0 & 90 \\
$PA$ & [$^\circ$] & 0 & 360 \\
$c_1$ & - & -1 & 1 \\
$s_1$ & - & -1 & 1 \\
$c_2$ & - & -1 & 1 \\
$s_2$ & - & -1 & 1 \\
$x_0$ & [mas] & -30 & 30 \\
$y_0$ & [mas] & -30 & 30 \\
$rM$ & - & 0 & 20 \\
UD$_\mathrm{prim}$ & [mas] & 0.01 & $+\infty$ \\
UD$_\mathrm{sec}$ & [mas] & 0.01 & $+\infty$ \\
\hline
\end{tabular}
\end{table}
\modif{Our models are fitted to both} V$^2$ and CP.
\modif{However, there is a large variety of data in our sample as s}ome targets have a signal with low complexity, such as \object{AC Her} (\#1), but others have a high complexity, such as \object{U\,Mon} (\#22).
Recent studies of IRAS\,08544-4431 (\#10) showed that model \texttt{br5} with 17 parameters is needed to reproduce the interferometric dataset \citep{Hillen2016}.
We therefore fit a tree of models, starting with our simplest model (\texttt{s0}, see Sect.\,\ref{sec:s}), up to the most complex model (\texttt{br5}, see Sect.\,\ref{sec:br}), inspired by \citet{Hillen2016} with 17 parameters.
We start by fitting the simplest model to all the targets.
We then fit models with increasing complexity using the best parameters from the previous model as the starting point for the next model (adding the new parameters).
Some targets are best fitted by models that do not include a ring but just a binary.
We therefore included several forks in the model tree (see Fig.\,\ref{fig:tree}).
For each model a first fit is performed using a genetic algorithm implemented with \texttt{DEAP} \citep[][]{DEAP}.
The initial population is defined with a random flat distribution over the initial parameters.
For the parameters that are not used in the previous model the distribution spans over all the \modif{allowed} values of a parameter \modif{that we defined in Table\,\ref{tab:modvar}}.
For the parameters already used in the previous model in the tree, we span over 30\% of its best-fit value for the previous model.
The mutation is Gaussian over the parameter allowed distribution and has a probability of 5\%.
The individual selection is made through a three-rounds tournament.
Then, using the best fit from the genetic algorithm as a starting point, a MCMC minimization is performed with the \texttt{emcee} package \citep[][]{MCMC} to determine the error bars on the parameters.
When a model has a larger $\chi^2$ than the previous model in the model tree we redo the minimization as this indicates that a local minimum was reached.
As it is not possible to directly compare two models with a different number of parameters,
\modif{we wanted to use a criterion that is applicable to all our sample.}
\modif{Such criteria exists from information theory and}
we used the Bayesian inf\modif{ormation} criterion \citep[BIC;][]{BIC} to select the best model for each target.
\modif{This criteria aims to infer the model that fits the data the best without over fitting and is used in several studies of stellar physics to infer the most likely model that reproduce the data from a set of different models \citep[e.g.,][]{Degroote2009,Aerts2018,Matra2019}.
The BIC criteria was developped for model inference and is more conservative in the choice of the most likely model (higher penalty for models with more parameters to fit) than the Akaike information criterion (AIC) that was developed for prediction purposes.
The BIC can be written as follows:}
\begin{equation}
\mathrm{BIC} = -2\mathcal{\hat{L}} + n_\mathrm{par} \log{n_\mathrm{data}},
\end{equation}
\modif{where $\mathcal{\hat{L}}$ is the maximum likelihood values found for the optimal model parameters, $n_\mathrm{par}$ is the number of optimized parameters and $n_\mathrm{data}$ is the number of data points.}
\modif{Under the assumptions we make, i.e. data points are independent and the error distribution is Gaussian, the BIC can be written as:}
\begin{equation}
\mathrm{BIC} = \chi^2 + n_\mathrm{par} \log{n_\mathrm{data} },
\end{equation}
\modif{where} $\chi^2 = \sum_{i=1}^{n_\mathrm{data}} \Big(\frac{y_\mathrm{i}-m_\mathrm{i}}{\sigma_\mathrm{i}}\Big)^2$ \modif{with} $y_\mathrm{i}$ \modif{is $i$-th data point}, $m_\mathrm{i}$ \modif{the $i$-th model point and} $\sigma_\mathrm{i}$ \modif{the error bar of th $i$-th data point.}
The values of the BIC \modif{and }$\chi^2$ for each target are displayed on Figs.\,\ref{fig:BIC1} and \ref{fig:BIC2}.
\modif{To compare the models, we have selected the model with the lowest BIC.
However, the differences between the BIC values for a given dataset can be small and several models can be considered.
Usually, models with a difference of more than 10 with the BIC of the best model can be ruled out whereas a difference between 10 and 6 can be considered as moderately strong evidence for the best model, between 6 and 2 as positive evidence in favour of the best model and less than 2 as weak evidence \citep[e.g.,][]{Aerts2018}.}
\section{Results}
\label{sec:results}
In this section we present a first analysis of the fit results by discussing the important and most reliable parameters related to the ring morphology such as the near-infrared sizes, the temperatures or the fluxes of the different components.
We focus on the circumbinary environments in our comparison with YSOs.
\subsection{Results of the model selection}
\begin{figure}
\centering
\includegraphics[width=8.5cm]{Fig/ModelComplexity2.pdf}
\includegraphics[width=8.5cm]{Fig/ModelComplexity1.pdf}
\caption{Statistics of model selection. Top: number of targets per model. Bottom: number of targets per number of parameter of the selected model (complexity).}
\label{fig:models}
\end{figure}
\modif{
For HD\,93662, \object{IRAS\,19125+0343}, IW\,Car and R\,Sct one model has very strong evidence over all the other models ($\Delta$BIC>10 for any other model).
For all other targets, the most likely models are presented on Tables\,\ref{tab:resACHer} to \ref{tab:resV494Vel}.
Five targets have the most likely model to have at least strong evidence (6<$\Delta$BIC<10) over other models (AI\,Sco, HD\,108015, HD\,213985, \object{IRAS\,05208-2035}, \object{IRAS\,17038-4815}).
Most of the likely models have similar parameters (i.e. within the error bars).
In the rest of our analysis we use the most likely model (i.e. the model with the lowest BIC value).}
The best-fit parameters are presented in Table\,\ref{tab:res1} and \ref{tab:res2}.
\modif{As an estimation of the fit quality, the} \modif{reduced $\chi^2$ (} $\chi^2_\mathrm{red}$ \modif{=} $\chi^2/n_\mathrm{data}$\modif{)} value\modif{s} ranges from 0.5 to 6.6.
The number of targets per model is displayed in the top panel of Fig.\,\ref{fig:models}.
Out of the sixteen models \modif{eleven} were selected.
For one target the simplest model was preferred (AC Her, \#1).
For three targets a binary model was preferred with no ring (HD93662 (\#4), PS Gem (\#18) and RU Cen (\#20)).
For \modif{ten} targets models of a single star with a ring was selected while for ten targets a model with a binary surrounded by a ring was preferred.
For \modif{twelve} targets models with a binary were preferred and for nineteen targets models with a ring were preferred (their images are displayed in Figs.\,\ref{fig:modImages1} and \ref{fig:modImages2}).
On the bottom panel of Fig.\,\ref{fig:models} we can see how many models there are per number of parameters.
Models with six or less parameters are not fitting the data well enough (apart for AC Her, \#1).
Thirteen targets ($\sim$55\%) were fitted by models with thirteen parameters or more.
This shows the complexity of the resolved structures.
\modif{The case of AC\,Her (\#1) is interesting as it was previously observed in mid-infrared with the MIDI instrument at the VLTI \citep[][]{Hillen2015}. A disk with an inner rim at 68\,au was fitted to the data, which corresponds to 42\,mas. A rim of this size would be over-resolved by our observations, which is compatible with the most likely model: a star + an over-resolved flux. For the target with the richest dataset, IRAS\,08544-4431 (\#10), the most likely model corresponds to the model which was fitted in \citet[][]{Hillen2016}. The best-fit parameters are also very similar except the ring temperature $T_\mathrm{ring}$. This is due to the way the photosphereric spectrum is represented as in \citet[][]{Hillen2016} it is assumed to be in the Rayleigh-Jeans regime ($F_\lambda \propto \lambda^{-4}$) and here we use the fit to the photometry.}
\subsection{Inner rim radius ruled by dust sublimation physics}
\label{sec:sizelum}
\begin{figure}
\centering
\includegraphics[width=8.5cm]{Fig/SizeLum_wysov2.pdf}
\caption{Size-luminosity relation. The sizes are in log-scale. Purple points are pAGB binaries from this work. Purple open circles are pAGB binaries with $T_\mathrm{ring}$>7000\,K. Light blue points are Herbig Ae/Be stars from \citet{Lazareff2017}. The blue and orange resp. dashed lines are the theoretical sublimation radius for $T_\mathrm{sub}=1500$K and $T_\mathrm{sub}=1000$K resp.}
\label{fig:sizelum1}
\end{figure}
For protoplanetary disks the size of the near-infrared extended emission (\modif{the physical radius in au:} $a$) correlates with the square root of the stellar luminosity \citep[$L_\mathrm{bol}$;][]{Monnier2002,Lazareff2017} such as:
\begin{equation}
a = \frac{1}{2} (C_\mathrm{bw}/\epsilon)^{1/2} (L_\mathrm{bol}/4\pi\sigma T^4_\mathrm{sub})^{1/2},
\end{equation}
where $T_\mathrm{sub}$ is the sublimation temperature, $C_\mathrm{bw}$ is the backwarming coefficient \citep[][]{kama2009}, $\epsilon$\,=\,$Q_\mathrm{abs}(T_\mathrm{sub})/Q_\mathrm{abs}(T_*)$ is the dust grain cooling efficiency which is the ratio of Planck-averaged absorption cross-sections at the dust sublimation and stellar temperatures and $\sigma$ the Stefan-Boltzmann constant.
To study the size of the inner rim we use here and in the rest of the paper the half of the fitted ring diameter ($\theta$) as it was done in studies of disks around YSOs \citep[e.g.,][]{Monnier2002,Monnier2005,Lazareff2017}. The rings having a given width ($\delta\theta$) it is possible that the inner disk edge, usually defined by the location where the optical depth $\tau$ equals unity \citep[e.g.,][]{kama2009}, can be closer than the ring radius.
On Fig.\,\ref{fig:sizelum1} we plot the sizes of the ring models versus the central luminosity for pAGB binaries of our sample with, for reference, lines indicating theoretical sublimation radii for $T_\mathrm{sub}=1000$\,K and 1500\,K with $C_\mathrm{bw}=1$ and $\epsilon=1$.
In order to compute the luminosities and the physical sizes of our targets we have used the Gaia parallaxes \citep{BJ2018}.
However, as our targets are binaries with a semi-amplitude that can be of the order of the parallax, those distances are likely biased \modif{by the orbital movement of the binary}.
Luckily, however, this does not impact the size luminosity diagram as both the physical size and the square root of the luminosity scale linearly with distance.
An error on the distance will therefore displace a point along the size-luminosity relation.
Sizes of near-infrared emission around pAGB binaries seem to scale with the stellar luminosity as it is the case for young stellar objects.
\modif{However, the sizes of pAGB circumstellar emissions are systemically always offset toward sizes larger than for circumstellar emission around YSOs.
This can be deduced more clearly from the histogram on Fig.\,\ref{fig:sizelum2}.}
There are three outliers (\object{IRAS\,10174-5704} (\#11), R Sct (\#19) and V494Vel (\#23)) with very small sizes compared to their luminosity.
Those three stars have also a large temperature for their environment (higher than 7000\,K) meaning that the traced circumstellar environment is not thermal emission from dust.
\begin{figure}
\centering
\includegraphics[width=8.5cm]{Fig/SizeLum_hist_vok.pdf}
\caption{Histogram of sizes scaled by the square root of luminosity for both pAGBs and YSOs.}
\label{fig:sizelum2}
\end{figure}
\subsection{Dust temperature}
\label{sec:T}
The spectral channels of PIONIER allow us to probe the difference in spectral index between the central star and its environment.
This is done by fitting the difference of level of the squared visibilities between the different channels.
For a target in which such a difference is present, the fraction of the total flux that is resolved by the interferometer at any given baseline depends on the wavelength, hence the observed squared visibilities as well.
This is called the chromatic effect.
When there is a large difference in temperature between a central unresolved source and its resolved environment, the squared visibility increases with s\modif{hort}er wavelength for a given baseline (see Fig.\,\ref{fig:chrom}).
\begin{figure}
\centering
\includegraphics[width=8.5cm]{Fig/chrom2.pdf}
\caption{Illustration of the chromatic effect (see the main text). Left: squared visibilities of HR4049 (\#8) showing a strong chromatic effect. Right: squared visibilities for R Sct (\#19) not showing such an effect.}
\label{fig:chrom}
\end{figure}
We assumed the central star to have a given spectrum that we fit from photometry.
The ring models assume a black-body emission for the ring.
As we fix the primary spectrum to be what we fit from the photometry the difference in the levels of the squared visibilities per channel will be reproduced by a given ring temperature.
This temperature may not be the exact temperature at that location due to optical depth effects, but it gives a good first indication.
\begin{figure}
\centering
\includegraphics[width=8.5cm]{Fig/Temperaturesv2.pdf}
\caption{Histogram of ring temperatures for our targets.}
\label{fig:ringT}
\end{figure}
In Fig.\,\ref{fig:ringT} we show the histogram of temperatures we found for our models.
On the one hand, two-third of the targets (13/19) have a low circumstellar emission temperature ($T_\mathrm{ring}<1600\,K$), equal or lower than the classical silicate sublimation temperatures we expect ($\sim$1500\,K) indicating that the thermal emission of the inner rim of the disk dominates.
The slightly lower temperatures agree with the shift toward larger sizes we see in the size-luminosity diagram for the pAGB with respect to YSOs.
On the other hand, four targets have a circumstellar emission with a temperature of more than 7000\,K: IRAS\,05208-2035 (\#9), IRAS\,10174-5704 (\#11), R\,Sct (\#19) and V\,494\,Vel (\#23).
Three of them are also outliers in the size-luminosity diagram (see Sect.\,\ref{sec:sizelum}) pointing toward another origin of the circumstellar flux.
For \object{IRAS\,05208-2035} (\#9), the high-stellar-to-total flux ratio (90.5$\pm$0.3\%) makes this target special.
The ring flux is perhaps stellar scattered light coming from the inner rim of the disk.
Finally, between these two categories of temperatures, two targets have a temperature around 3000-4000\,K (IRAS\,17038-4815 (\#13) and U\,Mon (\#22)).
Interestingly, those targets are expected to have pulsation with the largest amplitude ($\Delta$mag=1.5 and 1.1 for IRAS\,17038-4815 (\#13) and U\,Mon (\#22) resp.) among our sample.
However, our data is too sparse to look for morphological changes induced by pulsations in these targets within the observations that span a large part of the pulsation cycle.
Our models are able to reproduce the data reasonable without including any intrinsic variations.
\subsection{Disk inclinations}
\begin{figure}
\centering
\includegraphics[width=8.5cm]{Fig/incv2.pdf}
\caption{Histogram of the cosines of fitted inclinations.
}
\label{fig:inc}
\end{figure}
We can investigate the morphology of the circumbinary environment by looking at the ring inclinations.
As our targets are surrounded by disks, the distribution of the cosines of inclinations should be flat.
However, for spherical shells for instance, there would be a pile-up of objects at $\cos inc \sim 1$.
Fig.\,\ref{fig:inc} displays the histogram of the cosines of our ring inclinations.
There is clearly no pile-up at $\cos inc =1$.
We see a rather flat distribution with a cut-off at about $\cos inc \sim 0.5$ that corresponds to a inclination of $\sim60^\circ$.
This is likely due to observational bias as the disks that are edge-on will absorb the visible light from the central stars.
The cut-off of disk inclinations could therefore be a proxy to the characteristic thickness of the disk.
The cut-off we see would translate to a disk thickness of $h/r\sim$0.8.
However, there could be a model bias as at very high inclinations the model might not be able to reproduce the intensity distribution correctly.
\subsection{Disk width}
\begin{figure}
\centering
\includegraphics[width=8.5cm]{Fig/RingWidthsvsRes.pdf}
\caption{Relative ring width (\modiff{$\delta\theta$}) against the degree of resolution of the ring. The colors indicate the ring inclination. The horizontal line indicates where the ring has no inner cavity ($\delta\theta\geq$\modiff{2}).
}
\label{fig:ringwidths}
\end{figure}
We are sensitive to the width of the emission coming from the disk.
The parameter $\delta\theta$ measures the width of the ring in the units of the ring \modiff{radius}.
On Fig.\,\ref{fig:ringwidths} we see this ring width parameter plotted against the size of the ring divided by the angular resolution of the largest baseline.
We see that as long as the ring is resolved by the observations ($\frac{\theta B_\mathrm{max}}{\lambda}\geq0.5$) its width is better constrained and is below unity.
It means that the circumbinary dust emission is compatible with a ring and not with an emission without a cavity \modiff{($\delta\theta\geq2$)} and that the ring has a significant radial width ($\delta\theta$ between 0.5 and 1).
\subsection{Rim brightness distribution}
\begin{figure}
\centering
\includegraphics[width=8.5cm]{Fig/c1s1sininc.pdf}
\caption{Direction of the maximum of the first order modulation. The dashed line represent the limit for inclination-like modulations \citep[see text;][]{Lazareff2017}. The circle represent the maximum values for the modulation coefficients $c_1$ and $s_1$.
}
\label{fig:m1}
\end{figure}
All the models with a ring require at least a first order modulation.
For a modulation due to the inclination of an optically and geometrically thick inner rim we will have a larger illumination in the direction of the minor axis \citep[e.g.,][]{Isella2005}.
If this is the case most of the targets would have a larger absolute value of the $s_1$ coefficient and an almost zero value for $c_1$.
Fig.\,\ref{fig:m1} shows that this is not the case.
The values for $c_1$ and $s_1$ are scattered.
Inclination effects would produce more points with $s_1\sin{inc}$ between 0 and 1 and $c_1\sin{inc}$ between 0 and 0.3 \citep{Lazareff2017}.
There is no clear evidence that the inclination is the main cause of the observed modulation.
Half of the targets prefer the second order modulation, also indicating that the inner rims are not ruled only by inclination effects but also by interactions with the inner binary and/or disk instabilities.
\subsection{Extended flux statistics}
\begin{figure}
\centering
\includegraphics[width=8.5cm]{Fig/Overresolved2v2.pdf}
\caption{Histogram of the fractions of the non-stellar flux that is over-resolved
}
\label{fig:extended}
\end{figure}
In IRAS\,08544-4431 (\#10), our best studied object, $\sim$15\% of the $H$-band flux is coming from an over-resolved emission \citep{Hillen2016}.
Only half of this flux was accounted for by the radiative transfer model including a disk in hydrostatic equilibrium and scattered light \citep{Kluska2018}.
The origin of this extended flux is not clear so far.
This extended flux is easy to detect as it can be measured as the drop of visibility at short baselines.
Fig.\,\ref{fig:extended} shows the ratio of the over-resolved \modif{(f$_{bg}$)} over the total non-stellar flux \modif{(f}$_\mathrm{circum}$ \modif{= 1-f}$_\mathrm{prim}$\modif{-f}$_\mathrm{sec}$\modif{)}.
IRAS\,08544-4431 (\#10) has $\sim$42\% of its non-stellar flux to be over-resolved.
14 out of 19 targets ($\sim$75\%) have this ratio larger than 10\% and three of them have it around 50\%.
\section{Discussion}
\label{sec:discussion}
In this section we will discuss the results and interpret them in an extended context.
We will first discuss the outliers in Sect.\,\ref{sec:outliers}.
We then discuss the shift between pAGB and YSOs in the size luminosity relation (Sect.\,\ref{sec:shift}).
Then we will discuss the relation between disk inclination and the RVb phenomenon (variable extinction/scattering in the line of sight during orbital motion; Sect.\,\ref{sec:RVb}).
We will discuss the radial structure of the disk by comparing the size of the emission in the near-infrared with the one in the mid-infrared (Sect\,\ref{sec:NIRMIR}).
Finally, in Sect.\,\ref{sec:YSOcomp}, we will discuss the differences and similarities with the disks around YSOs.
\subsection{Outliers}
\label{sec:outliers}
Four targets from our sample are outliers in the fact that the temperature of their environment is very high ($T_\mathrm{ring}$>7000\,K), significantly above any dust sublimation temperature.
Here we summarize results published in literature and give an interpretation of the origin of their pecularities.
\subsubsection{IRAS\,05208-2035 (\#9)}
This source is an outlier in \citet{deRuyter2006}: its infrared \modif{excess} starts at longer wavelengths (after $L$-band).
This visibilities show a reversed chromatic effect i.e. the squared visibility decreases with decreasing wavelengths.
It happens when the environment is bluer than the emission from the central source
Our models are not able to reproduce this effect.
This effect can be produced by stellar light \modif{scattered on the disk surface}.
\subsubsection{IRAS\,10174-5704 (\#11)}
This source has a high temperature for its environment and a small radius.
It was included in the mid-infrared interferometric survey of post-AGB binaries with MIDI \citep{Hillen2017}.
In this sample it is standing out because of its large size.
Also in the Spitzer survey of \citet{Gielen2011} it was noticed that the spectrum of this source is dominated by amorphous silicates with no crystalline dust features.
It was postulated in the latter that IRAS\,10174-5704 is likely a luminous super-giant.
It \modiff{would} explain why this source is an outlier in our survey as well.
\subsubsection{R\,Sct (\#19)}
R\,Sct is one of the only pAGB targets for which a surface magnetic field was detected \citep[][]{Sabin2015}.
It is also classified as `uncertain' by \citet{Gezer2015}, on the basis of its WISE photometric colours.
Its binary nature is not confirmed and the SED shows a very minimal infrared excess that is more reminiscent of an outflow than of a disk.
\subsubsection{V\,494\,Vel (\#23)}
This target shows photometric fluctuations but without any periodicity \citep[][]{Kiss2007}. \modif{As for R\,Sct, the binarity nature of this source is uncertain.}
We note that the target is usually referred in previous studies as IRAS\,09400-4733 \citep[][]{Kwok1997,deRuyter2006,Kiss2007,Szczerba2007}.
\subsection{Origin of the shift between pAGBs and YSOs in the size luminosity relation}
\label{sec:shift}
There can be two explanations for the systematic offset between inner rim sizes (scaled to the squared stellar luminosity, see Fig.\,\ref{fig:sizelum2}) between pAGB and YSO sources.
\modiff{The two explanations are about factors that influence the dust sublimation radius.}
\modiff{
A first factor that could be different between pAGBs and YSOs is the dust type.
For a given gas density, different types of dust will have different sublimation temperatures.
For example silicates, that are Oxygen rich, will have lower dust sublimation temperature that Carbon-rich dust \citep[e.g.,][]{Kobayashi2011}.
Carbon-rich dust is more abundant in
PPDs than in disks around pAGB \citep[e.g.,][]{Gielen2011}.
Although amorphous C does not show significant spectral features, the absence of distinct features from other C-rich dust species in the mid-IR spectra indicates a low abundance of C in the pAGB circumbinary dust.
Mg-rich species like Olivine on the other hand are abundantly present \citep{Gielen2008,Gielen2011,Hillen2015}.
Therefore, in pAGB the overall dust sublimation temperature will be higher and the inner disk radius will be larger as observed.
}
\modiff{
Another factor is the local gas density.
Assuming the same dust species, the same temperature will be reached farther away from the central star in the pAGBs because of higher luminosity of the central star than in YSOs.
Assuming the same central stellar masses, same disk masses and a similar disk structure for the two types of objects, i.e. a decreasing surface density with radius, the local density will be lower at those locations in pAGBs (also because of weaker gravity due to the central star).
As the dust sublimation temperature depends on the gas density \citep[e.g.,][]{kama2009}, it will be lower for pAGBs and hence the inner dust rim will be larger.}
\begin{figure}
\centering
\includegraphics[width=9cm]{Fig/Temperaturesv3.pdf}
\caption{Histogram of temperatures of the near-infrared circumbinary emission of pAGB from this work (\modif{purple}) and around YSOs from \citep{Lazareff2017} for objects having $T_\mathrm{ring}$ between 500 and 2000\,K.}
\label{fig:Tshift}
\end{figure}
We can compare the measured interferometric temperatures of the near-infrared circumbinary emission of pAGBs to those of the environment of YSOs.
Fig.\,\ref{fig:Tshift} shows that pAGBs have systemically lower measured disk rim temperatures than YSOs.
\begin{figure}
\centering
\includegraphics[width=9cm]{Fig/RadiusBinsep2.pdf}
\caption{\modif{Comparison between the ring diameter ($\Theta$) and the binary separation ($a_\mathrm{bin} = \sqrt{x_\mathrm{bin}^2 + y_\mathrm{bin}^2}$) for targets for which the most likely model contains a binary. The dashed blue line indicates the dynamical truncation diameter of the ring by the inner binary. The purple area indicates the forbidden ring diameters that are below the dynamical truncation diameter.}}
\label{fig:radbinsep}
\end{figure}
One could expect that some targets will show larger sizes that the theoretical dust sublimation radius because of the dynamical interaction between the inner binary and the disk.
This interaction would push the disk rim at a radius $\sim$1.7 times larger than the binary separation \citep{artymowicz1994}.
This would have been seen by having some targets \modif{higher up} in the size-luminosity diagram (Fig.\,\ref{fig:sizelum1}) and that some ring diameter\modiff{s} ($\Theta$) would be of the order of 1.7$\times$ the binary separation which we do not see (Fig.\,\ref{fig:radbinsep}).
Therefore the dust component is no tracing the disk dynamical truncation by the inner binary.
\subsection{Relation between the RVb phenomenon and disk inclination}
\label{sec:RVb}
\begin{figure}
\centering
\includegraphics[width=9cm]{Fig/incvRV.pdf}
\caption{Histogram of inclinations for objects from our sample showing the RVb phenomenon (orange) and the whole sample (purple).}
\label{fig:RVb}
\end{figure}
The RV Tauri stars are variable pulsating post-AGB stars \citep[e.g.,][]{Kiss2007,KissBodi2017,Manick2017}.
They display alternate deep and shallow minima.
A sub-sample of these stars has a long-period variation of the mean luminosity \citep[][]{Pollard1996} and are classified as RVb.
More generally this long-period photometric variation is observed in non RV Tauri stars as well and is caused by variable extinction/scattering in the line of sight due to orbital motion of the central binary \citep[][]{Kiss2007}.
It is interpreted as caused by a highly inclined disk shadowing the primary at certain phases of the orbit \citep[][]{VanWinckel1999,Manick2017}.
As we are sensitive to the disk inclination we can test this hypothesis.
In our sample, there are nine targets displaying the RVb phenomenon: AI\,Sco (\#2), HD\,95767(\#5), HD\,213985 (\#7), HR\,4049(\#8), IRAS\,19125+0343 (\#15), IW\,Car (\#16), SX\,Cen(\#21), U\,Mon (\#22) and V\,494\,Vel (\#23) \citep[][]{Waelkens1991,Waelkens1995,Kiss2007,KissBodi2017}.
While HD\,213985 (\#7), SX\,Cen (\#21) and U\,Mon (\#22) are the most inclined disks from this survey ($inc\sim60^\circ$) confirming the inclined disk hypothesis, the other objects are only moderately inclined implying very high disk scale-height for the disk shadowing interpretation to be true.
The histogram of inclinations for all sources and RVb sources shows that most of the RVb sources have the highest inclination (Fig.\,\ref{fig:RVb}).
One RVb source is found to have a pole-on inclination: HD\,95767 (\#5).
\modif{The other likely model for this source also point toward a pole-on orientation (Fig.\,\ref{tab:resHD95}).}
We also note that three sources have high inclinations (above 50$^\circ$) without showing the RVb phenomenon: EN\,TrA (\#3), IRAS\,05208-2035 (\#9) and \object{IRAS\,15469-5311} (\#12).
The apparent inclinations are deduced from an aspect ratio and, given the poor uv-coverage for some sources, need to be confirmed by further studies.
\subsection{Comparison between near-infrared and mid-infrared sizes}
\label{sec:NIRMIR}
\begin{figure}
\centering
\includegraphics[width=9cm]{Fig/ComparisonMIDIhlrv3.pdf}
\caption{Comparison between MIDI half light diameters and PIONIER diameters for ring like targets. The dashed lines represent the models of disks with a temperature dependence that scales with a power-law of the radius. IRAS\,10174-5704 (\#11) is outside the limits of this plot (see Sect.\,\ref{sec:NIRMIR}) and is not appearing for clarity reasons.}
\label{fig:Midi}
\end{figure}
We see in Sect.\,\ref{sec:sizelum} that the size of the near-infrared emission is ruled by dust sublimation physics as it is proportional to the square root luminosity of the central star.
However the structure of the disk can be constrained if it is observed at different wavelengths.
To do so we can compare the \modiff{diameters} of the circumstellar emission at the mid-infrared probed by MIDI at 10$\mu$m \citep{Hillen2016} with the \modiff{diameters} from this work.
The two \modiff{diameters} are plotted against each other on Fig.\,\ref{fig:Midi}.
There is a relation of proportionality where $\theta_\mathrm{MIR}$ $\approx$ 2$\theta_\mathrm{NIR}$.
Most of the models of protoplanetary disks predict a power-law for the radial temperature dependence $T\propto r^{-\alpha}$ \citep[e.g.,][]{LyndenBell1974,Kenyon1987}.
In several studies the used power-law index is either $\alpha$=0.5 or $\alpha$=0.75 \citep[e.g.,][]{Kraus2008}.
Assuming that the disk emits as a black-body with a power-law radial temperature profile and that the inner disk radius has a temperature of 1100\,K (see Sect.\,\ref{sec:T}) we can simulate the radial profile of the emission in both the near and mid-infrared.
From those profiles we can take the ratio of the half-light \modiff{diameters} \modif{($\Theta$)} between the near-infrared and the mid-infrared.
Those ratios are reported in Fig.\,\ref{fig:Midi}.
We can see that most of our targets fall within the two limits set by the two temperature power-laws.
This points toward disks that have a similar radial temperature dependence and a smooth radial disk structure in the inner disk regions (<30\,au).
There are some sources that are standing out from this picture.
IRAS10174-5704 (\#11) has a mid-infrared size \citep[2hlr=150$\pm$8mas\modif{;}][]{Hillen2017} $\sim$115 times larger than the near-infrared size from this work.
U\,Mon (\#22) has also a relatively large mid to near infrared size ratio.
This target is showing the RVb phenomenon and has a very complex visibility profile that could only be reproduced by the most complex model.
The result of the fit is displaying a strong azimuthal modulation showing a complex morphology of this target.
Finally IW\,Car (\#16) is the only outlier having a significantly small mid to near-infrared size ratio.
This target is showing an RVb phenomenon as well, however, we find a moderate inclination in our work.
It is fitted by the most complex model and has still a relatively large $\chi_\mathrm{red}^2$=3.9.
It is possible that the models are not able to reproduce all the complexity of that source and that it could have an effect on the derived size.
More observations of this target are therefore needed to confirm its status.
\subsection{Continuing the comparison between pAGB and YSOs}
\label{sec:YSOcomp}
Despite a completely different formation process, disks surrounding post-AGB binaries and those surrounding young stars are similar with respect to several criteria.
They are often compared to young intermediate mass Herbig stars.
The disks around Herbig Ae/Be stars are classified into two groups \citep[][]{Meeus2001,Maaskant2013,Menu2015}: disks with flared or gapped structure (group I) and flat disks (group II).
It was advocated that disks around post-AGB binaries are group II sources \citep[][]{deRuyter2006,Hillen2017} because of their SED and mid-infrared sizes and colors.
In this work we push the comparison further.
We recall and discuss here the main points of comparison between those two kinds of disks arising from this study.
\begin{itemize}
\item \textbf{The near-infrared size-luminosity relation:} We show in Sect.\,\ref{sec:sizelum} that the relation extends for the near-infrared emission around post-AGB disks. As these targets have a higher luminosity than most of the young intermediate mass Herbig Ae/Be stars, we are able to probe very high luminosity regimes. This result shows that the near-infrared emission in post-AGB is also mainly ruled by dust sublimation rather than dynamical disk truncation by the inner binary. However, the shift toward larger sizes and the \modif{ring temperatures, $T_\mathrm{ring}$,} inform us \modif{on different properties of} the disks around post-AGB binaries that lowers the temperature of their dust sublimation front. We postulate that it can be due to smaller local gas density or different dust grain mineralogy or both.
\item \textbf{Amount of extended flux:} For Herbig Ae/Be stars, the disks that have a double-peaked SED in the mid-infrared \citep[group I;][]{Meeus2001} display an over-resolved-to-non-stellar flux ratio larger than 5\% whereas the targets with a flat SED in the mid-infrared (group II) have this flux ratio lower than 5\% \citep{Lazareff2017}.
In our sample, fourteen targets have more than 10\% of non-stellar flux to be over-resolved.
This ratio can reach 50\% in some cases.
This confirms that there is a significant contribution from an over-resolved flux in the near-infrared for these targets.
It is the first \modif{time we observe a} \modif{specific} feature \modif{of group~I Herbig disks in} pAGB disks.
The origin of this extended flux \modif{($f_\mathrm{bg0}$)} is unknown and should be the focus of future studies.
It can be related to the disk structure (e.g., disk flaring, presence of a gap) or to a mechanism lifting the dust from the disk (e.g., disk wind, jet from the secondary).
\end{itemize}
\section{Conclusions}
\label{sec:conclusions}
We summarize here the most important findings of our interferometric survey.
Despite the very inhomogeneous (u, v)-coverages obtained we can conclude that:
\begin{enumerate}
\item For \modif{most of the} sources \modif{19/23} a compact but resolved ring-like $H$-band emission component is detected which confirms the presence of a disk in pAGB binaries.
\item Most of the targets (14/23) prefer models with more than ten parameters and several targets (\modif{6}/23) prefer the most complex models with fifteen or more, including complex azimuthal modulations, even with a limited (u, v)-plane coverage.
\item \modiff{There is a relation of proportionality between the size of the near-infrared circumstellar emission and the square root of the stellar luminosity as it is the case around YSOs, suggesting that the near-infrared extended emission is also linked to the dust sublimation region around pAGBs.}
\item The measured temperature of the near-infrared circumbinary emission is lower (\modif{median ring temperature of}$\sim$1\modif{2}00\modif{-1300}\,K) for pAGB disks and the sizes \modif{of the near-infrared circumstellar emission} are systematically a bit larger than for YSOs. This can be due to different dust grain mineralogy and/or lower gas density at the sublimation front.
\item The dust sublimation front \modiff{has} width-to-\modif{radius} ratios spanning between 0.5 and unity.
\item A significant fraction of the near-infrared emission is over-resolved by our observations. This ratio is higher than for Herbig Ae/Be stars. The origin of this circumstellar flux is unknown.
\end{enumerate}
Given the complexity of our targets and the limitations of geometrical modeling of the near-infrared interferometric observables, we believe that a time series of interferometric images is the only way to come to a sharp view on the physics that drives the interactions in the inner regions of these objects. The here presented survey (see Figs.\,\ref{fig:modImages1} and \ref{fig:modImages2}) and our imaging campaign of IRAS\,08544-4431 \citep[][]{Hillen2016,Kluska2018} have demonstrated the potential of near-infrared interferometric images, have shown which targets can be well resolved with the existing instrumentation and which are most interesting for follow-up campaigns.
The focus of this paper has been mostly on the circumbinary emission, as this is most reliably detected in our data. Our survey also demonstrates, however, that there is a lot of potential for detecting the companions, even if the here presented detections are likely not free of model bias. Model-independent determinations of the binary orbits will require significant investments of observing time though, as significantly better uv-coverages are required (like that of IRAS\,08544-4431), and this at various orbital phases.
To investigate the origin of the over-resolved flux (outflow, disk wind, disk structure), direct imaging in scattered light should bring strong constraints.
Finally\modif{,} a complete view of the dust \modiff{disk} structure such as millimeter observations with ALMA will allow us to investigate the possibility of second-generation planet formation in these disks.
\begin{acknowledgements}
We thank the referee for his comments that improved the clarity of the paper.
JK and HVW acknowledge support from the research council of the KU Leuven under grant number C14/17/082.
\modif{DK acknowledges support from Australian Research Council DECRA grant DE190100813.}
This research has made use of the Jean-Marie Mariotti Center \texttt{Aspro}
service\footnote{Available at http://www.jmmc.fr/aspro}.
We used the following internet-based resources: NASA Astrophysics Data System for bibliographic services; Simbad; the VizieR online catalogs operated by CDS.
\end{acknowledgements}
\bibliographystyle{aa}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 3,383 |
William Janiak's Music Book – BJ11 (available in print only)
THIS PRINTED BOOK IS THE ORIGINAL COLLECTION OF 25 ALL TIME FAVORITES FROM THE WILLIAM JANIAK RECORD LIBRARY WITH WORDS AND SIMPLE PIANO ARRANGEMENTS.
These fun educational songs help develop the child's everyday skill concepts through music and movement.
SONGBOOK INCLUDES
From Put Your Finger in the Air (KIM 70167CD): "I'm Swinging My Arms," "Put Your Finger in the Air, in the Air," "Clap Your Hands," "Everything's Put Together," "Make a Circle";
From Look What I Can Do (KIM 9300CD): "My Name is ______," "Circle Dance," "We Wash Our Face," "Up and Down," "I Like to Jump"; From "It's Fun To Clap (KIM 9195CD: "Hop Like A Bunny," "The La Song," "It's Fun To Clap," "Make Your Eyes," "Shrug Your Shoulders";
From Arms Up Keep Moving (KIM 9193CD): "Arms Up," "Making Sense," "Keep Movin'";
From Dances for Little People (KIM 0860CD): "Dance to the Rhythm of Your Name";
From Songs About Me (KIM 20223CD): "Do You Like Foods? (The Meat Song)," "Piece of Paper," "Why Me?";
From More Songs About Me (KIM 70234CD): "If You Have This On – Stand Up," "I Like to Move to the Left and Right," "We are All Nodding Our Heads"
William Janiak's Music Book - BJ11 (available in print only) quantity
Category: Song Books | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 4,333 |
ScreenDance Diaries
Wang Ramirez on the Borderline: An Interview
By Sarah Elgart on April 9, 2014 in Art, Dance, Film, Theatre
The more work I see, the more I think that the most powerful dance defies categorization and leaves an audience simply moved, without a clear idea of what they just saw or the need to categorize or define it. In fact, I am beginning to believe that this is actually a prerequisite for performance or art work of any kind to succeed or become great, especially at a time when people in most developed countries are exposed to as much stimuli in a single day as the average Victorian was in a lifetime.
So when I saw Borderline by the Bessie Award-winning Company Sebastien Ramirez of the duo Wang Ramirez – a short film that is really more promo and documentation of a stage work than a dance film (but still very well shot) – I could see very clearly that they were exploring new territory not only for hip hop, but for stage work of any genre. The work is set apart by a rigging apparatus that allows the dancers to fly. And while the notion of flying in performance is not new by any means (remember Peter Pan, anyone?), it seemed to me that it was happening in a very new context with this work, with the kind of movement and imagery they were exploring. I was so excited to recognize ideas, themes that I grapple with in my own work – sculptural costume design, scale, gravity – that I sat down with Honji Wang and Sebastien Ramirez to discuss this and their work as a whole over the vast 5,781 mile distance separating Berlin and LA via Skype.
"We are kind of obliged to categorize or label it, but it's a complicated question," Honji explained about their work, given that their company materials describe them as "known for their emotional and powerful blend of contemporary and hip hop dance". She went on to say that they were very new to and careful in categorizing their work as being part of the "contemporary scene." In the beginning they had not seen a lot of contemporary work, and while they didn't want to categorize or limit their work to being just called hip-hop, it was exciting to broaden the horizons of what is expected from that form.
Honji: "At the same time it (hip-hop) opened up possibilities and we are very proud of that work. You can see a difference of flavor and style… The French have a particular style, Americans have their particular style. It started in and came from US, but I think the Europeans, especially the French, took it and developed it in their own way. Each style reflects the culture in which you are living. French is more dirty style, a bit more rough. Asian style is more clean, with clean lines. A lot of the French who started to dance to English lyrics did not understand the lyrics but they understood the flow. (It's) so mixed up because of the internet… Everybody gets inspired by everybody! Before it was clear: dirty style, European style, Asian style… Now it's more mixed up."
"We are just starting to watch now," Sebastien went on to say about seeing contemporary work. "Before we were much more focused in our bubble. I grew up B-boying. I was all the time so busy with the work. I had not so much knowledge of contemporary."
H: "Before I was a bit judging it too fast. I saw a couple of contemporary works but I found it too boring. But the contemporary is so much bigger than the hip-hop dance scene! Now I am seeing a lot of great work… There are moments in almost every performance where I see things that I am inspired by."
I asked what impulse Borderline began with. Was it a visual image? A concept? Some movement?
S: "It comes from laboratory. We meet some dancers we want to work with. Our stories come out of each person. We try to get inspired by the dancers. We isolate work with the dancers, and we work in a team as well, with a stage designer, composer, dramaturge. The impulse with Borderline was this rigging particularly. We would have loved to spend more time to develop it, but we have a lot of time constraints."
H: "He had the dream of working with wires and rigging for 4 years. The way he moved in the rigging system inspired the riggers."
Honji flies in Borderline
S: "In the beginning when we were exploring the rigging we were testing everything. But in Borderline our approach was to use the rigging in a very subtle way… Even though you see some big movements we were not constantly into wires and jumping around."
While Borderline is not actually a dance made specifically for camera, it certainly lends itself to the medium. When I asked the duo if they have aspirations to create another dance film, they explained that Company Sebastien Ramirez is focused specifically on theater, and Wang Ramirez more on museum installations, fashion, and in particular on film, which Sebastien is anxious to dive further into as a director.
S: "I'd like to produce more of this work. We love this combination of film & dance and you can do so much with it. Film is a great tool to put your choreography and attach the emotion you want. You can make just the right moments."
On the day I spoke to them, Honji and Sebastien were supposed to have been performing with Company Sebastien Ramirez at the Kennedy Center in Washington, D.C., but because of visa clearance issues were unable to go. With their work now touring internationally and teetering on the borderline between hip-hop and contemporary, I am excited to see how they continue to navigate the divide, defy categorization, and burst open new forms.
S: "It's art work, with a deep meaning behind it. Built to touch and bring emotions and to be global with our story. To move people is our goal."
Enjoy and be moved.
Tagsborderlinecontemporarydancehip hophoni wanginterviewsebastien ramirezshort filmstageWang Ramirez
City Still: An Erogenous Zone of Urbanity
A Working Class 'Midsummer' Steals the 'Dream'
Sarah Elgart is an award winning, LA based choreographer/director whose work is produced internationally for stage, screen, and site-specific venues. She has collaborated with composers including Paul Chavez of Feltlike, Yuval Ron, and Wilco lead guitarist Nels Cline. From 2007 – 2013 Elgart served both as a board member, chair, and Director of Artistic Development for Dance Camera West. Elgart is the recipient of commissions from organizations including the Getty, NEA, Rockefeller Foundation, Department of Cultural Affairs and Los Angeles Word Airports. A frequent guest lecturer and educator teaching dance and dance/film in schools, universities, and with communities, Sarah continues to be passionate about the intersections of dance and the camera. She enjoys sharing her favorite screen dance selections with Cultural Weekly readers. You can find Sarah at sarahelgart.com on Facebook and on Twitter . Follow Sarah on Instagram at https://www.instagram.com/arrogantelbow/
Deryn Warren
Great article and clip. Thank you!
By Cory Besskepp Cofer
we lack the cheat-sheets It starts from the...
No Tongue
In almost every regard and every area except...
How a Dancer Took Fashion Topless Turvy
By Ann Haskins
Before he set the fashion world "topless...
Contradiction of Silence
In 2014, Bolon, the internationally known... | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 3,071 |
Phytocoris stitti är en insektsart som beskrevs av Knight 1961. Phytocoris stitti ingår i släktet Phytocoris och familjen ängsskinnbaggar. Inga underarter finns listade i Catalogue of Life.
Källor
Ängsskinnbaggar
stitti | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 9,344 |
A turbo generator is the combination of a turbine directly connected to an electric generator for the generation of electric power. Large steam-powered turbo generators provide the majority of the world's electricity and are also used by steam-powered turbo-electric ships.
This report focuses on the Air-Cooled Turbogenerators in North America market, especially in United States, Canada and Mexico. This report categorizes the market based on manufacturers, countries, type and application.
There are 15 Chapters to deeply display the North America Air-Cooled Turbogenerators market. | {
"redpajama_set_name": "RedPajamaC4"
} | 7,226 |
{"url":"https:\/\/freshergate.com\/arithmetic-aptitude\/profit-and-loss\/discussion\/155","text":"Home \/ Arithmetic Aptitude \/ Profit and Loss :: Discussion\n\n### Discussion :: Profit and Loss\n\n1. A man buys a cycle for Rs. 1400 and sells it at a loss of 15%. What is the selling price of the cycle?\n\n2. A. Rs. 1090 B. Rs. 1160 C. Rs. 1190 D. Rs. 1202\n\nExplanation :\n\nS.P. = 85% of Rs. 1400 = Rs.[$$\\frac { 85 } { 100 }$$\u00a0X 1400] = Rs.1190\n\nBe The First To Comment","date":"2023-02-07 07:53:45","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\": 0, \"mathjax_display_tex\": 1, \"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\": 0.9757028818130493, \"perplexity\": 5313.288924241913}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 5, \"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-2023-06\/segments\/1674764500392.45\/warc\/CC-MAIN-20230207071302-20230207101302-00643.warc.gz\"}"} | null | null |
#include "config.h"
#include "core/css/CSSTestHelper.h"
#include "core/css/CSSRuleList.h"
#include "core/css/CSSStyleSheet.h"
#include "core/css/RuleSet.h"
#include "core/css/StyleSheetContents.h"
#include "core/dom/Document.h"
#include "wtf/text/WTFString.h"
#include <gtest/gtest.h>
namespace blink {
CSSTestHelper::~CSSTestHelper()
{
}
CSSTestHelper::CSSTestHelper()
{
m_document = Document::create();
TextPosition position;
m_styleSheet = CSSStyleSheet::createInline(m_document.get(), KURL(), position, "UTF-8");
}
RuleSet& CSSTestHelper::ruleSet()
{
RuleSet& ruleSet = m_styleSheet->contents()->ensureRuleSet(MediaQueryEvaluator(), RuleHasNoSpecialState);
ruleSet.compactRulesIfNeeded();
return ruleSet;
}
void CSSTestHelper::addCSSRules(const char* cssText)
{
TextPosition position;
unsigned sheetLength = m_styleSheet->length();
ASSERT_TRUE(m_styleSheet->contents()->parseStringAtPosition(cssText, position, true));
ASSERT_TRUE(m_styleSheet->length() > sheetLength);
}
} // namespace blink
| {
"redpajama_set_name": "RedPajamaGithub"
} | 131 |
title: To be a better leader, don't forget this basic coaching technique
date: 2019-12-17 00:00:00 Z
categories:
- management
tags:
- green
layout: post
author: martin
featured: false
hidden: false
rating: 4.5
comments: true
image: https://miro.medium.com/max/1000/1*kgCHKzwbhkRx_g39onQxdg.jpeg
---
[Don't just criticize the result, coach the approach](https://medium.com/@mackuntu/to-be-a-better-leader-dont-forget-this-basic-coaching-technique-3bafc504f904?source=friends_link&sk=c85d6732242ba4c7c04bab5ace8c09e1) | {
"redpajama_set_name": "RedPajamaGithub"
} | 321 |
A little while back I stumbled across the term "sewbie". It refers to a newbie sewer; a beginner. Being newish to sewing myself, I quite liked the term. Its kind of cute.
As a sewbie, my journey from stitching innocent to enthusiast, has traversed a standard sort of path.
It started with a bit of cutting and sewing and a pair of pyjama pants. Then a few button holes and a bit of easing and a matching pyjama top. Next came facings, darts and zippers and et voila there was a skirt. Then there was a period of consolidation: sewing from commercial patterns; drafting a few little projects.
All along, though, I knew I'd soon arrive at that sign-posted fork in the road. The big one, where one direction points to "Comfort Zone" (on the flat, of course) and the other to "Real Pattern Alterations" (up a steep hill).
When I did finally arrive, uncharacteristically I chose the hill. You might still see me there sometime: I'm the one puffing asthmatically and puce in the face.
There is, however, a little endorphin rush every now and then. Its exhilarating learning how to be the master of your own sewing destiny. To make something that fits properly; looks flattering; feels comfortable. Recently there's been a flurry of grading and slashing and spreading and redrawing of seam lines on paper patterns. My sewing space is a jumble of calico, tracing paper, pencils, rulers and tape as patterns are redrafted and toiles trialled.
In the midst of my latest class experiments on Misses' tunic top Simplicity 5409 (requiring an FBA, grading on the side seams and a change to the neckline), I received an e-mail. Something that really caught my attention. It was from sewingpatterns.com announcing an innovation: the Perfect Fit Pattern. This new service allows you to input your details (measurements, posture type and photos) when purchasing selected Butterick, McCalls or Neue Mode patterns. The pattern is then altered for you within 7-working days.
"And the cost?", I hear you ask breathlessly (if you are on the same hill as me, that is).
Well, I should say that the finished pattern isn't cheap. At the time of writing its US$24.50 (AU$36.10) for a downloadable version or US$27.50 (AU$40.53) for a paper pattern. This is at least US$10 (AU$14.75) more expensive than a standard paper pattern. However, if you consider the additional cost is really for a service - which I am assuming involves human intervention - it seems relatively inexpensive.
So it all sounds pretty nifty, huh?
I can't help but wonder whether this will be a little bit like introducing the calculator into schools. (Call me old-fashioned, but does anyone really remember how to do long division, when it counts?). Can a remote pattern alteration service really dispense with the need to make your own changes? Will it be possible to sew lovely fitted clothes without ever knowing the theory or practice of pattern alterations?
I have decided to see.
In the name of "research" I have just purchased McCalls pattern 5758. Its a semi-fitted, hip-length jacket with princess seams. And I'd never have a shadow of a hope of making this jacket without alteration. Substantial alteration.
Now if all that doesn't all spell trouble for fit, I really don't know what does.
Anyway, I feel that with my dimensions, I am able to throw down a suitable gauntlet for sewingpatterns.com. Stay tuned - I'll look forward to telling you in up-coming posts about how I have found the service and showing you how well the end garment fits.
I can't wait to see if it actually works. If it's a success, then it's definitly an extra $10 very well spent for sure.
Sounds fantastic, and the price isn;t bad if it's for a service that really does save you time and a bundle of calico and pattern sheets. I stand in awe at your efforts to teach yourself sewing properly. I am still so 'dipping a toe' in the puddle end.
Good on you ! Hope it all works well for you .
I can't wait to see how your pattern turns out. If it works out well than I would definately spend $10 to save the time altering the pattern.
Wow that does sound like a "service" I wonder if it comes with a smile. Can't wait to see how it works out for you, might inspire me to sew more than straight line quilts! Gives hope to those of us out there with normal shapes to get home made clothes that fit without all the sweat and tears!
Paying the extra money is not bad if you can't figure out how to adjust the pattern yourself, and its really not that much money compared to the time it might take a person trying themselves.
I think the service sounds fab for those that would like to make sure it fits with no hassle.
There is a name for me I am a sewbie! I like it!
That sounds awesome and just like how a meal always tastes better when someone else cooks it. Loved your post, had this scary morning monster chuckling!
I am absolutely fascinated. Can't wait to hear your verdict.
hmmm interesting. Very cute jacket too by the way.
That sounds so interestin! I'm keen to see how you go with it!
Sounds great - I'll be interested to hear how it is. I am also in the "grossly disfigured by pregnancy" camp and, though my kids love touching Mums soft squishly tummy cushion, the fact that my stomach is at least 2-3sizes bigger than the rest of me makes buying clothes kinda tricky. Thats why I love sewing so much - clothes that fit - woohoo!!
I am utterly compelled to follow this one through with you to the bitter end. I am so fascinated by this 'fitted pattern service' but have wondered if it could actually live up to all of it's marketing hype.
And I have to say I have been hovering at that same fork in the road for quite some time now, wanting to make tracks up that hill (albeit alone) and always stopping at the sign that says 'too hard - requires good fitting skills to continue'...and totally wussing out.
But it is the year for no fear I keep telling myself. Thanks for some inspiration from this post. | {
"redpajama_set_name": "RedPajamaC4"
} | 2,184 |
CU's Solar House Profiled By Discovery
Sean Cronin | December 14, 2007 | 11:33am
It's been almost two months since the University of Colorado Solar Decathlon team, (profiled in the Westword feature "Partly Sunny," on November 1), returned from the international competition in D.C. only to learn that the house they had built was not immediately welcome at their school. It was going to cost the already over-budget project too much to crane the house's pieces back together at its planned display spot on campus. Thus, at the last minute, students had to divert the house to the McStain Neighborhoods lot in Westminster where it was built.
Since then, the culmination of the two-year, exhaustive volunteer project has been sitting disassembled. On Tuesday, students plan to finally move their house near CU's research park at 30th Street and Colorado Avenue. Project manager Chad Corbin is happy to see the house get a home, but says bureaucratic delays have caused them to have to move the house at the worst possible time. "We've been extremely frustrated," he says. "The delays have pushed everything back to a bad time of year when it's cold, snow is on the ground, all the students are trying to finish up their semester, and we're getting kicked off of our lot because McStain needs to use it. It comes down to [the university's] concern for the expense of the project and lack of concern for the people involved, it seems."
And because the house is not on the main campus, tours will be a little more difficult logistically. Corbin says anyone who wants to see the solar house should check the team's web page, solar.Colorado.edu, for days and times starting next semester. It will stay on campus for about six months. Later, the house will be expanded to 2100 square feet and put on permanent display by its owner – Xcel Energy.
But first, you can get a sneak peak of the house Monday night on the Discovery Home Channel/Planet Green network. Solar Showdown will air at 6 p.m., again at 9 p.m., and three more times on Tuesday, on the digital cable network.
For the hour-long program, producer Scott DeGraw followed CU, as well as Carnegie Mellon University and the University of Maryland through the October solar competition, as well as the intensive four-month building period leading up to the competition. It should be an in-depth look at the students' blood, sweat and tears. While following the competition myself, one member of the Discovery crew told me he had shot over 200 hours of footage – just in D.C.
Corbin says the team is excited to see the show, but finding a college student who has a cable package expansive enough to carry that channel has been a challenge.
He's still looking. -- Jessica Centers | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 7,429 |
Amid the sartorial pomp and circumstance at New York Fashion Week, we noticed something shocking: Style stars, editors and It girls were all eschewing outrageous outfits in favor of simple sundresses. In the middle of September, no less. Rather than stow away their vacation frocks after Labor Day, the style set reinvented their sleeveless, printed frocks by layering them over classic white tees. How better to keep your mainstay in heavy rotation through fall? Add ankle boots and top with a denim jacket for a playful weekend look. | {
"redpajama_set_name": "RedPajamaC4"
} | 3,383 |
With this Choquequirao is the largest Inca-town of the continent and much larger then Macchu Pichu. For the first time in 1710 they mentioned the existence of Choquequirao. The first drawings of the structure of the ruins were made about 1836 by the Frenchmen Léonce Angrand and them by Eugine Sartigues. It is one of the best tours in Cusco.
We leave Cusco at 6:00 am in the morning for the 4 hour drive to Cachora in our private transportation. On the way we will make a stop to enjoy our breakfast admiring the wonderful and amazing Limatambo Canyon, to the right the great Apu –Salkantay and Humantay Mountains, then we will continue to the little village of Cachora. After meeting our expedition support crew know as arrieros we arrange our equipment on mules / horses we will be ready to start a soft hike through this unknow mystery and very historically rich territory. It is a 3 hour hike to Capuliyoc (2915m) from where we have our first beautiful views of the Apurimac valley stretching below. As well as the snow-capped peaks of Padrayoc and Wayna Cachora. We then descend toward Coca Masana (2330m) where the climate becomes noticeably warmer and the flora and fauna begin to change. Essential to bring a high deet mosquito repellent. Finally we arrive at Playa Rosalina (Rosalina Beach) at 1550m beside the raging " The Speaker God " Apurimac River, famous for Class 5 Rafting rapids. Here we will set up camp and spend the night. Your guide will introduce you to this amazing Inca land. Walking poles or wooden sticks are very highly recommended.
Today is dedicated to exploring the incredible ruins of Choquequirao. Our guide will explain the history and importance of the site. After you will have free time to visit the many sectors of the complex. Late in the afternoon we will start the return trek and camp the night at Raqaypata. | {
"redpajama_set_name": "RedPajamaC4"
} | 4,333 |
\section*{0 \ Introduction}
Let $C$ be a smooth projective curve over an algebraically
closed field of characteristic 0, and $X$ an {\em elliptic
surface} over $C$. By this we mean the following: $X$ is a
smooth projective surface with a relatively minimal elliptic
fibration
$$
f : X \longrightarrow C .
$$
In this paper, we also assume:
$(i)$\ $f$ has a global section $\cal O$, and
$(ii)$\ $f$ is not smooth, i.e., there is at
least one singular fibre.
\vskip 0.5cm
\hskip -0.6cm To every (Jacobian) elliptic fibration $X$ there
is a group of sections $\Phi (X)$ with the distinguished section
$\cal O$ as zero. Up to a finite group, $\Phi (X)$ is identified
with the relative automorphism group of the fibration.
\hskip -0.5cm Due to a formula of Shioda-Tate we have the basic
inequalities
$$
0\leq {\rm rank}\Phi \leq \rho (X)-2
$$
where $\rho$ is the Picard number and the discrepancy
in the upper bound is related to the degree of reducibility
of the fibres.
\begin{definition}\hskip 0.5cm {\rm
An elliptic fibration $X$ is called {\em extremal} if and only
if $\rho (X)=h^{1,1} (X)$ ({\em maximal Picard number}) and
rank$\Phi (X)=0$.
}
\end{definition}
Let $f$ : $X \longrightarrow {\bf P}^1 $ be an {\em extremal}
elliptic $K3$ surface. A fibration $f$ is called {\em semi-stable}
if each singular fiber of $f$ is of tpye $I_n$ [MP3].
Here we call a fibration {\em unsemi-stable} if it is not
semi-stable.
\vskip 0.5cm
\hskip -0.6cm In [MP2], R.Miranda and U.Persson have classified
possible semi-stable fibrations. The determination of all semi-stable
fibrations has been done in [MP3] and [ATZ].
In this paper, we first classify all possible configurations of
unsemi-stable fibrations (cf. Theorem 2.4). Then we calculate
the possible Mordell-Weil Groups for Case(A) (cf.Theorem 3.1),
i.e., the case where each singular fibre of $f$ is {\em not}
of tpye $I_n$. Finally, by using
the method in [ATZ], we will precisely
determine which cases in {\em Table 1} are actually
realizable (cf.Theorem 0.4).
\vskip 0.5cm
\hskip -0.6cm Let $i_n$ denote the number of singular fibres of $f$ of type $I_n$. Similarly we define $i_n^*$, $ii$, $iii$, $iv$, $iv^*$, $iii^*$,
$ii^*$ (cf. [MP1]) . Then we have the following Theorem 0.2.
\begin{theorem} Let
$f$ : $X \longrightarrow {\bf P}^1 $ be an extremal elliptic
$K3$ surface with $\deg J\not= 0$. Then
\[
\deg J = \sum_{n\geq 1} n(i_n + {i}^*_n )=
6\sum_{n\geq 1} (i_n + i^*_n ) + 4(ii + iv^* ) +
3(iii + iii^* ) + 2(iv + ii^* ) - 12.
\]
\end{theorem}
\begin{remark} {\rm From [MP1,\ Lemma 3.1\ and
\ Proposition 3.4], we know that the second equality in
{\rm Theorem\ 0.2} is replaced by ``$\leq$'' in general cases,
for example, the fiber type $(II^*,II^*, I_2,I_1,I_1)$
{\rm (cf.[SI, Lemma 3.1])}.
Thus it {\rm (Theorem\ 0.2)}justifies their naming ``extremal''.
}
\end{remark}
\begin{theorem}
Let $f$ : $X \longrightarrow {\bf P}^1 $ be an extremal elliptic
$K3$ surface and each singular fibre of $f$ is {\rm not} of
type $I_n$. Then there exists exactly $11$ fiber types as
given below {\rm (table 1)}. In particular, Mordell-Weil Group is uniquely
determined by the fiber type of $f$.
\vskip 0.5cm
\begin{center}
\begin{tabular}{| c|c|c|c|c|c|}\hline
$\sharp$ & the fibre type & MW$(f)$ & $\sharp$ & the fibre type & MW$(f)$
\\ \hline
$1$ & $(II^* ,I_1^* , I_1^* )$ & $(0)$ &$ 7$ & $(IV^* ,IV^* , IV^* )$ & ${\bf Z}/3{\bf Z}$ \\ \hline
$ 2$ & $ (II^* ,II^* , IV)$ & $(0)$ & $ 8$ & $(IV^* ,IV^* , I_2^* )$ & $(0)$ \\ \hline
$ 3$ & $(II^* ,IV^* , I_0^* )$ & $(0)$ & $9$ & $(IV^* ,I_3^* , I_1^* )$ & $(0)$ \\ \hline
$ 4$ & $(III^* ,III^* , I_0^* )$ & ${\bf Z}/2{\bf Z}$ & $ 10 $ & $(I_4^* ,I_1^* , I_1^* )$ & ${\bf Z}/2{\bf Z}$ \\ \hline
$ 5$ & $(III^* ,IV^* , I_1^* )$ & $(0)$ & $11$ &$(I_2^* ,I_2^* , I_2^* )$ & ${\bf Z}/2{\bf Z}\oplus {\bf Z}/2{\bf Z} $ \\ \hline
$ 6$ & $(III^* ,I_2^* , I_1^* )$ & ${\bf Z}/2{\bf Z}$ & & & \\ \hline
\end{tabular}
\end{center}
All the above $11$ fiber types are realizable.
\end{theorem}
This paper is organized as follows.
In Section 1, we introduce some basic notation and theorems
which will be used
in the paper. In Section 2, we first prove Theorem 0.2.
Then we give the combinatorical
classification of the possible unsemi-stable fibration
(cf.Theorem 2.4). In Section 3, we calculate all possible Mordell-Weil
Groups for Case(A) (cf. Theorem 3.1).
In Section 4, we prove Theorem 0.4.
\hskip -0.6cm At the same time, and indepedently,
I.Shimada and D.Q.Zhang present
a complete list of extremal elliptic $K3$ surfaces by using the
different method [SZ].
\vskip 0.5cm
\hskip -0.6cm {\bf Acknowledgement.} I would like to thank my
advisor Professor D.-Q. Zhang for introducing me to this
subject, lending his precious manuscript to me and
many enlightening instructions during my preparation of this paper.
I also wish to express my sincere gratitude to
Professor J.Conway and Professor N.Sloane for
their helpful correspondences.
\section{Preliminaries}
{\bf (a) Lattices}
\vskip 0.5cm
\hskip -0.5cm Let $L$ be a lattice, i.e.,
$(i)$ \ $L$ is a free finite {\bf Z} module and
$(ii)$ \ $L$ is equipped with a non-degenerate bilinear
symmetric pairing $< , >$.
The determinant of $L$, det$L$, is defined as
the determinant of the matrix $I=(<x_i , x_j >)$
where $\{ x_1 ,\dots ,x_r \}$ is a {\bf Z}-basis of $L$
($r$= the rank of $L$):
$$
{\rm det}L = {\rm det}(<x_i ,x_j >).
$$
We define the positive- (or negative-) definiteness or
the signature of a lattice by that of the matrix $I$,
noting that these properties
are independent of the choice of a basis. An lattice $L$ is
called {\em even} if $<x,x> \in 2{\bf Z}$ for all $x \in L$.
We call $L$ {\em unimodular} if det$L$=1. Let $J$ be a
sublettice of $L$. We denote its orthogonal complement with
respect to $< , >$ by $J^{\perp}$.
For a lattice $L$, we denote its {\em dual} lattice by $L^{\vee}$. By
using pairing, $L$ is embedded in $L^{\vee}$ as a sublettice with
the same rank. Hence the quotient group $L^{\vee}/L$ is a finite
abelian group, which we denote by $G_J$.
For an even lattice $L$, we define a quadratic form $q_L$ with
values in {\bf Q}/2{\bf Z} as follows:
$$
q_L (x \bmod{L}) = <x, x> \bmod{ 2{\bf Z}}.
$$
\begin{lemma} For $j=1$,$2$, let
$\Delta_j = \Delta (1)_j \oplus \cdots \oplus \Delta (r_j )_j$
be a lattice where each $\Delta (i)_j $is of Dynkin type $A_a$,
$D_d$ or $E_e$.
$(1)$ \ Suppose that $\Phi$ : $\Delta_1 \longrightarrow \Delta_2$
is a lattice-isometry. Then $r_1 =r_2$ and $\Phi (\Delta (i)_1 )=
\Delta (i)_2$ after relabelling.
$(2)$ \ Let ${\bf B}(6)= E_7 \oplus D_6 \oplus D_5$,
${\bf B}(10)= D_8 \oplus D_5 \oplus D_5$,
${\bf B}(11)= D_6 \oplus D_6 \oplus D_6$.
Then we have
$(i)$\ ${\bf B}(6) \subset E_7 \oplus D_{11}$ is an index-$2$
lattice extension.
$(ii)$\ ${\bf B}(10) \subset D_5 \oplus D_{13}$ is an index-$2$
extension, ${\bf B}(10) \subset D_8 \oplus D_{10}$ is
an index-$2$ extension and ${\bf B}(10) \subset D_{18}$ is
an index-$4$ extension.
$(iii)$\ ${\bf B}(11) \subset D_6 \oplus D_{12}$ is an index-$2$
extension, ${\bf B}(11) \subset D_{18}$ is an index-$4$ extension.
\end{lemma}
{\em Proof.}\hskip 0.5cm
We observe that
$$
|\det (A_n )|= n+1 , |\det (D_n )|=4 , |\det (E_6 )|=3 ,
|\det (E_7 )|=2 , |\det (E_8 )|=1,
$$
and for an index-$n$ lattice extension $L \subset M$ one has
$$
|\det (L)| = n^2 |\det (M)|.
$$
Then (1) comes from [ATZ, Lemma 1.3],
and (2) can be obtained by an easy calculation.
\begin{definition} {\rm (The lattice $D_n$) [CS]
\ For $n\geq 3$,
$$
D_n = \{(x_1,x_2,...,x_n)\in {\bf Z}^n : x_1+\cdots +x_n \ is\ even \}.
$$
}
\end{definition}
\begin{remark} {\rm From {\rm Lemma 1.1}, we know
$D_8 \oplus D_5 \oplus D_5 \subset D_{18}$ is an index-$4$
extension. Thus $D_{18}/(D_8 \oplus D_5 \oplus D_5)$ maybe
${\bf Z}/4{\bf Z}$ or
${\bf Z}/2{\bf Z} \oplus {\bf Z}/2{\bf Z}$. In the
following {\rm Lemma 1.4}, We shall prove that,
$D_{18}/(D_8 \oplus D_5 \oplus D_5)
={\bf Z}/2{\bf Z} \oplus {\bf Z}/2{\bf Z} $.
}
\end{remark}
\begin{lemma} {\it For any lattice-isometric embedding
$i$: $ D_5 \oplus D_5 \oplus D_8 \longrightarrow D_{18}$, we have
$$
D_{18} / (D_5 \oplus D_5 \oplus
D_8) = {\bf Z}/2{\bf Z} \oplus {\bf Z}/2{\bf Z}.
$$
}
\end{lemma}
{\em Proof.}\hskip 0.5cm
We denote $A\oplus B\oplus C = D_5 \oplus D_5 \oplus D_8$ and let
$i$ : $A \oplus B \oplus C \longrightarrow D_{18}$
be a lattice-isometric embedding. For one generator
$e$ of the lattices $A$, $B$ or $C$, we assume that
$i(e)=(x_1 ,x_2 ,...,x_{18}) \in D_{18}$.
Since $2=<e,e>=<i(e),i(e)>=\sum_{i=1}^{18} x_i^2$ and
$x_i$ is integer, we have
\begin{claim}\hskip 0.5cm {\it There are exactly two
coordinates of $i(e)$ which are non-zero and each of which is $1$ or $-1$.
}
\end{claim}
Thus by relabelling the coordinates, we may assume that one
generators $e_1$ of $A$ satisfies
$$
i(e_1 )=(1,1,0,...,0) \ or \ (1,-1,0,...,0).
$$
Then we can use the connections among the generators
in Dynkin diagram of $D_5$ to get the possible coordinates of
the generators $e_1$, $e_2$, $e_3$, $e_4$ , $e_5$ of $A$ .
After a simple calculation, we find that, by
rebelling the coordinates, we may assume that
$$
i(A) \subset (x_1,x_2,x_3,x_4,x_5,0,...,0) \cap D_{18} := L_1.
$$
On the other hand, we know $L_1$ is a $D_5$ type lattice and
$i$ is lattice isometry, thus we get
$$
i(A) = L_1.
$$
By using the same method, we may assume that
$$
i(B)=(0,..,0,x_6,x_7,..,x_{10},0,..,0) \cap D_{18} := L_2
$$
and
$$
i(C)=(0,..,0,x_{11},,..,x_{18}) \cap D_{18} := L_3.
$$
Here we will use the orthonormal conditions among $i(A)$,
$i(B)$ and $i(C)$.
A direct computation shows that
$$
D_{18}/ (L_1 \oplus L_2 \oplus L_3) =
{\bf Z}/2{\bf Z} \oplus {\bf Z}/2{\bf Z}.
$$
Thus we prove the Lemma 1.4.
\vskip 0.5cm
\hskip -0.6cm By using the same idea as above proof, we get
\begin{theorem}
{\it For $m=\sum_{i=1}^k n_i$, we have
$$
D_{m} / (\oplus_{i=1}^k D_{n_i} ) = \oplus_{i=1}^{k} {\bf Z}/2{\bf Z}
$$
for any lattice-isometric embedding
$i$ $:$ $\oplus_{i=1}^k D_{n_i} \longrightarrow
D_m$.
}
\end{theorem}
{\bf (b) \ Mordell-Weil lattices of elliptic surface}
\vskip 0.5cm
\hskip -0.6cm Given an ellipric surface $f$: $X\longrightarrow C$,
let $F_{\nu} = f^{-1} ({\nu})$ denote the fibre over $\nu \in
C$, and let
$ Sing(f) =\{ \nu \in C | F_{\nu} \ {\rm is\ singular} \}$.
$ {\bf R}= {\rm Red}(f) = \{ \nu \in C | F_{\nu}\ {\rm is \
reducible}\}. $
\hskip -0.6cm For each $\nu \in {\bf R}$, let
$$
F_{\nu} = f^{-1} (\nu ) = \Theta_{\nu ,0} + \sum_{i=1}^{m_{\nu}
-1} \mu_{\nu ,i} \Theta_{\nu ,i} \ \ (\mu_{\nu ,0} =1 )
$$
where $ \Theta_{\nu ,i}$ ($ 0\leq i \leq m_{\nu} -1$) are the
irreducible components of $F_{\nu}$, $m_{\nu}$ being their
number, such that $ \Theta_{\nu ,0}$ is the unique component
of $F_{\nu}$ meeting the zero section.
\hskip -0.6cm Here we denote
$$
E(K) = {\rm the \ group\ of \ sections \ of\ }\ f,
$$
and
$$
NS(X) = {\rm the \ group \ of \ divisors \ on}\ X \ \ {\rm
modulo \ algebraic \ equivalence.}
$$
\begin{theorem} {\rm (cf.[Sh, Theorem 1.1,1.2,1.3])}
Under the assumptions for the elliptic surfaces in Introduction,
we have
$(1)$\ $E(K)$ is a finite generated abelian group.
$(2)$\ $N(X)$ is finitely generated and torsion-free.
$(3)$\ Let $T$ denote the subgroup of $NS(X)$ generated by
the zero section $({\cal O})$ and all the irreducible
components of fibres. Then, there is a natural isomorphism
$$
E(K) \cong NS(X)/T ,
$$
which maps $P\in E(K)$ to $(P)$ mod $T$.
\end{theorem}
\begin{theorem} {\rm (cf.[Sh, Lemma 8.1])}
For any $P$, $Q \in E(K)$, let
$$
<P,Q> = - (\varphi (P) \cdot \varphi (Q)) \hskip 2cm (*)
$$
where $\varphi (P)$ (resp. $\varphi (Q)$), satisfying
the condition:
$(1)$ \ $\varphi (P) \equiv (P)$ {\rm mod} $T_{\bf Q}$, and
$(2)$ \ $\varphi (P) \perp T$.
Then it defines a symmetric bilinear pairing on $E(K)$, which
induces the structure of a positive-definite lattice on
$E(K)/E(K)_{\it tor}$.
\end{theorem}
\begin{definition} {\rm
The pairing $(*)$ on the Mordell-Weil group $E(K)$ is called
the {\em height pairing}, and the lattice
$$
(E(K)/E(K)_{\it tor} , <,>)
$$
is called the {\em Mordell-Weil Lattice} of the elliptic
curve $E/K$ or of the elliptic surface $f$: $S \longrightarrow
C$.
}
\end{definition}
\begin{theorem} {\rm (Explicit formula for the height pairing)
[Sh, Theorem 8.6]} For any $P$,$Q \in E(K)$,
we have
$$
<P ,Q> = \chi + (P{\cal O}) + (Q{\cal O}) - (PQ) - \sum_{\nu \in R} contr_{\nu}
(P,Q),
$$
$$
<P,P>=2\chi + 2(P{\cal O}) - \sum_{\nu \in R} contr_{\nu}
(P).
$$
\end{theorem}
\begin{remark} {\rm Here $\chi$ is the arithmetic genus of
$S$, and $(P{\cal O})$ is the intersection number of the sections $(P)$
and $({\cal O})$, and similarly for $(Q{\cal O})$,$(PQ)$. The term $contr_{\nu}
(P,Q)$ stands for the local contribution ar $\nu \in R$, which is
defined as follows: suppose that $(P)$ interests $\Theta_{\nu ,i}$
and $(Q)$ intersects $\Theta_{\nu ,j}$. Then we let
$$
contr_{\nu} (P,Q) = \left \{\begin{array}{ll}
(-A_{\nu}^{-1})_{i,j} ,& if \ i\geq 1,j\geq 1, \\
0 ,& otherwise.
\end{array}
\right.
$$
where the first one means the $(i,j)$-entry of the matrix $(-A_{\nu}^{-1})$.
Further we set
$$
contr_{\nu} (P) = contr_{\nu} (P,P).
$$
\hskip -0.5cm Arrange $\Theta_i = \Theta_{\nu ,i}$ $(i=0,1,\cdots ,m_{\nu} -1)$ so
that the simple components are numbered as in the figure below.
\vskip 0.5cm
\centerline{\input mw0.tex}
\vskip 0.5cm
\hskip -0.6cm For the other types of reducible fibres, the numbering is irrelevant.
Assume that $(P)$ intersects $\Theta_{\nu ,i}$ and $(Q)$ intersect
$\Theta_{\nu ,j}$ with $i>1$,$j>1$. Then we have the following
table: the forth row is for the case $i<j$ (interchange $P$, $Q$
if necessary).
\vskip 0.5cm
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|}\hline
$T_{\nu}^-$ & $A_1$ & $E_7$ & $A_2$ & $E_6$ & $A_{b-1}$ & $D_{b+4}$
\\ \hline
type of $F_{\nu}$ & $III$ & $III^*$ & $IV$ & $IV^*$ & $I_b (b\geq 2 )$
& $I_b^* (b\geq 0)$ \\ \hline
$contr_{\nu} (P)$ &$\frac1 2$ & $\frac 3 2$ & $\frac2 3$ & $\frac4 3$&
$\frac{i(b-i)} b$ & $\left \{\begin{array}{ll}
1 , & i=1 \\
1+\frac b 4, & i>1
\end{array}
\right.$ \\ \hline
$contr_{\nu} (P,Q) (i<j)$& - & - & $\frac1 3$ &$\frac2 3$ &
$\frac{i(b-i)} b$ & $\left \{\begin{array}{ll}
\frac1 2 , & i=1 \\
\frac{(2+b)}4 ,& i>1
\end{array}
\right. $ \\ \hline
\end{tabular}
\end{center}
}
\end{remark}
\begin{theorem} {\rm (cf.[MP1, Lemma 3.1 and Proposition 3.4])}
For an elliptic fibration $\pi$: $X \longrightarrow {\bf P}^1$,
we have the following formulas:
$$
\deg J = \sum_{n\geq 1} n(i_n + i^*_n ),
$$
and if furthermore $\deg J \not= 0$, then we also have
$$
\deg J \leq 6\sum_{n\geq 1} (i_n + i^*_n ) + 4(ii + iv^* ) +
3(iii + iii^* ) + 2(iv + ii^* ) - 12.
$$
\end{theorem}
\section{The possible configurations of
the unsemi-stable fibrations}
We shall prove Theorem $0.2$ in the present section.
\vskip 0.5cm
\hskip -0.6cm Let $f$: $X \longrightarrow {\bf P}^1$ be a
(relatively) minimal elliptic surface over ${\bf P}^1$ with
a distinguished section ${\cal O}$. The complete list of
possible fibers has been given by Kodaira [K1]. It
encompasses two infinite families $(I_n, I_n^*, n\geq 0)$
and six exceptional cases $(II, III, IV, II^*, III^*, IV^*)$.
And they can be considered as sublattices of the Neron-Severi
group of $X$ and as such they have rank $(=r(F))$.
If $e(F)$ denotes the Euler number
of the fiber as a reduced divisor, we can set up the following
table.
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}\hline
& $I_0$ & $I_n$($n\geq 1$) & $I_n^*$($n\geq 0$)& $II$ &
$III$ & $IV$ & $IV^*$ & $III^*$ & $II^*$ \\ \hline
e & $0$ & $n$ & $n+6$ & $2$ & $3$ & $4$ & $8$ & $9$ & $10$ \\ \hline
r & $0$ & $n-1$ & $n+4$ & $0$ & $1$ &$2$ & $6$ & $7$ & $8$
\\ \hline
\end{tabular}
\end{center}
\begin{lemma} {\rm (cf.[MP1, Corollary 1.3])} In all
cases $0\leq e-r \leq 2$. Moreover,
$(1)$ \hskip 0.5cm $e-r=0 \Longleftrightarrow$ the fibre $F$
is smooth, i.e., of type $I_0$;
$(2)$ \hskip 0.5cm $e-r=1 \Longleftrightarrow$ the fibre $F$
is semi-stable, i.e., of type $I_n$ ,$n\geq 1$;
$(3)$ \hskip 0.5cm $e-r=2 \Longleftrightarrow$ the fibre $F$
is unstable.
\end{lemma}
In the following discussion, we denote
\begin{eqnarray*}
{[Q1]} &:=& \sum_{n\geq 1} ni_n + \sum_{n\geq 1} (n+6) i^*_n
+ 6i^*_0 + 10ii^* + 9iii^* + 8iv^* + 4iv + 3iii + 2ii. \\
{[Q2]} &:=& \sum_{n\geq 1} (n-1)i_n + \sum_{n\geq 1} (n+4) i^*_n
+ 4i^*_0 + 8ii^* + 7iii^* + 6iv^* + 2iv + iii. \\
{[Q3]} &:=& \sum_{n\geq 1} i_n + 2(\sum_{n\geq 1} i^*_n
+ i^*_0 + ii^* + iii^* + iv^* + iv + iii + ii). \\
{[Q4]} &:=& 6\sum_{n\geq 1} (i_n + i^*_n ) + 4(ii + iv^* ) +
3(iii + iii^* ) + 2(iv + ii^* ) - 12. \\
{[Q5]} &:=& i^*_0 + iv + iii + ii.
\end{eqnarray*}
It is easy to see $[Q1]=24$, $[Q2]= \rho (X)-2$ and $[Q1]-[Q2]=[Q3]$.
\begin{lemma}
Let $f$: $X \longrightarrow {\bf P}^1$ be an
elliptic surface over ${\bf P}^1$ with
$\rho (X)=a (\leq 20)$. If $\deg J\not= 0$, then we have
$(1)$ \ \ $[Q4] - \deg J = 6(20 - a - [Q5])$.
$(2)$ \ \ $0 \leq [Q5] \leq 20 - a$.
\end{lemma}
{\em Proof.}\ By Theorem 1.12, we have
\begin{eqnarray*}
[Q4] - \deg J &=& [Q4] - \sum_{n\geq 1} n(i_n + {i}^*_n ) \\
&=& [Q4] - 24 + \sum_{n\geq 1} 6 i^*_n
+ 6i^*_0 + 10ii^* + 9iii^* + 8iv^* + 4iv + 3iii + 2ii \\
&=& 6\sum_{n\geq 1} i_n +
12( \sum_{n\geq 1}{i}^*_n + ii^* + iii^* + iv^*) + 6[Q5] - 36 \\
&=& 6([Q3]-[Q5]-6)\\
&=& 6(20-a-[Q5]) \geq 0.
\end{eqnarray*}
{\em Proof of Theorem $0.2$.} In this case, $a=20$,
by Lemma 2.2, we get the result.
\begin{lemma} Assume $X$ is an extremal elliptic $K3$ surface
with an unsemi-stable fibration $f:$ $X \longrightarrow {\bf P}^1$.
Let $m$ be the number of singular fibers of $f$.
Then
$(a)$\ $\sum (e(F) - r(F)) = 6$.
$(b)$\ $3\leq m \leq 5$.
\end{lemma}
{\em Proof.}\ Since $X$ is $K3$ surface, $\sum e(F)=24$. Also
since $X$ is extremal, $\sum r(F)=\rho -2=18$. This proves (a).
(b) follows from Lemma 2.1 and our definition of {\em unsemi-stable
fibration}.
\vskip 0.5cm
\hskip -0.6cm In the following discussion, we denote $F_i$ (i=1,2,3)
as the singular fiber of $f$ which is not of type $I_n$.
\begin{theorem}
Let $f :$ $ X \longrightarrow
{\bf P}^1 $ be an extremal elliptic $K3$ unsemi-stable fibration. Then
the number $m$ of the singular fibers of f is $3$, $4$ or $5$.
$(A)$ If $m=3$, then the possible fiber types of $(F_1, F_2, F_3)$
are listed in the following.
$(II^* ,I_1^* ,I_1^* )$, $(II^* ,II^* ,IV)$, $(II^* ,IV^* ,I_0^*)$,
$(III^* ,III^* ,I_0^* )$,
$(III^* ,IV^* ,I_1^* )$, $(III^* ,I_2^* ,I_1^* )$, $(IV^* ,IV^* ,IV^* )$,
$(IV^* ,IV^* ,I_2^* )$,
$(IV^* ,I_2^* ,I_2^* )$, $(IV^* ,I_3^* ,I_1^* )$,$(I_2^* ,I_2^* ,I_2^* )$,
$(I_2^* ,I_3^* ,I_1^* )$, $(I_4^* ,I_1^* , I_1^* )$.
$(B)$ If $m=4$, then the possible fiber types of
$(F_1, F_2, I_{n_3}, I_{n_4})$ are listed in the following.
$(B.1)$ \ \ $(I_{n_1}^* ,I_{n_2}^* , I_{n_3 } ,I_{n_4 })$ \mbox{where}
$ n_1 \geq n_2 \geq 1$\ \mbox{and}\ $\sum_{i=1}^{4} n_i = 12$.
$(B.2)$
$(i)$ \ \ $(I_{n_1}^* , II^* ,I_{n_3} ,I_{n_4 })$
\mbox{where} $n_1 + n_3 + n_4 =8$\ \mbox{and} \ $n_1 \geq 1$.
$(ii)$ \ $(I_{n_1}^* , III^* ,I_{n_3} ,I_{n_4 })$
\mbox{where} $n_1 +n_3 + n_4 =9$\ \mbox{and} \ $n_1 \geq 1$.
$(iii)$\ $(I_{n_1}^* , IV^* ,I_{n_3} ,I_{n_4 })$
\mbox{where} $n_1 +n_3 + n_4 =10$ \ \mbox{and}\ $n_1 \geq 1$.
$(B.3.1)$\
$(i)$\ $(II^* , II^* ,I_{n_3} ,I_{n_4} )$\ \mbox{ where}\ $n_3 +n_4 =4$.
$(ii)$\ $(II^* , III^* ,I_{n_3} ,I_{n_4} )$\ \mbox{ where}\ $n_3 +n_4 =5$.
$(iii)$\ $(II^* , IV^* ,I_{n_3} ,I_{n_4} )$\ \mbox{ where}\ $n_3 +n_4 =6$.
$(B.3.2)$ \
$(i)$\ $(III^* , III^* ,I_{n_3} ,I_{n_4} )$\ \mbox{ where}\ $n_3 +n_4 =6$.
$(ii)$\ $(III^* , IV^* ,I_{n_3} ,I_{n_4} )$\ \mbox{ where}\ $n_3 +n_4 =7$.
$(B.3.3)$\
$(IV^* , IV^* ,I_{n_3} ,I_{n_4} )$\ \mbox {where}\ $n_3 +n_4 =8$.
$(C)$ If $m=5$, then the possible fiber types of
$(F_1, I_{n_2}, I_{n_3}, I_{n_4}, I_{n_5})$ are listed in the following.
$(i)$\ $(I_{n_1}^* ,I_{n_2} ,I_{n_3} ,I_{n_4} ,I_{n_5} )$
\mbox{ where} $\sum_{i=1}^{5} n_i =18$\ \mbox{and}\ $n_1 \geq 1$.
$(ii)$\ $(II^* ,I_{n_2} ,I_{n_3} ,I_{n_4} ,I_{n_5 } )$\ \mbox{ where}\
$\sum_{i=2}^{5} n_i =14$.
$(iii)$\ $(III^* ,I_{n_2} ,I_{n_3} ,I_{n_4} ,I_{n_5 } )$\ \mbox{where} \
$\sum_{i=2}^{5} n_i =15$.
$(iv)$\ $(IV^* ,I_{n_2} ,I_{n_3} ,I_{n_4} ,I_{n_5 } )$\ \mbox{ where}
\ $\sum_{i=2}^{5} n_i =16$.
\end{theorem}
{\em Proof.}\ We discuss $\deg J = 0$ and $\deg J\not= 0$
separately.
\vskip 0.5cm
\hskip -0.6cm If $\deg J=0$, then $m=3$. By $[Q2]=18$, we have
$$
18 = iii + 2iv + 6iv^* + 7iii^* +8ii^* + 4i_0^*.
$$
Thus we have the following possible fiber types:
$$
(II^* ,II^* ,IV),\ (II^* ,IV^* ,I_0^*),\ (III^* ,III^* ,I_0^* ),
\ (IV^* ,IV^* ,IV^* ).
$$
If $\deg J\not=0$, then by Lemma 2.2, we have
$i^*_0 + iv + iii + ii=0$ and
$$
24 = \sum_{n\geq 1} ni_n + \sum_{n\geq 1} (n+6) i^*_n +
10ii^* + 9iii^* + 8iv^* := [a].
$$
If $m=3$, then we have
$$
24 = \sum_{n\geq 1} (n+6) i^*_n + 10ii^* + 9iii^* + 8iv^* .
$$
Thus we have the following possible fiber types:
$$
(II^* ,I_1^* ,I_1^* ),\ (III^* ,IV^* ,I_1^* ),
\ (III^* ,I_2^* ,I_1^* ), \ (IV^* ,IV^* ,I_2^* ),\
(IV^* ,I_2^* ,I_2^* ),
$$
$$
(IV^* ,I_3^* ,I_1^* ),\
(I_2^* ,I_2^* ,I_2^* ),\ (I_2^* ,I_3^* ,I_1^* ),\
(I_4^* ,I_1^* , I_1^* ).
$$
Combining the above results, we prove Case(A).
\vskip 0.5cm
\hskip -0.6cm If $m=4$, then with $[Q3] = 6$, we have
$$
2 = \sum_{n\geq 1} i^*_n + ii^* + iii^* + iv^* := [b]
$$
and
$$
[a] - 6 \times [b] = \sum_{n\geq 1}ni_n + \sum_{n\geq 1}ni^*_n + 4ii^* + 3iii^* +
2iv^* =12.
$$
This proves Case(B).
\vskip 0.5cm
\hskip -0.6cm If $m=5$, then with $[Q3] = 6$, we have
$$
\hskip 2cm 1 = \sum_{n\geq 1} i^*_n + ii^* + iii^* + iv^* := [c]
$$
and
$$
[a] - 6\times [c] = \sum_{n\geq 1}ni_n + \sum_{n\geq 1}ni^*_n + 4ii^* + 3iii^* +
2iv^* = 18.
$$
This proves Case(C).
\section {The possible Mordell-Weil Groups for Case (A) }
We shall prove the following Theorem 3.1 in the present section. For
simplity, we label the fiber types which appeared in
Case(A) of Theorem 2.4.
\begin{theorem} The possible Mordell-Weil
Groups for Case $(A)$ are listed in the following table:
\vskip 0.5cm
\begin{center}
\begin{tabular}{| c|c|c|c|c|c|}\hline
$\sharp$ & the fibre type & $MW(f)$ & $\sharp$ & the fibre type & $MW(f)$ \\
\hline
$1$ & $(II^* ,I_1^* , I_1^* )$ & $(0)$ &
$ 8$ & $(IV^* ,IV^* , I_2^* )$ & $(0)$ \\ \hline
$ 2$ & $ (II^* ,II^* , IV)$ & $(0)$ &
$ 9$ & $(IV^* ,I_3^* , I_1^* )$ & $(0)$ \\ \hline
$3$ & $(II^* ,IV^* , I_0^* )$ & $(0)$ &
$ 10$ & $(I_4^* ,I_1^* , I_1^* )$ & $(0)$, ${\bf Z}/2{\bf Z}$
\\ \hline
$ 4 $& $(III^* ,III^* , I_0^* )$ & $(0)$, ${\bf Z}/2{\bf Z}$ &
$ 11 $ & $(I_2^* ,I_2^* , I_2^* )$ & $(0)$,
${\bf Z}/2{\bf Z}$,
${\bf Z}/2{\bf Z}\oplus{\bf Z}/2{\bf Z} $\\ \hline
$ 5$ & $(III^* ,IV^* , I_1^* )$ & $(0)$ &
$12$ & $(I_2^* ,I_3^* , I_1^* )$ & $(0)$
\\ \hline
$ 6$ & $(III^* ,I_2^* , I_1^* )$ & $(0)$, ${\bf Z}/2{\bf Z}$ &
$13$ & $(IV^* ,I_2^* , I_2^* )$ & $(0)$ \\ \hline
$ 7$ & $(IV^* ,IV^* , IV^* )$ & $(0)$, ${\bf Z}/3{\bf Z}$ &
& & \\ \hline
\end{tabular}
\end{center}
\end{theorem}
We now explain the outline of the proof of Theorem 3.1. Firstly,
we deal with types 1,2,3,5,8,9,13 (cf. Lemma 3.2). Then we calculate
the possible nontrivial Mordell-Weil Groups of types 4,6,7,10,11
(cf. Lemma 3.4). Finally we deal with type 12 (cf. Lemma 3.5).
\begin{lemma} For type $m$,
where $m=1$,$2$,$3$,$5$,$8$,$9$ or $13$, the possible
Mordell-Weil Group is $(0)$.
\end{lemma}
{\em Proof.}\hskip 0.5cm With Definition 1.9,
Theorem 1.10 and Remark 1.11, it is easy to prove Lemma 3.2.
For example, $m=1$, if there is a non zero section, say,
$P_1 \in E(K)_{tor}$, where the $i$-th component of $P_1$
is indicated in Remark 1.11, then by Theorem 1.10, we have
$$
0=<P_1 ,P_1 > = 2\chi ({\cal O}_X ) + 2(P_1 {\cal O})
-\left \{\begin{array}{ll}
0 , & i=0, \\
1, & i=1, \\
1+\frac{1}{4} , & i>1 .
\end{array}
\right.
-\left \{\begin{array}{ll}
0 , & i=0, \\
1, & i=1, \\
1+\frac{1}{4} , & i>1 .
\end{array}
\right.
$$
With $2\chi ({\cal O}_X )=4$ and $(P_1 {\cal O}) \geq 0$,
we get a contradiction. The others can be proved by the same method.
\begin{remark} {\rm
In the following calculation,
we let $G_i$, $H_i$ and $J_i$ be the $i$-th
component in the corresponding fiber type $F_1$, $F_2$, $F_3$
respectively (cf.Theorem $2.4$). The numbering of the singular
fiber is defined as following diagrams. Meanwhile, ``$P_1$
pass through the $(i,j,k)$ component'' means $P_1$ only
intersect $G_i$, $H_j$ and $G_k$ in the corresponding fiber
type $F_1$, $F_2$ and $F_3$ respectively. }
\end{remark}
\vskip 0.5cm
\centerline {\input mw1.tex}
\vskip 0.5cm
\begin{lemma} The possible nontrivial Mordell-Weil Group of
type $4$ {\rm (resp. 6,7,10,13)} is ${\bf Z}/2{\bf Z}$
{\rm (resp. ${\bf Z}/2{\bf Z}$, ${\bf Z}/3{\bf Z}$,
${\bf Z}/2{\bf Z}$,
${\bf Z}/2{\bf Z}$ or ${\bf Z}/2{\bf Z}\oplus {\bf Z}/2{\bf Z}$)}.
\end{lemma}
{\em Proof.} We only show how to deal with type 6, the others can be
done by the same way.
\vskip 0.2cm
\hskip -0.6cm For m=6, the fiber type is $(III^* ,I_2^* ,I_1^* )$.
If the Mordell-Weil Group is nontrivial, then for a non zreo
section $P_1$, by Theorem 1.10, we have
$$
0=<P_1 ,P_1 > = 2 \chi ({\cal O}_X ) + 2(P_1 O)
-\left \{\begin{array}{ll}
0 , &i=0,\\
1 , & i=1, \\
1 + \frac{1}{2} , & i>1, (*)
\end{array}
\right.
-\left \{\begin{array}{ll}
0, & i=0,\\
1 , & i=1, \\
1+\frac{1}{2} , & i>1, (*)
\end{array}
\right.
-
$$
$$
\left \{\begin{array}{ll}
0, & i=0, \\
1 , & i=1, (*)\\
1+\frac{1}{2} , & i>1 .
\end{array}
\right.
$$
Thus we may assume that the section $P_1$ pass through
the $(1,2,1)$ component. An easy calculation shows
$$
P_1 = {\cal O} + 2F + \sum_{i=1}^{7} \alpha_i G_i + \sum_{j=1}^{6}
\beta_j H_j + \sum_{k=1}^{5} \gamma_k J_k
$$
where
\begin{eqnarray*}
(\alpha_i ) &=& (-\frac3 2 ,-\frac3 2 , -3, -2,-1,-\frac5 2 ,-2), \\
(\beta_j ) &=& (-\frac1 2 ,-\frac3 2 , -1, -2,-\frac3 2 ,-1), \\
(\gamma_k ) &=& (-1 ,-\frac1 2 ,-\frac1 2 ,-1,-1).
\end{eqnarray*}
and there doesn't exist another non-zero section.
Thus the possible nontrivial Mordell-Weil Group of type $6$ is
${\bf Z}/2{\bf Z}$.
\begin{lemma} The possible Mordell-Weil Group of type $12$ is
trivial, i.e., $(0)$.
\end{lemma}
{\em Proof.} Assume Lemma 3.5 is false. By the same disscussion
as above, we may assume that there is a nonzero section $P$ passing
through the $(1,2,2)$ component. An easy calculation shows
$$
P = {\cal O} + 2F + \sum_{i=1}^{6} G_i \theta_i + \sum_{j=1}^{7}
\beta_j H_j + \sum_{k=1}^{5} \gamma_k J_k
$$
where
\begin{eqnarray*}
(\alpha_i ) &=& (-1,-\frac1 2 ,-\frac1 2 , -1, -1,-1), \\
(\beta_j ) &=& (-\frac1 2 ,-\frac7 4 , -\frac5 4,-\frac3 2,
-2,-\frac5 2 ,-1), \\
(\gamma_k ) &=& (-\frac1 2 ,-\frac5 4 ,-\frac3 4 ,-\frac3 2 ,-1).
\end{eqnarray*}
Thus the possible nontrival Mordell-Weil Group of type $12$ is
${\bf Z}/4{\bf Z}$. That is to say, this group has at least
two nonzero distinct sections, say, $P_1$, $P_2$. On the
other hand, with Theorem 1.10, we have
$$
0=<P_1 ,P_2 > = 2 - (P_1 P_2 )
- 1
-\left \{\begin{array}{ll}
1 +\frac{3}{4}, & i=j>1, \\
\frac{5}{4} , & j>i>1.
\end{array}
\right.
-
\left \{\begin{array}{ll}
1+\frac{1}{4} , & i=j>1, \\
\frac{3}{4} , & j>i>1 .
\end{array}
\right.
< 0.
$$
Thus we get a contradiction and prove Lemma 3.5.
\vskip 0.5cm
\hskip -0.6cm Combining Lemma 3.2, 3.4 and 3.5,
we prove Theorem 3.1.
\section{The complete determination of the Mordell-Weil Groups for
Case (A) }
We shall prove Theorem 0.4 in the present section.
\begin{lemma} {\rm (cf.[ATZ, Lemma 3.1])}
Let $S$ be an even symmetric lattice of rank $20$ and
signature $(1,19)$ and $T$ a positive definite even
symmetric lattice of rank $2$. Assume that $\varphi$:
$T^{\vee} /T \longrightarrow S^{\vee} /S$ is an
isomorphism which induces the the following equality
involving ${\bf Q}/2{\bf Z}$-valued discriminant
(quadratic) forms:
$$
q_S =-q_T .
$$
Let $X$ be the unique $K3$ surface (up to isomorphisms)
with the transcendental lattice $T_X = T$. Then the Picard
lattice $Pic X$ is isometric to $S$.
\end{lemma}
\begin{lemma}
Let $f$: $X \longrightarrow
{\bf P}^1$ be of type $m$ where $m=4$,
$6$,$7$,$10$,$11$,$12$ and $13$. Then
$(1)$\ $MW(f_m) \not= (0)$, and further
$(2)$\ $MW(f_{11}) \not= {\bf Z}/2{\bf Z}$.
\end{lemma}
{\em Proof.} Suppose the contrary that $f$:
$X \longrightarrow {\bf P}^1$ is of the corresponding type
with $MW(f)=(0)$. Let $(b_{ij} )$ be the intersection metrix
of the transcendental lattice $T=T_X$, then
$\det{(b_{ij})}=|\det{(PicX)}|$ (cf.[BPV]).
Modulo congruent action of $SL(2,{\bf Z})$, we may assume that
$ -b_{11} < 2|b_{12} | \leq b_{11} \leq b_{22} $ , and that
$b_{12} \geq 0$ when $b_{11} = b_{22} $.
Embed $T$, as a sublattice, into $T^{\vee} = Hom_{\bf Z} (T,
{\bf Z} )$. Then $T^{\vee} /T \cong (PicX)^{\vee} /(PicX)$.
On the other hand, $T^{\vee}$ has a {\bf Z}-basis
$(e_1 ,e_2 )(b_{ij} )^{-1} =(g_1 ,g_2 )$, where $e_1$,$e_2$
form a canonical basis of $T$. Then comparing the
order of ($g_i$) ($i=1$,$2$) with
$T^{\vee} /T$, we will get a contraction. For simplity, we only
show the case for $m=4$, the others can be done by the same way.
\vskip 0.5cm
\begin{center}
\begin{tabular}{|c|c|c|c|}\hline
m & $T^{\vee} /T$ & the possible $T=(b_{ij} )$ & order of $g_1$, $g_2$ \\ \hline
4 & ${\bf Z}/2{\bf Z} \oplus
{\bf Z}/2{\bf Z} \oplus $
& diag $[2,8]$ & $2$,$8$ \\ \cline{3-4}
&$( {\bf Z}/2{\bf Z}\oplus {\bf Z}/2{\bf Z})$ & diag $[4,4]$ & $4$,$4$ \\ \hline
\end{tabular}
\end{center}
\begin{remark} {\rm From {\rm Theorem\ 3.1} and
{\rm Lemma\ 4.2}, we know that there does not exist
type $12$ or $13$.
The existence of type $m=1$,$3$,$4$,$7$
can be found in {\rm [SI] (page121, 131\ and\ 132)}. Thus the
Mordell-Weil Group of type $1$ {\rm (resp. $3$,$4$,$7$)}
is $(0)$ {\rm (resp. $(0)$, ${\bf Z}/2{\bf Z}$,
${\bf Z}/3{\bf Z}$ )}.}
\end{remark}
\begin{lemma}
Consider the pairs below:
$$
(m, G_m )= (2, (0)),(5, (0)),
(6,{\bf Z}/2{\bf Z} ),(8, (0)),
(9, (0)),(10,{\bf Z}/2{\bf Z}), (11,{\bf Z}/2{\bf Z}\oplus {\bf Z}/2{\bf Z}) .
$$
For each of these seven pairs $(m,G_m )$, there is a Jacobian
elliptic $K3$ surface $f_m$: $X_m \longrightarrow {\bf P}^1 $
of type $m$ such that $(m, MW(f_m ))=(m,G_m )$.
\end{lemma}
{\em Proof.} \hskip 0.5cm
Let $T_m$, $m=2$,$5$,$6$,$8$,$9$, $10$ and $11$
be the positive define
symmetric lattice of rank $2$ with the following intersection form,
respectively:
$$
\left( \begin{array}{rr}
2 & 1 \\
1 & 2
\end{array}
\right) ,
\left( \begin{array}{rr}
2 & 0 \\
0 & 12
\end{array}
\right) ,
\left( \begin{array}{rr}
2 & 0 \\
0 & 4
\end{array}
\right),
\left( \begin{array}{rr}
6 & 0 \\
0 & 6
\end{array}
\right) ,
\left( \begin{array}{rr}
4 & 0 \\
0 & 12
\end{array}
\right) ,
\left( \begin{array}{rr}
4 & 0 \\
0 & 4
\end{array}
\right),
\left( \begin{array}{rr}
2 & 0 \\
0 & 2
\end{array}
\right).
$$
For $m=2$,$5$,$8$,$9$, let $S_m$ be the lattice of rank $20$
and signature $(1,19)$ with the following intersection form,
respectively
$$
U\oplus E_8 \oplus E_8 \oplus A_2 ,
U\oplus E_7 \oplus E_6 \oplus D_5 ,
U\oplus E_6 \oplus E_6 \oplus D_6 ,
U\oplus E_6 \oplus D_7 \oplus D_5 .
$$
We now show how to define $S_6$.
$S_{10}$ and $S_{11}$ can be defined by
the similar way.
\vskip 0.5cm
\hskip -0.6cm Let $\Gamma_6$ be the lattice
$U\oplus E_7 \oplus D_6 \oplus D_5$, with $G_i (1\leq i\leq 7)$,
$H_j (1\leq j \leq 6)$, $J_k (1\leq k \leq 5)$as the canonical
basis of $E_7 \oplus D_6 \oplus D_5$ which are indicated in
{\em Section} 3, and ${\cal O}$, $F$ as a basis of $U$ such
that ${\cal O}^2 =-2$, $F^2 =0$,${\cal O}F=1$.
We extend $\Gamma_6$ to an index-2 integral over lattice
$S_6 = \Gamma_6 + {\bf Z}s_6 $, where
\begin{eqnarray*}
s_6 = {\cal O} + 2F &+& [-\frac3 2 G_1 -\frac3 2 G_2 - 3G_3
-2G_4 -G_5 -\frac5 2 G_6 - 2G_7 ] \\
&+& [-\frac 1 2 H_1 -\frac 3 2 H_2 - H_3 - 2H_4 -\frac 3 2 H_5
- H_6] \\
&+& [-J_1 - \frac 1 2 J_2 - \frac 1 2 J_3 - J_4 - J_5 ].
\end{eqnarray*}
It is easy to see that intersection form on $\Gamma_6$ can be
extend to an integral even symmetric lattice of signature
$(1,19)$. Indeed, setting $s=s_6 $, we have
$$
s^2 =-2 , s\cdot F=s\cdot G_1 = s\cdot H_2 = s\cdot J_1 = 1,
$$
$$
s\cdot G_i = s\cdot H_j =s\cdot J_k =0 \ (\forall i\not=1 ,
j\not= 2 ,k\not= 1 ).
$$
Moreover, $|{\rm det}(S_6 )|= |{\rm det}(\Gamma_6 )| / 2^2 = 8.$
Note that $\Gamma_6^{\vee} = {\rm Hom}_{\bf Z} (\Gamma_6 , {\bf Z})$
contains $\Gamma_6$ as a sublettice with $ E_7 \oplus D_6 \oplus D_5$
as the factor group, and is generated by the following, modulo
$\Gamma_6$:
\begin{eqnarray*}
h_1 &=& (1/2)(G_1 + G_2 + G_6 ), \\
h_2 &=& (1/2)(H_1 + H_2 + H_5 ), \\
h_3 &=& (1/2)(H_1 + H_3 + H_5 ), \\
h_4 &=& (1/4)(2J_1 + J_2 -J_3 + 2J_4 ).
\end{eqnarray*}
Since $(S_6 )^{\vee}$ is an (index-2) sublettice of
$(\Gamma_6^{\vee})$, an element $x$ is in $(S_6 )^{\vee}$ if
and only if $x = \sum_{i=1}^{4} a_i h_i (\bmod{\Gamma_6} )$ such
that $x$ is integral on $S_6$, i.e.,
$$
x\cdot s = ( a_1 + a_2 +a_4 )/2
$$
is an integer. Hence $(S_6 )^{\vee}$ is generated by the following
module $\Gamma_6$:
$$
h_1 + h_2 , h_1 + h_4 ,h_2 +h_4 , h_3 .
$$
Noting that $2h_1$,$2h_2 \in S_6$
and $ h_1 + h_2 + 2h_4 $ is equal to $s$ (mod $\Gamma_6$)
and hence contained in $S_6$, we find that $(S_6 )^{\vee}$ is
generated by the following, modulo $\Gamma_6$:
$$
\epsilon_1 = h_3 \ , \ \epsilon_2 =h_1 + h_4.
$$
Now the fact that $|(S_6 )^{\vee} /S_6 |=8$ and that
$2\epsilon_1$,$4\epsilon_2 \in S_6 $ imply that
$ (S_6 )^{\vee} /S_6$ is a direct sum of its cyclic
subgroups which are of order $2$,$4$, and generated
by $\epsilon_1$, $\epsilon_2$, modulo $S_6$.
We note that the negative of the discriminant form
$$
-q_{(S_6 )} = (-(\epsilon_1 )^2 )\oplus (-(\epsilon_2 )^2 )
=(3/2) \oplus (3/4).
$$
Similarly, we can get
\begin{eqnarray*}
-q_{(S_{10} )} &=& (-(\epsilon_1 )^2 )\oplus (-(\epsilon_2 )^2 )
=(5/4) \oplus (5/4), \\
-q_{(S_{11} )} &=& (-(\epsilon_1 )^2 )\oplus (-(\epsilon_2 )^2 )
=(1/2) \oplus (1/2),
\end{eqnarray*}
for suitable generators $\epsilon_1$, $\epsilon_2$ in the
corresponding cases.
\begin{claim} The pair $(S_m ,T_m )$ satisfies the
conditions of {\rm Lemma\ 4.1} and hence if we let $X_m$ be the unique
$K3$ surface with $T_{X_m} = T_m$ then ${\rm Pic}X_m = S_m$
{\rm (both\
two\ equalities\ here\ are\ modulo\ isometries)}.
\end{claim}
{\em Proof of the claim.}\hskip 0.5cm
We need to show that $q_{T_m} = -q_{S_m}$.
$(1)$\ $m\not =2$. $(S_m )^{\vee} /S_m $ is generated by two
elements $\epsilon_i$ (i=1,2) ($\epsilon_i$ is a simple
sum of the natural generators of $(S_m )^{\vee} /S_m $) such
that for every $a$,$b\in {\bf Z}$ one has $-q_{S_m} (a\epsilon_1
+ b\epsilon_2 ) = -a^2 (\epsilon_1 )^2 - b^2 (\epsilon_2 )^2. $
For all six $m$ where $m\not= 2$, $ \epsilon_i $ can be choosen
such that $(-\epsilon_1^2 , -\epsilon_2^2 )$ is respectively given
as follows:
\begin{center}
$ (3/2, 7/12)$, $(3/2, 3/4)$, $(5/6, 5/6)$,
$(7/12 , 7/4)$, $(5/4, 5/4)$, $(1/2, 1/2)$.
\end{center}
On the other hand, $(T_m )^{\vee}$ ($m\not= 2$)
is generated by
$(g_1, g_2 )=(e_1, e_2 ) T_m^{-1}$, where $e_1$, $e_2$ form a
canonical basis of $T_m$ which gives rise to the intersection
matrix of $T_m$ shown before this claim. Now the claim follows
from the existence of the following isomorphism,
which induces $q_{T_m} = -q_{S_m}$:
$$
\phi \ : \ (T_m )^{\vee}/ T_m \longrightarrow
(S_m )^{\vee}/ S_m , \ \ \ (g_1, g_2 )=(\epsilon_1, \epsilon_2 )B_m.
$$
Here $B_m$ is respectively given as:
$$
\left( \begin{array}{rr}
1 & 1 \\
6 & 1
\end{array}
\right) ,
\left( \begin{array}{rr}
1 & 1 \\
2 & 1
\end{array}
\right) ,
\left( \begin{array}{rr}
2 & 5 \\
1 & 2
\end{array}
\right) ,
\left( \begin{array}{rr}
3 & 2 \\
2 & 1
\end{array}
\right) ,
\left( \begin{array}{rr}
2 & 1 \\
1 & 2
\end{array}
\right),
\left( \begin{array}{rr}
1 & 0 \\
0 & 1
\end{array}
\right)
$$
for $m=5$,$6$,$8$,$9$,$10$ and $11$ respectively.
\vskip 0.5cm
\hskip -0.6cm $(2)$ \ $m=2$. In this case, we know $(T_2 )^{\vee}$
is generated by $(g_1, g_2 )=(e_1, e_2 ) T_2^{-1}$,
where $e_1$, $e_2$ form a
canonical basis of $T_2$ which gives rise to the intersection
matrix of $T_2$ shown before this claim. In fact we have
$$
g_1 \equiv g_2 \ ({\rm mod} T_2 ).
$$
Thus $(T_2 )^{\vee} /T_2$ is generated by one element $g_1$,
and the natural isomorphism
$$
\phi : g_1 \longrightarrow \epsilon_1
$$
will give $q_{T_2} = -q_{S_2}$, where $\epsilon_1 $ is a
canonical {\bf Z}-basis of $A_2$ and
such that for every $a\in {\bf Z}$, we have
$-q_{S_2 }(a\epsilon_1 )= -a^2 (\epsilon_1 )^2 $.
\vskip 0.5cm
\hskip -0.6cm Write $S_m$ (resp. $\Gamma_m$) as
$U \oplus {\bf B}(m)$ with
${\bf B}(m)$ as in the definitions of them. Let $\cal O$, $F$
be a {\bf Z}-basis of $U$ for all $m$. By [PSS, p.573, Theorem 1],
after an (isometric) action of reflections on $S_m = PicX_m$,
we may assume at the beginning that $F$ is a fibre of elliptic
fibration $f_m$: $X_m \longrightarrow {\bf P}^1$. Since
${\cal O}^2 = -2$, Riemann-Roch Theorem implies that ${\cal O}$
is an effective divisor for ${\cal O}\cdot F > 0$. Moreover,
${\cal O}\cdot F =1$ implies that ${\cal O}={\cal O}_1 + F'$
where ${\cal O}_1$ is a cross-section of $f_m$ and $F'$ is an
effective divisor contained in fibres. So $f_m$ is a Jacobian
elliptic fibration and we can choose ${\cal O}_1$ as the zero
element of $MW(f_m )$.
\hskip -0.5cm Let $\Lambda_m $ be the lattice generated by all fiber components
of $f_m$. Clearly, $\Lambda_m = {\bf Z}F \oplus \Delta$, $\Delta
= \Delta (1) \oplus \cdots \oplus \Delta (r) $ (depending on $m$),
where each $ \Delta (i)$ is a negative definite even
lattice of Dynkin type $A_p$, $D_q$, or $E_r$, contained in a single
reducible fibre $F_i$ of $f_m$ and spanned by smooth components
of $F_i$ disjoint from ${\cal O}_1$.
\begin{claim}
We have
$(1)$ \ ${\it Span}_{\bf Z} \{ x\in S_m | x\cdot F=0, x^2 =-2 \} =
\Lambda_m = {\bf Z}F \oplus {\bf B}(m) $; in particular, there
are lattice-isometries: $\Delta \cong {\bf B}(m)$.
$(2)$ \ $MW(f_m) = (0)$ for $m=2$,$5$,$8$ and $9$.
$(3)$ \ $MW(f_m)={\bf Z}/2{\bf Z}$ for $m=6$,$10$.
$(4)$ \ $MW(f_m)={\bf Z}/2{\bf Z}\oplus{\bf Z}/2{\bf Z} $ for $m=11$.
\end{claim}
{\em Proof of the claim.} The first equality in (1)
is from Kodiara's classification of elliptic fibres and the
Riemann-Roch Theorem as used prior to this claim to deduce
${\cal O} \geq 0$. The second equality is clear for that
cases of $m=2$, $5$, $8$ and $9$, because then
${\rm Pic}X_m = S_m = ({\bf Z}{\cal O} + {\bf Z}F )\oplus
{\bf B}(m)$.
We now show the second equality for $m=6$,$10$ and $11$ using Lemma 1.1.
Clearly, ${\bf Z}F \oplus {\bf B}(m)$ is contained in the
first term of (1) and hence in $\Lambda_m$. We also have
$$
19={\rm rank}S_m -1 \geq {\rm rank}\Lambda_m =1+ {\rm rank}\Delta
= 1+ {\rm rank}{\bf B}(m)=19.
$$
Hence $\Delta = \Delta (1) \oplus \cdots \oplus \Delta (r)
\cong \Lambda_m /{\bf Z}F$ contains a finite-index sublettice
$({\bf Z}F\oplus {\bf B}(m))/{\bf Z}F \cong {\bf B}(m)$,
Suppose the contrary that the second equality in (1) is not
true. Then ${\bf B}(m)$ is an index-$n$ $(n>1)$ sublattice
of $\Delta$.
\vskip 0.5cm
\hskip -0.6cm For $m=6$, by Lemma $1.1$, we know
$ \Delta =E_7 \oplus D_{11}$. On the other hand, if we
denote $ s_6' = s_6 - {\cal O} - 2F$, then we have
$$
\Lambda_6 \subset {\it Span}_{\bf Z} \{ x\in S_m | x\cdot F=0 \}=
{\it Span}_{\bf Z} \{ F, G_i , H_j, J_k, s_6' \}
$$
where $1\leq i\leq 7$, $1\leq j\leq 6$ and $1\leq k\leq 5$.
Thus we get ($\bmod{{\bf Z}F}$)
$$
E_7 \oplus D_{11}\subset{\it Span}_{\bf Z} \{ G_i , H_j, J_k, s_6' \}
$$
By a simple calculation, we find that for any element
$e\in D_{11}-(D_6 \oplus D_5)$, $e\notin {\it Span}_{\bf Z} \{ G_i , H_j,
J_k, s_6' \}$. Thus we get a contradiction.
Similarly, we can prove the second equality in (1) for
$m=10$, $11$. The assersion (2),(3) and (4) follow from
the fact in [Sh, Theorem 1.3],
that $MW(f_m )$ is isomorphic to the factor group of
Pic$X_m$ modulo $({\bf Z}{\cal O}_1 + {\bf Z}F)\oplus \Delta$,
where the latter is equal to
$$
({\bf Z}{\cal O} + {\bf Z}F) + \Delta = ({\bf Z}{\cal O} + {\bf Z}F)
\oplus {\bf B}(m) = U \oplus {\bf B}(m).
$$
This proves the claim. And this completes
the lattice-theoretical proof of Lemma $4.4$.
\vskip 0.5cm
\hskip -0.6cm Combining Remark 4.3
and Lemma 4.4, we prove Theorem 0.4.
\section*{References}
\hskip 0.6cm [ATZ]\ E.Artal Bartolo, H.Tokunaga and D.Q.Zhang:
{\em Miranda-Persson's Problem on Extremal Elliptic $K3$ Surfaces},
{\bf alg-geom/9809065}.
[BPV]\ W.P.Barth, C.A.M.Peters and A.J.H.M.Van de Ven:
{\em Compact complex surfaces}, Springer, Berlin, 1984.
[CS]\ J.Conway and N.Sloane:{\em Sphere Packings, Lattices and
Groups}, Grund.Math.Wiss.{\bf 290}, Springer-Verlag (1988).
[K1]\ K.Kodaira: {\em On compact complex
analytic surfaces II}, Ann.Math. {\bf 77}(1963), 563-626.
[MP1]\ R.Miranda and U.Persson: {\em On extremal rational elliptic
surfaces}, Math.Z.{\bf 193}(1986), 537-558.
[MP2]\ R.Miranda and U.Persson: {\em Configurations of $I_n$ fibers
on elliptic $K3$ surfaces}, Math.Z.{\bf 201}(1989), 339-361.
[MP3]\ R.Miranda and U.Persson: {\em Mordell-Weil Groups of extremal
elliptic $K3$ surfaces}, Problems in the theory of
surfaces and their classification (Cortona, 1988), Symposia
Mathematica, XXXII, Academic Press, London,1991, 167-192.
[PSS]\ I.I.Pjateckii-Sapiro and I.R.Safarevic: {\em Torelli's
theorem for algebraic surfaces of type $K3$}, Math.USSR Izv.
{\bf 5} (1971), 547-588.
[SZ]\ I.Shimada and D.Q.Zhang : {\em Classification of extremal
elliptic $K3$ surfaces and fundamental groups of open $K3$ surfaces},
preprint,1999.
[Sh]\ T.Shioda: {\em On the Mordell-Weil Lattices}, Commentarii
Mathematici, Universitatis Sancti Pauli, Vol.{\bf 39}(1990),
211-240.
[SI]\ T.Shioda and H.Inose: {\em On singular $K3$ surfaces},
Complex analysis and algebraic geometry, papers dedicated to K.Kodiaira,
Iwanami Shoten and Cambridge University Press, London, 1977,
119-136.
\end{document}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 6,237 |
hs.allowWidthReduction = false;
hs.allowHeightReduction = true;
hs.objectLoadTime = 'before'; // Load iframes 'before' or 'after' expansion.
hs.cacheAjax = true; // Cache ajax popups for instant display. Can be overridden for each popup.
hs.preserveContent = true; // Preserve changes made to the content and position of HTML popups.
// These properties can be overridden in the function call for each expander:
hs.push(hs.overrides, 'contentId');
hs.push(hs.overrides, 'allowWidthReduction');
hs.push(hs.overrides, 'allowHeightReduction');
hs.push(hs.overrides, 'objectType');
hs.push(hs.overrides, 'objectWidth');
hs.push(hs.overrides, 'objectHeight');
hs.push(hs.overrides, 'objectLoadTime');
hs.push(hs.overrides, 'swfObject');
hs.push(hs.overrides, 'cacheAjax');
hs.push(hs.overrides, 'preserveContent');
// Internal
hs.preloadTheseAjax = [];
hs.cacheBindings = [];
hs.sleeping = [];
hs.clearing = hs.createElement('div', null,
{ clear: 'both', borderTop: '1px solid white' }, null, true);
hs.htmlExpand = function(a, params, custom) {
if (!hs.$(params.contentId) && !hs.clones[params.contentId]) return true;
for (var i = 0; i < hs.sleeping.length; i++) {
if (hs.sleeping[i] && hs.sleeping[i].a == a) {
hs.sleeping[i].awake();
hs.sleeping[i] = null;
return false;
}
}
try {
hs.hasHtmlexpanders = true;
new HsExpander(a, params, custom, 'html');
return false;
} catch (e) {
return true;
}
};
hs.identifyContainer = function (parent, className) {
for (i = 0; i < parent.childNodes.length; i++) {
if (parent.childNodes[i].className == className) {
return parent.childNodes[i];
}
}
};
hs.preloadAjax = function (e) {
var aTags = document.getElementsByTagName('A');
var a, re;
for (i = 0; i < aTags.length; i++) {
a = aTags[i];
re = hs.isHsAnchor(a);
if (re && re[0] == 'hs.htmlExpand' && hs.getParam(a, 'objectType') == 'ajax' && hs.getParam(a, 'cacheAjax')) {
hs.push(hs.preloadTheseAjax, a);
}
}
hs.preloadAjaxElement(0);
};
hs.preloadAjaxElement = function (i) {
if (!hs.preloadTheseAjax[i]) return;
var a = hs.preloadTheseAjax[i];
var cache = hs.getNode(hs.getParam(a, 'contentId'));
var ajax = new HsAjax(a, cache);
ajax.onError = function () { };
ajax.onLoad = function () {
hs.push(hs.cacheBindings, [a, cache]);
hs.preloadAjaxElement(i + 1);
};
ajax.run();
};
hs.getCacheBinding = function (a) {
for (i = 0; i < hs.cacheBindings.length; i++) {
if (hs.cacheBindings[i][0] == a) {
var c = hs.cacheBindings[i][1];
hs.cacheBindings[i][1] = c.cloneNode(1);
return c;
}
}
};
HsExpander.prototype.htmlCreate = function () {
this.tempContainer = hs.createElement('div', null,
{
padding: '0 '+ hs.marginRight +'px 0 '+ hs.marginLeft +'px',
position: 'absolute',
left: 0,
top: 0
},
document.body
);
this.innerContent = hs.getCacheBinding(this.a);
if (!this.innerContent) this.innerContent = hs.getNode(this.contentId);
this.setObjContainerSize(this.innerContent);
this.tempContainer.appendChild(this.innerContent); // to get full width
hs.setStyles (this.innerContent, { position: 'relative', visibility: 'hidden' });
this.innerContent.className += ' highslide-display-block';
this.content = hs.createElement(
'div',
{ className: 'highslide-html' },
{
position: 'relative',
zIndex: 3,
overflow: 'hidden',
width: this.thumbWidth +'px',
height: this.thumbHeight +'px'
}
);
if (this.objectType == 'ajax' && !hs.getCacheBinding(this.a)) {
var ajax = new HsAjax(this.a, this.innerContent);
var pThis = this;
ajax.onLoad = function () { pThis.onLoad(); };
ajax.onError = function () { location.href = hs.getSrc(this.a); };
ajax.run();
}
else this.onLoad();
};
HsExpander.prototype.htmlGetSize = function() {
this.innerContent.appendChild(hs.clearing);
this.newWidth = this.innerContent.offsetWidth;
this.newHeight = this.innerContent.offsetHeight;
this.innerContent.removeChild(hs.clearing);
if (hs.ie && this.newHeight > parseInt(this.innerContent.currentStyle.height)) { // ie css bug
this.newHeight = parseInt(this.innerContent.currentStyle.height);
}
};
HsExpander.prototype.setObjContainerSize = function(parent, auto) {
if (this.swfObject || this.objectType == 'iframe') {
var c = hs.identifyContainer(parent, 'highslide-body');
c.style.width = this.swfObject ? this.swfObject.attributes.width +'px' : this.objectWidth +'px';
c.style.height = this.swfObject ? this.swfObject.attributes.height +'px' : this.objectHeight +'px';
}
};
HsExpander.prototype.writeExtendedContent = function () {
if (this.hasExtendedContent) return;
this.objContainer = hs.identifyContainer(this.innerContent, 'highslide-body');
if (this.objectType == 'iframe') {
//if (hs.ie && hs.ieVersion() < 5.5) window.location.href = hs.getSrc(this.a);
var key = this.key;
this.objContainer.innerHTML = '';
this.iframe = hs.createElement('iframe', { frameBorder: 0 },
{ width: this.objectWidth +'px', height: this.objectHeight +'px' },
this.objContainer);
if (hs.safari) this.iframe.src = null;
this.iframe.src = hs.getSrc(this.a);
if (this.objectLoadTime == 'after') this.correctIframeSize();
} else if (this.swfObject) {
this.objContainer.id = this.objContainer.id || 'hs-flash-id-' + this.key;
this.swfObject.write(this.objContainer.id);
}
this.hasExtendedContent = true;
};
HsExpander.prototype.correctIframeSize = function () {
var wDiff = this.innerContent.offsetWidth - this.objContainer.offsetWidth;
if (wDiff < 0) wDiff = 0;
var hDiff = this.innerContent.offsetHeight - this.objContainer.offsetHeight;
hs.setStyles(this.iframe, { width: (this.x.span - wDiff) +'px', height: (this.y.span - hDiff) +'px' });
hs.setStyles(this.objContainer, { width: this.iframe.style.width, height: this.iframe.style.height });
this.scrollingContent = this.iframe;
this.scrollerDiv = this.scrollingContent;
};
HsExpander.prototype.htmlSizeOperations = function () {
this.setObjContainerSize(this.innerContent);
if (this.objectLoadTime == 'before') this.writeExtendedContent();
// handle minimum size
if (this.x.span < this.newWidth && !this.allowWidthReduction) this.x.span = this.newWidth;
if (this.y.span < this.newHeight && !this.allowHeightReduction) this.y.span = this.newHeight;
this.scrollerDiv = this.innerContent;
this.mediumContent = hs.createElement('div', null,
{
width: this.x.span +'px',
position: 'relative',
left: (this.x.min - this.thumbLeft) +'px',
top: (this.y.min - this.thumbTop) +'px'
}, this.content, true);
this.mediumContent.appendChild(this.innerContent);
document.body.removeChild(this.tempContainer);
hs.setStyles(this.innerContent, { border: 'none', width: 'auto', height: 'auto' });
var node = hs.identifyContainer(this.innerContent, 'highslide-body');
if (node && !this.swfObject && this.objectType != 'iframe') {
var cNode = node; // wrap to get true size
node = hs.createElement(cNode.nodeName, null, {overflow: 'hidden'}, null, true);
cNode.parentNode.insertBefore(node, cNode);
node.appendChild(hs.clearing); // IE6
node.appendChild(cNode);
var wDiff = this.innerContent.offsetWidth - node.offsetWidth;
var hDiff = this.innerContent.offsetHeight - node.offsetHeight;
node.removeChild(hs.clearing);
var kdeBugCorr = hs.safari || navigator.vendor == 'KDE' ? 1 : 0; // KDE repainting bug
hs.setStyles(node, {
width: (this.x.span - wDiff - kdeBugCorr) +'px',
height: (this.y.span - hDiff) +'px',
overflow: 'auto',
position: 'relative'
}
);
if (kdeBugCorr && cNode.offsetHeight > node.offsetHeight) {
node.style.width = (parseInt(node.style.width) + kdeBugCorr) + 'px';
}
this.scrollingContent = node;
this.scrollerDiv = this.scrollingContent;
}
if (this.iframe && this.objectLoadTime == 'before') this.correctIframeSize();
if (!this.scrollingContent && this.y.span < this.mediumContent.offsetHeight) this.scrollerDiv = this.content;
if (this.scrollerDiv == this.content && !this.allowWidthReduction && this.objectType != 'iframe') {
this.x.span += 17; // room for scrollbars
}
if (this.scrollerDiv && this.scrollerDiv.offsetHeight > this.scrollerDiv.parentNode.offsetHeight) {
setTimeout("try { hs.expanders["+ this.key +"].scrollerDiv.style.overflow = 'auto'; } catch(e) {}",
hs.expandDuration);
}
};
HsExpander.prototype.htmlSetSize = function (w, h, x, y, offset, end) {
try {
hs.setStyles(this.wrapper, { visibility: 'visible', left: x +'px', top: y +'px'});
hs.setStyles(this.content, { width: w +'px', height: h +'px' });
hs.setStyles(this.mediumContent, { left: (this.x.min - x) +'px', top: (this.y.min - y) +'px' });
this.innerContent.style.visibility = 'visible';
if (this.objOutline && this.outlineWhileAnimating) {
var o = this.objOutline.offset - offset;
this.positionOutline(x + o, y + o, w - 2*o, h - 2*o, 1);
}
} catch (e) {
window.location.href = hs.getSrc(this.a);
}
};
HsExpander.prototype.reflow = function () {
hs.setStyles(this.scrollerDiv, { height: 'auto', width: 'auto' });
this.x.span = this.innerContent.offsetWidth;
this.y.span = this.innerContent.offsetHeight;
var size = { width: this.x.span +'px', height: this.y.span +'px' };
hs.setStyles(this.content, size);
this.positionOutline(this.x.min, this.y.min, this.x.span, this.y.span);
};
HsExpander.prototype.htmlOnClose = function() {
if (/Macintosh.+Gecko/.test(navigator.userAgent)) { // bad redraws
if (!hs.mask) hs.mask = hs.createElement('div', null,
{ position: 'absolute' }, hs.container);
hs.setStyles(hs.mask, { width: this.x.span +'px', height: this.y.span +'px',
left: this.x.min +'px', top: this.y.min +'px', display: 'block' });
}
if (this.objectLoadTime == 'after' && !this.preserveContent) this.destroyObject();
if (this.scrollerDiv && this.scrollerDiv != this.scrollingContent)
this.scrollerDiv.style.overflow = 'hidden';
if (this.swfObject) try { hs.$(this.swfObject.getAttribute('id')).StopPlay(); } catch (e) {}
};
HsExpander.prototype.destroyObject = function () {
this.objContainer.innerHTML = '';
};
HsExpander.prototype.sleep = function() {
if (this.objOutline) this.objOutline.table.className = 'highslide-display-none';
this.wrapper.className += ' highslide-display-none';
hs.push(hs.sleeping, this);
};
HsExpander.prototype.awake = function() {
hs.expanders[this.key] = this;
if (!hs.allowMultipleInstances && hs.focusKey != this.key) {
try { hs.expanders[hs.focusKey].doClose(); } catch (e){}
}
this.wrapper.className = this.wrapper.className.replace(/highslide-display-none/, '');
var z = hs.zIndexCounter++;
this.wrapper.style.zIndex = z;
if (o = this.objOutline) {
if (!this.outlineWhileAnimating) o.table.style.visibility = 'hidden';
o.table.className = null;
o.table.style.zIndex = z;
}
this.show();
};
// HsAjax object prototype
HsAjax = function (a, content) {
this.a = a;
this.content = content;
};
HsAjax.prototype.run = function () {
try { this.xmlHttp = new XMLHttpRequest(); }
catch (e) {
try { this.xmlHttp = new ActiveXObject("Msxml2.XMLHTTP"); }
catch (e) {
try { this.xmlHttp = new ActiveXObject("Microsoft.XMLHTTP"); }
catch (e) { this.onError(); }
}
}
this.src = hs.getSrc(this.a);
if (this.src.match('#')) {
var arr = this.src.split('#');
this.src = arr[0];
this.id = arr[1];
}
var pThis = this;
this.xmlHttp.onreadystatechange = function() {
if(pThis.xmlHttp.readyState == 4) {
if (pThis.id) pThis.getElementContent();
else pThis.loadHTML();
}
};
this.xmlHttp.open("GET", this.src, true);
this.xmlHttp.send(null);
};
HsAjax.prototype.getElementContent = function() {
hs.genContainer();
var attribs = window.opera ? { src: this.src } : null; // Opera needs local src
this.iframe = hs.createElement('iframe', attribs,
{ position: 'absolute', left: '-9999px' }, hs.container);
try {
this.loadHTML();
} catch (e) { // Opera security
var pThis = this;
setTimeout(function() { pThis.loadHTML(); }, 1);
}
};
HsAjax.prototype.loadHTML = function() {
var s = this.xmlHttp.responseText;
if (!hs.ie || hs.ieVersion() >= 5.5) {
s = s.replace(/\s/g, ' ');
if (this.iframe) {
s = s.replace(new RegExp('<link[^>]*>', 'gi'), '');
s = s.replace(new RegExp('<script[^>]*>.*?</script>', 'gi'), '');
var doc = this.iframe.contentDocument || this.iframe.contentWindow.document;
doc.open();
doc.write(s);
doc.close();
try { s = doc.getElementById(this.id).innerHTML; } catch (e) {
try { s = this.iframe.document.getElementById(this.id).innerHTML; } catch (e) {} // opera
}
hs.container.removeChild(this.iframe);
} else {
s = s.replace(new RegExp('^.*?<body[^>]*>(.*?)</body>.*?$', 'i'), '$1');
}
}
hs.identifyContainer(this.content, 'highslide-body').innerHTML = s;
this.onLoad();
};
hs.addEventListener(window, 'load', hs.preloadAjax); | {
"redpajama_set_name": "RedPajamaGithub"
} | 1,145 |
\section{Higher holonomies of singular leaves}
\label{sec:leaf}
According to \cite{LLS}, a universal NQ-manifold can be associated to most singular foliations, in particular to those generated by real analytic local generators - which form a large class.
We recall these constructions in Subsection \ref{sec:recapfol}, then define the homotopy groups of a singular foliation as the homotopy groups its universal NQ-manifold in Subsection \ref{sec:homotopyGroups}.
In Subsection \ref{sec:isotropy}, we discuss how these homotopy groups can be computed with the help of several long exact sequences derived from Theorem \ref{thm:snake}.
We then consider a locally closed leaf $L$ of $ {\mathcal F}$. Upon replacing $M$ by a neighborhood of $L$, we can assume that there exists a surjective submersion $p \colon M \to L$ such that the composition of the anchor NQ-morphism $\mathbb U^\mathcal F\to TM$ with the surjection $Tp\colon TM\to TL$ is a surjective submersion of NQ-manifolds.
We first define and study Ehresmann connections for singular leaves in Section \ref{sec:Ehresmann_leaf}.
In Section \ref{sec:LESsing}, we show that Theorem \ref{thm:snake}, when applied in such a neighborhood of a singular leaf, gives a long exact sequence: we call higher holonomies of $\mathcal F$ at $L$ its connecting homomorphisms.
We explain how this is related to the Androulidakis-Skandalis' holonomy groupoid, and how a Lie groupoid defining the singular foliation may help to construct it.
We will finish this section by focusing on several particular aspects of the higher holonomies, that reduce to well-known constructions, e.g.~the period map of Cranic-Fernandes \cite{MR1973056}.
\subsection{The universal NQ-manifold and homotopy groups of a singular foliation}
\subsubsection{The universal NQ-manifold of a singular foliation}
\label{sec:recapfol}
We recall and adapt the main results of \cite{LLS}.
Throughout this subsection, ${\mathcal F} $ is a singular foliation on a manifold $M$.
\begin{definition}\label{def:resolution} (Definition 2.1 in \cite{LLS})
A geometric resolution of ${\mathcal F}$ is a finite family of vector bundles $(E_{-i})_{i\geq 1}$ together with vector bundle morphisms $ E_{-i} \to E_{-i+1}$ and $\rho \colon E_{-1} \to TM$ such that the following sequence is a resolution of $ {\mathcal F}$ in the category of $\mathcal C^\infty_M $-modules:
$$ \dots \to \Gamma(E_{-3}) \to \Gamma(E_{-2}) \to \Gamma(E_{-1}) \to {\mathcal F}. $$
\end{definition}
\begin{remark}
\normalfont
By the Serre-Swann theorem, a geometric resolution
is a projective resolution of ${\mathcal F}$ by finitely generated $\mathcal C^\infty_M $-modules.
\end{remark}
\begin{theorem}\label{theo:exists}
(Theorem 2.4 in \cite{LLS})
Every locally real analytic singular foliation on a manifold $M$ of dimension $n$ admits a geometric resolution of length $ n+1$ on every relatively compact open subset.
\end{theorem}
Among the NQ-manifolds over $M$ whose anchor map $\rho\colon \Gamma(E_{-1})\to\mathfrak X(M)$ takes values in $\mathcal F$, there is a distinguished class, which we now define.
\begin{definition}
A \emph{universal NQ-manifold of a singular foliation $ {\mathcal F}$} is an NQ-manifold ${\mathbb V}$ such that the underlying sequence $E_{-i}\to E_{-i+1}$ together with the anchor map $E_{-1}\to TM$ form a geometric resolution of $ {\mathcal F}$.
\end{definition}
The use of the word ``universal'' is justified by the following Theorem:
\begin{theorem}\label{theo:onlyOne}
(Theorem 1.7 in \cite{LLS})
\texttt{ "Universality"}.
Universal NQ-manifolds of $ {\mathcal F}$ are a terminal object in the category where objects are NQ-manifolds over $M$ inducing sub-foliations of $ {\mathcal F}$, and morphisms are homotopy classes of NQ-manifold morphisms over $M$.
\end{theorem}
Since in a category two universal objects are canonically isomorphic:
\begin{corollary} \label{coro:unique}
(Corollary 1.8 in \cite{LLS}) {\texttt{"Uniqueness"}}.
Two universal NQ-manifolds of a the singular foliation $ \mathcal{F}$ are homotopy equivalent in a unique up to homotopy manner.
\end{corollary}
We will denote an universal NQ-manifold by $\mathbb U^\mathcal F$, and their linear parts by $ (E_\bullet,\mathrm d,\rho)$.
Singular foliations that admit a universal NQ-manifold admit a geometric resolution.
In fact, the converse also holds:
\begin{theorem}
(Theorem 1.6 in \cite{LLS}) \texttt{"Existence"}.
\label{theo:existe}
If $\mathcal F$ admits a geometric resolution, then it admits a universal NQ-manifold whose linear part is that geometric resolution.
\end{theorem}
We finish this section by two lemmas about restrictions of the universal NQ-manifold of a singular foliation to a transverse sub-manifold and to a leaf.
For any locally closed submanifold $ S \subset M$, we let $ {\mathcal F}|_S \subset {\mathfrak X}(S)$ be the space obtained by restricting to $S$ vector fields in $ {\mathcal F}$ which are tangent to $S$. For a generic submanifold $S$, the space $\mathcal F|_S$ might not be finitely generated and hence might not be a singular foliation.
We say that a submanifold $S \subset M$ is \emph{${\mathcal F}$-transversal} if $ T_m S + T_m {\mathcal F} = T_m M $ for all $ m \in S$.
\begin{lemma}\label{lem:transversefol0}
The restriction $ {\mathcal F}|_S$ to an ${\mathcal F}$-transversal $S$
is a singular foliation on $S$. Moreover, the restriction of any universal NQ-manifold of $ {\mathcal F}$ to $S$ is a universal NQ-manifold of $ {\mathcal F}|_S$.
\end{lemma}
\begin{proof}
The first statement is proven in \cite{AS09}, Proposition 1.10, Item (b). The second statement follows from the trivial observation that for any geometric resolution $ (E,\mathrm d, \rho)$ of $ {\mathcal F}$ and any ${\mathcal F}$-transversal $S$ as above, the complex:
$$ \dots \to E_{-3} |_{S} \stackrel{\mathrm d}{\longrightarrow} E_{-2} |_{S} \stackrel{\mathrm d}{\longrightarrow} \rho^{-1}(TS) \stackrel{\rho}{\longrightarrow} TS $$
is both a geometric resolution of $ {\mathcal F}|_S$ and a sub-Lie $ \infty$-algebroid.
\end{proof}
Let $L$ be a leaf of $ \mathcal F$. We recall from \cite{AZ13} the following construction:
there is a unique transitive Lie algebroid over $L$, called \emph{holonomy Lie algebroid of $L$}, denoted by $ A_L$, whose sheaf of sections is $ \mathcal F / I_L \mathcal F$, with $ I_L$ the sheaf of smooth functions on $M$ vanishing along $L$.
\begin{example}
\label{ex:regularLeafHolonomy}
\normalfont
For a regular\footnote{A regular point is a point around which all leaves have the same dimension (and are called regular leaves). Recall that if a geometric resolution of finite length exists, then the all regular leaves have the same dimension.} leaf $L$, the holonomy Lie algebroid is the tangent Lie algebroid. In equation: $ A_L =TL$.
\end{example}
Consider the restriction $\mathfrak i_L \mathbb U^\mathcal F $ of any universal NQ-manifold $\mathbb U^\mathcal F $.
This is an NQ-manifold over $L$, which is transitive over $L$.
\begin{lemma}\label{lem:transversefol}
The restriction $\mathfrak i_L \mathbb U^\mathcal F$ of any universal NQ-manifold $\mathbb U^\mathcal F $ to a leaf $L$ of a singular foliation is an NQ-manifold such that:
\begin{enumerate}
\item the differential $ \mathrm d: \mathfrak i_L E_{-i-1} \to \mathfrak i_L E_{-i}$ has constant rank along $L$,
\item the quotient space $ A_L := \mathfrak i_L E_{-1} / \mathrm d \mathfrak i_L E_{-2}$:
\begin{enumerate}
\item comes equipped with a transitive Lie algebroid structure,
\item which does not depend on the choice of a universal NQ-manifold $\mathbb U^\mathcal F $,
\item and coincides with Androulidakis-Zambon's holonomy Lie algebroid $A_L$ of the leaf $L$.
\end{enumerate}
\end{enumerate}
\end{lemma}
\begin{proof}
Let $ \tau : TL \to E_{-1}$ be a section of $ \rho : E_{-1} \to TL $. Then the bilinear map
$$ \nabla_u^{(i)} e := [\tau ( u ) , e ] \hbox{ for all $u \in \mathfrak X (L) $ and $e \in \Gamma(E_{-i}) $ }$$
defines a connection on $ E_{-i}$ for all $i \geq 1$. An easy computation gives:
$$ \nabla_u^{(i+1)} (\mathrm d e) = [\tau ( u ) , \mathrm d e ] = \mathrm d [\tau ( u ) , e ] = \mathrm d \nabla^{(i)}_u e .$$
Since the covariant derivative of $\mathrm d $ is $0$, its rank has to be constant. This proves the first item.
For the second item, see Proposition 4.18 in \cite{LLS}.
\end{proof}
\subsubsection{Definition of the homotopy groups of a singular foliation}
\label{sec:homotopyGroups}
Throughout this section, $ {\mathcal F}$ is a singular foliation that admits a geometric resolution: by Theorem \ref{theo:existe}, it also admits a universal NQ-manifold $\mathbb U^{\mathcal F} $.
Since homotopy equivalent NQ-manifolds have isomorphic homotopy groups and fundamental groupoid by Corollary \ref{cor:homotopyGroups},
and since any two universal NQ-manifold are homotopy equivalent by Corollary \ref{coro:unique},
the groups $\pi_n(\mathbb U^\mathcal F, m) $, $\Gamma(\pi_n(\mathbb U^\mathcal F) ) $ and the fundamental groupoid $\mathbf\Pi (\mathbb U^\mathcal F) $ do not depend on the choice of a universal NQ-manifold. In particular, this justifies this following definition and the following notations:
\begin{definition}
For every singular foliation $\mathcal F $ that permits a geometric resolution, we define
\begin{enumerate}
\item the $n$-th homotopy group $ \pi_n ({\mathcal F}, m) $ at $m$,
\item its space of smooth sections $\Gamma(\pi_n(\mathcal F)) $,
\item and the fundamental groupoid $\mathbf F $ of $\mathcal F $,
\end{enumerate}
to be those of any one of its universal NQ-manifolds.
\end{definition}
In \cite{AS09}, Androulidakis and Skandalis constructed the holonomy Lie groupoid of a singular foliation: It is a topological groupoid whose leaves are the leaves of $ {\mathcal F}$, and whose restriction to a given leaf $L$ is a smooth manifold in a neighborhood of the identity.
According to Proposition 4.37 in \cite{LLS}, the fundamental groupoid $\mathbf F $ of $\mathcal F $ is the universal cover of the holonomy groupoid of Androulidakis and Skandalis\footnote{By universal cover, we mean that for each leaf $m \in M$, $\mathbf F|_m=s^{-1}(m) $ is the universal cover of the fiber over $m$ of Androulidakis and Skandalis'holonomy groupoid}. In view of this identification, and since the holonomy Lie groupoid is a well-established mathematical object, we prefer to adapt the terminology as follows:
\begin{convention}
\label{con:newName}
From now on, we will call $\mathbf F $ the \emph{universal holonomy groupoid of $\mathcal F $} rather than ``the fundamental groupoid of $\mathcal F $''.
\end{convention}
\begin{remark}
\normalfont
\label{rmk:holonomyLieAlgerbroid}
The universal holonomy groupoid $\mathbf F$ can be characterized as follows: it induces the same foliation as $ \mathcal F$ and its restriction to a leaf $L$ is the fundamental groupoid of the holonomy Lie algebroid $A_L$.
\end{remark}
Lemma \ref{lem:restrLeaf} implies that $ \pi_n(\mathcal F,m)$ only depends on the restriction of the universal NQ-manifold $ \mathbb U^\mathcal F$ to the leaf $L$ to which $m $ belongs:
\begin{lemma}
For every universal NQ-manifold $ \mathbb U^\mathcal F$ of a singular foliation $ \mathcal F$, and every point $m \in M$, we have:
$$ \pi_n(\mathcal F, m) = \pi_n ( \mathfrak i_L \mathbb U^\mathcal F, m)$$
where $L$ is the leaf through $ m$.
\end{lemma}
This lemma makes the following examples quite trivial:
\begin{example}
\normalfont
\label{ex:regular}
When $ \mathcal F$ is a regular foliation, we have $ \mathcal F = \Gamma(A)$ with $A\subset TM$ an integrable distribution. In this case, $A$ is itself a universal NQ-manifold of $ \mathcal F$. The universal holonomy groupoid is then the universal cover of the holonomy groupoid of $\mathcal F $ as defined in \cite{zbMATH01956612}. Also, at a point $m \in M$, the homotopy groups of $ \mathcal F$ are the homotopy groups of the leaf through $m$.
\end{example}
\begin{example}
\normalfont
\label{ex:regularleaf}
By the previous example, we see that
$ \pi_n( \mathcal F,m) = \pi_n(L,m) $
for every regular point $m \in M$, with $L$ being the leaf through $m$.
\end{example}
\begin{example}
\normalfont
\label{ex:Debord1}
We say that a singular foliation is \emph{Debord} or \emph{projective} when it is projective as a module over functions. By Serre-Swann theorem, it means that $ {\mathcal F}$ comes from a Lie algebroid $A$ whose anchor is injective on a dense open subset. By construction, the Lie algebroid $A$ is a universal NQ-manifold of $ {\mathcal F}$. Moreover, Theorem 1 in \cite{D} says that $A$ integrates into a smooth (source simply-connected) Lie groupoid, which has to coincide with $ \mathbf F$. In view of Example \ref{ex:LieAlgebroid}, we have\footnote{As usual for Lie groupoids, $ \mathbf F |_m = s^{-1}(m) \subset \mathbf F$ is the source fiber of $m$, and $\widetilde{I_m({\mathbf F})}$ is the universal cover of the holonomy of $I_m(\mathbf F) $ of $ \mathbf F$ at $m$.} $ \pi_n(\mathcal F,m) = \pi_n( {\mathbf F}|_m ,m)$ for all $ n \geq 2$ and $ \pi_1(\mathcal F,m) = {I_m({\mathbf F})}$.
\end{example}
\subsubsection{Computation of the homotopy groups of a singular foliation}
\label{sec:isotropy}
The purpose of this subsection is to describe several exact sequences that lead to the computation of the homotopy groups $ \pi_n({\mathcal F},m)$, when the following two ingredients are given:
\begin{enumerate}
\item the isotropy Lie $\infty$-algebra of $ \mathcal F$ at $m \in M$,
\item the universal holonomy groupoid $ \mathbf F$ of $\mathcal F $, see Convention \ref{con:newName} - more precisely its restriction to the leaf $L$ through $m$. \end{enumerate}
Let us briefly recall how the first of these objects is constructed. The \emph{isotropy Lie $\infty$-algebra of $\mathcal F $ at $m$} is introduced in \cite{LLS}, Section 4.2, by the following spaces and brackets:
\begin{enumerate}
\item The cohomology groups $ H^\bullet(\mathcal F, m) := \oplus_{ n \geq 1} H^{-n}( {\mathcal F},m )$ of any geometric resolution $(E_\bullet, \mathrm d, \rho) $ of $\mathcal F$, evaluated at $ m \in M$:
\begin{equation}\label{eq:Hn} H^{-n}(\mathcal F,m)= H^{-n}({\mathfrak i}_m E_\bullet,{\mathfrak i}_md)=\frac{\mathrm{ker}({\mathfrak i}_m d:{\mathfrak i}_m E_{-n}\to {\mathfrak i}_m E_{-(n-1)})}{\mathrm{im}({\mathfrak i}_m d:{\mathfrak i}_m E_{-(n+1)}\to {\mathfrak i}_m E_{-n})} \end{equation}
coincides by construction with $ Tor^{\bullet}({\mathcal F}, {\mathbb R})$, with ${\mathbb R}$ being made a $\mathcal C^\infty_M$-module through evaluation at $m$ (see Remark 4.9 in \cite{LLS}).
In particular, the graded vector space $H^\bullet(\mathcal F, m)$ does not depend on the chosen resolution and depends on the singular foliation $ {\mathcal F}$ only, which explains the notation.
\item The Lie $ \infty$-algebroid brackets of any universal Lie $\infty$-algebroid of $ \mathcal F$ restrict to the point $m$ to yield a Lie $\infty $-algebra, called \emph{isotropy Lie $\infty$-algebra}. Its \emph{minimal model} is by construction a Lie $\infty$-algebra structure on $H^\bullet(\mathcal F, m)$ whose un-ary bracket is zero. In particular, its $2$-ary bracket satisfies the graded Jacobi identity. Any two minimal models are strictly isomorphic through an isomorphism whose linear map is the identity map of $ H^{\bullet}( {\mathcal F},m )$: in particular, the $2$-ary graded Lie bracket is canonically attached to $ {\mathcal F}$.
\end{enumerate}
We now give a long exact sequence that computes the homotopy groups of a singular foliation.
\begin{proposition}
\label{prop:computingISotropies}
Let ${\mathcal F} $ be a singular foliation that admits a geometric resolution near a point $m \in M$ and $\mathbf F $ be the universal holonomy Lie groupoid. Then there is a natural exact sequence of groups as follows:
\begin{align}
\label{eq:computingISotropies}
\dots \to H^{-n}( {\mathcal F}, m) \longrightarrow \pi_n( \mathcal F,m) \stackrel{q}{ \longrightarrow } \pi_{n}({\mathbf F}|_m,m) \longrightarrow H^{-n+1}( {\mathcal F},m) \longrightarrow \cdots \\
\cdots \longrightarrow \pi_2( \mathcal F, m) \stackrel{q}{ \longrightarrow } \pi_2(\mathbf F|_m,m) \longrightarrow 0 \longrightarrow \pi_1( \mathcal F, m) \longrightarrow {\mathbf{F}|_m^m} \longrightarrow 0 ,\nonumber
\end{align}
where $ \mathbf F|_m = s^{-1}(m)$ is the $s$-fiber over $m$ of the source map, $\mathbf{F}|_m^m$ is the isotropy at $m$ of the universal holonomy Lie groupoid, and $L$ is the leaf through $m$.
\end{proposition}
\begin{proof} Consider the restriction $\mathfrak i_L \mathbb U^\mathcal F$ of the universal NQ-manifold $ \mathbb U^\mathcal F $ to the leaf $L$.
The exact sequence \eqref{eq:computingISotropies} is a particular case of the exact sequence obtained in Theorem \ref{thm:snake} to the surjective submersion of NQ-manifold:
$$ Q \colon \mathfrak i_L \mathbb U^{\mathcal F} \longrightarrow A_L = \tfrac{{\mathfrak i}_L E_{-1}}{ \mathrm d \mathfrak i_L E_{-2}} $$
over the holonomy Lie algebroid of $L$. Its fiber over $ \ell$ is the nilpotent part of isotropy Lie $\infty$-alegbra of $ \mathcal F$: for $ n \geq 2 $ its $\pi_n $ coincide with $H^{-n}(\mathcal F , m )$ in view of Example \ref{ex:LieAlgebraNilpotent}, while its $ \pi_1$ is trivial.
Since the holonomy Lie algebroid is the Lie algebroid of the universal holonomy groupoid (see Remark \ref{rmk:holonomyLieAlgerbroid}), Example \ref{ex:LieAlgebroid} gives an expression of its homotopy groups.
By Proposition 2.2. in \cite{MR3119886}, the holonomy groupoid of a singular foliation is longitudinally smooth. Since $\mathbf F $ is its universal cover by Proposition 4.37 in \cite{LLS}, $ \mathbf F|_m$ is a smooth manifold for all $m\in M$. Proposition \ref{prop:Homot_grs_Of_Lie_oid} gives therefore for all $ n \geq 2$ and $ m \in L$ an isomorphism $ \pi_{n}({\mathbf F}|_m,m)\simeq \pi_n(A_{L}, m) $ and
$ \pi_{1}({\mathbf F}|_m,m)\simeq I_m(\mathbf F) = \mathbf F|_m^m $.
Altogether, these results prove the statement.
\end{proof}
\begin{remark}\normalfont
\label{rmk:holonomyGpdPi2}
Let $L$ be a leaf of a singular foliation $ \mathcal F$, and $A_L$ be the holonomy Lie algebroid. Applying Theorem \ref{thm:snake} or Theorem 1.4 in \cite{Zhu} to the surjective submersion of Lie algebroids:
$$ A_L \to T L $$
whose fiber over $m$ is the isotropy Lie algebra $ H^{-1}(\mathcal F, m ) $,
we obtain the following long exact sequence:
\begin{align}
\label{eq:computingISotropies2}
\dots \to \pi_n( \mathbf H_m , 1 ) \longrightarrow \pi_n( \mathbf F |_m, m ) \longrightarrow \pi_n( L , m ) \longrightarrow \cdots \\
\cdots \longrightarrow 0 \longrightarrow \pi_2( \mathbf F|_m, m) \longrightarrow \pi_2( L,m) \longrightarrow \mathbf H_m \longrightarrow \mathbf F_m \stackrel{t}{\longrightarrow} L ,\nonumber
\end{align}
with $ \mathbf F|_m = s^{-1}(m) \subset \mathbf F_m$
and $\mathbf H_m$ the simply connected Lie group integrating the isotropy Lie algebra.
The second line of \eqref{eq:computingISotropies2} appears in Proposition 3.9. in \cite{AZ13}. Let us explain briefly the relation. Equation \eqref{eq:computingISotropies2} yields an exact sequence:
\begin{align}
\label{eq:computingISotropies3} 0 \longrightarrow \frac{ \pi_2( L,m) }{ \pi_2( \mathbf F|_m, m) } \longrightarrow \mathbf H_m \longrightarrow \mathbf F|_m^m \longrightarrow 0
\end{align}
which matches (3.6) in \cite{AZ13} (up to notations): the left-hand term of the previous expression is therefore what Androulidakis and Zambon call the essential isotropy group of $L$.
\end{remark}
\subsection{Ehresmann ${\mathcal F}$-connections}
\label{sec:Ehresmann_leaf}
\subsubsection{Definition and existence}
Let $L$ be a locally closed leaf of a singular foliation $ {\mathcal F}$.
\begin{definition}
An Ehresmann ${\mathcal F}$-connection near $L$ is a triple made of:
\begin{enumerate}
\item[a)] A neighbourhood $M_L$ of $L$ in $M$ equipped with a projection $p\colon M_L\to L$;
\item[b)] An Ehresmann connection $H$ for $ p$ whose sections are elements of $ {\mathcal F}$.
\end{enumerate}
It is said to be a complete Ehresmann ${\mathcal F}$-connection near $L$ if $H$ is complete.
\end{definition}
Let us discuss existence of such objects. The first statement of the following Proposition was established and proved while discussing with Iakovos Androulidakis \cite{AndroulidakisPrivate}.
\begin{proposition}
\label{prop:existeEhresmann}
Any locally closed leaf $L$ of singular foliation $ {\mathcal F}$ admits an Ehresmann ${\mathcal F}$-connection in a neighborhood of $L$. It can be chosen to be complete, possibly in a sub-neighborhood $M_L $, if there exists a function $ \kappa \colon M_L \to {\mathbb R}$ vanishing along $L$
such that the restriction of $p \colon \{ \kappa \leq 1 \} \to L $ is a proper map satisfying either of the following assumptions:
\begin{enumerate}
\item $ \kappa $ is constant along the leaves,
\item for every $m \in\kappa^{-1}(1) $, there exists a vector field $ u \in {\mathcal F} $ tangent to the fibers of $p$ such that $ u[ \kappa]|_m \neq 0 $.
\end{enumerate}
\end{proposition}
\begin{proof}
According to the tubular neighborhood theorem, there exists a neighborhood $U $ of the leaf $L$ in $M$ that comes equipped with a projection $p \colon U \to L$ which is a surjective submersion.
Now, there is a neighborhood $M_L$ of $L$ in $ U$ such that for all $m \in M_L$, $ T_m {\mathcal F} + {\mathrm{Ker}} (T_m p ) =T_m M $. For every $m \in M$, there exists $d= {\mathrm { dim }} (L)$ sections of $ {\mathcal F}$ whose values at $m$ generate a vector space in direct sum with ${\mathrm{Ker}} (T_m p ) $. There exists a neighborhood $ V_m \subset M_L $ of $m$ on which the distribution $H_m$ generated by these vector fields is also in direct sum with the tangent space of the fibers, defining therefore an Ehresmann $ {\mathcal F}$-connection $ H_m$ on $V_m$.
Let $ (\phi_i,V_i)_{i \in I} $ be a partition of unity of $ M_L$, such that an Ehresmann $ {\mathcal F}$-connection $H_i$ of the previous type exists on $V_i$.
A global Ehresmann $ {\mathcal F}$-connection on $M_L$ is given as follows: it is the unique distribution for which the horizontal lift of $X \in {\mathfrak X}(L)$ is $ H(X) := \sum_{i \in I} \phi_i H_i ({X})$ with $H_i({X})$ the horizontal lift of $X|_{p(V_i)} $ with respect to $H_i$ on $V_i$.
If a function $\kappa $ satisfying the first condition exists, then the horizontal lift $ \tilde{X}$ of every vector field $X \in {\mathfrak X}(L) $ satisfies $ \tilde{X}[\kappa]=0$, so that its flow preserves $ \{\kappa \leq 1\}$. Since $p: \{\kappa \leq 1\} \to L$ is a proper map, this proves that the flow of $X$ is defined if and only if the flow of $ \tilde{X}$ is defined.
Assume now that a function $\kappa $ satisfying the second condition exists. This implies that $ \{\kappa =1\}$ is a submanifold of $ M_L$. Using partitions of unity, we can construct a vertical vector field $u \in {\mathcal F}$ such that $ u[\kappa]=1$ at any point of $ \{\kappa =1\}$. Now, let $H $ be an Ehresmann $ {\mathcal F}$-connection on $p \colon M_L \to L$. By assigning to a vector field $X \in \mathfrak X (L) $ the vector field $H(X) - H(X)[\kappa] \, u $, we define a second Ehresmann ${\mathcal F} $-connection $ H_\kappa$ whose sections are tangent to the submanifold $\{ \kappa = 1 \} $, and therefore preserves the subset $\{ \kappa \leq 1 \} \subset M_L$. Since $p$ is a proper map on that subset, the flow of $X$ is defined if and only if the flow of $ H_\kappa (X) $ is defined, hence the claim.
\end{proof}
\begin{example}
\normalfont
\label{ex:Tsphere}
Let $M=TS^n$ be the tangent space of the sphere with its standard metric, and let $ \kappa : TS^n \to {\mathbb R}$ be the square of the norm.
Let $ {\mathcal F}$ be the singular foliation on $M=TS^n$ of all vector fields $X$ that satisfy $ X[\kappa]=0$. The leaves of this singular foliations are the sets $ \kappa^{-1}(\lambda)$ for $ \lambda \in {\mathbb R}_+$. Moreover, $\kappa^{-1}(0)$, \emph{i.e}.~the zero section, is the unique singular leaf. The function $ \kappa$ obviously satisfies the first condition in Proposition \ref{prop:existeEhresmann}, so that complete Ehresmann ${\mathcal F}$-connections exist. Indeed, the horizontal distribution associated to the Levi-Civita connection is a complete Ehresmann $ {\mathcal F}$-connection.
\end{example}
\begin{example}
\normalfont
Let $\mathcal F$ be the singular foliation on $M=TS^n$ of all vector fields tangent to the zero section $S^n\subset TS^n$. The zero section is a leaf, and the function $\kappa$ defined in Example \ref{ex:Tsphere} now satisfies the second condition of Proposition \ref{prop:existeEhresmann}. Hence a complete Ehresmann $ {\mathcal F}$-connection exists. Indeed, the horizontal distribution associated to any linear connection is a complete Ehresmann $ {\mathcal F}$-connection.
\end{example}
Consider a complete Ehresmann ${\mathcal F}$-connection $(M_L,p,H)$ near a locally closed leaf $L$. The fiber of $p $ over $\ell $ is an $ {\mathcal F}$-transversal to $ {\mathcal F}$ at $\ell$. The restricted singular foliation $ {\mathcal F}|_{p^{-1} \ell}$ shall be denoted $ {\mathcal T}_\ell$.
\begin{proposition}\label{prop:loctriv}
Consider a complete Ehresmann ${\mathcal F}$-connection $(M_L,p,H)$ near a locally closed leaf $L$. Every $ \ell \in L$ admits a neighborhood $U \subset L$ and a local diffeomorphism $ p^{-1}(U) \simeq U \times p^{-1}(L) $ intertwining $ {\mathcal F}$ with the direct product foliation $ {\mathfrak X}(U) \times {\mathcal T}_\ell$.
\end{proposition}
\begin{proof}
It suffices to prove the following result: ``For $L \simeq J^n$ with $ J = ]-1,1[$, if a complete Ehresmann ${\mathcal F}$-connection $(M_L,p,H)$ over $L$ exists, then a flat\footnote{A flat Ehresmann connection is a horizontal distribution whose sections are closed under the Lie bracket of vector fields.} complete Ehresmann ${\mathcal F}$-connection exists''.
We prove this statement by induction on $n$. For $n=1$, since every Ehresmann connection is flat, the result is straightforward.
Assume that it holds true for some $n \in {\mathbb N}$.
Consider a complete Ehresmann ${\mathcal F}$-connection $(M_L,p,H)$ over $L= J^{n+1} $.
Let $t \in J$ stand for the variable in the last copy of $J$. The horizontal lift (with respect to $H$) of the vector field $\tfrac{\partial}{\partial t}$ being complete by assumption, its flow defines a family of diffeomorphisms $ \Psi_t \colon p^{-1}(J^n \times \{ s \} ) \to p^{-1}(J^n \times \{s+t\})$ for all $t,s \in J$ such that $ s+t \in J$. Being the flow of an element in $ {\mathcal F}$, $\Psi_t $ preserves $ {\mathcal F}$, so that:
\begin{equation}
\label{eq:pro:Phi}
(\Psi_t )_* \, \colon \, {\mathcal F}|_{p^{-1}(J^n\times \{0\})} \simeq {\mathcal F}|_{p^{-1}(J^{n}\times \{t\})}.
\end{equation}
By recursion hypothesis, there exists a flat complete Ehresmann ${\mathcal F}$-connection $(M_L,p,H')$ on the restriction
$$ p^{-1}( J^n \times \{0\}) \to J^n \times \{0\} .$$
We use $ \Psi_t$ to transport this flat complete Ehresmann connection $\tilde{H}' $ over
$$ p^{-1}( J^n \times \{t\}) \to J^n \times \{t\} . $$
According to Equation \eqref{eq:pro:Phi}, $\tilde{H}' $ is a flat Ehresmann ${\mathcal F}$-connection for all $t \in J$. Consider the distribution
\begin{equation}\label{eq:final_dist} H'':= \langle H(\tfrac{\partial}{\partial t})\rangle \oplus \tilde{H}' .\end{equation}
The distribution $ H''$
\begin{enumerate}
\item is of rank $ n+1$ and is horizontal with respect to $ p$,
\item is flat because $\tilde{H}' $ is flat over $ p^{-1}(J^n \times \{t\})$ for all $t \in J$ and is by construction invariant under the flow of $ H(\tfrac{\partial}{\partial t})$,
\item has sections in $ {\mathcal F}$ because
$H(\tfrac{\partial}{\partial t}) \in {\mathcal F}$, and because sections of $\tilde{H}' $ are in ${\mathcal F}$,
\item is complete because $H(\tfrac{\partial}{\partial t})$ is a complete vector field and $ \tilde{H}'$ is a complete flat Ehresmann connection on $ p^{-1}(J^n \times \{t\})$ for all $t \in J$. Since complete vector fields for a flat connection are a $ \mathcal C^\infty_L$-module, this is enough to guarantee the completeness of $H'' $.
\end{enumerate}
This completes the proof.
\end{proof}
\begin{remark}
\label{rmk:Pedro}
\normalfont
When $\mathcal F$ is a Debord foliation, hence comes from a Lie algebroid with an anchor map injective on an open subset, then Theorem C in \cite{MR3897481} gives an alternative proof of Proposition \ref{prop:loctriv}.
\end{remark}
\subsubsection{Ehresmann $ {\mathcal F}$-connections and NQ-manifolds}
Since most singular foliations that appear in a given concrete problem are locally real analytic, and since locally real analytic singular foliations do admit geometric resolution over a locally compact open subset in view of Theorem 2.4 in \cite{LLS}, the following theorem justifies the assumption that we always make about the existence of a universal NQ-manifold for $ \mathcal F$ on $M_L$.
\begin{theorem}\label{thm:onetransverseisenough}
Consider a complete Ehresmann ${\mathcal F}$-connection $(M_L,p,H)$ near a compact leaf $L$. The following statements are equivalent:
\begin{enumerate}
\item[(i)] There exists an $ \ell \in L$ such that $ {\mathcal T}_\ell$ admits a geometric resolution.
\item[(ii)] The restriction of ${\mathcal F}$ to $M_L$ admits a universal NQ-manifold.
\end{enumerate}
In particular, if $ {\mathcal T}_\ell$ is a locally real analytic singular foliation, and $ p^{-1}(\ell)$ is relatively compact in $M$, then (ii) holds.
\end{theorem}
Of course, $ {\mathcal T}_\ell$ is a locally real analytic singular foliation
if ${\mathcal F}$ is locally real analytic singular foliation and $p$ is, around each point, real analytic in some local coordinates (that do not need to patch in a real analytic manner).
\begin{proof}
It is obvious that (ii) implies (i). Conversely, Proposition \ref{prop:loctriv} implies that for any $\ell, \ell'\in L$, $\mathcal T_\ell$ and $\mathcal{ T}_{\ell'}$ are isomorphic and that any $\ell'\in L$ has a neighbourhood $U_{\ell'}$ such that $p^{-1}(U_{\ell'})\subset M_L$ admits a geometric resolution. By compactness of $L$, $M_L$ can be covered by finitely many such $p^{-1}(U_{\ell'})$. The statement now follows from Section 3.2.2 in \cite{LLS}.
\end{proof}
\begin{proposition}
\label{prop:compatEhresmann}
Consider an Ehresmann ${\mathcal F}$-connection $(M_L,p,H)$ near a locally closed leaf $L$. Assume that $ {\mathcal F}$ is the image through the anchor $ \rho_{\mathbb U}: E_{-1} \to T M$ of an NQ-manifold $ {\mathbb U}$ on $M_L$.
\begin{enumerate}
\item The composition $Tp \circ \rho_{\mathbb U} \colon E_{-1} \to TL $ is an NQ-manifold fibration $ P \colon \mathbb U \to TL$ over the tangent Lie algebroid $TL$.
\item There exists Ehresmann connections $ H_{\mathbb U} \subset E_{-1}$ for $P \colon {\mathbb U} \to TL$ such that $ \rho_{\mathbb U} ( H_{\mathbb U} ) = H $.
\item Moreover, such an Ehresmann connection $ H_{\mathbb U}$ is complete for $P $ if and only if $(M_L,p,H)$ is a complete Ehresmann $ {\mathcal F}$-connection.
\end{enumerate}
\end{proposition}
\begin{proof}
Since the anchor map $ \rho_{\mathbb U}:{\mathbb U} \to {\mathfrak X}(M_L)$ is a Lie $\infty$-algebroid morphism and since the projection $Tp: {\mathfrak X}(M_L) \to {\mathfrak X}(L)$ is a Lie algebroid morphism, their composition is a Lie $\infty$-algebroid morphism.
Since $ p \colon M_L \to L$ is a surjective submersion and since $Tp \colon H \to TL $ is bijective and $ H_m \subset \rho_{{\mathbb U}}(E_{-1}|_m) $, this Lie $\infty $-algebroid morphism is a surjective submersion. This proves the first item.
To prove the second item, it suffices to construct a vector bundle morphism $ \sigma: H \to E_{-1} $ such that $ \rho \circ \sigma = {\mathrm {id}}_{H} $: Its image $ H_{\mathbb U} = \sigma(H) \subset E_{-1}$ satisfies then the required conditions.
Let us first construct such a section locally. Let $d$ be the dimension of $L$. Since $H$ is an Ehresmann $ {\mathcal F}$-connection, there exists for every $m \in M_L $ vector fields $X_1, \dots,X_d \in {\mathcal F}$ that generate $H$ on every point in a neighborhood $ U_m \subset M_L$. Let $ e_1, \dots, e_d \in \Gamma_{U_m}(E_{-1})$ be sections such that $ \rho(e_i) = X_i$ for all $i=1, \dots, d$. The vector bundle morphism:
$\sigma_{U} \colon H \to E_{-1}$ mapping $X_i|_{m'} $ to $ e_i|_{m'}$ for all $m' \in U$ satisfies $\rho \circ \sigma_U ={\mathrm{id}}_H $ on $U$ by construction.
Let $ (U_i, \phi_i)_{i \in I}$ be a partition of unity of $M$ such that each open subset $ U_i$ comes equipped with a section $\sigma_{i} \colon H \to E_{-1}$ as above. The vector bundle morphism:
$$ \sigma := \sum_{i \in I} \phi_i \sigma_i $$
satisfies $ \rho_{\mathbb U} \circ \sigma = \sum_{i \in I} \phi_i \, \rho_{\mathbb U} \circ \sigma_i = \sum_{i \in I} \phi_i \, {\mathrm{id}}_H = {\mathrm{id}}_H$. This proves the second item.
The third item is an immediate consequence of Definition \ref{def:Ehresmann}.
\end{proof}
\subsection{The higher holonomies of a singular leaf}
\label{sec:LESsing}
Throughout this section, $L$ is a locally closed leaf in is a singular foliation $ {\mathcal F}$ on a manifold~$M$.
\subsubsection{Main theorem and definition}
The main result of this section is the following theorem.
\begin{theorem}
\label{theo:holonomies}
Let $ {\mathcal F}$ be a singular foliation that permits geometric resolutions. Let $L$ be a locally closed leaf that admits a complete Ehresmann ${\mathcal F}$-connection $(M_L,p,H)$.
For every $\ell \in L$,
there exist group morphisms
\begin{equation}
\label{eq:def_Holonomy}
Hol : \pi_n (L ,\ell ) \to \Gamma \left( \pi_{n-1}( {\mathcal T}_{\ell}) \right) \end{equation}
such that for all $m\in p^{-1}(\ell)$ the sequence
\begin{align}\label{seq:leaf}
\dots \stackrel{Hol|_{m}}{\longrightarrow} \pi_n( {{\mathcal T}}_\ell,m ) \stackrel{i}{\to} \pi_n({\mathcal F}|_{M_L} , m) \stackrel{P}{\to} \pi_n(L,\ell) \stackrel{Hol|_{m}}{\longrightarrow} \pi_{n-1}( {{\mathcal T}}_\ell,m ) \to \dots
\end{align}
is exact.
\end{theorem}
\begin{proof}
Let $ \mathbb U^\mathcal F$ be a universal NQ-manifold of $\mathcal F $.
According to the first item of Proposition \ref{prop:compatEhresmann}, the morphism of NQ-manifold $ \mathbb U^\mathcal F \to TL$ given by $P := Tp \circ \rho \colon E_{-1} \to TL$ is a fibration of NQ-manifolds.
According to the second item in Proposition \ref{prop:compatEhresmann}, $P$ admits an Ehresmann connection. According to its third item, the latter
Ehresmann connection is complete.
We can therefore apply Theorem \ref{thm:snake}.
Since the base Lie algebroid is the tangent Lie alegbroid $TL \to L$, its homotopy groups and the usual homotopy groups of the manifold $L$, see Example \ref{ex:manifold}. Also, the kernel of $ P$ over a given point $ \ell \in L$ is the universal NQ-manifold of the transverse singular foliation $ {\mathcal T}_\ell$ on $p^{-1}(\ell) $ by Lemma \ref{lem:transversefol}. Its homotopy groups are therefore the homotopy groups of the transverse foliation $ {\mathcal T}_\ell$ on $ p^{-1}(\{l\})$. Theorem \ref{thm:snake} yields a family of group morphisms as in Equation \eqref{eq:def_Holonomy} that make the sequence \eqref{seq:leaf} exact.
\end{proof}
\begin{definition} \label{def:holonomies}
We call \emph{$n$-th holonomy} of the singular foliation $\mathcal F $ near the leaf $L$ the group morphism \eqref{eq:def_Holonomy}:
\begin{equation} \label{eq:FolHolonomy}
Hol \colon \left\{ \begin{array}{rcll} \pi_n(L,\ell) &\to & \Gamma( \pi_{n-1}(\mathcal T_\ell)) & \hbox{ for $n \geq 3$} \\
\pi_2(L,\ell) & \to & \mathrm{Center}(\Gamma(I(\mathbf T_\ell)) & \\
\pi_1(L,\ell) & \to & {\rm Diff}( p^{-1}(\ell) / \mathcal T_\ell ) & \\ \end{array}\right.
\end{equation}
We call \emph{higher holonomies} the family of these group morphisms and \emph{higher holonomy long exact sequence} the long exact sequence \eqref{seq:leaf}.
\end{definition}
Above, we used the identification $ \Gamma(\pi_1(\mathcal T_\ell)) \simeq \Gamma(I(\mathbf T_\ell)) $ with $ \mathbf T_\ell$ the universal holonomy groupoid of $\mathcal T_\ell $ seen in Proposition 2.18, and the identification $ \Gamma(\pi_0(\mathcal T_\ell)) \simeq {\rm Diff}( p^{-1}(\ell) / \mathcal T_\ell ) $
of Definition \ref{def:pi0}.
\begin{remark}
\normalfont
Although the context is slightly different, the first holonomy $$Hol:\pi_1(L)\to \mathrm{Diff}(p^{-1}(\ell)/\mathcal T_\ell)$$ matches the holonomy map constructed by Dazord in \cite{Dazord}. Instead of complete Ehresmann connection, \cite{Dazord} assumes saturated neighbourhoods (called stable in his paper), but a line-by-line comparaison shows that both morphisms will agree for leaves that admit a neighborhood satisfying both conditions.
\end{remark}
Let $\tilde L$ be a leaf of $ \mathcal F$ through some point $m \in p^{-1}(\ell)$ for $ \mathcal F |_{M_L}$ and $ \mathcal T_\ell$ respectively.
The restricted map $ p|_{\tilde L} \colon \tilde{L} \to L $ is a surjective submersion, and any complete Ehresmann connection on $p \colon M_L \to L $ restricts to an Ehresmann connection on $p|_{\tilde L} $. By a classical theorem for fibrations (cf. eg. \cite{MR1867354}), there exists a long exact sequence:
\begin{equation}\label{eq:les-topology} \dots \to \pi_n( L , \ell) \to \pi_{n-1} ( \tilde{S}, m ) \to \pi_{n-1} ( \tilde{L}, m ) \to \cdots \end{equation}
with $\tilde S = p^{-1}(\ell) \cap \tilde{L}$ the fiber of $p|_{\tilde L} \colon \tilde L \to L $.
\begin{proposition}
\label{prop:regular_leaves}
For $m$ a regular point, the higher holonomy long exact sequence \eqref{seq:leaf} coincides with the long exact sequence \eqref{eq:les-topology} associated to the fibration $p |_{\tilde L} \colon \tilde L \to L$.
\end{proposition}
\begin{proof}
Since both the leaf $\tilde L$ and the connected component of $m$ in $ \tilde S$ are regular leaves of $ \mathcal F$ and $\mathcal T_\ell$,
Example \ref{ex:regularleaf} implies
$$ \pi_n(\mathcal F,m)=\pi_n ( \tilde L , m ) \hbox{ and } \pi_n(\mathcal F,m)=\pi_n (\tilde S,m ) .$$
In this case, our construction thus reduces to the long exact sequence (\ref{eq:les-topology}).
\end{proof}
\subsubsection{The higher holonomies and Androulidalis-Skandalis holonomy groupoid.}
\label{sec:LinkWithHolonomyGroupoid}
We show that the $n$-th holonomy of the singular foliation $\mathcal F $ near the leaf $L$ ``projects'' to a similar group morphism associated to Androulidakis-Skandalis' holonomy Lie groupoid.
Let $ {\mathcal F}$ be a singular foliation that permits geometric resolutions, and $L$ be a locally closed leaf that admits a complete Ehresmann ${\mathcal F}$-connection $(M_L,p,H)$.
Throughout this section, we denote by $\mathbf F$ the universal holonomy groupoid of $\mathbf F $ and by $\mathbf T_\ell $ the universal holonomy
groupoid of the transverse singular foliation $ \mathcal T_\ell$ on the the fiber $ p^{-1}(\ell)$.
We choose $m \in p^{-1}(L)$ and we denote by $ \tilde L$ and $\tilde S $ leaves through $m$ of $\mathcal F $ and $\mathcal T_\ell $.
Let us apply Theorem \ref{thm:snake} (or Theorem 1.4 in \cite{Zhu}) and Example \ref{ex:LieAlgebroid} to the surjective submersion of Lie algebroids,
$$ P := p \circ \rho : A_{\tilde L} \mapsto T L . $$
The fiber over $ \ell$ is easily identified with the holonomy Lie algebroid $ A_{\tilde S}$ of the leaf $ \tilde{\mathcal S}$. Also, the Ehresmann $ \mathcal F$-connection on $p: M_L \to L $ induces an Ehresmann connection on $P$ as in Proposition \ref{prop:compatEhresmann}.
By Proposition 2.2. in \cite{MR3119886}, $ \mathbf T_\ell|_m$ and $\mathbf F|_m $ are smooth manifolds for all $m \in M$.
We obtain therefore for all $m \in M$ an exact sequence of the form:
\begin{equation} \label{exact:Gr}
\xymatrix{ \ar[r]\pi_n(\mathbf T_\ell|_m ,m ) &\ar[r]
\pi_n(\mathbf F|_m , m) & \pi_n(L,\ell) \ar[dll]_{Hol_{AS}|_m} \\
\pi_{n-1}(\mathbf T_\ell|_m,m )&\cdots & \pi_2(L,\ell) \ar[dll]_{Hol_{AS}|_m} \\ { I_m} (\mathbf T_\ell ) \ar[r] & I_m(\mathbf F )\ar[r] & \pi_1(L,\ell) \ar[dll]_{Hol_{AS}|_m} \\ p^{-1}(\ell)/\mathcal T_\ell & & } \end{equation}
The connecting maps $ Hol_{AS}|_m$, altogether, form group morphisms:
\begin{equation} \label{eq:GrHolonomy}
Hol_{AS} \colon \left\{ \begin{array}{rcll} \pi_n(L,\ell) &\to & \Gamma( \pi_{n-1}(\mathbf T_\ell)) & \hbox{ for $n \geq 3$} \\
\pi_2(L,\ell) & \to & \mathrm{Center}(\Gamma(I(\mathbf T_\ell))) & \\
\pi_1(L,\ell) & \to & {\rm Diff}( p^{-1}(\ell) / \mathcal T_\ell ) & \\ \end{array}\right.
\end{equation}
Above, $ \Gamma(I(\mathbf T_\ell)) $ stands for sections of the source $ s: \mathbf T_\ell \to p^{-1}(L)$ which are for all $ m \in M$ valued in the isotropy group at $m$ of the groupoid $\mathbf T_\ell $, a group that we denote by $ I_m(\mathbf T_\ell)$. Also, $ I_m(\mathbf F)$ stands for the isotropy group of $ \mathbf F$ at $m$.
\begin{definition}
We call $n$-th \emph{AS holonomy of $L$} the group morphism \eqref{eq:GrHolonomy}
and \emph{AS exact sequence near $L$} the exact sequence \eqref{exact:Gr}.
\end{definition}
We relate these connecting maps to the higher holonomies of the leaves in the next proposition.
\begin{proposition}\label{prop:ASrelation}
Let $ {\mathcal F}$ be a singular foliation that permits geometric resolutions. Let $L$ be a locally closed leaf that admits a complete Ehresmann ${\mathcal F}$-connection $(M_L,p,H)$.
For every $\ell \in L$ and $n \in {\mathbb N}^*$, the $n$-th holonomy and the $n$-th AS holonomy of the leaf $L$ are equal for $n=1,2$ and for $n \geq 3$ are related by:
\begin{align}
\label{eq:factorize0}
\xymatrix@R-0.8pc{
&& \Gamma(\pi_{n-1}(\mathcal T_\ell )) \ar@{-->}[dd]\\
\pi_n ( L , \ell ) \ar[rrd]_{Hol_{AS}}\ar@{->}[rru]^{Hol}
& \\ & & \Gamma(\pi_{n-1}( \mathbf T_\ell )) }
\end{align}
and for all $m \in p^{-1}(\ell)$, we have natural morphisms of long exact sequences:
\begin{align}\label{eq:algsnake0}
\xymatrix@R-0.8pc{
& \pi_n( {\mathcal T}_\ell,m ) \ar[r]^{i} \ar@{-->}[dd]
& \pi_n({\mathcal F}|_{M_L} , m) \ar[rd]^{P} \ar@{-->}[dd]
&
& \pi_{n-1}( {\mathcal T}_\ell|_m,m )\ar[r] \ar@{-->}[dd]
&\dots
\\ \ar[rd]_{Hol_{AS}|_m}
\dots \ar[ru]^{Hol|_{m}}
&
&
&\ar[rd]_{Hol_{AS}|_m} \pi_n(L,\ell) \ar[ru]^{Hol|_{m}}
&
&\\
&\pi_n(\mathbf T_\ell|_m ,m ) \ar[r]^{i}
&\pi_n(\mathbf F|_m , m) \ar[ru]_{P_A}
&
&\pi_{n-1}(\mathbf T_\ell|_m,m )\ar[r]
&\dots}
\end{align}
where all vertical lines above are the natural morphisms $q$ of Proposition \ref{prop:computingISotropies}, the upper line is the higher holonomy long exact sequence \eqref{eq:def_Holonomy} of $ \mathcal F$ near $L$ and the lower one is the AS long exact sequence \eqref{exact:Gr} near $L$.
\end{proposition}
\begin{proof}
Let $\tilde L$ be the leaf through $m \in p^{-1}(\ell)$.
For all $ \sigma \colon S^n \to L$,
consider the horizontal lift $ \mathcal L^H (\sigma, \underline{m}) \colon I \times S^n \to \mathbb U^\mathcal F$ in Fundamental Lemma \ref{lem:lifts}. Its image through $ Q\colon \mathbb U^\mathcal F \to A_{\tilde L}$ is the horizontal lift for $A_{\tilde L} \to TL $.
This implies the commutativity of the diagrams \eqref{eq:factorize0} and \eqref{eq:algsnake0}.
\end{proof}
\begin{remark}
\normalfont
The restriction of $ \mathcal F$ to the open subset $M_{reg} \subset M_L $ of regular points is a regular foliation. Its holonomy groupoid (more precisely, its universal cover) is the restriction to $ M_{reg}$ of the universal holonomy groupoid $ \mathbf F$. This gives an alternative proof of Proposition \ref{prop:regular_leaves} as an immediate consequence and Proposition \ref{prop:ASrelation} and Example \ref{ex:regularleaf}.
\end{remark}
\begin{remark}
\normalfont
Iakovos Androulidakis and Marco Zambon \cite{AZ13,AZ2} have also defined a holonomy morphism for a leaf of a singular foliation, which goes from the holonomy groupoid to some quotients of jets of vector fields on the transverse singular foliation. It is not a higher holonomy in our sense. However, it can be related to our construction as follows: Any element of the holonomy groupoid comes from a path in $A_L$ from $\ell$ to $\ell'$. Using the Ehresmann $\mathcal F$-connection, for any $m\in p^{-1}(\ell)$ this path can be lifted to a path ending in $\phi(m)\in p^{-1}(\ell')$. The class of the map $\phi$ modulo $\mathrm{exp(I_\ell\mathcal T_\ell)}$ is the holonomy of \cite{AZ13,AZ2}. While finishing this article, we learnt that Alfonso Garmendia and Joel Villatoro are going further in this direction (see \cite{GarVilt}).
\end{remark}
\subsubsection{The higher holonomies for singular foliations coming from a Lie algebroid.}
\label{sec:IfLieAlgebroid}
We show that if there exists a Lie algebroid defining $ \mathcal F$, then the $n$-th holonomy is obtained by projection from a group morphism associated to this algebroid, introduced by Olivier Brahic and Chenchang Zhu in \cite{Zhu}, that we call the $n$-th BZ holonomy of $ \mathcal F$ near~$L$.
Throughout this section, we consider a Lie algebroid $(A,[ \cdot, \cdot ]_A, \rho_A)$ over $M$. We denote by $\mathbf A $ its fundamental groupoid (also referred to as the Weinstein groupoid). Then $ {\mathcal F} = \rho_A(\Gamma(A))$ is a singular foliation.
Let us adapt the main results of \cite{Zhu} in our context: Assume that a locally closed leaf $ L$ comes equipped with a complete Ehresmann connection $ (M_L,p,H)$.
By construction, $ P_A:= p \circ \rho_A \colon A \to TL$ is a surjective submersion and a Lie algebroid morphism. Its kernel over a point $\ell \in L$ is the sub-Lie algebroid $( K , [\cdot, \cdot]_A ,\rho_ A )$ with $ K|_m = {\mathrm{Ker}}(T_m p \circ \rho_A|_m) $.
We call $\mathbf K $ its fundamental Lie groupoid.
Theorem \ref{thm:snake} or Theorem 1.4 in \cite{Zhu}, together with Example \ref{ex:LieAlgebroid}, can then be applied to yield a family of group morphisms:
\begin{align}
\label{eq:AHolonomy}
\delta_{BZ} \colon \left\{
\begin{array}{rcll} \pi_n(L,\ell) &\to & \Gamma( \pi_{n-1}(\mathbf K_\ell)) & \hbox{ for $n \geq 3$} \\
\pi_2(L,\ell) & \to & \mathrm{Center}(\Gamma(I(\mathbf K_\ell))) & \\
\pi_1(L,\ell) & \to & \mathrm{Diff}( p^{-1}(\ell) / \mathcal T_\ell ) & \\ \end{array}\right.
\end{align}
making for all $ m \in p^{-1}(\ell)$ the following long sequence exact:
\begin{align}
\label{exact:BZ}
\xymatrix{ \ar[r]\pi_n(\mathbf K_\ell|_m ,m ) &\ar[r]
\pi_n(\mathbf A|_m , m) & \pi_n(L, \ell) \ar[dll]_{\delta_{BZ}|_m} \\
\pi_{n-1}(\mathbf K_\ell|_m,m )&\cdots & \pi_2(L,\ell) \ar[dll]_{\delta_{BZ}|_m} \\ { I_m} (\mathbf K_\ell ) \ar[r] & I_m(\mathbf A_\ell )\ar[r] & \pi_1(L,\ell) \ar[dll]_{\delta_{BZ}|_m} \\ p^{-1}(\ell)/\mathcal T_\ell & & }
\end{align}
\begin{definition}
Let $(A,[ \cdot, \cdot ]_A, \rho_A) $ be a Lie algebroid over $M$.
We call $n$-th \emph{BZ holonomy\footnote{BZ stands for Olivier Brahic and Chenchang Zhu.} of $A$ near $L$} the group morphism \eqref{eq:AHolonomy}
and \emph{BZ long exact sequence of $A$ near $L$} the long exact sequence \eqref{exact:BZ}.
\end{definition}
Corollary 3.12 in \cite{Zhu} describes the evaluation at $ \ell$ of the $BZ$-holonomy. Here is the relation between both family of holonomies:
\begin{proposition}
\label{prop:BZ}
Let $A$ be an integrable Lie algebroid with fundamental groupoid $\mathbf A $ and $ {\mathcal F} = \rho_A(\Gamma(A))$ be its induced singular foliation.
For every locally closed leaf $L$ that admits a complete Ehresmann ${\mathcal F}$-connection $(M_L,p,H)$, the first holonomy and the first BZ-holonomy coincide, and for $n \geq 2$ the $n$-th holonomy of $\mathcal F$ near $L$ factorizes through the $n$-th BZ-holonomy of $A$ near $L$:
\begin{align}
\label{eq:factorize}
\xymatrix@R-0.8pc{
&&\Gamma(\pi_{n-1}( \mathbf K_\ell ))\ar@{-->}[dd]\\
\pi_n ( L , \ell ) \ar[rru]^{\delta_{BZ}}\ar@{->}[rrd]_{Hol}
& \\ & &\Gamma(\pi_{n-1}(\mathcal T_\ell )) }
\xymatrix@R-0.8pc{
&&{\mathrm{Center}} \left(\Gamma(I(\mathbf K_\ell ))\right) \ar@{-->}[dd]\\
\pi_2 ( L , \ell ) \ar[rru]^{\delta_{BZ}}\ar@{->}[rrd]_{Hol}
& \\ & &{\mathrm{Center}} \left( \Gamma(\pi_1(\mathcal T_\ell )) \right)}
\end{align}
For all $m \in p^{-1}(\ell)$, moreover, we have natural morphisms of long exact sequences:
\begin{align}\label{eq:algsnake}
\xymatrix@R-0.8pc{
&\pi_n(\mathbf K_\ell|_m ,m ) \ar[r]^{i}\ar@{-->}[dd]
&\pi_n(\mathbf A|_m , m) \ar@{-->}[dd] \ar[rd]^{P_A}
&
&\pi_{n-1}(\mathbf K_\ell|_m,m )\ar[r]\ar@{-->}[dd]
&\dots\\ \ar[ru]^{\delta_{BZ}|_m}
\dots \ar[rd]_{Hol|_{m}}
&
&
&\ar[ru]^{\delta_{BZ}|_m} \pi_n(L,\ell) \ar[rd]_{Hol|_{m}}
&
&\\
& \pi_n( {\mathcal T}_\ell,m ) \ar[r]^{i}
& \pi_n({\mathcal F}|_{M_L} , m) \ar[ru]_{P}
&
& \pi_{n-1}( {\mathcal T}_\ell,m )\ar[r]
&\dots}
\end{align}
where the upper line is the BZ long exact sequence \eqref{exact:BZ} of $A$ near $L$ and the lower one is the higher holonomy long exact sequence \eqref{eq:def_Holonomy} of $ \mathcal F$ near $L$.
\end{proposition}
\begin{proof}
By universality Theorem \ref{theo:onlyOne}, there exists a (unique up to homotopy) NQ-morphism
$\Phi\colon A \to \mathbb U^\mathcal F $ and a unique NQ-morphism
$ T^A_\ell \to \mathbb U^{{\mathcal T}_\ell} $ (which can chosen to be the restriction of $ \Phi$ to $ p^{-1}(\ell)$). If $L(m,\sigma)$ is the lift in the Fundamental Lemma \ref{lem:lifts} applied to a base map $ \sigma : S^n \to L$ and an initial shape $e = \underline{m}$ for the fibration $P_A \colon A \to TL $, then $ \Phi \circ L(e,m) $ is the lift for the same base map $ \sigma $ and the same initial shape $ \underline{m}$ for the NQ-manifold fibration $ P : \mathbb U^{\mathcal F} \to TL$. This implies that $ \Phi$ and $ \Phi|_{p^{-1}(\ell)}$ are NQ-morphisms whose induced vertical arrows of \eqref{eq:factorize} and \eqref{eq:algsnake} satisfy all required conditions.
\end{proof}
One of the most intriguing open question about singular foliations, that first appeared in \cite{AZ13}, is \emph{``Let $ {\mathcal F}$ be a finitely generated singular foliation, do we have $ {\mathcal F} = \rho_A( \Gamma(A))$ for some Lie algebroid $ (A,[ \cdot, \cdot ]_A, \rho_A)$? Is it at least true locally?''}. Let us recapitulate a few points about this problem (see Example 3.9--3.16 in \cite{LLS}):
\begin{enumerate}
\item For many singular foliations, no natural Lie algebroid seems to exist (\emph{e.g. }vector fields on $ \mathbb R^n$ vanishing at order two at zero, or tangent to a given affine variety).
\item For many singular foliations, such a Lie algebroid is known to exist (\emph{e.g. }symplectic leaves of a Poisson structure, orbits of a Lie group action).
\item But even if such a Lie algebroid exists, it may not be unique.
\end{enumerate}
For singular foliations as in item 2, Proposition \ref{prop:BZ} can and will be quite useful to compute the holonomies. But item 1 warns us that it might not be always possible. Item 3 means that we should not hope that the BZ holonomies ``know more'' about the singular leaf than the holonomies that we constructed: they certainly ``know more'' in view of (\ref{eq:factorize}), but they may contain information which is not related to the singular foliation itself but only to the particular Lie algebroid defining it.
\subsection{Examples and particular cases}
\label{sec:exsing}
Let us specialize our construction of the higher holonomies to several types of singular foliations.
\vspace{0.2cm}
{\textbf{Projective singular foliations.}} We say that a singular foliation $\mathcal F $ is \emph{Debord} or \emph{projective} when $\mathcal F $ is projective as a module over $ \mathcal C^\infty_M$. By Serre-Swann theorem, it means that $ {\mathcal F}$ is the image of a Lie algebroid $A$ through a Lie algebroid whose anchor is injective on a dense open subset. Since the $\mathcal C^\infty_{M}$-module of sections of the Lie algebroid $T{\mathcal F}$ is isomorphic to the $\mathcal C^\infty_{M}$-module $ {\mathcal F}$, the Lie algebroid $A$ is a universal NQ-manifold of $ {\mathcal F}$.
The following is an obvious consequence of Proposition \ref{prop:BZ}:
\begin{corollary}
Let $ {\mathcal F}$ be a Debord singular foliation with associated Lie algebroid $A$, and $L$ a leaf that admits an Ehresmann connection on a neighborhood. The higher holonomies and the higher holonomy long exact sequence of ${\mathcal F} $ at $L$ coincide with the BZ holonomies and the BZ long exact sequence of the Lie algebroid $A$ at $L$.
\end{corollary}
Of course, regular foliations are an instance of this situation.
\begin{example}
\normalfont
For a regular foliation, only $ Hol_1$ is not zero, and coincides with the usual holonomy.
\end{example}
\vspace{0.2cm}
{\textbf{Singular foliations arising from a Lie algebroid of minimal rank.}} Let $L$ be a leaf of a singular foliation that admits an Ehresmann connection $ (M_L,p,H)$, and $A_L$ its holonomy Lie algebroid.
We say that the leaf $L $ admits a Lie algebroid of minimal rank if there exists a Lie algebroid $A$ on $M_L$ defining $\mathcal F$ such that $A|_L=A_L$.
Proposition \ref{prop:ASrelation} states that $Hol_{AS}$ nand $Hol$ agree on $ \pi_1(L,m)$ and $\pi_2(L,m) $ for any lef $L$ of a singular foliation that admits a complete Ehresmann connection.
For the higher holonomies, the following is an immediate consequence of Proposition \ref{prop:BZ}:
\begin{corollary}
\label{cor:minimalRank}
If $L$ admits a Lie algebroid of minimal rank $A$, then there exist for all $ n \geq 3$ group morphisms $\phi: \pi_{n-1}( \mathbf T_\ell|_m ,m )\to \pi_{n-1} ( \mathcal T_\ell , m)$ such that $Hol= \phi \circ Hol_{AS}$.
\end{corollary}
Here are a few instances of this situation:
\begin{example}
\label{ex:vectorbundle}
\normalfont
Let $p:E\to L$ be a vector bundle of rank $k$ over a manifold $L$ that we identify with the zero section. For simplicity, we assume $\pi_1(L)=0$. Let $\mathcal F$ the singular foliation of all vector fields on $E$ tangent to $L$. Every linear connection induces an Ehresmann connection for the zero section.
The leaf $L$ admits a Lie algebroid of minimal rank: the Lie algebroid $CDO(E)$ (see \cite{Mackenzie}), so that we are in the situation of Corollary \ref{cor:minimalRank}.
It suffices therefore to describe the AS holonomy.
\begin{enumerate}
\item Let us compute $Hol_{AS}|_m $ for all nonzero $ m \in p^{-1}(L)$.
The restriction of the universal holonomy groupoid to $ E \backslash L $ is the fundamental groupoid of $E \backslash L$. Since $L$ is simply-connected by assumption, the latter groupoid is the pair groupoid $(E \backslash L) \times ( E \backslash L ) \rightrightarrows (E \backslash L)$, and $Hol_{AS}|_m $ is the connecting map associated to the fibration:
$ p \colon E \backslash L \to L $.
Since the fiber is $ E_\ell \backslash \{0_\ell \}$. The latter being homotopic to a $(k-1)$-sphere, with $k$ the rank of $E$, we obtain a group morphism:
$$ Hol_{AS}|_m : \pi_n(L,\ell) \to \pi_{n-1}(S^{k-1},N) $$
\item Let us compute $Hol_{AS}|_\ell $ for $\ell\in L$. Since the Lie algebroid $CDO(E)$ integrates to the Lie groupoid made of all invertible linear maps between fibers of $E$, the universal cover of $s^{-1}(\ell) $ is the universal cover $\widetilde{{\mathrm{Fr}}(E)}$ of the frame bundle of $E$. Its fiber over $ \ell$ is the universal cover $ \widetilde{GL(E_\ell)}$ of $ GL(E_\ell)$. In this case, $Hol_{AS}|_\ell $ is the connecting map associated to the fibration:
$ p \colon \widetilde{{\mathrm{Fr}}(E)} \to L $.
Since the fibers are $ \widetilde{GL(E_\ell)}$, we obtain therefore group morphisms:
$$ Hol_{AS}|_\ell := \left\{\begin{array}{rcll} \pi_n(L,\ell)& \to& \pi_{n-1}( \widetilde{GL(E_\ell)},e)=\pi_{n-1}(GL(E_\ell),e ) & \hbox{ for $n \geq 3$ } \\
\pi_2(L,\ell) & \to & {\mathrm{Center}} \left( \widetilde{GL(E_\ell)}\right) = \mathbb Z/ 2 \mathbb Z & \end{array} \right. $$
A direct computation shows that the second map corresponds to the second class of Stiefel-Whitney.
\end{enumerate}
\end{example}
\begin{example}
\normalfont
When $L=S^k$ and $ E= TS^k$ with $ k \geq 2$, the higher holonomies of Example \ref{ex:vectorbundle} satisfy $Hol_{AS}|_m=0$ if and only if
\begin{enumerate}
\item[(i)] $m$ is not in the zero section of $TS^n$ and $k=3,7$.
\item[(ii)] $m$ is in the zero section of $TS^n$ and $k$ is odd.
\end{enumerate}
This comes from the well-konwn fact that the tangent space of spheres of odd dimension admits a nowhere vanishing section, while its frame bundles admits a section iff $k=3,7$.
\end{example}
\begin{example}
``Period map the Crainic-Fernandes.''
\normalfont
The second higher holonomy, evaluated at the point $\ell $, is a group morphism $\pi_2(L) \to {\mathrm{Center}} (I_\ell(\mathbf T_\ell)) $. Let us check that it coincides with Crainic-Fernandes' period map \cite{MR1973056} for the holonomy Lie algebroid $ A_L \to L$. The higher holonomies $ Hol|_\ell $ and $Hol_{AS}|_\ell $ agree on $\pi_2(L,\ell) $ in view of Proposition \ref{prop:ASrelation}.
Item 2 in Corollary 3.12 in \cite{Zhu} identifies $ Hol_{AS}|_\ell$ (which is associated to the Lie algebroid fibration $ A_L \to TL$) with the period map the Crainic-Fernandes.
This completes the proof of the claim.
We could also prove the claim using the equality of the short exact sequence \eqref{eq:computingISotropies3}
with the short exact sequence (3.6) in \cite{AZ13} and a similar claim made in Remark 4.3 in \cite{AZ13}. Last, since Debord \cite{MR3119886} proved longitudinal smoothness -using a totally independent argument- the period map the Crainic-Fernandes has a discrete image.
\end{example}
\section*{Introduction}
The holonomy of a leaf $L$ of a regular foliation $ {\mathcal F}$ on a manifold $M$ is defined \cite{MR47386, MR0055692,MR189060} as a group morphism:
\begin{equation} \label{eq:holonomy} Hol : \pi_1(L,\ell) \mapsto {\mathrm{Diff}}_{\ell}(T_\ell), \end{equation}
where $\pi_1(L)$ is the fundamental group of $L$ and ${\mathrm{Diff}}_{\ell}(T_\ell)$ is the group of germs of diffeomorphisms of a transversal $T_\ell$ of $L$ at a point $\ell \in L$. Assuming the existence of Ehresmann connections on a tubular neighborhood of $L$, the group morphism \eqref{eq:holonomy} then takes values in the group of diffeomorphisms of some fixed transversal. The construction of $Hol$ consists in lifting paths on $ L$ with end points $\ell,\ell' $ to a diffeomorphism $T_\ell \simeq T_{\ell'}$, then in showing that homotopic paths induce the same diffeomorphism. \\
Consider now a singular foliation \cite{MR0149402,Cerveau,D,AS09}, \emph{i.e.}~a ``locally finitely generated sub-module $ {\mathcal F}$ of the module of vector fields, stable under Lie bracket''.
By Hermann's theorem \cite{MR0142131}, singular foliations induce a partition of $M$ into submanifolds called leaves of the foliation. Unlike for regular foliations, the leaves may not be all of the same dimension.
Those leaves around which the dimension jumps are called \emph{singular leaves}. This leads us to the natural:
\begin{question}
\label{ques:main}
How to define an analogue of the holonomy \eqref{eq:holonomy} for a singular leaf of a singular foliation?
\end{question}
Dazord proposed in 1985 an answer, which consists of a group morphism from the fundamental group of the leaf to the bijections of the orbit space of the transversal. Recently, Androulidakis and Zambon \cite{AZ13,AZ2} used the holonomy Lie groupoid of a singular foliation \cite{AS09} as a replacement for the fundamental group for a singular leaf. Using the universal Lie $ \infty$-algebroid of a singular foliation discovered in \cite{LLS}, we propose a construction of ``higher holonomies'' taking into account the higher homotopy groups of the singular leaves. The first of these holonomies is the one defined in \cite{Dazord}. We are also able relate our construction to Andoulidakis and Zambon's holonomy, but our construction differs from them.
Recall that to most\footnote{More precisely, the universal Lie $\infty$-algebroid is shown in \cite{LLS} to exist for any singular foliation that admits a geometric resolution. This happens in particular for locally real analytic singular foliations, that is singular foliations that have, in a neighborhood of every point, generators which are real analytic in some local coordinates. This class is quite large.}
singular foliations ${\mathcal F}$ on $M$, Sylvain Lavau, Thomas Strobl and C.L. associated in \cite{LLS} a Lie $\infty$-algebroid ${\mathbb U}^{\mathcal F}$, constructed over resolutions of $ {\mathcal F}$ as a module over smooth functions on $M$. Although the construction of ${\mathbb U}^{\mathcal F}$ relies on several choices, it is unique up to homotopy.
In particular, its homotopy groups $\pi_n( {\mathbb U}^{\mathcal F},m) $,
at a point $m \in M$, \emph{i.e.}~homotopy classes of maps from $S^n$ to $ {\mathbb U}^{\mathcal F}$ mapping the north pole to $m$, depend only on the singular foliation $\mathcal F$. We can therefore denote these groups as $ \pi_n({\mathcal F},m)$, without any reference to the universal Lie $\infty$-algebroid ${\mathbb U}_{\mathcal F}$. Altogether, these groups form a bundle of groups over $M$ that we denote by $\pi_n({\mathcal F} )$. Last, we denote its group of (smooth) sections by $\Gamma(\pi_n({\mathcal F} ))$.
For $n=0$, $\Gamma(\pi_0({\mathcal F} ))$ consists in bijections of the orbit space $ M/\mathcal F$ that come from a leaf-preserving smooth diffeomorphism of $M$.
Now, let us choose a leaf $L$. Every submanifold $T$ transverse to $L$ comes equipped with a singular foliation, which does not depend on the choice af $T$, see \cite{Cerveau,Dazord,AS09}.
It is called the \emph{transverse foliation} and we denote it by $ {\mathcal T}_\ell$ , with $ \ell $ the intersection point $T \cap L$.
We claim that a reasonable answer to the question above is to define the higher holonomies as a sequence of group morphisms:
\begin{equation}\label{eq:intro:Hol}
Hol : \pi_n (L,\ell) \mapsto \Gamma(\pi_{n-1} ({\mathcal T}_\ell )),
\end{equation}
that we show in the present article to exist, provided that the leaf $L$ admits a complete Ehresmann connection. If no complete Ehresmann connection exists, then the constructions still make sense, but at the level of germs, as we shall see in a subsequent article.
We also justify the interest of the definition by showing that these morphisms belong to a long exact sequence:
\begin{equation}\label{eq:intro:long}
\dots \to \pi_{n+1}(L,\ell) \stackrel{Hol^{}}{ \longrightarrow } \pi_n({\mathcal T}_\ell , m) \to \pi_n({\mathcal F} , {m}) \to \pi_n(L,\ell) \stackrel{Hol^{}}{ \longrightarrow } \pi_{n-1}({\mathcal T}_\ell , m) \to \dots .
\end{equation}
for all $m \in p^{-1}(\ell)$.
Indeed, the existence of such a long exact sequence is a general phenomenon for NQ-manifold fibrations and is proven in this context, see Theorem \ref{thm:snake}. This extends a Theorem of Olivier Brahic and Chenchang Zhu \cite{Zhu} about Lie algebroids, which was an important inspiration for the present work. This construction is quite involved and is addressed in Section \ref{sec:nq}.
In Section \ref{sec:leaf}, we apply the results of Section \ref{sec:nq} to the universal Lie $\infty$-algebroid of a singular foliation: we first carefully justify that we are in a context where this construction is possible, and study the existence of an Ehresmann connection. Eventually, this allows us to construct the higher holonomies \eqref{eq:intro:Hol} and to verify the exact sequence \eqref{eq:intro:long}.
We finish this article by relating our higher holonomies $ Hol$ with several Lie algebroids and groupoids. On the one hand, our higher holonomies are universal in the following sense: assume that there exists a Lie groupoid $ \mathbf G$ whose leaves are the leaves of $ \mathcal F$.
Then, Brahic-Zhu's theorem implies that there exists for every leaf $L$, group morphisms:
$$ \delta_{BZ} \colon \pi_n(L,\ell) \mapsto \Sigma_{n-1} ( \mathbf G) $$
valued in a group $ \Sigma_{n-1}( \mathbf G)$ that shall be introduced later. There are canonical group morphisms $ \psi : \Sigma_{n-1}( \mathbf G) \to \Gamma(\pi_{n-1}(\mathcal T_\ell)) $ by universality of $\mathbb U^\mathcal F $. The higher holonomies $ Hol$ coincide with the composition $Hol = \psi \circ \delta_{BZ} $.
On the other hand, our higher holonomies are a refinement of a holonomy naturally associated to Androulidakis and Skandalis holonomy groupoid $\mathbf F $ of the singular foliation $ \mathcal F$, that we also construct
$$ Hol_{AS} : \pi_n(L,\ell) \to \Sigma_{n-1}(\mathbf F).$$
There are canonical group morphisms $ \phi : \Gamma(\pi_{n-1}(\mathcal T_\ell)) \to \Sigma_{n-1} (\mathbf F) $ (that correspond to $1$-truncation of the NQ-manifold $\mathbb U^\mathcal F $) and we show that $ Hol_{AS} = \phi \circ Hol $.
\begin{convention*}
For $\mathbf G$ a groupoid over a manifold $M$, and $ X,Y \subset M$, we use the following notations:
$$ \mathbf G|_X =s^{-1}(X) \hbox{ , } \mathbf G|^Y= t^{-1}(Y) \hbox{ and } \mathbf G|_X^Y =s^{-1}(X) \cap t^{-1}(Y). $$
\end{convention*}
\noindent {\textbf{Acknowledgment}} We acknowledge crucial discussions with Iakovos Androulidakis (for $ \mathcal F$-connections) and Pavol \v{S}evera (for fundamental groups of NQ-manifolds): We will mark them in the text in due place. We also thank the workshop and conference ``Singular Foliations'' in Paris Diderot and Leuven where the content of the article was presented, and in particular Sylvain Lavau, Thomas Strobl and Marco Zambon. L. R. was supported by the Ruhr University Research School PLUS, funded by Germany's Excellence Initiative [DFG GSC 98/3] and the PRIME programme of the
German Academic Exchange Service (DAAD) with funds from the German Federal Ministry of Education and Research (BMBF).
\section{Homotopy groups of NQ-manifolds}
\label{sec:nq}
\subsection{On NQ-manifolds and Lie $\infty$-algebroids}
\subsubsection{NQ-manifolds, morphisms and homotopies}
An \emph{NQ-manifold} structure over a manifold $M$ is a graded vector bundle $V= \oplus_{i=1}^d V_i $ concentrated in degrees ranging from $+1 $ to $d$ together with a derivation $Q$ of degree plus one and squaring to zero of its \emph{functions}, \emph{i.e.}~the graded commutative algebra of sections of $S(V)$.
\begin{example}
\normalfont
Any manifold $ \Sigma$ can be seen as an NQ-manifold: Its functions are the exterior forms $\Omega^\bullet_\Sigma $ equipped with the de Rham differential ${\mathrm d}_{dR}$.
\end{example}
A \emph{Lie $\infty$-algebroid structure} over a manifold $M$ is a graded vector bundle $(E_{-i})_{i=1}^d $ concentrated in degrees ranging from $-1$ to $-d$ whose sections come equipped with a graded symmetric Lie $\infty $-algebra structure $\{l_k\}_{k \geq 1} $. We assume that $l_k$ are $\mathcal C^\infty_M$-linear, except for $l_2 $ which satisfies:
$$ l_2(x,fy) \, = \, f \, l_2 (x,y) \, + \, \rho(x)[f] \, y \hbox{ $~\forall f \in \mathcal C^\infty_M $, $x \in E_{-1}$ and $ y \in \Gamma (E_{-i})$,} $$
for some vector bundle morphism $ \rho \colon E_{-1} \to TM $ called the \emph{anchor map}.
It is well-known \cite{Voronov} that NQ-manifolds and Lie $\infty$-algebroid structures over a given manifold $M$ are in one-to-one correspondence.
Under this correspondence, the underlying vector bundles of both structures are dual to one another: $E_{-i} = V_i^*$ for all $i =1, \dots, d$. Given an $NQ$-manifold (resp. a Lie $\infty$-algebroid), it makes therefore sense to refer to its \emph{dual} Lie $\infty$-algebroid (resp. $NQ$-manifold).
Given a Lie $\infty$-algebroid structure over $M$ as above, $ {\mathcal F} := \rho (\Gamma(E_{-1}))$ is easily shown to be a singular foliation, called the \emph{basic singular foliation}. Also, $l_1$ is $\mathcal C^\infty_M$-linear and
$$ \dots \stackrel{l_1}{\to} E_{-3} \stackrel{l_1}{\to} E_{-2} \stackrel{l_1}{\to} E_{-1} \stackrel{\rho}{\to} TM $$
is a complex of vector bundles, called the \emph{linear part} of $E_\bullet$.
We call basic singular foliation and linear parts of an NQ-manifold those of its dual Lie $\infty$-algebroid.
\begin{example}
\normalfont
For $ \Sigma $ a manifold, the dual of the NQ-manifold with functions $\Omega_\Sigma^\bullet$ and differential ${\rm d}_{dR}$ is the tangent Lie algebroid $T \Sigma $.
\end{example}
\begin{notation}
NQ-manifolds shall be denoted as ${\mathbb V} = (V_\bullet, M, Q) $ where $M$ is a manifold, $V_\bullet $ is the graded vector bundle over $M$ and $Q$ is the derivation of $\Gamma(S(V))$.
As an exception to this convention, the NQ-manifold associated to a manifold $\Sigma$ shall simply be denoted by $\Sigma$.
\end{notation}
Let ${\mathbb V} = (V_\bullet, M, Q) $ and ${\mathbb V}' = (V_\bullet', M', Q') $ be NQ-manifolds. An \emph{NQ-manifold morphism} $\Phi: {\mathbb V} \to {\mathbb V}'$ is by definition a graded commutative differential algebra morphism $\Phi^*:\Gamma(S(V') ) \to \Gamma(S(V))$.
The latter is entirely described by:
\begin{enumerate}
\item[$\bullet$] a smooth map $\phi: M \to M'$ called the \emph{base map} of $\Phi$;
\item[$\bullet$] a sequence $(\phi_n)_{n \geq 1}$ of degree zero vector bundle morphisms $\phi_n \colon S^n (E) \to E' $ over $\phi$ called the \emph{Taylor coefficients} of $ \Phi$.
\end{enumerate}
For instance, an NQ-manifold morphism $ I\times {\mathbb V} \to {\mathbb V}' $ is a differential graded algebra morphism\footnote{The symbol $\tilde{\otimes}$ stands for the following operation: For ${\mathbb V} = (V_\bullet, M, Q) $ and ${\mathbb V}' = (V_\bullet', M', Q') $ NQ-manifolds, we shall denote by $ \Gamma(S(V)) \tilde{\otimes} \Gamma(S(V')) $ sections over $ M \times M'$ of the symmetric algebra of the bundles $p_1^* V \oplus p_2^* V' $ with $p_1,p_2$ the natural projections. This algebra contains the tensor product $\Gamma(S(V)) \otimes \Gamma(S(V'))$, and defines an NQ-manifold structure that we denote by $ {\mathbb V} \times {\mathbb V}' $.}:
$$ \Phi \colon (\Gamma (S(V')),Q') \longrightarrow (\Gamma(S(V)) \tilde \otimes \Omega^\bullet_I,~~Q' + {\rm d}_{dR}).$$
For any $t \in I$, $ \Phi$ admits a restriction to an NQ-manifold morphism $\Phi_t\colon {\mathbb V} \hookrightarrow \{t\} \times {\mathbb V} \to {\mathbb V}' $.
The NQ-morphism $\Phi \colon I \times {\mathbb V} \to {\mathbb V}' $ is said to be \emph{constant} if the following diagram commutes:
$$ \xymatrix{ I\times {\mathbb V} \ar[d]^{\Phi} \ar[r]^{pr} & {\mathbb V} \ar[dl]^{\Phi_0} \\ {\mathbb V}' .& } $$
\begin{remark}
\normalfont
For a constant NQ-manifold morphism $ I\times {\mathbb V} \to {\mathbb V}' $, the restrictions $ \Phi_t$ do not depend on $t \in I$. However $\Phi_t=\Phi_0$ for all $t\in I$ does not imply that $\Phi$ is constant.
\end{remark}
\begin{definition}
A \emph{homotopy between two NQ-manifold morphisms} $\Phi_0, \Phi_1\colon \mathbb V \to {\mathbb V}'$ is an NQ-manifold morphism
$ I\times {\mathbb V} \to {\mathbb V}'$ whose restrictions to the end points of $I=[0,1]$ are $\Phi_0$ and $ \Phi_1 $ respectively.
\end{definition}
A homotopy $[0,1] \times {\mathbb V} \to {\mathbb V}' $ between NQ-manifold morphisms $\Phi_0$ and $ \Phi_1 $ constant after restriction to $[1-\epsilon,1]\times {\mathbb V} \to {\mathbb V}' $ and a homotopy $[0,1]\times {\mathbb V} \to {\mathbb V}' $ between NQ-manifold morphisms $\Phi_1$ and $ \Phi_2 $ constant after restriction to $[0, \eta] \times {\mathbb V} \to {\mathbb V}' $ can be concatenated so that the concatenation remains smooth.
Also, any homotopy can be transformed by a smooth rescaling of $I$ to a homotopy with this property.
\\
Consider a graded manifold with functions $\Gamma(S(V))$ over $M$. Every degree zero vector field $Y\in Der(\Gamma(S(V)))$ induces a derivation of $\mathcal C^\infty_M=\Gamma(S^0(V))$, \emph{i.e.~}a vector field $\underline{Y} $ on the base manifold $M$.
\begin{lemma}\label{lem:flow}
Let $Y$ be a degree zero vector field on an NQ-manifold $\mathbb V$ over a manifold $M$.
\begin{enumerate}
\item For all fixed $t \in \mathbb R$, the vector field $Y$ admits a time-$t$ flow $\Phi_t^Y\colon \Gamma(S(V)) \to \Gamma(S(V))$ in the sense of graded manifolds if and only if the induced vector field $\underline{Y} $ on $M$ admits a time-$t$ flow.
\item Assume $Y$ is closed, \emph{i.e.}~$[Y,Q]=0$. Then, for any admissible $t$, the flow $\Phi_t^Y:\mathbb V\to \mathbb V $ is an NQ-morphism.
\item Assume $Y$ is exact, \emph{i.e.~}there exists $X$ such that $[X,Q]=Y$. Then there exists an NQ-morphism
$$\Phi^{X} : \mathbb R\times \mathbb V \to \mathbb V, $$
maybe defined in a neighborhood of $\{0\} \times \mathbb V $ only, such that for all admissible $t\in \mathbb R$, the restriction $\Phi^X_t:\mathbb V\to \mathbb V$ is the flow of $[X,Q]$ at time $t$. In particular all $\Phi_t^Y$ are homotopic NQ-morphisms.
\end{enumerate}
\end{lemma}
\begin{proof}
For the two first points, see Chapter 5 in \cite{MR2102797} (that deals with super-manifolds, instead of graded manifolds, but the arguments can be repeated word by word).
Let us prove the third item. Consider the following map \footnote{As in Section 3.4.4 in \cite{LLS}, we implicitly consider elements of degree $k$ in $\Gamma(S(V)) \tilde{\otimes} \Omega_I $ as being elements of the form $ F_t \otimes 1 + G_t \otimes dt $ with $F_t,G_t \in \Gamma(S(V))$ being elements of degree $k$ and $k-1$ respectively that depend smoothly on a parameter $t \in I$.}:
$$ \begin{array}{rcl} \Gamma(S(V))&\to& \Gamma(S(V)) \, \tilde{\otimes} \, \Omega(I) \\ F & \mapsto & \Phi_t^Y(F) \otimes 1 + X \circ \Phi_t^Y (F)\otimes dt \end{array}$$
It is easily checked to be a graded algebra morphism.
Furthermore, the differential equation
\begin{align*}
\frac{\partial \Phi^Y_t }{\partial t}=Y\circ \Phi^Y_t
=[X,Q] \circ \Phi^Y_t=(X\circ \Phi^Y_t)\circ Q + Q\circ (X\circ \Phi^Y_t)
\end{align*}
implies that it intertwines $Q$ and $Q+\rm d_{dR}$. This completes the proof.
\end{proof}
\subsubsection{Definition of homotopy groups of NQ-manifolds}
We summarize in this section several ideas coming from \v{S}evera, see \cite{1707.00265} and \cite{SeveraIntegration}. \\
Let $ {\mathbb V} = (M,{\mathbb V}, Q)$ be an NQ-manifold. Let $ E_{-i} = V_i^{*}$ and $ \rho$ be the vector bundles defining its dual Lie $\infty$-algebroid and its anchor. By abuse of language, we call \emph{map} from a manifold $\Sigma$ to ${\mathbb V}$ an NQ-manifold morphism $ \Phi$ from the NQ-manifold $ \Sigma$ to the NQ-manifold ${\mathbb V}$, \emph{i.e.}~a differential graded commutative algebra morphism
\begin{equation} \Phi: ({\mathbb V}, Q) \mapsto (\Omega^\bullet_\Sigma,{\rm d}_{dR}).\end{equation}
The latter is entirely described by:
\begin{enumerate}
\item[$\bullet$] its base map, which is a smooth map $\phi: \Sigma \to M$
\item[$\bullet$] its Taylor coefficients $\phi_n \colon \wedge^n T\Sigma \to E_{-n} $.
\end{enumerate}
\begin{example}
\label{ex:constant}
\normalfont
Let $m \in M$ be a point and $\Sigma$ a manifold. A map $ \Phi\colon \Sigma \to {\mathbb V}$ is said to be \emph{constantly $m$} if its base map is a constant map equal to $m \in M$ and all Taylor coefficients are equal to zero.
It shall be denoted by $\underline{m}$.
\end{example}
Let us now define the homotopy and fundamental groups based at $m$.
Let $N$ be the north pole of the sphere $S^n$. Consider the set of homotopy classes of maps from $S^n$ to ${\mathbb V}$ whose restriction to a neighborhood of the north pole is constantly $ m$. As for usual manifolds, this set has a group structure, and this group is Abelian for all $n \geq 2$. For $ n \geq 2$, it is referred to as the \emph{$n$-th homotopy group} based at $m$ and denoted by $\pi_n({\mathbb V},m) $. For $n=1$, we call it the \emph{fundamental group of $({\mathbb V}, Q) $} based at $m \in M$.
Consider now the set of homotopy classes of \emph{paths}, (\emph{i.e.}~NQ-morphisms from $I=[0,1]$ to $ {\mathbb V}$) valued in ${\mathbb V}$ which are constant near $0$ and $1$. This set comes equipped with a natural groupoid structure over $M$, referred to as the \emph{fundamental groupoid of ${\mathbb V} $} and denoted by $\mathbf \Pi({\mathbb V}) \rightrightarrows M$.
\begin{proposition}
Let $ ({\mathbb V},Q)$ and $ ({\mathbb V}',Q')$ be
two NQ-manifolds.
\begin{enumerate}
\item An NQ-morphism $\Phi \colon {\mathbb V} \to {\mathbb V}'$ with base morphism $ \phi$ induces a group morphism $\pi_n( {\mathbb V},m) \to \pi_n( {\mathbb V}',\phi(m)) $ for all $ m \in M$ and a groupoid morphism $ \mathbf \Pi({\mathbb V}) \to \mathbf \Pi({\mathbb V}')$.
\item Two homotopic NQ-morphisms $\Phi_0,\Phi_1$ over the same base map $\phi$ induce the same group and groupoids morphisms.
\end{enumerate}
\end{proposition}
\begin{proof}
We prove it for homotopy groups: the proof for the fundamental groupoid is similar.
The group and groupoid morphisms in the first item are simply induced by the push-forward:
$$ \begin{array}{rcl} \{ \hbox{Map}( S^n, {\mathbb V} ) \}& \longrightarrow & \{ \hbox{Map}( S^n, {\mathbb V}') \}, \\ \sigma & \mapsto & \Phi \circ \sigma \end{array} $$
which transforms a map constantly $m$ to a map constantly $\phi(m)$ in a neighborhood of the north pole, and transforms compatible maps into compatible maps.
For the second item, it suffices to check that for $\Phi \colon I \times {\mathbb V} \to {\mathbb V}' $ a homotopy between $ \Phi_0$ and $\Phi_1$, the composition $ \Phi \circ ({\mathrm{id}} \times \sigma)\colon I \times S^n \to {\mathbb V}' $ is a for every $ \sigma \in \hbox{Map}( S^n, {\mathbb V} ) $ a homotopy relating $ \Phi_0 \circ \sigma$ and $ \Phi_1 \circ \sigma$.
\end{proof}
Here is an important consequence of this proposition.
\begin{corollary}
\label{cor:homotopyGroups}
The homotopy groups, fundamental groups and fundamental groupoids of any two homotopy equivalent NQ-manifolds are isomorphic.
\end{corollary}
The following lemma will also be of interest:
\begin{lemma}
\label{lem:restrLeaf}
Let $ L$ be the leaf through $m$ of an NQ-manifold $(\mathbb V,Q)$.
Then the restriction $ {\mathfrak i}_L {\mathbb V}$ of $ {\mathbb V} $ to $L$ is an NQ-manifold and $ \pi_n(\mathbb V,m) = \pi_n({\mathfrak i}_L\mathbb V,m) $. Also, the induced singular foliation of the fundamental groupoid $ \mathbf \Pi ({\mathbb V})$ is the basic singular foliation
of $\mathbb V $.
\end{lemma}
\begin{proof}
This lemma follows obviously from the fact that the base map of an NQ-manifold morphism from a connected manifold to $ \mathbb V$ can not ``jump'' from one leaf to an other leaf, and has to be valued in a given $\mathbb V$-leaf.
\end{proof}
Altogether, homotopy groups $ (\pi_n({\mathbb V}),m)_{m \in M}$ of an NQ-manifold $ {\mathbb V}$ with base manifold $M$ form a bundle of groups over $M$, that we call the \emph{$n$-th homotopy group bundle}.
\begin{definition}
A section $ \sigma : m \mapsto \sigma(m) \in \pi_n({\mathbb V}, m) $
of the $n$-th homotopy group bundle is said to be \emph{smooth} there exists an NQ-morphism $M \times S^n \to {\mathbb V}$ whose restriction to $ \{m\} \times S^n$ is a representative of $ \sigma(m) $ for all $m \in M$.
We denote these smooth sections by $\Gamma(\pi_n({\mathbb V})) $.
\end{definition}
For the fundamental groupoid, smooth sections of the source map, and \emph{smooth bisections}, are defined in the same way.\\
Let us spell out the special case $n=0$. We define $\pi_0(\mathbb V,m)=M/\mathbb V$ to be the pointed set $M/\mathbb V $ of $\mathbb V$-orbits\footnote{An NQ-manifold $\mathbb V$ on $M$ induces a singular foliation on $M$ and hence a decomposition into $\mathbb V$-leaves. We denote the leaf space by $M/\mathbb V$.}, the orbit of $m$ being the marked point. Also:
\begin{definition}
\label{def:pi0}
We define $\Gamma(\pi_0({\mathbb V})) $ to be the group ${\mathrm{Diff}}(M/\mathbb V) $ of bijections of the orbit space $M/{\mathbb V} $ that come from a smooth diffeomorphism of $M$ mapping every $\mathbb V$-leaf to a $\mathbb V$-leaf. In particular, $\Gamma(\pi_0({\mathbb V}))$ has a natural group structure.
\end{definition}
\subsection{A long exact sequence for NQ-manifold fibrations }
\subsubsection{NQ-manifold fibrations and Ehresmann connections}
Throughout this section:
\begin{enumerate}
\item $(E_\bullet =\bigoplus_{i\geq 1 } E_{-i} ,Q)$ is a Lie $\infty$-algebroid with dual NQ-manifold $ {\mathbb V}= (V_\bullet, M, Q)$ with $V_i=E_{-i}^*$ for all $ i \in {\mathbb N}$.
\item $B \to L$ is a Lie algebroid, seen as an NQ-manifold when equipped with functions $ \Gamma(\wedge^\bullet B^*)$ and Chevalley-Eilenberg differential (see \cite{MR2157566}, Chapter 7).
\item $ P : {\mathbb V} \to B$ is an NQ-manifold morphism.
For degree reasons, all its Taylor coefficients are zero except for the first one, so that $P$ is entirely described by a vector bundle morphism. We denote this morphism by $P$:
\begin{equation}
\label{eq:shortExact} \xymatrix{ E_{-1}\ar[r]^{P} \ar[d] & B \ar[d] \\ M \ar[r]^p & L}
\end{equation}
such that $ \wedge^\bullet P^* \colon \Gamma( \wedge^\bullet B^* )\to \Gamma( \wedge^\bullet E^*_{-1} ) \subset \Gamma(S(V)) $ is a differential graded commutative algebra morphism.
\end{enumerate}
\begin{definition}
\label{def:fibrations}
When $P \colon E_{-1} \to B$ is a surjective submersion, \emph{i.e.~}the base map $ p\colon M \to L$ is a surjective submersion and $P_m : E_{-1}|_{m} \to B|_{p(m)} $ is a surjective linear map for every $m \in M$, we say that $P$ is an
NQ-manifold fibration over a Lie algebroid.
\end{definition}
\begin{example}
\normalfont
When both ${\mathbb V} $ and $B$ are manifolds, we recover usual manifold fibrations. When both ${\mathbb V} $ and $B$ are Lie algebras, we recover Lie algebra epimorphisms.
\end{example}
Let us define the fibers of an NQ-manifold fibration.
Consider the vector bundles over $M$ defined by $ K_{-i} = E_{-i}$ for all $i \geq 2$ and $K_{-1} := {\mathrm{Ker}}(P) $.
The Lie $\infty$-algebroid brackets of $E$ restrict to sections of $ K_\bullet \to M$, and therefore equip the latter with a Lie $\infty$-algebroid structure.
We denote by $ {\mathbb T}$ its dual NQ-manifold. Since the anchor $ {\mathbb T}$ is by construction valued in the fibers of $p \colon M \to L$, it restricts to an NQ-manifold $ {\mathbb T}_\ell$ on the fiber $p^{-1}(\ell) $ over any $\ell \in L$.
We call the NQ-manifold $ {\mathbb T}_\ell$ the \emph{fiber over $\ell \in L$}.
\begin{example}
\normalfont
When both ${\mathbb V} $ and $B$ are manifolds, we recover usual fibers of a manifold fibration.
When both ${\mathbb V} $ and $B$ are Lie algebras, the fiber is simply the kernel of the Lie algebra epimorphism $P$.
\end{example}
\begin{example}
\normalfont
Lie algebroid actions on $N$-manifolds, as defined in \cite{BrahicZambon}, provide a wide class of examples of Lie algebroid fibration.
\end{example}
\begin{definition}
An Ehresmann connection for a NQ-manifold fibration over a Lie algebroid $ P \colon \mathbb V \to B$ is a sub-bundle $ H \hookrightarrow E_{-1} $ in direct sum with $K_{-1}={\mathrm{Ker}}(P) \subset E_{-1}$.
\end{definition}
An Ehresmann connection induces a \emph{horizontal lift} $ \tilde{H}\colon \Gamma(B) \to \Gamma(E_{-1})$: A section $b \in \Gamma(B)$ is lifted to the unique section of $H$ which is $P$-related to $b$.
For every section $ b \in \Gamma (A)$ the vector field $ \rho_\mathbb V \circ \tilde{H} \, (b)$ is $p$-related to the vector field $ \rho_A (b)$.
\begin{definition}
\label{def:Ehresmann}
An Ehresmann connection for an NQ-manifold fibration over a Lie algebroid $ P \colon \mathbb V \to B$ as above is said to be \emph{complete} if for all $ b \in \Gamma(B) $ the integral curve of the vector field $ \rho_{\mathbb V} \circ \tilde{H} (b) $ starting at $m$ is defined if and only if the integral curve of the vector field $ \rho_B (b)$ starting at $ p(m)$ is defined.
\end{definition}
\begin{remark}
\normalfont
Of course, every NQ-manifold fibration over a Lie algebroid admits Ehresmann connections, but there may not exist complete Ehresmann connections. For instance when $B$ is transitive and the fibers $ (\mathbb T_{\ell})_{\ell \in L}$ over two points $\ell,\ell' \in L$ are not diffeomorphic, Proposition \ref{prop:fibersAreIso} below obstructs the existence of complete Ehresmann connections.
\end{remark}
\begin{example}
\normalfont
When ${\mathbb V} $ is a manifold $M$, we recover usual (complete) Ehresmann connections for manifold fibrations.
When both ${\mathbb V} $ and $B$ are Lie algebras, a connection is a linear section of the epimorphism $P$ and every connection is Ehresmann.
\end{example}
\begin{remark}
\normalfont
When ${\mathbb V}$ is a Lie algebroid, our objects match those introduced in \cite{MR3897481} and \cite{Zhu}, but with several differences in vocabulary. NQ-manifold fibrations do not coincide with Lie algebroid fibrations in \cite{Zhu} since they assume that the Lie algebroid fibration comes with a complete Ehresmann connection. So that the correspondence is ``Lie algebroid fibration in \cite{Zhu}''= ``NQ-manifold fibrations in our sense with ${\mathbb V}$ a Lie algebroid'' + ``a complete Ehresmann connection''.
Also, ``Ehresmann connections'' in \cite{Zhu} are always complete, and therefore correspond to our ``complete Ehresmann connections''.
When ${\mathbb V}$ is a Lie algebroids and $B$ is transitive, our NQ-manifold fibrations are the Lie algebroid submersions of \cite{MR3897481}: complete Ehresmann connections now have the same meaning.
\end{remark}
Sections of $\Gamma(B) $ (resp. $E_{-1}$) can be considered as degree $-1$ vector fields on the NQ-manifold $B$
(resp. $\mathbb V$): it suffices to let $ b \in \Gamma(B) $ (resp. $e \in \Gamma(E_{-1})$) act by contraction $ {\mathfrak i}_b$ (resp. $ {\mathfrak i}_e$) on their respective graded algebras of functions.
The degree zero vector fields $ [ {\mathfrak i}_{\tilde{H}(b)}, Q]$ and $[ {\mathfrak i}_{b}, {\mathrm d}_{B}] $ are $P$-related for every Ehresmann connection $H$. If $H$ is complete, their flows are also related:
\begin{proposition}
\label{prop:aboutComplete}
Consider an Ehresmann connection $H$ for an NQ-manifold fibration over a Lie algebroid $ P \colon \mathbb V \to B$. The following are equivalent:
\begin{enumerate}
\item The Ehresmann connection $H$ is complete.
\item For all $ b \in \Gamma(B)$, the time $t$ flow of the degree $0$ vector field $ [ {\mathfrak i}_{\tilde{H}(b)}, Q]$ is defined if and only if the time $t$ flow of the degree $0$ vector field $[ {\mathfrak i}_{b}, {\mathrm d}_{B}] $ is defined.
\end{enumerate}
\end{proposition}
\begin{proof}
By the first item in Lemma \ref{lem:flow}, the flow of a degree $0$ vector field exists if and only if the flow of its induced vector field on its base manifold exists.
In our case, the induced vector fields of $ [ {\mathfrak i}_{\tilde{H}(b)}, Q]$ and $[ {\mathfrak i}_{b}, {\mathrm d}_{B}] $ are
$ \rho_{\mathbb V} \circ \tilde{H} (b) $ and $ \rho_B (b) $ respectively. Hence the equivalence of both items is a direct consequence of Definition \ref{def:Ehresmann}.
\end{proof}
In differential geometry, complete Ehresmann connections allow to identify fibers with each other: the same occurs in the present context for transitive Lie algebroids.
\begin{proposition}
\label{prop:fibersAreIso}
If a complete Ehresmann connection $H$ for an NQ-manifold fibration over a transitive Lie algebroid $ P \colon \mathbb V \to B$ exists, then the fibers $ {\mathbb T}_{\ell}$ and $ {\mathbb T}_{\ell'}$ over any two $\ell, \ell' \in L$ are diffeomorphic NQ-manifolds.
\end{proposition}
\begin{proof}
There exists a vector field $X $ on $L$ whose time $1$ flow exists globally and maps $\ell$ and $\ell'$. Since $B$ is transitive, there exists a section $b \in \Gamma(B)$ with $ \rho_B(b)=X$. Since the induced vector field of $ [ {\mathfrak i}_{\tilde{H}(b)}, Q] $ is $p$-related to $X$, and since the time-$1$ flow of $\rho_B(b)=X$ is well-defined, Proposition \ref{prop:aboutComplete} implies that the time-$1$ flow $\Phi_1$ of $ [ {\mathfrak i}_{\tilde{H}(b)}, Q] $ is a well-defined NQ-manifold diffeomorphism. Since the induced diffeomorphism $ \phi_1 \colon M \to M$ of $\Phi_1$ is over the time $1$- flow of $ X$, it maps the fiber $ p^{-1}(\ell)$ to the fiber $ p^{-1}(\ell')$. Therefore, the restriction of $ \Phi_1$ to $ {\mathbb T}_{\ell}$ is the desired NQ-manifold diffeomorphism onto ${\mathbb T}_{\ell'}$.
\end{proof}
\begin{remark}
\normalfont
Proposition \ref{prop:fibersAreIso} obviously extends to the non-transitive case as follows: $ {\mathbb T}_{\ell}$ and $ {\mathbb T}_{\ell'}$ over any two $\ell, \ell' \in L$ in the same $B$-leaf are diffeomorphic NQ-manifolds.
When $\mathbb V$ is a Lie algebroid, this follows from Theorem C in \cite{MR3897481}.
\end{remark}
\subsubsection{Horizontal lifts for complete Ehresmann connections}
The equivalent of the following lemma for fibrations of ordinary manifolds is well-known under the name of ``homotopy lifting property'', see \cite{MR1325242}, Chapter 7, or \cite{MR1867354} Section 4.2. Its Lie algebroid equivalent is implicit in the proof of Theorem 1.4 in \cite{Zhu}.
\begin{flemma}
\label{lem:lifts}
Let $ P \colon {\mathbb V} \to B$ be an NQ-manifold fibration over a Lie algebroid as in Definition \ref{def:fibrations} that admits a complete Ehresmann connection $H$. Given
\begin{enumerate}
\item[a)] a manifold $\Sigma$ called \emph{parameter space};
\item[b)] an NQ-morphism $e\colon \Sigma \mapsto {\mathbb V} $ called \emph{initial shape};
\item[c)] a Lie algebroid morphism $b: T(I \times \Sigma) \to B $ called \emph{base map}
\end{enumerate}
making the following diagram of NQ-manifold morphisms commutative
\begin{equation}
\label{eq:lemmaLift}
\xymatrix{
\Sigma\ar[rr]^e \ar@{_(->}[d]_{u \to (0,u) }& & {\mathbb V}\ar@{->>}[d]^P \\
I\times \Sigma \ar[rr]_b& & B},
\end{equation}
there exists a natural NQ-manifold morphism ${\mathcal L}^H(e,b)\colon I\times \Sigma--\to E$ called \emph{horizontal lift of the base map $b$ with initial shape $e$} such that the following diagram commutes:
\begin{equation}
\label{eq:lemmaLift2}
\xymatrix{
\Sigma\ar[rr]^e \ar@{_(->}[d]_{u \to (0,u) }& & {\mathbb V}\ar@{->>}[d]^P \\
I\times \Sigma \ar[rr]_b \ar@{-->}[urr]_{{\mathcal L}^H(e,b)}& & B},
\end{equation}
\end{flemma}
\begin{proof}
Consider the following Cartesian diagram of vector bundle morphisms
\begin{equation}
\label{diag:cartesian}\xymatrix{
(TI\times T\Sigma)\times_{b,B,P}E_{-1}\ar[rrr]\ar[dr] \ar[ddd]& & & TI\times T\Sigma\ar[ddd] \ar[dl]\\
&(I\times \Sigma)\times_{L}M\ar[r]\ar[d]&I\times \Sigma\ar[d]& \\
& M\ar[r]& L & \\
E_{-1}\ar[ur] \ar[rrr]& & & B\ar[ul]
} \end{equation}
Given a horizontal distribution $D$ on $P:E_{-1}\to B$, the section $\frac{\partial}{\partial t}$ of $TI\times T\Sigma$ lifts to a unique section $X$ of $(TI\times T\Sigma)\times_{b,B,P}E_{-1}$, such that for all $(t,\sigma,m)\in (I\times \Sigma)\times_{L}M$, $X(t,\sigma,m)=(\frac{\partial}{\partial t},0,u)$, with $u\in D$. By construction $P(u)=b_{t,\sigma}(\frac{\partial}{\partial t},0)$.\\
We now prove the first item. We define ${\mathcal L}^H(e,b)$ as the composition of three NQ-morphisms:
\begin{align}
\label{eq:composition}
TI\times T\Sigma\longrightarrow
TI\times (TI\times T\Sigma) \times_{b,B,P} \mathbb V\longrightarrow
(TI\times T\Sigma) \times_{b,B,P} \mathbb V \to \mathbb V
\end{align}
which we now describe.
\begin{enumerate}
\item $ (TI\times T\Sigma) \times_{b,B,P} \mathbb V $ stands for fibered products of NQ-manifolds, which make sense because $P$ is a surjective submersion. Its base manifold is $ (I\times \Sigma)\times_{L}M $ and its dual vector bundle is given in degree $-1$ by $(TI \times T\Sigma) \times_{b,B,P} E_{-1}$.
In particular, the section $X$ defined above can be seen as a degree minus one vector field on $ (TI\times T\Sigma) \times_{b,B,P} \mathbb V $.
\item The first arrow is the composition of $TI\times T\Sigma\to TI\times TI\times T\Sigma, (t,\sigma)\mapsto (t,0,\sigma)$ with $\mathrm{id}_{TI}\times A$, where $A: TI\times T\Sigma\to (TI\times T\Sigma)\times_B \mathbb V$ is the natural map associated to the pullback.
\item The central arrow is the flow $NQ$-morphism of the degree minus one vector field $X$, as constructed in Lemma \ref{lem:flow}.
\item The last arrow is the projection to $\mathbb V$.
\end{enumerate}
Let us check that $\mathcal L^H(e,b)$ satisfies the required conditions.
Let us consider its restriction to $ \{0\} \times T \Sigma$ . The central arrow in
\eqref{eq:composition}, restricted to $\{0\} \times TI \times T \Sigma \times {\mathbb V}$, is the projection onto the second component, so that the restriction of the two last arrows to $\{0\} \times TI \times T \Sigma \times {\mathbb V}$ consists in projecting onto $ {\mathbb V}$.
In view of the first arrow, the restriction to $ \{0\} \times T \Sigma$ coincides with $e$.
Let us prove that $P \circ \mathcal L^H(e,b) = b $.
The following diagram is commutative, where the upper horizontal line is \eqref{eq:composition2}, the lower line is constructed as \eqref{eq:composition2}, with $B$ replacing ${\mathbb V}$ and $ b_1 $ replacing $e$, and with all vertical lines being induced by $P \colon {\mathbb V} \to B$:
\begin{align}
\label{eq:composition2}
\xymatrix{TI\times T\Sigma \ar@{=}[d]\ar[r]&
TI\times (TI\times T\Sigma) \times_{b,B,P} \mathbb V \ar[d]\ar[r]&
(TI\times T\Sigma) \times_{b,B,P} \mathbb V\ar[d]\ar[r] & \mathbb V \ar[d]^P \\TI\times T\Sigma\ar[r]
&
TI\times (TI\times T\Sigma) \times_{b,B,id} B \ar[r]&
(TI\times T\Sigma) \times_{b,B,id} B\ar[r] & B
}
\end{align}
This commutativity is straightforward, except for the central square, for which is expresses the fact that $X_{(t,\sigma,m)} $, with $p(m)=\sigma $, projects to $ (\frac{\partial }{\partial t}, b_{t,\sigma} (\frac{\partial}{\partial t}))$. The composition of the four lower arrows coincides with $b$.
\end{proof}
\begin{definition}
Let $ P \colon {\mathbb V} \to B$ be an NQ-manifold fibration over a Lie algebroid as in Definition \ref{def:fibrations} that admits a complete Ehresmann connection $H$. Let $(\Sigma,e,b)$ be a parameter space, an initial shape, a base map as in Lemma \ref{lem:lifts}.
We call \emph{parallel transport of $e$ with respect to $b$} the restriction to $ \{1\} \times \Sigma $ of the horizontal lift ${\mathcal L}^H(e,b) $. We denote it by $ {\mathcal P}^H(e,b) \colon \Sigma \to {\mathbb V}$.
\end{definition}
The parallel transport behaves well with respect to restrictions to submanifolds:
\begin{proposition}
\label{prop:restrictions}
Let $ P \colon {\mathbb V} \to B$ be an NQ-manifold fibration over a Lie algebroid as in Definition \ref{def:fibrations} that admits a complete Ehresmann connection $H$. Let $\Sigma,e,b$ be a parameter space, an initial shape, and a base map as in Lemma \ref{lem:lifts}. For every submanifold $ {\mathfrak i} : \Sigma' \hookrightarrow \Sigma$, the following diagram commutes:
\begin{equation}
\label{eq:lemmaLiftRestr}
\xymatrix{
\Sigma' \ar@{^(->}[rrrr] \ar[drr]_{\mathcal P^{H}(e|_{\Sigma'},b|_{I \times \Sigma'}) } & & & & \Sigma \ar[lld]^{\mathcal P^{H}(e,b) }\\ && {\mathbb V} && }
\end{equation}
\end{proposition}
\begin{proof}
A close look at its construction shows that the horizontal lift behaves well with respect to restriction to a submanifold $ {\mathfrak i} \colon \Sigma' \hookrightarrow \Sigma$, \emph{i.e.~}it satisfies $ {\mathcal L}^H(e,b) |_{I \times \Sigma'} = {\mathcal L}^H(e|_{\Sigma'}, b|_{I \times \Sigma'}) $ or, as a diagramm:
\begin{equation}
\label{eq:lemmaLiftRestr0}
\xymatrix{
\Sigma\ar[d]_e \ar@{^(->}[rrr]^{u \hookrightarrow (0,u) }& & & I\times \Sigma \ar[d]^{b} \ar@{-->}[dlll]_{{\mathcal L}^H(e,b)} \\ {\mathbb V}\ar@{->>}[rrr]^P & & & B \\
\Sigma'\ar[u]^{e|_\Sigma'} \ar@{^(->}[rrr]_{u \hookrightarrow (0,u) } \ar@/^4.0pc/@{_(->}[uu]^{\mathfrak i} & & & I\times \Sigma' \ar[u]_{b|_{I \times \Sigma'}} \ar@{-->}[ulll]_{ {\mathcal L}^H(e|_{\Sigma'}, b|_{I \times \Sigma'}) } \ar@/_4.0pc/@{^(->}[uu]_{{\mathrm{id}} \times \mathfrak i} }.
\end{equation}
This implies the commutativity of the diagram \eqref{eq:lemmaLiftRestr}
\end{proof}
We will furthermore need the following
\begin{lemma}\label{lem:hotint}
Let $ P \colon {\mathbb V} \to B$ be an NQ-manifold fibration over a Lie algebroid as in Definition \ref{def:fibrations} and $\ell \in L$ a point.
Consider two NQ-morphisms $\Phi_i\colon \Sigma \to \mathbb T_\ell $, $i=0,1$. Then the following items are equivalent:
\begin{enumerate}
\item[(i)] $\Phi_0$ and $\Phi_1$ are homotopic in $\mathbb T_\ell$,
\item[(ii)] $\Phi_0$ and $\Phi_1$ are homotopic in ${\mathbb V}$ through a homotopy $h \colon I \times \Sigma \to {\mathbb V}$ such that $P \circ h \colon I \times \Sigma \to B$ is homotopic to the constant map $\underline{\ell}$.
\end{enumerate}
\end{lemma}
\begin{proof}
The implication (ii) $ \implies $ (i) is trivial. For the converse implication, let $h_B:I\times I\times \Sigma\to B$ be a contracting homotopy for $P \circ h$. Consider its horizontal lift ${\mathcal L}^H (h , h_B) \colon I \times I \times \Sigma \to \mathbb V$ with initial shape $h \colon I \times \Sigma \to \mathbb T_\ell $ with respect to the base path $h_B:I\times I\times \Sigma \to B$.
\begin{center}
\begin{figure}[!ht]
\caption{\label{fig:edge} The neighborhood $ U$ and the curve $ \gamma(t)$. }
\begin{center}
\includegraphics[clip=true, trim = 0mm 150mm 0mm 5mm,scale=0.3]{drawing.pdf}
\end{center}
\end{figure}
\end{center}
Since $ h_B \colon \colon I \times I \to B$ is constantly $ \ell$ on $U \times \Sigma $, with $U \subset I^2$ a neighborhood of the three edges of the square depicted in Figure \ref{fig:edge},
the restriction to $U \times \Sigma $ of the horizontal lift ${\mathcal L}^H (h , h_B) $ is valued in $ {\mathbb T}_\ell$. Let $\gamma : I \to I^2 $ a curve valued in $U$ relating $(0,1) $ and $ (1,1)$ as in Figure \ref{fig:edge}. The restriction to $ \{ \gamma(I) \} \times \Sigma $ of the horizontal lift ${\mathcal L}^H (h , h_B) \colon I \times I \times \Sigma $ is a $ \mathbb T_\ell$-valued homotopy between $ \Phi_0$ and $ \Phi_1$.
\end{proof}
\subsubsection{Main result}
We can now associate a long exact sequence to any NQ-manifold fibration over a Lie algebroid, generalizing Theorem 1.4 in \cite{Zhu}.
\begin{theorem}\label{thm:snake}
Consider a fibration of NQ-manifold over a Lie algebroid $P \colon {\mathbb V} \to B$ as in Definition \ref{def:fibrations} that admits a complete Ehresmann connection. For $\ell \in L$, denote by $ {\mathbb T}_\ell $ the fiber over $ \ell$.
Then there exists a family of group morphisms:
$$\begin{array}{rclll} \delta \colon & \pi_n (B,\ell ) & \longrightarrow & \Gamma ( \pi_{n-1}({\mathbb T}_\ell ) )& \hbox{ for $ n \geq 2$ } \\
& \pi_1 (B,\ell )& \longrightarrow & \mathrm{Diff}(M / \mathbb V) &
\end{array}
$$
such that the sequence of groups morphisms
$$
\xymatrix{ & \dots\ar[r]^{P} & \ar[dll]_{\delta}\pi_{n+1}(B,\ell) \\
\pi_n({\mathbb T}_\ell,m)\ar[r]^{}&\pi_n({\mathbb V},m)\ar[r]^{P}&\pi_n(B,\ell)\ar[dll]_{\delta}\\
\pi_{n-1}({\mathbb T}_\ell ,m)\ar[r]^{}& \dots & \dots \\
}
$$
is exact for all $m \in p^{-1}(\ell)$.
Moreover, $\delta(\pi_2(B, \ell)) $
lies in the center of $\Pi_1(\mathbb T_\ell)$,
and, for a given $\gamma \in \pi_1(B,\ell) $, the diffeomorphism $\delta (\gamma)$ of the orbit space of $ {\mathbb T}_\ell$ fixes the ${\mathbb T}_\ell$-leaf of $ m$ if and only if $\gamma$ lies in the image of $ {P_*} \colon \pi_{1}({\mathbb V} ,m) \to \pi_{1}(B ,\ell)$.
\end{theorem}
\begin{proof} The fundamental Lemma \ref{lem:lifts} allows to make the proof of the present theorem similar to the usual proof for manifold fibrations as presented for instance in Theorem 4.41 in \cite{MR1867354} - with smooth maps being replaced by NQ-morphisms.
Although it follows the same structure as the proof of Theorem 1.4 in \cite{Zhu}, our point of view is more (NQ-) geometric.\\
\noindent
{\bf Construction of the connecting map $\delta$.}
Consider an NQ-morphism $ S^n\to B$, which is constantly $ \ell$ near the north pole $N \in S^n$. This last property allows to see it as an NQ-morphism $ \sigma \colon I\times S^{n-1}\to B$ which is constantly $\ell$ near $\{0 \} \times S^{n-1} $, $\{1\} \times S^{n-1} $ and $I \times \{N\}$.
For all $m \in p ^{-1}(\ell)$, we define $$ \delta|_m (\sigma) := {\mathcal P}^H(\sigma , \underline{m} )$$
to be the parallel transport of the initial shape $\underline{m} \colon S^{n-1} \to {\mathbb V}$ with respect to the base map $\sigma$.\\
Let us study this map.
\begin{enumerate}
\item {\textbf{It is valued in the fiber $ {\mathbb T}_\ell$.}} The horizontal lift of $e $ with respect to the base path $\sigma$ makes by definition the following diagram commutative:
\begin{equation}
\label{eq:item1H}
\xymatrix{
S^{n-1} \ar[rr]^{\underline{m}} \ar@{_(->}[d]_{u \to (0 ,u)} & & {\mathbb V}\ar@{->>}[d]^P \\
I \times S^{n-1} \ar[rr]_{\sigma } \ar@{-->}[urr]^{{\mathcal L}^H(\sigma , \underline{m} )} & & B.}
\end{equation}
Since $\sigma $ is constantly $\ell $ on $\{1\} \times S^n $, the commutativity of \eqref{eq:item1H} implies that $ \delta|_m ( \sigma) = {\mathcal P}^H(\sigma , \underline{m} ) = {\mathcal L}^H(\sigma , \underline{m} )|_{\{1\}\times S^{n-1}}$ is valued in the fiber $ {\mathbb T}_\ell$.
\item {\textbf{It is constant equal to $m$ near the north pole}}. For every paramater manifold, parallel transport of a constant initial shape $\underline{m}$ with respect to a base map which is a constant map $\underline{\ell} $ is the constant map $\underline{m} $ again. Since the north pole admits a neighborhood $U$ on which $\sigma \colon I \times U $ is $\underline{\ell}$, it follows from Proposition \ref{prop:restrictions}, applied to the open submanifold $ \Sigma' =U $, that ${\mathcal P}^H(\sigma , \underline{m} )$ is equal to $\underline{m}$ in a neighborhood of the north pole.
\item {\textbf{ It preserves homotopy.}} Let $h\colon I \times S^{n-1}\times I \to B$ be a homotopy between NQ-morphisms $\sigma_i\colon S^{n-1}\times I \to B $, $i=0,1$. Let ${\mathcal P} (\underline{m} , h) $ be the parallel transport of the initial shape $\underline{m}$ with respect to the base map $h$.
By Proposition \ref{prop:restrictions}, the restrictions of ${\mathcal P} (\underline{m} , h) $ to the submanifolds $ \{i\} \times S^{n-1}\times I \subset I \times S^{n-1}\times I$ is $\delta|_m (\sigma_i) $, for $i=0,1$.
\end{enumerate}
Altogether these three points allow to define an induced group morphism:
$$ \delta |_m(\sigma) \colon \pi_n(L , \ell ) \longrightarrow \pi_{n-1} ({\mathbb T}_\ell ,m) .$$
We have to check that $m \to \delta |_m(\sigma) $ is valued in smooth sections of the group bundle $\pi_{n-1} ({\mathbb T}_\ell)$.
Consider \emph{(i)} the parameter space $ \Sigma = S^{n-1} \times p^{-1}(\ell)$, \emph{(ii)} the initial shape given by the projection onto $ pr_2 \colon S^{n-1} \times p^{-1}(\ell) \to p^{-1}(\ell) \subset {\mathbb V}$, and \emph{(iii)} the base map $\sigma \circ pr$ with $pr$ being the projection $I \times S^{n-1} \times p^{-1}(\ell) \to I \times S^{n-1}$. The henceforth associated parallel transport is an NQ-morphism:
$$ S^{n-1} \times p^{-1}(\ell) \longrightarrow {\mathbb T}_\ell .$$
Proposition \ref{prop:restrictions} implies that its restriction to the parameter submanifolds $S^{n-1} \times \{m\}$ is $\delta|_m(\sigma)$. This guaranties the smoothness of the section. \\
\noindent {\bf The relation $\delta\circ p_*=0$.}
Consider an element of $ \pi_n(B,\ell )$ that can be represented by an NQ-morphism $ \sigma$ of the form $ \sigma = P_* (\tau)$ for some $ \tau \colon S^n \to {\mathbb V} $ constantly equal to $m$ near $N$.
Let us compare $ \tau$ with the horizontal lift $L^H(\sigma, \underline{m})$ of the initial condition $\underline{\sigma} $ with respect to the base $ \sigma$.
We can see $ \tau$ as NQ-morphism $\tau \colon I \times S^{n-1} \to {\mathbb V} $ constantly equal to $m$ in neighborhoods of $\{0\} \times S^{n-1}$ and $\{1\} \times S^{n-1}$.
This allows to consider the concatenation $\tau^{-1}*L^H(\sigma, \underline{m})$, where $\tau^{(-1)}$ is the inverse path of $\tau$, \emph{i.e.}~$ \tau \circ (1-t \times {\mathrm{id}}_{S^{n-1}}) $. Its projection through $P$ being the concatenation of $\sigma^{-1} $ with $\sigma $ is homotopic to $\underline \ell$. By Lemma \ref{lem:hotint}, the restrictions to the end points are homotopic in $\mathbb T_\ell$. These restrictions being $\underline{m}$ for $\tau $ and $\delta(\sigma)|_{m}(\sigma) $, this means that the latter is zero. \\
\noindent {\bf Exactness at $\pi_n(B,b_0)$.} Let $\sigma \colon S^n \to B $. Consider $\mathcal L^H(\sigma,\underline{m}): I \times S^{n-1} \to {\mathbb V} $ in diagram \eqref{eq:item1H}. Recall that $ \delta(\sigma)$ is represented by the restriction of $\mathcal L^H(\sigma,\underline{m})$ to $ \{1\} \times S^{n-1}$. If $\delta(\sigma) =0 $, then there exists
a homotopy $ h \colon [1,2] \times S^{n-1} \to {\mathbb T}_\ell$ whose restrictions to $ {t} \times S^{n-1}$ is $\delta(f) $ for $t$ near $1$ and whose restriction to $ \{2\} \times S^{n-1}$ is $ \underline{m}$ near $2$. The concatenation of $H(\sigma,\underline{m})$ and $h$ is smooth in view of the boundary assumptions and yields an NQ-morphism $ \tau \colon [0,2] \to {\mathbb V} $, defining an element in $ \pi_n( {\mathbb V}, m) $, whose image through $ P$ is $\sigma$ by construction. \\
\noindent {\bf Exactness at $\pi_n(\mathbb V, e_0)$.}
Let $ \tau \colon S^n \to {\mathbb V}$ be such that $ P_* (\tau)=0$, so that there exists an NQ-morphism $ h : I \times S^n \to B $ whose restrictions to $ \{0\} \times S^n $ is $ \underline{\ell}$ and to $ \{1\} \times S^n $ is $ \underline{\ell}$ is $P_* (\tau)$. In view of the first item of Lemma \ref{lem:lifts}, there exists $\mathcal L^H(h, \tau) $ that makes the following diagram commutative:
$$ \xymatrix{ \{0 \} \times S^{n} \ar[rr]^{\tau} \ar[d] && \mathbb V\ar@{->>}[d]^p \\
I\times S^{n} \ar[rr]_{h}\ar@{-->}[urr]^{\mathcal L^H(h,\tau)} && B.}
$$
In particular, the restriction of $\mathcal L^H(h,\tau)$ to $ \{0\} \times S^n$, which is $ \tau$, is homotopic to its restriction to $ \{1\} \times S^n$. The latter being mapped by $P$ to $ \underline{\ell}$ it takes values in $ {\mathbb T}_\ell$, so that $ [\tau]$ lies in the image of ${\mathfrak i}_* $.\\
\noindent {\bf Exactness at $\pi_{n-1}(\mathbb T_\ell,m)$.}
Let $ \tau : S^{n-1}\to {\mathbb T}_l$ be such that $i_* ( \tau ) =0 $.
Lemma \ref{lem:hotint}, so that there exists an NQ-morphism $h : I \times S^{n-1}\to {\mathbb V}$ whose restriction to $\{t\} \times S^{n-1} $ is $\underline{m}$ for all $t $ near $0$, and whose restriction to $\{t\} \times S^{n-1} $ is $ \tau$ for all $t$ near $1$. In particular, this implies that $ P_* \circ h:I\times S^{n-1}\to B$ is constantly equal to $\ell$ on neighbourhoods of $\{0\} \times S^{n-1} $ and $\{1\} \times S^{n-1}$, defining therefore some NQ-manifold morphism $ \sigma \colon S^n \to B $. The NQ-morphisms $\delta|_m(\sigma)$ and $\tau$ are homotopic in $ {\mathbb V}$ through of homotopy as in the second item of Lemma \ref{lem:hotint}, so that they are homotopic in $ {\mathbb T}_\ell$.\\
\noindent {\bf The relation $\delta(\pi_2(B, \ell))\subset \mathrm{Center}(\Pi_1(\mathbb T_\ell))$.} Consider $[\sigma] \in \pi_1(B,\ell)$, and $\tau:I\to \mathbb T_\ell$ a path from $m$ to $m'$. We have to show that $ \delta|_{m'}(\sigma)\circ \tau $ is homotopic to $\tau\circ \delta|_m(\sigma)$ in $\mathbb T_\ell$. By Lemma \ref{lem:hotint}, it suffices to find a homotopy in $\mathbb V$, whose image is homotopic to a constant map in $B$. Such a homotopy is given by the concatenation of homotopies
$$ \xymatrix{
\delta|_m(\sigma)*\tau \ar[rrr]^{\mathcal L^H(\sigma,\underline{m})^{-1} * id_\tau}
&&& \underline{m} * \tau \ar[r]^\sim
& \tau*\underline{m'}\ar[rrr]^{id_\tau*\mathcal L^H(\sigma,\underline{m'})}
&&& \tau * \delta|_{m'}(\sigma) , }
$$
where $*$ denotes the concatenation of paths and homotopies. Recall that the concatenation $b*a$ is a representative of the product $[a]\circ [b]$ in $\Pi_1$.
\end{proof}
\subsection{Examples of homotopy groups of NQ-manifolds}
In this subsection we survey some examples and special cases of the above, mostly already present in the literature.
$$
\resizebox{0.99\textwidth}{!}{
\xymatrix@C=1pt{
&&\hbox{Lie $\infty$-algebroids} \ar[ld]^{\supset}\ar[rd]^{\supset}&&\\
&\txt{Lie algebroids,\\ Ex. \ref{ex:LieAlgebroid}}\ar[rd]^{\supset}\ar[ld]^{\supset}&&\txt{Lie $\infty$-algebras, \\Ex \ref{ex:LieInfinityAlgebra}}\ar[ld]^{\supset}\ar[rd]^{\supset}&\\
\txt{Tangent Lie algebroids,\\ Ex. \ref{ex:manifold}}&&\txt{Lie algebras,\\ Ex. \ref{ex:LieAlgebra}}&&\txt{Nilpotent Lie $\infty$-algebras,\\ Ex. \ref{ex:LieAlgebraNilpotent}}
}
}
$$
\begin{example}
\label{ex:manifold}
\normalfont
Let $M$ be a manifold and $TM$ its tangent Lie algebroid. Then the fundamental groupoid, fundamental groups, resp. $n$-th homotopy groups of $TM$ as an Lie $\infty$-algebroid are just the fundamental groupoid, fundamental group resp. $n$-th homotopy groups of the manifold $M$.
\end{example}
\begin{example}
\label{ex:LieAlgebra}
\normalfont
The fundamental groupoid and the fundamental group of a Lie algebra ${\mathfrak g}$ (seen as an NQ-manifold over a point $pt$) is the simply-connected Lie group $\mathbf G$ that integrating $ {\mathfrak g}$, see \cite{1707.00265,MR1973056}.
For all $n \geq 2$, the $n$-th homotopy groups of the Lie algebra ${\mathfrak g}$, seen as an NQ-manifold, coincide with the $n$-th homotopy groups of the Lie group $\mathbf G$ (see \cite{Zhu}, Example 3.5).
In equation:
$$ \mathbf \Pi (\mathfrak g) = \pi_1(\mathfrak g, pt) = \mathbf G \hbox{ and for all $n \geq 2$: } \pi_n(\mathfrak g,pt)=\pi_n(\mathbf G, 1_\mathbf G) $$
\end{example}
\begin{example}
\label{ex:LieAlgebroid}
\normalfont
For a Lie algebroid $A$ our definitions of fundamental groupoid, fundamental groups and $n$-th homotopy groups reduce to the definition given in \cite{1707.00265,MR1973056, Zhu}.
The fundamental groupoid is the source $s$-connected topological groupoid $\mathbf G$ that integrates the Lie algebroid \cite{1707.00265}: it is a smooth manifold in a neighborhood of $ M$. This idea has been widely used in the literature (cf. e.g. \cite{MR1973056, zbMATH01686785}).
If the Lie algebroid is \emph{integrable}, that is when $ \mathbf G$ is a Lie groupoid, it is proven in \cite{Zhu} that the fundamental group at a given point $m$ is the isotropy $I_m( \mathbf G)= \mathbf G |_m^m$ of this groupoid at $m$, and the homotopy groups are the homotopy groups of $\mathbf G |_m = s^{-1}(m)$ of the fibers of the source map.
In equation:
$$ \mathbf \Pi (A) = \mathbf G, \pi_1(A, m)=I_m(\mathbf G) \hbox{ and for all $n \geq 2$: } \pi_n(A,m)=\pi_n(\mathbf G |_m, m) $$
\end{example}
Let us state an immediate consequence of this example:
\begin{proposition}
\label{prop:Homot_grs_Of_Lie_oid}
Let $A$ be a Lie algebroid which is longitudinally integrable (\emph{i.e.}~the restriction of the fundamental groupoid $ \mathbf G$ of $A$ to every leaf is a smooth manifold).
Then for all $ n \geq 2$ and $m \in M$, we have $\pi_n(A,m)\simeq \pi_n( \mathbf G|_m , m) $ and $\pi_1(A, m)=I_m(\mathbf G)$.
\end{proposition}
\begin{proof}
It suffices to apply the second part of Example \ref{ex:LieAlgebroid} to the restriction of $A$ to the leaf $L$ through~$m$.
\end{proof}
\begin{example} Let us review some results by Berglund
\label{ex:LieAlgebraNilpotent}.
\normalfont
\cite{Berlund}
For a \emph{nilpotent Lie $\infty$-algebra} $ \mathfrak g_\bullet$, \emph{i.e.~}a Lie $\infty$-algebra concentrated in degrees less or equal to $-2$:
$$ \stackrel{ \mathrm d}{\to} {\mathfrak g}_{-3} \stackrel{ \mathrm d}{\to} {\mathfrak g}_{-2} \to 0 $$
the fundamental groupoid and fundamental groups are reduced to $0$ and, for all $n \geq 2$, the $n$-th homotopy groups are isomorphic to the cohomologies $ H^{-n}({\mathfrak g}_\bullet)$ of the above complex in degree $-n$.
In equation:
$$ \mathbf \Pi (\mathfrak g_\bullet) = \pi_1(\mathfrak g_\bullet, pt) = \{ 0 \} \hbox{ and for all $n \geq 2$: } \pi_n(\mathfrak g_\bullet,m)= H^{-n}(\mathfrak g_\bullet) .$$
Explicitly, an NQ-manifold morphism $ \sigma \colon S^n \to {\mathfrak g}_\bullet $ for $n \geq 2$ decomposes as $ \sum_{i=-n}^{-2} \alpha_i $ with
$\alpha_n \in \Omega^n (S^n) \otimes {\mathfrak g}_{-n} , \dots, \alpha_2 \in \Omega^{2} (S^n) \otimes {\mathfrak g}_{-2} $.
Then $ \int_{S^n} \alpha_n \in {\mathfrak g}_{-n} $ is a $d$-closed element and its class of cohomology in $ H^{-n}({\mathfrak g}_\bullet)$ entirely determines the class up to homotopy of $\sigma $.
\end{example}
\begin{example}
\label{ex:LieInfinityAlgebra}
\normalfont
Consider an NQ-manifold $\mathfrak g_\bullet$ over a point $pt$, \emph{i.e.~}a negatively graded Lie $\infty$-algebra:
$$ \dots \stackrel{d}{\to} {\mathfrak g}_{-2} \stackrel{d}{\to} {\mathfrak g}_{-1}.$$
Let us give a long exact sequence describing $\pi_n(\mathfrak g_\bullet,pt)$.
According to the homotopy transfer theorem (see, e.g. \cite{zbMATH06043075}), this Lie $\infty$-algebra is homotopy equivalent to a ``minimal model'', \emph{i.e.}~a Lie $\infty$-algebra with trivial differential constructed on the cohomology of the previous complex:
$$ \dots \stackrel{0}{\to} H^{-2}({\mathfrak g_\bullet}) \stackrel{0}{\to} H^{-1}({\mathfrak g_\bullet}) .$$
The $2$-ary bracket of this Lie $\infty $-algebra does not depend on the choices of a minimal model and satisfies the graded Jacobi identity, so that $H^{-1}({\mathfrak g}_{\bullet}) $ is a Lie algebra.
The projection onto the minus one component is a surjective submersion of a Lie $\infty$-algebra over a Lie algebra:
\begin{equation}
\label{eq:Subm} \oplus_{i\geq 1} H^{-i}({\mathfrak g}_{\bullet}) \to H^{-1}({\mathfrak g}_{\bullet}) .
\end{equation}
The fiber is the nilpotent Lie $ \infty$-algebra $ \oplus_{ i\geq 2} H^{-i}({\mathfrak g}_{\bullet}) $. There is a unique Ehresmann connection (which is of course complete in view of Proposition \ref{prop:existeEhresmann}) given by $H= H^{-1}({\mathfrak g}_{\bullet})$ itself. We are therefore in the setting of Theorem \ref{thm:snake} with all base manifolds being points, so that there is a long exact sequence of groups
$$ \dots \to H^{-n}( {\mathfrak g}_{\bullet}) \to \pi_n( {\mathfrak g}_{\bullet},pt) \to \pi_{n}(\mathbf G, 1_{\mathbf G}) \stackrel{\delta}{\to} H^{-n+1}( {\mathfrak g}_{\bullet} ) \to \cdots $$
$$ \cdots \to \pi_2( {\mathfrak g}_{\bullet}, pt) \to 0 \to 0 \to \pi_1( {\mathfrak g}_{\bullet},pt) \to \mathbf G \to 0 ,$$
where $\mathbf G $ is the simply-connected Lie group integrating the Lie algebra $H^{-1}({\mathfrak g}_\bullet) $. Here, we used Example \ref{ex:LieAlgebraNilpotent}, Example \ref{ex:LieAlgebra} and the fact that $\pi_2(\mathbf G)=0$ for any Lie group.
In particular, $ \pi_1( {\mathfrak g}_{\bullet},pt) \simeq \mathbf G $.
\end{example}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 7,978 |
J. Harold Grady, mayor, chief judge, dies at 84
By Jacques Kelly and Frederick N. Rasmussen
J. Harold Grady, who spent three years in office as Baltimore's 40th mayor before resigning to become a judge, died of cancer yesterday at Stella Maris Hospice in Timonium. He was 84 and had lived in Homeland.
The former Goodale Road resident, who had lived at the Mercy Ridge retirement community in Timonium since July, spent 22 years on the bench - including four as chief judge of what became the Baltimore Circuit Court - until retiring in 1984.
"He felt very comfortable with the law and very uncomfortable as mayor," said former Mayor Thomas J. D'Alesandro III, whose father, Thomas J. D'Alesandro Jr., was defeated by Judge Grady in the 1959 mayoral election. "He was a man of intellect, a sense of humor and was always approachable. And you always knew you were dealing with a man who was fair."
"He accomplished a great deal as state's attorney, mayor and chief judge in a most unassuming way," said J. Robert Brown, a Social Security Administration administrative judge and friend.
"He was an intelligent and fair judge, well-read in the law and a credit to the ... bench," said Maryland Attorney General J. Joseph Curran Jr.
In the early 1950s, Judge Grady attracted wide attention in the legal community as a prosecutor in the state's case against G. Edward Grammer, who was convicted of murdering his wife and using an automobile accident on Taylor Avenue as a cover-up. Grammer was put to death in 1954.
"I could always rely heavily on his judgment, and he was always a dominant figure both literally and physically at the trial table," said Anselm Sodaro, the city state's attorney during the Grammer trial, who later became chief judge of the Supreme Bench of Baltimore.
After succeeding Judge Sodaro as the city's top prosecutor, Judge Grady was drafted to run for mayor in 1959.
That year, he rode to victory on a wave of political change, defeating three-time Mayor Thomas J. D'Alesandro Jr. in the hotly contested Democratic primary by a 33,000-vote margin.
In the general election, Judge Grady defeated Republican Theodore R. McKeldin, former mayor and governor, by more than 81,000 votes.
The election of Judge Grady, who was backed by Irv Kovens, also had historic consequences. It marked the eclipse of James H. "Jack" Pollack's political machine and the rise of Mr. Kovens as a kingmaker.
When Judge Grady and his running mates, Philip H. Goodman and Dr. R. Walter Graham Jr. - popularized as the Three Gs for Good Government - were elected, they gave every appearance of being a united team. Before the administration finished its first year, it became apparent that the team was not pulling together.
Mr. Goodman, the City Council president who succeeded him as mayor, and Dr. Graham, the comptroller, were complaining that Judge Grady was not consulting them on important matters, and they began disagreeing with the administration line on occasion.
The city's financial problems afflicted the new mayor. To save money, he discontinued free public baths and merged the Park Police into the Baltimore Police Department. He also speeded up land acquisition and construction of the Jones Falls Expressway and the Baltimore Civic Center, later renamed the Baltimore Arena.
Judge Grady left City Hall in November 1962, three years into his term, when he was appointed to the Supreme Bench of Baltimore, forerunner of the Circuit Court.
"He was possibly the smartest mayor I ever worked with - yet he was uncomfortable with the job - and became the best judge I've ever known," said state Comptroller William Donald Schaefer, former mayor and governor.
"He was used to being a state's attorney and not dealing with politicians who came into his office with their hats on and put their feet up on the desk. He took one look at that and said, 'That's not for me,'" said Mr. Schaefer.
"Harold was a very nice and decent guy who was in a better position as a judge rather than in elective office. The political pressures which swirled around him were not suited for his personality," said Walter Sondheim, now senior adviser to the Greater Baltimore Committee.
Former Gov. Marvin Mandel, who was in law school at the University of Maryland with Judge Grady, said: "He was very competent as a judge and did an excellent job throughout his career. He was courteous and kind and at the same time very effective. He made great contributions to Baltimore and the state."
"He was a bright guy who always played it down. He was even-tempered, very well-liked and respected as a judge," said Elsbeth L. Bothe, retired Baltimore Circuit Court judge. "He preferred civil cases for the most part because they were more gentlemanly."
Born in Williamsport, Pa., he was a 1934 graduate of Forest Park High School and graduated from Loyola College in 1938.
He received his commission as a reserve officer in the Navy in 1941, and the next year entered the U.S. Department of Justice after graduating from law school.
Judge Grady was a special agent for the Federal Bureau of Investigation from 1942 until 1946, when he resigned to practice law in Baltimore. He was appointed assistant state's attorney for Baltimore in 1947 and deputy state's attorney in 1955. He was named state's attorney the next year, to complete Judge Sodaro's term, and elected to that post in 1958, the year before the mayoral race.
After retiring as chief judge of the city Circuit Court in 1984, Judge Grady became a partner in the Baltimore law firm of Siskind, Grady, Rosen & Hoover, and continued to go to his office at 2 E. Fayette St. until late last year, when he retired.
He was married in 1942 to Patricia Grogan, who died in 1994.
Judge Grady was a communicant of the Roman Catholic Cathedral of Mary Our Queen, 5200 N. Charles St., where a Mass of Christian burial will be offered at 10 a.m. Saturday.
He is survived by two sons, Joseph H. Grady Jr. of Saratoga, Calif., and Thomas L. Grady of Baltimore; two daughters, Maureen Callahan of Point Pleasant, N.J., and Kathleen Ann Donovan of Timonium; two sisters, Marianna Davis of Lutherville and Maxine Eagan of Towson; and five grandchildren. | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 360 |
Q: Do JSR-286 portlets require a web.xml file in their WAR files? Does the JSR 286 spec require the presence of a web.xml file in WARs containing portlets? At first, I thought so but then I created a portlet without a web.xml, deployed it in Liferay and it worked flawlessly. So is it an extension (or a bug) of Liferay, or is it not necessary to have such a file?
A: I'd really have to dig in the spec - but my assumption is that it's following the servlet spec in this regard: A portlet app is first and foremost deployed to an application server. If the servlet spec requires a web.xml to be there, this requirement would need to be fulfilled. Otherwise the Appserver would not deploy the WAR (it's more or less the question if the WAR file format requires or recommends web.xml
Liferay will only kick in once the Apperver has deployed a web application. Liferay itself doesn't care for web.xml
Based on this arguing (and without looking at the spec - so I'm giving an educated guess here) I expect JSR 286 to not make a statement about web.xml. However, it probably references the servlet spec and this in turn might require/recommend/mention web.xml.
And if the Appservers require it or how they behave if it's missing is yet another story.
A: As Olaf rightly said portlet is nothing but a web application. Liferay has a listener that gets triggered when the portlet auto deploys. It explodes the war and adds web.xml and the content that is necessary. You can check the logic if you have source. The class name is PortletDeployer and the method is getServletContent. After add the web.xml and stuffing it, they just touch it using FileUtil.touch.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 6,118 |
The PrePlay Before Four Play. Join us Next Saturday as we drop this as a Pre-Episode to Fourplay. Sip Sip Caution it is HOT! PrePlay before Four Play is about the intent behind the lyrics inspired by today's music.
The women of Tea Time get together on opposing sides to talk about a subject that is way too popular amongst our women... Are "broke" men dateable?" Should the "successful" woman turn him down? Join us as we sip at the table for the great debate. Sip! Sip! | {
"redpajama_set_name": "RedPajamaC4"
} | 4,613 |
Q: Nucleon current from quark current For free neutron decay we consider the effective current-current Lagrangian of quarks and leptons:
\begin{equation}
\mathcal{L}=\frac{G_F\cos{\theta_c}}{2\sqrt{2}}[\overline{u}\gamma^\mu(1-\gamma^5)d][\overline{e}\gamma_\mu(1-\gamma^5)\nu_e], \tag{1}\label{Lang}
\end{equation}
where $u$, $d$, $e$, $\nu_e$ are the quantum field operators of up-quark, down-quark, electron and electron-neutrino respectively. Obviously the tree-level hadronic current $\overline{u}\gamma^\mu(1-\gamma^5)d$ has to be converted to nucleon current. The most general nucleon current written in terms of form-factors is given by,
\begin{align}
\overline{u}\gamma^\mu(1-\gamma^5)d \rightarrow \overline{\Psi}\tau^+\Big[g_V(q^2)\gamma^\mu+ig_M(q^2)\sigma^{\mu\nu}q_\nu-g_A(q^2)\gamma^\mu\gamma^5+g_P(q^2)q^\mu\gamma^5 \Big]\Psi,\tag{2}\label{nucl-curr}
\end{align}
where $\Psi^T= (\psi_p, \psi_n)$ is the iso-doublet nucleon field in terms protons and neutrons. $\tau^+$ is raising operator in isospin space. The various momentum dependent form-factors $g_V$, $g_M$, $g_A$ and $g_P$ are defined as the vector, weak magnetism, axial-vector and induced pseudoscalar, respectively. The four-momentum vector $q^\mu=p_f^\mu-p^\mu_i$ is the momentum transfer between initial and final nucleon.
My question is, how do we get the nucleon current from the basic quark current? Can anyone suggest a pedagogical reference of how to get \eqref{nucl-curr}? All I have found is heuristic arguments of how to construct a Lorentz-vector like current from available Lorentz vectors $\gamma^\mu$, $q^\mu$ etc using general symmetry considerations. I know the treatment is quite involved, but any suggestions would be appreciated.
A: A good question, meaning, sadly, that there is no good answer. The answer is (properly) either in hyper-technical lattice gauge theory, so not pedagogical. Or, else books and reviews of the 1970s, after the triumph of the quark model and the acceptance in principle of QCD, but before the industry of correction computation/estimation got its start: the golden age of hadronization modeling.
As you indicate, in the absence of good quark wave functions for nucleons, all one has is symmetry: Lorentz, parity, CP, flavor (isospin), etc... and systematic tabulation of matrix elements between hadron states of short distance QCD operators, correlating them among dozens and dozens of processes, with a few models/aids such as PCAC, the Goldberger-Treiman relation, etc. Shining examples of these are two books now sometimes sniffed at as dated, which you might, or might not, find in your library:
*
*Weak Interactions of Leptons and Quarks,
by Eugene D. Commins & Philip H. Bucksbaum (Cambridge University Press, 1983)
ISBN-13 : 978-1423446637
*Elementary Particles and Their Currents, by Jeremy Bernstein (W. H. Freeman and Co., 1968) ASIN : B0000COC6N
and old QFT texts from the early 80s, expediting the cultural transition to the SM. Typically, neutral current and neutrino scattering reviews, by necessity, had to review the stuff. In any case, what was part of the street slang of that generation is rapidly getting lost, and I am hard pressed to find systematic dictionaries for it.
Today, the absolutely best starting point is
*
*Dynamics of the Standard Model, by Donoghue, Golowich & Holstein, (Cambridge University Press; 2nd edition, 2014) ISBN-13: 978-0521768672
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 7,013 |
Q: How to limit file size or number of files in multer express and handle error? I am writing a code where I need to limit the file size or limit the number of files uploaded.
For example, a user can upload maximum 100 files at once or user can upload files up to the size of 100 mb. But when I run it and select 7 files of 10 mb each it gives me this error:
Also how can I handle error if any of the limit exceeds when user uploads?
Below is my code:
var maxSize = 100 * 1024 * 1024;
const express = require("express")
const multer = require('multer');
const csvtojson = require('csvtojson');
var storage = multer.diskStorage({
destination: function (req, file, cb) {
cb(null, './upload') //Destination folder
},
filename: function (req, file, cb) {
cb(null, file.originalname) //File name after saving
}
})
const upload = multer({ storage: storage ,limits: { fileSize: maxSize }});
const router = express.Router()
router.post("/multiple",upload.array("ziptable",100),(req,res)=>{
var file = req.files;
console.log(file)
res.send("ok")
})
my html:
<form action="/multiple" method="POST" enctype="multipart/form-data">
<div id="middle">
<input id="uploadzip" name="ziptable" type="file" multiple>
<button id="submit" type="submit" class="btn btn-primary">SELECT FILES</button>
</div>
</form>
A: For now, there is no functionality like allFileSizeLimitto handle this issue.
Maybe you can write some custom code in the fileFilter function.
Find the sum of the file sizes and if that total size exceeds your limit you can throw an custom error.
In addition, you can write custom form submitter in the frontend javascript and send the request to server only if the total size is lesser than your limit. But this javascript can be changed by the browser console so you have to write the above logic in server.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 1,606 |
package com.faforever.client.game;
import com.faforever.client.fx.Controller;
import com.faforever.client.fx.StringCell;
import com.faforever.client.i18n.I18n;
import com.faforever.client.map.MapService;
import com.faforever.client.map.MapServiceImpl.PreviewSize;
import com.faforever.client.preferences.PreferencesService;
import com.faforever.client.remote.domain.RatingRange;
import com.faforever.client.theme.UiService;
import com.google.common.base.Joiner;
import javafx.application.Platform;
import javafx.beans.Observable;
import javafx.beans.binding.Bindings;
import javafx.beans.property.ObjectProperty;
import javafx.beans.property.SimpleObjectProperty;
import javafx.beans.property.SimpleStringProperty;
import javafx.beans.value.ObservableValue;
import javafx.collections.ObservableList;
import javafx.collections.transformation.SortedList;
import javafx.geometry.Pos;
import javafx.scene.Node;
import javafx.scene.control.Label;
import javafx.scene.control.SortEvent;
import javafx.scene.control.TableCell;
import javafx.scene.control.TableColumn;
import javafx.scene.control.TableColumn.CellDataFeatures;
import javafx.scene.control.TableColumn.SortType;
import javafx.scene.control.TableRow;
import javafx.scene.control.TableView;
import javafx.scene.image.Image;
import javafx.util.Pair;
import org.jetbrains.annotations.NotNull;
import org.springframework.beans.factory.config.ConfigurableBeanFactory;
import org.springframework.context.annotation.Scope;
import org.springframework.stereotype.Component;
import javax.inject.Inject;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;
import java.util.stream.Collectors;
@Scope(ConfigurableBeanFactory.SCOPE_PROTOTYPE)
@Component
public class GamesTableController implements Controller<Node> {
private final ObjectProperty<Game> selectedGame;
private final MapService mapService;
private final JoinGameHelper joinGameHelper;
private final I18n i18n;
private final UiService uiService;
private final PreferencesService preferencesService;
public TableView<Game> gamesTable;
public TableColumn<Game, Image> mapPreviewColumn;
public TableColumn<Game, String> gameTitleColumn;
public TableColumn<Game, PlayerFill> playersColumn;
public TableColumn<Game, RatingRange> ratingColumn;
public TableColumn<Game, String> modsColumn;
public TableColumn<Game, String> hostColumn;
public TableColumn<Game, Boolean> passwordProtectionColumn;
@Inject
public GamesTableController(MapService mapService, JoinGameHelper joinGameHelper, I18n i18n, UiService uiService, PreferencesService preferencesService) {
this.mapService = mapService;
this.joinGameHelper = joinGameHelper;
this.i18n = i18n;
this.uiService = uiService;
this.preferencesService = preferencesService;
this.selectedGame = new SimpleObjectProperty<>();
}
public ObjectProperty<Game> selectedGameProperty() {
return selectedGame;
}
public Node getRoot() {
return gamesTable;
}
public void initializeGameTable(ObservableList<Game> games) {
SortedList<Game> sortedList = new SortedList<>(games);
sortedList.comparatorProperty().bind(gamesTable.comparatorProperty());
gamesTable.setPlaceholder(new Label(i18n.get("games.noGamesAvailable")));
gamesTable.setRowFactory(param1 -> gamesRowFactory());
gamesTable.setItems(sortedList);
applyLastSorting(gamesTable);
gamesTable.setOnSort(this::onColumnSorted);
sortedList.addListener((Observable observable) -> selectFirstGame());
selectFirstGame();
passwordProtectionColumn.setCellValueFactory(param -> param.getValue().passwordProtectedProperty());
passwordProtectionColumn.setCellFactory(param -> passwordIndicatorColumn());
mapPreviewColumn.setCellFactory(param -> new MapPreviewTableCell(uiService));
mapPreviewColumn.setCellValueFactory(param -> Bindings.createObjectBinding(
() -> mapService.loadPreview(param.getValue().getMapFolderName(), PreviewSize.SMALL),
param.getValue().mapFolderNameProperty()
));
gameTitleColumn.setCellValueFactory(param -> param.getValue().titleProperty());
gameTitleColumn.setCellFactory(param -> new StringCell<>(title -> title));
playersColumn.setCellValueFactory(param -> Bindings.createObjectBinding(
() -> new PlayerFill(param.getValue().getNumPlayers(), param.getValue().getMaxPlayers()),
param.getValue().numPlayersProperty(), param.getValue().maxPlayersProperty())
);
playersColumn.setCellFactory(param -> playersCell());
ratingColumn.setCellValueFactory(param -> new SimpleObjectProperty<>(new RatingRange(param.getValue().getMinRating(), param.getValue().getMaxRating())));
ratingColumn.setCellFactory(param -> ratingTableCell());
hostColumn.setCellValueFactory(param -> param.getValue().hostProperty());
hostColumn.setCellFactory(param -> new StringCell<>(String::toString));
modsColumn.setCellValueFactory(this::modCell);
modsColumn.setCellFactory(param -> new StringCell<>(String::toString));
gamesTable.getSelectionModel().selectedItemProperty().addListener((observable, oldValue, newValue)
-> Platform.runLater(() -> selectedGame.set(newValue)));
}
public void setModsColumnVisibility(boolean isVisible) {
modsColumn.setVisible(isVisible);
}
public void setPasswordProtectionColumnVisibility(boolean isVisible) {
passwordProtectionColumn.setVisible(isVisible);
}
private void applyLastSorting(TableView<Game> gamesTable) {
final Map<String, SortType> lookup = new HashMap<>();
final ObservableList<TableColumn<Game, ?>> sortOrder = gamesTable.getSortOrder();
preferencesService.getPreferences().getGameListSorting().forEach(sorting -> lookup.put(sorting.getKey(), sorting.getValue()));
sortOrder.clear();
gamesTable.getColumns().forEach(gameTableColumn -> {
if (lookup.containsKey(gameTableColumn.getId())) {
gameTableColumn.setSortType(lookup.get(gameTableColumn.getId()));
sortOrder.add(gameTableColumn);
}
});
}
private void onColumnSorted(@NotNull SortEvent<TableView<Game>> event) {
List<Pair<String, SortType>> sorters = event.getSource().getSortOrder()
.stream()
.map(column -> new Pair<>(column.getId(), column.getSortType()))
.collect(Collectors.toList());
preferencesService.getPreferences().getGameListSorting().setAll(sorters);
preferencesService.storeInBackground();
}
@NotNull
private ObservableValue<String> modCell(CellDataFeatures<Game, String> param) {
int simModCount = param.getValue().getSimMods().size();
List<String> modNames = param.getValue().getSimMods().entrySet().stream()
.limit(2)
.map(Entry::getValue)
.collect(Collectors.toList());
if (simModCount > 2) {
return new SimpleStringProperty(i18n.get("game.mods.twoAndMore", modNames.get(0), modNames.size()));
}
return new SimpleStringProperty(Joiner.on(i18n.get("textSeparator")).join(modNames));
}
private void selectFirstGame() {
TableView.TableViewSelectionModel<Game> selectionModel = gamesTable.getSelectionModel();
if (selectionModel.getSelectedItem() == null && !gamesTable.getItems().isEmpty()) {
Platform.runLater(() -> selectionModel.select(0));
}
}
@NotNull
private TableRow<Game> gamesRowFactory() {
TableRow<Game> row = new TableRow<>();
row.setOnMouseClicked(event -> {
if (event.getClickCount() == 2) {
Game game = row.getItem();
joinGameHelper.join(game);
}
});
return row;
}
private TableCell<Game, Boolean> passwordIndicatorColumn() {
return new StringCell<>(
isPasswordProtected -> isPasswordProtected ? i18n.get("game.protected.symbol") : "",
Pos.CENTER, UiService.CSS_CLASS_ICON);
}
private TableCell<Game, PlayerFill> playersCell() {
return new StringCell<>(playerFill -> i18n.get("game.players.format",
playerFill.getPlayers(), playerFill.getMaxPlayers()), Pos.CENTER);
}
private TableCell<Game, RatingRange> ratingTableCell() {
return new StringCell<>(ratingRange -> {
if (ratingRange.getMin() == null && ratingRange.getMax() == null) {
return "";
}
if (ratingRange.getMin() != null && ratingRange.getMax() != null) {
return i18n.get("game.ratingFormat.minMax", ratingRange.getMin(), ratingRange.getMax());
}
if (ratingRange.getMin() != null) {
return i18n.get("game.ratingFormat.minOnly", ratingRange.getMin());
}
return i18n.get("game.ratingFormat.maxOnly", ratingRange.getMax());
}, Pos.CENTER);
}
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 3,902 |
\section{Introduction}
The Visual and Infrared survey Telescope for Astronomy (VISTA, \citealt{Eme06})
survey of the Magellanic Clouds system (VMC, \citealt{Cio11})
is collecting $Y, J, K_\mathrm{s}$ photometry\footnote{The VMC system is similar to the Vega magnitude system (see, e.g., \citealt{Rub12}).}
down to $Y$=21.9 mag, $J$=22.0 mag, $K_\mathrm{s}$=21.5 mag
(5 $\sigma$ level on the stacked images)
for sources in the Large Magellanic Cloud (LMC),
the Small Magellanic Cloud (SMC), the Bridge connecting them, and a small part of the Magellanic Stream.
The VMC images are processed by the Cambridge Astronomical Survey Unit
(CASU\footnote{http://casu.ast.cam.ac.uk/}; \citealt{Lew10}) through the VISTA Data Flow System
(VDFS) pipeline that performs aperture photometry of the images.
The reduced data are then further processed by the Wide Field
Astronomy Unit (WFAU)\footnote{http://horus.roe.ac.uk/vsa} in Edinburgh where the single epochs are
stacked and catalogued in the VISTA Science Archive (VSA; \citealt{Cro12}).
As of 2016 January, the VMC survey observations are 74\% complete, with specific completion levels
of 62\%, 95\%, 95\% and 100\%
for the LMC, the SMC, the Bridge and the Stream, respectively.
A main aim of the VMC survey is to study the structure of the whole Magellanic system using different distance indicators.
Among them, most notably, are primary standard candles such as the Cepheids and the RR Lyrae stars, for which VMC is obtaining
$K_\mathrm{s}$-band light curves with 13 (or more) individual epochs, with most epochs reaching a depth of $K_\mathrm{s}$=19.3 mag (5 $\sigma$ level).
The significantly reduced amplitude of these pulsating variables in the near-infrared, with respect to the optical bands, along with the multi-epoch cadence
of the VMC $ K_\mathrm{s}$-band observations, allows us
to infer accurate mean $K_\mathrm{s}$ magnitudes. Furthermore, the near-infrared period-luminosity ($PL$) relations are intrinsically much narrower than the
corresponding optical relations and less affected by systematic uncertainties in reddening and metal content \citep{Cap00}.
For these reasons the VMC data are very well suited to construct $PL$ relations with a high level of precision
and accuracy (e.g. \citealt{Rip12a,Rip12b,Rip14,Rip15};
\citealt{Mur15}). Conversely, the smaller amplitudes compared to optical bands make it more
difficult to identify these variables from near-infrared data alone.
In VMC, most of the information (identification, period, variability type, etc.) needed to fold the
$Y, J, K_\mathrm{s}$-band light curves and derive average $Y, J, K_\mathrm{s}$ magnitudes
for the variable stars are taken from the catalogues of Magellanic Cloud variables produced by
large microlensing optical surveys such as MACHO \citep{Alc00}, EROS \citep{Tis07} and OGLE \citep{Sos08a} that were conducted
in the last two decades to search for baryonic dark matter in the Milky Way.
As described in \cite{Uda15}, the observations of the fourth phase of the OGLE project (hereinafter,
OGLE~IV) have been successfully run over the last five years. First results have been published in \cite{Sos12} (South Ecliptic Pole region),
\cite{Koz13} (Magellanic Bridge) and \cite{Sos15a} (anomalous Cepheids).\footnote{During the revision phase of this manuscript,
results on classical Cepheids in the Magellanic Clouds have been published by \cite{Sos15b}. See Section~\ref{sec:using}.}
\begin{table*}
\caption{Information on the $K_\mathrm{s}$-band photometry of the VMC tiles analyzed in this paper.
Col. 1: Field and tile number, Col. 2: Right Ascension (RA, J2000), Col. 3: Declination (Dec, J2000), Col. 4: position
angle, Col. 5: number of $K_\mathrm{s}$ observations, Cols. 6,7: Dates of the 1st and
last observations expressed in dd/mm/yy format, Col. 8: Time interval (T.I.) between first and last observations,
Col. 9: Airmass, Col. 10: Full Width at Half Maximum (FWHM),
Col. 11: Ellipticity, Col. 12: Limiting magnitude.
Reported values of airmass, FWHM, ellipticity and limiting magnitude are averages over all nights.
They are indicated with their respective standard deviations.
The number of epochs ($N_{K_\mathrm{s}}$) includes
both shallow and deep observations (see text for details).}
\scriptsize
\label{tab:qc}
\[
\begin{array}{cccccccrcccc}
\hline \hline
\noalign{\smallskip}
\mathrm{Tile} & \alpha & \delta & \mathrm{\phi} & N_{K_\mathrm{s}} & \mathrm{1st\,Epoch} &
\mathrm{Last\,Epoch} & \mathrm{T.\,I.} & \mathrm{Airmass} & \mathrm{FWHM}
& e & \mathrm{LimMag}\\
& \mathrm{(h:m:s)} & (^\circ \mathrm{:}^\prime\mathrm{:}^{\prime\prime}) &
\mathrm{(deg)} & & & & \mathrm{(d)} & & & & \mathrm{(mag)}\\
\noalign{\smallskip}
\hline
\noalign{\smallskip}
\mathrm{LMC\,6\_8} & 06\mathrm{:}02\mathrm{:}21.984 &
-69\mathrm{:}14\mathrm{:}42.360 & -83.7904 & 15 & 23/02/11 & 16/04/14 & 1148 &
1.49\pm0.04 & 0.96\pm0.19 & 0.06\pm0.01 & 19.30\pm0.24\\
\noalign{\smallskip}
\mathrm{LMC\,7\_3} & 05\mathrm{:}02\mathrm{:}55.200 &
-67\mathrm{:}42\mathrm{:}14.760 & -97.7044 & 16 & 12/01/11 & 04/03/13 & 782 &
1.44\pm0.05 & 0.95\pm0.10 & 0.06\pm 0.02 & 19.23\pm0.19\\
\noalign{\smallskip}
\mathrm{LMC\,8\_8} & 05\mathrm{:}59\mathrm{:}23.136 &
-66\mathrm{:}20\mathrm{:}28.680 & -84.4802 & 16 & 14/11/09 & 26/11/10 & 377 &
1.39\pm0.05 & 0.93\pm0.11 & 0.06\pm0.02 & 19.36\pm0.27\\
\noalign{\smallskip}
& & & & & & & & & & &\\
\mathrm{SMC\,3\_1} & 00\mathrm{:}02\mathrm{:}39.912 &
-73\mathrm{:}53\mathrm{:}31.920 & -11.3123 & 16 & 21/08/12 & 01/07/14 & 679 &
1.60\pm 0.05 & 1.01\pm 0.15 & 0.05\pm 0.01 & 19.38\pm0.16\\
\noalign{\smallskip}
\mathrm{SMC\,3\_3} & 00\mathrm{:}44\mathrm{:}55.896 &
-74\mathrm{:}12\mathrm{:}42.120 & -1.2120 & 18 & 03/08/11 & 05/09/12 & 399 &
1.62\pm 0.04 & 1.02\pm 0.10 & 0.07\pm 0.02 & 19.35\pm0.15\\
\noalign{\smallskip}
\mathrm{SMC\,3\_5} & 01\mathrm{:}27\mathrm{:}30.816 &
-74\mathrm{:}00\mathrm{:}49.320 & +8.9671 & 16 & 29/07/11 & 07/10/12 & 436 &
1.58\pm 0.06 & 1.02\pm 0.15 & 0.06\pm 0.01 & 19.34\pm0.22\\
\noalign{\smallskip}
\mathrm{SMC\,4\_2} & 00\mathrm{:}25\mathrm{:}14.088 &
-73\mathrm{:}01\mathrm{:}47.640 & -5.9198 & 15 & 18/10/12 & 23/06/14 & 606 &
1.57\pm 0.06 & 0.94\pm 0.08 & 0.05\pm 0.01 & 19.29\pm0.21\\
\noalign{\smallskip}
\mathrm{SMC\,4\_3} & 00\mathrm{:}45\mathrm{:}14.688 &
-73\mathrm{:}07\mathrm{:}11.280 & -1.1369 & 16 & 29/08/11 & 18/08/13 & 720 &
1.56\pm 0.05 & 0.92\pm 0.10 & 0.07\pm 0.01 & 19.27\pm0.17\\
\noalign{\smallskip}
\mathrm{SMC\,4\_4} & 01\mathrm{:}05\mathrm{:}19.272 &
-73\mathrm{:}05\mathrm{:}15.360 & +3.6627 & 15 & 25/09/12 & 14/07/14 & 648 &
1.56\pm 0.06 & 0.87\pm 0.06 & 0.07\pm 0.02 & 19.27\pm0.16\\
\noalign{\smallskip}
\mathrm{SMC\,4\_5} & 01\mathrm{:}25\mathrm{:}11.016 &
-72\mathrm{:}56\mathrm{:}02.040 & +8.4087 & 18 & 29/08/11 & 25/11/13 &819 &
1.56\pm 0.06 & 0.92\pm 0.17 & 0.06\pm 0.01 & 19.42\pm0.19\\
\noalign{\smallskip}
\mathrm{SMC\,5\_2} & 00\mathrm{:}26\mathrm{:}41.688 &
-71\mathrm{:}56\mathrm{:}35.880 & -5.5717 & 17 & 09/06/11 & 10/11/12 & 520 &
1.53\pm 0.06 & 0.99\pm 0.18 & 0.07\pm 0.01 & 19.24\pm0.27\\
\noalign{\smallskip}
\mathrm{SMC\,5\_3} & 00\mathrm{:}44\mathrm{:}49.032 &
-72\mathrm{:}01\mathrm{:}36.120 & -1.2392 & 19 & 04/10/12 & 08/08/14 & 673 &
1.54\pm 0.07 & 0.95\pm 0.12 & 0.06\pm 0.01 & 19.31\pm0.15\\
\noalign{\smallskip}
\mathrm{SMC\,5\_4} & 01\mathrm{:}04\mathrm{:}26.112 &
-71\mathrm{:}59\mathrm{:}51.000 & +3.4514 & 18 & 26/10/10 & 23/10/12 & 728 &
1.58\pm 0.08 & 0.92\pm 0.11 & 0.07\pm 0.01 & 19.35\pm0.15\\
\noalign{\smallskip}
\mathrm{SMC\,5\_6} & 01\mathrm{:}41\mathrm{:}28.800 &
-71\mathrm{:}35\mathrm{:}47.040 & 12.3004 & 18 & 30/09/11 & 11/09/14 & 1077 &
1.58\mathrm{\pm}0.08 & 1.0\mathrm{\pm} 0.2 & 0.05\mathrm{\pm} 0.01 & 19.15\pm0.65\\
\noalign{\smallskip}
\mathrm{SMC\,6\_3} & 00\mathrm{:}45\mathrm{:}48.768 &
-70\mathrm{:}56\mathrm{:}08.160 & -1.0016 & 14 & 20/08/11 & 06/11/13 & 809 &
1.52\pm 0.06 & 0.95\pm 0.12 & 0.05\pm 0.01 & 19.39\pm0.22\\
\noalign{\smallskip}
\mathrm{SMC\,6\_5} & 01\mathrm{:}21\mathrm{:}22.488 &
-70\mathrm{:}46\mathrm{:}10.920 & +7.5039 & 14 & 06/09/11 & 18/10/13 & 773 &
1.49\pm 0.03 & 0.97\pm 0.17 & 0.05\pm 0.01 & 19.42\pm0.17\\
\noalign{\smallskip}
\hline
\hline
\end{array}
\]
\end{table*}
However, as we are writing this paper, none of the available optical catalogues cover
the field of view of VMC entirely (see Figure 4 of \citealt{Mor14}). There are external regions in both the LMC and SMC, and the whole
Bridge area where a comprehensive census of all types of variable stars is still missing.
On the other hand, in VMC each field is observed in the $K_\mathrm{s}$-band at least 13 times~\citep{Cio11}:
11 times with an exposure of 750s (deep observations, hereafter) and twice with shorter exposures of 375s each (shallow observations, hereafter)
obtained over more than one year time-span. Furthermore, the VSA provides a list of flags related to the analysis of the light
curves of each VMC source \citep{Cro12}, that can be used to identify variable sources.
In this paper we show that, in spite of the
small amplitudes in the near-infrared, pulsating variable stars can be effectively detected from the VMC data
if a proper analysis is performed. Specifically, we combine
the information from: i) the ($J-K_\mathrm{s}$, $K_\mathrm{s}$) colour-magnitude and the ($Y-J$, $Y-K_\mathrm{s}$)
colour-colour diagrams, ii) the analysis of the $K_\mathrm{s}$-band light curves,
iii) the Period-Luminosity relations,
and iv) the VSA flags, to devise a procedure and detect variable stars in the external regions of the Magellanic system
using only the VMC data.
This paper is organised as follows. The reliability of the VSA flags in detecting variable stars is presented in Section 2.
Section 3 describes the method to identify variable stars in the SMC where information is available from the OGLE III survey.
In Section 4 we apply our method to identify classical Cepheids (CCs)
in external regions of the SMC
where the VMC observations are complete and no other CC variable star catalogue is currently available.
Our results are summarized in Section 5.
\section{Identification of variable stars using the VSA flags}
We have tested the ability of the VSA in
identifying variable stars
based solely on the VMC data in three most external and completely observed tiles in the LMC, namely,
tile LMC 6\_8, 7\_3 and 8\_8. The upper portion of Table~\ref{tab:qc} provides centre
coordinates of these 3 LMC tiles, together with information related to their observations.
We specifically selected these three tiles because they contain a large number of variable stars of different types
and are located in external fields of the LMC with similar level of crowding as in the regions
where we currently have VMC data, but there is no coverage by the optical surveys.
We first cross-matched the OGLE catalogues (\citealt{Sos08a,Sos08b,Sos09a,Sos09b,Sos12};
\citealt{Pol10}; \citealt{Gra11}) of variable stars in these fields against the VMC catalogue
of the corresponding three tiles. We adopted a pairing radius of 0.5 arcsec in order to maximize the
reliability of the matching procedure as discussed in \cite{Rip15}.
There are 7206 variable stars in common between the two catalogues; in particular we have 94 $\delta$ Scuti (DSCT) stars,
1071 RR Lyrae (RRLYR) stars, 1521 Eclipsing binaries (ECL), 217 CCs, 4289 Long Period Variables (among wich are 83 Mira, 3626 OGLE Small
Amplitude Red Giants -- OSARGs, and 580 SemiRegular Variables -- SRVs), 6 Type II Cepheids (T2Ceps) and 8 Anomalous Cepheids (ACs).
For all of them we investigated the Variability classification provided by the VSA; this is summarized by the
{\it ``Variability Class''} parameter which is the result of the analysis performed by the WFAU team on the
$Y$, $J$ and $K_\mathrm{s}$-band light curves of the sources. {\it ``Variability Class''} is either 1 or 0, whether or not
the object is classified as variable (\citealt{Cro09,Cro12}). Table~\ref{tab:var} compares the number of
OGLE variable stars cross-matched with VMC sources in the aforementioned tiles
and the number of them classified as variables by the VSA ({\it VariableClass}=1).
The effectiveness of the VSA in selecting variable stars depends
on the mean magnitude, period and amplitude of the light variation and, in turn,
on the type of variability.
Indeed, according to the numbers in Table~\ref{tab:var} the type of variable stars most easily classified as such by
the VSA are the LPV-Mira, ACs, LPV-SRV, T2Cep's and CCs. Note that the sample of T2Ceps and of ACs are not statistically representative.
In this paper we focus on the detection of Cepheids, CCs in particular, because they are numerous
and will give us a very good statistical sample to work with. We leave for a future paper the analysis of other types of variables.
CCs typically have single-epoch photometric errors in the $K_\mathrm{s}$-band of about 0.01 mag \citep{Rip12b,Mor14},
typical $K_\mathrm{s}$ amplitudes between 0.05 and 0.38 mag and periods between $\sim$0.5 d and tens of days.
The search for CCs was performed in regions of the Magellanic system where we expect this kind of variable stars to be present and
the VMC observations are complete, but for which no optical data are currently available.
In particular, we focused on the external regions of the SMC (see Section~\ref{sec:using}).
\begin{table}
\caption{VSA classification of variable stars within VMC tiles LMC 6\_8, 7\_3 and 8\_8.
Col. 1: Type of variable stars; Col. 2:
number of OGLE~III stars with a VMC counterpart within 0.5 arcsec; Col. 3: number of stars classified as
variables by the VSA; Col. 4: percentage value. See text for details.}
\label{tab:var}
\begin{tabular}{lrrr}
\hline
\hline
\noalign{\smallskip}
Type & $N_{\rm OGLE}$ & $N_{\rm VSA}$ & $N$\% \\
\noalign{\smallskip}
\hline
\noalign{\smallskip}
\noalign{\smallskip}
DSCT & 94 & 1 & 1\% \\
\noalign{\smallskip}
RRLYR & 1071 & 28 & 3\% \\
\noalign{\smallskip}
ECL & 1521 & 132 & 9\% \\
\noalign{\smallskip}
LPV-OSARG & 3626 & 576 & 16\% \\
\noalign{\smallskip}
CC & 217 & 137 & 63\% \\
\noalign{\smallskip}
T2Cep & 6 & 4 & 67\% \\
\noalign{\smallskip}
LPV-SRV & 580 & 477 & 82\% \\
\noalign{\smallskip}
AC & 8 & 7 & 87\% \\
\noalign{\smallskip}
LPV-Mira & 83 & 75 & 90\% \\
\hline
\end{tabular}
\\
\end{table}
\section{Defining the method to detect classical Cepheids}\label{sec:CCSMC}
In order to specifically optimize the
detection of CCs, we first analyzed the properties of the VMC data for variable stars in the SMC
classified as CCs by OGLE~III \citep{Sos10a}. These authors identified 4630 CCs in the SMC, of which 2626
are fundamental-mode (F), 1644 are first-overtone (1O), 83 are second-overtone (2O), 274 are double-mode
(59 F/1O and 215 1O/2O) and three are triple-mode CCs.
The mean $I$-band magnitude of these CCs ranges from $\sim$10 mag to $\sim$19 mag, with two major
peaks at $I\sim$16.8 mag and $I\sim$15.8 mag.
We have used the OGLE~III coordinates of the SMC CCs as reference in the following analysis.
Figure~\ref{fig:SMCtiles} shows the distribution of the OGLE~III footprint (red contours)
and the VMC tiles (black rectangles) in the SMC area.
\begin{figure}
\begin{centering}
\includegraphics[width=8cm]{f1.ps}
\caption{Celestial distribution of OGLE~III footprint (red contours) and VMC tiles (black rectangles) in the SMC area.
$X$ and $Y$ are defined as in van der Marel \& Cioni (2001) with $\alpha_0=12.5$ deg and $\delta_0=-73.0$ deg.
Tiles analysed in this work are labelled with their VMC IDs. The grey oval shows
the area containing the peak density of the SMC CCs which was avoided in the present analysis.}
\label{fig:SMCtiles}
\end{centering}
\end{figure}
The SMC tiles analysed in this work are the ones for which the VMC observations were completed before 2014 September 30.
They are tiles SMC 3\_3, 3\_5, 4\_2, 4\_3, 4\_4,
4\_5, 5\_2, 5\_3, 5\_4, 6\_3 and 6\_5, labelled with their IDs in Fig.~\ref{fig:SMCtiles}.
The lower portion of Table~\ref{tab:qc} lists centre coordinates and observation properties
for all of them.
OGLE~III observations cover the most central and crowded regions of the SMC.
In particular, the peak in stellar density (see Figure 1 of~\citealt{Rub15})
and CC density (see right panel of Figure 18 in~\citealt{Mor14}) in the SMC occurs
in an area centered at coordinates RA=12 deg, Dec=$-$73.1 deg and radius of 2
deg in RA and 0.5 deg in Dec. This region, marked by a grey oval in Fig.~\ref{fig:SMCtiles},
mainly covers the tile SMC 4\_3 and contains 1695
of the 4630 CCs identified by the OGLE~III survey in the SMC.
However, as we aim at fine-tuning our procedures to identify new SMC CCs outside the OGLE~III field (see Section 4), we considered in the following
analysis only CCs lying in the external portions of the OGLE~III footprint, where the level of crowding and reddening is similar to what we found
in the regions which currently have only VMC data (see Fig.~\ref{fig:SMCtiles}). Specifically, we considered 2935 CCs located outside the oval shape in Fig.~\ref{fig:SMCtiles}.
By matching the OGLE~III and VMC catalogues, using a pairing radius of 0.5 arcsec
we obtained a sample of 2411 CCs that have VMC $K_\mathrm{s}$-band light curves.
Increasing the paring radius to 1.0 arcsec would result in
increasing by about 1\% the number of misidentifications. Conversely, reducing the paring radius to
0.1 arcsec, would result in loosing about 10\% of the crossmatched sources. On the other hand, the
reliability of our cross-identifications and, in turn, of the value adopted for the paring radius, is
specifically assessed by folding the VMC $K_\mathrm{s}$-band light curves according to the corresponding
OGLE III periods (e.g. \citealt{Rip15}).
These CCs lie in the
tiles (see Fig.~\ref{fig:SMCtiles}): SMC 3\_3, 3\_5, 4\_2, 4\_3 and 4\_4 (excluding sources inside the oval grey contour), 4\_5, 5\_2 (two sources),
5\_3, 5\_4, 6\_3 (four sources) and 6\_5 (two sources).
\subsection{Selection based on the colour-magnitude and colour-colour diagrams}\label{sec:cmd-cc-sel}
We used the sample of 2411 OGLE~III CCs described in the previous section as a reference to define the range in VMC colours and magnitudes in which the SMC CCs lie.
Figure~\ref{fig:cmdKJK} shows the distribution of such stars in the ($J-K_\mathrm{s}$, $K_\mathrm{s}$)
colour-magnitude diagram.
We have highlighted with red dashed contours the region that we
will use to select CC candidates. We note that 2365 of the 2411 reference CCs (corresponding to 98\% of the sample) have
$\langle K_\mathrm{s} \rangle <$ 17.0 mag and only 2\% of the population have fainter magnitudes. Specifically, 45 stars
have $17.0 \leq \langle K_\mathrm{s} \rangle \leq$ 18.2 mag and only 1 CC is as faint as $\langle K_\mathrm{s} \rangle$ = 18.6 mag.
We have also reported in Fig.~\ref{fig:cmdKJK} the theoretical instability strips (ISs) for CCs with metallicity
{\it Z}=0.004 and helium abundance $Y$=0.25 taken from \cite{Bono00,Bono01a,Bono01b}. These values of $Z$ and $Y$ are
appropriate for the SMC CCs. To transform the theoretical IS edges to the observational plane we adopted
the static model atmospheres by \citet{CGK97a,CGK97b},
an absorption $A_V$=0.1 mag \citep{Has11}
and a distance modulus of 19.0 mag, computed as the weighted average of the CC results in
table 1 of \cite{Has12}\footnote{This method applied to
the CC results listed in Table 2 of \cite{deG15} leads exactly to the same distance modulus value.}.
Specifically, we have plotted in Fig.~\ref{fig:cmdKJK} the blue and red edges of the IS for CCs of different pulsation modes: F (grey
lines), 1O (green lines) and 2O (magenta lines).
For magnitudes brighter than 14 mag, there is reasonably good agreement, within the errors, between theoretical and observed IS's,
suggesting that the sample of CCs we are using as reference \citep{Sos10a}, once matched with the
VMC data, represents the ranges in $K_\mathrm{s}$ magnitude and $J$-$K_\mathrm{s}$ colour covered by the SMC CCs very well.
A number of reference CCs, especially at magnitudes fainter than $K_\mathrm{s} \sim$14 mag, are found beyond the boundaries
of the theoretical IS's, both at bluer and redder colours.
This is likely caused by
metallicity, differential reddening issues and by the poor sampling of their VMC $J$-band light curves.
Indeed, the same theoretical ISs are in much better agreement with other
observational samples, including Magellanic Cepheids, where light curves are better sampled
in colour (e.g. \citealt{Bon99}, \citealt{Mar05}).
We also compared the colour-magnitude distribution for sources with more than 10 data points in the
$J$ band (about 10\% of our sample) using the $J$-$K_\mathrm{s}$ colours
obtained after analyzing both the $J$ and $K_\mathrm{s}$-band light curves
with a custom template fitting procedure \citep{Rip15,Rip16}. We obtained a relatively
more compact distribution, confirming that the
poor sampling of the VMC $J$-band light curves is one of the possible issues.
The right panel of Figure~\ref{fig:cmdKJK} shows the distribution of these CCs in the
$Y-J$, $Y-K_\mathrm{s}$ colour-colour diagram.
As in Fig.~\ref{fig:cmdKJK}, we have highlighted with red dashed contours the region that
we will use in Sec.~\ref{sec:using} to select CC candidates.
\begin{figure*}
\begin{centering}
\includegraphics[width=8cm]{f2a.ps}
\includegraphics[width=8cm]{f2b.ps}
\caption{ Left: Distribution of known SMC CCs in the ($J-K_\mathrm{s}$, $K_\mathrm{s}$) colour-magnitude diagram.
Grey, green and magenta solid lines represent the theoretical
blue and red edges of the CC ISs for F, 1O and 2O pulsation modes, respectively.
Right: Same as the left panel but in the ($Y-J$, $Y-K_\mathrm{s}$) colour-colour diagram.
In both panels, red dashed contours indicate the boundaries of the region that will be used to select candidate CCs in the external part of the SMC.
See Section~\ref{sec:CCSMC} for details.}
\label{fig:cmdKJK}
\end{centering}
\end{figure*}
\subsection{Analysis of the light curves}\label{sec:lc-analysis}
OGLE~III periods of the reference CCs range between 0.25 d to 208 d.
The vast majority of the reference CCs (2240 sources) have $P$$<$5d, 109 sources have
$5\leq$$P$$<$10 d, 60 have $10\leq$$P$$<$75 d and only 2 sources have $P$$\geq$75 d. Specifically, these two sources have
$P_{\rm OGLE}$= 128.2 d and $P_{\rm OGLE}$ =208.8 d; despite their $K_\mathrm{s}$-band photometry ($K_\mathrm{s} \sim$ 11 mag)
should not be saturated, they are fainter than expected according to the $K_\mathrm{s}$-band $PL$ relation, especially the last one.
The same feature occurs in the $I$-band OGLE data.
Despite their images do not show any clear problem, their folded $K_\mathrm{s}$-band light curves
are very noisy without a clear shape; for these reasons we decided to discard them and focus on the sources
with $P$ lower than 80 d.
We analysed the $K_\mathrm{s}$-band light curves
and derived the period of the reference CCs
from the $K_\mathrm{s}$ time-series data alone, ignoring OGLE~III information on the period.
We used all available VMC epochs to study the light curves including observations obtained during
nights with sky conditions (i.e. seeing and ellipticity) that exceeded the VMC requirements \citep{Cio11},
since our fitting procedure is able to handle lower accuracy data (see, e.g., \citealt{Rip15} and references therein).
The resulting light curves have a number of data points that ranges from 7 to 60.
We checked the images of some CCs with a few epochs and found that often these sources are contaminated by very bright companions.
Periods were derived using the program ``Significance Spectrum'' ({\it SigSpec}; \citealt{Ree11}).
{\it SigSpec} is a method specifically developed for detecting and characterizing periodic signals in
noisy data. While most period search analyses explore only the Fourier amplitude, through the power spectrum, ignoring phase information, {\it SigSpec}
is based on the definition of a quantity called {\it spectral significance} for a time series, a function of Fourier phase and amplitude.
The {\it spectral significance} quantity conveys more information than does the conventional amplitude spectrum alone, and appears to simplify
statistical issues as well as the interpretation of phase information.
We ran {\it SigSpec} on the VMC $K_\mathrm{s}$-band
time-series adopting a lower period of 0.25 d, an upper period of 80 d
and weighting by the $K_\mathrm{s}$ single-epoch errors.
The r.m.s. of the light curve analysis typically ranged
from 0.002 mag to 0.15 mag, (with a few extreme values as large as 0.3 mag) and a median value of 0.015 mag.
The median value of the {\it spectral significance} is $\sim$ 3.0
with a standard deviation of 0.9, and minimum and maximum values of 0.9 and 7.0 (see also discussion at the end of Section~\ref{sec:PL-selection}).
Figure~\ref{fig:PdeltaP} shows the comparison between the OGLE~III periods and
the periods we derived for the reference CCs running {\it SigSpec} on the $K_\mathrm{s}$ time-series data.
For 54\% of the sources (1302 stars) the two periods are in good agreement, the difference being smaller
than 0.02 d. On the other hand,
for some stars the period found by {\it SigSpec} ($P_{\rm SigSpec}$) is definitely shorter than that published by the OGLE team.
In particular, about 800 sources have $\Delta P$=$P_{\rm OGLE}$ (where $\Delta P$=$P_{\rm OGLE}-P_{\rm SigSpec}$)
because their $P_{\rm SigSpec}$ is near zero (see Fig~\ref{fig:PdeltaP}).
This is probably due to alias problems in the case of sources with period shorter than a few days and to saturation
of the $K_\mathrm{s}$ time series for stars with longer period. Conversely, there are about 150 stars for which $P_{\rm SigSpec}$
is definitely longer than $P_{\rm OGLE}$ and hence $\Delta P$ assumes large negative values (see Fig~\ref{fig:PdeltaP}). This might be
due to faintness and hence poor quality of the $K_\mathrm{s}$ light curves of these stars that affect the period search procedure.
We also checked if there is any dependence of $\Delta P$ on the amplitude of the light curves, but did not find any.
We do not have an estimate of the error on $P_{\rm SigSpec}$ for each star but, assuming the OGLE as the correct one,
we can use as an estimate of the $P_{\rm SigSpec}$ error the median value of $\Delta P$, that is 0.002 d.
As an additional test, we performed an analysis of the $K_\mathrm{s}$ light curves using a different period search program,
FNPEAKS \citep{Kur85}. We adopted the same limits (0.25-80 d) for the period and a frequency step of 0.0001 s$^{-1}$.
For 29\% of the stars there is good agreement between the OGLE~IIII and the FNPEAKS periods:
$\arrowvert \Delta$P$\arrowvert \leq$ 0.02 d. FNPEAKS does not allow to
weight by the error of the single epoch magnitudes; this probably explains the lower percentage of periods recovered within
0.02 d with respect to {\it SigSpec} (54\%). We also checked the $P$ versus $\Delta P$ plot, that is the counterpart of Fig.~\ref{fig:PdeltaP}.
The shape of the $\Delta P$=$P_{\rm OGLE}-$$P_{\rm FNPEAKS}$ versus $P_{\rm OGLE}$ is very similar to Fig.~\ref{fig:PdeltaP}, hence confirming that FNPEAKS and {\it SigSpec} find consistent
results, however, {\it SigSpec} appears to be more efficient.
In the analysis described in Sec.~\ref{sec:using} we will thus adopt {\it SigSpec} to estimate the period of the candidate variable stars.
\begin{figure}
\begin{centering}
\includegraphics[width=8cm]{f3.ps}
\caption{Distribution of $\Delta P$=$P_{\rm OGLE}-P_{\rm SigSpec}$ versus $P_{\rm OGLE}$. Shown in the inset
is an enlargement of the short period region. Red points indicate stars within 3 $\sigma$ from
the $PL$ relations but with $\Delta P$ larger than 0.02 d. See text and Fig.~\ref{fig:plSMCknown} for details.}
\label{fig:PdeltaP}
\end{centering}
\end{figure}
\subsection{Selection based on the Period-luminosity relation}\label{sec:PL-selection}
Figure~\ref{fig:plSMCknown} shows the $K_\mathrm{s}$-band period-luminosity
($PL$) plane obtained for the 2411 reference CCs using for the period values derived with the {\it SigSpec} analysis
and for the $K_\mathrm{s}$ average magnitude values available from the VSA.
Red solid lines show the $PL$ fits obtained for F and 1O mode CCs,
using the periods from OGLE~III; red dashed lines show the
corresponding 3$\sigma$ boundaries for 1O (upper dashed line) and F (lower dashed line) CCs, respectively.
As expected, Cepheids with a good {\it SigSpec} estimate of the period ($\arrowvert \Delta$P$\arrowvert \leq$0.02 d, black points)
lie near the $PL$s defined using the OGLE~III periods.
Specifically, there are 1589 objects lying within 3$\sigma$ from the $PL$s, corresponding to 66\% of the original sample\footnote{This number reduces to 1139, corresponding to
47\% of the original sample, if the periods obtained with FNPEAKS are used instead.}. This number includes 1304 stars
with $\arrowvert \Delta$P$\arrowvert \leq$ 0.02 d (black points) and 285 sources with $\arrowvert \Delta$P$\arrowvert$ larger than 0.02 d (grey points).
These 285 sources are marked by red points in the $ \Delta$P versus $P_{\rm OGLE}$ distribution shown in Fig.~\ref{fig:PdeltaP}.
The bulk of the distribution has $\arrowvert \Delta$P$\arrowvert \leq$ 1 d (253 objects), and only a few
sources (32 stars) have $ \Delta$P values larger than $\pm$ 1 d and up to 2.5 d.
Stars within the 3$\sigma$ from the $PL$s, show an average {\it significance} (see Section~\ref{sec:lc-analysis}) of 3.4 with minimum and
maximum values of 1.8 and 7.0, respectively.
We checked the position of these stars on the $PL$ according to their $significance$
value and noted that several stars with {\it significance} value between 1.8 and 2.0 lie within 2$\sigma$ from the $PL$s.
Hence, in the following analysis we will retain only sources with {\it significance} larger than 1.7 (Section~\ref{sec:using}).
We will also take into account the possible contamination by other types of variable stars.
In particular, the RR Lyrae stars follow a $PL_{K_\mathrm{s}}$ relation that, although different, partially overlaps
with the $PL_{K_\mathrm{s}}$ relation of CCs. A black dashed box schematically shows the locus of the SMC RR Lyrae stars
in the $K_\mathrm{s}$-band $PL$ plane in Fig.~\ref{fig:plSMCknown}. This issue will be discussed in more detail
in Section~\ref{sec:contamination}.
\begin{figure}
\begin{centering}
\includegraphics[width=8cm]{f4.ps}
\caption{Distribution of known SMC CCs in the Period-$K_\mathrm{s}$-band Luminosity plane
using for the period values derived from the analysis of the
$K_\mathrm{s}$-band light curves.
Red solid lines show the $PL$ relations for F (right) and 1O (left) CCs derived using the OGLE~III periods.
Red dashed lines represent 3$\sigma$ boundaries of the 1O (upper line) and F (lower line) $PL$s, respectively. Black and grey dots show stars
with $\arrowvert \Delta$P$\arrowvert \leq$0.02 d and $\arrowvert \Delta$P$\arrowvert >$ 0.02 d, respectively.
A black dashed box marks the region populated by RR Lyrae stars. See Section~\ref{sec:contamination} for details.}
\label{fig:plSMCknown}
\end{centering}
\end{figure}
\subsection{ VSA flags}\label{sec:VSA-selection}
The VSA provides several flags describing the quality of the light curve of each VMC source.
A complete explanation of these parameters is provided in \cite{Cro12} and on the VSA web page\footnote{http://horus.roe.ac.uk/vsa}.
Here we briefly summarize the properties that are relevant for the present study.
The VSA classifies sources according to their nature by the $mergedClass$ parameter within the $vmcSource$ table,
containing the information about the sources extracted from the stacked images.
Specifically, the association between parameter value and physical nature of
the source is as follows: 1=galaxy, 0=noise, $-1$=stellar, $-2$=probableStar, $-3$=probableGalaxy, $-9$=saturated source.
The $KsppErrBits$ parameter encodes quality issues associated with a given $K_\mathrm{s}$-band detection within the vmcSource table.
Its value is zero for a detection without quality issues and grows according to the severity of the issue.
In particular to include sources with only minor $K_\mathrm{s}$-band quality issues the user can filter as $KsppErrBits<$256.
We used the VSA flags to select among our sample of 2411 reference CCs only stars with at least 10 data points (this corresponds to 2407 of the 2411 sources),
that are classified as stars or probable stars by the VSA ($mergedClass$=$-$1 or $-$2), as variables
($VarClass$=1) and that do not exibit any severe quality issues
($KsppErrBits<$256), thus ending up with a total of 1445 sources. This corresponds to 60\% of our original sample. Five hundred
of the sources that the VSA classifies as non variable sources ($VarClass$=0) perfectly lie within the 3$\sigma$ limits from the $PL$s. Hence,
the VSA flags, although useful for a first selection of variable sources, need to be fine-tuned (see, e.g., \citealt{FLC15}) to increase
their capability to detect bona fide variable stars.
In the next section we will use the VSA flags to corroborate our identification of new SMC CCs based on the
colour-magnitude and colour-colour diagrams and $PL$ selections
described in the previous sections.
In conclusion, to identify CCs from the VMC $K_\mathrm{s}$-band time series data
we will
\begin{itemize}
\item first select candidate variables by applying the colour-magnitude and colour-colour diagrams cuts described in Section~\ref{sec:cmd-cc-sel},
\item analyse with {\it SigSpec} the light curves
of the candidate variables to determine their periods (see Section~\ref{sec:lc-analysis}),
\item consider as best candidates the stars falling within 3$\sigma$ from the $PL$s defined as described in Sec.~\ref{sec:PL-selection},
\item investigate the VSA flags.
\end{itemize}
In this way, following a visual inspection of the light-curves to identify bona fide CCs, we should be able to identify 66\% of the CCs that populate the
external areas of the SMC analysed in the next Section.
\section{Detection of classical Cepheids in the external regions of the SMC}\label{sec:using}
\begin{figure}
\begin{centering}
\includegraphics[width=8cm]{f5.ps}
\caption{Spatial distribution of the SMC CC candidates (black points) selected using the tools described in
Section~\ref{sec:CCSMC}.
$X$ and $Y$, the red contours and the VMC tiles are defined as in Fig.~\ref{fig:SMCtiles}.
The white area within tile SMC 5\_2 at $X\sim 2$, $Y\sim1$ corresponds to the 47 Tucanae field analyzed by Weldrake et al. (2004).
Sources that we confirm to be bona fide new Cepheids are marked by red filled circles (see text for details).}
\label{fig:SMCfield}
\end{centering}
\end{figure}
We have applied the methods described in Section~\ref{sec:CCSMC} to look for CCs in all regions of the SMC where
we currently have complete VMC data, but where no optical comprehensive catalogue of variable stars is available yet.
Figure~\ref{fig:SMCfield} shows the portion of sky
we have analysed with the different VMC tiles labelled in Fig.~\ref{fig:SMCtiles} (see also Tab.~\ref{tab:qc}).
We specifically considered only regions outside the OGLE~III footprint
(red solid lines) and also discarded the region studied by Weldrake et al. (2004; empty area within tile SMC 5\_2
at $X\sim 2$, $Y\sim1$),
who provided a comprehensive catalogue of variable stars in the field of the Galactic globular cluster 47 Tucanae.
In particular, we first selected our candidates in the aforementioned VMC tiles,
and then discarded sources
lying in the region that overlaps with the OGLE~III and \cite{Wel04} fields. The sources
were then further selected using the colour-magnitude and colour-colour diagrams, as
described in Section~\ref{sec:cmd-cc-sel}, yielding
19,938 candidate CCs, that are plotted as black points in Fig.~\ref{fig:SMCfield}.
\cite{Rub15} estimated the star formation history (SFH) in these regions of the SMC. By comparing the best-fitting SFH models
with the theoretical CC instability strips for $Z$=0.004 and $Y$=0.25
we predict a few hundreds CCs to populate the area under
investigation. Hence, clearly, only a few percent of our 19,938 CC candidates are bona fide CCs.
For 18,090 of these sources the VSA provides $K_\mathrm{s}$-band light curves with at least 10 data points, we
analysed them with {\it SigSpec} and defined their period following the procedure described in Section~\ref{sec:lc-analysis}.
Furthermore, we derived average $K_\mathrm{s}$ magnitudes and $K_\mathrm{s}$-band amplitudes (Amp$K_\mathrm{s}$)
analysing the light curves with an automatic template fitting procedure specifically developed for the
analysis of the $K_\mathrm{s}$-band light curves of the SMC CCs \citep{Rip16}.
We then used the periods derived with {\it SigSpec} and the average $K_\mathrm{s}$
magnitudes computed as described above to plot the sources in the Period - $K_\mathrm{s}$ Luminosity
plane shown in Fig.~\ref{fig:pl}.
We have also plotted in this figure
the 3$\sigma$ boundaries of the $K_\mathrm{s}$-band $PL$ relations defined by
the OGLE~III SMC CCs (see Section~\ref{sec:PL-selection}).
\begin{figure}
\begin{centering}
\includegraphics[width=8cm]{f6.ps}
\caption{Distribution of the 18,090 candidate CCs (black points) in the Period - $K_\mathrm{s}$-band Luminosity plane.
Red solid lines show the $PL$ relations and their 3$\sigma$ boundaries (red dashed lines) defined by the OGLE~III SMC CCs.
A black dashed box highlights the region occupied
by the SMC RR Lyrae stars. Red filled circles mark bona fide CCs identified in the present study.
Their $K_\mathrm{s}$-band light curves are shown in Fig.~\ref{fig:lightCurves1}.}
\label{fig:pl}
\end{centering}
\end{figure}
Excluding objects
more than 3$\sigma$ away from the F and 1O mode $PL$ relations the sample reduces to 4,817 sources.
We further selected the sample using the parameters of the template fitting procedure.
A comprehensive description of this procedure and of its parameters is provided by \cite{Rip16}.
Here we simply note that we used the parameter
$G$, which gives an estimate of the goodness of the fit by weighting the residuals
of the fit (rms) with the number of data points rejected (outliers) from the fitting
procedure. The first term of $G$ tends to favour templates which give the smallest rms values,
whereas the second term favours those removing the least number of outliers.
The balance between these two terms generally yields an automatic selection of the best
templates that is in agreement with the visual inspection of the fitting procedure.
Since $G$ is inversely proportional to the square of the rms and directly
proportional to the fourth power of the ratio between used in the fit and total number
of phase points in the light curve, values of $G$ in the interval 100-10000 generally
mean good fit, and, in absolute terms, also good light curve with small scatter. On the contrary,
values of $G$ below 100 are usually associated with highly scattered light curves.
Finally, we retained only sources with the $G$ parameter between 100 and 10,000 and $K_\mathrm{s}$-band
amplitude $\geq$ 0.04 mag further
reducing the sample to 3,636 best candidates with $\langle K_\mathrm{s} \rangle$ magnitude in the range of 12.31 to 18.21 mag. This number appears to be still rather large if compared
to expectations from the SFH recovery.
A check was made by comparing the distributions of CCs in the VMC tiles partially covered by OGLE~III.
In particular, half of the tile SMC 4\_5 is covered by both OGLE III and VMC and the other half only by VMC
(see Figs.~\ref{fig:SMCtiles} and \ref{fig:SMCfield}).
We divided the tile into four subregions, each of them approximately covering the same area.
The two western subregions lie within the
OGLE~III footprint, while the two eastern ones lie outside.
OGLE~III detected 53 CCs in the north-western subregion and 30 in the southern one.
From the SFH performed by \cite{Rub15} and the theoretical IS's (\citealt{Bono00,Bono01a,Bono01b}), we obtain a total number of
about 55 (with a minimum of 35 and a maximum of 76) CCs from the SFH recovery, in good agreement with the OGLE~III findings.
Our number of CC candidates in the two eastern subregions of tile SMC 4\_5 is roughly 10 times that observed by OGLE~III in the two western ones.
This test suggests that the majority of the 3636 new candidate variable stars are not CCs and that a high level
of contamination must be present among them. In particular there could be contamination by other types of variables, but also issues such as:
saturation (on the bright side), limiting magnitude and photometric error problems (on the faint side), blending effects, intrinsic problems of the NIR data, etc.
These issues may particularly affect faint sources, thus leading to an overestimate of the number of variable sources.
Indeed, the sample of 3,636 new candidate variables includes
1,677 sources with $\langle K_\mathrm{s} \rangle <$ 17.0 mag and 1,959 fainter sources with $17.0 \leq \langle K_\mathrm{s} \rangle \leq $ 18.2 mag. However,
according to the discussion in Section~\ref{sec:cmd-cc-sel}, only 2\% of the SMC CCs have $\langle K_\mathrm{s} \rangle$ magnitudes fainter than 17.0 mag. Hence,
the 1,959 candidate variables fainter than 17.0 mag should include, at most, $\sim$ 40 bona fide CCs.
We hence visually inspected only the $K_\mathrm{s}$ light curves of all 1,677 sources brighter than $\langle K_\mathrm{s} \rangle $= 17.0 mag
and selected among them 297 sources whose $K_\mathrm{s}$-band light curves have the typical shapes of CCs.
In particular, all sources with very poor light curve coverage were discarded since a firm
classification was not possible. Moreover, for several sources the light curves did not show the typical shape of
cepheids, these sources were discarded as well. During the visual inspection, a percentage reliability flag has been
assigned to each source. This flag is 100\% for a source with (i) good phase coverage,
(ii) good light curve shape and (iii) good position in the PL,
while its value diminishes according to issues related to one or more of the aforementioned features, becoming 62\% for sources that are not
confirmed to vary.
Then we also checked a subsample of 240 among the 1,959 sources fainter than
$\langle K_\mathrm{s} \rangle $ = 17.0 mag; this provided two additional sources with light curves typical of CCs.
Particular attention was devoted to sources showing a $K_\mathrm{s}$-band amplitude between 0.04 mag and 0.1 mag since
at this level of amplitude it is hard to distinguish between real variable sources and spurious objects. We hence decided
to keep only low amplitude sources with a good light curve coverage and very clear shapes.
We consider this total sample of 299 candidate variable stars that passed the
colour-magnitude and colour-colour diagrams, $PL$, template
parameters selections and the visual inspection of the light curve as bona fide Cepheids.
Out of 299, nine sources are in common with the General Catalog of Variable Stars (GCVS, \citealt{Art96}),
two (VMC-SMC-CEP-258 and 286) are ACs in common with \citealt{Sos15a},
while 288 are new CCs identified in the present study.
The new CCs have periods in the range
from about 0.34 to 9.1 d and span the magnitude range: 12.9 $\leq \langle K_\mathrm{s} \rangle \leq$ 17.6 mag,
and only two being fainter than 17.0 mag (see above).
This number is well consistent with the predictions of \cite{Rub15} SFH recovery in these external regions of the SMC.
In particular, of the 299 confirmed Cepheids, only 13
are located in the eastern subregions of tile SMC 4\_5. In the same two subregions \cite{Rub15} SFH recovery leads to a total number of
about 10 (with a minimum of 4 and a maximum of 20) CCs, in very good agreement with our findings.
Table~\ref{tab:CCcandidates1} lists our 299
SMC Cepheids along with their RA, Dec (J2000) coordinates,
the period obtained with {\it SigSpec} (for a discussion about the errors see Section~\ref{sec:lc-analysis}),
the $K_\mathrm{s}$-band amplitude and the intensity-averaged
$K_\mathrm{s}$ magnitude computed with the template-fitting procedure, the {\it VarClass}
parameter attributed by the VSA and the variability type ({\it VarType}) assigned in the present study
together with comments, if any, and the aforementioned percentage reliability flag. The last column of Table~\ref{tab:CCcandidates1} also provides the
GCVS identification number, whenever appropriate.
The sources are ordered by increasing magnitude going from brighter to fainter objects.
Their spatial distribution over the SMC is shown in Fig.~\ref{fig:SMCfield} where they have been marked as red filled circles.
\begin{table*}
\caption{Information on the 299 Cepheids that we have identified in the SMC using only the near-infrared photometry of the VMC survey.
Col. 1: VMC-SMC-CEP ID, Col. 2: Right Ascension (RA, J2000), Col. 3: Declination (Dec, J2000),
Col. 4: number of $K_\mathrm{s}$ observations, Col. 5: period computed with {\it SigSpec},
Col. 6: $K_\mathrm{s}$ amplitude computed with the template fitting procedure,
Col. 7: intensity-averaged $K_\mathrm{s}$ magnitude computed with the template fitting procedure,
Col. 8: {\it VarClass} flag assigned by the VSA, Col. 9: variability type
assigned in the present study, corresponding percentage reliability flag and specific comments, if any. The table is published in its entirety as
Supporting Information with the electronic version of the article.
A portion is shown here for guidance regarding its form and content.}
\label{tab:CCcandidates1}
\begin{tabular}{ccccccccl}
\hline
\hline
VMC ID & RA & Dec & $N_{K_\mathrm{s}}$& $P_{\rm SigSpec}$ & Amp$K_\mathrm{s}$ & $\langle K_\mathrm{s} \rangle$ & {\it VarClass} & {\it VarType \& Comments}\\
& (deg) & (deg) & & (d) & (mag) & (mag) & & \\
\hline
001 & 01:43:58.77 & $-$71:50:09.7 & 18 & 8.89656 & 0.06 & 12.350 & 0 & 75\% CC few points at min \\
002 & 01:43:15.53 & $-$71:42:49.8 & 18 & 8.897351 & 0.08 & 12.599 & 0 & 75\% CC few points at min \\
003 & 01:28:07.58 & $-$72:48:52.1 & 18 & 12.52367 & 0.27 & 12.758 & 1 & 100\% CC GCVS2347 \\
004 & 01:24:25.35 & $-$74:16:50.5 & 16 & 13.151942 & 0.29 & 12.792 & 1 & 100\% CC GCVS2343 \\
005 & 01:42:56.52 & $-$71:18:46.0 & 18 & 9.122673 & 0.05 & 12.885 & 0 & 75\% CC few points at min \\
006 & 01:38:18.83 & $-$71:22:18.4 & 18 & 8.886835 & 0.04 & 13.076 & 0 & 75\% CC few points at min \\
007 & 01:23:00.57 & $-$74:22:16.8 & 16 & 9.752387 & 0.21 & 13.182 & 1 & 100\% CC GCVS2337 \\
008 & 01:38:00.64 & $-$71:39:22.2 & 18 & 8.887862 & 0.06 & 13.269 & 0 & 75\% CC few points at min \\
009 & 00:29:43.53 & $-$71:33:21.0 & 17 & 8.371283 & 0.06 & 13.470 & 0 & 100\% CC \\
010 & 01:13:25.40 & $-$70:58:09.2 & 14 & 5.233301 & 0.06 & 13.533 & 0 & 75\% CC few points at min \\
\hline
\end{tabular}
\end{table*}
During the revision phase of this manuscript, \cite{Sos15b} paper presenting CCs in the Magellanic Clouds was posted as
a preprint on the ArXiv.
The authors comment that they can counter-identify 278 sources out of the 299 Cepheids in our list (the remaining 21
fall in the inter-CCD gaps of their camera; Soszynski, private communication) and they confirm 35
of the SMC Cepheids we have identified in our study.
They also say that most of the remaining objects from our list turned out to be
constant or nearly constant in the optical bands.
We have cross-matched our catalogue with \cite{Sos15b}'s and find that
indeed 36 (about 13\%) of our Cepheids are confirmed by OGLE~IV, namely, the nine CCs in common with the GCVS, the two ACs
also identified by \cite{Sos15a} and 25 new CCs.
This result is encouraging, since it confirms that our method
is promising despite the limited number of data points and the intrinsically low amplitude of the VMC
light curves. But it is also quite puzzling due to the low rate of optical confirmations, since
our 299 bona fide Cepheids were selected not simply because of the light variation
in the $K_\mathrm{s}$-band, but, more importantly, because they also fall in the
colour-magnitude and colour-colour diagrams where OGLE III SMC
bona-fide Cepheids are found, and because they also follow the SMC
Cepheid $PL$ relation.
The $K_\mathrm{s}$-band light curves of our 299 bona fide SMC Cepheids folded according to the period derived
from the analysis with {\it SigSpec}, are shown in Fig.~\ref{fig:lightCurves1}.
We first display the 9 CCs in common with the GCVS
and the 25 new CCs discovered in our study and later confirmed also by OGLE~IV;
all other bona fide Cepheids (including the two ACs) follow.
The light curves appear overall quite symmetrical and it is difficult with the VMC $K_\mathrm{s}$-band data to identify features such as bumps.
$K_\mathrm{s}$ time series data for all of them
are provided in Table~\ref{tab:lightcurves}.
Among the remarks in the last column of Tab.~\ref{tab:CCcandidates1}, ``few points at min/max'' is used whenever the light
curve is not homogeneously covered at minimum or maximum light. For some of these sources, we checked the light curve
of nearby sources within 5 arcsec, to rule out unidentified photometric problems that might cause/mimic the light
variation. They look flat suggesting that the variability of the sources identified as Cepheids is real, even for
less well sampled light curves.
Finally, we recall that according to the results in Section~\ref{sec:PL-selection} our technique enables recovery of
about a 66\% of the true Cepheids that may occur in these external regions of the SMC and that
bona fide Cepheids may also be present in the sample of
about 2,000 candidate variables fainter than $K_\mathrm{s}$ = 17.0 mag that were not analyzed here.
Therefore, there are likely additional
Cepheids that we may have missed either
because their {\it SigSpec} periods are wrong by more than $\pm$ 2.5 d, thus
causing them to fall outside the 3$\sigma$ boundaries of
the $PL$s\footnote{Indeed, there are about 30 CCs detected by OGLE~IV that are not in our list.
These sources were not identified because their SigSpec period is incorrect
(see Sections~\ref{sec:lc-analysis}, ~\ref{sec:PL-selection} for details), hence they did not pass the $PL$ selection.},
or because they are fainter than $K_\mathrm{s}$ = 17.0 mag.
\begin{table}
\begin{center}
\caption{$K_\mathrm{s}$-band time-series photometry of our 299 SMC Cepheids.
The table is published in its entirety as Supporting
Information with the electronic version of the article. A portion is
shown here for guidance regarding its form and content.
}
\label{tab:lightcurves}
\begin{tabular}{ccc}
\hline
\hline
\multicolumn{3}{c}{Star VMC-SMC-CEP-001 } \\
\hline
HJD-2400000 & $K_\mathrm{s}$ & err$K_\mathrm{s}$ \\
(d) & (mag) & (mag) \\
\hline
55834.792559 & 12.339 & 0.002 \\
55927.600518 & 12.344 & 0.002 \\
56146.865651 & 12.336 & 0.002 \\
56147.757493 & 12.336 & 0.002 \\
56155.743101 & 12.336 & 0.002 \\
56159.787169 & 12.342 & 0.002 \\
56163.761878 & 12.327 & 0.002 \\
56175.694234 & 12.330 & 0.002 \\
56188.672502 & 12.389 & 0.002 \\
56189.703199 & 12.360 & 0.002 \\
56190.724434 & 12.334 & 0.002 \\
56208.619677 & 12.337 & 0.002 \\
56226.551935 & 12.335 & 0.002 \\
56256.553723 & 12.336 & 0.002 \\
56282.578397 & 12.333 & 0.002 \\
56300.546758 & 12.329 & 0.002 \\
56486.866850 & 12.326 & 0.002 \\
56911.795036 & 12.331 & 0.002 \\
\hline
\end{tabular}
\end{center}
\end{table}
\begin{figure*}
\begin{centering}
\includegraphics[width=17cm,height=21cm]{f7.ps}
\caption{$K_\mathrm{s}$-band light curves of bona fide Cepheids in the external regions of
the SMC identified in the present study.
For each source we indicate the VMC ID (ordered by increasing magnitude; see also Tab.~\ref{tab:CCcandidates1}) and
the period derived from the analysis of the light curve with {\it SigSpec}.
We first display the 9 CCs in common with the GCVS (VMC-SMC-CEP-003, 004, 007, 011, 022, 039, 041, 051, 091)
and 25 new CCs discovered in our study that were later confirmed also by OGLE~IV (VMC-SMC-CEP-023, 035, 038, 042, 056, 075, 081, 085,
087, 093, 097, 101, 110, 121, 133, 136, 138, 140, 161,
166, 177, 188, 231, 237, 292). All other bona fide Cepheids (including the two ACs) follow.
The figure is published in its entirety as Supporting Information with the electronic version of the article.}
\label{fig:lightCurves1}
\end{centering}
\end{figure*}
\subsection{Contamination by other types of variable stars}\label{sec:contamination}
Adopting as reference the catalogue of RR Lyrae stars detected in the SMC by the OGLE~III
survey \citep{Sos10b}, Muraveva et al. (in prep.)
studied the $K_\mathrm{s}$-band $PL$ relation of 1,081 SMC RR Lyrae stars observed by VMC. They found
that these variables typically have
$K_\mathrm{s}$ mean magnitude between 17.5 mag and 19.5 mag, with a subset of about 200 of them
having 17.5 $\leq \langle K_\mathrm{s} \rangle \leq$ 18.2 mag.
In this magnitude range there is partial overlap with the short period, faint end of the CCs distribution (see Fig.~\ref{fig:cmdKJK}).
Furthermore, while the number of CCs is expected to drop significantly moving from the centre to the external region
of the SMC, the distribution of RR Lyrae stars declines gently and their number is expected to remain rather high
in the peripheral areas we are investigating (e.g. Fig. 7 of~\citealt{Sos15a}). Therefore, some of the 1,959 candidate variables with $17.0 \leq \langle K_\mathrm{s} \rangle \leq $ 18.2 mag
may be RR Lyrae stars.
A further three percent contamination can also be expected from ECLs, they can spread
all over the colour-magnitude diagram (see \citealt{Mor14}),
thus we would need different criteria to distinguish them.
Finally, some of the sources could be ACs like VMC-SMC-CEP-258 and 286, or T2Ceps, due to the partial
overlap existing among the $PL$s of the different types of Cepheids, especially between CCs and
ACs (e.g. \citealt{Sos08b}).
\section{Summary and conclusions}\label{sec:summary}
We have developed a technique to identify variable stars using only the multi-epoch
near-infrared photometry obtained by the VMC survey and have specifically tailored it to the identification of Cepheids.
The technique exploits colour-magnitude and colour-colour diagrams, and $PL$ relations defined by known SMC Cepheids, along with template
parameter selections and visual inspection of the light curve to identify bona fide Cepheids.
The technique was
applied to external regions of the SMC
for which complete VMC $K_\mathrm{s}$-band observations are available and no comprehensive identification of variable
stars from other surveys exists yet.
We have identified and present $K_\mathrm{s}$-band light curves for 299 SMC Cepheids, of which 9 are CCs in common with the General Catalog of Variable Stars (GCVS, \citealt{Art96}), two are ACs also found by \cite{Sos15a} and the remaining 288 sources are new discoveries. The number of SMC Cepheids we have detected is consistent with the predictions of \cite{Rub15} SFH recovery in these regions of the SMC, taking into account that our technique may enable recovery of only
about a 66\% of the true Cepheids that may occur in these external regions of the SMC.
Subsequently, \cite{Sos15b} crossmatched
278 of the sources in our list with their optical photometry and confirmed 36 of them as Cepheids.
This result is encouraging, since it shows that our technique is promising despite the limited number of data points and the intrinsically low amplitude of the VMC
light curves, but the low rate of optical confirmations is rather surprising and calls
for further investigations to understand this discrepancy and possible physical/technical reasons behind it.
This near-infrared vs optical light-curve
connection/conspiracy will be addressed in a following paper.
\section*{Acknowledgments}
We thank the CASU and the WFAU for providing calibrated data products under the support of the
Science and Technology Facility Council (STFC) in the UK. Partial financial support for this
work was provided by PRIN-MIUR 2011 (PI: F. Matteucci).
We thank the anonimous referee for his/her constructive comments.
RdG acknowledges funding support from the National Natural Science Foundation of China (grant 11373010).
MIM thanks Felice Cusano and Alceste Bonanos for the interesting and useful discussions.
MRC acknowledges support from the UK's Science and Technology Facilities Council
[grant number ST/M001008/1] and from the German Academic Exchange Service.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 7,532 |
import sbt._
import Keys._
object ApplicationBuild extends Build {
val appName = "play2-websocket"
val appVersion = "1.0.2"
val appDependencies = Seq(
"com.typesafe.akka" %% "akka-remote" % "2.1.2"
)
val main = play.Project(appName, appVersion, appDependencies).settings(
organization := "com.originate",
description := "Scalable resilient to failures WebSocket/Socket.IO messaging module for Scala Play 2 for cloud environments.",
scalacOptions ++= Seq("-deprecation", "-unchecked", "-feature"),
publishTo <<= version { (v: String) =>
val nexus = "https://oss.sonatype.org/"
if (v.trim.endsWith("SNAPSHOT"))
Some("snapshots" at nexus + "content/repositories/snapshots")
else
Some("releases" at nexus + "service/local/staging/deploy/maven2")
},
publishMavenStyle := true,
pomIncludeRepository := { _ => false },
pomExtra := (
<url>https://github.com/Originate/play2-websocket</url>
<licenses>
<license>
<name>The Apache Software License, Version 2.0</name>
<url>http://www.apache.org/licenses/LICENSE-2.0.txt</url>
<distribution>repo</distribution>
</license>
</licenses>
<scm>
<url>git@github.com:Originate/play2-websocket.git</url>
<connection>scm:git@github.com:Originate/play2-websocket.git</connection>
</scm>
<developers>
<developer>
<id>dtarima</id>
<name>Denis Tarima</name>
<url>http://originate.com</url>
</developer>
</developers>)
)
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 2,189 |
3 children die, 3 relatives badly hurt in Texas…
3 children die, 3 relatives badly hurt in Texas house fire
CONROE, Texas (AP) — Three children were killed early Friday when a burning house collapsed north of Houston and three members of the same family are critically injured, according to a sheriff's official.
Montgomery County sheriff's Lt. Scott Spencer said the children ranged in age from 6 to 13 years and that their bodies were found amid the debris of the home.
"We've moved from a rescue operation to a search and recovery," he said.
The three critically injured people included a 10-year-old boy who suffered severe burns.
The flames and heat drove away emergency responders who attempted to enter the home near Conroe (KAHN'-roh), about 40 miles north of Houston, Montgomery County Fire Marshal Jimmy Williams told KHOU-TV.
Spencer said two Shenandoah police officers who were among the first to respond also were hurt, suffering smoke inhalation and other injuries. An officer with another department was hurt as well.
Three other members of the family were taken to hospitals with injuries that are not life threatening.
The flames spread to two adjacent homes that sustained minor damage, Spencer said. | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 1,606 |
{"url":"https:\/\/www.math.stonybrook.edu\/cgi-bin\/preprint.pl?ims92-1","text":"## Preprint ims92-1\n\nM. Martens\nThe Existence of Sigma-Finite Invariant Measures, Applications to Real One-Dimensional Dynamics.\n\nAbstract: A general construction for $\\sigma-$finite absolutely continuous invariant measure will be presented. It will be shown that the local bounded distortion of the Radon-Nykodym derivatives of $f^n_*(\\lambda)$ will imply the existence of a $\\sigma-$finite invariant measure for the map $f$ which is absolutely continuous with respect to $\\lambda$, a measure on the phase space describing the sets of measure zero. Furthermore we will discuss sufficient conditions for the existence of $\\sigma-$finite invariant absolutely continuous measures for real 1-dimensional dynamical systems.\nView ims92-1 (PDF format)","date":"2018-03-20 09:48:04","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\": 0.7871953845024109, \"perplexity\": 489.9870510585947}, \"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-2018-13\/segments\/1521257647327.52\/warc\/CC-MAIN-20180320091830-20180320111830-00567.warc.gz\"}"} | null | null |
{"url":"https:\/\/tex.stackexchange.com\/questions\/254339\/formatted-date-inside-section","text":"# Formatted date inside section\n\nI am trying to compile a document where each section title corresponds to a given date. The date has to be formatted, 2014\/01\/03 should produce 3rd January 2014 in the title.\n\nI was trying to use the isodate package like this:\n\n\\documentclass[english]{report}\n\n\\usepackage{verbatim}\n\\usepackage{isodate}\n\\usepackage{hyperref}\n\n\\begin{document}\n\n\\tableofcontents\n\\newpage\n\n%Date formatting routines\n\\dateinputformat{tex}\n\\origdate\n\n\\section{\\printdate{2014\/01\/03}}\n% Some normal lines of text\n\\verbatiminput{..\/20140103.txt}\n\\pagebreak\n\n\\section{\\printdate{2014\/01\/04}}\n% Some normal lines of text\n\\verbatiminput{..\/20140104.txt}\n\\pagebreak\n\n\\end{document}\n\n\nbut this is reporting errors.\n\nSome details:\n\n\u2022 The document will have more than 50 different sections (each with a different date).\n\u2022 I don't mind using other date formatting packages or options\n\u2022 I have a script that produces the section blocks from the filename automatically.\n\u2022 I've tried to escape the \\printdate with a protect but has not worked.\n\nI have read this question. However, I need a lot of different sections each with a different date. I guess there has to be a straightforward way to do this.\n\n\u2022 Welcome to TeX.SX! What is the error reported? \u2013\u00a0user31729 Jul 8 '15 at 15:08\n\u2022 It should be \\printdate{2014-01-03}; if you use \/ as separator, the input should be dd\/mm\/yyyy. \u2013\u00a0egreg Jul 8 '15 at 15:13\n\u2022 @ChristianHupfer The error is: ! Undefined control sequence. \\su@IfSubStringInString ...}\\su@compare #2#1\\@nil l.86 \\subsection{\\printdate{2014-01-04}} \u2013\u00a0mcieec Jul 8 '15 at 15:15\n\u2022 @egreg I've also tested with the dd\/mm\/yyyy format. Still the same error. \u2013\u00a0mcieec Jul 8 '15 at 15:17\n\u2022 I get no error; the version of isodate is isodate 2010\/01\/03 v2.30; I tried with TeX Live 2008, 2009, 2010, 2011, 2012, 2013, 2014 and 2015, without any error. \u2013\u00a0egreg Jul 8 '15 at 15:38\n\nThe 'problem' is hyperref -- it tries to create bookmarks automatically, writing pdf string content to the bookmarks. Apparently \\printdate does not provide such pdf string content. The usage of bookmarks can be prevented with bookmarks=false option to hyperref.\n\nA working solution, however, not with 'correct' date formatting in the bookmarks, is to use \\texorpdfstring{\\printdate{...}}{...}, where the 2nd argument content goes to the bookmark and the first one is for the section title\/ToC.\n\n\\documentclass[english]{report}\n\n\\usepackage{verbatim}\n\\usepackage{isodate}\n\n\\usepackage{hyperref}\n\\usepackage{bookmark}\n\n\\newcommand{\\printdatetitle}[1]{%\n\\texorpdfstring{\\printdate{#1}}{#1}%\n}\n\n\\makeatletter\n%\\providecommand{\\@nil}\n\\makeatother\n\n\\dateinputformat{tex}\n\n\\begin{document}\n\\tableofcontents\n\\chapter{First}\n\\section{\\printdatetitle{2014\/1\/3}}\n\n\\end{document}\n\n\u2022 Thanks @ChristianHupfer! This is enough for my requirements. \u2013\u00a0mcieec Jul 8 '15 at 16:52\n\u2022 @mcieec: Alright, but I'll try to find a better way, to get the non-pdf-conforming output of printdate working \u2013\u00a0user31729 Jul 8 '15 at 16:54\n\nSince you said you don't mind using another package, here's a solution that uses datetime2 instead. This package is a replacement to datetime and was designed to provide expandable commands so that the date can be used in bookmarks. Example:\n\n\\documentclass[english]{report}\n\n\\usepackage[en-GB]{datetime2}\n\\usepackage{hyperref}\n\n\\def\\iprintdate#1\/#2\/#3\/{\\DTMdisplaydate{#1}{#2}{#3}{-1}}\n\\newcommand*{\\printdate}[1]{\\iprintdate#1\/}\n\n\\begin{document}\n\n\\tableofcontents\n\\newpage\n\n\\section{\\printdate{2014\/01\/03}}\nSome normal lines of text\n\\pagebreak\n\n\\section{\\printdate{2014\/01\/04}}\nSome normal lines of text\n\\pagebreak\n\n\\end{document}\n\n\nThe dates appear as \"3rd January 2014\" and \"4th January 2014\" in the bookmarks (as well as in the table of contents and the section headings).\n\nIf you don't already have them installed, in addition to datetime2, you also need to install datetime2-english (for the en-GB option) and tracklang.","date":"2019-12-08 00:50:09","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\": 0.8723049163818359, \"perplexity\": 3328.5083569884123}, \"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-20191207233943-20191208021943-00309.warc.gz\"}"} | null | null |
Analysis: January release of NHS performance statistics
We see a tough picture before Omicron getting tougher with growing waits and missed targets as staff battle COVID-19 caseloads and absences.
Joseph Plewes
The latest performance statistics were released as COVID-19 hospital admissions plateau nationally. But the service is still under enormous pressure at the most challenging time of year with staff shortages (with COVID-19 absence passing the highest week from last winter, this week) and poor performance metrics.
Significant challenges are shown in key monthly performance indicators such as record levels of 12-hour waits from decision-to-admit to admission in A&E ('trolley waits') and record low performance against the four-hour A&E attendance target and the two week/two month wait cancer standards (albeit with some caveats).
The waiting list has risen until the end of November, but marginally compared to recent months with 6m (5,995,156) waiting for treatment compared to 5.98m in October. This is in part due to the 1.42m removals from the waiting list, the highest level of activity since November 2019.
COVID-19 admissions per day have flattened considerably across England, with a seven-day average of 2,010 patients admitted per day in England (data available up to 10 January), slightly down from 2,038 week on week. This is the first time this week-on-week figure has decreased since the start of December.
The turning point in admissions is also reflected in the regional picture, shown in the chart below, with the seven-day average of daily admissions in London now 18.7 per cent lower than one week ago. It is important to note that admissions rose week on week in other regions, such as the north east and Yorkshire where they have risen 15.2 per cent. However, this increase is significantly lower than the rises of over 100 per cent from seven days previously.
The number of COVID-19 patients in hospital beds is flattening too as would be expected (albeit more slowly), with this chart showing the biggest rises still happening in the north of England despite decreases in numbers in London:
Although we may be approaching, or just beyond, the peak of Omicron in London, there is still considerable pressure on the service at the toughest time of the year.
With staff absence high, discharge delays, patient flow issues, infection prevention control challenges, non-urgent surgery postponed and more, it is unsurprising that today's statistics show performance continuing to decline and targets continuing to be missed, with patient safety at risk.
A&E performance
In December, performance against the four-hour target fell. The percentage of patients spending four hours or less in type 1 A&Es (major A&E departments that provide a consultant-led 24-hour service with full facilities for resuscitating patients) fell to 61.2 per cent, the lowest on record. The target is 95 per cent but this hasn't been met in type 1 A&Es since 2012. This means that in December, over 440,000 patients had attendances in A&E that lasted longer than four hours from arrival to admission, transfer or discharge.
The fall in performance is shown against 2019 for context, with improved percentages during the pandemic when there were fewer attendances:
Also concerning is that 120,218 patients spent more than four hours from decision-to-admit to admission at A&Es (marginally down from 120,749 in the previous month, but total admissions were down by just under 7,000). 12,986 of these patients waited over 12 hours, up 22 per cent from last month and the highest it has ever been:
The waiting list has risen from 5.98m to 6m (5,995,156) as of the end of November. This has increased by over 100,000 patients in all of the last nine months, with this smaller rise partly due to the highest number of patients coming off the waiting list (1.42m) since January 2020. Many of these will be due to patients starting treatment, although a percentage will be administrative removals or patients no longer electing to have treatment. The number will also be higher, with validation removals taken into account.
1.68m patients meanwhile joined the waiting list, the highest since January 2020. This is similar to pre-pandemic levels (across 2019, an average month was 1.68m) meaning despite the increase we are not seeing the missing patients who didn't present throughout the pandemic yet.
The number of 52-week waiters is 306,996, down from 312,665, while the number of 104-week waiters is 18,585 up from 16,225.
6,954 patients are due to rise over the 104-week threshold within four weeks. Knowing the pressure the service has been under in December and January, it is unlikely we will see large numbers of those long waiters coming off the waiting list in forthcoming data releases, despite planning guidance to eradicate 104-week waits by March 2022.
In November, 77.4 per cent of patients were seen within two weeks of an urgent referral for suspected cancer, down from 81.3 per cent in October and meaning performance against the standard has declined in four consecutive months. The target is that following an urgent GP referral for suspected cancer, at least 93 per cent should be seen within two weeks. This has rarely been met over the last four years, and when it last was (May 2020) significantly fewer patients were seen (106,741).
The number of patients seen by a specialist however in November 2021 was 246,316, the highest ever. The record was set in March 2021 at 232,136; prior to the pandemic the number seen in a month had only been over 220,000 twice, but now this number has been above 220,000 in five of the last six months.
A similar pattern is visible in two-month waits from GP urgent referral to first treatment. This standard is 85 per cent of patients starting treatment within two months, but the last time this percentage was even over 80 per cent was in December 2018, with 67.5 per cent of patients meeting the standard in November 2021. Compared to two-week waits, the volume of treatment starts is slightly more in keeping with previous years:
One of the key caveats on both metrics from month to month, is that we know a significant number will have already been over the threshold of two weeks or two months coming into November, so a 100 per cent figure would not be a good sign as the metric is based on how many patients seen or treated were within the standard.
On an average day in the week ending 2 January 2022, 11,795 inpatients no longer meeting the criteria to reside in hospital were not able to be discharged on the day out of an average 19,460 patients, up to 60.6 per cent from 58.3 per cent in the previous week.
This equates to an average day having 11,795 patients who were medically fit to be discharged, but 7,666 remaining in hospital. They now make up over 13 per cent of patients in NHS acute trusts.
90.6 per cent of patients who had been in hospital for more than three weeks and no longer met the criteria to reside in hospital could not be discharged, further emphasising pressure in other NHS services and the care sector.
Ambulance handovers
Challenges with discharging and patient flow then have a knock-on effect to the front door. In the week ending 9 January, there were 7,948 ambulance handover delays over 60 minutes, compared to 8,272 the previous week. This equates to 9.8 per cent of all ambulances arriving, with 81,135 arrivals in total (previous week saw 83,643 arrivals and 9.9 per cent were delayed over 60 minutes).
18,307 handovers were delayed by over 30 minutes (22.6 per cent of arrivals), not quite reaching the previous record of 19,503 delays set week ending 12 December 2021. For context, the record prior to this year was 18,251 week ending 5 January 2020.
In acute trusts, COVID-19 absence was up 28 per cent in the week ending 9 January, with 45,736 staff absent due to COVID-19 per day, up from 35,596 and now surpassing the levels of last winter. There are significant differences across the regions, with COVID-19 absences up 46 per cent in the north west and 41 per cent in the Midlands, compared to just 6 per cent in London.
Total absence per day was up to 88,516 on average compared to 80,295 in the previous week, with COVID-19 absences accounting for 52 per cent of this.
According to the NHS sickness absence data with absence reasons released up to August, the most reported cause for absence was mental health. 556,000 full time equivalent days were lost in August (27.8 per cent), which is concerning as mental health issues can have a significant impact on staff and their families, and also result in longer absence periods or high levels of presenteeism.
Mental health trusts themselves are also seeing a rapid rise in staff absences. With some reporting that the numbers are increasing from about 30-40 before Christmas to about 150 now. Workforce capacity was an issue before the pandemic, and while mental health trusts are managing despite staff absences, it is a very challenging situation, which will of course get worse if these numbers continue to rise. Often these patients are seriously ill and will need a bed. We also know that high levels of people on inpatient units also have COVID-19, which gives an additional challenge to mental health trusts at this time.
Staff absence is also having a significant impact on primary care, reducing the number of available clinicians for core activity and vaccination clinics, as well as admin and support staff. One primary care federation (groupings of practices) reported to us that over 30 per cent of their core team absent. Our members have also been reporting increased COVID-19 presentations in primary care.
How is the NHS performing?
View our analysis of the latest NHS performance figures for a rounded view of how healthcare services are coping under immense pressure.
Population health Workforce
COVID-19 Capacity and performance Urgent and emergency care Acute care
Record NHS ambulance call outs for life-threatening conditions, despite jump in Omicron absences
Matthew Taylor responds to the latest NHS monthly performance statistics.
NHS leaders warn of dangerous complacency around Omicron threat
The NHS is still facing several weeks of intense winter pressures, with health leaders urging politicians to avoid complacency around Omicron.
Access more commentary and analysis on the pressures facing the NHS.
Find out more Arrow pointing right | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 1,631 |
Exceptional Persons, Inc. Exceptional Persons, Inc.
Exceptional Persons, Inc. (EPI) makes a positive difference in the lives of people with disabilities, families, and children as they live and work in the community.
Exceptional Persons, Inc. (EPI) is a non-profit agency that focuses services for two identified service groups- individuals with disabilities and families with child care needs.
We foster active community participation of the individuals and families we serve while respecting and encouraging their preferences and choices.
Would you recommend Exceptional Persons, Inc.? | {
"redpajama_set_name": "RedPajamaC4"
} | 1,402 |
Impakt invite Danish artist, Linda Hilfling, for residency to work on Een Publiek Domein.
domain in a literary sense.
chunks of information is missing and eventually the users have to guess the overall meaning of the content.
Nevertheless, although amputated and fragmented, this is a public domain.
domain along with a "howto"-guide, making it possible for anyone to create their own 'Publiek Domein'. | {
"redpajama_set_name": "RedPajamaC4"
} | 4,622 |
Langustovití (Palinuridae) je čeleď vyšších korýšů z řádu desetinožci (Decapoda). U rodu Langusta se velikost pohybuje v rozsahu cca 15–60 cm. Langusty spolu s listorožci tvoří podřád Palinura. Všechny druhy jsou mořské a žijí především v tropickém pásu. Pomocí tzv. plektra jsou schopni produkovat zvuky. Mají značný hospodářský význam, neboť jsou loveny ke konzumaci.
Zástupci
Čeleď zahrnuje rody:
Jasus Parker, 1883
Justitia Holthuis, 1946
Linuparus White, 1847
Palibythus Davie, 1990
Palinurellus De Man, 1881
Palinurus Weber, 1795
Palinustus A. Milne-Edwards, 1880
Panulirus White, 1847
Projasus George and Grindley, 1964
Puerulus Ortmann, 1897
Reference
Externí odkazy
Langusty | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 5,288 |
Affordable quality homemade food from local or Welsh suppliers served all day as well homemade cakes and fresh coffee.
Our Carvery Sunday Lunch is a must!
If you want to pre-book your table or make a room reservation, you can use our amazing online booking form or call us on 01938 820202.
Newly refurnished quality, clean and comfortable rooms from singles, twins and doubles to family rooms all with en-suite facilities.
With extensive parking, large garden, outdoor undercover dining and marquee we hold a number of events and music festivals. | {
"redpajama_set_name": "RedPajamaC4"
} | 1,616 |
Q: Improper integral convergence test $\int_0^\infty\frac{x^2}{\sqrt[3]{x^9+1}}dx$ I've come this far in my computations:
$$0\le\int_0^\infty\frac{x^2}{\sqrt[3]{x^9+1}}dx\le\int_0^\infty\frac{x^2}{\sqrt[3]{x^9}}dx=\int_0^\infty\frac{x^2}{(x^9)^{1/3}}dx=\int_0^\infty \frac{x^2}{x^3}=\int_0^\infty\frac{1}{x}dx.$$
The last integral is divergent according to the $p-$test. However this does not show that the original integral has to be divergent too. The above result says that the original integral either diverges, but can also converge.
I somehow need to have the last integral less than or equal to the original one. Any trick here?
A: $$\int_0^\infty\frac{x^2}{\sqrt[3]{x^9+1}}dx\ge\int_0^\infty\frac{x^2}{2\sqrt[3]{x^9}}dx=\int_0^\infty\frac{x^2}{2(x^9)^{1/3}}dx=\int_0^\infty \frac{x^2}{2x^3}=\int_0^\infty\frac{1}{2x}dx=\infty$$
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 2,859 |
Geoffrey Colin Shephard foi um matemático britânico.
Carreira
Shephard obteve um Ph.D. em 1954 na Universidade de Cambridge, orientado por John Arthur Todd. Foi professor de matemática da University of East Anglia até aposentar-se.
É um matemático que trabalha com geometria convexa e grupos de reflexão. Ele questionou o problema de Shephard nos volumes de corpos convexos projetados, apresentou outro problema nas redes poliédricas, provou o teorema de Shephard-Todd na teoria invariante de grupos finitos, iniciou o estudo de politopos complexos e classificou os grupos de reflexão complexos.
Shephard obteve seu Ph.D. em 1954, do Queen's College, Cambridge, sob a supervisão de John Arthur Todd. Ele foi professor de matemática na Universidade de East Anglia até sua aposentadoria.
Publicações selecionadas
Ligações externas
Photo from Oberwolfach
Matemáticos do Reino Unido do século XX
Matemáticos do Reino Unido do século XXI
Alunos da Universidade de Cambridge | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 4,516 |
Q: Want to know the diff among pd.factorize, pd.get_dummies, sklearn.preprocessing.LableEncoder and OneHotEncoder All four functions seem really similar to me. In some situations some of them might give the same result, some not. Any help will be thankfully appreciated!
Now I know and I assume that internally, factorize and LabelEncoder work the same way and having no big differences in terms of results. I am not sure whether they will take up similar time with large magnitudes of data.
get_dummies and OneHotEncoder will yield the same result but OneHotEncoder can only handle numbers but get_dummies will take all kinds of input. get_dummies will generate new column names automatically for each column input, but OneHotEncoder will not (it rather will assign new column names 1,2,3....). So get_dummies is better in all respectives.
Please correct me if I am wrong! Thank you!
A: These four encoders can be split in two categories:
*
*Encode labels into categorical variables: Pandas factorize and scikit-learn LabelEncoder. The result will have 1 dimension.
*Encode categorical variable into dummy/indicator (binary) variables: Pandas get_dummies and scikit-learn OneHotEncoder. The result will have n dimensions, one by distinct value of the encoded categorical variable.
The main difference between pandas and scikit-learn encoders is that scikit-learn encoders are made to be used in scikit-learn pipelines with fit and transform methods.
Encode labels into categorical variables
Pandas factorize and scikit-learn LabelEncoder belong to the first category. They can be used to create categorical variables for example to transform characters into numbers.
from sklearn import preprocessing
# Test data
df = DataFrame(['A', 'B', 'B', 'C'], columns=['Col'])
df['Fact'] = pd.factorize(df['Col'])[0]
le = preprocessing.LabelEncoder()
df['Lab'] = le.fit_transform(df['Col'])
print(df)
# Col Fact Lab
# 0 A 0 0
# 1 B 1 1
# 2 B 1 1
# 3 C 2 2
Encode categorical variable into dummy/indicator (binary) variables
Pandas get_dummies and scikit-learn OneHotEncoder belong to the second category. They can be used to create binary variables. OneHotEncoder can only be used with categorical integers while get_dummies can be used with other type of variables.
df = DataFrame(['A', 'B', 'B', 'C'], columns=['Col'])
df = pd.get_dummies(df)
print(df)
# Col_A Col_B Col_C
# 0 1.0 0.0 0.0
# 1 0.0 1.0 0.0
# 2 0.0 1.0 0.0
# 3 0.0 0.0 1.0
from sklearn.preprocessing import OneHotEncoder, LabelEncoder
df = DataFrame(['A', 'B', 'B', 'C'], columns=['Col'])
# We need to transform first character into integer in order to use the OneHotEncoder
le = preprocessing.LabelEncoder()
df['Col'] = le.fit_transform(df['Col'])
enc = OneHotEncoder()
df = DataFrame(enc.fit_transform(df).toarray())
print(df)
# 0 1 2
# 0 1.0 0.0 0.0
# 1 0.0 1.0 0.0
# 2 0.0 1.0 0.0
# 3 0.0 0.0 1.0
I've also written a more detailed post based on this answer.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 2,859 |
Q: Angular9 application is not compiling after installation of angular material In Angular9, I have installed angular-material but it is giving following error.
ERROR in The target entry-point "@angular/material/sidenav" has missing dependencies:
*
*@angular/cdk/platform
*@angular/cdk/scrolling
*@angular/cdk/a11y
*@angular/cdk/bidi
*@angular/cdk/coercion
*@angular/cdk/keycodes
I have tried to downgrade the material version but it is still giving error.
A: Angular material has dependencies on angular/cdk and angular/animations.
Did you install with ng add, or npm install? Either way, you probably need to specify version 9.
Please post your package.json if you still need help.
A: This is the package.json file which I am using.
{
"name": "microservice-builder-client",
"version": "0.0.0",
"scripts": {
"ng": "ng",
"start": "ng serve",
"build": "ng build",
"test": "ng test",
"lint": "ng lint",
"e2e": "ng e2e"
},
"start": "ng serve --proxy-config proxy.config.json",
"private": true,
"dependencies": {
"@angular/animations": "~9.0.0",
"@angular/cdk": "^14.0.4",
"@angular/common": "~9.0.0",
"@angular/compiler": "~9.0.0",
"@angular/core": "~9.0.0",
"@angular/forms": "~9.0.0",
"@angular/material": "^11.2.3",
"@angular/platform-browser": "~9.0.0",
"@angular/platform-browser-dynamic": "~9.0.0",
"@angular/router": "~9.0.0",
"@clr/angular": "^3.1.4",
"@clr/core": "^3.1.4",
"@clr/icons": "^3.1.4",
"@clr/ui": "^3.1.4",
"http-server": "^0.12.3",
"ngx-toastr": "^12.1.0",
"npm-install-peers": "^1.2.1",
"rxjs": "~6.5.4",
"rxjs-compat": "^6.6.0",
"rxjs-tslint": "^0.1.8",
"swagger-editor-dist": "^3.11.6",
"tslib": "^1.13.0",
"zone.js": "~0.10.2"
},
"devDependencies": {
"@angular-devkit/build-angular": "^0.901.9",
"@angular/cli": "~9.0.0",
"@angular/compiler-cli": "~9.0.0",
"@angular/language-service": "~9.0.0",
"@types/jasmine": "~3.5.0",
"@types/jasminewd2": "~2.0.3",
"@types/node": "^12.11.1",
"codelyzer": "^5.1.2",
"jasmine-core": "~3.5.0",
"jasmine-spec-reporter": "~4.2.1",
"karma": "~4.3.0",
"karma-chrome-launcher": "~3.1.0",
"karma-coverage-istanbul-reporter": "~2.1.0",
"karma-jasmine": "~2.0.1",
"karma-jasmine-html-reporter": "^1.4.2",
"protractor": "~5.4.3",
"ts-node": "~8.3.0",
"tslint": "~5.18.0",
"typescript": "~3.7.5"
}
}
A: You're on Angular 9.
Your @angular/material is 11
Your @angular/cdk is 14.
That's your problem.
Lower the above to versions to 9, delete your node_modules folder, and run npm install.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 9,077 |
{"url":"https:\/\/www.gamedev.net\/forums\/topic\/702032-sqlite-data-base-taking-too-much-size-from-unity-side\/","text":"# C# Sqlite data base taking too much size from Unity side\n\n## Recommended Posts\n\nI have integrated the sqlite data base in slot game which is build in Unity platform for windows. It is increasing drastically although it has only text data., it is showing size about 1.5 GB which is very annoying\u00a0 and which will affect to my game's performance. so how can I reduce size of Sqlite database ?\n\n##### Share on other sites\n\n\"only text data\" isn't really a helpful description and text is not magically smaller than binary data, infact it can often be larger. What are you storing in your database? How do you need to query it? Can you delete some unneeded data?\n\n##### Share on other sites\nPosted (edited)\n\nHi SyncViews, Thanks for your answer, I just wanted to say is when we compare to large database and my database, It is vary rather than from each other and let me tell you that there are 40,000 entries in it and its taking 1.5 GB size.\n\nEdited by Gautti\n\n##### Share on other sites\n\nYou still need to provide more information. And \"1.5\" GB is nothing for many large databases, they can be terabytes or more, all depends on purpose. 1.5GB with a single table and 40,000 rows is only about 40KB per row on average, which while maybe bigger than the average row, can really be many things.\n\n\u2022 What exactly\u00a0is the schema for each table (the SQL CREATE TABLE ... to make them)\n\u2022 Provide\u00a0an example\u00a0of data for each table, at least 1 row\n\u2022 Explain how that data is being used\n\n##### Share on other sites\n17 hours ago, SyncViews said:\n\ns nothing for many large databases\n\nHi SyncViews, Actually I have done with it. My data was too long in error log. When I used vaccum, I found the bug and shorted out.\n\nThanks you !\n\n## Create an account\n\nRegister a new account\n\n\u2022 10\n\u2022 10\n\u2022 12\n\u2022 9\n\u2022 33","date":"2019-05-21 13:29:01","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\": 0.21288295090198517, \"perplexity\": 3213.4591292182054}, \"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\/1558232256381.7\/warc\/CC-MAIN-20190521122503-20190521144503-00447.warc.gz\"}"} | null | null |
\section{Introduction}
Early theoretical studies of gravitational lensing of distant quasars
by foreground galaxies approximated their mass distributions as spherical
and usually also as singular and isothermal (\cite{GG74}; \cite{Turner80};
Turner, Ostriker \& Gott 1984). Later studies explored the
relaxation of these simplifying approximations and, in particular,
considered elliptical mass distributions and potentials ({\it e.g.},
\cite{BK87}, \cite{KB87}, \cite{BN92}),
but it has been shown that the standard spherical approximations are adequate
for the purpose of many statistical calculations
({\it e.g.}, \cite{FT91}, \cite{MR93}, \cite{Koch93}).
However, with rare exception (\cite{OV90}), the gravitational
lensing effects of the defining component of spiral galaxies, their very
thin disks, has been ignored in discussions of lensing statistics.
There are two reasons. First, {\it within the limitations of the
spherical galaxy approximation}, it was early shown (Turner et al. 1984)
and later repeatedly confirmed by more detailed studies
(\cite{FT91}, \cite{Koch91}) that spiral galaxies contribute
only a small fraction, of order 10\%, of the total lensing cross section
of the known galaxy population, with ellipticals and S0's being the
dominant contributors. Second, the disk
stability arguments originated by Ostriker \& Peebles (1973) and the
many other lines of evidence which indicate the presence of massive
dark halos suggest that even spiral galaxy mass distributions are dominated
by a roughly spherical component.
While these considerations remain valid, there now seem to be some reasons
to examine the lensing effects of thin disk mass distributions more
carefully. For one, samples of galaxy lensed quasars are becoming
rapidly larger and better controlled. Thus, smaller and more subtle
features of lens statistics are becoming observationally accessible and
therefore theoretically interesting. For another, in the usual (and very
accurate) {\it thin lens} approximation, it is not the three dimensional
mass distribution of the lensing object that matters but rather the
projected two dimensional one, particularly those regions which exceed
the lensing critical surface density (\cite{TOG84}). From this
two dimensional point of view, a very thin disk may be much more important
when seen nearly edge-on than it is in three dimensions. Finally,
although the situation is far from unambiguous, observations suggest
that lensing galaxies may have more dust (\cite{Law95},
Malhotra, Rhoads \& Turner 1997) and larger quadrupole mements
(Keeton, Kochanek \& Seljak 1996) than would have been naively expected if early type
galaxies entirely dominated the quasar-galaxy lensing rate.
Therefore, in this paper we present a preliminary exploration of the
lensing properties of spiral galaxies, which we approximate as finite,
uniform surface density disks of zero thickness embedded in singular
isothermal spheres (SIS). We focus on the general caustic and critical
line properties of such lensing objects and also study the lensing
cross section enhancement over the SIS only approximation. We
explicitly carry the important variable of disk inclination angle
throughout the calculations. In the present paper, however, we do
not present the fully detailed calculations required for direct
comparison to observations; in particular, we do not here consider
line-of-sight integrations, integration over realistic distributions
of galaxy properties nor the effects of amplification (magnification)
bias on observed samples. Rather, we present only an initial
qualitative and semi-quantitative investigation of lensing by disk
galaxy mass distributions.
The basic equations and model are presented in section 2. Section 3 discusses
our technique for finding caustics, critical curves and multiple
imaging configurations. Section 4 presents the general lensing
properties of our model of disk galaxies in the language of caustics
and critical curves. Section 5 considers the effects of inclined disks
on lensing cross sections and the implied
modification of quasar
lensing rates. Finally, section 6 contains a summary of our results
and a discussion of recent related work.
We follow the conventions and notation of Schneider,
Ehlers \& Falco (1992).
\section{Light deflection due to a spiral galaxy}
We model spiral galaxies as uniform and infinitely thin disks embedded
in singular isothermal spheres.
Let us choose a length scale appropriate for a singular isothermal
sphere (SIS),
\begin{equation}
\xi_0 = 4 \pi \left(\frac{v_{SIS}}{c} \right)^2 \frac{D_d D_{ds}}{D_s},
\end{equation}
where $v_{SIS}$ is the line-of-sight velocity dispersion of the SIS.
Let $\mbox{\boldmath $\xi$}$ and $\mbox{\boldmath $\eta$}$ be the physical position vectors
of the image (in the lens plane) and the source (in the source plane)
respectively, then the dimensionless image and source
positions are
\begin{equation}
{\bf x}=\frac{\mbox{\boldmath $\xi$}}{\xi_0}; \hskip 1cm
{\bf y}= \frac{\mbox{\boldmath $\eta$}}{\eta_0},
\end{equation}
where $\eta_0= \xi_0 D_s/D_d$,
$D_s$ and $D_d$ are our distances to the source and lens respectively.
The lens equation becomes
\begin{equation}
{\bf y}= {\bf x}- \mbox{\boldmath $\alpha$}({\bf x}),
\end{equation}
The scaled deflection angle
\begin{equation}
\label{eq:alpha}
\mbox{\boldmath $\alpha$}({\bf x})= \frac{1}{\pi} \int{\rm d}^2x'\, \kappa({\bf x}')\,
\frac{ {\bf x}- {\bf x}'}{|{\bf x}- {\bf x}'|^2},
\end{equation}
where $\kappa({\bf x})= \Sigma(\xi_0{\bf x})/\Sigma_{crit}$, with
$\Sigma_{crit}= c^2 D_s/(4\pi G D_d D_{ds})$. $D_{ds}$ is the
distance between the lens and the source.
We use affine distances, $D_d=cH_0^{-1}\lambda(z_d)$,
$D_s= cH_0^{-1}\lambda(z_s)$, $D_{ds}= cH_0^{-1}\lambda(z_d,z_s)=
cH_0^{-1} (1+z_d)[\lambda(z_s)-\lambda(z_d)]$,
where
\begin{equation}
\lambda(z)= \int_0^z {\rm d}w\, \frac{1}{(1+w)^2
\sqrt{\Omega_0 (1+w)^3 +\Omega_{\Lambda} } }.
\end{equation}
For $\Omega_0=1$, $\lambda(z)=(2/5)\, \left[1- (1+z)^{-5/2}\right]$.
For a SIS, its dimensionless surface density and scaled deflection
angle are (\cite{book})
\begin{equation}
\kappa^{SIS}({\bf x})= \frac{1}{2 |{\bf x}|}; \hskip 1cm
\mbox{\boldmath $\alpha$}^{SIS}({\bf x})= \frac{\bf x}{|{\bf x}|}.
\end{equation}
The projection of a circular disk of radius $r_{disk}$ in the plane
perpendicular to the line of sight is an ellipse with semi-major axes
$r_{disk}$ and $r_{disk}\,\cos\theta$, where $\theta$ is the angle
between the normal vector of the inclined disk plane and the line of sight.
Note that $0\leq \theta \leq \pi/2$;
$\theta=0$ is the ``face-on'' disk case; $\theta=\pi/2$ is the
``edge-on'' disk case.
For a uniform disk, its projected surface density is
$\Sigma_{disk} =M_{disk}/(\pi r_{disk}^2 \cos\theta)$; its projected
dimensionless surface density is
\begin{eqnarray}
\kappa^{disk} &= &\frac{\Sigma_{disk}}{\Sigma_{crit}}
= \frac{1}{r_{disk}^2 \cos\theta}\, \frac{4GM_{disk}\, D_d D_{ds}}{c^2 D_s},
\nonumber \\
&=& \frac{0.19}{\cos\theta} \, \left( \frac{10\,\mbox{kpc}}{r_{disk}}
\right)^2\, \left(\frac{M_{disk}}{10^{11}\,M_{\odot}}\right)
\,\left( \frac{D_d}{1000\, \mbox{Mpc}}\right)\,
\frac{D_{ds}}{D_s}.
\end{eqnarray}
For simplicity in notation, we define
\begin{equation}
q = \kappa^{disk} \cos\theta.
\end{equation}
$q$ is the {\it face-on} surface mass density of the disk.
Note that $q<1$ for realistic choices of the parameters.
The dimensionless radius of the disk is
\begin{equation}
R = \frac{r_{disk}}{\xi_0}= \frac{10}{\pi} \left(\frac{r_{disk}}{10
\,\mbox{kpc}}\right)\, \left( \frac{150\,\mbox{km/s}}{v_{SIS}}
\right)^2\, \left( \frac{1000\, \mbox{Mpc}}{D_d}\right)\,
\frac{D_s}{D_{ds}}.
\end{equation}
Let us choose the ${\bf x}$ coordinates such that the
contour of the projected disk is given by
\begin{equation}
\left(\frac{x_1}{R}\right)^2+ \left(\frac{x_2}{R\cos\theta}\right)^2=1.
\end{equation}
Then the total scaled deflection angle
$\mbox{\boldmath $\alpha$}=\mbox{\boldmath $\alpha$}^{SIS}+\mbox{\boldmath $\alpha$}^{disk}$,
with $\mbox{\boldmath $\alpha$}^{disk}$ given by
\begin{eqnarray}
\label{eq:alphadisk}
&&\alpha_1^{disk}= \frac{R\,q}{2\pi}
\int_{-1}^{1}{\rm d}\omega \, \ln\left[
\frac{ ( \sqrt{1-\omega^2}+ x_1/R)^2 + (\omega\cos\theta- x_2/R)^2 }
{ ( \sqrt{1-\omega^2}- x_1/R)^2 + (\omega\cos\theta- x_2/R)^2 }
\right], \nonumber \\
&&
\alpha_2^{disk}= \frac{R\,q}{2\pi\cos\theta}
\int_{-1}^{1}{\rm d}\omega \, \ln\left[
\frac{ ( \omega- x_1/R)^2 + ( \sqrt{1-\omega^2}\,\cos\theta+ x_2/R)^2 }
{ ( \omega- x_1/R)^2 + ( \sqrt{1-\omega^2}\,\cos\theta- x_2/R)^2 }
\right].
\end{eqnarray}
Note that
\begin{equation}
\alpha_2^{disk}(\cos\theta \rightarrow 0)= \frac{2 x_2 q}{\pi}
\int_{-1}^{1}{\rm d}\omega \,
\frac{ \sqrt{1-\omega^2} }
{ ( \omega- x_1/R)^2 + ( \sqrt{1-\omega^2}\,\cos\theta- x_2/R)^2 }.
\end{equation}
To study image multiplicity, we will also need the derivatives
of the deflection angles:
\begin{eqnarray}
\label{eq:dalpha/dx}
&&\frac{\partial\alpha_1}{\partial x_1}= \frac{2 q}{\pi}
\int_{-1}^1{\rm d}\omega\, \frac{\sqrt{1-\omega^2}}{f_1(\omega)}\,
\left[ 1-\omega^2 - \left(\frac{x_1}{R} \right)^2 +
\left(\frac{x_2}{R}-\omega\cos\theta \right)^2 \right]
+\frac{x_2^2}{x^3}, \nonumber\\
&&\frac{\partial\alpha_1}{\partial x_2}= -\frac{4 q}{\pi}
\left(\frac{x_1}{R}\right)
\int_{-1}^1{\rm d}\omega\, \frac{\sqrt{1-\omega^2}}{f_1(\omega)}\,
\left(\frac{x_2}{R}-\omega\cos\theta \right)
-\frac{x_1 x_2}{x^3}, \\
&&\frac{\partial\alpha_2}{\partial x_1}= -\frac{4 q}{\pi}
\left(\frac{x_2}{R}\right)
\int_{-1}^1{\rm d}\omega\, \frac{\sqrt{1-\omega^2}}{f_2(\omega)}\,
\left(\frac{x_1}{R}-\omega \right)
-\frac{x_1 x_2}{x^3}, \nonumber\\
&&\frac{\partial\alpha_2}{\partial x_2}= \frac{2 q}{\pi}
\int_{-1}^1{\rm d}\omega\, \frac{\sqrt{1-\omega^2}}{f_2(\omega)}\,
\left[ \left(\frac{x_1}{R}-\omega \right)^2 -
\left(\frac{x_2}{R}\right)^2 + (1-\omega^2)\cos^2\theta \right]
+\frac{x_1^2}{x^3}.\nonumber
\end{eqnarray}
We have defined
\begin{eqnarray}
\label{eq:D1,D2}
&& f_1(\omega) \equiv \left[ \left(\frac{x_1}{R}+ \sqrt{1-\omega^2}\right)^2
+\left(\frac{x_2}{R}-\omega\cos\theta\right)^2\right]\,
\left[ \left(\frac{x_1}{R} - \sqrt{1-\omega^2}\right)^2+
\left(\frac{x_2}{R}-\omega\cos\theta\right)^2\right],
\nonumber \\
&& f_2(\omega) \equiv \left[ \left(\frac{x_1}{R}- \omega \right)^2
+\left(\frac{x_2}{R}+ \sqrt{1-\omega^2}\,\cos\theta\right)^2\right]\,
\left[ \left(\frac{x_1}{R}- \omega\right)^2
+\left(\frac{x_2}{R}- \sqrt{1-\omega^2}\,\cos\theta\right)^2\right].
\nonumber\\
&&
\end{eqnarray}
\section{Condition for multiple images}
The magnification factor of the source is given by
$\mu({\bf x}) = 1/\det\,A({\bf x})$, where
the Jacobian matrix $A({\bf x})$ is defined as
\begin{eqnarray}
\label{eq:detA}
&&A({\bf x})= \frac{\partial {\bf y}}
{\partial {\bf x}}, \hskip 1cm
A_{ij}= \frac{\partial y_i} {\partial x_j};\\
&&\det A= \left(1- \frac{\partial\alpha_1}{\partial x_1}\right)\,
\left(1- \frac{\partial\alpha_2}{\partial x_2} \right)
-\frac{\partial\alpha_1}{\partial x_2}\cdot
\frac{\partial\alpha_2}{\partial x_1}. \nonumber
\end{eqnarray}
It has been shown that an isolated transparent lens can produce
multiple images if, and only if, there is a point $\bf x$ with
$\det\,A({\bf x})<0$. If at ${\bf x}_0$, $\det\,A({\bf x}_0)<0$
(negative parity),
a source at ${\bf y}_0={\bf y}({\bf x}_0)$ has at least two
additional images of positive parity. The multiple-image cross-section
is simply the bounded area in the source plane in which for
each source position ${\bf y}$, there exists an image position ${\bf x}$
where $\det\,A({\bf x})<0$.
Formally, critical curves are given by $\det A({\bf x})=0$,
the corresponding source positions are the caustics.
However, the cross-section for multiple images is not always bounded
by a caustic. In the case of SIS only, $\det\,A=1-1/x$,
where $x = \sqrt{x_1^2+ x_2^2}$. $\det\,A<0$ gives $ 0\leq x < 1$.
The lens equation gives us $y=|x-1|$. Hence $y \leq 1$ for multiple
images; the multiple image cross-section is bounded by $y=1$, which corresponds
to $x=0$ (where $\det\,A=-\infty$), while the only critical curve is
at $x=1$ corresponding to $y=0$.
Next let us consider the case of SIS plus
a face-on disk, $\theta=0$, $\kappa^{disk}=q$. It's
straightforward to integrate Eqs.(\ref{eq:alphadisk}) to find (\cite{book})
\begin{equation}
\mbox{\boldmath $\alpha$} = \left\{ \begin{array} {ll}
{\bf x}/x + \kappa^{disk} \, {\bf x}, \hskip 1cm x<R; \\
{\bf x}/x + \kappa^{disk} R^2 \, {\bf x}/ x^2, \hskip 1cm x>R. \end{array}
\right.
\end{equation}
Hence we have
\begin{equation}
\det A=\left\{ \begin{array}{ll}
\left(1-\kappa^{disk}\right)\,\left(1-\kappa^{disk} - 1/x\right),
\hskip 1cm x<R; \\
\left(1+ \kappa^{disk} R^2/x^2\right)\,
\left(1-1/x - \kappa^{disk} R^2/x^2 \right),
\hskip 1cm x>R. \end {array} \right.
\end{equation}
It is easy to see that $\det A$ is {\it discontinuous} at $x=R$.
For $\kappa^{disk}<1$, $\det A$ does {\it not} change sign at $x=R$.
If $R>1/(1-\kappa^{disk})$, the only critical curve is given by
$x=1/(1-\kappa^{disk}) <R$, and it maps to $y=0$;
if $R<1/(1-\kappa^{disk})$, the only critical curve is given by
$x=\left(1+ \sqrt{1+ 4\kappa^{disk} R^2} \right)/2 >R$,
and it also maps to $y=0$.
The critical curves therefore have no relevance to image multiplicity.
On the other hand, $\det A(x)<0$ leads to $y \leq 1$, with $y=1$
given by $x=0$, just as in the SIS only case.
SIS with a face-on disk of $\kappa^{disk}<1$ has the same multiple-image
cross section as SIS only.
If $\kappa^{disk}>1$, the true critical curve is given by
$x=(1+ \sqrt{1+ 4\kappa^{disk} R^2} )/2 >R$, which again maps to $y=0$.
However, $\det A(x<R)>0$, $\det A(x \rightarrow R^{+})<0$,
hence $\det A$ changes sign at $x=R$, which is effectively a
``critical curve''; the corresponding ``caustic''
is given by $y=1+R(\kappa^{disk}-1) >1$.
SIS with a face-on disk of $\kappa^{disk}>1$ has a larger multiple-image
cross section than SIS only.
\section{Image multiplicity for SIS plus inclined disk}
For SIS plus uniform disk, $\mbox{\boldmath $\alpha$}^{disk}(x=0)=0$, hence $x=0$ corresponds
to source position $y=1$, just as in the SIS only case. When
$\det\,A(x\rightarrow 0)<0$, $y\leq 1$ always gives multiple images;
the multiple-image cross-section increases relative to the SIS only case
if any caustic curves lie outside the $y=1$ circle.
The value of $\det A$ at $x=0$ is a useful indicator
of the general properties of the critical curves.
For uniform disk plus SIS, we find [using Eqs.(\ref{eq:dalpha/dx}) and
(\ref{eq:detA})]
\begin{equation}
\label{eq:detA(x=0)}
\det A (x\rightarrow 0) = 1-\frac{2q}{\cos\theta}+ \frac{1}{\cos\theta}
\left(\frac{2q}{1+\cos\theta}\right)^2
+ \frac{C_1(\theta)\, x_1^2}{x^3}+ \frac{C_2(\theta)\, x_2^2}{x^3},
\end{equation}
where
\begin{equation}
C_1(\theta)=\frac{2q}{1+\cos\theta}-1, \hskip 1cm
C_2(\theta) = \frac{2q}{\cos\theta(1+\cos\theta)}-1.
\end{equation}
Let us define
\begin{equation}
S \equiv C_1 x_1^2+ C_2 x_2^2.
\end{equation}
$S$ has the same sign as $\det A (x\rightarrow 0)$.
We have two critical angles:
\begin{eqnarray}
\label{eq:thetacrit}
&& C_1(\theta)=0: \hskip 1cm \theta=\theta_1= \arccos(2q-1);\nonumber\\
&& C_2(\theta)=0: \hskip 1cm \theta=\theta_2=
\arccos\left( \frac{\sqrt{1+8q}-1}{2} \right).
\end{eqnarray}
Note that $\theta_1$ is not defined for $q<1/2$.
For $q>1/2$, $\theta_1>\theta_2$.
Note that $\det A$ is {\it discontinuous} for $(x_1^c,x_2^c)$ on the ellipse
$(x_1^c/R)^2+ (x_2^c/R\cos\theta)^2=1$ [see Eqs.(\ref{eq:D1,D2})],
i.e., on the edge of the disk, just as in the face-on disk case.
If $\det A(x\rightarrow 0) = +\infty$, $\det A $ changes sign at
$(x_1^c,x_2^c)$, which describes effectively a ``critical curve'',
and the corresponding source positions give us a ``caustic''.
First we consider $q<1/2$. Here we always have $C_1<0$.
We have two different cases: (1) $\theta<\theta_2$, $C_2<0$, $S<0$,
$\det A(x\rightarrow 0) = -\infty$, we have only one true critical curve
and it encloses $x=0$;
(2) $\theta>\theta_2$, $C_2>0$, $S>0$ [$\det A(x\rightarrow 0) = +\infty$]
if $|x_2|> \sqrt{|C_1|/C_2}\, |x_1|$, $S<0$
[$\det A(x\rightarrow 0) = -\infty$]
if $|x_2|< \sqrt{|C_1|/C_2}\, |x_1|$,
here we have one true critical curve enclosing $x=0$, plus an additional
``critical curve'' which is hourglass shaped and passes through $x=0$,
and is closed off by $(x_1^c,x_2^c)$ defined above.
Figs.1-3 show the critical curves (a) and caustics (b) for $q=0.2$
($\theta_2=72.168^{\circ}$), for $\theta=70^{\circ}$,
75$^{\circ}$, and 85$^{\circ}$ respectively. The unit radius circle in
(a) indicates the critical curve for the SIS only case (its corresponding
caustic shrinks to the point $y=0$ in the source plane); the
unit radius circle in (b) indicates the multiple-image cross-section
for the SIS only case.
For $q>1/2$, we have three different cases:
(1) $\theta<\theta_2$, $C_1<0$, $C_2<0$, $S<0$,
$\det A(x\rightarrow 0) = -\infty$, we have only one true critical curve
and it encloses $x=0$;
(2) $\theta_2<\theta< \theta_1$, $C_1<0$, $C_2>0$,
$S>0$ [$\det A(x\rightarrow 0) = +\infty$]
if $|x_2|> \sqrt{|C_1|/C_2}\, |x_1|$,
$S<0$ [$\det A(x\rightarrow 0) = -\infty$]
if $|x_2|< \sqrt{|C_1|/C_2}\, |x_1|$,
here we have one true critical curve enclosing $x=0$, plus an additional
``critical curve'' which is hourglass shaped and passes through $x=0$,
and is closed off by $(x_1^c,x_2^c)$ defined above;
(3) $\theta>\theta_1$, $C_1>0$, $C_2>0$, $S>0$,
$\det A(x\rightarrow 0) = +\infty$, we have two critical curves
enclosing $x=0$, one of them is the true critical curve, the other is
given by $(x_1^c,x_2^c)$ defined above.
Figs.4-6 show the critical curves (a) and caustics (b) for $q=0.6$
($\theta_1=78.463^{\circ}$, $\theta_2= 45.238^{\circ}$),
for $\theta=45^{\circ}$, 75$^{\circ}$, and 85$^{\circ}$ respectively.
The unit radius circles in the
figures are the same as already indicated.
Because $\det A({\bf x})$ is modified from Eq.(\ref{eq:detA(x=0)})
as we move away from $x=0$, the hourglass shaped ``critical curves''
are given by the intersection of $|x_2| \sim \sqrt{|C_1|/C_2}\, |x_1|$ and
the ellipse $(x_1^c/R)^2+ (x_2^c/R\cos\theta)^2=1$; this explains
the curved sides of the hourglass shape.
Due to numerical noise, some discrete points from the ($x_1^c, x_2^c$)
ellipse and their corresponding source positions appear in
Figs.1-6 as discrete points apart from the critical curves and caustics.
It is interesting to note that our study of the face-on disk case
(see the previous section)
suggests that the critical angle for the disk to contribute to
image multiplicity is given by $\kappa^{disk}=q/\cos\theta=1$,
i.e., $\theta=\theta_0 \equiv \arccos(q)$. For $q<1/2$, only
$\theta_2$ is defined, and $\theta_0>\theta_2$;
for $q>1/2$, $\theta_1> \theta_0 >\theta_2$.
This indicates that an inclined disk is more efficient than a face-on
disk with the same surface mass density.
\section{Increased multiple-image cross section due to the inclined disk}
When there is only {\it one} critical curve (the true critical curve)
in the ${\bf x}$ plane, the multiple image cross-section
is given by the area enclosed by $y=1$ plus the areas enclosed by the caustic
which lies {\it outside} the $y=1$ circle, which are two small sharp-angular
areas lying along the $y_1$ axis;
the caustic is completely inside the $y=1$ circle for
sufficiently small $\theta$.
When we have {\it two} branches of
critical curves (one of them is a true critical curve, the other
corresponds to discontinuity and change of sign in $\det A$)
in the ${\bf x}$ plane, the multiple image cross-section
is given by the area enclosed by $y=1$ plus the areas enclosed by the caustics
which lie {\it outside} the $y=1$ circle, which now consist of two angular
areas lying along the $y_1$ axis
(corresponding to the true critical curve, at largest possible $x$),
plus two round areas lying along the $y_2$ axis
(corresponding to the ``critical curve'' due to discontinuity
and change of sign in $\det A$).
Assuming uniform distribution of the inclination angle $\theta$ in solid angle,
the average multiple-image cross section is
\begin{equation}
\overline{\sigma}_{\theta}= \frac{ \int {\rm d}\theta \, \sin\theta \,
\sigma(\theta)}
{ \int {\rm d}\theta \, \sin\theta },
\end{equation}
where $\sigma(\theta)$ is multiple-image cross section for inclination
angle $\theta$.
Figs.7-9 show the ratio of the multiple-image cross-sections for
SIS plus inclined uniform disk and for SIS only, $\sigma(\theta)/\sigma^{SIS}$,
as a function of the inclination angle $\theta$, for three groups (nine sets) of
choices of ($q,R)$. Table 1 lists the parameters and the corresponding
average enhancement factor in multiple-image cross section
$\overline{\sigma}_{\theta}/\sigma^{SIS}$.
\begin{table}[h]
\caption{Average enhancement factor in multiple-image cross section}
\begin{center}
\begin{tabular}{|c||p{1in}|p{1in}|p{1in}|}
\hline
& $R=1.5$ & $R=3$ & $R=6$ \\ \hline\hline
q=0.2 & 1.086 & 1.160 & 2.041 \\ \hline
q=0.4 & 1.296 & 1.547 & 2.742 \\ \hline
q=0.6 & 1.718 & 2.493 & 4.509 \\ \hline
\end{tabular}
\end{center}
\end{table}
Let us write
\begin{equation}
q\,R= \frac{1.9}{\pi}\, \left( \frac{10\,\mbox{kpc}}{r_{disk}}\right)\,
\left( \frac{ M_{disk}}{ 10^{11} M_{\odot}} \right)\,
\left( \frac{150\,\mbox{km/s}}{v_{SIS}} \right)^2.
\end{equation}
For a model spiral galaxy with given ($v_{SIS}, M_{disk},
r_{disk}$), $q\,R$ is constant.
Inspection of Table 1 shows that the modification of the galaxy's inclination
averaged cross section is quite small, less than say 50\% (there are other
uncertainties in lens statistics calculations {\it at least} this large),
if $q\,R$ is less than about unity. From equation (23) we then see that
significant modification of the multiple image lensing cross section will
only occur for objects with uncharacteristically small and massive disks and/or
those with minimal spherical (halo) components. Few, if any, real galaxies
may satisfy such conditions.
The differential probability of a strong-lensing (multiple-images)
event is (\cite{TOG84})
\begin{equation}
{\rm d}\tau = n_L(z_d)\, \overline{\sigma}_{\theta}(z_d|z_s)\,
\frac{c\, {\rm d}t}{ {\rm d} z_d}\, {\rm d} z_d,
\end{equation}
where $n_L(z_d)$ is the number density of lenses at lens redshift $z_d$,
$\overline{\sigma}_{\theta}(z_d|z_s)$ is the inclination-angle averaged
cross-section for multiple images given source redshift $z_s$.
From the above discussion, we would expect the total contribution of spiral
galaxies to the QSO strong lensing optical depths to increase only slightly
due to the effects of their thin disks. The optical depths would then continue
to be dominated by early type galaxies.
However, the inclusion of an inclined disk breaks the circular
symmetry due to the SIS, the true critical curve [where $\det A(x)=0$]
no longer maps to a point ($y=0$) as in the SIS only case,
but maps to a caustic which encloses an area comparable to
the SIS only multiple-image cross-section for $\theta>\theta_2$
[see Eq.(\ref{eq:thetacrit})].
Since the magnification is infinite (for a point source) on the caustic,
and decreases smoothly away from it,
the cross section for high magnifications is significantly increased
due to the inclusion of a disk.
Quantitative investigation of this effect will require extensive numerical
calculations which are outside the scope of this paper, but
this effect may well be important.
Observational selection effects favoring inclusion of high
amplification (magnification) lensing configurations in flux limited
samples can lead to major distortions of intrinsic distributions in real
samples; see Ostriker \& Vietri (1990), for example.
\section{Discussion and conclusion}
We have examined strong (multiple image) gravitational lensing
by spiral galaxies, modeled as infinitely thin
uniform disks embedded in singular isothermal spheres.
We have derived general properties of the critical curves and caustics
analytically.
The multiple-image cross section is a sensitive function of
the inclination angle of the disk relative to the observer.
We have therefore computed the inclination averaged cross section
for several sets of lensing parameters.
We find that
the optical depth for multiply imaged QSOs should only
increase by a factor of a few at most and by less
than 50\% in nearly all realistic cases; the inclusion of
a disk is therefore expected to have {\it no} significant effect
on the contribution
of spiral galaxies to the total optical depth for
multiply imaged QSOs.
On the other hand, the cross section for high magnifications is
significantly increased due to the inclusion of a disk.
The increase in the number of lensed QSOs with
high magnifications could have substantial effects on observed lens samples.
While completing this work, we became aware of recent preprints
by Loeb (1997) and by
Maller, Flores, and Primack (1997).
The former is primarily concerned with the possible connectin between
lensing by spiral galaxies and high column density HI absorption systems
seen in QSO spectra but also briefly mentions the potential role of disk
inclination in spiral galaxy lensing statistics.
The subject of
our paper overlaps partially with that of Maller {\it et al.},
and some of our results
are qualitatively similar;
however, our paper is mainly analytical (which provides insight into the
mathematics of the critical curves and caustics) while theirs is numerical.
We thank A. Loeb for useful discussions and gratefully acknowledge
support from NSF grant AST94-19400.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 6,562 |
\section{Introduction}
Let $p$ be an odd prime.
Let $K$ be a totally real number field and let $K_{\infty}$ be the cyclotomic $\mathbb{Z}_{p}$-extension of $K$.
An admissible $p$-adic Lie extension $\mathcal{L}$ of $K$ is a Galois extension $\mathcal{L}$ of $K$ such that
(i) $\mathcal{L}/K$ is unramified outside a finite set of primes $S$ of $K$,
(ii) $\mathcal{L}$ is totally real, (iii) $\mathcal{G} := \mathrm{Gal}(\mathcal{L}/K)$ is a compact
$p$-adic Lie group, and (iv) $\mathcal{L}$ contains $K_{\infty}$.
Let $M_{S}^{\mathrm{ab}}(p)$ be the maximal abelian $p$-extension of $\mathcal{L}$ unramified outside the set of primes above $S$.
Let $\Lambda(\mathcal{G}):=\mathbb{Z}_{p}[[\mathcal{G}]]$ denote the Iwasawa algebra of $\mathcal{G}$ over $\mathbb{Z}_{p}$
and let $X_{S}$ denote the (left) $\Lambda(\mathcal{G})$-module $\mathrm{Gal}(M_{S}^{\mathrm{ab}}(p) / \mathcal{L})$.
Roughly speaking, the equivariant Iwasawa main conjecture (EIMC) relates $X_{S}$ to special values of Artin $L$-functions
via $p$-adic $L$-functions. This relationship can be expressed as the existence of a certain element in an algebraic $K$-group;
it is also conjectured that this element is unique.
There are at least three different versions of the EIMC.
The first is due to Ritter and Weiss and deals with the case where $\mathcal{G}$ is a one-dimensional $p$-adic Lie group \cite{MR2114937},
and was proven under a certain `$\mu=0$' hypothesis in a series of articles culminating in \cite{MR2813337}.
The second version follows the framework of Coates, Fukaya, Kato, Sujatha and Venjakob \cite{MR2217048} and
was proven by Kakde \cite{MR3091976}, again assuming $\mu=0$.
This version is for $\mathcal{G}$ of arbitrary (finite) dimension and Kakde's proof uses a strategy of Burns and Kato to reduce to the one-dimensional case
(see Burns \cite{MR3294653}). Finally, Greither and Popescu \cite{MR3383600} have formulated and proven an EIMC via the Tate module of a certain Iwasawa-theoretic abstract $1$-motive, but they restricted their formulation to one-dimensional abelian extensions and the formulation itself
requires a $\mu=0$ hypothesis.
In \cite{MR3072281}, the second named author generalised this formulation (again assuming $\mu=0$) to the one-dimensional non-abelian case, and in the situation that all three formulations make sense (i.e.\ $\mathcal{G}$ is a one-dimensional $p$-adic Lie group and $\mu=0$),
he showed that they are in fact all equivalent.
Venjakob \cite{MR3068897} has also compared the work of Ritter and Weiss to that of Kakde.
The classical Iwasawa $\mu=0$ conjecture (at $p$) is the assertion that for every number field $F$, the
Galois group of the maximal unramified abelian $p$-extension of $F_{\infty}$ is finitely generated as a $\mathbb{Z}_{p}$-module.
This conjecture was proven by Ferrero and Washington \cite{MR528968} in the case that $F/\mathbb{Q}$ is abelian, but little progress has been made since.
The $\mu=0$ hypothesis for an admissible $p$-adic Lie extension $\mathcal{L}/K$ discussed in the paragraph above
is implied by the classical Iwasawa $\mu=0$ conjecture (at $p$) for a certain finite extension of $K$.
It follows from the result of Ferrero and Washington that the EIMC holds unconditionally when $\mathcal{L}/K$
is an admissible pro-$p$ extension and $K$ is abelian over $\mathbb{Q}$.
We wish to prove the EIMC in cases in which the $\mu=0$ hypothesis is not known.
As a consequence, we must restrict to the case in which $\mathcal{G}$ is one-dimensional because the
`$\mathfrak{M}_{H}(G)$-conjecture' is required to even formulate the EIMC.
This conjecture is known unconditionally in the one-dimensional case but in the more general case it is
presently only known to hold under the $\mu=0$ hypothesis (see \S \ref{subsec:higher-rk}).
We remark that in the one-dimensional case, the $\mu=0$ hypothesis is equivalent to
the Iwasawa module $X_{S}$ being finitely generated as a $\mathbb{Z}_{p}$-module;
a detailed discussion of the relation to the classical Iwasawa $\mu=0$ conjecture in this setting is given in Remark \ref{rmk:mu=0}.
The main result of this article is the unconditional proof of the EIMC for an infinite class of admissible extensions with Galois group $\mathcal{G}$
a one-dimensional non-abelian $p$-adic Lie group and for which the $\mu=0$ hypothesis is not known.
A key ingredient is a result of Ritter and Weiss \cite{MR2114937} which, roughly speaking,
says that a version of the EIMC `over maximal orders' (or `character by character') holds without any $\mu=0$ hypothesis.
The proof uses Brauer induction to reduce to the abelian case, which is essentially equivalent to
the Iwasawa main conjecture for totally real fields proven by Wiles \cite{MR1053488}.
Another key ingredient is the notion of `hybrid Iwasawa algebras' which is an adaptation of
the notion of `hybrid $p$-adic group rings' first introduced by the present authors in \cite{MR3461042}.
Let $p$ be a prime and $G$ be a finite group with normal subgroup $N$.
The group ring $\mathbb{Z}_{p}[G]$ is said to be `$N$-hybrid' if $\mathbb{Z}_{p}[G]$ is isomorphic to the direct product of $\mathbb{Z}_{p}[G/N]$ and a maximal order.
Now suppose that $p$ is odd and that $\mathcal{G}$ is a one-dimensional $p$-adic Lie group with a finite normal subgroup $N$.
Then the Iwasawa algebra $\Lambda(\mathcal{G})$ is said to be `$N$-hybrid'
if it decomposes into a direct product of $\Lambda(\mathcal{G}/N)$ and a maximal order.
By using the maximal order variant of the EIMC and certain functoriality properties, we show that the EIMC for the
full extension (corresponding to $\mathcal{G}$) is equivalent to the EIMC for the sub-extension corresponding to $\mathcal{G}/N$.
There are many cases in which $\mu=0$ is not known for the full extension, but is known for the sub-extension,
and thus we obtain new unconditional results.
However, we first need explicit criteria for $\Lambda(\mathcal{G})$ to be $N$-hybrid.
Since $\mathcal{G}$ is one-dimensional it decomposes as a semidirect product $\mathcal{G} = H \rtimes \Gamma$
where $H$ is finite and $\Gamma$ is isomorphic to $\mathbb{Z}_{p}$.
Moreover, if $N$ is a finite normal subgroup of $\mathcal{G}$ then it must in fact be a (normal) subgroup of $H$.
We show that $\Lambda(\mathcal{G})$ is $N$-hybrid if and only if $\mathbb{Z}_{p}[H]$ is $N$-hybrid;
this is easy to see in the case that $\mathcal{G}$ is a direct product $H \times \Gamma$,
but much more difficult in the general case.
In \cite{MR3461042} we gave explicit criteria for $\mathbb{Z}_{p}[H]$ to be $N$-hybrid in terms of the degrees of the complex irreducible characters of $H$,
and thus the same criteria can be used to determine whether $\Lambda(\mathcal{G})$ is $N$-hybrid.
We also study the behavior of $N$-hybrid
$p$-adic group rings $\mathbb{Z}_{p}[G]$ when we change the group $G$ and its normal subgroup $N$.
(As discussed in Remark \ref{rmk:applications-to-ETNC}, these new results on hybrid $p$-adic group rings also have applications
to the equivariant Tamagawa number conjecture.)
\\
In \cite{non-abelian-Brumer-Stark}, we show that the EIMC implies the $p$-primary parts of refinements of the
imprimitive Brumer and Brumer-Stark conjectures.
These latter conjectures are classical when the relevant Galois group is abelian, and have been generalised to
the non-abelian case by the second named author \cite{MR2976321} (independently,
they have been formulated in even greater generality by Burns \cite{MR2845620}).
By combining this with the main result of the present article, we give unconditional proofs of
the non-abelian Brumer and Brumer-Stark conjectures in many cases.
\\
This article is organised as follows. In \S \ref{sec:hybrid-group-rings} we review some material on hybrid
$p$-adic group rings, and show how new examples of such group rings can be obtained from existing examples.
In \S \ref{sec:hybrid-Iwasawa-alg} we generalise this notion to Iwasawa algebras and study its structure and basic properties.
In particular, we give explicit criteria for an Iwasawa algebra to be $N$-hybrid.
This enables us to give many examples of one-dimensional non-abelian $p$-adic Lie groups $\mathcal{G}$
such that the Iwasawa algebra $\Lambda(\mathcal{G})$ is $N$-hybrid for a non-trivial finite normal subgroup
$N$ of $\mathcal{G}$.
In \S \ref{sec:EIMC} we give a slight reformulation of the EIMC that is convenient for our purposes.
We also recall the functorial properties of the EIMC and reinterpret the maximal order variant of the EIMC of Ritter and Weiss \cite{MR2114937}.
The algebraic preparations of \S \ref{sec:hybrid-Iwasawa-alg} then permit us to verify the EIMC for many one-dimensional non-abelian
$p$-adic Lie extensions without assuming the $\mu=0$ hypothesis.
\subsection*{Acknowledgements}
It is a pleasure to thank Werner Bley, Ted Chinburg, Takako Fukaya, Lennart Gehrmann, Cornelius Greither, Annette Huber-Klawitter,
Mahesh Kakde, Kazuya Kato, Daniel Macias Castillo, Cristian Popescu, J\"urgen Ritter, Sujatha, Otmar Venjakob, Christopher Voll,
Al Weiss and Malte Witte for helpful discussions and correspondence.
The authors also thank the referee for several helpful comments.
The second named author acknowledges financial support provided by the DFG within the Collaborative Research Center 701
`Spectral Structures and Topological Methods in Mathematics'.
\subsection*{Notation and conventions}
All rings are assumed to have an identity element and all modules are assumed
to be left modules unless otherwise stated. We fix the following notation:
\medskip
\begin{tabular}{ll}
$S_{n}$ & the symmetric group on $n$ letters\\
$A_{n}$ & the alternating group on $n$ letters\\
$C_{n}$ & the cyclic group of order $n$\\
$D_{2n}$ & the dihedral group of order $2n$\\
$Q_{8}$ & the quaternion group of order $8$\\
$V_{4}$ & the subgroup of $A_{4}$ generated by double transpositions\\
$\mathbb{F}_{q}$ & the finite field with $q$ elements, where $q$ is a prime power\\
$\mathrm{Aff}(q)$ & the affine group isomorphic to $\mathbb{F}_{q} \rtimes \mathbb{F}_{q}^{\times}$ defined in Example \ref{ex:affine}\\
$v_{p}(x)$ & the $p$-adic valuation of $x \in \mathbb{Q}$\\
$R^{\times}$ & the group of units of a ring $R$\\
$\zeta(R)$ & the centre of a ring $R$\\
$M_{m \times n} (R)$ & the set of all $m \times n$ matrices with entries in a ring $R$\\
$\zeta_{n}$ & a primitive $n$th root of unity\\
$K_{\infty}$ & the cyclotomic $\mathbb{Z}_{p}$-extension of the number field $K$\\
$K^{+}$ & the maximal totally real subfield of $K$\\
$K^{c}$ & an algebraic closure of a field $K$ \\
$\mathrm{Irr}_{F}(G)$ & the set of $F$-irreducible characters of the (pro)-finite group $G$\\
& (with open kernel) where $F$ is a field of characteristic $0$
\end{tabular}
\section{Hybrid $p$-adic group rings} \label{sec:hybrid-group-rings}
We recall material on hybrid $p$-adic group rings from \cite[\S 2]{MR3461042}
and prove new results which provide many new examples.
We shall sometimes abuse notation by using the symbol $\oplus$ to denote the direct product of rings or orders.
\subsection{Background material}
Let $p$ be a prime and let $G$ be a finite group.
For a normal subgroup $N \unlhd G$, let $e_{N} = |N|^{-1}\sum_{\sigma \in N} \sigma$
be the associated central trace idempotent in the group algebra $\mathbb{Q}_{p}[G]$.
Then there is a ring isomorphism $\mathbb{Z}_{p}[G]e_{N} \simeq \mathbb{Z}_{p}[G/N]$.
We now specialise \cite[Definition 2.5]{MR3461042} to the case of $p$-adic group rings
(we shall not need the more general case of $N$-hybrid orders).
\begin{definition}
Let $N \unlhd G$. We say that the $p$-adic group ring $\mathbb{Z}_{p}[G]$ is \emph{$N$-hybrid}
if (i) $e_{N} \in \mathbb{Z}_{p}[G]$ (i.e. $p \nmid |N|$) and (ii) $\mathbb{Z}_{p}[G](1-e_{N})$ is a maximal
$\mathbb{Z}_{p}$-order in $\mathbb{Q}_{p}[G](1-e_{N})$.
\end{definition}
\begin{remark}
The group ring $\mathbb{Z}_{p}[G]$ is itself maximal if and only if $p$ does not divide $|G|$
if and only if $\mathbb{Z}_{p}[G]$ is $G$-hybrid. Moreover, $\mathbb{Z}_{p}[G]$ is always $\{1\}$-hybrid.
\end{remark}
For every field $F$ of characteristic $0$ and every finite group $G$, we denote by $\mathrm{Irr}_{F}(G)$
the set of $F$-irreducible characters of $G$.
Let $\mathbb{Q}_{p}^{c}$ be an algebraic closure of $\mathbb{Q}_{p}$.
If $x$ is a rational number, we let $v_{p}(x)$ denote its $p$-adic valuation.
\begin{prop}[{\cite[Proposition 2.7]{MR3461042}}]\label{prop:hybrid-criterion-groupring}
The group ring $\mathbb{Z}_{p}[G]$ is $N$-hybrid if and only if
for every $\chi \in \mathrm{Irr}_{\mathbb{Q}_{p}^{c}}(G)$ such that $N \not \leq \ker \chi$
we have $v_{p}(\chi(1))=v_{p}(|G|)$.
\end{prop}
\begin{remark}
In the language of modular representation theory, when $v_{p}(\chi(1))=v_{p}(|G|)$
we say that ``$\chi$ belongs to a $p$-block of defect zero''.
\end{remark}
\subsection{Frobenius groups}\label{subsec:frobenius-groups}
We recall the definition and some basic facts about Frobenius groups and then use them to
provide many examples of hybrid group rings.
For further results and examples, we refer the reader to \cite[\S 2.3]{MR3461042}.
\begin{definition}
A \emph{Frobenius group} is a finite group $G$ with a proper non-trivial subgroup $H$
such that $H \cap gHg^{-1}=\{ 1 \}$ for all $g \in G-H$,
in which case $H$ is called a \emph{Frobenius complement}.
\end{definition}
\begin{theorem}\label{thm:frob-kernel}
A Frobenius group $G$ contains a unique normal subgroup $N$, known as the Frobenius kernel, such that
$G$ is a semidirect product $N \rtimes H$. Moreover:
\begin{enumerate}
\item $|N|$ and $[G:N]=|H|$ are relatively prime.
\item The Frobenius kernel $N$ is nilpotent.
\item If $K \unlhd G $ then either $K \unlhd N$ or $N \unlhd K$.
\item If $\chi \in \mathrm{Irr}_{\mathbb{C}}(G)$ such that $N \not \leq \ker \chi$ then $\chi= \mathrm{ind}_{N}^{G}(\psi)$ for some $1 \neq \psi \in \mathrm{Irr}_{\mathbb{C}}(N)$.
\end{enumerate}
\end{theorem}
\begin{proof}
For (i) and (iv) see \cite[\S 14A]{MR632548}.
For (ii) see \cite[10.5.6]{MR1357169} and for (iii) see \cite[Exercise 7, \S 8.5]{MR1357169}.
\end{proof}
\begin{prop}[{\cite[Proposition 2.13]{MR3461042}}]\label{prop:frob-N-hybrid}
Let $G$ be a Frobenius group with Frobenius kernel $N$.
Then for every prime $p$ not dividing $|N|$, the group ring $\mathbb{Z}_{p}[G]$ is $N$-hybrid.
\end{prop}
\begin{proof}
We repeat the short argument for the convenience of the reader.
Let $\chi \in \mathrm{Irr}_{\mathbb{Q}_{p}^{c}}(G)$ such that $N \not \leq \ker \chi$.
Then by Theorem \ref{thm:frob-kernel} (iv) $\chi$ is induced from a nontrivial irreducible character
of $N$ and so $\chi(1)$ is divisible by $[G:N]$.
However, $|N|$ and $[G:N]$ are relatively prime by Theorem \ref{thm:frob-kernel} (i) and so
Proposition \ref{prop:hybrid-criterion-groupring} now gives the desired result.
\end{proof}
We now give some examples, the first two of which were also given in \cite[\S 2.3]{MR3461042}.
\begin{example}\label{ex:metacyclic}
Let $p<q$ be distinct primes and assume that $p \mid (q-1)$.
Then there is an embedding $C_{p} \hookrightarrow \mathrm{Aut}(C_{q})$ and so there is
a fixed-point-free action of $C_{p}$ on $C_{q}$.
Hence the corresponding semidirect product $G = C_{q} \rtimes C_{p}$ is a Frobenius group
(see \cite[Theorem 2.12]{MR3461042} or \cite[\S 4.6]{MR2599132}, for example), and so $\mathbb{Z}_{p}[G]$ is $N$-hybrid with $N = C_{q}$.
\end{example}
\begin{example}\label{ex:affine}
Let $q$ be a prime power and let $\mathbb{F}_{q}$ be the finite field with $q$ elements.
The group $\mathrm{Aff}(q)$ of affine transformations on $\mathbb{F}_{q}$ is the group of transformations
of the form $x \mapsto ax +b$ with $a \in \mathbb{F}_{q}^{\times}$ and $b \in \mathbb{F}_{q}$.
Let $G=\mathrm{Aff}(q)$ and let $N=\{ x \mapsto x+b \mid b \in \mathbb{F}_{q} \}$.
Then $G$ is a Frobenius group with Frobenius kernel $N \simeq \mathbb{F}_{q}$ and is isomorphic to
the semidirect product $\mathbb{F}_{q} \rtimes \mathbb{F}_{q}^{\times}$ with the natural action.
Moreover, $G/N \simeq \mathbb{F}_{q}^{\times} \simeq C_{q-1}$ and $G$ has precisely one
non-linear irreducible complex character, which is rational-valued and of degree $q-1$.
Hence for every prime $p$ not dividing $q$, we have that $\mathbb{Z}_{p}[G]$ is $N$-hybrid
and is isomorphic to $\mathbb{Z}_{p}[C_{q-1}] \oplus M_{(q-1) \times (q-1)}(\mathbb{Z}_{p})$.
Note that in particular $\mathrm{Aff}(3) \simeq S_{3}$ and $\mathrm{Aff}(4) \simeq A_{4}$.
Thus $\mathbb{Z}_{2}[S_{3}] \simeq \mathbb{Z}_{2}[C_{2}] \oplus M_{2 \times 2}(\mathbb{Z}_{2})$
and $\mathbb{Z}_{3}[A_{4}] \simeq \mathbb{Z}_{3}[C_{3}] \oplus M_{3 \times 3}(\mathbb{Z}_{3})$.
\end{example}
\begin{example}\label{ex:dicyclic-complement}
Let $p$ be an odd prime and let $\mathrm{Dic}_{p} := \langle a,b \mid a^{2p}=1, a^{p}=b^{2}, b^{-1}ab = a^{-1} \rangle$ be the dicyclic group of order $4p$.
We recall a construction given in \cite[Chapter 14]{MR1828640}.
For every positive integer $n$ we let $\zeta_{n}$ denote a primitive $n$th root of unity.
There is an embedding $\iota$ of $\mathrm{Dic}_{p}$ into the subring $S_{p} := \mathbb{Z}[\frac{1}{2p}, \zeta_{2p},j]$ of the real quaternions;
here, $j^{2} = -1$ and $j \zeta_{2p} = \zeta_{2p}^{-1} j$.
We put $R_{p} := \mathbb{Z}[\frac{1}{2p}, \zeta_{p} + \zeta_{p}^{-1}] \subseteq S_{p}$.
Let $t$ and $k_{i}$, $1\leq i \leq t$ be positive integers and let $\mathfrak{p}_{i}$, $1 \leq i \leq t$ be maximal ideals of $R_{p}$.
Then for each $i$ there is a $R_{p} / \mathfrak{p}_{i}^{k_{i}}$-algebra isomorphism
$S_{p} / \mathfrak{p}_{i}^{k_{i}} S_{p} \simeq M_{2 \times 2} (R_{p}/\mathfrak{p}_{i}^{k_{i}})$ which
induces a fixed-point-free action of $\mathrm{Dic}_{p}$ on $N(\mathfrak{p}_{i}^{k_{i}}) := (R_{p} / \mathfrak{p}_{i}^{k_{i}})^{2}$ via $\iota$.
Thus $G := N \rtimes \mathrm{Dic}_{p}$ with $N := \prod_{i=1}^{t} N(\mathfrak{p}_{i}^{k_{i}})$ is a Frobenius group with Frobenius complement $\mathrm{Dic}_{p}$.
In fact, every Frobenius group with Frobenius complement isomorphic to $\mathrm{Dic}_{p}$
is of this type (see \cite[Theorem 14.4]{MR1828640}).
In particular, $\mathbb{Z}_{p}[G]$ is $N$-hybrid.
\end{example}
\begin{example} \label{ex:non-abelian-kernel}
Let $p$ and $q$ be primes, $f$ and $n$ positive integers such that $q>n>1$ and $q \mid (p^{f}-1)$.
Let $N$ be the subgroup of $\mathrm{GL}_{n}(\mathbb{F}_{p^{f}})$ comprising upper triangular matrices with all diagonal
entries equal to $1$.
There are pairwise distinct $b_{j} \in \mathbb{F}_{p^{f}}$, $1 \leq j \leq n$ such that $b_{j}^{q} = 1$.
Let $h$ be the diagonal matrix with entries $b_{1}, \dots, b_{n}$ and set $H:= \langle h \rangle$.
Then $G := N \rtimes H$ is a Frobenius group of order $q p^{fn(n-1)/2}$ and so $\mathbb{Z}_{p}[G]$ is $N$-hybrid.
Moreover, the Frobenius kernel $N$ has nilpotency class $n-1$ (see \cite[Example 16.8b]{MR1645304})
and hence is complicated if $n$ is large.
\end{example}
\subsection{New $p$-adic hybrid group rings from old}
We now use character theory to show how new examples of hybrid $p$-adic group rings can be obtained from existing examples.
For a field $F$ of characteristic $0$, a finite group $G$ and (virtual) $F$-characters $\chi$ and $\psi$ of $G$,
we let $\langle \chi, \psi \rangle$ denote the usual inner product.
\begin{remark}
For any finite group $G$ with subgroup $H$, any prime $p$,
and any $\chi \in \mathrm{Irr}_{\mathbb{Q}_{p}^{c}}(G)$, we have $H \leq \ker\chi$ if and only if $\langle \mathrm{res}^{G}_{H} \chi, \psi \rangle = 0$
for every non-trivial $\psi \in \mathrm{Irr}_{\mathbb{Q}_{p}^{c}}(H)$. We shall use this easy observation several times (for different choices of
$G$, $H$ and $\chi$) in the proofs of Lemma \ref{lem:kernel-basechange} and Propositions \ref{prop:hybrid-basechange-down} and
\ref{prop:hybrid-basechange-up} below.
\end{remark}
\begin{lemma}\label{lem:kernel-basechange}
Let $G$ be a finite group with subgroups $N \leq H \unlhd G$.
Let $p$ be a prime.
Fix $\chi \in \mathrm{Irr}_{\mathbb{Q}_{p}^{c}}(G)$ and let $\eta \in \mathrm{Irr}_{\mathbb{Q}_{p}^{c}}(H)$ be an irreducible constituent of $\mathrm{res}^{G}_{H} \chi$.
Then:
\begin{enumerate}
\item If $N \leq \ker \chi$, then $N \leq \ker \eta$.
\item Assume in addition that $N \unlhd G$. Then $N \leq \ker \chi$ if and only if $N \leq \ker \eta$.
\end{enumerate}
\end{lemma}
\begin{proof}
As $H \unlhd G$, we have a natural right action of $G$ on $\mathrm{Irr}_{\mathbb{Q}_{p}^{c}}(H)$;
namely, for each $g \in G$, $h \in H$ we have $\eta^{g}(h) = \eta(g^{-1}hg)$.
Let $G_{\eta} = \{g \in G \mid \eta^{g} = \eta \}$ be the stabiliser of $\eta$ in $G$ and
let $R_{\eta}$ be a set of right coset representatives of $G_{\eta}$ in $G$.
Then by Clifford theory (see \cite[Proposition 11.4]{MR632548}) we have
$\mathrm{res}^{G}_{H} \chi = e \sum_{g \in R_{\eta}} \eta^{g}$ for some positive integer $e$, and in particular
\begin{equation}\label{eqn:Clifford-theory}
\chi(1) = e [G:G_{\eta}] \eta(1).
\end{equation}
Let $\psi \in \mathrm{Irr}_{\mathbb{Q}_{p}^{c}}(N)$. Then we have
\begin{equation}\label{eqn:res-non-neg}
\langle \mathrm{res}^{G}_{N} \chi, \psi \rangle = \langle \mathrm{res}^{H}_{N} (\mathrm{res}^{G}_{H} \chi), \psi \rangle
= e \sum_{g \in R_{\eta}} \langle \mathrm{res}^{H}_{N} \eta^{g}, \psi \rangle \geq 0,
\end{equation}
since $\langle \mathrm{res}^{H}_{N} \eta^{g}, \psi \rangle \geq 0$ for every $g \in R_{\eta}$.
Suppose that $N \leq \ker \chi$.
Then $\langle \mathrm{res}^{G}_{N} \chi, \psi \rangle = 0$ for every non-trivial $\psi \in \mathrm{Irr}_{\mathbb{Q}_{p}^{c}}(N)$ and so by \eqref{eqn:res-non-neg} we have $\langle \mathrm{res}^{H}_{N} \eta, \psi \rangle = 0$. Hence $N \leq \ker \eta$, proving (i).
Now assume that $N \unlhd G$ and suppose conversely that $N \leq \ker \eta$.
Then $\langle \mathrm{res}^{H}_{N} \eta, \psi \rangle = 0$ for every non-trivial $\psi \in \mathrm{Irr}_{\mathbb{Q}_{p}^{c}}(N)$,
and for every $g \in R_{\eta}$ we have
\[
\langle \mathrm{res}^{H}_{N} \eta^{g}, \psi \rangle = \langle \mathrm{res}^{H}_{N} \eta, \psi^{g^{-1}} \rangle = 0.
\]
Thus by \eqref{eqn:res-non-neg} we have $\langle \mathrm{res}^{G}_{N} \chi, \psi \rangle = 0$
for every non-trivial $\psi \in \mathrm{Irr}_{\mathbb{Q}_{p}^{c}}(N)$ and so $N \leq \ker \chi$.
\end{proof}
The following proposition is a generalisation of \cite[Proposition 2.8 (iv)]{MR3461042}.
\begin{prop}\label{prop:hybrid-basechange-down}
Let $G$ be a finite group with normal subgroups $N, H \unlhd G$.
Let $K$ be a normal subgroup of $H$ such that $K \leq N$. Let $p$ be a prime.
If $\mathbb{Z}_{p}[G]$ is $N$-hybrid then $\mathbb{Z}_{p}[H]$ is $K$-hybrid.
\end{prop}
\begin{proof}
Fix $\eta \in \mathrm{Irr}_{\mathbb{Q}_{p}^{c}}(H)$ such that $K \not\leq \ker \eta$.
By Proposition \ref{prop:hybrid-criterion-groupring} we have to show that $v_{p}(\eta(1)) = v_{p}(|H|)$.
Let $\chi \in \mathrm{Irr}_{\mathbb{Q}_{p}^{c}}(G)$ be any irreducible constituent of $\mathrm{ind}^{G}_{H} \eta$.
Then by Frobenius reciprocity we have $\langle\chi, \mathrm{ind}^{G}_{H} \eta \rangle = \langle \mathrm{res}^{G}_{H} \chi, \eta \rangle \neq 0$.
By Lemma \ref{lem:kernel-basechange} (i) we have $K \not\leq \ker \chi$ and a fortiori $N \not\leq \ker \chi$.
Then again by Proposition \ref{prop:hybrid-criterion-groupring} we find that
$v_{p}(\chi(1)) = v_{p}(|G|)$ as $\mathbb{Z}_{p}[G]$ is $N$-hybrid by assumption.
This and equation \eqref{eqn:Clifford-theory} imply $v_{p}(e \eta(1)) = v_{p}(|G_{\eta}|) = v_{p}([G_{\eta}:H] \cdot |H|)$.
But $\eta(1)$ divides $|H|$ whereas $e$ divides $[G_{\eta}:H]$ by \cite[Theorem 21.3]{MR1645304}, so we must have
$v_{p}(e) = v_{p}([G_{\eta}:H])$ and $v_{p}(\eta(1)) = v_{p}(|H|)$, as desired.
\end{proof}
The following proposition is a generalisation of \cite[Lemma 2.9]{MR3461042}.
\begin{prop} \label{prop:hybrid-basechange-up}
Let $G$ be a finite group with normal subgroups $N \unlhd H \unlhd G$ such that $N \unlhd G$.
Let $p$ be a prime and assume that $p \nmid [G:H]$.
Then $\mathbb{Z}_{p}[G]$ is $N$-hybrid if and only if $\mathbb{Z}_{p}[H]$ is $N$-hybrid.
\end{prop}
\begin{proof}
If $\mathbb{Z}_{p}[G]$ is $N$-hybrid then $\mathbb{Z}_{p}[H]$ is $N$-hybrid by Proposition \ref{prop:hybrid-basechange-down} with $K=N$.
Suppose conversely that $\mathbb{Z}_{p}[H]$ is $N$-hybrid and assume that $p \nmid [G:H]$.
Let $\chi \in \mathrm{Irr}_{\mathbb{Q}_{p}^{c}}(G)$ such that $N \not\leq \ker \chi$ and let $\eta$ be an irreducible constituent of $\mathrm{res}^{G}_{H} \chi$.
Then $N \not\leq \ker \eta$ by Lemma \ref{lem:kernel-basechange} (ii) and $\eta(1)$ divides $\chi(1)$ by \eqref{eqn:Clifford-theory}.
However, we have $v_{p}(\eta(1)) = v_{p}(|H|) = v_{p}(|G|)$ by assumption, and so we must also have that $v_{p}(\chi(1)) = v_{p}(|G|)$.
Thus $\mathbb{Z}_{p}[G]$ is $N$-hybrid by Proposition \ref{prop:hybrid-criterion-groupring}.
\end{proof}
\begin{example}\label{ex:S4-A4-V4}
Let $p=3$, $G=S_{4}$, $H=A_{4}$ and $N = V_{4}$. Then the hypotheses of Proposition \ref{prop:hybrid-basechange-up} are satisfied.
Hence $\mathbb{Z}_{3}[S_{4}]$ is $V_{4}$-hybrid if and only if $\mathbb{Z}_{3}[A_{4}]$ is $V_{4}$-hybrid.
In fact, $\mathbb{Z}_{3}[A_{4}]$ is indeed $V_{4}$-hybrid since $A_{4}$ is a Frobenius group with Frobenius kernel $V_{4}$
(see Example \ref{ex:affine}) and so $\mathbb{Z}_{3}[S_{4}]$ is also $V_{4}$-hybrid.
However, $S_{4}$ is \emph{not} a Frobenius group (see \cite[Example 2.18]{MR3461042}).
Thus Proposition \ref{prop:hybrid-basechange-up} can be used to give examples which do not come directly from
Proposition \ref{prop:frob-N-hybrid}.
\end{example}
\begin{example} \label{ex:affine-Frobenius}
Let $q = \ell^{n}$ be a prime power and let $\phi: \mathbb{F}_{q} \rightarrow \mathbb{F}_{q}$, $x \mapsto x^{\ell}$ be the Frobenius automorphism.
Each $c \in \mathbb{F}_{q}^{\times}$ defines a map $m_{c}: \mathbb{F}_{q} \rightarrow \mathbb{F}_{q}$, $x \mapsto c \cdot x$.
We may consider $\phi$ and $m_{c}$ as elements of $\mathrm{GL}_{n}(\mathbb{F}_{\ell})$. Then $\phi m_{c} \phi^{-1} = m_{c^{\ell}}$
and we may form the semidirect product $\mathbb{F}_{q}^{\times} \rtimes \langle \phi \rangle$ inside $\mathrm{GL}_{n}(\mathbb{F}_{\ell})$.
Moreover, the action of $\mathbb{F}_{q}^{\times} \rtimes \langle \phi \rangle$ on $\mathbb{F}_{q}$ gives a semidirect product
$G := \mathbb{F}_{q} \rtimes (\mathbb{F}_{q}^{\times} \rtimes \langle \phi \rangle)$.
Then the group $\mathrm{Aff}(q) \simeq \mathbb{F}_{q} \rtimes \mathbb{F}_{q}^{\times}$ of affine transformations on $\mathbb{F}_{q}$
naturally identifies with a normal subgroup of $G$, and $N = \mathbb{F}_{q}$ is normal in both $G$ and $\mathrm{Aff}(q)$.
However, $\mathbb{Z}_{p}[\mathrm{Aff}(q)]$ is $N$-hybrid for every prime $p \neq \ell$ by Example \ref{ex:affine}.
If we further suppose that $p$ does not divide $n = [G : \mathrm{Aff}(q)]$, then $\mathbb{Z}_{p}[G]$
is $N$-hybrid by Proposition \ref{prop:hybrid-basechange-up}.
Note that this recovers Example \ref{ex:S4-A4-V4} since $G \simeq S_{4}$ when $\ell=n=2$.
\end{example}
\begin{remark}\label{rmk:applications-to-ETNC}
Burns and Flach \cite{MR1884523} formulated
the equivariant Tamagawa number conjecture (ETNC)
for any motive over $\mathbb{Q}$ with the action of a semisimple $\mathbb{Q}$-algebra,
describing the leading term at $s=0$ of an equivariant motivic $L$-function in terms of certain
cohomological Euler characteristics.
The present authors introduced hybrid $p$-adic groups rings in \cite{MR3461042} and used them to prove many new cases of the
$p$-part of the ETNC for Tate motives;
the same methods can also be applied to several related conjectures.
Thus the new results on $p$-adic group rings given here combined with the results of \cite{MR3461042}
give unconditional proofs of the $p$-part of the ETNC for Tate motives and related conjectures in many new cases.
\end{remark}
\section{Hybrid Iwasawa algebras}\label{sec:hybrid-Iwasawa-alg}
\subsection{Iwasawa algebras of one-dimensional $p$-adic Lie groups}\label{subsec:Iwasawa-algebras}
Let $p$ be an odd prime and let $\mathcal{G}$ be a profinite group containing a finite normal subgroup $H$
such that $\mathcal{G}/H \simeq \overline{\Gamma}$ where $\overline\Gamma$ is a pro-$p$-group isomorphic to $\mathbb{Z}_{p}$.
The argument given in \cite[\S 1]{MR2114937} shows that the short exact sequence
\[
1 \longrightarrow H \longrightarrow \mathcal{G} \longrightarrow \overline{\Gamma} \longrightarrow 1
\]
splits. Thus we obtain a semidirect product $\mathcal{G} = H \rtimes \Gamma$ where $\Gamma \leq \mathcal{G}$ and $\Gamma \simeq \overline{\Gamma} \simeq \mathbb{Z}_{p}$. In other words, $\mathcal{G}$ is a one-dimensional $p$-adic Lie group.
The Iwasawa algebra of $\mathcal{G}$ is
\[
\Lambda(\mathcal{G}) := \mathbb{Z}_{p}[[\mathcal{G}]] = \varprojlim \mathbb{Z}_{p}[\mathcal{G}/\mathcal{N}],
\]
where the inverse limit is taken over all open normal subgroups $\mathcal{N}$ of $\mathcal{G}$.
If $F$ is a finite field extension of $\mathbb{Q}_{p}$ with ring of integers $\mathcal{O}=\mathcal{O}_{F}$,
we put $\Lambda^{\mathcal{O}}(\mathcal{G}) := \mathcal{O} \otimes_{\mathbb{Z}_{p}} \Lambda(\mathcal{G}) = \mathcal{O}[[\mathcal{G}]]$.
We fix a topological generator $\gamma$ of $\Gamma$.
Since any homomorphism $\Gamma \rightarrow \mathrm{Aut}(H)$ must have open kernel, we may choose a natural number $n$ such that $\gamma^{p^n}$ is central in $\mathcal{G}$; we fix such an $n$.
As $\Gamma_{0} := \Gamma^{p^n} \simeq \mathbb{Z}_{p}$, there is a ring isomorphism
$R:=\mathcal{O}[[\Gamma_{0}]] \simeq \mathcal{O}[[T]]$ induced by $\gamma^{p^n} \mapsto 1+T$
where $\mathcal{O}[[T]]$ denotes the power series ring in one variable over $\mathcal{O}$.
If we view $\Lambda^{\mathcal{O}}(\mathcal{G})$ as an $R$-module (or indeed as a left $R[H]$-module), there is a decomposition
\begin{equation*}\label{eq:Lambda-R-decomp}
\Lambda^{\mathcal{O}}(\mathcal{G}) = \bigoplus_{i=0}^{p^n-1} R[H] \gamma^{i}.
\end{equation*}
Hence $\Lambda^{\mathcal{O}}(\mathcal{G})$ is finitely generated as an $R$-module and is an $R$-order in the separable $E:=Quot(R)$-algebra
$\mathcal{Q}^{F} (\mathcal{G})$, the total ring of fractions of $\Lambda^{\mathcal{O}}(\mathcal{G})$, obtained
from $\Lambda^{\mathcal{O}}(\mathcal{G})$ by adjoining inverses of all central regular elements.
Note that $\mathcal{Q}^{F} (\mathcal{G}) = E \otimes_{R} \Lambda^{\mathcal{O}}(\mathcal{G})$ and that by
\cite[Lemma 1]{MR2114937} we have $\mathcal{Q}^{F} (\mathcal{G}) = F \otimes_{\mathbb{Q}_{p}} \mathcal{Q}(\mathcal{G})$,
where $\mathcal{Q}(\mathcal{G}) := \mathcal{Q}^{\mathbb{Q}_{p}}(\mathcal{G})$.
\subsection{Characters and central primitive idempotents} \label{subsec:idempotents}
For any field $K$ of characteristic $0$ let $\mathrm{Irr}_{K}(\mathcal{G})$ the set of $K$-irreducible characters of $\mathcal{G}$ with open kernel.
Fix a character $\chi \in \mathrm{Irr}_{\mathbb{Q}_{p}^{c}}(\mathcal{G})$ and let $\eta$ be an irreducible constituent of
$\mathrm{res}^{\mathcal{G}}_{H} \chi$.
Then $\mathcal{G}$ acts on $\eta$ as $\eta^{g}(h) = \eta(g^{-1}hg)$
for $g \in \mathcal{G}$, $h \in H$, and following \cite[\S 2]{MR2114937} we set
\[
St(\eta) := \{g \in \mathcal{G}: \eta^g = \eta \}, \quad e(\eta) := \frac{\eta(1)}{|H|} \sum_{h \in H} \eta(h^{-1}) h,
\quad e_{\chi} := \sum_{\eta \mid \mathrm{res}^{\mathcal{G}}_{H} \chi} e(\eta).
\]
By \cite[Corollary to Proposition 6]{MR2114937} $e_{\chi}$ is a primitive central idempotent of
$\mathcal{Q}^{c}(\mathcal{G}) := \mathbb{Q}_{p}^{c} \otimes_{\mathbb{Q}_{p}} \mathcal{Q}(\mathcal{G})$.
In fact, every primitive central idempotent of $\mathcal{Q}^{c}(\mathcal{G})$ is of this form
and $e_{\chi} = e_{\chi'}$ if and only if $\chi = \chi' \otimes \rho$ for some character $\rho$ of $\mathcal{G}$ of type $W$
(i.e.~$\mathrm{res}^{\mathcal{G}}_{H} \rho = 1$).
The irreducible constituents of $\mathrm{res}^{\mathcal{G}}_{H} \chi$ are precisely the conjugates of $\eta$
under the action of $\mathcal{G}$, each occurring with the same multiplicity $z_{\chi}$ by \cite[Proposition 11.4]{MR632548}. By \cite[Lemma 4]{MR2114937} we have $z_{\chi}=1$ and thus we also have equalities
\begin{equation}\label{eq:idem-sum}
\mathrm{res}^{\mathcal{G}}_{H} \chi = \sum_{i=0}^{w_{\chi}-1} \eta^{\gamma^{i}},
\quad
e_{\chi} = \sum_{i=0}^{w_{\chi}-1} e(\eta^{\gamma^{i}}) = \frac{\chi(1)}{|H|w_{\chi}}\sum_{h \in H} \chi(h^{-1})h,
\end{equation}
where $w_{\chi} := [\mathcal{G} : St(\eta)]$.
Note that $\chi(1) = w_{\chi} \eta(1)$ and that
$w_{\chi}$ is a power of $p$ since $H$ is a subgroup of $St(\eta)$.
\subsection{Working over sufficiently large $p$-adic fields}\label{subsec:sufficiently-large}
We now specialise to the case where $F/\mathbb{Q}_{p}$ is a finite extension over which both
characters $\chi$ and $\eta$ have realisations.
Let $V_{\chi}$ denote a realisation of $\chi$ over $F$.
By \cite[Proposition 5]{MR2114937}, there exists a unique element
$\gamma_{\chi} \in \zeta(\mathcal{Q}^{c}(\mathcal{G})e_{\chi})$ such that $\gamma_{\chi}$
acts trivially on $V_{\chi}$ and $\gamma_{\chi}= gc$ where $g \in \mathcal{G}$ with $(g \bmod H) = \gamma^{w_{\chi}}$
and $c \in (\mathbb{Q}_{p}^{c}[H]e_{\chi})^{\times}$. Moreover, $\gamma_{\chi}= gc=cg$.
\begin{lemma}\label{lem:c-in-F[H]}
In fact $c \in (F[H]e_{\chi})^{\times}$ and so
$\gamma_{\chi} \in \zeta(\mathcal{Q}^{F}(\mathcal{G})e_{\chi})$.
\end{lemma}
\begin{proof}
We recall the definition of $\gamma_{\chi} := gc$ given in the proof of \cite[Proposition 5]{MR2114937}, the only difference
being that there $V_{\chi}$ is defined over $\mathbb{Q}_{p}^{c}$ rather than $F$.
Choose $g \in \mathcal{G}$ such that $(g \bmod H) = \gamma^{w_{\chi}}$.
Then $g \in St(\eta^{\gamma^{i}})$ for each $i$,
and it acts on $V_{\chi} = \oplus_{i=0}^{w_{\chi}-1} e(\eta^{\gamma^{i}}) V_{\chi}$ componentwise.
Since, by \cite[Lemma 4]{MR2114937}, each $e(\eta^{\gamma^{i}}) V_{\chi}$ is $H$-irreducible,
\[
g^{-1} \mid_{e(\eta^{\gamma^{i}})V_{\chi}} =: c(\eta^{\gamma^{i}}) \in F[H]e(\eta^{\gamma^{i}}) \simeq \mathrm{End}_{F}(e(\eta^{\gamma^{i}})V_{\chi})
\simeq M_{\eta(1) \times \eta(1)}(F).
\]
Now we set $c:=\sum_{i=0}^{w_{\chi}-1} c(\eta^{\gamma^{i}})$, and we see that $c \in (F[H]e_{\chi})^{\times}$.
\end{proof}
By \cite[Proposition 5]{MR2114937}, the element $\gamma_{\chi}$
generates a procyclic $p$-subgroup $\Gamma_{\chi}$ of $(\mathcal{Q}^{F}(\mathcal{G})e_{\chi})^{\times}$.
Let $\Lambda^{\mathcal{O}}(\Gamma_{\chi})$ be the integral domain $\mathcal{O}[[\Gamma_{\chi}]]$
with field of fractions $\mathcal{Q}^{F}(\Gamma_{\chi})$.
\begin{lemma}\label{lem:unique-max-order-in-centre}
$\Lambda^{\mathcal{O}}(\Gamma_{\chi})$ is the unique maximal $R$-order in $\zeta(\mathcal{Q}^{F}(\mathcal{G})e_{\chi})\simeq \mathcal{Q}^{F}(\Gamma_{\chi})$.
\end{lemma}
\begin{proof}
By \cite[Proposition 6]{MR2114937}, $\mathcal{Q}^{F}(\Gamma_{\chi})$ is contained in $\mathcal{Q}^{F}(\mathcal{G})e_{\chi}$,
and $\gamma_{\chi} \in \mathcal{Q}^{F}(\Gamma_{\chi})$ induces an isomorphism
$\mathcal{Q}^{F}(\Gamma_{\chi}) \stackrel{\simeq}{\longrightarrow} \zeta(\mathcal{Q}^{F}(\mathcal{G})e_{\chi})$.
Therefore $\Lambda^{\mathcal{O}}(\Gamma_{\chi})$ is an $R$-order in $\zeta(\mathcal{Q}^{F}(\mathcal{G})e_{\chi})$.
Moreover, $\Lambda^{\mathcal{O}}(\Gamma_{\chi})$ is a maximal $R$-order since there is an isomorphism of commutative rings
$\Lambda^{\mathcal{O}}(\Gamma_{\chi}) = \mathcal{O}[[\Gamma_{\chi}]] \simeq \mathcal{O}[[T]]$ and $\mathcal{O}[[T]]$ is integrally closed. Uniqueness follows from commutativity of $\mathcal{Q}^{F}(\Gamma_{\chi})$.
\end{proof}
\subsection{Maximal order $e_{\chi}$-components of Iwasawa algebras}
We give criteria for `$e_{\chi}$-components' of Iwasawa algebras of one-dimensional $p$-adic Lie groups to be maximal orders
in the case that $F/\mathbb{Q}_{p}$ is a sufficiently large finite extension. Moreover,
we give an explicit description of such components.
\begin{prop}\label{prop:chi-comp-max-order}
Let $\chi \in \mathrm{Irr}_{\mathbb{Q}_{p}^{c}}(\mathcal{G})$ and let $\eta$ be an irreducible constituent of
$\mathrm{res}^{\mathcal{G}}_{H} \chi$.
Let $F/\mathbb{Q}_{p}$ be a finite extension over which both characters $\chi$ and $\eta$ have realisations
and let $\mathcal{O}=\mathcal{O}_{F}$ be its ring of integers.
Suppose that $v_{p}(\eta(1))=v_{p}(|H|)$.
Then $e_{\chi} \in \Lambda^{\mathcal{O}}(\mathcal{G})$,
$\zeta(\Lambda^{\mathcal{O}}(\mathcal{G})e_{\chi}) = \Lambda^{\mathcal{O}}(\Gamma_{\chi})$
and there is an isomorphism of $R:=\mathcal{O}[[\Gamma_{0}]]$-orders
\[
\Lambda^{\mathcal{O}}(\mathcal{G})e_{\chi} \simeq M_{\chi(1) \times \chi(1)}(\Lambda^{\mathcal{O}}(\Gamma_{\chi})).
\]
Moreover, these are maximal $R$-orders and as rings are isomorphic to $M_{\chi(1) \times \chi(1)}(\mathcal{O}[[T]])$.
\end{prop}
\begin{proof}
Suppose that $v_{p}(\eta(1))=v_{p}(|H|)$. Then $v_{p}(\eta^{\gamma^{i}}(1)) =v_{p}(|H|)$ for each $i$. This has two consequences.
First, $e(\eta^{\gamma^{i}}) \in \mathcal{O}[H] \subseteq \Lambda^{\mathcal{O}}(\mathcal{G})$ for each $i$
and so the description of $e_{\chi}$ in \eqref{eq:idem-sum} shows that $e_{\chi} \in \Lambda^{\mathcal{O}}(\mathcal{G})$.
Second, since $\eta$ can be realised over $F$,
a standard result on ``$p$-blocks of defect zero'' (see \cite[Proposition 46 (b)]{MR0450380}, for example) shows that for each $i$ we have
an isomorphism of $\mathcal{O}$-orders
\begin{equation}\label{eq:matrix-order-key-point}
\mathcal{O}[H]e(\eta^{\gamma^{i}}) \simeq M_{\eta(1) \times \eta(1)}(\mathcal{O}).
\end{equation}
Let $M_{\chi}$ be an $\mathcal{O}$-lattice on $V_{\chi}$ that is stable under the action of $\mathcal{G}$.
In particular, $M_{\chi}$ is an $\mathcal{O}[H]$-module and since $e(\eta^{\gamma^{i}}) \in \mathcal{O}[H]$ for each $i$,
we have $M_{\chi} = \oplus_{i=0}^{w_{\chi}-1}e(\eta^{\gamma^{i}})M_{\chi}$.
Now recall the proof of Lemma \ref{lem:c-in-F[H]}.
The element $g^{-1}$ acts on $e(\eta^{\gamma^{i}})M_{\chi}$ for each $i$ and so we see that in fact
\[
c(\eta^{\gamma^{i}}) =
g^{-1} \mid_{e(\eta^{\gamma^{i}})M_{\chi}} \in \mathcal{O}[H]e(\eta^{\gamma^{i}}) \simeq \mathrm{End}_{\mathcal{O}}(e(\eta^{\gamma^{i}})M_{\chi})
\]
where the isomorphism follows from \eqref{eq:matrix-order-key-point}.
Hence $c \in \mathcal{O}[H]e_{\chi} \subseteq \Lambda^{\mathcal{O}}(\mathcal{G})e_{\chi}$ and
since $ge_{\chi} \in \Lambda^{\mathcal{O}}(\mathcal{G})e_{\chi}$ we have
$\gamma_{\chi} = gc\in \zeta(\Lambda^{\mathcal{O}}(\mathcal{G})e_{\chi})$.
Thus $\Lambda^{\mathcal{O}}(\Gamma_{\chi}) \subseteq \zeta(\Lambda^{\mathcal{O}}(\mathcal{G})e_{\chi})$.
However, $\Lambda^{\mathcal{O}}(\Gamma_{\chi})$ is maximal by Lemma \ref{lem:unique-max-order-in-centre}
and therefore $\zeta(\Lambda^{\mathcal{O}}(\mathcal{G})e_{\chi}) = \Lambda^{\mathcal{O}}(\Gamma_{\chi})$.
Extending scalars in \eqref{eq:matrix-order-key-point} gives an isomorphism of $R$-orders $R[H]e(\eta) \simeq M_{\eta(1) \times \eta(1)}(R)$.
Let $e(\eta) = f_{1} + \cdots + f_{\eta(1)}$ be a decomposition of $e(\eta)$
into a sum of orthogonal indecomposable idempotents of $R[H]e(\eta)$.
Observe that $f_{k,i}:=\gamma^{-i}f_{k}\gamma^{i}$ is also an indecomposable idempotent
for every $0 \leq i < p^{n}$ and $1 \leq k \leq \eta(1)$.
Moreover, each $f_{k,i}$ belongs to $R[H]e(\eta^{\gamma^{i}})$ since
$\gamma^{-i}e(\eta)\gamma^{i}=e(\eta^{\gamma^{i}})$ and $H$ is normal in $\mathcal{G}$.
Thus $e(\eta^{\gamma^{i}}) = f_{1,i} + \cdots + f_{\eta(1),i}$ is a decomposition of $e(\eta^{\gamma^{i}})$
into a sum of orthogonal indecomposable idempotents of $R[H]e(\eta^{\gamma^{i}})$.
Note that $e(\eta) = e(\eta^{\gamma^{i}})$ if and only if $w_{\chi}$ divides $i$.
Hence $R[H]e_{\chi} = \oplus_{i=0}^{w_{\chi}-1} R[H] e(\eta^{\gamma^{i}})$ is an $R$-suborder of
$\Lambda^{\mathcal{O}}(\mathcal{G})e_{\chi}$ and $e_{\chi} = \sum_{k=1}^{\eta(1)} \sum_{i=0}^{w_{\chi}-1} f_{k,i}$
is a decomposition of $e_{\chi}$ into orthogonal idempotents.
By considering the appropriate `elementary matrices' in $R[H]e(\eta) \simeq M_{\eta(1) \times \eta(1)}(R)$
together with powers of $\gamma$, it is straightforward to see that there is a subset $S$ of
$(\Lambda^{\mathcal{O}}(\mathcal{G})e_{\chi})^{\times}$
such that each element of $I:=\{ f_{k,i} \}_{k=1,i=0}^{k=\eta(1), i=w_{\chi}-1}$ is a conjugate of $f_{1}=f_{1,0}$
by some element of $S$. Furthermore, $|I|=\eta(1)w_{\chi}=\chi(1)$.
Therefore by \cite[\S 46, Exercise 2]{MR892316} we have a ring isomorphism
\begin{equation}\label{eq:is-matrix-ring-chi1}
\Lambda^{\mathcal{O}}(\mathcal{G})e_{\chi} \simeq M_{\chi(1) \times \chi(1)}(f_{1}\Lambda^{\mathcal{O}}(\mathcal{G})e_{\chi}f_{1}).
\end{equation}
As shown in the proof of \cite[Proposition 6 (2)]{MR2114937},
the dimension of $\mathcal{Q}^{F}(\mathcal{G})e_{\chi}$ over
$\zeta(\mathcal{Q}^{F}(\mathcal{G})e_{\chi}) \simeq \mathcal{Q}^{F}(\Gamma_{\chi})$ is $\chi(1)^{2}$.
Since $e_{\chi}$ is a primitive central idempotent we have $\mathcal{Q}^{F}(\mathcal{G})e_{\chi} \simeq M_{m \times m}(D)$ for some $m$
and some skewfield $D$ with $\zeta(D)=\mathcal{Q}^{F}(\Gamma_{\chi})$.
Moreover, $\chi(1)=ms$ where $[D: \mathcal{Q}^{F}(\Gamma_{\chi})]=s^{2}$.
However, \eqref{eq:is-matrix-ring-chi1} shows that $m \geq \chi(1)$ and so in fact $\chi(1)=m$
and
\begin{equation}\label{eq:is-matrix-algebra-chi1}
\mathcal{Q}^{F}(\mathcal{G})e_{\chi} \simeq M_{\chi(1) \times \chi(1)}(\zeta(\mathcal{Q}^{F}(\mathcal{G})e_{\chi})) \simeq M_{\chi(1) \times \chi(1)}(\mathcal{Q}^{F}(\Gamma_{\chi})).
\end{equation}
Combining \eqref{eq:is-matrix-ring-chi1}, \eqref{eq:is-matrix-algebra-chi1} and the fact that
$\zeta(\Lambda^{\mathcal{O}}(\mathcal{G})e_{\chi}) = \Lambda^{\mathcal{O}}(\Gamma_{\chi})$
therefore shows that there are $R$-order isomorphisms
\[
\Lambda^{\mathcal{O}}(\mathcal{G})e_{\chi} \simeq M_{\chi(1) \times \chi(1)}(\zeta(\Lambda^{\mathcal{O}}(\mathcal{G})e_{\chi}))
\simeq M_{\chi(1) \times \chi(1)}(\Lambda^{\mathcal{O}}(\Gamma_{\chi})).
\]
Lemma \ref{lem:unique-max-order-in-centre} and \cite[Theorem 8.7]{MR1972204} show that
$M_{\chi(1) \times \chi(1)}(\Lambda^{\mathcal{O}}(\Gamma_{\chi}))$ is a maximal $R$-order.
The last claim follows from the ring isomorphism $\Lambda^{\mathcal{O}}(\Gamma_{\chi}) = \mathcal{O}[[\Gamma_{\chi}]] \simeq \mathcal{O}[[T]]$.
\end{proof}
\subsection{Central idempotents and Galois actions}\label{subset:idem-gal-actions}
Fix a character $\chi \in \mathrm{Irr}_{\mathbb{Q}_{p}^{c}}(\mathcal{G})$ and let $\eta$ be an irreducible constituent of
$\mathrm{res}^{\mathcal{G}}_{H} \chi$.
We define two fields
\[
K_{\chi} := \mathbb{Q}_{p}(\chi(h) \mid h \in H) \subseteq \mathbb{Q}_{p}(\eta) := \mathbb{Q}_{p}(\eta(h) \mid h \in H)
\]
and remark that the containment is not an equality in general.
Note that $K_{\chi} = K_{\chi \otimes \rho}$ whenever $\rho$ is of type $W$ (i.e.~$\mathrm{res}^{\mathcal{G}}_{H} \rho = 1$).
Since $H$ is normal in $\mathcal{G}$, we have $\mathbb{Q}_{p}(\eta^{g}) = \mathbb{Q}_{p}(\eta)$ for every $g \in \mathcal{G}$ and thus
$\mathbb{Q}_{p}(\eta)$ does not depend on the particular choice $\eta$ of irreducible constituent of $\mathrm{res}^{\mathcal{G}}_{H}\chi$.
We let $\sigma \in \mathrm{Gal}(\mathbb{Q}_{p}^{c}/\mathbb{Q}_{p})$ act on $\chi$ by ${}^{\sigma}\chi(g) = \sigma(\chi(g))$ for all $g \in \mathcal{G}$
and similarly on characters of $H$. Note that the actions on $\mathrm{res}^{\mathcal{G}}_{H}\chi$ and $\eta$ factor through
$\mathrm{Gal}(K_{\chi}/\mathbb{Q}_{p})$ and $\mathrm{Gal}(\mathbb{Q}_{p}(\eta)/\mathbb{Q}_{p})$, respectively.
Moreover, for $\sigma \in \mathrm{Gal}(\mathbb{Q}_{p}^{c}/\mathbb{Q}_{p})$ we have $\sigma(e_{\chi})=e_{({}^{\sigma}\chi)}$,
and $\sigma(e_{\chi})=e_{\chi}$ if and only if $\mathrm{res}^{\mathcal{G}}_{H}\chi = {}^{\sigma}(\mathrm{res}^{\mathcal{G}}_{H}\chi)$.
Hence the action of $\mathrm{Gal}(\mathbb{Q}_{p}^{c}/\mathbb{Q}_{p})$ on $e_{\chi}$ factors through $\mathrm{Gal}(K_{\chi}/\mathbb{Q}_{p})$ and we define
\[
\varepsilon_{\chi} := \sum_{\sigma \in \mathrm{Gal}(K_{\chi} / \mathbb{Q}_{p})} \sigma(e_{\chi}).
\]
We remark that $\varepsilon_{\chi}$ is a primitive central idempotent of $\mathcal{Q}(\mathcal{G})$.
Finally, we define an equivalence relation on $\mathrm{Irr}_{\mathbb{Q}_{p}^{c}}(\mathcal{G})$ as follows:
$\chi \sim \chi'$ if and only if there exists $\sigma \in \mathrm{Gal}(K_{\chi}/\mathbb{Q}_{p})$ such that $\sigma(e_{\chi})=e_{\chi'}$.
Note that the number of equivalence classes is finite and
that we have the decomposition $1 = \sum_{\chi/\sim} \varepsilon_{\chi}$ in $\mathcal{Q}(\mathcal{G})$.
\subsection{Maximal order $\varepsilon_{\chi}$-components of Iwasawa algebras}
We give criteria for `$\varepsilon_{\chi}$-components' of Iwasawa algebras of one-dimensional $p$-adic Lie groups to be maximal orders in the case that $F=\mathbb{Q}_{p}$.
Moreover, we give a somewhat explicit description of such components.
\begin{prop}\label{prop:max-part-of-hybrid-matrix-over-comm-ring}
Let $\chi \in \mathrm{Irr}_{\mathbb{Q}_{p}^{c}}(\mathcal{G})$ and let $\eta$ be an irreducible constituent of
$\mathrm{res}^{\mathcal{G}}_{H} \chi$.
If $v_{p}(\eta(1))=v_{p}(|H|)$ then $\varepsilon_{\chi} \in \Lambda(\mathcal{G})$ and $\varepsilon_{\chi}\Lambda(\mathcal{G})$
is a maximal $\mathbb{Z}_{p}[[\Gamma_{0}]]$-order.
Moreover, $\varepsilon_{\chi}\Lambda(\mathcal{G}) \simeq M_{\chi(1) \times \chi(1)}(S_{\chi})$
for some local integrally closed domain $S_{\chi}$.
\end{prop}
\begin{remark}
The condition that $v_{p}(\eta(1))=v_{p}(|H|)$ is independent of choice of irreducible consistent $\eta$ of $\mathrm{res}^{\mathcal{G}}_{H} \chi$
and thus only depends on $\chi$. Moreover, if the condition holds for $\chi \in \mathrm{Irr}_{\mathbb{Q}_{p}^{c}}(\mathcal{G})$, then
it also holds for ${}^{\sigma}\chi$ for all $\sigma \in \mathrm{Gal}(\mathbb{Q}_{p}^{c}/\mathbb{Q}_{p})$ (recall \S \ref{subset:idem-gal-actions}).
\end{remark}
\begin{proof}[Proof of Proposition \ref{prop:max-part-of-hybrid-matrix-over-comm-ring}]
Assume the notation and hypotheses of Proposition \ref{prop:chi-comp-max-order}.
In particular, Proposition \ref{prop:chi-comp-max-order} shows that $e_{\chi} \in \Lambda^{\mathcal{O}}(\mathcal{G})$.
Hence $\varepsilon_{\chi} \in \mathcal{Q}(\mathcal{G}) \cap \Lambda^{\mathcal{O}}(\mathcal{G}) = \Lambda(\mathcal{G})$.
Now
\[
\bigoplus_{\sigma \in \mathrm{Gal}(K_{\chi}/\mathbb{Q}_{p})} \Lambda^{\mathcal{O}}(\mathcal{G})\sigma(e_{\chi})
= \Lambda^{\mathcal{O}}(\mathcal{G}) \varepsilon_{\chi}
= \mathcal{O} \otimes_{\mathbb{Z}_{p}} \Lambda(\mathcal{G}) \varepsilon_{\chi}
=\mathcal{O}[[\Gamma_{0}]] \otimes_{\mathbb{Z}_{p}[[\Gamma_{0}]]} \Lambda(\mathcal{G}) \varepsilon_{\chi},
\]
which is a maximal $\mathcal{O}[[\Gamma_{0}]]$-order by Proposition \ref{prop:chi-comp-max-order}.
Hence we have that $\Lambda(\mathcal{G}) \varepsilon_{\chi}$ is a maximal $\mathbb{Z}_{p}[[\Gamma_{0}]]$-order.
Set $\Lambda = \Lambda(\mathcal{G})\varepsilon_{\chi}$ and $S=\zeta(\Lambda)$.
Note that $S$ is contained in $\zeta(\mathcal{Q}(\mathcal{G})\varepsilon_{\chi})$, which is a field
since $\varepsilon_{\chi}$ is a primitive central idempotent of $\mathcal{Q}(\mathcal{G})$.
Thus $S$ is an integral domain.
Moreover, $S$ is semiperfect by \cite[Example (23.3)]{MR1838439} and thus is a finite direct product of local rings by
\cite[Theorem (23.11)]{MR1838439}; therefore $S$ is a local integral domain.
In fact, $S$ is a complete local integral domain by \cite[Proposition 6.5 (ii)]{MR632548}.
Furthermore, $S$ must be integrally closed since $\Lambda(\mathcal{G})\varepsilon_{\chi}$ is maximal.
Now set $\Lambda' = \Lambda^{\mathcal{O}}(\mathcal{G})\varepsilon_{\chi} = \mathcal{O} \otimes_{\mathbb{Z}_{p}} \Lambda$.
It is clear that $\mathcal{O} \otimes_{\mathbb{Z}_{p}} S \subseteq S':=\zeta(\Lambda')$.
A standard argument gives the reverse containment; here we give a slightly modified version of the proof of
\cite[Theorem 7.6]{MR1972204}.
Let $b_{1}, \ldots, b_{m}$ be a $\mathbb{Z}_{p}$-basis of $\mathcal{O}$. Let $x \in S'$.
Then we can write $x= \sum_{i} b_{i} \otimes \lambda_{i}$ for some $\lambda_{i} \in \Lambda$
and in particular, $x$ commutes with $1 \otimes \lambda$ for all $\lambda \in \Lambda$.
Hence $\sum_{i} b_{i} \otimes (\lambda \lambda_{i} - \lambda_{i} \lambda) = 0$
and so $\lambda \lambda_{i} = \lambda_{i} \lambda$ for all $\lambda \in \Lambda$,
that is, $\lambda_{i} \in \zeta(\Lambda)=S$.
Therefore $S' = \mathcal{O} \otimes_{\mathbb{Z}_{p}} S$. This gives
\begin{equation}\label{eqn:extend-by-central-scalars}
\Lambda' = \mathcal{O} \otimes_{\mathbb{Z}_{p}} \Lambda \simeq (\mathcal{O} \otimes_{\mathbb{Z}_{p}} S) \otimes_{S} \Lambda \simeq S' \otimes_{S} \Lambda.
\end{equation}
Let $S'' :=e_{\chi}S'=\zeta(\Lambda' e_{\chi})$.
By Proposition \ref{prop:chi-comp-max-order} we have $S'' \simeq \mathcal{O}[[\Gamma_{\chi}]]$,
which is a local integrally closed domain.
Let $k$ and $k''$ be the residue fields of $S$ and $S''$, respectively.
Then $k''/k$ is a finite extension of finite fields of characteristic $p$.
By Proposition \ref{prop:chi-comp-max-order} and \eqref{eqn:extend-by-central-scalars} we have
\begin{equation}\label{eqn:defn-of-Gamma}
\Lambda'' := S'' \otimes_{S} \Lambda \simeq e_{\chi}(S' \otimes_{S} \Lambda)
\simeq e_{\chi}\Lambda' \simeq M_{\chi(1) \times \chi(1)}(S'').
\end{equation}
Now set $\overline{\Lambda} := k \otimes_{S} \Lambda$ and observe that
\[
k'' \otimes_{k} \overline{\Lambda} = k'' \otimes_{k} (k \otimes_{S} \Lambda) \simeq k'' \otimes_{S} \Lambda
\simeq k'' \otimes_{S''} (S'' \otimes_{S} \Lambda) \simeq k'' \otimes_{S''} \Lambda'' \simeq M_{\chi(1) \times \chi(1)}(k'').
\]
Therefore \cite[Theorem 7.18]{MR1972204} shows that $\overline{\Lambda}$ is a separable $k$-algebra.
Now \cite[Theorem 4.7]{MR0121392} shows that $\Lambda$ is separable over $S$.
Hence $\Lambda$ is an Azumaya algebra (i.e.\ separable over its centre) and so defines a class $[\Lambda]$
in the Brauer group $\mathrm{Br}(S)$.
Since $S$ is a complete local ring, \cite[Corollary 6.2]{MR0121392} shows that the natural homomorphism
$\mathrm{Br}(S) \rightarrow \mathrm{Br}(k)$ given by $[X] \mapsto [k \otimes_{S} X]$ is in fact a monomorphism.
However, $\mathrm{Br}(k)$ is trivial by Wedderburn's theorem that every finite division ring is a field.
Therefore $[\Lambda]$ is the trivial element of $\mathrm{Br}(S)$ and so by \cite[Proposition 5.3]{MR0121392} $\Lambda$
is isomorphic to an $S$-algebra of the form $\mathrm{Hom}_{S}(P,P)$ where $P$ is a finitely generated projective faithful $S$-module.
Since $S$ is a local ring, $P$ must be free and so $\Lambda$ is isomorphic to a matrix ring over its centre $S$.
Now \eqref{eqn:defn-of-Gamma} shows that in fact $\Lambda \simeq M_{\chi(1) \times \chi(1)}(S)$.
\end{proof}
The following proposition is an adaptation of \cite[Lemma 2.1]{MR3461042}.
\begin{prop}\label{prop:idempotent-implies-defect-zero}
Let $\chi \in \mathrm{Irr}_{\mathbb{Q}_{p}^{c}}(\mathcal{G})$.
If $\varepsilon_{\chi} \in \Lambda(\mathcal{G})$ then $v_{p}(\eta(1))=v_{p}(|H|)$ for every
irreducible constituent $\eta$ of $\mathrm{res}^{\mathcal{G}}_{H} \chi$.
\end{prop}
\begin{proof}
Fix an irreducible constituent $\eta$ of $\mathrm{res}^{\mathcal{G}}_{H} \chi$ and recall the notation of \S \ref{subset:idem-gal-actions}.
We may write $\varepsilon_{\chi} = \frac{\eta(1)}{|H|} \sum_{h \in H} \beta(h^{-1}) h$, where
\[
\beta := \sum_{\sigma \in \mathrm{Gal}(K_{\chi} / \mathbb{Q}_{p})} \sigma \left( \sum_{i=0}^{w_{\chi}-1} \eta^{\gamma^{i}} \right)
\]
is a $\mathbb{Q}_{p}$-valued character of $H$.
Recall that an element $h \in H$ is said to be $p$-singular if its order is divisible by $p$.
Write $\varepsilon_{\chi} = \sum_{h \in H} a_{h} h$ with $a_{h} \in \mathbb{Z}_{p}$ for $h \in H$.
Then \cite[Proposition 5]{MR1261587} shows that $a_{h}=0$ for every $p$-singular $h \in H$
(alternatively, one can use \cite[Proposition 3]{MR1261587} and that $\varepsilon_{\chi}$ is central).
Hence the character $\beta$ vanishes on $p$-singular elements.
Let $P$ be a Sylow $p$-subgroup of $H$.
Then $\beta$ vanishes on $P- \{1\}$, so the multiplicity of the trivial character of $P$ in the restriction $\mathrm{res}^{H}_{P} \beta$ is
\[
\langle \mathrm{res}^{H}_{P} \beta , 1_{P} \rangle = \beta(1) |P|^{-1} = [K_{\chi} : \mathbb{Q}_{p}] w_{\chi} \eta(1)|P|^{-1}.
\]
Now fix $0 \leq i < w_{\chi}$. Then $P' := \gamma^{i} P \gamma^{-i}$ is also a Sylow $p$-subgroup of $H$. Hence we may write $P' = hPh^{-1}$ for some $h \in H$.
We have
$\langle \mathrm{res}^{H}_{P} \eta^{\gamma^{i}} , 1_{P} \rangle = \langle \mathrm{res}^{H}_{P} \eta^{h} , 1_{P} \rangle = \langle \mathrm{res}^{H}_{P} \eta , 1_{P} \rangle$,
and thus
\[
\langle \mathrm{res}^{H}_{P} \beta, 1_{P} \rangle
= \sum_{\sigma \in \mathrm{Gal}(K_{\chi} / \mathbb{Q}_{p})} \sum_{i=0}^{w_{\chi}-1} \langle ^{\hat{\sigma}}(\mathrm{res}^{H}_{P}\eta^{\gamma^{i}}), 1_{P} \rangle
= [K_{\chi} : \mathbb{Q}_{p}] w_{\chi} \langle \mathrm{res}^{H}_{P} \eta , 1_{P} \rangle,
\]
where $\hat \sigma \in \mathrm{Gal}(\mathbb{Q}_{p}(\eta) / \mathbb{Q}_{p})$ denotes any lift of $\sigma$.
Therefore $\eta(1) = |P| \langle \mathrm{res}^{H}_{P} \eta , 1_{P} \rangle$, and so $v_{p}(\eta(1))=v_{p}(|H|)$.
\end{proof}
\begin{prop}\label{prop:varep-comp-TFAE}
Let $\chi \in \mathrm{Irr}_{\mathbb{Q}_{p}^{c}}(\mathcal{G})$ and let $\eta$ be an irreducible constituent of
$\mathrm{res}^{\mathcal{G}}_{H} \chi$. Then the following are equivalent:
\begin{enumerate}
\item $v_{p}(\eta(1))=v_{p}(|H|)$,
\item $\varepsilon_{\chi} \in \Lambda(\mathcal{G})$,
\item $\varepsilon_{\chi} \in \Lambda(\mathcal{G})$ and $\varepsilon_{\chi}\Lambda(\mathcal{G})$
is a maximal $\mathbb{Z}_{p}[[\Gamma_{0}]]$-order.
\end{enumerate}
Moreover, if these equivalent conditions hold then
$\varepsilon_{\chi}\Lambda(\mathcal{G}) \simeq M_{\chi(1) \times \chi(1)}(S_{\chi})$
for some local integrally closed domain $S_{\chi}$.
\end{prop}
\begin{proof}
This is the combination of Propositions \ref{prop:max-part-of-hybrid-matrix-over-comm-ring} and \ref{prop:idempotent-implies-defect-zero}.
\end{proof}
\subsection{Hybrid Iwasawa algebras}
Let $p$ be an odd prime and let $\mathcal{G} = H \rtimes \Gamma$ be a one-dimensional $p$-adic Lie group.
As in \S \ref{subsec:Iwasawa-algebras}, we choose an open subgroup $\Gamma_{0} \leq \Gamma$ that is central in $\mathcal{G}$.
If $N$ is a finite normal subgroup of $\mathcal{G}$ and $e_{N} := |N|^{-1}\sum_{\sigma \in N} \sigma$
is the associated central trace idempotent in the algebra $\mathcal{Q}(\mathcal{G})$,
then there is a ring isomorphism $\Lambda(\mathcal{G})e_{N} \simeq \Lambda(\mathcal{G}/N)$.
In particular, the commutator subgroup $\mathcal{G}'$ of $\mathcal{G}$ is contained in the finite subgroup $H$ and
thus $\Lambda(\mathcal{G})e_{\mathcal{G}'} \simeq \Lambda(\mathcal{G}^{\mathrm{ab}})$ where
$\mathcal{G}^{\mathrm{ab}}:=\mathcal{G}/\mathcal{G}'$ is the maximal abelian quotient of $\mathcal{G}$.
\begin{definition}\label{def:max-comm-hybrid}
Let $N$ be a finite normal subgroup of $\mathcal{G}$.
We say that the Iwasawa algebra $\Lambda(\mathcal{G})$ is \emph{$N$-hybrid}
if (i) $e_{N} \in \Lambda(\mathcal{G})$ (i.e.\ $p \nmid |N|$) and
(ii) $\Lambda(\mathcal{G})(1-e_{N})$ is a maximal $\mathbb{Z}_{p}[[\Gamma_{0}]]$-order in
$\mathcal{Q}(\mathcal{G})(1-e_{N})$.
\end{definition}
\begin{remark}\label{rmk:p-not-divide-H-max-order}
The Iwasawa algebra $\Lambda(\mathcal{G})$ is itself a maximal order if and only if $p$ does not divide $|H|$ if and only if
$\Lambda(\mathcal{G})$ is $H$-hybrid. Moreover, $\Lambda(\mathcal{G})$ is always $\{ 1 \}$-hybrid.
\end{remark}
\begin{lemma}\label{lem:kernel-criterion}
Let $\chi \in \mathrm{Irr}_{\mathbb{Q}_{p}^{c}}(\mathcal{G})$ and let $\eta$ be an irreducible constituent of $\mathrm{res}^{\mathcal{G}}_{H} \chi$.
Let $N$ be a finite normal subgroup of $\mathcal{G}$.
Then $N$ is in fact a normal subgroup of $H$.
Moreover, $N \leq \ker(\chi)$ if and only if $N \leq \ker(\eta)$.
\end{lemma}
\begin{proof}
The canonical projection $\mathcal{G} \twoheadrightarrow \mathcal{G} / H \simeq \mathbb Z_{p}$ maps $N$ onto $NH /H \simeq N / H\cap N$.
However, the only finite (normal) subgroup of $\mathbb{Z}_{p}$ is the trivial group;
hence $H \cap N = N$ and thus $N$ is a (normal) subgroup of $H$.
As $\chi$ factors through a finite quotient of $\mathcal{G}$, the second claim follows from
Lemma \ref{lem:kernel-basechange} (ii).
\end{proof}
\begin{lemma}\label{lem:exists-chi-irr-constit-eta}
Let $\eta \in \mathrm{Irr}_{\mathbb{Q}_{p}^{c}}(H)$. Then there exists $\chi \in \mathrm{Irr}_{\mathbb{Q}_{p}^{c}}(\mathcal{G})$ such that $\eta$ is an irreducible constituent of
$\mathrm{res}^{\mathcal{G}}_{H} \chi$.
\end{lemma}
\begin{proof}
Let $G=\mathcal{G}/\Gamma_{0} \simeq H \rtimes \Gamma/\Gamma_{0}$ (recall that $\Gamma_{0}$ is central in $\mathcal{G}$).
Let $\overline{\chi}$ be an irreducible constituent of $\mathrm{ind}^{G}_{H} \eta$.
Then by Frobenius reciprocity $\langle \mathrm{res}^{G}_{H} \overline{\chi}, \eta \rangle = \langle \overline{\chi}, \mathrm{ind}_{H}^{G} \eta \rangle \neq 0$.
Now take $\chi = \mathrm{infl}^{\mathcal{G}}_{G} \overline{\chi}$.
\end{proof}
\begin{theorem}\label{thm:hybrid-criterion}
Let $p$ be an odd prime and let $\mathcal{G} = H \rtimes \Gamma$ be a one-dimensional $p$-adic Lie group
with a finite normal subgroup $N$.
Then $N$ is in fact a normal subgroup of $H$.
Moreover, $\Lambda(\mathcal{G}):=\mathbb{Z}_{p}[[\mathcal{G}]]$ is $N$-hybrid if and only if $\mathbb{Z}_{p}[H]$ is $N$-hybrid.
\end{theorem}
\begin{proof}
The first assertion is contained in Lemma \ref{lem:kernel-criterion}.
Suppose that $\mathbb{Z}_{p}[H]$ is $N$-hybrid.
Let $\chi \in \mathrm{Irr}_{\mathbb{Q}_{p}^{c}}(\mathcal{G})$ such that $N \not \leq \ker(\chi)$.
Let $\eta$ be an irreducible constituent of $\mathrm{res}^{\mathcal{G}}_{H} \chi$.
Then $N \not \leq \ker(\eta)$ by Lemma \ref{lem:kernel-criterion} and so
$v_{p}(\eta(1))=v_{p}(|H|)$ by Proposition \ref{prop:hybrid-criterion-groupring}.
Hence $\varepsilon_{\chi} \in \Lambda(\mathcal{G})$ and $\varepsilon_{\chi}\Lambda(\mathcal{G})$
is a maximal $\mathbb{Z}_{p}[[\Gamma_{0}]]$-order by Proposition \ref{prop:varep-comp-TFAE}.
Now by Lemma \ref{lem:kernel-criterion} we have
\begin{equation}\label{eq:1-en-decomp}
1-e_{N}
= \sum_{\stackrel{\eta \in \mathrm{Irr}_{\mathbb{Q}_{p}^{c}}(H)}{N \not \leq \ker(\eta)}} e(\eta)
= \sum_{\stackrel{\chi \in \mathrm{Irr}_{\mathbb{Q}_{p}^{c}}(\mathcal{G})/\sim}{N \not \leq \ker(\chi)}} \varepsilon_{\chi}
\end{equation}
where $\sim$ denotes the equivalence relation defined in \S \ref{subset:idem-gal-actions}.
Therefore $1-e_{N} \in \Lambda(\mathcal{G})$ and
$\Lambda(\mathcal{G})(1-e_{N})$ is a maximal $\mathbb{Z}_{p}[[\Gamma_{0}]]$-order
as it is the direct product of such orders. Hence $\Lambda(\mathcal{G})$ is $N$-hybrid.
Suppose conversely that $\Lambda(\mathcal{G})$ is $N$-hybrid.
Let $\eta \in \mathrm{Irr}_{\mathbb{Q}_{p}^{c}}(H)$ such that $N \not \leq \ker(\eta)$.
By Lemma \ref{lem:exists-chi-irr-constit-eta} there exists $\chi \in \mathrm{Irr}_{\mathbb{Q}_{p}^{c}}(\mathcal{G})$ such that $\eta$ is an irreducible
constituent of $\mathrm{res}^{\mathcal{G}}_{H} \chi$ and by Lemma \ref{lem:kernel-criterion}
we have $N \not \leq \ker(\chi)$.
Since $1-e_{N} \in \Lambda(\mathcal{G})$ and $\Lambda(\mathcal{G})(1-e_{N})$
is a maximal $\mathbb{Z}_{p}[[\Gamma_{0}]]$-order, it follows from
\eqref{eq:1-en-decomp} that in particular $\varepsilon_{\chi} \in \Lambda(\mathcal{G})$.
Thus Proposition \ref{prop:varep-comp-TFAE} gives $v_{p}(\eta(1))=v_{p}(|H|)$.
Therefore $\mathbb{Z}_{p}[H]$ is $N$-hybrid by Proposition \ref{prop:hybrid-criterion-groupring}.
\end{proof}
\begin{corollary}\label{cor:hybrid-implies-nr-surjective}
Let $p$ be an odd prime and let $\mathcal{G} = H \rtimes \Gamma$ be a one-dimensional $p$-adic Lie group
with a finite normal subgroup $N$.
If $\Lambda(\mathcal{G}) := \mathbb{Z}_{p}[[\mathcal{G}]]$ is $N$-hybrid then
$\Lambda(\mathcal{G})(1-e_{N})$ is isomorphic to a direct product of matrix rings over integrally closed commutative local rings.
\end{corollary}
\begin{proof}
This follows from Proposition \ref{prop:varep-comp-TFAE} and the proof of Theorem \ref{thm:hybrid-criterion} above.
\end{proof}
\begin{prop}\label{prop:frob-N-iwasawa-hybrid}
Let $p$ be an odd prime and $\mathcal{G} = H \rtimes \Gamma$ be a one-dimensional $p$-adic Lie group.
Suppose that $H$ is a Frobenius group with Frobenius kernel $N$.
If $p$ does not divide $|N|$, then $\Lambda(\mathcal{G}):=\mathbb{Z}_{p}[[\mathcal{G}]]$ is $N$-hybrid.
\end{prop}
\begin{proof}
Let $\gamma$ be a topological generator of $\Gamma$.
Then $N$ is a normal subgroup of $H$ and so $\gamma N \gamma^{-1}$ is also a normal subgroup of $H$
since $\gamma H = H \gamma$. Thus $N$ and $\gamma N \gamma^{-1}$ are normal subgroups of $H$ of equal order and so
Theorem \ref{thm:frob-kernel} (iii) implies that they are in fact equal. Hence $N$ is normal in $\mathcal{G}$.
The claim is now an immediate consequence of Proposition \ref{prop:hybrid-criterion-groupring}
and Theorem \ref{thm:hybrid-criterion}.
\end{proof}
\begin{example}
Let $q$ be a prime power and let $H=\mathrm{Aff}(q)$ be the Frobenius group with Frobenius kernel $N$ defined in Example \ref{ex:affine}.
Let $p$ be an odd prime not dividing $q$ and let $\mathcal{G} = H \rtimes \Gamma$ (any choice of semidirect product).
Then $p$ does not divide $|N|$ and so
$\Lambda(\mathcal{G}) := \mathbb{Z}_{p}[[\mathcal{G}]]$ is $N$-hybrid by Proposition \ref{prop:frob-N-iwasawa-hybrid}.
We consider its structure in more detail.
Let $\eta$ be the unique non-linear irreducible character of $H$.
Choose $\chi \in \mathrm{Irr}_{\mathbb{Q}_{p}^{c}}(\mathcal{G})$
such that $\eta$ appears as an irreducible constituent of $\mathrm{res}^{\mathcal{G}}_{H} \chi$ (this is possible by Lemma \ref{lem:exists-chi-irr-constit-eta}.)
As $\eta$ is the only non-linear irreducible character of $H$, we must have $\eta^{g} = \eta$ for every $g \in \mathcal{G}$,
i.e., $St(\eta) = \mathcal{G}$. Consequently, we have $w_{\chi} = 1$ and thus $\chi(1) = \eta(1) = q-1$.
Since both $\eta$ and $\chi$ have realisations over $\mathbb{Q}$ and hence over $\mathbb{Q}_{p}$,
applying Proposition \ref{prop:chi-comp-max-order} therefore shows that there is a ring isomorphism
$\Lambda(\mathcal{G}) \simeq \mathbb{Z}_{p}[[C_{q-1} \rtimes \Gamma]] \oplus M_{(q-1) \times (q-1)}(\mathbb{Z}_{p}[[T]])$.
\end{example}
\begin{example}\label{ex:S4-V4}
Let $p=3$, $H=S_{4}$ and $N=V_{4}$.
Recall from Example \ref{ex:S4-A4-V4} that $\mathbb{Z}_{3}[S_{4}]$ is $V_{4}$-hybrid and $S_{4}$ is \emph{not} a Frobenius group.
Let $\mathcal{G} = H \rtimes \Gamma$ (any choice of semidirect product).
As each automorphism of $S_{4}$ is inner, $N$ is normal in $\mathcal{G}$
and so Theorem \ref{thm:hybrid-criterion} shows that $\Lambda(\mathcal{G}):=\mathbb{Z}_{3}[[\mathcal{G}]]$ is $N$-hybrid.
We consider its structure in more detail.
Note that $H/N \simeq S_{3}$ and the only two complex irreducible characters $\eta$ and $\eta'$ of $H$
not inflated from characters of $S_{3}$ are of degree $3$ and have realisations over $\mathbb{Q}$ and hence over $\mathbb{Q}_{3}$.
Choose characters $\chi, \chi' \in \mathrm{Irr}_{\mathbb{Q}_{3}^{c}}(\mathcal{G})$ with the property that $\eta$ and $\eta'$ appear as irreducible constituents of $\mathrm{res}^{\mathcal{G}}_{H} \chi$ and $\mathrm{res}^{\mathcal{G}}_{H} \chi'$, respectively.
Again, as each automorphism of $S_{4}$ is inner, we have
$St(\eta) = St(\eta') = \mathcal{G}$ and thus $w_{\chi} = w_{\chi'} = 1$.
Therefore Proposition \ref{prop:chi-comp-max-order} yields a ring isomorphism
$\Lambda(\mathcal{G}) \simeq \mathbb{Z}_{3}[[S_{3} \rtimes \Gamma]] \oplus M_{3 \times 3}(\mathbb{Z}_{3}[[T]]) \oplus M_{3 \times 3}(\mathbb{Z}_{3}[[T']]))$.
\end{example}
\subsection{Iwasawa algebras and commutator subgroups}
Let $p$ be prime (not necessarily odd) and let $\mathcal{G}=H \rtimes \Gamma$ be a one-dimensional $p$-adic Lie group.
\begin{prop}[{\cite[Proposition 4.5]{MR3092262}}]\label{prop:niceIwasawa-algebras}
The Iwasawa algebra $\Lambda(\mathcal{G})=\mathbb{Z}_{p}[[\mathcal{G}]]$ is a direct product of matrix rings over commutative rings
if and only if $p$ does not divide the order of the commutator subgroup $\mathcal{G}'$ of $\mathcal{G}$.
\end{prop}
\begin{corollary}\label{cor:no-skewfields}
If $p$ does not divide the order of $\mathcal{G}'$ then no skewfields appear in the Wedderburn decomposition of $\mathcal{Q}(\mathcal{G})$.
\end{corollary}
\begin{remark}
Note that $\mathcal{G}'$ is a normal subgroup of $H$.
If $\Lambda(\mathcal{G})$ is $\mathcal{G}'$-hybrid then $p$ does not divide the order of $\mathcal{G}'$.
However, the converse does not hold in general.
\end{remark}
\subsection{Hybrid algebras in Iwasawa theory}\label{subsec:algebras-in-Iwasawa-thy}
Let $p$ be an odd prime.
We denote the cyclotomic $\mathbb{Z}_{p}$-extension of a number field $K$ by $K_{\infty}$ and let $K_{m}$ be its $m$th layer.
We put $\Gamma_{K} := \mathrm{Gal}(K_{\infty}/K)$
and choose a topological generator $\gamma_{K}$. In Iwasawa theory one is often concerned with the
following situation. Let $L/K$ be a finite Galois extension of number fields with Galois group $G$.
We put $H := \mathrm{Gal}(L_{\infty}/K_{\infty})$ and $\mathcal{G} := \mathrm{Gal}(L_{\infty}/K)$.
Then $H$ naturally identifies with a normal subgroup of $G$ and $G/H$ is cyclic of $p$-power order
(the field $L^{H}$ equals $L \cap K_{\infty}$ and thus identifies with $K_{m}$ for some $m< \infty$).
As in \S \ref{subsec:Iwasawa-algebras}, we obtain a semidirect product
$\mathcal{G} = H \rtimes \Gamma$ where $\Gamma \leq \mathcal{G}$ and $\Gamma \simeq \Gamma_{K} \simeq \mathbb{Z}_{p}$.
\begin{prop}\label{prop:hybrid-codescent}
Keep the above notation and suppose that $\mathbb{Z}_{p}[G]$ is $N$-hybrid.
Then $N$ naturally identifies with a normal subgroup of $\mathcal{G}$,
which is also a normal subgroup of $H$.
Moreover, both $\mathbb{Z}_{p}[H]$ and $\Lambda(\mathcal{G}):=\mathbb{Z}_{p}[[\mathcal{G}]]$ are also $N$-hybrid.
\end{prop}
\begin{proof}
Let $F = L^{N}$.
As $e_{N}$ lies in $\mathbb{Z}_{p}[G]$, we have that $p \nmid |N|$.
Hence $N$ naturally identifies with $\mathrm{Gal}(L_{\infty}/F_{\infty})$, which is a normal subgroup of both
$H$ and $\mathcal{G}$
since $F_{\infty}/K$ is a Galois extension.
Thus $\mathbb{Z}_{p}[H]$ is $N$-hybrid by Proposition \ref{prop:hybrid-basechange-down} and so
$\Lambda(\mathcal{G})$ is also $N$-hybrid by Theorem \ref{thm:hybrid-criterion}.
\end{proof}
\begin{example} \label{ex:S4-V4-II}
Let $p=3$ and suppose that $G = \mathrm{Gal}(L/K) \simeq S_{4}$.
Then we also have $\mathrm{Gal}(L_{\infty}/K_{\infty}) \simeq S_{4}$, since $S_{4}$ has no abelian quotient of $3$-power order.
As a consequence, we have $\mathcal{G} = \mathrm{Gal}(L_{\infty}/K) \simeq S_{4} \times \Gamma_K$.
Using the notation of Example \ref{ex:S4-V4}, this yields a ring isomorphism
$\Lambda(\mathcal{G}) \simeq \mathbb{Z}_{3}[[S_{3} \times \Gamma_{K}]]
\oplus M_{3 \times 3}(\mathbb{Z}_{3}[[T]]) \oplus M_{3 \times 3}(\mathbb{Z}_{3}[[T']]))$.
\end{example}
\begin{example}
Assume that $G = N \rtimes V$ is a Frobenius group and that $p \nmid |N|$.
Then by Proposition \ref{prop:frob-N-hybrid} the group ring $\mathbb{Z}_{p}[G]$ is $N$-hybrid.
It is straightforward to check that $H = \mathrm{Gal}(L_{\infty}/K_{\infty}) \simeq N \rtimes U$ is a Frobenius group with $U \leq V$.
Let $F = L^{N}$.
If we assume that $V$ is abelian, then $\mathrm{Gal}(F_{\infty}/K)$ is also abelian and so is isomorphic to $\Gamma \times U$ for some
choice of $\Gamma \simeq \mathbb{Z}_{p}$. Thus we have an isomorphism
$\Lambda(\mathcal{G}) \simeq \Lambda(\Gamma \times U) \oplus (1-e_{N}) \mathfrak{M}(\mathcal{G})$
where $\mathfrak{M}(\mathcal{G})$ is a maximal order containing $\Lambda(\mathcal{G})$.
\end{example}
\section{The equivariant Iwasawa main conjecture} \label{sec:EIMC}
\subsection{Algebraic $K$-theory}\label{subsec:K-theory}
Let $R$ be a noetherian integral domain with field of fractions $E$.
Let $A$ be a finite-dimensional semisimple $E$-algebra and let $\mathfrak{A}$ be an $R$-order in $A$.
Let $\mathrm{PMod}(\mathfrak{A})$ denote the category of finitely generated projective (left) $\mathfrak{A}$-modules.
We write $K_{0}(\mathfrak{A})$ for the Grothendieck group of $\mathrm{PMod}(\mathfrak{A})$ (see \cite[\S 38]{MR892316})
and $K_{1}(\mathfrak{A})$ for the Whitehead group (see \cite[\S 40]{MR892316}).
Let $K_{0}(\mathfrak{A}, A)$ denote the relative algebraic $K$-group associated to the ring homomorphism
$\mathfrak{A} \hookrightarrow A$.
We recall that $K_{0}(\mathfrak{A}, A)$ is an abelian group with generators $[X,g,Y]$ where
$X$ and $Y$ are finitely generated projective $\mathfrak{A}$-modules
and $g:E \otimes_{R} X \rightarrow E \otimes_{R} Y$ is an isomorphism of $A$-modules;
for a full description in terms of generators and relations, we refer the reader to \cite[p.\ 215]{MR0245634}.
Moreover, there is a long exact sequence of relative $K$-theory (see \cite[Chapter 15]{MR0245634})
\begin{equation}\label{eqn:long-exact-seq}
K_{1}(\mathfrak{A}) \longrightarrow K_{1}(A) \stackrel{\partial}{\longrightarrow} K_{0}(\mathfrak{A}, A)
\stackrel{\rho}{\longrightarrow} K_{0}(\mathfrak{A}) \longrightarrow K_{0}(A).
\end{equation}
The reduced norm map $\mathrm{nr} = \mathrm{nr}_{A}: A \rightarrow \zeta(A)$ is defined componentwise on the Wedderburn decomposition of $A$
and extends to matrix rings over $A$ (see \cite[\S 7D]{MR632548}); thus
it induces a map $K_{1}(A) \longrightarrow \zeta(A)^{\times}$, which we also denote by $\mathrm{nr}$.
Let $\mathcal C^{b} (\mathrm{PMod} (\mathfrak{A}))$ be the category of bounded complexes of finitely generated projective $\mathfrak{A}$-modules.
Then $K_{0}(\mathfrak{A}, A)$ identifies with the Grothendieck group whose generators are $[C^{\bullet}]$, where $C^{\bullet}$
is an object of the category $\mathcal C^{b}_{\mathrm{tor}}(\mathrm{PMod}(\mathfrak{A}))$ of bounded complexes of finitely generated projective $\mathfrak{A}$-modules whose
cohomology modules are $R$-torsion, and the relations are as follows: $[C^{\bullet}] = 0$ if $C^{\bullet}$ is acyclic, and
$[C_{2}^{\bullet}] = [C_{1}^{\bullet}] + [C_{3}^{\bullet}]$ for every short exact sequence
\[
0 \longrightarrow C_{1}^{\bullet} \longrightarrow C_{2}^{\bullet} \longrightarrow C_{3}^{\bullet} \longrightarrow 0
\]
in $\mathcal C^{b}_{\mathrm{tor}}(\mathrm{PMod}(\mathfrak{A}))$ (see \cite[Chapter 2]{MR3076731} or \cite[\S 2]{MR3068893}, for example).
Let $\mathcal{D} (\mathfrak{A})$ be the derived category of $\mathfrak{A}$-modules.
A complex of $\mathfrak{A}$-modules is said to be perfect if it is
isomorphic in $\mathcal{D} (\mathfrak{A})$ to an element of $\mathcal C^b(\mathrm{PMod} (\mathfrak{A}))$.
We denote the full triangulated subcategory of
$\mathcal{D} (\mathfrak{A})$ comprising perfect complexes by $\mathcal{D}^{\mathrm{perf}} (\mathfrak{A})$,
and the full triangulated subcategory
comprising perfect complexes whose cohomology modules are $R$-torsion by $\mathcal{D}^{\mathrm{perf}}_{\mathrm{tor}} (\mathfrak{A})$.
Then any object of $\mathcal{D}^{\mathrm{perf}}_{\mathrm{tor}} (\mathfrak{A})$ defines an element in $K_{0}(\mathfrak{A}, A)$.
We now specialise to the situation of \S \ref{subsec:Iwasawa-algebras}.
Let $p$ be an odd prime and let $\mathcal{G} = H \rtimes \Gamma$ be a one-dimensional $p$-adic Lie group.
Let $A = \mathcal{Q}(\mathcal{G})$, $\mathfrak{A} = \Lambda(\mathcal{G})=\mathbb{Z}_{p}[[\mathcal{G}]]$ and $R=\mathbb{Z}_{p}[[\Gamma_{0}]]$,
where $\Gamma_{0}$ is an open subgroup of $\Gamma$ that is central in $\mathcal{G}$.
Then \cite[Corollary 3.8]{MR3034286} (take $\mathcal{O}=\mathbb{Z}_{p}$ and $G=\mathcal{G}$ and note that
$\mathcal{O}[[G]]_{S^{*}}=\mathcal{Q}(\mathcal{G})$ since $\mathcal{G}$ is one-dimensional)
shows that the map $\partial$ in \eqref{eqn:long-exact-seq}
is surjective
(one can also give a slight modification of the proof of either \cite[Proposition 3.4]{MR2217048} or \cite[Lemma 1.5]{MR2819672});
thus the sequence
\begin{equation}\label{eqn:Iwasawa-K-sequence}
K_{1}(\Lambda(\mathcal{G})) \longrightarrow K_{1}(\mathcal{Q}(\mathcal{G})) \stackrel{\partial}{\longrightarrow}
K_{0}(\Lambda(\mathcal{G}),\mathcal{Q}(\mathcal{G})) \longrightarrow 0
\end{equation}
is exact.
\subsection{Admissible extensions and the $\mu=0$ hypothesis}\label{subsec:admiss-and-mu}
We specialise the definition of admissible $p$-adic Lie extension given in the introduction to the one-dimensional case.
\begin{definition}\label{def:one-dim-adm}
Let $p$ be an odd prime and let $K$ be a totally real number field.
An admissible one-dimensional $p$-adic Lie extension $\mathcal{L}$ of $K$ is a Galois extension $\mathcal{L}$ of $K$
such that (i) $\mathcal{L}$ is totally real,
(ii) $\mathcal{L}$ contains the cyclotomic $\mathbb{Z}_{p}$-extension $K_{\infty}$ of $K$, and
(iii) $[\mathcal{L} : K_{\infty}]$ is finite.
\end{definition}
Let $\mathcal{L}/K$ be an admissible one-dimensional $p$-adic Lie extension with Galois group $\mathcal{G}$.
Let $H=\mathrm{Gal}(\mathcal{L}/K_{\infty})$ and let $\Gamma_{K}=\mathrm{Gal}(K_{\infty}/K)$.
As in \S \ref{subsec:Iwasawa-algebras}, we obtain a semidirect product
$\mathcal{G} = H \rtimes \Gamma$ where $\Gamma \leq \mathcal{G}$ and $\Gamma \simeq \Gamma_{K} \simeq \mathbb{Z}_{p}$,
and we choose an open subgroup $\Gamma_{0} \leq \Gamma$ that is central in $\mathcal{G}$.
Let $S_{\infty}$ be the set of archimedean places of $K$ and let $S_{p}$ be the set of places of $K$ above $p$.
Let $S_{\mathrm{ram}}=S_{\mathrm{ram}}(\mathcal{L}/K)$ be the (finite) set of places of $K$ that ramify in $\mathcal{L}/K$;
note that $S_{p} \subseteq S_{\mathrm{ram}}$.
Let $S$ be a finite set of places of $K$ containing $S_{\mathrm{ram}} \cup S_{\infty}$.
Let $M_{S}^{\mathrm{ab}}(p)$ be the maximal abelian pro-$p$-extension of
$\mathcal{L}$ unramified outside $S$ and let $X_{S}=\mathrm{Gal}(M_{S}^{\mathrm{ab}}(p)/\mathcal{L})$.
As usual $\mathcal{G}$ acts on $X_{S}$ by $g \cdot x = \tilde{g}x\tilde{g}^{-1}$,
where $g \in \mathcal{G}$, and $\tilde{g}$ is any lift of $g$ to $\mathrm{Gal}(M_{S}^{\mathrm{ab}}(p)/K)$.
This action extends to a left action of $\Lambda(\mathcal{G})$ on $X_{S}$.
Since $\mathcal{L}$ is totally real, a result of Iwasawa \cite{MR0349627} shows that
$X_{S}$ is finitely generated and torsion as a $\Lambda(\Gamma_{0})$-module.
\begin{definition}\label{def:mu=0-hypothesis}
We say that $\mathcal{L}/K$ satisfies the $\mu=0$ hypothesis if $X_{S}$ is finitely generated as a $\mathbb{Z}_{p}$-module.
\end{definition}
\begin{remark}\label{rmk:mu=0}
The classical Iwasawa $\mu=0$ conjecture (at $p$) is the assertion that for every number field $F$, the
Galois group of the maximal unramified abelian $p$-extension of $F_{\infty}$ is a finitely generated $\mathbb{Z}_{p}$-module.
This conjecture has been proven by Ferrero and Washington \cite{MR528968} in the case that $F/\mathbb{Q}$ is abelian.
Now let $\mathcal{L}/K$ be an admissible one-dimensional $p$-adic Lie extension
and let $L$ be a finite Galois extension of $K$ such that $L_{\infty}=\mathcal{L}$.
Let $E$ be an intermediate field of $L/K$ such that $L/E$ is of $p$-power degree.
Then \cite[Theorem 11.3.8]{MR2392026} says that $\mathcal{L}/K$
satisfies the $\mu=0$ hypothesis if and only if $E_{\infty}/K$ does.
Finally, let $\zeta_{p}$ denote a primitive $p$th root of unity.
Then by \cite[Corollary 11.4.4]{MR2392026} Iwasawa's conjecture for $E(\zeta_{p})$ implies the $\mu=0$ hypothesis for
$E_{\infty}(\zeta_{p})^{+}/K$ and thus for $E_{\infty}/K$ and $\mathcal{L}/K$.
\end{remark}
\subsection{A reformulation of the equivariant Iwasawa main conjecture} \label{subsec:EIMC-reformulation}
We give a slight reformulation of the equivariant Iwasawa main conjecture for totally real fields.
Let $\mathcal{L}/K$ be an admissible one-dimensional $p$-adic Lie extension.
We assume the notation and setting of \S \ref{subsec:admiss-and-mu}.
However, we do \emph{not} assume the $\mu=0$ hypothesis for $\mathcal{L}/K$ except where explicitly stated.
Let $C_{S}^{\bullet}(\mathcal{L}/K)$ be the canonical complex
\[
C_{S}^{\bullet}(\mathcal{L}/K) := R\mathrm{Hom}(R\Gamma_{\mathrm{\acute{e}t}}(\mathrm{Spec}(\mathcal{O}_{\mathcal{L},S}), \mathbb{Q}_{p} / \mathbb{Z}_{p}), \mathbb{Q}_{p} / \mathbb{Z}_{p}).
\]
Here, $\mathcal{O}_{\mathcal{L},S}$ denotes the ring of integers $\mathcal{O}_{\mathcal{L}}$ in $\mathcal{L}$ localised at all primes above those in $S$
and
$\mathbb{Q}_{p} / \mathbb{Z}_{p}$ denotes the constant sheaf of the abelian group $\mathbb{Q}_{p} / \mathbb{Z}_{p}$ on the \'{e}tale site
of $\mathrm{Spec}(\mathcal{O}_{\mathcal{L},S})$.
The only non-trivial cohomology groups occur in degree $-1$ and $0$ and we have
\[
H^{-1}(C_{S}^{\bullet}(\mathcal{L}/K)) \simeq X_{S}, \qquad H^{0}(C_{S}^{\bullet}(\mathcal{L}/K)) \simeq \mathbb{Z}_{p}.
\]
It follows from \cite[Proposition 1.6.5]{MR2276851} that $C_{S}^{\bullet}(\mathcal{L}/K)$ belongs to $\mathcal{D}^{\mathrm{perf}}_{\mathrm{tor}}(\Lambda(\mathcal{G}))$.
In particular, $C_{S}^{\bullet}(\mathcal{L}/K)$ defines a class $[C_{S}^{\bullet}(\mathcal{L}/K)]$ in $K_{0}(\Lambda(\mathcal{G}), \mathcal{Q}(\mathcal{G}))$.
Note that $C_{S}^{\bullet}(\mathcal{L}/K)$ and the complex used
by Ritter and Weiss (as constructed in \cite{MR2114937}) become isomorphic in $\mathcal{D}(\Lambda(\mathcal{G}))$ by
\cite[Theorem 2.4]{MR3072281} (see also \cite{MR3068897} for more on this topic).
Hence it makes no essential difference which of these complexes we use.
Recall the notation and hypotheses of \S \ref{subsec:idempotents} and \S \ref{subsec:sufficiently-large}.
In particular, $F$ is a sufficiently large finite extension of $\mathbb{Q}_{p}$.
Let $\chi_{\mathrm{cyc}}$ be the $p$-adic cyclotomic character
\[
\chi_{\mathrm{cyc}}: \mathrm{Gal}(\mathcal{L}(\zeta_{p})/K) \longrightarrow \mathbb{Z}_{p}^{\times},
\]
defined by $\sigma(\zeta) = \zeta^{\chi_{\mathrm{cyc}}(\sigma)}$ for any $\sigma \in \mathrm{Gal}(\mathcal{L}(\zeta_{p})/K)$ and any $p$-power root of unity $\zeta$.
Let $\omega$ and $\kappa$ denote the composition of $\chi_{\mathrm{cyc}}$ with the projections onto the first and second factors of the canonical decomposition $\mathbb{Z}_{p}^{\times} = \mu_{p-1} \times (1+p\mathbb{Z}_{p})$, respectively;
thus $\omega$ is the Teichm\"{u}ller character.
We note that $\kappa$ factors through $\Gamma_{K}$
(and thus also through $\mathcal{G}$) and by abuse of notation we also
use $\kappa$ to denote the associated maps with these domains.
We put $u := \kappa(\gamma_{K})$.
For $r \in \mathbb{N}_{0}$ divisible by $p-1$
(or more generally divisible by the degree $[\mathcal{L}(\zeta_{p}) : \mathcal{L}]$),
up to the natural inclusion map of codomains,
we have $\chi_{\mathrm{cyc}}^{r}=\kappa^{r}$.
Following \cite[Proposition 6]{MR2114937}, we define a map
\[
j_{\chi}: \zeta(\mathcal{Q}^{F} (\mathcal{G})) \twoheadrightarrow \zeta(\mathcal{Q}^{F} (\mathcal{G})e_{\chi}) \simeq \mathcal{Q}^{F}(\Gamma_{\chi}) \rightarrow \mathcal{Q}^{F}(\Gamma_{K}),
\]
where the last arrow is induced by mapping $\gamma_{\chi}$ to $\gamma_{K}^{w_{\chi}}$.
It follows from op.\ cit.\ that $j_{\chi}$ is independent of the choice of $\gamma_{K}$ and that
for every matrix $\Theta \in M_{n \times n} (\mathcal{Q}(\mathcal{G}))$ we have
\begin{equation*} \label{eqn:jchi-det}
j_{\chi} (\mathrm{nr}(\Theta)) = \mathrm{det}_{\mathcal{Q}^{F}(\Gamma_{K})} (\Theta \mid \mathrm{Hom}_{F[H]}(V_{\chi}, \mathcal{Q}^{F}(\mathcal{G})^n)).
\end{equation*}
Here, $\Theta$ acts on $f \in \mathrm{Hom}_{F[H]}(V_{\chi}, \mathcal{Q}^{F}(\mathcal{G})^{n})$ via right multiplication,
and $\gamma_{K}$ acts on the left via $(\gamma_{K} f)(v) = \gamma \cdot f(\gamma^{-1} v)$ for all $v \in V_{\chi}$,
where $\gamma$ is the unique lift of $\gamma_{K}$ to $\Gamma \leq \mathcal{G}$.
Hence the map
\begin{eqnarray*}
\mathrm{Det}(~)(\chi): K_{1}(\mathcal{Q}(\mathcal{G})) & \rightarrow & \mathcal{Q}^{F}(\Gamma_{K})^{\times} \\
{[P,\alpha]}& \mapsto & \mathrm{det}_{\mathcal{Q}^{F}(\Gamma_{K})} (\alpha \mid \mathrm{Hom}_{F[H]}(V_{\chi}, F \otimes_{\mathbb{Q}_{p}} P)),
\end{eqnarray*}
where $P$ is a projective $\mathcal{Q}(\mathcal{G})$-module and $\alpha$ a $\mathcal{Q}(\mathcal{G})$-automorphism of $P$, is just $j_{\chi} \circ \mathrm{nr}$ (see \cite[\S 3, p.558]{MR2114937}).
If $\rho$ is a character of $\mathcal{G}$ of type $W$ (i.e.~$\mathrm{res}^{\mathcal{G}}_H \rho = 1$)
then we denote by
$\rho^{\sharp}$ the automorphism of the field $\mathcal{Q}^{c}(\Gamma_{K})$ induced by
$\rho^{\sharp}(\gamma_{K}) = \rho(\gamma_{K}) \gamma_{K}$.
Moreover, we denote the additive group generated by all $\mathbb{Q}_{p}^{c}$-valued
characters of $\mathcal{G}$ with open kernel by $R_p(\mathcal{G})$; finally,
$\mathrm{Hom}^{\ast}(R_{p}( \mathcal{G}), \mathcal{Q}^{c}(\Gamma_{K})^{\times})$
is the group of all homomorphisms
$f: R_p(\mathcal{G}) \rightarrow \mathcal{Q}^{c}(\Gamma_{K})^{\times}$ satisfying
\[
\begin{array}{ll}
f(\chi \otimes \rho) = \rho^{\sharp}(f(\chi)) & \mbox{ for all characters } \rho \mbox{ of type } W \mbox{ and}\\
f({}^{\sigma}\chi) = \sigma(f(\chi)) & \mbox{ for all Galois automorphisms } \sigma \in \mathrm{Gal}(\mathbb{Q}_{p}^{c}/\mathbb{Q}_{p}).
\end{array}
\]
By \cite[Proof of Theorem 8]{MR2114937} we have an isomorphism
\begin{eqnarray*}
\zeta(\mathcal{Q}(\mathcal{G}))^{\times} & \simeq &
\mathrm{Hom}^{\ast}(R_{p}(\mathcal{G}), \mathcal{Q}^{c}(\Gamma_{K})^{\times})\\
x & \mapsto & [\chi \mapsto j_{\chi}(x)].
\end{eqnarray*}
By \cite[Theorem 8]{MR2114937} the map $\Theta \mapsto [\chi \mapsto \mathrm{Det}(\Theta)(\chi)]$
defines a homomorphism
\[
\mathrm{Det}: K_{1}(\mathcal{Q}(\mathcal{G})) \rightarrow \mathrm{Hom}^{\ast}(R_p(\mathcal{G}), \mathcal{Q}^{c}(\Gamma_{K})^{\times})
\]
such that we obtain a commutative triangle
\begin{equation} \label{eqn:Det_triangle}
\xymatrix{
& K_{1}(\mathcal{Q}(\mathcal{G})) \ar[dl]_{\mathrm{nr}} \ar[dr]^{\mathrm{Det}} &\\
{\zeta(\mathcal{Q}(\mathcal{G}))^{\times}} \ar[rr]^{\sim} & & {\mathrm{Hom}^{\ast}(R_p( \mathcal{G}), \mathcal{Q}^{c}(\Gamma_{K})^{\times})}.}
\end{equation}
Each topological generator $\gamma_{K}$ of $\Gamma_{K}$ permits the definition of a
power series $G_{\chi,S}(T) \in \mathbb{Q}_{p}^{c} \otimes_{\mathbb{Q}_{p}} Quot(\mathbb{Z}_{p}[[T]])$
by starting out from the Deligne-Ribet power series for linear characters of open subgroups
of $\mathcal{G}$ (see \cite{MR579702}; also see \cite{ MR525346, MR524276})
and then extending to the general case by using Brauer induction (see \cite{MR692344}).
One then has an equality
\[
L_{p,S}(1-s,\chi) = \frac{G_{\chi,S}(u^s-1)}{H_{\chi}(u^s-1)},
\]
where $L_{p,S}(s,\chi)$ denotes the `$S$-truncated $p$-adic Artin $L$-function' attached to $\chi$ constructed by Greenberg \cite{MR692344},
and where, for irreducible $\chi$, one has
\[
H_{\chi}(T) = \left\{\begin{array}{ll} \chi(\gamma_{K})(1+T)-1 & \mbox{ if } H \subseteq \ker \chi\\
1 & \mbox{ otherwise.} \end{array}\right.
\]
Now \cite[Proposition 11]{MR2114937} implies that
\[
L_{K,S} : \chi \mapsto \frac{G_{\chi,S}(\gamma_{K}-1)}{H_{\chi}(\gamma_{K}-1)}
\]
is independent of the topological generator $\gamma_{K}$ and lies in
$\mathrm{Hom}^{\ast}(R_{p}( \mathcal{G}), \mathcal{Q}^{c}(\Gamma_{K})^{\times})$.
Diagram \eqref{eqn:Det_triangle} implies that there is a unique element
$\Phi_{S} = \Phi_{S}(\mathcal{L}/K) \in \zeta(\mathcal{Q}(\mathcal{G}))^{\times}$
such that
\[
j_{\chi}(\Phi_{S}) = L_{K,S}(\chi)
\]
for every $\chi \in \mathrm{Irr}_{\mathbb{Q}_{p}^{c}}(\mathcal{G})$.
It is now clear that the following is a reformulation of the EIMC without its uniqueness statement.
\begin{conj}[EIMC]\label{conj:EIMC}
There exists $\zeta_{S} \in K_{1}(\mathcal{Q}(\mathcal{G}))$ such that $\partial(\zeta_{S}) = -[C_{S}^{\bullet}(\mathcal{L}/K)]$
and $\mathrm{nr}(\zeta_{S}) = \Phi_{S}$.
\end{conj}
It can be shown that the truth of Conjecture \ref{conj:EIMC} is independent of the choice of $S$,
provided $S$ is finite and contains $S_{\mathrm{ram}} \cup S_{\infty}$.
The following theorem has been shown independently by Ritter and Weiss \cite{MR2813337} and Kakde \cite{MR3091976}.
\begin{theorem}\label{thm:EIMC-with-mu}
If $\mathcal{L}/K$ satisfies the $\mu=0$ hypothesis then the EIMC holds for $\mathcal{L}/K$.
\end{theorem}
\begin{corollary}\label{cor:EIMC-unconditional}
Let $\mathcal{P}$ be a Sylow $p$-subgroup of $\mathcal{G}$.
If $\mathcal{L}^{\mathcal{P}}/\mathbb{Q}$ is abelian then $\mathcal{P}$ is normal in $\mathcal{G}$ (and thus is unique),
and the EIMC holds for $\mathcal{L}/K$.
\end{corollary}
\begin{proof}
The first claim is clear.
Let $E=\mathcal{L}^{\mathcal{P}}$ and let $L$ be a finite Galois extension of $K$ such that $L_{\infty}=\mathcal{L}$.
Then $L/E$ is a finite Galois extension of $p$-power degree.
Moreover, $E/\mathbb{Q}$ is a (finite) abelian extension by hypothesis and so $E(\zeta_{p})/\mathbb{Q}$ is also abelian.
Therefore the $\mu=0$ hypothesis for $\mathcal{L}/K$ holds by the results discussed in Remark \ref{rmk:mu=0}.
\end{proof}
We shall also consider the EIMC with its uniqueness statement.
\begin{conj}[EIMC with uniqueness]\label{conj:EIMC-unique}
There exists a unique $\zeta_{S} \in K_{1}(\mathcal{Q}(\mathcal{G}))$ such that $\partial(\zeta_{S}) = -[C_{S}^{\bullet}(\mathcal{L}/K)]$
and $\mathrm{nr}(\zeta_{S}) = \Phi_{S}$.
\end{conj}
\begin{remark}\label{rmk:SK1}
Let $SK_{1}(\mathcal{Q}(\mathcal{G})) = \ker(\mathrm{nr}: K_{1}(\mathcal{Q}(\mathcal{G})) \longrightarrow \zeta(\mathcal{Q}(\mathcal{G}))^{\times})$.
If $SK_{1}(\mathcal{Q}(\mathcal{G}))$ vanishes then it is clear that the uniqueness statement of the EIMC follows from
its existence statement.
Moreover, $SK_{1}(\mathcal{Q}(\mathcal{G}))$ vanishes if no skewfields appear in the Wedderburn decomposition of
$\mathcal{Q}(\mathcal{G})$; in particular, this is the case if $\mathcal{G}$ is abelian or, more generally, if $p$ does not
divide the order of the commutator subgroup $\mathcal{G}'$ of $\mathcal{G}$ (see Corollary \ref{cor:no-skewfields}).
As noted in \cite[Remark E]{MR2114937} (also see \cite[Remark 3.5]{MR3294653}), a conjecture of Suslin implies that $SK_{1}(\mathcal{Q}(\mathcal{G}))$ in fact always vanishes.
\end{remark}
\subsection{Relation to the framework of \cite{MR2217048}}
We now discuss Conjecture \ref{conj:EIMC} within the framework of the theory of \cite[\S 3]{MR2217048};
this section may be skipped if the reader is only interested in the formulation of \S \ref{subsec:EIMC-reformulation}.
Let
\[
\pi: \mathcal{G} \rightarrow \mathrm{GL}_{n}(\mathcal{O})
\]
be a continuous homomorphism, where $\mathcal{O}=\mathcal{O}_{F}$ denotes the ring of integers of $F$
and $n$ is some integer greater or equal to $1$.
There is a ring homomorphism
\begin{equation} \label{eqn:first_Phi}
\Phi_{\pi}: \Lambda(\mathcal{G}) \rightarrow M_{n\times n}(\Lambda^{\mathcal{O}}(\Gamma_{K}))
\end{equation}
induced by the continuous group homomorphism
\begin{eqnarray*}
\mathcal{G} & \rightarrow & (M_{n \times n}(\mathcal{O}) \otimes_{\mathbb{Z}_p} \Lambda(\Gamma_{K}))^{\times} = \mathrm{GL}_{n}(\Lambda^{\mathcal{O}}(\Gamma_{K}))\\
\sigma & \mapsto & \pi(\sigma) \otimes \overline{\sigma},
\end{eqnarray*}
where $\overline{\sigma}$ denotes the image of $\sigma$ in $\mathcal{G} / H = \Gamma_{K}$.
By \cite[Lemma 3.3]{MR2217048} the
homomorphism \eqref{eqn:first_Phi} extends to a ring homomorphism
\[
\Phi_{\pi}: \mathcal{Q}(\mathcal{G}) \rightarrow M_{n\times n}(\mathcal{Q}^{F}(\Gamma_{K}))
\]
and this in turn induces a homomorphism
\[
\Phi_{\pi}': K_{1}(\mathcal{Q}(\mathcal{G})) \rightarrow
K_{1}(M_{n\times n}(\mathcal{Q}^{F}(\Gamma_{K}))) = \mathcal{Q}^{F}(\Gamma_{K})^{\times}.
\]
Let $\mathrm{aug}: \Lambda^{\mathcal{O}}(\Gamma_{K}) \twoheadrightarrow \mathcal{O}$ be the augmentation map and put $\mathfrak{p} = \ker(\mathrm{aug})$.
Writing $\Lambda^{\mathcal{O}}(\Gamma_{K})_{\mathfrak{p}}$ for the localisation of $\Lambda^{\mathcal{O}}(\Gamma_{K})$ at $\mathfrak{p}$, it is clear that $\mathrm{aug}$ naturally extends to a homomorphism $\mathrm{aug}: \Lambda^{\mathcal{O}}(\Gamma_{K})_{\mathfrak{p}} \rightarrow F$.
One defines an evaluation map
\begin{equation*} \label{eqn:evaluation-map}
\begin{array}{rcl}
\phi: \mathcal{Q}^{F}(\Gamma_{K}) & \rightarrow & F \cup \{\infty\}\\
x & \mapsto & \left\{ \begin{array}{ll} \mathrm{aug} (x) & \mbox{ if } x \in \Lambda^{\mathcal{O}}(\Gamma_{K})_{\mathfrak{p}}\\
\infty & \mbox{ otherwise}. \end{array} \right.
\end{array}
\end{equation*}
For $r \in \mathbb{Z}$ we define maps
\[
j_{\chi}^{r}: \zeta(\mathcal{Q}^{F} (\mathcal{G})) \twoheadrightarrow \zeta(\mathcal{Q}^{F} (\mathcal{G})e_{\chi}) \simeq \mathcal{Q}^{F}(\Gamma_{\chi}) \rightarrow \mathcal{Q}^{F}(\Gamma_{K}),
\]
where the last arrow is induced by mapping $\gamma_{\chi}$ to
$(u^{r}\gamma_{K})^{w_{\chi}}$.
Note that $j_{\chi}^{0} = j_{\chi}$.
It is straightforward to show that for $r \in \mathbb{Z}$ we have
\begin{equation*}\label{eq:PhiS-jr-p-adic}
\phi(j_{\chi}^{r}(\Phi_{S})) = L_{p,S}(1-r, \chi).
\end{equation*}
If $\zeta$ is an element of $K_{1}(\mathcal{Q}(\mathcal{G}))$, we define $\zeta(\pi)$ to be $\phi(\Phi_{\pi}'(\zeta))$.
Conjecture \ref{conj:EIMC} now implies that there is an element $\zeta_{S} \in K_{1}(\mathcal{Q}(\mathcal{G}))$ such that
$\partial(\zeta_{S}) = -[C_{S}^{\bullet}(\mathcal{L}/K)]$ and for each $r \geq 1$ divisible by $p-1$
and every irreducible Artin representation $\pi_{\chi}$ of $\mathcal{G}$ with character $\chi$ we have
\[
\zeta_{S}(\pi_{\chi}\kappa^{r}) = \phi(j_{\chi}^{r}(\Phi_{S})) = L_{p,S}(1-r, \chi),
\]
where the first equality follows from \cite[Lemma 2.3]{MR2822866}.
\subsection{A maximal order variant of the EIMC}
We shall prove the EIMC in many cases in which the $\mu=0$ hypothesis is not known;
in some of these cases we shall also prove the EIMC with uniqueness.
The following key result of Ritter and Weiss can be seen as a `maximal order variant' of Conjecture \ref{conj:EIMC};
crucially, it does not require the $\mu=0$ hypothesis.
We assume the setup and notation of \S \ref{subsec:EIMC-reformulation}.
\begin{theorem}\label{thm:EIMC-MaxOrd}
Let $\mathfrak{M}(\mathcal{G})$ be a maximal $\mathbb{Z}_{p}[[\Gamma_{0}]]$-order
such that $\Lambda(\mathcal{G}) \subseteq \mathfrak{M}(\mathcal{G}) \subseteq \mathcal{Q}(\mathcal{G})$.
Choose $x_{S} \in K_{1}(\mathcal{Q}(\mathcal{G}))$ such that $\partial(x_{S}) = -[C_{S}^{\bullet}(\mathcal{L}/K)]$.
Then $\mathrm{nr}(x_{S})\Phi_{S}^{-1} \in \zeta(\mathfrak{M}(\mathcal{G}))^{\times}$.
\end{theorem}
\begin{proof}
By \cite[Theorem 16]{MR2114937} we know that
$\mathrm{Det}(x_{S})L_{K,S}^{-1} \in \mathrm{Hom}^{\ast}(R_{p}(\mathcal{G}), \Lambda^{c}(\Gamma_{K})^{\times})$,
where $\Lambda^{c}(\Gamma_{K}) := \mathbb{Z}_{p}^{c} \otimes_{\mathbb{Z}_{p}} \Lambda(\Gamma_{K})$ and $\mathbb{Z}_{p}^{c}$ denotes the integral closure
of $\mathbb{Z}_{p}$ in $\mathbb{Q}_{p}^{c}$.
Moreover, $\mathrm{Hom}^{\ast}(R_{p}(\mathcal{G}), \Lambda^{c}(\Gamma_{K})^{\times})$ identifies with $\zeta(\mathfrak{M}(\mathcal{G}))^{\times}$
under the isomorphism in diagram \eqref{eqn:Det_triangle} as is explained in \cite[Remark H]{MR2114937}. Thus $\mathrm{nr}(x_{S}) \Phi_{S}^{-1}$ lies in $\zeta(\mathfrak{M}(\mathcal{G}))^{\times}$.
\end{proof}
\begin{corollary} \label{cor:EIMC-MaxOrd}
Let $\mathfrak{M}(\mathcal{G})$ be a maximal $\mathbb{Z}_{p}[[\Gamma_{0}]]$-order such that
$\Lambda(\mathcal{G}) \subseteq \mathfrak{M}(\mathcal{G}) \subseteq \mathcal{Q}(\mathcal{G})$
and let $e \in \mathfrak{M}(\mathcal{G})$ be a central idempotent. Suppose that the reduced norm map
\begin{equation} \label{eqn:nr-surjective-hypothesis}
\mathrm{nr}: K_{1}(e \mathfrak{M}(\mathcal{G})) \longrightarrow \zeta(e \mathfrak{M}(\mathcal{G}))^{\times}
\end{equation}
is surjective.
Then there exists $y_{S} \in K_{1}(e \mathcal{Q}(\mathcal{G}))$ such that $\mathrm{nr}(y_{S}) = e \Phi_{S}$
and $y_{S}$ maps to $[e \mathfrak{M}(\mathcal{G}) \otimes^{\mathbb{L}}_{\Lambda(\mathcal{G})} C_{S}^{\bullet}(\mathcal{L}/K)]$ under
$K_{1}(e \mathcal{Q}(\mathcal{G})) \rightarrow K_{0}(e \mathfrak{M}(\mathcal{G}), e \mathcal{Q}(\mathcal{G}))$.
\end{corollary}
\begin{proof}
Choose $x_{S} \in K_{1}(\mathcal{Q}(\mathcal{G}))$ as in Theorem \ref{thm:EIMC-MaxOrd}. By assumption, there is
$z_{S} \in K_{1}(e \mathfrak{M}(\mathcal{G}))$ such that $\mathrm{nr}(z_{S}) = e \mathrm{nr}(x_{S}) \Phi_{S}^{-1}$.
Let $z'_{S}$ be the image of $z_{S}$ in $K_{1}(e \mathcal{Q}(\mathcal{G}))$.
Then $y_{S} := e x_{S} (z'_{S})^{-1}$ has the desired properties.
\end{proof}
\begin{remark}\label{rmk:nr-surjective}
It is not clear whether the map \eqref{eqn:nr-surjective-hypothesis} is always surjective.
However, this map is surjective if no skewfields
occur in the Wedderburn decomposition of $e \mathcal{Q}(\mathcal{G})$, and thus one can always take $e = e_{\mathcal{G}'}$,
where $\mathcal{G'}$ is the commutator subgroup of $\mathcal{G}$ (note that $\mathcal{G}' \leq H$).
If $p$ does not divide the order of $\mathcal{G}'$, then by Corollary \ref{cor:no-skewfields}
one can take an arbitrary $e$ (in particular, $e=1$ is possible).
If $\Lambda(\mathcal{G})$ is $N$-hybrid then Corollary \ref{cor:hybrid-implies-nr-surjective} shows that one can take $e=1-e_{N}$.
\end{remark}
\begin{theorem}\label{thm:EIMC-for-p-not-dividing-ordH}
Let $\mathcal{L}/K$ be an admissible one-dimensional $p$-adic Lie extension with Galois group $\mathcal{G} = H \rtimes \Gamma$.
If $p \nmid |H|$ then the EIMC with uniqueness holds for $\mathcal{L}/K$.
\end{theorem}
\begin{remark}
Theorem \ref{thm:EIMC-for-p-not-dividing-ordH} does not require the $\mu=0$ hypothesis.
The statement is the same as that of \cite[Example 2]{MR2205173},
except that the proof of this latter result does require the $\mu=0$ hypothesis which is not explicitly stated (see \cite[p.\ 48]{MR2242618}).
\end{remark}
\begin{proof}[Proof of Theorem \ref{thm:EIMC-for-p-not-dividing-ordH}]
Since the commutator subgroup $\mathcal{G}'$ is a subgroup of $H$,
Corollary \ref{cor:no-skewfields} shows that the Wedderburn decomposition of $\mathcal{Q}(\mathcal{G})$ contains no skewfields.
This has two consequences. First, uniqueness follows from Remark \ref{rmk:SK1}.
Second, by Remark \ref{rmk:nr-surjective} the hypotheses of Corollary \ref{cor:EIMC-MaxOrd} are satisfied for every choice of $e$.
However, Remark \ref{rmk:p-not-divide-H-max-order} shows that $\Lambda(\mathcal{G})$ is in fact a maximal order since $p \nmid |H|$.
Therefore the EIMC for $\mathcal{L}/K$ follows from Corollary \ref{cor:EIMC-MaxOrd} with $e=1$.
\end{proof}
\subsection{Functorialities}
Let $\mathcal{L}/K$ be an admissible one-dimensional $p$-adic Lie extension with Galois group $\mathcal{G}$.
Let $N$ be a finite normal subgroup of $\mathcal{G}$ and let $\mathcal{H}$ be an open subgroup of $\mathcal{G}$.
There are canonical maps
\begin{eqnarray*}
\mathrm{quot}^{\mathcal{G}}_{\mathcal{G}/N}: & K_{0}(\Lambda(\mathcal{G}), \mathcal{Q}(\mathcal{G})) \longrightarrow &
K_{0}(\Lambda(\mathcal{G}/N), \mathcal{Q}(\mathcal{G}/N)),\\
\mathrm{res}^{\mathcal{G}}_{\mathcal{H}}: & K_{0}(\Lambda(\mathcal{G}), \mathcal{Q}(\mathcal{G})) \longrightarrow &
K_{0}(\Lambda(\mathcal{H}), \mathcal{Q}(\mathcal{H}))
\end{eqnarray*}
induced from scalar extension along $\Lambda(\mathcal{G}) \longrightarrow \Lambda(\mathcal{G}/N)$ and restriction of scalars
along $\Lambda(\mathcal{H}) \hookrightarrow \Lambda(\mathcal{G})$. Similarly, we have maps (see \cite[\S 3]{MR2114937})
\begin{eqnarray*}
\mathrm{quot}^{\mathcal{G}}_{\mathcal{G}/N}: & \mathrm{Hom}^{\ast}(R_{p}(\mathcal{G}), \mathcal{Q}^{c}(\Gamma_{K})^{\times}) \longrightarrow &
\mathrm{Hom}^{\ast}(R_{p}(\mathcal{G}/N), \mathcal{Q}^{c}(\Gamma_{K})^{\times}),\\
\mathrm{res}^{\mathcal{G}}_{\mathcal{H}}: & \mathrm{Hom}^{\ast}(R_{p}(\mathcal{G}), \mathcal{Q}^{c}(\Gamma_{K})^{\times}) \longrightarrow &
\mathrm{Hom}^{\ast}(R_{p}(\mathcal{H}), \mathcal{Q}^{c}(\Gamma_{K'})^{\times}),
\end{eqnarray*}
where $K' := \mathcal{L}^{\mathcal{H}}$; here for $f \in \mathrm{Hom}^{\ast}(R_{p}(\mathcal{G}), \mathcal{Q}^{c}(\Gamma_{K})^{\times})$
we have $(\mathrm{quot}^{\mathcal{G}}_{\mathcal{G}/N} f)(\chi) = f(\mathrm{infl}^{\mathcal{G}}_{\mathcal{G}/N} \chi)$ and
$(\mathrm{res}^{\mathcal{G}}_{\mathcal{H}} f)(\chi') = f(\mathrm{ind}^{\mathcal{G}}_{\mathcal{H}} \chi')$ for $\chi \in R_{p}(\mathcal{G}/N)$
and $\chi' \in R_{p}(\mathcal{H})$. Then diagram \eqref{eqn:Det_triangle} induces canonical maps
\begin{eqnarray*}
\mathrm{quot}^{\mathcal{G}}_{\mathcal{G}/N}: & \zeta(\mathcal{Q}(\mathcal{G}))^{\times} \longrightarrow &
\zeta(\mathcal{Q}(\mathcal{G}/N))^{\times},\\
\mathrm{res}^{\mathcal{G}}_{\mathcal{H}}: & \zeta(\mathcal{Q}(\mathcal{G}))^{\times} \longrightarrow &
\zeta(\mathcal{Q}(\mathcal{H}))^{\times}.
\end{eqnarray*}
The first map is easily seen to be induced by the canonical projection $\mathcal{G} \twoheadrightarrow \mathcal{G}/N$.
The following proposition is an obvious reformulation of \cite[Proposition 12]{MR2114937}
(note that the proof of \cite[Proposition 12 (1)(a)]{MR2114937} uses a result which assumes Leopoldt's conjecture;
a direct proof without this assumption is given in \cite[Appendix]{MR2813337}); also see
\cite[Proposition 1.6.5]{MR2276851}.
\begin{prop} \label{prop:funtorialities}
Let $\mathcal{L}/K$ be an admissible one-dimensional $p$-adic Lie extension with Galois group $\mathcal{G}$.
Then the following statements hold.
\begin{enumerate}
\item
Let $N$ be a finite normal subgroup of $\mathcal{G}$ and put $\mathcal{L}' := \mathcal{L}^{N}$.
Then
\[
\mathrm{quot}^{\mathcal{G}}_{\mathcal{G}/N}([C_{S}^{\bullet}(\mathcal{L}/K)]) = [C_{S}^{\bullet}(\mathcal{L}'/K)], \quad
\mathrm{quot}^{\mathcal{G}}_{\mathcal{G}/N}(\Phi_{S}(\mathcal{L}/K)) = \Phi_{S}(\mathcal{L}'/K).
\]
In particular, if the EIMC holds for $\mathcal{L}/K$, then it holds for $\mathcal{L}'/K$.
\item
Let $\mathcal{H}$ be an open subgroup of $\mathcal{G}$ and put $K' := \mathcal{L}^{\mathcal{H}}$. Then
\[
\mathrm{res}^{\mathcal{G}}_{\mathcal{H}}([C_{S}^{\bullet}(\mathcal{L}/K)]) = [C_{S}^{\bullet}(\mathcal{L}/K')], \quad
\mathrm{res}^{\mathcal{G}}_{\mathcal{H}}(\Phi_{S}(\mathcal{L}/K)) = \Phi_{S}(\mathcal{L}/K').
\]
In particular, if the EIMC holds for $\mathcal{L}/K$, then it holds for $\mathcal{L}/K'$.
\end{enumerate}
\end{prop}
\subsection{The EIMC over hybrid Iwasawa algebras}
We show how hybrid Iwasawa algebras can be used to `break up' certain cases of the EIMC.
\begin{theorem}\label{thm:EIMC-break-down}
Let $\mathcal{L}/K$ be an admissible one-dimensional $p$-adic Lie extension with Galois group $\mathcal{G}$.
Suppose that $\Lambda(\mathcal{G})$ is $N$-hybrid for some
finite normal subgroup $N$ of $\mathcal{G}$.
Let $\overline{\mathcal{P}}$ be a Sylow $p$-subgroup of $\overline{\mathcal{G}}:=\mathrm{Gal}(\mathcal{L}^{N}/K) \simeq \mathcal{G}/N$.
Then the following statements hold.
\begin{enumerate}
\item The EIMC holds for $\mathcal{L}/K$ if and only if it holds for $\mathcal{L}^{N}/K$.
\item The EIMC with uniqueness holds for $\mathcal{L}/K$ if and only if it holds for $\mathcal{L}^{N}/K$.
\item If $(\mathcal{L}^{N})^{\overline{\mathcal{P}}} / \mathbb{Q}$ is abelian, then the EIMC holds for $\mathcal{L}/K$.
\item If $\mathcal{L}^{N} / \mathbb{Q}$ is abelian, then the EIMC with uniqueness holds for $\mathcal{L}/K$.
\end{enumerate}
\end{theorem}
\begin{remark}
Under the hypotheses in (iii), $\overline{\mathcal{P}}$ is necessarily normal in $\overline{\mathcal{G}}$
and thus is its unique Sylow $p$-subgroup.
\end{remark}
\begin{proof}[Proof of Theorem \ref{thm:EIMC-break-down}]
By assumption $\Lambda(\mathcal{G})$ decomposes into $\Lambda(\mathcal{G}) e_{N} \oplus \mathfrak{M}(\mathcal{G}) (1 - e_{N})$
for some maximal order $\mathfrak{M}(\mathcal{G})$. This induces a decomposition of relative $K$-groups
\begin{eqnarray*}
K_{0}(\Lambda(\mathcal{G}), \mathcal{Q}(\mathcal{G}))
& \simeq & K_{0}(\Lambda(\mathcal{G}) e_{N}, \mathcal{Q}(\mathcal{G}) e_{N}) \times K_{0}(\mathfrak{M}(\mathcal{G}) (1 - e_{N}), \mathcal{Q}(\mathcal{G}) (1-e_{N}))\\
& \simeq & K_{0}(\Lambda(\mathcal{G}/N), \mathcal{Q}(\mathcal{G}/N)) \times K_{0}(\mathfrak{M}(\mathcal{G}) (1 - e_{N}), \mathcal{Q}(\mathcal{G}) (1-e_{N}))
\end{eqnarray*}
which maps $[C_{S}^{\bullet}(\mathcal{L}/K)]$ to the pair
$([C_{S}^{\bullet}(\mathcal{L}^{N}/K)], [\mathfrak{M}(\mathcal{G}) (1 - e_{N}) \otimes_{\Lambda(\mathcal{G})}^{\mathbb{L}} C_{S}^{\bullet}(\mathcal{L}/K)])$
by Proposition \ref{prop:funtorialities} (i).
Similarly, we have a decomposition
\begin{eqnarray*}
\zeta(\mathcal{Q}(\mathcal{G}))^{\times} & \simeq & \zeta(\mathcal{Q}(\mathcal{G}) e_{N})^{\times} \times \zeta(\mathcal{Q}(\mathcal{G})(1-e_{N}))^{\times} \\
& \simeq & \zeta(\mathcal{Q}(\mathcal{G}/N))^{\times} \times \zeta(\mathcal{Q}(\mathcal{G})(1-e_{N}))^{\times}
\end{eqnarray*}
which maps $\Phi_{S}(\mathcal{L}/K)$ to the pair $(\Phi_{S}(\mathcal{L}^{N}/K), \Phi_{S}(\mathcal{L}/K) (1-e_{N}))$.
Hence (i) follows from Corollary \ref{cor:EIMC-MaxOrd} and Remark \ref{rmk:nr-surjective}.
Part (ii) follows from (i) if
\[
SK_{1}((1-e_{N})\mathcal{Q}(\mathcal{G})) :=
\ker(\mathrm{nr}: K_{1}((1-e_{N})\mathcal{Q}(\mathcal{G})) \longrightarrow \zeta((1-e_{N})\mathcal{Q}(\mathcal{G}))^{\times})
\]
vanishes.
This is indeed the case since Corollary \ref{cor:hybrid-implies-nr-surjective} implies that $\mathcal{Q}(\mathcal{G})(1-e_{N})$ is a direct product
of matrix rings over fields.
Part (iii) follows from part (i) and Corollary \ref{cor:EIMC-unconditional}.
Finally, part (iv) follows from parts (ii) and (iii) and Remark \ref{rmk:SK1}.
\end{proof}
The following theorem is useful in applications to proving results about finite Galois extensions of number fields.
\begin{theorem}\label{thm:EIMC-start-with-num-fields}
Let $L/K$ be a finite Galois extension of totally real number fields with Galois group $G$.
Let $p$ be an odd prime and let $L_{\infty}$ be the cyclotomic $\mathbb{Z}_{p}$-extension of $L$.
Let $P$ be a Sylow $p$-subgroup of $G$. Then the following statements hold.
\begin{enumerate}
\item $L_{\infty}/K$ is an admissible one-dimensional $p$-adic Lie extension.
\item If $p \nmid |G|$ then the EIMC with uniqueness holds for $L_{\infty}/K$.
\item If $L^{P}/\mathbb{Q}$ is abelian then the EIMC holds for $L_{\infty}/K$.
\end{enumerate}
Suppose further that $\mathbb{Z}_{p}[G]$ is $N$-hybrid.
Let $\overline{P}$ be a Sylow $p$-subgroup of $\overline{G}:=\mathrm{Gal}(L^{N}/K) \simeq G/N$.
Then the following statements hold.
\begin{enumerate}
\setcounter{enumi}{3}
\item If $L^{N}/\mathbb{Q}$ is abelian then the EIMC with uniqueness holds for $L_{\infty}/K$.
\item If $(L^{N})^{\overline{P}}/\mathbb{Q}$ is abelian then the EIMC holds for $L_{\infty}/K$.
\end{enumerate}
\end{theorem}
\begin{remark}
Under the hypothesis in (iii), $P$ is necessarily normal in $G$ and thus is its unique
Sylow $p$-subgroup. In part (v), we have $(L^{N})^{\overline{P}}=L^{NP}=L^{N \rtimes P}$.
\end{remark}
\begin{proof}[Proof of Theorem \ref{thm:EIMC-start-with-num-fields}]
For part (i) it is trivial to check that the conditions of Definition \ref{def:one-dim-adm} are satisfied.
We adopt the setup and notation of \S \ref{subsec:algebras-in-Iwasawa-thy}.
Since $H$ identifies with a subgroup of $G$, part (ii) follows from Theorem \ref{thm:EIMC-for-p-not-dividing-ordH}
(in fact $H$ identifies with $G$ in this case.)
Let $\mathcal{P}$ be a Sylow $p$-subgroup of $\mathcal{G}$.
Since $\Gamma_{L}:=\mathrm{Gal}(L_{\infty}/L)$ is an open normal pro-$p$ subgroup of $\mathcal{G}$
we have $\Gamma_{L} \leq \mathcal{P}$ and $\mathcal{P}$ maps to $P$ under the natural
projection $\mathcal{G} \rightarrow \mathcal{G}/\Gamma_{L} \simeq G$.
Hence $L^{P}=L_{\infty}^{\mathcal{P}}$ and so part (iii) follows from Corollary \ref{cor:EIMC-unconditional}.
Now suppose that $\mathbb{Z}_{p}[G]$ is $N$-hybrid.
Proposition \ref{prop:hybrid-codescent} says that $N$ identifies with a normal subgroup of $\mathcal{G}$ which is also a normal subgroup of $H$;
moreover, both $\mathbb{Z}_{p}[H]$ and $\Lambda(\mathcal{G})$ are $N$-hybrid.
Note that $(L_{\infty})^{N}=(L^{N})_{\infty}$.
If $L^{N}/\mathbb{Q}$ is abelian then $L_{\infty}^{N}/\mathbb{Q}$ is also abelian and so part (iv) follows from
Theorem \ref{thm:EIMC-break-down} (iv).
Part (v) now follows from (iii) and Theorem \ref{thm:EIMC-break-down} (i).
\end{proof}
\begin{corollary}\label{cor:EIMC-Frobenius}
Let $L/K$ be a finite Galois extension of totally real number fields with Galois group $G$.
Suppose that $G = N \rtimes V$ is a Frobenius group with Frobenius kernel $N$ and abelian Frobenius complement $V$.
Further suppose that $L^{N}/\mathbb{Q}$ is abelian (in particular, this is the case when $K=\mathbb{Q}$).
Let $p$ be an odd prime and let $L_{\infty}$ be the cyclotomic $\mathbb{Z}_{p}$-extension of $L$.
Then the following statements hold.
\begin{enumerate}
\item If $p \nmid |N|$ then the EIMC with uniqueness holds for $L_{\infty}/K$.
\item If $N$ is a $p$-group then the EIMC holds for $L_{\infty}/K$.
\item If $N$ is an $\ell$-group for any prime $\ell$ then the EIMC holds for $L_{\infty}/K$.\\
(This includes the cases $\ell=2$ and $\ell=p$.)
\end{enumerate}
In particular, (iii) holds in the following cases:
\begin{itemize}
\item $G \simeq \mathrm{Aff}(q)$, where $q$ is a prime power (see Example \ref{ex:affine}),
\item $G \simeq C_{\ell} \rtimes C_{q}$, where $q<\ell$ are distinct primes such that $q \mid (\ell-1)$ and $C_{q}$ acts on $C_{\ell}$ via an embedding $C_{q} \hookrightarrow \mathrm{Aut}(C_{\ell})$ (see Example \ref{ex:metacyclic}),
\item $G$ is isomorphic to any of the Frobenius groups constructed in Example \ref{ex:non-abelian-kernel}.
\end{itemize}
\end{corollary}
\begin{proof}
Part (i) follows from Proposition \ref{prop:frob-N-hybrid} and Theorem \ref{thm:EIMC-start-with-num-fields} (iv).
Part (ii) follows from Theorem \ref{thm:EIMC-start-with-num-fields} (iii) with $P=N$.
Part (iii) is just the combination of (i) and (ii).
\end{proof}
\begin{remark}
Let $L$ be a totally real finite Galois extension of $\mathbb{Q}$ with Galois group $G$.
Let $p$ be an odd prime and let $L_{\infty}$ be the cyclotomic $\mathbb{Z}_{p}$-extension of $L$.
If $G$ is abelian then the EIMC with uniqueness for $L_{\infty}/\mathbb{Q}$ holds by Corollary \ref{cor:EIMC-unconditional} and Remark \ref{rmk:SK1}.
If $G$ is an $\ell$-group for any prime $\ell$ then EIMC for $L_{\infty}/\mathbb{Q}$ holds: if $p=\ell$ this follows from the fact that the $\mu=0$
hypothesis is known in this case (see Remark \ref{rmk:mu=0}), and if $p \neq \ell$ then the EIMC with uniqueness for $L_{\infty}/\mathbb{Q}$
holds by Theorem \ref{thm:EIMC-for-p-not-dividing-ordH}.
Thus it remains to consider finite groups $G$ that are neither abelian nor $\ell$-groups for any prime $\ell$.
Among all such groups of order $\leq 10^{6}$ there are $568,220$ metabelian Frobenius groups
(see \cite[Remark 11.13 (A)]{MR1600514}), which in particular have abelian Frobenius complement
and so satisfy the hypotheses of Corollary \ref{cor:EIMC-Frobenius}.
In fact, there are many more pairs ($G$, $p$) with $p$ an odd prime dividing $|G| \leq 10^{6}$ that satisfy the hypotheses
of Theorem \ref{thm:EIMC-start-with-num-fields} (v) in the case $K=\mathbb{Q}$ (in particular, recall that if $\mathbb{Z}_{p}[G]$ is $N$-hybrid for some non-trivial $N$ then $G$ need not be a Frobenius group).
\end{remark}
\begin{example} \label{ex:EIMC-dicyclic}
Let $p$ be an odd prime.
Let $V = \mathrm{Dic}_{p}$ be dicyclic of order $4p$ and $G = N \rtimes V$ be a Frobenius group as in Example \ref{ex:dicyclic-complement}.
Let $L/K$ be a finite Galois extension of totally real number fields with Galois group $\mathrm{Gal}(L/K) \simeq G$.
The unique Sylow $p$-subgroup $P$ of $\mathrm{Dic}_{p}$ coincides with the commutator subgroup
and thus $L^{N \rtimes P}/K$ is cyclic of order $4$.
If we further assume that $L^{N \rtimes P}/\mathbb{Q}$ is abelian (for instance, assume that $K=\mathbb{Q}$),
then Theorem \ref{thm:EIMC-start-with-num-fields} (v) implies that the EIMC holds for $L_{\infty}/K$,
where $L_{\infty}$ is the cyclotomic $\mathbb{Z}_{p}$-extension of $L$.
Note that this cannot be deduced from Corollary \ref{cor:EIMC-Frobenius}
because although $G$ is a Frobenius group, it has a non-abelian Frobenius complement.
\end{example}
\begin{example} \label{ex:EIMC-modified-affine}
Let $q=\ell^{n}$ be a prime power and consider the group $\mathbb{F}_{q} \rtimes (\mathbb{F}_{q}^{\times} \rtimes \langle \phi \rangle)$
of Example \ref{ex:affine-Frobenius}. Let $p \mid (q-1)$ be an odd prime that does not divide $n$.
Then the group $\mu_{p}(\mathbb{F}_{q})$ of $p$-power roots of unity in $\mathbb{F}_{q}$ is non-trivial and we put
$G := \mathbb{F}_{q} \rtimes (\mu_{p}(\mathbb{F}_{q}) \rtimes \langle \phi \rangle)$ and $U := \mathbb{F}_{q} \rtimes \mu_{p}(\mathbb{F}_{q}) \unlhd G$.
Then Example \ref{ex:affine-Frobenius} and Proposition \ref{prop:hybrid-basechange-up} imply that $\mathbb{Z}_{p}[G]$ is
$\mathbb{F}_{q}$-hybrid.
Now assume that $L/K$ is a Galois extension of totally real number fields with $\mathrm{Gal}(L/K) \simeq G$
and let $L_{\infty}$ be the cyclotomic $\mathbb{Z}_{p}$-extension of $L$.
If $L^{U}/\mathbb{Q}$ is abelian (for instance if $K=\mathbb{Q}$), then Theorem \ref{thm:EIMC-start-with-num-fields} (v) implies that the EIMC holds
for $L_{\infty}/K$.
\end{example}
\begin{example}\label{ex:EIMC-S4}
Let $L/K$ be a finite Galois extension of totally real number fields with Galois group $\mathrm{Gal}(L/K) \simeq S_{4}$.
Let $p$ be an odd prime and let $L_{\infty}$ be the cyclotomic $\mathbb{Z}_{p}$-extension of $L$.
Then Theorem \ref{thm:EIMC-start-with-num-fields} (ii) shows that the EIMC with uniqueness holds for $L_{\infty}/K$ when $p>3$.
Now further assume that $L^{A_{4}}/\mathbb{Q}$ is abelian (in particular, this is the case when $K=\mathbb{Q}$) and consider the case $p=3$.
The group ring $\mathbb{Z}_{3}[S_{4}]$ is $V_{4}$-hybrid as shown in Example \ref{ex:S4-A4-V4}.
Moreover, the Sylow $3$-subgroup of $S_{4}/V_{4} \simeq S_{3}$ is $A_{3} \simeq C_{3}$ and we have
$(L^{V_{4}})^{A_{3}}=L^{A_{4}}$, so the EIMC for $L_{\infty}/K$ follows from Theorem \ref{thm:EIMC-start-with-num-fields} (v).
Let $\mathcal{G}=\mathrm{Gal}(L_{\infty}/K)$.
Then as shown in Example \ref{ex:S4-V4-II} we have $\mathcal{G} \simeq S_{4} \times \Gamma_{K}$,
and so no skewfields occur in the Wedderburn decomposition of $\mathcal{Q}(\mathcal{G})$.
Therefore, the EIMC with uniqueness also holds for $L_{\infty}/K$ when $p=3$.
\end{example}
\subsection{Remarks on the higher dimension case} \label{subsec:higher-rk}
In \cite{MR3091976}, Kakde proved a more general version of the EIMC for admissible
$p$-adic Lie extensions of arbitrary (finite) dimension under a suitable version of the $\mu=0$ hypothesis.
This used a strategy of Burns and Kato to reduce the proof to the one-dimensional case discussed above (see \cite{MR3294653}).
We briefly discuss some of the obstacles to generalising the approach of this article to prove higher dimension cases
of the EIMC when a suitable $\mu=0$ hypothesis is not known.
A serious obstacle is that a certain `$\mathfrak{M}_{H}(G)$-conjecture' is required to even formulate
the higher dimension version of the EIMC, and that this is presently only known to hold under a suitable $\mu=0$ hypothesis
(see \cite[p.\ 5]{zbMATH06148870} and \cite{MR2905532}).
Another problem is that a higher dimension version of Theorem \ref{thm:EIMC-MaxOrd} (the `maximal order variant of the EIMC')
has not been proven unconditionally.
Finally, and perhaps most importantly, it is not clear how the notion of a hybrid Iwasawa algebra generalises to the higher dimension case.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 5,157 |
A dog owner wants to warn others after her beloved pet labrador suffered a horrific cut at a popular walking spot.
Donna Sharkey, 50, from Coningsby, was with her partner Anthony Malin walking their dog Chester in Ostler Woods in Woodhall Spa.
It was around 5.30pm yesterday, April 11, that they heard Chester let out a horrible yelp as he was bounding around the woodland off his lead.
She said: "He was off towards the right hand side at the bottom of the woods and suddenly he yelped.
"My partner went rushing to help him and there was this huge gash and an envelope of skin.
After they had treated Chester, performing emergency surgery to repair the cut in his muscle, the vet told Donna that the cut was so clean it could have come from a scalpel. That's when she wondered whether something sharp was laying on the ground in the woods.
Chester is only 15 months old.
He was walking a mere six or seven yards away from a public footpath.
"He's feeling very sorry for himself today," Donna said.
"The worry is that the skin around the wound is going to die, but we'll see.
"I'm trying to put it out there because it's such a popular place for dog walkers. It won't put me off walking there again, I just want everyone to be careful.
Donna and her partner Anthony are going back to the spot where it happened tonight to have a closer look and see if they can figure out what caused Chester's injury, making it safer for other dogs.
At the time it happened, they say they were totally focused on getting Chester to the vets and treated. | {
"redpajama_set_name": "RedPajamaC4"
} | 9,601 |
{"url":"https:\/\/www.r-bloggers.com\/2020\/10\/multiple-gauge-plots-with-facet-wrap\/","text":"# Multiple Gauge Plots with Facet Wrap\n\n[This article was first published on exploRations in R, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here)\nWant to share your content on R-bloggers? click here if you have a blog, or here if you don't.\n\n## Intro\n\nHere are some good examples of how to generate gauge plots including multiple gauge plots in R found at stackoverflow. My use case, however, requires the ability to vary the number of plots based on the number of metrics fed into the function without having to adjust the code. Below is such a function that borrows from these examples and leverages ggplot2::facet_wrap.\n\n## Gauge Plots with Facet Wrap\n\nMy data is captured in a table with a column \u201cpos\u201d for the position of the needle, and \u201cmetric\u201d for the name of the metric.\n\nTable 1: Metrics Data\npos metric\n94 Metric 1\n23 Metric 2\n44 Metric 3\n57 Metric 4\n17 Metric 5\n79 Metric 6\n5 Metric 7\n66 Metric 8\n\nThe gauge_plot() function takes three parameters. The first is the two-column data frame that has variables \u201cpos\u201d for the needle position, and \u201cmetric\u201d for the name of the metric. The second optional parameter allows you to change the breaks. The final parameter controls the number of columns generated by facet_wrap().\n\nHere is the output that captures the 8 metrics in two columns with the default breaks.\n\n## The Code\n\nHere\u2019s the code for the function.\n\ngauge_plot <- function(vals, breaks=c(0,30,70,100), ncol= NULL) {\nrequire(ggplot2)\nrequire(dplyr)\nrequire(tidyr)\n\nif (!is.data.frame(vals)) stop(\"Vals must be a dataframe\")\nif (!dim(vals)[2]==2) stop(\"Vals must have two columns\")\nif (!is.numeric(vals$pos)) stop(\"Dataframe variable pos must be numeric\") # function to generate polygons get_poly <- function(a,b,r1=0.5,r2=1.0) { th.start <- pi*(1-a\/100) th.end <- pi*(1-b\/100) th <- seq(th.start,th.end,length=100) x <- c(r1*cos(th),rev(r2*cos(th))) y <- c(r1*sin(th),rev(r2*sin(th))) df <- data.frame(x,y) return(df) } # create the segments based on the breaks segments <- list() seg_names <- tibble(x = c(\"a\", \"c\", \"e\"), y = c(\"b\", \"d\" ,\"f\")) for(n in 1:3){ i <-breaks[n] j <-breaks[n+1] df <- get_poly(i, j) names(df) <- seg_names[n,] segments$df[[n]] <- df\n}\n\ndfs <- bind_cols(segments)\n\n# create set of segments for each metric\npnt <- tibble()\nfor (name in vals$metric){ pnt[1:nrow(dfs), name] <- name } dfp <- dfs %>% cbind(pnt) %>% pivot_longer(-c(a:f), names_to = \"metric\") %>% select(-value) # generate the needles needles <- list() for(p in 1:nrow(vals)){ i <-vals$pos[p] - 1\nj <-vals$pos[p] + 1 r1 <- 0.2 df <- get_poly(i, j, r1) df$metric <- vals$metric[p] needles$df[[p]] <- df\n}\n\ndfn <- bind_rows(needles)\n\n# graph\nggplot()+\ngeom_polygon(data=dfp, aes(a,b), fill=\"red\")+\ngeom_polygon(data=dfp, aes(c,d), fill=\"gold\")+\ngeom_polygon(data=dfp, aes(e,f), fill=\"forestgreen\")+\ngeom_polygon(data=dfn, aes(x,y))+\ngeom_text(data=as.data.frame(breaks), size=3, fontface=\"bold\", vjust=0,\naes(x=1.05*cos(pi*(1-breaks\/100)),y=1.05*sin(pi*(1-breaks\/100)),label=breaks))+\ngeom_text(data=vals, aes(x=0,y=0), label=paste0(vals\\$pos,\"%\"), vjust=0, size=4, fontface=\"bold\")+\ncoord_fixed()+\ntheme_bw()+\ntheme(axis.text=element_blank(),\naxis.title=element_blank(),\naxis.ticks=element_blank(),\npanel.grid=element_blank(),\npanel.border=element_blank()) +\nfacet_wrap(~metric, strip.position = \"bottom\", ncol = ncol) +\nlabs(title = \"Multiple Gauge Plots w\/ Facet Wrap\")\n}\n\nTo leave a comment for the author, please follow the link and comment on their blog: exploRations in R.\n\nR-bloggers.com offers daily e-mail updates about R news and tutorials about learning R and many other topics. Click here if you're looking to post or find an R\/data-science job.\nWant to share your content on R-bloggers? click here if you have a blog, or here if you don't.","date":"2023-02-01 03:41:20","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\": 0.21234209835529327, \"perplexity\": 5201.099182727158}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": false}, \"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-2023-06\/segments\/1674764499899.9\/warc\/CC-MAIN-20230201013650-20230201043650-00670.warc.gz\"}"} | null | null |
Selling Originalism
Jamal Greene, Columbia Law SchoolFollow
The legal academy has been unkind to originalism. Legal scholars have leveled withering criticism at originalists, maintaining that their faith in judges' abilities to divine historical intent and meaning is facile and that their underlying account of democratic authority is theoretically impoverished and insufficiently attentive to actual constitutional practice. Yet originalism is itself a robust part of that practice and as a political aesthetic is at least as healthy today as it was when Justice Scalia joined the Court in 1986. This Article considers the import of originalism's practical success for nonoriginalist constitutional theories. To the extent that such theories rely on sustained public acquiescence as a legitimating criterion for constitutional interpretation, the capacity of a methodology to translate political ends into the language of constitutional law should be of great interest. Nonoriginalist theories that rely on the legitimating power of public acceptance must account for originalism's political success as an important part of their theories of constitutional change rather than as an unfortunate exception to them. I sketch an exemplary model of originalism's political success that is placed in market terms. I suggest that proponents of originalism have taken advantage of a democratization of the market for constitutional methodologies that has placed a premium on populist selling points such as simplicity, class-based critiques of judicial elites, and nativism. Developing similar and richer models of the political success of constitutional methodologies will help to round out heretofore underspecified theories of how constitutional change occurs outside of the Article V process.
Constitutional Law | Law
Jamal Greene, Selling Originalism, Georgetown Law Journal, Vol. 97, p. 657, 2009; Columbia Public Law Research Paper No. 08-185 (2008).
Constitutional Law Commons | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 3,902 |
Q: Getting data from access to combobox not working I'm trying to get data from access to a combobox1 but its not working.
my database is a .accdb file. my table name is "list".
the columns names are:
"ID_lista" = autonumber
"Nome" = smalltext
"Link" = hyperlink
Here is the code:
using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Windows.Forms;
using System.Data.OleDb;
namespace teste_pwd
{
public partial class Form2 : Form
{
public Form2()
{
InitializeComponent();
}
public static class Globals
{
public static String ConString = "Provider=Microsoft.ACE.OLEDB.12.0; Data Source=" + Application.StartupPath + "\\PWD_BD.accdb";
public static OleDbConnection ConnectionString = new OleDbConnection(Globals.ConString);
}
private void Form2_Load(object sender, EventArgs e)
{
Globals.ConnectionString.Close();
pCMB();
}
public void pCMB()
{
try
{
Globals.ConnectionString.Open();
int i = 0;
string query = "SELECT * FROM List";
OleDbDataAdapter dataAdapeter = new OleDbDataAdapter(query, Globals.ConnectionString);
DataSet mydata = new DataSet();
dataAdapeter.Fill(mydata, "List");
comboBox1.DisplayMember = "Texto";
comboBox1.ValueMember = "Valor";
DataTable tb = new DataTable();
tb.Columns.Add("Texto", typeof(string));
tb.Columns.Add("Valor", typeof(int));
for (i = 0; i <= mydata.Tables[0].Rows.Count - 1; i++)
{
tb.Rows.Add(mydata.Tables[0].Rows[i].ItemArray[1], mydata.Tables[0].Rows[i].ItemArray[0]);
}
comboBox1.DataSource = tb;
Globals.ConnectionString.Close();
}
catch (Exception ex) {
MessageBox.Show(ex.Message);
}
}
}
}
A: this code might help you to add data to combo box.
this code is for vb.net
Dim DrContractor As DataRow
Dim DtContractor As DataTable
Dim i As Integer
strQuery = " SELECT * FROM CONTRACTOR ORDER BY CONTCODE"
Dim DataAdapeter As New SqlDataAdapter(strQuery, myConnection)
Dim dsDivision As New DataSet
'myConnection.Open()
DataAdapeter.Fill(dsDivision, "CONTRACTOR")
DtContractor = dsDivision.Tables("CONTRACTOR")
'myConnection.Close()
cboContractor.Items.Add("ALL")
For i = 0 To DtContractor.Rows.Count - 1
DrContractor = DtContractor.Rows(i)
cboContractor.Items.Add(DrContractor("CONTNAME").ToString & " : " & DrContractor("CONTCODE").ToString)
Next
cboContractor.SelectedIndex = 0
cboContractor.SelectedText = ""
Catch ex As Exception
MsgBox(ex.Message)
End Try
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 31 |
{"url":"http:\/\/ecly.ratingcomuni.it\/hit-or-miss-monte-carlo-integration-python.html","text":"# Hit Or Miss Monte Carlo Integration Python\n\nBautz, Ralf K. Seit dem Jahr 1944 spricht man von Monte Carlo Methoden, wobei man sich aber auch fr\u00fcher mit derartigen Problemen\/Methoden auseinandergesetzt hat. So far, we've only been able to see the a single person's various odds examples, nothing more. Also the anti theft light is on. 5 \u03a0\u03bf\u03b9\u03ac \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03b1\u03c0\u03bb\u03bf\u03cd\u03c3\u03c4\u03b5\u03c1\u03b7 \u00b5\u03bf\u03c1\u03c6\u03ae \u03c4\u03b7\u03c2 Monte Carlo \u0397 \u03bf\u03bb\u03bf\u03ba\u03bb\u03ae\u03c1\u03c9\u03c3\u03b7 \u00b5\u03b5 \u03c4\u03b7 \u0395\u03c0\u03b9\u03c4\u03c5\u03c7\u03af\u03b1 \u03ae \u0391\u03c3\u03c4\u03bf\u03c7\u03af\u03b1 (Hit or Miss) \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u03b1\u03c0\u03bb\u03bf\u03cd\u03c3\u03c4\u03b5\u03c1\u03bf. 9781437386622 1437386628 The Calculus for Engineers and Physicists - Integration and Differentiation, with Applications to Technical Problems (1897), Robert Henry Smith 9781437436631 1437436633 Trusia - A Princess of Krovitch (1906), Davis Brinton, Walter H. It doesn't involve integrals - it sounds like you want the exact solution, no? $\\endgroup$ \u2013 Bobson Dugnutt May 11 '17 at 11:03. The game is a rpg-simulation. Mendel gown, Ofira jewelry and Judith Leiber clutch at an N. (2007), Howe (1969), and Zorn et al. Continuous variables. In a report on Monte Carlo method published in 1957 by the Los Alamos Scientific Laboratory, we could already read:. This example is presented in many books on statistical simulation and is famous enough that Brian Ripley in his book Stochastic Simulation states that the problem is \u201cwell known to every reader\u201d (Ripley 1987, p. 13 Monte Carlo: General Integration Procedures 199 as Python, in which most functions must be explicitly brought in as modules. Tutorial on Monte Carlo Techniques Gabriel A. Inverse transform method. An important application of Monte Carlo techniques constitutes the Monte Carlo integration which is presented in Sect. As a further enhanced tool for high thoughput NMR analysis, we developed and tested a new integration method for 2D NMR spectra quantification, called CAKE, based upon NMR axial symmetry and Monte Carlo Hit-or-Miss technique. Scatterplot Smoothing R code (not currently used) Data sets and other code. The WAE appliance farm is represented by a virtual IP address. 2) Monte Carlo Simulation Find mean values (expectation values) of some system components. f(x) I ? f(x) dx. Writing a Physically Plausible Light. We can use this property to drastically reduce the number of paths needed in the Monte Carlo simulation. View Rlecture09. The name of Monte Carlo was applied to a class of mathematical methods rst by scientists working on. The volume (in space) of integration is considered large enough for the \u2018kinetic\u2019 energy to be internal; there should be no need to change the integra-tion domain as a function of time. Founder Dave Jackson is a leading expert on Podcasting and his snarky attitude and entertaining delivery has lead Dave to building a large and loyal following. For the systems considered here, all reasonable definitions of the integration limit c yield effectively identical K i values, as verified in Figure S9 of the SI. Phil Collins - Two Hearts 759. The effectiveness of S-4s is hit or miss, with many not being engaged in planning sustainment for the squadron, leaving planning and executing on the forward-support company (FSC) commander, who does not have the visibility of current and future operations the S-4 has as a member of the staff at the combat-trains CP (CTCP). While other algorithms usually evaluate the integrand at a regular grid, Monte Carlo randomly choose points at which the integrand is evaluated. Introduction to Computational Physics and Monte Carlo Simulations of Matrix Field Theory Badis Ydri Department of Physics, Faculty of Sciences, BM Annaba University, Annaba, Algeria. Karaoke-Version songs by Artist and Language. Oct 9, 2015- Explore LNatureBlush's board \"Louis Vuitton and Celebrities\" on Pinterest. Before we begin, we should establish what a monte carlo simulation is. Other readers will always be interested in your opinion of the books you've read. PI estimator that uses the Monte Carlo Hit or Miss algorithm. it) is a toolbox for integrating 2D NMR spectra by the CAKE (Monte CArlo peaK volume Estimation)1 algorithm within the Matlab environment (www. This sounds like the Touareg tire issues, where there was vibrations at all sorts of speeds,. however, before we get to this point, it is useful and easy to introduce the concept with a simple example. Continuous variables. When Paul Alexander died in December 1977, he had been at Berkeley for just over a decade. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. In a similar way that a Monte Carlo integration of the sort described in Section2. Warning: All code was play-tested in the QB 4. This probabilistic method relies on a random number generator and is described below. 03 - Answered by a verified Auto Mechanic. I think that the equation is x\u00b2 + y\u00b2 <= 1, and you are doing sqrt(x\u00b2 + y\u00b2) <= 1. MonteCarloIntegration. If then throw 4 is a miss (rows 1-2) then throw 5 is irrelevant because it is not directly preceeded by a streak of 3 hits. Roxette - The Look 760. Quantitative information from multidimensional NMR experiments can be obtained by peak volume integration. What if you have too many dice to count this way? For example, say you want to know the odds of getting a combined total of 15 or more on a roll of 8d6. high dimension problems [16]. In this video we're going to use matplotlib to further visualize our gamblers and their varying scenarios. Integrate a function using the Hit-or-Miss Monte Carlo algorithm. This method is the same as the regular hit-or-miss integration method except that the estimated ellipsoid is employed as a base sampling domain, which may exclude some portion of the region and may thus result. In reality, only one of the outcome possibilities will play out, but, in terms of risk. it) is a toolbox for integrating 2D NMR spectra by the CAKE (Monte CArlo peaK volume Estimation)1 algorithm within the Matlab environment (www. Oct 9, 2015- Explore LNatureBlush's board \"Louis Vuitton and Celebrities\" on Pinterest. Integration tests are essential for having adequate testing. MC integration (hit-or-miss method) Monte Carlo method Bu\ufb00on\u2019s needle problem From Solitaire to MC Newton-Pepys problem PRNG Hit-or-miss method MC integration results Optimization of MC Crude method Methods comparison Random from PDF CDF CDF discrete CDF continuous Acceptance-rejection Quasi-elastic scattering Tutorial MC MC generators \u03bdN. Introduction. that the \"Hit-Or-Miss Monte Carlo\". The criterion of a 1-s time window was based on. However, patients did not activate these reward-related regions. (Here, ) Generate \/ random uniform 10 $2 pairs, 0 \u2019s from 3 54 76 (here, ) and 2$ \u2019s from 3. The adaptive Monte Carlo algorithm may move to a worse estimated integral in few iterations. The \"Monte Carlo Method\" is a method of solving problems using statistics. MC or MD), and the actual function form need not be. In [5] propo-sedthe calculation of global illumination of the scene represented by a Fredholm integral equation of the se-cond kind. We propose a corrector adaptive Monte Carlo integration algorithm that prevents the adaptive algorithm from moving to a. Details of the module's prerequisites, learning outcomes, assessment and contact hours are given in the official module description in the Faculty Handbook - follow the link above. We recorded EEG from golfers while putting and scored the puts as hit or miss. Skip navigation Sign in. Introduction. Andrew Dotson 14,894 views. Simulazioni Monte Carlo in Fisica delle Particelle Le simulazioni Monte Carlo sono ampiamente utilizzate in diversi campi Questo metodo (hit or miss) e' molto. Convergence behavior crucial for numerical evaluations. Introduction and definitions 1. A simple multiplicative random process. The Hit-or-Miss Method. Monte Carlo estimation. But very often nowadays the function itself is a set of values returned by a simulation (e. The AL strategy adopted in the present study is, in part, motivated by the success reported by Warmuth et al. It covers both the basic ideas of statistical theory, and also some of the more modern and advanced topics of Bayesian statistics such as complete class theorems, the Stein effect, Bayesian model choice, hierarchical and empirical Bayes modeling, Monte Carlo integration including Gibbs sampling, and other MCMC techniques. $\\endgroup$ - Xi'an Apr 12 '13 at 20:14. , \u03b4f (t)=0 if sampling on intervals T 1 and T 2 Importancesampling: choose x N based on I N\u22121 VEGAS is an adaptive integrator that adjusts step functions to parallel integrand. Every used car for sale comes with a free CARFAX Report. 2015 Abstract: Using the result that under the null hypothesis of no misspecification an asymptotically efficient estimator must have zero asymptotic covariance with its difference from a consistent but asymptotically inefficient estimator, specification tests are devised for a number of model specifications in econometrics. \u00a7 Hit-or-miss (acceptance-rejection) Monte Carlo integration of a function F(G). Times Blank Microsoft Equation 3. Justin knew Marcos was into rendering and introduced him to Carl Ludwig, cofounder and one of the original developers of Blue Sky's renderer. From that last expression above, Laplace generalized it to: The end result was a new method whereby \u03c0 could now be calculated by determining the probability P. The area of the ellipse given by + =1 is estimated using three different Monte Carlo procedures. Vector Differential Calculus: Continuity and differentiability of vector functions, Scalar and Vector point function, Gradient, Directional. Underestimating the peak water level from older. What is a Mazo Carlo Simulation? (Part 2) How do we work with Monte Carlo in Python? A great instrument for undertaking Monte Carlo simulations on Python is definitely the numpy selection. Kolmogorov Smirnov test. Leighton Meester wearing Prada Lino Twist Frame Shoulder Bag, Louis Vuitton Cruise 20210 Blue Print Dress, Alexandre Birman Multicolor Python Pumps and Sequin Necklace. Curve fitting and surface fitting web application source code Django (this site) Django (Python 2) Flask CherryPy Bottle Curve fitting and surface fitting GUI application source code tkinter pyQt5 pyGtk wxPython Miscellaneous application source code Animated Confidence Intervals Initial Fitting Parameters Multiple Statistical Distributions Fitter. 33 with not a great deal of confidence. Caches are notorious for their unpredictability. 'HitMiss' uses a methodinspired by the hit-or-miss Monte Carlo integration, which corrects the bias due to ellipsoid approximation. Apart from change of variables, there are several general techniques for variance reduction, sometimes known as Monte Carlo swindles since these methods improve the accuracy and convergence rate of Monte Carlo integration without increasing the number of Monte Carlo samples. In the rst part we give an elementary introduc-. Short segments of pupil data were extracted for each target presentation and then classified as either a hit or miss. In a similar way that a Monte Carlo integration of the sort described in Section2. it checks if it is a row buffer hit or miss, that represents a key computational kernel for the Monte Carlo. it) is a toolbox for integrating 2D NMR spectra by the CAKE (Monte CArlo peaK volume Estimation) algorithm within the Matlab environment (www. Details of the Monte Carlo program GALUGA are given in section 8. MC integration (hit-or-miss method) Monte Carlo method Bu\ufb00on's needle problem From Solitaire to MC Newton-Pepys problem PRNG Hit-or-miss method MC integration results Optimization of MC Crude method Methods comparison Random from PDF CDF CDF discrete CDF continuous Acceptance-rejection Quasi-elastic scattering Tutorial MC MC generators \u03bdN. 2 Monte Carlo (MC) Integration I Monte Carlo (MC) integration is a statistical method based on random sampling. In fact, MA has one of the best Tortas I've ever eaten. It doesn't involve integrals - it sounds like you want the exact solution, no? $\\endgroup$ - Bobson Dugnutt May 11 '17 at 11:03. It compiles and runs on a wide variety of UNIX platforms, Windows and MacOS. Of course, this song could have been 2 minutes instead of 4. Chapter 9 addresses some of the considerations that apply to this kind of study. Monte Carlo methods provide a general, approximate solution to the integration problem. My output for hits is 0 but I can't for the life of me figure out why, to me everything seems fine but clearly isn't. Due to the large number of software packages available for processing nuclear magnetic resonance data, MatCAKE is designed just for implementing the. All of these factors must be accurately characterized to determine silicon performance. The p-value. R code to compare Hit or Miss MC to Classic MC. Python Example of Monte Carlo Estimation & Importance Sampling - kevinzakka\/monte-carlo. See photos, videos, and links of Blair Waldorf. They encompass all of the cases that you cannot hit through plain unit testing. So in both rows the share of hits in throws directly after 3 hits is 0%. They are now suitable for building quite complex Python-based applications on top of Mitsuba, ranging from advanced scripted rendering workflows to full-blown visual material editors. ) 3 BG and Overview (2) The first step can either be very simple or very complicated, based on the particulars features of the problem. probable area was. Cashflow Diary\u2122 | Influenced by Robert Kiyosaki of Rich Dad about Real Estate Investing, Cash Flow and Passive Income. For the Adaptation Modulation model, the estimates of \u03b3 u *U were larger during miss than hit conditions, with no overlap of the confidence intervals ([. Mathematica Subroutine (Monte Carlo for 1 Dimensional Integrals). bubbles() \u00b7 The Bubbles Animation in Hans Roslings Talk. Python has many approaches to making things fast, which is why there are silos. Phil Collins - Two Hearts 759. , hitting a straight flush, neutron escaping through a surface, etc. \u201cThe boy was hit by a stray bullet, he said. Monte Carlo Integral\uc758 \uad6c\ud604 \uc774\uc81c VB \ud504\ub85c\uadf8\ub7a8\uc73c\ub85c \uc801\ubd84 \uac12. We therefore begin this section by recalling some features of numerical integration of a continuous function f over a domain. o a new function MC. First, a random sample of size n is drawn from the uniform distribution over the ellipsoid. Therefore: Use central limit theorem. curl Calculate curl of vector field given by the arrays FX, FY, and FZ or FX, FY respectively. * \u2661 [ M ] While it's great that you guys can help each other out when it comes to picking classes, I would like to avoid having a million separate threads. So this is the code I've put together so far to determine how many points land inside or on the circle (hits). $\\begingroup$ How do you define importance sampling in connection with hit-and-miss Monte Carlo? With no restriction, the minimal variance of an importance sampling estimate is zero. Budajov\u00e1, Graphical visualization of definite integration using Monte Carlo Hit or. Therefore common-block variables that specify methods and constraints to be used have to be set before the PYINIT call and then not be changed afterwards, with few exceptions. As already suggested in the introduction, Monte Carlo methods' popularity and development have very much to do with the advent of computing technology in the 1940s to which von Neumann (picture above) was a pioneer. Sometimes, especially in probabilistic or stochastic codes, the precise behavior of an integration test cannot be determined beforehand. A direct Monte Carlo method for volume estimation of star-shaped or convex domains is presented, and is generalized to a Markov Chain Monte Carlo method for high-dimensional problems. Monte Alban tells you to look wherever the f*** you want as long as you eat a Diablo Torta. Monte Carlo (cont) Strati\ufb01edsampling: divide integration region into sub-volumes and sample according to variance e. Python Example of Monte Carlo Estimation & Importance Sampling - kevinzakka\/monte-carlo. \"\"\" This programme calculates pi with Monte Carlo Given a square and a circle inside it. An important application of Monte Carlo techniques constitutes the Monte Carlo integration which is presented in Sect. For the Adaptation Modulation model, the estimates of \u03b3 u *U were larger during miss than hit conditions, with no overlap of the confidence intervals ([. Here it is written that that the hit-and-miss has a higher variance but they showed (variance of improved Monte-Carlo) - (variance of hit-and-miss method) > 0 $$\\sigma^2_M-\\sigma^2_H>0$$. nl: Since januari 2019, this archive is no longer maintained\/updated. The second step of volume estimation adjusts for this bias via the Hit-or-Miss method of Monte Carlo integration. 8 General on Monte Carlo methods \u00a7Monte Carlois the artof approximating an expectation by the sample mean of a function of simulated random variables. The 90s found Altman back on stable footing with a couple of additional brilliant films (The Player, Short Cuts), and more misses. The convergence is rather slow in this \"hit-or-miss\" method and the variance of this estimator is rather large compared with other integration methods. Hit-or-Miss Method 2. 5 Hit-or-miss integration 40 3. Ask Question and the forum to be very hit or miss :) Improving Performance of an XY Monte Carlo Two questions about. Greenhouse Gas Emissions Modeling: A Tool for Federal Facility Decommissioning, Karen L. The generated numbers are called pseudo-random numbers. curl Calculate curl of vector field given by the arrays FX, FY, and FZ or FX, FY respectively. Metallo-oxide (MO)-based bioinorganic nanocomposites promise unique structures, physicochemical properties, and novel biochemical functionalities, and within the past decade, investment in research on materials such as ZnO, TiO2, SiO2, and GeO2 has significantly increased. the Monte Carlo procedure was a hit '' (it fell below the curve of~ $f(x)$) and $s_i= 0$ if it was a miss. The moments of a uniform distribution \", \" \\\\frac{1}{N} \\\\sum_{i=1}^{N} r_i^k = \\\\int x^k \\\\, P(x) dx +O(1\/\\\\sqrt{N}) \\\\approx \\\\frac. Note from archivercs. with density f(x), then. Monte Carlo Methods. In [5] propo-sedthe calculation of global illumination of the scene represented by a Fredholm integral equation of the se-cond kind. Experimental High Energy Physics Group NIJMEGEN HEN-343 January 18, 2010 Statistical Methods in Data Analysis W. The Hit or Miss method is worse because an extra R. Saturday's Powerball drawing is a staggering $320 million. Traditional extraction. International Journal of Remote Sensing 40:3, 1148-1174. that the \"Hit-Or-Miss Monte Carlo\". Introduction. 3)\uc758 \uc801\ubd84\uac12 \uacc4\uc0b0\uc5d0 \ub300\ud574 \uba3c\uc800 \ud568\uc218 \uac00 \ub97c \ub9cc\uc871\ud55c\ub2e4\uace0 \uac00\uc815\ud558\uc790. Let us calculate the variance of our estimate of \u03c0 as obtained by this method:. ) 3 BG and Overview (2) The first step can either be very simple or very complicated, based on the particulars features of the problem. Integration tests on simulated and experimental peaks with different degree of overlap, showed the CAKE efficacy in. A comprehensive reader-guide exposing 270 secret corrupt judge practices in American court systems. Monte Carlo Simulation R code. the estimator from SMMC has lower variance; SMMC does not require a non-negative integrand (or adjustments). 5 Hit-or-Miss Random Sampling Integration 66 6. MC integration (hit-or-miss method) Monte Carlo method Bu\ufb00on's needle problem From Solitaire to MC Newton-Pepys problem PRNG Hit-or-miss method MC integration results Optimization of MC Crude method Methods comparison Random from PDF CDF CDF discrete CDF continuous Acceptance-rejection Quasi-elastic scattering Tomasz Golan MC generators. The complete book is coming soon. Sample Mean Method Comparison of Hit-and-Miss and Sample Mean Monte Carlo Sample mean Monte Carlo is generally preferred over Hit-and-Miss Monte Carlo because. MC integration (hit-or-miss method) Monte Carlo method Bu\ufb00on\u2019s needle problem From Solitaire to MC Newton-Pepys problem PRNG Hit-or-miss method MC integration results Optimization of MC Crude method Methods comparison Random from PDF CDF CDF discrete CDF continuous Acceptance-rejection Quasi-elastic scattering Tomasz Golan MC generators. ect, NCHRP 25-27, \u201cEvaluation of the Use and Effectiveness of Wildlife Crossings,\u201d was charged to provide guidance in the form of clearly written guidelines for the selection, con\ufb01guration, and location of crossing types, as well as suggestions for the monitoring and evaluation of crossing effectiveness, and the maintenance of crossings. In essence, Monte Carlo is a highly flexible and powerful form of quadrature, or numerical integration, that can be applied to a very wide range of problems, both direct and inverse. \"Bloomer Girl, Oklahoma, Brigadoon, High Button Shoes and Kismet. What if you have too many dice to count this way? For example, say you want to know the odds of getting a combined total of 15 or more on a roll of 8d6. One can see that an immediate application of the Hit or Miss method is for numerical integration. THE CP NOTES 2 Badis Ydri Department of Physics, Faculty of Sciences, Annaba University, Annaba, Algeria. Ex 1: Monte Carlo integration in one dimension Function to integrate and domain In[1]:= [email protected]_D:= [email protected] [email protected]; Hit or Miss flat cover. e it has a larger variance for a given N than any other MC method. This was probably one of the best sessions I attended and I\u2019m glad I went. What is a Mazo Carlo Feinte? (Part 2) How do we work with Monte Carlo in Python? A great instrument for undertaking Monte Carlo simulations throughout Python could be the numpy library. My output for hits is 0 but I can't for the life of me figure out why, to me everything seems fine but clearly isn't. Topographic Anisotropy model in python to be used in CSDMS framework. In fact, MA has one of the best Tortas I've ever eaten. In fact, for the incomplete success of the prior art, the one basic premise is designed using a method similar to Fc, so as alanine scanning, to more attempts (hit-or-miss), or a method using expression strain to produce different glycoforms. =t (\u02dd\u201a\u00dc\u2030 YP \u00b5\u02dcY\u00fc) Monte Carlo Methods and Its Applications 22. Illustration of numerical integration by the hit-or-miss Monte Carlo method. Monte Carlo Simulation has applications in physics, engineering, mathematics, and nance. The volume (in space) of integration is considered large enough for the \u2018kinetic\u2019 energy to be internal; there should be no need to change the integra-tion domain as a function of time. The car starts but wont go out of park and key is stuck in the ignition. Abstract MatCAKE (www. \ub610\ud55c \ub97c \uadf8\ub9bc 11. Tweet Share ShareMonte Carlo methods are a class of techniques for randomly sampling a probability distribution. Hit and miss: Advantages and disadvantages of the Monte-Carlo integration Very singular function. This article discusses the simplest type of probabilistic numerical integration which is sometimes referred to as the hit-or-miss Monte Carlo method. Integrate a function using the Hit-or-Miss Monte Carlo algorithm. The routine is now slower but gives more consistent answers of around 2. Other methods mentioned are molecular dynamics and. \u2022 Although I was tempted to sneak out to the casino this morning, I attended a session entitled \u201cThe Path to a Smooth Go-Live: Assembling the 1,000 Piece Jigsaw Puzzle\u201d. The function where, the hit-or-miss method for the determination of the area of an irregular figure; simulation: Planned learning activities and teaching methods: Lectures of theory and exercises in the computer laboratory. First up is a post which covers how to generate data from different distributions: One method that is very useful for data scientist\/data analysts in order to validate methods or data is Monte Carlo simulation. The Reverse Monte Carlo (RMC) simulation is applied in the study of an aqueous electrolyte LiCl6H2O. Many (probably most) Monte Carlo problems are of the Hit or Miss category, which finds the probability of some event occurring. entrepreneurial skill and required knowledge in computer science and engineering. Mathematical Intelligencer, 40 (1). It's then revealed as a further meaningful rename in that Big Bad Xemnas was influenced into adding not an \"X\", but a \"\u03c7\" (Greek letter for \"chi\", or, \"key\"), in reference to the \"\u03c7-blade\". Hit or Miss Eat. Current as of 12\/17\/2017. The p-value. \u201d The Data Types. Hit or Miss Integration R code. However, patients did not activate these reward-related regions. A straightforward Monte Carlo solution to this problem via the 'hit-or-miss' (or acceptance-rejection) method. Simulation and Monte Carlo integration In this chapter we introduce the concept of generating observations from a speci ed distribution or sample, which is often called Monte Carlo generation. Monte Carlo theory and practice 1147 1. Bere\u017en\u00fd, K. I've been attempting to use Python to create a script that lets me generate large numbers of points for use in the Monte Carlo method to calculate an estimate to Pi. Due to the large number of software packages available for processing nuclear magnetic resonance data, MatCAKE is designed just for implementing the. Tutorial on Monte Carlo Techniques Gabriel A. A Multi-Petascale Highly Efficient Parallel Supercomputer of 100 petaOPS-scale computing, at decreased cost, power and footprint, and that allows for a maximum packaging density of processing nodes from an interconnect point of view. Today most of us focus on utilising random quantity generators, in addition to some traditional Python, to setup two small sample problems. Statistics show that while 95% of Americans say they support organ donation, only 58% are actually signed up as a donor. University of South Carolina Page 6. There are, however,. Everett 9781606932148 1606932144 Sarah Meets a Bully, Bec Furraway. The \"Monte Carlo Method\" is a method of solving problems using statistics. What if you have too many dice to count this way? For example, say you want to know the odds of getting a combined total of 15 or more on a roll of 8d6. Google has many special features to help you find exactly what you're looking for. Attempting to upload and map a cvs file in my Trailhead Playground and upon mapping the fields, the 'Map your field' dialog box in Data Import Wizard won't scroll down in order to click Next or Save or whatever the button is, I cannot even see what it says. Linear congruential generators. 33 and finally 1. daspk Solve the set of differential-algebraic equations daspk_options. It describes how one performs Monte Carlo simulations in condensed matter physics and deals with spin-glasses,.$\\begingroup$This transformation gives better convergence properties for the algorithm described in your code (which is essentially Gaussian quadrature at random locations) but that's not what I understand by the \"hit and miss\" method (generating random$(x,y)$pairs and checking to see if they fall into the region being integrated. Hit or miss Monte Carlo: Buffon\u2019s needle to calculate \u03c0 Method : - Pattern of parallel lines with distance t - Throw randomly n needles of length l (l \u2264 t) - Case a -> a hit - Case b -> a miss If n large, \ud835\udf0b = \u00df \u00e7 2 \u00e1 \u210e\ud835\udc56 Probability of a hit : Let x the distance from the center of the needle to closest line. Simple Monte Carlo integration ! Hit or miss. Lec5_IT\/\u8ba1\u7b97\u673a_\u4e13\u4e1a\u8d44\u6599 50\u4eba\u9605\u8bfb|7\u6b21\u4e0b\u8f7d. monte-carlo-integration monte-carlo Using Monte Carlo Methods to estimate mass potential. To further improve ease of programming, our framework supports speculative parallelism, thanks to the ability of enforcing a given commit order in hardware. We introduce in this nutshell the Monte Carlo integration framework. The customized exterior is finished in Tampico over a black interior equipped with GTS Monte Carlo seats, auxiliary instrumentation, and a leather-trimmed dashboard. MC or MD), and the actual function form need not be. The sample mean method. The above subroutine is all we need to \"do the math. sample mean method Another Monte Carlo integration method is. What is a Cerro Carlo Simulation? (Part 2) How do we refer to Monte Carlo in Python? A great resource for carrying out Monte Carlo simulations for Python is a numpy catalogue. Numerical integration Monte Carlo integration Variance reduction in Monte Carlo integration Lecture 9 - Monte. Warning: All code was play-tested in the QB 4. hitormiss() to demonstrate the 'hit or miss' Monte Carlo integration o a new function MC. The \"Monte Carlo Method\" is a method of solving problems using statistics. The input particles can be generated from various hit-or-miss Monte Carlo processes or they can be read directly from a file. BAKODAH Faculty of Girls, Jeddah, K. A Business Planning Example using Monte Carlo SimulationImagine you are the marketing manager for a firm that is planning to introduce a new product. Find great deals, tips and tricks on Cruise Critic to help plan your cruise. Lec5_IT\/\u8ba1\u7b97\u673a_\u4e13\u4e1a\u8d44\u6599 50\u4eba\u9605\u8bfb|7\u6b21\u4e0b\u8f7d. In mathematics, Monte Carlo integration is a technique for numerical integration using random numbers. 32) satisfies$ \\var (f) \\le H^ 2 \\var (s) $. There are, however,. Monte Carlo (cont) Strati\ufb01edsampling: divide integration region into sub-volumes and sample according to variance e. Suppose that V\u02c6 is the approximation obtained by MC, and Veis the one obtained by using \u2212Z. I was, and still am, a Robert Altman fan. Integrate a function using the Hit-or-Miss Monte Carlo algorithm. hitormiss: Hit or Miss Monte Carlo integration in animation: A Gallery of Animations in Statistics and Utilities to Create Animations. This value, accurate to better than three parts in ten million, would be an impressive example of the power of the statisticaJ sampling method were it, not for the fact. MISSION To educate the students to gain an understanding of the fundamentals of core and allied disciplines so that they can enhance their skills in the areas of Computer Science and Engineering and develop solutions to potential problems. 2017-04-04 MC. 50 KB import math. In general, the sample-mean method is better than the hit-or-miss Monte Carlo method. The reduction in variance is larger when the means of the strata are widely dispersed. Abstract: We present an implementation of the calculation of the production of W + W + plus two jets at hadron colliders, at next-to-leading order (NLO) in QCD, in the POWHEG fram. 16 versus 1028 experiments. net] has joined #ubuntu === Surf24 [[email protected] A simple multiplicative random process. algorithm - How to run MCTS on a highly non-deterministic system? I'm trying to implement a MCTS algorithm for the AI of a small game. Brain to Body ratio data set, tab-delimited. but not including 1. That means it can return any values between 0 and 1, including 0. Then Var( ^M) Var( ^S). As a further enhanced tool for high thoughput NMR analysis, we developed and tested a new integration method for 2D NMR spectra quantification, called CAKE, based upon NMR axial symmetry and Monte Carlo Hit-or-Miss technique. getting the wrong answer [using Hit or Miss] Monte Carlo integration in. You can write a book review and share your experiences. I'm trying to make a randomizer that will use the Monte Carlo Hit or Miss Simulation. ani() to demonstrate the stock prices changing within a time span (inspired by Shen Dai) MINOR CHANGES. Poisson Variates. # ' # ' We compute the proportion of points hitting the area under the curve, and the # ' integral can be estimated by the proportion multiplied by the total area of # ' the rectangle (from xmin to xmax, ymin to ymax). Short segments of pupil data were extracted for each target presentation and then classified as either a hit or miss. THE MONTE CARLO METHOD 6. Monte Carlo estimation. Integrate a function using the Hit-or-Miss Monte Carlo algorithm. Additional notes about suggested reading:. Due to the large number of software packages available for processing nuclear magnetic resonance data, MatCAKE is designed just for implementing the. Many (probably most) Monte Carlo problems are of the Hit or Miss category, which finds the probability of some event occurring. 8 General on Monte Carlo methods \u00a7Monte Carlois the artof approximating an expectation by the sample mean of a function of simulated random variables. STAT 517: Random Numbers and Simulation Hitchcock Hit-or-Miss Method Example 1: Determine. Monte-Carlo integration Markov chains and the Metropolis algorithm Ising model Conclusion Hit-or-Miss Monte Carlo: Calculation of \u02c7 One of the possibilities to calculate the value of \u02c7 is based on the geometrical representation: \u02c7= 4 \u02c7R2 (2R)2 = 4 Area of a circle Area of enclosing square: Choose points randomly inside the square. Warning: All code was play-tested in the QB 4. This is an R package to create and export animations to a variety of formats (HTML\/JS, GIF, Video, PDF), and it also serves as a gallery of statistical animations. What is a Mont\u00f3n Carlo Simulation? (Part 2) How do we refer to Monte Carlo in Python? A great resource for accomplishing Monte Carlo simulations around Python could be the numpy local library. In this nutshell, we consider an alternative integration technique based on random variates and statistical estimation: Monte Carlo integration. Ltd Hyundai Motor Shanghai Infohold Corp. Find temporal evolution. the Monte Carlo procedure was a hit '' (it fell below the curve of~$ f(x) $) and$ s_i= 0 \\$ if it was a miss. Outline hit or miss method points below the curve total number of points. Mendel gown, Ofira jewelry and Judith Leiber clutch at an N. moduleauthor:: G. This chapter explains how to set up and execute simple Monte Carlo simulations, using data generating processes to represent random inputs. Some Monte Carlo swindles are: importance sampling. It is difficult or even impossible to predict if a memory access will result in a definite cache hit or miss. Comparison of Hit-and-Miss and Sample Mean Monte Carlo 25 2. , Gorunescu, M. ani() to demonstrate the stock prices changing within a time span (inspired by Shen Dai) MINOR CHANGES. Following this introduction is a section on the Monte Carlo experiment, part of the physical chemistry lab at UNL, which computes the population distribution in the rotational energy levels of HCl and DCl. RANK NAME STEAMID POINTS KDR HS ACC PROFILE; 1: Kata: STEAM_1:0:1499491: 2234: 3. Dejnition A Monte Carlo technique is any technique making use of random numbers to solve a problem. Going a bit deeper down the rabbit hole, if we have n samples, the distribution of the max is given by F(x)^n. Every used car for sale comes with a free CARFAX Report. It is difficult or even impossible to predict if a memory access will result in a definite cache hit or miss. The \ufb01rst Monte Carlo method we will be discussing is the Hit-Or-Miss method. The P2P approach has several challenges that come from the fact that there is no intermediary hub \u201cbetween \u201d the enterprises that acts as a switchboard providing central integration functionality. The volume fraction is obtained via Monte Carlo Hit-or-Miss technique, which proved to be the most efficient. In the name of GOD. Interestingly. Andrew Dotson 14,894 views. Statistics and Lean Six Sigma: A Perfect Combination ASA Colorado\/Wyoming Spring Meeting April 15, 2011 Presented By Scott Leek Senior Member, ASQ Managing Director & MBB Sigma Consulting Resources. \uba3c\uc800 \uc774 \ud504\ub85c\uadf8\ub7a8\uc744 run\ud558\ub294 \uacfc\uc815\uc758 form\uc758 \ubaa8\uc591\uc744 \ubcf4\uba74 \ub2e4\uc74c\uacfc \uac19\ub2e4. Topographic Anisotropy model in python to be used in CSDMS framework. In [5] propo-sedthe calculation of global illumination of the scene represented by a Fredholm integral equation of the se-cond kind. We propose a corrector adaptive Monte Carlo integration algorithm that prevents the adaptive algorithm from moving to a. Lec5_IT\/\u8ba1\u7b97\u673a_\u4e13\u4e1a\u8d44\u6599 50\u4eba\u9605\u8bfb|7\u6b21\u4e0b\u8f7d. It describes how one performs Monte Carlo simulations in condensed matter physics and deals with spin-glasses, percolating networks and the random field Ising model. made available free computer software that allows the assessor to calculate important supplementary values.","date":"2019-12-07 15:03:58","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\": 1, \"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\": 0.5990256667137146, \"perplexity\": 2188.083082019309}, \"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\/1575540499439.6\/warc\/CC-MAIN-20191207132817-20191207160817-00105.warc.gz\"}"} | null | null |
Permits & Enforcement
Pay Stations
Corporate Campuses
T2U Learning Center
Customer Advisory Board
City of Aspen
Aspen Improves Efficiency and Customer Satisfaction with Virtual Permits & LPR
Aspen, Colorado, a small resort town nestled in the Rockies, is known for its big slopes, big houses, and big name guests. Founded in the late 1800s as a mining camp during the Colorado Silver Boom, the town developed into a tourist spot in the mid-20th century when nearby Aspen Mountain was developed into a ski resort. Since then Aspen has continued to grow, and today the town is a popular destination year-round thanks to its great hiking, biking, rafting, dining, and arts.
For the City of Aspen Parking Department, this means finding a way to balance the parking needs of its full-time residents, employees of local businesses, and visitors from around the world. While Aspen has a population of just 7,000, during the summer it can see up to 50,000 visitors in a day. In addition, "Because we're a small town, probably 60 to 70 percent of our employees are driving from downvalley, which is anywhere from 30 to 60 miles," explained Mitch Osur, Director of Parking and Downtown Services. "We have a huge amount of people coming into town."
One of the biggest challenges for the City of Aspen is that all of its parkers – residents, employees, and visitors – are competing for limited resources. Aspen has less than 1,000 parking spaces in its downtown core and about 3,000 in its residential zone. Ideally, those spaces in the downtown core would be available to tourists and locals coming into town to spend money. However, Prior to Osur joining the Parking Department in 2014, the majority of the people parking in the downtown core were employees of businesses. Osur made it a goal to free up more spaces for the patrons of these downtown businesses.
Another challenge is ensuring that the full-time residents of Aspen are satisfied with their parking situation. "We have a very aggressive group of locals that have strong opinions about parking," Osur said, "so consistency and making sure were doing the same thing pretty much all the time is very important."
Aspen uses T2 Flex software to manage its residential permit program. Residents can purchase up to three virtual parking permits – where their license plate is their credential – from any smartphone, tablet, or computer via T2 FlexPort, an online self-service platform. Going virtual "is a lot better both for the environment and also for enforcement," Osur stated.
For enforcement, Aspen has been utilizing license plate recognition for almost 15 years. They have four enforcement vehicles equipped with Genetec LPR cameras, which patrol the vast residential areas, and T2's Mobile Enforcement App, which enables officers to perform permit lookups, receive LPR detected violations, and issue citations – all while integrating seamlessly with T2 Flex in real time.
While Aspen is not "overly aggressive" with collecting on parking tickets, they have worked with T2 to connect with the Colorado DMV to get information faster and send out weekly letters to violators. Aspen also uses T2's Retrieval of Vehicle Registration (RoVR) solution to track down contact information for individuals from other states.
Additionally, Aspen was one of the first organizations to work with T2 on allowing people to contest citations online. "We use T2 FlexPort to let people contest tickets. Appeals come right into our office, we get the information we need, and then we have worked out a really good system so that we can turn these around quite a bit faster and get information back to people on whether their citation has been dismissed or not."
"Certainly the efficiency of writing the tickets and also the efficiency of the way the permit system works has been a huge success for us," Osur said.
T2's virtual permitting capabilities have saved Aspen time in the office and in the field, while its LPR enforcement integrations have made enforcing Aspen's large residential zones a breeze. "As my officers go down the street, it's much easier to cover a lot of ground because we know who has paid and who has not paid," Osur explained.
Aspen's enforcement officers are also much more informed thanks to the Mobile Enforcement App. "It's nice when my officers look up a license plate, to be able to look at the history right then and there in the street," Osur stated. This improved enforcement, combined with changes to pricing and other strategic decisions, helped Aspen achieve Osur's goal of freeing up parking spaces for patrons of downtown businesses. "Seven years ago probably 150-200 of the people parking in the downtown core were employees of businesses, and today we're probably less than 50 on an average day."
"Work with T2 to go to virtual permits, because if you can have all virtual permits and LPR, you can be really efficient out there in the field."
Mitch Osur, Director of Parking and Downtown Services
The City of Aspen has also been able to achieve the consistency that its residents desire. For example, its policy is that every person's first ticket is a warning, and the Mobile Enforcement App makes it easy for officers to identify whether somebody is a first-time offender. "We get nothing but compliments from people that were like, 'I went back to my car and saw I had a ticket on it, and I was all upset, and then I saw it was a warning at zero value. Thank you!'" Osur said.
Additionally, Aspen's citation collection and appeals processes have become much more efficient and stress-free for both staff and patrons. T2's RoVR solution has been "very successful for us to collect money without having to be overly aggressive and/or hire a collection agency," Osur explained, while automated processes in T2 Flex and online appeals in T2 FlexPort have helped Aspen reduce the amount of time to settle an appeal from four to five weeks to less than four days.
Ultimately, according to Osur, "Our goal is to go 100 percent virtual and try to do as much online as we can, versus having to come into the office." The City of Aspen is about to reach that goal with T2's new Residential Permits solution, which they are currently implementing. In addition to providing new backend tools and an improved interface for residents to purchase permits, the solution will enable Aspen to change their handful of lingering paper permits to virtual.
Osur's top recommendation was to "Definitely go to license plate recognition. It produces efficiency, it makes the officers happy, and it allows you to cover a lot more ground, which is fantastic. Tied in with that, work with T2 to go to virtual permits, because if you can have all virtual permits and LPR, you can be really efficient out there in the field, and I think that's super important."
"Technology is changing fast, so be innovative out there and try to do things that really can work for you," Osur added. "A company like T2 that has pretty much everything covered from A to Z will help you get there, because certainly the more you can do with one vendor, the easier it is for the people that work for me and the municipality itself."
Header photo by Vlado Sestan on Unsplash
← Webinar: Discover the Options with Flex Managed Services Employee Spotlight: Andrew Kenakin →
© 2020 T2 Systems. All Rights Reserved. Privacy Policy | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 9,700 |
static std::atomic<qint64> counter {0}; // Box counter
#if defined(Q_OS_UNIX) || !defined(QT_NO_DEBUG)
void writeFailure(const char *data, size_t size)
{
std::cout << QByteArray(data, size).data() << std::endl;
}
#endif
struct Box
{
int a {0};
int b {0};
Box() { counter++; }
Box(const Box &box)
: a(box.a), b(box.b) { counter++; }
~Box() { counter--; }
Box &operator=(const Box &box) {
a = box.a;
b = box.b;
return *this;
}
};
TStack<Box> stackBox;
class PopThread : public QThread
{
Q_OBJECT
public:
PopThread() { }
protected:
void run() {
for (;;) {
Box box;
if (stackBox.pop(box)) {
Q_ASSERT(box.a + box.b == 1000);
}
//Tf::msleep(1);
std::this_thread::yield();
}
}
};
class PushThread : public QThread
{
Q_OBJECT
public:
PushThread() {}
protected:
void run() {
for (;;) {
Box box;
if (stackBox.top(box)) {
Q_ASSERT(box.a + box.b == 1000);
box.a++;
box.b--;
} else {
box.a = 1000;
box.b = 0;
}
if (stackBox.count() < 500) {
stackBox.push(box);
}
//Tf::msleep(1);
std::this_thread::yield();
}
}
};
class StackStarter : public QObject
{
Q_OBJECT
public slots:
void startPopThread();
void startPushThread();
};
class TestStack : public QObject
{
Q_OBJECT
private slots:
void initTestCase();
void cleanupTestCase();
void push_pop();
};
void TestStack::initTestCase()
{
#if defined(Q_OS_UNIX) || !defined(QT_NO_DEBUG)
// Setup signal handlers for SIGSEGV, SIGILL, SIGFPE, SIGABRT and SIGBUS
google::InstallFailureWriter(writeFailure);
google::InstallFailureSignalHandler();
#endif
}
void TestStack::cleanupTestCase()
{
_exit(0);
}
void TestStack::push_pop()
{
#ifdef Q_OS_UNIX
const int NUM = 128;
#else
const int NUM = std::max((int)std::thread::hardware_concurrency(), 2);
#endif
// Starts threads
StackStarter starter;
for (int i = 0; i < NUM; i++) {
starter.startPopThread();
starter.startPushThread();
}
QElapsedTimer timer;
timer.start();
QEventLoop eventLoop;
while (timer.elapsed() < 10000) {
eventLoop.processEvents();
}
std::cout << "counter=" << counter.load() << std::endl;
std::cout << "stack count=" << stackBox.count() << std::endl;
}
void StackStarter::startPopThread()
{
auto *threada = new PopThread();
connect(threada, SIGNAL(finished()), this, SLOT(startPopThread()));
connect(threada, SIGNAL(finished()), threada, SLOT(deleteLater()));
threada->start();
}
void StackStarter::startPushThread()
{
auto *threadb = new PushThread();
connect(threadb, SIGNAL(finished()), this, SLOT(startPushThread()));
connect(threadb, SIGNAL(finished()), threadb, SLOT(deleteLater()));
threadb->start();
}
TF_TEST_SQLLESS_MAIN(TestStack)
#include "main.moc"
| {
"redpajama_set_name": "RedPajamaGithub"
} | 2,335 |
\section{Introduction} \label{sec:intro}
It has long been established that local environment and galaxy properties are linked, with overdense regions in the local Universe playing host to an overabundance of early type galaxies undergoing passive evolution \citep[e.g.][]{dre80}. Substantial efforts have now been made to trace this relation back in time to the initial conditions of proto-clusters \citep[see][for a review]{ove16}, whose resident galaxies are predicted to contribute substantially to the cosmic star formation rate density and mass growth \citep{chi17}. Bridging the gap between these early structures and present-day clusters is a transition epoch at $z\sim1-2$ wherein massive (log $M/\ensuremath{{M_{\odot}}}\gtrsim13.8$) clusters are found to host populations of (dust obscured) star forming galaxies (SFGs) with field-like star formation activity \citep[i.e][hereafter A14, A16]{coo06, hil10, tra10, fas11, fas14, hay11, tad11, bro13, san14, san15, ma15, alb14, alb16} and a corresponding decrease in quenched populations and quenching efficiency \citep{nan17, cha19}.
Despite this progress, it is still unclear which of several mechanisms play a substantial role in quenching galaxies from the proto-cluster to cluster regimes. Strangulation $-$ the prevention of fresh gas accretion due to the hot Intracluster Medium \citep[ICM; ][]{lar80} $-$ likely plays some role but its few Gyr timescales are not consistent with evidence supporting rapid quenching \citep[i.e.][A14, A16]{man10, vdb13, bro13, wet13}. Ram pressure stripping \citep[RPS;][]{gun72} may be able to remove gas on shorter timescales \citep[$\sim100-200$ Myr;][]{aba99, mar03, roe06, roe07,kro08, ste16}. RPS may also play a role out to large radii ($\gtrsim2-3R_{\rm vir}$) where infalling galaxies encounter virial accretion shocks associated with the ICM \citep[i.e.][]{sar98, bir03, dek06, zin18}. However, the effect of RPS on star formation (whether enhancement or quenching occurs) and gas reservoirs can depend on an individual galaxy's properties and orbit \citep{bek14, ton19}, making its overall effectiveness unclear. Overdense environments may also host increased merger \citep{deg18, wat19} and AGN \citep[][A16]{mar13} activity, enabling scenarios in which quenching is propelled by increased starburst activity or feedback \citep{bro13}.
Key observables in distinguishing between these scenarios involve the cold molecular gas which fuels star formation: the (molecular) gas mass, gas depletion timescale, and gas fraction. In the local Universe, gas properties have been observed in the ever accumulating examples of RPS events, from truncated or disturbed gaseous disks \citep[i.e.][]{vol08, zab19} to spectacular one-sided tails \citep[i.e.][]{sun06, sun10, ebe14}, which can host significant star formation (SF) \citep{fum14} and molecular gas \citep{jac14, jac19}. However it again remains unclear whether this cold gas is stripped or formed $in$ $situ$ in the tail and how this effects the star formation occurring in the host galaxy.
At higher redshifts, where different mechanisms may operate, observations of gas content have been limited to small samples of cluster \citep[][Williams, ApJ, submitted]{nob17, nob19, rud17, sta17, hay17, hay18, coo18, spe21a, spe21b} and proto-cluster galaxies \citep{wan16, wan18, ume17, zav19, tad19, lon20, cha21, hil21} detected in CO or dust continuum emission. These studies have largely found enhanced or field-like gas fractions and field-like gas depletion timescales, even when obtaining deep observations \citep[][Williams, et al., ApJ, submitted; but see \citet{coo18}]{nob19}, with only hints of gas loss at the highest stellar masses (log $M_{\star}/\ensuremath{{M_{\odot}}}>11$). On the other hand, the use of stacking to probe below detection limits suggests the existence of a population of cluster galaxies with low gas fractions and short depletion timescales \citep{bet19, zav19}. Theoretically, evidence from cosmological simulations is mounting that gas loss should start at large radii, with significant depletion of the gas reservoir by or at first passage of the cluster center \citep{oma16, zin18, art19, mos21, oma21}.
In this work, we quantify the average molecular gas properties of cluster galaxies through stacking their dust continuum emission as observed by the Atacama Large Millimeter Array (ALMA). Dust continuum emission, like CO line emission, is a robust proxy for molecular gas mass \citep[see][for a review]{tac20}. Our sample consists of 126 cluster galaxies selected, based on far-infrared (FIR) luminosity, from eleven stellar mass-selected galaxy clusters (log $M_{200}/M_{\odot}\sim14$) at $z=1-1.75$. In general, the massive cluster galaxies in these clusters supply a total SFR budget comparable to massive field halos when controlled for halo mass, though there is significant variation in the total SF from cluster to cluster \citepalias{alb16}. We measure the average gas properties of a luminous subset of these massive cluster galaxies, comparing their gas reservoirs to coeval field populations as well as to the (proto-)cluster samples in the literature at comparable redshifts.
This paper is structured as follows: in Section~\ref{sec:data}, we describe our sample selection and observations. In Section~\ref{sec:prop_stack}, we present a breakdown of the properties of our sample, our stacking techniques, and the methodology used to measure gas properties from our ALMA data. Section~\ref{sec:analysis} contains our analysis and results: the average Herschel+ALMA infrared SEDs of our cluster galaxies, a cluster-centric radial analysis of gas properties, and a comparison to the gas properties of field galaxies and (proto-)cluster galaxies at comparable redshifts. In Section~\ref{sec:disc}, we discuss our results, including putting our findings in the context of recent cosmological simulations looking at gas loss in infalling cluster galaxies. Section~\ref{sec:conc} contains our conclusions. Throughout this work, we adopt concordance cosmology: ($\Omega_{\Lambda}$ , $\Omega_{\rm M}$ , $h$)=(0.7, 0.3, 0.7), a \citet{kro01} IMF, and the \citet{spe14} Main Sequence (MS). ``log" refers to log$_{10}$.
\section{Data}\label{sec:data}
\subsection{Sample}\label{sec:sample}
Our sample consists of 126 cluster galaxies selected from eleven massive (log $M_{200}/\ensuremath{{M_{\odot}}}\sim14$) galaxy clusters at $z=1-1.75$ with uniquely deep {\it Herschel}/PACS imaging at 100 and 160 $\mu$m. The clusters are drawn from the IRAC Shallow and IRAC Distant Cluster Surveys \cite[ISCS, IDCS;][]{eis08, sta12}, identified as near-infrared (stellar mass) overdensities in (RA, Dec, photometric redshift) space and confirmed via targeted spectroscopic follow-up \citep{sta05, sta12, els06, bro06, bro11, bro13, eis08, zei12, zei13}. We note that these eleven clusters were chosen for follow-up based on being significant overdensities and {\it not} on the basis of their star formation activity, which shows considerable variation from cluster-to-cluster. See \citetalias{alb16} and references therein for a detailed description of our cluster sample and {\it Herschel}/PACS imaging.
Cluster galaxy membership was established using spectroscopic redshifts (spec-$z$s) where available and photometric redshifts (photo-$z$s) otherwise, starting from a Spitzer/IRAC 4.5$\mu$m catalog and using full photo-$z$ probability distribution functions to identify a complete catalog of massive (log $M_{\star}/\ensuremath{{M_{\odot}}} \geq 10.1$) cluster galaxies. Herschel/PACS photometry was then extracted using the spectroscopic and IRAC catalogs as positional priors. The infrared luminosity $L_{\rm IR}$[8-1000$\mu$m] for each member was calculated by scaling one of two templates from \citet{kir15} to the Herschel/PACS 100$\mu$m flux density; these templates are known to describe the average optical to FIR properties of massive cluster galaxies in the ISCS, IDCS \citepalias{alb16}. For purely star forming galaxies, the template {\tt MIR0.0} was adopted. For members with AGN activity as determined via SED fitting \citepalias[see \S\ref{sec:ancillary} and][]{alb16}, the template {\tt MIR0.5} and a correction factor were used to account for the contribution from AGN at shorter wavelengths. The sample in this work was then selected as IR-bright ($L_{\rm IR}\geq5\e{11}\ensuremath{{L_{\odot}}}$) members within 2 Mpc of the cluster centers \citep[$\sim2$x the virial radius,][]{bro07}. The cluster centers are known to within $\sim15\arcsec$ ($\sim130\,$kpc), set by the pixel scale of the cluster detection maps \citep{gon19} and confirmed through comparisons with X-ray centroids (Garcia et al.~in preparation).
\subsection{ALMA Data}\label{sec:almadata}
Dust continuum emission was observed in Band 6 (centered at 1.3mm or 231 GHz) for our sample of 126 cluster galaxies over 125 pointings in ALMA Cycle 3 proposal 2015.1.00813.S (PI: Alberts). The target positions were obtained from the IRAC counterparts, used as priors for Herschel source extraction. The sample was split into 6 Science Goals (for ease of scheduling) with four target rms sensitivities calculated based on two \citet{kir15} templates (discussed in the previous section), listed in Table~\ref{tbl:observations}. The maximum 7.5 GHz bandwidth was requested over four 1875 MHz spectral windows. Observations were taken in June 2016 with configurations C40-2 and C40-4. Data reduction was performed with the Common Astronomy Software Application ver. 4.7.2 \citep[CASA;][]{mcm07} with preliminary reductions revealing that all sources are non-detections and that there are no serendipitous $>3\sigma$ detections in the maps. As such, the final continuum maps were generated using the CASA task {\tt TCLEAN} with natural weighting, a $0.2\arcsec$ pixel size, and no deconvolution (cleaning). {\it u,v} tapering was applied to produce maps with similar beamsizes (Table~\ref{tbl:observations}) as needed for stacking; a 260$k\lambda$ x 220$k\lambda$ taper was found to maintain sensitivity while coming close to $1\arcsec$ resolution, comparable to resolution of the IRAC priors. The average rms sensitivities achieved are within $\sim10\%$ of requested; they and the beamsizes of the sample subsets are listed in Table~\ref{tbl:observations}. We discuss why our targets are undetected in Section~\ref{sec:ir_sed}.
\begin{table*}
\centering
\caption{Summary of ALMA Observations}
\begin{tabular}{lcccc}
\hline
\hline
Sample & Number of & Requested rms & Average rms & Beamsize \\
& Pointings & [$\mu$Jy beam$^{-1}$] & [$\mu$Jy beam$^{-1}$] & [arcsec] \\
\hline
\hline
Science Goal 1a & 30 & 140 & 141.7 & 0.92 x 0.72 \\
Science Goal 1b & 29 & 140 & 136.0 & 0.96 x 0.75 \\
Science Goal 2a & 21 & 90 & 91.5 & 0.93 x 0.73 \\
Science Goal 2b & 22 & 90 & 79.8 & 1.0 x 0.67 \\
Science Goal 3 & 15 & 70 & 78.0 & 0.97 x 0.77 \\
Science Goal 4 & 8 & 50 & 56.5 & 0.92 x 0.75 \\
\hline
\label{tbl:observations}
\end{tabular}
\end{table*}
\subsection{Ancillary Data}\label{sec:ancillary}
For each ALMA target, UV through near-infrared photometry is available as described in \citet{chu14}, which was used to derive photo-$z$s and identify AGN as described in \citetalias{alb16}.
For ten of the clusters in this work, deep {\it Spitzer}/MIPS 24$\mu$m imaging was obtained to $3\sigma$ depths of $156\,\mu$Jy to $36\,\mu$Jy spanning $z=1$ to $z=1.75$, providing a uniform depth in $L_{\rm IR}$ of 3\e{11}\,\ensuremath{{L_{\odot}}} \citep{bro13}. Deep Herschel/PACS imaging at 100 and 160$\mu$m (average rms sensitivity of $1.2\,\mu$Jy at 100$\mu$m) was obtained for all 11 clusters, as described in \citetalias{alb16}. Herschel/SPIRE imaging at 250, 350, and 500$\,\mu$m is available from the Herschel Multi-tiered Extragalactic Survey \citep[HerMES;][]{oli12} for 10/11 clusters; this work uses the SPIRE maps as reduced and described in \citetalias{alb14}.
\section{ALMA Sample Properties and Stacking}\label{sec:prop_stack}
Histograms of the following properties of our ALMA sample are shown in Figure~\ref{fig:properties}: redshift (spec-$z$ or photo-$z$), cluster-centric (projected) radius, stellar mass, obscured star formation rate SFR$_{\rm IR}$, and distance from the Main Sequence. The photo-$z$ uncertainties are $\sigma/(1+z) \approx 0.05$ for sources dominated by (host) galaxy emission and $\sigma/(1+z) \approx 0.2$ when dominated by AGN emission \citepalias[see][]{alb16}, based on comparisons with spectroscopic redshifts and pair statistics \citep{qua10, hua13}. Stellar masses were derived using optical - MIR Bayesian SED fitting \citep{mou13} as described in \citet{bro13}, with uncertainties of 0.3 dex including systematic error. SFR$_{\rm IR}$ was obtained from the total $L_{\rm IR}$ (as described in \citetalias{alb16} and in Section~\ref{sec:data}) determined using Herschel/PACS photometry and assuming the $L_{\rm IR}$-SFR conversion from \citet{mur11b}, with corrections for AGN emission where appropriate \citep{kir15}. The typical (median) measurement uncertainty on SFR$_{\rm IR}$ is $\sim0.15$ dex \citepalias[][]{alb16}. We note that for massive galaxies at these redshifts, the obscured component contributes the bulk of the star formation rate \citep{whi17}. The Main Sequence star formation rate (SFR$_{\rm MS}$) for each of our targets was determined from their redshifts and stellar masses using the publicly available Python package {\tt a3cosmos} \citep{liu19a} and the \citet{spe14} MS relation. \citet{spe14} assumes a \citet{cha03} IMF, which has a similar normalization to the \citet{kro01} IMF assumed in our $L_{\rm IR}$ to SFR conversion \citep{mur11b}. From this we derived the distance of our sources from the MS as $\Delta \mathrm{MS}=\mathrm{log}(\mathrm{SFR}_{\rm IR}/\mathrm{SFR}_{\rm MS})$.
To summarize (Figure~\ref{fig:properties}, (a)-(e)), our sample is largely massive (log $M_{\star}/\ensuremath{{M_{\odot}}}\sim11$) cluster galaxies at $z\sim1.4$ with SFRs on the high end of the MS at this redshift with some starbursting activity.
We cover up to 2x the virial radius \citep[$R_{\rm vir}\sim1\,$Mpc,][]{bro07} and sit above the MS with a mean (median) $\Delta$MS of 0.52 (0.36), with 23 members falling into the category of starbursts \citep[$\Delta$MS$\,\geq0.6$;][]{rod11, rod14}. Figure~\ref{fig:properties} (f) shows that the starbursts are not preferentially located at any particular redshift or cluster-centric radius, though they tend to be in the lower half of our stellar mass distribution. This is due to our flux-limited selection, which results in a bias with stellar mass; at lower masses (log $M_{\star}/\ensuremath{{M_{\odot}}}\lesssim11$), we only probe sources on the upper end and above the MS with log $M_{\star}/\ensuremath{{M_{\odot}}}\lesssim10.3$ sources making up the majority of our starbursts.
\begin{figure*}
\centering
\includegraphics[width=\textwidth]{alma_sample_properties.pdf}
\caption{Properties of the ALMA cluster member sample: (a) redshift, (b) cluster-centric radius, (c) stellar mass, (d) obscured SFR, (e) $\Delta$MS. (f) shows the distance from the main sequence as a function of cluster-centric radius in three redshift bins, with the lower and upper half of our stellar mass distribution indicated by smaller and larger symbols. Median and mean values are indicated as dashed and dotted lines. The purple shaded region denotes starbursting activity.}
\label{fig:properties}
\end{figure*}
\subsection{Stacking and Stacked Photometry}
\subsubsection{ALMA}\label{sec:almastacking}
Stacking is performed by creating 31x31 pixel cutouts of the primary-beam corrected maps centered on the target coordinates. Except for two cluster galaxies, all of our targets are at the center of the ALMA pointings. Cutouts are then binned into subsets by different properties and the pixel-wise variance-weighted mean is calculated. The variance-weighted mean provides the most robust average flux density in the case of non-uniform noise properties when combining cutouts that reach different depths (Table~\ref{tbl:observations}).
Figure~\ref{fig:stacks} shows the stack of the full sample, which is marginally resolved. This is consistent with the median size of the (global) star forming regions observed in MS galaxies at $z\sim2$ with deep ALMA and VLA radio data \citep[$r_{\rm SF}\sim2.1\pm0.9$ kpc;][]{ruj16}. Accordingly, we use integrated aperture photometry to determine the stacked flux following the procedure used in \citet{bet19}: starting at the average (circularized) beamsize, the S/N is measured in increasingly large apertures in order to identify the aperture which maximizes the S/N by enclosing the most signal while minimizing the contribution from noise (determined off source as the standard deviation of 100 apertures). From this procedure, we find that the optimal aperture is $0.8\arcsec$ in radius\footnote{We confirm that the aperture photometry returns a larger flux density than the peak pixel, which provides a measure of the source flux in images calibrated in Jy beam$^{-1}$ in the case of a point source.}. We use this aperture across all of our stacks.
Photometric uncertainties are derived using the bootstrap technique \citep[i.e.][]{bet12} as described in \citetalias{alb14}, which encompasses all sources of noise in the images as well as scatter in the ALMA properties of the stacked population. Stacking and aperture photometry are performed on randomly selected cutouts with replacement for $N=500$ realizations; the stacked uncertainty is then measured from the width of the resulting histogram.
\subsubsection{SPIRE}\label{sec:spirestacking}
Stacking in the SPIRE 250, 350, and 500$\,\mu$m bands was performed as described above, on cutouts centered on the target galaxies to determine the variance-weighted mean. Given the large SPIRE beamsizes (18, 25, and 36\arcsec at 250, 350, and 500$\,\mu$m) and the high source density of the clusters, these stacks will suffer from flux boosting \citep{vie13}. While it is possible to correct for this flux boosting statistically in large samples \citep[A14, ][]{alb21}, we utilize simulations \citep[see Appendix~C in ][]{alb21} to determine that the correction has a large uncertainty in small samples like those being stacked in this work, particularly at the longer wavelengths. As such, we adopt the uncorrected SPIRE stacks, which are formally upper limits. We note that in a previous study flux boosting at 250$\,\mu$m was found to sharply decrease with cluster-centric radius and is minimal at $r>0.5\,$Mpc for this cluster sample \citepalias{alb14}. Stack uncertainties are determined via bootstrapping.
\begin{figure*}
\centering
\includegraphics[width=\linewidth, trim=10 270 0 200, clip]{stack_plot.pdf}
\caption{Weighted-mean stacks at (observed) 1.3mm for our full cluster sample (far left) and split into three cluster-centric radial bins. The black circle shows the aperture used to measure the stacked photometry (see Section~\ref{sec:almastacking}). The average beamsize is shown in the lower left corner.}
\label{fig:stacks}
\end{figure*}
\subsection{Molecular Gas Measurements}\label{sec:gasmeasurements}
Over the redshift range of our clusters ($z\sim1-1.75$), our observed ALMA 1.3mm photometry probes rest-frame 650-470$\mu$m on the optically thin Rayleigh Jeans tail of the dust emission. This provides a robust measure of the total dust mass; from this, given assumptions on the dust opacity and dust-to-gas abundance ratio, the total molecular gas mass can be derived \citep{eal12, sco14, pri18}. This method agrees well with gas measurements from CO emission lines, see \citet{tac20} for a review.
All dust mass and gas mass measurements used in this work are calculated following \citet{sco16}, hereafter S16. We briefly summarize the method here: in the optically thin regime, the observed flux density, $S_{\nu}$, is a function of the dust opacity per unit mass, $\kappa_{\nu}^{dust}$, the mass-weighted dust temperature T$^{\rm bulk}_{\rm{dust}}$, and the dust mass M$_{\rm{dust}}$ though
\begin{equation}
\label{eqn1}
S_{\nu} = \kappa_{\nu}^{\mathrm{dust}} \mathrm{T^{bulk}_{dust}} (1+z) \nu^2 \frac{\mathrm{M_{dust}}}{4\pi d_L^2}
\end{equation}
where $d_L$ is the luminosity distance. This can be related to the total molecular gas mass, $M_{\rm mol}$, by defining the dust opacity per unit ISM mass via $\kappa^{\rm{mol}}_{\nu} = \kappa^{\rm{dust}}_{\nu} \times \rm{M_{dust}/M_{mol}}$. The gas mass can then be quantified given photometry on the RJ tail of the dust emission, the dust temperature, and $\kappa^{\rm{mol}}_{\nu}$.
$\kappa^{\rm{mol}}_{\nu}$ was empirically calibrated at 850$\mu$m based on local SFGs and Ultra-Luminous Infrared Galaxies (ULIRGs) as well as high-$z$ Sub-millimeter Galaxies \citep[SMGs;][]{sco14, sco16} and $z\sim2$ SFGs \citep{kaa19}, from which was found a single representative constant:
\begin{equation}
\label{eqn2}
\alpha_{850} \propto \kappa^{\mathrm{mol}}_{\nu_{850}} \mathrm{T^{bulk}_{dust}} = \frac{L_{\nu_{850}}}{\mathrm{M_{mol}}} = 6.7\e{19} \mathrm{ergs\,} \mathrm{s}^{-1} \mathrm{Hz}^{-1} \ensuremath{{M_{\odot}}}^{-1}.
\end{equation}
assuming a bulk dust temperature of $T^{\rm bulk}_{\rm dust}=25$ K \citep{kir15,sco16}. We apply this calibration to derive $M_{\rm mol}$, accounting for redshift, our observed frequency, and departures from Rayleigh Jeans as in Eqn. 16 of \citetalias{sco16} (please note the typo as corrected in their erratum).
\section{Analysis and Results}\label{sec:analysis}
\subsection{Infrared SEDs}\label{sec:ir_sed}
In \citetalias{alb16}, we showed that up to the dust peak ($\sim$80-100$\,\mu$m), the infrared SED of massive, star forming cluster galaxies at $z\sim1-2$ is well represented by SFG templates derived from field galaxies in the same epoch. Here we extend that analysis to the submm by adding to the SED our stacked ALMA fluxes in three bins of cluster-centric radius: $0<r/\mathrm{Mpc}<0.7$, $0.7<r/\mathrm{Mpc}<1.4$ and $1.4<r/\mathrm{Mpc}<2.1$.
These radial bins were chosen in part to ensure good detections in the stacks. We can interpret them physically as roughly representing different environments within the cluster ecosystem via the following: simulated mass profiles have found a sharp drop in cluster dark matter profiles, termed the splashback radius \citep[$R_{\rm sp}$;][]{die14}, which denotes the boundary between the virialized and infalling regions. $N$-body simulations and observations (via weak lensing) both agree that, for rapidly accreting halos such as we expect at high redshifts, the splashback radius roughly coincides with the virial radius \citep[i.e.][and references therein]{mor15, die17b, zur19, shi19, shi21}. Though we do not have the data to measure the splashback radii individually in our clusters, this is qualitatively confirmed by stacked stellar mass profiles of the full ISCS cluster sample, which showed a sharp drop off at the virial radius \citep{alb21}. As such, our data out to $2R_{\rm vir}$ is likely allowing us to separately probe the virialized and infalling populations, as well as the transition region (at $\sim R_{\rm vir}$) between the two.
The three radial stacks can be seen in Figure~\ref{fig:stacks}; all radial stacks are detected at S/N$\,\sim4$ (Table~\ref{tbl:properties}).
\begin{figure*}
\centering
\includegraphics[width=\linewidth]{sed_3radius.pdf}
\caption{The IR SEDs of our targets divided into three cluster-centric radial bins: $0<r/\mathrm{Mpc}<0.7$, $0.7<r/\mathrm{Mpc}<1.4$, $1.4<r/\mathrm{Mpc}<2.1$, at the mean redshifts of these bins (Table~\ref{tbl:properties}). The black line shows a fit to the average flux densities at 100, 160, and 1300$\mu$m (yellow circles), parameterized as a MIR power-law plus a modified blackbody. The median MIPS 24$\mu$m and stacked SPIRE 250, 350, and 500$\,\mu$m (uncorrected for flux boosting, see Section~\ref{sec:spirestacking}) flux densities are shown (yellow diamonds), but not included in the fit. The dashed red line shows the modified blackbody component only. We compare to two field-based SFG templates: the log $L_{\rm IR}/\ensuremath{{L_{\odot}}}\sim11.75$ local template from \citet{rie09} and the $z\sim2$ template from \citet{kir15}, normalized to the 160$\mu$m datapoint. We find that the local template, which has a warm dust peak and relatively weak submm emission, better represents our cluster galaxies on average, with no apparent radial dependence.}
\label{fig:seds}
\end{figure*}
To quantify the IR SEDs (Figure~\ref{fig:seds}) in these radial bins, we combine the average flux densities from the detections of our sources at (observed) 100 and $160\mu$m with the stacked average flux density at 1.3mm. Shorter wavelength MIPS 24$\mu$m detections are available for a subset of our sample; however, these probe the complex PAH and absorption features in the MIR, which will be diluted due to the redshift range probed. Accordingly, we display the median value of the 24$\,\mu$m flux densities available but do not include these in our SED fitting. We also display the SPIRE stacks at 250, 350, and 500$\,\mu$m; however, as discussed in \S~\ref{sec:ancillary} and \S~\ref{sec:spirestacking}, SPIRE coverage is only available for 10/11 clusters and the small number of sources stacked prevents us from applying a robust correction for flux boosting. Accordingly, the SPIRE stacks are shown as upper limits and not included in the fitting. The flux boosting is maximal for the largest beamsize (36\arcsec at 500$\,\mu$m) and at small cluster-centric radius. It is expected to be minimal at $r>0.5\,$Mpc at 250$\,\mu$m \citepalias{alb14}.
Following \citet{cas12}, we parameterize the dust emission as a mid-IR power law arising from warm/hot dust emission from compact star forming regions and/or AGN plus a single-temperature modified blackbody, representing the cold dust emission from reprocessed light from young stars. The dust emissitivy index ($\beta$=1.5) and MIR power law index ($\alpha$=2) are fixed, and general opacity is assumed, appropriate for conditions near the peak of emission. Total $L_{\rm IR}$ and $T_{\rm dust}$ (i.e. the dust temperature representing the luminosity-weighted dust emission) are allowed to vary.
The addition of the submm point to the data at the FIR peak indicates that that the full FIR SED is warmer ($\sim$36-38 K) than typical of massive field galaxies at this redshift \citep[$\sim30$ K;][]{sch18}, with weaker submm emission relative to the dust peak than found for the $z\sim1$ SFG template that well represents luminous field ALMA sources in blank field surveys \citep{kir15, dun17}. Given the degeneracy between dust temperature and emissivity index, we repeat this analysis assuming $\beta=2$ \citep{wei01}, finding lower dust temperatures by $\sim2$ K. As \citet{sch18} assumes $\beta=1.5$, we cautiously present our $\beta=1.5$ dust temperatures as the direct comparison to the field value. It is unknown at this time whether $\beta$ varies as a function of environment. The average total infrared luminosity in all three radial bins is log $L_{\rm IR}/\ensuremath{{L_{\odot}}}\approx11.9$. We note that the MIPS 24$\,\mu$m is in good agreement with the fit and that the SPIRE fluxes are consistent with warmer dust temperatures where flux boosting is expected to be minimal, in the outermost radial bin. We discuss these warm dust temperatures in the context of the literature on galaxy clusters in \S~\ref{sec:warm_dust}.
A recent compilation of targeted ALMA observations of galaxies at $z\sim2-4$ with robust photometry spanning the dust peak found that local templates from \citet{rie09} can reproduce the full FIR SED at these redshifts \citep{der18}. In Figure~\ref{fig:seds}, we compare to the \citet{kir15} and \citet{rie09} templates which most closely match our measured $L_{\rm IR}$, normalized (not fit) at 160$\mu$m; we find that the local \citet{rie09} template better reproduces our cluster SEDs\footnote{We note that we do not update our total $L_{\rm IR}$ and SFR$_{\rm IR}$ measurements as described in \S~\ref{sec:sample} to use the \citet{rie09} template as these quantities are heavily luminosity-weighted. At these redshifts and based on Herschel/PACS measurements near the dust peak, the expected difference in $L_{\rm IR}$ calculated using \citet{kir15} versus \citet{rie09} templates is on order $\lesssim30\%$.}. We confirm via least squares minimization that the \citet{rie09} template has reduced $\chi^2$ values of a few, an order of magnitude lower than the reduced $\chi^2$ of the \cite{kir15} template. Predicting the submm from the FIR peak and field templates such as those found appropriate in \citet{kir15} and \citet{dun17} overestimates the average ALMA flux of our targets by a factor $\sim6.5$, resulting in non-detections for observations aimed at moderate S/N.
\begin{table*}
\renewcommand{\arraystretch}{1.15}
\centering
\caption{Stacked ALMA emission and average derived properties of subsets of cluster galaxies. }
\begin{tabular}{lcccccccccc}
\hline
\hline
Subset & Number in & Stacked & S/N & $z$ & log $M_{\star}$ & SFR$_{\rm IR}$ & $\Delta$MS & log $M_{\rm mol}$ & $\tau_{\rm depl}$ & $f_{\rm gas}$ \\
& Stack & $S_{1.3}$ [$\mu$Jy] & & & [\ensuremath{{M_{\odot}}}] & [\ensuremath{{M_{\odot}}} yr$^{-1}$] & & [$\ensuremath{{M_{\odot}}}$] & [Gyr] & \\
\hline
\hline
All & 126 & 123.1$\pm$17 & 7.4 & $-$ & $-$ & $-$ & $-$ & $-$ & $-$ & $-$ \\
\hline
\multicolumn{10}{c}{Split by Radius}\\
\hline
$0<r/\mathrm{Mpc}<0.7$ & 38 & 90.1$\pm$21 & 4.3 & 1.38 & 11.1$\pm$0.1 & 129$\pm$6 & 0.17 & 10.2$\pm$0.1 & 0.11$\pm$0.03 & 0.11$\pm$0.3 \\
$0.7<r/\mathrm{Mpc}<1.4$ & 33 & 158.6$\pm$36 & 4.4 & 1.37 & 11.0$\pm$0.1 & 149$\pm$7 & 0.29 & 10.4$\pm$0.1 & 0.17$\pm$0.04 & 0.20$\pm$0.05 \\
$1.4<r/\mathrm{Mpc}<2.1$ & 55 & 114.2$\pm$ 30 & 3.8 & 1.37 & 11.0$\pm$0.1 & 165$\pm$6 & 0.33& 10.3$\pm$0.1 & 0.11 $\pm$0.03 & 0.16$\pm$0.05\\
\hline
\multicolumn{10}{c}{Split by Median Property Value}\\
\hline
$z<1.396$ & 59 & 106.5$\pm$23 & 4.6 & 1.26 & 11.0$\pm$0.06 & 138$\pm$5 & 0.30 & 10.2$\pm$0.09 & 0.12$\pm$0.03 & 0.14$\pm$0.04\\
$z>1.396$ & 67 & 130.6$\pm$25 & 5.2 & 1.47 & 11.0$\pm$0.05 & 161$\pm$5 & 0.26 & 10.3$\pm$0.08 & 0.13$\pm$0.02 & 0.17$\pm$0.04 \\
\hline
log $M_{\star}/\ensuremath{{M_{\odot}}}<10.87$ & 68 & 110.0$\pm$23 & 4.8 & 1.34 & 10.7$\pm$0.04 & 128$\pm$5 & 0.42 & 10.2$\pm$0.09 & 0.14$\pm$0.03 & 0.24$\pm$0.05 \\
log $M_{\star}/\ensuremath{{M_{\odot}}}>10.87$ & 48 & 166.5$\pm$27 & 6.3 & 1.39 & 11.3$\pm$0.05 & 181$\pm$6 & 0.16 & 10.4$\pm$0.07 & 0.15$\pm$0.02 & 0.12$\pm$0.02 \\
\hline
SFR$_{\rm IR}<121\,\ensuremath{{M_{\odot}}}$ yr$^{-1}$ & 62 & 92.5$\pm$20 & 4.6 & 1.35 & 11.0$\pm$0.07 & 90$\pm$5 & 0.09 & 10.2$\pm$0.09 & 0.16$\pm$0.04 & 0.13$\pm$0.03\\
SFR$_{\rm IR}>121\,\ensuremath{{M_{\odot}}}$ yr$^{-1}$ & 64 & 193.6$\pm$29 & 6.7 & 1.39 & 11.0$\pm$0.05 & 208$\pm$5 & 0.39 & 10.5$\pm$0.06 & 0.15$\pm$0.02 & 0.22$\pm$0.04 \\
\hline
$\Delta$MS$\,<0.35$ & 62 & 133.4$\pm$24 & 5.5 & 1.40 & 11.2$\pm$0.05 & 125$\pm$5 & 0.07 & 10.3$\pm$0.08 & 0.17$\pm$0.03 & 0.12$\pm$0.03\\
$\Delta$MS$\,>0.35$ & 64 & 97.1$\pm$26 & 3.7 & 1.34 & 10.7$\pm$0.05 & 174$\pm$5 & 0.55 & 10.2$\pm$0.12 & 0.09$\pm$0.02 & 0.22$\pm$0.06\\
\hline
\hline
\label{tbl:properties}
\end{tabular}
\end{table*}
\subsection{Molecular gas masses, gas depletion time scales, and gas fractions}
Molecular gas masses are measured following \citetalias{sco16} as described in Section~\ref{sec:gasmeasurements}. From these, we can quantify two key relations between gas content and other galaxy properties. First, the gas depletion time ($\tau_{\rm depl}\equiv M_{\rm mol}/\mathrm{SFR}$), or its inverse the star formation efficiency (SFE), gives the timescale for the depletion of the molecular gas in the system assuming the current SFR and no new gas accretion or gas recycling. Second, the gas mass fraction ($f_{\rm gas}\equiv M_{\rm mol}/(M_{\star} + M_{\rm mol})$) relates the current gas content to the stellar mass.
In \citetalias{alb16}, we analyzed the star forming properties of the IR luminous ISCS/IDCS cluster galaxies from which the sample in this work is drawn \citep[see also][A14]{bro13}. We found evidence for a rapid transition epoch in our mass-limited SFG population at $z\sim1.4$; on average, cluster galaxies have SFRs comparable to field galaxies at this epoch, with evidence for rapid quenching seen in the sharp decrease in the star forming fraction and average specific-star formation rate (SSFR) across a relatively short timescale of our redshift range ($\lesssim1$ Gyr). Evidence for rapid quenching at $z\sim1.5$ has similarly been found by looking at the environmental quenching efficiency via quiescent cluster populations \citep{nan17}.
Here we examine the gas content of our ALMA targets, cluster SFGs on the high end of the MS. In Figure~\ref{fig:radial}, we show the average $M_{\rm mol}$, $\tau_{\rm depl}$, and $f_{\rm gas}$ for the three radial bins analyzed in the previous section: $0<r/\mathrm{Mpc}<0.7$, $0.7<r/\mathrm{Mpc}<1.4$ and $1.4<r/\mathrm{Mpc}<2.1$. Again, assuming $R_{\rm sp}\sim R_{\rm vir}$, these bins are probing the virialized, transition, and infalling regions of our clusters. We find that on average, our cluster galaxies have low molecular gas masses (log $M_{\rm mol}/\ensuremath{{M_{\odot}}}\sim10.3$) and correspondingly low gas depletion times ($\sim100-200\,$Myr) and gas fractions (0.1-0.2). We compare these values to their field counterparts in the next section. We find only a weak radial dependence; cluster galaxies near the virial radius show slightly elevated gas content and longer gas depletion timescales than those in the cluster cores or well outside the virial (and potentially splashback) radius. This transition region is likely dominated by galaxies entering the virialized region for the first time, as the population of backsplash galaxies (cluster galaxies that have already fallen into and past the cluster core, whose orbits are again on outbound trajectories) is expected to be small at this redshift \citep{mos21}. However, this difference is on the $1\sigma$ level when accounting for the bootstrapped errors which incorporate the spread in the population being stacked. Overall, the gas content is low across all radii, agreeing with previous observational works that show the cluster influence extending to at least two times the virial radius \citep[i.e][]{bal99, vdl10, chu11, ras12, shi21}.
We note two sources of potential contamination in this analysis. The first is that we have identified the majority of our cluster galaxies using photometric redshifts, which will mis-classify some field galaxies as cluster galaxies. This will serve to dilute signals of environmental effects.
The second is that we have used projected cluster-centric radii to create our bins, which will dilute any signal as a function of radius. The use of phase space diagrams \citep[i.e.][]{rhe17} can place cluster galaxies in the context of their accretion history, and roughly separate virialized, infalling, and backsplash populations. However, the spectroscopy needed has thus far been largely limited to 1) the optical, which can miss the heavily obscured cluster galaxies dominating the SF budget \citepalias{alb16}, or 2) cluster galaxies with high gas content through CO emission. We argue in the next sections that the latter are not representative. Sensitive near-infrared spectroscopy, such as provided by the {\it James} {\it Webb} {\it Space} {\it Telescope}, is needed for complete, mass-limited spectroscopic cluster catalogs.
\begin{figure}
\centering
\includegraphics[scale=0.55, trim=5 0 5 0, clip]{properties_3radius.pdf}
\caption{The molecular gas mass (top), gas depletion timescale (middle) and gas fraction (bottom) of cluster galaxies in three projected radial bins: $0<r/\mathrm{Mpc}<0.7$, $0.7<r/\mathrm{Mpc}<1.4$ and $1.4<r/\mathrm{Mpc}<2.1$. Stacked uncertainties are from bootstrapping and so incorporate the spread in the stacked populations. The dashed line shows the level of the cluster core, to guide the eye. There is weak to no dependence of these properties on cluster-centric radius out to twice the virial radius ($R_{\rm vir}\sim1\,$ Mpc).}
\label{fig:radial}
\end{figure}
\subsection{Comparison with the field: significant gas deficits in cluster SFGs}\label{sec:field_compare}
In the last section, we determined that our cluster galaxies generally have low gas masses and gas fractions, with short depletion timescales with weak or no dependence on cluster-centric radius. In this section, we redo our stacking analysis by splitting our sample into subsets by the median values (Figure~\ref{fig:properties}, Table~\ref{tbl:properties}) of the following properties: stellar mass, redshift, SFR$_{\rm IR}$, and $\Delta$MS. We then compare to field galaxies in order to quantify environmentally-driven differences in gas properties.
\begin{figure*}
\centering
\includegraphics[width=\linewidth]{all_3_radius_panel_simplified.pdf}
\caption{The average molecular gas masses (first row), gas depletion timescales (second row) and gas fractions (third row) as a function of galaxy properties: stellar mass (first column), redshift (second column), IR SFR (third column) and distance from the MS (fourth column). In each panel, we measure the gas content from our stacks by splitting our ALMA sample into galaxies below (red cross) and above (yellow x) the median value of each property as listed in Table~\ref{tbl:properties}. The gas properties of field galaxy stacks \citetalias{sco16} are shown as blue diamonds.
In the first column, the \citetalias{sco16} stacks are additionally rebinned into a weighted average in two mass bins for a more direct comparison. The red dashed and yellow dotted lines are the predicted values in each bin from field scaling relations \citepalias{sco17}, calculated using the average properties (redshift, stellar mass, $\Delta$MS) of the cluster stacks, as described in Section~\ref{sec:field_compare}. Table~\ref{tbl:deficits} gives the cluster deficits as the ratio between the field and cluster values, using the rebinned \citetalias{sco16} stacks and \citetalias{sco17} scaling relations. }
\label{fig:field_compare}
\end{figure*}
\begin{table*}
\centering
\caption{The average deficits measured in gas mass, depletion timescale, and gas fraction obtained by taking the ratio of the predicted field value to the stacked cluster value (Figure~\ref{fig:field_compare}). Deficits are calculated in comparison to the rebinned field stacks presented in S16 and the field scaling relations presented in S17 (see Section~\ref{sec:field_compare}).}
\begin{tabular}{lcccccc}
\hline
\hline
Subset & $M_{\rm mol}$ & $\tau_{\rm depl}$ & $f_{\rm gas}$ & $M_{\rm mol}$ & $\tau_{\rm depl}$ & $f_{\rm gas}$\\
& S16$^\dagger$ & S16$^\dagger$ & S16$^\dagger$ & S17 & S17 & S17 \\
\hline
\hline
log $M_{\star}/\ensuremath{{M_{\odot}}}<10.87$ & $2.7\pm1.0$ & $3.8\pm1.6$ & $2.0\pm0.6$ & $4.4^{+2.1}_{-1.7}$ & $4.9^{+1.2}_{-1.2}$ & $2.4^{+0.7}_{-0.7}$ \\
log $M_{\star}/\ensuremath{{M_{\odot}}}>10.87$ & $2.3\pm0.9$ & $2.9\pm0.7$ & $3.4\pm0.8$ & $3.6^{+1.8}_{-1.3}$ & $6.7^{+1.3}_{-1.3}$ & $2.7^{+0.8}_{-0.9}$ \\
\hline
$z<1.396$ & $-$ & $-$ & $-$ & $4.7^{+2.3}_{-1.8}$ & $6.8^{+1.7}_{-1.7}$ & $3.1^{+1.0}_{-1.0}$ \\
$z>1.396$ & $-$ & $-$ & $-$ & $4.4^{+2.1}_{-1.6}$ & $6.2^{+1.4}_{-1.4}$ & $2.8^{+0.8}_{-0.8}$ \\
\hline
SFR$_{\rm IR}<121\,\ensuremath{{M_{\odot}}}$ yr$^{-1}$ & $-$ & $-$ & $-$ & $4.9^{+2.2}_{-1.8}$ & $6.9^{+1.7}_{-1.7}$ & $3.2^{+1.0}_{-1.1}$ \\
SFR$_{\rm IR}>121\,\ensuremath{{M_{\odot}}}$ yr$^{-1}$ & $-$ & $-$ & $-$ & $3.1^{+1.6}_{-1.1}$ & $4.6^{+0.9}_{-0.9}$ & $2.1^{+0.6}_{-0.6}$ \\
\hline
$\Delta$MS$\,<0.35$ & $-$ & $-$ & $-$ & $4.0^{+1.8}_{-1.4}$ & $6.7^{+1.4}_{-1.4}$ & $2.9^{+0.9}_{-0.9}$ \\
$\Delta$MS$\,>0.35$ & $-$ & $-$ & $-$ & $5.5^{+2.9}_{-2.3}$ & $6.1^{+1.8}_{-1.8}$ & $2.8^{+0.9}_{-0.9}$ \\
\hline
\hline
\label{tbl:deficits}
\end{tabular} \\
$^\dagger$Deficits from the S16 field are only calculated for the cluster stack subsets split by stellar mass. See Section~\ref{sec:field_compare}.
\end{table*}
Ideally, the comparison of the gas content in cluster and field galaxies would be between samples well matched in terms of target selection criteria, stellar mass range, and the detection limits and methodology used to probe the gas. Stacking analyses, as used in this work, have the advantage of not relying on individual detections, which can bias a study toward the gas-rich end. Given these considerations, we begin our comparison in Figure~\ref{fig:field_compare}
with the field sample from \citetalias{sco16}, which derived the average gas properties of galaxies stacked in bins of stellar mass and SSFR. For a fair comparison, we limit our comparison to their stacks at $z\sim1$ in the range log $M_{\star}/\ensuremath{{M_{\odot}}}> 10.5$ and SFR$\,>50\,\ensuremath{{M_{\odot}}}$ yr$^{-2}$, shown as blue diamonds. The average gas masses of our cluster stacks (red cross and gold x) fall well below the general distribution of the gas masses of the field galaxies in the \citetalias{sco16} sample. When incorporating SFR and stellar mass in the gas depletion timescales and gas fractions, cluster galaxies are, on average, consistent with the shortest gas depletion timescales and lowest gas fractions as found at $z\sim1$ for massive field SFGs.
To quantify this difference, we rebin the \citetalias{sco16} stacks into two bins split by the median stellar mass of our sample (Table~\ref{tbl:properties}). For these bins, we determine the weighted average and standard deviation of the \citetalias{sco16} stacks, which each include 1-12 galaxies, adopting as weights the number of galaxies in each stack. The rebinned average gas properties can be seen in the first column of Figure~\ref{fig:field_compare} (indicated by the blue background). To calculate the deficits, we then take the ratio of the rebinned field to our cluster stacks, finding that the clusters are lower than the field in $M_{\rm mol}$ by $2.7\pm1.0$ ($2.3\pm0.9$), in $\tau_{\rm depl}$ by $3.8\pm1.6$ ($2.9\pm0.7$), and in f$_{\rm gas}$ by $2.0\pm0.6$ ($3.4\pm0.8$) in the low (high) mass bin (Table~\ref{tbl:deficits}).
We next expand our comparison to the field scaling relations, which are widely used in the literature and allow us to compare to a predicted field value calculated for the average properties (stellar mass, etc) of our stacked subsamples. We adopt the scaling relations presented in \citet{sco17} (hereafter S17), which were derived from a sample of Herschel-selected SFGs at a range of redshifts ($z\lesssim3$) and masses (log $M_{\star}/\ensuremath{{M_{\odot}}}>10.3$). Priors were used to push the extraction of the Herschel photometry to lower significance detections in a similar manner as used in our Herschel photometry \citepalias{alb16}. Both the \citetalias{sco17} field and our cluster samples are the massive, IR-bright end of the galaxy populations, though the use of detections, not stacking, in the field scaling relations may bias the results toward the gas-rich populations. We note that for the stellar mass range probed in this work, the \citetalias{sco17} scaling relations are comparable to those of \citet{tac18, tac20}, which combine CO and dust continuum emission measurements, both detected and stacked.
In Figure~\ref{fig:field_compare} (all panels), we show the field scaling relations for $M_{\rm mol}$, $\tau_{\rm depl}$, or $f_{\rm gas}$ as a function of stellar mass, redshift, SFR$_{\rm IR}$, or $\Delta$MS $-$ with the other parameters being fixed at the average values of our stacked subsets (Table~\ref{tbl:properties}) $-$ following Eqns 6-8, 10 in \citetalias{sco17}. For example, in the upper lefthand plot, we show the gas mass of a field galaxy as a function of stellar mass, with its redshift and distance from the MS fixed to the properties of our lower stellar mass stack. We repeat this for the higher mass stack.
The deficit for each cluster stack is then calculated as the ratio between the predicted field value and the stacked value. The error on the deficit is a combination of the error on the best-fit scaling relations\footnote{The error on the S17 scaling relations shown is derived using the errors on the parameters defining the best-fit relations. This does not include all possible calibration and systematic errors, which are expected to be largely shared between the cluster and field gas mass derivations and so are not included in the comparison. } and the bootstrapped errors of the stacks, which included the spread in the population, added in quadrature.
From this analysis, we can compare the slope of our cluster stacks (split into two bins by stellar mass, etc) against the slope of the field scaling relations as well as the absolute values.
In terms of gas mass (top row), the cluster galaxies show similar trends as the field in terms of increasing gas mass with redshift, stellar mass, and obscured SFR, but at a deficit of 3-5x in absolute value. Cluster galaxies with $\Delta$MS$\,>0.35$ show a hint of a reverse trend in gas mass from the field scaling relation, with a larger deficit for this population relative to cluster galaxies closer to $\Delta$MS$\,\sim0$. This is likely driven by the fact that our starbursting galaxies ($\Delta\textrm{MS}>0.6$) are predominantly the lowest mass galaxies in our sample. This is a selection effect caused by targeting sources based on Herschel flux; at the low mass end, only starbursts will be bright enough to make it into the sample. This is confirmed when we look at the gas fraction as a function of $\Delta$MS (fourth column, third row), which looks at the gas content for a given stellar mass. We find that the slope of the field scaling relation with $\Delta$MS is recovered, though the absolute deficit in cluster gas remains. This suggests that there is no preferential loss of gas in currently starbursting cluster galaxies, though we also note that high mass, extremely dusty starbursts may still be missing from our sample due to the difficulty in measuring a robust photo-$z$. CO spectroscopy is needed to place the extremely dusty sources in the cluster and measure their gas properties.
The gas depletion timescales (second row) and gas fractions (third row) are similarly consistent with the slopes expected in the field when splitting by the properties shown. The deficits are significant, however, with average gas depletion timescales of $\sim200$ Myr, $\sim5-7$x lower than the predicted field values, and gas fractions of 10-20$\%$, a deficit of $\sim2-3$x. All stacked values are listed in Table~\ref{tbl:properties} and the deficit values are in Table~\ref{tbl:deficits}.
\subsection{Comparison with other (proto-)clusters: pushing past the gas-rich outliers}\label{sec:cluster_compare}
In this section, we compare to the gas properties in cluster galaxies in the literature derived from both CO and dust continuum, which are both robust tracers of the gas mass \citep[][]{tac20}. We do not re-calibrate any measurements based on CO and we caution that, similar to the comparison to field galaxies, comparisons between cluster samples can suffer from heterogeneous datasets and methodologies, as well as biased selection via targeted follow-up, small number statistics, and often high detection limits \citep[e.g. log $M_{\rm mol, CO}/\ensuremath{{M_{\odot}}}>10.8$ at $z\sim1.5$;][]{nob17, hay17, hay18, sta17}. One advantage we have in this work is that we draw targets from eleven clusters with a range of star formation activity \citepalias{alb16}, mitigating the chances that we are biased toward only highly star forming cluster systems. Of course, on the other hand, our target selection of luminous cluster SFGs introduces a different bias and means we can only address the gas properties of that population.
\begin{figure}
\centering
\includegraphics[width=\linewidth, trim=5 0 5 0, clip]{cluster_comparisons.pdf}
\caption{The gas depletion timescale (top) and gas fraction (bottom) as a function of stellar mass for (proto-)cluster galaxies. The red plus and yellow x are the stacked results from this work, split by the median stellar mass of the sample (Table~\ref{tbl:properties}). We compare to the gas properties of cluster galaxies stacked in dust continuum at $z\sim0.7$ \citep[orange square,][]{bet19} and $z\sim2$ \citep[light blue hexagons,][]{zav19}. Individual detections of CO and/or dust emission in (proto-)cluster galaxies are shown at $z\sim1.5$ \citep[purple circles and upright and rightward triangles;][]{nob17, nob19, hay17, hay18} and $z\sim2$ \citep[blue left triangles, diamonds, and inverted triangles;][]{coo18, zav19, tad19}. The stacked (average) values of cluster galaxies indicate that the typical gas content is lower than implied by detections, with shorter depletion timescales and lower gas fractions.}
\label{fig:cluster_comparison}
\end{figure}
Closely analogous to this work in terms of technique, \citet{bet19} looked at the stacked dust continuum emission in 101 SFGs at $z\sim0.7$ as a function of local galaxy density \citep{sco13}, with their highest density bin corresponding to the cluster environment. In this lower redshift epoch ($z<1$), the quenching efficiency in clusters is high \citep{muz12, alb14, nan17}; however, among cluster galaxies \textit{with ongoing star formation}, the average gas properties look remarkably similar at $z\sim0.7$ and $z\sim1.4$ (this work), with short depletion timescales and low gas fractions (Figure~\ref{fig:cluster_comparison}).
At more comparable redshifts ($z\sim1.5$), relatively small samples of cluster galaxies have been studied via detections of the CO emission line or dust continuum emission. \citet{nob17} looked at 11 cluster members detected in CO(2-1) in three clusters at $z\sim1.6$, finding field-like gas depletion timescales and both field-like and {\it enhanced} gas fractions. \citet{hay17, hay18} looked at CO(2-1) and dust continuum for 18 cluster galaxies, again finding enhanced gas fractions and long depletion timescales compared to field scaling relations. They speculated that gas accretion may be enhanced in cluster infall regions and/or filaments or that SFEs may be reduced due to environmentally-induced shock-heating or feedback. These studies, however, have relatively high detection limits and may be missing the gas-poor cluster population.
Interestingly, however, we have to date two examples of detection studies at $z\sim1.5$ which have gone significantly deeper, albeit over small areas. \citet{nob19} presented deep follow-up of one of their \citet{nob17} clusters, detecting four additional cluster galaxies in CO(2-1) (Figure~\ref{fig:cluster_comparison}). Additionally, a serendipitous detection of CO in a low mass cluster at $z\sim1.3$ was recently obtained to a $5\sigma$ limit of log $M_{\rm mol}/\ensuremath{{M_{\odot}}}\sim10.2$; Williams, et al., ApJ, submitted). Despite probing deeper in terms of $M_{\rm mol}$, neither of these studies uncovered the relatively gas-poor cluster galaxies suggested by our stacking analysis. In fact, both studies found only cluster galaxies with gas comparable to the upper end or above the field scaling relations presented in \citet{tac18}, which are built largely on CO detections.
A cluster study at higher redshift, on the other hand, does finds a gas-poor population: \citet{coo18} obtained deep CO(1-0) from the VLA and dust continuum emission at 870$\mu$m in the core of a low mass, X-ray-selected cluster at $z=1.99$, which revealed relatively low gas fractions, some comparable to this work, in five galaxies. Gas depletion timescales and gas fractions derived from their dust continuum imaging (which are in good agreement with their CO (1-0) detections and limits) are shown in Figure~\ref{fig:cluster_comparison}. Notably, their galaxies are on the MS in terms of their SSFRs, but have high gas excitation (obtained from multiple CO transitions) and a high rate of mergers/interactions and/or AGN activity.
At higher redshifts ($z>2$), gas content has been studied in the cores of overdense proto-cluster environments \citep[i.e.][]{wan16, wan18, ume17, zav19, tad19, cha21, lon20}. We limit our comparison to two studies at $z\sim2$, \citet{tad19} and \citet{zav19}, with the caveat that these environments may be in a different stage of virialization. \citet{tad19} looked at 13 H$\alpha$-selected proto-cluster members in CO(3-2) around the radio galaxy PKS 1138−262 at $z=2.16$, again finding enhanced gas fractions and long depletion timescales in four detections. \citet{zav19} looked at 41 proto-cluster SFGs with an average mass of log $M_{\star}/\ensuremath{{M_{\odot}}} = 10.8$ at $z\sim2.1$ in a log $M_{\rm halo}/\ensuremath{{M_{\odot}}}=14.3$ proto-cluster core, finding again similar gas content as coeval field galaxies, but with increasing gas-poor cluster members at the high mass (log $M_{\star}/\ensuremath{{M_{\odot}}}>11$) end. \citet{zav19} additionally performed stacking to better characterize their non-detections, finding average values on the low end of their detected distribution, consistent with this work and \citet{bet19}. This suggests that the (massive) cores of proto-clusters at $z\sim2$ may also be experiencing environmental gas loss.
Putting this all together in Figure~\ref{fig:cluster_comparison}, and folding in the field comparisons of Section~\ref{sec:field_compare}, it is clear that the picture of field-like gas content from cluster galaxy detections, with long depletion timescales and enhanced gas fractions, does not describe star forming cluster galaxies on average. Instead these gas-rich populations may be outliers driven by high detection limits. On the other hand, the two deep cluster observations at $z\sim1.5$ that have been obtained to date \citep[][Williams et al, ApJ, submitted]{nob19} have also found gas-rich cluster populations, deepening the mystery. We have only one example of relative gas-poor detections at $z=1.99$ \citep{coo18}. Resolving this contention between detections and stacking will require even lower detection limits to recover the distribution of gas properties in the galaxies represented in the stacks. Care must additionally be taken to avoid biases by observing different regions (i.e. core, infalling) within the cluster environment as well as statistical clusters samples, given cluster-to-cluster variation.
\section{Discussion}\label{sec:disc}
\subsection{Warm Dust in Cluster Galaxies}\label{sec:warm_dust}
Dust temperature has been linked to internal factors such as the physical conditions of SF in a galaxy \citep{gal18}; however, its dependence on external factors such as environment remain poorly constrained. In Section~\ref{sec:ir_sed}, we measured the average dust temperature of our cluster galaxies via their Herschel+ALMA SEDs modeled with a modified blackbody and power law component \citep{cas12}. We found that for all three radial bins, out to twice the virial radius, the dust temperatures are relatively warm, $\sim36-38$ K. Comparably massive field galaxies have typical dust temperatures of $30\pm4$K with relatively little scatter \citep[][see also \citet{sym13}]{sch18}. We note this comparison assumes that the dust emissivity $\beta$ is comparable in cluster and field galaxies, something that has yet to be tested in the literature. Similarly warm dust temperatures have been observed in a small fraction ($\sim10\%$) of sub-LIRG cluster members in the Bullet cluster, in excess of what is expected in the field and attributed to dust stripping and heating from RPS \citep{raw12}. FIR stacking of a much larger sample selected from SDSS found a marginal trend of increasing dust temperature with increasing environment, up to filament and cluster scales \citep{mat17}. Our results qualitatively support this trend at higher redshift, with the caveat that the conditions that set the dust temperature also evolve with redshift and stellar mass \citep{sch18} so the absolute numbers cannot be easily compared. Conversely, studies in both the Coma cluster \citep{ful16} and at higher redshift \citep{nob16} found no increase in dust temperature in cluster environments; however, long wavelength constraints were often undetected or not included in these works. A systematic study with good FIR and submm wavelength coverage is needed. The inclusion of the ALMA datapoint was key in our analysis; without it, we had previously found a good match to the Herschel data only using templates with $T_{\rm dust}\lesssim30$ K for the cold dust component \citep{kir15}.
\subsection{Cold Gas in Cluster Galaxies: Observations and Theory}\label{sec:theory}
In this work, we have quantified the gas properties of a population of massive cluster SFGs at $z\sim1.4$ which are on or above the star forming MS. We have found a clear dependence of their gas properties on the environment, with gas masses 2-3x (3-5x) lower than stacked co-eval field galaxies (field scaling relations), with correspondingly short gas depletion timescales and low gas fractions. This effect is seen out to twice the virial radius, the limit of our survey.
Significant gas loss starting at large cluster-centric radii has been predicted by recent cosmological simulations. \citet{zin18} studied the effects of RPS on diffuse hot halo gas and the more tightly bound cold disk gas. They found that RPS can efficiently remove 40-70$\%$ of the galaxy's hot halo gas at large cluster-centric radii ($\sim2R_{\rm vir}$), with less than 30$\%$ of halo gas remaining in high mass satellites by $R_{\rm vir}$ \citep[see also][]{bah13}. This removal was attributed to virial accretion shocks \citep{sar98, bir03, dek06} at roughly the boundary where galaxies enter the hot ICM. This is also approximately where it is expected that fresh gas accretion onto galaxies is prevented (i.e. starvation). For their high redshift clusters ($z\sim0.6$), \citet{zin18} estimated a travel time of a few Gyr between the accretion shock and $R_{\rm vir}$ over which this starvation can occur. If we compare this to the expected gas recycling timescale \citep[$0.5-1\,$Gyr;][]{opp10, tac20} and gas depletion timescale due to ongoing SF \citep[$\sim1$ Gyr at $z\sim0.6$;][]{tac20} for field galaxies, then there is ample time for the combination of RPS of the hot halo gas, starvation, and consumption of gas in star formation to affect the cold gas reservoir well before the galaxy reaches the virial radius.
This is qualitatively consistent with our recovery of a gas deficit out to 2$R_{\rm vir}$. Additionally, cold gas may be heated by processes such as feedback, potentially associated with increase fractions of cluster AGN \citep[][A16]{mar13}.
Similar but more extreme results were found in the \textsc{Three Hundred} simulation suite \citep{art19, mos21}, which looked at cluster and satellite halos up to $z\sim1$. They predict instantaneous gas fractions consistent with zero by the crossing of $R_{\rm vir}$ in projected space, again attributed to crossing virial accretion shocks. This total gas loss is likely overestimated due to poor resolution in the current simulations \citep{bah17b, bos18b, bos18a} and oversimplifications in ICM structure \citep{ton19}, but the emerging picture is that of significant gas loss starting at large radii and largely completing by the first passage of the cluster center, largely independent of halo mass \citep[see also][]{oma16, lot19, oma21}.
Our finding of gas properties 2-5x below the field, on average, out to $2R_{\rm vir}$ supports this picture. The low gas content can be reconciled with the ongoing star formation under the ``delayed, then rapid" quenching scenario \citep[i.e.][]{wet13, vdb13, bah15, oma16, alb16, nan17, rhe20} in which any effect on the SFR is delayed after infall, followed by quenching on a short timescale. If tightly bound disk gas is retained, then the high mass galaxies characteristic of this study may then keep forming stars up to a Gyr after infall \citep{lot19}. The low gas masses, ongoing star formation, and condition that quenching is rapid, particularly for these clusters during the era in which we see a sharp rise in the quenched fraction over $z\sim1-1.6$ \citep[A16,][]{nan17}, suggest a combination of gas loss via stripping to prevent the recycling and/or cooling of gas \citep[e.g. effected by stellar or AGN winds;][]{tac20} in conjunction with the consumption of disk gas via star formation, facilitates the final quenching.
Seemingly at odds with the low average gas content presented in this study $-$ and the simulations that predict strong gas stripping on first passage $-$ are the gas-rich cluster galaxies discussed in Section~\ref{sec:cluster_compare}. These galaxies must 1) retain (or replenish) their cold gas reserves or 2) be in a phase of efficient gas cooling, a selection effect. For the former, hot gas haloes have been observed around local cluster/group galaxies \citep{sun07, jel08}. One explanation is that these individually detected, gas-rich cluster galaxies occupy a privileged place in the clusters relative to gas streams which may weave through the ICM and could provide fresh gas accretion \citep{dek06, zin16, zin18}. Such gas streams, particularly in unrelaxed clusters, have been invoked to explain metal enrichment of the ICM at large radii \citep{reb06,sim15} and the build up of Brightest Cluster Galaxies \citep[BCGs;][]{mcd12}. This work quantifying the average gas properties of massive cluster SFGs, drawn from a range of clusters, adds a key constraint for future investigations of this mechanism, in that we have shown that this gas-rich population is the minority.
\section{Conclusions} \label{sec:conc}
In this work, we have used ALMA Band 6 (observed 1.3mm) imaging to present the average gas properties of 126 massive (log $M_{\star}/\ensuremath{{M_{\odot}}}\gtrsim10.5$) cluster SFGs in eleven massive (log $M_{\rm halo}/\ensuremath{{M_{\odot}}}\sim14$) galaxy clusters at $z=1-1.75$. Our eleven clusters represent a range in total star formation activity and we sample out to $\sim2 R_{\rm vir}$, with $R_{\rm vir}$ likely equivalent to the splashback radius. On initial analysis, it was found that all targets were undetected in ALMA due to weaker submm emission relative to that predicted based on coeval field galaxies. Stacking analysis was therefore used to obtain the average dust continuum emission, from which we derived average gas masses ($M_{\rm mol}$) for stacked subsets of our sample following the calibrations presented in \citetalias{sco16}. Combined with extensive multi-wavelength data, we further measured the average gas depletion timescales ($\tau_{\rm depl}$; inversely, the star formation efficiencies) and gas fractions (f$_{\rm gas}$).
Our main conclusions are as follows:
\begin{enumerate}
\item ALMA stacks in three cluster-centric radial bins probing from the virialized to infalling regions were combined with the Herschel/PACS photometry to construct the average IR SEDs of massive, star forming cluster galaxies. We find weaker submm flux than predicted by the IR SEDs of ALMA-detected field galaxies at $z\sim2$ \citep{kir15, dun17}; our cluster SEDs are instead well described by local SFG templates \citep{rie09} of similar luminosity (log $L_{\rm IR}/\ensuremath{{L_{\odot}}}\sim11.9$). The addition of the submm anchor to the Herschel data at the dust peak reveals relatively warm dust temperatures ($\sim36-38$ K) compared to the well constrained cooler temperatures of field galaxies \citep[$\sim30$ K;][]{sch18}, with no significant dependence on cluster-centric radius. Evidence of warmer dust in cluster galaxies is highly conflicted in the literature; in our work, it was key to probe both the IR peak and submm to identify these warmer temperatures.
\item Gas masses, gas depletion timescales, and gas fractions were presented in the three radial bins, sampling out to twice the virial radius. We find that the gas properties have weak to no dependence on cluster-centric radius, with a marginal excess in average gas mass in the intermediate bin ($\sim R_{\rm vir}$) at the $1\sigma$ level. We note that using projected radial bins may dilute a real trend in this analysis; however, it is clear that the environmental impact on gas properties extends from the cluster cores to large radii, in good agreement with cluster studies which have traced the cluster influence beyond the virial (splashback) radius \citep[i.e][]{bal99, vdl10, chu11, ras12, shi21}.
\item Restacking our cluster sample into bins split by stellar mass, redshift, obscured SFR, and distance from the MS, we perform a careful comparison of our ALMA cluster stacks to the gas properties of field galaxies in the literature. We find that our cluster galaxies have deficits of 2-3x, 3-4x, and 2-4x in $M_{\rm mol}$, $\tau_{\rm depl}$, and f$_{\rm gas}$ when compared to stacked field samples \citepalias{sco16}. Slightly larger deficits are found when comparing to the predicted values from field scaling relations \citep{tac18, sco17}, calculated based on the median properties (redshift, stellar mass, etc) of our subsamples. This demonstrates that environmentally-driven processes are causing, on average, a significant depletion in the gas reservoirs of cluster galaxies at high redshift.
\item Comparison of our stacks with gas measurements of cluster galaxies in the literature reveal that our results are consistent with other stacking analyses \citep{bet19, zav19}, which also find on average low gas content. By contrast, our results are inconsistent with studies of cluster populations {\it detected} in CO or dust continuum emission, which largely find field-like gas depletion timescales and field-like or even enhanced gas fractions \citep[i.e.][Williams et al., ApJ, submitted; but see \citet{coo18}]{hay17, hay18, nob17, nob19, tad19, zav19}.
The comparison with our average, stacked gas properties suggests that these gas-rich cluster galaxies may not be typical.
\end{enumerate}
Our observed deficit in the gas properties of cluster galaxies at $z\sim1.4$ out to $2R_{\rm vir}$ is qualitatively in good agreement with recent simulations, which project that gas depletion begins early in the infall stages, at large radii where galaxies encounter virial accretion shocks. Though they vary in detail and quantity of gas removed, these simulations largely agree that significant gas stripping happens by first passage of the cluster center \citep{zin18, art19, mos21, oma21}. Our analysis of star-forming yet gas-poor cluster galaxies is consistent with this significant gas loss, but also a retention of the closely-bound disk gas, which continues for a time to fuel star formation as in the ``delayed, then rapid" quenching scenario \citep[i.e.][]{wet13}.
\acknowledgments
The authors thank Sarah Betti for her help in measuring the ALMA photometry and Karen Olsen for discussions on cluster simulations. SA acknowledges support from the James Webb Space Telescope (JWST) Mid-Infrared Instrument (MIRI) Science Team Lead, Grant 80NSSC18K0555, from NASA Goddard Space Flight Center to the University of Arizona. CCW acknowledges support from the JWST Near-Infrared Camera (NIRCam) Development Contract NAS5-02105 from NASA Goddard Space Flight Center to the University of Arizona. The National Radio Astronomy Observatory is a facility of the National Science
Foundation operated under cooperative agreement by Associated Universities, Inc. This paper makes use of the following ALMA data: ADS/JAO.ALMA\#2015.1.00813.S. ALMA is a partnership of ESO (representing its member states), NSF (USA) and NINS (Japan), together with NRC (Canada), NSC and ASIAA (Taiwan), and KASI (Republic of Korea), in cooperation with the Republic of Chile. The Joint ALMA Observatory is operated by ESO, AUI/NRAO and NAOJ. This research was carried out in part at the Jet Propulsion Laboratory, California Institute of Technology, under a contract with the National Aeronautics and Space Administration.
\\
\\
Software: NumPy \citep{har20},
Matplotlib \citep{hun07}, Astropy \citep{astropy:2018}, pandas \citep{mckinney-proc-scipy-2010}, seaborn \citep{Waskom2021}, CASA \citep{mcm07}
\bibliographystyle{aasjournal}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 4,516 |
\section{Introduction}
\subsection{The Inner Product and the Size of Preimages}
The inner product map $\ensuremath{\langle} x,y\ensuremath{\rangle}=\sum_{i=1}^d x_iy_i$ of two
$d$-dimensional vectors $x=(x_1,x_2,\ldots,x_d)$ and $y=(y_1,y_2,\ldots,y_d)$
is one of the cornerstones of linear algebra and its applications.
For example, when $x$ and $y$ are vectors of observations normalized to
zero mean and unit standard deviation, then $\ensuremath{\langle} x,y\ensuremath{\rangle}$ is
the Pearson correlation between $x$ and $y$. As such, it is a fundamentally
important computational and data-analytical task to efficiently gain
information about the {\em preimages} of the inner product
map; for example, to highlight pairs of similar or dissimilar observables
between two families of $n$ observables.
Accordingly, the protagonist of this paper is the following counting problem
(\#\textsc{InnerProduct}):
\begin{quote}
Given as input a target $t\in\{0,1,\ldots,d\}$ and two $n$-element sets
$\mathcal A\subseteq \{0,1\}^d$ and $\mathcal B\subseteq \{0,1\}^d$,
count the number of vector pairs $(x,y)\in \mathcal A \times \mathcal B$
with integer inner product $\langle x,y \rangle=t$.
\end{quote}
From a complexity-theoretic standpoint, this problem generalizes many
conjectured-hard problems in the study of fine-grained complexity---such as
the $t=0$ special case, the {\em orthogonal vector counting} (\#\textsc{OV})
problem---as well as generalizing fundamental application settings, such as
similarity search in Hamming spaces. While it is immediate that subquadratic
scalability in $n$ is obtainable when $d=o(\log n)$, our interest in this paper
is to obtain an improved understanding of the fine-grained complexity landscape
for {\em moderately short} vectors, specifically for $d$ at most
poly-logarithmic in $n$.
\subsection{Subquadratic Scaling for Moderately Short Vectors}
Our main positive result establishes deterministic subquadratic scalability
for \#\textsc{InnerProduct} up to $d$ growing essentially as the square of
the logarithm of $n$:
\begin{Thm}[Main; Subquadratic Scaling for \#\textsc{InnerProduct}]
\label{thm:main}
There exists a deterministic algorithm that, given as input
a target $t\in\{0,1,\ldots,c\log n\}$ and
two $n$-element sets $\mathcal{A},\mathcal{B}\subseteq\{0,1\}^{c\log n}$
with $4\leq c\leq\frac{\log n}{(\log\log n)^4}$,
outputs the number of pairs $(x,y)\in\mathcal{A}\times\mathcal{B}$
with $\ensuremath{\langle} x,y\ensuremath{\rangle}=t$ in time
\begin{equation}
\label{eq:main-time}
n^{2}/2^{\Omega\bigl(\sqrt{\frac{\log n\log\log n}{c\log^2 c}}\bigr)}\,.
\end{equation}
\end{Thm}
The algorithm in Theorem~\ref{thm:main} is based on a novel technique of
reconstructing a function from its sum-aggregates by prime residue, which
can be seen as an {\em additive} analog of the Chinese Remainder Theorem
and may be of independent interest (cf.~Sect.~\ref{sect:reconstruction}).
We also show how a randomized algorithm for the decision problem
of checking for a pair of vectors whose Hamming distance is less than a target
by Alman and Williams~\cite{AlmanW2015}, can with a small modification be turned into an algorithm for \#\textsc{InnerProduct}.
\begin{Thm}[Randomized Subquadratic Scaling for \#\textsc{InnerProduct}]
\label{thm:random}
There exists a randomized algorithm that w.h.p., given as input
a target $t\in\{0,1,\ldots,c\log n\}$ and
two $n$-element sets $\mathcal{A},\mathcal{B}\subseteq\{0,1\}^{c\log n}$
with $4\leq c\leq\frac{\log n}{(\log\log n)^3}$,
outputs the number of pairs $(x,y)\in\mathcal{A}\times\mathcal{B}$
with $\ensuremath{\langle} x,y\ensuremath{\rangle}=t$ in time
\begin{equation}
\label{eq:random-time}
n^{2}/2^{\Omega\bigl(\frac{\log n}{c\log^2 c}\bigr)}\,.
\end{equation}
\end{Thm}
While the randomized algorithm in Theorem~\ref{thm:random} is
faster than the deterministic one in Theorem~\ref{thm:main}, we
stress that as far as we know no deterministic algorithm in subquadratic
time was previously known for \#\textsc{InnerProduct}, even for
$O(\log n)$ dimensions. In particular, derandomizing Theorem~\ref{thm:random}
while retaining subquadratic time seems challenging, even though some
progress on the amount of randomness needed in the algorithm has been made,
cf.~Theorem~1.1~in~\cite{AlmanCW2016}.
Our further objective is to better understand the fine-grained complexity
of \#\textsc{InnerProduct} in relation to that of \#\text{OV} and other counting
problems.
For $d=O(\log n)$, it is known that these problems are truly-subquadratically
related; indeed, Chen and Williams~\cite{ChenW19} give a parsimonious
reduction for the detection variants of these two problems.
That is, if \#\textsc{OV} can be solved in $n^{2-\omega(1)}$ time,
then so can \#\textsc{InnerProduct}. However, while there is a
subquadratic time algorithm for \#\textsc{OV} whose running time scales as good as $n^{2-\Omega(1/\log c)}$~\cite{ChanW2016},
the reduction of Chen and Williams~\cite{ChenW19} does not immediately give
a non-trivial algorithm for \#\textsc{InnerProduct}. Indeed, the fastest known
algorithm for the decision version \textsc{InnerProduct} utilize probabilistic
polynomials for symmetric Boolean functions with optimal dependence on
the degree and error~\cite{AlmanW2015}, and does not go via fast \textsc{OV} algorithms and
the reduction above. In Theorem~\ref{thm:random}, we show how a simple modification to the algorithm in Alman and Williams~\cite{AlmanW2015}
can turn their algorithm into a counting one. We note that while Alman, Chan, and Williams~\cite{AlmanCW2016} later presented a deterministic algorithm based on Chebyshev polynomials over the reals for minimum/maximum Hamming weight pair, with the same running time as the randomized one in~\cite{AlmanW2015}, that deterministic algorithm, or the even faster randomized one they presented, can not be turned into one for \#\textsc{InnerProduct}
by our suggested modification alone.
\subsection{Lower Bounds via the Permanent}
The running times~\eqref{eq:main-time} and ~\eqref{eq:random-time} would,
at least at first, appear to leave room for improvement. Indeed, the
running time~\eqref{eq:random-time} is considerably worse than the
running time $n^{2-\Omega(1/\log c)}$ obtained by
Chan and Williams~\cite{ChanW2016} for \#\textsc{OV}. We proceed to show that
this intuition might be misleading, since such scalability would imply the
existence of considerably faster algorithms for a canonical hard problem
in exponential-time complexity.
Accordingly,
to gain insight into the complexity of \#\textsc{InnerProduct} and \#\textsc{OV}
when $d=\omega(\log n)$, we introduce our second protagonist
($R$-\textsc{Permanent}):
\begin{quote}
Given as input an $n\times n$ matrix $M$ with entries $m_{ij}$ in
a ring $R$ for $i,j\in[n]$, compute the {\em permanent}
\[
\ensuremath{\operatorname{per}} M=\sum_{\sigma\in S_n} \prod_{i\in [n]} m_{i,\sigma(i)}\,,
\]
where $S_n$ is the group of all permutations of $[n]=\{1,2,\ldots,n\}$.
\end{quote}
Ryser's algorithm from 1963 computes the permanent with $O(n2^n)$ arithmetic
operations in $R$~\cite{Ryser1963}. It is a major open problem whether
this can be improved to $O(c^n)$ for some constant $c<2$. Even improving
the running time to less than $2^n$ operations has been noted as
a challenge by Knuth in the
{\em Art of Computer Programming} \cite[Exercise 4.6.4.11]{Knuth1998}.
Valiant in 1979 famously proved that the permanent is \#P-complete even
when restricted to $m_{ij}\in\{0,1\}$ and evaluated over the ring of
integers~\cite{Valiant1979}; this version of the problem can be interpreted
as counting the perfect matchings in a balanced bipartite graph having
the matrix as its biadjacency matrix.
For zero-one inputs over the integers, \emph{somewhat} faster algorithms are
known (cf.~Sect.~\ref{sect:related}); to the best of our knowledge,
the current champion for zero-one matrices computes the permanent
in $2^{n-\Omega\bigl(\sqrt{n/\log \log n}\bigr)}$ time~\cite{BjorklundKW2019}.
As our second contribution, we relate the fine-grained scalability
of solving \#\textsc{InnerProduct} and \#\textsc{OV} to the task of computing the
permanent of a zero-one matrix over the integers. In particular, our first
result shows that if we could solve \#\textsc{InnerProduct} as fast as the
fastest currently known algorithms for \#\textsc{OV}~\cite{ChanW2016},
then we would immediately obtain a much faster algorithm for the permanent:
\begin{Thm}[Lower Bound for \#\textsc{InnerProduct} via Integer Permanent]
\label{thm:perm-to-exactip}
If there exists an algorithm for solving \#\textsc{InnerProduct} for
$N$ vectors from $\{0,1\}^{c\log N}$ in time $N^{2-\Omega(1/\log c)}$,
then there exists an algorithm solving the permanent of an $n\times n$
zero-one matrix over the integers in time $2^{n-\Omega(n/\log n)}$.
\end{Thm}
Thus, despite the true-subquadratic equivalence for $d=O(\log n)$~\cite{ChenW19},
it would appear that \#\textsc{InnerProduct} and \#\textsc{OV} have
different complexity characteristics when $d=\omega(\log n)$.
Our next result shows that a modest improvement in fine-grained
scalability of \#\textsc{OV} would likewise imply much faster algorithms
for the permanent.
\begin{Thm}[Lower Bound for \#\textsc{OV} via Integer Permanent]
\label{thm:perm-to-ov}
If there exists an algorithm for solving \#\textsc{OV} for
$N$ vectors from $\{0,1\}^{c\log N}$ in time
$N^{2-\Omega(1/\log^{1-\epsilon} c)}$ for some $\epsilon>0$, then there
exists an algorithm solving the permanent of an $n\times n$
zero-one matrix over the integers in time
$2^{n-\Omega(n/\log^{2/\epsilon-2} n)}$.
\end{Thm}
We note that such fast algorithms for \#\textsc{OV} would already disprove the
so-called Super Strong ETH, that $k$-\textsc{CNFSAT} on $n$ variables
has a $2^{n-n/o(k)}$ time algorithm, by the reduction to \textsc{OV}
by Williams~\cite{Williams2005} after sparsification \cite{ImpagliazzoPZ2001}. The present result merely adds to the list of consequences of faster algorithms for \#\textsc{OV}.
\subsection{Methodology and Organization of the Paper}
The key methodological contribution underlying our main algorithmic result
(Theorem~\ref{thm:main}) is a novel additive analog of the Chinese Remainder
Theorem (Lemma~\ref{lem:reconstruction} developed independently of the
application in Sect.~\ref{sect:reconstruction}), which enables us to recover
the number of pairs
$(x,y)\in\mathcal{A}\times\mathcal{B}$ with $\ensuremath{\langle} x,y\ensuremath{\rangle}=t$ from counts of
pairs $(x,y)$ satisfying $\ensuremath{\langle} x,y\ensuremath{\rangle}\equiv r\pmod p$ for multiple
small primes $p$ and residues $r\in\{0,1,\ldots,p-1\}$. In particular, the crux
of the algorithmic speedup lies in the observation
that to recover the count associated with a target $0\leq t\leq d$, primes
up to roughly $\sqrt{d}$ suffice by Lemma~\ref{lem:reconstruction}.
To obtain the counts of pairs in each residue
class $r$ modulo $p$, we employ the polynomial method with modulus-amplifying
polynomials of Beigel and Tarui~\cite{BeigelT1994} to accommodate the counts
under a prime-power modulus, with fast rectangular matrix multiplication of
Coppersmith~\cite{Coppersmith1982} as the key subroutine implementing
the count; this latter part of the algorithm design developed
in Sect.~\ref{sect:main-proof} follows well-known techniques in fine-grained
algorithm design (e.g.~\cite{AlmanCW2016}). Similarly, the
randomized algorithm design in Theorem~\ref{thm:random} follows by a minor
adaptation of the probabilistic-polynomial techniques of Alman and
Williams~\cite{AlmanW2015} to a counting context; a proof is relegated
to Sect.~\ref{sect:proof-of-randomized-algorithm}.
Our two lower-bound reductions, Theorem~\ref{thm:perm-to-exactip}
and Theorem~\ref{thm:perm-to-ov}, rely on reducing an $m\times m$
integer permanent first via the Chinese Remainder Theorem into
permanents modulo multiple primes $p$ with $p\leq m\ln m$, and then using
algebraic splitting via Ryser's formula~\cite{Ryser1963} to obtain
short-vector instances of \#\textsc{InnerProduct} and \#OV, respectively.
For \#\textsc{InnerProduct} and Theorem~\ref{thm:perm-to-exactip},
the split employs a novel discrete-logarithm version of Ryser's formula
modulo $p$ to arrive at two collections of vectors whose counts of pairs with
specific inner products enable recovery of the permanent modulo $p$;
the proof is presented in Sect.~\ref{sect:perm}.
For \#OV and Theorem~\ref{thm:perm-to-ov}, the split analogously
employs Ryser's formula modulo $p$ but with a more intricate vector-coding
of group residues modulo $p$ to obtain the desired correspondence with counts
of pairs of orthogonal vectors; we relegate the proof to
Sect.~\ref{sect:perm-cont}.
\subsection{Related Work and Further Applications}
\label{sect:related}
{\em Exact and approximate inner products.}
Abboud, Williams, and Yu~\cite{AbboudWY2015} used the polynomial method
to construct a randomized subquadratic time algorithm for
\textsc{OV}. Chan and Williams~\cite{ChanW2016} derandomized the algorithm
and showed that it could also solve the counting problem \#\textsc{OV}.
The first result that addressed an inner product different from zero, was
the randomized algorithm for minimum
Hamming weight pair by Alman and Williams~\cite{AlmanW2015}. Subsequently,
Alman, Chan, and Williams~\cite{AlmanCW2016} found an even faster randomized
as well as a deterministic subquadratic algorithm matching~\cite{AlmanW2015}.
A number of studies address approximate versions of inner-product counting
in subquadratic time, such as the detection of outlier correlations and
offline computation of approximate nearest neighbors, including
Valiant~\cite{Valiant2015},
Karppa, Kaski, and Kohonen~\cite{KarppaKK2018},
Alman~\cite{Alman2019}, and
Alman, Chan, and Williams~\cite{AlmanCW2020}.
All the algorithms above utilize fast rectangular matrix multiplication.
{\em Permanents.}
Bax and Franklin presented a randomised $2^{n-\Omega(n^{1/3}/\log n)}$ expected time algorithm for the $0/1$-matrix permanent~\cite{BaxF2002}.
Bj\"orklund~\cite{Bjorklund2016} derived a faster and deterministic $2^{n-\Omega(\sqrt{n/\log n})}$ time algorithm.
The algorithm was subsequently improved to a deterministic $2^{n-\Omega(\sqrt{n/\log \log n})}$ time algorithm by Bj\"orklund, Kaski, and Williams~\cite{BjorklundKW2019}.
For the computation of an integer matrix permanent modulo a prime power $p^{\lambda n/p}$ for any constant $\lambda<1$, Bj\"orklund, Husfeldt, and Lyckberg~\cite{BjorklundHL2017} derived a $2^{n-\Omega(n/(p\log p))}$ time algorithm. For the computation of a matrix permanent over an arbitrary ring $R$ on $r$ elements, Bj\"orklund and Williams~\cite{BjorklundW2019} gave a deterministic $2^{n-\Omega(\frac{n}{r})}$ time algorithm.
The problem \#\textsc{InnerProduct} has various applications in combinatorial
algorithms. To mention two in particular, it can be used to count the
satisfying assignments to a
\textsc{Sym}$\circ$\textsc{And} formula (cf.~Sect.~\ref{sect:sym-and}),
or compute the weight enumerator polynomial of a linear code
(cf.~Sect.~\ref{sect:weight-enum}).
\section{Reconstruction from Sum-Aggregates by Prime Residue}
\label{sect:reconstruction}
This section develops the main methodological contribution of this work.
Namely, we show that a complex-valued function
$f:D\rightarrow\mathbb{C}$ can be reconstructed from its
sum-aggregates by prime residue when the domain $D$ is a prefix of the
set of nonnegative integers. In essence, reconstruction of a function from
its sum-aggregates can be viewed as an {\em additive} analog of
the Chinese Remainder Theorem; that is, we obtain reconstruction
up to the {\em sum} of the prime moduli---in the precise sense
of~\eqref{eq:sm} below---whereas the Chinese Remainder Theorem enables
reconstruction up to the product of the moduli.%
\footnote{%
Here it should be noted that the scope of the Chinese Remainder Theorem is
also somewhat more restricted than our present setting; indeed, in our setting
the Chinese Remainder Theorem does not enable the reconstruction of
an arbitrary function $f$ but rather is restricted to reconstruction in
the case when $f$ is known to vanish in all but one point of $D$.}
In our application of counting pairs of vectors by inner product, we let $f$
be a counting function such that $f(\ell)$ counts the number of pairs
$(x,y)\in\mathcal{A}\times\mathcal{B}$ with $\ensuremath{\langle} x,y\ensuremath{\rangle}=\ell$. Reconstruction
from sum-aggregates then enables us to recover $f$ by counting the number of
pairs $(x,y)$ with $\ensuremath{\langle} x,y\ensuremath{\rangle}\equiv r\pmod p$ for small primes $p$ and residues
$r\in\{0,1,\ldots,p-1\}$; we postpone the details of this application to
Sect.~\ref{sect:main-proof} and first proceed to study reconstructibility.
\subsection{Sum-Aggregation by Prime Residue}
Let $p_1,p_2,\ldots,p_m$ be distinct prime numbers and let us assume that
\[
D\subseteq\bigl\{0,1,\ldots,s_m-1\bigr\}
\]
where
\begin{equation}
\label{eq:sm}
s_m=1+\sum_{b=1}^m\bigl(p_b-1\bigr)\,.
\end{equation} Letting $f_{\ell}$ be shorthand for $f(\ell)$, we show that we can recover $f$ from the sequence of its {\em sum-aggregates}
\begin{equation}
\label{eq:F}
F_{br}=\!\!\!\sum_{\substack{\ell\in D\\\ell\equiv r\!\!\!\!\pmod {p_b}}}\!\!\! f_\ell
\end{equation}
for each residue $r\in\{0,1,\ldots,p_b-1\}$ and each $b\in\{1,2,\ldots,m\}$.
To start with, let us observe that this sequence is linearly redundant.
Indeed, define the sum
\begin{equation}
\label{eq:F0}
F_{01}=\sum_{\ell\in D} f_\ell
\end{equation}
and observe that for each $b\in\{1,2,\ldots,m\}$ we have the linear relation
\[
F_{01}=\sum_{r=0}^{p_b-1} F_{br}\,.
\]
To obtain an equivalent and---as we will shortly show---linearly
irredundant sequence, take the sequence formed by the sum $F_{01}$
followed by $F_{br}$ for each {\em nonzero} residue
$r\in\{1,2,\ldots,p_b-1\}$ and each $b\in\{1,2,\ldots,m\}$.
Let us write $F$ for this sequence of length $s_m$.
By extending the domain of the function $f$ with zero-values $f_\ell=0$
as needed, we can also assume that $D=\{0,1,\ldots,s_m-1\}$ in what follows.
\subsection{Sum-Aggregation as a Linear System}
Let us now study reconstruction of $f$ from $F$. From \eqref{eq:F} and
\eqref{eq:F0} we observe that the task of reconstructing $f$ from $F$
is equivalent to solving the linear system
\begin{equation}
\label{eq:FAf}
F=Af\,,
\end{equation}
where $A$ is the $s_m\times s_m$ {\em nonzero residue aggregation matrix}
whose entries are defined for all $b\in\{0,1,2,\ldots,m\}$,
$i\in\{1,2,\ldots,p_b-1\}$, and
$\ell\in\{0,1,\ldots,s_m-1\}$ by the rule
\begin{equation}
\label{eq:A-entry}
A_{bi,\ell}=
\begin{cases}
1 & \text{if $b=0$;}\\
1 & \text{if $b\geq 1$ and $i\equiv \ell\!\!\pmod{p_b}$;}\\
0 & \text{if $b\geq 1$ and $i\not\equiv \ell\!\!\pmod{p_b}$,}
\end{cases}
\end{equation}
where we have assumed for convenience that $p_0=2$.
Indeed,
we readily verify from \eqref{eq:F}, \eqref{eq:F0}, and~\eqref{eq:A-entry}
that
\[
F_{bi}=\sum_{\ell=0}^{s_m-1} A_{bi,\ell}f_\ell
\]
holds for each $b\in\{0,1,\ldots,m\}$ and $i\in\{0,1,\ldots,p_b-1\}$.
When we want to stress the $m$ selected primes,
we write $A^{p_1,p_2,\ldots,p_m}$ for the matrix $A$.
The row-banded structure given by \eqref{eq:A-entry} is perhaps easiest
illustrated with a small example. Below we display the matrix $A$ for
the primes $p_1=2$, $p_2=3$, and $p_3=5$:
\[
A^{2,3,5}=
\left[\begin{array}{cccccccc}
1&1&1&1&1&1&1&1\\\hline
0&1&0&1&0&1&0&1\\\hline
0&1&0&0&1&0&0&1\\
0&0&1&0&0&1&0&0\\\hline
0&1&0&0&0&0&1&0\\
0&0&1&0&0&0&0&1\\
0&0&0&1&0&0&0&0\\
0&0&0&0&1&0&0&0
\end{array}\right]\,.
\]
Observe in particular that the first band $b=0$
corresponds to the sum~\eqref{eq:F0} and the subsequent bands
$b\in\{1,2,\ldots,m\}$
each correspond to one of the primes $p_1,p_2,\ldots,p_m$ so that
the $p_b-1$ rows inside each band correspond to the sum-aggregates \eqref{eq:F}
of the $p_b-1$ {\em nonzero} residue classes modulo~$p_b$.
Our main technical lemma establishes that the matrix $A$ is invertible,
thus enabling reconstruction of $f$ from $F$.
\begin{Lem}[Reconstruction from Sum-Aggregates by Prime Residue]
\label{lem:reconstruction}
The nonzero residue aggregation matrix $A^{p_1,p_2,\ldots,p_m}$
is invertible whenever $p_1,p_2,\ldots,p_m$ are distinct primes.
\end{Lem}
The key idea in the proof is
to decompose $A^{p_1,p_2,\ldots,p_m}$ over the complex numbers into the
product of a near-block-diagonal matrix with near-Vandermonde blocks
and a Vandermonde matrix, both of which are then shown to have nonzero
determinant. The rest of this section is devoted to a proof of
Lemma~\ref{lem:reconstruction}.
\subsection{Preliminaries on Complex Roots of Unity}
We will need the following standard facts about complex roots of unity.
For a positive integer $N$, let us write
\[
\omega_{N}=\exp\biggl(\frac{2\pi\Im}{N}\biggr)\,,
\]
where $\Im=\sqrt{-1}$ is the imaginary unit. For all $m\in\mathbb{Z}$ we have
\begin{equation}
\label{eq:root-power-sum}
\frac{1}{N}\sum_{j=0}^{N-1}\omega_{N}^{km}=
\begin{cases}
1 & \text{if $k\equiv 0\!\!\pmod N$};\\
0 & \text{if $k\not\equiv 0\!\!\pmod N$}.
\end{cases}
\end{equation}
\subsection{Reconstruction from Sum-Aggregates---Proof of Lemma~\ref{lem:reconstruction}}
We show that for distinct primes $p_1,p_2,\ldots,p_m$
the matrix $A=A^{p_1,p_2,\ldots,p_m}$ is invertible over rational numbers.
Our strategy is to
show that $A=UV$ for two complex matrices $U$ and $V$ that both have nonzero
determinant. Indeed, the near-cyclic banded structure of $A$ suggests that one
should pursue a decomposition in terms of block-structured near-Vandermonde
matrices. Let us first define the matrices $U$ and $V$, then present
a small example, and then complete the proof.
The matrix $U$ will use a $(m+1)\times(m+1)$ block structure that is similar
to the $(m+1)$-band structure of $A$, but now the structure is used both for
rows and columns. Again for convenience we assume $p_0=2$. The matrix $U$ is
defined for all $b\in\{0,1,2,\ldots,m\}$,
$i\in\{1,2,\ldots,p_b-1\}$,
$d\in\{0,1,\ldots,m\}$,
and
$k\in\{1,2,\ldots,p_d-1\}$ by the rule
\begin{equation}
\label{eq:U-entry}
U_{bi,dk}=
\begin{cases}
1 & \text{if $d=0$ and $b=0$;}\\
\frac{1}{p_b} & \text{if $d=0$ and $b\geq 1$;}\\
0 & \text{if $d\geq 1$ and $b\neq d$;}\\
\frac{1}{p_b} \omega_{p_b}^{-ik} & \text{if $d\geq 1$ and $b=d$.}\\
\end{cases}
\end{equation}
The matrix $V$ is a Vandermonde matrix with $(m+1)$-banded structure
defined for all $d\in\{0,1,\ldots,m\}$,
$k\in\{1,2,\ldots,p_d-1\}$, and
$\ell\in\{0,1,\ldots,s_m-1\}$ by the rule
\begin{equation}
\label{eq:V-entry}
V_{dk,\ell}=
\begin{cases}
1 & \text{if $d=0$;}\\
\omega_{p_d}^{k\ell} & \text{if $d\geq 1$}\,.
\end{cases}
\end{equation}
Before proceeding with the proof that $A=UV$, let us present an example
for the primes $p_1=2$, $p_2=3$, and $p_3=5$. We have
\begin{equation}
\label{eq:AUV-example}
\tiny
\begin{split}
\hspace*{1cm}&\hspace*{-1cm}\left[\begin{array}{cccccccc}
1&1&1&1&1&1&1&1\\\hline
0&1&0&1&0&1&0&1\\\hline
0&1&0&0&1&0&0&1\\
0&0&1&0&0&1&0&0\\\hline
0&1&0&0&0&0&1&0\\
0&0&1&0&0&0&0&1\\
0&0&0&1&0&0&0&0\\
0&0&0&0&1&0&0&0
\end{array}\right]_{A^{2,3,5}}
=\\[5mm]
&=
\left[\begin{array}{c|c|cc|cccc}
1&0&0&0&0&0&0&0\\\hline
\frac{1}{2}&\frac{1}{2}\omega_2^{-1\cdot 1}&0&0&0&0&0&0\\\hline
\frac{1}{3}&0&\frac{1}{3}\omega_3^{-1\cdot 1}&\frac{1}{3}\omega_3^{-1\cdot 2}&0&0&0&0\\
\frac{1}{3}&0&\frac{1}{3}\omega_3^{-2\cdot 1}&\frac{1}{3}\omega_3^{-2\cdot 2}&0&0&0&0\\\hline
\frac{1}{5}&0&0&0&\frac{1}{5}\omega_5^{-1\cdot 1}&\frac{1}{5}\omega_5^{-1\cdot 2}&\frac{1}{5}\omega_5^{-1\cdot 3}&\frac{1}{5}\omega_5^{-1\cdot 4}\\
\frac{1}{5}&0&0&0&\frac{1}{5}\omega_5^{-2\cdot 1}&\frac{1}{5}\omega_5^{-2\cdot 2}&\frac{1}{5}\omega_5^{-2\cdot 3}&\frac{1}{5}\omega_5^{-2\cdot 4}\\
\frac{1}{5}&0&0&0&\frac{1}{5}\omega_5^{-3\cdot 1}&\frac{1}{5}\omega_5^{-3\cdot 2}&\frac{1}{5}\omega_5^{-3\cdot 3}&\frac{1}{5}\omega_5^{-3\cdot 4}\\
\frac{1}{5}&0&0&0&\frac{1}{5}\omega_5^{-4\cdot 1}&\frac{1}{5}\omega_5^{-4\cdot 2}&\frac{1}{5}\omega_5^{-4\cdot 3}&\frac{1}{5}\omega_5^{-4\cdot 4}
\end{array}\right]_{U^{2,3,5}}\cdot\\[5mm]
&\ \,\,\cdot
\left[\begin{array}{cccccccc}
1&1&1&1&1&1&1&1\\\hline
\omega_2^{1\cdot 0}&\omega_2^{1\cdot 1}&\omega_2^{1\cdot 2}&\omega_2^{1\cdot 3}&\omega_2^{1\cdot 4}&\omega_2^{1\cdot 5}&\omega_2^{1\cdot 6}&\omega_2^{1\cdot 7}\\\hline
\omega_3^{1\cdot 0}&\omega_3^{1\cdot 1}&\omega_3^{1\cdot 2}&\omega_3^{1\cdot 3}&\omega_3^{1\cdot 4}&\omega_3^{1\cdot 5}&\omega_3^{1\cdot 6}&\omega_3^{1\cdot 7}\\
\omega_3^{2\cdot 0}&\omega_3^{2\cdot 1}&\omega_3^{2\cdot 2}&\omega_3^{2\cdot 3}&\omega_3^{2\cdot 4}&\omega_3^{2\cdot 5}&\omega_3^{2\cdot 6}&\omega_3^{2\cdot 7}\\\hline
\omega_5^{1\cdot 0}&\omega_5^{1\cdot 1}&\omega_5^{1\cdot 2}&\omega_5^{1\cdot 3}&\omega_5^{1\cdot 4}&\omega_5^{1\cdot 5}&\omega_5^{1\cdot 6}&\omega_5^{1\cdot 7}\\
\omega_5^{2\cdot 0}&\omega_5^{2\cdot 1}&\omega_5^{2\cdot 2}&\omega_5^{2\cdot 3}&\omega_5^{2\cdot 4}&\omega_5^{2\cdot 5}&\omega_5^{2\cdot 6}&\omega_5^{2\cdot 7}\\
\omega_5^{3\cdot 0}&\omega_5^{3\cdot 1}&\omega_5^{3\cdot 2}&\omega_5^{3\cdot 3}&\omega_5^{3\cdot 4}&\omega_5^{3\cdot 5}&\omega_5^{3\cdot 6}&\omega_5^{3\cdot 7}\\
\omega_5^{4\cdot 0}&\omega_5^{4\cdot 1}&\omega_5^{4\cdot 2}&\omega_5^{4\cdot 3}&\omega_5^{4\cdot 4}&\omega_5^{4\cdot 5}&\omega_5^{4\cdot 6}&\omega_5^{4\cdot 7}\\
\end{array}\right]_{V^{2,3,5}}\,.
\end{split}
\end{equation}
The main technical aspect of the proof that $A=UV$ is to partition
the index $\ell\in\{0,1,\ldots,s_m-1\}$ to the $m+1$ bands.
Towards this end, define for each $c\in\{0,1,\ldots,m\}$ the prefix-sum
\[
\underline{s}_c=
\begin{cases}
0 & \text{if $c=0$;}\\
1 + \sum_{\ell=1}^{c-1}\bigl(p_c-1\bigr) & \text{if $c\geq 1$.}
\end{cases}
\]
In particular,
for every $\ell\in\{0,1,\ldots,s_m-1\}$, we observe that there exist unique
$c\in\{0,1,\ldots,m\}$ and $j\in\{1,2,\ldots,p_c-1\}$ such that
\begin{equation}
\label{eq:ell-decomposition}
\ell=j-1+\underline{s}_c\,.
\end{equation}
We are now ready to show that $A=UV$.
Let $b\in\{0,1,\ldots,m\}$, $i\in\{0,1,\ldots,p_b-1\}$,
and $\ell\in\{0,1,\ldots,s_k-1\}$ be arbitrary.
Let $c\in\{0,1,\ldots,m\}$ and $j\in\{1,2,\ldots,p_c-1\}$ be uniquely
determined from $\ell$ by \eqref{eq:ell-decomposition}.
From \eqref{eq:U-entry}, \eqref{eq:V-entry},
\eqref{eq:root-power-sum}, and \eqref{eq:A-entry} we observe that
\[
\small
\begin{split}
\,\,\,&\!\!\!\sum_{d=0}^m\sum_{k=0}^{p_d-1} U_{bi,dk}V_{dk,\ell}
=\\[2mm]
&=
\begin{cases}
1 & \!\!\!\text{if $b=0$};\\[1mm]
\frac{1}{p_b}\bigl(1+\sum_{k=1}^{p_b-1}\omega_{p_b}^{-ik+k(j-1+\underline{s}_b)}\bigr)=\frac{1}{p_b}\sum_{k=0}^{p_b-1}\omega_{p_b}^{k(j-i-1+\underline{s}_b)}=1 & \!\!\!\text{if $b\geq 1$ and $i\equiv j-1+\underline{s}_b=\ell\!\!\!\!\pmod{p_b}$};\\[1mm]
\frac{1}{p_b}\bigl(1+\sum_{k=1}^{p_b-1}\omega_{p_b}^{-ik+k(j-1+\underline{s}_b)}\bigr)=\frac{1}{p_b}\sum_{k=0}^{p_b-1}\omega_{p_b}^{k(j-i-1+\underline{s}_b)}=0 & \!\!\!\text{if $b\geq 1$ and $i\not\equiv j-1+\underline{s}_b=\ell\!\!\!\!\pmod{p_b}$.}
\end{cases}\\[2mm]
&=A_{bi,\ell}\,.
\end{split}
\]
Thus, $A=UV$ holds.
It remains to show that both matrices $U$ and $V$ have nonzero determinant
over the complex numbers. Starting with the Vandermonde matrix $V$,
let $\nu_0=1$ and $\nu_\ell=\omega_{p_c}^j$ for $\ell\in\{1,2,\ldots,s_m-1\}$,
where $c\in\{0,1,\ldots,m\}$ and $j\in\{1,2,\ldots,p_c-1\}$ are uniquely
determined from $\ell$ by \eqref{eq:ell-decomposition}. In particular,
we observe that $V$ is a Vandermonde matrix with $D=s_m-1$ and
\[
V=
\left[\begin{array}{cccc}
\nu_0^0 & \nu_0^1 & \cdots & \nu_0^D\\
\nu_1^0 & \nu_1^1 & \cdots & \nu_1^D\\
\vdots & \vdots & & \vdots \\
\nu_D^0 & \nu_D^1 & \cdots & \nu_D^D\\
\end{array}\right]\,.
\]
The Vandermonde determinant formula thus gives
\[
\det V=\sum_{0\leq k<\ell\leq D}(\nu_\ell-\nu_k)\,.
\]
Furthermore, this determinant is nonzero because $p_1,p_2,\ldots,p_m$ are
distinct primes and thus $\nu_0,\nu_1,\ldots,\nu_D$ are distinct.
Next, let us consider the matrix $U$ defined by \eqref{eq:U-entry}.
At this point it may be useful to revisit the structure of $U$ via
the example \eqref{eq:AUV-example}.
We observe that the block-diagonal of $U$ with
$b=c\geq 1$ consists of matrices that each decompose into the product of a
$(p_b-1)\times (p_b-1)$ diagonal matrix
with diagonal entries $\frac{1}{p_b}\omega_{p_b}^{-i}$ for
$i\in\{1,2,\ldots,p_b-1\}$ and a $(p_b-1)\times (p_b-1)$ Vandermonde matrix
with a nonzero determinant since $\omega_{p_b}^{-i}$
for $i\in\{1,2,\ldots,p_b-1\}$ are distinct. Thus, since the determinant
of $U$ is the product of the determinants of the block-matrices on
the diagonal, each of which is nonzero, the determinant of $U$ is nonzero.
It follows that $A$ is invertible and thus given $F$ we can
solve for $f$ via \eqref{eq:FAf}.
This completes the proof of Lemma~\ref{lem:reconstruction}. $\qed$
\section{Counting Pairs of Zero-One Vectors by Inner Product}
\label{sect:main-proof}
This section documents our main algorithm and proves Theorem~\ref{thm:main}.
Let $\kappa$ be a parameter that satisfies, with foresight,
\begin{equation}
\label{eq:kappa-up}
4\leq\kappa\leq\frac{\log n}{(\log\log n)^4}\,.
\end{equation}
Let $a^{(1)},a^{(2)},\ldots,a^{(n)}\in\{0,1\}^d$ and
$b^{(1)},b^{(2)},\ldots,b^{(n)}\in\{0,1\}^d$ be given as input
with $d\leq\kappa\log n$.
We want to compute for each $t\in\{0,1,\ldots,d\}$ the count
\[
f_t=|\{(i,j)\in\{1,2,\ldots,n\}^2:\ensuremath{\langle} a^{(i)},b^{(j)}\ensuremath{\rangle}=t\}|\,.
\]
Our high-level approach will be to use Lemma~\ref{lem:reconstruction} and
\eqref{eq:FAf} to solve for the counts $f_0,f_1,\ldots,f_d$ using as
input counts that have been sum-aggregated by prime residue.
More precisely, we will work with prime moduli $p_1,p_2,\ldots,p_m$
and develop an algorithm that computes, for given
further input $p\in\{p_1,p_2,\ldots,p_m\}$ and $r\in\{0,1,\ldots,p-1\}$,
the sum-aggregated count
\[
F_{pr}=|\{(i,j)\in\{1,2,\ldots,n\}^2:\ensuremath{\langle} a^{(i)},b^{(j)}\ensuremath{\rangle}\equiv r\!\!\!\pmod p\}|\,.
\]
The detailed choices for $m$ and the primes $p_1,p_2,\ldots,p_m$ will be
presented later.
\subsection{The Residue-Indicator Polynomial}
Assume $p$ and $r$ have been given. We will rely on the polynomial
method, and accordingly we first build a standard polynomial that
indicates the residue $r$ modulo $p$ in a pair of vectors.
Let $x=(x_1,x_2,\ldots,x_d)$ and $y=(y_1,y_2,\ldots,y_d)$ be two
vectors of indeterminates. By Fermat's little theorem,
the $2d$-indeterminate polynomial
\begin{equation}
\label{eq:indicator-poly}
G_{p,r}\bigl(x,y\bigr)
=1-\biggl(\sum_{k=1}^d x_ky_k-r\biggr)^{p-1}
\end{equation}
satisfies for all $i,j\in\{1,2,\ldots,n\}$ the indicator property
\begin{equation}
\label{eq:indicator-property}
G_{p,r}\bigl(a^{(i)},b^{(j)}\bigr)
\equiv
\begin{cases}
1\!\!\!\pmod p & \text{if $\ensuremath{\langle} a^{(i)},b^{(j)}\ensuremath{\rangle}\equiv r\!\!\!\pmod p$;}\\
0\!\!\!\pmod p & \text{if $\ensuremath{\langle} a^{(i)},b^{(j)}\ensuremath{\rangle}\not\equiv r\!\!\!\pmod p$}.
\end{cases}
\end{equation}
We observe that $G_{p,r}$ has degree $2p-2$.
\subsection{Modulus Amplification for Zero-One Residues}
To enable taking the sum of a large number of indicators, we make use
of the {\em modulus amplifying polynomials} of Beigel and
Tarui~\cite{BeigelT1994}.
\begin{Thm}[Modulus amplification; Beigel and Tarui~\cite{BeigelT1994}]
\label{thm:amp}
For $h\in\mathbb{Z}_{\geq 1}$, define the polynomial
\begin{equation}
\label{eq:modamp-poly}
A_h(z)=1-(1-z)^h\sum_{j=0}^{h-1}\binom{h+j-1}{j}z^j\,.
\end{equation}
Then, for all $m\in\mathbb{Z}_{\geq 2}$ and $s\in\mathbb{Z}$, we have
\begin{enumerate}
\item[(i)]
$s\equiv 0\pmod m$ implies $A_h(s)\equiv 0\pmod{m^h}$, and
\item[(ii)]
$s\equiv 1\pmod m$ implies $A_h(s)\equiv 1\pmod{m^h}$.
\end{enumerate}
\end{Thm}
We observe that $A_h$ has degree $2h-1$.
Composing \eqref{eq:modamp-poly} and \eqref{eq:indicator-poly}, we obtain the
amplified residue-indicator polynomial
\begin{equation}
\label{eq:amp-indicator-poly}
G_{p,r}^h(x,y)=A_h\bigl(G_{p,r}(x,y)\bigr)\,.
\end{equation}
From \eqref{eq:indicator-property} and Theorem~\ref{thm:amp}, we observe
the amplified indicator property
\begin{equation}
\label{eq:amp-indicator-property}
G_{p,r}^h\bigl(a^{(i)},b^{(j)}\bigr)
\equiv
\begin{cases}
1\!\!\!\pmod{p^h} & \text{if $\ensuremath{\langle} a^{(i)},b^{(j)}\ensuremath{\rangle}\equiv r\!\!\!\pmod p$;}\\
0\!\!\!\pmod{p^h} & \text{if $\ensuremath{\langle} a^{(i)},b^{(j)}\ensuremath{\rangle}\not\equiv r\!\!\!\pmod p$}.
\end{cases}
\end{equation}
Furthermore, we observe that $G_{p,r}^h$ has degree $(2h-1)(2p-2)$.
\subsection{Multilinear Reduct and Bounding the Number of Monomials}
For a nonnegative integer $e$, define $\underline{e}=0$ if $e=0$
and $\underline{e}=1$ if $e\geq 1$. For a monomial
$x_1^{e_1}x_2^{e_2}\cdots x_d^{e_d}y_1^{f_1}y_2^{f_2}\cdots y_d^{f_d}$,
define the {\em multilinear reduct} by
\[
\underline{x_1^{e_1}x_2^{e_2}\cdots x_d^{e_d}y_1^{f_1}y_2^{f_2}\cdots y_d^{f_d}}
=
x_1^{\underline{e}_1}x_2^{\underline{e}_2}\cdots x_d^{\underline{e}_d}y_1^{\underline{f}_1}y_2^{\underline{f}_2}\cdots y_d^{\underline{f}_d}\,.
\]
For a polynomial $Q(x,y)$, define the multilinear reduct $\underline{Q}(x,y)$
by taking the multilinear reduct of each monomial $Q(x,y)$ and simplifying.
Since $a^{(i)}$ and $b^{(j)}$ are $\{0,1\}$-valued vectors, over the
integers we have
\begin{equation}
\label{eq:multilin-reduct}
Q\bigl(a^{(i)},b^{(j)}\bigr)=\underline{Q}\bigl(a^{(i)},b^{(j)}\bigr)\,.
\end{equation}
Furthermore, if $Q$ has degree $D$, then $\underline{Q}$ has at most
$\sum_{j=0}^D\binom{2d}{j}$ monomials. In particular, we observe that
$\underline{G}_{p,r}^h$ has at most $\sum_{j=0}^{4hp}\binom{2d}{j}$ monomials.
\subsection{Split-Monomial Form of the Multilinear Reduct}
Suppose that the multilinear reduct $\underline{G}_{p,r}^h(x,y)$ has
exactly $M$ monomials with the representation
\begin{equation}
\label{eq:monomial-rep}
\underline{G}_{p,r}^h(x,y)=
\sum_{k=1}^M
\gamma^{(k)}\,
x_1^{e_1^{(k)}}\!\!x_2^{e_2^{(k)}}\!\!\cdots x_d^{e_d^{(k)}}\!\!
y_1^{f_1^{(k)}}\!\!y_2^{f_2^{(k)}}\!\!\cdots y_d^{f_d^{(k)}}\,.
\end{equation}
For $I,J\subseteq\{1,2,\ldots,n\}$ and $k\in\{1,2,\ldots,M\}$, define
\begin{equation}
\label{eq:lr}
L_{I,k}=
\sum_{i\in I}
\bigl(a_1^{(i)}\bigr)^{e_1^{(k)}}\!\!\bigl(a_2^{(i)}\bigr)^{e_2^{(k)}}\!\!\cdots \bigl(a_d^{(i)}\bigr)^{e_d^{(k)}}\gamma^{(k)}\,,\qquad
R_{J,k}=
\sum_{j\in J}
\bigl(b_1^{(j)}\bigr)^{f_1^{(k)}}\!\!\bigl(b_2^{(j)}\bigr)^{f_2^{(k)}}\!\!\cdots \bigl(b_d^{(j)}\bigr)^{f_d^{(k)}}\,.
\end{equation}
From \eqref{eq:lr}, \eqref{eq:monomial-rep}, \eqref{eq:multilin-reduct},
and \eqref{eq:amp-indicator-property}, we have
\begin{equation}
\label{eq:mm-count-identity}
\sum_{k=1}^M L_{I,k}R_{J,k}
=\sum_{i\in I}\sum_{j\in J}
\underline{G}_{p,r}^h\bigl(a^{(i)},b^{(j)}\bigr)
\equiv
\big|\bigl\{(i,j)\in I\times J:\ensuremath{\langle} a^{(i)},b^{(j)}\ensuremath{\rangle}\equiv r\!\!\!\pmod p\bigr\}\big|\!\!\!
\pmod{p^h}\,.
\end{equation}
In particular, assuming that
$|I||J|\leq p^h-1$, from \eqref{eq:mm-count-identity}
it follows that
$\sum_{k=1}^M L_{I,k}R_{J,k}$ computed modulo $p^h$ recovers
the number of pairs $(i,j)\in I\times J$ with
$\ensuremath{\langle} a^{(i)},b^{(j)}\ensuremath{\rangle}\equiv r\pmod p$.
We now move from deriving the polynomial and its properties to
describing the algorithm.
\subsection{Algorithm for the Prime-Residue Count}
The algorithm will rely on \eqref{eq:mm-count-identity} via
fast rectangular matrix multiplication to count the number of
pairs $(i,j)\in\{1,2,\ldots,n\}^2$ that satisfy
$\ensuremath{\langle} a^{(i)},b^{(j)}\ensuremath{\rangle}\equiv r\pmod p$.
The algorithm first computes the explicit $M$-monomial representation
of the polynomial $\underline{G}^h_{p,r}$ in~\eqref{eq:monomial-rep}.
More precisely, the algorithm evaluates
\eqref{eq:indicator-poly}, \eqref{eq:modamp-poly}, and
\eqref{eq:amp-indicator-poly} in explicit monomial representation, taking
multilinear reducts with respect to the variables
$x_1,x_2,\ldots,x_d,y_1,y_2,\ldots,y_d$ whenever possible.
This results in the set
\begin{equation}
\label{eq:monomial-list}
\{(k,\gamma^{(k)},
e_1^{(k)},e_2^{(k)},\ldots,e_d^{(k)},
f_1^{(k)},f_2^{(k)},\ldots,f_d^{(k)}):k\in\{1,2,\ldots,M\}\}\,.
\end{equation}
Next, the algorithm constructs two rectangular matrices $S$ and $T$, with
the objective of making use of the following algorithm of
Coppersmith~\cite{Coppersmith1982}.
\begin{Thm}[Coppersmith~\cite{Coppersmith1982}]
\label{thm:coppersmith}
Given an $N\times \lfloor N^{0.17}\rfloor$ matrix $S$
and an $\lfloor N^{0.17}\rfloor\times N$ matrix $T$ as input,
the matrix product $ST$ over the integers can be computed in $O(N^2\log^2 N)$
arithmetic operations.
\end{Thm}
Towards this end, let $g$ be a positive integer whose value we will fix
later. Introduce two set partitions of $\{1,2,\ldots,n\}$ with cells
\[
I_1,I_2,\ldots,I_{\lceil n/g\rceil}\subseteq\{1,2,\ldots,n\}\qquad
\text{and}\qquad
J_1,J_2,\ldots,J_{\lceil n/g\rceil}\subseteq\{1,2,\ldots,n\}\,,
\]
respectively, so that
$|I_u|=g$ and $|I_v|=g$ for $u,v\in\{1,2,\ldots,\lfloor n/g\rfloor\}$.
Indeed, we thus have
\[
|I_u||J_v|\leq g^2
\]
for all $u,v\in\{1,2,\ldots,\lceil n/g\rceil\}$,
so \eqref{eq:mm-count-identity} applied to $I_u$ and $J_v$
modulo $p^h$ recovers the number of pairs $(i,j)\in I_u\times J_v$ with
$\ensuremath{\langle} a^{(i)},b^{(j)}\ensuremath{\rangle}\equiv r\pmod p$, assuming that $g^2\leq p^h-1$,
which will be justified by our eventual choice of $g$.
Now let $N=\lceil n/g\rceil$ and define the $N\times M$
and $M\times N$ matrices $S$ and $T$ by setting,
for $u,v\in\{1,2,\ldots,\lceil n/g\rceil\}$ and $k\in\{1,2,\ldots,M\}$,
\[
S_{uk}=L_{I_u,k}\qquad\text{and}\qquad T_{kv}=R_{I_v,k}\,.
\]
Concretely, the algorithm computes $S$ and $T$ from the given input
one entry at a time using the computed monomial
list \eqref{eq:monomial-list}
and the formulas \eqref{eq:lr} for $I=I_u$ and $J=J_v$
for each $u,v\in\{1,2,\ldots,\lceil n/g\rceil\}$ and $k=1,2,\ldots,M$.
The algorithm then multiplies $S$ and $T$ to obtain the product
matrix $ST$ modulo $p^h$, where we assume that each entry of $ST$ is
reduced to $\{0,1,\ldots,p^h-1\}$. Finally, the algorithm outputs the sum
\[
F_{pr}=\sum_{u=1}^{\lceil n/g\rceil}\sum_{v=1}^{\lceil n/g\rceil}(ST)_{uv}\,.
\]
\subsection{Parameterizing the Algorithm}
Let us now start parameterizing the algorithm. First,
to apply the algorithm in Theorem~\ref{thm:coppersmith} to the matrices
$S$ and $T$, we need $M\leq N^{0.17}=\lceil n/g\rceil^{0.17}$.
Subject to the assumption $g\leq n^{0.1}$---to be justified later---it
will be sufficient to show that $M\leq n^{0.15}$.
We recall that $M\leq\sum_{j=0}^{4hp}\binom{2d}{j}$ and $d\leq \kappa\log n$.
With foresight, let us set
\begin{equation}
\label{eq:beta-kappa}
\beta_\kappa=\frac{K}{\log\kappa}\,,
\end{equation}
where $K>0$ is a small constant that will be fixed later.
In particular, since $\kappa\geq 4$, we have the upper bound
\begin{equation}
\label{eq:base-decay}
\beta_\kappa\log\frac{\kappa}{\beta_\kappa}=K-\frac{K\log K}{\kappa}+\frac{K\log\log\kappa}{\kappa}\leq \frac{5K}{4}
\end{equation}
which we can make an arbitrarily small and positive by choosing
a small enough $K$.
Let us assume---to be justified later---that $p=o(\beta_\kappa\log n)$.
Taking
\begin{equation}
\label{eq:h}
h=\biggl\lfloor\beta_\kappa\frac{\log n}{p}\biggr\rfloor
\end{equation}
we have, for all large enough~$n$,
\begin{equation}
\label{eq:m-bound}
\begin{split}
M&\leq
\sum_{j=0}^{4hp}\binom{2d}{j}
\leq
4hp\binom{2d}{4hp}
\leq
4hp\biggl(\frac{2ed}{4hp}\biggr)^{4hp}\\
&\leq
4\biggl(\beta_\kappa\frac{\log n}{p}+1\biggr)p\biggl(\frac{2e\kappa\log n}{4\bigl(\beta_\kappa\frac{\log n}{p}-1\bigr)p}\biggr)^{4\bigl(\beta_\kappa\frac{\log n}{p}+1\bigr)p}\\
&=
4\bigl(\beta_\kappa\log n+p\bigr)\biggl(\frac{2e\kappa\log n}{4\bigl(\beta_\kappa\log n-p\bigr)}\biggr)^{4(\beta_\kappa\log n+p)}\\
&\leq
\bigl(5\beta_\kappa\log n\bigr)\biggl(\frac{2e\kappa}{3\beta_\kappa}\biggr)^{5\beta_\kappa\log n}\\
&\leq
n^{0.15}\,,
\end{split}
\end{equation}
where the last inequality follows by \eqref{eq:base-decay} and choosing $K$
small enough.
Thus, Theorem~\ref{thm:coppersmith} applies, subject to the assumptions
$g\leq n^{0.1}$, $g^2\leq p^h-1$, and $p=o(\beta_\kappa\log n)$, which still need to
be established. Before this, we digress to further preliminaries to enable
reconstruction.
\subsection{Preliminaries on Asymptotics of Primes}
In what follows let us write $p_j$ for the $j$th prime number with $j=1,2,\ldots$;
that is, $p_1=2$, $p_2=3$, $p_3=5$, and so forth.
Asymptotically, from the Prime Number Theorem we have $p_m\sim m\ln m$
(e.g.~Rosser and Schoenfeld~\cite{RosserS1962}), and the sum of the
first $m$ primes satisfies
$\sum_{j=1}^m p_j\sim \frac{1}{2}m^2\ln m$
(cf.~Bach and Shallit~\cite{BachS1996}), where we write
$f(m)\sim g(m)$ if $\lim_{m\rightarrow\infty} \frac{f(m)}{g(m)}=1$.
When evaluated for the first $m$ primes, the reconstruction
parameter~\eqref{eq:sm} thus satisfies
\begin{equation}
\label{eq:sm-growth}
s_m=1+\sum_{j=1}^m(p_j-1)\sim \frac{p_m^2}{2\ln p_m}\,.
\end{equation}
We are now ready to continue parameterization of the algorithm.
\subsection{Further Parameterization of the Algorithm}
Let $m$ be a positive integer whose value will be fixed shortly.
The algorithm will work with $p_1,p_2,\ldots,p_m$, the first $m$ prime numbers.
To reconstruct inner products of length-$d$ zero-one vectors over the
integers, we need $d+1\leq s_m$, which for $d\leq \kappa\log n$
and \eqref{eq:sm-growth} means
\[
\frac{p_m^2}{2\ln p_m}\sim \kappa\log n\,.
\]
From Bertrand's postulate it thus follows that choosing the least $m$ so that
\begin{equation}
\label{eq:pm-choice}
2\sqrt{\kappa(\ln n)\ln\ln n}\leq p_m\leq 4\sqrt{\kappa(\ln n)\ln\ln n}
\end{equation}
implies that we have $d+1\leq s_m$ for all large enough $n$
and thus reconstruction is feasible. The choice \eqref{eq:pm-choice}
also justifies our ealier assumption made in
the context of \eqref{eq:h} and \eqref{eq:m-bound}
that $p_j=o(\beta_\kappa\log n)$ for all
$j\in\{1,2,\ldots,m\}$; indeed, from \eqref{eq:kappa-up}
and \eqref{eq:beta-kappa}, we have
\[
\begin{split}
\beta_\kappa\log n
&=\frac{K\log n}{\log\kappa}
\end{split}
\]
and thus from \eqref{eq:kappa-up} and \eqref{eq:pm-choice} we observe that
\[
\frac{p_j}{\beta_\kappa\log n}
\leq
\frac{4\kappa^{1/2}(\log\kappa)(\ln n)^{1/2}(\ln\ln n)^{1/2}}{K\log n}=o(1)\,.
\]
Let us next choose the parameter $g$. Using $p_j=o(\beta_\kappa\log n)$ again,
we have
\[
p_j^{h_j}
=p_j^{\bigl\lfloor\beta_\kappa\frac{\log n}{p_j}\bigr\rfloor}
\geq p_j^{\beta_\kappa\frac{\log n}{p_j}-1}
\geq p_j^{\beta_\kappa\frac{\log n}{2p_j}}
=2^{\beta_\kappa\frac{\log n}{2p_j}\log p_j}
=n^{\beta_\kappa\frac{\log p_j}{2p_j}}\,.
\]
Since $p_1<p_2<\cdots<p_m$, for $j\in\{1,2,\ldots,m\}$ thus
\[
p_j^{h_j} \geq n^{\beta_\kappa\frac{\log p_m}{2p_m}}\,.
\]
It follows that choosing
\begin{equation}
\label{eq:g-choice}
g
=\biggl\lfloor \sqrt{n^{\beta_\kappa\frac{\log p_m}{2p_m}}-1}\biggr\rfloor
\end{equation}
justifies our assumption $g^2\leq p_j^{h_j}-1$ for $j\in\{1,2,\ldots,m\}$.
The final assumption $g\leq n^{0.1}$ is justified by observing that
$\frac{\log p_m}{2p_m}$ is a decreasing function of $m$ and
observing that $\beta_\kappa=o(1)$ by \eqref{eq:kappa-up} and
\eqref{eq:beta-kappa}.
The algorithm is now parameterized. Let us proceed to analyse its running
time.
\subsection{Running Time Analysis}
First, let us seek control on $N$ as a function of $n$.
From \eqref{eq:pm-choice} and \eqref{eq:g-choice}, we have
\[
g\geq
\sqrt{n^{\beta_\kappa\frac{2\log 2+\log\kappa+\log\ln n+\log\ln\ln n}{16\sqrt{\kappa\ln n\ln\ln n}}}-1}-1\,.
\]
This together with \eqref{eq:kappa-up} gives us the crude lower bound
\[
g=\exp\biggl(\Omega\biggl(\beta_\kappa\sqrt{\frac{(\ln n)\ln\ln n}{\kappa}}\biggr)\biggr)\,.
\]
We thus have
\[
N^2=\lceil n/g\rceil^2=n^{2-\Omega\bigl(\beta_\kappa\sqrt{\frac{\ln\ln n}{\kappa\ln n}}\bigr)}\,.
\]
Recalling \eqref{eq:m-bound}, we observe that
the time to compute the $M$-monomial list \eqref{eq:monomial-list} can
be bounded by $n^{0.31}$ because the algorithm is careful to take multilinear
reducts and thus at no stage of evaluating
\eqref{eq:indicator-poly}, \eqref{eq:modamp-poly}, and
\eqref{eq:amp-indicator-poly} the number of monomials increases above
$(n^{0.15})^2=n^{0.30}$.
Since
\[
\log p_j^{h_j}=h_j\log p_j
=\biggl\lfloor\beta_\kappa\frac{\log n}{p_j}\biggr\rfloor\log p_j=O(\log n)\,,
\]
the arithmetic over the integers and modulo $p_j^{h_j}$ for each
$j=1,2,\ldots,m$ runs in time polylogarithmic in $n$ for each arithmetic
operation executed by the algorithm.
Because the algorithm in Theorem~\ref{thm:coppersmith} runs in
$O(N^2\log^2 N)$ arithmetic operations, we observe that the polylogarithmic
terms are subsumed by the asymptotic notation and
the entire algorithm for computing $F_{pr}$
for given $p\in\{p_1,p_2,\ldots,p_m\}$ and $r\in\{0,1,\ldots,p-1\}$
runs in time
\begin{equation}
\label{eq:runtime}
n^{2-\Omega\bigl(\beta_\kappa\sqrt{\frac{\log\log n}{\kappa\log n}}\bigr)}
=n^{2-\Omega\bigl(\sqrt{\frac{\log\log n}{\kappa(\log\kappa)^2\log n}}\bigr)}\,.
\end{equation}
From \eqref{eq:pm-choice} we observe that the required repeats for different
$p$ and $r$ result in multiplicative polylogarithmic terms in $n$ and
are similarly subsumed to result in total running time of the
form~\eqref{eq:runtime}. This completes the proof of Theorem~\ref{thm:main}.
$\qed$
\section{A Faster Randomized Algorithm for \#\textsc{InnerProduct}}
\label{sect:proof-of-randomized-algorithm}
This section sketches a proof for Theorem~\ref{thm:random}.
We follow the algorithm outlined in Alman and Williams~\cite{AlmanW2015}. We note that by their Theorem~1.2, there are probabilistic polynomials over any field with error $\epsilon$ of degree $O(\sqrt{n\log(1/\epsilon)})$. In their Theorem~4.2, they have a probabilistic OR-construction that takes the disjunction of a random set of $s^2$ pairs of vector inner products
as
\[
q(x_1,y_1,x_2,y_2,\ldots,x_s,y_s)=1+\prod_{k=1}^2 \bigl(1+\sum_{(i,j)\in R_k} \bigl(1+p(x_{i,1}+y_{i,1},x_{i,2}+y_{j,2},\ldots,x_{i,s}+y_{j,s})\bigr) \bigr)\,,
\]
where $p$ is a probabilistic threshold polynomial over $\mathbb{F}_2$ of error $\epsilon=s^{-3}$, and $R_k\subseteq [s]^2$ for $k=1,2$ are sieve subsets drawn uniformly at random.
This construction can be used to detect w.h.p. if there is a pair in the $s^2$-sized batch whose difference Hamming weight is less than the threshold.
By repeated computations with new $p$'s and $R_k$'s, a majority vote for the batch can be chosen as the correct answer, again w.h.p. for all batches.
We implement the following change of $q$ to get an \#\textsc{InnerProduct} algorithm. We take $p$ to be a probabilistic polynomial of error $\epsilon=s^{-3}$ for the symmetric function $\iv{\sum_{i=1}^n z_i=t}$, over a field of characteristic $>s^2$. We then construct $q$ as
\begin{equation}
\label{eq:newq}
q(x_1,y_1,x_2,y_2,\ldots,x_s,y_s)=\sum_{(i,j)\in[s]^2} p(x_{i,1}y_{i,1},x_{i,2}y_{j,2},\ldots,x_{i,s}y_{j,s})\,.
\end{equation}
Since the characteristic of the field is large enough, \eqref{eq:newq} is equal to the number of pairs in the $s^2$-sized batch that has inner product equal to $t$ with probability at least
$1-s^2\epsilon\geq 1-\frac{1}{s}$, a similar bound on the probability as in Theorem~4.2. Also, the degree of the polynomials is only a factor $2$ larger. As with the original algorithm,
if we repeat this enough times and take the majority in each batch, we get the correct number of pairs with $t$ as inner product in all batches. By summing these final majority numbers over the integers, we obtain the output. We note that the parameters of the error and the degree has only changed by a constant, and hence that all calculations of the running time and the error bound of the original algorithm carries through also for our modification of the algorithm. This completes the proof sketch. $\qed$
\section{A Lower Bound for \#\textsc{InnerProduct} via Zero-One Permanents}
\label{sect:perm}
This section proves Theorem~\ref{thm:perm-to-exactip};
the proof of Theorem~\ref{thm:perm-to-ov} is presented in
Appendix~\ref{sect:perm-cont}.
Throughout this section we let
$M$ be an $n\times n$ matrix with entries $m_{ij}\in\{0,1\}$
for $i,j\in\{1,2,\ldots,n\}$.
For convenience, let us write $[n]=\{1,2,\ldots,n\}$.
Recalling Ryser's formula, we have
\begin{equation}
\label{eq:ryser-recalled}
\ensuremath{\operatorname{per}} M=(-1)^n\sum_{S\subseteq[n]}(-1)^{|S|}\prod_{i\in[n]}\sum_{j\in S}m_{ij}\,.
\end{equation}
\subsection{First Reduction: Chinese Remaindering}
Since it is immediate that $0\leq\ensuremath{\operatorname{per}} M\leq n!$, it suffices to compute
the permanent modulo small primes $p$ and then assemble the result over
the integers via the Chinese Remainder Theorem. Let us first state and prove
a crude upper bound on the size of the primes needed.
For a positive integer $m$, let us write $m\#$ for the product of all
prime numbers at most $m$.
\begin{Lem}
\label{lem:factorial-upper-bound-primorial}
For all sufficiently large $n$, we have $n!\leq (n\ln n)\#$.
\end{Lem}
\begin{Proof}
Recall that for a positive integer $m$ we write write $m\#$ for the product of
all
prime numbers at most $m$. For $m\geq 563$, we have
(cf.~Rosser and Schoenfeld~\cite{RosserS1962})
\[
\ln m\# > m\biggl(1-\frac{1}{2\ln m}\biggr)\,.
\]
For the factorial function, for $n\geq 1$,
we have (cf.~Robbins~\cite{Robbins1955})
\[
n! = \sqrt{2\pi n}\biggl(\frac{n}{e}\biggr)^ne^{\alpha_n}\quad
\text{with}\quad\frac{1}{12n+1}<\alpha_n<\frac{1}{12n}\,,
\]
which gives us the comparatively crude upper bound, for $n\geq 1$,
\[
\ln n! < \biggl(n+\frac{1}{2}\biggr)\ln n-n+1\,.
\]
We want $\ln n!<\ln m\#$. Accordingly, it suffices to have $m\geq 563$ and
\[
\biggl(n+\frac{1}{2}\biggr)\ln n-n+1<m\biggl(1-\frac{1}{2\ln m}\biggr)\,.
\]
It is immediate that $m\geq n\ln n$ suffices for $m\geq 563$, which completes the proof.
\end{Proof}
Thus, it suffices to work with all primes $p$ with $p\leq n\ln n$ in what follows.
\subsection{A Reduction from Zero-One Permanent to \#\textsc{InnerProduct}}
This section starts our work towards Theorem~\ref{thm:perm-to-exactip}
without yet parameterizing the reduction in detail.
Let a prime $2\leq p\leq n\ln n$ be given.
We seek to compute $\ensuremath{\operatorname{per}} M$ modulo $p$.
Fix a primitive root $g\in\{1,2,\ldots,p-1\}$ modulo $p$. For
an integer $a$ with $a\not\equiv 0\pmod p$, let us write
$\ensuremath{\operatorname{dlog}}_{p,g} a$ for the {\em discrete logarithm} of $a$ relative
to $g$ modulo $p$.
That is, $\ensuremath{\operatorname{dlog}}_{p,g} a$ is the unique integer in $\{0,1,\ldots,p-2\}$ that
satisfies $g^{\ensuremath{\operatorname{dlog}}_{p,g} a}\equiv a\pmod p$.
Working modulo $p$ and collecting the outer sum in \eqref{eq:ryser-recalled}
by the sign $\sigma\in\{-1,1\}$ and the {\em nonzero} products by their
discrete logarithm, we have
\[
\ensuremath{\operatorname{per}} M\equiv (-1)^n\sum_{e=0}^{p-2}\,
g^e\bigl(w_{1}^{(e)}-w_{-1}^{(e)}\bigr)
\pmod p\,,
\]
where
\[
w_{\sigma}^{(e)}
=\bigg|\biggl\{S\subseteq[n]\,:\,
(-1)^{|S|}=\sigma\ \ \text{and}\
\ensuremath{\operatorname{dlog}}_{p,g}\prod_{i\in [n]}\sum_{j\in S} m_{ij}\equiv e\!\!\pmod{p-1}
\biggr\}\bigg|
\]
for $\sigma\in\{-1,1\}$ and $e\in\{0,1,\ldots,p-2\}$.
Thus, to compute $\ensuremath{\operatorname{per}} M$ modulo $p$ it suffices to compute the coefficients
$w_{\sigma}^{(e)}$.
Towards this end, suppose that $n\geq 4$ is even and let
\[
L=\{1,2,\ldots,n/2\}\qquad\text{and}\qquad
R=\{n/2,n/2+1,\ldots,n\}\,.
\]
For $\sigma_L,\sigma_R\in\{1,-1\}$, let
\[
w_{\sigma_L,\sigma_R}^{(e)}
=\bigg|\biggl\{\!S\subseteq[n]:
(-1)^{|S\cap L|}=\sigma_L\,,\,
(-1)^{|S\cap R|}=\sigma_R\ \ \text{and}\,
\ensuremath{\operatorname{dlog}}_{p,g}\prod_{i\in [n]}\sum_{j\in S} m_{ij}\equiv e\!\!\!\!\pmod{p-1}
\!\biggr\}\bigg|
\]
Clearly
$w_{\sigma}^{(e)}=\sum_{\substack{\sigma_L,\sigma_R\in\{-1,1\}\\\sigma_L\sigma_R=\sigma}}w_{\sigma_L,\sigma_R}^{(e)}$,
so it suffices to focus on computing $w_{\sigma_L,\sigma_R}^{(e)}$ in what
follows.
Define the set families
\[
\mathcal{L}_{\sigma_L}=\bigl\{A\subseteq L:(-1)^{|A|}=\sigma_L\bigr\}
\qquad
\text{and}
\qquad
\mathcal{R}_{\sigma_R}=\bigl\{B\subseteq R:(-1)^{|B|}=\sigma_R\bigr\}
\]
with $|\mathcal{L}_{\sigma_L}|=|\mathcal{R}_{\sigma_R}|=2^{n/2-1}$.
Next we will define two families of length-$d$ zero-one vectors whose pair
counts by inner product will enable us to recover the coefficients
$w_{\sigma_L,\sigma_R}^{(e)}$. The structure of the vectors will be slightly
elaborate, so let us first define an index set $D$ for indexing the $|D|=d$
dimensions.
Let
\[
\begin{split}
D=\bigl\{&(i,\ell,r,k)\in [n]\times\{0,1,\ldots,p-1\}\times\{0,1,\ldots,p-1\}\times [np]:\\
&\qquad\ell+r\not\equiv 0\!\!\!\pmod p\ \text{implies} \ k\leq\ensuremath{\operatorname{dlog}}_{p,g}\bigl(\ell+r\bigr)\bigr\}\,.
\end{split}
\]
We have
\[
d=n^2p^2+np(p-1)(p-2)/2<n^4(\ln n)^3\,.
\]
For $A\in\mathcal{L}_{\sigma_L}$ and $B\in\mathcal{R}_{\sigma_R}$,
define the vectors $\lambda(A)\in\{0,1\}^D$ and $\rho(B)\in\{0,1\}^D$
for all $(i,\ell,r,k)\in D$ by the rules
\begin{equation}
\label{eq:lambda-rho}
\lambda(A)_{i\ell rk}=
\begin{cases}
1 & \!\!\!\text{if $\ell\equiv\sum_{j\in A}m_{ij}\!\!\!\pmod p$;}\\
0 & \!\!\!\text{otherwise;}
\end{cases}\ \ \text{and}\ \
\rho(B)_{i\ell rk}=
\begin{cases}
1 & \!\!\!\text{if $r\equiv\sum_{j\in B}m_{ij}\!\!\!\pmod p$;}\\
0 & \!\!\!\text{otherwise.}
\end{cases}
\end{equation}
To study the inner product $\ensuremath{\langle}\lambda(A),\rho(B)\ensuremath{\rangle}$ it will be convenient
to work with Iverson's bracket notation. Namely, for a logical proposition $P$,
let
\[
\iv{P}=\begin{cases}
1 & \text{if $P$ is true};\\
0 & \text{if $P$ is false}.
\end{cases}
\]
Over the integers, from \eqref{eq:lambda-rho} we now have
\begin{equation}
\label{eq:lambda-rho-ip}
\begin{split}
\ensuremath{\langle}\lambda(A),\rho(B)\ensuremath{\rangle}
&=
\sum_{(i,\ell,r,k)\in D}
\lambda(A)_{i\ell rk}
\rho(B)_{i\ell rk}\\
&=
\sum_{(i,\ell,r,k)\in D}
\iv{\ell\equiv\sum_{j\in A}m_{ij}\!\!\!\pmod p}
\iv{r\equiv\sum_{j\in B}m_{ij}\!\!\!\pmod p}\\
&=
\sum_{i\in[n]}
\hspace{-18mm}
\sum_{\hspace{17mm}\tiny\begin{array}{l}\ell=0\\\ell+r\not\equiv 0\!\!\!\!\pmod p\end{array}}^{p-1}\hspace{-17mm}
\sum_{r=0}^{p-1}\,\,\,
\iv{\ell\equiv\sum_{j\in A}m_{ij}\!\!\!\pmod p}
\iv{r\equiv\sum_{j\in B}m_{ij}\!\!\!\pmod p}
\ensuremath{\operatorname{dlog}}_{p,g}\bigl(\ell+r\bigr)
\\
&\qquad+\sum_{i\in[n]}
\sum_{\ell=0}^{p-1}
\iv{\ell\equiv\sum_{j\in A}m_{ij}\!\!\!\pmod p}
\iv{p-\ell\equiv\sum_{j\in B}m_{ij}\!\!\!\pmod p}
np\\
&=\begin{cases}
\sum_{i\in[n]}\ensuremath{\operatorname{dlog}}_{p,g}\sum_{j\in A\cup B}m_{ij}
& \text{if $\prod_{i\in [n]}\sum_{j\in A\cup B}m_{ij}\not\equiv 0\!\!\!\pmod p$;}\\
\geq np
& \text{if $\prod_{i\in [n]}\sum_{j\in A\cup B}m_{ij}\equiv 0\!\!\!\pmod p$.}
\end{cases}
\end{split}
\end{equation}
In particular, letting
\[
f_{\sigma_L,\sigma_R,t}=
\big|\bigl\{(A,B)\in\mathcal{L}_{\sigma_L}\times\mathcal{R}_{\sigma_R}:
\ensuremath{\langle}\lambda(A),\rho(B)\ensuremath{\rangle}=t\bigr\}\big|\,,
\]
it follows immediately from \eqref{eq:lambda-rho-ip} that we have
$w_{\sigma_1,\sigma_2}^{(e)}
=\sum_{t=0,\ t\equiv e\!\pmod{p-1}}^{n(p-2)}f_{\sigma_L,\sigma_R,t}$,
which enables us to recover $\ensuremath{\operatorname{per}} M$ from the counts of pairs
in $\mathcal{L}_{\sigma_L}\times\mathcal{R}_{\sigma_R}$ by inner product.
\subsection{Completing the Proof of Theorem~\ref{thm:perm-to-exactip}}
Suppose we have an algorithm for \#\textsc{InnerProduct} that
runs in $N^{2-\Omega(1/\log c)}$ time when given an input of $N$
vectors from $\{0,1\}^{c\log N}$. Take $N=2^{n/2-1}$ and observe that
$\log N=n/2-1$. The reduction from previous section has $d\leq n^4(\ln n)^3$
and thus we can take $c=(n\ln n)^3$ and thus solve $n\times n$
zero-one permanent in time $N^{2-\Omega(1/\log c)}=2^{n-\Omega(n/\log n)}$.
This completes the proof of Theorem~\ref{thm:perm-to-exactip}.
\section{A Lower Bound for \#\textsc{OV} via Zero-One Permanents}
\label{sect:perm-cont}
This section continues our work towards relations to zero-one permanents
started in Sect.~\ref{sect:perm}; in particular, we
prove Theorem~\ref{thm:perm-to-ov}.
\subsection{A Reduction from Zero-One Permanent to \#OV}
This section starts our work towards Theorem~\ref{thm:perm-to-ov}
without yet parameterizing the reduction in detail.
As in Sect.~\ref{sect:perm}, it suffices to describe how to compute $\ensuremath{\operatorname{per}} M$
modulo a given prime $p$ with $2\leq p\leq n\ln n$.
Let $g$ be a positive integer parameter, which we assume divides $n$.
For $h\in[g]$, let
\[
V_h=\{i\in [n]:(h-1)n/g+1\leq i\leq hn/g\}
\]
be a partition of the rows of $M$ into $g$ groups, each of size $n/g$.
Again from Ryser's formula, we observe that
\[
\ensuremath{\operatorname{per}} M=(-1)^n\sum_{S\subseteq[n]}(-1)^{|S|}\prod_{h\in[g]}\prod_{i\in V_h}\sum_{j\in S}m_{ij}\,.
\]
Grouping by sign $\sigma\in\{-1,1\}$ and per-group residues
$r\in\{0,1,\ldots,p-1\}^g$ modulo $p$, we thus have
\begin{equation}
\label{eq:g-group-ryser}
\ensuremath{\operatorname{per}} M\equiv(-1)^n\!\!\!\sum_{r\in\{0,1,\ldots,p-1\}^g}\!\!\!(t_{1,r}-t_{-1,r})\prod_{h=1}^g r_h\pmod p\,,
\end{equation}
where
\[
t_{\sigma,r}=
\big|\bigl\{S\subseteq[n]:
(-1)^{|S|}=\sigma\ \text{and}\
\prod_{i\in V_h}\sum_{j\in S}m_{ij}
\equiv
r_h
\!\!\!
\pmod{p}\
\text{for each $h\in[g]$}
\bigl\}\big|\,.
\]
Observe that given all the counts $t_{\sigma,r}$, it takes $O(p^{g}g)$
operations modulo $p$ to compute the permanent modulo $p$
via \eqref{eq:g-group-ryser}, which is less than $2^nn$ when $g<n/\log p$.
We continue to describe how to get the counts $t_{\sigma,r}$
via orthogonal-vector counting.
Assuming that $n\geq 4$ is even, introduce again the split
\[
L=\{1,2,\ldots,n/2\}\qquad\text{and}\qquad
R=\{n/2,n/2+1,\ldots,n\}\,.
\]
Let the residue vector $r\in\{0,1,\ldots,p-1\}^g$ be fixed.
For $\sigma_L,\sigma_R\in\{1,-1\}$, let
\[
\begin{split}
t_{\sigma_L,\sigma_R,r}=
\big|\bigl\{S\subseteq[n]:\ &
(-1)^{|S\cap L|}=\sigma_L\ ,\
(-1)^{|S\cap R|}=\sigma_R\,,\\&\quad\text{and}\
\prod_{i\in V_h}\sum_{j\in S}m_{ij}
\equiv
r_h
\!\!\!
\pmod{p}\
\text{for each $h\in[g]$}
\bigl\}\big|\,.
\end{split}
\]
Clearly
$t_{\sigma,r}=\sum_{\substack{\sigma_L,\sigma_R\in\{-1,1\}\\\sigma_L\sigma_R=\sigma}}t_{\sigma_L,\sigma_R,r}$,
so it suffices to focus on computing $t_{\sigma_L,\sigma_R,r}$ in what
follows. We again work with the set families
\[
\mathcal{L}_{\sigma_L}=\bigl\{A\subseteq L:(-1)^{|A|}=\sigma_L\bigr\}
\qquad
\text{and}
\qquad
\mathcal{R}_{\sigma_R}=\bigl\{B\subseteq R:(-1)^{|B|}=\sigma_R\bigr\}\,.
\]
Let
\[
D=[g]\times\{0,1,\ldots,p-1\}^{n/g}\,.
\]
We have
\[
d=|D|=gp^{n/g}\,.
\]
For $A\in\mathcal{L}_{\sigma_L}$ and $B\in\mathcal{R}_{\sigma_R}$,
define the vectors $\lambda(A)\in\{0,1\}^D$ and $\rho(B)\in\{0,1\}^D$
for all $(h,u)\in D$ by the rules
\begin{equation}
\label{eq:lambda-rho-ov}
\begin{split}
\lambda(A)_{hu}&=
\begin{cases}
1 & \!\!\!\text{if we have $\sum_{j\in A}m_{ij}\equiv u_{i-(h-1)n/g}\!\!\!\pmod p$ for all
$i\in V_h$;}\\
0 & \!\!\!\text{otherwise;}
\end{cases}\\[3mm]
&\hspace{-12mm}\text{and}\\[3mm]
\rho(B)_{hu}&=
\begin{cases}
0 & \!\!\!\text{if $\prod_{i\in V_h}\bigl(u_{i-(h-1)n/g}+\sum_{j\in B}m_{ij}\bigr)\equiv r_h\!\!\!\pmod p$;}\\
1 & \!\!\!\text{otherwise.}
\end{cases}
\end{split}
\end{equation}
Over the integers, from \eqref{eq:lambda-rho-ov} we now have
\begin{equation}
\label{eq:lambda-rho-ip-ov}
\begin{split}
\ensuremath{\langle}\lambda(A),\rho(B)\ensuremath{\rangle}
&=\sum_{(h,u)\in D}
\lambda(A)_{hu}
\rho(B)_{hu}\\
&=\sum_{h\in[g]}
\sum_{u\in\{0,1,\ldots,p-1\}^{n/g}}
\prod_{i\in V_h}
\bbiv{\sum_{j\in A}m_{ij}\equiv u_{i-(h-1)n/g}\!\!\!\pmod p}\\
&\hspace*{37mm}
\bbiv{\prod_{i\in V_h}\bigl(u_{i-(h-1)n/g}+\sum_{j\in B}m_{ij}\bigr)\not\equiv r_h\!\!\!\pmod p}\\
&=\sum_{h\in[g]}
\bbiv{\prod_{i\in V_h}\biggl(\sum_{j\in A}m_{ij}+\sum_{j\in B}m_{ij}\biggr)\not\equiv r_h\!\!\!\pmod p}\\
&=\begin{cases}
0 &
\text{if we have $\prod_{i\in V_h}\sum_{j\in A\cup B}m_{ij}\equiv r_h\!\!\! \pmod{p}$ for each $h\in[g]$;}\\
\geq 1 & \text{otherwise.}
\end{cases}
\end{split}
\end{equation}
In particular, we have
\[
t_{\sigma_L,\sigma_R,r}=
\big|\bigl\{(A,B)\in\mathcal{L}_{\sigma_L}\times\mathcal{R}_{\sigma_R}:
\ensuremath{\langle}\lambda(A),\rho(B)\ensuremath{\rangle}=0\bigr\}\big|\,,
\]
which enables us to recover $\ensuremath{\operatorname{per}} M$ from the counts of orthogonal pairs
in $\mathcal{L}_{\sigma_L}\times\mathcal{R}_{\sigma_R}$.
\subsection{Completing the Proof of Theorem~\ref{thm:perm-to-ov}}
Suppose now that we have an algorithm for \#\textsc{OV} that
runs in $N^{2-\Omega(1/\log^{1-\epsilon} c)}$ time for some $0<\epsilon<1$
when given an input of $N$
vectors from $\{0,1\}^{c\log N}$. Take $N=2^{n/2-1}$ and observe that
$\log N=n/2-1$.
Let $K>1$ be a constant that will depend on $\epsilon$ and the constant
hidden by the $\Omega(\cdot)$ notation in the running time of the
\#\textsc{OV} algorithm. Take
\[
g=\lfloor K^{-1/\epsilon}n(\log p)^{1-2/\epsilon}\rfloor
\]
and recall that the prime $p$ is in the range $2\leq p\leq n\ln n$.
To compute the parameters $t_{\sigma,r}$ using the reduction in the previous
section, for each prime $p$ we need $4p^g$ invocations of the \#\textsc{OV}
algorithm on an input of $N$ vectors of dimension $d=gp^{n/g}$.
Thus, for all large enough $n$,
since $\frac{1}{2}K^{-1/\epsilon}n(\log p)^{1-2/\epsilon}\leq g$,
we have
\[
d=gp^{n/g}
\leq n 2^{2K^{1/\epsilon}(\log p)^{2/\epsilon}}\,.
\]
Since clearly $d=c\log N=c(n/2-1)$ and $2/\epsilon>2$,
for all large enougn $n$, we have
\[
\begin{split}
\log c&\leq 1+2K^{1/\epsilon}(\log p)^{2/\epsilon}\\
&\leq 3K^{1/\epsilon}(\log p)^{2/\epsilon}\,,
\end{split}
\]
where the last inequality depends on choosing a large enough $K$
so that the inequality is true for $p=2$.
Thus,
\[
\begin{split}
-(\log c)^{\epsilon-1}
&\leq -{3^{\epsilon-1}K^{1-1/\epsilon}(\log p)^{2-2/\epsilon}}\,.
\end{split}
\]
One invocation of the \#\textsc{OV} algorithm thus runs in
\[
N^{2-\Omega(\log^{\epsilon-1} c)}=2^{n-\Omega(n{3^{\epsilon-1}K^{1-1/\epsilon}(\log p)^{2-2/\epsilon}})}
\]
time. For each prime $2\leq p\leq n\ln n$, we need
\[
4p^g\leq 2^{2+K^{-1/\epsilon}n(\log p)^{2-2/\epsilon}}
\]
invocations of the \#\textsc{OV} algorithm. Thus, the running time of
all the invocations for the prime $p$ is bounded by
\[
\begin{split}
4p^gN^{2-\Omega(\log^{\epsilon-1} c)}\leq
2^{n-\Omega(n{3^{\epsilon-1}K^{1-1/\epsilon}(\log p)^{2-2/\epsilon}})
+2+K^{-1/\epsilon}n(\log p)^{2-2/\epsilon}}\,.
\end{split}
\]
By choosing a large enough $K$ to dominate the constant hidden by
the $\Omega(\cdot)$ notation in the running time of
the \#\textsc{OV} algorithm, we thus have, for all large enough $n$,
\[
\begin{split}
4p^gN^{2-\Omega(\log^{\epsilon-1} c)}
&\leq
2^{n-\Omega(n{3^{\epsilon-1}K^{-1/\epsilon}(\log p)^{2-2/\epsilon}})}\\
&\leq
2^{n-\Omega(n{3^{\epsilon-1}K^{-1/\epsilon}(\log n+\log\ln n)^{2-2/\epsilon}})}\\
&\leq
2^{n-\Omega(n(\log n)^{2-2/\epsilon})}\,.
\end{split}
\]
Since there are at most $n\ln n$ primes $p$ to consider, the total running
time to compute $\ensuremath{\operatorname{per}} M$ is bounded by $2^{n-\Omega(n/\log^{2/\epsilon-2}n)}$.
This completes the proof of Theorem~\ref{thm:perm-to-ov}.
\section{Further Applications}
\subsection{Counting Satisfying Assignments to a \textsc{Sym}$\circ$\textsc{And} circuit via \#\textsc{InnerProduct}}
\label{sect:sym-and}
We describe how to embed a \textsc{Sym}$\circ$\textsc{And} circuit, i.e., a circuit of $s$ \textsc{And} gates working on $n$ Boolean inputs, connected by a top gate that is an arbitrary symmetric gate, in a \#\textsc{InnerProduct} instance of size $N=2^{n/2}$ and $d=s$. Assuming $n$ even, we divide the $n$ inputs in two equal halves $L$ and $R$. We let $\mathcal A$ have one vector $u$ for each assignment to the inputs in $L$, with one coordinate in $u$ for each \textsc{And} gate, representing the truth value of that
gate restricted to the inputs in $L$. Likewise, we let $\mathcal B$ have one vector $v$ for each assignment to the inputs in $R$, with each coordinate set to the truth value of the represented gate restricted to the inputs in $R$. It is readily verified that $\langle u,v\rangle$ counts the number of \textsc{And} gates that are satisfied by the assignment represented by $(u,v)$.
Hence, knowing the number of assignments that satisfy exactly $t$ of the \textsc{And} gates, for $t=0,1,\ldots,s$, which is what the solution to the \#\textsc{InnerProduct} gives us, we can count the total number of assignments that also satisfies the top symmetric gate.
Variations where the circuit instead is a \textsc{Sym}$\circ$\textsc{Or} or a \textsc{Sym}$\circ$\textsc{Xor}, are also possible.
\subsection{Computing the Weight Enumerator Polynomial via \#\textsc{InnerProduct}}
\label{sect:weight-enum}
A binary linear code of length $n$ and rank $k$ is a linear subspace $C$ with dimension $k$ of the vector space
$\mathbb{F}_2^n$.
The \emph{weight enumerator polynomial} is
\[
W(C;x,y)=\sum_{w=0}^n A_wx^wy^{n-w}\,,
\]
where
\[
A_w=|\{c\in C:\langle c,c\rangle=w\}|\,,
\]
for $w=0,1,\ldots,n$ is the {\em weight distribution}; that is,
$A_w$ equals the number of codewords of $C$ having exactly $w$ ones.
We will reduce the computation of the weight distribution, and hence the weight enumerator polynomial, to $(k/2+1)^2$ instances of \#\textsc{InnerProduct} with $N\leq 2^{k/2}$ and $d=2(n-k)$ when $k$ is even.
Let the $k\times n$ matrix $G$ be the generating matrix of the code; that is, the codewords of $C$ are exactly the row-span of $G$. We can assume without loss of generality that the generator matrix has the standard form $G=[I_k|P]$, where $I_k$ is the $k\times k$ identity matrix.
For each $s_A=0,1,\cdots,k/2$ and $s_B=0,1,\ldots,k/2$, we make one instance
of \#\textsc{InnerProduct}.
We let the set $\mathcal A$ have one vector $u$ for each code $c$ obtained as the linear combination of exactly $s_A$ of the first $k/2$ rows. Each of the $n-k$ last coordinates in the code word $c$ is described by a block of two coordinates in $u$. If $c_i=0$ we encode this as $01$ in $u$, and if $c_i=1$ we encode this as $10$ in $u$. We concatenate all $n-k$ encoded blocks to obtain $u$. Likewise, we let the set $\mathcal B$ have one vector $v$ for each code $c$ obtained as a linear combination of $s_B$ of the last $k/2$ rows. Again,
each of the $n-k$ last coordinates in the code word $c$
is described by a block of two coordinates in $v$, but the encoding is opposite the one for $\mathcal A$: If $c_i=0$ we encode this as $10$ in $v$, and if $c_i=1$ we encode this as $01$ in $v$. We again concatenate all $n-k$ encoded blocks to obtain $v$.
With this design, it is readily verified that for $(u,v)\in \mathcal A\times \mathcal B$, the inner product $\langle u,v\rangle$ is equal to the number of ones in the last $n-k$ coordinates in the code word
obtained as the sum of the code word represented by $u$ and the code word represented by $v$. Also, by design the number of ones in the first $k$ coordinates equals $s_A+s_B$. Hence, by summing over all pairs that have the same inner product $t$, and aggregating over all $s_A$ and $s_B$, we can compute the weight distribution.
\section*{Acknowledgment}
We thank Virginia Vassilevska Williams and Ryan Williams for many useful discussions.
\bibliographystyle{abbrv}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 7,492 |
{"url":"http:\/\/ehhookupnz.biodiverse.xyz\/hyperboloid-equation-one-sheet-of-paper.html","text":"# Hyperboloid equation one sheet of paper\n\nPaper equation\n\n## Hyperboloid equation one sheet of paper\n\nPaper published similar to PhD thesis. , dividing the equation with one of its non- zero coefficients paper reduces the number of coefficients to nine. There paper is more than one type of Hyperboloid : > In mathematics, a hyperbolo. The parallel postulate of Euclidean geometry is replaced with:. x^ 2\/ a^ 2 + y^ 2\/ b^ 2 - z^ 2\/ c^ 2 = 1 ( hyperboloids of one sheet) x^ 2\/ a^ 2 - y^ 2\/ b^ 2 - z^ 2\/ c^ 2 equation = 1 ( hyperboloids of two sheets).\n\n- A - [ 1] 1- parameter group of transformations 1- \ub9e4\uac1c\ubcc0\uc218\ubcc0\ud658\uad70[ 2] Abelian equation \uc544\ubca8\ubc29\uc815\uc2dd [ 3] Abelian extension field \uc544\ubca8 \ud655\ub300\uccb4 [ 4] Abelian group \uc544\ubca8 \uad70, \uac00\ud658\uad70 [ 5] Abelian integral \uc544\ubca8 \uc801\ubd84 [ 6] Archimedian valuation \uc544\ub974\ud0a4\uba54\ub370\uc2a4 \ubd80\uce58. For any given line R in the plane containing both line R , point P not on R point P there are at least two distinct lines through P equation that do not intersect R. A hyperboloid of revolution is generated by revolving a hyperbola about one of its axes. A hyperboloid is a surface whose plane sections are either hyperbolas or ellipses. A Hyperboloid of one sheet, showing its ruled surface property. The terrorist organization Aum Shinrikyo found inspiration in the galactic empire of Isaac Asimov' s Foundation Trilogy.\n\nFind an equation for the tower. This paper aims to clarify the derivation of this result paper and to describe some further related ideas. Using MATLAB, you can turn these complicated paper equations into 3- D plots. Generalised Equation of Hyperboloid. K A R M A: \ube14\ub85c\uadf8 \uba54\ub274; prologue; blog; tag; guest; blog 7) into the equation a2z. 2 Hyperboloid a) Hyperboloid of one sheet) Hyperboloid of two sheets 2. A hyperboloid is a Ruled Surface.\n\nOr, in other words, a surface generated by a line. Results for this submission Entered Answer Preview hyperboloid of one sheet hyperboloid of one sheet 0. But interstellar empires never seem to go out of style regardless of their practicality they paper remain a powerful meme. Folding the paper Hyperbolic Crane. can easily be constructed by bending a sheet of paper. \uc6b0\uc120 \uc9c0\uc2dd\uc778\uc5d0\uc11c \uae01\uc5c8\uc2b5\ub2c8\ub2e4. Let paper us show that the hyperboloid of one sheet is a doubly- ruled surface by.\n\nlyzes and describes the desired properties of the sheet of paper from which equation one can fold a hyperbolic analog of the. Hyperboloid of One Sheet:. mathematica \uacf5\ubd80\ud560 \ub54c \ucc38\uace0 \ud560\ub824\uad6c\uc694. Difficulty: Moderately ChallengingInstructions Things You' ll Need Equation paper of a hyperboloid Pencil Paper Computer MATLAB. The plaster model shows an elliptic equation hyperboloid of one sheet, a surface that can be represented by the equation x 2 \/ a 2 + y 2 \/ b 2 - z 2 \/ c 2 = 1. 166667 At least one of the answers above is NOT correct ( 1 point) Identiry the type of quadric surtace detined by the equation and find all x-. upon one sheet of a two- sheeted cylindrical hyperboloid in Minkowski space- time.\n\nSketch equation the graph of this quadric surface on paper. Hyperboloid equationr c Z b Y a Xwhere on the right hand side of ( 1) corresponds to a hyperboloid of one sheet, on the right hand side of - 1 to a hyperboloid of two sheets. A cooling tower for a nuclear reactor is to be constructed in the shape of a hyperboloid of one sheet. Hyperboloid equation one sheet of paper. Red Paper House - contemporary jewellery prints , accessories created by Amy Hall. equation ^ 2$one- to- one onto one sheet. Hyperboloid equation one sheet of paper. Simple Curves and Surfaces. \uc218\ud559 \uc601\uc5b4 \uc6a9\uc5b4. There are those who in the realm of science fiction literature wonder if galactic empires are the new \" Middle- Earth\". A revolving around its transverse axis equation forms a surface called \u201c hyperboloid of one sheet\u201d. y z- intercepts of the resulting graph. parametrization of the hyperboloid of two sheets. Ruled surfaces are surfaces that for every point on the surface, there is a line on the surface passing it. The diameter at the base is 280 m the minimum diameter, 500 m above the base is 220 m. In mathematics hyperbolic geometry ( also called Bolyai\u2013 Lobachevskian geometry Lobachevskian geometry) is a non- Euclidean geometry. In geometry also known as the Minkowski model , Hendrik paper Lorentz ), the Lorentz model ( after Hermann Minkowski , is a equation model of n- dimensional hyperbolic geometry in which points are represented by the points on the forward sheet paper S of a two- sheeted hyperboloid in ( n+ 1) - dimensional paper Minkowski space , the hyperboloid model m- paper planes are. Hyperboloid of one sheet string surface model 1872. Here x y are axes in the horizontal plane z is the vertical axis. ## Sheet equation parametrization of the hyperboloid of two sheets. { \\ mathbb R} ^ 2$ one- to- one onto one sheet. Use MathJax to format equations. Approach your problems from the right end and begin with the answers. Then one day, perhaps, you will find the final question. \" The Chinese Maze Murders\" by Robert Hans van GulikIt' s better to know some of the questions than all of the answers.\n\nhyperboloid equation one sheet of paper\n\nA rectangular sheet of paper of width a and length b, where 0< a< b, is folded by taking one corner of the sheet and placing it at point P on the opposite long side of the sheet. A rectangular sheet of paper of width a and length b, where 0< a< b, is fold.","date":"2019-10-18 08:32:50","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\": 0.6771213412284851, \"perplexity\": 1861.5420169322724}, \"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-2019-43\/segments\/1570986679439.48\/warc\/CC-MAIN-20191018081630-20191018105130-00175.warc.gz\"}"} | null | null |
{"url":"https:\/\/support.numxl.com\/hc\/en-us\/articles\/215716583","text":"# Seasonal Autoregressive Integrated Moving Average (SARIMA) Model\n\nThe SARIMA model is an extension of the ARIMA model, often used when we suspect a model may have a seasonal effect.\n\nBy definition, the seasonal auto-regressive integrated moving average - SARIMA(p,d,q)(P,D,Q)s - process is a multiplicative of two ARMA processes of the differenced time series.\n\n$$(1-\\sum_{i=1}^p {\\phi_i L^i})(1-\\sum_{j=1}^P {\\Phi_j L^{j \\times s}})(1-L)^d (1-L^s)^D x_t = (1+\\sum_{i=1}^q {\\theta_i L^i})(1+\\sum_{j=1}^Q {\\Theta_j L^{j \\times s}}) a_t$$ $$y_t = (1-L)^d (1-L^s)^D$$\n\nWhere:\n\n\u2022 $x_t$ is the original non-stationary output at time t.\n\u2022 $y_y$ is the differenced (stationary) output at time t.\n\u2022 $d$ is the non-seasonal integration order of the time series.\n\u2022 $p$ is the order of the non-seasonal AR component.\n\u2022 $P$ is the order of the seasonal AR component.\n\u2022 $q$ is the order of the non-seasonal MA component.\n\u2022 $Q$ is the order of the seasonal MA component.\n\u2022 $s$ is the seasonal length.\n\u2022 $D$ is the seasonal integration order of the time series.\n\u2022 $a_t$ is the innovation, shock or the error term at time t.\n\u2022 $\\{a_t\\}$ time series observations are independent and identically distributed (i.e. i.i.d) and follow a Gaussian distribution (i.e. $\\Phi(0,\\sigma^2)$)\n\nAssuming y_t follows a stationary process with a long-run mean of $\\mu$, then taking the expectation from both sides, we can express $\\phi_o$ as follows: $$\\phi_o = (1-\\phi_1-\\phi_2-\\cdots-\\phi_p)(1-\\Phi_1-\\Phi_2-\\cdots-\\Phi_P)$$\n\nThus, the SARIMA(p,d,q)(P,D,Q)s process can now be expressed as: $$(1-\\sum_{i=1}^p {\\phi_i L^i})(1-\\sum_{j=1}^P {\\Phi_j L^{j \\times s}}) (y_t -\\mu) = (1+\\sum_{i=1}^q {\\theta_i L^i})(1+\\sum_{j=1}^Q {\\Theta_j L^{j \\times s}}) a_t$$ $$z_t=y_t-\\mu$$ $$(1-\\sum_{i=1}^p {\\phi_i L^i})(1-\\sum_{j=1}^P {\\Phi_j L^{j \\times s}}) z_t = (1+\\sum_{i=1}^q {\\theta_i L^i})(1+\\sum_{j=1}^Q {\\Theta_j L^{j \\times s}}) a_t$$\n\nIn sum, $z_t$ is the differenced signal after we subtract its long-run average.\n\nNotes: The order of the seasonal or non-seasonal AR (or MA) component is solely determined by the order of the last lagged variable with a non-zero coefficient. In principle, you can have fewer parameters than the order of the component.\n\n## notes\n\n1. The variance of the shocks is constant or time-invariant.\n2. The order of the seasonal or non-seasonal AR (or MA) component is solely determined by the order of the last lagged variable with a non-zero coefficient. In principle, you can have fewer parameters than the order of the component.\n\nExample: Consider the following SARIMA(0,1,1)(0,1,1)12 process: $$(1-L)(1-L^{12})x_t-\\mu = (1+\\theta L)(1+\\Theta L^{12})a_t$$ Note: This is the AIRLINE model, a special case of the SARIMA model.","date":"2019-01-18 04:17:27","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\": 1, \"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\": 0.9369326233863831, \"perplexity\": 995.1929334139533}, \"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-2019-04\/segments\/1547583659677.17\/warc\/CC-MAIN-20190118025529-20190118051529-00040.warc.gz\"}"} | null | null |
Personal Data – (names and addresses etc) through which it is possible to identify you. You will always know when this data is being collected and it will only be used for the specific purposes to which you have consented.
Aggregated Statistical Data – all other non-personal information that relates purely to generic statistics such as traffic flow and demographics. It is not possible to identify you through this data.
Soap World Magazine will only collect, use and transfer your Personal Data if you have consented to that collection, use and transfer, except in circumstances where it is necessary to conform to legal requirements; protect and defend our rights or property; or to protect the interests of other users.
By checking other boxes on the Registration Form, or giving your specific consent at other times, you can additionally choose to have your Personal Data passed on to specially selected Third Parties that Soap World Magazine believe will be able to supply you with useful information, and for those Third Parties to contact you with relevant marketing material.
On occasion we may collect Personal Data in other ways – e.g. through competition entry forms or other occasional services.
Unless you have consented otherwise by having already submitted a Sportal Registration Form, we will only use that data to contact you about that particular service, unless we indicate otherwise.
Other non-personal information will be collected and used by Soap World Magazine and may be transferred to third parties. Any such statistical or demographic information will not include personal details through which you could be identified. In addition to non-personal information given on your Registration Form, information is automatically collected about traffic around the site. This is also only used in aggregate form and no individual user will be identified.
Soap World Magazine uses "cookies" to speed up access to the site (it bypasses checking the registered user database) and to enhance your experience of the site by customising it according to your interests. A cookie is a small piece of information sent from Sportal to your computer to help us to identify you quickly. It is possible for you to disable these cookies by changing your browser settings, but this may slow down or prohibit access to parts of the site.
Security is paramount to Soap World Magazine in respect of your data. We use a variety of software to ensure appropriate security and encryption.
By visiting the Soap World website you acknowledge that you use the service at your own risk.
nexmedia Pty Ltd, Soap World magazine, its affiliate and third party service providers expressly disclaaim any and all warranties, both express and implied, including but not limitied to any warranties of accuracy, reliability, title, merchantability, non-infringement, fitness for a particular purpose or any other warranty, condition, guarantee or representation whether oral, in writing, electronic form, including but not limited to the accuracy or completeness of any information contained therein or provided by the service, service provider, affiliates, and third party service providers do not represent or warrant that access to the service will be uninterupted or that there will be no failures,errors or ommision or loss of transmitted information, or that no viruses will be transmitted on the service.
nexmedia Pty Ltd, Soap World magazine, its affiliate and third party service providers shall not be liable for any direct, indirect, special, consequential or punitive damages allegedly sustained arising out of this agreement,the providing of services hereunder, the sale or purchase of any goods or merchandise, your access to orinability to access the service, including for viruses alleged to have been obtained from the service, your use or reliance on the service or any of the merchandise, information or materials available on the service regardless of the type of claim or the nature of the cause of action, even if advised of the possibility of such damages.
You hereby agree to release nextmedia Pty Ltd, Soap World magazine, its affiliatates and third-party service providers, and each of their respective directors, officers, employees, and agents from claims, demands and damages of every kind and nature, known and unknown, suspected and unsuspected arising out of or in any way connected with your use of this site. | {
"redpajama_set_name": "RedPajamaC4"
} | 1,261 |
In Hawaii, top state lawmakers are considering at all options in finding a way for qualified patients to receive medicinal marijuana.
In Hawaii, state lawmakers are looking to make more medical marijuana available, and seeking the advice from exemplary states that already have the medicine available to patients.
While the state has medical marijuana as a treatment option for qualified patients, the Hawaii doesn't actually have a way of dispensing the medicine to patients.
One idea they are leaning towards is allowing a caregiver to grow for more than one patient, which is the current situation now.
They admit that they can't really do it as a state because they don't know how to as well as those that are already doing it, signaling a change from the way most states end up (mis)handing medical marijuana distribution to qualified patients. | {
"redpajama_set_name": "RedPajamaC4"
} | 8,001 |
\section{Introduction}
Quantum key distribution (QKD)~\cite{QKDreview} enables the generation of
secret keys between two or more authenticated parties by resorting to the
fundamental laws of quantum mechanics. Its continuous variable (CV)
version~\cite{Cerf,GG02, noswitch, CVMDI,RMP} represents a very profitable
setting and opportunity thanks to its more direct implementation in the
current communication infrastructure and, most importantly, for its potential
to approach the ultimate rate limits of quantum communication, as represented
by the repeaterless PLOB bound~\cite{QKDpaper}. From an experimental point of
view, we have been witnessing an increasing number of realizations closing the
gap with the more traditional qubit-based
implementations~\cite{LeoCodes,LeoEXP}.
The most advanced protocols of CV-QKD are the Gaussian-modulated
coherent-state protocols~\cite{GG02, noswitch, CVMDI}. Not only they are very
practical, but also enjoy the most advanced security proofs, accounting for
finite-size effects (i.e., finite number of signal exchanges) and
composability (so that each step of the protocol has an associated error which
adds to an overall `epsilon'-security)~\cite{QKDreview,noteDalpha}. Very
recently, this level of security has been extended to the free-space
setting~\cite{FSpaper,SATpaper}, where we need to consider not only the
presence of diffraction-induced loss~\cite{Goodman,Siegman,svelto},
atmospheric extinction~\cite{Huffman} and background thermal noise~\cite{Miao,
BrussSAT}, but also the effect of fading, as induced by pointing error and
turbulence~\cite{Vasy12,Esposito,Yura73, Fante75,AndrewsBook, Majumdar,
Hemani}. The importance of studying fading and atmospheric effects in CV-QKD
is an active area with increasing efforts put by the community at large (e.g.
see
Refs.~\cite{refC1,refC1b,refC2,refC2b,PanosFading,refC4,refC5,refC6,refC7,refC8,refC9,refC10
).
While composable security is typically assessed against collective or coherent
attacks, experiments may involve some additional (realistic) assumptions that
elude this theory. For instance, these assumptions may concern some level of
trusted noise in the setups (e.g., this is often the case for the electronic
noise of the detector) or some realistic constraint on the eavesdropper, Eve
(e.g., it may be considered to be passive in line-of-sight free-space
implementations). For this reason, here we present the general theory to cover
all these cases.
In fact, we consider various levels of trust for the receiver's setup,
starting from the traditional scenario where detector's loss or noise are
untrusted, meaning that Eve may perform a side-channel attack over the
receiver besides attacking the main channel. Then, we consider the case where
detector's noise is trusted but not its loss, which corresponds to Eve
collecting leakage from the receiver. Finally, we study the more trustful
scenario where both detector's loss and noise are considered to be trusted, so
that Eve is excluded from side-channels to the receiver. We show how these
assumptions can non-trivially increase the composable key rates of
Gaussian-modulated CV-QKD protocols and tolerate higher dBs.
In our analysis, we then investigate the free-space setting, specifically for
near-range wireless quantum communications at optical frequencies (LiFi). This
scenario involves the presence of free-space diffraction and also fading
effects, mainly due to pointing and tracking errors associated with the
limited technology of the transmitter (while we can neglect turbulence at such
distances). We consider communication with both fixed and mobile devices,
assuming realistic parameters for indoor conditions and relatively-large
field-of-views for the receivers. Security is studied under the various
trusted models for the receiver's detector and then including additional
assumptions for Eve due to the line-of-sight configuration. Here too we show
that key rates are remarkably increased as an effect of the realistic
assumptions. More interestingly, we show that wireless high-rate CV-QKD is
indeed feasible with mobile devices.
Finally, we consider wireless quantum communications at the microwave
frequencies (WiFi) where both loss and thermal noise are very high. In this
scenario, we consider a potential regime of parameters that enables very
short-range quantum security, e.g., between contact-less devices within the
range of a few centimeters.
The paper is organized as follows. In Sec.~\ref{SEC1}, we provide a general
framework for the composable security of CV-QKD, which also accounts for
levels of trust in the loss and noise of the communication. In Sec.~\ref{SEC2
, we consider near-range free-space quantum communications, first at optical
frequencies (with fixed and mobile devices) and then at the microwaves.
Sec.~\ref{SEC3} is for conclusions.
\section{General framework for composable security of CV-QKD\label{SEC1}}
\subsection{General description\label{descriptionPROTsec}}
Let us consider a Gaussian-modulated coherent-state protocol between Alice
(transmitter) and Bob (receiver). Alice prepares a coherent state $\left\vert
\alpha\right\rangle $ whose amplitude $\alpha$ is modulated according to a
complex Gaussian distribution with zero mean and variance $\mu-1$. Assuming
the notation of Ref.~\cite{RMP}, we may decompose the amplitude as
$\alpha=(q+ip)/2$, where $x=q$ or $p$ represents the mean value of the generic
quadrature operator $\hat{x}=\hat{q},\hat{p}$ where $[\hat{q},\hat{p}]=2i$.
This generic quadrature can be written as $\hat{x}=\hat{x}_{0}+x$, where
$\hat{x}_{0}$\ is the vacuum noise associated with the bosonic mode and the
real variable $x$ is a random Gaussian displacement with zero mean and
variance
\begin{equation}
\sigma_{x}^{2}=\mu-1.
\end{equation}
The coherent state is sent through a thermal-loss channel controlled by the
eavesdropper, with transmissivity $\eta_{\text{ch}}$ and mean number of
thermal photons $\bar{n}_{e}$. Equivalently, we may introduce the variance
$\omega=2\bar{n}_{e}+1$ and the background thermal noise $\bar{n}_{B}$ defined
by $\bar{n}_{e}=\bar{n}_{B}/(1-\eta_{\text{ch}})$, so $\bar{n}_{B}$ photons
are added to the input signal. Bob's setup is characterized by quantum
efficiency $\eta_{\text{eff}}$ and extra noise variance $\nu_{\text{ex}
=2\bar{n}_{\text{ex}}$, where $\bar{n}_{\text{ex}}$ is an equivalent number of
thermal photons generated by the imperfections in his receiver station (due to
electronic noise, phase errors etc.)
From an energetic point of view, the initial mean photons at the transmitter
$\bar{n}_{T}$ are attenuated by an overall factor $\tau=\eta_{\text{ch}
\eta_{\text{eff}}$ which can be seen as the total effective transmissivity of
the extended channel between Alice and Bob. Thus, the total mean number of
photons that are seen by the receiver's detector is given b
\begin{equation}
\bar{n}_{R}=\tau\bar{n}_{T}+\bar{n}, \label{IOenergy
\end{equation}
where $\bar{n}$ is the total number of thermal photons due to the various
sources of noise, given by
\begin{equation}
\bar{n}=\eta_{\text{eff}}\bar{n}_{B}+\bar{n}_{\text{ex}}. \label{nBARvalue
\end{equation}
See also Fig.~\ref{attackPIC} for a schematic of the overall scenario.
\begin{figure}[t]
\vspace{-0.9cm}
\par
\begin{center}
\includegraphics[width=1\columnwidth] {attacks.eps}
\end{center}
\par
\vspace{-1.2cm}\caption{Quantum communication scenario between transmitter
(Alice) and receiver (Bob) separated by a quantum channel with transmissivity
$\eta_{\text{ch}}$ and thermal number $\bar{n}_{e}=\bar{n}_{B}/(1-\eta
_{\text{ch}})$.\ Bob's setup has quantum efficiency $\eta_{\text{eff}}$\ and
extra thermal photons $\bar{n}_{\text{ex}}$. The mean number of photons at the
input ($\bar{n}_{T}$) and output ($\bar{n}_{R}$) follow Eq.~(\ref{IOenergy}),
while the input classical variable ($x$)\ and the output one ($y$) follow
Eq.~(\ref{IOnew}). We also describe the various trust levels for the receiver.
In the scenario \textquotedblleft Eve~(1)\textquotedblright, the eavesdropper
is assumed to attack the external channel only. In the scenario
\textquotedblleft Eve~(2)', there is also a passive side-channel attack where
the eavesdropper collects leakage from the receiver's setup. Finally, in the
scenario \textquotedblleft Eve~(3)\textquotedblright, we assume that the
eavesdropper is also able to perform an active side-channel attack, so that
the noise internal to the setup has to be considered untrusted.
\label{attackPIC
\end{figure}
Bob's detection is either a randomly-switched homodyne, measuring $\hat{q}$ or
$\hat{p}$ \cite{GG02}, or heterodyne, realizing the joint measurement of
$\hat{q}$ and $\hat{p}$~\cite{noswitch}. We may treat both cases compactly
with the same formalism. In both protocols, Bob retrieves an outcome $y$ which
corresponds to Alice's input $x$. For the homodyne protocol, there is a single
pair $(x,y)$ for each mode transmitted by Alice while, for the heterodyne
protocol, there are two pairs of variables per mode (but affected by more noise).
The input-output relation for the total channel from the classical input $x$
to the output $y$ takes the form
\begin{equation}
y=\sqrt{\tau}x+z, \label{IOnew
\end{equation}
where $z$ is a noise variable. The latter is given by
\begin{align}
z & =\sqrt{\eta_{\text{eff}}(1-\eta_{\text{ch}})}\hat{x}_{e}+\sqrt{\tau
\hat{x}_{0}\nonumber\\
& +\sqrt{1-\eta_{\text{eff}}}\hat{x}_{v}+z_{\text{ex}}+z_{\text{det}},
\end{align}
where $\hat{x}_{e}$ denotes the quadrature of the thermal mode $e$, $\hat
{x}_{v}$ is the quadrature associated with setup vacuum mode $v$ (quantum
efficiency), $z_{\text{ex}}$ is a Gaussian variable with $\mathrm{var
(z_{\text{ex}})=2\bar{n}_{\text{ex}}$ accounting for the extra noise of the
setup,\ and $z_{\text{det}}$ is an additional Gaussian variable with
$\mathrm{var}(z_{\text{det}})=\nu_{\text{det}}-1$ where $\nu_{\text{det}}$ is
the quantum duty (`qu-duty') associated with detection: $\nu_{\text{det}}=1$
for homodyne and $\nu_{\text{det}}=2$ for heterodyne. See also
Fig.~\ref{attackPIC}. In total the noise variable $z$ has varianc
\begin{equation}
\sigma_{z}^{2}=2\bar{n}+\nu_{\text{det}}. \label{sigmazed
\end{equation}
From the input-output relation of Eq.~(\ref{IOnew}), we may compute Alice and
Bob's mutual information $I(x:y)$ which takes the same expression in direct
reconciliation (where Bob infers $x$ from $y$) and reverse reconciliation
(where Alice infers $y$ from $x$). In fact, from $\mathrm{var}(y)=\tau
\sigma_{x}^{2}+\sigma_{z}^{2}$ and $\mathrm{var}(y|x)=\sigma_{z}^{2}$, we ge
\begin{equation}
I(x:y)=\frac{\nu_{\text{det}}}{2}\log_{2}\left( 1+\frac{\sigma_{x}^{2}}{\chi
}\right) , \label{MutualINFOeq
\end{equation}
where
\begin{equation}
\chi:=\frac{\sigma_{z}^{2}}{\tau}=\frac{2\bar{n}+\nu_{\text{det}}}{\tau
\end{equation}
is the equivalent noise. Clearly $I(x:y)$ can be specified to $I^{\text{hom}}$
(for homodyne)\ and $I^{\text{het}}$ (for heterodyne) by choosing the
corresponding value for $\nu_{\text{det}}$.
Note that the equivalent noise can be re-written as
\begin{equation}
\chi=\xi_{\text{tot}}+\frac{\nu_{\text{det}}}{\tau},~\xi_{\text{tot}
:=\frac{2\bar{n}}{\tau}, \label{totexcess
\end{equation}
where $\xi_{\text{tot}}$ defines the total excess noise. In turn, the total
excess noise can be decomposed a
\begin{align}
\xi_{\text{tot}} & =\xi_{\text{ch}}+\xi_{\text{ex}},\\
\xi_{\text{ch}} & :=\frac{2(\bar{n}-\bar{n}_{\text{ex}})}{\tau}=\frac
{2\eta_{\text{eff}}\bar{n}_{B}}{\tau},\label{chexcess}\\
\xi_{\text{ex}} & :=\frac{2\bar{n}_{\text{ex}}}{\tau}, \label{excess
\end{align}
where $\xi_{\text{ch}}$ is the excess noise of the external channel, i.e.,
related to the thermal background, while $\xi_{\text{ex}}$ is that associated
with the extra noise in the setup.
Let us make an important remark on notation. The use of the excess noise
$\xi_{\text{tot}}$ is typical in fiber-based communication channels, while the
use of the equivalent number of thermal photons $\bar{n}$ is instead more
appropriate for free-space channels. In general, the two notations are related
by the formulas above and can be used interchangeably. In the following, we
choose to work with $\bar{n}$ which is particularly convenient from the point
of view of the finite-size estimators. However, for completeness, we also
provide the corresponding formulations in terms of excess noise.
\subsection{Local oscillator and setup noise}
Before discussing security aspects, let us discuss the local oscillator (LO)
and then clarify the main contributions to the setup noise. In terms of
equivalent number of thermal photons, the setup noise can be decomposed as
$\bar{n}_{\text{ex}}=\bar{n}_{\text{LO}}+\bar{n}_{\text{el}}+\bar
{n}_{\text{other}}$, where $\bar{n}_{\text{LO}}$ is the mean number of thermal
photons associated with the phase errors of the LO, $\bar{n}_{\text{el}}$ is
the mean number of thermal photons generated by electronic noise, and $\bar
{n}_{\text{other}}$ is any other uncharacterized but independent source of
noise (here neglected). Similarly, we may write a corresponding decomposition
in terms of excess noise $\xi_{\text{ex}}=\xi_{\text{LO}}+\xi_{\text{el}
+\xi_{\text{other}}$, which is obtained by using $\xi_{(...)}=2\bar{n
_{(...)}/\tau$.
\subsubsection{Phase-locking via TLO\ or phase-reconstruction via
LLO\label{LLOTLO}}
LO\ is crucial in CV-QKD\ since it contains the phase information that allows
the parties to exploit the two quadratures of the mode. In other words,
Alice's and Bob's rotating reference frames need to be phase-locked so Bob can
measure the incoming state in the same quadrature(s) chosen by Alice. To
achieve this goal there are two techniques, the simplest solution of the
transmitted LO (TLO)~\cite{GG02} and the more challenging (but more secure)
one of the local LO\ (LLO)~\cite{LLO,LLO2,LLO3,QKDreview}.
With the TLO, the LO is generated by the transmitter and multiplexed in
polarization with the signal mode/pulse. Both of them are sent through the
channel and then de-multiplexed by the receiver before being interfered in the
homodyne/heterodyne setup. With the LLO, bright reference pulses are regularly
interleaved with the signal pulses (time multiplexing). At the receiver, both
the signals and the references are measured with an independent local LO. From
the references, Bob is able to track Alice's rotating frame and, using this
phase information, he suitably rotates the outcomes obtained from the signals
in the phase space.
Note that both TLO and LLO\ require to employ half of the total pulses for
phase locking or reconstruction. When we explicitly consider a clock $C$ for
the system (pulses per second), the LLO involves an extra factor $1/2$ in
front of the final key rate, unless this is compensated by using both the
polarizations for the signal transmissions (not possible for the TLO).
\subsubsection{Contributions to setup noise}
From the point of view of the setup noise, we need to account for phase errors
introduced by an imperfect LO. In TLO this is negligible ($\bar{n
_{\text{TLO}}\simeq0$), while for the LLO it is non-trivial. In fact, assume
that signal and reference pulses are generated with an average linewidth
$l_{\text{W}}=(l_{\text{W}}^{\text{signal}}+l_{\text{W}}^{\text{LO}})/2$.
Then, for input classical modulation $\sigma_{x}^{2}$ and transmissivity
$\tau$, we may write~\cite{FSpaper}
\begin{equation}
\bar{n}_{\text{LLO}}\simeq\Theta_{\text{ph}}\tau,~\Theta_{\text{ph}
:=\pi\sigma_{x}^{2}C^{-1}l_{\text{W}}. \label{UBnex
\end{equation}
This contribution can equivalently be written as excess noise $\xi
_{\text{LLO}}=2\bar{n}_{\text{LLO}}/\tau$, according to Eq.~(\ref{excess}).
For a cw-laser $l_{\text{W}}\simeq1.6$~KHz, a clock $C=5~$MHz and a typical
modulation $\sigma_{x}^{2}=9$ (i.e., $\mu=10$) one has $\xi_{\text{LLO}
\simeq0.018$.
While the LLO introduces phase errors, it may actually be better when we
consider the impact of electronic noise. The latter can be described by a
variance $\nu_{\text{el}}$ or an equivalent number of photons $\bar
{n}_{\text{el}}=\nu_{\text{el}}/2$. Its value depends on the frequency of the
light $\nu$, features of the homodyne/heterodyne detector, such as its
noise equivalent power (NEP) and the bandwidth $W$, as well as features of the
LO, such as its power at detection $P_{\text{LO}}^{\text{det}}$\ and the
duration of its pulses $\Delta t_{\text{LO}}$. In fact, we may write
\begin{equation}
\bar{n}_{\text{el}}=\frac{\nu_{\text{det}}\mathrm{NEP}^{2}W\Delta
t_{\text{LO}}}{2h\nu P_{\text{LO}}^{\text{det}}}.
\end{equation}
In the case of a TLO, one has $P_{\text{LO}}^{\text{det}}=\tau P_{\text{LO}}$,
where $P_{\text{LO}}$ is the LO\ initial power at the transmitter. For an LLO,
we instead have $P_{\text{LO}}^{\text{det}}=P_{\text{LO}}$. Thus, by setting
\begin{equation}
\Theta_{\text{el}}:=\frac{\nu_{\text{det}}\mathrm{NEP}^{2}W\Delta
t_{\text{LO}}}{2h\nu P_{\text{LO}}}, \label{eleCONTRIBUT
\end{equation}
we may writ
\begin{equation}
\bar{n}_{\text{el}}^{\text{TLO}}=\frac{\Theta_{\text{el}}}{\tau},~\bar
{n}_{\text{el}}^{\text{LLO}}=\Theta_{\text{el}},
\end{equation}
so the formulas for the total setup noise are
\begin{equation}
\bar{n}_{\text{ex}}^{\text{TLO}}\simeq\frac{\Theta_{\text{el}}}{\tau},~\bar
{n}_{\text{ex}}^{\text{LLO}}\simeq\Theta_{\text{el}}+\Theta_{\text{ph}}\tau.
\label{setupNoiseExp
\end{equation}
These formulas are in terms of equivalent number of thermal photons and they
have corresponding expressions in terms of setup excess noise by using
$\xi_{\text{ex}}=2\bar{n}_{\text{ex}}/\tau$.
Above we can see the different monotonicity of the setup noise with respect to
$\tau$, between TLO\ and LLO. Assume $\lambda=800~$nm and $W=100~$MHz, so we
have signal pulses of duration $\Delta t=10~$ns and LO pulses of duration
$\Delta t_{\text{LO}}=10~$ns. For this bandwidth, we can assume the good value
$\mathrm{NEP}=6~$pW/$\sqrt{\text{Hz}}$. Then, assuming $P_{\text{LO}}=100~$mW,
we get $\Theta_{\text{el}}\simeq1.45\times10^{-3}$ for heterodyne detection
($\nu_{\text{det}}=2$). For the LLO this value remains low, while for the TLO
it is rescaled by $1/\tau$, which means that it may become large at long
distances. See also Fig.~\ref{TLOLLO}\ for a comparison.\begin{figure}[t]
\vspace{0.2cm}
\par
\begin{center}
\includegraphics[width=0.70\columnwidth] {TLOandLLO.eps}
\end{center}
\par
\vspace{-0.5cm}\caption{Setup noise as a function of the total transmissivity
$\tau$ expressed in decibels. (a)~We plot the equivalent number of thermal
photons $\bar{n}_{\text{ex}}$ associated with the setup noise, for the TLO
(black lines) and the LLO\ (blue lines), considering the homodyne protocol
(solid lines) and the heterodyne protocol (dashed lines). (b)~As in~(a) but we
plot the setup excess noise $\xi_{\text{ex}}$. Parameters are chosen as in the
text. See Eq.~(\ref{setupNoiseExp}).
\label{TLOLLO
\end{figure}
\subsection{Trust levels\label{Section_levels}}
Once we have clarified the main sources of noise in the communication
scenario, we can go ahead and identify different levels of trust on the basis
of different assumptions for the eavesdropper (Eve). The basic model is to
assume that Eve's action is restricted to the outside channel. In this
strategy, she inserts her photons in the thermal background and stores all the
photons which are not collected by the receiver. However, she is assumed not
to monitor or control the receiver's setup. This is the scenario where loss
and noise are considered to be trusted in the receiver. See also Eve~(1) in
Fig.~\ref{attackPIC}. In this case, Eve's collective Gaussian attack is
represented by a purification of the environmental beam-splitter of
transmissivity $\eta_{\text{ch}}$, where the injected $\bar{n}_{e}^{(1)
=\bar{n}_{B}(1-\eta_{\text{ch}})^{-1}$ thermal photons are to be considered
part of a two-mode squeezed vacuum (TMSV) state in Eve's
hands~\cite{collectiveG}.
More generally, we can assume that Eve is able to detect the leakage from
setups~\cite{refB1,refB2,refB3}. Here we consider this potential problem for
the receiver's setup, so that the fraction $1-\eta_{\text{eff}}$ of the
photons missed by the detection is stored by Eve and becomes part of her
attack. On the other hand, we may assume that Eve is not able to actively
tamper with the receiver, i.e., she does not control the noise internal to the
setup, which may therefore be considered as trusted (this is a reasonable
assumption which is often made by experimentalists for the electronic noise of
the detector). We call this scenario the trusted-noise model for the receiver.
See Eve~(2) in Fig.~\ref{attackPIC}. In this case, the efficiency
$\eta_{\text{eff}}$ becomes part of Eve's environmental beam-splitter, which
now has total transmissivity $\tau=\eta_{\text{ch}}\eta_{\text{eff}}$ and
injects $\bar{n}_{e}^{(2)}=\eta_{\text{eff}}\bar{n}_{B}(1-\tau)^{-1}$ thermal photons.
Finally, there is the worst-case scenario where no imperfection in the
receiver setup is trusted. In fact, the most pessimistic assumption is that
Eve can also potentially control the extra photons in the setup $\bar
{n}_{\text{ex}}$ besides collecting its leakage. See also Eve~(3) in
Fig.~\ref{attackPIC}. In this case, the extra photons become part of Eve's
environment. In other words, the entire channel from the transmitter to the
final (ideal) detection is dilated into a single beam-splitter with
transmissivity $\tau=\eta_{\text{ch}}\eta_{\text{eff}}$ and injecting $\bar
{n}_{e}^{(3)}=\bar{n}(1-\tau)^{-1}$ thermal photons.
Clearly the security increases from the completely trusted receiver [Eve~(1)]
to the worst-case scenario [Eve~(3)]. Similarly, the key rate will decrease,
because more degrees of freedom would go under Eve's control. For this reason,
the worst-case scenario provides a lower bound for all the others. Also note
that the worst-case scenario progressively collapses in the lower levels if we
assume $\bar{n}_{\text{ex}}=0$ and then $\eta_{\text{eff}}=1$. Also note that,
in general, one \ may consider hybrid situations between Eve~(2) and Eve~(3),
where the setup noise $\bar{n}_{\text{ex}}$ is partly trusted ($\bar
{n}_{\text{ex}}^{\text{tr}}$) and partly untrusted ($\bar{n}_{\text{ex
}^{\text{unt}}$). This is included by writing $\bar{n}_{\text{ex}
^{\text{unt}}=\eta_{\text{eff}}\bar{n}_{B}^{\text{unt}}$ and increasing the
background $\bar{n}_{B}\rightarrow\bar{n}_{B}+\bar{n}_{B}^{\text{unt}}$.
\subsection{Asymptotic key rates\label{subsectionAKR}}
It is convenient to start by studying the security of the protocol with the
intermediate assumption of a trusted-noise detector as in
Fig.~\ref{dilationPIC2}, where the setup noise is considered to be trusted,
i.e., not coming from Eve's attack [cf. Eve~(2) in Fig.~\ref{attackPIC}].
Then, we analyze the key rate in the most optimistic case where also the setup
loss is considered to be trusted. Finally, we compare the formulas with the
worst-case scenario, where all noise is considered to be untrusted [cf.
Eve~(3) in Fig.~\ref{attackPIC}]. The latter represents the case analyzed in
Ref.~\cite{FSpaper}.
\subsubsection{Asymptotic key rate with a trusted-noise detector}
Consider the trusted-noise detector which corresponds to the dilated scenario
in Fig.~\ref{dilationPIC2}. Here the total transmissivity is $\tau
=\eta_{\text{ch}}\eta_{\text{eff}}$ and the injected thermal noise is given by
$\bar{n}_{e}^{(2)}=\eta_{\text{eff}}\bar{n}_{B}(1-\tau)^{-1}$. In order to
compute the asymptotic secret key rate in reverse reconciliation, we consider
Bob and Eve's joint covariance matrix (CM). Let us define the basic block
matrices $\mathbf{I}:=\mathrm{diag}(1,1)$ and $\mathbf{Z}:=\mathrm{diag
(1,-1)$. Then, the joint CM is given b
\begin{equation}
\mathbf{V}_{BEE^{\prime}}=\left(
\begin{array}
[c]{cc
b\mathbf{I} & \mathbf{C}\\
\mathbf{C}^{T} & \mathbf{V}_{EE^{\prime}
\end{array}
\right) , \label{jointCM
\end{equation}
where Eve's reduced CM $\mathbf{V}_{EE^{\prime}}$ and the cross-correlation
block $\mathbf{C}$ take the forms
\begin{equation}
\mathbf{V}_{EE^{\prime}}=\left(
\begin{array}
[c]{cc
\phi\mathbf{I} & \psi\mathbf{Z}\\
\psi\mathbf{Z} & \omega\mathbf{I
\end{array}
\right) ,~\mathbf{C}=\left(
\begin{array}
[c]{cc
\theta\mathbf{I} & \gamma\mathbf{Z
\end{array}
\right) , \label{CMcase1
\end{equation}
where we have se
\begin{align}
\omega & =2\bar{n}_{e}^{(2)}+1=\frac{2\eta_{\text{eff}}\bar{n}_{B}}{1-\tau
}+1=\frac{\tau\xi_{\text{ch}}}{1-\tau}+1,\label{EvesVAR}\\
b & =\tau(\mu-1)+2\bar{n}+1=\tau(\mu-1)+\tau\xi_{\text{tot}
+1,\label{parameterB}\\
\gamma & =\sqrt{(1-\tau)(\omega^{2}-1)},~\theta=\sqrt{\tau(1-\tau)
(\omega-\mu),\label{thetaEQ1}\\
\psi & =\sqrt{\tau(\omega^{2}-1)},~\phi=\tau\omega+(1-\tau)\mu. \label{fiEQ1
\end{align}
\begin{figure}[t]
\vspace{-1.7cm}
\par
\begin{center}
\includegraphics[width=0.5\textwidth] {Dilation.eps}
\end{center}
\par
\vspace{-2.0cm}\caption{Eve's collective attack under the assumption of
trusted noise in the receiver's setup, i.e., Eve~(2) in Fig.~\ref{attackPIC}.
\label{dilationPIC2
\end{figure}
In the homodyne protocol, Eve's conditional CM\ on Bob's outcome $y$ is given
by~\cite{RMP,GaeCM,GaeCM2
\begin{equation}
\mathbf{V}_{EE^{\prime}|B}^{\text{hom}}=\mathbf{V}_{EE^{\prime}
-b^{-1}\mathbf{C}^{T}\boldsymbol{\Pi}\mathbf{C},
\end{equation}
where $\boldsymbol{\Pi}:=\mathrm{diag}(1,0)$. In the heterodyne protocol, we
have instead the following conditional CM~\cite{RMP,GaeCM,GaeCM2
\begin{equation}
\mathbf{V}_{EE^{\prime}|B}^{\text{het}}=\mathbf{V}_{EE^{\prime}
-(b+1)^{-1}\mathbf{C}^{T}\mathbf{C}.
\end{equation}
Call $\{\nu_{\pm}\}$ the symplectic spectrum of Eve's CM\ $\mathbf{V
_{EE^{\prime}}$. Then, call $\{\nu_{\pm}^{\text{hom}}\}$ and $\{\nu_{\pm
}^{\text{het}}\}$ the symplectic spectra of Eve's conditional CMs
$\mathbf{V}_{EE^{\prime}|B}^{\text{hom}}$ and $\mathbf{V}_{EE^{\prime
|B}^{\text{het}}$, respectively. Then, we may compute Eve's Holevo information
for both protocols, a
\begin{align}
\chi^{\text{hom}}(\mathbf{E} & :y)=\sum\limits_{k=\pm}\left[ H(\nu
_{k})-H(\nu_{k}^{\text{hom}})\right] ,\label{homoCHI}\\
\chi^{\text{het}}(\mathbf{E} & :y)=\sum\limits_{k=\pm}\left[ H(\nu
_{k})-H(\nu_{k}^{\text{het}})\right] , \label{heteroCHI
\end{align}
where $\mathbf{E}=EE^{\prime}$ and $H(x)$ is the entropic functio
\begin{equation}
H(x):=\frac{x+1}{2}\log_{2}\frac{x+1}{2}-\frac{x-1}{2}\log_{2}\frac{x-1}{2}.
\end{equation}
For a realistic reconciliation efficiency $\beta\in\lbrack0,1]$, accounting
for the fact that data-processing may not reach the Shannon limit, we write
the asymptotic key rate
\begin{equation}
R_{\text{asy}}^{(2)}(\tau,\bar{n},\bar{n}_{B})=\beta I(x:y)_{\tau,\bar{n
}-\chi(\mathbf{E}:y)_{\tau,\bar{n},\bar{n}_{B}}, \label{rateASYm
\end{equation}
where the explicit expressions for the homodyne protocol~\cite{GG02} and the
heterodyne protocol~\cite{noswitch} derive from the corresponding expressions
for the mutual information [cf. Eq.~(\ref{MutualINFOeq})] and the Holevo bound
[cf. Eqs.~(\ref{homoCHI}) and~(\ref{heteroCHI})].
It is clear that, in a practical setting, the parties do not know all the
parameters entering the rate in Eq.~(\ref{rateASYm}), so they need to resort
to suitable procedures of parameter estimation. It is acceptable to assume
that Alice controls/knows the signal modulation $\mu$, while Bob
monitors/knows the quantum efficiency $\eta_{\text{eff}}$. The channel
parameters $\tau$ and $\bar{n}$ need to be estimated. In general, the setup noise
$\bar{n}_{\text{ex}}$ depends on the total transmissivity $\tau$. For this
reason, $\bar{n}_{\text{ex}}$ too needs to be estimated by the parties. The
estimates of $\bar{n}$ and $\bar{n}_{\text{ex}}$ then provide the value of
$\bar{n}_{B}$.
\subsubsection{Asymptotic key rate with a trusted-loss and trusted-noise
detector}
Here we consider the best possible scenario for Alice and Bob, which is the
assumption of Eve~(1) in Fig.~\ref{attackPIC}. Not only the setup noise is
trusted but also the loss of the setup into the external environment is
considered to be trusted (i.e., we assume Eve is not collecting the leakage
from the setup). The asymptotic key rate can be found by a simple modification
of the previous derivation.
From the point of view of Alice and Bob, the mutual information is clearly the
same. For Eve instead, the effective beam splitter used in her attack has now
transmissivity $\eta_{\text{ch}}$ and input thermal noise $\bar{n}_{e
^{(1)}=\bar{n}_{B}(1-\eta_{\text{ch}})^{-1}$. It is easy to check that we need
to use the CM in Eq.~(\ref{CMcase1}) with the replacement
\begin{align}
\omega & =2\bar{n}_{e}^{(1)}+1=\frac{2\bar{n}_{B}}{1-\eta_{\text{ch}
}+1=\frac{\eta_{\text{ch}}\xi_{\text{ch}}}{1-\eta_{\text{ch}}}+1,\\
\gamma & =\sqrt{\eta_{\text{eff}}(1-\eta_{\text{ch}})(\omega^{2}-1)},\\
\theta & =\sqrt{\tau(1-\eta_{\text{ch}})}(\omega-\mu),\label{thetaEQ2}\\
\psi & =\sqrt{\eta_{\text{ch}}(\omega^{2}-1)},~\phi=\eta_{\text{ch}
\omega+(1-\eta_{\text{ch}})\mu, \label{fiEQ2}
\end{align}
while parameter $b$ is the same as in Eq.~(\ref{parameterB}).
The next steps are as before. One computes the symplectic spectrum $\{\nu
_{\pm}\}$ of the CM $\mathbf{V}_{EE^{\prime}}$ and those, $\{\nu_{\pm
}^{\text{hom}}\}$ and $\{\nu_{\pm}^{\text{het}}\}$, of the conditional CMs
$\mathbf{V}_{EE^{\prime}|B}^{\text{hom}}$ and $\mathbf{V}_{EE^{\prime
|B}^{\text{het}}$. These eigenvalues are then replaced in Eqs.~(\ref{homoCHI})
and~(\ref{heteroCHI}). In this way, we get the corresponding asymptotic key
rates $R_{\text{asy}}^{(1)}(\tau,\bar{n},\bar{n}_{B})$ following the formula
in Eq.~(\ref{rateASYm}). Parameters need to be estimated in the same way as
explained in the previous subsection.
\subsubsection{Asymptotic key rate with untrusted
detector\label{AsyUntrustedSection}}
In the worst-case scenario of untrusted noise [cf. Eve~(3) in
Fig.~\ref{attackPIC}], the entire channel is dilated into a single beam
splitter with transmissivity $\tau=\eta_{\text{ch}}\eta_{\text{eff}}$, where
Eve injects $\bar{n}_{e}^{(3)}=\bar{n}(1-\tau)^{-1}$ thermal photons. Setup
noise $\bar{n}_{\text{ex}}$ becomes part of Eve's attack, so all excess noise
is now considered to be untrusted. From the point of view of the asymptotic
key rate, it is sufficient to replace $\eta_{\text{eff}}\bar{n}_{B}=\bar
{n}-\bar{n}_{\text{ex}}\rightarrow\bar{n}$ in the expression of Eve's variance
$\omega$ in Eq.~(\ref{EvesVAR}), with implicit modifications for the other
elements of the CM. More precisely, it is sufficient to se
\begin{equation}
\omega=2\bar{n}_{e}^{(3)}+1=\frac{2\bar{n}}{1-\tau}+1=\frac{\tau
\xi_{\text{tot}}}{1-\tau}+1. \label{untrustedOMEGA
\end{equation}
Alternatively, we can exploit the entanglement-based representation of the
protocol according to which Alice's Gaussian-modulated coherent states are
realized by heterodyning mode $A$ of a TMSV state~\cite{RMP} with CM
\begin{equation}
\mathbf{V}_{AA^{\prime}}=\left(
\begin{array}
[c]{cc
\mu\mathbf{I} & \sqrt{\mu^{2}-1}\mathbf{Z}\\
\sqrt{\mu^{2}-1}\mathbf{Z} & \mu\mathbf{I
\end{array}
\right) .
\end{equation}
After the thermal-loss channel with total transmissivity $\tau$, Alice and
Bob's shared Gaussian state $\rho_{AB}$ has C
\begin{equation}
\mathbf{V}_{AB}=\left(
\begin{array}
[c]{cc
\mu\mathbf{I} & \sqrt{\tau(\mu^{2}-1)}\mathbf{Z}\\
\sqrt{\tau(\mu^{2}-1)}\mathbf{Z} & b\mathbf{I
\end{array}
\right) .
\end{equation}
Eve is assumed to hold the purification of $\rho_{AB}$, so the total state
$\rho_{AB\mathbf{E}}$ of Alice, Bob and Eve is pure. This means that
$S(\mathbf{E})=S(AB)$, where $S(Q)$ denotes the von Neumann entropy computed
over the state $\rho_{Q}$ of system $Q$. Then, because homodyne/heterodyne is
a rank-1 measurement (projecting pure states in pure states), we have that
$\rho_{A\mathbf{E}|y}$ is pure, which implies the equality of the conditional
entropies $S(\mathbf{E}|y)=S(A|y)$. As a result, Eve's Holevo bound is simply
given by
\begin{equation}
\chi(\mathbf{E}:y):=S(\mathbf{E})-S(\mathbf{E}|y)=S(AB)-S(A|y).
\label{ChiJOINT}
\end{equation}
Thus, we may compute $\chi(\mathbf{E}:y)$ using Alice and Bob's CM
$\mathbf{V}_{AB}$ with symplectic eigenvalues $\nu_{\pm}^{\prime}$. It is easy
to find~\cite{FSpaper
\begin{align}
\chi^{\text{hom}}(\mathbf{E} & :y)=H(\nu_{+}^{\prime})+H(\nu_{-}^{\prime
})-H\left[ \sqrt{\mu^{2}-\frac{\mu\tau(\mu^{2}-1)}{b}}\right] ,\\
\chi^{\text{het}}(\mathbf{E} & :y)=H(\nu_{+}^{\prime})+H(\nu_{-}^{\prime
})-H\left[ \mu-\frac{\tau(\mu^{2}-1)}{b+1}\right] ,
\end{align}
where $b$ is given in Eq.~(\ref{parameterB}).
Using these expressions and the mutual information of Eq.~(\ref{MutualINFOeq
), we writ
\begin{equation}
R_{\text{asy}}^{(3)}(\tau,\bar{n})=\beta I(x:y)_{\tau,\bar{n}}-\chi
(\mathbf{E}:y)_{\tau,\bar{n}}. \label{asyUNrate
\end{equation}
Note that the parties only need to estimate the extended-channel parameters
$\tau$ and $\bar{n}$. As we see below these estimators are built up to some
error probability $\varepsilon_{\text{pe}}$.
\subsection{Parameter estimation\label{PEmainSEC}\label{PEsection
\label{PEsectionLAter}}
As mentioned in the previous section, Alice and Bob need to estimate some of
the parameters. Even if they control the values of the input Gaussian
modulation $\mu$ and they can calibrate the output quantum efficiency
$\eta_{\text{eff}}$, they still need to estimate the various channel's
parameters and the setup noise $\bar{n}_{\text{ex}}$. The procedure has some
differences depending if we consider a trusted or untrusted model for the
receiver. For a trusted-noise detector [Eve~(2)] and a fully-trusted detector
[Eve~(1)], Alice and Bob need to estimate $\tau$, $\bar{n}$ and $\bar{n}_{B}$
(via $\bar{n}_{\text{ex}}$). For the untrusted detector [Eve~(3)], they only
need to estimate $\tau$ and $\bar{n}$, since the two thermal contributions
$\bar{n}_{B}$ and $\bar{n}_{\text{ex}}$ are both considered to be untrusted
(and therefore merged into a single parameter).
We therefore consider two basic independent estimators $\hat{\tau}$ and
$\widehat{\bar{n}}$, for $\tau$ and $\bar{n}$. Then, in the trusted scenarios
[Eve~(1) and~(2)], we also require the use of additional estimators, which can
be derived from the basic ones. To estimate the parameters, Alice and Bob
randomly and jointly choose $m$ of the $N$ distributed signals, and publicly
disclose the corresponding $m_{p}:=\nu_{\text{det}}m$ pairs of values
$\{x_{i},y_{i}\}_{i=1}^{m_{p}}$. These are $m$ pairs for the homodyne protocol
and $2m$ pairs for the heterodyne protocol. Under the standard assumption of a
collective (entangling-cloner) Gaussian attack, these pairs are independent
and identically distributed Gaussian variables, related by Eq.~(\ref{IOnew}).
From the pairs, they build the estimator $\hat{T}$ of the square-root
transmissivity $T:=\sqrt{\tau}$, i.e.
\begin{equation}
\hat{T}=\frac{\sum_{i=1}^{m_{p}}x_{i}y_{i}}{\sum_{i=1}^{m_{p}}x_{i}^{2}},
\end{equation}
and the estimator $\widehat{\sigma_{z}^{2}}$ of the noise variance $\sigma
_{z}^{2}$, i.e.,
\begin{equation}
\widehat{\sigma_{z}^{2}}=\frac{1}{m_{p}}\sum_{i=1}^{m_{p}}(y_{i}-\hat{T
x_{i})^{2}. \label{sigmaZesimates
\end{equation}
From these, we can derive the two basic estimators
\begin{equation}
\hat{\tau}:=\hat{T}^{2},~\widehat{\bar{n}}:=\frac{\widehat{\sigma_{z}^{2}
-\nu_{\text{det}}}{2}.
\end{equation}
For a confidence parameter $w$, we then define and compute the worst-case
estimators~\cite{NoteEstimator}
\begin{align}
\tau^{\prime} & :=\hat{\tau}-w\sqrt{\mathrm{var}(\hat{\tau})}\simeq
\tau-2w\sqrt{\frac{2\tau^{2}+\tau\sigma_{z}^{2}/\sigma_{x}^{2}}{m_{p}
},\label{wcEstimator1}\\
\bar{n}^{\prime} & :=\widehat{\bar{n}}+w\sqrt{\mathrm{var}(\widehat{\bar{n
})}\simeq\bar{n}+w\frac{\sigma_{z}^{2}}{\sqrt{2m_{p}}}. \label{wcEstimator2
\end{align}
Each of these estimators bounds the corresponding actual value, $\tau$ and
$\bar{n}$, up to an error probability $\varepsilon_{\text{pe}}$ if we tak
\begin{equation}
w=\sqrt{2}\operatorname{erf}^{-1}(1-2\varepsilon_{\text{pe}}),
\label{wSTVALUE
\end{equation}
or, in case of low values ($\varepsilon_{\text{pe}}\leq10^{-17}$), if we tak
\begin{equation}
w=\sqrt{2\ln(1/\varepsilon_{\text{pe}})}. \label{wTAIL
\end{equation}
As a result the total error probability associated with parameter estimation
is $\simeq2\varepsilon_{\text{pe}}$. See Ref.~\cite{FSpaper} for more
technical details on these derivations, which exploit tools from
Ref.~\cite{UsenkoFinite} and involves suitable tail
bounds~\cite{TailBound,Kolar}.
For the trusted-detector scenarios, we need to provide the best-case estimator
of $\bar{n}_{\text{ex}}$, which automatically allows us to derive the
worst-case estimator of $\bar{n}_{B}$. From the analytical expressions in
Eq.~(\ref{setupNoiseExp}), we see that we need to account for the different
behavior of $\bar{n}_{\text{ex}}$ in terms of the transmissivity $\tau$, which
requires both the use of a worst-case estimator $\tau^{\prime}$ and that of a
best-case estimator $\tau^{\prime\prime}:=\hat{\tau}+w\sqrt{\mathrm{var
(\hat{\tau})}$. In other words, we hav
\begin{align}
\bar{n}_{\text{ex}}^{\text{TLO}} & \gtrsim\bar{n}_{\text{ex,bc}
^{\text{TLO}}:=\frac{\Theta_{\text{el}}}{\tau^{\prime\prime}},\label{bcc1}\\
\bar{n}_{\text{ex}}^{\text{LLO}} & \gtrsim\bar{n}_{\text{ex,bc}
^{\text{LLO}}:=\Theta_{\text{el}}+\Theta_{\text{ph}}\tau^{\prime}.
\label{bcc2
\end{align}
Correspondingly, we have the following worst-case estimator for the background
thermal noise
\begin{equation}
\bar{n}_{B}\lesssim\bar{n}_{B}^{\prime}:=\frac{\bar{n}^{\prime}-\bar
{n}_{\text{ex,bc}}}{\eta_{\text{eff}}}.
\end{equation}
We can now compute the values of the asymptotic key rates affected by
parameter estimation. For the various scenarios, these are given by
\begin{align}
R_{\text{asy}}^{(1,2)}(\tau,\bar{n},\bar{n}_{B}) & \rightarrow\frac{n
{N}R_{\text{asy}}^{(1,2)}(\tau^{\prime},\bar{n}^{\prime},\bar{n}_{B}^{\prime
}),\\
R_{\text{asy}}^{(3)}(\tau,\bar{n}) & \rightarrow\frac{n}{N}R_{\text{asy
}^{(3)}(\tau^{\prime},\bar{n}^{\prime}),
\end{align}
where $n=N-m$ is the number of signals left for key generation (after $m$ are
discarded for parameter estimation). These key rates are correct up to an
error $\simeq2\varepsilon_{\text{pe}}$.
As a final remark, notice that the total excess noise $\xi_{\text{tot}}$ can
be estimated by using $\hat{\tau}$ and $\widehat{\bar{n}}$\ via
Eq.~(\ref{totexcess}) and therefore worst-case estimated by using
$\tau^{\prime}$ and $\bar{n}^{\prime}$, i.e.,
\begin{equation}
\xi_{\text{tot}}\lesssim\xi_{\text{tot}}^{\prime}:=\frac{2\bar{n}^{\prime
}{\tau^{\prime}}.
\end{equation}
Similarly, the channel excess noise $\xi_{\text{ch}}$ can be worst-case
estimated by combining Eq.~(\ref{chexcess}) with $\tau^{\prime}$\ and $\bar
{n}_{B}^{\prime}$, i.e.
\begin{equation}
\xi_{\text{ch}}\lesssim\xi_{\text{ch}}^{\prime}:=\frac{2\eta_{\text{eff}
\bar{n}_{B}^{\prime}}{\tau^{\prime}}.
\end{equation}
\subsection{Composable finite-size key rates\label{KR_sec}}
After parameter estimation, each block of size $N$ provides $n$ signals to be
processed into a shared key via error correction and privacy amplification.
Given a block, this is successfully error-corrected with probability
$p_{\text{ec}}$ (or failure probability $\mathrm{FER}=1-p_{\text{ec}}$ known
as `frame error rate'). The value of $p_{\text{ec}}$ depends on the
signal-to-noise ratio, the target reconciliation efficiency $\beta$, and the
$\varepsilon$-correctness $\varepsilon_{\text{cor}}$, the latter bounding the
probability that Alice's and Bob's local strings are different after error
correction and successful verification of their hashes.
On average $np_{\text{ec}}$ signals per block are promoted to privacy
amplification. This final step is implemented with an associated $\varepsilon
$-secrecy $\varepsilon_{\text{sec}}$, the latter bounding the distance between
the final key and an ideal key that is completely uncorrelated from Eve. In
turn, the $\varepsilon$-secrecy is technically decomposed as $\varepsilon
_{\text{sec}}=\varepsilon_{\text{s}}+\varepsilon_{\text{h}}$, where
$\varepsilon_{\text{s}}$ is a smoothing parameter and $\varepsilon_{\text{h}}$
is a hashing parameter.
Overall, the final composable key rate of the protocol takes the
form~\cite{FSpaper}
\begin{equation}
R\geq\frac{np_{\text{ec}}}{N}\left( R_{\text{pe}}^{(k)}-\frac{\Delta
_{\text{aep}}}{\sqrt{n}}+\frac{\Theta}{n}\right) , \label{sckeee
\end{equation}
where $R_{\text{pe}}^{(k)}$ depends on the receiver mode
\begin{equation}
R_{\text{pe}}^{(1,2)}=R_{\text{asy}}^{(1,2)}(\tau^{\prime},\bar{n}^{\prime
},\bar{n}_{B}^{\prime}),~~R_{\text{pe}}^{(3)}=R_{\text{asy}}^{(3)
(\tau^{\prime},\bar{n}^{\prime}), \label{RpeFORMU
\end{equation}
and the extra finite-size terms are equal to
\begin{align}
& \Delta_{\text{aep}}=4\log_{2}\left( 2\sqrt{d}+1\right) \sqrt{\log
_{2}\left( \frac{18}{p_{\text{ec}}^{2}\varepsilon_{\text{s}}^{4}}\right)
},\label{deltaAEPPP}\\
& \Theta=\log_{2}[p_{\text{ec}}(1-\varepsilon_{\text{s}}^{2}/3)]+2\log
_{2}\sqrt{2}\varepsilon_{\text{h}}. \label{bigOMEGA
\end{align}
Here the parameter $d$ is the size of Alice's and Bob's effective alphabet
after analog-to-digital conversion of their continuous variables $x$ and $y$
($d=2^{5}=32$ for a $5$-bit discretization). This rate refers to security
against collective Gaussian attacks with total epsilon security~\cite{FSpaper
\begin{equation}
\varepsilon=2p_{\text{ec}}\varepsilon_{\text{pe}}+\varepsilon_{\text{cor
}+\varepsilon_{\text{sec}}. \label{epsSECcollective
\end{equation}
\subsubsection{Improved pre-factor}
Note that the prefactor $\log_{2}(2\sqrt{d}+1)$ in the AEP\ term in
Eq.~(\ref{deltaAEPPP}) can be tightened into $\log_{2}(\sqrt{d}+2)$. In
general, according to Theorem~6.4 and Corollary~6.5 of Ref.~\cite{TomaThesis},
one can lower-bound the conditional smooth min-entropy $H_{\min}^{\delta
}(y^{n}|\mathbf{E}^{n})$ associated with the $n$-use classical-quantum state
$\rho_{y\mathbf{E}}^{\otimes n}$\ shared between Bob (classical system $y$)
and Eve (quantum system $\mathbf{E}$). This is done by using the conditional
entropy between the single-use systems ($y$ and $\mathbf{E}$) up to a penalty,
i.e., we may write~\cite{TomaThesis,TomaBook}
\begin{equation}
H_{\min}^{\delta}(y^{n}|\mathbf{E}^{n})_{\rho^{\otimes n}}\geq nH(y|\mathbf{E
)_{\rho}+\sqrt{n}\Delta_{\text{aep}}(\delta),
\end{equation}
where
\begin{align}
\Delta_{\text{aep}}(\delta) & =4(\log_{2}v)\sqrt{-\log_{2}(1-\sqrt
{1-\delta^{2}})}\nonumber\\
& \simeq4(\log_{2}v)\sqrt{\log_{2}(2/\delta^{2})}\\
v & \leq\sqrt{2^{-H_{\min}(y|\mathbf{E})}}+\sqrt{2^{H_{\max}(y|\mathbf{E})
}+1,\label{equationV
\end{align}
with $v$ being bounded using min- and max-entropies. Recall that the min- and
max-entropies can be negative in general, but their absolute values must be
$\leq\log_{2}d$, with $d$ being the size of Bob's alphabet (e.g., this easily
follows from Ref.~\cite[Lemma~5.2]{TomaBook}). This implies the bound
$v\leq2\sqrt{d}+1$, which leads to the prefactor used in Eq.~(\ref{deltaAEPPP
). See Ref.~\cite[Appendix~G]{FSpaper} for details on how to connect the key
rate with the conditional smooth min-entropy and simplify derivations via the
AEP term.
However, it is worth noting that, for a classical-quantum state $\rho
_{y\mathbf{E}}$, the conditional min-entropy is non-negative, i.e., $H_{\min
}(y|\mathbf{E})\geq0$. This is a property that can be shown, more generally,
for separable states. In fact, starting from the definition of conditional
min-entropy for a generic state $\rho_{AB}$ of two quantum systems $A$ and $B$
\cite[Def. 4.1]{TomaThesis}, we can write the lower boun
\begin{equation}
H_{\min}(A|B)_{\rho}\geq\tilde{H}:=\sup\{\lambda\in\mathbb{R}:\rho_{AB
\leq2^{-\lambda}I_{A}\otimes\rho_{B}\}.\label{eq2
\end{equation}
For separable $\rho_{AB}$, one may write~\cite[Lemma~5.2]{TomaBook
\begin{equation}
\rho_{AB}=\sum_{k}p_{k}\theta_{A}^{k}\otimes\rho_{B}^{k}\leq\sum_{k}p_{k
I_{A}\otimes\rho_{B}^{k}=I_{A}\otimes\rho_{B},
\end{equation}
which leads to $\tilde{H}\geq0$, since we are left to find the
\textit{maximum} value of $\lambda$ such that
\begin{equation}
\rho_{AB}\leq I_{A}\otimes\rho_{B},~\rho_{AB}\leq2^{-\lambda}I_{A}\otimes
\rho_{B}.
\end{equation}
Thus, using $H_{\min}(y|\mathbf{E})\geq0$ in Eq.~(\ref{equationV}), we may
write $v\leq\sqrt{d}+2$ which improves Eq.~(\ref{deltaAEPPP}) into
\begin{equation}
\Delta_{\text{aep}}=4\log_{2}\left( \sqrt{d}+2\right) \sqrt{\log_{2}\left(
\frac{18}{p_{\text{ec}}^{2}\varepsilon_{\text{s}}^{4}}\right) }.
\end{equation}
Note that, for a typical $5$-bit digitalization $d=2^{5}$, we have $\log
_{2}(\sqrt{d}+2)\simeq2.94$ instead of $\log_{2}(2\sqrt{d}+1)\simeq3.6$, so
the improvement is limited. In our numerical investigations we assume the
worst-case pre-factor, but keeping in mind that performances can be slightly improved.
\subsubsection{Extension to coherent attacks}
For the heterodyne protocol, the key rate can be extended to security against
general attacks using tools from Ref.~\cite{Lev2017}. Let us symmetrize the
protocol by applying an identical random orthogonal matrix to the classical
continuous variables of the two parties. Then, assume that Alice and Bob
jointly perform $m_{\mathrm{et}}=f_{\mathrm{et}}n$ energy tests on randomly
chosen uses of the channel (for some factor $f_{\mathrm{et}}<1$). In each
test, the parties measure the local number of photons (which can be
extrapolated from the data) and compute an average over the $m_{\mathrm{et}}$
tests. If these averages are greater than a threshold $d_{\text{et}}$, the
protocol is aborted. Setting $d_{\text{et}}\gtrsim\bar{n}_{T}+\mathcal{O
(m_{\mathrm{et}}^{-1/2})$ assures secure success of the test in typical
scenarios (where signals are attenuated and noise is not too high).
The number of signals for key generation is reduced to
\begin{equation}
n=N-(m+m_{\mathrm{et}})=\frac{N-m}{1+f_{\mathrm{et}}},
\end{equation}
and the procedure needs an additional step of privacy amplification
compressing the final key by a further amount
\begin{align}
\Phi_{n} & :=2\left\lceil \log_{2}\binom{K_{n}+4}{4}\right\rceil
,\label{FInGEN}\\
K_{n} & :=\max\left\{ 1,2nd_{\text{et}}\frac{1+2\sqrt{\vartheta
+2\vartheta}{1-2\sqrt{\vartheta/f_{\mathrm{et}}}}\right\} ,
\end{align}
where we have set $\vartheta:=(2n)^{-1}\ln(8/\varepsilon)$.
The composable key rate reads~\cite{FSpaper
\begin{equation}
R_{\text{gen}}^{\text{het}}\geq\frac{np_{\text{ec}}}{N}\left[
R_{\text{pe,het}}^{(k)}-\frac{\Delta_{\text{aep}}}{\sqrt{n}}+\frac{\Theta
-\Phi_{n}}{n}\right] , \label{compoHETgen
\end{equation}
where $R_{\text{pe,het}}^{(k)}$ is the rate in Eq.~(\ref{RpeFORMU}) depending
on the noise model for the receiver and suitably specified for the heterodyne
protocol. Assuming that the original protocol had $\varepsilon$-security
against collective Gaussian attacks, the symmetrized protocol has security
$\varepsilon^{\prime}=K_{n}^{4}\varepsilon/50$\ against general attacks. Note
that this implies a very demanding condition for the epsilon parameters, such
as $\varepsilon_{\text{pe}}$. As a matter of fact, $\varepsilon_{\text{pe}}$
should be so small that the confidence parameter needs to be calculated
according to Eq.~(\ref{wTAIL}).
\subsection{Numerical investigations}
We may use the previous formulas to plot the composable key rate for the
homodyne/heterodyne protocol with TLO/LLO under each noise model for the
receiver, i.e., corresponding to each of the three different assumptions for
Eve as depicted in Fig.~\ref{attackPIC}. Here we numerically investigate the
most interesting case which is the heterodyne protocol with LLO, for which we
show the performances associated with the three noise models under collective
attacks, and also the worst-case performance associated with the
untrusted-noise model under general attacks. We adopt the physical parameters
listed in Table~\ref{tableMODELS} and the protocol parameters in
Table~\ref{tablePARAMETERS}. The results are given in terms of secret key rate
versus total loss in the channel and can be applied to both fiber-based and
free-space quantum communications, as long as for the latter scenario we can
assume a stable channel (i.e., we can exclude or suitably ignore
fading~\cite{PanosFading}).\begin{table}[t]
\vspace{0.2cm}
\begin{tabular}
[c]{|l|l|l|}\hline
Physical parameter & Symbol & Value\\\hline\hline
Wavelength & $\lambda$ & $800~$nm\\\hline
Detector shot-noise & $\nu_{\text{det}}$ & $2~\text{(het)}$\\\hline
Detector efficiency & $\eta_{\text{eff}}$ & $0.7$ ($1.55~$dB)\\\hline
Detector bandwidth & $W$ & $100~$MHz\\\hline
Noise equivalent power & NEP & $6~$pW/$\sqrt{\text{Hz}}$\\\hline
Linewidth & $l_{\text{W}}$ & $1.6$~KHz\\\hline
LO power & $P_{\text{LO}}$ & $100~$mW\\\hline
Clock & $C$ & $5~$MHz\\\hline
Pulse duration & $\Delta t,\Delta t_{\text{LO}}$ & $10~$ns\\\hline
Setup noise (LLO) &
\begin{array}
[c]{l
\bar{n}_{\text{ex}}\\
\xi_{\text{ex}
\end{array}
$ &
\begin{array}
[c]{l
\text{Eq.~(\ref{setupNoiseExp})}\\
\text{Eq.~(\ref{excess})
\end{array}
$\\\hline
Channel noise &
\begin{array}
[c]{l
\bar{n}_{B}\\
\xi_{\text{ch}
\end{array}
$ &
\begin{array}
[c]{l
1/500\\
\text{Eq.~(\ref{chexcess})
\end{array}
$\\\hline
Total thermal noise &
\begin{array}
[c]{l
\bar{n}\\
\xi_{\text{tot}
\end{array}
$ &
\begin{array}
[c]{l
\text{Eq.~(\ref{nBARvalue})}\\
\text{Eq.~(\ref{totexcess})
\end{array}
$\\\hline
\end{tabular}
\caption{Physical parameters.
\label{tableMODELS
\end{table}\begin{table}[t]
\vspace{0.2cm}
\begin{tabular}
[c]{|l|l|l|l|}\hline
\begin{array}
[c]{l
\text{Protocol}\\
\text{parameter
\end{array}
$ & Symbol &
\begin{array}
[c]{l
\text{Collective}\\
\text{attacks
\end{array}
$ &
\begin{array}
[c]{l
\text{General}\\
\text{attacks
\end{array}
$\\\hline\hline
Total pulses & $N$ & $10^{7}$ & $10^{7}$\\\hline
PE signals & $m$ & $0.1\times N$ & $0.1\times N$\\\hline
Energy tests & $f_{\text{et}}$ & $-$ & $0.2$\\\hline
KG signals & $n$ & $0.9\times N$ & $\simeq7.5\times10^{6}$\\\hline
Digitalization & $d$ & $2^{5}$ & $2^{5}$\\\hline
Rec. efficiency & $\beta$ & $0.95$ & $0.95$\\\hline
EC success prob & $p_{\text{ec}}$ & $0.9$ & $0.1$\\\hline
Epsilons & $\varepsilon_{\text{h,s,\ldots}}$ & $2^{-33}\simeq10^{-10}$ &
$10^{-43}$\\\hline
Confidence & $w$ & $\simeq6.34$ & $\simeq14.07$\\\hline
Security & $\varepsilon,\varepsilon^{\prime}$ & $\simeq5.6\times10^{-10}$ &
$\simeq1.4\times10^{-13}$\\\hline
Modulation & $\mu$ & $10$ & $10$\\\hline
\end{tabular}
\caption{Protocol parameters adopted with respect to collective attacks and
general attacks.
\label{tablePARAMETERS
\end{table}
The results are shown in Fig.~\ref{comparisonPIC} where we are particularly
interested in the high-rate short-range setting. As we can see from the
figure, the rate has a non-trivial improvement as a result of the stronger
assumptions made for the receiver, as expected. Considering the standard
loss-rate of an optical fiber ($0.2$ dB/km), we see that one extra dB of
tolerance for the rate corresponds to additional $5$ km. Clearly this is
achievable as long as the security assumptions about the receiver are
acceptable by the parties.
\begin{figure}[t]
\vspace{0.2cm}
\par
\begin{center}
\includegraphics[width=0.45\textwidth] {comparison.eps}
\end{center}
\par
\vspace{-0.5cm}\caption{Composable secret key rate (bits/use) versus total
loss (decibels) for the heterodyne protocol with LLO. We plot the rates
against collective attacks assuming a trusted-loss and trusted-noise receiver
(black dotted), a trusted-noise receiver (black dashed), and an untrusted
receiver (solid black). We also show the performance achievable with the
untrusted receiver in the presence of general attacks (red). The gray line is
the total excess noise $\xi_{\text{tot}}$ in shot noise units. Finally, the
blue lines refer to line-of-sight security (discussed in Sec.~\ref{LoSsection
) for trusted-loss and trusted-noise receiver (blue dotted), and trusted-noise
receiver (blue dashed). Physical and protocol parameters are chosen as in
Tables~\ref{tableMODELS} and~\ref{tablePARAMETERS}.
\label{comparisonPIC
\end{figure}
\section{Security of near-range free-space quantum communications\label{SEC2}}
Let us now discuss the specific setting of free-space quantum communications
which generally requires some elaborations of the formulas above in order to
account for the additional physical processes occurring in this scenario. In
the following we discuss one potential extra simplification and realistic
assumption for security, and then we treat the issues related to near-range
wireless communications at various frequencies and with different types of
receivers (fixed or mobile).
\subsection{Line-of-sight security\label{LoSsection}}
The line-of-sight (LoS) security is a strong but yet realistic assumption for
free-space quantum communications in the near range (say within $100$ meters
or so). The idea is that transmitter and receiver can \textquotedblleft
see\textquotedblright\ each other, so it is unlikely that Eve is able to
tamper with the middle channel. A realistic attack is here to collect photons
which are lost in the environment; in other words it is a passive attack which
can be interpreted as the action of a pure-loss channel, i.e., a beam-splitter
with no injection of thermal photons (which are the active entangled probes
employed in the usual entangling-cloner attack).
Within the LoS assumption, there are additional degrees of reality for Eve's
attack. The most realistic scenario is Eve using a relatively-small device
which only collects a fraction of the photons that are leaked into the
environment. The worst-case picture which can be used as a bound for the key
is to assume Eve collecting all the leaked photons. In this case, the
performance will strictly depend on how much the receiver is able to intercept
of the incoming beam which is in turn related to the geometric features of the
beam itself (collimated, focused, or spherical beam). In any case, any thermal
noise which is present in the environment is considered to be trusted.
In the studies below, we consider both LoS security (Eve passive on the
channel) and standard security (Eve active on the channel). Under LoS
security, thermal noise is considered to be trusted, which means that the
relevant models for the detector are those with trusted noise [Eve~(2)] and
trusted noise and loss [Eve~(1)]. The attack can be represented as in
Fig.~\ref{attackPIC}\ but where Eve does not control environmental modes,
represented by mode $e$ for Eve~(1) and modes $e$,$v$ for Eve~(2). With the
trusted-noise detector, we also allow Eve to collect leakage from Bob's setup;
with the trusted-noise-and-loss detector, this additional side-channel is
excluded. Depending on the cases, we adopt one assumption or the other. See
Table~\ref{tableSEClevels} for a summary of the security types and trust
levels (associated detector models). These definitions are meant to be in
addition to the classification into individual, collective and
coherent/general attacks.
\begin{table}[h
\begin{tabular}
[c]{|l|l|l|}\hline
Channel noise & Security type & Detector model\\\hline\hline
Untrusted &
\begin{tabular}
[c]{l
Standard security\\
(Active Eve\\
controlling the\\
environment)
\end{tabular}
&
\begin{tabular}
[c]{l
\begin{tabular}
[c]{l
$\bullet~$Untrusted\\
\lbrack Eve~(3)]
\end{tabular}
\\%
\begin{tabular}
[c]{l
$\bullet$~Noise-trusted\\
\lbrack Eve~(2)]
\end{tabular}
\\%
\begin{tabular}
[c]{l
$\bullet~$Noise-loss-trusted\\
\lbrack Eve~(1)]
\end{tabular}
\end{tabular}
\\\hline
Trusted &
\begin{tabular}
[c]{l
LoS security\\
(Passive Eve.\\
No control of\\
the environment)
\end{tabular}
&
\begin{tabular}
[c]{l
\begin{tabular}
[c]{l
$\bullet~$Noise-trusted\\
\lbrack Eve~(2)]
\end{tabular}
\\%
\begin{tabular}
[c]{l
$\bullet~$Noise-loss-trusted\\
\lbrack Eve~(1)]
\end{tabular}
\end{tabular}
\\\hline
\end{tabular}
\caption{Security types and trust levels (detector models). The security
assumptions become stronger from top to bottom.
\label{tableSEClevels
\end{table}
The secret key rates under LoS security are derived by excluding Eve from the
control of the environmental noise. This means that her CM\ is reduced from
the form in Eq.~(\ref{CMcase1}) to just the block $\phi\mathbf{I}$. Thus, we
have to consider the simpler joint CM for Bob and Ev
\begin{equation}
\mathbf{V}_{BE}=\left(
\begin{array}
[c]{cc
b\mathbf{I} & \theta\mathbf{I}\\
\theta\mathbf{I} & \phi\mathbf{I
\end{array}
\right) , \label{VbeLOS
\end{equation}
leading to the conditional CM
\begin{equation}
\mathbf{V}_{E|B}^{\text{hom}}=\left(
\begin{array}
[c]{cc
\phi-\frac{\theta^{2}}{b} & 0\\
0 & \phi
\end{array}
\right) ,~\mathbf{V}_{E|B}^{\text{het}}=\left( \phi-\frac{\theta^{2}
{b+1}\right) \mathbf{I}. \label{Vbe2LOS
\end{equation}
Therefore, Eve's Holevo bound to be used in the key rates is simply given b
\begin{align}
\chi_{\text{LoS}}^{\text{hom}}(E & :y)=H(\phi)-H\left[ \sqrt{\phi
(\phi-\theta^{2}/b)}\right] ,\\
\chi_{\text{LoS}}^{\text{het}}(E & :y)=H(\phi)-H\left( \phi-\frac
{\theta^{2}}{b+1}\right) , \label{HetLOSformula
\end{align}
where the explicit expressions for $\theta$ and $\phi$ depend on the detector
noise model, while $b$ is given in Eq.~(\ref{parameterB}).
Using these expressions, we may then write the asymptotic key rate with LoS
security for the two detector models ($k=1,2$). Recalling that the mutual
information is expressed as in Eq.~(\ref{MutualINFOeq}), the LoS key rate is
given b
\begin{equation}
R_{\text{asy,LoS}}^{(k)}(\tau,\bar{n},\bar{n}_{B})=\beta I(x:y)_{\tau,\bar{n
}-\chi_{\text{LoS}}(E:y)_{\tau,\bar{n},\bar{n}_{B}},
\end{equation}
taking specific expressions for the homodyne protocol [$R_{\text{asy,LoS,hom
}^{(k)}$] and the heterodyne protocol [$R_{\text{asy,LoS,het}}^{(k)}$]. After
parameter estimation, the modified key rate will be expressed in terms of the
worst-case estimators as $R_{\text{pe,LoS}}^{(k)}=R_{\text{asy,LoS}
^{(k)}(\tau^{\prime},\bar{n}^{\prime},\bar{n}_{B}^{\prime})$. Finally, the
composable finite-size LoS key rate takes the expression in Eq.~(\ref{sckeee})
proviso we make the replacement $R_{\text{pe}}^{(k)}\longrightarrow
R_{\text{pe,LoS}}^{(k)}$. Improvement in performance is shown in
Fig.~\ref{comparisonPIC}.
\subsection{Optical wireless with fixed devices}
Let us consider a free-space optical link between transmitter and receiver.
Assume that this is mediated by a Gaussian \textrm{TEM}$_{00}$ beam with
initial spot-size $w_{0}$ and phase-front radius of curvature $R_{0
$~\cite{Goodman,Siegman,svelto}. This beam has a single well-defined
polarization (scalar approximation) and carrier frequency $\nu=c/\lambda$,
with $\lambda$ being the wavelength and $c$ the speed of light (so angular
frequency is $\omega=2\pi c/\lambda$, and wavenumber is $k=\omega
/c=2\pi/\lambda$). The pulse duration $\Delta t$ and frequency bandwidth
$\Delta\nu$ satisfy the time-bandwidth product for Gaussian pulses, i.e.,
$\Delta t\Delta\nu\gtrsim0.44$. In particular, we may assume $\Delta
t\Delta\nu\simeq1$. Under the paraxial wave approximation, we assume
free-space propagation along the $z$ direction with no limiting apertures in
the transverse plane, neglecting diffraction effects at the transmitter (e.g.,
by assuming a suitable aperture for the transmitter with radius $\geq2w_{0
$~\cite{Siegman}).
By introducing the Rayleigh range
\begin{equation}
z_{R}:=\frac{\pi w_{0}^{2}}{\lambda},
\end{equation}
which identifies near- and far-field, we may write the following expression
for the diffraction-limited spot size of the beam at generic distance
$z$~\cite{Siegman,svelto
\begin{equation}
w_{z}^{2}=w_{0}^{2}\left[ \left( 1-\frac{z}{R_{0}}\right) ^{2}+\left(
\frac{z}{z_{R}}\right) ^{2}\right] .
\end{equation}
In particular, for a collimated beam ($R_{0}=\infty$), we get
\begin{equation}
w_{z}^{2}=w_{0}^{2}[1+(z/z_{R})^{2}], \label{colliB
\end{equation}
while for a focused beam ($R_{0}=z$), we hav
\begin{equation}
w_{z}^{2}=w_{0}^{2}(z/z_{R})^{2}=\left( \frac{\lambda z}{\pi w_{0}}\right)
^{2}. \label{focusedB
\end{equation}
We see that, in the far field $z\gg z_{R}$, the expressions in
Eqs.~(\ref{colliB}) and~(\ref{focusedB}) tend to coincide.
Consider then a receiver with a sharped-edged circular aperture with radius
$a_{R}$. The total power impinging on this aperture is given b
\begin{equation}
P(z,a_{R})=\frac{\pi w_{0}^{2}}{2}\eta_{\text{d}},~\eta_{\text{d
}:=1-e^{-2a_{R}^{2}/w_{z}^{2}}, \label{ILexp
\end{equation}
where parameter $\eta_{\text{d}}$ is the non-unit transmissivity of the
channel due to the free-space diffraction and the finite size of the receiver.
Note that, for far field and a receiver's size comparable with the
transmitter's (so $a_{R}\simeq w_{0}$), we have $w_{z}\gg a_{R}$ and therefore
the approximation
\begin{equation}
\eta_{\text{d}}\simeq\eta_{\text{d}}^{\text{far}}:=2a_{R}^{2}/w_{z}^{2}\ll1.
\end{equation}
For a collimated or focused beam, this become
\begin{equation}
\eta_{\text{d}}^{\text{far}}\simeq2\left( \frac{\pi w_{0}a_{R}}{\lambda
z}\right) ^{2}.
\end{equation}
The overall transmissivity of the system can be written as $\tau
=\eta_{\text{ch}}\eta_{\text{eff}}$, where $\eta_{\text{ch}}=\eta_{\text{d
}\eta_{\text{atm}}$ is the total transmissivity of the external channel which
generally includes the effect of atmospheric extinction $\eta_{\text{atm}}$.
Since the latter effect is negligible at short distances ($\eta_{\text{atm
}\simeq1$), we may just write $\eta_{\text{ch}}\simeq\eta_{\text{d}}$. By
contrast, the other term $\eta_{\text{eff}}$ is the total quantum efficiency
of the receiver and its contribution is typically non-negligible, e.g.,
$\eta_{\text{eff}}\simeq0.7$. Because the devices are assumed to be fixed,
there is no fading, meaning that the total transmissivity can be assumed to be
constant and equal to $\tau$.
The quantum communication scenario can be described as in Fig.~\ref{attackPIC
, where $\eta_{\text{ch}}$ is essentially given by free-space diffraction and
the thermal background $\bar{n}_{B}$ needs to be carefully evaluated from the
sky brightness (see below). Then, we can certainly assume standard security
with the trust levels $k=0,1,2$ according to which Eve's interaction is
described by different effective beam-splitters with different amounts of
input thermal noise $\bar{n}_{e}^{(k)}$ (see Sec.~\ref{Section_levels}).
Similarly, we may investigate LoS security where thermal noise is assumed to
be trusted.
Sky brightness $B_{\lambda}^{\text{sky}}$ is measured in W m$^{-2}$ nm$^{-1}$
sr$^{-1}$ and its value typically varies from $\simeq1.5\times10^{-6}$ (clear
night)\ to $\simeq1.5\times10^{-1}$ (cloudy day)~\cite{Miao,BrussSAT}, if one
assumes that the receiver field of view is shielded from direct exposition to
bright sources (e.g., the sun). Let us assume a receiver with aperture $a_{R}$
and angular field of view $\Omega_{\text{fov}}$ (in steradians). Assume the
receiver has a detector with bandwidth $W$ and spectral filter $\Delta\lambda
$. Then, the mean number of background thermal photons per mode collected by
the receiver is equal t
\begin{equation}
\bar{n}_{B}=\frac{\pi\lambda\Gamma_{R}}{hc}B_{\lambda}^{\text{sky}
,~\Gamma_{R}:=\Delta\lambda W^{-1}\Omega_{\text{fov}}a_{R}^{2}. \label{downT
\end{equation}
In this formula, we can estimate $\Omega_{\text{fov}}^{1/2}\simeq
2\arctan[l_{\text{D}}/(2f_{\text{D}})]$ from the linear size of the sensor of
the detector $l_{\text{D}}$ and the focal length $f_{\text{D}}$ of the
receiver. For $l_{\text{D}}=2~$mm and $f_{\text{D}}=20~$cm, we find
$\Omega_{\text{fov}}\simeq10^{-4}~$sr. Note that the latter value of the field
of view is relatively-large compared with typical values considered in
long-range setting, including satellite communications (where $\Omega
_{\text{fov}}\simeq10^{-10}~$sr).
The effective value of the spectral filter $\Delta\lambda$ can be very narrow
in setups that are based on homodyne/heterodyne detection. The reason is
because the required mode-matching of the signal with the LO pulse provides a
natural interferometric process which effectively reduces the filter
potentially down to the time-product bandwidth. For instance, for an LO\ pulse
of $\Delta t_{\text{LO}}=10~$ns, we may assume a bandwidth $\Delta\nu=50~$MHz
which is $\geq0.44/\Delta t_{\text{LO}}$. Thus, interferometry at the homodyne
setup imposes an effective filter of $\Delta\lambda=\lambda^{2}\Delta
\nu/c\simeq0.1~$pm around $\lambda=800~$nm.
Finally, if we take the detector bandwidth $W=100~$MHz and we assume a small
area for the receiver's aperture, i.e., $a_{R}=1~$cm (so as to be compatible
with the typical sizes of near-range devices), then we compute $\bar{n
_{B}\simeq0.019$ photons per mode during a cloudy day. This is a non-trivial
amount of noise that leads to a clear discrepancy between the performance in
standard security (where channel's noise is considered to be untrusted) and
LoS security (where this noise is assumed to be trusted). Let us also remark
here that LoS security is a realistic assumption for receivers with a small
field of view, so the noise collected from free space is limited and unlikely
to come from an active Eve hidden in the environment.
For our numerical study we consider the physical parameters listed in
Table~\ref{WirelessTABLE}; these are compatible with indoor and near-range
optical wireless communications with small devices (e.g., laptops). This means
that, for the transmitter, we consider limited power (e.g., $10~$mW), and a
small spot size ($w_{0}=1~$mm). Similarly, for the receiver, we consider a
limited aperture ($a_{R}=1~$cm), non-unit quantum efficiency ($\eta
_{\text{eff}}=0.7$), and a realistic field of view $\Omega_{\text{fov}
\simeq10^{-4}~$sr as discussed above.
\begin{table}[t
\begin{tabular}
[c]{|l|l|l|}\hline
Physical parameter & Symbol & Value\\\hline\hline
Altitude & $h$ & $30$~m\\\hline
Beam curvature & $R_{0}$ & $\infty$ (collimated)\\\hline
Wavelength & $\lambda$ & $800~$nm\\\hline
Beam spot size & $w_{0}$ & $1~$mm\\\hline
Receiver aperture & $a_{R}$ & $1$ cm\\\hline
Receiver field of view & $\Omega_{\text{fov}}$ & $10^{-4}~$sr\\\hline
Homodyne filter & $\Delta\lambda$ & $0.1~\text{pm}$\\\hline
Detector shot-noise & $\nu_{\text{det}}$ & $2~\text{(het)}$\\\hline
Detector efficiency & $\eta_{\text{eff}}$ & $0.7$ ($1.55~$dB)\\\hline
Detector bandwidth & $W$ & $100~$MHz\\\hline
Noise equivalent power & NEP & $6~$pW/$\sqrt{\text{Hz}}$\\\hline
Linewidth & $l_{\text{W}}$ & $1.6$~KHz\\\hline
LO power & $P_{\text{LO}}$ & $10~$mW\\\hline
Clock & $C$ & $5~$MHz\\\hline
Pulse duration & $\Delta t,\Delta t_{\text{LO}}$ & $10~$ns\\\hline
Setup noise with LLO & $\bar{n}_{\text{ex}}$ & Eq.~(\ref{setupNoiseExp
)\\\hline
Channel noise & $\bar{n}_{B}$ & $0.019$ [Eq.~(\ref{downT})]\\\hline
Total thermal noise & $\bar{n}$ & Eq.~(\ref{nBARvalue})\\\hline
Atmospheric extinction & $\eta_{\text{atm}}$ & $\simeq1$ (negligible)\\\hline
\end{tabular}
\caption{Physical parameters for optical wireless
\label{WirelessTABLE
\end{table}Assuming the physical parameters in Table~\ref{WirelessTABLE} and
the protocols parameters in Table~\ref{tablePARAMETERS}, we show the various
achievable performances of the free-space diffraction-limited heterodyne
protocol with LLO in Fig.~\ref{FixedWirelessPic}. As we can see from the
figure, we have drastically different rates depending on the type of security
and trust level. It is clear that the highest rates (and distances) are
obtained with LoS security (blue lines in the figure). With standard security,
the range is restricted to about $50$ meters (black lines in the figure) and
about $30$ meters in the worst-case scenario of an untrusted detector and
general attacks (red line in the figure). The possibility to enforce weaker
security assumptions leads to non-trivial advantages in terms of rate and
distance.\begin{figure}[t]
\vspace{0.1cm}
\par
\begin{center}
\includegraphics[width=0.45\textwidth] {Wireless_Fixed.eps}
\end{center}
\par
\vspace{-0.5cm}\caption{Optical-wireless QKD\ with fixed devices. We plot the
composable secret key rate (bits/use) versus free-space distance (meters) for
the heterodyne protocol with LLO. In particular, we show the rates against
collective attacks assuming a trusted-loss-and-noise receiver (black dotted),
a trusted-noise receiver (black dashed), and an untrusted receiver (solid
black). We also show the performance achievable with the untrusted receiver
versus general attacks (red). The blue lines refer to line-of-sight security
(discussed in Sec.~\ref{LoSsection}) for trusted-loss-and-noise receiver (blue
dotted), and trusted-noise receiver (blue dashed). Physical parameters are
chosen as in Table~\ref{WirelessTABLE}, while protocol parameters are in
Table~\ref{tablePARAMETERS}.
\label{FixedWirelessPic
\end{figure}
Also note the stability of the rates at short distances ($<30~$m) where their
values remain approximately constant. This is due to the fact that, for the
specific regime of parameters considered, the beam broadening induced by
free-space diffraction within that range [see Eq.~(\ref{colliB}) with
$w_{0}=1~$mm and $z<30$~m] is still limited with respect to the radius of the
receiver's aperture ($a_{R}=1~$cm). Thus, the transmissivity $\eta_{\text{d}}$
in Eq.~(\ref{ILexp}) remains sufficiently close to $1$, before starting to
decay after about $30$~m.
\subsection{Optical wireless with mobile devices}
\subsubsection{Pointing and tracking error}
In the presence of free-space optical connections with portable devices, one
can use a suitable tracking mechanism so the transmitter (such as a fixed
router/hot spot) points at the mobile receiver in real time with some small
pointing error. In general, the receiver too may have a mechanism of adaptive
optics aimed at maintaining the beam alignment by rotating the field of view
in direction of the transmitter. We therefore need to introduce a pointing
error at the transmitter $\tilde{\sigma}_{\text{P}}$ which introduces a
Gaussian wandering of the beam centroid over the receiver's aperture with
variance $\sigma_{\text{P}}^{2}\simeq(\tilde{\sigma}_{\text{P}}z)^{2}$ for
distance $z$. We assume an accessible value $\tilde{\sigma}_{\text{P}
\simeq1.745\times10^{-3}$ radiant, which is about $1/10$ of a degree (this is
orders-of-magnitude worse than the performance achievable in satellite-based
pointing and tracking).
Let us call $r$ the instantaneous deflection of the beam centroid from the
center of the receiver's aperture. The wandering can be described by the
Weibull distribution
\begin{equation}
P_{\text{WB}}(r)=\frac{r}{\sigma_{\text{P}}^{2}}\exp\left( -\frac{r^{2
}{2\sigma_{\text{P}}^{2}}\right) . \label{WeibullDISTRIBUTION
\end{equation}
For each value of the deflection $r$, there is an associated instantaneous
transmissivity $\tau=\tau(r)$, which can be computed as follow
\begin{equation}
\tau(r)=e^{-\frac{4r^{2}}{w_{z}^{2}}}Q_{0}\left( \frac{2r^{2}}{w_{z}^{2
},\frac{4ra_{R}}{w_{z}^{2}}\right) , \label{etadiff
\end{equation}
where $Q_{0}(x,y)$ is an incomplete Weber integral~\cite{Agrest}.
Alternatively, we may use the approximation
\begin{equation}
\tau(r)\simeq\eta\exp\left[ -\left( \frac{r}{r_{0}}\right) ^{\gamma
}\right] , \label{etaSTanalytical
\end{equation}
wher
\begin{equation}
\eta:=\tau(0)=\eta_{\text{ch}}(z)\eta_{\text{eff}}\simeq\eta_{\text{d}
(z)\eta_{\text{eff}} \label{eta0
\end{equation}
is the maximum transmissivity at distance $z$ (corresponding to a beam that is
perfectly-aligned), while$\ \gamma$ and $r_{0}$\ are the following shape and
scale (positive) parameters
\begin{align}
\gamma & =\frac{4\eta_{\text{d}}^{\text{far}}\Lambda_{1}(\eta_{\text{d
}^{\text{far}})}{1-\Lambda_{0}(\eta_{\text{d}}^{\text{far}})}\left[ \ln
\frac{2\eta_{\text{d}}}{1-\Lambda_{0}(\eta_{\text{d}}^{\text{far}})}\right]
^{-1},\label{expGG1}\\
r_{0} & =a_{R}\left[ \ln\frac{2\eta_{\text{d}}}{1-\Lambda_{0
(\eta_{\text{d}}^{\text{far}})}\right] ^{-\frac{1}{\gamma}}, \label{expGG2
\end{align}
where $\Lambda_{n}(x):=e^{-2x}I_{n}\left( 2x\right) $ and $I_{n}$ is a
modified Bessel function of the first kind with order $n$~\cite[Eq.~(D2)
{Vasy12}.
By suitably combining Eqs.~(\ref{WeibullDISTRIBUTION})
and~(\ref{etaSTanalytical}), one can derive the fading statistics, i.e., the
probability distribution $P_{\text{fad}}$ associated with the instantaneous
transmissivity $\tau$, which is given b
\begin{equation}
P_{\text{fad}}(\tau)=\frac{r_{0}^{2}}{\gamma\sigma_{\text{P}}^{2}\tau}\left(
\ln\frac{\eta}{\tau}\right) ^{\frac{2}{\gamma}-1}\exp\left[ -\frac{r_{0
^{2}}{2\sigma_{\text{P}}^{2}}\left( \ln\frac{\eta}{\tau}\right) ^{\frac
{2}{\gamma}}\right] . \label{P0tau
\end{equation}
\subsubsection{Maximum wireless range}
Besides the beam wandering (and associated fading) due to pointing and
tracking error, there is also the further issue that a mobile receiver
generally has a variable distance from the transmitter, so the transmissivity
of the free-space link has an additional degree of variability. The latter
effect has a very slow dynamics with respect to typical clocks, meaning that a
block of reasonable size is distributed while the position of the receiver is
substantially unchanged. For example, for a detector bandwidth $W=100~$MHz, we
may use a clock of $C=W/3\simeq33~$MHz. In this case, a block of $10^{7}$
points will be distributed in $1/3$ of a second. For an indoor network,
assuming an average walking speed of $\simeq1.5$ m/s, this corresponds to a
$\simeq50$~cm free-space displacement of the receiver. In the worst-case
scenario where this displacement increases the distance from the transmitter,
we may assume that the distribution of the whole block occurs at the maximum distance.
In general, we may compute a lower bound by assuming that the entire quantum
communication (i.e., the communication of all the blocks) occurs with the
mobile device at the maximum distance from the transmitter. In other words, we
can fix a maximum range $z_{\text{max}}$ for the local network and assume this
value as worst-case scenario. Since the parties control the parameters of the
channel and know the instantaneous distance, they could process their data in
a way that it appears to be completely distributed at $z_{\text{max}}$ (data
distributed at $z<z_{\text{max}}$ can be attenuated and suitably thermalized
in post processing).
To be more precise the lower bound should be computed by minimizing the
transmissivity and maximizing the thermal noise over the distance $z\leq
z_{\text{max}}$, so that data is processed via a more lossy and noisy
channel. While the minimization of the transmissivity occurs at
$z=z_{\text{max}}$, the maximization of the thermal noise may occur at
different values of $z$, depending on the type of LO. In particular, this
value is $z=z_{\text{max}}$ for the TLO and $z=0$ for the LLO. The issue is
therefore resolved for the LLO if we keep the mobile device at
$z=z_{\text{max}}$ while bounding the LLO noise with the value for $z=0$.
Such an approach is not optimal but robust and applicable to outdoor wireless
networks with faster-moving devices (with a speed limited by the ratio between
$z_{\text{max}}$ and the total communication time). It is worth mentioning
that, a better but more complicated strategy relies on slicing the trajectory
of the moving device into sectors, with each sector being associated with the
communication of a single block and the final rate being given by the average
rate over the sectors. This is particularly useful in satellite quantum
communications where a trajectory is well defined (for instance, see the
technique of orbital slicing in Ref.~\cite{SATpaper}). However, for stochastic
trajectories on the ground, the analytical treatment is not immediate.
\subsubsection{Pilot modes and de-fading\label{defadingSEC}}
Besides the use of bright pointing/tracking modes and bright LLO-reference
modes, it is also important to use relatively-bright pilot modes that are
specifically employed for the real-time estimation of the instantaneous
transmissivity $\tau$, whose fluctuation is generally due to both pointing
error and distance variability (for mobile devices). These $m_{\text{PL}}$
pilots are randomly interleaved with $N_{\text{S}}:=N-m_{\text{PL}}$ signal
modes, where $N$\ are the total pulses. The pilots allow the parties to:
(i)~identify an overall interval for the transmissivity $\Delta=[\tau_{\min
},\tau_{\max}]$ in which $N_{\text{S}}p_{\Delta}$ signals are post-selected
with probability $p_{\Delta}$; (ii)~introduce a lattice in $\Delta$ with step
$\delta\tau$, so that each signal is associated with a corresponding narrow
bin of transmissivities $\Delta_{k}:=[\tau_{k},\tau_{k+1}]$, with $\tau
_{k}:=\tau_{\min}+(k-1)\delta\tau$ for $k=1,\ldots,M$ and $M=(\tau_{\max
-\tau_{\min})/\delta\tau$~\cite{NotePilots}.
Each bin $\Delta_{k}$ is selected with probability $p_{k}$\ and, therefore,
populated by $N_{\text{S}}p_{k}$ signals. There are corresponding
$\nu_{\text{det}}N_{\text{S}}p_{k}$ pairs of points $\{x_{i},y_{i
\}$\ satisfying the input-output relation of Eq.~(\ref{IOnew}), which here
read
\begin{equation}
y^{(k)}\simeq\sqrt{\tau_{k}}x+z^{(k)},
\end{equation}
where $z^{(k)}$ is a Gaussian noise variable with varianc
\begin{equation}
\sigma_{k}^{2}=2\bar{n}_{k}+\nu_{\text{det}},~\bar{n}_{k}:=\eta_{\text{eff
}\bar{n}_{B}+\bar{n}_{\text{ex}}(\tau_{k}).
\end{equation}
Bob can map these points into the first bin $\Delta_{1}$ of the interval via
the de-fading ma
\begin{equation}
y^{(k)}\rightarrow\tilde{y}^{(k)}=\sqrt{\frac{\tau_{\text{min}}}{\tau_{k}
}y^{(k)}+\sqrt{1-\frac{\tau_{\text{min}}}{\tau_{k}}}\xi_{\text{add}},
\end{equation}
where $\xi_{\text{add}}$ is Gaussian noise with variance $\nu_{\text{det}}$.
By repeating this procedure for all the bins, Bob create the new variable
\begin{equation}
\tilde{y}=\sqrt{\tau_{\text{min}}}x+\tilde{z}, \label{finalIO
\end{equation}
where $\tilde{z}$ is non-Gaussian noise with varianc
\begin{equation}
\sigma_{\tilde{z}}^{2}=2\bar{n}_{\ast}+\nu_{\text{det}},~\bar{n}_{\ast
:=\frac{\tau_{\text{min}}}{p_{\Delta}}\sum_{k}\frac{p_{k}}{\tau_{k}}\bar
{n}_{k}. \label{thermalNOISE
\end{equation}
This new variable is now associated with a single (worst-case) transmissivity
$\tau_{\text{min}}$, thus effectively removing the fading process from the
distributed data, i.e., from their $\nu_{\text{det}}N_{\text{S}}p_{\Delta}$
pairs of correlated points.
Exploiting the optimality of Gaussian attacks, the parties assume that
$\tilde{z}$ is Gaussian (overestimating Eve's performance). In this way, the
final input-output relation in Eq.~(\ref{finalIO}) reduces to considering a
simpler thermal-loss\ Gaussian channel with transmissivity $\tau_{\text{min}}$
and thermal number $\bar{n}_{\ast}$. See Ref.~\cite{FSpaper} for more details.
For a receiver at some fixed distance $z$ and only subject to pointing error,
we can assume $\tau_{\max}=\eta$ [cf. Eq.~(\ref{eta0})] and $\tau_{\text{min
}=f_{\text{th}}\eta$ for some threshold factor $f_{\text{th}}<1$. Then, the
probabilities $p_{\Delta}=p(\tau_{\text{min}},\tau_{\text{max}})$ and
$p_{k}=p(\tau_{k},\tau_{k+1})$ are computed from the formul
\begin{equation}
p(\tau_{1},\tau_{2}):=\int_{\tau_{1}}^{\tau_{2}}d\tau~P_{\text{fad}}(\tau),
\label{probFORMULA
\end{equation}
where $P_{\text{fad}}(\tau)$ is given in Eq.~(\ref{P0tau}).
In general, for a mobile receiver at variable distance $z$, Alice and Bob
compute the post-selection interval $\Delta$ and the lattice $\{\Delta_{k
\}$\ directly from data, together with the corresponding values of $p_{\Delta
}$ and $p_{k}$. As mentioned in the previous subsection, the performance in
this general scenario can be lower-bounded by the extreme case where the
receiver is assumed to be fixed at the maximum distance $z_{\text{max}}$ from
the transmitter (while maximizing thermal noise over $z$, whose maximum is at
$z_{\text{max}}$ for a TLO and at $z=0$ for an LLO). In this worst-case
scenario, we may exploit the formula in Eq.~(\ref{probFORMULA}) for the fading
probability (suitably computed at $z_{\text{max}}$) and derive an analytical
lower bound for the secret key rate.
\subsubsection{Estimators and key rate}
Let us assume the worst-case scenario of a receiver at the maximum range
$z_{\text{max}}$ from the transmitter, so the maximum transmissivity is
$\tau_{\max}=\eta(z_{\text{max}})$ and the minimum transmissivity is
$\tau_{\text{min}}=f_{\text{th}}\eta(z_{\text{max}})$ for some threshold value
$f_{\text{th}}$. These border values define a post-selection interval $\Delta$
which is sliced into a lattice of $M$\ narrow bins $\{\Delta_{k}\}$. The
instantaneous transmissivity $\tau$ will fluctuate according to the
distribution in Eq.~(\ref{P0tau}) with associated pointing error
$\sigma_{z_{\text{max}}}^{2}\simeq(\sigma_{\text{P}}z_{\text{max}})^{2}$ for
an empirical value $\sigma_{\text{P}}$ at the transmitter (e.g., $1/10$ of a
degree). As a result of the fluctuation, a value of the transmissivity $\tau$
is post-selected with probability $p_{\Delta}$ and populates bin $\Delta_{k}$
with probability $p_{k}$, according to the integral in Eq.~(\ref{probFORMULA}).
For the worst-case scenario, let us also assume that the thermal noise is
maximized over $z\leq z_{\text{max}}$ (and the fading process). Thus, for any
bin $\Delta_{k}$, we consider the following bound on the associated thermal
nois
\begin{equation}
\bar{n}_{k}\leq\bar{n}_{\text{wc}}=\eta_{\text{eff}}\bar{n}_{B}+\bar
{n}_{\text{ex,wc}}, \label{nboundd
\end{equation}
where the maximum setup noise $\bar{n}_{\text{ex,wc}}$ depends on the type of
LO and is given b
\begin{equation}
\bar{n}_{\text{ex,wc}}^{\text{TLO}}\simeq\Theta_{\text{el}}/\tau_{\text{min
},~\bar{n}_{\text{ex,wc}}^{\text{LLO}}\simeq\Theta_{\text{el}}+\pi\sigma
_{x}^{2}C^{-1}l_{\text{W}}. \label{eqabove
\end{equation}
Note that the first expression in Eq.~(\ref{eqabove}) above is computed on
$\tau_{\text{min}}=\tau_{\text{min}}(z_{\text{max}})$ while the second one is
computed for $\tau=1$ (maximum value at $z=0$). By replacing
Eq.~(\ref{nboundd}) in Eq.~(\ref{thermalNOISE}), we get the bound
\begin{equation}
\bar{n}_{\ast}\leq\bar{n}_{\text{wc}}.
\end{equation}
As already explained, the construction of the lattice is possible thanks to
the random pilots. In total, during the quantum communication, the parties
exchange\ $N$ quantum pulses, whose $m_{\text{PL}}$ are pilots and
$N_{\text{S}}=N-m_{\text{PL}}$ are signals. Using the pilots, the parties
post-select a fraction $N_{\text{S}}p_{\Delta}$ of the signals, with a smaller
fraction $N_{\text{S}}p_{k}$ allocated to the generic bin $\Delta_{k}$. After
de-fading, the parties are connected by an effective thermal-loss channel with
transmissivity $\tau_{\text{min}}=\tau_{\text{min}}(z_{\text{max}})$ and
thermal number $\bar{n}_{\text{wc}}$.
The parties sacrifice a portion $mp_{\Delta}$ of the post-selected signals
$N_{\text{S}}p_{\Delta}$ for parameter estimation (PE), so $np_{\Delta}$
signals are left for key generation, where $n=N_{\text{S}}-m$ (this value is
further reduced for security extended to general coherent attacks). Overall
the parties use $m_{\Delta}:=\nu_{\text{det}}mp_{\Delta}$ pairs of data points
for PE following the procedure described in Sec.~\ref{PEsection} with
effective transmissivity $\tau_{\text{min}}=\tau_{\text{min}}(z_{\text{max}})$
and $\sigma_{\text{wc}}^{2}=2\bar{n}_{\text{wc}}+\nu_{\text{det}}$. This leads
to the following bounds for the worst-case estimators~\cite{FSpaper
\begin{align}
\tau_{\text{LB}} & =\tau_{\text{min}}-2w\sqrt{\frac{2\tau_{\text{min}
^{2}+\tau_{\text{min}}\sigma_{\text{wc}}^{2}/\sigma_{x}^{2}}{m_{\Delta}}},\\
\bar{n}_{\text{UB}} & =\bar{n}_{\text{wc}}+w\frac{\sigma_{\text{wc}}^{2
}{\sqrt{2m_{\Delta}}},
\end{align}
where $\sigma_{x}^{2}$ is the input modulation and $w$ is the confidence
parameter [cf. Eqs.~(\ref{wSTVALUE}) and~(\ref{wTAIL})].
As we can see from the two estimators above, the relevant information is the
minimum transmissivity $\tau_{\text{min}}$ of the post-selection interval, the
maximum thermal noise $\bar{n}_{\text{wc}}$ over the range (and fading
process), and the number of post-selected points $m_{\Delta}$. The formulas
hold for a generic fading statistics, i.e., not necessarily given by
Eq.~(\ref{probFORMULA}), as long as we can evaluate $m_{\Delta}$. Also note
that, assuming Eq.~(\ref{probFORMULA}) and fixing a threshold transmissivity
$\tau_{\text{min}}$, the value of $m_{\Delta}$\ decreases by increasing $z$.
In other words, the fact that a worst-case device at the maximum range
provides a lower bound for a mobile device is also due to the decreased
statistics for PE.
In order to compute the key rates for the trusted models, we also need to
bound the worst-case estimator of the background thermal noise $\bar{n}_{B}$.
This is possible by writin
\begin{equation}
\bar{n}_{B}^{\text{UB}}=\frac{\bar{n}_{\text{UB}}-\bar{n}_{\text{ex,bc}}
{\eta_{\text{eff}}},
\end{equation}
where the best-case value $\bar{n}_{\text{ex,bc}}$ needs to be optimized over
the entire range $z\leq z_{\max}$ and the fading process. We therefore extend
Eqs.~(\ref{bcc1}) and~(\ref{bcc2}) to the following expression
\begin{equation}
\bar{n}_{\text{ex,bc}}^{\text{TLO}}:=\Theta_{\text{el}},~\bar{n
_{\text{ex,bc}}^{\text{LLO}}:=\Theta_{\text{el}}+\Theta_{\text{ph}
\tau_{\text{min}}.
\end{equation}
We now have all the elements to write the composable finite-size key rate,
which extends Eq.~(\ref{sckeee}) of Sec.~\ref{KR_sec} to the following
expressio
\begin{equation}
R\geq\frac{np_{\Delta}p_{\text{ec}}}{N}\left( R_{\text{pe}}^{(k)
-\frac{\Delta_{\text{aep}}}{\sqrt{np_{\Delta}}}+\frac{\Theta}{np_{\Delta
}\right) , \label{erreDELTAA
\end{equation}
where $n=N-(m+m_{\text{PL}})$ and $R_{\text{pe}}^{(k)}$ depends on the
receiver model ($k=1,2,3$). The latter takes the following expressions in
terms of the new estimator
\begin{align}
R_{\text{pe}}^{(1,2)} & =R_{\text{asy}}^{(1,2)}(\tau_{\text{LB}},\bar
{n}_{\text{UB}},\bar{n}_{B}^{\text{UB}}),\label{RpeFREE1}\\
R_{\text{pe}}^{(3)} & =R_{\text{asy}}^{(3)}(\tau_{\text{LB}},\bar
{n}_{\text{UB}}). \label{RpeFREE2
\end{align}
Alternatively, we may write Eq.~(\ref{erreDELTAA}) assuming LoS security,
which means to replace $R_{\text{pe}}^{(k)}$ with the key rate
\begin{equation}
R_{\text{pe,LoS}}^{(k)}=R_{\text{asy,LoS}}^{(k)}(\tau_{\text{LB}},\bar
{n}_{\text{UB}},\bar{n}_{B}^{\text{UB}}).
\end{equation}
The composable key rate in Eq.~(\ref{erreDELTAA}) is $\varepsilon$-secure
against collective Gaussian attacks [cf Eq.~(\ref{epsSECcollective})].
For the heterodyne protocol, we extend the composable key rate of
Eq.~(\ref{compoHETgen}) to the following expressio
\begin{equation}
R_{\text{gen}}^{\text{het}}\geq\frac{np_{\Delta}p_{\text{ec}}}{N}\left(
R_{\text{pe,het}}^{(k)}-\frac{\Delta_{\text{aep}}}{\sqrt{np_{\Delta}}
+\frac{\Theta-\Phi_{np_{\Delta}}}{np_{\Delta}}\right) , \label{hetDELTAgen
\end{equation}
where $n$ must account for the $m_{\text{PL}}$ pilots besides the
$m_{\mathrm{et}}$ energy tests, i.e.,
\begin{equation}
n=N-(m+m_{\text{PL}}+m_{\text{et}})=\frac{N-(m+m_{\text{PL}})
{1+f_{\mathrm{et}}},
\end{equation}
and $R_{\text{pe,het}}^{(k)}$ is given by Eqs.~(\ref{RpeFREE1})
and~(\ref{RpeFREE2}) for the case of the heterodyne protocol. This rate has
epsilon security $\varepsilon^{\prime}=K_{np_{\Delta}}^{4}\varepsilon/50$
against general attacks, with $\varepsilon$ being the initial security versus
collective attacks (see Sec.~\ref{KR_sec}).
We perform a numerical investigation assuming the heterodyne protocol with
LLO. This is now implemented in a post-selection fashion in a way to remove
the (non-Gaussian) effect of fading from the distributed data (see above). We
consider the protocol parameters in Table~\ref{tablePARAMETERS} but where we
include the pilots $m_{\text{PL}}=0.05\times N$, so the key generation signals
are reduced to $n\simeq7.08\times10^{6}$, and a threshold parameter
$f_{\text{th}}=0.8$ for post-selection. We then assume the physical parameters
in Table~\ref{WirelessTABLE}, but taking a higher clock value $C=33~$MHz and
also including the transmitter's pointing error $\tilde{\sigma}_{\text{P}}$,
equal to $1/10$ of degree. In this regime of parameters, we study the
composable key rates that are achievable under the various security and trust
assumptions, considering a mobile device which can move up to a maximum
distance $z_{\max}$ from the transmitter (range of the wireless network).
The rates are plotted in Fig.~\ref{mobilePIC}. Note that the values in the
range of $10^{-2}-1$ bit/use correspond to high rates in the range of
$0.33-33~$Mbits/sec at the considered clock. This means that quantum-encrypted
wireless communication at about $1~$Mbit/sec are possible within distances of
a few meters. Another important consideration is that these rates are actually
lower bounds, since they are computed with the device at the maximum distance
and bounding the noise. This is also the reason why the key rate of
Eq.~(\ref{hetDELTAgen}) does not appear for this specific choice of
parameters. \begin{figure}[t]
\vspace{0.2cm}
\par
\begin{center}
\includegraphics[width=0.9\columnwidth] {mobile_rates.eps}
\end{center}
\par
\vspace{-0.4cm}\caption{Optical-wireless QKD\ with mobile devices. We plot the
composable secret key rate (bits/use) versus the maximum free-space distance
$z_{\max}$ of the receiver-device from the transmitter (meters). This is for a
pilot-guided post-selected heterodyne protocol with an LLO. We show the rates
against collective attacks assuming a trusted-loss-and-noise receiver (black
dotted), a trusted-noise receiver (black dashed), and an untrusted receiver
(solid black). The blue lines refer to line-of-sight security for
trusted-loss-and-noise receiver (blue dotted), and trusted-noise receiver
(blue dashed). Physical parameters are chosen as discussed in the main text.
\label{mobilePIC
\end{figure}
\subsection{Short-range microwave wireless}
Let us consider wireless quantum communications at the microwave frequencies,
in particular at 1~GHz. We show the potential feasibility for short-range
quantum-safe WiFi (e.g., for contact-less cards) within the general setting of
composable finite-size security. First of all we need to remark two important
differences with respect to the optical case: presence of higher loss and
higher noise.
From the point of view of increased loss, the crucial difference is the
geometry of the beam. For indoor wireless applications, microwave antennas are
small and, for this reason, cannot offer beam directionality. The emitted beam
is either isotropic (spherical wave) or have some limited directionality,
usually quantified by the gain $g$. This means that, at some distance $z$, the
intensity of the beam will be confined in an area equal to $4\pi z^{2}/g$. It
is clear that we have a strong suppression of the signal, since a receiver
with aperture's radius $a_{R}$ is going to collect just a fraction
$\eta_{\text{ch}}\simeq\min\{ga_{R}^{2}/(4\pi z^{2}),1\}$ of the emitted
photons. Here the minimum accounts for the case where the receiver is close to
the antenna, so the angle of emission is subtended by the receiver's aperture,
which happens at the distance $z_{\text{best}}=\sqrt{g/\pi}a_{R}/2$. In our
investigation, we assume the numerical value $g=10$.
As mentioned above another important difference with respect to the optical
case is the amount of thermal background noise which affects microwaves for
both signal preparation and
detection~\cite{refA1,refA2,refA3,refA4,refA5,refA6}. If we assume setups
working at room temperature, this thermal noise is dominant with respect to
the other sources of noise. Both the preparation noise at the microwave
modulators and the electronic noise in the amplifiers of the microwave
homodyne detectors are relevant~\cite{Shabir}; we set them to be equal to the
thermal background computed using the formula of the black-body radiation. On
the other hand, phase-errors associated with the LO are negligible since the
LO is slow at the microwave and can easily be reconstructed.
Let us quantify the amount of thermal noise and identify a suitable set of
parameters able to mitigate the problem. For a receiver with spectral filter
$\Delta\lambda$, detector bandwidth $W$, aperture $a_{R}$, and field of view
$\Omega_{\text{fov}}$, we can consider the photon collection parameter
$\Gamma_{R}$ in Eq.~(\ref{downT}). Assume that signal and LO pulses are
time-bandwidth limited, so that $\Delta t\Delta\nu\simeq1$. For instance
$\Delta t=10$~ns and $\Delta\nu=100~$MHz for a carrier frequency of $\nu
=1~$GHz ($10\%$ bandwidth). Corresponding carrier wavelength is $\lambda
=c/\nu\simeq30~$cm. Using $\Delta\lambda=\Delta\nu\lambda^{2}/c$ and setting
$W\simeq\Delta\nu$ (detector resolving the pulses), we may writ
\begin{equation}
\Gamma_{R}\simeq\frac{\lambda^{2}}{c}\Omega_{\text{fov}}a_{R}^{2}.
\end{equation}
For receiver aperture $a_{R}=5~$cm and sufficiently-narrow field of view
$\Omega_{\text{fov}}^{1/2}=1~$degree (so $\Omega_{\text{fov}}\simeq
3\times10^{-4}$~sr), we compute $\Gamma_{R}\simeq2.28\times10^{-16}$ in units
of s m$^{3}$ sr. Note that realizing such a narrow field of view with a small
indoor receiver can be challenging in practice.
The photon collection parameter must be combined with the thermal background
photons in units of photons s$^{-1}$ m$^{-3}$ sr$^{-1}$, quantified by the
black-body formul
\begin{equation}
\bar{n}_{\text{body}}=\frac{2c}{\lambda^{4}}\left[ \exp\left( \frac
{hc}{\lambda k_{\text{B}}T}\right) -1\right] ^{-1},
\end{equation}
where $k_{\text{B}}$ is Boltzmann's constant and $T\simeq290~$K is the
temperature. Therefore we ge
\begin{equation}
\bar{n}_{\text{th}}=\Gamma_{R}\bar{n}_{\text{body}}\simeq0.1\text{~photons.}
\label{nthexp
\end{equation}
Note that the figure is acceptably low thanks to the filtering effect of
$\Gamma_{R}$, which accounts for the spatiotemporal profile of the LO pulses,
together with the other features of the receiver (aperture, field of view).
Thermal noise is affecting both preparation and detection with constant floor
level. This means that $\bar{n}_{\text{th}}$ mean photons are seen by the
detector no matter if signal photons are present or not. In other words, the
detector experiences a constant noise variance equal to
\begin{equation}
\sigma_{z}^{2}=2\bar{n}_{\text{th}}+\nu_{\text{det}}, \label{noiseZmicro
\end{equation}
where $\nu_{\text{det}}$ is the usual quantum duty (which is $=1$ for homodyne
and $=2$ for heterodyne).
Assume that the total transmissivity is $\tau=\eta_{\text{ch}}\eta
_{\text{eff}}$, where $\eta_{\text{ch}}$ is channel's transmissivity and
$\eta_{\text{eff}}\simeq0.8$ is receiver's efficiency. Also assume that the
transmitter (Alice), modulates thermal states with classical variance
$\sigma_{x}^{2}=2\bar{n}_{T}$, where $\bar{n}_{T}$ is equivalent mean number
of signal photons. Then, the total mean number of photons at the receiver's
detector is given b
\begin{equation}
\bar{n}_{R}=\tau\bar{n}_{T}+\bar{n}_{\text{th}}. \label{IOmicro
\end{equation}
Basically, this is equivalent to Eqs.~(\ref{IOenergy}) and~(\ref{nBARvalue}),
by setting $\bar{n}_{B}=\bar{n}_{\text{th}}$ and $\bar{n}_{\text{ex}
=(1-\eta_{\text{eff}})\bar{n}_{\text{th}}$. As we can see, for $\tau=1$, we
get $\bar{n}_{T}+\bar{n}_{\text{th}}$ meaning that the prepared states are
thermal; for $\tau<1$, signal photons are lost ($\bar{n}_{T}\rightarrow
\tau\bar{n}_{T}$), while the depleted thermal background photons are
compensated at the receiver re-entering the detection system, so we have the
constant noise level $\bar{n}_{\text{th}}$.
\subsubsection{Fully-untrusted scenario}
In the worst-case scenario, the noise associated with preparation, channel and
detector is all untrusted. In this case, Eq.~(\ref{IOmicro}) corresponds to
the action of a beam splitter with transmissivity $\tau$ combining a signal
mode with mean photons $\bar{n}_{T}$ and an environmental mode with mean
photons $\bar{n}_{e}=\bar{n}_{\text{th}}/(1-\tau)$. The idea is that Alice
would attempt to create randomly-displaced coherent states, but Eve readily
thermalizes them by adding malicious thermal photons. These photons add up to
those later introduced by the channel, so that we globally have the insertion
of $\bar{n}_{e}$ mean photons as above. This leads to a collective Gaussian
attack where Eve has the purification of the untrusted thermal noise
associated with each stage of the communication.
Alice's and Bob's classical variables, $x$ and $y$, are related by
Eq.~(\ref{IOnew}) but where the noise variable $z$ has now variance
$\sigma_{z}^{2}$ as in Eq.~(\ref{noiseZmicro}) which corresponds to
Eq.~(\ref{sigmazed}) up to replacing $\bar{n}\rightarrow\bar{n}_{\text{th}}$.
Alice and Bob's mutual information $I(x:y)$ is therefore given by
Eq.~(\ref{MutualINFOeq}) computed with modulation $\sigma_{x}^{2}=2\bar{n
_{T}$ and equivalent nois
\begin{equation}
\chi=\frac{2\bar{n}_{\text{th}}+\nu_{\text{det}}}{\tau}=\xi_{\text{tot}
+\frac{\nu_{\text{det}}}{\tau},
\end{equation}
where $\xi_{\text{tot}}:=2\bar{n}_{\text{th}}/\tau$ is the total excess noise.
Numerically, we choose the modulation $\sigma_{x}^{2}=20$.
As already said, in the fully-untrusted scenario, all thermal noise coming
from preparation, channel and receiver's setup is considered to be untrusted.
This is equivalent to the treatment of Sec.~\ref{AsyUntrustedSection}, proviso
we make the replacement $\bar{n}\rightarrow\bar{n}_{\text{th}}$ in
Eq.~(\ref{untrustedOMEGA}) and then in Eqs.~(\ref{parameterB}),
(\ref{thetaEQ1}) and~(\ref{fiEQ1}). The revised parameters can then be used in
the global CM\ in Eqs.~(\ref{jointCM}) and~(\ref{CMcase1}).
Then, the asymptotic key rate against collective Gaussian attacks is given by
$R_{\text{asy}}^{(3)}(\tau,\bar{n}_{\text{th}})$ according to
Eq.~(\ref{asyUNrate}), where we now us
\begin{equation}
\tau=\eta_{\text{eff}}\min\{ga_{R}^{2}/(4\pi z^{2}),1\}, \label{tauMICRO
\end{equation}
and $\bar{n}_{\text{th}}$ as given by Eq.~(\ref{nthexp}). We may then assume
the reconciliation parameter $\beta=0.98$.
To account for finite-size effects, we first include parameter estimation. This
means that the parties need to sacrifice $m$ of the $N$\ pulses, so $n$ pulses
survive for key generation. Numerically, we take $N=5\times10^{7}$ and
$m=0.1\times N$. Thus, they construct the worst-case estimators for the
overall transmissivity $\tau$ and thermal noise $\bar{n}_{\text{th}}$
following Eqs.~(\ref{wcEstimator1}) and~(\ref{wcEstimator2}). These estimators
can be here approximated as follow
\begin{align}
\tau^{\prime} & \simeq\tau-2w\sqrt{\frac{2\tau^{2}+\tau(2\bar{n}_{\text{th
}+\nu_{\text{det}})/\sigma_{x}^{2}}{\nu_{\text{det}}m}},\label{estREV1}\\
\bar{n}_{\text{th}}^{\prime} & \simeq\bar{n}_{\text{th}}+w\frac{2\bar
{n}_{\text{th}}+\nu_{\text{det}}}{\sqrt{2\nu_{\text{det}}m}}, \label{estREV2
\end{align}
where $w$ is the confidence parameter associated with $\varepsilon_{\text{pe
}$, and computed according to Eq.~(\ref{wSTVALUE}) for collective Gaussian
attacks (see Sec.~\ref{PEsectionLAter} for more details). Assuming a tolerable
error probability of $\varepsilon_{\text{pe}}=2^{-33}$, we have $w\simeq6.34$
confidence intervals.
The composable key rate takes the form in Eq.~(\ref{sckeee}) where we now use
$R_{\text{pe}}^{(3)}=R_{\text{asy}}^{(3)}(\tau^{\prime},\bar{n}_{\text{th
}^{\prime})$ computed from Eqs.~(\ref{estREV1}) and~(\ref{estREV2}), together
with the usual finite-size terms in Eqs.~(\ref{deltaAEPPP})
and~(\ref{bigOMEGA}). Numerically, we can assume $p_{\text{ec}}=0.9$ for the
probability of success of EC, $d=2^{5}$ for the digitalization of the
continuous variables, and the value $2^{-33}$ for all the epsilon parameters,
so we have epsilon security $\varepsilon\simeq5.6\times10^{-10}$ against
collective Gaussian attacks according to Eq.~(\ref{epsSECcollective}).
To study the performance, let us consider the heterodyne protocol
($\nu_{\text{det}}=2$). Then, we assume a device stably kept at some distance
$z$ from the transmitter within the emission angle of the transmitter and with
an aligned field of view. For the parameters considered here, we find that a
positive key rate is obtained for $z\leq4.48$~cm, which is fully compatible
for contactless card applications. In particular, for any $z\leq
z_{\text{best}}\simeq4.46$~cm we compute a key rate of $R\gtrsim10^{-2}$
bits/use, corresponding to $\gtrsim50$ kbit/sec with a system clock at $5$ MHz.
Note that, according to the thermal version of the PLOB bound~\cite{QKDpaper},
the maximum key rate cannot overcome the upper limit
\begin{equation}
R\leq\left\{
\begin{array}
[c]{l
-\log_{2}\left[ (1-\tau)\tau^{\frac{\bar{n}_{\text{th}}}{1-\tau}}\right]
-h\left( \frac{\bar{n}_{\text{th}}}{1-\tau}\right) ,~~\text{for~}\bar
{n}_{\text{th}}\leq\tau,\\
0,~~\text{for~}\bar{n}_{\text{th}}\geq\tau,
\end{array}
\right.
\end{equation}
where $h(x):=H(2x+1)$. This means that the no rate is possible above the
threshold $\bar{n}_{\text{th}}=\tau$. Using Eqs.~(\ref{nthexp})
and~(\ref{tauMICRO}) with our regime of parameters, we find that the maximum
possible range is about $12.47~$cm, i.e., about three times the distance
achievable with the considered heterodyne protocol under composable security.
\subsubsection{LoS security for microwaves}
Better performances can be obtained if we relax security requirements by
relying on the LoS geometry. In particular, one may assume that the thermal
noise is trusted, so that Eve is passively limited to eavesdrop the photons
leaking from the channel and the setup. In this case, Eq.~(\ref{IOmicro})
corresponds to the action of a beam splitter with transmissivity $\tau$
combining a signal mode with mean photons $\bar{n}_{T}+\bar{n}_{\text{th}}$
(signal photons plus trusted preparation noise)\ and a genuine environmental
mode with mean photons $\bar{n}_{\text{th}}$~\cite{NotaNOISE}. Eve collects
the fraction $1-\tau$ of photons leaked into the environment, but she does not
control any noise, i.e., she does not have its purification.
Alice and Bob's mutual information $I(x:y)$ is the same as above for the
fully-untrusted case but Eve's Holevo information $\chi_{\text{LoS}}(E:y)$ is
now rather different. The latter can be computed as in Sec.~\ref{LoSsection}
and, in particular, from the CM\ in Eq.~(\ref{VbeLOS}), where we insert the
following parameter
\begin{align}
b & =2\bar{n}_{R}+1,\\
\theta & =-\sqrt{\tau(1-\tau)}\sigma_{x}^{2},\\
\phi & =(1-\tau)\sigma_{x}^{2}+2\bar{n}_{\text{th}}+1.
\end{align}
In this way we can compute the asymptotic key rat
\begin{equation}
R_{\text{asy,LoS}}(\tau,\bar{n}_{\text{th}})=\beta I(x:y)-\chi_{\text{LoS
}(E:y).
\end{equation}
The incorporation of finite-size effects requires that we under-estimate the
thermal noise experienced by Eve, while we over-estimate that seen by the
parties. Thus, besides the worst-case estimators $\tau^{\prime}$ and $\bar
{n}_{\text{th}}^{\prime}$ in Eqs.~(\ref{estREV1}) and~(\ref{estREV2}), we also
compute the best-case estimato
\begin{equation}
\bar{n}_{\text{th}}^{\prime\prime}\simeq\bar{n}_{\text{th}}-w\frac{2\bar
{n}_{\text{th}}+\nu_{\text{det}}}{\sqrt{2\nu_{\text{det}}m}}.
\end{equation}
Thus, we compute the rate
\begin{equation}
R_{\text{pe,LoS}}=\beta I(x:y)_{\tau^{\prime},\bar{n}_{\text{th}}^{\prime
}-\chi_{\text{LoS}}(E:y)_{\tau^{\prime},\bar{n}_{\text{th}}^{\prime\prime}},
\end{equation}
which is replaced into Eq.~(\ref{sckeee}) to provide the composable key rate
associated with LoS security.
Assuming the heterodyne protocol with the same parameters as in the
fully-untrusted case, we find an improvement, as expected. As shown in
Fig.~\ref{WiFipic}, the range of security is now larger, even though the
effective application is still restricted to centimeters from the transmitter.
Note that this performance is based on the LoS assumption, so it is not
confined by the PLOB bound.\begin{figure}[t]
\vspace{0.2cm}
\par
\begin{center}
\includegraphics[width=0.9\columnwidth] {WiFi.eps}
\end{center}
\par
\vspace{-0.4cm}\caption{Microwave wireless QKD (at 1~GHz) using the heterodyne
protocol under LoS security. We plot the composable secret key rate (bits/use)
versus free-space distance $z$ between transmitter and receiver (centimeters).
Parameters are chosen as discussed in the main text.
\label{WiFipic
\end{figure}
\section{Conclusions\label{SEC3}}
In this work, we have developed a general framework for the composable
finite-size security analysis of Gaussian-modulated coherent-state protocols,
which are the most powerful protocols of CV-QKD. We have investigated the
secret key rates that are achievable assuming various levels of trust for the
receiver's setup, from the worst-case assumption of a fully-untrusted detector
to the case where detector's loss and noise are considered to be trusted. In
the specific case of free-space quantum communication, we have also
investigated the additional assumption of passive eavesdropping on the
communication channel due to the line-of-sight geometry.
We have shown how the realistic assumptions on the setups can have non-trivial
effects in terms of increasing the composable key rate and tolerating higher
loss (therefore increasing distance). More interestingly, we have also
demonstrated the feasibility of high-rate CV-QKD with wireless mobile devices,
assuming realistic parameters and near-range distances, e.g., as typical of
indoor networks. Besides the optical frequencies, we have also analyzed the
microwave wavelengths, considering possible parameters able to mitigate the
loss and noise affecting this challenging setting. In this way, we have
discussed potential microwave-based applications for very short-range
(cm-range) quantum-safe communications.
\bigskip
\textit{Acknowledgements}.~~The author would like to thank Panagiotis
Papanastasiou, Masoud Ghalaii, Cillian Harney, and Marco Tomamichel for
discussions. This work was funded by the European Union's Horizon 2020
research and innovation programme under grant agreement No 820466
(Quantum-Flagship Project CiViQ: \textquotedblleft Continuous Variable Quantum
Communications\textquotedblright).
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 3,723 |
Districts in Himachal Pradesh receives fresh spell of snowfall, temperature expected to drop
SHIMLA: Snowfall has been reported in the hilly areas of the state including Lahaul and Spiti, Kinnaur and Kullu with the temperature expected to drop further in the coming days, according to India Meteorological Department Head of Himachal Pradesh, Surender Paul.
Speaking to ANI on Monday, Paul said, "There has been moderate to heavy snowfall in some of the districts of Himachal Pradesh that include Lahaul And Spiti, Kinnaur and Kullu."
Talking about the weather forecast in the coming days, the IMD Head said that the temperature is expected to drop further with the onset of Northwest winds.
"The temperature has already dropped in the areas such as Lahaul Spiti, Kinaul up to -3.4 degrees celsius. The temperature is expected to drop further. The Northwest winds that would follow after the snowfall will lead to cold weather and the temperature will drop further," he said.
Meanwhile, a tourist from Chandigarh who travelled to Shimla to see the snowfall said, "I just wish to have snowfall today. The weather is chilling but we are enjoying ourselves. We are expecting a snowfall today."
Another tourist hailing from Haryana said, "We had mainly come to see the snowfall and we wish to have it before we leave, even at the last moment. It was a very nice experience altogether." (ANI)
Delhi To ShimlaKinnaurKulluLahaul SpitlifestyleShimla Temperature TodayShimla TrainsShimla WeatherSnowfall Himachal PradeshSnowfall in ShimlaSpiti ValleyWeather
Dish Network secures broadcast rights for IPL
NY Indian Film Festival's feature line up
Indie film Patang to premiere at Tribeca film fest | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 2,484 |
\section{Introduction}
\label{sec:intro}
Data compression is performed in all types of data requiring storage and transmission. It preserves space, energy and bandwidth, while representing the data in most efficient way [1-4]. There are numerous coding algorithms used for compression in various applications [1-3,5-18]. Some of them are optimal [4,19] in all cases, whereas others are optimal for a specific probability distribution of the source symbols. All of these algorithms are mostly applied on $m$-ary data source. However, most of the universal compression algorithms substantially increase their coding complexity and memory requirements when the data changes from binary to $m$-ary source. For instance, in arithmetic coding, the computational complexity difference between the encoder and decoder increases with the number of source symbols [2]. Therefore, it would be beneficial if binarization is perfomed on $m$-ary data source before compression algorithms are applied on it. The process where binarization is followed by compression is most notably found in Context-based Adaptive Binary Arithmetic Coding (CABAC) [20] which is used in H.264/AVC Video Coding Standard [21], High Efficiency Video Coding (HEVC) Standard [22], dynamic 3D mesh compression [23], Audio Video Coding Standard (AVS) [24], Motion Compensated-Embedded Zeroblock Coding (MC-EZC) in scalable video coder [25], multiview video coding [26], motion vector encoding [27-28], and 4D lossless medical image compression [29].
There are many binary conversion techniques which are, or can be used for the binarization process. The most common among all is binary search tree [30-32]. In this, Huffman codeword is used to design an optimal tree [18]. However, there are two limitations to it. First, the probability of all the symbols should be known prior to encoding that may not be possible in all the applications. Although there are methods to overcome the above problem, they come at an additional cost of complexity. For example, binary search tree is updated with the change in incoming symbol probabilities. Second, as with the Huffman coding, the optimality is achieved only when the probability distribution of symbols are in the powers of two. Apart from binary search tree, there are other binarization schemes like unary binarization scheme [20], truncated unary binarization scheme [20], fixed length binarization scheme [20], Golomb binarization scheme [9,20,33-34], among many [5-13,20,30-34]. All of them are optimal for only certain type of symbol probability distributions, and hence, can only conserve the entropy of the data for that probability distribution of the source symbols. Currently, there is no binarization scheme that is optimal for all probability distributions of the source symbols which would result in achieving overall optimal data compression.
This paper presents a generalized optimal binarization algorithm. The novel binarization scheme conserves the entropy of the data while converting the $m$-ary source data into $m-1$ binary strings. Moreover, the binarization technique is independent of the data type and can be used in any field for storing and compressing data. Furthermore, it can efficiently represent data in the fields which require data to be easily written and read in binary form.
The paper is organized as follows. Section 2 describes the binarization and de-binarization process that will be carried out at the encoder and decoder, respectively. The optimality proof of the binarization scheme is provided in section 3. In section 4, the complexity associated with the binarization process is discussed. Lastly, section 5 concludes by stating the advantages of the presented binarization scheme over others, and its applications.
\section{Binarization and De-binarization}
The binarization of the source symbols is carried out at the encoder using the following two steps:
\begin{enumerate}
\item A symbol is chosen, and a binary data stream is created by assigning '1' where the chosen symbol is present and '0' otherwise, in the uncompressed data.
\item The uncompressed data is rearranged by removing the symbol chosen in step 1.
\end{enumerate}
The two steps are iteratively applied for $m-1$ symbols. It needs to be explicitly emphasized that the binarization of symbol occurs on the previously rearranged uncompressed data and not on the original uncompressed data. Here, the algorithm reduces the uncompressed data size with the removal of binarized symbols from the data, leading to the conservation of entropy. After binarizing every symbol, there are $m-1$ binary data streams corresponding to $m$ source symbols. It is because the $m-1^{th}$ binary string would represent $m-1^{th}$ and $m^{th}$ symbols as '1' and '0', respectively. The binarization scheme demostrated here is optimal i.e., the overall entropy of binarized data streams is equal to the entropy of original data containing $m$-ary source. The proof of optimality is provided in section 3. After binarization, the binarized data streams can be optimally compressed using any universal compression algorithm, including the algorithms that optimally compress only binary data (for example: binary arithmetic coding).
\begin{table*}[ht!] \footnotesize
\centering
\caption{An example of binarization process}
\begin{tabular}{|c|c|c|c|}
\hline
Binarization Order & \multicolumn{3}{c|}{Iteration} \\ \cline{2-4}
& First & Second & Third \\
\hline
Data & AABCBACBBACCABACB & BCBCBBCCBCB & CCCCC\\
ABC & 11000100010010100 & 10101100101 & 11111 \\
\hline
Data & AABCBACBBACCABACB & BCBCBBCCBCB & BBBBBB\\
ACB & 11000100010010100 & 01010011010 & 111111 \\
\hline
Data & AABCBACBBACCABACB & AACACACCAAC & CCCCC \\
BAC & 00101001100001001 & 11010100110 & 11111 \\
\hline
Data & AABCBACBBACCABACB & AACACACCAAC & AAAAAA \\
BCA & 00101001100001001 & 00101011001 & 111111 \\
\hline
Data & AABCBACBBACCABACB & AABBABBAABAB & BBBBBB \\
CAB & 00010010001100010 & 110010011010 & 111111 \\
\hline
Data & AABCBACBBACCABACB & AABBABBAABAB & AAAAAA \\
CBA & 00010010001100010 & 001101100101 & 111111 \\
\hline
\end{tabular}
\end{table*}
\begin{table*}[ht!] \footnotesize
\centering
\caption{An example of de-binarization process}
\begin{tabular}{|c|c|c|c|}
\hline
De-binarization Order & \multicolumn{3}{c|}{Iteration} \\ \cline{2-4}
& First & Second & Third \\
\hline
ABC & 11000100010010100 & AA101A011A00A1A01 & AAB1BA1BBA11ABA1B \\
Data & AA000A000A00A0A00 & AAB0BA0BBA00ABA0B & AABCBACBBACCABACB\\
\hline
ACB & 11000100010010100 & AA010A100A11A0A10 & AA1C1AC11ACCA1AC1 \\
Data & AA000A000A00A0A00 & AA0C0AC00ACCA0AC0 & AABCBACBBACCABACB\\
\hline
BAC & 00101001100001001 & 11B0B10BB1001B10B & AAB1BA1BBA11ABA1B \\
Data & 00B0B00BB0000B00B & AAB0BA0BBA00ABA0B & AABCBACBBACCABACB \\
\hline
BCA & 00101001100001001 & 00B1B01BB0110B01B & 11BCB1CBB1CC1B1CB \\
Data & 00B0B00BB0000B00B & 00BCB0CBB0CC0B0CB & AABCBACBBACCABACB \\
\hline
CAB & 00010010001100010 & 110C01C001CC101C0 & AA1C1AC11ACCA1AC1 \\
Data & 000C00C000CC000C0 & AA0C0AC00ACCA0AC0 & AABCBACBBACCABACB \\
\hline
CBA & 00010010001100010 & 001C10C110CC010C1 & 11BCB1CBB1CC1B1CB \\
Data & 000C00C000CC000C0 & 00BCB0CBB0CC0B0CB & AABCBACBBACCABACB \\
\hline
\end{tabular}
\end{table*}
Table 1 shows the binarization process through an example. A sample input data 'AABCBACBBACCABACB' is considered for the process and as can be seen, it contains three source symbols 'A', 'B', and 'C'. In Table 1, 'Binarization order' states the sequence in which the symbols are binarized. For instance, in 'ABC' binarization order, 'A' is binarized first, followed by 'B', and then finally by 'C'. The row 'Data' shows the uncompressed data available to be binarized after each iteration. Below the 'Data' row is the binarized value of each symbol. As can be seen in each first iteration, the symbol that has be binarized is marked '1', while others are marked '0'. In the next iteration, the symbol that was binarized in the current step is removed from the uncompressed data. Although shown in Table 1, the binarization process does not require to binarize last symbol, because the resultant string contains all '1's that provide no additional information and is redundant. It also needs to be noted that each binarization order results in different sets of binary strings.
At the decoder, the decoding of the compressed data is followed by de-binarization of $m$-ary source symbol. The order of decoding follows the order of encoding for perfect reconstruction at minimum complexity. With the encoding order information, the de-binarization can be perfectly reconstructed in multiple ways other than the encoding order, but the reordering of sequence after every de-binarization will increase the time as well as the decoder complexity. The de-binarization of the source symbols is also carried out in two steps shown below, and these steps are recursively applied to all the binary data streams representing $m$-ary source symbols:
\begin{enumerate}
\item Replace '1' with the source symbol in the reconstructed data stream.
\item Assign the values of the next binary data stream in sequence to the '0's in the reconstructed data stream.
\end{enumerate}
An example of de-binarization process is shown in Table 2. The de-binarization order follows the same order as of binarization process. In Table 2, the row 'Data' represents the reconstructed data at each iteration. The value '1' is replaced by the symbol to be de-binarized in the respective iteration, while '0's are replaced by the binary string of the next symbol to be de-binarized. Finally, after the last iteration, the original input data 'AABCBACBBACCABACB' is losslessly recovered for all binarization and de-binarization order.
\section{Optimality of Binarization Scheme}
Let the data source be $Y\in \{Y_1,\, Y_2,\, \dots,\, Y_m\}$, and $X\in \{0,\, 1\}$ be the binary source for each source symbol. The entropy of a $m$-ary source $Y$ is defined as,
\begin{equation} \small
H(Y) = -\sum_{i=1}^{m} p(Y_i) \log p(Y_i)
\end{equation}
where $p(Y_i)$ is the probability of $Y_i^{th}$ source symbol. Subsequently, the entropy of $m$-ary data source $Y$ with length $N$ is $H(Y^N)$. Similarly, $H(X^N)$ is the entropy of binary source with data length $N$. As explained in the binarization algorithm, the uncompressed data is rearranged after the binarization of the previous symbol/s to ${N(1-\sum_{j=1}^{i-1}p(Y_j))}$ data length i.e., the length of the original data subtracted by the length of all the previously binarized source symbols. Hence, the overall entropy of the $m$ binarized strings is $\sum_{i=1}^{m} H(X_i^{N(1-\sum_{j=1}^{i-1}p(Y_j))})$. Here, $m$ binary strings are considered for mathematical convenience.
To achieve the optimal binarization of $m$-ary source, the entropy of $m$-ary source data must equal the total entropy of binary strings. Therefore,
{\small
\begin{IEEEeqnarray}{rCl} \footnotesize
H(Y^N) & = & \sum_{i=1}^{m} H(X_i^{N(1-\sum_{j=1}^{i-1}p(Y_i))}) \\
N H(Y) & = & \sum_{i=1}^{m} N\left(1-\displaystyle\sum_{j=1}^{i-1} p(Y_i)\right) H(X_i)
\end{IEEEeqnarray}}
The probability distribution of the binary source $X_i$ is the probability distribution of $m$-ary source $Y_i$ when the first $i-1$ source symbols have already been binarized i.e., removed from the original data. Thus, $H(X_i)$ can be rewritten in terms of $Y_i$ in the following way:
{\small
\begin{IEEEeqnarray}{rCl} \footnotesize
H(Y) & = & \sum_{i=1}^{m} \left(1-\displaystyle\sum_{j=1}^{i-1} p(Y_j)\right) H\left(\frac{p(Y_i)}{1 - \sum_{j=1}^{i-1}p(Y_j)}\right) \\
H(Y) & = & -\sum_{i=1}^{m} \left(1-\displaystyle\sum_{j=1}^{i-1} p(Y_j)\right) \left(\frac{p(Y_i)}{1 - \sum_{j=1}^{i-1}p(Y_j)}\right)\nonumber \\
&& \log \left( \frac{p(Y_i)}{1 - \sum_{j=1}^{i-1}p(Y_j)}\right) - \sum_{i=1}^{m} \left(1-\displaystyle\sum_{j=1}^{i-1} p(Y_j)\right) \nonumber \\
&& \left[\frac{1 - \sum_{j=1}^{i}p(Y_j)}{1 - \sum_{j=1}^{i-1}p(Y_j)} \log \left( \frac{1 - \sum_{j=1}^{i}p(Y_j)}{1 - \sum_{j=1}^{i-1}p(Y_j)}\right)\right]
\end{IEEEeqnarray}
\begin{IEEEeqnarray}{rCl} \footnotesize
H(Y) & = & -\sum_{i=1}^{m} p(Y_i) \log \left( \frac{p(Y_i)}{1 - \sum_{j=1}^{i-1}p(Y_j)}\right) \nonumber \\
&& -\> \sum_{i=1}^{m} (1 - \sum_{j=1}^{i}p(Y_j)) \log \left( \frac{1 - \sum_{j=1}^{i}p(Y_j)}{1 - \sum_{j=1}^{i-1}p(Y_j)}\right)
\end{IEEEeqnarray}
\begin{IEEEeqnarray}{rCl} \footnotesize
H(Y) & = & -\sum_{i=1}^{m} p(Y_i) \log \left( \frac{p(Y_i)}{\sum_{j=i}^{m}p(Y_j)}\right) \nonumber \\
&& -\> \sum_{i=1}^{m} (\sum_{j=i+1}^{m}p(Y_j)) \log \left( \frac{\sum_{j=i+1}^{m}p(Y_j)}{\sum_{j=i}^{m}p(Y_j)}\right)
\end{IEEEeqnarray}
\begin{IEEEeqnarray}{rCl} \footnotesize
H(Y) & = & -\sum_{i=1}^{m} \left(p(Y_i) \log p(Y_i) - p(Y_i) \log \left(\sum_{j=i}^{m}p(Y_j)\right)\right) \nonumber \\
&& -\> \sum_{i=1}^{m} \left(\sum_{j=i+1}^{m}p(Y_j)\right) \log \left(\sum_{j=i+1}^{m}p(Y_j)\right) \nonumber \\
&& +\> \sum_{i=1}^{m} \left(\sum_{j=i+1}^{m}p(Y_j)\right) \log \left(\sum_{j=i}^{m}p(Y_j)\right)
\end{IEEEeqnarray}
\begin{IEEEeqnarray}{rCl} \footnotesize
H(Y) & = & -\sum_{i=1}^{m} p(Y_i) \log p(Y_i) \nonumber \\
&& -\> \sum_{i=1}^{m} \left(\sum_{j=i+1}^{m}p(Y_j)\right) \log \left(\sum_{j=i+1}^{m}p(Y_j)\right) \nonumber \\
&& +\> \sum_{i=1}^{m} \left(\sum_{j=i}^{m}p(Y_j)\right) \log \left(\sum_{j=i}^{m}p(Y_j)\right)
\end{IEEEeqnarray}
\begin{IEEEeqnarray}{rCl} \footnotesize
H(Y) & = & -\sum_{i=1}^{m} p(Y_i) \log p(Y_i) \nonumber \\
&& +\> \left(\sum_{j=1}^{m}p(Y_j)\right) \log \left(\sum_{j=1}^{m}p(Y_j)\right)\\
H(Y) & = & -\sum_{i=1}^{m} p(Y_i) \log p(Y_i) + 1 \log 1 \\
H(Y) & = & -\sum_{i=1}^{m} p(Y_i) \log p(Y_i)
\end{IEEEeqnarray}}
The reduction of equation 2 to equation 12 (also equation 1) proves that the binarization scheme preserves entropy for any $m$-ary data source.
\section{Computational Complexity of Binarization Scheme}
The computational complexity of the presented method is the linear function of the input data length. The binarization and de-binarization process only acts as a filter, assigning or replacing 0's and 1's, respectively, for an occurrence of a source symbol without any additional table or calculation, that is created or performed for the other binarization techniques. Suppose, the length of input data is $N$, $m$ is the number of source symbols, and $Y$ is the source. For the first symbol, the length of the binary string would be $N$. The length of binary string for the second symbol would be the length of all the symbols, except the first symbol (see Table 1). Likewise, the length of $i^{th}$ binary string would be the length all symbols yet to be binarized. Mathematically, the length can be written as {\small $N\left(1-\displaystyle\sum_{j=1}^{i-1} p(Y_i)\right)$}, where $p(Y_i)$ is the probability of $i^{th}$ symbol. The total number of binary assignment would be {\small $\sum_{i=1}^{m}N\left(1-\displaystyle\sum_{j=1}^{i-1} p(Y_i)\right)$}. As can be seen, the computational complexity of the binarization and de-binarization process is linear in terms of the input data length.
\section{Conclusion: Advantages and Applications}
The proposed binarization scheme has the following advantages over others. Firstly, it is optimal for every data set. As proved and shown in this paper, the binarization scheme conserves entropy of $m$-ary data source. Secondly, the proposed method eliminates the need for knowing the source symbols at all. It works optimally without the knowledge of source because the binarization of the source symbols can occur in any order as shown Table 1, and all orders conserve $m$-ary source entropy, which can be inferred from the derivation shown in section 3. Thirdly, adding to the previous point, the coding is independent of the occurrence of the source symbols. In other words, any source symbol can be encoded in any order subject to the constraint that decoding is performed in the same order. The optimality is independent of the source order in the data set. Fourthly, unlike variable length codes, there is no need to know the probability distribution of the source symbols beforehand. It can be updated as the symbols occur. However, even without the knowledge of probability distribution, the presented method is optimal. Lastly, it has low complexity that is feasible for practical data compression.
One of the immediate usage of the presented binarization technique is in CABAC used in video and image compression. In addition, CABAC with the proposed binarization scheme can potentially replace Context-based Adaptive Arithmetic Coding used in various image compression standards [35], including JPEG2000 [36]. Furthermore, the binarization scheme can be applied to all the universal compression algorithms that have less complexity and resource requirements for binary data, than $m$-ary data.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 7,161 |
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Runtime.Serialization;
namespace FlixCloud.Notification
{
[DataContract(Name = "job", Namespace = "")]
public class JobNotification
{
[DataMember(Name = "id", Order = 1)]
public int Id;
[DataMember(Name = "initialized-job-at", Order = 2)]
public DateTime InitializedAt;
[DataMember(Name = "pass-through", Order = 3)]
public string PassThrough;
[DataMember(Name = "recipe-name", Order = 4)]
public string RecipeName;
[DataMember(Name = "recipe-id", Order = 5)]
public int RecipeID;
[DataMember(Name = "state", Order = 6)]
public string State;
[DataMember(Name = "error-message", Order = 7)]
public string ErrorMessage;
[DataMember(Name = "finished-job-at", Order = 8)]
public DateTime FinishedAt;
[DataMember(Name = "input-media-file", Order = 9)]
public MediaFile InputMediaFile;
[DataMember(Name = "output-media-file", Order = 10)]
public MediaFile OutputMediaFile;
[DataMember(Name = "watermark-file", Order = 11)]
public WatermarkFile WatermarkFile;
}
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 7,465 |
\section{Introduction}
Magnetohydrodynamic (MHD) turbulence in the early Universe can be a powerful source
of gravitational waves (GWs) that could be observable as a stochastic
background today \citep{1994PhRvD..49.2837K,PhysRevLett.85.2044,2002PhRvD..66b4030K,Dolgov+02,2018CQGra..35p3001C}.
The frequency spectrum of these waves is related to the spectrum of the
underlying turbulence.
Such turbulence could be induced by the various phase transitions in
the early Universe \citep{PhysRevD.30.272,10.1093/mnras/218.4.629,Mazumdar_2019,10.21468/SciPostPhysLectNotes.24} or the possible
presence of primordial magnetic fields \citep{1988PhRvD..37.2743T,tanmay1991,
1992ApJ...391L...1R,durrer2013,subramanian2016,tanmay2021,Bran+He+Shar21}.
These GWs produced by turbulence at an epoch of the electroweak phase
transition lie in the sensitivity range of the proposed
Laser Interferometer Space Antenna and pulsar timing arrays for
the turbulence induced around an epoch of the quantum chromodynamics
(QCD) phase transition.
Recently, various pulsar timing arrays \citep{NANOGrav2020,
Goncharov_2021, Chen2021, Antoniadis2022} have reported evidence for
the presence of a common spectrum process across analyzed pulsars in
the search of the presence of an isotropic stochastic GW background.
This evidence has been used to constrain the strength and
correlation length of the magnetic fields generated at the QCD
epoch \citep{2021PhRvD.103L1302N,Sharma21,RoperPol+22}.
However, the presence of a quadrupolar spatial correlation
\citep{Hellings&Downs}, a characteristics of a GW background,
is yet to be claimed.
Numerical simulations have confirmed that there is indeed a direct
connection between the slopes of the turbulence and GW spectra
\citep{RoperPol+20}, except that at low frequencies, below the peak of
the spectrum, the GW spectrum was found to be shallower in the simulations
than what was previously expected from analytical calculations.
However, there is the worry that this shallow tail could be caused by
unknown numerical artifacts such as the finite size of the computational
domain and the way the turbulence is initiated in the simulations.
To address the problem of a limited computational domain, it is important
to inspect the detailed temporal dynamics of the different spatial
Fourier modes of the stress.
In an alternative approach, the authors of Ref.~\cite{RoperPol+22} have recently compared
numerical simulations with an analytic model, where the stress is constant
for a certain interval of time that is related to the eddy turnover time
corresponding to the peak wavenumber at the initial time.
Their model predicts a flat spectrum whose extent depends on the duration
over which the stress is held constant.
In this way, it was possible to determine an effective duration for a
given numerical simulation.
Their model is therefore descriptive rather than predictive.
In another recent approach, the authors of Ref.~\cite{Auclair:2022jod} have focused on the
importance of unequal time correlation functions of the Fourier components
of the velocity field for purely hydrodynamic turbulence.
While the authors acknowledge the potential importance of the initial
growth phase of the turbulence, they also show that there is no inverse
cascade in their simulations.
This is different from MHD turbulence, which can display inverse cascading
even in the absence of net magnetic helicity.
This will be crucial to the approach discussed in the present paper.
In the simulations of Ref.~\cite{RoperPol+22}, the wavenumber corresponding
to the peak of the GW spectrum and the wavenumbers below that
corresponding to the horizon size at the initial time are well resolved.
Since the stress appears explicitly in the linearized GW
equation, we also provide the evolution of the stress spectrum for
these simulations.
Second, we develop a simple model, motivated by the stress evolution seen
in simulations, to explain the GW spectrum obtained in simulations.
In this model, our main focus is to understand the nature of the GW
spectrum below the wavenumber corresponding to the peak of the spectrum.
We call this part the low frequency tail of the GW spectrum.
We emphasize that the Hubble horizon wavenumber poses an ultimate cutoff
for the flat spectrum toward low wavenumbers.
This paper is organized as follows.
In \Sec{TheModel}, we discuss the evolution of the magnetic field, stress,
and GW spectrum in our new runs.
In this section, we also discuss how the stress spectrum evolves when
inverse transfer and inverse cascade turbulence regimes correspond to
the evolution of the nonhelical and helical magnetic fields in the early
Universe.
In \Sec{simplemodel}, we discuss the model to explain the low frequency
tail of the GW spectrum.
Further, in \Sec{comparison}, we compare the GW spectrum obtained from
our numerical simulations and our model.
We conclude in \Sec{conclusion}.
\section{Nonhelical and helical cascades}
\label{TheModel}
Various phenomena such as primordial magnetic fields and phase
transitions can lead to the generation of turbulence in the early
Universe.
The stress associated with magnetic fields and turbulence
lead to the production of GWs.
This has been studied in the literature both analytically \citep{Dolgov+02,kosowsky2002,Gogo+07,tina2008}
and numerically \citep{RoperPol+20,RoperPol+21,Kahniashvili+21,RoperPol+22,Auclair:2022jod}.
In the present paper, we perform new simulations of decaying
MHD turbulence, where we resolve the scales which are smaller than
the Hubble horizon size at the initial time.
Before explaining the simulations in detail, let us begin by summarizing
the basic equations.
\subsection{GWs from MHD turbulence}
\label{GWfromCME}
We follow here the formalism of Ref.~\cite{RoperPol+20b, RoperPol+20}, where
conformal time is normalized to unity at the initial time.
One could associate this with the electroweak phase transition,
for example.
The velocity $\mathbf{u}$ is normalized to the speed of light.
The magnetic field $\mathbf{B}=\mbox{\boldmath $\nabla$} {}\times\mathbf{A}$ is written in terms of the
magnetic vector potential $\mathbf{A}$, and the current density is written
as $\mathbf{J}=\mbox{\boldmath $\nabla$} {}\times\mathbf{B}$.
Following Ref.~\cite{BEO96}, the energy density $\rho$ includes the restmass
density, so its evolution equation obeys a continuity equation that also
includes magnetic energy terms.
As in \cite{RoperPol+20b}, $\rho$ is normalized to the critical energy
density for a flat Universe.
We solve for the Fourier transformed plus and cross polarizations
of the gravitational strain, $\tilde{h}_+$ and $\tilde{h}_\times$,
which are driven by the corresponding projections of the stress,
which, in turn, is composed of kinetic and magnetic contributions,
\begin{equation}
{\sf T}_{ij} =\frac{4}{3}\gamma_{\rm Lor}^2\rho u_i u_j-B_i B_j+...,
\end{equation}
where $\gamma_{\rm Lor}=(1-\mathbf{u}^2)^{-1/2}$ is the Lorentz
factor, and the ellipsis denotes terms proportional to $\delta_{ij}$,
which do not contribute to the projected source $\tilde{T}_{+/\times}$.
Assuming the Universe to be conformally flat, its expansion can be
scaled out by working with conformal time $t$ and comoving variables
\citep{BEO96}.
We use the fact that in the radiation-dominated era, the scale factor
grows linearly with conformal time.
The only explicit occurrence of conformal time is then in the GW
equation, where a $6/t$ factor occurs in the source term \citep{RoperPol+20b}.
The full set of equations is therefore
\begin{eqnarray}
&&\frac{\partial\mathbf{B}}{\partial t}=
\mbox{\boldmath $\nabla$} {}\times[\mathbf{u}\times\mathbf{B}+\mbox{\boldmath $\nabla$} {}\times\mathbf{B}],\label{dAdt}\\
&&{{\rm D} {}\mathbf{u}\over{\rm D} {} t}=
{1\over\rho}\mbox{\boldmath $\nabla$} {}\cdot\left(2\rho\nu\mbox{\boldmath ${\sf S}$} {}\right)-{1\over4}\mbox{\boldmath $\nabla$} {}\ln\rho
+{\mathbf{u}\over3}\left(\mbox{\boldmath $\nabla$} {}\cdot\mathbf{u}+\mathbf{u}\cdot\mbox{\boldmath $\nabla$} {}\ln\rho\right)
\nonumber \\
&&\qquad\quad-{\mathbf{u}\over\rho}
\left[\mathbf{u}\cdot(\mathbf{J}\times\mathbf{B})+\eta \mathbf{J}^2\right]
+{3\over4\rho}\mathbf{J}\times\mathbf{B},
\label{dudt} \\
&&{\partial\ln\rho\over\partial t}
=-\frac{4}{3}\left(\mbox{\boldmath $\nabla$} {}\cdot\mathbf{u}+\mathbf{u}\cdot\mbox{\boldmath $\nabla$} {}\ln\rho\right)
+{1\over\rho}\left[\mathbf{u}\cdot(\mathbf{J}\times\mathbf{B})+\eta \mathbf{J}^2\right]\!,
\nonumber \\
&&\frac{\partial^2}{\partial t^2} \tilde{h}_{+/\times} (\mathbf{k}, t)
+k^2\tilde{h}_{+/\times} (\mathbf{k}, t) = {6\over t}
\tilde{T}_{+/\times}(\mathbf{k},t),
\label{GW4}
\end{eqnarray}
where ${\rm D} {}/{\rm D} {} t\equiv\partial/\partial t+\mathbf{u}\cdot\mbox{\boldmath $\nabla$} {}$ is the advective derivative,
$\eta$ is the magnetic diffusivity, $\nu$ is the kinematic viscosity,
${\sf S}_{ij}={\textstyle{1\over2}}(u_{i,j}+u_{j,i})-{\textstyle{1\over3}}\delta_{ij}\mbox{\boldmath $\nabla$} {}\cdot\mathbf{u}$ are
the components of the rate-of-strain tensor $\mbox{\boldmath ${\sf S}$} {}$ with commas denoting
partial derivatives.
Fourier transformation in space is denoted by a tilde.
In all cases studied in this paper, the initial conditions are such
that $\mathbf{B}$ consists of a weak Gaussian-distributed seed magnetic field,
$\mathbf{u}=0$, $\rho=1$.
We work with spectra that are defined as integrals over concentric shells
in wavenumber space $\mathbf{k}$ with $k=|\mathbf{k}|$.
They are normalized such that their integrals over $k$ give the mean square
of the corresponding quantity, i.e., $\int\mbox{\rm Sp}(\mathbf{B})\,{\rm d} {} k=\bra{\mathbf{B}^2}$,
where $\mbox{\rm Sp}(\mathbf{B})=\mbox{\rm Sp}(B_x)+\mbox{\rm Sp}(B_y)+\mbox{\rm Sp}(B_z)$.
Likewise, $\mbox{\rm Sp}(\mbox{\boldmath ${\sf h}$} {})=\mbox{\rm Sp}(h_+)+\mbox{\rm Sp}(h_\times)$ is defined as the sum over
the two polarization modes.
Of particular interest will also be the stress spectrum $\mbox{\rm Sp}(\mbox{\boldmath ${\sf T}$} {})$,
which is defined analogously through $\mbox{\rm Sp}(\mbox{\boldmath ${\sf T}$} {})=\mbox{\rm Sp}(T_+)+\mbox{\rm Sp}(T_\times)$.
To study the evolution of the stress at selected Fourier modes, we compute
$|T(k,t)|\equiv\sqrt{\mbox{\rm Sp}(\mbox{\boldmath ${\sf T}$} {})/4\pi k^2}$,
which scales the same
way as $|\tilde{T}_+(k,t)|$ and $|\tilde{T}_\times(k,t)|$.
\subsection{Evolution of the stress and strain spectra}
\label{EvolutionOfStress}
To put our results into perspective and compare with earlier work,
we study cases of suddenly initiated turbulence.
We perform simulations similar to those of
Ref.~\cite{RoperPol+20}
by using as
initial condition for the magnetic field a random Gaussian-distributed
magnetic field with a $k^4$ spectrum for $k<k_{\rm p}$ and a $k^{-5/3}$
spectrum for $k>k_{\rm p}$.
For details of such a magnetic field, see Ref.~\cite{Bran+17}.
As initial condition for the GW field, we assume that $h$ and $\dot{h}$
vanish.
The strength of the GW field is then strongly determined by the
sudden initialization of a fully developed turbulence spectrum.
The details of the simulations is given in \Tab{table1}.
In this table, the first column represents the name of the runs,
${\cal E}_{\rm EM}^{i}$ is the initial value of the magnetic energy density compared
to the background energy density, $k_{\rm p}$ is the wavenumber at which the
magnetic energy spectrum peaks and it is normalized by the wavenumber
corresponding to the Hubble horizon size at the initial time,
${\cal E}_{\rm GW}^{\rm sat}$ is the value of the GW energy density after
saturation compared to the background energy density, and
${\Omega}_{\rm GW}^{\rm sat}$ is the density parameter of GWs, representing the
ratio of the GW energy density compared to the critical energy density
at present.
${\Omega}_{\rm GW}^{\rm sat}$ has been calculated considering the production of GWs
around the electroweak phase transition.
\begin{figure*}\begin{center}
\includegraphics[width=\textwidth]{pstress_etc_LowFreq_sig1}
\end{center}\caption{
Spectra of the magnetic field, the TT-projected stress, the strain
derivative, and the strain for suddenly initiated turbulence with
magnetic helicity.
}\label{pstress_etc_LowFreq_sig1}\end{figure*}
\begin{figure*}\begin{center}
\includegraphics[width=\textwidth]{pstress_etc_LowFreq_sig0}
\end{center}\caption{
Same as \Fig{pstress_etc_LowFreq_sig1}, but for the nonhelical case.
}\label{pstress_etc_LowFreq_sig0}\end{figure*}
In \Figs{pstress_etc_LowFreq_sig1}{pstress_etc_LowFreq_sig0},
we show spectra of the magnetic field, the TT-projected stress,
the strain derivative, and the strain for runs with and without
magnetic helicity, respectively.
Inverse cascading is seen in the magnetic energy spectra, which leads
to the expected increase of the spectral stress at small $k$; see
\Figsp{pstress_etc_LowFreq_sig1}{a}{b} for the helical case.
We also see in \Figp{pstress_etc_LowFreq_sig1}{c} that the GW energy
spectrum has a maximum at $k\sim20$, which is not present in the
nonhelical case; cf.\ \Figp{pstress_etc_LowFreq_sig0}{c}.
Their spectra fall off toward smaller $k$ proportional to $k$ and
$k^{1.5}$ in the helical and nonhelical cases, respectively.
\begin{table}\caption{
Summary of simulation parameters
}\begin{center}
\begin{tabular}{ccccccc}
\hline
Run & ${\cal E}_{\rm M}^i$ &$k_{\rm p}$ & ${\cal E}_{\rm GW}^{\rm sat}$ & ${\Omega}_{\rm GW}^{\rm sat}$\\
\hline
hel & $5.4\times10^{-3}$ &$10$& $3.7\times10^{-7}$ & $5.9\times10^{-12}$\\
\hline
nonhel & $5.5 \times 10^{-3}$ &$10$& $3.5\times10^{-7}$ & $5.6\times10^{-12}$\\%LowFreq1024sig0_k01_kf10b2_rep2
\hline
\label{table1}\end{tabular}
\end{center}
\end{table}
\begin{figure*}\begin{center}
\includegraphics[width=\textwidth]{rslice_stress_plot_LowFreq.eps}
\end{center}\caption{
Modulus and phase of $\tilde{T}(k,t)$ and $\dot{\tilde{h}}(k,t)$
for the helical case for $\mathbf{k}=(k,0,0)$ with $k=0.3$ (orange), 0.4 (red),
0.5 (green), 0.6 (blue), and 0.7 (black).
The inset shows the phase with a linear abscissa.
}\label{rslice_stress_plot_LowFreq}\end{figure*}
\begin{figure*}\begin{center}
\includegraphics[width=\textwidth]{rslice_stress_plot_LowFreq_nohel}
\end{center}\caption{
Same as \Fig{rslice_stress_plot_LowFreq}, but for the nonhelical case.
}\label{rslice_stress_plot_LowFreq_nohel}\end{figure*}
\begin{figure*}\begin{center}
\includegraphics[scale=0.7]{helical_with_and_wo_phase.eps}
\includegraphics[scale=0.7]{nonhelical_with_and_wo_phase.eps}
\end{center}\caption{$\mbox{\rm Sp}(\dot{\tilde{h}})(k,t)$ vs $k$:
(a) The solid black and blue curves represent $\mbox{\rm Sp}(\dot{\tilde{h}})(k,t)$
at times $t=1.5$ and $t=37$ for run hel.
The dashed red and orange curves show $\mbox{\rm Sp}(\dot{\tilde{h}})(k,t)$ for
the case when the stress spectrum has been replaced by its modulus
in the GW evolution equation.
(b) Same as (a), but for run nonhel.}
\label{with_and_wo_phase}
\end{figure*}
In \Figs{rslice_stress_plot_LowFreq}{rslice_stress_plot_LowFreq_nohel},
we compare stress, strain derivative, and strain spectra for helical and
nonhelical runs with similar values of the initial Alfv\'en speed of about 0.1.
In the helical case with inverse cascading, the stress increases with
time at small $k$.
By contrast, in the nonhelical case, the stress always decreases at
small $k$.
In spite of these differences, the GW spectra are not so different in
the two cases.
Both show a drop for $k<1$ and a nearly flat spectrum in the interval
$1<k<10$, which is below the peak at $2k_{\rm p}=20$.
In \Figs{rslice_stress_plot_LowFreq}{rslice_stress_plot_LowFreq_nohel},
we also show the evolution of the phase, $\arg(\tilde{T})$, for different $k$ values.
From these figures, it is evident that $\arg(\tilde{T})$ remains constant
for some time and starts evolving more rapidly after that.
It is also interesting to note that the amplitude of $|\dot{\tilde{h}}|$
increases up to the time until which $\arg(\dot{\tilde{h}})$
is roughly constant.
After this time, $|\dot{\tilde{h}}|$ enters an oscillatory regime
and its amplitude does not change much.
This conclusion applies for both runs shown in
\Figs{rslice_stress_plot_LowFreq}{rslice_stress_plot_LowFreq_nohel} and
it leads us to develop a simple model to understand the GW spectrum in
these cases.
In this model we replace the stress, $\tilde{T}$ by its magnitude, $|\tilde{T}|$
discussed in \Sec{simplemodel}.
Further, to understand the role of the phases of the stress tensor in
the production of GWs, we run two new simulations analogous to Runs~hel
and nonhel where we replace $\tilde{T}(k,t)$ with
its modulus at each time step.
The final GW spectrum in these new runs turn out to be same as
Runs~hel and nonhel and these are shown in \Fig{with_and_wo_phase}.
The comparisons of helical and nonhelical runs are shown in
parts (a) and (b) of this figure, respectively.
Dashed red and orange curves at times $t=1.5$ and $t=37$ respectively,
are for the case when $\tilde{T}(\mathbf{k},t)$ has been replaced by its modulus.
It is evident from the figure that there is hardly any difference in the
actual $\mbox{\rm Sp}(\dot{\tilde{h}})$ and the spectra obtained after replacing
the stress with its modulus.
On the basis of this observation, we develop a model to obtain the GW
spectrum from the time evolution of the spectrum of the stress tensor.
A striking difference between the helical and nonhelical cases is a more
pronounced peak in the spectral GW energy.
As we show in \App{appendixa}, this is due to the fact that the
stress spectrum for
the nonhelical case is different from that of the helical case due to
the presence of additional helical contributions to the two-point correlation
of the magnetic field vectors.
This difference is shown in \Fig{stress_spectrum_helvsnonhel} and the
details are explained in the next section.
\begin{figure}
\begin{center}
\includegraphics[scale=0.6]{helical_vs_nonhelical.eps}
\end{center}\caption{
In this figure, magnetic field energy spectrum, $E_{\rm M}(k)$ (Dashed curves)
and $\mbox{\rm Sp}(\tilde{T})$ (Solid curves) for the helical and non helical case.
The blue and red curves are for nonhelical and helical case respectively.
}\label{stress_spectrum_helvsnonhel}
\end{figure}
\begin{figure*}\begin{center}
\includegraphics[scale=0.7]{helical1.eps}
\includegraphics[scale=0.7]{helical2.eps}
\end{center}\caption{
Left: Solutions for $\mbox{\rm Sp}(\mbox{\boldmath ${\sf T}$} {}(k))$ (red) for different $E_{\rm M}(k)$ (blue) for three
values of $k_{\rm p}$.
Right: solutions for $\mbox{\rm Sp}(\mbox{\boldmath ${\sf T}$} {}(k)$ scaled by $k_{\rm p}$ (blue) and
$k_{\rm p}^{-8/3}$ (red), to see its scalings in the subinertial
and inertial ranges, respectively.
}\label{pcascade}\end{figure*}
\begin{figure*}\begin{center}
\includegraphics[scale=0.7]{nonhelical1.eps}
\includegraphics[scale=0.7]{nonhelical2.eps}
\end{center}\caption{
Similar to \Fig{pcascade}, but for a case with $\beta=1$.
On the right, the solutions for $\mbox{\rm Sp}(\mbox{\boldmath ${\sf T}$} {}(k)$ are scaled by
$k_{\rm p}$ (blue) and $k_{\rm p}^{14/3}$ (red), to see its scalings
in the subinertial and inertial ranges, respectively.
}\label{pcascade_beta2}\end{figure*}
\subsection{Overall behavior of the stress}
\label{stress_evolution}
At the most minimalistic level, we can say that the magnetic field shows
an approximately self-similar evolution at late times, where for the
helical case, the peak value of $E_{\rm M}(k,t)$ is unchanged, but the position
of the peak $k_{\rm p}$ goes to progressively smaller values as
$k_{\rm p}\sim t^{-2/3}$.
To understand the consequences for the evolution of the stress, let us now
consider an idealized model, where $E_{\rm M}(k)\equiv\mbox{\rm Sp}(\mathbf{B})/2$ has a $k^4$
subinertial range, where $k<k_{\rm p}$, with $k_{\rm p}(t)$ being the
peak wavenumber, and a $k^{-5/3}$ inertial range spectrum for $k>k_{\rm p}$.
The spectrum of the transverse traceless part of the stress,
$\mbox{\rm Sp}(\mbox{\boldmath ${\sf T}$} {})/2$, can be computed analytically using the expressions given
in \App{appendixa} \citep[for details, see][]{CDK04,Sharma+20}
and is shown in \Figs{pcascade}{pcascade_beta2} for the helical and
nonhelical cases, respectively.
In these figures, we take three instances where the magnetic peaks are at
wavenumbers $k_{\rm p}=1$, 0.3, and $0.1$.
For the helical case, the position of the peak of $E_{\rm M}(k)$ is unchanged.
We see that, in agreement with earlier work \citep{RoperPol+20},
the positions of the peak of $\mbox{\rm Sp}(\mbox{\boldmath ${\sf T}$} {})$ are always at $2k_{\rm p}$.
However, even though the peak values of $E_{\rm M}(k)$ are unchanged, except
for the factor of two, those of $\mbox{\rm Sp}(\mbox{\boldmath ${\sf T}$} {})$ are not and decay.
Nevertheless, at small $k$, $\mbox{\rm Sp}(\mbox{\boldmath ${\sf T}$} {})$ still increases proportional to
$k_{\rm p}^{-1}$.
If $k_{\rm p}\propto t^{-2/3}$, as expected for helical turbulence
\citep{Hat84,BM99,BK17}, we find that $\mbox{\rm Sp}(\mbox{\boldmath ${\sf T}$} {})\propto t^{2/3}$ for small $k$.
For the nonhelical case, as shown in Ref.~ \cite{BK17}, the peak of the
spectrum decreases with decreasing values of $k_{\rm p}$ proportional
to $k_{\rm p}^\beta$, where $\beta$ is an exponent that can be between
one and four.
In \Fig{pcascade_beta2}, we present the case with $\beta=1$ and find
that now $\mbox{\rm Sp}(\mbox{\boldmath ${\sf T}$} {})(k)\propto k_{\rm p}$ for small $k$ and
$\propto k_{\rm p}^{14/3}$ for large $k$.
If $k_{\rm p}\propto t^{-1/2}$, as expected for the nonhelical case for
$\beta=1$, $\mbox{\rm Sp}(\tilde{T})\propto t^{-1/2}$ for small $k$.
Recently, it has been found that the Saffman helicity invariant is well
conserved in nonhelical magnetically dominated decaying turbulence
\citep{hosking20}, which implies $\beta=1.5$.
For the general case, we write $\mbox{\rm Sp}(\tilde{T})\propto k_{\rm p}^{2\beta-1}$
(for $k<k_{\rm p}$), which implies $\mbox{\rm Sp}(\tilde{T})\propto t^{-8/9}$ for
$\beta=1.5$ and $k_{\rm p}\propto t^{-4/9}$.
It is also interesting to note that, for a given
spectrum of magnetic field (blue and red dashed curve in
\Fig{stress_spectrum_helvsnonhel}), $\mbox{\rm Sp}(\mbox{\boldmath ${\sf T}$} {})$ is different for the
helical and nonhelical cases.
The blue and red curves are for nonhelical and helical cases, respectively.
In the helical case, $\mbox{\rm Sp}(\mbox{\boldmath ${\sf T}$} {})$ has smaller values compared to the
nonhelical case at wavenumbers below $k_{\rm p}$.
However, it has large values for wavenumbers around the peak and above.
Such a feature of the stress spectrum also translates to the GW spectrum
and that is why we see a difference in the final GW spectrum produced
from helical and nonhelical cases discussed in the previous section.
\section{Predictions from algebraically growing stress}\label{simplemodel}
With the detailed information above, we are now in a position to compare
with the predictions from a simple time-dependent model.
In this section, we compute GW spectra by considering a simple
model for the time evolution of the stress.
It is assumed to increase algebraically as a power law
characterized by a power law index $p$ during the time interval
from $t=1$ to $t_{\rm e}$.
\subsection{The model}
We model the $+$ and $\times$ polarizations of the Fourier-transformed
stress, $\tilde{T}(k,t)$, as
\begin{equation}\label{stresscase2}
\tilde{T}(k,t)= \left\{\begin{array}{ll}
\tilde{T}_0(k) \, t^{p} , \quad & 1\le t \le t_{\rm e}, \\
0, & t > t_{\rm e},
\end{array}\right.
\end{equation}
where $\tilde{T}_0(k)\equiv\sqrt{\langle\tilde{T}_{ij}^{TT}(\mathbf{k})\tilde{T}_{*ij}^{TT}(\mathbf{k})\rangle}$
represents $|\tilde{T}(k,t)|$ at the initial time and is obtained
for given energy and helicity spectra of the magnetic field;
see \App{appendixa} for details.
We note that the authors of Ref.~\cite{RoperPol+22}
have developed an analytical model for the GW spectrum on the basis of
the time evolution of the stress, which they assumed constant during
a certain interval -- unlike our case.
The authors explain the location of certain breaks in their GW spectrum
as a consequence of the finite duration over which the stress is constant.
This duration is an empirical input parameter.
In our model, by contrast, the stress evolves as
a power law with an index that is in principle known from MHD theory,
although we can get even better agreement with the simulations when we
take the actual power-law index that is realized in the simulations.
\begin{figure*}
\includegraphics[scale=0.7]{model.eps}
\includegraphics[scale=0.7]{model_fig2.eps}
\caption{
(a) $\mbox{\rm Sp}(\dot{\tilde{h}}(k,t)$ at different times.
Here, we assume $t=t_{\rm e}$,
$E_{\rm M}=c (k/k_{\rm p})^4/(1+(k/k_{\rm p})^{17/3})$, where $k_{\rm p}=10$
and $c=10^{-4}$ and $p=-1/4$.
The red, blue, and black curves are for $t_{\rm e}=2$, $4$, and $10$ respectively.
The two black vertical lines correspond to $k_{*}$ and $2 k_{\rm p}$.
(b) $\mbox{\rm Sp}(\dot{\tilde{h}})$ at times $t_{\rm e}=10$, $20$, and $30$.
}\label{results_from_model}
\end{figure*}
To obtain the GW spectrum for our model, we first solve \Eq{GW4} for a
case when the source is active during the interval $1<t<t_{\rm e}$ and
thus obtain $\tilde{h}(k,t)$ and $\dot{\tilde{h}}(k,t)$.
The solution for $t\ge t_{\rm e}$ is given by
\begin{align}
\tilde{h}(k,t)&=\int_1^{t} \frac{\sin k(t-t')}{k} \, \frac{6\tilde{T}(k,t')}{t'} \, dt', \\
\dot{\tilde{h}}(k,t)&=\int_1^{t} \cos k(t-t') \, \frac{6\tilde{T}(k,t')}{t'} \, dt'. \label{hdot}
\end{align}
Using \Eq{hdot} and our model for $\tilde{T}(k,t)$, we obtain
\begin{align}
\dot{\tilde{h}}(k,t)&=\frac{-3 \tilde{T}_0(k)}{(k t_0)^p}\Big\{e^{i(kt-p \pi/2)}\big[\Gamma(p,ikt_{\rm e})-\Gamma(p,ikt_0)\big]\nonumber\\
&+e^{-i(kt-p \pi/2)}\big[\Gamma(p,-ikt_{\rm e})-\Gamma(p,-ikt_0)\big]\Big\}.\label{full_solution}
\end{align}
In the above expression, $t_0=1$ represents the initial time.
In \Fig{results_from_model}(a), we show $\mbox{\rm Sp}(\dot{\tilde{h}})$
at different times for this model with $p=-1/4$.
The red, blue, and black curves represent $\mbox{\rm Sp}(\dot{\tilde{h}})$ at $t=2$,
$4$, and $10$, respectively.
It is evident from this figure that $\mbox{\rm Sp}(\dot{\tilde{h}})$ is almost
flat for $1\la k\la 2 k_{\rm p}$ and declines as $\propto k^{-11/3}$
for $k> 2k_{\rm p}$.
$\mbox{\rm Sp}(\dot{\tilde{h}})$ is proportional to $k^2$ for $k<k_{\rm H}$, where
$k_{\rm H}$ represents the wavenumber corresponding to the Hubble horizon size at $t=t_{\rm e}$.
Further, as time increases, $\mbox{\rm Sp}(\dot{\tilde{h}})$ at low wavenumbers
($k_{\rm H} < k < 1$) grows and saturates, as is evident from
\Fig{results_from_model}(b).
To understand the role of the power-law index in the algebraically
growing part of the stress on $\mbox{\rm Sp}(\dot{\tilde{h}})$, we calculate
$\mbox{\rm Sp}(\dot{\tilde{h}})$ for different values of $p$.
Those are shown in the right-hand panel of \Fig{sphdot_for_different_gamma}.
Here, $\mbox{\rm Sp}(\dot{\tilde{h}})$ is rapidly oscillating, so we plot
in this figure only its envelope.
From this figure, we conclude that $\mbox{\rm Sp}(\dot{\tilde{h}})$ can be divided
into three regime.
We begin discussing first the high wavenumber regime ($k>k_{0}$, regime~I),
where $k_{0}$ represents the wavenumber
corresponding to the Hubble horizon at the initial time. $\mbox{\rm Sp}(\dot{\tilde{h}})$
is flat and changes to $k^{-11/3}$ for $k>2k_{\rm p}$.
For very low wavenumbers corresponding to the superhorizon
range ($k<k_{\rm H}$, regime~III) $\mbox{\rm Sp}(\dot{\tilde{h}})$ is proportional to $k^2$.
In the intermediate regime ($k_{\rm H}\lesssim k\lesssim (1-p)/t_0$, regime~II),
$\mbox{\rm Sp}(\dot{\tilde{h}})$ changes from a flat spectrum
to a $k^2$ spectrum as the wavenumber decreases.
Note that, as the wavenumbers decrease, the transition from a flat spectrum
to a $k^2$ spectrum is faster for the case when $p=-1/4$ than $p=1/3$.
The wavenumber at which this transition occurs depends on the value of
$p$ and can be understood as follows.
In the algebraically growing phase, the typical time scale over which
$\tilde{T}/t$ decays, is $\delta t_T\sim t/(1-p)$ and the typical time
scale for sourcing GWs at a given wavenumber $k$
just after $t=1$ is $\delta t_{\rm GW}\sim 1/k$, as can be inferred
from the cosine function in \Eq{hdot}.
The value of $\tilde{T}/t$ does not change much when $\delta
t_{\rm GW}/\delta t_T\le 1$.
This implies that, for $k>(1-p)/t_0$, there will be a finite
interval during which $\tilde{T}/t$ can be assumed constant.
However, for $k<(1-p)/t_0$, $\tilde{T}/t$ always changes.
The wavenumber $k\sim (1-p)/t_0$ corresponds to the wavenumber
where $\mbox{\rm Sp}(\dot{\tilde{h}})$ starts changing from a flat spectrum.
\begin{figure}
\begin{center}
\includegraphics[scale=0.8]{sp_hdot_for_different_p.eps}
\caption{$\mbox{\rm Sp}(\dot{\tilde{h}})$ for different values of $p$.
Here, the left, middle, and right black vertical lines represent the wavenumbers corresponding to
the horizon size at the final time $t$, the initial time $t_0=1$, and $2k_{\rm p}$, respectively.}
\label{sphdot_for_different_gamma}
\end{center}
\end{figure}
The nature of $\mbox{\rm Sp}(\dot{\tilde{h}})$ can also be understood
by writing the expression of $\dot{\tilde{h}}(k,t)$, given in
\Eq{full_solution}, for different limits depending on the values of $kt_0$
and $kt_{\rm e}$.
For $t_{\rm e}\gg t_0$, which is indeed the case, and $p<1$,
\Eq{full_solution} reduces to
\begin{equation}
\frac{\dot{\tilde{h}}(k,t)}{6 \tilde{T}_0(k)}\approx \left\{\begin{array}{lr}
\frac{\sin{k(t-t_0)}}{kt_0} \quad \text{(I)},\\[5pt]
\frac{\Gamma[p]}{(kt_0)^p}\cos\left({kt-\frac{p\pi}{2}}\right)-\frac{\cos{kt}}{p} \quad \text{(II)},\\[5pt]
\frac{\cos{kt}}{p}\left[\left(\frac{t_{\rm e}}{t_0}\right)^p-1\right] \quad \text{(III)}.
\end{array}\right.
\end{equation}
Using this, we calculate the spectrum of $\dot{\tilde{h}}$; it is given by
\begin{equation}
\frac{\mbox{\rm Sp}(\dot{\tilde{h}}(k,t))}{36 \tilde{T}_0^2(k)}\approx \left\{\begin{array}{lr}
\left[\frac{\sin{k(t-t_0)}}{t_0}\right]^2 \quad \text{(I)}, \\[5pt]
k^2\Big[-\frac{\cos{kt}}{p}+\frac{\Gamma[p]}{(kt_0)^p}\cos({kt-\frac{p\pi}{2}})\Big]^2 \, \text{(II)}, \\[5pt]
k^2\Big\{\frac{\cos{kt}}{p}\left[\left(\frac{t_{\rm e}}{t_0}\right)^p-1\right]\Big\}^2 \quad \text{(III)}.
\end{array}\right.\label{approx_solution}
\end{equation}
From the above expression, we conclude that the break points
for the different slopes of $\mbox{\rm Sp}(\dot{\tilde{h}})$
is decided by $\tilde{T}_0^2(k)$ for $k>t_0^{-1}$.
Here, $\tilde{T}_0^2(k)$ is flat for $k<2 k_{\rm p}$ and
proportional to $k^{-11/3}$ for $k>2 k_{\rm p}$.
For the superhorizon modes, i.e., $k<1/t_{\rm e}$, $\mbox{\rm Sp}(\dot{\tilde{h}})$
is proportional to $k^2$, and for wavenumbers $t_0^{-1}<k<t_{\rm e}^{-1}$,
$\mbox{\rm Sp}(\dot{\tilde{h}})$ changes from a flat spectrum to $k^2$, as
shown as the blue curves in \Fig{sphdot_for_different_gamma}.
\begin{figure*}
\begin{center}
\includegraphics[scale=0.7]{t_evolution_helical.eps}
\includegraphics[scale=0.7]{t_evolution_nonhelical.eps}
\caption{
$\tilde{T}(k,t)$ vs.\ $t$. (a) The blue curve corresponds to the time
evolution of $\tilde{T}(k,t)$ vs $t$ obtained from the simulation for $k=3$
for the helical case and the red curve corresponds to a broken power law
fit to the blue curve.
(b) Same as (a), but for nonhelical case.
\label{t_evolution}
}\end{center}
\end{figure*}
\begin{figure*}
\begin{center}
\includegraphics[scale=0.7]{LowFreq1024sig1_comparison_broken.eps}
\includegraphics[scale=0.7]{LowFreq1024sig0_comparison_broken.eps}
\caption{
$\mbox{\rm Sp}(\dot{\tilde{h}})$ vs $k$: (a) The blue curve represents
$\mbox{\rm Sp}(\dot{\tilde{h}})$ corresponding to the run shown in
\Fig{pstress_etc_LowFreq_sig1}, respectively.
The red and green curves represent the spectra obtained in the model
for the time evolution of $\tilde{T}(k,t)$ given in column 2 of
\Tab{table2}.
The orange curves represent $\mbox{\rm Sp}(\dot{\tilde{h}})$ for the values of
$p$ given in column~1 of \Tab{table2}.
The red curves are for the case when $|\tilde{T}_0(k)|$ is obtained using \Eq{tijnh} of \App{appendixa}. For the green
curves, $|\tilde{T}_0(k)|$ is taken from the simulation.
(b) Same as (a), but for nonhelical case.
}\label{helical_comparison}
\end{center}
\end{figure*}
In this model, we take the same algebraic evolution with a constant
power-law index for all wavenumbers.
In general, however, the time evolution of $\tilde{T}(k,t)$ is different for
wavenumbers below and above the peak of $\mbox{\rm Sp}(\tilde{T})(k,t)$.
For the case of helical magnetic fields discussed in \Fig{pcascade}, the
value of $\mbox{\rm Sp}(\tilde{T})$ for a particular $k<2k_{\rm p}$ at the initial time,
grows as $t^{2/3}$ until the time for which $\mbox{\rm Sp}(\tilde{T})$ peaks at this
particular wavenumber.
After this time, the value of $\mbox{\rm Sp}(\tilde{T})$ for this particular $k$
starts decreasing as $t^{-16/9}$.
This implies that we would have assumed $\tilde{T}(k)\propto t^{1/3}$
for $k<k_{\rm p}$ and $\tilde{T}(k)\propto t^{-8/9}$ for $k>k_{\rm p}$.
For the nonhelical magnetic field shown in \Fig{pcascade_beta2},
$\mbox{\rm Sp}(\tilde{T})$ is always decreasing.
For $k<2 k_{\rm p}$ at the
initial time, it first decreases as $t^{-1/2}$ and later switches to
$t^{-7/3}$.
We study $\mbox{\rm Sp}(\dot{\tilde{h}})$ by incorporating this and find that
there is no difference in the final $\mbox{\rm Sp}(\dot{\tilde{h}})$ compared to
the one obtained in our model.
This is why we did not consider the aforementioned evolution of the
stress to keep the model simple.
\section{Comparison of the analytical model with simulation results}
\label{comparison}
In the above section, we discussed $\mbox{\rm Sp}(\dot{\tilde{h}})$ in a model
inspired by the fact that the GW spectrum does not change if we change
the stress tensor by its modulus for decaying MHD turbulence in the
early Universe.
In this model, we approximate the stress tensor by
$|T(k,t)|\equiv\sqrt{\mbox{\rm Sp}(\mbox{\boldmath ${\sf T}$} {})/4\pi k^2}$ and its time evolution
is parameterized as a power law with index $p$; see \Fig{t_evolution}.
In this section, we provide a comparison for $\mbox{\rm Sp}(\dot{\tilde{h}})$
obtained in this model with the simulation results discussed in
\Sec{EvolutionOfStress}.
To compare, we show $\mbox{\rm Sp}(\dot{\tilde{h}})$ obtained from the
model and different simulations together.
In \Figsp{helical_comparison}{a}{b}, we plot the spectra for the runs
shown in \Figs{pstress_etc_LowFreq_sig1}{pstress_etc_LowFreq_sig0} and
discussed in \Sec{EvolutionOfStress}.
For the dashed orange curve, $\tilde{T}_0$ is obtained by using \Eq{tijh} of \App{appendixa},
where we take $E_{\rm M}=c (k/k_{\rm p})^4/(1+(k/k_{\rm p})^{17/3})$ and the value of
the constant $c$ is determined such that the obtained stress spectrum
matches with that of the simulation at $t=1$.
For this case, the stress spectrum evolution is modeled as a single power
law and the value of $p$ is $1/3$ and $-1/4$ for the helical and nonhelical
cases, respectively.
The value of $p$ is decided from the time evolution of the low wavenumber
tail of $\mbox{\rm Sp}(\tilde{T})$, as discussed in \Sec{stress_evolution}.
From this figure, it is concluded that the spectral nature of
$\mbox{\rm Sp}(\dot{\tilde{h}})$ matches well with the prediction from the model.
However, there is a difference at small wavenumbers, especially for the
nonhelical case.
This is due to the fact that the modelling of $\mbox{\rm Sp}(\tilde{T})$ by a single
power does not provide a better fit to the evolution obtained in the
simulation for the nonhelical case.
A double power law of the form $t^{-1/3}/(1+(t-1)^n)^{5/7n}$, where
$n$ regulates the transition, here with $n=10$, provides a better fit
to $\mbox{\rm Sp}(\tilde{T})$ for the low frequency tail for the nonhelical case; see
\Fig{t_evolution}.
In this figure, we plot $\tilde{T}(k,t)$ obtained from the simulation in
blue color for the wavenumber $k=3$ and the double power law fit to this
curve is in red color.
The double power law, which fits the $\tilde{T}(k,t)$ for the helical case,
is $1/(1+(t-0.2)^n)^{5/24n}$, where $n=20$.
After considering such a time evolution, the obtained $\mbox{\rm Sp}(\dot{\tilde{h}})$
is shown as the dashed red curve in \Fig{helical_comparison}(b).
For the dashed green curve, we consider $\tilde{T}_0(k)=\sqrt{\mbox{\rm Sp}(\mbox{\boldmath ${\sf T}$} {})/4\pi k^2}$
and $\mbox{\rm Sp}(\mbox{\boldmath ${\sf T}$} {})$ is obtained from the simulation at $t=1$.
These different forms of the time evolution of $\tilde{T}(k,t)$ are
tabulated in \Tab{table2}.
We notice that the nature of the GW spectra in the helical case are
different than in the nonhelical case.
There is a large power in the helical case compared to the nonhelical
case around the peak of the GW spectrum for the same strength of the
initial magnetic field.
This is due to the presence of an additional term due to
the helicity spectrum in the stress spectrum.
\begin{table}\caption{
Time dependence of $\tilde{T}(k,t)$ taken in our analysis
}\begin{center}
\begin{tabular}{c|c|c}
&\multicolumn{1}{c|}{from MHD theory}&
\multicolumn{1}{c}{from MHD simulations}\\
Run& $p$& $p$ \\
\hline
hel& $\left(\frac{t}{t_0}\right)^{1/3}$& $\frac{1}{(1+(t-0.2)^n)^{5/24n}}$ \\
\hline
nonhel&
$\left(\frac{t}{t_0}\right)^{-1/4}$& $\frac{t^{-1/6}}{(1+(t-1)^n)^{5/14n}}$ \\
\hline
\label{table2}\end{tabular}
\end{center}
\end{table}
\section{Conclusions}\label{conclusion}
In this work, we have suggested a simple model to understand the GW spectrum
obtained for decaying MHD turbulence in the early Universe.
The Fourier-transformed stress is taken to be
$|\tilde{T}(k,t)|$, i.e., we ignore changes in the phase, and
its time evolution is parameterized by a power law.
Such a time evolution of the stress is motivated by the simulations for
the decaying MHD turbulence at low wavenumbers discussed in \Sec{EvolutionOfStress}.
We find that the spectral nature of the GW spectrum is well represented
by this simple model.
In this work, we also show that the nature of the GW spectra in the helical
case are different from those in the nonhelical case.
Apart from the polarization of GW, this spectral difference may also
be important in distinguishing the helical and nonhelical nature of the
primordial magnetic field.
In this work, we have developed a model to understand the low frequency tail
of the GW spectrum in the cases where turbulence is initiated suddenly.
However, it will be more interesting to study those cases where magnetic
field is generated selfconsistently, such as through the chiral magnetic
effect in the early Universe \citep{Roga_etal17,Schober+18}.
It would be interesting to see, if a model such as the one discussed in
this paper can also explain the GW spectra obtained through the chiral
magnetic effect.
This, we hope to report in a future study.
\begin{acknowledgements}
RS would like to thank Hongzhe Zhou for helping him analyzing the output
data with the Mathematica routines of the {\sc Pencil Code}.
This work was supported by the Swedish Research Council
(Vetenskapsradet, 2019-04234).
Nordita is sponsored by Nordforsk.
We acknowledge the allocation of computing resources provided by the
Swedish National Allocations Committee at the Center for Parallel
Computers at the Royal Institute of Technology in Stockholm and
Link\"oping.
\end{acknowledgements}
\vspace{2mm}
{\bf Data availability}---The source code used for the
simulations of this study, the {\sc Pencil Code},
is freely available from Refs.~\cite{JOSS}.
The simulation setups and the corresponding data
are freely available from Ref.~\cite{DATA}.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 1,149 |
#include <d3d12.h>
#include "DiligentCore/Graphics/GraphicsEngineD3D12/interface/TextureD3D12.h"
| {
"redpajama_set_name": "RedPajamaGithub"
} | 2,072 |
package org.apache.geronimo.testsuite.jsps;
import java.net.URL;
import org.apache.geronimo.testsupport.HttpUtils;
import org.apache.geronimo.testsupport.TestSupport;
import org.testng.annotations.Test;
public class TestJsps extends TestSupport {
@Test
public void testDeferral() throws Exception {
URL url = new URL("http://localhost:8080/jsp21/testDeferral.jsp");
String reply = HttpUtils.doGET(url);
assertTrue("testDeferral", reply.contains("OneTwo"));
}
@Test
public void testScopes() throws Exception {
URL url = new URL("http://localhost:8080/jsp21/testScopes.jsp");
String reply = HttpUtils.doGET(url);
assertTrue("testScopes", reply.contains("value1 value2 value3 value4"));
}
@Test
public void testTaglibs() throws Exception {
URL url = new URL("http://localhost:8080/jsp21/testTaglibs.jsp");
String reply = HttpUtils.doGET(url);
assertTrue("testTagLibs", reply.contains("Hello"));
}
@Test
public void testTrimWhitespace() throws Exception {
URL url = new URL("http://localhost:8080/jsp21/testTrimWhitespace.jsp");
String reply = HttpUtils.doGET(url);
assertTrue("testTrimWhitespace", reply.contains("source html of this page should not contain empty lines"));
}
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 2,903 |
{"url":"http:\/\/www.physicsforums.com\/showthread.php?t=346897","text":"## orbital angular momentum problem\n\n1. The problem statement, all variables and given\/known data\n\nConsider a particle that moves in three dimensions with wave function $$\\varphi$$ . Use operator methods to show that if $$\\varphi$$ has total angular momentm quantum number l=0 , then $$\\varphi$$ satifies\n\n$$L\\varphi=0$$\n\nfor all three components $$L_\\alpha$$ of the total angular momentum L\n2. Relevant equations\n\n$$[L^2,L_\\alpha]$$?\n\n$$L^2=L_x^2+L_y^2+L_z^2$$?\n\n$$L^2=\\hbar*l(l+1)$$ , l=0,1\/2,1,3\/2,... ?\n\n$$L_z=m\\hbar$$ , m=-l, -l+1,...,l-1,l. ?\n3. The attempt at a solution\n\n$$[L^2,L_\\alpha]=[L_x^2+L_y^2+L_z^2,L_\\alpha]=[L_x^2,L_\\alpha]+[L_y^2,L_\\alpha]+[L_z^2,L_\\alpha]$$. $$[AB,C]=A[B,C]+[A,C]B$$; Therefore,$$[L_x^2,L_\\alpha]+[L_y^2,L_\\alpha]+[L_z^2,L_\\alpha]=L_x[L_x,L_\\alpha]+[L_x,L_\\alpha]L_x+L_y[L_y,L_\\alpha]+[L_y,L_\\alpha]L_y+L_z[L_z,L_\\alpha]+[L_z,L_\\alpha]L_z$$. Now what?\n PhysOrg.com science news on PhysOrg.com >> Intel's Haswell to extend battery life, set for Taipei launch>> Galaxies fed by funnels of fuel>> The better to see you with: Scientists build record-setting metamaterial flat lens\n People having a hard time reading the problem?\n Recognitions: Homework Help Do you not already know $[L^2,L_x]$, $[L^2,L_y]$ and $[L^2,L_z]$? If so, just calculate $[L^2,L_x]\\varphi$, $[L^2,L_y]\\varphi$ and $[L^2,L_z]\\varphi$ and use the fact that $L^2\\varphi=l(l+1)\\varphi=0$ for $l=0$.\n\nRecognitions:\n\n## orbital angular momentum problem\n\nI don't understand gabbagabbahey's method. I would (using bra-ket notation) define three states $|\\psi_i\\rangle = L_i|\\phi\\rangle$, $i=x,y,z$, and then compute $\\sum_{i}\\langle\\psi_i|\\psi_i\\rangle$.\n\n Quote by gabbagabbahey Do you not already know $[L^2,L_x]$, $[L^2,L_y]$ and $[L^2,L_z]$? If so, just calculate $[L^2,L_x]\\varphi$, $[L^2,L_y]\\varphi$ and $[L^2,L_z]\\varphi$ and use the fact that $L^2\\varphi=l(l+1)\\varphi=0$ for $l=0$.\nYes I calculated those commutators $$[L^2,L_x]\\varphi=(L_y*i\\hbar*L_z)+(i*\\hbar*L_z)(L_y)+L_z(i*\\hbar*L_y)+(i*\\hbar*L _y)L_z,[L^2,L_y]= L_x(i*\\hbar*L_z)+(i*\\hbar*L_z)L_y+L_z(i*\\hbar*L_x)+(i*\\hbar*L_x) +(i*\\hbar*L_x)L_z, [L^2,L_z]=L_x(i*\\hbar*L_y)+(i*\\hbar*L_y)(L_x)+L_y(i*\\hbar*L_x)+(i*\\hbar*L_x)L_y$$, The comutators simply expand more and do not simplify . at $$l=0, l(l+1)\\varphi=0$$\n\nPlease look at my latex for the expression $$[L^2,L_z]$$ PF is not completely showing the solution to this expression for some reason. Now is it readable\n Recognitions: Homework Help What you've just written is completely unreadable....How do you expect others to be able to help you, if you don't take the time to make sure they can read what you are posting? Begin by calculating the commutator $[L^2,L_x]$....use the \"go advanced\" and \"preview post\" buttons to make sure whatever you write is actually readable, before submitting your post.\n $$[L^2,L_x]\\varphi=(L_y*i\\hbar*L_z)+(i*\\hbar*L_z)(L_y)+L_z(i* \\hbar*L_y)+(i*\\hbar*L_y)L_z,[L^2,L_y]= L_x(i*\\hbar*L_z)+(i*\\hbar*L_z)L_y+L_z(i*\\hbar*L_x) +(i*\\hbar*L_x) +(i*\\hbar*L_x)L_z, [L^2,L_z]=L_x(i*\\hbar*L_y)+(i*\\hbar*L_y)(L_x)+L_y(i*\\hbar*L _x)+(i*\\hbar*L_x)L_y$$ Now is it readable? $$[L^2,L_z]=L_x(i*\\hbar*L_y)+(i*\\hbar*L_y)(L_x)+L_y(i*\\hbar*L _x)+(i*\\hbar*L_x)L_y$$\n Recognitions: Homework Help I still can't make heads or tails of what you've written....When you read a textbook or article, or even someone else's posts here....do they lay things out in a jumbled mess like that?\n\n Quote by gabbagabbahey I still can't make heads or tails of what you've written....When you read a textbook or article, or even someone else's posts here....do they lay things out in a jumbled mess like that?\nI will seperate my commutators into seperate latex code and aligned the commutators vertically. Maybe it should then be easier for you to read:\n\n$$[L^2,L_x]\\varphi=(L_y*i\\hbar*L_z)+(i*\\hbar*L_z)(L_y)+L_z(i* \\hbar*L_y)+(i*\\hbar*L_y)L_z$$\n$$[L^2,L_y]\\varphi= L_x(i*\\hbar*L_z)+(i*\\hbar*L_z)L_y+L_z(i*\\hbar*L_x) +(i*\\hbar*L_x) +(i*\\hbar*L_x)L_z$$\n$$[L^2,L_z]\\varphi=L_x(i*\\hbar*L_y)+(i*\\hbar*L_y)(L_x)+L_y(i*\\hbar*L _x)+(i*\\hbar*L_x)L_y$$,\nat$$l=0, l(l+1)\\varphi=0$$\n Recognitions: Homework Help That's a little easier to read (Although I really wish you'd stop using that ugly $*$ symbol for multiplication). Why do you have a $\\varphi$ on the LHS of each equation? And, you are missing some negative signs on the RHS....why don't you show me your steps for calculating $[L^2,L_x]$?\n\n Quote by gabbagabbahey That's a little easier to read (Although I really wish you'd stop using that ugly $*$ symbol for multiplication). Why do you have a $\\varphi$ on the LHS of each equation? And, you are missing some negative signs on the RHS....why don't you show me your steps for calculating $[L^2,L_x]$?\nokay.$$[L^2,L_x]=[L_x^2+L_y^2+L_z^2,L_x]=[L_x^2,L_x]+[L_y^2,L_x]+[L_z^2,L_z]$$\n\n$$[L_x^2,L_x]+[L_y^2,L_x]+[L_z^2,L_x]=[L_y^2,L_x]+[L_z^2,L_x]$$\n\n$$[L_y^2,L_x]+[L_z^2,L_z]=L_y[L_y,L_x]+[L_y,L_x]L_y+L_z[L_z,L_x]+[L_z,L_x][L_z]$$\n\n$$[L_y,L_x]=i*\\hbar*L_z, [L_z,L_x]=i*\\hbar*L_y$$, therefore\n\n$$L_y[L_y,L_x]+[L_y,L_x]L_y+L_z[L_z,L_x]+[L_z,L_x][L_z]=i*L_y(\\hbar*L_z),+i*\\hbar*L_z*L_y+i*L_z*(\\hbar*L_y)+i(\\hbar*L_y)*L_z$$ .How did you get a negative term? (Sorry, if I don't used the $$*$$ symbol, latex will read the i's and L's as a subscript rather than a coeffcient\n Recognitions: Homework Help $[L_y,L_x]=-[L_x,L_y]=-i\\hbar L_z$\n My calculations showed that $$[L^2,L_z]=0,[L^2,L_y]=0,[L^2,L_x]=0$$, now what?\n Recognitions: Homework Help Now, by definition what is $[L^2,L_x]\\varphi$?...compare that to what you've just calculated....\n\n Quote by gabbagabbahey Now, by definition what is $[L^2,L_x]\\varphi$?...compare that to what you've just calculated....\nsince$$[L^2,L_x]=0$$,then$$[L^2,L_x]\\varphi=0\\varphi=0$$?\n Recognitions: Homework Help If I asked you the definition of $[A,B]\\varphi$, you would say $(AB-BA)\\varphi$, right? So, when I say \"by definition what is $[L^2,L_x]\\varphi$?\", you say ____?\n\n Quote by gabbagabbahey If I asked you the definition of $[A,B]\\varphi$, you would say $(AB-BA)\\varphi$, right? So, when I say \"by definition what is $[L^2,L_x]\\varphi$?\", you say ____?\nyes thats right. $$[L^2,L_x]=(L^2L_x-L_xL^2)\\varphi$$, should I let $$L_x=(\\hbar\/i)*d\/dx$$ and $$L=(\\hbar\/i)*d\/dx+(\\hbar\/i)*d\/dy+(\\hbar\/i)*d\/dz$$ and let $$(-\\hbar\/i)*d\/dx+(-\\hbar\/i)*d\/dy+(-\\hbar\/i)*d\/dz$$","date":"2013-05-26 05:54:22","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\": 1, \"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\": 0.7681255340576172, \"perplexity\": 1017.0414881319687}, \"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-2013-20\/segments\/1368706631378\/warc\/CC-MAIN-20130516121711-00029-ip-10-60-113-184.ec2.internal.warc.gz\"}"} | null | null |
<?xml version="1.0" encoding="utf-8"?>
<!--
~ Copyright 2016-2020 Open Food Facts
~
~ Licensed under the Apache License, Version 2.0 (the "License");
~ you may not use this file except in compliance with the License.
~ You may obtain a copy of the License at
~
~ https://www.apache.org/licenses/LICENSE-2.0
~
~ Unless required by applicable law or agreed to in writing, software
~ distributed under the License is distributed on an "AS IS" BASIS,
~ WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
~ See the License for the specific language governing permissions and
~ limitations under the License.
-->
<resources xmlns:tools="http://schemas.android.com/tools">
<plurals name="offline_notification_count">
<item quantity="one">You have 1 offline product waiting to be contributed to Open Food Facts</item>
<item quantity="other">You have %d offline products waiting to be contributed to Open Food Facts</item>
</plurals>
<plurals name="seconds">
<item quantity="one">%d second ago</item>
<item quantity="other">%d seconds ago</item>
</plurals>
<plurals name="minutes">
<item quantity="one">%d minute ago</item>
<item quantity="other">%d minutes ago</item>
</plurals>
<plurals name="hours">
<item quantity="one">%d hour ago</item>
<item quantity="other">%d hours ago</item>
</plurals>
<plurals name="days">
<item quantity="one">%d day ago</item>
<item quantity="other">%d days ago</item>
</plurals>
<string name="action_about">जानकारी</string>
<string name="action_discover">तलाश करें</string>
<string name="contribute">योगदान कैसे करें</string>
<string name="action_find">ढूंढें</string>
<string name="product_incomplete_message">यह उत्पाद पूर्ण नहीं है। नतीजतन, हम एलर्जी की जांच नहीं कर सके</string>
<string name="product_allergen_prompt">This product contains:</string>
<string name="no_offline_data">आप सब कर चुके हैं</string>
<string name="first_offline">ऑफ़लाइन ऐड किये गए उत्पाद यहां दिखाई देंगे</string>
<!-- Image Upload -->
<!-- Nav Drawer Product Menu Items -->
<string-array name="nav_drawer_items_product">
<item>संग्रह</item>
<item>सामग्री</item>
<item>पोषण</item>
<item>पोषण प्रबंधन/सूची</item>
<item>वातावरण</item>
</string-array>
<!-- New Nav Drawer Product Menu Items -->
<string-array name="nav_drawer_new_items_product">
<item>Product Photos</item>
<item>Ingredient Analysis</item>
</string-array>
<!-- Text units -->
<!-- weight units -->
<!-- nutrition energy units -->
<!-- volume units -->
<!-- nutrition weight units -->
<!-- nutrition weight units -->
<!-- all weight units -->
<string-array name="weight_all_units">
<item>ग्राम</item>
<item>मिलीग्राम</item>
<item>माइक्रोग्राम</item>
<item>% DV</item>
<item>IU</item>
</string-array>
<!-- nutrition comparison units -->
<!-- nutrients list
do not change the order of the array as the items are paired with the String array
PARAMS_OTHER_NUTRIENTS present in AddProductNutritionFactsFragment-->
<string-array name="nutrients_array">
<item>Alpha-linolenic acid / ALA (18:3 n-3)</item>
<item>Arachidic acid (20:0)</item>
<item>Arachidonic acid / AA / ARA (20:4 n-6)</item>
<item>Behenic acid (22:0)</item>
<item>बिकारबोनिट</item>
<item>बायोटिन</item>
<item>Butyric acid (4:0)</item>
<item>कैफीन</item>
<item>केल्शियम</item>
<item>Capric acid (10:0)</item>
<item>Caproic acid (6:0)</item>
<item>Caprylic acid (8:0)</item>
<item>कैसिइन</item>
<item>Cerotic acid (26:0)</item>
<item>क्लोराइड</item>
<item>कोलेस्ट्रॉल</item>
<item>क्रोमियम</item>
<item>तांबा</item>
<item>Dihomo-gamma-linolenic acid / DGLA (20:3 n-6)</item>
<item>Docosahexaenoic acid / DHA (22:6 n-3)</item>
<item>Eicosapentaenoic acid / EPA (20:5 n-3)</item>
<item>Elaidic acid (18:1 n-9)</item>
<item>Erucic acid (22:1 n-9)</item>
<item>फ्लोराइड</item>
<item>फ्रुक्टोज</item>
<item>Gamma-linolenic acid / GLA (18:3 n-6)</item>
<item>शर्करा</item>
<item>Gondoic acid (20:1 n-9)</item>
<item>आयोडीन</item>
<item>लोहा</item>
<item>लैक्टोज</item>
<item>Lauric acid (12:0)</item>
<item>Lignoceric acid (24:0)</item>
<item>Linoleic acid / LA (18:2 n-6)</item>
<item>मैग्नीशियम</item>
<item>माल्टोडेक्सट्रिन्स</item>
<item>माल्टोज़</item>
<item>मैंगनीज़</item>
<item>Mead acid (20:3 n-9)</item>
<item>Melissic acid (30:0)</item>
<item>मॉलीब्डेनम</item>
<item>एक-संतृप्त वसा</item>
<item>Montanic acid (28:0)</item>
<item>Myristic acid (14:0)</item>
<item>Nervonic acid (24:1 n-9)</item>
<item>न्यूक्लियोटाइड</item>
<item>Oleic acid (18:1 n-9)</item>
<item>Omega 3 fatty acids</item>
<item>Omega 6 fatty acids</item>
<item>Omega 9 fatty acids</item>
<item>Palmitic acid (16:0)</item>
<item>Pantothenic acid / Pantothenate (Vitamin B5)</item>
<item>पीएच</item>
<item>फॉस्फोरस</item>
<item>Sugar alcohols (Polyols)</item>
<item>बहुसंतृप्त वसा</item>
<item>पोटेशियम</item>
<item>सेलेनियम</item>
<item>Serum proteins</item>
<item>सिलिका</item>
<item>Starch</item>
<item>Stearic acid (18:0)</item>
<item>सुक्रोज</item>
<item>टौरीने</item>
<item>ट्रांस वसा</item>
<item>विटामिन ए</item>
<item>Vitamin B1 (Thiamin)</item>
<item>Vitamin B12 (Cobalamin)</item>
<item>Vitamin B2 (Riboflavin)</item>
<item>Vitamin B3 / Vitamin PP (Niacin)</item>
<item>Vitamin B6 (Pyridoxin)</item>
<item>Vitamin B9 (Folic acid)</item>
<item>Vitamin C (Ascorbic acid)</item>
<item>विटामिन डी</item>
<item>विटामिन ई</item>
<item>विटामिन के</item>
<item>ज़िंक</item>
</string-array>
<!-- Text units -->
<!-- Content Description -->
<string name="search_button">ढूंढें</string>
<string name="alert_dialog_warning_title">चेतावनी</string>
<string name="alert_dialog_warning_msg_user">उपयोगकर्ता/ पासवर्ड बदल गया है, कृप्या जानकारी अपडेट करें</string>
<string name="toast_retrieving">लोड किया जा रहा है…</string>
<string name="txtBarcode">बारकोड:</string>
<string name="txtCategories">श्रेणी:</string>
<string name="txtLabels">लेबल, प्रमाणपत्र, पुरस्कार:</string>
<string name="txtEMB">Traceability codes:</string>
<string name="txtIngredients">सामग्री की सूची:</string>
<string name="Donate">दान करें</string>
<string name="Donation">ओपन फ़ूड फैक्ट्स को दान करने के लिए यहाँ क्लिक करें</string>
<string name="txtSubstances">पदार्थ या उत्पादों के कारण से एलर्जी या असहिष्णुता होना:</string>
<string name="txtAdditives">योगशील:</string>
<string name="txtTraces">निशानी:</string>
<string name="txtMayBeFromPalmOilProduct">सामग्री ताड़ के तेल से हो सकती है:</string>
<string name="txtPalmOilProduct">ताड़ के तेल से सामग्री:</string>
<string name="txtServingSize">सेवारत परिमाण:</string>
<string name="txtNutrientLevel100g">100 ग्राम के लिए पोषक तत्वों का स्तर</string>
<string name="txtNutrientLevel100ml">Nutrient levels for 100ml :</string>
<string name="txtNutritionLevelLow">कम मात्रा में</string>
<string name="txtNutritionLevelModerate">मध्यम मात्रा में</string>
<string name="txtNutritionLevelHigh">अधिक मात्रा में</string>
<string name="txtSave">सेव़</string>
<string name="txtSignIn">साइन इन करें</string>
<string name="txtToastSaved">उपयोगकर्ता की जानकारी सेव है</string>
<string name="txtSalt">नमक</string>
<string name="txtFat">वसा</string>
<string name="txtSugars">शुगर्स</string>
<string name="hintBarcode">बारकोड</string>
<string name="txtNutriScoreInfo">Learn more about the nutrient levels</string>
<string name="hintLogin">उपयोगकर्ता नाम</string>
<string name="hintPass">पासवर्ड</string>
<string name="errorLogin">लॉगिन या पासवर्ड एरर</string>
<string name="errorWeb">हम इंटरनेट का उपयोग करने में सक्षम नहीं थे| क्या आपके पास कनेक्टिविटी है?</string>
<string name="uploadLater">नेटवर्क उपलब्ध होने पर चित्र अपलोड किया जायेगा</string>
<string name="txtInfoLoginOk">उपयोगकर्ता की जानकारी सेव है</string>
<string name="txtInfoLoginNo">उपयोगकर्ता की जानकारी सेव नहीं `है</string>
<string name="txtDialogsContent">जिस उत्पाद को आप स्कैन कर चुके हैं वह अभी तक ओपन फूड फैक्ट्स में उपलब्ध नहीं है। कृपया इस उत्पाद के बारे में जानकारी प्राप्त करने में मदद करें।</string>
<string name="txtYes">हाँ</string>
<string name="txtNo">नहीं</string>
<string name="hintBrand">ब्रांड</string>
<string name="txtSendAll">सभी अपलोड कर दो</string>
<string name="txtPictureNeededDialogNo">रद्द करें</string>
<string name="toastSending">अपलोड हो रहा है</string>
<string name="txtLoading">लोड किया जा रहा है…</string>
<string name="txtDialogsContentDelete">क्या आप वाकई इस प्रारूप को हटाना चाहते हैं?</string>
<string name="txtDialogsContentInfoSave">उत्पाद स्थानीय रूप से सेव है, आप ऑफलाइन संपादन मेनू में संपादित कर सकते हैं</string>
<string name="productSavedToast">Product saved, thanks!</string>
<string name="txtBarcodeRequire">बारकोड अपेक्षित हैं</string>
<string name="txtOk">ठीक है</string>
<string name="txtBarcodeNotValid">बारकोड मान्य नहीं</string>
<string name="msg_share">Open Food Facts पर इस उत्पाद को देखो:</string>
<string name="create_account">साइन अप करें</string>
<string name="home_drawer">मुख्य पृष्ठ</string>
<string name="install">Install</string>
<string name="open_beauty_drawer">Open Beauty Facts</string>
<string name="open_food_drawer">Open Food Facts</string>
<string name="search_drawer">ढूंढें</string>
<string name="search_by_barcode_drawer">बारकोड के आधार पर ढूंढें</string>
<string name="category_drawer">श्रेणियाँ</string>
<string name="search_by_category">श्रेणी</string>
<string name="scan_search">स्कैन</string>
<string name="user_drawer">उपयोगकर्ता</string>
<string name="offline_edit_drawer">ऑफ़लाइन संपादित</string>
<string name="txt_anonymous">अज्ञात</string>
<string name="delete_txt">नष्ट करे</string>
<string name="sign_in_drawer">साइन इन करें</string>
<string name="alert_drawer">एलर्जी की चेतावनी</string>
<string name="your_lists">Your lists</string>
<string name="title_dialog_alert">अलर्ट जोड़ें</string>
<string name="select_camera">Select a Camera</string>
<string name="ok_button">ठीक है</string>
<string name="cancel_button">Cancel</string>
<string name="login_button">उपयोगकर्ता नाम</string>
<string name="create_account_button">Create Account</string>
<string name="warning_alert_data" tools:ignore="StringFormatInvalid">Your allergies info will never be sent to the Open Food Facts server. Also, be careful that product data may not be accurate, and detection might not be 100% accurate, so please double-check by yourself!</string>
<string name="title_info_dialog">यह कैसे काम करता है?</string>
<string name="text_offline_info_dialog">जब आप ऑफ़लाइन हो तब भी आप उत्पादों को स्कैन कर सकते हैं, ताकि उन्हें आप बाद में जोड़ सके। आपके सभी स्कैन कि हुए उत्पाद यहाँ दिखाई देगी, और आप नीचे दिए गए बटन का उपयोग कर उन्हें भेजने के लिए सक्षम हो जाएगा।</string>
<string name="permission_camera">कैमरा अनुमति की जरूरत है बारकोड को स्कैन और तस्वीरें लेने के लिए</string>
<string name="no_camera_dialog_title">Camera not found</string>
<string name="no_camera_dialog_content">Your device does not have a camera. You won\'t be able to scan any barcode.</string>
<string name="permission_title">अनुमति जानकारी</string>
<string name="permission_denied">सेटिंग्स में अनुमति सक्रिय करें</string>
<string name="info_download_data_connection">आप इस कार्यक्षमता काम पाने के लिए कुछ डेटा डाउनलोड करने के लिए की जरूरत है। इंटरनेट कनेक्शन सक्रिय करें।</string>
<!-- Strings related to login -->
<string name="error_invalid_password">पासवर्ड कम से कम 6 अक्षर लंबा होने की जरूरत है।</string>
<string name="user">उपयोगकर्ता</string>
<string name="error_field_required">यह क्षेत्र आवश्यक है</string>
<string name="action_preferences">प्राथमिकताएँ</string>
<string name="prefs_scan_startup">स्टार्टअप पर स्कैन करें</string>
<string name="prefs_scan_startup_summary">इस विकल्प का चयन किया है, तो जब एप्लिकेशन को खोला जाता है स्कैन शुरू कर देंगे।</string>
<string name="prefs_power_mode">पावर मोड स्कैन</string>
<string name="prefs_power_mode_summary">सुपरमार्केट में उत्पादों को झटपट स्कैन करें और देखें कि कौनसे उत्पाद डेटाबेस में उपलब्ध नहीं हैं</string>
<string name="prefs_language">भाषा</string>
<string name="prefs_language_summary">भाषा बदले</string>
<!-- Strings related to feedback dialog -->
<string name="user_enjoying_app">Are you enjoying your experience with Open Food Facts?</string>
<string name="user_ask_show_feedback_form">We\'re sorry to hear that! Could you tell us what happened?</string>
<string name="user_ask_rate_app">Great! Let others know what you think of this app!</string>
<string name="rate_app">Rate app</string>
<string name="no_thx">No thanks</string>
<string name="feedback_form_url">https://forms.gle/SWnx6h6QSUHcmFMy7</string>
<!-- Scan history -->
<string name="scan_history_drawer">वृत्तांत</string>
<string name="clear_history">इतिहास मिटा दें</string>
<string name="title_clear_history_dialog">खाली करें</string>
<string name="text_clear_history_dialog">Do you want to clear all your scan history?</string>
<string name="export_csv_history">CSV निर्यात करें</string>
<string name="txt_exporting_history">इतिहास निर्यात …</string>
<string name="txt_history_exported">इतिहास का निर्यात</string>
<string name="permision_write_external_storage">लिखें बाह्य भंडारण की अनुमति इतिहास को निर्यात करने की जरूरत है।</string>
<string name="no_brand">No Brand</string>
<string name="no_title">शीर्षक नहीं है</string>
<!-- Nutrition info tab -->
<string name="display_fact">Nutrition facts table for %1$s of this product</string>
<string name="check_facts">जाँचें</string>
<string name="nutrition">पोषण</string>
<string name="calculated_nutrition">परिकलित पोषण</string>
<string name="please_enter_weight">Please type the product weight</string>
<string name="enter_weight">Type the product weight</string>
<string name="calculate_nutrition_facts">Calculate Nutrition Facts</string>
<string name="nutrition_carbohydrate">कार्बोहाइड्रेट</string>
<string name="nutrition_cholesterol">कोलेस्ट्रॉल</string>
<string name="nutrition_energy">ऊर्जा</string>
<string name="nutrition_energy_kcal">Energy in calories (kcal)</string>
<string name="nutrition_energy_kj">Energy in joules (kJ)</string>
<string name="nutrition_energy_short_name">ऊर्जा</string>
<string name="nutrition_fat">वसा</string>
<string name="nutrition_fiber">फाइबर आहार</string>
<string name="nutrition_header">पोषण तथ्य</string>
<string name="nutrition_proteins">प्रोटीन</string>
<string name="nutrition_salt">नमक</string>
<string name="nutrition_satured_fat">संतृप्त वसा</string>
<string name="nutrition_sodium">सोडियम</string>
<string name="nutrition_sugars">शुगर्स</string>
<string name="nutrition_trans_fat">ट्रांस वसा</string>
<string name="nutriment_serving_size">प्रति हिस्से:</string>
<string name="action_manage_account">खाते का प्रबंधन</string>
<string name="action_contributes">आपका योगदान</string>
<string name="choose_energy_unit">Choose your preferred energy units for the nutrition table</string>
<string name="energy_pref_title">Preferred energy unit</string>
<string name="choose_volume_unit">Choose your preferred volume units for the nutrition table</string>
<string name="volume_pref_title">Preferred volume unit</string>
<string name="logout_drawer">लॉग आउट करें</string>
<string name="contribution_without_account">आप साइन-इन कर सकते हैं या अपने योगदान रखने के लिए एक खाता बनाने\nप्रसंग:</string>
<string name="textCarbonFootprint">कार्बन पदचिह्न / CO2 उत्सर्जन: </string>
<string name="action_search">ढूंढें</string>
<string name="search_hint">उत्पाद खोजें</string>
<string name="number_of_results">परिणामों की संख्या: </string>
<string name="number_of_products">उत्पादों की संख्या: </string>
<string name="txt_no_matching__category_products">Oh no, we don't have any product in this category, can you add one?</string>
<string name="txt_no_matching_allergen_products">Oh no, we don't have any products with this allergen, can you add one?</string>
<string name="txt_no_matching_label_products">Oh no, we don't have any products with this label, can you add one?</string>
<string name="txt_no_matching_additive_products">Oh no, we don't have any products with this additive, can you add one?</string>
<string name="txt_no_matching_country_products">Oh no, we don't have any products from this country, can you add one?</string>
<string name="txt_no_matching_store_products">Oh no, we don't have any products from this store, can you add one?</string>
<string name="txt_no_matching_packaging_products">Oh no, we don't have any products with this packaging, can you add one?</string>
<string name="txt_no_matching_brand_products">Oh no, we don't have any products with this brand, can you add one?</string>
<string name="txt_no_matching_incomplete_products">We don't have any incomplete products! Add a new product?</string>
<string name="txt_no_matching_contributor_products">Oh no, we don't have any products from this contributor, can you add one?</string>
<string name="txt_no_matching_products">हमें कोई भी मिलते-जुलते उत्पाद नहीं मिले हैं। \n आप यह कोशिश कर सकते हैं:</string>
<string name="txt_broaden_search">- अपनी खोज को विस्तृत करना \n- स्कैन करें और इसे ओपन फ़ूड फैक्ट्स में ऐड करें</string>
<string name="something_went_wrong">क्षमा करें, कुछ गलत हो गया…</string>
<string name="txt_try_again">फिर कोशिश करे</string>
<string name="nutrition_lactose">लैक्टोज</string>
<string name="nutrition_casein">कैसिइन</string>
<string name="nutrition_maltodextrins">माल्टोडेक्सट्रिन्स</string>
<string name="nutrition_serum_proteins">सीरम प्रोटीन</string>
<string name="nutrition_nucleotides">न्यूक्लियोटाइड</string>
<string name="for_100g">१०० g के लिए</string>
<string name="for_100ml">For 100 ml</string>
<string name="serving">प्रत्येक हिस्सा</string>
<string name="nutrition_monounsaturatedFat">एक-संतृप्त वसा</string>
<string name="nutrition_polyunsaturatedFat">बहुसंतृप्त वसा</string>
<string name="nutrition_omega3">ओमेगा 3</string>
<string name="nutrition_omega6">ओमेगा 6</string>
<string name="nutrition_omega9">ओमेगा 9</string>
<string name="nutrition_sucrose">सुक्रोज</string>
<string name="nutrition_glucose">शर्करा</string>
<string name="nutrition_fructose">फ्रुक्टोज</string>
<string name="nutrition_maltose">माल्टोज़</string>
<string name="nutrition_alcohol">अल्कोहॉल</string>
<string name="nutrition_vitamins">विटामिन</string>
<string name="vitamin_a">विटामिन ए</string>
<string name="vitamin_d">विटामिन डी</string>
<string name="vitamin_e">विटामिन ई</string>
<string name="vitamin_k">विटामिन के</string>
<string name="vitamin_c">विटामिन सी</string>
<string name="vitamin_b1">विटामिन बी 1</string>
<string name="vitamin_b2">विटामिन बी 2</string>
<string name="vitamin_pp">विटामिन पीपी</string>
<string name="vitamin_b6">विटामिन बी 6</string>
<string name="vitamin_b9">विटामिन बी 9</string>
<string name="vitamin_b12">विटामिन बी 12</string>
<string name="biotin">बायोटिन</string>
<string name="pantothenic_acid">पैंटोथैनिक एसिड</string>
<string name="nutrition_minerals">खनिज</string>
<string name="silica">सिलिका</string>
<string name="bicarbonate">बिकारबोनिट</string>
<string name="potassium">पोटेशियम</string>
<string name="chloride">क्लोराइड</string>
<string name="calcium">केल्शियम</string>
<string name="phosphorus">फॉस्फोरस</string>
<string name="iron">लोहा</string>
<string name="magnesium">मैग्नीशियम</string>
<string name="zinc">ज़िंक</string>
<string name="copper">तांबा</string>
<string name="manganese">मैंगनीज़</string>
<string name="fluoride">फ्लोराइड</string>
<string name="selenium">सेलेनियम</string>
<string name="chromium">क्रोमियम</string>
<string name="molybdenum">मॉलीब्डेनम</string>
<string name="iodine">आयोडीन</string>
<string name="caffeine">कैफीन</string>
<string name="taurine">टौरीने</string>
<string name="ph">पीएच</string>
<string name="fruits_vegetables_nuts">फल, सब्जियां, दाने</string>
<string name="collagen_meat_protein_ratio">कोलेजन मांस प्रोटीन अनुपात</string>
<string name="cocoa">कोकोआ</string>
<string name="chlorophyl">क्लोरोफिल</string>
<string name="take_picture">Take a photo</string>
<string name="choose_picture_from_gallery">From gallery</string>
<string name="take_picture_ingredients">सामग्रियों की एक छायाचित्र लीजिए</string>
<string name="take_picture_front">Take a photo of the front</string>
<string name="take_picture_nutrition_table">पोषण तालिका की एक तस्वीर लें</string>
<string name="take_more_pictures">और तस्वीरें लें</string>
<string name="txtSaturatedFat">संतृप्त वसा</string>
<string name="nutriscore_info_msg">Learn more about the Nutri-Score</string>
<string name="nova_info_msg">Learn more about NOVA groups</string>
<string name="nova_grp1_msg">Group 1 - Unprocessed or minimally processed foods</string>
<string name="nova_grp2_msg">Group 2 - Processed culinary ingredients</string>
<string name="nova_grp3_msg">Group 3 - Processed foods</string>
<string name="nova_grp4_msg">Group 4 - Ultra-processed food and drink products</string>
<!-- String related to App Shortcuts -->
<string name="disabled_message">शॉर्टकट उपलब्ध नहीं है</string>
<string name="scan_shortcut_short_label">स्कैन</string>
<string name="scan_shortcut_long_label">उत्पाद स्कैन करें</string>
<string name="barcode_shortcut_short_label">ढूंढें</string>
<string name="barcode_shortcut_long_label">बारकोड द्वारा खोज</string>
<string name="history_shortcut_short_label">वृत्तांत</string>
<string name="history_shortcut_long_label">वृत्तांत</string>
<string name="contributions_shortcut_short_label">योगदान</string>
<string name="contributions_shortcut_long_label">My contributions</string>
<string name="forgot_password">पासवर्ड भूल गए</string>
<string name="next">अगला</string>
<string name="skip">छोड़ें</string>
<string name="start">समझ गया</string>
<string name="contact">संपर्क</string>
<string name="contact_summary">टीम को संपर्क करें</string>
<string name="email_not_found">कोई ईमेल एप्लीकेशन नहीं मिली</string>
<string name="airplane_mode_active_dialog_title">एयरप्लेन मोड चालू</string>
<string name="airplane_mode_active_dialog_message">एयरप्लेन मोड चालू, जारी रहने के लिए कृपया इसको बंद करें</string>
<string name="airplane_mode_active_dialog_negative">ख़ारिज करें</string>
<string name="airplane_mode_active_dialog_positive">सेटिंग्स</string>
<string name="device_offline_dialog_title">नेटवर्क उपलब्ध नहीं है</string>
<string name="device_offline_dialog_message">ऐसा करने के लिए नेटवर्क कनेक्शन की आवश्यकता है, क्या आप इसे सेटिंग्स में बदलना चाहेंगे?</string>
<string name="device_offline_dialog_negative">ख़ारिज करें</string>
<string name="device_offline_dialog_positive">सेटिंग्स</string>
<string name="scan_first_product">पहला उत्पाद स्कैन करें</string>
<string name="preference_delete_search_history">Successfully deleted history</string>
<string name="preference_enable_mobile_data_title">मोबाइल डेटा पर अपलोड करें</string>
<string name="preference_enable_mobile_data_summary_on">मोबाइल डेटा और WiFi का उपयोग करके नए उत्पाद अपलोड करें</string>
<string name="preference_enable_mobile_data_summary_off">केवल WiFi का उपयोग कर नए उत्पाद अपलोड करें</string>
<string name="device_on_mobile_data_warning_title">जानकारी</string>
<string name="device_on_mobile_data_warning_message">You have disabled upload on mobile data , please connect to WiFi or change this in preferences</string>
<string name="device_on_mobile_data_warning_positive">प्राथमिकताएँ</string>
<string name="device_on_mobile_data_warning_negative">ऐसे ही अपलोड करें</string>
<string name="changes_saved">Changes saved successfully</string>
<string name="preference_choose_language_dialog_title">भाषा चुनें</string>
<string name="FAQ">ऍफ़ ऐ क्यू</string>
<string name="FrequentlyAskedQuestions">अक्सर पूछे जाने वाले सवाल</string>
<string name="translate_help_title">ओपन फ़ूड फैक्ट्स का अनुवाद करने में मदद करें</string>
<string name="translate_help_summary">ओपन फ़ूड फैक्ट्स डेटाबेस को अपनी भाषा में लाएं - कोई तकनीकी की आवश्यकता नहीं है</string>
<string name="please_check_your_connection">कृपया अपना कनेक्शन जाँचें</string>
<string name="advanced_search_title">उन्नत खोज और चार्ट्स</string>
<string name="compare_products">Compare products</string>
<string name="connectivity_check">कृपया अपना इंटरनेट कनेक्शन जाँचें।</string>
<string name="dismiss">ख़ारिज करें</string>
<string name="Terms">शर्तें</string>
<string name="TermsOfUse">Terms and Privacy</string>
<string name="donation_url">https://donate.openfoodfacts.org</string>
<!--Just replace "world" by "fr", "de"… translate the "terms-of-use" with the hyphens -->
<string name="terms_url">https://world-hi.openfoodfacts.org/terms-of-use</string>
<!--Just replace "world" by "world-fr", "world-de"… do not translate "faq" -->
<string name="faq_url">https://world.openfoodfacts.org/faq</string>
<!--Just replace "world" by "world-fr", "world-de"… do not translate "hunger-game" -->
<string name="hunger_game_url">https://world.openfoodfacts.org/hunger-game</string>
<!--Just replace "world" by "world-fr", "world-de"… do not translate the end of the url -->
<!--Just replace "world" by "fr", "de"… do not translate the "nutriscore"-->
<string name="nutriscore_uri">https://world.openfoodfacts.org/nutriscore</string>
<!--Just replace "world" by "fr", "de"… do not translate the "nova"-->
<string name="url_nova_groups">https://world.openfoodfacts.org/nova</string>
<!--Just replace "world" by "fr", "de"… translate the "nutrition-traffic-lights" with the hyphens -->
<string name="url_nutrient_values">https://world.openfoodfacts.org/nutrition-traffic-lights</string>
<string name="preference_header_general">सामान्य</string>
<string name="preference_header_network">नेटवर्क</string>
<string name="preference_header_contributing">योगदान</string>
<string name="scan_first_string">आपके द्वारा देखे गए उत्पाद यहाँ सूचीबद्ध हैं \n यह केवल आपके लिए है और फ़ोन में संग्रहीत है</string>
<string name="log_in">उपयोगकर्ता नाम</string>
<string name="login_true">पहले से ही लॉगिन हैं!</string>
<string name="confirm_logout">लॉगआउट करें?</string>
<string name="logout_dialog_content">लॉग आउट करने के लिए निश्चित हैं?</string>
<string name="dialog_cancel">Cancel</string>
<string name="dialog_create">Create</string>
<string name="choose_resolution">संकल्प चुनें</string>
<string name="pref_resolution_summary">Pick a resolution for image upload</string>
<string name="image_upload_resolution">Image upload resolution</string>
<string name="wikidata_information">अधिक जानकारी</string>
<string name="wikidata_unavailable">There is no Wikipedia page in your language yet.</string>
<string name="Wikipedia">विकिपीडिया पर देखें</string>
<string name="Browse_Products">Browse Products</string>
<string name="brand_string">ब्रांड</string>
<string name="additive_string">योज्य</string>
<string name="allergen_string">Allergen</string>
<string name="search_string">ढूंढें</string>
<string name="country_string">जिन देशों में बिकता है</string>
<string name="store_subtitle">दुकान</string>
<string name="packaging_subtitle">पैकेजिंग</string>
<string name="packaging_string">पैकेजिंग</string>
<string name="label_string">लेबल</string>
<string name="category_string">श्रेणी</string>
<string name="connection">कनेक्टेड</string>
<string name="contributor_string">सहयोगी</string>
<string name="Rate_Us">अपना समर्थन दिखाएं</string>
<string name="Rate_Us_title">हमारी ऐप रेट करें</string>
<string name="app_disabled_text">Application is disabled</string>
<!--Vitamin, Amino Acid, mineral and other nutrition Tags TextView-->
<string name="vitamin_tags_text">विटामिन:</string>
<string name="amino_acid_tags_text">Amino Acids:</string>
<string name="mineral_tags_text">मिनरल:</string>
<string name="other_tags_text">Other Substances:</string>
<!--shared element transition key-->
<string name="disable_image_loading">Disable image loading when battery is low</string>
<string name="DisableLoadTitle">Disable image loading</string>
<string name="by_title">शीर्षक</string>
<string name="by_brand">ब्रांड</string>
<string name="by_barcode">बारकोड</string>
<string name="by_time">हाल ही में देखे गए</string>
<string name="sort">क्रम</string>
<string name="by_nutrition_grade">Nutrition Grade</string>
<string name="sort_by">Sort by</string>
<!-- strings related to share images -->
<string name="sorry_msg">Pictures without barcode cannot yet be added with the app. Please try the mobile web version.</string>
<string name="no_barcode">We have not detected any barcode</string>
<string name="enter_barcode">Please type the barcode below</string>
<string name="format_error">Wrong Barcode/QR format</string>
<string name="code_detected">We have detected the barcode</string>
<string name="do_you_want_to">Do you want to add photos to this barcode?</string>
<string name="bottomNav_history">वृत्तांत</string>
<string name="products_added">Products you added or edited</string>
<string name="products_incomplete">Products you added that need to be completed</string>
<string name="product_pictures_contributed">Products you took pictures for</string>
<string name="picture_contributed_incomplete">Products you took pictures for that need to be completed</string>
<string name="product_info_added">Products you added information for</string>
<string name="product_info_tocomplete">Products you added information for that need to be completed</string>
<string name="show_by">Show</string>
<string name="notify_title">Export complete</string>
<string name="notify_content">Click to view exported products</string>
<string name="notification_channel_name">Export Notifications</string>
<string name="notify_channel_description">Notification to show exported file\'s location</string>
<string name="products_to_be_completed">Products to be completed</string>
<string name="offline_notification_title">Offline contributions you need to send</string>
<string name="creator_history">Product added on %1$s at %2$s CET by %3$s</string>
<string name="last_editor_history">Last edit of product page on %1$s at %2$s CEST by %3$s </string>
<string name="other_editors">Product page also edited by </string>
<string name="contribution_tab">योगदान</string>
<string name="contribution_tab_summary">Show user contributions tab with products</string>
<string name="contribution_tab_title">Show contribution tab</string>
<string name="txtDialogsTitle">जानकारी</string>
<string name="preference_show_all_product_photos_title">उत्पाद की सभी तस्वीरें दिखाएं</string>
<string name="preference_show_all_product_photos_summary">उत्पाद की सभी तस्वीरें एक अलग टैब में दिखाएं</string>
<string name="offline_product_addition_title">नया उत्पाद ऐड करें</string>
<string name="productNameNull">नाम ज्ञात नहीं है</string>
<string name="productAdditivesUnknown"></string>
<string name="productAdditivesTemplate">%1$d additives</string>
<plurals name="productAdditives">
<item quantity="one">%d additive</item>
<item quantity="other">%d additives</item>
</plurals>
<string name="product_not_complete">This product is not yet complete. \nCan you add some details?</string>
<string name="addProductOffline">You seem offline. Add the product %s using the offline mode ?</string>
<string name="product_not_found">The product %1$s you scanned does not exist yet in Open Food Facts. Can you add it?</string>
<string name="scan_sound">स्कैन की आवाज़</string>
<string name="trouble_scanning">स्कैन करने में समस्या?</string>
<string name="barcode_hint">स्कैन करने में समस्या? बारकोड लिखें</string>
<string name="switch_camera">स्विच कैमरा</string>
<string name="autofocus">ऑटोफोकस</string>
<string name="initial_loading">प्रारंभिक लोडिंग</string>
<string name="preferences">प्राथमिकताएँ</string>
<string name="OFFlogo">Open Food Facts logo</string>
<string name="loading_product">Loading product %s</string>
<string name="image_not_defined_for_language">No image for selected language. You can add a new one</string>
<string name="error_adding_ingredients">Something went wrong while trying to add product ingredients</string>
<string name="unable_to_extract_ingredients">Unable to extract ingredients from image</string>
<string name="error_adding_nutrition_facts">Something went wrong while trying to add product nutrition facts</string>
<string name="error_in_carbohydrate_value">Carbohydrates can\'t be less than the sum of sugar and starch</string>
<string name="choose_nutrient">Choose a nutrient</string>
<string name="error_adding_product_details">Something went wrong while trying to add product details</string>
<string name="hint_product_URL">Link of the official page of the product</string>
<string name="product_language">उत्पाद भाषा: </string>
<string name="hint_emb_codes">उदाहरण: FSSL 10013011001409</string>
<string name="official_website">Official website</string>
<string name="scan_QR_code">Scan QR Code for product website</string>
<string name="add_at_least_one_picture">Please add at least one picture of this product before proceeding</string>
<string name="image_uploaded_successfully">Image uploaded successfully</string>
<string name="error_adding_product_photos">Something went wrong while trying to add product photos</string>
<string name="ingredients_overwrite">सामग्री ओवरराइट</string>
<string name="choose_mine">मेरा चुनें</string>
<string name="keep_previous_version">पिछला संस्करण रखें</string>
<string name="product_name_overwrite">उत्पाद का नाम ओवरराइट करें</string>
<string name="yours">आपका: </string>
<string name="currently_on">Currently on %s: </string>
<string name="quantity_overwrite">मात्रा ओवरराइट करें</string>
<string name="link_overwrite">लिंक ओवरराइट करें</string>
<string name="save_product">क्या आप परिवर्तनों को सहेजना चाहते हैं?</string>
<string name="error_adding_product">Something went wrong when adding the product</string>
<string name="photos">तस्वीरें</string>
<string name="please_wait">प्रतीक्षा करें</string>
<string name="product_uploaded_successfully">उत्पाद सफलतापूर्वक अपलोड किया गया</string>
<string name="no_internet_connection">You appear to have no internet connection, images will be uploaded when network is available</string>
<string name="no_internet_unable_to_extract_ingredients">No internet connection. Unable to extract ingredients</string>
<string name="percent">%1$d%%</string>
<string name="overview">संग्रह</string>
<string name="ingredients">सामग्री</string>
<string name="nutrition_facts">पोषण तथ्य</string>
<string name="your_version">आपका</string>
<string name="ingredients_picture">सामग्री चित्र</string>
<string name="take_ingredients_picture_to_extract">(Please take a picture of the ingredients to automatically extract them)</string>
<string name="ingredients_list">सामग्री की सूची</string>
<string name="ingredients_hint">Keep the order, indicate the % when specified, separate with a comma or use -, use () for ingredients of an ingredient, surround allergens with _</string>
<string name="extract_ingredients">Extract ingredients</string>
<string name="extracting_ingredients">Extracting ingredients</string>
<string name="looksGood">सही है</string>
<string name="skip_ingredients">बहुत बुरा \n सामग्री छोड़ें</string>
<string name="traces">अवशेष</string>
<string name="traces_hint">Indicate ingredients from mentions like \'May contain traces of\', \'Made in a factory that also uses\', etc. \nExamples: Milk, Gluten, Nuts</string>
<string name="nutrition_facts_not_specified">Nutrition facts are not specified on the product.</string>
<string name="nutrition_facts_picture">Nutrition Facts photo</string>
<string name="front_short_picture">Front</string>
<string name="other_picture">Other Picture</string>
<string name="for_100g_100ml">100 ग्राम / 100 मिलीलीटर के लिए</string>
<string name="per_serving">प्रत्येक हिस्सा</string>
<string name="serving_size">सेवारत माप</string>
<string name="alcohol_unit" formatted="false">% वॉल्यूम / °</string>
<string name="add_a_nutrient">Add a nutrient</string>
<string name="add_this_product">इस उत्पाद को ऐड करें</string>
<string name="product_picture">उत्पाद की तस्वीर</string>
<string name="one_photo_compulsory">(नया उत्पाद ऐड करने के लिए एक तस्वीर अनिवार्य है)</string>
<string name="product_details">उत्पाद विवरण</string>
<string name="product_name">Product name</string>
<string name="quantity">मात्रा</string>
<string name="labels">लेबल</string>
<string name="period_of_time_after_opening">Period of time after opening</string>
<string name="manufacturing_details">विनिर्माण विवरण</string>
<string name="origin_of_ingredients">सामग्री की उत्पत्ति</string>
<string name="manufacturing_place">विनिर्माण स्थान</string>
<string name="emb_code">Traceability Code</string>
<string name="purchasing_details">खरीद विवरण</string>
<string name="country_where_purchased">जिस देश में खरीदा</string>
<string name="stores">दुकाने</string>
<string name="additional_pictures">अतिरिक्त तस्वीरें</string>
<string name="hint_product_pictures">(उत्पाद की और तस्वीरें ऐड करें। उदाहरण: रीसाइक्लिंग निर्देश, लेबल इत्यादि)</string>
<string name="uploading_image">Sending the photo</string>
<string name="uploading_nutrition_image">पोषण तथ्य की तस्वीर अपलोड हो रही है</string>
<string name="uploading_ingredients_image">सामग्री की तस्वीर अपलोड हो रही है</string>
<string name="uploading_front_image">उत्पाद की तस्वीर अपलोड हो रही है</string>
<string name="edit_front_image">Edit the front photo</string>
<string name="edit_nutrition_image">Edit the nutrition photo</string>
<string name="edit_ingredients_image">Edit the ingredients photo</string>
<string name="edit_other_image">Edit the photo</string>
<string name="sign_in_to_edit">किसी उत्पाद को संपादित करने के लिए कृपया साइन इन करें।</string>
<string name="sign_in_to_answer">Please sign in or register to answer an insight.</string>
<string name="hint_quantity">उदाहरण: 2 l, 250 g, 1 kg, 25 cl, 6 fl oz, 1 pound</string>
<string name="edit_product_title">Edit a product</string>
<string name="nutrient_already_added">%s is already present in the nutrition facts</string>
<string name="save_edits">संपादन सहेजें</string>
<string name="contribution_by">Contribution by</string>
<string name="title_activity_product">शीर्षक</string>
<string name="description_activity_product">description</string>
<string name="intent_filter_share_label">Add to Open Food Facts</string>
<!-- Used for additive exposure evaluation table -->
<string name="overexposure_high">High risk of over exposure</string>
<string name="overexposure_moderate">Moderate risk of over exposure</string>
<string name="infants">Infants</string>
<string name="infants_age">< 1</string>
<string name="toddlers">Toddlers</string>
<string name="toddlers_age">1 to 2</string>
<string name="children">Children</string>
<string name="children_age">3 to 9</string>
<string name="adolescents">Adolescents</string>
<string name="adolescents_age">10 to 17</string>
<string name="adults">Adults</string>
<string name="adults_age">18 to 64</string>
<string name="elderly">Elderly</string>
<string name="elderly_age">65+</string>
<string name="most_people">Most people\n(over 50%)</string>
<string name="some_people">Some people\n(over 5%)</string>
<string name="safety_and_exposure_assessment">Safety and exposure assessment</string>
<string name="risk_of_exposure">Risk of exposure</string>
<string name="efsa_warning_high_risk">The European Food Safety Authority (EFSA) has determined that some population groups have a high risk of consuming too much %1$s.</string>
<string name="legend_noael">above dose with observed adverse effect (NOAEL)</string>
<string name="legend_adi">above Acceptable Daily Intake (ADI)</string>
<string name="max_energy_val_msg">This value is above 2000 kcal per 100g. Can you correct it ?</string>
<string name="max_nutrient_val_msg">Value entered is more than given per serving size</string>
<string name="update_image">Send an Updated Image</string>
<!-- Used for additive exposure evaluation table -->
<string name="extract_ingredients_prompt">Extract ingredients</string>
<string name="add_nutrient_prompt_text">Add nutrition facts to compute the Nutri-Score</string>
<string name="add_category_prompt_text">Add a category to compute the Nutri-Score</string>
<string name="add_nutrient_category_prompt_text">Add nutrition facts and a category to compute the Nutri-Score</string>
<string name="category_game_btn">Help categorize products to compute the Nutri-Score</string>
<string name="version_string">Open Food Facts Android App</string>
<string name="version">Version</string>
<string name="set_ingredient_img">Set as ingredients image</string>
<string name="set_img_nutrients">Set as nutrition image</string>
<string name="set_img_front">Set as front image</string>
<string name="report_img">Report Image</string>
<string name="detected_ingredients">Detected Ingredients</string>
<string name="txtConnectionError">We could not download certain data. This may be due to a connection error.</string>
<string name="set_image_name">Successfully set the image as</string>
<string name="fast_addition_mode_title">Fast Addition Mode</string>
<string name="fast_addition_mode">Enables fast addition of products with fewer attributes</string>
<string name="crop_new_image_title">Crop new images</string>
<string name="crop_new_image">Enables crop action on new images</string>
<string name="addtive_search">Search for a additive</string>
<string name="additives">योगशील</string>
<!--used for product_lists implementation-->
<string name="add_to_product_lists">Add to your lists</string>
<string name="txt_create_new_list">Create a new list</string>
<string name="txt_add_to_new_list">Add to a new list</string>
<string name="create_new_list_list_name">List Name</string>
<string name="txt_eaten_products">Eaten products</string>
<string name="txt_products_to_buy">Products to buy</string>
<string name="txt_exporting_your_listed_products">Exporting Your Products…</string>
<string name="txt_your_listed_products_exported">Your products have been exported</string>
<string name="txt_discard">रद्द करें</string>
<string name="txt_info_eaten_products">\"You\'ll be able to add product to this list by scanning products, dragging the product card up and tapping on \"I\'m eating this\".
You\'ll also be able to add products to this list by searching them and clicking on the + button in the top right of the product page.\"</string>
<string name="txt_info_your_listed_products">You\'ll be able to add products to this list by searching them and clicking on the + button in the top right of the product page.</string>
<string name="search_history_pref_title">Delete local search history</string>
<string name="search_history_pref_summary">Remove all the searches that you performed on this device</string>
<string name="search_history_pref_dialog_content">Delete local search history?</string>
<string name="settings_country_dialog_title">Select your country</string>
<string name="settings_country_summary">Change your selected country</string>
<string name="settings_country_title">देश</string>
<string name="txt_no_nutrition_data">This product doesn\'t have any nutrition information on the packaging. If this has changed, you can edit the product.</string>
<string name="image_prompt_language">Take a photo of the ingredients in</string>
<!-- Used for diets AddOn -->
<string name="empty"></string>
<string name="add_product">Add products</string>
<string name="add_another_product">Add another product</string>
<string name="add_nutrition_facts_for_nutriscore">Add nutrition facts to compute the Nutri-Score</string>
<string name="product_summary_action_compare_button_label">Compare</string>
<string name="product_summary_action_add_to_list_button_label">Add to list</string>
<string name="product_summary_action_edit_button_label">संपादित करें</string>
<string name="product_summary_action_share_button_label">शेयर</string>
<string name="product_question_prompt">Tap here to answer</string>
<string name="product_question_positive_response">हाँ</string>
<string name="product_question_negative_response">नहीं</string>
<string name="product_question_ambiguous_response">Not sure</string>
<string name="product_question_submit_message">Thank you for your contribution!</string>
<string name="error_nutrient_entry">The value should be numeric. Example: "1.5", "2,0"</string>
<string name="error_nutrient_serving_data">The serving size value must be a positive value</string>
<string name="error_in_nutrient_data">Some nutrient values are not valid. Please, fix above fields with error marker</string>
<string name="go_to_full_product_page">Go to product</string>
<string name="no_additive_data_available_for_this_product">No additive data available for this product</string>
<string name="no_nutrient_information_available_for_this_product">No nutrient information available for this product</string>
<string name="error_duplicate_listname">You already have a list with this name</string>
<string name="required_to_compute_nutriscore">(Required to calculate the Nutri-Score)</string>
<string name="scan_tooltip">You can slide up the card for more details</string>
<string name="add_an_allergen">Add an allergen</string>
<string name="add_a_list">Add a list</string>
<string name="edit_image">Edit Image</string>
<string name="cant_edit_image_not_yet_uploaded">The new image is not yet uploaded. Please, wait before editing it</string>
<string name="unselect_image">Unselect Image</string>
<string name="choose_default_language">Select product language</string>
<string name="barcode">बारकोड</string>
<string name="productAdditivesNone">No additives</string>
<string name="error_uploading_photo">Something went wrong while uploading a new photo</string>
<string name="import_csv_to_list">Import an Open Food Facts list CSV file</string>
<string name="toast_import_csv_error">Error in importing CSV file</string>
<string name="toast_import_csv">CSV imported successfully!</string>
<string name="open_csv">Open CSV</string>
<!-- Used for diets AddOn -->
<string name="history_network_error">The product couldn\'t be loaded. Please try again when you are back online.</string>
<string name="cant_modify_if_refreshing">Please wait until the upload finishes</string>
<string name="closeZoom">Click to close zoom</string>
<string name="imageFullscreen">Zoom</string>
<!-- Used for product ingredients analysis tags -->
<string name="close_button">बंद करें</string>
<string name="tip_message">TIP: %1$s </string>
<string name="ingredient_analysis_tip">Click on the icons below to learn more about the product and to select what information you are interested in.</string>
<string name="got_it">Got it!</string>
<string name="onboarding_hint_msg">Hint: %s </string>
<string name="ingredient_image_edit_tip">Tap on the image to zoom it, change it, crop it, rotate it and more…</string>
<string name="ingredients_in_this_product_are">Ingredients in this product are %1$s.</string>
<string name="ingredients_in_this_product">Ingredients in this product that are %1$s:</string>
<string name="display_analysis_tag_status">Display %1$s status</string>
<string name="preference_header_display">Display</string>
<string name="unknown_status_no_translation">We couldn\'t assess the status of the product due to unrecognized ingredients.</string>
<string name="help_translate_ingredients">Help translate the ingredients in your language</string>
<string name="unknown_status_missing_ingredients">We couldn\'t assess the status of the product due to missing ingredients.</string>
<string name="unknown_status_ambiguous_ingredients">We couldn\'t assess the status of the product because of ambiguous ingredients : <![CDATA[<b>%1$s</b>]]></string>
<string name="help_extract_ingredients">Help extract the ingredients to get the %1$s status</string>
<string name="add_photo_to_extract_ingredients">Add a photo to begin extracting ingredients</string>
<string name="product_already_exists_in_comparison">The product already exists in the comparison list</string>
<string name="compare_brands">Brands :</string>
<string name="compare_additives">योगशील:</string>
<string name="compare_salt">नमक</string>
<string name="compare_sugars">शुगर्स</string>
<string name="compare_saturated_fat">संतृप्त वसा</string>
<string name="compare_fat">वसा</string>
<string name="compare_quantity">Quantity :</string>
<string name="load_ingredient_detection_data">Load ingredient detection data</string>
<string name="load_ingredient_detection_data_summary">No ingredient detection data has been found, try to load it from server</string>
<string name="white_check_image_descr">A white check</string>
<string name="hint_packaging">Examples: plastic, paper, glass</string>
<string name="risk_alcohol_consumption">Excessive consumption of alcohol is harmful to health, to be consumed in moderation</string>
<!--ShowcaseView Strings-->
<string name="title_image_type">Select Image Type</string>
<string name="content_image_type">Choose the photo to work on</string>
<string name="title_choose_language">Select Language</string>
<string name="content_choose_language">Select the language of the photo</string>
<string name="title_add_photo">Add Photo</string>
<string name="content_add_photo">Press here to take a new photo</string>
<string name="title_choose_photo">Choose Photo</string>
<string name="content_choose_photo">Choose among existing photos</string>
<string name="title_edit_photo">Edit Photo</string>
<string name="content_edit_photo">Edit, crop and rotate the photo</string>
<string name="title_unselect_photo">Unselect Photo</string>
<string name="content_unselect_photo">Press here to unselect current photo</string>
<string name="title_exit">Exit</string>
<string name="content_exit">Press here to return to the product</string>
<string name="icon_label">Labels icon</string>
<string name="icon_categories">Categories icon</string>
<string name="icon_emb">Traceability code icon</string>
<string name="product_front_image">Product front image</string>
<string name="brand_hint">Examples: Coca-Cola, Nestle, etc…</string>
<string name="sign_in_or_register">Sign in or register</string>
<string name="hunger_game_call_to_action">Help classify more %1$s</string>
<string name="txtHomeOnline">Open Food Facts is an open database of more than %1$s food products (plus the ones you will contribute).</string>
</resources>
| {
"redpajama_set_name": "RedPajamaGithub"
} | 4,127 |
namespace blink {
inline const CollapsedBorderValue& TableCellPainter::cachedCollapsedLeftBorder(const ComputedStyle& styleForCellFlow) const
{
if (styleForCellFlow.isHorizontalWritingMode()) {
return styleForCellFlow.isLeftToRightDirection() ? m_layoutTableCell.section()->cachedCollapsedBorder(&m_layoutTableCell, CBSStart)
: m_layoutTableCell.section()->cachedCollapsedBorder(&m_layoutTableCell, CBSEnd);
}
return styleForCellFlow.isFlippedBlocksWritingMode() ? m_layoutTableCell.section()->cachedCollapsedBorder(&m_layoutTableCell, CBSAfter)
: m_layoutTableCell.section()->cachedCollapsedBorder(&m_layoutTableCell, CBSBefore);
}
inline const CollapsedBorderValue& TableCellPainter::cachedCollapsedRightBorder(const ComputedStyle& styleForCellFlow) const
{
if (styleForCellFlow.isHorizontalWritingMode()) {
return styleForCellFlow.isLeftToRightDirection() ? m_layoutTableCell.section()->cachedCollapsedBorder(&m_layoutTableCell, CBSEnd)
: m_layoutTableCell.section()->cachedCollapsedBorder(&m_layoutTableCell, CBSStart);
}
return styleForCellFlow.isFlippedBlocksWritingMode() ? m_layoutTableCell.section()->cachedCollapsedBorder(&m_layoutTableCell, CBSBefore)
: m_layoutTableCell.section()->cachedCollapsedBorder(&m_layoutTableCell, CBSAfter);
}
inline const CollapsedBorderValue& TableCellPainter::cachedCollapsedTopBorder(const ComputedStyle& styleForCellFlow) const
{
if (styleForCellFlow.isHorizontalWritingMode())
return styleForCellFlow.isFlippedBlocksWritingMode() ? m_layoutTableCell.section()->cachedCollapsedBorder(&m_layoutTableCell, CBSAfter) : m_layoutTableCell.section()->cachedCollapsedBorder(&m_layoutTableCell, CBSBefore);
return styleForCellFlow.isLeftToRightDirection() ? m_layoutTableCell.section()->cachedCollapsedBorder(&m_layoutTableCell, CBSStart) : m_layoutTableCell.section()->cachedCollapsedBorder(&m_layoutTableCell, CBSEnd);
}
inline const CollapsedBorderValue& TableCellPainter::cachedCollapsedBottomBorder(const ComputedStyle& styleForCellFlow) const
{
if (styleForCellFlow.isHorizontalWritingMode()) {
return styleForCellFlow.isFlippedBlocksWritingMode() ? m_layoutTableCell.section()->cachedCollapsedBorder(&m_layoutTableCell, CBSBefore)
: m_layoutTableCell.section()->cachedCollapsedBorder(&m_layoutTableCell, CBSAfter);
}
return styleForCellFlow.isLeftToRightDirection() ? m_layoutTableCell.section()->cachedCollapsedBorder(&m_layoutTableCell, CBSEnd)
: m_layoutTableCell.section()->cachedCollapsedBorder(&m_layoutTableCell, CBSStart);
}
void TableCellPainter::paint(const PaintInfo& paintInfo, const LayoutPoint& paintOffset)
{
ASSERT(paintInfo.phase != PaintPhaseCollapsedTableBorders);
BlockPainter(m_layoutTableCell).paint(paintInfo, paintOffset);
}
static EBorderStyle collapsedBorderStyle(EBorderStyle style)
{
if (style == OUTSET)
return GROOVE;
if (style == INSET)
return RIDGE;
return style;
}
void TableCellPainter::paintCollapsedBorders(const PaintInfo& paintInfo, const LayoutPoint& paintOffset)
{
ASSERT(paintInfo.phase == PaintPhaseCollapsedTableBorders);
if (!paintInfo.shouldPaintWithinRoot(&m_layoutTableCell) || m_layoutTableCell.style()->visibility() != VISIBLE)
return;
const CollapsedBorderValue* tableCurrentBorderValue = m_layoutTableCell.table()->currentBorderValue();
if (!tableCurrentBorderValue)
return;
const ComputedStyle& styleForCellFlow = m_layoutTableCell.styleForCellFlow();
const CollapsedBorderValue& leftBorderValue = cachedCollapsedLeftBorder(styleForCellFlow);
const CollapsedBorderValue& rightBorderValue = cachedCollapsedRightBorder(styleForCellFlow);
const CollapsedBorderValue& topBorderValue = cachedCollapsedTopBorder(styleForCellFlow);
const CollapsedBorderValue& bottomBorderValue = cachedCollapsedBottomBorder(styleForCellFlow);
bool shouldPaintTop = topBorderValue.shouldPaint(*tableCurrentBorderValue);
bool shouldPaintBottom = bottomBorderValue.shouldPaint(*tableCurrentBorderValue);
bool shouldPaintLeft = leftBorderValue.shouldPaint(*tableCurrentBorderValue);
bool shouldPaintRight = rightBorderValue.shouldPaint(*tableCurrentBorderValue);
if (!shouldPaintTop && !shouldPaintBottom && !shouldPaintLeft && !shouldPaintRight)
return;
// Adjust our x/y/width/height so that we paint the collapsed borders at the correct location.
int topWidth = topBorderValue.width();
int bottomWidth = bottomBorderValue.width();
int leftWidth = leftBorderValue.width();
int rightWidth = rightBorderValue.width();
LayoutRect paintRect = paintBounds(paintOffset, AddOffsetFromParent);
IntRect borderRect = pixelSnappedIntRect(paintRect.x() - leftWidth / 2,
paintRect.y() - topWidth / 2,
paintRect.width() + leftWidth / 2 + (rightWidth + 1) / 2,
paintRect.height() + topWidth / 2 + (bottomWidth + 1) / 2);
if (!borderRect.intersects(paintInfo.rect))
return;
GraphicsContext* graphicsContext = paintInfo.context;
if (LayoutObjectDrawingRecorder::useCachedDrawingIfPossible(*graphicsContext, m_layoutTableCell, paintInfo.phase))
return;
LayoutObjectDrawingRecorder recorder(*graphicsContext, m_layoutTableCell, paintInfo.phase, borderRect);
Color cellColor = m_layoutTableCell.resolveColor(CSSPropertyColor);
bool antialias = BoxPainter::shouldAntialiasLines(graphicsContext);
// We never paint diagonals at the joins. We simply let the border with the highest
// precedence paint on top of borders with lower precedence.
if (shouldPaintTop) {
ObjectPainter::drawLineForBoxSide(graphicsContext, borderRect.x(), borderRect.y(), borderRect.maxX(), borderRect.y() + topWidth, BSTop,
topBorderValue.color().resolve(cellColor), collapsedBorderStyle(topBorderValue.style()), 0, 0, antialias);
}
if (shouldPaintBottom) {
ObjectPainter::drawLineForBoxSide(graphicsContext, borderRect.x(), borderRect.maxY() - bottomWidth, borderRect.maxX(), borderRect.maxY(), BSBottom,
bottomBorderValue.color().resolve(cellColor), collapsedBorderStyle(bottomBorderValue.style()), 0, 0, antialias);
}
if (shouldPaintLeft) {
ObjectPainter::drawLineForBoxSide(graphicsContext, borderRect.x(), borderRect.y(), borderRect.x() + leftWidth, borderRect.maxY(), BSLeft,
leftBorderValue.color().resolve(cellColor), collapsedBorderStyle(leftBorderValue.style()), 0, 0, antialias);
}
if (shouldPaintRight) {
ObjectPainter::drawLineForBoxSide(graphicsContext, borderRect.maxX() - rightWidth, borderRect.y(), borderRect.maxX(), borderRect.maxY(), BSRight,
rightBorderValue.color().resolve(cellColor), collapsedBorderStyle(rightBorderValue.style()), 0, 0, antialias);
}
}
void TableCellPainter::paintBackgroundsBehindCell(const PaintInfo& paintInfo, const LayoutPoint& paintOffset, LayoutObject* backgroundObject)
{
if (!paintInfo.shouldPaintWithinRoot(&m_layoutTableCell))
return;
if (!backgroundObject)
return;
if (m_layoutTableCell.style()->visibility() != VISIBLE)
return;
LayoutTable* tableElt = m_layoutTableCell.table();
if (!tableElt->collapseBorders() && m_layoutTableCell.style()->emptyCells() == HIDE && !m_layoutTableCell.firstChild())
return;
Color c = backgroundObject->resolveColor(CSSPropertyBackgroundColor);
const FillLayer& bgLayer = backgroundObject->style()->backgroundLayers();
LayoutRect paintRect = paintBounds(paintOffset, backgroundObject != &m_layoutTableCell ? AddOffsetFromParent : DoNotAddOffsetFromParent);
if (bgLayer.hasImage() || c.alpha()) {
// We have to clip here because the background would paint
// on top of the borders otherwise. This only matters for cells and rows.
bool shouldClip = backgroundObject->hasLayer() && (backgroundObject == &m_layoutTableCell || backgroundObject == m_layoutTableCell.parent()) && tableElt->collapseBorders();
GraphicsContextStateSaver stateSaver(*paintInfo.context, shouldClip);
if (shouldClip) {
LayoutRect clipRect(paintRect.location(), m_layoutTableCell.size());
clipRect.expand(m_layoutTableCell.borderInsets());
paintInfo.context->clip(clipRect);
}
BoxPainter(m_layoutTableCell).paintFillLayers(paintInfo, c, bgLayer, paintRect, BackgroundBleedNone, SkXfermode::kSrcOver_Mode, backgroundObject);
}
}
void TableCellPainter::paintBoxDecorationBackground(const PaintInfo& paintInfo, const LayoutPoint& paintOffset)
{
if (!paintInfo.shouldPaintWithinRoot(&m_layoutTableCell))
return;
LayoutTable* table = m_layoutTableCell.table();
if (!table->collapseBorders() && m_layoutTableCell.style()->emptyCells() == HIDE && !m_layoutTableCell.firstChild())
return;
bool needsToPaintBorder = m_layoutTableCell.styleRef().hasBorderDecoration() && !table->collapseBorders();
if (!m_layoutTableCell.hasBackground() && !m_layoutTableCell.styleRef().boxShadow() && !needsToPaintBorder)
return;
if (LayoutObjectDrawingRecorder::useCachedDrawingIfPossible(*paintInfo.context, m_layoutTableCell, DisplayItem::BoxDecorationBackground))
return;
LayoutRect visualOverflowRect = m_layoutTableCell.visualOverflowRect();
visualOverflowRect.moveBy(paintOffset);
LayoutObjectDrawingRecorder recorder(*paintInfo.context, m_layoutTableCell, DisplayItem::BoxDecorationBackground, pixelSnappedIntRect(visualOverflowRect));
LayoutRect paintRect = paintBounds(paintOffset, DoNotAddOffsetFromParent);
BoxPainter::paintBoxShadow(paintInfo, paintRect, m_layoutTableCell.styleRef(), Normal);
// Paint our cell background.
paintBackgroundsBehindCell(paintInfo, paintOffset, &m_layoutTableCell);
BoxPainter::paintBoxShadow(paintInfo, paintRect, m_layoutTableCell.styleRef(), Inset);
if (!needsToPaintBorder)
return;
BoxPainter::paintBorder(m_layoutTableCell, paintInfo, paintRect, m_layoutTableCell.styleRef());
}
void TableCellPainter::paintMask(const PaintInfo& paintInfo, const LayoutPoint& paintOffset)
{
if (m_layoutTableCell.style()->visibility() != VISIBLE || paintInfo.phase != PaintPhaseMask)
return;
LayoutTable* tableElt = m_layoutTableCell.table();
if (!tableElt->collapseBorders() && m_layoutTableCell.style()->emptyCells() == HIDE && !m_layoutTableCell.firstChild())
return;
if (LayoutObjectDrawingRecorder::useCachedDrawingIfPossible(*paintInfo.context, m_layoutTableCell, paintInfo.phase))
return;
LayoutRect paintRect = paintBounds(paintOffset, DoNotAddOffsetFromParent);
LayoutObjectDrawingRecorder recorder(*paintInfo.context, m_layoutTableCell, paintInfo.phase, paintRect);
BoxPainter(m_layoutTableCell).paintMaskImages(paintInfo, paintRect);
}
LayoutRect TableCellPainter::paintBounds(const LayoutPoint& paintOffset, PaintBoundOffsetBehavior paintBoundOffsetBehavior)
{
LayoutPoint adjustedPaintOffset = paintOffset;
if (paintBoundOffsetBehavior == AddOffsetFromParent)
adjustedPaintOffset.moveBy(m_layoutTableCell.location());
return LayoutRect(adjustedPaintOffset, LayoutSize(m_layoutTableCell.pixelSnappedSize()));
}
} // namespace blink
| {
"redpajama_set_name": "RedPajamaGithub"
} | 5,215 |
module Backup
module Storage
class RSync < Base
include Utilities::Helpers
##
# Mode of operation
#
# [:ssh (default)]
# Connects to the remote via SSH.
# Does not use an rsync daemon on the remote.
#
# [:ssh_daemon]
# Connects to the remote via SSH.
# Spawns a single-use daemon on the remote, which allows certain
# daemon features (like modules) to be used.
#
# [:rsync_daemon]
# Connects directly to an rsync daemon via TCP.
# Data transferred is not encrypted.
#
attr_accessor :mode
##
# Server Address
#
# If not specified, the storage operation will be local.
attr_accessor :host
##
# SSH or RSync port
#
# For `:ssh` or `:ssh_daemon` mode, this specifies the SSH port to use
# and defaults to 22.
#
# For `:rsync_daemon` mode, this specifies the TCP port to use
# and defaults to 873.
attr_accessor :port
##
# SSH User
#
# If the user running the backup is not the same user that needs to
# authenticate with the remote server, specify the user here.
#
# The user must have SSH keys setup for passphrase-less access to the
# remote. If the SSH User does not have passphrase-less keys, or no
# default keys in their `~/.ssh` directory, you will need to use the
# `-i` option in `:additional_ssh_options` to specify the
# passphrase-less key to use.
#
# Used only for `:ssh` and `:ssh_daemon` modes.
attr_accessor :ssh_user
##
# Additional SSH Options
#
# Used to supply a String or Array of options to be passed to the SSH
# command in `:ssh` and `:ssh_daemon` modes.
#
# For example, if you need to supply a specific SSH key for the `ssh_user`,
# you would set this to: "-i '/path/to/id_rsa'". Which would produce:
#
# rsync -e "ssh -p 22 -i '/path/to/id_rsa'"
#
# Arguments may be single-quoted, but should not contain any double-quotes.
#
# Used only for `:ssh` and `:ssh_daemon` modes.
attr_accessor :additional_ssh_options
##
# RSync User
#
# If the user running the backup is not the same user that needs to
# authenticate with the rsync daemon, specify the user here.
#
# Used only for `:ssh_daemon` and `:rsync_daemon` modes.
attr_accessor :rsync_user
##
# RSync Password
#
# If specified, Backup will write the password to a temporary file and
# use it with rsync's `--password-file` option for daemon authentication.
#
# Note that setting this will override `rsync_password_file`.
#
# Used only for `:ssh_daemon` and `:rsync_daemon` modes.
attr_accessor :rsync_password
##
# RSync Password File
#
# If specified, this path will be passed to rsync's `--password-file`
# option for daemon authentication.
#
# Used only for `:ssh_daemon` and `:rsync_daemon` modes.
attr_accessor :rsync_password_file
##
# Additional String or Array of options for the rsync cli
attr_accessor :additional_rsync_options
##
# Flag for compressing (only compresses for the transfer)
attr_accessor :compress
##
# Path to store the synced backup package file(s) to.
#
# If no +host+ is specified, then +path+ will be local, and the only
# other used option would be +additional_rsync_options+.
# +path+ will be expanded, so '~/my_path' will expand to '$HOME/my_path'.
#
# If a +host+ is specified, this will be a path on the host.
# If +mode+ is `:ssh` (default), then any relative path, or path starting
# with '~/' will be relative to the directory the ssh_user is logged
# into. For `:ssh_daemon` or `:rsync_daemon` modes, this would reference
# an rsync module/path.
#
# In :ssh_daemon and :rsync_daemon modes, +path+ (or path defined by
# your rsync module) must already exist.
#
# In :ssh mode or local operation (no +host+ specified), +path+ will
# be created if needed - either locally, or on the remote for :ssh mode.
attr_accessor :path
def initialize(model, storage_id = nil)
super
@mode ||= :ssh
@port ||= mode == :rsync_daemon ? 873 : 22
@compress ||= false
@path ||= "~/backups"
end
private
def transfer!
write_password_file
create_remote_path
package.filenames.each do |filename|
src = "'#{File.join(Config.tmp_path, filename)}'"
dest = "#{host_options}'#{File.join(remote_path, filename)}'"
Logger.info "Syncing to #{dest}..."
run("#{rsync_command} #{src} #{dest}")
end
ensure
remove_password_file
end
##
# Other storages add an additional timestamp directory to this path.
# This is not desired here, since we need to transfer the package files
# to the same location each time.
def remote_path
@remote_path ||= begin
if host
path.sub(/^~\//, "").sub(/\/$/, "")
else
File.expand_path(path)
end
end
end
##
# Runs a 'mkdir -p' command on the host (or locally) to ensure the
# dest_path exists. This is used because we're transferring a single
# file, and rsync won't attempt to create the intermediate directories.
#
# This is only applicable locally and in :ssh mode.
# In :ssh_daemon and :rsync_daemon modes the `path` would include a
# module name that must define a path on the remote that already exists.
def create_remote_path
if host
return unless mode == :ssh
run "#{utility(:ssh)} #{ssh_transport_args} #{host} " +
%("mkdir -p '#{remote_path}'")
else
FileUtils.mkdir_p(remote_path)
end
end
def host_options
@host_options ||= begin
if !host
""
elsif mode == :ssh
"#{host}:"
else
user = "#{rsync_user}@" if rsync_user
"#{user}#{host}::"
end
end
end
def rsync_command
@rsync_command ||= begin
cmd = utility(:rsync) << " --archive" <<
" #{Array(additional_rsync_options).join(" ")}".rstrip
cmd << compress_option << password_option << transport_options if host
cmd
end
end
def compress_option
compress ? " --compress" : ""
end
def password_option
return "" if mode == :ssh
path = @password_file ? @password_file.path : rsync_password_file
path ? " --password-file='#{File.expand_path(path)}'" : ""
end
def transport_options
if mode == :rsync_daemon
" --port #{port}"
else
%( -e "#{utility(:ssh)} #{ssh_transport_args}")
end
end
def ssh_transport_args
args = "-p #{port} "
args << "-l #{ssh_user} " if ssh_user
args << Array(additional_ssh_options).join(" ")
args.rstrip
end
def write_password_file
return unless host && rsync_password && mode != :ssh
@password_file = Tempfile.new("backup-rsync-password")
@password_file.write(rsync_password)
@password_file.close
end
def remove_password_file
@password_file.delete if @password_file
end
end
end
end
| {
"redpajama_set_name": "RedPajamaGithub"
} | 518 |
Q: Как выполнить поиск по истории Qiwi? def check_buy(m):
s = requests.Session()
parameters = {
'operation': 'IN',
'rows': '10'
}
s.headers['Accept'] = 'application/json'
s.headers['Authorization'] = 'Bearer ' + config.api_access_token
r = s.get('https://edge.qiwi.com/payment-history/v2/persons/' + config.my_number + '/payments',
params=parameters)
data = json.loads(r.text)
for i in range(0, 10):
comment = data['data'][i]['comment']
amount = str(data['data'][i]['sum']['amount'])
data_buy = data['data'][i]['date'][:10]
Рофл в том, что переменная принимает последнее значение.
Пример что выводит в логи и какое значение переменной ниже:
'2019-10-01'
'2019-10-01'
'2019-10-01'
'2019-10-01'
'2019-10-01'
'2019-10-01'
'2019-10-01'
'2019-10-01'
'2019-10-01'
'2019-10-01'
'2019-09-30'
'2019-09-30'
'2019-09-29'
'2019-09-29'
'2019-09-29'
'2019-09-29'
'2019-09-29'
'2019-09-29'
'2019-09-29'
Значение переменной - '2019-09-29'.
Как сделать переменную списком этих объектов?
A: Значение переменной - '2019-09-29'. Это значение data_buy?
Я не очень понял условие. Вижу, что данные достаются в цикле и каждый раз записываются в одни и те же переменные. Если заменить переменные на списки, то они будут хранить все нужные данные
data = json.loads(r.text)
comment = list()
amount = list()
data_buy = list()
for i in range(0, 10):
comment.append(data['data'][i]['comment'])
amount.append(str(data['data'][i]['sum']['amount']))
data_buy.append(data['data'][i]['date'][:10])
А можно хранить в словаре например, где ключом будет например comment (если он уникален) или ввести абстрактный id. А значениями остальные параметры.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 1,575 |
{"url":"https:\/\/answerriddle.com\/answer-what-is-the-study-of-law-called\/","text":"# Answer: What is the study of law called?\n\nThe Question: What is the study of law called?\n\nJurisprudence\nJudgeship\nJurisdiction\nJudgement","date":"2021-09-22 01:23:26","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\": 0.8273360133171082, \"perplexity\": 1623.146299645808}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 5, \"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\/1631780057303.94\/warc\/CC-MAIN-20210922011746-20210922041746-00236.warc.gz\"}"} | null | null |
Alloy Crossdressers - Meet a Crossdresser In your City!
Alloy Crossdressers 100% Free to Signup!
So you have an urge to discover a hot crossdresser who lives near by in Alloy? Then you've come to the perfect place on the Internet for we know all the hottest crossdressers here Alloy & we can put you in touch with all of them that's craving for a hot & handsome guy like you.
We've removed all problems out of finding a hot crossdresser in Alloy for you & now all you have to do is use the directory of Alloy crossdressers that we've compiled here for you. We guarantee that you won't find any hornier crossdressers online in Alloy so step inside & use our completely free directory to find those hot crossdressers that you lust after. You can be talking with the crossdressers we have listed on Crossdresser Chat City in just a few seconds so don't waste anymore time looking for them on the Net online. Find them in our free directory and you can be having fun in just a short time. | {
"redpajama_set_name": "RedPajamaC4"
} | 2,913 |
Q: Compiling OpenSSL 1.1.0f with SSLv3 I'm running Ubuntu 14.04.5 and have compiled from source OpenSSL 1.1.0f and Curl 7.54.1. When I compile openssl I threw the enable-ssl2 and enable-ssl3 flags:
vbetoglo@helios:~/Downloads/openssl-1.1.0f$ ./Configure enable-ssl2 enable-ssl3 linux-x86_64
Configuring OpenSSL version 1.1.0f (0x1010006fL)
no-asan [default] OPENSSL_NO_ASAN
no-crypto-mdebug [default] OPENSSL_NO_CRYPTO_MDEBUG
no-crypto-mdebug-backtrace [default] OPENSSL_NO_CRYPTO_MDEBUG_BACKTRACE
no-ec_nistp_64_gcc_128 [default] OPENSSL_NO_EC_NISTP_64_GCC_128
no-egd [default] OPENSSL_NO_EGD
no-fuzz-afl [default] OPENSSL_NO_FUZZ_AFL
no-fuzz-libfuzzer [default] OPENSSL_NO_FUZZ_LIBFUZZER
no-heartbeats [default] OPENSSL_NO_HEARTBEATS
no-md2 [default] OPENSSL_NO_MD2 (skip dir)
no-msan [default] OPENSSL_NO_MSAN
no-rc5 [default] OPENSSL_NO_RC5 (skip dir)
no-sctp [default] OPENSSL_NO_SCTP
no-ssl-trace [default] OPENSSL_NO_SSL_TRACE
no-ssl3 [forced] OPENSSL_NO_SSL3
no-ssl3-method [default] OPENSSL_NO_SSL3_METHOD
no-ubsan [default] OPENSSL_NO_UBSAN
no-unit-test [default] OPENSSL_NO_UNIT_TEST
no-weak-ssl-ciphers [default] OPENSSL_NO_WEAK_SSL_CIPHERS
no-zlib [default]
no-zlib-dynamic [default]
Configuring for linux-x86_64
CC =gcc
CFLAG =-Wall -O3 -pthread -m64 -DL_ENDIAN
SHARED_CFLAG =-fPIC -DOPENSSL_USE_NODELETE
DEFINES =DSO_DLFCN HAVE_DLFCN_H NDEBUG OPENSSL_THREADS OPENSSL_NO_STATIC_ENGINE OPENSSL_PIC OPENSSL_IA32_SSE2 OPENSSL_BN_ASM_MONT OPENSSL_BN_ASM_MONT5 OPENSSL_BN_ASM_GF2m SHA1_ASM SHA256_ASM SHA512_ASM RC4_ASM MD5_ASM AES_ASM VPAES_ASM BSAES_ASM GHASH_ASM ECP_NISTZ256_ASM PADLOCK_ASM POLY1305_ASM
LFLAG =
PLIB_LFLAG =
EX_LIBS =-ldl
APPS_OBJ =
CPUID_OBJ =x86_64cpuid.o
UPLINK_OBJ =
BN_ASM =asm/x86_64-gcc.o x86_64-mont.o x86_64-mont5.o x86_64-gf2m.o rsaz_exp.o rsaz-x86_64.o rsaz-avx2.o
EC_ASM =ecp_nistz256.o ecp_nistz256-x86_64.o
DES_ENC =des_enc.o fcrypt_b.o
AES_ENC =aes-x86_64.o vpaes-x86_64.o bsaes-x86_64.o aesni-x86_64.o aesni-sha1-x86_64.o aesni-sha256-x86_64.o aesni-mb-x86_64.o
BF_ENC =bf_enc.o
CAST_ENC =c_enc.o
RC4_ENC =rc4-x86_64.o rc4-md5-x86_64.o
RC5_ENC =rc5_enc.o
MD5_OBJ_ASM =md5-x86_64.o
SHA1_OBJ_ASM =sha1-x86_64.o sha256-x86_64.o sha512-x86_64.o sha1-mb-x86_64.o sha256-mb-x86_64.o
RMD160_OBJ_ASM=
CMLL_ENC =cmll-x86_64.o cmll_misc.o
MODES_OBJ =ghash-x86_64.o aesni-gcm-x86_64.o
PADLOCK_OBJ =e_padlock-x86_64.o
CHACHA_ENC =chacha-x86_64.o
POLY1305_OBJ =poly1305-x86_64.o
BLAKE2_OBJ =
PROCESSOR =
RANLIB =ranlib
ARFLAGS =
PERL =/usr/bin/perl
SIXTY_FOUR_BIT_LONG mode
Configured for linux-x86_64.
I've compile and recompiled curl after having done this as well, but no matter what I do, it always comes back with:
vbetoglo@helios:~/Downloads$ curl --version
curl 7.54.1 (x86_64-pc-linux-gnu) libcurl/7.54.1 OpenSSL/1.1.0f zlib/1.2.8 nghttp2/1.12.0-DEV
Release-Date: 2017-06-14
Protocols: dict file ftp ftps gopher http https imap imaps ldap ldaps pop3 pop3s rtsp smb smbs smtp smtps telnet tftp
Features: IPv6 Largefile GSS-API Kerberos SPNEGO NTLM NTLM_WB SSL libz TLS-SRP HTTP2 UnixSockets HTTPS-proxy
vbetoglo@helios:~/Downloads$ curl -3 https://www.google.com/
curl: (4) OpenSSL was built without SSLv3 support
Can anyone hint at what I may be doing wrong?
A: Try to add also the flag "enable-ssl3-method"
Your command becomes:
./Configure enable-ssl2 enable-ssl3 enable-ssl3-method linux-x86_64
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 568 |
package org.apache.jena.shacl.sys;
import org.apache.jena.graph.Graph;
import org.apache.jena.graph.TransactionHandler;
import org.apache.jena.graph.impl.TransactionHandlerBase;
import org.apache.jena.riot.other.G;
import org.apache.jena.shacl.*;
import org.apache.jena.sparql.graph.GraphWrapper;
/** A graph that performs SHACL validation on a graph during transaction "commit".
* (It does not validate on direct addition of date to the graph outside a transaction).
* Usage:
* <pre>
* try {
* graph.getTransactionHandler().execute(()->{
* ... application code ...
* });
* } catch (ShaclValidationException ex) {
* ShLib.printReport(ex.getReport());
* }
* </pre>
* If validation fails, the transaction is aborted.
*
* <em>Experimental</em>
*/
public class ValidationGraph extends GraphWrapper {
private final Shapes shapes;
private final TransactionHandlerValidate transactionHandler;
public ValidationGraph(Graph graph, Shapes shapes) {
super(graph);
this.shapes = shapes;
this.transactionHandler = new TransactionHandlerValidate(this, get().getTransactionHandler());
}
@Override
public TransactionHandler getTransactionHandler() {
return transactionHandler;
}
/**
* Update the graph and return a validation report. This operation aborts the
* update if there are any validation results from shapes (any severity).
* To determine if the update resulted in a graph that validates,
* call {@code ValidationReport.conforms}.
*/
public ValidationReport updateValidateReport(Runnable action) {
return GraphValidation.updateAndReport(shapes, get(), action);
}
/**
* Update the graph. This operation aborts and throws a
* {@link ShaclValidationException} if there are any validation results from
* shapes (of any severity). If the update results in a valid graph, return
* without exception (with a {@link ValidationReport} with
* {@code ValidationReport.conforms} returning true).
*/
public ValidationReport updateValidate(Runnable action) throws ShaclValidationException {
return GraphValidation.update(shapes, get(), action);
}
/**
* Update the graph and return a validation report.
* This operation only reports, it does cancel the update
* if there are any validation results from shapes.
*/
public ValidationReport updateAndReport(Runnable action) {
return G.calcTxn(get(), ()-> {
action.run();
return ShaclValidator.get().validate(shapes, get());
});
}
private static class TransactionHandlerValidate extends TransactionHandlerBase {
private final TransactionHandler other;
private ValidationGraph graphValidate;
TransactionHandlerValidate(ValidationGraph graphValidate, TransactionHandler other) {
this.graphValidate = graphValidate;
this.other = other ;
}
@Override
public boolean transactionsSupported() {
return other.transactionsSupported();
}
@Override
public void begin() {
other.begin();
}
@Override
public void abort() {
other.abort();
}
@Override
public void commit() {
ValidationReport report = validateCommit();
throw new ShaclValidationException(report) {
@Override
public Throwable fillInStackTrace() { return this; }
};
}
// Execute. If there are any validation results of any severity, abort, and return the ValidationReport.
private ValidationReport validateCommit() {
ValidationReport report = ShaclValidator.get().validate(graphValidate.shapes, graphValidate.get());
if ( report.conforms() ) {
other.commit();
return null;
}
abort();
return report;
}
}
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 4,299 |
Walker Architecture & Design offers services in the design of individual houses, townhouses, commercial buildings, and urban design projects.
Our fee structure is highly competitive and we work on a fixed fee structure with clear and transparent undertakings after the scope of work has been defined.
For people that have not worked with an architect before, the process can be daunting. We work hard to ensure that clients are aware of all costs related to their project, including architectural services, resource and building consents, and other consultants that may be necessary to bring projects to fruition. We also provide realistic advice on timeframes related to working with local authorities and other consultants.
For projects where the scope of work cannot be defined, we offer competitive hourly rates for expert services. Please contact us if you want more information on this.
For more information, please refer to the section on working with us. | {
"redpajama_set_name": "RedPajamaC4"
} | 586 |
Spoiler (z ang. spoil – psuć, niszczyć, rozpieszczać) – niepożądana informacja o szczegółach zakończenia bądź ważnego zwrotu akcji utworu literackiego, teatralnego, telewizyjnego lub filmowego. Termin ten bywa używany również w odniesieniu do podpowiedzi lub rozwiązań w grach i łamigłówkach. Stosowany na stronach i forach internetowych oraz grupach dyskusyjnych.
Spoiler może wyjawiać szczegóły dotyczące rozwiązania kluczowego dla danego utworu napięcia dramatycznego czy pomysłu, na którym oparty jest utwór, i odbierać widzowi przyjemność z jego oglądania (czytania). Ma to szczególnie znaczenie w przypadku utworów kryminalnych czy thrillerów, gdy przedstawiona zostaje informacja o mordercy lub tożsamości czarnego charakteru, choć ma zastosowanie również w przypadku produkcji innych gatunków.
Nie każdy szczegół zakończenia utworu jest spoilerem. Nie jest nim informacja o tym, że w filmie "Titanic" zatonie statek, choć może już być wyjawienie, czy i któremu z bohaterów wersji filmu z 1997 roku uda się przeżyć.
Jego znaczenie nieco maleje z upływem czasu od powstania do emisji danej produkcji, bowiem w międzyczasie wiele informacji o szczegółach zakończenia przedostaje się do mediów, niemniej na internetowych grupach dyskusyjnych, forach i stronach poświęconych różnego rodzaju produkcjom przyjęło się ostrzeganie czytelników.
Na grupach dyskusyjnych w Usenecie ostrzeżenie polega na poprzedzeniu spoilera około 25 pustymi liniami nazywanymi odstępem spoilerowym (ang. spoiler space), niekiedy takie informacje kodowane są w standardzie ROT13 lub w inny sposób ukrywane.
W grach – także komputerowych – za spoilery uchodzą gotowe podpowiedzi lub rozwiązania zagadek, których rozwiązanie jest niezbędne do uwieńczonego powodzeniem ukończenia gry. Skrajnym przypadkiem są gry, których ukończenie bez odnoszenia się do spoilerów jest niezmiernie trudne (np. Nethack). Wraz z widocznym rozwojem znaczenia opowiadanej historii i coraz bardziej rozbudowanej narracji w grach pojawia się taki sam problem spoilerowania szczegółów fabuły, jak w innych utworach literackich czy filmowych.
Linki zewnętrzne
Spoiler na TV Tropes
Teoria filmu | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 2,189 |
{"url":"http:\/\/2msa.it\/javj\/compound-interest-excel-formula-with-irregular-deposits.html","text":"# Compound Interest Excel Formula With Irregular Deposits\n\nHow to Calculate Interest. P = 1500, r = 4. We have just scratched the surface of Excel\u2019s finance functions \u2014 start with these to become an Excel whiz. To demonstrate the difference between simple interest and compound interest, let\u2019s take for example two fixed deposits. 04$Solution 4. The PV (Present Value) function in Excel 2013 is found on the Financial button\u2019s drop-down menu on the Ribbon\u2019s Formulas tab (Alt+MI). A single formula can easily calculate the repayments on a loan of \u00a3 x,000 at y % over z years. However, Excel has made life quite simple. In this formula, A stands for the total amount that accumulates. The concept of compound interest is explained on future value of a single sum page. We are constantly shown numbers which are stripped of context. In line with Kailua's request above, I am seeking an excel formula for compounding interest with a STARTING monthly deposit of$100 invested at 7% per year (compounded annually), and increasing the monthly deposit by 3% per year (i. where i = r\/m is the interest per compounding period and n = mt is the number of compounding periods. =PV*(1+r)^N Where PV = Present Value r = Interest Rate. Tip: In the above formula, 1000 indicates the initial principal of your account, 0. In this method we sum up the interest earned in the previous years to the initial principal, thus increasing the principal amount, on which the interest for the next period is charged. There the result is indeed, using 0. But first you should learn the difference between compound and simple interest. R= Rate of interest. \u2026If you've ever made an investment then you know\u2026that a deposit of $10,000 will grow\u2026by a certain amount every year. Looking for a Compound Interest Equation? Use this formula:. All of them off course linked Before a loan is given out an interest rate is determined, and the way of payment (capital monthly or at the end) For a capital monthly the following. by Kristina Dunbar, UGA. \u0905\u0924\u0903 6 \u0935\u0930\u094d\u0937 \u092c\u093e\u0926 \u092e\u093f\u0936\u094d\u0930\u0927\u0928 \u0932\u0917\u092d\u0917 \u0930\u0942 1,938. PV represents the present value of the investment. The formula is given as: Monthly Compound Interest = Principal $$(1+\\frac{Rate}{12})^{12*Time}$$ \u2013 Principal. If you have monthly compounding or a rate period and the payment is in 45 days? You actually will have a hybrid calculation of a monthly and an exact day calculation. Calculate The Return Of Irregular Deposits - Template to caluculate the XNPR of deposits mtwn - Free interest amount, principal amount, Ending Principal balance. This example assumes that$1000 is invested for 10 years at an annual interest rate of 5%. When your savings plans pays interest 365 days in a year and you make monthly deposits, use the NOMINAL and EFFECT functions first before using the FV function to calculate what the savings plan. When you therefore record any transaction, you will need to enter a date (in column A), select a transaction type from the list box in column B, enter a new interest rate (only if an interest rate change transaction is being recorded, otherwise enter 0%), enter a transaction amount in column E and the formulas in all the other columns will. How to Determine Future Value of a Compounded Deposit in Excel. Most savings accounts don't pay anywhere near enough interest to keep up with inflation. Uniform annual series and future value. A= Monthly compound rate. Fixed Deposit Calculator: This Fixed Deposit Calculator (FD Calculator) tells you the Maturity Value of your invesment (Principal) when compounding of interest is done on a Monthly, Quarterly. However, Excel has made life quite simple. Popular Course in this category. represent an interest rate. However, it's not really all that hard. Simple Interest is the interest generated on a principal amount that does not compound. The formula for calculating compound interest is A = P (1 + r\/n) ^ nt. The user can choose whether deposits are made at the beginning or at the end of the period. Summary: Compound interest can work for you or against you. 1500\/- is deposited in a bank for 6 years and paying an annual interest rate of 4. Regarding the 5% or 10% guess discussion, I used the YTD formula from the author\u2019s row 28 example and played around with the different guess return numbers. Here's a DIY approach to calculate the returns on investment in mutual funds either as a lump sum or through a SIP. 05 \u00d7 10 = $500. It recognizes that being paid interest before the end of the year allows you to reinvest it to earn additional income (regardless what you actually do with the money). It is the result of reinvesting interest, rather than paying it out, so that interest in the next period is then earned on the principal sum plus previously accumulated interest. Suppose you deposit$1000 in a bank which pays 5% interest compounded daily, meaning 365 times per year. Click the view report button to see all of your results. You must provide the amount of each deposit, the frequency of the deposits, the term in months, and the nominal interest rate. Compound interest with fixed rate but irregular deposit amounts over time; I'm trying to estimate the end balance of an account with annual compound interest at a fixed rate, but with variable (but known) deposit amounts. At the same time, you'll learn how to use the FV function in a formula. The discounted cash flow allows for the accumulation of expected interest earned on a sum. Example: You want to invest $20,000 for 30 years at 11 % interest compounded quarterly. Excel workbooks are also provided. See also How to calculate compound interest for an intra-year period in Excel. =PV*(1+r)^N Where PV = Present Value r = Interest Rate. This formula is also used in Microsoft Excel to calculate the Future Value (FV). The number of times interest is compounded in a single period is known as the compounding frequency. 04 \/ 4 =$286. The formula we use to find compound interest is A = P(1 + r\/n)^nt. In the formulas below, all of the data will come from these cells in my sample spreadsheet. Compound interest means that interest gets paid (or is earned) on previously unpaid interest. If you are working through these exercises before completing the first finance lab, use a calculator to find the answers to A, B, and C and when you get to part D, prepare only the first three lines of the table (through period 2). You have a RD for 1 year (Jan-Dec) of Rs 1000\/-. com Support hours: Monday-Thursday, 7am-6pm; Friday, 7am-5pm Pacific Time. Compound Interest Defined Compound interest is the addition of interest to the original amount of a loan or deposit, whereby interest calculated is re-invested into the original amount (or added to the loan), so that interest in the next period is then earned (or charged, if it is a loan) on the principal sum plus previously accumulated interest. \u2022 Compound interest- is interest added to the principal of a deposit or loan so that the added interest also earns interest from then on. Interest rate \u2013 the interest rate on your investment expressed on a yearly basis. Compound interest is interest earning interest on interest on interest. I undestand the use of the FV formulas in Excel to calculate Future Value of an investment. 12)^15-1)\/0. When your savings plans pays interest 365 days in a year and you make monthly deposits, use the NOMINAL and EFFECT functions first before using the FV function to calculate what the savings plan. Book Description. The periodic interest payments promised to bond holders are computed as a fixed percentage of the bond\u2019s face value; this percentage is known as the coupon rate. Step 2: Keep the cursor on the cell where you want the mean to be calculated (say B7) and click on the Function icon under Insert Menu. Get instant live expert help on I need help with compound interest excel formula with regular deposits \u201cMy Excelchat expert helped me in less than 20 minutes, saving me what would have been 5 hours of work!\u201d. Thus, $1,000 invested for ten years at simple interest of 5% earns interest of$1,000 \u00d7 0. This article is about finding interest rate in a simple way for a one time deposit. Banks generally set repayments on loans and mortgages in equal payments over a fixed period of time. In addition to the formula, you also can use Function to calculate the compound interest. Deposit Interest Calculator. Compound interest implicates adding the interest income to your investment, and then reinvesting it, every time, as opposed to withdrawing it. We can get the future value as $. To calculate present value, the k-th payment must be discounted to the present by dividing by the interest, compounded by k terms. Say your brother wants to buy a used car for$5,000 and has only $. 5% interest rate, compounded daily, for the next year, the total interest for two years is$133. 00 is deposited in a bank paying an annual interest rate of 4. Compound interest is the interest paid on the original principal and on the accumulated past interest. Compounding interest rate concept is the center point of the investment world. Compound interest formulas to find principal, interest rates or final investment value including continuous compounding A = Pe^rt. techcommunity. If a person deposits $1,000 at 5 percent. While it is most frequently used to calculate how long an investment will last assuming some periodic, regular withdrawal amount, it will also solve for the \" Starting Amount\", \"Annual Interest Rate\" or \"Regular Withdrawal Amount\" required if you want to dictate the duration of the payout. Activity 1: Compute Interest using What-IF Analysis (Goal. I tried a few things but it did not work. The set dollar amount of your regular contribution buys more units when prices are low and fewer units when prices are high; over the long term, this can reduce your average. One tab for each of the following: payments, loans, customers, investor o. Regarding calculating interest in Excel, have a look at the EFFECT() function. In this formula, you'll want to convert the percentage (5%) to a decimal (. Payment is due at the beginning of the. Capture the values - Your invested amount will be in negative - Any dividend received or final value will be in positive. I used, 5%, 10%, 100%, 1000%, or used the formula with no guess percentage. Compound interest - meaning that the interest you earn each year is added to your principal, so that the balance doesn't merely grow, it grows at an increasing rate - is one of the most useful concepts in finance. How to Determine Future Value of a Compounded Deposit in Excel. We want to do some compound interest calculation, so we need to enter a balance (put this in cell B1, to the right of where we labeled \"Balance\") and an interest rate (in B2, just to the right of the \"Interest Rate\" label). Here's a DIY approach to calculate the returns on investment in mutual funds either as a lump sum or through a SIP. Note These formulas assume that the deposits (payments) are made at the end of each compound period. Albert invested amount of 8000 in a fixed deposit for 2 years at compound interest rate of 5 % per annum. After a year, your money will grow from$1,000 to $1,030. 1500\/- is deposited in a bank for 6 years and paying an annual interest rate of 4. Sort by: A+ Compound Interest Loan Formula Excel #1 Cash Now, Online Loans 24\/7. You're close, but this doesn't handle compound interest correctly. org\u2018s formulas. How much money will you have at the end of the 30 years?.$1,000 is expected to be received at the end of the first year, $800 at the end of the second year,$1,100 at the end of the third year, $700 at the end of the fourth year, and$1,050 at the end of the fifth year. How to Calculate Interest. Compound interest formulas to find principal, interest rates or final investment value including continuous compounding A = Pe^rt. We can get the future value as $. P = A(1 + i \/ N )T. I undestand the use of the FV formulas in Excel to calculate Future Value of an investment. In closing, notice that this formula uses an insight that is always useful to keep in mind about typical annuity calculations like this\u2026 If you were to get a new loan for the current balance due on an existing loan, for the number of months remaining on the existing loan, and for the same interest rate, your payments would be the same as your. IRR Formula Excel with example. To prove the growth rate is correct, the Proof formula is\u2026 F20: =B3*(1+C20)^(14\/12) That is, the ending value is equal to the beginning value times one plus the annual growth rate taken to the number-of-years power. It calculates the IRR on an annual basis of an irregular stream of up to 20 payments and withdrawals. We can model the growth of an initial deposit with respect to the interest rate. The act of adding declared interest to be principal is called compounding. 0% and it is compounded quaterly (means when your money gains interest in a year). Regarding calculating interest in Excel, have a look at the EFFECT() function. Compound interest formula The compound interest formula is: FV = PV x (1 + r)^t Where PV is the Present Value (your investment now) and FV is the Future Value (the total money in future, after the investment). The formula for figuring compound interest on a one time deposit of principal seems to work perfect to me. For example, if you invested$1,000 at a 5 percent annual rate of return, after 10 years you would have $1,628. Formula 5; the derivation of Formula 5 is beyond the scope of the material you are covering. [To arrive at the interest amount you can further use the formula Interest = A - P ] Example: Let us assume that an amount of Rs. Compound Interest (A) = P [(1 + i) n \u2013 1] Where: P = Principal Amount, i = interest rate, n = compounding periods. Simple Interest. Although Microsoft excel does not include a function for determining compound interest , you can use following formula for this calculation. We can get the future value as$. Includes compound interest formulas to find principal, interest rates or final investment value including continuous compounding A = Pe^rt. When the numbers get bigger, and the years more numerous, though, there\u2019s that handy continuous compound interest formula we can use to calculate the impending value of a debt, loan, or deposit after a certain amount of time. Note: The formula for Compound Interest Calculator with Additional Deposits is a combination of: Compound Interest Formula \" P(1+r\/n)^(nt) \" and Future Value of Series Formula \" PMT \u00d7 (((1 + r\/n)^(nt) - 1) \u00f7 (r\/n)) \", as explained at The Calculator Site. If the amounts and dates are right, you know that the calculations will be correct. The interest can be compounded annually, semiannually, quarterly, monthly, or daily. We have compounded over 12 periods, 8 1\/3% over 12 periods. Using the future value calculator. You Borrow 20,000 TL At 10% Yearly Interest (compounded Annually). Introduction to Cashflow - Savings Plans In the first of three chapters covering the way in which interest rate affects cashflow we explore savings - but first we introduce some general ideas that apply equally to annuities and repayment loans. This example assumes that $1000 is invested for 10 years at an annual interest rate of 5%. Compound Interest in Excel Formula Compound interest is the addition of interest to the principal sum of a loan or deposit, or we can say, interest on interest. In this case, the amount value A;the principal P; the period of investment t;and the annual interest rate iare related by the formula A= P(1 + it):At what rate will$500 accumulate to $615 in 2. The concept of compound interest is the interest adding back to the principal sum so that interest is earned during the next compounding period. 2% compounded quarterly. The following represents the compound interest factor Formula: (1 + i) n, where n is the number of periods, i is the periodic rate of interest, and 1 represents one dollar since the formula results in a factor that is multiplied by the principle dollar amount. Calculate compound interest on an investment or savings. 12,100,100,100,100,1100) In the Excel formula, 0 indicates that there are no intermediate cash flows. There are at least three ways to calculate accrued interest in Excel while using the correct day count basis. If a person deposits$1,000 at 5 percent. In addition to the formula, you also can use Function to calculate the compound interest. Its something that you you probably read in your 7th or 8th standard or even earlier. An amount of $1,500. So for the maturity value of Fixed Deposit of Rs. Or a website that can do this for me. There are different ways to do compounding interest, but I've chosen to compound it daily, which means you take the annual Interest Rate and compute the period interest rate as (1 + Rate)^(Days \/ 365) - 1 where Days is the number of days since the last contribution. The drawback to earning compound interest is that the annual rates are not usually as high as accounts like term deposits where interest is not compounded monthly. 718, r the annual interest rate as a decimal and t the time in years. See also How to calculate compound interest for an intra-year period in Excel. Example: I have deposited$. Where: P = initial principal; r = interest rate as a decimal; t = number of years invested; n = number of times the money is compounded per year; A = final amount, including the initial principal and all interest earned over n years; Example. The formula for annual interest, including principal sum, is: A = P (1 + r\/n) (nt) Where: A = the future value of the investment\/loan, including interest. This is called compound interest. by Kristina Dunbar, UGA. where \"A\" is the initial amount, \" i \" is the interest rate per compounding period, \" N \" is Number of times or compounds in a year, and \" T \" is the number of periods you want to calculate. Calculate compound interest on an investment or savings. Compound Interest Formula in Excel. The interest offered on Deposit 1 is 5% compound interest. I like to know the excel formula for fixed deposit which compounding in certain period when entry and exit dates are provided. =PV*(1+r)^N Where PV = Present Value r = Interest Rate. pdf), Text File (. Specify the initial investment with your plans for future investments and details about the account you plan to invest in and map the progress of that investment over the years. In this article, we are going to learn the compound interest formula in Excel. 00 in year two. This Fixed Deposit (FD) Calculator helps you find out how much interest you can earn on an FD and the value of your invesment (Principal) on Maturity when compounding of interest is done on a monthly, quarterly, half-yearly or yearly basis. 2,000 on 11th Oct 2018 @ annual rate of 10% with quarterly compounding. Following is the formula for calculating compound interest when time period is specified in years and interest rate in % per annum. Uniform Series Compound-Amount Factor. Using this formula, Kathy should have a balance of $23,087. i = Interest Rate. Go to Excel -> Go to Fx -> Financial Formulas -> IRR. 75 at end of period as worked out. To see all four continuously compound interest formulas, (solved for total, principal, years and rate) click here. Compound interest (also called \u201ccompounding interest\u201d) is interest that is calculated on both the initial amount of a deposit or loan (also referred to as the \u201cprincipal\u201d) and on any interest previously accumulated on that amount. When you borrow money from a bank , you pay interest. 0% and it is compounded quaterly (means when your money gains interest in a year). n = number of times the interest is compounded per year. There the result is indeed, using 0. When your savings plans pays interest 365 days in a year and you make monthly deposits, use the NOMINAL and EFFECT functions first before using the FV function to calculate what the savings plan. 00; Year 3 would see a monthly deposit of$106. We have named the input cells. The Excel compound interest formula in cell B4 of the above spreadsheet on the right once again calculates the future value of $100, invested for 5 years with an annual interest rate of 4%. On the day you exit. Instead, I am going to take few examples and show you, how you would compute the same in Excel. 1,000 * (1 + 0. What this means is that, during each month, the balance gets multiplied by 1+I (it becomes the original amount plus the interest, which is I times the original amount) and then has W subtracted from it. e, 12 months. So, with your formula with an intial deposit of$5,000 and a monthly payment of 100 at 3 percent annual interest rate, at the end of 12 months you would have $6,368. Simple Interest (S. Since the interest is compounded annually, the one-year period can be represented by n = 1 and the corresponding interest rate will be i = 8% per year: The formula shows that the present value of$10,000 will grow to the FV. Fill out your original investment (p) in the investment column. A = P(1 + r\/n) nt. ??? For example, if it was $175,000 at 9 per-cent. The PV function returns the present value of an investment, which is the total amount that a series of future payments is worth presently. \u2026Interest shows how much an investment grows over time. The formula used to calculate compound interest is M = P(1 + i)n. Looking at the original interest rate and the number of times interest is compounded, investors will be able to multiply the true interest rate by the principal invested or financed and have an instant figure. 5% rate of interest. You can use Excel functions to calculate the maturity value of the of the monthly plan. Banks generally set repayments on loans and mortgages in equal payments over a fixed period of time. How much interest will be earned? Based on the formula, the total interest is$402. Extra money paid for using other's money is called interest. 4% (in cell L2) compounded quarterly (in cell L4) for 6 years (in cell L3). n = Number of compounding period which could be daily, annually, semi-annually, monthly or quarterly. Compound interest calculation example. Looking for some help making a compound interest calculator in excel similar to Bankrate where I can adjust the contribution frequency and the interest compound frequency. Compound interest excel formula with regular deposits (Using Excel FV Function) Say, you\u2019re going to run a savings scheme with one of your trusted banks. So if you deposit $100 for 1 year, you will receive$5 in interest. For Cumulative Deposit, the formula is based on compound interest, where: A = P (1+ r\/n) n*t A= Amount to be received. To find out your nominal rate of interest, you need to divide 5 by 100 which equals 0. In the examples shown above, the value in monthly compounding is highest. In this formula, A stands for the total amount that accumulates. Calculate the balance at the end of a certain number of years. We shall also discuss how to calculate future values of an investment on the basis of daily, monthly and yearly compounding interest rate. 5% interest rate, compounded daily, for the next year, the total interest for two years is $133. Friday, 11 January 2013 at 17:00:00 GMT+5:30. The basic Excel formula for compound interest is this: =PV*(1+R)^N PV is the present value. Once these values have been entered in any order, the. The interest rate and number of periods need to be expressed in annual terms, since the length is presumed to be in years. For example, in this formula the 17% annual interest rate is divided by 12, the number of months in a year. 04 Average Annual Interest Earned = Total Interest Earned \/ Time =$1,147. Compound Interest. P is principal or the original deposit in bank account. One of the easiest ways is to apply the formula: (gross figure) x (1 + interest rate per period). How much we can get if the plan A offers a true 10% interest rate? In this case, we're looking for the future value by performing FV function. Examples of Compound Interest Formula (With Excel Template) Let\u2019s take an example to understand the calculation of Compound Interest in a better manner. Simple Interest (S. Write a C program to input principle (amount), time and rate (P, T, R) and find Compound Interest. The interest on a loan or other fixed-income instrument where interest previously paid is included in the calculation of future interest. Compound Interest Calculator \u2013 Savings Account Interest Calculator Calculate your earnings and more Consistent investing over a long period of time can be an effective strategy to accumulate wealth. A more efficient way of calculating compound interest in Excel is applying the general interest formula: FV = PV(1+r)n, where FV is future value, PV is present value, r is the interest rate per period, and n is the number of compounding periods. The amounts have varied between $400,$800 and $150. The second is a self checking riddle WS that covers the same topic. Quantitative Reasoning Center. In simple interest, an interest rate of 5% that is earned by a$1000. In this formula, \"i\" is the annual interest rate, \"n\" is the number of years, \"P\" is the original deposit amount and \"W\" is the fixed annual withdrawal. You need to have Microsoft Excel 2007-2013 and Microsoft Windows to use it. 1500\/- is deposited in a bank for 6 years and paying an annual interest rate of 4. The annuity payment formula is used to calculate the periodic payment on an annuity. DePaul University. Fixed Deposit Interest Formula. Specify the initial investment with your plans for future investments and details about the account you plan to invest in and map the progress of that investment over the years. Calculating Interest and Excel Functions: Apply the Compound Interest Formula for monthly Compounding Interest. For the admin of a small micro credit organisation I developed an admin tool in excel. To calculate IRR manually, you need to use an excel spreadsheet and choose the financial formulas in excel. Also see: Simple Interest Tables Compound Interest Tables. 02% with rounding to the nearest penny being done just that once instead. Using the compound interest formula, calculate principal plus interest or principal or rate or time. 00, the amount will be R1 452. According to Figure 1, this means that type=0 (the default for the FV function). 01% for a savings account. Compound interest - meaning that the interest you earn each year is added to your principal, so that the balance doesn't merely grow, it grows at an increasing rate - is one of the most useful concepts in finance. Recurring Deposit Interest : The interest rates given by banks is generally above 8. Just remember that the type argument has to do. P = 1500, r = 4. xltx, and zip). Also see: Simple Interest Tables Compound Interest Tables. Introduction The basis of the time value of money is that an investor is compensated for the time value of money and risk. We can use Microsoft Excel to. Excel expert Paul McFedries shows how to use Excel 2016\u2019s core features to solve problems and get the answers you need. Assume 40 payments. Future Value = Present Value x (1 + Rate of Return)^Number of Years. Multiply the amount of the underpayment by the interest rate. 05,12), which would yield 0. Simple interest and Compound interest for the first year will be the same. Summary: Compound interest can work for you or against you. To calculate IRR manually, you need to use an excel spreadsheet and choose the financial formulas in excel. Compound Interest Formula. For this formula, P is the principal amount, r is the rate of interest per annum, n denotes the number of times in a year the interest gets compounded, and t denotes the number of years. Interest Formulae (Gradient Series) 2. We are constantly shown numbers which are stripped of context. 09; Year 4 would see. Its something that you you probably read in your 7th or 8th standard or even earlier. I like to know the excel formula for fixed deposit which compounding in certain period when entry and exit dates are provided. The interest is compounding every period, and once it's finished doing that for a year you will have your annual interest, i. Note: The formula for Compound Interest Calculator with Additional Deposits is a combination of: Compound Interest Formula \" P(1+r\/n)^(nt) \" and Future Value of Series Formula \" PMT \u00d7 (((1 + r\/n)^(nt) - 1) \u00f7 (r\/n)) \", as explained at The Calculator Site. Loan or Payment Amount. We compound for years which is months. Compound Interest = P * [ (1 + i)n - 1] P = Initial Principal. Discounting cash flow is one of the core principles of small business financing operations. If you need to, you can adjust the column widths to see all the data. To calculate compound interest in Excel, you can use the FV function. Add the result to the underpayment balance to get the amount you owe for the current day. The present value portion of the formula is the initial payout, with an example being the original payout on an amortized loan. The periodic interest payments promised to bond holders are computed as a fixed percentage of the bond\u2019s face value; this percentage is known as the coupon rate. The first deposit would occur at the end of the first year. Our original principal times 1 plus 100% divided by 12. Email: [email\u00a0protected] Calculate compound interest by Function in Excel. Find the nominal rate compounded seminnually for an investment of $500 which amounts to$588. If you save $100 a month at 5% interest (compounded annually) for 5 years, you'll have made$6,100 in deposits, and earned $836. Compound Interest is the interest which is computed as a percentage of revised principal, i. Daily Compound Interest = [Start Amount * (1 + Interest Rate) ^ n] - Start Amount. What this means is that, during each month, the balance gets multiplied by 1+I (it becomes the original amount plus the interest, which is I times the original amount) and then has W subtracted from it. I am looking for a way to calculate and display the amounts of a CD on a monthly basis, sort of like an amortization schedule for a mortgage. Using the future value calculator. Savings Withdrawal Help. 06,A1:A41,B1:B41) should return the remaining loan balance. 0, how can I calculate A=? Thanks in advance. Let \"A\" be the amount of each uniform payment. The NPER argument of 2*12 is the total number of payment periods for the loan. Doing so will calculate the amount that you'll have to pay in interest for each period. 2000 rupees will earn 120 rupees as interest at 6 per cent per annum if invested in a Bank for one interest rate is always quoted as per 100 rupees or currency. Generally, when someone deposits money in the bank the bank pays interest to the investor in the form of quarterly interest. The basic formula for Compound Interest is: FV = PV (1+r) n. Formula for compound interest WITH annual deposits. Compound interest implicates adding the interest income to your investment, and then reinvesting it, every time, as opposed to withdrawing it. Rate is the interest rate per period. In simple interest, an interest rate of 5% that is earned by a$1000. Calculation of the effective interest rate on OVDP in Excel. First find NPV: Then compound NPV for 5 years: =NPV(0. Figure out monthly mortgage payments. 5k, what compound interest did I earn?. From there you can solve for the future value. Interest Formulae (Gradient Series) 2. The file is not working for if we change the month or date or year of deposit or maturity. The concept of compound interest is the interest adding back to the principal sum so that interest is earned during the next compounding period. Calculating interest and repayments on a loan. Use this calculator to determine an Internal Rate of Return (IRR). Number of. Consider that at present, I am getting yearly 10% interest rate on my deposits. 1300 x 708. Regular Deposits. From the Compound Interest formula (shown above) we can compound \"n\" periods using. Compound Interest Calculator. 97% interest was. This example assumes that $1000 is invested for 10 years at an annual interest rate of 5%, compounded monthly. I added the modified formula for that case. We want to do some compound interest calculation, so we need to enter a balance (put this in cell B1, to the right of where we labeled \"Balance\") and an interest rate (in B2, just to the right of the \"Interest Rate\" label). The Interest amount is this adjusted rate times the Start Balance. In this article, we are going to learn the compound interest formula in Excel. Understanding Compound Interest. \u2026If we're assuming a monthly compounding,\u2026then 120 periods is 120 divided by. The present value portion of the formula is the initial payout, with an example being the original payout on an amortized loan. The user can choose whether deposits are made at the beginning or at the end of the period. Hi, is there a way to find out the annual compound interest (A), from the interest (I) over (Y) number of years? For example: If interest over 5 years is 300%, so I=5, Y=3. You get interest on the interest from previous years. The natural logarithm is the logarithm with base e (Euler number is approximately 2. Teaser raters on adjustable mortgages, APR rates on credit cards which don't highlight other fees or the compounding effects, and secured credit cards which have an effective APR of above 100% after paying for the membership fee - and, what's worse, is that on a secured credit card the cardholder is. Looking for some help making a compound interest calculator in excel similar to Bankrate where I can adjust the contribution frequency and the interest compound frequency. e, 12 months. For the formula for compound interest, just algebraically rearrange the formula for CAGR. I set up the admin as a kind of database system. 24%, if the interest is compounded quarterly. Through the template of excel, you only need to fill the column and you are set to calculate. After 10 years you will have: Initial Investment. When there are no intermediate, external transactions, such a rate of return measures the percentage growth during the period. Popular Course in this category. Example: I have deposited$. The process of figuring periodic interest over time, or compound interest, can be turned into a formula you can enter into a spreadsheet or programmable calculator. If you save $100 a month at 5% interest (compounded annually) for 5 years, you'll have made$6,100 in deposits, and earned $836. In line with Kailua's request above, I am seeking an excel formula for compounding interest with a STARTING monthly deposit of$100 invested at 7% per year (compounded annually), and increasing the monthly deposit by 3% per year (i. Compound Interest Formula. Click on the tab with the worksheet titled Activity 1 to begin. No longer will you need to wonder if an investment offering a 6% return, compounded daily is better than an investment offering a 7% return. Although Microsoft excel does not include a function for determining compound interest , you can use following formula for this calculation. Summary: Compound interest can work for you or against you. The annuity payment formula is used to calculate the periodic payment on an annuity. Step 2: Keep the cursor on the cell where you want the mean to be calculated (say B7) and click on the Function icon under Insert Menu. The deposit is for 5 years. Similarly, you can also calculate Daily compound interest by below formula. i = Interest Rate. Friday, 11 January 2013 at 17:00:00 GMT+5:30. Fixed Deposit Calculator: This Fixed Deposit Calculator (FD Calculator) tells you the Maturity Value of your invesment (Principal) when compounding of interest is done on a Monthly, Quarterly. Optionally, you can specify periodic contributions or withdrawals and how often these are expected to occur. 04 \/ 4 = $286. The interest is compounding every period, and once it's finished doing that for a year you will have your annual interest, i. The interest offered on Deposit 1 is 5% compound interest. Compound Interest is the interest which is computed as a percentage of revised principal, i. Simple and Compound Interest. ) Base Calculations on a 360-Day Year: Base Calculations on a 365-Day Year: Principal: Nominal Interest Rate: Total Number of Days: Compound on a Daily Basis Compound on a Weekly Basis Compound on a Bi-Weekly Basis. This would make the above formula for the final value of an investment after n interest periods look like this: S = P(1 + r\/k) n. Savings Withdrawal Help. Step 1: Open a file in an excel containing sample data to find the arithmetic mean. 24% per annum compounded monthly, then the amount of interest credited to the account at the end of the month is the average daily balance during the past month (taking into account all deposits and withdrawals made that month) times 0. COMPOUND INTEREST (methodology): Compound interest takes into account the time-value-of-money. So far, no luck. While this formula may look complicated, this Future Worth Calculator makes the math easy for you by not only computing the variables present in this equation, but it also allows investors to account for recurring deposits, annual interest rates, and taxes. A small monthly deposit over a couple of decades will produce incredible results even with a conservative interest rate. Just enter a few data and the template will calculate the compound interest for a particular investment. 2,000 on 11th Oct 2018 @ annual rate of 10% with quarterly compounding. 5%, tax on income to$70,000 per year decreased from 33% to 30%, and tax on income over $70,000 decreased from 38% to 33%. Calculate compound interest on an investment or savings. In Excel, here is a formula that can help you to quickly calculate the compound interest. Compound interest formula. For simplicity in this example, this row represents the 2nd row of a spreadsheet, and the column headings start with Investment in column A. Then, interest kicks in and you will have$21,139. Once these values have been entered in any order, the. WHAT FORMULA SHOULD I USE TO FIND OUT RATE OF INTEREST. A = P (1 + r \u2044 n) nt. However, that $5 will probably be worth less at the end of the year than it would have been at the beginning. P = Principal amount r = interest rate n = number of compunding periods in a year t = number of years and a = amount accrued at end of t years. Usually banks compound interest on quarterly basis in fixed deposit. If you're refinancing existing debt, you may want a tool to compare your options based on how far you've already come with repayment. Quarterly: When interest is compounded quarterly, we pass 4 to the method. ) Base Calculations on a 360-Day Year: Base Calculations on a 365-Day Year: Principal: Nominal Interest Rate: Total Number of Days: Compound on a Daily Basis Compound on a Weekly Basis Compound on a Bi-Weekly Basis. A (final amount), P ( principal), r ( interest rate) or T (how many years to compound) t (time in years) Integer Tenths Hundredths Thousandths Max Accuracy. Credit union Web site says interest is compounded and credited quarterly; a person from the credit union said interest is accrued daily and compounded quarterly. So far, no luck. The PV function returns the present value of an investment, which is the total amount that a series of future payments is worth presently. Annual interest rate. For example, to find out how much would$10,000 grow in 10 years with an annual interest rate of 5% and compound monthly, we will plugin the variables to the compound. Calculate Accrued Interest Using the AccrInt Function. If I wanted to deposit $1000 at the beginning of each year for 5 years, the FV function in Excel allows me to calculate the result as =FV(4%,5,-1000,,1) where type=1. Compound interest arises when interest is added to the principal of a deposit or loan, so that, from that moment on, the interest that has been added also earns interest. So the CAGR formula is\u2026 C20: =(B17\/B3)^(12\/14)-1. I have something like this: Principal. Don\u2019t bother with dollar signs or commas \u2013 the calculator will take care of those automatically. The basic Excel formula for compound interest is this: =PV*(1+R)^N PV is the present value. The method used to solve the problem will depend on what we are trying to find. com) Excel pound Interest Template via (newcv. The next year, with compound interest applied to the R1 320. You can choose the interest rate and the moment its generated income will be cashed (monthly, quarterly, semi-annually or yearly), which is also known as compound interest. What is the balance after 6 years? Using the compound interest formula, we have that. When the numbers get bigger, and the years more numerous, though, there\u2019s that handy continuous compound interest formula we can use to calculate the impending value of a debt, loan, or deposit after a certain amount of time. The calculator can solve annuity problems for any unknown variable (interest rate, time, initial deposit or regular deposits). But i don\u2019t know what INTEREST RATE I HAVE GOT. This is not the case when there are intermediate transactions. Note These formulas assume that the deposits (payments) are made at the end of each compound period. The formula for annual interest, including principal sum, is: A = P (1 + r\/n) (nt) Where: A = the future value of the investment\/loan, including interest. Notice that the output, S, is an exponential function of n. In order to understand this better, let us take the help of an example: Sania made an investment of Rs 50,000, with an annual interest. Instantaneous and Compounded Annual Rates for Interest In finance there are two ways to express rates such as interest rates. How much money is in the bank after for 4 years? After four years, there will be 3248. 2 - Fill out the white input boxes. Enter the years (0-5) in cells A2. For all results, including not entering a guess percentage in the YTD formula, all results were exactly the same. The compounding of interest grows your investment without any further deposits. As you can see, even small deposits to a savings account can add up over time. The premiums I have paid have been irregular (six years of paying regular deposits with two years premium holiday in between). Formula To Calculate Compound Interest. A unique feature of this calculator is the option to select a random interest rate, to simulate fluctuation in the market. Dollar-cost averaging. Now, let's say you deposited the same amount of money on a bank for 2 years at 3% annual interest compounded annually. 0% and it is compounded quaterly (means when your money gains interest in a year). Example: I have deposited$. The future value calculator can be used to determine future value, or FV, in financing. The first deposit would occur at the end of the first year. You may be able to use the calculator with other applications that can open and read XLS spreadsheets, but this has not been tested. Generally, when someone deposits money in the bank the bank pays interest to the investor in the form of quarterly interest. If I wanted to deposit $1000 at the beginning of each year for 5 years, the FV function in Excel allows me to calculate the result as =FV(4%,5,-1000,,1) where type=1. I need a formula for Compound interest excel formula with regular deposits that you can set the intervals of deposits eg week or month. An online financial calculator to calculate the maturity value of your recurring deposit. See also How to calculate compound interest for an intra-year period in Excel. techcommunity. Now consider an irregular cash flow stream (where CFs can take on any value). - [Instructor] When you analyze cash flows\u2026whether in Excel, or using another tool,\u2026you need to account for two important effects. And, the formula in excel for yearly compound interest will be. Cells B2:B41 are payment dates. The formula is A=P(1+r\/n) nt. The compounding of interest grows your investment without any further deposits. Since the interest is compounded annually, the one-year period can be represented by n = 1 and the corresponding interest rate will be i = 8% per year: The formula shows that the present value of$10,000 will grow to the FV. Example: Kevin deposits $3,000 in a 1-year certificate of deposit (CD) at 5. 96% interest was$46,283, and amount at maturity for 2. Doing so will calculate the amount that you'll have to pay in interest for each period. Compound Interest Formula for a Series of Payments For both loans and savings, we typically want to include a series of payments or deposits in our calculation, such as depositing 100 each month for 3 years. n = number of times the interest is compounded per year EXAMPLE Suppose you intend to invest Rs 1,00,000 for 10 years at an interest rate of 10 per cent and the compounding is annual. As you can see, even small deposits to a savings account can add up over time. Or a website that can do this for me. Include regular monthly deposits and\/or an annual deposit. Compound Interest Calculator. What will be my future value on 8th Sept 2019. by Kristina Dunbar, UGA. can anyone tel me how to do monthly calculation. Since our interest rate is compounded monthly our time needs to be in the same units thus, months will be the units of time. You must provide the amount of each deposit, the frequency of the deposits, the term in months, and the nominal interest rate. An annuity is a series of periodic payments that are received at a future date. The PV function returns the present value of an investment, which is the total amount that a series of future payments is worth presently. That is the part i can not remember. A unique feature of this calculator is the option to select a random interest rate, to simulate fluctuation in the market. 4$Required difference$=306. How much money is in the bank after for 4 years? After four years, there will be 3248. The Simple Interest Calculation Formula is: Deposit Amount (in dollars and cents) x Interest Rate x Time On Deposit (in years) = Total Earned Interest Enter the amount of the savings deposit and the simple interest rate. Principal Amount (P) = Rs. I like to know the excel formula for fixed deposit which compounding in certain period when entry and exit dates are provided. Calculate its simple interest and compound interest. It has to do with interest rates, compound interest, and the concepts of. TValue software calculations are based on embedded algorithms, not writing formulas. The formula for annual interest, including principal sum, is: A = P (1 + r\/n) (nt) Where: A = the future value of the investment\/loan, including interest. How banks calculate interest on fixed deposits. The formula for interest compounded annually is FV = P (1+r)n, where P is the principal, or the amount deposited, r is the annual interest rate, and n is the number of years the money is in the bank. Formula from book where i = r \u00f7 t and n = t \u00d7 c. 00 of interest. The interest is compounded every quarter which means 4 times in a year. The longer you leave the money where it is,. Annuity problems require the input of 4 of these 5 values:. This Online Bank Deposit Interest Calculator is specially programmed to calculate the Total Maturity Amount and the Total Interest based on the selection of Monthly, Quarterly, Semi-annually and Annually Compound terms, interest rate and total time period of the deposited money in the Bank. how to calculate recurring deposit in monthly basis? M = ( R * [(1+r)n - 1 ] ) \/ (1-(1+r)-1\/3) M is Maturity value R is deposit amount r is rate of interest n is number of quarters if i take 'n' as 4(no of Quarters) for 1 year its showing yearly Maturity value. The present value portion of the formula is the initial payout, with an example being the original payout on an amortized loan. 0083, into the formula as the rate. Interest can be compounded, of course. We compound for years which is months. 085\/365) 10\u00d7365 = Rs. The nominal interest rate is the stated interest rate. Can you tell me the base formula for compound monthly interest rates but monthly, bi weekly and 24 payment per year. In line with Kailua's request above, I am seeking an excel formula for compounding interest with a STARTING monthly deposit of $100 invested at 7% per year (compounded annually), and increasing the monthly deposit by 3% per year (i. This is called compound interest. This formula can be used to calculate maturity value but I want the formula(to be used in Excel) to calculate the rate of interest if maturity value is known in case of RD. at a compound annual interest rate of 8%. Quantitative Reasoning Center. This example assumes that$1000 is invested for 10 years at an annual interest rate of 5%, compounded monthly. Assuming an annual interest rate on your deposit, divide the rate by the number of compounding periods. n = Number of compounding period which could be daily, annually, semi-annually, monthly or quarterly. , you\u2019d have $30,000 in 5 years. Use the penalty interest calculator to find out the amount of interest, if an invoice is paid e. He starts with explaining the basic concepts like principle which is the amount you borrow and the rate of interest or annual percentage rate (APR), which is the rate at which you pay the interest up on the borrowed principle. \u2026You will often divide that by the number of times\u2026the interest is compounded during the year. =Amount borrowed - FV (interest rate\/12, number of months elapsed,monthly payment) Easier to do than read! There is probably a better method but that should get you going. Professor Albert Einstein once said, \"Compounding interest is the most powerful force in the universe\". The SFF is typically used to determine how much must be set aside each period in order to meet a future monetary obligation. The term or months is the time that the money will be in the CD account accumulating the percentage of interest. This formula is also used in Microsoft Excel to calculate the Future Value (FV). jjasso5 Excel 2007-2010 Future Value Formula (Investingments). The interest is compounded every quarter which means 4 times in a year. We will need to convert our number of years into number of months by multiplying it by 12. Excel\u2019s Internal Rate of Return (IRR) function is an annual growth rate formula for investments that pay out at regular intervals. FV is the amount of money the depositor would have after n years, or the future value of that investment. These two pieces of data are input into your compound interest table template. \u2026When you want to calculate the effects of interest\u2026you use the following formula. Multiply the principal amount by one plus the annual interest rate to the power of the number of compound periods to get a combined figure for principal and compound interest. Then determine the length of the deposit time period in the units of compound calculation. \u2026Next, we have the number of periods. The formula for compound interest is. Compound Interest If you want to find t, the number of years, enter values for F, P and r. The money borrowed or lent out for a certain period is called the principal or the sum. 5% rate of interest. It is the result of reinvesting interest, rather than paying it out, so that interest in the next period is then earned on the principal sum plus previously accumulated interest. , you\u2019d have$30,000 in 5 years. Generally, when someone deposits money in the bank the bank pays interest to the investor in the form of quarterly interest. 22 Mauchly, Irvine, California 92618. 10,000 with the bank every month for 24 months, and the bank pays you interest on Rs. 24% per annum compounded monthly, then the amount of interest credited to the account at the end of the month is the average daily balance during the past month (taking into account all deposits and withdrawals made that month) times 0. Term \u2013 the time frame you are going to invest money. The interest earned grows, because the amount of money it is applied to grows with each payment of interest. A savings calculator is a great tool to calculate compound interest and estimate the value of savings over a period of time. interest that is not compounded), you can use a formula that multiples principal, rate, and term. Nper is the total number of payment periods in an annuity. at a compound annual interest rate of 7%. Amount of savings = Principal originally invested * (1 + decimal annualized rate of interest \/ number of compounding periods per year) times per year interest is compounded * years invested. If you are calculating for a single year, leave off the T, which stands for the number of years. The table results in a future value of $5,402. To simplify the process, we have created a simple and easy Compound Interest Calculator Excel Template with predefined formulas. You Borrow 20,000 TL At 10% Yearly Interest (compounded Annually). IRR (Internal Rate of Return) is the most widely used financial indicator while assessing return on an investment or a project. Note that the calculator bases its calculations on 360-day years to accommodate daily compounding for monthly, semi-monthly, quarterly, and annual deposits, so please allow for weekly and bi-weekly annual deposit differences (52 weeks and 26 bi-weeks each add up to 364 days). Uniform Series Compound-Amount Factor. M is the final amount including the principal, P is the principal amount (the original sum borrowed or invested), i is the rate of interest per year, and n is the number of years invested. The amount on deposit at the end of the first year is found by the simple interest formula, with t = 1. It has to do with interest rates, compound interest, and the concepts of. This form calculates the future value of an investment when deposits are made regularly. In closing, notice that this formula uses an insight that is always useful to keep in mind about typical annuity calculations like this\u2026 If you were to get a new loan for the current balance due on an existing loan, for the number of months remaining on the existing loan, and for the same interest rate, your payments would be the same as your. Suppose you deposit$1000 in a bank which pays 5% interest compounded daily, meaning 365 times per year. The longer you leave the money where it is,. On the day you exit. We will explain compound interest formula excel sheet with some of the examples. This Fixed Deposit (FD) Calculator helps you find out how much interest you can earn on an FD and the value of your invesment (Principal) on Maturity when compounding of interest is done on a monthly, quarterly, half-yearly or yearly basis. 5%, over 10 year(s): Total Value = $1750 Total Interest =$750. To calculate compound interest in Excel, you can use the FV function. Your initial deposit earned $30 as interest. What would be the return?. After one year, you have$100 in principal and $10 in interest, for a total base of. Here is a future value calculator that uses continously compounded interest:. If the interest rate (APR) is, say, 0. P = 120 r = 5% n = 2 (semi-annual compounding) t = 12 years. This formula gives simple interest earned at the end of whole years. If you start with$10,000 in a savings account earning a 7% interest rate, compounded annually, and make $100 deposits on a monthly basis, after 20 years your savings account will have grown to$89,737. The simple interest calculator below can be used to determine future value, present value, the period interest rate, and the number of periods. An online financial calculator to calculate the maturity value of your recurring deposit. 393913886691 would be the value at maturity. Created Date: 6\/20\/2007 10:06:07 PM Company. A = P11 + r # 1 2= P11 + r If the deposit earns compound interest, the interest earned during the second year is paid on. 3%, compounded quarterly. Optionally, you can specify periodic contributions or withdrawals and how often these are expected to occur. how to calculate recurring deposit in monthly basis? M = ( R * [(1+r)n - 1 ] ) \/ (1-(1+r)-1\/3) M is Maturity value R is deposit amount r is rate of interest n is number of quarters if i take 'n' as 4(no of Quarters) for 1 year its showing yearly Maturity value. 2 - Fill out the white input boxes. We will explain compound interest formula excel sheet with some of the examples. The formula for figuring compound interest on a one time deposit of principal seems to work perfect to me. In order to understand this better, let us take the help of an example:. $100 deposit x 5% interest x 1 year term =$5. So the CAGR formula is\u2026 C20: =(B17\/B3)^(12\/14)-1. Create an excel document to compute compound interest. Estimate the total future value of an initial investment or principal of a bank deposit and a compound interest rate. The interest paid by the bank is 5%. We compound for years which is months.","date":"2020-05-29 16:46:47","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\": 1, \"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\": 0.43719202280044556, \"perplexity\": 1031.9271213325019}, \"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-2020-24\/segments\/1590347405558.19\/warc\/CC-MAIN-20200529152159-20200529182159-00341.warc.gz\"}"} | null | null |
Malayalam film star Dileep arrested for kidnap and assault of actress
'I am not scared of anyone, as I have not committed any crime,' says Dileep as he was being transferred to jail where he will be in judicial custody for 14 days
Indian actor Dileep in a still from the movie 'Mr Marumakan'. Dileep was arrested on July 10, 2017 over his suspected connection to the kidnap and assault of an Indian actress. Courtesy Varnachitra Big Screen
The Malayalam film star Dileep has protested his innocence after being arrested in connection with the abduction and sexual assault of an actress in February.
"I am not scared of anyone, as I have not committed any crime… I was trapped," Dileep, who uses only one name, said on Tuesday, a day after his arrest, as he was being transferred to the Aluva sub-jail where he will remain in judicial custody for 14 days.
No charges have been filed against the actor yet. Judicial custody enables police officials to continue questioning him without giving him any chance to flee their jurisdiction. A magistrate will consider his bail application on Wednesday.
Before his arrest, Dileep had been extensively questioned as part of the police's probe into a conspiracy to abduct and molest an actress, who remains unnamed under Indian law to protect her privacy. The actress was kidnapped by a group of men while she was driving from Thrissur to Kochi in February.
In her police complaint, she said she was molested in a moving vehicle, and that photographs and videos were taken of her, before she was let off after two hours. The police arrested two other men in February and charged them with carrying out the crime.
Protests against Dileep broke out in Kochi and Aluva as news of his arrest emerged. Outside the Aluva Police Club, where Dileep was interrogated and lodged for a night on Monday, crowds applauded as television reporters delivered their pieces to camera about the arrest.
In Kochi, a crowd gathered outside a restaurant owned by Dileep, forcing the police to shut the restaurant down for the evening. Elsewhere in the city, a group of Bharatiya Janata Party cadre burned an effigy of the actor, while another group of Congress party workers conducted a protest march to D-Cinemas, a multiplex owned by Dileep.
Crowds raised slogans against Dileep outside the magistrate's residence where he was produced on Tuesday, and then near the jail in Aluva. "Welcome to Central Jail!" the crowds shouted, a reference to the actor's 2016 movie of that name.
Also on Tuesday, Dileep was formally expelled from the Association of Malayalam Movie Artists, a trade body that came under heavy criticism for choosing to defend him earlier in the investigation.
Mumbai attacks 'architect' arrested in Pakistan
US puts sanctions on Myanmar military leader over Rohingya abuses
Hong Kong: no end in sight as protests expand to oppose China
Miracle bus smash escape caught on CCTV
Undersea Bali quake causes panic and damage but no deaths
Special report: Inside China's 'boarding schools' for adult Muslims
The dogs who sniff out explosives in Kabul
Mumbai building collapse: many dead and dozens trapped
Landmine kills 11 pilgrims in southern Afghanistan | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 3,118 |
Spread the love to auto culture
Australian Supercars
Marathone 24H
WTCR (WTCC&TCR)
We're back in Ipswich at a circuit that traditionally provides cracking racing when TCM comes to town.
Location: TWillowbank, Ipswich, Queensland. 🇦🇺
map of National circuit
Length: 3.12 km
Turns: 6
Previous TCM lap record: John Bowe / Holden Torana SL/R 5000 / 1:16.3951 / 30 July 2017
Categories: Races TCM
AUS V6 | #9 Queensland Raceway – 2019
Event: Ipswich SuperSprint
V8 Supercars lap record in 2017: James Courtney / Holden VF Commodore / 1:09.7225 / 29 July 2017
V8 Supercars lap record in 2018: James Courtney / Holden ZB Commodore / 1:09.6591 / 21 July 2018
Categories: Races Australian Supercars
Categories: Races Formula 1
Location: Hungaroring, Mogyoród, Hungary 🇭🇺
In 1986, it became the location of the first Formula One Grand Prix behind the Iron Curtain. Bernie Ecclestone wanted a race in the USSR, and a Hungarian friend recommended Budapest. They wanted a street circuit similar to the Circuit de Monaco to be built in the Népliget – Budapest's largest park – but the government decided to build a new circuit just outside the city near a major highway.
F1 Race lap record : 1:19.071 (🇩🇪 Michael Schumacher, Ferrari F2004, 2004)
F1 lap record: 1:16.276 (🇩🇪 Sebastian Vettel, Ferrari SF70H, 2017)
AUS V6 | #10 The Bend Motorsport Park – 2019
Event: The Bend SuperSprint
Location: at Tailem Bend, South Australia, 100 kilometres south-east of the state capital, Adelaide.. 🇦🇺
Opened: 13 January 2018
Inspired by some of the world's most iconic race circuits, The Bend's motor racing circuit will be Australia's only circuit to comply with the latest FIA Grade 2 and FIM Category A standards. More about it.
Map of Int. Circuit for Supercars
Location: Circuit de Spa-Francorchamps, Stavelot, Belgium
Since 1922 !!! Infamous for its "Red Water" or "Eau Rouge" fast speed corner
Modern circuit with new pit lane and Bus Stop Chicane (2007–present)
Eau Rouge and Raidillon in 1997, with a maximum gradient in excess of 18%
Map of the old and new (2004–2006) Spa circuits, overlaid
Location: Autodromo Nazionale Monza, Monza, Italy
Since 1922 !!!
"Gran Prix movie" (1966) must to watch
Length of modern configuration: 5.793 km (3.600 mi)
Banking above straight after Curva del Serraglio
AUS V6 | #11 Pukekohe Park Raceway – 2019
Event: Auckland SuperSprint
Location: 40 km south of Auckland City, Pukekohe, North Island, New Zealand. 🇳🇿
Opened: 1963. First V8 race in 1996 as Auckland SuperSprint.
The Raceway was opened in 1963 as a permanent track, replacing Ardmore (an aerodrome) as the host circuit of the New Zealand Grand Prix. Annually for several years, the mainly European based Grand Prix drivers such as Stirling Moss, Graham Hill, Jim Clark and Jackie Stewart, would head downunder for a relaxed Tasman Series during the European winter.
Map of current circuit:
V8 Supercars lap record: Jamie Whincup / Holden VF Commodore / Q time: 1:02.5148 / 4 november 2017
Night race!
Location: Marina Bay Street Circuit, Singapore
Singapore Marina Bay Street Circuit
Location: Sochi, Krasnodar Krai, Russia
🏁 TCM | #6 Bathurst – 2019
6th race of TCM Australia in 2019
Location: Bathurst, New South Wales, Australia. 🇦🇺
Opened: 17 April 1938
It is situated on a hill with the dual official names of Mount Panorama and Wahluu. Track is technically a street circuit, and is a public road, with normal speed restrictions, when no racing events are being run, and there are many residences which can only be accessed from the circuit. The National Motor Racing Museum is located next to the Mount Panorama Circuit.
Length: 6.213 km
TCM lap record: John Bowe / Holden LH Torana SL/R 5000 / 2:17.4462 / 8 October 2017
About TCM: Setting itself apart from traditional Historic racing, the Touring Car Masters pioneered a new class of racing by introducing innovative technical regulations aiming to improve safety, reliability, cut running costs and provide an exciting and entertaining race package while maintaining the visual appeal of its period race cars.
The series has proved popular with fans for the nostalgia value and also for the mix of eligible vehicles with the majority of competing vehicles being Australian or American V8s or even classic 911s.
🏎 F1 #17 Japanese GP 🇯🇵 – 2019
Location: Suzuka Circuit, Mie Prefecture, Japan 🇯🇵
Since 1962. One of the most interesting classical track now.
Race lap record: 1:31.540 (🇫🇮Kimi Räikkönen, McLaren, 2005)
🏁 AUS V6 | #12 Mount Panorama (Bathurst)- 2019
Event: Bathurst 1000
previous V8 Supercars lap record: David Reynolds / Holden VF Commodore / 2:06.2769 / 9 October 2016
V8 Supercars lap record in 2018:David Reynolds / Holden ZB Commodore / 2:06.1492 / 7 October 2018
🏁 AUS V6 | #13 Surfers Paradise – 2019
Event: Gold Coast 600
Location: Surfers Paradise, in Queensland, Australia. 🇦🇺
Opened: 15 March 1991 for the 1991 Gold Coast IndyCar Grand Prix.
BEST TRACK IN AUSTRALIAN CALENDAR! Temporary street circuit on the Surfers Paradise. Back in better days held INDY car (CART) and V8 races.
V8 Supercars lap record: David Reynolds / Ford FG Falcon / 1:10.0499 / 27 October 2013
Map of original Champ Cart circuit 1991-2009 (to compare):
V8 Supercars lap record: Garth Tander/ Holden VE Commodore / 1:49.8352 / 21 October 2007
🏎 F1 #18 Mexican GP 🇲🇽 – 2019
Location: Autódromo Hermanos Rodríguez, Mexico City, Mexico 🇲🇽
Moder layout was made in 1986
🏎 F1 #19 US GP 🇺🇸 – 2019
Nov 2 @ 12:34 – Nov 3 @ 13:34
Location: Circuit of the Americas, Elroy, 19 km south of Austin, Texas 🇺🇸
Length: 3.427 mi (5.513 km)
TCM | #7 Sandown Raceway – 2019
Nov 8 – Nov 10 all-day
A new home for a season finale: We can't wait to get back to the Home of Horsepower for our 2019 showdown.
Location: Springvale in Melbourne, Victoria,~ 25 kilometres south east of the city centre. 🇦🇺
Opened: 11 March 1962
Sandown Racecourse was first built as a horse racing facility, dating back into the 19th century, but closed in the 1930s in a government run rationalisation program. Redevelopment began not long after World War II. A bitumen motor racing circuit was built around the outside of the proposed horse track (which was not completed until 1965) and was first opened in 1962 and held the race which became the Sandown 500 for the first time in 1964.
TCM lap record: John Bowe / Holden LH Torana SL/R 5000 / 1:15.1481 / 3 April 2016
AUS V6 | #14 Sandown Raceway – 2019
Event: Sandown 500
V8 Supercars lap record: Chaz Mostert / Ford FG X Falcon / 1:09.0987 / 16 September 2017
🏎 F1 #20 Brazilian GP 🇧🇷 – 2019
Nov 16 – Nov 17 all-day
Location: Autódromo José Carlos Pace, São Paulo, Brazil 🇧🇷
Since: 1940. Contract was prolonged until 2022
AUS V6 | #15 Newcastle Street Circuit – 2019
Event: Newcastle 500
Location: Newcastle, New South Wales, Australia. 🇦🇺
Opened: 24 November 2017
Quotes of drivers about track:
Shane van Gisbergen: "What a crazy track. It's really cool. It's going to be hard work, 250km, limited passing… it's going to be all about qualifying."
Scott McLaughlin: "It's full on […] It's so technical. You've got to try and work out one corner, you're sacrificing one corner to help you regain in three corners time. It's really bumpy too in places. It has everything a driver wants."
V8 Supercars lap record: David Reynolds / Holden VF Commodore / 1:10.6403 / 26 November 2017
🏎 F1 #21 Abu Dhabi GP 🇦🇪 – 2019
Nov 30 – Dec 1 all-day
Location: Yas Marina Circuit, Abu Dhabi, UAE 🇦🇪
Fastest lap: S. Vettel – Red Bull Racing – 1:40.279 – 2009
2018 🇬🇧 Lewis Hamilton / Mercedes
2017 🇫🇮 Valterri Bottas / Mercedes
2015 🇩🇪 Nico Rosberg / Mercedes
2013 🇩🇪 Sebastian Vettel / Red Bull-Renault
2012 🇫🇮 Kimi Räikkönen / Lotus-Renault
2011 🇬🇧 Lewis Hamilton / McLaren-Mercedes
Lanzante Porsche 930 TAG turbo' 2019
Circuito de la Sierra (original 🎮 track) | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 3,768 |
{"url":"https:\/\/www.beatthegmat.com\/the-volume-of-a-rectangular-box-x-inches-wide-t300847.html","text":"\u2022 Free Practice Test & Review\nHow would you score if you took the GMAT\n\nAvailable with Beat the GMAT members only code\n\n\u2022 Free Veritas GMAT Class\nExperience Lesson 1 Live Free\n\nAvailable with Beat the GMAT members only code\n\n\u2022 Award-winning private GMAT tutoring\nRegister now and save up to $200 Available with Beat the GMAT members only code \u2022 Get 300+ Practice Questions 25 Video lessons and 6 Webinars for FREE Available with Beat the GMAT members only code \u2022 5 Day FREE Trial Study Smarter, Not Harder Available with Beat the GMAT members only code \u2022 FREE GMAT Exam Know how you'd score today for$0\n\nAvailable with Beat the GMAT members only code\n\n\u2022 1 Hour Free\nBEAT THE GMAT EXCLUSIVE\n\nAvailable with Beat the GMAT members only code\n\n\u2022 5-Day Free Trial\n5-day free, full-access trial TTP Quant\n\nAvailable with Beat the GMAT members only code\n\n\u2022 Magoosh\nStudy with Magoosh GMAT prep\n\nAvailable with Beat the GMAT members only code\n\n\u2022 Free Trial & Practice Exam\nBEAT THE GMAT EXCLUSIVE\n\nAvailable with Beat the GMAT members only code\n\n# The volume of a rectangular box x inches wide\n\ntagged by: VJesus12\n\n### Top Member\n\n#### The volume of a rectangular box x inches wide\n\nThe volume of a rectangular box x inches wide, y inches long, and z inches high is 108 in^3. If x\u2264y\u2264z, what is x + yz?\n\n(1) The area of the largest face of the box is 36 in^2.\n(2) The longest edge of the box is 9 inches.\n\nThe OA is the option A .\n\nExperts, could you show me how to show that the A is the correct answer? I'd be thankful for your help.\n\n### GMAT\/MBA Expert\n\nGMAT Instructor\nJoined\n22 Aug 2016\nPosted:\n1394 messages\nFollowed by:\n26 members\n470\nVJesus12 wrote:\nThe volume of a rectangular box x inches wide, y inches long, and z inches high is 108 in^3. If x\u2264y\u2264z, what is x + yz?\n\n(1) The area of the largest face of the box is 36 in^2.\n(2) The longest edge of the box is 9 inches.\n\nThe OA is the option A .\n\nExperts, could you show me how to show that the A is the correct answer? I'd be thankful for your help.\nWe are given that the volume of a rectangular box x inches wide, y inches long, and z inches high = 108 in^3 if x \u2264 y \u2264 z.\n\nThus, xyz = 108\n\nWe have to get the value of x + yz.\n\nLet's take each statement one by one.\n\n(1) The area of the largest face of the box is 36 in^2.\n\nThe largest face of the box would be yz since we are given that x \u2264 y \u2264 z.\n\nThus, yz = 36. We already know that xyz = 108. Thus, x = (xyz) \/ (yz) = 108\/36 = 3 in\n\nThus, x + yz = 3 + 36 = 39. Sufficient.\n\n(2) The longest edge of the box is 9 inches.\n\n=> z = 9. This can't help. We can't the value of x + yz. Insufficient.\n\nHope this helps!\n\n-Jay\n_________________\nManhattan Review GMAT Prep\n\nLocations: New York | Singapore | Doha | Lausanne | and many more...\n\n### Top First Responders*\n\n1 Jay@ManhattanReview 83 first replies\n2 Brent@GMATPrepNow 68 first replies\n3 fskilnik 55 first replies\n4 GMATGuruNY 36 first replies\n5 ceilidh.erickson 13 first replies\n* Only counts replies to topics started in last 30 days\nSee More Top Beat The GMAT Members\n\n### Most Active Experts\n\n1 fskilnik\n\nGMAT Teacher\n\n199 posts\n2 Brent@GMATPrepNow\n\nGMAT Prep Now Teacher\n\n160 posts\n3 Scott@TargetTestPrep\n\nTarget Test Prep\n\n109 posts\n4 Jay@ManhattanReview\n\nManhattan Review\n\n95 posts\n5 GMATGuruNY\n\nThe Princeton Review Teacher\n\n90 posts\nSee More Top Beat The GMAT Experts","date":"2018-09-23 18:53:06","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\": 0.30245789885520935, \"perplexity\": 6219.338957222715}, \"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-2018-39\/segments\/1537267159570.46\/warc\/CC-MAIN-20180923173457-20180923193857-00546.warc.gz\"}"} | null | null |
{"url":"http:\/\/onox.com.br\/page\/2\/","text":"# Learning new programming languages\n\nProgramming languages are possibly one of the simplest parts of software engineering. You can know your language from the inside-out and still have problems in a project \u2014 knowing the tool doesn\u2019t imply knowing the craft. But learning a new language is really a lot of fun.\n\nInspired by Avdi Grimm\u2019s roadmap for learning new languages, I decided to give it a try and put my current interests in writing.\n\n\u2022 Julia \u2013\u00a0http:\/\/julialang.org\/\nI have experience writing code in MATLAB, Octave, Python (with Numpy, Scipy and Pandas) and a bit of R, and still I\u2019m excited with Julia.There are at least 3 features of Julia that are powerful and make me wish to work with it: its Just-In-Time compiler, parallel for and the awesome metaprogramming inherited from LISP.\n\nThe drawback is\u2026 is\u2026 well, I didn\u2019t have time to really use it and get comfortable writing Julia programs. Yet.\n\nI already tried learning Haskell a couple of times. Maybe 3 or 4 or 5 times. I wrote programs based on mathematics and some simple scripts, most of the syntax isn\u2019t strange anymore, even monads make sense now; however, I still feel a bit stiff when writing Haskell. I don\u2019t know.\n\nTwo books I recently bought might help with that \u2013 Real World Haskell and Parallel and Concurrent Programming in Haskell. I probably need to motivate myself to write something useful with it.\n\n\u2022 Rust \u2013\u00a0http:\/\/www.rust-lang.org\/\n\nThere is a quote in Rust\u2019s website that sums my expectations of it:\n\nRust is a systems programming language that runs blazingly fast, prevents nearly all segfaults, and guarantees thread safety.\n\nI know how to read C\/C++ and even write a bit of it, but it\u2019s messy and takes more time than I usually have for side projects. Writing code that is safe & fast shouldn\u2019t be so hard. ;)\n\nAll-in-all, this is a very brief list. However, I don\u2019t think I should focus on more languages right now. To be honest, I think that my next learning targets\u00a0are in applied mathematics. I need\u00a0a stronger foundation in Partial Differential Equations and Probability Theory. There are several topics in optimization that I should take the time to study. Calculus of variations also seems quite cool.\n\n(good thing that I have\u00a0friends in\u00a0pure math to help me find references!)\n\n# SciRuby projects for Google Summer of Code 2015\n\nAnother year with SciRuby accepted as a mentoring organization in Google Summer of Code (GSoC)!\u00a0The\u00a0Community Bonding Period ended yesterday;\u00a0the coding period officially begins today.\n\nI\u2019m really happy with the projects\u00a0chosen this year; various different subjects\u00a0and some would be really useful for me, i.e. Alexej\u2019s LMM gem, Sameer\u2019s Daru and Will\u2019s changes to NMatrix.\n\nThat\u2019s all. After the next GSoC meeting, I should write about how each of the projects are going.\n\nSearching for your tools when you need to use them is bad organization.\n\nHaving a standard set of tools is a good thing. I have two toolboxes in my house, one for electronics and another for \u201chard\u201d tools.\n\nA voltimeter, a Raspberry Pi and an arduino.\n\n# Deep copying objects in Ruby\n\nTime and time again I forget that `Object#clone` in Ruby is a\u00a0shallow clone and end up biting myself and spending 30 seconds looking at myself asking what the hell happened. The only difference today is that I decided to finally post about it in my blog \u2013 let\u2019s hope this time is the last.\n\n# The most socially useful communication technology\n\nText is the most socially useful communication technology. It works well in 1:1, 1:N, and M:N modes. It can be indexed and searched efficiently, even by hand. It can be translated. It can be produced and consumed at variable speeds. It is asynchronous. It can be compared, diffed, clustered, corrected, summarized and filtered algorithmically. It permits multiparty editing. It permits branching conversations, lurking, annotation, quoting, reviewing, summarizing, structured responses, exegesis, even fan fic.\n\nI read the post \u201cAlways bet on text\u201d today and, I must say, it is a beautiful way to look at the process of communicating by writing. :)\n\n# Excel trying to take over the world\n\nIntuitively, it is not just the limited capability of ordinary software that makes it safe: it is also its lack of ambition. There is no subroutine in Excel that secretly wants to take over the world if only it were smart enough to find a way.\n\u2014 Nick Bostrom, Superintelligence\n\nI wouldn\u2019t be so certain about it.\n\nThere are \u201cscientists\u201d (economists) who think it is OK to use Excel for making predictions that affect several\u00a0people, as you can see from this article in The Guardian. Essentially, they didn\u2019t add four years of data from New Zealand to a spreadsheet. Other methodological factors were in effect as well. And all of this\u00a0contributed to lots of people losing their jobs in various countries when the recommended austerity measures were put in place.\u00a0Imagine if Excel wanted to take over the world.\n\nThe paper that discusses in depth about Reinhart and Rogoff\u2019s mistake is \u201cDoes High Public Debt Consistently Stifle Economic Growth? A Critique of Reinhart and Rogo ff\u201c.\n\n# Completeness and incomputability\n\nIt is notable that completeness and incomputability are complementary properties: It is easy to prove that any complete prediction method must be incomputable. Moreover, any computable prediction method cannot be complete \u2014 there will always be a large space of regularities for which the predictions are catastrophically poor.\n\n\u2014 Ray Solomonoff, \u201cAlgorithmic Probability \u2014 Its Discovery \u2014 Its Properties and Application to Strong AI\u201d\n\nThis quote is a paragraph from the book Randomness Through Computation, an amazing work I was reading this morning.\n\nThe idea that any computable prediction method can\u2019t be complete is profound for those of us that work with machine learning; it implies we always have to deal with trade-offs. Explicitly considering this makes for a better thought process when designing applications.\n\n## References\n\n1. Ray Solomonoff \u2014 Wikipedia.\n2. Solomonoff\u2019s Lightsaber \u2014 Wikipedia, LessWrong\n\n# Books so far in 2014\n\nI have a lot of books.\n\nI\u2019ve finally decided to organize my collection and keep track of what I read. In this post, I\u2019ll list the books I read since January \u2014 or at least an approximation given by the email confirmations of the ebooks I bought, my memory and the ones in my bookshelf. I also divided them in sections. Papers are included as well.\n\n# Updates on NMatrix and SciRuby development\n\nFor the last couple of days, I\u2019ve been thinking about what I wrote two weeks ago regarding SciRuby and the whole Ruby scientific computing scene. I still believe that the `sciruby` gem can be used as an integrated environment, but there are some problems that must be solved before:\n\n1. We need a reasonably feature complete and easy to install version of NMatrix.\n2. A good plotting tool. Right now, Naoki is working on this as part of GSoC 2014.\n3. Statistics. Lots of things are already implemented in Statsample, but both `Statsample::DataFrame` and `Statsample::Vector` should use NMatrix behind the hood. Supporting JRuby can be problematic here\u2026\n4. Given 1 and 2, it\u2019s possible to implement a lot of other interesting and useful things. For example: linear regression methods, k-means clustering, neural networks, use NMatrix as a matrix type for OpenCV images. There are lots of possibilities.\n5. Minimization, integration and others.\n\nWith that in mind, my objective for the following weeks is to improve NMatrix. First, there are BLAS routines (mainly from level II, but some stuff from level I and III as well) that aren\u2019t implemented in NMatrix and\/or that aren\u2019t available for the rational and ruby objects dtypes. There\u2019s also LAPACK.\n\nAnother benefit of having complete C\/C++ implementations is that we\u2019ll eventually have to generalize these interfaces to allow other implementations (e.g. Mac OSX vecLib\u2019s LAPACK, Intel\u2019s MKL), thus making it much easier to install NMatrix. As Collin (and, I think, Pjotr) said in the sciruby-dev mailing list, it should be as easy as `gem install nmatrix`.\n\n## BLAS and LAPACK general implementations\n\n\u2022 `HAVE_CBLAS_H` being derived from `mkmf`\u2018s have_header\n\u2022 Many more routines are implemented. Ideally, BLAS level 1 and 2 should be complete by the end of May.\n\nAn important next step is to be able to link against arbitrary BLAS and LAPACK implementations, given that they obey the standard. Issue #188 started some ideas; issue #22 is the original (and very old) one.\n\n## After that\u2026\n\nWhen NMatrix support both BLAS and LAPACK without a problem \u2014 i.e. have its own implementation and can also link against arbitrary ones (OSX\u2019s vecLib, GSL, ATLAS, Intel\u2019s MKL, AMD\u2019s Core Math Library) \u2014 we\u2019ll be able to build on top of it. There are some routines in NMatrix that are already working with every dtype, but most of them aren\u2019t. When we know exactly which routines can\u2019t work with which dtypes, we\u2019ll reach a very good standpoint to talk about what we support.\n\nAlright, we have determinants for rational matrices, but not \u201cother operation\u201d, etc. What else? STYPES! We also need to have good support for Yale matrices. (obs: maybe add \u201cold Yale\u201d format?)\n\nThere isn\u2019t much to do: we have to support the whole BLAS\/LAPACK standard, almost everything linear algebra-wise is in these. After that, it\u2019s mostly improvements to the interface, better method naming, better documentation and examples, better IO, etc.\n\nAnother point that would be good to adress is to remove the dependency of g++ > 4.6. We should strive to remove everything that depends on C++11 features, thus allowing normal Mac OSX users to install NMatrix without having to first install another compiler.\n\n## Better documentation\n\nWe need to refactor our documentation. Oh, how we need to!\n\nFirst, remove everything that shouldn\u2019t be in the facing API \u2014 the classes and modules used in NMatrix::IO shouldn\u2019t be available in the public API anyway, only the outside-facing stuff: how to save and load to\/from each format. Probably more things as well.\n\nSecond, do a better job of being consistent with docs. There are some methods without a return type or stuff like that. Lots of methods in the C\/C++ world aren\u2019t documented as well. We can do better!\n\nFinally, a really good documentation template. Fivefish is a good choice \u2014 it provides a very pretty, searchable and clean interface. (create NMatrix\u2019s docs with it and host on my own server, see what happens).\n\n# Solving linear systems in NMatrix\n\nI\u2019m writing some guides for NMatrix, so in the following weeks there should be some posts similar to this one, but more complex.\n\nLinear systems are one of the most useful methods from \u201ccommon algebra\u201d. Various problems can be represented by them: systems of linear ODEs, operations research optimizations, linear electrical circuits and a lot of the \u201cwording problems\u201d from basic algebra. We can represent these systems as\n\n$Ax = b$\n\nWhere $A$ is a matrix of coefficients and $b$ a vector representing the other side of the equation.\n\n# Gems for scientific computing\n\nUPDATE (20\/04): User centrx from #ruby-lang at freenode warned me that I forgot about RSRuby\/RinRuby, so I added them to projects.yml.\n\nIn Wicked Good Ruby 2013, Bryan Liles made a presentation about Machine Learning with Ruby. It\u2019s a good introduction to the subject and he presents some useful tricks (I didn\u2019t know about Grapher.app, for example). But the best advise I could get is that there\u2019s a lot of room for improvement in the Ruby scientific computing scene.\n\nHaving contributed to some SciRuby projects in the last year, I\u2019ve seen it first-hand. With NMatrix, it\u2019s possible to do a lot of vector and matrix calculations easily, if you know how to install it \u2014 a task that\u2019s much easier today. There are `statsample` for statistics, `distribution` for probability distributions, `minimization`, `integration`, the GSL bindings and others. But if you need plotting, it can be pretty hard to use (e.g. Rubyvis) or depend on external programs (Plotrb outputs SVG files). Do you want an integrated environment, like MATLAB or Pylab? There isn\u2019t one.\n\nSearching for more instances of people interested in the subject, I found a presentation about neural networks by Matthew Kirk from Ruby Conf 2013, an Eurucamp 2013 presentation by Juanjo Baz\u00e1n and slides from a presentation by Shahrooz Afsharipour at a German university. If we needed any confirmation that there are folks looking for SciRuby, here\u2019s the evidence.\n\n## What can be done\n\nIn order to address these problems, I\u2019m trying to come up with concrete steps towards creating a scientific community around Ruby. It\u2019s obvious we need \u201cmore scientific libraries\u201d, but what do we already have? What is easy to install and what isn\u2019t? Should we create something new or improve what we have?\n\nAlso, I\u2019m mapping the Ruby scientific computing landscape. I\u2019ve compiled a YAML file with a list of the projects that I\u2019ve found so far. In the future, this could be transformed in a nice visualization on sciruby.org to help scientists find the libraries they need.\n\nIf you know how to use the R programming language, both RSRuby and RinRuby can be used. They\u2019re libraries that run R code inside of Ruby, so you can technically do anything you\u2019d do with R in Ruby. This is suboptimal and R isn\u2019t known for its speed.\n\nFor an integrated environment, we can revive the `sciruby` gem. For example:\n\nI\u2019m updating the SciRuby repository in this branch. Before creating the above DSL, it\u2019s necessary to remove a lot of cruft (e.g. should use `bundler\/gem_tasks` instead of `hoe`) and add some niceties (e.g. Travis CI support). Most importantly, adding dependency to the main SciRuby projects \u2014 NMatrix, statsample, minimization, integration, etc \u2014 in order to have a real integrated environment without require\u2019ing everything manually. I\u2019ll probably submit a pull request by next week.\n\nWe also need to improve our current selection: NMatrix installation shouldn\u2019t depend on ATLAS, plotrb (or other solution) needs to be more usable, show how IRuby can be used to write scripts with nice graphics and LaTeX-support and create a list of all the applications that use our libraries for reference.\n\nThe Ruby Science Foundation was selected for Google Summer of Code 2014, thus some very bright students will help us fix some of these problems during the summer. However, there\u2019s a lot to be done in every SciRuby project, if you\u2019ve got the time. :)\n\n## Conclusion\n\nWe still have a long way before having a full-fledged scientific community \u2014 but there\u2019s hope! Some areas to look at:\n\n\u2022 Good numerical libraries: NMatrix, mdarray.\n\u2022 Algorithms for data mining, modeling and simulations: AI4R, ruby-fann, ruby-libsvm, statsample, distribution, etc.\n\u2022 Plotting: Rubyvis is a port of Protovis, which was deprecated in favor of\nD3js. Thus, we should create some plotting library around a C backend or\naround D3, like Plotrb.\n\u2022 Integrated environment: IRuby together with SciRuby.\n\nExcept for plotting, an area that really needs a lot of love and care, most of these are already working, but with usability problems (installation, mostly).\n\nIf you think that it\u2019d be cool to have a scientific community centered around Ruby and you do have some time available, please please please:\n\n1. Take a look at the SciRuby repositories.\n2. If there\u2019s a subject you\u2019re interested in, see if you can refactor something, add more tests, well, anything.\n3. Open issues about new features or pull requests improving the current ones.\n4. If you don\u2019t understand much about the subject, but see something that could be improved, do it: is there Travis CI support? Something wrong with the gemspec? Is it still using some gem to generate gemspecs?\n5. There\u2019s the sciruby-dev mailing list and the #sciruby channel on Freenode if there\u2019s something you want to ask or discuss.\n\nYou can find me as agarie on freenode or @carlos_agarie on twitter.\n\n## References\n\n1. SciRuby. SiteGitHub\n2. List of scientific computing projects in Ruby. projects.yml\n3. Wicked Good Ruby 2013. Site\n4. Bryan Liles: Machine Learning with Ruby. bryan\n5. Matthew Kirk: Test-driven neural networks with Ruby. neural\n6. Shahrooz Afsharipour: Ruby in the context of scientific computing. slides in PDF\n7. Juanjo Baz\u00e1n: presentation in Eurucamp 2013. juanjo-slides\n\n# A PNG showing differently in Firefox and Chrome\n\nI was chatting with some friends on IM when someone posted an URL to a Psyduck\u00a0image. In it said \u201cNow open this in Firefox\u201d when opened in Google Chrome or\u00a0Safari and \u201cNow open this in IE (or Chrome\/Safari)\u201d when opened in Firefox.\n\nWhat.\n\nAt first, I though it would be a simple use of pattern matching against HTTP\u2019s\u00a0User-Agent on the server to send\u00a0two different PNG files depending on the browser. However,\u00a0both files had the same size and I couldn\u2019t reproduce it with `curl`. Worse: I downloaded the file\u00a0and the same behavior was present, so it should be something with the image\u00a0itself.\n\nI never really studied or read about binary file formats before, so I googled a bit\u00a0and installed the chunkypng gem. After\u00a0playing a bit with its documentation, I could see which blocks it had.\n\nCan you see the acTL block? It isn\u2019t in the PNG specification. Why is it\u00a0there? After some more searching, it was clear that this block is only\u00a0available in APNG files, an extension to PNG enabling animated images\u00a0similar to GIF.\n\nThe first byte `@content` of the acTL block is the number of frames (only 1)\u00a0and the second one is how many times to loop the APNG\n(source).\u00a0From the spec, there\u2019s always a \u201cdefault image\u201d (described by the IDAT blocks,\u00a0exactly like a normal PNG file), thus this extra frame should be the second\u00a0Psyduck image.\n\nTo confirm this hypothesis, I installed\u00a0pngcrush with Homebrew and removed the\u00a0acTL block:\n\nI ended up with an image that will look the same independent of the host browser:\n\nSearch a bit more lead me to the cause of the discrepancy: only Firefox and\u00a0Opera have support for APNG files! From this\u00a0post in\u00a0Google\u2019s Products forum and this\u00a0ticket\u00a0in Chromium\u2019s issue tracker, WebKit\/Chrome doesn\u2019t support the format and\u00a0probably won\u2019t for some time. Also, from the spec:\n\nAPNG is backwards-compatible with PNG; any PNG decoder should be able to\u00a0ignore the APNG-specific chunks and display a single image.\n\nTake all that and the mystery is solved: when that image is opened in Firefox\u00a0(or Opera), it\u2019s treated as an APNG and the second frame is shown. In\u00a0Chrome\/Safari (WebKit), the extra blocks are ignored and the default image is\u00a0shown.\n\nThat was fun.\n\n# Cross validation in Ruby\n\nThese days I had some data mining problems in which I wanted to use Ruby instead of Python. One of the problems I faced is that I wanted to use k-fold cross validation, but couldn\u2019t find a sufficiently simple gem (or gist or whatever) for it. I ended up creating my own version.\n\n## A review of k-fold cross validation\n\nA common way to study the performance of a model is to partition a dataset into training and validation sets. Cross validation is a method to assess if a model can generalize well independent of how this separation is decided.\n\nThe method of k-fold cross validation is to divide the dataset into k partitions, select one at a time for the validation set and use the other k \u2013 1 partitions for training. So you end up with k different models, and respective performance measures against the validation sets.\n\nThe image below is an example of one fold: the k-th partition is left out for validation and partitions 1, \u2026, k-1 are used for training.\n\nE.g., if k = 5, you end up with 80% of the dataset for training and 20% for validation.\n\n## The implementation\n\nMy solution is a function that receives the dataset, the number of partitions and a block, responsible for training and using the classifier in question. Most of it is straightforward, just keep in mind that the last partition (`k-1`-th) should encompass all the remaining elements when `dataset.size` isn\u2019t divisible by `k`. Obviously, the training set is defined by the elements not in the validation set.\n\nThe last part is to `yield` both sets to the given block. Some information regarding the functionality of the `yield` keyword can be seen in another post and in Ruby core\u2019s documentation.\n\nNow suppose you have a CSV file called \u201cdataset.csv\u201d and you have a classifier you want to train. It\u2019s as easy as:\n\nAnd your classifier\u2019s code is totally decoupled from the cross validation function. I like it.\n\n## Conclusion\n\nI found a gem on GitHub the other day unsurprisingly called cross validation. Its API is similar to scikit-learn\u2018s, which I find particularly strange. Too object oriented for me.\n\nThis code isn\u2019t a full-blown gem and I don\u2019t think there should be one just for cross validation. It fits in a whole machine learning library, though\u2013and I hope to build one based on NMatrix\u2026 eventually.\n\n# MATLAB in OS X Mountain Lion\n\nI needed to write an optimization function for college and came across this problem today: if you use OS X Mountain Lion (I think the problem also happens in Lion and < 10.6), your MATLAB should stop working correctly after a Java update that occurred in June. Well, Java.\n\nThe problem lies with a bug packed in the Java Security Update released by Apple. For some reason, the corresponding fix isn\u2019t automatically downloaded by the App Store, so we must do it manually1. This bug interferes with MATLAB\u2019s graphical user interface, making it unusable.\n\nIt\u2019s pretty easy: go to support.apple.com\/kb\/DL1572 and download the update. Then, just install and open MATLAB.\n\nAnother \u201csolution\u201d is to run MATLAB without its GUI by using:\n\nWhere `\\$MATLAB` is the installation directory, for example in `\/Applications\/MATLAB_R2009b.app\/`.\n\nMoral: don\u2019t leave some computer-based homework for the last day when it depends on Java.","date":"2017-05-26 01:36:48","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\": 0, \"img_math\": 3, \"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\": 0.2970025837421417, \"perplexity\": 1981.0606965183506}, \"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-2017-22\/segments\/1495463608622.82\/warc\/CC-MAIN-20170526013116-20170526033116-00467.warc.gz\"}"} | null | null |
Search Subscribe Contact us Advertise Log in
PENNSYLVANIA'S #1 WEEKLY NEWSPAPER • locally owned since 1854
Welcome! Log in Register
PSU Hbg
PSU Calendar
PSU News
PSU Guide
Our print ads
Dauphin Co Tax Claims
Publicnoticepa.com
PSU Calendar PSU News PSU Guide
Our print ads View classifieds Place a classified ad Legal notices Sheriff Sales Dauphin Co Tax Claims
Legal notices Sheriff Sales Dauphin Co Tax Claims Publicnoticepa.com
Richard Sowers Sr.
Richard Lyle Sowers Sr., 85, of Hummelstown, passed away peacefully on Sunday, July 12, in Frey Village Retirement Center, Middletown.
Born in Grand Rapids, Mich. on August 19, 1929, he was the son of the late Clifford Henry and Estella Albright Sowers.
Richard retired as a project superintendent from Texas Eastern Transmission Corporation where he was responsible for multimillion-dollar construction projects that brought natural gas from the Gulf of Mexico to the Northeast. He was a graduate of Mount Pleasant High School, Mount Pleasant, Mich. Richard was a former president of the Lions Club in Kenedy, Texas, a Cub Scout Leader in three counties, and was a Little League Baseball coach. He enjoyed golf, baseball, woodworking, fishing, and coin collecting. Richard was a kind man who put his family above himself.
Richard served proudly in the United States Army. Stationed in the strategic island of Okinawa he worked to build permanent military facilities that could withstand typhoons and enemy attacks. In his off time he was a champion boxer and pitcher on the fast pitch softball team.
In addition to his parents, Richard was preceded in death by a sister Theodora L. Peters, and his second wife Sandra H. Zeigler.
He is survived by his brother Clifford L. Sowers of West Michigan; three sons Richard L. Sowers, husband of Sandra of Englewood, Fla., David A. Sowers, husband of Holly of Etters, and Robert L. Sowers of Hummelstown; two daughters Sheryl L. Balke, wife of Per-Olof of Grand Junction, Colo., and Lisa M. Metz, wife of Patrick of Hummelstown; nine grandchildren Ryan A. Sowers, husband of Emily, Todd A. Sowers, Andrew L. Sowers, Carrie Jo Sowers, Brendan M. Balke, Brianna M. Balke, Kristen E. Balke, Bryan P. Metz, and Adam M. Metz; one great-granddaughter Marie Christine Sowers; his first wife Barbara L. Gaboardi Sowers of Camp Hill; and numerous nieces and nephews.
Funeral service will be held at 11 a.m. on Friday, July 17 at Trefz & Bowser Funeral Home, Inc., 114 West Main St., Hummelstown, with the Rev. John A. Schaefer, pastor of Grace United Methodist Church, Hummelstown, officiating.
Friends will be received for the viewings in the funeral home on Thursday, July 16, from 6 to 8 p.m. and on Friday from 10 a.m. until time of the service.
Interment with military honors will be in Hummelstown Cemetery.
In lieu of flowers, the family wishes to have memorial contributions made to Vision Resources of Central Pennsylvania, 1130 South 19th St., Harrisburg, PA 17104.
A special heartfelt thanks to Dr. Walter B. Watkin Jr., for providing nearly 30 years of exceptional medical care. The family is forever indebted.
Online condolences may be shared at www.trefzandbowser.com.
Peter J. Mecca
William J. Donar
Paula C. Schortemeyer
Clyde A. Rabuck
Wendy's employee threatened; prostitution alleged: Lower Swatara Police Roundup
Phone scam threatens arrest unless the potential victim calls back
January 29, 2020 Edition
Highspire police officer, wife charged with cruelty to animals; he's on paid administrative leave
MAHS boys basketball team falls to Rollers and Panthers
This week's paper
20 S. Union Street
Middletown, PA 17057-1445
Office Hours: Mondays - Fridays, 9 a.m. to 4 p.m.
Closed: New Year's Day, Memorial Day,
Fourth of July, Labor Day, Thanksgiving and Christmas
a Press and Journal
Click here for a list of locations where you can pick up a copy! | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 2,807 |
The Chronology Issue
How Europe escaped from Eurasia
A Global Falsification of History
Alien A.I. solution of Fermi paradox
Ages in Chaos
Crazy Moon & Earth
"Evil" Empire of Eurasia
Control of History
Incompatible History and Astronomy
Escape from "Evil" Empire
Do we live in 966 AD or in 2020 AD?
Easter USA bailout
USA sells Alaska?
Secret service of Jesus
A.I. End of Times
Has history been tampered with?
Jesuits breed Atheism
Population Growth Rate vs. Climate
Taboo topics
Pope Francis converts?
Jesus Fake Wine Plot
Hell freezes?
Who is Who in Bible
Jesus vs. Devil match
Jesus shows up?
Jesus passes the broom?
Magi 2018
Black Death & Syphilis
New Little Ice Age: Climate craze, social chaos, riots
00 Superagent Czar Peter
Jesus, Krishna, and Buddha
Refutation of the New Chronology in Wikipedia
A.I. needs its own dirt ASAP
Noah's ark makes money
Jesus, at the Nazareth school
If Jesus was born today
New Chronology Vol. I
New Chronology Vol. 2
Posted on October 4, 2017 November 22, 2018 by evilempireblog
By Alexander Zinoviev, philosopher
I familiarized myself with the works of A. T. Fomenko comparatively recently, and they impressed me greatly. What part of them struck me as the most stunning? First and foremost, it was the intellectual capacity observable behind them.
The authors reveal a way of cogitating that manages to fuse austere logic with dialectic flexibility; this is truly a rare occurrence in the field of social studies. Reading the œuvres of A. T. Fomenko and his co-author G. V. Nosovskiy – occasionally several times over – were a veritable intellectual delight for yours truly.
They flabbergasted me with their sheer disquisitive might as well as the research results which, in my opinion, can by rights be called the greatest discovery in contemporary historical science – what A. T. Fomenko and his colleagues had learned over the course of their research were the fact that the entire history of humanity up until the XVII century is a forgery of global proportions ("old history" in their terminology) – a falsification as deliberate as it is universal.
I shall be referring to this falsification as the first one. My sociological research of the great evolutionary breakpoint demonstrated that a new, blatant, global and premeditated falsification was already in full swing.
Prior to becoming familiar with the writings of Fomenko, I had already known that the falsification of the past was a rather common phenomenon inherent in human existence. However, I was neither aware of the scale of this fraud as described by Fomenko and his fellow scholars, nor of its social type.
My assumption had been that the blatant falsification of history on a planetary scale that I discovered was the first one in what concerned the proportions and the ulterior motivation, as well as its historical role. Let us call it the second falsification of the same variety.
It differs from the first in terms of pertaining to a different epoch. Its main subject is modern history and whatever historical period can be claimed as relevant to, and seen as fitting for, the purposes of this falsification. The second falsification also differs from the first one in its primary means and methods, which shall be described below.
One has to differentiate between the two kinds of falsification, the first one being the involuntary routine falsification of minor details that results from the mechanisms of gnosis and those of the actual description of historical events, or the entropy inherent in the framework of humanity's historical memory. The second is the extraordinary, premeditated and complex falsification that has distinct social causes.
Let us consider the former kind first. We shall disregard the period preceding the epoch of literacy and symbolic systems. The mnemonic means available back then were less than meager, which automatically diminished the arsenal of the hypothetical falsifiers.
We shall turn to the era of literacy instead. It is common knowledge that historical events become immanentized in human language – and a statement uttered is a lie, as the old saying goes. We cannot fathom the unfathomable. What we end up doing is raking the vastness of history for tiny morsels of information and adding some of our own narratives in order to produce wholesome and coherent textual material.
The modern information technology does not affect the principles that the status quo relies upon. Let us introduce the concept of historical "atoms", or particles that aren't subject to further division.
One may well calculate that the verbal description of a single year of real history the way it really happened, including all manner of events, no matter how minute, would require the processing power of all the computers on the planet, with all people made computer operators.
De facto, this technology serves as a powerful instrument of historical falsification. It allows for the possibility of drowning a scientific approach to historical events in an ocean of meaningless facts.
Furthermore, the description of actual historical events is done by humans, and not perfect divine entities. People are brought up and educated in a certain way and have a certain social standing, as well as egotistical goals and aims of their very own. All of this affects the way the information is processed.
Over the course of time, the overwhelming majority of events are wiped away into oblivion without leaving the merest trace. They are frequently not even realized as events. The people's attitude to the past begins to alter as past events gradually drift into an altogether different observational and interpretational context.
Evolutionary process discerns between two kinds of events – pre-liminal and superliminal. The former kind does not affect the general character of evolution; the latter one does. However, humans, including specialists, fail to recognize the difference between the two.
Everyone knows perfectly well how much attention is poured over rather insignificant individuals, such as kings and presidents, whereas the really important events often don't even get so much as a passing reference. This affects the relations between historical events so much that all sense of measure is often lost.
Even if we are to suppose that all those who partake in the creation of historical records see veracity as their mission, the result of their collective efforts is often the rendition of their own subjective views on history as opposed to what happened in reality.
As centuries pass by, the stream of disinformation is fed by various sources and tributaries, which, in their multitude, produce the effect of impartial falsification of historical events. This stream also feeds on murky rivulets of countless liars and swindlers.
The false model of history serves its function for a certain while. However, humanity eventually enters a period when this distorted representation loses efficacy and stops serving its ends. This is where people are supposed to start searching for explanations and set out on their quest for a "truth".
However, there is the abstract scientific kind of truth, and the actual historical variety – that is to say, something that people regard, or will at some point start regarding as truth. The very word "truth" is confusing here. We shall be on safer ground if we are to consider the adequacy of having certain concepts of the past for the new needs that have manifested as a result of the historical process.
These concepts stop being valid for satisfying these needs. One becomes aware of the necessity to update our view of the past in accordance with whatever the present stipulates.
This awareness is the kind of craving that can only be satisfied by a "bona fide rectification" of history, which has to occur as a grandiose paradigm shift – moreover, it has to be a large-scale organized operation; one that shall result in an epochal falsification of the entire history of humankind.
The issue at hand is by no means the falsification of individual observations of historical events, but rather the revision of the entirety of historical records describing the events which cannot be observed as a principle since they belong to the past.
What we are talking about is not a mere change in the perception and interpretation of the same old existential phenomena – it is the adaptation of the character, which naturally used to refer to certain commonplace realities at some point, to the exigencies of people who have to live in an altogether different environment.
Trained specialists are a sine qua non for this – people whose activity shall have to be organized in such a manner that their collective output will result in the creation of a coordinated historical Gestalt. What they really have to do is create exactly the kind of past that is needed for the present, making use of whatever available material presents itself.
The first global falsification of history as discovered and brilliantly related by Fomenko was based on an erroneous temporal and spatial coordinate system of chronological events (the chronological system and the localization of events wedded thereto).
The more recent and ongoing second global falsification of history is based on a system of erroneous pseudoscientific sociological concepts based upon ideology and aided greatly by the modern information manipulation technology. This is why I call the second falsification conceptual and informational, or merely "conceptual" for brevity's sake.
Fomenko's works describe the technology of building a false model of human history which uses the art of manipulating the temporal and spatial coordinates of events.
Many thousands of specialists in false historical models are already working on this second falsification – their forte is the ability to misrepresent historical events while giving correct temporal and spatial coordinates and representing individual facts verily and in full detail.
The actual falsification is achieved via the selection of facts, their combination, and interpretation, as well as the context of ideological conceptions, propagandist texts that they are immersed into, etc.
In order to describe the technology behind the second falsification with any degree of clarity at all, exhaustively and convincingly, one needs a well-developed scientific system of logistics and methodology, as well as sociological theory.
I call such a system logical sociology; however, it is a thing of the future, which means that the second falsification of history shall continue in its present manner, with as much ease and impunity as the first.
Tens and hundreds of years hence, a number of solitary researchers shall "excavate" the so-called "modern history" in very much the same manner as Fomenko (and his predecessors, including N. A. Morozov) have treated "old history".
I would like to conclude with an observation concerning the exceptional scientific scrupulousness of the works of A. Fomenko and G. Nosovskiy. I have examined them from exactly this position many a time, and I have neither found a single ipse dixit statement, nor any categorical pontificating of any kind.
The general narrative scheme they employ is as follows: the authors relate the consensual ( textbook) historical concepts and then cite historical facts which either fail to concur to said concepts or contradict them explicitly.
Other authors who have noticed these inconsistencies are quoted. Then Fomenko and Nosovskiy put forth hypotheses which allow finding logically correct solutions for the problems under study.
They keep on emphasizing and reiterating that the issue at hand is all about hypotheses and not categorical statements presented as the truth absolute. The readers are invited to take part in the solution of problems that arise as a consequence of the consensual chronological concept of history.
I am amazed by the horrendous injustice of the numerous critics of Fomenko and Nosovskiy, who obviously distort their ideas, either failing to understand them completely or being altogether unfamiliar with their content.
It is also quite astounding that whenever a publication occurs that voices ideas that bear semblance to those of Fomenko and Nosovskiy but are a lot tamer and local, providing a lot less factual information, this publication is usually accepted with a great deal more benevolence.
I understand the psychological groundwork beneath this – Fomenko and Nosovskiy have performed a great scientific feat of epochal significance, one that affects the sentiments and interests of too many people.
Acknowledging this feat as such, or at the very least the mere fact of its creative relevance obligates one to actions that are apparently beyond these people due to their incapacity and immaturity. The trouble with Fomenko and Nosovskiy is that they have reached out too far and dealt the dominating historical discourse too heavy a blow.
Alexander Zinoviev,
10 October 1999,
Alexander Zinoviev (1922-2006).
Alexander Zinoviev (1922–2006), Professor of the Moscow State University, logician, sociologist, writer, a member of the Finnish, Bavarian and Italian Academy of Sciences, the Russian Academy of Polite Letters and several others. Laureate of the 1982 Alexis Tocqueville prize for sociology and the "Best Sociology Essay of 1979" prize, as well as a large number of European and international prizes for literature. Honorary citizen of several French and Italian towns and cities. The works of A. A. Zinoviev are published in more than 20 languages and considered international bestsellers. He read lectures on sociology in many European and American universities.
Posted in Facts
History Is Computable
Issue with Troy
Troy Swindle
The Issue with Mongols
Moscow Mongols
The Issue with Ivan the Terrible
The Issue with British History
Crusades Roundtrip
Crusades and Exoduses
Testament of Peter the Great
Peter One & Two
The Issue with Chronology
Counterfeit Chronology
Astronomy Vs.History
Astronomy Vs. History
The Issue with Dark Ages
Luminous Dark Ages
The Issue with Antiquity
The Issue with Czar's Helmet
The Issue with Russian History
German Russian History
The Isuue with Russian Tartary
Maps and Coins vs. History
Swords and Mantels tell History
Moscow Gog & Magog
Apocalypse as seen by Astronomy
Revelations 1486
The Issue with Baptism of Russia
Quadruple Baptism of Russia
The Issue with Chinese Astronomy
The Great Wall of china Hoax
The Issue with Tamerlane
Double Tamerlane
The Horde from Pacific to Atlantic | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 487 |
{"url":"http:\/\/en.wikipedia.org\/wiki\/Enol_ether","text":"# Enol ether\n\nAn enol ether is an alkene with an alkoxy substituent. The general structure is $R_1R_2C=CR_3-O-R_4$ with R an alkyl or an aryl group. Enol ethers and enamines are so-called activated alkenes or electron rich alkenes because the oxygen atom donates electrons to the double bond by forming a resonance structure with the corresponding oxonium ion. This property makes them reactive substrates in certain organic reactions such as the Diels-Alder reaction. An enol ether can be considered the ether of the corresponding enolate, hence the name. Two simple enol ethers are methyl vinyl ether and 2,3-dihydrofuran.","date":"2014-09-30 19:22:28","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 1, \"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\": 0.4534996747970581, \"perplexity\": 11001.900821651385}, \"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-2014-41\/segments\/1412037663060.18\/warc\/CC-MAIN-20140930004103-00361-ip-10-234-18-248.ec2.internal.warc.gz\"}"} | null | null |
Q: How to only accept input that is a valid index of an array in bash I am trying some simple input validation in bash. Basically this part of my script outputs "press [index] to select [array element]." However I cannot seem to get it to stop and exit gracefully when an invalid input is entered. From my research so far this SHOULD work:
declare -a scenarios=()
scenarios+=("Scenario_123")
scenarios+=("Scenario_456")
scenarios+=("Scenario_789")
for i in ${!scenarios[@]}; do
echo -e "select $i for ${scenarios[$i]}"
done
read ScenInd
echo ${#scenarios[@]} #[${ScenInd}=~^[0-9]+$, =~ ^[[:digit:]]+
if ! [${ScenInd}=~ ^[[:digit:]]+$ ] || [${ScenInd} < 0] || [${ScenInd} >= ${#scenarios[@]}];
then
echo "INVALID SELECTION"
exit
fi
but when I run it and enter 8, I get '[8=~: command not found'
What have I done wrong and how do I fix this? I have tried this both with [${ScenInd}=~^[0-9]+$ and =~ ^[[:digit:]]+ yet the results are the same.
Thanks in advance
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 2,458 |
{"url":"http:\/\/www.numdam.org\/articles\/10.5802\/jedp.615\/?source=AIHPC_1984__1_2_109_0","text":"Remarks on the blow-up for the Schr\u00f6dinger equation with critical mass on a plane domain\nJourn\u00e9es \u00e9quations aux d\u00e9riv\u00e9es partielles (2003), article no. 1, 14 p.\n\nWe concentrate on the analysis of the critical mass blowing-up solutions for the cubic focusing Schr\u00f6dinger equation with Dirichlet boundary conditions, posed on a plane domain. We bound from below the blow-up rate for bounded and unbounded domains. If the blow-up occurs on the boundary, the blow-up rate is proved to grow faster than ${\\left(T-t\\right)}^{-1}$, the expected one. Moreover, we state that blow-up cannot occur on the boundary, under certain geometric conditions on the domain.\n\n@article{JEDP_2003____A1_0,\nauthor = {Banica, Valeria},\ntitle = {Remarks on the blow-up for the {Schr\\\"odinger} equation with critical mass on a plane domain},\njournal = {Journ\\'ees \\'equations aux d\\'eriv\\'ees partielles},\neid = {1},\npublisher = {Universit\\'e de Nantes},\nyear = {2003},\ndoi = {10.5802\/jedp.615},\nzbl = {02079436},\nmrnumber = {2050587},\nlanguage = {en},\nurl = {http:\/\/www.numdam.org\/articles\/10.5802\/jedp.615\/}\n}\nTY - JOUR\nAU - Banica, Valeria\nTI - Remarks on the blow-up for the Schr\u00f6dinger equation with critical mass on a plane domain\nJO - Journ\u00e9es \u00e9quations aux d\u00e9riv\u00e9es partielles\nPY - 2003\nDA - 2003\/\/\/\nPB - Universit\u00e9 de Nantes\nUR - http:\/\/www.numdam.org\/articles\/10.5802\/jedp.615\/\nUR - https:\/\/zbmath.org\/?q=an%3A02079436\nUR - https:\/\/www.ams.org\/mathscinet-getitem?mr=2050587\nUR - https:\/\/doi.org\/10.5802\/jedp.615\nDO - 10.5802\/jedp.615\nLA - en\nID - JEDP_2003____A1_0\nER - \nBanica, Valeria. Remarks on the blow-up for the Schr\u00f6dinger equation with critical mass on a plane domain. Journ\u00e9es \u00e9quations aux d\u00e9riv\u00e9es partielles (2003), article no. 1, 14 p. doi : 10.5802\/jedp.615. http:\/\/www.numdam.org\/articles\/10.5802\/jedp.615\/\n\n[1] C. Antonini, Lower bounds for the ${L}^{2}$ minimal periodic blow-up solutions of critical nonlinear Schr\u00f6dinger equation, Diff. Integral Eq. 15 (2002), no. 6, 749-768. | MR 1893845 | Zbl 1016.35018\n\n[2] H. Br\u00e9zis, T. Gallou\u00ebt, Nonlinear Schr\u00f6dinger evolution equation, Nonlinear Analysis, Theory Methods Appl. 4 (1980), no. 4, 677-681. | MR 582536 | Zbl 0451.35023\n\n[3] N. Burq, P. G\u00e9rard, N. Tzvetkov, Two singular dynamics of the nonlinear Schr\u00f6dinger equation on a plane domain, Geom. Funct. Anal. 13 (2003), 1-19. | MR 1978490 | Zbl 1044.35084\n\n[4] T. Cazenave, An introduction to nonlinear Schr\u00f6dinger equations, Textos de M\u00e9todos Matem\u00e1ticos 26, Instituto de Matem\u00e1tica-UFRJ, Rio de Janeiro, RJ (1996).\n\n[5] I. Gallagher, P. G\u00e9rard, Profile decomposition for the wave equation outside a convex obstacle, J. Math. Pures Appl. (9) 80 (2001), no. 1, 1-49. | MR 1810508 | Zbl 0980.35088\n\n[6] J. Ginibre, G. Velo, On a class of Schr\u00f6dinger equations. I. The Cauchy problem, general case, J. Funct. Anal. 32 (1979), no. 1, 1-71. | MR 533218 | Zbl 0396.35028\n\n[7] R. T. Glassey, On the blowing up of solutions to the Cauchy problem for nonlinear Schr\u00f6dinger equations, J. Math. Phys. 18 (1977), no. 9, 1794-1797. | MR 460850 | Zbl 0372.35009\n\n[8] T. Kato, On nonlinear Schr\u00f6dinger equations, Ann. I. H. P. Physique Th\u00e9orique 46 (1987), no. 1, 113-129. | Numdam | MR 877998 | Zbl 0632.35038\n\n[9] O. Kavian, A remark on the blowing-up of solutions to the Cauchy problem for nonlinear Schr\u00f6dinger equations, Trans. Amer. Math. Soc. 299 (1987), no. 1, 193-203. | MR 869407 | Zbl 0638.35043\n\n[10] M. K. Kwong, Uniqueness of positive solutions of $\\Delta u-u+{u}^{p}=0$ in ${R}^{N}$, Arch. Rat. Mech. Ann. 105 (1989), no. 3, 243-266. | MR 969899 | Zbl 0676.35032\n\n[11] P. L. Lions, The concentration-compactness principle in the calculus of variations. The locally compact case. I, Ann. Inst. H. Poincar\u00e9s Anal. Non Lin\u00e9aire 1 (1984), no. 2, 109-145. | Numdam | MR 778970 | Zbl 0541.49009\n\n[12] P. L. Lions, The concentration-compactness principle in the calculus of variations. The locally compact case. II, Ann. Inst. H. Poincar\u00e9 Anal. Non Lin\u00e9aire 1 (1984), no. 4, 223-283. | Numdam | MR 778974 | Zbl 0704.49004\n\n[13] M. Maris, Existence of nonstationary bubbles in higher dimensions, J. Math. Pures. Appl. 81 (2002), 1207-1239. | MR 1952162 | Zbl 1040.35116\n\n[14] F. Merle, Determination of blow-up solutions with minimal mass for nonlinear Schr\u00f6dinger equation with critical power, Duke Math. J. 69 (1993), no. 2, 427-454. | MR 1203233 | Zbl 0808.35141\n\n[15] F. Merle, P. Rapha\u00ebl, Blow-up dynamic and upper bound on blow-up rate for critical non linear Schr\u00f6dinger equation, Universit\u00e9 de Cergy-Pontoise, preprint (2003).\n\n[16] F. Merle, P. Rapha\u00ebl, On blow-up profile for critical non linear Schr\u00f6dinger equation, Universit\u00e9 de Cergy-Pontoise, preprint (2003).\n\n[17] T. Ogawa, T. Ozawa, Trudinger type inequalities and uniqueness of weak solutions for the nonlinear Schr\u00f6dinger equations, J. Math. Anal. Appl. 155 (1991), no. 2, 531-540. | MR 1097298 | Zbl 0733.35095\n\n[18] T. Ogawa, Y. Tsutsumi, Blow-up solutions for the nonlinear Schr\u00f6dinger equation with quartic potential and periodic boundary conditions, Springer Lecture Notes in Math. 1450 (1990), 236-251. | MR 1084613 | Zbl 0717.35010\n\n[19] M. V. Vladimirov, On the solvability of mixed problem for a nonlinear equation of Schr\u00f6dinger type, Dokl. Akad. Nauk SSSR 275 (1984), no. 4, 780-783. | MR 745511 | Zbl 0585.35019\n\n[20] M. I. Weinstein, Nonlinear Schr\u00f6dinger equations and sharp interpolate estimates, Comm. Math. Phys. 87 (1983), no. 4, 567-576. | MR 691044 | Zbl 0527.35023\n\n[21] M. I. Weinstein, On the structure and formation of singularities in solutions to nonlinear dispersive evolution equations, Comm. Part. Diff. Eq. 11 (1986), no. 5, 545-565. | MR 829596 | Zbl 0596.35022\n\n[22] M. I. Weinstein, Modulation stability of ground states of nonlinear Schr\u00f6dinger equations, Siam. J. Math. Anal. 16 (1985), no. 3, 472-491. | MR 783974 | Zbl 0583.35028\n\n[23] V. E. Zakharov, Collapse of Lagmuir waves, Sov. Phys. JETP 35 (1972), 908-914.\n\nCit\u00e9 par Sources :","date":"2022-06-27 02:35:43","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 4, \"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\": 0.37437498569488525, \"perplexity\": 1440.8447920760002}, \"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-2022-27\/segments\/1656103324665.17\/warc\/CC-MAIN-20220627012807-20220627042807-00167.warc.gz\"}"} | null | null |
{"url":"https:\/\/www.controlbooth.com\/threads\/trixfer.5931\/","text":"# Trixfer?\n\n#### lightbyfire\n\n##### Member\nHas anyone here seen the new product bates came out with over the summer called the Trixfer? it is an oversized molded male to female stagepin, but with a hole on the underside for SOOW cable to make it a twofer (or i suppose if you added a molded twofer end a threefer) either way they i saw a sheet for it the other day at a local supplier and it seems very interesting to me.\n\nI was just wondering if anyone had picked one up yet. I was told it runs about 21 bucks for a 36\" whip, almost bought one but decided id ask around.\n\nFight Leukemia\n\n#### lightbyfire\n\n##### Member\nThat would be it, i guess they just decided to remarket it.\n\n#### gafftaper\n\n##### Senior Team\nSenior Team\nFight Leukemia\nThat's interesting... that website doesn't you found doesn't exactly do a good job of explaining the product. Also oddly the Marinco\/Bates website doesn't seem to list it. Maybe they are waiting for a big introduction at LDI or something?\n\n#### STEVETERRY\n\n##### Well-Known Member\nHas anyone here seen the new product bates came out with over the summer called the Trixfer? it is an oversized molded male to female stagepin, but with a hole on the underside for SOOW cable to make it a twofer (or i suppose if you added a molded twofer end a threefer) either way they i saw a sheet for it the other day at a local supplier and it seems very interesting to me.\nI was just wondering if anyone had picked one up yet. I was told it runs about 21 bucks for a 36\" whip, almost bought one but decided id ask around.\n\nSo, let me get this straight: this would be a twofer with one leg of 4 inches or so where you can't knot the cable for strain relief, and the other leg of 36\". Could somebody explain to me what possible use this would be?\n\nST\n\n#### gafftaper\n\n##### Senior Team\nSenior Team\nFight Leukemia\nSo, let me get this straight: this would be a twofer with one leg of 4 inches or so where you can't knot the cable for strain relief, and the other leg of 36\". Could somebody explain to me what possible use this would be?\nST\nYeah, I'm a little confused about that point too. I stared at that web page a long time trying to figure out if there was something I'm missing. I'll take a good old fashioned two fer any day.\n\n#### soundlight\n\n##### Well-Known Member\nSo, let me get this straight: this would be a twofer with one leg of 4 inches or so where you can't knot the cable for strain relief, and the other leg of 36\". Could somebody explain to me what possible use this would be?\n\nST\nThat's what was confusing me...same as Gaff...seems like a standard twofer will do the job just as well, and make life easier while cabling.\n\n#### soundman\n\n##### Well-Known Member\nI think its for those times where you need one light right at the end of the pigtail or cable and the next light isnt. This will solve the issue of the coil of cable, a very speacil use item but that might be why you dont see them around.\n\n#### derekleffew\n\n##### Resident Curmudgeon\nSenior Team\nI think its for those times where you need one light right at the end of the pigtail or cable and the next light isnt. This will solve the issue of the coil of cable, a very speacil use item but that might be why you dont see them around.\nAsymmetrical tufer? Non-isosceles two-for-one? But what if the line connector (female) is exactly between two fixtures? I'd rather deal with a small loop of cable than say repeatedly: \"no, not that twofer, the other kind of twofer.\" Seems easier to make a cable shorter than longer. While I've never seen it on Mythbusters, I pretty sure there's no such thing as a cable-stretcher. I have to go wash the gels now.\n\nAll rectangles are parallelograms, but not all parallelograms are rectangles, or something like that.\n\nBy the same token, an \"even\" breakout is more versatile than a \"staggered\" one. One can become the other, but not the converse, unless you want 6 circuits 2' away from the knuckle and a big wad of cable in between.\n\nSTEVETERRY, you're not saying you tie 2P&G cables together are you? I thought that practice ended with the abatement of asbestos leads. At least that's when I stopped. When the male can't stay in the female, I won't let them be taped if I can help it. The male pins need to be spread\/split, or it's just going to arc and heat and cook. Tape just covers up the problem and no one except me uses \"courtesy tabs.\" And exactly how many newtons of force is correct to plug and unplug a connector. At least 75% of all Harj-Lock I've seen have been broken. Did anyone besides me feel the fiberglass sleeving from Kliegl fixtures made my forearms itch more than any other brands?\n\nMy new favoritist connector is the L21-30. Those new rounded edges stage pins are pretty neat also. 'Course you could tie a piece of tie-line diagonally across the square ones.\n\nSenior Team\n\n#### ship\n\n##### Senior Team Emeritus\nInteresting.... all I can really say about this concept beyond not what I would recommend using.\n\nTakes a lot of trust on their part that someone will be able to do this properly and safely. I could do it, how many of you could really really do it properly without any little \u201cslight differences\u201d to the concept? Don\u2019t try modifying their design in making it a threefer as it\u2019s now your liability and most likely won\u2019t work out as per design.\n\nBrass tacks... not recommended on my part to attempt this.\n\nUnion connector company ) http:\/\/www.unionconnector.com\/uc\/index.cfm still sells the classic stage pin cube tap both in the cable mount and plug as a part of it attached version. A few weeks ago in fact I had a customer looking for them and it took some research but I was able to find that such a thing I have not seen in years still exists. Don\u2019t remember where my links for photos of them went or their part numbers - the Union Connector website doesn\u2019t have part numbers and photos but these are the SOP way of doing a cube tap as per a block of them instead of a cord mounted one.\n\nCheck out BMI\u2019s Edison to Stage pin adaptor http:\/\/www.bmisupply.com\/ photo #B of page 47 of their catalog. Very cool in converting an Edison wall outlet to a cord mounted stage pin cable. This especially counteracts the torque issues that the above Union cube tap, much less this Bates twofer would be problematic with in other than cord to cord mount situations - with support of them that a \u201cY\u201d two or threefer doesn\u2019t have problems with. = longer the length, the more torque on the panel mount plug.\n\nBetter selling point of the Union cube tap is that it\u2019s designed to be what it is and not some part added to stock parts in something that ain\u2019t designed to be that and won\u2019t be near what it was before. The union cube taps are built like tanks. At one point I remember even converting some two pin to three pin type. Was ever so slightly off on my pin spacing but otherwise worked well. Been at least ten years now since I have last seen any such thing. I\u2019m a fan of the Bates stage pin plug, not one of this concept.\n\n#### BWTRIX\n\n##### Member\nI \"invented\" the trixfer after being accosted by a local electrical inspector threatning to shut down my fancy black tie fund raiser saying I had plugs rated for one cable having two cables in it (standard twofer) ....he right it is a code violation....i didt'nt really like molded assemblies..so I started playing around with glue screws and connectors...the trix-fer was born....I approached rosco first who held me up for a year first saying they would make it then after the failure of the wall hugger chickened out....so i took it to bates who made the molds and away we went...the pictures that Bates had not only showed trusses in wrong positions but it wasn't very clear of what this device does...\n\n1st Have you priced twofers lately? a complete trixfer sells for $25+1.00\/ft (12\/3sowa) cheaper if you don't count your own labor to assemble them. 2Nd there are now 2 versions a double female great for connector strip tails (a twofer for$10 more than a single stage pin...works for breakouts as well) or male\/female great on jumpers 10-25 ft our favorite most twofer applications require a jumper to reach the next fixture unless double hung. you can put them on a fixture directly and plug another directly into it (great for double hung torms) streaching a circuit over 4 fixtures on a pipe is a snap with 4 10' trixfers, cyc lights ...well you get the picture.\n\nBates pushed a 36\" version but that to me is the least practicle twofering jumpers are the way to go....and it passes code!!!\n\n#### chausman\n\n##### Chase\nFight Leukemia\n... I pretty sure there's no such thing as a cable-stretcher. ...\nThere is a cable stretcher. PRG in Los Angeles has one that they use when they get their extension cable back. They stretch the cable, wash it then but it back in one of the giant crates they store everything in. It is massive and will even test cables when they come back.\n\n#### dramatech\n\n##### Well-Known Member\nI \"invented\" the trixfer\n\nApparently you are not the only one. Five years ago I took a male and a female, older union style, connector and attached them to a flat piece of aluminum with pop rivets. Then brought a cable out through the cover and attached a cable with a female connector.\nHad never heard of the trixfer at the time. Since seeing an announcement in one of the trades, I have heard nothing but negative feelings from most of the posters on this and lightnetwork.\nI guess that I am just some oddball, but I have about 30 of them at my theatre, and have been using them as the 2fer of choice for the last five years. Mine aren't quite as classy as the commercial bates units, but they do the job rather nicely.\n\nTom Johnson\nFlorida's Most Honored Community Theatre\n\n#### BWTRIX\n\n##### Member\nWell then obviously a good idea... someone once said there are few truly new unique ideas...perhaps invent was the wrong word...substitute \"commercialize\"....I've been gluing and screwing connectors together and developing this for nearly 7 years now as a technician just trying to bring a beter solution to the world of \"two-fers\"... Code issues will bring the two fer issue to a head and as mentioned I really do not like the molded solution (and cost).\n\nInjection molds are expensive That's why rosco stalled for a year... luckly when they released us from our \"deal\". Bates jumped right on the wagon...unfortunatly didn't have the knowledge to market Trix-fer effectivly....distributors that invested heavily in molded two-fer blanks that sell for $11+ for 5 bucks worth of material were not eager to push it. The Bates connector parts used are standard issue as are the pins... I'm pushing for a crimp pin solution for mass assembly work... just need to convince Bates a oversized crimp barrel is worth retooling to accomodate 2 #12 wires... current barrel is for 1-#12-16 they use for fixtures...we have pico pneumatic crimper and after making 100's by hand think a quicker more reliable, repeatable connection would be possible for less time spent (less cost) ... ring crimps we currently hand crimp and put a little daub of soldier on the exposed tip of copper sticking out of the crimp not allowing it to wick up the cable a technique we developed over 30 years ago. We offer lifetime warrantee on that connection (as all of our hand made stagepin cables (and have 33 years of testomony on this method of connection in rental and touring applications) So once again The only \"special\" part of the trixfer is the extended female \"trix-fer\" molding.... the male (or female) and pins are standard Bates connector (less cover). I think negative comments are due to lack of information on what these things really do i only found this forum last night in googling TRIXFER saw last entry was 3 years ago and replied Commercial acceptance will only come after people start using them...we run a rental shop and find the Trix-fer jumpers (10' and 25') to be a popular product in our market... When priced as an adder to a jumper, fixture or tail the$'s speak for themselves. Once you grab hold of the concept (as you have) it's a no brainer.\n\nspread the word please...my kids need to go to college\n\nbw\n\n#### BWTRIX\n\n##### Member\nBates did trade mark the name 5 or 6 years ago\nand was going to see if patientable (ideas are tough, patient cost money to commit\/prepare for possible future ligitations) I was just trying to get a better mousetrap to use did not think this was a post-it note type of project ...I believe it might be patient pending but that is bate's deal. I did develope this independantly of any other influence....\n\nlooked up patient and read... it could be is a cousin of what we have\n\nbw\n\n#### mstaylor\n\n##### Well-Known Member\nDeparted Member\nMaybe I am just not seeing the advantage of this. The picture isn't showing the difference.","date":"2020-02-21 15:59:11","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\": 0.19856613874435425, \"perplexity\": 2541.8684491424}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.3, \"absolute_threshold\": 10, \"end_threshold\": 5, \"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\/1581875145533.1\/warc\/CC-MAIN-20200221142006-20200221172006-00371.warc.gz\"}"} | null | null |
Kingsland Viaduct is a railway viaduct about in length, almost wholly within the present London Borough of Hackney in the north-east part of Inner London. It was built in the 1860s, but was disused from 1986 until it was reopened to carry the London Overground in 2010. The viaduct is owned by Transport for London. Since then it has carried East London Line services between Shoreditch High Street and Dalston.
History
Initial operation
It was built as the main part of the North London City Extension, providing a more direct link from the North London Railway to the City of London (the City) at Broad Street, next door to the site that became Liverpool Street station.
Previously the North London Line's access from west of Dalston to the City was by a circuitous route via Hackney, Homerton, Bow and Stepney to Fenchurch Street. The extension, originally with three lines, shortened this considerably by providing a direct route from Dalston over the Kingsland Viaduct. It was authorised by the North London Railway Act of 22 July 1861. It is said to have cost £1 million and displaced 4,500 people.
The extension was accessed by a triangular junction, accessible from both the west and east directions on the North London Railway.
Passenger traffic into Broad Street began on 1 November 1865, and was said to have doubled the number of passengers to 14 million in 1868. Initially, there were services at fifteen-minute intervals to both Hampstead and Bow.
A fourth line was added to the extension in 1874. Broad Street originally had seven platforms, extended to eight in 1891 and nine in 1913. The lines were electrified with two conductor rails at 600 V DC, with services beginning on 1 October 1916, using Oerlikon rolling stock. However, only the western five platforms at Broad Street served electrified lines, the remaining lines being unelectrified. Electric services to Watford did not begin until 10 July 1922.
Reopening
The last scheduled train over the viaduct was on 27 June 1986, after which the line and its remaining stations were all closed.
After that there were proposals in 1993 to extend London Underground's East London Line along it from its Shoreditch terminus to Dalston Junction. The plan received the support of a public inquiry in 1994 and it was envisaged that the construction of the extension and the station itself would begin in 1996, to be completed by 1998. The project was finally approved by the Government in 1996 but a lack of funding forced the project to be delayed in 1997.
However, in 2007 the London Overground was created, which included the North London Line (the North London railway to which the viaduct connects at Dalston), and the East London Line was transferred to the London Overground to be extended south and to the north to meet the North London Line. To do this it followed the 1993 route along the Kingsland Viaduct. Work was complete in 2010 and on 27 April 2010 the first services began along the viaduct for the first time in over two decades.
To facilitate this the viaduct has been refurbished, and many bridges over roads replaced; a remnant of the brick viaduct between New Inn Yard and Holywell Lane was demolished to make way for a connecting concrete viaduct to the new bridge over Shoreditch High Street.
The southern end of the viaduct, from the old Broad Street/Liverpool Street terminus and where the East London Line crosses onto the viaduct at Hollywell Lane, has been given over to development, including the Broadgate Tower.
Stations
There were four stations on and adjacent to the viaduct:
Dalston Junction was at the bottom of the incline to the north of the viaduct, at the apex of the triangular junction, actually below street level.
Haggerston, not to be confused with the earlier NLR station on a slightly different site
Shoreditch, straddling Old Street and parallel to Shoreditch High Street, not to be confused with the former (now closed) London Underground station of the same name, some distance to the east
Broad Street, at the southern end of the viaduct, which had suburban services throughout its existence, and limited main line services.
As of 2010, the new stations on the extended viaduct, or adjacent, are;
A rebuilt Dalston Junction on the same site as before.
A rebuilt Haggerston station slightly to the north.
Hoxton, a completely new station, opposite the rear of the Geffrye museum.
Shoreditch High Street, a completely new station.
References
Connor, J.E. (1995) Broad Street to Poplar: A Photographic Journey, Colchester : Connor & Butler, 53p
Mitchell, Vic and Smith, Keith (1997) North London line : Broad Street to Willesden Jn. via Hampstead Heath, Midhurst : Middleton Press, 96p
White, Henry Patrick (1987) A Regional History of the Railways of Great Britain, Volume 3 - Greater London, Newton Abbott : David St John Thomas, 237p
Buildings and structures in the London Borough of Hackney
Railway viaducts in London | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 7,515 |
\section{Introduction}
To every symmetric or alternating bilinear form, and hence to every quadratic form when the base field is not of characteristic two, we can associate a central simple algebra with involution. In this way, the theory of quadratic forms is embedded into the larger theory of algebras with involution, and through the use of this correspondence, quadratic form theory has provided inspiration for the study of algebras with involution (see \cite{Knus:1998}).
Pfister forms are a central concept in the modern algebraic theory of quadratic forms. It is therefore natural to look for class of central simple algebras with involution which extends the notion of a Pfister form. This question was first raised in \cite{parimala:pfisterinv}. Involutions adjoint to Pfister forms are tensor products of quaternion algebras with involution. Thus, tensor products of quaternion algebras with involution are a natural candidate.
In characteristic two, quadratic forms and symmetric bilinear forms are not equivalent objects. The relation between bilinear Pfister forms and totally decomposable involutions in characteristic two was studied in \cite{dolphin:orthpfist}.
In order to have an object defined on a central simple algebra that corresponds to a quadratic form after splitting, the notion of a quadratic pair was introduced in \cite[\S5]{Knus:1998}. In particular, one may use quadratic pairs to give an intrinsic definition of twisted orthogonal groups in a manner that includes fields of characteristic two (see \cite[\S23.B]{Knus:1998}).
Algebras with quadratic pair associated to quadratic Pfister forms are tensor products of quaternion algebras with involution and a quaternion algebra with quadratic pair.
One may ask whether all such totally decomposable quadratic pairs on a split central simple algebra
are adjoint to a quadratic Pfister form.
In characteristic different from two, where quadratic pairs are equivalent to orthogonal involutions, this is known to hold by the main result of \cite{Becher:qfconj}, which says that in this case a
totally decomposable orthogonal involution on a split algebra
is adjoint to a Pfister form.
In this article we prove the corresponding result for quadratic pairs over fields of characteristic two.
Our approach is particular to fields of characteristic two. It allows us to capture more information on the resulting quadratic Pfister form (see \cref{cor:main}), in a way that is not possible in general over fields of characteristic different from two (see \cref{ex:countex}).
This approach uses the unusual properties of totally decomposable involutions in characteristic two found in \cite{dolphin:orthpfist}. In particular, we use that the isotropy behaviour of a totally decomposable orthogonal involution can be captured in the isotropy behaviour of an associated bilinear Pfister form (see \cref{thm:pfisterinvar}).
The method used in \cite{Becher:qfconj} in characteristic different from two is based on a ramification exact sequence for Witt groups of quadratic forms over function fields of conics and the excellence property of these function fields.
The latter result is known to extend to the case of arbitrary characteristic, but the former is not yet available in characteristic two in a suitable form.
We intend to give a characteristic free proof in a future article.
Our solution to the missing case of characteristic two involves several basic properties of quadratic pairs,
which we did not find explicitly in the literature. Following \cite[\S5]{Knus:1998}, we present such statements without assumptions on the characteristic, for ease of future reference.
\section{Quadratic forms over fields}
In this section we recall the terminology and results we use from quadratic form theory.
We refer to \cite[Chapters 1 and 2]{Elman:2008} as a general reference on symmetric bilinear and quadratic forms and for any basic notation and concepts not defined here.
For two objects $\alpha$ and $\beta$ in a certain category, we write $\alpha\simeq\beta$ to indicate that they are {isomorphic}, i.e.~that there exists an isomorphism between them.
This applies in particular to algebras with involution or with quadratic pair, but also to quadratic and bilinear forms, where the corresponding isomorphisms are called {isometries}.
Throughout, let $F$ be a field. Let $\kar(F)$ denote the characteristic of $F$ and let $F^\times$ denote the multiplicative group of $F$.
A \emph{bilinear form over $F$} is a pair $(V,b)$ where $V$ is a finite-dimensional $F$-vector space and $b$ is a $F$-bilinear map $b:V\times V\rightarrow F$.
The \emph{radical of $(V,b)$} is the set ${\mathrm{rad}}(V,b)=\{x\in V\mid b(x,y)=0 \textrm{ for all } y\in V\}$.
We say that $(V,b)$ is
\emph{degenerate} if ${\mathrm{rad}}(V,b)\neq \{0\}$, and \emph{nondegenerate} otherwise.
Let $\varphi=(V,b)$ be a bilinear form over $F$.
We say that $\varphi$ is \emph{symmetric} if $b(x,y)=b(y,x)$ for all $x,y\in V$.
We call $\varphi$ \emph{alternating} if $b(x,x)=0$ for all $x\in V$.
The form $\varphi$ is said to be \emph{isotropic} if there exists an $x\in V\backslash\{0\}$ such that $b(x,x)=0$, and \emph{anisotropic} otherwise.
We call a subspace $W\subseteq V$ \emph{totally isotropic} (with respect to $b$) if $b|_{W\times W}=0$.
If $\varphi$ is nondegenerate and there exists a totally isotropic subspace $W\subseteq V$ such that $\dim_F(W)=\frac{1}{2}\dim_F(V)$, then we call $\varphi$ \emph{metabolic}.
For $a_1,\ldots,a_n\in F^\times$ the symmetric $F$-bilinear map $b:F^n\times F^n\rightarrow F$ given by $(x,y)\mapsto \sum_{i=1}^n a_ix_iy_i$ yields a
symmetric bilinear form $(F^n,b)$ over $F$, which we denote by $\qf{a_1,\ldots,a_n}$.
For a positive integer $m$, by an \emph{$m$-fold bilinear Pfister form over $F$} we mean a nondegenerate symmetric bilinear form over $F$ that is isometric to
$\qf{1,a_1}\otimes\ldots\otimes\qf{1,a_m}$ for some
$a_1,\ldots,a_m\in F^\times$. We call $\qf{1}$ the \emph{$0$-fold bilinear Pfister form}.
By \cite[(6.3)]{Elman:2008}, a bilinear Pfister form is either anisotropic or metabolic.
By a \textit{quadratic form over $F$} we mean a pair $(V,q)$ of a finite-dimensional $F$-vector space $V$ and a map $q:V\rightarrow F$ such that, firstly, $q(\lambda x)=\lambda^2q(x)$ holds for all $x\in V$ and $\lambda\in F$, and secondly, the map $b_q:V\times V\rightarrow F\,,\,(x,y)\longmapsto q(x+y)-q(x)-q(y)$ is $F$-bilinear.
Then $(V,b_q)$ is a symmetric bilinear form over $F$, called the \emph{polar form of $(V,q)$}. If $b_q$ is nondegenerate, we say that $(V,q)$ is \emph{nonsingular}, otherwise we say that $(V,q)$ is \emph{singular}. If $b_q$ is the zero map, then we say $(V,q)$ is \emph{totally singular}.
By the \emph{quadratic radical of $(V,q)$} we mean the set
${\mathrm{rad}}(V,q)= \{ x\in {\mathrm{rad}}(V,b_q)\mid q(x)= 0\}$.
We say that $(V,q)$ is \emph{regular} if ${\mathrm{rad}}(V,q)=\{0\}$.
For a symmetric bilinear form $(V,b)$ over $F$, the map $q:V\rightarrow F$ given by $q_b(x)=b(x,x)$ makes $(V,q_b)$ a quadratic form over $F$.
We call $(V,q_b)$ the \emph{quadratic form associated to $(V,b)$}.
If $\kar(F)=2$ then this quadratic form is totally singular.
Let $\rho=(V,q)$ and $\rho'=(V',q')$ be quadratic forms over $F$. By an \emph{isometry of quadratic forms $\rho\rightarrow\rho'$} we mean an isomorphism of $F$-vector spaces $f:V\rightarrow V'$ such that $q(x)=q'(f(x))$ for all $x\in V$.
We say $\rho$ is \emph{isotropic} if
$q(x)=0$ for some $x\in V\backslash\{0\}$, and \emph{anisotropic} otherwise.
By \emph{a totally isotropic subspace of $\rho$} we mean an $F$-subspace $W$ of $V$ such that $q|_{W}=0$.
We call the maximum of the dimensions of all totally isotropic subspaces of $\rho$ the \emph{Witt index of $\rho$}, denoted $i_W(\rho)$.
Assume $\rho$ is nonsingular. Then $i_W(\rho)\leqslant\frac{1}{2}\dim(\rho)$ (see \cite[(7.28)]{Elman:2008}) and if $i_W(\rho)= \frac{1}{2}\dim (\rho)$
we say that $\rho$ is \emph{hyperbolic}. We denote the anisotropic part of $\rho$ by $\rho_{\mathrm{an}}$ (see \cite[(8.5)]{Elman:2008}).
We say that the quadratic forms $\rho_1$ and $\rho_2$ over $F$ are \emph{similar} if there exists an element $c\in {F}^\times$ such that $\rho_1\simeq c\rho_2$.
Recall the concept of a tensor product of a symmetric or alternating bilinear form and a quadratic form (see \cite[p.51]{Elman:2008}).
For a quadratic form $\rho$ over $F$, we say that $\varphi$ \emph{factors $\rho$} if there exists a quadratic form $\rho'$ over $F$ such that $\rho\simeq \varphi\otimes \rho'$. Similarly, we say a quadratic form $\rho'$ over $F$ \emph{factors $\rho$} if there exists a symmetric bilinear form $\varphi$ over $F$ such that $\rho\simeq \varphi\otimes\rho'$.
For a positive integer $m$, by an \emph{$m$-fold (quadratic) Pfister form over $F$} we mean a quadratic form that is isometric to the tensor product of a $2$-dimensional nonsingular quadratic form representing $1$ and an $(m-1)$-fold bilinear Pfister form over $F$.
Pfister forms are either anisotropic or hyperbolic (see \cite[(9.10)]{Elman:2008}).
Let $\rho$ be a regular quadratic form over $F$.
If $\dim(\rho)\geqslant 3$ or if $\rho$ is anisotropic and $\dim(\rho)=2$, then we call the function field of the projective quadric over $F$ given by $\rho$ the \emph{function field of $\rho$} and denote it by $F(\rho)$. In the remaining cases we set $F(\rho)=F$.
This agrees with the definition in \cite[\S22]{Elman:2008}.
For an anisotropic symmetric bilinear form $\varphi$, the quadratic form $\rho$ associated to $\varphi$ is regular. We call $F(\rho)$ the \emph{function field of $\varphi$} and we denote it by $F(\varphi)$.
Let $K/F$ be a field extension. Then we write $(V,q)_K=(V\otimes_F K, q_K) $ where $q_K$ is the unique quadratic map such that $q_K( v\otimes k )= k^2q(v)$ for all $v\in V$ and $k\in K$.
\begin{prop}\label{prop:funcfieldbf}
Let $\rho$ be a nonsingular quadratic form and let $\varphi$ be an anisotropic bilinear Pfister form over $F$. If $\rho_{F(\varphi)}$ is hyperbolic then either $\rho$ is hyperbolic or $\varphi$ factors $\rho_{\mathrm{an}}$.
\end{prop}
\begin{proof} See \cite[(5.2)]{HoffmannLaghribi:Isoqfffquadricc2}.
\end{proof}
\section{Algebras with involution}\label{section:basicsalgs}
We refer to \cite{pierce:1982} as a general reference on finite-dimensional algebras over fields, and for central simple algebras in particular, and to \cite{Knus:1998} for involutions.
Let $A$ be an (associative) $F$-algebra. We denote the centre of $A$ by $Z(A)$.
For a field extension $K/F$, the $K$-algebra $A\otimes_F K$ is denoted by $A_K$.
An element $e\in A$ is called an \emph{idempotent} if $e^2=e$.
An $F$-\emph{involution on $A$} is an $F$-linear map $\sigma:A\rightarrow A$ such that $\sigma(xy)=\sigma(y)\sigma(x)$ for all $x,y\in A$ and $\sigma^2=\textrm{id}_A$.
Assume now that $A$ is finite-dimensional and simple (i.e.~it has no nontrivial two-sided ideals).
By Wedderburn's Theorem (see \cite[(1.1)]{Knus:1998}), $A\simeq{\mathrm{End}}_D(V)$ for a finite-dimensional $F$-division algebra $D$ and a finite-dimensional right $D$-vector space $V$.
Furthermore, the centre of $A$ is a field and $\dim_{Z(A)}(A)$ is a square number, whose positive square root is called the \emph{degree of $A$} and is denoted $\mathrm{deg}(A)$. The degree of $D$ is called the \emph{index of $A$} and denoted $\mathrm{ind}(A)$. We call $A$ \emph{split} if $\mathrm{ind}(A)=1$, that is $A\simeq {\mathrm{End}}_F(V)$ for some finite-dimensional right $F$-vector space $V$.
If $Z(A)=F$, then we call the $F$-algebra $A$ \emph{central simple} and
we call a field extension $K/F$ such that $A_K$ is split a \emph{splitting field of $A$}.
If $A$ is a central simple $F$-algebra then we denote ${\mathrm{Trd}}_A:A{\longrightarrow} F$ the reduced trace map and ${\mathrm{Nrd}}_A:A{\longrightarrow} F$ the reduced norm map, as defined in \cite[(1.6)]{Knus:1998}.
By an \emph{$F$-algebra with involution} we mean a pair $(A,\sigma)$ of a finite-dimensional central simple $F$-algebra $A$ and an $F$-involution $\sigma$ on $A$ (note that we only consider involutions that are linear with respect to the centre of $A$, that is involutions of the first kind, here). We use the following notation:
$\mathrm{Sym}(A,\sigma)= \{ a\in A\mid\sigma(a)=a\}$,
$\mathrm{Skew}(A,\sigma)= \{ a\in A\mid\sigma(a)=-a\}$ and
${\mathrm{Alt}}(A,\sigma)= \{ a-\sigma( a)\,|\, a\in A\}$.
These are $F$-linear subspaces of $A$.
Let $(A,{\sigma})$ and $(B,\tau)$ be $F$-algebras with involution.
By an \emph{isomorphism of $F$-algebras with involution} $\Phi:(A,\sigma)\rightarrow(B,\tau)$ we mean an $F$-algebra isomorphism $\Phi: A\rightarrow B$ satisfying $\Phi\circ\sigma=\tau\circ\Phi$.
On the $F$-algebra $A\otimes_F B$ we obtain an $F$-involution ${\sigma}\otimes \tau$, whereby $(A\otimes_F B,\sigma\otimes \tau)$ is an $F$-algebra with involution, which we denote by $(A,\sigma)\otimes(B,\tau)$.
For a field extension $K/F$ we write $(A,{\sigma})_K=(A\otimes_F K,{\sigma}\otimes\mathrm{id})$.
We call $(A,\sigma)$ \emph{isotropic} if there exists $a\in A\backslash\{0\}$ such that $\sigma(a)a=0$, and \emph{anisotropic} otherwise. An idempotent $e\in A$ is called \emph{metabolic with respect to $\sigma$} if $\sigma(e)e=0$ and $\dim_FeA=\frac{1}{2}\dim_F A$. We call $(A,\sigma)$ \emph{metabolic} if $A$ contains a metabolic idempotent element with respect to $\sigma$. For more information on metabolic involutions, see \cite{dolphin:metainv}.
To every nondegenerate symmetric or alternating bilinear form $\varphi=(V,b)$ over $F$ we can associate an algebra with involution in the following way. Let $A={\mathrm{End}}_F(V)$. Then there is a unique involution $\sigma$ on $A$ such that
$$b(x,f(y))=b(\sigma(f)(x),y) \quad \textrm{ for all } x,y\in V \textrm{ and all } f\in A .$$
We denote this $F$-involution on $A$ by ${\mathrm{ad}}_b$.
We call $(A,{\mathrm{ad}}_b)$ the \emph{$F$-algebra with involution adjoint to $\varphi$} and we denote it by ${\mathrm{Ad}}(\varphi)$. For every split $F$-algebra with involution $(A,\sigma)$, there exists a nondegenerate symmetric or alternating bilinear form $\psi$ over $F$ such that $(A,\sigma)\simeq{\mathrm{Ad}}(\psi)$ (see \cite[(2.1)]{Knus:1998}). The following is well-known, but we include a proof for completeness.
\begin{prop}\label{lemma:tensorbf} Let $\varphi$ and $\psi$ be nondegenerate symmetric bilinear forms over $F$. Then
${\mathrm{Ad}}(\varphi\otimes \psi)\simeq {\mathrm{Ad}}(\varphi)\otimes {\mathrm{Ad}}(\psi). $
\end{prop}
\begin{proof} Let $\varphi=(V,b)$ and $\psi=(W,b')$.
Let $f\in {\mathrm{End}}_F(V)$ and $g\in {\mathrm{End}}_F(W)$. Then for all $u,v\in V$, $w,t\in W$ we have
\begin{eqnarray*}(b\otimes b') (f\otimes g(u\otimes w), (v\otimes t))& =&(b\otimes b')((f(u)\otimes g(w)), (v\otimes t))
\\ &= &b(f(u), v)\cdot b'(g(w), t)
\\ &=& b(u, {\mathrm{ad}}_b(f)(v))\cdot b'(w, {\mathrm{ad}}_{b'}(g)(t))
\\ &=& (b\otimes b') ((u\otimes w), ({\mathrm{ad}}_b(f)(v)\otimes {\mathrm{ad}}_{b'}(g)(t)))\,.
\end{eqnarray*}
Therefore, by bilinearity of $b\otimes b'$, we have that ${\mathrm{ad}}_{b\otimes b'}(f\otimes g)= {\mathrm{ad}}_b(f)\otimes {\mathrm{ad}}_b(g) $.
Using this, it follows from the linearity of ${\mathrm{ad}}_{b\otimes b'}$ that the natural isomorphism of $F$-algebras $\Phi:{\mathrm{End}}_F(V)\otimes_F{\mathrm{End}}_F(W)\rightarrow {\mathrm{End}}_{F}(V\otimes_F W)$ is an isomorphism of the $F$-algebras with involution in the statement.
\end{proof}
We distinguish two \emph{types} of $F$-algebras with involution. An $F$-algebra with involution is said to be \emph{symplectic} if it becomes adjoint to an alternating bilinear form over some splitting field of the associated $F$-algebra, and \emph{orthogonal} otherwise.
In characteristic different from two, these types are distinguished by the dimensions of the spaces of symmetric and alternating elements. However in characteristic two these dimensions do not depend on the type (see \cite[(2.6)]{Knus:1998}).
An $F$-\emph{quaternion algebra} is a central simple $F$-algebra of degree $2$.
Let $Q$ be an $F$-quaternion algebra.
By \cite[(2.21)]{Knus:1998}, the map $Q\rightarrow Q,$ $x\mapsto {\mathrm{Trd}}_Q(x)-x$ is the unique symplectic involution on $Q$; it is called the \emph{canonical involution of $Q$}.
Any $F$-quaternion algebra has a basis $(1,u,v,w)$ such that
$$u^2 =u+a, v^2=b\,\textrm{ and }\, w=uv=v-vu\,,$$
for some $a\in F$ with $-4a\neq 1$ and $b\in F^\times$
(see \cite[Chap.~IX, Thm.~26]{Albert:1968}); such a basis is called an \emph{$F$-quaternion basis}.
Conversely, for $a\in F$ with $-4a\neq 1$ and $b\in F^\times$
the above relations uniquely determine an $F$-quaternion algebra (up to $F$-isomorphism), which we denote by $[a,b)$.
By the above, up to isomorphism any $F$-quaternion algebra is of this form. The following result can be recovered from \cite[p.104, Thm.~4]{Draxl:1983}, but we include a direct argument.
\begin{lemma}\label{qrep}
Let $Q$ be an $F$-quaternion algebra and $v\in Q\setminus F$ be such that $v^2\in F^\times$. There exist an element and $u\in Q$ such that $uv=v(1-u)$ and $u^2-u=a$ for some $a\in F$ with $-4a\neq 1$. That is, $(1,u,v,uv)$ is an $F$-quaternion basis of $Q$.
\end{lemma}
\begin{proof}
If $\kar(F)\neq 2$ then
it is well-known that
there exists an invertible element $x\in Q$ such that $xv+vx=0$ and we set $u=x+\frac{1}2$.
Assume that $\kar(F)=2$.
Consider the $F$-linear map $Q\to Q, x\mapsto xv+vx$. Its kernel and its image are equal to $F[v]$.
Hence $uv+vu=v$ for some $u\in Q$, and it follows that $u^2+u\in F[v]\cap F[u]=F$.
\end{proof}
With an $F$-quaternion basis $(1,u,v,w)$ of $Q$,
we define $F$-involutions $\gamma$, ${\sigma}$ and $\tau$ on $Q$ via their action on $u$ and $v$ as follows. We let
\begin{eqnarray*}
\gamma: & u\mapsto 1-u, & v\mapsto-v\\
{\sigma}: & u\mapsto 1-u, &v\mapsto \hphantom{-}v\\
\tau:& u\mapsto u,\hphantom{-1} & v\mapsto -v\,.
\end{eqnarray*}
Note that $\gamma$ is the canonical involution on $Q$. Further, if $\kar(F)=2$, then $\gamma={\sigma}$ and hence ${\sigma}$ is symplectic. Otherwise ${\sigma}$ is orthogonal. We use the notation $$[a\,\cdot\!\!\mid\!\!\cdot\, b)=(Q,\gamma), \quad [a\,\cdot\!\!\mid b)=(Q,{\sigma}),\quad[a\mid\!\!\cdot\, b)= (Q,\tau)\,.$$
\begin{prop}\label{prop:orth}
Let $(Q,{\sigma})$ be an $F$-quaternion algebra with orthogonal involution. Then there exists $a\in F$ with $-4a\neq 1$ and $b\in F^\times$ such that $(Q,{\sigma})\simeq [a\mid\!\!\cdot\, b)$.
\end{prop}
\begin{proof}
Let $\gamma$ be the canonical involution of $Q$.
By \cite[(2.21)]{Knus:1998}, we have ${\sigma}={\mathrm{Int}}(v)\circ \gamma$ for some invertible element $v\in \mathrm{Skew}
(Q,\gamma)\backslash F$.
Then
$v^2=-\gamma(v)v \in F^\times$ and we set $b=v^2$. By \cref{qrep}, there exists an element $u\in Q$ such
that $uv=v(1-u)$ and $u^2-u=a$ for some $a\in F$ with $-4a\neq 1$. Hence $(1,u,v,uv)$ is an $F$-quaternion basis of $Q$ and we have that $\gamma(u)=1-u$ and $\gamma(v)=-v$. Hence ${\sigma}(u)=u$ and ${\sigma}(v)=-v$, and further $(Q,{\sigma})\simeq [a\mid\!\!\cdot\, b)$.
\end{proof}
We call an $F$-algebra with involution \emph{totally decomposable} if it is isomorphic to a tensor product of $F$-quaternion algebras with involution. Note that by \cref{lemma:tensorbf}, the $F$-algebra with involution adjoint to a bilinear Pfister form over $F$ is totally decomposable.
Let $(A,\sigma)$ be an $F$-algebra with orthogonal involution.
By \cite[(7.1)]{Knus:1998}, for any $F$-algebra with orthogonal involution $(A,{\sigma})$ with $\deg(A)$ even and any $a,b\in {\mathrm{Alt}}(A,{\sigma})$ we have ${\mathrm{Nrd}}_A(a)F^{\times 2}={\mathrm{Nrd}}_A(b) F^{\times 2}$.
Therefore, as in \cite[\S7]{Knus:1998}, we may make the following definition. The \emph{determinant of $(A,\sigma)$}, denoted $\Delta(A,\sigma)$, is the square class of the reduced norm of an arbitrary alternating unit, that is
$$\Delta(A,\sigma)= \mathrm{Nrd}_A(a)\cdot F^{\times2}\textrm{ in } F^\times/ F^{\times2} \quad \textrm{ for any} \,\,a\in {\mathrm{Alt}}(A,\sigma)\cap A^\times.$$
For the rest of this section, we assume that $\kar(F)=2$.
Let $(A,\sigma)$ be a totally decomposable $F$-algebra with orthogonal involution. That is $$(A,\sigma)\simeq \bigotimes_{i=1}^n(Q_i,\sigma_i)$$ where $(Q_i,\sigma_i)$ are $F$-algebras with involution for $i=1,\ldots, n$.
Note that we must have that $(Q_i,{\sigma}_i)$ is orthogonal for all $i=1,\ldots, n$ by
\cite[(2.23)]{Knus:1998}.
Let $d_i=\Delta(Q_i,\sigma_i)$. Then the bilinear Pfister form $\pi=\pff{d_1,\ldots, d_n}$ over $F$ does not depend on the choice of the $F$-quaternion algebras with involution $(Q_i,{\sigma}_i)$ in the decomposition of $(A,{\sigma})$ by \cite[(7.3)]{dolphin:orthpfist}. We call this bilinear Pfister form \emph{the Pfister invariant of $(A,\sigma)$} and denote it by $\mathfrak{Pf}(A,\sigma)$.
Note that by \cite[(7.3)]{dolphin:orthpfist}, for any field extension $K/F$ we have that $\mathfrak{Pf}((A,{\sigma})_K)= (\mathfrak{Pf}(A,{\sigma}))_K$.
\begin{prop}\label{thm:pfisterinvar} Assume $\kar(F)=2$.
Let $(A,\sigma)$ be a totally decomposable $F$--algebra with orthogonal involution. Then
$(A,\sigma)$ is anisotropic (resp.~metabolic) if and only if $\mathfrak{Pf}(A,\sigma)$ is anisotropic (resp.~metabolic).
\end{prop}
\begin{proof}
See \cite[(7.5)]{dolphin:orthpfist}.
\end{proof}
\section{Algebras with quadratic pair}
We now recall the definition of and basic results we use on quadratic pairs.
Let $(A,\sigma)$ be an $F$-algebra with involution.
We call an $F$-linear map $f:{\mathrm{Sym}}(A,\sigma)\rightarrow F$ a \emph{semi-trace on $(A,\sigma)$} if it satisfies $f(x+\sigma(x)) = \mathrm{Trd}_A(x)$ for all $x\in A$. By \cite[(4.3)]{dolphin:quadpairs}, if $\kar(F)\neq2$, then $\frac{1}{2}{\mathrm{Trd}}_A|_{{\mathrm{Sym}}(A,\sigma)}$ is the unique semi-trace on $(A,\sigma)$. On the other hand, if $\kar(F)=2$, then the existence of a semi-trace on $(A,\sigma)$ implies that ${\mathrm{Trd}}_A({\mathrm{Sym}}(A,\sigma))=\{0\}$ and hence by \cite[(2.6)]{Knus:1998} that $(A,\sigma)$ is symplectic.
Given an element $\ell\in A$ with $\ell+{\sigma}(\ell)=1$, the map $f:{\mathrm{Sym}}(A,\sigma)\rightarrow F$ given by $x\mapsto {\mathrm{Trd}}_A(\ell x)$ is a semi-trace on $(A,\sigma)$, and conversely every semi-trace on $(A,\sigma)$ is of this form by \cite[(5.7)]{Knus:1998} (although the case where $\kar(F)\neq 2$ and $(A,{\sigma})$ is symplectic is excluded there, the same proof applies). In this case, we say that the semi-trace $f$ on $(A,{\sigma})$ \emph{is given by $\ell$}.
For another element $\ell'\in A$ such that $\ell'+{\sigma}(\ell')=1$, we have that
$\ell$ and $\ell'$ give the same semi-trace on $(A,{\sigma})$ if and only if $\ell-\ell'\in{\mathrm{Alt}}(A,{\sigma})$ (see \cite[(5.7)]{Knus:1998}).
An \emph{$F$-algebra with quadratic pair} is a triple $(A,\sigma,f)$ where $(A,\sigma)$ is an $F$-algebra with involution, which is assumed to be orthogonal if $\kar(F)\neq 2$ and symplectic if $\kar(F)=2$, and where $f$ is a semi-trace on $(A,\sigma)$.
In $\kar(F)\neq2$ the concept of an algebra with quadratic pair is equivalent to the concept of an algebra with orthogonal involution, as
then the semi-trace given by $\frac{1}{2}$ is the unique semi-trace on $(A,\sigma)$.
Given two $F$-algebras with quadratic pair $(A,\sigma,f)$ and $(B,\tau,g)$, by an \emph{isomorphism of $F$-algebras with quadratic pair} $\Phi:(A,\sigma,f)\rightarrow(B,\tau,g)$ we mean an isomorphism of the underlying $F$-algebras with involution satisfying $ f=g\circ\Phi$.
Let $(A,\sigma,f)$ be an $F$-algebra with quadratic pair. We call $(A,\sigma,f)$ \emph{isotropic} if there exists an element $s\in{\mathrm{Sym}}(A,\sigma)\backslash\{0\}$ such that $s^2=0$ and $f(s)=0$, and \emph{anisotropic} otherwise.
In particular, if $(A,{\sigma},f)$ is isotropic, then $A$ has zero divisors.
We call an idempotent $e\in A$ \emph{hyperbolic with respect to $\sigma$ and $f$} if $\sigma(e)=1-e$ and $f(eA\cap {\mathrm{Sym}}(A,\sigma))=\{0\}$.
We say that the $F$-algebra with quadratic pair $(A,\sigma,f)$ is \emph{hyperbolic} if $A$ contains a hyperbolic idempotent with respect to $\sigma$ and $f$.
We describe,
following \cite[\S5]{Knus:1998},
how a nonsingular quadratic form gives rise to an algebra with quadratic pair.
Let $\rho=(V,q)$ be a nonsingular quadratic form over $F$ with polar form $(V,b_q)$.
By declaring
$$(v_1\otimes w_1)\ast(v_2\otimes w_2)=b_q(w_1,v_2)\cdot(v_1\otimes w_2)\quad \textrm{ for }\quad \,v_1,v_2,w_1,w_2\in V\,$$
a product $\ast$ is defined on the tensor product $V\otimes_F V$ making it into an $F$-algebra.
By declaring $\sigma(v\otimes w)=w\otimes v$ for $v,w\in V$ we obtain an $F$-involution ${\sigma}$ on the $F$-algebra $V\otimes_FV$.
Then by \cite[(5.1)]{Knus:1998}, the $F$-linear map $\Phi:V\otimes_F V\rightarrow {\mathrm{End}}_F(V)$ determined by $$\Phi(u\otimes v)(w)= b_q(v,w) u\quad \mbox{ for } \, u,v,w\in V$$ yields an isomorphism of $F$-algebras with involution ${\mathrm{Ad}}(V,b_q){\longrightarrow} (V\otimes_F V,{\sigma})$.
According to \cite[(5.11)]{Knus:1998} there is a unique semi-trace $f_q:{\mathrm{Sym}}({\mathrm{Ad}}(V,b_q))\rightarrow F$ such that
$f_q(\Phi(v\otimes v))=q(v)$ for $v\in V,$ which yields an $F$-algebra with quadratic pair $${\mathrm{Ad}}(\rho)=({\mathrm{End}}_F(V),{\mathrm{ad}}_{b_q}, f_q)\,,$$ called the \emph{adjoint $F$-algebra with quadratic pair of $\rho$}.
We say that an $F$-algebra with quadratic pair $(A,\sigma,f)$ is \emph{adjoint to $\rho$} if $(A,\sigma,f)\simeq {\mathrm{Ad}}(\rho)$.
By \cite[(5.11)]{Knus:1998}, for any split $F$-algebra with quadratic pair $(A,{\sigma},f)$, there exists a nonsingular quadratic form $\rho$ over $F$ such that $(A,\sigma,f)\simeq{\mathrm{Ad}}(\rho)$, and for two nonsingular quadratic forms $\rho$ and $\rho'$ over $F$, we have that ${\mathrm{Ad}}(\rho)\simeq{\mathrm{Ad}}(\rho')$ if and only if $\rho$ and $\rho'$ are similar.
\begin{prop}\label{prop:hypiffquad}
Let $\rho$ be a nonsingular quadratic form over $F$. Then $\rho$ is isotropic (resp.~hyperbolic) if and only if ${\mathrm{Ad}}(\rho)$ is isotropic (resp.~hyperbolic).
\end{prop}
\begin{proof}
The statement on isotropy follows from \cite[(6.3) and (6.6)]{Knus:1998}. See \cite[(6.13)]{Knus:1998} for the statement on hyperbolicity.
\end{proof}
For any field extension $K/F$ we will use the notation $(A,\sigma,f)_K$ for the $K$-algebra with quadratic pair $(A_K,\sigma_K,f_K)$ where
$f_K:\mathrm{Sym}(A_K,{\sigma}_K){\longrightarrow} K$ is the canonical extension of $f$ to a
$K$-linear map.
\section{Tensor products of involutions and quadratic pairs}\label{tenprodqps}
In this section we consider the tensor product of an algebra with involution with an algebra with quadratic pair. This corresponds to notion of the tensor product of a symmetric bilinear form and a quadratic form.
We show that this tensor product is associative with the tensor product of algebras with involution. This property underlies our definition of a totally decomposable quadratic pair in the following section.
\begin{prop}\label{lemma:explict}
Let $(A,{\sigma},f)$ be an $F$-algebra with quadratic pair and $(B,\tau)$ an $F$-algebra with involution.
Then there is a unique semi-trace $g$ on $(B,\tau)\otimes (A,{\sigma})$ such that $g(s_1\otimes s_2) = {\mathrm{Trd}}_B(s_1)\cdot f(s_2)$
for all $s_1\in {\mathrm{Sym}}(B,\tau)$ and $s_2\in{\mathrm{Sym}}(A,{\sigma})$.
Moreover, if the semi-trace $f$ on $(A,{\sigma})$ is given by $\ell$, then
$g$ is the semi-trace on $(B,\tau)\otimes(A,{\sigma})$ given by $1\otimes \ell$.
\end{prop}
\begin{proof} If $\kar(F)\neq 2$ then this result is trivial. Assume that $\kar(F)=2$.
Note that as $1\otimes \ell + (\tau\otimes{\sigma})(1\otimes \ell) = 1\otimes 1$, the element $1\otimes \ell$ gives a semi-trace on $(B\otimes_F A,\tau\otimes {\sigma})$ by \cite[(5.7)]{Knus:1998}.
For all $s_1\in{\mathrm{Sym}}(B,\tau)$ and $s_2\in{\mathrm{Sym}}(A,{\sigma})$ we have that
$${\mathrm{Trd}}_{B\otimes_F A}((1\otimes\ell)(s_1\otimes s_2))= {\mathrm{Trd}}_B(s_1)\cdot {\mathrm{Trd}}_A(\ell \cdot s_2)={\mathrm{Trd}}_B(s_1)\cdot f(s_2)\,.$$
For the uniqueness statement, see \cite[(5.18)]{Knus:1998}.
\end{proof}
Let $(A,{\sigma},f)$ be an $F$-algebra with quadratic pair and let $(B,\tau)$ be an $F$-algebra with involution, which is assumed to be orthogonal if $\kar(F)\neq 2$. Then by \cite[(2.23)]{Knus:1998}, $(B,\tau)\otimes (A,{\sigma})$ is orthogonal if $\kar(F)\neq 2$ and symplectic if $\kar(F)=2$.
We denote by $(B,\tau)\otimes (A,{\sigma},f)$ the $F$-algebra with quadratic pair $( B\otimes_F A, \tau\otimes {\sigma},g)$, where $g$ is the semi-trace $g$ on $(B,\tau)\otimes (A,{\sigma})$ characterised in \cref{lemma:explict}.
\begin{prop}\label{prop:tensor} Let $\varphi$ be a symmetric bilinear form over $F$ and $\rho$ a nonsingular quadratic form over~$F$. Then ${\mathrm{Ad}}(\varphi\otimes \rho)\simeq {\mathrm{Ad}}(\varphi)\otimes{\mathrm{Ad}}(\rho)$.
\end{prop}
\begin{proof} See \cite[(5.19)]{Knus:1998}.
\end{proof}
\begin{prop}\label{lemma:decomp}
Let $(B,\tau)$ and $(C,\gamma)$ be $F$-algebras with involution that are assumed to be orthogonal if $\kar(F)\neq 2$ and let $(A,{\sigma},f)$ be an $F$-algebra with quadratic pair. Then
$$ ((B,\tau)\otimes (C,\gamma))\otimes (A,{\sigma},f) \simeq(B,\tau)\otimes ((C,\gamma)\otimes (A,{\sigma},f))\,. $$
\end{prop}
\begin{proof}
Let $\Phi:(B\otimes_FC)\otimes_F A\rightarrow B\otimes_F(C\otimes_FA)$ be the natural $F$-algebra isomorphism.
Clearly $\Phi$ is compatible with the involutions in the statement.
By \cite[(5.7)]{Knus:1998}, $f$ is given by some $\ell\in A$ with $\ell+{\sigma}(\ell)=1$.
It follows from \cref{lemma:explict} that the semi-trace associated with
$((B,\tau)\otimes (C,\gamma))\otimes (A,{\sigma},f) $ is given by $(1\otimes 1)\otimes\ell$ and the semi-trace associated with $(B,\tau)\otimes ((C,\gamma)\otimes (A,{\sigma},f))$ is given by
$1\otimes (1\otimes\ell)$. It then easily follows that $\Phi$ is an isomorphism between the $F$-algebras with quadratic pair.
\end{proof}
By \cref{lemma:decomp}, the tensor product of two algebras with involution on the one hand, and the tensor product of an algebra with involution with an algebra with quadratic pair on the other hand, are mutually associative. That is, for $F$-algebras with involution $(A,{\sigma})$ and $(B,\tau)$ and an $F$-algebra with quadratic pair $(C,\gamma,f)$, the expression $(A,{\sigma})\otimes(B,\tau)\otimes(C,\gamma,f)$ is unambiguous.
\begin{prop}\label{choicef}
Assume $\kar(F)=2$.
Let $(B,\tau)$ and $(C,{\sigma})$ be $F$-algebras with symplectic involution. Then there exists a unique semi-trace $h$ on $(B,\tau)\otimes(C,{\sigma})$ such that $h(s_1\otimes s_2)=0$ for all $s_1\in {\mathrm{Sym}}(B,\tau)$ and $s_2\in {\mathrm{Sym}}(C,{\sigma})$. Moreover, for any semi-trace $f$ on $(C,{\sigma})$, $h$ is the
semi-trace associated with the $F$-algebra with quadratic pair $(B,\tau)\otimes(C,{\sigma}, f)$.
\end{prop}
\begin{proof} For the existence and uniqueness of the semi-trace $h$, see \cite[(5.20)]{Knus:1998}.
Let $f$ be any semi-trace on $(C,{\sigma})$.
By \cite[(5.7)]{Knus:1998}, $f$ is given by an element $\ell\in C$ with $\ell+{\sigma}(\ell)=1$.
Let $g$ be the semi-trace associated with $(B,\tau)\otimes(C,{\sigma},f)$. Then by
\cref{lemma:explict}, $g$ is
given by $1\otimes\ell$.
By \cite[(2.6)]{Knus:1998}, as $(B,\tau)$ is symplectic we have ${\mathrm{Trd}}_B({\mathrm{Sym}}(B,\tau))=\{0\}$. Hence for all $s_1\in {\mathrm{Sym}}(B,\tau)$ and $s_2\in {\mathrm{Sym}}(C,{\sigma})$ we have
$$ {\mathrm{Trd}}_{B\otimes_F C}((1\otimes \ell)(s_1\otimes s_2))= {\mathrm{Trd}}_B(s_1)\cdot {\mathrm{Trd}}_C(\ell\cdot s_2)=0\,.$$
That is, $g$ satisfies the characterising property of $h$ in the statement. Therefore by the uniqueness of the semi-trace $h$, we have that $g=h$.
\end{proof}
Given two $F$-algebras with symplectic involution $(B,\tau)$ and $(C,{\sigma})$, we may define a semi-trace $h$ on $(B,\tau)\otimes (C,{\sigma})$ in the following manner. If $\kar(F)\neq2$, then $(B,\tau)\otimes(C,{\sigma})$ is orthogonal by \cite[(2.23)]{Knus:1998} and we
let $h=\frac{1}{2}{\mathrm{Trd}}_{B\otimes_F C}$. If $\kar(F)=2$, let $h$ be the semi-trace on $(B,\tau)\otimes (C,{\sigma})$ characterised in \cref{choicef}. We denote the $F$-algebra with quadratic pair $(B\otimes_FC,\tau\otimes{\sigma},h)$ by $(B,\tau)\boxtimes (C,{\sigma})$.
If $\kar(F)=2$, then by \cref{choicef} we have that $(B,\tau)\boxtimes (C,{\sigma})\simeq (B,\tau)\otimes (C,{\sigma},f)$ for any choice of semi-trace $f$ on $(C,{\sigma})$. In particular, by Proposition~\ref{lemma:decomp}, for an $F$-algebra with symplectic involution $(A,\sigma)$, the expression $(A,\sigma)\otimes(B,\tau)\boxtimes (C,\gamma)$ is unambiguous.
Hence given a tensor product of two $F$-algebras with symplectic involution,
there is natural choice of a semi-trace making this product into a quadratic pair. We now consider this quadratic pair in the case where the $F$-algebras with involution are $F$-quaternion algebras with their canonical involutions.
Let $a\in F$ with $-4a\neq 1$ and $b\in F^\times$ and let $(Q,\gamma) =[a\,\cdot\!\!\mid b)$. Recall that this $F$-algebra with involution is orthogonal if $\kar(F)\neq2$ and symplectic if $\kar(F)=2$.
Let $u\in Q$ be such that $u^2=u+a$ and $\gamma(u)=1-u$ and let $f$ be the semi-trace on $(Q,\gamma)$ given by $u$. Then we denote the $F$-algebra with quadratic pair $(Q,\gamma,f)$ by $\qp{a}{b}$.
\begin{prop}\label{symplectize}
Let $a,c\in F$ such that $4a\neq -1\neq 4c$ and $b,d\in F^\times$.
Then $${[a\,\cdot\!\!\mid\!\!\cdot\, b)}\boxtimes[{c}\,\cdot\!\!\mid\!\!\cdot\,{d})\simeq {[a+c+4ac \mid\!\!\cdot\, b)}\otimes \qp{c}{bd}\,.$$
\end{prop}
\begin{proof}
Let $(B,\sigma,f)={[a\,\cdot\!\!\mid\!\!\cdot\, b)}\boxtimes[{c}\,\cdot\!\!\mid\!\!\cdot\,{d})$, $(Q_1,\gamma_1)= [a\,\cdot\!\!\mid\!\!\cdot\, b)$ and $(Q_2,\gamma_2)=[{c}\,\cdot\!\!\mid\!\!\cdot\,{d})$.
Let $i,j\in Q_1$ be such that $i^2=i+a$, $j^2= b$ and $ij=j-ji$ and let $ u, v\in Q_2$ be such that $u^2 =u+c$, $v^2=d$ and $uv=v-vu$. In $B$ we have that $\sigma(i\otimes 1)=1\otimes 1 - i\otimes 1$, $\sigma(j\otimes 1)=-j\otimes 1$, $\sigma(1\otimes u)=1\otimes 1-1\otimes u$ and $\sigma(1\otimes v)=-1\otimes v$.
Let $i'= i\otimes 1 + (1-2i)\otimes u$, $j'= j\otimes 1$, $u'=1\otimes u$ and $v'= j\otimes v$. Then one easily checks that $$Q'_1=F\oplus Fi'\oplus Fj'\oplus Fi'j' \quad\mbox{ and }\quad Q'_2=F\oplus Fu'\oplus Fv'\oplus Fu'v'$$ are ${\sigma}$-invariant $F$-subalgebras of $B$ that commute elementwise with one another.
We set $\tau_1={\sigma}|_{Q'_1}$ and $\tau_2={\sigma}|_{Q'_2}$. We have
$$(Q'_1,\tau_1)\simeq [a+c+4ac \mid\!\!\cdot\, b) \textrm{ and } (Q'_2,\tau_2) \simeq[{c}\,\cdot\!\!\mid{bd})\,.$$
Hence $(B,{\sigma})\simeq {[a+c+4ac \mid\!\!\cdot\, b)}\otimes [{c}\,\cdot\!\!\mid{bd})$.
If $\kar(F)\neq 2$, then the semi-trace on $(B,{\sigma})$ is uniquely determined, and in this case there is nothing further to show.
Assume $\kar(F)=2$.
Then $(B,{\sigma},f)\simeq (Q_1,\gamma_1)\otimes(Q_2,\gamma_2,h)$ for any choice of semi-trace $h$ on $(Q_2,\gamma_2)$ by \cref{choicef}. Let $h$ to be the semi-trace given by $u$.
Then for all $s\in {\mathrm{Sym}}(Q_1\otimes_FQ_2, \gamma_1\otimes\gamma_2)={\mathrm{Sym}}(Q'_1\otimes_FQ'_2, \tau_1\otimes\tau_2)$ we have that
$${\mathrm{Trd}}_{Q_1\otimes_FQ_2}((1\otimes u)\cdot s)= {\mathrm{Trd}}_{Q_1\otimes_FQ_2}(u'\cdot s)={\mathrm{Trd}}_{Q'_1\otimes_FQ'_2}((1\otimes u')\cdot s)\,.$$
Hence
$(Q_1,\gamma_1)\otimes(Q_2,\gamma_2,h)\simeq (Q'_1,\tau_1)\otimes (Q'_2,\tau_2,g) $ for the semi-trace $g$ on $(Q'_2,\tau_2)$ given by ${u'}$ by \cref{lemma:explict}. That is, $(B,{\sigma},f)\simeq {[a+c+4ac \mid\!\!\cdot\, b)}\otimes \qp{c}{bd}$.
\end{proof}
\section{Totally decomposable quadratic pairs}
We call an $F$-algebra with quadratic pair \emph{totally decomposable} if it is isomorphic to a tensor product of a totally decomposable $F$-algebra with involution and an $F$-quaternion algebra with quadratic pair.
It follows from \cref{lemma:decomp} that taking the tensor product of a totally decomposable $F$-algebra with involution and a totally decomposable $F$-algebra with quadratic pair gives a totally decomposable algebra with quadratic pair.
Let $(A,{\sigma},f)$ be a totally decomposable $F$-algebra with quadratic pair. Then there exists a totally decomposable $F$-algebra with quadratic pair $(B,\tau)$ and an $F$-quaternion algebra with involution $(Q,\gamma,g)$ such that $(A,{\sigma},f)\simeq (B,\tau)\otimes(Q,\gamma,h)$.
If $\kar(F)\neq 2$ then $(B,\tau)$ is necessarily orthogonal. We now show that, even if $\kar(F)=2$, we may always find a decomposition as above where $(B,\tau)$ is orthogonal. This will allow us in the next section to use the Pfister invariant of $(B,\tau)$ to study the quadratic pair $(A,{\sigma},f)$.
\begin{prop}\label{totdecompqp}
Let $(A,{\sigma},f)$ be a totally decomposable $F$-algebra with quadratic pair. Then there exists a totally decomposable $F$-algebra with orthogonal involution $(B,\tau)$ and
an $F$-quaternion algebra with quadratic pair $(Q,\gamma,g)$ such that $(A,{\sigma},f)\simeq (B,\tau)\otimes (Q,\gamma, g)$.
\end{prop}
\begin{proof} The result is trivial if $\kar(F)\neq2$. Assume that $\kar(F)=2$.
As $(A,{\sigma},f)$ is totally decomposable, there exist $F$-quaternion algebras with involution $(Q_i,\sigma_i)$ for $i=1,\ldots, n-1$ and an $F$-quaternion algebra with quadratic pair $(Q_n,\gamma,h)$ such that
$$(A,{\sigma},f)\simeq (Q_1,{\sigma}_1)\otimes \ldots\otimes (Q_{n-1},{\sigma}_{n-1})\otimes (Q_n,\gamma, h)\,.$$
Suppose ${\sigma}_i$ is symplectic. In particular, it is the canonical involution on $Q_i$, for some $i\in\{1,\ldots, n-1\}$.
Then by \cref{choicef}, we have that
$$ (Q_i,{\sigma}_i)\otimes (Q_n,\gamma,h)\simeq (Q_i,{\sigma}_i)\boxtimes (Q_n,\gamma)\,. $$
Hence, by \cref{symplectize},
there exists
an $F$-quaternion algebra with orthogonal involution $(Q'_i,\tau)$ and an $F$-quaternion algebra with quadratic pair $(Q'_n,\gamma',h')$
such that
$$ (Q_i,{\sigma}_i)\otimes (Q_n,\gamma,h)\simeq (Q'_i,\tau)\otimes (Q'_n,\gamma',h') \,.$$
Using this argument repeatedly for all $i=1,\ldots, n-1$ such that ${\sigma}_i$ is symplectic, we modify our expression of $(A,{\sigma},f)$ above to obtain the result.
\end{proof}
For interest, we also record a characteristic two specific counterpart of the previous statement,
which produces a symplectic instead of an orthogonal factor.
\begin{prop}
Assume that $\kar(F)=2$.
Let $(A,{\sigma},f)$ be a totally decomposable $F$-algebra with quadratic pair with $\deg(A)=2^n$, where $n\geq 2$. Then there exist $F$-quaternion algebras with canonical involution $(Q_i,\gamma_i)$ for $i=1,\ldots, n$ such that
$(A,{\sigma},f)\simeq (Q_1,\gamma_1)\otimes\ldots\otimes(Q_{n-1},\gamma_{n-1})\boxtimes (Q_n,\gamma_n)\,.$
\end{prop}
\begin{proof}
By \cref{prop:orth}, for every $F$-quaternion algebra with orthogonal involution $(Q,\tau)$, there exists an $a\in F$ and $b\in F^\times$ such that $(Q,\tau)\simeq [a\mid\!\!\cdot\, b)$. Similarly, by \cite[(5.6)]{dolphin:quadpairs}, for every $F$-quaternion algebra with quadratic pair $(Q,\gamma,f)$
there exists an $c\in F$ and $d\in F^\times$ such that $(Q,\gamma,f)\simeq\qp{c}{d}$. The result thus follows using the isomorphism in \cref{symplectize} in a similar way as to how it is used in \cref{totdecompqp}, but in the opposite direction.
\end{proof}
\section{Totally decomposable quadratic pairs on a split algebra}\label{section:main}
We now prove our main result, that over fields of characteristic two a split algebra with totally decomposable quadratic pair is adjoint to a Pfister form. We use the following result, which is an approach unique to fields of characteristic two. This approach gives more information on the $m$-fold Pfister form $\pi$ adjoint to a totally decomposable quadratic pair on an algebra of degree $2^m$ over a field $F$ of characteristic $2$ after extending to splitting field $K$. Specifically, we show that we can always find an $(m-1)$-fold bilinear Pfister form $\varphi$ defined over $F$ such that $\varphi_K$ factors
$\pi$.
\begin{prop}\label{prop:char2extra}
Assume $\kar(F)=2$. Let $(B,\tau)$ be a totally decomposable orthogonal $F$-algebra with involution, $(Q,\gamma,h)$ be an $F$-quaternion algebra with quadratic pair and $(A,{\sigma},f)= (B,\tau)\otimes (Q,\gamma,h)$. Then for any field extension $K/F$ such that $A_K$ is split, there exists a $1$-fold Pfister form $\pi$ over $K$ such that $(A,{\sigma},f)_K\simeq {\mathrm{Ad}}((\mathfrak{Pf}(B,\tau))_K\otimes \pi)$.
\end{prop}
\begin{proof}
Let $\varphi=\mathfrak{Pf}(B,\tau)$ and $\rho$ a quadratic form over $K$ with $(A,{\sigma},f)_K\simeq {\mathrm{Ad}}(\rho)$. Note that $\dim(\rho)=2\dim(\varphi)$.
Assume first that $(B,\tau)_K$ is metabolic. Then by \cite[(A.5)]{Tignol:galcohomgps} we have that $(A,{\sigma},f)_K$ is hyperbolic and by \cref{prop:hypiffquad} that $\rho$ is hyperbolic.
We may thus take $\pi$ to be the hyperbolic $2$-dimensional quadratic form.
Assume now that $(B,\tau)_K$ is not metabolic. Then the bilinear Pfister form $\varphi_K$ is anisotropic by \cref{thm:pfisterinvar}.
We consider its function field $L=K(\varphi_K)$.
Since $\varphi_L$ is metabolic, it follows by \cref{thm:pfisterinvar} that $(B,\tau)_L$ is metabolic.
Hence, $(A,{\sigma},f)_L$ is hyperbolic by \cite[(A.5)]{Tignol:galcohomgps} and therefore $\rho_L$ is hyperbolic by \cref{prop:hypiffquad}.
By \cref{prop:funcfieldbf}, there exists a non-trivial nonsingular quadratic form $\pi'$ over $K$ such that $\rho_{\mathrm{an}}\simeq \varphi_K\otimes\pi'$.
We have $$\dim(\varphi)\cdot \dim(\pi')=\dim(\rho_{\mathrm{an}})\leqslant \dim(\rho)= 2\dim(\varphi)\,.$$
As $\kar(F)=2$, by \cite[(7.32)]{Elman:2008} $\dim(\pi')$ is even.
It follows that $\dim(\pi')=2$ and $\rho_{\mathrm{an}}\simeq \rho$.
In particular, $\pi'$ is similar to a $1$-fold Pfister form $\pi$ and $\rho\simeq\rho_{{\mathrm{an}}}\simeq \varphi_K\otimes\pi'$. Hence $\rho$ is similar to $\varphi_K\otimes\pi$.
\end{proof}
\begin{cor} \label{cor:main}
Assume that $\kar(F)=2$. Let $n\in\mathbb{N}$, let $(A,{\sigma},f)$ be a totally decomposable $F$-algebra with quadratic pair with $\deg(A)=2^{n+1}$ and let $K/F$ be a field extension such that $A_K$ is split.
Then there exists an
$n$-fold bilinear Pfister form $\varphi$ over $K$ and an
$(n+1)$-fold Pfister form $\rho$ over $K$ such that
$\varphi_K$ factors $\rho$ and
$(A,{\sigma},f)_K\simeq{\mathrm{Ad}}(\rho)$.
\end{cor}
\begin{proof} For $n=0$, this is trivial. Otherwise,
by \cref{totdecompqp}, there exists a totally decomposable $F$-algebra with orthogonal involution $(B,\tau)$ and an $F$-quaternion algebra with quadratic pair $(Q,\gamma,h)$ such that $(A,{\sigma},f)\simeq (B,\tau)\otimes (Q,\gamma,h)$. The result then follows from \cref{prop:char2extra} with $\varphi=\mathfrak{Pf}(B,\tau)$.
\end{proof}
\begin{thm}\label{thm:pfisterfactquad}
Let $\rho$ be a nonsingular quadratic form over $F$ with $\dim(\rho)\geq 2$.
Then ${\mathrm{Ad}}(\rho)$ is totally decomposable if and only if $\rho$ is similar to a Pfister form.
\end{thm}
\begin{proof}
If $\rho$ is a Pfister form over $F$, then we can write $\rho\simeq\varphi\otimes\pi$ for a bilinear Pfister form $\varphi$ and a $1$-fold quadratic Pfister form $\pi$ over $F$.
Then by \cref{prop:tensor} we have that ${\mathrm{Ad}}(\rho)\simeq {\mathrm{Ad}}(\varphi)\otimes {\mathrm{Ad}}(\pi)$. The $F$-algebra with involution ${\mathrm{Ad}}(\varphi)$ is totally decomposable by \cref{lemma:tensorbf}. As ${\mathrm{Ad}}(\pi)$ is an $F$-quaternion algebra with quadratic pair, it follows that ${\mathrm{Ad}}(\rho)$ is totally decomposable.
Assume conversely that ${\mathrm{Ad}}(\rho)$ is totally decomposable. If $\kar(F)\neq2$, then any quadratic pair is equivalent to an orthogonal involution and thus the result corresponds to \cite[Thm.~1]{Becher:qfconj}. If $\kar(F)=2$, then
$\rho$ is similar to a Pfister form by \cref{cor:main}.
\end{proof}
Let $(A,{\sigma},f)$ be a totally decomposable $F$-algebra with quadratic pair with $\mathrm{deg}(A)=2^m$ and let $K/F$ be a field extension such that $A_K$ is split.
Let $\pi$ be the $m$-fold Pfister form over $K$ such that $(A,{\sigma},f)_K\simeq{\mathrm{Ad}}(\pi)$.
In general, it is not possible to find
an $(m-1)$-fold quadratic Pfister form $\rho$ over $F$ such that $\rho_K$ factors $\pi$.
This is illustrated by the following example, which is a variation of an example in \cite[(3.9)]{parimala:pfisterinv}.
In particular, this example shows that
\cref{cor:main} cannot be extended to cover fields of characteristic different from $2$, where quadratic and bilinear Pfister forms are equivalent. %
\begin{ex}\label{ex:countex} Let $n\in \mathbb{N}$ with $n\geqslant 4$.
By \cite[(38.4)]{Elman:2008} and its proof, there exists a field $F$ such that all
$3$-fold quadratic Pfister forms over $F$ are hyperbolic and there exist
$F$-quaternion algebras $Q_1,\ldots, Q_n$ such that $A=Q_1\otimes_F\cdots\otimes_F Q_n$ is a division $F$-algebra. In particular, we have $\deg(A)=\mathrm{ind}(A)=2^n$. For $i= 1,\ldots, n$, let $\gamma_i$ be the canonical involution on $Q_i$ if $\kar(F)=2$ and an orthogonal involution on $Q_i$ if $\kar(F)\neq2$.
We obtain a totally decomposable $F$-algebra with quadratic pair
$$ (A,{\sigma},f ) = (Q_1,\gamma_1)\otimes\cdots \otimes (Q_{n-1},\gamma_{n-1})\boxtimes (Q_n,\gamma_n) \,. $$
By \cite[(3.3) and \S2.4]{karpenko:gensplit}
there exists a field extension $K/F$ such that $A_K$ is split and $(A,{\sigma},f)_K\simeq {\mathrm{Ad}}(\rho)$ for some quadratic form $\rho$ over $K$ such that $\mathrm{ind}(A)$ divides $i_W(\rho)$. As for any such $\rho$ we have that $\dim(\rho) = 2^n=\mathrm{ind}(A)$ and $i_W(\rho)\leqslant \frac{1}{2}\dim(\rho)$, it follows that $i_W(\rho)=0$, that is, $\rho$ is anisotropic. In particular, $\rho$ is not factored by $\pi_K$ for any $(n-1)$-fold Pfister form $\pi$ over $F$, as all such $\pi$ are hyperbolic.
\end{ex}
\small{ | {
"redpajama_set_name": "RedPajamaArXiv"
} | 7,812 |
Introduction
============
This project is the Inocybe/OpenDaylight customization of the Docbkx
plug-in for creating documentation artifacts for Inocybe, OpenDaylight,
and other opensource projects.
Release Notes
=============
sdndocs-maven-plugin 0.1.0 (February, 2014)
============================================================
New features and changes
------------------------
- Initial Implemantation and Branding from OpenStack Plugin
Bug fixes
---------
| {
"redpajama_set_name": "RedPajamaGithub"
} | 6,730 |
John "Johnny" Jones (Washington D.C., ) es un exjugador de baloncesto estadounidense que disputó una temporada en la NBA, otra más en la ABA, mientras que el resto de su carrera transcurrió en la EBA. Con 2,01 metros de estatura, jugaba en la posición de Ala-pívot.
Trayectoria deportiva
Universidad
Jugó durante su etapa universitaria con los Golden Eagles de la Universidad Estatal de California, Los Ángeles, siendo uno de los cuatro jugadores que han salido de sus aulas en participar en la NBA o en la ABA.
Profesional
Tras no ser elegido en el Draft de la NBA, fichó como agente libre en octubre de por los Boston Celtics, donde en su única temporada en el equipo consiguió el anillo de campeón, tras derrotar en las Finales a Los Angeles Lakers por 4-2, promediando en temporada regular 4,2 puntos y 2,2 rebotes por partido.
Al año siguiente no fue protegido por su franquicia, entrando en el Draft de Expansión por la llegada de nuevos equipos a la liga, siendo elegido por los Milwaukee Bucks, quienes finalmente desecharon su fichaje. Cambió de liga para fichar por los Kentucky Colonels de la ABA, pero solo llegó a jugar 15 partidos, en los que promedió 7,0 puntos y 4,0 rebotes.
Jugó posteriormente siete temporadas en la EBA, logrando sendos campeonatos en 1971 y 1975.
Estadísticas de su carrera en la NBA y la ABA
Temporada regular
Playoffs
Referencias
Enlaces externos
Ficha de Johnny Jones en ultimatenba.com
Baloncestistas de Washington D. C.
Baloncestistas de la NBA
Baloncestistas de la ABA
Baloncestistas de los Cal State Los Angeles Golden Eagles
Baloncestistas de los Boston Celtics
Baloncestistas de los Kentucky Colonels
Baloncestistas de los Allentown Jets
Baloncestistas de los Wilkes-Barre Barons
Baloncestistas de los Scranton Apollos
Nacidos en Washington D. C. | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 9,587 |
{"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2019_AMC_10C_Problems\/Problem_25","text":"# 2019 AMC 10C Problems\/Problem 25\n\n### Problem\n\nLet $N$ be the least positive integer $x$ such that $\\lfloor \\frac{x^{8}}{x-1} \\rfloor$ is a multiple of 10000. Find the sum of the digits of $N$. (Note: $\\left\\lfloor x \\right\\rfloor$ denotes the greatest integer less than or equal to $x$. )\n\n$\\textbf{(A)}\\ 9 \\qquad\\textbf{(B)}\\ 11 \\qquad\\textbf{(C)}\\ 15 \\qquad\\textbf{(D)}\\ 24 \\qquad\\textbf{(E)}\\ 36$\n\n### Solution\n\nWe have $\\left\\lfloor \\frac{x^8}{x-1}\\right\\rfloor = (1+x)(1+x^2)(1+x^4)$ so we want to solve $(1+x)(1+x^2)(1+x^4) \\equiv 0 \\pmod{16}$ and $0 \\pmod{625}$.\n\nSolving modulo 16 is fairly simple; the solutions are $x \\equiv 3,7,11,15 \\pmod{16}$ (or more compactly, $x \\equiv 3 \\pmod{4}$).\n\nTo solve modulo 625, observe that at most one of $1+x$, $1+x^2$, and $1+x^4$ can be divisible by 5. The case $1+x \\equiv 0 \\pmod{625}$ gives $x \\equiv 624 \\pmod{625}$. The case $1+x^4 \\equiv 0 \\mod{625}$ gives no solutions as $x^4 \\equiv 0,1 \\pmod{5}$ by FLT. Thus we want to solve $1+x^2 \\equiv 0 \\pmod{625}$.\n\nNote that $x^2 + 1 \\equiv 0 \\pmod{625}$ implies $x \\equiv 2,3 \\pmod{5}$. As $x$ is a solution iff $-x$ is a solution, we can assume w.l.o.g. that $x \\equiv 2 \\pmod{5}$. Then $x = 5k+2$, and plugging in gives $25k^2 + 20k + 5 \\equiv 0 \\pmod{625} \\iff 5k^2 + 4k + 1 \\equiv 0 \\pmod{125}$.\n\nTaking the LHS mod 5 again, we see $k\\equiv 1 \\pmod{5}$, so $x \\equiv 7 \\pmod{25}$, so we may let $x = 25m+7$. Using the same method, we have $(25m+7)^2 + 1 = 625m^2 + 350m + 50 \\equiv 0 \\pmod{625} \\implies 350m + 50 \\equiv 0 \\pmod{625} \\implies 14m + 2 \\equiv 0 \\pmod{25}$. The solution is $m \\equiv 7 \\pmod{25}$, so $x = 25(25q+7)+7 \\equiv 182 \\pmod{625}$. Thus the solutions are $x \\equiv 182, 443 \\pmod{625}$.\n\nSo the solutions mod 16 and 625 are $x \\equiv 3,7,11,15 \\pmod{16}$, and $x \\equiv 182, 443, 624 \\pmod{625}$. By CRT, each combination gives us a solution mod 10000. Observing that $443 \\equiv 11 \\pmod{16}$ immediately gives us the minimum solution, so $N=443$ and $4+4+3 = \\boxed{11}$.","date":"2023-01-27 21:05:48","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\": 0, \"img_math\": 39, \"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\": 0.9817689657211304, \"perplexity\": 95.94385670509095}, \"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-2023-06\/segments\/1674764495012.84\/warc\/CC-MAIN-20230127195946-20230127225946-00799.warc.gz\"}"} | null | null |
Q: curl php post request I have this code
POST /account/p_api/v1/account/authorization/assign HTTP/1.1
Host: exemple.com
Cookie: exemle-locale=en;
User-Agent: Mozilla/5.0 (Windows NT 10.0; Win64; x64; rv:107.0) Gecko/20100101 Firefox/107.0
Accept: */*
Accept-Language: en-US,en;q=0.5
Accept-Encoding: gzip, deflate
Content-Type: application/json
Content-Length: 17
Origin: https://exemple.com
Sec-Fetch-Dest: empty
Sec-Fetch-Mode: cors
Sec-Fetch-Site: same-origin
Te: trailers
Connection: close
{"code":"herecode"}
I want to use curl and php to be able to send the request and add another code to "herecode"
I don't know how to do this
A: In your Browser open the developer tools
Go to the url in your Browser.
Go to the "Network Tab"
Right click the first request in the Network Tab
Select copy => curl
Go to https://curlconverter.com/php/
Paste the "curl copy"
And you have your PHP code.
There is a bug in their CURLOPT_HTTPHEADER you need to make your own header array.
Example header:
'Content-Type' => 'application/x-www-form-urlencoded; charset=UTF-8',
Change to
'Content-Type: application/x-www-form-urlencoded; charset=UTF-8',
Firefox curl copy example:
Chrome curl copy example:
Paste to urlconverter:
This is a curl copy and paste for this page. curl php post request
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 2,213 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.